Objective Mathematics. Volume1 9789325299122

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COMPLETE STUDY PACK FOR

ENGINEERING

ENTRANCES

OBJECTIVE MATHEMATICS Volume 1

COMPLETE STUDY PACK FOR

ENGINEERING

ENTRANCES

OBJECTIVE MATHEMATICS Volume 1 Amit M. Agarwal

ARIHANT PRAKASHAN (SERIES), MEERUT

Arihant Prakashan (Series), Meerut All Rights Reserved

© AUTHOR

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PREFACE Engineering offers the most exciting and fulfilling of careers. As a Engineer you can find satisfaction by serving the society through your knowledge of technology. Although the number of Engineering colleges imparting quality education and training has significantly increased after independence in the country, but simultaneous increase in the number of serious aspirants has made the competition difficult, it is no longer easy to get a seat in a prestigious Engineering college today. For success, you require an objective approach of the study. This does not mean you 'prepare' yourself for just 'objective questions'. Objective Approach means more than that. It could be defined as that approach through which a student is able to master the concepts of the subject and also the skills required to tackle the questions asked in different entrances such as JEE Main & Advanced, as well other regional Engineering entrances. These two-volume books on Mathematics ‘Objective Mathematics (Vol.1 & 2)’ fill the needs of such books in the market in Mathematics and are borne out of my experience of teaching Mathematics to Engineering aspirants.

The plan of the presentation of the subject matter in the books is as follows — The whole chapter has been divided under logical topic heads to cover the syllabi of JEE Main &

Advanced and various Engineering entrances in India. — The Text develops the concepts in an easy going manner, taking the help of the examples from the

day-to-day life. — Important points of the topics have been highlighted in the text. Under Notes, some extra points

regarding the topics have been given to enrich the students. — The Solved Examples make the students learn the basic problem solving skills in Mathematics. Very

detailed explanations have been provided to make the students skilled in systematically tackling the problems. — The answers / solutions to all the questions have been provided. — The Objective Questions have been divided according to their types Single correct option, More

than One, Assertion-Reason, Matching Type, Integer Type, Passage Based, etc. which can take the students to a level required for various Engineering entrances in the present scenario. — Entrance Corner includes the Previous Years' Questions asked in JEE Main & Advanced and other

various Engineering entrances. At the end of the book, JEE Main & Advanced & Other Regional Entrances Solved Papers have been given.

I would open-heartedly welcome the suggestions for the further improvements of this book (Vol.1) from the students and teachers. Amit M. Agarwal

CONTENTS 1. SETS 1-19 Ÿ Introduction Ÿ Set Ÿ Notations Ÿ Representation of Sets Ÿ Types of Sets Ÿ Venn Diagram Ÿ Operations on SetsLaws of Algebra of Sets Ÿ Formulae to Solve Practical Problems on Union and Intersection of Sets 2. FUNDAMENTALS OF RELATION AND FUNCTION 20-40 Ÿ Ordered Pair Ÿ Cartesian Product of Sets Ÿ Properties of Cartesian Product of Sets Ÿ Relation Ÿ Representation of Relation Ÿ Domain and Range of Relations Ÿ Some Particular Types of Relations Ÿ Inverse Relation Ÿ Composition of Relations Ÿ Functions or Mappings Ÿ Difference between Relation and Function Ÿ Domain, Codomain and Range of a Function Ÿ Equal Functions Ÿ Classification of Functions Ÿ Algebra of Real Functions Ÿ Composition of Functions 3. SEQUENCE AND SERIES Ÿ Introduction Ÿ Arithmetic Progression (AP) Ÿ Geometric Progression (GP)

41-108

Ÿ Harmonic Progression (HP) Ÿ Arithmetico-Geometric Progression (AGP) Ÿ Some Special Series

4. COMPLEX NUMBERS 109-181 Ÿ The Real Number System Ÿ Modulus of a Real Number Ÿ Imaginary Number Ÿ Complex Number Ÿ Algebra of Complex Numbers Ÿ Conjugate of a Complex Number Ÿ Modulus of a Complex Number Ÿ Argument (or Amplitude) of a Complex Number Ÿ Various Forms of a Complex Number Ÿ De-Moivre’s Theorem Ÿ Roots of Unity Ÿ Geometrical Applications of Complex Numbers Ÿ Loci in Complex Plane Ÿ Logarithm of Complex Numbers 5. INEQUALITIES AND QUADRATIC EQUATION 182-268 Ÿ Inequality Ÿ Generalised Method of Intervals for Solving Inequalities by Wavy Curve Method (Line Rule) Ÿ Absolute Value of a Real Number Ÿ Logarithms Ÿ Arithmetico-Geometric Mean Inequality Ÿ Quadratic Equation with Real Coefficients Ÿ Formation of a Polynomial Equation from Given Roots Ÿ Symmetric Function of the Roots

Ÿ Transformation of Equations Ÿ Common Roots Ÿ Quadratic Expression and its Graph Ÿ Maximum and Minimum Values of Rational

Expression Ÿ Location of the Roots of a Quadratic Equation Ÿ Algebraic Interpretation of Rolle’s Theorem Ÿ Condition for Resolution into Linear Factors Ÿ Some Application of Graphs to Find the Roots

of Equations 6. PERMUTATION AND COMBINATION 269-332 Ÿ Fundamental Principles of Counting (FPC) Ÿ Factorial Ÿ Exponent of Prime p in Factorial n n n Ÿ Representation of Symbols Pr and Cr Ÿ Some Basic Arrangements and Selections Ÿ Summation of Numbers (3 different ways) Ÿ Permutations under Certain Conditions Ÿ Circular Permutations n Ÿ Geometrical Applications of Cr Ÿ Selection of One or More Objects Ÿ Number of Divisors and the Sum of the Divisors of a Given Natural Number Ÿ Division of Objects into Groups Ÿ Dearrangements Ÿ Number of Integral Solutions of Linear Equations and Inequations 7. MATHEMATICAL INDUCTION 333-347 Ÿ Introduction Ÿ Statement Ÿ Principle of Mathematical Induction Ÿ Algorithm for Mathematical Induction Ÿ Types of Problems

8. BINOMIAL THEOREM 348-417 Ÿ Binomial Theorem for Positive Integral Index Ÿ Multinomial Theorem Ÿ Greatest Coefficient Ÿ Greatest Term Ÿ R-f Factor Relation Ÿ Divisibility Problems Ÿ Properties of Binomial Coefficients Ÿ Binomial Theorem for any Index Ÿ Approximation Ÿ Exponential Series Ÿ Logarithmic Series 9. TRIGONOMETRIC FUNCTIONS AND EQUATIONS 418-511 Ÿ Introduction Ÿ Measure of Angles Ÿ Systems of Measurement of Angles Ÿ Trigonometric Ratios Ÿ Trigonometric Function Ÿ Graph of Trigonometric Functions Ÿ Trigonometrical Identities Ÿ Trigonometric Ratios of Allied Angles Ÿ Trigonometrical Ratios of Compound Angles Ÿ Trigonometric Ratios of Multiples of an Angle Ÿ Maximum and Minimum Values of Trigonometrical Expressions Ÿ Trigonometric Equations Ÿ General Solution of Trigonometric Equations Ÿ Solution of Trigonometric Inequality 10. PROPERTIES OF TRIANGLES, HEIGHTS AND DISTANCES 512-589 Ÿ Introduction Ÿ Relation between the Sides and Angles of Triangle

Ÿ Trigonometric Ratios of Half Angles of a Ÿ Ÿ Ÿ Ÿ Ÿ Ÿ Ÿ Ÿ Ÿ Ÿ

Triangle Area of a Triangle Conditional Identities Solution of Triangles Circles Connected with Triangle The Orthocentre and the Pedal Triangle Cyclic Quadrilateral Regular Polygon Heights and Distances Some Important Properties of Triangles Some Properties Related to Circle

11. CARTESIAN SYSTEM OF

RECTANGULAR COORDINATES

590-626

Ÿ Introduction Ÿ Coordinate System Ÿ Distance Formulae Ÿ Applications of Distance Formula Ÿ Section Formulae Ÿ Area of a Triangle Ÿ Area of a Quadrilateral Ÿ Some Standard Points of a Triangle Ÿ Locus Ÿ Transformation of Axes

12. STRAIGHT LINE AND PAIR OF STRAIGHT LINES 627-694 Ÿ Straight Line Ÿ Angle between Two Lines Ÿ Point of Intersection of Two Lines Ÿ Image of a Point with Respect to a Line Ÿ Family of Lines through the Intersection of Two Given Lines

Ÿ Locus and its Equation Ÿ Combined Equation of a Pair of Straight Lines Ÿ Bisectors of the Angle between the Lines

Given by a Homogeneous Equation Ÿ General Equation of Second Degree Ÿ Equations of the Angle Bisectors Ÿ Distance between the Pair of Parallel Lines 13. CIRCLE 695-791 Ÿ Introduction Ÿ Standard Equation of a Circle Ÿ Circle Passing through Three Points Ÿ Position of a Point with respect to a Circle Ÿ Intersection of a Straight Line and a Circle Ÿ Equation of Tangent Ÿ Normal to a Circle Ÿ Pair of Tangents Ÿ Director Circle Ÿ Pole and Polar Ÿ Diameter of a Circle Ÿ Angle of Intersection of Two Circles Ÿ Family of Circles Ÿ Coaxial System of Circles Ÿ Limiting Points 14. PARABOLA 792-853 Ÿ Conic Sections Ÿ Parabola Ÿ Other Standard Forms of Parabola Ÿ Position of a Point with respect to a Parabola Ÿ Intersection of a Line and a Parabola Ÿ Equation of Tangent Ÿ Angle of Intersection of Two Parabolas Ÿ Equation of Normal to Parabola

Ÿ Number of Normals and Conormal Points

Ÿ Conjugate points

Ÿ Combined Equation of Pair of Tangents

Ÿ Conjugate Lines

Ÿ Director Circle

Ÿ Diameter

Ÿ Diameter of a Parabola

Ÿ Asymptotes

Ÿ Pole and Polar of a Parabola

Ÿ Rectangular Hyperbola

Ÿ Lengths of Tangent, Subtangent, Normal

and Subnormal 15. ELLIPSE 854-918 Ÿ Introduction Ÿ Position of a Point with respect to an Ellipse Ÿ Equation of the Chord Ÿ Intersection of a Line and an Ellipse Ÿ Tangent Ÿ Combined Equation of the Pair of Tangents Ÿ Director Circle Ÿ Normal Ÿ Number of Normals and Conormal Points Ÿ Pole and Polar Ÿ Conjugate Lines Ÿ Diameter 16. HYPERBOLA 919-971 Ÿ Introduction Ÿ Conjugate Hyperbola Ÿ Position of a Point with respect to a Hyperbola Ÿ Intersection of a Line and a Hyperbola Ÿ Tangent to a Hyperbola Ÿ Director Circle Ÿ Normals to a Hyperbola Ÿ Equation of the Pair of Tangents Ÿ Equations of Chord Ÿ Pole and Polar

Ÿ Tangent to a Rectangular Hyperbola Ÿ Normals to a Rectangular Hyperbola

17. INTRODUCTION TO THREE DIMENSIONAL (3D) GEOMETRY 972-986 Ÿ Coordinate Axes and Coordinate Planes in Three Dimensional Space Ÿ Coordinates of a Point in Space Ÿ Distance between Two Points Ÿ Section Formulae Ÿ Centroid of a Triangle 18. INTRODUCTION TO LIMITS & DERIVATIVES 987-1040 Ÿ Limits Ÿ Existence of Limit Ÿ Algebra of Limits Ÿ Evaluation of Limits by Using L' Hospital’s Rule Ÿ Evaluation of Algebraic Limits Ÿ Evaluation of Trigonometric Limits Ÿ Evaluation of Exponential and Logarithmic Limits Ÿ Evaluation of Exponential Limits of the Form 1¥ Ÿ Sandwich Theorem for Evaluating Limits Ÿ Some Useful Expansions Ÿ Use of Newton-Leibnitz’s Formula in

Evaluating the Limits Ÿ Derivative

Ÿ Geometrical Meaning of a Derivative Ÿ Derivative from First Principle Ÿ Differentiation of Some Important Functions Ÿ Algebra of Derivative of Functions Ÿ Chain Rule Ÿ Logarithmic Differentiation

19. MATHEMATICAL REASONING 1041-1057 Ÿ Statements or Propositions Ÿ Use of Venn Diagrams in Checking Truth and Falsity of Statements Ÿ Truth Table Ÿ Logical Connectives/Operators Ÿ Quantifiers and Quantified Statements Ÿ Negation of a Quantified Statement Ÿ Logical Equivalence Ÿ Negation of a Compound Statement Ÿ Converse, Inverse and Contrapositive of an Implication Ÿ Tautologies and Contradictions Ÿ Algebra of Statements Ÿ Duality

20. STATISTICS 1058-1094 Ÿ Measures of Central Tendency Ÿ Measures of Dispersion Ÿ Skewness Ÿ Some Results to be Remembered Ÿ Correlation Analysis Ÿ Characteristics of Correlation Coefficient Ÿ Regression Analysis Ÿ Properties of Regression Coefficients Ÿ Properties of Lines of Regression 21. FUNDAMENTALS OF PROBABILITY 1095-1132 Ÿ Introduction Ÿ Some Basic Definitions Ÿ Event Ÿ Important Events Ÿ Algebra of Events Ÿ Probability Ÿ Geometrical Probability Ÿ Addition Theorem of Probability Ÿ Independent Events Ÿ Booley’s Inequality

JEE Advanced Solved Paper 2015 JEE Main & Advanced Solved Papers 2016 JEE Main & Advanced/ BITSAT/Kerala CEE/ KCET/AP & TS EAMCET/ VIT/MHT CET Solved Papers 2017 JEE Main & Advanced/ BITSAT/ KCET/AP & TS EAMCET/ VIT/MHT CET Solved Papers 2018 JEE Main & Advanced/ BITSAT/ AP & TS EAMCET/ MHT CET/WB JEE Solved Papers 2019-20

1135-1140 1-12 1-32 1-35 1-31

1 Sets Introduction In our mathematical language, everything in this universe whether living or non-living is called an object. If we consider a collection of objects given in such a way that it is possible to tell beyond doubt, whether a given object is in the collection under consideration or not, then such a collection of objects is called a well-defined collection of objects.

Set A set is a well-defined collection of objects. By ‘well-defined collection of objects’, it means that we can definitely decide whether a given particular object belongs to a given collection or not. The objects, elements and members of a set are synonymous terms. X

Example 1. The set of intelligent students in a class is (a) a null set (b) a well-defined collection (c) a finite set (d) not a well-defined collection Sol. (d) Since, the criterion for determining the intelligence of a student may vary from person to person, so this is not a well-defined collection of objects.

X

Example 2. Which of the following is the collection of first five prime numbers? (a) {1, 2, 3, 5, 7} (b) {2, 3, 5, 7, 11} (c) {3, 5, 7, 11, 13} (d) {1, 2, 3, 4, 5} Sol. (b) It is clear that first five prime numbers are {2, 3, 5, 7, 11}.

Notations Sets are usually denoted by capital letters A, B, C etc., and their elements by small letters a, b, c, etc. Let A be any set of objects and let a be a member of A, then we write a ∈ A and read it as ‘a belongs to A’ or ‘a is an element of A’ or ‘a is member of A’. If a is not an object of A, then we write a ∉ A and read as ‘a does not belong to A’ or ‘a is not an element of A’ or ‘a is not a member of A’.

Representation of Sets A set is often represented in one of the following two forms : (i) Roster form or Tabular form (ii) Set builder form

Chapter Snapshot ●

Introduction

●

Set

●

Notations

●

Representation of Sets

●

Types of Sets

●

Venn Diagram

●

Operations on Sets

●

Laws of Algebra of Sets

●

Formulae to Solve Practical Problems on Union and Intersection of Sets

Objective Mathematics Vol. 1

1

Roster Form or Tabular Form In this form, a set is described by listing its elements, separated by commas, within braces { }. e.g. The set of months of a year which have thirty days, in roster form can be described as {April, June, September, November} X

Example 3. The roster form of set A = {x : x is a positive integer and divisor of 9} (a) {3, 6, 9} (b) {1, 5, 9} (c) {1, 3, 9} (d) {2, 3, 9}

Ø

●

●

A set consisting of only one element is called a singleton set. X

can take values 1, 3, 9. ∴Roster form of the set A is {1, 3, 9}.

Set Builder Form

X

Example 4. If set A = {1, 2, 3, 4}, then it can be written in set builder form as (a) A = {x : x ∈ N and x ≤ 5} (b) A = {x : x ∈ N and x < 5} (c) A = {x : x ∈ N and 1 < x < 5} (d) A = {x : x ∈ N and x < 4} Sol. (b) Clearly, A = { x : x ∈ N and x < 5} describes the set A = {1, 2, 3, 4}

Types of Sets

Finite Set A set which is empty or having a definite number of elements is called a finite set. X

Sol. (c) Since, x ∈ N and 3 < x < 4, so the given set has no element. X

Example 6. The set {x : x 2 + 1 = 0 and x ∈ R } is equal to (a) {−1, 1} (b) {1} (c) {−1} (d) φ Sol. (d) We know that there is no real number x such that

2

x2 + 1 = 0.

Example 8. A set {x | x ∈ N and 1 ≤ x ≤ 10} is called (a) infinite set (b) finite set (c) singleton set (d) null set Sol. (b) Since, elements of the given set is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Cardinal Number of a Finite Set The number of distinct elements contained in a finite set A is called its cardinal number and is denoted by n( A ). e.g. If A = {1, 3, 5, 7}, then n( A ) = 4. X

A set which does not contain any element is called an empty set or null set or void set and is denoted by the symbol φ or {}.

Example 5. The set {x : x ∈ N , 3 < x < 4} is equal to (a) {3, 4} (b) {4} (c) φ (d) {3}

Example 7. A set A = {x : x ∈ N and 2 < x < 4} is called (a) null set (b) infinite set (c) singleton set (d) None of the above Sol. (c) {3}; Since, x ∈ N and 2 < x < 4 ⇒ x = 3

Empty Set

X

The set {0} is not an empty set as it contains the element 0 (zero). The set {φ} is not a null set. It is a set containing one element φ. A set which has atleast one element is called a non-empty set.

Singleton Set

Sol. (c) Since, x is a positive integer and a divisor of 9. So, x

In this form, instead of listing all the elements of a set we describe the set by some special property (properties) satisfied by all of its elements and write it as A = {x : P ( x ) holds} = {x | x has the property P ( x )} and read it as ‘A is the set of all elements of x such that x has the property P’. The symbol ‘:’ or ‘|’ stands for ‘such that’.

●

Example 9. The number of elements in a set {x : x ∈all vowels} is (a) 6 (b) 26 (c) 5 (d) 2 Sol. (c) A = {a, e, i , o, u} So,

n( A) = 5

Infinite Set A set which is not finite is called an infinite set. In other words, a set in which the process of counting of elements does not come to an end is called an infinite set. X

Example 10. A set of all points in a plane is called (a) empty set (b) finite set (c) infinite set (d) None of these Sol. (c) Points in a plane cannot be counted.

Two sets A and B are said to be equal, if every element of A is also an element of B and every element of B is also an element of A. Thus, if x ∈ A ⇒ x ∈ B and y ∈ B ⇒ y ∈ A, then A and B are equal sets and we can write A = B . X

Example 11. The set A = {x : x ∈ N and 1 < x < 7} is equal to (a) {2, 3, 4, 5, 6} (b) {1, 2, 3, 4, 5, 6} (c) {2, 3, 4, 5, 6, 7} (d) {1, 2, 3, 4, 5, 6, 7}

We write it as B ⊂ A and read it as B is a proper subset of A. Thus, B is a proper subset of A, if every element of B is an element of A and there is atleast one element in A which is not in B. Observe that A ⊆ A i.e. every set is a subset of itself, but not a proper subset. X

Equivalent Sets Two finite sets A and B are said to be equivalent, if n( A ) = n( B ). Clearly, equal sets are equivalent, but equivalent sets need not to be equal. Equivalence of two sets is denoted by the symbol ‘~’. Thus, if A and B are equivalent sets, we write A ~ B which is read as ‘A is equivalent to B’. X

Example 12. The set {1, 2, 3, 4, 5} is equivalent to (a) {x : x ∈all vowels} (b) {2, 3, 4} (c) {1, 2, 4, 5} (d) {1, 2, 3}

Intervals as Subsets of R

Let a, b ∈ R and a < b, then (i) An open interval denoted by ( a, b) is the set of all real numbers such that ( a, b) = {x ∈ R : a < x < b} (ii) A closed interval denoted by [ a, b] is the set of all real numbers such that [ a, b] = {x ∈ R : a ≤ x ≤ b} (iii) Interval closed at one end and open at the other are given by [ a, b) = {x ∈ R a ≤ x < b} and ( a, b] = {x ∈ R : a < x ≤ b} X

∴ X

The set B is said to be subset of set A, if every element of set B is also an element of set A. Symbolically, we write it as, B ⊆ A or A ⊇ B , where A is superset of B. (i) B ⊆ A is read as B is contained in A or B is subset of A or A is superset of B. (ii) A ⊇ B is read as A contains B or B is a subset of A. Evidently, if A and B are two sets such that A ∈ B ⇒ x ∈ A, then B is subset of A. The symbol ‘⇒’ stands for ‘implies’. We read it as ‘x belongs to B’ implies that ‘x belongs to A’. X

Example 13. The set A = {1, 2, 3} is the subset of (a) {1, 2, 4, 5} (b) {1, {2, 3}, 4, 5} (c) {1, 2, 3, 7, 8} (d) {{1, 2}, 3, 4} Sol. (c)

Proper Subset The set B is said to be a proper subset of set A, if every element of set B is an element of A, whereas every element of A is not an element of B.

Example 15. The subset of R as intervals {x : x ∈ R , − 12 < x < − 10} is (a) ( −12, − 10] (b) [ −12, − 10] (c) ( −12, − 10) (d) [ −12, − 10) Sol. (c) The given subset is belong to open interval.

Sol. (a)

Subset and Superset

Example 14. The set {1, 2} is the proper subset of (a) {1, 2} (b) {1, 3, 4} (c) {1, 2, 3} (d) {{1, 2}, 3, 4} Sol. (c)

Sol. (a) Whenever, we have to show that two sets A and B are equal, show that A ⊆ B and B ⊆ A, then A = B.

1 Sets

Equal Sets

{ x : x ∈ R, − 12 < x < − 10} is (−12, − 10).

Example 16. Let R be set of points inside a rectangle of sides a and b ( a, b >1) wth two sides along the positive direction of X-axis and Y-axis, then (a) R = {( x, y) : 0 ≤ x ≤ a, 0 ≤ y < b} (b) R = {( x, y) : 0 ≤ x < a, 0 ≤ y ≤ b} (c) R = {( x, y) : 0 ≤ x ≤ a, 0 < y < b} (d) R = {( x, y) : 0 < x < a, 0 < y < b} Sol. (d) Since, ( x, y) both lies in first quadrant. ∴ x, y > 0 and less than side a and b because R lies inside the rectangle. ∴ R = {( x, y) : 0 < x < a, 0 < y < b}

Power Set The set formed by all the subsets of a given set A is called the power set of A. It is usually denoted by P ( A ). X

Example 17. If set A = {1, 2, 3}, then P ( A ) is equal to (a) {φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}} (b) {{1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}} (c) {φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}, {1, 2, 3}} (d) {{1}, {2}, {3}, {1, 2, 3}} Sol. (c)

3

Objective Mathematics Vol. 1

1

Some Results on Subsets (i) Every set is a subset of itself. (ii) The empty set is a subset of every set. (iii) The total number of subset of finite set containing n elements is 2 n .

Operations on Sets Now, we introduce some operations on sets to construct new sets from the given ones.

i.

Comparable Sets Two sets A and B are said to be comparable, if one of them is a subset of the other i.e. either A ⊆ B or B ⊆ A. X

U

Example 18. The set {1, 2, 3} is comparable with (a) {1, 2, 4, 5} (b) {{1, 2}, 4, 3} (c) {{1, 3}, 2, 4} (d) {1, 2, 3, 4}

A

Universal Set In any discussion of set theory, there is always a set that contains all the sets under consideration i.e. it is a superset of each of the given sets. Such a set is called the universal set and is denoted by U .

Example 19. Which of the following set is the universal set of sets A = {2, 4, 5}, B = {1, 3, 5}? C = {3, 5, 7, 11}, D = {2, 4, 8, 10} (a) {1, 2, 3, 5, 6, 7, 8, 9, 10} (b) {1, 2, 3, 4, 5, 6, 7, 8, 10} (c) {1, 2, 3, 4, 5, 7, 8, 10, 11} (d) {1, 2, 3, 4, 6, 7, 8, 10, 11}

X

ii.

Venn Diagram Venn diagrams are the diagrams which represent the relationship between sets.

Intersection of sets The intersection of two sets A and B, denoted by A ∩ B is the set of all elements, common to both A and B. U

Example 20. The set of natural numbers is a subset of whole numbers which is a subset of integers. Then, correct representation of it through Venn diagram is U

A

(a)

N

(b)

Z

W

N Z

U

(c)

N

ZW

(d) None of these X

Sol. (a)

U

W N

Z

B

Thus, A ∩ B = {x : x ∈ A and x ∈ B } Clearly, x ∈A ∩ B ⇒ x ∈ A and x ∈ B and x ∉ A ∩ B ⇒ x ∈ A or x ∈ B In the figure, the shaded part represents A ∩ B . It is evident that A ∩ B ⊆ A, A ∩ B ⊆ B .

U

W

4

Example 21. If two sets A = {1, 2, 3} and B = {1, 3, 5, 7}, then A ∪ B is equal to (a) {1, 2, 3, 7} (b) {1, 2, 3, 5, 7, 8} (c) {1, 2, 3, 5, 7} (d) {1, 2, 3, 4, 5, 6, 7} Sol. (c)

Sol. (c)

X

B

Thus, A ∪ B = {x : x ∈ A or x ∈ B } Clearly x ∈A ∪ B ⇒ x ∈A or x ∈B and x∉A∪B ⇒ x ∉ A and x ∉ B In the figure, the shaded part represents, A ∪ B . It is evident that A ⊆ A ∪ B , B ⊆ A ∪ B .

Sol. (d) Since, {1, 2, 3} is the subset of {1, 2, 3, 4}.

X

Union of sets The union of two sets A and B, denoted by A ∪ B is the set of all those elements, each one of which is either in A or in B or in both A and B.

Example 22. If two sets A = {1, 2, 3, 4} and B = {2, 4, 5}, then A ∩ B is equal to (a) {1, 3} (b) {2, 4} (c) {3, 4} (d) {1, 6} Sol. (b) Since, 2 and 4 are the common elements in sets A and B.

v.

Disjoint sets Two sets A and B are said to be disjoint, if A ∩ B = φ. If A ∩ B ≠ φ, then A and B are said to be intersecting or overlapping sets.

Symmetric difference of two sets The symmetric difference of two sets A and B is the set ( A − B ) ∪ ( B − A ) and is denoted by A∆B . U

U

A–B A

A

X

X

B

Example 25. If set A = {1, 3, 5, 7, 9} and set B = {2, 3, 5, 7, 11}, then A ∆ B is equal to (a) {3, 5, 7} (b) {1, 2} (c) {9, 11} (d) {1, 2, 9, 11} Sol. (d) A ∆ B = ( A − B) ∪ (B − A) = {1, 9} ∪ {2, 11} = {1, 2, 9, 11}

vi.

Sol. (a)

iv.

B–A

Thus, A∆B = ( A − B ) ∪ ( B − A ) = {x : x ∉ A ∩ B } The shaded part represents A ∆B .

B

Example 23. Given that, A = {1, 2, 3, 4, 5, 6} B = {7, 8, 9, 10, 11} C = {6, 8, 10, 12, 14}. Which of the following pair of sets are disjoint sets? (a) A and B (b) B and C (c) C and A (d) None of the above

1 Sets

iii.

Difference of sets If A and B are two sets, then their difference A − B is the set of all those elements of A which do not belong to B.

Complement of a set Let U be the universal set and A ⊂ U , then the complement of A, denoted by A′ or U − A is defined as A ′ = {x : x ∈U and x ∉ A} Clearly, x ∈ A ′ ⇔ x ∉ A The shaded part represents A′ . A'

U

U A–B

A

A

B X

Thus, A − B = {x : x ∈ A and x ∉ B } Clearly, x ∈A − B ⇒ x ∈ A and x ∉ B In the figure, the shaded part represents A − B . Similarly, the difference B − A is the set of all those elements of B that do not belong to A i.e. U B–A

A

B

B − A = {x : x ∈ B and x ∉ A} X

Example 24. Given that two sets A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 7, 11}, then A − B is (a) {1, 9} (b) {2, 11} (c) {3, 5, 7} (d) {1, 2, 3, 5, 7, 9, 11} Sol. (a) {1, 9}, since 1, 9 ∉B

Example 26. Given that U = {x : x is a letter in English alphabet} and A = {x : x is a vowel}, then A′ is equal to (a) φ (b)U (c) {x : x ∈ all consonants} (d) None of these Sol. (c)

vii.

Some results on complement The following results are the direct consequences of the definition of the complement of the set. (a) U ′ = φ (b) φ′ = {x ∈U : x ∉φ } = U (c) ( A ′ ) ′ = {x ∈U : x ∉ A ′ } = {x ∈U : x ∈ A} = A (d) A ∩ A ′ = {x ∈U : x ∈ A} ∩ {x ∈U : x ∉ A} = φ (e) A ∪ A ′= {x ∈U : x ∈ A} ∪ {x ∈U : x ∉ A} = U

Laws of Algebra of Sets In this article, we shall state and prove some fundamental laws of algebra of sets. 1. Idempotent laws For any set A, we have (i) A ∪ A = A (ii) A ∩ A = A

5

Objective Mathematics Vol. 1

1

2. Identity laws For any set A, we have φ and U as identity elements for union and intersection, respectively. Proof (i) A ∪ φ = {x : x ∈ A or x ∈φ} = {x : x ∈ A} = A (ii) A ∩ U = {x : x ∈ A and x ∈U } = {x : x ∈ A} = A 3. Commutative laws For any two sets A and B, we have (i) A ∪ B = B ∪ A (ii) A ∩ B = B ∩ A i.e. union and intersection are commutative. Proof Recall that two sets X and Y are equal iff X ⊆ Y if every element of X belongs to Y . (i) Let x be an arbitrary element of A ∪ B . Then, x ∈A ∪ B ⇒ x ∈ A or x ∈ B ⇒ x ∈ B or x ∈ A ⇒ x ∈B ∪ A ∴ A∪B ⊆B ∪ A Similarly, B ∪ A⊆A∪B Hence, A∪B =B ∪ A (ii) Let x be an arbitrary element of A ∩ B . Then, x ∈A ∩ B ⇒ x ∈ A and x ∈ B ⇒ x ∈ B and x ∈ A ⇒ x ∈B ∩ A A∩B ⊆B ∩ A Similarly, B ∩ A⊆A∩B Hence, A∩B =B ∩ A 4. Associative laws If A, B and C are any three sets, then (i) ( A ∪ B ) ∪ C = A ∪ ( B ∪ C ) (ii) A ∩ ( B ∩ C ) = ( A ∩ B ) ∩ C i.e. union and intersection are associative.

6

Proof (i) Let x be an arbitrary element of ( A ∪ B ) ∪ C . Then, x ∈( A ∪ B ) ∪ C ⇒ x ∈ ( A ∪ B ) or x ∈C ⇒ ( x ∈ A or x ∈ B ) or x ∈C ⇒ x ∈ A or ( x ∈ B or x ∈C ) ⇒ x ∈ A or x ∈ ( B ∪ C ) ⇒ x ∈ A ∪ (B ∪ C ) ∴ ( A ∪ B ) ∪ C ⊆ A ∪ (B ∪ C ) Similarly, A ∪ ( B ∪ C ) ⊆ ( A ∪ B ) ∪ C Hence, ( A ∪ B ) ∪ C = A ∪ (B ∪ C )

(ii) Let x be an arbitrary element of A ∩ ( B ∩ C ). Then, x ∈ A ∩ (B ∩ C ) ⇒ x ∈ A and x ∈ ( B ∩ C ) ⇒ x ∈ A and ( x ∈ B and x ∈C ) ⇒ ( x ∈ A and x ∈ B ) and x ∈C ⇒ x ∈ ( A ∩ B ) and x ∈C ⇒ x ∈( A ∩ B ) ∩ C ∴ A ∩ (B ∩ C ) ⊆ ( A ∩ B ) ∩ C Similarly, ( A ∩ B ) ∩ C ⊆ A ∩ ( B ∩ C ) Hence, A ∩ (B ∩ C ) = ( A ∩ B ) ∩ C 5. Distributive laws If A, B and C are any three sets, then (i) A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C ) (ii) A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) i.e. union and intersection are distributive over intersection and union, respectively. Proof (i) Let x be an arbitrary element of A ∪ ( B ∩ C ). Then, x ∈ A ∪ (B ∩ C ) ⇒ x ∈ A or x ∈ ( B ∩ C ) ⇒ x ∈A or ( x ∈ B and x ∈C ) ⇒ ( x ∈ A or x ∈ B ) and ( x ∈ A or x ∈C ) ⇒ x ∈( A ∪ B ) and x ∈( A ∪ C ) ⇒ x ∈ (( A ∪ B ) ∩ ( A ∪ C )) ∴ A ∪ (B ∩ C ) ⊆ ( A ∪ B ) ∩ ( A ∪ C ) Similarly,( A ∪ B ) ∩ ( A ∪ C ) ⊆ A ∪ ( B ∩ C ) Hence, A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C ) (ii) Let x be an arbitrary element of A ∩ ( B ∪ C ). Then, x ∈ A ∩ ( B ∪ C ) ⇒ x ∈A and x ∈ (B ∪ C ) ⇒ x ∈A and ( x ∈ B or x ∈C ) ⇒ ( x ∈ A and x ∈ B ) or

( x ∈ A and x ∈C )

⇒

x ∈( A ∩ B )

or

x ∈( A ∩ C )

⇒

x ∈( A ∩ B ) ∪ ( A ∩ C )

∴

A ∩ (B ∪ C ) ⊆ ( A ∩ B ) ∪ ( A ∩ C )

Similarly, ( A ∩ B ) ∪ ( A ∩ C ) ⊆ A ∩ ( B ∪ C ) Hence,

A ∩ (B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C )

Proof (i) Let x be an arbitrary element of ( A ∪ B ) ′. Then, x ∈ ( A ∪ B )′

Formulae to Solve Practical Problems on Union and Intersection of Sets Let A, B , C be any finite sets and U be the finite universal sets, then (i) n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B )

⇒

x ∉ (A ∪ B)

⇒

x∉A

and

x ∉B

(iii) n ( A − B ) = Number of elements which belong to only A of sets A and B

⇒

x ∈ A′

= n( A ) − n( A ∩ B ) = n( A ∪ B ) − n( B )

and

x ∈B ′

⇒

x ∈ A′ ∩ B ′

∴

(ii) If ( A ∩ B ) = φ, then n ( A ∪ B ) = n ( A ) + n ( B )

(iv) n ( A∆B ) = Number of elements which belong to exactly one of A or B

( A ∪ B )′ ⊆ A′ ∩ B ′

= n ( A ) + n ( B ) − 2n ( A ∩ B ) (v) n ( A ∪ B ∪ C ) = n ( A ) + n ( B ) + n (C )

Again, let y be an arbitrary element of A ′ ∩ B ′ . Then, y ∈ A ′ ∩ B ′ ⇒ and

y ∈ A′ y ∈B ′

⇒

y∉ A

and

y∉B

⇒

y∉ A ∩ B

⇒

− n ( A ∩ B ) − n (B ∩ C ) − n ( A ∩ C ) + n (A ∩ B ∩ C)

y ∈ ( A ∪ B )′

∴

A′ ∩ B ′ ⊆ ( A ∪ B )′

Hence,

( A ∪ B )′ = A′ ∩ B ′

(ii) Let x be an arbitrary element of ( A ∩ B ) ′ . Then, x ∈ ( A ∩ B )′ ⇒ x ∉ (A ∩ B) ⇒

x∉A

and

x ∉B

⇒

x ∈ A′

or

x ∈B ′

⇒

x ∈ A′ ∪ B ′

(vi) Number of elements in exactly two of the sets A, B , C = n ( A ∩ B ) + n ( B ∩ C ) + n (C ∩ A ) − 3n ( A ∩ B ∩ C ) (vii) Number of elements in exactly one of the sets A, B , C = n ( A ) + n( B ) + n (C ) − 2n ( A ∩ B ) − 2n ( B ∩ C ) − 2n ( A ∩ C ) + 3n ( A ∩ B ∩ C ) (viii) n( A ′ ∪ B ′ ) = n {( A ∩ B )′ } = n (U ) − n ( A ∩ B ) (ix) n ( A ′∩ B ′ ) = n {( A ∪ B ) ′ } = n (U ) − n ( A ∪ B )

Again, let y be an arbitrary element of A ′ ∪ B ′. Then,

Example 27. In a group of 50 people, 35 speak Hindi, 25 speak both English and Hindi and all the people speak atleast one of two languages. The number of people who speak ‘English’, ‘English but not Hindi’ are respectively (a) 40, 15 (b) 15, 40 (c) 35, 15 (d) 25, 15

⇒

y ∈ ( A ′∪B ′ )

Sol. (a) Let H denotes the set of people speaking Hindi and

⇒

y ∈ A′

or

y ∈B ′

⇒

⇒

X

( A ∩ B ) ′ ⊆ A ′∪B ′

y ∉ A or

E denotes the set of people speaking English. We have, n (H ∪ E ) = 50, n (H) = 35, n(H ∩ E ) = 25

y∉B

⇒

y∉ (A ∩ B)

⇒

y ∈ ( A ∩ B )′

∴

1 Sets

6. De-Morgan’s laws If A and B are any two sets, then (i) ( A ∪ B ) ′ = A ′ ∩ B ′ (ii) ( A ∩ B ) ′ = A ′ ∪ B ′

A′ ∪ B ′ ⊆ ( A ∩ B )′

Hence, ( A ∩ B ) ′ = A ′ ∪ B ′

Also, we have n (H ∪ E ) = n(H) + n(E ) − n (H ∩ E ) ⇒ 50 = 35 + n(E ) − 25 ⇒ n(E ) = 75 − 35 = 40 which is the number of people who speak English. We have to find the people speak only English i.e.

n (E − H) = n (H ∪ E ) − n(H) = 50 − 35 = 15

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Objective Mathematics Vol. 1

1

Work Book Exercise 1 Let A = {1, 2, 3}, B = { 3, 4} C = { 4, 5, 6}. Then, A ∪ (B ∩ C ) is a {3} c {1, 2, 5, 6}

b {1, 2, 3, 4} d {1, 2, 3, 4, 5, 6}

2 If A = { a, b, d , e }, B = {c , d , f , l , m} and C = { a, l , m, o}, then C ∩ ( A ∪ B) will be given by a b c d

{a, d , l, m} { b, c, l, m, o} { a, l, m} { a, b, c, d , f, l, m, o}

contain b 2 n elements d None of these

4 If A and B are two sets, then A ∩ ( A ∪ B) equals a A c φ

b B d None of these

5 If A = { φ, { φ}}, then the power set of A is a A c { φ, { φ}, {{ φ}}, A}

b { φ, { φ}, A} d None of these

6 Subtraction of integers is an operation that is a b c d

8

with entries 0 or 1 only. Let B be the subset of A consisting of all determinants with value 1. Let C be the subset of the set of all determinants with value −1. Then, a C is empty b B has as many elements as C c A = B∪C d B has twice as many elements as C

8 In a town of 840 persons, 450 persons read

3 If a set contains n elements, then power set will a n elements c n2 elements

7 Consider the set A of all determinants of order 3

commutative and associative not commutative but associative neither commutative nor associative commutative but not associative

Hindi, 300 read English and 200 read both, then the number of persons who read neither is a 210

b 290

c 180

d 260

9 Let S = { x : x is a positive multiple of 3 less than

100}, P = { x : x is a prime number less than 20}. Then, n(S ) + n(P ) is a b c d

34 41 33 30

10 If A and B are two disjoint sets, then n ( A ∪ B) is equal to a b c d

n ( A ) + n ( B) n ( A ) + n ( B) − n ( A ∩ B) n ( A ) − n ( B) n ( A ) ⋅ n ( B)

WorkedOut Examples Type 1. Only One Correct Option Ex 1. If A = {1, 3, 5, 7, 9, 11, 13, 15, 17}, B = {2, 4, ....., 18} and N is the universal set, then A ′ ∪ {( A ∪ B ) ∩ B ′ } is (a) A (c) B

(b) N (d) None of these

Now, A ∩ B = {2, 4 , 6, 8, 12, 20} ∩ {3, 6, 9, 12, 15} = {6, 12} B ∩ C = {3, 6, 9, 12, 15} ∩ {5, 10, 15, 20} = {15} C ∩ A = {5, 10, 15, 20} ∩ {2, 4 , 6, 8, 12, 20} = {20} and A ∩ B ∩C =φ A

Sol. We have, ( A ∪ B ) ∩ B′ = A − ( A ∩ B )

2 8

Since, A and B are disjoint sets. ∴ A∩ B =φ ⇒ ( A ∪ B ) ∩ B′ = A ∴ {( A ∪ B ) ∩ B′ } ∪ A′ = A ∪ A′ = N

(a)

A

2 8

(c)

U

A

(b)

2 8

4 6

12

1 20 5

5 10

5 10

C

C

4 6

12

20 15 5 10

3 B 9

B

3

U

9

U

(d) None of these

C

Sol. Given,

1 20 5 5 10

3

U

B

9

Hence, (b) is the correct answer.

Ex 2. Let A, B and C are subsets of universal setU . If A = {2, 4, 6, 8, 12, 20}, B = {3, 6, 9, 12, 15}, C = {5, 10, 15, 20} and U is the set of all whole numbers. Then, the correct Venn diagram is B 3 2 4 6 12 8 9 1 20 5

12

C

Hence, (b) is the correct answer.

A

4 6

A = {2, 4 , 6, 8, 12, 20} B = {3, 6, 9, 12, 15} C = {5, 10, 15, 20}

Ex 3. Let A and B have 3 and 6 elements, respectively. What can be minimum number of elements in A ∪ B ? (a) 3 (c) 9

(b) 6 (d) 18

Sol. Note that A ∪ B will contain minimum number of elements, if A ⊂ B So that, n ( A ∪ B) = n ( B) = 6 Hence, (b) is the correct answer.

Ex 4. Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than total number of subsets of second set. The values of m and n are (a) 7 and 6 (c) 5 and 1

(b) 6 and 3 (d) 8 and 7

Sol. Given, 2m − 2n = 56 By hit and trial method, we get m = 6 and n = 3 Hence, (b) is the correct answer.

Type 2. More than One Correct Option Ex 5. Let A and B be any two sets. Which of the following are correct? (a) ( A − B ) ∩ B = φ (b) ( A − B ) ∪ B = φ (c) ( A − B ) ∩ A = A ∩ B ′ (d) ( A − B ) ∪ B = A ∪ B

Sol. We have,

(a) ( A − B ) ∩ B = ( A ∩ B′ ) ∩ B [QA − B = A ∩ B′ ] = A ∩ ( B′ ∩ B ) [Q B′ ∩ B = φ] =A∩φ =φ (b) ( A − B ) ∪ B = ( A ∩ B′ ) ∪ B = ( A ∪ B) ∩ ( B′ ∪ B) [Q B ′ ∪ B = U ] = ( A ∪ B) ∩ U = A∪ B (c) ( A − B ) ∩ A = ( A ∩ B′ ) ∩ A = A ∩ B ′ (d) ( A − B ) ∪ B = A ∪ B Hence, (a), (c) and (d) are the correct answers.

Ex 6. If A and B are two sets such that n( A ) = 70, n( B ) = 60, n( A ∪ B ) =110, then (a) n( A − B ) = 40 (b) n( B − A ) = 40 (c) n( A ∩ B ) = 20 (d) n( A − B ) = n( A ∩ B ′ )

Sol. We have, n( A ) = 70, n( B ) = 60, n( A ∪ B ) = 110 ∴ ⇒

n( A ∪ B ) = n( A ) + n( B ) − n( A ∩ B ) n( A ∩ B ) = n( A ) + n( B ) − n( A ∪ B ) = 70 + 60 − 110 = 20 (a) n( A − B ) = n( A ∩ B′ ) = n( A ) − n( A ∩ B ) = 70 − 20 = 50 (b) n( B − A ) = n( B ∩ A′ ) = n( B ) − n( A ∩ B ) = 60 − 20 = 40 Hence, (b), (c) and (d) are the correct answers.

9

Objective Mathematics Vol. 1

1 Type 3. Assertion and Reason Directions (Ex. Nos. 7-9) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 7. Statement I If A and B are disjoint sets, then ( A ′ ∪ B ′ ) ′ = φ. Statement II ( A ∪ B ) ′ = A ′ ∩ B ′ [De-Morgan’s law] Sol. ( A′ ∪ B′ )′ = ( A′ )′ ∩ ( B′ )′ = A ∩ B = φ, as A and B are disjoint. ∴ Statement I is true, Statement II is true. Also, Statement II is a correct explanation of Statement I. Hence, (a) is the correct answer.

Ex 8. Statement I If A and B are any two non-empty sets, then ( A − B ) ∪ (B − A) = ( A ∪ B ) − ( A ∩ B ) Statement II ( A − B ) = A ∩ B ′ and ( A ∩ B ) ∪ C = ( A ∪ C ) ∩ (B ∪ C ) Sol. ( A − B ) ∪ ( B − A ) = ( A ∩ B′ ) ∪ ( B ∩ A′ )

[Q ( A − B ) = A ∩ B′] [distributive = {( A ∩ B′ ) ∪ B} ∩ {( A ∩ B′ ) ∪ A′ } law] = {( A ∪ B ) ∩ ( B′ ∪ B )} ∩ {( A ∪ A′ ) ∩ ( B′ ∪ A′ )} = {( A ∪ B ) ∩ U } ∩ {U ∩ B′∪ A′ )}[U = Universal set] = ( A ∪ B ) ∩ ( B′ ∪ A′ ) = ( A ∪ B ) ∩ ( A ∩ B )′ = ( A ∪ B) − ( A ∩ B) So, both statements are true and Statement II is correct explanation of Statement I. Hence, (a) is the correct answer.

Ex 9. Statement I If A ′ ∪ B = U , then A ⊆ B . Statement II A ⊂ B ↔ B C ⊂ A C Sol. A′ ∪ B = U , iff B = A ⇒ Also,

A=B ⇒ A⊆B A⊆B ⇒ BC ⊂ AC

[Q A ∪ A′ = U ]

So, both statements are true but Statement II is not the correct explanation of Statement I. Hence, (b) is the correct answer.

Type 4. Linked Comprehension Based Questions Passage (Ex. Nos. 10-12) In a town of 10000 families, it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C. 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three news papers, then Ex 10. The number of families which buy newspaper A only, is (a) 3300

(b) 4000

(c) 1400

(d) 2000

Ex 11. The number of families which buy newspaper none of A, B and C, is (a) 4400

(b) 4000

(c) 1000

(d) 3300

Ex 12. The number of families which buy newspaper B only, is (a) 1400 (c) 3000

(b) 2500 (d) 2000

Sol. (Ex. Nos. 10-12) Let P, Q and R be the sets of families

10

buying newspapers A, B and C respectively. Let U be the universal set. Then, n(U ) = 10000, n(P ) = 40% of 10000 = 4000 n(Q ) = 20% of 10000 = 2000, n(R ) = 10% of 10000 = 1000 n(P ∩ Q ) = 5% of 10000 = 500 n(Q ∩ R ) = 3% of 10000 = 300 n(R ∩ P ) = 4% of 10000 = 400 n(P ∩ Q ∩ R ) = 2% of 10000 = 200

10. The number of families which buy newspaper A only

= n (P ∩ Q′ ∩ R′ ) = n (P ∩ (Q ∪ R )′ ) = n(P ) − n [ P ∩ (Q ∪ R )] [Q n( A ∩ B′ ) = n( A ) − n( A ∩ B )] = n (P ) − n[ P ∩ Q ] ∪ (P ∩ R ) = n(P ) − [ n(P ∩ Q ) + n(P ∩ R ) − n(P ∩ Q ∩ R )] = 4000 − [ 500 + 400 − 200 ] = 3300 Hence, (a) is the correct answer.

11. The number of families which buy newspaper none of A , B and C = n(P ∪ Q ∪ R )′ = n(U ) − n(P ∪ Q ∪ R ) = n(U ) − [ n(P ) + n(Q ) + n(R ) − n(P ∩ Q ) − n(Q ∩ R ) − n(P ∩ R ) + n(P ∩ Q ∩ R )] = 10000 − [ 4000 + 2000 + 1000 − 500 − 300 − 400 + 200] = 10000 − 6000 = 4000 Hence, (b) is the correct answer.

12. The number of families which buy newspaper B only = n(P′ ∩ Q ∩ R′ ) = n(Q ∩ P ′ ∩ R ′ ) = n[ Q ∩ (P ∪ R )′ ] = n(Q ) − n[ Q ∩ (P ∪ R )] n(Q ) − n [(Q ∩ P ) ∪ (Q ∩ R )] = n(Q ) − [ n(P ∩ Q ) + n(Q ∩ R ) − n(P ∩ Q ∩ R )] = 2000 − [ 500 + 300 − 200 ]= 2000 − 600 = 1400 Hence, (a) is the correct answer.

B. Since, A ⊂ B ∴ A ∩ B′ = φ ⇒ ( A ∩ B′ )′ = φ′ ⇒ A′ ∪ B = U C. ( A − B ) ∪ A = A [Q ( A − B ) ⊂ A] D. ( A − B ) ∪ B = ( A ∩ B ′ ) ∪ B = ( A ∪ B) ∩ ( B′ ∪ B) = ( A ∪ B) ∩ U =A∪ B A → q; B → r, C → s; D → p

Ex 13. Match the following sets for all sets A and B. Column I

Column II

A.

If A ⊂ B, then A ∩ B′

p.

B.

If A ⊂ B, then A ′ ∪ B

q.

A∪B φ

C.

( A − B) ∪ A

r.

U

D.

( A − B) ∪ B

s.

A

Sol. A. Since, A ⊂ B, so there is no element in A, which does not belong to B. ∴ A − B = φ ⇒ A ∩ B′ = φ

1 Sets

Type 5. Match the Column

Type 6. Single Integer Answer Type Questions Ex 14. If U = {x : x 5 − 6x 4 + 11x 3 − 6x 2 = 0}, A = {x : x 2 − 5x + 6 = 0} and B = {x : x 2 − 3x + 2 = 0}, then n( A ∩ B ) ′ is equal to ________ .

Ex 15. If A = {x : x is an even natural number} and B = {x : x is a prime number }, then n( A ∩ B ) is equal to ________ . Sol. (1) We have,

Sol. (3) We have, U = {x : x 5 − 6x 4 + 11x 3 − 6x 2 = 0} = {0, 1, 2, 3} A = {x : x 2 − 5x + 6 = 0} = {2, 3} and B = {x : x 2 − 3x + 2 = 0} = {1, 2} ∴ A ∩ B = {2} Hence, ( A ∩ B )′ = U − ( A ∩ B ) = {0, 1, 2, 3} − {2} = {0, 1, 3} ∴ n( A ∩ B )′ = 3

A = {x : x is an even natural number} = {2, 4 , 6, 8, …} and

B = {x : x is a prime number} = {2, 3, 5, 7, 11, …}

A ∩ B = {2} n( A ∩ B ) = 1

11

Target Exercises Type 1. Only One Correct Option 1. The set {x : x is a positive integer less than 6 and 3 − 1 is an even number} in roster form is x

(a) {1, 2, 3, 4, 5} (c) {2, 4, 6}

(b) {1, 2, 3, 4, 5, 6} (d) {1, 3, 5}

2. The set A = {x : x 4 − x 3 − x 2 = 0and x ∈ N }represents (a) a null set (c) an infinite set

(b) a singleton set (d) None of these

3. If A = {1, 2, 3}, B = {x ∈ R : x 2 − 2x + 1 = 0}, C = {1, 2, 3} and D = {x ∈ R : x 3 − 6x 2 + 11x − 6 = 0}, then the equal sets are (a) A and B (c) A, B and C

(b) A and C (d) A, C and D

Ta rg e t E x e rc is e s

4. Which of the following is true? (a) a ∈{{a}, b} (c) {a, b} ⊂ {a,{b, c}}

(b) {b, c} ⊂ {a, {b, c}} (d) None of these

5. Let P be the set of prime numbers and S = { t : 2 − 1 is a prime}. Then, t

(a) S ⊂ P (c) S = P

(b) P ⊂ S (d) None of these

6. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}, then ( A ∪ B ) ′, ( A ′ ∩ B ′ ), ( A∆B ) is equal to (a) {1, 9}, {2, 8}, {3, 4, 5, 6, 7, 8} (b) {1, 9}, {1, 9}, {3, 4, 5, 6, 7, 8} (c) {1, 9}, {1, 9}, {5, 6, 7, 8} (d) None of the above

(b) A ∩ B

(c) A ∩ B

(d) A ∩ B

8. Which of the following is not correct? (a) A ⊆ A′ if and only if A = φ (b) A′ ⊆ A if and only if A = X , where X is the universal set (c) If A ∪ B = A ∪ C , then B = C (d) B = C if and only if A ∪ B = A ∪ C and A ∩ B = A ∩C

9. The set ( A ∪ B ∪ C ) ∩ ( A ∩ B ′ ∩ C ′ )′ ∩ C ′ is equal to (a) B ∩ C ′ (c) B′ ∩ C ′

(b) A ∩ C (d) None of these

10. If A ∪ B = A ∪ C and A ∩ B = A ∩ C, then

12

(a) B = C only when A ⊆ B (b) B = C only when A ⊆ C (c) B = C (d) None of the above

(a) b = cd (c) d = bc

where

(b) c = bd (d) None of these

12. If A = { θ : 2cos 2 θ + sin θ < 2} and

3π   π B = θ : ≤ θ ≤ , then A ∩ B is equal to 2  2

5π   π (a) θ : < θ <  6   2 3π   (b) θ : π < θ <  2   5π 3π   π (c) θ : < θ < or π < θ <  6 2   2 (d) None of the above

13. Let A = {x : x is a digit in the number 3591}, B = {x : x ∈ N , x < 10}. Which of the following is false? (a) A ∩ B = {1, 3, 5, 9} (c) B − A = {2, 4 , 6, 7, 8}

(b) A − B = φ (d) A ∪ B = {1, 2, 3, 5, 9}

14. Let A and B be two sets, such that A ∪ B = A. Then, A ∩ B is equal to (a) φ (c) A

(b) B (d) None of these

15. Let A and B be two sets, then ( A ∪ B )′ ∪ ( A ′ ∩ B ) is equal to

7. If A = {x : x is a multiple of 3} and B = { x : x is a multiple of 5}, then A − B is equal to (a) A ∩ B

11. If aN = { an : n ∈ N }, bN = { nb : n ∈ N }, cN = { cN : n ∈ N } and bN ∩ cN = d N , a, b, c ∈ N and b, c are coprime, then

(a) A′ (c) B′

(b) A (d) None of these

16. LetU be the universal set and A ∪ B ∪ C = U. Then, {( A − B ) ∪ ( B − C ) ∪ (C − A )}′ is equal to (a) A ∪ (b) A ∪ (c) A ∩ (d) A ∩

B ∪C (B ∩ C ) B ∩C (B ∪ C )

17. If A and B are two sets, ( A − B ) ∪ ( B − A ) ∪ ( A ∩ B ) is equal to (a) A ∪ B (c) A

then

(b) A ∩ B (d) B′

18. If A = { y : y = 2x, x ∈ N }, B = { y : y = 2x − 1, x ∈ N }, then ( A ∩ B )′ is (a) A

(b) B

(c) φ

(d) U

19. If n( A ) = 4 and n( B ) = 7, then the minimum and maximum value of n( A ∪ B ) are, respectively (a) 4 and 11 (c) 7 and 11

(b) 4 and 7 (d) None of these

(b) 700 (d) 900

21. Suppose A1 , A 2 , ... , A 30 are thirty sets each with five elements and B1 , B 2 , ... , B n are n sets each with three elements.

30

n

i=1

j =1

Let ∪ A i = ∪ B j = S . Assume that

each elements of S belongs to exactly 10 of the A i ’ s and exactly 9 of B j ’s. The value of n must be (a) 30 (c) 45

(b) 40 (d) 50

22. If there are three atheletic teams in a school, 21 are in the basketball team, 26 in hockey team and 29 in the football team. 14 play hockey and basketball, 15 play hockey and football, 12 play football and basketball and 8 play all the games. The total number of members is (a) 42 (c) 45

(b) 43 (d) None of these

I. 10% families own both a car and a phone. II. 35% families own either a car or a phone. III. 40000 families live in the town. Which of the above statements are correct? (a) I and II (c) II and III

(b) I and III (d) I, II and III

24. 20 teachers of a school either teach Mathematics or Physics. 12 of them teach Mathematics, while 4 teach both the subject. Then, the number of teachers teaching Physics only is (a) 12 (c) 16

(b) 8 (d) None of these

25. A survey shows that 63% of the Americans like cheese whereas 76% like apples. If x% of the Americans like both cheese and apples, then we have (a) x ≥ 39 (c) 39 ≤ x ≤ 63

(b) x ≤ 63 (d) None of these

Type 2. More than One Correct Option 26. If A = {1, 5, 7, 9}, B = {2, 5}, then A − B is equal to (a) {1, 7, 9} (b) {1, 7, 9, 2} (c) A ∩ B (d) A ∩ B

27. If X ∪ {1, 2} = {1, 2, 3, 5, 9}, then (a) the smallest set of X is {3, 5, 9} (b) the smallest set of X is {2, 3, 5, 9} (c) the largest set of X is {1, 2, 3, 5, 9} (d) the largest set of X is {2, 3, 4, 9}

1 Sets

(a) 600 (c) 800

23. In a certain town, 25% families own a phone and 15% own a car, 65% families own neither a phone nor a car. 2000 families own both a car and a phone. Consider the following statements in this regard :

28. In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows French = 17; English = 13; Sanskrit = 15; French and English = 09; English and Sanskrit = 4; French and Sanskrit = 5; English, French and Sanskrit = 3. The number of students who study (a) French only is 6 (b) Sanskrit only is 8 (c) French and Sanskrit but not English is 2 (d) atleast one of the three language is 30

Targ e t E x e rc is e s

20. Let X be the universal set for sets A and B. If n( A ) = 200, n( B ) = 300 and n( A ∩ B ) = 100, then n( A ′ ∩ B ′ ) is equal to 300, provided (X) is equal to

Type 3. Assertion and Reason Directions (Q. Nos. 29-31) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

29. Statement I If A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21}, B = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} and N is the universal set, then A ′ ∪ (( A ∪ B ) ∩ B ′ ) = A Statement II ( A ∪ B ) ∩ B ′ is always A. 30. Statement I If A, B and C are any three non-empty sets, then A − ( B ∪ C ) = ( A − B ) ∩ ( A − C ) Statement II A ∩ ( B ∆ C ) = ( A ∩ B ) ∆ ( A ∩ C ) 31. Statement I If A, B and C any three sets and U is universal set, such that n(U ) = 700, n( A ) = 200, and then n( B ) = 300 n( A ∩ B ) = 100, n( A ′∩ B ′ ) = 300. Statement II n( A ′ ∩ B ′ ) = n(U ) − n( A ∩ B )

13

Objective Mathematics Vol. 1

1

Type 4. Linked Comprehension Based Questions 33. The number of students only passed in Mathematics is

Passage (Q. Nos. 32-34) Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science, 6 in English and Mathematics, 7 in Mathematics and Science, 4 in English and Science, 4 in all the three passed.

(a) 5 (c) 3

34. The number of students only passed in more than one subject is

32. The number of students passed in English and Mathematics but not in Science is (a) 3

(b) 2

(c) 4

(b) 4 (d) 2

(a) 9 (c) 2

(d) 5

(b) 3 (d) 1

Type 5. Match the Column 35. Match the following sets for all sets A, B and C : Column I p.

A−B

B. [B′ ∪ ( B′ − A )]′

q.

A

A − ( A ∩ B)

r.

B

D. ( A − B) ∩ (C − B)

s.

(A ∩ C) − B

C.

Ta rg e t E x e rc is e s

Column II

A. [( A ′ ∪ B′ ) − A ]′

Type 6. Single Integer Answer Type Questions 36. If A = {a, b, c}, then the number of proper subsets of A is equal to________ . 37. If A = {x ∈ C : x 2 = 1} and B = {x ∈ C : x 4 = 1}, then number of elements in ( A − B ) is ________ . 38. If A and B are two sets containing 3 and 6 elements, respectively. The maximum number of elements in A ∪ B is equal to ________ . 39. A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men and only three men got medals in all the three sports, how many received medals in exactly two of three sports?

Entrances Gallery JEE Advanced/IIT JEE 1. Let P = {θ :sin θ − cos θ = 2 cos θ } and Q = {θ :sin θ + cos θ = 2 sin θ} be two sets. Then, (a) P ⊂ Q and Q − T ≠ φ (c) P ⊄ Q

(b) Q ⊄ P (d) P = Q

[2011]

2. Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to (a) 25 (c) 42

(b) 34 (d) 41

(a) N (c) X

(b) Y − X (d) Y

[2010]

JEE Main/AIEEE 3. Let A and B be two sets containing four and two elements respectively. Then, the number of subsets of the set A × B, each having atleast three elements, is [2015] (a) 219

(b) 256

14

(c) 275

(d) 510

4. If X = {4 − 3n − 1: n ∈ N }and Y = {9 ( n − 1) : n ∈ N }, where N is the set of natural numbers, then X ∪ Y is equal to [2014] n

5. If A, B and C are three sets, such that A ∩ B = A ∩ C and A ∪ B = A ∪ C, then [2009] (a) A = C (b) B = C (c) A ∩ B = φ (d) A = B

(a) 45

(b) 0

(c) 25

[Karnataka CET 2014] (a) D = {x : 0 < (x + 5) < 7} (b) B = {x : − 3 < x < 7} (c) E = {x : − 7 < x < 7} (d) C = {x : 13 < 2x < 4}

8. There is a group of 265 persons who like either singing or dancing or painting. In this group 200 like singing, 110 like dancing and 55 like painting. If 60 persons like both singing and dancing. 30 like both singing and painting and 10 like all three activities, then the number of persons who like only dancing and painting is [WB JEE 2014] (b) 20

(a) 500 ≤ p ≤ 1000 (c) 1000 ≤ p ≤ 1499 (e) 1000 ≤ p ≤ 1498

(d) 35

7. The set A = { x : 2x + 3 < 7} is equal to the set

(a) 10

11. If n( A ) = 1000, n( B ) = 500 and if n( A ∩ B ) ≥ 1 and [Kerala CEE 2012] n( A ∪ B ) = p, then

(c) 30

(d) 40

1 2  9. Let X n =  z = x + iy : z <  for all integers n > 1. n  ∞

Then, ∩ X n is (a) a singleton set (b) not a finite set (c) an empty set (d) an finite set with more than one element

13. There are 100 students in a class. In the examination, 50 of them failed in Mathematics, 45 failed in Physics, 40 failed in Biology and 32 failed in exactly two of the three subjects. Only one student passed in all the subjects. Then, the number of students failing [WB JEE 2012] in all the three subjects is (a) 12 (c) 2

(b) 4 (d) Cannot be determined

14. The set A = {x : x ∈ R , x 2 = 16 and 2x = 6} is equal to (a) φ

(b) {14, 3, 4} (c) {3}

(a) A′ (c) B′

U

(b) A (d) None of these

16. Let A = {1, 2}, B = {{1}, {2}}, C = {{1, 2}}. Then which of the following relation is true? [J&K CET 2011] (a) A = B (c) A ∈ C

B

(c) B − A

[BITSAT 2011] (d) {4}

15. Let A and B be two sets, then ( A ∪ B )′ ∩ ( A ′ ∩ B ) is [GGSIPU 2011] equal to

[Kerala CEE 2014]

(a) A ∩ B (b) A ∪ B (e) ( A − B ) ∪ ( B − A )

[OJEE 2012] (b) 19 (d) None of these

(a) 18 (c) 20

10. The shaded region in the figure represents

A

(b) 1001 ≤ p ≤ 1498 (d) 999 ≤ p ≤ 1499

12. Out of 64 students, the number of students taking Mathematics is 45 and number of students taking both Mathematics and Biology is 10. Then, the number of students taking only Biology is

[WB JEE 2014]

n=1

(d) A − B

1

(b) B ⊆ C (d) D ⊂ C

Targ e t E x e rc is e s

6. In a class of 60 students, 25 students play cricket and 20 students play tennis and 10 students both the games, then the number of students who play neither is [Karnataka CET 2014]

Sets

Other Engineering Entrances

17. 25 people employed for programme A, 50 people for programme B, 10 people for both. So, number of employee employed for only A is [OJEE 2011] (a) 15

(b) 20

(c) 35

(d) 40

Answers Work Book Exercise 1. (b)

2. (c)

3. (b)

4. (a)

5. (c)

6. (c)

7. (b)

8. (b)

9. (b)

10. (a)

10. (c)

Target Exercises 1. (a)

2. (a)

3. (d)

4. (d)

5. (a)

6. (b)

7. (b)

8. (c)

9. (a)

11. (c)

12. (c)

13. (d)

14. (b)

15. (a)

16. (c)

17. (a)

18. (d)

19. (c)

20. (b)

21. (c)

22. (b)

23. (c)

24. (b)

25. (c)

26. (a,d)

27. (a,c)

28. (a,c,d)

29. (c)

30. (b)

31. (c)

32. (b)

33. (c)

34. (a)

35. (*)

36. (7)

37. (0)

38. (9)

39. (9)

8. (a)

9. (a)

*A → q; B → r; C → p; D → s

Entrances Gallery 1. (d)

2. (d)

3. (a)

4. (d)

5. (b)

6. (c)

7. (a)

11. (c)

12. (b)

13. (c)

14. (a)

15. (d)

16. (c)

17. (a)

10. (e)

15

Explanations Target Exercises 1. Since, 3 x − 1 is an even number for all x ∈ Z + . So, the given set in roster form is {1, 2, 3, 4, 5}.

2. Given equation can be written as

1± 5 2 ∴There is no natural value of x, which satisfies x4 − x3 − x2 = 0 x 2 ( x 2 − x − 1) = 0 ⇒

x = 0,

7. A − B = A ∩ B is a general result. 8. Options (a) and (b) are trivially true. Option (c) is incorrect because if A = {2, 3, 4}, B = { 3}, C = { 4}, then A ∪ B = A ∪ C but B ≠ C. Option (d) is also correct.

9. ( A ∪ B ∪ C ) ∩ ( A ∩ B′ ∩ C′ )′ ∩ C′ = ( A ∪ B ∪ C ) ∩ ( A′ ∪ B ∪ C ) ∩ C′ = [( A ∩ A′ ) ∪ (B ∪ C )] ∩ C′ = (φ ∪ B ∪ C ) ∩ C′ = (B ∩ C′ ) ∪ (C ∩ C′ ) = (B ∩ C′ ) ∪ φ = B ∩ C′

3. A = {1, 2, 3} B = { x ∈ R : x 2 − 2 x + 1 = 0} = {1} C = {1, 2, 3} D = { x ∈ R : x 3 − 6 x 2 + 11x − 6 = 0}= {1, 2, 3} ∴A, C and D are equal.

4. a is not an element of {{ a}, b}

Ta rg e t E x e rc is e s

∴ a ∈{{ / a}, b} { b, c } is the element of { a, { b, c }} ∴ { b, c } ∈ { a, { b, c }} b ∈{ a, b} but b ∈{ / a, { b, c }} ∴ { a, b} ⊄ { a, { b, c }}

5. Let x ∈ / P

⇒ x is a composite number. Let us now assume that x ∈ S ⇒ 2 x − 1 = m [where, m is a prime number] ⇒ 2x = m + 1 which is not true for all composite numbers, say for x = 4 because 2 4 = 16 which cannot be equal to the sum of any prime number m and 1. Thus, we arrive at a contradiction ⇒ x∈ /S So, S ⊂P U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7} Now, A′ = U − A = {1, 2, 3, 4, 5, 6, 7, 8, 9 } − {2, 4, 6, 8} = {1, 3, 5, 7, 9} and B′ = U − B = {1, 2, 3, 4, 5, 6, 7, 8, 9} − {2, 3, 5, 7} = {1, 4, 6, 8, 9} (i) A ∪ B = {2, 4, 6, 8} ∪ {2, 3, 5, 7} = {2, 3, 4, 5, 6, 7, 8} ∴ ( A ∪ B)′ = U − A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9} − {2, 3, 4, 5, 6, 7, 8} = {1, 9} (ii) ( A′ ∩ B′ ) = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9} (iii) Now, A − B = {2, 4, 6, 8} − {2, 3, 5, 7} = { 4, 6, 8} and B − A = {2, 3, 5, 7} − {2, 4, 6, 8} = { 3, 5, 7} ∴ A ∆ B = ( A − B ) ∪ ( B − A) = { 4, 6, 8} ∪ { 3, 5, 7} = { 3, 4, 5, 6, 7, 8}

6. Given,

16

10. Let x ∈ B ⇒ x ∈ A ∪ B ⇒ x ∈ A ∪ C Case I ∴ ∴ Case II ∴ Similarly, ∴

x ∈A x ∈ A ∩ B or x ∈ A ∩ C or x ∈ C B ⊆C x ∈C x ∈ B ⇒ x ∈ C or B ⊆ C C⊆B B =C

11. We have, bN = { nb : n ∈ N}, cN = {cn : n ∈ N} and dN = {dn : n ∈ N} Also, bN ∩ cN = dN ∴ bc ∈ bN ∩ cN or bc ∈ dN [as b and c are coprime] ∴ bc = d π 3π 2 12. Let 2 cos θ + sin θ ≤ 2 and ≤ θ ≤ 2 2 ⇒ 2 − 2 sin 2 θ + sin θ ≤ 2 ⇒ 2 sin 2 θ − sin θ ≥ 0 ⇒ sin θ(2 sin θ − 1) ≥ 0 π 5π 3π or π ≤ θ ≤ ≤θ ≤ ⇒ 2 6 2 5π 3π   π A ∩ B = θ : ≤ θ ≤ or π ≤ θ ≤ ∴  6 2   2

13. We have, A = {1, 3, 5, 9} and B = {1, 2, 3, 4, 5, 6, 7, 8, 9} Now, find A ∪ B, A ∩ B, A − B, B − A ∴ Option (d) is false.

14. A ∪ B = A ⇒ B ⊆ A

⇒

A∩B=B

15. ( A ∪ B)′ ∪ ( A ′ ∩ B) = ( A′ ∩ B ′ ) ∪ ( A′ ∩ B) = ( A′ ∪ A′ ) ∩ ( A′ ∪ B) ∩ (B′ ∪ A′ ) ∩ (B′ ∪ B) = A′ ∩ { A′ ∪ (B ∩ B′ )} ∩ U = A′ ∩ ( A′ ∪ φ ) ∩ U = A′ ∩ A′ ∩ U = A′ ∩ U = A′

16. ( A − B) ∪ (B − C ) ∪ (C − A) is represented by the shaded portion in the figure. The unshaded portion is A ∩ B ∩ C. A

B

C

∴

{( A − B) ∪ (B − C ) ∪ (C − A)}′ = A ∩ B ∩ C

18. A = {2, 4, 6, 8, K }, B = {1, 3, 5, 7, K } A∩B=φ ( A ∩ B )′ = U

19. n( A ∪ B) is minimum when A ⊆ B. In this case, A∪B=B and we have n( A ∪ B) = n(B) = 7 n( A ∪ B) is maximum when A ∩ B = φ. In this case, n( A ∪ B) = n( A) + n(B) = 4 + 7 = 11

20. We have, n( A ∪ B) = n( A) + n(B) − n( A ∩ B) ∴ n( A ∪ B) = 200 + 300 − 100 = 400 Also, n( A ′ ∩ B′ ) = n(( A ∪ B)′ ) = n( X ) − n( A ∪ B) ⇒ 300 = n( X ) − 400 ⇒ n( X ) = 700 30

21. Since, S = ∪ A i and each element of S is in 10 A i’s. i =1

We have, n(S ) =

1 30 1 n( A i ) = (30 × 5) = 15 ∑ 10 i = 1 10

30

Also, S = ∪ B j and each element of S is in 9 B j’s. j =1

We have, ⇒

n(S ) =

1 n n(B j ) 9∑ j =1

1 15 = (n × 3) ⇒ n = 45 9

22. Let B, H, F be the sets of members in the basketball team, hockey team, football team, respectively. ∴ n(B) = 21, n(H ) = 26, n(F ) = ,29 , n(H ∩ B) = 14, n(H ∩ F ) = 15 n(F ∩ B) = 12, n(B ∩ H ∩ F ) = 8 ∴ n(B ∪ H ∪ F ) = n(B) + n(H ) + n(F ) − n(B ∩ H ) − n(H ∩ F ) − n(B ∩ F ) + n(B ∩ H ∩ F ) = 21 + 26 + 29 − 14 − 15 − 12 + 8 = 43

23. Let X, P, C denote the sets of all families, families owning phone, families owning car, respectively. Let total number of families be k. k×5 k k Since, n(P ∩ C ) = = ⇒ 2000 = 20 100 20 ⇒ k = 40000 ∴ Statement III is correct. 2000 Now, n(C ∩ P ) = 2000 = × 100% = 5% 40000 ∴ Statement I is incorrect.

n(P) n(C) = 25% – 5% 5% = 15% – 5% = 20% = 10%

n (P∩C) n (P ∪C) = 65%

∴ It is clear from the Venn diagram that Statement II is correct. ∴ The correct answer is (c).

24. n(M ∪ P ) = 20, n(M ) = 12, n(M ∩ P ) = 4 Q n(M ∪ P ) = n(M ) + n(P ) − n(M ∩ P ) ⇒ 20 = 12 + n(P ) − 4 ∴ n(P ) = 12 So, the required number = n(P ) − n(M ∩ P ) = 12 − 4 = 8

1 Sets

( A − B ) ∪ ( B − A) ∪ ( A ∩ B ) = A ∪ B

25. n( X ) = 100, n(C ) = 63, n( A) = 76, n(C ∩ A) = x Now, n(C ∪ A) = n(C ) + n( A) − n(C ∩ A) ∴ n(C ∪ A) = 63 + 76 − x = 139 − x ∴ 139 − x ≤ 100 or x ≥ 39 Also, C ∩ A ⊆ C and C ∩ A ⊆ A ∴ n(C ∩ A) ⊆ n(C ) and n(C ∩ A) ⊆ n( A) ⇒ n(C ∩ A) ≤ 63 and n(C ∩ A) ≤ 76 ⇒ n(C ∩ A) ≤ 63 i.e. x ≤ 63 ∴ 39 ≤ x ≤ 63

26. Given, A = {1, 5, 7, 9}, B = {2, 5} A − B = Set of all those elements of A, which do not belong to B = {1, 7, 9} Also, A − B = A ∩ B is a general result. Hence, options (a) and (d) are correct.

27. Given,

X ∪ {1, 2} = {1, 2, 3, 5, 9}

The smallest set of X = {1, 2, 3, 5, 9} − {1, 2} = { 3, 5, 9} The largest set of X = The set which contains atleast elements 3, 5, 9 = {1, 2, 3, 5, 9}

28. Given, n(F ) = 17, n(E ) = 13, n(S ) = 15, n(F ∩ E ) = 9, n(E ∩ S ) = 4, n(F ∩ S ) = 5, n(E ∩ F ∩ S ) = 3 (a) Number of students study only French = n(F ) − n(F ∩ E ) − n(F ∩ S ) + n(F ∩ E ∩ S ) = 17 − 9 − 5 + 3 = 6

Targ e t E x e rc is e s

17. Draw the Venn diagram. From the Venn diagram,

(b) Number of students study only Sanskrit = n(S ) − n(S ∩ F ) − n(S ∩ E ) + n(F ∩ E ∩ S ) = 15 − 5 − 4 + 3 = 9 (c) n(F ∩ S ∩ E ) = n(F ∩ S ) − n(F ∩ E ∩ S ) = 5− 3=2 (d) ∴ n(F ∪ E ∪ S ) = n(F ) + n(E ) + n(S ) − n(F ∩ E ) − n(E ∩ S ) − n(S ∩ F ) + n(F ∩ E ∩ S ) = 17 + 13 + 15 − 9 − 4 − 5 + 3 = 30

29. In Statement I, A ∩ B = φ ⇒

( A ∪ B) ∩ B′ = A A′ ∪ (( A ∪ B) ∩ B′ ) = A ′ ∪ A = N Statement I is true and Statement II is false. Option (c) is true.

30. A − (B ∪ C ) = A ∩ (B ∪ C )′ = A ∩ (B′ ∩ C′ ) = ( A ∩ B′ ) ∩ ( A ∩ C′ ) = ( A − B) ∩ ( A − C ) Statement I is true. A ∩ (B ∆ C ) = A ∩ {(B − C ) ∪ (C − B)} = { A ∩ (B − C )} ∪ { A ∩ (C − B)} = {( A ∩ B) − ( A ∩ C )} ∪ {( A ∩ C ) − ( A ∩ B)} = ( A ∩ B) ∆ ( A ∩ C ) Statement II is true. Option (b) is true.

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Objective Mathematics Vol. 1

1

31. Since, n( A′ ∩ B′ ) = n(U ) − n( A ∪ B) = n(U ) − { n( A) + n(B) − n( A ∩ B)} = 700 − {200 + 300 − 100} = 300 ∴Statement I is true and Statement II is false. Option (c) is true. Solutions (Q. Nos. 32-34) Let U , E, M and S be denote the total number of students passed in English, passed in Mathematics and passed in Science, respectively. Here, n(U ) = 100, n(E ) = 15, n(M ) = 12, n(S ) = 8, n(E ∩ M ) = 6, n(M ∩ S ) = 7, n(E ∩ S ) = 4 and n(E ∩ M ∩ S ) = 4

C. A − ( A ∩ B) = A ∩ ( A ∩ B)′ = A ∩ ( A′ ∪ B′ ) = ( A ∩ A′ ) ∪ ( A ∩ B′ ) = φ ∪ ( A ∩ B′ ) = A ∩ B′ = A − B D. ( A − B) ∩ (C − B) = ( A ∩ B′ ) ∩ (C ∩ B′ ) = ( A ∩ C ) ∩ B′ = ( A ∩ C ) − B

36. We know that the total number of proper subsets of a finite set containing n elements is 2 n − 1. ∴ Number of proper subsets of A = 23 − 1 = 7

37. A = { −1, 1}, B = { −i , i , − 1, 1}

32. The number of students passed in English and Mathematics but not in Science = n(E ∩ M ∩ S ) = n(E ∩ M ) − n(E ∩ M ∩ S ) = 6 − 4 = 2

33. The number of students only passed in Mathematics = n(M ∩ E ∩ S ) = n(M ) − n(M ∩ E ) − n(M ∩ S ) + n(M ∩ E ∩ S ) = 12 − 6 − 7 + 4 = 16 − 13 = 3

Ta rg e t E x e rc is e s

34. The number of students only passed in more than one subject = n(M ∩ E ) + n(M ∩ S ) + n(S ∩ E ) − 2n (M ∩ E ∩ S ) = 6 + 7 + 4 − 2(4) = 17 − 8 = 9

35. A. [( A′ ∪ B′ ) − A]′ = [( A ∩ B)′ ∩ A′ ]′ = ( A ∩ B) ∪ A = A B. [B′ ∪ (B′ − A)]′ = B′ ∪ (B′ ∩ A′ )]′ = [B′ ∪ (B ∪ A)′ ]′ = [B ∩ (B ∪ A)] = B

A−B=φ n( A − B) = 0

38. The maximum number of elements in A ∪ B iff ∴

n( A ∩ B) = 0 n( A ∪ B) = n( A) + n(B) − n( A ∩ B) = 3+ 6= 9

39. We have, n(F ) = 38, n(B) = 15 ,n(C ) = 20, n(F ∪ B ∪ C ) = 58, n(F ∩ B ∩ C ) = 3 Now, n(F ∪ B ∪ C ) = n(F ) + n(B) + n(C ) − n(F ∩ B) − n(F ∩ C ) − n(B ∩ C ) + n(F ∩ B ∩ C ) ⇒ n(F ∩ B) + n(B ∩ C ) + n(F ∩ C ) = 38 + 15 + 20 + 3 − 58 = 18 Now, number of men who received medals in exactly two of the three sports = n(F ∩ B) + n(B ∩ C ) + n(F ∩ C ) − 3 n (F ∩ B ∩ C ) = 18 − 3 × 3 = 18 − 9 = 9

Entrances Gallery 1. Since, cos θ ( 2 + 1) = sin θ ⇒ and ⇒ ∴

tan θ = 2 + 1 sin θ ( 2 − 1) = cos θ 1 2 +1 tan θ = × = ( 2 + 1) 2 −1 2 +1 P =Q

2. Let A ∩ B = φ, A, B ⊂ S ∴Total number of unordered pairs of disjoint subsets of 34 + 1 S = = 41 2

3. Given, n( A) = 4, n(B) = 2

18

∴

⇒ n( A × B) = 8 Total number of subsets of set ( A × B) = 2 8 Number of subsets of set ( A × B) having no element (i.e. φ ) = 1 Number of subsets of set ( A × B) having one element = 8C1 Number of subsets of set ( A × B) having two elements = 8C2 ∴Number of subsets having atleast three elements = 2 8 − (1 + 8C1 + 8C2 ) = 2 8 − 1 − 8 − 28 = 2 8 − 37 = 256 − 37 = 219

4. We have, X = { 4n − 3n − 1 : n ∈ N} X = { 0, 9, 54, 243, K } Y = { 9 (n − 1) : n ∈ N} Y = { 0, 9, 18, 27, K } It is clear that X ⊂ Y ∴ X∪Y=Y

[put n = 1, 2, 3, K] [put n = 1, 2, 3, K]

5. Since, A ∩ B = A ∩ C and

A∪ B = A∪C ⇒ B =C

6. Let the number of students play cricket = C Number of students play tennis = T and total number of students = S ∴ n(S ) = 60, n(C ) = 25, n(T ) = 20 and n(C ∩ T ) = 10 Now, n(C ∪ T ) = n (C ) + n(T ) − n(C ∩ T ) = 25 + 20 − 10 = 35 ∴ The number of students who play neither game = n(C ∪ T )′ = n(S ) − n(C ∪ T ) = 60 − 35 = 25

7. Given, set A = { x :|2 x + 3| < 7} Now, ⇒ ⇒ ⇒ ⇒ ⇒

|2 x + 3| < 7 − 7 < 2x + 3 < 7 − 7 − 3 < 2x < 7 − 3 − 10 < 2 x < 4 − 5< x 0

Sol. (c) We have, f( x) =

Sol. (c) We have, f( x) = 2 x2 − 1 ⇒

4 − x2 2

X

Example 21. The domain for which the functions f ( x ) = 2x 2 − 1 and g ( x ) = 1 − 3x are equal, is 1  (a) {1, 2} (b) {−2, 2} (c) −2,  (d) {0, 1} 2 

1

Sol. (c) We have, f( x) =

⇒

X

y 1− y

y ≥0 1− y

y ≤ 0 ⇒ 0≤ y< 1 y−1

∴ Range of f = [0, 1)

Equal Functions Two functions f and g are said to be equal iff (i) domain of f = domain of g. (ii) codomain of f = codomain of g. (iii) f ( x ) = g ( x ) for every x belonging to their common domain. If two functions f and g are equal, then we write f = g. e.g. Let A = {1, 2}, B = {3, 6} and f : A → B given by f ( x ) = x 2 + 2 and g : A → B given by g ( x ) = 3x. Then, we observe that f and g have the same domain and codomain. Also, we have f (1) = 3 = g (1) and f (2) = 6 = g (2) Hence, f =g ∴ f and g are equal functions.

2 Fundamentals of Relation and Function

X

Classification of Functions Constant Function

Y

A function which does not (0) change as its parameters vary, i.e. the function whose rate of X′ X change is zero. In short, a Y′ constant function is a function that always gives or returns the same value. or Let k be a constant, then function f ( x ) = k , ∀ x ∈ R is known as constant function. Domain of f ( x ) = R and Range of f ( x ) = {k }

Polynomial Function The function y = f ( x ) = a 0 x n + a1 x n −1 + K + a n , where a 0 , a1 , a 2 ,…,a n are real coefficients and n is a non-negative integer, is known as a polynomial function. If a 0 ≠ 0, then degree of polynomial function is n. Domain of f ( x ) = R and range varies from function to function.

Rational Function If P ( x ) and Q ( x ) are polynomial functions, P (x ) is known as Q ( x ) ≠ 0, then function f ( x ) = Q (x ) rational function. Domain of f ( x ) = R − [ x : Q ( x ) = 0] and range varies from function to function.

Irrational Function The function containing one or more terms having non-integral rational powers of x is called irrational function. 5x 3/ 2 − 7x 1/ 2 e.g. y = f (x ) = x 1/ 2 − 1 Domain varies from function to function.

25

Objective Mathematics Vol. 1

2 Identity Function

Modulus Function Y

Function f ( x ) = x, ∀ x ∈ R is known as identity function. It is a straight line passing through origin and having slope unity. Domain of f ( x ) = R and range of f ( x ) = R

y=x

X′

X

O

Function y = f ( x ) = | x | is known as modulus function.  x, x ≥ 0  y = f (x ) = | x | =  −x, x < 0  Domain of f ( x ) = x ∈ R and range of f ( x ) = [0, ∞ )

Y′

Square Root Function

Y

The function that associates every positive real number x to + x is called the square root function, i.e. f (x ) = + x .

y = x, if x > 0

y = –x, if x < 0

Y y = √x

X′

X′

Y′

X

O

Properties of Modulus Function

Y′

Range of f ( x ) = [0, ∞ )

i.

Exponential Function

ii.

A function of the form f ( x ) = a x , a is a positive real number, is an exponential function. The value of the function depends upon the value of a. For 0 < a < 1, function is decreasing and for a >1, function is increasing. Domain of f ( x ) = R and range of f ( x ) = [0, ∞ ) Y

Y

y = ax

y = ax (0, 1)

X'

X

O

X

O

X′

Y′ 0 0) and a ≠1, is known as logarithmic function. Domain of f ( x ) = (0, ∞ ) and range of f ( x ) = R Y

For any real number x, x 2 = | x |.

Y 3

Y

2 1

(1, 0) X′

26

O

Y′ 0 0 

Let f( x) = y ∴

x + y = 1, y = 1, and y − x = 1, ∴ Graph of this function is

x< 0 x=0 x> 0

Y

y=1 x>0

1

(0, 1)

X′

X X′

y = –1 x 2 ), then 2 If the function f ( x ) = 2 f ( x + y ) + f ( x − y ) is equal to a 2 f( x )f( y) f( x ) c f( y)

3 Domain and range of f ( x ) = 28

respectively

b f( x )f( y) d None of these

| x − 3| are, x−3

a R, [−1, 1] c RT , R

b R − { 3}, {1, − 1} d None of these

4 The domain of the function f ( x ) = (2 − 2 x − x 2 ) is a − 3≤ x≤ 3 b ( −1 − 3 ) ≤ x ≤ ( −1 + 3 ) c −2 ≤ x ≤ 2 d ( −2 − 3 ) ≤ x ≤ ( −2 + 3 )

5 If f ( x ) = sin 2 x and the composite function

g { f ( x )} = |sin x |, then the function g( x ) is equal to a

x−1

b

x

c

x+1

d − x

WorkedOut Examples Type 1. Only One Correct Option Ex 1. The composite mapping fog of the maps f : R → R , f ( x ) = sin x, g : R → R , g ( x ) = x 2 is (a) sin x + x 2 (b) (sin x ) 2 (c) sin x 2 (d)

sin x x2

Sol. We have, ( fog ) (x ) = f (g (x )) = f (x 2 ) = sin x 2 Hence, (c) is the correct answer.

Ex 2. If R is a relation ‘ −x It is true for all R + . [R + means set of all positive real numbers] ∴Domain of the given function = R + Hence, (b) is the correct answer.

Sol. Given, f (x ) =

Type 2. More than One Correct Option Ex 10. Let f , g : R → R be defined, respectively by f ( x ) = x +1 and g ( x ) = 2x − 3. Then, (a) ( f + g ) ( x ) = 3x − 2 (b) ( f − g ) ( 2) = 6  f (c)   ( 2) = 3 (d) ( f − g ) ( x ) = 4 − x g

Sol. We have, f (x ) = x + 1and g (x ) = 2x − 3

( f + g ) (x ) = x + 1 + 2x − 3 = 3x − 2 ( f − g ) (x ) = x + 1 − 2x + 3 = 4 − x  f x+1   (x ) =  g 2x − 3 f 3 (2) = =3 g 4−3

( f − g ) (2) = 4 − 2 = 2 Hence, (a), (c) and (d) are the correct answers.

Ex 11. If n( A ) =10 and n(B ) = 5, then (a) n( A × B ) = 50 (b) n( A × B ) = n( B × A ) (c) number of relations from A to B is 2 10 (d) All of the above Sol. Given, n( A ) = 10, n( B ) = 5 n( A × B ) = n( A ) × n( B ) = 10 × 5 = 50 n( B × A ) = n( B ) × n( A ) = 5 × 10 = 50 ∴Total number of relations from A to B is 2n(A × B) = 250 Hence, (a) and (b) are the correct answers.

Type 3. Assertion and Reason Directions (Ex. Nos. 12-13) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 12. Let A = {1, 2, 3}, we define R1 = {(1, 3), (1, 4), (2, 1), (2, 2), (2, 3)} and R 2 = {(1, 1), (1, 2), (1, 3)}. Then, Statement I R1 is not a relation and R 2 is relation on A. Statement II If R is relation on A, then R ⊆ A × A. Sol. If R1 is a relation on A, then R1 ⊆ A × A 30

Here, R1 contains (1, 4) ∉ A × A ⇒ R1 ⊆/ A × A

∴ R1 is not a relation. Whereas R2 ⊂ A × A ⇒ R2 is relation. Hence, (a) is the correct answer.

1 , x ≠ 0, 1, then the 1− x graph of the function y = f ( f ( f ( x ))), x >1 is a straight line.

Ex 13. Statement I

Statement II ⇒

f (x ) =

If f ( x ) =

1 1− x

f [ f { f ( x )}] = x, ∀ x ∈ R

x −1 1 1 , ∀ x ∈ R − {0,1} = = 1 x 1 − f (x ) 1 − 1− x 1 1 ∴ f [ f { f (x )}] = = 1 − f { f (x )} 1 − x − 1 x = x , ∀ x ∈ R − {0, 1} ⇒ Statement I is true, Statement II is false. Hence, (c) is the correct answer.

Sol. f { f (x ) =

Sol. It is obvious that the relation R is ‘x is the square of y’.

Passage (Ex. Nos. 14-15) The figure shows a relation

In set builder form, R = {(x , y) : x is the square of y, x ∈ P , y ∈ Q}

R between the sets P and Q. Q 5 3 2 1 –2 –3 –5

P 9 4 25

Hence, (c) is the correct answer.

Ex 15. The relation R in roster form is (a) {(9, 3), (4, 2), (25, 5)} (b) {(9, −3), (4, −2), (25, −5)} (c) {( 9, − 3), ( 9, 3), (4, − 2), (4, 2), ( 25, − 5), ( 25, 5)} (d) None of the above

Ex 14. The relation R in set builder form is

(a) {( x, y ) : x ∈ P , y ∈ Q} (b) {( x, y ) : x ∈ Q , y ∈ P} (c) {( x, y ) : x is the square of y, x ∈ P , y ∈ Q} (d) {( x, y ) : y is the square of x, x ∈ P , y ∈ Q}

Sol. In roster form,

2 Fundamentals of Relation and Function

Type 4. Linked Comprehension Based Questions

R = {(9, 3), (9, − 3), (4 , 2), (4 , − 2), (25, 5), (25, − 5)} Hence, (c) is the correct answer.

Type 5. Match the Columns Ex 16. Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Then, match the following in Column I with the sets of ordered pairs in Column II. Column I A.

Column II

A × (B ∩ C )

p. {(1, 4), (2, 4), (3, 4)}

B. ( A × B) ∩ ( A × C ) q. {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)} C.

A × (B ∪ C )

r. {(3, 4)}

D. ( A × B) ∪ ( A × C ) E. ( A ∩ B) × ( B ∩ C )

Sol. Given, A = {1, 2, 3}, B = {3, 4} and C = {4 , 5, 6}

A. B ∩ C = {4} ∴ A × ( B ∩ C ) = {(1, 4 ), (2, 4 ), (3, 4 )} B. A × B = {(1, 3), (1, 4 ), (2, 3), (2, 4 ), (3, 3), (3, 4 )} A × C = {(1, 4 ), (1, 5), (1, 6), (2, 4 ), (2, 5), (2, 6), (3, 4 ), (3, 5), (3, 6)} ∴ ( A × B ) ∩ ( A × C ) = {(1, 4 ), (2, 4 ), (3, 4 )} C. B ∪ C = {3, 4 , 5, 6} ∴ A × ( B ∪ C ) = {(1, 3), (1, 4 ), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)} D. ( A × B ) ∪ ( A × C ) = {(1, 3), (1, 4 ), (1, 5), (1, 6) (2, 3), (2, 4 ), (2, 5), (2, 6), (3, 3), (3, 4 ), (3, 5), (3, 6)} E. A ∩ B = {3}, B ∩ C = {4} ∴ ( A ∩ B ) × ( B ∩ C ) = {3, 4} A → p; B → p; C → q; D → q; E → r

Ex 17. Let A = {1, 2, 3, 4, ..., 14}. A relation R from A to A is defined by R = {( x, y) :3x − y = 0, where x, y ∈ A}. Then, match the terms of Column I with the terms of Column II and choose the correct option from the codes given below: Column I A.

Column II

In roster form, the relation R is

p.

{1, 2, 3, 4}

B.

The domain of R is

q.

{3, 6, 9, 12}

C.

The range of R is

r.

{1, 2, 3, 4, ..., 14}

D.

The codomain of R is

s.

{(1, 3), (2, 6), (3, 9), (4, 12)}

Sol. We have, 3x − y = 0 ⇒ y = 3x

x = 1, y = 3 ∈ A x = 2, y= 6 ∈A x = 3, y= 9 ∈A x = 4 , y = 12 ∈ A x = 5 , y = 15 ∉ A A. In roster form, R = {(1, 3), (2, 6), (3, 9), (4 , 12)} B. Domain of R = Set of first elements of ordered pairs in R = {1, 2, 3, 4} C. Range of R = Set of second elements of ordered pairs in R = {3, 6, 9, 12} D. Codomain of R is the set A. A → s; B → p; C → q; D → r

For

Type 6. Single Integer Answer Type Questions Ex 18. If n( A ) = 4, n(B ) = 3 and n( A × B × C ) = 24, then n(C ) is equal to________. Sol. (2) We have, n( A ) = 4, n( B ) = 3 and ⇒

n( A × B × C ) = 24 n( A ) × n( B ) × n(C ) = 24 4 × 3 × n(C ) = 24 n(C ) = 2

Ex 19. Let A = {x, y, z} and B = {1, 2}. Then, the number of relations from A to B is 2 k , then k is equal to________. Sol. (6) We have, n( A ) = 3, n( B ) = 2

n ( A × B) = 6 Total number of relations from A to B = 2n(A × B) = 26 ∴ k =6

31

Target Exercises Type 1. Only One Correct Option 1. If two sets A and B are having 99 elements in common, then the number of elements common to each of the sets A × B and B × A are (a) 299

(b) 992

(c) 100

(d) 18

2. Let n( A ) = m and n( B ) = n. Then, the total number of non-empty relations that can be defined from A to B is (a) mn

(b) nm − 1

(c) mn − 1

(d) 2mn − 1

3. Let X = {1, 2, 3, 4, 5} and Y = {1, 3, 5, 7, 9}. Which of the following is not a relation from X to Y ?

Ta rg e t E x e rc is e s

(a) (b) (c) (d)

R1 = {(x , y) : y = x + 2, x ∈ X , y ∈ Y} R2 = {(1, 1), (2, 1), (3, 3), (4 , 3), (5, 5)} R3 = {(1, 1), (1, 3), (3, 5), (3, 7), (5, 7)} R4 = {(1, 3), (2, 5), (2, 4 ), (7, 9)}

4. A relation R from C to R is defined by xR y iff | x | = y. Which of the following is correct? (a) (2 + 3i ) R 13 (c) (1 + i ) R 2

(b) 3R (−3) (d) i R1

5. The relation R defined on the set A = {1, 2, 3, 4, 5} by R = {( x, y ) : | x 2 − y 2 | < 16} is given by (a) (b) (c) (d)

R = {(0, 3), (0, − 3), (3, 0), (−3, 0)} Domain of R = {−3, 0, 3} Range of R = {− 3, 0, 3} None of the above

(a) {2, 4, 6} (c) {1, 2, 3, 4, 6}

by

(b) {1, 2, 3} (d) None of these

9. Let A = {1, 2, 3}, B = {1, 3, 5}. If relation R from A to B is given by {(1, 3), (2, 5), (3, 3)}, then R −1 is

32

(a) (b) (c) (d)

{(3, 3), (3, 1) , (5, 3)} {(1, 3), (2, 5) , (3, 3)} {(1, 3), (5, 2) } None of the above

11. Let a relation be defined by R R = {( 4, 5), (1, 4 ), ( 4, 6), ( 7, 6), ( 3, 7)}. The relation R −1 oR is given by (a) (b) (c) (d)

{(1, 1), (4, 4), (7, 4), (4, 7), (7, 7)} {(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)} {(1, 5), (1, 6), (3, 6)} None of the above

12. If R is a relation on the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} given by xRy ⇔ y = 3x, then RoR −1 is (a) (b) (c) (d)

{(1, 3), (2, 6), (3, 9)} {(1, 1), (2, 2), (3, 3)} {(3, 3), (6, 6), (9, 9)} None of the above

13. If R ⊆ A × B and S ⊆ B × C be two relations, then ( SoR ) −1 is equal to (b) RoS (d) None of these

14. If R is a relation from a set A to the set B and S is a relation from B to C, then the relation SoR (a) is from C to A (c) does not exist

(b) is from A to C (d) None of these

15. Let A = [ − 1, 1], B = [ − 1, 1], C = [ 0, ∞ ). If R 2 = {( x, y ) ∈ A × C : x 2 + y 2 = 1}, then

(b) {−2, − 1, 0} (d) None of these

8. Let R be a relation in N defined R = {( x, y ) : x + 2 y = 8}. The range of R is

{(11, 8), (13, 10)} {(8, 11), (10, 13)} {(8, 11), (9, 12), (10, 13)} None of the above

R1 = {( x, y ) ∈ A × B : x 2 + y 2 = 1}and

7. If R = {( x, y ) : x, y ∈ Z , x 2 + y 2 ≤ 4} is a relation in Z, then domain of R is (a) {0, 1, 2} (c) {−2, − 1, 0, 1, 2}

(a) (b) (c) (d)

(a) S −1 oR −1 (c) R −1 oS −1

{(1, 1) , (2, 1) , (3, 1) , (4, 1) , (2, 3)} {(2, 2) , (3, 2) , (4, 2) , (2, 4)} {(3, 3) , (4, 3) , (5, 4) , (3, 4)} None of the above

6. A relation R is defined in the set Z of integers as follows ( x, y ) ∈ R iff x 2 + y 2 = 9. Which of the following is false? (a) (b) (c) (d)

10. R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x − 3. The relation R −1 is

(a) (b) (c) (d)

R1 defines a function from R2 defines a function from R1 defines a function from R2 defines a function from

A into B A into C A onto B A onto C

x 2 , 0 ≤ x ≤ 3 16. The relation f is defined by f ( x ) =   3x, 3 ≤ x ≤ 10 and the relation is defined by g 2 x , 0 ≤ x ≤ 2 . Which of the following g (x ) =   3x, 2 ≤ x ≤ 10 relation is a function? (a) (b) (c) (d)

f g f,g None of the above

x 2 + 3x + 2

(a) R − {− 1, − 2} (c) R − {− 1, − 2, − 3}

26. If f ( x ) = x 2 + 1 , then the value of ( fof ) ( x ) is equal to

is

(b) (− 2, ∞ ) (d) (− 3, ∞ ) − {−1 , − 2}

18. The domain and range of the function f given by f ( x ) = 2 − | x − 5 | is (a) (b) (c) (d)

Domain = R + , Range = (− ∞ , 1] Domain = R, Range = (− ∞ , 2 ] Domain = R, Range = (− ∞ , 2) Domain = R + , Range = (− ∞ , 2 ]

19. The range of the function f ( x ) = (a) R

(b) R − {1}

x−2 when x ≠ 2 is 2− x

(c) {− 1}

(d) R − {− 1}

x ≤ 0 ; g ( x ) = 3, x ≥ 0, then the domain of f + g is (b) [ 0, ∞ )

(c) (− ∞ , ∞ ) (d) (− ∞ , 0)

22. If [ x] 2 − 5[ x] + 6 = 0, where [⋅] denotes the greatest integer function, then (a) x ∈[ 3, 4 ] (b) x ∈ (2, 3 ] (c) x ∈[ 2, 3 ] (d) x ∈[ 2, 4 )

1   1  23. If f ( x ) = 1 − , then f  f    is x   x  (a)

1 x

24. Let f ( x ) = (a)

1 x

(b)

1 1+ x

(c)

x x −1

(d)

1 x −1

x−1 , then f { f ( x )}is x+1 (b) −

1 x

(c)

4

1 x+1

2

27. If f ( x ) is defined on [0, 1] by the rule if x ∈ Q  x, , then for x ∈[ 0, 1] f (x ) =  1 − x, if x ∉Q (b) ( fof ) (x ) = x (d) None of these

 x3 + 1 x < 0  ( x − 1)1/ 3 , x < 1 28. If f ( x ) =  2 , g (x ) =  , 1/ 2  x + 1, x ≥ 0  ( x − 1) , x ≥ 1 then ( gof )( x ) is equal to (a) x, ∀ x ∈ R (c) x + 1, ∀ x ∈ R

2

(b) x − 1, ∀ x ∈ R (d) None of these

29. If f ( x ) = [ x] and g ( x ) = x − [ x] , then which of the following is the zero function?

(b) (− ∞ , 0) (d) None of these

21. If f and g are two functions defined as f ( x ) = x + 2 , (a) {0}

(b) x + 2x − 2 (d) None of these

2

(a) ( fof ) (x ) = 1 + x (c) ( fof ) (x ) = 1 − x

20. The range of the function f ( x ) = | x | is (a) (0, ∞ ) (c) [ 0, ∞ )

(a) x + 2x + 2 (c) x 4 + x 2 + 1 4

Fundamentals of Relation and Function

log 2 ( x + 3)

(a) ( f + g ) (x ) (c) ( f − g ) (x )

30. If f ( x ) =

(a)

x x −1

(b) ( fg ) (x ) (d) ( fog ) (x )

x , x ≠ 1, then ( fofo K of ) ( x ) is equal to 14243 x−1 19 times 19

 x  19x (b)   (c) x −1  x − 1

(d) x

1 1  31. Let f  x +  = x 2 + 2 , x ∈ R − { 0}, then f ( x ) is  x x equal to (a) x 2 (c) x 2 − 2 , when | x | ≥ 2

(b) x 2 − 1 (d) None of these

Targ e t E x e rc is e s

17. The domain of F ( x ) =

f :[ 0, 1] → [ 0, 1] and g:[ 0, 1] → [ 0, 1] be two 1− x functions defined by and f (x ) = 1+ x g ( x ) = 4x (1 − x ), then ( fog ) ( x ) is equal to

32. Let (d)

1 x −1

25. Two functions f and g are said to commute, if ( fog ) ( x ) = ( gof ) ( x ), ∀ x, then which one of the following functions are commute? (a) f (x ) = x 3 , g (x ) = x + 1 (b) f (x ) = x , g (x ) = cos x (c) f (x ) = x m , g (x ) = x n , m ≠ n, m, n ∈ I (I is the set of all integers) (d) f (x ) = x − 1, g (x ) = x 2 + 1

8x (1 − x ) (1 + x )2 1 − 4 x + 4 x2 (c) 1 + 4 x − 4 x2

(a)

(b)

4 (1 − x ) 1+ x

(d) None of these

33. Let f ( x ) = | x − 1 | , x ∈ R, then (a) f (x 2 ) = ( f (x ))2 (c) f (| x |) = | f (x ) |

(b) f (x + y) = f (x ) + f ( y) (d) None of these

Type 2. More than One Correct Option 34. If A = {a, b, c, d}, B = {1, 2, 3}, which of the following sets of ordered pairs are not relation from A to B? (a) (b) (c) (d)

{(a, 1), (a, 3)} {(b, 1), (c, 2), (d , 1)} {(a, 2), (b, 3), (3, b)} {(a, 1), (b, 2), (3, c)}

35. The relation R defined in A = {1, 2, 3} by aRb, if | a 2 − b 2 | ≤ 5. Which of the following is true? (a) (b) (c) (d)

R = {(1, 1), (2, 2), (3, 3), (2, 1), (1, 2),(2, 3), (3, 2)} R −1 = R Domain of R = {1, 2, 3} Range of R = {5}

33

Objective Mathematics Vol. 1

2

Type 3. Assertion and Reason Directions (Q. Nos. 36-37) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are

36. Statement I The domain of the real function f defined by f ( x ) = x − 1 is R − {1}.

(a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

37. Statement I Let f and g be two real functions given by f = {( 0, 1), ( 2, 0), ( 3, − 4 ), ( 4, 2), ( 5, 1)} and g = {(1, 0), ( 2, 2), ( 3, − 1), ( 4, 4 ), ( 5, 3)}.

(b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

Statement II The range of the function defined by f ( x ) = x − 1 is [ 0, ∞ ).

Then, domain of f ⋅ g is given by {2, 3, 4, 5}. Statement II Let f and g be two real functions. Then, ( f ⋅ g )( x ) = f {g ( x )}.

(c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Type 4. Linked Comprehension Based Questions Passage (Q. Nos. 38-39) The figure shows a relation R

38. The domain of the relations R is

between the sets P and Q.

(a) (b) (c) (d)

Q P

Ta rg e t E x e rc is e s

9 4 25

5 3 2 1 –2 –3 –5

{4, 9, 25} {2, 3, 5} {−2, − 3, − 5} {−4 , − 9, − 25}

39. The range of the relation R is (a) (b) (c) (d)

{2, 3, 4} {−2, − 3, − 5} {−5, − 3, − 2, 2, 3, 5} {−5, − 3, − 2, 1, 2, 3, 5}

Type 5. Match the Column 40. Match the terms of Column I with the terms of Column II. Column I

Column II

A.

If P = { x : x < 3, x ∈ N} and Q = { x : x ≤ 2, x ∈ N}, then ( P ∪ Q ) × ( P ∩ Q ) is equal to

p. {(0, 8), (8, 0), ( 0, − 8), ( −8, 0)}

B.

If A = { x : x ∈ W , x < 2}, B = { x : x ∈ N, 1 < x < 5} and C = { 3, 5}, where W is the set of whole numbers, then A × ( B ∩ C ) is equal to

q. {( −2, 3), ( −1, 5), (0, 7), (1, 9) , (2, 11)}

C.

If R = {( x, y) : x and y are integers and { x 2 + y 2 = 64} is a relation, then R in roster form is

r. {(0, 3) , (1, 3)}

D. If R = {( x, y) : y = 2 x + 7, x ∈ Z and −2 ≤ x ≤ 2} is a relation, then R is

s. {(1, 1), (1, 2), (2, 1), (2, 2 )}

Type 6. Single Integer Answer Type Questions 41. Let A and B be two sets having 3 elements in common. If n( A ) = 5 and n( B ) = 4, then n[( A × B ) ∩ ( B × A )] is equal to _______. 42. Let f = {(1, 1), ( 2, 3), ( 0, − 1), ( −1, − 3)} be a linear function from Z to Z , such that f ( x ) = kx − 1, then k is equal to _______.

34

Entrances Gallery AIEEE (a) (0, ∞ ) (c) (−∞ , ∞ ) − {0}

is | x | − x [2011]

(b) (−∞ , 0) (d) (−∞ , ∞ )

2. Let for a ≠ a1 ≠ 0, f ( x ) = ax 2 + bx + c, g ( x ) = a1 x 2 + b1 x + c1 and p( x ) = f ( x ) − g ( x ). If p( x ) = 0 only for x = − 1 and p( − 2) = 2 , then the value of p( 2) is [2011] (a) 18

(b) 3

(c) 9

y−3 4 y+ 3 (c) g ( y) = 4 + 4

where a is a given constant and f ( 0) = 1, f ( 2a − x ) is [2005] equal to (a) f (− x ) (c) f (x )

(b) f (a) + f (a − x ) (d) − f (x )

5. If f : R → R satisfies f ( x + y ) = f ( x ) + f ( y ) , n

∀ x, y ∈ R and f (1) = 7, then

(d) 6

3. Let f : N → Y be a function defined as f ( x ) = 4x + 3 where Y = { y ∈ N : y = 4x + 3 for some x ∈ N }. [2008] If f is invertible, then its inverse is (a) g ( y) =

4. A real valued function f ( x ) satisfies the functional equation f ( x − y ) = f ( x ) f ( y ) − f ( a − x ) f ( a + y ) ,

3y + 4 3 y+ 3 (d) g ( y) = 4

(b) g ( y) =

Σ r =1

f ( r ) is

[2003]

7n 2 7(n + 1) (b) 2 (c) 7n(n + 1) 7n(n + 1) (d) 2 (a)

Other Engineering Entrances 6. Let A = {1, 2, 3, 4} and R be the relation on A defined by {( a, b ) : a, b ∈ A, a × b is an even number}, then [J&K CET 2014] the range of R is (a) {1, 2, 4} (c) {2, 3, 4}

(b) {2, 4} (d) {1, 2, 4}

7. The domain of the function f ( x ) =

( x 2 + 1) x 2 − 3x + 3

is

[J&K CET 2014] (a) R − {1, 2} (b) R − {1, 4} (c) R (d) R − {1}

8. Let A = {1, 2, 3, 4, 5}. The domain of the relation on A defined by R = {( x, y ) : y = 2x − 1} is [J&K CET 2014] (a) {1, 2, 3} (b) {1, 2} (c) {1, 3, 5} (d) {2, 4}

9. If f ( x ) = x and g ( x ) = 2x − 3, then domain of [Kerala CEE 2014] ( fog ) ( x ) is (a) (− ∞ , − 3) 3  (b)  − ∞ , −   2  3  (c) − , 0  2   3 (d) 0,   2 3  (e)  , ∞ 2 

10. If f ( x ) =

x+2 , then f { f ( x )}is 3x − 1

(a) x 1 (c) x (e) 0

[Kerala CEE 2014]

(b) −x 1 (d) − x

11. The range of the function f ( x ) = log e ( 3x 2 + 4 ) is [Kerala CEE 2012] equal to (a) [loge 2, ∞ ) (c) [ 2 loge 3, ∞ ) (e) [ 2 loge 2, ∞ )

Targ e t E x e rc is e s

1

1. The domain of the function f ( x ) =

(b) [loge 3, ∞ ) (d) [ 0, ∞ )

12. Let R be the set of real numbers and the functions f : R → R and g : R → R be defined by f ( x ) = x 2 + 2x − 3 and g ( x ) = x + 1. Then, the value [WB JEE 2012] of x for which g ( f ( x )) = f ( g ( x )) is (a) −1 (c) 1

(b) 0 (d) 2

13. Let A = {x, y, z}and B = {a, b, c, d}. Which one of the following is not a relation from A to B? [Kerala CEE 2011] (a) {(x , a), (x , c)} (b) {( y, c), ( y, d )} (c) {(z, a), (z, d )} (d) {(z, b), ( y, b), (a, d )} (e) {(x , c)}

14. If f ( x ) = 3 − x, − 4 ≤ x ≤ 4, then the domain of [J&K CET 2011] log e ( f ( x )) is (a) [ −4 , 4 ] (c) (−∞ , 3)

(b) (−∞ , 3 ] (d) [ −4 , 3)

35

Objective Mathematics Vol. 1

2

15. Let f = {( 0, − 1), ( −1, 3), ( 2, 3), ( 3, 5)} be a function from Z to Z defined by f ( x ) = ax + b. Then, [AMU 2011]

(a) a = 1, b = − 2 (b) a = 2, b = 1 (c) a = 2, b = − 1 (d) a = 1, b = 2

17. Let R be the set of real numbers and the mapping f : R → R and g : R → R be defined by f ( x ) = 5 − x 2 and g ( x ) = 3x − 4, then the value of ( fog ) ( −1) is (a) − 44

16. If f ( x ) = x − 1and g ( x ) = ( x + 1) , then ( gof )( x ) is 2

2

[Kerala CEE 2011] (a) (x + 1)4 − 1 (b) x 4 − 1 (c) x 4 (d) (x + 1)4 (e) (x − 1)4 − 1

(b) − 54

(c) − 32

[WB JEE 2010] (d) − 64

18. Let A = {1, 0, 1, 2}, B = {4, 2, 0, 2} and f , g : A → B be and functions defined by f (x ) = x 2 − x 1  [AMU 2010] g ( x ) = 2 x − − 1. Then, 2  (a) f = g (b) f = 2g (c) g = 2 f (d) None of the above

Answers Work Book Exercise 2.1

Ta rg e t E x e rc is e s

1. (c)

2. (b)

3. (c)

4. (c)

5. (a)

4. (b)

5. (b)

6. (a)

7. (c)

8. (c)

Work Book Exercise 2.2 1. (b)

2. (a)

3. (b)

Target Exercises 1. (b)

2. (d)

3. (d)

4. (d)

5. (d)

6. (d)

7. (c)

8. (b)

9. (d)

10. (b)

11. (b)

12. (b)

13. (c)

14. (b)

15. (d)

16. (a)

17. (d)

18. (b)

19. (c)

20. (c)

21. (a)

22. (d)

23. (c)

24. (b)

25. (c)

26. (a)

27. (b)

28. (a)

29. (d)

30. (a)

31. (c)

32. (c)

33. (d)

34. (c,d)

35. (a,b,c)

36. (d)

37. (c)

38. (a)

39. (c)

40. (*)

41. (9)

42. (2)

9. (e)

10. (a)

* A → s; B → r; C → p; D → q

Entrances Gallery

36

1. (b)

2. (a)

3. (a)

4. (d)

5. (d)

6. (b)

7. (c)

8. (a)

11. (e)

12. (a)

13. (d)

14. (d)

15. (c)

16. (c)

17. (a)

18. (a)

Explanations Target Exercises = n( A ∩ B) × n(B ∩ A) = 99 × 99 = 992

2. Given, n( A) = m and n(B) = n Now, total number of relations from A to B = 2 mn ∴Total number of non-empty relations from A to B = 2 mn − 1

3. R1 is a relation from X to Y because R1 ⊆ X × Y. R2 is a relation from X to Y because R2 ⊆ X × Y. R3 is a relation from X to Y because R3 ⊆ X × Y. R4 is not a relation from X to Y because (2, 4), (7, 9) ∉ X × Y.

4. | i | = 0 2 + (1)2 = 1 = 1 ∴ iR1 is correct.

5. We have, R = {( x, y ):| x 2 − y 2| < 16} ∴ R = {(1, 1), (1, 2 ), (1, 3), (1, 4), (2, 1), (2, 2 ), (2, 3),(2, 4), (3, 1), (3, 2 ), (3, 3), (3, 4), (4, 1), (4, 2 ), (4, 3), (4, 4), (4, 5), (5, 4), (5, 5)} y = ± 9− x

7. We have, R = {( x, y ) : x, y ∈ Z, x + y ≤ 4} 2

2

R = {(0, 0 ), (0, − 1), (0, 1), (0, − 2 ), (0, 2 ),(−1, 0 ), (1, 0 ), (1, 1), (1, − 1), (−1, 1), (−1, − 1), (2, 0 ), (−2, 0 )} ∴ Domain of R = { x : ( x, y ) ∈ R} = { 0, − 1, 1, − 2, 2}

8. R = {( x, y ) : x + 2 y = 8, x, y ∈ N} 8− x 2 R = {(2, 3), (4, 2 ), (6, 1)} Range of R = { y : ( x, y ) ∈ R} = {1, 2, 3}

∴

⇒

y=

9. We have, R = {(1, 3), (2, 5), (3, 3)} ∴

R −1 = {( y, x ) : ( x, y ) ∈ R} = {(3, 1), (5, 2 ), (3, 3)}

10. We have, R = {( x, y ) : y = x − 3, x = 11or 12 or 13, y = 8 or 10 or 12} ∴

ROR −1 = {(1, 1), (2, 2 ), (3, 3)}

13. SoR is a relation from A to C.

∴ (SoR )−1 is a relation from C to A. R −1 is a relation from B to A. S −1 is a relation from C to B. ∴R −1oS −1 is a relation from C to A. Let (c , a) ∈ (SoR )−1 ∴ (a, c ) ∈ SoR ⇒ ∃ b ∈ B : (a, b) ∈ R and (b, c ) ∈ S ⇒ (b, a) ∈ R −1 and (c , b) ∈ S −1 ⇒ (c , a) ∈ R −1oS −1 ∴ (SoR )−1 ⊆ R −1oS −1 Conversely, let (c , a) ∈ R −1oS −1 ⇒ ∃ b ∈ B : (c , b) ∈ S −1 and (b, a) ∈ R −1 ⇒ (b, c ) ∈ S and (a, b) ∈ R ⇒ (a, c ) ∈ SoR ⇒ (c , a) ∈ (SoR )−1 ∴

R −1oS −1 ⊆ (SoR )−1

Combining, we get (SoR )−1 = R −1oS −1 ∴ SoR is a relation from A to C.

2

∴ R = {(0, 3), (0, − 3), (3, 0 ), (−3, 0 )} Domain of R = { x : ( x, y ) ∈ R} = { 0, 3, − 3} Range of R = { y : ( x, y ) ∈ R} = { 3, − 3, 0}

x + 2y = 8

∴

R −1 = {(3, 1), (6, 2 ), (9, 3)}

14. Since, R ⊆ A × B and S ⊆ B × C, we have SoR ⊆ A × C

6. x 2 + y 2 = 9 ⇒ y = 9 − x 2 ⇒

and

15. R2 = {( x, y ) ∈ A × C : x 2 + y 2 = 1} x2 + y2 = 1 ⇒

y = ± 1 − x2

⇒

y = 1 − x2 ∴R2 is a function from A onto C.  x 2, 0 ≤ x ≤ 3 16. Given, f ( x ) =  3 x, 3 ≤ x ≤ 10

[Q y ≥ 0 ]

Targ e t E x e rc is e s

1. n[( A × B) ∩ (B × A)] = n[( A ∩ B) × (B ∩ A)]

Q f ( x ) = x 2 is well-defined in 0 ≤ x ≤ 3 and f ( x ) = 3 x is also well-defined in 3 ≤ x ≤ 10 At x = 3, f ( x ) = x 2 ⇒ f(3) = 32 = 9 At x = 3, f ( x ) = 3 x ⇒ f(3) = 3 × 3 = 9 ⇒ f ( x ) is defined at x = 3. Hence, f is a function.  x 2, 0 ≤ x ≤ 2 g( x ) =  3 x, 2 ≤ x ≤ 10 2 Q g( x ) = x is well-defined in 0 ≤ x ≤ 2 and g( x ) = 3 x is also well-defined in 2 ≤ x ≤ 10

= {(11, 8), (13, 10)} R −1 = {( y, x ) : ( x, y ) ∈ R}

2

= {(8, 11), (10, 13)}

4 6

11. We have, R = {(4, 5), (1, 4), (4, 6), (7, 6), (3, 7 )} R −1 = {(5, 4), (4, 1), (6, 4), (6, 7 ), (7, 3)} ∴

R −1oR = {(4, 4), (1, 1), (4, 7 ), (7, 4), (7, 7 ), (3 ,3)}

12. We have, set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} ∴

xRy ⇒ y = 3 x R = {(1, 3), (2, 6), (3, 9)}

But at x = 2, g( x ) = x 2 ⇒

g(2 ) = 2 2 = 4

At x = 2, g( x ) = 3 x ⇒ g(2 ) = 3 × 2 = 6 Therefore, g( x ) is not defined at x = 2. Hence, g is not a function.

37

Objective Mathematics Vol. 1

2

17. We have, F( x ) =

log 2 ( x + 3) x 2 + 3x + 2

28. Let x < 0 ∴

∴ F( x ) is defined, if x + 3 > 0 and x 2 + 3x + 2 ≠ 0

= [( x 3 + 1) − 1]1/ 3 [Q x < 0 ⇒ x 3 + 1 < 1]

⇒ F( x ) is defined, if x > − 3 and x ≠ − 1, − 2 Domain of F( x ) = (−3, ∞ ) − { − 1, − 2} ⇒ Domain of f ( x ) is defined for all real values of x. Q | x − 5| ≥ 0 ⇒ − | x − 5| ≤ 0 ⇒ 2 − | x − 5| ≤ 2 ⇒ f ( x ) ≤ 2 Hence, range of f ( x ) is (−∞, 2 ]. x −2 x −2 19. f ( x ) = =− = − 1, if x ≠ 2 x −2 2−x ∴

Range of f = { −1}

21. D(f + g ) = D(f ) ∩ D(g ) (−∞, 0 ] ∩ [0, ∞ ) = { 0}

22. Given, [ x ] − 5[ x ] + 6 = 0 2

⇒

Ta rg e t E x e rc is e s

= ( x 3 )1/ 3 = x Let x ≥ 0 ∴ (gof )( x ) = g(f ( x )) = g( x 2 + 1) = [( x 2 + 1) − 1]1/ 2 [Q x ≥ 0 ⇒ x 2 + 1 ≥ 1]

18. Given, f ( x ) = 2 −| x − 5|

20. f ( x ) =| x | ⇒ R(f ) = [0, ∞)

[ x ]2 − 3[ x ] − 2[ x ] + 6 = 0

⇒ [ x ]([ x ] − 3) − 2([ x ] − 3) = 0 ⇒ ([ x ] − 3) ([ x ] − 2 ) = 0 ⇒ [ x ] = 3 or [ x ] = 2 ⇒ x ∈[3, 4) or x ∈[2, 3) ∴ x ∈[2, 4)  1   1  1 23. f f    = f 1 −  = f (1 − x ) = 1 −  1− x (1 / x )   x  1− x − 1 x = = 1− x x −1 x −1 −1  x − 1 x+1 24. f { f ( x )}d = f  =  x −1  x + 1 +1 x+1 x − 1 − x − 1 −2 1 = = =− x − 1+ x + 1 2x x f( x) = x As

f { g( x )} = ( x ) = x

and

g{ f ( x )} = ( x m )n = x nm

∴

f { g( x )} = g{ f ( x )}

n m

nm

26. (fof )( x ) = f { f ( x )} = f ( x + 1) = ( x + 1) + 1 2

∴ (gof ) ( x ) = x, ∀ x ∈ R

29. (f + g )( x ) = f ( x ) + g( x ) = [ x ] + ( x − [ x ]) = x ≠ 0 (fg )( x ) = f ( x )⋅ g( x ) = [ x ]( x − [ x ]) ≠ 0 (f − g )( x ) = f ( x ) − g( x ) = [ x ] − ( x − [ x ]) = 2 [x] − x ≠ 0 (fog )( x ) = f (g( x )) = f ( x − [ x ]) = [ x − [ x ]] = 0 [Q x − [ x ] ∈ [0, 1)] x  x  x −1 30. (fof )x = f  =x  =  x − 1  x    −1  x − 1 x ⇒ (fofof )x = f (fof )( x ) = f ( x ) = x −1 x ∴ (fofof…19 times )( x ) = x −1 2 1 1 1 31. f  x +  = x 2 + 2 =  x +  − 2



x

Let ∴



x 1 z=x+ x f ( z) = z2 − 2 x

Replacing z by x, we get f ( x ) = x 2 − 2,

when| x | ≥ 2

1 − g( x ) 1 − 4 ( x )(1 − x ) 32. f { g( x )} = = 1 + g( x ) 1 + 4 x(1 − x ) =

1 − 4x + 4x 2 1 + 4x − 4x 2

(a)| x 2 − 1| = (| x − 1|)2 , which is not true.

m

g( x ) = x n

and

[Q x ≥ 0]

= ( x 2 )1/ 2 = | x | = x

33. We have, f ( x ) =| x − 1|

25. (fog ) ( x ) = (gof ) ( x ) only when

2

2

= x4 + 2 x2 + 2

27. Let x ∈[0, 1]

38

(gof )( x ) = g{ f ( x )} = g( x 3 + 1)

Case I x ∈ Q ∴ f( x) = x and (fof ) ( x ) = f { f ( x )} = f ( x ) = x [Q x ∈ Q ] Case II x ∉ Q ∴ f( x) = 1 − x and (fof )( x ) = f { f ( x )} = f (1 − x ) = 1 − (1 − x ) = x [Q 1 − x ∉ Q ] Now, in both cases, (fof )( x ) = x

(b)| x + y − 1| = | x − 1| + | y − 1|, which is not true. (c)|| x | − 1| = |(| x − 1|)|, which is not true.

34. {(a, 1), (a, 3)} ⊆ A × B ∴This is a relation. {(b, 1), (c , 2 ), (d , 1)} ⊆ A × B ∴This is a relation. (3, b) ∉ A × B ∴ (a, 2 ), (b, 3), (3, b) is not a relation from A to B. (3, c ) ∈ / A×B ∴ {(a, 1), (b, 2 ), (3, c )} is not a relation from A to B.

35. Q R = {(1, 1), (1, 2 ), (2, 1), (2, 2 ), (2, 3), (3, 2 ), (3, 3)} R −1 = {( y, x ) : ( x, y ) ∈ R} = {(1, 1), (2, 1), (1, 2 ), (2, 2 ), (3, 2 ), (2, 3), (3, 3)}= R Domain of R = { x : ( x, y ) ∈ R} = {1, 2, 3} and range of R = { y : ( x, y ) ∈ R} = {1, 2, 3}

= { −5, 5, − 3, 3, − 2, 2}

f ( x ) is defined, if x − 1 ≥ 0 i.e. x ≥1 Domain of f = [1, ∞ ) ∴ Hence, Statement I is false. Let f( x) = y Then, y = x −1 ⇒ y2 = x − 1 ⇒

40. A. P = {1, 2}, Q = {1, 2} P ∪ Q = {1, 2}, P ∩ Q = {1, 2} ∴ (P ∪ Q ) × (P ∩ Q ) = {(1, 1), (1, 2 ), (2, 1), (2, 2 )} B. A = { 0, 1}, B = {2, 3, 4}, C = { 3, 5} B ∩ C = { 3} A × (B ∩ C ) = {(0, 3), (1, 3)} C. R = {(0, 8), (0, − 8), (8, 0 ), (−8, 0 )} D. R = {(−2, 3), (−1, 5), (0, 7 ), (1, 9), (2, 11)}

x = y2 + 1

∴

Range of f = [0, ∞ )

[Q x ≥ 1]

37. Domain of f = { 0, 2, 3, 4, 5} Domain of g = {1, 2, 3, 4, 5} Domain of f ⋅ g = Domain of f ∩ Domain of g = {2, 3, 4, 5} Hence, Statement I is true. If f and g are two real functions. Then, (f ⋅ g )( x ) = f ( x )⋅ g( x ) Hence, Statement II is false.

41. Given, n( A) = 5, n(B) = 4, n( A ∩ B) = 3 n[( A × B) ∩ (B × A)] = n[( A ∩ B) × (B ∩ A)] = n( A ∩ B) × n(B ∩ A) = 3 × 3 = 9

2 Fundamentals of Relation and Function

39. Range of R = Set of second elements of ordered pairs in R

36. We have, f ( x ) = x − 1

42. Since, f ( x ) is a linear function, f ( x ) = mx + c f (1) = m + c = 1 f (0 ) = c = − 1 m=2 f( x) = 2 x − 1 k =2

Also, and This gives and ∴

38. Domain of R = Set of first elements of ordered pairs in R = { 4, 9, 25}

1. y =

For domain,| x | − x > 0 ⇒ | x| > x This is possible, only if x < 0. ∴ x ∈ (−∞, 0 )

2. Given, p( x ) = f ( x ) − g( x ) has only one root −1. ...(i) ∴ p( x ) = (a − a1 ) x 2 + (b − b1 )x + (c − c1 ) − (b − b1 ) …(ii) ⇒ = −2 (a − a1 ) (c − c1 ) ...(iii) and =1 (a − a1 ) Also, p(−2 ) = 2 ⇒ 4 (a − a1 ) − 2 (b − b1 ) + (c − c1 ) = 2 ⇒ 4(a − a1 ) − 4(a − a1 ) + (a − a1 ) = 2 [using Eqs. (ii) and (iii)] ⇒ (a − a1 ) = 2 Substituting (a − a1 ) = 2 in Eq. (i) and Eq. (ii), (b − b1 ) = 4, (c − c1 ) = 2 ∴ p(2 ) = 4 (a − a1 ) + 2 (b − b1 ) + (c − c1 ) [from Eq. (i)] = 8 + 8 + 2 = 18

3. Since, Y = { y ∈ Y : y = 4 x + 3 for some x ∈ N} ∴ Now,

Y = {7, 11, K ∞} y = 4x + 3 y−3 ⇒ x= 4 y−3 Inverse of f ( x ) is g( y ) = . 4

4. Given, f ( x − y ) = f ( x ) f ( y ) − f (a − x ) f (a + y ) Let ⇒

⇒

1 | x| − x

x=0=y f (0 ) = { f (0 )} 2 − { f (a)} 2

1 = 1 − { f (a)} 2

⇒ ∴

[f(0 ) = 1, given]

f (a) = 0 f (2 a − x ) = f { a − ( x − a)} = f (a)f ( x − a) − f (a − a)f ( x ) = 0 − f ( x )⋅ 1 = − f ( x )

n

5.

∑ f (r ) = f (1) + f (2 ) + f (3) + K + f (n )

r =1

= f (1) + 2 f (1) + 3f (1) + K + nf (1) [since, f ( x + y ) = f ( x ) + f ( y )] = (1 + 2 + 3 + K + n ) f (1) = f (1) ∑ n 7 n(n + 1) [Q f(1) = 7, given] = 2

Targ e t E x e rc is e s

Entrances Gallery

6. Since, A = {1, 2, 3, 4} and R be a relation on A defined by {(a, b); a, b ∈ A, a × b is an even number}. Thus, R be the set of even numbers. Range of R = {2, 4} ∴

7. Q f ( x ) =

( x 2 + 1) ( x − 3 x + 3) 2

⇒ ( x 2 + 1) is defined for all x ∈ R. Also, x 2 − 3 x = − 3 has imaginary roots. ∴x 2 − 3 x + 3 is also defined for all real numbers, i.e. x ∈ R. Hence, domain of the function is R.

8. Since, A = {1, 2, 3, 4, 5} ∴Relation from A to A R = {( x, y ) : y = 2 x − 1} = {(1, 1), (2, 3), (3, 5)} Domain = {1, 2, 3} ⇒

9. Given, f ( x ) = x and ∴

g( x ) = 2 x − 3 (fog ) ( x ) = f { g( x )} = f (2 x − 3) = 2 x − 3

39

Objective Mathematics Vol. 1

2

For domain of (fog )( x ), 2x − 3 ≥ 0 ⇒ 2x ≥ 3 3 ⇒ x≥ 2 3  ∴ x ∈  , ∞ 2 

10. Given, f ( x ) = ∴

x+2 3x − 1

 x + 2 f { f ( x )} = f   =  3 x − 1

x+2 +2 3x − 1  x + 2 3  −1  3 x − 1

x + 2 + 6x − 2 3 x + 6 − (3 x − 1) 7x = =x 7 =

11. Given, f ( x ) = loge (3 x + 4) 2

y = loge (3 x + 4)

⇒

3x 2 + 4 = e y

⇒

x=

Ta rg e t E x e rc is e s

⇒ ⇒

f( x) = x2 + 2 x − 3 g( x ) = x + 1 f { g( x )} = g{ f ( x )} f ( x + 1) = g( x 2 + 2 x − 3)

and Q ⇒

14. Given, f ( x ) = 3 − x, here −4 ≤ x ≤ 4 To find domain of loge { f ( x )} = loge (3 − x ) For loge { f ( x )} to be defined, 3− x > 0 ⇒ 3> x ⇒ x defined by a n = 4n + 5 is an AP. Also, find its common difference.

X

Sol. We have, an = 4n + 5 On replacing n by (n + 1,) we get an + 1 = 4(n + 1) + 5 = 4n + 9 Now, an + 1 − an = (4 n + 9) − (4 n + 5) Clearly, an + 1 − an = 4, ∀ n ∈ N

Example 5. If the sequence < a n > is defined by a n = 3n + 2, then difference of any two consecutive terms is (a) 2 (b) 3 (c) 4 (d) 5 Sol. (b) Since, nth term is of the form An + B , therefore it is an AP and the common difference is coefficient of n i.e. 3.

iii.

So, the given sequence is an AP with common difference 4.

Properties of an Arithmetic Progression i.

X

X

If a is the first term and d is the common difference of an AP, then its nth term or general term a n is given by a n = a + ( n − 1) d

Example 3. If the sequence 9, 12, 15, 18, ... is an AP, then its general term is (a) 3n + 1 (b) 3n + 2 (c) 3n + 4 (d) 3n + 6

term of the given AP is being added by a constant number 3.

As the given sequence is an AP with common difference 3 and first term 9. ∴General term = nth term = a + (n − 1)d = 9 + (n − 1) 3 = 3n + 6

iv.

Example 4. Let Tr be the rth term of an AP for r =1, 2, 3,K , if for some positive integers m, n, we 1 1 have Tm = and Tn = , then Tmn is equal to n m 1 1 1 (a) (b) + (c)1 (d) 0 mn m n Sol. (c) Let a be the first term and d be the common 1 n 1 and Tn = a + (n − 1) d = m On subtracting Eq. (ii) from Eq. (i), we get 1 1 m− n (m − n)d = − = n m mn 1 ⇒ d= mn Again, Tmn = a + (mn − 1)d = a + (mn − n + n − 1)d = a + (n − 1)d + (mn − n)d 1 = Tn + n(m − 1) mn 1 (m − 1) = + =1 m m

ii. 42

Tm = a + (m − 1) d =

X

If each term of a given AP is multiplied or divided by a non-zero constant k, then the resulting sequence is also an AP with common difference kd or d / k respectively, where d is the common difference of the given AP.

Example 7. If a, b and c are in AP, then prove that  1 1 1 1   1 1 a  +  , b  +  , c  +  are in AP.  b c  c a   a b Sol. Since, a, b and c are in AP.

difference. ∴

Example 6. If a, b and c are in AP, then which of the following will also be in AP? (a) 3 + a, 3 + b, 3 + c (b) a − 1, b − 2, c − 3 (c) a + 1, b + 2, c + 3 (d) a − k , b − 2k , c − 3k Sol. (a) a + 3, b + 3 and c + 3 are in AP because each

Sol. (d) We have, d = (12 − 9) = 3

X

If a constant is added to or subtracted from each term of an AP, then the resulting sequence is also an AP with the same common difference.

...(i) ...(ii)

On dividing each term by abc, the resulting sequence a b c are in AP. , and abc abc abc 1 1 1 are in AP. , and ⇒ bc ac ab ab + bc + ca ab + bc + ca ab + bc + ca are in AP ⇒ , , bc ac ab [on multiplying each term by ab + bc + ca] ab + bc + ca ab + bc + ca ab + bc + ca − 1, − 1, −1 ⇒ bc ac ab are in AP. [on adding − 1 to each term] ab + ac ab + bc bc + ca are in AP. ⇒ , , bc ca ab 1 1 1 1 1 1 ⇒ a  +  , b  +  , c  +  are in AP.  b c   c a  a b  Hence proved.

v. A sequence is an AP iff its nth term is of the form An + B , i.e. a linear expression in n. The common difference in such a case is A, i.e. the coefficient of n.

In a finite AP, the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of first and last term, i.e. a k + a n ( k −1) = a1 + a n for all k = 1, 2, 3, K, ( n − 1).

Example 8. If an AP is given by 3, a 2 , ..., 20, then which of the following is the sum of second and second last terms? (a) 20 (b) 3 (c) 23 (d) 22

X

Sol. (c) As we know, ak + an − ( k − 1) = a1 + an

Sol. (d) Let A1, A2 , A3 and A4 be four numbers in AP.

On putting k = 2, we get a2 + an − 1 = a1 + an = 3 + 20 = 23 X

Given, A1 + A4 = 8 and A2 A3 = 15 We know, A2 + A3 = A1 + A4 ∴ A2 + A3 = 8 On solving Eqs. (ii) and (iii), we get A2 = 3, A3 = 5 or A2 = 5, A3 = 3 Q 2 A2 = A1 + A3 ⇒ 2 × 3 = A1 + 5 or 2 × 5 = A1 + 3 ⇒ A1 = 1 or A1 = 7 Similarly, A4 = 7 or A4 = 1 Hence, least number of sequence is 1.

Example 9. If a1 , a 2 , a 3 , K , a 2n + 1 are in AP, then ( a 2n + 1 + a1 ) + ( a 2n + a 2 ) + K + ( a n + 2 + a n ) is equal to (b) 2na n − 1 (a) 2na n (c) 2na n + 1 (d) None of these Sol. (c) Clearly, a1 + a2 n + 1 = a2 + a2 n = a3 + a2 n −1 = K = an + 2 + an and ar = a1 + (r − 1)d ∴ (a2 n + 1 + a1 ) + (a2 n + a2 ) + K + (an + 2 + an ) = n(a2 n + 1 + a1 ) = n(2 a1 + 2 nd ) = 2 n(a1 + nd ) = 2 nan + 1

vi. X

Three numbers a, b and c are in AP iff 2b = a + c

X

[by property]

Selecting Odd Number of Terms of AP i. ii. X

Selecting 3 terms of AP a − d, a, a + d Selecting 5 terms of AP a − 2d, a − d, a, a + d, a + 2d

Example 13. If the sum of three numbers in AP is − 3 and their product is 8, then the numbers are (a) 2, 1, 4 (b) 2, −1, − 4 (c) 4, 2, 1 (d) None of these Sol. (b) Let the numbers be a − d , a and a + d . Given, sum = − 3 ⇒ (a − d ) + a + (a + d ) = − 3 ⇒ 3a = − 3 ⇒ a = − 1 and product = 8 ⇒ ( a − d ) a( a + d ) = 8 ⇒ a(a2 − d 2 ) = 8

If the terms of an AP are chosen at regular intervals, then they form an AP.

Example 11. If T1 , T2 , T3 , T4 , T5 , T6 , T7 ,… are in AP with common difference 2. Then, the difference of any two consecutive terms of the sequence T1 , T4 , T7 , ... is (a) 3 (b) 4 (c) 6 (d) 8

⇒

(− 1)(1 − d 2 ) = 8

⇒

d2 = 9

⇒ d=± 3 If d = 3, then the numbers are − 4, − 1, 2. If d = − 3, then the numbers are 2, − 1, − 4.

If a n , a n +1 and a n + 2 are any three consecutive

Example 14. In a triangle, the lengths of the two larger sides are 10 and 9, respectively. If the angles are in AP, then the length of the third side can be

terms of an AP, then 2a n +1 = a n + a n + 2

(a) 91 (c) 5

Sol. (c) Clearly, T1, T4 , T7,... are in AP. Now,

viii.

...(iii)

Sometimes, it is required to select a finite number of terms in AP. It is always convenient, if we select the terms in the following manner:

bc ca ab  b + 1  −  a + 1  = c + 1  −  b + 1           ca   bc   ab   ca  ( b − a) (c − b ) = (c − b ) + ⇒ ( b − a) + abc abc ⇒ (b − a){abc + 1} = (c − b ){abc + 1} ⇒ b− a=c − b ⇒ 2b = a + c which is true, as given a, b and c are in AP.

vii.

…(i) ...(ii)

T4 − T1 = T1 + 3 × 2 − T1 = 6

3

Selection of Terms in an AP

Example 10. If a, b and c are in AP, then 1 1 1 are in and c + a + ,b+ bc ca ab (a) HP (b) AP (c) GP (d) None of these Sol. (b) a + 1 , b + 1 and c + 1 are in AP, if and only if

Example 12. Four numbers are in arithmetic progression. The sum of first and last term is 8 and the product of both middle terms is 15. The least number of the sequence is (a) 4 (b) 3 (c) 2 (d) 1

Sequence and Series

X

X

(b) 3 3 (d) None of these

43

3

Sol. (d) Since, angles of a triangle are in AP.

Objective Mathematics Vol. 1

∴ ⇒

α + (α − δ) + (α + δ) = 180° α = 60° A

9 B

(α + δ )

x

(α – δ )

α 10

C

Using cosine law in ∆ACB, (10)2 + ( x)2 − 92 cos 60° = 2 ⋅ (10) ⋅ ( x)

Sum of n Terms of an AP The sum S n of n terms of an AP with first term a and common difference d is given by n S n = [2a + ( n − 1) d ] 2 n Also, S n = [ a + l] 2 where, l = last term = a + ( n − 1) d X

1 19 + x2 = 2 20 x x2 − 10 x + 19 = 0

⇒ ⇒

x = 5±

⇒

6

Sol. (b) Let a be the first term and d be the common

Selecting Even Number of Terms of AP i. ii. Ø

●

●

X

difference of the given AP. Then, a = 1and d = 3. We have to find the sum of 20 terms of the given AP. n On putting a = 1, d = 3 and n = 20 in S n = [2 a + (n − 1)d ], 2 we get 20 [2 × 1 + (20 − 1) × 3] S 20 = 2 = 10 × 59 = 590

Selecting 2 terms of AP a − d, a + d Selecting 4 terms of AP a − 3d, a − d, a + d, a + 3d

It should be noted that in case of odd number of terms, the middle term is a and the common difference is d while in case of an even number of terms the middle terms are a − d , a + d and the common difference is 2d. If sum of the numbers is not given, then the numbers can be taken as a, a + d , a + 2 d , a + 3 d ,... .

Example 17. Which of the following is the sum of 20 terms of an AP 1, 4, 7, 10,...? (a) 690 (b) 590 (c) 591 (d ) 691

X

Example 15. If the sum of four numbers of AP is 20, then which of the following is first term of the given AP? (a) 5 (b) 4 (c) 6 (d) 10

Example 18. If S r denotes the sum of the first S 3r − S r − 1 is equal to r terms of an AP, then S 2r − S 2r − 1 (a) 2r − 1 (b) 2r + 1 (c) 4r + 1 (d) 2r + 3 Sol. (b)

a + 3d .

and ⇒ ⇒ ⇒

44

⇒

Sum = 32 (a − 3d ) + (a − d ) + (a + d ) + (a + 3d ) = 32 4 a = 32 ⇒ a = 8 (a − 3d ) (a + 3d ) 7 = 15 (a − d ) (a + d ) a2 − 9d 2 a −d 2

2

64 − 9d 2 64 − d 2

=

7 15

7 = 15

128d = 512 2

d =4 ⇒ d=±2 2

Thus, the four parts are 2, 6, 10, 14.

−1

Ø A sequence is an AP if and only if the sum of its n terms is of the

form An2 + Bn, where A, B are constants. In such a case, the common difference of the AP is 2A.

Sol. (a) Let the four parts be a − 3d , a − d , a + d and a + 3d . Q ⇒ ⇒

−1

=

Now, a − 3d + a − d + a + d + a + 3d = 20 ⇒ 4 a = 20 ⇒ a = 5

Example 16. Divide 32 into four parts which are in AP such that the ratio of product of extremes to the product of means is 7 :15. (a) 2, 6, 10, 14 (b) 3, 9, 15, 18 (c) 4, 12, 20, 28 (d) 5, 15, 25, 35

S2r − S2r

(r − 1) 3r [2 a + (3r − 1) d ] − [2 a + (r − 2 )d ] 2 2 (2 r − 1) 2r [2 a + (2 r − 1)d ] − [2 a + (2 r − 2 )d ] 2 2 2 a (2 r + 1) + d (8r 2 − 2 ) = 2 a + d(4 r − 2 ) (2 r + 1) [2 a + 2(2 r − 1) d ] = (2 r + 1) = [2 a + 2(2 r − 1)d ]

Sol. (a) Let the four numbers in AP be a − 3d , a − d , a + d ,

X

S3r − Sr

X

Example 19. If the sum of n terms of an AP is 3n 2 + 5n, then the common difference is (a) 2 (c) 5

(b) 3 (d) 6

Sol. (d) Since, S n = 3 n2 + 5 n is of the form An2 + Bn. Therefore, the common difference is 2 A = 6.

Arithmetic Mean of Two Given Numbers Let a and b be two given numbers, then the arithmetic mean of a and b is a number A such that a, A, b are in AP.

a+b ⇒ 2A = a + b ⇒ A = 2 Thus, the arithmetic mean of two numbers a and b a+b is given by A = . 2

X

Arithmetic Mean of n Given Numbers

Sol. (b) (1 − 1 + 2 − 2 + ... + 5 − 5)2

Let a, a 2 , ..., a n be n given numbers, then the arithmetic mean of a1 , a 2 , a 3 , ..., a n is given by a + a 2 + a 3 + ... + a n A= 1 n

where, S is the required number. or 0 = 2(12 + 2 2 + 32 + 42 + 52 ) + 2S − 5(5 + 1) (10 + 1) ∴ = − 55 2S = 6 X

If between two given quantities a and b, we have to insert n quantities A1 , A2 , ..., An such that a, A1 , A2 , ... An , b form an AP, then we say that A1 , A2 , ..., An are n arithmetic means between a and b. Let d be the common difference of this AP. Clearly, it contains ( n + 2) terms. ∴ b = ( n + 2) th term b−a ⇒ b = a + ( n + 1) d ⇒ d = n +1

A1 + A 2 + .... + A n = nA. X

a + b  2 

(c) ( n 2 − n + 1) + ( n 2 − n + 3) + ( n 2 − n + 5) + ... + ( n 2 + n − 1) (d) None of the above

i.e.

Sol. (c) 13 = 1⋅ (1 − 1) + 1, 2 3 = (2 ⋅ 1 + 1) + 5

Example 20. Three arithmetic means between 3 and 19 are (a) 7, 11, 15 (b) 7, 10, 13 (c) 7, 12, 17 (d) None of these

33 = (3 ⋅ 2 + 1) + 9 + 11 43 = (4 ⋅ 3 + 1) + 15 + 17 + 19, etc. ∴

Example 21. If (2n + r )r, n ∈ N , r ∈ N is expressed the sum of k consecutive odd natural numbers, then k is equal to (a) r (b) n (c) r +1 (d) n +1

Example 24. Along a road lies an odd number of stones placed at intervals of 10 m. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried out the job starting with the stone in the middle, carrying stones in succession, thereby covering a distance of 4.8 km. Then, the number of stones is (a)15 (b) 29 (c) 31 (d) 35

Sol. (a) (2 n + r )r = (n + r )2 − n2

Sol. (c) Let there be 2 n + 1 stone i.e. n stones on each side

∴

A1 = 3 + d = 3 + 4 = 7 A2 = 3 + 2d = 3 + 8 = 11 A3 = 3 + 3d = 3 + 12 = 15

Hence, the required AM’s are 7,11,15. X

n3 = {n ⋅ (n − 1) + 1} + ... next term being 2 more than the previous = (n2 − n + 1) + (n2 − n + 3) + ... + (n2 + n − 1)

Sol. (a) Let A1, A2 , A3 be three AM’s between 3 and 19. Then, 3, A1, A2 , A3 19 are in AP whose common difference is 19 − 3 =4 d= 3+1

Example 23. Observe that 13 = 1, 2 3 = 3 + 5, 3 3 = 7 + 9 + 11, 4 3 = 13 + 15 + 17 + 19. Then, n 3 as a similar series is   n( n − 1)     ( n + 1) n   (a) 2  + 1 + 1 + 1 − 1 + 2        2   2    ( n + 1) n  + ... + 2  + 1 + 2n − 3     2 (b) ( n 2 + n + 1) + ( n 2 + n + 3) + ( n 2 + n + 5) + ... + ( n 2 + 3n − 1)

Now, A1 = a + d, A2 = a + 2d, ... and An = a + nd of n AM's between a and b is n  

3

= 12 + 12 + 2 2 + 2 2 + ... + 52 + 52 + 2S

Insertion of n Arithmetic Means between a and b

Ø Sum

Example 22. The sum of the products of the numbers ± 1, ± 2, ± 3, ± 4, ± 5 taking two at a time is (a)165 (b) − 55 (c) 55 (d) None of the above

Sequence and Series

Now, a, A, b are in AP.

= {1 + 3 + 5 + ... to (n + r ) terms} − {1 + 3 + 5 + ... to n terms} as (n + r )2 − n2 = {1 + 3 + 5 + ... (n + r ) terms) − {1 + 3 + 5 + ... n terms) ∴ Sum of r consecutive odd natural numbers, k=r

X

of the middle stone. The man will run 20 m, to pick up the first stone and return, 40 m for the second stone and so on. So, he runs (n/2 ) {2 × 20 + (n − 1)20} = 10n(n + 1) m to pick up the stones on one side and hence 20n(n + 1) m to pick up all the stones. ∴ 20n(n + 1) = 4800 or n = 15 Hence, there are 2 n + 1 = 31 stones.

45

3

Work Book Exercise 3.1

Objective Mathematics Vol. 1

1 If the nth term of an AP 5, 8, 11, ... is 320, then n

9 The sum of the first ten terms of an AP is four times the sum of the first five terms, then the ratio of the first term to the common difference is

is equal to 105

104

106

112

1/2

2 If the 5th term of an AP is 11 and the 9th term is 2

1

0

b −4

4

c 2

− 6 pqr 2 pqr

d −2

x − 63

63 − x

middle term is 8

3

a = 2d 2a = d

2

6 The sum of the integers from 1 to 100 that are

a1 + a5 + a10 + a15 + a20 + a24 = 225, then a1 + a2 + a3 + .... + a23 + a24 is equal to

3050 None of these

909

7 If N which is the set of natural numbers, is

8 If

1

2

1 + 2+ 5

1 + 5+ 8

3n + 2 − 2

is 37

900

0

16c

14 Sum of n terms of the series

3 + 5 + 7 + ... + n terms = 7, then the value of n 5 + 8 + 11 + ...+ 10 terms 36

750

value of a − 4 b + 6 c − 4 d + e is

62525 62555

35

75

13 If the numbers a, b, c , d , e are in AP, then the

partitioned into subsets S1 = {1},S 2 = {2, 3}, S 3 = { 4, 5, 6}, ... , then the sum of the members in S 50 is 62255 65225

a=d None of these

12 If a1, a2 , a3 ,... are in AP such that

divisible by 2 or 5, is 3000 3600

4 pqr None of these

difference d. Let S k denotes the sum of first k S terms. If kx is independent of x, then Sx

63 + x

5 If the sum of three numbers of an AP is 24, then 6

3

11 Consider an AP with first term a and common

4 The 15th term of sequence x −7, x − 2 and x + 3 is x + 63

3

to

3 If p − 1, p + 3 and 3 p − 1 are in AP, then p is equal to a

4

10 If p, q and r are in AP, then p + r − 8 q is equal 3

7, then the 14th term is −1

1/4

2

40

3n + 2 +

2

1 + ... is 8 + 11 1 ( 3n + 2 − 2 ) 3 None of these

Geometric Progression (GP) A sequence of non-zero numbers is called a geometric progression, if the ratio of a term and the term preceding to it is always a constant quantity. The constant ratio is called the common ratio of the GP. In other words, a sequence a1 , a 2 , a 3 , ..., a n , ... is called a geometric progression, if an+ 1 = Constant, ∀ n ∈ N an X

Example 25. Show that the sequence given by a n = 3 (2 n ), ∀ n ∈ N is a GP. Also, find its common ratio. Sol. We have, an = 3 (2 n ) and an + 1 = 3 (2 n + 1 ) an + 1

Now,

3 (2 n + 1 ) =2 3 (2 n )

= 2 (constant), ∀ n ∈ N. So, the given an sequence is a GP with common ratio 2. Clearly,

46

an an + 1

=

X

Example 26. Which of the following is a GP? (a) 2, 4, 8, 16, ... 1 −1 1 −1 (b) , , , ,... 9 27 81 243 (c) 0.01, 0.0001, 0.000001,... (d) All of the above Sol. (d) a2 a a = 2 , 3 = 2, 4 = 2 and so on a1 a2 a3 1 a2 −1 (b) We observe, a1 = , = , 9 a1 3 a3 −1 a4 −1 and so on. = , = a2 3 a3 3 a (c) We have, a1 = 0.01, 2 = 0.01, a1 a3 a = 0.01, 4 = 0.01 and so on. a2 a3 It is observed that in each case, every term except the first term bears a constant ratio to the term immediately preceding it. Thus, options (a), (b) and (c) all are in GP. (a) We have, a1 = 2 ,

If a is the first term of a GP and r is the common ratio, then its nth term (general term) t n is given by t n = ar n − 1 X

Example 27. The nth and 10th terms of a GP 5, 25, 125, ..., are respectively (a) 5 n − 1 , 5 9 (b) 5 n − 2 , 5 8 (d) None of these (c) 5 n , 510

X

Sol. (c) Here, a = 5 and r = 5. Thus, the nth term of the given GP is given by an = ar n − 1

an = 5 (5)n − 1 = 5n

⇒

Sol. (c) Let the four numbers be a, ar, ar 2 and ar 3 .

Now, for n = 10, we have a10 = 510 X

It is given that,

Example 28. The third term of a geometric progression is 4. The product of the first five terms is (a) 4 3 (b) 4 5 4 (c) 4 (d) None of these

and ⇒ and

Selection of Terms in GP

3 4 5

Common ratio

Properties of Geometric Progressions i.

X

r r2

Example 29. If the sum of three numbers in GP is 38 and their product is 1728, then numbers are (a) 8, 12, 18 (b) 8, 16, 32 (c) 18, 12, 8 (d) None of these Sol. (a,c) Let the three numbers be a , a and ar. r

⇒

product = 1728 a. . a ar = 1728 r a3 = 1728

⇒ Q

a = 12 Sum = 38

Then, ⇒

∴ ⇒

1 a + a + ar = 38 ⇒ a  + 1 + r r  1 + r + r2   = 38 12  r  

If all the terms of a GP is multiplied or divided by the same non-zero constant, then it remains a GP with the same common ratio.

Example 31. If 5, 25, 125,... are in GP, then 1, 5, 25,... are in (a) AP (b) GP (c) HP (d) None of these Sol. (b)Q5, 25, 125,... are in GP.

r

On dividing each term by 5, we get the following sequence 1, 5, 25 ,..., which is again a GP as here a a a1 = 1, 2 = 5, 3 = 5 and so on. a1 a2

If the product of the numbers is not given, then the numbers can be taken as a, ar, ar 2 , ar 3 , ... . X

...(i) ...(ii)

Hence, the numbers are 3, − 6, 12, − 24.

Sometimes, it is required to select a finite number of terms in GP. It is always convenient, if we select the terms in the following manner: Terms a , a, ar r a a , , ar, ar 3 r3 r a a , , a, ar, ar 2 r2 r

ar − ar 3 = 18 a (r 2 − 1) = 9 ar (1 − r 2 ) = 18

On putting r = − 2 in a (r 2 − 1) = 9, we get a = 3

∴Product of first five terms = a. ar. ar 2 . ar 3 . ar 4 = a5 r10 = (ar 2 )5 = 45

ar 2 − a = 9

On dividing Eq. (ii) by Eq. (i), we get ar (1 − r 2 ) 18 = ⇒r = − 2 9 a (r 2 − 1)

Sol. (b) Here, t 3 = 4 ⇒ ar 2 = 4

Number of terms

Example 30. If four numbers in GP are such that the third number is greater than the first by 9 and the second number is greater than the fourth by 18, then the numbers are (a) 3, 6, 12, 24 (b) − 3, 6, − 12, 24 (c) 3, − 6, 12, − 24 (d) None of these

3 Sequence and Series

⇒ 6 + 6r + 6r 2 = 19 r ⇒ 6r 2 − 13r + 6 = 0 ⇒ (3r − 2 ) (2 r − 3) = 0 ⇒ r = 3/2 and 2/3 Hence, on putting the values of a and r, the required numbers are 8, 12, 18 or 18, 12, 8.

General Term of a GP

ii. X

The reciprocals of the terms of a given GP form a GP.

1 1 1 1 If , 2 , 3 , 4 ,... are in GP, 2 2 2 2 then 2, 2 2 , 2 3 ,... are in (a) AP (b) GP (c) HP (d) None of these

Example 32.

Sol. (b) On reciprocating each term of a GP, we get the

r  = 38 

following sequence 2, 2 2 , 2 3 , 2 4 ,... which is again a GP as a a here a1 = 2, 2 = 2, 3 = 2 and so on. a1 a2

iii.

If each term of a GP is raised to the same power, then the resulting sequence also form a GP.

47

Objective Mathematics Vol. 1

3

X

Example 33. If a, b and c are in GP, then a 1/ 2 , b1/ 2 , c1/ 2 are in (a) AP (b) GP (c) HP (d) None of these

X

Sol. (a) Since, 3, 9 and 27 are in GP.

Sol. (b) Since, a, b and c are in GP. b c = ⇒ a b Hence, a1/ 2 , b1/ 2

iv.

X

 b    a

1/ 2

c =    b

1/ 2

⇒

b1/ 2 a1/ 2

=

c1/ 2 b1/ 2

and c1/ 2 are in GP.

In a finite GP, the product of the terms equidistant from the beginning and the end is always same and is equal to the product of the first and the last terms.

X

n

∑ n,

n =1

n

vi.

∴ ⇒

X

is 1, 25, 625 ,..., which is a GP as every term except the first term bears a constant ratio to the term immediately preceding it.

If a1 , a 2 , a 3 , ... , a n , ... are in GP of non-zero non-negative terms, then log a1 , log a 2 , ..., log a n K are in AP and vice-versa.

 5c 

 3b 

 3b  5c a .   = a 3b  5c  3b = 5c b 2 = ac 9ac = 25c 2 9a = 25c 9a = 5c = 3b 5 a b c = = 5 3 9/ 5

⇒ b+ c< a Since, sum of two sides is smaller than the third side. So, triangle is not formed.

2

Sol. (b) Clearly, the sequence of odd terms of the given GP

48

⇒

n =1

Example 36. If 1, 5, 25, 125, 625,... forms a GP, then the sequence consisting odd term of the given GP, is (a) AP (b) GP (c) HP (d) None of these

vii.

⇒

3 ∑ n are in GP.

If the terms of a given GP are chosen at regular intervals, then the new sequence so formed also form a GP.

 3b   5c  If log   , log   and  5c  a

5c 3b a are in GP. , and a 5c 3b

⇒ Also,

n

2

Example 38.

2

n

n (n + 1) 10 n (n + 1) (2 n + 1) n (n + 1) , are in GP. , . 2 3 6 4 2 2 2 2 . 10 n (n + 1) (2 n + 1) n (n + 1) n (n + 1)2 = ⋅ ⇒ 9 .36 2 4 ⇒ 20 (2 n + 1)2 = 81n (n + 1) ⇒ n2 + n − 20 = 0 ⇒ n = 4

⇒

2 log 9 = log 3 + log 27

⇒

10 If ∑ n, ⋅ ∑ n 2 and ∑ n 3 3 n =1 n =1 n =1

10 n 2 ⋅ ∑ n and 3 n =1

⇒

⇒

are in GP, then the value of n is (a) 2 (b) 3 (c) 4 (d) None of these Sol. (c) Since,

log 92 = log(3 × 27 )

 a

Three non-zero numbers a, b and c are in GP iff b 2 = ac.

Example 35.

⇒

  Sol. (d) Since, log  5c  , log  3b  and log  a  are in AP.

∴ Product of a and b = Product of first and last terms = 2 ⋅ 2 5 = 2 6 = 64

X

92 = 3 × 27

a log   are in AP, where a, b and c are in GP,  3b  then a, b, c are the lengths of sides of (a) an isosceles triangle (b) an equilateral triangle (c) a scalene triangle (d) None of these

Sol. (b) Since, 2, a, 2 3 , b and 2 5 are in GP..

n

∴

Hence, log 3, log 9 and log 27 are in AP.

Example 34. If 2, a, 2 3 , b and 2 5 are in GP, then ab is (a) 32 (b) 64 (c) 128 (d) 55

v.

Example 37. If 3, 3 2 , 3 3 are in GP, then log 3, log 9 and log 27 are in (a) AP (b) GP (c) HP (d) None of these

X

Example 39. A circle of radius r is inscribed in a square. The mid-points of sides of the square have been connected by line segment and a new square resulted. The sides of the resulting square were also connected by segment so that a new square was obtained and so on, then the radius of the circle inscribed in the nth square is  1−n  (a) 2 2  r   3 − 3 n   (b) 2 2  r    −n (c) 2 2  r    − 5 − 3n  (d) 2 2  r  

Sol. (c) S n =

Side of square S 2 = r 2

r

S2

S1

r

r√2

r

2r  1    2  2

1  Side of square S 3 = 2 r    2 1  Side of square S n = 2 r    2 ∴

Radius = r(2

2 −1

3 −1

1 = 2 r   2 

2 −1

X

2

1  = 2 r   and so on.  2

n −1

− 1/ 2 n − 1

)

X

( 1− n )/ 2

= r(2

) and so on.

| r| > 1 r =1

X

If | r| r represents exterior of the circle | z − z 0 | = r. (iv) General equation of a circle The general equation of the circle is zz + az + az + b = 0, where a is complex number and b ∈ R . ∴ Centre and radius are −a and | a | 2 − b, respectively. (v) Equation of circle in diametric form If end points of diameter represented by A ( z1 ) and B ( z 2 ) and P ( z ) is any point on the circle, then ( z − z1 )( z − z 2 ) + ( z − z 2 )( z − z1 ) = 0 which is required equation of circle in diametric form. 129

X

Objective Mathematics Vol. 1

4

Example 64. A circle whose radius is r and centre z 0 , then the equation of the circle is (a) zz − zz 0 − zz 0 + z 0 z 0 = r 2 (b) zz + zz 0 − zz 0 + z 0 z 0 = r 2 (c) zz − zz 0 + zz 0 − z 0 z 0 = r 2 (d) None of the above

 z − z1  (x) arg   = 0 or π  z − z2  ⇒ Locus of z is a straight line passing through z1 and z 2 . X

Sol. (a) Equation of circle| z − z0|2 = r 2 ⇒

( z − z0 )( z − z0 ) = r 2 ⇒ ( z − z0 )( z − z0 ) = r 2 zz − zz0 − zz0 + z0 z0 = r 2

X

.

Example 65. The set of values of k for which the equation zz + ( −3 + 4i) z − (3 + 4i) z + k = 0 represents a circle is (a) ( −∞ , 25) (b) (25 , ∞ ) (c) (5, ∞ ) (d) ( −∞, 5)

Example 66. The complex numbers z = x + iy z − 5i which satisfy the equation = 1, lie on z + 5i (a) the X-axis (b) the straight line y = 5 (c) a circle passing through the origin (d) None of the above Sol. (a) Given, z − 5 i = 1 ⇒ | z − 5 i| = | z + 5 i|

z + 5i [if| z − z1| = | z – z2|, then it is a perpendicular bisector of z1 and z2 ]

Sol. (a) We have, zz + (−3 + 4i ) z − (3 + 4i ) z + k = 0 This equation represents a circle with centre

Y

a + (3 − 4i ) and Radius = (−3 − 4 i ) − k = 25 − k 2

(0, 5)

For circle to exist, we must have 25 − k > 0 ⇒ k < 25 Hence, the given equation will represent a circle if k < 25.

X¢

130

(vi) | z − z1 | 2 + | z − z 2 | 2 = | z1 − z 2 | 2 ⇒ Locus of z is a circle with z1 and z 2 as the extremities of diameter. (vii) | z − z1 | = k | z − z 2 |, ( k ≠ 1) ⇒ Locus of z is a circle.  z − z1  (viii) arg   = α (fixed)  z − z2  ⇒ Locus of z is a segment of circle.  z − z1  (ix) arg   = ± π /2  z − z2  ⇒ Locus of z is a circle with z1 and z 2 as the vertices of diameter.

X

(0, –5)

Loci in Complex Plane If z is a variable point and z1 , z 2 are two fixed points in the argand plane, then (i) | z − z1 | = | z − z 2 | ⇒ Locus of z is the perpendicular bisector of the line segment joining z1 and z 2 . (ii) | z − z1 | + | z − z 2 | = k , if | k | > | z1 − z 2 | ⇒ Locus of z is an ellipse. (iii) | z − z1 | + | z − z 2 | = | z1 − z 2 | ⇒ Locus of z is the line segment joining z1 and z 2 (iv) | z − z1 | − | z − z 2 | = | z1 − z 2 | ⇒ Locus of z is a straight line joining z1 and z 2 but z does not lie between z1 and z 2 . (v) | z − z1 | – | z − z 2 | = k , where k < | z1 − z 2 | ⇒ Locus of z is a hyperbola.

O

Y¢

∴ Perpendicular bisector of (0, 5) and (0, − 5) is X-axis. X

Example 67. The equation | z − i | + | z + i | = k, k > 0 can represent an ellipse, if k 2 is (a) 4 (d) None of these Sol. (c)| z − z1 | + | z – z2 | = k represents ellipse, if | k | > | z1 − z2| Thus,| z − i | + | z − i | = k represents ellipse, if | k | > |i + i| or | k | > |2 i| ∴ | k | > 2 or k 2 > 4

X

Example 68. The equation | z + i | − | z − i | = k represents a hyperbola if (a) −2 < k < 2 (b) k >2 (c) 0 < k < 2 (d) None of these Sol. (a)| z − z1| − | z − z2| = k, represents hyperbola, if| k | < |z1 − z2| Thus, | z + i| − | z − i| = k, represents hyperbola, if | k | < | i + i | or| k | < 2 ⇒ −2 < k < 2

X

Example 69. If z = 1 − t + i t 2 + t + 2, where t is a real parameter. The locus of z in the argand plane is (a) a hyperbola (b) an ellipse (c) a straight line (d) None of these

z−i

x = 1− t , y = t + t + 2

⇒

2

Eliminating t, y2 = t 2 + t + 2 = (1 − x)2 + 1 − x + 2

2

3 7 or y2 −  x −  =  2 4 which is a hyperbola.

X

Example 70. Identify the locus of z, if r2 z=a + , > 0. z−a Sol. z = a + r

2

z−a

⇒ z−a=

⇒

( z − a) ( z − a ) = r 2

⇒

| z − a|2 = r 2

2

r z−a

X

⇒

| z − z1|2 + | z − z2|2 = | z1 − z2|2

⇒

k = | z1 − z2|2 = |2 + 3i − 4 − 3i|2 = | − 2 |2 = 4

Example 72. If | z + 1| = 2 | z – 1,| then the locus described by the point z in the argand diagram is a (a) straight line (b) circle (c) parabola (d) None of these z = x + iy ⇒ x + iy + 1 = 2| x + iy − 1| | ( x + 1) + iy| = 2 |( x − 1) + iy| ( x + 1)2 + y2 = 2 [( x − 1)2 + y2 ] x 2 + y2 − 6 x + 1 = 0

which is the equation of a circle.

Example 73. The locus of z given by is (a) a circle (c) a straight line

| ( x − 2 ) + i ( y + 1)| = | ( x − 3) + i ( y − 1)|

⇒

( x − 2 )2 + ( y + 1)2 = ( x − 3)2 + ( y − 1)2

Example 75. If z = z 0 + A ( z − z 0 ), where A is a constant, then prove that locus of z is a straight line. …(i) Az − z − Az0 + z0 = 0 ⇒ …(ii) A z − z − A z0 + z0 = 0 ⇒ Adding Eqs. (i) and (ii), we get ( A − 1) z + ( A − 1) z − ( Az0 + Az0 ) + z0 + z0 = 0 This is of the form az + az + b = 0, where a = A − 1 and b = − ( Az0 + Az0 ) + z0 + z0 ∈R

z −1 =1 z−i

X

Example 76. Plot the region represented by  z + 1  2π π in the argand plane. ≤ arg  ≤ 3  z −1  3 Sol.

Let us take  z + 1 2 π , clearly z arg   = 3  z − 1 (1, 0) lies on the minor arc of the (–1, 0) 2π/3 circle passing through (1, 0) and (−1 , 0). π/3  z + 1 π Similarly, arg   =  z − 1 3 means that z is lying on the major arc of the circle passing through (1, 0) and (−1, 0). Now, if we take any point in the region included between the two arcs, say P1( z1 ), we get  z + 1 2 π π ≤ arg  ≤ 3 3  z − 1

Thus,

(b) an ellipse (d) a parabola

2 x + 4y − 5 = 0

Hence, locus of z is a straight line.

Sol. (b)| z + 1| = 2 | z − 1|

X

⇒

Sol. z = z0 + A ( z − z0 )

Sol. (b) As z1 and z2 are the extremities of diameter.

⇒

x2 + ( y − 1)2

Example 74. If z = x + iy and | z − 2 + i | = | z − 3 − i |, then locus of z is (a) 2x + 4 y − 5 = 0 (b) 2x − 4 y − 5 = 0 (c) x + 2 y = 0 (d) x − 2 y + 5 = 0

⇒

Example 71. If the equation | z − z1 | 2 + | z − z 2 | 2 = k represents the equation of a circle, where z1 = 2 + 3 i, z 2 = 4 + 3 i are the extremities of a diameter, then the value of k is 1 (b) 4 (a) 4 (c) 2 (d) None of these

⇒ ⇒

( x − 1)2 + y 2 =

⇒ x 2 + 4 − 4 x + y2 + 1 + 2 y = x 2 + 9 − 6 x + y2 + 1 − 2 y

X

Putting

⇒

4

Sol. (a) | z − 2 + i| = | z − 3 − i|

⇒ | z − a| = r Hence, locus of z is circle having centre a and radius r.

X

|( x − 1) + iy| = | x + i ( y − 1)|

⇒ 2x = 2y or x− y=0 which is the equation of a straight line.

2

3 7 =  x −  +  2 4

X

⇒

Complex Numbers

Sol. (c) z − 1 = 1⇒ | z − 1| = | z − i|

Sol. (a) x + iy = 1 − t + i t 2 + t + 2

 z + 1 2 π π represents the shaded region ≤ arg  ≤ 3  z − 1 3

excluding the points (1, 0) and (−1, 0).

131

i.

Objective Mathematics Vol. 1

4

Some Important Results Four points z1 , z 2 , z 3 and z 4 in anti-clockwise order will be concyclic, if and only if  z − z4   z2 − z3  θ = arg  2  = arg    z1 − z 4   z1 − z 3 

⇒

  z − z 4   z1 − z 3   ⇒ arg   2    = 2nπ   z1 − z 4   z 2 − z 3    z − z 4   z1 − z 3  ⇒ 2   is real and positive.  z1 − z 4   z 2 − z 3 

Sol.

R (z2) S(0)

β

X

Q(z1) P(z3)

⇒

132

2 1 1 = + z1 z2 z3 z2 − z1 z1 − z3 = z1 z2 z3 z1

2

4 = a, then find the z greatest and least values of | z | .

Example 79. If z +

z 2 a z + = . Now, put = w 2 z 2 2 1 a w+ = ∴ w 2

Sol.

a + 2 Now, greatest value of|w| =

a2 + 4 4 and| z| = 2|w| 2

a a2 + + 4 2 4 Similarly, we can find least value of a a2 | z| = − + + 4 2 4 Greatest value of| z| =

X

Example 80. Let z1 and z 2 be two non-real complex cube roots of unity and 2 2 z − z1 + z − z 2 = λ be the equation of a circle with z1 , z 2 as ends of a diameter, then the value of λ is (a) 4 (b) 3 (c) 2 (d) 2 2

Sol. (b) z − ω 2 + z − ω2 − λ could be shows as given below : ⇒λ = ω−ω =

α

2

−a + a + 4

and

Example 77. Show that the points 3 + 4i, 3 − 4i, −4 + 3i, −4 − 3i are concyclic.

Example 78. If z1 , z 2 and z 3 are complex 2 1 1 numbers, such that = + , then show that z1 z 2 z 3 the points represented by z1 , z 2 and z 3 lie on a circle passing through the origin.

a + a2 + 4

2

Sol. Let z1 = 3 + 4i , z2 = 3 − 4i , z3 = − 4 + 3i

X

1 = a, then the greatest and least z

values of z are respectively

z − z   z − z  if  2 4  1 3  is positive or negative.  z1 − z4   z 2 − z3 

= a real number Hence, the given points are concyclic.

α = π −β ⇒ α + β = π

If z +

ii.

Ø Points z1, z 2, z3 and z4 (not necessarily in order) will be concyclic,

and z4 = − 4 − 3i Then,  z2 − z4   z1 − z3  (7 − i )(7 + i )    = − 7 − ( + 7 i )(7 − 7 i ) z z z z  1 4  2 3 25 50 = = 49 + 49 49

 z3     z2 

Hence, the points are concyclic.

 z − z4   z2 − z3  ⇒ arg  2  −arg   = 2nπ , n ∈ I z − z  1  z1 − z 3  4

X

 z − z1  arg  2  = π − arg  z3 − z1 

⇒

2

z–ω

2

− 1+ i 3 − 1− i 3 − 2 2

− 1+ i 3 + 1+ i 3 = 2

z

z – ω2

2

2

O ω – ω2

= |i 3|2 = 3 ⇒ ⇒

1 1 1 1 − = − z1 z2 z3 z1 z2 − z1 z2 =− z3 − z1 z3

⇒

 z − z1   z  arg  2  = arg  − 2  − z z  3  z3  1

⇒

 z − z1  z  arg  2  = π + arg  2   z3 − z1   z3 

Logarithm of Complex Numbers Let

z = α + iβ = re i ( θ + 2nπ )

log z = log( re i ( θ + 2nπ ) ) = log r + i(θ + 2nπ ) = log | z | + i arg z + 2nπi If we put n = 0, we get principal value of log z. ∴ Principal value of log z = log| z | + i arg z.

Sol. (a) Let z = i , loge z = loge i = i loge i = i loge e i

2

ei

= iπ / 2

2

log e ( 2 −

elog e ( 2 −

=

π π = i 2 loge e = − 2 2 z = e −π / 2

⇒ X

i

4

iz − iz Sol. (a) We know that, cos z = e + e

Example 81. The value of i i is (a) e −π / 2 (b) e π / 2 (c) e π / 4 (d) None of these

(2 −

=

+ e −i 2

2

log e ( 2 −

3)

3 ) −1

+ elog e ( 2 − 3 ) 2 3 )−1 + (2 − 3 ) 2  1 2 + 3 1 + (2 − + 2 − 3 =  − 3  2 4− 3

1 2  2 1 = [2 + 2 =

Example 82. If z = i log e (2 − 3 ), then the value of cos z is (a) 2 (b) −2 (c) 2i (d) −2i

3)

Complex Numbers

X

3+2−

 3 ) 

3]= 2

Work Book Exercise 4.4 z −1 = 1, represents z+1

1

a a circle c a straight line

9 The figure formed by four points 1 + 0 i, −1 + 0i, 3 + 4i and b an ellipse d None of these

2 | z − 4| 0 c Re ( z) > 2

b Re ( z) < 0 d None of these

3 If 2 z1 − 3 z2 + z3 = 0, then z1, z2 , z3 are

a b c d

25 on the argand plane is −3 − 4i

a parallelogram but not a rectangle a trapezium which is not equilateral a cyclic quadrilateral None of the above

10 If i = −1 , then define a sequence of complex number by z1 = 0, zn + 1 = zn 2 + i for n ≥ 1. In the

represented by

complex plane, how far from the origin is z111 ?

a b c d

a 1

three vertices of a triangle three collinear points three vertices of a rhombus None of the above

4 If z = x + iy, such that| z + 1| =| z − 1| and  z − 1 π amp   = , then  z + 1 4 a b c d

x= x= x= x=

3

d

110

parts are integers and satisfy the relation zz 3 + z 3 z = 350, forms a rectangle on the argand plane, the length of whose diagonal is b 10 units

c 15 units

d 25 units

2π 3 π b same as the locus of z for arg z = 3 c the part of the straight line 3 x + y = 0 with y < 0, x> 0 d the part of the straight line 3 x + y = 0 with y > 0, x < 0 a same as the locus of z for arg z =

collinear concyclic the vertices of irregular polygon None of the above

6 The equation zz + az + az + b = 0, b ∈ R represents a circle, if a | a|2 = b

b | a|2 ≥ b

c | a|2 < b

d None of these

7 If z = (λ + 3) + i 3 − λ2 , where| λ| < 3, then locus of z is b parabola d None of these

8 If a point P denoting the complex number z moves on the complex plane such that |Re ( z )| + |Im ( z )| = 1, then the locus of z is a a square c two intersecting lines

c

π 12 The locus of z, for arg z = − is 3

2 + 1, y = 0 0, y = 2 + 1 0, y = 2 − 1 2 − 1, y = 0

a circle c line

2

11 The complex numbers whose real and imaginary

a 5 units

5 If z 8 = ( z − 1)8 , then the roots of this equation are a b c d

b

b a circle d a line

13 If z1 and z2 are two complex numbers and

z1 + z2 π = but| z1 + z2| ≠ | z1 − z2|, then the z1 − z2 2 figure formed by the points represented by 0, z1, z2 and z1 + z2 is

arg

a b c d

a parallelogram but not a rectangle or a rhombus a rectangle but not a square a rhombus but not a square a square

14 z1 and z2 are two distinct points in an argand plane. If a| z1| = b| z2|, where a, b ∈ R, then the az bz point 1 + 2 is a point on the bz2 az1 a b c d

line segment [−2, 2] of the real axis line segment [−2, 2] of the imaginary axis unit circle| z| = 1 the line with arg z = tan−1 2

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Objective Mathematics Vol. 1

4

15 The roots z1, z2 , z3 of the equation z 3 + 3 αz 2 + 3 βz + γ = 0 correspond to the points A, B and C on the complex plane. Then, the triangle is equilateral, if a α =β c α 2 = 3β 2 2

b α =β d 3α 2 = β 2

z22

a equilateral triangle c isosceles triangle

b right angled triangle d None of these

17 Complex numbers z1, z2 , z3 are the vertices A, B and C, respectively of an isosceles right angled triangle, right angled at C. Then, ( z1 − z2 )2 is equal to a 2( z1 − z3 )( z3 − z2 ) c ( z1 − z3 )( z3 − z2 )

complex plane, has intercept on the imaginary axis equal to a 5

1 ( z1 + z3 )( z3 − z1 ) 2 d None of these b

c

z1 z2 z3 zz z=− 1 2 z3 z=

b d

z3 z2 z1 z3 z2 z= z1 z=−

19 If z1 and z2 are the roots of the equation

z 2 + pz + q = 0, where the coefficients p and q may be complex number. Let A and B represent z1 and z2 in the complex plane. If ∠AOB = α ≠ 0 and OA = OB, where O is the origin, then p2 is equal to

α a 4 q cos   2 α c q cos 2 4 2

b 2q cos α 2

d None of these

 z + 4 1 20 If Re   = , then z is represented by a 2z − i 2 point lying on a a circle c a straight line

b an ellipse d None of these

21 Suppose z1, z2 and z3 are the vertices of an

equilateral triangle inscribed in the circle| z| = 2. If z1 = 1 + 3 i and z1, z2 and z3 are in the clockwise sense, then a z2 = 1 − 3i , z3 = 2 b z2 = 2 , z3 = 1 − 3i c z2 = − 1 + 3i , z3 = − 2 d None of the above

22 Suppose z1, z2 and z3 are the vertices of an equilateral triangle circumscribing in the circle | z| = 2. If z1 = 1 + 3 i and z1, z2 and z3 are in the anti-clockwise sense, then z2 is

134

a 1− 3i b 2 1 c (1 − 3 i ) 2 d None of the above

c −

5 2

d −5

| z| ≥ 3, then the least value of z + 5 3 11 c 3

b

a

1 is equal to z

8 3

d None of these

a b , z2 = , z3 = a − bi , for a, b ∈ R and 1− i 2+i z1 − z2 = 1, then the centroid of the triangle formed by the points z1, z2 , z3 in the argand plane is given by

25. If z1 =

a

numbers z1, z2 and z3 . If the circumcentre of the ∆ABC is at the origin and the altitude AD of the triangle meets the circumcircle again at P, then the complex number representing point P is

5 2

24 If the complex number z satisfies the condition

18 A, B and C are points represented by complex

a

b

2

+ + 2 z1 z2 cos θ = 0, then the points represents by z1, z2 and the origin form

16 If

z12

23 The straight line (1 + 2 i )z + (2 i − 1)z = 10 i on the

1 (1 + 7 i ) b 9

1 (1 + 7 i ) c 3

1 (1 − 3i ) d 3

1 (1 − 3i ) 9

26. Let λ ∈ R. If the origin and the non-real roots of 2 z 2 + 2 z + λ = 0 form three vertices of an equilateral triangle in the argand plane, then λ is a 1 c 2

b 2 /3 d −1

27 If a, b, c and u, v, w are complex numbers representing the vertices of two triangles, such that c = (1 − r ) a + rb and w = (1 − r ) u + r v . When r is a complex number, then the two triangles a have the same area c are congruent

b are similar d None of these

28. On the complex plane, ∆OAP and ∆OQR are similar and I(OA) = 1. If the points P and Q denote the complex numbers z1 and z2 , then the complex numbers z denoted by the point R is given by

Y R(z)

Q(z2) θ

P(z1)

θ

O

A (1, 0)

a

z1 z2

b

c

z2 z1

d

X

z1 z2 z1 + z2 z2

29 Intercept made by the circle zz + αz + αz + r = 0 on the real axis on complex plane, is a

(α + α ) − r

b

(α + α )2 − 2 r

c

(α + α )2 + r

d

(α + α )2 − 4r

30 The equation of the radical axis of the two circles represented by the equations,| z − 2| = 3 and | z − 2 − 3 i | = 4 on the complex plane is a b c d

3y + 1 = 0 3y − 1 = 0 2y − 1= 0 None of the above

WorkedOut Examples Type 1. Only One Correct Option b = cos β + i sin β, a b c + + =1, then c = cos γ + i sin γ and b c a cos (α − β) + cos (β − γ ) + cos ( γ − α ) is equal to 3 3 (b) − (a) 2 2 (c) 0 (d) 1

Ex 1. If

Sol.

a = cos α + i sin α,

a b c + + =1 b c a cis α cis β cis γ ⇒ + + = 1, cis β cis γ cis α where, cis θ represents cos θ + i sin θ. ⇒ cis (α − β ) + cis(β − γ ) + cis( γ − α ) = 1 Equation real parts of both sides cos (α + β ) + cos (β − γ ) + cos (γ − α ) = 1 Hence, (d) is the correct answer.

Ex 2. The locus of the centre of a circle which touches the circles | z − z1 | = a and | z − z 2 | = b externally (z, z1 and z 2 are complex numbers) will be (a) an ellipse (c) a circle

α  α tan α − i  sin + cos   2 2 Ex 4. If is α 1 + 2i sin 2 imaginary, then α is given by π π (b) nπ − (a) nπ + 4 4 π (c) ( 2n + 1) π (d) 2nπ + 4 α    α  tan α − i  sin + cos   2 2  =0 Sol. Re  α   1 + 2i sin   2  

 α  α   α  tan α − i  sin + cos   1 − 2i sin      2 2 2   ⇒ Re   =0 α   2 + 1 4 sin   2  

⇒

Sol. Let A (z1 ), B (z2 ) be the centres of given circles and P be

⇒

the centre of the variable circle which touches given circles externally, then | AP | = a + r and | BP | = b + r,

⇒ ⇒

⇒ sin α (1 − cosα ) = cosα (1 − cosα ) ⇒ sin α = cosα , cosα = 1 π ⇒ α = nπ + 4 or α = 2nπ, n ∈ I Hence, (a) is the correct answer.

  a − ib   Ex 5. The expression tan  i log    reduces to  a + ib    (a)

Ex 3. If arg ( z1 ) = arg ( z 2 ), then (a) z 2 = kz1−1 , k > 0 (c) | z 2 | = | z1 |

Sol. z1 = ⇒

α α α    tan α − 2 sin 2  cos 2 + sin 2     α α α  Re  − i  tan α ⋅ 2 sin + sin + cos   = 0 2 2 2  −  α   1 + 4 sin 2   2 2α tan α = 2 sin + sin α 2 sin α = sin α + 1 − cosα cosα sin α = sin α ⋅ cosα + cosα − cos2 α

(b) a hyperbola (d) None of these

where r is the radius of the variable circle. On subtraction, we get | AP | − | BP | = a − b ⇒ || AP | − | BP || = | a − b | , a constant. Hence, locus of P is (i) right bisector AB, if a = b. (ii) a hyperbola, if | a − b | < | AB | = | z2 − z1 |. (iii) an empty set, if | a − b | > | AB | = | z2 − z1 |. (iv) set of all points on line AB except those which lie between A and B, if | a − b | = | AB | ≠ 0. Hence, (d) is the correct answer.

(b) z 2 = kz1 , k > 0 (d) None of these

z1z1 = | z1 |2 z1−1 z1 arg (z1−1 ) = arg (z1 ) = arg (z2 )

⇒ z2 = kz1−1 (k > 0) Hence, (a) is the correct answer.

purely

(c)

ab a +b ab 2

(b)

2

(d)

a2 − b2 

2ab a − b2 2ab 2

a2 + b2

 a − ib  

Sol. Given, tan i log    a + ib   Let a + ib = reiθ , r2 = a2 + b2 b ⇒ a − ib = re− iθ , tan θ = a

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Objective Mathematics Vol. 1

4

⇒

a − ib = e−2 i θ a + ib  a − ib  −2iθ i log   = i log(e ) = 2 θ  a + ib

  a − ib   ⇒ tan i log   = tan 2θ  a + ib   2 tan θ = 1 − tan 2 θ 2b / a 2ab = = b2 a2 − b2 1− 2 a Hence, (b) is the correct answer.

Ex 6. If z1 = a + ib and z 2 = c + id are complex numbers such that | z1 | = | z 2 | = 1 and Re ( z1 z 2 ) = 0, then the pair of complex numbers a + ic = w1 and b + id = w2 satisfies (a) | w1 | ≠ 1 (c) Re ( w1 w 2 ) = 0

(b) | w 2 | ≠ 1 (d) None of these

Sol. z1 = a + ib, z2 = c + id ⇒

a2 − c2 = d 2 − b2

⇒ a2 − c2 + 2i ac = d 2 − b2 − 2 ibd [Q as ac = − bd ] ⇒ ⇒ ⇒ ⇒

(a + ic)2 = (d − ib)2 a + ic = (d − ib) or − d + ib a = d and c = − b or a = − d , b = c c2 + d 2 = b2 + d 2 a2 + c2 = a2 + b2

⇒

136

⇒ | z1 | = | z2 | ⇒ | z1 | = | z2 | = r1 Now, arg (z1 , z2 ) = 0 ⇒ arg(z1 ) + arg (z2 ) = 0 ⇒ arg (z2 ) = − θ 1

Hence, (b) is the correct answer.

Ex 8. If z is a complex number, then z 2 + z 2 = 2 represents (a) a circle (c) a hyperbola

(b) a straight line (d) an ellipse

Sol. Let z = x + iy, then ⇒ ⇒

z2 + z 2 = 2 (x + iy)2 + (x − iy)2 = 2 x 2 − y2 = 1

which represents a hyperbola. Hence, (c) is the correct answer.

1 − iα = A + iB , then A 2 + B 2 equals 1 + iα (a) 1 (b) α 2 (c) −1 (d) − α 2

Ex 9. If

1 − iα 1 + iα ⇒ A − iB = 1 + iα 1 − iα (1 − iα )(1 + iα ) ⇒ ( A + iB ) ( A − iB ) = =1 (1 + iα )(1 − iα ) ⇒

A2 + B2 = 1

Hence, (a) is the correct answer.

Ex 10. If | z1 | = | z 2 | and arg ( z1 ) + arg ( z 2 ) = π / 2, then (a) z1 z 2 is purely real (b) z1 z 2 is purely imaginary (c) ( z1 + z 2 ) 2 is purely real π (d) arg ( z1−1 ) + arg ( z 2−1 ) = 2 Sol. Let | z1 | = | z2 | = r ⇒

and

z1 = r (cosθ + i sin θ )  π  π  z2 = r cos  − θ + i sin  − θ      2 2 

z1z2 = r2i, which is purely imaginary.

which is purely imaginary. Also,

(b) | z 2 |2 = z1 z 2 (d) None of these

Sol. Let z1 = r1 (cos θ 1 + i sin θ 1 ) , then

z2 = (z1 ) = z1 ⇒ | z2 |2 = z1z2

z1 + z2 = r [(cosθ + sin θ ) + i (cosθ + sin θ )] ⇒ (z1 + z2 )2 = 2r2 ⋅ (cosθ + sin θ )2 ⋅ i

z1 = 1 and arg ( z1 z 2 ) = 0, then z2

(a) z1 = z 2 (c) z1 z 2 = 1

⇒

⇒

a2 + c2 = 1, b2 + d 2 = 1

⇒ | w1 | = | w2 | = 1 Also, ab + cd = − cd + cd = 0 ⇒ Re (w1w2 ) = 0 Hence, (c) is the correct answer.

Ex 7. If

z2 = r1[cos (− θ 1 ) + i sin (− θ 1 )] = r1 (cos θ 1 − i sin θ 1 ) = z1

Sol. A + iB =

| z1 | = | z2 | = 1 a2 + b2 = c2 + d 2 = 1

w1 = a + ic, w2 = b + id z1z2 = (a + ib) (c − id ) = (ac + bd ) + i (bc − ad ) As Re (z1z2 ) = 0 ⇒ ac + bd = 0 ⇒ ac = − bd w1w2 = (a + ic) (b − id ) = (ab + cd ) + i (bc − ad ) We have, a2 + b2 = c2 + d 2 ⇒

Therefore,

z1 =1 z2

arg (z1−1 ) + arg (z2−1 ) = −

π 2

Hence, (b) is the correct answer.

Ex 11. The value of the expression 1  1  1  1    2 1 +  1 + 2  + 3  2 +   2 + 2     ω  ω   ω ω  1  1   + 4  3 +   3 + 2  +… + ( n + 1)  ω  ω  1  1   n +  n + 2  ,    ω ω  where ω is an imaginary cube root of unity, is

(b)

n( n 2 − 2) 3

Sol. If z1 is the new complex number, then

(d) None of these

     Sol. tn = (n + 1) n +   n + 2  n3 + n2  2 + + 1     ω  ω ω ω 1

1

1

1

1 1  + n 1 + 2 +  + 1  ω ω = n3 + n2 (ω + ω 2 + 1) + n(ω + ω 2 + 1) + 1 = n3 + 1 ∴ Sn =

n

n

r=1

r=1

∑ tr = ∑ (r3 + 1) =

n2 (n + 1)2 +n 4

Hence, (c) is the correct answer.

Ex 12. If z1 and z 2 are two complex numbers z  z + iz 2   = 1, then 1 satisfying the equation  1 z2  z1 − iz 2  is (a) purely real (b) of unit modulus (c) purely imaginary (d) None of the above Sol. (z1 + iz2 )(z1 − iz2 ) = (z1 − iz2 )(z1 + iz2 ) ⇒

⇒

z1z2 = z1z2 z1 z1 = z2 z2

z ⇒ 1 is purely real. z2 Hence, (a) is the correct answer.

| z1 | = | z| + 2 = 2 2 z |z | Also, 1 = 1 ⋅ ei 3π / 2 z | z| 3π 3π   ⇒ z1 = z ⋅ 2  cos + i sin   2 2 = 2(1 + i )(0 − i ) = − 2i + 2 = 2(1 − i ) Hence, (d) is the correct answer.

Ex 15. If 2 cos θ = x +

Sol. z = − 2 + 2 3i = 4ω ∴z2n + 22n zn + 24n = 4 2nω 2n + 22n ⋅ 4 n ⋅ ω n + 24n = 4 2n[ω 2n + ω n + 1] = 0, if n is not a multiple of 3. = 3 ⋅ 4 2n, if n is a multiple of 3. Hence, (c) is the correct answer.

Ex 14. The complex number z = 1 + i is rotated through an angle 3π / 2 in anti-clockwise direction about the origin and stretched by additional 2 units, then the new complex number is (a) − 2 − 2 i (b) 2 − 2 i (c) 2 − 2 i (d) None of the above

1 1 and 2 cos φ = y + , then x y

x y + = 2cos (θ + φ ) y x 1 (b) x m y n + m n = 2cos ( mθ + nφ ) x y (a)

xm yn

+

yn

= 2cos ( mθ + nφ ) xm 1 (d) xy + = 2cos (θ − φ ) xy

(c)

1 x

Sol. 2 cosθ = x + , 2 cos φ = y +

1 y

⇒ x 2 − 2x cos θ + 1 = 0 2 cos θ ± 4 cos2 θ − 4 2 x = cosθ ± i sin θ = e± iθ

⇒

x=

⇒

y = e± iφ x y + = 2cos (θ − φ ) y x 1 x m yn + m n = 2cos (mθ + nφ ) x y

Similarly, ⇒

Ex 13. If z = − 2 + 2 3i, then z 2n + 2 2n z n + 2 4n may be equal to (a) 22n (b) 0 (c) 3⋅ 4 2n , n is multiple of 3 (d) None of the above

4 Complex Numbers

n( n 2 + 2) 3 n 2 ( n + 1) 2 + 4n (c) 4

(a)

xm yn + m = 2cos (mθ − nφ ) n y x 1 xy + = 2cos (θ + φ ) xy Hence, (b) is the correct answer.

Ex 16. The complex numbers z1 and z 2 are such that z1 ≠ z 2 and | z1 | = | z 2 |. If z1 has positive real part and z 2 has negative imaginary part, then  z1 + z 2    may be  z1 − z 2  (a) zero (b) real and positive (c) real and negative (d) purely imaginary Sol. Given, | z1 | = | z2 |

Re (z1 ) > 0, Im (z2 ) < 0

 z + z2  1  z1 + z2 z1 + z2  Re  1 +  =    z1 − z2  2  z1 − z2 z1 − z2  =

1  (z1 + z2 )(z1 − z2 ) + (z1 + z2 )(z1 − z2 )    2 (z1 − z2 )(z1 − z2 )

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Objective Mathematics Vol. 1

4

 z1z1 − z1z2 + z2z1 − z2z2 + z1z1 + z1z2    1 − z z − z2z2  =  21  2 | z1 − z2 |2     =

1  2 | z1 |2 − | z2 |2   =0  2  | z1 − z2 |2 

Hence, (d) is the correct answer.

 z z − z2   = k , ( z1 , z 2 ≠ 0), then Ex 17. If 1  z1 z + z 2 

  = k  

Ex 18. If A ( z1 ), B ( z 2 ) and C ( z 3 ) are the vertices of a π AB ∆ABC in which ∠ABC = and = 2, then 4 BC the value of z 2 is equal to (a) z 3 + i( z1 + z 3 ) (b) z 3 − i( z1 − z 3 ) (c) z 3 + i( z1 − z 3 ) (d) None of the above

(a) 1 − ccos θ (c) 1 + ccos θ

then

(b) 1 + 2ccos θ (d) 1 − 2ccos θ

1 − c2 = n2c2 − 2nc + 1 c 1 …(i) ∴ = 2n 1 + n2 c n 1  1    or 1 + n2 + n  z +   (1 + nz) 1 +  = 2     2n z z  1+ n  1 {1 + n2 + n ⋅ (2 cosθ )} = 1 + n2 (1 + n2 ) + 2n cos θ = 1 + n2  2n  =1+   cos θ  1 + n2  = 1 + ccos θ

[using Eq. (i)]

Hence, (c) is the correct answer.

AB = 2 BC Considering the rotation about B, we get z1 − z2 | z1 − z2 | iπ / 4 AB iπ / 4 = ⋅e = ⋅e z3 − z2 | z3 − z2 | BC i   1 = 2 +  =1+ i  2 2

Sol. Given,

⇒ z1 − z2 = (1 + i )(z3 − z2 ) ⇒ z1 − (1 + i ) z3 = z2 (1 − 1 − i ) ⇒ iz2 = − z1 + (1 + i )z3 ⇒ z2 = iz1 − i (1 + i )z3 ∴ z2 = z3 + i (z1 − z3 ) Hence, (c) is the correct answer.

(c) 10

Ex 21. Consider an ellipse having its foci at A ( z1 ) and B ( z 2 ) in the argand plane. If the eccentricity of the ellipse is e and it is known that origin is an interior point of the ellipse, then  | z + z2 |  (a) e ∈  0, 1   | z1 | + | z 2 |  | z − z2 |  (b) e ∈  0, 1   | z1 | + | z 2 |  | z + z2 |  (c) e ∈  0, 1   | z1 | − | z 2 | (d) Can’t be determined Sol. If P (z) is any point on the ellipse. Then, equation of the ellipse is

| z1 − z2 | …(i) e If we replace z by z1or z2, Eq. (i) becomes | z1 − z2 |. For P (z) to lie on ellipse, we have | z − z2 | | z − z1 | + | z − z2 | < 1 e | z − z1 | + | z − z2 | =

Ex 19. If z1 z 2 ∈C , z12 + z 22 ∈ R , z1 ( z12 − 3 z 22 ) = 2 and z 2 (3 z12 − z 22 ) = 11, then the value of z12 + z 22 is (b) 6

Ex 20. 1 − c 2 = nc − 1 and z = e iθ , c  n (1 + nz ) 1 +  is equal to  2n z

⇒

If k = 0, z represents a point. Hence, (b) is the correct answer.

(a) 5

Hence, (a) is the correct answer.

Sol. Here, 1 − c2 = nc − 1

Clearly, if k ≠ 0, 1 then z would lie on a circle. If k = 1, z would lie on a perpendicular bisector of the z z line segment joining 2 and − 2 . z1 z1

138

On multiplying Eq. (ii) by i and adding it to Eq. (i), we get z13 − 3z22z1 + i (3z12z2 − z23 ) = 2 + 11i …(iii) ⇒ (z1 + iz2 ) = 2 + 11i

On multiplying Eqs. (iii) and (iv), we get (z12 + z22 )3 = 4 + 121 ⇒ z12 + z22 = 5

(a) for k ≠ 1, locus z is a straight line (b) for k ∉{1, 0}, z lies on a circle (c) for k ≠ 0, z represents a point (d) for k ≠ 1, z lies on the perpendicular bisector z z of the line segment joining 2 and − 2 z1 z1 z2 z1 z2 z1

…(i) …(ii)

Again multiplying Eq. (ii) by i and subtracting it from Eq. (i), we get z13 − 3z22z1 − i (3z12z2 − z23 ) = 2 − 11i …(iv) ⇒ (z1 − iz2 )3 = 2 + 11i

 z + z2  ⇒ 1  is purely imaginary.  z1 − z2 

z −  z1z − z2     = k ⇒ Sol. Given,  z1z + z2  z + 

z1 (z12 − 3z22 ) = 2 z2 (3z12 − z22 ) = 11

Sol. Here,

(d) 12

Hence, (b) is the correct answer.

 π  Ex 22. If | z − 2 − i | = | z |  sin  − arg ( z ) ,   4  locus of z is

then

(a) a pair of straight lines (b) a circle (c) a parabola (d) an ellipse Sol. We have, |(x − 2) + i ( y − 1)| = | z|

1 1 cosθ − sin θ   2  2

Ex 25. Locus of z, if 3π , when | z | ≤ | z − 2 | 4 , is arg[ z − (1 + i)] =  −π  , when | z | > | z − 2 |  4 (a) straight line passing through (2, 0) (b) straight line passing through (2, 0), (1, 1) (c) a line segment (d) a set of two rays Sol. The given equation is written as  3π , when x ≤ 2 4 arg [ z − (1 + i )] =   − π , when x > 2  4 The locus is a set of two rays.

where, θ = arg (z) 1 (x − 2) + ( y − 1) = |x − y | 2 which is a parabola. Hence, (c) is the correct answer. 2

2

Ex 23. α 1 , α 2 , α 3 , …, α 100 are all the 100th roots of unity. The numerical value of 5 ∑ ∑ (α i α j ) is

(0, 2)

(1, 1)

1 ≤ i < j ≤ 100

(a) 20 (c) ( 20)1/ 20 Sol.

∑∑

(b) 0 (d) None of these

α 5i α 5j 1 ≤ i < j ≤ 100

=

(α 51

+

α 52

+

α 53

+

4

By the given conditions, the area of ∆ABC is 1 | z1 − z2 | r 2 Hence, (b) is the correct answer.

Complex Numbers

It is given that origin is an interior point of the ellipse. | z − z2 | ⇒ |0 − z1 | + |0 − z2 | < 1 e  | z1 − z2 |  ⇒ e ∈  0,   | z1 | + | z2 |

α 54

+

α 55

+ …)

(2, 0) 2

Re (z) = 1

10 10 10 10 − (α 10 1 + α 2 + α 3 + α 4 + α 5 + …)

=0−0=0

100, if r = 100k r Because (α 1r + α 2r + … + α 100 )=  0 , if r ≠ 100k Hence, (b) is the correct answer.

Ex 24. The maximum area of the triangle formed by the complex coordinates z, z1 and z 2 , which satisfy the relations | z − z1 | = | z − z 2 |, z1 + z 2     z −   ≤ r, where r >| z1 − z 2 |, is  2   1 | z1 − z 2 | 2 2 1 (c) | z1 − z 2 |2 r 2 2

(a)

Sol.

1 | z1 − z 2 | r 2 1 (d) | z1 − z 2 | r 2 2 (b)

Hence, (d) is the correct answer.

π  A =  z | arg ( z ) =  4  2π   B =  z | arg ( z − 3 − 3i) = , z ∈C . 3  n( A ∩ B ) is equal to (a) 1 (b) 2 (c) 3 (d) 0

Ex 26. If

and Then,

Sol. We can observe that, 3 + 3i ∈ A but ∉ B. A

B Y

A(z)

Y

(3, 3)

B(z1)

r z 1 +z 2 2

O

O

X

C(z2)

X

∴ n( A ∩ B ) = 0 Hence, (d) is the correct answer.

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Objective Mathematics Vol. 1

4

Ex 27. Dividing f ( z ) by z − i, we obtain the remainder i and dividing it by z + i, we get the remainder 1 + i. The remainder upon the division of f ( z ) by z 2 + 1, is 1 ( z + 1) + i 2 1 (c) ( iz − 1) + i 2

1 ( iz + 1) + i 2 1 (d) ( z + i ) + 1 2

(b)

(a)

Sol. f (z) = g (z) (z − i )(z + i ) + az + b; where a, b ∈ C f (i ) = i ⇒ ai + b = i f (− i ) = 1 + i ⇒ a(− i ) + b = 1 + i From Eqs. (i) and (ii), we get i 1 a= ,b= + i 2 2

…(i) …(ii)

Hence, required remainder = az + b =

1 1 iz + + i 2 2

Hence, (b) is the correct answer.

Type 2. More than One Correct Option Ex 28. If z1 = a1 + ib1 and z 2 = a 2 + ib2 are complex numbers such that | z1 | = 1, | z 2 | = 2 and Re ( z1 z 2 ) = 0, then the pair of complex ia numbers ω1 = a1 + 2 and ω 2 = 2b1 + ib2 , 2 satisfy (a) |ω 1 | = 1 (c) Re (ω 1ω 2 ) = 0

(b) |ω 2 | = 2 (d) Im (ω 1ω 2 ) = 0

Sol. a12 + b12 = 1, a22 + b22 = 4 and a1a2 = b1b2 a22 + b22 = 4 a12 + 4 b12 (a2 + 2ia1 )2 = (2b1 + ib2 )2 ⇒ a2 = ± 2b1 a2 |ω 1 |2 = a1 + a12 + 2 = a12 + b12 = 1 4 ⇒ |ω 1 | = 1 and |ω 2 |2 = 4 b12 + b22 = 4 ⇒ |ω 2 | = 2 ab Re (ω 1ω 2 ) = 2a1b1 − 2 2 = 0 2 Im (ω 1ω 2 ) = a1b2 + a2b1 = 2a12 + 2b12 = 2

z + z2 + z3  (b) 1 = 1 ⇒ | z1 + z2 + z3 |2 = 9 3   ⇒ (z1 + z2 + z3 ) (z1 + z2 + z3 ) = 9 4 1 1  4 1 1 ⇒ +  + +  =9  +  z1 z2 z3   z1 z2 z3  (c) ∠QOR = 120° Hence, (a), (b), (c) and (d) are the correct answers.

Ex 30. One vertex of the triangle of maximum area that can be inscribed in the curve | z − 2i | = 2, is 2 + 2i, remaining vertices is/are (a) −1 + i( 2 + 3 ) (c) −1 + i( 2 − 3 )

(b) −1 − i( 2 + 3 ) (d) −1 − i( 2 − 3 )

Sol. Clearly, the inscribed triangle is equilateral. z2

Hence, (a), (b) and (c) are the correct answers.

Ex 29. If from a point P representing the complex number z1 on the curve | z | = 2, pair of tangents are drawn to the curve | z | =1, meeting at point Q ( z 2 ) and R ( z 3 ), then (a) complex number

z1 + z 2 + z 3 will lie on the 3

curve | z | = 1 4 1 1  4 1 1 (b)  + +  + +  =9  z1 z 2 z 3   z1 z 2 z 3   z  2π (c) arg  2  = 3  z3  (d) orthocentre and circumcentre of ∆PQR will coincide

Sol. Options (a) and (d) are true as PQR is an equilateral triangle, so orthocentre, circumcentre and centroid will coincide. Y P(z1)

Q(z2)

X¢

X

O

R(z3)

140

Y¢

z0(2i) z1(2 + 2i )

z3 2π

⇒ ⇒

i −i z2 − z0 z − z0 =e 3, 3 =e z1 − z0 z1 − z0

z2 = − 1 + i (2 +

2π 3

3)

and z3 = − 1 + i (2 − 3 ) Hence, options (a) and (c) are the correct answers.

Ex 31. Let z be a complex number and a be a real parameter, such that z 2 + az + a 2 = 0, then (a) locus of z is a pair of straight lines (b) locus of z is a circle 2π (c) arg ( z ) = ± 3 (d) | z | = | a | Sol. z2 + az + a2 = 0

⇒ z = aω, aω 2 where, ω is non-real root of cube unity. ⇒ Locus of z is a pair of straight lines and arg(z) = arg (a) + arg (ω ) or arg (a) + arg (ω 2 ) 2π arg (z) = ± ⇒ 3 Also, | z | = | a | |ω | or | a | |ω 2 | ⇒ | z | = | a | Hence, (a), (c) and (d) are the correct answers.

Directions (Ex. Nos. 32-36) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true,Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 32. Statement I If A ( z1 ), B ( z 2 ), C ( z 3 ) are the vertices of an equilateral ∆ABC, then  z + z 3 − 2 z1  π arg  2 =  z3 − z2  4 Statement II If ∠B = α,  z1 − z 2  z1 − z 2 AB iα = e or arg   =α z 3 − z 2 BC  z3 − z2 

then

α

Now, for x = − ω , p = ω 4000 + Similarly, for x = − ω 2 , p = − 1 For n > 1, 2n = 4 k

C(z3)

  z2 + z3 − z1    z2 + z3 − 2z1  2 Sol. Q arg    = arg   z3 − z2   z3 − z2   

1 ω

4000

=ω +

1 = −1 ω

n

∴ (2)2 = 24k = (16)k = a number with last = 6 ⇒ q = 6 + 1= 7 Hence, p + q = −1 + 7 = 6 Hence, (d) is the correct answer.

Ex 34. Statement I If z1 , z 2 , z 3 are complex numbers represent the points A, B , C such that 2 1 1 = + . Then, A, B , C passes through z1 z 2 z 3 origin. Statement II If 2 z 2 = z1 + z 3 , then z1 , z 2 , z 3 are concyclic. Sol.

A(z1)

B(z2)

1 Sol. x + = 1 ⇒ x 2 − x + 1 = 0 x ∴ x = − ω, − ω2

Complex Numbers

4

Type 3. Assertion and Reason

2 1 1 1 1 = + ⇒ − z1 z2 z3 z1 z2 z2 − z1 ⇒ z1z2 z2 − z1 ⇒ z3 − z1  z2 − z1  ⇒ arg    z3 − z1  ⇒

1 1 − z3 z1 z1 − z3 = z1z3 z =− 2 z3  z = arg  − 2   z3  =

 z −z  z  z  arg  2 1  = π − arg  2  = π − arg  2   z3 − z1   z3   z3 

⇒ ∠CAB = π − ∠COB ⇒ ∠CAB + ∠COB = π ⇒ Points O , A , B , C are concyclic. Hence, (b) is the correct answer.

A(z1)

Ex 35. Statement I 3 + ix 2 y and x 2 + y + 4i are complex conjugate numbers, then x 2 + y 2 = 4. B(z2)

C(z3)

D z2+z3 2

π 2 Hence, (d) is the correct answer. =

Ex 33. Statement I If p=x

4000

+

x+

Statement II If sum and product of two complex numbers is real, then they are conjugate complex number. [ as AD ⊥ BC ]

1 =1 x

and

1

and q is the digit at unit place x 4000 n in the number 2 2 + 1, n ∈ N and n >1, then the value of p + q = 8. Statement II ω, ω 2 1 1 x + = − 1, x 3 + 3 = 2. x x

are the roots of

Sol. If 3 + ix 2 y and x 2 + y + 4 i are conjugate, then x 2 y = − 4 and x 2 + y = 3 ⇒ x 2 = 4 , y = − 1 ⇒ x 2 + y2 = 5 Hence, (d) is the correct answer.

Ex 36. Statement I If | z| < 2 − 1, then | z 2 + 2 z cos α | < 1 Statement II | z1 + z 2 | ≤ | z1 | + | z 2 |, |cos α| ≤1.

also

Sol. | z2 + 2z cos α | < | z2 | + |2z cos α | < | z|2 + 2| z| |cos α | < ( 2 − 1)2 + 2( 2 − 1) < 1 [Q |cos α | ≤ 1] Hence, (a) is the correct answer.

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Objective Mathematics Vol. 1

4 Type 4. Linked Comprehension Based Questions Passage I (Ex. Nos. 37-39) In argand plane, | z | represents the distance of a point z from the origin. In general, | z1 − z 2 | represents the distance between two points z1and z 2 . Also, for a general moving point z in argand plane, if arg ( z ) = θ, then z = | z | ei θ , where ei θ = cos θ + i sin θ.

Ex 37. The equation | z − z1 | + | z − z 2 | = 10, z1 = 3 + 4i and z 2 = − 3 − 4i represents (a) point circle (c) ellipse

if

z − z1 + z − z2 = z1 − z2 ∴ ⇒ z lies between z1 and z2. i.e. a line segment. Hence, (d) is the correct answer.

(b) hyperbola (d) None of these

then

1 1 cos θ + sin θ , where θ = arg (z) 2 2 1 | x + y| ⇒ (x − 3)2 + ( y − 2)2 = 2 which is a parabola. Hence, (b) is the correct answer.

Passage II (Ex. Nos. 40-42) The complex slope of a line passing through two points represented by complex z − z1 numbers z1 and z 2 is defined by 2 and we shall z 2 − z1 denote by ω. If z 0 is complex number and c is a real number, then z 0z + z 0z = 0 represents a straight line. Its z complex slope is − 0 . Now ,consider two lines z0 α z + αz + i β = 0…(i) and a z + az + b = 0 …(ii) where, α, β and a, b are complex constants and let their complex slopes be denoted by ω1 and ω 2 , respectively.

Ex 40. If the lines are inclined at an angle of 120° to each other, then (d) ω 1 + 2ω 2 = 0

Hence, (b) is the correct answer.

(a) a must be pure imaginary (b) β must be pure imaginary (c) a must be real (d) β must be imaginary

Ex 42. If line (i) makes an angle of 45° with real axis,  2α  then (1 + i)  −  is  α π ±i e 2

(b) 2 2i

(c) 2(1 − i ) (d) − 2(1 + i )

=±i

 2α  (1 + i )  −  = ± 2 (− 1 + i )  α Hence, (c) is the correct answer.

∴

= |z |

(c) ω 12 = ω 22

ω 2 ω12 = ω 1 ω22

∴

α α

Sol. Given, | z − (3 + 2i )| = |(x − 3) + i ( y − 2)|

142

2

⇒ω 1 ω12 = ω 2 ω12 , as ω 1 = ω 2

Sol. − =

(b) parabola (d) hyperbola

(b) ω 2 ω12 = ω 1 ω 22

2

ω 1 ω 1 ω22 = ω 2 ω 2 ω12

⇒

(a) 2 2

| z1 − z2 | = 10 ⇒ z lies on hyperbola. Hence, (b) is the correct answer.

(a) ω 2 ω1 = ω 1 ω 2

⇒ ω 31 = ω 32 ⇒ω13 ω12 ω22 = ω 32 ω12 ω22

∴ β is pure imaginary. Hence, (b) is the correct answer.

Sol. Given equation can represent a hyperbola, since

(a) circle (c) ellipse

4π 3

Sol. Since, i β is real.

Ex 38. z − z1 | − | z − z 2 = t, t >10 where t is a real parameter, always represents

π  Ex 39. If | z − (3 + 2i)| = z cos  − arg ( z ) , 4  locus of z is a/an

i

Ex 41. Which of the following must be true?

(b) ordered pair (0, 0) (d) None of these

Sol. As z1 − z2 = 6 + 8i = 10

(a) ellipse (c) circle

Sol. ω 1 = ω 2 e

Passage III (Ex. Nos. 43-45) Consider ∆ABC in argand plane. Let A(0), B (1) and C (1 + i ) be its vertices and M be the mid-point of CA. Let z be a variable complex number in the plane. Let us another variable complex number defined as, u = z 2 + 1.

Ex 43. Locus of u, when z is on BM, is a/an (a) circle (c) ellipse

(b) parabola (d) hyperbola

Ex 44. Axis of locus of u, when z is on BM, is (a) real axis (c) z + z = 2

(b) imaginary axis (d) z − z = 2i

Ex 45. Directrix of locus of z, when z is on BM, is (a) real axis (c)z + z = 2

(b) imaginary axis (d) z − z = 2i

Sol. (Ex. Nos. 43-45) ∴

BM ≡ y − 0 = − 1(x − 1) x + y=1 u − 1 = t + i (1 − t )

u = 2t + 2i t (1 − t ) x = 2t and y = 2t (1 − t ), where u = x + iy 1  (x − 1)2 = − 2  y −  , which is parabola.  2 Axis is x = 1, i.e. z + z = 2 Directrix is y = 1, i.e. z − z = 2i

43. (b)

44. (c)

45. (d)

75 × 25 = z1 + 1 ⇒ ∴ z1 − (− 1) = 1875 ⇒ z1 lies on circle.

Ex 46. Match the following : Column I

Column II p. circle

A. Locus of the point z satisfying the

Ex. 47. If z1 , z 2 , z 3 , z 4 are the roots of the equation z 4 + z 3 + z 2 + z + 1 = 0, then

equationRe ( z2 ) = Re( z + z), is a/an q. straight line

B. Locus of the point z satisfying the equation| z − z1| + | z − z2| = λ, λ ∈ R + and λ | z1 − z2 | and for straight line, λ = | z1 − z2 | 2z − i C. Q =m z+1 i z− m 2 = ⇒ 2 z+1

⇒

q.

i =1 4

Sol. A. Put z = x + iy

For m = 2,

0

∑

A.

B.

Column II p.

4

+

Complex Numbers

4

Type 5. Match the Columns

A. z14 + z24 + z34 + z44 + 14 = 0 4

∑ zi4 = 1

⇒

i=1

B.

z15

+

+

z25

z35

+

z45

+ 1= 5 4

∑ zi5 = 4

⇒

i=1

C. z4 + z3 + z2 + z + 1 = (z − z1 ) (z − z2 ) (z − z3 ) (z − z4 ) Putting z = − 2 on both the sides, we get

1 2 =1 z+1

z−

4

∏ (zi + 2) = 11 i=1

i ⇒ z − = | z + 1|, i.e. straight line 2

D. | z1 + z2 | = 2 + 2 cos144 ° for minimum 5 −1 2 whose greatest integer is 0. A → r; B → q; C → s; D → p = 2 cos 72° =

D. Given, z = 25 Let z1 = − 1 + 75 z ∴ 75 z = z1 + 1 or 75 z = z1 + 1

Type 6. Single Integer Answer Type Questions Ex 48. Consider an equilateral triangle having vertices at points  2 iπ   2 − iπ  A e 2 , B  e 6  and    3  3      2 − i 5π C e 6  3 

Sol. (5) A (z1 ) =

A

  . If P ( z ) is any point on its  

incircle, then AP + BP + CP _______ . 2

2

2

2i 3

O

is equal to

r

B

D

C

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Objective Mathematics Vol. 1

4

B (z2 ) =

i 2  3 1 − i  =1−  2 3 3  2

= 1 − cos

2

∴

| A1 A2 |2 =

n

∑

r=1

rπ    2 − 2 cos   n

= 2 (n − 1) − 2 (1) [since, {1 + α + α 2 + K + α n− 1 = 0}]

AP 2 + BP 2 + CP 2 = 5

∴

Ex 49. Let A1 , A2 , ..... An be the vertices of a regular polygon of n sides in a circle of radius unity and a = | A1 A2 | 2 + | A1 A3 | 2 + .... + | A1 An | 2 , a b = | A1 A2 | | A1 A3 | ....| A1 An |, then = ____. b Sol. (2) Let us assume that O is the centre of the polygon and z0 , z1 , … , zn − 1 represent the affixes of A1 , A2 , … , An such that

144

2

2π 4π 2(n − 1)π   = 2 (n − 1) cos + cos + ... + cos  n n n   2 n− 1 = 2 (n − 1) − 2 ⋅ Real part of (α + α + ... + α )

Solving for | AP |2 + | BP |2 + |CP |2 , we get

| A1 Ar |2 = |α r − 1|2 = |1 − α r |2

n

∑

r=1

Hence, any point on incircle i.e. P (z) is 1 1 1 cos α , sin α i.e. (cosα + i sin α ) 3 3 3

Now,

2

2rπ  2rπ  2rπ   = 1 − cos  = 2 − 2 cos  +  sin   n  n  n

i 2  3 i C (z3 ) = −  = −1− − 2 3  2 3 1 unit. Radius of incircle of ∆ABC , i.e. r = 3

z0 = 1, z1 = α , z2 = α 2 ,..., zn− 1 = α n− 1, where α = e

2rπ 2rπ − i sin n n

i 2π n

| A1 A2 |2 + | A1 A3 |2 + ...+ | A1 An |2 = 2n

Also, let E = | A1 A2 | | A1 A3 | K | A1 An | = |1 − α | |1 − α 2 | |1 − α 3 | ... |1 − α n− 1 | = (|1 − α )(1 − α 2 )(1 − α 3 ) K (1 − α n− 1 )| Since, 1 α ,α 2 ,… , α n− 1 are the roots of zn − 1 = 0. zn − 1 ⇒ (z − 1) (z − α ) (z − α 2 ) ... (z − α n−1 ) = z −1 ⇒ (z − α ) (z − α 2 ) (z − α n− 1 ) = zn − 1 = 1 + z + z2 + … + zn −1 Substituting z = 1, we have (1 − α ) (1 − α 2 )... (1 − α n− 1 ) = n |1 − α | |1 − α 2 |... |1 − α n− 1 | = n a 2n Hence, the value of = =2 b n

Target Exercises Type 1. Only One Correct Option i 592 + i 590 + i 588 + i 586 + i 584 i

582

+i

580

+i

(a) −1 (c) −3

2. i 57 +

578

+i

576

+i

574

− 1 is

(a) I quadrant (c) III quadrant

(b) −2 (d) −4

1

i

12. The value of

, when simplified has the value 125

(a) 0 (c) − 2i

(b) 2i (d) 2

(a) −

(b) −1

7. For

a

positive

(d) −i

(c) i

integer

n,

the

8. If the number

(1 − i ) n (1 + i ) n − 2

(a)

9. The smallest positive n  1 + i  = − 1 is   1 − i (a) 1 (c) 3

10. The smallest positive (1 + i ) 2n = (1 − i ) 2n is (a) 4 (c) 2

a − ib 1− c

6i 17. If 4

(d) 4 n

(b)

integer

n

for

18. which

1 10

(d) −

1 10

(1+ b + ia ) is equal to (1 + b − ia )

a − ib 1+ c

(c)

1

3i

−1 = x + iy, then

3

a + ib 1− c

(d)

1 + iz 1 − iz

a + ib 1+ c

i

(a) x = 3, y = 1 (c) x = 0, y = 3 (b) 2λ (d) None of these

(c)

−3i

20

is real and positive, then n is

(a) any integer (c) 4 λ + 1

1 5

(b) 2 (d) a + ib

equals

n

(c) 2n

(b)

16. If b + ic = (1 + a ) z and a 2 + b 2 + c 2 = 1, then

expression

 1 (1 − i ) 1 −  equals  i (b) 2i n

1 5

(a) 1 (c) b + ia

n

(a) 0

(d) − 2

(b) − 2i (d) 2

15. If a 2 + b 2 = 1, then

(d) − 2

6. If n is any positive integer, then the value of i 4n + 1 − i 4n − 1 equals 2 (a) 1

(c) 1

2

 (1 + i ) 2  14. The value of Re   is equal to  3 − i 

2

25   1  5. The value of i19 +    is  i   

(c) 2

is equal to

(a) 2i (c) − 2

(b) negative (d) Can’t be determined

(b) − 4

1− i3

 1 − i  1 + i 13.   is equal to  +  1 + i  1 − i

4. 1 + i 2 + i 4 + i 6 + K + i 2n is

(a) 4

(1 − i ) 3

2

(b) − 1 (d) None of these

(a) positive (c) 0

(b) II quadrant (d) IV quadrant

(b) − 1

(a) i

3. i n + i n + 1 + i n + 2 + i n + 3 is equal to (a) 1 (c) 0

1 + 2i lies in the 1− i

11. The complex number

Targ e t E x e rc is e s

1. The value of

(b) x = 1, y = 3 (d) x = 0, y = 0

2i equals (a) 1 + i (c) − 2i

(b) 1 − i (d) None of these

19. If z is a complex number such that | z | ≠ 0 and Re ( z ) = 0, then (a) Re (z2 ) = 0 (c) Re (z2 ) = Im (z2 )

(b) 2 (d) 4

number (b) 8 (d) 12

n for

which

(b) Im (z2 ) = 0 (d) None of these

20. If ( x + iy )1/ 3 = a + ib, then (a) 2 (a2 − b2 ) (c) 8 (a2 − b2 )

x y + is equal to a b

(b) 4 (a2 − b2 ) (d) None of these

145

Objective Mathematics Vol. 1

4

21. If 8iz 3 + 12z 2 − 18z + 27i = 0, then (a) | z | =

3 2

(b) | z | =

2 3

32. If x + iy = (d) | z | =

(c) | z | = 1

3 4

22. If z = 1 + i, then the multiplicative inverse of z is 2

(a) 1 − i

(b)

i 2

3

(c) −

i 2

(d) 2i

3

 1 − i  1 + i 23. If   = x + iy, then ( x, y ) is equal to  −  1 + i  1 − i (a) (0, 2) (c) (0, −2)

(b) (− 2, 0) (d) None of these

24. The multiplicative inverse of ( 6 + 5i ) 2 is 11 60 (a) − i 60 61 9 60 (c) − i 61 61

8 31 + i 25 25 8 31 (c) − − i 25 25

Ta rg e t E x e rc is e s

(a) −

(b)

8 31 − i 25 25

z also lies in z

1 2

(b)

2− i (1 − 2i ) 2

is

2 11 (b) − i 25 25 2 11 (d) − − i 25 25

1 2

(c) 1

(b) 2

(c) 0

2a (1 + a)2 + b2 2a (c) (1 + b)2 + a2

31. The modulus of 146

(a)

13 5

(b)

(d) − 1

(d) 2 2

2b (1 + a)2 + b2 2b (d) (1 + b)2 + a2

(d)

5π 12

π π  (b) 2  cos + i sin   6 6 π π  (d) 4  cos + i cos   6 6

(b) 2 (d) None of these

38. If z is any complex number such that | z + 4 | ≤ 3, then the least value and greatest value of | z + 1| are (a) 1, 6 (c) 2, 8

(b) 0, 6 (d) None of these

(b) | w + 1| < | w − 7| (d) | w + 5| < | w − 4 |

40. For any complex number z, the minimum value of | z | + | z − 1| is (a) 1

(b) 0

(c)

1 2

(d)

3 2

41. The greatest positive argument of complex number satisfying | z − 4| = Re ( z ), is π 3 π (c) 2 (a)

(a) ± (1 + 3 i ) (c) ± (− 1 + 3 i )

2π 3 π (d) 4 (b)

(b) ± (1 − 3 i ) (d) None of these

43. If z = a + ib, where a > 0, b > 0, then

( 3 + 2i ) 2 is ( 4 − 3i ) (c)

5π 12

42. The square roots of −2 + 2 3 i are

(b)

11 5

(c) −

37. If z = x + iy, x and y are real, then | x | + | y | ≤ k | z | , where k is equal to

(a) | w + 1| < | w − 8| (c) | w + w | > 7

1 − ix 30. If = a − ib and a 2 + b 2 = 1, where a and b are 1 + ix real, then x is equal to (a)

7π 12

39. If z satisfies | z + 1| < | z − 2|, then w = 3z + 2 + i satisfies

29. The value of | 2i − − 2i | is (a) 2

(b) −

(a) 1 (c) 3

 2 + i 28. (1 + i )   is equal to  3 + i (a) −

19π 12

(a) | z2 | > | z |2 (b) | z2 | = | z |2 (c) | z2 | < | z |2 (d) | z2 | ≥ | z |2

(b) x < y < 0 (d) y > x > 0

2 11 (a) + i 25 25 2 11 (c) − + i 25 25

34. The principal argument of the complex number (1 + i ) 5 (1 + 3i ) 2 is −2i ( − 3 + i )

36. If z is a complex number, then

III quadrant, if

27. The conjugate complex number of

1 (a − b) 2 1 (c) | z | < (a + b) 2

(a) | z | ≥ 9 5

(d)

then

(b) α 2 − β 2 (d) None of these

(a)

(d) None of these

(a) x > y > 0 (c) y < x < 0

(a) α − iβ (c) α 2 + β 2

π 1  π  sin + i cos  6 6 2  1 π π (c)  sin + i cos  2 6 6

3 + 4i is 4 − 5i

26. If z = x + iy lies in III quadrant, then

33. If (1 + i ) (1 + 2i ) (1 + 3i )... (1 + ni ) = α + iβ, 2⋅ 5⋅ 10 ... (1 + n 2 ) is equal to

35. The complex number i + 3 in polar form can be written as

(d) None of these

25. The multiplicative inverse of

(b) −1 (d) None of these

(a) 1 (c) 0

(a)

11 60 (b) − i 61 61

u + iv , then x 2 + y 2 is equal to u − iv

7 5

(b) | z | ≥

1 (a + b) 2

(d) None of these

the

complex

numbers

z1

z2,

and

|1 − z1 z 2 | − | z1 − z 2 | = k (1 − | z1 | ) (1 − | z 2 | ), then k is equal to 2

2

(b) −1

(a) 1

2

2

(c) 2

(d) 4

45. The range of real number α for which the equation z + α| z − 1| + 2i = 0 has a solution, is  5 5 (a) − ,   2 2   5 (c) 0,   2 

 3 (b) − , 2   (d)  −∞ , − 

3 2 

5  5  , ∞ ∪ 2   2 

46. If | z | = 2 and locus of 5z − 1 is the circle having radius a and z12 + z 22 − 2z1 z 2 cos θ = 0, then | z1 |: | z 2 | is equal to (a) a : 1 (c) a : 10

(b) 2a : 1 (d) None of these

47. The maximum value of | z | when z satisfies the 2 condition z + = 2 is z (a) 3 − 1 (c) 3

(b) 3 + 1 (d) 2 + 3

48. If z1 ≠ z 2 and | z1 + z 2 | =

1 1 , then + z1 z 2

(a) atleast one of z1 , z2 is unimodular (b) z1 ⋅ z2 is unimodular (c) z1 ⋅ z2 is non-unimodular (d) None of the above

n −1

∑ | z1 + ω k z 2 |2 is equal to (b) (n − 1) [| z1 |2 + | z2 |2 ] (d) None of these

50. If z satisfies the equation | z | − z = 1 + 2i, then z is equal to 3 + 2i 2  3 (c) 2 −   i  2

51. If

3 − 2i 2  3 (d) 2 +   i  2

(b)

3 + i = ( a + ib ) ( c + id ), then tan

is equal to

π + 2 π (b) − 3 π (c) + 6 π (d) − 6 (a)

2nπ , for some integer n + nπ , for some integer n nπ , for some integer n + 2nπ , for some integer n

π 3 5π (c) 3

(a)

(b)

−1

2π 10π or 2π or 3 3

 z − z1  π 54. If z1 = 8 + 4i, z 2 = 6 + 4i and arg   = , then  z − z2  4 z satisfies (b) | z − 7 − 5i | = 2 (d) | z − 7i | = 18

(a) | z − 7 − 4 i | = 1 (c) | z − 4 i | = 8

55. If arg ( z ) < 0, then arg ( − z ) − arg (z) is equal to (a) π

(b) − π

(c) −

π 2

(d)

π 2

56. If z is a point on the argand plane such that | z − 1| = 1 z−2 , then is equal to z (a) tan (arg z) (c) i tan (arg z)

(b) cot (arg z) (d) None of these

1 + C + iS is equal to 1 + C − iS

(b) C − iS

(c) S + iC

(d) S − iC

58. The imaginary part of ( z − 1) (cos α − i sin α) + ( z − 1) −1 × (cos α + i sin α ) is zero, if

59. The value of

(a) −1

60. If a = cos b d + tan −1 a c

4

(d) π

(a) | z − 1| = 2 (c) arg (z − 1) = α

k =0

(a)

53. If z = − 1, then the principal value of arg ( z 2/ 3 ) is equal to

(a) C + iS

n−1

(a) n[| z1 |2 + | z2 |2 ] (c) (n + 1) [| z1 |2 + | z2 |2 ]

(a) arg (z − 1) = 2 arg (z) (b) 2 arg (z) = 2/ 3 arg (z2 − z) (c) arg (z − 1) = arg (z + 1) (d) arg (z) = 2 arg (z + 1)

57. If C 2 + S 2 = 1, then

are the n, nth roots of unity and 49. If 1, ω , ω , K , ω z1 , z 2 are any two complex numbers, then 2

52. If z is any complex number satisfying | z − 1| = 1, then which of the following is correct?

Complex Numbers

for

(b) arg (z − 1) = 2α (d) | z | = 1

π  π sin + i cos   8 8

8

π π  sin − i cos   8 8

8

(b) 0

(c) 1

is

(d) 2i

4π 4π  1+ a , then the value of  + i sin   2  3 3

(a) (−1)n

(b)

(−1)n 23n

(c)

1 23n

8

 1 + i  1 − i 61. The value of   +   2  2 (a) 4

Targ e t E x e rc is e s

44. If

(b) 6

3n

is

(d) (−1)n + 1

8

(c) 8

is equal to (d) 2

2πk 2πk   − i cos sin  is  11 11  k =1 10

62. The value of (a) 1 (c) i

∑

(b) −1 (d) − i

147

Objective Mathematics Vol. 1

4

63. The value of

4 (cos 75° + i sin 75° ) is 0.4 (cos 30° + i sin 30° )

10 (1 + i ) 2 5 (c) (1 + i ) 2

(a)

(b)

72. If ω ( ≠ 1) is cube root of unity satisfying 1 1 1 + + = 2ω 2 and a+ω b+ω c+ω

10 (1 − i ) 2

1 1 1 + + = 2 ω, then the value of a + ω2 b + ω2 c + ω2 1 1 1 is equal to + + a+1 b+1 c+1 (a) 2 (b) − 2 (c) − 1 + ω 2 (d) None of these

(d) None of these

64. The value of (cos 2θ − i sin 2θ ) 7 (cos 3θ + i sin 3θ ) −5 (cos 4θ + i sin 4θ )12 (cos 5θ + i sin 5θ ) −6 (a) cos 33θ + i sin 33θ (c) cos 47θ + i sin 47θ

is

73. If α and β are the roots of the equation x 2 − 2x + 4 = 0, then the value of α 6 + β 6 is

(b) cos 33θ − i sin 33θ (d) cos 47θ − i sin 47θ

65. (cos 2θ + i sin 2θ ) −5 (cos 3θ − i sin 3θ ) 6 (sin θ − i cos θ ) 3 is equal to

(a) 64 (c) 256

74. If z is any complex number such that z +

(a) cos 25θ + i sin 25θ (b) cos 25θ − i sin 25θ (c) sin 25θ + i cos 25θ (d) sin 25 θ − i cos 25θ

the value of z 99 +

66. The principal value of the arg ( z ) and | z | of the

Ta rg e t E x e rc is e s

complex number z =

(1 + cos θ + i sin θ) 5 (cos θ + i sin θ ) 3

θ θ , 32 cos5 2 2 θ 4 θ (c) − , 16 cos 2 2

68. If z = cos θ + i sin θ, then

1 (b) z + n = 2n cos nθ z 1 (d) zn − n = (2i )n sin nθ z n

z 2n − 1 , where n is an z 2n + 1

integer, is equal to (b) i tan nθ

(c) tan nθ

(d) cot nθ

69. If a = cos 2α + i sin 2α , b = cos 2β + i sin 2β , c = cos 2γ + i sin 2γ and d = cos 2δ + i sin 2δ, then 1 is equal to abcd + abcd (a) 2 cos (α + β + γ + δ ) (c) cos (α + β + γ + δ )

70. e 2mi cot (a) 0 (c) − 1

71. If

−1

(b) 2cos (α + β + γ + δ) (d) None of these

 pi + 1 ⋅  is equal to  pi − 1

 8π   ,  11

Re (α + α 2 + α 3 + α 4 + α 5 ) is equal to 148

1 (a) 2 (c) 0

1 (b) − 2 (d) None of these

(b) 264 (d) 0

(a) 4 (c) − 4

(b) 0 (d) None of these 6

6

 1 − i 3  1 + i 3  is  + 77. The value of   1+ i 3   1− i 3  (a) 2

(b) −2

(c) 1

(d) 0

78. If z1 = 3 + i 3 and z 2 = 3 + i , then the complex z  number  1   z2 

50

(a) I

(b) II

lies in the quadrant number (c) III

(d) IV

79. If x = a + b, y = aω + bω 2 , z = aω 2 + bω, then xyz is equal to (b) a3 + b3 (d) (a + b)3 + 3ab(a + b)

(a) (a + b)3 (c) a3 − b3

(a) 2ω

(b) 1 (d) None of these

 8π  α = cos   + i sin  11

(b) −1 (d) − 2

(a) 1 (c) 2

80. If 1, ω and ω 2 are the cube roots of unity, then the value of (1 + ω ) 3 − (1 + ω 2 ) 3 is

m

p

is

76. If ω is a cube root of unity, then ( 3 + 5ω + 3ω 2 ) 2 ( 3 + 3 ω + 5 ω 2 ) 2 is equal to

67. If z = cos θ + i sin θ, then 1 (a) z + n = 2cos nθ z 1 (c) zn − n = 2n i sin nθ z

z 99

(a) 262 (c) − 2 62

(d) None of these

n

1

1 = 1, then z

75. The value of ( − 1 + −3 ) 62 + ( − 1 − − 3 ) 62 is

are

θ θ (b) , 32 cos5 2 2

(a) −

(a) i cot nθ

(b) 128 (d) None of these

then

(b) 2

(c) −2

(d) 0

81. If 1, ω and ω 2 are the three cube roots of unity and α , β and γ are the roots of p, p < 0, then for any x, y xα + yβ + zγ and z, the expression equals xβ + yγ + zα (a) 1 (b) ω (c) ω 2 (d) None of the above

2

2

 n(n + 1) (b)   −n  2 

2

 n(n + 1) (c)   +n  2 

(d) None of these

83. If α and β are the complex cube roots of unity, then

n−1

1

∑ 2 − α i is equal to

i=1

(a)

(n − 2)2n − 1 + 1 2n − 1

(b) (n − 2) ⋅ 2n

(c)

(n − 2) ⋅ 2n − 1 2n − 1

(d) None of these

92. The triangle formed by the points 1, vertices in the argand diagram is a/an

(a) 0 (c) − 3

(a) scalene (c) isosceles

85. If

(a) ab′ + a′ b = 0 (c) ab + a′ b′ = 0

(b) 14 (d) None of these

x = ω − ω 2 − 2,

then

the

value

of

x 4 + 3x 3 + 2x 2 − 11 x − 6 is (b) − 1 (d) None of these

(a) 1 (c) 2

86. If 1, ω and ω 2 are the three cube roots of unity, then (1 + ω ) (1 + ω 2 ) (1 + ω 4 ) (1 + ω 8 ) … to 2n factors is equal to (b) − 1 (d) None of these

(a) 1 (c) 0

87. If x = a + b + c, y = aα + bβ + c and z = aβ + bα + c where, α and β are complex cube roots of unity, then xyz is equal to (a) 2 (a + b + c ) (c) a3 + b3 + c3 − 3abc 3

3

3

(b) 2 (a − b − c ) (d) a3 − b3 − c3 3

3

3

88. The common roots of the equations z 3 + 2z 2 + 2z + 1 = 0 and z 1985 + z 100 + 1 = 0 are (a) −1, ω (b) −1, ω 2 (c) ω , ω 2 (d) None of the above

89. The values of (16)1/ 4 are (a) ± 2, ± 2i (b) ± 4 , ± 4 i (c) ± 1, ± i (d) None of the above

90. If p, q and r are positive integers and ω is an imaginary cube root of unity and f ( x ) = x 3 p + x 3q + 1 + x 3r + 2 , then f (ω ) is equal to (a) ω (b) −ω 2 (c) 0 (d) None of the above

(b) equilateral (d) right angled

93. If the points represented by complex numbers z1 = a + ib, z 2 = a′ + ib′ and z1 − z 2 are collinear, then

84. If ω is a complex cube root of unity, then (1 − ω + ω 2 ) (1 − ω 2 + ω 4 ) (1 − ω 4 + ω 8 ) (1 − ω 8 + ω 16 ) is equal to (a) 12 (c) 16

and i as

2

α 3 + β 3 + α −2 β −2 is equal to (b) 3 (d) None of these

1+ i

4

(b) ab′ − a′ b = 0 (d) ab − a′ b′ = 0

94. If α + iβ = tan −1 z , z = x + iy and α is constant, then the locus of z is (a) x 2 + y2 + 2x cot 2α = 1 (b) (x 2 + y2 ) cot 2α = 1 + x (c) x 2 + y2 + 2 y tan 2α = 1 (d) x 2 + y2 + 2x sin 2α = 1

95. The equation | z + 1 − i | = | z + i − 1| represents (a) a straight line (c) a parabola

(b) a circle (d) a hyperbola

96. If the roots of ( z − 1) n = i ( z + 1) n are plotted in the argand plane, they as (a) on a parabola (c) collinear

(b) concyclic (d) the vertices of a triangle

Targ e t E x e rc is e s

 n(n + 1) (a)    2 

91. If 1, α , α 2 , K , α n − 1 are the n, nth roots of unity, then

Complex Numbers

82. The value of the expression 1 ( 2 − ω ) ( 2 − ω 2 ) + 2( 3 − ω ) ( 3 − ω 2 ) + .... + ( n − 1) ( n − ω ) ( 2 − ω 2 ), where ω is an imaginary cube root of unity, is

97. Locus of the point z satisfying the equation | iz − 1| + | z − i | = 2 is (a) a straight line (c) an ellipse

(b) a circle (d) a pair of straight lines

98. For x1 , x 2 , y1 , y 2 , ∈ R , if 0 < x1 , < x 2 , y1 = y 2 and 1 z1 = x1 + iy1 , z 2 = x 2 + iy 2 and z 3 = ( z1 + z 2 ), 2 then z1 , z 2 , z 3 satisfy (a) | z1 | = | z2 | = | z3 | (c) | z1 | > | z2 | > | z3 |

(b) | z1 | < | z2 | < | z3 | (d) | z1 | < | z3 | < | z2 |

99. The equation not representing a circle is given by  1 + z (a) Re   =0  1 − z (b) zz + iz − iz + 1 = 0  z − 1 π (c) arg   =  z + 1 2 (d)

z −1 =1 z+1

100. If z1 , z 2 and z 3 are three complex numbers in AP, then they lie on (a) a circle (c) a parabola

(b) a straight line (d) an ellipse

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Objective Mathematics Vol. 1

4

101. If the area of the triangle on the argand plane formed by the complex numbers − z , iz , z − iz is 600 sq units, then | z | is equal to (a) 10 (c) 30

(b) 20 (d) None of these

102. The complex numbers given by 1 − 3i, 4 + 3i and 3 + i represent the vertices of (a) a right angled triangle (b) an isosceles triangle (c) an equilateral triangle (d) None of the above

(b) (− 2, 1) (d) (2, 1)

Ta rg e t E x e rc is e s

(a) x 2 + y2 + 2x + 2 y = 0 (b) x 2 + y2 − 2x = 0 (c) x + y + 2 = 0 (d) None of the above

(c)

Re (a) |a |

|a | 2 Im (a) (d) |a | (b)

1 |1 − a | 1 (b) | z − 1| = |1 − a | 1 1 (c) z − = 1− a |1 − a | (d) None of these (a) | z − a | =

105. In the argand plane, all the complex numbers satisfying | z − 4i| + | z + 4i| = 10 lie on (b) a circle (d) a parabola

106. If z = x + iy and a is a real number such that | z − ai| = | z + ai|, then locus of z is (a) X -axis (b) Y-axis (c) x = y (d) x 2 + y2 = 1

107. If P represents z = x + iy in the argand plane and | z − 1|2 + | z + 1|2 = 4, then the locus of P is (a) x 2 + y2 = 2 (b) x 2 + y2 = 1 (c) x 2 + y2 = 4 (d) x + y = 2

(a) 1 1 (b) 3 2 (c) 3 (d) None of the above

109. The region of argand | z − 1| + | z + 1| ≤ 4 is

113. If z1 , z 2 , z 3 and z 4 are the four complex numbers represented by the vertices of a quadrilateral taken in order such that and z1 − z 4 = z 2 − z 3  z 4 − z1  π amp   = , then the quadrilateral is a  z 2 − z1  2 (a) square (b) rhombus (c) cyclic quadrilateral (d) None of the above

114. The locus of z satisfying Im ( z 2 ) = 4 is

108. All the roots of the equation a1 z 3 + a 2 z 2 + a 3 z + a 4 = 3, where | a i | ≤ 1, i = 1, 2, 3, 4, lie outside the circle with centre origin and radius

150

(a) 1

112. Let a be a complex number such that | a| < 1 and z1 , z 2 , … , z n be the vertices of a polygon such that z k = 1 + a + a 2 +… a k , then the vertices of the polygon lie within the circle

 z + 2i 104. If Im   = 0, then z lies on the curve  z + 2

(a) a straight line (c) an ellipse

(a) a circle (b) interior of a circle (c) exterior of a circle (d) None of the above

111. The closest distance of the origin from a curve given as az + az + aa = 0 (a is a complex number), is

 z + 2i 103. If Re   = 0, then z lies on a circle with centre  z + 4 (a) (− 2, − 1) (c) (2, − 1)

110. The locus of the complex number z in an argand plane satisfying the inequality  | z − 1| + 4  2  log 1/ 2   > 1  where,| z − 1| ≠  is  3  3| z − 1| − 2

diagram

(a) interior of an ellipse (b) exterior of a circle (c) interior and boundary of an ellipse (d) None of the above

defined

by

(a) a circle (b) a rectangular hyperbola (c) a pair of straight lines (d) None of the above

115. The curve represented by Re ( z 2 ) = 4 is (a) a parabola (b) an ellipse (c) a circle (d) a rectangular hyperbola

116. A point z is equidistant from three distinct points z1 , z 2 , z 3 in argand plane. If z , z1 and z 2 are  z − z1  collinear, then arg  3  will be (z1 , z 2 , z 3 in  z3 − z2  anti-clockwise sense) π 2 π (c) 6 (a)

π 4 2π (d) 3 (b)

126. Let complex number z of the form x + iy satisfy  3z − 6 − 3i π and | z − 3 + i| = 3. Then, the arg  =  2z − 8 − 6i 4 ordered pairs ( x, y ) are

a 0 z 4 + a1 z 3 + a 2 z 2 + a 3 z + a 4 = 0 where, a 0 , a1 , a 2 , a 3 and a 4 are real, then (a) z1 , z2 , z3 , z4 are also roots of equation (b) z1 is equal to atleast one of z1 , z2 , z3 , z4 (c) − z1 , − z2 , − z3 , − z4 are also roots of equation (d) None of the above

4 2  (a)  4 − ,1+   5 5 (c) (6, − 1)

118. If z 3 + ( 3 + 2i ) z + ( − 1 + ia ) = 0 has one real root, then the value of a lies in the interval ( a ∈ R )

(a) (− 2, 1) (c) (0, 1)

(b) (−1, 0) (d) (−2, 3)

119. The real value of θ for which the expression i + i cos θ is a real number, is 1 − 2i cos θ π (a) 2nπ + , n ∈ I 2 π (c) 2nπ ± , n ∈ I 2

π (b) 2nπ − , n ∈ I 2 π (d) 2nπ ± , n ∈ I 4

(a) α + α = 1 (b) α + α = 0 (c) α + α = − 1 (d) the absolute value of the real root is 1

(b) | z1 − z2 |min = 10 (d) | z1 − z2 |max = 25

(c) | z|max

(d) | z|min =

5 −1 2 5 −1 2

satisfying

−

z 22 | =

| z12

+

z 22

129. If z is complex number satisfying z + z −1 = 1, then z n + z − n , n ∈ N has the value (a) 2(−1)n, when n is a multiple of 3 (b) (−1)n − 1, when n is not a multiple of 3 (c) (−1)n + 1 , when n is a multiple of 3 (d) 0, when n is not a multiple of 3

− 2z1 z 2 | , then

z (a) 1 is purely imaginary z2

z (b) 1 is purely real z2

(c) | arg z1 − arg z2 | = π

(d) | arg z1 − arg z2 | =

125. If | z − 1| = 1, then (a) arg {(z − 1 − i )/ z} can be equal to − π /4 (b) (z − 2)/ z is purely imaginary number (c) (z − 2)/ z is purely real number (d) If arg (z) = θ, where z ≠ 0 and θ is acute, then 1 − 2 / z = i tan θ

(b) min | z1 − z2 | = 1 (d) None of these

(a) B represents the complex number iz (b) D represents the complex number iz (c) B represents the complex number iz (d) D represents the complex number − iz

124. If z1 and z 2 are two complex numbers ( z1 ≠ z 2 ) | z12

(b) z1z2 = 1 (d) None of these

131. ABCD is a square, vertices being taken in the anti-clockwise sense. If A represents the complex number z and the intersection of the diagonals is the origin, then

123. If | z − (1/ z )| = 1, then (b) | z|min =

128. If amp ( z1 z 2 ) = 0 and | z1 | = | z 2 | = 1, then

(a) max |2z1 + z2 | = 4 1 (c) z2 + ≤3 z1

122. If | z1 | = 15 and | z 2 − 3 − 4i| = 5, then

1+ 5 2 5−2 = 2

(b) z = 0 (d) z = − 1

points on the circle | z | = 1 and | z | = 2 respectively, then

(a) z1z2 = 1 (b) z1 + z2 = 0 (c) z1 = z2 (d) None of the above

(a) | z|max =

(a) z = 1 (c) z = − 2

130. Let z1 , z 2 be two complex numbers represented by

121. If | z1 | = | z 2 | = 1 and arg z1 + arg z 2 = 0, then

(a) | z1 − z2 |min = 5 (c) | z1 − z2 |max = 20

127. If z = ω , ω 2 , where ω is a non-real complex cube root of unity, are two vertices of an equilateral triangle in the argand plane, then the third vertex may be represented by

(a) z1 + z2 = 0 (c) z1 = z2

120. If α is a complex constant such that αz + z + a = 0has a real root, then

4 2  (b)  4 + ,1−   5 5 (d) (0, − 1)

Targ e t E x e rc is e s

117. If z1 , z 2 , z 3 and z 4 are roots of the equation

Complex Numbers

4

Type 2. More than One Correct Option

π 2

132. If z 0 , z1 represent point P , Q on the locus | z − 1| = 1 π and the line segment PQ subtend an angle at the 2 point z = 1, then z1 is equal to (a) 1 + i (z0 − 1) (c) 1 − i (z0 − 1)

i z0 − 1 (d) i (z0 − 1)

(b)

133. If | z1 | = | z 2 | = | z 3 | = 1 and z1 , z 2 and z 3 are represented by the vertices of an equilateral triangle, then (a) z1 + z2 + z3 = 0 (c) z1z2 + z2z3 + z3z1 = 0

(b) z1z2z3 = 1 (d) None of these

151

Objective Mathematics Vol. 1

4

Type 3. Assertion and Reason Direction (Q. Nos. 134-144) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I. (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true.

134. Consider z1 = 1, z 2 = 2 and z 3 = 3. Statement I

If z1 + 2z 2 + 3z 3 = 6, then the value

of z 2 z 3 + 8z 3 z1 + 27z1 z 2 is 36. Statement II

z1 + z 2 + z 3 ≤ z1 + z 2 + z 3

135. Statement I 7 + i > 5 + i

Ta rg e t E x e rc is e s

Statement II Cancellation laws does not hold true in complex numbers. 136. Statement I If z1 and z 2 are two complex numbers z  such that | z1 | = | z 2 | + | z1 − z 2 |, then Im  1  = 0.  z2 

139. Statement I If cos (1− i ) = a + ib, where a, b ∈ R and 1 1 1 1 i = −1, then a =  e +  cos 1, b =  e −  sin 1. e 2 2 2 Statement II

140. Statement I The product of all values of (cos α + i sin α ) 3/ 5 is cos 3a + i sin 3a. Statement II The product of fifth roots of unity is 1. 141. Statement I The locus of the centre of a circle which touches the circles | z − z1 | = a and | z − z 2 | = b externally ( z , z1 and z 2 are complex numbers) will be hyperbola. Statement II

1 = 1 ( z ≠ 0 is a complex z 5 +1 number), then the maximum value of | z | is . 2 1 Statement II On the locus z + = 1, the farthest z 5 +1 . distance from origin is 2

138. Statement I The number of complex numbers z satisfying | z |2 + a | z | + b = 0 ( a, b ∈ R ) is atmost 2. Statement II A quadratic equation in which all the coefficients are non-zero can have atmost two roots.

| z − z1 | − | z − z 2 | < | z 2 − z1 |

⇒ z lies on hyperbola. 142. Statement I Consider an ellipse having its foci at A ( z1) and B ( z 2) in the argand plane. If the eccentricity of the ellipse is e and it is known that origin is an interior point of the ellipse, then  | z + z 2 | | z1 + z 2 |  e ∈ 1 ,   | z1 | + | z 2 | | z1 | + | z 2 | Statement II If z 0 is the point interior to curve | z − z1 | + | z − z 2 | = λ | z 0 − z1 | + | z 0 − z 2 | < λ

Statement II arg ( z ) = 0 ⇒ z is purely real. 137. Statement I If z +

e iθ = cos θ + i sin θ

143. Statement I The equation | z − i| + | z + i| = k , k > 0, can represent an ellipse, if k > 2i. Statement II | z − z1 | + | z − z 2 | = k , represents an ellipse, if | k | > | z1 − z 2 |. 144. Statement I The equation zz + az + az + λ = 0 , where a is a complex number represents a circle in argand plane, if λ is real. Statement II The radius zz + az + az + λ = 0 is aa − λ .

of

the

circle

Type 4. Linked Comprehension Based Questions Passage I (Q. Nos. 145-147) Consider the complex numbers z1 and z 2 2 2 2 z1 + z1 = z1 + z 2 .

satisfying

the

(a)

145. Complex number z1 z 2 is (a) purely real (c) zero

146. Complex number z1 / z 2 is

152

(a) purely real (b) purely imaginary (c) zero (d) None of the above

relation

147. One of the possible argument of complex number i( z1 / z 2 ) is

(b) purely imaginary (d) None of these

π 2

(b) −

π 2

(c) 0

(d) None of these

Passage II (Q. Nos. 148-150) If z satisfy the relation | z − {(α 2 − 7α + 11) + i }| = 1, α ∈R. π Also, arg ( z ) ≥ is satisfied by atleast one z. 2

148. The values of α lies in the interval (a) [ 2, 6 ]

 7 (b) 1,  2 

(c) [ 2, 5 ]

(d) [ − 4 , − 2 ]

(a)

7 2

(b)

4 9

(c)

5 9

(d)

AB and points A and E lies in the opposite side of line BC. If A, B and C are represent by the complex numbers 1, ω and ω 2 respectively.

9 4

150. Maximum value of arg ( z ), for which | z − i| is maximum, is  4 (a) π − tan −1    9  1 (c) π − tan −1    9

 9 (b) π − tan −1    4  4 (d) π − tan −1    5

151. Angle between AC and DE is equal to (a)

π 3

(b)

π 6

(c)

π 4

(d)

π 2

152. The length of DE is 3 2

(a)

Passage III (Q. Nos. 151-153) On the sides AB and BC of a ∆ABC, squares are drawn with centres D and E such that points C and D lies on the same side of line AB

(b) 3

(c) 2

(d) 6

(c) 3 − 3

(d) 3 +

4 Complex Numbers

149. Maximum value of | z − i| is

153. The length of AE is (a)

3− 3 2

(b)

3+ 3 2

3

Type 5. Match the Columns 154. Match the following: Column I  z2 − 1  A. arg   = 0; z ≠ ± i , ± 1 2  z + 1

P. Portions of a line

B. z − cos −1 cos 12 |−| z − sin−1 sin 12

q. Point of intersection of hyperbola

= 8 ( π − 3)

Column I

Column II

C. z2 + k1 = i | z1|2 + k 2; k1 ≠ k 2 ∈ R − { 0}

r. Pair of open rays

D. z − 1 − sin−1 1 + z + cos −1 1 − π = 1 s. line segment 3 3 2

A. z4 − 1 = 0

p.

B. z4 + 1 = 0

q.

C. i z4 + 1 = 0

r.

D. i z4 − 1 = 0

s.

π 8 π 8 π 4

z = cos 0 + i sin 0

157. Match the following: Column I A.

155. For the equation z − 6z + 20 = 0, match the items of Column I with that of Column II. 6

Column I

Column II π z = cos + i sin 8 π z = cos − i sin 8 π z = cos + i sin 4

B.

−1 − −1 − ... ∞ is If zi, i = 1, 2, ..., 6 are the vertices of the six sided regular polygon inscribed in a circle| z| = 2, then

Column II

ColumnI

6

∑ zi

2

p.

0

q.

n ω−1

is equal to

i =1

A.

The number of the roots in the first quadrant can be

p.

1

B.

The number of the roots in the second quadrant can be

q.

2

C.

The number of the roots in the third quadrant can be

r.

3

D.

The number of the roots in the fourth quadrant can be

s.

4

C.

If ω k, k ∈ I; 0 ≤ k ≤ n − 1 are the nth roots of unity, then the maximum value of ( n!)1/ n ω( n−1)/ 2 is

r.

ω

D.

If z1, z2, z3 are the affixes points and A, B and C are lying on circle centred at origin. If the altitude is drawn from vertex A to base BC, such that it meets the circumcircle at P( z), then zz1 + z2 z3 is

s.

ω2

Targ e t E x e rc is e s

156. Match the following:

Type 6. Single Integer Answer Type Questions 158. If x = a + bi is a complex number such that x 2 = 3 + 4i and x 3 = 2 + 11i, where i = −1, then ( a + b ) equals _______.

161. If z is a complex number satisfying 4 z + z 3 + 2z 2 + z + 1 = 0, then | z | is equal to _____ .

159. If the complex numbers z is simultaneously satisfy z − 12 5 z − 4 equations = 1, then Re ( z ) is ___. = , z − 8i 3 z − 8

162. ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD = 2AC. Its points D and M represent the complex numbers 1 + i and 2 − i respectively. Then, the complex number represented by A is z, has Re ( z ) = {λ 1 , λ 2 }, then the value of λ 1 + λ 2 is __________ .

160. If | z | ≥ 3, then the least value of z + is __________ .

1 λ is , where λ z 3

153

Entrances Gallery JEE Advanced/IIT JEE 1. Match the following :

[2014]

 2kπ   2kπ  Let zk = cos   ; k = 1, 2, ..., 9.  + i sin   10   10  Column I A. For each zk , there exists a z j such zk ⋅ z j = 1

p. True

B. There exists k ∈ {1, 2, ..., 9}, such that z1 ⋅ z = zk has no solution z in the set of complex numbers

q. False

C. 1 − z1 1 − z2 ... 1 − z9 equals 10 9  2 kπ  D. 1 − ∑ cos  10  equals k=1

r. 1

Codes A (a) p (c) p

Ta rg e t E x e rc is e s

Column II

B q q

C s r

D r s

1 2

s. 2

A (b) q (d) q

(c)

B p p

C r s

D s r

1 7

2

(d)

= r2 + 2 , [2013] 1 3

3+i and P = {w n : n = 1, 2, 3, ...}. Further 2 1 and H1 =  z ∈ C : Re ( z ) >  2  − 1  H 2 =  z ∈ C : Re ( z )  , where C is the set of all 2  complex numbers. Ifz1 ∈ P ∩ H1 , z 2 ∈ P ∩ H 2 and O represents the origin, then ∠z1Oz 2 equals [2013]

3. Let w =

(a)

π 2

(b)

π 6

(c)

2π 3

(d)

5π 6

16π (c) 3

z ∈S

(a)

154

2− 3 2

(b)

2+ 3 2

(c)

3− 3 2

(d)

x a

2

+ y

2

+ z

2

2

+ b

2

+ c

2

is ________ .

[2011]

9. Let z1 andz 2 be two distinct complex numbers and z = (1 − t ) z1 + tz 2 for some real number t with 0 < t 0  1− 3 i    and S 3 { z ∈ C : Re z > 0}. 20π (b) 3

π 3

p. an ellipse with 4 eccentricity 5

where,

10π (a) 3

7. If z is any complex number satisfying z − 3 − 2i ≤ 2,

A. The set of point z satisfying z − i z = z + i z is contained in or equal to

Passage (Q. Nos. 4-5) Let S = S1 ∩ S 2 ∩ S 3

4. The area of S equals

1 2

2i

z 0 = x 0 + iy 0 satisfies the equation 2 z 0 then α equals (b)

(b)

[2011]. then the minimum value of 2z − 6 + 5i is________

1 lie on the circles α and ( x − x0 ) 2 + ( y − y0 ) 2 = r 2 2 2 2 respectively. If ( x − x0 ) + ( y − y0 ) = 4r ,

1 2

1 3 3 (d) 4

(a) − 1 (c)

2. Let complex numbers α and

(a)

6. Let z be a complex number such that the imaginary part of z is non-zero and a = z 2 + z + 1is real. Then, a cannot take the value [2012]

3+ 3 2

t. the set of point z satisfying Im| z| ≤ 5

11. A complex number z is said to be unimodular, if | z | = 1. Suppose z1 and z 2 are complex numbers such z1 − 2 z 2 that is unimodular and z 2 is not 2 − z1 z 2 unimodular. Then, the point z1 lies on a [2015] (a) straight line parallel to X -axis (b) straight line parallel to Y-axis (c) circle of radius 2 (d) circle of radius 2.

1 2



∑ sin

(d) 0

2kπ 2kπ  + i cos  is 11 11 

(b) − 1

(c) − i

[2006]

(d) i

23. If z + z + 1 = 0, where z is complex number, then 2

2

(d) π − θ

1 1 1   the value of  z +  +  z 2 + 2  +  z 3 + 3     z z  z 

2

2

1  + ... +  z 6 + 6  is  z  (a) 54

(b) 6

[2006] (c) 12

(d) 18 2

(b) β = 1 (d) β ∈ (0, 1)

(1 + ω ) 7 = A + Bω. Then, (A , B) equals

of the equation ( x − 1) 3 + 8 = 0, are (a) − 1, 1 + 2ω, 1 + 2ω (c) −1, − 1, − 1

[2011]

17. The number of complex numbers z such that [2010] z − 1 = | z + 1| = z − i equals (b) 1 (d) ∞

18. If α and β are the roots of the equation x 2 − x + 1 = 0, [2010] then α 2009 + β 2009 is equal to

2

[2005] 2

(b) −1, 1 − 2ω, 1 − 2ω (d) −1, − 1 + 2ω , − 1 − 2ω 2

25. If z1 and z 2 are two non-zero complex numbers such that z1 + z 2 = z1 + z 2 , then arg ( z1 ) − arg ( z 2 ) is equal to [2005] (a) −

π 2

26. If w =

(b) 0

z z−

i 3

(c) − π

(d)

π 2

and w = 1, then z lies on

(a) a parabola (c) a circle

(b) (1, 0) (d) (0, 1)

(b) − 1 (d) 2

24. If the cube roots of unity are 1, ω, ω , then the roots

Targ e t E x e rc is e s

(c) θ

(c) 6

2

16. If ω (≠ 1) is a cube root of unity and

(a) − 2 (c) 1

[2007] (b) 10

(a) 1

15. Let α and β be real and z be a complex number. If z 2 + α z + β = 0 has two distinct roots on the line [2012] Re ( z ) = 1, then it is necessary that

(a) 0 (c) 2

1 . Then, i−1

21. If z + 4 ≤ 3, then the maximum value of z + 1 is

k =1

(a) either on the real axis or on a circle passing through the origin (b) on a circle with centre at the origin (c) either on the real axis or on a circle not passing through the origin (d) on the imaginary axis

(a) (1, 1) (c) (−1, 1)

2

[2008] 1 (d) i −1

1 (c) − i+1

1 (b) i+1

10

z2 is real, then the point represented by 14. If z ≠ 1and z −1 [2012] the complex number z lies

(a) β ∈ (− 1, 0) (c) β ∈ (1, ∞ )

1 (a) − i −1

22. The value of

13. If z is a complex number of unit modulus and 1 + z argument θ, then arg  [2013]  equals 1 + z π −θ 2

(d) 2 +

(c) 2

20. The conjugate of a complex number is

(a) 4

5 2 (b) lies in the interval (1, 2) 5 (c) is strictly greater than 2 3 5 (d) is strictly greater than but less than 2 2

(b)

(b) 5 + 1

[2014]

(a) is equal to

(a) − θ

(a) 3 + 1

that complex number is

12. If z is a complex number such that z ≥ 2, then the minimum value of z +

4 19. If z − = 2 , then the maximum value of z is z [2009] equal to

Complex Numbers

4

JEE Main/AIEEE

[2005]

(b) a straight line (d) an ellipse

27. Let z and w be two complex numbers such that z + iw = 0 and arg ( zw ) = π. Then, arg ( z ) equals [2004] π (a) 4 3π (c) 4

π (b) 2 5π (d) 4

28. If and z = x − iy z 1/ 3 = p + iq,  x y 2 2  +  / ( p + q ) is equal to  p q (a) 1

(b) − 1

(c) 2

(d) − 2

then [2004]

155

Objective Mathematics Vol. 1

4

29. If | z 2 − 1| = | z |2 + 1, then z lies on (a) the real axis (c) a circle

[2004]

(b) the imaginary axis (d) an ellipse

30. Let z1 and z 2 be two roots of the equation z 2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z 2 form an equilateral triangle. Then, [2003] (a) a2 = b

(b) a2 = 2b

(c) a2 = 3b

(d) a2 = 4 b

31. If z and w are two non-zero complex numbers such π that zw = 1, and arg ( z ) − arg ( w ) = , then z ( w ) is 2 equal to [2003] (b) − 1

(a) 1

(d) − i

(c) i

x

 1 + i 32. If   = 1, then  1 − i

[2003]

(a) x = 4 n, where n is any positive integer (b) x = 2n, where n is any positive integer (c) x = 4 n + 1, where n is any positive integer (d) x = 2n + 1, where n is any positive integer

33. If ω is an imaginary cube root of unity, then [2002] (1 + ω − ω 2 ) 7 equals (a) 128 ω (b) −128 ω (c) 128 ω 2 (d) −128 ω 2

Other Engineering Entrances 34. If α and β are two different complex numbers with β = 1, then

β −α 1 − αβ

is equal to

(a) 1/2 (c) −1

Ta rg e t E x e rc is e s

35. If z =

[Karnataka CET 2014]

(b) 0 (d) 1

( 3 + i ) 3 ( 3i + 4 ) 2

(a) 8 (d) 4

( 8 + 6i ) 2 (b) 2 (e) 10

, then z is equal to [Kerala CEE 2014] (c) 5

36. Let w ≠ ± 1 be complex number. If w = 1 and w−1 , then Re ( z ) is equal to [Kerala CEE 2014] z= w+1 (a) 1

(b)

1 w+1

(c) Re (w)

(d) 0

(e) w + w

37. The value of z when z equals

2

2

+ z−3 + z−i

2 i 3 i (c) 1 + 3 (a) 2 −

2

is minimum, [WB JEE 2014]

(b) 45 + 3i (d) 1 −

i 3

π π 38. Convert ( i + 1) / cos − i sin  in polar form.  4 4 [J&K CET 2014] (a) cos (π /4) + i sin (π / 4 ) (b) cos (π / 2) − i sin (π / 2) (c) 2 [cos (π /4) + i sin (π /4 )] (d) 2 [cos (π / 2) + i sin (π / 2)]

39. Let z1 ≠ z 2 and z1 = z 2 . If z1 has positive real part z + z1 and z 2 has negative imaginary part. Then, 2 z1 − z 2 may be [Manipal 2014] (a) 0 (c) real and negative

(b) real and positive (d) None of these

40. If z = e 2π / 3 , then 1 + z + 3z 2 + 2z 3 + 2z 4 + 3z 5 is equal to [Kerala CEE 2014] 156

(a) − 3eπ i / 3 (b) 3eπ i / 3 (d) − 3e2π i / 3 (e) 0

(c) 3e2π i / 3

41. If ω, ω 2 are the cube roots of unity, then roots of [ RPET 2014] equation ( x − 1) 3 + 5 = 0 are (a) −5, − 5 ω, − 5 ω 2 (b) −4 , 1 − 5 ω, 1 − 5 ω 2 (c) 6, 1 − 5 ω, 1 + 5 ω 2 (d) None of the above

42. The complex number z = x + iy, which satisfies the z − 3i [BITSAT 2014] equation = 1, lie on z + 3i (a) the X -axis (b) the straight line y = 3 (c) a circle passing through the origin (d) None of the above

43. If the complex numbers z1 , z 2 and z 3 denote the vertices of an isosceles triangle, right angled at z1 , then ( z1 − z 2 ) 2 + ( z1 − z 3 ) 2 is equal to [Kerala CEE 2014] (a) 0 (d) 3

(b) (z2 + z3 )2 (e) (z2 − z3 )2

(c) 2

44. Let z1 and z 2 be two fixed complex numbers in the argand plane and z be an arbitrary point satisfying z − z1 + z − z 2 = 2 z1 − z 2 . Then, the locus of z will be [WB JEE 2014] (a) an ellipse (b) a straight line joining z1 and z2 (c) a parabola (d) a bisector of the line segment joining z1 and z2

45. Let z1 be a fixed point on the circle of radius 1 centred at the origin in the argand plane and z1 ≠ ± 1. Consider an equilateral triangle inscribed in the circle with z1 , z 2 and z 3 as the vertices taken in the counter clockwise direction. Then, z1 z 2 z 3 is equal to [WB JEE 2014] (a) (b) (c) (d)

z12 z13 z14 z1

[WB JEE 2014] (a) the vertices of an equilateral triangle (b) the vertices of an isosceles triangle (c) collinear (d) the vertices of a scalene triangle

(a) a square (b) an equilateral triangle (c) a rhombus (d) a rectangle

50.

(b) − 2

(c) 0

π (b) 4

[UP SEE 2013]

1 1 1 + + = 1, z1 z 2 z 3

z1 + z 2 + z 3 is (a) 3 (c) greater than 3

then [AMU 2013]

(b) 3 (d) 0

53. The value of (1 + 3i ) 4 + (1 − 3i ) 4 is (a) − 16 (c) 14

[RPET 2013]

(b) 16 (d) − 14

54. If the fourth roots of unity are z1 , z 2 , z 3 and z 4 , then [Karnataka CET 2013] z12 + z 22 + z 32 + z 42 is equal to (a) 0 (c) 3

(b) 2 (d) None of these

55. Among the complex number z satisfying condition z + 1 − i ≤ 1, the number having the least positive [OJEE 2013] argument is (a) 1 − i (b) 1 + i (c) − i (d) None of the above

(a) (− 3, 0)

(b) (0, 3)  1 3  (d)  , 2 2 

(c) (0, − 3)

59. If z − z + z + z = 2, then z lies on (b) a square

[AMU 2012]

(c) an ellipse (d) a line

60. If 2x = 3 + 5i, then the value of 2x 3 + 2x 2 − 7x + 72 is [MP PET 2011]

(b) − 4 (d) − 8

(a) 4 (c) 8 2

= z1

2

2

+ z 2 , then

z1 is [MP PET 2011] z2

(a) purely real (b) purely imaginary (c) zero of purely imaginary (d) neither real nor imaginary

62. The value of

(b) 1 (d) less than 1

52. If ( 3 i + 1)100 = 299 ( a + ib ), then a 2 + b 2 is equal to [RPET 2013] (a) 4 (c) 2

3 3 25 58. If  + i  = 3 ( x + iy ), where x and y are real, 2 2   [WB JEE 2012] then the ordered pair ( x, y ) is

61. If z1 + z 2

(d) None of these

51. If z1 , z 2 and z 3 are complex numbers such that z1 = z 2 = z 3 =

[Karnataka CET 2012] 1− i (b) x + iy = 1 − 2i 1 (d) x = − 5

1− i (a) x − iy = 1 − 2i 1 (c) x = 5

(a) a circle

(d) − 4

If z ( 2 − i2 3 ) 2 = i( 3 + i ) 4 , then amplitude of z is −π (a) 6 π (c) 6

(d) None of these

1+ i 3  1   1 +  i + 1

(a) 20 5 (c) 4

2

is

Targ e t E x e rc is e s

(c) − 64i

49.The value of (1 + i ) 3 + (1 − i ) 3 is equal to [J&K CET 2013] (a) 1

z  (b) Re  1  = 0  z2 

4

50

48. If z1 = 2 2 (1 + i ) and z 2 =1 + i 3, then z12 z 23 is equal to [Kerala CEE 2013] (b) 64i (e) 256

z  (a) Im  1  = 0  z2  z  z  (c) Re  1  = Im  1   z2   z2 

57. If the conjugate of ( x + iy ) (1 − 2i ) is 1 + i, then

47. In the argand plane, the distinct roots of 1 + z + z 3 + z 4 = 0 (z is a complex number) [WB JEE 2014] represent vertices of

(a) 128i (d) −128i

56. If z1 and z 2 are two complex numbers such that [Manipal 2012] z1 = z 2 + z1 − z 2 , then

Complex Numbers

46. Suppose that z1 , z 2 and z 3 are three vertices of an equilateral triangle in the argand plane. Let 1 α = ( 3 − i ) and β be a non-zero complex 2 numbers. The points αz1 + β, αz 2 + β , αz 3 + β will be

[Karnataka CET 2011]

(b) 9 4 (d) 5

63. If z1 and z 2 are two non-zero complex numbers such z  that z1 + z 2 = z1 + z 2 , then arg  1  is  z2  (a) 0 (d)

π 2

(b) − π

π (c) − 2

[Kerala CEE 2011]

(e) π

64. If ω ≠ 1 is a cube root of unity, then the sum of the series S = 1 + 2ω + 3ω 2 + ... + 3 nω 3n − 1 is [WB JEE 2011] 3n ω −1 ω− 1 (c) 3n

(a)

(b) 3n (ω − 1) (d) 0

157

Objective Mathematics Vol. 1

4

  θ 1 + cos  2 − i sin 65.  1 + cos  θ + i sin   2

 θ     2   θ     2 

(a) cos nθ − i sin nθ (c) cos 2nθ − i sin 2nθ

4n

is equal to

[UP SEE 2011]

− 3+ i 3 is a complex number, then the value 2 of ( x 2 + 3x ) 2 ( x 2 + 3x + 1) is [UP SEE 2011] 9 8 (c) − 18

(d) 36

67. If ( x + iy ) (a) − 20 (d) 60

= 2 + 3i, then 3x + 2 y is equal to (b) − 60 (e) 156

(c) −120

Ta rg e t E x e rc is e s 158

(c) 9

69. If z1 , z 2 , .... , z n are complex numbers such that z1 = z 2 = ... = z n = 1, then z 2 + z 2 + ... + z n is equal to [Kerala CEE 2010] (a) z1z2z3... zn (c)

1 1 1 + + ... + z1 z2 zn

(e) n

(b) z1 + z2 + ... + zn (d) n

3

72. If z =

z z + is z z

[Kerala CEE 2010] (b) 2 cos 2 θ (d) 2 sin θ

4 , then z is (where, z is complex conjugate 1− i

of z)

[Kerala CEE 2010] (b) 2 (e) 3

(b) 2 (d) 3

71. If z = r (cos θ + i sin θ ), then the value of

[Kerala CEE 2010]

68. The modulus of the complex number z such that z + 3 − i = 1 and arg ( z ) = π, is equal to (a) 1 (d) 4

(a) 6 (c) 6 (e) 2 +

(a) cos 2 θ (c) 2 cos θ (e) 2 sin 2 θ

(b) 6

1/ 3

and

[Kerala CEE 2010]

(b) cos nθ + i sin θ (d) cos 2nθ + i sin 2nθ

66. If x =

(a) −

π π z1 = 2 cos + i sin   4 4 π π z 2 = 3 cos + i sin  , then z1 z 2 is  3 3

70. If

[WB JEE 2010]

(a) 2(1 + i ) 2 (c) 1− i

(b) (1 + i ) 4 (d) 1− i

π 73. If − π < arg ( z ) < − , then arg ( z ) − arg ( − z ) is 2 [WB JEE 2010] (a) π (b) − π (c) π /2 (d) − π / 2

74. The value of (a) i 1+ (c)

3i 2

cos 30° + i sin 30° is equal to cos 60° − i sin 60° [VITEEE 2010] (b) − i 1 − 3i (d) 2

Answers Work Book Exercise 4.1 1. (b)

2. (a)

3. (c)

4. (b)

11. (d)

12. (d)

13. (c)

14. (c)

5. (a)

6. (d)

7. (b)

8. (d)

9. (a)

10. (c)

6. (b)

7. (c)

8. (b)

9. (d)

10. (d)

Work Book Exercise 4.2 1. (a)

2. (b)

3. (a)

4. (c)

5. (a)

11. (a)

12. (a)

13. (d)

14. (b)

15. (b)

Work Book Exercise 4.3 1. (c)

2. (b)

3. (b)

4. (c)

5. (a)

6. (d)

7. (c)

8. (b)

9. (a)

10. (d)

11. (a)

12. (b)

13. (b)

14. (c)

15. (b)

16. (c)

17. (c)

18. (a)

19. (a)

20. (c)

21. (c)

22. (a)

23. (c)

24. (a)

25. (d)

26. (c)

27. (b)

Work Book Exercise 4.4 1. (c)

2. (d)

3. (b)

4. (b)

5. (a)

6. (b)

7. (a)

8. (a)

9. (c)

10. (b)

11. (b)

12. (c)

13. (c)

14. (a)

15. (a)

16. (c)

17. (a)

18. (b)

19. (a)

20. (c)

21. (a)

22. (d)

23. (a)

24. (b)

25. (a)

26. (b)

27. (b)

28. (a)

29. (d)

30. (b)

Target Exercises 2. (a)

3. (c)

4. (d)

5. (b)

6. (c)

7. (c)

8. (c)

9. (b)

12. (d)

13. (c)

14. (a)

15. (c)

16. (d)

17. (d)

18. (a)

19. (b)

10. (c) 20. (b)

21. (a)

22. (c)

23. (c)

24. (d)

25. (c)

26. (c)

27. (d)

28. (c)

29. (a)

30. (b)

31. (a)

32. (a)

33. (c)

34. (c)

35. (b)

36. (b)

37. (b)

38. (b)

39. (a)

40. (a)

41. (d)

42. (a)

43. (b)

44. (a)

45. (a)

46. (c)

47. (b)

48. (b)

49. (a)

50. (b)

51. (c)

52. (a)

53. (b)

54. (b)

55. (a)

56. (c)

57. (a)

58. (c)

59. (c)

60. (b)

61. (d)

62. (c)

63. (a)

64. (d)

65. (c)

66. (a)

67. (a)

68. (b)

69. (b)

70. (b)

71. (b)

72. (a)

73. (b)

74. (d)

75. (c)

76. (c)

77. (a)

78. (a)

79. (b)

80. (d)

81. (c)

82. (b)

83. (b)

84. (c)

85. (a)

86. (a)

87. (c)

88. (c)

89. (a)

90. (c)

91. (a)

92. (c)

93. (b)

94. (a)

95. (a)

96. (c)

97. (a)

98. (d)

99. (d)

100. (b)

101. (b)

102. (d)

103. (a)

104. (c)

105. (c)

106. (a)

107. (b)

108. (c)

109. (c)

110. (c)

111. (b)

112. (c)

113. (c)

114. (b)

115. (d)

116. (a)

117. (a,b)

118. (a,b,d)

119. (a,b,c) 120. (a,c,d)

121. (a,c) 122. (a,d) 123. (a,b) 124. (a,d) 125. (a,b,d) 126. (a,b) 127. (a,c)

128. (b,c)

129. (a,b)

130. (a,b,c)

131. (a,d) 132. (a,c)

133. (a,b) 134. (b)

135. (d)

136. (a)

137. (a)

138. (d)

139. (a)

140. (b)

141. (d)

142. (d)

143. (d)

144. (a)

145. (b)

146. (b)

147. (c)

148. (c)

149. (d)

150. (a)

151. (c)

152. (a)

153. (b)

154. (*)

155. (**)

156. (***) 157. (****) 158. (3)

159. (6)

160. (8)

161. (1)

162. (4)

Targ e t E x e rc is e s

1. (b) 11. (b)

* A → r; B → p; C → q; D → s ** A → p, q; B → p, q; C → p, q; D → p, q *** A → s; B → r; C → p; D → q **** A → r,s; B → p; C → q; D → p

Entrances Gallery 1. (c)

2. (c)

4. (b)

5. (c)

6. (d)

7. (5)

8. (3)

11. (c)

12. (b)

13. (c)

3. (c,d)

14. (a)

15. (c)

16. (a)

17. (b)

18. (c)

19. (b)

9. (a,c,d)

10. (*) 20. (c)

21. (c)

22. (c)

23. (c)

24. (b)

25. (b)

26. (b)

27. (c)

28. (d)

29. (b)

30. (c)

31. (d)

32. (a)

33. (d)

34. (d)

35. (b)

36. (d)

37. (c)

38. (d)

39. (a)

40. (a)

41. (b)

42. (a)

43. (a)

44. (a)

45. (b)

46. (a)

47. (b)

48. (d)

49. (d)

50. (a)

51. (b)

52. (a)

53. (a)

54. (a)

55. (d)

56. (a)

57. (b)

58. (d)

59. (a)

60. (a)

61. (b)

62. (d)

63. (a)

64. (a)

65. (c)

66. (c)

67. (c)

68. (e)

69. (c)

70. (c)

71. (b)

72. (d)

73. (a)

74. (a)

* A → q; B → p, t; C → p; D → s,q

159

Explanations Target Exercises i 584 (i 8 + i 6 + i 4 + i 2 + 1) i 584 − 1 = 574 − 1 574 8 6 4 2 (i + i + i + i + 1) i i = i 10 − 1 = − 1 − 1 = −2 1 1 1 2. i 57 + 125 = (i 4 )14 i + 4 31 = i + = i − i = 0 i i (i ) i

1.

3. We have, i n + i n + 1 + i n + 2 + i n + 3 = i n (1 + i + i 2 + i 3 ) [Q i 3 = i 2 ⋅ i = (− 1)⋅ i = − i ] = i n (1 + i − 1 − i ) n = i (0 ) = 0

2

1+ i 1− i  +  1− i 1+ i

=

− 2i (1 + i )2 (1 − i )2 2i = + + 2i (1 − i )2 (1 + i )2 − 2 i Q (1 + i )2 = 1 + i 2 + 2 i = 2 i    2 2 and (1 − i ) = 1 + i − 2 i = − 2 i 

= −1− 1= −2

= 1 − 1 + 1 − 1 + ... + (− 1)n which cannot be determined unless n is known. 

1

25  2

1 



2

Ta rg e t E x e rc is e s

2

1  = i2 ⋅ i + = (− i − i )2  i  = (− 2 i )2 = 4i 2 = − 4 i 4n + 1 − i 4n − 1 2 1 i− 4n 4n − 1 i ⋅i −i ⋅i i = = 2 2 i2 − 1 − 1− 1 − 2 1 = = = =i = −i 2i 2i 2i

15. Given that, a2 + b2 = 1 ∴

6. We have,

[Q i 4 n = (i 4 )n = (1)n = 1]

∴

n

1 7. (1 − i ) 1 −  = (1 − i ) n (1 + i ) n  i = (1 + 1) n = 2 n

8.

⇒

n−2

(1 − i ) (1 − i ) (1 − i ) × = n−2 n−2 (1 + i ) (1 + i ) (1 − i ) n − 2 n

n −1

= 2 (− i ) n − 1 = 2 (− 1) 2 n −1 This is positive and real, if is even. 2 n −1 Let = 2λ 2 ∴ n = 4λ + 1 n

n

 1 − 1 + 2i  1+ i 1+ i 1+ i 9.  ×   =  =  1+ 1  1− i 1− i 1+ i =i

n

b + ic ib − c ⇒ iz = 1+ a 1+ a 1 + iz 1 + a − c + ib = 1 − iz 1 + a + c − ib (1 + a − c ) + ib (1 + a + c ) + ib = × (1 + a + c ) − ib (1 + a + c ) + ib a2 + a + iab + ib = 1 + a + ac + c (a + 1) (a + ib) a + ib = = (a + 1) (1 + c ) 1 + c z=

17. x + iy = 6 i (3 i 2 + 3) + 3 i (4 i + 20 ) + 1(12 − 60 i ) n

= − 1, if n = 2

10. (1 + i ) = (1 − i )2 n ⇒ 1 = (− 1)n 2n

∴The smallest value of n is 2. 1 + 2i 1 + 2i 1 + i 1 + 3 i − 2 1 3 11. = × = =− + i 1− i 1− i 1+ i 2 2 2 1 + 2i lies in II quadrant ∴ 1− i

160

1 + b + ia (1 + b + ia)(1 + b + ia) = 1 + b − ia (1 + b − ia) (1 + b + ia) (1 + b)2 − a2 + 2 ia (1 + b) = 1 + b2 + 2 b + a2 (1 − a2 ) + 2 b + b2 2 ia (1 + b) = 2 (1 + b) 2 b2 + 2 b + 2 ia (1 + b) = 2 (1 + b) = b + ia

16. We have, b + ic = (1 + a) z

n

n

 2 i   3 + i    3 − i   3 + i      6i − 2  = Re    9+ 1 6  1  2 i =− = Re − +  10 10  5

(1 + i )2   = Re  3− i 

14. Re 

4. Given expression = 1 + i 2 + i 4 + i 6 + ... + i 2 n

5. We have, i 19 +    = i 16 ⋅ i 3 + 24   i i ⋅i   

2

13. We have, 

(1 − i )3 1 − 3 i + 3 i 2 − i 3 − 2 − 2 i 12. = = = −2 1+ i 1+ i 1− i3

∴

= 0 − 12 + 60 i + 12 − 60 i = 0 + 0 i x = 0, y = 0

18. 2 i = 1 + i 2 + 2 i = (1 + i )2 = 1 + i 19. Let z = x + iy, where x, y ∈ R ∴ ∴ ⇒ ⇒

x 2 + y 2 ≠ 0 and x = 0 z = 0 + iy = iy, y ≠ 0 z2 = − y2 Im ( z 2 ) = 0

20. We have, ( x + iy )1/ 3 = a + ib On cubing both sides, we get x + iy = (a + ib)3 = a3 + 3 a2 (ib) + 3 a (ib)2 + (ib)3

⇒ ∴

21. We have, 8 iz 3 + 12 z 2 − 18 z + 27 i = 0 ⇒ 4 z (2 iz + 3) + 9 i (2 iz + 3) = 0 2

⇒

(2 iz + 3) (4 z 2 + 9 i ) = 0

⇒

2 iz + 3 = 0 or

∴

z =

4z2 + 9 i = 0 3 2

22. Multiplicative inverse of z 2 =

23. Since,

1 1 1 i = = =− 2 2 2 i 2 (1 + i ) z

1+ i 1− i = i and =−i 1− i 1+ i 3

∴

3

1+ i 1− i   −  = x + iy 1− i 1+ i

⇒ ⇒ ⇒

i 3 − (− i )3 = x + iy − i + i 3 = x + iy − i − i = x + iy x + iy = 0 − 2 i Hence, ( x, y ) = (0, − 2 ).

z lies in III quadrant, if x 2 − y 2 < 0 and − 2 xy < 0 z x < 0, y < 0 ⇒ − 2 xy < 0 Also, x 2 − y 2 < 0, if ( x + y ) ( x − y ) < 0 If x − y > 0, if x > y [Q x, y < 0 ⇒ x + y < 0] z ∴ lies in III quadrant, if y < x < 0. z 2−i 27. Let z = (1 − 2 i )2 − 2 + i 3 − 4i 2−i z= = × ∴ − 3 − 4i 3 + 4i 3 − 4i 2 11 =− + i 25 25 2 11 z=− − i ∴ 25 25

=

30.

3 + 4i 3 + 4i 4 + 5i = × 4 − 5i 4 − 5i 4 +5 i 12 + 15 i + 16 i + 20 i 2 = 16 − 25 i 2 12 + 31 i − 20 = 16 + 25 8 31 + i = 41 41 8 31 z=− − i 41 41 2

 8  31 z = −  + −   41  41 64 + 961 = (41)2 =

= (1 + i ) − (1 − i ) = 2 i 2i − − 2i = 2i = 0 + 4 = 2

1 − ix = a − ib 1 + ix (1 − ix ) (1 − ix ) ⇒ = a − ib (1 + ix ) (1 − ix ) 1 − x 2 − 2 ix a − ib = ⇒ 1 1 + x2 1 − x2 ⇒ = a and 1 + x2

2x =b 1 + x2

Now, x can be written as 2x 2x 1 + x2 1 + x2 = x= 2 1 − x2 +1 2 1+ x 1 + x2 b 2b = = 1 + a 1 + 1 + 2a 2b 2b = = 2 2 1 + (a + b ) + 2 a (1 + a)2 + b2

2

1025 25 × 41 5 = = 41 (41)2 (41)2

2× 5 =1 10

2 i − − 2 i = (1 + i )2 − (1 − i )2

29. ∴

= 121 + 3600 = 3721 = 61 Multiplicative inverse of z 11 − 60 i 11 60 z= 2 = = − i 2 2 (61) (61) (61)2 z

and

z x − iy ( x − iy )2 x 2 − y 2  2 xy  − = = = 2  i z x + iy ( x + y 2 ) x 2 + y 2  x 2 + y 2 

1+ i × 2 + i 2 + i  =  3 + i 3+ i

= 36 + 60 i − 25 = 11 + 60 i Then, z = 11 − 60 i and z = (11)2 + (60 )2

Then,

∴

28. (1 + i ) 

24. Let z = (6 + 5i )2 = 36 + 2(6)(5 i ) + 25 i 2

25. Let z =

26. Let z = x + iy, where x < 0, y < 0

4 Complex Numbers

∴ and

∴ Multiplicative inverse of 8 31 − − i z 41 41 = − 8 − 31 i z= 2 = 25 25 z 41 8 31 =− − i 25 25

Targ e t E x e rc is e s

= a3 + 3 a2 bi + 3 ab2 i 2 + i 3 b3 = a3 + 3 a2 bi − 3 ab2 − ib3 [Q i 3 = i 2 ⋅ i = − i ] = a (a2 − 3 b2 ) + ib (3 a2 − b2 ) x = a (a2 − 3 b2 ) y = b(3 a2 − b2 ) x y = a2 − 3 b2 and = 3 a2 − b2 a b x y + = 4 (a2 − b2 ) a b

2

31.

3 + 2i (3 + 2 i )2 = (4 − 3 i ) 4 − 3i =

( 9 + 4 )2 ( 16 + 9 )

=

13 5

161

u + iv u − iv

…(i)

We know that, when two complex numbers are equal their conjugates are also equal. u − iv …(ii) ∴ x − iy = u + iv

Objective Mathematics Vol. 1

4

32. We have, x + iy =

On multiplying Eqs. (i) and (ii), we get u + iv u − iv ( x + iy ) ( x − iy ) = × u − iv u + iv u 2 − i 2v 2 ⇒ x2 + y2 = 2 2 2 u −i v u 2 + v2 2 2 =1 ∴ x + y = 2 u + v2 2

2

2

2

⇒ ⇒

Ta rg e t E x e rc is e s

34.

 3 i π π  +  = 2 cos + i sin   2 6 6  2

36. z 2 = z z = z z = z

2

37. For every a ∈ R, a = a2 ∴

a

⇒

2

= a2

( x − y )2 ≥ 0

Now, 2

2

x + y −2x y ≥ 0 2

⇒

2x y ≤ x + y 2

2

2

2

2

⇒

x + y + 2x y ≤2 x + 2y

⇒

( x + y )2 ≤ 2 ( x 2 + y 2 )

⇒

( x + y )2 ≤ 2 z

∴

Y

39. z + 1 < z − 2 3z + 3 < 3z − 6 w + 1− i < w − 8 − i w+1< w−8

40. z + z − 1 ≥ z − z + 1 ≥ 1

X

O

(4, 0)

2

⇒ ⇒

x − 8 x + 16 + y 2 = x 2 y 2 = 8 ( x − 2) 2

The given relation represents the part of parabola with focus (4, 0) lying above X-axis and the imaginary axis as the directrix. The two tangents from directrix are at right angle. Hence, the greatest positive argument π of z is . 4

42. Let a + bi be a square root of − 2 + 2 3 i. ∴ ⇒ ⇒ ∴ and Now,

∴

(a + bi )2 = − 2 + 2 3 i a + 2 abi + i 2 b2 = − 2 + 2 3 i (a2 − b2 ) + 2 abi = − 2 + 2 3 i …(i) a2 − b2 = − 2 …(ii) 2 ab = 2 3 (a2 + b2 )2 = (a2 − b2 )2 + 4 a2 b2 = (− 2 )2 + (2 3 )2 = 4 + 12 = 16 …(iii) a2 + b2 = 16 = 4 2

On solving Eqs. (i) and (iii), we get 2 a2 = 2 or a2 = 1 ∴ a=±1 ⇒ 2 b2 = 6 or b2 = 3 b = ± 3. ∴ From Eq. (ii), ab = 3, which is positive. Either a = 1, b = 3 or a = − 1, b = − 3 Hence, the two square roots are 1 + 3 i and − 1 − 3 i i.e. ± (1 + 3 i ). Aliter  z +a z − a  z = ±  +i 2 2    4−2 4 + 2 =± +i  = ± (1 + i 3 ) 2 2  

x + y ≤ 2z

∴ Greatest value of z + 1 = 6 Also, z + 1 ≥ z + 4 − − 3 ≥ 0 ∴ 0≤ z+1 ≤6 ⇒ ⇒ ⇒

( x − 4) + y 2 = x

2

38. z + 1 = ( z + 4) − 3 ≤ z + 4 + − 3 ≤ 3 + 3 = 6

162

2

(1 + i )5 (1 + 3 i )2 − 2 i (− 3 + i )

35. i + 3 = 2 

⇒

2

Y′

(1 + 1)⋅ (1 + 4)⋅ (1 + 9) ... (1 + n 2 ) = α 2 + β 2 2 ⋅ 5 ⋅ 10 ... (1 + n 2 ) = α 2 + β 2

5 3 1  2 1  1 ( 2 )5  +   2  +i  2 2  2 2 =  3 i 2i 2  −  2  2 5π 2 π π π 9π ⇒ Argument = + − + = 4 3 2 6 12 5π Therefore, the principal argument is − . 12

z − 4 = Re ( z )

Now,

X′

33. Taking modulus and squaring on both sides, we get 1 + i ⋅ 1 + 2 i ⋅ 1 + 3 i .... 1 + ni = α + i β

41.

43. As,(a − b)2 ≥ 0, a2 + b2 ≥ 2 ab z = a +b 2

But

.... (i) 2

From Eq. (i), we get| z 2| ≥ 2 ab ∴ | z|2 + a2 + b2 ≥ a2 + b2 + 2 ab 2

2

⇒

z + z ≥ (a + b)2

⇒

2 z ≥ (a + b)2

⇒ ∴

2

2 z ≥ a + b, as z is positive 1 ( a + b) z≥ 2

= (1 − z1 z2 ) (1 − z1 z2 ) − ( z1 − z2 ) ( z1 − z2 ) [Q z1 − z2 = z1 − z2 and 1 = 1] =(1 − z1 z2 ) (1 − z1 z2 ) − ( z1 − z2 ) ( z1 − z2 ) [Q ( z1 ) = z1 ] = 1 − z1 z2 − z1 z2 + z1 z1 z2 z2 − z1 z1 + z1 z2 + z1 z2 − z2 z2 = 1 + z1 ∴

2

2

2

2

2

z2 − z1 − z2 = (1 − z1 ) (1 − z2 ) k =1

x + i ( y + 2 ) + α ( x − 1)2 + y 2 = 0

x + α ( x − 1)2 + 4 = 0

α = 5 z − 1 ⇒ α + 1 = 5 z = 5 (2 ) = 10 α + 1 = 10

2

⇒ ∴

 z1  z   − 2 1 cos θ + 1 = 0  z2  z2 2 z1 2 cos θ ± 4 cos θ − 4 = = cos θ ± i sin θ 2 z2 z1 a = cos θ ± i sin θ = 1 = z2 10

z1 : z2 = a : 10 2 2 2 2 − ≤ z+ + z z z z 2 2 z ≤2 + ⇒ z ≤2 z + 2 z

47. We have, z = z + ⇒ ⇒ ⇒ ⇒

k=0 n −1

2

z − 2 z + 1≤ 1+ 2 ( z − 1)2 ≤ 3 ⇒ − 3 ≤ z − 1 ≤ 3 1− 3 ≤ z ≤ 1+

3

So, the maximum value of z is 1 +

3.

48. We have, z1 + z2 =

z + z2 1 1 + = 1 z1 z2 z1 z2

⇒

 1  z1 + z2 1 −  =0 z1 z2  

⇒

z1 z2 = 1

∑

∴ ⇒

∴Locus of α, i.e. 5 z − 1 is the circle having centre at − 1 and radius 10. ∴ a = 10 Again, z12 + z22 − 2 z1 z2 cos θ = 0

⇒

∑ [ z1 z1 + 2

z1 +

z2 }

z1 z2 (ω )k + z1 z2ω k + z2 z2 (ω )k (ω )k ] n −1

n −1

n −1

∑ z1 z2 (ω )k + ∑ z1 z2ω k + ∑

k=0

k=0

2

z2

2

k=0

2

2

50. Let z = x + iy

46. Given, z = 2

⇒

n −1

∑ ( z1 + ω k z2 ) { z1 + (ω)k

k=0

2

∴ x 2 = α 2 ( x 2 − 2 x + 5) or (1 − α 2 )x 2 + 2α 2 x − 5 α 2 = 0 Since, x is real. ∴ D = B 2 − 4 AC ≥ 0 4 ⇒ 4 α + 20 α 2 (1 + α 2 ) ≥ 0 ⇒ − 4 α 4 + 5α 2 ≥ 0 5  4 α 2 α 2 −  ≤ 0 ⇒  4 − 5 5 ∴ ≤α ≤ 2 2 Let ⇒

z1 + ω z2 =

k=0

=

∑ (ω)k = 0

k=0 n −1

2

k

n −1

= n z1 + 0 + 0 + n z2 = n ( z1 + z2 )

On equating real and imaginary parts y + 2 = 0 ⇒ y = −2 and

∑

Now,

k=0

z + α z − 1 + 2i = 0

We have,

∑ ω k = 0 and

k=0 n −1

=

45. Let z = x + iy ⇒

n −1

∴

Complex Numbers

2

[Q z z = z ]

[Q z1 ≠ − z2 ]

z − z = 1 + 2i x 2 + y 2 − ( x + iy ) = 1 + 2 i

⇒ ( x 2 + y 2 − x ) + i (− y ) = 1 + 2 i ⇒ and If ⇒

x2 + y2 − x = 1 y = −2 y = − 2,

x2 + 4 − x = 1

x + 4 = (1 + x )2 2

⇒

2 x = 3 or x =

∴

z = x + iy =

3 2

3 − 2i 2

51. ( 3 + i ) = (a + ib) (c + id ) On taking argument both sides, we get 1 b d tan − 1 = tan − 1 + tan − 1 a c 3 π −1 b −1 d + tan = nπ + , n ∈ Z ⇒ tan a c 6

52. Let z − 1 = e iθ ⇒

[Q z − 1 = 1]

z = 1 + cos θ + i sin θ θ θ θ = 2 cos 2 + 2 i sin cos 2 2 2 θ  θ θ = 2 cos cos + i sin  2 2 2  θ 1 ∴ arg ( z ) = = arg ( z − 1) or arg ( z − 1) = 2 arg ( z ) 2 2 2 53. arg ( z 2 / 3 ) = arg (− 1) 3 2 2 2 2π 10 π = π , ⋅ 3π , ⋅ 5π = , 2 π, 3 3 3 3 3 z − z1 ( x + iy ) − (8 + 4i ) ( x − 8) + i ( y − 4) 54. = = z − z2 ( x + iy ) − (6 + 4i ) ( x − 6) + i ( y − 4)  z − z1  π We have, arg   =  z − z2  4 π arg ( z − z1 ) − arg ( z − z2 ) = ⇒ 4 π −1 y − 4 −1 y − 4 tan − tan = ⇒ x−8 x−6 4 2( y − 4) ⇒ =1 2 2 x + y − 14 x − 8 y + 64 ⇒ x 2 + y 2 − 14 x − 10 y + 72 = 0 ⇒ ( x −7 )2 + ( y − 5)2 = ( 2 )2 ∴ z − (7 + 5 i ) = 2

Targ e t E x e rc is e s

=(1 − z1 z2 ) (1 − z1 z2 ) − ( z1 − z2 ) ( z1 − z2 )

2

4

49. Since, 1, ω, ω 2 , ..., ω n − 1 are the nth roots of unity.

44. We have, 1 − z1 z2 2 − z1 − z2 2

163

Objective Mathematics Vol. 1

4

8

55. arg (− z ) = π − θ = π + (− θ ) = π + arg ( z ) ∴

1− i 1+ i   +  2  2 

arg (− z ) − arg ( z ) = π

8

and

z − 1 = cos θ + i sin θ z − 2 = cos θ + i sin θ − 1 θ θ θ = − 2 sin 2 + 2 i sin cos 2 2 2 θ θ θ = 2 i sin cos + i sin  2 2 2 z = 1 + cos θ + i sin θ θ θ θ = 2 cos 2 + 2 i sin cos 2 2 2 θ θ θ = 2 cos cos + i sin  2 2 2

10

…(i)

r =1 ⇒ z −1 =1 sin (θ − α ) = 0 ⇒ (θ − α ) = 0 θ = α ⇒ arg ( z − 1) = α

 1 + a ∴    2 

 1 = −   2

Ta rg e t E x e rc is e s

i

2 πk 11

k =1

(cos θ + i sin θ )− 14 (cos θ + i sin θ)− 15 (cos θ + i sin θ)48 (cos θ + i sin θ )− 30 = (cos θ + i sin θ )− 47 = cos 47θ − i sin 47θ =

sin θ − i cos θ = − i 2 sin θ − i cos θ = − i (cos θ + i sin θ ) ∴The given expression, using De-Moivre’s theorem = (− i )3 [cos (− 25 θ) + i sin (− 25 θ )] = i [cos 25 θ − i sin 25 θ] = sin 25 θ + i cos 25 θ

66. We have, z =

2π 2π   + i sin  cos  3 3

(1 + cos θ + i sin θ)5 (cos θ + i sin θ)3

θ θ θ cos + i sin  2 2 2 cos 3 θ + i sin 3 θ

5

5

32 cos 5 =

5θ 5θ  + i sin  {cos 3 θ − i sin 3 θ} cos 2 2   θ 5 θ  5 θ  = 32 cos 5 cos  − 3 θ  + i sin  − 3 θ  2 2  2 

1+ a 1  4π 4π  = 1 + cos + i sin   2 2 3 3 1 2π  2π 2π  = . 2 cos + i sin  cos  2 3 3 3 1 2π 2π  = − cos + i sin   2 3 3 3n

10

∑e

  10 i 2 πk = − i  ∑ e 11 − 1   k = 1 = − i (Sum of 11th roots of unity − 1) = − i (0 − 1) = i 4 (cos 75° + i sin 75° ) 63. We have, 0.4 (cos 30 ° + i sin 30 ° ) 10 {cos (75° − 30 ° ) + i sin (75° − 30 ° )} = cos 2 30 ° + sin 2 30 ° 10 (1 + i ) = 10 (cos 45° + i sin 45° ) = 2

− 16

π π  sin − i cos   8 8 = cos 2 π − i sin 2 π = 1

3n

k =1

2 πk 2 πk  + i sin  = −i 11 11 

θ θ  2 θ − 1 + 2 i sin cos  1 + 2 cos 2 2 2  = cos 3 θ + i sin 3 θ

8

8



65. We have,

1 ∴ Given expression = re iθ ⋅ e − iα + iθ e iα re 1 = re i( θ − α ) + e −i( θ − α) r Since, imaginary part of given expression is zero, we have 1 r sin (θ − α ) − sin (θ − α ) = 0 r r2 − 1= 0 ⇒ r2 = 1

60.

164

= cos θ + i sin θ

π π  = cos + i sin   8 8

10

∑ cos

2 πk 2 πk  − i cos  11 11 

64. Using De-Moivre’s theorem, the given expression

2

58. Let z − 1 = r (cos θ + i sin θ ) = re iθ

59.



∑  − i 2 sin

=−i

…(ii)

Let C = cos θ, S = sin θ 1 + C + iS 1 + cos θ + i sin θ = ∴ 1 + C − iS 1 + cos θ − i sin θ θ θ θ cos cos + i sin  2 2 2 = θ θ θ cos cos − i sin  2 2 2

π π  sin + i cos   8 8

k =1 10

2 πk 2 πk  − i cos  11 11 

k =1

57. We have, C 2 + S 2 = 1

⇒ or ⇒



∑ sin

62. We have, =

From Eqs. (i) and (ii), we get θ z −2 θ   = i tan = i tan (arg z ) Q arg z = , from Eq. (ii)   2 z 2

θ θ  cos + i sin   2 2 = 1 = C + iS

8

π π π π   + cos − i sin = cos + i sin    4 4 4 4   = cos 2 π + i sin 2 π + cos 2 π − i sin 2 π [using De-Moivre’s theorem] = 2 cos 2 π = 2 (1) = 2

56. Since, z − 1 = 1 Let Then,

8

61. We have, 

3n

=

(− 1) 23n

n

= 32 cos 5

θ 2

= 32 cos 5

θ 2

  θ  θ  cos  − 2  + i sin  − 2    

θ  Hence, the modulus and argument of z are 32 cos 5  2   θ and  −  , respectively.  2

68. We have, z 2 n − 1 (cos θ + i sin θ)2 n − 1 = z 2 n + 1 (cos θ + i sin θ)2 n + 1 cos 2 nθ + i sin 2 nθ − 1 = cos 2 nθ + i sin 2 nθ + 1

1

73. x 2 − 2 x + 4 = 0 ⇒ x = 1 + 3 i = 2  ± 2

= − 2ω − 2ω 2

[using De-Moivre’s theorem] (1 − 2 sin 2 nθ) + 2 i sin nθ cos nθ − 1 = (2 cos 2 nθ − 1) + 2 i sin nθ cos nθ + 1 i sin nθ cos nθ + i 2 sin 2 nθ [Q i 2 = − 1] cos 2 nθ + i sin nθ cos nθ i sin nθ (cos nθ + i sin nθ ) = i tan nθ = cos nθ (cos nθ + i sin nθ ) =

74. z + ⇒

1 =1 z

∴

+ i sin (2α + 2β + 2 γ + 2δ ) abcd = [cos (2α + 2β + 2 γ + 2δ ) + i sin (2α + 2β + 2 γ + 2δ )]1/ 2 or abcd = cos (α + β + γ + δ ) + i sin (α + β + γ + δ ) …(i) [using De-Moivre’s theorem] 1 ∴ = cos (α + β + γ + δ) abcd − i sin (α + β + γ + δ) …(ii) On adding Eqs. (i) and (ii), we get 1 abcd + = 2 cos (α + β + γ + δ) abcd

70. Let cot − 1 p = θ, then p = cot θ ∴

 pi + 1 ⋅   pi − 1

m

 i (cot θ − i ) = e 2 miθ.    i (cot θ + i )

m

e

2 mi cot −1 p

=e

2 mi θ

 i cot θ + 1 ⋅   i cot θ − 1

 cot θ − i  = e 2 miθ ⋅    cot θ + i 

 cos θ − i sin θ  = e 2 miθ ⋅    cos θ + i sin θ 

m

 e − iθ  = e 2 miθ ⋅  iθ   e 

m

z 99

= e 2 miθ (e − 2 iθ )m = e 0 = 1 8π 8π 71. We have, α = cos   + i sin  

8π i = e 11

 11  11 Re (α + α 2 + α 3 + α 4 + α 5 ) α + α2 + α3 + α4 + α5 + α + α2 + α3 + α4 + α5 = 2 − 1 + (1 + α + α 2 + α 3 + α 4 + α 5    + α + α 2 + α 3 + α 4 + α 5 )  = 2 − 1+ 0 [sum of 11 and 11th roots of unity] = 2 1 =− 2

⇒ α 6 + β 6 = 128

or − ω 2 1 1 + 99 = (− 1) + = −2 (− 1) z

75. (− 1 + − 3 )62 + (− 1 − − 3 )62 = 2 62 ω 62 + 2 62 (ω 2 )62 = 2 62 [(ω 3 )20 ω 2 + (ω 3 )41ω] = 2 62 [(ω 2 + ω] = − 2 62

76. Given expression = { 3 (1 + ω 2 ) + 5 ω} 2 + { 3 (1 + ω ) + 5 ω 2 } 2 = (− 3 ω + 5 ω)2 + (− 3 ω 2 + 5 ω 2 )2 = 4ω2 + 4ω = − 4 6

6  ω2  ω  +  2  = 2, ω  ω

77. 

ω2 =

using

ω=

− 1+ i 3 2

− 1− i 3 2 2

 z1  − 3i ω  =  z2  2

and

78. Here, 

z  ⇒  1  z2 

m

m

3  i 2 

z = −ω

69. We have, abcd = cos (2α + 2β + 2 γ + 2δ ) ∴

4

1 1 1 2 + + = 2ω 2 = a+ω b+ω c +ω ω 1 1 1 2 and + + = 2ω = 2 2 2 ω a+ω b+ω c +ω ∴ ω and ω 2 are roots of the equation 1 1 1 2 + + = . a+ x b+ x c + x x When x = 1, 1 1 1 2 + + = =2 a+1 b+1 c +1 1

72. We have,

Complex Numbers

1 1 = = cos θ − i sin θ z cos θ + i sin θ ∴ z n = (cos θ + i sin θ )n = cos nθ + i sin nθ , 1 and = (cos θ − i sin θ )n = cos nθ − i sin nθ zn 1 1 Hence, z n + n = 2 cos nθ and z n − n = 2 i sin nθ z z

Targ e t E x e rc is e s

67. We have,

z  ∴  1  z2 

50

25

25 25  z  2   3  − 3 iω  =  1   =   = −   iω  2  2   z2   25  3  3 i  =   +   2  2 2

50

lies in I quadrant.

79. Q xyz = (a + b) (aω + bω 2 ) (aω 2 + bω) = (a + b) (a2 − ab + b2 ) = a3 + b3

80. (1 + ω )3 − (1 + ω 2 )3 = (− ω 2 )3 − (− ω)3 = − ω6 + ω3 = − 1 + 1 = 0

81. p1/ 3 = (− p)1/ 3 (− 1)1/ 3 = − (− p)1/ 3 , − (− p)1/ 3 ω, − (− p)1/ 3 ω 2 = − q, − qω, − qω 2, where q = (− p)1/ 3 Let α = − q, β = − qω and γ = − qω 2 xα + yβ + z γ − q ( x + yω + zω 2 ) = ∴ xβ + yγ + zα − q ( x ω + yω 2 + z ) xω 3 + yω 4 + zω 2 = ω2 = x ω + yω 2 + z

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82. Given expression = =

z3 + 2 z2 + 2 z + 1 = 0 ⇒ ( z + 1) ( z 2 + z + 1) = 0 Its roots are − 1, ω and ω 2 . The root z = − 1 does not satisfy the equation, z1985 + z100 + 1 = 0 but z = ω and z = ω 2 satisfy it. Hence, ω and ω 2 are the common roots.

88. We have,

n

∑ (k − 1)(k − ω ) (k − ω 2 )

k=2 n

n

k=2

k=2

∑ (k − 1) (k 2 + k + 1) = ∑ (k 3 − 1)

= (2 3 + 33 + ... + n 3 ) − (n − 1) = (Σ n 3 − 13 ) − n + 1 2

 n (n + 1) =  −n  2 

83. Let α = ω and β = ω 2 1 1 ⋅ ω2 ω4 1 1 = ω 3 + (ω 3 )2 + 3 2 = 1 + (1)2 + 2 = 3 [Qω 3 = 1] (ω ) (1)

Now, α 3 + β 3 + α − 2β − 2 = ω 3 + ω 6 +

89. We have, (16)1/ 4 = (2 4 )1/ 4 = 2(1)1/ 4 = 2 (cos 0 + i sin 0 )1/ 4 1 1   = 2 cos (2 kπ + 0 ) + i sin (2 kπ + 0 ), k = 0, 1, 2, 3 4 4   = 2 × 1, 2 × i , 2 × − 1, 2 × − i , = ± 2, ± 2 i

90. We have,

f (ω ) = ω 3 p + ω 3 q + 1 + ω 3 r + 2 = ω 3p + ω 3q ⋅ ω + ω 3r ⋅ ω 2 = (ω 3 ) p + (ω 3 )q ⋅ ω + (ω 3 ) r ⋅ ω 2 = 1 + ω + ω2 = 0

Ta rg e t E x e rc is e s

84. We have, (1 − ω + ω 2 ) (1 − ω 2 + ω 4 ) (1 − ω 4 + ω 8 ) (1 − ω 8 + ω16 ) = (1 + ω 2 − ω ) (1 − ω 2 + ω ) (1 − ω + ω 2 ) (1 − ω 2 + ω ) [Qω 4 = ω 3 ⋅ ω = ω; ω 8 = (ω 3 )2 ⋅ ω 2 ; ω16 = (ω 3 )5 ⋅ ω = ω and ω 3 = 1] 2 2 = (− ω − ω ) (− ω − ω ) (− ω − ω ) (− ω 2 − ω 2 ) = (− 2ω ) (− 2ω 2 )(− 2ω ) (− 2ω 2 ) = 16 ⋅ ω 6 = 16 (ω 3 )2 = 16 (1)2 = 16

85. We have, or

x =ω −ω −2 x + 2 = ω − ω2 2

On squaring, x 2 + 4 x + 4 = ω 2 + ω 4 − 2ω 3 = ω 2 + ω 3 ⋅ ω − 2ω 3 [Qω 2 = 1] = ω2 + ω − 2 = − 1− 2 =−3 ⇒ x 2 + 4x + 7 = 0 On dividing x 4 + 3 x 3 + 2 x 2 − 11x − 6 by x 2 + 4 x + 7, we get x 4 + 3 x 3 + 2 x 2 − 11x − 6 = ( x 2 + 4 x + 7 ) ( x 2 − x − 1) + 1 = (0 ) ( x 2 − x − 1) + 1 = 0 + 1=1

86. We have, (1 + ω ) (1 + ω )2 (1 + ω 4 ) (1 + ω 8 ) ... to 2n factors = (1 + ω ) (1 + ω 2 ) (1 + ω 3 ⋅ ω ) (1 + ω 6 ⋅ ω 2 ) ... to 2n factors

= (1 + ω ) (1 + ω 2 ) (1 + ω ) (1 + ω 2 ) ... to 2n factors [Qω 3 = ω 6 = ... = 1] = [(1 + ω ) (1 + ω ) ... to n factors] [(1 + ω 2 ) (1 + ω 2 ) ... to n factors] = (1 + ω ) n (1 + ω 2 ) n = [(1 + ω ) (1 + ω 2 )] n = (1 + ω + ω 2 + ω 3 ) n = (0 + 1) n = 1 [Q1 + ω + ω 2 = 0, ω 3 = 1]

87. We have, α = ω and β = ω 2

166

Then, xyz = (a + b + c ) (aω + bω 2 + c ) (aω 2 + bω + c ) = (a + b + c ) (a2 + b2 + c 2 − ab − bc − ca) = a3 + b3 + c 3 − 3 abc

[Qω 3 = 1]

91. Since 1, α, α 2 , ..., α n − 1 are n and nth roots of unity. x n − 1 = ( x − 1) ( x − α) ( x − α 2 ) ...( x − α n − 1 )

∴

⇒ log ( x n − 1) = log ( x − 1) + log ( x − α ) + log ( x − α 2 ) + ... + log ( x − α n − 1 ) On differentiating both sides w.r.t. x, we get 1 1 1 1 nx n − 1 = + + + ... + xn − 1 x − 1 x − α x − α2 x − αn −1 Putting x = 2, we get n2 n − 1 1 1 1 1 = + + ... + + n 2 1 2 − α 2 −1 2 −α 2 − αn −1 ∴

n ⋅ 2n − 1 − 1= 2n − 1 n −1

Hence,

n −1

1

∑ 2 − αi i =1

∑ 2 − αi =

n ⋅ 2 n − 1− 2 n + 1 2n − 1

=

(n − 2 )2 n − 1 + 1 2n − 1

1

i =1

92. The vertices of the triangle are A (1, 0 ), B (1 / 2 , 1 / 2 ) and C (0, 1.) ∴

∴

AB 2 = 2 − 2 BC 2 = 2 − 2 AC 2 = 1 + 1 = 2 AB = BC

93. By definition, z1 = x1 + iy1, z2 = x2 + iy2 and z3 = collinear, if x1 y1 1 a b x 2 y2 1 = 0 ⇒ a′ b′ x 3 y3 1 a − a′ b − b′ ⇒

x3 + iy3

are

1 1=0 1

ab′ = a′ b

94. Let z = x + iy, where x, y ∈ R ∴ ⇒ ⇒

α + iβ = tan − 1 ( x + iy ) α − iβ = tan − 1 ( x − iy ) 2α = tan − 1 ( x + iy ) + tan − 1 ( x − iy ) 2x = tan − 1 1 − x2 − y2

∴ x 2 + y 2 + 2 x cot 2α = 1

96. We have, ( z − 1)n = i ( z + 1) n ∴

z −1 = i z + 1 n

n

n

⇒ z −1 = z + 1 ⇒ z −1 = z + 1 2 2 z −1 = z + 1 ⇒ ⇒ ( x − 1)2 + y 2 = ( x + 1)2 + y 2 ⇒ 4x = 0 ⇒ x = 0 ∴The roots lies on the Y-axis.

97. Let z = x + iy, where x, y ∈ R We have, ⇒

iz − 1 + z − i = 2 ix − y − 1 + x + iy − i = 2

⇒ (− y − 1)2 + x 2 +

x 2 + ( y − 1)2 = 2

⇒ x 2 = 0 or x = 0 ∴The locus of z is a straight line.

98. Clearly, z1 < z3 < z2 , as z3 is mid-point of z1 and z2 .  1 + z π  1 + z 99. (a) Re   =  = 0 ⇒ arg   1 − z 2  1 − z This represents a circle. (b) z z + iz − iz + 1 = 0 represent a circle.  z − 1 π (c) arg   =  z + 1 2 This represents a circle. (d)

z −1 = 1⇒ z − 1 = z + 1 z+1

This represents a straight line. z + z3 100. We have, z2 = 1 2 The point representing z2 divided the line segment joining points representing z1 and z3 in the ratio 1 : 1. ∴The points lies on a line.

101. Area of the triangle on the argand plane formed by 3 2 complex numbers − z, iz, z − iz is z . 2 3 2 ∴ z = 600 ⇒ z = 20 2

102. Let z1 = 1 − 3 i , z2 = 4 + 3 i and z3 = 3 + i . Then, z3 divides the line segment joining z1 and z2 in the ratio 2 : 1 internally. So, the points z1, z2 and z3 are collinear.

103. Let z = x + iy Then,

z + 2 i x + iy + 2 i x + i ( y + 2 ) = = z+4 x + iy + 4 ( x + 4) + iy [ x + i ( y + 2 )][( x + 4) − iy ] = ( x + 4)2 + y 2

( x 2 + 4 x + y 2 + 2 y ) + i (2 x + 4 y + 8) ( x + 4)2 + y 2  z + 2i  2 2 Since, Re   = 0 ⇒ x + y + 4 x + 2 y = 0,  z + 4 which represents a circle with centre (− 2, − 1).

Since,

which represents a straight line.

105. We have, z − 4 i + z + 4 i = 10 ⇒ z − (0 + 4 i ) + z − (0 − 4 i ) = 10 This represents an ellipse. 4 i + 4 i < 10 i.e. 10 > 8

106. We have, z − a i = z + a i 2

⇒ x + i ( y − a) = x + i ( y + a) ⇒ x 2 + ( y − a)2 = x 2 + ( y + a)2 ⇒ 4ay = 0; y = 0, which is X-axis.

107. We have, z − 1

Then,

z + 2 i x + iy + 2 i x + ( y + 2 )i = = z+2 x + iy + 2 ( x + 2 ) + iy

2

+ z+1

2

2

=4

2

2

⇒ ( x − 1) + iy + ( x + 1) + iy = 4 ⇒ ( x − 1)2 + y 2 + ( x + 1)2 + y 2 = 4 ⇒ 2 ( x 2 + 1) + 2 y 2 = 4 2 ∴The locus of P is x + y 2 = 1.

108. We have, a1 z 3 + a2 z 2 + a3 z + a4 = 3 ⇒

3 = a1 z 3 + a2 z 2 + a3 z + a4

⇒

3 ≤ a1 z 3 + a2 z 2 + a3 z + a4

⇒

3 ≤ a1 z 3 + a2 z 2 + a3 z + a4

⇒

3≤ z + z + z + 1

3

2

2

3

[Q ai ≤ 1] 2

3

⇒ 3 ≤ 1 + z + z + z < 1 + z + z + z + ... ∞ 2

⇒

3

3 < 1 + z + z + z + ... ∞ 1 1 3< ⇒ 1− z < 1− z 3 2 − z 3

⇒ ⇒ ∴

109. We have, z − 1 + z + 1 ≤ 4 ( x − 1)2 + y 2 + ( x + 1)2 + y 2 ≤ 4

⇒ where, ⇒

z = x + iy = A+ B≤4

where,

A = ( x − 1)2 + y 2

…(i)

B = ( x + 1)2 + y 2

and

But [( x − 1) + y 2 ] − [( x + 1)2 + y 2 ] = − 4 x i.e. A2 − B 2 = − 4 x On dividing Eq. (ii) by Eq. (i), we get 2

…(ii)

( x − 1)2 + y 2 − ( x + 1)2 + y 2 ≤ − x

=

104. Let z = x + iy

4

Targ e t E x e rc is e s

n

[ x + ( y + 2 ) i ][( x + 2 ) − i y ] ( x + 2 )2 + y 2 ( x 2 + y 2 + 2 x + 2 y ) + i (2 x + 2 y + 4) = ( x + 2 )2 + y 2  z + 2i  Im   =0 ⇒ x+ y+2=0  z + 2 =

Complex Numbers

95. z − z1 = z − z2 represents a perpendicular bisector.

⇒

2 ( x − 1)2 + y 2 ≤ 4 − x

⇒

3 x 2 + 4 y 2 ≤ 12 [squaring and simplifying] x2 y2 or + ≤1 4 3 which represents the interior and boundary of an ellipse.

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4

 z −1 + 4  1  > 1 = log1/ 2    2  3 z − 1 − 2 z−1 + 4 1 < 6 ⇒ z −1 > 3 which is an exterior of a circle.

111. The closest distance = length of the perpendicular from the origin on the line az + az + aa = 0 2 a (0 ) + a 0 + aa a a = = = 2 a 2 a 2

112. We have, zk = 1 + a + a2 + ... + ak = ⇒

zk −

ak + 1 1 − 1− a 1− a

⇒

zk −

a 1 1 = < 1− a 1− a 1− a

1 − ak + 1 1− a

k+1

[Q a < 1]

Ta rg e t E x e rc is e s

∴Vertices of the polygon z1, z2 , ... , zn lie within the circle 1 1 z− = 1− a 1− a

113. We have, z1 − z4 = z2 − z3 or

z1 + z3 z2 + z4 i.e. the = 2 2

diagonals bisect each other. ∴It is a parallelogram.  z − z1  π Also, amp  4  =  z2 − z1  2 ⇒ Angle at z1 is a right angle. ∴ It is a rectangle and hence, a cyclic quadrilateral.

114. We have, Im ( z 2 ) = 4 ⇒ Im [( x 2 − y 2 ) + 2 ixy ] = 4 [putting z = x + iy ] ⇒ 2 xy = 4 or xy = 2, which is a rectangular hyperbola. Re ( z 2 ) = 4 ⇒ Re [( x 2 − y 2 ) + 2 ixy ] = 4 ⇒ x2 − y2 = 4 which is a rectangular hyperbola.

118. Let z = α be a real root. Then,

α 3 + (3 + 2 i )α + (− 1 + ia) = 0 ⇒ (α 3 + 3 α − 1) − i (a + 2 α ) = 0 ⇒ α 3 + 3α − 1 = 0 and α = − a / 2 a3 3a ⇒ − − − 1= 0 8 2 3 ⇒ a + 12 a + 8 = 0 Let f (a) = a3 + 12 a + 8 ∴ f(− 1) < 0, f(0 ) > 0, f(− 2 ) < 0, f(1) > 0 and f(3) = 0

119. We have,

1 + i cos θ (1 + i cos θ ) (1 + 2 i cos θ ) = 1 − 2 i cos θ (1 − 2 i cos θ ) (1 + 2 i cos θ ) (1 − 2 cos 2 θ ) + i 3 cos θ = 1 + 4 cos 2 θ (1 + i cos θ ) is a real number, if cos θ = 0 Thus, (1 − 2 i cos θ ) π θ = 2 nπ ± ⇒ 2 where, n is an integer.

120. Let z = c be a real root. Then, αc 2 + c + α = 0 Let α = p + iq Then, ( p + iq )c 2 + c + p − iq = 0 ⇒ pc 2 + c + p = 0 and qc 2 − q = 0 ⇒ c =±1 From Eq. (i), we get α ± 1 + α = 0 Also, c = 1

116. z − z1 = z − z2 = z − z3 ∴ z must be the mid-point of z1 and z2 and since z − z1 = z − z2 and z, z1 and z2 are collinear. ⇒ z is circumcentre of ∆ formed by z1, z2 and z3 .  z − z1  π [neglecting − ve value] arg  3 ∴  =±  z3 − z2  2

z2 = r2 ⇒ Also, arg ( z1 ) + arg ( z2 ) = 0 ⇒ arg ( z1 ) = − arg ( z2 ) = − θ 2 ∴

z1 = r2 [cos (− θ 2 ) + i sin (− θ 2 )] = r2 [cos θ 2 − i sin θ 2 ] z1 = z2 1 z1 = z2

⇒ ⇒ ∴

z1 z2 = 1

122. We have, z1 = 15, z2 − 3 − 4i = 5 z2 – 3 – 4 i = 5 B A

z1 = 15

117. a0 z 4 + a1 z 3 + a2 z 2 + a3 z + a4 = 0 ⇒

a0 z 4 + a1 z 3 + a2 z 2 + a3 z + a4 = 0 [taking conjugate on both sides] ⇒ a0 ( z 4 ) + a1( z )3 + a2 ( z )2 + a3 ( z )2 + a3 z + a4 = 0

168

∴z is a root of the equation if z is a root. So, option (a) is correct. Also, if z1 is real, z1 = z1. If z1 is non-real complex, then z1 is also a root because imaginary root occurs in conjugate pairs. So, option (b) is also correct.

[Qq ≠ 0]

121. Let z2 = r2 (cos θ 2 + i sin θ 2 )

115. We have,

[putting z = x + iy ]

…(i)

O C

Minimum value of z1 − z2 , AB = OB − OA = 15 − 10 = 5 Maximum value of z1 − z2 , CA = OA + OC = 10 + 15 = 25

1 1≥ z − z

⇒

3 z − 6 − 3 i 3( x + iy ) − 6 − 3 i = 2 z − 8 − 6 i 2( x + iy ) − 8 − 6 i [3 x − 6 + i (3 y − 3)][(2 x − 8) − i (2 y − 6)] = [(2 x − 8) + i (2 y − 6)][(2 x − 8) − i (2 y − 6)]

Now,

1 ⇒ − 1≤ z − ≤1 z 2

⇒

− z ≤ z − 1≤ z

6 x 2 − 36 x + 48 + 6 y 2 − 24 y + 18 + i [6 xy

2

From z − 1 ≥ z , we get =

2

z + z − 1≥ 0 ⇒

z ≥

− 1+ 2

5

…(i)

2

From z − 1 ≤ z , we get 2

z − z − 1≤ 0 1− 5 1+ 5 ≤ z ≤ 2 2 From Eqs. (i) and (ii), we get − 1+ 5 1+ 5 ⇒ ≤ z ≤ 2 2 5 −1 1+ 5 z min = , z max = ⇒ 2 2 ⇒

…(ii)

124. z12 − z22 = z12 − z22 − 2 z1 z2 z1 − z2 z1 + z2 = z1 − z2

⇒ ⇒

⇒ ⇒ ⇒ ⇒ ∴

2

z1 + z2 = z1 + z2 z z z1 + z2 = z1 − z2 ⇒ 1 + 1 − 1 − 1 z2 z2 z1 lies on ⊥ bisector of 1 and − 1. z2 z1 lies on imaginary axis. z2 z1 is purely is imaginary. z2 z  π arg  1  = ±  z2  2 π arg ( z1 ) − arg( z2 ) = 2

125. Since, arg (( z − 1 − i ) / z ) is the angle subtended by the chord joining the points O and 1 + i at the circumference of the circle z − 1 = 1, so it is equal to − π /4. The line joining the points z = 0 and z = 2 + 0 i is the diameter. Y P(z)

O

θ 1

2 X A

− 6 x − 24 y + 24 − 6 xy + 12 y + 18 x − 36 (2 x − 8)2 + (2 y − 6)2

6 x 2 + 6 y 2 − 36 x − 24 y + 66 (2 x − 8)2 + (2 y − 6)2 (12 x − 12 y − 12 ) = a + ib [say] + (2 x − 8)2 + (2 y − 6)2 π Since, arg (a + ib) = 4 π b ∴ tan = ⇒ a=b 4 a 2 2 ⇒ 6 x + 6 y − 36 x − 24 y + 66 = 12 x − 12 y − 12 ⇒ 6 x 2 + 6 y 2 − 48 x − 12 y + 78 = 8 …(i) ⇒ x 2 + y 2 − 8 x − 2 y + 13 = 0 =

Again, z − 3 − i = 3 ⇒ x + iy − 3 + i = 3 ⇒ ( x − 3)2 + ( y + 1)2 = 9 …(ii) ⇒ x 2 + y 2 − 6x + 2 y + 1 = 0 On subtracting Eq. (ii) from Eq. (i), W get − 2 x − 4 y + 12 = 0 …(iii) ⇒ x = − 2y + 6 Putting the value of x in Eq. (ii), we get (− 2 y + 6)2 + y 2 − 6 (− 2 y + 6) + 2 y + 1= 0 ⇒ 5 y 2 − 10 y + 1 = 0 10 ± 4 5 ∴ y= 10 2 = 1± 5 4 ∴ x = − 2y + 6 = 4 m 5 4 2   z = x + iy = 4 m + i 1 ± ∴   5 5 Thus, the order pairs ( x, y ) satisfying the given 4 2   equations such that x = x + iy and  4 − ,1+   5 5 4 2   and  4 + ,1− .  5 5

127. Clearly, we have to find it for real z. Let z = x. Then,

x − w = x − w2 = w − w2 2

⇒ z −2 π z −2 is purely imaginary. =± ⇒ z 2 z−0 π We have, ∠POA = 2  2 − z π z − 2 AP ⇒ arg  ⇒ = i  =  0 − z 2 OP 2 AP Thus, z − 2 = 0 Now, in ∆OAP, tan θ = OP

∴ arg

4 Complex Numbers

126. Given, z = x + iy

1 =1 z

Targ e t E x e rc is e s

123. Given, z −

⇒ ⇒

1 3 − 1+ 3 i − 1− 3 i  − x +  + =  2 4 2 2 1 3 x+ =± 2 2 x = 1, − 2

2

=3

128. amp ( z1 z2 ) = 0 ⇒ amp ( z1 ) + amp ( z2 ) = 0 ∴ Q So, Also,

amp ( z1 ) = − amp ( z2 ) = amp ( z2 ) z1 = z2 , we get z1 = z2 . z1 = z2 2 z1 z2 = z2 z2 = z2 = 1, because z2 = 1

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4

129. z + z − 1 = 1⇒ z 2 − z + 1 = 0 ∴

z=

∴z + z n

or

or

−n

1±

132. Q z1 − 1 = 1, z0 − 1 = 1

3i

= − ω, − ω 2 = (− ω ) n + (− ω)− n or (− ω 2 ) n + (− ω 2 )− n 1 = (− 1) n ω n + (− 1)− n ⋅ n ω 1 = (− 1) n ω 2 n + (− 1)− n . 2 n ω 1  = (− 1) n ω n + n   ω  1   = (− 1)n ω 2 n + 2 n   ω  1  = (− 1) n ω n + n , because ω 3 n = 1  ω  = (− 1) n ⋅ (ω n + ω 2 n ) = (− 1) n ⋅ (1 + 1) or (− 1) n ⋅ (− 1) 2

Dependening on whether n is a multiple of 3 or not.

130. 2 z1 + z2 ≤ 2 z1 + z2 = 2 z1 + z2 = 2 × 1 + 2 = 4 From the figure, z1 − z2 is the least, when 0, z1, z2 are collinear. z1 – z2

Y

z –1 = 1

P (z0)

1 C

X

Q (z1)

z1 − 1 =1 z0 − 1 π ∠QCP = 2  z1 − 1 π π ⇒ amp   = ,−  z0 − 1 2 2 z1 − 1 ±π ± π  = 1⋅ cos + i sin ∴  = i, − i  z0 − 1 2 2  ∴ z1 − 1 = ( z0 − 1) i , z1 − 1 = − ( z0 − 1) i ∴

133. Given, z1 = z2 = z3 A(z1)

z2

Ta rg e t E x e rc is e s

z1 O

A 1

1

B 1

1

O

z =1 B(z2)

z1 − z2 = 1

Then, Again,

z2 +

1 1 ≤ z2 + z1 z1 =2 +

1 1 =2 + = 3 z1 1

131. OB = OD = OA = z B(z2)

A(z)

C(z3)

⇒ The vertices are at equal distances from the origin z = 0. ∴The origin is at the centroid of the equilateral triangle. z1 + z2 + z3 ∴ =0 3 ∴ z1 + z2 + z3 = 0 with OA as real axis, z1 = 1, z2 = 1 (cos 120 ° + i sin 120 ° ), z3 = 1 (cos 120 ° − i sin 120 ° )  1 3  1 3 z1 z2 z3 = 1⋅  − + i ∴  − − i  2   2 2   2 2  3  1 = −  +    2  2 

2

=1 C

D (z 1 )

π If amp ( z ) = θ, then amp ( z1 ) = θ − , 2 π amp ( z2 ) = θ + 2 and z = z {cos θ + i sin θ} ∴

π π     z1 = z cos θ −  + i sin θ −      2 2   = z {sin θ − i cos θ} = z (− i ) (cos θ + i sin θ ) = − iz

170

π π     z2 = z cos θ +  + i sin θ +   = iz   2 2  

1 8 27  + +   z1 z2 z3 

134. z2 z3 + 8 z3 z1 + 27 z1 z2 = z1 z2 z3  = z1 z2 z3

1 8 27 + + z1 z2 z3

= z1 z2 z3

z1

= z1 z2 z3 = z1 z2 z3

z1

2

+

8 z2 z2

2

+

27 z3 z3

z1 8 z2 27 z3 + + 1 4 9 z1 + 2 z2 + 3 z3

= z1 z2 z3 z1 + 2 z2 + 3 z3 = 1⋅ 2 ⋅ 3 ⋅ 6 = 36

2

136. Given that, arg ( z ) = 0 ⇒ z is purely real. ∴ Statement II is true. Also, z1 = z2 + z1 − z2 ∴

( z2 − z1 ) = ( z1 − z2 ) 2

⇒

2

2

2

2

2

= z1 + z2 − 2 z1 z2 cos (θ1 − θ 2 ) = z1 + z2 − 2 z1 z2

⇒ cos(θ1 − θ 2 ) = 1 ⇒ θ1 − θ 2 = 0 ⇒ arg( z1 ) − arg ( z2 ) = 0 z  z1 is purely real. ⇒ arg  1  = 0 ⇒  z2  z2 z  Im  1  = 0  z2 

⇒

Hence, Statement I and Statement II are true and Statement II is correct explanation of Statement I.

137. We will show that Statement I is true and follows from Statement II. Indeed z = z+ ⇒

1 1 1 1 1 − ≤ z+ + ≤ 1+ z z z z z 2

z − z − 1≤ 0 1 − 5 1 + 5  z ∈ , ⇒ 2   2  1 + 5 But as z > 0, we have z ∈  0,  2   1+ 5 . ⇒ Maximum value of z is 2 −a±

a2 − 4 b 138. Statement I is false, since z = , if a2 > 4b 2 and z is a positive number c, then z = c ⇒ z = c (cos θ + i sin θ ) ⇒ Infinite complex numbers satisfy the given equation. Statement II is true. A quadratic can have more than two roots, if all coefficients are zero. e iθ = cos θ + i sin θ ⇒ e − iθ = cos θ − i sin θ e iθ + e − iθ ∴ cos θ = 2 − (1 + i ) e i(1 − i) + e − i(1− i) e (1 + i) + e Now, cos (1 − i ) = = 2 2 e (e i ) + e − 1(e − i ) = 2 e (cos 1 + i sin 1) + e − 1 (cos 1 − i sin 1) = 2 i  1 1 1 = e +  cos 1 + e −  sin 1 e e 2 2 1 1 1 1 a = e +  cos 1, b = e −  sin 1 ∴ e e 2 2

139. Q

4

140. We have, (cos θ + i sin θ )3 / 5 = (cos 3 θ + i sin 3 θ )1/ 5 = [cos (2 rπ + 3 θ ) + i sin (2 rπ + 3 θ )]1/ 5 where, r = 0, 1, 2, 3, 4

Complex Numbers

set of complex numbers. Cancellation laws, a + c > b + c ⇒ a > b does not hold true in complex numbers, therefore Statement II is true.

 2 πr + 3 θ  i    5

, r = 0, 1, 2, 3, 4 =e Hence, product of all values of (cos θ + i sin θ )3 / 5 i

3θ 4

 2π + 3θ  i    5

 6π + 3θ  i    5

 8π + 3θ  i    5

=e e e e = e i 3 θ + 4 πi = e 4 iπ e i 3 θ = cos 3 θ + i sin 3 θ Also, product of roots of the equation x 5 − 1 = 0 is 1. Hence, Statement II is true, but it is not a correct explanation of Statement I.

141. Let A( z1 ) and B( z1 ) be the centres of the given circles and P be the centre of the variable circle, which touches given circles externally, then AP = a + r and BP = b + r where, r is the radius of the variable circle. On subtraction, we get AP − BP = a − b ⇒ AP − BP = | a − b is a constant. Hence, locus of P is (i) a right bisector of AB, if a = b (ii) a hyperbola if a − b < AB = z2 − z1 (iii) an empty set, if a − b > AB = z2 − z1 (iv) set of all points on line AB except those which lie between A and B, if a − b = AB ≠ 0 Thus, Statement I is false and Statement II is true.

142. If P( z ) is any point on the ellipse, then equation of the ellipse is z − z1 + z − z2 =

z1 − z2 e

…(i)

Targ e t E x e rc is e s

135. Statement I is false, since there is no order relation in the

For P( z )to lie in ellipse, we have z1 − z2 e It is given that origin is an interior point of the ellipse. z − z2 ⇒ 0 − z1 + 0 − z2 < 1 e  z − z2  e ∈  0, 1 ∴   z1 + z2  z − z1 + z − z2
z1 − z2 . Thus, z − i + z + i = k represents an ellipse, if k > i + i or k > 2. ∴ Statement I is false and Statement II is true.

144. The equation can be rewritten as zz + az + az + aa = aa − λ ⇒ ( z + a) ( z + a) = aa − λ z + a = aa − λ ⇒ Since, aa is real, 1 should be real. aa − λ represents radius of the circle. Hence, Statement I and Statement II both are true and Statement II is the correct explanation for Statement I.

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4

Solutions (Q. Nos. 145-147) Given that, | z1 + z2|2 2 2 ⇒| z1| + | z2| + z1 z2 + z1 z2 z1 z2 + z1 z2 ⇒ z1 z + 1 ⇒ z2 z2 z1  z1  +  ⇒ z2  z2 

152. DE = = | z1| + | z2| = | z1|2 + | z2|2 =0 2

153. zE = …(i)

=0

[dividing by z2 z2 ]

=0

…(ii)

arguments are 0 and π.

148. Clearly, according to the least possibility α 2 − 7α + 11 ≤ 1 ⇒ α ∈[2, 5]

149. ( z − i ) − (α 2 − 7α + 11 ) = 1 ∴ ( z − i ) − (α 2 − 7α + 11) ≥ z − i − α 2 − 7α + 11 [using z1 − z2 ≥ z1 − z2 ]

Ta rg e t E x e rc is e s

α 2 − 7α + 11 + 1 ≥ z − i [since, α 2 − 7α + 11 ≥ − 5/ 4] or z − i ≤ 9/ 4

X

−1

4 9 Y

B – 9 ,1 4

151. zD =

X

zB − zA i z − zC i , zE = B 1− i 1− i

zB − zC i ω − ω 2 i = 1− i 1− i (ω + ω 2 ) + i (ω − ω 2 ) − 1 − 3 = = 2 2 1+ 3 3 + 3 AE = 1 + = 2 2

Since, cos − 1 cos 12 − sin − 1 sin 12 = 8 (π − 3) ∴ Locus of z is portion of a line joining z1 and z2 except the segment between z1 and z2 . 2

C. z 2 − i z1 = k2 − k1 x 2 − y 2 + 2 ixy − iλ 1 = λ 2

λ1 2 ∴ Locus of z is point of intersection of hyperbola. 1 1 π D. Given, z − 1 − sin − 1 + z + cos − 1 − =1 3 3 2 1 1 π since, 1 + sin − 1 + cos − 1 − =1 3 3 2 ∴ z − z1 + z − z2 = z1 + z2 ⇒ Locus of z is the segment joining z1 and z2 . x2 − y2 = λ 2

and

xy =

If x + iy is a root, then x − iy is also a root. Let the roots be x1 ± iy1, x2 ± iy2 , x3 ± iy3 Sum of the roots = 2 ( x1 + x2 + x3 ) = 0 ⇒ x1 + x2 + x3 = 0 ⇒ One of x1, x2 , x3 is negative and other two are positive or vice-versa. ⇒ The number of roots in each of the quadrant is either 1 or 2.

156. A. z 4 − 1 = 0 ⇒ z 4 = 1 = cos 0 + i sin 0

A

⇒

D B

C

E

172

3 2

155. There are no real roots of the equation z 6 − 6 z + 20 = 0.

A (0, 1)

O – 5 ,1 4

=

z2 − 1 z 2 − 1 = z2 + 1 z 2 + 1 ⇒ z − z = 0, z + z = 0, y = 0, x = 0 Locus of z is portion of pair of lines xy = 0   z 2 − 1   > 0 Q  2   z + 1  B. Given, z − cos − 1 cos 12 − z − sin − 1 sin 12| = 8 (π − 3)

⇒

9 and arg ( z ) = π − tan 4

2

=

∴

Y

150. AB =

1 − ω2

 z 2 − 1  = 0;z ≠ ± i  z 2 + 1

147. Also, i ( z1 / z2 ) is purely real. Hence, its possible

z − i ≤ 1 + 5/ 4

∴

=

154. A. arg 

146. From Eq. (ii), z1 / z2 is purely imaginary

∴

1− i

2

145. From Eq. (i),z2 z2 is purely imaginary.

⇒

( zA − zC )i

∴Angle between AC and DE  z − zA   z − zA 1 − i  π = arg  C  − arg  C =  zE − zD   zA − zC i  4

z = (cos 0 + i sin 0 )1/ 4 = cos 0 + i sin 0 B. z 4 + 1 = 0 ⇒ z 4 = − 1 = cos π + i sin π ⇒ z = (cos π + i sin π)1/ 4 π π = cos + i sin 4 4 π π 4 4 C. iz + 1 = 0 ⇒ z = i = cos + i sin 2 2 1/ 4 π π  ⇒ z = cos + i sin   2 2 π π = cos + i sin 8 8

⇒

⇒ y = 17, 8 Thus, the required numbers are z = 6 + 17 i , 6 + 8 i. Hence, the value of Re ( z ) is 6. 1 1 1 1 ≥ z − = z − = z − as z ≥ 3 z −z z z 1 1 Let f ( x ) = x − for x ≥ 3,f ' ( x ) = 1 + 2 > 0 x x ⇒ f ( x ) is increasing function, so f ( x )min at x = 3 is

160. z +

157. A. Let z = 1 − z ⇒ z 2 + z + 1 = 0 ⇒

z = ω , ω2

B. Here, z1 + z2 + z3 + .... + z6 = 0 ... (i) z2 = z1 e ( i2 π )/ n, if e ( i2π )/ n = α ⇒ z2 = z1 α Similarly, z3 = z1α 2 , z4 = z2 α 3 , z5 = z2 α 4 , z6 = z1α 5 On squaring and adding and then using Eq. (i), we get z12 + z22 + z32 + z42 + z52 + z62 = 0 1 + 2 ω + 3 ω 2 + 4 ω 3 + ... + nω n − 1  C. ≥ n ! ω n 

n( n + 1) 1/ n  2

 

[Q AM ≥ GM] Now, E = 1 + 2 ω + 3 ω 2 + 4 ω 3 + ... + nω n − 1 + .... ω + 2 ω 2 + 3 ω 3 + ... + (n − 1)ω n − 1 + n ω n ωE = (1 − ω)E = 1 + ω + ω 2 + ... + ω n − 1 − n ω n 0 − n(1) E= ⇒ 1− ω n ∴ E= ω −1 D. Let x 2 + y 2 = 1 where, z1 − i , z2 = 1, z3 = − 1 ∴ z=−i ⇒ zz1 + z2 z3 = 1 − 1 = 0

158. Consider, x 3 2 + 11 i 3 − 4 i 50 + 25 i = × = 3 + 4i 3 − 4i 25 x2 =2 + i a + b = 2 + 1= 3 x=

∴

159. Here,

z−4 =1 ⇒ z−8

⇒ ⇒

x − 4 + iy =1 x − 8 + iy

( x − 4)2 + y 2 = ( x − 8)2 + y 2 x=6 z − 12 5 With x = 6, = z − 8i 3 ⇒ y 2 − 25 y + 136 = 0

9 (36 + y 2 ) = 25 [36 + ( y − 8)2 ]

z+

1 1 = z − z min z

= z =3

4 Complex Numbers

⇒

⇒

π z 4 = − i = cos − i sin π /2 2 1/ 4 π   z = cos − i sin π /2   2 π π = cos − i sin 8 8

8 3

Hence, λ is equal to 8.

161. Consider ⇒ ⇒ ⇒ ∴

z4 + z3 + 2 z2 + z + 1 = 0 z + z3 + z2 + z2 + z + 1 = 0 2 z ( z 2 + z + 1) + ( z 2 + z + 1) = 0 ( z 2 + z + 1) ( z 2 + 1) = 0 z = i, − i, ω, ω 2, for each z = 1 4

162. Let A be ( x, y ). D(1,1)

A (x, y)

M (2, –1)

Targ e t E x e rc is e s

D. iz 4 − 1 = 0

C

B

It is given that, BD = 2 AC ⇒ MD = 2 AM Also, DM is perpendicular to AM. ⇒ (1 − 2 )2 + (1 + 1)2 = 4 [( x − 2 )2 + ( y + 1)2 ] y + 1 1+ 1 and ⋅ = −1 x − 2 1− 2

…(i)

⇒ 2 ( y + 1) = x − 2 With x − 2 = 2( y + 1) , Eq. (i) can be written as 1 ( y + 1)2 = 4 1 3 y =− ,− ⇒ 2 2 ⇒ x = 3, 1 ∴ λ1 + λ 2 = 4

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4

Entrances Gallery 1. A. zk is 10th root of unity ⇒ zk will also be 10th root of unity. Take z j as zk . z B. z1 ≠ 0, take z = k , we can always find z. z1

OC =

r

In ∆OCD,

1 − 4r 2 |α|2 1 2| z0| |α|

| z0|2 + |α|2 − r 2 = ⇒ 2| z0||α | ⇒

3. w =

|α | = 3+i 2

D

θz 0

O

X 60° x2 + y2 < 16 Y′

5. Distance of

(1,

− 3) from y +

3x = 0 −3+ 3 ×1 3− 3 > > 2 2

6. Given, is z 2 + z + 1 − a = 0

C (3, 2)

1 − 4r 2 |α | 2 1 2| z0| |α |

| z0|2 +

X¢

B (3, 0)

(0, 0) O

X

A(3, –5/2)

1 7

Y¢

iπ =e 6

∴ Minimum value = 5

8. The expression may not attain integral value for all a, b,c. A3 π/6

B2

A2

O B3 x = –1/2

174

60°

X

B1

A1

If we consider a = b = c , then x = 3a y = a (1 + ω + ω 2 ) = 0 z = a (1 + ω 2 + ω ) = 0 ∴ | x|2 + | y|2 + | z|2 = 9| a|2 ∴

x = 1/2  nπ  i   6 

wn = e nπ 1 Now, for z1, cos > 6 2 nπ 1 and for z2 , cos , b).

Properties of Inequalities

●

●

●

We shall learn about some elementary properties of inequalities which will be used in the subsequent discussion.

i.

Chapter Snapshot

If a > b and b > c, then a > c. Generally, if a1 > a 2 , a 2 > a 3 ,...,a n − 1 > a n , then a1 > a n

●

●

●

ii.

If a > b, then a ± c > b ± c, ∀c ∈ R

iii.

If a > b, then for m > 0, am > bm,

a b b a > and for m < 0, bm > am, > m m m m

●

●

iv.

If a > b > 0, then (a) a 2 > b 2

(c)

(b) | a| > | b |

1 1 (c) > a b

and if a < b < 0 , then (a) a 2 > b 2

v.

1 1 < a b

(b) | a| > | b |

(a) a 2 > b 2 , if | a| > | b |

vii.

●

If a < x < b and a, b are positive real numbers, then

(a) 0 ≤ x < b , if | b | > | a| 2

2

●

(b) 0 ≤ x ≤ a , if | a | > | b | 2

2

Arithmetico-Geometric Mean Inequality Quadratic Equation with Real Coefficients Formation of a Polynomial Equation from Given Roots Symmetric Function of the Roots Common Roots

●

If a < x < b and a is negative number and b is positive number, then

Logarithms

Transformation of Equations

(b) a 2 < b 2 , if | a| < | b |

a 2 < x 2 < b2

Absolute Value of a Real Number

●

●

vi.

Generalised Method of Intervals for Solving Inequalities by Wavy Curve Method (Line Rule)

●

●

If a < 0 < b, then

Inequality

●

Quadratic Expression and its Graph Maximum and Minimum Values of Rational Expression Location of the Roots of a Quadratic Equation Algebraic Interpretation of Rolle’s Theorem Condition for Resolution into Linear Factors Some Application of Graphs to Find the Roots of Equations

ix. x. xi.

a > 0, then b (a) a > 0, if b > 0

⇒

3 a2 a2 + b 2 + c 2 3c 2 < < a+ b+c 3c 3a

X

If

(b) a < 0, if b < 0

If a i > bi > 0, where i = 1, 2, 3, …, n, then a1 a 2 a 3K a n > b1 b2 b3K bn

⇒

If a i > bi , where i = 1, 2, 3, …, n, then a1 + a 2 + a 3 + K + a n > b1 + b2 + K + bn

If a and b are positive real numbers such that a < b and if n is any positive rational number, then (a) a n < b n (b) a − n > b − n (c) a 1/ n < b1/ n

xiv.

If a >1 and n is any positive rational number, then (b) 0 < a − n < 1 (a) a n >1 If 0 < a < 1 and m, n are positive rational numbers, then (a) m > n ⇒ a m < a n (b) m < n ⇒ a m > a n If a >1 and m, n are positive rational numbers, then (a) m > n ⇒ a m > a n (b) m < n ⇒ a m < a n

Example 1. If a, b and c are positive real numbers such that a < b < c, then show that a 2 a 2 + b2 + c2 c2 < < . c a +b+c a Sol. We have,

c > b> a> 0

⇒

c >b >a >0

Now, ⇒

a< b1 (a) 0 < a n < 1

xvi.

…(ii)

From Eqs. (i) and (ii), we get

xii.

xv.

3 a2 < a2 + b 2 + c 2 < 3c 2

Inequalities and Quadratic Equation

viii.

a2 < b 2 < c 2

and

 1 1 1   x ∈ b , a ,   if a and b have same sign  1  1  1  If x ∈[ a, b] ⇒ ∈ − ∞,  ∪  , ∞  , x  a b    if a and b have opposite  signs 

12 x < 6 ⇒

x
4 So, there is no real number x satisfying both the inequalities (i) and (ii). Hence, the given system of inequalities has no solution. X

Example 3. The value of x for which x −3 x −1 x − 2 , 2 − x > 2x − 8 −x< − 4 2 3  10   10  (b)  −1,  (a)  −1,   3 3  (c) R (d) None of these Sol. (b) We have, x − 3 − x < x − 1 − x − 2 4

2

3

⇒ ⇒ ⇒ and ⇒

3 x − 9 − 12 x < 6 x − 6 − 4 x + 8 −11x < 11 ⇒ − x < 1 …(i) x> −1 2 − x> 2x − 8 −3 x > − 10 10 …(ii) ⇒ x< 3 From Eqs. (i) and (ii), the solution of the given system 10 of inequalities is given by x ∈  −1,  .  3

Generalised Method of Intervals for Solving Inequalities by Wavy Curve Method (Line Rule) Let F ( x ) = ( x − a1 ) k 1 ( x − a 2 ) k 2 K ( x − a n − 1 ) k n−1

2

(x − a n ) k n …(i)

where, k1 , k 2 , …, k n ∈Z and a1 , a 2 , …, a n are fixed real numbers satisfying the condition a1 < a 2 < a 3 < K < a n − 1 < a n

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Objective Mathematics Vol. 1

5

For solving F ( x ) > 0 or F ( x ) < 0, consider the following algorithm: We mark the numbers a1 , a 2 ,…, a n on the number axis and put the plus sign in the interval on the right of the largest of these numbers, i.e. on the right of a n . Then, we put the plus sign in the interval on the left of a n , if k n is an even number and the minus sign, if k n is an odd number. In the next interval, we put a sign according to the following rule: When passing through the point a n − 1 the polynomial F ( x ) changes sign, if k n − 1 is an odd number.Then, we consider the next interval and put a sign in it using the same rule. Thus, we consider all the intervals. The solution of the inequality F ( x ) > 0 is the union of all intervals in which we have put the plus sign and the solution of the inequality F ( x ) < 0 is the union of all intervals in which we have put the minus sign.

X

Example 4. Solve (x − 1)(x − 2)(1 − 2x ) > 0. Sol. We have, ( x − 1)( x − 2 )(1 − 2 x) > 0

■

⇒ ⇒

− ( x − 1)( x − 2 )(2 x − 1) > 0 ( x − 1)( x − 2 )(2 x − 1) < 0 1 On number line mark x = , 1, 2 2

■

–

■

X

Solution of Rational Algebraic Inequation

Example 5. Solve

2x − 3 − 3≥ 0 3x − 5 2 x − 3 − 9 x + 15 ≥0 3x − 5 −7 x + 12 ≥0 3x − 5 7 x − 12 ≤0 3x − 5

⇒ ⇒ ⇒

are known as rational algebraic inequations. To solve these inequations, we use the sign method as explained in the following algorithm :

⇒

⇒ (7 x − 12 )(3 x − 5) ≤ 0 Sign scheme of (7 x − 12) (3 x − 5) is as follows :

Algorithm

+

Obtain P ( x ) and Q ( x ).

Step IV Obtain critical points by equating all factors to zero. Plot the critical points on the number line. If there are n critical points, then they divide the number line into ( n +1) regions.

Step VI In the right most region the expression P (x ) bears positive sign and in other Q (x ) regions the expression bears positive and negative signs depending on the exponents of the factors.

184

– 5/3

+ 12/7

5 12 x ∈  ,   3 7 

⇒

Step II

Step V

2x − 3 ≥ 3. 3x − 5

3x − 5

P (x ) ≤0 Q (x )

Factorise P ( x ) and Q ( x ) into linear factors. Step III Make the coefficient of x positive in all factors.

+ 2

Sol. 2 x − 3 ≥ 3

If P ( x ) and Q ( x ) are polynomial in x, then the inequation P (x ) P (x ) P (x ) > 0, < 0, ≥0 Q (x ) Q (x ) Q (x )

Step I

– 1

When x > 2, all factors ( x − 1,) (2 x − 1) and ( x − 2 ) are positive. Hence, ( x − 1)( x − 2 )(2 x − 1) > 0 for x > 2. Now, put positive and negative sign alternatively as shown in figure. Hence, solution set of ( x − 1)( x − 2 )(1 − 2 x) > 0 or 1 ( x − 1)( x − 2 ) (2 x − 1) < 0 is  −∞,  ∪ (1, 2 ).  2

■

and

+ 1/2

5 5 is not included in the solutions as at x = denominator 3 3 becomes zero.

Ø x=

X

Example 6. The value of x for which ( x − 2) 3 ( x − 3) < 0, is (a) (2, 3)

(b) [2, 3)

(c) (0, 3)

(d) (2, 3]

Sol. (a) ( x − 2 )3 ( x − 3) < 0 ⇒

( x − 2 )( x − 3) < 0 [as ( x − 2 )2 is positive for all real values of x ≠ 2] –∞

+

– 2

+ 3

∞

Here, interval is open as x = 2, 3 do not satisfy inequality. i.e. 2 < x < 3 or x ∈(2, 3)

1 The value of x satisfying the inequalities 1 x + ≥ 2 hold x ( 0, ∞ )

φ

R

4 If c < d , x 2 + (c + d )x + cd < 0 (− d , − c ] (− d , − c ) R φ

[0, ∞ )

x2 2 ≥0 x −1

5

(1, ∞ ) { 0} ∪ (1, ∞ )

[1, ∞) None of these

(2 x − 1)( x − 1)4 ( x − 2 )4 ≤0 ( x − 2 )( x − 4)4  1, 2   2 

3 ( x − 2 ) ( x − 3) ( x − 4) (1 − x ) ≤ 0 4

3

2

R φ  1, 2   2 

(1, 3) ( − ∞, 1) ∪ ( 3, ∞ ) ( −∞, 1] ∪ [3, ∞ ) None of the above

Absolute Value of a Real Number The absolute value of a real number ‘x’ is denoted x, if x ≥ 0 by | x| and defined by | x| =  −x, if x < 0 Ø

●

(i) | x| is also defined as x 2

e.g.

x2 = − x

9 = (−3)2 = 32 = 3

●

(i) a ≤ | a| (ii) | ab| = | a||b| a | a| (iii) = b |b| (iv) | a + b| ≤ | a| + |b|

[triangle inequality]

(v) | a − b| ≥ | a| − |b|

[triangle inequality]

(vi) | a + b| = | a| + |b| , iff ab ≥ 0 (vii) | a + b| = | a| − |b| , iff ab ≤ 0

Inequations Containing Absolute Values By definition, | x| < a | x| ≤ a | x| > a | x| ≥ a a ≤ | x| ≤ b where, a, b > 0

Form 1 The inequation of the form f (| x| ) < g ( x ) is equivalent to the collection of systems  f ( x ) < g ( x ), if x ≥ 0   f ( −x ) < g ( x ), if x < 0

Form 3 The inequation of the form | f ( x )| > g ( x ), is equivalent to the systems  f ( x ) > g ( x ), if g ( x ) < 0  − f ( x ) > g ( x ), if g ( x ) < 0

9 = ±3

or

Forms of the Inequations Containing Absolute Values

Form 2 The inequation of the form | f ( x )| < g ( x ), is equivalent to the systems  f ( x ) < g ( x ), if g ( x ) > 0  − f ( x ) < g ( x ), if g ( x ) > 0

(ii) If x is positive, then x 2 = x (iii) If x is negative, then

Inequalities and Quadratic Equation

5

Work Book Exercise 5.1

⇒ − a < x < a ( a > 0) ⇒ −a ≤ x ≤ a ⇒ x < − a and x > a ⇒ x ≤ − a and x ≥ a ⇒ x ∈[ −b, − a ] ∪ [ a, b]

Form 4 The inequation of the form | f (| x| )| ≥ g ( x ) is equivalent to the collection of systems | f ( x )| ≥ g ( x ), if x ≥ 0  | f ( −x )| ≥ g ( x ), if x < 0 Form 5 The inequation of the form | f ( x )| ≥ | g ( x )| is equivalent to the collection of system f 2 (x ) ≥ g 2 (x ) Form 6

The inequation of the form h( x, | f ( x )| ) ≥ g ( x )

is equivalent to the collection of systems h{x, f ( x )} ≥ g ( x ), if f ( x ) ≥ 0  h{x, − f ( x )} ≥ g ( x ), if f ( x ) < 0 185

X

Objective Mathematics Vol. 1

5

Example 7. The solution set of |3 − 4x| ≥ 9 is 3  (a)  −∞,  ∪ [3, ∞ )  2 3  (b)  −∞, −  ∪ [3, ∞ )  2 (c) ( −∞, 2) ∪ [2, ∞ ) (d) None of the above

X

Example 9. Solve the inequation  1 − | x|   ≥ 1. 1 + | x |   2 Sol. The given inequation is equivalent to the collection of systems.  x  1 ≥ , if x ≥ 0 1 − 1 + x 2  ⇒ ⇒  1 + x ≥ 1 , if x < 0     1 − x 2  1 ≥ 1 , if x ≥ 0 1 + x 2 ⇒  1 1 ≥ , if x < 0  1 − x 2  1 − x ≥ 0, if x ≥ 0 1 + x ⇒ ⇒  1 + x ≥ 0, if x < 0  1 − x x−1 For ≤ 0, if x ≥ 0, then x+1 0≤ x≤ 1

Sol. (b) We have, |3 − 4 x| ≥ 9 ⇒ 3 − 4 x ≤ − 9 or 3 − 4 x ≥ 9 ⇒ ⇒ ⇒

X

[since,| x| ≥ a ⇒ x ≤ − a or x ≥ a] −4 x ≥ 6 −3 [dividing both sides by −4] x ≥ 3 or x ≤ 2 −3 x ∈  −∞,  ∪ [3, ∞ )  2 

−4 x ≤ − 12

or

Example 8. The solution set of (a) [ −5, − 2] ∪ [ −1, ∞ ) (c) ( −5, − 2) ∪ ( −1, ∞ )

| x + 3| + x > 1 is x +2

(b) [ −5, − 2) ∪ [ −1, ∞ ) (d) None of these

⇒

∴

⇒

…(ii)

x > − 1 or

⇒

x ∈ (−1, ∞ ) or

⇒

x ∈ (−3, − 2 ) ∪ (−1, ∞ )

–

1

From Eqs. (i) and (ii), the solution of the given equation is x ∈ [−1, 1.] X

⇒

+ –1

⇒ {x + 1 > 0 and x + 2 > 0} or {x + 1 < 0 and x + 2 < 0} {x > − 1 and x > − 2} or {x < − 1 and x < − 2} ⇒

Example 10. The value of x, | x + 3| > |2x − 1| is  −2   2  (b)  − , ∞  (a)  , 4  3   3  (c) (0, 1) (d) None of these Sol. (a) Given,| x + 3| > |2 x − 1|

x< −2

On squaring both sides, we get

x ∈ (−∞, − 2 )

| x + 3|2 > |2 x − 1|2

[Q x ≥ − 3] …(i)

Case II When x + 3 < 0, i.e. x < − 3 | x + 3| − 2 >0 ∴ x+2 −x − 3 − 2 >0 ⇒ x+2 −( x + 5) ⇒ >0 x+2 ( x + 5) ⇒ 0 x+2

x+1 >0 x+2

1

x+1 ≤ 0, if x < 0 x−1

For

∴

x+ 3−2 >0 x+2

...(i) +

–

–1

| x + 3| + x − 1> 0 x+2 x+ 3 −2 >0 x+2

⇒

 x − 1 ≤ 0, if x ≥ 0  x + 1   x + 1 ≤ 0, if x < 0  x − 1

+

Sol. (c) We have, | x + 3| + x > 1 x+2 ⇒

 1 ≥ 1 , if x ≥ 0 |1 + x| 2   1 ≥ 1 , if x < 0 |1 − x| 2

⇒ {x + 5 < 0 and x + 2 > 0} or {x + 5 > 0 and x + 2 < 0} ⇒ (x < − 5 and x > − 2), which is not possible or {x > − 5 and x < − 2} …(ii) ⇒ x ∈ (−5, − 2 ) From Eqs. (i) and (ii), we get x ∈ (−5, − 2 ) ∪ (−1, ∞ )

⇒ ⇒

{( x + 3) − (2 x − 1)} {( x + 3) + (2 x − 1)} > 0 {(− x + 4)(3 x + 2 )} > 0 –

⇒

–

+ –2/3

4

−2 x ∈  , 4  3 

Logarithms Let there be a number a > 0, ≠1. A number p is called the logarithm of a number x to the base a, if a p = x and is written as p = log a x. Obviously x must be positive. Ø

● ●

If x < 0, log a x is imaginary and if x = 0, log a x does not exist . log a x exists if and only if x, a> 0 and a ≠ 1.

i. ii.

a log a x = x; a ≠ 0, ±1, x > 0

log a a =1, log a 1 = 0; a > 0, ≠1

iv.

1 ; x, a > 0, ≠1 log a x = log x a log a x = log b x ⋅ log a b =

log b x ; a, b > 0, ≠1, log b a

X

x >0

vi.

For m, n > 0, a > 0, ≠1 (a) log a ( m ⋅ n) = log a m + log a n  m (b) log a   = log a m − log a n  n

viii. ix.

xi.

If 0 < a < 1, then (a) log a x > p ⇒ 0 < x < a p (b) 0 < log a x < p ⇒ a p < x 0 ⇒ x> 3 Also, e > 1 given inequality may written as x − 3 < (e )1 ⇒ x < 3 + e

(c) log a ( m x ) = x log a m

vii.

If a >1, then (a) log a x > p ⇒ x > a p (b) 0 < log a x < p ⇒ 0 < x < a p

a log b x = x log b a ; a > 0, b > 0, ≠1, x > 0

iii.

v.

x.

Inequalities and Quadratic Equation

5

Properties of Logarithms

⇒

For x > 0, a > 0, ≠1  1 (a) log a n (x ) =   log a x  n  m (b) log a n ( x m ) =   log a x  n

X

…(i)

x ∈ (3, 3 + e )

Example 12. The value of x, log 1/ 2 x ≥ log 1/ 3 x is (a) (0, 1] (b) (0, 1) (c) [0, 1) (d) None of these Sol. (a) Case I When x ≠ 1and x > 0

(a) log a x > 0, iff x >1, a >1 or 0 < x < 1, 0 < a < 1 (b) log a x < 0, iff x >1, 0 < a < 1 or 0 < x < 1, a >1 For x > y > 0, (a) log a x > log a y, if a >1 (b) log a x < log a y, if 0 < a < 1

log1/ 2 x ≥ log1/ 3 x log x 2 ≥ log x 3, when x ≠ 1

⇒

which is possible, only if 0 < x < 1. Case II When x = 1 log1/ 2 x = log1/ 3 x, i.e. equality holds. Combining the above cases, 0 < x ≤ 1 or

x ∈(0, 1]

Work Book Exercise 5.2 7 ( x − 1)2 > 9

Solve for x,

1 x+4 >5

( −∞, − 2 ) ∪ ( 4, ∞ )

( −∞, 1) R − [−9, 1]

( 4, ∞ )

( −∞, 9) None of these

( −∞, 2 )

2 | x + 2| < 4 ( −6, 2 )

( −∞, − 2 ] ∪ ( 4, ∞ ) ( −6, 0)

( −6, 2 ]

( 0, 2 )

3 log 2 x > 0 ( 0, ∞ )

( − ∞ , 0)

(1, ∞ )

( 4, ∞ )

( −4, ∞ )

( 0, 1)

( −∞, 4)

( −5, 4)

( 0, 4)

( 0, 5)

( −∞, 1)

( −1, 0)

( −1, 1)

( −∞, 1)

( −1, 0)

( −1, 1)

x +6 ≥1 5x 2

10

6 log x 0.5 > 2  1  , 1   2 

( −6, 4) ( −4, 6)

9 log x ( x + 7 ) < 0

5 log 0. 2 ( x + 5) > 0 ( −5, − 4)

( −6, 4] ( 4, 6)

( 4, ∞ )

4 log 4 x > 1 ( 0, ∞ )

8 ( x + 1)2 < 25

( −∞, − 3) ( −∞, − 3) ∪ ( 3, ∞ ) R ( −∞, − 3] ∪ [−2, 0) ∪ ( 0, 2 ] ∪ [3, ∞ )

187

Objective Mathematics Vol. 1

Mean 5 Arithmetico-Geometric Inequality If a1 , a 2 , …, a n are n distinct positive real numbers, then a1 + a 2 + K + a n > ( a1 a 2K a n )1/ n n i.e. AM > GM

Example 15. If the equation x 4 − 4x 3 + ax 2 + bx + 1 = 0 has four positive roots, then a, b are (a) a = 4, b = 6 (b) a = − 4, b = 6 (c) a = 2, b = 3 (d) a = 6, b = − 4 Sol. (d) Let α, β, γ and δ be four roots of the given equation.

Ø If a1 = a2 = K = an , then AM = GM. Thus, equality occurs when

all quantities are equal. X

X

Example 13. If a, b and c are three distinct positive real numbers, then  1 1 1 (a) ( a + b + c)  + +  > 9  a b c  1 1 1 (b) ( a + b + c)  + +  < 9  a b c  1 1 1 (c) ( a + b + c)  + +  ≥ 9  a b c  1 1 1 (d) ( a + b + c)  + +  ≤ 9  a b c

Then, α + β + γ + δ = 4 and αβγδ = 1 ⇒ AM of α, β, γ, δ = GM of α,β, γ, δ ⇒ α = β = γ = δ ⇒ α = β = γ = δ = 1 [Qα + β + γ + δ = 4, also αβ + αγ + αδ + βγ + βδ + γδ = a, αβγ + αβδ + βγδ + αγδ = − b] ∴ x4 − 4 x3 + ax2 + bx + 1 = x4 − 4 x3 + 6 x2 − 4 x + 1 ⇒

Two Important Inequalities i.

m

a1m + a 2m + ... + a nm  a1 + a 2 + K + a n   1  > (abc ) and ∴   abc 3 3 1 1 1 3  1/ 3 ⇒ a + b + c > 3 (abc ) and  + +  >  a b c  (abc )1/ 3 1 1 1 ⇒ (a + b + c )  + +  > 9  a b c

Weighted AM and GM Inequalities If a1 , a 2 , K , a n are n positive real numbers and m1 , m2 , …, mn are n positive rational numbers, then m1 a1 + m2 a 2 + K + mn a n m1 + m2 +... + mn

m

 b a + b2 a 2 + ... + bn a n  > 1 1  , b1 + b2 + K + bn   if m < 0 or m >1 and b1 a1m + b2 a 2m + K + bn a nm b1 + b2 +... + bn

1 m1 m2 mn m1 + m2 + K + mn ( a1 ⋅ a 2 K a n )

i.e. weighted AM > weighted GM X

Example 14. If a, b and c are distinct positive integers, then ax b − c + bx c − a + cx a − b is (a) > 2 ( a + b + c) (b) R (c) > ( a + b + c) (d) < ( a + b + c) Sol. (c) We have, a⋅ x b − c + b⋅ xc − a + c ⋅ x a − b > a+ b+c {( x ⇒

b −c a

) (x

) (x

a − b c 1/ a + b + c

) }

ax b − c + bx c − a + cx a − b > a+ b+c { xa( b − c ) +

188

c −a b

⇒ ax

b −c

+ bx

c −a

+ cx

a−b

b( c − a ) + c ( a − b ) 1/ a + b + c

> (a + b + c )

}

If a1 , a 2 , ..., a n are n positive distinct real numbers, then m a1m + a 2m + a nm  a1 + a 2 +K + a n  (a) >  ,   n n if m < 0 or m >1 (b)

Sol. (a) We have, AM > GM

>

a = 6 and b = − 4

m

 b a + b2 a2 + K + bn an   n 

  a1q + a 2q +K + a nq      n   r r r  a1 + a 2 +K + a n    n  

(a) ≥ ( a12 + a 22 + K + a n2 ) ( b1−2 + b2−2 + K + bn−2 )

Example 16. If m >1 and n ∈ N , then 1m + 2 m + 3 m + K + n m  n + 1 >   2  n

m

1m + 2 m + 3 m + K + n m  n + 1 ( a12 + a 22 + K + a n2 ) ( b1−2 + b2−2 + K + bn−2 ) (d) < ( a12 + a 22 + K + a n2 ) ( b1−2 + b2−2 + K + bn−2 )

(b)

Sol. (b) By Cauchy-Schwartz’s inequality, we have (a1b1−1 + a2 b2−1 + K + an bn−1 )2 ≤ (a12 + a22 + K + an2 )(b1−2 + b2−2 + K + bn−2 ) a a a  ⇒ 1 + 2 + K+ n b b b  1 n 2

Sol. (a) We know that for m > 1,

 1 1 1 ≤ (a12 + a22 + K + an2 )  2 + 2 + K + 2  b2 bn   b1

AM of mth power > mth power of AM

∴ ⇒

1m + 2 m + 3m + K + nm  1 + 2 + 3 + K + n  >    n n m 1m + 2 m + 3m + K + nm  n + 1 >   2  n

m

Applications of Inequalities to Find the Greatest and Least Values i.

Some Other Standard Inequalities i.

ii.

Weierstrass Inequality (a) If a1 , a 2 , K , a n are n positive real numbers, then for n ≥ 2 , (1 + a1 ) (1 + a 2 ) K (1 + a n ) > 1 + a1 + a 2 +L + a n (b) If a1 , a 2 ,K , a n are positive numbers less than unity, then (1 − a1 )(1 − a 2 ) K (1 − a n ) > 1 − a1 − a 2 −K − a n Cauchy-Schwartz Inequality If a1 , a 2 , K , a n and b1 , b2 , K , bn are 2n real numbers, then ( a1 b1 + a 2 b2 + ... + a n bn ) 2 ≤ ( a12 + a 22 +K + a n2 ) ( b12

+

b22

+K +

If x1 , x 2 , K , x n are positive variables such that x1 x 2 K x n = c (constant), then the sum x1 + x 2 + K + x n is least when x1 = x 2 = K = x n = c1/ n and the least value of the sum is n( c1/ n ).

iii.

If x1 , x 2 , K, x n are variables and m1 , m2 , K, mn are positive real numbers such that x1 + x 2 +… + x n = c (constant), then m m x1 1 ⋅ x 2 2 … x nmn is greatest, when x1 x x x + x 2 +… + x n = 2 =… = n = 1 m1 m2 mn m1 + m2 +… + mn

bn2 )

X

Tchebychef’s Inequality If a1 , a 2 , K, a n and b1 , b2 , K, bn are real numbers such that a1 ≤ a 2 ≤ a 3 ≤K ≤ a n and b1 ≤ b2 ≤ K ≤ bn , then n ( a1 b1 + a 2 b2 + K + a n bn ) ≥ ( a1 + a 2 + K + a n ) ( b1 + b2 + K + bn ) a1 b1 + a 2 b2 + K + a n bn or n  a1 + a 2 + K + a n   b1 + b2 + K + bn  ≥      n n

Example 17. If none of b1 , b2 , K , bn is zero, 2

a a a  then  1 + 2 + K + n  is b b bn   1 2

If x1 , x 2 , K , x n are n positive variables such that x1 + x 2 + K + x n = c (constant), then the product x1 x 2 K x n is greatest when c x1 = x 2 =K = x n = and the greatest value is n n  c   .  n

ii.

with the equality holding if and only if a1 a 2 a = =K = n b1 b2 bn

iii.

2

5 Inequalities and Quadratic Equation

X

X

Example 18. The greatest value of x 2 y 3 when 3x + 4 y = 5, is 3 3 6 8 (b) (c) (d) (a) 8 16 5 3 Sol. (b) Let P = x2 y3 . Clearly, P is the product of 5 factors such that two of them are equal to x and the remaining 3 are equal to y. Now,

3x + 4y = 5

3x 4y 2   + 3   = 5  3 2  3x 3x 4y 4y 4y ⇒ + + + + =5 2 2 3 3 3 5 3  16 x2 y3  ≤  5  ⇒ x 2 y3 ≤ ⇒      3   5 16 or maximum of x2 y3 = 3/16 ⇒

189

5

Work Book Exercise 5.3

Objective Mathematics Vol. 1

1 If a, b and c are all positive real numbers, which

3 If a and b are two positive quantities whose sum is  λ, then the minimum value of 1 + 

one of the following is true? ( b + c )(c + a )( a + b ) ≥ 8abc ( b + c )(c + a )( a + b ) < 8abc b c e f g a +  < 9 +   +  + e f g  a b c 1 1  ( a + b ) +  < 4 a b

λ−

c3 2 ab

2 λ

1+

2 λ

1+

1 λ

c3 4ab

c3 2 ab

c3 2 ab

5 Which of the following is true, where a, b, c > 0?

positive real numbers, then a22

λ−

4 If a2 x 4 + b2 y 4 = c 6 , then the greatest value of xy is

2 If a1, a2 , K, an and b1, b2 , K, bn are any two sets of  a12

1 λ

1  1  1 +  is   a b

2 an2 

2 ( a 3 + b 3 + c 3 ) ≥ bc( b + c ) + ca(c + a ) + ab( a + b )

+ + ... +   bn   b1 b2

a3 + b 3 + c 3 (a + b + c ) ⋅ (a2 + b 2 + c 2 ) > 3 9 bc ca ab 1 + + < (a + b + c ) b+c c+ a a+ b 2

≤ ( a14 + a24 + K + an4 )( b1−2 + b 2−2 + K + b n−2 ) ≥ ( a14 + a24 + K + an4 )( b1−2 + b 2−2 + K + b n−2 ) < ( a14 + a24 + K + an4 )( b1−2 + b 2−2 + K + b n−2 )

2 2 2 1 1 1 + + < + + b+c c+ a a+ b a b c

> ( a14 + a24 + K + an4 )( b1−2 + b 2−2 + K + b n−2 )

Quadratic Equation with Real Coefficients An equation of the form ax 2 + bx + c = 0

…(i)

where, a ≠ 0, a, b, c ∈ R is called a quadratic equation with real coefficients.

Ø

●

The quantity D = b 2 − 4ac is known as the discriminant of the quadratic equation in (i) whose roots are given by α=

−b + b 2 − 4ac

and β =

190

●

−b − b 2 − 4ac

2a 2a The nature of the roots is as given below: (i) If a quadratic equation in x has more than two roots, then it is an identity in x that is a = b = c = 0. (ii) If a =1 and b, c ∈ I and the roots are rational numbers, then these roots must be integers. (iii) The roots are of the form p ± q ( p, q ∈Q ), iff a, b and c are rational and D is not a perfect square. (iv) The roots are real and distinct, iff D > 0. (v) The roots are real and equal, iff D = 0. (vi) The roots are complex with non-zero imaginary part, iff D < 0. (vii) The roots are rational, iff a, b and c are rational and D is a perfect square.

●

●

●

●

X

If α is a root of the equation f (x) = 0, then the polynomial f (x) is exactly divisible by (x − α) or (x − α) is a factor of f (x) and conversely. Every equation of nth degree (n≥ 1) has exactly n roots and if the equation has more than n roots, it is an identity. If the coefficients of the equation f (x) = 0 are all real and α + iβ is its root, then α − iβ is also a root, i.e. imaginary roots occur in conjugate pairs. If the coefficients in the equation are all rational and α + β is one of its roots, then α − β is also a root, where α, β ∈ Q and β is not a perfect square. If there is any two real numbers ‘a’ and ‘b’ such that f (a) and f (b) are of opposite signs, then f (x) = 0 must have atleast one real root between ‘a’ and ‘b’. Every equation f (x) = 0 of odd degree has atleast one real root of a sign opposite to that of its last term.

Example 19. The number of values of a for which ( a 2 − 3a + 2) x 2 + ( a 2 − 5a + 6) x + a 2 − 4 = 0 is an identity in x, is (a) 0 (b) 2 (c)1 (d) 3 Sol. (b) It is an identity in x, if a2 − 3 a + 2 = 0, a2 − 5 a + 6 = 0, a2 − 4 = 0 ⇒ a = 1, 2 and a = 2, 3 and a = 2, −2 ∴ Equation is identity, if a = 2

Sol. (d) Since, ax2 + x + b = 0 has real roots

Example 20. If cos θ, sin φ and sin θ are in GP, then roots of x 2 + 2 cot φ ⋅ x + 1 = 0 are always (a) equal (b) real (c) imaginary (d) greater than 1

⇒ ⇒ or

sin2 φ = cos θ ⋅ sin θ

⇒

cos 2 φ = 1 − sin 2θ

Also, discriminant, D = 4cot 2 φ − 4 = =

4(1 − sin2 θ ) sin φ 2

4cos 2 φ

…(i)

sin2 φ

X

> 0; as sin 2θ < 1 and sin2 φ > 0

Example 21. If l, m and n are real, l ≠ m, then the roots of the equation ( l − m) x 2 − 5( l + m) x − 2( l − m) = 0 are (a) real and equal (b) complex (c) real and unequal (d) None of these

Let

f( x) = a2 x2 + 2 bx + 2c

∴

f(α ) = a2α 2 + 2 bα + 2c

and

X

Example 24. If the roots of the equation ax 2 + x + b = 0 are real, then the roots of the equation x 2 − 4 ab x + 1 = 0 will be (a) rational (b) irrational (c) real (d) imaginary

[from Eq. (ii)]

α< γ 7 possible. Hence, no such a exists.

Sol. (c) Given α and β are the roots of ( x − a)( x − b) − c = 0

( x − α) ( x − β) + c = 0

f(β ) = a β + 2 bβ + 2c

∴

Example 23. Let α and β be the roots of the equation ( x − a ) ( x − b) = c, c ≠ 0. Then, the roots of the equation ( x − α ) ( x − β) + c = 0 are (a) a, c (b) b , c (c) a, b (d) a + c, b + c

⇒ a and b are the roots of equation

[from Eq. (i)]

2 2

⇒ f(α ) f(β ) < 0 ∴f( x) must have a root lying in the open interval (α, β ).

⇒ (| x| − 1)(| x| − 2 ) = 0 ⇒ | x| = 1, 2 ∴ x = 1, −1, 2, − 2 Hence, four real solutions exist.

( x − a)( x − b ) = ( x − α )( x − β ) + c

…(ii)

= a2β 2 + 2 a2β 2 = 3a2β 2

Sol. (a) Since,| x|2 − 3| x| + 2 = 0

X

a2β 2 − bβ − c = 0

= a2α 2 − 2 a2α 2 = − a2α 2

(l + m) ≥ 0

( x − a)( x − b ) − c = ( x − α )( x − β )

…(i)

and β is a root of a x − bx − c = 0 ⇒

2

Example 22. The number of real solutions of the equation | x | 2 − 3 | x | + 2 = 0 is (a) 4 (b) 1 (c) 3 (d) 2

⇒

a2α 2 + bα + c = 0 2 2

Thus, D> 0 Hence, roots are real and unequal.

⇒

D< 0

⇒

As, l ≠ m, (l − m)2 > 0.

X

D = 16 ab − 4

From Eq. (i),

Sol. (d) Since, α is a root of a2 x2 + bx + c = 0

D = 25(l + m)2 + 8(l − m)2

X

∴

Example 25. Let a, b and c be real numbers, a ≠ 0. If α is a root of a 2 x 2 + bx + c = 0, β is the root of a 2 x 2 − bx − c = 0 and 0 < α < β, then the equation a 2 x 2 + 2bx + 2c = 0 has a root γ that always satisfies α +β β (b) γ = α + (a) γ = 2 2 (c) γ = α (d) α < γ < β

Sol. (c) Discriminant of the given equation is

Also,

…(i)

Hence, roots are imaginary.

⇒ Roots are real. Also, when cos θ = sin θ ⇒ Three numbers are equal which is a special case i.e. the series forms both AP and GP. X

− 4 ab ≥ − 1 4 ab ≤ 1

Now, second equation is x2 − 4 a b x + 1 = 0

Sol. (b) As cos θ, sin φ and sin θ are in GP. ⇒

5

(1)2 − 4 ab ≥ 0

Inequalities and Quadratic Equation

X

X

Example 27. The roots of ax 2 + bx + c = 0, were a ≠ 0 and coefficients are real, are non-real complex and a + c < b. Then, (a) 4a + c > 2b (b) 4a + c < 2b (c) 4a + c = 2b (d) None of these Sol. (b) Since, ax2 + bx + c = 0 has non-real. Thus, either f( x) = ax2 + bx + c ⇒ As

f( x) > 0 or f( x) < 0, for all x a + c < b ⇒ f(− 1) < 0

191

⇒ a − b + c < 0 or ∴ f( x) < 0, for all x.

5

a+ c< 0

and product of roots =

Objective Mathematics Vol. 1

a −b

Hence, the quadratic equation

Y

 2a   x + x2 −  2  a − b X′

⇒

X

O

X

Y′

Thus, for all x ∈ R, ax2 + bx + c < 0 for x = − 2, 4a − 2 b + c < 0 4a + c < 2 b

or ⇒

Formation of a Polynomial Equation from Given Roots If α 1 , α 2 , α 3 , … , α n are the roots of an nth degree equation, then the equation is x n − S 1 x n − 1 + S 2 x n − 2 − S 3 x n − 3 + ... + ( −1) n S n = 0 where, S k denotes the sum of the products of roots taking k roots at a time.

Particular Cases

i.

Quadratic equation If α and β are the roots of a quadratic equation, then the equation is x 2 − S1 x + S 2 = 0 i.e.

ii.

x 2 − (α + β) x + αβ = 0

Cubic equation If α, β and γ are the roots of a cubic equation, then the equation is x 3 − S1 x 2 + S 2 x − S 3 = 0 i.e. x 3 − (α + β + γ ) x 2 + (αβ + βγ + γα ) x − αβγ = 0

X

Example 28. If a and b are rational and b is not a perfect square, then the quadratic equation 1 with rational coefficients whose one root is , a+ b is (a) x 2 − 2ax + ( a 2 − b) = 0 (b) ( a 2 − b) x 2 − 2ax + 1 = 0 (c) ( a 2 − b) x 2 − 2bx + 1 = 0 (d) None of the above Sol. (b) As irrational roots alway occur in pairs, i.e. the roots of given equation are

192

1 2

Thus, sum of roots =

2a a2 − b

1 1 , . a+ b a− b

 1   2  = 0  a − b

(a2 − b )x2 − 2 ax + 1 = 0

Example 29. The quadratic equation whose roots are the AM and HM of the roots of the equation x 2 + 7x − 1 = 0 is (a)14x 2 + 14x − 45 = 0

(b)14x 2 − 14x + 14 = 0

(c)14x 2 + 45x − 14 = 0

(d) None of these

Sol. (c) Here, x2 + 7 x − 1 = 0 has α + β = − 7 and αβ = − 1 QQuadratic equation’s roots are AM and HM of the roots of the equation x2 + 7 x − 1 = 0 α+β 2αβ and β ′ = α′ = ⇒ α+β 2 7 −2 2 and β′ = or = α′ = − 2 −7 7 ∴Required equation is x2 − (α ′ + β ′ ) x + α ′β ′ = 0 45  x2 −  − x − 1 = 0 ⇒  14  or

14 x2 + 45 x − 14 = 0

Symmetric Function of the Roots A function of α and β is said to be a symmetric function, if it remains unchanged when α and β are interchanged. e.g. α 2 + β 2 + 2αβ is a symmetric function of α and β whereas α 2 − β 2 + 3αβ is not a symmetric function of α and β. In order to find the value of a symmetric function of α and β, express the given function in terms of α + β and αβ. The following results may be useful:

i. ii.

α 2 + β 2 = (α + β) 2 − 2αβ α 3 + β 3 = (α + β) 3 − 3αβ(α + β)

iii.

α 4 + β 4 = (α 3 + β 3 )(α + β) − αβ(α 2 + β 2 )

iv.

α 5 + β 5 = (α 3 + β 3 )(α 2 + β 2 ) − α 2 β 2 (α + β)

v.

| α − β| = (α + β) 2 − 4αβ

vi.

α 2 − β 2 = (α + β)(α − β)

vii.

α 3 − β 3 = (α − β)[(α + β) 2 − αβ] = (α − β) 3 + 3αβ(α − β)

viii.

α 4 − β 4 = (α + β)(α − β)(α 2 + β 2 )

Example 30. If α and β are the roots of α β ax 2 + 2bx + c = 0, then + is equal to β α 4b 2 − 2ac 4b 2 − 4ac (b) ac ac 2b 2 − 2ac 2b 2 − 4ac (c) (d) ac ac Sol. (a) Since, α and β are the roots of equation

v.

To obtain an equation whose roots are reciprocals of the roots of a given equation is obtained by replacing x by 1/ x in the given equation.

vi.

To transform an equation to another equation whose roots are negative of the roots of a given equation, replace x by −x.

vii.

To transform an equation to another equation whose roots are square of the roots of a given equation, replace x by x.

viii.

To transform an equation to another equation whose roots are cubes of the roots of a given equation replace x by x 1/ 3 .

(a)

ax2 + 2 bx + c = 0 −2b c and αβ = α+β= a a α β α 2 + β 2 (α + β )2 − 2αβ + = = β α αβ αβ

∴ Now,

α β (4b 2 / a2 ) − 2c / a 4b 2 − 2 ac = + = β α ac c/a

⇒ X

Example 31. If α and β are the roots of the equation px 2 + qx + r = 0, then the value of α 3β + β 3α is (a) (c)

X

(b)

p3 r ( q 2 + 2 pr )

(d)

p3

r ( q − 2 pr )

a( x1/ 3 )3 + b( x1/ 3 )2 + c( x1/ 3 ) + d = 0

p3 ( q 2 − 2 pr )

⇒

px2 + qx + r = 0, therefore α + β = −

⇒ a x + 3 a2 d x2 + 3 ad 2 x + d 3

p3

= − {b 3 x2 + c 3 x + 3bcx (bx2 / 3 + cx1/ 3 )} ⇒ a x + 3 a dx + 3 ad 2 x + d 3 3 3

q r and αβ = . p p

2

2

= − {b 3 x2 + c 3 x − 3 bcx (ax + d )} ⇒ a x + x (3 a d − 3 abc + b 3 ) 3 3

2

2

α β + β α = αβ(α + β ) 3

3

2

2

+ x (3 ad 2 − 3bcd + c 3 ) + d 3 = 0

= (αβ ) [(α + β ) − 2αβ ]  r   q 2 2r  r(q 2 − 2 pr ) =    2 −  =  p  p p p3

Transformation of Equations Let α and β be the roots of the equation ax + bx + c = 0, then the equation 2

i.

Whose roots are α + k, β + k is obtained by replacing x by x − k in the given equation.

ii.

Whose roots are α − k, β − k is obtained by replacing x by x + k in the given equation.

iv.

(a x + d )3 = − (b x2 / 3 + c x1/ 3 )3 3 3

2

iii.

a x + d = − (b x2 / 3 + cx1/ 3 )

⇒

Sol. (b) Since, α and β are the roots of the equation ∴

Example 32. Form an equation whose roots are cubes of the roots of equation ax 3 + bx 2 + cx + d = 0. Sol. Replacing x by x1/ 3 in the given equation, we get

2

r

Whose roots are αk, βk is obtained by multiplying the coefficients of x 2 , x and constant term by k 0 , k 1 , k 2 respectively in the given equation. α β Whose roots are , is obtained by k k multiplying the coefficients of x 2 , x and constant term by k 2 , k 1 , k 0 respectively in the given equation.

5 Inequalities and Quadratic Equation

X

which is the required equation. X

Example 33. If the roots of a1 x 2 + b1 x + c1 = 0 are α 1 , β1 and that of a 2 x 2 + b2 x + c2 = 0 are α 2 , β 2 such that α 1α 2 = β1β 2 = 1, then a b c (a) 1 = 1 = 1 a 2 b2 c2 a b c (b) 1 = 1 = 1 c2 b2 a 2 (c) a1 a 2 = b1 b2 = c1 c2 (d) None of the above Sol. (b) Roots of the second equation are reciprocal of the first. As and ∴ Replacing x by

and

1 α1 1 β2 = β1

α2 =

1 in a1 x2 + b1 x + c1 = 0, we get x c1 x2 + b1 x + a1 = 0 a2 x2 + b2 x + c 2 = 0

Eqs. (i) and (ii) are same equations. c1 b a = 1 = 1 a2 b2 c 2

∴

…(i)

[given]…(ii)

193

Objective Mathematics Vol. 1

5

X

Example 34. If α, β and γ are the roots of the equation x (1 + x 2 ) + x 2 (6 + x ) + 2 = 0, then the value of α −1 + β −1 + γ −1 is 1 (a) −3 (b) 2 1 (c) − (d) None of these 2

Common Roots Let a1 x 2 + b1 x + c1 = 0 and a 2 x 2 + b2 x + c2 = 0 be two quadratic equations such that a1 , a 2 ≠ 0 and a1 b2 ≠ a 2 b1 . When one common root If α is the common root of these equations, then ( b1 c2 − b2 c1 ) ( a1 b2 − a 2 b1 ) = ( c1 a 2 − c2 a1 ) 2 which is the condition for roots of two quadratic equations to be common.

i.

Sol. (c) Since, 2 x3 + 6 x2 + x + 2 = 0 has roots α, β and γ. So, 2 x3 + x2 + 6 x + 2 = 0 has roots α −1, β −1 and γ −1 [writing coefficients in reverse order, since roots are reciprocal] Coefficient of x2 Hence, sum of the roots = − Coefficient of x3 1 α −1 + β −1 + γ −1 = − ∴ 2

ii.

X

Descartes Rule of Signs for the Roots of a Polynomial The maximum number of positive real roots of a polynomial equation

Rule 1

f ( x ) = a 0 x n + a1 x n − 1 + a 2 x n − 2 + K + an − 1 x + an = 0 is the number of changes of the signs of coefficients from positive to negative and negative to positive. For instance, in the equation x 3 + 3x 2 + 7x − 11 = 0 the signs of coefficients are + + + − As there is just one change of sign, the number of positive roots of x 3 + 3x 2 + 7x − 11 = 0 is atmost 1. The maximum number of negative roots of the polynomial equation f ( x ) = 0 is the number of changes from positive to negative and negative to positive in the signs of coefficients of the equation f ( − x ) = 0.

Rule 2

When two common roots The required a b c condition is 1 = 1 = 1 a 2 b2 c2

Example 35. If ax 2 + bx + c = 0 and bx 2 + cx + a = 0 have a common root and a, b and c are non-zero real numbers, then find the value of ( a 3 + b 3 + c 3 ) / abc. Sol. Given that, ax2 + bx + c = 0 and bx2 + cx + a = 0 have a common root. Hence, (bc − a2 )2 = (ab − c 2 ) (ac − b 2 ) ⇒ b 2c 2 + a4 − 2 a2 bc = a2 bc − ab 3 − ac 3 + b 2c 2 ⇒

a4 + ab 3 + ac 3 = 3a2 bc a3 + b 3 + c 3 =3 abc

⇒ X

Example 36. If the equation x 2 + 5x + 8 = 0 and ax 2 + bx + c = 0; a, b, c ∈ R have a common root, then a : b : c is (a)1 : 5 : 6 (b)1 : 5 : 8 (c) 8 : 5 :1 (d) None of these Sol. (b) Since, the quadratic equation x2 + 5 x + 8 = 0 has imaginary roots. So, equation ax2 + bx + c = 0 will have both roots same as the equation x2 + 5 x + 8 has. a b c Thus, = = ⇒ a = λ, b = 5λ, c = 8λ 1 5 8 Hence, the required ratio is 1 : 5 : 8.

Work Book Exercise 5.4 1 If α and β are the roots of the equation

2 x − 3 x − 6 = 0, then equation whose roots are α 2 + 2 , β 2 + 2 is 2

4 x 2 + 49 x + 118 = 0 4 x 2 − 49 x − 118 = 0

4 x 2 − 49 x + 118 = 0 x 2 − 49 x + 118 = 0

2 If 2 + i 3 is a root of x 2 + px + q = 0, where p and q are real, then ( p, q ) is equal to ( −4, 7 ) ( −7 4)

194

( 4, − 7 ) ( 4, 7 )

3 The equations ax 2 + bx + a = 0,

x 3 − 2 x 2 + 2 x − 1 = 0 have two roots in common. Then, a + b must be equal to 1 0

−1 None of these

4 If| x 2| + | x| − 2 = 0, then the value of x is equal to 2 −2 ±1 None of the above

is {2, 9}

(2, 7)

{2}

d [2, 9]

6 If a, b and c are three distinct positive real numbers, then the number of real roots of ax 2 + 2 b| x| + c = 0 is 0

1

2

4

7 If sin α, sin β and cos α are in GP, then roots of x + 2 x cot β + 1 = 0 are always 2

equal imaginary

real greater than 1

8 If x 2 + ax + b is an integer for every integer x, then d is always an integer but b need not be an integer b is always an integer but d need not be an integer Cannot be discussed a and b are always integers

9 Let a > 0, b > 0 and c > 0. Then, both the roots of the equation ax 2 + bx + c = 0 are real and negative are rational numbers

have negative real part None of these

10 If p, q and r are real and p ≠ q, then the roots of the equation ( p − q ) x 2 + 5 ( p + q )x − 2( p − q ) = 0 are real and equal real and unequal

complex None of these

11 If α and β are the roots of x 2 + px + q = 0 and α 4 , β 4 are the roots of x 2 − r x + s = 0, then the equation x 2 − 4 qx + 2q 2 − r = 0 has always two real roots two positive roots two negative roots one positive and one negative root

12 The real roots of the equation | x|3 − 3 x 2 + 3| x| − 2 = 0 are ±1

0, 2

±2

1, 2

13 If α and β are the roots of the quadratic equation 6 x 2 − 6 x + 1 = 0, then +

1 (a + bα + cα 2 + dα 3 ) 2

5 Inequalities and Quadratic Equation

5 If| x − 2 | + | x − 9| = 7, then the set of values of x

1 (a + bβ + cβ 2 + dβ 3 ) is equal to 2

12d + 6 c + 4b + a 12 12 a + 6b + 4 c + 9 d 1 (12 a + 6b + 4 c + 3 d ) 12 None of these

14 Both the roots of the equation ( x − b) ( x − c ) + ( x − a) ( x − c ) + ( x − a) ( x − b) = 0 are always positive real

negative None of these

Quadratic Expression and its Graph Let a, b and c be real numbers and a ≠ 0. Then, f ( x ) = ax 2 + bx + c is known as a quadratic expression or a quadratic polynomial in x. In this section, we shall discuss the graph of the quadratic polynomial, i.e. the curve whose equation is y = ax 2 + bx + c, a ≠ 0. Consider, y = ax 2 + bx + c 2  4ac − b 2  b = a  x +  +  2a  4a 2    2  b D  = a  x +  − 2  2a  4a   

b D   ⇒  y +  = a x +     4a 2a 

Y

2

X′

When a < 0 X-axis a>0 X-axis

In this case, the parabola cuts X-axis at (α, 0) −b − D −b + D and β = and (β, 0), where α = 2a 2a For a > 0, we have −b −D at x = , y max → ∞ y min = 2a 4a y = ax2 + bx + c

Thus, y = f ( x ) represents a parabola. The parabola opens upwards or downwards accordingly as a > 0 or a < 0 When a > 0

Thus, we may draw the following conclusions: (i) The parabola will intersect the X -axis in two distinct points iff D > 0

a 0, if 

α < x β

For a < 0, we have –D −b at x = , y min → − ∞ y max = 2a 4a

195

Objective Mathematics Vol. 1

5

and

< 0, if  y = f ( x ) is = 0, if > 0, if 

x < α or x > β x = α, β α < x 0, we have y min = 0 −b at x = , y max → ∞ 2a > 0, if x ≠ α and y = f ( x ) is  = 0, if x = α

a 0, ∀ x ∈R, iff a> 0 and D < 0 and 2

a>0 O

ax 2 + bx + c < 0, ∀ x ∈R , iff a< 0 and D < 0.

X X

Y′

−b y max = 0 at x = , y min → − ∞ 2a < 0, if x ≠ α y = f (x ) =   0, if x = α Y

X′

X

O a 0 or a < 0. For a > 0, we have −D −b at x = , y max → ∞ y min = 2a 4a and y = f ( x ) > 0, ∀ x

Example 37. What is the minimum height of any point on the curve y = x 2 − 4x + 6 above the X-axis? Sol. Here, a = 1 > 0 and D = 16 − 4(6) = − 8 < 0 Therefore, the parabola remains completely above X-axis and minimum value of any point on the curve is −D . given by 4a

For a < 0, we have

and

X

O

ymin =

i.e.

−D 8 = =2 4a 4

Hence, the minimum height is 2. X

Example 38. The roots of ax 2 + bx + c = 0, where a ≠ 0 and coefficients are real, are non-real complex and a + c < b. Then, (a) 4a + c > 2b (b) 4a + c < 2b (c) 4a + c = 2b (d) None of the above Sol. (b) Since, f( x) = ax2 + bx + c = 0 has non-real roots. Thus, either f( x) > 0 or f( x) < 0, ∀ x a + c < b ⇒ f(−1) < 0

As

Y

X′

O

Y y = ax2 + bx + c Y¢

a>0 X′

196

O

Y′

X

i.e. a − b + c < 0 or a + c < b ∴ f( x) < 0, ∀ x Thus, for all x ∈ R; ax2 + bx + c < 0. Now, for x = − 2; 4 a − 2 b + c < 0 ⇒ 4a + c < 2b

X

Let f ( x ) = ax 2 + bx + c, where a, b, c ∈ R and a ≠ 0. Then, f ( x ) ≥ 0, f ( x ) < 0, f ( x ) ≤ 0, f ( x ) > 0 are quadratic inequations. The set of all real values of x, which satisfy an inequation is called its solution set. The following example will make it clear: X

X

Example 39. Solve the inequation −x 2 + 3x + 4 < 0. Sol. We have,

2 Sol. Let y = x2 − 3 x + 4. Then,

− x + 3x + 4 < 0 2

⇒

x − 3x − 4 > 0

⇒

( x − 4) ( x + 1) > 0

x + 3x + 4

x ( y − 1) + 3 x( y + 1) + 4( y − 1) = 0 2

2

+

– –1

⇒ or ⇒

Now, if ⇒

+ 4

x< −1 x> 4 x ∈ (− ∞, − 1) ∪ (4, ∞ )

● ●

(x − a)(x − b) < 0 ⇒ x ∈(a, b) , where a < b (x − a)(x − b) > 0 ⇒ x ∈(−∞ , a) ∪ (b , ∞), where a < b x 2 ≤ a2 ⇒ x ∈[− a, a]

●

x 2 ≥ a2 ⇒ x ∈(−∞ , − a] ∪ [a, ∞)

●

If ax 2 + bx + c < 0, (a> 0) ⇒ x ∈(α , β) , where α and β (α < β) are roots of the equation ax 2 + bx + c = 0

●

If ax 2 + bx + c > 0, (a> 0) ⇒ x ∈(−∞ , α ) ∪ (β , ∞) , where α and β(α < β) are roots of the equation ax 2 + bx + c = 0.

Maximum and Minimum Values of Rational Expression Here, we shall find the values attained by a rational a x 2 + b1 x + c1 expression of the form 1 2 for real values a 2 x + b2 x + c2 of x. The following algorithm will explain the procedure:

Algorithm Step I

Equate the given rational expression to y.

Step II

Obtain a quadratic equation in x by simplifying the expression in Step I.

D≥ 0 9( y + 1)2 − 16( y − 1)2 ≥ 0

⇒

− 7 y2 + 50 y − 7 ≥ 0

⇒

7 y2 − 50 y + 7 ≤ 0

⇒

(7 y − 1)( y − 7 ) ≤ 0 1 ⇒ ≤ y≤ 7 7 1 Hence, the given expression lies between and 7. 7

Ø Frequently used inequalities ●

Example 40. If x is real, then prove that the x 2 − 3x + 4 1 values of 2 lie between and 7. 7 x + 3x + 4

5 Inequalities and Quadratic Equation

Step III Obtain the discriminant of the quadratic equation in Step II. Step IV Put discriminant ≥ 0 and solve the inequation for y. The values of y so obtained determines the set of values attained by the given rational expression.

Solution of Quadratic Inequations

X

Example 41. If f (x ) =

x 2 −1

for every real x 2 +1 number x, then the minimum value of f (a) does not exist because f is unbounded (b) is not attained even though f is bounded (c) is equal to 1 (d) is equal to −1

2 Sol. (d) Let y = x2 − 1

x +1

⇒ ⇒

yx2 + y − x2 + 1 = 0 x2 ( y − 1) + y + 1 = 0

Now, for real values of x consider D ≥ 0 ⇒ − 4( y2 − 1) ≥ 0 ⇒

y2 − 1 ≤ 0

⇒ − 1≤ y ≤ 1 Thus, minimum value of f( x) is − 1.

Location of the Roots of a Quadratic Equation Here, we shall discuss various conditions satisfied by the coefficients of a quadratic equation when the location of its roots is given. Let f ( x ) = ax 2 + bx + c, a, b, c ∈ R , a ≠ 0 and α, β be the roots of f ( x ) = 0. Suppose k , k1 , k 2 ∈ R and k1 < k 2 .

197

5

Then, the following hold good :

Objective Mathematics Vol. 1

i.

ii.

Conditions for a number k (If both the roots of f ( x ) = 0 are less than k)

Conditions for a number k (If both the roots of f ( x ) = 0 are greater than k) a>0

f (k ) a>0

α

– b 2a

f (k )

– b 2a

α

k

X

X

β k

– b ,– D 2a 4a

– b ,– D 2a 4a

– b ,– D 2a 4a

– b ,– D 2a 4a α

β

β

k α β k

– b 2a

X

X

– b 2a

f (k )

a 0 b (c) k > − , where α ≤ β 2a X

X

Example 42. Find the values of m for which both roots of equation x 2 − mx + 1 = 0 are less than unity. Sol. Let f( x) = x2 − mx + 1, as both roots of f( x) = 0 are less than 1, we can take D ≥ 0, af(1) > 0 and −

b < 1. 2a

⇒

Sol. Let f( x) = x2 − 6mx + 9m2 − 2 m + 2. as both roots of f( x) = 0 are greater than 3, we can take b > 3. D ≥ 0, af(3) > 0 and − 2a Case I Consider, D ≥ 0 (−6 m)2 − 4 ⋅ 1⋅ (9m2 − 2 m + 2 ) ≥ 0 ⇒ 8m − 8 ≥ 0 ∴ m ≥ 1 or m ∈ [1, ∞ ) Case II Consider, af (3) > 0 1⋅ (9 − 18m + 9m2 − 2 m + 2 ) > 0

Case I Consider, D ≥ 0 (− m)2 − 4 ⋅ 1⋅ 1 ≥ 0 (m + 2 ) (m − 2 ) ≥ 0

⇒ m ∈ (− ∞, − 2 ] ∪ [2, ∞ ) Case II Consider, af (1) > 0 1⋅ (1 − m + 1) > 0 ⇒ m− 2 < 0 ⇒ m< 2 ⇒ m ∈ (− ∞, 2 ) b Case III Consider, − 0

⇒

(9m − 11) (m − 1) > 0  m − 11 (m − 1) > 0 ⇒    9 11 ⇒ m ∈ (− ∞, 1) ∪  , ∞  9  b Case III Consider, − >3 2a +

+

+ – –2

2

m 3 ⇒ m> 1 2 …(iii) ⇒ m ∈ (1, ∞ ) Hence, the values of m satisfying Eqs. (i), (ii) and (iii) at 11 the same time are m ∈  , ∞  . 9 

Conditions for a number k (If k lies between the roots of f ( x ) = 0)

X

Sol. Let f( x) = x2 − 2 mx + m2 − 1as exactly one root of

a>0 k

α

Example 45. Find the values of m for which exactly one root of the equation x 2 − 2mx + m 2 − 1 = 0 lies in the interval ( − 2, 4). f( x) = 0 lies in the interval, we can take D > 0 and f(−2 ) f(4) < 0.

X

β

Case I Consider, D > 0 (− 2 m)2 − 4 ⋅ 1⋅ (m2 − 1) > 0

f (k)

⇒ 4> 0 ∴ m∈R Case II Consider, f(−2 ) f(4) < 0

– b ,– D 2a 4a – b ,– D 2a 4a f (k ) α

+

+ k

X

β

…(i)

–3

–

–1

5 Inequalities and Quadratic Equation

iii.

+ 3

–

5

(4 + 4m + m2 − 1) (16 − 8m + m2 − 1) < 0 a 0 X

⇒

⇒ (m + 1) (m + 3) (m − 3) (m − 5) < 0 ⇒ (m + 3) (m + 1) (m − 3)(m − 5) < 0 ...(ii) ∴ m ∈ (−3, − 1) ∪ (3, 5) Hence, the values of m satisfying Eqs. (i) and (ii) at the same time are m ∈ (−3, − 1) ∪ (3, 5).

(b) af ( k ) < 0, where α < β

Example 44. Find all values of p, so that 6 lies between roots of the equation x 2 + 2 ( p − 3) x + 9 = 0. Sol. Let f( x) = x2 + 2 ( p − 3) x + 9 +

v. +

as 6 lies between the roots of – 0 6 f( x) = 0, we can take D > 0 and af(6) < 0. Case I Consider, D > 0 {2 ( p − 3)}2 − 4 ⋅ 1⋅ 9 > 0 ⇒ ( p − 3)2 − 9 > 0 ⇒ p( p − 6) > 0 …(i) ⇒ p ∈ (− ∞, 0) ∪ (6, ∞ ) Case II Consider, af (6) < 0 1 ⋅ {36 + 12 ( p − 3) + 9} < 0 3 ⇒ 12 p + 9 < 0 ⇒ p + < 0 4 …(ii) ⇒ p ∈ (−∞, − 3/ 4) Hence, the values of p satisfying Eqs. (i) and (ii) at the same time are p ∈ (− ∞, − 3 / 4).

iv.

(m2 + 4m + 3) (m2 − 8m + 15) < 0

Conditions for numbers k 1 and k 2 (If both roots of f ( x ) = 0 are confined between k1 and k 2 )

a>0 f (k1)

f (k2)

k1 α

β k2

X

– b,– D 2a 4a

– b,– D 2 a 4a

Conditions for numbers k 1 and k 2 (If exactly one root of f ( x ) = 0 lies in the interval ( k1 , k 2 )).

β k2

k1 α

X

f (k2)

f (k1) a0 f (k1) k1

α

k2

β

(a) D ≥ 0 (roots may be equal) (b) af ( k1 ) > 0 (c) af ( k 2 ) > 0 b (d) k1 < − < k 2 , where α ≤ β and k1 < k 2 2a

X

f(k2)

f (k2) k1 f (k1)

α

k2

β

a 0 (b) f ( k1 ) f ( k 2 ) < 0, where α < β

f( x) = 0 lie between (−1, 1,) we can take D ≥ 0, af (−1) > 0 , 1 af (1) > 0 and −1 < < 1. 4

199

Case I Consider, D ≥ 0

5

X

1 4 1 a ∈  − ∞,   4 

Objective Mathematics Vol. 1

(−2 )2 − 4 ⋅ 4 ⋅ a ≥ 0 ⇒ a ≤ ⇒

…(i)

Consider, af (−1) > 0 4(4 + 2 + a) > 0 …(ii) ⇒ a > − 6 ⇒ a ∈ (−6, ∞ ) Case III Consider, af (1) > 0 4(4 − 2 + a) > 0 …(iii) ⇒ a > − 2 ⇒ a ∈ (−2, ∞ ) Hence, the values of a satisfying Eqs. (i), (ii) and (iii) at 1 the same time are a ∈  −2, .  4  Case II

vi.

Conditions for numbers k 1 and k 2 (If k1 and k 2 lie between the roots of f ( x ) = 0) a>0 k2

k1 α

f (k2) f (k1) f (k1) f (k2)

α

k1

β

β

between the roots of f( x) = 0, we can take D > 0, (a − 5)f(1) < 0 and (a − 5) f(2 ) < 0.

(a − 3) (a + 4) < 0 −4 < a < 3

…(ii)

β

3

⇒

(−2 a)2 − 4 ⋅ 1 (a2 + a − 3) ≥ 0

⇒

4a2 − 4a2 − 4a + 12 ≥ 0 −4a + 12 ≥ 0 a≤ 3 af(3) > 0

⇒

1[(3)2 − 2 a(3) + a2 + a − 3] > 0

⇒

9 − 6 a + a2 + a − 3 > 0

⇒

a2 − 5 a + 6 > 0

⇒

(a − 2 ) (a − 3) > 0

∴

a ∈ (−∞, 2 ) ∪ (3, ∞ )

…(iii)

…(iv)

From Eqs. (i), (ii), (iii) and (iv), we get a ∈ (−4, 2 )

Case I Consider, D > 0 (−2 a)2 − 4 ⋅(a − 5) ⋅ (a − 4) > 0

200

⇒ ⇒

⇒ ⇒ Again,

Sol. Let f( x) = (a − 5)x2 − 2 ax + a − 4(a ≠ 5) as 1 and 2 lie

…(i)

…(ii)

+ –

a2 + a − 12 < 0

Again, D = B − 4 AC ≥ 0

Example 47. Find the values of a for which one root of equation ( a − 5) x 2 − 2ax + a − 4 = 0 is smaller than 1 and the other greater than 2.

5

⇒

…(i)

2

(a) D > 0 (b) af ( k1 ) < 0 (c) af ( k 2 ) < 0, where α < β

+

Since, both roots are less than 3. i.e. α < 3, β < 3 Sum (S ) = α + β < 6 α+β 2a 0 ⇒ a ∈ (5, ∞ ) Case III Consider, (a − 5) f(2 ) < 0

Sol. (a) Let f( x) = x2 − 2 ax + a2 + a − 3

X

a< 0

X

Example 48. If the roots of the equation x 2 − 2ax + a 2 + a − 3 = 0 are real and less than 3, then (a) a < 2 (b) 2 ≤ a ≤ 3 (c) 3a < a ≤ 4 (d) a > 4

24

(a − 5) {4(a − 5) − 4a + a − 4} < 0 ⇒ (a − 5)(a − 24) < 0 …(iii) ⇒ a ∈(5, 24) Hence, the values of a satisfying Eqs. (i), (ii) and (iii) at the same time are a ∈(5, 24).

Algebraic Interpretation of Rolle’s Theorem Let f ( x ) be a polynomial having α and β as its roots such that α < β. Then, f (α ) = f (β) = 0. Also, a polynomial function is everywhere continuous and differentiable. Thus, f ( x ) satisfies all the three conditions of Rolle’s theorem. Consequently, there exists γ ∈(α, β) such that f ′ ( γ ) = 0, i.e. f ′ ( x ) = 0 at x = γ. In other words, x = γ is a root of f ′ ( x ) = 0. Thus, algebraically Rolle’s theorem can be interpreted as follows: Between any two roots of a polynomial f ( x ), there is always a root of its derivative f ′ ( x ).

Example 49. If a + b + c = 0, then the quadratic equation 3ax 2 + 2bx + c = 0 has (a) atleast one root in (0, 1) (b) both imaginary roots (c) one root in (1, 2), other in ( −1, 0) (d) None of the above

x 3 − [ x] = 3 ⇒ x 3 − 3 = [ x] Let

f( x) = x3 − 3 and

3 2

f( x) being a polynomial function is continuous and differentiable, ∀ x ∈ R

1

⇒

X′

[given]

i.e.

X

b f

f =0 c

Y′

Solving the given equations means finding the x-coordinates of the points of intersection of the curves y = f( x) and y = g ( x). If 1 < x < 2, we have g ( x) = 1 and f( x) = x3 − 3 At the point of intersection of the two curves, we have x3 − 3 = 1 ⇒ x = 41/ 3 1/ 3 Hence, x = 4 is the solution of the equation x 3 − [ x] = 3 X

Example 50. For what values of m can the expression 2x 2 + mxy + 3 y 2 − 5 y − 2 be expressed as the product of two linear factors?

abc + 2 hxy − af 2 − bg 2 − ch2 = 0  m2  25  = 0 + 2  2  4 

⇒

m2 = 49

∴

m= ±7

Some Application of Graphs to Find the Roots of Equations

X

Example 51. If x 3 − [x ] = 3, where [x ] denotes the greatest integer less than or equal to x, then (a) x ∈{41/ 3 } (b) x ∈ R (c) Cannot be discussed (d) None of the above

y=x y = cos x

O

π/2

X

Y′

Example 53. If b > a, then the equation ( x − a )( x − b) − 1 = 0 has (a) both roots in ( a, b) (b) both roots in ( −∞, a ) (c) both roots in ( b, + ∞ ) (d) one root in ( −∞, a ) and the other in ( b, ∞ ) Sol. (d)

Y y = (x – a)(x – b) – 1

Here, we shall discuss some examples to find the roots of equations with the help of graphs. X

Y

y=–x

Thus, to find value of x for which both curves have point of X′ intersections. –π/2 Clearly, there are two points of intersection of y =| x| and y = cos x. Hence, there are two real solutions.

ax2 + 2 hxy + by2 + 2 gx + 2 fy + c = 0, we have m 5 a = 2, h = , b = 3, c = − 2, f = − and g = 0 2 2 The given expression is resolvable into linear factors, if

−12 −

Example 52. The number of solutions of the equation | x | = cos x are (a) one (b) two (c) three (d) zero Sol. (b) Here,| x| = cos x

Sol. Comparing the given equation with

⇒

3

–3

− bg 2 − ch 2 = 0 h g

2

–2

ax 2 + 2hxy + by 2 + 2gx + 2 fy + c is resolvable into linear rational factors iff g

1 –1

The quadratic function

h

(4,1/3 1) X

–1

Condition for Resolution into Linear Factors

a

y = g ( x)

O –2

Thus, f( x) satisfies all conditions of Rolle’s theorem. So, there exists a point k ∈(0, 1) such that f ′(k ) = 0.

2

y = f (x)

2

f(0) = 0, f(1) = a + b + c = 0 f(0) = f(1)

∆ = abc + 2 fgh − af

g ( x) = [ x]

Y

Sol. (a) Consider the function f( x) = ax + bx + cx 3

5

Sol. (a) We have,

Inequalities and Quadratic Equation

X

α

a

b

β

X

1

From graph, it is clear that one of the roots of ( x − a) ( x − b ) − 1 = 0 lies in (−∞, a) and other lies in (b, ∞ ).

201

Objective Mathematics Vol. 1

5

Work Book Exercise 5.5 1 The number of values of k for which { x 2 − (k − 2 ) x + k 2 }{ x 2 + kx + (2 k − 1)} is a perfect square, is a c

1 0

b 2 d None of these

2 If x 2 − 4 x + log1/ 2 a = 0 does not have two distinct real roots, then maximum value of a is 1 4 −

1 16 1 4

a root in the interval b (2, 4) (−2, 0) 2

less than zero for atleast one positive x are a c

a − 3 > 0, b < 0 a, b ∈ R

b a − 3 > 0, b > 0 d Cannot be discussed

5 The set of possible values of λ for which

x 2 − (λ2 − 5λ + 5)x + (2 λ2 − 3λ − 4) = 0 has roots, whose sum and product are both less than 1, is  −1, 5     2 1, 5   2 

(1, 4)  1, 5     2

6 If ax 2 + bx + 6 = 0 does not have two distinct real roots, then the least value of 3 a + b is −2 −1

2 1

7 If x is real, then the least value of the expression x 2 − 6x + 5 is x2 + 2 x + 1 −1 −

1 3

a b c d

exactly two positive real roots exactly two negative real roots exactly two real roots None of the above

ax 2 + bx + c = 0 are real and of opposite sign. Then, the roots of the equation α ( x − β )2 + β( x − α )2 = 0 are a b c d

positive negative real and of opposite sign imaginary

12 The graph of curve x 2 = 3 x − y − 2 is 3 2 b between the lines x = 1 and x = 2 c strictly below the line 4 y = 1 d None of the above a

between the lines x = 1 and x =

x2 + 5 = x − 2 cos(a + bx ) 2 is satisfied for atleast one real x, then the greatest value of a + b is

13 If 0 < a < 5,0 < b < 5 and

π π 2 c 3π d 4π a

b −

1 2

None of these

8 If the quadratic equation α x 2 + β x + a2 + b2 + c 2 − a b − bc − ca = 0 has imaginary roots, then 2(α − β ) + ( a − b )2 + ( b − c )2 + (c − a )2 > 0 2(α − β ) + ( a − b )2 + ( b − c )2 + (c − a )2 < 0 2(α − β ) + ( a − b )2 + ( b − c )2 + (c − a )2 = 0 None of the above

202

a ( −∞, 0) 1 b  −2, −   2 1 1 c  − ,   2 4 d null set

11 The roots α and β of the quadratic equation are

4 Conditions on a and b for which x − ax − b is 2

which the roots of the equation x 2 − ax − a2 = 0 exceeds d, then S equals

10 x 4 − 4 x − 1 = 0 has

None of these

3 If c > 0 and 4 a + c < 2 b, then a x 2 − bx + c = 0 has a (0, 2) c (0, 1)

9 Let S denotes the set of real values of d for

14 If the roots of ax 2 − bx − c = 0 change by the same quantity, then the expression in a, b and c that does not change, is a b c

b 2 − 4ac a2 b − 4c a b 2 + 4ac

a2 d None of the above

WorkedOut Examples Type 1. Only One Correct Option Ex 1. If 0x 2 + x + | x | + 1 ≤ 0, then x lies in (a) ( 0, ∞ ) (c) R

(b) ( − ∞ , 0) (d) φ

Sol. x 2 + x + | x | + 1 ≤ 0 gives two cases due to | x | . Case I When x ≥ 0

…(i)

x + 2x + 1 ≤ 0 or 2

(x + 1) ≤ 0 2

⇒ x + 1= 0 ∴ x = − 1 but x ≥ 0 ∴ No value of x ≥ 0.

Case II

[from Eq. (i)] …(ii)

x2 + x − x + 1 ≤ 0

(d) ( n + 1) ( 2 c )1/ n

which is not true for any value of x < 0 ∴ No value of x < 0 From above two cases no value of x is possible i.e. x ∈ φ. Hence, (d) is the correct answer.

Sol. We have, 1 (a + a2 + ... + an −1 + 2an ) n 1 ≥ [ a1 a2 ... an −1 (2a n )]1 / n = (2c)1 / n [using AM ≥ GM]

Ex 2. If | x − 1| + | x| + | x + 1| ≥ 6, then x lies in (b) ( − ∞ , − 2] ∪ [ 2, ∞ ) (d) φ

Sol. | x − 1| + | x | + | x + 1| ≥ 6, gives four cases When x < − 1 Then, − (x + 1) − x − (x − 1) ≥ 6 ⇒ −x − 1 − x − x + 1 ≥ 6 ⇒ −3x ≥ 6 or x ≤ − 2 From Eqs. (i) and (ii), x ≤ − 2

Case I

Case II When −1 ≤ x ≤ 0 ⇒ ⇒ ∴

…(ii) …(iii)

Thus, minimum value of a1 + a2 + ... + a n −1 + 2a n is n (2c)1 / n .

Ex 4. Solution set of 3  log 3 ( x 2 − 2) < log 3  | x | − 1 is 2  (a) ( − 2 , − 1)

(b) ( − 2, − 2 )

(c) ( − 2 , 2)

(d) None of these

Sol. We have, x 2 − 2 > 0,

…(v)

⇒ x + 1 + x − (x − 1) ≥ 6 ⇒ x≥4 No solution, using Eqs. (v) and (vi).

…(vi)

Case IV When x > 1

…(vii)

⇒ x + 1+ x + x −1≥ 6 ⇒ 3x ≥ 6 or x ≥ 2 From Eqs. (vii) and (viii), x ≥ 2 Thus, from above four cases, x ≤ − 2 or x ≥ 2 ⇒ x ∈ (− ∞ , − 2 ] ∪ [ 2, ∞ )

⇒ a1 + a2 + ... + an − 1 + 2an ≥ n (2c)1 / n

Hence, (a) is the correct answer. …(i)

x + 1 − x − (x − 1) ≥ 6 …(iv) 1− x + 1≥ 6 ⇒ x ≤ −4 No value of x. [from Eqs. (iii) and (iv)]

Case III When 0 ≤ x ≤ 1

(b) ( n + 1) c1/ n (c) 2 nc1/ n

x2 + 1 ≤ 0

(a) ( − ∞ , 2] (c) R

Ex 3. If a1 , a 2 , ..., a n are positive real numbers whose product is a fixed real number c, then the minimum value of a1 + a 2 + ... + a n − 1 + 2a n is (a) n ( 2c )1/ n

When x < 0

⇒

⇒ −3x ≥ 6 or 3x ≥ 6 ⇒ x ≤ −2 or x≥2 ⇒ x ∈ (− ∞ , − 2 ] ∪ [ 2, ∞ ) Hence, (b) is the correct answer.

x2 − 2
0 2

3 |x| − 1 2

⇒

2 3 and | x |2 − | x | − 1 < 0 3 2 | x | > 2 and | x | < 2

⇒

x ∈ (−2, − 2 ) ∪ ( 2, 2)

⇒ x 2 > 2, | x | >

Hence, (d) is the correct answer.

Ex 5. Solution set of the inequality 1 1 is > x 2 − 1 1 − 2 x− 1 (a) (1, ∞ ) (c) ( −1, ∞ )

  4  (b) 0, log 2     3     4  (d) 0, log 2    ∪ (1, ∞ )  3  

203

Objective Mathematics Vol. 1

5

Sol. Put 2x = t, then t > 0. The given inequality becomes 1 2 > t −1 2 − t 1 2 − >0 t −1 2− t 2 − t − 2t + 2 >0 (t − 1) (2 − t ) 4 − 3t >0 (t − 1) (2 − t ) (3t − 4 ) >0 (t − 1) (t − 2)

⇒ ⇒ ⇒ ⇒ –

–∞

are positive, then (a) b = 8 = c (c) b = 24, c = − 32

–

+

4 — 3

∞

2

4 or t > 2 3 4 or 2x > 2 1 < 2x < 3  4 0 < x < log2   or x > 1  3

⇒ ⇒

Ex 9. If a, b and c are three positive real numbers such that a + b ≥ c, then

Hence, (d) is the correct answer.

(a) ( n + 1) 4 < 10n + 1

(b) ( n + 1) 4 > 10n + 1

(c) ( n − 1) 4 < 10n + 2

(d) None of these

Sol. We have, 4

4

1   ≤ 1 +  n

4

4

1  3  =   < 10 [Q n ≥ 2]  2  2

⇒ (n + 1)4 ≤ n4 ⋅ 10 < 10n ⋅ 10 = 10n + 1 Hence, (a) is the correct answer.

Ex 7. If x1 , x 2 , x 3 and x 4 are four positive real 1 1 numbers such that x1 + = 4, x 2 + =1, x2 x3 1 1 x 3 + = 4 and x 4 + = 1, then x4 x1 (a) x1 = x 3 and x 2 = x 4 (b) x 2 = x 4 but x1 ≠ x 3 (c) x1 x 2 = 1, x 3 x 4 ≠ 1 (d) x 3 x 4 = 1, x1 x 2 ≠ 1 Sol. Using AM ≥ GM, x1 +

204

x 1 1 ≥ 2 1 ; x2 + ≥2 x2 x2 x3 x3 x4

x2 x3

x 1 ≥2 4 x1 x1

x3 +

1 ≥2 x4

⇒

 1  1  1 1  x1 +   x2 +   x3 +   x4 +  ≥ 24 x2   x3   x4   x1  

; x4 +

a b c a b c (b) + ≥ + < 1+ a 1+ b 1+ c 1+ a 1+ b 1+ c a b c (d) None of these (c) + > 1+ a 1+ b 1+ c

(a)

Ex 6. Given, n 4 < 10 n for a fixed positive integer n ≥ 2, then

  n + 1  = 1 +    n 

x1 + x2 + x3 + x4 = 8 and x1 x2 x3 x4 = 16 As AM ≥ GM, we get 1 2 = (x1 + x2 + x3 + x4 ) ≥ (x1 x2 x3 x4 )1 / 4 = 2 4 i.e. AM = GM Thus, x1 = x2 = x3 = x4 = 2 ∴ x 4 − 8x 3 + bx 2 + cx + 16 = (x − 2)4 ⇒ b = 24 and c = − 32 Hence, (c) is the correct answer.

1< t
0 for i =1, 2, ..., n and a1 a 2 ... a n = 1, then minimum value of (1 + a1 ) (1 + a 2 ) ... (1 + a n ) is (a) 2n/ 2 (c) 22n

(b) 2n (d) 1

Sol. We have, 1 (1 + a1 ) ≥ 1⋅ a1 = a1 [using A.M ≥ G.M.] 2 1 (1 + a2 ) ≥ 1⋅ a2 = a2 2 M M M 1 (1 + an ) ≥ 1⋅ an = an 2 On multiplying above inequalities, we get 1 (1 + a1 ) (1 + a2 ) ... (1 + an ) ≥ a1 a2 ... an = 1 2n ⇒ (1 + a1 ) (1 + a2 ) ... (1 + an ) ≥ 2n Hence, (b) is the correct answer.

(b) ≥ 3 (d) ≤ 2

Similarly, a2 + b2 + c2 − bc − ca − ab ≥

Sol. As a, b and c are sides of a triangle, b + c − a, c + a − b, a + b − c > 0. Let x = b + c − a, y = c + a − b and z = a + b − c ⇒ y + z = 2a, z + x = 2b and x + y = 2c Thus, LHS of the inequality y+ z z+ x x + y = + + 2x 2y 2z =

1 y x x z y z  + + + + +  2 x y z x z y

≥

6  y x x z y z  ⋅ ⋅ ⋅ ⋅ ⋅  2  x y z x z y

1/ 6

= 3 [Q AM ≥ GM]

Hence, (b) is the correct answer.

Ex 12. If a + b + c = 6, then (a) ≤ 9

4a + 1 + 4b + 1 + 4c + 1 (b) ≥ 9 (c) > 9 (d) < 9

Sol. By the Cauchy-Schwartz’s inequality, ( 4 a + 1) +

⇒

4b + 1 +

4 c + 1) 2

≤ (1 + 1 + 1) (4 a + 1 + 4 b + 1 + 4 c + 1) = 3 [ 4 (a + b + c) + 3 ] = (3) (27) 4a + 1 + 4b + 1 + 4c + 1 ≤ 9

Hence, (a) is the correct answer.

Ex 13. If a, b and c ∈ R, then the square root of a 2 + b 2 + c 2 − bc − ca − ab is greater than or equal to 3 (a) max{| b − c|, | c − a|, | a − b|} 2 3 (b) max{| b − c|, | c − a|, | a − b|} 2 (c) max{| b − c|, | c − a|, | a − b|} 3 (d) max{| b − c|, | c − a|, | a − b|} 4 Sol. We have, a2 + b2 + c2 − bc − ca − ab

Also,

1 = [(b2 + c2 − 2bc) + (c2 + a2 − 2ca) 2 + (a2 + b2 − 2ab)] 1 = [(b − c)2 + (c − a)2 + (a − b)2 ] ≥ 0 2 3 a2 + b2 + c2 − bc − ca − ab − (b − c)2 4 1 = [ 4 a2 + 4 b2 + 4 c2 − 4 bc − 4 ca − 4 ab 4 −3 (b2 + c2 − 2bc)] 1 = [ 4 a2 + b2 + c2 + 2bc − 4 a(c + b)] 4

and a2 + b2 + c2 − bc − ca − ab ≥

3 | c − a| 2

3 | a − b| 2

a2 + b2 + c2 − ab − bc − ca 3 ≥ max{| b − c |, | c − a |, | a − b |} 2 Hence, (a) is the correct answer.

⇒

Inequalities and Quadratic Equation

(a) ≤ 3 (c) ≥ 2

5

1 [ 4 a2 + (b + c)2 − 4 a (b + c)] 4 1 = [ 2a − (b + c)]2 ≥ 0 4

=

Ex 11. If a, b and c are the sides of a triangle, then a b c + + b+c−a c+a −b a +b−c

Ex 14. If a, b and c are three distinct positive real numbers, then the number of real roots of ax 2 + 2b | x| − c = 0 is (a) 4 (b) 2 (c) 0 (d) None of the above Sol. a| x |2 + 2b| x | − c = 0 ∴

|x| =

−2b ±

= −b±

4 b2 + 4 ac 2 b2 + ac

Since, a, b and c are positive. So, | x | = − b +

b2 + ac

∴ x has two real values, neglecting | x | = − b − b2 + ac, as | x | ≥ 0 Hence, (b) is the correct answer.

Ex 15. If the ratio of the roots of λx 2 + µx + ν = 0 is equal to the ratio of the roots of x 2 + x + 1 = 0, then λ, µ and ν are in (a) AP (b) GP (c) HP (d) None of the above Sol. As ratio of roots for λ x 2 + µ x + v = 0

and

x 2 + x + 1 = 0 are equal. α α′ ∴ = , where α and β are roots of λx 2 + µx + v = 0 β β′ and α ′ , β′ are roots of x 2 + x + 1 = 0 Now, ω and ω 2 are the roots of x 2 + x + 1 = 0 α ω 1 = 2 = ⇒ β = ωα ∴ ω β ω µ ν and α + β = − , αβ = λ λ µ ν ⇒ α (1 + ω ) = − , α 2 ω = λ λ −µ 2 ν − αω 2 = ,α ω = ⇒ λ λ µ2 ν or µ 2 = λν = ⇒ λ2 λ Hence, (b) is the correct answer.

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Ex 16. If x 2 − 2r ⋅ pr x + r = 0; r =1, 2, 3 are three quadratic equations of which each pair has exactly one root common, then the number of solutions of the triplet ( p1 , p2 , p3 ) is (a) 2

(b) 1

(c) 9

Ex 18. If a ∈ R and the equation ( a − 2) ( x − [ x ]) 2 + 2( x − [ x ]) + a 2 = 0 (where, [ x ] denotes the greatest integer ≤ x) has no integral solution and has exactly one solution in (2, 3), then a lies in the interval

x − 6 p3 x + 3 = 0 has roots β , γ ⇒α + β = 2 p1 , α + γ = 4 p2 , β + γ = 6 p3 , αβ = 1, αγ = 2, βγ = 3 ∴ γ − β = 4 p2 − 2 p1 and α (γ − β ) = 1 1 α= ⇒ 4 p2 − 2 p1 β 1 Also, = ⇒ 2β = γ γ 2 2

⇒

β=±

(a) ( − 1, 2) (c) ( − 1, 0)

 1 3 ±  = 1 4 p2 − 2 p1  2 1 1 1 1 Again, + = 2 p1 , + = 2 p2 α β α γ 1 and = − p1 + p2 + p3 γ 1 1 1 ∴ + + = p1 + p2 + p3 α β γ 1 ⇒ = p1 + p2 − p3 α 1 = p1 − p2 + p3 β 1 = − p1 + p2 + p3 γ p − p2 + p3 γ So, =2= 1 − p1 + p2 + p3 β

…(i)

− p1 + p2 + p3 α =3= γ p1 + p2 − p3

p1 p2 P3 = = = t. 7 4 9 Using this in Eq. (i), we get two values of x. Hence, (a) is the correct answer.

(a − 2)(x − [ x ])2 + 2 (x − [ x ]) + a2 = 0

…(i)

can be written as f ( y) = (a − 2) y2 + 2 y + a2 = 0

…(ii)

As x cannot be an integer, y = x − [x ] ≠ 0 When 2 < x < 3, [ x ] = 2 ⇒ 0 < x − [ x ] < 1 i.e. 0 < y < 1 Since, Eq. (i) has exactly one solution in the interval (2, 3), Eq. (ii) has exactly one solution in the interval (0, 1). This is possible, if f (0) f (1) < 0 For otherwise the Eq. (ii) has either no or two solutions in (0, 1). ⇒ a2 {a − 2 + 2 + a2 } < 0 ⇒

a(a + 1) < 0

Ex 17. If f (x ) = x 2 + 2bx + 2c 2 and g ( x ) = − x 2 − 2cx + b 2 are such that min f ( x ) > max g ( x ), then relation between b and c is (c) | c| < 2 | b|

b 2 (d) | c| > 2 | b| (b) 0 < c
0]

⇒ −1 < a < 0 or a ∈ (−1, 0) Hence, (c) is the correct answer.

Ex 19. If a,b and c are all distinct, then ( x − b)( x − c) ( x − c) ( x − a ) a +b ( a − b) ( a − c) ( b − c)( b − a ) +c

Thus, we get

(a) no relation

(b) ( 0, 1) (d) ( 2, 3)

Sol. Let y = x − [ x ]. Then, equation

3 2

∴

and

| c| > 2 | b| Hence, (d) is the correct answer.

x 2 − 4 p2 x + 2 = 0 has roots α , γ

β ⋅ 2β = 3

c2 > 2b2

⇒

(d) 27

Sol. x 2 − 2 p1 x + 1 = 0 has roots α , β

⇒

Thus, min f (x ) > max g (x ) ⇒ 2c2 − b2 > b2 + c2

(a) 0 (c) a + b + c Sol. Let f (x ) = a

( x − a ) ( x − b) − x is equal to ( c − a ) ( c − b) (b) abc (d) None of these

(x − b) (x − c) (x − c) (x − a) +b (a − b) (a − c) (b − c) (b − a) (x − a) (x − b) +c −x (c − a) (c − b)

We have, f (a) = 0, f (b) = 0 and f (c) = 0, i.e. f is a trinomial which has three distinct zeroes but f (x ) is quadratic, so it could not have more than two zeroes. ⇒ f (x ) ≡ 0 Hence, (x − b) (x − c) (x − a) (x − c) (x − a) (x − b) a +c +b (a − b) (a − c) (b − a) (b − c) (c − a) (c − b) ≡ 0, ∀ x ∈ R Hence, (a) is the correct answer.

(a) a 2 + b 2

(b) a 2

(c) a 2 + b 2

(d) a a 2 + b 2

1 = t x+

∴

t−

2

1 x 2 + b2

b = 2x and t

Thus, 2(a − x )(x +

=

x 2 + b2 − x b2

(a) 15

2

b = 2 x 2 + b2 t  b2  x 2 + b2 ) =  2a − t +  (t ) t  t+

∴ Product of the roots < 0 p ( p − 1) < 0 ⇒ p ( p − 1) < 0 ⇒ ⇒ 3 Hence, (b) is the correct answer.

D ≥ 0 ⇒ q2 − 4 p ≥ 0 ⇒ q2 ≥ 4 p If p = 1, then q2 ≥ 4 p ⇒ q2 ≥ 4 ⇒ q = 2, 3, 4

If p = 4, then q2 ≥ 4 p ⇒ q2 ≥ 16 ⇒ q = 4

= a2 + b2 − (a − t )2

Thus, we see that there are 7 cases. Hence, (c) is the correct answer.

x 2 + b2  ≤ a2 + b2

Ex 24. If p and q are the roots of the equation x 2 + px + q = 0, then (a) (b) (c) (d)

Ex 21. If a > 0, a ≠1, then the equation 2 log x a + log ax a + 3 log a 2x a = 0 has

x 2 + px + q = 0 ⇒ p + q = − p and pq = q pq = q ⇒ q = 0 or p = 1 If q = 0, then p = 0 and if p = 1, then q = − 2 Hence, (a) is the correct answer.

Sol. The equation 2 logx a + logax a + 3 loga 2 x a = 0

Ex 25. If x 2 + 6x − 27 > 0 and x 2 − 3x − 4 < 0, then

where, b = log a and y = log x ⇒ 2 (b + y) (2b + y) + y (2b + y) + 3 y (b + y) = 0 ⇒ 4 b2 + 11by + 6 y2 = 0 − 11b ± 121b − 96b 4b b ,− =− 12 3 2 4 1 log x = − log a or − log a 3 2 x = a− 4 / 3 , a − 1 / 2 2

Hence, (b) is the correct answer.

Ex 22. The set of values of p for which the roots of the equation 3x 2 + 2x + p ( p − 1) = 0 are of opposite sign, is (a) ( − ∞ , 0) (c) (1, ∞ )

(b) ( 0, 1) (d) ( 0, ∞ )

(a) x > 3

…(i)

As a > 0 and a ≠ 1, log a ≠ 0, Eq. (i) can be written as 2 1 3 + + =0 y b + y 2b + y

2

p = 1, q = − 2 p = 0, q = 1 p = − 2, q = 0 p = − 2, q = 1

Sol. Since, p and q are roots of the equation

(a) exactly one real root (b) two real roots (c) no real root (d) infinite number of real roots

can be written as 2 log a log a 3 log a + + =0 log x log (ax ) log (a2 x )

5

If p = 2, then q2 ≥ 4 p ⇒ q2 ≥ 8 ⇒ q = 3, 4

= a2 + b2 − (a2 − 2at + t 2 )

a2 − b2 . 2a Hence, (a) is the correct answer.

⇒

(d) 8

Sol. For real roots, we have

attained at t = a or x =

⇒

(c) 7

If p = 3, then q2 ≥ 4 p ⇒ q2 ≥ 12 ⇒ q = 4

So, the maximum value of y is a2 + b2 . This value is

⇒ y=

(b) 9

= 2at − t 2 + b2

Therefore, y = 2 (a − x )  x +

p ∈ (0, 1)

Ex 23. If p, q ∈{1, 2, 3, 4}, then the number of equations of the form px 2 + qx + 1 = 0 having real roots is

Sol. Let t = x + x 2 + b2 ⇒

Sol. Since, the roots of the given equation are of opposite sign.

Inequalities and Quadratic Equation

Ex 20. If x is real, then the maximum value of y = 2 ( a − x ) ( x + x 2 + b 2 ) is

(c) 3 < x < 4 Sol.

(b) x < 4 7 (d) x = 2

x 2 + 6x − 27 > 0 and x 2 − 3x − 4 < 0 ⇒ (x + 9)(x − 3) > 0 and (x − 4 )(x + 1) < 0 ⇒ ∈ (−∞ , − 9) ∪ (3, ∞ ) and (−1 < x < 4 ) ⇒ 3 a]

Ex 31. The number of real solutions of the equation x 9 2   = − 3 + x − x is  10  (b) 1 (d) None of these

Ex 32. Roots of the equation ax 3 + bx 2 + cx + d = 0 remain unchanged by increasing each coefficient by one unit, then (b) a = b ≠ c = d ≠ 0 (d) a = b = c ≠ d ≠ 0

…(i)

= sin 4 θ − sin 2 θ + 1 …(ii)

From Eqs. (i) and (ii), we get 3 ≤ sin 2 θ + cos4 θ ≤ 1 4 Hence, (b) is the correct answer.

Ex 30. If 0 < a < b < c and the roots α and β of the equation ax 2 + bx + c = 0 are imaginary, then (b) | α | < 1 (d) None of these

(a) [1, ∞ ) (c) ( 0, 1] Sol. y =

2

1 3 3  =  sin 2 θ −  + ≥   2 4 4

and (a + 1)x 3 + (b + 1) x 2 + (c + 1) x + (d + 1) = 0 must be identical. a+1 b+1 c+1 d +1 = = = ⇒ a b c d ⇒ a=b=c=d ≠0 Hence, (a) is the correct answer.

Ex 33. Complete set of values of a such that x2 − x attains all real values, is 1 − ax

0 ≤ cos2 θ ≤ 1

Also, sin 2 θ + cos4 θ = sin 2 θ + (1 − sin 2 θ )2

208

2

Sol. ax 3 + bx 2 + cx + d 4

sin 2 θ + cos4 θ ≤ sin 2 θ + cos2 θ ≤ 1

(a) | α | = | β | (c) | b |< 1

4a

(a) a = b = c = d ≠ 0 (c) a ≠ b = c = d ≠ 0

cos4 θ ≤ cos2 θ

⇒

b2

coefficient of x 2 < 0 and D < 0. Thus, LHS of the given equation is always positive whereas the RHS is always less than zero. The given equation has no solution. Hence, (a) is the correct answer.

Ex 29. For real θ, value of sin θ + cos θ lies in the interval

Sol. Since,

|α | =

Sol. Let f (x ) = − 3 + x − x 2 . Then, f (x ) < 0, ∀ x, because

[where, x + 2 ≠ 0]

2

1 5  (c)  ,   4 16

Further,

(a) 0 (c) 2

Hence, (b) is the correct answer.

(a) [1, 2]

Therefore,

− b − i 4 ac − b2 2a |α | = |β | β=

and

x 2 + 5 | x | + 6 = 0 does not have any real root.

− b + i 4 ac − b2 2a

⇒

(b) ( 0, 4] (d) None of these

x2 − x ⇒ x 2 − x = y − axy 1 − ax x 2 + x (ay − 1) − y = 0

Since, x is real ⇒ (ay − 1)2 + 4 y ≥ 0 ⇒

a2 y2 + 2 y (2 − a) + 1 ≥ 0, ∀ y ∈ R

⇒

a2 > 0, 4 (2 − a)2 − 4 a2 ≤ 0

⇒

4 + a2 − 4 a − a 2 ≤ 0

⇒

a2 > 0, 4 a ≥ 4

⇒ a≥1 ⇒ a ∈ [1, ∞ ) Hence, (a) is the correct answer.

(a) a = b = c (c) a ≠ b = c

(b) a ≠ b ≠ c (d) a = b ≠ c

b c c a a b > 0, > 0, > 0, > 0, > 0, > 0, a a b b c c Thus, a, b and c all have same signs. Also, b2 ≥ ac, c2 ≥ ab, a2 ≥ bc

D = (a − 3)2 − 4 a = a2 − 10a + 9 D≥0 ⇒ (a − 9) (a − 1) ≥ 0 ⇒ a ∈ (− ∞ , 1] ∪ [ 9, ∞ ) Case I

Sol. Clearly,

These last three inequalities can hold simultaneously if and only if a = b = c. Hence, (a) is the correct answer.

Ex 35. If a ∈ R − and a ≠ − 2, then the equation x 2 + a| x| + 1 = 0 (a) cannot have any real root (b) must have exactly two real roots (c) must have either exactly two real roots or no real root (d) must have either four real roots or no real root Sol. x 2 + a| x | + 1 = 0 − a ± a2 − 4 |x| = 2 a < 0 and not equal to −2.

⇒

−a +

a −4 2 2

−a − a − 4 both are 2 2

and

positive. Thus, there are four roots, if a < − 2, else no real root. Hence, (d) is the correct answer.

Ex 36. If x 2 + ax + 1 is a factor of ax 3 + bx + c , then (b) b − a + a 2 = 0, a = c (c) b + a − a 2 = 0, a = c (d) None of the above Sol. x 2 + ax + 1 must divide ax 3 + bx + c. ax + bx + c 3

x 2 + ax + 1 = a (x − a) +

(b − a + a3 )x + c + a2 x 2 + ax + 1

⇒ (b − a + a3 )x + (c + a2 ) = 0, ∀ x ⇒

…(i)

Case II When exactly one root is positive. ⇒ a≤0 Thus, from Eqs. (i) and (ii), a ∈ (− ∞ , 0 ] ∪ [ 9, ∞ ) Hence, (c) is the correct answer.

…(ii)

Ex 38. If roots of x 2 − (a − 3)x + a = 0 are such that both of them is greater than 2, then (a) a ∈[ 7, 9] (c) a ∈[ 9, 10)

(b) a ∈ [ 7, ∞ ) (d) a ∈[ 7, 9)

Sol. x 2 − (a − 3)x + a = 0 ⇒

D = (a − 3)2 − 4 a

When both roots are greater than 2, then B D ≥ 0, f (2) > 0, − >2 2A ⇒ (a − 1) (a − 9) ≥ 0; 4 − (a − 3) 2 + a > 0; ⇒ a ∈ (− ∞ , 1] ∪ [ 9 , ∞ ); a < 10; a > 7 a ∈[ 9, 10) Hence, (c) is the correct answer.

b − a + a3 = 0 and c + a2 = 0

Ex 39. If the quadratic equation ax 2 + bx + 6 = 0 does not have real roots and b ∈ R + , then

b2  (c) a > min  , b − 6  24   b2 (d) a < min  , b − 6   24

Sol. ax 2 + bx + 6 = 0, roots are not real. ⇒

D < 0 ⇒ b2 − 24 a < 0 ⇒ a >

Hence, (d) is the correct answer.

Ex 37. If x 2 − (a − 3)x + a = 0 has atleast one positive root, then (a) a ∈ ( −∞ , 0) ∪ [ 7, 9] (b) a ∈ ( − ∞ , 0) ∪ [ 7, ∞ ) (c) a ∈ ( −∞ , 0] ∪ [ 9, ∞ ) (d) None of the above

a−3 >2 2

 b2 (a) a > max  , b − 6   24 b2  (b) a < max  , b − 6  24 

(a) b + a + a 2 = 0, a = c

Now,

⇒ ⇒ ⇒

When both roots are positive. D≥0 (a − 9) (a − 1) ≥ 0 D ≥ 0, a > 0, a > 3 a ∈ [ 9, ∞ )

= a2 − 10a + 9 = (a − 1) (a − 9)

⇒ | x |2 + a| x | + 1 = 0

Q

5

Sol. x 2 − (a − 3) x + a = 0,

Inequalities and Quadratic Equation

Ex 34. If the equations ax 2 − 2bx + c = 0, 2 2 bx − 2cx + a = 0 and cx − 2ax + b = 0 all have only positive roots, then

b2 24

i.e. a is positive. …(i) Also, f (− 1) > 0 ⇒ a−b+ 6>0 …(ii) ⇒ a>b−6  b2  a > max  , b − 6 [from Eqs. (i) and (ii)] ⇒  24  Hence, (a) is the correct answer.

209

Objective Mathematics Vol. 1

5

Ex 40. Consider the equation x 2 + 2x − n = 0, where n ∈ N and n ∈[5, 100]. Total number of different values of n so that the given equation has integral roots, is (a) 4

(b) 8

(c) 3

(d) 6

Sol. x 2 + 2x − n = 0 ⇒ (x + 1)2 = n + 1 ⇒

x = −1±

Thus, n + 1should be a perfect square. Since, n ∈[ 5, 100 ] ⇒ n + 1 ∈ [ 6, 100 ] Number of perfect squares between 1 and 100 is 10. Thus, n can take 10 − 2, i.e. 8 different values. Hence, (b) is the correct answer.

Ex 41. If log 2 (ax 2 + x + a ) ≥ 1, ∀ x ∈ R , then exhaustive set of values of a is

 5 5 (b) 1 − ,1+  2 2  

 5 (c)  0, 1 −  2 

 5  (d) 1 + , ∞ 2  

ax 2 + x + a ≥ 2, ∀ x ∈ R

⇒

ax 2 + x + (a − 2) ≥ 0, ∀ x ∈ R

⇒

x + 16 − x 2 > a > − 16 + x 2 + x 2

65  1 −  x −  > a > − 16 + x 2 + x; x ≤ 0 4  2 65  Since, x ∈ R − ⇒ a ∈  − 16,   4 Hence, (a) is the correct answer.

⇒

(b) n( x1 ) n − 1

(d) n( x1 ) n−1 + b

Sol. x n + x + b = (x − x1 ) (x − x2 ) K (x − xn ) ⇒

(x − x2 ) (x − x3 ) K (x − xn ) =

x n + ax + b (x − x1 ) x n + ax + b x→ x1 x − x1

⇒ (x1 − x2 ) (x1 − x3 ) K (x1 − xn ) = lim

= n (x1 )n −1 + a Hence, (c) is the correct answer.

Hence, (d) is the correct answer.

Ex 42. If x − x + a − 3 < 0 for atleast one negative value of x, then complete set of values of ‘a’ is 2

(a) ( − ∞ , 4 ) (b) ( − ∞ , 2) (c) ( − ∞ , 3) (d) ( − ∞ , 1)

Ex 45. If where x 2 + ( a − b) x + (1 − a − b) = 0, a, b ∈ R , then the values of a for which equation has unequal roots for all values of b, is given by (a) a < 1 (c) a ∈ R

(b) a > 1 (d) None of these

⇒

D>0 (a − b)2 − 4 (1 − a − b) > 0

⇒

b2 + b (4 − 2a) + a2 + 4 a − 4 > 0

Sol. For unequal roots,

Sol. The equation x 2 − x + a − 3 = 0 must have atleast one

210

x 2 − 16 < x − a < 16 − x 2 x − 16 − x < − a < 16 − x 2 − x 2

(c) n( x1 ) n−1 + a

 5  a ∈ 1 + , ∞ 2  

negative root. For real roots, D ≥ 0 ⇒ 1 − 4 (a − 3) ≥ 0 13  13  ⇒ a≤ ⇒ a ∈  − ∞,  4  4

⇒ ⇒

(a) nx1 + b

⇒ a > 0, 1 − 4 a (a − 2) ≤ 0 ⇒ 4 a2 − 8a − 1 ≥ 0   5 5  , ∞ ⇒ a > 0, a ∈  − ∞ , 1 −  ∪  1+ 2 2    ⇒

65  (b)  − 8,   2 (d) ( − 8, 8)

Ex 44. If x1 , x 2 , x 3 , ..., x n are the roots of then the value of x n + ax + b = 0, ( x1 − x 2 ) ( x1 − x 3 ) ( x1 − x 4 ) ... ( x1 − x n ) is

Sol. log2 (ax 2 + x + a) ≥ 1, ∀ x ∈ R ⇒

65  (a)  − 16,   4 (c) ( − 16, 16)

Sol. 16 − x 2 > | x − a|

n+1

 5 (a)  0, 1 +  2  

Ex 43. If 16 − x 2 > | x − a| is to be satisfied by atleast one negative value of x, then complete set of values of a is

...(i)

Both root will be non-negative, if D ≥ 0, a − 3 ≥ 0 13 ⇒ a≤ ,a≥3 4  13  ...(ii) a ∈ 3, ⇒  4  Thus, equation will have atleast one negative root, if 13   13   [from Eqs. (i) and (ii)] a ∈  − ∞, ∪ 3,  4   4  ⇒ a ∈ (− ∞ , 3) Hence, (c) is the correct answer.

For the above quadratic inequation to be true, ∀ b ∈ R, the discriminant of its corresponding equation should be less than zero. i.e. (4 − 2a)2 − 4 (a2 + 4 a − 4 ) < 0 ⇒ − 32a + 32 < 0 ⇒ a>1 Hence, (b) is the correct answer.

Ex 46. Let a, b and c be real numbers with a ≠ 0 and let α, β be the roots of the equation ax 2 + bx + c = 0. Then, a 3 x 2 + abcx + c 3 = 0 has roots (a) α 2β , β 2α

(b) α , β 2

(c) α 2β , βα

(d) α 3β , β 3α

2

 ax   ax  get a   + b   + c = 0  c  c ax = α , β are roots ⇒ c c c ⇒ x = α and x = β a a ⇒ x = α 2 β and αβ 2 are roots of above equation. Hence, (a) is the correct answer.

Ex 47. The values of ‘a’ for which x 2 + ax + sin −1 ( x 2 − 4x + 5) + cos −1 ( x 2 − 4x + 5) = 0 has atleast one solution, is (a) − 2 π (c) − 4

(b) − 2 + π π (d) − 2 − 4

−1

Sol. Since, sin θ is defined only when ⇒

−1≤θ ≤1 − 1 ≤ x2 − 4 x + 5 ≤ 1

⇒

x 2 − 4 x + 4 ≤ 0 and

x2 − 4 x + 6 ≥ 0

⇒

(x − 2)2 ≤ 0 and

(x − 2)2 + 2 ≥ 0

⇒ x = 2 is only solution Putting x = 2 , we get π 4 + 2a + = 0 2

π 4 Hence, (d) is the correct answer.

Ex 48. The set of values of x satisfying log | sin x | |cos x|  π + log | cos x| |sin x| = 2, when x ∈  0,  , is  2  π (a)    4

 π (b)    3

Sol. Here, log|sin x | |cos x | + ⇒

 π  π π (c)  0,  (d)  ,   4  4 2 1 log|sin x | |cos x |

=2

1 y + = 2, where y = log|sin x | |cos x | y

⇒

y2 − 2 y + 1 = 0

⇒ ∴ ⇒ ⇒

y=1 log|sin x | |cos x | = 1 |cos x | = |sin x | |tan x | = 1 π x= ⇒ 4 Hence, (a) is the correct answer.

(a) ≥

∴ For any λ ∈ Q , f (λ ) ≠ 0 ⇒ f (λ ) > 0 or f (λ ) < 0 , but f (λ ) ≠ 0 ∴

| f (λ )| > 0

or

a

p

2

q2

+b

p +c >0 q

1

| ap2 + bpq + cq2 | > 0 q2 As a, b, c, p, q ∈integers ∴ | ap2 + bpq + cq2 | ≥ 1 1 1 ⇒ | f (λ )| = 2 | ap2 + bpq + cq2 | ≥ 2 q q or

...(i)

...(ii)

5

[from Eqs. (i) and (ii)] Hence, (a) is the correct answer.

Ex 50. The equation x 2 ⋅ 2| x − 3| +4 + 2 x −1 = x 2 ⋅ 2 x +1 + 2| x − 3| +2 has (a) no solution 1 (c) x = ± and x ≥ 3 2

(b) two solutions (d) x ≥ 3

If ⇒

x−3≥0 x 2 ⋅ 2x − 3 + 4 + 2 x − 1 = x 2 ⋅ 2 x + 1 + 2 x − 3 + 2

⇒

x 2 ⋅ 2x +1 + 2x −1 = x 2 ⋅ 2x +1 + 2x −1

which is always true for x ≥ 3. Again, if x − 3 < 0 x 2 ⋅ 2− (x − 3) + 4 + 2x −1 = x 2 ⋅ 2x +1 + 2− (x − 3) + 2 ⇒

x 2 ⋅ 2 7 − x + 2x − 1 = x 2 ⋅ 2 x + 1 + 2 5 − x

⇒

x 2 ⋅ 27 − x − 25 − x = x 2 ⋅ 2x + 1 − 2x − 1

⇒

25 − x (x 2 ⋅ 22 − 1) = 2x −1 (x 2 ⋅ 22 − 1)

⇒ (25 − x − 2x −1 ) (4 x 2 − 1) = 0 1 and 5 − x = x − 1 ⇒ x2 = 4 ⇒ x = ± 1/ 2 and x = 3 1 and x ≥ 3 ∴Solution set is x = ± 2 Hence, (c) is the correct answer.

Ex 51. For any real value of θ ≠ π , the value of the cos 2 θ − 1 expression y = is cos 2 θ + cos θ

Ex 49. If a, b, c ∈ I and ax 2 + bx + c = 0 has irrational p root and λ ∈Q = , where f ( x ) = ax 2 + bx + c, q then | f ( λ )| 1

Since, coefficients are integers and one root is irrational, then both the roots are irrational.

Sol. Here, x 2 ⋅ 2|x − 3 |+4 + 2x −1 = x 2 ⋅ 2x +1 + 2|x − 3 |+2

a=−2−

∴

Sol. Let f (x ) = ax 2 + bx + c

Inequalities and Quadratic Equation

Sol. Dividing the equation, a3 x 2 + abcx + c3 = 0 by c2 , we

1

(b) < 2 q2 q (c) Cannot be discussed (d) None of these

(a) − 1≤ y ≤ 2 (c) − 1≤ y ≤ 1 Sol. Here, y =

(b) y < 0 and y > 2 (d) y ≥ 1

cos2 θ − 1

cos2 θ + cos θ ⇒ ( y − 1)cos2 θ + y cos θ + 1 = 0 ⇒

∴

cosθ = − 1 and −1
2 Hence, (b) is the correct answer. ⇒

Ex 52. If the roots of the equation x 2 + ax + b = 0 are c and d, then one of the roots of the equation x 2 + (2c + a ) x + (c 2 + ac + b 2 ) = 0 is (b) d − c

(a) c

(c) 2d

(d) 2c

Sol. Here, f (x ) = x 2 + ax + b, then = x 2 + (2c + a) x + c2 + ac + b which shows roots of f (x ) are transformed to (x − c), i.e. roots of f (x + c) = 0 are c − c and d − c. Thus, x 2 + (2c + a) x + c2 + ac + b2 = 0 has roots 0 and (d − c). Hence, (b) is the correct answer.

Ex 53. Number of integers, which satisfy the (16)1/ x inequality x+ 3 > 1, is equal to (2 )

Sol. Here,

(16)1 / x 2

x+3

⇒ 2x

4/x

−x−3

(d) 4

– –4

212

2

K − (−2)   = 0  2 K 2 − 49 = 0 ⇒ K = 7, − 7

⇒

Hence, (d) is the correct answer.

Ex 56. The number of triangles formed by the lines represented by x 3 − x 2 − x − 2 = 0 and xy 2 + 2xy + 4x − 2 y 2 − 4 y − 8 = 0 is (a) 1 (c) 3

(b) 2 (d) None of these

4 −x−3>0 x − (x + 4 ) (x − 1) or >0 x

(c) 3

(d) infinite

4

but x is non-negative. y4

8 − 6 y2

⇒

(x − 2) ( y2 + 2 y + 4 ) = 0

...(ii) ⇒ x=2 Both the equation represent same line. So, number of triangles is zero. Hence, (d) is the correct answer.

Ex 57. If x 2 − (a + b + c) x + (ab + bc + ca ) = 0 has imaginary roots, where a, b, c ∈ R + , then a , b and c

1

(b) 2 8 − 6 y2

...(i)

2

–

+ 0

y

(x − 2) (x 2 + x + 1) = 0 ⇒ x = 2

and x y + 2x y + 4 x − 2 y − 4 y − 8 = 0

>1

Ex 54. Number of non-negative integral solution of equation y 4 + 6xy 2 − 8x = 0 is equal to

⇒

 5 K  5 2(3) (−2) + 2  −  (0)   − 2  −  − 3 (0)2  2  2  2

2

⇒ x ∈ (− ∞ , − 4 ) ∪ (0, 1) Hence, (a) is the correct answer.

So,

be

⇒ 2x 2 + K x y + 3 y2 − 5 y − 2 can be expressed as (a1 x + b1 y + c1 ) (a 2 x + b2 y + c2 ) , if

⇒

⇒

(x 2 + 3x − 4 ) >0 x Using number line rule,

Sol. Here, x =

ax 2 + 2hx y + by2 + 2g x + 2 f y + c can expressed as product of two linear factors, if abc + 2 f g h − af 2 − bg 2 − ch2 = 0

Sol. As,

⇒ y2 (x − 2) + 2 y (x − 2) + 4 (x − 2) = 0

> 20

+

4 −x−3 2x

> 1 or

⇒−

(a) 1

(b) 2, 3 (d) 7, − 7

Sol. Here, x 3 − x 2 − x − 2 = 0

2x + 3 4

i.e.

(c) 1

>1 2

(a) −3, − 4 (c) 3, 4

2

f (x + c) = (x + c)2 + a (x + c) + b

(a) infinite (b) 0

Ex 55. For what values of K ∈ R the expression 2x 2 + K x y + 3 y 2 − 5 y − 2 can be expressed as ( a1 x + b1 y + c1 ) ⋅ ( a 2 x + b2 y + c2 )?

≥0

4 3 [as yis non-negative integers]

8 − 6 y2 > 0 or

y2
b ( a+

c )2 > b or

a+

Similarly, b +

c> a

⇒

c can be sides of triangle.

a , b and

and

Hence, (a) is the correct answer.

a+

c> b b> c

has

(a) {2}

(b) {0, 2}

(c) {0}

Y

(d) [0, 2]

Sol. As we know, AM ≥ GM and equality holds only when values are equal,

⇒

( 2+

2 )x + ( 2 − 2 )x ≥ (2x / 2 )1 / 2 2

( 2+

2 ) x + ( 2 − 2 ) x ≥ 2 ⋅ 2x / 4

X¢

X –1

Here, equality holds only, if ( 2+

Hence, (c) is the correct answer.

Ex 59. If c < a < b < d, then roots of the equation bx 2 + {1 − b ( c + d )} x + bcd − a = 0 are (a) real and distinct in which one lie between c and a (b) real and distinct in which one lie between c and d (c) real and distinct in which one lie between a and b (d) complex roots Sol. Here, f (x ) = bx 2 + {1 − b (c + d )} x + bcd − a

(a) Always integer (b) Always irrational (c) Always rational but never integer (d) Nothing can be said

...(i) ...(ii)

From Eqs. (i) and (ii), q = 2k which is not possible as q is prime. ∴ D is not a perfect square but positive. Hence, (b) is the correct answer.

Ex 61. Number of solutions for x − 2 − 2 [x ] = 0, (where, [ ] denotes the greatest integer function) is 2

(a) 0 (b) 1 (c) 2 (d) infinite

x = 0, not possible x2 = 4 x = ± 2, not possible x2 = 6

⇒ x=±

6 , only one possible value x= 6

i.e.

(b) ( − 5, − 7) (d) ( 5, − 7)

(a) (5, 7) (c) ( − 5, 7)

Sol. Since, two common roots must satisfy − 2x 2 + q − r = 0

where, sum of common roots = 0 If x 3 + 5x 2 + px + q = 0 and x 3 + 7x 2 + px + r = 0

If D is perfect square, then p2 + 4 q = d 2 d − p = 8k ; k ∈integer

⇒ If [ x ] = 2, then

i.e.

⇒ D = 4 p2 + 16q > 0; as p, q ∈odd prime numbers

2

⇒ If [ x ] = 0, then ⇒

(x 3 + 5x 2 + px + q) − (x 3 + 7x 2 + px + r) = 0

Sol. Here, x 2 + 2 px − 4 q = 0

4 q = d 2 − p2

∴ 2 + 2 [x ] ≥ 0 ⇒ [x ] ≥ − 1 If [ x ] = − 1, then x2 − 2 = − 2

Ex 62. The equations x 3 + 5x 2 + px + q = 0 and x 3 + 7x 2 + px + r = 0 have two roots in common. If their third roots are γ 1 and γ 2 respectively, then the ordered pair ( γ 1 , γ 2 ) is

Ex 60. If p and q are odd prime numbers, then which of the following statements about the roots of the equation x 2 + 2 px − 4q = 0 is correct?

But

3

Hence, (b) is the correct answer.

or f (x ) = b(x − c) (x − d ) + (x − a) where, f (c) = (c − a) = − ve f (d ) = (d − a) = + ve So, one root lies between c and d. Hence, (b) is the correct answer.

2

2

Y¢

2 )x = ( 2 − 2 )x , i.e. x = 0

⇒

1

5 Inequalities and Quadratic Equation

Sol. As x 2 = 2 + 2 [ x ], where LHS is always positive.

Ex 58. The solution set for ( 2 + 2 ) x + ( 2 − 2 ) x = 2 ⋅ 2 x / 4 is

have roots α , β , γ1 and α , β , γ 2 , respectively. ∴ α + β + γ1 = − 5 and α + β + γ 2 = − 7, where α+β=0 ∴ γ1 = − 5 or γ 2 = − 7 Hence, (b) is the correct answer.

Ex 63. Let P (x ) = 0 be the polynomial equation of least possible degree with rational coefficients having 3 7 + 3 49 as a root. Then, the product of all the roots of P ( x ) = 0 is (a) 56

(b) 42

(c) 343

(d) 7

Sol. If x = 7 + 49 3

3

⇒

x 3 = 7 + 49 + 3 ⋅ 3 7 ⋅ 3 49 (3 7 +

⇒

x = 56 + 21x

3

49 )

3

⇒ x 3 − 21x − 56 = 0 ; where, P (x ) is least possible degree 3. Product of all roots = 56 ∴ Hence, (a) is the correct answer.

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Ex 64. If α and β are the roots of the equation 1 x 2 + px − 2 = 0, where p ∈ R , then the 2p minimum value of α 4 + β 4 is (a) 2

(b) 2 + 2

(c) 2 2

Sol. If the given inequalities hold, then a (1 − b) b(1 − c) c(1 − a) >

(d) 2 − 2

2

=

 1 1 =  p2 + 2  − 4  p  2p

 =  p2 − 

... (ii)

Ex 67. x 4 − 4x − 1 = 0

+2 2

1   + 2+ 2 p2 

Thus, minimum value of α 4 + β 4 is 2 +

2 ≥2+

2

(a) all real roots (b) all imaginary roots (c) exactly two real roots (d) None of these Sol. Let f (x ) = x 4 − 4 x − 1 f ′ (x ) = 4 x 3 − 4 = 4 (x − 1)(x 2 + x + 1)

2.

Hence, (b) is the correct answer.

Ex 65. If α and β are roots of the equation x 2 + ax + b = 0, then maximum value of the (α − β) 2 will be expression −( x 2 + ax + b) − 4 a 2 − 4b (a) 4 (c) 0

2   1  Q  − x ≥ 0 2  

⇒ a(1 − a) b(1 − b) c(1 − c) ≤

2

1

1 1 1  −  − x ≤  4 2 4

1 64 Eqs. (i) and (ii) cannot hold simultaneously. Hence, (a) is the correct answer.

= {(α + β )2 − 2αβ}2 − 2 (αβ )2

2 p4

... (i)

For x > 0, we have x (1 − x ) = x − x 2

Sol. Here, α 4 + β 4 = (α 2 + β 2 )2 − 2α 2 β 2

= p4 +

1 64

b 2 − 4a (b) 4 (d) None of these

Sol. Here, y = − (x 2 + ax + b) D (a2 − 4 b) =− 4a 4 4 b − a2 (α − β )2 ∴ ymax = =− 4 4 ( α − β )2 ⇒ Maximum value of − (x 2 + ax + b) − =0 4 Hence, (c) is the correct answer. y is maximum i.e. ymax = −

1 1 Ex 66. If a, b, c > 0 and a (1 − b) > , b (1 − c) > and 4 4 1 c(1 − a ) > , then 4 (a) never possible (b) always true (c) Cannot be discussed (d) None of these

So, f (x ) decrease in (−∞ , 1) and increases in (1, ∞ ). But f (1) = − 4, so f (x ) = 0 has only two real roots in which one will be positive and one will be negative. (as f(0) is negative). Hence, (c) is the correct answer.

Ex 68. If an unknown polynomial is divided by (x −1) and ( x − 2), we obtain the remainder 2 and 1 respectively, then the remainder resulting from the division of this polynomial by ( x − 1) ( x − 2) is (a) 3

(b) −3

(c) 3 − x

(d) 3 − 2x

Sol. Let the polynomial be P (x ), when divided by (x − 1) leaves remainder 2 ⇒ P (1) = 2 and when P (x ) divided by (x − 2) leaves remainder 1 ⇒ P(2) = 1. When P (x ) is divided by (x − 1) (x − 2), could be given by P (x ) = {q(x )} (x − 1) (x − 2) + r(x ) where, q (x ) is quotient and r(x ) is remainder. i.e. r (x ) = ax + b ... (i) ⇒ P (x ) = (x − 1) (x − 2) q (x ) + (ax + b) ∴ P (1) ⇒ 0 + a + b = 2 and P (2) ⇒ 0 + 2a + b = 1 On solving, we get a = − 1, b = 3 Remainder = − x + 3 ∴ Hence, (c) is the correct answer.

Type 2. More than One Correct Option Ex 69. If 3 x + 2 − 9 −1/ x > 0, then the interval of x can be (a) x ∈ ( 0, ∞ ) (c) x ∈ R

2 x + 2x + 2 ⇒ >0 x x (x + 1)2 + 1 >0 x ⇒ x>0 Hence, (a) and (b) are the correct answers.

Sol. 3x + 2 > 9−1 / x ⇒ x + 2 > −

214

(b) x ∈ ( 0, 250) (d) x ∈ ( − 250, 250) 2

Ex 70. If ax 2 − bx + c = 0 has two distinct roots lying in the interval (0, 1), a, b, c ∈ N . Then, (a) log 5 abc = 1 (b) log 6 abc = 2 (c) log 5 abc = 3 (d) log 6 abc = 4

b c a a 0 < (1 − α ) ⋅ (1 − β ) < 1

Sol. α + β = , αβ = , 0 < αβ < 1 ⇒

⇒ ⇒

0 < α (1 − α ) ≤

a2 0 < c (a − b + c) < 16 a = b, c = 1 a2 > 16

⇒ a>4 ⇒ a=5 ⇒ b=5 ⇒ min (abc) = 25 Hence, (b), (c) and (d) are the correct answers.

Ex 71. If ax 2 + bx + c = 0 and cx 2 + bx + a = 0 ( a, b, c ∈ R ) have a common non-real root, then (a) | b| < 2 | a| (b) | b| < 2 | c| (c) a = ± c (d) a = c

Sol. D1 = b2 − 4 ac < 0, D2 = b2 − 4 ac < 0, as the root is non-real. ⇒ Both roots will be common. a b c ⇒ = = =1 c b a ⇒ a=c Now, b2 − 4 ac < 0 ⇒

b2 − 4 a2 ( 4 c2 ) < 0

⇒ | b| < 2 | a|(2| c |) Hence, (a), (b) and (d) are the correct answers.

Ex 72. Consider the equation x 2 + x − a = 0, a ∈ N . If equation has integral roots, then

5 Inequalities and Quadratic Equation

1 4 1 and 0 < β (1 − β ) ≤ 4 1 0 < αβ (1 − (α + β ) + αβ ) ≤ 16 c b c 1 0 < 1 − +  ≤ a a a 16 ⇒

(a) a = 2 (b) a = 6 (c) a = 12 (d) a = 20

Sol. Discriminant, D = 1 + 4 a ⇒1 + 4 a should be a perfect square. As 1 + 4 a is always odd. ⇒ 1 + 4 a = (2λ + 1)2 , λ ∈ I + ⇒ a = λ (λ + 1) Hence, (a), (b), (c) and (d) are the correct answers.

Type 3. Assertion and Reason Directions (Ex. Nos. 73-76) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 73. Statement I The quadratic equation ( a − b) x 2 + ( b − c) x + ( c − a ) = 0 has one root x =1. Statement II If sum of the coefficients in a quadratic equation vanishes, then its one root is x =1. Sol. Statement I a − b + b − c + c − a = 0 If quadratic equation ax 2 + bx + c = 0 has one root x = 1, then a + b + c = 0 Sum of coefficients = 0 Hence, (a) is the correct answer.

Ex 74. Statement I The equation (x − p) (x − r ) + λ ( x − q ) ( x − s) = 0, p < q < r < s has non-real roots, if λ > 0.

Statement II The equation ax 2 + bx + c = 0, a, b, c ∈ R has non-real roots, if b 2 − 4ac < 0. Sol. Statement I Let f (x ) = (x − p) (x − r) + λ (x − q) (x − s) If λ > 0, then f ( p) > 0, f (q) < 0, f (r) < 0 and f (s) > 0 f (x ) = 0 has one real root between p and q and other real root between r and s. Statement II Obviously true. Hence, (d) is the correct answer.

Ex 75. Statement I If roots of the equation x 2 − bx + c = 0 are two consecutive integers, then b 2 − 4c < 1. Statement II If a, b and c are odd integers, then the roots of the equation 4abc x 2 + ( b 2 − 4ac) x − b = 0 are real and distinct. Sol. Statement I Given equation x 2 − bx + c = 0 Let α and β be two roots such that |α − β | = 1. ⇒

(α + β )2 − 4αβ = 1 ⇒ b2 − 4 c = 1

Statement II Given equation 4 abc x 2 + (b2 − 4 a) x − b = 0 ∴ D = (b2 − 4 ac)2 + 16ab2 c ⇒ D = (b2 + 4 ac)2 > 0 So, roots are real and unequal. Hence, (d) is the correct answer.

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Ex 76. Statement I If one root is 5 − 2, then the equation of lowest degree with rational coefficient is x 4 − 14x 2 + 9 = 0. Statement II For a polynomial equation with rational coefficient irrational roots occur in pairs. Sol. Given a root of the equation is x= 5− 2 On squaring both sides, we get x 2 = 5 + 2 − 2 10 ⇒ 7 − x 2 = 2 10 Again squaring both sides, we get 49 + x 4 − 14 x 2 = 40 ⇒ x 4 − 14 x 2 + 9 = 0 For polynomial equation with rational coefficients irrational roots occurs in pairs. Hence, (a) is the correct answer.

Ex 77. Statement I The number of values of a for which ( a 2 − 3a + 2) x 2 + ( a 2 − 5a + 6) x + a 2 − 4 = 0 is an identity in x, is 2. Statement II If a = b = c = 0, then equation ax 2 + bx + c = 0 is an identity in x. Sol. Statement I a2 − 3a + 2 = 0 ⇒

a = 1, 2 a2 − 5a + 6 = 0

⇒

a = 2, 3 a −4=0 2

⇒ a=± 2 a = 2 is the only solution. Hence, Statement I is false. Statement II is true by definition. Hence, (d) is the correct answer.

Type 4. Linked Comprehension Based Questions Passage I (Ex. Nos. 78-80) Let f ( x ) = x 2 + b1x + c1, g( x ) = x 2 + b2 x + c2 , real roots of f ( x ) = 0 be α, β and real roots of g( x ) = 0 be α + δ, β + δ. Least value of f ( x ) 1 7 be − . Least value of g( x ) occurs at x = . 4 2

Ex 78. The least value of g (x ) is

(c) −

1 4

(b) −7 (d) 0

Ex 80. The roots of g (x ) = 0 are (b) −3, 4 (d) −3, − 4

(a) 3, 4 (c) 3, − 4 Sol. (Q. Nos. 78-80)

(β − α ) = (β + δ ) − (α + δ )(β + α )2 − 4αβ = [(β + δ ) + (α + δ )]2 − 4 (β + δ ) (α + δ ) (− b1 )2 − 4 c1 = (− b2 )2 − 4 c2 ...(i) D1 = D2 D1 1 Least value of f (x ) is − = − ⇒ D1 = 1 4 4 and D2 = 1 D 1 Least value of g (x )is − 2 = − ∴ 4 4 b2 7 Least value of g (x ) occurs at x = − = ⇒ b2 = − 7 2 2 2 b2 − 4 c2 = D2 ⇒ 49 − 4 c2 = 1 48 ⇒ = c2 ⇒ c2 = 12 4 2 x − 7x + 12 = 0 ⇒ x = 3, 4

216

78. (c)

79. (b)

Ex 81. Value of b is (a) − 54

(b) 54

(c) 27

(d) − 27

(a) 108

(b) −108

(c) 54

(d) −54

Ex 83. Root of equation 2bx + c = 0 is (a) −1/ 2

(b) 1/ 2

(c) 1

(d) − 1

Sol. (Q. Nos. 81-83)

Ex 79. The value of b2 is (a) 6 (c) 8

x 4 − 12 x 3 + bx 2 + cx + 81 = 0 are positive.

Ex 82. Value of c is

1 (b) − 2 1 (d) − 3

(a) −1

Passage II (Ex. Nos. 81-83) If roots of the equation

80. (a)

Let roots of x 4 − 12x 3 + bx 2 + cx + 81 = 0 α , β , γ , δ. ⇒ α + β + γ + δ = 12and αβγδ = 81 α+β+γ+δ ≥ (αβγδ )1 / 4 4 α+β+γ+δ Since, = 3 and (αβγδ )1 / 4 = 3 4 ∴ α =β = γ =δ ⇒ α =β = γ =δ = 3 Now, b = ∑ αβ = 6 × 9 = 54 c = − ∑ αβγ = − 4 × 27 = − 108 ⇒ 2b + c = 0 ∴ 1 is a root of equation 2bx + c = 0.

81. (b)

82. (b)

be

83. (c)

Passage III (Ex. Nos. 84-86) In the given figure vertices of ∆ABC lie on y = f ( x ) = ax 2 + bx + c. The ∆ABC is right angled isosceles triangle whose hypotaneous AC = 4 2 units, then Y

A

O

B

y = f(x)= ax2 + bx + c

C

X

(a) y =

x

2

(a) 2 2

x −2 2 (d) y = x 2 − 2 2

−2 2

(b) y =

2 2 (c) y = x 2 − 8

∴

AB = BC =

(c) 2

(d) −2

x2 at x = 0 is −2 2. 2 2 Hence, (b) is the correct answer.

Ex 86. Number of integral value of k for which k /2 lies between the roots of f ( x ) = 0, is

4 2 = 4 units 2

(a) 9

and OB = 4 2 − (2 2 )2 = 2 2

(b) 10

(c) 11

(d) 12

Sol.Q Roots of f (x ) = 0

A (− 2 2 , 0), B (2 2 , 0), C (0, − 2 2 )

∴

(b) −2 2

Sol. Minimum value of y =

AC = 4 2

Sol. Q

5

Ex 85. Minimum value of y = f (x ) is 2

x2 −2 2=0 ⇒ x=±2 2 2 2

Q y = ax 2 + bx + c passes through A , B and C, we get

i.e.

x2 −2 2 2 2 Hence, (a) is the correct answer.

∴ Number of integral value, of k for which k /2 lies in (−2 2 , 2 2 ) is 11. Hence, (c) is the correct answer.

y=

Inequalities and Quadratic Equation

Ex 84. y = f (x ) is given by

Type 5. Match the Columns Ex 87. Match the statements of Column I with values of Column II. Column I A.

B.

C.

D.

The equation x 3 − 6 x 2 + 9 x + λ = 0 has exactly one root in (1, 3), then [λ + 1] is (where, [ ] denotes the greatest integer function) If − 3
2 is x (b) [ − 1, 0) (d) None of these

(a) (0, 3 ] (c) (− 1, 0) ∪ (0, 3)

17. Solution of | x 2 − 10 | ≤ 6 is (b) (− 4 , − 2) (d) [ − 4 , − 2 ] ∪ [ 2, 4 ]

1 > 2 is x

(a) R − {0} (c) R − {1}

19. The solution set of

(b) R − {− 1, 0, 1} (d) R − {− 1, 1}

( x + 1) 2 x+1 is + | x + 1| = | x| x

(a) {x | x ≥ 0} (c) {− 1, 1}

(b) {x | x > 0} ∪ {− 1} (d) {x | x ≥ 1 or x ≤ − 1}

20. The set of real values of x satisfying | x − 1 | ≤ 3 and | x − 1 | ≥ 1is (a) [ 2, 4 ] (c) [ − 2, 0 ] ∪ [ 2, 4 ]

(b) (− ∞ , 2 ] ∪ [ 4 , ∞ ) (d) None of these

21. Solution of the inequality x > (1 − x ) is given by

10. Solution of | 3 − x | = x − 3 is (a) x < 3

13. Solution of | x − 1 | ≥ | x − 3 | is

18. Solution of x +

7. Solution of ( 2x + 1) ( x − 3) ( x + 7) < 0 is (a) (− ∞ , − 7) ∪ (− 1/ 2, 3) (c) (− ∞ , 7) ∪ (− 1/ 2, 3)

(a) (− 4 / 9, − 2/ 9) (b) [ − 4 / 9, − 2/ 9 ] (c) (− 4 / 9, − 2/ 9) − {− 1/ 3} (d) [ − 4 / 9, − 2/ 9 ] − {− 1/ 3}

(a) (2, 4) (c) (− 4 , − 2) ∪ (2, 4 )

(b) (− ∞ , − 4 ) (d) (1, 4)

1 is 3

Targ e t E x e rc is e s

1. The least integer satisfying

(b) (1/ 3, 5) (d) (− ∞ , 1 / 3) ∪ (5, ∞ )

(a) ] − ∞ , (− 1 − 5 )/ 2 [ (b) ]( 5 − 1) / 2, ∞[ (c) ] − ∞ , ( − 1 − 5 ) / 2 [ ∪ ] ( 5 − 1) / 2, ∞[ (d) ]( 5 − 1) / 2, 1 [

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22. The product of all the solutions of the equation |( x − 2)|2 − 3 | x − 2 | + 2 = 0 is (b) − 4 (d) None of these

(a) 2 (c) 0

23. Which of the following statement(s) is/are correct? (a) a > b ⇒ ax > bx , x ≠ 0, (a, b, x ∈ R ) (b) | x | > | y | ⇒ x > y (x , y ∈ R ) 1 1 (c) a > b ⇒ > a b (d) a > b ⇒ a + c > b + c (b) (− ∞ , − 2) ∪ [ 4 , ∞ ) (d) None of these

1 2a + 1

1 2b + 1 1 (d) 2ab − 1 (b)

(c) 2ab + 1

Ta rg e t E x e rc is e s

(b) 0 (d) None of these

28. The solution set of log 2 | 4 − 5x | > 2 is  8 (a)  , + ∞  5

 4 8 (b)  ,   5 5

8  (c) (− ∞ , 0) ∪  , + ∞ 5 

(d) None of these

5  (a)  − ∞ , −  ∪ (0, + ∞ )  2 (c) (− ∞ , − 2) ∪ (0, + ∞ )

31. x

log x a × log a y × log y z

(a) x (c) z

x+2 ≤ 1is x

5  (b) , + ∞  2  (d) None of these

(b) 1 (d) None of these

is equal to (b) y (d) None of these

32. The number of zeroes coming immediately after the decimal point in the value of ( 5) 25 is (given, log 10 2 = 0.30103)

220

(a) 16 (b) 17 (c) 18 (d) None of the above

(a) [ − 1, 1] (c) (− ∞ , − 1]

(b) [1, + ∞ ) (d) (− ∞ , − 1] ∪ [1, + ∞ )

36. If x n > x n−1 > K > x 2 > x1 > 1, then the value of

37. 4 sin

2

(b) 1 (d) None of these x

+ 4 cos

2

x

is equal to

(a) ≤ 4 (c) ≤ 2

(b) ≥ 4 (d) ≥ 2

38. If y = 3x − 1 + 3− x −1 , then the least value of y is (a) 2 2 (c) 3

(b) 6 3 2

(d)

b+c c+a a+b (for real + + a b c positive numbers a, b, c) is

39. Minimum value of

30. The number of real values of the parameter k for which (log 16 x ) 2 − log 16 x + log 16 k = 0, with real coefficients will have exactly one solution, is (a) 2 (c) 4

(b) [ − 1, 2 ] (d) (− ∞ , + ∞ )

35. Let f ( x ) = log 10 x 2 . The set of all values of x for which f ( x ) is real, is

(a) 0 (c) 2

27. The number of solutions of log 2 ( x + 5) = 6 − x is

29. The set of real values of x for which log 0. 2

(d) None of the above

xKx1

(b) a2 (d) None of these

(a) 2 (c) 3

(c) x x ⋅ y y ⋅ z z = 1

log x1 log x2 log x3 ... log xn x nn− 1 is

26. If log 5 a ⋅ log a x = 2, then x is equal to (a) 125 (c) 25

(b) x x ⋅ y y ⋅ z z = 1

(a) (− ∞ , − 2) ∪ (1, + ∞ ) (c) (− 2, 1)

25. If log 4 5 = a and log 5 6 = b, then log 3 2 is equal to (a)

(a) x y ⋅ y z ⋅ z x = 1

34. The solution set of the inequation log 1/ 3 ( x 2 + x + 1) + 1> 0 is

24. Solution of | x − 1 | + | x − 2 | + | x − 3 | ≥ 6 is (a) [0, 4] (c) (− ∞ , 0 ] ∪ [ 4 , ∞ )

33. If log x log y log z = ( y − z ) ( z − x ) ( x − y ) , then

(a) 1 (c) 4

(b) 2 (d) 6

40. If a, b and c are different positive real numbers such that b + c − a, c + a − b and a + b − c are positive, then ( b + c − a ) ( c + a − b ) ( a + b − c ) − abc is (a) positive (c) non-positive

(b) negative (d) non-negative

41. If the product of n positive numbers is unity, then their sum is (a) a negative integer (b) divisible by n (c) never less than n 1 (d) equal to n + n

42. For positive numbers a, b and c, the least value of 1 1  1 ( a 2 + b 2 + c 2 )  2 + 2 + 2  is a b c  (a) 3 (b) 9 27 (c) 4 (d) None of the above

(a) a + b + c ≥ bc + ca + ab (b) (b + c) (c + a) (a + b) ≤ 8abc a b c (c) + + ≤ 3 b c a (d) a3 + b3 + c3 ≥ abc 2

2

(a) four real zeroes (b) atleast two real zeroes (c) atmost two real zeroes (d) no real zeroes

44. For positive real numbers a, b and c, which of the following holds? (a) a + b + c > 3 ⇒ a2 + b2 + c2 > 3

(a) rational (c) irrational

(b) a + b ≤ 12a b − 64 1 1 1 9 (c) a + b + c = α ⇒ + + ≤ a b c α (d) None of the above 6

6

2 2

8 3 p 27 (b) ( p − a) ( p − b) ( p − c) ≥ 8abc bc ca ab (c) + + ≥p a b c (d) None of the above (a) ( p − a) ( p − b) ( p − c) ≥

(b) 6abc (d) 4abc

of

P = a2b3c4,

(a) P = 4 4 ⋅ 64 ⋅ 84

(b) P = 4 2 ⋅ 63 ⋅ 84

(c) P = 32 ⋅ 63 ⋅ 84

(d) P = 4 2 ⋅ 62 ⋅ 82

when

 n  (a) n ∑ xi2 <  ∑ xi    i = 1  i=1 n

(c)

∑

xi2 i=1

2

 n  ≥ n  ∑ xi    i = 1 

 n  (b) n ∑ xi2 ≥  ∑ xi    i = 1  i=1 n

2

(d) None of these

(a) (2 − x ) (2 − y) (2 − z) ≥ 8xyz (b) x −1 + y−1 + z−1 ≥ 1/ 2 (c) (2 − x ) (2 − y) (2 − z) < 8xyz (d) None of the above

1 is a root of ax 2 + bx + 1 = 0, where a and b 4 − 3i are real, then

(a) a = 25, b = − 8 (c) a = 5, b = 4

(a) real and different (c) complex

50. For non-negative real numbers such a1 + a 2 + K + a n = p and q = ∑ a i a j , then 1 2 p 4 p2 (d) q > 2

(b) q >

(b) a = 25, b = 8 (d) None of these

that

(b) real and identical (d) None of these

57. The number of positive integral values of k for which (16 x 2 + 12 x + 39) + k ( 9x 2 − 2 x + 11) is a perfect square, is (b) zero (d) None of these

58. For the equation 3x 2 + px + 3 = 0, p > 0, if one of the roots is the square of the other, then p is equal to (a)

1 3

(c) 3

i< j

(b) p > 4 (d) 0 ≤ p < 4

55. If

(a) two (c) one

2

49. If x, y and z are positive real numbers such that x + y + z = 2, then

1 (a) q ≤ p2 2 p (c) q < 2

(a) p < − 1or p > 4 (c) − 1 < p < 4

56. If a ∈ R , b ∈ R, then the factors of the expression a ( x 2 − y 2 ) − bxy are

48. If x1 , x 2 , ... , x n are any real numbers and n is any positive integer, then n

(a) all rational values of a except a = − 2 (b) all real values of a except a = − 2 1 (c) rational values of a > 2 (d) None of the above

54. If the absolute value of the difference of roots of the equation x 2 + px + 1 = 0 exceeds 3 p , then

46. The minimum value of P = bcx + cay + abz , when xyz = abc, is

47. The greatest value a + b + c = 18, is

(b) imaginary (d) None of these

53. The equation ( a + 2) x 2 + ( a − 3) x = 2a − 1, a ≠ − 2 has roots rational for

45. For positive real numbers a, b and c, such that a + b + c = p, which one holds?

(a) 3abc (c) abc

52. If a, b and c are non-zero, unequal rational numbers, then the roots of the equation abc 2 x 2 + ( 3a 2 + b 2 ) cx − 6a 2 − ab + 2b 2 = 0 are

5

Targ e t E x e rc is e s

2

51. The polynomial ( ax 2 + bx + c ) ( ax 2 − dx − c ), ac ≠ 0, has

Inequalities and Quadratic Equation

43. For positive real numbers a, b and c, which one of the following holds ?

(b) 1 (d)

2 3

59. Let α and β be the roots of the quadratic equation x 2 + px + p 3 = 0 ( p ≠ 0). If (α , β ) is a point on the parabola y 2 = x, then the roots of the quadratic equation are (a) 4 , − 2 (b) − 4 , − 2 (c) 4, 2 (d) − 4 , 2

221

Objective Mathematics Vol. 1

5

60. If a, b and c are positive real numbers, then the number of real roots of the equation ax 2 + b | x | + c = 0 is (a) 2 (c) 0

(b) 4 (d) None of these

q 2 − 4 pr  p (a)    a

is equal to

2

(b) 1

 a (c)    p

2

(b) b2 (m + n) = mn

(c) m + n = b mn

(d) c (mn) = ab (m + n) 2

Ta rg e t E x e rc is e s

(a) 0 (c) − 5

1  x

(b) 5 (d) 10

64. The value of 6 + 6 + 6 + K ∞ is (a) 3 (c) − 2

(b) 6 (d) − 4

65. The sum of the roots of the equation x 2 + px + q = 0 is equal to the sum of their squares, then (a) p2 − q2 = 0

(b) p2 + q2 = 2q

(c) p2 + p = 2q

(d) None of these

66. The roots of the equation ( q − r ) x 2 + ( r − p ) x + ( p − q ) = 0 are r− p ,1 q−r q−r (c) ,1 p−q (a)

p−q ,1 q−r r− p (d) ,1 p−q

(b)

67. A lad was asked his age by his friend. The lad said, ‘The number you get when you subtract 25 times my age from twice the square of my age will be thrice your age.’ If the friend’s age is 14, then age of the lad is (a) 21 (c) 14

(b) 28 (d) 25 2

x − bx λ − 1 are such = λ +1 ax − c that α + β = 0, then the value of λ is

68. If the roots of the equation a−b a+ b 1 (c) c (a)

222

(b) c (d)

a+ b a−b

p −7 0} ∪ {− 1} (d) {x | x ≥ 1 or x ≤ − 1}

94. { x ∈ R : | x − 2 | = x 2 } is equal to (a) [ − 1, 2 ] (c) [ − 1, − 2 ]

(b) [1, 2 ] (d) {− 2, 1}

95. Number of solutions | x |2 − 3 | x | + 2 = 0 will be (a) 4

(b) 1

of

an

(c) 3

equation (d) 2

96. In the equation 4 x + 2 = 2x + 3 + 48 , the value of x will be (b) − 2 (d) 1

97. The real values of x which satisfy the equation (5 + 2 6 )x

2

−3

+ (5 − 2 6 )x

2

−3

= 10 are

(b) ±

(a) ± 2

(a) α −1 , β −1

(c) {2, 4}

92. If the roots of the equation

(c) ± 2, ±

87. If α and β are the roots of ax 2 + c = bx, then the equation ( a + cy ) 2 = b 2 y in y has the roots (c) α β , α β (d) α , β

(b) {2, 3}

(a) −

(a) AP (b) GP (c) HP (d) None of the above

−2

(a) {1, 2}

3 2 (c) − 3

(c) (bc′ − b′ c)2 = (ab′ − a ′ b) (ac′ − a ′ c) (d) None of the above

−1

91. If 9x − 4 × 3x + 2 + 35 = 0, then the solution set is

(a) {x | x ≥ 0} (c) {− 1, 1}

84. A car travels 25 km an hour faster than a bus for a journey of 500 km. If the bus takes 10 h more than the car, then the speeds of the bus and the car are

86. If a ( b − c ) x + b ( c − a ) xy + c ( a − b ) y perfect square, then a, b and c are in

(b) 4 , − 5 (d) 0, 0

(d) None of the above

83. If sin θ and cos θ are the roots of the equation ax 2 + bx + c, then

2

90. If x 2 − 3x + 2 is a factor of x 4 − px 2 + q, then p and q are

(c) product of roots =

(b) GP (d) None of these

(a) 25 km/h, 40 km/h (c) 25 km/ h, 60 km/h

(b) 4 (d) 8

(a) 2, 3 (c) 5, 4

81. The equation x + 3 − 4 x − 1 + x + 8 − 6 x − 1 = 1 has

−1

(b) GP (d) None of these

5 Inequalities and Quadratic Equation

(a)

q + p

88. If the sum of the roots of the quadratic equation ax 2 + bx + c = 0 is equal to the sum of squares of a b c their reciprocals, then , , are in c a b

Targ e t E x e rc is e s

79. If the ratio of the roots of l x 2 + nx + n = 0 is p : q, then

2

2

(d) 2, 2

98. The equation 125x + 45x = 2⋅ ( 27) x has (a) no solution (b) one solution (c) two solutions (d) more than two solutions

223

Objective Mathematics Vol. 1

5

99. If

α and β

are

the

roots

of

the

equation

x − px + q = 0 and α > 0, β > 0, then the value of 2

α 1/ 4 + β 1/ 4 is( p + 6 q + 4q1/ 4

p + 2 q ) k , where

k is equal to (a) 1

(b)

1 2

(c)

1 3

(d)

1 4

100. The solutions of the equation ( 3 | x | − 3) 2 = | x | + 7 which belongs to the domain of definition of the function y = x ( x − 3), are given by 1 (a) , − 2 9

(b) −

1 ,2 9

(c) ±

1 1 , ± 2 (d) − , − 2 9 9 3x

101. The roots of the equation 2x + 2 ⋅ 3 x −1 = 9 are given by  2 (a) log2   , − 2  3

(b) 3, − 3

(c) − 2, 1 − log2 3

(d) 1 − log2 3, 2

Ta rg e t E x e rc is e s

(b) 1

(c) 2

(d) infinite

103. The number of real solutions of the equation 271/ x + 121/ x = 2⋅ 81/ x is (a) 1

(b) 2

(c) infinite

(d) 0

104. If α and β (α < β ) are the roots of the equation x 2 + bx + c = 0, where c < 0 < b, then (a) 0 < α < β (c) α < β < 0

(b) α < 0 < β < | α | (d) α < 0 < |α | < β

105. If x 2 − 2x + sin 2 α = 0, then (a) x ∈ [ − 1, 1] (c) x ∈ [ − 2, 2 ]

106. The set of values of m for which the roots of the equation x 2 − ( m + 1) x + m + 4 = 0 are real and negative consists of all m such that (b) − 4 < m ≤ − 3 (d) − 3 ≥ m or m ≥ 5

107. If a ≤ 0 , then the roots of the equation x 2 − 2a | x − a | − 3a 2 = 0 are (a) (1 − 2 ) a, ( 6 − 1) a (b) (− 1 + (c) (1 +

2) a

2 ) a, 0

(d) − (1 ±

108. For

6 ) a, (1 + 6 ) a, (1 ±

a ≥ 0,

the

of

the

x 2 − 2a | x − a | − 3a 2 = 0 are given by 224

(a) − a ( 6 + 1)

(b) a ( 6 + 1)

(c) a (1 +

(d) a (1 +

2 ), a (− 1 − 6 )

2)

which

(b) 0 (d) None of these

110. If [ x] 2 = [ x + 2], where [ x] is the greatest integer less than or equal to x, then x must be such that (a) x = 2, − 1 (c) x ∈ [ − 1, 0)

(b) [ − 1, 0) ∪ [ 2, 3) (d) None of these

111. Number of real roots e sin x − e − sin x − 4 = 0 is (a) 1 (c) 2

of

the

equation

(b) 0 (d) None of these

112. The roots of the equation x

x

= x x are

(b) 0 and 1 (d) 1 and 4

113. The quadratic equation 8sec 2 θ − 6sec θ + 1 = 0 has (a) infinitely many roots (c) exactly four roots

(b) exactly two roots (d) no roots

10 = log 2 y + log y 2 and x ≠ y, 3 then x + y 3 is equal to

114. If log 2 x + log x 2 =

(a) 10 (c)

(b)

37 6

65 8

(d) None of these

115. The number of solutions sin ( e x ) = 5x + 5− x is (a) 0 (c) 2

of

the

equation

(b) 1 (d) infinitely many

(a) always an even integer (b) always an odd integer (c) always irrational (d) Can’t be discussed

117. The constant n  1  ∑  x − k − 1 k =2

term of the quadratic expression 1   x −  , as n → ∞ is  k

(a) − 1 (b) 0 (c) 1 (d) None of the above

118. The system y (x

2) a

roots

for

116. Let D = a 2 + b 2 + c 2 , where a, b being consecutive integers and c = ab, then D is

(b) x ∈[ 0, 2 ] (d) x ∈[1, 2 ]

(a) − 3 < m ≤ − 1 (c) − 3 ≤ m ≤ 5

(a) 1 (c) 2

(a) 0 and 4 (c) 0, 1 and 4

102. If f ( x ) = x − [ x], x ( ≠ 0) ∈ R, where [ x] is greatest integer less than or equal to x, then the number of  1 solutions of f ( x ) + f   = 1is  x (a) 0

109. The value of x x/ 2 x x/ 2 2 + ( 2 + 1) = ( 5 + 2 2 ) , is

equation

2

+ 7 x +12)

(a) no solution (c) two solutions

= 1 and x + y = 6, y > 0 has (b) one solution (d) more than two solutions

119. The curve y = ( λ + 1) x 2 + 2 intersects the curve y = λx + 3 in exactly one point, if λ equals (a) {− 2, 2}

(b) {1}

(c) {− 2}

(d) {2}

(b) a = ± b = ± c (d) None of these

121. Sum the non-real roots of ( x 2 + x − 2)( x 2 + x − 3) = 12 is (a) − 1

(c) − 6

(b) 1

(d) 6

122. The number of irrational roots of the equation 4x / ( x 2 + x + 3) + 5x / ( x 2 − 5x + 3) = − 3 / 2 is (a) 4

(b) 0

(c) 1

(d) 2

123. If α and β are the roots of the equation x − 2x + 3 = 0. Then, the equation whose roots are P = α 3 − 3α 2 + 5α − 2 and Q = β 3 − β 2 + β + 5, is 2

(a) x 2 + 3x + 2 = 0

(b) x 2 − 3x − 2 = 0

(c) x 2 − 3x + 2 = 0

(d) None of these

+ 1 = ( 32 + 8 15 ) x

(a) 3 (c) 2

126. If

2

−3

is

(b) 0 (d) None of these

α and β

are

the

roots

(b)

of

the

equation

b2 − 4 ac 2a

(d) None of these

127. If α and β are the roots of the equation x 2 + px + q = 0 and α 4 , β 4 are the roots of x 2 − rx + q = 0, then the roots of x 2 − 4qx + 2q 2 − r = 0 are always (a) both non-real (c) both negative

(b) both positive (d) opposite in sign

128. Suppose A , B and C are defined as A = a 2 b + ab 2 − a 2 c − ac 2 , B = b 2 c + bc 2 − a 2 b − ab 2 and C = a 2 c + ac 2 − b 2 c − bc 2 , where a > b > c > 0 and the equation Ax 2 + Bx + C = 0 has equal roots, then a, b and c are in (a) AP

(b) GP

(b) 0 (d) 2

(c) HP

(d) AGP

129. If α and β are the roots of ax 2 + c = bx, then the equation ( a + cy ) 2 = b 2 y in y has the roots (a) αβ −1 , α −1β

(b) α −2 , β −2

(c) α −1 , β −1

(d) α 2 , β 2

(b) 0 (d) None of these

x, y, z and t

are

real

numbers

x + y = 9, z + t = 4 and xt − yz = 6. Then, the greatest value of P = xz is 2

2

2

2

(b) 3

(c) 4

(d) 6

135. If x + ax − 3x − ( a + 2) = 0 has real and distinct roots, then minimum value of ( a 2 + 1) / ( a 2 + 2) is 2

( aα 2 + c ) / ( aα + b ) + ( aβ 2 + c ) /( aβ + b ) is b (b2 − 2ac) 4a b (b2 − 2ac) (c) a2c

(b) negative (d) non-real

132. The number of integral values of a for which the quadratic equation ( x + a ) ( x + 1991) + 1 = 0 has integral roots, is

(a) 2

ax 2 + bx + c = 0, then the value of

(a)

(a) real and of opposite sign (c) positive

134. If

125. The sum of values of x satisfying the equation −3

2

131. If x1 and x 2 are the roots of ax 2 + bx + c = 0 and then the roots of x1 x 2 < 0, x1 ( x − x 2 ) 2 + x 2 ( x − x1 ) 2 = 0 are

(a) 2 (c) 1

(b) 1 (d) None of these 2

(d) 2 +

133. The number of real solutions of the equation ( 9 / 10) x = − 3 + x − x 2 is

2x 2 − 35x + 2 = 0, then the value of ( 2α − 35) 3 ( 2 β − 35) 3 is

( 31 + 8 15 ) x

(b) 2 − 2

(c) 2

(a) 3 (c) 1

124. If α and β are the roots of the equation

(a) 8 (c) 64

(a) 2 2

5

(a) 1 1 (c) 2

(b) 0 1 (d) 4

Targ e t E x e rc is e s

(a) a = b = c (c) a = b ≠ c

130. If α and β are the roots of the equation x 2 + px − 1/ ( 2 p 2 ) = 0, where p ∈ R. Then, the minimum value of α 4 + β 4 is

Inequalities and Quadratic Equation

120. If the expression x 2 + 2 ( a + b + c ) x + 3 ( bc + ca + ab ) is a perfect square, then

136. The value of the expression x 4 − 8x 3 + 18x 2 − 8x + 2, when x = 2 + 3, is (a) 2 n

(b) 1

(c) 0

(d) 3

n

137. x 3 + y 3 is divisible by x + y, if (a) n is any integer ≥ 0 (b) n is an odd positive integer (c) n is an even positive integer (d) n is a rational number

138. If z 0 = α + iβ, i = −1, then the roots of the cubic equation x 3 − 2 (1 + α ) x 2 + ( 4α + α 2 + β 2 ) x 2 2 − 2 (α + β ) = 0 are (a) 2, z0 , z0 (c) 2, z0 , − z0

(b) 1, z0 , − z0 (d) 2, − z0 , z0

139. If α , β and γ are the roots of x 3 + 64 = 0, then the 2

2

 α  α equation whose roots are   and   , is γ β (a) x 2 − 4 x + 16 = 0

(b) x 2 + x + 1 = 0

(c) x 2 + 4 x + 16 = 0

(d) x 2 − x + 1 = 0

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Objective Mathematics Vol. 1

5

140. If ( x − 1) 3 is a factor of x 4 + ax 3 + bx 2 + cx − 1, then the other factor is (a) x − 3 (c) x + 2

(b) x + 1 (d) None of these

(a) − 160 (c) 0

141. Let α + iβ , α , β ∈ R be a root of the equation x 3 + qx + r = 0, q, r ∈ R. The cubic equation is independent of α and β whose one root is 2α, is (a) x 3 + qx − r = 0

(b) x 3 + qx + 2r = 0

(c) x 3 + qx + r = 0

(d) x 3 − qx + r = 0

142. If α , β are the roots of x + px + q = 0 and α β are the roots of x 2n + p n x n + q n = 0 and , β α x n + 1 + ( x + 1) n = 0, then n must be 2

(a) even integer (c) rational but non-integer

(a) x 2 + b (b2 − 3c) x − c3 = 0

Ta rg e t E x e rc is e s

the

roots

of

the

equation

ax + bx + c = 0, then the equation whose roots are 1 1 and , is aα + b aβ + b 2

(a) cax − bx + 1 = 0

(b) cax + bx + 1 = 0

(c) cax + bx − 1 = 0

(d) None of these

2

2

2

145. If x = 2 + 21/ 3 + 22/ 3 , then the value of x 3 − 6x 2 + 6x is (a) 3 (c) 1

(b) 2 (d) None of these

146. If roots of the equation x n − 1 = 0 are 1, a1 , a 2 , a 3 , ... , a n−1 , then the value of (1 − a1 ) (1 − a 2 ) (1 − a 3 ) K (1 − a n−1 ) will be (a) n

(b) n

147. If α and β

2

are

(c) n

the

roots

n

(d) 0

of

the

quadratic

x 2 − 2cos θx + 1 = 0 , then equation whose roots are α n , β n , is (a) x 2 − (2 cos nθ ) x + 1 = 0 (b) 2x 2 − (2 cos nθ ) x − 1 = 0 (d) x 2 + (2 cos nθ ) x − 1 = 0

148. If 1, ω , ω , ... , ω

n−1

are the nth roots of unity, then (1 + ω ) (1 + ω ) ... (1 + ω n−1 ) equals 2

226

(a) 0, n is even (c) n, n is even

(b)

1  ,1  2 

 1 (c) 0,  2 

(d) [ − 1, 1]

151. If α , β , γ and σ are the roots of the equation x 4 + 4x 3 − 6x 2 + 7x − 9 = 0, then the value of (1 + α 2 ) (1 + β 2 ) (1 + γ 2 ) (1 + σ 2 ) is (a) 6 (c) 13

(b) 11 (d) 5

where x 3 − ( a + 1) x 2 + ( b − a ) x − b = 0, θ1 + θ 2 + θ 3∈ ( 0, π ), then θ1 + θ 2 + θ 3 is equal to (a) π/2

(b) π/4

(d) π

(c) 3π/4

(a) − 5 (c) − 7

2

(b) − 6 (d) − 2

154. The equation formed by decreasing each root of a x 2 + b x + c = 0 by 1 is 2 x 2 + 8 x + 2 = 0, then (a) a = − b (c) c = − a

(b) b = − c (d) b = a + c

155. If the equations ax 2 + bx + c = 0 and x 2 + x + 1 = 0 have a common root, then (a) a + b + c = 0 (c) a = b or b = c or c = a

(b) a = b = c (d) a − b + c = 0

156. If a, b, c ∈ R and the equations ax 2 + bx + c = 0 and x 3 + 3 x 2 + 3 x + 2 = 0 have two roots in common, then (a) a = b ≠ c (c) a = b = c

(b) a = b = − c (d) None of these

and a x2 + b x + c = 0 2 c x + b x + a = 0, a ≠ c have a negative common root, then the value of a − b + c is

157. If

(a) 0 (c) 1

the

equations

(b) 2 (d) None of these

158. If the equations x 2 + i x + a = 0 and x 2 − 2 x + ia = 0, a ≠ 0 have a common root, then

(c) x 2 + (2 cos nθ ) x + 1 = 0

2

(a) [0, 1]

153. If α , β and γ are the roots of x − x − 1 = 0, then (1 + α ) / (1 − α ) + (1 + β ) / (1 − β ) + (1 + γ ) / (1 − γ ) is equal to

(c) x 2 − b (b2 − 3c) x + c3 = 0 (d) None of the above

are

150. If p ∈ [ − 1, 1] , then the value of x for which 4x 3 − 3x − p = 0 has a root lies in

3

(b) x 2 + b (b2 − 3c) x + c3 = 0

α and β

(b) 160 (d) None of these

152. If tan θ1 , tan θ 2 and tan θ 3 are the real roots of the

(b) odd integer (d) None of these

143. The quadratic equation whose roots are cubes of the roots of x 2 + bx + c = 0, is

144. If

149. If f ( x ) = x 4 + 9x 3 + 35x 2 − x + 4, then f ( − 5 + 2 −4 ) is equal to

(b) 1, n is even (d) n2 , n is even

(a) a is real 1 (b) a = + i 2 1 (c) a = − i 2 (d) the other root is also common

160. If the equations ax 2 + bx + c = 0 and x 2 + 2x + 3 = 0 have a common root, then a : b : c is equal to (a) 2 : 4 : 5 (c) 1 : 2 : 3

(b) 1 : 3 : 4 (d) None of these

161. If the equations x − px + q = 0 and x − ax + b = 0 have a common root and the other root of the second equation is the reciprocal of the other root of the first, then ( q − b ) 2 is equal to 2

2

(a) aq ( p − b)2

(b) bq ( p − a)2

(c) bq ( p − b)2

(d) None of these

162. Let f ( x ) = 1 + 2x + 3x 2 + K + ( n + 1) x n , where n is even. Then, the number of real roots of the equation f ( x ) = 0 is (a) 0 (c) n

(b) 1 (d) None of these

163. The value of x 2 + 2bx + c is positive, if (a) b2 − 4 ac > 0

(b) b2 − 4 ac < 0

(c) c2 < b

(d) b2 < c

(b) − 2 (d) None of these

165. If − a 2 x 2 + 2x + 3a 2 > 0, ∀ x ∈ ( 2, 4 ), then the values of a lie in the interval  1 1 (a)  − ,   3 3

 2 2 (b)  − ,   7 7  2 2 2 2 (d)  − ,   13 13 

 4 4 (c)  − ,   7 7

166. The values of a which make the expression x 2 − ax + 1 − 2a 2 always positive for real values of x, are 2 2 0, then the number of possible values of x is (a) 3 (c) 2

168. If

(b) 4 (d) infinite

6x 2 − 5x − 3

x 2 − 2x + 6 of 4x 2 are

(a) 0, 81

(b) (1, 2) (d) (− 4 , 3)

170. The set of real values of x satisfying | x − 1 | − 1 ≤ 1is (a) [ − 1, 3 ] (c) [ − 1, 1]

(b) [ 0, 2 ] (d) None of these

171. If 5x + ( 2 3 ) 2x ≥ 13x , then the solution set for x is (a) [ 2, + ∞ ] (c) (4 , + ∞ )

(b) (2, + ∞ ) (d) None of these

172. If log 10 x + log 10 y ≥ 2, then the smallest possible value of x + y is (a) 10 (c) 20

173. If x ∈ R, then

(b) 30 (d) None of these

x2 − x + 1 x2 + x + 1

takes values in the interval

1  (a)  , 3 3 

(b)

(c) (0, 3)

(d) None of these

1  ,3  3 

174. If y = f ( x ) = tan x cot 3x, then (a)

164. If ax 2 + bx + 10 = 0 does not have two distinct real roots, then the least value of 5a + b is (a) − 3 (c) 3

(a) (3, 4) (c) (− 1, 2)

1 < y 0, ∀ x ∈ R, is (a) 5 (c) 3

for

which

(b) 4 (d) None of these

178. The values of a for which exactly one root of the equation e a x 2 − e 2a x + e a − 1 = 0 lies between 1 and 2 are given by (5 − 17 ) 5 + 17 < a < log 4 4 (b) 0 < a < 100 5 10 (c) log < a < log 4 3 (d) None of the above (a) log

< 4, then the least and highest values

(b) 0, 36

(c) − 10, 3

(d) 10, − 3

227

Objective Mathematics Vol. 1

5

179. If the roots of the equation x 2 + 2ax + b = 0 are real and distinct and they differ by atmost 2 m , then b lies in the interval (a) (a2 , a2 + m2 )

(b) (a2 − m2 , a2 )

(c) [ a − m , a ]

(d) None of these

2

2

2

180. If a, b, c and d are four consecutive terms of an increasing AP, then the roots of the equation ( x − a ) ( x − c ) + 2 ( x − b ) ( x − d ) = 0 are (a) non-real complex (c) integers

(b) real and equal (d) real and distinct

181.The interval of a for which the equation tan 2 x − ( a − 4 ) tan x + 4 − 2a = 0 has atleast one solution, ∀ x ∈ [ 0, π / 4], is (a) a ∈ (2, 3) (b) a ∈[ 2, 3 ] (c) a ∈ (1, 4 ) (d) a ∈[1, 4 ]

Ta rg e t E x e rc is e s

(b) 0

(c) 1/4

(d) 1/2

183. If a, b and c are distinct positive numbers, then the nature of roots of the equation 1/ ( x − a ) + 1/ ( x − b ) + 1/ ( x − c ) = 1/ x is (a) all real and distinct (b) all real and atleast two are distinct (c) atleast two real (d) all non-real

184. If a, b, c and d are four consecutive terms of an increasing AP, then the roots of the equation ( x − a ) ( x − c ) + 2 ( x − b ) ( x − d ) = 0 are (a) real and distinct (b) non-real complex (c) real and equal (d) integers 2

(a) p ≤ 1 (c) p ≥

(b)

3 4

3 ≤ p≤1 4

(d) None of these

186. If the roots of x + x + a = 0 exceed a, then 2

(a) 2 < a < 3 (c) − 3 < a < − 3

(b) a > 3 (d) a < − 2

187. The necessary and sufficient condition for the equation (1 − a 2 ) x 2 + 2ax − 1 = 0 to have roots lying in the interval (0, 1) is

228

(b) (0, ∞ ) (d) None of these

189. Let a, b and c be real and ax 2 + bx + c = 0 has two real c b roots α and β, where α < − 1and β > 1, then 1 + + a a (a) < 2 (c) < 0

(b) < 1 (d) None of these

190. If 2 a + 3b + 6 c = 0, a, b, c ∈ R, then the quadratic equation a x 2 + b x + c = 0 has (a) atleast one root in [0, 1] (c) atleast one root in (3, 4]

(b) atleast one root in [2, 3] (d) None of these

(a) a ≠ 0 (b) a > 0 (c) a < 2 or a > 2 (d) None of the above

(a) no real root (c) two equal roots

(b) 1 and 2 as real roots (d) two distinct real roots

192. If x + λy − 2 and x − µy + 1are factors of the expression 6x 2 − xy − y 2 − 6x + 8 y − 12, then 1 1 ,µ = 3 2 1 1 (c) λ = , µ = − 3 2

(a) λ =

(b) λ = 2, µ = 3 (d) None of these

193. If the expression y 2 + 2xy + 2x + my − 3 can be resolved into two rational factors, then m must be (a) any positive real number (b) any negative real number (c) − 2 (d) 3

194. If the expression ax 2 + by 2 + cz 2 + 2ayz + 2bzx +2cxy can resolved into rational factors, then

185. If cos x + sin x − p = 0, p ∈ R has real solutions, then 4

(a) (− 3, 0) (c) (− ∞ , − 3) ∪ (0, ∞ )

191. Let f ( x ) = ax 2 + bx + c and b, c ∈ R , a ≠ 0, satisfying f (1) + f ( 2) = 0. Then, the quadratic equation f ( x ) = 0 has

182. If the expression [ mx − 1 + (1/ x )] is non-negative for all positive real x, then the minimum value of m must be (a) − 1/ 2

188. The values of a for which the equation 2x 2 − 2 ( 2a + 1) x + a ( a − 1) = 0 has roots α and β satisfying the condition α < a < β, are

(a) a + b + c = 0 (b) a3 + b3 + c3 = 3abc (c) ab + ac + bc = 0 (d) None of the above

195. If the equation x 2 + 9 y 2 − 4x + 3 = 0 is satisfied for real values of x and y, then (a) 1 ≤ x ≤ 3 1 (c) − < y < 1 3

(b) 2 ≤ x ≤ 3 2 (d) 0 < y < 3

196. If ( mr , 1/ mr ), r = 1, 2, 3, 4 are four pairs of values of x and y that satisfy 2 2 x + y + 2gx + 2 fy + c = 0, m1 m2 m3 m4 is (a) 0 (b) 1 (c) − 1 (d) None of the above

the equation then the value of

(a) 3 (b) 2 (c) infinitely many (d) no value of k satisfies the requirements

199. The number of real solutions of the equation log 0. 5 x = | x | is (a) 1 (c) 0

(b) 2 (d) None of these

200. If (log 5 x ) 2 + log 5 x < 2, then x belongs to the interval

198. The number of real solutions of the equation e x = x is

 1  (a)  , 5  25  1 1  (b)  ,   5 5 (c) (1, ∞ ) (d) None of the above

(a) 1 (b) 2 (c) 0 (d) infinite

5 Inequalities and Quadratic Equation

197. The number of values of k for which the equation x 2 − 3x + k = 0 has two distinct roots lying in the interval (0, 1), is

Type 2. More than One Correct Option roots

+ (a − b )x (a) ± 3

of 2

−15

the

(a + b )x

equation

2

−15

= 2a, where a 2 − b = 1, are

(b) ± 4

(c) ± 14

(d) ±

5

202. If 0 < a < b < c and the roots α , β of the equation ax 2 + bx + c = 0 are imaginary, then (a) | α | = | β | (b) | α | = 1 (c) | β | < 1 (d) None of the above

(a) sin 2 (α + β ) + p sin (α + β )cos (α + β ) + q cos2 (α + β ) = q p q−1 (c) cos (α + β ) = − p (d) sin (α + β ) = 1 − q (b) tan (α + β) =

204. If ( ax + c ) y + ( a ′x + c ′ ) = 0 and x is a rational function of y, then 2

a a (b) = ′ c c′ (d) None of these

a c (a) = a′ c′ (c) ac′ − a′ c = 0 2

2

 x   x  205. The equation   = a ( a − 1) has  +  x − 1  x + 1

206. If α and β are the roots of the quadratic equation ax 2 + bx + c = 0, then which of the following expression will be the symmetric function of roots α β

207. If the quadratic equation ax 2 + bx + c = 0 ( a > 0) has sec 2 θ and cosec 2θ as its roots, then which of the following must hold good? (a) b + c = 0 (c) c ≥ 4 a

(b) b2 − 4 ac ≥ 0 (d) 4 a + b ≥ 0

(a) p + q = 0 (b) p ∈ (− 3, ∞ ) (c) one of the roots is unity (d) one root is smaller than 1 and one root is greater than 1

that α , γ are roots of the equation Ax − 4x + 1 = 0 and β , δ are the roots of the equation of Bx 2 − 6x + 1 = 0 such that α , β , γ and δ are in HP, then

209. Given 2

(a) A = 3 (c) B = 2

(b) A = 4 (d) B = 8

210. If (1 + k ) tan 2 x − 4 tan x − 1 + k = 0 has real roots tan x1 and tan x 2 , then (a) k 2 ≤ 5 (b) tan (x1 + x2 ) = 2 (c) for k = 2, x1 = π / 4 (d) for k = 1, x1 = 0

211. If the equation x 2 + px + q = 0 has roots u and v, where p and q are non-zero constants, then

(a) four real roots, if a > 2 (b) two real roots, if 1 < a < 2 (c) no real root, if a < − 1 (d) four real roots, if a < − 1

(a) log

1  (d)  log  + (log β )2  α

208. If the roots of the equation x 3 + px 2 + qx – 1 = 0 form an increasing GP, where p and q are real, then

203. If tan α and tan β are roots of the equation x 2 + px + q = 0 with p ≠ 0, then

2

2

(c) tan (α − β )

Targ e t E x e rc is e s

201. The

(b) α 2 β 5 + β 2α 5

(a) qx 2 + px + 1 = 0 has roots 1/u and 1/v (b) (x + p) (x − q) = 0 has roots u + v and uv (c) x 2 + p2x + q2 = 0 has roots u2 and v 2 u v (d) x 2 + px + p = 0 has roots and v u

212. For real x, the function real values provided (a) a > b > c

(x − a ) (x − b ) will assume all x−c 229

Objective Mathematics Vol. 1

5

(b) b ≤ c ≤ a (c) a > c > b (d) a ≤ c ≤ b

(d) µ = 0

215. If x, y ∈ R and 2x 2 + 6xy + 5 y 2 = 1, then

213. Let P ( x ) = x 2 + bx + c, where b and c are integers. If P ( x ) is a factor of both x 4 + 6x 2 + 25 and 3x 4 + 4x 2 + 28x + 5, then (a) P (x ) = 0 has imaginary roots (b) P (x ) = 0 has roots of opposite sign (c) P(1) = 4 (d) P(1) = 6

214. If the equations and 4x 2 − x − 1 = 0 3x 2 + ( λ + µ ) x + λ − µ = 0 have a common root, then the rational values of λ and µ are −3 4 (b) λ = 0 3 (c) µ = 4

(a) | x | ≤ 5

(b) | x | ≥ 5

(c) y ≤ 2

(d) y2 ≤ 4

2

216. If cos x − y 2 − (a) y ≥ 1

y − x 2 − 1 ≥ 0, then

(b) x ∈ R

(c) y = 1

(d) x = 0

217. If the following figure shows the graph of f ( x ) = ax 2 + bx + c, then Y

X

(a) λ =

(a) ac < 0 (c) ab > 0

(b) bc > 0 (d) abc < 0

Ta rg e t E x e rc is e s

Type 3. Assertion and Reason ax 2 + bx + c = 0

Directions (Q. Nos. 218-219) In the following

Statement

questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has four choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

a1 x + b1 x + c1 = 0 have a common root and a b c are in AP, then a1 , b1 and c1 are in , and a1 b1 c1

a b c are in AP, then , and a1 b1 c1 a1 , b1 and c1 are in GP.

218. Statement I If

II

If

and

2

GP. 219. Statement I If roots of the equation x 2 − bx + c = 0 are two consecutive integers, then b 2 − 4c = 1. Statement II If a, b and c are odd integers, then the roots of the equation 4abc x 2 + ( b 2 − 4ac ) x − b = 0 are real and distinct.

Type 4. Linked Comprehension Based Questions Passage I (Q. Nos. 220-222) Let the roots of f ( x ) = x be α and β, where f ( x ) is quadratic polynomial ax 2 + bx + c. α and β are also the roots of f (f ( x )) = x . Let the other two roots of f (f ( x )) = x be λ and δ.

220. The correct statements are I. If α and β are real and unequal, then λ and δ are also real. II. If α and β are imaginary, then λ and δ are also imaginary. (a) Only I (c) both I and II

(b) Only II (d) neither I nor II

221. If α + β = λ + δ, then the correct statements are

230

I. α and β are real II. λ and δ are real III. α and β cannot be real IV. λ and δ cannot be real

(a) I and II (c) II and III

(b) III and IV (d) I and IV

222. If α and β are real and equal, then (a) λ and δ are imaginary (b) λ and δ are real (c) λ and δ may be real or imaginary (d) Cannot say

Passage II (Q. Nos. 223-225) If f ( x ) is a differentiable function wherever it is continuous and f ′(c1) = f ′(c2 ) = 0. f ′′(c1) ⋅ f ′′(c2 ) < 0, f (c1) = 5, f (c2 ) = 0 and (c1 < c2 ).

223. If

f (x )

is

continuous

in

[ c1 , c 2 ]

and

f ′′ ( c1 ) − f ′′ ( c 2 ) > 0, then minimum number of roots of f ′ ( x ) = 0 in [ c1 − 1, c 2 + 1] is (a) 2 (c) 4

(b) 3 (d) 5

f (x )

is

continuous

in

[ c1 , c 2 ]

f ′′ ( c1 ) − f ′′ ( c 2 ) < 0, then minimum number of roots of f ′ ( x ) = 0 in [ c1 − 1, c 2 + 1] is (a) 1

(b) 2

(c) 3

(b) 3

(c) 4

(d) 5

Passage III (Q. Nos. 226-228) Let f ( x ) = x 2 + b1x + c1 and g( x ) = x + b2 x + c2 . If real roots of f ( x ) = 0 are α, β and real roots of g( x ) = 0 are α + δ, β + δ and least value of 1 7 f ( x ) is − . Then, least value of g( x ) occurs at x = . 4 2

Passage IV (Q. Nos. 229-231) Let x 1, x 2, x 3 and x 4 be the roots (real or complex) of the equation x 4 + ax 3 + bx 2 + cx + d = 0 and x 1 + x 2 = x 3 + x 4 and a, b, c, d ∈ R .

229. If a = 2, then the value of b − c is (a) − 1

1 (c) − 4

(c) − 2

(b) 1

(d) 2

230. If b < 0, then how many different values of a we may have? (a) 3

(b) 2

(c) 1

(d) 0

231. If b + c = 1and a ≠ − 2 , then for real values of a, c ∈

1 2 1 (d) − 3 (b) −

(a) − 1

(b) − 3, 4 (d) − 3, − 4

(a) 3, 4 (c) 3, − 4

2

226. The least value of g ( x ) is

5

(d) 0

228. The roots of g ( x ) = 0 are

(d) 4

225. If f ( x ) is continuous in [ c1 , c 2 ] and f ′′ ( c1 ) − f ′′ ( c 2 ) > 0, then minimum number of roots of f ( x ) = 0 in [ c1 − 1, c 2 + 1] is (a) 2

(c) 8

and

Inequalities and Quadratic Equation

224. If

1  (a)  − ∞ ,   4 (c) (− ∞ , 1)

(b) (− ∞ , 3) (d) (− ∞ , 4 )

227. The value of b2 is

Type 5. Match the Columns 232. Match the statements of Column I with the values of Column II. Column I

Column II a≥ 6

The real values of a for which the quadratic equation 2 x 2 − ( a 3 + 8a − 1) x + a 2 − 4a = 0 possesses roots of opposite signs are given by

p.

If the equation x 2 + 2 ( a + 1) x + 9a − 5 = 0 has only negative roots, then

q.

C.

The value of a for which the inequality x 2 − 2 ( 4a − 1) x + 15a 2 − 2 a − 7 > 0 is valid for all x ∈ R, is

r.

0< a< 4

D.

If x ∈ R, then

s.

2 < a< 4

A.

B.

x2 + 2 x + a x 2 + 4 x + 3a

233. Match the statements of Column I with the values of Column II.

can take all

Column I A.

p. If the equation x + 2 ( k + 1) x + ( 9k − 5) = 0 has only negative roots, then

2< k< 4

B.

If the inequality x 2 − 2 ( 4k − 1) x 2 + 15k − 2 k − 7 > 0 is valid for all x, then

k≥ 6

C.

If x 2 − 2 ( k − 1) x + (2 k + 1) = 0 has both r. roots positive, then

k< −1 k> 0

D.

If 2 x 2 − 2 (2 k + 1) x + k ( k + 1) = 0 has one s. root less than k and other root greater than k, then

k≥ 4

E.

The graph of the curve x 2 = 3 x − y − 2 is t. strictly below the line y equal to

1 4

0≤ a≤ 1

real values, if

Column II

2

q.

Targ e t E x e rc is e s

(b) −7

(a) 6

or

Type 6. Single Integer Answer Type Questions 234. Let satisfying the equation x1 , x 2 , x 3 x 3 − x 2 + βx + γ = 0 are in GP, where ( x1 , x 2 , x 3 > 0), then the maximum value of [β ] + [ γ ] + 2 is ______, where [ ] denotes the greatest integer function. 235. Consider the equation x 3 − ax 2 + bx − c = 0 , where a, b, c ∈ Q ( a ≠ 1). It is given that x1 , x 2 and x1 x 2 are

the real roots of the equation. If b + c = 2 ( a + 1), then x1 x 2 is equal to ______. 236. If α and β are the roots of x 2 + px − q = 0 and γ , δ are the roots of x 2 + px + r = 0, q + r ≠ 0, then (α − γ ) (α − δ ) is equal to ______. (β − γ ) (β − δ ) 231

Entrances Gallery JEE Advanced/IIT JEE 1. Let a ∈ R and let f : R → R be given by f ( x ) = x 5 − 5x + a, then (a) (b) (c) (d)

7. Let ( x 0 , y 0 ) be solution of the following equations

f (x ) has three real roots, if a > 4 f (x ) has only one real root, if a > 4 f (x ) has three real roots, if a < − 4 f (x ) has three real roots, if − 4 < a < 4

3ln x = 2ln y.

Then, x 0 is equal to

2. The quadratic equation p( x ) = 0 with real coefficients has purely imaginary roots. Then, the equation p( p( x )) = 0 has [2014]

1 (a) 6 1 (c) 2

[2011] 1 (b) 3 (d) 6

8. Let α and β be the roots of x 2 − 6 x − 2 = 0 with

(a) only purely imaginary roots (b) all real roots (c) two real and two purely imaginary roots

α > β. If a n = α n − β n for n > 1, then the value of a10 − 2a 8 is [2011] 2a 9

(d) neither real nor purely imaginary roots

(a) 1

3. The number of points in (− ∞, ∞), for which x − x sin x − cos x = 0, is 2

(a) 6

Ta rg e t E x e rc is e s

(2x )ln 2 = (3 y)ln 3 and

[2014]

(b) 4

[2013]

(a)

2 log3 2 2 log3 2 − 1

(b)

2 2 − log2 3

(c)

1 1 − log4 3

(d)

2 log2 3 2 log2 3 − 1

( 3 1 + a − 1) x 2 + ( 1 + a − 1) x + ( 6 1 + a − 1) = 0, where a > − 1. Then, lim α( a ) and lim β ( a ) are 5 and 1 2 7 (c) − and 2 2

a → 0+

[2012]

1 and − 1 2 9 (d) − and 3 2

(a) −

(b) −

6. The value of   1 1 1 1 6 + log 3/ 2  4− 4− 4− K  3 2  3 2 3 2 3 2   [2012] is (a) 4

(b) 3

(c) 1

x 2 + x + b = 0,

have one root in common, is (a) − 2

(b) − i 3

[2011]

(c) i 5

(d) 2

10. Consider the polynomial f ( x ) = 1+ 2x + 3x 2 + 4x 3 . Let s be the sum of all distinct real roots of f ( x ) and let t = | s |. The real number s lies in the interval [2010]

5. Let α( a ) and β( a ) be the roots of the equation

a → 0+

(d) 4

x 2 + bx − 1 = 0 and

(d) 0

4. If 3x = 4 x − 1 , then x is equal to

(c) 3

9. A value of b for which the equations [2013]

(c) 2

(b) 2

 1  (a)  − , 0  4  1  3 (c)  − , −   4 2

 (b)  − 11, − 

3  4

 1 (d)  0,   4

11. Let p and q be real numbers such that p ≠ 0, p 3 ≠ q and p 3 ≠ − q. If α and β are non-zero complex numbers satisfying α + β = − p and α 3 + β 3 = q, β α then a quadratic equation having and as its α β roots, is [2010] (a) ( p3 + q)x 2 − ( p3 + 2q)x + ( p3 + q) = 0 (b) ( p3 + q)x 2 − ( p3 − 2q)x + ( p3 + q) = 0 (c) ( p3 − q)x 2 − (5 p3 − 2q) x + ( p3 − q) = 0 (d) ( p3 − q)x 2 − (5 p3 + 2q)x + ( p3 − q) = 0

(d) 0

JEE Main/AIEEE 12. Let α and β be the roots of equation x 2 − 6x − 2 = 0. If a n = α − β , for n ≥ 1, then the value of a10 − 2a 8 is equal to 2a 9 n

232

(a) 6 (c) 3

n

(b) −6 (d) −3

[2015]

13. If a ∈ R and the equation − 3( x − [ x]) 2 + 2( x − [ x]) + a 2 = 0 (where, [x] denotes the greatest integer < x) has no integral solution, then all possible values of a lie in the [2014] interval (a) (− 1,0) ∪ (0, 1) (c) (− 2, − 1)

(b) (1, 2) (d) (− ∞ , − 2) ∪ (2, ∞ )

(b) 3 : 2 : 1

15. The equation e

sin x

−e

(c) 1 : 3 : 2

− sin x

(d) 3 : 1 : 2

− 4 = 0 has

[2012]

(a) 1

16. Let α , β be real and z be a complex number. If z 2 + αz + β = 0 has two distinct roots on the line Re z = 1, then it is necessary that [2011] (b) β = 1 (d) β ∈ (0, 1)

[2011] (b) 6, 1

(c) 4, 3

(d) − 6, − 1

18. If α and β are the roots of the equation x 2 − x + 1 = 0 [2010] , then α 2009 + β 2009 is equal to (a) − 2 (c) 1

(b) − 1 (d) 2

19. If the roots of the equation bx 2 + cx + a = 0 are imaginary, then for all real values of x, the expression 3b 2 x 2 + 6bcx + 2c 2 is [2009] (a) greater than 4ab (c) greater than − 4ab

(b) less than 4ab (d) less than − 4ab

20. The quadratic equations x 2 − 6x + a = 0 and x 2 − cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 [2008] : 3. Then, the common root is (a) 2

(b) 1

(c) 4

(d) 3

21. If the difference between the roots of the equation x 2 + ax + 1 = 0 is less than 5, then the set of [2007] possible values of a is (b) (− 3, ∞) (d) (− ∞ , − 3)

(a) (− 3, 3) (c) (3, ∞ )

22. If the roots of the quadratic equation x 2 + px + q = 0 are tan 30° and tan 15° respectively, then the value of 2 + q − p is [2006] (a) 3

(b) 0

(c) 1

(c) 0

(b) 2

(d) 1

(d) 2

23. All the values of m for which both roots of the equation x 2 − 2mx + m2 − 1 = 0 are greater than − 2 but less than 4 lie in the interval [2006] (a) m > 3

(b) − 1 < m < 3

(c) 1 < m < 4

(d) − 2 < m < 0

(c) 3

(d) − 2

26. If both the roots of the quadratic equation x 2 − 2 kx + k 2 + k − 5 = 0 are less than 5, then k lies in the interval [2005] (a) [4, 5]

17. Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4, 3). Rahul made a mistake in writing down coefficient of x to get roots (3, 2). The correct roots of equation are (a) − 4 , − 3

(b) 3

25. If the roots of the equation x 2 − bx + c = 0 are two consecutive integers, then b 2 − 4c is equal to [2005]

(a) infinite number of real roots (b) no real roots (c) exactly one real root (d) exactly four real roots

(a) β ∈ (− 1, 0) (c) β ∈ (1, ∞ )

(a) 2

(b) (− ∞, 4)

(c) (6, ∞)

5 Inequalities and Quadratic Equation

(a) 1 : 2 : 3

24. The value of a for which the sum of the squares of the roots of the equation x 2 − ( a − 2) x − a − 1 = 0 assume the least value, is [2005]

(d) (5, 6]

27. If the equation a n x n + a n − 1 x n − 1 + ... + a1 x = 0, a1 ≠ 0, n > 2 , has a positive root x = α, then the equation na n x n − 1 + ( n − 1)a n − 1 x n − 2 + ... + a1 = 0 has a positive root, which

[2005]

(a) is equal to α (b) is greater than or equal to α (c) belongs to (0, α ) (d) is greater than α

28. If (1 − p ) is a root of quadratic equation x 2 + px + (1 − p ) = 0, then its roots are (a) 0, 1

(b) − 1, 1

(c) 0, − 1

[2004] (d) − 1, 2

29. If one root of the equation x 2 + px + 12 = 0 is 4, while the equation x 2 + px + q = 0 has equal roots, [2004] then the value of q is 49 4 (c) 3 (a)

Targ e t E x e rc is e s

14. If the equations x 2 + 2x + 3 = 0 and ax 2 + bx + c = 0, a, b, c ∈ R, have a common root, then a : b : c is [2013]

(b) 12 (d) 4

30. Let two numbers have arithmetic mean 9 and geometric mean 4. Then, these numbers are the roots of the quadratic equation [2004] (a) x 2 + 18x + 16 = 0

(b) x 2 − 18x + 16 = 0

(c) x + 18x − 16 = 0

(d) x 2 − 18x − 16 = 0

2

31. Let z1 and z 2 be two roots of the equation z 2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z 2 form an equilateral triangle. Then, [2003] (a) a2 = b

(b) a2 = 2b

(c) a2 = 3b

(d) a2 = 4 b

32. If the sum of the roots of the quadratic equation ax 2 + bx + c = 0 is equal to the sum of the squares of their reciprocals, then (a) AP

(b) GP

a b c , and are in c a b (c) HP

[2003] (d) AGP

233

Objective Mathematics Vol. 1

5

33. The number of the real solutions of the equation x 2 − 3| x| + 2 = 0 is [2003] (a) 2

(b) 4

(c) 1

(d) 3

34. The value of a for which one root of the quadratic equation ( a 2 − 5a + 3)x 2 + ( 3a − 1)x + 2 = 0 is twice as large as the other, is (b) −2/ 3

(a) 2/ 3

(c) 1/ 3

[2003] (d) −1/ 3

[WB JEE 2014]

(a) 602 (b) 301 (c) 392 (d) 391

38. The number of solutions of the equation [WB JEE 2014] log 101 log 7 ( x + 7 + x ) = 0 is (a) 3 (c) 9

Ta rg e t E x e rc is e s

equation having α /β and β / α as its roots, is [2002] (a) 3x 2 + 19x + 3 = 0

(b) 3x 2 − 19x + 3 = 0

(c) 3x − 19x − 3 = 0

(d) x 2 − 16x + 1 = 0

2

36. The number of real roots of 32x (a) 0

(b) 2

2

− 7x + 7

(c) 1

= 9 is [2002] (d) 4

Other Engineering Entrances 37. The number of digits in 20301 (given, log 10 2 = 0.3010) is

(b) 7 (d) 49

39. The value of a, so that the sum of squares of the roots of the equation x 2 − ( a − 2) x − a + 1 = 0 [BITSAT 2014] assume the least value, is (a) 2 (c) 3

(b) 0 (d) 1

40. If the roots of x 2 − ax + b = 0 are two consecutive odd integers, then a 2 − 4b is equal to[Kerala CEE 2014] (a) 3 (d) 6

(b) 4 (e) 7

(c) 5

41. If α and β are the roots of x 2 − ax + b 2 = 0, then [Kerala CEE 2014] α 2 + β 2 is equal to (a) a2 + 2b2

(b) a2 − 2b2

(d) a + 2b

(e) a − b

2

2

(c) a2 − 2b

2

−3 4 4 (d) 3

3 4 3 (e) 2

(b)

(a)

(c)

−4 3

43. If the roots of the equation x 2 + 2bx + c = 0 are α and β, then b 2 − c is equal to [Kerala CEE 2014] (α − β ) (a) 4 (b) (α + β )2 − αβ (c) (α + β )2 + αβ (α − β )2 (d) + αβ 2 (α + β )2 (e) + αβ 2 2

44. If a, b and c are positive numbers in a GP, then the roots of the quadratic equation (log e a ) x 2 − ( 2 log e b ) x + (log e c ) = 0 are loge c (a) −1and loge a

[WB JEE 2014] loge c (b) 1 and − loge a

(c) 1 and loga c

(d) −1and logc a

45. If α , β are the roots of ax 2 + bx + c = 0 ( a ≠ 0) and α + h, β + h are the roots of px 2 + qx + r = 0 ( p ≠ 0), then the ratio of the squares of their discriminants is (a) a2 : p2

(b) a : p2

[WB JEE 2014] (d) a : 2 p

(c) a2 : p

46. Let f ( x ) = 2x 2 + 5x + 1. If we write f ( x ) is f ( x ) = [ a ( x + 1)( x − 2) + b( x − 2) ( x − 1) + c( x − 1)( x + 1)] for real numbers a, b and c, then

[WB JEE 2014]

(a) there are infinite number of choices for a, b and c (b) only one choice for a but infinite number of choices for b and c (c) exactly one choice for each of a, b and c (d) more than one but finite number of choices for a, b and c

47. If α and β are the roots of the equation ax 2 + bx + c = 0 and

42. If α and β are the roots of the equation 1 1 x 2 + 3x − 4 = 0, then + is equal to α β [Kerala CEE 2014]

234

35. If α ≠ β and α 2 = 5α − 3, β 2 = 5β − 3, then the

then aS n + 1 + bS n + cS n − 1

S n = αn + βn, is equal to [AMU 2014]

(a) 0 (b) abc (c) a + b + c (d) None of the above

48. If the roots of ax 2 + bx + c = 0 are sin α and cos α for some α, then which one of the following is correct? [RPET 2014] (a) a + b = 2ac 2

2

(b) b2 − c2 = 2ab (c) b2 − a2 = 2ac (d) b2 + c2 = 2ab

49. The value of x such that 32x − 2( 3x + 2 ) + 81 = 0 is [Kerala CEE 2014] (a) 1 (d) 4

(b) 2 (e) 5

(c) 3

(a) 4 x 2 + 5x + 1 = 0

[Kerala CEE 2014] (b) 4 x 2 − x + 1 = 0

(c) 4 x 2 − 5x − 1 = 0

(d) 4 x 2 − 5x + 1 = 0

(e) 4 x + 5x − 1 = 0

59. The number of solutions of the inequation [Kerala CEE 2014] | x − 2| + | x + 2| < 4 is (a) 1 (d) 0

(b) 2 (e) 5

60. Solve the inequality 2x − 5 ≤ ( 4x − 7)/ 3.

2

51. In a ∆ABC, tan A and tan B are the roots of pq( x 2 + 1) = r 2 x. Then, ∆ABC is [WB JEE 2014] (a) a right angled triangle (b) an acute angled triangle (c) an obtuse angled triangle (d) an equilateral triangle

52. If α and β are the roots of the quadratic equation x 2 + px + q = 0, then the values of α 3 + β 3 and α 4 + α 2β 2 + β 4 are respectively 4

2

[WB JEE 2014]

2

(c) pq − 4 and p − q

4

(d) 3 pq − p3 and ( p2 − q) ( p2 − 3q)

53. The number of solution(s) of the equation [WB JEE 2014] x + 1 − x − 1 = 4x − 1 is (a) 2

(b) 0

(c) 3

(d) 1

54. If 9 x = 12 + 147, then the value of x is [RPET 2014] (a) −2

(b) 2

(d) −3

(c) 3

55. Solve the equation x 2 − 3x + 1 = 0.

[J&K CET 2014]

(b) x = (− 3 ± i ) / 2

(c) x = (− 3 ± i )

(d) x = ( 3 ± i ) / 2

2π 2π + i sin , then the quadratic equation 7 7 whose roots are α = a + a 2 + a 4 and

56. If a = cos

[Manipal 2014]

(a) x 2 − x + 2 = 0

(b) x 2 + 2x + 2 = 0

(c) x 2 + x + 2 = 0

(d) x 2 + x − 2 = 0

57. Let p and q be real numbers. If α is the root of x 2 + 3 p 2 x + 5q 2 = 0, β is the root of x 2 + 9 p 2 x + 15q 2 = 0 and 0 < α < β, then the equation x 2 + 6 p 2 x + 10q 2 = 0 has a root γ that always satisfies α (a) γ = + β 4 α (c) γ = + β 2

[WB JEE 2014] (b) β < γ (d) α < γ < β

58. If the roots of the equation λ 2 + 8λ + µ 2 + 6 µ = 0 are real, then µ lies between [RPET 2014] (a) −2 and 8 (c) −8 and 2

(b) −3 and 6 (d) −6 and 3

61. Solve the inequality 3x + 2 > − 16 and 2x − 3 ≤ 11 . [J&K CET 2014] (a) (− 6, 7 ]

(b) [ − 6, 7)

(c) (− 6, 7)

(d) [ − 6, 7 ]

62. The number of real solutions of the equation [Karnataka CET 2013] sin e x = 5x + 5− x is (a) 1 (c) 2

(b) 0 (d) None of these

(a) γ = α

(b) α < β < γ α+β (d) γ = 2

(c) α < γ < β

64. If the arithmetic mean of the roots of a quadratic equation is 8 and the geometric mean is 5, then the equation is [BITSAT 2013] (a) x 2 − 16x − 25 = 0

(b) x 2 + 16x − 25 = 0

(c) x − 16x + 25 = 0

(d) x 2 − 8x + 5 = 0

2

(a) x = (− 3 ± 2i ) / 2

β = a 3 + a 5 + a 6 , is

[J&K CET 2014] (b) x ∈ (− ∞ , 4 ] (d) x ∈ (− ∞ , − 4 ]

63. Let a, b and c be real numbers, a ≠ 0. If α is a root of a 2 x 2 + bx + c = 0, β is a root of a 2 x 2 − bx − c = 0 and 0 < α < β. Then, the equation a 2 x 2 + 2bx + 2c = 0 has a root γ that always satisfies [UP SEE 2013]

(b) − p(3q − p2 ) and ( p2 − q) ( p2 + 3q) 4

(a) x ∈ (− ∞ , 4 ) (c) x ∈ (− ∞ , 8 ]

5

65. If the roots of the equation ax 2 + bx + c = 0 are of k +1 k+2 the form , then ( a + b + c ) 2 is equal and k k +1 to [Manipal 2013] (a) b2 − 4 ac

(b) b2 − 2ac

(c) 2b2 − ac

(d) Σa2

Targ e t E x e rc is e s

(a) 3 pq − p and p − 3 p q + 3q 3

(c) 4

Inequalities and Quadratic Equation

50. The equation whose roots are the squares of the roots of the equation 2x 2 + 3x + 1 = 0, is

66. If the equations 2ax 2 − 3bx + 4c = 0 and 3x 2 − 4x + 5 = 0 have a common root, then ( a + b ) / ( b + c ) is equal to ( a, b, c ∈ R ) [Kerala CEE 2013] (a) 1/2 (d) 29/23

(b) 3/35 (c) 34/31 (e) None of these

67. If log 2 ( 9x −1 + 7) − log 2 ( 3x −1 + 1) = 2 , then the values of x are [Karnataka CET 2012] (a) 0, 2 (c) 1, 4

(b) 0, 1 (d) 1, 2

68. If a, b and c are in arithmetic progression, then the roots of the equation ax 2 − 2bx + c = 0 are c (a) 1 and a (c) −1and −

c a

[WB JEE 2012] 1 (b) − and −c a c (d) −2 and − 2a

235

Objective Mathematics Vol. 1

5

69. The harmonic mean of the roots of the equation ( 5 + 2 ) x 2 − ( 4 + 5 ) x + 8 + 2 5 = 0 is (a) 2

(b) 4

(c) 6

[Manipal 2012] (d) 8

70. If α , β and γ are the roots of the equation x 3 + 4x + 2 = 0, then α 3 + β 3 + γ 3 is equal to (a) 2

(b) 6

1 1 1 1 (a)  +  and  −  β β α α

Ta rg e t E x e rc is e s

(

3 +1+ (

3 − 1) 2 ( 3 − 2 )

3 + 1) 2 − (

3 − 1) 2

is

[UP SEE 2011] (a)

3+ 2 3

(b) 1

(c) 3

(d)

1 3

72. If the product of the roots of the equation x 2 − 2 2 kx + 2e 2 log k − 1 = 0 is 31, then the roots of the equation are real for k, is equal to [AMU 2012] (b) 1 (d) 0

73. If α and β are the roots of the equation x 2 − 2x + 4 = 0, then the value of α n + β n will be [BITSAT 2011]

(a) i 2n+ 1 sin (nπ / 3)

[Kerala CEE 2011] (b) (5, 9) 2  (d)  −8,   3

2  (e)  −5,   3

(d) ( α + β ) and ( α − β )

80. If α and β are the roots of the equation ax 2 + bx + c = 0 , then the equation whose roots are k k [MP PET 2011] and , is α β (a) cx 2 + kbx + k 2a = 0

(b) cx 2 + k 2bx + ka = 0

(c) kcx 2 + bx + k 2a = 0

(d) k 2cx 2 + bx + ka = 0

81. If 3 ≤ 3t − 18 ≤ 18, then which one of the following is correct? [Kerala CEE 2010] (a) 15 ≤ 2t + 1 ≤ 20 (c) 8 ≤ t + 1 ≤ 13 (e) t ≤ 7 or t ≥ 12

(b) 2n+ 1 cos (nπ / 3) (c) i 2n− 1 sin (nπ / 3) (d) 2n− 1 cos (nπ / 3)

(b) 8 ≤ t < 12 (d) 21 ≤ 3t ≤ 24

82. The solution set of the inequation

74. If α , β and γ are the roots of x 3 − 2x + 1 = 0, then the   1 [Karnataka CET 2011] value of Σ   is α + β − γ  (b) −1 (d) 1 / 2

75. The number of integral values of b, for which the equation x 2 + bx − 16 has integral roots, is [Kerala CEE 2011] (b) 3 (d) 5

a + 2b + 4c = 0. Then, the equation ax 2 + bx + c = 0 [WB JEE 2011] (a) has both the roots complex (b) has its roots lying within −1 < x < 0 1 (c) has one of roots equal to 2 (d) has its roots lying within 2 < x < 6

x + 11 > 0 is x−3 [Kerala CEE 2010]

(a) (− ∞ , − 11) ∪ (3, ∞ ) (b) (− ∞ , − 10) ∪ (2, ∞ ) (c) (− 100, − 11) ∪ (1, ∞ ) (d) (− 5, 0) ∪ (3, 7) (e) (0, 5) ∪ (− 1, 0)

83. In a right angled triangle, the sides are a, b and c, with c as hypotenuse and c − b ≠ 1, c + b ≠ 1 . Then, the value of (log c + b a + log c − b a ) / ( 2 log c + b a × log c − b a ) [WB JEE 2010] will be (b) −1

(a) 2

76. If a, b and c are three real numbers such that

236

78. The value of

[UP SEE 2011] (d) −3

(c) 8/49

(a) (−3, 5)  −2  (c)  , 8  3 

 1  1 1 1 (c)  + −  and   β β  α  α

(a) 2 (c) 4 (e) 6

(b) 49/8

79. If |2x − 3| < | x + 5|, then x lies in the interval

1 1  1  1 (b)  +  and  −   α  α β β

(a) −1 / 2 (c) 0

(a) 3

[Karnataka CET 2012] (c) −2 (d) −6

71. If (α + β ) and (α − β ) are the roots of the equation x 2 + px + q = 0, where α , β , p and q are real, then the roots of the equation ( p 2 − 4q ) ( p 2 x 2 + 4 px ) − 16q = 0 are [WB JEE 2012]

(a) −4 (c) 4

77. If one of the roots of 2x 2 − cx + 3 = 0 is 3 and another equation 2x 2 − cx + d = 0 has equal roots, where c and d are real numbers, then d is equal to

(c) 1/2

(d) 1

84. If α and β are the roots of the equation ax 2 + bx + c = 0, ( c ≠ 0) , then the equation whose 1 1 and , is roots are [BITSAT 2010] aα + b aβ + b (a) acx 2 − bx + 1 = 0

(b) x 2 − acx + bc + 1 = 0

(c) acx + bx − 1 = 0

(d) x 2 + acx − bc + 11 = 0

2

(b) 4 , − 1

(c) 2, 3

86. If x 2 + px + q = 0 has the roots α and β, then the value of (α − β ) 2 is [VITEEE 2010] (a) p2 − 4 q

(b) ( p2 − 4 q)2

(c) p2 + 4 q

(d) ( p2 + 4 q)2

87. If a and b are the roots of the equation x 2 + ax + b = 0, a ≠ 0, b ≠ 0 , then the values of a and b are, respectively [Kerala CEE 2010] (a) 2 and − 2 (b) 2 and −1 (d) 1 and 2 (e) −1and 2

(c) 1 and − 2

88. If the roots of the equation x 2 − bx + c = 0 are two consecutive integers, then b 2 − 4c is equal to [Kerala CEE 2010] (a) −1 (d) 2

(b) 0 (e) 3

(c) 1

89. If α and β are the roots of the quadratic equation x 2 + x + 1 = 0, then the equation whose roots are [WB JEE 2010] α 19 and β 7 , is (a) x 2 − x + 1 = 0

(b) x 2 − x − 1 = 0

(c) x 2 + x − 1 = 0

(d) x 2 + x + 1 = 0

[BITSAT 2010] (a) only positive solutions (b) only negative solutions (c) no solution (d) both positive and negative solutions

91. The value of a for which the equation 2x 2 + 2 6x + a = 0 has equal roots, is [VITEEE 2010] (a) 3 (b) 4 (c) 2 (d) 3

5 Inequalities and Quadratic Equation

(a) − 4 , 1

[VITEEE 2010] (d) − 2, − 3

90. The quadratic equation x 2 + 15 | x| + 14 = 0 has

3 7 + i is a solution of the equation 2 2 ax 2 − 6x + b = 0, where a and b are real numbers, then the value of a + b is [Kerala CEE 2010]

92. If

(a) 10 (d) 29

(b) 22 (e) 31

(c) 30

93. The roots of the quadratic equation x 2 − 2 3x − 22 = 0 are (a) imaginary (b) real, rational and equal (c) real, irrational and unequal (d) real, rational and unequal

[WB JEE 2010]

Targ e t E x e rc is e s

85. If one root of the equation x 2 + px + q = 0 is 2 + 3, then the value of p and q are, respectively

237

Answers Work Book Exercise 5.1 1. (a)

2. (c)

3. (d)

4. (b)

5. (a)

3. (d)

4. (b)

5. (a)

3. (c)

4. (d)

5. (a,b,c,d)

3. (c) 13. (c)

4. (c) 14. (c)

4. (c) 14. (c)

Work Book Exercise 5.2 1. (c)

2. (a)

6. (a)

7. (a)

8. (b)

9. (a)

10. (d)

5. (d)

6. (a)

7. (b)

8. (d)

9. (b)

10. (c)

5. (d)

6. (b)

7. (c)

8. (a)

9. (d)

10. (c)

Work Book Exercise 5.3 1. (a)

2. (a)

Work Book Exercise 5.4 1. (b) 11. (d)

2. (a) 12. (c)

Work Book Exercise 5.5 1. (a) 11. (c)

2. (b) 12. (c)

3. (a) 13. (c)

Ta rg e t E x e rc is e s

Target Exercises 1. 11. 21. 31. 41. 51. 61. 71. 81. 91. 101. 111. 121. 131. 141. 151. 161. 171. 181. 191. 201. 211. 221. 231. * **

(b) 2. (b) 3. (a) 4. (b) 12. (c) 13. (b) 14. (d) 22. (c) 23. (d) 24. (c) 32. (b) 33. (b) 34. (c) 42. (b) 43. (a) 44. (b) 52. (a) 53. (a) 54. (c) 62. (a) 63. (a) 64. (b) 72. (c) 73. (b) 74. (d) 82. (a) 83. (d) 84. (b) 92. (b) 93. (b) 94. (c) 102. (d) 103. (d) 104. (b) 112. (d) 113. (d) 114. (a) 122. (d) 123. (c) 124. (a) 132. (d) 133. (b) 134. (a) 142. (a) 143. (b) 144. (c) 152. (b) 153. (a) 154. (b) 162. (a) 163. (d) 164. (d) 172. (c) 173. (b) 174. (b) 182. (c) 183. (a) 184. (d) 192. (a) 193. (c) 194. (b,c) 202. (a,b) 203. (a,b) 204. (a,b) 212. (b,d) 213. (a,c) 214. (b) 222. (a) 223. (c) 224. (a) 232. (*) 233. (**) 234. A → r; B → p; C → s; D → q A → q; B → p; C → s; D → r; E → t

(c) (d) (c) (c) (a) (b) (a) (b) (b) (d) (b) (a) (c) (b) (a) (b) (b) (b) (a) (b) (a,b,c) (a,d) (b) (1)

5. 15. 25. 35. 45. 55. 65. 75. 85. 95. 105. 115. 125. 135. 145. 155. 165. 175. 185. 195. 205. 215. 225. 235.

(a) (d) (d) (d) (b) (a) (c) (c) (a) (a) (b) (a) (b) (c) (b) (b) (d) (c) (b) (a) (a,b,d) (a,c) (a) (2)

6. 16. 26. 36. 46. 56. 66. 76. 86. 96. 106. 116. 126. 136. 146. 156. 166. 176. 186. 196. 206. 216. 226. 236.

(b) (c) (c) (b) (a) (a) (b) (a) (c) (d) (b) (b) (c) (b) (a) (c) (a) (d) (d) (b) (a,b,d) (c,d) (c) (1)

7. 17. 27. 37. 47. 57. 67. 77. 87. 97. 107. 117. 127. 137. 147. 157. 167. 177. 187. 197. 207. 217. 227.

(a) (d) (d) (b) (b) (c) (c) (b) (d) (c) (a) (c) (d) (a) (a) (a) (a) (a) (c) (d) (a,b,c) (a,b,d) (b)

8. 18. 28. 38. 48. 58. 68. 78. 88. 98. 108. 118. 128. 138. 148. 158. 168. 178. 188. 198. 208. 218. 228.

(d) (b) (c) (c) (b) (c) (a) (a) (c) (b) (c) (d) (c) (a) (a) (c) (a) (a) (c) (c) (a,c,d) (d) (a)

9. 19. 29. 39. 49. 59. 69. 79. 89. 99. 109. 119. 129. 139. 149. 159. 169. 179. 189. 199. 209. 219. 229.

(c) (c) (c) (c) (c) (c) (a) (b) (c)

9. 19. 29. 39. 49. 59. 69. 79. 89.

(c) (b) (a) (d) (a) (a) (d) (b) (b) (d) (c) (c) (b) (b) (a) (b) (b) (c) (c) (a) (a,d) (b) (b)

10. 20. 30. 40. 50. 60. 70. 80. 90. 100. 110. 120. 130. 140. 150. 160. 170. 180. 190. 200. 210. 220. 230.

(c) (c) (a) (b) (a) (c) (a) (a) (c) (d) (b) (a) (d) (b) (b) (c) (a) (d) (a) (a) (a,b,c,d) (c) (c)

Entrances Gallery

238

1. 11. 21. 31. 41. 51. 61. 71. 81. 91.

(b,d) (b) (a) (c) (b) (a) (a) (a) (c) (a)

2. 12. 22. 32. 42. 52. 62. 72. 82. 92.

(d) (c) (a) (c) (b) (d) (d) (c) (a) (e)

3. 13. 23. 33. 43. 53. 63. 73. 83. 93.

(c) (a) (b) (b) (a) (b) (c) (b) (d) (c)

4. 14. 24. 34. 44. 54. 64. 74. 84.

(a,b,c) (a) (d) (a) (c) (c) (c) (b) (a)

5. 15. 25. 35. 45. 55. 65. 75. 85.

(b) (d) (a) (b) (a) (d) (a) (a) (a)

6. 16. 26. 36. 46. 56. 66. 76. 86.

(a) (c) (b) (b) (c) (c) (c) (c) (a)

7. 17. 27. 37. 47. 57. 67. 77 87.

(c) (b) (c) (c) (a) (d) (d) (b) (c)

8. 18. 28. 38. 48. 58. 68. 78. 88.

(b) (c) (a) (d) (b) (d) (b) (c) (d)

10. 20. 30. 40. 50. 60. 70. 80. 90.

(c) (a) (b) (b) (d) (a) (d) (a) (c)

Explanations Target Exercises 1. The given inequation reduces to

1 3 1 1 − < 3x + 1 < ⇒ 3 3 4 2 − < 3x < − ⇒ 3 3 4 2 − − 1, 9 a + 3 b + c < − 4 ⇒

a− b+ c 0 b + 1> 0

or ⇒

3. (a4 + b4 ) − (a3 b + ab3 ) = (a4 − a3 b) − (ab3 − b4 ) = a3 (a − b) − b3 (a − b) = (a − b)(a3 − b3 ) = (a − b)2 (a2 + ab + b2 )

13. | x − 1| is the distance of x from 1 | x − 3| is the distance of x from 3 The points x = 2 is equidistant from 1 and 3. Hence, the solution consists of all x ≥ 2.

which is positive as both factors are positive. Thus, (a4 + b4 ) − (a3 b + ab3 ) > 0 ⇒

a4 + b4 > a3 b + ab3

14. || x| − 1| < |1 − x|

4. Using number line rule, 2  x ∈ (− ∞, 0 ) ∪  , ∞ 3 

Case I x ≥ 0 ∴ Eq. (i) becomes| x − 1| < x − 1or|1 − x| < 1 − x which is not satisfied by any x, because | a| ≥ a, ∀ a ∈ R Case II − 1≤ x < 0 ∴ Eq. (i) becomes |− 1 − x| < 1 − x or | x + 1| < 1 − x ⇒ x + 1 < 1 − x or x < 0 Thus, Eq. (i) is satisfied for − 1 ≤ x < 0 Case III x < −1 Eq. (i) becomes |− 1 − x| < 1 − x ⇒ |1 + x| < 1 − x ⇒ − (1 + x ) < 1 − x ⇒ − 2 < 0, which is true So, solution set is (− ∞, 0 ).

5. Using number line rule, x ∈ (− ∞, 3) ∪ (10, ∞ )

6. Using number line rule, x ∈ (− ∞, − 4)

7. Using number line rule, – ve

+ ve

– ve

+ ve

–— 1 2

–7

3

 1  x ∈ (− ∞, − 7 ) ∪  − , 3  2 

8. 5 x + 2 < 3 x + 8 and ⇒

x+2 2 x

3 3 [clearly x > 0] >2 ⇒ >1 x x ⇒ 3 > x or x < 3 3 3 [hence x < 0] Case II 1 + < − 2 ⇒ − 3x ⇒ − 1 < x or x > − 1 Hence, either 0 < x < 3 or − 1 < x < 0 Case I

( x − 5)(3 x − 1) < 0 –∞

1 1 −2 < 4 ⇒ − 4< −2 < 4 x x 1 −2 < < 6 ⇒ x 1 1   Hence, x ∈  − ∞, −  ∪  , ∞ . 6   2

16. 1 +

+

…(i)

∞

Targ e t E x e rc is e s

8x or 2< 10 ∴ Least integer is 3.

1+

239

Objective Mathematics Vol. 1

5

17. | x 2 − 10| ≤ 6 ⇒

25. Here, 5 = 4a and 6 = 5b

− 6 < x − 10 ≤ 6 2

⇒

Let Then,

4 ≤ x ≤ 16 2

⇒ 2 ≤ | x| ≤ 4 ∴ Either 2 ≤ x ≤ 4 or − 4≤ x ≤ −2 ∴ x ∈ [− 4, − 2 ] ∪ [ 2, 4] 1 [clearly x ≠ 0] 18. x + > 2 x x2 + 1 >2 ⇒ x

⇒ ⇒

x2 + 1 >2 | x|

| x|2 − 2| x| + 1 > 0

⇒

(| x| − 1)2 > 0

⇒ ∴

| x| ≠ 1 ⇒ x ≠ − 1, 1 x ∈ R − { − 1, 0, 1}

Now,

6 = 5b = (4a )b = 4ab

or

3 = 2 2 ab − 1

∴

2 = (2 2 ab − 1 )x = 2 x( 2 ab − 1)

⇒ x(2 ab − 1) = 1 log a log x 26. ⋅ =2 log 5 log a ⇒

[Q x 2 + 1 > 0 ]

log 5 x = 2 ⇒

x = 52 = 25

27. x + 5 = 2 6 − x .

x 2 + 1 > 2| x|

⇒

log 3 2 = x 2 = 3x

Its graph is as follows 2 6–x Y (x+5) O

| x + 1| | x + 1| + | x + 1| = | x| | x|

2

19.

Ta rg e t E x e rc is e s

Since,| a| + | b| = | a + b| iff ab ≥ 0 ( x + 1) ( x + 1)2 ⋅ ( x + 1) ≥ 0 ⇒ ∴ ≥0 x x ⇒ x > 0 or { − 1}

Clearly, there cannot be more than one solution and by trial, the solution is x = 3.

28. |4 − 5 x| > 2 2 = 4

⇒ (− 2 ≤ x ≤ 4) and (x ≤ 0 or x ≥ 2) ⇒ − 2 ≤ x ≤ 0 or 2 ≤ x ≤ 4

∴

⇒

x2 > 1 − x

and 1 − x > 0, x > 0

x + x − 1 > 0 and 2

29.

0 < x  2 4 5 −1 and 0 < x < 1 x> 2 5 −1 < x b>0 ⇒

x = 3, 1, 4, 0

x > y > 0, 1 1 < a b

Option (c) is false a > b ⇒ a + c > b + c , for c ∈ R ∴Option (d) is true.

24. For x ≤ 1, the given inequation gives x ≤ 0 and for x ≥ 3, the given inequation gives x ≥ 4. For 1 < x ≤ 2 or 2 < x < 3, the given inequation gives no solution. ∴ Solution is (− ∞, 0 ] ∪ [4, ∞ ).

x0 x ⇒ x( x + 2 ) > 0 ∴ x < − 2 or x > 0 5  ∪ (0, + ∞ ). ∴The solution set is  − ∞, −  2  Also,

23. Here, a > b ⇒ ax > bx, only when x > 0 and a>b>0 Option (a) is false | x| > | y| ⇒ Option (b) is false

5x − 1< − 1 4

x+2 x+2 1 ≥ (0.2 )1 or ≥ x x 5 5 x( x + 2 ) ≥ x 2 or ∴

22. | x − 2|2 − 3| x − 2| + 2 = 0 ⇒ (| x − 2| − 1)(| x − 2| − 2 ) = 0 ⇒ ∴ Product of roots = 0

5x − 1 > 1 or 4 8 or x> 5

 8 So, the solution set = (− ∞, 0 ) ∪  , + ∞  5

21. Here, x > 1 − x i.e. greater than zero

⇒

240

⇒

20. (− 3 ≤ x − 1 ≤ 3) and (x − 1 ≤ − 1or x − 1 ≥ 1)

∴

X

O

1 ± 1 − 4 log16 k 2 For exactly one solution, 4 log16 k = 1 ∴ k 4 = 16

30. log16 x =

Hence, k = 2, − 2, 2 i , − 2 i But k is positive and real.

31. xlog x a ⋅log a y ⋅log y z ⇒

x

log a log y log z ⋅ ⋅ log x log a log y

⇒ xlog x z = z

32. log10 x = 25 log10 (5) = 25 {1 − log10 2} = 25 {1 − 0. 30103} = 25 { 0.69897} = 17.469 ∴ Number of zeroes coming immediately after the decimal place is 17.

x log x + y log y + z log z = 0 ⇒ log( x x y y z z ) = 0 ⇒ x x y y z z = 1 1 34. log1/ 3 ( x 2 + x + 1) < − 1 = log1/ 3  

 1  ≥ (a2 b2c 2 )1/ 3  2 2 2  a b c  1 1 1   ⇒ (a2 + b2 + c 2 )  2 + 2 + 2  ≥ 9 a b c  ∴ Least value is 9.

−1

 3

⇒

 1 x2 + x + 1 <    3

⇒

x + x −2 < 0

−1

a2 + b2 b2 + c 2 ≥ a2 b2 = ab, ≥ bc 2 2 2 2 c +a and ≥ ca 2 2 2 On adding, we get a + b + c 2 ≥ ab + bc + ca

∴

x ∈ (− 2, 1)

35. log10 x 2 ≥ 0 ⇒ log10 x 2 ≥ log10 1 ⇒

x2 ≥ 1 ⇒

x ≥1 ⇒

x ≤ −1 x ...

⇒ Option (a) holds. a3 + b3 + c 3 Again, ≥ (a3 b3c 3 )1/ 3 3 ⇒ a3 + b3 + c 3 ≥ 3 abc

x1

36. Value = log x1log x2log x3 K log xn − 1 ( xn n−−1 2 log xn xn ) x...

x1

= log x1log x2log x3 K log n − 1 xn n−−1 2 = K = log x1 x1 = 1 sin 2 x

37. 4

+

4 4sin

2

sin 2 x

≥2 4

x

sin 2 x

⇒

4

⋅

⇒ Option (d) does not hold. b+c c+a a+ b Next, ≥ bc , ≥ ca, ≥ ab 2 2 2  b + c   c + a  a + b 2 2 2 ⇒  ≥ a bc     2  2  2 

4 4sin

cos 2 x

+4

2

x

≥4

x

⇒ (b + c )(c + a)(a + b) ≥ 8 abc ⇒ Option (b) does not hold. 1/ 3 1  a b c  a b c Again,  + +  ≥ ⋅ ⋅  3  b c a  b c a a b c + + ≥3 ⇒ b c a ⇒ Option (c) does not hold.

3 1 + x , we know AM ≥ GM 3 3 ⋅3 3x 1 + x x 3 ⋅3 ≥ 3 . 1 3 ⇒ 2 3 3x ⋅ 3 2 3 x − 1 + 3− x − 1 ≥ ⇒ 3

38. y =

39. Since, AM ≥ GM

44. 3 (a2 + b2 + c 2 ) − (a + b + c )2

a b b c c + + + + ≥6 b a c b a b+c c + a a+ b + + ≥6 ⇒ a b c Hence, the least value is 6.

= 2(a2 + b2 + c 2 − bc − ca − ab) = (b − c )2 + (c − a)2 + (a − b)2 ≥ 0 ⇒ ⇒

40. Since, AM > GM for different numbers. 1

∴ ⇒

(b + c − a) + (c + a − b) > [(b + c − a)(c + a − b)]2 2 c > [(b + c − a)(c + a − b)]1/ 2

Similarly,

b > [(b + c − a)(a + b − c )]

and

a > [(a + b − c )(c + a − b)]1/ 2

1/ 2

41. Here, x1 ⋅ x2 … xn = 1 1

⇒

x1 + x2 + K + xn ≥ ( x1 ⋅ x2 K xn )n n x1 + x2 + K + xn ≥ n

42. Since, AM ≥ GM ∴ and

a2 + b2 + c 2 ≥ (a2 b2c 2 )1/ 3 3 1/ 3 1 1 1 1  1   2 + 2 + 2 ≥  2 2 2 3a b c  a b c 

3(a2 + b2 + c 2 ) ≥ (a + b + c )2 > 9 a2 + b2 + c 2 > 3

⇒ Option (a) holds. Now, a6 + b6 ≥ 12 a2 b2 − 64 If

a6 + b6 + 64 ≥ 12 a2 b2

i.e. if

a6 + b6 + 2 6 ≥ 3 ⋅ 2 2 ⋅ a2 b2

a6 + b6 + 2 6 ≥ (2 6 a6 b6 )1/ 3 3 ⇒ Option (b) does not hold. Again, AM ≥ HM a+ b+c 3 ≥ ∴ 1 1 1 3 + + a b c α 3 ⇒ ≥ ⇒ 3 1+ 1+ 1 a b c ⇒ Option (c) does not hold.

[Q AM ≥ GM]

i.e. if

On multiplying, we get abc > (b + c − a)(c + a − b)(a + b − c ) ⇒ (b + c − a)(c + a − b)(a + b − c ) − abc < 0

∴

1/ 3

43. Since, AM ≥ GM

2

⇒

5

On multiplying, we get 1 2 1 1 1 1 ( a + b2 + c 2 ) ⋅  2 + 2 + 2  3 3a b c 

Inequalities and Quadratic Equation

log x log y log z = = y−z z−x x−y

Targ e t E x e rc is e s

33.

1 1 1 9 + + ≥ a b c α

45. Using AM ≥ GM, one can show

(b + c )(c + a)(a + b) ≥ 8 abc ⇒ ( p − a)( p − b)( p − c ) ≥ 8 abc ⇒ Option (b) holds.

241

Objective Mathematics Vol. 1

5

( p − a) + ( p − b) + ( p − c ) 3 ≥ [( p − a)( p − b)( p − c )]1/ 3 3 p − (a + b + c ) ⇒ ≥ [( p − a)( p − b)( p − c )]1/ 3 3 2p ⇒ ≥ [( p − a)( p − b)( p − c )]1/ 3 3 8 p3 ⇒ ( p − a)( p − b)( p − c ) ≤ 27 Option (a) does not hold. ⇒ 1  bc ca  bc ca Again, +  ≥  ⋅  , etc.   a b 2 a b

Also,

On adding the inequalities, we get bc ca ab + + ≥a+ b+c = p a b c ⇒ Option (c) does not hold. bcx + cay + abz ≥ (a2 b2c 2 ⋅ xyz )1/ 3 3 bcx + cay + abz ≥ 3 xyz or bcx + acy + abz ≥ 3abc c  b  a 2   + 3  + 4   4 1/ 9 2 3  4  3  2  a  b  c   47. ≥         9   2   3  4 

Ta rg e t E x e rc is e s

or

a2 b3c 4 ≤ 42 ⋅ 63 ⋅ 84

n

n∑

⇒

xi2

i =1

 n  ≥  ∑ xi   i =1 

⇒ D=0 ∴Only one possible integral solution for k. p 58. α + α 2 = − 3 3 and α ⋅α 2 = ⇒ α 3 = 1 3 ⇒ α = 1, ω, ω 2 2

2

59. α + β = − p, αβ = p3 and β 2 = α Hence, the roots are 4 and − 2.

50. p = (a1 + a2 + K + an ) = ∑ 2

−1

n

n

2

ai2

i =1

+ 2 ∑ ai aj

n

⇒ p2 − 2q = ∑ ai2 ≥ 0 ⇒ q ≤ i =1

51. D = b − 4ac, D′ = d + 4 ac ⇒

242

If α = 1, then p = − 6 If α = ω, then p = 3 But p>0 ⇒ p≠−6 Hence, p=3

60. Since,

 2 x −1 + y −1 + z −1 ≥ 3 ⋅    3 9 x −1 + y −1 + z −1 ≥ 2

2

a = 25 −b α+β= a 8 b =− 25 25 b=−8

57. (16 + 9 k )x 2 + 2(6 − k )x + (39 + 11k )

Now, (2 − x )(2 − y )(2 − z ) = ( y + z )( z + x )( x + y ) We know, ( y + z ) ≥ 2 yz , ( x + z ) ≥ 2 xz , ( x + y ) ≥ 2 xy ∴ ( y + z )( z + x )( x + y ) ≥ 8 xyz ⇒ (2 − x ) (2 − y ) (2 − z ) ≥ 8 xyz  x −1 + y −1 + z −1   x + y + z  − 1 Also,   ≥   3 3   

⇒

⇒ p2 − 3 p − 4 > 0 and p > 0 ⇒ ( p − 4)( p + 1) > 0 and p > 0 ⇒ p>4 1 1 1 55. α = ,β = ⇒ αβ = 4 − 3i 4 + 3i a

56. For ax 2 − bxy − ay 2 , D = b2 + 4a2 > 0

49. x + y + z = 2

⇒

54. |α − β| > 3p ⇒ (α + β)2 − 4 αβ > 3p

⇒

⇒ Maximum value of P is 42 ⋅ 63 ⋅ 84 . ( x 2 + x22 + K + xn2 )  x1 + x2 + K + xn  48. Here, 1 ≥    n n

= (a + 2 ) + (a − 3) − (2 a − 1) = 0 x = 1is a root. 2a − 1 ∴ Other root = − = Rational, ∀ a, a ≠ − 2 a+2 So,

⇒

 a2 b3c 4  29 ≥  2 3 4  2 ⋅ 3 ⋅ 4  a2 b3c 4 ≤ 2 9 ⋅ 2 2 ⋅ 33 ⋅ 44

53. The sum of the coefficients

Also,

⇒

or

= c 2 (3 a2 − b2 + 4 ab)2

or

46. AM ≥ GM

⇒

52. D = c 2 (3 a2 + b2 )2 − 4 abc 2 (− 6 a2 − ab + 2 b2 )

2

D + D′ = b2 + d 2 > 0

∴Atleast one of D, D′ is positive.

a, b and ax 2 + b| x| + c > 0

are

positive,

therefore

Hence, the equation ax 2 + b| x| + c = 0 has no real root. b c 61. α + β = − , αβ = a a ⇒ [{(α + k ) + ( β + k )} 2 − 4 (α + k )( β + k )] = (α + β )2 − 4 αβ 2 q r b2 4 c ⇒ −4 = 2 − 2 p a a p q 2 − 4 pr b2 − 4 ac ⇒ = p2 a2 b2 − 4 ac  a  =  q 2 − 4 pr  p

⇒

i< j

1 2 p 2

c

α β

62. Since, = ⇒ ⇒

m n

2

α +β m+ n = α −β m− n (α + β )2 (m + n )2 = 2 (α + β ) − 4 αβ (m + n )2 − 4 mn

⇒

ac (m + n )2 = mnb2

= 125 − 125 − 2(5) + (10 ) =0

71. Given, ⇒

7 x 2 = 175

⇒

x 2 = 25

⇒

α + β =α + β 2

2

− p = p2 − 2q p + p = 2q 2

∴

1+ β = − 1

⇒

β = −2

74. α and β are roots of 8 x 2 − 3 x + 27 = 0.

(2 x + 3)( x − 14) = 0 3 x = − , 14 ⇒ 2 But x cannot be negative. Hence, x = 14

(λ + 1)( x − bx ) = (λ − 1)(ax − c ) 2

⇒ (λ + 1)x 2 − [λb + b + λa − a]x + (λ − 1)c = 0 α+β=0 λb + b + λa − a =0 λ+1 λ(a + b) = a − b a−b λ= a+ b

⇒ 4(ac + bd )2 − 4(a2 + b2 )(c 2 + d 2 ) = 0 ⇒

2 abcd − a2d 2 − b2c 2 = 0

⇒

(ad − bc )2 = 0

⇒

1

1

1

 α 2  3  β 2  3 (α 3 )3 + (β 3 )3   +  = (αβ )1/ 3  β α 3/ 8 α+β = = (αβ)1/ 3 (27 / 8)1/ 3 3/ 8 1 = = 3/ 2 4

∴ ⇒ or

α =β =1 αβ = 1 c 2 (a2 − b2 ) = a2 (b2 − c 2 )

⇒

b2 (a2 + c 2 ) = 2 a2c 2

⇒

b2 =

2 a2c 2 a2 + c 2

Hence, a2 , b2 and c 2 are in HP.

76. We have,(1 + m)x 2 − 2(1 + 3 m)x + (1 + 8 m) = 0

69. Discriminant = 0

⇒

3 27 , αβ = 8 8

75. Since, f(1) = 0 and f ( x ) is perfect square.

x 2 − bx λ − 1 = ax − c λ+1

⇒

α+β=

⇒

2 x 2 − 25 x − 42 = 0

⇒

α + β = − α , αβ = β ⇒ α = 1 [Qβ = 0 does not satisfy the equation]

∴

∴

p−q . q−r

⇒

⇒

2 ac a+c

c (a − b) = a(b − c ) ⇒ b =

⇒

67. 2 x 2 − 25 x = 3 (14)

Since,

α =β =1

1

66. Here, q − r + r − p + p − q = 0

⇒

x=±5

∴

= (α + β )2 − 2αβ

68. Given,

2 x + 1 x 2 + 7 2 x + 17 = + 2 9 9 x −7

73. Since, α and β are roots of equation x 2 + x α + β = 0.

65. α + β = − p, αβ = q

⇒

r −7 ≥ 4 3 p

Hence, a, b and c are in HP.

⇒ x = − 2, 3 But x cannot be negative. Hence, x = 3

∴Its roots are 1,

r r ∴   − 14   + 1 ≥ 0 ⇒  p  p

72. x = 1, satisfy a(b − c )x + b(c − a)x + c (a − b) = 0

Squaring on both sides, we get x2 = 6 + x

⇒

 p + r q 2 − 4 pr ≥ 0 ⇒   − 4 pr ≥ 0  2 

2

64. Let x = 6 + 6 + 6 + K ∞ = 6 + x

⇒

2

2

1 1 1 63.  x 3 + 3  − 5  x 2 + 2  +  x +    x x  x   3 2    1 1 1 1    =  x +  − 3  x +  − 5  x +  − 2  +  x +         x x x x   3 2 1 1 1    =  x +  − 5  x +  − 2  x +  + 10    x x x

Given,

The roots of px 2 + qx + r = 0 are real, if

Targ e t E x e rc is e s

⇒

5

70. Given, p, q and r are in AP ⇒ 2q = p + r

Inequalities and Quadratic Equation

4 αβ 4 mn = (α + β )2 (m + n )2 c/a mn = 2 2 b /a (m + n )2

⇒

ad = bc a c = b d

...(i)

Let D be the discriminant of Eq. (i). Roots of Eq. (i) will be equal, if D = 0 ⇒ m2 − 3 m = 0 ⇒ m(m − 3) = 0 ∴

m = 0, 3

77. Let the other root be β. 2 1 =− 4 2 1 β = − −α ⇒ 2 and 4 α 2 + 2α − 1 = 0 Then,

Now,

α+β=−

...(i)

4 α 3 − 3 α = α (4 α 2 − 3) = α (1 − 2 α − 3) [Q 4 α 2 + 2 α − 1 = 0]

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Objective Mathematics Vol. 1

5

= − 2α 2 − 2α = −

1 (4 α 2 ) − 2α 2

1 (1 − 2α ) − 2α 2 1 = − −α =β 2 Hence, 4 α 3 − 3 α is the other root. =−

[Q 4 α 2 = 1 − 2α ] [from Eq. (i)]

83. Since, sin θ and cos θ are the roots of the equation

∴

B − 4 AC = 0

⇒

(c − a)2 − 4(a − b)(b − c ) = 0

ax 2 + bx + c = 0.

2

∴

⇒ c 2 + a2 − 2ca − 4ab + 4ac + 4b2 − 4bc = 0 ⇒

c + a + 2 ac + 4b − 4b(c + a) = 0

⇒

(c + a)2 + (2 b)2 − 2 ⋅ 2 b(c + a) = 0

⇒

[(c + a) − (2 b)]2 = 0

2

2

⇒ ⇒ Hence, a, b and c are in AP.

c + a − 2b = 0 2b = a + c

Ta rg e t E x e rc is e s

79. Let the roots be α and β.

Now,

n α and = l β n α = + l β

p q β + αβ α n n − + α + β + αβ l =0 = = l n αβ l

80. The given equation can be written as (ab − 1)x 2 + (a + b)x − ab = 0 Here,

85. The given equation can be written as

D = (a + b)2 + 4ab(ab − 1) = (a + b) − 4ab + 4(ab) 2

∴Roots are always real.

81. Put x − 1 = t ⇒ x − 1 = t 2 ⇒ x = t 2 + 1 The given equation reduces to t 2 + 1 + 3 − 4 t + t 2 + 1 + 8 − 6 t = 1, where t ≥ 0 ⇒|t − 2| + |t − 3| = 1, where t ≥ 0. This equation will be satisfied, if 2 ≤ t ≤ 3 ∴ 2 ≤ x − 1 ≤ 3 or 5 ≤ x ≤ 10 Hence, the given equation is satisfied for all values of lying in [5, 10].

82. Let α and β be the roots of the given equation. b c and αβ = a a 1 1 α 2 + β 2 (α + β )2 − 2αβ Given, α + β = 2 + 2 = = α β (αβ )2 (αβ )2

244

(ay + a′ )x 2 + (by + b′ )x + (cy + c ′ ) = 0

2

= (a − b)2 + (2 ab)2 ≥ 0, ∀ a, b

Then, α + β = −

500 h x 500 h and time taken by the car to cover 500 km = ( x + 25) 500 500 50 50 Given, = + 10 ⇒ = +1 x ( x + 25) x ( x + 25) 50 50 ⇒ − =1 x ( x + 25) 50( x + 25 − x ) =1 ⇒ x( x + 25) ⇒ 1250 = x 2 + 25 x ⇒ x 2 + 25 x − 1250 = 0 2 ⇒ x + 50 x − 25 x − 1250 = 0 ⇒ ( x + 50 )( x − 25) = 0 Q x ≠ − 50 or x = 25 Hence, the speed of the bus = 25 km/h and speed of the car = (25 + 25) = 50 km/h. Time taken by the bus to cover 500 km =

2b = a + c −n , αβ = l p q + + q p

Now, (sin θ + cos θ )2 b2 ∴ a2 ⇒ b2 2 ⇒ b + c2 Hence, (a + c )2 car = ( x + 25) km/h

As x = 1satisfy given equation. ∴ α = 1is the root ⇒ α =β =1 a−b αβ = 1 = ∴ b−c

Then, α + β =

b c and sin θ cos θ = a a = 1 + 2 sin θ cos θ 2c a + 2 c = 1+ = a a = a(a + 2c ) = a2 + 2 ac = a2 + 2 ac + c 2 = (a + c )2 = b2 + c 2

sin θ + cos θ = −

84. Let the speed of the bus = x km/h, then the speed of the

Aliter

⇒

⇒

Hence, bc 2 , ca2 and ab2 are in AP.

78. Since, the roots are equal.

2

⇒

b2 2 c − 2 b a2 a = b − 2 ca − = a c2 c2 2 a 2ca2 = bc 2 + ab2

The condition that x may be rational function of y is, (by + b′ )2 − 4(ay + a′ )(cy + c ′ ) is a perfect square i.e. (b2 − 4ac )y 2 + (2 bb′ − 4ac ′ − 4a′ c )y + b′ 2 − 4a′ c ′ is a perfect square ⇒ D = 0 i.e. 4 (bb′ − 2 ac ′ − 2 a′ c )2 − 4(b2 − 4ac )(b′ 2 − 4a′ c ′ ) = 0 or (ac ′ + a′ c )2 − 4aa′ cc ′ = abb′ c + a′ bb′ c − a′ bb′ c − a′ c ′ b2 − acb′ 2 2 ⇒ (ac ′ − a′ c ) = (ab′ − a′ b)(bc ′ − b′ c )

86. The corresponding quadratic equation of the given expression is a(b − c )x 2 + b(c − a)xy + c (a − b)y 2 = 0 2

⇒

x  x a(b − c )  + b(c − a) + c (a − b) = 0  y y

x =X y Then, a(b − c )X 2 + b(c − a)X + c (a − b) = 0

Let

⇒ r ( x + q + x + p) = ( x + p)( x + q ) or x 2 + x( p + q − 2 r ) + pq − r ( p + q ) = 0

(bc + ab − 2 ac )2 = 0

⇒

Let its roots be α and − α. Then, α + (− α ) = − ( p + q − 2 r ) ⇒ p + q = 2r Also, product = α (− α ) = pq − r ( p + q ) 1 ( p + q )2 = pq − = − ( p2 + q 2 ) 2 2

b(a + c ) = 2 ac 2 ac b= a+c

⇒ Hence, a, b and c are in HP.

87. ax 2 − bx + c = 0 b a c αβ = a 1 y= 2 x

⇒

α+β=

and Clearly,

93.

b 1 − +c =0 y y a − b y + cy = 0 ⇒ ⇒ b2 y = (a + cy )2 1 1 ⇒ (a + cy )2 = b2 y has roots 2 , 2 . α β

88. α + β =

b b − 2 ac = a c2 2 2 − bc = ab − 2 a2c b c 2a + = c a b −

⇒ ⇒

2

c a b are in AP. , and ⇒ a b c a b c are in HP. i.e. , and c a b

89. α + β =

4+ 5 5+ 2

and αβ =

8+ 2 5 5+ 2

2αβ HM = α+β =

16 + 4 5 =4 4+ 5

90. Since, x 2 − 3 x + 2 i.e. ( x − 2 )( x − 1) is a factor of x 4 − px 2 + q. ∴

x = 2, x = 1are roots of x 4 − px 2 + q = 0

⇒ 16 − 4 p + q = 0 and 1 − p + q = 0 On solving these equations, we get p = 5, q = 4

91. (3 x )2 − 36 ⋅ 3 x + 243 = 0 Put 3 = y, the equation becomes y 2 − 36 y + 243 = 0 ⇒ ( y − 9)( y − 27 ) = 0 ⇒ y = 9, 27 ⇒ 3x = 9 or 3 x = 27 ⇒ x =2 or x=3 ∴Solution pair is {2, 3}. x

∴

| x + 1| = 0 ⇒ 1 + | x | − | x + 1| = 0 | x + 1| = 0 ⇒

∴

x + 1 > 0 and

x≠0

x = − 1, x > 0

Aliter ( x + 1)2 >0 x x > 0 ∪ { − 1}

⇒

1 1 + α 2 β2

⇒

| x + 1| | x + 1|2 + | x + 1| = | x| | x|  1 ( x + 1) ⇒| x + 1|  + 1− =0 | x|  | x|

a⋅

⇒

5

1 1 1 + = x+ p x+q r

94. Given,| x − 2| = x 2 On taking x = 1and x = − 2, the equation is satisfied. Hence, its solution is { − 2, 1}.

95. Given equation is| x|2 − 3| x| + 2 = 0 ⇒ (| x| − 1) (| x| − 2 ) = 0 Either| x| = 1 or | x| = 2 ⇒ x = ± 1, x = ± 2 Hence, the given equation has four solutions.

96. Put 2 x = y, the equation becomes 16 y 2 = 8 y + 48 ⇒

2 y2 − y − 6 = 0

Targ e t E x e rc is e s

⇒

92. We have,

Inequalities and Quadratic Equation

The given expression will be perfect square, if the discriminant of the corresponding equation of the given expression is zero. ∴ b2 (c − a)2 − 4a(b − c )⋅ c (a − b) = 0

⇒

( y − 2 ) (2 y + 3) = 0 3 ⇒ y = 2, − 2 But 2 x cannot be negative. Hence, 2 x = 2 ⇒ x = 1

97. 5 − 2 6 =

1 5+ 2 6

∴Given equation (5 + 2 6 )x Becomes y + ⇒

2

−3

+ (5 − 2 6 )x

⇒

−3

y = 5± 2 6 (5 + 2 6 )x

2

−3

= (5 + 2 6 )1

(5 + 2 6 )x

2

−3

= (5 + 2 6 )−1

⇒

x − 3=1

or

x2 − 3 = − 1

⇒ Hence, or

= 10

2 1 = 10, where y = (5 + 2 6 )x − 3 y y 2 − 10 y + 1 = 0

⇒ ⇒

2

2

x 2 = 4 or x=±2 x=± 2

x2 = 2

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Objective Mathematics Vol. 1

5

98. The given equation can be written as  5    3

3x

⇒

x

 5 +   =2  3

⇒

 5 On putting   = t, the equation becomes  3

⇒ k 2 + 2 k − 3 ≥ 0 ⇒ (k + 3)(k − 1) ≥ 0 ⇒ k ≤ −3 or k ≥ 1 Therefore, number of solutions is infinite.

t3 + t −2 = 0 t − 1 + ( t − 1) = 0 3

103. The given equation can be written as

⇒ ( t − 1) ( t + t + 1) + ( t − 1) = 0 2

⇒

( t − 1) ( t + t + 2) = 0

⇒

t = 1 or t + t + 2 = 0

 3    2

2

2

 3 Put    2

But t + t + 2 = 0 does not have real solutions. 2

x

 5 Therefore, t = 1 ⇒   = 1 ⇒  3

⇒

x=0

x 2 − px + q = 0

⇒

α+β=p αβ = q

⇒

= [α 1/ 2 + β1/ 2 + 2(αβ)1/ 4 ]2 = [ α + β + 2 αβ + 2(αβ )

Ta rg e t E x e rc is e s

= [ p + 2 q + 2(q )

]

= p + 6 q + 4q

p+2 q

1/ 4

α 1/ 4 + β1/ 4 = [ p + 6 q + 4 q1/ 4

p + 2 q ]1/ 4

100. The function y = x ( x − 3) is defined for x ( x − 3) ≥ 0 x≥3

...(i)

The given equation can be written as 9| x|2 − 19| x| + 2 = 0 ⇒ ⇒

101. We have, 2 x + 2 ⋅ 33 x /( x − 1) = 9 = 32 3x log 3 = 2 log 3 x −1

 3x  ⇒ ( x + 2 ) log 2 +  − 2 log 3 = 0  x −1 

x = − 2 or

x = 1−

log 3 log 2

x

246

⇒

1  1 − =1 x  x  1  1 x + − 1 = [x] +  x  x x − [x] +

=1

[Qb > 0]

sin 2 α = 2 x − x 2 0 ≤ sin 2 α ≤ 1

Since, ⇒

⇒

0 ≤ 2 x − x2 ≤ 1 x − 2 x ≤ 0 and 2

x2 − 2 x + 1 ≥ 0

x ∈[0, 2 ] and x ∈ R x ∈[0, 2 ] (m − 1)2 ≥ 42

⇒ | m − 1| ≥ 4 ⇒ m − 1 ≤ − 4 or m − 1 ≥ 4 ⇒ m ≤ − 3 or m ≥ 5 ∴Their product > 0, i.e. m + 4 > 0 ⇒ m>−4 Hence, −4 < m ≤ − 3

107. If x − a < 0,| x − a| = − ( x − a) ⇒

1 102. We have, f ( x ) + f   = 1   ⇒

 3    2

[Q t = 1]

1/ x

∴Equation becomes x 2 + 2 a( x − a) − 3 a2 = 0

  1 ( x + 2 ) log 2 + log 3 = 0   x −1

⇒

(t − 1) (t 2 + t + 2 ) = 0

106. (m + 1)2 − 4 (m + 4) ≥ 0

1 In the domain (i), the required solutions are −2, − . 9

⇒

= t, the equation becomes t 3 + t − 2 = 0

⇒

⇒ ⇒

(9| x| − 1) (| x| − 2 ) = 0 1 | x| = 2 or | x| = 9

( x + 2 ) log 2 +

1/ x

105. Given, x 2 − 2 x + sin 2 α = 0

⇒

1 ∴Solutions of the given equation are ± 2, ± . 9

⇒

=2

[Qc < 0] αβ = c < 0 Since, αβ < 0 and α < β ⇒ α is a negative root. α + β < 0 and α < 0 ⇒ β 0

⇒

λ2 + 4λ + 4 = 0

⇒

(λ + 2 )2 = 0 λ = −2

This quadratic expression will be a perfect square, if the discriminant of its corresponding equation is zero. Hence, 4 (a + b + c )2 − 4 × 3 (bc + ca + ab) = 0 ⇒

(a + b + c )2 − 3 (bc + ca + ab) = 0

⇒ a2 + b2 + c 2 + 2 ab + 2 bc + 2ca

But sec θ ≤ − 1or sec θ ≥ 1

x + y 3 = 8 + 2 = 10

D=0 λ2 − 4 (λ + 1) (−1) = 0

x 2 + 2 (a + b + c ) x + 3 (bc + ca + ab) .

6 ± 36 − 32 6 ± 2 1 1 = = , 16 16 2 4

Hence, the given equation has no roots. 1 1 114. Given, log 2 x + = 3+ log 2 x 3 1 = log 2 y + log 2 y 1 ⇒ log 2 x = 3 and log 2 y = 3 ⇒ x = 2 3 and y = 21/ 3

⇒

∴

Either x = 1 x or x = ,x>0 2 Hence, x = 1and 4

Hence,

[when y ≠ 1]

120. Given quadratic expression is

⇒

113. sec θ =

=1

119. As (λ + 1) x 2 + 2 = λx + 3 has only one solution.

5

So, the given equation has no real solution. x

1  + ... ∞ 4

∴ Solutions are (− 3, 9) (− 4, 10 ) and (5, 1.)

e sin x > e

112. Given, x

+ 7 x + 12

1  1  + − 3  3

⇒ x 2 + 7 x + 12 = 0 ⇒ x = − 3, − 4 ⇒ y = 9, 10 Again, when y = 1, x = 5

1 − 4= 0 t t =2 ± 5 e sin x = 2 + 5, 2 − 5

But e sin x cannot be negative. ∴ e sin x = 2 +

2

1 k ⋅ (k − 1)

 1   1   1  1 =  +  +  +… k(k − 1)  1 ⋅ 2   2 ⋅ 3  3 ⋅ 4

110. [ x ] − [ x ] − 2 = 0 [ x ] = 2, − 1 x ∈ [− 1, 0 ) ∪ [2, 3)

5

∑

2

⇒

[Q 5 x > 0 ]

⇒ sin (e x ) ≥ 2, which is not possible for any real x.

x = a (1 ± 2 )

2

1  1 = t −  +2 t  t

Targ e t E x e rc is e s

x 2 − 2 a ( x − a) − 3 a2 = 0

Inequalities and Quadratic Equation

115. sin (e x ) = 5 x + 5 − x, let 5 x = t , then

108. For x ≥ a, the given equation becomes

⇒

− 3 (bc + ca + ab) = 0 a2 + b2 + c 2 − ab − bc − ca = 0

1 [2 a2 + 2 b2 + 2c 2 − 2 ab − 2 bc − 2ca] = 0 2 1 ⇒ [(a2 + b2 − 2 ab) + (b2 + c 2 − 2 bc ) 2 + (c 2 + a2 − 2ca)] = 0 1 [(a − b)2 + (b − c )2 + (c − a)2 ] = 0 ⇒ 2 which is possible only when (a − b)2 = 0 , (b − c )2 = 0 and (c − a)2 = 0 i.e. a = b = c

⇒

[Q x ≠ y ]

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Objective Mathematics Vol. 1

5

121. ( x 2 + x − 2 )( x 2 + x − 3) = 12

…(i)

Put x + x = y, so that Eq.(i) becomes ( y − 2 )( y − 3) = 12 ⇒ y 2 − 5y − 6 = 0 ⇒ ( y − 6) ( y + 1) = 0 ⇒ y = 6, − 1 When y = 6, we get x2 + x − 6 = 0 ⇒ ( x + 3) ( x − 2 ) = 0 ⇒ x = − 3, 2 When y = − 1, we get x2 + x + 1 = 0 which has non-real roots and sum of roots is − 1. 2

122. Here, x = 0 is not a root. Divide both the numerator and

As product of roots is negative, so the roots must be real.

128. A = a(b − c ) (a + b + c ) B = b(c − a) (a + b + c ) C = c ( a − b) ( a + b + c ) Now, Ax 2 + Bx + C = 0 ⇒ (a + b + c ) [a (b − c ) x 2 + b (c − a) x + c (a − b)] = 0 Given that roots are equal. ∴ D=0 ⇒ b2 (c − a)2 − 4ac (b − c ) (a − b) = 0

x + 3/ x = − 5 has two irrational roots and x + 3/ x = 3 has imaginary roots.

⇒ b2c 2 − 2 ab2c + b2 a2 − 4a2 bc + 4acb2 ⇒ b c + b a + 4a c + 2 ab c − 4a2 bc − 4abc 2 = 0

x − 2x + 3 = 0 …(i) α 2 − 2α + 3 = 0 …(ii) β 2 − 2β + 3 = 0 α 2 = 2α − 3 α 3 = 2α 2 − 3 α P = (2α 2 − 3 α ) − 3 α 2 + 5 α − 2

⇒ and ⇒ ⇒ ⇒

+ 4a2c 2 − 4abc 2 = 0 2 2

2

Ta rg e t E x e rc is e s

= − (α 2 − β 2 )2 < 0

denominator by x and put x + 3/ x = y to obtain 4 5 3 + = ⇒ y = − 5, 3 y + 1 y−5 2

123. Given α and β are roots of equation

= − α + 2α − 2 = 3 − 2 = 1

⇒

[using Eq. (i)]

Similarly, we have Q = 2 Now, sum of roots is 3 and product of roots is 2. Hence, the required equation is x 2 − 3 x + 2 = 0.

124. Since, α and β are the roots of the equation

3

3

2

−3

⇒ (31 + 8 15 )x ⇒

2

+ 1 = (32 + 8 15 )x −3

+ 1x

x − 3=1 2

2

or

−3

2

bc + ab = 2 ac 1 1 2 + = ⇒ a c b Hence, a, b and c are in HP.

129. Q ax 2 − bx + c = 0 ∴

b a c αβ = a (a + cy )2 = b2 y α+β=

and Also,

c 2 y 2 − (b2 − 2 ac ) y + a2 = 0

2   b 2 c c  ⇒   y 2 −   − 2    y + 1 = 0  a  a  a  

2

125. (31 + 8 15 )x

2 2

⇒

⇒

2 x − 35 x + 2 = 0. ∴ 2α 2 − 35 α = − 2 2 ⇒ 2α − 35 = − α and 2 β 2 − 35 β = − 2 −2 ⇒ 2β − 35 = β

2 2

(bc + ab − 2 ac )2 = 0

2

 − 2  − 2 Now, (2α − 35)3 (2β − 35)3 =    α   β  8 × 8 64 = 3 3 = = 64 [Qα β = 1] 1 α β 2

−3

= (32 + 8 15 )x

2

−3

x = ± 2 [Q a + b = (a + b)n ] n

n

c α Hence, the given expression is b 2 b (b2 − 2 ac ) (α + β 2 ) = c a2c

126. aα 2 + c = − bα, aα + b = −

127. Since, α and β are roots of x 2 + px + q = 0.

248

Now, for equation x 2 − 4 qx + 2 q 2 − r = 0, product of roots is 2q 2 − r = 2 (αβ)2 − (α 4 + β 4 )

⇒

(αβ)2 y 2 − (α 2 + β 2 )y + 1 = 0

⇒

y 2 − (α −2 + β −2 )y + α −2β −2 = 0

⇒

( y − α −2 ) ( y − β −2 ) = 0

Hence, the roots are α −2 and β −2 .

130. Here, α 4 + β 4 = (α 2 + β 2 )2 − 2α 2β 2 = [(α + β )2 − 2αβ ]2 − 2 (αβ )2 2

 1 1 =  p2 + 2  −  2 p4 p  = p4 +

1 +2 2 p4 2

 1  =  p2 −  + 2 + 2 ≥2 + 2  2 p2 

131. x1( x − x2 )2 + x2 ( x − x1 )2 = 0 ⇒ x 2 ( x1 + x2 ) − 4 x x1 x2 + x1 x2 ( x1 + x2 ) = 0 D = 16 ( x1 x2 )2 − 4 x1 x2 ( x1 + x2 )2 > 0

[Q x1 x2 < 0 ]

∴ α + β = − p and αβ = q 4 Now, α and β 4 are roots of x 2 − rx + q = 0.

The product of roots is x1 x2 < 0.

∴

Thus, the roots are real and of opposite signs.

α +β =r 4

4

and α β = q 4 4

γ is its real root. ⇒ ( x − γ )( x 2 − 2αx + α 2 + β 2 ) = x 3 + qx + r

( x + a)( x + 1991) = − 1 ( x + a) = 1 x + 1991 = − 1 a = 1993 x + a = −1 x + 1991 = 1 a = 1989

On comparing coefficients, we get − γ − 2α = 0 ⇒ γ = − 2α Since, γ 3 + qγ + r = 0 ⇒ (− 2α )3 + q(− 2α ) + r = 0

133. Let f ( x ) = − 3 + x − x 2 . Then, f ( x ) < 0 for all x, because coefficient of x 2 is less than 0 and D < 0. Thus, LHS of the given equation is always positive, whereas the RHS is always less than zero. Hence, there is no solution.

134. x = 3 cos θ and y = 3 sin θ z = 2 cos φ, t = 2 sin φ ∴ 6 cos θ sin φ − 6 sin θ cos φ = 6 ⇒ sin (φ − θ ) = 1 ⇒ φ = 90 ° + θ ⇒ P = xz = − 6 sin θ cos θ = − 3 sin 2θ ⇒ Pmax = 3

135. D > 0 ⇒ (a − 3)2 + 4 (a + 2 ) > 0 ⇒

a2 − 6 a + 9 + 4 a + 8 > 0

⇒

a2 − 2 a + 17 > 0

⇒

α + β = − p, αβ = q α 2 n + pnα n + q n = 0 β 2 n + pn . β n + q n = 0 On subtraction, α n + β n = − pn α Now, is a root of x n + 1 + ( x + 1)n = 0. β n

αn α  + 1 +  + 1 = 0 β  βn α n + β n (α + β )n + =0 βn βn − pn + (− p)n = 0

∴ ⇒

Then,

α + β = − b, αβ = c

Roots of the required equation are α 3 and β 3 . So, the equation is x 2 − (α 3 + β 3 )x + α 3β 3 = 0

From Eq. (i), we get x 4 − 8 x 3 + 18 x 2 − 8 x + 2 = 1

⇒ x 2 − {(α + β)3 − 3 αβ (α + β)} x + (αβ)3 = 0 ⇒

x 2 − (− b3 + 3 cb)x + c 3 = 0

⇒

x 2 + b(b2 − 3 c )x + c 3 = 0

144. Since,α and β are roots of the equation ax 2 + bx + c = 0.

137. x 3 + (− x )3 = 0, if n ≥ 0 and n is an integer. 138. At x = 2, x 3 − 2(1 + α )x 2 + (4α + α 2 + β 2 )x − 2(α 2 + β 2 ) = 8 − 8 (1 + α ) + 8 α + 2 α 2 + 2β 2 − 2α 2 − 2β 2 = 0 ∴ x = 2, z0 and z0 are three roots.

139. Let α = − 4, β = − 4 ω and γ = − 4ω 2 2

∴Required equation is x 2 − (ω + ω 2 )x + ω 3 = 0 ⇒

2

α 1 2   = 4 =ω  γ ω x2 + x + 1 = 0

140. Let f ( x ) = x 4 + ax 3 + bx 2 + cx − 1 Since, ( x − 1)3 is a factor. ∴ f(1) = 0, f′ (1) = 0, f ″ (1) = 0 ⇒ a + b + c = 0, 3 a + 2 b + c = − 4 and 3a + b = − 6 ⇒ a = − 2, b = 0, c = 2 So, other factor is ( x + 1.)

b a

α+β=−

∴

αβ =

and

n

1 α   = 2 = ω and  β ω

…(i)

This is true only, if n is an even integer.

⇒ ( x − 2 )2 = 3 2 ...(i) ⇒ x − 4x + 1 = 0 ⇒ ( x − 2 )4 = 9 ⇒ x 4 − 8 x 3 + 24 x 2 − 32 x + 16 = 9 ⇒ x 4 − 8 x 3 + 18 x 2 − 8 x + 2 + 6 ( x 2 − 4 x + 1) − 1 = 0

∴

∴

143. Let α and β be the roots of x 2 + bx + c = 0.

136. x = 2 + 3

n

142. Since, α and β are the roots of x 2 + px + q = 0.

⇒

a ∈R 1 1 a2 + 1 = 1− 2 ≥ 2 a +2 a +2 2

∴

i.e. 2α is root of cubic equation x 3 + qx − r = 0.

5 Inequalities and Quadratic Equation

⇒ ⇒ and ⇒ or and ⇒

141. α ± iβ are complex roots of equation x 3 + qx + r = 0 and

Targ e t E x e rc is e s

132. ( x + a)( x + 1991) + 1 = 0

c a

The required equation is  1 1  1 1 x2 −  + ⋅ =0 x+  aα + b aβ + b aα + b aβ + b ⇒ { a2αβ + ab (α + β) + b2 } x 2 ⇒

− { a (α + β ) + 2 b} x + 1 = 0 (ca − b2 + b2 )x 2 − (2 b − b)x + 1 = 0

⇒

cax 2 − bx + 1 = 0

145. x = 2 + 21/ 3 + 2 2 / 3 ⇒

x − 2 = 21/ 3 + 2 2 / 3

⇒

( x − 2 )3 = (21/ 3 + 2 2 / 3 )3

⇒

x 3 − 8 − 6x ( x − 2) = (21/ 3 )3 + (2 2 / 3 )3 + 3 (21/ 3 )⋅ (2 2 / 3 ) (21/ 3 + 2 2 / 3 )

On putting 21/ 3 + 2 2 / 3 = x − 2 in above equation, we get x 3 − 6 x 2 + 12 x − 8 = 6 + 6 ( x − 2 ) ⇒

x 3 − 6x 2 + 6x = 2

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Objective Mathematics Vol. 1

5

146. 1, a1, a2 ,..., an − 1 are roots of x n − 1 = 0 x −1 = ( x − a1 ) ( x − a2 )...( x − an − 1 ) x −1 ⇒ x n − 1 + x n − 2 + x n − 3 + ... + x + 1 = ( x − a1 ) ( x − a2 )... ( x − an − 1 ) n

⇒

Put x = 1on the both sides, we get 1 + 1 + 1 + ... + 1 = (1 − a1 ) (1 − a2 ) ... (1 − an − 1 ) ⇒ (1 − a1 ) (1 − a2 ) ... (1 − an − 1 ) = n 2 cos θ ± 4 cos θ − 4 = cos θ ± i sin θ 2 Take α = cos θ + i sin θ, then β = cos θ − i sin θ ⇒ α n = (cos θ + i sin θ)n = cos nθ + i sin nθ Also, β n = (cos θ − i sin θ)n

On putting x = i and then x = − i , we get 1 − 4i + 6 + 7 i − 9 = (i − α ) (i − β )(i − γ )(i − σ ) and 1 + 4i + 6 − 7 i − 9 = (− i − α )(− i − β )(− i − γ )(− i − σ ) On multiplying these two equations, we get (−2 + 3i ) (−2 − 3i ) = (1 + α 2 )(1 + β 2 ) (1 + γ 2 ) (1 + σ 2 ) ⇒

152. tan θ1 + tan θ 2 + tan θ 3 = (a + 1)

2

147. x =

= cos nθ − i sin nθ Now, α n + β n = 2 cos nθ and α nβ n = 1

13 = (1 + α 2 )(1 + β 2 ) (1 + γ 2 ) (1 + σ 2 )

∴

⇒

Σ tan θ1 tan θ 2 = (b − a) tan θ1 tan θ 2 tan θ 3 = b Σ tan θ1 − ∏ tan θ1 tan(θ1 + θ 2 + θ 3 ) = 1 − Σ tan θ1 tan θ 2 ( a + 1 − b) = =1 1 − (b − a) π θ1 + θ 2 + θ 3 = 4

153. Σα = 1, Σαβ = 0, αβγ = 1 Σ

∴Required equation is x 2 − (α n + β n )x + α nβ n = 0 ⇒

Ta rg e t E x e rc is e s

148.

x 2 − (2 cos nθ) x + 1 = 0

xn − 1 = ( x − a1 ) ( x − a2 )...( x − an − 1 ) x −1 ⇒

x n − 1 + x n − 2 + ... + x + 1

= ( x − a1 ) ( x − a2 ) ...( x − an − 1 ) Put x = − 1, we get (−1)n − 1 (1 + a1 ) (1 + a2 )...(1 + an − 1 ) = 1 − 1 + 1 − ... + 1 0, if n is even = 1, if n is odd Thus, (1 + ω) (1 + ω )...(1 + ω 2

n −1

)= n

149. Let y = − 5 + 4i ⇒ y + 10 y + 41 = 0 2

∴

f ( x ) = x 4 + 9 x 3 + 35 x 2 − x + 4 = x 2 ( x 2 + 10 x + 41) − x ( x 2 + 10 x + 41) +4 ( x 2 + 10 x + 41) − 160 = − 160 ∴f (− 5 + 4i ) = − 160

150. Let x = cos θ, we get 4 cos 3 θ − 3 cos θ = p ⇒

p = cos 3 θ 1 −1 ⇒ θ = cos ( p) and x = cos θ 3 1  ...(i) ⇒ x = cos  cos − 1 ( p) 3  Since, − 1 ≤ p ≤ 1 ⇒ 0 ≤ cos − 1 p ≤ π 1 π or 0 ≤ cos −1 p ≤ , as we know cos x is decreasing. 3 3 π  1 …(ii) cos 0 ≥ cos  cos − 1 p ≥ cos ∴  3 3 From Eqs. (i) and (ii), we get 1 1  ≤ x ≤ 1 ⇒ x ∈ ,1 2  2

151. Since, α, β, γ and σ are the roots of the given equation, 250

therefore x 4 + 4 x 3 − 6 x 2 + 7 x − 9 = ( x − α ) ( x − β) ( x − γ ) ( x − σ )

Now, ⇒

 2  1+ α −α + 1 − 2 = −Σ =Σ − 1  1− α  1− α 1− α 1 = 2Σ −3 1− α 1 1 1 3x 2 − 2 x + + = 3 ( x − α ) ( x − β) ( x − γ ) x − x2 − 1 1 1 1 3−2 + + = = −1 1− α 1− β 1− γ 1−1−1 1+ α = −5 1− α

⇒

154. Here, x → x + 1 ⇒

ax 2 + (b + 2 a)x + a + b + c = 0

But equation given is ⇒ ⇒ ⇒

2 x 2 + 8x + 2 = 0

a = 2, b + 2 a = 8, a + b + c = 2 b = 4, c = − 4 b= −c

155. x 2 + x + 1 = 0 has roots ω, ω 2 . ∴

ax 2 + bx + c = 0

∴Both roots common. a b c = = ⇒ 1 1 1

156. We have, x 3 + 3 x 2 + 3 x + 2 = 0 ⇒

( x + 1)3 + 1 = 0

⇒ ( x + 1 + 1) {( x + 1)2 − ( x + 1) + 1} = 0 ⇒ ⇒ ⇒

( x + 2 ) ( x 2 + x + 1) = 0 −1 ± 3 i 2 x = − 2, ω, ω 2 x = − 2,

Since, a, b, c ∈ R, ax 2 + bx + c = 0 cannot have one real and one imaginary root. Therefore, two common roots of ax 2 + bx + c = 0 and x 3 + 3 x 2 + 3 x + 2 = 0 are ω and ω 2 . b Thus, − = ω + ω2 = − 1 a c = ω ⋅ ω2 = 1 ⇒ c = a ⇒ a = b and a ⇒ a= b=c

ax + bx + c = 0 are α and β, while the roots of 2

So, the given equation has no roots. ⇒ D < 0 ⇒ 4b2 − 4 c < 0

1 1 cx 2 + bx + a = 0 are and . α β

⇒

roots. ⇒ f (5) ≥ 0 ⇒ 25 a + 5 b + 10 ≥ 0 ⇒ 5 a + b ≥ − 2

158. Let α be the common root.

165. f (2 ) > 0 and f (4) > 0

Then, α 2 + iα + a = 0, α 2 − 2α + ia = 0 1 α2 α = ⇒ = a a(1 − i ) − 2 − i ⇒ a2 (1 − i )2 = a (− 2 − i ) 1 a= −i ⇒ 2

⇒

∴ Roots of x + 2 x + 3 = 0 are imaginary. Since, the equations x 2 + 2 x + 3 = 0 and ax 2 + bx + c = 0 are given to have a common root, therefore both roots will be common. Hence, both the equations are identical. ⇒ a : b : c = 1: 2 : 3 1 161. Let α and β be the roots of x 2 − px + q = 0 and α and β be the roots of x 2 − ax + b = 0. Then, α + β = p and αβ = q α 1 Also, α + = a and =b β β 2

 1   (α + β) − α + β    = bq ( p − a)2 α = ⋅ βα β

f( x) = 0 Now, f(0 ) = 1 ≠ 0, so x = 0 is not a root f ( x ) = 1 + 2 x + 3 x 2 + K + (n + 1)x n

⇒

(− a)2 − 4 (1 − 2 a2 ) < 0

⇒

9 a2 − 4 < 0 (3 a + 2 ) (3 a − 2 ) < 0 2 2 − 0 ⇒ 1≤ x ≤ 5 and x < 0 or x > 2 ⇒ 2 0, ∀x ⇒

2 x 2 + 3 x − 27 = (2 x + 9)( x − 3) ≤ 0 9 which gives − ≤ x ≤ 3 leading to 0 ≤ 4 x 2 ≤ 81 2 1< x < 2 x ∈(1, 2 )

170. | x − 1| − 1 ≤ 1 ⇒

0 ≤ | x − 1| ≤ 2

⇒

| x − 1| ≤ 2

⇒ …(i)

(1 − x )f ( x ) = (1 + x + x 2 + K + x n ) − (n + 1)x n + 1 1 − xn + 1 = − (n + 1)x n + 1 1− x 1 − x n + 1 (n + 1)x n + 1 f( x) = − (1 − x ) (1 − x )2 1 − (n + 2 )x n + 1 + (n + 1)x n + 2 = (1 − x )2

So, the equation f ( x ) > 0, ∀x. Hence, no solution.

2 2 2 2 0, so there is no positive real root of

⇒

−

⇒

2

2

⇒

⇒

Discriminant = (2 )2 − 4 ⋅ 1 ⋅ 3 < 0

α 1   (q − b)2 = αβ −  = α 2 β −    β β

− 2 < a < 2 and

Targ e t E x e rc is e s

α α 1 = = λ λ −1 α = 1 and α = − λ λ = −1 2

160. For the equation x 2 + 2 x + 3 = 0,

Now,

⇒

∴The given expression is positive for all real values of x, if D < 0.

159. α + 2α + 3λ = 0 and 2α + 3 α + 5λ = 0

⇒ ⇒

a2 < 4 and 13 a2 < 8

5

166. Since, the coefficient of x 2 is 1 which is positive.

2

⇒

b2 < c

164. f (0 ) = 10 > 0, as f ( x ) = 0 does not have distinct real

One negative root is common. 1 α= 0

Inequalities and Quadratic Equation

157. As the coefficients are in reverse order, the roots of

− 1≤ x ≤ 3 x

x

5 12 171.   +   ≥ 1  13  13 ∴ where, and

cos x α + sin x α ≥ 1 5 cos α = 13 12 sin α = 13

Equality holds for x = 2. If x < 2, both cos α and sin α increase (being positive fractions). So,

cos x α + sin x α > 1

If x < 2. Thus, x ≤ 2.

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Objective Mathematics Vol. 1

5

172. Here, xy ≥ 100

179. Let the roots be α and β. x+ y ≥ xy 2 x + y ≥ 2 ⋅ 10 x + y ≥ 20

⇒ ⇒

∴ Given, ⇒

x2 − x + 1 173. Let 2 =m x + x+1

⇒

(α + β ) − 4ab ≤ 4m2

⇒

4a2 − 4b ≤ 4m2

and discriminant

D>0

⇒

2

181. tan x =

Ta rg e t E x e rc is e s

⇒

− 12 + λ > 0, λ < 16, λ > 12 and 16 (16 − λ ) > 0 ⇒ 12 < λ < 16 ∴ λ ∈{13, 14, 15} Hence, three integral solutions.

177. k − 2 > 0 and 64 − 4 (k − 2 ) (k + 4) < 0 k > 2 and 16 − (k 2 + 2 k − 8) < 0

⇒

(k < − 6 or k > 4)

∴ Least integral value of k = 5

178. (e a − e 2 a + e a − 1) (4 e a − 2 e 2 a + e a − 1) < 0 ⇒ Let

(e 2 a − 2e a + 1) (2e 2 a − 5 e a + 1) < 0 x = ea

⇒

( x − 1)2 (2 x 2 − 5 x + 1) < 0  5 − 17   5 + 17  ⇒ ( x − 1)2  x −  x −  0, f(2 ) < 0, f(3) > 0 and D > 0

k > 2 and k>4

4a2 − 4b > 0 b < a2

f ( x ) = ( x − a)( x − c ) + 2( x − b)( x − d ) ⇒ f (b) = (b − a)(b − c ) < 0 and f (d ) = (d − a)(d − c ) > 0 Hence, f ( x ) = 0 has one root in (b, d ). Also, f (a)f (c ) < 0. So, the other root lies in (a, c ). Hence, roots of the equation are real and distinct.

x≥0 The given inequation can be written in the form 372 − x − x > 1

k 2 + 2 k − 24 > 0

or

a − m ≤ b and 2

180. Given, a < b < c < d . Let

175. The given inequation is valid only when

k > 2 and

a2 − m2 ≤ b

Hence, b ∈ [a2 − m2 , a2 ]

tan x tan x (1 − 3 tan 2 x ) 174. y = = tan 3 x 3 tan x − tan 3 x 1 or y > 3 y< ⇒ 3 1 or 3 < y < ∞ ⇒ −∞ 0 x + x − 72 < 0 ( x + 9) ( x − 8) < 0 x + 9 > 0, ∀x ≥ 0 x − 8< 0 ⇒ x 0 and b2 − 4ac ≤ 0. 1 mx − 1 + ≥ 0 ∴ x mx 2 − x + 1 ⇒ ≥0 x ⇒ mx 2 − x + 1 ≥ 0 as x > 0 Now, mx 2 − x + 1 ≥ 0, if m > 0 and 1 − 4m ≤ 0 1 ⇒ m > 0 and m ≥ 4 1 Thus, the minimum value of m is . 4

183. The equation on simplifying gives x ( x − b) ( x − c ) + x ( x − c ) ( x − a) + x ( x − a) ( x − b) ...(i) − ( x − a) ( x − b) ( x − c ) = 0 Let f ( x ) = x ( x − b) ( x − c ) + x ( x − c ) ( x − a) + x ( x − a) ( x − b) − ( x − a) ( x − b) ( x − c ) We can assume without loss of generality that a < b < c . Now, f (a) = a (a − b) (a − c ) > 0 f (b) = b (b − c ) (b − a) < 0 f (c ) = c (c − a) (c − b) > 0 So, one root of Eq. (i) lies in (a, b) and one root in (b, c ). Obviously the third root must also be real.

184. As a < b < c < d are in AP. ∴f ( x ) = ( x − a)( x − c ) + 2( x − b)( x − d ) ⇒ f (a) = +ve, f (b) = − ve, f (c ) = − ve and f (d ) = + ve ⇒ One root lies in (a, b) and other in (c , d ).

i.e. f (1)⋅ f (2 ) < 0, Hence, the equation must have a root x ∈(1, 2 ). ∴Two distinct real roots.

and ⇒

and ∴ and ⇒

α ≥ 0, β ≥ 0 α − 1≤ 0 β − 1≤ 0 D≥0 α+β≥0 αβ ≥ 0, α + β − 2 ≤ 0, αβ − (α + β ) + 1 ≥ 0 D≥0 1 ≥ 0, 1 − p ≥ 0, 1 − 2 ≤ 0, 1 − p − 1 + 1 ≥ 0 1 − 4 (1 − p) ≥ 0 3 ≤ p≤1 4

192. 6 x 2 − xy − y 2 − 6 x + 8 y − 12 = 6 ( x + λy − 2 )( x − µy + 1) ⇒

193. The given equation can be written as y 2 + y(2 x + m) + (2 x − 3) = 0 = 4 x + 4 xm − 8 x + m + 12 2

= [4 x 2 + 4 x(m − 2 ) + (m − 2 )2 ] + (m2 + 12 ) − (m − 2 )2 = [2 x + (m − 2 )]2 + [12 + 4m − 4]

…(i) …(ii)

For rational factors, D should be a perfect square. ∴ 8 + 4m = 0 ⇒ m = − 2 Aliter Use abc + 2 fgh − af 2 − bg 2 − ch 2 = 0

…(iii)

⇒

2   x 2  x  y  y = z 2 a   + b   + c + 2 a   + 2 b    z  z  z  z 

_ 2

2 f (a) < 0 f (a) < 0 f ( x ) = 2 x 2 − 2(2 a + 1) x + a(a − 1)

∴ ⇒ ⇒ ⇒

f (a) = 2 a2 − 2(2 a + 1) a + a(a − 1) < 0 −a − 3 a < 0 2

a (a + 3) > 0 a ∈ (− ∞, − 3) ∪ (0, ∞ )

189. Since, α < − 1and β > 1 ∴ α + λ = − 1and β = 1 + µ, where λ , µ > 0 c b Now, 1 + + = 1 + αβ + |α + β | a a = 1 + (− 1 − λ ) (1 + µ) + |− 1 − λ + 1 + µ| = 1 − 1 − µ − λ − λµ + |µ − λ| − µ − λ − λµ + µ − λ , if µ > λ = − µ − λ − λµ + λ − µ, if λ > µ c b 1+ + = − 2 λ − λµ or − 2µ − λµ ∴ a a c b In both cases, 1 + + [Q λ , µ > 0] 0

0

2

= 4 x 2 + 4 x(m − 2 ) + (m2 + 12 )

Discriminant > 0 f (a) > 0 1 − >a 2 On solving Eqs. (i), (ii) and (iii), we get a< −2

+

D = (2 x + m)2 − 4(2 x − 3)

Here,

186. Since, both roots of f ( x ) = 0 exceed a. Therefore,

(− 1) (1 − a2 + 2 a − 1) > 0 ⇒ a2 − 2 a > 0 ⇒ a(a − 2 ) > 0 ⇒ a2

1 1 ,µ = 3 2

λ=

…(i) = z 2 [aX 2 + bY 2 + c + 2 aY + 2 bX + 2cXY ] x y where, = X and =Y z z The given expression can be resolved into rational factors when the expression under the brackets given in Eq. (i) can be resolved into rational factors. The condition for this is abc + 2 ⋅ a ⋅ b ⋅ c − a ⋅ a2 − b ⋅ b2 − c ⋅ c 2 = 0 ⇒

Targ e t E x e rc is e s

∴

5

191. Since, f (1) = − f (2 ),

The roots of y 2 − y + 1 − p = 0 lie between 0 and 1.

Inequalities and Quadratic Equation

185. cos 4 x − cos 2 x + 1 − p = 0; 0 ≤ cos 2 x ≤ 1,

[Q here, f = a, g = b, h = c ] a3 + b3 + c 3 = 3abc

195. Given equation is x 2 + 9y 2 − 4x + 3 = 0 or

x 2 − 4x + 9y 2 + 3 = 0

...(i)

Since, x is real. ∴ (− 4)2 − 4(9 y 2 + 3) ≥ 0 ⇒

16 − 4 (9 y 2 + 3) ≥ 0

⇒

4 − 9y 2 − 3 ≥ 0

⇒

9y 2 − 1 ≤ 0

⇒

(3 y − 1) (3 y + 1) ≤ 0 −1 1 ≤y≤ ⇒ 3 3 Eq. (i) can also be written as 9y 2 + 0 y + x 2 − 4x + 3 = 0 Since, y is real. ∴ 0 2 − 4 × 9 ( x 2 − 4 x + 3) ≥ 0 ⇒ ⇒ ⇒

x 2 − 4x + 3 ≤ 0 ( x − 1) ( x − 3) ≤ 0 1≤ x ≤ 3

253

 1 

Case I y = a +

r   x 2 + y 2 + 2 gx + 2 fy + c = 0, then 1 2f mr2 + 2 + 2 gmr + +c =0 mr mr

Objective Mathematics Vol. 1

5



196. If mr ,    satisfy the given equation m 

⇒

mr4

+

+

2 gmr3

cmr2

+ 2 fmr + 1 = 0

Now, roots of given equation are m1, m2 , m3 and m4 . The product of roots, Constant term 1 m1m2 m3 m4 = = =1 4 Coefficient of mr 1

197. Let f ( x ) = 0, where f ( x ) = x 2 − 3 x + k has two roots α and β in (0, 1) for some value of k, then by Rolle’s theorem f ′ ( x ) = 2 x − 3 = 0 has a root in (0, 1)[Q α , β ∈ (0, 1]) but 3 f ′ ( x ) = 0 gives x = ∉ (0, 1), which is contradiction. 2

⇒

Ta rg e t E x e rc is e s

X′

⇒

(a +

Y′

⇒

Y

X′

x = ± 14

∴

b2 − 4ac < 0

Roots are Let

|α| =

= cos 2 (α + β ) [tan 2 (α + β ) + p tan (α + β ) + q ]

y = log0.5 x

⇒

a2 + a − 2 < 0

=

then a2 + a < 2

(a + 2 ) (a − 1) < 0 − 2 < a < 1 or − 2 < log 5 x < 1 1 1 ∴Option (c) is false.

∴ No solution.

⇒ ⇒

=a+

2

Also,

X

O

−15

b )−1

y = (a +

⇒

∴

y=x

2

x − 15 = 1 ⇒ x = ± 4 1 Case II y = a − b ⇒ y = a+ b

and

y = ex

Y

b )x

2

Hence, f ( x ) = 0 has no root in (0, 1) for any value of k.

198.

b ⇒ (a +

[Q a2 − b = 1]

∴

(ay + a′ ) x 2 + (cy + c ′ ) = 0

Since, x is a rational function, the discriminant must be a perfect square. Discriminant = 0 − 4 (ay + a′ ) (cy + c ′ ) = − 4 [acy 2 + (ac ′ + a′ c )y + a′ c ′ ] ∴ Roots of acy 2 + (ac ′ + a′ c )y + a′ c ′ = 0 must be equal. ∴ Discriminant = 0, i.e.(ac ′ + a′ c )2 − 4ac ⋅ a′ c ′ = 0 ⇒ ⇒ ⇒ ⇒

(ac ′ − a′ c )2 = 0 ac ′ − a′ c = 0 ac ′ = a′ c a c a a′ = ⇒ = a′ c ′ c c′

2

 x  x   x  x  ⇒  +  −2     = a (a − 1)  x + 1 x − 1  x + 1  x − 1 2

 2 x2  2 x2 = a (a − 1)  − 2  2 x −1  x − 1

⇒

⇒ z 2 − z − a (a − 1) = 0, where z = ⇒

z=a

or

z=a ⇒ ⇒ ⇒

2 x2 x2 − 1

209. Since α, β, γ and δ are in HP. So, 1 / α, 1 / β, 1 / γ and 1 / δ are in AP and they may be taken as a − 3 d , a − d , a + d , a + 3 d . Replacing x by 1/ x, we get the equation whose roots are 1 / α , 1 / β, 1 / γ, 1 / δ. Therefore, equation x 2 − 4 x + A = 0 has roots a − 3 d , a + d and equation x 2 − 6 x + B = 0 has roots a − d , a + 3 d . Sum of the roots is

a−1 2 x2 =a x2 − 1

2 x 2 = ax 2 − a x=±

a a−2

⇒ ∴

∴

x=± x=±

a ,± a−2

(a − 3 d ) (a + d ) = A = 3 (a − d ) (a + 3 d ) = B = 8

210. (1 + k ) tan 2 x − 4 tan x − 1 + k = 0

a−1 a+1

Since, roots are real, we have (− 4)2 − 4 (1 + k ) (− 1 + k ) ≥ 0

⇒ a < −1 ⇒ All roots are real a a−1 1< a < 2 ⇒ x = ± i, ± a+1 2−a ⇒ Only two roots are real. ⇒ a > 2 ⇒ All roots are real.

206. Symmetric functions are those which do not change by interchanging α and β.

207. sec 2 θ + cosec 2θ = sec 2 θ ⋅ cosec 2θ Sum of the roots is equal to their product and the roots are real. b c − = ⇒ b+c =0 ∴ a a Also, b2 − 4ac ≥ 0 c 2 − 4ac ≥ 0

⇒ ⇒ Further

c (c − 4a) ≥ 0 c − 4a ≥ 0 b2 + 4ab ≥ 0

[Qc > 0]

b + 4a ≤ 0

[Q b < 0]

208. Let the roots be a/r, a, ar, where a > 0, r > 1. Now,

a/ r + a + ar = − p

a(a/ r ) + a(ar )(ar )(a/ r ) = q (a/ r ) (a) (ar ) = 1 ⇒ a3 = 1 ⇒ a = 1

− p− 3> 0 ⇒ p< −3 Hence, option (b) is not correct. Also, 1/ r + 1 + r = − p

16 − 4 (k 2 − 1) ≥ 0

⇒

k2 ≤ 5

 −4 4 We have, tan x1 + tan x2 = −   = 1 + k 1 + k − 1+ k and tan x1 ⋅ tan x2 = 1+ k 4 4 1+ k tan ( x1 + x2 ) = = =2 ∴  − 1 + k 2 1−    1+ k 

...(i)

For k = 2, from Eq. (i), 3 tan 2 x − 4 tan x + 1 = 0 tan x = 1,

1 3

1 π , x2 = tan − 1 4 3 For k = 1, from Eq. (i), 2 tan 2 x − 4 tan x = 0 x1 =

∴

⇒

tan x = 0, 2 ⇒

x1 = 0, x2 = tan −1 2

211. We have, u + v = − p and uv = q ...(i) ...(ii) ...(iii)

Hence, option (c) is correct. From Eq. (i), putting a = 1, we get

⇒

⇒

⇒

⇒

a = 5 / 2, d = 1 / 2

Product of the roots is

2 x 2 = (1 − a) x 2 − 1 + a a−1 a+1

5

2(a − d ) = 4 and 2(a + d ) = 6

2 x2 = 1− a z = 1− a ⇒ 2 x −1 ⇒

…(v)

Inequalities and Quadratic Equation

From Eq. (ii), putting a = 1, we get 1/r + r + 1 = q From Eqs. (iv) and (v), we have − p=q ⇒ p+ q = 0 Hence, option (a) is correct. Now, as r > 1 a/ r = 1 / r < 1 and ar = r > 1 Hence, option (d) is correct.

2

Targ e t E x e rc is e s

2

 x   x   = a (a − 1)  +  x − 1  x + 1

205. We have, 

1   Qr + >2   r

(a)

1 1 u+v −p 1 1 1 1 and ⋅ = + = = = u v uv q u v uv q ∴ The required equation is 1  p x2 −  −  x + = 0  q q ⇒ qx 2 + px + 1 = 0

(b) (u + v ) + uv = − p + q and (u + v ) uv = − pq ∴ The required equation is x 2 − (− p + q ) x + (− pq ) = 0 ⇒ x 2 + ( p − q ) x − pq = 0

...(iv)

⇒

( x + p) ( x − q ) = 0

255

5

(c) u 2 + v 2 = (u + v )2 − 2uv = p2 − 2q u v = (uv ) = q 2 2

Objective Mathematics Vol. 1

and

(d)

216. cos x − y 2 − y − x 2 − 1 ≥ 0

u v u 2 + v 2 p2 − 2q u v and ⋅ =1 + = = v u uv q v u ∴ The required equation is  p2 − 2q  x2 −   x + 1= 0  q 

From Eq. (i),

y ≥ x + 1. So, minimum value of y is 1. 2

cos x − y 2 ≥

y − x2 − 1

where, cos x − y 2 ≤ 0 [as when cos x is maximum (= 1) and y 2 is minimum (= 1,) so cos x − y 2 is maximum]. Also,

qx 2 − ( p2 − 2q ) x + q = 0

Hence,

y − x2 − 1 ≥ 0 cos x − y 2 =

⇒

y − x2 − 1 = 0

y = 1 and cos x = 1, y = x 2 + 1

⇒ x = 0, y = 1

( x − a) ( x − b) − k ( x − c ) = 0 x 2 − (a + b + k ) x + ab + kc = 0

217.

⇒ { − (a + b + k )} 2 − 4 ⋅ 1⋅ (ab + kc ) ≥ 0 ⇒

y − x 2 − 1 is defined when y − x 2 − 1 ≥ 0 or

Now,

( x − a) ( x − b) =k x −c ⇒ ⇒

...(i)

2

∴ The required equation is x 2 − ( p2 − 2q ) x + q 2 = 0

⇒

212.

2

Y

[Q x is real]

k 2 + 2 (a + b − 2c ) + (a − b)2 ≥ 0

X

This is true for all real k.

Ta rg e t E x e rc is e s

∴ Discriminant = 4 (a + b − 2c )2 − 4 ⋅ 1 (a − b)2 ≤ 0 ⇒

(a + b − 2c + a − b) (a + b − 2c − a + b) ≤ 0

⇒

2 (a − c ) 2 (b − c ) ≤ 0

⇒

(a − c ) (b − c ) ≤ 0

⇒

(c − a) (c − b) ≤ 0

⇒

a ≤ c ≤ b or

From the graph,

b≤c ≤ a

213. Since, P( x ) divides both of them. So, P( x ) also divides (3 x 4 + 4 x 2 + 28 x + 5) − 3 ( x 4 + 6 x 2 + 25)

(a1c − ac1 )2 = 4 (ab1 − ba1) (bc1 − b 1c )

P(1) = 4

common implies both roots are common. 4 −1 −1 −3 = ⇒ λ= ,µ = 0 ∴ = 3 λ + µ λ −µ 4

2 x 2 + (6 y )x + 5 y 2 − 1 = 0 Since, x is real. 36 y 2 − 8 (5 y 2 − 1) ≥ 0 ⇒

y2 ≤ 2

− 2≤y≤ 2 ⇒ Eq. (i) can also be rewritten as 5 y + (6 x )y + 2 x − 1 = 0 2

2

Since, y is real. ∴ 36 x 2 − 20 (2 x 2 − 1) ≥ 0 ⇒

36 x 2 − 40 x 2 + 20 ≥ 0

⇒

− 4 x 2 ≥ − 20

⇒

x2 ≤ 5

⇒

On putting these values in Eq. (i), we have c1a1 = b 12 ...(i)

Eq. (i) can be rewritten as

− 5≤x≤ 5

a b c are in AP. , , a1 b 1 c 1 ba1 − ab 1 cb 1 − c 1 b = =k a1b 1 b 1c 1 cb 1 − ac1 and = 2k a1c1

If

which is a quadratic. Hence, P( x ) = x 2 − 2 x + 5

215. 2 x 2 + 6 xy + 5 y 2 = 1

...(ii) ...(iii)

218. For ax 2 + bx + c = 0 and a1 x 2 + b1 x + c1 = 0 have a

= − 14( x 2 − 2 x + 5)

⇒

...(i)

common root

= − 14 x 2 + 28 x − 70

214. Roots of 4 x 2 − x − 1 = 0 are irrational. So, one root

256

f (0 ) = c > 0 Also, the graph is concave downward. ∴ a 0

219. Statement I Given equation is x 2 − bx + c = 0 Let α and β be two roots such that lα − βl = 1 ⇒ (α + β)2 − 4αβ = 1 ⇒

b2 − 4 c = 1

Statement II Given equation is 4abc x 2 + (b2 − 4ac ) x − b = 0. ∴

D = (b2 − 4ac )2 + 16 ab2c = (b2 + 4ac )2 > 0

Hence, roots are real and unequal.

Solutions (Q.Nos. 220-222) Let the other roots of f { f ( x )} = x be λ and δ. ⇒ f { f (λ )} = λ Let f(λ ) = γ ⇒ f (γ ) = γ

...(i)

b−c =1

230. Investigating the nature of the cubic equation of a let f (a) = a3 − 4ab + 8 c f ′ (a) = 3a2 − 4b

B Y

If b < 0, then f ′ (a) > 0

(β, β) (α, α)

O

∴The equation a3 − 4ab + 8 c = 0 has only one real root. X

231. On substituting c = 1 − b in Eq. (i), we get

⇒ If α and β are real so are λ and δ. If α + β = λ + δ ⇒ middle points of AB and CD become same. This is not possible ⇒ α, β and λ , δ cannot be real. Also, if α and β are equal to then λ , δ cannot be real.

223. f ′′ (c 2 ) f ′′ (c1 ) < 0 and f ′ (c1 ) = f ′ (c 2 ) = 0 ⇒ ⇒ and

(a + 2 ) [(a − 1)2 + 3 − 4b] = 0 ⇒

4b − 3 > 0 3 b> 4 1 c< 4

⇒ ⇒

232. A. Roots of the given equation will be opposite sign, if D ≥ 0 and product of roots < 0. (a3 + 8a − 1)2 − 8 (a2 − 4a) ≥ 0

f ′′ (c1 ) − f ′′(c 2 ) > 0 f ′′ (c1 ) > 0 f ′′ (c 2 ) > 0

⇒ c 2 is local maximum and c1 is local minimum for f ( x ) ⇒ f ′ ( x ) = 0 has atleast four times in [c1 − 1, c 2 + 1].

a2 − 4a 0 ∴ Discriminant < 0 ⇒

...(i)

2

a2 − 6 a + 8 < 0

⇒ D. Let y =

[Q coefficient of x 2 > 0]

4 (4a − 1) − 4 (15 a − 2 a − 7 ) < 0 2

⇒

2

a3 − 4ab + 8 c = 0

a≥6

From Eqs. (i), (ii) and (iii), we get a≥6

x + ax + bx + cx + d = ( x − x1 ) ( x − x2 ) ( x − x3 ) ( x − x4 ) Let ( x − x1 ) ( x − x2 ) = x 2 + px + q and ( x − x3 ) ( x − x4 ) = x 2 + px + r ∴ q = x1 x2 and r = x3 x4 x 4 + ax 3 + bx 2 + cx + d = x 4 + 2 px 3 + ( p2 + q + r )x 2 + p(q + r )x + qr ∴ a = 2 p; b = p2 + q + r ; c = p(q + r ); d = qr Clearly,

a ≤ 1 or α+β 0 a> −1 f(0 ) > 0 9a − 5 > 0 5 a> 9

⇒

Solutions (Q. Nos. 229-231) Let

a2 − 7 a + 6 ≥ 0

⇒ ⇒ and

Least value of f ( x ) is D 1 − 1 =− ⇒ D1 = 1 ⇒ D2 = 1 4 4 ∴Least value of g( x ) is D 1 − 2 =− 4 4 Least value of g( x ) accurs at b 7 x=− 2 = ⇒ b2 = − 7 2 2 b22 − 4 c 2 = D2 ⇒ 49 − 4 c 2 = 1 48 ⇒ = c 2 ⇒ c 2 = 12 4 x 2 − 7 x + 12 = 0 ⇒ x = 3, 4 3

⇒ ⇒

D1 = D2

4

⇒ 0 0

Targ e t E x e rc is e s

A

5

229. If a = 2, then

Inequalities and Quadratic Equation

Other roots γ and δ lie on the line y=−x+c There must be two points C and D on the parabola y = ax 2 + bx + c which are image of each other in the line y = x

2 −1 and f(0 ) > 0 ⇒ 9k − 5 > 0 5 ⇒ k> 9 From Eqs. (i), (ii) and (iii), we get k≥6 B. f ( x ) > 0 ⇒ Discriminant < 0 ⇒ 4(4k − 1)2 − 4(15k 2 − 2 k − 7 ) < 0

…(i)

...(i)

f ( x ) = 0 has three positive real roots in GP. ...(ii)

⇒ f ′ ( x ) = 0 will have two distinct real roots. ⇒ 3 x 2 − 2 x + β = 0 has two distinct roots.

…(iii)

∴

D>0 4 − 12β > 0 ⇒ β
0 both

roots

⇒ Also, ⇒

[Q x22 = x1 x3 ]

r >0 x2 ( x1 + x3 ) + x22 = β x2 (1 − x2 ) + x22 = β

⇒

x2 = β > 0

From Eqs. (ii) and (iv), we get 0 < β
0 ⇒ 2(k − 1) > 0 ⇒ k >1 and f(0 ) > 0 ⇒ (2 k + 1) > 0 1 ⇒ k>− 2 From Eqs. (i), (ii) and (iii), we get k≥4 D. Since, k lies between the roots of f ( x ) = 2 x 2 − 2(2 k + 1)x + k(k + 1) ⇒ Discriminant > 0 and f (k ) < 0 ⇒ 4 (2 k + 1)2 − 8k(k + 1) > 0 and

2 k 2 − 2 k(2 k + 1) + k(k + 1) < 0

[ β ] + [γ ] + 2 = [0 ] + [− 1] + 2 = 1

235. x1 + x2 + x1 x2 = a

Now, discriminant ≥ 0 ⇒ 4 (k − 1)2 − 4 (2 k + 1) ≥ 0

258

This is always true and k > 0 or k < − 1. E. y = − x 2 + 3 x − 2 9  9 = −  x 2 − 3x +  − 2  4 4 1  x − 3 1 y= −  ≤ 4  2  4

234. Let f ( x ) = x 3 − x 2 + βx + γ

k 2 − 6k + 8 < 0

⇒ 2 0 (iii) f(0 ) > 0

1  8 k2 + k +  ≥ 0  2 2 −k −k 4 or a < − 4 and f ( x ) has three real roots, if − 4 < a < 4. (–1,4)

g(x) = 4

–1

1 (1,– 4)

g(x) = – 4

2. P( x ) = ax 2 + b with a, b of same sign. P(P( x )) = a (ax 2 + b)2 + b If x ∈ R or ix ∈ R ⇒

1 1 ... = y 4− 4− 3 2 3 2 1 So, y = y2 [Q y > 0 ] 4− 3 2 1 8 y2 + y − 4= 0 ⇒ y = ⇒ 3 2 3 2 8   1 So, the required value is 6 + log 3 / 2  ×   3 2 3 2 4 = 6 + log 3 / 2 = 6 − 2 = 4 9

1 6. Let 4 − 3 2

7. (2 x )In 2 = (3 y )In 3

…(i) 3 In x = 2 In y

x ∈R 2

⇒ P( x ) ∈ R ⇒ P { P( x )} ≠ 0 Hence, real or purely imaginary number cannot satisfy P(P( x )) = 0.

3. Let f ( x ) = x − x sin x − cos x ⇒ f ′ ( x ) = 2 x − x cos x 2

lim f ( x ) → ∞

x→ ∞

lim f ( x ) → ∞

x → −∞

f(0 ) = − 1

⇒

(log x ) (log 3) = (log y ) log 2 (log x ) (log 3) log y = log 2

⇒

Multiply with α 8 on both sides, we get α 10 − 6 α 9 − 2α 8 = 0

x log 2 3 = 2 x − 2 2 x= 2 − log 2 3 x=

α 10 − β10 − 6(α 9 − β 9 ) = 2 (α 8 − β 8 ) a − 2 a8 ⇒ a10 − 6 a9 = 2 a8 ⇒ 10 =3 2 a9

9. x 2 + bx − 1 = 0

5. Let 1 + a = y + ( y1/ 2 − 1) x + y1/ 6 − 1 = 0  y1/ 2 − 1 y1/ 6 − 1 =0   x+ y −1  y −1 

Taking lim on both the sides, we get

⇒ ∴

x2 + x + b = 0

and

1 log 3 4 1 log 4 3 x= = = 1 log 3 4 − 1 1 − log 4 3 −1 log 4 3

y →1

...(ii)

From Eqs. (i) and (ii), we get 2

Rearranging, again

⇒

…(i)

β10 − 6 β 9 − 2β 8 = 0

Similarly,

1 2− log 3 2 2 log 3 2 = 2 log 3 2 − 1

( y1/ 3 − 1) x 2  y1/ 3 − 1 2   x +  y −1 

1 2

α 2 − 6α − 2 = 0

After rearranging, we get

⇒

x0 =

8. an = α n − β n

4. log 2 3 x = ( x − 1) log 2 4 = 2( x − 1) ⇒

− log 2 = log x 1 x= ⇒ 2

⇒

Hence, 2 solutions. ⇒

…(iii)

In Eq. (i), taking log on both sides, we get (log 2 ) {log 2 + log x} = log 3 {log 3 + log y} (log 3)2 (log x ) ⇒ (log 2 )2 + (log 2 ) (log x ) = (log 3)2 + log 2 [from Eq. (iii)] (log 3)2 − (log 2 )2 ⇒ (log 2 )2 − (log 3)2 = (log x ) log 2 ⇒

(0,–1)

…(ii)

Targ e t E x e rc is e s

1. Let y = x 5 − 5 x and g( x ) = − a. Then, it is clear from the

Inequalities and Quadratic Equation

5

Entrances Gallery

1 2 1 1 x + x+ =0 3 2 6 2 x 2 + 3x + 1 = 0 x = − 1, −

Common root is (b − 1) x − 1 − b = 0 ⇒

b+1 b−1

This value of x satisfies Eq. (i). (b + 1)2 b + 1 ∴ + + b=0 b−1 (b − 1)2 b = 3 i, − 3 i, 0 ⇒ 1 3 3 1 10. Since,f  −  ⋅ f  −  < 0, so lies in  − , −  .  2   4  4 2

11. α 3 + β 3 = q ⇒ ⇒

1 2

x=

…(i)

⇒

(α + β)3 − 3αβ (α + β) = q − p3 + 3 pαβ = q q + p3 αβ = 3p

259

Objective Mathematics Vol. 1

5

α β α β x2 −  +  x + ⋅ = 0  β α β α (α 2 + β 2 ) 2 x − x + 1= 0 ⇒ αβ  (α + β)2 − 2αβ x2 −  ⇒  x + 1= 0 αβ    2  p3 + q    p −2   3p   ⇒ x2 −  x + 1= 0   p3 + q   3p   3 2 3 3 ⇒ ( p + q ) x − (3 p − 2 p − 2q ) x + ( p3 + q ) = 0 ⇒ ( p3 + q ) x 2 − ( p3 − 2q )x + ( p3 + q ) = 0

14. Given equations are

⇒

and

⇒ Now, let

x − 6 x − 2 = 0. Q an = α n − β n for n ≥ 1 ∴ a10 = α 10 − β10 8 a8 = α − β 8 ⇒ a9 = α 9 − β 9

∴ ⇒

α 8 (α 2 − 2 ) − β 8 (β 2 − 2 ) 2(α 9 − β 9 ) Q α and β are the roots of   2   x − 6x − 2 = 0  So,  x 2 = 6x + 2  α 8 ⋅ 6 α − β8 ⋅ 6 β  2 ⇒  α = 6α + 2 = 9 9 2(α − β )   2 α − 2 = 6α ⇒  and  β2 = 6 β = 2   ⇒ β 2 − 2 = 6 β  6 α 9 − 6 β9 6 = =3 = 2(α 9 − β 9 ) 2 =

Ta rg e t E x e rc is e s

e sin x − e − sin x = 4 1 e sin x − sin x = 4 e y = e sin x 1 y− =4 y y 2 − 4y − 1 = 0

4 ± 16 + 4 2 y =2 ± 5 ⇒ On substituting the value of y, we get e sin x = 2 ± 5 ⇒

Now, consider a10 − 2 a8 α 10 − β10 − 2(α 8 − β 8 ) = 2 a9 2(α 9 − β 9 )

β10 = 6 β 9 + 2 β 8

…(ii)

15. Given equation is

2

Similarly,

…(i)

ax 2 + bx + c = 0

Since, Eq. (i) has imaginary roots. So, Eq. (ii) will also have both roots same as Eq. (i). a b c Thus, = = 1 2 3 Hence, a : b : c is 1 : 2 : 3

12. Given, α and β are the roots of the equation

Aliter Since, α and β are the roots of the equation x 2 − 6x − 2 = 0 2 ⇒ x = 6x + 2 ⇒ α 2 = 6α + 2 ⇒ α 10 = 6 α 9 + 2α 8

x2 + 2 x + 3 = 0

y=

Now, sine is a bounded function, i.e. − 1 ≤ sin x ≤ 1. ∴ e − 1 ≤ e sin x ≤ e 1  ⇒ e sin x ∈ , e e 

16. Let z = x + iy, given Re( z ) = 1 ∴ x =1 ⇒ z = 1 + iy Since, the complex roots are conjugate of each other. ∴ z = 1 + iy and 1 − iy are two roots of z 2 + α z + β = 0 Product of roots = β ⇒ (1 + iy ) (1 − iy ) = β ∴ β = 1 + y2 ≥ 1 ⇒

β ∈ (1, ∞ )

17. Let the quadratic equation be ax 2 + bx + c = 0 …(i) …(ii)

On subtracting Eq. (ii) from Eq. (i), we get α 10 − β10 = 6(α 9 − β 9 ) + 2(α 8 − β 8 ) ⇒

a10 = 6 a9 + 2 a8 [Q an = α n − β n ] a − 2 a8 ⇒ a10 − 2 a8 = 6a9 ⇒ 10 =3 2 a9

13. Q a2 = 3 t 2 − 2 t

Sachin made a mistake in writing down constant terms. ∴Sum of roots is correct. i.e. α + β =7 Rahul made mistake in writing down coefficient of x. ∴Product of roots is correct. i.e. αβ = 6 ⇒ Correct quadratic equation is x 2 − (α + β) x + αβ = 0 ⇒ x 2 − 7 x + 6 = 0 having roots 1 and 6.

18. Since, α and β are roots of the equation x 2 − x + 1 = 0. 1 (0,0)

(2/3,0)

For non-integral solution, 0 < a2 < 1 ⇒ a ∈ (− 1, 0 ) ∪ (0, 1)

260

Ø It is assumed that a real solution of given equation exists.

⇒ ⇒ ⇒ ⇒

α + β = 1, αβ = 1 1 ± 3i x= 2 1 + 3i 1 − 3i or x= 2 2 x = − ω or − ω 2

Hence, α

2009

+β

2009

then β = − ω = (− ω )

2009

[where, ω = 1] 3

+ (− ω )

2 2009

= − [(ω 3 )669 ⋅ ω 2 + (ω 3 )1339 ⋅ ω ] = − [ω 2 + ω] = − (− 1) = 1

19. Given bx 2 + cx + a = 0 has imaginary roots ⇒

c 2 − 4ab < 0 ⇒

⇒

greater than − 2 but less than 4. b < 4, ∴ D ≥ 0, − 2 < − 2a f(4) > 0 and f(− 2 ) > 0 Now, D ≥ 0 ; 4m2 − 4m2 + 4 ≥ 0 ⇒

− c 2 > − 4ab

…(i)

Let

f ( x ) = 3b2 x 2 + 6bcx + 2c 2

⇒

Here,

3b > 0

⇒

2

So, the given expression has a minimum value. −D ∴Minimum value = 4a 4ac − b2 4 (3b2 ) (2c 2 ) − 36b2c 2 = = 4a 4(3b2 ) 2 2 12 b c [from Eq. (i)] =− = − c 2 > − 4 ab 12 b2

20. Let the roots of x 2 − 6 x + a = 0 be α, 4 β and that of x 2 − cx + 6 = 0 be α, 3 β. ∴ and ⇒ ∴

α + 4β = 6 α + 3β = c a 4 = 6 3 2 x − 6x +

and and

4α β = a 3 αβ = 6

⇒ a=8 8= 0

⇒ ( x − 4) ( x − 2 ) = 0 ⇒ x = 2, 4 and x 2 − cx + 6 = 0 If x = 2, then 2 2 − 2c + 6 = 0 ⇒ c = 5 ∴ x 2 − 5x + 6 = 0 ⇒ x = 2, 3 Hence, the common root is 2.

21. Letα andβ be the roots of equation x 2 + ax + 1 = 0, then α + β = − a and αβ = 1. Now, α − β = (α + β)2 − 4αβ ⇒

α − β = a2 − 4

⇒

−2 < m< 4 f(4) > 0 16 − 8m + m2 − 1 > 0

⇒

m2 − 8m + 15 > 0

⇒ ⇒ and ⇒ ⇒

⇒

a2 − 4 < 5

⇒

a2 < 9

⇒ ⇒

a 0 − ∞ < m < 3 and 5 < m < ∞ f(− 2 ) > 0 4 + 4m + m2 − 1 > 0

…(i)

…(ii)

…(iii)

m2 + 4m + 3 > 0

⇒ (m + 3) (m + 1) > 0 …(iv) ⇒ − ∞ < m < − 3 and − 1 < m < ∞ From Eqs. (i), (ii), (iii) and (iv), we get m lies between − 1 and 3.

24. Let α and β be the roots of equation x 2 − (a − 2 )x − a − 1 = 0 Then, Now,

α + β = a − 2 and αβ = − a − 1 α 2 + β 2 = (α + β)2 − 2αβ

⇒

α 2 + β 2 = (a − 2 )2 + 2(a + 1)

⇒

α 2 + β 2 = a2 − 2 a + 6

⇒

α 2 + β 2 = (a − 1)2 + 5

The value of α 2 + β 2 will be least, if a − 1 = 0 ⇒

a=1

Aliter Since, α + β = (a − 2 ) and αβ = − a − 1 Let f (a) = α 2 + β 2 = (α + β)2 − 2αβ

According to given condition, a2 − 4 < 5

⇒

4 > 0, ∀ m ∈ R b −2 < − 0 ⇒ k < 4 and k > 5 From Eqs. (i), (ii) and (iii), we get k 0, then x = x ∴ x 2 − 3x + 2 = 0 ⇒ ( x − 1) ( x − 2 ) = 0 ⇒ x = 1, 2 Case II When x < 0, then x = − x ∴ x 2 + 3x + 2 = 0 ⇒ ( x + 1) ( x + 2 ) = 0 ⇒ x = − 1, − 2 Hence, four solutions are possible.

34. Since,

one root of the quadratic equation (a2 − 5a + 3)x 2 + (3a − 1) x + 2 = 0 is twice as large as the other, then let their roots be α and 2α. (3 a − 1) ∴ α + 2α = − 2 (a − 5 a + 3) (3 a − 1) 3α = − 2 ⇒ (a − 5 a + 3) 2 and α ⋅ 2α = 2 (a − 5 a + 3) 2 2 2α = 2 ⇒ (a − 5 a + 3) ⇒

31. Since, origin z1 and z2 are the vertices of an equilateral z12

b  b 2a =  − a c c 2a b  b c  =  +  c c  c a 2a b c = + b c a c a b are in AP. , and a b c a b c are in HP. , and c a b −

⇒

So, (1 − p) satisfied the above equation. ∴ (1 − p)2 + p (1 − p) + (1 − p) = 0

∴ ⇒

2

2 b  − b / a  −  =    a c/a c/a 2

So, the other equation is x 2 − px + q = 0 whose roots are equal. Let the roots be α and α. 7 Sum of roots = α + α = ∴ 1 7 α= ⇒ 2 and product of roots = α ⋅ α = q 2 49  7 ⇒   =q ⇒ q =  2 4

262

Let α and β be the roots of the equation. b c Then, and αβ = α+β=− a a 1 1 α 2 + β2 Also given, α + β = 2 + 2 = α β α 2β 2

⇒

28. Since, (1 − p) is a root of quadratic equation

⇒

a2 = 3b

2

f (0 ) = 0 and f (α) = 0 According to the Rolle’s theorem, f′ ( x) = 0 has atleast one root between (0, α ). ⇒ f ′ ( x ) = 0 has a positive root less than α. x 2 + px + (1 − p) = 0

⇒

32. Given equation is ax 2 + bx + c = 0

⇒ 25 − 10 k + k 2 + k − 5 > 0 ⇒

z1 + z2 = − a z1 z2 = b On putting these values in Eq. (i), we get (− a)2 = 3 b Then, and

⇒ ⇒

…(i)

⇒ ⇒

(3 a − 1)2 1 = 2 2 2 9 (a − 5 a + 3) (a − 5 a + 3) (3 a − 1)2 = 9 (a2 − 5 a + 3) 9 a2 − 6 a + 1 = 9 a2 − 45 a + 27 45 a − 6 a = 27 − 1 26 2 a= = 39 3

39. Let α and β be the roots of the equation

α 2 − 5 α + 3 = 0 and β 2 = 5 β − 3

⇒

β − 5β + 3 = 0 2

These two equations show that α and β are the roots of the equation x 2 − 5x + 3 = 0 α + β = 5 and αβ = 3 α β α 2 + β2 Now, + = β α αβ (α + β)2 − 2αβ = αβ 25 − 6 19 = = 3 3 α β and ⋅ =1 β α α β Thus, the equation having roots and is given by β α α β α β x2 −  +  x + ⋅ = 0  β α β α 19 2 x − x + 1= 0 ⇒ 3 ⇒ 3 x 2 − 19 x + 3 = 0

x 2 − (a − 2 ) x − a + 1 = 0 ∴ and ⇒

α + β = a−2 αβ = − (a − 1) s = α 2 + β 2 = (α + β )2 − 2αβ = (a − 2 )2 + 2 (a − 1) = a2 − 4a + 4 + 2 a − 2

∴

2 x2 − 7x + 7

=3 2 x2 − 7 x + 7 = 2 2 x2 − 7 x + 5 = 0 D = b2 − 4ac = (− 7 )2 − 4 × 2 × 5 = 49 − 40 = 9 > 0 Hence, it has two real roots.

36. Given, 3

2

⇒ ⇒ Now,

37. Let y = 20 301 Number of digits = Integral part of (301 log10 20 ) + 1 = Integral part of [301 (log10 10 + log10 2 )] + 1 = Integral part of [301 (1 + 0.3010 )] + 1 = Integral part of [301 × 13010 . ]+ 1 = Integral part of [391601 . ]+ 1 = 391 + 1 = 392

38. Given, log101 log 7 ( x + 7 +

x)= 0

log 7 ( x + 7 +

x ) = (101)

⇒

log 7 ( x + 7 +

x)= 1

( x+7 +

⇒

∴

∴Sum of roots, α + 1 + α + 3 = + a

a−4 2 and product of roots, (α + 1) (α + 3) = b ⇒ α 2 + 4α + 3 = b

⇒

⇒ ⇒

49 x = 441 441 x= 49 x=9

 a − 4  a − 4  + 3= b   + 4  2   2  a2 + 16 − 8 a 4 a − 16 + + 3= b 4 2 2 a + 16 − 8 a + 8 a − 32 + 12 = 4b

∴

a2 − 4b = 4

41. Since, α and β are the roots of x 2 − ax + b2 = 0. Then, and Now,

α+β=a αβ = b2 2 α + β 2 = (α + β )2 − 2αβ = (a)2 − 2 (b)2 = a2 − 2 b2

42. Since, α and β are the roots of the equation Then,

x 2 + 3 x − 4 = 0. α + β = − 3 and αβ = − 4 1 1 β+α −3 3 + = = = α β αβ −4 4

43. Since, α and β are the roots of x 2 + 2 bx + c = 0.

Now,

α + β = − 2b αβ = c 2  α + β b2 − c =   − αβ  −2  α 2 + β 2 + 2αβ − 4αβ 4 α 2 + β 2 − 2αβ (α − β )2 = = 4 4

=

2 x − 42 = − 2 x + 7 x x − 21 = −

2α = a − 4 ⇒ α =

2

⇒

2

x2 + 7 x

x 2 + 441 − 42 x = x 2 + 7 x ⇒

40. Let two consecutive odd integers be(α + 1)and(α + 3).

Then, and

x ) = 71

Again, squaring on both sides, we get ⇒

Hence, at a = 1, s will have minimum value.

0

⇒ x + 7 + x =7 On squaring both sides, we get ( x + 7 ) + x + 2 x 2 + 7 x = 49 ⇒

ds = 2a − 2 da For maximum and minimum, put ds / da = 0 ⇒ 2a − 2 = 0 ⇒ a = 1 d 2s Also, =2 > 0 da2 Now,

Now,

∴ ⇒

= a2 − 2 a + 2

5 Inequalities and Quadratic Equation

⇒

Targ e t E x e rc is e s

35. Since, α 2 = 5 α − 3

44. Since, a, b and c are in GP. ∴

b2 = ac

Given equation is (loge a) x 2 − (2 loge b) x + (loge c ) = 0 Put x = 1, we get loge a − 2 loge b + loge c = 0

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Objective Mathematics Vol. 1

5

⇒ ⇒

2 loge b = loge a + loge c loge b2 = loge ac

⇒ b2 = ac , which is true.

⇒

Hence, one of the root of given equation is 1. Let another root be α. 2 loge b loge b2 ∴ Sum of roots, 1 + α = = loge a loge a loge ac α= −1 ⇒ loge a (loge a + loge c ) = −1 loge a loge c = = log a c loge a

∴ aS n + 1 + bS n + cS n − 1 = 0

Hence, roots are 1 and log a c.

45. Given α , β are the roots of ax 2 + bx + c = 0 and

Ta rg e t E x e rc is e s

α + h, β + h are the roots of px 2 + qx + r = 0. b c ∴ α + β = − , αβ = a a q r and α + h + β + h = − , (α + h ) (β + h ) = p p Now, (α + h ) − (β + h ) = α − β ⇒ [(α + h ) − (β + h )]2 = (α − β )2 ⇒ [(α + h ) + (β + h )]2 − 4 (α + h ) ( β + h ) = (α + β )2 − 4αβ ⇒ ⇒ ∴

q 2 4r b2 4 c − = − p a2 a p2 q − 4 pr p − 4ac = p2 a2 2 2 b − 4ac a = 2 q 2 − 4 pr p 2

2

Hence, the ratio of the square of their discriminants is a2 : p2 .

46. Given, f ( x ) = 2 x 2 + 5 x + 1

…(i) Also, f ( x ) = a ( x + 1) ( x − 2 ) + b ( x − 2 ) ( x − 1) + c ( x − 1) ( x + 1) = a ( x 2 − x − 2 ) + b ( x 2 − 3 x + 2 ) + c ( x 2 − 1) ⇒ f ( x ) = (a + b + c ) x 2 + (− a − 3b) x

+ (− 2 a + 2 b − c ) …(ii) On equating the coefficients of x 2 , x and constant terms in Eqs. (i) and (ii), we get a + b + c = 2, − a − 3b = 5 and− 2 a + 2 b − c = 1 On solving above equations, we get a = − 4, b = − 1/ 3 and c = 19 / 3 Hence, exactly one choice for each of a, b and c.

47. Given, α and β are the roots of equation ax 2 + bx + c = 0. b c and αβ = α+β=− ∴ a a Now, S n + 1 = α n + 1 + β n + 1 = α n + 1 + β n + 1 + α nβ + β nα − α nβ − β nα = α n (α + β ) + β n (α + β ) − αβ (α n − 1 + β n − 1 )

264

b c Sn − Sn − 1 a a − bS n − cS n − 1

= −

= (α + β ) (α n + β n ) − αβ (α n − 1 + β n − 1 )

Sn + 1 =

a

48. Since,sin α andcos α are the roots of ax 2 + bx + c = 0. b c and sin α cos α = a a 2  b 2 (sin α + cos α ) =  −   a

∴ sin α + cos α = − ⇒

⇒ sin 2 α + cos 2 α + 2 sin α cos α =

b2 a2

c b2 = a a2 a (a + 2c ) = b2 1+ 2⋅

⇒ ⇒ ⇒

b2 − a2 = 2 ac

49. Given, 32 x − 2 (3 x + 2 ) + 81 = 0 ∴

(3 x )2 − 2 (3 x ) 32 + 81 = 0 3x = y

Let ∴

y − 18 y + 81 = 0

⇒

( y − 9)2 = 0

2

⇒

y = 9 ⇒ 3 x = 32

∴

[Q 3 x = y]

x =2

50. Let the roots of the equation 2 x 2 + 3 x + 1 = 0 beα andβ. Then, and ∴

−3 2 1 αβ = 2 2 α + β 2 = (α + β )2 − 2αβ α+β=

2

…(i) …(ii)

5  1 9  − 3 =  − 2   = − 1=  2 4  2  4 2

1  1 α 2β 2 =   =  2 4 ∴ Required equation is x 2 − (α 2 + β 2 ) x + α 2β 2 = 0 5 1 ⇒ x2 − x + = 0 4 4 ⇒ 4x 2 − 5x + 1 = 0 and

51. Given equation can be rewritten as x2 −

r2 x + 1= 0 pq

r2 and tan A tan B = 1 pq We know that, A + B + C = 180 ° ⇒ ( A + B) = 180 ° − C ⇒ tan ( A + B) = tan (180 ° − C ) tan A + tan B ⇒ = − tan C 1 − tan A tan B r 2 / pq = − tan C ⇒ 1− 1 ⇒ ∞ = − tan C ∴ C = 90 °

∴ tan A + tan B =

Hence, ∆ABC is a right angled triangle.

= (− p) − 3 q (− p) = − p + 3 pq 3

and

3

α 4 + α 2β 2 + β 4 = (α 4 + β 4 ) + (αβ )4 = (α 2 + β 2 )2 − (αβ )2 2

2

= p4 − 4 p2q + 4 q 2 − q 2 = p4 − 4 p2q + 3 q 2 = p2 ( p2 − 3 q ) − q ( p2 − 3 q ) = ( p − q ) ( p − 3q ) 2

...(i)

On squaring both sides, we get ( x + 1) + ( x − 1) − 2 x 2 − 1 = 4 x − 1 2 x − 2 x − 1 = 4x − 1

⇒

− 2 x − 1 = 2x − 1

54. Given, 9 x = 12 + 147 On squaring both sides, we get 81x = 12 + 147 + 2 12 147 ⇒ 81x = 159 + 84 ⇒ 81x = 243 ∴ x=3 Aliter 9 x = 12 + 147 9 x =2 3 + 7 3 ⇒ ⇒ 9 x =9 3 x = 3 ⇒ ∴ x=3

2π 2π + i sin 7 7 a7 = cos 2 π + i sin 2 π

=

3±i 2

56. Given, a = cos ∴

∴

[Q e

2

2

β 2 + 9 p2β + 15 q 2 = 0

Let

f ( x ) = x 2 + 6 p2 x + 10 q 2

Then,

f (α ) = α 2 + 6 p2α + 10 q 2

…(ii)

iθ

f (α ) > 0 and

[from Eq. (i)]

f ( β ) = β 2 + 6 p2β + 10 q 2

= ( β 2 + 9 p2 β + 15 q 2 ) − (3 p2β + 5 q 2 ) = 0 − (3 p2β + 5 q 2 ) ⇒ f ( β) < 0 Thus, f ( x ) is a polynomial such that f (α ) > 0 and f ( β ) < 0. Therefore, there exists γ satisfying α < γ < β such that f (γ ) = 0.

58. Given equation is λ2 + 8λ + µ 2 + 6 µ = 0. Since, roots are real. ∴ b2 − 4ac ≥ 0 ⇒

(8)2 − 4 ( µ 2 + 6 µ ) ≥ 0

⇒

64 − 4 ( µ 2 + 6 µ ) ≥ 0

⇒

16 − ( µ 2 + 6 µ ) ≥ 0

⇒

µ 2 + 6 µ − 16 ≤ 0

⇒

µ 2 + 8 µ − 2 µ − 16 ≤ 0

⇒ ⇒ ∴

µ ( µ + 8) − 2 ( µ + 8) ≤ 0 ( µ + 8) ( µ − 2 ) ≤ 0 − 8≤µ ≤2

59. Given,| x − 2| + | x + 2| < 4 Case I When x < − 2 ⇒ − ( x − 2) − ( x + 2) < 4 ⇒ − 2x < 4 ⇒ − x < 2 ⇒ x > − 2, which is not true. Hence, no value of x exist. Case II When − 2 < x < 2 ⇒ − ( x − 2) + x + 2 < 4 ⇒ 4 < 4, which is not true. Hence, no value of x exist.

55. x 2 − 3 x + 1 = 0 3− 4

…(i)

and β is a root of x + 9 p x + 15 q = 0, then 2

⇒

Hence, no value of x satisfy the given equation.

2

[Q a7 = 1]

6

= 0 + 3 p2α + 5 q 2

Again, squaring on both sides, we get 4 ( x 2 − 1) = 4 x 2 + 1 − 4 x ⇒ − 4 = + 1 − 4x ⇒ 4x = 5 ⇒ x = 5 / 4 5 But when we put x = in Eq. (i), we get 4 5 5 5 + 1− −1= 4× −1 4 4 4 9 1 ⇒ − = 5−1 4 4 3 1 ⇒ − = 2 ⇒ 1 = 2, which is not true. 2 2

3±

5

= (α 2 + 3 p2α + 5 q 2 ) + 3 p2α + 5 q 2

2

x=

4

α 2 + 3 p2α + 5 q 2 = 0

2

⇒

3

57. Since, α is a root of x 2 + 3 p2 x + 5 q 2 = 0, then

= p4 − 3 p2q − p2q + 3 q 2

⇒

[Q a7 = 1]

= 3 + (a + a + a + a + a + a ) = 3 − 1 = 2 Hence, the required quadratic equation is x 2 + x + 2 = 0. 2

= [ p2 − 2q ]2 − 3 q 2

53. Given equation is x + 1 − x − 1 = 4 x − 1

⇒

= a4 + a5 + 1 + a6 + 1 + a2 + 1 + a + a3

= [(− p)2 − 2q ]2 − 3 (q )2

2

a (1 − a6 ) a − a7 a − 1 S = = = = −1 1− a 1− a 1− a

Product of the roots, P = αβ = (a + a2 + a4 ) (a3 + a5 + a6 )

= [(α + β ) − 2αβ ] − (αβ ) 2

5

Then, sum of the roots, S = α + β = a + a2 + a3 + a4 + a5 + a6

(α 3 + β 3 ) = (α + β )3 − 3 αβ (α + β )

Targ e t E x e rc is e s

∴

Inequalities and Quadratic Equation

52. Q Sum of roots, α + β = − p and αβ = q

= cos θ + i sin θ ]

=1 Also, α = a + a2 + a4 and β = a3 + a5 + a6

Case III When x > 2 ⇒ x −2 + x + 2 < 4 ⇒ 2 x < 4 ⇒ x < 2, which is not true. So, no value of x exist. Hence, the number of solutions of the inequation is 0.

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Objective Mathematics Vol. 1

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60. 2 x − 5 ≤

4x − 7 3

⇒ ⇒

Now, on substituting the value of k in Eq. (iii), we get c −a 1 b + =− 2+ 2 a 2a a +1 c −a c −a c −a b 2+ + =− ⇒ 2a a+ c a 2 (2 a) (a + c ) + (c − a) (c + a) + 2 a (c − a) b ⇒ =− 2 a (a + c ) a

6 x − 15 ≤ 4 x − 7 2x ≤ 8 ⇒ x ≤ 4

61. 3 x + 2 > − 16 ⇒ and ⇒ ∴

x>−6 2 x − 3 ≤ 11 x ≤7 x ∈ (− 6, 7 ] 1 62. We can write, 5 x + 5− x = 5 x + x 5 1 = 5x + x − 2 + 2 5 2 1  x = 5 − x + 2  5 

⇒

a2 + c 2 + 6ac = − 2 ab − 2 bc

⇒ a2 + b2 + c 2 + 2 ab + 2 bc + 2ca = b2 − 4ac ∴

(a + b + c )2 = b2 − 4ac

66. Since, the second equation has imaginary roots. 2 a − 3b 4 c = = =k 3 −4 5 3k 4k 5k a= ,b= ,c = 2 3 4 3k 4k + a+ b 3 = 34 = 2 b + c 4k + 5k 31 3 4

∴

This means 5 x + 5− x > 1

⇒

Thus, sin (e ) > 1, which is not possible. x

63. By hypothesis, a2α 2 + bα + c = 0

…(i)

a β − bβ − c = 0

…(ii)

2 2

and

0

∴

2 3 0

log a log a + log(c + b) log(c − b) = log a log a ⋅ 2⋅ log(c + b) log(c − b) log a{log(c − b) + log(c + b)} = 2(log a)2

268

Given, ⇒ ⇒

⇒

⇒ 7 ≤ t ≤ 12 ⇒ 8 ≤ t + 1 ≤ 13 x + 11 82. >0 x−3 ⇒ ( x − 3) ( x + 11) > 0 ⇒ x < − 11, x > 3 ⇒ x ∈ (− ∞, − 11) ∪ (3, ∞ ) logc + b a + logc − b a

log(c 2 − b2 ) log a2 = 2 log a log a2

From Eq. (ii), From Eq. (i),

ab = b a=1 b= −2

88. Let α and β be the roots, then

k k  α + β  − b / a + = k   = k    c/a  α β αβ

=

(α − β )2 = α 2 + β 2 − 2αβ = ( p2 − 2q ) − 2q = p2 − 4 q

81. 3 ≤ 3 t − 18 ≤ 18 ⇒ 21 ≤ 3 t ≤ 36

83.

= p2 − 2q ⇒

=

⇒

− 3. 3=− p p=−4 3) = q q =1

α 2 + β 2 = (α + β )2 − 2αβ

…(ii)

3 nCr + 1 > nCr + 2 > ... > nCn , then r is equal to

n−1

C r − 1 d r ! nC r = nPr

b 19

4 The exponent of 7 in a 0 c 4

c 18 100

b (−∞, −2) d ( 3, 2 )

6 If n is even and n C0 < nC1 < nC2 < ... < nCr

n +1

C8 : n − 2 P4 = 57 : 16, then the value of n is

a 20

Cr = (k 2 − 3) nCr + 1, then k belongs to

a [− 3, 3 ] c (2, ∞)

d 8

2 Which of the following is incorrect? a

n −1

a

d 17

n 2

b

n−1 2

c

n−2 2

d

n+2 2

7 The number of positive integers satisfying the inequality

C50 is b 2 d None of these

a 9 c 5

Permutation and Combination

6

Work Book Exercise 6.3

n+1

Cn − 2 − n + 1Cn − 1 ≤ 100, are

b 8 d None of these

Some Basic Arrangements and Selections Combinations Each of the different selections made by taking some or all of a number of distinct objects or items, irrespective of their arrangements or order in which they are placed is called a combination.

We find that the sub-job J1 can be completed in n C r ways and sub-job J 2 can be completed in r! ways. Therefore, number of ways of arranging n distinct items, taking r at a time = n C r × r!

Permutations Each of the different arrangements which can be made by taking some or all of a number of distinct objects is called a permutation. 1. Let r and n be positive integers such that 1 ≤ r ≤ n. Then, the number of all permutations of n distinct items or objects taken r at a time is n ( n − 1) ( n − 2) ( n − 3) ... ( n − r − 1)

Clearly, it is consistent with the result stated in point (1), because n

X

e.g. We have, n ( n − 1) ( n − 2) K ( n − r − 1 ) n ( n − 1) ( n − 2)... ( n − r − 1) ( n − r )! ( n − r )! n! = = n Pr ( n − r )! So, the total number of arrangements (permutations) of n distinct items, taking r at time is n Pr or P ( n , r ). =

2. The number of all permutations (arrangements) of n distinct objects taken all at a time is n!. 3. The number of ways of selecting r items or objects from a group of n distinct items or objects is n! = nCr ( n − r )! r ! e.g. Let there be n distinct objects which are to be arranged in a row by taking r at a time. The job of arranging n items by taking r at a time can be divided into two ordered sub-jobs J1 and J 2 given by J1 : Selecting r items from n distinct items. J 2 : Arranging r selected items.

C r × r ! = n ( n − 1) ( n − 2)... ( n − r − 1) = n Pr

Example 25. Seven atheletes are participating in a race. In how many ways can the first three prizes be won? (a) 420 (b) 210 (c) 35 (d) 105 Sol. (b) The total number of ways in which first three prizes can be won is equivalent to the number of arrangements of seven different things taken 3 at a time 7! 7! = 7P3 = = = 210 (7 − 3)! 4!

X

Example 26. How many numbers between 400 and 1000 can be made with the digits 2, 3, 4, 5, 6 and 0? (a) 120 (b) 720 (c) 60 (d) None of the above Sol. (c) Any number between 400 and 1000 will be three digits. Since, the number should be greater than 400, therefore hundred’s place can be filled up by any one of the three digits 4, 5 and 6 in 3 ways. Remaining two places can be filled up by remaining five digits in 5 P2 ways. ∴

Required number = 3 × 5 P2 5! =3× = 60 3!

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Objective Mathematics Vol. 1

6

X

Sol. (a) The numbers are of five digits having 4 or 5 in ten

(n − 2 )(n − 3) = 78 1⋅2

⇒

Example 27. The number of numbers lying between 40000 and 70000 that can be made from the digits 0, 1, 2, 4, 5, 7, if digits can be repeated in the same number, is (a) 2591 (b) 932 (c) 766 (d) None of these

⇒ ⇒ ⇒ ∴ X

thousand’s place and the remaining digits are any of the given six digits. ∴Number of ways to fill ten thousand’s place = 2 P1 = 2 Number of ways to fill other four places = 64

n2 − 5n + 6 = 156 n2 − 5n − 150 = 0 (n − 15) (n + 10) = 0 n = 15

Example 31. The maximum number of points of intersection of 8 circles are (a) 56 (b) 28 (c) 24 (d) 16 Sol. (a) Maximum number of points

∴Required number of numbers = 2 × 64 − 1 = 2591 X

Example 28. In a network of railways, a small island has 15 stations. The number of different types of tickets to be printed for each class, if every station must have tickets for other station, are (a) 230 (b) 210 (c) 340 (d) None of the above

= 8 P2 = 8 × 7 = 56 X

Sol. (b) For each pair of stations, two different types of

X

2 !13!

X

Example 29. In a certain test, a i students gave wrong answers to atleast i questions, where i =1, 2, 3,...,k. No student gave more than k wrong answers. The total number of wrong answers given is (a) a1 + a 2 + ... + a k (b) a1 + a 2 + ... + a k − 1 (c) a1 + a 2 + K + a k + 1 (d) None of the above

Sol. (b) Suppose, the two players did not play at all, so that the remaining (n − 2 ) players played

276

n−2

C 2 matches.

Since, these two players played 3 matches each, hence the total number of matches [given] = n − 2C 2 + 3 + 3 = 84

x+ 2

Also,

and ⇒ Q ∴ ⇒ ⇒ and ∴ X

Px + x

2

x − 11

= a ⇒ a = ( x + 2 )!

P11 = b b=

⇒

= 1 ⋅ (a1 − a2 ) + 2 ⋅ (a2 − a3 ) + K + (k −1) (ak − 1 − ak ) + kak = a1 + a2 + a3 + K + ak

Example 30. In a class tournament, the participants were to play one game with another, two class players fell ill, having played 3 games each. If the total number of games played is 84, then the number of participants at the beginning, was (a) 22 (b) 15 (c) 17 (d) None of these

Example 33. If a denotes the number of permutations of ( x + 2) things taken all at a time,b denotes the number of permutations of x things taken 11 at a time and c denotes the number of permutations of ( x −11) things taken all at a time such that a =182 bc, then the value of x is (a) 15 (b) 12 (c) 10 (d) 19 Sol. (b) We have,

Sol. (a) Total number of wrong answers

X

Example 32. On a railway route, there are 15 stations. The number of tickets required in order that it may be possible to book a passenger from every station to every other, is 15! 15! (a) (b) 13!2! 13! 15! (c)15! (d) 2! Sol. (a) Required number of tickets = 15C 2 = 15!

tickets are required. Now, the number of selections of 2 stations from 15 stations = 15 C 2 . ∴Required number of types of tickets 15! = 2 ⋅ 15C 2 = 2 ⋅ = 15 × 14 = 210 2 !13!

[Q n ≠ − 10 ]

x! ( x − 11)!

Px − 11 = c c = ( x − 11)! a = 182 bc

x! ⋅ ( x − 11)! ( x − 11)! ( x + 2 ) ( x + 1) = 182 = 14 × 13 x + 1 = 13 x + 2 = 14 x = 12 ( x + 2 )! = 182 ⋅

Example 34. In a certain test, there are n questions. In this test, 2 n − i students gave wrong answers to atleast i questions, where i =1, 2, 3, ..., n. If the total number of wrong answers given is 2047, then n is equal to (a) 10 (b) 11 (c) 12 (d) 13

i (1 ≤ i ≤ n − 1) questions wrongly is 2 n − i − 2 n − i −1. The

number of students answering all n questions wrongly is 2 0 . Hence, the total number of wrong answer is n −1

∑ i (2

n−i

− 2 n − i −1 ) + n(2 0 ) = 2047

i =1

⇒ 2 n −1 + 2 n − 2 + 2 n − 3 + K + 2 0 = 2047 ⇒ 2 n − 1 = 2047 n ⇒ 2 = 2048 ⇒ 2 n = 211 ∴ n = 11 X

Example 35. The number of numbers of 9 different non-zero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle, is (a) 2 (4!) (b) ( 4!) 2 (c) 8! (d) None of these

Method 3 Applicable only if the digits used are such that they have the same common difference. (Valid even, if the digits are repeating) Writing all the numbers in ascending order of magnitude, S = (13579 + 13597) + ... + 97513 + 97531 S = (13579 + 99531) + (13597 + 97513) + ... = (111110) 60 times = 6666600 n S = ( l + L), where n = Number of numbers, 2 l = Smallest, L = Largest X

Sol. (b) The required number of numbers = 4 P4 × 4 P4 = (4!)2

Summation of Numbers (3 different ways)

Example 36. The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time, is equal to (a) 93324 (b) 93328 (c) 92324 (d) None of the above Sol. (a) The total number of numbers formed with the digits 2, 3, 4, 5 taken all at a time = Number of arrangements of 4 digits taken all at a time = 4 P4 = 4! = 24

Sum of all the numbers greater than 10000 formed by the digits 1, 3, 5, 7, 9, no digit being repeated.

To find the sum of these 24 numbers, we will find the sum of digits at unit’s, ten’s, hundred’s and thousand’s places in all these numbers.

Method 1

Consider the digits in the unit’s place in all these numbers. Each of the digits 2, 3, 4, 5 occur in 3! i.e. 6 times in the unit’s place.

All possible numbers = 5! = 120 If one occupies the unit’s place, then total number = 24 Hence,1 enjoys unit’s place 24 times.

So, total for the digits in the unit’s place in all the numbers = (2 + 3 + 4 + 5) × 3! = 84. Since, each of the digits 2, 3, 4, 5 occurs 3! times in anyone of the remaining places. So, the sum of the digits in the ten’s, hundred’s and thousand’s places in all the numbers = (2 + 3 + 4 + 5) × 3! = 84 Hence, the sum of all the numbers = 84 (100 + 101 + 102 + 103 ) = 93324

1

Similarly, 1 enjoys each place 24 times. Sum due to 1 = 1 × 24 (1 + 10 + 10 2 + 10 3 + 10 4 ) Similarly, sum due to the digit 3 = 3 × 24 (1 + 10 + 10 2 + 10 3 + 10 4 ) M M M M M M M M 2 Required total sum = 24 (1 + 10 + 10 + 10 3 + 10 4 )

Ø Sum of all the numbers that can be formed with n (non-zero)

digits, when repetition is not allowed and each digit being used (10 n − 1) (sum of digits) once = (n − 1)! (10 − 1) = n − 1! (sum of digits) × (111 … n times)

(1 + 3 + 5 + 7 + 9) X

Method 2 In 1st column, there are twenty four 1’s, twenty four 3’s and so on and their sum = 24 × 25 = 600 Hence, add in vertical column normally, we get Sum = 6666600 5th X X

X X

X X

2nd 1st X X X X

120 Number X 666

X 6

X 6

X 0

X 0 = 6666600

6 Permutation and Combination

Sol. (b) The number of students answering exactly

Example 37. The sum of all the numbers greater than 10000 formed by using digits 0, 2, 4, 6, 8, no digit repeated in any number, is equal to (a) 5199960 (b) 5209960 (c) 5199980 (d) 5299960 Sol. (a) The total number of numbers formed by arranging the digits 0, 2, 4, 6, 8 taken all at a time is 5! = 120. Some of these numbers are less than 10000 and some are greater than 10000. Infact all those numbers having 0 at ten thousand’s place are four-digit numbers.

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6

∴ Required sum = Sum of the numbers formed by using digits 0, 2, 4, 6, 8 − Sum of the numbers formed by using digits 2, 4, 6, 8  105 − 1  = (0 + 2 + 4 + 6 + 8)(5 − 1)!   10 − 1 

X

 104 − 1  − (2 + 4 + 6 + 8) (4 − 1)!   10 − 1 

Sol. (d) Total number of persons = 5 + 3 = 8 When there is no restriction, then they can be seated in a row in 8! ways. But when all the three girls sit together, consider the three girls as one person, we have only (5 + 1) = 6 persons. These 6 persons can be arranged in a row in 6! ways. But the three girls can be arranged among themselves in 3! ways. ∴Number of ways when three girls are together = 6! 3! ∴Required number of ways in which all the three girls do not sit together = 8! − 6! 3! = 6! (56 − 6) = 6! ⋅ 50 = 50 ⋅ 6!

 104 − 1  105 − 1   − 120  = 480   10 − 1   10 − 1  = 5333280 − 133320 = 5199960

Permutations under Certain Conditions The number of all permutations (arrangements) of n different objects taken r at a time, 1. when a particular object is to be always included in each arrangement, is n − 1 C r − 1 × r !.

X

2. when a particular object is never taken in each arrangement, is n − 1 C r × r !. 3. when two specified objects always occur together is n − 2 C r − 2 × ( r − 1)! × 2!.

278

2nd

3rd

when there is no restriction = 10! When the best and the worst papers come together, consider the two as one paper, we have only 9 papers. These 9 papers can be arranged in 9! ways. But these two papers can be arranged among themselves in 2! ways. ∴Number of arrangement when the best and the worst paper do not come together = 10! − 9! 2 ! = 8 ⋅ 9!

nth

Let us first put aside the specified item and select ( r −1) items from the remaining ( n −1) items. This can be done in n − 1 C r − 1 ways. Now, we have r items, namely, one specified item and ( r −1) selected items. These r items can be arranged in r! ways. Hence, the required number of arrangements is n − 1 C r − 1 × r !. 2. If a particular objects is never taken in each arrangement, then we first select r objects from the remaining ( n −1) objects and then they are arranged. So, required number of arrangements is n −1 C r × r !. 3. In order to arrange n distinct items by taking r at a time when two specified objects always occur together, we first select ( r − 2) objects from the remaining ( n − 2) objects. This can be done in n−2 C r − 2 ways. Now, we consider two specified objects temporarily as a single object and mix it with ( r − 2) selected objects. Now, these ( r −1) objects can be arranged in ( r −1)! ways. But two specified objects can be put together in 2! ways. Hence, required number of arrangements is n−2 C r − 2 × ( r − 1)! × 2!

Example 39. In how many ways can 10 examination paper be arranged, so that the best and the worst papers never come together? (a) 9 ⋅ 8! (b) 8 ⋅ 9! (c)10 ⋅ 9! (d) 9 ⋅ 10! Sol. (b) The number of permutation of 10 papers

Proof 1. In order to arrange n distinct items by taking r at a time when a particular object is to be always included in each arrangement, M M M … M 1st

Example 38. There are 5 boys and 3 girls. In how many ways can they be seated in a row, so that all the three girls do not sit together? (a) 8! (b) 8!5! (c) 8! − 5! (d) 50 ⋅ 6!

X

Example 40. The number of ways in which a committee of 5 can be chosen from 10 candidates so as to exclude the youngest, if it includes the oldest, is (a) 196 (b) 178 (c) 202 (d) None of the above Sol. (a) There are two different ways of forming the committee: (i) oldest may be included (ii) oldest may be excluded (i) If oldest is included, then youngest has to be excluded and we are to select 4 candidates out of 8. This can be done in 8×7 × 6× 5 8! 8 = 70 ways C4 = = 4! 4! 4× 3×2 (ii) If oldest is excluded, then we are to select 5 candidates from 9 which can be done in 9× 8×7 × 6 9! 9 = 126 ways C5 = = 5! 4! 4 × 3 × 2 × 1 Hence, the total number of ways in which committee can be formed = 126 + 70 = 196

Sol. (b) There are 8 chairs on each side of the table. Let the

Example 41. The number of ways in which the letters of the word VOWEL can be arranged so that the letters O, E occupy only even places, is (a) 12 (b) 18 (c) 24 (d) None of these

sides be represented by A and B. Let four persons sit on side A, then number of ways of arranging 4 persons on 8 chairs on side A = 8 P4 and then two persons sit on side B. The number of ways of arranging 2 persons on 8 chairs on side B = 8 P2 and the remaining 10 persons can be arranged in remaining 10 chairs in 10! ways.

Sol. (a) V

O W E L 2 3 4 5 × × QO, E occupy only even places. ∴O and E can be arranged in two × marked places in 2 P2 ways = 2 ! = 2.

Hence, the total number of ways in which the person can be arranged 8! 8!10! = 8 P4 × 8 P2 × 10! = 4! 6!

1

X

Also, the remaining 3 letters can be arranged in themselves in 3! ways = 3 × 2 × 1 = 6 ∴Required number of words = 3! = 2 × 6 = 12 X

Example 42. The number of ways in which the letters of word FRACTION be arranged so that no two vowels are together, is (a) 17330 (b) 14400 (c) 16440 (d) None of these

Sol. (a) As, p things are excluded. ∴Number of things left = (n − p), out of which r are to be arranged. Now, the number of permutations = P(n − p, r )

Sol. (b) The word ‘FRACTION’ consists of 8 different letters out of which A, I, O are 3 vowels and F, R, C, T, N are 5 consonants. First of all, we arrange the consonants among themselves. This can be done in P (5, 5) = 5! = 120 ways. Let one such way be × F × R × C × T × N × Now, no two vowels are together, if they are put at places marked (× ). The 3 vowels can fill up these 6 places in P (6, 3) ways. Hence, the total number of words = 120 × P(6, 3) = 120 × 6 × 5 × 4 = 120 × 120 = 14400 X

Example 43. A boat is to be managed by eight men of whom 2 can only row on bow side and 3 can only row on stroke side, the number of ways in which the crew can be arranged, is (a) 4360 (b) 5760 (c) 5930 (d) None of these Sol. (b) First, we have to select 2 men for bow side and 3 for stroke side. ∴ The number of selections of the crew for two sides = 5C 2 × 3C 3 For each selection, there are 4 persons, each on both sides who can be arranged in 4! × 4! ways. ∴Required number of arrangements 5×4 × 1 × 24 × 24 = 5760 = 5C 2 × 3C 3 × 4! × 4! = 2

X

Example 44. A teaparty is arranged for 16 people along two sides of a large table with 8 chairs on each side. Four men want to sit on one particular side and two on the other side. The number of ways in which they can be seated, is 6!8!10! 8!8!10! (b) (a) 4! 6! 4! 6! 8! 8! 6! (c) (d) None of these 6! 4!

Example 45. The number of permutations of n different things taken r at a time, in which p ( r ≤ n − p) particular things will never occur, is (a) P ( n − p, r ) (b) P ( n, r ) − P ( n, p) (c) P ( n, r ) × P ( n, p) (d) P ( n − p, r ) × P ( n, n − p)

6 Permutation and Combination

X

X

Example 46. The number of permutations of n different objects taken r ( ≥ 3) at a time which include 3 particular objects, is (b) n Pr − 3 ⋅ 3! (a) n Pr ⋅ 3! (c)

n−3

Pr − 3 ⋅ 3!

(d) r P3 ⋅

n−3

Pr − 3

Sol. (d) First, we arrange 3 particular things in r places. This can be done in r P3 ways. Then, remaining n − 3 things can be arranged taken r − 3 at a time in n − 3 Pr − 3 ways. ∴Total number of ways = r P3 ⋅n − 3 Pr X

−3

Example 47. We are required to form different words with the help of the letters of the word INTEGER. Let m1 be the number of words in which I and N are never together and m2 be the number of words which begin with I and end with R, then m1 is given by m2 1 (a) 42 (b) 30 (c) 6 (d) 30 Sol. (b) The number of words in which I and N are together 6! × 2 ! = 6! = 720 2! ∴The number of words in which I and N are never together, 7! m1 = − 720 = 1800 2! The number of words which begin with I and end with R, 5! = 60 m2 = 2! m1 1800 = = 30 ∴ m2 60 =

279

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6

Example 48. The number of six-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7, so that digits do not repeat and the terminal digits are even, is (a) 144 (b) 72 (c) 288 (d) 720

X

Sol. (d) Terminal digits are the first and last digits.

Sol. (b) Total number of coins = 2 + 2 + 3 + 1 = 8

Since, terminal digits are even. ∴1st place can be filled in 3 ways and last place can be filled in 2 ways and remaining places can be filled in 5 P4 = 120 ways.

Q2 coins are of 10 paise, 2 coins are of 20 paise, 3 coins are of 25 paise and 1 coin is of 50 paise. ∴Required number of ways 8×7 × 6× 5× 4× 3×2 ×1 8! = = 2 ! × 2 ! × 3! × 1! 2 × 1 × 2 × 1 × 3 × 2 × 1 × 1

Hence, the number of six digit numbers, so that the terminal digits are even = 3 × 120 × 2 = 720.

Permutations of Objects not all Distinct 1. The number of mutually distinguishable permutations of n things taken all at a time, of which p are alike of one kind, q alike of second n! kind such that p + q = n, is . p! q ! 2. The number of permutations of n things, of which p1 are alike of one kind, p2 are alike of second kind, p3 are alike of third kind,..., pr are alike of rth kind such that n! . p1 + p2 + ... + pr = n, is p1 ! p2 ! p3 !... pr !

= 8 × 7 × 6 × 5 = 1680 X

X

three books can be borrowed. (i) When Chemistry part II is selected, then Chemistry part I is also borrowed and the third book is selected from the remaining 6 books = C(6, 1) = 6. (ii) When Chemistry part II is not selected, in this case, he has to select the three books from the remaining 7 books. First choice can be made in 7 ×6×5 = 35 ways. C(7, 3) = 1× 2 × 3

in the 1st, 4th, 7th places and the 4 consonants in their places. The number of ways to arrange 3 vowels in 1st, 4th 3! and 7th places = [Qthere are 2A’s] 2! The number of ways to arrange 4 consonants in their places = 4!

280

∴ Total number of ways in which he can choose the three books to be borrowed = 6 + 35 = 41. X

Example 49. The number of arrangements which can be made out of the letters of the word ALGEBRA, without changing the relative order (positions) of vowels and consonants, is (a) 72 (b) 54 (c) 36 (d) 62 Sol. (a) We have to arrange the 3 vowels in their places, i.e.

∴Required number of arrangements 3! = × 4! = 3 × 24 = 72 2!

Example 51. A boy has 3 library tickets and 9 books of his interest in the library. Out of these 8, he does not want to borrow Chemistry part II, unless Chemistry part I is also borrowed. The number of ways in which he can choose the three books to be borrowed, is (a) 41 (b) 32 (c) 51 (d) None of these Sol. (a) The following are the different possibilities in which

3. The number of permutations of n things, of which p are alike of one kind, q are alike of second kind n! and remaining all are distinct, is . p! q ! 4. Suppose there are r things to be arranged, allowing repetitions. Let further p1 , p2 , ..., pr be the integers such that the first object occurs exactly p1 times, the second occurs exactly p2 times, etc. Then, the total number of permutations of these r objects to the above condition is ( p1 + p2 + ... + pr )! p1 ! p2 ! p3 !... pr !

Example 50. The number of ways in which two 10 paise, two 20 paise, three 25 paise and one 50 paise coins can be distributed among 8 children, so that each child gets only one coin, is (a) 1720 (b) 1680 (c) 1570 (d) None of these

Example 52. There are three copies, each of 4 different books. The number of ways in which they can be arranged on a shelf, is 12! 11! (b) (a) 4 (3!) (3!) 2 9! (c) (d) None of these (3!) 2 Sol. (a) Total number of books = 3 × 4 = 12 in which each of 4 different books is repeated 3 times. Hence, the required number of arrangements 12 ! 12 ! = = 3! × 3! × 3! × 3! (3!)4

X

Example 53. A library has a copies of one book, b copies of two books, c copies of each of three books and single copy of d books. The total number of ways in which these books can be distributed, is ( a + 2b + 3c + d )! ( a + b + c + d )! (a) (b) a ! b! c! a ! b! c! ( a + 2b + 3c + d )! (c) (d) None of these a ! ( b!) 2 ( c!) 3

a6

∴Total number of arrangements (a + 2 b + 3c + d )! = a! (b !)2 (c !)3 X

X

9! 2 ! 3! 4!

Example 55. The number of ways in which the six faces of a cube is painted with six different colours, is (a) 6 (b) 6! (c) 6 C 2 (d) None of the above Sol. (d) Number of ways = 6! = 1 6!

X

a3

an–1

Sol. (b) A 2 B3C 4 written in full is AA BBB CCCC. ∴Required number of ways =

a4

an–2

Example 54. How many different arrangements can be made out of the letters in the expansion A 2 B 3C 4 , when written in full? 9! 9! (a) (b) 2! + 3! + 4! 2 ! 3 ! 4! (c) 2! + 3! + 4!(2!3! 4!) (d) 2!3! − 4

[all faces are alike]

Example 56. The number of numbers greater than a million that can be formed with the digits 2, 3, 0, 3, 4, 2 and 3, is (a) 360 (b) 340 (c) 370 (d) None of these Sol. (a) Any number greater than a million must be of 7 or more than 7 digits. Here, number of given digits is seven, therefore we have to form numbers of seven digits only. Now, there are seven digits of which 3 occurs thrice and two occurs twice. 7! = 420 ∴Number of numbers formed = 2 ! 3! But this also includes those numbers of seven digits, whose first digit is zero and so infact, they are only six digit numbers. Number of numbers of seven digits having zero in the 6! first place = 1 × = 60 3!2 ! Hence, required number = 420 − 60 = 360

Circular Permutations The number of circular permutations of n distinct objects is ( n −1)!. Proof Let a1 , a 2 , a 3 ,…,a n − 1 , a n be n distinct objects. Let the total number of circular permutations be x. Consider one of these x permutations as shown in figure.

6

a5

Since, there are b copies each of two books, c copies each of three books and single copy of d books.

an

a1

a2

Clearly, this circular permutation provides n linear permutations as given below: a1 , a 2 , a 3 , ..., a n − 1 , a n a 2 , a 3 , a 4 , ..., a n , a1 a 3 , a 4 , a 5 , ..., a n , a1 , a 2 a 4 , a 5 , a 6 , ..., a n , a1 , a 2 , a 3 … … … … … … … … a n , a1 , a 2 , a 3 , ..., a n − 1 Thus, each circular permutation gives n linear permutations. But there are x circular permutations, so that number of linear permutations is xn. But the number of linear permutations of n distinct objects is n!. n! ∴ xn = n! ⇒ x = = ( n − 1)! n

Permutation and Combination

Sol. (c) Total number of books = a + 2 b + 3c + d

Ø In the above theorem, anti-clockwise and clockwise orders of

arrangements are considered as distinct permutations.

Difference between Clockwise and Anti-clockwise Arrangements Consider the following circular permutations: We observe that in both the order of the circular arrangements is a1 , a 2 , a 3 a 4 . In Fig. (i), the order is anti-clockwise whereas in Fig. (ii), the order is clockwise. a3

a3

a2

a4

a1 (i)

a2

a4

a1 (ii)

Thus, the number of circular permutation of n things in which clockwise and anti-clockwise arrangements give rise to different permutations is ( n −1)!. e.g. The number of permutations of 5 persons seated around a table is (5 − 1)! = 4!, because with respect to the table, the clockwise and anti-clockwise arrangements are distinct. If anti-clockwise and clockwise orders of arrangements are not distincts, then the number of 1 circular permutations of n distinct items is {( n − 1)!}. 2 e.g. Arrangements of beads in a necklace, arrangements of flowers in a garland.

281

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Objective Mathematics Vol. 1

6

Example 57. The Chief Minister of 11 states of India meet to discuss the language problem. The number of ways they can seat themselves at a round table so that Punjab and Chennai Chief Ministers sit together, is (a) 9! × 2! (b) 9! (c)10! (d) None of the above

Sol. (c) The two particular persons can be arranged among themselves in 2 P2 ways i.e. 2! ways. Taking them as one person and keeping him fixed, we can arrange the remaining 5 persons among themselves in 5! ways. Hence, the required number of ways of which 2 particular persons come together = 5! × 2 ! = 120 × 2 = 240 X

Sol. (a) Treat the Punjab and Chennai Chief Ministers as one (P, C) + 9 others. So, we have to arrange 10 persons around round table. This can be done in (10 − 1)! = 9! ways. Corresponding to each of these 9! ways, Punjab and Chennai Chief Ministers can interchange their places in 2! ways. ∴ Associating the two operations, total number of ways = 9! × 2 !. X

Sol. (a) As shown in figure, 1, 2 and X are the three boys

Example 58. The number of ways in which n things of which r are alike, can be arranged in a circular order, is ( n −1)! (a) r! n! (b) r! n! (c) ( r −1)! (d) None of the above

and 3, 4 and Y are three girls. Boy X will have neighbours as boys 1 and 2 and the girl Y will have neighbours as girls 3 and 4. X

X

282

Example 59. The number of ways in which 7 people can be arranged at a round table, so that 2 particular persons may be together, is (a) 132 (b) 148 (c) 240 (d) None of the above

2

1

4

3

Sol. (a) Let the required number of arrangements = x Consider anyone of these x arrangements. If in this arrangements, r alike things are replaced by r different things, then these r things can be arranged among themselves in r! ways. Since, one arrangement gives rise to r! arrangements. ∴x arrangements will give rise to xr! arrangements. But if all the n things are different, then the number of circular arrangements = (n − 1)! Thus, x ⋅ r ! = (n − 1)! (n − 1)! Hence, x= r!

Example 60. Three boys and three girls are to be seated around a table, in a circle. Among them, the boy X does not want any girl neighbour and the girl Y does not want any boy neighbour. The number of such arrangements possible is (a) 4 (b) 6 (c) 8 (d) None of the above

Y

1 and 2 can be arranged in P(2, 2 ) = 2 ! = 2 × 1 = 2 ways Also, 3 and 4 can be arranged in P(2, 2 ) = 2 ! = 2 × 1 = 2 ways Hence, required number of permutations = 2 ×2 = 4 X

Example 61. There are n seats round a table numbered 1, 2, 3,..., n. The number of ways in which m ( ≤ n) persons can take seats, is (a) n Pm (b) n C m ⋅ ( m − 1)! 1 (d) n − 1 Pm (c) ⋅ n Pm 2 Sol. (a) Since, the seats are numbered. So, the arrangement is not circular. Hence, the required number of ways = The number of arrangements of n things taken m at a time = n Pm

1 If the letters of the word ‘VARUN’ are written in all possible ways and then are arranged as in a dictionary, then the rank of the word ‘VARUN’ is 98

99

100

101

2 5 Indian and 5 American couples meet at a party and shake hands. If no wife shakes hands with her own husband and no Indian wife shakes hands with a male, then the number of hand shakes that takes place in the party, is 95

110

135

150

3 If m denotes the number of 5-digit numbers, if each successive digits are in their descending order of magnitude and n is the corresponding figure, when the digits are in their ascending order of magnitude, then (m − n ) has the value 10

C4

9

C5

10

C3

9

C3

4 The number of ways in which 9 different prizes be given to 5 students, if one particular boy receives 4 prizes and the rest of the students can get any numbers of prizes, is C 4 ⋅ 2 10 4 ⋅ 45

C 5 ⋅ 54 None of these

9

9

5 The number of ways in which 8 non-identical apples can be distributed among 3 boys such that every boy should get atleast 1 apple and atmost 4 apples is k ⋅ 7P3 , where k has the value equal to 88

66

44

22

6 A rack has 5 different pairs of shoes. The number of ways in which 4 shoes can be chosen from it so that there will be no complete pair, is 1920 110

200 80

7 The number of ways in which 7 people can occupy six seats, 3 seats on each side in a first class railway compartment, if two specified persons are to be always included and occupy adjacent seats on the same side, is (5 !)⋅ k, then k has the value equal to 2 8

4 None of these

8 The number of different ways in which five ‘dashes’ and eight ‘dots’ can be arranged, using only seven of these 13 ‘dashes’ and ‘dots’, is 1287 120

119 1235520

9 There are n identical red balls and m identical green balls. The number of different linear arrangements consisting of n red balls but not necessarily all the green balls is x Cy , then x x x x

= = = =

m+ m+ m+ m+

n, y = m n + 1, y = m n + 1, y = m + 1 n, y = n

10 In a unique hockey series between India and Pakistan, they decide to play on till a team wins 5 matches. The number of ways in which the series can be won by India, if no match ends in a draw, is 126 225

252 None of these

11 The number of ways in which n things of which r alike and the rest different can be arranged in a circle distinguishing between clockwise and anti-clockwise arrangement, is ( n + r − 1)! r! ( n − 1) ! ( r − 1) !

Permutation and Combination

6

Work Book Exercise 6.4

( n − 1) ! r ( n − 1) ! r!

12 A gentleman invites a party of m + n (m ≠ n ) friends to dinner and places m at one table T1 and n at another table T2 , the table being round. If not all people shall have the same neighbour in any two arrangement, then the number of ways in which he can arrange the guests, is ( m + n) ! 4mn ( m + n) ! 2⋅ mn

1 ( m + n) ! ⋅ 2 mn

None of these

13 There are 12 guests at a dinner party. Supposing that the master and mistress of the house have fixed seats opposite one another and that there are two specified guests who must always, be placed next to one another, the number of ways in which the company can be placed, is 20 ⋅ 10 ! 22 ⋅ 10 ! 44 ⋅ 10 ! None of the above

14 The number of signals that can be generated by using 6 differently coloured flags, when any number of them may be hoisted at a time, is 1956 1958

1957 1959

15 The sum of all five-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 when repetition of digits is not allowed, is 366000 360000

660000 3999960

16 By using the digits 0, 1, 2, 3, 4 and 5 (repetitions not allowed), any number of digits being used, the number of non-zero numbers that can be formed, is 1030

1630

1200

1530

17 The total number of numbers of not more than 20 digits that are formed by using the digits 0, 1, 2, 3 and 4, is 5 20 5 20 − 1 5 20 + 1 None of the above

283

Objective Mathematics Vol. 1

6

18 Six boys and six girls sit along a line alternately in

24 The number of ways in which we can select four

x ways and along a circle (again alternatively in y ways), then x= y

y = 12 x

x = 10 y

numbers from 1 to 30, so as to exclude every selection of four consecutive numbers, is

x = 12 y

27378 27399

19 The number of times the digit 3 will be written

25 The number of ways in which 10 candidates

when listing the integers from 1 to 1000, is 269

300

271

A1, A 2 , …, A 10 can be ranked, so that A 1 is always above A 2 , is

302

20 The number of ways in which a mixed double

10! 10 ! 2 9! None of the above

game can be arranged from amongst 9 married couples, if no husband and wife play in the same game, is 756 3024

1512 None of these

26 There are n white and n black balls marked

21 The number of different seven digit-numbers that

1, 2, 3, …, n. The number of ways in which we can arrange these balls in a row, so that neighbouring balls are of different colours, is

can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number, is P2 ⋅ 2 5 7 C 2 ⋅ 52

n!

C2 ⋅25 None of these

2

7

2( n!)

2990

dinner and places m at one round table and n at another. The number of ways of arranging the guests is

3120

( m + n)! m! n ! ( m + n)! ( m − 1)!( n − 1)! ( m − n)!( n − 1)! None of the above

23 Given that n is odd, the number of ways in which three numbers in AP can be selected from 1, 2, 3, …, n, is ( n − 1)2 2

( n + 1)2 4

( n + 1)2 2

( n!)2

27 A man invites in a party of (m + n ) friends to

in 5 different boxes such that atmost three boxes is empty, is 3010

(2 n)! (2 n)!

2

22 The number of ways one can put 5 different balls

3000

27405 None of these

( n − 1)2 4

Geometrical Applications of n Cr Some basic Geometrical Applications on n C r are as

8. Number of parallelogram formed by two system of parallel lines (when 1st set contains m parallel lines and 2nd set contains n parallel lines) = n C 2 × mC 2

follow: 1. Out of n non-concurrent and non-parallel straight lines, points of intersection are n C 2 .

9. Number of squares formed by two system of parallel lines in which 1st set is perpendicular 2nd sets of lines (when 1st set contains m parallel lines and 2nd set contains n parallel lines)

2. Out of n points, the number of straight lines (when three points are not collinear) are n C 2 . 3. If out of n points, m are collinear, then Number of straight lines = n C 2 − mC 2 + 1 4. In a polygon, total number of diagonals out of n points (when three points are not collinear) n ( n − 3) = 2 5. Number of triangles formed from n points (when three points are not collinear) are n C 3 . 6. Number of triangles out of n points in which m are collinear, are n C 3 − mC 3 .

284

7. Number of triangles that can be formed out of n points (when none of the side is common to the sides of polygon) are n C 3 − n C1 − n C1 ⋅ n − 4C1 .

=

m−1

∑ (m − r ) (n − r ); m < n

r =1

X

Example 62. The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines, is (a) 6 (b) 18 (c) 12 (d) 9 Sol. (b) Required number of parallelograms = 4C 2 × 3C 2 = =

4! 3! × 2 ! 2 ! 2 ! 1!

4×3 3 × = 18 2 ×1 1

Sol. (b) 4 lines intersect each other in 4 C 2 = 6 points and

Example 63. The number of diagonals that can be drawn by joining the vertices of an octagon, is (a) 28 (b) 48 (c) 20 (d) None of these Sol. (c) The total number of lines obtained by joining 8 vertices of octagon = 8C 2 = 28 ways

4 circles intersect in 4 P2 = 12 points. Each line cuts 4 circles into 8 points. ∴ 4 lines cut 4 circles into 32 points. ∴ Required number = 6 + 12 + 32 = 50 X

Out of these 28 lines, 8 are sides and remaining diagonals. Hence, the number of diagonals = 28 − 8 = 20 X

Example 64. There are n concurrent lines and another line parallel to one of them. The number of different triangles that will be formed by the ( n +1) lines, is ( n −1) n ( n − 1)( n − 2) (a) (b) 2 2 n( n +1) ( n + 1)( n + 2) (c) (d) 2 2

(b)

X

selections of 4 lines, two from each set Q parallelogram has two   sets of parallel sides    n ( n − 1) m (m − 1) = nC 2 × mC 2 = × 2 2 mn (m − 1)(n − 1) = 4 X

Example 66. The maximum number of points into which 4 circles and 4 straight lines intersect, is (a) 26 (b) 50 (c) 56 (d) 72

C 3 − mC 3 − n C 3 − k C 3

Sol. (b) Total number of points are m + n + k. The triangles formed by these points =

m+ n+ k

C3 .

Joining 3 points on the same line gives no triangle, such triangles are m C 3 + nC 3 + kC 3 . ∴ Required number of triangles = m + n + kC 3 − mC 3 − nC 3 − kC 3 X

Example 65. In a plane, there are two sets of parallel lines, one of m lines and the other of n lines. If the lines of one set cut those of the other, then the number of different parallelograms that can be formed is mn ( m − 1) ( n − 1) (a) 2 m ( m − 1)( n − 1) (b) 6 nm ( m − 1) ( n − 1) (c) 4 (d) None of the above Sol. (c) The number of parallelograms = Number of

m+ n + k

(c) m C 3 + n C 3 + k C 3 (d) None of the above

Sol. (b) The number of triangles = Number of selections of 2 lines from the (n − 1) concurrent lines which are cut by the another line (n − 1)! = n − 1C 2 = 2 ! (n − 3)! (n − 1) (n − 2 ) = 2

Example 67. The straight lines I 1 , I 2 , I 3 are parallel and lie in the same plane. A total number of m points are taken on I 1 , n points on I 2 , k points on I 3 . The maximum number of triangles formed with vertices at these points, are (a) m + n + k C 3

6 Permutation and Combination

X

Example 68. Two straight lines intersect at a point O. Points A1 , A2 , …, An are taken on one line and points B1 , B 2 , …, B n on the other. If the point O is not to be used, then the number of triangles that can be drawn using these points as vertices, is (a) n ( n −1) (b) n ( n −1) 2 (c) n 2 ( n − 1) (d) n 2 ( n − 1) 2 Sol. (c) Number of triangles =

C 3 − nC 3 − nC 3 2 n(2 n − 1)(2 n − 2 ) 2 n(n − 1)(n − 2 ) = − 6 6 1 2 = n(n − 1)(3n) = n (n − 1) 3

X

2n

Example 69. There are 10 points in a plane, of which no 3 points are collinear and 4 points are concyclic. The number of different circles that can be drawn through atleast 3 of these points, is (a) 116 (b) 120 (c) 117 (d) None of the above Sol. (c) The required number of circles = (10 C 3 − 4C 3 ) + 1 = 117

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6 Objective Mathematics Vol. 1

Work Book Exercise 6.5 m+ n−2 m+ n m+ n−2 m+ n+2

1 The interior angles of a regular polygon measure 150° each. The number of diagonals of the polygon is 35 54

44 78

m+ n−2 m+ n−1 m ( n − 1) ( m + 1) ( n + 1)

4 Let there be 9 fixed points on the circumference

2 18 points are indicated on the perimeter of a

of a circle. Each of these points is joined to every one of the remaining 8 points by a straight line and the points are so positioned on the circumference that atmost 2 straight lines meet in any interior point of the circle. The number of such interior intersection points is

∆ ABC (see figure). How many triangles are there with vertices at these points? A

126 351 756 None of the above B

331

5 The greatest possible number of points of

C

408

710

intersection of 9 different straight lines and 9 different circles in a plane, is

711

117 153 270 None of the above

3 There are m points on a straight line AB and n points on the line AC, none of them being the point A. Triangles are formed with these points as vertices, when (i) A is excluded. (ii) A is included.

6 The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the sides of the octagon, is

Then, the ratio of number of triangles in the two cases is

24 48

52 16

Selection of One or More Objects Selection from Different Items

X

The number of ways of selecting one or more items from a group of n distinct items is 2 n − 1. Out of n items, 1 item can be selected in n C1 ways, 2 items can be selected in n C 2 ways, 3 items can be selected in n C 3 ways and so on. Hence, the required number of ways = n C1 + n C 2 + n C 3 + K + n C n = ( n C 0 + n C1 + n C 2 + K + n C n ) − n C 0 = 2 n − 1 X

Example 70. Nishi has 5 coins, each of the different denomination. The number of different sums of money, she can form, is (a) 32 (b) 25 (c) 31 (d) None of the above Sol. (c) Required number of ways = 5C1 + 5C 2 + 5C 3 + 5C 4 + 5C 5 = 2 5 − 1 = 31

286

Example 71. A man has 6 friends. In how many ways he can invite one or more of them at dinner? (a) 64 (b) 65 (c) 63 (d) 61 Sol. (c) The man has to select some or all of his 6 friends. So, required number of ways = 2 6 − 1 = 63

Ø

●

●

●

The number of ways of selecting r items out of n identical items is 1. The total number of ways of selecting zero or more i.e. items from a group of n identical items is (n + 1) . The total number of selections of one or more out of p + q + r items, where p are alike of one kind, q are alike of second kind and rest are alike of third kind is [(p + 1)(q + 1)(r + 1)] − 1.

Selection from Identical and Distinct Objects The total number of ways of selecting one or more items from p identical items of one kind, q identical items of second kind, r identical items of third kind and n different items is ( p + 1) ( q + 1) ( r + 1) 2 n − 1

n

n

n

n

n = p1 1 ⋅ p2 2 ⋅ p3 3 K pk k

Let

m

m

The total number of divisors is same as the number of ways selecting same or all out of four 2’s, three 3’s and one 5’s. The number of ways = (4 + 1)(3 + 1)(1 + 1) − 1 = 39

…(i)

where, p1 , p2 , …, pk are distinct prime numbers and n1 , n2 , …, nk are positive integers. Clearly, any divisor of n is of the form m

2160 = 2 4 × 33 × 51

X

m

…(ii) d = p1 1 ⋅ p2 2 ⋅ p3 3 … pk k where, m1 , m2 , …, mk are natural numbers such that 0 ≤ mi ≤ ni , i =1, 2, 3, …, k Therefore, the total number of divisors of Eq. (i) will be equal to the number of ways of selecting atleast one from n1 identical primes p1 , n2 identical primes p2 and so on, finally nk identical primes pk . The number of such ways is ( n1 + 1) ( n2 + 1) K ( nk + 1) Hence, the total number of divisors of n

n

n

Sol. (c) We have, 7875 = 32 × 53 × 71 The total number of factor = (2 + 1)(3 + 1)(1 + 1) − 1 = 23 But this includes the given number itself. ∴ Number of proper factors = 23 − 1 = 22 X

n

n = p1 1 ⋅ p2 2 ⋅ p3 3 K pk k is ( n1 + 1) ( n2 + 1) ( n3 + 1) K ( nk + 1) This includes 1 and n as divisors. Therefore, number of divisors other than 1 and n is ( n1 + 1) ( n2 + 1) ( n3 + 1) K ( nk + 1) − 2 The sum of all divisors of Eq.(i) is given by n1

n2

n3

∑ ∑ ∑

r1 = 0 r2 = 0 r3 = 0

nk

K

∑

rk = 0

r p11

⋅

r p22

⋅

r p33

K

 pknk + 1 − 1 K   pk − 1 

is equal to number of ways of selecting atleast one from seven a’s, four b’s, three c’s, d, e, f are different. ∴ The number of ways = (7 + 1)(4 + 1)(3 + 1)2 3 − 1 = 1279 X

10800 = 2 4 × 33 × 52

 1 (n1 + 1)(n2 + 1)…(nk + 1), if nis not perfect square  = 2 1  [(n1 + 1)(n2 + 1)…(nk + 1) + 1], if nis perfect square 2

X

Example 72. The total number of factors (excluding 1) of 2160 is (a) 40 (b) 39 (c) 41 (d) 38

Example 75. The number of even divisors of 10800 is (a) 12 (b) 24 (c) 36 (d) 48 Sol. (d) We have,

The number of ways of expressing n as a product of two natural numbers

e.g. To find number of ways in which the number 94864 can be resolved as a product of two factors. 94864 = 24 × 7 2 × 112 So, 94864 is a perfect square. 1 Hence, the number of ways = [(4 + 1)(2 + 1)(2 + 1) + 1] = 23 2

Example 74. The number of factors (excluding 1 and the expression itself) of the product a 7 b 4 c 3 def , where a, b, c, d, e and f are all prime numbers, is (a)1279 (b)1278 (c)1280 (d)1277 Sol. (a) The total number of factors of the product a7b 4c 3def

r pkk

 p n1 + 1 − 1  p2n2 + 1 − 1  p3n3 + 1 − 1 = 1     p1 − 1   p2 − 1   p3 − 1 

Example 73. The total number of proper factors of 7875 is (a) 23 (b) 24 (c) 22 (d) 21

6 Permutation and Combination

Sol. (b) We have,

Number of Divisors and the Sum of the Divisors of a Given Natural Number

For a divisor to be even, we must have atleast one 2 out of four 2’s any number of 3’s and 5’s. ∴ Total number of even divisors = 4(3 + 1)(2 + 1) = 48 X

Example 76. The sum of divisors of 2520 is (a) 9360 (b) 5040 (c) 9600 (d) None of these Sol. (a) We have, 2520 = 2 3 × 32 × 5 × 7 The sum of divisors of 2520 = (1 + 2 + 2 2 + 2 3 )(1 + 3 + 32 )(1 + 5)(1 + 7 ) = 15 × 13 × 6 × 8 = 9360

287

Objective Mathematics Vol. 1

6 Division of Objects into Groups Division of Items into Groups of Unequal Size Number of ways in which ( m + n) distinct items can be divided into two unequal groups containing m ( m + n)! and n items is . m! n!

The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of groups is important, is  ( mn)! 1  ( mn)! ×  m! =  m m!   ( n!) ( n!) m X

Proof The number of ways in which ( m + n) items are divided into two groups containing m and n items is same as the number of combinations of ( m + n) things taken m at a time. Thus, the required number ( m + n)! . = m + nCm = m! n! Ø

●

●

The number of ways in which (m + n + p) items can be divided into unequal groups containing m, n, p items, is (m + n + p) ! m+ n + p C m⋅ n + pC n = m !n ! p ! The number of ways to distribute (m + n + p) items among 3 persons in the groups containing m, n and p items

Sol. (b) Here, 52 cards are to be divided in four equal groups and order of the group is not important. 52 ! So, required number of ways = (13!)4 × 4! X

= (Number of ways to divide) × (Number of groups)! (m + n + p) ! = ×3! m !n ! p ! X

Example 77. The number of ways to 16 different things to three persons A, B, C so that B gets one more than A and C gets two more than B, is 16 ! 16! (b) (a) 3 !5 !8 ! 4 !5 ! 7 ! (c) 4 !5 ! 7 ! (d) None of these

Division of Objects into Groups of Equal Size

288

The number of ways in which mn different objects can be divided equally into m groups, each containing n objects and the order of groups is not important, is  ( mn)!  1    ( n!) m  m!

Example 79. The number of ways 12 different books can be distributed equally among 4 persons, is 12! 12! 12! 12! (b) (c) (d) (a) 3 4 4 ( 4!) (3!) × 3 (3!) ( 4!) 4 Sol. (c) Here, 12 different books can be divided equally among 4 persons that each person will get 3 books. Hence, required number of distribution 12 ! = 12C 3 ⋅ 9C 3 ⋅ 6C 3 ⋅ 3C 3 = (3!)4

Division of Identical Objects into Groups i.

Sol. (a) Suppose, A gets x things, then B gets ( x + 1) and C gets ( x + 3) things. ∴ x + x + 1 + x + 3 = 16 ⇒ x=4 Thus, we have to distribute 16 things to A, B and C in such that A gets 4, B gets 5 and C gets 7 things. (4 + 5 + 7 )! ∴Required number of ways = 4! 5! 7 ! 16! = 4! 5! 7 !

Example 78. The number of ways in which a pack of 52 cards can be divided equally into four groups, is 52! 52! (b) (a) 4 (13!) (13!) 4 × 4! 52! (d) None of these (c) (13!) × 4!

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0, 1, 2 or more items ( ≤ n) is n + r − 1 C r − 1 . Or The total number of ways of dividing n identical objects into r groups, if blank groups are allowed, is n + r − 1 C r − 1 .

X

Example 80. The total number of ways can 15 identical Mathematics books be distributed among six students, is 15! 20! (b) (a) 6! 9! 5!15! (c) 15P6 (d) None of these Sol. (b) We know that the total number of ways in which n things all alike can be distributed into r different boxes is n + r −1 C r − 1, when blank box is allowed. Hence, number of ways in which 15 identical books be distributed among six students 20! = 15 + 6 − 1C 6 − 1 = 20C 5 = 5!15!

The total number of ways of dividing n identical items among r persons, each one of whom, receives atleast one item is n − 1 C r − 1 . Or The number of ways in which n identical items can be divided into r groups such that blank groups are not allowed is n − 1 C r − 1 .

X

If r (0 ≤ r ≤ n) objects occupy the places assigned to them i.e. their original places and none of the remaining ( n − r ) objects occupies its original places, then the number of such ways is D (n − r ) = n C r ⋅ D ( n − r )  1  1 1 1 = nC r ⋅(n − r )! 1 − + − + K + (−1)n − r  1! 2 ! 3! (n − r )! 

X

Example 81. The number of ways in which 10 identical balls can be distributed in 4 distinct boxes, so that none of the boxes remain empty, is (b) 10 P4 (a) 10 C 4 (c) 9 P3

(d) None of these

Sol. (d) The required number of ways of distributing 10 identical items to three persons such that each person receives atleast one item. So, required number of ways 9! = 10 − 1C 4 − 1 = 9C 3 = = 84 3! 6!

Sol. (a) The required number of ways 1 1 1 1 = 4! 1 − + − +   1! 2 ! 3! 4! = 12 − 4 + 1 = 9 X

iii. The number of ways in which n identical items can be divided into r groups so that no group contains less than m items and more than k ( m < k ) items is coefficient of x n in the expansion of ( x m + x m + 1 + K + x k ) r . X

Sol. (a) The required number of ways is equal to the

Sol. (c) For 1 ≤ i ≤ 4, let xi ( ≥ 3) be the number of blanks

X

= Coefficient of t in (1 + C1 + C 2 t + C 3 t + K ) 3

4

5

2

6

3

But 5 letters can be permuted in 5! = 120 ways. Thus, the required number of arrangements = (120) (20) = 2400

Dearrangements If n distinct objects are arranged in a row, then the number of ways in which they can be dearranged, so that none of them occupies its original place is 1  1 1 1 1 n!1 − + − + − K + (−1) n  n!  1! 2! 3! 4! and it is denoted by D ( n).

number of dearrangements of 10 objects. Hence, required number of ways 1 1 1 1  = 10! 1 − + − + K+  1! 2 ! 3! 10! 

…(i)

= Coefficient of t 3 in (1 − t )−4

Example 84. The number of ways can 10 letters be placed in 10 marked envelopes, so that no letter is in the right envelope, are 1  1 1 1 (a)10! 1 − + − + K +  10!  1! 2! 3! 1  1 1 1 (b)10! 1 + − + − K −  1 ! 2 ! 3 ! 10 !  1  1 1 1 (c) 1 + − + − K −  10!  1! 2! 3! 1  1 1 1 (d) 9! 1 + − + − K −  10!  1! 2! 3!

Example 82. Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the letters with atleast three between every two. The number of ways in which this can be done, is (a) 1200 (b) 1800 (c) 2400 (d) 3000 between ith and (i + 1) th letters. Then, x1 + x2 + x3 + x4 = 15 The number of solutions of Eq. (i) = Coefficient of t 15 in (t 3 + t 4 + K )4

Example 83. There are four balls of different colours and four boxes of colours, same as those of the balls. The number of ways in which the balls, one each in a box, could be placed such that a ball does not go to a box of its own colour is (a) 9 (b) 13 (c) 8 (d) None of the above

6 Permutation and Combination

ii.

Example 85. Ajay writes letters to his five friends and addresses the corresponding. The number of ways can the letters be placed in the envelopes, so that atleast two of them are in the wrong envelopes, are (a)120 (b)125 (c)119 (d) None of the above Sol. (c) Required number of ways = =

5

∑

r =2 5

5

C 5 − r D(r ) 5!



1

1

1

∑ r !(5 − r )! r ! 1 − 1! + 2 ! − 3! + K +

r =2



(−1)r   r! 

289

Objective Mathematics Vol. 1

6

(−1)r   r!   r =2 5! 1 5! 1 1 5! 1 1 1 =   +  −  +  − +  3!  2 ! 2 !  2 ! 3! 1!  2 ! 3! 4! 5! 1 1 1 1 +  − + −  0!  2 ! 3! 4! 5! = 10 + 20 + (60 − 20 + 5) + (60 − 20 + 5 − 1) = 10 + 20 + 45 + 44 = 119 =

5

5!

1

3

 1 − xn + 1   1 − x2 n + 1  = Coefficient of x3 n in      1− x   1− x 

1

∑ (5 − r )! 2 ! − 3! + K +

= Coefficient of x3 n in (1 − xn + 1 )3 (1 − x2 n + 1 ) (1 − x)−4 = Coefficient of x3 n in [(1 − 3 xn + 1 + 3 x2 n + 2 − x3 n + 3 ) (1 − x2 n + 1 ) (1 − x)−4 ] = Coefficient of x3 n in [(1 − 3 xn + 1 − x2 n + 1 + 3 x2 n +

− 3 Coefficient of x2 n − 1 in (1 − x)−4 − Coefficient of xn − 1 in (1 − x)−4 + 3 Coefficient of xn − 2 in (1 − x)−4

Consider the equation

=

●

●

●

The total number of non-negative integral solutions of the equation x1 + x 2 + K + x r = n is n + r − 1C r − 1 and the total number of solutions of the same equation in the set N of natural numbers is n − 1C r − 1. If the upper limit of a variable in solving an equation of the form x1 + x 2 + K + x m = n subject to the conditions ai ≤ x ≤ b i ; i = 1, 2 , K , m is more than or equal to the sum required and lower limit of all the variables are non-negative, then upper limit of that variable can be taken as infinite. In order to solve inequations of the form x1 + x 2 + K + x m≤ n We introduce a dummy (artificial) variable x m+ 1 such that x1 + x 2 + K + x m + x m+ 1 = n, where x m+ 1 ≥ 0. The number of solutions of this equation are same as the number of solutions of Eq. (i).

X

Example 86. In an examination, the maximum marks for each of three papers is n and that for fourth paper is 2n. Then, the number of ways in which a candidate can get 3n marks, is 1 (a) ( n − 1) (5n 2 + 10n + 6) 6 1 (b) ( n + 1) (5n 2 + 10n + 6) 6 1 (c) ( n + 1) (5n 2 + n + 6) 6 (d) None of the above

290

C4 − 1 −

n −1+ 4 −1

C4 − 1 n − 2 + 4 −1

C4 − 1

C 3 − 3 × 2 n + 2C 3 − n + 2C 3 + 3 × n + 1C 3 (3n + 3)(3n + 2 )(3n + 1) (2 n + 2 )(2 n + 1)(2 n) = − 3⋅ 6 6 (n + 2 )(n + 1)(n) (n + 1)(n)(n − 1) − + 3⋅ 6 6 1 2 = (n + 1)(5n + 10n + 6) 6 =

X

3n + 3

Example 87. The number of ways of selecting n objects out of 3n objects, of which n are alike and rest are different, is (2n − 1)! (2n − 1)! (a) 2 2n − 1 + (b) 2 2n − 1 − n!( n − 1)! n!( n − 1)! ( 2 + 1 )! n (c) 2 2n + 1 + (d) None of these n!( n + 1)! Sol. (a) The required number of ways = Coefficient of xn in ( x0 + x1 + x2 + K + xn )( x0 + x)2 n 1 − x n + 1 2n = Coefficient of x n in   (1 + x)  1− x  = Coefficient of x n in (1 − x)−1 (1 + x)2 n  ∞  = Coefficient of x n in  ∑ xr  (1 + x)2 n   r = 0   n  = Coefficient of x n in  ∑ xr  (1 + x)2 n   r = 0  = Coefficient of x n in

n

∑x

(1 + x)2 n

r

r =0

=

n

∑

Coefficient of x

n−r

in (1 + x)2 n

r =0

=

n

∑

2n

Cn − r

r =0

=

Cn +

2n

Cn + 1 + K +

2n

1 = [2 n C n + {2 n C 0 + 2

C1 +

2n

=

2n

C0

C1 + K +

Cn +

2n

+ K+

Sol. (b) The total number of ways of getting 3n marks = Coefficient of x3 n in ( x0 + x1 + x2 + K + xn )3 × ( x0 + x1 + K + x2 n )

2n − 1 + 4 − 1

C4 − 1 − 3 ×

+ 3×

where, x1 , x 2 , x 3 , x 4 , …, x r and n are non-negative integers.

Ø

3n + 4 − 1

…(i)

This equation may be interpreted as that n identical objects are to be divided into r groups, where a group may contain any number of objects. Therefore, Total number of solutions of Eq. (i) = Coefficient of x n in ( x 0 + x 1 + K + x n ) r = n + r − 1C n or n + r − 1 C r − 1

+ … ) (1 − x)−4 ]

= Coefficient of x3 n in (1 − x)−4

Number of Integral Solutions of Linear Equations and Inequations x1 + x 2 + x 3 + x 4 + K + x r = n

2

2n

2n + 1

Cn + 1

C 2 n }] [ Q nC r = nC n − r ]

2n

 (2 n − 1)! 1  2 n! 1 2n [ Cn + 2 2n ] =  + 2 2n  = 2 2n − 1 + n! (n − 1)! 2  n! n! 2 

 1 − x5  = Coefficient of x4 in    1− x 

= Coefficient of x in [(1 − x ) (1 − x) ]

Sol. (a) There are 11 letters in the word EXAMINATION viz.

= Coefficient of x4 in (1 − x)−4 = 7C 3 =

4

2 A’s, 2 I’s, 2 N’s, E, X, M, T, O. Therefore, Number of ways of selecting 4 letters from these letters = Coefficient of x4 in ( x0 + x1 + x2 )3 (1 + x)5 3

−3

= Coefficient of x in (1 − x ) (1 − x) 4

3 3

= (1 + x)

5

= Coefficient of x in (1 − 3 x + 3 x − x9 )(1 + x)5 (1 − x)−3 4

3

6

= Coefficient of x4 in (1 − 3 x3 ) (1 + 5C1 x + 5C 2 x2 + 5C 3 x3 + 5C 4 x4 ) (1 − x)−3 4 = Coefficient of x in (1 + 5 x + 10 x2 + 7 x3 − 10 x4 )(1 − x)−3 =

4 + 3 −1

C3 − 1 + 5 ×

3 + 3 −1

C 3 − 1 + 10 ×

2 + 3 −1

C3 − 1

+7×

1+ 3 −1

C 3 − 1 − 10

= 6C 2 + 5 × 5C 2 + 10 × 4C 2 + 7 × 3C 2 − 10 = 15 + 50 + 60 + 21 − 10 = 136

Example 89. The number of ways in which 16 identical things can be distributed among 4 persons, if each person gets atleast 3 things, is (a) 33 (b) 35 (c) 38 (d) None of these Sol. (b) Required number = Coefficient of x

16

in ( x + x + K + x ) 3

4

7 4

= Coefficient of x4 in (1 + x + K + x4 )4

−4

5 4

4⋅ 5⋅ 6⋅7 = 35 4!

Aliter x1 + x2 + x3 + x4 = 16, where x1, x2 , x3 , x4 ≥ 3 or x1 − 3 = a1, x2 − 3 = a2 , x3 − 3 = a3 , x4 − 3 = a4 ∴ a1, a2 , a3 , a4 ≥ 0 or a1 + a2 + a3 + a4 = 8 Number of solutions = 4 + 4 − 1C 4 − 1 = 7C 3 ⇒

 1 − x3  5 = Coefficient of x4 in   (1 + x)  1− x 

X

6

4

Example 88. The number of ways of selecting 4 letters from the word EXAMINATION, is (a)136 (b)140 (c)126 (d)102

X

7 ×6×5 = 35 3×2

Permutation and Combination

X

Example 90. If N is the number of positive integral solution of x1 x 2 x 3 x 4 = 770. Then, (a) N is 250 (b) N is 252 (c) N is 254 (d) N is 256 Sol. (d) We have, 770 = 2 ⋅ 5 ⋅ 7 ⋅ 11 We can assign 2 to x1 or x2 or x3 or x4 i.e. 2 can be assigned in 4 ways. Similarly, each of 5, 7 or 11 can be assigned in 4 ways. Thus, the number of positive integral solution of x1 x2 x3 x4 = 770 is 44 = 2 8 = 256 Aliter x1 x2 x3 x4 = 21 ⋅ 51 ⋅ 71 ⋅ 111 ∴ a1 + a2 + a3 + a4 = 1 i.e. 21 can be assigned to a1, a2 , a3 , a4 . 1+ 4 −1

⇒

C 4 − 1 = 4C 3 = 4

Similarly, for 51, 71, 111 ∴ Total number of ways = 4 ⋅ 4 ⋅ 4 ⋅ 4 = 256

Work Book Exercise 6.6 1 The total number of ways of dividing 15 different 15 ! 7 ! 3 !2 ! 15 ! 8! 4! 3!

having 6 one rupee coins, 7 one rupee coins and 8 one rupee coins respectively can donate 10 one rupee coins collectively, is

None of these

a

7

c

10

2 The number of ways of distributing 50 identical things among 8 persons in such a way that three of them get 8 things each, two of them get 7 things each and remaining 3 get 4 things each, is 8!

8!

( 3 !)2 (2 !) 8!

(2 !)2 3 !

( 4 !)2 3 !

None of these

3 In how many ways can we put 12 different balls in three different boxes such that the first box contains exactly 5 balls? C 3 ⋅ 57

12

C 5 ⋅ 52

None of these

12

c

12

4 The number of ways, 3 persons A, B and C

15 ! 9! 5! 4!

things into groups of 8, 4 and 3 respectively, is

C5 ⋅27

C3 C3

8

C3

None of these

5 In an examination, the maximum marks for each of the three papers are 50. Maximum marks for the fourth paper is 100. The number of ways in which a candidate can score 60% marks on the whole, is a c

100550 110551

101551 d None of these

6 The number of ways of dividing 20 persons into 10 couples, is a c

20 ! 2 10 20 ! (2 !)9

20

C10

None of these

291

Objective Mathematics Vol. 1

6

7 The number of integer solutions for the equation x + y + z + t = 20, where x, y, z, t ≥ − 1, is a

20

C4

23

c

27

C4

27

C3 C3

8 The number of ways in which mn students can be distributed equally among m sections, is a c

5382

136

c 6062

8055

c 455

105

b 18 d None of these

12 If x, y, z, … are (m + 1) distinct prime numbers, then then number of factors of x n y z K is a

m( n + 1)

c ( n + 1)2

m

b 2 nm d n⋅2

m

13 There are m copies of each of n different books in a university library. The number of ways in which one or more than one book can be selected, is a

mn + 1

c ( n + 1) − m n

b ( m + 1)n − 1 d m

14 The number of ways in which one or more balls can be selected out of 10 white, 9 green and 7 blue balls, is a c

892 891

b 881 d 879

15 The number of all three element subsets of the set { a1, a2 , a3 , K , an } which contain a3 , is a

n

c

n−1

C3 C2

b

n−1

C3

d None of these

16 If one quarter of all three element subsets of the set A = { a1, a2 , a3 , K , an } contains the element a3 , then n is equal to a b c d

10 12 14 None of the above

30 8

n− 3

C2

d None of these

C3

20 There are three piles of identical red, blue and green balls and each pile contains atleast 10 balls. The number of ways of selecting 10 balls, if twice as many red balls as green balls are to be selected, is a c

3 6

b 4 d 8

21 Two classrooms A and B having capacity of 25 and (n − 25) seats, respectively. Let A n denotes the number of possible sitting arrangements of room A, when n students are to be seated in these rooms, starting from room A which is to be filled up full to its capacity. If A n − A n − 1 = 25!(49 C25 ), then n equals a

49

b

48

c

50

d 51

22 Between two junction stations A and B, there are 12 intermediate stations. The number of ways in which a train can be made to stop at 4 of these stations, so that no two of these halting stations are consecutive, is a

8

c

12

9

C4

b

C4 − 4

d None of these

C4

23 The sum of the divisors of 2 5 ⋅ 34 ⋅ 52 is a b c d

3 2 ⋅ 7 1 ⋅ 112 63 ⋅ 121 ⋅ 31 3 ⋅ 7 ⋅ 11 ⋅ 31 None of the above

24 The number of ways arranging m positive and n < m + 1negative signs in a row, so that no two negative signs are together, is a

m +1

b

n+1

c

m +1

d

n +1

Pn

Pm Cn

Cm

sign, then the minimum value of Σ

abc = 30 is a c

b

C3

25 If ai, i = 1, 2, 3, 4 is four real numbers of the same

17 The number of positive integral solutions of

292

19 If n objects are arranged in a row, then the

n− 3

be made from a throw by three persons, each throwing a single die once, is 45 27

d None of these

C5

c

11 The number of ways in which a score of 11 can

a c

C4

n− 2

x + y + z = 10, with x ≥ − 5, y ≥ − 5, z ≥ − 5, is 135

n +4

a

10 The number of integral solutions of a

c

n +5

b

C3

( mn)m

10 6 and which have the sum of the digits equal to 12, is 8550

n +3

( n!)m

9 The number of integers which lie between 1 and

a

x + y + z ≤ n, where n ∈ N is a

number of ways of selecting three of these objects, so that no two of them are next to each other, is

( m n)!

( mn)! n! ( mn)! m! n !

18 The number of non-negative integral solutions of

b 27 None of these

i, j ∈{1, 2, 3, 4}, i ≠ j , is a b c d

8 6 12 None of the above

ai , aj

WorkedOut Examples Type 1. Only One Correct Option Ex 1. Number of positive integer n less than 17, for which n! + ( n + 1)! + ( n + 2)! is an integral multiple of 49, is (a) 0 (c) 5

(b) 3 (d) 2

Sol. Here, n! + (n + 1)! + (n + 2)! = n![1 + (n + 1) + (n + 2)(n + 1)] = n!(n + 2)2 Either 7 divides (n + 2) or 49 divides n !. i.e. n = 5 ,12, 14, 15, 16

[Qn < 17]

Thus, number of solutions is five. Hence, (c) is the correct answer.

2 2 1 2a + + = , where a, b ∈ N , then 9! 3! 7! 5!5! b! the ordered pair ( a, b) is (a) (9, 10) (b) (10, 9) (c) (7, 10) (d) (10, 7) 2 2 1 + + 9! 3!7! 5!5! 1 1 1 1 1 = + + + + 1!9! 3!7! 5!5! 3!7! 9!1! 1  10! 10! 10! 10! 10!  = + + + +   10!  9!1! 7!3! 5!5! 3!7! 9! 1 ! 1 10 = { C 1 + 10C 3 + 10C 5 + 10C 7 + 10C 9 } 10! 1 29 2a , but it is = ⋅ (210 − 1 ) = . (10)! 10! b! ⇒ a = 9 and b = 10 Hence, (a) is the correct answer.

Sol. Here,

Ex 3. Number of ordered triplets (x, y, z ) such that x, y, z are primes and x y + 1 = z, is (a) 0 (b) 1 (c) 3 (d) None of the above

is equal to

Sol. As

(b) n n + 1C 4 (d) n ( n + 1)

∑ ∑ ∑ ∑n = n∑ ∑ ∑ ∑

0≤ i < j < k< l ≤ n

represents

0≤ i < j < k< l ≤ n

selection of four terms from {0, 1, 2, ..., n} i.e. (n + 1) terms. Thus, ∑ ∑ ∑ ∑ n = n ⋅n + 1 C 4 Hence, (b) is the correct answer.

Ex 5. Let A denotes the property that two elements of A = {1, 5, 9, 13, ..., 1093} add upto 1094. Then, maximum number of elements in A can be (a) 273 (c) 137

(b) 136 (d) 138

Sol. Since, A = {1, 5, 9, 13, ..., 1093} is an arithmetic progression. ∴

1093 − 1 +1 4 = 274

Number of terms =

Since, sum of equidistant terms in AP is equal to sum of first and last terms = 1 + 1093 = 1094. ∴ Maximum number of elements in A that add upto 1094 274 = = 137 2 Hence, (c) is the correct answer.

Ex 6. Let n1 = x1 x 2 x 3 and n2 = y1 y2 y3 be two 3-digit numbers, then the pairs of n1 and n2 can be formed, so that n1 can be subtracted from n2 without borrowing, is (a) 55⋅ 54 (c) 45⋅ ( 55) 2

(b) 45⋅ 55 (d) 55⋅ ( 45) 2

Sol. Here, n1 = x1x2x3 and n 2 = y1 y2 y3

Sol. Here, x y + 1 = z, where x , y, z are prime. Thus, y cannot be odd, if y is prime. ⇒ x y + 1is divisible by ( x + 1). [Qx y = z − 1]

Thus, only even value (i.e. prime) is x = 2. ⇒ x = 2, y = 2, z = 5 So, there is only one such triplet (2, 2, 5). Hence, (b) is the correct answer.

∑∑∑

0≤i < j 6), then S n − 7  n   7  r =1 (where, [ ] denotes the greatest integer function) is equal to  n  n (a)   (b) n ! − 7    7  7 (c) 5 (d) 3

Ex 12. Let S n =

n

∑ r! ;

Sol. All the number r! 1; (r ≥ 7) will be multiple of 7. So, for remainder, consider S6 = 1 + 2 + 6 + 24 + 120 + 720 = 873 which leaves the remainder 5, when divided by 7. Hence, (c) is the correct answer.

Ex 13. The number of ways of arranging m numbers out of 1, 2, 3, ..., n, so that maximum is ( n − 2) and minimum is 2 (repetitions of numbers is allowed) such that maximum and minimum both occur exactly once ( n > 5, m > 3), is (a)

n−3

Cm − 2

(b)

m

C 2 ( n − 3) m − 2

(c) m( m − 1)( n − 5) m − 2 (d) n C 2 ⋅ n C m Sol. First, we take one number as 2 and one as (n − 2) and put

them in m(m − 1) ways. Now, remaining (m − 2), numbers can be anyone from, 3, 4, ..., (n − 4 ), (n − 3). Which we can do in (n − 5)(m − 2) .

∴ Total number of ways = m(m − 1)(n − 5)m − 2 Hence, (c) is the correct answer.

(a) 666 (b) 667 (c) 665 (d) None of the above Sol. Q 36 = 22 ⋅ 32 Total number of integers ∈[ 2, 2000 ] that are divisible by 2 is  2000  = 1000  2  The integer ∈[ 2, 2000 ] that are divisible by 3 is  2000  = 666  3  Integers that are divisible by 2 and 3 and contained in  2000  = 333 [ 2, 2000 ] =  6  Thus, total number of integers that are neither divisible by 2 nor by 3 = 1999 − (1000 + 666 − 333) = 666 Hence, (a) is the correct answer.

1 2 3 4 30 31 Ex 15. If E = ⋅ ⋅ ⋅ K ⋅ = 8 x , then value of 4 6 8 10 62 64 x is (a) − 7 (c) − 10

(b) − 9 (d) − 12 31! 1 1 = = 231 (32)! 231 (32) 236

= 2−36 = (23 )−12 = 8−12, but it is 8x. Thus, x = − 12 Hence, (d) is the correct answer.

Ex 16. The number of rational numbers lying in the interval (2002, 2003) all whose digits after the decimal points are non-zero and are in decreasing order, is 9

(a)

∑

9

10

(b)

Pi

∑

9

Pi

i=1

i=1

(c) 29 − 1

(d) 210 − 1

Sol. A rational number of the desired category is of the form 2002. x1x2K xk, where 1 ≤ k ≤ 9 and 9 ≥ x1 > x2 > K > xk ≥ 1. We can choose k digits out 9 in 9 C k ways and arrange them in decreasing order in just one way. Thus, the desired number of rational numbers is 9 C 1 + 9C 2 + ... + 9C 9 = 29 − 1

Hence, (c) is the correct answer.

Ex 17. The number of ways of dividing 15 men and 15 women into 15 couples, each consisting of a man and a woman, is (a) 1240

(b) 1840

(15C 1 )(15C 1 ) = 152, the number of ways of choosing 2nd couple is (14C 1 )(14C 1 ) = 14 2 and so on. Thus, the number of ways of choosing the couple is 152 + 14 2 + 132 + K + 22 + 12 15 × (15 + 1)[ 2(15) + 1] = = 1240 6 Hence, (a) is the correct answer.

Ex 18. The number of ordered pairs (m, n), m, n ∈ {1, 2, ..., 50} such that 6 n + 9 m is a multiple of 5, is (a) 2500

(b) 1250

(c) 625

6

(d) 500

n

Sol. All the numbers of the form 6 will end with 6 and 9m will end with 9, if m is odd and will end with 1, if m is even, so 6n + 9m will end with 5, if nis any number and m is odd. So, ordered pairs will be 50 × 25 = 1250. Hence, (b) is the correct answer.

Ex 19. n is selected from the set {1, 2, 3, ..., 100} and the number 2 n + 3 n + 5 n is formed. Total number of ways of selecting n, so that the formed number is divisible by 4, is equal to (a) 50 (c) 48

(b) 49 (d) None of these

Sol. If n is odd, 3n = 4 λ 1 − 1, 5n = 4 λ 2 + 1

Sol. We have, E=

Sol. The number of ways of choosing first couple is

Permutation and Combination

Ex 14. Total number of integer n such that 2 ≤ n ≤ 2000 and HCF of n and 36 is one, is equal to

(c) 1820

(d) 2005

⇒ 2n + 3n + 5n is not divisible by 4, as 2n + 3n + 5n will be in the form of 4 λ + 2. Thus, total number of ways of selecting n is 49. Hence, (b) is the correct answer.

Ex 20. The number of ordered pairs (m, n), m, n ∈{1, 2, ..., 100} such that 7 m + 7 n is divisible by 5, is (a) 1250

(b) 2000

(c) 2500

(d) 5000

Sol. Note that 7 (r ∈ N ) ends in 7, 9, 3 or 1 (corresponding to r

r = 1, 2, 3 and 4, respectively). Thus, 7m + 7n cannot end in 5 for any values of m, n ∈ N . In other words, for 7m + 7n to be divisible by 5, it should end in 0. For 7m + 7n to end in 0, the forms of m and n should be as follows: S.No.

m

n

1.

4r

4s + 2 4s + 3

2.

4r + 1

3.

4r + 2

4s

4.

4r + 3

4s + 1

Thus, for a given value of m, there are just 25 values of n for which 7m + 7n ends in 0. [for instance, if m = 4 r, then n = 2, 6, 10, ..., 98] ∴ There are 100 × 25 = 2500 ordered pairs (m, n) for which 7m + 7n is divisible by 5. Hence, (c) is the correct answer.

295

Objective Mathematics Vol. 1

6

Ex 21. If the number of ways of selecting K coupons out of an unlimited number of coupons bearing the letters A, T, M, so that they cannot be used to spell the word MAT is 93, then K is equal to (a) 3 (c) 7

(b) 5 (d) None of these

⇒ a + b + c = 14, a, b, c ≥ 1and a, b and care distinct. Let a < b < c and x1 = a, x2 = b − a, x3 = c − b So, 3x1 + 2x2 + x3 = 14; x1 , x2 , x3 ≥ 1 ∴ The number of solutions = Coefficient of t 14 in {(t 3 + t6 + t 9 + ...) (t 2 + t 4 + ...) (t + t 2 + ...)}

Sol. Here, A, T, M bears unlimited number of coupons. Thus, selecting K in which word MAT is not spelled. If atleast one letter is not selected. i.e. 3C 1. ...(i)

= Coefficient of t 8 in {(1 + t 3 + t6 + K )

From the remaining two we have to select K i.e. for each selection of K coupons we have 2 ways. For K coupons, we have 2K ways but to select atleast one, we have …(ii) ⇒ 2K − 1

= Coefficient of t 8 in {(1 + t 2 + t 3 + t 4 + t 5 + 2t6

Words cannot be used to spell MAT ⇒ 3(2K − 1) = 93 ⇒ K = 5 Hence, (b) is the correct answer.

Ex 22. In a shooting competition, a man can score 5, 4, 3, 2 or 0 points for each shot. Then, the number of different ways in which he can score 30 in seven shots, is (a) 419

(b) 418

(c) 420

(d) 421

(1 + t 2 + t 4 + ...) (1 + t + t 2 + ...)} + t 7 + 2t 8 ) (1 + t + t 2 + K + t 8 )} = 1 + 1 + 1 + 1 + 1 + 2 + 1 + 2 = 10 Since, three distinct numbers can be assigned to three boys in 3! ways. So, total number of ways = 10 × 3! = 60 Hence, (c) is the correct answer.

Ex 25. Total number of positive integral solutions of 15 < x1 + x 2 + x 3 ≤ 20 is equal to (a) 1125 (c) 1245

(b) 1150 (d) 685

Sol. 15 < x1 + x2 + x3 ≤ 20

= Coefficient of x 30 in {(x 0 + x 2 + x 3 ) + x 4 (x + 1)}7

⇒ x1 + x2 + x3 = 16 + r; r = 0, 1, 2, 3, 4 Now, number of positive integral solutions of x1 + x2 + x3 = 16 + r is 13 + r + 3 − 1C 13 + r

= Coefficient of x 30 in

i.e.

Sol. Number of ways of making 30 in 7 shots = Coefficient of x in (x + x + x + x + x ) 0

30

2

3

4

5 7

{x 28 (x + 1)7 + 7C 1x 24 (x + 1)6 (1 + x 2 + x 3 ) 20

5

3

2

= C 5 + C 1 ( C 3 + C 2 + C 0 ) + C 2 ( C 1 + 2) 7

7

6

6

6

7

5

= 21 + 252 + 147 = 420 Hence, (c) is the correct answer.

Ex 23. The number of ways in which the sum of upper faces of four distinct dices can be six, is (a) 10 (c) 6

(b) 4 (d) 7

Sol. The number of required ways will be equal to the number of solutions of x1 + x2 + x3 + x4 = 6 where, x1 , x2 , x3 , x4 ≥ 1 Let, x1 − 1 = a1, x2 − 1 = a2, x3 − 1 = a3, x4 − 1 = a4 ≥ 0 ⇒ (a1 + 1) + (a2 + 1) + (a3 + 1) + (a4 + 1) = 6 where, a1 , a2 , a3 , a4 ≥ 0 ⇒ a1 + a2 + a3 + a4 = 2 ∴ Number of solutions = 2 + 4 − 1C 4 − 1 = 5C 3 = 10 Hence, (a) is the correct answer.

Ex 24. The number of ways can 14 identical toys be distributed among three boys, so that each one gets atleast one toy and no two boys get equal number of toys, is (a) 45 (c) 60

(b) 48 (d) None of these

15 + r

C 13 + r =

15 + r

C2

Thus, total solutions

+ C 2x (x + 1) (x + x + 1) + K} 7

296

Sol. Let the boys gets a, b and c toys, respectively.

=

4

∑

15 + r

C 2 = 15C 2 +

C2 +

16

C2 +

17

C2 +

18

19

C2

r=0

= 20C 3 − 15C 3 = 685 Hence, (d) is the correct answer.

Ex 26. Ten persons numbered 1, 2, ..., 10 play a chess tournament, each player playing against every other player exactly one game. It is known that no game ends in a draw. Let w1 , w2 , ..., w10 be the number of games won by players 1, 2, 3, ..., 10 respectively and l1 , l2 , ..., l10 be the number of games lost by the players 1, 2, ..., 10, respectively. Then, (a) Σw i2 = 81 − Σli2

(b) Σw i2 = 81 + Σli2

(c) Σw i2 = Σli2

(d) None of these

Sol. Clearly, each player will play 9 games and total number of games = 10C 2 = 45 Clearly, and ⇒ ⇒ ⇒

wi + li = 9 Σwi = Σli = 45 wi = 9 − li wi2 = 81 + li2 − 18li Σwi2 = 81⋅ 10 + Σli2 − 18Σli = 810 + Σli2 − 18 ⋅ 45 = Σli2

Hence, (c) is the correct answer.

(a) 2n − 1

(b) 2n − 1

(c) 2n − 2

(d) 2n

Sol. For each of the first (n − 1) elements a1 , a2 , K , an − 1, we

have two choices : either ai (1 ≤ i ≤ n − 1) lies in the subset or ai does not lie in the subset. For the last element we have just one choice. If even number elements have already been taken, we do not include an, in the subset, otherwise (when odd number of elements have been added), we include it in the subset. Thus, the number of subsets of A = {a1 , a2 , ..., an}

which contain even number of elements is equal to 2n − 1. Hence, (a) is the correct answer.

Ex 28. A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the elements of P. A subset Q of A is again chosen. The number of ways of choosing P and Q, so that P ∩ Q = φ, is (a) 22n −

2n

(b) 2n

Cn

(c) 2n − 1

(d) 3n

Sol. Let A = {a1 , a2 , ..., an}. For ai ∈ A, we have the following choices: (i) ai ∈ P and ai ∈ Q (ii) ai ∈ P and ai ∉ Q (iii) ai ∉ P and ai ∈ Q (iv) ai ∉ P and ai ∉ Q Out of these, only (ii), (iii) and (iv), ai ∉ P ∩ Q. Therefore, the number of required subsets is 3n. Hence, (d) is the correct answer.

Ex 29. The number of integral points that lie exactly in the interior of the triangle with vertices (0, 0), (0, 21), (21, 0), is (a) 133 (c) 233

(b) 190 (d) 105

Sol. The integral points lying on the line x = 1which also lie in the interior of ∆OAB are (1, 1), (1, 2), ..., (1, 19). ∴ There are 19 points lying on the line x = 1.

Similarly, the number of integral points x = 2is 18 and on x = 3 is 17 and so on. The number of integral points on x = 19 is just 1. Thus, number of integral points lying in the interior of the triangle is 1 19 + 18 + K + 1 = (19)(19 + 1) = 190 2 Hence, (b) is the correct answer.

Ex 30. Let S = {1, 2, 3, ..., n}. If X denotes the set of all subsets of S containing exactly two elements, then the value of ∑ (min A ) is (a)

n+1

A ∈X

C3

(c) n C 2

(b) n C 3 (d) None of these

Sol. There are exactly (n − 1) subsets of S containing two

elements having 1 as least element, exactly (n − 2) subsets of S having 2 as least element and so on. Thus, ∑ min ( A ) = 1(n − 1) + 2(n − 2) + K + (n − 1)(1)

A ∈X

=

n− 1

n− 1

n− 1

r=1

r=1

r=1

∑ r(n − r) = n ∑ r −

∑ r2

1  1 = n  n(n − 1) − n(n − 1)(2n − 1) 2  6 1 = n(n − 1){3n − 2n + 1} 6 1 = (n + 1)n(n − 1) = n + 1C 3 6 Hence, (a) is the correct answer.

6 Permutation and Combination

Ex 27. The number of subsets of the set A = {a1 , a 2 , ..., a n } which contain even number of elements, is

Ex 31. The sum of factors of 8! which are odd and are of the form 3m + 2, where m is a natural number, is (a) 40 (b) 8 (c) 45 (d) 35 Sol. Here, 8! = 27 ⋅ 32 ⋅ 51 ⋅ 71 Obviously, the factors are not multiple of either 2 or 3. So, the factors may be 1, 5, 7, 35 of which 5 and 35 are of the form 3m + 2. So, the sum = 40 Hence, (a) is the correct answer.

Ex 32. The number of positive integral solutions of the equation x1 x 2 x 3 x 4 x 5 = 420 is (a) 1875 (b) 1600 (c) 1250 (d) None of the above Sol. Since, x1 x2 x3 x4 x5 = 420 ∴

x 1 x 2 x 3 x 4 x 5 = 22 × 3 × 5 × 7

Each of 3, 5 or 7 can take 5 places and 22 can be disposed in 15 ways. So, required number of solutions = 53 (5C 1 + 5C 2 ) = 53 × 15 = 1875 Hence, (a) is the correct answer. Aliter x 1 x 2 x 3 x 4 x 5 = 22 × 3 × 5 × 7 Here, 22 can be assigned to a1 , a2 , a3 , a4 , a5 ⇒ ⇒

a1 + a2 + a3 + a4 + a5 = 2. 2+ 5− 1 C 5 − 1 = 6C 4 = 15.

Also, 31 can be assigned to b1 , b2 , b3 , b4 , b5 ⇒ ⇒

b1 + b2 + b3 + b4 + b5 = 1 1+ 5− 1 C 5 − 1 = 5C 4 = 5

Similarly, for 5 and 7. ∴Total number of positive integral solutions = 15 × 5 × 5 × 5 = 1875

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Ex 33. The number of numbers lying in (2, 3), where all the digits after decimal are non-zero and in increasing order, is 9

(a)

∑

9

10

Pi (b) 511

i=1

(c)

∑

10

Pi (d) 1023

i=1

Sol. Since, the digits required is non-zero. ∴ To select amongst {1, 2, 3, 4, 5, 6, 7, 8, 9} Thus, any number is of the form 2x1 x2 K xk 3, where 1 ≤ x1 < x2 < x3 < K < x9 ≤ 9 i.e. number of numbers lying between (2, 3) = 9C 1 + 9C 2 + 9C 3 + K + 9C 9 = 29 − 1 = 1023

Ex 34. The number of divisors of n = 2 7 ⋅ 3 5 ⋅ 5 3 , which are of the form 4t + 1( t ∈ N ), is (b) 10 (d) 14

contain any power of 2 but contain even number of odds of the form 4 k + 3 and any number of odds of the form 4 k + 1, so total factors = 1 × 3 × 4 = 12 But one will be 1, also so required number = 12 − 1 = 11 Hence, (c) is the correct answer. Aliter Odd numbers of the type 4 t + 1 or 4 t + 3 and they are symmetrical. Number of odd numbers = (1 + 5)(1 + 3) = 24 [any number of 5’s and 3’s and no 2’s] Required number = 12 but 1 is included. ∴Required number of numbers = 12 − 1 = 11

Ex 35. Total number of divisors of n = 2 5 ⋅ 3 4 ⋅ 510 ⋅ 7 6 that are of the form 4λ + 2, λ ≥ 1, is equal to (b) 55

(c) 384

(d) 54

Sol. Q 4 λ + 2 = 2 (2λ + 1) = odd multiple of 2

Thus, total divisors = 1⋅ 5 ⋅ 11⋅ 7 − 1 = 384 [one is subtracted because there will be case when selected powers of three, five and seven are zero each and this will make λ = 0] Hence, (c) is the correct answer.

Ex 36. Total number of divisors of n = 3 5 ⋅ 5 7 ⋅ 7 9 that are of the form 4λ + 1, λ ≥ 0, is equal to (a) 240 Sol. Q 3 and

298

n1

(b) 30

(c) 120

n1

(d) 15

= (4 − 1) = 4 λ 1 + (− 1) , 5n2 = (4 + 1)n2 = 4 λ 2 + 1 7

n3

= (8 − 1)

n1

n3

= 4 λ 3 + (− 1)

(b) 54 (d) None of these

Sol. Q x1 ⋅ x2 ⋅ x3 = 22 ⋅ 3 ⋅ 5 Total number of positive integral solutions = 3 ⋅ 3 ⋅ (3C 1 + 3C 2 ) = 54 Hence, (b) is the correct answer. Aliter x 1 ⋅ x 2 ⋅ x 3 = 22 ⋅ 3 ⋅ 5 Here, 22 can be assigned to a1 , a2 , a3. a1 + a2 + a3 = 2 C 3 − 1 = 4C 2 = 6

2+ 3− 1

and 31 can be assigned to b1 , b2 , b3. ⇒

Hence, any positive integer power of 5 will be in the form of 4 λ 2 + 1. Even powers of 3 and 7 will be in the form of 4 λ + 1 and odd powers of 3 and 7 will be in the form of 4 λ − 1. Thus, required number of divisors = 8 ⋅ (3 ⋅ 5 + 3 ⋅ 5) = 240 Hence, (a) is the correct answer.

C3 − 1 = 3

x1 ⋅ x2 ⋅ x3 = 2 ⋅ 3 ⋅ 5 2

1

1

Total number of positive integral solutions = 6 × 3 × 3 = 54

Ex 38. The number of positive integral solutions of the equation x1 x 2 x 3 x 4 x 5 = 1050, is (a) 1800 (c) 1400

(b) 1600 (d) None of these

Sol. Using prime factorisation of 1050, we can write the given equation as x1x2x3x4x5 = 2 × 3 × 52 × 7 We can assign 2, 3 or 7 to any of variables. We can assign entire 52 to just one variable in 5 ways or can assign 52 = 5 × 5 to two variables in 5C 2 ways. Thus, 52 can be assigned in 5 C 1 + 5C 2 = 5 + 10 = 15 ways Hence, required number of solutions = 5 × 5 × 5 × 15 = 1875 Hence, (d) is the correct answer. Aliter x 1 ⋅ x 2 ⋅ x 3 ⋅ x 4 ⋅ x 5 = 2 × 3 × 52 × 7 For 5 , the number of ways =

5+ 2− 1

For 21, the number of ways =

1+ 5− 1

2

C 5 − 1 = 6C 4 = 15

C 5 − 1 = 5C 4 = 5

∴Positive integral solutions for x1 ⋅ x2 ⋅ x3 ⋅ x4 ⋅ x5 = 2 × 3 × 52 × 7 is 15 × 5 × 5 × 5 = 1875

Ex 39. Let a be a factor of 120, then the number of positive integral solutions of x1 x 2 x 3 = a is (a) 160

n3

1+ 3− 1

b1 + b2 + b3 = 1 ⇒

∴

Sol. Any number will be of the form 4 t + 1, if it does not

(a) 385

(a) 27 (c) 64

⇒ ⇒

Hence, (d) is the correct answer.

(a) 12 (c) 11

Ex 37. Total number of positive integral solutions of x1 ⋅ x 2 ⋅ x 3 = 60 is equal to

(b) 320

(c) 480

(d) 960

120 , then the number of positive a integral solutions of x1x2x3 = a is same as that of number of positive integral solutions of x1x2x3x4 = 120 = 23 × 3 × 5

Sol. Let x4 be such that x4 =

We can assign 3 and 5 to unknown quantities in 4 × 4 ways.

Hence, the number of required solutions = 4 × 4 × [ 4C 1 + (4C 2 )(2) + 4C 3 ] = 4 × 4 × 20 = 320 Hence, (b) is the correct answer. Aliter x1 ⋅ x2 ⋅ x3 ⋅ x4 = 120 x 1 ⋅ x 2 ⋅ x 3 ⋅ x 4 = 23 × 3 × 5

⇒

For 2 , the number of ways =

3+ 4− 1

For 3 , the number of ways =

1+ 4− 1

3

1

C 4 − 1 = C 3 = 20 6

C 4 − 1 = 4C 3 = 4

Ex 40. The number of ways of choosing triplets ( x, y, z ) such that z ≥ max {x, y} and x, y, z ∈{1, 2, ..., n, n + 1}, is n+1

C3 +

n+2

C3

(b) n( n + 1)( 2n + 1)

(c) 11 + 22 +K+ ( n − 1) 2 (d) None of these Sol. Triplets with x = y < z, x < y < z, y < z < x can be n+ 1

n+ 1

n+ 1

chosen in C2, C3, C 3 ways. ∴There n + 1C 2 + 2(n + 1C 3 ) = n + 2C 3 +

n+ 1

C 3 ways

Hence, (a) is the correct answer.

Ex 41. The number of maps (functions) from the set A = {1, 2, 3} into the set B = {1, 2, 3, 4, 5, 6, 7} such that f ( i) ≤ f ( j) whenever i < j, is (a) 84 (c) 88

(b) 90 (d) None of these

Sol. If the function is one-one, then select any three from the set B in 7 C 3 ways i.e. 35 ways

.

If the function is many-one, then there are two possibilities. All three corresponds to same element number of such functions = 7C 1 = 7 ways. Two corresponds to same element. Select any two from the set B. The larger one corresponds to the larger and the smaller one corresponds to the smaller the third may corresponds to any two. Number of such functions = 7C 2 × 2 = 42 So, the required number of maps = 35 + 7 + 42 = 84 Hence, (a) is the correct answer.

Ex 42. If f :{1, 2, 3, 4, 5} → {x, y, z, t}, then the total number of onto functions is equal to (a) 242

(b) 245

(c) 102

(d) 240

Sol. Total functions = 4 5 Total functions when any one element is the left out = 4C 1 ⋅ 35 Total functions when any two elements are left out = 4C 2 ⋅ 25 ∴ Total number of onto functions = 4 5 − (4C 1 ⋅ 35 − 4C 2 ⋅ 25 + 4C 3 ) = 240 Hence, (d) is the correct answer.

(a) 9 (b) 44 (c) 16 (d) None of the above Sol. Total number of required functions

∴Total number of integral solutions = 20 × 4 × 4 = 320

(a)

Ex 43. The function f :{1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} that are onto and f ( i) ≠ i, is equal to

= Number of dearrangement of 5 objects 1 1 1 1 = 5!  − + −  = 44  2! 3! 4 ! 5!

6 Permutation and Combination

We can assign all 2 to one unknown in 4C 1 ways, to two unknown in 4C 2 and to three unknown in 4C 3 ways.

Hence, (b) is the correct answer.

Ex 44. The total number of functions f from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that f ( i) ≤ f ( j), ∀ i < j, is equal to (a) 35 (c) 50

(b) 30 (d) 60

Sol. Let 1 is associated with r, r ∈{1, 2, 3, 4 , 5}, then 2 can be associated with r, r + 1, ..., 5.

Let 2 is associated with j, then 3 can be associated with j, j + 1, ..., 5. Thus, required number of functions 5  5 5  (6 − r)(7 − r) = ∑  ∑ (6 − j ) = ∑   2  r=1 r = 1 j = r =

 1  5 (42 − 13r + r2 ) ∑   2 r = 1 

1 6 ⋅ 5 5 ⋅ 6 ⋅ 11 +  42 ⋅ 5 − 13 ⋅  2 2 6  = 35 Hence, (a) is the correct answer. =

Ex 45. The number of functions f from the set A = {0, 1, 2} into the set B = {0, 1, 2, 3, 4, 5, 6, 7} such that f ( i) ≤ f ( j) for i < j and i, j ∈ A, is (a) 8 C 3 (b) 8 C 3 + 2( 8 C 2 ) (c) 10 C 3 (d) None of the above Sol. A function f : A → B such that f (0) ≤ f (1) ≤ f (2) falls in one of the following four categories.

Case I f (0) < f (1) < f (2) There are 8 C 3 functions in this category. Case II f (0) = f (1) < f (2) There are 8 C 2 functions in this category.

Case III f (0) < f (1) = f (2) There are again 8 C 2 functions in this category.

Case IV f (0) = f (1) = f (2) There are 8 C 1 functions in this category. Thus, the number of desired functions is 8 C 3 + 8C 2 + 8C 2 + 8C 1 = 9C 3 + 9C 2 = 10C 3 Hence, (c) is the correct answer.

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Ex 46. The integers from 1 to 1000 are written in order around a circle. Starting at 1, every fifteenth number is marked (i.e. 1, 16, 31 etc.) This process is continued until a number is reached which has already been marked, then unmarked numbers are (a) 200

(b) 400

(c) 600

(d) 800

Sol. In first round all the integers, which leaves the remainder 1 when divided by 15, will be marked. Last number of this category is 991. Next number to be marked is (991 + 15 − 1000) = 6 Again, second round of integers which leaves the remainder 6 when divided by 15, will be marked. Last number of this category is 996. Next number to be marked is (996 + 15 − 1000) = 11 Thus, third round of integers which leaves the remainder 11 when divided by 15, will be marked last number of this category is 986. Next number to be marked is 986 + 15 − 1000 = 1, which is already been marked. Hence, the numbers of the form 15λ + 1, 15λ + 6, 15λ + 11are marked. All these numbers can also be put in the form of 5λ 1 + 1(λ 1 ≥ 0). Hence, total number of marked numbers  999  =1+ = 200  5  ∴Unmarked numbers = 800 Hence, (d) is the correct answer.

Ex 47. Let f (n) denotes the number of different ways the positive integer n can be expressed as the sum of 1’s and 2’s. For example f ( 4) = 5, since 4 = 1 + 1 + 1 + 1, 1 + 1 + 2, 1 + 2 + 1, 2 + 1 + 1, 2 + 2 (Note Order of 1’s and 2’s is important). Then, f { f (6)} is (a) f ( 6) (c) f (13)

(b) f (12) (d) None of these

Sol. As f (4 ) = 5

[given]

300

(b) 292 (d) 296

Sol. Either we have to choose

or

(i) Every square of (2 by 2)

will contribute four

shaped figures by removing anyone square out of it. Number of ways to choose these squares = 7 × 7 = 49 So, total

shaped figures = 49 × 4 = 196

(ii) In every line it is possible to have 6

shaped

figures. So, total number of such figures = 6 × 8 × 2 = 96 ∴Total number of cases = 196 + 96 = 292 Hence, (b) is the correct answer.

Ex 49. A man on the verge of dying. He has unlimited amount of money and intends to distributed some of it among his n relatives, so that total money that is to be distributed is a positive multiple of four and no relative gets more than ` ( 4n − 1), then the number of ways in which the rich man can write his will, is (a) 4 n − 1 − 1 (b) n n − 1 (c) 4 n − 1 ⋅ n n (d) 4 n − 1 ⋅ n n − 1 4K; (K ∈Natural numbers) and the relative xi receives `pi .

⇒

Number of arrangements

Number of 1’s

Number of 2’s

0

3

3! =1 3!

2

2

4! =6 2 !2 !

4

1

5! =5 4!

6

0

6! =1 6!

∴ f (6) = 13 ∴ f { f (6)} = f (13) Hence, (c) is the correct answer.

(a) 290 (c) 294

Sol. Let us assume the total money that is to be distributed is

∴ f (6) can be written using 1’s and 2’s as

Total

Ex 48. The number of ways in which we can choose 3 squares on a chessboard such that one of the squares has its two sides common to other two squares, is

13

n

∑ pi = 4 K

and

0 ≤ pi ≤ 4 n − 1

i=1

Thus, to obtain the number of integral solutions of p1 + p2 + p3 + ... + pn = 4 K , K = 1, 2, K is Coefficient of x 4K in (1 + x + x 2 + K + x 4n − 1 )n ...(i) Let (1 + x + x 2 + K + x 4n − 1 )n =

4n2 − n

∑

ar x r

...(ii)

r=0

Now, required coefficients = a4 + a8 + a12 + K Putting x = 1, − 1, i and − i in Eq. (ii) and adding, we get 4 (a0 + a4 + a8 + ...) = (4 n)n + 0 + 0 + 0 ⇒

a0 + a4 + a8 + K = 4 n − 1 ⋅ nn

...(iii)

From Eqs. (i) and (iii), we get Coefficient of x 4K in (1 + x + x 2 + K + x 4n − 1 )n is (4 n − 1 ⋅ nn − 1). Hence, (d) is the correct answer.

(a) 4 (b) 8 (c) 2 (d) None of the above Sol. We have, 8, 7, 6, 4, 2, x and y Any number is divisible by 3, if sum of digits is divisible by 3. i.e. x + y + 27 is divisible by 3. x and y can take values from 0, 1, 3, 5, 9. ∴ Possible pairs are (5, 1), (3, 0), (9, 0), (9, 3), (1, 5), (0, 3), (0, 9) and (3, 9). Hence, (b) is the correct answer.

Ex 51. 10 different toys are to be distributed among 10 children. Total number of ways of distributing these toys so that exactly 2 children do not get any toy, is equal to  1 1   (a) (10!) 2  +  3! 2! 7! ( 2!) 5 6!  1 1   (b) (10!)  +  3! 2! 7! ( 2!) 4 6! 2

 1 1   (c) (10!)  +  3! 7! ( 2!) 5 6! 2

 1 1   (d) (10!)  +  3! 7! ( 2!) 4 6! 2

Sol. It is possible in two mutually exclusive cases. Case I 2 children get none, one child gets three and all remaining children get one each. Case II 2 children get none, 2 children get 2 each and all remaining children get one each. 10! In first case, number of ways = ⋅ 10! 3!2! 7! 10! In second case, number of ways = ⋅ 10! (2!)4 ⋅ 6! Thus, total number of ways  1 1  = (10!)2  +   3!2! 7! (2!)4 6!

6 Permutation and Combination

Ex 50. A seven-digit number made up of all distinct digits 8, 7, 6, 4, 2, x and y is divisible by 3. Then, possible number of order pair ( x, y) is

Hence, (b) is the correct answer.

Ex 52. A delegation of five students is to be formed from a group of 10 students. If three particular students want to remain together whereas two particular students do not want to remain together, then the number of selection is (a) 10 (c) 30

(b) 20 (d) None of these

Sol. Let A , B and C want to remain together and D, E do not want to remain together. In

Out

A, B, C , D

E

5

D

5

D

A, B, C , E

5

E

A, B, C , D

5

A, B, C , E

Number of selections C1 = 5 C1 = 5

C4 = 5 C4 = 5

∴ Required number of selections = 5C 1 + 5C 1 + 5C 4 + 5C 4 = 20 Hence, (b) is the correct answer.

Type 2. More than One Correct Option Ex 53. If m and n are positive integers more than or equal to 2, m > n, then ( mn)! is divisible by (a) ( m!) n (b) ( n !) m (c) ( m + n )! (d) ( m − n )! Sol.

(mn)! is the number of ways of distribution mn distinct (m!)n objects in n persons equally. (mn)! Hence, is an integer ⇒ (m!)n /(mn)! (m!)n Similarly, (n!)m /(mn)! Further m + n < 2m ≤ mn ⇒ (m + n)!/(mn)! and m − n < m < mn Hence, (a), (b), (c) and (d) are the correct answers.

Ex 54. Let n be a positive integer with f ( n) = 1! + 2! + 3! + K + n! and P ( x ), Q ( x ) be polynomials in x such that f ( n + 2) = P ( n) f ( n +1) + Q ( n) f ( n) for all n ≥1. Then, (a) P ( x ) = x + 3 (b) Q ( x ) = − x − 2 (c) P ( x ) = − x − 2 (d) Q ( x ) = x + 3

Sol. Q f (n) = 1! + 2 ! + 3! + K + n!

f (n + 1) = 1! + 2! + 3! + K + (n + 1)! f (n + 2) = 1! + 2! + 3! + K + (n + 2)!

∴ f (n + 2) − f (n + 1) = (n + 2)! = (n + 2)(n + 1)! = (n + 2)[ f (n + 1) − f (n)] ⇒ f (n + 2) = (n + 3) f (n + 1) − (n + 2) f (n) ⇒ P (x ) = x + 3, Q (x ) = − x − 2 Hence, (a) and (b) are the correct answers.

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Ex 55. The number of ways in which we can choose 2 distinct integers from 1 to 100, such that the difference between them is atmost 10, is (a)

100

C2 −

(b)

100

C 98 −

(c)

100

C2 −

90

C2

90 90

C 88

C 88

(d) None of the above Sol. Let the chosen integers be x1 and x2. Let a be an integer before x1 , b be integer between x1 and x2 and c be integer after x2. ∴ a + b + c = 98, where a ≥ 0, b ≥ 10, c ≥ 0. Now, if we consider the choices, where difference is atleast 11, then the number of solutions is 88 + 3 − 1 C 3 − 1 = 90C 2 ∴ The number of ways in which b is less than 10 is 100 C 2 − 90C 2 which is equal to options (a), (b) and (c). Hence, (a), (b) and (c) are the correct answers.

Ex 56. n locks and n corresponding keys are available. But the actual combination is not known. The maximum numbers of trials that are needed to assign the keys to their corresponding locks, are (a) n C 2 n

(b)

∑ ( k − 1)

k =2

(c) n ! (d) n + 1 C 2 Sol. First key will be tried for atmost (n − 1) locks. Second key will be tried for atmost (n − 2) locks and so on. Thus, the maximum number of trials needed = (n − 1) + (n − 2) + K + 1 n(n − 1) n = = C2 2 Hence, (a) and (b) are the correct answers.

Type 3. Assertion and Reason Directions (Ex. Nos. 57-61) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 57. Statement I The maximum number of points of intersection of 8 circles of unequal radii is 56. Statement II The maximum number of points into which 4 circles of unequal radii and 4 non-coincident straight lines intersect is 50. Sol. Statement I Two circles intersect in 2 points. ∴Maximum number of points of intersection = 2 × Number of selection of two circles from 8 circles = 2 × 8C 2 = 2 × 28 = 56 Statement II 4 lines intersect each other in 4 C 2 = 6 points and 4 circles intersect each other in 2 × 4C 2 = 12 points. Further, one line and one circle intersect in two points. So, 4 lines will intersect four circles in 32 points. ∴Maximum number of points = 6 + 12 + 32 = 50 Hence, (b) is the correct answer.

302

Ex 58. Statement I If there are six letters, L1 , L2 , L3 , L4 , L5 , L6 and their corresponding six envelopes are E1 , E 2 , E 3 , E 4 , E 5 , E 6 . Letters having odd value can be put into odd value envelopes and even value letters can be put into even value envelopes, so that no letter go into the right envelopes, the number of arrangements will be equal to 4. Statement II If Pn is the number of ways in which n letter can be put in n corresponding envelopes, such that no letters goes to correct envelopes, then  1 1 ( −1) n   Pn = n! 1 − + − K + n!   1! 2! Sol. L1 L3 L5 L2 L4 L6 E1 E3 E5 E2 E4 E6 ∴ Number of ways 1 1 1 1 1 1   = 3!  1 − + −  ⋅ 3!  1 − + −  =4   1! 2! 3! 1! 2! 3! Hence, (a) is the correct answer.

Ex 59. Statement I The maximum value of k, such that (50) k divides 100! is 2. Statement II If p is any prime number, then power of p in n! is equal to n  n   n   p  +  2  +  3  + ... , where [ ] represents   p  p  the greatest integer function.

Power of 2 in 100! 100  100  100  100  = + + + +  2   22   23   24  = 50 + 25 + 12 + 6 + 3 + 1 = 97 So, power of 5 in 100! is 24. ∴ Exponent of 25 in (100)! = 12

100  +  25 

100   26 

Statement II 3 m + 7 n has last digit zero, when m is of 4k − 2 type and n is of 4l type where k , l ∈W. Sol. 3m gives 1 at unit place, if m = 4 k and these are 5 in numbers. 7n gives 9 at unit place, if n = 4 l + 2 and these are 5 in numbers. Number of ways = 25 ∴ 3m gives 3 at unit place, if m = 4 k + 1

⇒ Maximum value of k is 12. Hence, (d) is the correct answer.

Ex 60. Statement I A five-digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4 and 5 with repetition.The total number formed are 216. Statement II If sum of digits of any number is divisible by 3, then the number must be divisible by 3. Sol. Number of numbers form by using 1, 2, 3, 4, 5 = 5! = 120 Number of numbers formed by using 0, 1, 2, 4 , 5 4 4 3 2 1 = 4 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 96 Statement I is incorrect and Statement II is correct. Hence, (d) is the correct answer.

Ex 61. Statement I The number of ordered pairs (m, n); m, n ∈{1, 2, 3, ..., 20}, such that 3 m + 7 n is a multiple of 10, is equal to 100.

and these are 5 in numbers. 7n gives 7 at unit place, if n = 4 l + 1

6 Permutation and Combination

n

n

Sol. If power of p is l in n!, then l =   +  2  +...  p  p 

and these are 5 in numbers. Number of ways = 25 ∴ 3m gives 7 at unit place, if m = 4 k + 3 and these are 5 in numbers. 7n gives 3 at unit place, if n = 4 l + 3 and these are 5 in numbers. Number of ways = 25 ∴ 3m gives 9 at unit place, if m = 3k + 2 and these are 5 in numbers. 7n gives 1 at unit place, if n = 4 l and these are 5 in numbers. Hence, (c) is the correct answer.

Type 4. Linked Comprehension Based Questions Passage I (Ex. Nos. 62-64) One of the most important techniques of counting is the principle of exclusion and inclusion. Let A1, A2 , ..., A n be m sets and n( A i ) represents the cardinality of the set A i (the number of elements in the set A i ), then according to the principle of exclusion and inclusion n( A1 ∪ A2 ∪ K ∪ A n ) =

m

∑ n( Ai ) − ∑ n( Ai

i =1

i ≠j

+

∑

∩ Aj )

n( Ai ∩ Aj ∩ Ak )

i < j z, x − z = a, a ≥ 1 ⇒ x=z+ a So, 2x + z = 12 ⇒ 3z + 2a = 12 , a ≥ 1, z ≥ 0 i.e. coefficient of x 12 in (1 + x 3 + x6 + K )(x 2 + x 4 + K ) = 2 (b) x < z, z − x = a ⇒ z=x+ a So, 3x + a = 12, a ≥ 1, x ≥ 0 i.e. coefficient of x 12 in (1 + x 3 + x6 + x 9 + ...)(x + x 2 + ...) = 4

Case III x + y + z = 12, x ≠ y ≠ z Taking only different combinations of x , y and z. Let x > y > z ⇒ y−z=a ⇒ y = z + a, a ≥ 1 x − y=b ⇒ x = z + a + b, b ≥ 1 x + y + z = 12 ⇒ 3z + 2a + b = 12 ; a, b ≥ 1, z > 0

306

Coefficient of x 12 in (1 + x 3 + x6 + ...)(x 2 + x 4 + x6 + ...)(x + x 2 + x 3 + ...) = 12 So, total ways = 1 + 2 + 4 + 12 = 19 Hence, λ = 1

Ex 76. If the number of ordered triplet (a, b, c), such that LCM ( a, b) =1000, LCM ( b, c) = 2000 and LCM ( c, a ) = 2000 is 10 k, then k is ________. Sol. (7)Q 1000 = 23 ⋅ 53 , 2000 = 24 ⋅ 53 So,

a = 2A ⋅ 5R , b = 2B ⋅ 5S , c = 2C ⋅ 5T

max( A , B ) = 3, max( A , C ) = max ( B , C ) = 4 max(R , S ) = max(R , T ) = max (S , T ) = 3 There are 10 ways of choosing R , S , T . 1 with all 3, 3 with one 2, 3 with one 1, 3 with one 0. C must be 4. There are 7 ways of choosing A , B. 1 with both 3, 3 with A = 3, B not, 3 with B = 3 and A not thus, 7 ways of choosing A , B , C , hence 7 ⋅ 10 = 70 ways ⇒ k =7

Ex 77. N is the number of ways in which a person can walk up a stairway which has 7 steps. If he can take 1 or 2 steps up the stairs at a time, then the value of N /3 is _________ . Sol. (7) If x denotes the number of times, he can take unit step and y denotes the number of times, he can take 2 steps, then x + 2 y = 7. We must have x = 1, 3, 5. If x = 1, then steps will be 1, 2, 2, 2. 4! Number of ways = ⇒ =4 3! If x = 3, then steps will be 1, 1, 1, 2, 2. 5! Number of ways = = 10 ⇒ 3! 2! If x = 5, then steps will be 1, 1, 1, 1, 1, 2. Number of ways = 6C 1 = 6 ⇒ If x = 7, then steps will be 1, 1, 1, 1, 1, 1, 1. ⇒

Number of ways = 7C 0 = 1

Hence, total number of ways (N ) = 4 + 10 + 6 + 1 = 21 N 21 ⇒ = 3 3 =7

Target Exercises Type 1. Only One Correct Option (a)

m+ 1

C4

(b)

C 2 is equal to

m−1

2. If n C 3 + nC 4 >

C4

n+1

∑ nCr / ( nCr

r=0

(a) n + 1 (c) n + 2

∑

r =1

5. If

2n + 1

+ nC r + 1 ) equals

Pr is r!

Pn − 1 :

2n − 1

(b) 6

(c) 2n − 1

(d) 2n + 1

Pn = 3 : 5, then n is equal to (c) 3

(d) 8

6. The number less than 1000 that can be formed using the digits 0, 1, 2, 3, 4, 5 when repetition is not allowed is equal to (a) 130 (c) 156

(b) 131 (d) 155

7. The numbers greater than 1000 but not greater than 4000, which can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is allowed), are (a) 350 (c) 450

(b) 375 (d) 576

8. A variable name in certain computer language must be either an alphabet or an alphabet followed by a decimal digit. The total number of different variable names that can exist in that language is equal to (a) 280 (c) 286

(b) 290 (d) 296

9. The number of five-digit numbers that contain 7 exactly once is (a) (41) (93 ) (c) (7) (94 )

(b) (37) (93 ) (d) (41) (94 )

10. The total number of flags with three horizontal strips in order, which can be formed by using 2 identical red, 2 identical green and 2 identical white strips, is equal to (a) 4! (b) 3 × (4 !) (c) 2 × (4 !) (d) None of these

11. Let A be a set of n (≥ 3) distinct elements. The number of triplets (x, y, z) of the elements of A in which atleast two coordinates same, is (a) n P3

n

(b) 2n − 1

(a) 4

C4

(b) n / 2 (d) None of these n

(a) 2n

(d) 3

(b) n > 7 (d) None of these n−1

4. The value of

C4

m+ 1

C 3 , then

(a) n > 6 (c) n < 6

3. The value of

(c) 3

m+ 2

(b) n3 − nP3

(c) 3n2 − 2n

(d) 3n2 (n − 1)

12. The total number of six-digit numbers that can be formed, having the property that every succeeding digit is greater than the preceding digit, is (a) 9 C 3

(b) 10C 3

(c) 9 P3

(d)

10

P3

13. The number of four-digit numbers that can be made with the digits 1, 2, 3, 4 and 5 in which atleast two digits are identical, is (a) 4 5 − 5! (c) 600

(b) 505 (d) None of these

14. The total number of five-digit numbers of different digits in which the digit in the middle is the largest, is 9

(a)

∑ n P4

(b) 33(3!)

n= 4

(c) 30(3!)

(d) None of these

15. The number of nine non-zero digit numbers such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle, is (a) 2(4 !)

(b) 3(7!) / 2

(c) 2(7!)

Targ e t E x e rc is e s

α

1. If α = mC 2 , then

(d) 4 P4 × 4P4

16. The total number of three-digit numbers, the sum of whose digits is even, is equal to (a) 450

(b) 350

(c) 250

(d) 325

17. The permutations of n objects taken (i) atleast r objects at a time (ii) atmost r objects at a time (if repetition of the objects is allowed) are respectively nn − r + 1 nr + 1 − 1 and n−1 n−1 nn− r + 1 nr (nr + 1 − 1) (b) and n −1 n−1 r n− r n (n nr + 1 − 1 − 1) (c) and n−1 n−1 (d) None of these (a)

18. How many ten-digit numbers can be written by using the digits 1 and 2? (a) (c)

C 1 + 9C 2 C2

10

10

(b) 210 (d) 10!

307

Objective Mathematics Vol. 1

6

19. A five-digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways this can be done is (a) 216

(b) 240

(c) 600

(d) 3125

20. The total number of 9-digit numbers which have all different digits is (a) 10!

(b) 9!

(c) 9 ⋅ 9!

(d) 10 ⋅ 10!

21. Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 and then the men select the chairs from amongst the remaining. The number of possible arrangements is (a) C 3 × C 2 (c) 4 P2 × 4P3 4

(b) C 2 × P3 (d) None of these

4

4

4

22. From 4 Officers and 8 Jawans, a committee of 6 is to be chosen to include exactly one officer. The number of such committee is

Ta rg e t E x e rc is e s

(a) 160

(b) 200

(c) 224

(d) 300

23. Seven different lecturers are to deliver lectures in seven periods of a class on a particular day. A, B and C are three of the lecturers. The number of ways in which a routine for the day can be made such that A delivers his lecture before B and B before C, is (a) 420 (c) 210

(b) 120 (d) None of these

24. The number of arrangements of letters of the word ‘BANANA’ in which the two N’s do not appear together is (a) 40

(b) 60

(c) 80

(d) 100

25. How many different nine-digit numbers can be formed from the number 223355888 by rearranging its digits, so that odd digits occupy even positions? (a) 16

(b) 36

(c) 60

(d) 180

26. Let A = {x | x is a prime number and x < 30}. The number of different rational numbers whose numerator and denominator belong to A, is (a) 90 (c) 91

(b) 180 (d) None of these

27. Let S be the set of all functions from the set A to the set A. If n ( A ) = k, then n ( S ) is (a) k !

(b) k k

(c) 2k − 1

(d) 2k

28. Let A be the set of 4-digit numbers a1 a 2 a 3 a 4 , where a1 > a 2 > a 3 > a 4 , then n ( A ) is equal to (a) 126 (c) 210

308

(b) 84 (d) None of these

29. An n-digit number is a positive number with exactly n-digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is (a) 6

(b) 7

(c) 8

(d) 9

30. The number of 5-digit telephone numbers having atleast one of their digits repeated is (a) 90000

(b) 100000

(c) 30240

(d) 69760

31. If letters of the word ‘KUBER’ are written in all possible orders and arranged as in a dictionary, then rank of the word ‘KUBER’ will be (a) 67

(b) 68

(c) 65

(d) 69

32. The total number of 4-digit numbers that are greater than 3000, that can be formed using the digits 1, 2, 3, 4, 5, 6 (no digit is being repeated in any number) is equal to (a) 120

(b) 240

(c) 480

(d) 80

33. In a country, no two persons have identical set of teeth and there is no person without a tooth, also no person has more than 32 teeth. If shape and size of tooth is disregarded and only the position of tooth is considered, then maximum population of that country can be (a) 232 (c) Cannot be determined

(b) 232 − 1 (d) None of these

34. n different toys have to be distributed among n children. Total number of ways in which these toys can be distributed, so that exactly one child gets no toy, is equal to (a) n! (c) (n − 1) ! ⋅ nC 2

(b) n !⋅n C 2 (d) n !⋅ n − 1C 2

35. The total number of permutations of k different things, in a row, taken not more than r at a time (each thing may be repeated any number of times) is equal to (a) k r − 1

(b) k r

(c)

kr − 1 k −1

(d)

k (k r − 1) (k − 1)

36. A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions, so that exactly 10 predictions are correct, is equal to (a) (c)

C 10 ⋅ 210 20 C 10 ⋅ 310

20

(b) (d)

C 10 ⋅ 320 20 C 10 ⋅ 220

20

37. A team of four students is to be selected from a total of 12 students. Total number of ways in which team can be selected such that two particular students refuse to be together and other two particular students wish to be together only, is equal to (a) 220 (c) 226

(b) 182 (d) None of these

38. The total number of numbers that are less than 3⋅ 108 and can be formed using the digits 1, 2, 3, is equal to 1 9 (3 + 4 ⋅ 38 ) 2 1 (c) (7 ⋅ 38 − 3) 2 (a)

1 9 (3 − 3) 2 1 (d) (39 − 3 + 38 ) 2

(b)

(b) 12

(c) 14

(d) 16

40. In an examination of 9 papers, a candidate has to pass in more papers than the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful, is (a) 255

(b) 256

(c) 193

(d) 319

41. The number of 5-digit numbers that can be made using the digits 1 and 2 and in which atleast one digit is different, is (a) 30 (c) 32

(b) 31 (d) None of these

42. In a club election, the number of contestants is one more than the number of maximum candidates for which a voter can vote. If the total number of ways in which a voter can vote be 62, then the number of candidates is (a) 7 (c) 6

(b) 5 (d) None of these

43. Two players P1 and P2 play a series of ‘2n’ games. Each game can result in either a win or loss for P1 . Total number of ways in which P1 can win the series of these games, is equal to 1 2n 2n (2 − C n ) 2 1 (c) (2n − 2nC n ) 2 (a)

1 2n (2 − 2 ⋅ 2nC n ) 2 1 (d) (2n − 2 ⋅ 2nC n ) 2 (b)

44. In the decimal system of numeration, the number of 6-digit numbers in which the sum of digits is divisible by 5, is (a) 180000 (c) 5 × 105

(b) 540000 (d) None of these

45. The number of possible outcomes in a throw of n ordinary dice in which atleast one of the dice shows an odd number, is (a) 6n − 1 (c) 6n − 3n

(b) 3n − 1 (d) None of these

46. The number of different matrices that can be formed with elements 0, 1, 2 or 3, each matrix having 4 elements, is (a) 3 × 24 (c) 3 × 4 4

(b) 2 × 4 4 (d) None of these

47. There are 20 questions in a question paper. If no two students solve the same combination of questions but solve equal number of questions, then the maximum number of students who appeared in the examination, is (a) (c)

20

C9 20 C 10

(b) 20C 11 (d) None of these

(a) 828 (c) 396

(b) 1260 (d) None of these

49. A shopkeeper sells three varieties of perfumes and he has a large number of bottles of the same size of each variety in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varieties of perfumes in the showcase is (a) 6 (c) 150

6 Permutation and Combination

(a) 10

48. The number of different pairs of words (****, ***) that can be made with the letters of the word ‘STATICS’ is

(b) 50 (d) None of these

50. The number of ways in which 3 boys and 3 girls (all are of different heights) can be arranged in a line, so that boys as well as girls among themselves are in decreasing order of height from left to right, is (a) 1 (c) 20

(b) 6! (d) None of these

51. The number of words of four letters containing equal number of vowels and consonants, repetition allowed, is (b) 210 × 243 (d) None of these

(a) 1052 (c) 105 × 243

52. There are three teams each of a chairman, a supervisor and a worker three different companies. There are nine bonus of different denominations to be paid to these nine persons in all. In how many ways can this be done with due respect to superiority is given in every team? (a) 865 (c) 1680

(b) 129 (d) None of these

Targ e t E x e rc is e s

39. Four couples (husband and wife) decide to form a committee of four members. The number of different committees that can be formed in which no couple finds a place, is

53. If two numbers are selected from numbers 1 to 25, then the number of ways that their difference does not exceed 10 is (a) 105 (c) 15C 2

(b) 195 (d) None of these

54. The number of ways in which a mixed double game can be arranged amongst 9 married couples, if no husband and wife play in the same game, is (a) 756 (c) 3024

(b) 1512 (d) None of these

55. In a certain test, there are n questions. In this test 2n − i students gave wrong answers to atleast i question, where i = 1, 2, ... , n. If the total number of wrong answers given is 2047, then n is equal to (a) 10

(b) 11

(c) 12

(d) 13

56. The number of natural numbers which are less than 2 ⋅ 108 and which can be written by means of the digit 1 and 2, is (a) 772

(b) 870

(c) 900

(d) 766

57. The number of times digit 3 will be written when listing the integer from 1 to 1000, is (a) 269

(b) 300

(c) 271

(d) 302

309

Objective Mathematics Vol. 1

6

58. Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. The number of ways in which we can place the balls in the boxes (order is not considered in the box) so that no box remains empty is (a) 150 (c) 200

(b) 300 (d) None of these

59. The number of permutations of the letters of the word ‘HINDUSTAN’ such that neither the pattern ‘HIN’ nor ‘DUS’ nor ‘TAN’ appear, are (a) 166674

(b) 168474

(c) 166680

(d) 181434

60. Two teams are to play a series of 5 matches between them. A match ends in a win or loss or draw for a team. A number of people forecast the result of each match and no two people make the same forecast for the series of matches. The smallest group of people in which one person forecasts correctly for all the matches will contain n people, where n is

Ta rg e t E x e rc is e s

(a) 81 (c) 486

(b) 243 (d) None of these

61. In a class of 20 students, every student had a hand shake with every other student. The total number of hand shakes were (a) 180

(b) 190

(c) 200

(d) 210

62. There are 10 persons among whom two are brother. The total number of ways in which these persons can be seated around a round table so that exactly one person sit between the brothers, is equal to (a) (2!) (7!) (c) (3!) (7!)

(b) (2!) (8!) (d) (3!) (8!)

63. The number of ways in which a couple can sit around a table with 6 guests, if the couple take consecutive seats, is (a) 1440 (c) 5040

(b) 720 (d) None of these

64. The number of different garlands, that can be formed using 3 flowers of one kind and 3 flowers of other kind, is (a) 60 (c) 4

(b) 20 (d) 5

65. In the next world cup of cricket, there will be 12 teams, divided equally in two groups. Teams of each group will play a match against each other. From each group, 3 top teams will qualify for the next round. In this round each team will play against others once. Four top teams of this round will qualify for the semifinal round, where each team will play against the other three. Two top teams of this round will go to the final round, where they will play the best of three matches. The minimum number of matches in the next world cup will be 310

(a) 54 (c) 38

(b) 53 (d) None of these

66. The value of 10 C 5 / 11 C 6 , when numerator and denominator takes their greatest value, is (a)

6 11

(b)

5 11

(c)

10 6

(d)

10 5

67. The number of subsets {1, 2, 3, ..., n} having least element m and greatest element k, 1≤ m < k ≤ n, is (a) 2n − (k − m)

(b) 2k − m − 2

(c) 2k − m− 1

(d) 2k − m + 1

68. A class has 21 students. The class teacher has been asked to make n groups of r students each and go to zoo taking one group at a time. The size of group zoo i.e. the value of r) for which the teacher goes to the maximum number of times is (no group can go to the zoo twice) (a) 9 or 10

(b) 10 or 11

(c) 11 or 12

(d) 12 or 13

69. A class has n students. We have to form a team of the students including atleast two students and also excluding atleast two students. The number of ways of forming the team is (a) 2n − 2n (c) 2n − 2n − 4

(b) 2n − 2n − 2 (d) None of these

70. From 4 gentlemen and 6 ladies, a committee of five is to be selected. The number of ways, in which the committee can be formed so that gentlemen are in majority, is (a) 66 (c) 60

(b) 156 (d) None of these

71. If the total number of m elements subsets of the set A = {a1 , a 2 , a 3 , ... , a n } is k times the number of m elements subsets containing a 4 , then n is (a) (m − 1) k (c) (m + 1) k

(b) mk (d) None of these

72. An urn contains 3 red pens, 4 green pens and 6 yellow pens. The number of ways of drawing 4 pens from the urn, if atleast one red pen is to be included in the draw, is (all the pens are different from each other) (a) 500 (c) 510

(b) 505 (d) None of these

73. Two numbers are chosen from 1, 3, 5, 7, ... , 147, 149 and 151 and multiplied together in all possible ways. The number of ways which will give us the product a multiple of 5, is (a) 1710 (c) 1700

(b) 2900 (d) None of these

74. Let Tn denotes the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn + 1 − Tn = 21, then n equals (a) 5

(b) 7

(c) 6

(d) 4

75. If a polygon has 44 diagonals, then the number of its sides are (a) 11 (c) 8

(b) 7 (d) None of these

(b) 145

(c) 178

(d) 205

77. ABCD is a convex quadrilateral 3, 4, 5 and 6 points are marked on the sides AB, BC, CD and DA, respectively. The number of triangles with vertices on different sides is (a) 270 (c) 282

(b) 220 (d) None of these

78. In a polygon, no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon is 70, then the number of diagonals of the polygon is (a) 20 (c) 8

(b) 28 (d) None of these

79. n lines are drawn in a plane such that no two of them are parallel and no three of them are concurrent. The number of different points at which these lines will cut, is n

(a)

∑k

(b) n (n − 1)

k=1 2

(c) n

ab + bc + ca 2 (c) Σ ab (a + b)

(b) Σ ab (a + b − 2)

(a)

(d) None of these

86. The number of points in the cartesian plane with integral coordinates satisfying the inequalities x ≤ k , y ≤ k , x − y ≤ k, is (b) (k + 2)3 − (k + 1)3 (d) None of these

(a) (k + 1)3 − k 3 (c) (k 2 + 1)

87. Given 5 different green dyes, 4 different blue dyes and 3 different red dyes. The number of combinations of dyes, which can be chosen taking atleast one green and one blue dye, is (a) 3600 (c) 3800

(b) 3720 (d) None of these

88. Six X ′s have to be placed in the squares of the figure given below such that each row contains atleast one X ′. The number of ways in which this can be done is

(d) None of these

80. If n objects are arranged in a row, then the number of ways of selecting three objects, so that no two of them are next to each other, is (n − 2) (n − 3) (n − 4 ) 3 (c) n − 2C 3 + n − 3C 2

(a)

(b)

n− 2

C3

(d) None of these

81. There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum number of triangles with vertices at these points is (a) 3 p2 ( p − 1) + 1 (c) p2 (4 p − 3)

(b) 3 p2 ( p − 1) (d) None of these

82. In a plane, there are two families of lines y = x + r, y = − x + r, where r ∈{0, 1, 2, 3, 4}. The number of squares of diagonals of length 2 formed by the lines, is (a) 9 (c) 25

(b) 16 (d) None of these

83. Line L1 contains l1 point and line L2 contains l2 point. If the points on L1 are joined to the points on L2 , then number of points of intersection of new lines, is (a) l1 C 2 × l2 C 2 (b) 4 ⋅ l1C 2 × l2 C 2 (c) 2 ⋅ l1C 2 × l2 C 2 (d) None of the above

(a) 16

(b) 28

(a) 26 (c) 22

(c) 56

(d) 70

(b) 27 (d) None of these

89. The total number of six-digit numbers x1 x 2 x 3 x 4 x 5 x 6 having the property x1 < x 2 ≤ x 3 < x 4 < x 5 ≤ x 6 , is (a) (c)

(b) 12C6 (d) None of these

10

C6 C6

11

90. A class contains three girls and four boys. Every Saturday five students go on a picnic, a different group of students is being sent each week. During the picnic, each girl in the group is given doll by the accompanying teacher. All possible groups of five have gone once, the total number of dolls the girls have got, is (a) 21

(b) 45

(c) 27

(d) 24

91. The total number of ways of selecting two numbers from the set {1, 2, 3, 4, ...., 3n}, so that their sum of divisible by 3, is (a)

84. The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the octagon, is

6 Permutation and Combination

(a) 135

85. The sides AB, BC and CA of a ∆ABC have a, b and c interior points on them respectively, then the number of triangles that can be constructed using these interior points as vertices, is

Targ e t E x e rc is e s

76. The sides AB, BC, CA of a ∆ABC have 3, 4, 5 interior points, respectively on them. Total number of triangles that can be formed using these points as vertices, is equal to

2n2 − n 2

(b)

3n2 − n 2

(c) 2n2 − n

(d) 3n2 − n

92. If r > p > q, the number of different selections of p + q things taking r at a time, where p things are identical and q other things are identical, is (a) p + q − r (c) r − p − q + 1

(b) p + q − r + 1 (d) None of these

311

Objective Mathematics Vol. 1

6

93. The number of proper divisors of 2 p ⋅ 6q ⋅ 15r is (a) ( p + q + 1) (q + r + 1) (r + 1) (b) ( p + q + 1) (q + r + 1) (r + 1) − 2 (c) ( p + q) (q + r) r − 2 (d) None of the above

94. The number of even proper divisors of 1008 is (a) 23 (c) 22

(b) 24 (d) None of these

95. The number of ways to fill each of the four cells of the table with a distinct natural number such that the sum of the numbers is 10 and the sum of the numbers places diagonally are equal, is

102. A person writes letters to 6 friends and addresses the corresponding envelopes. The number of ways in which all 5 letters can be placed in wrong envelopes, is (a) 264 (c) 206

(b) 210 (d) None of these

103. The number of ways in which m + n ( n ≤ m + 1) different things can be arranged in a row such that no two of the n things may be together, is (m + n)! m ! n! m ! n! (c) (m − n + 1)! these (a)

(b)

m! (m + 1)! (m + n)!

(d) None of

104. Number of divisors of the form 4n + 2 ( n ≥ 0) of the integer 240, is

Ta rg e t E x e rc is e s

(a) 2! × 2! (c) 2(4 !)

(a) 4 (c) 10

(b) 4! (d) None of these

96. In the figure, two 4-digit numbers are to be formed by filling the places with digits. The number of different ways in which the places can be filled by digits so that the sum of the numbers formed is also a 4-digit number and in no place the addition is with carrying, is Th

H

T

U

105. Total number of non-negative integral solutions of x1 + x 2 + x 3 = 10 is equal to (a) (c)

(b) 10C 3 (d) 10C 2

12

C3 C2

12

106. Total number of positive integral solutions of x1 + x 3 + 2x 3 = 15 is equal to (a) 42 (c) 32

(b) 36 ⋅ (55)3 (d) None of these

(a) 554 (c) 454

97. A bag contains 2 apples, 3 oranges and 4 bananas. The number of ways in which 3 fruits can be selected, if atleast one banana is always in the combination (assume fruit of same species to be alike) is (b) 10

(c) 29

(d) 7

98. The number of divisors of the number 38808 (excluding 1 and the number itself), is (a) 70 (c) 71

(b) 72 (d) None of these

99. The number of integral solutions of x1 + x 2 + x 3 = 0, with x i ≥ − 5, is (a) 15C 2

(b)

16

C2

(c)

17

C2

(d)

18

C2

100. If 33! is divisible by 2n , then the maximum value of n is equal to (a) 30

(b) 31

(c) 32

(d) 33

101. Let p be a prime number such that p ≥ 3. Let n = p ! + 1. The number of primes in the list n + 1, n + 2, n + 3, ...., n + p − 1is (a) p − 1 (c) 1

(b) 2 (d) None of these

(b) 70 (d) None of these

107. Total number of ways in which 15 identical blankets can be distributed among 4 persons, so that each of them gets atleast two blankets, is

+

(a) 6

312

(b) 8 (d) 3

(a) (c)

10

(b) 9 C 3 (d) None of these

C3 11 C3

108. Total number of 3 letter words that can be formed from the letter of the word ‘AAAHNPRRS’ is equal to (a) 210

(b) 237

(c) 247

(d) 227

109. 15 identical balls have to be put in 5 different boxes. Each box can contain any number of balls. Total number of ways of putting the balls into box, so that each box contains atleast 2 balls, is equal to (a) 9 C 5 (c) 6C 5

(b) (d)

10

C5 C6

10

110. The minimum marks required for clearing a certain screening paper is 210 out of 300. The screening paper consists of 3 sections each of Physics, Chemistry and Mathematics. Each section has 100 as maximum marks. Assuming, there is no negative marking and marks obtained in each section are integers, the number of ways in which a student can qualify the examination, assuming no cut-off limit, is (a) 210C 3 − 90C 3 (b) 93C 3 (c) 213C 3 (d) (210)3

(b) 108

(c)

17

C7

(d)

10

C8

112. A bag contains 3 black, 4 white and 2 red balls, all the balls being different. The number of selections of atmost 6 balls containing balls of all the colours, is

(a)

n− 1

P2 (n + 1) (n + 2) (n + 3) (b) 6 (c) n − 1C n − 4 (d) None of these

116. The number of points ( x, y, z ) in space, whose each coordinate is a negative integer such that x + y + z + 12 = 0, is

(a) 42 (4 !) (b) 26 × 4 ! (c) (26 − 1) (4 !) (d) None of the above

113. The number of ways to give 16 different things to three persons, so that B gets 1 more than A and C gets 2 more than B, is 16! 4 ! 5! 7! (b) 4 ! 5! 7! 16! (c) 3! 5! 8! (d) None of the above (a)

(a) 385 (c) 110

(b) 55 (d) None of these

117. If a, b, c are three natural numbers in AP and a + b + c = 21, then the possible number of values of the ordered triplet (a, b, c) is (a) 15 (b) 14 (c) 13 (d) None of the above

114. The number of positive integral solutions of x + y + z = n, n ∈ N , n ≥ 3, is (a) n − 1C 2 (b) n − 1 P2 (c) n (n − 1) (d) None of the above

118. If a, b, c, d are odd natural numbers such that a + b + c + d = 20, then the number of values of the ordered 4-tuple (a, b, c, d) is (a) 165 (b) 455 (c) 310 (d) None of the above

Type 2. More than One Correct Option 119. Number of ways in which three numbers in AP can be selected from 1, 2, 3,..., n, is 2

 n − 1 (a)   , if n is even0  2  n (n − 2) (b) , if n is even 4 (n − 1)2 , if n is odd (c) 4 (d) None of the above

123. If n C α = nCβ , then it may be true that

(b) 2 × 8C 3 (d) None of these

121. A forecast is to be made of the results of five cricket matches, each of which can be a win or a draw or a loss for Indian team. Let p = number of forecasts with exactly 1 error q = number of forecasts with exactly 3 errors and r = number of forecasts with all five errors Then, the correct statement(s) is/are (a) 2q = 5r (c) 8 p = 5r

122. If x is the number of 5-digit numbers sum of whose digits is even and y is the number of 5-digit numbers sum of whose digits is odd, then (a) x = y (b) x + y = 90000 (c) x = 45000 (d) x < y

120. Kanchan has 10 friends among whom two are married to each other. She wishes to invite five of them for a party. If the married couples refuse to attend separately, then the number of different ways in which she can invite five friends, is (a) 8 C 5 (c) 10C 5 − 2 × 8C 4

6

Targ e t E x e rc is e s

(a) 810

115. The number of non-negative integral solutions of a + b + c + d = n, n ∈ N , is

Permutation and Combination

111. Number of ways of distributing 10 identical objects among 8 persons (one or many persons may not be getting any object), is

(b) 8p = q (d) 2( p + r) > q

(a) α = β (b) α + β = 2n (c) α + β = n (d) All of the above

124. There are n married couples at a party. Each person shakes hand with every person other than her or his spouse. The total number of hand shakes must be (a) 2nC 2 − n (b) 2nC 2 − (n − 1) (c) 2n (n − 1) (d) 2nC 2

125. There are n lines in a plane, no two of which are parallel and no three of which are concurrent. If plane is divided in u n parts, then (a) u4 = 11 (b) u3 = 7 (c) u2 = 3 (d) un = un − 1 + 4

313

Objective Mathematics Vol. 1

6

Type 3. Assertion and Reason Directions (Q. Nos. 126-133) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ta rg e t E x e rc is e s

126. Statement I The number of positive integral solutions of abc = 30 is 27.

Statement II prime factors.

1400 is divisible by exactly three

130. Statement I Let E = {1, 2, 3, 4} and F = {a, b}. Then, the number of onto functions from E to F is 14. Statement II Number of ways in which four distinct objects can be distributed into two different boxes, if no box remains empty is 14. 131. Statement I Number of ways in which India can win the series of 11 matches, if no match is drawn is 210 .

Statement II Number of ways in which three prizes can be distributed among three persons is 33 .

Statement II For each match, there are two possibilities, either India wins or loses.

127. Statement I Number of ways in which 10 identical toys can be distributed among three students, if each receives at least two toys, is 9 C 2 .

132. Statement I Number of ways in which Indian team (11 players) can bat, if Yuvraj wants to bat before Dhoni and Pathan wants to bat after Dhoni is 11!/3!.

Statement II Number of positive integral solutions of x + y + z + w = 7 is 6 C 3 .

Statement II Yuvraj, Dhoni and Pathan can be arranged in batting order in 3! ways.

( n 2 )! / ( n !) n is a natural number,

133. Statement I When number of ways of arranging 21 objects of which r objects are identical of one type and remaining are identical of second type is maximum, then maximum value of 13 C r is 78.

128. Statement I ∀ n ∈ N.

Statement II Number of ways in which n 2 objects can be distributed among n persons equally is ( n 2 )!/( n !) n .

Statement II

2n + 1

C r is maximum, when r = n.

Type 4. Linked Comprehension Based Questions Passage I (Q. Nos. 134-136) We have to choose 11 players for cricket team from eight batsmen, six bowlers, four all rounders and two wicket keepers in the following questions.

Passage II (Q. Nos. 137-139) 12 persons are to be

134. The number of selections, when atmost one all rounder and one wicket keeper will play, is

137. Number of ways in which these 12 persons can be arranged is

(a) 4C 1 × 14C 10 + 2C 1 × 14C 10 + 4C 1 × 2C 1 × 14C 9 + (b) 4C 1 × 15C 11 + 15C 11 (c) 4C 1 × 15C 10 + 15C 11 (d) None of the above

14

C 11

135. Number of selections when two particular batsmen do not want to play when a particular bowler will play, is (a)

17

C 10 +

19

(c)

17

C 10 +

20

C 11

C 11

(b)

17

C 10 +

19

(d)

19

C 10 +

19

C 11 +

17

C 11

C 11

136. Number of selections when a particular batsman and a particular wicket keeper do not want to play together, is 314

129. Statement I The number of ways of writing 1400 as a product of two positive integers is 12.

(a) 218 C 10 (c) 19 C 10 +

19

C 11

(b) 19 C 11 + 18C 10 (d) None of these

arranged around two round tables such that one table can accommodate seven persons and another five persons only.

(a) (c)

12

C 5 6! 2! C 5 6! 4 !

12

(b) 6!4 ! (d) None of these

138. Number of ways of arrangements, if two particular persons A and B do not want to be on the same table, is (a) (c)

10

C 4 6! 4 ! C6 6! 4 !

11

(b) 210C6 6! 4 ! (d) None of these

139. Number of ways of arrangement, if two particular persons A and B want to be together and consecutive, is (a) 10C 7 6! 3!2! + 10C 5 4 ! 5! 2! (b) 10C 5 6! 3! + 10C 7 4 ! 5! (c) 10C 7 6! 2! + 10C 5 5! 2! (d) None of the above

140. 4 candidates are competing for two managerial posts. In how many ways can the candidates be selected? (a) 4 2 (c) 24

(b) 4C 2 (d) None of these

141. 8 different balls can be distributed among 3 children so that every child receives atleast 1 ball, is 8

8

(a) 3 (c) 83

(b) C 3 (d) None of these

142. 5 letters can be posted into 3 letter boxes in (a) 35 ways (b) 53 ways (c) 5C 3 ways (d) None of the above

Type 5. Match the Columns 143. Match the following. Column I

Column II

A. Number of straight lines joining any two of 20 points of which four points are collinear, is

p.

56

B. Maximum number of points of intersection of 20 straight lines in the plane, is

q.

60

C. Maximum number of points of intersection of 8 circles in the plane, is

r.

185

D. Maximum number of points intersection of six parabolas, is

s.

190

of

144. A function is defined as f :{a1 , a 2 , a 3 , a 4 , a 5 , a 6 } → {b1 , b2 , b3 }. Then, match the following. Column I

Column I

Column II

A. The value of 1 ⋅ 1! + 2 ⋅ 2 ! + 3 ⋅ 3 ! + ... + n ⋅ n! is

p. ( n + 2 ) 2 n − 1 − ( n + 1)

B. The value of n ⋅ n C1 + ( n − 1) ⋅ n C 2 + ( n − 1) ⋅ nC 3 + ... + 1 ⋅ nC n is

q.

2n

Cn

Column II p. divisible by 6

B. Number of f( ai ) ≠ b i , is

q. divisible by 2

functions

145. Match the statements of Column I with values of Column II.

r. ( n + 1)! − 1 C. The value of 2 C 2 + 3C 2 + 4C 2 + ... + nC 2 is

A. Number of surjective functions is in

which

C. Number of invertible functions is

r. divisible by 4

D. Number of many-one functions is

s. divisible by 3

6 Permutation and Combination

distributing n different things into r different groups is r n when blank groups are taken into account and is r n − r C1 (r − 1)n + r C 2 (r − 2) 2 − .. + ( − 1)r − 1 r Cr − 1 when blank groups are not permitted.

n

D. The value of

∑( C r ) n

2

is

s.

n +1

C3

r=0

t.

Cn − 1 − 1

2n

Targ e t E x e rc is e s

Passage III (Q. Nos. 140-142) Number of ways of

t. not possible

Type 6. Single Integer Answer Type Questions 146. A class has three teachers, Mr. P, Ms. Q and Mrs. R and six students A, B, C, D, E, F. Number of ways in which they can be seated in a line of 9 chairs, if between any two teachers, there are exactly two students, is k ! (18), then the value of k is ____________ . 147. Consider, the five points comprising the vertices of a square and the intersection point of its diagonals. How many triangles can be formed using these points? 148. If number of selections of 6 different letters that can be made from the words ‘SUMAN’ and ‘DIVYA’, so that each selection contains 3 letters from each word is N 2 , then the value of N is _________ .

149. There are n distinct white and n distinct black balls. If the number of ways of arranging them in a row, so that neighbouring balls are of different colours, is 1152, then the value of n is _________ . 150. There are 4 oranges, 5 apples and 6 mangoes in a fruit basket. If (21k + 20) ways can a person make a selection of fruits from among the fruits in the basket, when all fruits of the same type, are identical, then k refers to _________ . 151. The number of ways of filling three boxes (named A, B and C) by 12 or less number of identical balls, if no box is empty, box B has atleast 3 balls and box C has atmost 5 balls, is 55 λ, where λ r refers to _______ . 152. The number of non-negative integral solutions 2x + y + z = 21is 22( k ), where k refers to _______ . 315

Entrances Gallery JEE Advanced/IIT JEE 1. Let n1 < n 2 < n 3 < n 4 < n 5 be positive integers such that n1 + n 2 + n 3 + n 4 + n 5 = 20. Then, the number of such distinct arrangements (n1 , n 2 , n 3 , n 4 , n 5 ) [2014] is _________ . 2. Let n ≥ 2 be an integer. Take n distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of n is _________ . [2014]

3. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then, the number of ways it [2014] can be done is (a) 264

(b) 265

(c) 53

(d) 67

4. Consider the set of eight vectors Three V = {a$i + b$j + ck$ ; a, b, c ∈ {−1, 1}}. non-coplanar vectors can be chosen from V in 2 p [2013] ways. Then, p is _________ .

JEE Main/AIEEE

Ta rg e t E x e rc is e s

5. The number of integers greater than 6000 that can be formed, using the digits 3, 5, 6, 7 and 8 without [2015] repetition, is (a) 216 (c) 120

(b) 192 (d) 72

6. Let A and B be two sets containing 2 elements and 4 elements, respectively. The number of subsets of [2013] A × B having 3 or more elements, is (a) 256 (c) 219

(b) 220 (d) 211

(b) 150

(c) 210

(d) 243

Directions (Q. Nos. 8-9) Let an denotes the number of all n-digit positive integers formed by the digits 0, 1 or both, such that no consecutive digits in them are 0. Let bn be the number of such n digit integers ending with digit 1 and cn be the number of such n digit integers ending with digit 0. [2012]

8. Which of the following is correct? (a) a17 (b) c17 (c) b17 (d) a17

= a16 + a15 ≠ c16 + c15 ≠ b16 + c16 = c17 + b16

(b) 8

(c) 9

(d) 11

10. Statement I The number of ways distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9 C 3 . 316

(b) N ≤ 100 (d) 140 < N ≤ 190

12. There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn, two balls are taken out at random and then transferred to the other. The number of ways in which this can be [2010] done is (a) 3 (c) 66

(b) 36 (d) 108

13. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then, the number of such [2009] arrangements is (a) atleast 500 but less than 750 (b) atleast 750 but less than 1000 (c) atleast 1000 (d) less than 500

9. The value of b6 is (a) 7

11. There are 10 points in a plane, out of these 6 are collinear. If N is the number of triangles formed by joining these points, then [2011] (a) N > 190 (c) 100 < N ≤ 140

7. The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets atleast one ball, is [2012] (a) 75

(a) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (b) Statement I is true, Statement II is false (c) Statement I is false, Statement I is true (d) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

Statement II The number of ways of choosing any 3 places from 9 different places is 9 C 3 . [2011]

14. In a shop, there are five types of ice-creams available. A child buys six ice-creams. Statement I The number of different ways the child can buy the six ice-creams is 10 C 5 . Statement II The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging 6A’s and 4B’s [2008] in a row.

15. How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in which no two S are adjacent? [2008] (a) 8 ⋅ 6C 4 ⋅ 7C 4 (b) 6 ⋅ 7 ⋅ 8C 4 (c) 6 ⋅ 8 ⋅ 7C 4 (d) 7 ⋅ 6C 4 ⋅ 8C 4

(b) 385

(c) 600

(d) 601 7−x

19. The range of the function f ( x ) = (a) {1, 2, 3} (c) {1, 2, 3, 4}

(a) 120

Px − 3 is [2004]

(b) {1, 2, 3, 4, 5, 6} (d) {1, 2, 3, 4, 5}

(b) 12!/ 3! (3!) (d) 12!/ (3!)4

(c) 1110

(b) 240

(c) 360

(d) 480

21. The number of ways of distributing 8 identical balls in 3 distinct boxes, so that none of the boxes is empty, is [2004] (a) 5

(c) 38

(b) 21

(d) 8 C 3

4

17. At an election, a voter may vote for any number of candidates not greater than the number to be elected. There are 10 candidates and 4 are to be elected. If a voter votes for atleast one candidate, then the number [2006] of ways in which he can vote, is (a) 6210

(b) 603

(d) 5040

22. The number of ways in which 6 men and 5 women can dine at a round table, if no two women are to sit together, is given by [2003] (a) 6! × 5!

(c) 5! × 4 !

(b) 30

(d) 7! × 5!

n

23. If C r denotes the number of combinations of n things taken r at a time, then the expression n [2003] C r + 1 + nC r − 1 + 2 × nC r equals (a)

n+ 2

(b)

Cr

n+ 2

C r + 1 (c)

n+ 1

Cr

(d)

n+ 1

Cr + 1

Other Engineering Entrances 24. How many 5-digit telephone numbers can be constructed using the digits 0 to 9, if each number starts with 67 and no digit appears more than once? (a) 335

(b) 336

[Karnataka CET 2014] (c) 338 (d) 337

28. 10 different toys are to be distributed among 10 children. Total number of ways of distributing these toys, so that exactly two children do not get any toy, is [Manipal 2014] 10! 10! (b) 2! 3! 7! (2!)4 × 6!  1 1  10! × 10!  25  (d) (c) (10!)2  +  4 (2!)2 × 6!  84   (2!) × 6! 2! × 3! (a)

25. A student is allowed to select atmost n books from a collection of ( 2n + 1) books. If the number of ways in which he can do this, is 64, then the value of n is (a) 6 (c) 3

[BITSAT 2014] (b) n (d) None of these

26. S = {1, 2, 3, ... , 20} is to be partitioned into four sets A, B , C and D of equal size. The number of ways it can be done, is [Manipal 2014] 20! 4 ! × 5! 20! (c) (5!)4

(a)

20! 45 20! (d) (4 !)5

29. Out of 7 consonants and 4 vowels, the number of words (not necessarily meaningful) that can be made, each consisting of 3 consonants and 2 vowels, is (a) 24800

30.

m

Cr + 1 +

(b)

[WB JEE 2014] (d) 25400

n

[AMU 2014]

k =m

(c) C r

31.

(c) 25200

∑ k C r is equal to

n

[Manipal 2014] (b) 36 (d) 37

(b) 25100

(a) nC r + 1

27. In how many ways can a student choose a program of 5 courses, if 9 courses are available and 2 specific courses are compulsory for every student? (a) 34 (c) 35

6

Targ e t E x e rc is e s

(a) 12!/ 3! (4 !) (c) 12!/ (4 !)3

(a) 602

20. How many ways are there to arrange the letters in the word ‘GARDEN’ with the vowels in alphabetical [2004] order?

16. The set S :{1, 2, 3,... , 12} is to be partitioned into three sets A, B, C of equal size. Thus, A ∪ B ∪ C = S , A ∩ B = B ∩ C = A ∩ C = φ. [2007] The number of ways to partition S is 3

18. If the letters of the word ‘SACHIN’ are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number is [2005]

Permutation and Combination

(a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

(b)

n+ 1

Cr + 1

(d) None of these

C1 C C Cn is equal to + 2 2 + 3 3 + ... + n C0 C1 C2 Cn − 1 [AMU 2014] n (n − 1) 2 (n + 1) (n + 2) (c) 2

(a)

(b)

n (n + 1) 2

(d) None of these

317

Objective Mathematics Vol. 1

6

32. If n is an integer with 0 ≤ n ≤ 11, then the minimum value of n ! (11 − n )! is attained, when a value of n is [EAMCET 2014] (a) 11 (c) 7

(b) 5 (d) 9

33. Find n C 21 , if n C10 = nC11 . (a) 1 (c) 11

[J&K CET 2014]

(b) 0 (d) 10

34. Determine n, if

2n

C 2 : nC 2 = 9 : 2.

(a) 5 (c) 3

35. Out of thirty points in a plane, eight of them are collinear. The number of straight lines that can be formed by joining these points, is [EAMCET 2014] (b) 540 (d) 348

36. The number of positive integers which can be formed by using any number of digits from 0, 1, 2, 3, 4, 5 but using each digit not more than once in each number [Karnataka CET 2013] is

Ta rg e t E x e rc is e s

(a) 1200 (c) 1600

(b) 1500 (d) 1630

37. The total number of ways in which five ‘+’ and three ‘−’ signs can be arranged in a line such that no two ‘−’ sign occur together, is [AMU 2013] (a) 10 (b) 20 (c) 15 (d) None of the above

(b) 196 (d) 346

39. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if atleast one black ball is to be included in the draw? [AMU 2013] (a) 64

(b) 24

(c) 3

(d) 12

40. Four speakers will address a meeting, where speaker Q will always speak after speaker P. Then, the number of ways in which the order of speakers can be prepared is [WB JEE 2012] (a) 256 (b) 128 (c) 24 (d) 12

41. Sum of digits in the unit’s place formed by the digits 1, 2, 3 and 4 taken all at a time is [OJEE 2012]

318

(a) 30 (b) 60 (c) 59 (d) 61

43. A student is allowed to select atmost n books from a collection of ( 2n + 1) books. If the total number of ways in which he can select atleast one book is 225, [J&K CET 2011] then the value of n is (a) 6 (c) 4

(b) 5 (d) 3

44. In how many ways can 5 prizes be distributed among four students, when every student can take one or more prizes? [BITSAT 2011] (a) 1024 (c) 120

(b) 625 (d) 600

45. The number of ways in which the digits 1, 2, 3, 4, 3, 2, 1 can be arranged so that the odd digits always occupy the odd places, is [MP PET 2011] (a) 6 (c) 18

(b) 12 (d) 24

46. The number of integers greater than 6000 that can be formed with 3, 5, 6, 7 and 8, where no digit is repeated, is [Kerala CEE 2011]

38. A student is to answer 10 out of 13 questions in an examination such that he must choose atleast 4 from the first five questions. The number of choices available to him is [Manipal 2013] (a) 140 (c) 280

(a) 12 (b) 10 (c) 60 (d) None of the above

[J&K CET 2014]

(b) 4 (d) 2

(a) 296 (c) 408

42. A teacher takes 3 children from her class to the zoo at a time as often as she can, but she does not take the same three children to the zoo more than once. She finds that she goes to the zoo 84 times more that a particular child goes to the zoo. The number of [GGSIPU 2011] children in her class is

(a) 120 (c) 216 (e) 202

47. If

56

Pr + 6 :

(b) 192 (d) 72 54

Pr + 3 = 30800 : 1, then the value of r is

[Kerala CEE 2010]

(a) 40 (c) 41 (e) 101

(b) 51 (d) 510

48. The number of permutations by taking all letters and keeping the vowels of the word ‘COMBINE’ in the [WB JEE 2010] odd places, is (a) 96 (b) 144 (c) 512 (d) 576

49. From 12 books, the difference between number of ways of selection of 5 books when one specified book is always excluded and one specified book is [Kerala CEE 2010] always included, is (a) 64 (c) 132 (e) 462

50. If n − 1 C 3 + integer (a) 5 (c) 4

(b) 118 (d) 330 n−1

C 4 > nC 3 , then n is just greater than [WB JEE 2010] (b) 6 (d) 7

Answers Work Book Exercise 6.1 1. (d)

2. (b)

3. (d)

4. (b)

5. (a)

3. (a)

4. (a)

5. (c)

3. (b)

4. (a)

5. (d)

6. (d)

7. (a)

6. (a)

7. (b)

8. (d)

9. (a)

10. (c)

Work Book Exercise 6.2 1. (c)

2. (c)

Work Book Exercise 6.3 1. (d)

2. (b)

Work Book Exercise 6.4 1. (c)

2. (c)

3. (b)

4. (a)

5. (d)

6. (d)

7. (c)

8. (c)

9. (b)

10. (a)

11. (d)

12. (a)

13. (a)

14. (a)

15. (d)

16. (b)

17. (a)

18. (d)

19. (b)

20. (b)

21. (b)

22. (d)

23. (d)

24. (a)

25. (b)

26. (c)

27. (d)

4. (a)

5. (c)

6. (d)

Work Book Exercise 6.5 1. (c)

2. (d)

3. (a)

1. (c)

2. (a)

3. (b)

4. (a)

5. (c)

6. (d)

7. (d)

8. (b)

9. (c)

10. (b)

11. (c)

12. (c)

13. (b)

14. (d)

15. (c)

16. (b)

17. (b)

18. (a)

19. (a)

20. (b)

21. (c)

22. (b)

23. (b)

24. (c)

25. (c)

Target Exercises 1. (d)

2. (a)

3. (b)

4. (b)

5. (a)

6. (b)

7. (b)

8. (c)

9. (a)

10. (a)

11. (b)

12. (a)

13. (b)

14. (d)

15. (d)

16. (a)

17. (d)

18. (b)

19. (a)

20. (c)

21. (d)

22. (c)

23. (d)

24. (a)

25. (c)

26. (c)

27. (b)

28. (c)

29. (b)

30. (d)

31. (a)

32. (b)

33. (b)

34. (b)

35. (d)

36. (a)

37. (c)

38. (c)

39. (d)

40. (b)

41. (a)

42. (c)

43. (a)

44. (a)

45. (c)

46. (c)

47. (c)

48. (b)

49. (c)

50. (c)

51. (d)

52. (d)

53. (d)

54. (b)

55. (b)

56. (d)

57. (b)

58. (a)

59. (b)

60. (b)

61. (b)

62. (b)

63. (a)

64. (d)

65. (b)

66. (a)

67. (d)

68. (b)

69. (b)

70. (a)

71. (b)

72. (b)

73. (d)

74. (b)

75. (a)

76. (d)

77. (d)

78. (a)

79. (d)

80. (b)

81. (c)

82. (a)

83. (c)

84. (a)

85. (d)

86. (a)

87. (b)

88. (a)

89. (c)

90. (b)

91. (b)

92. (b)

93. (b)

94. (a)

95. (a)

96. (b)

97. (a)

98. (a)

99. (c)

100. (b) 110. (b)

101. (d)

102. (a)

103. (d)

104. (a)

105. (c)

106. (a)

107. (a)

108. (c)

109. (a)

111. (c)

112. (a)

113. (a)

114. (a)

115. (b)

116. (b)

117. (c)

118. (a)

119. (b,c)

120. (b,c)

123. (a,c)

124. (a,c)

125. (a,b)

126. (a)

127. (d)

128. (a)

129. (b)

130. (a)

121. (a,b,d) 122. (a,b,c) 131. (b)

132. (a)

133. (d)

134. (a)

135. (b)

136. (b)

137. (c)

138. (b)

139. (a)

140. (b)

141. (d)

142. (a)

143. (*)

144. (**)

145. (***)

146. (6)

147. (8)

148. (8)

149. (4)

150. (9)

151. (2)

152. (6)

Targ e t E x e rc is e s

Work Book Exercise 6.6

* A → r; B → s; C → p; D → q ** A → p,q,r,s; B → p,q,r,s; C → p,q,r,s,t; D → s *** A → r; B → p; C → s; D → q

Entrances Gallery 1. (7)

2. (5)

3. (c)

4. (5)

5. (b)

6. (c)

7. (b)

8. (a)

9. (b)

10. (a)

11. (b)

12. (d)

13. (c)

14. (d)

15. (d)

16. (c)

17. (b)

18. (d)

19. (a)

20. (c)

21. (b)

22. (a)

23. (b)

24. (b)

25. (c)

26. (c)

27. (c)

28. (d)

29. (c)

30. (b)

31. (b)

32. (b)

33. (a)

34. (a)

35. (c)

36. (d)

37. (b)

38. (b)

39. (a)

40. (d)

41. (b)

42. (d)

43. (c)

44. (a)

45. (c)

46. (b)

47. (c)

48. (d)

49. (c)

50. (d)

319

Explanations Target Exercises

8. Total number of variables, if only alphabet is used is 26.

m(m − 1) 1. We have, α = C2 ⇒ α = 2 α (α − 1) 1 m(m − 1)  m(m − 1)  α C2 = = ⋅ − 1 ∴  2 2 2 2   1 = m(m − 1) (m − 2 ) (m + 1) 8 1 = (m + 1) m (m − 1) (m − 2 ) = 3 m + 1C4 8

Total number of variables, if alphabets and digits both are used is 26 × 10. Hence, the total number of variables is 26 (1 + 10 ) = 286.

m

⇒ ⇒

C4 >1 ⇒ C3

n+1

3. We have,

∑

r =0

[Q Cr + Cr + 1 = n

C3

n+1

n −1

n

is 3 ! = 6 . When two strips are of same colour, then number of flags is 3 C1 × (3 !/ 2 ) × 2C1 = 18. ∴Total number of flags = 6 + 18 = 24 = 4!

n+1

Cr + 1]

n −2 >1 ⇒ n > 6 4

n

n

10. If all strips are of different colours, then number of flags

C3

n+1

C4 >

and for any other four places, it is 8 × 93 . Now, required number of numbers = 94 + 4 × 8 × 93 = 93 (9 + 32 ) = 41 (93 )

n+1

2. Given, n C3 + nC4 > n+1

9. If 7 is used at first place, the number of numbers is 94

n −1

Cr = ∑ Cr + nCr + 1 r = 0

1 Cr + 1

11. Total number of triplets without restriction is n × n × n. The number of triplets with all different coordinates is n P3 . Therefore, the required number of triplets is n 3 − n P3 .

n

1+

n −1

n

Cr

1 ∑ n−r r = 0 1+ r +1 n −1 n −1 r +1 1 (r + 1) = ∑ = n + 1 n + 1 r∑ r =0 =0 1 n = [1 + 2 + K + n ] = (n + 1) 2

12. x1 < x2 < x3 < x4 < x5 < x6 , when the number is x1 x2 x3 x4 x5 x6 . Clearly, no digit can be zero. Also, all the digits are distinct. So, let us first select six digits from the list of digits 1, 2, 3, 4, 5, 6, 7, 8, 9 which can be done in 9 C6 ways. After selecting these digits, they can be put only in one order. Thus, total number of such numbers = 9C6 × 1 = 9C3

Ta rg e t E x e rc is e s

=

The number of numbers when digits cannot be repeated is 5 P4 . Therefore, the required number of numbers is 54 − 5 ! = 505.

n

4. We know that, ∴

n

Pr n = Cr r!

∑ nCr

= 2n − 1

[1 corresponding to n C0 ]

r =1

(2 n + 1)! (n − 1)! 3 5. Given, = (n + 2 )! (2 n − 1)! 5 (2 n + 1) (2 n ) 3 = (n + 2 ) (n + 1) n 5 ⇒ 10 (2 n + 1) = 3 (n 2 + 3 n + 2 ) 2 ⇒ 3 n − 11 n − 4 = 0 ⇒ (n − 4) (3 n + 1) = 0 ∴ n=4

6. The number of one-digit numbers = 0

The number of two-digit numbers = 5 × 5 = 25. The number of three-digit numbers = 5 × 5 × 4 = 100. Hence, the total numbers are 131.

7. Numbers greater than 1000 and less than or equal to

320

13. The number of numbers when repetition is allowed is 54 .

4000 will be of 4-digits and will have either 1 or 2 or 3 in the 1st place. After fixing 1 at first place, then 2nd place can be filled by any of the 5-digits. Similarly, the 3rd place can be filled up in 5 ways and 4th place can be filled up in 5 ways. Thus, there will be 5 × 5 × 5 = 125 ways in which 1 will be in first place but this also includes 1000. Hence, there will be 124 numbers having 1 in the first place. Similarly, 125 for each 2 or 3. One number will be there in which 4 will be in the first place i.e. 4000. Hence, the required number of numbers = 124 + 125 + 125 + 1 = 375

14. Digits available Middle for remaining four digit places

Pattern

Number of ways of filling remaining four places

4

0, 1, 2, 3

4

3 × 3P3

5

0, 1,...,4

5

4 × 4 P3

6

0, 1,...,5

…

5 × 5P3

7

0, 1,...,6

…

6 × 6P3

8

0, 1,...,7

…

7 × 7P3

9

0, 1,...,8

…

8 × 8P3

Thus, number of required numbers = 5292

15. According to the given conditions, numbers can be formed by the following format:

Filled with 1, 2, 3, 4

Filled with 6, 7, 8, 9

∴ Number of required numbers = 4 P4 × 4 P4

permutations of n objects (taken r at a time + taken (r + 1) at a time + taken (r + 2 ) at a time ) + .... + taken n at a time] = n r + n r + 1 + n r + 2 + ....+ n n =

r

n (n

n−r +1

− 1)

n −1

(ii) Required number of permutations = Number of permutations of n objects (taken 0 at a time + taken 1 at a time + K + taken r at a time) 1 × (n r + 1 − 1) = n 0 + n1 + n 2 + K + n r = n −1

18. Total number of required numbers

is made by taking any two in any order. So, the required number of rational numbers [including 1] = 10 P2 + 1 = 9

27. Each element of the set A (domain) can be give the image in the set A (codomain) in k ways. So, the required number of functions. i.e. n(S ) = k × k × ....(k times) = k k

28. Any selection of four digits from the ten digits 0, 1, 2, 3, ...,9 gives one such number. So, the required number of numbers = 10C4 = 210

29. Number of distinct n digit numbers using 2, 5, 7 = 3n . Now consider, 3n > 900 ⇒ n = 7

Number of numbers using 0, 1, 2, 4, 5 = 5 ! − 4 ! = 120 − 24 = 96 [Q there are 4! numbers beginning with ‘0’] ∴Total number of required numbers = 120 + 96 = 216

20. Number of ways of filling first place = 9 Number of ways of filling remaining 8 places = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9! ∴Total number of numbers = 9 × 9!

21. Number of ways for women = 4 P2 Number of ways for men = 6 P3 ∴ Total number of ways = 4 P2 × 6 P3

22. Number of committees = 4C1 × 8C5 = 4 × 56 = 224

[for minimum]

30. Number of numbers with repetition = 10 × 10 × 10 × 10 × 10 = 100000

= 2 × 2 × ... 10 times = 210

19. Number of numbers using 1, 2, 3, 4, 5 = 5 ! = 120

Number of numbers without repetition = 10 × 9 × 8 × 7 × 6 = 30240 ∴Number of numbers with atleast one digit repeated = 100000 − 30240 = 69760

31. Total words starting with B = 4 ! = 24 Total words starting with E = 4 ! = 24 Total words starting with KB = 3 ! = 6 Total words starting with KE = 3 ! = 6 Total words starting with KR = 3 ! = 6 The first word starting with KU will be KUBER. So, rank of the word KUBER = 24 + 24 + 18 + 1 = 67

32. Total number of such numbers

= 4 ⋅ 5 ⋅ 4 ⋅ 3 = 240

33. Maximum population = 2 32 − 1[as each position of tooth has two options]

23. As the order of A, B and C is not to change, therefore number of ways of assigning three periods to them is 7 C3 . Now, we left with 4 periods for 4 lecturers that can assign in 4! ways.

34. Total number of ways

∴Required number of ways = 7C3 × 4 ! 7! 7! = × 4! = = 840 3! × 4! 3!

35. Total number of permutations = 0 + k + k 2 + k 3 +K+ k r

24. Total number of arrangements of the letters of the given word 6! = = 60 3! 2 ! Number of arrangements in which two N’s are together 5! = = 20 3! ∴ Number of arrangements in which two N’s are not together

= nC2 × nC1 × (n − 1) ! ⋅

=

5! 2, 2, 8, 8, 8 in = 10 ways and remaining 4 even 2 ! 3! 4! places can be filled by using 3, 3, 5, 5 in = 6 ways 2! × 2! ∴Total number of numbers = 10 × 6 = 60

n! ⋅ n ! = n !⋅ nC2 2 ! (n − 2 )! k (k r − 1) (k − 1)

36. Matches whose predictions are correct can be selected in 20 C10 ways. Now, each wrong prediction can be made in 2 ways. Thus, total number of ways =

C10 ⋅ 210 .

20

37. LetS1 andS 2 refuse to be together andS 3 andS 4 want to be together only. Total number of ways, when S 3 and S 4 are selected = (8 C2 + 2C1 ⋅ 8C1 ) = 44 Total number of ways when S 3 and S 4 are not selected

= 60 − 20 = 40

25. There are 5 odd places and that can be filled by using

6 Permutation and Combination

17. (i) Required number of permutations = Number of

26. A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}. A rational number

Targ e t E x e rc is e s

16. Total number of digits (sum of digits is even) = 450

= (8 C2 + 2C1 ⋅ 8C3 ) = 182 Thus, total number of ways = 44 + 182 = 226

38. Formed number can be atmost of nine digits. Total number of such numbers = 3 + 32 + 33 + K + 38 + 2 ⋅ 38 =

39 − 3 + 4 ⋅ 38 7 ⋅ 38 − 3 3 (38 − 1) = + 2 ⋅ 38 = 2 2 3−1

321

Objective Mathematics Vol. 1

6

39. The number of committees of 4 gentlemen = 4C4 = 1 The number of committees of 3 gentlemen, 1 wife after selecting 3 gentlemen, only 1 = 4C3 × 1C1 wife is left, which can be included    The number of committees of 2 gentlemen, 2 wives = 4C2 × 2C2 The number of committees of 1 gentleman, 3 wives = 4C1 × 3C3 The number of committees of 4 wives = 1 ∴The required number of committees = 1 + 4 + 6 + 4 + 1 = 16

∴The number of ways to be unsuccessful = 9C9 + 9C8 + 9C7 + 9C6 + 9C5 = 1 + 9 + 36 + 84 + 126 = 256 Two numbers have all the digits equal. So, the required number of numbers = 2 5 − 2 = 32 − 2 = 30

42. 62 = nC1 + nC2 + nC3 + K + nCn − 1 = 2 − C0 − Cn = 2 − 2 62 + 2 = 2 n 2 n = 64 = 2 6 ⇒ n = 6

Ta rg e t E x e rc is e s

n

n

43. Total number of ways =

n

∑

r =1

= ×

×

×

Cn + r =

2n

1 2n (2 − 2 ×

×

Cn + 1 + K +

2n

2n

C2 n

Cn )

×

44. Ways 9 10 10 10 10 2 0 cannot be placed in the first place. In the other four places, any digit can go. After filling the first five places, the last place can be filled by 0 or 5. ∴Required number of numbers = 9 × 10 4 × 2 = 18000

45. Total number of outcomes = 6 × 6 × ... to n times = 6n Total number of outcomes which show only even numbers = 3 × 3 × 3 ...to n times = 3n

46. The matrix will be of the order 4 × 1 or 1 × 4 or 2 × 2. For each order, the number of different matrices = The number of ways to fill four places by 0, 1, 2, 3 = 4× 4× 4× 4 47. If r questions are solved by each student, then the number of possible selections of questions is 20 Cr . ∴The number of students = 20Cr Q each student has solved different   combinations of questions    ∴The maximum number of students = Maximum value of 20 Cr = 20C10 because 20 C10 is the largest among 20 C0 , being the middle one.

20

C1,...,20 C20

48. From every arrangement of 7 letters, we get a pair by 322

∴The required number of pairs =

Two pairs, one different

3

C1 × 2C 2

C 2 × 1C1

3

Arrangements 5! C1 × 2C 2 × = 60 3!

C 2 × 1C1 ×

3

5! = 90 2 !2 !

∴Required number of ways = 150 6! 50. Total number of ways = = 20 3! × 3!

∴Required number of words 4! = 5C1 × 21C1 × + 5C2 × 21C2 × 4 ! 2!2!

53. If x = 1, y ∈{1, 2,...,11} x = 2, y ∈ {1, 2,...,12} ... ... ... ... x = 11, y ∈ {1, 2,..., 21} ∴Number of ways = 11 + 12 + K + 21

54. We can choose two men out of 9 in 9 C2 ways. Since, no husband and wife are to play in the same game, two women out of the remaining 7 can be chosen in 7 C2 ways. If M1, M2 , W1 and W2 are chosen, then a team may consist of M1 and W1 or M1 and W2 . Thus, the number of ways of arranging the game is 9× 8 7 × 6 9 C2 ⋅7 C2 = × × 2 = 1512 2 2

55. The number of students answering exactly i (1 ≤ i ≤ n − 1)

2n

putting a sign (,).

3

52. Required number of ways = 3 ! ⋅ 3 ! ⋅ 3 ! = 216

41. Total number of numbers of without restriction = 2 5

⇒ ∴

Selections

consonants = 5C1 × 21C1

6 or 5 papers.

n

Possibilities One triplet, two different

51. The number of selections of 1 pair of vowels and 1 pair of

40. The candidate is unsuccessful, if he fails in 9 or 8 or 7 or

n

49.

7! = 1260 2 !2 !

questions wrongly is 2 n − i − 2 n − i − 1. The number of students answering all n questions wrongly is 2 0 . Thus, the total number of wrong answers is 1 (2 n − 1 − 2 n − 2 ) + 2 (2 n − 2 − 2 n − 3 ) + K + (n − 1) (21 − 2 0 ) + n(2 0 ) n −1 n−2 =2 +2 + 2n − 3 + K + 20 = 2n − 1 Now, 2 n − 1 = 2047 ⇒ 2 n = 2048 = 211 ∴ n = 11

56. The required numbers are 1, 2, 11, 12, 21, 22, ...,122222222 Let us calculate how many numbers are these. There are 2 one-digit such numbers. There are 2 2 two-digit such numbers and so on. There are 2 8 eight-digit such numbers. Also, the 9-digit numbers beginning with 1 and written by means of 1 and 2 are smaller than 2 ⋅ 10 8 . Thus, there are 2 8 such nine-digit numbers. Hence, the required number of numbers = 2 + 2 2 + 2 3 + ....+ 2 8 + 2 8 2 (2 8 − 1) = + 28 2 −1 = 29 − 2 + 28 = 766

65. The number of matches in the first round = 6C2 + 6C2

number of times 3 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz, where 0 ≤ x, y, z ≤ 9. Let us first count the numbers in which 3 occurs exactly once. Since, 3 can occur at one place in 3 C1 ways, there are 3 C1 (9 × 9) = 3 × 92 such numbers. Next, 3 can occur exactly two places in (3 C2 ) (9) = 3 × 9 such numbers.

The number of matches in the next round = 6C2 The number of matches in the semifinal round = 4C2 So, the required number of matches = 6C2 + 6C2 + 6C2 + 4C2 + 2 = 53

58. One possible arrangements is 2

2

3

Three such arrangements are possible. Therefore, the number of ways is (5 C2 ) (3 C2 ) (1C1 ) (3) = 90 The other possible arrangements is 1 1 3 Three such arrangement are possible. In this case, the number of ways is (5 C1 ) (4 C1 ) (3 C3 ) (3) = 60 Hence, the total number of ways = 90 + 60 = 150 9! 59. Total number of permutations = 2! Number of permutation of those containing ‘HIN’ = 7 ! 7! Number of permutation of those containing ‘DUS’ = 2! Number of permutation of those containing ‘TAN’ = 7 ! Number of permutation of those containing ‘HIN’ and ‘DUS’ = 5! Number of permutation of those containing ‘HIN’ and ‘TAN’ = 5! Number of permutation of those containing ‘TAN’ and ‘DUS’ = 5! Number of permutation of those containing ‘HIN’, ‘DUS’ and ‘TAN’ = 3! 9!  7 ! ∴Required number = − 7 ! + 7 ! +  − 3 × 5 ! − 3 !  2! 2 = 168474

60. The smallest number of people = Total number of possible forecast = Total number of possible results = 3 × 3 × 3 × 3 × 3 = 243

61. Since, every pair of the students gives us a hand shake ∴Total number of hand shakes = Total number of pairs of students = Number of ways of choosing two students out of 20 ! = 20C2 = 2 !⋅ 18 ! 20 × 19 × 18 ! = = 190 2 × 18 !

62. Total number of ways = 8 ! 2 ! 63. A couple and 6 guests can be arranged in (7 − 1)! ways. But the two people forming the couple can be arranged among themselves in 2 ! ways. ∴ Required number of ways = 6 ! × 2 ! = 1440 5! 64. Required number of garlands = =5 2 ⋅ 2 !⋅ 3 !

C5 6 = C6 11

11

67. Required set is { m, m + 1, ..., k} ⇒ Number of subsets = 2 k − ( m − 1)

68. Number of groups each having r students = 21Cr Let G = Number of times the teacher can go to the zoo Z = Number of groups she can form = 21Cr ∴ G is maximum, when Z is maximum. 21 − 1 21 + 1 i.e. or r= 2 2 i.e. ] r = 10 or 11

69. Required number of ways = nC2 + nC3 + ... + nCn − 3 + nCn − 2 = 2 n − (n C0 + nC1 + nCn − 1 + nCn ) = 2 n − (1 + n + n + 1) = 2 n − 2(n + 1)

70. In a committee of 5 persons, the gentlemen will be in majority, if the number of gentlemen ≥ 3. Total number of ways of forming committee = 4C3 × 6C2 + 4C4 × 6C1 = 4 × 15 + 1 × 6 = 66

71. Total number of m elements subsets of A = nCm Number of m elements subsets of A each containing the element a4 = Number of ways of choosing remaining (m − 1) elements of the subset, out of remaining (n − 1) elements of A = n − 1Cm − 1 Given that,

n

Cm = k ×

Targ e t E x e rc is e s

Lastly, 3 can occur in all three-digits in one number only. Hence, the number of times 3 occurs is 1 × (3 × 92 ) + 2 × (3 × 9) + 3 × 1 = 300

10

66.

6 Permutation and Combination

57. Since, 3 does not occur in 1000, we have to count the

n −1

Cm − 1 n! (n − 1)! =k⋅ m ! (n − m)! (m − 1)! (n − m)! n = km

⇒

72. Required number of ways = Total number of ways of choosing 4 pens − Number of ways of choosing 4 non-red pens =

3+ 4+ 6

=

13 ⋅ 12 ⋅ 11 ⋅ 10 10 ⋅ 9 ⋅ 8 ⋅ 7 − 4⋅ 3⋅2 ⋅1 4⋅ 3⋅2 ⋅1

C4 −

4+ 6

C4 = 13C4 − 10C4

= 715 − 210 = 505

73. Out of the given numbers, the numbers which are multiple of 5, are 5, 10, 15, 20,...,145. This is an AP whose first term a = 5, Common difference = 5 nth term, t n = 145 ∴ 145 = 5 + (n − 1) × 5 ⇒ n = 29

323

Objective Mathematics Vol. 1

6

Now, if total number of numbers is m, then 151 = 1 + (m − 1) × 2 151 − 1 m − 1= ⇒ m = 76 ⇒ 2

81. The number of triangles with vertices on different lines = pC1 × pC1 × pC1 = p3 The number of triangles with two vertices on one line and the third vertex on any one of the other two lines p ( p − 1) = 3C1 { p C2 × 2 pC1} = 6 p ⋅ 2 So, the required number of triangles

So, number of ways in which product is a multiple of 5 = Both two numbers from 5, 10, 15, 25,...,145 + One number from 5, 10, 15, 25,...,145 and one from remaining numbers = 29C2 + 29 C1 × 76C1 = 29 × 14 + 29 × 76= 29 × 90 = 2610

= p3 + 3 p2 ( p − 1) = p2 (4 p − 3)

82. There are two sets of five parallel lines at equal distances.

74. Q n + 1C3 − nC3 = 21

m5

(n + 1) (n − 1) n n(n − 1) (n − 2 ) − = 21 6 6 n(n − 1) ⇒ [(n + 1) − (n − 2 )] = 21 6 ⇒ n(n − 1) = 42 ⇒ n 2 − n − 42 = 0 ⇒ n = − 6, 7 ∴ n =7 n(n − 1) 75. Q n C2 − n = 44 ⇒ − n = 44 2 ∴ n 2 − 3 n − 88 = 0 ⇒ n = − 8 , 11 ∴ n = 11

Ta rg e t E x e rc is e s

∴

76. Total number of triangles =12 C3 − 3C3 − 4C3 − 5C3 = 205 77. The number of triangles with vertices on sides AB, BC, CD = C1 × C1 × C1 3

4

5

4

l3

l2 O

l1

Clearly, lines like l1, l 3 , m1, m3 form a squares whose diagonals length is 2. So, the number of required squares = 3 × 3 = 9. Since, choices are (l1, l 3 ),(l 2 , l 4 ),(l 3 , l 5 ) for one set.

83.

1

l1

2 3 4 5

L1

L2 1

2 3 4 5

l2

Required number of points on intersection of new lines = 2 × Number of choosing 1 point on L1 and 2 point on L2 = 2 ⋅ l1C1 × l 2C2

84. Required number of triangles = Total number of triangles − Number of triangles having two sides common − Number of triangles having one side common = 8C3 − 8 − 8 × 4 = 16

85.

A

chosen, x1 be the number of objects between the first and the second, x2 be the number of objects between the second and the third and x3 be the number of objects to the right of the third object. We have, x0 , x3 ≥ 0, x1, x2 ≥ 1 and ...(i) x0 + x1 + x2 + x3 = n − 3

324

l5 l4

1

∴ The number of diagonals = 8 C2 − 8 = 20

The number of solutions of Eq. (i) = Coefficient of y n − 3 in (1 + y + y 2 + ... ) (1 + y + y 2 + ... ) ( y + y 2 + y 3 + ... ) ( y + y 2 + y 3 + ... ) n−3 in y 2 (1 + y + y 2 + y 3 + ... )4 = Coefficient of y n−5 in (1 − y )−4 = Coefficient of y = Coefficient of y n − 5 in (1 + 4C1 y + 5C2 y 2 + 6C3 y 3 + ... ) (n − 2 ) (n − 3) (n − 4) = n − 5 + 3Cn − 5 = n − 2C3 = 6

m2 m1

2

78. A selection of four vertices of the polygon gives an

79. n C2 is point of intersection. 80. Let x0 be the number of objects to left of the first object

m3

3

Similarly, for other cases, the total number of triangles = 3C1 × 4C1 × 5C1 + 3C1 × 4C1 × 6C1 + 3C1 × 5C1 × 6C1 + 4C1 × 5C1 × 6C1 = 342 interior intersection. ∴The number of sides = n n ⇒ C4 = 70 ⇒ n (n − 1) (n − 2 ) (n − 3) = 24 × 70 = 8×7 × 6× 5 ∴ n=8

m4

c 1 2 3

3

2 1

a B

1 2

3

b

C

Required number of triangles = Total number of ways of choosing 3 points − Number of ways of choosing all the 3 points from AB or BC or CA =

a+ b+c

C3 − (a C3 + bC3 + c C3 )

...(i)

| y| ≤ k ⇒ − k ≤ y ≤ k | x − y| ≤ k ⇒ | y − x| ≤ k −k≤y−x≤k x−k≤y≤x+k

∴The required number of even proper divisors = Total number of selections of atleast one 2 and any number of 3’s or 7’s = 4 × (2 + 1) × (1 + 1) − 1 = 23

...(ii)

...(iii)

Y

95. The natural numbers are 1, 2, 3 and 4. Clearly, in one diagonal we have to place 1, 4 and in the other 2, 3. The number of ways = 2 ! × 2 ! = 4

k 3 2 1 X¢

–k

–3 –2 –1

96. If 0 is placed in the unit’s place of the upper number,

If 1 is placed in the units place of the upper number, then the unit’s place of the lower number can be filled in 9 ways (filling by any one of 0, 1, 2, ..., 8) etc. ∴The unit’s column can be filled in 10 + 9 + 8 + .... + 1, i.e. 55 ways. Similarly, for the second and the third columns. The number of ways for the fourth column = 8 + 7 + ... + 1 = 36 ∴The required number of ways = 55 × 55 × 55 × 36

,– (k k)

Y¢

k) ,– (2 –k) , (1

–k

X

k

1 2 3 –1 –2 –3

then the unit’s place of the lower number can be filled in 10 ways (filling by any one of 0, 1, 2, ..., 9).

∴Number of points having integral coordinates = (2 k + 1)2 − 2[k + (k − 1) + ...+ 2 + 1] = (3 k 2 + 3 k + 1)

97. Coefficient of x 3 in ( x 0 + x1 + x 2 ) ( x 0 + x1 + x 2 + x 3 )( x1 + x 2 + x 3 + x 4 ) =6

87. Total number of ways = 31 × 15 × 8 = 3720 88. Total number of squares = 8. We are to choose 6

squares for six X’s and this can be done is 8 C6 = 28 ways. This also include the possibility that either the top horizontal row does not have any X or the bottom horizontal row have no X. These possibilities are not required.

∴Total number of ways = 28 − 2 = 26

89. Total number of such numbers = 9C6 + 9C5 + 9C5 + 9C4 = 10C6 +

6

C5 = 11 C6

10

90. The number of times, a particular girl goes on picnic = 6C4 ∴Total number of dolls given to the girls 6! ⋅ 3 = 45 = 6 C4 ⋅ 3 = 4! 2 !

91. Case I 1 4 7 ... 3 n − 2 Case II 2 5 8 ... 3 n − 1 Case III 3 6 9 ... 3 n That means we must take 2 numbers from last row or one number each from first and second rows. Total number of ways = nC2 + nC1 ⋅ nC1z n (n − 1) 3 n2 − n = + n2 = 2 2

92. The number of selections of p things from q identical things are (r − p) things from q identical things = 1 × 1 p→ p p − 1... r − q q → r − p r − p + 1... q Similarly, in all other cases, total number of ways = p − (r − q ) + 1or q − (r − p) + 1 = p + q − r + 1

93. 2 p ⋅ 6 q ⋅ 15r = 2 p + q ⋅ 3 q + r ⋅ 5r ∴The number of proper divisors = {Total number of selections from ( p + q ) twos, (q + r ) threes and r fives} − 2 = ( p + q + 1)(q + r + 1)(r + 1) − 2

98. 38808 = 2 3 × 32 × 7 2 × 111 ∴

99.

Divisors = 4 × 3 × 3 × 2 − 2 = 70

15 + 3 − 1

C3 − 1 = 17C2

33 33 33 33 33 100.   +   +   +   +   2 

4

 8   16   32  = 16 + 8 + 4 + 2 + 1 = 31

101. For 1 ≤ k ≤ p − 1, n + k = p ! + k + 1, is clearly divisible by k + 1. Therefore, there is no prime number in the given list.

Targ e t E x e rc is e s

and and ⇒ ⇒

94. 1008 = 2 4 × 32 × 7

Permutation and Combination

86. | x| ≤ k ⇒ − k ≤ x ≤ k

102. The number of ways in which all 5 letters be placed in wrong envelopes

1 1 1 1 1  − + −  = 6 C5 . 5 ! 1 − +  1! 2 ! 3 ! 4 ! 5 ! 1 1  1 1 = 720  − + −  = 264  2 6 24 120 

103. There is (m + 1) space between them (including the beginning and the end) and n things can be placed in all possible spaces in m + 1 Pn ways. ∴Required number of ways m !⋅ (m + 1)! = m ! ⋅ m + 1Pn = (m + 1 − n )!

104. We have, 240 = 2 4 × 31 × 51 ∴ There are 4 divisors of 240 which of the form 4n + 2, n ≥ 0.

105. Number of non-negative integral solutions = 12C10 = 12C2

106. Q x1 + x2 + 2 x3 = 15 ⇒ x1 + x2 = 15 − 2 x3 where, x3 can be 1, 2, ..., 7. Let x3 = r ⇒ x1 + x2 = 15 − 2 r

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Objective Mathematics Vol. 1

6

Number of positive integral solutions of this equation = (14 − 2 r ) ∴Total number of solutions 7  7 ⋅ 8 = ∑ (14 − 2 r ) = 14 ⋅ 7 − 2 ⋅   = 42  2  r =1

107. x1 + x2 + x3 + x4 = 15 and xi ≥ 2 ⇒ ( x1 − 2 ) + ( x2 − 2 ) + ( x3 − 2 ) + ( x4 − 2 ) = 7 ⇒ y1 + y2 + y3 + y4 = 7, where yi = xi − 2 ≥ 0 ⇒ 10 C7 = 10C3 are number of non-negative. 3

108. Coefficient of x in {( x 0 + x1 + x 2 + x 3 ) ( x 0 + x1 + x 2 ) ⋅ ( x 0 + x1 )4 } = 247

109. Let the balls put in the box be x1, x2 , x3 , x4 and x5 . We have, x1 + x2 + x3 + x4 + x5 = 15, xi ≥ 2 ⇒ ( x1 − 2 ) + ( x2 − 2 ) + ( x3 − 2 ) + ( x4 − 2 ) + ( x5 − 2 ) = 5 ⇒ y1 + y2 + y3 + y4 + y5 = 5, yi = xi − 2 ≥ 0 = 5 + 5 − 1C5 − 1 = 9C4

110. Coefficient of x 210 in{( x 0 + x1 + x 2 + x 3 + ... + x100 )3 }

Ta rg e t E x e rc is e s

=

93

C3

111. Required number of ways = Number of non-negative integral solution of x1 + x2 + x3 + x4 + ... + x8 = 10 = 10 + 8 − 1C8 − 1 = 17C7

112. The required number of selections = 3 C1 × 4C1 × 2C1 (6 C3 + 6C2 + 6C1 + 6C0 )

113. x + y + z = 16, y = x + 1 and z = y + 2 ∴ x = 4, y = 5, z = 7 Now, number of ways to give 16 different things =16 C4 × 12C5 × 7C7

114. x + y + z = n x, y, z ≥ 1or x − 1= a, y − 1 = b, z − 1 = c ≥ 0 ⇒ a+ b+c =n−3 The required number of solutions = n − 3 + 3 − 1C3 − 1 = n − 1C2

115. The number of non-negative integral solutions of a+ b+c +d =n (n + 3)! (n + 3) (n + 2 ) (n + 1) n+ 3 ⇒ Cn = = n ! 3! 6

116. The required number of solutions = The number of positive integral solution of (a + b + c = 12 ) (12 − 1) (12 − 2 ) = = 55 2

117. a + a + d + a + 2d = 21or a + d = 7 ∴ a + c = 14 and b=7 The number of positive integral solutions of (a + c = 14) is 13.

118. Let a = 2 p + 1, b = 2q + 1, c = 2 r + 1, d = 2 s + 1, where p 326

q, r and s are non-negative integers. ∴ 2 p + 1 + 2q + 1 + 2 r + 1 + 2 s + 1 = 20

p+q + r + s=8

or

The required number of solutions = The number of non-negative integral solutions of ( p + q + r + s = 8) (8 + 3) (8 + 2 ) (8 + 1) = = 165 6

119. If a, b and c are in AP, then a and c both are odd or both are even. Case I

When n is even.

The number of ways of selection of two even numbers, a and c is n / 2 C2 . Number of ways of selection of two odd numbers is n / 2 C2 . Hence, the number of AP’s n 2 n/ 2 2 ⋅ C2 = 2 ⋅ Case II

n   − 1  2 2

=

n (n − 2 ) 4

When n is odd.

The number of ways of selection of two odd numbers a and c is ( n + 1)/ 2 C2 . The number of ways of selection of two even numbers a and c is ( n − 1)/ 2 C2 . Hence, the number of AP’s = ( n + 1)/ 2C2 + ( n − 1)/ 2C2  n − 1  n − 1   n + 1  n + 1  − 1 − 1        2   2   2   2  + = 2 2 1 (n − 1)2 = (n − 1)[(n + 1) + (n − 3)] = 4 8

120. The number of ways of inviting, when the couple not included is 8 C5 . The number of ways of inviting with the couple included is 8 C3 . Therefore, the required number of ways = 8C5 + 8C3 = 8C3 + 8C3 [Q 8C5 = 8C3 ] 10 ! 8! Also, 10 C5 − 2 × 8C4 = −2 × 5! 5! 4! 4! 10 × 9 × 8 × 7 × 6 8×7 × 6× 5 = −2 × 120 24 8! = 9 × 4 × 7 − 140 = 112 = 2 × 3! 5!

121. Q p = 5C4 × 2C1 = 10 and ∴ and

q = 5C2 (2 C1 )3 = 80 r = 5C0 (2 C1 )5 = 32 2q = 5r , 8 p = q 2( p + r ) > q

122. By principle of homogenuity, x = y is evident and since sum of digits is either or odd. ⇒ x + y = Total 5-digit numbers = 9 × 10 × 10 × 10 × 10 ⇒ 2 x = 90000 ∴ x = 45000

123. Since, n Cr = nCn − r , hence options (a) and (c) follow. 124. Total hand shakes possible =

2n

C2 . This will include n hand shakes in which a person shakes hand with her or his spouse. ∴Required number of hand shakes = 2 n C2 − n = 2 n (n − 1)

Hence, options (a) and (b) are correct and option (c) is false.

126. We have, 30 = 2 × 3 × 5. So, 2 can be assigned to either a or b or c i.e. 2 can be assigned in 3 ways. Similarly, each of 3 and 5 can be assigned in 3 ways. Thus, the number of solutions is 3 × 3 × 3 = 27. Also, the number of ways in which three prizes can be distributed among three persons = 3 × 3 × 3 = 33

127. As each student receives atleast two toys. So, let us first give each student one toy. Now, we are left with 7 toys, which can be distributed among three students such that each receives atleast one toy, which is equivalent to number of positive integral solutions of the equation x + y + z + w = 7, which is given by 7−1C4 − 1 = 6C3 . Hence, Statement I is false and Statement II is correct.

128. Number of ways of dividing n 2 objects into n groups of 2

same size is

(n ) ! . (n !)n n !

131. India must win atleast 6 matches out of 11 matches. Then, number of ways in which India can win the series = 11C6 + 11C7 + ... + 11C11 = 210 Thus, both the statements are true, but Statement II is not a correct explanation of Statement I.

132. The batting order of 11 players can be decided in 11! ways. Now, Yuvraj, Dhoni and Pathan can be arranged in 3! ways. But the order of these three players is fixed i.e. Yuvraj-Dhoni-Pathan. Now, 11! answer is 3! times more, hence the required answer is 11!/3!.

133. Number of ways of arranging 21 identical objects, when r is identical of one type and remaining are identical of 21! second type, is = 21Cr , which is maximum r ! (21 − r )! when r = 10 or 11. Therefore, 13 Cr = 13C10 or 13 C11, hence maximum value of 13 Cr is 13 C10 = 286. Hence, Statement I is false. Obviously, Statement II is true.

134. When one all rounder and ten players from bowlers and batsman play, number of ways is 4 C1 14 C10 . When one wicketkeeper and 10 players from bowlers and batsman play, number of ways is 2 C114C10 .

Now, number of ways of distributing these n groups among n persons is  (n 2 )!  (n 2 )! = n , which is always an integer. !   n (n !)n (n !) n !

When one all rounder, one wicketkeeper and nine from batsman and bowlers play, number of ways is 4 C1 2C1 14C9 .

Also, we know that product of r is divisible by r ! . Now, (n 2 ) ! = 1 × 2 × 3 × 4 × ... × n 2 = 1 × 2 × × 3.... n × (n + 1) (n + 2 )... 2 n × (2 n + 1) (2 n + 2 )... 3 n × { n 2 − (n 2 − 1)}

Total number of selections

{ n 2 − (n 2 − 1)} ... n 2 2

Thus, in n ! , there are n rows each consisting product of n integers. Each row is divisible by n ! . (n 2 )! is a natural Hence, (n 2 )! is divisible by (n !)n or (n !)n number. Hence, both statements are correct and Statement II is the correct explanation of Statement I.

129. Q1400 = 2 3 527 The number of ways in which 1400 can be expressed as a product of two positive integers is (3 + 1) (2 + 1) (1 + 1) = 12 2 Statement II is correct but does not explain Statement I as it just gives the information of prime factor about which 1400 is divisible.

130. In onto functions, each image must be assigned to atleast one pre-image. Now, if we consider the images a and bas two different boxes, then four distinct objects 1, 2, 3 and 4 (pre-images) can be distributed, such that no box remains empty, in 2 4 − 2C1 (2 − 1)4 = 14 ways. Hence, both statements are correct and Statement II is the correct explanation of Statement I.

6 Permutation and Combination

parts are already there. ⇒ un = un − 1 + n ⇒ Option (d) is incorrect. Again, u 0 = 1, u1 = 2 We easily get u 2 = 4, u 3 = 7, u 4 = 11

When all eleven players play from bowlers and batsmen then number of ways is 14 C11. = 4C1 14C10 + 2C1 14C10 + 4C1 2C1 14C9 +

14

C11

Targ e t E x e rc is e s

125. The nth line will rise to n additional parts, when u n − 1

135. If the particular bowler plays, then two batsmen will not play. So, rest of 10 players can be selected from 17 other players. Number of such selections is 17C10 . If the particular bowler does not play, then number of selections = 19C11 If all the three players do not play, then number of selections = 17C11 ∴Total number of selections = 17C10 +

C11 + 17C11

19

136. If the particular batsman is selected, then rest of 10 players can be selected in 18 C10 ways. If particular wicketkeeper is selected, then rest of 10 players can be selected in 18 C10 ways. If both are not selected, then number of ways is 18 C11. Hence, total number of ways = 2 ⋅18 C10 + 18C11 = 19C11 +

18

C10

137. Seven persons can be selected for first table in

12

C7 ways. Now, these seven persons can be arranged in 6! ways. The remaining five persons can be arranged on the second table in 4! ways.

Hence, total number of ways = 12C5 6 ! 4 !

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Objective Mathematics Vol. 1

6

138. Here, A can sit on first table and B on the second or A on

Let C = (n − 1)⋅ nC1 + (n − 2 )⋅ nC2 + ....+ 1⋅ nCn − 1 ...(i)

second table and B on the first table. If A is on the first table, then remaining six for first table can be selected in 10 C6 ways. Now, these seven

We use n Cr = nCn − r and rewrite the terms in the

Ta rg e t E x e rc is e s

C = 1⋅ nC1 + 2 ⋅ nC2 + K+ (n − 1)⋅ nCn − 1

persons can be arranged in 6! ways. Remaining five can be arranged on the other table in 4! ways. Hence, total number of ways is 2

10

= n (2 n − nC0 − nCn ) = n(2 n − 2 )

139. If A, B are on the first table, then remaining five can be selected in

10

⇒

C5 ways.

Now, seven persons including A and B can be arranged on the first table in which A and B are together in 2 ! 5 ! ways. Remaining five can be arranged on the second table in 4! ways. Total number of ways is 10 C5 4!5!2!.

...(ii)

On adding Eqs. (i) and (ii), we get 2C = n (n C1 + nC2 + ....+ n Cn − 1 )

C6 6 ! 4 ! .

C = n (2 n − 1 − 1)

Thus, S = n(2 n − 1 − 1) + (2 n − 1)= (n + 2 ) 2 n − 1 − (n + 1) C. We have, 2 C2 + 3C2 + 4C2 + K + nC2 = n

n

r =0

r =0

n+1

C3

∑ (nCr )2 = ∑ (nCr ) (nCn − r )

If A and B are on the second table, then remaining three can be selected in 10 C3 ways.

D. We have,

Now, five persons including A and B can be arranged on the second table in which A and B are together in 2!3! ways. Remaining seven can be arranged on the first table in 6! ways. Total number of ways for first table is 10 C7 6!3!2!.

∴ The number of ways of selecting n persons out on n men and n women = 2nCn

∴Total number of ways = 10C7 6 ! 3 ! 2 ! +

10

C5 4 ! 5 ! 2 !

140. 2 candidates can be selected out of 4 in 4C2 ways. 141. Required number of ways = 38 − 3C1 ⋅ 2 8 + 3C2 142. Each letter has 3 choices of letter boxes in 35 ways. 143. A. 20 C2 − 4C2 + 1 = 190 − 6 + 1 = 185 B. 1 ×

C2 = 190

20

146. Let T and S denote teacher and student, respectively. Then, we have following possible patterns according to questions (i) T S S T S S T S S (ii) S T S S T S S T S (iii) S S T S S T S S T Hence, total number of arrangements are 3 ⋅ (3 !)6 ! = 18 × 6 ! ⇒ k = 6

147. To form a triangle, 3 points out of 5 can be chosen in 5

C. 2 × C2 = 56 8

D. 6 C2 × 4 = 60

144. A. Number of surjective functions is 36 − 3C1 (3 − 1)6 + 3C2 (3 − 2 )6 = 729 − 192 + 3 = 540 B. If f (ai ) ≠ bi, then pre-image a1, a2 , a3 cannot be assigned to images b1, b2 , b3 , respectively. Hence, each of a1, a2 , a3 can be assigned to images in 2 ways. a4 , a5 , a6 can be assigned to images in 3 ways each. Hence, number of functions is 2 3 33 = 216 C. One-one functions are not possible, as pre-images are more than images. D. Number of many one functions = Total number of functions − Number of one-one functions = 36 − 0 = 729

145. A. For k ≥ 1, we have k ⋅ k ! = [(k + 1) − 1] k ! = (k + 1)! − k ! Thus, 1⋅ 1! + 2 ⋅ 2 ! + 3 ⋅ 3 ! + K + n ⋅ n ! = (2 ! − 1) + (3 ! − 2 !) + (4 ! − 3 !) + K+ {(n + 1)! − n !} = (n + 1) ! − 1 B. Let S = n ⋅ nC1 + (n − 1)⋅ nC2 + (n − 2 )⋅ nC3 + K+ 1⋅ nCn = (n − 1) ⋅n C1 + (n − 2 )⋅ nC2 + K + 1 ⋅ nCn − 1

328

reverse order, to obtain

+ (n C1 + nC2 + K + nCn )

C3 = 10 ways.

But of these, the three points lying on the 2 diagonals will be collinear. So, 10 − 2 = 8 triangles can be formed.

148. Here, A is common letter in words SUMAN and DIVYA. Now, for selecting six different letters, we must select A either from word SUMAN or from word DIVYA. Hence, for possible selections, we have A excluded from SUMAN + A included in SUMAN = 4C3 ⋅ 5C3 + 4C2 ⋅ 4C3 = 40 + 24 = 64 Hence, N 2 = 64 ⇒ N = 8

149. Number of arrangements are 2 n ! n !. Given that, ⇒ ∴

2 n ! n ! = 1152 (n !)2 = 576 n ! = 24 ⇒ n = 4

150. Zero or more oranges can be selected out of 4 identical oranges in 4 + 1 = 5 ways. Zero or more apples can be selected out of 5 identical apples in 5 + 1 = 6 ways. Zero or more mangoes can be selected out of 6 identical mangoes in 6 + 1 = 7 ways. ∴Total number of selections when all the three types of fruits are selected = 5 × 6 × 7 = 210 But in one of these selections number of each type of fruit is zero and hence this selection must be excluded. ∴ Required number of ways = 210 − 1 = 209

has x3 balls. Then, x1 + x2 + x3 ≤ 12, x1 ≥ 1, x2 ≥ 3, 1 ≤ x3 ≤ 5 Let x4 = 12 − ( x1 + x2 + x3 ). Then, x1 + x2 + x3 + x4 = 12 [1 ≤ x1 ≤ 8, 3 ≤ x2 ≤ 10, 1 ≤ x3 ≤ 5 and 0 ≤ x 4 ≤ 7 ] ∴Required number of ways = Coefficient of x12 in ( x1 + x 2 + ... + x 8 ) ( x 3 + x 4 + ... + x10 ) ( x1 + x 2 + ... + x 5 ) ( x 0 + x1 + ... + x 7 ) = Coefficient of x in (1 + x + x + ...+ x ) 7

2

= Coefficient of x 7 in (1 − x 5 ) (1 + 4C1 x + 5C2 x 2 + 6C3 x 3 + . .. ) = C7 − C2 = 110 10

5

152. Clearly, x = 0, 1, 2, 3, ..., 10. Let x = k, then 0 ≤ k ≤ 10 When x = k, y + z = 21 − 2 k The number of non-negative integral solutions in above equation = Number of ways to distribute (21 − 2 k ) identical things (each thing is number 1) among 2 persons =

21 − 2 k + 2 − 1

7

(1 + x + x 2 + ... + x 7 ) (1 + x + x 2 + x 3 + x 4 )

C2 − 1 =

22 − 2 k

C1 = 22 − 2 k 10

∴Required number of solutions =

∑ (22 − 2 k )

6 Permutation and Combination

151. Suppose box Ahas x1 balls, box B has x2 balls and boxC

k=0

(1 + x + x 2 + ... + x 7 )

= 22 + 20 + 18 + ... + 2 = 2 (1 + 2 + 3 + ... + 11) 11 × 12 =2 × = 132 2

= Coefficient of x 7 in (1 − x )− 4 (1 − x 5 )(1 − x 8 ) = Coefficient of x 7 in (1 − x )−4 (1 − x 5 )

Entrances Gallery moves from 0 to 4 which can be arranged in 4 4 4 C1 4 C2 C3 C4 4 C0 + + + + =7 4 3 2 1 Aliter Possible solutions are 1, 2, 3, 4, 10 1, 2, 3, 5, 9 1, 2, 3, 6, 8 1, 2, 4, 5, 8 1, 2, 4, 6, 7 1, 3, 4, 5, 7 2, 3, 4, 5, 6 Hence, there are 7 solutions.

2. Number of red lines = nC2 − n Number of blue lines = n n Hence, C2 − n = n n ⇒ C2 = 2 n n (n − 1) = 2n ⇒ 2 ⇒ n − 1= 4 ∴ n=5

3. Required number of ways = 5 ! − { 4 ⋅ 4 ! − 4C2 ⋅ 3 ! + 4C3 ⋅ 2 ! − 1} = 53

4. Let (1, 1, 1), (− 1, 1, 1), (1, − 1, 1), (−1, − 1, 1) be vectors a , b, c, d and rest of the vectors be − a , − b, − c, − d and let us find the number of ways of selecting coplanar vectors. Observe that out of any three coplanar vectors, two will be collinear (anti-parallel). Number of ways of selecting the anti-parallel pair = 4 Number of ways of selecting the third vector = 6 Total = 24 ∴Number of non-coplanar selections = 8C3 − 24 = 32 = 2 5 , p = 5

Aliter

8× 6× 4 3! p=5

Required value = ∴

5. The integer greater than 6000 may be of 4 digit or 5 digit. So, here two cases arise. Case I When number is of 4 digit. Four-digit number can starts from 6, 7 or 8. 6,7 or 8

3

4

3

2

Thus, total number of 4-digit number, which are greater than 600 = 3 × 4 × 3 × 2 = 72 Case II When number is of 5 digit. Total number of five-digit number which are greater than 6000 = 5 ! = 120 ∴Total number of integers = 72 + 120 = 192

Targ e t E x e rc is e s

1. When n5 takes value from 10 to 6 the carry forward

6. Given, n( A) = 2, n(B) = 4

∴ n ( A × B) = 8 Now, the number of subsets of A × B having 3 or more elements = 8C3 + 8C4 + ... + 8C8 = (8 C0 + 8C1 + ... + 8C8 ) − (8 C0 + 8C1 + 8C2 ) = 2 8 − 8C0 − 8C1 − 8C2 [Q n C0 + nC1 + ... + nCn = 2 n ] = 256 − 1 − 8 − 28 = 219

7. Number of ways = 35 − 3C1 ⋅ 2 5 + 3C2 ⋅ 15 = 243 − 96 + 3 = 150

8. As, an = bn + c n ⇒ an = 1 ... (1 or 0)

⇒ an = 1_ _ _ _ _ _ _1 = an − 1 + 1 _ _ _ _ _ _ _ 1 0 = an − 2 14 4244 3 ( n − 1) places

∴

14 4244 3 ( n − 2 ) places

an = an −1 + an − 2 a17 = a16 + a15

329

⇒

Objective Mathematics Vol. 1

6

9. b6 = Six-digit numbers ending with 1. 1 – – – – 1

Now, the four places are to be filled. Case I 1 ................... 1 1 For 3 places, all 1’s are used in 1 way. One zero is used = 3C1 = 3 Two zeros are used = 1 way 0 1 0 1... Total = 5 ways Case II 1 ................. 0 1 For 3 places, All 1’s are used = 1 way One zero is used = 2C1 = 2 ways ∴ Total ways = 3 ways Thus, b6 = 5 + 3 = 8

10. The number of ways of distributing n identical objects among r persons such that each person gets atleast one object is same as the number of ways of selecting (r − 1)places out of(n − 1)different places, i.e. is n −1Cr −1.

11. If out of n points, m are collinear.

Ta rg e t E x e rc is e s

7! 4 !2 ! 7 × 6! = 8C4 × 4 !2 ! = 7 ⋅ 8C4 ⋅ 6C4

∴Required number of words = 8C4 ×

Number of triangles = nC3 − mC3 ∴ Number of triangles = 10 C3 − 6C3 = 120 − 20 ⇒ N = 100

12.

16. Required number of ways = 12C4 × 8C4 × 4C4 =

12 ! 8! 12 ! × ×1 = 8! × 4! 4! × 4! (4 !)3

17. Total number of ways =10 C1 + 10C2 + 10C3 + 10C4 = 10 + 45 + 120 + 210 = 385

18. In the word SACHIN, order of alphabets is A, C, H, I, N, S. Number of words start with A = 5! so with C, H, I, N. Now, words start with S and after that ACHIN are in ascending orders of position, so 5 ⋅ 5 ! = 600 words are in dictionary before words with start S. Thus, position of given word is 601.

19. Given that, f ( x ) =

Px − 3 The above function is defined, if 7 − x ≥ 0 and x − 3 ≥ 0 and 7−x≥x−3

⇒

x ≤ 7 and

f (3) = 4 P0 = 1

Now, 9 distinct blue balls

Urn A

Urn B

The number of ways in which two balls from urn A and two balls from urn B can be selected = 3C2 × 9C2 = 3 × 36 = 108

13. The number of ways in which 4 novels can be selected = C4 = 15 The number of ways in which 1 dictionary can be selected = 3C1 = 3 4 novels can be arranged in 4! ways. [since, dictionary is fixed in the middle] ∴ Total number of ways = 15 × 3 × 4! = 15 × 3 × 24 = 1080 6

14. The number of different ways the child can buy the six ice-creams = Total number of non-negative integral

x≥3

x≤5 Df = { 3, 4, 5}

and ∴ 3 distinct red balls

7− x

f (4) = 3 P1 = 3 f (5) = 2 P2 = 2 Rf = {1, 2, 3}

and ∴

20. Total number of ways in which all letters can be arranged = 6! There are two vowels (A, E ) in the word GARDEN. Total number of ways in which these two vowels can be arranged = 2 ! ∴ Total number of required ways 6! = = 360 2!

21. Required number of ways = =

8 −1

C3 −1 = 7C2 =

7⋅6 = 21 2 ⋅1

22. First, we fix the position of men, the number of ways in which men can sit = 5!. Now, the number of ways in which women can sit = 6 P5 M

solution of the equation x1 + x2 + x3 + x4 + x5 = 6 = 6 + 5 − 1C5 −1 = 10C4 and number of ways to arrange 6A’s and 4B’s in a row 10 ! 10 = = C4 6! 4! ∴Statement I is false and Statement II is true.

15. Given word is MISSISSIPPI. 330

Here, I = 4 times, S = 4 times, P = 2 times, M = 1time _M_I_I_I_I_P_P_

7! 2 ! 5!

M

M

M

M

M

∴ Total number of ways = 5 ! × 6 P5 = 5! × 6!

n

n

n

already fixed. Now, we have to arrange three digits from remaining eight digits i.e. 0, 1, 2, 3, 4, 5, 8, 9 in 8 P3 ways 8! = = 8 × 7 × 6 = 336 ways 5! Thus, required number of telephone numbers = 336

25. Number of ways =

2n + 1

C0 +

2n + 1

C1 +

2n + 1

C2 + ... +

2n + 1

Cn

1 2n + 1 (2 ) = 22n 2 Thus, 2 2n = 64 ⇒ 2 2 n = 2 6 On comparing, 2 n = 6 ∴ n=3 =

26. Set S = {1, 2, 3, ..., 20} is to be partitioned into four sets of equal size i.e. having 5-5 numbers. The number of ways in which it can be done =

C5 × 15C5 × 10C5 × 5C5 20 ! 15 ! 10 ! = × × ×1 15 ! × 5 ! 10 ! × 5 ! 5 ! × 5 ! (20 !) = (5 !)4 20

27. Total number of available courses = 9 Out of these, 5 courses can be chosen. But it is given that 2 courses are compulsory for every student i.e. you have to choose only 3 courses instead of 5, out of 7 instead of 9. 7 × 6× 5 It can be done in 7C3 ways = 6 = 35 ways

28. There may be two cases. Case I Two children get none and one get three and others get one each. Then, total number of ways 10 ! = × 10 ! 2 ! × 3! × 7 ! Case II Two get none and two get 2 each and the others get one each. Then, total number of ways 10 ! = × 10 ! (2 !)4 × 6 ! Hence, total number of ways (10 !)2 (10 !)2 = + 2 ! × 3 ! × 7 ! (2 !)4 × 6 ! =

(10 !)2 × 25 (2 !)2 × 6 ! × 84

29. 3 consonants can be selected from 7 consonants in 7C3 ways. 2 vowels can be selected from 4 vowels in 4C2 ways. ∴ Required number of words = 7C3 × 4C2 × 5 ! selected 5! letters can be arranged  in 5! ways, to get different words    = 35 × 6 × 120 = 25200

k=m

m+1

m+1

=

m+ 2

m+ 2

=

n+1

=(

24. Since, telephone number start with 67, so two digits is

n

∑ k Cr

Cr + 1 +

Cr + 1 +

M Cr + 1

31. We have,

= (mCr + 1 + mCr ) +

Cr ) +

m+1

Cr

+ m + 2Cr + ... + nCr Cr + ... + nCr

m+ 2

Cr + ... + nCr M

M

C1 C C C + 2 2 + 3 3 + ... + n n C0 C1 C2 Cn − 1

n n n C C1 C C + 2 n 2 + 3 n 3 + ... + n n n Cn −1 C0 C1 C2 n (n − 1) (n − 2 ) n (n − 1) n 1 3×2 2 = +2× + 3× + ... + n × − n n ( 1 ) 1 n n 2 n (n + 1) = n + (n − 1) + (n − 2 ) + ... + 1 = Σ n = 2

=

n

6 Permutation and Combination

30. We have, mCr + 1 +

= Cr + 1 + Cr + Cr −1 + Cr = n +1Cr + 1 + n + 1Cr = n + 2Cr + 1 n

n

32. We know, 11Cn is maximum when n = 5. 11! n ! (11 − n )! ∴ n ! (11 − n )! is minimum, when n = 5.

∴

Cn =

11

33. Given, n C10 = nC11 n! n! = 10 ! (n − 10 )! 11! (n − 11)!

⇒ ⇒ ⇒ Now,

11 × 10 ! (n − 11)! = 10 ! (n − 10 ) (n − 11)! 11 = n − 10 ⇒ n = 11 + 10 = 21 n C21 = 21C21 = 1

⇒ ⇒ ⇒ ⇒ ∴

2n

C2 9 = C2 2 (2 n )! 2 ! (n − 2 )! 9 × = 2 ! (2 n − 2 )! n! 2 (2 n ) (2 n − 1) (2 n − 2 )! 2(n − 2 )! 9 × = 2(2 n − 2 )! n (n − 1) (n − 2 )! 2 2(2 n − 1) 9 = ⇒ 4 (2 n − 1) = 9 (n − 1) n −1 2 8n − 4 = 9n − 9 ⇒ 9n − 8n = − 4 + 9 n=5

34. Given,

2n

C2 : nC2 = 9 : 2 ⇒

n

Targ e t E x e rc is e s

23. Now, n Cr + 1 + nCr − 1 + 2 nCr

35. Total number of points in a plane is 30. Out of them, 8 points are collinear. ∴ Total number of straight lines formed 30 × 29 8 × 7 − +1 = 30C2 − 8C2 + 1 = 2 2 ×1 = 435 − 28 + 1= 408

36. We have, 0, 1, 2, 3, 4, 5 i.e. six numbers, one zero and 5 positive numbers. One-digit number = 5 P1 = 5 Two-digit numbers = 6 P2 − 5 P1 = 25 Three-digit numbers = 6 P3 − 5 P2 = 100 Four-digit numbers = 6 P4 − 5 P3 = 300 Five-digit numbers = 6 P5 − 5 P4 = 600 Six-digit numbers = 6 P6 − 5 P5 = 600 ∴ Total possible numbers = 5 + 25 + 100 + 300 + 600 + 600 = 1630

331

Objective Mathematics Vol. 1

6

37. First, we fix the 5 plus ‘+’ in five alternate positions. ... + ... + ... + ... + ... + ... Now, select the three positions for three ‘minus’ out of 6 blank positions. ∴ Required number of ways 6 P3 6× 5× 4 = 1× = 1 × 6C3 = 1 × = 1 × 5 × 4 = 20 3! 3×2 ×1

∴

43. Since, the student is allowed to select atmost n books out of (2 n + 1) books. Therefore, in order to select atleast one book, he has the choice to select one, two, three, ..., n books. Thus, if T is the total number of ways of selecting atleast one book, then

38. Case I When he chooses 4 questions from first five

T=

questions. Now,

Now, the number of choices = 5C4 × 8C6 8×7 = 5× = 140 2 ×1 Case II When he chooses 5 questions from first five questions. Now, the number of choices 8×7 × 6 = 5C5 × 8C5 = 1 × 8C3 = = 56 3×2 ×1 Now, the total number of choices available = 140 + 56 = 196 = 3C1 ⋅ 2C2 ⋅ 4C0 + 3C1 ⋅ 2C1 ⋅ 4C1 + 3C1 ⋅ 4C2 ⋅ 2C0 = 3 ⋅ 1 + 3 ⋅ 2 ⋅ 4 + 3 ⋅ 6 = 3 + 24 + 18 = 45 Case II When 3 drawn balls contains 2 black balls = 3C2 ⋅ 2C1 ⋅ 4C0 + 3C2 ⋅ 4C1 ⋅ 2C0 = 3 ⋅ 2 ⋅ 1 + 3 ⋅ 4 ⋅ 1 = 6 + 12 = 18 Case III When 3 drawn balls contain all black balls = 3C3 ⋅ 2C0 ⋅ 4C0 = 1 On adding all three cases, we get required ways = 45 + 18 + 1 = 64 Aliter Case I

When 3 drawn balls contains 1 black ball 6× 5 = 6C2 × 3C1 = × 3 = 45 2 ×1

When 3 drawn balls contain 2 black balls 3× 2 = 6C1 × 3C2 = 6 × = 18 2 ×1 Case III When 3 drawn balls contain all black balls = 3C3 = 1 On adding all three cases, we get Required ways = 45 + 18 + 1 = 64

different ways in which half number of cases will be such that P speaks before Q and half number of cases will be such that P speaks after Q. 24 ∴Required number of ways = = 12 2

41. Required sum = 3 ! (1 + 2 + 3 + 4) = 6 (10 ) = 60

...(i) C1 + 2 n + 1C2 + ... + 2 n + 1Cn = 255 C0 + 2 n + 1C1 + ... + 2 n + 1Cn + 2 n + 1Cn + 1 2n + 1

= (1 + 1) 2n + 1

∴

C0 + 2(

2n + 1

Cn ) 2n + 1 C2 n + 1

1 + 2 (255) + 1 = 2 2 n + 1

⇒

255 + 1 = 2

∴

C2 n + 1

2n + 1

C1 + ... +

⇒

2n

2n + 1

2n + 1

1 + 2T + 1 = 2 2 n + 1 ⇒ 256 = 2

2n

= 22n + 1

[from Eq. (i)]

⇒ 4 = 4n 4

n=4

44. Clearly, first prize can be distributed among 4 students in 4 C1 = 4 ways and second prize can be distributed among 4 students in 4 C1 = 4 ways. Similarly, 3rd, 4th and 5th prizes can be distributed in 4 ways each. Hence, by fundamental principle of counting, the total number of ways = 4 × 4 × 4 × 4 × 4 = 45 = 1024

45. Required number of ways =

4! 3! × = 18 2 !2 ! 2 !

46. The number of integers greater than 6000 = 5 ! + (3 × 4 × 3 × 2 ) = 192 56

47. Q 54

Pr + 6 Pr + 3

∴ ⇒ ⇒ ∴

= 30800 56 ! (51 − r )! × = 30800 54 ! (50 − r )! 56 × 55 × (51 − r ) = 56 × 55 × 10 51 − r = 10 r = 41

48. The vowels in the word COMBINE are O, I, E which can be arranged at 4 places in 4 P3 ways and other words can be arranged in 4! ways. Hence, total number of ways = 4 P3 × 4 ! = 4 ! × 4 ! = 576

49. Required number of ways = 11C5 − 11C4 =

n

332

Cn + 2 + ... +

⇒

42. Number of ways of selecting 3 children from n children = C3 Since, the number of times she will go to the garden = 84 × (Number of ways that a particular child goes to the zoo) (n − 1)(n − 2 ) n n −1 n ⇒ C3 = 84 × (1 × C2 ) ⇒ C3 = 84 × 2 n(n − 1)(n − 2 ) (n − 1)(n − 2 ) = 84 × ⇒ 3×2 ×1 2

=2

2n + 1

+

Case II

40. Four speakers will address the meeting in 4! ways = 24

2n + 1

2n + 1

+

39. Case I When 3 drawn balls contain 1 black ball

Ta rg e t E x e rc is e s

n = 84 3 n = 252

⇒

50.

n −1

C3 +

⇒ ⇒ ⇒ ∴

11! 11! − = 132 5 ! 6 ! 4 !7 !

n −1

C4 > nC3 [Q n Cr + nCr + 1 = C4 > nC3 n! n! > (n − 4)! 4 ! (n − 3)! 3 ! (n − 3) (n − 4)! > (n − 4)! 4 n >7 n

n+1

Cr + 1]

7 Mathematical Induction Introduction There are two basic processes of reasoning. Deduction (result is deduced from general to particular) and induction (result is generalised from a particular result). Deduction is the application of a general case to a particular case i.e. given a statement to be proven, often called a conjecture or a theorem in Mathematics, valid deductive steps are derived and a proof may or may not be established. In contrast to deduction, induction depends on working with each case and developing a conjecture by observing incidences till each and every case is observed. Induction is a key aspect of scientific reasoning, where collecting and analysing data is the norm. Thus, induction means the generalisation from particular cases or facts. In algebra or in other disciplines of Mathematics, there are certain results or statements that are formulated in terms of n (where, n is a positive integer). To prove such statements, the well suited principles that is used based on the specific technique, is the principle of mathematical induction.

Statement A sentence or description which can be judged to be true or false, is called a statement. e.g. (i) 3 divides 9. (ii) Jaipur is the capital of Rajasthan. A statement involving mathematical relations is known as the mathematical statement. e.g. 3 divides 9.

Principle of Mathematical Induction Suppose there is a given statement P ( n) involving the natural number n, such that (i) statement is true for n =1 i.e. P (1) is true and (ii) if the statement is true for n = k (where, k is some positive integer), then statement is also true for n = k +1 i.e. truth of P ( k ) implies the truth of P ( k +1).

Chapter Snapshot ●

Introduction

●

Statement

●

●

●

Principle of Mathematical Induction Algorithm for Mathematical Induction Types of Problems

7

Now, show that the result is true, for n = k + 1. P(k + 1): 1 + 2 + 3 + ... + k + (k + 1) (k + 1) (k + 2 ) = 2 Then, LHS = 1 + 2 + 3 + ... + k + (k + 1) k(k + 1) [from Eq. (i)] = + (k + 1) 2 k + 1 =   (k + 2 ) = RHS  2 

Objective Mathematics Vol. 1

Then, P ( n) is true for all natural numbers n.

i.

It is simply a statement of fact. There may be a situation when a statement is true for all n ≥ 4. In this case, (i) will start from n = 4 and we shall verify the result for n = 4 i.e. P ( 4).

ii.

It is a conditional property, it does not assert that the given statement is true for n = k but only that, if it is true for n = k , then it is also true for n = k +1. So, to prove that property holds, only prove that conditional proposition.

Ø The step ‘if the statement is true for n = k, then it is also true for

Therefore, P(k + 1) is true, whenever P(k ) is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers. X

n = k + 1 ’ is sometimes referred as the inductive step and the assumption in this step that statement is true for n = k is called the inductive hypothesis.

(b)13 + 2 3 + 3 3 +… + n 3 > (1 + 2 + 3 +… + n) 2 (c)13 + 2 3 + 3 3 +… + n 3 < (1 + 2 + 3 +… + n) 2

Algorithm for Mathematical Induction

(d)13 + 2 3 + 3 3 +… + n 3 ≠ (1 + 2 + 3 +… + n) 2

The steps used to prove that a statement is true for all natural numbers by using principle of mathematical induction are given below: Step I Consider the given statement as P ( n). Step II Put n =1, then verify that it is a true statement. Step III Suppose that the statement is true, for n = k (any arbitrary number). Step IV Show that the statement is true for n = k +1. Combining the results of steps II and IV conclude by principle of mathematical induction that P ( n) is true for all n ∈ N . Ø This algorithm for principle of mathematical induction is used to

solve problems which can be categorised into the following types.

Types of Problems Identity Type Problems In this type of problems, we use principle of mathematical induction to show LHS is equal to RHS. X

Example 1. 1 + 2 + 3 + ... + n is equal to n( n −1) n( n +1) n( n +1) n( n −1 ) (a) (b) (c) (d) 2 2 3 3 Sol. (b) By the help of AP, we have P(n) : 1 + 2 + 3 + ... + n = Put n = 1, LHS = 1 and

n(n + 1) 2

Sol. (a) We have, 13 + 2 3 + 33 + … + n3 =  n(n + 1)

2

 2  = (1 + 2 + 3 + … + n)2

Aliter Let P(n) : 13 + 2 3 + 33 + … + n3 = (1 + 2 + 3 + … + n)2 For n = 1, we have P(1) : 13 = 12 ⇒ 1 = 1, which is true and for n = 2, we have P(2 ) : 13 + 2 3 = (1 + 2 )2 ⇒ 9 = 9, which is true. Similarly, we can prove that P(n) is true for all n ∈ N.

Divisibility Type Problems In this type of problems, we use principle of mathematical induction to show that given statement P ( n) is divisible by given number or P ( n) is a multiple of given number. This process can also be explained with the following examples: X

Example 3. The greatest positive integer, which divides ( n + 2) ( n + 3) ( n + 4) ( n + 5) ( n + 6) for all n ∈ N , is (a) 4 (b) 120 (c) 240 (d) 24 Sol. (b) Let P(n) : (n + 2 ) (n + 3) (n + 4) (n + 5) (n + 6) Put n = 1,

P(1) = (1 + 2 )(1 + 3)(1 + 4)(1 + 5)(1 + 6) = 3 × 4 × 5 × 6 × 7 = 120 × 21

Put n = 2,

P(2 ) = (2 + 2 )(2 + 3)(2 + 4)(2 + 5)(2 + 6) = 4 × 5 × 6 × 7 × 8 = 120 × 56 and so

11 ( + 1) RHS = = 1 = LHS 2

on. Hence, P(n) is always divisible by 120.

Therefore, P(1) is true.

334

Example 2. For a natural number n, which one is the correct statement? (a)13 + 2 3 + 3 3 +… + n 3 = (1 + 2 + 3 +… + n) 2

Suppose that the statement is true, for n = k, k ∈ N. k(k + 1) P(k ) : 1 + 2 + 3 + ... + k = ∴ 2

Aliter Qn consecutive positive integer is divided by n!. …(i)

∴5 consecutive positive integer is divided by 5! = 5 × 4 × 3 × 2 × 1= 120

Example 4. If n ∈ N , then 11n + 2 + 12 2n + 1 is divisible by (a) 113 (b) 123 (c) 133 (d) None of these

X

Example 7.

If un + 1 = 3un − 2un − 1 and u0 = 2,

u1 = 3, then un is equal to (a)1 − 2 n

Sol. (c) Putting n = 1in 11n + 2 + 12 2 n + 1, we get

(b) 2 n + 1

…(i)

Step I Given, u1 + 1 = 3 = 2 + 1 = 2 + 1 1

Inequality Type Problems

which is true, for n = 1.

In this type of problems, we have inequality symbols like >, 2n (c) n 4 < 10 n (d) 2 3n > 7n + 1

Step II Assume it is true for n = k, then it is also true for n = k − 1. Then, ...(ii) uk = 2 k + 1 u k − 1 = 2 k −1 + 1

and

…(iii)

Step III On putting n = k in Eq. (i), we get u k + 1 = 3 u k − 2u k − 1 = 3(2 k + 1) − 2(2 k − 1 + 1) [from Eqs. (ii) and (iii)]

Sol. (c) Let n = 1, then options (a), (b) and (d) do not satisfy.

= 3⋅2k + 3 − 2 ⋅2k − 1 − 2

Only option (c) satisfied. X

(d) 2 n + 2

Sol. (b) We have, u n + 1 = 3u n − 2u n − 1

113 + 12 3 = 3059, which is divisible by 133.

X

(c) 2 n − 1

7 Mathematical Induction

X

= 3 ⋅ 2 k + 3 − 2 k − 2 = (3 − 1)2 k + 1 = 2 ⋅2k + 1 = 2k + 1 + 1

Example 6. For any natural number n, 2 n ( n − 1)! < n n , if (a) n < 2 (b) n > 2 (c) n ≥ 2 (d) Never Sol. (b) Check through option, the condition 2 n (n − 1)! < nn is satisfied for n > 2.

Recursion Type Problems In this type of problems, we get a sequence in which later terms are deduced from earlier ones by using the principle of mathematical induction. This process can be explained by following examples:

This shows that the result is true for n = k + 1, hence by the principle of mathematical induction, the result is true for all n ∈ N. X

Example 8. If a1 = 2, a n = 5 a n − 1 , for all natural numbers n ≥ 2, then the value of a 4 is equal to (a) 150 (b) 50 (c) 200 (d) None of these Sol. (d) We have, a1 = 2 Then, ∴

a2 = 5 a2 − 1 = 5 a1 = 10 a3 = 5 a3 − 1 = 5 a2 = 5 × 10 = 50 a4 = 5 a4 − 1 = 5 a3 = 5 × 50 = 250

Work Book Exercise 1 Let P(n ) be a statement and let P(n ) → P(n + 1) is true for all natural numbers n, then P(n ) is true a b c d

for all n for all n > 1 for all n > m, m being a fixed integer Nothing can be said

2 The value of the natural numbers n such that the inequality 2 n > 2 n + 1 is valid, is a for n ≥ 3 c for n ∈ N

3 If n ∈ N, then 7

b for n < 3 d for any n 2n

+2

3n − 3

n −1

⋅3

is always divisible

a n> 2

b n≥ 3

c n≥ 4

d n< 4

5 For every positive integer n, 2 n < n ! when a n< 4 c n< 3

b n≥ 4 d None of these

6 For every natural number n, n(n + 1) is always a even c multiple of 3

b odd d multiple of 4 n

n + 1  > n ! is true,  2 

7 If n is a natural number, then  when

by a 25 c 45

4 For every positive integral value of n, 3 n > n 3 when

b 35 d None of these

a n< 1 c n≥ 2

b n≥ 1 d n< 3

335

WorkedOut Examples Type 1. Only One Correct Option Ex 1. The value of sum in the nth bracket of (1) + (2 + 3) + ( 4 + 5 + 6 + 7) + (8 + 9 + 10 +… 15) +… , is (a) 2n ( 2n + 2n − 1 − 1) (b) 2n − 1 ( 2n + 2n − 1 − 1)

Ex 4. Sum of nth bracket of

(c) 2n − 2 ( 2n + 2n − 1 − 1) (d) None of the above Sol. If n = 1, then 2n (2n + 2n − 1 − 1) = 2 (2 + 1 − 1) = 2(2) = 4, 2n − 1 (2n + 2n − 1 − 1) = 20 (2 + 20 − 1) = (2 + 1 − 1) = 2 and 2n − 2 (2n + 2n − 1 − 1) = 2− 1 (21 + 20 − 1) = 2− 1 (2) = 1 ∴

Sum of nth bracket = 2n − 2 (2n + 2n − 1 − 1)

Hence, (c) is the correct answer.

Ex 2. If (1) + (2 + 3 + 4) + (5 + 6 + 7 + 8 + 9)+… , then the sum of terms in the nth bracket is (a) ( n − 1) 3 + n 3 (b) ( n + 1) 3 + 8n 2 ( n + 1) ( n + 2) (c) 6n (d) ( n + 1) 3 + n 3 (n + 1)3 + 8n2 = 8 + 8 = 16,

(a)

(n + 1) + n = 8 + 1 = 9 3

If n = 2, then (n − 1)3 + n3 = 1 + 8 = 9,

∴

Sum of nth bracket = (n − 1) + n

3

Hence, (a) is the correct answer.

Ex 3. (1) + (2 + 3) + (4 + 5 + 6) + ... n brackets is equal to n( n + 1) ( n 2 + n + 2) 8 n ( n + 1) ( n 2 − n + 2) (b) 8 n( n − 1) ( n 2 + n + 2) (c) 8 n( n − 1) ( n 2 − n + 2) (d) 8 (a)

336

n ( n + 1) ( n + 2) 6

(c) n ( n + 1) ( n + 2)

(n + 1) (n + 2) 12 = =1 6n 12 3

(3n − 1)3 (2)3 = = 4; 2 2⋅ 4n − 1 (3n − 1) 2 3n + 1 4 and = = 1 = 2 ⋅ 3(n − 1) (n + 2) / 2 2 ⋅ 30 3 ⋅ 7n − 1 3 (3n − 1) ∴ Sum of nth bracket = (n − 1) (n + 2) / 2 2⋅ 3 Aliter nth bracket 1 − 1/ 3n  1 1 1 = n(n − 1) / 2 + … + ( n + 1) n/ 2 − 1 = n(n − 1) / 2   3 3 3  1 − 1/ 3 

Sol. If n = 1, then

Ex 5. The sum of (12 ) + (12 + 2 2 ) + (12 + 2 2 + 3 2 ) +… n brackets is equal to

(n + 1) (n + 2) 6 = = 1, 6n 6 and

1 1 1 1   1 (1) +  + 2  +  3 + 4 + 5  +… is 3 3  3 3 3  n 3 ( 3 − 1) ( 3n − 1) (b) (a) 2⋅ 4 n − 1 2⋅ 3(n − 1)(n + 2)/ 2 n ( 3 + 1) (d) None of these (c) 3⋅ 7n − 1

 3(3n − 1)  3n − 1 = =  n n (n − 1) / 2 (n − 1) (n + 2) / 2  2⋅ 3  2⋅ 3 3 Hence, (b) is the correct answer.

Sol. If n = 1, then (n − 1)3 + n3 = 0 + 1 = 1,

3

n (n + 1) (n2 + n + 2) 2(4 ) = =1 8 8 Hence, (a) is the correct answer.

Sol. If n = 1, then

n( n + 1) 2 ( n + 2) 12 n( n + 1) ( 2n + 1) (d) 6

(b)

Sol. Q tn = 12 + 22 + … + n2 n (n + 1) (2n + 1) n(2n2 + 3n + 1) = 6 6 n3 n2 n = + + 3 2 6  n3 n2 n ∴S n = ∑  + +  2 6  3 =

n2 (n + 1)2 n(n + 1)(2n + 1) n(n + 1) + + 12 12 12  n(n + 1)  [ n(n + 1) + 2n + 1 + 1] =  12  n(n + 1) (n2 + 3n + 2) n(n + 1)2 (n + 2) = = 12 12 Hence, (b) is the correct answer. =

n( n + 1) ( n + 2) 6 n( n + 1) ( 3n 2 + 23n + 46) (b) 12 n( 27n 3 + 90n 2 + 45n − 50) (c) 4 n( n + 1) ( 2n + 1) (d) 6

Ex 9. 2 ⋅ 12 + 3 ⋅ 2 2 + 4 ⋅ 3 2 +… + (n + 1)n 2 is equal to n( n + 1) ( n + 2) ( 3n + 5) 12 n( n + 1) ( n + 2) ( n + 3) (b) 4 (c) 2n( n + 1) ( n + 2) ( n + 3) n( n + 1) ( n + 2) ( 3n + 1) (d) 12

(a)

(a)

Sol. Q tn = (n + 1)n2 = n3 + n2 n2 (n + 1)2 n(n + 1) (2n + 1) + 4 6 n(n + 1) = [ 3n(n + 1) + 2(2n + 1)] 12 n(n + 1) (3n2 + 7n + 2) = 12 n(n + 1) (n + 2) (3n + 1) = 12 Hence, (d) is the correct answer. =

Sol. n th bracket = 1 + 3 + 5 + ... + (2n − 1) = n2 ∴

S n = Σ n2 =

n(n + 1) (2n + 1) 6

Hence, (d) is the correct answer.

Ex 7. (1) + (1 + 2) + (1 + 2 + 3) +… n brackets is equal to n( n + 1) ( n + 2) 6 n( n + 1) ( 3n 2 + 23n + 46) (b) 12 n( 27n 3 + 90n 2 + 45n − 50) (c) 4 n( n + 1) ( 2n + 1) (d) 16 (a)

Ex 10. 1 ⋅ 2 2 + 2 ⋅ 3 2 + 3 ⋅ 4 2 +… + n(n + 1) 2 is equal to n( n + 1) ( n + 2) ( 3n + 5) 12 n( n + 1) ( n + 2) ( n + 3) (b) 4 (c) 2n( n + 1) ( n + 2) ( n + 3) n ( n + 1) ( n + 2) ( 3n + 1) (d) 12

(a)

Sol. If n = 1, then the sum of the terms = (1) = 1, ∴

Sol. Q tn = n (n + 1)2 = n(n2 + 2n + 1) = n3 + 2n2 + n

n(n + 1) (n + 2) 1 × 2 × 3 = 6 6

∴ S n = Σn3 + 2Σn2 + Σn n2 (n + 1)2 2n(n + 1)(2n + 1) n(n + 1) + + 4 6 2 n(n + 1) = [ 3n(n + 1) + 4 (2n + 1) + 6 ] 12 n(n + 1) (3n2 + 11n + 10) = 12 n(n + 1) (n + 2) (3n + 5) = 12 Hence, (a) is the correct answer.

=1

=

Hence, (a) is the correct answer.

Ex 8. 1 ⋅ 3 ⋅ 4 + 2 ⋅ 4 ⋅ 5 + 3 ⋅ 5 ⋅ 6 +… upto equal to

n terms

n( n + 1) ( n + 2) 6 n( n + 1) ( 3n 2 + 23n + 46) (b) 12 3 n( 27n + 90n 2 + 45n − 50) (c) 4 n( n + 1) ( 2n + 1) (d) 6

is

(a)

Sol. Given expression Σn(n + 2) (n + 3) = Σ (n3 + 5n2 + 6n) n2 (n + 1)2 5n(n + 1) (2n + 1) 6n(n + 1) + + 4 6 2 n(n + 1) = [ 3n (n + 1) + 10(2n + 1) + 36 ] 12 n(n + 1) (3n2 + 23n + 46) = 12 =

Hence, (b) is the correct answer.

7 Mathematical Induction

Ex 6. (1) + (1 + 3) + (1 + 3 + 5) + ... n brackets is equal to

Ex 11.

12 12 + 2 2 12 + 2 2 + 3 2 + + +… upto n terms is 3 5 7 equal to n( n + 1) ( n + 2) n( n + 1) ( n + 2) (b) (a) 18 6 n( n + 1) ( n + 2) 2n ( n + 1) ( n + 2) (c) (d) 3 3

Sol. Q tn =

12 + 22 + … + n2 n(n + 1) = 2n + 1 6

(n2 + n) 1  n(n + 1)(2n + 1) n(n + 1)  + =  6 6 2  6 n(n + 1) (n + 2) = 18 Hence, (a) is the correct answer.

∴ Sn = Σ

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7

Ex 12. 1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 +… upto n terms is equal to n( n + 1) ( n + 5) 3 n( 4n 2 + 6n − 1) (c) 3 (a)

(b)

n( n + 1) ( n + 2) 3

(d) n( n + 1) ( n + 2)

Sol. If n = 1, then the sum of the terms = 1⋅ 2 = 2 n(n + 1) (n + 2) 1 × 2 × 3 = =2 3 3 Hence, (b) is the correct answer.

Ex 13. 1 ⋅ 3 + 2 ⋅ 3 2 + 3 ⋅ 3 3 +… + n ⋅ 3 n is equal to ( 2n − 1)3n + 1 + 3 4 ( 2n + 1)3n + 1 − 3 (c) 4

(a)

( 2n + 1)3n + 1 + 3 4 ( 2n − 1)3n + 1 − 3 (d) 4 (b)

Sol. If n = 1, then (2n − 1)3n + 1 + 3 (2n + 1)3n + 1 + 3 15 = 3, = , 4 4 2 (2n + 1)3n + 1 − 3 (2n − 1)3n + 1 − 3 3 = 6, = 4 4 2 (2n − 1)3n + 1 + 3 ∴ Sum = 4 Hence, (a) is the correct answer.

Ex 14. 1 ⋅ 1! + 2 ⋅ 2! + 3 ⋅ 3! +… + n ⋅ n! is equal to (a) ( n + 1)! − 1 (b) ( n − 1)! + 1 (c) ( n + 1)! + 1 (d) ( n − 1)! − 1 Sol. If n = 1, then (n + 1)! − 1 = 2 − 1 = 1, (n − 1)! + 1 = 2, (n + 1)! + 1 = 2 + 1 = 3, (n − 1)! − 1 = 1 − 1 = 0 Sum = (n + 1)! − 1 ∴ Hence, (a) is the correct answer.

Ex 15. 1 ⋅ 3 + 3 ⋅ 5 + 5 ⋅ 7 +… upto n terms is equal to n( n + 1) ( n + 5) 3 n( n + 1) ( n + 2) (b) 3 n( 4n 2 + 6n − 1) (c) 3 (d) n( n + 1) ( n + 2) (a)

n(n + 1) (n + 5) n(n + 1) (n + 2) = 4, = 2, 3 3 n(4 n2 + 6n − 1) = 3, n(n + 1) (n + 2) = 6 3 Hence, (c) is the correct answer.

Ex 16. 1 + 4 + 13 + 40 +… upto n terms is equal to 3n + 1 − 2n 2n 3n − 1 + 3n (c) 9

338

⇒ ⇒ ∴

S n = 1 + 4 + 13 + 40 + … + tn Sn = 1 + 4 + 13 + … + tn + 1 + tn 0 = 1 + 3 + 9 + 27 + … − tn (3n − 1) (3n − 1) tn = = (3 − 1) 2 (3n − 1) 2 3(3n − 1) n = − 2(3 − 1) 2

Sn = Σ

(3n + 1 − 3 − 2n) 4 Hence, (b) is the correct answer. =

Ex 17. 1 + 4 + 10 + 19 +… + n( n + 1) ( 2n + 1) 6 n 2 ( n + 1) 2 (b) 4 n ( n + 1) ( n + 2) (c) 6 n( n 2 + 1) (d) 2

(3n 2 − 3n + 2) is equal to 2

(a)

(3n2 − 3n + 2) 2 3 2 3 = Σn − Σn + Σ1 2 2 3n (n + 1) (2n + 1) 3n (n + 1) = − +n 12 4 n = [ 2n2 + 3n + 1 − 3n − 3 + 4 ] 4 n n(n2 + 1) = [ 2n2 + 2 ] = 4 2 Hence, (d) is the correct answer.

Sol. Sum = Σ

1   1  1  1   is equal to Ex 18. 1 −  1 −  1 −  … 1 −  2   3   4   n + 1 1 n+1 n (b) n+1 n (c) 2n + 1 n (d) 3n + 1 (a)

Sol. If n = 1, then

(a)

Sol.

( 3n + 1 − 2n − 3) 4 ( 3n − 1 + 2n 2 ) (d) 8 (b)

1 1 n 1 = , = , n+1 2 n+1 2 n 1 n 1 = , = 2n + 1 3 3n + 1 4 1 1 n 2 If n = 2, then = , = n+1 3 n+1 3 n Sum = ∴ n+1 Hence, (b) is the correct answer.

Sol. If n = 1, then

Target Exercises Type 1. Only One Correct Option 3 15 63 + + +K n terms is equal to 4 16 64 1 n 1 4 − 3 3 1 n 1 (c) n + 4 − 3 3

9. 1 +

1 −n 4 − 3 1 (d) n − 4 − n + 3

(a) n −

(b) n +

1 3 1 3

a1 a 2… a n

3. Σ

n2 (n + 1)2

(b)

n3 (n + 1)3

(x + a1 )(x + a2 )… (x + an ) a1a2… an (x − a1 )(x − a2 )… (x − an ) (b) a1a2… an (c) (x + a1 )(x + a2 ) … (x + an ) (d) (x − a1 )(x − a2 ) … (x − an )

2

2

(d)

1 n+1

defined by a 0 = 1, a n + 1 = 3n 2 + n + a n , ( n ≥ 0). Then, a n is equal to

(a) n3 + n2 + 1 (b) n3 − n2 + 1 (c) n3 − n2

n2 − 2n (b) 6

11. For n ∈ N ,

(d) None of these

1⋅ 22 + 2⋅ 32 + 3⋅ 4 2 + … + n( n + 1) 2 12 ⋅ 2 + 22 ⋅ 3 + 32 ⋅ 4 + … + n 2 ( n + 1) 3n + 5 3n + 1 (c) (3n + 1)(3n + 5)

is equal to

3n + 1 3n + 5 (d) None of these

7.

(b)

n n+1

1 n+1

(b)

n n+1

sin 2 θ 2n sin θ cos 2nθ (c) n 2 cosθ (a)

(d) y − 1

(a) positive integer (c) odd positive integer

(b) integer (d) None of these

14. x n + 1 + ( x + 1) 2n − 1 is divisible by (b) x + 1

(c) x 2 + x + 1 (d) x 2 − x + 1

15. If n ∈ N , then 2n (a) = n (c) < n

(c)

n 2n + 1

(c)

n 2n + 1

sin 2 θ sin θ cos 2nθ (d) n 2 sin θ n

(b)

(b) > n (d) None of these

16. If n ∈ N , then 1+ (d)

(d)

n 3n + 1

n 3n + 1

1 2

+

1

(a) = n (c) ≤ n

17.

8. cos θ cos 2θ cos 4θ …cos 2n − 1 θ is equal to n

(c) x + y

n

(a) x

1 1 1 + + + K n terms is equal to 1⋅ 3 3⋅ 5 5⋅ 7 (a)

(b) x − y

13. x + a is divisible by x + a for n is any n

1 1 1 + + + K n terms is equal to 1⋅ 2 2⋅ 3 3⋅ 4 1 n+1

(b) a natural number (d) None of these

12. If n ∈ N , then x n − y n is divisible by (a) x − 1

14 24 34 n4 is equal to + + +K+ 1⋅ 3 3⋅ 5 5⋅ 7 ( 2n + 1)( 2n − 1)

(a)

(d) n3 + n2

1 5 1 3 7 n + n + n is 5 3 15

(a) an integer (c) a positive fraction

(b)

n(n + 1)(n + 2) (a) 6n n(n + 1)(n2 + n + 1) (b) 6(2n + 1) (c) n(n + 2)(n + 3)2 (d) None of the above

6.

10. Using mathematical induction, the numbers a n ’s are

2

(a)

5.

n n+1

1 + 2 + 3 + ... + n is equal to 1+ 2 + 3 + K + n 2

(n2 + 2n) (a) 3 n2 + 11 (c) 12n

4.

(c)

is equal to

(a)

1 2 2 3 3 4 ⋅ ⋅ ⋅ 2. 2 32 + 32 2 3 + 3 2 32 3 + ... n terms is equal to 1 1 +2 1 +2 +3 (a)

x x( x + a1 ) + +…+ a1 a1 a 2 x( x + a1 ) ( x + a 2 )… ( x + a n − 1 )

Targ e t E x e rc is e s

1.

( 2n )! 2n

2 ( n !) 2 (a)

3

+…+

1 n

(b) ≥ n (d) None of these

≤

1 3n + 1

(b)

1 3n + 2

(c)

1 (d) 3n + 4

(b)

(2n)! 4n < 2 2n + 1 (n!)

1 3n + 5

18. If n > 1, then (2n)! 4n = 2 2n + 1 (n!) (2n)! 4n (c) > (n!)2 2n + 1

(a)

(d) None of these

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Objective Mathematics Vol. 1

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Type 2. More than One Correct Option 19. If a1 = 1 and a n = na n−1 + 5 for all positive integer n ≥ 2 , then a 5 is equal to

(a) n ≥ 1(n ∈ Z ) (c) all n ∈ N

(b) n > 2 (d) None of these

21. If p is prime number, then n p − n is divisible by p, when n is

(a) 550 (b) 120 (c) 5a4 + 5 (d) 24

(a) natural number greater than 1 (b) irrational number (c) complex number (d) positive integer greater than or equal to 2

20. x( x n−1 − na n−1 ) + a n ( n − 1) is divisible by ( x − a ) 2 for

Type 3. Assertion and Reason Directions (Q. Nos. 22-24) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

Ta rg e t E x e rc is e s

(b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

22. Statement I Sum of 20th group of {1}, {1, 2}, {1, 2, 3},... is 110. n( n + 1) Statement II Sum of n natural numbers is . 2 23. Statement I If n is a positive integer, then 32n + 7 is divisible by 8. Statement II HCF of 16 and 88 is 8. 24. Statement I If n ∈ N , product of n( n + 1)( n + 2) is divisible by 6. Statement II Product of 3 consecutive positive integers is divisible by 3!.

Type 4. Linked Comprehension Based Questions Passage I (Q. Nos. 25-26) If an = 7 + 7 + 7 + …

27. S ( k ) can be expressed as (a) k 2 (b) 2k 2 (c) 2 + k 2 (d) None of the above

having n radical signs, then by methods of mathematical induction, answer the following questions.

25. Which of the following is true? (a) an > 7, ∀ n ≥ 1 (b) an > 3, ∀ n ≥ 1 (c) an < 4 , ∀ n ≥ 1 (d) an < 3, ∀ n ≥ 1

26. The value of a1 + a 2 + a 3 is (a) < 6 (b) < 4 (c) < 12 (d) None of the above

Passage II (Q. Nos. 27-28) Consider S(k) is the sum of first k odd natural numbers expressed as S (k ): 1 + 3 + 5 + … + (2k − 1)

340

28. If S ( k ) expressed as 3 + k 2 , then (a) S (1) is true (c) S (k ) ⇒ S (k + 1)

(b) S (k ) ⇒ / S (k + 1) (d) None of these n

Passage III (Q. Nos. 29-30) Consider an = 2 2 + 1, if n > 1.

29. The last digit of a 2 is (a) 1 (c) 3

(b) 2 (d) None of these

30. The last digit of a n is (a) 2

(b) 3

(c) 7

(d) 4

7 Mathematical Induction

Type 5. Match the Column 31. Match the following: Column II

Column I n +1

< 4 holds is n

p.

1

If x − 1is divisible by x − k. Then, the least positive integral value of k is

q.

4

C.

If 49 + 16n + k is divisible by 64 for n ∈ N. Then, the numerically least negative integral value of k is

r.

2

D.

For all n ∈ N, 2 4 n − 15n − 1 is divisible by 15 k . Then, the maximum integer value of k is

s.

3

A.

The smallest positive integer for which the statement 3

B.

n

n

Type 6. Single Integer Answer Type Questions 32. If

( n + 2)! divisible by n, n ∈ N and 1≤ n ≤ 9, then n is___________ . 6( n − 1)!

33. If n ∈ N and 1< n ≤ 9, then n 3 + 2n is divisible by___________ .

Entrances Gallery JEE Main/AIEEE 1. 23n − 7n − 1 is divisible by (a) 64 (c) 49

[2014] (b) 36 (d) 25

2. Statement I The sum of the series 1 + (1 + 2 + 4 ) + ( 4 + 6 + 9) + ( 9 + 12 + 16) + .... + ( 361 + 380 + 400) is 8000. n

Statement II

∑ [ k 3 − ( k − 1)3 ] = n 3 , for any

k =1

natural number n.

[2012] (a) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (b) Statement I is true, Statement II is false (c) Statement I false, Statement II is true (d) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

3. Statement I For each natural number n, ( n + 1) 7 − n 7 − 1is divisible by 7. Statement II For each natural number n, n 7 − n is [2011] divisible by 7. (a) Statement I is incorrect, Statement II is correct (b) Statement I is correct, Statement II is correct; Statement II is correct explanation for Statement I (c) Statement I is correct, Statement II is correct; Statement II is not a correct explanation for Statement I (d) Statement I is correct, Statement II is incorrect

4. Statement I

For every natural number n ≥ 2 1 1 1 + +K+ > n. n 1 2

Statement II For every natural number n ≥ 2

n( n + 1) < n + 1.

[2008]

(a) Statement I is correct, Statement II is correct; Statement II is a correct explanation for Statement I (b) Statement I is correct, Statement II is correct; Statement II is not a correct explanation for Statement I (c) Statement I is correct, Statement II is incorrect (d) Statement I is incorrect, Statement II is correct

Targ e t E x e rc is e s

34. The minimum value of n for which 10n + 3⋅ 4 n+ 2 + 5 is an integral multiple of 97, is___________ .

1 0 1 0 and I =  5. If A =  , then which one of the  0 1 1 1 following holds, ∀ n ≥ 1 by the principle of [2005] mathematical induction? (a) A n = 2n − 1 A + (n − 1)I (c) A n = 2n − 1 A − (n − 1)I

(b) A n = nA + (n − 1)I (d) A n = nA − (n − 1)I

6. Let S ( k ) = 1 + 3 + 5 + ... + ( 2k − 1) = 3 + k 2 .

Then, which of the following is true? [2004] (a) S (1) is correct (b) Principle of mathematical induction can be used to prove the formula (c) S (k ) ⇒ / S (k + 1) (d) S (k ) ⇒ S (k + 1) 7. 1⋅ 2⋅ 3 + 2⋅ 3⋅ 4 + 3⋅ 4 ⋅ 5 + ... n terms is equal to n(n + 1)(n + 2)(3n + 5) (a) 12 (c) 2n(n + 1)(n + 2)(n + 3)

[2002] n(n + 1)(n + 2)(n + 3) (b) 4 n(n + 1)(n + 2)(3n + 1) (d) 12

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Objective Mathematics Vol. 1

7

Other Engineering Entrances 8. If n is a positive integer, then n 3 + 2n is divisible by (a) 2

(b) 6

[BITSAT 2014] (d) 3

(c) 15

(a) a + b (b) a − b (c) a3 + b3 (d) a2 + b2

9. 10n × 3( 4 n + 2 ) + 5 is divisible by ( n ∈ N ) (a) 7

(b) 5

[BITSAT 2014] (d) 17

(c) 9

10. Let n ≥ 2 be an integer,

13. Which of the following result is valid?

[AMU 2011] (a) (1 + x ) > (1 + nx ), for all natural numbers n (b) (1 + x )n ≥ (1 + nx ), for all natural numbers n, where x > − 1 (c) (1 + x )n ≤ (1 + nx ), for all natural numbers n (d) (1 + x )n < (1 + nx ), for all natural numbers n

   2π   2π   cos   sin   0  n n   2 π 2 π     A = − sin   cos   0   n  n   0 0 1   and I is the identity matrix of order 3. Then,

n

14. If n is a natural number, then

1 1 1 kn , then + + + ... n terms = 2× 4 4 × 6 6× 8 n+1 [EAMCET 2012] k is equal to

11. If

(a)

1 4

2

2

15. If a1 = 1 and a n = na n − 1 for all positive integers n ≥ 2, then a 5 is equal to

1 2 1 (d) 8

(b)

(c) 1

[AMU 2011,13]

n3 (a) 1 + 2 + ... + n < 3 n3 2 2 2 (b) 1 + 2 + ...+ n = 3 (c) 12 + 22 + ...+ n2 > n3 n3 (d) 12 + 22 + ...+ n2 > 3 2

[WB JEE 2014]

(a) A n = I and A n − 1 ≠ I (b) A m ≠ I for any positive integer m (c) A is not invertible (d) A m = 0 for a positive integer m

Ta rg e t E x e rc is e s

12. If a, b and n are natural numbers, then a 2n − 1 + b 2n − 1 is divisible by [EAMCET 2011]

(a) 125 (c) 100 (e) None of these

[Kerala CEE 2010] (b) 120 (d) 115

Answers Work Book Exercise 1. (d)

2. (a)

3. (a)

4. (c)

5. (b)

6. (a)

7. (c)

Target Exercises 1. (b)

2. (c)

3. (a)

4. (a)

5. (b)

6. (b)

7. (c)

8. (a)

9. (a)

10. (b)

11. (b)

12. (b)

13. (c)

14. (c)

15. (b)

16. (b)

17. (a)

18. (c)

19. (a,c)

20. (a,c)

21. (a,d)

22. (d)

23. (a)

24. (a)

25. (c)

26. (c)

27. (a)

28. (c)

29. (d)

30. (c)

31. (*)

32. (1)

33. (3)

34. (2)

9. (c)

10. (a)

* A → q; B → p; C → p; D → r

Entrances Gallery

342

1. (c)

2. (d)

3. (b)

4. (b)

5. (d)

11. (a)

12. (a)

13. (b)

14. (d)

15. (b)

6. (d)

7. (b)

8. (d)

Explanations Target Exercises 1 1 4 1 −2 n − 4n − = 1 − − = , 3 3 3 3 3 1 1 1 1 9 3 n + 4− n − = 1 + − = = , 3 3 12 3 12 4 1 1 4 1 n + 4n − = 1 + − = 2, 3 3 3 3 1 1 1 1 15 5 and n − 4− n + = 1 − + = = 3 3 12 3 12 4 1 −1 Sum = n + 4−n ∴ 3 3

n(n + 1)(n + 2 )(3 n + 1) 12 n (n + 1)(n + 2 )(3 n + 5) Σn (n + 1)2 12 ∴ = Σn 2 (n + 1) n(n + 1)(n + 2 )(3 n + 1) 12 3n + 5 = 3n + 1 =

n4 (2 n + 1)(2 n − 1) n4 =Σ 2 4n − 1   n2 1 1 =Σ + +  16 16 (2 n + 1)(2 n − 1)  4 n(n + 1)(2 n + 1) n (2 n + 1) − (2 n − 1) = + +Σ 24 16 32(2 n + 1)(2 n − 1) n(n + 1)(2 n + 1) n 1  1  = + + 1 −  24 16 32  2 n + 1 n(n + 1)(2 n + 1) n n = + + 24 16 16 (2 n + 1) n(n + 1)(2 n + 1) n(2 n + 2 ) = + 24 16 (2 n + 1) 2 n(n + 1) (2 n + 1) + 3  = 24  2 n + 1  n(n + 1)(n 2 + n + 1) = 6 (2 n + 1)

5. S n = Σ

2. If n = 1, then 1 n2 = , 2 4 (n + 1) 1 n3 = , (n + 1)3 8 n 1 = , n+1 2 1 1 = n+1 2

and If n = 2, then

2 n = n+1 3 1 1 = n+1 3

∴

3. We have,

Sum =

n n+1

12 + 2 2 + 32 + K + n 2 1+ 2 + 3 + K + n n(n + 1)(2 n + 1)2 (2 n + 1) = = n(n + 1) 6 3 (Σn ) n n(n + 1) n + = Sn = 2 + 3 3 3 3 2 (n + 2 n ) = 3

4. Σ n(n + 1)2 = Σ (n 3 + 2 n 2 + n ) n 2 (n + 1)2 2 n(n + 1)(2 n + 1) n(n + 1) = + + 4 6 2  n(n + 1) = [3 n(n + 1) + 4(2 n + 1) + 6]  12  n(n + 1)(3 n 2 + 11 n + 10 ) = 12 n(n + 1)(n + 2 )(3 n + 5) = 12 Σn 2 (n + 1) = Σ(n 3 + n 2 ) n (n + 1) n(n + 1)(2 n + 1) + 4 6  n(n + 1) = [3 n(n + 1) + 2(2 n + 1)]  12  n(n + 1)(3 n 2 + 7 n + 2 ) = 12 =

2

2

1 n(n + 1) 2 −1 3−2 4− 3 (n + 1) − n + + + ... + ∴ Sn = n(n + 1) 1⋅ 2 2⋅3 3⋅ 4 1 1 1 1 1 = 1 − + − + ... + − 2 2 3 n n+1 1 n = 1− = n+1 n+1

6. t n =

Targ e t E x e rc is e s

1. If n = 1, then

7. If n = 1, then

and ∴

1 1 = , n+1 2 n 1 = , n+1 2 n 1 = , 2n + 1 3 n 1 = 3n + 1 4 Sum =

n 2n + 1

8. If n = 1, then sin 2 n θ sin 2θ = 2 n sin θ 2 sin θ 2 sin θ cos θ = = cos θ 2 sin θ

343

Objective Mathematics Vol. 1

7

9. If n = 1, then given sum = 1 +

x x + a1 = a1 a1

10. Given, a0 = 1, an + 1 = 3 n 2 + n + an ⇒

a1 = 3 (0 ) + 0 + a0 = 1

⇒

a2 = 3 (1)2 + 1 + a1 = 3 + 1+ 1= 5

From option (b), Let P(n ) = n 3 − n 2 + 1 ∴

P(0 ) = 0 − 0 + 1 = 1 = a0 P(1) = 13 − 12 + 1 = 1 = a1

and ∴

P(2 ) = (2 )3 − (2 )2 + 1 = 5 = a2 an = n 3 − n 2 + 1

11. If n = 1, then 1 5 1 3 7 n + n + n 5 3 15 1 1 7 = + + 5 3 15 3+ 5+ 7 = 15 =1

12. If n = 1, then

Ta rg e t E x e rc is e s

⇒

19. Given, an = nan − 1 + 5 For

∴

n = 2, then a2 = 2 a1 + 5 = 7 a3 = 3 a2 + 5 = 21 + 5 = 26 a4 = 4a3 + 5 = 104 + 5 = 109 a5 = 5a4 + 5 = 545 + 5 = 550

x n + an = x 3 + a3 = ( x + a)( x 2 + a2 + ax ) ∴ x n + an is divisible for odd positive integer n.

14. If n = 1, then x n + 1 + ( x + 1)2 n − 1 = x 2 + x + 1, divisible by x 2 + x + 1. 2 n = 2, > 1 = n 2n > n

⇒

1 2 2 +1 2+ 2 = = 2 2 3.414 = 2 = 1.707 2 = 1.414 Sum ≥ n

16. If n = 2, then given sum = 1 +

17. If n = 1, then (2 n )! 2 1 = = 2 4 2 2 (n !) 1 1 1 = = , 3n + 1 4 2 1 1 , = 5 3n + 2 1 1 = , 3n + 4 7 2n

[Q a1 = 1]

20. Check through options, the condition is satisfied, ∀ n ∈ N.

21. n p − n is divisible by p for any natural number greater

x n + an = x + a

15. If n = 1, then

344

and

(2 n )! 24 = =6 4 (n !)2 4n 8 = 2n + 1 3 (2 n )! 4n > (n !)2 2 n + 1

Aliter Let Then,

If n = 3, then

Now,

18. If n = 2, then

than 1. It is Fermet’s theorem.

13. If n = 1, then

⇒ ∴

∴

1 1 = 3n + 5 8 (2 n )! 1 ≤ 2n 2 3n + 1 2 (n !)

n

x − y = x − y, divisible by x − .y. n

and

n = 4 and p = 2 (4)2 − 4 = 16 − 4 = 12

It is divisible by 2. So, it is true for any natural number greater than 1. 20 ⋅ 21 22. Sum of 20th group = = 210 2 n(n + 1) Sum of n natural numbers = 2 1⋅ 2 It is true, for n = 1, =1 2 Assume, it is true, for n = k. k(k + 1) So, 1 + 2 + 3 + ... + k = 2 Now, for k + 1, 1 + 2 + 3 + ... + k + k + 1 k(k + 1) (k + 1)(k + 2 ) = +k+1= 2 2 So, Statement II is true, for all n ∈ N.

23. Let statement P(n ) = 32 n + 7 For P(n = 1) = 16 For P(n = 2 ) = 88 ∴HCF of 16 and 88 is 8.

24. Let P(n ) ≡ n(n + 1)(n + 2 ) P(n = 1) ≡ 1⋅ 2 ⋅ 3 = 3 ! P(n = 2 ) ≡ 2 ⋅ 3 ⋅ 4 = 2 ⋅ (1⋅ 2 ⋅ 3) = 2 ⋅ 3 ! P(n = 3) ≡ 3 ⋅ 4 ⋅ 5 = 2 ⋅ 5 ⋅ (1⋅ 2 ⋅ 3) = 10 ⋅ 3 ! So, Statement I is true, Statement II is also true and Statement II is a correct explanation of Statement I.

25. We have, a1 = 7 < 4. Assume ak < 4 for some natural number k. We have, ∴

ak + 1 = 7 + ak < 7 + 4 < 4 an < 4 , ∀ n ≥ 1

a1 + a2 + a3 < 4 + 4 + 4 = 12 k 27. Consider 1 + 3 + 5 + ... + (2 k − 1) = [1 + (2 k − 1)] 2 Q if a and l are the first term and last    n  term of an AP, then S n = (a + l )    2 ⇒ 1 + 3 + 5 + ... + (2 k − 1) = k 2 Thus, S (k ) can be expressed as k 2 . i.e. S (k ): 1 + 3 + 5 + ...+ (2 k − 1) = k 2

28. Consider S (k ): 1 + 3 + 5 + ... + (2 k − 1) = 3 + k 2 Then, S(1) : 1 = 3 + 12 , which is not true. Suppose S (k ) is true, then 1 + 3 + 5 + ... + (2 k − 1) = 3 + k 2 On adding (2 k + 1) both sides, we get 1 + 3 + 5 +…+ (2 k − 1) + (2 k + 1) = 3 + k 2 + (2 k + 1) = 3 + (k + 1)2 which is S (k + 1.) Thus, S (k ) ⇒ S (k + 1) 2

29. For n = 2, a2 = 2 2 + 1 = 2 4 + 1 = 16 + 1 = 17 Thus, the last digit of a2 is 7.

where, k > 1and m is some positive integer. k +1 k Now, ak + 1 = 2 2 + 1 = (2 2 )2 + 1 = (10 m + 6)2 + 1 = 10(10 m2 + 12 m+ 3)+ 7 Thus, last digit of an is 7, ∀ n > 1.

31. A. If n = 1, then 32 k +

1 = k+1

k(k + 1) + 1

(k + 1) k+1 [Q k(k + 1) > k, ∀ k ≥ 0] > k+1 = (k + 1)

∴

P(k + 1) > k + 1

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Objective Mathematics Vol. 1

7

By mathematical induction, Statement I is true, ∀ n ≥ 2. Now, let α(n ) = n(n + 1) ∴ α(2 ) = 2(2 + 1) = 6 − 1] [Q k > 0 and x > − 1]

= 1 + (k + 1)x Thus,

(1 + x )k + 1 ≥ 1 + (k + 1)x, if x > − 1

7 Mathematical Induction

2π   sin 2 n − 2 ×   n 2π   cos 2 n − 2 ×   n 0

14. By taking option (d), 1 3 8 When n = 2, then 5 > , 3 When n = 3, then 14 > 9, 64 When n = 4, then 30 > = 2133 . , 3 When n = 1, then 1 >

[true] [true] [true] [true]

15. Given, an = nan − 1 ⇒

∴

a2 = 2 a1 = 2 a3 = 3 a2 = 3 (2 ) = 6 a4 = 4(a3 ) = 4 (6) = 24 a5 = 5 (a4 ) = 5 (24) = 120

[Q a1 = 1]

Targ e t E x e rc is e s

  n − 2 2π  ×   cos 2 n  2π   and An − 1 = − sin 2 n − 2 ×   n   0  

347

8

Binomial Theorem Binomial Theorem for Positive Integral Index If x and a are real numbers, then for all n Î N , ( x + a ) n = n C 0 x n a 0 + n C1 x n -1 a 1 + n C 2 x n - 2 a 2 +K + n C r x n - r a r + K + n C n -1 x 1 a n -1 + n C n x 0 a n

Chapter Snapshot ●

n

(x + a ) n = å n C r x n - r a r

i.e.

.

r =0

X

Example 1. x 10 + 10x 8 a + 40x 6 a 2 + 80x 4 a 3 + 80x 2 a 4 + 32a 5 is equal to (a) ( x 2 + a ) 5 (c) ( x 2 + 2a ) 5 (b) ( x + a )10 (d) ( x 2 - 2a ) 5 ( x2 + 2 a)5 = 5C 0 ( x2 )5 (2 a)0 + 5C1( x2 )4 (2 a)1 + 5C 2 ( x2 )3 (2 a)2 + 5C 3 ( x2 )2 (2 a)3 5

2

4

●

Multinomial Theorem

●

Greatest Coefficient

●

Sol. (c) Using binomial theorem, we have 2 0

R-f Factor Relation

●

Divisibility Problems

5

+ C 4 ( x )(2 a) + C 5 ( x ) (2 a) 10

= x

10

= x

8

6

+ 5( x )(2 a) + 10( x )(4a ) + 10( x )(8a ) + 5( x2 )(16a4 ) + 32 a5 8

6 2

2

4

4 3

3

2 4

+ 10 x a + 40 x a + 80 x a + 80 x a + 32 a

●

5

Properties of Binomial Expansion i. The above expansion is also valid when x and a are complex numbers. n ii. We have, (x + a ) n = å n C r x n - r a r . Since, r can have values from 0 to n, the r =0

total number of terms in the expansion is ( n +1).

iii.

The sum of the indices of x and a in each term is n. 9

e.g. (2x + 3 y) 9 = å 9 C r (2x ) 9- r (3 y) r and it has (9 + 1) i.e. 10 terms. r =0

Also, the sum of indices of 2x and 3y in each term is 9.

iv.

Replacing a by -a, in the expansion of ( x + a ) n , we get ( x - a ) n = n C 0 x n a 0 - n C1 x n -1 a 1 + n C 2 x n - 2 a 2 - n C 3 x n - 3 a 3 n

i.e.

( x - a ) n = å ( -1) r r =0

+ K + ( -1) r n

C r x n -r a r

n

C r x n - r a r + K + ( -1) n

n

Cn x 0a n

Greatest Term

●

●

5

Binomial Theorem for Positive Integral Index

Properties of Binomial Coefficients Binomial Theorem for any Index

●

Approximation

●

Exponential Series

●

Logarithmic Series

Putting x =1 and a = x in the expansion of ( x + a ) n , we get

X

(1 + x ) n = n C 0 + n C1 x + n C 2 x 2 +K + nCr x r +K + nCn x n

Sol. (c) Since, here n = 9 is odd, therefore the expansion of

n

i.e. (1 + x ) n = å n C r x r

9 + 1ö (1 + 5 2 x)9 + (1 - 5 2 x)9 has æç ÷ = 5 terms. è 2 ø

r =0

known as the expansion of (1 + x ) n in ascending powers of x.

vi.

X

Putting a =1 in the expansion of ( x + a ) n , we get ( x + 1) n = n C 0 x n + n C1 x n -1 + n C 2 x n - 2 + K + n C r x n - r + K + n C n -1 x + n C n n

n

n

i.e. ( x + 1) = å C r x

Example 4. Using binomial theorem, expand {( x + y) 5 + ( x - y) 5 } and hence find the value of {( 2 + 1) 5 + ( 2 - 1) 5 }. ( x + y)5 + ( x - y)5 = 2 [5 C 0 x5 + 5C 2 x3 y2 + 5C 4 x1 y4 ] = 2( x5 + 10 x3 y2 + 5 xy4 )

n -r

Putting x = 2 and y = 1, we get ( 2 + 1)5 + ( 2 - 1)5 = 2 {( 2 )5 + 10( 2 )3 + 5 2 }

known as the expansion of ( x +1) n in descending powers of x.

= 2(4 2 + 20 2 + 5 2 ) = 58 2

x.

Putting x =1 and a = - x in the expansion of ( x + a ) n , we get (1 - x ) n = n C 0 - n C1 x + n C 2 x 2 - n C 3 x 3 + K + ( -1) r n C r x r + K + ( -1) n n C n x n n

i.e. (1 - x ) n = å ( -1) r

n

Here, ( r +1)th term is given by n C r x n - r a r . If Tr +1 denotes the ( r +1)th term, then Tr +1 = n C r x n - r a r

Cr x r

This is called the general term, because by giving different values to r we can determine all terms of the expansion. Also, the general term can be expressed as n! s r x a , where r + s = n. r ! s!

Example 2. Expand (2x - 3 y) 4 by binomial theorem. Sol. Using binomial theorem, we have (2 x - 3 y)4 = {2 x + (- 3 y)}4 =

4

r 4

å(- 1)

C r (2 x)4 - r (3 y)r

r =0 1 4

= 4C 0 (2 x)4 (3 y)0 - 4C1(2 x)3 (3 y) + C 2 (2 x)2 (3 y)2

xi.

- 4C 3 (2 x)(3 y)3 + 4C 4 (2 x)0 (3 y)4 = 16 x4 - 4(8 x3 )(3 y) + 6(4 x2 )(9 y2 ) - 4(2 x)(27 y3 ) + 81y4 = 16 x4 - 96 x3 y + 216 x2 y2 - 216 xy3 + 81y4

viii.

ix.

We have, (x + a ) n + (x - a ) n = 2 [ n C 0 x n a 0 + n C 2 x n - 2 a 2 + K] and ( x + a ) n - ( x - a ) n = 2 [ n C1 x n -1 a 1 + n C 3 x n - 3 a 3 + K] If n is odd, then {( x + a ) n + ( x - a ) n } and {( x + a ) n - ( x - a ) n } both have the same æ n + 1ö number of terms equal to ç ÷ whereas, if n è 2 ø æn ö is even, then {( x + a ) n + ( x - a ) n } has ç + 1÷ è2 ø terms and æ nö {( x + a ) n - ( x - a ) n } has ç ÷ terms. è 2ø

We have, ( x + a ) n = n C 0 x n a 0 + n C1 x n -1 a 1 + n C 2 x n - 2 a 2 + K+ n C r x n - r a r + K + n C n x 0 a n

r =0

X

8

Sol. We have,

r =0

vii.

Example 3. The number of terms in the expansion of (1 + 5 2x ) 9 + (1 - 5 2x ) 9 are (a) 3 (b) 6 (c) 5 (d) None of these

Binomial Theorem

v.

xii.

In the binomial expansion of ( x - a ) n , the general term is given by Tr +1 = ( -1) r n C r x n - r a r In the binomial expansion of (1 + x ) n , we have Tr +1 = n C r x r

xiii.

In the binomial expansion of (1 - x ) n , we have Tr +1 = ( -1) r

n

Cr x r

Ø To find the terms free from radicals or rational indicies in the

expansion of (a1/ p + b1/ q )n , where a and b are prime numbers, first N

1/ p N - r 1/ q r

N

N- r r p q

find the general termTr + 1 = C r(a ) (b ) = C r a b , then put the values of r, where 0 £ r £ N, for which indicies of a and b become integers. X

Example 5. If the rth term in the expansion of 10 æx 2 ö 4 ç - 2 ÷ contains x , then r is equal to è3 x ø (a) 2 (b) 1 (c) 3 (d) 5

349

8

Sol. (a) Let the general term i.e. (r + 1) th contains x11.

Sol. (c) Clearly, the general term in the binomial expansion 10

Objective Mathematics Vol. 1

x 2 of æç - 2 ö÷ è3 x ø

r

is given by

We have, Tr 10 - r

Tr + 1 =

x C r æç ö÷ è 3ø

10

æ- 2ö ç 2 ÷ èx ø

r

-2 C r ( x3 )12 - r æç 2 ö÷ èx ø = 12C r x36 - 3 r - 2 r (- 1)r 2 r

=

12

=

12

C r (- 1)r 2 r x36 - 5 r

10 - r

=

1 C r æç ö÷ è 3ø

10

(- 2 ) x

to find the coefficient of x , put 36 - 5r = 11

(- 2 )r x10 - 3 r

Þ 5r = 25 Þ r = 5 On substituting the value of r in Eq. (i), we get T6 = 12C 5 (- 1)5 2 5 x11

10 - r

=

1 C r æç ö÷ è 3ø

10

Now, replace r by r - 1, we get the rth term i.e. 11 - r 1 (- 2 )r -1 x13 - 3 r Tr = 10C r -1 æç ö÷ è 3ø

X

Thus, the coefficient of x11 is - 12 ´ 11 ´ 10 ´ 9 ´ 8 12 ´ 32 C 5 (- 1)5 2 5 = 5´ 4´ 3´2 = - 25344 X

Example 6. The number of integral terms in 642 1ö æ 1 the expansion of ç 5 2 + 7 6 ÷ is ç ÷ è ø (a) 106 (b) 108 (c) 103 (d) 109

ç è

+

1ö 76 ÷

÷ ø

(1 + x)p

642

æ 1ö Cr ç52 ÷ ç ÷ è ø

642 - r

642

æ 1ö ç7 6 ÷ ç ÷ è ø

r

The coefficient of x k in the expansion of ( x + a ) n can be obtained by equating the power of x in the general term to k. This will give value of r and on substituting this value in general term we can find the coefficient of x k .

Ø If r is not a positive integer, then there will be no term containing

xvii.

X

X

p +q

C P and

p +q

Cp =

The coefficient of ( r +1)th term in the expansion of (1 + x ) n is n C r . The coefficient of x r in the expansion of (1 + x ) n is n C r .

Example 7. The coefficient of x 11 in the 12 æ 3 2 ö expansion of ç x - ÷ is è x2 ø (a) - 25344 (b) 25344 (c) 23544 (d) - 23544

p +q

p +q

Cq

Cq =

( p + q )! p! q !

Independent term or constant term of a binomial expansion is the term in which exponent of the variable is zero. i.e. the coefficient of x 0 .

Example 9. The ratio of the coefficient of x 15 15 2ö æ to the term independent of x in ç x 2 + ÷ is è xø (a) 12 : 32 (b) 1 : 32 (c) 32 : 12 (d) 32 : 1 Sol. (b) Clearly, the general term in the expansion of 15

æ x2 + 2 ö ç ÷ è xø

x k and hence coefficient of x k will be zero.

xvi.

are

is given by

Obviously, r should be a multiple of 6. 642 Total numbers = = 107, but first term for r = 0 is 6 also integral. Hence, total terms are 107 + 1 = 108

xv.

+q

and

Tr + 1 =

xiv.

Example 8. The coefficient of x p and x q (p and q are positive integers) in the expansion of (1 + x ) p + q are (a) equal (b) equal with opposite signs (c) reciprocal of each other (d) None of the above Sol. (a) Coefficient of xp and xq in the expansion of

Sol. (b) Clearly, the general term in binomial expansion of æ 1 ç52

…(i)

11

r 10 - r - 2 r

Since, the r th term contains the x4 , therefore we have 13 - 3 r = 4 Þ 3r = 9 Þ r = 3

350

+1

is given by 2 C r ( x2 )15 - r æç ö÷ è xø

Tr + 1 =

15

=

15

C r (2 )r x30 - 3 r

r

…(i)

Now, for the coefficient of term containing x15 , 30 - 3r = 15, i.e. r = 5 Therefore, 15 C 5 (2 )5 is the coefficient of x15 [from Eq.(i)] Now, to find the term independent of x, put 30 - 3 r = 0 Thus, 15 C10 210 is the term independent of x [from Eq.(i)] 15

Now, the ratio is

xviii.

15

C52 5

C10 210

=

1 25

=

1 32

In the binomial expansion of ( x + a ) n , the rth term from the end is ( n - r + 2)th term from the beginning.

Example 10. The 4th term from the end in the 9 æ x3 2 ö expansion of ç - ÷ is è 2 x2 ø 670 671 (b) (a) 3 x x3 672 (c) (d) None of these x3

ii.

The above expansion has =

iii.

Sol. (c) Clearly, the 4th term from the end is 9 - 4 + 2 i.e. 7th 3

xix.

xx.

6

Middle term in a binomial expansion If n is an even natural number, then in the binomial æn ö expansion of ( x + a ) n , ç + 1÷ th term is the è2 ø middle term.

Example 11. The middle term (terms) in the 9 æ p xö expansion of ç + ÷ is/are è x pø 126 p 125 p 126x 125x (a) (c) (b) (d) p p x x

+

4

n

5

(x + a ) = å C r x

n -r

a , nÎ N

r =0 n

n! x n -r a r ( n r )! r ! r =0

=

n! s r x a , where s = n - r r + s= n r ! s!

å

16

C16 ( y - z)16

\ Number of terms in ( x + y - z)16 18 ´ 17 = 16 + 3 -1C 3 -1 = 18C 2 = = 153 2

r

=å

C r x16 - r ( y - z)r + .... +

16

Clearly, all the terms are distinct. \ The number of distinct terms = 1 + 2 + 3 + K + 17 17 ´ 18 = 153 = 2 Aliter Number of terms in ( x1 + x2 + K + xm )n = n + m -1C m -1

We know, n

n

Sol. (b) ( x + y - z)16 = 16C 0 x16 + 16C1 x15 ( y - z) + …

p æ xö x 126 x T6 = 9C 5 æç ö÷ ç ÷ = 9C 5 = è x ø è pø p p

n

p + q + r + s= n C 4-1 = n + 3C 3

Example 12. The number of distinct terms in the expansion of ( x + y - z )16 is (a) 136 (b) 153 (c) 16 (d) 17

Multinomial Theorem i.

n! x p yq z r us p! q ! r ! s!

X

p æ xö p 126 p T5 = 9C 4 æç ö÷ ç ÷ = 9C 4 = è x ø è pø x x 4

å

The coefficient of x1 1 × x 2n2 K x nmm in the expansion of n! (x1 + x 2 + K + x m )n is n1 ! n2 ! n3 ! K nm!

have two middle terms which are 5th and 6th terms. These are given by

and

( x + y + z + u) n =

Ø

Sol. (a, b) Since, the power of binomial is odd, therefore we 5

C 2 terms

The number of terms in the above expansion is equal to the number of non-negative integral solution of the equation r1 + r2 + K + rk = n, because each solution of this equation gives a term in the above expansion. The number of such solutions is n + k -1 C k -1 .

the binomial expansion of ( x + a ) n .

X

C 3-1

The general term in the above expansion is n! x1r1 x 2r2 x 3r3 K x krk r1 ! r2 ! r3 !K rk !

If n is odd natural number, then æ n + 1ö æ n + 3ö ç ÷ th and ç ÷ th are the middle terms in è 2 ø è 2 ø

When there are two middle terms in the expansion, then their binomial coefficients are equal.

n + 3-1

8

There are terms in the above expansion. Above result can be generalized in the following form (known as multinomial theorem) ( x1 + x 2 + K + x k ) n n! r r r = å r ! r !K r ! x11 x 22 K x kk k r1 + r2 +K+ rk = n 1 2

9 æ -2 ö = 9C x × 64 ç 2÷ 3 èx ø 8 x12 64 672 ´ 3 = 3 x x

Ø

n +2

n + 4-1

term from the beginning, which is given by æ x3 ö T7 = 9C 6 çç ÷÷ è2 ø 9´ 8´7 = 3´2 ´1

Some another form of binomial expansion n! (x + y + z ) n = å x r ys z t ! ! ! r s t r + s+ t = n

Binomial Theorem

X

X

Example 13. The coefficient of a 8 b 6 c 4 in the expansion of ( a + b + c)18 is (b) 18 C14 (a) 18 C14 × 14C 8 (c)

12

C8

(d) None of these

351

Objective Mathematics Vol. 1

8

Sol. (a) In (a + b + c )18 , general term is given by (18)! (a)r1 (b )r 2 (c )r 3 (r1 )!(r2 )!(r3 )!

Sol. (d) (1 - x + x2 )5 = {1 + x( x - 1)}5 = 5C 0 + 5C1 x( x - 1) + 5C 2 x2 ( x - 1)2 + 5C 3 x3 ( x - 1)3 + K \ The coefficient of x3 = - 2 ×5 C 2 - 5C 3 = - 30

…(i)

\ Coefficient of a8 b 6c 4 is obtained by putting r1 = 8, r2 = 6, r3 = 4 in Eq.(i) (18)! 8 6 4 i.e. a bc 8! × 6!× 4! (18)! 14! 18 Thus, required coefficient = ´ = C14 ´14 C 8 8! × 6! × 4! 14! X

Aliter General term of (1 - x + x2 )5 is 5! (1)r1 (- x)r 2 ( x2 )r 3 (r1 )! (r2 )!(r3 )! \ For coefficient of x3 , r2 + 2 r3 = 3 where r1 + r2 + r3 = 5 Let r3 = 0, then r2 = 3 and r1 = 2 r3 = 1, then r2 = 1 and r1 = 3 If r3 = 2, 3, 4, 5, then r2 is not possible as negative. \ Only two terms contains x5 i.e. (r1 = 2, r2 = 3, r3 = 0) and (r1 = 3, r2 = 1, r3 = 1) Thus, from Eq.(i), 5! 5! = (- 1)3 + (- 1)1 = - 10 - 20 = - 30 2 ! 3! 0! 3!1!1!

Example 14. The coefficient of x 3 y 4 z 2 in the expansion of (2x - 3 y + 4 z ) 9 is (a) 12063600 (b) 13063680 (c) 11063600 (d) None of these Sol. (b) The general term in the expansion of (2 x - 3 y + 4 z)9 9! = × (2 x)a1 (- 3 y)a 2 (4 z)a 3 a1 ! a2 ! a3 ! 9! = × 2 a1 (- 3)a 2 4a 3 × xa1 ya 2 za 3 a1 ! a2 ! a3 ! \ Coefficient of x3 y4 z2 9! = × 2 3 (- 3)4 × 42 [taking a1 = 3, a2 = 4, a3 = 2] 3! 4!2 ! = 1260 ´ 10368 = 13063680

X

Example 15. The coefficient of x 3 in the expansion of (1 - x + x 2 ) 5 is (a) 10 (b) 8 (c) - 50 (d) - 30

…(i)

X

Example 16. The total number of terms in the expansion of ( a + b + c + d ) n , n Î N is n( n + 1)( n + 2) n( n + 1)( n + 2) ( n + 3) (a) (b) 6 6 ( n + 1)( n + 2) ( n + 3) (c) (d) None of these 6 Sol. (c) The number of terms in the expansion of (a + b + c + d )n =

n + 4 -1

C4 - 1 =

n+ 3

C3 =

(n + 1) (n + 2 ) (n + 3) 6

Work Book Exercise 8.1 1 If the second, third and fourth terms in the

expansion of ( x + y )n are 135, 30 and 10 / 3 respectively, then

a n=7 c n=6

b n=5 d None of these

2 The total number of dissimilar terms in the expansion of ( x1 + x2 + K + xn )3 is 3

a n3 c

b

n( n + 1) ( n + 2 ) 6

d

n + 3n 4 n2( n + 1)2 4

c 14

2 10

d 25 2

then a1 equals b 20

c 210

d 110

5 The number of terms which are free from radical signs in the expansion of ( y1/ 5 + x1/ 10 )55 is a 5

b 6

c 7

d 4

6

6 The coefficient of x in the expansion of (1 + x 2 - x 3 )8 is a 80

352

b 84

c 88

51

C5 C6 -

31

21

C6

b d

9

C5 C5 +

30

20

C5

8 If the nineth term in the expansion of 25 x - 1 + 7

a log10 15 c loge 15

4 If (1 + 2 x + 3 x ) = a0 + a1 x + a2 x + ¼ + a20 x 20 , a 10

a c

log

expansion of (3 + 7 x )29 are equal, then r equals b 21

(1 + x )21 + (1 + x )22 + ¼ + (1 + x )30 is

x -1

+ 1) 10 + 3-1/ 8 log 3( 5 [3 3 ] is equal to 180 and x > 1, then x equals

2

3 If the coefficients of rth and (r + 1) th term in the a 15

7 The coefficient of x 5 in the expansion of

d 92

b log 5 15 d None of these

9 The expression ìæ 1 + 4 x + 1 ö 7 æ1 - 4 x + 1 ö 7 ü 1 ï ï ÷÷ ý ÷÷ - çç íç 2 2 4 x + 1 ï çè ø ï è ø î þ

is a polynomial in x of degree a 7 c 4

b 5 d 3

ì æ 1 + 4x + 1ö n æ 1 - 4x + 1ö nü 1 ï ï ÷ -ç ÷ ý 10 If íç 2 2 4x + 1 ï è ø è ø ï î þ 5 = a0 + a1 x + ¼ + a5 x , then n is equal to a 11 c 10

b 9 d 8

In the binomial expansion of ( x + a ) n , the coefficient n C 0 , n C1 , n C 2 , …, n C n are known as the binomial coefficients. For given value of n, (a) If n is even, then greatest coefficient = n C n / 2 (b) If n is odd, then greatest coefficient are n

Cn - 1 2

Ø

●

●

X

n

and

Cn + 1 2

.

Example 17. Find the value of r for which 200 C r is the greatest. Sol. Coefficient of terms in the expansion of ( x + y)200 are C0,

200

C1,

C 2 , …, 200 C 200 ×

200

Since, middle term has greatest coefficient. \ Greatest coefficient = Coefficient of middle term = Coefficient of 101st term = 200C100 Hence, X

200

C r is the greatest, when r = 100

Example 18. The greatest coefficient in the expansion of ( x + y + z + w)15 is 15! 15! (a) (b) 3 3!( 4!) (3!) 3 4! 15! (d) None of these (c) 2!( 4!) 2 Sol. (a) The greatest coefficient is = =

n! (q !)k - r [(q + 1)!]r 15! (3!)4 - 3 [(3 + 1)!]3

[here, n = 15, q = 3, r = 3, k = 4] =

15! 3!(4!)3

Greatest Term Let Tr + 1 and Tr be ( r +1)th and rth terms, respectively in the expansion of ( x + a ) n . Then, Tr + 1 = n C r x n - r a r and Tr = n C r - 1 x n - r + 1 a r - 1 n Tr + 1 Cr x n - r a r \ = n Tr Cr - 1x n - r + 1a r - 1 ( r - 1)!( n - r + 1)! a n! ´ × ( n - r )! r ! n! x n - r +1 a = × r x =

Þ Þ

Þ

Tr

> = = = < 1 íç îè r ø þ x

n +1 n +1 x xö æ -1 > = < Þ > = < ç1 + ÷ è aø r a r n +1 n + 1ö > = < r Þ r > = = < Tr according as

Þ

Binomial coefficient of middle term is the greatest binomial coefficient. The greatest coefficient in the expansion of n! , where q and r are (x1 + x 2 + ¼ + x m )n is m- r (q !) [(q + 1)!]r the quotient and remainder respectively when n is divided by m.

200

Now, Tr + 1 > = < Tr

Tr + 1

Binomial Theorem

8

Greatest Coefficient

æn + 1 ö ÷ …(i) r>=< ç ç x ÷ 1ç + ÷ è aø Now, two cases arise. n +1 is an integer. Case I When x 1+ a n +1 Let = m. Then, from Eq. (i), we have x 1+ a mth and ( m +1)th terms are greatest terms. n +1 Case II When is not an integer. x 1+ a n +1 . Then, from Let m be the integral part of x 1+ a Eq. (i), we have ( m +1)th term is the greatest term. Above discussion suggests the following algorithm to find the greatest term in a binomial expansion.

Algorithm Write Tr + 1 and Tr from the given expansion. Tr + 1 Step II Find Tr Tr + 1 Step III Put >1 Tr Step I

Step IV Solve the inequality in step III for r to get an inequality of the form r < m or r > m. If m is an integer, then mth and ( m +1)th terms are equal in magnitude and these two are the greatest terms. If m is not an integer, then obtain the integral part of m, say k. In this case, ( k +1) th term is the greatest term.

353

Objective Mathematics Vol. 1

8

Shortcut Method

X

To find the greatest term (numerically) in the expansion of (1 + x ) n . | x | ( n + 1) . (a) Calculate | x | +1 (b) If m is integer, then Tm and Tm + 1 are equal and both are greatest term. (c) If m is not integer, there T[ m] + 1 is the greatest term, where [ ] denotes the greatest integral part. n

n

næ

Ø To find greatest term in (x + y) , express (x + y) = x ç1 +

è

Sol. (c) We have, 9

n

yö ÷ . xø

n

æ yö Then, find the greatest term in ç1 + ÷ . è xø X

Example 19. The greatest term (numerically) 1 in the expansion of (3 - 5x )15 , when x = is 5 15 10 15 11 (a) C 3 ´ 3 (b) C 3 ´ 3 (c)

15

C 3 ´ 312

Example 20. The greatest term (numerically) 3 in the expansion of (2 + 3x ) 9 , when x = , is 2 5 ´ 311 5 ´ 313 (a) (b) 2 2 13 7´3 (c) (d) None of these 2

6

9 9 = 2 9 × T6 + 1 = 2 9 ×9 C 6 æç ö÷ = 2 9 ×9 C 3 æç ö÷ è 4ø è 4ø 9 × 8 × 7 312 7 ´ 313 × = 29 × = 1× 2 × 3 212 2

Sol. (c) Let Tr + 1 and Tr denote the (r + 1th ) and rth terms

Þ

Tr

+1

15

=

Tr Þ

Tr

+1

Tr

+1

Tr

+1

Tr

15 - r + 1 (-5 x)r - 1 - 13

C r 315 - r (-5 x)r 16 - r (-5 x)r - 1 - 13

15 - r + 1 æ -5 x ö ç ÷ è 3 ø r

=

16 - r æ 5 1 ö 1 ´ ç - ´ ÷, when x = è 3 3ø 5 r

=

16 - r 1 ´ , numerically 3 r

Tr Þ

Cr

Cr

=

Tr Þ

15

15

[neglecting minus sign] Now, Tr

+1

Tr

> 1 (numerically)

16 - r 1 ´ >1 r 3 Þ 16 > 4r Þ r cos x sin 3 x , then complete set of values of x is æ p ö æ p 3p ö (a) x Î ç 0, ÷ È ç , ÷ è 4ø è 2 4 ø æ p p ö æ 3p ö (b) x Î ç , ÷ È ç , p÷ è 4 2ø è 4 ø p p æ ö (c) x Î ç , ÷ è 4 2ø (d) None of the above

(a)

n=0

42. If

(a)

51. (b) xyz = xy + y (d) xyz = yz + x

y

=

2p ö æ cos çq - ÷ è 3 ø x + y + z is equal to

(a) 1 (c) -1

43. If cos q =

z 2p ö æ cos çq + ÷ è 3 ø

,

then

1æ 2 1ö 1ö ÷, then ç x + 2 ÷ is equal to ø è 2 x x ø

(a) sin 2q (c) tan 2q

(b) cos 2q (d) None of these

44. If - p £ x £ p, - p £ y £ p and cos x + cos y = 2 , then the value of cos ( x - y ) is (a) -1 (c) 1

(b) 0 (d) None of these

and cos x + cos y + cos a = 0 æ x + yö sin x + sin y + sin a = 0, then cot ç ÷ is equal to è 2 ø

45. If

(a) sin a (c) cot a

(b) cos a æx + (d) sin ç è 2

yö ÷ ø

46. If a + b - g = p, then sin 2 a + sin 2 b - sin 2 g is equal to (a) 2sin a sin b cos g (c) 2sin a sin b sin g

2 3

(b) 2cos a cosb cos g (d) None of these

47. If a + b + g = 2p , then a b g a b g + tan + tan = tan tan tan 2 2 2 2 2 2 a b b g g a (b) tan tan + tan tan + tan tan = 1 2 2 2 2 2 2 a b g a b g (c) tan + tan + tan = - tan tan tan 2 2 2 2 2 2 (d) None of the above

p -a 4

(b)

3p -a 4

(c)

3 2

(d)

3 5

(c)

p -a 8

(d)

3p a 8 2

3cos q + cos 3q is equal to 3sin q - sin 3q (a) 1 + cot 2 q (b) cot 4 q

(b) 0 (d) None of these

1æ çx + 2è

(b)

1 1 p and sin b = , where 0 < a, b < , then 7 2 10 2b is equal to

å cos 2n f sin 2n f, then

x = cos q

5 3

50. If tan a =

¥

(a) xyz = xz + y (c) xyz = x + y + z

(d) None of these

tan ( A + B ) 1 49. If sin B = sin ( 2A + B ) , then is equal tan A 5 to

¥ ¥ p 41. For 0 < f < , if x = å cos 2n f, y = å sin 2n f and 2 n=0 n=0

z=

7-2 3

(b)

9

(c) cot 3 q

(d) 2cot q

a +b 2 is equal to 52. If 3sin a = 5sin b, then a -b tan 2 tan

(a) 1

53. If cos 2B =

(b) 2

(c) 3

Targ e t E x e rc is e s

(a) sin (x - y) (c) cos (x + y)

7 x é pù , where x Î ê 0, ú, then tan is 2 2 ë 4û

48. If sin x + cos x =

Trigonometric Functions and Equations

æp ö æp ö æp ö æp ö 39. cos ç - x÷ cos ç - y÷ - sin ç - x÷ sin ç - y÷ is è4 ø è4 ø è4 ø è4 ø equal to

(d) 4

cos ( A + C ) , then cos ( A - C )

(a) tan A , tan B and t an C are in AP (b) tan A , tan B and tan C are in GP (c) tan A , tan B and tan C are in HP (d) None of the above

b 54. If tan x = , then the value of a cos 2x + b sin 2x is a (a) a (c) a + b

(b) a - b (d) b

55. The graph of 2 cos x cos ( x + 2) - cos ( x + 1) is

the

function

(a) a straight line passing through (0, - sin 2 q) with slope 2 (b) a straight line passing through (0, 0) (c) a parabola with vertex (1, - sin 2 1) æp ö (d) a straight line passing through the point ç , - sin 2 1÷ è2 ø and parallel to the X-axis

56. If a cos 2 3q + b cos 4 q = 16cos 6 q + 9cos 2 q is an identity, then (a) a = 1, b = 18 (c) a = 3, b = 24

(b) a = 1, b = 24 (d) a = 4, b = 2

(a) tan

n

57. Let n be an odd integer. If sin n q =

å br sin r q for r=0

every value of q, then (a) b0 = 1, b1 = 3 (c) b0 = - 1, b1 = n

(b) b0 = 0, b1 = n (d) b0 = 0, b1 = n2 - n + 3

467

Objective Mathematics Vol. 1

9

p and x = X cos q + Y sin q, 2 such that y = X sin q - Y cos q x 2 + 2xy + y 2 = aX 2 + bY 2 , where a and b are constants, then 0£q £

58. If

(a) a = - 1, b = - 3 (c) a = 3, b = - 1

p 2 p (d) q = 3

(b) q =

(b) -1

a Î ( 0, p / 2),

60. If

x2 + x +

tan 2 a x2 + x

(a) 2tan a (c) 1

(c) 2

then

(d) 1

the

expression

is always greater than or equal to (b) 2cos a (d) sec2 a

A B C , tan and tan are in HP, then the 2 2 2 A C value of cot cot is 2 2

Ta rg e t E x e rc is e s

(b) 2 (d) 4

62. The expression cos 2 f + cos 2 ( a + f ) - 2cos a cos f cos ( a + f ) is independent of (a) f (c) both a and f

(b) a (d) None of these

63. If cos x - sin a cot b sin x = cos a, then tan a b tan 2 2 a b (c) tan tan 2 2

x is equal 2

(d) None of these

(b) |cos A - cos B | =

468

(a) 7 (c) 9

(b) 8 (d) None of these

70. If f (q ) = sin 4 q + cos 4 q + 1, then range of f (q ) is é 3ù (b) ê1, ú ë 2û (d) None of these

71. If f ( x ) = sin 6 x + cos 6 x + 1, then range of f (q ) is é1 ù (a) ê , 1ú ë4 û é3 ù (c) ê , 1ú ë4 û

é 1 3ù (b) ê , ú ë4 4û (d) None of these

pö pö æ æ 72. The maximum value of sin ç x + ÷ + cos ç x + ÷ in è è 6ø 6ø æ pö the interval ç 0, ÷ is attained at è 2ø p 12

(b)

p 6

(c)

p 3

(d)

p 2

73. The number of integral value of k for which the equation 7cos x + 5 sin x = 2 k + 1 has a solution, is (b) 8

(c) 10

cos 2 q x sin 2 q

(d) 12

x sin 2 q , q Î ( 0, p / 2), then cos 2 q

roots of f ( x ) = 0 are 2 3

(b) I (n + 2) (d) I (n + 2) × I (2)

(d) [ 0, p ]

69. If f (q ) = (sin q + cosec q ) 2 + (cos q + sec q ) 2 , then minimum value of f (q ) is

74. If f ( x ) = cos q x

(d) None of these

(b) [ - p , 0 ]

(b) f (q) Î [ 0, 2 ] (d)

2

66. The equation (cos p - 1) x 2 + (cos p ) x + sin p = 0 in x has real roots. Then, the set of values of p is (a) [ 0, 2p ] é p pù (c) ê - , ú ë 2 2û

equation

(b) a2 + b2 = p2 + q2 (d) None of these

(a) f (q) Î [ 0, 2 ] (c) f (q) Î [ 0, 1] f (q) Î [1, 2 ]

sin 2 q

65. Let I ( n ) = 2cos nx, " n Î N , then I (1) × I ( n + 1) - I ( n ) is equal to (a) I (n + 3) (c) I (n + 1) × I (2)

(a) 1 + b + a2 = p2 - q (c) b + q = a2 + p2 - 2

(a) 4

b a (b) cot tan 2 2

p 64. If A + B = and cos A + cos B = 1, then 3 1 (a) cos ( A - B ) = 3 2 (c) cos ( A - B ) = 3

the

cos x + a cos x + b = 0 and sin x + p sin x + q = 0, then the relation between a, b, p and q is

(a)

to (a) cot

both 2

é3 ù (a) ê , 2ú ë2 û (c) [1, 2 ]

61. In DABC, tan

(a) 1 (c) 3

satisfies

2

68. If f (q ) = | sin q | + |cos q | , q Î R, then

59. If q 1 and q 2 are two values lying in [0, 2p] for which q q tan q = l , then tan 1 tan 2 is equal to 2 2 (a) 0

x=a , b

67. If

(a) 1/ 2, - 1, 0 (c) -1/ 2, 1, - 1

75. If f (q ) =

(b) 1/ 2, - 1 (d) -1/ 2, - 1, 1

1 , then range of f (q ) is 2 - 3cos q

1ù æ (a) ç - ¥ , ú È [1, ¥ ) è 5û æ1 ù (c) (- ¥ , - 1] È ç , ¥ ú è5 û

1ù é (b) ê -1, ú 5û ë (d) None of these

pö æ 76. If f (q ) = 5cos q + 3cos çq + ÷ + 3, then range of è 3ø f (q ) is (a) [ - 5, 11] (c) [ - 2, 10 ]

(b) [ - 3, 9 ] (d) [ - 4 , 10 ]

(b)

1 d 2 - a2 2| a |

(b) zero (d) None of these

p 79. If A ³ 0, B ³ 0, A + B = and y = tan A tan B, then 3 (b) 0 £ y £

1 3

(b) | a | ³ 1 (d) None of these

83. If sin q 1 - sin q 2 = a and cos q 1 + cos q 2 = b, then (b) a2 + b2 £ 4 (d) a2 + b2 £ 2

0£ b £ 3

and

the

equation

x 2 + 4 + 3cos ( ax + b ) = 2x has atleast one solution, then the value of a + b is p 2 (d) None of these

(a) 0

(b)

(c) p

85. If tan 2 x + sec x - a = 0 has atleast one solution, then the complete set of values of a is (a) (- ¥ , 1] (c) [ - 1, 1]

86. sec 2 q =

(b) [1, 1] (d) [ -1, ¥ )

4xy (x + y )2

(a) x + y ¹ 0 (b) x = y, x ¹ 0 (c) x = y (d) x ¹ 0, y ¹ 0

11p 6

89. The number of solutions of tan 2 x - sec10 x + 1 = 0 in ( 0, 10), is (a) 3 (c) 10

the

equation

(b) 8 (d) None of these

(b) 4

(b) 2

7p 2 9p (c) 2

82. If sin x + a sin x + 1 = 0 has no real number solution, then

0 £ a £ 3,

(d) 2np +

(c) 6

(d) 8

(c) 3

(b)

(a)

2

84. If

7p 6

(d) 4

92. If and the equation a, b Î [ 0, p ] x 2 + 4 + 3 sin ( ax + b ) - 2x = 0 has atleast one solution, then the value of ( a + b ) can be

1 2 1 (d) a £ 2

(a) a2 + b2 ³ 4 (c) a2 + b2 ³ 3

(c) 2np +

7p 6

(b) np + (-1)n

(a) 1

(b) a ³

(a) | a | ³ 2 (c) | a | < 2

p 6

91. Total number of solutions of the equation 5p in x Î[ 0, 2p ], is 3x + 2 tan x = 2

a , " x Î R ~ {0}, then x

1 4 1 (c) a £ 4

(a) np ±

(a) 2

(a) | a | ³ | b | (b) | b | = | a | (c) | a | £ | b | (d) None of the above

(a) a ³

(b) x ¹ y and x , y Î R (d) x ¹ y and x ¹ 0, y ¹ 0

90. Total number of solutions of tan x + cot x = 2 cosec x in [ - 2p , 2p ] is

(d) None of these

80. If ( a sin 4 q + b ) ( a + b ) = ab cos 4q , then

81. If sin q = x +

9

, then

88. The most general values of q satisfying the equation (1 + 2sin q ) 2 + ( 3 tan q - 1) 2 = 0 is/are given by

78. The minimum value of the expression sin a + sin b + sin g , where a , b and g are real numbers satisfying a + b + g = p, is

1ù æ (a) y Î ç -¥ , ú È [3, ¥) è 3û 1 (c) y > 3

x2 - y2

(a) x ¹ y (c) x = 0 or y = 0

(d) None of these

(a) positive (c) negative

x2 + y2

is true if and only if

3p 2

(d) None of these

1 93. If sin q, cos q and tan q are in GP, then the general 6 value(s) of q is/are p , n ÎI 3 p (c) 2np + (-1)n , n Î I 3

(a) 2np ±

Targ e t E x e rc is e s

1 d 2 - a2 2| b | 1 (c) d 2 - a2 2| d | (a)

87. If sin q =

Trigonometric Functions and Equations

77. If a sin x + b cos ( x + q ) + b cos ( x - q ) = d], then the minimum value of |cos q | is

p , n ÎI 6 p (d) np + , n Î I 3

(b) 2np ±

1 94. The general values of x for which cos 2x, and sin 2x 2 are in AP, are given by (a) np, np + (c) np +

p 2

p 4

95. The

(b) np, np + (d) np

most general values æ 3p ö tan q + tan ç + q ÷ = 2 are è 4 ø

p , n ÎI 3 p (b) 2np ± , n Î I 6 p (c) 2np ± , n Î I 3 p (d) 2np + (-1)n , n Î I 3

p 4

of

q

satisfying

(a) np ±

469

Objective Mathematics Vol. 1

9

96. The most general solutions of the equation sec x - 1 = ( 2 - 1) tan x are given by p , np 8 (c) 2np, np

p 4 (d) None of these

(a) np +

(a) 1

(b) 2np , 2np +

97. If n tan y = tan x, n Î R + , then the maximum value of sec 2 ( x - y ) is equal to (n + 1)2 (b) n (n + 1)2 (d) 4n

(n + 1)2 (a) 2n (n + 1)2 (c) 2

98. Let a , b be any two positive values of x for which 2 cos x, |cos x | and 1 - 3cos 2 x are in GP. The minimum value of |a - b | is p 3 p (c) 2

Ta rg e t E x e rc is e s

(a)

(b)

p 4

p 4

p 4 (d) None of these

(b) np +

(c) np

100. The

number of solutions x 1 + sin x sin 2 = 0 in [ - p , p ] is 2 (a) 0 (c) 2

of

the

equation

101. The number of solutions of cos x = | 1 + sin x | , 0 £ x £ 3p is (b) 2 (d) None of these

102. The most general solutions of 21 + |cos x | + cos

2

x + |cos x |3 + ¼ ¥

p ,n Î I 3 2p (c) 2np ± ,n Î I 3

(a) np ±

= 4 are given by (b) 2np ±

p ,n Î I 3

(d) None of these

2 np and (cos x ) sin x - 3 sin x + 2 = 1, then all 2 solutions of x is/are given by

103. If x ¹

p (a) 2np + 2 (c) 2np + (-1)n

p (b) (2n + 1) p 2 p 2

(d) None of these

104. If 0 £ x £ 3p, 0 £ y £ 3p and cos x sin y = 1, then the possible number of values of the ordered pair ( x, y ) is 470

(a) 6 (c) 8

(b) 12 (d) 15

(c) 5

106. The number of values sin 2x + cos 4x = 2, is (a) 0

(b) 1

(d) 7

of

x

(c) 2

for

which

(d) infinite

107. If |tan x | £ 1and x Î [ - p , p ] , then the solution set for x is 3p ù é p p ù é 3p ù é (a) ê - p , È - , È , pú 4 úû êë 4 4 úû êë 4 ë û é p p ù é 3p ù (b) ê - , ú È ê , pú ë 4 4û ë 4 û é p pù (c) ê - , ú ë 4 4û (d) None of the above

é pù (a) ê 0, ú ë 6û

é 5p ù (b) ê 0, ú ë 6û

é 5p é p 5p ù ù (c) ê , 2p ú (d) ê , ë 6 ë 6 6 úû û

109. The set of values of x for sin x cos 3 x > cos x sin 3 x, 0 £ x £ 2p, is (a) (0, p)

æ pö (b) ç 0, ÷ è 4ø

æp ö (c) ç , p ÷ è4 ø

(d) None of these

which

110. If x1 and x 2 are two positive values of x for which 2cos x, | cos x | and 3sin 2 x - 2 are in GP. The minimum value of | x1 - x 2 | is

(b) 1 (d) 3

(a) 3 (c) 4

(b) 3

108. If 4 sin 2 x - 8sin x + 3 £ 0, 0 £ x £ 2p, then the solution set for x is

(d) None of these

99. The value(s) of x Î [ -2p , 2p ] such that sin x + i cos x , i = -1 is purely imaginary, is/are 1+ i given by (a) np -

105. If x, y Î [ 0, 2p ] , then total number of ordered pairs ( x, y ) satisfying sin x cos y = 1is

(a)

4p 3

(b)

æ 2ö (c) 2 cos-1 ç ÷ è 3ø

p 3

æ 2ö (d) cos-1 ç ÷ è 3ø

111. If 3sin x + 4 cos ax = 7 has atleast one solution, then a has to be necessarily a/an (a) odd integer (b) even integer (c) rational number (d) irrational number

112. If the equation a1 + a 2 cos 2x + a 3 sin 2 x = 1 is satisfied by every real value of x, then the number of possible values of the triplet ( a1 , a 2 , a 3 ) is (a) 0

(b) 1

(c) 3

(d) infinite

2

113. If sin q - 2 sin q - 1 = 0 is to be satisfied for exactly 4 distinct values of q Î [ 0, np ], n Î N, then the least value of n is (a) 2

(b) 6

(c) 4

(d) 15

114. If 2 tan 2 x - 5sec x is equal to 1 for exactly 7 distinct é np ù values of x Î ê 0, ú, n Î N , then the greatest value ë 2û of n is (a) 6

(b) 12

(c) 13

(d) 15

116. If

(b) np + (-1)n - 1

p , n ÎI 2

(d) None of these

max{5sin q + 3sin (q - a )} = 7, then set of q ÎR

possible values of a is p ü ì (a) í x | x = 2np ± ; n Î I ý 3 þ î 2p ü ì (b) í x | x = 2np ± ; n Î Iý 3 þ î é p 2p ù (c) ê , ë 3 3 úû (d) None of the above

117.

The number 0 £ x £ 4 p is

of

solutions

(a) 8 (c) 2

124. If cos x -

of |cos x | = sin x,

(b) 4 (d) None of these

118. The values of a for which the equation cos 4 x - ( a + 2)cos 2 x - ( a + 3) = 0 possesses a solution, lies in (a) [ - 3, - 2 ] (b) (- 3, - 2) (c) (0, 2)

(d) (0, 3)

119. The least positive non-integral solution of sin p ( x 2 + x ) - sin px 2 = 0 is (a) rational (b) irrational of the form (c) irrational of the form (d) irrational of the form

q Î[ 0, 5p ]

p -1 , where p is an odd integer 4 p+1 , where pis an even integer 4

and

121. Total number of solutions of [sin x] = cos x, where [ ] denotes the greatest integer function in [ 0, 3p ] , is (a) 2 (b) 6 (c) 5 (d) None of the above

3 p x - + a has no solution, where 2 2

a Î R + , then (a) a Î R + 3 p (b) a > 2 3 p ö æ 3 (c) a Î ç 0, + ÷ è 2 3ø (d) None of the above

b (a) tan tan 15° 2 b (b) tan 2 (c) tan15° (d) None of the above

125. If the expression n sin 2 q + 2n cos (q + a ) sin a sin q + cos 2(a + q ) is independent of q, then the value of n is (a) 1 (b) 2 (c) 3 (d) 4

126. For any real value of q ¹ p, the value of the

p

(a) 8 (b) 10 (c) 6 (d) None of the above

122. If 1 - sin x =

cot b sin x 3 x = , then the value of tan is 2 2 2

expression y =

r Î R such that 4 2 2sin q = r - 2r + 3, then the maximum number of values of the pair ( r, q ) is

120. If

1 2 1 (b) | m | ³ 2 (c) | m | ³ 1 (d) | m | £ 1 (a) | m | £

9 Trigonometric Functions and Equations

p (a) np + (-1)n , n Î I 2 (c) 2np , n Î I

123. If sin 2 (q - a ) cos a = cos 2 (q - a )sin a = m sin a cos a , then

cos 2 q - 1 cos 2 q + cos q

is

(a) 1 £ y £ 2 (b) y < 0 and y > 2 (c) - 1 £ y £ 1 (d) y ³ 1

127. Solution of equation [sin x] = [1 + sin x] + [1 - cos x], 0 £ x £ 2p is

Targ e t E x e rc is e s

115. If sin a, 1 and cos 2a are in GP, then a is equal to

3p 2 3p (b) x = 4 (c) no real solution (d) None of the above (a) x =

128. The general solution sin 50 x - cos 50 x = 1 is

of

the

equation

p 2 p (b) 2np + 3 p (c) np + 2 p (d) np + 3 (a) 2np +

129. Number of ordered pairs ( a, x ) satisfying the equation sec 2 ( a + 2) x + a 2 - 1 = 0 , - p < x < p is (a) 2 (b) 1 (c) 3 (d) infinite

471

Objective Mathematics Vol. 1

9

Type 2. More than One Correct Option q (1 + sec q ) (1 + sec 2q )(1 + sec 4q ) ¼ 2 (1 + sec 2n q ) , then

130. If f n (q ) = tan

æpö (a) f2 ç ÷ = 1 è 16 ø æ p ö (c) f4 ç ÷ = 1 è 64 ø

æpö (b) f3 ç ÷ = 1 è 32 ø æ p ö (d) f5 ç ÷ =1 è 128 ø

p and x = X cos q + Y sin q , 2 y = X sin q - Y cos q such that x 2 + 4xy + y 2 = aX 2 + bY 2 , where a and b are constants. Then,

131. Let 0 £ q £

p 4 p (d) q = 3

(b) q =

(a) a = - 1, b = 3 (c) a = 3, b = - 1 2

132.

2

sin x + a - 2 a has a solution, if = cos 2x 1 - tan 2 x

Ta rg e t E x e rc is e s

(b) a ³ 1 (d) a is any real number

133. If cos 3 x sin 2x =

n

139. If y = sin 2 x + cos 4 x , then for all real x (a) the maximum value of y is 2 3 (b) the minimum value of y is 4 (c) y £ 1 1 (d) y ³ 4

å a m sin mx is identity in x. Then,

æ pö (a) k 1 - k 2 > 0, x Î ç 0, ÷ è 4ø æ pö (b) k 1 - k 2 < 0, x Î ç 0, ÷ è 4ø æ pö (c) k 1 + k 2 > 0, x Î ç 0, ÷ è 2ø æ pö (d) k 1 + k 2 < 0, x Î ç 0, ÷ è 4ø

m=1

3 (a) a3 = , a2 = 0 8 1 (c) n = 5, a1 = 4

(b) n = 6, a1= (d) San =

141. If a Î [ - 2p , 2p ] and a a cos + sin = 2 (cos 36° - sin 18° ) , then a value 2 2 of a is

1 2

3 4

134. If 0 £ x, y £ 180° and sin ( x - y ) = cos ( x + y ) = 1/ 2, then x and y are given by (a) x = 45° , y = 15° (c) x = 165° , y = 15°

(b) x = 45° , y = 135° (d) x = 165° , y = 135°

x x 135. If tan = cosec x - sin x, then tan 2 is equal to 2 2 (a) 2 - 5 (c) (9 - 4 5 ) (2 +

136. If y =

5)

(b) 5 - 2 (d) (9 + 4 5 ) (2 - 5 )

1 - sin 4 A + 1 , then one of the values of y is 1 + sin 4 A - 1

(a) - tan A æp ö (c) tan ç + A ÷ è4 ø

(b) cot A æp ö (d) - cot ç + A ÷ è4 ø

137. If x = a 2 cos 2 a + b 2 sin 2 a + a 2 sin 2 a + b 2 cos 2 a 2

2

2

, then x = a + b + 2

2

2

2

p ( a + b ) - p , where p

is equal to

472

(a) sin 2 (a + b ) + p sin (a + b )cos (a + b ) + q cos2 (a + b ) = q p (b) tan (a + b ) = q-1 (c) cos (a + b ) = 1 - q (d) sin (a + b ) = - p

140. If k1 = sin x cos 3 x and k 2 = cos x sin 3 x, then

2

(a) a £ - 1 (c) a = 1/ 2

138. If tan a and tan b are the roots of the equation x 2 + px + q = 0 with p ¹ 0, then

(a) a2 cos2 a + b2 sin 2 a (b) a2 sin 2 a + b2 cos2 a 1 (c) [ a2 + b2 + (a2 - b2 ) cos 2a ] 2 1 (d) [ a2 + b2 - (a2 - b2 )cos 2a ] 2

7p 6 5p (c) 6 (a)

(b)

p 6

(d) -

p 6

142. If cos x = 1- sin 2x , where 0 £ x £ p, then a value of x is (a) p (b) 0 (c) tan -1 2 (d) None of the above

143. If sin q + 3 cos q = 6x - x 2 - 11, 0 £ q £ 4p, x Î R, then holds for (a) no value of x and q (b) one value of x and two values of q (c) two values of x and two values of q (d) two pairs of values of (x , q)

144. If 0 £ x £ 2p and | cos x | £ sin x, then ép pù (a) the set of values of x is ê , ú ë 4 2û (b) the number of solutions that are integral multiples of is three (c) the sum of the largest and the smallest solution is é p p ù é p 3p ù (d) x Î ê , ú È ê , ë 4 2 û ë 2 4 úû

3p 4

p 4

questions, each question contains Statement I (Assertion) and Statement II ( Reason). Each question has 4 choices (a), (b), (c) and (d ) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

145. Statement I If A , B and C are the angles of a triangle such that ÐA is obtuse, then tan B tan C > 1. tan B + tan C Statement II In any triangle, tan A = tan B tan C - 1 q = 1 + sin q + 1 - sin q , 2 p 3p q then lies between 2np + and 2np + . 4 4 2

146. Statement I

Statement II

If

If

2sin

q p 3p q runs from to , then sin > 0. 2 4 4 2

147. Statement I The numbers sin 18° and - sin 54° are equations of such quadratic equations with integer coefficients. Statement II If x = 18°, then 5x = 90° and if y = - 54°, then 5 y = - 270°. 148. Statement I cos 36° > tan 36° Statement II cos 36° > sin 36° 149. Statement I The minimum value of the expression sin a + sin b + sin g , where a , b and g are real numbers such that a + b + g = p, is negative. Statement II

a , b and g are angles of a triangle.

sin x + cosec x = 2 5p ù é pù é possesses unique solution, if x Î ê 0, ú È ê 2p , ú. 2û ë 2û ë

150. Statement I

The

Statement II x +

equation

1 ³ 2, x > 0 equality holds, only x

when x = 1. If (sin x ) 2 + (tan x ) 2 = 0, then x = np,

151. Statement I Statement II

If a 2 + b 2 = 0, then a, b = 0.

Type 4. Linked Comprehension Based Questions Passage I (Q. Nos. 152-154) For each positive real number k, let Ck denotes the circle with centre at origin and radius k units. On a circle Ck , particle a moves k units in the counter-clockwise direction. After completing its motion on Ck , the particle moves onto the circle Ck + g in some well defined manner, where g > 0. The motion of the particle continues in this manner.

152. Let k Î I + and g = 1, particle moves in the radical direction from circle C k to C k + 1 . If particle starts from the point P ( - 1, 0), then (a) it will cross the X-axis again at (3, 0) (b) it will cross the X-axis again at (4, 0) (c) it will cross the positive Y-axis at (0, 4) (d) it will cross the positive Y-axis at (0, 6)

153. Let k Î I + and g Î R, particle moves tangentially from the circle C k to C k + y , such that length of tangent is equal to k unit itself. If particle starts from the point Q (1, 0), then (a) the particle will cross positive X-axis again for 2 2 3 3 (c) k < 3 3 (d) 0 < k < 3 3

1 9

(c)

1 81

(d)

1 243

162. Number of roots of the equation cos 7 x + sin 4 x = 1in the interval [ 0, 2p ] is (a) 0

160. There exist three or more non-similar isosceles triangles for

Ta rg e t E x e rc is e s

(b)

(a) no value of k (b) 0 < k < 3 3 (c) k > 3 (d) k > 0

(b) 1

(c) 2

(d) 4

163. The smallest positive number p for which the equation cos ( p sin x ) = sin ( p cos x ) has a solution in [ a, 2p ] , is p 4 p (b) 3 p (c) 4 2 p (d) 2 2 (a)

Passage IV (Q. Nos. 161-163) Use boundedness for solving trigonometric equations. Consider an equation …(i) sin x + sin y = 2 We know that, sin x £ 1, sin y £ 1 and sin x + sin y £ 2, " x and y .

Type 5. Match the Columns 164. Match the items of Column I with that of Column II. Column I

165. Match the items of Column I with that of Column II. Column I

Column II

A.

The number of pairs ( x, y) satisfying the equation sin x + sin y = sin( x + y) | x | + | y | = 1, is

p.

3

B.

The number of values of x for which f( x ) is valid, where 1 - | x |ö f( x ) = sec -1 æç ÷ , is è 2 ø

q.

8

C.

If x, y Î [0, 2 p ], then total number of ordered pairs ( x, y) satisfying sin x cos y = 1, is

r.

¥

D.

f( x ) = sin x - cos x - kx + b decreases for all values of real values of x when 4 2 k is always greater than

s.

6

A.

If in a D ABC, then sin2 A + sin2 B = sin ( A + B), the triangle must be

p. right angled

B.

If in a D ABC, bc = b 2 + c 2 - 2 bc cos A, then 2 cos A the triangle must be

q. equilateral

If in a D ABC, A B C + tan + tan = 2 2 2 triangle must be

r. isosceles

C.

474

Column II

tan

D.

3, then the

If in a triangle, the sides and the altitudes are in AP, then the triangle must be

s. obtuse angles

Column I

Column II

A. The ratio of greatest value of p. k 2 - cos x + sin2 x to its least value is , 4 then k equals

0

B. The equation cos 2 x + p sin x = 2 p - 7 q. possesses a solution, then number of integral values of p satisfying equation is

12

C. Ifcos a = 3, then value of16 sin a sin 3a is 2 2 4

r.

13

D. If a is a real constant and A, B, C are s. variable angles and

5

Column II p.

1

B. The number of solutions of equation sin-1 (| x 2 - 1|) + cos -1 p (|2 x 2 - 5 |) = is 2

q.

0

C. The number of solutions p x 4 - 2 x 2 sin2 x + 1 = 0 is 2

r.

A. The number of solutions x sin x p + = in [-p, p ], is 2 cos x 4

of

of

3

a 2 - 4 tan A + a tan B D. The number of solutions of x 2 + 2 x + 2 sec 2 px + tan2 px = 0 is

s.

2

167. Match the items of Column I with that of Column II. Column I p.

sin3 x, 0 £ x £ 2 p, is

169. Match the items of Column I with that of Column II.

2

é 3p , 2 p ù È { 0} êë 2 úû

C. |tan x | £ 1and x Î [- p, p ], is

r.

æ 0, p ö ç ÷ è 4ø

D. cos x - sin x ³ 1 0 £ x £ 2 p, is

s.

é p , 5p ù êë 6 6 úû

4 sin x - 8 sin x + 3 £ 0, 0 £ x £ 2 p, is

and

Column I

é - p,- 3p ù êë 4 úû p p È é- , ù êë 4 4 úû 3p È é , pù êë 4 úû

q.

B.

+ a 2 - 4 tanC = 6a, then least value of tan2 A + tan2 B + tan2 C is

Column II

A. sin x cos 3 x > cos x

9 Trigonometric Functions and Equations

Column I

168. Match the items of Column I with that of Column II.

3 , 4 A 5A is sin sin 2 2

Column II p.

0

q.

11 32

then

r.

1 p cot 2 14

tan( A + B) 1 sin(2 A + B), then tan A 5 is equal to

s.

3 2

A. If

cos A =

B. The value of sin

then

the

value

of

p 2p 3p is + sin + sin 7 7 7

C. If tan2 q = 2 tan2 f + 1, cos 2 q + sin2 f equals D. If sin B =

Targ e t E x e rc is e s

166. Match the items of Column I with that of Column II.

170. Match the items of Column I with that of Column II. Column I

Column II

A.

The number of solutions of the equation|tan2 x | = sin x, x Î[0, p ] is

p.

1

B.

p p p p The value of 4 tan - 4 tan3 + 6 tan2 - tan4 - 1 is 16 16 16 16

q.

4

r.

3

s.

2

4 p C. If the equation tan ( p cot x ) = cot ( p tan x ) has a solution in ( 0, p ) - ìí üý, then pmax is equal to p î2 þ 2 2 2 2 2x in [0, 2 p ], if 5cos 2x + 2sin x + 5 2cos x + sin 2x = 126 has a solution, is D. The value of p

Type 6. Single Integer Answer Type Questions 171. The number of real solutions of [ x] 2 - 5[ x] + 6 - sin x = 0, where [ ] denotes the . greatest integer function, is 172. If f ( nq ) =

2sin 2q and cos 2q - cos 4nq

f (q ) + f ( 2q ) + f ( 3q ) + ¼ + f ( nq ) = then the value of m - l is

.

sin lq , sin q sin mq

p 7 = l cos p , then l is ________ . 173. If p 7 1 - tan 2 7 3 - tan 2

174. The number of integral values of a such that sin x cos 3x - a cos x sin 3x = 0 does not have any real p root other than ( 2n + 1) , n Î I for any real value of 2 x, is ______.

475

Objective Mathematics Vol. 1

9

175. The expression n sin 2 q + 2n cos (q + a ) sin a sin q + cos 2 (a + q ) is independent of q, the value of n is ________. 176. For any real value of q ¹ p, the value of the 2

expression y =

cos q - 1 cos 2 q + cos q

, y Î R - [ a, b] , then

the value of ( b - a ) is _______.

182. If the values of x and y satisfy the equation tan 2 ( x + y ) + cot 2 ( x + y ) = 1 - 2x - x 2 , then the value of x 2 - 3x + 2 is ______. 183. If the smallest positive values of x and y satisfying mp p lp and y = , x - y = and cot x + cot y = 2are x = 6 4 12 then l + m is equal to ______. 184. The system of simultaneous equations for x and y, 1

177. The number of solutions of equation [sin x] = [1 + sin x] + [1 - cos x], 0 £ x £ 2p, is _____. p and sin A sin C = l, then the set 3 of all possible values of l Î[ a, b], then ( b - a ) is _______. A B C 179. In a DABC, cot cot cot ³ l 3, then the value 2 2 2 of l is _____. A B C 180. If in a DABC, tan , tan , tan are in HP and 2 2 2 B cot ³ l , then the value of l is _______. 2

æ pö where x, y Î ç 0, ÷ , are 4 sin x + 3 cos y = 11 and è 2ø 1

Ta rg e t E x e rc is e s

178. In a DABC, ÐB =

185. The number of solutions for pö 3p ö ü æ æ sin ç x - ÷ - cos ç x + ÷ = 1ï è è 4ø 4 ø ï in (0, 2p ) is _____ . ý 2cos 7x cos 2x ï >2 ïþ cos 3 + sin 3 186. The integral value of p for which

181. If a and b satisfy the equation 12sin a + 5cos a = 2 b 2 - 8 b + 21, then the value of b 2 - 2 b is _______.

p cos x - 2sin x = 2 + 2 - p has a solution, is _______.

Entrances Gallery JEE Advanced/IIT JEE 1. For x Î ( 0, p ), the equation sin x + 2sin 2x - sin 3x = 3 has [2014] (a) infinitely many solutions (b) three solutions (c) one solution (d) no solution

2. The number of points in ( - ¥ , ¥ ) , for which x 2 - x sin x - cos x = 0, is [2013] (a) 6 (c) 2

(b) 4 (d) 0

be such that f Î[ 0, 2p ] q qö æ 2 2cos q (1 - sin f ) = sin q ç tan + cot ÷ cos f - 1, è 2 2ø

3. Let

q,

tan ( 2p - q ) > 0 and

- 1 < sin q < -

3 . Then, f 2

cannot satisfy

476

5× 16sin x - 2× 3 cos y = 2 . p If x + y = , then l is equal to _______. l

p (a) 0 < f < 2 4p 3p (c) 3 satisfying the 1 1 1 is ______. equation = + æpö æ 2p ö æ 3p ö sin ç ÷ sin ç ÷ sin ç ÷ ènø è nø è nø [2011] 6. Let f ( x ) = x 2 and g ( x ) = sin x, for all x Î R. Then, the set of all x satisfying ( fogogof) ( x ) = (gogof) ( x ), where ( fog) ( x ) = f [ g ( x )] , is [2011] (a) ± np , n Î {0, 1, 2, ¼} (b) ± np , n Î { 1, 2, ¼} p (c) + 2np , n Î {..., - 2, - 1, 0, 1, 2, ¼} 2 (d) 2np , n Î {..., - 2, - 1, 0, 1, 2, ¼}

9. The

( xyz )sin 3q = ( y + 2z )cos 3q + y sin 3q have a solution ( x 0 , y 0 , z 0 ) with y 0 z 0 ¹ 0, is

.

maximum 1

value

of

sin 2 q + 3sin q cos q + 5cos 2 q

the

expression .

is

[2010]

[2010]

JEE Main/AIEEE 1 (sin k x + cos k x ) , where, x Î R and k [2014] k ³ 1. Then, f 4 ( x ) - f 6 ( x ) equals

10. Let f k ( x ) =

(a)

1 6

1 3

(b)

11. The expression

(c)

1 4

(d)

1 12

as

[2013] (b) sec A cosec A + 1 (d) sec A + cosec A

5p 6

(b)

p 6

(c)

p 4

(d)

3p 4

13. If A = sin 2 x + cos 4 x, then for all real x 13 (a) £ A £1 16 3 13 (c) £ A £ 4 16

[2011]

3 £ A £1 4

2p p 4 p p 3p 8p , , , , , 9 4 9 2 4 9 p 5p p 2p 3p 8p (b) , , , , , 4 12 2 3 4 9 2p p p 2p 3p 35p (c) , , , , , 9 4 2 3 4 36 2p p p 2p 3p 8p (d) , , , , , 9 4 2 3 4 9 (a)

0 £ a ,b £ 25 (a) 16

19 (c) 12

16. Let A and B denote A :cos a + cos b + cos g = 0 B :sin a + sin b + sin g = 0

the

(c) px 2

3 (d) x 2 2

(4 + 7 ) (b) 3 (1 - 7 ) (d) 4

(4 - 7 ) (a) 3 (1 + 7 ) (c) 4

a and b be such that p < a - b < 3p. If 21 27 and cos a + cos b = - , then sin a + sin b = 65 65 æa - b ö [2004] the value of cos ç ÷ is è 2 ø 3 130

3 130 6 (d) 65

(b)

6 65

21. If f : R ® S , defined by f ( x ) = sin x - 3 cos x + 1, is [2004] onto, then the interval of S is where [2010]

20 (d) 7

statements

If cos (b - g ) + cos ( g - a ) + cos (a - b ) = (a) A is true and B is false (c) both A and B are true

1 (b) x 2 2

1 19. If 0 < x < p and cos x + sin x = , then tan x is equal to 2

(c)

4 5 and sin (a - b ) = , 5 13

56 (b) 33

x3 8

(a) -

p . Then, tan 2a is equal to 4

(d) 4

20. Let

14. The possible values of q Î ( 0, p ) such that [2011] sin q + sin 4q + sin 7q = 0 are

15. Let cos (a + b ) =

(c) 2

[2006]

(b) 1 £ A £ 2 (d)

(b) 1

18. A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length x. The [2006] maximum area enclosed by the park is (a)

12. In a DPQR, if 3sin P + 4 cos Q = 6 and [2012] 4 sin Q + 3cos P = 1, then ÐR is equal to (a)

[2006] (a) 6

tan A cot A can be written + 1 - cot A 1 - tan A

(a) sin A cos A + 1 (c) tan A + cot A

17. The number of values of x in the interval [ 0, 3p ] satisfying the equation 2sin 2 x + 5sin x - 3 = 0 is

9 Trigonometric Functions and Equations

( y + z ) cos 3q = ( xyz ) sin 3q 2cos 3q 2sin 3q x sin 3q = + y z

æ p pö 8. The number of values of q in the interval ç - , ÷ è 2 2ø np such that q ¹ for n = 0 , ±1, ± 2 and tan q = cot 5q 5 . [2010] as well as sin 2q = cos 4q , is

Targ e t E x e rc is e s

7. The number of all possible values of q, where 0 < q < p, for which the system of equations

3 , then 2 [2009]

(b) A is false and B is true (d) both A and B are false

(b) [ -1, 1] (d) [ -1, 3 ]

(a) [ 0, 3 ] (c) [ 0, 1]

22. sin 2 q =

4xy (x + y )2

is true if and only if

[2002]

(a) x - y ¹ 0 (b) x = - y (c) x + y ¹ 0 (d) x ¹ 0, y ¹ 0

23. The value of (a) 1

1 - tan 2 15° 1 + tan 2 15° (b) 3

is (c) 3 / 2

[2002] (d) 2

477

Objective Mathematics Vol. 1

9

4 24. If tan q = - , then sin q is 3

[2002]

(a) y = 0 (c) y ³ - 2

4 4 but not 5 5 4 4 (b) - or 5 5 4 4 (c) but not 5 5 (d) None of the above (a) -

sin (a + b ) = 1

[2002]

(b) y £ 2 (d) y ³ 2

27. The equation a sin x + b cos x = c, where |c | > a 2 + b 2 has

1 sin (a - b ) = , then 2 [2002] tan (a + 2b ) tan ( 2a + b ) is equal to

25. If

26. If y = sin 2 q + cosec 2 q, q ¹ 0, then

and

(a) 1 (b) -1 (c) 0 (d) None of the above

[2002]

(a) a unique solution (b) infinite number of solutions (c) no solution (d) None of the above

28. If a is a root of 25cos 2 q + 5cos q - 12 = 0, p [2002] < a < p , then sin 2a is equal to 2 (a)

24 25

(b) -

24 25

(c)

13 18

(d) -

13 18

Other Engineering Entrances ¥

29. The sum of the series

æn !pö

å sin çè 720 ÷ø is

[WB JEE 2014]

34. tan 81° - tan 63° - tan 27° + tan 9° equals [EAMCET 2014] (d) 4

Ta rg e t E x e rc is e s

n=1

æ p ö æ p ö æ p ö (a) sin ç ÷ + sin ç ÷ + sin ç ÷ è 180 ø è 360 ø è 540 ø æpö æpö æ p ö æ p ö (b) sin ç ÷ + sin ç ÷ + sin ç ÷ + sin ç ÷ è 6ø è 30 ø è 120 ø è 360 ø æpö æpö æ p ö (c) sin ç ÷ + sin ç ÷ + sin ç ÷ è 6ø è 30 ø è 120 ø æ p ö æ p ö + sin ç ÷ + sin ç ÷ è 360 ø è 720 ø æ p ö æ p ö (d) sin ç ÷ + sin ç ÷ è 180 ø è 360 ø

30. If cos x = tan y, cot y = tan z and cot z = tan x, then sin x is equal to [EAMCET 2014] (a)

5+1 4

(b)

5 -1 4

(c)

5+1 2

(d)

5 -1 2

31. The value 2cot 2 ( p / 6) + 4 tan 2 ( p / 6) - 3cosec ( p / 6) is (a) 2

(b) 4

(c) 4/3

32. Find the value of cos (29 p / 3). (a) 1

(b) 0

(c) 3 / 2

of

[J&K CET 2014] (d) 3/4 [J&K CET 2014] (d) 1 / 2

(a) 6

(b) 0

(c) 2

35. If sin A + cos A = 2, then the value of cos 2 A is (a) 2

(b) 1/2

[RPET 2014] (d) -1

(c) 4

b a and tan are the roots of 8x 2 - 26x + 15 = 0 , 2 2 [Manipal 2014] then cos (a + b ) is equal to

36. If tan

627 726 (c) -1

627 725 (d) None of these

(b) -

(a)

3 3p x 37. If tan x = , p < x < , then the value of cos is 4 2 2 1 (a) 10

38. cos

(b)

3 10

[Karnataka CET 2014] 3 1 (c) (d) 10 10

2p 4p 6p + cos + cos 7 7 7

(a) is equal to zero (c) is a negative number

39. The value of tan

[WB JEE 2014] (b) lies between 0 and 3 (d) lies between 3 and 6

p 2p 4p is + 2 tan + 4 cot 5 5 5

33. If sin q = sin a , then

478

[Karnataka CET 2014] q+ a p q-a (a) is any odd multiple of and is any 2 2 2 multiple of p q+ a q-a is any odd multiple of p and is any (b) 2 2 multiple of p q+ a p q-a (c) is any multiple of and is any even 2 2 2 multiple of p q+ a p q-a (d) is any even multiple of and is any odd 2 2 2 multiple of p

p (a) cot 5

2p (b) cot 5

4p (c) cot 5

[WB JEE 2014] 3p (d) cot 5

40. If x and y are acute angles such that cos x + cos y =

3 2

3 and sin x + sin y = , then sin ( x + y ) equals 4 [Karnataka CET 2014] 2 (a) 5 3 (c) 5

3 (b) 4 4 (d) 5

(b) 5 / 26

(c) 5 / 13

42. Let

n be a positive p p n , then sin + cos = 2n 2n 2 (a) 6 £ n £ 8 (c) 4 £ n < 8

(d)

integer

such

that

[AMU 2014]

(b) 4 < n £ 8 (d) 4 < n < 8

(b) p

(c) 2p

(d) 3p

44. Which one of the following is not correct? [AMU 2013]

(b) 1 (d) None of these

46. The most general solutions of the equation sec 2 x = 2 (1 - tan 2 x ) are given by [BITSAT 2013] p (b) 2np + 4

47. The number of solutions of the equation (1 - cos 2x ) (a) 3

(b) 2

(c) 1

[Manipal 2013] (d) 0

48. If log cos x tan x + log sin x cot x = 0, then the most [UP SEE 2013] general solution of x is 3p , n ÎZ 4 p (c) np - , n Î Z 4

(a) 2np -

(b) 2np +

p , n ÎZ 4

(d) None of these

49. The number of real solutions of the equation x 3 + x 2 + 4x + 2sin x = 0 in 0 £ x £ 2p is [OJEE 2013] (a) 4

(b) 2

(c) 1

(d) 0

50. The general solution of sin 3x + sin x - 3sin 2x [AMU 2013] = cos 3x + cos x - 3cos 2x is np p + for n integer 2 8 p (c) np + for n integer 6

(a)

np p - for n integer 2 8 p (d) np - for n integer 6

(b)

51. If then sin A + sin B + sin C = 3, [MP PET 2012] cos A + cos B + cos C is equal to (a) 3

(b) 2

(c) 1

m m -1 m+1 (c) m -1

(b)

m+1 m

(d) None of these

54. The least value of 3sin 2 q + 4 cos 2 q is (a) 2

(b) 3

(c) 0

[OJEE 2012] (d) 1

55. If sin 2x = 4 cos x, then x is equal to np p ± , n ÎZ 2 4 p (c) np + (-1)n , n Î Z 4

(d) 0

(b) no value (d) 2np ±

56. Number of solutions of tan x + sec x = 2cos x, x Î[ 0, p ] is (a) 0 (c) 2

p , n ÎZ 2

the

equation

[WB JEE 2012]

(b) 1 (d) 3

57. The value of q satisfying cos q + 3 sin q = 2 is [OJEE 2012] (a) 30° (c) 45°

(d) None of these

(cos 2x + cos 2 x ) = 0, 0 £ x £ 2p is

[Karnataka CET 2012] (d) 9

æ A + Bö æB - Aö 53. If cos A = mcos B and cot ç ÷ = l/tan ç ÷, è 2 ø è 2 ø [Manipal 2012] then l is equal to

(a)

p 3 4 45. If 0 < b < a < , cos (a + b ) = and cos (a - b ) = , 4 5 5 [Karnataka CET 2013] then sin 2a is equal to

p (a) np ± 4 p (c) np ± 8

(c) 3

[Karnataka CET 2012]

(a) |sin x | £ 1 (b) - 1 £ cos x £ 1 (c) |sec x | < 1 (d) cosec x ³ 1 or cosec x £ - 1

(a) 0 (c) 2

(b) 0

(a)

43. The sum of the solutions in ( 0, 2p ) of the equation æp ö æp ö 1 cos x cos ç - x÷ cos ç + x÷ = is [EAMCET 2014] è3 ø è3 ø 4 (a) 4p

(a) -18

9

(b) 60° (d) 90°

58. If sin q =

2t

1+ t 2 cos q is equal to

and q lies in second quadrant, then [WB JEE 2011]

1 - t2 (a) 1 + t2 - |1 - t 2 | (c) 1 + t2

t2 - 1 (b) 1 + t2 1 + t2 (d) |1 - t 2 |

59. If sin q = 3sin(q + 2a ), tan(q + a ) + 2 tan a is (a) 3 (d) 0

Targ e t E x e rc is e s

(a) 5 13

1 26

52. If cos a + 2cos b + 3cos g = 0, sin a + 2sin b and then + 3sin g = 0 a + b + g = p, sin 3 a + 8sin 3 b + 27 sin 3g is equal to

Trigonometric Functions and Equations

41. Find the value of cos ( x / 2), if tan x = 5 / 12 and x lies in third quadrant. [J&K CET 2014]

then

the

value

of

[Kerala CEE 2011] (c) -1

(b) 2 (e) 1

60. If a , b , g Î [ 0, p ] and a , b , g are in AP, then sin a - sin g is equal to [Kerala CEE 2011] cos g - cos a (a) sinb (d) cosec b

(b) cosb (e) 2cosb

(c) cotb

61. If A , B and C are the angles of a triangle such that sec ( A - B ), sec A and sec( A + B ) are in AP, then B (a) cosec A = 2cosec 2 B (c) 2cosec2 A = cosec2 2 2

2

[UP SEE 2011] B (b) 2sec A = sec2 2 A (d) 2sec2 B = sec2 2 2

479

Objective Mathematics Vol. 1

9

1 1 and tan q = , " n Î I, then most 2 3 [BITSAT 2011] general value of q is sin q =

62. If

p (a) 2np + ," n Î I 6 p (c) 2np + ," n Î I 3

p (b) 2np + ," n Î I 4 p (d) 2np + ," n Î I 5

63. A value of q satisfying sin 5q - sin 3q + sin q = 0 p such that 0 < q < is [Karnataka CET 2011] 2 (a)

p 12

(b)

p 6

(c)

p 4

(d)

p 2

64. The number of solutions of 2sin x + cos x = 3 is [WB JEE 2011] (b) 2 (d) no solution

(a) 1 (c) infinite

65. If sin q + cos q = 0 and 0 < q < p, then q is equal to p (b) 4

Ta rg e t E x e rc is e s

(a) 0

p (c) 2

[WB JEE 2011] 3p (d) 4

66. The solution of trigonometric equation cos 4 x + sin 4 x = 2cos ( 2x + p )cos ( 2x - p ) is [UP SEE 2011] (a) x = (b) x = (c) x = (d) x =

np 2 np 4 np 2 np 4

æ 1ö ± sin -1 ç ÷ è 5ø 2ö ÷ ø

(d) 4

p cos A cos B 1 p = = - < A < 0 and - < B < 0, 2 3 4 5 2 [BITSAT 2010] then the value of 2sin A + 4 sin B is

68. If

(a) 4

69. The value of (a) 0 (c) 3

(b) -2

(c) -4

cot 54° tan 20° is + tan 36° cot 70°

(d) 0 [VITEEE 2010]

(b) 2 (d) 1

70. The value of tan 40° + tan 20° + 3 tan 20° tan 40° [BITSAT 2010] is (a) 12

480

(c) 2

(b) 2 (d) 2

4 12 æ pö and cos (a + b ) = - , 72. If a , b Î ç 0, ÷, sin a = è 2ø 5 13 then sin b is equal to [Kerala CEE 2010] 63 65 3 (c) 5 8 (e) 65

61 65 5 (d) 13 (b)

(a)

b , a > b > 0 and a a-b is equal to a+b

tan a =

73. If

a+b a-b 2 sin a cos 2a 2 sin a (c) sin 2a 2 tan a (e) cos 2a

1 3 (d) 3 (b)

p 0 cos x sin 3 x ⇒ sin x cos x(cos 2 x − sin 2 x ) > 0 ⇒ sin x cos x ⋅ cos 2 x > 0 ⇒ cos x cos 2 x > 0 –

+ π — 4

0

3π — 2

 x ∈  0, 

∴

41. We have, x=

–

+ π — 4

π

π  π 3π   ∪ ,   2 4  4

∞

∑ cos 2n φ = 1 + cos 2 φ + cos 4 φ + ...

n=0

Targ e t E x e rc is e s

π 2

35. Since, α + β = , then

1 1 = 1 − cos 2 φ sin 2 φ 1 1 Similarly, y = = 2 1 − sin φ cos 2 φ 1 and z= 1 − sin 2 φ cos 2 φ 1 xy = = ⇒ xyz = xy + z 1 1 xy − 1 1− ⋅ x y 1 1 1 Also, x + y = + = = xy sin 2 φ cos 2 φ sin 2 φ cos 2 φ Thus, xyz = xy + z = x + y + z x y z 42. We have, = = 2π  2π   cos θ  cos θ −   cos θ +   3 3 =

Therefore, each ratio x+ y+ z = 2π  2π    cos θ + cos θ −  + cos θ +    3 3 x+ y+ z x+ y+ z = =0 = 2π 0 cos θ + 2 cos θ cos 3

485

Objective Mathematics Vol. 1

9

1 1 43. Given that, cos θ =  x +  2

⇒

x+

48. sin x + cos x =

x

1 = 2 cos θ x 2

∴

1  1 = x +  −2 x x2  = (2 cos θ )2 − 2 = 4 cos 2 θ − 2

[from Eq. (i)] = 2 cos 2θ 1 2 1 1  x + 2  = × 2 cos 2θ = cos 2θ 2 x  2

44. Maximum value of cos θ = 1

Ta rg e t E x e rc is e s

...(i) ...(ii)

From Eqs. (i) and (ii), we get  x − y  x + y 2 cos    cos   2  cos α  2  =  x − y  x + y sin α 2 sin   cos    2   2   x + y cot  ⇒  = cot α  2 

46. We have, α + β − γ = π Now, sin 2 α + sin 2 β − sin 2 γ = sin α + sin (β − γ )sin (β + γ ) 2

= sin 2 α + sin (π − α )sin (β + γ ) [Qα + β − γ = π ] = sin 2 α + sin α sin (β + γ ) = sin α [sin α + sin (β + γ )] = sin α [sin { π − (β − γ )} + sin (β + γ )] = sin α [sin (β − γ ) + sin (β + γ )] = sin α [2 sin β cos γ ] = 2 sin α sin β cos γ

47. We have, α + β + γ = 2 π ⇒ ⇒ ⇒

⇒ ⇒

α β γ + + =π 2 2 2 α β γ + =π− 2 2 2 γ   α β tan  +  = tan  π −    2 2 2 α β tan + tan 2 2 = − tan γ α β 2 1 − tan tan 2 2 α β γ α β γ tan + tan + tan = tan tan tan 2 2 2 2 2 2

1 − tan 2

⇒

tan

1 x 4 ± 16 − 4(7 − 4) x π as = = < 2 2 8 7 +2 2( 7 + 2 )

⇒

tan

x 7 −2 = 2 3

sin(2 A + B) =5 sin B sin(2 A + B) + sin B 6 ⇒ = cos(2 A + B) − sin B 4 sin( A + B)cos A 3 = ⇒ cos( A + B)sin A 2 tan( A + B) 3 ⇒ = tan A 2 1 1 50. If tan α = and sin β = 7 10 1 3 3 sin 2β = 2 sin β cos β = 2 ⋅ ⋅ = 10 10 5 1 3 + tan α + tan 2β 25 tan(α + 2β ) = = 7 4 = =1 1 − tan α tan 2β 1 − 1 ⋅ 3 25 7 4 π ∴ 2β = − α 4 1 5

45. Given equations may be written as cos x + cos y = − cos α and sin x + sin y = − sin α  x − y  x + y ⇒ 2 cos   = − cos α  cos   2   2   x − y  x + y and 2 sin   = − sin α  cos   2   2 

x 2

49. sin B = sin(2 A + B) ⇒

So, the equation can have solution only when cos x = 1, cos y = 1 ⇒ x = 0, y = 0 ∴ cos( x – y ) = cos 0 = 1

486

x 2 = 7 + ⇒ 2 2 x 2 x 1 + tan 1 + tan 2 2 x x ⇒ ( 7 + 2 ) tan 2 − 4 tan + ( 7 − 2 ) = 0 2 2 2 tan

...(i)

We know that, x2 +

7 2

51.

3 cos θ + cos 3 θ 4 cos 3 θ = = cot 3 θ 3 sin θ − sin 3 θ 4 sin 3 θ sin α 5 = sin β 3  α + β tan    2  sin α + sin β 8 =4 = ⇒ ⇒  α − β sin α − sin β 2 tan    2 

52. 3 sin α = 5 sin β ⇒

cos( A + C ) cos( A − C ) 1 − cos 2 B cos( A − C ) − cos( A + C ) = 1 + cos 2 B cos( A − C ) + cos( A + C )

53. cos 2B = ⇒ ⇒ ⇒

2 sin 2 B 2 sin A sin C = 2 cos 2 B 2 cos A cos C tan A tan C = tan 2 B

Hence, tan A, tan B, tan C are in GP.

54. a cos 2 x + b sin 2 x = a

1 − tan 2 x 2 tan x +b 1 + tan 2 x 1 + tan 2 x

b2 2b 2  a2 − b2  2 ab2 a = a⋅ + b ⋅ a 2 = a 2  + 2 2 2 b b  a + b  a + b2 1+ 2 1+ 2 a a  a  2 = 2  ( a − b2 + 2 b2 ) = a  a + b2  1−

parallel to X-axis and clearly passes through the point π 2   , − sin 1 .  2

56. From the given identity, we have α [4 cos 3 θ − 3 cos θ ]2 + β cos 4 θ = 16 cos 6 θ + 9 cos 2 θ ⇒

16 α cos 6 θ + (β − 24 α ) cos 4 θ + 9 α cos 2 θ = 16 cos 6 θ + 9 cos 2 θ

⇒

α = 1and β − 24 α = 0 ⇒ α = 1 and β = 24

57. Given that, sin nθ =

n

∑ br sinr θ = b0 + b1 sin θ + b2 sin2 θ

r =0

+ K + bn sin n θ

Putting θ = 0 in Eq. (i), we get 0 = b0

...(i)

n

Again, Eq. (i) can be written as sin nθ =

∑ br sinr θ

r =0

sin nθ = ∑ br sin r − 1 θ sin θ r =1 n

⇒

Taking limit as θ → 0, we get sin nθ lim = b1 θ → 0 sin θ  sin nθ   θ  lim n  ⇒  = b1  θ → 0  nθ   sin θ  ⇒ ⇒

62. The given expression = cos 2 φ + cos 2 (a + φ ) − [cos(a + φ ) = cos 2 φ − cos(a + φ )cos(a − φ )

2

2

which is independent of φ.

63. cos x − sin α cot β sin x = cos α

2

xy = ( X 2 − Y 2 )sin θ ⋅ cos θ − XY(cos 2 θ − sin 2 θ ) x 2 + 4 xy + y 2 = X 2 + Y 2 + 2( X 2 − Y 2 )sin 2θ − 2 XY cos 2θ = (1 + 2 sin 2θ )X + (1 − 2 sin 2θ )Y − 2 cos 2θ ⋅ XY 2

2

According to the question, a = 1 + 2 sin 2θ, b = 1 − 2 sin 2θ, cos 2θ = 0 ⇒

+ cos(a − φ )]cos(a + φ )

9

= cos 2 φ − (cos 2 φ − sin 2 a) = sin 2 a

n = b1 b0 = 0, b1 = n

58. x + y = X + Y , 2

tan 2 α  tan 2 α 1 2 x2 + x × ≥  x + x + 2 2 x + x  x2 + x  =|tan α| 2 1 1 61. = + B A C tan tan tan 2 2 2 A C B C B A 2 tan tan = tan tan + tan tan ⇒ 2 2 2 2 2 2 A C B A C 2 tan tan = tan  tan + tan  ⇒  2 2 2 2 2 A C A C  2 tan tan tan + tan  B 2 2 2 2  ⇒ = tan  C A C A 2  1 − tan tan 1 − tan tan   2 2 2 2 B  A C = tan tan  +   2 2 2 B  π B = tan tan  −   2 2 2 A C 2 tan tan 2 2 =1 ⇒ A C 1 − tan tan 2 2 A C A C 2 tan tan = 1 − tan tan ⇒ 2 2 2 2 A C 1 tan tan = ⇒ 2 2 3 A C cot cot = 3 ⇒ 2 2

Trigonometric Functions and Equations

60. Since, AM ≥ GM

1 = [cos(2 x + 2 ) + cos 2 ] − cos 2 ( x + 1) 2 1 = [2 cos 2 ( x + 1) − 1 + cos 2 ] − cos 2 ( x + 1) 2 1 1 = − (1 − cos 2 ) = − (2 sin 2 1) = − sin 2 1 2 2 This shows that y = − sin 2 1 is a straight line which is

Targ e t E x e rc is e s

55. Let y = cos x cos( x + 2 ) − cos 2 ( x + 1)

cos 2θ = 0 π θ = , then 4

π π a = 1 + 2 sin , b = 1 − 2 sin 2 2 ∴ a = 3, b = − 1 2 tan θ / 2 59. tan θ = λ ⇒ =λ 1 − tan 2 θ / 2 θ θ λ tan 2 + 2 tan − λ = 0 ⇒ 2 2 θ1 θ ⇒ tan ⋅ tan 2 = − 1 2 2

⇒ ⇒

sin β cos x − sin α cos β sin x = cos α sin β x x  sin β 1 − tan 2  − sin α cos β ⋅ 2 tan  2 2

x  = cos α sin β 1 + tan 2   2 x x ⇒ tan 2 (− sin β − cos α sin β ) − sin α cos β ⋅ 2 tan 2 2 + sin β(1 − cos α ) = 0

⇒

tan

x = 2

4 [sin 2 α cos 2 β + sin 2 β (1 + cos α )(1 − cos α )] 2 sin β(1 + cos α )

− 2 sin α cos β ±

− sin α cos β ± sin 2 α (sin 2 β + cos 2 β ) sin β(1 + cos α ) − sin α cos β ± sin α = sin β(1 + cos α ) sin α (− cos β ± 1) = sin β(1 + cos α ) β α α β = tan tan or − tan cot 2 2 2 2

=

487

Objective Mathematics Vol. 1

9

A + B 3  A − B  A − B ⋅ cos   = 1 or 2 ⋅  ⋅ cos   =1  2   2   2  2 1  A − B ∴ cos   =  2  3 2 1  A − B ∴ cos( A − B) = 2 cos 2   − 1= − 1= −  2  3 3

64. 2 cos 

 A − B  A + B |cos A − cos B| = 2 sin    ⋅ sin   2   2  = 2⋅

1 1 2 1− = 2 3 3

− 74 ≤ 7 cos x + 5 sin x ≤ 74 ∴ So, − 74 < 2 k + 1 < 74 Therefore, 2 k + 1 = ± 8, ± 7 ± 6, ..., ± 1, 0 So, k = − 4, ± 3, ± 2, ± 1, 0 i.e. 8 values of k.

= 4 cos x cos(n + 1)x − 2 cos nx = 2[2 cos(n + 3)x ⋅ cos x − cos nx ] = 2[cos(n + 2 )x + cos nx − cos nx ] = 2 cos(n + 2 )x = I(n + 2 )

1 cos 2 θ x 2 2 74. f ( x ) = (sin θ + cos θ + x ) 1 x sin 2 θ 2 1 sin θ cos 2 θ

66. D ≥ 0 ⇒ cos 2 p − 4 sin p ⋅ (cos p − 1) ≥ 0 cos 2 p + 4 sin p(1 − cos p) ≥ 0

As,

cos p ≥ 0 and 1 − cos p ≥ 0

6 6 5π  π π     = 2 sin  x + +   = 2 sin  x +  ≤ 2    6 4  12   5π π π Equality hods, when x + = i.e. x = 12 2 12 Therefore, maximum value of given expression is π attained at x = . 12

73. Since,− a2 + b2 ≤ a sin x + b cos x ≤ a2 + b2

65. I(n ) = 2 cos nx ⇒ I(1)⋅ I(n + 1) − I(n )

or

π π 72. sin  x +  + cos  x +     

...(i)

= ( x + 1) 0

Eq. (i) will hold for all sin p ≥ 0. So, p ∈[0, π ]

67. cos α + cos β = − a, cos α ⋅ cos β = b

Ta rg e t E x e rc is e s

and sin α + sin β = − p, sin α ⋅ sin β = q ∴

a2 + p2 = 2 + cos(α − β )

and ∴

b + q = cos(α − β ) a2 + p2 = 2 + b + q

68. f(θ ) = |sin θ| + |cos θ|, ∀θ ∈ R. Clearly, f(θ ) > 0. Also,

f 2 (θ ) = sin 2 θ + cos 2 θ + |2 sin θ ⋅ cos θ|

⇒

= 1 + |sin 2θ| 0 ≤ |sin 2θ| ≤ 1 1 ≤ f 2 (θ ) ≤ 2

⇒

1 ≤ f(θ ) ≤ 2 2

= sin 2 θ + cos 2 θ + cosec 2θ + sec 2 θ + 4 1 1 = 5+ + sin 2 θ cos 2 θ 4 = 5+ 2 4 sin θ cos 2 θ 4 = 5+ ≥9 sin 2 2θ

70. f(θ ) = sin 4 θ + cos 4 θ + 1 = (sin 2 θ + cos 2 θ )2 − 2 sin 2 θ ⋅ cos 2 θ + 1 1 = 2 − sin 2 2θ 2 3  ⇒ f(θ ) ∈ , 2 2 

71. f ( x ) = cos x + sin x 6

6

= (cos 2 x + sin 2 x )(sin 4 x + cos 4 x − cos 2 x ⋅ sin 2 x )

488

x

x − cos 2 θ

sin 2 θ − x

0 sin 2 θ − cos 2 θ cos 2 θ − x = ( x + 1)[( x − cos 2 θ )(cos 2 θ − x ) − (sin 2 θ − cos 2 θ )(sin 2 θ − x )] 2 4 = ( x + 1)[− x − cos θ + 2 x cos 2 θ − x cos 2 θ + x sin 2 θ − sin 4 θ + sin 2 θ cos 2 θ] = (−1/ 2 )( x + 1)[( x − sin 2 θ )2 + ( x − cos 2 θ )2 + (sin 2 θ − cos 2 θ )2 ] So f ( x ) = 0, if x = − 1 or x = sin 2 θ = cos 2 θ π If sin 2 θ = cos 2 θ, then θ = ⇒ x = 1/ 2 4 Hence x = − 1, 1/ 2

75. 3 ≥ 3 cos θ ≥ − 3 ⇒ − 3 ≤ − 3 cos θ ≤ 3

69. f(θ ) = (sin θ + cosecθ ) + (cos θ + sec θ ) 2

cos 2 θ

1

2

= [(sin 2 x + cos 2 x )2 − 3 sin 2 x ⋅ cos 2 x ] 3 = 1 − sin 2 2 x 4 1  ⇒ f( x) ∈ , 1  4 

1 , − 1 ≤ 2 − 3 cos θ ≤ 5 and ≠ 0 2 − 3 cos θ For − 1 ≤ 2 − 3 cos θ < 0, − 1 ≥ f(θ ) > − ∞ 1 For 0 < 2 − 3 cos θ ≤ 5, ∞ > f(θ ) ≥ 5 1  Thus, the range of f(θ ) is (− ∞, − 1] ∪  , ∞ .  5  π 76. Q f(θ ) = 5 cos θ + 3 cos θ +  + 3  3 3 3 3 = 5 cos θ + cos θ − sin θ + 3 2 2 13 3 3 = cos θ − sin θ + 3 2 2 For f(θ ) =

As, − a2 + b2 ≤ a sin θ + b cos θ ≤ a2 + b2 Thus, range of f(θ ) is [− 4, 10 ].

77. a sin x + b cos( x + θ ) + b cos( x − θ ) = d ⇒ a sin x + 2 b ⋅ cos x ⋅ cos θ = d ⇒ ⇒ ⇒

|d| ≤ a2 + 4 b2 ⋅ cos 2 θ d 2 − a2 ≤ cos 2 θ 4 b2 |cos θ| ≥

d 2 − a2 2| b|

As tan A is real, D ≥ 0 ⇒ 3 ( y − 1)2 − 4 y ≥ 0 1 ∴ 3 y 2 − 10 y + 3 ≥ 0 ⇒ y ≤ or y ≥ 3 3

⇒

2

1 1 5  sec x +  = 1 + a + = + a   2 4 4 Now, sec x ≥ 1or ≤ − 1 1 3 1 or ≤ − sec x + ≥ ⇒ 2 2 2 2 5 1 1 1  ⇒ + a≥ ⇒ sec x +  ≥  4 4 2 4 ⇒ a ≥ − 1, a ∈ [− 1, ∞ ) ⇒

86. Since, the value of sec 2 θ is always ≥ 1. ∴

4 xy ≥ 1 ⇒ ( x + y )2 − 4 xy ≤ 0 ( x + y )2

⇒

( x − y )2 ≤ 0 ⇒ ( x − y )2 = 0

80. We have, (a sin 4 θ + b)(a + b) = ab cos 4 θ ⇒ a sin θ + ab sin θ + ba + b = ab cos θ

As, ( x − y )2 can never be negative.

⇒ a sin θ + ab + b = ab (cos θ − sin θ )

⇒

2

2

4

4

4

2

2

4

4

4

⇒ a2 sin 4 θ + ab + b2 = ab (cos 2 θ + sin 2 θ ) × (cos 2 θ − sin 2 θ ) [Q a2 − b2 = (a + b) (a − b)] ⇒ a sin θ + ab + b = ab cos 2θ 2

4

2

[Q cos 2θ = cos 2 θ − sin 2 θ ] ⇒ a sin θ + ab + b = ab (1 − 2 sin 2 θ ) 2

4

2

x=y 4 xy Here, the fraction is not defined for ( x + y )2

x+ y=0 ∴ x+ y≠0 ⇒ x+ x≠0 ⇒ 2 x ≠ 0 or x ≠ 0 Hence, x = y, x ≠ 0

87. sin θ =

⇒ a2 sin 4 θ + ab + b2 = ab − 2 ab sin 2 θ ⇒ a2 sin 4 θ + 2 ab sin 2 θ + b2 = 0

⇒

−b b ≤ 1⇒ ≤1 ⇒ a a

−b a

From

b ≤ a

sin 2 θ − 4a ≥ 0 ⇒ 4a ≤ sin 2 θ ≤ 1 ⇒

82. sin 2 x + a sin x + 1 = 0 Let t = sin x, then we have ⇒ t 2 + at + 1 = 0 Clearly, t = 0 does not satisfy this equation. 1 Dividing throughout by t, we get t + = − a t 1 | a| = t + ≥ 2 ⇒ | a| ≥ 2 ⇒ t Thus, for no real solution,| a| < 2.

83. sin θ1 − sin θ 2 = a, cos θ1 + cos θ 2 = b ⇒

a2 + b2 = 2 + 2 cos(θ1 + θ 2 )

⇒

0 ≤ a2 + b2 ≤ 4

84. x 2 − 2 x + 4 = − 3 cos(ax + b) ( x − 1)2 + 3 = − 3 cos(ax + b)

As, − 1 ≤ cos(ax + b) ≤ 1 and ( x − 1)2 ≥ 0, the above is possible only, if cos(ax + b) = − 1, when x = 1 So, a + b = π, 3π, 5π, etc., and 3π > 6

a≤

1 4

x2 + y2 ≤1 x2 − y2

y 2 = 0 or

x + y Also, − 1 ≤ 2 x − y2

As, D ≥ 0 for real roots.

x2 + y2 ≤1 x2 − y2

x2 + y2 2 y2 ≤ 1, we get 2 ≤0 2 2 x − y2 x −y 2

81. x − x sin θ + a = 0

⇒

⇒ − 1≤

⇒

2

⇒

x2 + y2 x2 − y2

⇒

⇒ (a sin 2 θ + b)2 = 0 ⇒ a sin 2 θ + b = 0 ⇒ sin 2 θ =

9

sec 2 x + sec x − 1 − a = 0

Trigonometric Functions and Equations

π π , β = − and γ = 2 π 2 2 ∴ sin α + sin β + sin γ = − 1 − 1 + 0 = − 2 But sin α + sin β + sin γ ≥ − 3 for any α , β, γ. Hence, the minimum value of sin α + sin β + sin γ is negative. π 79. y = tan A ⋅ tan  − A  3 2 tan A + 3( y − 1) tan A + y = 0 ⇒ Taking α = −

85. tan 2 x + sec x − a = 0

⇒

2

x2 − y2 ≤ 0

⇒

2 x2 ≥0 2 x − y2

x 2 = 0 or

x2 − y2 > 0

...(i)

Targ e t E x e rc is e s

78. Given that, α + β + γ = π

...(ii)

From Eqs. (i) and (ii), we get x 2 = 0 and y 2 = 0 Thus, x = 0 or y = 0 is one only possibility. 1 1 88. Here, sin θ = − and tan θ = 2 3 a2 + b2 = 0 ⇒ a = b = 0

As,

1  π ⇒ θ = mπ + (− 1)n  −   6 2 1 π tan θ = ⇒ θ = mπ + 6 3

sin θ = − and

For common values, m must be odd. m = 2n + 1 ⇒

i.e.

θ = 2 nπ +

7π 6

89. sec 2 x − sec10 x = 0 ⇒

sec 2 x(1 − sec 8 x ) = 0

⇒

1 − sec 8 x = 0 as sec 2 x ≥ 1

⇒ sec x = 1 ⇒ cos 8 x = 1 ⇒ cos x = ± 1 8

⇒ x = nπ, n ∈ I ⇒ x = { π , 2 π , 3π } ∈ (0, 10 ) So, the required number is 3.

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Objective Mathematics Vol. 1

9

sin x cos x 2 + = cos x sin x sin x 1 2 = ⇒ sin x ⋅ cos x sin x 1 π 4π 5π ⇒ cos x = ⇒ x =± ,± ,± 2 3 3 3 Thus, there are six solutions. 5π 5π 3 x 91. 3 x + 2 tan x = − ⇒ tan x = 2 4 2 5π 3 x and y = tan x, meet exactly three Graphs of y = − 4 2 times in [0, 2π ].

90. tan x + cot x = 2 cosec x ⇒

97. tan x = n tan y, cos( x − y ) = cos x cos y + sin x sin y ⇒ cos( x − y ) = cos x cos y(1 + tan x ⋅ tan y ) = cos x cos y(1 + n tan 2 y ) ⇒ sec 2 ( x − y ) = =

π

3π 2

2π

X

Y'

Thus, there are 3 solutions.

Ta rg e t E x e rc is e s

92. x 2 + 4 − 2 x + 3 sin (ax + b) = 0 ( x − 1)2 + 3 + 3 sin (ax + b) = 0 ⇒ ⇒

x = 1 and sin (ax + b) = − 1 3π 2

sin (a + b) = − 1 ⇒ a + b = 1 6

93. cos 2 θ = sin θ ⋅ tan θ or 6 cos 3 θ + cos 2 θ − 1 = 0 1 satisfies the equation (by trial) 2 (2 cos θ − 1) (3 cos 2 θ + 2 cos θ + 1) = 0 1 cos θ = [other values of cos θ are imaginary] 2 π θ = 2 nπ ± , n ∈ I 3

As, cos θ =

⇒ So,

94. 1 = cos 2 x + sin 2 x

π 1  sin 2 x +  =   4 2 π π n π ⇒ 2 x + = nπ + (− 1) ⇒ x = nπ , nπ + 4 4 4 − 1 + tan θ 95. Here, tan θ + =2 1 + tan θ ⇒ tan θ + tan 2 θ − 1 + tan θ = 2 + 2 tan θ π ⇒ tan 2 θ = 3 ⇒ θ = nπ ± , n ∈ I 3 ⇒

96. 1 − cos x = ( 2 − 1) sin x ⇒ ∴ ⇒

490

∴

x x  sin − ( 2 − 1) cos  = 0 2  2 x x 45° sin = 0 or tan = 2 − 1 = tan 2 2 2 x x π = nπ or = nπ + 2 2 8 π x = 2 nπ , 2 nπ + 4

sin

x 2

 1 + n tan 2 y   ≥ n tan 2 y  2  

[Q AM ≥ GM]

1 tan 2 y ≤ 2 2 4n (1 + n tan y )

(n − 1)2 (n + 1)2 = 4n 4n 2 2 98. cos x = 2 cos x ⋅ (1 − 3 cos x ) ⇒ cos x { 6 cos 2 x + cos x − 2} = 0 1 2 ∴ cos x = 0, , − 2 3 π π The two smallest positive values of x are and . 3 2 π |α − β| = ⇒ 6 sin x + i cos x (1 − i ) (sin x + i cos x ) 99. = (1 + i ) (1 − i ) 1+ i sin x + cos x + i (cos x − sin x ) = 2 which is purely imaginary. ⇒ sin x + cos x = 0 ⇒ tan x = − 1 π ∴ x = nπ − 4 2 x 100. Since, 1 + sin x sin =0 2  1 − cos x  ∴ 1 + sin x   =0   2 ⇒

π 2

(n − 1)2 tan 2 y (1 + n tan 2 y )2 2

Now, ⇒

0

(1 + n 2 tan 2 y )(1 + tan 2 y ) (1 + n tan 2 y )2

= 1+

Y

X'

sec 2 x sec 2 y (1 + tan 2 x )(1 + tan 2 y ) = (1 + n tan 2 y )2 (1 + n tan 2 y )2

sec 2 ( x − y ) ≤ 1 +

⇒ 2 + sin x − sin x cos x = 0 ⇒ sin 2 x − 2 sin x = 4 which is not possible for any x in [−π , π ].

101. Clearly, 1 + sin x ≥ 0 So, the equation is

π 1  cos x − sin x = 1 ⇒ cos  x +  =  4 2 π π 7 π 9π 15π 3π 7π , ... , , ⇒ x = 0, , 2 π, ⇒ x+ = , 4 4 4 4 4 2 2 3π As 0 ≤ x ≤ 3π ⇒ x = 0, , 2π 2 Thus, there are 3 solutions. 1

1 1 = 2 ⇒|cos x | = 1 − |cos x | 2 1 cos x = ± 2 π π  x = 2 nπ ± , 2 nπ ±  π −   3 3 π π π = 2 nπ ± , (2 n ± 1) π m = nπ ± 3 3 3

102. 21 − |cos x | = 2 2 ⇒ ⇒ ∴

nπ , we get cos x ≠ 0, 1, − 1 2 2 sin x − 3 sin x + 2 = 0

108. Here, (2 sin x − 1) (2 sin x − 3) ≤ 0. But 2 sin x − 3 is

103. As, x ≠

always negative. 1

⇒

sin x = 1, 2 nπ which is not possible as x ≠ . 2 ∴ No solution.

1 2

1 2

0

104. Max cos θ = 1. So, cos x ⋅ sin y = 1 cos x = 1, sin y = 1 cos x = − 1, sin y = − 1 cos x = 1, sin y = 1 ⇒ x = 0, 2 π π 5π and [from the equation] y= , 2 2 cos x = − 1, sin y = − 1 ⇒ x = π , 3π 3π and [from the equation] y= 2 ∴ Required number of ordered pair = 2 × 2 + 2 × 1= 6

0

⇒ or

105. sin x ⋅ cos y = 1⇒ sin x = 1, cos y = 1

π 5π ≤x≤ 6 6

1 2

⇒

1 sin 2 x ⋅ cos 2 x > 0 2 sin 4 x > 0 0 < 4x < π π 0 0

⇒

106. sin 2 x + cos 4 x = 2 2

2 sin x − 1 ≥ 0

From the figure,

⇒ ∴

or sin x = − 1, cos y = − 1 If sin x = 1, then cos y = 1 π x = , y = 0, 2 π ⇒ 2 If sin x = − 1, then cos y = − 1 3π ⇒ x= ,y =π 2 Thus, possible ordered pairs are   3π  π  π , π .  , 0 ,  , 2 π  ,     2 2  2 ⇒ ∴ 1 − 2 sin

–1

∴

9 Trigonometric Functions and Equations

So,

⇒ cos x = 0, which is not possible. or 6 cos 2 x + cos x − 2 = 0 1 2 cos x = , − ⇒ 2 3 π 5π x= , ⇒ 3 3  2  2 or x = π − cos − 1   , π + cos − 1    3  3 2 4π   or | x1 − x2 | = 2 cos − 1   ⇒ | x1 − x3 | =  3 3 2   ⇒ | x1 − x2|min = 2 cos − 1    3

111. 3 sin x + 4 cos ax = 7 ⇒ 0

0

⇒ ⇒ ⇒

1

–1 –∞

sin x = 1, cos ax = 1 π x = (4n1 + 1) , ax = 2 n2 π 2 π 2n π (4n1 + 1) = 2 a 2 4 n2 a= = Rational 4 n1 + 1

112. a1 + a2 (2 cos 2 x − 1) + a3 (1 − cos 2 x ) = 1 or (2 a2 − a3 ) cos 2 x + (a1 − a2 + a3 − 1) = 0

From the figure, −

This can hold for all x, if 2 a2 − a3 = 0 and a1 − a2 + a3 − 1 = 0

or

As there are two equations in three unknowns, so the number of solutions is infinite.

or ∴

π π ≤x≤ 4 4 3π −π ≤ x ≤ − 4 3π ≤x≤π 4 3π   π π   3π   ,π ∪ ∪ − , x ∈ − π, −   4   4 4   4

113. Here, sin θ = 1 ± 2; but |sin θ | ≤ 1. So, sin x = 1 − 2, which gives two values [0, 2 π ], (2 π , 4π ], (4π , 6π ] , etc. Thus, the least value of n is 4.

of

θ

in

each

of

491

Objective Mathematics Vol. 1

9

114. Here, (2sec x + 1) (sec x − 3) = 0; but | sec x | ≥ 1. So, sec x = 3, which gives two values of θ in each of 3π   . [0, 2 π ], (2 π , 4π ], (4π , 6π ]and one value in  6π , 6π +  2  Thus, n = 15

120. 2 sin θ = (r 2 − 1)2 + 2 ≥ 2. But max sin θ = 1 ∴ The equation implies sin θ = 1, r 2 = 1 π 5π 9π [from the equation] sin θ = 1 ⇒ θ = , , 2 2 2 2 r = 1 ⇒ r = 1, − 1 ∴ Number of values of pair (r , θ ) = 2 × 3 = 6

115. Here, 1 = sin α ⋅ cos 2α

121. [sin x ] = cos x ⇒ cos x = − 1, 0 , 1

sec 2α = sin α ⇒ which is only possible, if both are 1 or − 1. π α = nπ + (− 1)n − 1 ⇒ 2

π x = − π , , 2π ⇒ [sin x ] = 0, 1, 0 2 Thus, [sin x ] = cos x has no solution.

116. 5 sin θ + 3 (sin θ ⋅ cos α − cos θ ⋅ sin α )

122. Slope of y = 1 − sin x is − cos x.

= (5 + 3 cos α ) sin θ − 3 sin α ⋅ cos θ ∴ θ ∈ R min { 5 sin θ + 3 sin (θ − α )}

Slope of y =

π 3 π 3 . x −  + a for x > is 2 2  2 2 Y

= (5 + 3 cos α )2 + 9 sin 2 α = 34 + 30 cos α ∴The given equation is 34 + 30 cos α = 49 1 cos α = ⇒ 2 π α = 2 nπ ± ⇒ 3

Ta rg e t E x e rc is e s

117.

P

O π/2

y = cos x

0

2π

Y'

4π

X

y = sin x

Clearly, from the graph, there are four solutions.

118. We have, cos 4 x − (a + 2 )cos 2 x − (a + 3) = 0 ⇒ ⇒ ⇒

(a + 2 ) ± (a + 2 )2 + 4 (a + 3) 2 (a + 2 ) ± (a + 4) 2 [Qcos 2 x ≠ − 1] cos x = 2 cos 2 x = a + 3 cos 2 x =

We know, ⇒ ⇒ or

0 ≤ cos x ≤ 1 0 ≤ a+ 3≤1 − 3≤ a≤ −2 a ∈ [− 3, − 2 ] π ( x 2 + x ) = nπ + ( − 1)n πx 2

sin 2 (θ − α ) cos 2 (θ − α ) = =m sin α cos α sin 2 (θ − α ) + cos 2 (θ − α ) 1 = = sin α + cos α sin α + cos α 1 1 π  ⇒ sin α +  = ⇒ sin α + cos α =  4 m 2m  1  ≤ 1 ⇒ | m| ≥ 1 ⇒  2m  2

⇒

⇒

⇒

x 2 + x = 2 m + x 2 , where n = 2 m

⇒

⇒ or

x = 2m ∈ Z x 2 + x = p′ − x 2

⇒

For non-integral solution, − 1 ± 1 + 8 p′ − 1 + 1 + 8 p′ x= = 4 4 [taking the positive value] p −1 , where p = (8 p′ + 1) [an odd integer] = 4

cot β sin x 3 = 2 2 x x cot β tan 1 − tan 2 2 − 2 = 3 x x 2 1 + tan 2 1 + tan 2 2 2 3 x x 2 x 1 − tan 2 − cot β tan = 1 + tan   2 2 2 2 x x (2 + 3 ) tan 2 + 2 cot β tan + ( 3 − 2 ) = 0 2 2 2 x − 2 cot β ± 4 cot β + 4 tan = 2 2 (2 + 3 ) x − 2 cot β ± 2 cosec β tan = 2 2 (2 + 3 ) x − cot β + cosec β tan = 2 (2 + 3 ) x − cot β − cosecβ tan = 2 (2 + 3 ) x β tan = tan tan 15° 2 2

124. cos x −

∴

where, p′ is an odd integer. ⇒ x 2 + x = p′ − x 2

492

123. sin 2 (θ − α ) cos α = cos 2 (θ − α ) sin α = m sin α cos α

2

119. sin π ( x 2 + x ) = sin πx 2 ⇒

7π 3  7 π 3 ,  ⇒ P≡ ⇒ x=  6 2 2 6 3 π If y = x −  + a passes through P, then 2  2 3 3 π π a= − ⇒ a> − 2 2 3 3

⇒ cos x = −

Y

X'

X

⇒ ⇒ or ⇒

= n sin θ + n cos(θ + α ){cos(θ − α ) − cos (θ + α )} + 2 cos (θ + α ) − 1

sin θ sin 2 θ ⋅ 2n ⋅ n 2 sin θ cos 2 n θ  π  π ∴ f2   = tan  4 ⋅  = 1, f3  16  16 and so on.

2

+ 2 cos 2 (α + θ ) − 1 = n sin θ + n cos θ − n sin α + (2 − n ) 2

2

cos 2 (θ + α ) − 1 = (n − 1) − n sin α + (2 − n ) cos (θ + α ) ⇒ n = 2 2

2

cos θ = − 1 1 cos θ = 1− y 1 − 1< 2

⇒ 0 = 2 + [− cos x ] ⇒ [− cos x ] = − 2 Since, − 1 ≤ cos x ≤ 1 ⇒ − 1 ≤ − cos x ≤ 1 ⇒ [− cos x ] ≠ − 2 Hecne, [− cos x ] = − 2 has no real solution.

132.

128. We have, sin 50 x − cos 50 x = 1 ⇒

sin

x = 1 + cos

and ∴

x

∴

cos 50 x = 0 π , n ∈I 2

tan 2 (a + 2 ) x + a2 = 0

⇒

tan 2 (a + 2 ) x = 0

and ⇒

2

cos 3 x + 3 cos x ⋅ sin 2 x 4 1 3 = (sin 5 x − sin x ) + (sin 3 x + sin x ) 8 8 1 3 1 = sin x + sin 3 x + sin 5 x 4 8 8 1 3 1 n = 5, a1 = , a2 = 0, a3 = , a4 = 0, a5 = 4 8 8

134. sin( x − y ) = 1/ 2 ⇒ x − y = 30 ° or150°

a=0 tan 2 2 x = 0

π π x = 0, , − ⇒ 2 2 π  π  Hence, (0, 0 ),  0,   0, −  are ordered  2  2 satisfying the equation. θ 2 cos 2 1 + cos θ 2 130. 1 + sec θ = = cos θ cos θ Similarly, for others θ 2 cos 2 2 cos 2 θ 2 cos 2 2 n − 1θ θ 2 fn (θ ) = tan ⋅ ⋅ ... 2 cos θ cos 2θ cos 2 n θ

9

…(i)

and cos( x + y ) = 1/ 2 ⇒ x + y = 60 ° or 300° …(ii) Since, x and y lie between 0° and 180°, Eqs. (i) and (ii) are simultaneously true when x = 45°, y = 15° or x = 165°, y = 135°. But for the values given by (b) or (c), Eqs. (i) and (ii) do not hold simultaneously.

129. Given equation is sec 2 (a + 2 ) x + a2 − 1 = 0 ⇒

2

133. cos 3 x sin 2 x = 50

50

x = nπ +

2

(sin 2 x + α 2 − 2 )(1 + tan 2 x ) α2 = 1 − tan 2 x 1 − tan 2 x 2 2 2 ⇒ α cos x = sin x + α 2 − 2 ⇒ 2 = sin 2 x(1 + α 2 ) 2 2 ⇒ ⇒ 0≤ ≤1 sin 2 x = 1+ α2 1+ α2 ⇒ α 2 ≥ 1 ⇒ α ≤ − 1 or α ≥ 1

Since, sin x ≤ 1 and cos x ≥ 1, therefore the two sides are equal only, if sin 50 x = 1 = 1 + cos 50 x i.e. sin 50 x = 1 50

2

According to the question, a = 1 + 2 sin 2θ, b = 1 − 2 sin 2θ, cos 2θ = 0 π cos 2θ = 0 ⇒ θ = 4 π π Then, a = 1 + 2 sin , b = 1 − 2 sin 2 2 ⇒ a = 3, b = − 1

127. [sin x ] = [1 + sin x ] + [1 − cos x ]

50

π  π    = tan  8 ⋅  = 1  32   32 

− XY (cos 2 θ − sin 2 θ ) x + 4 xy + y = X + Y + 2( X − Y 2 )sin 2θ − 4 XY cos 2θ = (1 + 2 sin 2θ ) X 2 + (1 − 2 sin 2θ ) Y 2 − 4 cos 2θ ⋅ XY 2

⇒ ( y − 1) cos 2 θ + y cos θ + 1 = 0 and

= tan 2 n θ

131. x 2 + y 2 = X 2 + Y 2 , xy = ( X 2 − Y 2 ) sin θ ⋅ cos θ

cos 2 θ − 1 126. Given that, y = cos 2 θ + cos θ ⇒

n

=

= n sin 2 θ + n (cos 2 θ − sin 2 α ) − n cos 2 (θ + α ) 2

2 sin θ ⋅ cos θ ...cos 2 n − 1θ  sin θ ⋅ 2n   = ... n 2 sin θ cos 2 θ  

Trigonometric Functions and Equations

=

2

Targ e t E x e rc is e s

125. n sin 2 θ + 2 n cos (θ + α ) sin α sin θ + cos 2(α + θ )

135. The given relation can be written as x 1 = − sin x 2 sin x x x 2 tan 1 + tan 2 x 2 2 − tan = ⇒ x x 2 1 + tan 2 2 tan 2 2 2 x  x x x  ⇒ 2 tan 2 1 + tan 2  = 1 1 + tan 2  − 4 tan 2  2  2 2 2 x  where, y = tan 2 ⇒ 2 y(1 + y ) = (1 + y )2 − 4 y  2  ⇒ y 2 + 4y − 1 = 0 − 4 ± 16 + 4 y= = −2 ± 5 ⇒ 2 Since, y > 0, we get ( 5 − 2 )2 2 + 5 = (9 − 4 5 ) (2 + 5 ) ⋅ y = 5 −2 = 5 −2 2 + 5 tan

pair

θ cos 2 θ n+ 1 n− 1 2 = tan 2 ⋅ (cos θ cos 2θ ...cos 2 θ ) 2 cos 2 n θ

493

Objective Mathematics Vol. 1

9

136. y = ⇒

(cos 2 A − sin 2 A)2 + 1 (cos 2 A + sin 2 A)2 − 1 ± (cos 2 A − sin 2 A) + 1 y= ± (cos 2 A + sin 2 A) − 1

2 cos 2 A − 2 sin A cos A − 2 sin 2 A + 2 sin A cos A cos A (cos A − sin A) = = cot A sin A(cos A − sin A) − (cos 2 A − sin 2 A) + 1 y2 = − (cos 2 A + sin 2 A) − 1 (1 − cos 2 A) + sin 2 A = − (1 + cos 2 A) − sin 2 A

Ta rg e t E x e rc is e s

2

2 sin A − 2 sin A cos A − 2 sin 2 A + 2 sin A cos A cos A + sin A = cos A − sin A 1 + tan A  π = = tan  + A  4 1 − tan A 2

x 2 = a2 + b2 + 2

2

2

(a sin α + b cos α ) 2

2

2

= a2 + b2 + k where, k 2 =

× (a2 sin 2 α + b2 cos 2 α )

∴

x = a2 + b2 + 2 (a2 + b2 )p − p2

where,

p = a2 sin 2 α + b2 cos 2 α

⇒

494

[(a2 + b2 ) − (a2 sin 2 α + b2 cos 2 α )]

a2 b2 (1 − cos 2α ) + (1 + cos 2α ) p= 2 2 1 = [a2 + b2 − (a2 − b2 )cos 2α ] 2

2

4

3 1  ∴ ymin = and y is maximum when cos 2 x −   4 2 the maximum. 1 3 ∴ ymax = + = 1 4 4

2

is

140. k1 − k2 = sin x cos x(cos 2 x − sin 2 x ) 1 1 sin 2 x cos 2 x = sin 4 x 2 4 k1 − k2 > 0, 0 < 4 x < π π 0 0 0 < 2x < π π 0 0, x = − x, if x < 0 ] This will lead to θ π θ π sin  +  > 0, cos  +  < 0  2 4  2 4

4 cos 2 x − 3 = 2 sin x, 4 cos 2 y − 3 = 2 sin y

9 Trigonometric Functions and Equations

cos x = sin x − cos x,

⇒ 4 sin 2 x + 2 sin x − 1 = 0, 4 sin 2 y + 2 sin y − 1 = 0 ⇒ sin x (i.e. sin 18°) and sin y (− sin 54° ) are roots of the same quadratic equation.

148. The Statement I is true but does not follow from Statement II. Indeed cos 36° > tan 36°, if cos 2 36° > sin 36° ...(i) At this stage, we may think when cos 36° is greater than sin 36° , cos 2 36° should also be greater than sin 36° but this is not true, since cos 2 36° < cos 36°. Eq. (i) is proved as follows: 1 + cos 72 ° > 2 sin 36° ⇒ 1 + sin 18° ≥ 2 sin (30 ° + 6° ) ⇒ 1 + sin 18° ≥ 2 (sin 30 ° cos 6° + cos 30 ° sin 6° ) 1 + 2 sin 9° cos 9° ≥ cos 6° + 2 cos 30 ° sin 6° The last inequality is true in the light of the facts 1 > cos 6° , sin 9° > sin 6° , cos 9° > cos 30 ° 149. The minimum value of sum of three sines can be − 3 provided it is attainable through some angle. Indeed value − 3 is possible, if sin α = − 1, sin β = − 1, sin γ = − 1 π π π ⇒ α = (4l − 1) , β = (4m − 1) , γ = (4n − 1) 2 2 2 Now, α+β+γ=π π [4 (l + m + n ) − 3] = π ⇒ 2 ...(i) ⇒ 4 (l + m + n ) = 5 Since, l , m and n are integers, so Eq. (i) is impossible. ⇒ Minimum cannot be − 3. 3π 2π But for α= ,β = , γ = − 2 π, α + β + γ = π 2 2 and sin α + sin β + sin γ = − 2 Thus, sin α + sin β + sin γ can attain negative values, hence the Statement I is true. The Statement II is false, since α,β and γ are arbitrary real numbers. Though their sum is π but they can be of the type π π , , 0. 2 2

Targ e t E x e rc is e s

and

150. Statement I sin x + cosec x = 2 ⇒ sin x = cosec x = 1 π 5π , ⇒ Two solutions 2 2 ⇒ Statement I is false. 1 Statement II x + ≥ 2, only when x ≥ 0. x ∴ Statement II is false. ∴

x=

495

Objective Mathematics Vol. 1

9

151. sin 2 x + tan 2 x = 0 iff sin x = tan x = 0

155. f(θ, α) = 2 sin 2 θ + 4 cos θ (θ + α) sin θ sin α

⇒ x = nπ So, Statement I is true. But Statement II is not always true. i.e. if a, b ∈complex numbers, say a = 2 and b = − 2 i ⇒ a2 + b2 = 0

+ 2 cos 2 (θ + α) − 1 = 2 sin 2 θ + 2 cos (θ + α) [2 sin θ sin α + cos (θ + α )] − 1 = 2 sin 2 θ + 2 cos (θ + α) [sin θ sin α + cos θ cos α] − 1 = 2 sin 2 θ + 2 cos (θ + α ) cos (θ − α) − 1

152. Particle moves 1 radian on each circle starts from P (− 1, 0 ).

= 2 sin 2 θ + 2 cos 2 θ − 2 sin 2 α − 1

Y

∴

P (–1, 0)

X'

X

= cos 2 α π   π π f  ,  = cos 2 ×  = 0   3 4 4

156. f(θ, α ) = 0 ⇒

π 2 π α = (2 n + 1) , n ∈ I 4

⇒

Y'

To cross X-axis, it should cover π radian, hence will be on 4th circle. So, to cross positive Y-axis, it should cover a distance of 3π radian i.e. 5th circle. 2

153. Angular distance travelled from starting of motion on

Ta rg e t E x e rc is e s

cos 2α = 0

π  one circle to reach to other is 1 +  .  4 Y

2α = (2 n + 1)

∴ 1 π 157. f θ,  = 

6

2

 2π  1 cos   =  6 2 1 1 = ⇒ 2 2 ⇒ θ is any real. Solutions ( Q. Nos. 158-160) k = 0, graph will be shown as given below: Y

π/4 X'

X

O

X'

O

X

Y'

Y'

⇒ On 4th circle, particle will cross the positive X-axis again ⇒ x ∈(2 2 , 4) π 154. Angular distance travelled is . 4 Y A

2 tan A + tan C = k

⇒ 2 tan A + tan (180 ° − 2 A) = k ⇒

π/4

X

O

B C2 C3 Y'

496

Also,

...(i)

⇒ 2 tan A + tan (180 ° − 2 A) = k

P C1

X'

So, no such triangle is possible. Let A = B, then 2 A + C = 180 ° and 2 tan A + tan C = k Now, 2 A + C = 180 ° tan 2 A = − tan C

⇒ Coordinates of point C are ( 2 , 2 ). ⇒ Equation of tangent AB is x + y = 2. It intersect the circle x 2 + y 2 = 9 at A and B. 9   9 ⇒ , P=  2 2 2 2

⇒ ⇒

2 tan A − tan 2 A = k − 2 tan 3 A = k (1 − tan 2 A) 2 tan 3 A − k tan 2 A + k = 0

Let tan A = x, x > 0 Consider, f ( x ) = 2 x 3 − kx 3 + k

...(ii)

f ′ ( x ) = 6 x 2 − 2 kx = 0 k x = ,0 3 Following cases are arise: Case I If k < 0, then f(0 ) < 0 and the equation has only one positive solution.

Y

X k/3

X

O

9

Consider the positive sign, we have π p (sin x + cos x ) = (4n + 1) 2 (4n + 1)π sin x + cos x = ⇒ 2p cos x + sin x ≤ 2 Q ⇒

Y

(4 n + 1)π ≤ 2 2p

or

p≥

(4n + 1) π 2 2

So, p is smallest positive, if n = 0, giving again p = X

Thus, f ( x ) = 0 iff x = 0 which is not possible. k Case III If k > 0, then 0, are critical points. 3 3  k  27 k − k f (0 ) = k > 0 and f   =  3 27 Y

Y

k/3 O

O

k/3

X

X

Y

k/3

O

X

 k There is one isosceles triangle, if f   = 0 i.e. k = 3 3  3  k There are two isosceles triangle, if f   < 0 i.e. k > 3 3  3 There can never be three isosceles triangles.

161. Let y = 27 cos 2x ⋅ 81 sin 2 x = 33

cos 2 x + 4 sin 2 x

Now, − 32 + 42 ≤ 3 cos 2 x + 4 sin 2 x ≤ 32 + 42 ⇒

− 5 ≤ 3 cos 2 x + 4 sin 2 x ≤ 5

∴ 3

−5

1 1 ( x + y ) cos ( x − y ) 2 2 1 1 − 2 sin ( x + y ) cos ( x + y ) = 0 2 2 1 1 1 or 2 sin ( x + y )⋅ 2 sin x sin y = 0 2 2 2 1 1 Either sin ( x + y ) = 0, sin x = 0 ∴ 2 2 1 or sin y = 0 2 ∴ x + y = 0, x = 0, y = 0 x + y =1 i.e. x + y = 1, x − y = 1, x + y = − 1, x − y = − 1 When x + y = 0, reject x + y = 1 and x + y = −1 Solve with x − y = 1 or x − y = − 1 1  1  1 1 which gives  ,−  or  − , −   2  2 2 2 Again, solving with x = 0, we get (0, ± 1) and solving with y = 0, we get (± 1, 0 ) ∴ There are six set of solution. B. For f ( x ) to be obtained, 1− x ≤ − 1 ⇒ 1− x ≤ − 2 ⇒ x ≥ 3 2 ...(i) ⇒ x ≤ − 3 or x ≥ 3 1− x or ≥ 1 ⇒ 1 − x ≥ 2 or x ≤ − 1 2 which is not possible. 1− x  Also, sec − 1   ≥ 0 is always true as  2  π 0 ≤ sec − 1 x ≤ π ⇒ sec − 1 x ≠ 2 Hence, value of x lies between (− ∞, − 3] ∪ [3, ∞ ). ∴ Infinite values of x. C.sin x cos y = 1 ⇒ sin x = 1, cos y = 1 3π sin x = − 1, cos y = − 1⇒ x = ,y=π 2 Thus, ordered pairs are three. D. f ( x ) = sin x − cos x − kx + b ⇒ f ′ ( x ) = cos x + sin x − k Maximum value of cos x + sin x = 2 If k > 2, then f ′ ( x ) < 0, ∀ x ∈ R and hence f ( x ) will decreasing function, ∀ x ∈ R. ∴ k> 2 4 2k > 8 ⇒

164. A. 2 sin

Case II If k = 0, then f ( x ) = 2 x 3

≤ 3 3 cos 2 x + 4 sin 2 x ≤ 35

162. cos 7 x ≤ cos 2 x and sin 4 x = sin 2 x ⇒ cos 7 x + sin 4 x ≤ cos 2 x + sin 2 x = 1 So, the given equation is satisfies if and only if cos 7 x = 0 and sin 2 x = 1 cos 7 x = 1and sin 2 x = 0 π ⇒ x = (2 n + 1) or x = 2 mπ 2 π 3π Q 0 ≤ x ≤ 2 π, so x = 0, , , 2π 2 2 π 163. cos ( p sin x ) = sin ( p cos x ) = cos  − p cos 2  π ⇒ p sin x = 2 nπ ±  − p cos x , n ∈ I  2 or

 x 

π . 2 2

Trigonometric Functions and Equations

Y

Targ e t E x e rc is e s

∴ There is only one triangle.

497

Objective Mathematics Vol. 1

9

165. A. sin 2 A + sin 2 B = sin ( A + B) cos 2 A + cos 2 B 1− = sin C ⇒ 2 ⇒ 1 − cos ( A + B) cos ( A − B) = sin ( A + B) ⇒ cos 2 C cos 2 ( A − B) + 2 cos C cos ( A − B) = cos 2 C

B. Here, (2 sin x − 1) (2 sin x − 3) ≤ 0 But 2 sin x − 3 is always negative. ∴ 2 sin x − 1 ≥ 0 1 sin x ≥ ⇒ 2

⇒ cos C = 0 Hence, triangle is right angled. bc B. = b2 + c 2 − 2 bc cos A 2 cos A bc ⇒ cos A = 2 ⇒ b2c 2 = a2 (b2 + c 2 − a2 ) 2a ⇒ (a2 − b2 ) (a2 − c 2 ) = 0

1

1 2

⇒ a = b or a = c Hence, triangle is isosceles. A B C C. tan + tan + tan = 3 2 2 2 2 2 A B B C   ⇒  tan − tan  +  tan − tan    2 2 2 2

–1

From the figure,

π 5π ≤x≤ 6 6

C. − 1 ≤ tan θ ≤ 1 The value of scheme for this is shown below –∞ +∞

2

Ta rg e t E x e rc is e s

C A  +  tan − tan  = 0  2 2 Hence, triangle is equilateral. D. If a, b and c are in AP and ha , hb, hc are in AP, where ha , hb, hc are the altitudes, then a = b = c Hence, the triangle is equilateral. π x x π 166. A. + tan x = ⇒ tan x = − 4 2 2 4 Y

1 2

1

–1

0

0

1

–1

y = tan x

+∞ –∞

π/4

π π ≤x≤ 4 4 3π 3π or or − π ≤ x ≤ − ≤x≤π 4 4 3π   π π   3π   ∴ x ∈ − π, − ∪ − , ∪ ,π   4   4 4   4 π 1  . D.cos  x +  ≥  4 2 From the figure, −

X'

–π/2 0

π/2

π

3π/2

X

π x y =2 – 2 Y'

From the graph, the equation has 3 solutions in[− π, π]. π B. sin − 1 x 2 − 1 + cos − 1 2 x 2 − 5 = 2 ⇒ x2 − 1 = 2 x2 − 5 ⇒

x =2 ⇒ 2

The value for this is shown below: 0 1 √2

x = ± 2 (two solutions)

πx + 1= 0 2 2 πx   4 πx ⇒  x 2 − sin 2 =0  + 1 − sin   2 2

C. x 4 − 2 x 2 sin 2

⇒

x = (2 n + 1), n ∈ I and x 2 = sin 2

⇒

x = ± 1is the solution

⇒ No solution.

167. A. sin x ⋅ cos x (cos 2 x − sin 2 x ) > 0 498

∴

1 √2

πx =1 2

D. Let y = x 2 + 2 x + 2 sec 2 πx + tan 2 πx ⇒ y = ( x + 1)2 + (2 sec 2 πx − 1) + tan 2 πx > 0

⇒

1

–1

1 sin 2 x ⋅ cos 2 x > 0 ⇒ sin 4 x > 0 2 0 < 4x < π

0

π π π ≤x+ ≤ 4 4 4 π π π and in general, 2 nπ − ≤ x + ≤ 2 nπ + 2 4 4 π ∴ 2 nπ − ≤ x ≤ 2 nπ 2 π 3π For n = 0, − ≤ x ≤ 0; for n = 1, ≤ x ≤ 2π 2 2 From the figure,

−

B. 1 − 2 sin 2 x + p sin x = 2 p − 7 i.e. 2 sin 2 x − p sin x + 2 p − 8 = 0 It has solution, if (i) p2 − 8 (2 p − 8) ≥ 0 i.e.( p − 8)2 ≥ 0 which is always true. (ii) sin x = iff − 1 ≤

p±

p2 − 8 p + 64 p−4 = 2, 4 2

C. tan 2 θ = 2 tan 2 φ + 1 1 + tan 2 θ = 2(1 + tan 2 φ) ⇒ sec 2 θ = 2 cos 2 θ = 1 + cos 2θ ⇒ cos 2 θ = − sin 2 θ cos 2θ + sin 2 φ = 0 D.

p−4 ≤ 1, iff 2 ≤ p ≤ 6 2

∴ p has 5 integral values. 3 C. cos α = 4 3α 3α  α α  16 sin sin = 8 2 sin sin   2 2 2 2 

sin (2 A + B) =5 sin B Using componendo and dividendo, sin (2 A + B) + sin B 6 = sin (2 A + B) − sin B 4 2 sin (2 A + B) cos A 3 ⇒ = 2 cos (2 A + B) sin A 2 tan ( A + B) 3 = ∴ tan A 2

170. A. Clearly, number of solutions of|tan 2 x| = sin x in [0, π ]

= 8 (cos α − cos 2 α ) = 8 (cos α − 2 cos 2 α + 1)

are 4. Y

 3 18  = 8 − + 1 = 5  4 16  D. Let A = a2 − 4 $i + a $j +

a2 − 4 k$

X'

B = tan A $i + tan B$j + tan Ck$

π 2

π 4

O

3π 2

π

X

A ⋅ B =| A|| B|cos θ or a2 − 4 tan A + a tan B +

a2 + 4 tan C

Y'

= a2 − 4 + a2 + a2 + 4 tan 2 A + tan 2 B + tan 2 C cos θ = 3 a tan A + tan B + tan C cos θ 2

or cos θ =

or

169. A. sin

B. sin

2

a2 − 4 tan A + a tan B +

Since, cos θ ≤ 1 ∴

2

a2 + 4 tan C

3 a tan A + tan B + tan C 2

2

6a

3 a tan 2 A + tan 2 B + tan 2 C

2

≤1

tan 2 A + tan 2 B + tan 2 C ≥ 12 5A 1 A sin = (cos 2 A − cos 3 A) 2 2 2 1 = (2 cos 2 A − 1 − 4 cos 3 x + 3 cos A) 2 1 9 27 3 11 = 2 × − 1− 4 × + 3×  = 2 16 64 4 32 π 2π 3π 1 + sin + sin = π 7 7 7 2 sin 7 π π 2π  2π 3π  sin  + 2 sin 1 − cos  + 2 sin sin 7 7 7 7 7 

9

2 tan 2 A B. tan 4 A = = 1 − tan 2 2 A

tan 4 A =

Targ e t E x e rc is e s

1 13  = − cos x +  +   2 4 13 and least value = 1 ∴Greatest value = 4 13 Ratio = ∴ 4 ⇒ k = 13

1

=

2

Trigonometric Functions and Equations

π 2π 3π  1 − cos + cos − cos π  7 7 7 2 sin 7 2π 4π  + cos − cos  7 7  π 1 π 1  1 + cos  = cot = π  7 2 14  2 sin 7

168. A. f ( x ) = 2 − cos x + sin 2 x = 3 − cos x − cos 2 x

4 tan A 1 − tan 2 A  2 tan A  1−    1 − tan 2 A

2

4 tan A(1 − tan 2 A) 1 + tan 4 A − 6 tan 2 A

4 tan A − 4 tan 3 A + (6 tan 2 A − tan 4 A − 1) tan 4 A = 0 π If A= 16 π π π π 4 tan − 4 tan 3 + 6 tan 2 − tan 4 − 1= 0 16 16 16 16 ∴ Required value is 2. C. tan ( p cot x ) = cot ( p tan x ) π  tan( p cot x ) = tan  − p tan x 2  π p cot x = nπ + − p tan x 2 π nπ + π 2 p= = sin x cos x [Q x ∈ [0, π]] tan x + cot x 2 ⇒

Pmax =

π 4

⇒

4Pmax =1 π

499

Objective Mathematics Vol. 1

9

D. 5 cos

2

2 x + 2 sin 2 x

+ 52 cos

2

x + sin 2 2 x

= 126

5 y + 53 − y = 126 125 5 y + y = 126 5 5y = 125, 1

⇒ ⇒ ⇒

y = 3, 0 ⇒

cos 2 2 x + 2 sin 2 x = 3 π 3π x= , 2 2 cos 2 2 x + 2 sin 2 x ≠ 0 2x = 1, 3 π

⇒ ∴ ⇒

171. [ x ]2 − 5 [ x ] + 6 = sin x ⇒

RHS must be an integer. Now, integer values of sin x are 0, + 1 and − 1. Case I Let sin x = 1 [ x ]2 − 5 [ x ] + 5 = 0

Ta rg e t E x e rc is e s

[x] =

5± 5 2

⇒ sin x = − 1 ⇒ [ x ]2 − 5 [ x ] + 7 = 0 [this equation has no real roots] Case III Let sin x = 0 ⇒

[ x ]2 − 5 [ x ] + 6 = 0

⇒

[ x ] = 2, [ x ] = 3

⇒ 4 cos θ(2 cos 2 θ − 1) = 4 cos 2 θ − 1 ⇒ 4 cos θ =

⇒

4 cos 2 θ − 1 3 − tan 2 θ = 2 cos 2 θ − 1 1 − tan 2 θ π 3 − tan 2 π 7 4 cos = 7 1 − tan 2 π 7 λ=4 ...(i)

⇒

sin x cos 3 x α= cos x sin 3 x

⇒

α=

tan x tan 3 x

⇒

α=

tan x (1 − 3 tan 2 x ) 3 tan x − tan 3 x

⇒

α=

1 − 3 tan 2 x 3 − tan 2 x

...(ii)

1  For real value of x, RHS of 2nd never lies between  , 3 . 3  So, the number of integral values is { 0, 1, 2} i.e.3

175. n sin 2 θ + 2 n cos (θ + α ) sin α sin θ + cos 2 (α + θ) = n sin 2 θ + n cos (θ + α) [cos(θ − α) − cos (θ + α)] + 2 cos 2 (θ + α) − 1

⇒ x = π is the only solution. 2 sin 2θ 172. Given that, f (n θ) = cos 2θ − cos 4 nθ 2 sin 2θ 2 sin (2 n + 1) θ sin(2 n − 1)θ

sin {(2 n + 1) θ − (2 n − 1) θ} = sin (2 n + 1) θ sin(2 n − 1) θ sin (2 n + 1) θ cos (2 n − 1) θ  − cos (2 n + 1) θ sin (2 n − 1) θ   = sin (2 n + 1) θ sin (2 n − 1) θ = cot (2 n − 1) θ − cot (2 n + 1) θ ⇒ f (θ) + f (2 θ) + f (3 θ ) + ...+ f (nθ ) = (cot θ − cot 3 θ ) + (cot 3 θ − cot 5 θ ) + (cot 5 θ − cot 7θ ) + ...+ cot(2 n − 1) θ − cot (2 n + 1) θ cos θ cos (2 n + 1)θ = cot θ − cot (2 n + 1) θ = − sin θ sin (2 n + 1) θ = ⇒ ⇒ ⇒

8 cos 3 θ − 4 cos θ = 4 cos 2 θ − 1

x ∈[2, 4) and sin x = 0 for x ∈[2, 4) at x = π

=

500

⇒

3 − 4 sin 2 θ = 4 cos θ (2 cos 2 θ − 1)

[not possible]

Case II Let sin x = − 1

⇒

⇒

174. sin x cos 3 x − α cos x sin 3 x = 0

LHS is always represents an integer.

⇒

⇒ ⇒

Hence,

([ x ] − 1) = sin x

⇒

π ⇒ 3θ = π − 4θ 7 sin 3 θ = sin 4 θ 3 sin θ − 4 sin 3 θ = 4 sin θ cos θ cos 2 θ

173. Let θ =

sin 2 nθ sin θ sin (2 n + 1)θ µ = 2n + 1 λ = 2n µ − λ =1

= n sin θ + n (cos θ − sin α) − n cos 2 (θ + α) 2

2

2

+ 2 cos 2 (α + θ) − 1 = n sin θ + n cos θ − n sin α + (2 − n ) cos 2 (θ + α) − 1 2

2

2

= (n − 1) − n sin 2 α + (2 − n ) cos 2 (θ + α) ∴

n =2

176. Given that, y =

cos 2 θ − 1 cos 2 θ + cos θ

⇒

( y − 1) cos 2 θ + y cos θ + 1 = 0

⇒

cos θ = − 1 and cos θ =

⇒

− 1
0 for all x, so 5 u + 12 > 0 ]

Ta rg e t E x e rc is e s

⇒ ⇒

u = 2 ⇒ 4sin x = 2 2 = 21 ⇒ 2 sin x = 1 1 sin x = ⇒ 2 π π ⇒ sin x = sin ⇒ x= 6 6 Putting the value of u in Eq. (i), we get v=9 2 sin x

⇒ ⇒ ⇒ ∴

3cos y = 32 1 =2 cos y

⇒ cos y =

cos y = cos

π 3

  4 4 π π   ⇒ sin  x −  + sin  x +  = 1 [Q sin θ = sin (π − θ)]   4 4 π ⇒ 2 sin x cos = 1 4 1 π 3π ⇒ x= , sin x = ⇒ 4 4 2 We know that, 3 radians ≅ 171° 22′. Therefore, sin 3 > 0, cos 3 < 0 and|cos 3| > sin 3 ∴ cos 3 + sin 3 < 0 2 cos 7 x Now, > 2cos 2 x cos 3 + sin 3 ⇒ 2 cos 7 x < 2cos 2 x (cos 3 + sin 3) [Q cos 3 + sin 3 < 0 ] ⇒ 2 cos 7 x < 0 [Q 2cos 2 x > 0 ] ⇒ cos 7 x < 0 3π satisfies this equation. Clearly, x = 4 3π Hence, x = is the required solution. 4 So, only one solution.

186. The given equation will have a solution, if

1

⇒

π 3π 185. We have, sin  x −  − cos  x +  = 1

1 2

and ⇒ and ⇒ and

p ≥ 0, 2 − p ≥ 0 | p cos x − 2 sin x| ≤ p + 4 p ≥ 0, 2 − p ≥ 0 | 2 + 2 − p| ≤ p + 4 p ≥ 0, p ≤ 2 ( 2 + 2 − p )2 ≤ p + 4

⇒ and

p ≥ 0, p ≤ 2 p2 + 2 p − 4 ≥ 0

⇒ p ≥ 0, p ≤ 2 and p ∈ (− ∞, − 5 − 1) ∪ [ 5 − 1, ∞ ) ⇒ p ∈ ( 5 − 1, 2 ] So, the integral value of p is 2.

π 3 π x+ y= ⇒ λ =2 2 y=

Entrances Gallery lim f ( x ) → ∞

1. sin x + 2 sin 2 x − sin 3 x = 3 ⇒ sin x + 4 sin x cos x − 3 sin x + 4 sin x = 3 3

⇒

sin x [−2 + 4 cos x + 4(1 − cos 2 x )] = 3

⇒

sin x [2 − (4 cos 2 x − 4 cos x + 1) + 1] = 3

⇒

sin x [3 − (2 cos x − 1)2 ] = 3

⇒

sin x = 1 and 2 cos x − 1 = 0 π π x = and x = ⇒ 2 3 which is not possible at same time. Hence, no solution.

2. Let f ( x ) = x 2 − x sin x − cos x ⇒

f ′( x ) = 2 x − x cos x

(0, –1)

lim f ( x ) → ∞

502

x→ ∞

x→ −∞

f(0 ) = − 1 Hence, there are two solutions. 2 sin 2 θ 3. 2 cos θ (1 − sin φ ) = cos φ − 1 sin θ = 2 sin θ cos φ − 1 ⇒ 2 cos θ − 2 cos θ sin φ = 2 sin θ cos φ − 1 …(i) ⇒ 2 cos θ + 1 = 2 sin (θ + φ ) …(ii) tan(2 π − θ) > 0 ⇒ tan θ < 0 3 and − 1 < sin θ < − 2  3π 5π  …(iii) ⇒ ,  θ ∈  2 3 1 < sin (θ + φ ) < 1 [from Eqs. (i) and (iii)] ∴ 2 5π π ⇒ 2 π + − θmax < φ < 2 π + − θmin 6 6 π 4π c sin B . There are further following cases: (a) B is an acute angle ⇒ cos B is positive. In this case two values of a will exists if and only if c cos B > b 2 − ( c sin B ) 2 or c > b

X

⇒ Two such triangle is possible. If c < b, only one such triangle is possible. (b) B is an obtuse angle ⇒ cos B is negative. In this case triangle will exists if and only if b 2 − ( c sin B ) 2 > | c cos B | ⇒ b > c.

Sol. (a) Let A = 60° ⇒ B + C = 120° Then, tan  

●

●

X

⇒ ⇒ ⇒

If one side a and angles B and C are given, then a sin B a sin C and c = A = 180 ° − (B + C), b = sin A sin A If the three angles A, B and C are given, we can only find the ratios of the sides a, b and c by using sine rule (since, there are infinite similar triangles possible).

Also,

=

5×9 = 15

3

A = 60° 2 A = 120°

⇒ ∴ Aliter

b 2 + c 2 − a2 2 bc 100 + 36 − 196 = 120 1 =− 2 A = 120°

By cosine rule, cos A =

⇒

3+1 cot 30° = 1 3 −1

( 3 + 1)sin 60° sin 60° cos 45° + cos 60° sin 45° 2 ( 3 + 1) 3 = 3+1

6

Aliter By cosine rule, a2 = b 2 + c 2 − 2 bc cos A

Sol. (c) Let a = 14, b = 10, c = 6 ⇒ s = 14 + 10 + 6 = 15

=

3 + 1− 3 + 1+

B−C = 45° 2 B − C = 90° B = 105°, C = 15° sin A ( 3 + 1)sin 60° ( 3 + 1)sin 60° = a=b = sin B sin105° sin(60° + 45° ) =

Example 9. In any ∆ABC, the sides are 6 cm, 10 cm and 14 cm. Then, the triangle is obtuse angled with the obtuse angle equals (a)180° (b) 90° (c)120° (d)160° 2 The largest angle is opposite to the longest side. Then, (s − b )(s − c ) A tan = 2 s(s − a)

B − C b − c A cot  = 2  b+c 2 =

So, in this case only one such triangle is possible. If b < c, there exists no such triangle. Ø

Example 10. Two sides of a triangle are 3 − 1 and 3 + 1 units and their included angle is 60°. Then, the third side of the triangle is (b) 8 (a) 6 (c)1 + 6 (d) 5

= ( 3 + 1)2 + ( 3 − 1)2 − 2( 3 + 1)( 3 − 1)cos 60° 1 = 8 − 4⋅ = 6 2 a= 6

⇒ X

Example 11. If a, b and A are given in a triangle and c1 , c2 are the possible values of the third side, then c12 + c22 − 2c1 c2 cos 2 A is equal to (a) a 2 cos 2 A (c) 4a 2 cos 2 A

(b) 4b 2 cos 2 A (d) None of these

2 2 2 Sol. (c) cos A = b + c − a

2 bc

⇒ c − 2 bc cos A + b 2 − a2 = 0 2

⇒

c1 + c 2 = 2 b cos A and c1c 2 = b 2 − a2

⇒ c12 + c 22 − 2c1c 2 cos 2 A = (c1 + c 2 )2 − 2c1c 2 (1 + cos 2 A) = 4 b 2 cos 2 A − 2(b 2 − a2 ) 2 cos 2 A = 4 a2 cos 2 A

Work Book Exercise 10.1 1 In ∆ABC,

b a+ b b b c c c b d None of the above a

516

sin B is equal to sin( A + B)

2 In ∆ABC,c cos( A − α ) + a cos(C + α ) is equal to a acosα

b bcosα

c c cosα

d 2 bcosα

3 If in ∆ABC, a = 6 cm, b = 8 cm, c = 10 cm, then the value of sin 2 A is 6 25 10 c 25

a

8 25 24 d 25 b

a 2b

b 3b

c

2b

8 In a ∆ABC, if

d b

5 In a ∆ABC, AD is the altitude from A. Given b > c ,

a 1

abc , then ∠B is equal to b2 − c 2

C = 23° and AD = a 23° c 67°

b 113° d 90°

6 If 4 sin A = 4 sin B = 3 sin C in a ∆ABC, then cos A 1 3 1 c 27

a equilateral c Right angled

3

b isosceles d None of these

B(c , d ) subtends an angle θ at the origin, then cos θ is equal to

1 9 1 d 18

ab + cd

a

7 In an acute angled ∆ABC, the least value of sec A + sec B + sec C is b 3

d

10 If the line segment joining the points A(a, b) and

b

a 6

c 3 2

9 In ∆ABC, a2 + b2 + c 2 = ac + ab 3, then triangle is

is equal to a

b 2

b

( a 2 + b 2 )(c 2 + d 2 ) ac + bd

c

( a 2 + b 2 )(c 2 + d 2 ) ac − bd ( a 2 + b 2 )(c 2 + d 2 )

c 9

d 4

10 Properties of Triangles, Heights and Distances

cos A cos B cos C and the sides = = a b c a = 2, then area of triangle is

4 If A = 45°, B = 75°, then a + c 2 is equal to

d None of the above

Circles Connected with Triangle i.

In ∆ABC, ∆ (a) r = s

Circumcircle of a triangle The circle passing through the vertices of a ∆ABC is called circumcircle. Its radius R is called the circumradius and its centre is known as circumcentre. Circumcentre is the point of intersection of perpendicular bisectors of the sides.

(b) r = ( s − a ) tan

A B = ( s − b) tan 2 2 = ( s − c) tan

A

C 2

B C A C b sin sin sin 2 2 = 2 2 (c) r = A B cos cos 2 2 B A c sin sin 2 2 = C cos 2 A B C (d) r = 4R sin sin sin 2 2 2 a sin

c

O

b

R B

C

a

In ∆ABC, a b c = = 2 sin A 2 sin B 2 sin C abc (b) R = 4∆

(a) R =

ii.

Incircle of a triangle The circle touching all the sides of a triangle internally is known as an incircle of the triangle. Its centre is called incentre and its radius r is called the inradius of the circle. Incentre is the point of intersection of bisectors of the angles of the triangle. A

iii.

Escribed circles of a triangle The circle touching BC and the two sides AB and AC produced of ∆ABC externally is called the escribed circle opposite to A. Its radius is denoted by r1 . The centre of this circle is known as excentre and denoted by I. It is the point, where the external bisectors of the ∠B and ∠C and the internal bisector of the ∠A meet. B

c b O

r1 O I1

r B

a

C

A

C

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10 Objective Mathematics Vol. 1

Similarly, r2 and r3 denote the radii of the escribed circles opposite to the angles B and C, respectively and excentres are denoted by I 2 and I 3 . r1 , r2 and r3 are called the exradii of ∆ABC. Here, A ∆ (a) r1 = = s tan s−a 2 A B C = 4R sin cos cos 2 2 2 B C a cos cos 2 2 = A cos 2 B ∆ (b) r2 = = s tan s−b 2 B C A = 4R sin cos cos 2 2 2 A C b cos cos 2 2 = B cos 2 ∆ C (c) r3 = = s tan s−c 2 A B c cos cos C A B 2 2 = 4R sin cos cos = C 2 2 2 cos 2 Ø

X

●

r1 + r2 + r3 = 4R + r

●

r1 r2 + r2 r3 + r3 r1 = s 2 =

●

1 1 1 s 1 A B C  + + = = and r1 r2 r3 = r 2  cot cot cot   r1 r2 r3 ∆ r 2 2 2

r1 r2 r3 r

Sol. (d) cos A + cos B + cos C

A + B  A − B  + cos C = 2 cos  cos    2   2  A − B π C 2C = 2 cos  −  cos   + 1 − 2 sin 2 2  2  2 A − B C  2C = 1 + 2 sin cos   − 2 sin  2  2 2

A − B C  C cos   − sin   2  2  2 A − B C    A + B  = 1 + 2 sin cos   − cos    2   2   2  A B C = 1 + 4 sin sin sin 2 2 2 r  Q r = 4R sin A sin B sin C  = 1 +    R 2 2 2  = 1 + 2 sin

X

Sol. (a)Q r1 − r = r2 + r3

⇒ ⇒ ⇒

s(− a + b + c ) = bc ( b + c + a) ( b + c − a) = bc 2 ⇒ (b + c )2 − a2 = 2 bc ⇒

518

Example 13. In a ∆ABC, cos A + cos B + cos C is equal to r r r R (c) (b)1 − (d)1 + (a) R R R r

b 2 + c 2 + 2 bc − a2 = 2 bc

⇒

2

= 2 R sin A cos A + 2 R sin Bcos B + 2R sinC cos C [Qa = 2 R sin A, b = 2 R sin B and c = 2 R sinC] = R (sin 2 A + sin 2 B + sin 2C ) = R[2 sin( A + B)cos( A − B) + 2 sinC cos C ] = 2R[sin( π − C ) cos ( A − B) + sinC cos { π − ( A + B)}] = 2 R [sinC {cos ( A − B) − cos ( A + B)}] = 4 R sin A sin BsinC

s 2 − (b + c ) s + bc = s 2 − as

⇒

⇒

Example 12. In any ∆ABC, a cos A + b cos B + c cos C is equal to (a) 4 R (b) sin A sin B sin C (c) R sin A sin B sin C (d) 4R sin A sin B sin C

∆ ∆ ∆ ∆ − = + s−a s s− b s−c s −s + a s − c + s − b = s(s − a) (s − b ) (s − c ) a a = s(s − a) (s − b ) (s − c )

⇒

b 2 + c 2 = a2

∴ ∠A = 90° Thus, the triangle is right angled triangle. X

Sol. (d) acos A + b cos B + c cos C

X

Example 14. If r1 = r2 + r3 + r, then triangle is (a) right angled (b) obtuse angled (c) equilateral (d) None of the above

Example 15. If in a triangle, R and r are the circumradius and inradius respectively, then the HM of the exradii of the triangle is (a) 3r (b) 2R (c) R + r (d) None of these Sol. (a) Σ 1 = Σ s − a = 3s − (a + b + c ) = s

∆ ∆ 1 3 3∆ = = = 3r ∴ HM of exradii = Σ1 / r1 s/∆ s 3 r1

X

∆

Example 16. If the exradii of a triangle are in HP, then the corresponding sides are in (a) AP (b) GP (c) HP (d) None of the above

⇒ ⇒ ⇒ ⇒ X

Orthocentre is a point P at which all the three altitudes, drawn from the vertices to the opposite sides of the triangle, meet.

Example 17. In a ∆ABC, the inradius and three exradii are r, r1 , r2 and r3 , respectively. In usual notations, the value of r ⋅ r1 ⋅ r2 ⋅ r3 is equal to (a) 2∆ (b) ∆ 2 abc (c) (d) None of these 4R Sol. (b) r ⋅ r1 ⋅ r2 ⋅ r3 = ∆ ⋅ ∆ ⋅ ∆ ⋅ ∆

s s −a s − b s − c

= X

∆4 ∆2

A N M P

B

= ∆2

Ø

●

●

●

Sol. (b) In a right angled triangle, the circumcentre lies on

●

the hypotenuse.

Also,

⇒ ⇒ X

a , where A = 90° 2 A r = (s − a)tan 2 = (s − a) tan 45° = ( s − a) = s − 2 R r + 2R = s s−r R= 2 R=

●

X

Example 19. In a ∆ABC, a =1 and the perimeter is six times the AM of the sines of the angles. The measure of ∠A is π π π π (a) (b) (c) (d) 3 2 6 4

Orthocentre of the triangle is the incentre of the pedal triangle. If l1 , l 2 and l3 are the centres of escribed circles which are opposite to A , B and C, respectively and l is the centre of incircle, then ∆ABC is the pedal triangle of ∆ l1 l 2 l3 and l is the orthocentre of ∆ l1 l 2 l3 . The centroid of the triangle lies on the line joining the circumcentre to the orthocentre and divides it in the ratio 1:2 . Circle circumscribing the pedal triangle of a given triangle bisects the sides of the given triangle and also the lines joining the vertices of the given triangle to the orthocentre of the given triangle. This circle is known as nine point circle. Circumcentre of the pedal triangle of a given triangle bisects the line joining the circumcentre of the triangle to the orthocentre.

Example 20. The distance of the orthocentre from the vertices of the triangle are (a) 2R cos A, 2R cos B , 2R cos C (b) 2R sin A, 2R sin B , 2R sin C (c) 2R cos A, R cos B , R cos C (d) 2R cosec A, 2R cosec B , 2R cosec C Sol. (a) Let L, M and N be the feet of the altitudes from the vertices A, B and C of the triangle, respectively.

Sol. (c) 1 + b + c = 6  sin A + sin B + sinC  

⇒ Q ⇒ ⇒

3

1 b c  = 2  + +   2R 2R 2R  1 = (1 + b + c ) R R =1 1 = 2R = 2 sin A 1 sin A = 2 π A= 6

C

L

The ∆ LMN which is formed by joining the feet of these three altitudes is called the pedal triangle.

Example 18. In a right angled triangle, R is equal to s+r s−r (a) (b) 2 2 s+r (c) s − r (d) a

⇒

10

The Orthocentre and the Pedal Triangle

r1 r2 r3 s− a s− b s−c are in AP. , , ∆ ∆ ∆ s − a, s − b, s − c are in AP. − a, − b, − c are in AP. a, b, c are in AP.

Properties of Triangles, Heights and Distances

Sol. (a) 1 , 1 and 1 are in AP.

A



N M P

[Q1 + b + c ≠ 0]

B

In ∆ABL, In ∆BPL,

L

C

BL = c cos B

BP = BL sec PBL = c cos B sec (90° − C ) = c cos B cosec C = 2 R cos B Similarly, PC = 2 R cos C and PA = 2 R cos A

519

X

Objective Mathematics Vol. 1

10

Sol. (a) Let O and H be the circumcentre and the

Example 21. sin A, sin B and sin C are in AP for the ∆ABC. Then, (a) the altitudes are in AP (b) the altitudes are in HP (c) the medians are in GP (d) the medians are in AP

orthocentre, respectively. If OF is the perpendicular to AB. A

F

AP.

a2 + b 2 c 2 − 2 4

These are neither in AP nor GP. X

= R 2 + 4R 2 cos 2 A − 4R 2 cos A cos (C − B)

Example 22. The ratio of the distances of the orthocentre of an acute angled ∆ABC from the sides BC, AC and AB is (a) cos A :cos B :cos C (b) sin A :sin B :sin C (c) sec A :sec B :sec C (d) None of the above Sol. (c) HD = BD ⋅ tan ∠EBC = c cos B ⋅ tan (90° − C )

= R 2 + 4R 2 cos A [cos A − cos (C − B)] = R 2 − 4R 2 cos A [cos (B + C ) + cos (C − B)] = R 2 − 8R 2 cos A cos B cos C Hence, OH = R 1 − 8cos A cos Bcos C

Cyclic Quadrilateral A cyclic quadrilateral is a quadrilateral which can be circumscribed by a circle.

A

Ø

● ●

E

●

H

●

B

=

D

C

c cos B ⋅ cos C sin C

= 2 R cos B cos C 2R cos A cos B cos C = cos A Similarly, for others. So, the ratio of the distances of the 1 1 1 . orthocentre from the sides = : : cos A cos B cos C X

520

C

∠OAF = 90° − ∠AOF = 90° − C Also, ∠HAL = 90° − C Hence, ∠OAH = A − ∠OAF − ∠HAL = A − 2 (90° − C ) = A + 2C − ( A + B + C ) =C − B Also, OA = R and PA = 2 R cos A Now, in ∆AOH, OH2 = OA 2 + PA 2 − 2OA ⋅ HA cos OAH

and c are in AP. 2∆ are in AP. and h3 1 are in AP. and h3

a2 + c 2 b 2 − , 2 4

K

We have,

⇒ h1, h2 and h3 are in HP. Now, medians are c 2 + b 2 a2 − , 2 4

L H

B

Sol. (b) sin A, sin B, sin C are in AP. a b c are in , , ⇒ 2R 2R 2R ⇒ a, b 2∆ 2∆ , ⇒ h1 h2 1 1 ⇒ , h1 h2

O

Example 23. The distance between the circumcentre and the orthocentre of a ∆ABC is (a) R 1 − 8 cos A cos B cos C

X

Sum of the opposite angles of a cyclic quadrilateral is 180°. In a cyclic quadrilateral, sum of the products of the opposite sides is equal to the product of the diagonals. This is known as Ptolemy's theorem. If sum of the opposite sides of a quadrilateral is equal, then a circle can be inscribed in the quadrilateral. The area of the cyclic quadrilateral with sides a, b , c and d is (s − a)(s − b)(s − c)(s − d).

Example 24. The area of a cyclic (3 3 ) quadrilateralABCD is . The radius of the 4 circle circumscribing cyclic quadrilateral is 1. If AB =1 and BD = 3, then BC⋅ CD is equal to 1 (a) 2 (b) 3 − 3 (c) 2 3 + 1 (d) None of these Sol. (a) Clearly, ∠BOD = 2C cos 2C =

12 + 12 − ( 3 )2 2 ⋅ 1⋅ 1

(b) R 1 + 8 cos A cos B cos C

∴

(c) R 1 − 4 cos A cos B cos C

⇒

C = 60°

(d) None of the above

∴

A = 180° − C = 120°

AD2 + 12 − ( 3 )2 2 ⋅ AD ⋅ 1

X

A

D 1

√3 O

C

1 B

1 AD2 − 2 = 2 2 AD ⇒ AD2 + AD − 2 = 0 ⇒ AD = 1 Now, ar ( ABCD) = ar (∆ADB) + ar (∆BCD) 1 3 3 1 = ⋅ 1⋅ 1⋅ sin 120° + ⋅ BC ⋅ CD ⋅ sin 60° ⇒ 2 4 2 ⇒ BC ⋅ CD = 2

∴

−

Example 25. If the area of circle is A1 and area of regular pentagon inscribed in the circle is A A2 . Then, 1 is equal to A2 2π (a) 5 2π π (b) sec 5 10 π (c) sec 10 (d) None of the above Sol. (b) In ∆OAB, OA = OB = r and ∠AOB = 72 ° D

E

R

π n

B

a M 2

a 2

C

1 2 π na cot n 4 π = nr 2 tan n n 2 2π = R sin 2 n

Area of a regular polygon =

Ø

●

●

Sum of the exterior angles of a polygon taken in one direction (clockwise or anti-clockwise) remains constant and it is equal to 360°. In the regular polygon, the centroid, the circumcentre and the incentre are same.

54°

54° B

Area of ∆AOB =

⇒

A1 = πr 2

…(i)

...(ii)

A1 πr = 5 2 A2 r cos 18° 2 2π π = sec 5 10 2

Thus,

X

r

r

1 r ⋅ r sin 72 ° 2 1 = r 2 cos 18° 2 Area of pentagon = 5 (Area of ∆AOB) 5 A2 = r 2 cos 18° ⇒ 2 Also, we know that, Area of circle = πr 2

∴

D

E

r

A

A regular polygon is a polygon which has all its sides as well as all its angles equal. If the polygon has n sides, sum of its internal angles is ( n − 2) π and each ( n − 2) π angle is ⋅ The circle passing through all the n vertices of a regular polygon is called circumscribed circle and the circle which touches all sides of a regular polygon is called its inscribed circle. π a Radius of circumcircle, R = cosec and radius 2 n a π of incircle, r = cot , where a is the length of the side 2 n of a regular polygon. A

C

O

Regular Polygon

10 Properties of Triangles, Heights and Distances

cos 120° =

So,

Example 26. Let A0 , A1 , A2 , A3 , A4 and A5 be the consecutive vertices of a regular hexagon inscribed in a unit circle. The product of the lengths of A0 A1 , A0 A2 and A0 A4 is 3 (b) 3 3 (a) 4 3 3 (c) 3 (d) 2 2 2 2 Sol. (c) cos 60° = OA0 + OA1 − A0 A1

⇒ ⇒ ⇒

2 ⋅ OA0 ⋅ OA1 1 × 2 × 1 × 1 = 2 − A0 A12 2 A0 A12 = 2 − 1 = 1 A0 A1 = A1 A2

521

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Objective Mathematics Vol. 1

10

A3

O

A5

A2

cos120° = =

X

10x. Then, each side of the pentagon is 2 x and its area π …(i) = 5 x2 cot 5 [Q n = 5 and a =2 x]

A1

A0

∴ ⇒

Sol. (b) Let the perimeter of the pentagon and decagon be

O

A0 A12 + A1 A22 − A0 A22 2 ⋅ A0 A1 ⋅ A1 A2 1 + 1 − A0 A22 2 ⋅ 1⋅ 1

A0 A2 = 3 = A0 A4 A0 A1 × A0 A2 × A0 A4 = 1 × 3 × 3 = 3

Example 27. If regular pentagon and a regular decagon have the same perimeter, then the ratio of their areas is (a)1 : 5 (b) 2 : 5 (c) 5 : 2 (d) 5 :1

A

B

P

Also, as each side of decagon is x, so its area π 5 [Q n = 10 and a = x] = x2 cot 2 10 Area of pentagon 2 cot 36° 2 cos 36° sin18° = = ⇒ sin 36° cos 18° cot 18° Area of decagon 2 cos 36° sin18° cos 36° = = 2 sin18° cos 18° cos 18° cos 2 18° 2 cos 36° 2 cos 36° = = 2 2 cos 18° 1 + cos 36° =

2( 5 + 1) 5 + 1  4 

 4 1 + 

=

2( 5 + 1) 2 = 5 5+ 5

Work Book Exercise 10.2 1 If the radius of the circumcircle of an isosceles ∆ABC is equal to AB = AC, then the angle A is a c

π 6 π 2

π b 3 2π d 3

2 If a2 , b2 , c 2 are in AP, then cot A, cot B and cot C are in a b c d

b 12 − 7 3 d 4π

4 A regular polygon of nine sides, each of length 2, is inscribed in a circle. The radius of the circle is π 9

π b sin 9 π d tan 9

5 In a ∆ABC, sin A + sin B + sin C = 1 + 2, cos A + cos B + cos C = 2, if the triangle is

522

a b c d

equilateral isosceles right angled right angle isosceles

c

a2 4 a2 9

a2 6 2 a2 d 3 b

3 2 3 c 4 a

120° and radius of its incircle is 3. Then, the area of the triangle (in sq units) is

c cosec

a

incentre is parallel to BC, then cos B + cos C is equal to

3 Given an isosceles triangle, whose one angle is

π a sec 9

side a. The area of any square inscribed in the circle is

7 In a ∆ABC the line joining the circumcentre to the

AP GP HP None of the above

a 7 + 12 3 c 12 + 7 3

6 A circle is inscribed in an equilateral triangle of

b 1 d

1 2

8 In a triangle a = 13, b = 14, c = 15, then r is equal to a 4

b 8

c 2

d 6

9 The exradii of a triangle r1 , r2 and r3 are in harmonic progression, then the sides a, b and c are in a AP c HP

b GP d None of these

10 In an equilateral triangle, the inradius and the circumradius are connected by a r = 4R c r=

R 3

b r=

R 2

d None of these

Angle of elevation If O is the observer’s eye and OX is the horizontal line through O. If the object P is at a higher level than eye, then ∠POX is called the angle of elevation. P

Dimensions (a) If the total actual figure is located in one plane, the problem is of two dimensions. For directions in two dimensional figures, cross vertically as in given figure. N

Line of sight θ

O

X

Horizontal line

W

E

Angle of depression If the object P is at a lower level than O, then ∠POX is called the angle of depression. S

Horizontal line

O

X

θ

Properties of Triangles, Heights and Distances

10

Heights and Distances

(b) If the total actual figure is located in more than one plane, the problem will be of three dimensions. For directions in three dimensional figures, cross obliquely as shown. Clearly, this oblique cross represents the horizontal plane.

Line of sight

P

Bearing If the observer and the object i.e. O and P are on the same level, then bearing is defined. To measure the bearing the four standard directions East, West, North and South are taken as the cardinal directions. Angle between the line of observation i.e. OP and any one standard direction East, West, North or South is measured.

W

N 90° E

S

(c) If OP is a vertical tower perpendicular to the plane, then it will be represented as shown in the figure. Clearly, ∠POA = 90°.

N P

P

W

θ

W

E

O

N 90°

O

A S

S

Thus, ∠POE is called the bearing of the point P with respect to O measured from East to North. In other words, the bearing of P as seen from O is the direction in which P is seen from O.

B

If the observer at A moves towards East, we draw a line AB parallel to OE to represent this movement. Clearly, ∠OAB = 90° (angle between North and East).

N P1 NE

W

O

P2 ENE 45° 45° 1° 222 E 45°

Some Important Properties of Triangles i.

P3

E

In a ∆ABC, if DE || AB , then

AB BC AC = = DE DC EC A

S

SE

North-East means equally inclined to North and East. South-East means equally inclined to South and East. ENE means equally inclined to East and North-East.

E

C

D

B

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Objective Mathematics Vol. 1

10

In any ∆ABC, if AD is the angle bisector of ∠BAC, then BD AB c = = DC AC b

viii.

m-n theorem If in a ∆ABC , D is a point on the line BC such that BD : DC = m : n and ∠ADC = θ, ∠BAD = α, ∠DAC = β, then A

A A/2 A/2

c

α β θ

b

B B

iii.

In any ∆ABC, the exterior angle is equal to the sum of interior opposite angles. θ =α +β

X

A α

B

iv.

θ C

β

In an isosceles triangle, the median is perpendicular to the base. ∆ABC is isosceles and AD ⊥ BC .

n

D

C

( m + n)cot θ = m cot α − n cot β ( m + n)cot θ = n cot B − m cot C

(a) (b)

C

D

m

Example 28. P is a point on the segment joining the feet of two vertical poles of heights a and b. The angles of elevation of the tops of the poles from P are 45° each. Then, the square of the distance between the tops of the poles is a 2 + b2 (a) 2 2 (b) a + b 2 (c) 2( a 2 + b 2 ) (d) 4( a 2 + b 2 )

A

Sol. (c) In ∆APD, tan45° = a ⇒ AP = a AP

C

D

E b

B

D

a

C

A

v. vi.

In similar triangle, the corresponding sides are proportional.

45°

45° P

B

b PB PB = b DE = a + b CE = b − a DC 2 = DE 2 + EC 2

and in ∆BPC, tan45° =

In a right angled ∆ABC, the mid-point D of hypotenuse AC is equidistant from its vertices A, B and C.

⇒ ∴ and In ∆DEC,

A

= (a + b )2 + (b − a)2 = 2 (a2 + b 2 ) D X

C

B

AD = BD = CD

i.e.

vii.

Apollonius theorem ∆ABC, then

If AD is median of the

A

Example 29. From the top of a cliff 50 m high, the angles of depression of the top and bottom of a tower are observed to be 30° and 45°. The height of tower is (a) 50 m (b) 50 3 m  3 m (d) 50 1 − (c) 50 ( 3 − 1) m 3   Sol. (d) Let the height of the cliff be BD = 50 m and height of the tower be AE = h m. In ∆DEC,

B

524

D

C

AB + AC = 2 ( AD 2 + BD 2 ) 2

2

⇒

50 − h x 50 − h x= 1 3

tan30° =

x=

3 (50 − h)

…(i)

D

30° 45°

X

50–h 30°

E

C

x

50 m h 45°

A

B

x

and in ∆BAD,

Example 31. The angle of elevation of the top of a hill from a point is α. After walking b m towards the top up, a slope inclined at an angle β to the horizon, the angle of elevation of the top becomes γ. Then, the height of the hill is bsin α sin ( γ − β) bsin α sin ( γ − α ) (b) (a) sin ( γ − α ) sin ( γ − β) bsin ( γ − β) sin ( γ − β) (c) (d) sin ( γ − α ) b sin α sin ( γ − α ) Sol. (a) In ∆ABE, sinβ = BE

50 tan45° = x ⇒ x = 50 m From Eq. (i), 50 = 3 (50 − h) 3 h = 50 3 − 50 ⇒ 50 ( 3 − 1) m h= ⇒ 3  3 h = 50  1 − ⇒ m 3  

b BE = h1 = b sin β

⇒

α

γ−

α A

X

…(i) D

Example 30. A man standing 30 m South of the tower of height h, walks 60 m to the East and finds the angle of elevation of the top of the tower to be 30°. Then, the height of tower is equal to (b)15 3 m (a)15 10 m (c) 10 15 m (d) None of these

α−β β

10 Properties of Triangles, Heights and Distances

⇒

h2

γ

α E

b

F h1

B

C

Using sine rule in ∆AED, sin(α − β ) sin(γ − α ) b sin(α − β ) ⇒ ED = = ED b sin(γ − α ) h2 Now, in ∆FED, sinγ = ED b sin(α − β )sin γ …(ii) ⇒ h2 = sin(γ − α ) ∴Total height,

Sol. (c) Let the lines East-West and North-South meet at the foot of the tower T.

CD = h1 + h2 = bsinβ +

N

bsin(α − β )sin γ sin(γ − α )

[from Eqs. (i) and (ii)] b[sinβ sin(γ − α ) + sin(α − β )sin γ ] sin(γ − α )  b{sinβ (sin γ cos α − cos γ sinα)   + sin γ (sinα cos β − sinβ cos α )}  =  sin(γ − α )  b(sinβ sin γ cos α − sinβ cos γ sinα   + sin γ sinα cos β − sin γ sinβcos α )  =  sin(γ − α ) bsinα sin(γ − β ) = sin(γ − α ) =

T

W

E

30 m P

60 m

Q

S Base view

Let the man be standing at P, 30 m South of the tower. The man walks 60 m to Q parallel to the line WE. The angle of elevation of the tower at Q is 30°. R h W P

N

T E 30° Q

S 3-D view

Hence, TQ = hcot 30° = 3 h From the ∆PQT,QT 2 = PT 2 + PQ 2 ⇒

3 h2 = (30)2 + (60)2

⇒

= 4500 h = 10 15 m

X

Example 32. From the top of a hill h m high, the angle of depressions of the top and the bottom of the pillar are α and β, respectively. The height (in metre) of the pillar is h(tan β − tan α ) (a) tan β h(tan α − tan β) (b) tan α h (tan β + tan α ) (c) tan β h(tan β + tan α ) (d) tan α

525

10

Sol. (a) Let AB be a hill whose height is h m and CD be a pillar of height h′ m.

h − h′ ED h h tan β = = AC ED

Objective Mathematics Vol. 1

In ∆EDB,

tan α =

and in ∆ACB, B

…(i)

Some Properties Related to Circle i.

…(ii)

Angles in the same segment of a circle are equal. i.e. ∠APB = ∠AQB = ∠ARB Q

P

α β

R

h – h′ α

E

D

h h′ A β

A

C

ii.

From Eqs. (i) and (ii), tan α h − h′ = h tan β tan α ⇒ h⋅ = h − h′ tan β h(tan β − tan α ) h′ = ∴ tan β X

Angles in the alternate segments of a circle are equal. A α B α

Example 33. ABCD is a rectangular field. A vertical lamp post of height 12 m stands at the corner A. If the angle of elevation of its top from B is 60° and from C is 45°, then the area of the field is (a) 48 2 sq m (b) 48 3 sq m (c) 48 sq m (d)12 2 sq m Sol. (a) In ∆ABE, tan60° = 12 AB ⇒ AB = 4 3 m

C

O

A α

B C

iii.

E

α

T

O

If the line joining two points A and B subtends the greatest angle α at a point P, then the circle will touch the straight line XX ′ at the point P. A

12 m 90°

A

B

D

B α 60°

X′

45°

B

C

and in ∆ACE, 12 AC AC = 12 m

tan45° = ⇒ In ∆ABC,

BC =

526

iv.

The angle subtended by any chord at the centre is twice the angle subtended by the same on any point on the circumference of the circle. P α

AC 2 − AB2

= 144 − 48 = 4 6 m ∴ Area of rectangular field = AB × BC =4 3×4 6 = 48 2 sq m

X

P

2α A

B

Example 34. A statue on the top of a pillar subtends the same angle α at distances 9 m and 11 m from the pillar in the horizontal plane through the 1 foot of the pillar. If tan α = , then the height of 10 the pillar and height of the statue are (a) 9 m, 2 m (b) 1 m, 9 m (c) 10 m, 9 m (d) 1 m, 10 m Sol. (a) Let PQ be the pillar and QR the statue. Let A be a point 9 m from P and B be the point 11 m from P. Since, ∠RAQ = ∠RBQ, the four points A, B, R and Q are concyclic.

(a)

10

2csin α sin β cos β − cos α

2csin α sin β cos α − cos β 2csin α sin β (c) cos α + cos β (d) None of the above

(b)

Sol. (c) Let O be the initial position of the man and P and Q be the positions of the objects. Since, PQ subtends the greatest angle at R, a circle will passes through P, Q, R and OR will be tangential to this circle at R. Q θ

x

Properties of Triangles, Heights and Distances

X

P β

R

C

O

α

L

Q α

α B

D

A

P

Let D be the mid-point of AB and L be the mid-point of RQ. Draw perpendiculars from D on AB and from L on RQ respectively. Let they meet at C. Then, C is the centre of the circle passing through A, B, R and Q. ⇒ ∠RCQ = 2 ∠RAQ = 2α ⇒ ∠RCL = α. Also, CL = DP = 10 m 1 RL = LC tan α = 10 ⋅ = 1m ⇒ 10 Hence, the height of the statue is 2 m. Also, AB = 2 m [radii of the circle] ⇒ LP = CD and CR = CA ⇒ ∆ABC and ∆RQC are congrunent. ⇒ LP = CD = CL = 10 m ∴ PQ = LP − LQ = 10 − 1 = 9 m X

Example 35. A person walking along a canal observes that two objects are in the same line which is inclined at an angle α to the canal. He walks a distance c further and observes that the objects subtend their greatest angle β. Then, the distance between the object is

α

θ C

R

Let ∠PQR =∠PRO = θ [from the property of circle] We have, to find PQ i.e. x Clearly, ∠OPR = θ + β Applying sine rule in ∆PRQ, we get x PR = sin β sin θ PR sin β sinθ Applying sine rule in ∆POR, we get PR c = sin α sin (θ + β ) c sin α PR = ⇒ sin (θ + β ) ⇒

x=

From Eqs. (i) and (ii), we have 2c sin α sin β c sin α sin β x= = sin (θ + β ) sin θ 2 sin (θ + β ) sin θ 2c sin α sin β = cos β − cos (2θ + β )

…(i)

…(ii)

…(iii)

From ∆ORQ, α + θ + β + θ = 180° ⇒ ⇒

2θ + β = 180° − α cos (2θ + β ) = cos (180° − α ) = − cos α

…(iv)

From Eqs. (iii) and (iv), we get 2c sin α sin β x= cos α + cos β

527

Objective Mathematics Vol. 1

10

Work Book Exercise 10.3 1 A flagstaff stands vertically on a pillar, the height of the flagstaff being double the height of the pillar. A man on the ground at a distance finds that both the pillar and the flagstaff subtend equal angles at his eyes. Then, ratio of the height of the pillar and the distance of the man from the pillar, is a 3 :1 c 1: 3

b 1:3 d 3 :2

2 A circular ring of radius 3 cm is suspended horizontally from a point 4 cm vertically above the centre by 4 strings attached at equal intervals to its circumference. If the angle between two consecutive strings is θ, then cos θ is 4 5 4 b 25 16 c 25 d None of the above

a

3 a 5 4 b 5 3 c 4 d None of the above

3 m 2 d None of these

b

7 A man standing between two vertical posts find that the angle subtended at his eyes by the tops of the posts is a right angle. If the heights of the two posts are two times and four times the height of the man and the distance between them is equal to the length of the longer post, then the ratio of the distances of the man from the shorter and the longer post is b 2:3 d 1: 3

the top of a 24 ft high pillar and the angle of depression of its foot are complementary angles. The distance of the man from the pillar is a 2 3 ft c 6 3 ft

b 8 3 ft d None of these

9 A rocket of height h m is fired vertically upwards.

4 The shadow of a tower standing on a level ground is x metre long when the Sun’s altitude is 30°, while it is y metre long when the altitude is 3 60°. If the height of the tower is 45 × m, then 2 x − y is b 45 3 m 45 3 d m 2

5 A vertical lamp-post of height 9 m stands at the corner of a rectangular field. The angle of elevation of its top from the farthest corner is 30°, while from another corner it is 45°. The area of the field is

528

5 m 2 c 4m

a

8 A 6 ft tall man finds that the angle of elevation of

circle of radius 10 cm and of angle 216° just covers the lateral surface of a right circular cone of vertical angle 2θ. The sin θ is

a 81 2 sq m c 81 3 sq m

distance of 2 m from a wall, 4 m high. A 15 . m tall man starts to walk away from the wall on the other side of the wall, in line with the lamp-post. The maximum distance to which the man can walk remaining in the shadow is

a 3:5 c 3:2

3 A piece of paper in the shape of a sector of a

a 45 m 45 c m 3

6 A vertical lamp-post, 6 m high, stands at a

b 9 2 sq m d 9 3 sq m

Its velocity at time t seconds is (2 t + 3) m/s. If the angles of elevation of the top of the rocket from a π point on the ground after 1 s of firing is and 6 π after 3 s it is , then the distance of the point 3 from the rocket is

a b c d

14 3 m 7 3m 2 3m Cannot be found without the value of h

10 The angle of elevation of the top of a vertical pole when observed from each vertex of a regular π hexagon is . If the area of the circle 3 circumscribing the hexagon is A sq m, then the area of the hexagon is a c

3 3 A sq m 8 3 3 A sq m 4π

b d

3 A sq m π 3 3 A sq m 2π

WorkedOut Examples Type 1. Only One Correct Option Ex 1. If the data given to construct a ∆ABC is 3 a = 5, b = 7, sin A = , then it is possible to 4 construct (a) only one triangle (b) two triangle (c) infinitely many triangles (d) no triangle sin A sin B 3/ 4 sin B = = ⇒ a b 5 7 21 sin B = > 1, which is not possible ⇒ 20 Thus, no triangle is possible. Hence, (d) is the correct answer.

Sol. We have,

Ex 2. In ∆ABC, a = 5, b = 4 and 31 cos ( A − B ) = , then side c is 32 (a) 6 (b) 7 (c) 9 (d) None of the above Sol. We have, tan

A−B 1 − cos ( A − B ) = 2 1 + cos ( A − B )

31 1 32 = = 31 63 1+ 32 a−b C 1 cot = a+ b 2 63 1−

⇒

 A− B a−b C Q tan 2 = a + b cot 2    1 1 C cot = ⇒ 9 2 63 C 7 ⇒ tan = 2 3 C 1 − tan 2 2 Now, cos C = 2 C 1 + tan 2 7 1− 9 =1 cos C = ⇒ 7 8 1+ 9 Since, c2 = a2 + b2 − 2ab cos C 1 ⇒ c2 = 25 + 16 − 40 × = 36 8 ⇒ c=6 Hence, (a) is the correct answer.

Ex 3. In an acute angled ∆ABC, r + r1 = r2 + r3 and π ∠B > , then 3 (a) b + 2c < 2a < 2b + 2c (b) b + 4c < 4a < 2b + 4c (c) b + 4c < 4a < 4b + 4c (d) b + 3c < 3a < 3b + 3c

Sol. r − r2 = r3 − r1

∆ ∆ ∆ ∆ − = − s s− b s − c s − a −b c−a ⇒ = s(s − b) (s − a)(s − c) (s − a)(s − c) a − c = ⇒ s(s − b) b B a c − ⇒ tan 2 = 2 b B  π π But ∈ ,  2  6 4 B 1  ⇒ tan 2 ∈  , 1 2 3  1 a−c < 0. The triangle is (a) right angled (c) obtuse angled

(b) equilateral (d) None of these

Sol. Let a = 3x + 4 y, b = 4 x + 3 y and c = 5x + 5 y Clearly, cis the largest side and then the largest angleC is given by −2xy a2 + b2 − c2 3  

of

(a) A ≤ (c) A >

s

2

(b) A ≥

3 3 s2

s

2

3 3

(d) None of these

3

Sol. We have, 2s = a + b + c, A 2 = s(s − a) (s − b) (s − c) AM ≥ GM s + (s − a) + (s − b) + (s − c) 4 ≥ [ s (s − a) (s − b) (s − c)]1/ 4 4 s − 2s ≥ [ A 2 ]1/ 4 4 s s2 ≥ A 1/ 2 ⇒ A ≤ 4 2

Now, ⇒

⇒ ⇒ Also,

(s − a) + (s − b) + (s − c) ≥ [(s − a) (s − b) (s − c)]1/ 3 3 1

s  A2  3 ≥ 3  s 

or

or

A 2 s3 ≤ 27 s

s2 3 3 Hence, (a) is the correct answer. A≤

⇒

 A Ex 12. If in a ∆ABC , ∠B = 90°, then tan 2   is equal 2 to b−c (a) b+c b+c (b) b−c b − 2c (c) b+c

A 2 = c A b 1 + tan 2 2 A   − 1 − tan 2   2  (b − c) = A  (b + c)  + 1 − tan 2   2

 2 A  1 + tan  2  2 A  1 + tan  2

[by componendo and dividendo rule] A b−c tan 2 = 2 b+ c

Hence, (a) is the correct answer.

(a)

1 1 − 2R r

(c) r − 2R

1 r 1 1 2 −  = − 2r  R  r 2R Hence, (d) is the correct answer.

=

Ex 14. If A, B and C are angles of triangle such that ∠A is obtuse, then tan B tan C will be less than (a)

1

3 2 (d) None of these (b)

⇒ 90° > B + C > 0 ⇒ B + C < 90° ⇒ B < 90° − C ⇒ tan B < tan (90° − C ) ⇒ tan B < cot C ⇒ tan B tan C < 1 Hence, (c) is the correct answer.

1 − tan 2

Ex 13. In a ∆ABC,

10

Sol. Since, A is obtuse angle. Then, 90° < A < 180°

c b

∴

1 (1 − cos A + 1 − cos B + 1 − cosC ) 2r 1 = [ 3 − (cos A + cos B + cosC )] 2r 1   A   B   C   = 3 − 1 + 4 sin   sin   sin     2   2   2   2r   =

3

⇒

⇒

 A  A sin 2   sin 2    2  2 = = r  A  B  C  4 R sin   sin   sin    2  2  2 So, 1 r1 r r C    B  A + 2 + 3 = sin 2   + sin 2   + sin 2     2   2  2 bc ca ab r 

(c) 1

(d) None of the above Sol. cos A =

 A  B C  4 R sin   cos   cos    2  2  2 2R sin B ⋅ 2R sinC  A sin    2 =  B C  4 R sin   sin    2  2

r Sol. 1 = bc

Properties of Triangles, Heights and Distances

Ex 11. If A is the area and 2sis the sum of the sides of a triangle, then

r1 r r + 2 + 3 is equal to bc ca ab (b) 2R − r (d)

1 1 − r 2R

Ex 15. The sides of two angles of a triangle are equal 5 99 and . The cosine of the third angle is to 13 101 255 1315 255 (c) 1313

(a)

(b)

251 1313

(d) None of these

5 99 and sin C = 13 101 12 20 and cos C = cos B = 13 101 cos A = cos {180° − ( B + C )} = − cos ( B + C ) = − (cos B cos C − sin B sin C ) = sin B sin C − cos B cos C 5 99 12 20 255 = × − × = 13 101 13 101 1313 Hence, (c) is the correct answer.

Sol. Let sin B =

531

Objective Mathematics Vol. 1

10

Ex 16. If a, b, c and d are the sides of a quadrilateral, a 2 + b2 + c2 is then the minimum value of d2 1 2 1 (d) 4

(a) 1 (c)

(b)

1 3

⇒ 3 (a2 + b2 + c2 ) ≥ a2 + b2 + c2 + 2ab + 2bc + 2ca ⇒ 3 (a + b + c ) ≥ (a + b + c) > d 2

⇒ ⇒

⇒

⇒ 2 (a2 + b2 + c2 ) ≥ 2ab + 2bc + 2ca 2

sin C =

Now,

Sol. Since, (a − b)2 + (b − c)2 + (c − a)2 ≥ 0

2

1 − cos2 A sin 2 A = =1 sin 2 A sin 2 A C = 90° A = B = 45° sin A sin B sin C = = a b c 1 1 1 = = 2a 2b c a : b : c = 1: 1: 2

⇒

2

2

⇒ 3 (a2 + b2 + c2 ) > d 2 a2 + b2 + c2 1 > 3 d2 a2 + b2 + c2 1 So, minimum value of is . 3 d2 Hence, (c) is the correct answer. ⇒

⇒

Hence, (a) is the correct answer.

Ex 19. If D is a point on the side BC of a ∆ABC such that BD : DC = m : n and ∠ADC = θ, ∠BAD = α and ∠DAC = β. Then, ( m + n) cot θ is equal to (a) m cot α + n cot β (c) m cot α − n cot β

Sol. Given,

BD m = and ∠ADC = θ DC n A

Ex 17. If b = 2, c = 3 and ∠A = 30°, then inradius of ∆ABC is 3−1 (a) 2 3−1 (c) 4

(d) None of these

Sol. b = 2, c = 3 and ∠A = 30° a = b + c − 2bc cos A 2

3 = 1=1 2 A  A b + c − a tan r = (s − a) tan   =  2 2 2 = 4 + 3 − 2⋅ 2 3 ⋅

3+1 tan 15° 2 3 + 1 3 −1 = = ⋅ 2 3+1

=

3 −1 2

Hence, (a) is the correct answer.

Ex 18. If in a ∆ABC, cos A cos B + sin A sin B sin C =1, then a : b : c is equal to (a) 1: 1: 2 (c) 1: 2 : 1

(b) 1: 1: 3 (d) 1: 3 : 1

1 − cos A cos B ≤1 Sol. Here, sin C = sin A sin B 1 − cos A cos B ⇒ −1≤ 0 sin A sin B

532

α β

3+1 2

(b)

2

(b) n cot β − m cot α (d) m cot β − n cot α

⇒ 1 − cos A cos B − sin A sin B ≤ 0 ⇒ 1 − cos ( A − B ) ≤ 0 ⇒ cos ( A − B ) ≥ 1 But cos ( A − B ) cannot be greater than 1. ⇒ cos ( A − B ) = 1 ⇒ A−B=0 ⇒ A=B

B

θ m

D

n

C

Since, ∠ADB = (180° − θ ), ∠BAD = α and ∠DAC = β, ∠ABD = 180° − (α + 180° − θ ) = θ − α and ∠ACD = 180° − (θ + β ) BD AD From ∆ABD , = sin α sin(θ − α ) DC AD From ∆ADC, = sin β sin[180° − (θ + β )] DC AD or = sin β sin(θ + β )

…(i)

…(ii)

On dividing Eq. (i) by Eq. (ii), we get BD sin β sin(θ + β ) = DC sin α sin(θ − α ) m sin β sin θ cosβ + cosθ sin β = ⇒ n sin α sin θ cosα − cosθ sin α ⇒

m sin β (sin θ cosα − cosθ sin α ) = nsin α (sin θ cosβ + cosθ sin β )

⇒

m cot α − m cot θ = n cot β + n cot θ [dividing both sides by sinα sinβ sinθ]

⇒ (m + n)cot θ = m cot α − n cot β Hence, (c) is the correct answer.

Ex 20. The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60°. If the area of the quadrilateral is 4 3, then the remaining two sides are (a) 2, 3 (c) 3 + 1, 3

(b) 4, 6 (d) 3, 3

Sol. Let the points P and Q divide the side BC in three equal parts.

quadrilateral is cyclic. ∴ ∠BCD = 120°

BP = PQ = QC = x A

B

α β γ

y C x

60° A

θ

120°

5

D

2

1 3 ⋅ 2 ⋅ 5 sin 60° = 5 ⋅ 2 2 Area of ∆BCD = Area of quadrilateral ABCD − Area of ∆ABD 5 3 3 3 =4 3− = 2 2 Let CD = x and BC = y 1 Now, area of ∆BCD = ⋅ x ⋅ y ⋅ sin 120° 2 3 3 1 3 or = ⋅ xy ⋅ 2 2 2 …(i) ⇒ xy = 6 Area of ∆ABD =

On applying cosine rule in ∆BAD, we get AD 2 + AB 2 − BD 2 cos 60° = 2 AD ⋅ AB 1 22 + 52 − BD 2 ⇒ BD 2 = 19 = ⇒ 2 2⋅ 2⋅ 5 On applying cosine rule in ∆BCD, we get x 2 + y2 − 19 cos120° = 2xy 1 x 2 + y2 − 19 ⇒ − = 2 2xy 2 2 ⇒ x + y + xy = 19 Now, x 2 + y2 + 2xy = 13 + 12 = 25 ⇒ (x + y)2 = 25 …(ii) ⇒ x + y=5 and x 2 + y2 − 2xy = 13 − 12 …(iii) ⇒ x − y=±1 On solving, we get x = 3, y = 2 or x = 2, y = 3 Hence, (a) is the correct answer.

Ex 21. The base of a triangle is equal parts. If t1 , t 2 and t 3 the angles subtended by 1 opposite vertex, then   t1 equal to

divided into three are the tangents of these parts at the 11 1 +   +  is t2   t2 t3 

 1 (a) 4 1 − 2   t2 

 1 (b) 1 + 2   t2 

 1 (c) 4 1 + 2   t1 

 1 (d) 4 1 + 2   t2 

B

P

Q

C

Also, let ∠BAP = α, ∠PAQ = β, ∠QAC = γ and ∠AQC = θ From equation, tanα = t1, tanβ = t2 , tan γ = t3 On applying m : n rule in ∆ABC, we get …(i) (2x + x )cot θ = 2x cot(α + β ) − x cot γ From ∆APC, we get …(ii) (x + x )cot θ = x cot β − x cot γ On dividing Eq. (i) by Eq. (ii), we get 3 2 cot(α + β ) − cot γ = 2 cot β − cot γ ⇒ 3 cot β − 3 cot γ = 4 cot(α + β ) − 2 cot γ 4 (cot α cot β − 1) ⇒ 3 cot β − cot γ = cot β + cot α ⇒ 3 cot 2 β − cot β cot γ + 3 cot α cot β − cot α cot γ = 4 cot α cot β − 4 ⇒ 4 + 4 cot 2 β = cot 2 β + cot α cot β + cot β cot γ + cot γ cot α ⇒ 4 (1 + cot 2 β ) = (cot β + cot α ) (cot β + cot γ)  1  1 1  1 1 ⇒ 4 1 + 2  =  +   +   t2   t1 t2   t2 t3 

10 Properties of Triangles, Heights and Distances

Sol. Let AD = 2, AB = 5, ∠DAB = 60°. Since, the

Hence, (d) is the correct answer.

Ex 22. In the ∆ABC, if (a 2 + b 2 ) sin ( A − B ) = ( a 2 − b 2 ) sin ( A + B ), then the triangle is (a) either isosceles or right angled (b) only right angled (c) only isosceles triangle (d) None of the above a b c [say] = = =k sin A sin B sin C ⇒ a = k sin A , b = k sin B , c = k sin C Now, the given relation is (a2 + b2 ) sin ( A − B ) = (a2 − b2 ) sin ( A + B ) ⇒ k 2 [sin 2 A + sin 2 B ] sin ( A − B ) = k 2 [sin 2 A − sin 2 B ] sin ( A + B ) 2 2 ⇒[sin A + sin B ] sin ( A − B ) = sin 2 ( A + B ) sin ( A − B ) 2 ⇒ sin ( A − B ) [sin A + sin 2 B − sin 2 C ] = 0 Hence, either the first factor = 0 or the second factor = 0 If sin ( A − B ) = 0 ⇒ A − B = 0 ⇒ A = B ⇒ The triangle is isosceles. If sin 2 A + sin 2 B − sin 2 C = 0 a2 b2 c2 ⇒ + − =0 k2 k2 k2 2 2 2 ⇒ a + b − c = 0 ⇒ a2 + b2 = c2 So, the triangle is right angled.

Sol. Let

Hence, (a) is the correct answer.

533

Objective Mathematics Vol. 1

10

Ex 23. The sides of a ∆ABC are in AP, if the angles A and C are the greatest and the smallest angles respectively, then 4 (1 − cos A ) (1 − cos C ) is equal to A +C 2 2 A −C (c) 4 cos 2

(a) cos 2

(b) 4 cos 2 (d) 4 cos 2

A +C 2 A −C 4

Sol. Since, A is the greatest and C is the smallest angle, a is the greatest and c is the smallest side. ⇒ a, b and c are in AP. ⇒ 2b = a + c ⇒ 4 R sin B = 2R (sin A + sin C ) B B A+C A −C cos ⇒ 2 ⋅ 2 sin cos = 2 sin 2 2 2 2 B A −C  A + C B ⇒ 2 sin = cos Q = 90° −  2 2 2 2  A+C A −C …(i) ⇒ 2 cos = cos 2 2 A C Now, 4 (1 − cos A ) (1 − cos C ) = 4 ⋅ 2 sin 2 ⋅ 2 sin 2 2 2 2 2 A C A −C A+C   = 4 cos − cos ⇒ 4 2 sin ⋅ sin   2 2  2 2  2

A+C A+C A+C  = 4 cos2 − cos ⇒ 4 2 cos  2 2  2 [from Eq. (i)] Hence, (b) is the correct answer.

Ex 24. The sides of a triangle inscribed in a given circle subtend angles α, β and γ at the centre. Then, the minimum value of the arithmetic mean of π π π    cos  α +  , cos  β +  , cos  γ +  is    2 2 2 2π (b) 3

3 (a) 2

2 (c) 3

3 (d) − 2

Sol. Here, α + β + γ = 2π A

b c γ

β α

B

534

Now, arithmetic mean π   cos α +  + cos β +   2 = 3 − sin α − sin β − sin γ = 3  sin α + sin β + sin γ  =−    3

C

π π   + cos  γ +    2 2

For this arithmetic mean to be minimum, sin α + sin β + sin γ must be maximum. Clearly, sin α + sin β + sin γ will be maximum when α = β = γ. Also, α + β + γ = 2π 2π ⇒ α =β = γ = 3 Now, (sin α + sin β + sin γ)max 2π 2π 2π 3 3 = sin + sin + sin = 3 3 3 2 3 3 3 ∴ (AM)min = − =− 3⋅ 2 2 Hence, (d) is the correct answer.

Ex 25. In a triangle of base a, the ratio of the other sides is r ( 20 ⇒ BD = 72 m Hence, (c) is the correct answer.

Ex 54. Two persons who are 500 m apart, observe the direction and the angle of elevation of a balloon at the same instant. One find the π elevation to be and direction being 3 South-West, while the other find the elevation π to be and direction being West. Height of the 4 balloon is (a) 500

B

Now, using sine rule in ∆CAD, we get CA CD = sin β sin α CA ⋅ sin α asin α ⋅ sin γ = ⇒ CD = sin β sin β ⋅ sin(α + β + γ ) Hence, (b) is the correct answer.

Ex 53. The vertical poles, AB of the length 2 m and CD of length 20 m are erected with base B and D respectively. It is given that distance between the poles is more than 20 m and 2 tan( ∠ACB ) = , then distance between the 77 poles is (a) 68 m (b) 24 m (c) 72 m (d) None of the above

10

BD BD = CD 20 AE BD and tan (θ + α ) = = CE 18 2 BD + BD 77 20 ⇒ = 18 1 − 2 BD 77 ⋅ 20 ⇒ BD 2 − 77 BD + 360 = 0 tanα =

⇒

(c) 250

3 4+ 6 3 4+ 6

m

(b) 500

m

(d) 250

3 4− 6 3 4− 6

Properties of Triangles, Heights and Distances

Ex 52. A and B are two points on one bank of a straight river and C, D are two other points on the other bank of river. If direction from A to B is same as that from C to D and AB = a, ∠CAD = α, ∠DAB = β, ∠CBA = γ, then CD is equal to

m m

Sol. Let P represents the position of balloon at the time of observations and A and B are the points of observations. π We have, ∠PAQ = , 4 π π ∠PBQ = , ∠AQB = and AB = 500 m 3 4 π π h AQ = PQ cot = h, BQ = PQ ⋅ cot = 4 3 3 P

h A

Sol. We have, ∠ACB = θ

SW

π 4 θ

π 3

π 4

Q

W

B

2 where, tanθ = , AB = 2,CD = 20. 77 Let ∠BCD = α

Using cosine rule in ∆AQB, we get π AQ 2 + BQ 2 − AB 2 cos = 4 2 AQ ⋅ BQ

C θ

⇒

α

h2 − 250000 3 2h2 3 3 m h = 500 4− 6

1 = 2

h2 +

A

E

⇒

B

D

Hence, (b) is the correct answer.

543

Objective Mathematics Vol. 1

10

Ex 55. A man stands due East of the top of the tower to be θ 1 . He, then walks a m North-West and finds the angle of elevation to be θ 2 . Height of the tower is the root of the equation (a) t 2 (cot 2 θ 2 − cot 2 θ1 ) + t 2a cot θ1 + a 2 = 0 (b) t 2 (cot 2 θ1 − cot 2 θ 2 ) − t 2a cot θ1 + a 2 = 0 (c) t 2 (cot 2 θ1 − cot 2 θ 2 ) − t 2 cot θ 2 + a 2 = 0 (d) t 2 (cot 2 θ 2 − cot 2 θ1 ) − t 2 cot θ 2 + a 2 = 0 Sol. Let PQ be the tower of height h and A, B be the points of π observation. We have, AB = a, ∠PAB = , ∠QAP = θ 1, 4 and ∠QBP = θ 2

(a) 5(1 + 3 ) m 5 (b) (1 + 3 ) m 2 (c) 5( 3 − 1) m 5 (d) ( 3 − 1) m 2 Sol. Let PQ be the tower of height h and A, B be the points of observations. Then, π π ∠QAP = , ∠BAP = , AB = 10 m and BQ = 10 m 4 6 Q

Q

θ2

B

a θ1

π 4

E

A

P

PA = PQ ⋅ cot θ 1 = h cot θ 1 PB = PQ ⋅ cot θ 2 = h cot θ 2 Using cosine rule in ∆PAB, we get π PA 2 + AB 2 − PB 2 cos = 4 2PA ⋅ AB ⇒ 2ha cot θ 1 = h2 cot 2 θ 1 + a2 − h2 cot 2 θ 2 ⇒

B2

B π 6

A

P

B1

π = ∠AQB 12 π 5π ABQ = π − = ⇒ 6 6 Applying cosine rule in ∆ABQ, we get ∠QAB =

h2 (cot 2 θ 1 − cot 2 θ 2 ) − h 2a cot θ 1 + a2 = 0

AQ 2 = AB 2 + BQ 2 − 2 AB ⋅ BQ cos

⇒ h is the root of t 2 (cot 2 θ 1 − cot 2 θ 2 ) − t 2a cot θ 1 + a2 = 0

= 100 + 100 + 200 ⋅

Hence, (b) is the correct answer.

5π 6

3 2

= 200 + 100 3

Ex 56. The angle of elevation of the top of vertical tower from a point A on the horizontal ground π is found to be . From A, a man walks 10 m up 4 π a path sloping at an angle . After this, the 6 slope becomes steeper and after walking up another 10 m, the man reaches the top of the tower. The distance of A from the foot of the tower is

= 100(2 + ⇒

AQ = 10 2 +

3) 3

AP = AQ ⋅ cos

π 10 2 + = 4 2

3

=5 4+2 3 = 5 ( 3 + 1)2 ⇒

AP = 5(1 +

3) m

Hence, (a) is the correct answer.

Type 2. More than One Correct Option Ex 57. If in ∆ABC, a 4 + b 4 + c 4 = 2a 2 (b 2 + c 2 ), then ∠A is equal to (a) 45°

(b) 60°

(c) 90°

(d) 135°

Sol. We have, a + b + c = 2a (b + c ) 4

4

4

2

2

2

⇒ a4 + b4 + c4 − 2a2b2 − 2a2 c2 + 2b2c2 = 2b2c2 ⇒

(b2 + c2 − a2 )2 = 2b2c2

⇒

b2 + c2 − a2 = 2bc

or

544

⇒

b2 + c2 − a2 = − 2bc b + c − a2 1 = 2bc 2 2

2

b2 + c2 − a2 1 =− 2bc 2 1 cos A = = cos 45° ⇒ 2 π 1  or cos A = − = cos  π −   4 2 3π = cos 4 ⇒ A = 45° or A = 135° Hence, (a) and (d) are the correct answers. or

(b) − 2

(a) − 1

(c) 2

(d) 2

Sol. We are given, cosθ =

2 cos (θ − α ) cos(θ + α ) 2 (cos2 θ − sin 2 α) = cos (θ − α ) + cos (θ + α ) 2 cosθ cosα

⇒ cos2 θ cos α = cos2 θ − sin 2 α sin 2 α ⇒ cos θ = 1 − cos α 2

⇒ cos θ =

α α cos2 2 2 2 α 2 sin 2

4 sin 2

α α = 2 ⇒ cos θ sec = ± 2 2 Hence, (b) and (c) are the correct answers. ⇒ cos2 θ sec2

2

Ex 59. In ∆ABC, which of the following statements are true? (a) Maximum value of sin 2A + sin 2B + sin 2C is same as the maximum value of sin A + sin B + sin C (b) R ≥ 2r, where R is circumradius and r is inradius abc (c) R 2 ≥ a+b+c (d) ∆ABC is right angled, if r + 2R = s, where s is semi-perimeter

Sol. (a)Q sin 2 A + sin 2 B + sin 2C = 4 sin A sin B sin C Maximum value of sin 2 A + sin 2 B + sin 2C = 4 ⋅ 1⋅ 1⋅ 1 = 4 A B C and sin A + sin B + sin C = 4 cos cos cos 2 2 2 ∴ Maximum value of sin A + sin B + sin C = 4 ⋅ 1⋅ 1⋅ 1 = 4 (b) We know that A B C 1 sin sin sin ≤ 2 2 2 8 A B C ⇒ 8 sin sin sin ≤ 1 2 2 2 A B C  ⇒ 2  4 R sin sin sin  ≤ R  2 2 2 ⇒ 2r ≤ R ∴ R ≥ 2r (c) From the above relation R ≥ 2r, 2∆R R 2 ≥ 2rR ⇒ R 2 ≥ s 2abc abc 2 2 ⇒ R ≥ ⇒ R ≥ (a + b + c)  a + b + c 4    2 (d) Q ∠ABC = 90° Q ∠B = 90° ⇒ sin B = 1 ⇒ 2R sin B = 2R ⇒ b = 2R B ⇒ (s − b) + 2R = s ⇒ r tan + 2R = s 2 ⇒ r ⋅ 1 + 2R = s ⇒ r + 2R = s Hence, (a), (b), (c) and (d) are the correct answers.

Ex 60. There exists a ∆ABC satisfying the conditions π 2 π (b) b sin A > a, A > 2 π (c) b sin A > a, A < 2 π (d) b sin A < a, A < , b > a 2

(a) b sin A = a, A
a ⇒ a sin B > a ⇒ sin B > 1 Therefore, ∆ABC is not possible. (d) b sin A < a ⇒ a sin B < a ⇒ sin B < 1 ⇒ ∠B exists Now, b>a⇒ B> A π Since, A< 2 Therefore, the triangle is possible. Hence, (a) and (d) are the correct answers.

Ex 61. If a right angled of ∆ABC of maximum area is inscribed within a circle of radius R, then 1 1 1 + + = r1 r2 r3

(a) ∆ = 2R 2

(b)

(c) r = ( 2 − 1) R

(d) s = (1 + 2 ) R

2+1 R

Sol. For a right angled inscribed in a circle of radius R, the length of the hypotenuse is 2R. ∴The area is maximum when the triangle is isosceles with each side = 2 R. C

A

B

1 (2 2 + 2) R = ( 2 + 1) R 2 1 ∆= ⇒ 2R ⋅ 2R = R 2 2 1 2+1 = ⇒ r R 1 1 1 s−a s−b s−c Now, + + = + + r1 r2 r3 ∆ ∆ ∆ s 1 2+1 = = = ∆ r R Hence, (b) and (d) are the correct answers. ⇒

s=

545

Objective Mathematics Vol. 1

10 Type 3. Assertion and Reason Directions (Ex. Nos. 62-66) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 62. Statement I In a ∆ABC, if a < b < c and r is in radius and r1 , r2 , r3 are the exradii opposite to angles A, B , C respectively, then r < r1 < r2 < r3 . For ∆ABC, rrr r1 r2 + r2 r3 + r3 r1 = 1 2 3 . r

Statement II

A 2 = s(s − a) Sol. Q a abc 2 A cos 2 2 = s ∴ Σ a abc Hence, (c) is the correct answer. cos2

Ex 65. Statement I If I is incentre of ∆ABC and I 1 is excentre opposite to A and P is the intersection of II 1 and BC, then IP ⋅ I 1 P = BP ⋅ PC . Statement II In a ∆ABC, I is incentre and I 1 is excentre opposite to A, then IBI 1 C must be square. Sol. Q ICI 1 =

π 2 A

Sol. Statement I a < b < c s−a>s−b>s−c s>s−a>s−b>s−c ∆ ∆ ∆ ∆ < < < s s−a s−b s−c r < r1 < r2 < r3 ∆ ∆ ∆ ∆ Statement II r1 = , r2 = , r3 = ,r= s−a s−b s−c s 1 1 1 s 1 + + = = r1 r2 r3 ∆ r Hence, (b) is the correct answer.

Ex 63. Statement I If the sides of a triangle are 13, 14 and 15, then the radius of incircle is 4. Statement II In a ∆ABC, a +b+c ∆ = s ( s− a ) ( s − b) ( s − c), where s = 2 ∆ and r = . s Sol. s = 21,∆ = 21⋅ 8 ⋅ 7 ⋅ 6 = 3 ⋅ 7 ⋅ 24 ⋅ 7 ⋅ 3 = 3 ⋅ 7 ⋅ 4 = 84 ∆ 84 = =4 s 21 Hence, (a) is the correct answer. r=

Ex 64. Statement I

In a ∆ABC, Σ

s2 . value equal to abc

cos 2 a

A 2 has the

l B

546

(s − b ) (s − c ) (s − a ) (s − c ) B , , cos = bc 2 ac

A = 2

cos

( s − a ) ( s − b) C . = 2 ab

C

l1

π 2 ∴BICI 1 is cyclic quadrilateral BP ⋅ PC = IP ⋅ I 1 P Hence, (c) is the correct answer. IBI 1 =

Ex 66. All the notations used in Statement I and Statement II are usual. Statement I In ∆ABC, if cos A cos B cos C , then value of = = a b c r1 + r2 + r3 is equal to 9. r Statement II In ∆ABC, a b c = = = 2R , where R is sin A sin B sin C circumradius. Sol. Statement II is true. Statement I

Statement II In a ∆ABC, cos

P

∴

tan A = tan B = tan C

A = B = C i.e. a = b = c r1 = r2 = r3 ∆ r1 + r2 + r3 r1 s−a = 3⋅ = 3⋅ ∆ 3 r s a+ b+ c = 3⋅ =9 b+ c−a

Hence, (a) is the correct answer.

Passage I (Ex. Nos. 67-69) Let a, b and c be the sides

Passage II (Ex. Nos. 70-72) Consider a ∆ABC, where

opposite to angles A, B and C respectively in a ∆ABC A−B a−b a b c C . If and = = = tan cot sin A sin B sin C 2 2 a+b 4 a = 6, b = 3 and cos ( A − B ) = . 5

x , y and z are the length of perpendicular drawn from the vertices of the triangle to the opposite sides a, b and c respectively, let the letters R, r, S and ∆ denote the circumradius, inradius, semi-perimeter and area of the triangle, respectively.

Ex 67. ∠C is equal to

bx cy az a 2 + b 2 + c 2 , then the value + + = c a b k of k is (a) R (b) S 3 (c) 2R (d) R 2

Ex 70. If

π 4 π (b) 2 3π (c) 4 2π (d) 3 (a)

Sol.

Ex 68. Area of the triangle is equal to (a) 8

(b) 9

(c) 10

(d) 11

Ex 69. The value of sin A is (a) (b) (c) (d)

bx cy az + + = b sin B + c sin C + a sin A c a b b2 + c2 + a2 = 2R ∴ k = 2R Hence, (c) is the correct answer.

 1 1 1  cot A + cot B + cot C = k  2 + 2 + 2  , y z  x then the value of k is (a) R 2 (b) rR (c) ∆ (d) a 2 + b 2 + c 2

1

Ex 71. If

5 2 5 1 2 5 1

Sol.

1 1 1 a2 c2 a2 + b2 + c2 + 2+ 2+ + = 2 2 2 x y z 4∆ 4∆ 4 ∆2

3

cot A + cot B + cot C =

Sol. (Ex. Nos. 67-69) cos ( A − B ) =

A−B 2 =4 ⇒ −B 5 A 1 + tan 2 2 A−B 2 tan 2 2 = 1 ⇒ tan A − B = 1 ⇒ 2 9 2 3 1 6−3 C ⇒ = cot 3 6+ 3 2 C cos = 1 ⇒ C = 90° ⇒ 2 1 Area of triangle = ab sin C 2 1 = × 6 × 3 ×1= 9 2

⇒ ⇒

67. (b)

a2 + b2 a = 1 sin A 6 = 45 sin A 2 sin A = 5

68. (b)

R abc

(b2 + c2 − a2 + c2 + a2 − b2 + a2 + b2 − c2 )

4 5

1 − tan 2

∴

Properties of Triangles, Heights and Distances

10

Type 4. Linked Comprehension Based Questions

=

R R  4 ∆2 4 ∆2 4 ∆2  + 2 + 2  (b2 + c2 + a2 ) =  abc abc  x 2 y z  =

=

4 ∆2R  1 1 1  + 2 + 2 abc  x 2 y z 

 1  1 4 ∆R 1 1 1 1 ⋅ ∆  2 + 2 + 2 = ∆  2 + 2 + 2 abc x x y z  y z 

∴ k =∆ Hence, (c) is the correct answer.

Ex 72. The value of c sin B + b sin C a sin C + c sin A + x y b sin A + a sin B is equal to + z R r (c) 2 (a)

S R (d) 6

(b)

c sin B + b sin C x+x ,Σ =6 x x Hence, (d) is the correct answer.

Sol. Σ

69. (b)

547

D. 2R (− cos A + cos B )  b = 2R  − +  2 a + b2 

Ex 73. Match the statements of Column I with values of Column II. A.

Column II

If in ∆ABC, ∠B = 60°, then a + b + c is p. equal to 2

2

2

a−b

  a +b  a

B

2

a

b

2

+

r.

2 b 2 + ac

D. In a right angled ∆ABC, A+B ( A − B) π ∠C = , then 4R sin ⋅ sin 2 2 2

s.

abc

 a−b  = a2 + b2   =a−b A  a2 + b 2  A → r; B → p; C → s; D → q

C

b

Ex 74. Match the statements of Column I with values of Column II.

1 c2 + a2 − b2 = cos B = 2 2ca ⇒ c2 + a2 = b2 + ca

Column I

⇒ a2 + b2 + c2 = 2b2 + ca l B. sinC = b

…(i)

n a m sin A = c  bl   cm  an 2R   + 2R   + 2R    c  a  b sin B =

=

 b ∴2R  − +  2 a + b2 

a2 + b2   2 

2

C. In a ∆ABC, R ( b 2 sin 2C + c 2 sin2 B) equals

∴

 Q R = 

a+ b+c

B. If l, m and n are perpendicular drawn from the q. vertices of triangle having sides a, b andc, then bl cm an  2 R  + +  + 2 ab + 2 bc + 2ca is c a b equal to

Sol. A.

2

2

Column I

  a +b  a

2

√a

Objective Mathematics Vol. 1

10 Type 5. Match the Columns

c  bl  a  cm b  an   +   +   sin c  c  sin A  a  sin B  b 

…(ii) …(iii)

A.

In a ∆ABC, ( a + b + c ) ( b + c − a ) = λbc, where λ ∈ I, then greatest value of λ is

p.

3

B.

In a ∆ABC, tan A + tan B + tanC = 9 and if tan2 A + tan2 B + tan2 C = k, then least value of k satisfying is

q.

9( 3)1/ 3

C. In a ∆ABC, if line joining the circumcentre to the incentre is parallel to BC, then value of cos B + cos C is

r.

1

D. If the sides of ∆ABC are in AP (order being and satisfy a, b, c) 2! 2 1 8a , then the value of + + = 1! 9 ! 3 !7 ! 5 ! 5 ! (2 b )! cos A + cos B is

s.

12 /7

Sol. A. (b + c)2 − a2 = λbc ⇒

A

Column II

b2 + c2 − a2 = (λ − 2)bc

b2 + c2 − a2 λ − 2 = 2bc 2 λ−2 cos A = 1 (c) cos2 A + cos2 B + cos2 C > 1 (d) None of the above

(b) a2 − b2 = c2 (d) None of these

32. In a ∆ABC, a = 5, b = 4 and tan (a) 6 (c) 2

24. If the ∆ABC is acute angled at C, then

C = 2

7 . The side c is 9

(b) 3 (d) None of these

3 5 and cos B = . The value of 5 13

cos C can be 7 13 33 (c) 65 (a)

25. In a ∆ABC,cos B ⋅ cos C + sin B ⋅ sin C ⋅ sin 2 A = 1 Then, the triangle is (b) isosceles whose equal angles are greater than

A B and tan satisfy 6x 2 − 5x + 1 = 0. 2 2

(a) a2 + b2 > c2 (c) a2 + b2 = c2

(b) k > 27 (d) k ≤ 27

(a) right angled isosceles

(b) GP (d) None of these

then

(b) l = r 2 = d 2 (d) 2l = r = d

23. In a ∆ABC, tan A ⋅ tan B ⋅ tan C = 9. For such triangles, if tan 2 A + tan 2 B + tan 2 C = k, then

(c) equilateral (d) None of the above

(d) None of these

(a) AP (c) HP

C

(a) 9 ⋅ 3 3 < k < 27 (c) k < 9 ⋅ 3 3

2 13 , 17 17

(b)

30. In a ∆ABC, if 3 tan

O

(a) 3l = r = d (c) l = r 3 = d 3

A and 2

(a) (a + b + c)cos

(d) None of these

B

1 15 , 17 17 2 11 (c) , 17 17

(a)

21. In ∆ABC,cos 2 A + cos 2 B − cos 2 C = 1, then (a) ∠A =

(d) − 1

(c) 2

cos A are equal to

20. In a ∆ABC, if a, b and A are given, then there are two triangles with third sides c1 and c 2 , then ( c1 − c 2 ) is (b) a2 − b2 sin 2 A

1 2

27. In a ∆ABC, if a = 8, b = 15, c = 17, then sin

(b) 0 (d) None of these

(a) ab sin A

(b)

10

Targ e t E x e rc is e s

(a) λ < 0 (c) λ > 0

26. In a ∆ABC, if a, b and c are in AP, then the value of A C sin sin 2 2 is B sin 2

Properties of Triangles, Heights and Distances

17. If in a ∆ABC, ( a + b + c )( b + c − a ) = λbc, then

π 4

(b)

12 13

(d) None of these

34. In a ∆ABC, the sides a, b and c are in AP.Then,  tan A + tan C  :cot B is equal to    2 2 2 (a) 3 : 2 (c) 3 : 4

(b) 1 : 2 (d) None of these

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Objective Mathematics Vol. 1

10

35. In ∆ABC,1 − tan

B C ⋅ tan is equal to 2 2

2a a+ b+ c 2c (c) a+ b+ c

(a)

(b)

44. In a ∆ABC, if

2b a+ b+ c

(a) scalene (b) isosceles (c) right angled (d) equilateral

(d) None of these

36. In ∆ABC, a : b : c = (1 + x ) : 1: (1 − x ), where x ∈ ( 0, 1). π If ∠A = + ∠C , then x is equal to 2 1 6 1 (c) 7

1 2 3 1 (d) 2 7

(a)

R ≤ 2, then the triangle is r

(b)

45. In the given figure, P is any arbitrary interior point of the ∆ABC, H a , H b and H c are the length of altitudes drawn from vertices A, B and C respectively. If x a , x b and x c represent the distance of P from sides BC, AC and AB respectively, then xa x x + b + c is always equal to Ha Hb Hc A

37. If a, b and c are the sides of a triangle such that b ⋅ c = λ 2 , for some positive λ, then

Ta rg e t E x e rc is e s

C (a) c ≥ 2λ sin 2 A (c) a ≥ 2λ sin 2

(d) None of these

B

38. ABC is a triangle, B = 60°, C = 30° . BC is produced 8 to D so that ∠ADB, then given 3 = 1 and 11 BC × CD = 23 × 32 × 19 × 11, the square of the altitude from A to BC is (a) 3250 (c) 3249

(b) 3248 (d) None of these

39. If c 2 = a 2 + b 2 , then 4s ( s − a )( s − b )( s − c ) is equal to (b) c2a2 (d) s4

(a) a2b2 (c) b2c2

(a) non-negative (c) negative (d) positive

area

of

a

be

λ,

then

2

(a) 2λ (c) 4λ

(b) λ (d) None of these

42. In a ∆ABC, R = circumradius and r = inradius. The a cos A + b cos B + c cos C is equal to value of a+b+c R r r (c) R (a)

R 2r 2r (d) R

(b)

43. In a ∆ABC, 2s = perimeter and R = circumradius. s Then, is equal to R 552

(a) sin A + sin B + sin C A B C (c) sin + sin + sin 2 2 2

(b) cos A + cos B + cosC (d) None of these

(b) 2 (d) None of these

π 46. In a ∆ABC, a = 2b and A − B= . The measure of 3 ∠C is π 4 π (c) 6 (a)

(b)

π 3

(d) None of these

47. If the sides of a right angled triangle are in AP, then tangents of the acute angled triangle are 3+

(c) 3,

1 , 2

3−

1 2

1 3

(b)

5−

1 , 2

5+

1 2

3 4 (d) , 4 3

48. In a ∆ABC, the angles A and B are two values of θ satisfying 3 cos θ + sin θ = k ,k< 2. The triangle

(b) non-positive

∆ABC a sin 2B + b sin 2A is equal to the

2

C

(a) 3 (c) 1

(a)

40. In ∆ABC, ( a + b − c )( b + c − a )( c + a − b ) − abc is always

41. If

P

B (b) b ≥ 2λ sin 2

(a) is an acute angled (c) is an obtuse angled

(b) is a right angled π (d) has one angle 3

3π 4 and the other two angles are equal to two values of θ satisfying a tan θ + b sec θ = c, where b≤ a 2 + c 2 ,

49. If in an obtuse angled triangle the obtuse angle is

then a 2 − c 2 is equal to (a) ac a (c) c

(b) 2ac (d) None of these

50. The length of the sides of a triangle are x, y and x 2 + y 2 + xy. The measure of the greatest angle is (a) 120° (b) 150° (c) 135° (d) None of the above

(b) 90° (d) None of these

C

52. In ∆ABC, internal angle bisector of ∠A makes an angle θ with side BC. Value of sin θ is equal to  B −C (a) sin    2   B −C (c) cos    2 

 B (b) sin  − C   2  B (d) cos  − C   2

A

53. In a ∆ABC, a : b : c = 4 : 5 : 6. The ratio of the radius of the circumcircle to that of the incircle is (a)

16 9

(b)

16 7

(c)

11 7

(d)

7 16

54. If r1 , r2 and r3 are exradii of any triangle, then r1 r2 + r2 r3 + r3 r1 is equal to ∆ r r (c) ∆

∆2 r2 r2 (d) 2 ∆ (b)

(a)

(a) 1 : 3 : 2 (c) 2 : 3 : 4

(b) a < b < c (d) a < b and b > c

57. If x, y and z are perpendicular drawn from the vertices of triangle having sides a, b and c, then the value of bx cy az will be + + c a b a2 + b2 + c2 2R a2 + b2 + c2 (c) 4R

a2 + b2 + c2 R 2(a2 + b2 + c2 ) (d) R (b)

58. If the sides of a triangle are in the ratio 3 : 7 : 8, then R : r is equal to (b) 7 : 2

(c) 3 : 7

(d) 7 : 3

59. If the length of median AA1 , BB1 and CC1 of ∆ABC are ma , mb and mc , respectively. Then, 3 (a) Σ ma > s 2 5 (c) Σ ma > s 2

(b) Σ ma > 3s (d) Σ ma > 2s

60. In a ∆ABC, B = 90° , AC = h and the length of the perpendicular from B to AC is p such that h = 4 p. If AB < BC, then ∠C has the measure 5π (a) 12 π (c) 12

3 cm 7

7 cm 3

(b)

3 cm 7

(d) None of these

62. If α, β and γ are the altitudes of a ∆ABC and 2s denotes its perimeter, then α − 1 + β − 1 + γ − 1 is equal to s ∆ (d) None of these (b)

63. If in a ∆ABC, a 2 + b 2 + c 2 = 8R 2 , where R = circumradius, then the triangle is (a) equilateral (b) isosceles (c) right angled (d) None of the above

56. In a triangle, if r1 > r2 > r3 , then

(a)

(c)

B

D 5 cm

∆ s (c) s ⋅ ∆

(b) 1 : 1 : 1 (d) 1 : 2 : 3

(a) a > b > c (c) a > b and b < c

(a) 5

3 cm

(a)

55. In an equilateral triangle, the ratio of the inradius, circumradius and exradius are

(a) 2 : 7

60°

π (b) 6

(d) None of these

10 Properties of Triangles, Heights and Distances

(a) 60° (c) 120°

61. In the given figure, ABC is a triangle in which C = 90° and AB = 5 cm. D is a point on AB such that AD = 3 cm and ∠ACD = 60°. Then, the length of AC is

64. A ∆ABC is right angled at B. Then, the diameter of the incircle of the triangle is (a) 2(c + a − b) (c) c + a − b

(b) c + a − 2b (d) None of these

s s s 65. In any ∆ABC, 4  − 1  − 1  − 1 is equal to a  b  c  r R 3r (c) R

(b)

(a)

66. In any ∆ABC,

Targ e t E x e rc is e s

51. The greatest angle of a triangle whose sides are x 2 + x + 1, 2x + 1 and x 2 − 1, is

2r R

(d) None of these

1 r12

+

1 r22

+

a2 + b2 + c2 2∆2 2 a + b2 + c2 (c) 3∆2

(a)

1 r32

+

(b)

1 r2

is equal to

a2 + b2 + c2 ∆2

(d) None of these

67. In any ∆ABC, rr1 + r2 r3 is equal to (a) ab (b) ac (c) bc (d) None of the above

68. In an equilateral triangle, (a) r1 = r2 = r3 = 2r (b) r1 = r2 = r3 = r (c) r1 = r2 = r3 = 3r (d) None of the above

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Objective Mathematics Vol. 1

10

69. If in a ∆ABC, AD, BE and CF are the altitudes and R is the circumradius, then the radius of the circle DEF is R 2 (c) R (a)

(d) None of these

70. If R is the radius of the circumcircle of the ∆ABC and its area is ∆, then a+ b+ c ∆ abc (c) R = 4∆

(a) R =

(a) 4, 6, 8 (c) 6, 8, 10

(a)

a+ b+ c 4∆ abc (d) R = ∆

B

(b) 3, 9, 11 (d) None of these

4 3

6 5 5 (d) 2

(a) 3 (b) 3 (c) 3 3 (d) None of the above

(a) 2r

(b) 2R

r1 + r2 is always equal to 1 + cos C (c)

2r2 R

(d)

2R 2 r

75. If A, A1 , A 2 , A 3 are the area of the incircle and 1 1 1 excircles, then is equal to + + A1 A2 A3 1 A 3 (c) A

(b)

(a)

2 A

(d) None of these

76. If H is the orthocentre of ∆ABC, then AH is equal to (a) c cot A

(b) b cot A

(c) a cot B

(d) a cot A

77. ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to then the triangle has perimeter BC, 2 P = 2( 2hr − h + 2hr ) and area A is equal to (a) h 3rh − h2 (c) 2h 2rh − h

(b) h 2rh − h2 2

(b) 3

(c)

3 2

(d) 2 3

79. In ∆ABC, base BC and area of triangle ‘∆’ are fixed. Locus of the centroid of ∆ABC is a straight line that is (a) parallel to side BC (b) right bisector of side BC (c) right angle of BC

 ∆ (d) inclined at an angle sin − 1   to side BC  BC 

73. In ∆ABC, line joining the circumcentre and orthocentre is parallel to side AC, then the value of tan A ⋅ tan C is equal to

74. In ∆ABC, the value of

C

If x a , x b and x c represent the distance of P from the sides BC,CA and AB respectively, then x a + x b + x c is equal to (a) 6

(b)

(c) 2

P

(b) R =

72. In ∆ABC, BB1 is the bisector of ∠B. Altitude drawn from A to BB1 meets side BC in D1 and BB1 in D2 . AD1 is Value of AD2

Ta rg e t E x e rc is e s

A

(b) 2R

71. The radii r1 , r2 , r3 of escribed circles of the ∆ABC are in HP. If its area is 24 sq cm and its perimeter is 24 cm, then the lengths of its sides are

554

78. In the given figure, P is any interior point of the equilateral ∆ABC of side length 2 units.

(d) None of these

π , AA1 and AA 2 are the median and 2 altitude respectively. If A A ∠BAA1 = ∠A1 AA 2 = ∠A 2 AC, then sin 3 ⋅ cos is 3 3 equal to

80. In ∆ABC, ∠A =

3a3 16b2c 3a2 (c) 2 4b c (a)

3a3 64 b2c 3a3 (d) 128b2c (b)

81. At a point A, the angle of elevation of a tower is such 5 that its tangent is , on walking 120 m nearer the 12 3 tower the tangent of the angle of elevation is . The 4 height of the tower is (a) 225 m (b) 200 m (c) 230 m (d) None of the above

82. A observer on the top of a cliff 200 m above the sea level, observes the angles of depression of two ships on opposite sides of the cliff to be 45° and 30°, respectively. The distance between the ships if the line joining them points to the base of cliff, is (a) 100( 3 + 1) m (b) 200( 3 + 1) m (c) 150( 3 + 1) m (d) None of the above

(b) 200 2 m (d) None of these

84. The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of 100 m from its base is 45°. If the angle of elevation of the top of the complete pillar at the same point is to be 60°, then the height of the incomplete pillar is to be increased by (a) 50 2 m (c) 100 ( 3 − 1) m

(b) 100 m (d) 100 ( 3 + 1) m

(c) 10 (1 +

(b) 20 m  3 (d) 10 1 + m 2 

2) m

86. If a flagstaff 6 m high placed on the top of a tower throws a shadow of 2 3 m along the ground, then the angle (in degree) that the Sun makes with the ground, is (a) 30° (c) 45°

(b) 60° (d) None of these

87. The angle of elevation of the top of a tower from the top and bottom of a building of height a are 30° and 45°, respectively. If the tower and the building stand at the same level, the height of the tower is a(3 + 3 ) 2 (c) a 3

(b) a(3 +

(a)

3)

(d) a( 3 − 1)

88. At a point 15 m away from the base of a 15 m high house, the angle of elevation of the top is (a) 90° (c) 30°

(b) 60° (d) 45°

3 + 1 m 3 − 1 3 + 1 m 3 − 1

 (b) 30   (d) 60 

3 − 1 m 3 + 1 3 − 1 m 3 + 1

h 3

(b)

h 3d

(c) 3h

(a) (150 − 20 3 ) m (c) 216 m

(a) 75 3 ft (c) 150 3 ft

(d)

70 m 2

(d) 35 m

(b) 200 m (d) None of these

(b) 200 3 ft (d) None of these

94. An aeroplane flying at a height of 300 m above the ground passes vertically above another plane at an instant when the angle of elevation of the two planes from the same point on the ground are 60° and 45°, respectively. Then, the height of the lower plane from the ground (in metre) is 100 3 (d) 150 ( 3 + 1)

(a) 100 3

(b)

(c) 50

95. A person walking alone a straight road observes that at two points 1000 m apart, the angles of elevation of π 5π . The a vertical tower in front of him are and 6 12 height of the tower is equal to (a) 250 ( 3 + 1) m (c) 250 ( 2 + 1) m

3h d

(b) 250 ( 3 − 1) m (d) 250 ( 2 − 1) m

96. A chimney of 20 m high, standing vertically on the 1 top of a building, subtends an angle tan −1   at a  6 distance of 70 m from the foot of the building. The height of the building is (b) 90 m

(c) 50 m

(d) 30 m

97. A and B are two points 30 m apart in a line on the horizontal plane through the foot of a tower lying on opposite sides of the tower. The distances of the top of the tower from A and B are 20 m and 15 m respectively. If the angle of elevation of the top of the tower at A is θ and the height of the tower is h, then (a) cos θ = 28/ 36 (c) h = 5 5 / 3

90. A tower subtends an angle of 30° at a point distant d from the foot of the tower and on the same level as the foot of the tower. At a second point, h vertically above the first, the angle of depression of the foot of the tower is 60°. The height of the tower is (a)

(c)

92. The horizontal distance between two tower is 60 m. The angular elevation of the top of the taller tower as seen from the top of the shorter one is 30°. If the height of the taller tower is 150 m, the height of the shorter one is

(a) 25 m

89. From the top of a light house 60 m high with its base at sea level, the angle of depression of a boat is 15°. The distance of the boat from the light house is  (a) 30   (c) 60 

(b) 70 2 m

93. If the elevation of the Sun is 30°, then the length of the shadow cast by a tower of 150 ft height, is

85. A tree is broken by wind, its upper part touches the ground at a point 10 m from the foot of the tree and makes an angle of 45° with the ground. The entire length of the tree is (a) 15 m

(a) 70 m

10 Properties of Triangles, Heights and Distances

(a) 100 (2)1/ 4 m (c) 200 (2)1/ 4 m

91. The angle of depression of a point situated at a distance of 70 m from the base of a tower is 45°. The height of the tower is

Targ e t E x e rc is e s

83. A tower is observed from two stations A and B, where B is East of A at a distance of 200 m. The tower is due North of A and due West of B. The angles of elevation of the tower from A and B are complementary. The height of the tower is

(b) cos θ = 43/ 48 (d) h = 5 455 /12

98. The longer side of a parallelogram is 10 cm and the shorter is 6 cm. If the longer diagonal makes an angle 30° with the longer side, then length of the longer diagonal is (a) (5 3 + 11) cm (c) (5 3 + 13 ) cm

(b) (4 3 + 11) cm (d) None of these

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Objective Mathematics Vol. 1

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99. PQ is a post of height a and AB is a tower at some distance, α and β are the angles of elevation of B, the top of the tower, at P and Q, respectively. The height of the tower is a sin α cos β (a) sin (α − β ) a sin α sin β (c) sin (α − β )

a cos α sin β (b) sin (α − β )

(d) None of these

100. A man observes that when he move up a distance c m on a slope, the angle of depression of a point on the horizontal plane from the base of the slope is 30° and when he moves up further a distance c m the angle of depression of that point is 45°. The angle of inclination of the slope with the horizontal is (a) 60°

(b) 45°

(c) 75°

Ta rg e t E x e rc is e s

 a 2 + b2  (b) b  2  2 a −b   a 2 + b2  (d) a  2  2 a −b 

102. AB is a vertical pole and C is its middle point. The end A is on the level ground and P is any point on the level ground other than A. The portion CB subtends an angle β at P. If AP : AB = 2 : 1, then β is equal to (a) tan −1 (c) tan −1

4 9 5 9

(b) tan −1 (d) tan −1

1 9 2 9

103. Each side of an equilateral triangle subtends an angle of π/ 3 at the top of a tower of height h, located at the centre of the triangle. If a is the length of the side of the triangle, then (a) 3a2 = h2 (c) 2a2 = h2

(b) 2a2 = 3h2 (d) None of these

104. A house of height 100 m subtends a right angle at the window of an opposite house. If the height of the window is 64 m, then the distance between the two houses is (a) 48 m (c) 54 m

(b) 36 m (d) 72 m

105. The angle of elevation of a cloud from a point h m above a lake is θ1 and the angle of depression of its reflection is θ 2 . Then, height of the cloud above the lake is tan (θ 1 + θ 2 ) tan (θ 2 − θ 1 ) sin (θ 1 + θ 2 ) (c) h sin (θ 2 − θ 1 )

(a) h

556

cos (θ 1 + θ 2 ) cos (θ 1 − θ 2 ) cot (θ 1 + θ 2 ) (d) h cot (θ 1 − θ 2 ) (b) h

l

(a)

cot y − cot x 2l 2

(c)

2

cot 2 y − cot 2 x

(b)

l tan y − tan 2 x 2

(d) None of these

107. A pole stands vertically on the centre of a square. When α is the elevation of the Sun, its shadow just reaches the side of the square and is at a distance x and y from the ends of that side. The height of the pole is x 2 + y2 tanα 2 x 2 − y2 tan α 2

(a)

(d) 30°

101. A tower of height b subtends an angle at a point O on the level of the foot of the tower and at a distance a from the foot of the tower. If the pole mounted on the tower also subtends on equal angle at O, then height of the pole is  a2 − b2  (a) b  2  2 a + b   a2 − b2  (c) a  2  2 a + b 

106. The angle of elevation of a tower from a point A due South of it is x and from a point B due East of A is y. If AB = l, then the height h of the tower is given by

(c)

(b)

x 2 + y2 cotα 2

(d) None of these

108. If a flagstaff subtends equal angles at the points A , B , C and D on the horizontal ground through the foot of flagstaff, then the points A, B, C and D necessarily form a (a) rectangle (c) square

(b) parallelogram (d) None of these

109. The angles of elevation of the top of a tower from two points (collinear with foot of tower) on the ground at a distance a and b ( b > a ) from its foot are found to be α and β. If h is the height of tower, then (a) (b − a) = h (tan α − tan β ) (b) (b − a) = h (cot α − cot β ) (c) (b − a) = h (tan β − tan α ) (d) (b − a) = h (cot β − cot α )

110. A flagstaff standing vertically at the vertex A of an π isosceles ∆ABC, subtends angle at the mid-point of 3 π the base BC and an angle at the vertex B. If BC = a, 6 then height of flagstaff is (a)

a 2

(c) a

3 2 3 2

a 3 3 2 a 3 (d) 4 2

(b)

111. A tower of height h standing vertically at the centre of a square of side length a subtends the same angle θ at all the corner points of the square. Then, (a) 2h2 = a2 tan 2 θ (c) a2 = 2h2 cot 2 θ

(b) a2 = 2h2 tan 2 θ (d) 2a2 = h2 cot 2 θ

112. The angle of elevation of the top of a vertical tower PQ, Q being its foot standing on the horizontal ground, from two points A and B on ground are found to be θ A and θ B , respectively. If Q, A and B are collinear and QA = 4 m, QB = 9 m and θ A , θ B are complementary angles, then height of tower is (a) 3 m

(b) 6 m

(c) 4 m

(d) 16 m

114. Three vertical tower standing at A, B, C subtends the angle θ A , θ B and θC , respectively at the circumcentre of ∆ABC, then tan θ A , tan θ B and tan θC are in (a) AP (b) GP (c) HP (d) None of the above

115. On the top of a hemispherical dome of radius r, there stands a flag of height h. From a point on the ground the elevation of the top of the flag is 30°. After moving a distance d towards the dome, when the flag is just visible , then elevation is 45°, then r and h are

116. A spherical ball of radius r subtends an angle θ1 at a point P on the ground. If the angle of elevation of centre of the ball at P is θ 2 , then height of the centre of the ball from the ground is (a) r sin θ 2 ⋅ cosec (c) r sin

θ1 2

θ1 ⋅ cosec θ 2 2

θ2 ⋅ cosec θ 1 2 θ (d) r sinθ 1 ⋅ cosec 2 2

(b) r sin

117. A circular ring of radius r m is suspended from a point h m vertically above the centre of ring by four strings of equal lengths attached at equal intervals to the circumference of the ring. If θ is the angle between two consecutive strings, then cos θ is equal to (a)

h2 h + r2 2

(b)

r2 h + r2 2

(c)

2h2 h + r2 2

(d)

2r2 h + r2 2

Type 2. More than One Correct Option 118. If in a ∆ABC, sin C + cos C + sin ( 2B + C ) − cos ( 2B + C ) = 2 2, then ∆ABC is (a) equilateral (b) isosceles (c) right angled (d) obtuse angled

119. If r1 , r2 and r3 are the radii of the escribed circles of a ∆ABC and r is the radius of its incircle, then the root(s) of the equation x 2 − r ( r1 r2 + r2 r3 + r3 r1 ) x + r1 r2 r3 − 1 = 0 is/are (a) 1 (c) r

(b) r1 + r2 + r3 (d) r1r2r3 − 1

120. Let ABC be an isosceles triangle with base BC. If r is the radius of the circle inscribed in the ∆ABC and r1 be the radius of the circle escribed opposite to the angle A, then the product r1 r can be equal to (a) R 2 sin 2 A (b) R 2 sin 2 2 B 1 (c) a2 2 a2 (d) 4

Properties of Triangles, Heights and Distances

(a) x 2 = (a + b + c) / abc (b) x 2 = 1 / (a + b + c) (c) tan α + tan β + tan γ = (a + b + c)1/ 2 / abc (d) tan α tan β tan γ = (a + b + c)3/ 2 / abc

10

d d ( 3 + 1) , 2 2 2 d ( 3 + 1) d ( 3 + 1) ( 2 − 1) (b) , 2 2 2 2 d ( 3 + 1) d ( 2 − 1) (c) , 2 2 2 2 (d) None of the above (a)

121. If cos B cos C + sin B sin C sin 2 A = 1, then ABC is (a) isosceles (b) right angled (c) equilateral (d) None of the above

122. A pole of length h stands inside a triangular plot ABC and subtends equal angles α at its vertices, then (a) 2 h cos α (b) 2 h cos α (c) 2 h cos α (d) 2 h cos α

Targ e t E x e rc is e s

113. A monument ABCD stands at a point A on a level ground. At a point P on the ground, the portions AB , AC , AD subtend angles α , β , γ respectively. If and AB = a, AC = b, AD = c, AP = x α + β + γ = 180° , then

sin A = a sin α sin C = c sin α sin B = b sin α =R

123. The area of a regular polygon of n sides is (where, r is inradius, R is circumradius and a is side of the triangle) nR 2  2π  sin    n 2  π 2 (b) nr tan    n na2 π (c) cot 4 n  π 2 (d) nR tan    n (a)

557

Objective Mathematics Vol. 1

10

Type 3. Assertion and Reason Directions (Q. Nos. 124-128) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Satement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

124. In acute angled ∆ABC , a > b > c Statement I r1> r2 > r3 Statement II cos A < cos B < cos C 125. Statement I In ∆ABC, if tan A : tan B : tan C = 1: 2 : 3, then A = 45°.

Ta rg e t E x e rc is e s

Statement II

If p : q : r = 1: 2 : 3, then p = 1.

126. Statement I If in a triangle, orthocentre, circumcentre and centroid are rational points, then its vertices must also be rational points.

Statement II If the vertices of a triangle are rational points, then the centroid, circumcentre and orthocentre are also rational points. 127. Let the equations of the sides BC , CA and AB of the ∆ABC be Li : a i x + bi y + c i = 0 for i = 1, 2, 3 respectively such that a i2 + bi2 = k for i = 1, 2, 3. Let the equations of bisectors of internal angles B and C be L1 + L3 = 0 and L1 + L2 = 0. Statement I The bisector of internal angle A is L2 − L3 = 0. Statement II Any line passing through the points of intersection of the lines L1 and L2 can be written as L1 + λL2 = 0. 128. In the ∆ABC, AD is the angle bisector of ∠BAC meeting BC at D. Circumcircles S 1 and S 2 are drawn to circumscribe ∆ABD and ∆ACD, respectively. If S 1 cuts AC at E and S 2 cuts AB at F , then Statement I

BF = EC

Statement II If from an exterior point P, two secants are drawn to cut a circle at A , B and C , D, then PA × PB = PC × PD.

Type 4. Linked Comprehension Based Questions Passage I (Q. Nos. 129-131) Given that ∆ = 6, r1 = 2, r2 = 3, r3 = 6

129. Circumradius R is equal to (a) 2.5 (b) 3.5 (c) 1.5 (d) 4.2

132. Ratio of area of ∆DEF to ∆ABC is (a) 2cos A cos B cosC (b) 2sin A sin B sin C (c) 2cos A cos B sin C (d) None of the above

133. Radius of the circumcircle of ∆DEF is R 2 (b) R R (c) 4 (d) None of the above (a)

130. Inradius r is equal to (a) (b) (c) (d)

2 1 1.5 2.5

131. Difference between the greatest and the least angles is 4 (a) cos−1 5 3 (b) tan −1 4 3 (c) cos−1 5 (d) None of the above

Passage II (Q. Nos. 132-134) Let ABC be a triangle,

558

from vertices A, B and C altitudes AD, BE and CF are drawn to opposite sides BC, CA and AB respectively, which meet at a point O. Now, ∆DEF is completed, also length OA, OB and OC, respectively are p, q and r.

134. Radius of the incircle of ∆DEF is (a) 2R cos A cos B cosC (b) R cos A cos B cosC abc cos A cos B cosC (c) ∆ABC (d) None of the above

Passage III (Q. Nos. 135-137) O is the circumcentre of ∆ABC. R1, R 2, R 3 are the radii of the circumcircles of ∆OBC,OCA and ∆OAB, respectively. r1, r2, r3 are the radii of the circles drawn on the altitudes OD, OE and OF of these triangles respectively. ∆1, ∆ 2, ∆ 3 being the respective areas of these triangles. R, r are respectively the circumradius and inradius of ∆ABC, ∆ being its area and a, b, c are the lengths of the sides BC, CA and AB, respectively.

4 ∆1 (a) R2 4 (c) 2 (a + b + c) R

Passage IV (Q. Nos. 138-140) The two adjacent 4∆ (b) 2 R

sides AB and BC of a cyclic quadrilateral ABCD are 2 and 5 units, respectively and the angle between them is 60°. If the area of quadrilateral is 4 3 sq units.

(d) None of these

138. The area of ∆ADC is

(a) 4 tan A tan B tan C (c) 4 cot A cot B cot C

3 2 3 (d) 2 (b)

(a) 3 3

a b c 136. + + equals r1 r2 r3

(c) (b) 4 sin A sin B sin C (d) 4 cos A cos B cos C

R R R 137. 1 + 2 + 3 equals r1 r2 r3

3 3 2

139. The area of circle circumscribing the quadrilateral ABCD is (a) 48π 19π (b) 3 (c) 12π 19π (d) 12

 sin A sin B sin C  (a) R 2  + + ∆2 ∆ 3   ∆1  cos A cos B cosC  (b) R 2  + + ∆2 ∆ 3   ∆1  tan A tan B tan C  (c) R 2  + + ∆2 ∆ 3   ∆1  cot A cot B cot C  (d) R 2  + + ∆2 ∆ 3   ∆1

140. For given quadrilateral AD may be equal to (a) AB (b) BC (c) DC (d) None of the above

Type 5. Match the Columns 141. Match the statements of Column I with the values of Column II. Column I

Column II

A.

In a ∆ABC, if p. a 4 − 2( b 2 + c 2 )a 2 + b 4 + b 2c 2 + c 4 = 0, then ∠A is

30°

B.

In a ∆ABC, if a 4 + b 4 + c 4 = a 2b 2 + 2 b 2c 2 + 2c 2a 2, then ∠C is

q.

60°

r.

90°

C. In a ∆ABC, if a 4 + b 4 + c 4 + 2 a 2c 2 = 2 a 2b 2 + 2 b 2c 2, then ∠B is

10 Properties of Triangles, Heights and Distances

a b c equals + + R1 R 2 R 3

142. If r1 , r2 , r3 represent the radii of the escribed circles opposite to angles A, B and C, respectively of a ∆ABC r and R represent the inradius and circumradius of the same triangle respectively, then match the following: Column I In a ∆ABC, if R = 40, then r can be equal to

p.

1 2009

B.

If

r1 + r2 + r3 = λ, then λ can be equal to r rr r If 1 23 3 = µ, then µ can be equal to r

q.

25

r.

10

In ∆ABC, angle A is obtuse, then tan B ⋅ tanC can be equal to

s.

2009

C. s.

120°

t.

150°

Column II

A.

D.

Targ e t E x e rc is e s

135.

Type 6. Single Integer Answer Type Questions 143. In a ∆ABC, if a is the arithmetic mean and b, c are two geometric means between any two positive number. Then,

sin 3 B + sin 3 C is equal to sin A sin B sin C

.

9 4 and a = 2, then the value of 3∆, where ∆ is area of triangle, is ______.

146. In ∆ABC, sin A sin B + sin B sin C + sin C sin A =

b c is equal to + c+a a+b

147. In any ∆ABC, if sin A, sin B, and sin C are in AP and B λ the maximum value of tan = , then λ + µ is __. 2 µ

145. In a triangle of the base a the ratio of the other sides is r(< 1) and the altitude of the triangle is less than or ar . , then λ − µ is equal to λ − µr 2

148. The incircle of ∆ABC touches BC at P, AC at Q and AB at R. If G is foot of perpendicular from P to QR, RG CQ is equal to . then ⋅ QG BR

144. In a ∆ABC, if A = 60°, then .

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Entrances Gallery JEE Advanced/IIT JEE 1. In a triangle, the sum of two sides is x and the product of the same two sides is y. If x 2 − c 2 = y, where c is the third side of the triangle, then the ratio of the inradius to the circumradius of the triangle is [2014] (a)

3y 3y 3y (b) (c) 2x (x + c) 2c(x + c) 4 x (x + c)

(d)

3y 4 c(x + c)

1 2. In a ∆PQR, P is the largest angle and cos P = . 3 Further, the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then, possible lengths(s) of the side(s) of [2013] the triangle is/are (a) 16 (c) 24

(b) 18 (d) 22

Ta rg e t E x e rc is e s

3. Let PQR be a triangle of area ∆ with a = 2 , b =

7 and 2

5 c = , where a, b and c are the lengths of the sides of 2 the triangle opposite to the angles at P, Q and R 2sin P − sin 2P respectively. Then, [2012] equals 2sin P + sin 2P (a)

3 4∆

(b)

45 4∆

 3 (c)    4 ∆

2

 45  (d)    4 ∆

2

(a)

1 2

(b)

3 2

(c) 1

(d)

3

π and a, b 6 and c denote the lengths of the sides opposite to A, B and C, respectively. The value(s) of x for which a = x 2 + x + 1, b = x 2 − 1and c = 2x + 1is(are) [2010]

5. Let ABC be a triangle such that ∠ACB =

(a) − (2 + 3 ) (b) 1 + 3 (c) 2 + 3 (d) 4 3

6. Consider a ∆ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C, respectively. Suppose a = 6, b = 10and the area of the triangle is 15 3. If ∠ACB is obtuse and if r denotes the radius of the incircle of the triangle, then r 2 is equal to . [2010]

JEE Main/AIEEE 7. If the angles of elevation of the top of a tower from three collinear points A, B and C on a line leading to the foot of the tower are 30°, 45° and 60° [2015] respectively, then the ratio AB : BC is (a) 3 : 1 (c) 1 : 3

(b) 3 : 2 (d) 2 : 3

8. A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then, the speed (in m/s) of the bird is [2014] (a) 40( 2 − 1) (b) 40( 3 − 2 ) (c) 20 2 (d) 20( 3 − 1)

(a) 2 +

2

(b) 2 − 2(c) 1 +

10. ABCD is a trapezium such that AB and CD are parallel and BC ⊥ CD. If ∠ADB = θ, BC = p and [2013] CD = q, then AB is equal to ( p2 + q2 )sin θ p cosθ + q sin θ p2 + q2 cosθ (b) p cosθ + q sin θ p2 + q 2 (c) 2 p cosθ + q2 sin θ ( p2 + q2 )sin θ (d) ( p cosθ + q sin θ )2 (a)

11. For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement among the following is [2010] r R r (b) there is a regular polygon with R r (c) there is a regular polygon with R r (d) there is a regular polygon with R (a) there is a regular polygon with

9. The x-coordinates of the incentre of the triangle that has the coordinates of mid-points of its sides as (0, 1), (1, 1) and (1, 0) is [2013] 560

4. If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C, respectively, then the value of the expression a c [2010] sin 2C + sin 2A is c a

2

(d) 1 − 2

1 2 1 = 2 2 = 3 3 = 2 =

7 3 1   m 2  3 + 1

(b)

7 3 1   m 2  3 − 1

(c)

7 3 ( 3 + 1) m 2

(d)

7 3 ( 3 − 1) m 2

[2007] (b) 2a 3

a (c) 3

(d) 3

(c) c = a + b (d) a = b + c

π 15. In a ∆ABC, let ∠C = , if r is the inradius and R is the 2 circumradius of the ∆ABC, then 2 ( r + R ) equals (a) c + a

[2005] (d) b + c

(b) a + b + c (c) a + b

16. If in a ∆ABC, the altitudes from the vertices A , B and C on opposite sides are in HP, then [2005] sin A , sin B and sin C are in (a) HP

(b) AGP

(c) AP

(d) GP

17. The sides of a triangle are sin α , cos α and π 1 + sin α cos α for some 0 < α < . Then, the greatest 2 angle of the triangle is [2004] (a) 60°

(b) 90°

(c) 120°

(d) 150°

18. A person standing on the bank of a river, observes that the angle of elevation of the top of a tree on the opposite bank of the river is 60° and when he retreats 40 m away from the tree the angle of elevation becomes 30°. The breadth of the river is [2004] (a) 20 m

(b) 30 m

(a) are in AP (b) are in GP (c) are in HP (d) satisfy a + b = c

21. In a ∆ABC, medians AD and BE are drawn. If AD = 4, ∠DAB = π / 6 and ∠ABE = π/ 3, then the area [2003] of the ∆ABC is 8 sq units 3 32 sq units (c) 3 3

16 sq units 3 64 (d) sq units 3

(c) 40 m

(d) 60 m

3 22. The upper   th portion of a vertical pole subtends  4 3 and angle tan −1   at a point in the horizontal plane  5 through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is [2003] (a) 20 m

10

(b)

(a)

π P  Q 14. In a ∆PQR , ∠R = . If tan   and tan   are the    2 2 2 2 [2005] roots of ax + bx + c = 0, a ≠ 0, then (a) b = a + c (b) b = c

a π (b) cot    2n 2 a π (d) cot    2n 4

 π (a) a cot    n π (c) a cot    2n

3b C A 20. If in a ∆ABC, a cos 2   + c cos 2   = , then the  2  2 2 [2003] sides a, b and c

13. A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB ( = a ) subtends an angle of 60° at the foot of the tower and the angle of elevation of the top of the tower from A or B is 30°. The height of the tower is 2a (a) 3

[2003]

Properties of Triangles, Heights and Distances

(a)

19. The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a, is

(b) 40 m

(c) 60 m

(d) 80 m

 A − B + C 23. In a ∆ABC , 2ca sin   is equal to   2 (a) a2 + b2 − c2 (c) b2 − c2 − a2

[2002]

(b) c2 + a2 − b2 (d) c2 − a2 − b2

Targ e t E x e rc is e s

12. AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to point D such that CD = 7 m. From D, the angle of elevation of the point A is 45°. Then, the height of the pole is [2008]

24. In a ∆ABC , a = 4, b = 3, ∠A = 60° , then c is the root of the equation [2002] (a) c2 − 3c − 7 = 0 (c) c2 − 3c + 7 = 0

25. In a ∆ABC , tan

(b) c2 + 3c + 7 = 0 (d) c2 + 3c − 7 = 0

A 5 C 2 = , tan = , then 2 6 2 5

[2002]

(a) a, c, b are in AP (b) a, b, c are in AP (c) b, a, c are in AP (d) a, b, c are in GP

Other Engineering Entrances 26. In a triangle, the lengths of two larger sides are 10 cm and 9 cm. If the angles of the triangle are in AP, then [Manipal 2014] the length of the third side is (a) 5 − 6

(b) 5 +

6 (c) 5 ±

6 (d) 5 ±

6

27. In a ∆ABC , a [ b cos C − c cos B ] is equal to (a) 0

(b) a2

[Karnataka CET 2014] (c) b2 − c2 (d) b2

28. In a ∆ABC , a, b and c are the sides of the triangle opposite to the angles 2222222222dA , B , C , respectively. Then, the value of a 3 sin ( B − C ) + b 3 sin (C − A ) + c 3 sin ( A − B ) is [WB JEE 2014] (a) 0 (c) 3

(b) 1 (d) 2

561

Objective Mathematics Vol. 1

10

29. In any ∆ABC, (a + b + c ) (b + c − a ) (c + a − b ) (a + b − c )

equal to

4b 2 c 2 equals (a) sin 2 B

(b) cos2 A

(c) cos2 B

[EAMCET 2014] (d) sin 2 A

30. If the angles of a triangle are in the ratio 1: 1: 4, then the ratio of the perimeter of the triangle to its largest [EAMCET 2014] side is (a) 2 + 2 : 3 (c) 3 + 2 : 2

(b) 3 : 2 (d) 3 + 2 : 3

31. If angles A , B and C of a ∆ABC are in AP and b : c = 3 : 2 , then ∠A is given by [EAMCET 2014] (a) 45°

(b) 60°

(c) 75°

(d) 90°

32. If in a ∆ABC , r1 = 2, r2 = 3 and r3 = 6, then a equals [EAMCET 2014] (a) 4 (c) 2

33. If in a ∆ABC, a tan A + b tan B = ( a + b ) tan

Ta rg e t E x e rc is e s

(A + B) , 2

(b) C = A (d) B = C

34. a 3 cos ( B − C ) + b 3 cos (C − A ) + c 3 cos ( A − B ) is equal to [Manipal 2013] (b) 3 (a + b + c) (d) 0

(a) 3abc (c) abc (a + b + c)

35. If in a ∆ABC ,

a b = , then cos A cos B [Karnataka CET 2013]

(a) sin 2 A + sin 2 B = sin 2 C (b) 2sin A cos B = sin C (c) 2 sin A sin B sin C = 1 (d) None of the above

π π : n n π π (c) sin : n n

π π : n n π π (d) cot : n n (b) cos

(a) tan

37. ABCD is a square plot. The angle of elevation of the top of a pole standing at D from A or C is 30° and that from B is θ, then tan θ is equal to (b)

1 6

[Karnataka CET 2013] 3 2 (c) (d) 2 3

38. A verticle pole PO is standing at the centre O of a square ABCD. If AC subtends an ∠90° at the top P of the pole, then the angle subtended by a side of the square at P is [UP SEE 2013] 562

(a) 30° (c) 60°

40. The base BC of a ∆ABC is 6 cm and ∠B = 112.5° , ∠C = 22.5° , then its altitude is [MP PET 2012] (a) 12 cm (c) 1.5 cm

(b) 6 cm (d) 3 cm

41. Let p, q and r be the sides opposite to the angles P , Q and R, respectively in a ∆PQR. If r 2 sin P sin Q = pq, [WB JEE 2012] then the triangle is (a) equilateral (b) acute angled but not equilateral (c) obtuse angle (d) right angled

is (a) a − b 2

2

(b) 45° (d) None of these

1 (b) 2 a − b2

cos 2A a

2

−

cos 2B b2

[Karnataka CET 2011] 1 1 (c) 2 − 2 (d) a2 + b2 a b

43. If in a ∆ABC , sin A , sin B and sin C are in AP, then (a) the altitudes are in AP (c) the angles are in AP

[WB JEE 2011] (b) the altitudes are in HP (d) the angles are in HP

44. Let C be right angle sin 2 A cos 2 A is equal to − sin 2 B cos 2 B a −b ab 2

(a)

36. The area of a circle and the area of a regular polygon of n sides and of perimeter equal to that of the circle are in the ratio [OJEE 2013]

(a) 6

[BITSAT 2012] (b) cot A cot B cot C (d) 0

(a) 1 (c) −1

[BITSAT 2013]

(a) A = B = C (c) A = B

cot A + cot B + cot C is cot A cot B cot C

42. In any ∆ABC, the simplified form of

(b) 1 (d) 3

then

39. If A + B + C = 180°, then

a −b a2b2 4

2

(b)

4

of

a +b a2b2 4

(c)

a

4

∆ABC, then [J&K CET 2011] a2 + b 2 (d) ab

45. In a ∆ABC, a = 8 cm, b = 10 cm, c = 12 cm. Then, relation between angles of the triangle is [J&K CET 2011] (a) C = A + B (c) C = 2 A

(b) C = 2 B (d) C = 3 A

46. A ladder rests against a wall, so that its top touches the roof of the house. If the ladder makes an angle of 60° with the horizontal and height of the house be 6 3 m, then the length of the ladder is [BITSAT 2011] (a) 12 3 m 12 (c) m 3

(b) 12 m (d) None of these

47. A man from the top of a 100 m high tower sees a car moving towards the tower at an angle of depression of 30°. After sometime, the angle of depression becomes 60°. The distance travelled by the car during [MP PET 2011] this is (a) 100 3 m (c)

100 3 m 3

(b)

200 3 m 3

(d) 200 3 m

(a) ab

(b) ab

(c)

a b

[MP PET 2011] b (d) a

49. Angles of elevation of the top of a tower from three points (collinear) A , B and C on a road leading to the foot of the tower are 30° , 45° and 60°, respectively. [Karnataka CET 2011] The ratio of AB to BC is (a) 3 : 1

(b) 3 : 2

(c) 1 : 2

50. If angles A , B and C are in AP, then

a+c is equal to b [VITEEE 2010]

 A −C (a) 2 sin    2   A −C (b) 2 cos    2   A −C (c) cos    2   A −C (d) sin    2 

(d) 2 : 3

Answers

10 Properties of Triangles, Heights and Distances

48. The angles of elevation of the top of a tower at two points, which are at distances a and b from the foot in the same horizontal line and on the same sides of the tower, are complementary. The height of the tower is

Work Book Exercise 10.1 1. (b)

2. (b)

3. (d)

4. (a)

5. (b)

6. (b)

7. (a)

8. (d)

9. (c)

10. (b)

4. (c)

5. (d)

6. (b)

7. (b)

8. (a)

9. (a)

10. (b)

4. (a)

5. (d)

6. (a)

7. (d)

8. (c)

9. (b)

10. (d)

10. (b)

1. (d)

2. (a)

3. (c)

Work Book Exercise 10.3 1. (c)

2. (c)

3. (a)

Target Exercises 1. (a)

2. (a)

3. (c)

4. (c)

5. (c)

6. (c)

7. (c)

8. (d)

9. (d)

11. (c)

12. (b)

13. (b)

14. (c)

15. (b)

16. (a)

17. (d)

18. (b)

19. (b)

20. (c)

21. (c)

22. (c)

23. (b)

24. (a)

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26. (b)

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31. (c)

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39. (a)

40. (b)

41. (c)

42. (c)

43. (a)

44. (d)

45. (c)

46. (b)

47. (d)

48. (c)

49. (b)

50. (a)

51. (c)

52. (c)

53. (b)

54. (b)

55. (d)

56. (a)

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58. (b)

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60. (c)

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73. (b)

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76. (d)

77. (b)

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80. (d)

81. (a)

82. (b)

83. (c)

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88. (d)

89. (c)

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96. (c)

97. (d)

98. (a)

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100. (c)

101. (b)

102. (d)

103. (b)

104. (a)

105. (c)

106. (a)

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108. (d)

109. (d)

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111. (c)

112. (b)

113. (d)

114. (d)

115. (b)

116. (a)

117. (a)

118. (b,c)

119. (a,d)

120. (a,b)

121. (a,b)

122. (a,b,c) 123. (a,b,c)

124. (b)

125. (c)

126. (d)

127. (b)

128. (a)

129. (a)

130. (b)

139. (b)

140. (a)

131. (c)

132. (a)

133. (a)

134. (a)

135. (b)

136. (a)

137. (c)

138. (c)

141. (*)

142. (**)

143. (2)

144. (1)

145. (0)

146. (3)

147. (4)

148. (1)

Targ e t E x e rc is e s

Work Book Exercise 10.2

* A → q, s; B → p, t; C → r ** A → p, r; B → q, r, s; C → s; D → p

Entrances Gallery 3. (c)

4. (d)

5. (b)

6. (3)

7. (a)

8. (d)

9. (b)

10. (a)

11. (c)

1. (b)

12. (c)

2. (b,d)

13. (c)

14. (c)

15. (c)

16. (c)

17. (c)

18. (a)

19. (b)

20. (a)

21. (c)

22. (b)

23. (b)

24. (a)

25. (b)

26. (d)

27. (c)

28. (a)

29. (d)

30. (d)

31. (c)

32. (d)

33. (c)

34. (a)

35. (b)

36. (a)

37. (b)

38. (c)

39. (a)

40. (d)

41. (d)

42. (c)

43. (b)

44. (b)

45. (c)

46. (b)

47. (b)

48. (b)

49. (a)

50. (b)

563

Explanations Target Exercises = a2 + b2 − (a2 + b2 − c 2 )  a2 + b2 − c 2  = c2  Q cos C = 2 ab  

1. Here, ratio of angles are 4 : 1 : 1. ⇒ ⇒ ∴

4 x + x + x = 180 ° x = 30 ° ∠ A = 120 °, ∠B = ∠C = 30 °

30° a b 120° 30° c

Ta rg e t E x e rc is e s

A

B

Thus, the ratio of longest side to the perimeter a = a+ b+c Let b=c = x ∴ a2 = b2 + c 2 − 2 bc cos A ⇒ a2 = 2 x 2 − 2 x 2 cos A = 2 x 2 (1 − cos A) A a2 = 4 x 2 sin 2 ⇒ 2 A ⇒ a = 2 x sin 2 a = 2 x sin 60 ° = 3 x ⇒ Thus, required ratio is 3x 3 a = = a + b + c x + x + 3x 2 + 3

2. In ∆ABD,

2 2 + 52 − BD 2 2(2 )(5) BD 2 = 19 D

C 120° 3

2

1 1 cos A cos B = 4 ⋅ bc sin A ⋅ + 4 ⋅ ca sin B ⋅ 2 2 sin A sin B 1 cos C + 4 ⋅ ab sin C ⋅ 2 sin C = 2 bc cos A + 2ca cos B + 2 ab cos C b2 + c 2 − a2 c 2 + a2 − b2 = 2 bc ⋅ + 2ca ⋅ 2 bc 2ca a2 + b2 − c 2 + 2 ab ⋅ 2 ab = b2 + c 2 − a2 + c 2 + a2 − b2 + a2 + b2 − c 2 = a2 + b2 + c 2 cos A cos B cos C = = a b c cos A cos B cos C ⇒ = = k sin A k sin B k sin C ⇒ cot A = cot B = cot C ⇒ A = B = C = 60 ° ∴ The triangle is equilateral.

6. We have,

7. We have,

60° A

cos C =

5. We have, 4∆ (cot A + cot B + cot C )

cos 60 ° = ⇒

5

Now, in ∆BCD,

B

CD 2 + 9 − 19 (2 )(3)(CD ) ⇒ CD 2 + 3 CD − 10 = 0 ⇒ CD = − 5, 2 ⇒ CD = 2 [Q CD ≠ − 5] 2 2 C 2 2 C 3. (a − b) cos + (a + b) sin 2 2 C 2 2 2 C = (a + b − 2 ab)cos + (a2 + b2 + 2 ab)sin 2 2 2 C C  = a2 + b2 + 2 absin 2 − cos 2   2 2 cos 120 ° =

= a2 + b2 − 2 ab cos C

564

a2 + b2 − c 2 2 ab a2 + b2 − c 2 cos 60 ° = ⇒ 2 ab 2 ⇒ a + b2 − c 2 = ab ⇒ b2 + bc + a2 + ac = ab + ac + bc + c 2 ⇒ b(b + c ) + a(a + c ) = (a + c ) (b + c ) On dividing by (a + c ) (b + c ) and add 2 on both sides, we get b a 1+ + 1+ =3 a+c b+c 1 1 3 ⇒ + = a+c b+c a+ b+c

4. We have,

C

A + B + C = 90 °, A − B = 30 ° and A + C = 60 ° B = 30 °, A = 60 °, C = 0 ° cos 76° cos 16° 3+ 3 + cot 76° cot 16° sin 76° sin 16° 8. = cos 76° cos 16° cot 76° + cot 16° + sin 76° sin 16° 3 sin 76° sin 16° + cos 76° cos 16° = cos 76° sin 16° + sin 76° cos 16° 2 sin 76° sin 16° + cos (76° − 16° ) = sin(76° + 16° ) 1 1 2 sin 76° sin 16° + cos 60 ° − cos 92 ° + 2 2 = = sin 92 ° sin 92 ° 1 − cos 92 ° 2 sin 2 46° = = = tan 46° = cot 44° 2 sin 46° cos 46° sin 92 ° ⇒

⇒ Order of AP can be b, c , a or c , b, a. Case I When 2c = a + b b2 + c 2 − a2 b2 + c 2 − (2c − b)2 cos A = = 2 bc 2 bc 4bc − 3 c 2 4b − 3 c = = 2 bc 2b Case II 2b = a + c b2 + c 2 − a2 b2 + c 2 − (2 b − c )2 cos A = = 2 bc 2 bc 4 bc − 3 b2 4 c − 3 b = = 2 bc 2c

= sin C ⋅ cos A + cos C ⋅ sin A ⇒ cos A sin C − sin A cos C = 0 ⇒ sin (C − A) = 0 ⇒ C=A 2 cos A cos B 2 cos C a b 16. We have, + + = + a b c bc ca 2(b2 + c 2 − a2 ) c 2 + a2 − b2 2(a2 + b2 − c 2 ) ⇒ + + 2 bca 2 bac 2 abc a2 + b2 = abc ⇒ 2 b2 + 2c 2 − 2 a2 + c 2 + a2 − b2 + 2 a2 + 2 b2 − 2c 2 = 2 a2 + 2 b2

10. Clearly, a + b > c b ) − 2 ab > ( c ) ( a + b )2 > ( c )2 a+ b> c a+ b− c >0

2

13. cos θ =

2

2

2

∴ ⇒

π  − B 2 

18. cos 2 B + cos 2 C = cos 2 B + cos 2 

2

= cos 2 B + sin 2 B = 1

19. We have,

6

b2 − c 2 c 2 − a2 a2 − b2 ⋅ cos A + ⋅ cos B + ⋅ cos C a b c b2 − c 2 b2 + c 2 − a2 c 2 − a2 c 2 + a2 − b2 = ⋅ + ⋅ a 2 bc b 2 ac a2 − b2 a2 + b2 − c 2 + ⋅ c 2 ab =0

20. We have, a2 = b2 + c 2 − 2 bc cos A ⇒

⇒

2

b2 + c 2 − a2 λ − 2 = 2 bc 2 λ −2 = cos A 2 − 1 < cos A < 1 λ −2 − 1< 2 b − c ⇒ 2c > b b b ⇒ c = 2 a − c > b ⇒ 2b > b b 2  Thus, ∈  , 2 c 3 

2

3 b + c + a = 2a + 2b

Hence,

11. a + b > c ⇒ 2(b − c ) + b > 0 ⇒ 3b > 2c

12. Here, a + b = 2 3, ab = 2 and C =

⇒

2

Targ e t E x e rc is e s

( a+

⇒ ⇒ ⇒ ⇒

2

10 Properties of Triangles, Heights and Distances

15. 2 cos A sin C = sin (C + A)

9. Sides are in AP and a < minimum { b, c }.

C

θ = 30 ° 1 sin θ = ⇒ 2 cos A cos B cos C 14. + + a b c 2 2 b + c − a2 c 2 + a2 − b2 a2 + b2 − c 2 = + + 2 abc 2 abc 2 abc 2 2 2 a +b +c = 2 abc (a + b + c )2 − 2 (ab + bc + ca) = 2 abc 112 − 2 ⋅ 38 9 = = 2 ⋅ 40 16

c 2 − 2 bc cos A + (b2 − a2 ) = 0

Clearly, c1, c 2 are the roots of this equation such that c1 + c 2 = 2 ab cos A and c1c 2 = b2 − a2 ∴

c1 − c 2 = (c1 + c 2 )2 − 4 c1c 2 = 4b2 cos 2 A − 4 (b2 − a2 ) = 2 a2 − b2 (1 − cos 2 A) = 2 a2 − b2 sin 2 A

21. cos 2 A + cos 2 B − cos 2 C = 1 ⇒ 1 − sin 2 A + 1 − sin 2 B − 1 + sin 2 C = 1 ⇒ sin 2 A + sin 2 B = sin 2 C ⇒ a2 + b2 = c 2 Thus, triangle is right angled at C.

565

Objective Mathematics Vol. 1

10

22. AC = d , OA = OB = r , CD = BD = l , ∠COA = B

l °

C A

⇒

AC 2 = OA2 + OC 2 − 2OA ⋅ OC ⋅cos 1 = r2 2 2π π ∠BOD = ∠COD = = 3⋅ 2 3 π BD 1 tan = = ⇒l = r 3 =d 3 3 OB r

⇒

π 3

d 2 = 2r 2 − 2r 2 ⋅

Also, ⇒

Ta rg e t E x e rc is e s

23. 3 (tan 2 A + tan 2 B + tan 2 C ) − (tan A + tan B + tan C )2 = (tan A − tan B)2 + (tan B − tan C )2 + (tan C − tan A)2 > 0 ⇒ For, here tan A = tan B = tan C = 3 is not true. ⇒ 3 k − (tan A ⋅ tan B ⋅ tan C )2 > 0 because in ∆ABC, tan A + tan B + tan C = tan A ⋅ tan B ⋅ tan C ⇒ 3 k − 81 > 0 ⇒ k > 27

24. In ∆ABC, cos 2 A + cos 2 B − cos 2C = 1 − 4 sin A sin B cos C < 1 for all A, B cos 2 A + cos 2 B + cos 2C = − 1 − 4 cos A cos B cos C < − 1 for acute A, B cos 2 A + cos 2 B + cos 2 C = 1 − 2 cos A cos B cos C < 1 for acute A, B

25. cos B ⋅ cos C + sin B ⋅ sin C ≥ 1 because sin B ⋅ sin C ⋅ sin 2 A is positive and sin 2 A ≤ 1 ⇒ cos (B − C ) ≥ 1. But cos (B − C ) >/ 1 So, cos (B − C ) = 1 Therefore, B = C and then sin A = 1 π π A= ,B =C = ∴ 2 4 2b = a + c

⇒

3 b = 2s ⇒ s = A C sin 2 2 = B sin 2

sin Now,

3b 2

ac (s − b) (s − c ) (s − b) (s − a) (s − a) (s − c ) × bc × ab

3b −b s−b 1 = = 2 = b b 2 a + b + c 8 + 15 + 17 27. Here, s = = = 20 2 2 A  ( s − b) ( s − c )  sin =   2 bc

566

∴

(s − a) (s − b) (s − c ) s ⇒ (s − a) + (s − c ) = 2 (s − b) ⇒ 2b = a + c ⇒ a, b and c are in AP. A C 30. We have, 3 tan tan = 1 2 2 (s − b) (s − c ) (s − a) (s − b) ⇒ 3 ⋅ =1 s (s − a) s(s − c ) s−b 3⋅ =1 ⇒ s ⇒ 3s − 3b = s ⇒ 3 s − s = 3 b; 2 s = 3 b ⇒ a + b + c = 3b ⇒ a + c = 2b ∴ a, b and c are in AP. A B 5 A B 1 31. tan + tan = , tan ⋅ tan = 2 2 6 2 2 6 A B tan + tan  A + B 2 2 =1 tan  ⇒  = A B  2  1 − tan tan 2 2 π A+ B =2 × ⇒ 4 Triangle is right angled at angle C. ∴ c 2 = a2 + b2 On multiplying both sides by

32. tan 2

26. Since, a, b and c are in AP. ⇒

B C cos 2 2 s(s − b) s(s − c ) 2 a ⋅ s (s − b) (s − c ) = 2a ⋅ = ca ab a bc A A = 2 s ⋅ sin = (a + b + c )sin 2 2 A B C 29. We have, cot , cot , cot are in AP. 2 2 2 A C B ⇒ cot + cot = 2 cot 2 2 2 s(s − b) s(s − a) s(s − c ) ⇒ = 2⋅ + (s − c ) (s − a) ( s − b) ( s − c ) (s − a) (s − b)

28. We have, 2 a cos

D

O 60

π 3

(20 − 15) (20 − 17 )  5× 3  1 =   = 15 × 17  = 17 × 15 17     1 15 2 A cos A = 1 − 2 sin = 1− 2⋅ = 2 17 17

C (s − a) (s − b) = 2 s (s − c ) =

⇒ ∴

(9 + c − 10 ) (9 + c − 8) c 2 − 1 = (9 + c ) (9 − c ) 81 − c 2 c2 − 1 7 = 9 81 − c 2

c =6 4 12 33. tan A = and tan B = . Clearly, tan C should be such 3 5 that tan A + tan B + tan C = tan A tan B tan C 4 12 4 12 ∴ + + tan C = ⋅ ⋅ tan C 3 5 3 5 56 16 + tan C = tan C ⇒ 15 5 56 ⇒ tan C = 33 33 cos C = ∴ 65

⇒

⇒

2 sin B = sin A + sin C B  A − C ⇒ 2 sin = cos    2  2 A C A C ⇒ cos ⋅ cos = 3 sin ⋅ sin 2 2 2 2 A C 1 tan ⋅ tan = ⇒ 2 2 3 A C B B tan + tan cos sin 2 2 2 2 = ⋅ ∴ B A C B cot cos ⋅ cos cos 2 2 2 2  A + C cos    2  = A C cos ⋅ cos 2 2 1 2 A C = 1 − tan ⋅ tan = 1 − = 2 2 3 3 B C 35. 1 − tan ⋅ tan 2 2 (s − a) (s − c ) (s − a) (s − b) = 1− ⋅ s ( s − b) s (s − c ) s−a a 2a = 1− = = s s a+ b+c

From Eqs. (i) and (ii), we have a ≥ 2 λ sin

= ( 3 + 1/ 3 )x = (4 / 3 )x CD = x (cot 15° − cot 30 ° ) = x ( 2 + 3 − 3) = 2 x A

x 60° B

30° L

15° C

10

D

[given] BC × CD = (4 / 3) ⋅ 2 x 2 = 2 3 × 32 × 19 × 11 19 2 2 2 x = × 3 × 19 × 11 = (19 × 3) = 3249 ⇒ 11

39. Since, c 2 = a2 + b2 ∴ ∆ABC is right angled with ∠C = 90 ° . 1 Area of triangle = ab ∴ 2 1 ⇒ s(s − a) (s − b) (s − c ) = ab 2 ⇒ 4s (s − a) (s − b) (s − c ) = a2 b2

s2 (s − a) (s − c ) bc ⋅ ab

40. (a + b − c ) (b + c − a) (c + a − b) − abc 1 2 1 2

37. Since, AM ≥ GM

b+c ≥ bc ⇒ b + c ≥ 2λ 2 a b c b+c Also, = = = sin A sin B sin C sin B + sin C a b+c = ⇒ A A B + C  B − C  2 sin cos 2 sin    cos   2   2  2 2

A 2

38. BC = x (cot 60 ° + cot 30 ° )

A C π − = 2 2 4 A C A C 1 cos ⋅ cos + sin ⋅ sin = 2 2 2 2 2

(s − b) (s − c ) (s − a) (s − b) + ⋅ = bc ab s (s − a) (s − c ) (s − b) (s − a) (s − c ) ⇒ + = b ac b ac 1  2 s − b (s − a) (s − c ) = ⇒   b  ac 2 a + c (s − a) (s − c ) 1 = ⇒ b ac 2 2 ac  a + c ⇒ 2  =  b  (s − a) (s − c ) a + c = 2λ , b = λ a+c a + b + c 3λ = 2, s = = ⇒ b 2 2 (1 − 2 x ) λ (1 + 2 x ) λ , (s − c ) = s−a= 2 2 2 1 (1 − x ) 4 ⇒ ⇒ x= 8= 7 (1 − 4 x 2 ) ∴

...(ii)

...(i)

= 8 (s − c ) (s − a) (s − b) − abc ∆2  2∆  = 8⋅ − R ⋅ 4∆ = 4∆  − R  s  s = 4∆ (2 r − R ) ≤ 0

41. a2 sin 2 B + b2 sin 2 A

Targ e t E x e rc is e s

⇒

(b + c )sin A / 2  B − C cos    2 

 B − C Since, cos   is always positive.  2 

36. a = (1 + x )λ, b = λ, c = (1 − x )λ, ⇒

a=

Properties of Triangles, Heights and Distances

34. 2b = a + c

= 2 a2 sin B ⋅ cos B + 2 b2 sin A ⋅ cos A a2 b b2 a = cos B + cos A R R ab abc = = 2 bc sin A (a cos B + b cos A) = R R  1 = 4  bc sin A = 4λ  2

42.

a cos A + b cos B + c cos C a+ b+c 2 R sin A cos A + 2 R sin B cos B + 2 R sin C cos C = 2s R = (sin 2 A + sin 2 B + sin 2C ) 2s R = ⋅ 4 sin A sin B sin C 2s abc 2 R abc = ⋅ 3 = s 8R 4sR 2 ∆ abc But ,r = R= 4∆ s r 4∆R So, the value = = ∆ 4 ⋅ ⋅ R2 R r

567

Objective Mathematics Vol. 1

10

s (a + b + c ) a b c = = + + R 2R 2R 2R 2R = sin A + sin B + sin C r A B C 44. As, = 4 sin sin sin 2 2 2 R r C   A + B   A − B ⇒ = 2 cos    ⋅ sin  − cos      2 2 2 R   r C   A − B  C  = 2 cos  ⇒  − sin    ⋅ sin  2   2  2 R  r C C  ≤ 2 1 − sin  sin ⇒  2 2 R r C  C ⇒ ≤ 2 sin − sin 2   2 2 R

sin A − sin B = 3 cos B − cos A  A − B  A + B 2 cos    ⋅ sin   2   2  A+ B 3= ∴ = cot  A + B  A − B 2 2 sin   ⋅ sin    2   2  π π 2π A+ B π ⇒C = π − = ⇒A+ B= = ⇒ 3 3 3 2 6 So, triangle is an obtuse angled. ⇒

49. c cos θ − a sin θ = b Therefore, c cos α − a sin α = c cos β − a sin β, where α and β are the other two angles of the triangle. ∴ c (cos α − cos β ) = a (sin α − sin β )  α − β  α + β 2 cos    sin   2   2  c  α + β ⇒ = − cot  =   2  α β β − α +     a 2 sin    sin   2   2   α + β −a tan  ⇒  =  2  c 2 × (− a/ c ) 2 ac ⇒ = tan (α + β ) = 1 − a2 / c 2 a2 − c 2 3π  2 ac  ⇒ tan  π −  =  4  a2 − c 2 ∴ a2 − c 2 = 2 ac

2 1  1 C  − − sin     2 2  4 r 2 R R or ≤ ≥ 2 but ≤2 ⇒ R 4 r r R ⇒ =2 r A−B 1 C i.e. if = 0 and − sin = 0 2 2 2 ⇒ A = B and C = 60 ° ⇒ A = B = C = 60 ° So, triangle is equilateral.

r ≤2 R

⇒

Ta rg e t E x e rc is e s

48. 3 cos A + sin A = 3 cos B + sin B

43.

45. We have, ∆ = ∆ BPC + ∆ APC + ∆ APB 1 1 1 axa + bxb + cxc 2 2 2 1 1 1 Also, ∆ = aHa = bHb = cHc 2 2 2  xa xa x x xb xc  + + + b + c =1 ⇒ ∆=∆  ⇒  Ha Hb Hc  Ha Hb Hc ⇒

∆=

46. Clearly, A > B C  A − B a − b Now, tan  cot  =  2  a+ b 2 1 C ⇒ tan 30 ° = cot 3 2 C π C π ∴ 3 = cot ⇒ = ⇒C= 2 3 2 6

[Q a > b]

47. Let sides of the given triangle be a − d , a, a + d . Now, (a + d )2 = (a − d )2 + a2 C

50. Let a = x, b = y and c = x 2 + y 2 + xy Then, a2 = x 2 , b2 = y 2 and c 2 = x 2 + y 2 + xy ∴ c is the length of the greatest side and hence ∠C is the greatest angle ∴ By cosine rule, 1 a2 + b2 − c 2 x 2 + y 2 − x 2 − y 2 − xy cos C = =− = 2 ab 2 xy 2 ∴ ∠C = 120 °

51. Since, the sides are positive. ∴x > 1or else c will be negative. a = x2 + x + 1 > x + x + 1 = 2 x + 1 = b ∴ a>b Also, a > c . Hence, a is the greatest side, so A is the greatest angle. 1 b2 + c 2 − a2 ∴ cos A = = − ⇒ A = 120 ° 2 bc 2 A A 52. θ = B + = π − C +   2 2 A

a

+

d a–d

90° A

or ⇒ Now,

568

a

B

4ad = a a = 4d a − d 4d − d 3 tan A = = = a 4d 4 a 4d 4 tan C = = = a − d 4d − d 3 2

B

⇒

θ D

C

A A   sin θ = sin  B +  = sin C +    2 2 π B + C  = sin  B + −   2 2  π B + C   B − C = sin C + −  = cos     2  2 2 

[say]

54. ∴ r1r2 + r2 r3 + r3 r1

∆ ∆ ∆ ∆ ∆ ∆ × + × + × (s − a) (s − b) s − b s − c s − c s − a ∆2 ∆2 ∆2 = + + (s − b) (s − a) (s − b) (s − c ) (s − a) (s − c ) ∆2 [(s − c ) + (s − a) + (s − b)] = (s − a) (s − b) (s − c ) ∆2 [3 s − (a + b + c )] s ∆2 s2 ∆2 = × = 2 = s2 = 2 (s − a) (s − b) (s − c ) s ∆ r =

3 2 55. The area of an equilateral triangle = a , where a is 4 side. a + a + a 3a Also, s= = 2 2 2 a ∆ 3a ×2 Inradius, r = = ∴ = s 4 × 3a 2 3 abc a3 a Circumradius, R = = = 4∆ 3 3 a2 2 3 ∆ ( 3 / 4) a and exradius, r1 = a = = s−a a/ 2 2 a a 3 Hence, r : R : r1 = : : a = 1: 2 : 3 2 2 3 3 ∆ ∆ ∆ 56. We know that, r1 = , r2 = , r3 = s−a s−b s −c r1 > r2 > r3 ∆ ∆ ∆ > > s − a s − b s −c 1 1 1 > > s − a s − b s −c s − a< s − b< s −c [Q s − a, s − b, s − c are positive] −a < − b < − c ⇒ a > b > c

Given that, ⇒ ⇒ ⇒ ⇒

57. Let area of triangle be ∆, then according to the question,

1 1 1 ax = by = cz 2 2 2 bx cy az b  2 ∆  c  2 ∆  a  2 ∆  ∴ + + =   +   +   c a b c  a a b b c  2 ∆(b2 + c 2 + a2 ) = abc abc  2 (a2 + b2 + c 2 ) abc  = ⋅ Q∆ = 4R  4R  abc 2 2 2 a +b +c = 2R ∆=

2s = a + b + c ⇒ s = 9 x ∆ = s(s − a) (s − b) (s − c ) = 9x ⋅ 6x ⋅ 2 x ⋅ x = 6 3x 2 abc ∆ and r = R= 4∆ s R sabc 9 x ⋅ 3 x ⋅ 7 x ⋅ 8 x 7 = = = 2 r 4∆2 4 ⋅ 108 ⋅ x 4

Then, ∴ Now, ∴

59. If G is the centroid, then we have the following 2 2 mb + mc > a 3 3 2 2 2 2 ma + mb > c , ma + mc > b 3 3 3 3

BG = CG > a ⇒ Similarly, ⇒ ⇒ ⇒

4 (ma + mb + mc ) > a + b + c 3 3 ma + mb + mc > (a + b + c ) 4 3 ma + mb + mc > s 2

10 Properties of Triangles, Heights and Distances

58. Let the sides be a = 3 x, b = 7 x, c = 8 x.

60. Here, CD = p cot C and AD = p cot A = p tan C Adding,4p = p cot C + p tan C A D p

h = 4p

B

⇒ ⇒ ⇒ ⇒

C

tan C + cot C = 4 1 =4 sin C ⋅ cos C 1 sin 2C = 2 C = 15°

Targ e t E x e rc is e s

abc ∆ and r = 4∆ s R abc s abc ∴ = ⋅ = r 4∆ ∆ 4 (s − a) (s − b) (s − c ) Since, a: b :c = 4: 5: 6 a b c ⇒ = = =k 4 5 6 (4k )(5k )(6k ) R Thus, =   15k   15k   15k r 4 − 4k  − 5k  − 6k   2   2   2 120 k 3 ⋅ 2 16 = = 2 k ⋅ 7 ⋅ 5⋅ 3 7

53. We have, R =

61. Using (m + n )cot θ = n cos β − m cot α, we get (3 + 2 )cot ∠CDA = 2 cot 30 ° − 3 cot 60 ° 5 3 ⇒ sin ∠CDA = ⇒ cot ∠CDA = 5 2 7 AC 3 ∴ = sin ∠CDA sin 60 ° 3 3⋅ 2 5 ⇒ AC = ⋅ =5 7 3 2 7 1 62. ⋅ α ⋅ a = ∆ 2 1 a ∴ = α 2∆ 1 b 1 c Similarly, = , = β 2∆ γ 2∆ 1 s ∴ α −1 + β −1 + γ −1 = (a + b + c ) = 2∆ ∆

63. a2 + b2 + c 2 = 8 R 2 ⇒ ⇒ ∴ ⇒

sin 2 A + sin 2 B + sin 2 C = 2 cos 2 A + cos 2 B + cos 2C + 1 = 0 − 4 cos A cos B cos C = 0 π A or B or C = 2

569

Ta rg e t E x e rc is e s

Objective Mathematics Vol. 1

10

64. r =

ac ∆ (1 / 2 )⋅ ac = = s (1 / 2 )⋅ (a + b + c ) a + b + c ac (c + a − b) ac (c + a − b) = = (c + a)2 − b2 c 2 + 2ca + a2 − b2 ac (c + a − b) c + a − b [Q a2 + c 2 = b2 ] = = 2 2ca + b2 − b2

∠FBE = 90 ° − A FE = 2 R = sin A = a cos A ∠FDE = 180 ° − 2 A

Also, and

A

s s s 65. We have, 4  − 1  − 1  − 1 a

  b  c  (s − a) (s − b) (s − c ) = 4⋅ abc s (s − a) (s − b) (s − c ) = 4⋅ s ⋅ abc ∆2 = 4⋅ s ⋅ abc 4∆ ∆ r = ⋅ = abc s R 1 1 1 1 66. We have, 2 + 2 + 2 + 2 r1 r2 r3 r (s − a)2 (s − b)2 (s − c )2 s2 = + + + 2 ∆2 ∆2 ∆2 ∆ 1 2 2 = 2 [4s − 2 s (a + b + c ) + a + b2 + c 2 ] ∆ 1 = 2 [4s2 − 2 s ⋅ 2 s + a2 + b2 + c 2 ] ∆ a2 + b2 + c 2 = ∆2

67. We have, rr1 + r2 r3   ∆ ∆ ∆ ∆ 1 1 + ⋅ + ⋅ = ∆2   s ( s − a ) ( s − b ) ( s − c ) s s − a s − b s −c     − − + − ( s b ) ( s c ) s ( s a ) = ∆2    s (s − a) (s − b) (s − c )  =

2s2 − s (a + b + c ) + bc  = ∆2   ∆2   = 2 s2 − s ⋅ 2 s + bc = bc

F

E

B

∴ Radius of circle DEF FE a cos A = = 2 sin ∠FDE 2 sin (180 ° − 2 A) a cos A 2 R sin A cos A = = 2 sin 2 A 2 ⋅ 2 sin A cos A R a   = Q =R  2 sin A  2 ∆ ⋅2 70. We have, a = 2 R sin A = 2 R ⋅ bc abc 4R∆ a= ⇒ R= ∴ 4∆ bc s−a s−b s −c 71. r1, r2 and r3 are in HP or are in AP. and , ∆ ∆ ∆ or s − a, s − b, s − c are in AP or a, b, c are in AP. Now, a + b + c = 24 ⇒ b = 8, s = 12, c = 4 − a or a2 − 16 a + 60 = 0 or a = 10, 6 If a = 10, b = 6 and if a = 6, b = 10 ∴ Sides of the triangle are 6, 8, 10 cm. B 72. In ∆ABD1, ∠ABD2 = ∠D2 BD1 = 2 A

B1 D2

68. In an equilateral triangle, a = b = c ∴

∴

a + a + a 3a = 2 2 ∆ ∆ 2∆ r= = = s 3a 3a 2 ∆ ∆ r1 = = s − a a/ 2 2∆ 2∆ = = 3⋅ = 3r 3a a ∆ ∆ r2 = = = 3r s−b s−a ∆ ∆ r3 = = = 3r s −c s − a s=

r1 = r2 = r3 = 3 r

69. Circle on BC as diameter passes through E, F.

570

[since, ∠BEC = ∠CFB = 90 °] a and radius of this circle = 2

C

D

B

A1

D1

C

⇒ ∆ABD2 and ∆D1D2 B are congruent. AD1 2 = =2 ⇒ AD2 1

73. Distance of circumcentre from side AC = R cos B and distance of orthocentre from side AC = 2 R cos A ⋅ cos C ⇒ R cos B = 2 R cos A ⋅ cos C ⇒ − cos ( A + C ) = 2 cos A ⋅ cos C ⇒ sin A ⋅ sin C = 3 cos A ⋅ cos C ⇒ tan A ⋅ tan C = 3 A C A B C B 74. r1 + r2 = 4 R sin ⋅ cos ⋅ cos + sin ⋅ cos ⋅ cos   2 2 2 2 2 2 C A B B A   = 4 R cos sin ⋅ cos + sin ⋅ cos    2 2 2 2 2 C   = 4 R cos 2  = 2 R (1 + cos C )  2 r1 + r2 ⇒ = 2R 1 + cos C

πr12 ,

∴ A = πr , A1 = 1 1 ∴ + + A1 A2 2

=

1 π

1 π 1 = π =

A2 = A3 = 1 1 1 1 = + + A3 r1 π r2 π r3 π

1 1 r + r + 1 2  3s − (a +  ∆ s 1 ⋅ = ∆ r π

πr22 ,

πr32

1 1 s − a s − b s − c  = + + r3  ∆ ∆  π  ∆ b + c ) 1 3s − 2 s = ⋅  ∆ π 1 1 = = A πr 2

AH c = sin (90° − A) sin ( A + B) c cos A c cos A a AH = = = cos A sin (180 ° − C ) sin C sin A a c   Q =  sin A sin C  = a cot A

76. From ∆AHB, ⇒

10

1 2

79. ∆ = (BC )h where, h is the distance of vertex A from side BC. ∆ (BC )h ∴ ,where G is centroid. ∆GBC = = 3 6 2∆ ⇒ h= = constant BC Thus, distance of vertex A from side is fixed. This is turn implies that distance of centroid from side BC will be fixed, hence locus of G will be a line parallel to BC. A 80. ∠BAA1 = ∠A1 AA2 = ∠A2 AC = 3 Clearly, ∆AAC 1 is isosceles. ⇒

AA1 = AC = b A

Properties of Triangles, Heights and Distances

75. Area of circle = π × (Radius) 2

b

c

77. In ∆ABC, AB = BC and altitude AD = h B

A1

a/4

C

A2

A

A1 A2 = A2C = r

Now, sin 3

P r B

r C

D

⇒ ⇒

PA = PB = PC = r PD = h − r

Now,

BD = BP 2 − PD 2

A A ⋅ cos 3 3 1 A A A = sin 2 ⋅ 2 sin ⋅ cos 2 3 3 3 1 2A A = sin 2 ⋅ sin 2 3 3 2 1  A2C  BA 2 1 a2 3a 1 = ⋅ ⋅ ⋅ = ⋅  ⋅ 2  b  AB 2 16b2 4 c =

= r 2 − (h − r )2 = 2 rh − h 2 ∴

BC = 2 BD = 2 rh − h 1 Thus, area of ∆ABC = × DC × DA 2 1 = × 2 2 rh − h 2 × h = h 2 rh − h 2 2

a 4

Targ e t E x e rc is e s

Ler r be the radius of circumscribed circle.

3a3 128b2c

81. Let BC be the tower and tan ∠BAC =

2

5 . 12 C

78. Let A1, B1 and C1 be the foot of altitudes drawn from P to A

sides BC, CA and AB, respectively. We have, PA1 = xa , PB1 = xb and PC1 = xc ar (∆ABC ) = ar (∆PBC ) + ar (∆CPA) + ar (∆APB) A

AD = 240 m Now, ⇒

C1

B

⇒ ⇒

B1

A1

⇒ C

3 1 ⋅ 4 = ⋅ 2 ( xa + xb + xc ) 4 2 xa + xb + xc = 3

B

D 240 m

and

tan ∠BDC =

3 4

BC 5 = AB 12 BC 5 = 240 + BD 12 tan ∠BAC =

BC = (240 + BD )

5 12

BC 3 3 = ⇒ BC = BD BD 4 4 From Eqs. (i) and (ii), we get 3 5 BD = (240 + BD ) 4 12 ⇒ 9 BC = 1200 + 5 BD ⇒ 4BD = 1200

Again, tan ∠ BDC =

…(i) …(ii)

571

Objective Mathematics Vol. 1

10

1200 = 300 m 4 3 ∴ BC = BD 4 3 = × 300 4 = 225 m Hence, the height of the tower is 225 m. ⇒

84. Let OP be the incomplete and OQ be the complete pillar.

BD =

If A is a point on the horizontal plane, such that OA = 100 m. Q x P

82. Let AB = 200 m be the cliff and P, M be the two ships on opposite sides of the cliff such that PBM is a straight line. A 30°

45°

60° 45° 100 m

A

200 m M

Ta rg e t E x e rc is e s

45°

30° B

∠OAP = 45° ∠OAQ = 60 ° OP tan 45° = OA OP = OA = 100 m OQ tan 60° = OA (100 + x ) 3= 100 x = 100 ( 3 − 1)m

Then, and

P

Distance between this ships = PM In right angled ∆ABP, BP = cot 45° 200 ⇒ BP = 200 × cot 45° = 200 × 1 = 200m In right angled ∆ABM, BM = cot 30 ° 200 ∴ BM = 200 × cot 30 ° = 200 × 3 Hence, PM = PB + BM = 200 + 200 3 = 200 ( 3 + 1)m

In ∆OAP, ∴ In ∆OAQ, ⇒ ⇒

85. Let AQ (= PQ ) be the broken part of the tree OP , whose upper part touches the ground at A such that P

83. Let PQ be the tower. Let θ be the angle of elevation of Q as seen from A. Then, the angle of elevation ofQ as seen from B is 90° − θ.

Q

Q h P

45°

N 200 E

θ °–

A

10 m

A

90

θ

B

From the figure, we have ∠QAP = θ and ∠QBP = 90 ° − θ Let h be the height of tower PQ. Now, ∠APQ = 90 ° ∴ h = AP = tan θ Also, ∠BPQ = 90 ° ∴ h = BP tan (90 ° − θ ) = BP cot θ From Eqs. (i) and (ii), we get h 2 = AP × BP In right angled ∆BAP, AB = AP = 200 m ∴ BP = 200 2 From Eqs. (iii) and (iv), we get h 2 = 200 × 200 2 ⇒ h = 200 (2 )1/ 4 m

572

O

∴ Now, in ∆OAQ,

O

∠QAO = 45° OA = 10 m

[QOQ = 10 m]

AQ = (OA)2 + (OQ )2

…(i)

= 10 2 m Hence, the entire length of the tree = OP = OQ + PQ = OQ + AQ = 10 (1 + 2 ) m

86. Let OP be the tower of height, say h, m and PQ be the …(ii)

flagstaff of height 6 m. Q

…(iii)

6m P

…(iv)

h θ B 2√3 m A

θ

x

O

⇒ ∴

90. Let CD be the tower. A 60°

87. CD is tower of height h, AB is building of height a.

h

h−a LB

C H

D B

B

45°

A

C

In ∴ ⇒

D

H = tan 30 ° d h and from ∆ABD, = tan 60 ° d H tan 30 ° d ∴ = h tan 60 ° d h H= ⇒ 3

a

∴

d

From ∆BCD,

30°

hL

60°

30°

(h − a) LB = tan 30 ° = 3(h − a) h ∆ACD, tan 45° = ⇒ h ( 3 − 1) = 3a LB 3a 3 ( 3 + 1) a h= = 2 3 −1  3 + 3 h=  a  2 

…(i) …(ii)

Targ e t E x e rc is e s

In ∆BLD, tan 30° =

10

1 1− 1 − tan 30 ° 3 = = 1 + tan 30 ° 1 + 1 3 AC 3 −1 = x 3+1  3 + 1 x = 60  m  3 − 1

Properties of Triangles, Heights and Distances

Let the Sun makes an angle θ with the ground and OA and AB = 2 3 m be the shadows of the tower and the flagstaff, respectively. Let OA = x h …(i) In ∆OAP, tan θ = x (h + 6) and in ∆OBQ, tan θ = ( x + 2 3) (h + 6) h …(ii) = ∴ x ( x + 2 3) From Eqs. (i) and (ii), we get hx + 2 3 h = xh + 6 x ⇒ 2 3 h = 6 x h 6 ⇒ tan θ = = = 3 ⇒ θ = 60 ° x 2 3

91. AB is tower of height h. AB = tan 45° BC h = 70 tan 45° h = 70 m

In ∆ABC, ⇒ ⇒

92. From figure,

88. Let MP denotes the house. Let θ be the angle of

In

∆ADE,

elevation. ∴

P

150 − h 1 = tan 30 ° = 60 3 60 h = 150 − 3 = (150 − 20 3 ) m

93. Given that height of the tower AB = h = 150 ft 15 m

A θ

M

150 ft

O

15 m

MP 15 = =1 OM 15 θ = 45°

∴

tan θ =

⇒

30 °

B

89. Here, B is the position of boat and AC is light house. AC = tan 15° = tan(45° − 30 ° ) x 15°

B

x

C

60 m

Now,

A

C

Angle of elevation of the Sun = 30 ° Let BC be the length of the shadow cast by the tower. Then, from right angled ∆ABC, we have AB h tan 30° = = BC BC 150 1 ⇒ = BC 3 ⇒ BC = 150 3 ft

573

96. Let RQ be the chimney, PQ be the vertical tower and A be the point of observation. We have,

C

Objective Mathematics Vol. 1

AP = 70 m, RQ = 20 m 300

R

B

Q

60° 45° A

O

x =1 OA OA = x OC tan 60° = OA 300 = 3 OA 300 OA = 3 = 100 3 = x

∴ In ∆OAC, ⇒ ∴

θ α

A

P

∠ QAP = α

and PQ = H H + 20 H and , tan α = tan (θ + α ) = 70 70 H 1 + H + 20 70 6 = 6H + 70 = ⇒ H 70 6 ⋅ 70 − H 1− 6 ⋅ 70 ⇒ 6 ⋅ 70 H − H 2 + 120 ⋅ 70 − 20 H = 70 ⋅ 6 H + 70 2 If

95. Let PQ be the vertical tower of height h and A and B be the points of observation. We have, AB = 1000 m

Ta rg e t E x e rc is e s

1 6

x

⇒

⇒

H 2 + 20 H − 70 ⋅ 50 = 0

⇒ P

H 2 + 20 H + 100 = 3500 + 100 = 3600

⇒

(H + 10 )2 = (60 )2

⇒

H = 50 m

97. Let OP be the tower of height h, then AB = 30, AP = 20 and BP = 15.

A

P

5π 12

π 6

Q

B

AQ = h cot

π =h 3 6

5π 12  π π = h cot  +   6 4 π π cot ⋅ cot − 1 6 4 BQ = h π π cot + cot 6 4 ( 3 − 1) =h ( 3 + 1) h( 3 − 1)2 = 2 AB = AQ − BQ   4 − 2 3  =h 3−  2    BQ = h cot

⇒

Now,

⇒ ⇒

15 m

20 h

In ∆AQP, In ∆BQP,

= h( 3 − 2 + 1000 = 2 h( 3 − 1) 500 h= ( 3 − 1)

A

θ

B

30 m O

From ∆ABP,

20 2 + 30 2 − 152 2 × 20 × 30 1075 43 = = 1200 48

cos θ =

⇒

 43 sin θ = 1 −    48 =

2

2304 − 1849 = 48

h = 20 sin θ = 20 ×

and

455 48 455 5 455 = 48 12

98. Draw BP ⊥ AC, the longer diagonal. From right angled ∆APB, we have D

3)

= 250( 3 + 1)m

574

tan θ =

and

P

30°

C

6 cm

10

OB OA

94. In ∆OAB,tan 45° =

A

30° 10 cm

B

99. Let AB = h be the height of tower at a distance x from the post PQ = a.

10

101. Let AB be the tower and AC be pole of height h. C h A

α

O

2α

b B a

b = tan α a b+ h = tan 2α a b+ h 2 tan α = a 1 − tan 2 α 2 a × (b / a) b+ h= 1 − (b2 / a2 )  a2 + b2  h=b 2   a − b2 

From ∆ABO, From ∆CBO, ⇒ ⇒ ⇒

…(i)

102. Given that, AC = CB = h

Properties of Triangles, Heights and Distances

AP = cos 30 ° AB ⇒ AP = AB cos 30 ° 3 ∴ AP = 10 × 2 = 5 3 cm BP Also, = sin 30 ° AB ∴ BP = AB sin 30° 1 = 10 × 2 = 5cm From right angled ∆BPC, we have BP 2 + PC 2 = BC 2 ⇒ 52 + PC 2 = 62 ⇒ 62 − 52 = 11 ⇒ PC = 11cm ∴ Longer diagonal, AC = AP + PC = (5 3 + 11) cm

[say]

B B

h Tower β

Q a P

x

R

h

A

A

x

α

AB subtends angles α and β at P and Q, respectively. We have, to determine h and x in terms of known quantities α, β and a. AB = x tan α, BR = x tan β PQ = AR = AB − BR = x (tan α − tan β ) PQ a a cos α cos β = = ∴ x= tan α − tan β tan α − tan β sin(α − β ) a cos α cos β sin α ∴ h = AB = x tan α = ⋅ sin(α − β ) cos α asin α cos β = sin(α − β )

100. Applying m-n theorem of trigonometry, we get ⇒ ⇒

(c + c )cot (θ − 30 ° ) = c cot 15° − c cot 30 ° 1 sin(30 ° − 15° ) cot(θ − 30 ° ) = ⋅ 2 sin 15° sin 30 ° 1 1 cot(θ − 30 ° ) = ⋅ 2 sin 30 ° C

β α 4h

P

∠CPA = α, ∠BPC = β AP 2 and = ⇒ AP = 2 AB = 4 h AB 1 h 1 In ∆APC, tan α = ⇒ tan α = 4h 4 2h 1 In ∆ABP, tan (α + β ) = = 4h 2 1 + tan β 1 tan α + tan β 1 = ⇒ = ⇒ 4 1 2 1 − tan α ⋅ tan β 2 1 − tan β 4 1 + 4 tan β 1 ⇒ = ⇒ 2 + 8 tan β = 4 − tan β 4 − tan β 2 2 ⇒ 9 tan β = 2 ⇒ β = tan −1 9

Targ e t E x e rc is e s

C

h

103. Let ABC be the equilateral triangle having its centre at Q and PQ be the vertical tower. We have, a a AQ = QC = = π 3 2 ⋅ sin 3 B

P

A

c Q

P

θ–

15°

30

°

B

30°

c θ

A C

⇒ ∴

θ − 30 ° = 45° θ = 75°

Now,

PA = PC = PQ 2 + AQ 2 = h 2 + 2

2

a2 3

575

Objective Mathematics Vol. 1

10

π 3 π PA2 + PC 2 − AC 2 1 ⇒ cos = = 3 2 PA ⋅ PC 2 2  2 a2  a ⇒ 2  h +  − a2 = h 2 + 3 3  ∠APC =

Since,

⇒

3 h 2 = 2 a2

104. 64 cot θ = d D

OA = cot x h …(i) ⇒ OA = h cot x In right angled ∆OBP, ∠OBP = y OB ∴ = cot y h …(ii) ⇒ OB = h cot y In right angled ∆OAB, AB 2 + OA2 = OB 2 ∴ l 2 + h 2 cot 2 x = h 2 cot 2 y 2 ⇒ h (cot 2 y − cot 2 x ) = l 2 l h= ⇒ 2 cot y − cot 2 x

107. Let O be the centre of the square, OP the pole. Shadow A

90°

100 m

of the poleOP isOQ. From question, BQ = y andCQ = x. Then, BC = x + y P

64 m D

θ

B

C

d

Ta rg e t E x e rc is e s

Also,(100 − 64) tan θ = d or (64) (36) = d 2 ∴ d = 8 × 6 = 48m reflection of cloud. We have, AD = h, ∠PAB = θ1, ∠P1 AB = θ 2 Let PC = H P

θ1

B

θ2

D

x

O

105. LetCD be the surface of take P be the cloud and P1 be the

A

C h

C

α

R Q y B

A

OR ⊥ BC x+ y x+ y and BR = 2 2 x+ y x−y RQ = −y= 2 2 Let h be the right of the pole. From right angled ∆POQ, OQ cot α = ⇒ OQ = h cot α h Now, from right angled ∆ORQ, OQ 2 = OR 2 + RQ 2 Let ∴

OR =

2

P1

Now, Also, ⇒ ⇒

 x − y  x + y h 2 cot 2 α =    +  2   2 

⇒

AB = PB ⋅ cot θ1 = (H − h )cot θ1 AB = P1B ⋅ cot θ 2 = (H + h )cot θ 2 (H − h )cot θ1 = (H + h )cot θ 2 h(cot θ1 + cot θ 2 ) H= (cot θ1 − cot θ 2 ) h sin (θ1 + θ 2 ) = sin (θ 2 − θ1 )

106. Let OP be the tower of height h.

⇒ h 2 cot 2 α =

2( x 2 + y 2 ) ⇒ h= 4

2

x2 + y2 tan α 2

108. Let PQ be the vertical tower, having Q as its foot. We have, AQ = BQ = CQ = DQ, which implies that the points A, B, C and D are concyclic having Q as its centre.

109. Let PQ be the tower and A, B be the points of observations. We have, PQ = h, AQ = a, BQ = b P

In right angled ∆OAP,∠OAP = x P

h

South

O

B

576

B

y

l

East

α

β

x

A

Now, and ⇒

A

Q

AQ = PQ ⋅ cot α ⇒ a = h cot α BQ = PQ ⋅ cot β ⇒ b = h cot β (b − a) = h(cot β − cot α )

a = AB = AP tan α = x tan α b = AC = x tan β and c = AD = x tan γ, so that a + b + c = x (tan α + tan β + tan γ ) [Q α + β + γ = 180 °] = x tan α tan β tan γ and abc = x 3 tan α tan β tan γ

A1

π 6

B

D

A

π 3

C B

D

a C

h π = 3 3 π and AB = AA1 ⋅ cot = h 3 6 In ∆ABD, AB 2 = AD 2 + BD 2 h 2 a2 8 h 2 a2 ⇒ 3 h2 = + ⇒ = 3 4 3 4 a 3 a 3 h= = ⇒ 4 2 4 2 AD = AA1 ⋅ cot

Now,

111. Let ABCD be the square of side length a and OP be the vertical tower. We have, OP = h, BC = AB = CD = AD = a, ∠PCO = θ

P

αβγ

x P

abc , a+ b+c so that tan α + tan β + tan γ = tan α tan β tan γ = (a + b + c )/ x = (a + b + c )3 / 2 / abc

⇒

x2 =

114. Let AA1, BB1 and CC1 be the towers and O be the circumcentre of ∆ABC. We have, ∠AOA = θ A , ∠B1OB = θ B , ∠C1OC = θC 1 A1 B1

P

A

B B

A C1

O

O

C

C

D

Now, OA = AA1 ⋅ cot θ A , OB = BB1 ⋅ cot θ B OC = CC1 ⋅ cot θC Since, OA = OB = OC ⇒ AA1 cot θ A = BB1 cot θ B = CC1 cot θC tan θ A tan θ B tan θC = ⇒ = CC1 AA1 BB1

OC = h cot θ a = h cot θ ⇒ a2 = 2 h 2 cot 2 θ 2

Now, ⇒

112. Let PQ = h. We have, QA = 4, QB = 9, θ A + θ B =

10 Properties of Triangles, Heights and Distances

113. We have,

π π We have, BC = a, ∠A1DA = , ∠A1BA = 3 6

Targ e t E x e rc is e s

110. Let AA1 be the flagstaff of height h.

π 2 P

Any other relationship between tan θ A , tan θ B , tan θC cannot be establish.

115. Let PQ = h be a pole standing on a hemispherical dome, of radius r. The elevation of topQ as seen from A is 30°. Q

θB B

Now,

⇒

R

θA A

Q

QA = PQ ⋅ cot θ A QB = PQ ⋅ cot θ B QA ⋅ QB = PQ 2 ⋅ cot θ A ⋅ cot θ B = PQ 2 ⋅ cot θ A ⋅ tan θ A = PQ 2 PQ = QA ⋅ QB = 6m

A

30° d B

r 45°

r 45° h P r O

C

…(i) ∴ r + h = OA tan 30 ° On walking AB = d , the top is just visible from B which means that BQ is a tangent touching at R. OR is normal.

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Objective Mathematics Vol. 1

10

If ∠APB = θ

∠OBQ = 45° = ∠OQB ∴

OB = OQ = r + h

and ∴ and

OA = r + h + d r = OR = RQ = RB BQ = BR + RQ = 2 r

⇒

118. We have,

Hence, from right angled ∆OBQ, we have (2 r )2 = (r + h )2 + (r + h )2 h = r( 2 − 1)

⇒

…(ii)

1 From Eq. (i), (r + h ) = (r + h + d ) ⋅ 3 ⇒ ⇒

( 3 − 1) (r + h ) = d ( 3 − 1) ⋅ 2 r = d

[from Eq. (ii)] d d ( 3 + 1) r= = 2 ( 3 − 1) 2 2 d ( 3 + 1) ( 2 − 1) h = r ( 2 − 1) = 2 2

⇒ Also,

PA2 + PB 2 − AB 2 2 PA ⋅ PB 2 h2 2(h + r 2 ) − 2 r 2 = 2 = 2 2 h + r2 2(h + 4 )

cos θ =

116. Let O be the centre of spherical ball. We have, ∠APB = θ1, ∠OPQ = θ 2 , OA = OB = r

sin C + cos C + sin(2 B + C ) − cos(2 B + C ) = 2 2 ⇒ [sin(2 B + C ) + sin C ] + [cos C − cos(2 B + C )] = 2 2 ⇒ 2 sin(B + C )cos B + 2 sin(B + C )sin B = 2 2 ⇒ sin(180 ° − A)cos B + sin(180 ° − A)sin B = 2 ⇒ sin A(cos B + sin B) = 2 1  1  sin A  cos B + sin B = 1 ⇒  2  2 π  ⇒ sin A sin  + B = 1, is possible only when 4   π sin A = 1 and sin  + B = 1  4 ∴

∠A = 90 ° and

r (r1r2 + r2 r3 + r3 r1 ) =

∆3 (s − a)(s − b)(s − c )

and

r1r2 r3 =

∆3 (s − a)(s − b)(s − c )

∴

x 2 − r (r1r2 + r2 r3 + r3 r1 )x + r1r2 r3 − 1 = 0

A

Ta rg e t E x e rc is e s

O

It satisfied by x = 1. So, one root is x = 1and therefore other root is r1r2 r3 − 1. ∆ ∆ 120. r = , r1 = s s−a

B P

Q

Clearly, ∠APO = ∠OPB = Now, and

θ1 2

r1r =

OP = OA ⋅ cosec (∠APO ) = r cosec OQ = OP ⋅ sin θ 2 = r sin θ 2 ⋅ cosec

∠B = 45°, then ∠C = 45°

119. We have,

θ1 2

θ1 2

117. Let O be the centre of ring and P be the point of suspension. If PA and PB are two consecutive string, π then ∠AOB = 2

s(s − a)(s − b)(s − c ) ∆2 = s(s − a) s(s − a) [Q b = c ]

= (s − b)(s − c ) = (s − b)2 (2 s − 2 b) ( a + b + c − 2 b) = 4 4 a2 4R 2 sin 2 A 2 = = = R sin 2 A 4 4 Also, if ∠B = θ ⇒ ∠A = π − 2θ =

∴

2

2

[Q b = c ]

r1r = R 2 sin 2 (π − 2θ ) = R 2 sin 2 2θ = R 2 sin 2 2 B

A

O

B P

We have, OA = OB = r , OP = h In ∆AOB, AB 2 = OA2 + OB 2 ⇒ In ∆OPB,

AB = r 2 PB 2 = OP 2 + OB 2 = h 2 + r 2

578

⇒

PA = PB = h 2 + r 2

121. We are given that cos B cos C + sin B sin C sin 2 A = 1 and sin 2 A ≤ 1

...(i) ...(ii)

Now, sin B, sin C are positive in a triangle. ∴ sin B sin C sin 2 A ≤ sin B sin C

...(iii)

⇒ 1 − cos B cos C ≤ sin B sin C [from Eq. (i)] ⇒ 1 ≤ cos(B − C ) or cos(B − C ) ≥ 1 But cos θ cannot be > 1, hence we must have sign of equality throughout. ⇒ cos(B − C ) = 1 ⇒ B −C = 0 ∴ B =C π Also, sin 2 A = 1, i.e. A = , hence the triangle both right 2 angled and isosceles.

h A

128. We have, BA × BF = BD × BC and CE × CA = CD × CB

α

Q C

But

BQ = h cot α = CQ = AQ ∴Q is the circumcentre of ∆ABC. a Hence, = 2 R = 2 h cot α sin A ⇒ a sin α = 2 h cos α sin A

⇒

abc 60 = = 2.5 4∆ 24 ∆ 6 130. r = = = 1 s 6

129. R =

123. Area of polygon, ∆ = n times the area of ∆OBC = A

na2 π cot 4 n

...(i)

D

C R R π n r B

Now, ⇒ Also, ⇒

BA × BF BD = CE × CA CD AB BD BD BF BD = ⇒ ⋅ = AC DC DC CE CD BF = CE

Dividing, we have

B

L a

C

π n  π 2 ∆ = nr tan    n π a = 2 R sin n nR 2  2π  ∆= sin    n 2 a = 2 r tan

[from Eq. (i)]

131. Since, the triangle is right angled, greatest angle is 90°. Also, the least angle is opposite to side a, which is 3 sin −1 . 5 3 3 ⇒ 90 ° − sin −1 = cos −1 5 5 1 132. ∆ DEF = a cos Ab cos B sin(π − 2C ) a 2 ab cos A cos B sin 2C = 2 ab sin C = 2 cos A cos B cos C ⋅ 2 sin C  ∆ DEF  ⇒ Q ∆ ABC = ab = 2 cos A cos B cos C  2  ∆ ABC

133. Let R be the radius of the circumcircle of ∆DEF, then [from Eq. (i)]

124. a > b > c ⇒ s − a > s − b > s − c

∆ ∆ ∆ > > ⇒ r1 > r2 > r3 s − a s − b s −c Also, cosine is a decreasing function for x ∈[0, π / 2 ] For a > b > c ⇒ A > B > C ⇒ cos A < cos B < cos C Hence, both statements are true but Statement II does not explain Statement I. ⇒

125. Statement II is false. Because if p = 2, q = 4, r = 6 , then p : q : r = 1 : 2 : 3 but p ≠ 1. For Statement I, let tan A = k, tan B = 2 k, tan C = 3k, then from tan A + tan B + tan C = tan A tan B tan C (in a triangle), we get 6k = 6k 3 ⇒ k = 0, 1, − 1, but k = 0, − 1 is not possible, so k = 1. ⇒ tan A = 1 ⇒ A = 45°

(a cos A)(b cos B)(c cos C ) 4∆ DEF (abc )(cos A cos B cos C ) 1 abc 1 = = ⋅ = R 4 ⋅ 2(cos A cos B cos C )∆ ABC 2 4∆ ABC 2 [here, R is circumradius of ∆ABC] ∆ DEF 134. Radius of ∆DEF = S DEF 1 (2 cos A cos B cos C ) ab sin C 2 =  a cos A + b cos B + c cos C      2 R′ =

10 Properties of Triangles, Heights and Distances

of intersection of the two lines L1 + L3 = 0 and L1 + L2 = 0 and also passes through the point A, hence it has to be the bisector of internal ∠A. Statement II is also correct but is not a correct explanation of Statement I.

P

Targ e t E x e rc is e s

127. Since, (L1 + L3 ) − (L1 + L2 ) = 0 passes through the point

122. From the figure (if PQ is the pole of height h),

= 2R cos A cos B cos C Solutions (Q. Nos. 135-137) We know that, ∆ = abc / 4R OB × OC × BC a R 2 ∴ ∆1 = = 4 R1 4 R1 A

So, Statement I is correct.

126. Statement II is obviously correct. For Statement I, let the circumcentre be at (0, 0) and the vertices of the triangle be ( x1, y1 ), ( x2 , y2 ) and ( x3 , y3 ), then orthocentre of the triangle becomes ( x1 + x2 + x3 , y1 + y2 + y3 ) This implies that, if the centroid is rational, then orthocentre is also rational and x1 + x2 + x3 can be rational even, if x1, x2 , x3 are not all rational.

O R

B

⇒

a/ 2

A 2p1 D

C

a 4∆1 b 4∆ c 4∆ , similarly 2 = 22 , = = 23 R1 R 2 R R R R 3

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135.

136.

137.

a b c 4 4∆ + + = (∆1 + ∆ 2 + ∆ 3 ) = 2 R1 R2 R3 R 2 R 1 ∠BOD = ∠BOC = ∠A 2 a/ 2 From ∆OBD, = tan A 2 r1 a b c ⇒ = 4 tan A, similarly = 4 tan B, = 4 tan C r1 r2 r3 a b c + + = 4 (tan A + tan B + tan C ) r1 r2 r3 = 4 tan A tan B tan C [Q A + B + C = 180 °] 4∆ R a = 4r1 tan A = 12 1 R R1 R 2 tan A = ⇒ r1 ∆1 R2 R 2 tan B R3 R 2 tan C Similarly, = , = r2 r3 ∆2 ∆3  tan A tan B tan C  R1 R2 R3 + + = R2  + + r1 r2 r3 ∆2 ∆ 3   ∆1 1 2

3 5 3 = 2 2 5 3 3 3 Area of ∆ADC = 4 3 − = 2 2

Ta rg e t E x e rc is e s

138. Area of ∆ABC = ⋅ 2 ⋅ 5 ⋅

139. cos 60 ° = ⇒ ⇒ ∴

29 − AC 2 1 = 20 2 AC 2 = 19 19 R= 3 19π Area of circle = 3

3 3 , let AD = x and DC = y 2 1 3 3 xy sin 120 ° = ⇒ xy = 6 2 2 2 AC = 19 x 2 + y 2 + xy = 19 x 2 + y 2 = 13 ( x, y ) = (2, 3) or (3, 2)

140. Area of ∆ADC = ⇒ ⇒ ⇒ ⇒

Hence, AD may be equal to AB = 2

141. A. Q a4 − 2(b2 + c 2 ) a2 + b4 + b2c 2 + c 4 = 0 ⇒(b2 + c 2 − a2 )2 − (b2 + c 2 )2 + b4 + b2c 2 + c 4 = 0 ⇒

⇒

3 a2 b2 = 4a2 b2 cos 2 C [Q a4 + b4 + c 4 = a2 b2 + 2 b2c 2 + 2c 2 a2 ] 3 ⇒ cos 2 C = 4 3 ⇒ cos C = ± 2 ∴ C = 30 ° or 150° C. Q a4 + b4 + c 4 + 2 a2c 2 = 2 a2 b2 + 2 b2c 2 ⇒ b4 − 2(a2 + c 2 )b2 + (a2 + c 2 )2 = 0 ⇒ { b2 − (a2 + c 2 )} 2 = 0 2 ⇒ a + c 2 = b2 ⇒ cos B = 0 ∴ B = 90 °

142. A. R ≥ 2 r , if R = 40, r ≤ 20

1 1 1 1 r + r + r3 ≥9 + + = , AM ≥ HM 1 2 r1 r2 r3 r r C. Q GM ≥ HM 3∆ 3 = 3r ⇒ (r1r2 r3 )1/ 3 ≥ ∴(r1r2 r3 )1/ 3 ≥ 1 1 1 s + + r1 r2 r3 r1r2 r3 ⇒ ≥ 27 r3 D. tan B ⋅ tan C < 1 x+ y 143. Let a = and b = xr , c = xr 2 2 Then, y = xr 3 sin 3 B + sin 3 C b3 + c 3 Now, = sin A sin B sin C abc 2 r 3 (1 + r 3 ) x 3r 3 + x 3r 6 =2 = = r 3 (1 + r 3 ) x + xr 3 ⋅ xr ⋅ xr 2 2 B.

b2 + c 2 − a2 2 bc ...(i) b2 + c 2 = a2 + bc 2 2 b c ab + b + c + ac + = c + a a+ b (a + b)(c + a) ab + a2 + bc + ac [from Eq. (i)] = 2 a + ac + ab + bc =1

144. cos 60 ° = ⇒ ∴

145. Let D be the foot of the altitude from A and BC = a, AB = c , AC = cr , AD = L = c sin B ar ac 2 r abc Now, = 2 = 2 1− r c − r 2c 2 c 2 − b2

(b2 + c 2 − a2 )2 = b2c 2

A

2

 b2 + c 2 − a2  1  =  2 bc 4   1 1 ⇒ cos 2 A = ⇒ cos A = ± 4 2 ∴ A = 60 ° or 120° a2 + b2 − c 2 B. Q cos C = 2 ab ⇒ a2 + b2 − c 2 = 2 ab cos C On squaring both sides, then a4 + b4 + c 4 + 2 a2 b2 − 2 a2c 2 − 2 b2c 2 ⇒

580

= 4a2 b2 cos 2 C

b = cr c L

B

D

C

a sin B sin C a sin B sin C = = 2 2 sin C − sin B sin(C − B)sin(C + B) a sin B sin C c sin B L = = = sin A sin(C − B) sin(C − B) sin(C − B) ar ar ⇒ = ⇒ λ −µ = 0 L≤ 1 − r 2 λ − µr 2

λ 1 = µ 3 λ+µ=4

Q

148. Since, BP = BR ⇒ ∠BPR = ∠BRP = 90 ° −

R

90°– B/ 2

⇒

⇒

PR = 2 BP sin

Similarly, ,

⇒

⇒

2 k ≤ 1 + k 2 ⇒ 4k 2 ≤ 1 + k 2 1 3k2 − 1≤ 0 ⇒ k ≤ 3

⇒

α

B

[say]

B 2

A

147. 2 sin B = sin A + sin C = sin A + sin( A + B) = sin A + sin A cos B + cos A sin B ⇒ sin B(2 − cos A) = sin A(1 + cos B) sin B sin A = ⇒ 1 + cos B 2 − cos A B sin A ⇒ tan = =k 2 2 − cos A ⇒ sin A = 2 k − k cos A ⇒ sin A + k cos A = 2 k 2k sin( A + α ) = ⇒ 1+ k2 k where, tan α = 1 2k ⇒ ≤1 1+ k2

10

B is 2

⇒

B 2

G β

Q α C

P 90°– α

⇒ RG = 2 BP sin

B cos α 2

C cos β 2 B sin cos α RG BP 2 = ⋅ QG PC sin C cos β 2 B C  sin cos  90 ° −   RG BP 2 2 = ⋅ C B  QG CQ sin cos  90 ° −   2 2 RG BR RG CQ = ⇒ ⋅ =1 QG CQ QG BR QG = 2 PC sin

Entrances Gallery 1. x = a + b y = ab x2 − c 2 = y a2 + b2 − c 2 1 = − = cos 120 ° ⇒ 2 ab 2 ∆ 2π abc ⇒ ∠C = ,r = ⇒ R= 4∆ s 3 2 1  2π   4 ab sin   2  3   4∆2 r r 3y ⇒ = = =  ⇒ x+c R s (abc ) R 2 c ( x + c) ⋅ y ⋅c 2

2. Let s − a = 2 k − 2, s − b = 2 k, s − c = 2 k + 2, k ∈ I, k > 1 Adding, we get s = 6k P s–a

Properties of Triangles, Heights and Distances

4 sin A sin B + 4 sin B sin C + 4 sin C sin A = 9 ⇒ 2 cos ( A − B) − 2 cos ( A + B) + 2 cos (B − C ) −2 cos(B + C ) + 2 cos (C − A) − 2 cos (C + A) = 9 ⇒ 2 [cos ( A − B) + cos (B − C ) + cos (C − A)] 3 = 9 − 2 [cos A + cos B + cos C ] ≥ 9 − 2 × = 6 2 ⇒ cos ( A − B) + cos (B − C ) + cos (C − A) ≥ 3 But cos ( A − B) ≤ 1, cos (B − C ) ≤ 1, cos (C − A) ≤ 1 ⇒ cos ( A − B) = 1, cos (B − C ) = 1, cos (C − A) = 1 ⇒ A = B = C, therefore ∆ABC is equilateral. Hence, ∆ = 3

⇒ Maximum value of tan

Targ e t E x e rc is e s

146. The given equation is

1 3 b2 + c 2 − a2 1 ⇒ = 2 bc 3 ⇒ 3 [(4k )2 + (4k − 2 )2 − (4k + 2 )2 ] = 2 × 4k (4k − 2 ) ⇒ 3 [16k 2 − 4 (4k ) × 2 ] = 8k (4k − 2 ) ⇒ 48k 2 − 96k = 32 k 2 − 16k ⇒ 16k 2 = 80 k ⇒ k=5 So, sides are 22, 20, 18. P 2 sin 2 2 sin P − 2 sin P cos P 1 − cos P 2 3. = = 2 sin P + 2 sin P cos P 1 + cos P 2 cos 2 P 2 cos P =

Now,

P M

N

c = 5/2

s–c

Q

s–b

L

So, a = 4k + 2, b = 4k, c = 4k − 2

b = 7/2

R

Q

a=2

R

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Objective Mathematics Vol. 1

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In ∆EDB,

P 2 ( s − b) ( s − c ) = s (s − a) [(s − b) (s − c )]2 = ∆2 2  1  3      2  2   2    = 3 =   2  4∆  ∆ = tan 2

⇒ In ∆EDC,

⇒ Now,

a c ∴ sin 2C + sin 2 A = 2 sin A cos C + 2 sin C cos A c a = 2 sin ( A + C ) = 2 sin B 3 =2 × = 3 2

5. Using cosine rule for ∠C,

Ta rg e t E x e rc is e s

h CD h CD = 3

tan 60° =

4. B = 60 °

3 ( x 2 + x + 1)2 + ( x 2 − 1)2 − (2 x + 1)2 = 2 2 ( x 2 + x + 1) ( x 2 − 1) 2 x2 + 2 x − 1 ⇒ 3= 2 x + x+1 2 ⇒ ( 3 − 2 )x + ( 3 − 2 )x + ( 3 + 1) = 0 (2 − 3 ) ± 3 x= ⇒ 2 ( 3 − 2) ⇒ x = − (2 + 3 ), 1 + 3 x = 1+ 3 ⇒ 1 6. ∆ = ab sin C 2 2 ∆ 2 × 15 3 3 ⇒ sin C = = = ab 6 × 10 2 ⇒ C = 120 ° c = a2 + b2 − 2 ab cos C ⇒ ∴

h BD BD = h

tan 45° =

[as x > 0 ]

= 62 + 10 2 − 2 × 6 × 10 × cos 120 ° = 14 ∆ 225 × 3 =3 r= ⇒ r2 = 2 s  6 + 10 + 4     2

7. According to the given information, the figure should be as follows.

AB AD − BD = BC BD − CD AB h 3 − h ⇒ = h BC h− 3 AB h( 3 − 1) ⇒ = BC h ( 3 − 1) 3 3 −1 AB ⇒ × 3 = BC ( 3 − 1) AB 3 ⇒ = BC 1 AB : BC = 3: 1 ∴ 20 1 8. tan 30 ° = = 20 + x 3 20 m=h 20 m π/6

π/4

x

20 m

0

20 + x = 20 3 x = 20 ( 3 − 1) ⇒ Speed is 20 ( 3 − 1) m/s.

9. Given, mid-points of a triangle are (0, 1), (1, 1) and (1, 0). Plotting these points on a graph paper and make a triangle. So, the sides of a triangle will be 2, 2 and 2 2 + 2 2 i.e. 2 2. Y

E C (0, 2)

h

30° A

C

Let the height of tower be h. In ∆EDA, ED tan 30° = AD ED h 1 ⇒ = = 3 AD AD AD = h 3 ⇒

582

X′

60°

45° B

(0,1)

D

B (0, 0)

(1, 1)

(1, 0)

A(2, 0)

X

Y′

2⋅0 + 2 2 ⋅ 0 2+2+2 2 2− = × 2+ 2 2− =2 − 2

x-coordinate of incentre =

+ 2 ⋅2 2 2 2

11. θ

C

a π = tan 2r n

and

α 2+

√p

p

p 2

q

π–(θ+α) A

x–q

B

q M

∴

x

p x −q p tan (θ + α ) = q−x q − x = p cot (θ + α ) x = q − p cot (θ + α)  cot θ cot α − 1 =q − p   cot α + cot θ   q  cot θ − 1 p  =q − p  q + cot θ      p

In ∆DAM, tan (π − θ − α ) = ⇒ ⇒ ⇒

⇒ ⇒

r 1 = R 2 r 1 n = 4 gives = R 2 r 3 n = 6 gives = R 2

q  Q cot α = p 

12. In ∆ABC,

BC = h cot 60 ° and in ∆ABD, BD = h cot 45° Since, BD − BC = DC ⇒ h cot 45° − h cot 60 ° = 7 A

h

D

⇒

Aliter Applying sine rule in ∆ABD, θ

A

q

45° 7m

60° C

B

7 7 h= = 1 cot 45° − cot 60 °  1 −   3 =

C

α

√p 2 +q

r π = cos R n n = 3 gives

 q cot θ − p  =q − p   q + p cot θ   q cos θ − p sin θ  =q − p   q sin θ + p cos θ  2 q sin θ + pq cos θ − pq cos θ + p2 sin θ x= p cos θ + q sin θ ( p2 + q 2 ) sin θ AB = p cos θ + q sin θ

D

r

π n

R 2π n

7 3 × 3 −1

3+1 7 3 = ( 3 + 1) m 2 3+1

13. Let h be the height of a tower. p

2

π–(θ+α)

Since, ∠AOB = 60 ° So, ∆OAB is an equilateral. ∴ OA = OB = AB = a

B

C h

p +q AB = sin θ sin { π − (θ + α)} 2

⇒ ⇒

10

Targ e t E x e rc is e s

q

D

a π = sin 2R n

Q

Properties of Triangles, Heights and Distances

10. Let AB = x

2

O

p2 + q 2 AB = sin θ sin (θ + α ) p2 + q 2 sin θ AB = sin θ cos α + cos θ sin α ( p2 + q 2 ) sin θ = q sin θ + p cos θ ( p2 + q 2 ) sin θ = p cos θ + q sin θ q   Q cos α =  2 2  p +q     p and sin α =  2 2 p + q  

a

60°

⇒

30° A

a

B

In ∆OAC, tan 30° =

90° a

h a

⇒ h=

1 h = 3 a a 3

P Q and tan are the roots of equation 2 2 ax 2 + bx + c = 0. P Q b tan + tan = − 2 2 a ...(i) ∴  P Q c  and tan tan = 2 2 a 

14. Since, tan

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Objective Mathematics Vol. 1

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Also, ⇒ ⇒ ⇒ ⇒

⇒ ⇒ ⇒ ⇒

P Q R π + + = 2 2 2 2 P+Q π R = − 2 2 2 P+Q π = 2 4  P Q tan  +  = 1 2 2 P Q tan + tan 2 2 =1 P Q 1 − tan tan 2 2 b − a =1 c 1− a b c − = 1− a a − b= a−c c =a+ b

[Q P + Q + R = π ]

π   Q ∠R = , given   2

[from Eq. (i)]

π π ⇒ ∠P + ∠Q = 2 2 ∠P π ∠Q ⇒ = − 2 4 2 P  π Q tan = tan  −  ∴  4 2 2 π Q tan − tan 4 2 = π Q 1 + tan tan 4 2 P P Q Q ⇒ tan + tan tan = 1 − tan 2 2 2 2 P Q P Q tan + tan = 1 − tan tan ⇒ 2 2 2 2 b c ⇒ − = 1− ⇒ − b= a −c a a ⇒ c =a+ b c 15. We know that, = 2R sin C [QC = 90 °]…(i) ⇒ c = 2R C r π r and tan = ⇒ tan = 2 s −c 4 s −c a+ b+c −c ∴ r = s −c ⇒ r = 2 ...(ii) ⇒ a + b − c = 2r On adding Eqs. (i) and (ii), we get 2(r + R ) = a + b

Ta rg e t E x e rc is e s

Here, we see that the greatest side is c. a2 + b2 − c 2 cos C = ∴ 2 ab sin 2 α + cos 2 α − 1 − sin α cos α ⇒ cos C = 2 sin α cos α sin α cos α cos C = − ⇒ 2 sin α cos α 1 cos C = − = cos 120 ° ⇒ 2 ⇒ ∠C = 120 °

18. Let CD (= h ) be the height of the tree and BC (= x ) be the width of the river. D

h

A

c F

E

⇒ and in ∆CAD,

b

B

C

CD BC h x h=x 3

3=

CD AC 1 h = 3 40 + x h 3 = 40 + x 3 x = 40 + x 2 x = 40 x = 20 m

...(i)

tan 30° = 90° –

90°– C D

60°

30°

In ∆CBD, tan 60° =

90

°–B

A

⇒

17. Let a = sin α, b = cos α, c = 1 + sin α cos α

⇒

16. In ∆BAD,

584

1 × BC × AD 2 1 ∆ = × a × AD ⇒ 2 2∆ ⇒ AD = a 2∆ 2∆ Similarly, BE = and CF = b c Q AD, BE and CF are in HP. 1 1 1 ∴ , and are in HP. a b c ⇒ a, b and c are in AP. ⇒ sin A, sin B and sin C are in AP. ar (∆ABC ) =

∠R =

B

BE = a sin C and CF = b sin A Since, AD, BE and CF are in HP. ∴ c sin B, a sin C, b sin A are in HP. 1 1 1 are in AP. ⇒ , , sin C sin B sin A sin C sin B sin A ⇒ sin A, sin B and sin C are in AP. Aliter

Aliter

Since,

Similarly,

A

C a

AD cos (90° − B) = c AD = c sin B

⇒ ⇒ ⇒ ⇒ ⇒

[from Eq. (i)]

⇒

O

ON = AN cot a π cot 2 2 π AN sin = n OA

π n

A

a/ 2 N a/2

=

and

B

...(i)

B

π a π ...(ii) ⇒ OA = AN cosec = cosec n 2 n Now, sum of the radii = ON + OA a π a π [from Eqs. (i) and (ii)] = cot + cosec 2 n 2 n π   cos a 1  2 =  + π 2 sin π sin    n n π π    1 + cos 2 cos 2 a a    n n 2 =  = π  2 π π  2 cos   sin  2 sin   n  2n 2n  a π = cot 2 2n C A 3b 20. Given, a cos 2 + c cos 2 = 2 2 2  s(s − a) 3 b  s(s − c ) ⇒a = +c  bc  2  ab  s(s − c + s − a) 3 b = ⇒ b 2 ⇒ 2 s (2 s − c − a) = 3 b2 ⇒ 2 s (a + b + c − c − a) = 3 b2 ⇒ (a + b + c )b = 3 b2 ⇒ a + b + c = 3b ⇒ 2b = a + c Hence, a, b and c are in AP.

21. Given, AD = 4 and BD = DC Since, the centroid G divides the line AD in the ratio 2 : 1. 8 4 ∴AG = and DG = 3 3 A π/6

8/3

E

In ∆ABG,

D

π AG = 3 BG π BG = AG cot 3 8 1 8 BG = × = 3 3 3 3 tan

⇒ ⇒

1h 4

θ2

O

θ1

40 m

A

3 5 AC tan θ1 = AO 1 h h tan θ1 = 4 = 40 160

tan θ 2 =

⇒ In ∆AOC, ⇒

...(i)

...(ii)

and in ∆AOB, AB h = AO 40 tan θ1 + tan θ 2 h = 1 − tan θ1 tan θ 2 40 3 h + 160 5 = h 3 40 h 1− × 160 5 5 [h + 96] h = 800 − 3 h 40

tan (θ1 + θ 2 ) = ⇒

⇒

⇒ ⇒

[from Eqs. (i) and (ii)]

200 [h + 96] = 800 h − 3 h 2

⇒

3 h 2 − 600 h + 19200 = 0

⇒

h 2 − 200 h + 6400 = 0

23. Since, A + B + C = π ⇒

4/3

B

h

C

⇒ (h − 160 ) (h − 40 ) = 0 ⇒ h = 160 or h = 40 Hence, height of the vertical pole is 40 m.

G π/3

3h 4

10 Properties of Triangles, Heights and Distances

In ∆AON,

ON ⊥ AB AN = BN π AN tan = n ON

π/n

and

1 1 8 16 × AD × BG = × 4 × = 2 2 3 3 3 3 Since, median divides a triangle into two triangles of equal area. Therefore, Area of ∆ABC = 2 × Area of ∆ADB 16 32 sq units =2 × = 3 3 3 3 3 22. Given, θ 2 = tan − 1 5 Area of ∆ADB =

Targ e t E x e rc is e s

19. Let AB = a,

C

A+C=π −B A−B+C π = −B ⇒ 2 2  A − B + C π  ∴ 2 ca sin   = 2 ca sin  − B   2  2 = 2 ac cos B  a2 + c 2 − b2  = 2 ac   2 ac   = a2 + c 2 − b2

585

Ta rg e t E x e rc is e s

Objective Mathematics Vol. 1

10

24. Given that, a = 4, b = 3 ∠A = 60 ° c 2 + 9 − 16 cos 60 ° = 2 × 3×c

and Now,

1 c2 − 7 = 2 2 × 3c

⇒ ⇒

c 2 − 3c − 7 = 0

25. Given that, A 5 = 2 6 C 2 and tan = 2 5 A C 5 2 Now, tan tan = × 2 2 6 5 (s − b) (s − c ) (s − a) (s − b) 1 ⇒ ⋅ = s (s − a) s (s − c ) 3 s−b 1 = ⇒ s 3 ⇒ 2s = 3 b ⇒ a + c = 2b [Q2s = a + b + c ] ⇒ a, b and c are in AP. tan

26. Let in the figure, ∠A > ∠B > ∠C Since, angles are in AP. C b=9

a = 10

28. Σ a3 sin (B − C ) = Σk 3 sin 3 A sin (B − C )   a b c Q sin A = sin B = sin C = k    = Σk 3 [sin 2 A sin (B + C ) sin (B − C )] 1   = k 3 sin 2 A × (cos 2C − cos 2 B) 2   1 2 + {sin B × (cos 2 A − cos 2C )} 2 1  2 + {sin C × (cos 2 B − cos 2 A)}  2 3 k [sin 2 A (1 − 2 sin 2 C − 1 + 2 sin 2 B) = 2 + sin 2 B (1 − 2 sin 2 A − 1 + 2 sin 2 C ) + sin 2 C (1 − 2 sin 2 B − 1 + 2 sin 2 A)] =

k3 [− 2 sin 2 A sin 2 C + 2 sin 2 A sin 2 B 2 − 2 sin 2 B sin 2 A + 2 sin 2 B sin 2 C − 2 sin 2 C sin 2 B + 2 sin 2 C sin 2 A] 3

k [0 ] = 0 2 (a + b + c ) (b + c − a) (c + a − b) (a + b − c ) 29. Consider, 4b2c 2 2 s (2 s − 2 a) (2 s − 2 b) (2 s − 2c ) [Q a + b + c = 2 s] = 4b2c 2 16 [s (s − a) (s − b) (s − c )] = 4b2c 2 2 2 2∆  4∆  2∆   = 2 2 =   = sin 2 A Q sin A =   bc  bc  bc =

30. Given the ratio of angles of a triangle is1 : 1 : 4. A

c=x

B

∴ ∠C = θ − d , ∠B = θ and ∠A = θ + d As, ∠A + ∠B + ∠C = 180 ° θ + d + θ + θ − d = 180 ° ⇒ 3θ = 180 ° ⇒ θ = 60 ° Now, in ∆ABC, a2 + c 2 − b2 cos B = 2 ac (10 )2 + x 2 − 92 ⇒ cos 60 ° = 2 × 10 × x ⇒ ⇒ ∴

1 100 + x 2 − 92 = 2 20 x x 2 − 10 x + 19 = 0 10 ± 100 − 76 2 10 ± 24 = = 5± 6 2

x=

Let angles of a triangle be A, B and C. ∴ A : B : C = 1: 1: 4 Let A = x, B = x and C = 4 x Q A + B + C = 180 ° ∴ x + x + 4 x = 180 ° ⇒ 6 x = 180 ° ⇒ x = 30 ° ∴ A = 30 °, B = 30 ° and C = 120 ° Hence, the largest angle is 120°. So, the largest side of triangle is c. ∴Perimeter of triangle : Largest side of a triangle = (a + b + c ) : c = (sin 30 ° + sin 30 ° + sin 120 ° ) : sin 120° 1 1 3 3  3 3 = + + = 1 + =2 + 3: 3 : : 2  2 2  2 2 2 

31. Since, A, B and C are in AP. 2B = A + C A + B + C = 180 ° 3B = 180 ° ⇒ B = 60 °

We have, Also, ⇒

A

27. Consider, a [b cos C − c cos B]

c

 a2 + c 2 − b2   a2 + b2 − c 2  = ab    − ac  2 ab 2 ac    

586

=

a2 + b2 − c 2 (a2 + c 2 − b2 ) − = b2 − c 2 2 2

B

b

a

C

...(i) ...(ii)

r1 =

33. Rewrite the given relation as

    A + B   A + B a tan A − tan    = b tan   − tan B     2 2     A + B  A + B  b sin a sin A − −B   2  2  = ⇒  A + B  A + B cos A cos   cos B cos    2   2   A − B  A − B a sin    b sin   2   2  ⇒ = cos A cos B a b ⇒ = cos A cos B sin A sin B = ⇒ cos A cos B

⇒ ⇒

tan A = tan B A=B

35. We have,

a b = cos A cos B sin A sin B = cos A cos B

 a b c  Q sin A = sin B = sin C    ...(i) ⇒ sin A cos B = sin B cos A ⇒ 2 sin A cos B = cos A sin B + cos A sin B = cos A sin B + sin A cos B [from Eq. (i)] = sin ( A + B) = sin (π − C ) = sin C ⇒

10 Properties of Triangles, Heights and Distances

∆ ∆ ⇒ 2= s−a s−a s−a 1 ...(i) = ⇒ ∆ 2 ∆ ∆ ⇒ 3= r2 = s−b s−b s−b 1 ...(ii) = ⇒ ∆ 3 ∆ ∆ s −c 1 and r3 = ...(iii) ⇒ 6= ⇒ = s −c s −c ∆ 6 On adding Eqs. (i), (ii) and (iii), we get s − a s − b s −c 1 1 1 + + = + + ∆ ∆ ∆ 2 3 6 3s − (a + b + c ) 3 + 2 + 1 ⇒ = ∆ 6 3s − 2 s 6 ⇒ = =1 ∆ 6 s ...(iv) ⇒ =1 ∆ 2 Q s = r1r2 + r2 r3 + r3 r1 = 2 × 3 + 3 × 6 + 6 × 2 = 6 + 18 + 12 = 36 ⇒ s2 = 36 ⇒ s =6 6 From Eq. (iv), = 1, ∆ = 6 ∆ ∆ 6 Now, r1 = ,2 = s−a 6− a ⇒ 6− a= 3 ⇒ a= 3 Q

= k 3 sin 2 A sin (B + C ) cos (B − C ) 1 = k 3 [sin 2 A (sin 2 B + sin 2C )] 2 1 = k 3 [2 sin 2 A sin B cos B + 2 sin 2 A sin C cos C ] 2 ⇒ Σa3 cos (B − C ) = Σk 3 sin A sin B (sin A cos B + cos A sin B) = Σk 3 sin A sin B sin ( A + B) = Σk 3 sin A sin B sin C = Σabc = 3 abc

36. Area of circle, A1 = πr 2 Perimeter of a circle, P = 2πr If a is side of regular polygon of n sides, then na = 2πr 2πr ⇒ a= n Now, if A2 = Area of regular polygon of n sides 1  π Then, A2 = na2 cot    n 4 2

...(i)

1  2 πr   π [from Eq. (i)] n  cot    n 4  n  2 2 1  4π 2 r 2   π π r  π cot   = n  2  cot   =  n  n n 4  n 

=

∴

Area of circle πr 2 = 2 2 Area of regular polygon π r  π cot    n n  π tan    n  π π = = tan   : π  n n n

Targ e t E x e rc is e s

32. Given, r1 = 2, r2 = 3 and r3 = 6

34. By sine rule, T1 = k 3sin 3 A cos (B − C )

37. Let PD be a pole of height h. From ∆PAD or ∆PCD, h = a tan 30 ° =

a 3

P h C

30°

D

30°

Now, by using sine rule, we have a b c = = sin A sin B sin C c ⇒ sin C = sin B b 2 sin 60 ° = 3 2 3 1 = × = = sin 45° 2 3 2 ...(iii) ⇒ C = 45° [from Eqs. (i), (ii) and (iii)] ⇒ A = 75°

θ B

a

A

587

Objective Mathematics Vol. 1

10

From ∆PBD, diagonal BD = a 2 a ∴ = a 2 tan θ 3 1 tan θ = ⇒ 6 π 38. Let ∠OPA = ∠OPC = and OP = n 4

⇒ sin 2 R = 1 ⇒ sin R = 1 [Q 0 < R ≤ 90 ° ] ⇒ R = 90 ° ⇒ ∆PQR is a right angled triangle. cos 2 A cos 2 B 42. We have, − a2 b2 2 1 − 2 sin A (1 − 2 sin 2 B) = − a2 b2 2 1 1 2 sin A 2 sin 2 B = 2 − 2 − + a b a2 b2 2 sin B sin 2 A  1 1 = 2 − 2 +2 2 −  a b a2   b 1 1  sin A sin B  = 2 − 2 Q =  a b  a b

P π 4 D

A

O C

2∆ 2∆ 2∆ sin A + sin C andsin B = ,b= ,c = p1 p2 p3 2 [Q sin A, sin B and sin C are in AP] a+c ⇒ b= 2 2 ∆ 2 ∆ / p1 + 2 ∆ / p3 ⇒ = 2 p2 2 p1 p3 p2 = ⇒ p1 + p3 Hence, altitudes are in HP.

43. Since, a =

B

OA = tan 45° = 1 OP ∴ OA = n, AC = 2 n OP π Also, = cos PA 4 ⇒ PA = 2 n Also, AC = 2 AB ⇒ 2 n = 2 ⋅ AB ⇒ AB = 2 n ∴ PA = AB = PB So, ∆BPA is equilateral. Hence, the required angle is 60°.

Ta rg e t E x e rc is e s

So,

44. Q ∠C = 90 ° ∴

39. Since, A + B + C = 180 °

Σ cot A cot B − 1 1 = cot A cot B cot C − Σ cot A 0 cot A cot B cot C − Σ cot A = 0 cot A + cot B + cot C =1 cot A cot B cot C

⇒ cot ( A + B + C ) = ⇒ ⇒

40. Applying sine rule in ∆ABC, A 45°

67

h

.5°

112.5°

D x B

⇒

22.5° 6 cm

C

sin 45° sin 22.5° = 6 AB 2− 2 ×6 =3 2 2− 2 AB = 1 2× 2

∠A + ∠B = 90 ° sin 2 A cos 2 A sin 2 A cos 2 (90 ° − B) Consider, − = − sin 2 B cos 2 B sin 2 B cos 2 (90° − A) sin 2 A sin 2 B a2 b2 a4 − b4 = − = − = 2 2 a b sin 2 B sin 2 A b2 a2

45. Clearly, a2 + b2 − c 2 64 + 100 − 144 1 = = …(i) 2 ab 2 × 8 × 10 8 b2 + c 2 − a2 100 + 144 − 64 3 and cos A = = = 2 bc 2 × 10 × 12 4 9 1 2 Also, cos 2 A = 2 cos A − 1 = 2 × …(ii) − 1= 16 8 Now, from Eqs. (i) and (ii), cos 2 A = cos C ⇒ C = 2 A cos C =

46. Given, height of house is 6 3 and the angle of elevation is 60°. In

∆ABC, sin 60° =

h l

3 h = 2 l

⇒

A

2+ 2 h h In ∆ABD,sin 67.5° = ⋅ = 2 AB 3 2 2− 2 ⇒

4−2 × 3 2 2 3×2 = = 3 cm 2

60° C

41. We have, r 2 sin P sin Q = pq ⇒

588

sin 2 R sin P sin Q = sin P sin Q  p q r  Q sin P = sin Q = sin R   

h = 6√3

l

h=

⇒ ∴

B

3l = 6 3 ×2 6 3 ×2 l= = 12 m 3

⇒ In ∆PDB, ⇒ In ∆PDC,

60° B

y

⇒

P

x

tan 30 ° =

and

…(ii)

tan 60° =

100 x+ y

h DC

P

x + y = 100 3

∴The required distance = 100 3 − =

48. In ∆ABC, tan θ =

h

100 3

200 3 m 3

D

h b

...(i)

⇒ Now,

h

B

90° – θ a b

θ D

C

h a [Qangles of elevation at C and D are complement] h cot θ = ⇒ a a ...(ii) ⇒ tan θ = h On comparing Eqs. (i) and (ii), we get h a = b h 2 ⇒ h = ab In ∆ABD,

tan (90° − θ ) =

h = ab

60°

45° C

A

∴

h h ⇒ 1= DB DB DB = h

tan 45° =

100 m

A

…(i)

30° B

A

h h …(iii) ⇒ DC = DC 3 AB AD − DB 3h − h = = h BC DB − DC h− 3 [from Eqs. (i), (ii) and (iii)] h( 3 − 1) 3 = 3 :1 = h( 3 − 1) 3=

50. We have, A + B + C = π and 2B = A + C ⇒ ∴

Properties of Triangles, Heights and Distances

⇒ Q Top of tower

30°

h AD h 1 = 3 AD AD = 3 h

tan 30° =

3 B = π ⇒ B = π /3 a + c sin A + sin C = b sin B  A − C  A + C 2 sin    cos   2   2  = π sin 3 π  A − C 2 sin cos    2  3 = π sin 3  A − C = 2 cos    2 

Targ e t E x e rc is e s

⇒

10

49. In ∆PDA,

100 x 100 100 x= = tan 60 ° 3

47. Q tan 60 ° =

589

11 Cartesian System of Rectangular Coordinates Introduction Coordinate geometry is the branch of Mathematics which includes the study of different curves and figures by representing points in a plane by ordered pairs of real number called cartesian coordinates. Also, it involves the methods of algebra to solve geometrical problems which is known as analytical geometry. Since, to study geometrical figure, we deal with coordinates that’s why it is called coordinate geometry.

Coordinate System i.

●

Introduction

●

Coordinate System

●

Distance Formulae

●

Cartesian coordinate system Let XOX ′ and YOY ′ be two perpendicular straight lines drawn through any fix point O on the plane of the paper. Then,

Y

X-axis The horizontal line XOX ′ is called X-axis. X′ X O (0, 0) Y-axis The vertical line YOY ′ is called Y-axis. Coordinate axes X-axis and Y-axis together are called axes of coordinate or axes of reference. Y′ Origin The point O is called the origin of coordinates or the origin. Oblique axes If both the axes are not perpendicular, then they are called as oblique axes as shown in the figure: Y

θ X′

Chapter Snapshot

O (0, 0)

Y′

X

Applications of Distance Formula

●

Section Formulae

●

Area of a Triangle

●

Area of a Quadrilateral

●

Some Standard Points of a Triangle

●

Locus

●

Transformation of Axes

X

x P(x, y) y

(b) (1, 3 )

(c) (1, − 3 )

(d) None of these

iii.

X

O (0, 0)

(a) ( 3, 1)

Sol. (b) Clearly, x = 2 cos π and y = 2 sin π 3 3 ⇒ x = 1 and y = 3 ∴ Cartesian coordinates of P are (1, 3 ).

Y

X′

Example 2. If polar coordinates of any point  π are  2,  , then its cartesian coordinates are  3

Y′

Ø

● ● ●

Coordinates of the origin are (0, 0). y-coordinate of a point on X-axis is zero. x-coordinate of a point on Y-axis is zero.

Relation between polar and cartesian coordinates system Let P be any point in a plane whose cartesian coordinates are P ( x, y) and let the same point when polar system is used have coordinates ( r, θ ). Then, …(i) OA = x = r cos θ and …(ii) OB = y = r sin θ Y x r θ

B X

Example 1. One of the vertices of a square is origin and adjacent sides of the square are coincident with positive axes. If length of a side is 6 units, then which will not be its vertex? (a) (0, 6) (b) (6, 0) (c) ( − 6, − 6) (d) (0, 0)

X'

Again, dividing Eq. (ii) by Eq. (i), we get  y θ = tan −1    x

(6, 6)

X

6

Ø

Polar coordinates system Let P be a point whose distance be r from origin and makes an angle θ with X -axis. y x ∴ = cos θ and = sin θ r r or x = r cos θ and y = r sin θ Thus, to represent ( x, y) in ( r, θ ) is called polar representation as shown in the figure:

X

On adding 360° (or any multiple of 360°) to the vectorial angle does not alter the final position of revolving lines so (r , θ) is always the same point as (r , θ + 2nπ), where n ∈I. On adding 180° (or any odd multiple of 180°) to the vectorial angle and changing the sign of radius vector gives the same point as before. Thus, the point [ − r , θ + (2n + 1) π] is same as [− r , θ + π].

Example 3. The polar coordinates of a point having cartesian coordinates ( − 1, − 1) are π 3π    (b)  2, (a)  2,      4 4  5π  π   (c)  2,  (d)  2, −    4 4 Sol. (c) The cartesian coordinates of P are (− 1, − 1.)

Y

,y P(x

)

r y θ O

●

●

Y′

ii.

X

OP = r = x 2 + y 2

Y

O

A

On squaring and adding Eqs. (i) and (ii), we get

square. Since, the square lies in the first quadrant as shown below:

X′

O

P(x, y) y

Y'

Sol. (c) Obviously, (− 6, − 6) will not be the vertex of the

6

11 Cartesian System of Rectangular Coordinates

Cartesian coordinates The ordered pair of perpendicular distances from both axes of a point P lying in the plane is called cartesian coordinates of P. If the cartesian coordinates of a point P are ( x, y), then x is called the abscissa or x-coordinate of point P and y is called the ordinate or y-coordinate of point P.

x

X

So, x = − 1 = r cos θ and y = − 1 = r sin θ r = 1+ 1 = 2 ⇒  − 1 −1 and θ = tan− 1   = tan (1)  − 1 Since, the point lies in the third quadrant. 5π 3π or − θ= ∴ 4 4 5π  Hence, the polar coordinates of P(− 1, − 1) are P  2 , .  4 

591

Objective Mathematics Vol. 1

11

X

Example 4. Point P (5, 60°) having polar coordinates is equivalent to (a) ( − 5, 60° ) (b) (5, 240°) (c) (5, 420°) (d) (5, 300°) Sol. (c) Since, point having coordinates (r, θ) is equivalent to (r, 2 nπ + θ), where n ∈ I. Hence, (5, 60° ) = (5, 2 nπ + 60° ) = (5, 420° )

X

[take n = 1]

Sol. (a) PQ = (3)2 + (9)2 + 2 × 3 × (− 9) cos 60°

Distance Formulae i.

Example 6. The distance between points P ( − 3, − 2) and Q ( − 6, 7), the axes being inclined at 60°, is (b) 7 (a) 3 7 (d) None of these (c) 2 7

9 + 81 − 54 ×

⇒

When coordinates of two points are given in cartesian form Let P ( x1 , y1 ) and Q ( x 2 , y2 ) be the two points.

iii.

Q (x2, y2)

Y

1 = 2

63 = 3 7

When coordinates are in polar form If ( r1 , θ 1 ) and ( r2 , θ 2 ) are given points, then distance between them is r12 + r22 − 2r1 r2 cos (θ 1 − θ 2 ) Here, OA = r1 , OB = r2 and AB = d

Proof

y

1)

(y2 – y1)

B (r2,θ2)

P

(x

1,

Y

d

(x2 – x1 )

X′

M

r2

X

Y′

X′

Then, the distance between P and Q will be given by (Difference of abscissae) 2 + (Difference of ordinates) 2

PQ = ∴ X

Sol. (c) Given, 2 = ( x − 3)2 + (2 − 4)2 ∴

x=3

18

0°–

ω

d ω

(x1 − x2)

ω

ω

O (0, 0)

Y′

PQ = d = 592

( x1 − x 2 ) 2 + ( y1 − y2 ) 2 + 2 ( x1 − x 2 )( y1 − y2 )cos ω

X θ2

θ1

| AB | = r12 + r22 − 2r1 r2 cos (θ 1 − θ 2 )

Example 7. The distance between the points P (2, 60° ) and Q(5, 270° ) is 29 + 10 3

(b) 29 − 10 3 (c) 25 − 10 3 (d)

25 + 10 3

Sol. (a) PQ = (2 )2 + (5)2 − 2 (2 ) (5) cos (60° − 270° )

P (x1,y1)

(x2,y2) Q

X′

X

(a)

When coordinate axes are inclined at an angle ω If the coordinate system is oblique, i.e. if the coordinate axes are inclined at an angle ω. In this case, the distance between two points P and Q will be given by Y

∴

= 4 + 25 − 20 cos (− 210° ) = 29 − 20 cos (180° + 30° )

(y1 − y2)

ii.

A (r1,θ1)

By cosine law, d 2 = r12 + r22 − 2r1 r2 cos (θ 1 − θ 2 )

Example 5. If the distance between the points ( x, 2) and (3, 4) is 2, then the value of x is (a) 2 (b) 1 (c) 3 (d) 4 4 = ( x − 3)2 + 4 ⇒ x − 3 = 0

− θ2 r 1

Y′

d = ( x 2 − x1 ) 2 + ( y2 − y1 ) 2

⇒

O

θ1

= 29 + 20 cos 30° = 29 + 10 3

Applications of Distance Formula X

i.

Collinearity of three given points The three given points A, B and C are collinear i.e. lie on the same straight line, if any of the three points (say B) lie on the straight line joining the other two points. i.e. AB + BC = AC

Example 8. The points (1, 1), (− 2, 7) and (3, − 3) (a) form a right angled triangle (b) form an isosceles triangle (c) are collinear (d) None of the above

Example 11. The triangle joining the points P ( − 2, 7), Q ( − 4, − 1) and R (2, 6) is (a) an equilateral triangle (b) a right angled triangle (c) an isosceles triangle (d) a scalene triangle

Sol. (c) Let A(1 ,1), B (− 2, 7 ) and C (3, − 3) be the given

Sol. (b) Since, PQ = 68, PR = 17 and QR = 85

X

∴

points, then

i.e. right angled triangle.

AB = (− 2 − 1)2 + (7 − 1)2 =

9 + 36 = 3 5

iii.

BC = (3 + 2 ) + (− 3 − 7 ) 2

2

= 25 + 100 = 5 5 and

AC = (3 − 1)2 + (− 3 − 1)2 =

4 + 16 = 2 5

Clearly, BC = AB + AC Hence, A, B and C are collinear.

ii.

X

Types of triangles If A, B and C are vertices of triangle, then it would be (a) equilateral triangle, when AB = BC = CA. (b) isosceles triangle, when any two sides are equal. (c) right angled triangle, when sum of square of any two sides is equal to square of the third side.

Example 9. The points A (2a, 4a ), B (2a, 6a ) and C (2a + 3a, 5a ) (when a > 0) are vertices of (a) an obtuse angled triangle (b) an equilateral triangle (c) an isosceles obtuse angled triangle (d) a right angled triangle

Ø

●

●

= ( 3 a)2 + a2 = 2 a

●

3 a)2 + (4 a − 5 a)2

●

= ( 3 a) + a = 2 a 2

2

Since, AB = BC = CA, A, B and C form an equilateral triangle. X

Example 10. The triangle formed by P ( − 2, 3), Q (3, − 2) and R(0, 0) is (a) an equilateral (b) an isosceles (c) a right angled (d) None of these Sol. (b) PQ = 25 + 25 = 5 2 RQ = 9 + 4 = 13 PR = 4 + 9 = 13 RQ = PR Hence, ∆PQR is isosceles.

●

●

3 a − 2 a)2 + (5 a − 6 a)2

and CA = (2 a − 2 a −

Position of four points Let A, B , C and D be the four given points in a plane. By joining these points following figures are formed : (a) Parallelogram, if AB = DC , BC = AD, AC ≠ BD (b) Rectangle, if AB = CD, BC = DA, AC = BD (c) Rhombus, if AB = BC = CD = DA and AC ≠ BD (d) Square, if AB = BC = CD = DA and AC = BD Quadrilateral Diagonals

Sol. (b) AB = (2 a − 2 a)2 + (4 a − 6 a)2 = 2 a BC = (2 a +

PQ 2 + PR 2 = QR 2

X

Parallelogram

Not equal

Rectangle

Equal

Rhombus

Not equal

Square

Equal

11 Cartesian System of Rectangular Coordinates

X

Angles between diagonals π θ≠ 2 π θ≠ 2 π θ= 2 π θ= 2

Diagonals of square, rhombus, rectangle and parallelogram always bisect each other. Diagonals of rhombus and square bisect each other at right angle. Every parallelogram is not a rectangle but every rectangle is a parallelogram. Every rhombus is not a square but every square is a rhombus. Every parallelogram is not a rhombus but every rhombus is a parallelogram. Every rectangle is not a square but every square is a rectangle.

Example 12. The figure formed by four points (1, − 2), (3, 6), (5, 10) and (3, 2) is a (a) square (b) rectangle (c) rhombus (d) parallelogram Sol. (d) Let A ≡ (1, − 2 ), B ≡ (3, 6), C ≡ (5, 10) and Then,

D ≡ (3, 2 ) AB = (1 − 3)2 + (− 2 − 6)2 = 2 17 BC = (3 − 5)2 + (6 − 10)2 = 2 5 CD = (5 − 3)2 + (10 − 2 )2 = 2 17 AD = (1 − 3)2 + (−2 − 2 )2 = 2 5

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Objective Mathematics Vol. 1

11

AC = (1 − 5)2 + (− 2 − 10)2 = 4 10 and

●

BD = (3 − 3)2 + (6 − 2 )2 = 4

Parallelogram Show that diagonals AC and BD bisect each other.

Clearly, AB = CD, BC = AD and AC ≠ BD Hence, ABCD is a parallelogram. X

Example 13. If vertices of a quadrilateral are A (0, 0), B(3, 4), C (7, 7) and D(4, 3), then quadrilateral ABCD is (a) parallelogram (b) rectangle (c) square (d) rhombus Sol. (d) Obviously, AB = BC = CD = AD = 5 and also AC ≠ BD. Therefore, it is a rhombus.

●

(x1,y1)

ii.

m1

(x,y)

X

(x2,y2)

(x1,y1)

m1

Q (x2,y2)

m2

X

R (x, y)

●

●

●

●

594

If λ is positive, then it divides internally and if λ is negative, then it divides externally.  x + x 2 y1 + y 2  The mid-point of (x1 , y1) and (x 2 , y 2) is  1 , .  2 2 

The coordinates of the points of trisection P and Q of the line 

2x1 + x2 2y 1 + y2  ,  3 3 

Example 14. The coordinates of point which divides the line segment joining points A (0, 0) and B(9, 12) in the ratio 1 : 2, are (a) ( − 3, 4) (b) (3, 4) (c) (3, − 4) (d) None of these

Example 15. The ratio in which the line 3x + y − 9 = 0 divides the segment joining the points (1, 3) and (2, 7), is (a) 1 : 2 (b) 4 : 3 (c) 3 : 4 (d) None of these joining A (1, 3) and B (2, 7) in the ratio k : 1 at point C, then  2k + 1 7k + 3 the coordinates of C are  , .  k + 1 k +1 

Coordinates of any point which divide the line segment joining two points P (x1 , y1) and Q (x 2 , y 2) in the ratio λ :1is given by  x1 + λx 2 y1 + λy 2  ,   ,(λ ≠ − 1) λ +1   λ +1 Line formed by joining (x1 , y1) and (x 2 , y 2) is divided by y1 ã X-axis in the ratio − . y2 x1 ã Y-axis in the ratio − . x2 If ratio is positive, then the axis divides it internally and if ratio is negative, then the axis divides externally. Line Ax + By + C = 0 divides the line joining the points (x1 , y1) and (x 2 , y 2) in the ratio λ :1, then  Ax + By1 + C  λ =− 1   Ax 2 + By 2 + C 

Show that diagonals AC andBDare equal and bisect each other.

Sol. (c) Let the line 3 x + y − 9 = 0 divides the line segment

When m1 and m2 are of opposite sign, then the division is external. Ø

Rectangle

and B (9, 12) internally in the ratio 1 : 2 9 × 1+ 0 × 2 9 x= = =3 ∴ 3 3 12 × 1 + 2 × 0 12 and y= = =4 3 3 Hence, the coordinates of point P are (3, 4).

 m1 x 2 − m2 x1 m1 y2 − m2 y1  ,   (external division) m1 − m2   m1 − m2 P

Show that diagonals AC andBDare equal and bisect each other and adjacent sides AB andBC are equal.

Sol. (b) Let P divides the line segment joining points A (0, 0)

Q m2

Square

 x1 + 2x2 y 1 + 2y2  ,  respectively. 3   3

 m1 x 2 + m2 x1 m1 y2 + m2 y1  ,   (internal division) m1 + m2   m1 + m2 R

Show that diagonals AC and BD bisect each other and adjacent sides AB and BC are equal.

and 

Coordinates of a point which divides the line segment joining two points P ( x1 , y1 ) and Q ( x 2 , y2 ) in the ratio m1 : m2 are

P

Rhombus

segment joining A (x1 , y1) and B (x 2 , y 2) are 

Section Formulae

i.

Another conditions to prove that A , B , C and D are vertices of

Q C lies on line 3 x + y − 9 = 0, so it satisfies the equation of line.  2 k + 1  7 k + 3  3 3 ∴  +   −9=0 ⇒ k= 4  k + 1  k + 1  Hence, the required ratio is 3 : 4 (internally). Aliter ( Ax1 + By1 + C ) ( Ax2 + By2 + C ) (3 + 3 − 9) 3 = =− (6 + 7 − 9) 4

Required ratio = −

X

Example 16. The ratio in which the line segment joining the points (3, − 4) and ( − 5, 6) is divided by the X -axis, is (a) 2 : 3 (b) 3 : 2 (c) 6 : 4 (d) None of these Sol. (a) Let P be the point on X-axis, which divides the line segment joining points in the ratio λ : 1. − 5λ + 3 6λ − 4 Then, and y = x= λ+1 λ+1

Area of ∆ABC 1 = |[( x1 ( y2 − y3 ) + x 2 ( y3 − y1 ) + x 3 ( y1 − y2 )]| 2

Aliter

This can also be written as x1 1 ∆ = x2 2 x3

Required ratio = −

X

y1 − (− 4) 2 = = y2 6 3

Example 17. Prove that the points ( − 2, − 1), (1, 0), ( 4, 3) and (1, 2) are the vertices of a parallelogram. Is it a rectangle?

Ø If area of triangle is zero. Then, the points will be collinear and if

the points are collinear, then

Sol. Let the given points be A, B, C and D, respectively. Then, the coordinates of the mid-point of AC are  − 2 + 4 , − 1 + 3  = (1, 1)    2 2  and coordinates of the mid-point of BD are  1 + 1, 0 + 2  = (1, 1)    2 2 

x1 y1 1 Area of triangle = 0 or x 2 y 2 1 = 0 x3 y3 1 X

Thus, AC and BD have the same mid-point. Hence, ABCD is a parallelogram. Now, we shall see whether ABCD is a rectangle or not. We have, AC = {4 − (− 2 )}2 + {3 − (− 1)}2 = 2 13 BD = (1 − 1)2 + (0 − 2 )2 = 2

and

Clearly, AC ≠ BD So, ABCD is not a rectangle. X

y1 1 y2 1 y3 1

11 Cartesian System of Rectangular Coordinates

But as it is known for a point on X-axis, its y-coordinate = 0 6λ − 4 2 ⇒ =0 ⇒ λ= λ+1 3

Example 19. If the vertices of a triangle are (1, 2), ( 4, − 6) and (3, 5), then its area is 23 sq units (a) 2 25 sq units (b) 2 (c) 12 sq units (d) None of the above Sol. (b) Area of triangle, whose vertices are given,

Example 18. The points of trisection of line segment joining the points A(2, 1) and B (5, 3), are  5  7  7  5  (a)  3,  ,  , 4 (b)  4,  ,  3,   3  3   3 3   5  7  7  7 (d)  4,  ,  3,  (c)  3,  ,  4,   3  3   3  3 Sol. (c) Let P1 and P2 be the points, which trisect the line segment joining the points A(2, 1) and B(5, 3).  1 × 5 + 2 × 2 1 × 3 + 2 × 1  5  , P1( x, y) =   =  3,  1+ 2 1+ 2    3

 2 × 5 + 1 × 2 2 × 3 + 1 × 1  7  , P2 ( x, y) =   =  4,  1+ 2 1+ 2    3

1 ∆ = |[ x1( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )]| 2 1 = |[1(− 6 − 5) + 4(5 − 2 ) + 3(2 + 6)]| 2 25 sq units = 2 X

Example 20. The points (k , 2 − 2k ), (− k + 1, 2k ) and ( −4 − k , 6 − 2k ) are collinear for (a) all values of k 1 (b) k = −1 or 2 (c) k = −1 (d) no values of k Sol. (c) The given points are collinear, if area of ∆ = 0

Area of a Triangle Let A, B and C be the vertices of ∆ABC, whose coordinates are given by A ( x1 , y1 ), B ( x 2 , y2 ) and C ( x 3 , y3 ), then Y

C (x 3 , y 3 ) (x1, y1)

O Y'

L

On applying R 2 → R 2 − R1, R 3 → R 3 − R1, we get k 2 − 2k 1 − 2 k + 1 4k − 2 0 = 0 − 4 − 2k ⇒

M

N

X

k 2 − 2k 1 −k+1 2k 1 =0 − 4 − k 6 − 2k 1

B (x2, y2)

A

X'

⇒

4

0

− 2 k + 1 4k − 2 − 4 − 2k

4

=0

⇒ (1 − 2 k ) (k + 1) = 0 1 1 1 ⇒ k = − 1 or k = , neglecting as when k = , points 2 2 2 are same.

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Objective Mathematics Vol. 1

11 Area of a Quadrilateral

Sol. (b) Area of quadrilateral 1 = |[(1 × 4) − (3 × 1) + {3 × (− 2 )} − (5 × 4) 2 + {5 × (− 7 )} − {4 × (− 2 )} + (4 × 1) − {1 × (− 7 )}]| 41 41 sq units = − = 2 2

If ( x1 , y1 ), ( x 2 , y2 ), ( x 3 , y3 ) and ( x 4 , y4 ) are vertices of a quadrilateral, then its area will be given by Y C D

X

B A X'

O

L N

R M

X

Sol. (a) Area of the pentagon ABCDE

Y'

Area of quadrilateral = Area of trapezium ALND + Area of trapezium DNRC + Area of trapezium CRMB − Area of trapezium ALMB On substituting the value of area of trapezium 1 = (Sum of parallel sides) × Distance between them 2 ∴ Area of quadrilateral 1 = |[( x1 y2 − x 2 y1 ) + ( x 2 y3 − x 3 y2 ) 2 + ( x 3 y4 − x 4 y3 ) + ( x 4 y1 − x1 y4 )]| Ø

●

ã

●

X

596

∴ ⇒ ⇒ ⇒ X

Area of polygon can be calculated as ã Row Method When vertices of polygon are given, then area of polygon will be

x x x 1  x1 x 2 + 2 3 + ... + n  y 2 y3 yn 2  y1 y 2

Example 22. The values of t, if the area of the 45 pentagon ABCDE is sq units, where 2 A = (1, 3), B = ( − 2, 5), C = ( − 3, − 1), D (0, − 2) and E = (2, t ) and here the points are given in order, are (a) −1, 89 (b) − 2, 89 (c) 1, 89 (d) 2, 89

x1   y1 

Stair Method Again, when vertices of polygon are given say (x1, y1), (x2 , y2) , . . . , (xn , yn). Then, its area is x1 y1 x2 y2 y3 1 x3 . ... 2 .. xn yn x1 y1

1 = |{(x1 y 2 + x 2 y3 + ... + x n y1) − (y1x 2 + y 2x3 + ... + y n x1)}| 2 If the two opposite vertices of a square are A (x1 , y1) and C (x 2 , y 2), then its area 1 1 = AC 2 = [(x 2 − x1)2 + (y 2 − y1)2] 2 2

Example 21. The area of quadrilateral, whose vertices are (1, 1), (3, 4), (5, − 2) and ( 4, − 7), is (a) 41 sq units 41 (b) sq units 2 31 (c) sq units 2 (d) 7 sq units

−3 −2 5  1 3 + +  0 − 3 −1 1− 2 5 =  2 0 −2 + +  2 t  1 [(5 + 6) + (2 + 15) + (6 − 0) 45 = 2 2 + (0 + 4) + (6 − t )]

−1  − 2   2 t  1 3 

45 1 1 = (11 + 17 + 6 + 4 + 6 − t ) = |44 − t| 2 2 2 44 − t = ± 45 t = 44 m 45 = − 1, 89

Example 23. The area of the pentagon, whose vertices are A (1, 1), B ( 7, 21), C ( 7, − 3), D (12, 2) and E (0, − 3), is 137 137 (a) sq units (b) sq units 2 3 137 sq units (d) None of these (c) 4 x1

y1

x2

y2

1 7

1 21

−3 12 2 2 0 −3

x y 7 Sol. (a) Required area = 1 3 3 = 1 2 x4 x5

y4 y5

x1

y1 1 1 1 = |[(21 − 21 + 14 − 36 + 0) − (7 + 147 − 36 + 0 − 3)]| 2 137 sq units = 2 X

Example 24. If the coordinates of two opposite vertices of a square are (6, − 3) and ( − 2, 3), then area of square is (a) 50 sq units (b) 25 sq units (c) 35 sq units (d) None of these Sol. (a) Area of square = 1 d 2

2 1 {(6 + 2 )2 + (− 3 − 3)2 } 2 1 = (64 + 36) = 50 sq units 2

=

i.e. ( x − x1 ) 2 + ( y − y1 ) 2 = ( x − x 2 ) 2 + ( y − y2 ) 2 = ( x − x 3 ) 2 + ( y − y3 ) 2

In this section, we will discuss problems on finding centroid, circumcentre, incentre and orthocentre of a triangle.

i.

The coordinates of the circumcentre are also given by  x1 sin 2 A + x 2 sin 2B + x 3 sin 2 C ,  sin 2 A + sin 2 B + sin 2 C 

Centroid The centroid of a triangle is the point of intersection of its medians. It divides the medians in the ratio 2 : 1. If A ( x1 , y1 ), B ( x 2 , y2 ) and C ( x 3 , y3 ) are the vertices of a ∆ABC, then the coordinates of its centroid G are

y1 sin 2 A + y2 sin 2B + y3 sin 2C   sin 2 A + sin 2B + sin 2 C 

iv.

A (x1, y1) c F

E

b

G

(x 2 , y2 ) B

C (x3, y3)

D a

 x1 + x 2 + x 3 y1 + y2 + y3  ,     3 3

ii.

iii.

Incentre The point of intersection of the internal bisectors of the angles of a triangle is called its incentre. If A ( x1 , y1 ), B ( x 2 , y2 ) and C ( x 3 , y3 ) are the vertices of a ∆ABC such that BC = a, CA = b and AB = c, then the coordinates of the incentre are  ax1 + bx 2 + cx 3 ay1 + by2 + cy3  , .  a +b+c a +b+c   Circumcentre The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of its sides. A ( x1 , y 1 )

O B (x2, y2)

C (x 3 , y 3 )

It is the centre of the circle passing through the vertices of a triangle and so it is equidistant from the vertices of the triangle. Thus, if O is the circumcentre of a ∆ABC, then OA = OB = OC Let A ( x1 , y1 ), B ( x 2 , y2 ) and C ( x 3 , y3 ) be the vertices of ∆ABC and let O ( x, y) be its circumcentre. Then, the coordinates of O are obtained by solving

Ø

●

●

●

Orthocentre The orthocentre of a triangle is the point of intersection of its altitudes. In order to find the coordinates of the orthocentre of a triangle, we first find the equations of its altitudes and then we find the coordinates of the point of intersection of any two of them. If A ( x1 , y1 ), B ( x 2 , y2 ) and C ( x 3 , y3 ) are the vertices of a ∆ABC, then the coordinates of its orthocentre are  x1 tan A + x 2 tan B + x 3 tan C  ,  tan A + tan B + tan C    y1 tan A + y2 tan B + y3 tan C    tan A + tan B + tan C  

11 Cartesian System of Rectangular Coordinates

OA 2 = OB 2 = OC 2

Some Standard Points of a Triangle

The orthocentre of a right angled triangle is at the vertex forming the right angle. The orthocentre O′, circumcentre O and centroid G of a triangle are collinear and G divides O′ O in the ratio 2 : 1 i.e. O ′ G : OG = 2 : 1. The circumcentre of a right angled triangle is the mid-point of its hypotenuse.

v.

Nine point centre Let ABC be a triangle such that AD, BE and CF are its altitudes drawn from A, B and C on the opposite sides; H , I , J are the mid-points of the sides BC , CA and AB, respectively and K , L, M are the mid-points of the line segments joining the orthocentre O′ to the angular points A, B , C . These nine points ( D, E , F , H , I , J , K , L, M ) are concyclic and the circle passing through these nine points is called the nine point circle. Its centre is known as the nine point centre. The nine point centre of a triangle is collinear with the circumcentre O and the orthocentre O′ and bisects the segment joining them. The radius of nine point circle of a triangle is half the radius of the circumcircle.

Ø The orthocentre, nine point centre, centroid and circumcentre are

collinear.

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11 Objective Mathematics Vol. 1

vi.

Excentre of a triangle Let A ( x1 , y1 ), B ( x 2 , y2 ) and C ( x 3 , y3 ) be the vertices of a ∆ABC. Let a, b and c be the lengths of the sides BC , CA and AB or a, b and c be the lengths of the sides opposite to ∠A, ∠B and ∠C respectively, then the circle which touches the side BC and two sides AB and AC produced is called the escribed circle opposite to the ∠A. The bisector of the external angles B and C meet at E1 which is the centre of the escribed circle such that  − ax1 + bx 2 + cx 3 − ay1 + by2 + cy3  , E1   −a +b+c −a +b+c  

X

Sol. (c) Let P( x, y) be the coordinates of the centre of circumcircle. ∴ ∴ and

Similarly,  ax − bx 2 + cx 3 ay1 − by2 + cy3  E2  1 ,  a −b+c a −b+c    ax + bx 2 − cx 3 ay1 + by2 − cy3  and E 3  1 ,  a +b−c a +b−c  

E2

A

E3 C

B

E1

Example 27. If the vertices A, B and C of a triangle are (2, 1), (5, 2) and (3, 4), respectively. Then, the circumcentre is  13 − 9  (a)  ,  4 4   − 13 9  (b)  ,   4 4  13 9  (c)  ,   4 4  13 9  (d)  ,   2 4

PA 2 = PB2 and PB2 = PC 2 3 x + y = 12 x− y=1

…(i) …(ii) 13 9 On solving Eqs. (i) and (ii), we get the centre at  ,  .  4 4

X

Example 28. The orthocentre of the triangle formed by the line 2x 2 − xy − 6 y 2 = 0 and the line x + y + 1 = 0, is 4  4 (b) ( 4, 4) (a)  − , −   3 3 (c) ( − 4, − 4) (d) None of these Sol. (a) We have, 2 x2 − xy − 6 y2 = 0 ⇒ ⇒

(2 x + 3 y)( x − 2 y) = 0 2 x + 3 y = 0 and x − 2 y = 0 Y

X

Example 25. Centroid of the triangle, whose vertices are (0, 0), (2, 5) and (7, 4), is (a) (3, 3) (b) (2, 3) (c) (3, 2) (d) (2, 2) Sol. (a) Centroid of the triangle 0 + 2 + 7 0 + 5 + 4 =  ,  = (3, 3)   3 3

X

Example 26. Incentre of triangle, whose vertices are A ( − 36, 7), B (20, 7) and C (0, − 8), is (a) (0, − 1) (b) (0, 0) (c) ( −1, 0) (d) None of the above Sol. (c)Qa = BC = (20 − 0)2 + (7 + 8)2 = 25 b = CA = (− 36 − 0)2 + (7 + 8)2 = 39 and

598

c = AB = (36 + 20)2 + (7 − 7 )2 = 56  25(− 36) + 39(20) + 56(0),    25 + 39 + 56  = (− 1, 0) I=  25(7 ) + 39(7 ) + 56(− 8)    25 + 39 + 56  

B (–3, 2) 2x + 3y = 0 X′

X x+y+1=0 x – 2y = 0 A –2,–1 3 3 Y′

Thus, the equations of sides of the triangle are 2 x + 3 y = 0, x − 2 y = 0 and x + y + 1 = 0. The equation of the altitude through the vertex O(0, 0) is y − 0 = 1( x − 0) ...(i) ⇒ y= x The equation of altitude through the vertex B(− 3, 2 ) is y − 2 = − 2( x + 3) ...(ii) ⇒ 2x + y + 4 = 0 On solving Eqs. (i) and (ii), we get 4 4 and y = − x=− 3 3 Hence, the coordinates of the orthocentre are  − 4, − 4.    3 3

It is the path or curve traced by a moving point satisfying the given condition.

i.

X

Equation of the locus of a point The equation of the locus of a point is the algebraic relation which is satisfied by the coordinates of every point on the locus of the point. Steps taken to find the equation of locus of a point (a) Assumes the coordinates of the point say ( h, k ), whose locus is to be find. (b) Write the given condition involving ( h, k ). (c) Eliminate the variable (s), if any. (d) Replace h → x and k → y. The equation so obtained is the locus of the point which moves under some definite conditions.

Transformation of Axes i.

Example 29. The sum of squares of the distances of a moving point from two fixed points ( a, 0) and ( − a, 0) is equal to a constant. Then, the equation of its locus is (b) x 2 + y 2 = C − a 2 (a) x 2 − y 2 = C + a 2 (c) x 2 + y 2 = C 2 + a 2 (d) x 2 − y 2 = C − a 2

To alter the origin of coordinates without altering the direction of the axes Let origin O (0, 0) be shifted to a point ( a, b) by moving the X -axis and Y -axis parallel to themselves. If the coordinates of point P with reference to old axis are ( x1 , y1 ), then coordinates of this point with respect to new axis will be ( x1 − a, y1 − b). Y Y′ P(x1, y1)

Sol. (b) Let P (h, k ) be any position of the moving point and

(a, b)O′

let A (a, 0) and B (− a, 0) be the given points. Then, [given] PA 2 + PB2 = λ ⇒ ⇒

(h − a)2 + (k − 0)2 + (h + a)2 + (k − 0)2 = λ 2

X

Example 30. A rod of length l slides with its ends on two perpendicular lines. Then, the locus of its mid-point is l2 l2 (b) x 2 + y 2 = (a) x 2 + y 2 = 4 2 2 l (c) x 2 − y 2 = (d) None of these 4 Sol. (a) Let the two perpendicular lines be the coordinate axes. Let AB be rod of length l and the coordinates of A and B be (a, 0) and (0, b), respectively. Y B (0, b) P (h, k)

X′

O

h − 2 ah + a2 + k 2 + h2 + 2 ah + a2 + k 2 = λ λ ∴ Locus of (h, k) is x2 + y2 = C − a2 , where C = . 2 X

11 Cartesian System of Rectangular Coordinates

Let P (h, k ) be the mid-point of rod AB in one of the infinite position it attains, then a+ 0 0+ b and k = h= 2 2 a b and k = ...(i) ⇒ h= 2 2 From ∆OAB, we have AB2 = OA 2 + OB2 2 ⇒ a + b2 = l 2 2 ⇒ (2 h) + (2 k )2 = l 2 ⇒ 4h2 + 4k 2 = l 2 l2 h2 + k 2 = ⇒ 4 l2 So, the equation of locus is x2 + y2 = . 4

Locus

X

Example 31. If the axes are transformed from origin to the point (3, 4), then new coordinates of (7, 8) are (a) (4, 3) (b) (4, 2) (c) (4, 4) (d) None of these Sol. (c) We know that, if the origin is shifted to (h, k ), then new coordinates of ( x, y) becomes ( x − h, y − k ). Therefore, the new coordinates of (7, 8) with respect to new origin are (4, 4).

X

Example 32. If the axes are shifted to the point (1, − 2) without rotation. What do the following equations become? (i) 2x 2 + y 2 − 4x + 4 y = 0 (ii) y 2 − 4x + 4 y + 8 = 0 Sol. (i) Substituting x = X + 1, y = Y + (− 2 ) = Y − 2 in the equation 2 x2 + y2 − 4 x + 4 y = 0, we get

X′

O

(a, 0)A

X

2( X + 1)2 + (Y − 2 )2 − 4( X + 1) + 4(Y − 2 ) = 0 ⇒ 2 X 2 + Y 2 = 6, where ( X, Y ) are new coordinates

Y′

and ( x, y) are old coordinates.

599

11

(ii) Substituting x = X + 1, y = Y − 2 in the equation y2 − 4 x + 4 y + 8 = 0, we get

Sol. (d) X ′ = 2 cos(− 30° ) + 1 sin(− 30° ) 2 3 −1 2 Y ′ = − 2 sin 30° + cos 30° −2 + 3 = 2 =

Objective Mathematics Vol. 1

(Y − 2 )2 − 4 ( X + 1) + 4 (Y − 2 ) + 8 = 0 ⇒

ii.

Y 2 = 4X

To change the direction of the axes of coordinates without changing origin Let OX and OY be the old axes and OX ′ and OY ′ be the new axes obtained by rotating the old OX and OY through an angle θ, then the coordinates of P ( x, y) with respect to new coordinate axes will be given by x = ON − NL = x ′ cos θ − y′ sin θ y = PQ + QL = y′ cos θ + x ′ sin θ Y (x', y')

X

Example 34. If the axes be turned through an angle tan − 1 2. What does the equation 4xy − 3x 2 = a 2 become? (a) X 2 − 4Y 2 = a 2 (b) X 2 + 4Y 2 = a 2 (c) X 2 + 4Y 2 = − a 2 (d) None of the above Sol. (a) Here, tan θ = 2 1 5 2 and sin θ = 5 For x and y, we have x = X cos θ − Y sin θ X − 2Y = 5

θ

y′

Y

′

X′ Q

y

M

y = X sin θ + Y cos θ =

x′

and

θ O

cos θ =

So,

P (x, y)

2X + Y 5

The equation 4 xy − 3 x2 = a2 reduces to

θ x

L

X

N

2

X − 2Y  4( X − 2 Y ) (2 X + Y ) 2 ⋅ − 3   =a  5 5 5  ⇒ 4(2 X 2 − 2 Y 2 − 3 XY ) − 3( X 2 − 4 XY + 4Y 2 ) = 5 a2

Thus, if the axes are rotated through an angle θ. Then, the coordinates of a point with respect to new axis will be x ′ = x cos θ + y sin θ and y′ = − x sin θ + y cos θ The above relation between ( x, y) and ( x ′ , y′ ) can also be obtained from the table. x↓

y↓

x′→

cosθ

sinθ

y′→

− sinθ

cosθ

⇒ ⇒

iii.

5 X 2 − 20Y 2 = 5 a2 X 2 − 4Y 2 = a2

To change the direction of the axes of coordinates with changing origin If P ( x, y) and the axes are shifted parallel to the original axes so that new origin is O′ (α, β) and then the axes are rotated about the new origin (α, β) by angle φ in the anti-clockwise direction, then the coordinates of P will be given by Y

Y

′ X′

●

X

600

x and y are old coordinates, x′ and y′ are new coordinates. The axes rotation in anti-clockwise is positive and clockwise rotation of axes is negative.

Example 33. Keeping the origin constant axes are rotated at an angle 30° in negative direction, then coordinates of (2, 1) with respect to new axes are  2 3 +1 2 3   2 3 +1 − 2 + 3    (a)  (b)  , , 2  2  2  2  2 + 3 3  (c)  , 2   2

(d) None of these

y)

●

P( x,

Ø

φ O′

O

X

x = α + x ′ cos φ − y′ sin φ y = β + x ′ sin φ + y′ cos φ where, x and y are old coordinates, x ′ and y′ are new coordinates.

bx − ay + d = 0 are taken as new axes such that new coordinates of P are (x ′ , y ′ ) , then Y

bx –a y+ d= 0

X′

ax

+

by

+

c

=

0

1 2 C 1C 2C 3

O′ X

O

X

bx − ay + d b +a 2

2

and y ′ =

ax + by + c a +b 2

2

Example 35. What does the equation 2x 2 + 4xy − 5 y 2 + 20x − 22 y − 14 = 0 become when referred to rectangular axes through the point ( − 2, − 3), the new axes being inclined at an angle of 45° with the old axes? Sol. Let O′ be (− 2, − 3). Since, the axes are rotated about O′ by an angle 45° in anti-clockwise direction, let ( x′, y′ ) be the new coordinates with respect to new axes and ( x, y) be the coordinates with respect to old axes. Then, we have x = − 2 + x′cos 45° − y′sin 45° x′ − y′  = − 2 +    2  x′ + y′  y = − 3 + x′sin 45° + y′cos 45° = − 3 +    2 

b1 b2

c1 c2

a3

b3

c3

2

5. Area of parallelogram (i) Whose sides are a and b and angle between them is θ, is given by area of parallelogram ABCD = ab sin θ

2

x′ − y′    x′ − y′    2 − 2 +    + 4 − 2 +      2  2    − 3 +  x′ + y′       2   2

x′ + y′    x′ − y′    − 5 − 3 +    + 20 − 2 +     2  2    x′ + y′   − 22 − 3 +    − 14 = 0  2   x′2 − 14 x′ y′ − 7 y′2 − 2 = 0

Hence, new equation of curve is x2 − 14 xy − 7 y2 − 2 = 0

Points to be Remembered 1. Reflection (or image) of a point Let ( x, y) be any point, then its image with respect to (i) X -axis is ( x, − y) (ii) Y -axis is ( − x, y) (iii) origin is ( − x, − y) (iv) line y = x is ( y, x )

a

D

A

11

C

b

b

θ

B

a

(ii) Whose length of perpendicular from one vertex to the opposite sides are P1 and P2 and angle between sides is θ, is given by PP area of parallelogram ABCD = 1 2 sin θ D

C P1 P2

The new equation is

⇒

a1 a2

where, C1 , C 2 and C 3 are the cofactors of c1 , c2 and c3 in the determinant.

Y′

x′ =

2. A triangle is isosceles, if any two of its median are equal. 3. Triangle having integral coordinates can never be equilateral. 4. If a r x + br y + cr = 0 ( r = 1, 2, 3) are the sides of a triangle, then the area of the triangle is given by

Cartesian System of Rectangular Coordinates

Ø If P (x , y) and two mutually perpendicular lines ax + by + c = 0 and

A

θ

6. A triangle having vertices

B

( at12 , 2at1 ),

( at 22 , 2at 2 )

and ( at 32 , 2at 3 ), then area of triangle = a 2 [( t1 − t 2 ) ( t 2 − t 3 ) ( t 3 − t1 )] 7. Area of triangle formed by coordinate axes and c2 . the line ax + by + c is 2 | ab | 8. Area of rhombus formed by | ax | + | by | + | c | = 0 is 2c 2 . | ab | 9. Three points ( x1 , y1 ), ( x 2 , y2 ) and ( x 3 , y3 ) are y − y1 y3 − y2 . collinear, if 2 = x 2 − x1 x 3 − x 2 10. To remove the term of xy in the equation ax 2 + 2hxy + by 2 = 0. The angle θ through which the axes must be turned, is  2h  1 θ = tan − 1   2  a − b

601

Objective Mathematics Vol. 1

11

Work Book Exercise 1 If the line segment joining (2, 3) and (− 1, 2 ) is divided internally in the ratio 3 : 4 by the line x + 2 y = k, then k is a c

41 7 36 7

5 7 31 d 7

b

2 The polar coordinates of the vertices of a triangle  π  π are (0, 0 ),  3,  and  3,  . Then, the triangle is  2  2

a right angled c equilateral

b isosceles d None of these

3 If the points (− 2, 0 ),  − 1,

1  and (cos θ, sin θ ) 3 are collinear, then the number of values of θ ∈[0, 2 π ] is 

a 0 c 2

b 1 d infinite

4 The incentre of the triangle formed by the axes and the line

x y + = 1is a b

a b a  ,  2 2    ab ab b  ,  a + b + ab a + b + ab   a b c  ,   3 3   ab ab  d  ,  a + b + a2 + b 2 a + b + a2 + b 2   

5 The diagonals of a parallelogram PQRS are along the lines x + 3 y = 4 and 6 x − 2 y = 7. Then, PQRS must be a a b c d

rectangle square cyclic quadrilateral rhombus

6 The coordinates of three consecutive vertices of a parallelogram are (1, 3), (− 1, 2 ) and (2, 5). The coordinates of the fourth vertex are a b c d

602

(6, 4) (4, 6) ( − 2, 0) None of the above

7 In a ∆ABC, the coordinates of B are (0, 0),

π and the middle point of BC 3 has the coordinates (2, 0). The centroid of the triangle is AB = 2, ∠ABC =

 1 3 a  ,  2 2  4+ 3 c  , 3 

5 1  b  ,   3 3 1  3

d None of these

8 The points (α, β), (δ, γ ), (α, δ ) and (β, γ ) taken in order, where α , β, γ, δ are different real numbers, are a collinear c vertices of a rhombus

b vertices of a square d concyclic

9 The limiting position of the point of intersection of the lines 3 x + 4 y = 1 and (1 + c ) x + 3 c 2 y = 2 as c tends to 1, is a b c d

( − 5, 4) ( 5, − 4) ( 4, − 5) None of the above

10 A point moves in the xy-plane such that the sum of its distances from two mutually perpendicular lines is always equal to 3. The area enclosed by the locus of the point is a 18 sq units c 7 sq units

9 sq units 2 d None of these b

11. Three vertices of a quadrilateral in order are (6, 1), (7, 2) and (− 1, 0 ). If the area of the quadrilateral is 4 sq units, then the locus of fourth vertex has the equation a x − 7y = 1 b x − 7 y + 15 = 0 c ( x − 7 y)2 + 14 ( x − 7 y) − 15 = 0 d None of the above

12. If a vertex of an equilateral triangle is the origin and the side opposite to it has the equation x + y = 1, then the orthocentre of triangle is 1 1 a  ,   3 3 2 2 c  ,   3 3

 2 2 b  ,  3   3 d None of these

WorkedOut Examples Type 1. Only One Correct Option Ex 1. The

cartesian form of the equation θ 1/ 2 r = a 1/ 2 cos is 2 (a) ( 2x 2 + 2 y 2 − ax ) 2 = a 2 ( x 2 + y 2 ) (b) ( x 2 + y 2 − ax ) 2 = 2a 2 ( x 2 − y 2 ) (c) ( x 2 + y 2 − 2ax ) 2 = a 2 ( x 2 − y 2 ) 2 (d) None of the above

θ …(i) 2 On squaring Eq. (i) both sides, we get θ a r = a cos2 = (1 + cos θ ) 2 2 Now, by using the relation between cartesian and polar coordinates, the equation becomes 2r2 = ar + ar cosθ

Sol. Given, r1/ 2 = a1/ 2 cos

2(x 2 + y2 ) = a x 2 + y2 + ax

i.e.

⇒

= a2 + b2 + c2 + d 2 − 2 a2 + b2 c2 + d 2 cos θ ⇒ 2(ac + bd ) = 2 a2 + b2 c2 + d 2 cos θ ac + bd cos θ = ∴ a2 + b2 c2 + d 2 Hence, (a) is the correct answer.

Ex 3. If the point P (x, y) is equidistant from the points A ( a + b, b − a ) and B ( a − b, a + b), then (a) ax = by (b) bx = ay and P can be ( a, b ) (c) x 2 − y 2 = 2( ax + by ) (d) None of the above

Sol. We have, PA = PB ⇒ (PA )2 = (PB )2 ⇒ [ x − (a + b)]2 + [ y − (b − a)]2 = [ x − (a − b)]2 + [ y − (a + b)]2

i.e. (2x 2 + 2 y2 − ax )2 = a2 (x 2 + y2 )

⇒ [(x − a) − b ]2 + [( y − b) + a ]2

Hence, (a) is the correct answer.

Ex 2. If the segments joining the points A (a, b) and B ( c, d ) subtend an angle θ at the origin, then (a) cos θ = (b) sin θ = (c) cos θ =

= [(x − a) + b ]2 + [( y − b) − a ]2 ⇒ [(x − a) + b ]2 − [(x − a) − b ]2 = [( y − b) + a ]2 − [( y − b) − a ]2

ac + bd

⇒ 4 b(x − a) = 4 a( y − b) ⇒ bx = ay

( a 2 + b 2 )( c 2 + d 2 ) ac + bd

Ex 4. If t1 , t 2 and t 3 are distinct and the points ( t1 , 2at1 + at13 ), ( t 2 , 2at 2 + at 23 ), ( t 3 , 2at 3 + at 33 ) are collinear, then

( a 2 + b 2 )( c 2 + d 2 )

Sol. Let O be the origin. Then, OA 2 = a2 + b2, OB 2 = c2 + d 2 AB 2 = (c − a)2 + (d − b)2 Y

(a) t 1 t 2 t 3 = 1 (b) t 1 + t 2 + t 3 = t 1 t 2 t 3 (c) t 1 + t 2 + t 3 = 0 (d) t 1 + t 2 + t 3 = − 1 Sol. The given points are collinear, if

B (c, d)

t1 2at1 + at13 1 t2 2at2 + at23 1 = 0 t3 2at3 + at33 1

A (a, b) θ X′

X

O

⇒ Y′

Using cosine formula in ∆OAB, we have AB 2 = OA 2 + OB 2 − 2 ⋅ OA ⋅ OB cos θ ⇒

…(i)

Also, P (a, b) satisfies Eq. (i), so that P can be (a, b). Hence, (b) is the correct answer.

( a 2 + b 2 )( c 2 + d 2 ) ac − bd

(d) None of the above

and

c2 + a2 − 2ac + d 2 + b2 − 2bd

(c − a)2 + (d − b)2 = a2 + b2 + c2 + d 2 − 2 a2 + b2 c2 + d 2 cos θ

t1 2t1 + t13 1 a t2 2t2 + t23 1 = 0 t3 2t3 + t33 1

On applying R2 → R2 − R1, R3 → R3 − R1, we get t1 2t1 + t13 1 3 3 t2 − t1 2(t2 − t1 ) + (t2 − t1 ) 0 = 0 t3 − t1 2(t3 − t1 ) + (t33 − t13 ) 0

603

Objective Mathematics Vol. 1

11

t1 2t1 + t13 1 2 2 ⇒ (t2 − t1 )(t3 − t1 ) 1 2 + t2 + t1 + t2t1 0 = 0 1

Ex 7. The area of the triangle with vertices [( a + 1)( a + 2), ( a + 2)] [( a + 2)( a + 3), ( a + 3)] and [( a + 3)( a + 4), ( a + 4)] is

2 + t32 + t12 + t3t1 0

⇒

(t2 − t1 )(t3 − t1 )(t3 − t2 )(t3 + t2 + t1 ) = 0 [Qt1 ≠ t2 ≠ t3] ⇒ t1 + t2 + t3 = 0 Hence, (c) is the correct answer.

(a) independent of a (b) dependent on a (c) Cannot be discussed (d) None of these Sol. Area of the given triangle is given by ∆=

Ex 5. If the vertices of a triangle have integral coordinates, then the triangle cannot be (a) equilateral (c) scalene

(b) isosceles (d) None of these

Sol. Let A = (x1 , y1 ), B = (x2 , y2 ), and C = (x3 , y3 ) be the vertices of a triangle and x1 , x2 , x3 , y1 , y2 , y3 be integers. ∴BC 2 = (x2 − x3 )2 + ( y2 − y3 )2, it is a positive integer. If the triangle is equilateral, then AB = BC = CA = a [say] and ∠A = ∠B = ∠C = 60°  1  1 ∴Area of the triangle =   bc sin A =   a2 sin 60°  2  2  a2   3   3  =     =   a2  2 2   4  [Qa2 is a positive integer]

which is irrational.

Now, area of the triangle in terms of the coordinates 1 = | x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )| 2 which is a rational number.

1 2

(a + 1)(a + 2) a + 2 1 (a + 2)(a + 3) a + 3 1 (a + 3)(a + 4 ) a + 4 1

On applying R3 → R3 − R2 and R2 → R2 − R1, we get (a + 1)(a + 2) a + 2 1 1 ∆= 2(a + 2) 1 0 = | −1| =1 2 2(a + 3) 1 0 which is independent of a. Hence, (a) is the correct answer.

Ex 8. If the coordinates of mid-points of the sides of a triangle are (1, 2), (0, −1) and (2, −1). Then, the coordinates of its vertices are (a) (1, − 4), (3, 2), ( − 1, 2) (b) (1, 4), (3, − 2), (1, 2) (c) (1, 4), (3, 2), (1, 2) (d) None of the above

Sol. Let A (x1 , y1 ), B (x2 , y2 ) and C (x3 , y3 ) be the vertices of ∆ ABC. Let D (1, 2), E (0, − 1) and F (2, − 1) be the mid-points of BC, CA and AB, respectively. A (x1, y1)

This contradicts that the area is an irrational number, if the triangle is equilateral. Hence, (a) is the correct answer. F (2,–1)

Ex 6. If the coordinates of two points A and B are (3, 4) and (5, − 2) respectively. Then, the coordinates of any point P, if PA = PB and area of ∆PAB =10, are (a) (7, 5), (1, 0) (c) (7, 2), (− 1, 0)

(b) (7, 2), (1, 0) (d) None of these

∴

Sol. Let the coordinates of P be (x , y). Then,

⇒

PA = PB ⇒ PA 2 = PB 2 ⇒

(x − 3) + ( y − 4 ) = (x − 5) + ( y + 2) 2

2

2

⇒ x − 3y −1= 0 Now, area of ∆PAB = 10 x y 1 1 3 4 1 = ± 10 ⇒ 2 5 −2 1

2

...(i)

⇒ 6x + 2 y − 26 = ± 20 ⇒ 6x + 2 y − 46 = 0 or 6x + 2 y − 6 = 0 ...(ii) ⇒ 3x + y − 23 = 0 or 3x + y − 3 = 0 On solving x − 3 y − 1 = 0 and 3x + y − 23 = 0, we get x = 7 and y = 2 On solving x − 3 y − 1 = 0 and 3x + y − 3 = 0, we get x = 1 and y = 0

604

Thus, the coordinates of P are (7, 2) or (1, 0). Hence, (b) is the correct answer.

B (x2, y2)

E (0,–1)

D (1, 2)

x2 + x3 = 1 and 2 x2 + x3 = 2 and

Similarly, E and F are the respectively. x1 + x3 = 0 and 2 ⇒ x1 + x3 = 0 and x1 + x2 and = 2 and 2 ⇒ x1 + x2 = 4 and

C (x 3 , y 3 )

y2 + y3 =2 2 y2 + y3 = 4

...(i)

mid-points of CA and AB,

y1 + y3 = −1 2 …(ii) y1 + y3 = − 2 y1 + y2 = −1 2 …(iii) y1 + y2 = − 2 From Eqs. (i), (ii) and (iii), we get (x2 + x3 ) + (x1 + x3 ) + (x1 + x2 ) = 2 + 0 + 4 and ( y2 + y3 ) + ( y1 + y3 ) + ( y1 + y2 ) = 4 − 2 − 2 ...(iv) ⇒ x1 + x2 + x3 = 3 and y1 + y2 + y3 = 0 From Eqs. (i) and (iv), we get x1 + 2 = 3 and y1 + 4 = 0 ∴ x1 = 1 and y1 = − 4

Ex 10. A line L intersects the three sides BC, CA and AB of a ∆ABC at P, Q and R respectively. BP CQ AR Then, is equal to ⋅ ⋅ PC QA RB

So, the coordinates of B are (3, 2). From Eqs. (iii) and (iv), we get x3 + 4 = 3 and y3 − 2 = 0 ∴ x3 = − 1 and y3 = 2 So, coordinates of C are (− 1, 2). Hence, the vertices of the ∆ABC are A(1, − 4 ), B(3, 2) and C (− 1, 2). Hence, (a) is the correct answer.

Ex 9. If ∆1 is the area of the triangle with vertices (0, 0), ( a tan α, b cot α ), ( a sin α, b cos α ), ∆ 2 is the area of the triangle with vertices ( a, b), ( a sec 2 α, b cosec 2α ), ( a + a sin 2 α, b + b cos 2 α ) and ∆ 3 is the area of the triangle with vertices (0, 0), ( a tan α, − b cot α ), ( a sin α, b cos α ). Then, (a) ∆ 1 , ∆ 2 and ∆ 3 are in GP (b) ∆ 1 , ∆ 2 and ∆ 3 are not in GP (c) Cannot be discussed (d) None of the above

and ∆ 2 =

0 1 a tan α 2 a sin α 1 2

a + a sin α 2

A (x1, y1) R

P

1 1

b b cosec2α

⇒

b + b cos α 1

1 = ab | sin 2 α − cos2 α 2 0 0 1 ∆ 3 = a tan α − b cot α 2 a sin α b cosα

| 1 1 1

1 ab|sin α + cosα | 2 1 Clearly, ∆ 1∆ 3 = ab∆ 2. 2 Now, ∆ 1 , ∆ 2 and ∆ 3 are in GP, if 1 1 ∆ 1∆ 3 = ∆22 ⇒ ab∆ 2 = ∆22 ⇒ ∆ 2 = ab 2 2 1 1 2 2 ab(sin α − cos α ) = ab ⇒ 2 2 ⇒ sin 2 α − cos2 α = 1 =

π Then, α = (2m + 1) , m ∈ I. But for this value of α, the 2 vertices of the given triangles are not defined. ∆ 1, ∆ 2 and ∆ 3 cannot be in GP for any value of α. Hence, (b) is the correct answer.

C (x3, y3)

Also, as P lies on L, we have  λy + y2   λ x + x2  l 3  + n=0  + m 3  λ+1   λ+1 

2

On applying C 1→ C 1 − aC 3 and C 2→ C 2 − bC 3, we get 0 0 1 1 ∆ 2 = ab tan 2 α cot 2 α 1 2 sin 2 α cos2 α 1

and

∆ABC and let lx + my + n = 0 be the equation of line, if P divides BC in the ratio λ : 1, then the coordinates of P are  λx3 + x2 λy3 + y2  ,   λ+1   λ+1

B (x2, y2)

0 1 1 b cot α 1 = ab|sin α − cosα | 2 b cosα 1

a a sec2 α

Sol. Let A (x1 , y1 ), B (x2 , y2 ) and C (x3 , y3 ) be the vertices of

Q

Sol. We have, ∆1 =

(a) 1 (b) 0 (c) − 1 (d) None of the above

11 Cartesian System of Rectangular Coordinates

So, the coordinates of A are (1, − 4). From Eqs. (ii) and (iv), we get x2 + 0 = 3 and y2 − 2 = 0 ∴ x2 = 3 and y2 = 2

−

lx2 + my2 + n BP =λ= lx3 + my3 + n PC

Similarly, we obtain CQ =− QA AR and =− RB

lx3 lx1 lx1 lx2

+ + + +

my3 my1 my1 my2

+ + + +

n n n n

…(i)

…(ii) …(iii)

On multiplying Eqs. (i), (ii) and (iii), we get BP CQ AR ⋅ ⋅ = −1 PC QA RB Hence, (c) is the correct answer.

Ex 11. Vertices of a triangle are A (x1 , x1 tan α 1 ), B ( x 2 , x 2 tan α 2 ) and C ( x 3 , x 3 tan α 3 ). If the circumcentre coincides with the origin and the then orthocentre H = ( x , y ), y(cos α 1 + cos α 2 + cos α 3 ) is equal to (a) x(sin α 1 + sin α 2 + sin α 3 ) (b) x(sin α 1 − sin α 2 − sin α 3 ) (c) x(sin α 3 − sin α 2 + sin α 1 ) (d) None of the above

Sol. Here, the circumcentre O = (0, 0) So, OA = OB = OC ∴ x12 + x12 tan 2 α 1 = x22 + x22 tan 2 α 2 = x32 + x32 tan 2 α 3 ⇒

x12 sec2 α 1 = x22 sec2 α 2 = x32 sec2 α 3

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Objective Mathematics Vol. 1

11

Since, vertex A (a, a tan α ) lies on Eq. (i), therefore a2 (1 + tan 2 α ) = r2 H (x , y )

(0, 0)

G

O C

B

⇒

x1 x2 x3 = = =k cosα 1 cosα 2 cosα 3

[say]

∴The vertices of the triangle become A = (k cosα 1 , k sin α 1 ) B = (k cosα 2 , k sin α 2 ) C = (k cosα 3 , k sin α 3 ) ∴The centroid G  cosα 1 + cosα 2 + cosα 3 sin α 1+ sin α 2 + sin α 3  =k ,k    3 3 We know that the orthocentre H , centroid G and circumcentre O are collinear. So, m of HO = m of GO y−0 y = ∴ m of HO = x −0 x sin α 1 + sin α 2 + sin α 3 −0 k 3 and m of GO = cosα 1 + cosα 2 + cosα 3 −0 k 3 sin α 1 + sin α 2 + sin α 3 = cosα 1 + cosα 2 + cosα 3 y sin α 1 + sin α 2 + sin α 3 ∴ = x cosα 1 + cosα 2 + cosα 3 ⇒ y(cosα 1 + cosα 2 + cosα 3 ) = x (sin α 1 + sin α 2 + sin α 3 ) Hence, (a) is the correct answer.

Ex 12. The circumcentre of a triangle having vertices A ( a, a tan α ), B ( b, b tan β), C ( c, c tan γ ) is at the origin, where α + β + γ = π. If the orthocentre lies on the line β γ α   4 cos cos cos  x − λy = 0, then λ is  2 2 2 β α γ sin sin 2 2 2 β α γ (b) 2sin sin sin 2 2 2 β α γ (c) 1 + 4 sin sin sin 2 2 2 (d) None of the above

⇒ a = r cosα ∴ A ≡ (r cosα , r sin α ) Similarly, B ≡ (r cosβ , r sin β ), C ≡ (r cos γ , r sin γ ) ∴Centroid G is  r(cosα + cosβ + cos γ ) r(sin α + sin β + sin γ )  ,   3 3   Circumcentre O′(0, 0) and orthocentre (H )(h, k ) (let) Since, we know thatO′,G and H are collinear, therefore Slope of O′ G = Slope of O′ H sin α + sin β + sin γ k ⇒ = cosα + cosβ + cos γ h ⇒ ⇒

x(sin α + sin β + sin γ ) − y (cosα + cosβ + cos γ ) = 0 α β γ  x  4 cos cos cos   2 2 2 α β γ  − y 1 + 4 sin sin sin  = 0  2 2 2

and hence the result becomes α + β + γ = π α β γ Hence, λ = 1 + 4 sin sin sin 2 2 2 Hence, (c) is the correct answer.

Ex 13. The mid-point of line segment joining (3, − 1) and (1, 1) is shifted by two units (in the sense of increasing y) perpendicular to the line segment. Then, the coordinates of the point in the new position are (a) ( 2 − 2 , 2) (c) ( 2 + 2 , 2 )

(b) ( 2, 2 − 3 ) (d) None of these

Sol. Let P be the mid-point of the line segment joining A(3, − 1) and B(1, 1).  3 + 1 − 1 + 1 Then, P≡ ,  = (2, 0)  2 2 

Let P be shifted to Q, where PQ = 2 and y-coordinate of Q is greater than that of P (from question). Q

(a) 4 sin

A(3, –1)

equation of the circumcircle may be x 2 + y2 = r2 A

x O' xG xH

606

B(1, 1)

1 − (− 1) 2 = = −1 −2 1− 3 1 m of PQ = − =1 (−1)

Now, m of AB =

Sol. Since, circumcentre O′ is the origin and therefore

B

P

…(i)

∴

Coordinates of Q by distance formula are (2 ± 2 cosθ , 0 ± 2 sin θ ), where tanθ = 1 1 1  =  2 ± 2⋅ , 0 ± 2⋅  = (2 ± 2 , ±  2 2

2)

As y -coordinate of Q is greater than that of P. Q = (2 + 2 , 2 ) ∴ C

This is the required point. Hence, (c) is the correct answer.

(a) 3( x 2 + y 2 ) − 2x − 4 y − 1 = 0 (b) 3 ( x 2 + y 2 ) − 2x − 4 y + 1 = 0 (c) 3( x 2 + y 2 ) + 2x + 4 y − 1 = 0 (d) 3( x 2 + y 2 ) + 2x + 4 y + 1 = 0

On adding Eqs. (i) and (ii), we get k  3(x1 + x2 ) = 3  h −   4 k c k  c  ⇒ 3 −  = 3 h −  ⇒h − = −  4  4 4 4

k  (x1 − x2 )2 = 9  h +   4

2

 1 + cos t + sin t 2 + sin t − cos t  Then, (α , β ) =  ,    3 3 1 + cos t + sin t 2 + sin t − cos t ,β= α= ∴ 3 3 or 3α − 1 = cos t + sin t …(i) and 3β − 2 = sin t − cos t …(ii)

c2 k  + 1 = 9 h +   16 4

2

⇒

On squaring and adding Eqs. (i) and (ii), we get (3α − 1)2 + (3β − 2)2 = (cos t + sin t )2 + (sin t − cos t )2

⇒ 16h2 + 10hk + k 2 − 2 = 0

Sol. Let G (α , β) be the centroid in any position.

∴The equation of locus of the centroid is (3x − 1)2 + (3 y − 2)2 = 2 ⇒ 9(x 2 + y2 ) − 6x − 12 y + 3 = 0 ⇒ 3(x 2 + y2 ) − 2x − 4 y + 1 = 0 Hence, (b) is the correct answer.

Ex 15. A variable straight line of slope 4 intersects the hyperbola xy =1 at two points. Then, the locus of the point which divides the line segment between these two points in the ratio 1 : 2, is (a) 16x 2 (b) 16x 2 (c) 16x 2 (d) 16x 2

− 10xy + + 10xy + + 10xy − + 10xy +

y2 y2 y2 y2

− 2= 0 + 2= 0 + 2= 0 − 2= 0

Sol. Let equation of the line be y = 4 x + c, where c is parameter. It intersects the hyperbola xy = 1 at two points, for which x (4 x + c) = 1. ⇒ 4 x 2 + cx − 1 = 0 Let x1 and x2 be the roots of this equation. Then, c 1 and x1x2 = − x1 + x2 = − 4 4 If A and B are the points of intersection of the lines and  1 the hyperbola, then the coordinates A are  x1 ,  and x  1  1 that of B are  x2 ,  . x2   Let R (h, k ) be the point which divides AB in the ratio 1 : 2, then 2 1 + 2x + x2 x x2 2x2 + x1 and k = 1 h= 1 = 3 3x1x2 3 ...(i) ⇒ 2x1 + x2 = 3h  1  3 …(ii) and x1 + 2x2 = 3  −  k =  −  k  4  4

...(iii)

On subtracting Eq. (ii) from Eq. (i), we get k  x1 − x2 = 3  h +   4 ⇒

= 2(cos2 t + sin 2 t ) = 2

11

⇒ ⇒

 h − 

2

2

k k   + 1 = 9 h +    4 4 2  1 k 1 k 2 h2 − hk + + 1 = 9  h2 + hk +  2 16 2 16  

Cartesian System of Rectangular Coordinates

Ex 14. ABC is a variable triangle with the fixed vertex C (1, 2) and A, B having the coordinates (cos t, sin t ), (sin t, − cos t ) respectively, where t is a parameter. The locus of centroid of the ∆ABC is

So, the locus of R (h, k ) is 16x 2 + 10xy + y2 − 2 = 0. Hence, (d) is the correct answer.

Ex 16. The equation of a pair of straight lines is ax 2 + 2hxy + by 2 = 0. By what angle must the axes be rotated so that the term containing xy in the equation may be removed?  2h  1 tan −1   2  a + b 1  2h  (c) tan −1    ab  2

(a)

(b)

 2h  1 tan −1   2  a − b

(d) None of these

Sol. Here, the origin remains fixed. Let the axes be turned about the fixed origin through an angle φ in the anti-clockwise sense and new coordinates of the point (x , y) become (x′ , y′ ).

Then, the equations of transformation will be x = x′ cos φ − y′ sin φ and y = x′ sin φ + y′ cos φ ∴The changed equation will be a(x′ cos φ − y′ sin φ )2 + 2h(x′ cos φ − y′ sin φ ) (x′ sin φ + y′ cos φ ) + b(x′ sin φ + y′ cos φ )2 = 0 ⇒ (a cos2 φ + 2h sin φ cos φ + b sin 2 φ )x′ 2 + (− a sin 2 φ + 2h cos 2φ + b sin 2φ )x′ y′ + (a sin 2 φ − 2h sin φ cos φ + b cos2 φ ) y′ 2 = 0 ∴The transformed equation of the pair of lines for the new axes will be (a cos2 φ + h sin 2φ + b sin 2 φ )x 2 + {(b − a)sin 2 φ + 2h cos 2φ}xy + (a sin 2 φ − h sin 2φ + b cos2 φ ) y2 = 0 This equation will not contain xy, if (b − a)sin 2φ + 2h cos 2φ = 0 2h tan 2φ = ⇒ a−b 1 2h ⇒ φ = tan −1 2 a−b which is the required angle. Hence, (b) is the correct answer.

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Objective Mathematics Vol. 1

11 Type 2. More than One Correct Option Ex 17. If the coordinates of the vertices of a triangle are rational numbers, then which of the following points of the triangle have rational coordinates? (a) centroid (c) circumcentre

(b) incentre (d) orthocentre

Sol. Let A (x1 , y1 ), B (x2 , y2 ) and C (x3 , y3 ) be the vertices of

∆ABC such that x1 , x2 , x3 , y1 , y2 , y3 are rational x + x2 + x3 y + y2 + y3 and 1 are also numbers. Then, 1 3 3 rational numbers.

So, the coordinates of the centroid are always rational numbers. Let P be the circumcentre of ∆ABC. Then, PA = PB = PC ⇒ PA 2 = PB 2 = PC 2 ⇒

PA 2 = PB 2

and

PB 2 = PC 2

These two relations provide two linear equations in terms of the coordinates of point P such that the coefficients are rational numbers. So, the coordinates of P are also rational numbers. Since, the centroid G divides the line segment joining circumcentre P and orthocentre H in the ratio 1 : 2. Therefore, coordinates of H will also have rational values. The coordinates of the incentre are  ax1 + bx2 + cx3 ay1 + by2 + cy3  ,   a+ b+ c   a+ b+ c where, a = BC , b = CA and c = AB Clearly, a = (x2 − x3 )2 + ( y2 − y3 )2

b = (x3 − x1 )2 + ( y3 − y1 )2 and

c = (x2 − x1 )2 + ( y2 − y1 )2

may not be rational number. Therefore, coordinates of the incentre may not be rational numbers. Hence, (a), (c) and (d) are the correct answers.

Ex 18. If a > 0, b > 0, then the maximum area of the triangle formed by the points O(0, 0), A ( a cos θ, b sin θ ) and B ( a cos θ, − b sin θ ) is (in sq units) π 3ab , when θ = 4 4 π ab (b) , when θ = 4 2 π ab (c) , when θ = − 4 2 (d) a 2 b 2

(a)

Sol. Let ∆ be the area of ∆ABC. Then, ∆ = Absolute value of

0 0 1 1 a cosθ b sin θ 1 2 a cosθ − b sin θ 1

= Absolute value of (− ab sin θ cosθ ) = ab|sin θ cosθ | ab ∴∆ = |sin 2θ | 2 Clearly, ∆ is maximum when sin 2θ = ± 1 i.e. θ = ±

π 4

ab . 2 Hence, (b) and (c) are the correct answers. and the maximum value of ∆ is

Type 3. Assertion and Reason Directions (Ex. Nos. 19-20) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 19. Statement I If point of intersection of the lines 4x + 3 y = λ and 3x − 4 y = µ, ∀ λ, µ ∈R is ( x1 , y1 ), then the locus of (λ, µ) is x + 7 y = 0, ∀ x1 = y1 . Statement II If 4λ + 3µ > 0 and 3λ − 4µ > 0, then ( x1 , y1 ) is in first quadrant. Sol. On solving 4 x + 3 y = λ and 3x − 4 y = µ, we get 608

x1 =

4 λ + 3µ 3λ − 4µ and y1 = 25 25

Q ⇒ ⇒

x1 = y1 4 λ + 3µ 3λ − 4µ = 25 25 λ + 7µ = 0

∴ Locus of (λ, µ) is x + 7 y = 0 For first quadrant, x1 > 0 and y1 > 0 4 λ + 3µ 3λ − 4µ i.e. > 0 and >0 25 25 or 4 λ + 3µ > 0 and 3λ − 4µ > 0 Hence, (b) is the correct answer.

Ex 20. Statement I If centroid and circumcentre of a triangle are known, then its orthocentre and nine point centre can be found. Statement II Orthocentre, nine point centre, centroid and circumcentre are collinear. Sol. Orthocentre (O), nine point centre (N ), centroid (G) and circumcentre (C) are collinear and G divides O and C in the ratio 2 : 1 (internally) and N is the mid-point of O and C. Hence, (a) is the correct answer.

Passage (Ex. Nos. 21-23) Suppose we define the distance between two points P ( x 1, y 1) and Q( x 2, y 2 ) as d (P, Q) = max {| x 2 − x 1|, | y 2 − y 1 |}.

Ex 21. For a point P along the line y = 3x, d (0, P ) is equal to (a) x (c) x, if x < 0

(b) y (d) y, if y ≥ 0

Sol. Clearly, | y | < | x | , if θ
| x |, if θ < 4 Y

Sol. Consider a square with vertices at (− 1, 0), (1, 0), (0, 1)

P

and (0, − 1) as shown in following figure:

y

Y L

θ

X x

and ⇒

Cartesian System of Rectangular Coordinates

11

Type 4. Linked Comprehension Based Questions

| y | = | x |, if θ =

C (0, 1)

π 4

M

d (0, P ) = | y| as θ >

⇒ d (0, P ) = y, y ≥ 0 Hence, (d) is the correct answer.

π 4

X B (1, 0)

(–1, 0)A

D (0, –1)

Ex 22. The area of the region bounded by the locus of a point P satisfying d ( P , A ) = 4, where A is (1, 2), is (a) 64 sq units (c) 16π sq units

X′

(b) 54 sq units (d) None of these

Sol. We have, max {| x − 1|, | y − 2|} = 4

If {| x − 1| ≥ | y − 2|}, then | x − 1| = 4 i.e. if (x + y − 3) (x − y + 1) ≥ 0, then x = − 3 or 5 If | y − 2 | ≥ | x − 1 |, then |y− 2|= 4 i.e. (x + y − 3) (x − y + 1) ≤ 0 Then, y = − 2 or 6

Y′

If we select a point P1 in the region BCLM, then d ( A, P) = 1 + x d ( B, P) = y ∴ d ( A, P) + d ( B, P) = 1 + x + y > 2 However, for points P2 and P3 lying respectively on the line BC and below the line BC, d (P , A ) + d (P , B ) = 2 The same argument holds for other quadrant also. Hence, d (P , A ) + d (P , B ) = 2 represents the regions lying inside the square ABCD. Hence, (c) is the correct answer.

Type 5. Match the Column Ex 24. Match the following: Column I

Column II

A.

If the point { x1 + t ( x2 − x1 ), y1 + t ( y2 − y1 )}, divides the join of ( x1, y1 ) and ( x2, y2 ) internally, then

p.

−

B.

Set of values of t for which point P (t , t 2 − 2 ) lies inside the triangle formed by the lines x + y = 1, y = x + 1 and y = − 1, is

q.

1 < |t|
0. ⇒

⇒

0< t 0 ⇒ | t | > 1 Case II

4t 2t   + 8+ − 15 < 0 2 +   2 2 3t   6t   4 +   −5 +  0

…(i)

Case II B and P lies on same side of AC.

⇒ ⇒

…(iii)

(0 + 0 − 4 ) (1 + t − 4 ) > 0 (t − 3) < 0 ⇒ t < 3

…(ii)

Case III C and P lies on same side of AB. ⇒ (4 − 0) (1 − t ) > 0 ⇒ t < 1 From Eqs. (i), (ii) and (iii), we get 0< t

1 8

(d) None of these

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Objective Mathematics Vol. 1

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12. If the coordinates of a point (4, 5) become (− 3, 9), then the point of origin be shifted at (a) (7, − 4 ) (c) (− 7, 4 )

(b) (7, 4 ) (d) None of these

becomes a 2 X 2 + 2h2 XY + b2Y 2 , then

13. If the axes are rotated through an angle of 45° in the clockwise direction, the coordinates of a point in the new system are (0, − 2), then its original coordinates are (a) ( 2 , 2 )

(b) (− 2 , 2 )

(c) ( 2 , − 2 )

(d) (− 2 , − 2 )

3  (b)  , − 2 4 

(a) (a1 − b1 )2 + 4 h12 = (a2 − b2 )2 + 4 h22 (b) (a1 − b1 )2 − 4 h12 = (a2 − b2 )2 + 4 h22 (c) (a1 − b1 )2 + 4 h12 = (a2 − b2 )2 − 4 h22 (d) None of the above

19. If by rotating the coordinate axes without translating the origin, the expression a1 x 2 + 2h1 xy + b1 y 2

14. Shift the origin to a suitable point so that the equation y 2 + 4 y + 8x − 2 = 0 will not contain term in y and the constant term, then the origin is 3  (a)  , 2 4 

18. If by rotating the coordinate axes without translating the origin, the expression a1 x 2 + 2h1 xy + b1 y 2

 3  3   (c)  − , − 2 (d)  − , 2  4  4  

becomes a 2 X 2 + 2h2 XY + b2Y 2 , then (a) a1b1 + h12 = a2b2 + h22 (b) a1b1 + h12 = a2b2 − h22 (c) a1b1 − h12 = a2b2 − h22 (d) None of the above

Ta rg e t E x e rc is e s

15. If ( x, y ) and ( X , Y ) are the coordinates of the same point referred to two sets of rectangular axes with the same origin and if ax + by becomes pX + qY , where a, b are independent of x, y, then (a) a2 − b2 = p2 + q2

(b) a2 − b2 = p2 − q2

(c) a2 + b2 = p2 − q2

(d) a2 + b2 = p2 + q2

16. The angle through which the coordinates axes be rotated, so that xy-term in the equation 5x 2 + 4 3xy + 9 y 2 = 0 may be missing, is (a) −

π 6

(b)

π 4

(c)

π 2

(d)

2π 3

17. If by rotating the coordinate axes without translating the origin, the expression a1 x 2 + 2h1 xy + b1 y 2 becomes a 2 X 2 + 2h2 XY + b2Y 2 , then (a) a1 − b1 = a2 − b2 (c) a1 + b2 = a2 + b1

(b) a1 + b1 = a2 + b2 (d) a1− b1 = a2 + b2

20. The axes be shifted without rotation, so that the equation ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 does not contain terms in x, y and constant term, then the point is  hf − bg hg − af   (a)  − ,  ab − h2 ab − h2 

 hf − bg 2 hg − af   (b)  ,  ab − h2 ab − h2 

 hf − bg hg − af   (c)  − ,− 2 ab − h2   ab − h

 hf −bg hg − af   (d)  ,  ab − h2 ab − h2 

21. The angle through which the axes may be turned, so that the equation Ax + By + C = 0 may be reduced to the form x = constant, is  B (a) tan − 1    A (c)

1  A tan − 1    B 2

 A (b) tan − 1    B (d)

1  B tan − 1    A 2

Type 2. More than One Correct Option 22. All points lying inside the triangle formed by the points (1, 3), (5, 0) and (− 1, 2) satisfy (a) 3x + 2 y ≥ 0 (c) 2x − 3 y − 12 ≤ 0

(b) 2x + y − 13 ≥ 0 (d) − 2x + y ≥ 0

23. If ( − 6, − 4 ), ( 3, 5) and ( − 2, 1) are the vertices of a parallelogram, then remaining vertex can be

(a) (0, − 1) (c) (− 1, 0)

(b) (7, 9) (d) (− 11, − 8)

24. If ( − 4, 0) and (1, − 1) are two vertices of a triangle of area 4 sq units, then its third vertex lies on (a) y = x (c) x + 5 y − 4 = 0

(b) 5x + y + 12 = 0 (d) x + 5 y + 12 = 0

Type 3. Assertion and Reason Directions (Q. Nos. 25-29) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

612

(b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

(c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

25. Statement I If the point ( 2a − 5, a 2 ) is on the same side of the line x + y − 3 = 0 as that of the origin, then a ∈ ( 2, 4 ). Statement II The points ( x1 , y1 ) and ( x 2 , y 2 ) lie on the same or opposite sides of the line ax + by + c = 0, as ax1 + by1 + c and ax 2 + by 2 + c have the same or opposite signs.

triangle formed by the lines 2x + 3 y − 1 = 0, x + 2 y − 3 = 0 and 5x − 6 y − 1 = 0 for every  3  1  α ∈  − , − 1 ∪  , 1 .  2  2  Statement II Two points ( x1 , y1 ) and ( x 2 , y 2 ) lie on the same side of straight line ax + by + c = 0, if ax1 + by1 + c and ax 2 + by 2 + c are of opposite signs. 27. Statement I Area of the triangle formed by 4x + y + 1 = 0 with the coordinate axes is 1/8 sq unit. Statement II Area of the triangle made by the line c2 . ax + by + c = 0 with the coordinate axes is 2 ab

28. Statement I The incentre of a triangle formed by π π 8π the lines x cos + y sin = π, x cos + 9 9 9 8π 13π 13π y sin = π and x cos + y sin = π is ( 0, 0). 9 9 9 Statement II The point ( 0, 0) is equidistant from the π π 8π 8π lines x cos + y sin = π, x cos + y sin =π 9 9 9 9 13π 13π and x cos + y sin = π. 9 9 29. Statement I The area of triangle formed by the points and A (1000, 1002), B (1001, 1004 ) C(1002, 1003) is same as the area formed by A ′ ( 0, 0), B ′ (1, 2) and C ′ ( 2, 1). Statement II The area of the triangle is constant with respect to translation of axes.

11 Cartesian System of Rectangular Coordinates

26. Statement I The point (α, α 2 ) lies inside the

Type 4. Linked Comprehension Based Questions sides of unit length. Points E and F are taken on sides AB and AD respectively, so that AE = AF . Let P be a point inside the square ABCD.

30. The maximum possible area of quadrilateral CDFE is (a) 1/8 (b) 1/4 (c) 5/8 (d) 3/8

31. The value of ( PA ) 2 − ( PB ) 2 + ( PC ) 2 − ( PD ) 2 is equal to (a) 3 (c) 1

(b) 2 (d) 0

32. Let a line passing through point A divides the square ABCD into two parts so that area of one portion is double the other, then the length of portion of line inside the square is (a) 10 / 3

(b) 13 / 3

(c) 11 / 3

(d) 2 / 3

Type 5. Match the Column 33. Consider the lines represented by equation ( x 2 + xy − x )( x − y ) = 0 , forming a triangle. Then, match the following: Column I

Targ e t E x e rc is e s

Passage (Q. Nos. 30-32) Let ABCD be a square with

Column II

A.

Orthocentre of triangle

p.

(1/6, 1/2)

B.

Circumcentre

q.

{1/(2 + 2 2 ), 1/2}

C.

Centroid

r.

( 0, 1 / 2 )

D.

Incentre

s.

(1 / 2, 1 / 2 )

Type 6. Single Integer Answer Type Questions 34. If the area of triangle formed by the points ( 2a, b ), ( a + b, 2b + a ) and ( 2b, 2a ) is 2 sq units, then the area of the triangle whose vertices are ( a + b, a − b ), ( 3b − a, b + 3a ) and ( 3a − b, 3b − a ) will be _______ . 35. The equations of three sides of a triangle are x = 2, y + 1 = 0 and x + 2 y = 4. The coordinates of the

circumcentre of the triangle are (λ, µ), then λ + µ is _____. 36. The distance between the circumcentre and orthocentre of the triangle, whose vertices are (0, 0), 2 (6, 8) and (− 4, 3), is L, then the value of Lis ____. 5

613

Entrances Gallery JEE Main/AIEEE 1. The number of points having both coordinates as integers that lie in the interior of the triangle with [2015] vertices (0, 0), (0, 41) and (41, 0), is (a) 901 (c) 820

(b) 861 (d) 780

2. Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d = 0 lies in the fourth quadrant and is [2014] equidistant from the two axes, then (a) 2bc − 3ad = 0 (c) 3bc − 2ad = 0

(b) 2bc + 3ad = 0 (d) 3bc + 2ad = 0

Ta rg e t E x e rc is e s

3. The x-coordinate of the incentre of the triangle that has the coordinates of mid-points of its sides as (0, 1), (1, 1) and (1, 0), is [2013] (a) 2 +

2

(b) 2 − 2

(c) 1 +

2

(d) 1 − 2

29 5

(b) 5

(c) 6

(d)

11 5

5. If A( 2, − 3) and B( − 2, 1) are two vertices of a triangle and third vertex moves on the line 2x + 3 y = 9, then [2011] the locus of the centroid of the triangle is (a) 2x − 3 y = 1 (c) 2x + 3 y = 1

(b) x − y = 1 (d) 2x + 3 y = 3

6. Three distinct points A , B and C are given in the 2-dimensional coordinate plane such that the ratio of the distance of anyone of them from the point (1, 0) to the distance from the point ( − 1, 0) is equal to 1/3. Then, the circumcentre of the ∆ABC is at the point (a) (0, 0)

5  (b)  , 0 4 

5  (c)  , 0 2 

[2009] 5  (d)  , 0 3 

7. Let A ( h, k ), B (1, 1) and C( 2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which k can take is given by [2007] (a) {1, 3} (c) {− 1, 3}

(b) {0, 2} (d) {− 3, − 2}

8. If ( a, a 2 ) falls inside the angle made by the lines x y = , x > 0and y = 3x, x > 0, then a belongs to [2006] 2

614

 (b) 1, 

7  3 7  (d)  − 1,   3

1 (a)  , 3

7  3  1 7 (c)  − ,   3 3

10. Let A( 2, − 3) and B( − 2, 1) be the vertices of a ∆ABC. If the centroid of this triangle moves on the line 2x + 3 y = 1, then the locus of the vertex C is the line (a) 2x + 3 y = 9 (c) 3x + 2 y = 5

(b) 2x − 3 y = 7 (d) 3x − 2 y = 3

[2004]

11. If the equation of the locus of a point equidistant from is and the points ( a 2 , b2 ) ( a1 , b1 ) ( a1 − a 2 )x + ( b1 − b2 ) y + c = 0 , then the value of c is [2003]

4. If the line 2x + y = k passes through the point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2 , then k equals [2012] (a)

9. If a vertex of a triangle is (1, 1) and the mid-points of two sides through this vertex are ( − 1, 2) and ( 3, 2), then the centroid of the triangle is [2005]

1 2 (a2 + b22 − a12 − b12 ) 2 1 (c) (a12 + a22 + b12 + b22 ) 2 (a)

(b) a12 − a22 + b12 − b22 (d) a12 + b12 − a22 − b22

12. Locus of centroid of the triangle whose vertices are ( a cos t , a sin t ), ( b sin t , − b cos t ) and (1, 0), where t [2003] is a parameter, is (a) (3x − 1)2 + (3 y)2 = a2 − b2 (b) (3x − 1)2 + (3 y)2 = a2 + b2 (c) (3x + 1)2 + (3 y)2 = a2 + b2 (d) (3x + 1)2 + (3 y)2 = a2 − b2

13. A triangle with vertices ( 4, 0),( − 1, − 1) and ( 3, 5) is [2002] (a) isosceles and right angled (b) isosceles but not right angled (c) right angled but not isosceles (d) neither right angled nor isosceles

14. The incentre of the triangle with vertices (1, 3 ), ( 0, 0) and ( 2, 0) is [2002]  3  (a) 1,  2

2 1  (b)  ,   3 3

 2 3  (c)  , 3 2 

 1 (d) 1,   3

15. The locus of the mid-point of the distance between the axes of the variable line x cos α + y sin α = p, where p is constant, is [2002]

(a) (3, ∞ )

1  (b)  , 3 2 

(a) x 2 + y2 = 4 p2

1  (c)  − 3, −   2

 1 (d)  0,   2

(c) x 2 + y2 =

4 p2

1 1 4 + 2= 2 x2 y p 1 1 2 (d) 2 + 2 = 2 x y p (b)

( 8, − 2) and ( 2, − 2) is at the point (a) (2, − 1) (d) (2, 5)

(b) (1, − 2) (e) (4 , 5)

[Kerala CEE 2014]

(c) (5, 2)

17. The triangle joining the points P ( 2, 7), Q ( 4, − 1) and [UP SEE 2013] R( − 2, 6) is (a) isosceles triangle

(b) equilateral triangle

(c) right angled triangle

(d) None of these

18. ABC is a triangle with vertices A ( − 1, 4 ), B ( 6, − 2) and C ( − 2, 4 ). D, E and F are the points which divide each AB, BC and CA, respectively in the ratio 3 : 1 internally. Then, the centroid of the ∆DEF is

25. The vertices of a triangle are A ( 0, 0), B ( 0, 2) and C ( 2, 0), then find the distance between its orthocentre [Guj CET 2011] and circumcentre. (a) 0 1 (c) 2

(b) 2 (d) None of these

26. Orthocentre of the triangle formed by the lines [Guj CET 2011] x − y = 0, x + y = 0 and x = 3 is (a) (0, 0) (c) (0, 3)

(b) (3, 0) (d) Cannot be determined

[Kerala CEE 2013]

27. If A = ( − 3, 4 ), B = ( − 1, − 2), C = ( 5, 6) and D = ( x, − 4 ) are vertices of a quadrilateral such that [GGSIPU 2011] ∆ABD = 2∆ACD. Then, x is equal to

19. A point moves such that the sum of its distances from two fixed points (ae, 0) and ( − ae, 0) is always 2a. Then, equation of its locus is [WB JEE 2013]

28. The equation of the locus of the point of intersection of the straight lines x sin θ + (1 − cos θ) y = a sin θ and [WB JEE 2011] x sin θ − (1 + cos θ ) y + a sin θ = 0 is

(a) (3, 6) (d) (− 3, 6)

(b) (1, 2) (c) (4 , 8) (e) None of these

(a)

x2 y2 + 2 =1 2 a a (1 − e2 )

(b)

(c)

x2 y2 + 2 =1 2 a (1 − e ) a

(d) None of these

2

x2 y2 − 2 =1 2 a a (1 − e2 )

20. The line joining and A ( b cos α, b sin α ) B ( a cos β, a sin β ), where a ≠ b, is produced to the point M ( x, y ), so that AM : MB = b : a. Then, α + β α + β x cos   + y sin   is equal to [WB JEE 2012]  2   2  (d) a2 + b2

(a) 6 (c) 69

(b) 9 (d) 96

(a) y = ± ax (c) y2 = 4 x

(b) x = ± ay (d) x 2 + y2 = a2

29. A line segment of 8 units in length moves so, that its end points are always on the coordinate axes. Then, the equation of locus of its mid-point is [J&K CET 2011] (a) x 2 + y2 = 4

(b) x 2 + y2 = 16

(c) x 2 + y2 = 8

(d) x + y = 8

30. The distance between the points ( a cos α, a sin α) and [Kerala CEE 2011] ( a cos β, a sin β) is

21. If the mid-points of the sides of a triangle are ( − 2, 3), ( 4, − 3) and ( 4, 5), then the centroid of the triangle is

 α − β (a) a cos    2 

 α − β (b) 2a cos    2 

 α − β (c) a sin    2 

 α − β (d) 2a sin    2 

[Karnataka CET 2012] (c) (2, 5/ 3) (d) (1, 5/ 6)

(e) None of these

(a) 0

(a) (5/ 3, 2)

(b) 1

(b) (5/ 6, 1)

(c) − 1

22. The vertices of ∆ABC are A ( 2, 2), B ( − 4, − 4 ) and C ( 5, − 8). Find the length of a median of a triangle, which is passing through the point C. [Guj CET 2011] (a) 65

(b) 117

(c) 85

(d) 116

23. The line x + y = 4 divides the line joining the points [BITSAT 2011] ( −1, 1) and ( 5, 7) in the ratio (a) 2 : 1 (c) 1 : 2 externally

(b) 1 : 2 internally (d) None of these

24. The points A (1, 2), B ( 2, 4 ) and C( 4, 8) form a/an [Karnataka CET 2011] (a) isosceles triangle (b) equilateral triangle (c) straight line (d) right angled triangle

Targ e t E x e rc is e s

16. The circumcentre of the triangle with vertices ( 8, 6),

Cartesian System of Rectangular Coordinates

11

Other Engineering Entrances

31. The vertices of a rectangle ABCD are A( − 1, 0), B( 2, 0), C ( a, b ) and D( − 1, 4 ). Then, the length of the diagonal AC is [Kerala CEE 2011] (a) 2 (e) 6

(b) 3

(c) 4

(d) 5

32. The points ( 0, 8 / 3), (1, 3) and ( 82, 30) are the vertices of [Karnataka CET 2011] (a) an equilateral triangle (b) an isosceles triangle (c) a right angled triangle (d) None of the above

33. The locus of a point which moves so that its distance from X -axis is double of its distance from Y -axis, is (a) x = 2 y (c) x = 5 y + 1

[Karnataka CET 2011] (b) y = 2x (d) y = 2x + 3

615

Objective Mathematics Vol. 1

11

34. The three distinct points A ( at 12 , 2at 1 ), B ( at 22 , 2at 2 ) and C ( 0, a ) (where, a is a real number) are collinear, if [J&K CET 2011]

(a) t1t2 = − 1 (c) 2t1t2 = t1 + t2

(b) t1t2 = 1 (d) t1 + t2 = a

35. If the distance between (2, 3) and ( − 5, 2) is equal to the distance between ( x, 2) and (1, 3), then the values of x are [BITSAT 2010] (a) − 6, 8 (c) − 8, 6

37. If the three points ( 3q, 0), ( 0, 3 p ) and (1, 1) are collinear, then which one is true? [WB JEE 2010] 1 1 + =0 p q 1 1 (c) + = 3 p q

36. If the three points (0, 1), (0, − 1) and ( x, 0) are vertices of an equilateral triangle, then the values of x are (a) 3 , 2

[Kerala CEE 2010] (b) 3 , − 3

(c) − 5 , 3

(d) 2 , − 2

(e) 5 , − 5

(b)

38. The area (in sq units) of the triangle formed by the points (2, 2), (5, 5) and (6, 7) is [Kerala CEE 2010] (a) 9/2 (e) 14

(b) 6, 8 (d) − 7, 7

1 1 + =1 p q 1 3 (d) + = 1 p q

(a)

(b) 5

(c) 10

(d) 3/2

39. Q, R and S are points on the line joining the points P ( a, x ) and T ( b, y ) such that PQ = RS = ST , then  5a + 3b 5x + 3 y ,  is the mid-point of the line   8 8  segment [MP PET 2010] (a) RS

(b) ST

(c) PQ

(d) QR

Answers Ta rg e t E x e rc is e s

Work Book Exercise 1. (a)

2. (c)

11. (c)

12. (a)

3. (b)

4. (d)

5. (d)

6. (b)

7. (b)

8. (b)

9. (a)

10. (a)

Target Exercises 1. (a)

2. (a)

3. (d)

4. (a)

5. (b)

6. (b)

7. (c)

8. (a)

9. (a)

10. (a)

11. (a)

12. (a)

13. (d)

14. (b)

15. (d)

16. (a)

17. (b)

18. (a)

19. (c)

20. (d)

21. (a)

22. (a,c)

23. (c,d)

24. (c,d)

25. (d)

26. (c)

27. (a)

28. (b)

29. (a)

30. (c)

31. (d)

32. (b)

33. (*)

34. (8)

35. (4)

36. (5)

* A → s; B → r; C → p; D → q

Entrances Gallery

616

1. (d)

2. (c)

3. (b)

4. (c)

5. (c)

6. (b)

7. (c)

8. (b)

9. (b)

10. (a)

11. (a)

12. (b)

13. (a)

14. (d)

15. (b)

16. (c)

17. (c)

18. (b)

19. (a)

20. (a)

21. (c)

22. (c)

23. (b)

24. (c)

25. (b)

26. (a)

27. (c)

28. (d)

29. (b)

30. (d)

Explanations Target Exercises

Now,

∴

cos θ1 − cos θ 2 sin θ1 − sin θ 2 1 ab cos θ 2 − cos θ 3 sin θ 2 − sin θ 3 2 cos θ 3 sin θ 3

c = AB = (4 − 1) + (5 − 1) = 5 2

2

b2 + c 2 − a2 2 bc 169 + 25 − 68 63 = = 2 × 13 × 5 65

cos A =

⇒ ⇒

[using Eqs. (i) and (ii)] 4b1b2 2(c1a2 + c 2 a1 ) = a1a2 a1a2 ac + a c = 2 b b 1 2 2 1 1 2

3. Let A(− 1, 0 ), B(3, 1,) C(2, 2 ) and D( x, y )be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other. ∴Coordinates of the mid-point of AC = Coordinates of the mid-point of BD  − 1 + 2 0 + 2  3 + x 1 + y  ⇒ , ,    =  2 2   2 2   1   3 + x y + 1 , ⇒   , 1 =  2   2 2  3+ x 1 y+1 and = =1 ⇒ 2 2 2 ⇒ x = − 2 and y = 1 Hence, the fourth vertex of the parallelogram is (−2, 1.)

4. Let ∆ be the area of the triangle. Then, ∆=

a cos θ1 1 a cos θ 2 2 a cos θ 3

b sin θ1 1 b sin θ 2 1 b sin θ 3 1

1

 θ − θ2   θ + θ2  − 2 sin  1   sin  1  2   2  1  θ + θ3   θ − θ3  = ab − 2 sin  2   sin  2   2   2 2 cos θ 3  θ − θ2  2 sin  1  cos  2   θ − θ3  2 sin  2  cos  2  sin θ 3

2. Since, α, β are the roots of a1 x 2 + 2 b1 x + c1 = 0 and γ, δ are the roots of a2 x 2 + 2 b2 x + c 2 = 0, therefore 2b c …(i) α + β = − 1 , αβ = 1 a1 a1 Suppose C divides AB in the ratio λ : 1 and D divides AB in the ratio µ : 1. Then, λβ + α µβ + α and δ = γ= λ+1 µ+1 γ −α δ −α and µ = …(ii) ⇒ λ= β−γ β −δ γ −α δ −α λ+µ= + ⇒ β − γ β −δ (γ − α ) (β − δ ) + (δ − α ) (β − γ ) ⇒ 0= (β − γ) (β − δ ) [Q λ + µ = 0] ⇒ (γ − α ) (β − δ ) + (δ − α ) (β − γ ) = 0  2 b1   2 b2  2c1 2c 2 − =0 ⇒ −  × −  −  a1   a2  a1 a2

0 0

[applying R1 → R1 − R2 , R2 → R2 − R3 ]

a = BC = (4 − 6)2 + (5 − 13)2 = 68 b = CA = (6 − 1)2 + (13 − 1)2 = 169 = 13

and

=

 θ1 + θ 2     2   θ2 + θ3     2 

0 0 1

 θ + θ3   θ − θ3    θ − θ2  = 2 ab sin  1   × − cos  2  sin  2  2   2    2   θ + θ3    θ + θ2   θ + θ2  sin  1   sin  2  + cos  1  2    2   2   θ − θ3   θ − θ2  = 2 ab sin  1   sin  2  2   2 

 θ + θ 3 θ1 + θ 2  sin  2 −   2 2   θ − θ2   θ − θ1   θ − θ3  = 2 ab sin  1   sin  3  sin  2  2   2   2 

5. Let the coordinates of the third vertex be ( x, y ). x + 3−7 =2 3 y − 5+ 4 and =−1 3 ⇒ x − 4 = 6 and y − 1 = − 3 ⇒ x = 10 and y = − 2 Thus, the coordinates of the third vertex are (10, − 2).

Targ e t E x e rc is e s

b2 + c 2 − a2 , where a = BC, 2 bc b = CA and c = AB are the lengths of sides of the ∆ABC.

1. We know that, cos A =

Then,

6. Since, α,β and γ are the roots of the equation x 3 − 3 px 2 + 3 qx − 1 = 0 Therefore, α + β + γ = 3 p, αβ + βγ + γα = 3 q   and αβγ = 1 

…(i)

Let G ( x, y ) be the centroid of ∆ABC. Then, 1  1 1 1 α+β+γ and y =  + +  x= 3  α β γ 3 α+β+γ αβ + βγ + γα and y = x= ⇒ 3 3 αβγ 3p 3q and y = [using Eq. (i)] x= ⇒ 3 3 ⇒ x = p and y = q Hence, the coordinates of G are ( p, q )0.

7. It is given that α and β are the roots of the equation x 2 − 6 p1 x + 2 = 0. ∴ α + β = 6 p1, αβ = 2

…(i)

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Objective Mathematics Vol. 1

11

β and γ are the roots of the equation x 2 − 6 p2 x + 3 = 0. …(ii) ∴ β + γ = 6 p2 , βγ = 3 γ and α are the roots of the equation x 2 − 6 p3 x + 6 = 0. …(iii) ∴ γ + α = 6 p3 , γα = 6 From Eqs. (i), (ii) and (iii), we get [Qα, β, γ > 0] αβγ = 6 Now, αβ = 2 and αβγ = 6 ⇒ γ = 3 βγ = 3 and αβγ = 6 ⇒ α = 2 γα = 6 and αβγ = 6 ⇒ β = 1 1 ∴ α + β = 6 p1 ⇒ 3 = 6 p1 ⇒ p1 = 2 2 β + γ = 6 p2 ⇒ 4 = 6 p2 ⇒ p2 = 3 5 and γ + α = 6 p3 ⇒ 5 = 6 p3 ⇒ p3 = 6 The coordinates of the centroid of triangle are α + β + γ 1  1 1 1  ,  + +   3 3  α β γ   1   11 6 1  1 =  ,  + 1 +   = 2,   3 3 2 3   18 

Ta rg e t E x e rc is e s

8. Since, tan α, tan β and tan γ are the roots of the equation x 3 − 3 ax 2 + 3 bx − 1 = 0 Therefore, …(i) tan α + tan β + tan γ = 3 a …(ii) tan α tan β + tan β tan γ + tan γ tan α = 3 b and …(iii) tan α tan β tan γ = 1 Let ( x, y ) be the coordinates of the centroid G of ∆ABC. tan α + tan β + tan γ Then, x= = a [using Eq. (i)] 3 cot α + cot β + cot γ y= 3 tan α tan β + tan β tan γ + tan γ tan α = 3 tan α tan β tan γ 3b [using Eqs. (ii) and (iii)] = =b 3 Hence, the coordinates of the centroid G are (a, b).

9. Let P (1, 1), Q (2, − 3) and R (3, 4) be the mid-points of sides AB, BC and CA, respectively of ∆ABC. Let A ( x1, y1 ), B ( x2 , y2 )andC( x3 , y3 )be the vertices of ∆ABC. Then, P is the mid-point of AB. x1 + x2 =1 2 y1 + y2 and =1 2 ⇒ x1 + x2 = 2 and y1 + y2 = 2 Q is the mid-point of BC. x2 + x3 =2 ⇒ 2 y2 + y3 and =−3 2 ⇒ x2 + x3 = 4 and y2 + y3 = − 6 R is the mid-point of AC. x1 + x3 ⇒ =3 2 y1 + y3 and =4 2 ⇒ x1 + x3 = 6 and y1 + y3 = 8 ⇒

618

…(i)

From Eqs. (i), (ii) and (iii), we get x1 + x2 + x2 + x3 + x1 + x3 = 2 + 4 + 6 and y1 + y2 + y2 + y3 + y1 + y3 = 2 − 6 + 8 ⇒ x1 + x2 + x3 = 6 and y1 + y2 + y3 = 2 From Eqs. (i) and (iv), we get x3 + 2 = 6 and 2 + y3 = 2 ∴ x3 = 4 and y3 = 0 So, the coordinates of C are (4, 0 ). From Eqs. (ii) and (iv), we get x1 + 4 = 6 and y1 − 6 = 2 ∴ x1 = 2 and y1 = 8 So, the coordinates of A are (2, 8). From Eqs. (iii) and (iv), we get x2 + 6 = 6 and y2 + 8 = 2 ∴ x2 = 0 and y2 = − 6 So, the coordinates of B are (0, − 6).

…(iv)

Now, a = BC = (4 − 0 )2 + (0 + 6)2 = 52 = 2 13 b = AC = (4 − 2 )2 + (0 − 8)2 = 68 = 2 17 and c = AB = (2 − 0 )2 + (8 + 6)2 = 200 = 10 2 The coordinates of the incentre of the ∆ABC are  ax1 + bx2 + cx3 ay1 + by2 + cy3  ,    a+ b+c a+ b+c  or

 2 13 + 20 2 2 13 − 6 17  ,   13 + 17 + 5 2 13 + 17 + 5 2  

10. We know that the centroid of a triangle divides the segment joining the orthocentre and circumcentre internally in the ratio 2 : 1. Therefore, 2α + u x= 2+1 2β + v and y= 2+1 3 x = 2α + u ⇒ and 3 y = 2β + v a 11. Let a ≤ sin A ⇒ ≤1 sin A 1 ⇒ 2R ≤ 1 ⇒ R ≤ 2 So, for any point ( x, y ) inside the circumcircle, 1 x2 + y2 < . 4   x2 + y2 1 Using AM ≥ GM,  ≥ xy  ⇒ xy < 2 8  

12. Let (h, k ) be the point to which the origin shifted.

…(ii)

Then, x = 4, y = 5, X = − 3,Y = 9 ∴ x = X + h and y = Y + k ⇒ 4 = − 3 + h and 5 = 9 + k ⇒ h = 7 and k = − 4 Hence, the origin must be shifted to (7, − 4).

13. Here, X = 0, Y = − 2 and θ = − 45°

…(iii)

Let ( x, y ) be the original coordinates of the point. Then, x = X cos θ − Y sin θ and y = X sin θ + Y cos θ ⇒ x = 2 sin (− 45° ) and y = − 2 cos (− 45° ) x = − 2 and y = − 2 ⇒ Hence, the original coordinates are (− 2 , − 2 ).

For this equation to be free from the term containing Y and the constant term, we must have 4 + 2 k = 0 and k 2 + 4 k + 8 h − 2 = 0 3 ⇒ k = − 2 and h = 4  3 Hence, the origin is shifted at the point  , − 2 .  4

15. Suppose one set of rectangular axes is obtained by rotating the coordinate axes of the other set by an angle θ in anti-clockwise sense. Let ( x, y ) be the coordinates of the point with respect to old axes and ( X, Y ) be the coordinates with respect to the new axes. Then, x = X cos θ − Y sin θ, y = X sin θ + Y cos θ ∴ ax + by = a( X cos θ − Y sin θ ) + b( X sin θ + Y cos θ ) = (a cos θ + b sin θ )X + (b cos θ − a sin θ )Y It is given that ax + by becomes pX + qY. Therefore, p = a cos θ + b sin θ and q = b cos θ − a sin θ ⇒ p2 + q 2 = (a cos θ + b sin θ )2 + (b cos θ − a sin θ )2 ⇒ p2 + q 2 = a2 + b2

16. In order to remove xy-term from the equation ax 2 + 2 hxy + by 2 + 2 gx + 2 fy + c = 0, the axes are rotated through an angle θ given by 2h tan 2θ = a−b Here, a = 5, b = 9 and h =2 3 2h ∴ tan 2θ = a−b 4 3  π ⇒ tan 2θ = = − 3 = tan  −   3 5− 9 π ⇒ 2θ = nπ − , n ∈ I 3 nπ π θ= − , n ∈I ⇒ 2 6

17. Let the axes be rotated through an angle θ in anti-clockwise sense. Then, x = X cos θ − Y sin θ

and

y = X sin θ + Y cos θ

∴ a1 x + 2 h1 xy + b1 y = a1 ( X cos θ − Y sin θ )2 2

2

+ 2 h1( X cos θ − Y sin θ ) ( X sin θ + Y cos θ) + b1( X sin θ + Y cos θ)2 = (a1 cos θ + b1 sin θ + h1 sin 2θ )X 2

2

2

+ XY (2 h1 cos 2θ − a1 sin 2θ + b1 sin 2θ) + (a1 sin 2 θ − h1 sin 2θ + b1 cos 2 θ) y 2 …(i) It is given that the expression a1 x 2 + 2 h1 xy + b1 y 2 transforms to a2 X 2 + 2 h2 XY + b2 Y 2 …(ii) by rotating the axes. Therefore, Eqs. (i) and (ii) are the same. …(iii) ∴ a2 = a1 cos 2 θ + b1 sin 2 θ + h1 sin 2θ

and

2 h2 = 2 h1 cos 2θ − a1 sin 2θ + b1 sin 2θ b2 = a1 sin 2 θ − h1 sin 2θ + b1 cos 2 θ

…(iv) …(v)

On adding Eqs. (iii) and (v), we get a2 + b2 = a1 + b1

18. We have, a2 − b2 = a1 cos 2θ − b1 cos 2θ + 2 h1 sin 2θ ⇒ ⇒

a2 − b2 = (a1 − b1 )cos 2θ + 2 h1 sin 2θ (a2 − b2 )2 = (a1 − b1 )2 cos 2 2θ + 4 h12 sin 2 2θ + 4 (a1 − b1 ) h1 sin 2θ cos 2θ …(vi)

From Eq. (iv), we have 2 h2 = 2 h1 cos 2θ − (a1 − b1 ) sin 2θ ⇒

4h22 = 4h12 cos 2 2θ + (a1 − b1 )2 sin 2 2θ

− 4 (a1 − b1 ) h1 sin 2θ cos 2θ …(vii) On adding Eqs. (vi) and (vii), we get (a2 − b2 )2 + 4h22

11

= (a1 − b1 )2 (cos 2 2θ + sin 2 2θ ) + 4h12 (cos 2 2θ + sin 2 2θ ) = (a1 − b1 )2 + 4h12 Hence, (a2 − b2 )2 + 4h22 = (a1 − b1 )2 + 4h12

19. We have,

(a1 − b1 )2 + 4h12 = (a2 − b2 )2 + 4h22

and a1 + b1 = a2 + b2 ∴ (a1 − b1 )2 + 4h12 − (a1 + b1 )2 = (a2 − b2 )2 + 4h22 − (a2 + b2 )2 ⇒

− 4 a1b1 + 4h12 = − 4 a2 b2 + 4 h22

⇒

h12 − a1b1 = h22 − a2 b2

⇒

a1b1 − h12 = a2 b2 − h22

20. Let the origin be shifted to ( x1, y1 ). Then, x = X + x1 and y = Y + y1 On substituting x = X + x1 and y = Y + y1 in ax 2 + 2 hxy + by 2 + 2 gx + 2 fy + c = 0, we get a( X + x1 )2 + 2 h ( X + x1 ) (Y + y1 ) + b(Y + y1 )2 ⇒

Targ e t E x e rc is e s

x = X + h and y = Y + k On substituting x = X + h, y = Y + k in the equation y 2 + 4 y + 8 x − 2 = 0, we get (Y + k )2 + 4 (Y + k ) + 8 ( X + h ) − 2 = 0 2 ⇒ Y + (4 + 2 k )Y + 8 X + (k 2 + 4k + 8 h − 2 ) = 0

Cartesian System of Rectangular Coordinates

14. Let the origin be shifted to (h, k ). Then,

+ 2 g( X + x1 ) + 2 f (Y + y1 ) + c = 0 aX 2 + 2 hXY + bY 2 + 2 X (ax1 + hy1 + g ) + 2 Y (hx1 + by1 + f ) + ax12 + 2 hx1 y1

+ by12 + 2 gx1 + 2 fy1 + c = 0 This equation will be free from the terms containing X, Y and constant term, if …(i) ax1 + hy1 + g = 0 …(ii) hx1 + by1 + f = 0 and ax12 + 2 hx1 y1 + by12 + 2 gx1 + 2 fy1 + c = 0 …(iii) Now, ax12 + 2 hx1 y1 + by12 + 2 gx1 + 2 fy1 + c = 0 ⇒

x1(ax1 + hy1 + g ) + y1(hx1 + by1 + g )

+ (gx1 + fy1 + c ) = 0 x1 × 0 + y1 × 0 + gx1 + fy1 + c = 0 [using Eqs. (i) and (ii)] …(iv) ⇒ gx1 + fy1 + c = 0 On solving Eqs. (i) and (ii) by cross-multiplication, we get hg − af hf − bg , y1 = x1 = ab − h 2 ab − h 2  hf − bg hg − af  Hence, the origin must be shifted at  , .  ab − h 2 ab − h 2  ⇒

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11

21. Let the axes be turned through an angleα. If the axes are turned through an angle α, then x = X cos α − Y sin α and y = X sin α + Y cos α On substituting these values in Ax + By + C = 0, we get A( X cos α − Y sin α ) + B ( X sin α + Y cos α) + C = 0 ⇒ X ( A cos α + B sin α ) + Y (B cos α − A sin α) + C = 0 …(i) If this equation is of the form X = constant, then [coefficient of Y = 0] B cos α − A sin α = 0 B  B tan α = ⇒ α = tan − 1   ⇒  A A

22. Substituting the coordinates of points (1, 3), (5, 0) and

Ta rg e t E x e rc is e s

(− 1, 2 ) in 3 x + 2 y, we obtain the values 9, 15 and 1 which are all positive. Therefore, all the points lying inside the triangle formed by given points satisfy 3 x + 2 y ≥ 0. Substituting the coordinates of the given points in 2 x + 3 y − 13, we find the values − 2, − 3 and − 9 which are all negative. So, (b) is not correct. Again, substituting the given points in 2 x − 3 y − 12, we get −19, − 2, − 20 which are all negative. It follows that all points lying inside the ∆ formed by given point satisfy 2 x − 3 y − 12 ≤ 0. So, (c) is the correct answer. Finally, substituting the coordinates of the given point in − 2 x + y, we get 1, − 10 and 4 which are not all positive. So, (d) is not correct.

23. If the remaining vertex is (h, k ), then

24. Let the third vertex be ( x, y ). Then, x y 1 1 −4 0 1 =±4 2 1 −1 1 ⇒ ⇒ or

x + 5y + 4 = ± 8 x + 5 y + 12 = 0 x + 5y − 4 = 0

25. According to given data, 2 a − 5 + a2 − 3 < 0 ⇒ a2 + 2 a − 8 < 0 ⇒(a − 2 ) (a + 4) < 0 ⇒ a ∈ (− 4, 2 ) 3 26. A and P lie on same side of BC ⇒α ∈  − , − 1 on same   2 side of CA 1  1  α ∈  − ∞,  ∪  , ∞ on same side of AB, if  2  3 1  α ∈ (− ∞, − 1) ∪  , ∞ taking intersection, we get 3  result.

27. Put in formula

c2 1 1 = = sq unit. 2 ab 2 4 × 1 8

28. Since, the point (0, 0 ) is equidistant from the lines,

Y

therefore the circumcentre is (0, 0 ) and in equilateral triangle, circumcentre and incentre coincides.

(3, 5)

(–2, 1)

h − 6 = 3 − 2, k − 4 = 5 + 1 h = 7, k = 10

⇒

29. Area of triangle is unaltered by shifting origin to any

X'

point. If origin is shifted to (1000, 1002). A, B, C become A′ (0, 0 ), B′ (1, 2 ) and C′ (2, 1). Both are true. 1 1 30. Area of CDFE A = 1 − x 2 − (1 − x ) 2 2

X

O (h, k) (– 6, – 4)

Y

Y'

h − 2 = − 6 + 3, k + 1 = 5 − 4 h = − 1, k = 0

⇒

D (0,1)

C (1, 1)

Y F(0, x)

(–2, 1)

(3, 5)

X'

(0, 0) A

X

X

E (x, 0) B (1,0)

O

2 − x2 − 1 + x 1 + x − x2 = 2 2 1 1 1+ − 2 4 = 5 at x = 1 = 2 8 2

= (– 6, – 4)

Y'

A max

( h , k)

h + 3 = − 6 − 2, k + 5 = − 4 + 1 h = − 11, k = − 8

⇒

31.

Y D (0,1)

(h, k) Y

β P

(3, 5) X'

(–2

,

1)

α A

X

O

(0, 0)

⇒

(– 6, – 4)

620

C (1, 1)

Y'

γ

δ

B (1,0)

X

(PA)2 − (PB)2 + (PC )2 − (PD )2 = (α 2 + γ 2 ) − (α 2 + δ 2 ) + (δ 2 + β 2 ) − (γ 2 + β 2 ) = 0

Also, OA = 1, OB = OC = 1 / 2 . Hence, the incentre is   1 1  1 1  1  1   0   + (1) + 0   0   + (1) + 1          2  2 2 2 2 2  , 1 1 1 1   + 1+ + 1+   2 2 2 2  

C

Q (1, y) A

1 1 y(1) = (1) ⇒ 2 3

y=

2 3

34. We know that the area of the triangle formed by joining

 2 Length of AQ = (1)2 +    3

∴

33.

 1 1 = ,   2 + 2 2 2

X

B

2

=

13 3

A (0, 1) B (1/2, 1/2)

X

O Y'

35. One side of the triangle is parallel to the Y-axis and another side is parallel to the X-axis. So, the triangle is a right angled triangle. Hence, the mid-point of the hypotenuse is the circumcentre. Solving, x = 2, x + 2 y = 4, we get one end of the hypotenuse and solving y + 1 = 0, x + 2 y = 4, we get the other end. Their coordinates are (2, 1) and (6, − 1).  2 + 6 1 − 1 Circumcentre =  ∴ ,   2 2 

Y

X'

the mid-points of any triangle is one-fourth of area of that triangle. Therefore, required area is 8.

⇒

The given lines are x( x + y − 1) ( x − y ) = 0. In other words, lines x = 0, x + y − 1 = 0 and x − y = 0 form ∆OAB as shown in the above diagram. The triangle is right angled at point B, hence orthocentre is (1 / 2, 1 / 2 ). Also, circumcentre is mid-point of OA which is (0, 1 / 2 ). The centroid is 1 1    0 + + 0 0 + + 1  1 1 2 2 ,  = ,   3 3   6 2   

λ+µ=4

36. Given, vertices of triangle are

O (0, 0 ), B(6, 8) and

C(− 4, 3). Clearly, BC 2 = OB 2 + OC 2 ∴ ∆OBC is right angled at O. Now, circumcentre = mid-point of hypotenuse  11 BC = 1,  and orthocentre = vertex O(0, 0 )  2 121 5 5  units ∴ Required distance = 1 +  =  4  2

Entrances Gallery 1. Required points ( x, y ) are such that it satisfy x + y < 41 and x > 0, y > 0 Number of positive integral solution of the equation x + y + k = 41will be number of integral coordinates in the bounded region.

11 Cartesian System of Rectangular Coordinates

Y D

Targ e t E x e rc is e s

32.

Aliter Consider the following figure: Y (0,41) (1,40) (2,39)

Y

1 point

(0,41)

(40,1) X'

(0,0)O x = 1 x = 2 x= 40 (41,0)

39 points X

Y' X'

X

(0,0) O

(41,0)

Y'

∴Total number of integral coordinates =

41−1

=

40 ! = 780 2 ! 38 !

C3 −1 =

40

C2

Clearly, the number of required points 39 (39 + 1) = 780 = 1 + 2 + 3 + ... + 39 = 2

2. Let point of intersection be (h, − h). ⇒ So,

4 ah − 2 ah + c = 0  5 bh − 2 bh + d = 0 c d − =− ⇒ 3 bc − 2 ad = 0 2a 3b

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Objective Mathematics Vol. 1

11

3. Given, mid-points of sides of a triangle are (0, 1), (1, 1) and (1, 0). Plotting these points on a graph paper and making a triangle. So, the sides of a triangle will be 2, 2

( x − 1)2 + y 2

( x + 1)2 + y 2 1 x = ,2 ⇒ 2 1   + 2  5  2 ∴Circumcentre =  , 0 =  , 0  2  4   

Y C(0, 2)

(1, 1)

2 (0, 1)

9k = 3 − 6h 2 h + 3k = 1 2 x + 3y = 1

6. According to the question,

22 + 22 = 2 2

and

⇒ ⇒ or

=

1 3

7. Since, A (h, k ), B(1, 1) and C(2, 1) are the vertices of a right X′

B (0, 0)

A(2, 0)

(1, 0) 2

X

angled ∆ABC. Y A (h, k)

Y′

x-coordinate of incentre = =

2 × 0 + 2 2 ⋅ 0 + 2 ⋅2 2+2+2 2 2 2− 2 =2 − 2 × 2+ 2 2− 2

O

AB = (1 − h )2 + (1 − k )2

Now,

BC = (2 − 1)2 + (1 − 1)2 = 1

Ta rg e t E x e rc is e s

4. Given line L : 2 x + y = k passes through point (say P) which divides a line segment (say AB) in the ratio 3 : 2, where A(1, 1) and B(2, 4). Using section formula, the coordinates of the point P, which divides AB internally in the ratio 3 : 2, are  3 × 2 + 2 × 1 3 × 4 + 2 × 1  8 14 P ,  ≡P ,  5 5   3+ 2 3+ 2 Also, since the line L passes through P, hence  8 14 substituting the coordinates of P  ,  in the 5 5 equation of L : 2 x + y = k, we get  8  14 2  +  =k  5  5  ⇒

CA = (h − 2 )2 + (k − 1)2

and

From Pythagoras theorem, AC 2 = AB 2 + BC 2 2 ⇒ 4 + h − 4h + k 2 + 1 − 2 k = h2 + 1 − 2h + k 2 + 1 − 2k + 1 …(i) ⇒ 5 − 4h = 3 − 2 h ⇒ h = 1 Now, given that area of triangle is 1. 1 Then, area of ∆ABC = × AB × BC 2 1 ⇒ 1 = × (1 − h )2 + (1 − k )2 × 1 2 ... (ii) 2 = (1 − h )2 + (1 − k )2 ⇒ [from Eq. (i)] 2 = (k − 1)2 ⇒ ⇒ 4 = k 2 + 1 − 2k ⇒ k 2 − 2 k − 3 = 0 ⇒ (k − 3) (k + 1) = 0 ∴ k = − 1, 3 Thus, the set of values of k is { − 1, 3}.

k=6

5. The third vertex lies on 2 x + 3 y = 9  9 − 2 x   x,  3 

i.e.

C (2, 1) X

B (1, 1)

A (2, –3)

8. The graph of equations x − 2 y = 0 and 3 x − y = 0 is

shown in the figure. Since, given point (a, a2 ) lies in the shaded region. a Then, a2 − > 0 and a2 − 3 a < 0 2 Y

B (–2, 1)

622

(

C x, 9 – 2x 3

)

∴ Locus of centroid is 9 − 2x   − 3 + 1+ 2 − 2 + x 3  = (h, k ) ,   3 3     x 3−2x h= ,k = ∴ 3 9 ⇒ 9 k = 3 − 2 (3 h )

y = 3x y= X'

O

x 2 X

Y'

⇒ and

1  a ∈ (− ∞, 0 ) ∪  , ∞ 2  1  a ∈(0,3) ⇒ a ∈  , 3 2 

The coordinates of B are 1 + x2 − 1= 2 i.e. x2 = − 3 and coordinates of C are

1 + y2 and 2 = 2 and y2 = 3

1 + x3 2 1 + y3 2= 2 x3 = 5 and y3 = 3 3=

and i.e. A (1, 1)

(–1, 2)D

12. Since, the vertices of the triangle are (a cos t , a sin t ), (b sin t , − b cos t ) and (1, 0). Let the coordinate of centroid be a cos t + b sin t + 1 x= 3 …(i) ⇒ 3 x − 1 = a cos t + b sin t a sin t − b cos t + 0 and y= 3 …(ii) ⇒ 3y = a sin t − b cos t On squaring both sides and adding Eqs. (i) and (ii), we get (3 x − 1)2 + (3 y )2 = a2 (cos 2 t + sin 2 t ) + b2 (sin 2 t + cos 2 t ) 2 2 2 2 ⇒ (3 x − 1) + (3 y ) = a + b

13. Let the vertices of ∆ABC be A(4, 0 ), B(−1, − 1)andC(3, 5).

E (3, 2)

AB = (− 1 − 4)2 + (−1 − 0 )2

Now,

11 Cartesian System of Rectangular Coordinates

9. Let D and E be the mid-points of AB and AC.

= 25 + 1 = 26

∴ Centroid of triangle  x + x2 + x3 y1 + y2 + y3  , = 1    3 3  1 − 3 + 5 1 + 3 + 3  7  = ,  = 1,     3 3 3

10. Let ( x, y ) be the coordinates of vertex C and ( x1, y1 ) be the

BC = (3 + 1)2 + (5 + 1)2 = 42 + 62 = 16 + 36 = 52 CA = (4 − 3)2 + (0 − 5)2 = 1 + 25 = 26

and ∴

AC 2 + AB 2 = ( 26 )2 + ( 26 )2

= 26 + 26 = 52 = BC 2 Hence, CA + AB 2 = BC 2 Thus, the triangle is isosceles and right angled triangle. 2

coordinates of centroid of the triangle. x + 2 −2 y − 3+ 1 and y1 = x1 = ∴ 3 3 x y −2 and y1 = …(i) ⇒ x1 = 3 3 Since, the centroid lies on the line 2 x + 3 y = 1. So, point ( x1, y1 ) satisfies the equation of line. ∴ 2 x1 + 3 y1 = 1 2 x 3( y − 2) [from Eq. (i)] ⇒ + =1 3 3 ⇒ 2 x + 3y − 6 = 3 ⇒ 2 x + 3y = 9 Hence, this is the required equation of locus of the vertex C.

14. Let the vertices of ∆ABC be A(1, 3 ), B(0, 0 ) and C(2, 0 ).

11. Let P (α, β) be the point which is equidistant to A(a1, b1 )

  p at the point A  , 0 and the Y-axis at the point  cos α  p   B  0,  . Let (h, k ) be the coordinates of the  sin α  mid-point of the line segment AB. p Then, h= 2 cos α p and k= 2 sin α p ⇒ cos α = 2h p and sin α = 2k p2 p2 =1 sin 2 α + cos 2 α = 2 + ⇒ 4h 4k 2 Hence, locus of the point (h, k ) is 1 1 4 + 2 = 2. x2 y p

and B(a2 , b2 ). ∴ PA = PB ⇒ (α − a1 )2 + (β − b1 )2 = (α − a2 )2 + (β − b2 )2 ⇒

α 2 + a12 − 2αa1 + β 2 + b12 − 2βb1 = α 2 + a22 − 2αa2 + β 2 + b22 − 2βb2

⇒

2(a2 − a1 ) α + 2(b2 − b1 )β + (a12 + b12 − a22 − b22 ) = 0 Thus, the equation of locus (α, β) is 1 (a2 − a1 )x + (b2 − b1 )y + (a12 + b12 − a22 − b22 ) = 0 2 But the given equation is (a2 − a1 )x + (b2 − b1 )y − c = 0 1 c = − (a12 + b12 − a22 − b22 ) ∴ 2 1 = (a22 + b22 − a12 − b12 ) 2

Again, let a = BC = (2 − 0 )2 + (0 − 0 )2 = 2 b = AC = (2 − 1)2 + (0 − 3 )2 = 2 and

c = AB = (0 − 1)2 + (0 − 3 )2 = 2

Since, all sides of a triangle are equal, therefore the triangle is an equilateral triangle. Also, incentre is same as centroid of the triangle. Hence, coordinates of incentre are  1 + 0 + 2 3 + 0 + 0  1 ,   i.e. 1, .  3 3 3  

Targ e t E x e rc is e s

B (–3, 3)

C (5, 3)

15. The straight line x cos α + y sin α = p meets the X-axis

623

Similarly,

x2 = 0, y2 =

and

x3 = −

C (2, − 2 ). A (8, 6)

Objective Mathematics Vol. 1

11

16. Let the vertices of a triangle be A (8, 6), B(8, − 2 ) and

B (8, –2)

19. Let A (ae , 0 ) and B (−ae , 0 ) be two given points and (h, k )

C (2, –2)

be the coordinates of the moving point P. Now, PA + PB = 2 a

AB = (8 − 8)2 + (6 + 2 )2 = 0 + 82 = 8

⇒

BC = (2 − 8) + (− 2 + 2 ) 2

and

2

= 36 + 0 = 6 CA = (8 − 2 )2 + (6 + 2 )2

[(h − ae )2 + k 2 ] − [(h + ae )2 + k 2 ] = − 2eh 2 [(h − ae )2 + k 2 ] = 2(a − eh )

Hence, the required circumcentre is (5, 2).

AM b = MB a (b cos α, b sin α) (a cos β, a sin β)

QR = (− 2 − 4)2 + (6 + 1)2 = 85 RP = (− 2 − 2 )2 + (6 − 7 )2 = 17

So, using external division, ba cos β − ab cos α x= b−a

18. Let ( x1, y1 ), ( x2 , y2 ) and ( x3 , y3 ) be coordinates of the points D, E and F which divide each AB, BC and CA, respectively in the ratio 3 : 1 internally. A (–1, 4)

F (x3, y3)

E (x2, y2)

C (–2, 4)

3 × 6 − 1 × 1 17 = 4 4 −2 × 3+ 4×1 y1 = 4 2 1 =− =− 4 2

x1 =

B

(x, y) M

ab (cos β − cos α ) b−a ba sin β − ab sin α ab and y = = (sin β − sin α) b−a b−a ab  α + β  α + β ∴ x cos  cos   (cos β − cos α )  =  2   2  b−a =

ab  α + β  α + β and y sin  sin   (sin β − sin α )  =  2   2  b−a ∴

D ( x 1 , y1 )

and

20. Q

A

∴ PQ 2 + RP 2 = QR 2 So, the triangle formed by joining the given points is right angled triangle, right angled at P .

∴

On squaring both sides, we get (h − ae )2 + k 2 = (a − eh )2 h2 k2 ⇒ + 2 =1 2 a a (1 − e 2 ) x2 y2 Hence, locus of P is 2 + 2 = 1. a a (1 − e 2 )

PQ = (4 − 2 )2 + (− 1 − 7 )2 = 2 17

B (6, –2)

…(iii)

On adding Eqs. (i) and (iii), we get

= 64 + 36 = 100 = AC 2 So, ABC is a right angled triangle and right angled at B. We know that, in a right angled triangle, the circumcentre is the mid-point of hypotenuse.  8 + 2 6 − 2 , ∴ Mid-point of AC =   = (5, 2 )  2 2 

and

…(i)

On dividing Eq. (ii) by Eq. (i), we get

2

= 36 + 64 = 100 = 10 Now, AB 2 + BC 2 = (8)2 + (6)2

17. Q

(h − ae )2 + k 2 + (h + ae )2 + k 2 = 2 a

But we know that, [(h − ae )2 + k 2 ] − [(h + ae )2 + k 2 ] = − 4aeh …(ii)

= 6 +8 2

Ta rg e t E x e rc is e s

5 , y3 = 4 4

17 5 − +0 12 x1 + x2 + x3 For centroid, x = = 4 4 = =1 3 12 3 1 5 − + +4 6 y + y2 + y3 and y= 1 = 2 2 = =2 3 3 3 Hence, ( x, y ) = (1, 2 )

Now,

624

5 2

ab  α + β  α + β x cos   =  + y sin   2  b−a  2 

 β − α  β + α  α + β  cos  2  ⋅ − 2 sin  2  sin  2       β − α   β + α  α + β  + sin    sin   2 cos   2    2   2   ab  β − α = − sin(α + β) sin    2  b − a  β − α  + sin(α + β) sin   =0  2  

C( x3 , y3 ), then

BC = (4 − 2 )2 + (8 − 4)2 = 20 = 2 5 A (x1, y1)

and CA = (4 − 1)2 + (8 − 2 )2 = 45 = 3 5 Now, AB + BC = CA [Q 5 + 2 5 = 3 5] Hence, A, B and C are forming a straight line.

E (4, –3)

(4, 5) F

25. Given, vertices of ∆ABC are the vertices of right angled (x2, y2)B

and

triangle, right angled at A. In a right angled triangle, A is orthocentre and mid-point of BC is D (1, 1) which is the circumcentre.

C (x3, y3)

D (–2, 3)

x1 + x2 = 8 y1 + y2 = 10 x2 + x3 = − 4 y2 + y3 = 6 x3 + x1 = 8 y3 + y1 = − 6

…(i) …(ii) …(iii) …(iv) …(v) …(vi)

∴Required distance = AD = (1 − 0 )2 + (1 − 0 )2 = 2

26. From the intersection of given lines, we get the vertices of ∆ABC as A (0, 0 ), B (3, − 3) and C (3, 3). Y C (3, 3)

On solving these equations, we get x1 = 10, x2 = − 2, x3 = − 2, y1 = − 1, y2 = 11 and y3 = − 5  5 Hence, the centroid is 2,  .  3 Aliter As we know that the centroid of the ∆ABC and that of the triangle formed by joining the mid-points of the sides of ∆ABC is same. Therefore, required centroid  4 + 4 − 2 5 − 3 + 3  5 = ,  ≡ 2,     3 3 3

22. Let D be the mid-point of AB, then coordinates of D are (−1, − 1).

A (2, 2) D (–1, –1)

∴

X'

B (– 4, – 4)

CD = ( − 1 − 5)2 + (−1 + 8)2 = 36 + 49 = 85

23. Let the line divides the line joining the points (− 1, 1) and (5, 7) in the ratio k : 1and at ( x1 , y1 ). 5k − 1 x1 = ∴ k+1 7k + 1 and y1 = k+1 Since, ( x1, y1 ) satisfies the equation x + y = 4. 5k − 1 7 k + 1 ∴ + =4 k+1 k+1 ⇒ 5k − 1 + 7 k + 1 = 4k + 4 ⇒ 8k = 4 ⇒ k = 1/ 2 ∴Required ratio = 1 : 2, internally

X

(0,0) A

B (3, –3) Y'

Now, and Now,

AB = 9 + 9 = 18, BC = 36 = 6 CA = 9 + 9 = 18 CA2 + AB 2 = BC 2

∴ ∆ABC is right angled at A. Hence, orthocentre is (0, 0). 1 27. Area of ∆ACD = 2

C (5, –8)

11 Cartesian System of Rectangular Coordinates

24. AB = (2 − 1)2 + (4 − 2 )2 = 5

−3

4

1

5

6

1

x

−4 1

1 = |[− 3 (6 + 4) − 4 (5 − x ) + 1 (− 20 − 6 x )]| 2 1 = |[− 30 − 20 + 4 x − 20 − 6 x ]| 2 1 = |[−2 x − 70 ]| = | x + 35| 2 Similarly, area of ∆ABD = |3 x + 1| (given) Area of ∆ABD 2 ∴ = Area of ∆ACD 1 3x + 1 2 =± ⇒ x + 35 1 −71 ⇒ x = 69 or x = 5 28. The equations of the locus of the point of intersection of given straight lines are …(i) x sin θ + (1 − cos θ ) y − a sin θ = 0 …(ii) x sin θ − (1 + cos θ ) y + a sin θ = 0 On subtracting Eq. (ii) from Eq. (i), we get y (1 − cos θ + 1 + cos θ ) = 2 a sin θ ⇒ 2 y = 2 a sin θ ⇒ y = a sin θ ∴ x sin θ + a sin θ (1 − cos θ − 1) = 0 ⇒ x sin θ = a sin θ cos θ ⇒ x = a cos θ

Targ e t E x e rc is e s

21. Let the vertices of the triangle are A( x1, y1 ), B( x2 , y2 ) and

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Objective Mathematics Vol. 1

11

y x and cos θ = a a y2 x2 + 2 =1 2 a a x 2 + y 2 = a2

∴

sin θ =

⇒ ⇒

29. Let P (h, k ) be the mid-point of AB, whose length is 8 units. Y B (0, 2k) P (h, k) X'

(0, 0) O

X A (2h, 0)

Y'

∴ (2 h − 0 ) + (0 − 2 k )2 = 82 ⇒ 4h 2 + 4k 2 = 64 ⇒ h 2 + k 2 = 16 So, locus of the required point is x 2 + y 2 = 16. 2

30. Distance

Ta rg e t E x e rc is e s

= a2 (cos α − cos β )2 + a2 (sin α − sin β )2 sin 2 α + cos 2 α + cos 2 β + sin 2 β − 2 cos α cos β  =a   − 2 sin α sin β    α − β = a 2{1 − cos (α − β )} = 2 a sin    2  Aliter Put a = 1, α =

π π and β = , then the points will be (0, 1) 2 6

 3 1 and  ,  . Obviously, the distance between these  2 2 two points is 1 which is given by option (d). 1  α −β (π /2 ) − (π / 6)  Q 2 a sin = 2 × 1 × sin =2 × =1   2 2 2

31. In a rectangle ABCD, the diagonals are equal. So, AC = BD Now, ⇒

length of BD = 32 + 42 = 5 AC = 5

32. Since, the area of triangle formed by these points is zero, therefore the points are collinear.

33. Clearly, x = 2 y satisfy the given condition.

626

34. The given points are collinear, if t12 2 t1 1 at12 2 at1 1 2 2 2 at 2 2 at 2 1 = 0 ⇒ a t 2 2 t 2 1 = 0 0 1 1 0 a 1 ⇒

t12 (2 t 2 − 1) − 2 t1 (t 22 ) + 1 (t 22 ) = 0

⇒

2 t12 t 2 − t12 − 2 t1 t 22 + t 22 = 0

⇒ ⇒ ⇒ or

2 t1 t 2 (t1 − t 2 ) + (t 2 − t1 ) (t 2 + t1 ) = 0 (t1 − t 2 ) (2 t1 t 2 − t1 − t 2 ) = 0 t1 = t 2 t1 + t 2 = 2 t1 t 2

35. (2 + 5)2 + (3 − 2 )2 = ( x − 1)2 + (2 − 3)2 ⇒ ⇒ ⇒

49 + 1 = ( x − 1)2 + 1 x − 1= ± 7 x = − 6, 8

36. Since, A (0, 1), B (0, − 1) and C ( x, 0 ) are the vertices of an equilateral ∆ABC. ∴ ⇒ ⇒ ⇒ 3q

37. 0 1 ⇒

AB = BC 0 + 4 = x2 + 1 x2 = 3 x=± 3

0

1

3 p 1 = 0 ⇒ 3 q (3 p − 1) + 1 (0 − 3 p) = 0 1 1 1 1 9 pq = 3 p + 3 q ⇒ + =3 p q

38. Area of triangle =

1 2

2 2 1 5 5 1

6 7 1 1 = |[2 (5 − 7 ) − 2 (5 − 6) + 1 (35 − 30 )]| 2 1 3 = |(−4 + 2 + 5)| = sq units 2 2

39. Given, PQ = RS = ST P (a, x) Q

R

S

T (b, y)

 a + b x + y Mid-point of PT is R  , .  2 2   3 a + b 3x + y ∴Point Q is mid-point of PR i.e. Q  , .  4 4   5 a + 3 b 5x + 3y Hence, mid-point of QR is  , .  8 8 

12 Straight Line and Pair of Straight Lines Straight Line When we say that a first degree equation in x, yi.e. ax + by + c = 0 represents a line it means that all points ( x, y) satisfying a x + by + c = 0 lie along a line. Thus, a line is also defined as the locus of a point satisfying the condition ax + by + c = 0, where a, b and c are constants. It follows from the above discussion that a x + by + c = 0 is the general equation of a line. It should be noted that, in general equation of a line, there are only two unknowns, because equation of every straight line can be put in the form a x + by + 1 = 0, where a and b are two unknowns. Note that x and y are not unknowns. Infact, these are the coordinates of any point on the line and are known as the current coordinates. Thus, to determine a line, we will need two conditions to determine the two unknowns. In the further discussion on straight line, you will find that whenever it will be asked to find a straight line, there will always be two conditions connecting the two unknowns.

Chapter Snapshot ●

Straight Line

●

Angle between Two Lines

●

●

●

●

●

Slope (Gradient) of a Line The trigonometrical tangent of the angle that a line makes with the positive direction of the X -axis in anti-clockwise sense is called the slope or gradient of the line. The slope of a line is generally denoted by m. Thus, m = tan θ

Y

Y

●

B

B

●

θ

X′ A

X

θ

X′ O

O

●

X

A ●

Y′

Y′

Point of Intersection of Two Lines Image of a Point with Respect to a Line Family of Lines through the Intersection of Two Given Lines Locus and its Equation Combined Equation of a Pair of Straight Lines Bisectors of the Angle between the Lines Given by a Homogeneous Equation General Equation of Second Degree Equations of the Angle Bisectors Distance between the Pair of Parallel Lines

Objective Mathematics Vol. 1

12

Since, a line parallel to X -axis makes an angle of 0° with X -axis. Therefore, its slope is tan 0° = 0. A line parallel to Y -axis i.e. perpendicular to X -axis makes an π angle of 90° with X -axis, so its slope is tan = ∞. 2 Also, the slope of a line equally inclined with axes is 1 or −1 as it makes an angle of 45° or 135° with X -axis.

X

Sol. Let θ be the angle between the given lines. Then, m1 =Slope of the line joining (0, 0) and (2, 3) 3−0 3 = = 2−0 2 and m2 = Slope of the line joining (2, − 2 ) and (3, 5) 5+2 =7 = 3−2

Ø The angle of inclination of a line with the positive direction of

X-axis in anti-clockwise sense always lies between 0° and 180°. X

Example 3. Find the angle between the lines joining the points (0, 0), (2, 3) and the points (2, − 2), (3, 5).

Example 1. Slope of a line making an angle of 30° with Y -axis and lie in Ist quadrant and passing through origin, is 1 1 3 (b) (c) (d) 3 (a) 2 2 3

 m − m1  tan θ = ±  2   1 + m1m2 

∴

 3   7−  2   =±  1 + 7 3     2   11 11 =± 2 =± 23 23 2 11 θ = tan−1  ±   23 

Sol. (d) Since, line AB makes an angle of 30° with Y-axis. Hence, the inclination of line with X-axis is 60°. Y B

⇒ 30° 60° O(0, 0)

X′

Y′

A

∴

Condition of Parallelism and Perpendicularity of Two Lines

X

Slope = m = tan 60° =

i. 3

Slope of a Line in Terms of Coordinates of any Two Points on it Let ( x1 , y1 ) and ( x 2 , y2 ) be coordinates of any two points on a line, then its slope m is given by y − y1 Difference of ordinates m= 2 = x 2 − x1 Difference of abscissae X

Example 2. The value of x, so that 2 is the slope of the line through (2, 5) and ( x, 3), is (a) 1 (b) 2 (c) −1 (d) −2

⇒ m2 = m1 Thus, when two lines are parallel, their slopes are equal. Ø Three points A(x1 , y1), B(x 2 , y 2) and C (x3 , y3 ) are collinear, if slope of

AB = slope of BC , i.e.

Sol. (a) Slope of the line through (2, 5) and ( x, 3), m= ⇒

3−5 3−5 ⇒ 2= x−2 x−2

x=1

Angle between Two Lines The angle θ between the lines having slopes m1 and m2 is given by  m − m1  tan θ = ±  2   1 + m1 m2 

628

and the acute angle between the lines is given by m − m1 tan θ = 2 1 + m1 m2

Condition of parallelism of lines If two lines of slopes m1 and m2 are parallel, then the angle θ between them is of 0°. ∴ tan θ = tan 0° = 0 m2 − m1 ⇒ =0 1 + m1 m2

X

y 2 − y1 y3 − y 2 . = x 2 − x1 x3 − x 2

Example 4. The value of y, so that the line through (3, y) and (2, 7) is parallel to the line through ( −1, 4) and (0, 6), is (a) 5 (b) −5 (c) 9 (d) −9 Sol. (c) Let P(3, y), Q(2, 7 ), R(−1, 4) and S(0, 6) be the given points. Then,

7− y = y−7 2−3 6−4 m2 = Slope of the line RS = =2 0 − (−1)

m1 = Slope of the line PQ =

Since, PQ and RS are parallel, therefore m1 = m2 ⇒ y − 7 = 2 ⇒ y=9

The equation of a line parallel to X -axis at a distance b from it, is y = b.

Example 5. If three points A (h, 0), P (a, b) and a b B (0, k ) lie on a line, show that + =1. h k

Y y=b P (x,y)

Sol. Since, the points A(h, 0 ), P (a, b)and B(0, k ) lie on a line.

b

∴The points A(h, 0), P(a, b ) and B(0 , k ) are collinear. Slope of PA = Slope of PB ⇒ b−0 k−b ⇒ = a−h 0−a ⇒ ⇒ ⇒

Condition of perpendicularity of lines If two lines of slopes m1 and m2 are perpendicular, then the angle θ between them is of 90°. ∴ cot θ = 0 1 + m1 m2 =0 ⇒ m1 − m2

Since, X -axis is parallel to itself at a distance 0 from it. Therefore, the equation of X -axis is y = 0. If a line is parallel to X-axis at a distance b and below X -axis, then its equation is y = − b. X

Example 7. Determine the equation of line through the point ( −4, − 3) and parallel to X -axis. Sol. We know, the equation of line which is parallel to X-axis is of the form y = k. Here, line is passing through (−4, − 3). Therefore, we have −3 = k Thus, the required equation of line is y = − 3 or y + 3 = 0

⇒ m1 m2 = − 1 Thus, when two lines are perpendicular, the product of their slopes is −1. If m is the slope of a line, then the slope of a line  1 perpendicular to it is  −  .  m X

ii.

Example 6. A(4, 4), B (3, 5) and C(−1, − 1) are the vertices of (a) an equilateral triangle (b) a right angled triangle (c) an isosceles triangle (d) None of the above

Line parallel to Y -axis Let AB be a line parallel to Y -axis and at a distance a from it. Then, the abscissa of every point on AB is a. So, it can be treated as the locus of a point at a distance a from Y-axis. Thus, if P ( x, y) is any point on AB, then x = a. The equation of a line parallel to Y -axis at a distance a from it, is x = a. Since, Y-axis is parallel to itself at a distance 0 from it, therefore the equation of Y-axis is x =0. Y P (x, y)

Sol. (b) In ∆ABC,

5−4 = −1 3−4 −1 − 4 =1 m2 = Slope of AC = −1 − 4

a

m1 = Slope of AB =

i.

Line parallel to X -axis Let AB be a straight line parallel to X -axis at a distance b from it. Then, the ordinate of each point on AB is b. Thus, AB can be considered as the locus of a point at a distance b from X -axis and if P ( x, y) is any point on AB, then y = b.

X

X' x=a

Clearly, m1m2 = − 1, this shows that AB ⊥ AC π i.e. ∠CAB = 2 Hence, the given points are the vertices of a right angled triangle.

Lines Parallel to the Coordinate Axes

X

O

Y'

ab = (a − h)(b − k ) ab = ab − ak − bh + hk ak + bh = hk a b + =1 h k

∴

ii.

X'

12 Straight Line and Pair of Straight Lines

X

Y'

If a line is parallel to Y -axis at a distance a and to the left of Y-axis, then its equation is x = − a. X

Example 8. Find the equation of line, which is parallel to Y-axis and at a distance 3 units from left of the origin. (a) x = 3 (b) y = − 3 (c) x = − 3 (d) y = 3 Sol. (c) We know, the equation of line which is parallel to Y-axis is of the form x = k. Here, we take k = − 3, because it is left side of the origin. ∴ x= −3

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Objective Mathematics Vol. 1

Forms of the Equation 12 Different of a Straight Line

Thus, for the straight line Ax + By + C = 0, Coefficient of x A Slope ( m) = − = − B Coefficient of y

The equation of a line in the general form can also be written as a x + by = c. In this section, we shall discuss some standard forms of a line. These standard forms will also be compared with the general form.

i.

The slope intercept form of a line The equation of a line with slope m and making an intercept c on Y -axis is y = mx + c

and intercept on Y -axis Constant term C =− =− B Coefficent of y X

Sol. Let m be the slope of the line

Y

(k − 3)x − (4 − k 2 )y + k 2 − 7 k + 6 = 0 Coefficient of x k−3 −(k − 3) Then, = m= − = Coefficient of y −(4 − k 2 ) 4 − k 2

c X′

θ

As the line is parallel to X-axis, therefore we have k−3 =0 4 − k2

X

O

Example 10. Find the value of k for which the line is ( k − 3) x − ( 4 − k 2 ) y + k 2 − 7k + 6 = 0 parallel to X -axis.

⇒

iii.

and the equation of a line with slope m and x-intercept d is y = m( x − d ).

Example 9. The equation of a straight line which cut off an intercept of 5 units on negative direction of Y -axis and make an angle of 120° with positive direction of X -axis, is (a) y + 3x − 5 = 0 (b) y + 3x + 5 = 0 (c) y − 3x + 5 = 0 (d) y − 3x − 5 = 0 Sol. (b) Here, given that m = tan120° = − 3 and c = − 5 So, the equation of the line is given by y = − 3 x − 5 ⇒ y + 3 x + 5 = 0.

ii.

630

Reduction of general form to slope intercept form The general form of the equation of a line is Ax + By + C = 0 ⇒ By = − Ax − C  A  C y = −  x + −  ⇒  B  B This is of the form y = mx + c, A C where, m = − and c = − ⋅ B B

Example 11. Find the equation of the perpendicular bisector of the line segment joining the points A(2, 3) and B ( 4, − 5). Sol. The slope of AB is given by m=

−5 − 3 = −4 4−2

∴Slope of a line perpendicular to AB =

−1 1 = m 4

Let P be the mid-point of AB. Then, the coordinates of P 2 + 4 3 − 5 are  ,  i.e. (3, − 1).  2 2  Thus, the required line passes through P(3, − 1) and has 1 slope . So, its equation is 4 1 y + 1 = ( x − 3) ⇒ 4 y + 4 = x − 3 ⇒ x − 4 y − 7 = 0 4

iv.

The two point form of a line The equation of a line passing through two points ( x1 , y1 ) and ( x 2 , y2 ) is  y − y1  y − y1 =  2  ( x − x1 ) ⋅  x 2 − x1  Y B (x2, y2)

X' 1)

X

X

O

y

●

If the line passes through the origin, then 0 = m0 + c ⇒ c = 0 Therefore, the equation of a line passing through the origin is y = mx , where m is the slope of the line. If the line is parallel to X-axis, then m = 0, therefore the equation of a line parallel to X-axis is y = c.

1,

●

(x

Ø

The point-slope form of a line The equation of a line which passes through the point ( x1 , y1 ) and has the slope m is y − y1 = m( x − x1 ).

A

Y′

k−3=0 ⇒ k=3

Y'

X

Sol. (a) Let the equation of the line be

Example 12. Find the equation of the line joining the points ( at12 , 2at1 ) and ( at 22 , 2at 2 ).

x y + =1 a 2a Since, it passes through the point (1, 2). So, 2 + 2 = 2a ⇒ a=2 ∴ The required equation is 2 x + y = 4.

Sol. Here, x1 = at 12 , y1 = 2 at 1, x2 = at 22 , y2 = 2 at 2 So, the equation of the required line is 2 at 2 − 2 at 1 y − 2 at 1 = ( x − at 12 ) at 22 − at 12 2 ( y − 2 at 1 ) = ( x − at 12 ) ⇒ t1 + t 2 ⇒ y (t 1 + t 2 ) − 2 at 12 − 2 at 1 t 2 = 2 x − 2 at 12 ⇒

v.

y (t 1 + t 2 ) = 2 x + 2 at 1 t 2

The equation of a line passing through two points ( x1 , y1 ) and ( x 2 , y2 ) can also be written in the determinant form as x y 1 y1 1 = 0 y2 1

x1 x2 X

Example 13. The equation of the line passing through the points P (2, 3) and Q( 4, 5) is (a) 2x + y − 1 = 0 (b) 2x − y − 1 = 0 (c) x + y + 1 = 0 (d) x − y + 1 = 0 Sol. (d) The equation of the line through P(2, 3) and Q(4, 5) is x

y 1

2 3 1 =0 4 5 1

Example 15. The equation of the line through (2, 3), so that the segment of the line intercepted between the axes is bisected at this point, is (a) 3x − 2 y = 12 (b) 3x + 2 y = 12 (c) x − 2 y = 12 (d) 3x − y = 12 Sol. (b) Let the equation of the line be x + y = 1, which meets

a b the X and Y-axes at A (a, 0) and B (0, b ). The coordinates of a b the mid-point of ABare  ,  . Since, the point (2, 3) bisects 2 2  AB. Therefore, a b = 2 and =3 2 2 ⇒ a=4 and b=6 x y Hence, the equation of the line is + = 1 4 6 or 3 x + 2 y = 12

vii.

⇒ x (3 − 5) − y (2 − 4) + 1 (10 − 12 ) = 0 ⇒ −2 x + 2 y − 2 = 0 ⇒ −x + y − 1= 0 ⇒ x − y + 1= 0

vi.

X

The intercept form of a line The equation of a line which cuts off intercepts a and b x y respectively on the X and Y -axes is + =1. a b

Reduction of general equation of a line to intercept form The general equation of a line is Ax + By+C = 0. y x ⇒ + =1 C  C  −  −   A B Proof ⇒

Y B

⇒

b X'

A a

O

X

12 Straight Line and Pair of Straight Lines

X

Ax + By = − C By Ax + =1 −C −C y x + =1  C  C − −      A  B 

x y + =1. a b Thus, for the straight line Ax + By + C = 0, we have Constant term C Intercept on X -axis = − = − A Coefficient of x Constant term C Intercept on Y -axis = − = − B Coefficient of y This is of the form

Y' X

Example 14. The intercept made by a line on Y -axis is double to the intercept made by it on X -axis, if it passes through (1, 2), then its equation is (a) 2x + y = 4 (b) 2x + y + 4 = 0 (c) 2x − y = 4 (d) 2x − y + 4 = 0

Ø The intercept made by a line on X-axis can also be obtained by

putting y = 0 in its equation. Similarly, y-intercept is the value of y obtained from the line when x is replaced by zero.

631

X

Objective Mathematics Vol. 1

12

Example 16. Transform the equation of the line 3x + y − 8 = 0 to intercept form and find intercepts on the coordinate axes.

ix.

3 x+ y−8=0 3x + y = 8 y 3 ⇒ x+ =1 8 8 x y + =1 ⇒ 8 8 3 This is the intercept form of the given line. 8 and y-intercept = 8 So, x-intercept = 3

Sol. We have, ⇒

viii.

Reduction of the general equation of a line to the normal form The general equation of a line is Ax + By + C = 0 Let its normal form be x cos α + y sin α = p. A B C = = cos α sin α − p Ap Bp and sin α = − ⇒ cos α = − C C 2 2 Q cos α + sin α = 1 Therefore,

⇒

The normal or perpendicular form of a line The equation of the straight line upon which the length of the perpendicular from the origin is p and this perpendicular makes an angle α with X -axis is x cos α + y sin α = p.

⇒

Y

A 2 p2 C2

p=

α O

X

A

Y' X

Example 17. The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150° with the positive direction of Y -axis. Then, the equation of the line is (a) 3 x + y = 14 (b) 3 x − y = 14 (c) 3x − y = 14 (d) None of the above

| C| ⋅ C

A2 + B 2

and

sin α = −

| C| ⋅ C

A + B2

150° Q

Y′

⇒

632

⇒

x 3 y + =7 2 2 3 x + y = 14

B 2

 A   A2 + B 2 

  B  x+  A2 + B 2   

 y      

(Coefficient of x ) 2 + (Coefficient of y) 2

60° O

A

This is the normal form of the line Ax + By + C = 0 It follows from the above discussion that to reduce the general equation of a line to normal form, we first shift the constant term on the RHS and make it positive if it is not, so and then divide both sides by

Y

X′

A2 + B 2

 −C =  A2 + B 2 

∴ Equation of the required line is xcos 30° + y sin 30° = 7

7 30°

=1

cos α = −

⇒

Sol. (a) Given that, p = 7, α = 30°

30°

C2 | C|

On putting the values of cos α, sin α and p in x cos α + y sin α = p, we get  | C|  By | C| Ax  − = −    A2 + B 2 A2 + B 2  C A2 + B 2 By −C Ax ⇒ + = A2 + B 2 A2 + B 2 A2 + B 2

p X'

B 2 p2

∴

B P

+

X

X

Example 18. The normal form of the equation x − 3 y + 8 = 0 is (a) x cos 120° − y sin 120° = 4 (b) x cos 120° + y sin 120° = 4 (c) x cos 120° + y sin 120° = 8 (d) None of the above

x− ⇒

x−

⇒ Now, divide by

X

3 y+ 8=0

−x +

3 y=−8 3 y=8

(Coefficient of x)2 + (Coefficient of y)2

Example 19. A line is drawn from the point P (α, β) making an angle θ with the positive direction of X -axis to meet the line a x + by + c = 0 at Q. The length of PQ is (a) −

= (−1)2 + ( 3 )2 = 1 + 3 = 2,we get 1 3 8 x+ y= 2 2 2 ⇒ − cos 60° x + sin 60° y = 4 [Qconvert in form of x cos α + y sinα = p] ⇒ x cos (180° − 60° ) + y sin (180° − 60° ) = 4 Qcos x is negative and sin x is positive,  which is possible in second quadrant    −

⇒

x.

(c)

Y

x, P(

y)

r y 1) x 1, θ ( A

X'

aα + bβ + c a cos θ + b sin θ

(b)

aα + bβ + c a 2 + b2

(d) None of these

Sol. (a) Equation of a straight line passing through the point P(α, β ) and making an angle θ with positive direction of X-axis is x−α y−β [say] = =r cos θ sinθ

x cos 120° + y sin120° = 4

The distance form of a line The equation of the straight line passing through ( x1 , y1 ) and making an angle θ with the positive direction of X -axis is x − x1 y − y1 = =r cos θ sin θ where, r is the distance of the point ( x, y) on the line from the point ( x1 , y1 ).

aα + bβ + c a cos θ + b sin θ

12 Straight Line and Pair of Straight Lines

Sol. (b) Given equation of line is

Now, coordinates of any point at a distance r from P(α,β ) on this line are (α + r cos θ, β + r sin θ), If it lies on the line ax + by + c = 0, then a (α + r cos θ) + b ( β + r sin θ) + c = 0 aα + bβ + c ⇒ r=− a cos θ + b sin θ Thus,

X

PQ = r = −

aα + bβ + c a cos θ + b sin θ

Example 20. The equation of the line is passing through P ( 4, 5) and making 30° with X -axis. Then, coordinates of point which is at a distance 4 units on either side of P, are (a) ( 4 − 2 3, 7) (b) ( 4 ± 2 3, 7) (c) ( 4 + 2 3, 7), ( 4 − 2 3, 3)

θ

X

O Y'

(d) ( 4 − 2 3, 7), ( 4 − 2 3, 7) Sol. (c) Since, the equation of the line passing through P(4, 5) and making an angle 30°.

Ø

●

●

If the equation of the line is x − x1 y − y1 = =r cos θ sin θ Then, x − x1 = r cosθ and y − y1 = r sinθ ⇒ x = x1 + r cosθ and y = y1 + r sinθ Thus, the coordinates of any point on the line at a distance r from the given point (x1 , y1) are (x1 + r cos θ , y1 + r sin θ). If P is on the right side of (x1 , y1), then r is positive and if P is on the left side of (x1 , y1), then r is negative. Since, different values of r determine different points on the line, therefore the above form of the line is also called parametric form or symmetric form of a line. In the above form, we can determine the coordinates of any point on the line at a given distance from the given point through which it passes. At a given distance r from the point x − x1 y − y1 , there are two points viz. (x1 , y1) on the line = cos θ sin θ (x1 + r cos θ , y1 + r sin θ) and (x1 − r cos θ , y1 − r sin θ)

∴ Point on the line at a distance 4 from (4, 5) is given by x−4 y−5 = =± 4 cos 30° sin 30° x = 4 ± 4 cos 30° and

y = 5 ± 4 sin 30°

⇒

x=4±2 3

and y=5±2 The coordinates of point are (4 + 2 3, 7 ) and (4 − 2 3, 3).

Position of Two Points Relative to a Given Line In this section, we shall see how to check whether two given points are on the same side or opposite sides of a given line. The two points ( x1 , y1 ) and ( x 2 , y2 ) are on the same (or opposite) sides of the straight line ax + by + c = 0 according as the quantities ax1 + by1 + c and ax 2 + by2 + c have the same (or opposite) signs.

633

Objective Mathematics Vol. 1

12

Ø

●

●

X

A point (x1 , y1) will lie on the side of the origin relative to a line ax + by + c = 0, if ax1 + by1 + c and c have the same sign. A point (x1 , y1) will lie on the opposite side of the origin relative to the line ax + by + c = 0, if ax1 + by1 + c and c have the opposite sign.

Sol. (a) The essential condition for the points on the same side or opposite side of the line is a x1 + by1 + c and a x2 + by2 + c are of same sign or opposite sign, respectively. Let c = 3 x − 4 y − 8, then value of c1 at (2, 3) is given by c1 = 3 × 2 − 4 × 3 − 8 ⇒ c1 = 6 − 12 − 8 ⇒ c1 = − 14 and c 2 at (−4, 5) is given by c 2 = 3 × (−4) − 4 × 5 − 8 = − 40 Since, c1 and c 2 are of same sign. Therefore, the two points are on the same sides of the given line.

Example 21. The given line 3x − 4 y = 8, the points (2, 3), ( −4, 5) are on the (a) same sides (b) opposite sides (c) lie on the line (d) None of the above

Work Book Exercise 12.1 1 Consider the equation y − y1 = m( x − x1 ) . In this

6 A square is constructed on the portion of the line x + y = 5, which is intercepted between the axes on the side of the line away from origin. The equations to the diagonals of the square are

equation, if m and x1 are fixed and different lines are drawn for different values of y1, then a b c d

the line will pass through a single point there will be one possible line only there will be a set of parallel lines None of the above

a c

c

b 13 3

10 4

 13 , 0    5  c ( −7, 0)

d None of these

8 The larger of the two angles made with X-axis of a straight line drawn through (1, 2), so that it 6 intersects x + y = 4at a distance from (1, 2), is 3

vertices of a square, then the diagonal through B is 3)

a

105°

b 75°

c 60°

d 15°

9 If two vertices of a triangle are (5, − 1) and (− 2, 3)

4 If each of the points ( x1, 4) and (−2, y1 ) lies on the

and its orthocentre lies at the origin, then the coordinates of the third vertex are

line joining the points (2, − 1 ) and (5, − 3), then the point P( x1, y1 ) lies on the line

a

a x = 3y b x = − 3y c y = 2x + 1 d 2 x + 6y + 1 = 0

(4, 7)

b ( − 4, − 7 ) c (2, − 3)

d ( 5, − 1)

10 The equations of the line on which the perpendiculars from the origin make 30° angle 50 3 with axes, are

with X-axis and which form a triangle of area

5 All points lying inside the triangle formed by the points (1, 3), (5, 0 ) and (−1, 2 ) satisfy a 3x + 2 y ≥ 0 c 2 x − 3 y + 12 ≤ 0

5 b  , 0  13  d None of these

a

3 If A(1, 1), B( 3 + 1, 2 ) and C( 3, 3 + 2 ) are three a y = ( 3 − 2 ) x + (3 − b y=0 c y= x d None of the above

x = 5, y = 5 x − y = 5, x − y = − 5

reflected at a point A on the X-axis and then passes through the point (5, 3). The coordinates of the point A are

and (7, 6), then the length of the portion of the line AB intercepted between the axes is 5 4

b d

7 A ray of light coming from the point (1, 2) is

2 If the coordinates of the points A and B are (3, 3)

a

x − 5, y = − 5 x = − 5, y = 5

b 2 x + y − 13 ≥ 0 d −2 x + y ≥ 0

a c

x+ x±

3 y ± 10 = 0 3 y − 10 = 0

b 3 x + y ± 10 = 0 d None of these

Point of Intersection of Two Lines Let the equations of two lines be …(i) a1 x + b1 y + c1 = 0 and …(ii) a 2 x + b2 y + c2 = 0 Then, the coordinates of the point of intersection of Eqs. (i) and (ii) are  b1 c2 − b2 c1 c1 a 2 − c2 a1  ,  ⋅  a1 b2 − a 2 b1 a1 b2 − a 2 b1  634

Ø To find the coordinates of the point of intersection of two

non-parallel lines, we solve the given equations simultaneously and the values of x and y, so obtained determine the coordinates of the point of intersection. X

Example 22. Find the value(s) of m for which the lines and mx + (2m + 3) y + m + 6 = 0 (2m + 1) x + ( m − 1) y + m − 9 = 0 intersect at a point on Y -axis.

...(i) m x + (2 m + 3) y + m + 6 = 0 and ...(ii) (2 m + 1) x + (m − 1) y + m − 9 = 0 Solving these two by cross-multiplication method, we have x (2 m + 3) (m − 9) − (m − 1) (m + 6) y = (2 m + 1) (m + 6) − m (m − 9) 1 = m (m − 1) − (2 m + 1) (2 m + 3) x y 1 = = ⇒ 2 m − 20 m − 21 m2 + 22 m + 6 −3 (m2 + 3m + 1) ⇒

x=

and

y=

m2 − 20 m − 21 −3 (m2 + 3 m + 1)

This is the required condition of concurrency of three lines. Another condition Three lines L 1 ≡ a1 x + b1 y + c1 = 0 L2 ≡ a 2 x + b2 y + c2 = 0 and L3 ≡ a 3 x + b3 y + c3 = 0 are concurrent iff there exist constants λ 1 , λ 2 and λ 3 not all zero such that λ 1 L1 + λ 2 L2 + λ 3 L3 = 0 i.e. λ 1 ( a1 x + b1 y + c1 ) + λ 2 ( a 2 x + b2 y + c2 ) + λ 3 ( a 3 x + b3 y + c3 ) = 0 X

m2 + 22 m + 6 −3 (m2 + 3 m + 1)

So, given lines intersect at point  m2 − 20m − 21 m2 + 22 m + 6    , 2 2  −3 (m + 3m + 1) −3 (m + 3 m + 1) It lies on Y-axis. ∴

X

m2 − 20m − 21 −3 (m2 + 3m + 1)

=0

⇒

m2 − 20 m − 21 = 0

⇒ ⇒

(m − 21) (m + 1) = 0 m = − 1, 21

Example 23. The number of integer values of m for which the x-coordinate of the point of intersection of the lines 3x + 4 y = 9 and y = mx +1 is also an integer, is (a) 2 (b) 0 (c) 4 (d) 1

Sol. (c) Given lines x + 2 y − 3 = 0 and 3 x + 4 y − 7 = 0 intersect at (1, 1), which does not satisfy 2 x + 3 y − 4 = 0 and Also, 4 x + 5 y − 6 = 0. 3 x + 4 y − 7 = 0 and 2 x + 3 y − 4 = 0 intersect at (5, − 2 ), which does not satisfy x + 2 y − 3 = 0 and 4 x + 5 y − 6 = 0. Lastly, intersection point of x + 2 y − 3 = 0 and 2 x + 3 y − 4 = 0 is (−1, 2 ) which satisfy 4 x + 5 y − 6 = 0. Hence, three lines are concurrent. X

Sol. (a) On solving equation 3 x + 4 y = 9 and y = m x + 1, we get x=

5 3 + 4m

1

⇔

b1 b2

c1 c2 = 0

a3

b3

c3

0 −k − m

0 1 m −1 ⇒

2 0

=0

m2 + km + 2 = 0

...(i)

Since, m is real, the discriminant of Eq. (i) is greater than or equal to zero (0). ⇒ k2 ≥ 8

Condition of Concurrency of Three Lines

a1 a2

Example 25. The least value of | k |, so that the lines x = k + m, y = − 2 and y = mx are concurrent, is (b) 2 (a) 2 2 (c) 4 2 (d) None of these Sol. (a) Since, the given lines are concurrent, therefore

Now, for x to be an integer, 3 + 4 m = ± 5 or ± 1 The integral values of m satisfying these conditions are −2 and −1.

Three lines are said to be concurrent, if they pass through a common point i.e. they meet at a point. Thus, if three lines are concurrent the point of intersection of two lines lies on the third line. ...(i) Now, the lines a1 x + b1 y + c1 = 0 ...(ii) a 2 x + b2 y + c2 = 0 are concurrent and a 3 x + b3 y + c3 = 0 ...(iii)

Example 24. Given the four lines with the equations x + 2 y − 3 = 0, 3x + 4 y − 7 = 0, 2x + 3 y − 4 = 0, 4x + 5 y − 6 = 0, then (a) they are all concurrent (b) they are the sides of a quadrilateral (c) three lines are concurrent (d) None of the above

12 Straight Line and Pair of Straight Lines

Sol. The equations of the lines are

or |k | ≥ 2 2 Thus, the least value of|k | is 2 2 . X

Example 26. Show that the following lines are concurrent L1 ≡ ( a − b) x + ( b − c) y + ( c − a ) = 0, L2 ≡ ( b − c) x + ( c − a ) y + ( a − b) = 0 and L3 ≡ ( c − a ) x + ( a − b) y + ( b − c) = 0. Clearly, λ1L 1 + λ 2 L2 + λ 3 L3 = 0 where, λ1 = λ 2 = λ 3 = 1 Hence, the given lines are concurrent.

Sol.

635

Objective Mathematics Vol. 1

Parallel and Perpendicular to a 12 Lines Given Line i.

Distance of a Point from a Line

Equation of a line parallel to a given line The equation of a line parallel to a given line a x + by + c = 0 is a x + by + λ = 0, where λ is a constant.

Ø To write a line parallel to a given line, we keep the expression

The length of the perpendicular from a point ( x1 , y1 ) to a line ax + by + c = 0 is ax1 + by1 + c a 2 + b2 Ø

●

containing x and y same and simply replace the given constant by a new constant λ. The value of λ can be determined by some given conditions. ●

X

Example 27. The equation of a straight line parallel to 2x + 3 y + 11 = 0 and which is such that the sum of its intercept on the axes is 15, is (a) 2 x + 3 y − 18 = 0 (b) 2 x − 3 y − 18 = 0 (c) 2 x + 3 y + 18 = 0 (d) 2 x − 3 y + 18 = 0

X

Sol. (a) The equation of the line parallel to the line 2 x + 3 y + 11 = 0 is 2 x + 3y + λ = 0 where, λ is a constant.

...(i)

λ The intercept by the line on X-axis is given by  −  and  2 λ on the Y-axis is given by  −  .  3 It is given that the sum of the intercepts on the axes is 15.  − λ  +  − λ  = 15 ∴      2   3  5λ ⇒ − = 15 ⇒ λ = − 18 6 On putting λ = − 18 in Eq. (i), we get 2 x + 3 y − 18 = 0 Hence, the equation of the required line is 2 x + 3 y − 18 = 0.

ii.

X

Equation of a line perpendicular to a given line The equation of a line perpendicular to a given line ax + by + c = 0 is bx − ay + λ = 0, where λ is a constant and its value can be determined by some given condition.

Example 28. The equation of the line perpendicular to the line 2x + 3 y + 5 = 0 and passing through (1, 1), is (a) 3x + 2 y + 1 = 0 (b) 3x − 2 y − 2 = 0 (c) 3x − 2 y − 1 = 0 (d) 3x − y − 1 = 0 Sol. (c) Equation of line perpendicular to the given line is

636

3x − 2 y + λ = 0 Since, it passes through point (1, 1). ∴ 3−2 + λ = 0 ⇒ λ = −1 On putting the value of λ in Eq. (i), we get 3x − 2 y − 1 = 0

...(i)

.

The length of the perpendicular from the origin to the line | c| ax + by + c = 0 is . 2 a + b2 Distance between parallel lines ax + by + c1 = 0 and |c − c | ax + by + c2 = 0 is 2 1 . a2 + b 2

Example 29. The distance of point (−1, 1) from the line 12x − 5 y + 9 = 0 is 13 8 (b) (a) 8 13 5 8 (d) − (c) 13 13 Sol. (b) Required distance = −12 − 5 + 9 = 8 (12 )2 + (5)2

X

13

Example 30. The distance between the parallel lines 3x − 4 y + 9 = 0 and 6x − 8 y − 15 = 0, is 33 10 33 33 (a) − (b) (c) (d) 10 33 10 20 Sol. (c) Given equations of lines are ...(i) 3x − 4y + 9 = 0 and ...(ii) 6 x − 8 y − 15 = 0 Eq. (ii) can be rewritten as 15 ...(iii) =0 3x − 4y − 2 On comparing Eqs. (i) and (iii) with ax + by + c1 = 0 and ax + by + c 2 = 0, we get 15 a = 3, b = − 4, c1 = 9, c 2 = − 2 Now, distance between parallel lines 33 15 − −9 − |c 2 − c1| 33 2 = = = 2 = 2 2 2 2 25 10 (3) + (−4) a + b

Area of a Parallelogram Let ABCD be a parallelogram. The equations whose sides are AB , BC , CD and DA are a1 x + b1 y + c1 = 0, and a1 x + b1 y + d1 = 0 a 2 x + b2 y + c2 = 0, a 2 x + b2 y + d 2 = 0. Then, Area =

( c1 − d1 ) ( c2 − d 2 ) a1 b1 a 2 b2

The image of ( x1 , y1 ) on ax + by + c = 0 be ( x 2 , y2 ), x − x1 y2 − y1 2 (ax1 + by1 + c) given by 2 = =− ⋅ a b a 2 + b2

mx − y = 0, nx − y = 0, mx − y + 1 = 0 and nx − y + 1 = 0

Required area =

X

Particular Cases

(c1 − d1 ) (c 2 − d 2 ) , we get a1 b1 a2

i.

The image of the point P ( x1 , y1 ) with respect to X-axis is ( x1 , − y1 ).

ii.

The image of the point P ( x1 , y1 ) with respect to Y-axis is ( −x1 , y1 ).

iii.

The image of the point P ( x1 , y1 ) with respect to the line mirror y = x is Q ( y1 , x1 ).

iv.

The image of the point P ( x1 , y1 ) with respect to the line mirror y = x tan θ is given by x = x1 cos 2θ + y1 sin 2θ, y = x1 sin 2θ − y1 cos 2θ.

v.

The image of a point P ( x1 , y1 ) with respect to the origin is the point ( − x1 , − y1 ).

b2

(0 − 1) (0 − 1) 1 1 = = m −1 |− m + n| |m − n| n −1

Example 32. Prove that the four straight lines x y x y x y x y + = 1, + = 1, + = 2 and + = 2 form a a b b a a b b a rhombus. Find its area. Sol. The equations of the four sides are x + a x + b x + a x + b

and

y =1 b y =1 a y =2 b y =2 a

...(i) ...(ii) ...(iii) ...(iv)

Clearly, Eqs. (i), (iii) and Eqs. (ii), (iv) form two sets of parallel lines. So, the four lines form a parallelogram. Let p1 be the distance between parallel Eqs. (i) and (iii) and p2 be the distance between parallel Eqs. (ii) and (iv). Then, p1 =

2 −1 = 1 1 + a2 b2

ab a2 + b 2

 Q d =  

  a2 + b 2 

c1 − c 2

2 −1 ab = 2 1 1 a + b2 + 2 2 a b Clearly, p1 = p2 . So, the given lines form a rhombus. Area of the rhombus (2 − 1) (2 − 1) a2 b 2 1 = = = 2 1 1 1 1   |b − a2|  2 − 2 a a b b  1 1 b a and

p2 =

= c + by B (x 2, y 2)

Sol. (d) The given equation of lines can be rewritten as

Now, using the formula

0

A (x 1, y 1)

12 Straight Line and Pair of Straight Lines

Image of a Point with Respect to a Line

+

Example 31. Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = nx and y = nx +1 equals | m + n| (a) ( m − n) 2 2 (b) | m + n| 1 (c) | m + n| 1 (d) | m − n|

ax

X

X

Example 33. Find the image of the point (3, 8) with respect of the line x + 3 y = 7 assuming the line to be plane mirror. (a) (1, 4) (b) ( −1, − 4) (c) ( −1, 4) (d) (1, − 4) Sol. (b) Given point ( x1, y1 ) is (3, 8) and line x + 3 y − 7 is 0. By using the relation x − x1 y − y1 −2(ax1 + by1 + c ) , we get = = a b a2 + b 2 x−3 y−8 = 1 3 −2 (1 × 3 + 3 × 8 − 7 ) = 12 + 32 x − 3 y − 8 −2(20) ⇒ = = 1 3 10 x−3 y−8 = ⇒ 1 3 =−4 ⇒ x = − 4 + 3, y = − 12 + 8 ⇒ x = − 1, y = − 4

637

diagonals, so equation of the sides through (1, 1) are given by −1 ± tan 45° ( x − 1) y − 1= 1 m (−1)tan 45°

with a Given Line

pass through a point ( x1, y1 ) and made a given angle α with the given straight line y = mx + c are m ± tan α ( x − x1 ) y − y1 = 1 m m tan α Here, x1 = 7, y1 = 9, α = 60° and m = Slope of the line x − 3 y − 2 3 = 0 1 So, m= 3 Now, the equations of the required lines are 1 + tan 60° ( x − 7) y−9= 3 1 tan 60° 1− 3 1 − tan 60° 3 and ( x − 7) y−9= 1 tan 60° 1+ 3 ⇒

1 1 ( y − 9)  1 − tan 60°  =  + tan 60°  ( x − 7 )    3  3

and

1 1 ( y − 9)  1 + tan 60°  =  − tan 60°  ( x − 7 )    3  3 1 0 =  +  3

⇒ ⇒ ⇒

3  ( x − 7 ) 

1 − x − 7 = 0 and ( y − 9) (2 ) =   3 x+

3  ( x − 7 ) 

3 y=7 + 9 3

Hence, the required lines are x = 7 and x + 3 y = 7 + 9 3 X

Example 35. If (1, 1) and (−3, 5) are vertices of a diagonal of a square, then the equations of its sides through (1, 1) are (a) 2x − y = 1, y − 1 = 0 (b) 3x + y = 4, x − 1 = 0 (c) x = 1, y = 1 (d) None of these

x = 1, y = 1

or

Slope of a Line Equally Inclined to Two Given Lines The slope m of a line which is equally inclined with two lines of slopes m1 and m2 , is given by m − m1 m −m = 2 . 1 + mm1 1 + mm2 Ø The above equation in m always gives two values of m which are

the slopes of the lines parallel to the bisectors of the angles between the two given lines. X

Example 36. A line 4x + y = 1 through the point A (2, − 7) meets the line BC whose equation is 3x − 4 y + 1 = 0 at the point B. If AB = AC , then the equation of the line AC is (a) 52x − 89 y + 519 = 0 (b) 52x + 89 y − 519 = 0 (c) 52x + 89 y + 519 = 0 (d) None of the above Sol. (c) Since, AB = AC ∴

∠ABC = ∠ACB = θ

[say]

A (2, – 7)

1

Sol. (a) We know that the equations of two straight lines which

⇒

y=

Example 34. The equations of the two straight lines through (7, 9) and making an angle of 60° with the line x − 3 y − 2 3 = 0, are (a) x = 7 and x + 3 y = 7 + 9 3 (b) x = − 7 and x + 3 y = 7 + 9 3 (c) x = 7 and x − 3 y = 7 + 9 3 (d) x = 7 and x + 3 y = 7 + 3

[ Qslope of diagonal, m = − 1] −1 ± 1 ( x − 1) ⇒ x − 1 = 0, y − 1 = 0 y − 1= 1± 1

+

The equations of the straight lines which pass through a given point ( x1 , y1 ) and make a given angle α with the given straight line y = mx + c are m ± tan α ( x − x1 ) y − y1 = 1 m m tan α X

638

Sol. (c) Since, each side of a square makes 45° angle with its

4x

Objective Mathematics Vol. 1

of Straight Lines Passing through 12 Equation a Given Point and Making a Given Angle

B

3x – 4y + 1 = 0

C

3 and slope of AB = − 4 4 Let slope of AC be m. Since, BC is equally inclined with AB and AC. 3 3 + 4 m− 4 4 = 3 3 1+ m 1 − 4⋅ 4 4 19 3 − 4m ⇒ = 8 4 + 3m Now, slope of BC =

⇒

76 + 57 m = 24 − 32 m 52 ⇒ m= − 89 52 ( x − 2) ∴ Equation of AC is ( y + 7 ) = − 89 ⇒ 52 x + 89 y + 519 = 0

1 The image of the point (−8, 12 ) with respect to the line mirror 4 x + 7 y + 13 = 0 is

a (16, − 2 )

b ( −16, 2 )

c (16, 2 )

8 The equation of the line passing through the intersection of and x − 3y + 3 − 1 = 0 x + y − 2 = 0 and making an angle of 15° with the first line, is

d ( −16, − 2 )

2 The straight line x + 2 y − 9 = 0, 3 x + 5 y − 5 = 0 and ax + by − 1 = 0 are concurrent, if the straight line 35 x − 22 y + 1 = 0 passes through the point a ( a, b )

b ( b, a )

c ( a, − b )

d ( −a, b )

3 The equations of the lines through (−1, − 1) and making angle 45° with the line x + y = 0 are given by a c

x + 1 = 0, y + 2 = 0 x − 1 = 0, y − 1 = 0

b d

a b c d

3 sq units 9 sq units

2 x + 3 y − 14 = 0 measured parallel to the line x − 2 y = 1is

x − 1 = 0, y − 2 = 0 x + 1 = 0, y + 1 = 0

a c

b 12 sq units d 24 sq units

a b c d

a

d

Ax1 + By1 + C A + B Ax1 + By1 + C A cos α Bsinα 2

b 3 x + 2 y − 63 = 0 d None of these

from A( x1, y1 ), B( x2 , y2 ) and C( x3 , y3 ) to a variable line is zero, then the line passes through

b

7 13 13

the X-axis, to meet Ax + By + C = 0 at Q. Then, the lenght PQ is equal to

3 2 concurrent with the lines 4 x + 3 y − 7 = 0 and 8 x + 5 y − 1 = 0, is

6 The algebraic sum of the perpendicular distances

7 5 5

10 A line is drawn from P( x1, y1 ) in the direction α with

5 The equation of the line with gradient − , which is

a 3x + 2 y − 2 = 0 c 2 y − 3x − 2 = 0

3=0

9 The distance of the point (3, 5) from the line

4 The area enclosed 2 | x | + 3| y | ≤ 6 is a c

x− y=0 x − y + 1= 0 y=1 3x − y + 1 −

c

2

b −

Ax1 + By1 + C a cos α + Bsinα

d −

Ax1 + By1 + C A sinα + Bcos α

11 The image of the point P(α, β) with respect to the line y = − x is the point Q and the image of Q with respect to the line y = x is R. Then, the mid-point of R is

the orthocentre of ∆ABC the centroid of ∆ABC the circumcentre of ∆ABC None of these

a (α + β, β + α ) c (α − β, β − α )

b (0, 0) d None of these

7 If the algebraic sum of the perpendicular

12 The equation of a straight line passing through

distances from the points (2, 0 ), (0, 2 ) and (1, 1) to a variable straight lines be zero, then the line passes through the point

(1, 2) and having intercept of length 3 between the straight lines 3 x + 4 y = 24 and 3 x + 4 y = 12 is

a ( −1, 1)

b (1, 1)

c (1, − 1)

Straight Line and Pair of Straight Lines

12

Work Book Exercise 12.2

a 7 x + 24 y − 55 = 0 c 24 x − 7 y − 10 = 0

d ( −1, − 1)

b 24 x + 7 y − 38 = 0 d 7 x − 24 y + 41 = 0

Family of Lines through the Intersection of Two Given Lines The equation of the family of lines passing through the intersection of the lines a1 x + b1 y + c1 = 0 and a 2 x + b2 y + c2 = 0 is ( a1 x + b1 y + c1 ) + λ ( a 2 x + b2 y + c2 ) = 0 where, λ is a parameter. Ø The equation L1 + λL2 = 0 represents a line passing through the

intersecting of the line L1 = 0 and L2 = 0 , which is a fixed point. Hence, L 1 + λ L2 = 0 represents a family of straight lines for different values of λ, which pass through a fixed point.

X

Example 37. The equation of the straight line through the intersection of line 2x + y = 1 and 3x + 2 y = 5 and passing through the origin, is (a) 7x + 3 y = 0 (b) 7x − y = 0 (c) 3x + 2 y = 0 (d) x + y = 0

Sol. (a) The family of line through the intersection of L 1 and L2 is given by L1 + λL2 = 0, where λ is constant. ∴The equation of required line is given by (2 x + y − 1) + λ(3 x + 2 y − 5) = 0

…(i)

Since, it passes through the origin O (0, 0). ∴ ⇒

− 1 − 5λ = 0 λ=−

1 5

On putting value of λ in Eq. (i), we get 2 3 x  2 −  + y  1 −  + (−1 + 1) = 0   5 5 ⇒

7 3 x+ y=0 5 5

⇒

7 x + 3y = 0

639

Objective Mathematics Vol. 1

12 Equation of the Bisectors

Sol. (a) First make the constant terms (c1 and c 2 ) positive in

The equation of the bisectors of the angles between the lines ...(i) a1 x + b1 y + c1 = 0 and ...(ii) a 2 x + b2 y + c2 = 0 a1 x + b1 y + c1 a 2 x + b2 y + c2 are given by =± a12 + b12 a 22 + b22 Acute bisector L1

the two equation i.e. write equation as 12 x − 5 y + 7 = 0 and 4x − 3y + 1 = 0 Now, check a1a2 + b1b2 = 12 × 4 + (−5) × (−3) = 63, which is positive. 12 x − 5 y + 7 4x − 3y + 1 Now, =+ (12 )2 + (−5)2 (4)2 + (−3)2 ⇒ 5 (12 x − 5 y + 7 ) = 13 (4 x − 3 y + 1) So, 4 x + 7 y + 11 = 0 is the required obtuse angle bisector.

Bisector of the Angle Containing the Origin Let the equations of two lines be a1 x + b1 y + c1 = 0 and a 2 x + b2 y + c2 = 0 where, c1 and c2 are positive.

Obtuse bisector L2

...(i) ...(ii)

Y

where, if

i.

ii.

c1 , c2 > 0 anda1 a 2 + b1 b2 > 0, then the bisector corresponding to + sign gives the obtuse angle bisector and the bisector corresponding to ‘−’ sign gives the acute angle bisector. c1 , c2 > 0 and a1 a 2 + b1 b2 < 0, then the bisector corresponding to ‘+’ and ‘−’ signs gives the acute and obtuse angle bisector, respectively.

Ø A line which is equally inclined to given two lines is parallel to the

angle bisectors of the given lines. X

Example 38. The equation of the bisector of the angle between the straight lines 3x − 4 y + 7 = 0 and 12x − 5 y − 8 = 0, is (a) 21x + 27 y + 131 = 0 (b) x + 27 y − 131 = 0 (c) 21x − 27 y + 131 = 0 (d) 21x + 27 y − 131 = 0 Sol. (d) The equation of the bisectors of the angles between 3 x − 4 y + 7 = 0 and 12 x − 5 y − 8 = 0 is 3x − 4y + 7 12 x − 5 y − 8 =± 2 2 3 + (−4) (12 )2 + (−5)2 ⇒ 39 x − 52 y + 91 = ± (60 x − 25 y − 40) Taking the positive sign, 21x + 27 y − 131 = 0 bisector.

X

640

a1x + b1y + c1 = 0

A L

a2x + b2y + c2 = 0

M

X'

X

O P(h, k) Y'

The equation of the bisector containing the origin is given by | a1 x + b1 y + c1 | | a 2 x + b2 y + c2 | = a12 + b12 a 22 + b 22 a1 x + b1 y + c1

⇒

=

a 2 x + b2 y + c2

+ a 22 + b22 since a1 x + b1 y + c1 and a 2 x + b2 y + c2   are either both positive or both negative    a12

b12

Thus, we have the following algorithm to find the bisector of the angle containing the origin :

Algorithm as one

Example 39. The equation of the obtuse angle bisector of lines 12x − 5 y + 7 = 0 and 3 y − 4x − 1 = 0 is (a) 4x + 7 y + 11 = 0 (b) 4x + 7 y − 11 = 0 (c) 7x + 4 y + 11 = 0 (d) 4x + y + 11 = 0

Step I Obtain the two lines, let the lines be a1 x + b1 y + c1 = 0 and +a 2 x + b2 y + c2 = 0. Step II Rewrite the equation of the two lines, so that their constant terms are positive. Step III The bisector corresponding to the positive sign i.e. a1 x + b1 y + c1 a x + b2 y + c2 =+ 2 a12 + b12 a 22 + b22 is the bisector of the angle that contains the origin.

a1 x + b1 y + c1

i.e.

a12

+ b12

=−

Locus and its Equation

a2x + b 2 y + c 2 a22

+ b 22

It is the path or curve traced by a moving point satisfying the given condition.

is the bisector of the angle not containing the origin. X

i.

Example 40. For the straight lines 4x + 3 y − 6 = 0 and 5x + 12 y + 9 = 0, find the equation of the bisector of the angle which contains the origin. Sol. When constants have been made positive, then equation of lines become −4 x − 3 y + 6 = 0 and 5 x + 12 y + 9 = 0 Equation of bisector of the angle containing the origin will be Y

B (0, 2) X

c X′

3 A 2, 0 O D 0, –3 C 4 –9 , 5 0

X

Y′

⇒

Example 42. If P = (1, 0), Q = (−1, 0) and R = (2, 0) are three given points, then locus of the point S satisfying the relation SQ 2 + SR 2 = 2SP 2 , is (a) a straight line parallel to X -axis (b) a circle passing through the origin (c) a circle with the centre at the origin (d) a straight line parallel to Y -axis Sol. (d) Let the coordinates of S be (h, k ).

−4 x − 3 y + 6 5 x + 12 y + 9 = 5 13 7 x + 9y − 3 = 0

⇒ (h + 1) + k + (h − 2 )2 + k 2 = 2 [(h − 1)2 + k 2 ] 2

⇒

Sol. (a) Since, the origin and the point (1, − 3) lie on the same side of x + 2 y − 11 = 0 and on the opposite side of 3 x − 6 y − 5 = 0. Therefore, the bisector of the angle containing(1, − 3) is the bisector of that angle which does not contain the origin and is given by −3 x + 6 y + 5  − x − 2 y + 11 = −     45 5 i.e.

3 x = 19

2

h2 + 2 h + 1 + k 2 + h2 − 4h + 4 + k 2 = 2 (h2 − 2 h + 1 + k 2 )

that the origin lies in the acute angle. Hence, the equation of the bisector of the acute angle is the bisector of the angle between the given lines containing the origin, which is 7 x + 9y − 3 = 0 .

Example 41. The equation of the bisector of that angle between the lines x + 2 y − 11 = 0, 3x − 6 y − 5 = 0 which contains the point (1, − 3) is (a) 3x = 19 (b) 3 y = 7 (c) 3x = 19 and 3 y = 7 (d) None of the above

SQ 2 + SR 2 = 2SP 2

Q

Ø On drawing the graphs of the given straight lines it will be found

X

Equation of the locus of a point The equation of the locus of a point is the algebraic relation which is satisfied by the coordinates of every point on the locus of the point. Step taken to find the equation of locus of a point (a) Assumes the coordinates of the point say ( h, k ) whose locus is to be find. (b) Write the given condition involving ( h, k ). (c) Eliminate the variable(s) (if any). (d) Replace h → x and k → y The equation, so obtained is the locus of the point which moves under some definite condition.

12 Straight Line and Pair of Straight Lines

Ø The bisector corresponding to the negative sign

⇒

2h + 3 = 0

⇒

h=−

−3 3 ⇒ x= 2 2 Hence, it is a straight line parallel to Y-axis.

X

Example 43. If sum of distances of a point from the origin and lines x = 2 is 4, then its locus is (a) x 2 − 12 y = 36 (b) y 2 + 12x = 36 (c) y 2 − 12x = 36 (d) x 2 + 12 y = 36 Sol. (b) Let point be P(h, k ). So, distance from the origin = ∴

h2 + k 2 and distance from the line = (h − 2 ) h2 + k 2 + (h − 2 ) = 4

⇒

h2 + k 2 = (− h + 6)

⇒

h2 + k 2 = h2 + 36 − 12 h

⇒ ⇒

k 2 + 12 h = 36 y2 + 12 x = 36 is the required locus of point.

641

12

Work Book Exercise 12.3

Objective Mathematics Vol. 1

1 The equation of the bisector of the acute angle between the lines 12 x + 5 y − 2 = 0 is a b c d

3x − 4y + 7 = 0

and

21x + 77 y − 101 = 0 11x − 3 y + 9 = 0 31x + 77 y + 101 = 0 11x − 3 y − 9 = 0

a b c d

2 It the family of lines x(a + 2 b) + y(a + 3b) = a + b passes through the point for all values of a and b, then the coordinates of the point are a b c d

5 If u = a1 x + b1 y + c1 = 0 and v = a2 x + b2 y + c 2 = 0

and

a1 b1 c1 = ≠ , then u + kv = 0 represents a2 b2 c 2 a

7 The locus of a point which moves, so the difference of its distance from two fixed straight lines which are at right angles is equal to the distance from another fixed straight line is

a x ( ab ′ − a ′ b ) + (cb ′ − c ′ b ) = 0 b x ( ab ′ + a ′ b ) + (cb ′ + c ′ b ) = 0 c y ( a ′ b − ab ′) + ( a ′c − ac ′) = 0 d None of the above

a c

4 The equation of line bisecting the obtuse angle is

a family of concurrent lines

b a family of parallel lines c u = 0 or v = 0 d None of the above

3 The straight lines through the point of intersection

between and y − x =2 2y + x = 5 y − x − 2 x + 2y − 5 , where n is equal to = n 2

u =0 a family of concurrent lines a family of parallel lines None of the above

6 If u ≡ a1 x + b1 y + c1 = 0, v ≡ a2 x + b2 y + c 2 = 0 and

(2, 1) (2, − 1) ( −2, 1) None of the above

of ax + by + c = 0 and a′ x + b′ y + c ′ = 0 are parallel to the Y-axis has the equation

a1 b1 c1 ≠ = , then u + kv = 0 represents a2 b2 c 2

a straight line a parabola

b a circle d an ellipse

8 A and B are fixed points. The vertex C of ∆ABC moves such that cot A + cot B = constant. The locus of C is a straight line

a 5 b 2 c 5 d None of the above

a b c d

perpendicular to AB parallel to AB inclined at an angle ( A − B) to AB None of the above

Combined Equation of a Pair of Straight Lines The combined equation of the straight line a1 x + b1 y + c1 = 0 and a 2 x + b2 y + c2 = 0 is ( a1 x + b1 y + c1 ) ( a 2 x + b2 y + c2 ) = 0 .

It can be written as 2 x2 − ( y − 9) x − ( y2 + 3 y − 10) = 0

e.g. The combined equation of lines 2x + y + 3 = 0 and x − y + 4 = 0 is (2x + y + 3) ( x − y + 4) = 0 or 2x 2 − xy − y 2 + 11x + y + 12 = 0. Ø

●

●

X

The equation ax 2 + 2hxy + by 2 + 2 gx+ 2 fy + c = 0 is known as the general equation of second degree. Thus, the joint equation of a pair of straight lines is general equation of second degree. In order to find the separate equation of a pair of lines when their joint equation is given, write down the equation of pair of lines as a quadratic equation in x (or y) and from this write down the value of x in terms of y (or value of y in terms of x). Do cross-multiplication and bring all the terms on LHS making RHS equal to zero. The two equations, thus obtained will be the equations of the two lines.

Example 44. Find the separate equation of the lines represented by 2x 2 − xy − y 2 + 9x − 3 y + 10 = 0. Sol. Given equation is

642

2 x2 − xy − y2 + 9 x − 3 y + 10 = 0

...(i)

∴

x=

...(ii)

( y − 9) ± ( y − 9)2 + 4 ⋅ 2 ⋅ ( y2 + 3 y − 10) 4

⇒

4 x = ( y − 9) ±

9 y2 + 6 y + 1

⇒

4 x = y − 9 ± (3 y + 1)

⇒

4 x = 4 y − 8 and

4 x = − 2 y − 10

⇒ x − y + 2 = 0 and 2 x + y + 5 = 0 Hence, the separate equations of the lines represented by Eq. (i) are x− y+2=0 and 2x + y + 5 = 0

Pair of Straight Lines through the Origin Homogeneous equation of second degree A rational, integral, algebraic equation in two variables x and y is said to be a homogeneous equation of the second degree, if the sum of the indices (exponents) of x and y in each term is equal to 2. The general form of the homogeneous equation of the second degree in x and y is ax 2 + 2hxy + by 2 = 0, where a, b and h are constants.

Ø

●

●

●

The homogeneous equation of second degree ax 2 + 2hxy + by 2 = 0 represents a pair of straight lines passing through the origin. These lines are real and distinct, if h2 > ab and coincident, if h2 = ab and the lines are imaginary i.e. they do not exist, if h2 < ab . To find the separate equation of lines proceed as follows : y 2 ➣ Divide by x to obtain quadratic equation in . x ➣ Solve the quadratic equation to get the required lines. If y = m1x and y = m2x are lines represented by the equation ax 2 + 2hxy + by 2 = 0, then 2h Coefficient of xy m1 + m2 = − = − b Coefficient of y 2 m1m2 =

and X

X

Example 46. If the slope of one of the line represented by ax 2 + 2hxy + by 2 = 0 is the square of the other, then a − b 8h 2 a + b 8h 2 (b) (a) + =6 + =6 h ab h ab a − b 8h 2 (c) (d) None of these − =6 h ab Sol. (b) Let mand m2 be the slopes of the lines represented by ax2 + 2 hxy + by2 = 0. Then, m + m2 = −

a Coefficient of x 2 = b Coefficient of y 2

⇒

⇒

⇒

⇒

On comparing the given equation with ax2 + 2 hxy + by2 = 0, we get a = 6, b = 1 and 2 h = − 5 1 25 −6= >0 h2 − ab = ∴ 4 4 ⇒ h2 > ab

⇒ ⇒

Hence, the given equation represents a pair of distinct lines passing through the origin. Now consider, 6 x2 − 5 xy + y2 = 0

⇒ or

1/ 3

a +    b

2/ 3

=−

 a    b

1/ 3

2h b

3   a  1/ 3  a  2 / 3  2h +    =  −     b   b   b   2

1/ 3

2/ 3

a  a a a +   + 3      b  b b  b 1/ 3 2/ 3 8h3 a   a  +    = − 3    b  b  b   2 3 2h 8h a a a + + 3    −  = − 3  b  b  b b2 b   a  1/ 3  a  2 / 3 2h =− +    Q       b b b   8h3 6ah a a2 + 2 + 3 = 2 b b b b a (a + b ) 8h3 6ah + 3 = 2 b2 b b a + b 8h2 + =6 h ab  b2   multiplying both sides by ah  

2

 y − 5  y + 6 = 0      x  x

⇒

or

 a    b

2h and m = b

3

⇒

second degree. So, it represents a pair of straight lines passing through the origin.

⇒

m + m2 = −

⇒

Sol. (a) The given equation is a homogeneous equation of

⇒

a b

Q m + m = − 2 h and m m = a  1 2 1 2  b b 

Example 45. If the equation 6x 2 − 5xy + y 2 = 0 represents a pair of distinct straight lines each passing through the origin. Then, the separate equation of these lines are (a) y − 3x = 0 and y − 2x = 0 (b) y + 3x = 0 and y + 2x = 0 (c) y − 3x = 4 and y − 2x = 0 (d) y − 3x = 0 and y − 2x = 4

⇒

m ⋅ m2 =

and

2h b

12 Straight Line and Pair of Straight Lines

e.g. the equation 2x 2 − xy + 5 y 2 = 0 is a homogeneous equation of second degree, but the equation 2x 2 + 5xy + y 2 − 2x = 0 is not homogeneous equation of second degree.

2

 y − 3  y − 2      x  x  y − 3   x 

[dividing by x2 ]

 y + 6 = 0    x  y − 2 = 0   x  y −3=0 x y −2 = 0 x y − 3x = 0 y − 2x = 0

So, the given equation represents the straight lines y − 3 x = 0 and y − 2 x = 0.

Angle between the Pair of Lines Given by ax 2 + 2 hxy + by 2 = 0 The angle θ between the pair of lines represented by ax + 2hxy + by 2 = 0 is given by 2

tan θ = ±

2 h 2 − ab a+b

and acute angle between the lines is given by tan θ =

2 h 2 − ab a+b 643

X

Objective Mathematics Vol. 1

12

Example 47. The angle between the lines whose joint equation is 2x 2 − 3xy − 6 y 2 = 0, is  53   (a) θ = tan −1   4 

(c) θ = tan

−1

comparing

the

given

equation

2

⇒ ⇒

 57  θ = tan−1  −  4  

a+ b 57 4

2 h 2 − ab a+b

Example 49. Find the value of a for which the lines represented by ax 2 + 5xy + 2 y 2 = 0 are mutually perpendicular.

Bisectors of the Angle between the Lines Given by a Homogeneous Equation The joint equation of the bisectors of the angles between the lines represented by the equation 2 2 ax + 2hxy + by = 0 is x 2 − y 2 xy = a−b h ⇒

=0

h 2 − ab = 0 h = ab 2

The lines represented by ax 2 + 2hxy + by 2 = 0 are coincident iff h 2 = ab.

Example 48. If the lines represented by 2x 2 + 4xy + by 2 = 0 are coincident, then find the value of b. Sol. Since, the lines represented by 2 x2 + 4 xy + by2 = 0 are coincident, therefore we have 22 = 2 ⋅ b b=2

Condition of the Lines to be Perpendicular 644

a + b=0

perpendicular, if a + 2 = 0, i.e. a = − 2.

The lines are coincident, if the angle between them is zero. Thus, lines are coincident. ⇔ θ =0 ⇔ tan θ = 0

⇒

=0

Sol. The lines given by ax2 + 5 xy + 2 y2 = 0 are mutually

Condition for the Lines to be Coincident

X

X

2 h2 − ab

9 2 + 12 4 =m tan θ = ± 2−6

⇔

2 h 2 − ab

The lines represented by ax 2 + 2hxy + by 2 = 0 are perpendicular iff a + b = 0 i.e. coefficient of x 2 + coefficient of y 2 = 0

tan θ = ±

⇔

a+b

π 2

⇔ Coefficient of x + Coefficient of y 2 = 0

Then,

⇔

cot θ = cot

2

with 3 ax + 2 hxy + by = 0, we get a = 2, b = − 6 and h = − 2 Let θ be the angle between the given lines. 2

⇔

⇔

 3    4 

 57  1  (d) θ = tan −1  2  4  Sol. (b) On

θ=

⇔

 57   (b) θ = tan −1  − 4  

π 2

⇔

The lines are perpendicular, if the angle between them is π . Thus, lines are perpendicular. 2

hx 2 − hy 2 = ( a − b) xy

⇒ hx 2 − ( a − b) xy − hy 2 = 0 X

Example 50. The equation of the lines bisecting the angles between the pair of lines 3x 2 + xy − 2 y 2 = 0 is (a) x 2 − 10xy + y 2 = 0 (b) x 2 − xy − y 2 = 0 (c) x 2 + 10xy − 5 y 2 = 0 (d) x 2 − 10xy − y 2 = 0 Sol. (d) On comparing the given equation with ax2 + 2 hxy + by2 = 0, we get 1 and b = − 2 a = 3, h = 2 x 2 − y2 xy Using the formula = , we obtain the equation a−b h of the bisectors as x 2 − y2 xy = 3 − ( −2 ) 1 / 2 ⇒

x2 − 10 xy − y2 = 0

Angle between the Pair of Lines Given by ax 2 + 2 hxy+ by 2 + 2 gx + 2 fy + c = 0

The necessary and sufficient condition for ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 to represent a pair of straight lines is abc + 2 fgh − af 2 − bg 2 − ch 2 = 0

The angle θ between pair of lines represented by ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 can be obtained by using the formula

or

X

a h

h b

g f =0

g

f

c

tan θ = ±

tan θ =

Ø

●

●

Sol. (a) The given equation is of the form ax2 + 2 hxy + by2 + 2 gx + 2 fy + c = 0

●

On comparing the given equation with ax2 + 2 hxy + by2 + 2 gx + 2 fy + c = 0, we obtain 5 5 7 a = 3, h = , b = 2, g = , f = and c = 2 2 2 2 Now, abc + 2 fgh − af 2 − bg 2 − ch2 = 12 + 2 ×

5 5 7 5 × × − 3   − 2  2 2 2 2

2

 5 − 2   2

X

2 h 2 − ab a+b

The lines represented by ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 are perpendicular iff a + b = 0. i.e. coefficient of x 2 + coefficient of y 2 = 0 The lines are coincident, if g2 = ac or f 2 = bc .

Example 53. Find the angle between the pair of straight lines x 2 − 5xy + 4 y 2 + 3x − 4 = 0. Sol. Given equation is x2 − 5 xy + 4 y2 + 3 x − 4 = 0

7 = 0   2

Here,

2 Since, the given equation represents a pair of straight line, therefore ...(i) abc + 2 fgh − af 2 − bg 2 − ch2 = 0 2907 20h2 − 341h + =0 ⇒ 2  h − 17  (20h − 171) = 0 ⇒    2  171 17 or Hence, h= 20 2

…(i)

a = Coefficient of x2 = 1 b = Coefficient of y2 = 4

2h = Coefficient of xy = − 5 5 h=− 2 If θ is the acute angle between the pair of line (i), then 25 2 −4 2 h2 − ab 3 4 = tan θ = = 1+ 4 5 a+ b

∴

Example 52. The value of h, so that the equation 6x 2 + 2hxy + 12 y 2 + 22x + 31 y + 20 = 0 represent two straight lines, is 171 (a) 20 171 (b) 40 170 (c) 21 17 (d) 20 Sol. (a) Here, a = 6, b = 12, g = 11, f = 31 and c = 20

.

The lines represented by ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 a h g are parallel iff h2 = ab and af 2 = bg2 or = = . h b f

2

Hence, the given equation represents a pair of straight lines. X

a+b

and the acute angle between the lines is given by

Example 51. The equation 3x 2 + 7xy + 2 y 2 + 5x + 5 y + 2 = 0 represents (a) a pair of straight lines (b) a circle (c) an ellipse (d) a hyperbola

2

2 h 2 − ab

12 Straight Line and Pair of Straight Lines

General Equation of Second Degree

∴

3 θ = tan−1    5

3 Thus, acute angle between pair of lines (i) is tan−1    5 3 and obtuse angle is π − tan−1   .  5 X

Example 54. If ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 represents parallel straight lines, then (a) hf = bg (b) h 2 = bc (c) a 2 f = b 2 g (d) None of the above Sol. (a) The given equation represents a pair of parallel straight lines, if h2 = ab and bg 2 = af 2 , which on simplification gives hf = bg .

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12 Equations of the Angle Bisectors

Sol. (d) First check for parallel lines a h g = = h b f 3/2 1 −3 = = ⇒ −9 / 2 −3 9 which is true, hence lines are parallel. ∴ Distance between the pair of parallel straight lines i.e.

The equation of the bisectors of the angles between the lines represented by ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 are given by ( x − x1 ) 2 − ( y − y1 ) 2 ( x − x1 ) ( y − y1 ) = a−b h

=2

where, ( x1 , y1 ) is the point of intersection of the lines represented by the given equation.

g 2 − ac a (a + b)

9 + 4 5 =2 4 = 10 2

Ø Two pairs of straight lines are said to be equally inclined, if both

have the same pair of bisectors. X

Example 55. To the lines a x 2 + 2h xy + by 2 = 0 the lines a 2 x 2 + 2h ( a + b) xy + b 2 y 2 = 0 are (a) equally inclined (b) perpendicular (c) bisector of the angle (d) None of the above

X

Sol. Given equation of lines can be rewritten as x2 + (4 y + 3) x + (4 y2 + 6 y − 4) = 0 ⇒ ⇒

Sol. (a) If the two pairs of straight lines have the same bisectors, then the two pairs are equally inclined. The equation of the bisectors of the angle between the lines given by ax2 + 2 hxy + by2 = 0 is x 2 − y2 xy = a−b h

⇒ ⇒

...(i)

⇒ ⇒

The equation of the bisectors of the angle between the lines given by a2 x2 + 2 h (a + b ) xy + b 2 y2 = 0 is x − y 2

⇒

x 2 − y2 xy = a−b h

...(ii)

Clearly, Eqs. (i) and (ii) are the same. The two pairs of straight lines are equally inclined.

Distance between the Pair of Parallel Lines If ax 2 + 2h xy + by 2 + 2g x + 2 f y + c = 0 represents a pair of parallel straight lines, then the distance between them is given by 2

g 2 − ac f 2 − bc or 2 . a ( a + b) b( a + b)

Ø The distance between parallel lines can be found, by finding their

separate equations and then find the distance between them. X

646

∴

2

xy = a2 − b 2 h ( a + b )

Example 56. Distance between two lines represented by the pair of straight lines x 2 − 6xy + 9 y 2 + 3x − 9 y − 4 = 0 is 1 5 (a) (b) 2 2 5 (c) 10 (d) 2

Example 57. Find the distance between the pair of parallel line x 2 + 4xy + 4 y 2 + 3x + 6 y − 4 = 0.

x= x=

− (4 y + 3) ±

(4 y + 3)2 − 4(4 y2 + 6 y − 4

− (4 y + 3) ±

16 y2 + 9 + 24 y − 16 y 2 − 24 y + 16

2

2 − (4 y + 3) ± 25 − 4y − 3 ± 5 ⇒ x= x= 2 2 − 4y − 3 + 5 − 4y − 3 − 5 or x = x= 2 2 x = − 2 y + 1 or x = − 2y − 4 x + 2 y − 1 = 0 or x + 2 y + 4 = 0 Required distance =

|4 − (−1)| 5 = = 5 1+ 4

5

Point of Intersection of Straight Lines If the equation ax 2 + 2h x y + by 2 + 2g x + 2 fy + c = 0 represents a pair of straight lines, then they intersect at  hf − bg gh − af  . the point  ,  ab − h 2 ab − h 2  Ø The point of intersection of the two non-parallel lines can be

found, by finding their separate equations and solve them. X

Example 58. Find the point of intersection of lines represented by 2x 2 − xy − y 2 + 9x − 3 y + 10 = 0. Sol. On comparing the given equation with a x2 + 2 hxy + by2 + 2 gx + 2 f y + c = 0, we get 1 9 3 a = 2, h = − , b = − 1, g = , f = − and c = 10 2 2 2  hf − bg gh − af   , we get Now, by using the formula  , 2 2  ab − h ab − h   3 + 9 −9 + 3    Point of intersection =  4 2 , 4   −2 − 1 − 2 − 1   4 4  21 3    −7 −1 =  4 , 4  =  ,  9 9 − −   3 3   4 4 

 lx + my   lx + my  ax 2 + 2hxy + by 2 − 2g   x −2f   y  n   n 

Example 59. Prove that the equation 2x 2 + 5xy + 3 y 2 + 6x + 7 y + 4 = 0 represents a pair of straight lines. Find the coordinates of their point of intersection. Sol. Given equation is 2 x2 + 5 xy + 3 y2 + 6 x + 7 y + 4 = 0

…(i)

Writing the Eq. (i) as a quadratic equation in x, we have 2 x2 + (5 y + 6) x + 3 y2 + 7 y + 4 = 0 x= = =

−(5 y + 6) ±

(5 y + 6)2 − 4 ⋅ 2 (3 y2 + 7 y + 4)

−(5 y + 6) ±

25 y2 + 60 y + 36 − 24 y2 − 56 y − 32

−(5 y + 6) ±

y2 + 4 y + 4

4 4

4 −(5 y + 6) ± ( y + 2 ) = 4 −5 y − 6 + y + 2 −5 y − 6 − y − 2 ∴ x= , 4 4 ⇒ 4 x + 4 y + 4 = 0 and 4 x + 6 y + 8 = 0 ⇒ x + y + 1 = 0 and 2 x + 3 y + 4 = 0 Hence, Eq. (i) represents a pair of straight lines whose equations are …(ii) x + y + 1= 0 and ...(iii) 2 x + 3y + 4 = 0 On solving Eqs. (ii) and (iiii), we get x = 1 and y = − 2 Hence, coordinates of the point of intersection of Eqs. (ii) and (iii) is (1, − 2 ).

Lines Joining the Origin to the Points of Intersection of a Line and a Curve The combined equation of the straight lines joining origin to the points of intersection of a second degree curve a x 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 and a straight line lx + my + n = 0 is

2

 lx + my  +c  =0  n  X

Example 60. The angle between the lines joining the origin to the points of intersection of the line y = 3x +2 and the curve x 2 + 2xy + 3 y 2 + 4x + 8 y = 11, is 2 2 1  (a) tan −1  2  3   2 (b) tan −1    3  2 2  (c) tan −1   3   3  (d) tan −1  2 2 Sol. (c) The equation of the given straight line is

y − 3x =1 2 The equation of the given curve is x2 + 2 xy + 3 y2 + 4 x + 8 y − 11 = 0 y = 3x + 2 ⇒

...(ii)

2

y − 3x − 11   =0  2  ⇒

4 x2 + 8 xy + 12 y2 + 2 (8 y2 − 12 x2 − 20 xy) − 11 ( y2 − 6 xy + 9 x2 ) = 0 7 x2 − 2 xy − y2 = 0

This is the required equation. On comparing the equation with ax2 + 2 hxy + by2 = 0, we obtain a = 7, b = − 1 and h = − 1. Let θ be the required angle. Then,

A

tan θ = B

⇒ O

...(i)

The combined equation of the straight lines joining the origin to the points of intersection of Eqs. (i) and (ii) is a homogeneous equation of second degree obtained with the help of Eqs. (i) and (ii). Making Eq. (ii) homogeneous of the second degree in x and y with the help of Eq. (i), we get y − 3x  y − 3x x2 + 2 xy + 3 y2 + 4 x   + 8y    2   2 

⇒

Y

12 Straight Line and Pair of Straight Lines

X

X

⇒

tan θ =

2 h2 − ab a+ b 2 1+ 7 7 −1

=

2 2 3

2 2  θ = tan−1    3 

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Objective Mathematics Vol. 1

12

Work Book Exercise 12.4 1 The equation 3 x 2 + 2 hxy + 3 y 2 = 0 represents a pair of straight lines passing through the origin. The two lines are a b c d

real and distinct, if h2 > 3 real and distinct, if h2 > 9 real and coincident, if h2 = 3 real and coincident, if h2 > 3

+ 13 y + 15 = 0 represents a pair of straight lines. The distance between them is

c

equation ax 2 + by 2 + cx + cy = 0, represents a pair of straight lines, if

c ≠0

a+ b=0 a– b=0 b+c=0 None of the above

b

7 2 5

d None of these

7 One bisectors of the angle between the lines given by a( x − 1)2 + 2 h( x − 1)y + by 2 = 0 2 x + y − 2 = 0. The other bisector is a x − 2y − 1= 0 c 2x − y − 1= 0

is

b x + 2y + 1= 0 d None of these

8 Two

lines represented by the equation 9 x 2 + 24 xy + 16 y 2 + 21x + 28 y + 6 = 0 are

lines represented by the equation will a x 4 + b x 3 y + c x 2 y 2 + d x y 3 + ay 4 = 0 bisect the angles between the other two, if

a b c d

a b c d

3 The

parallel coincident perpendicular None of the above

4 If the angle between the pair of straight lines represented

by

the

x 2 − 3 xy + λy 2 + 3 x − 5 y + 2 = 0

is

equation  1 tan −1   ,  3

b 0

c 3

d 1

5 The equation y − x + 2 x − 1 = 0, represents 2

a b c d

2

a pair of straight lines a circle a parabola an ellipse

c + 6a = 0 and ab + d = 0 b + d = 0 and a + c = 0 c + 6a = 0 and b + d = 0 None of the above

9 The coordinates of the orthocentre of the triangle formed by the lines 2 x 2 + 3 xy + y 2 = 0 and x + y = 1are a (1, 1)

where λ is a non-negative real number. Then, λ is a 2

648

7 15 7 5

a

2 The a b c d

6 The equation 8 x 2 + 8 xy + 2 y 2 + 26 x

1 1 b  ,  2 2

1 1 c  ,   3 3

1 1 d  ,   4 4

10 The image of the pair of lines represented by ax 2 + 2 hxy + by 2 = 0 by the line mirror y = 0 is a b c d

a x 2 + 2 hx y + by 2 = 0 a x 2 − 2 hx y − b y 2 = 0 bx 2 + 2 hxy + ay 2 = 0 a x 2 − 2 h x y + by 2 = 0

WorkedOut Examples Type 1. Only One Correct Option Ex 1. Given, a straight lines and a fixed point O, a straight line is drawn through O meeting lines in the points R1 , R 2 , R 3 , ..., R n and on it a point R is n 1 1 1 , taken such that = + +K + OR OR1 OR 2 OR n then the locus of R is (a) a straight line (c) a parabola

(b) a circle (d) None of these

Sol. Let equations of the given lines be a1x + b1 y + c1 = 0,

i = 1, 2, ..., n and the point O be the origin (0, 0). Then, equation of the line through O can be written as x y = = r, where θ is the angle made by the line cosθ sin θ with the positive direction of X-axis and r is the distance of any point on the line from the origin O. Let r, r1 , r2 ,… , rn be distance of the points R , R1 , R2 ,… , Rn from O ⇒ OR = r and OR1 = r1 ... Then, coordinates of R are (r cos θ , r sin θ ) and of Ri are (ri cosθ , ri sin θ ), where i = 1, 2, …, n. Since, Ri lies on ai x + bi y + ci = 0 ⇒ ai ri cos θ + bi ri sin θ + ci = 0, for i = 1, 2, ... , n a b 1 − i cos θ − i sin θ = , i = 1, 2, ..., n ⇒ ci ci ri  n a  n b = −  ∑ i  cos θ −  ∑ i  sinθ i = 1 ci  i = 1 ci  i=1 i  n b  n a n = −  ∑ i  cos θ −  ∑ i  sin θ ⇒ r i = 1 ci  i = 1 ci  Hence, the locus of R is  n a  n   ∑ i  x +  ∑ bi  y + n = 0      i = 1 ci   i = 1 ci  ⇒

n

1

∑r

which is a straight line. Hence, (a) is the correct answer.

Ex 2. A variable straight line drawn through the x y point of intersection of lines + =1 and a b x y + =1, meets the coordinate axes in A and b a B. Then, the locus of the mid-point of AB is (a) xy( a + b ) = ab( x + y ) (b) 2xy( a − b ) = ab( x + y ) (c) 2xy( a − b ) = ab( x − y ) (d) 2xy( a + b ) = ab( x + y ) Sol. The equation of the variable line through the point of intersection of the given lines is of the type y  y  x x  + − 1 + k  + − 1 = 0 a b  b a 

⇒ ⇒

 1 k  + x +  a b (ak + b)x +

∴ Points

A

 1 k  +  y = (1 + k )  b a (bk + a) y = ab(1 + k )  ab(1 + k )  and B are  , 0  ak + b 

and

 ab(1 + k ) .  0,  (bk + a)  Let P (x1 , y1 ) be the mid-point of AB. ab(1 + k ) ab(1 + k ) Then, x1 = and y1 = 2(ak + b) 2(bk + a)

…(i)

Therefore, we write Eq. (i) in the following form 1 2(ak + b) 1 2(bk + a) = = , x1 ab(1 + k ) y1 ab(1 + k ) 1 1 2(a + b)(1 + k ) + = ⇒ x1 y1 ab(1 + k ) x1 + y1 2(a + b) = ⇒ x1 y1 ab Hence, the locus of P (x1 , y1 ) is 2xy (a + b) = ab(x + y). Hence, (d) is the correct answer.

Ex 3. The equation of the line passing through the point P(1, 2) and cutting the lines x + y − 5 = 0 and 2x − y = 7 at A and B respectively such that the harmonic mean of PA and PB is 10, is 11 14   (a) y − 2 = tan cos −1 − cos −1  ( x − 1)  146 5 146  11 14   (b) y + 2 = tan cos −1 + cos −1  ( x + 1)  146 5 146  11 14   (c) y − 2 = tan cos −1 + cos −1  ( x − 1)  146 5 146  (d) None of the above Sol. Let the equation of the line passing through P(1, 2) be y − 2 = m(x − 1) Let PA = r1, PB = r2 Then, A = (1 + r1 cos θ , 2 + r1 sin θ ) and B = (1 + r2 cos θ , 2 + r2 sin θ ), where, tanθ = m As A is on the line x + y − 5 = 0. ⇒1 + r1 cos θ + 2 + r1 sin θ − 5 = 0 ⇒ r1 (cos θ + sin θ ) = 2 1 cos θ + sin θ = ∴ r1 2 As B is on the line 2x − y − 7 = 0. ⇒ 2(1 + r2 cos θ ) − (2 + r2 sin θ ) − 7 = 0 ⇒ r2 (2 cos θ − sin θ ) = 7 1 2 cos θ − sin θ ∴ = r2 7

…(i)

…(ii)

…(iii)

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Objective Mathematics Vol. 1

12

Now, HM of PA and PB is 10. 2PA ⋅ PB ∴ 10 = PA + PB 2r r 10 = 1 2 ⇒ r1 + r2 r1 + r2 1 1 1 1 ⇒ = ⇒ + = r1r2 5 r1 r2 5 2 cos θ − sin θ cos θ + sin θ 1 + = ⇒ 7 2 5 [from Eqs. (i) and (iii)] ⇒ 10(2 cos θ − sin θ ) + 35(cos θ + sin θ ) = 14 ⇒ 55 cos θ + 25 sin θ = 14 55 where, cos α = 552 + 252 14 ∴ cos (θ − α ) = 5 146 14 θ = α + cos−1 ⇒ 5 146 11 14 = cos−1 + cos−1 146 5 146 From Eq. (i), the equation of the line is 11 14   y − 2 = tan  cos−1 + cos−1  (x − 1)  146 5 146  Hence, (c) is the correct answer.

Ex 4. A triangle has the line y = m1 x and y = m2 x for two of its sides with m1 and m2 being roots of the equation bx 2 + 2hx + a = 0. If H ( a, b) is the orthocentre of the triangle, then the equation of the third side is (a) ( a + b )( ax + by ) = ab( a + b − h ) (b) ( a − b )( ax − by ) = ab( a − b − 2h ) (c) ( a − 2b )( ax + by ) = ab( a + b − h ) (d) ( a + b )( ax + by ) = ab( a + b − 2h ) Sol. The given lines y = m1x and y = m2x intersect at the origin O (0, 0). Thus, one vertex of the triangle is at the origin O. Therefore, let OAB be the triangle and OA, OB be the lines respectively …(i) y = m1 x and …(ii) y = m2 x Let the equation of the third side AB be …(iii) y = mx + c Given that, H (a, b) is the orthocentre of the ∆OAB. ∴ OH is perpendicular to AB.  b ⇒   × m = −1  a a …(iv) m=− ⇒ b On solving Eq. (iii) with Eqs. (i) and (ii), the coordinates of A are  c cm1  ,    m1 − m m1 − m  c cm2  and those of B are  ,   m2 − m m2 − m

650

Now, equation of the line through A and perpendicular to OB is 1  cm1 c  y− =−  x − m1 − m m2  m1 − m x c(m1m2 + 1) …(v) ⇒ y=– + m2 m2 (m2 − m) Similarly, equation of line through perpendicular to OA is x c(m1m2 + 1) y=− + m1 m1 (m2 − m)

B

and

…(vi)

The point of intersection of Eqs. (v) and (vi) is the orthocentre H (a, b). On subtracting Eq. (vi) from Eq. (v), we get − cm(m1m2 + 1) x=a= (m1 − m)(m2 − m) ⇒

c=

− [ m1m2 − m(m1 + m2 ) + m2 ]a m(m1m2 + 1)

…(vii)

Since, m1 and m2 are the roots of the equation bx 2 + 2hx + a = 0. a 2h and m1m2 = ∴ m1 + m2 = − b b  a 2hm 2 − + + m a b b  c=  a  m + 1 b  =

− [ a + 2hm + bm2 ]a m(a + b)

From Eq. (iii), the equation of third side AB is (a + 2hm + bm2 )a y = mx − m(a + b) ⇒

(a − 2ha / b + ba2 / b2 )a  a y = −  x −  b  a  −  (a + b)  b

⇒ (ax + by)(a + b) = ab(a + b − 2h) Hence, (d) is the correct answer.

Ex 5. The centroid (x , y ) of the triangle with sides ax 2 + 2hxy + by 2 = 0 and lx + my =1, is y x = = bl − hm am − hl 3( am2 y x (b) = = bl + hm am − hl 3( am2 y x (c) = = bl + hm am + hl 3( am2 y x (d) = = bl − hm am + hl 3( am2 (a)

2 − 2hlm + bl 2 ) 2 − 2hlm − bl 2 ) 2 + 2hlm − bl 2 ) 2 − 2hlm − bl 2 )

Sol. Let y = m1x and y = m2x be the lines represented by ax 2 + 2hxy + by2 = 0. Then, 2h m1 + m2 = – b a and m1m2 = b

ax 2 + 2hxy + by2 = 0, then

y = m2x

Y

B A

X′

1 , m2 l + mm2 l + mm2 y = m1x 1 , m1 A X l + mm1 l + mm1

X

O G (x, y)

y = m 2x B

y = m1x

Ix + my = 1 X′

lx + my = 1

Y′

On solving y = m1x and y = m2x respectively with lx + my = 1, we obtain that the coordinates of A and B which are   1 m1  1 m2  , , A  and B  ⋅  l + mm1 l + mm1   l + mm2 l + mm2  Since, C (x , y) is the centroid of ∆OAB. Therefore, 1 1 1  x=  ,  3  l + mm1 l + mm2  and

y=

1  m1 m2  ,   3  l + mm1 l + mm2 

⇒

x=

 1 2l + m(m1 + m2 )   3  l 2 + lm (m1 + m2 ) + m2m1m2 

and

y=

 1  l (m1 + m2 ) + 2mm1m2   2 2 3  l + lm (m1 + m2 ) + m m1m2 

  2hm 2l −  1  b ⇒ x=   3  2 2hlm am2  l − +  b b  a  2hl  − + 2m 1 b b  and y=  hlm a 2 3  l2 − + m2  b b  2 bl − hm ⇒ x= ⋅ 2 3 am − 2hlm + bl 2 2 am − hl and y= ⋅ 2 3 am − 2hlm + bl 2 x y 2 ⇒ = = bl − hm am − hl 3(am2 − 2hlm + bl 2 ) Hence, (a) is the correct answer.

Ex 6. The orthocentre of the triangle formed by the lines ax 2 + 2hxy + by 2 = 0 and lx + my =1 is a+b x y = = 2 l m am + 2hlm + bl 2 a+b x y (b) = = 2 l m am + 2hlm − bl 2 a+b x y (c) = = 2 l m am − 2hlm − bl 2 a+b x y (d) = = 2 l m am − 2hlm + bl 2 (a)

Y′

12 Straight Line and Pair of Straight Lines

Sol. Let y = m1x and y = m2x be the lines represented by

Y

a 2h and m1m2 = …(i) b b Let OAB be the triangle formed by the lines y = m1x, y = m2 x and lx + my = 1 The coordinates of the vertices of ∆OAB are O (0, 0),   1 m1  1 m2  , , A  and B  .  l + mm1 l + mm1   l + mm2 l + mm2 

m1 + m2 = −

The equation of the altitude through A is 1  1  m1 y− =−  x − l + mm1 m2  l + mm1  1 + m1m2 ⇒ x + m2 y = 1 + mm1

…(ii)

Similarly, the equation of the altitude through B is 1 + m1m2 …(iii) x + m1 y = l + mm2 From Eqs. (ii) and (iii), we get x y = l (m1 − m2 )(1 + m1m2 ) m(m1 − m2 )(1 + m1m2 ) (l + mm1 )(l + mm2 ) (l + mm1 )(l + mm2 ) = ⇒ ⇒

1 m1 − m2

1 + m1m2 x y = = l m l 2 + lm(m1 + m2 ) + m2m1m2 x y a+ b [using Eq. (i)] = = 2 l m am − 2hlm + bl 2

Hence, the coordinates (x , y) of the orthocentre of ∆OAB are given by x y a+ b = = 2 l m am − 2hlm + bl 2 Hence, (d) is the correct answer.

Ex 7. The straight lines represented by ( y − mx ) 2 = a 2 (1 + m 2 ) and ( y − nx ) 2 = a 2 (1 + n 2 ), is (a) square (b) rhombus (c) rectangle (d) None of these Sol. We have, ( y − mx )2 = a2 (1 + m2 ) and

( y − nx )2 = a2 (1 + n2 )

⇒ y − mx = ± a 1 + m2 and y − nx = ± a 1 + n2

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⇒

y = mx ± a 1 + m2 and y = nx ± a 1 + n2

Thus, the equation of the line represented by the given equation are y = mx + a 1 + m2

…(i)

y = mx − a 1 + m

2

…(ii)

2

…(iii)

y = nx + a 1 + n

y = nx − a 1 + n2

Ex 8. The equation λ(x 3 − 3xy 2 ) + y 3 − 3x 2 y = 0 represents (a) three lines equally inclined to the other (b) two lines equally inclined to each other (c) no line equally inclined to each other (d) None of the above Sol. The given equation is a homogeneous cubic equation in x, y. So, it represents three lines passing through the origin. Let one of the three lines be y = mx, where m = tanθ, θ is the angle made by the line with X-axis. Then, λ (x 3 − 3m2x 3 ) + m3x 3 − 3mx 3 = 0

⇒

λ (1 − 3m2 ) + m3 − 3m = 0 λ=

3m − m 1 − 3m2

3

2λ = 2

(h2 − ab)(α 2 + β 2 ) (a − b)2 + 4 h2

⇒ λ2[(a − b)2 + 4 h2 ] = (α 2 + β 2 )(h2 − ab) Hence, the locus of (α , β ) is (x 2 + y2 )(h2 − ab) = λ2[(a − b)2 + 4 h2 ] Hence, (b) is the correct answer.

Ex 10. The condition that the pair of straight line joining the origin to the intersections of the line y = mx + c and the circle x 2 + y 2 = a 2 is at the right angles, is (a) 2c 2 = a 2 (1 − m2 ) (c) c 2 = a 2 (1 + m2 )

(b) 2c 2 = a 2 (1 + m2 ) (d) None of these

Sol. The equations of the given straight line and the given curve are

y − mx =0 c

y = mx + c or x 2 + y2 = a2

and

…(i) …(ii)

The combined equation of the straight lines joining the origin to the points of intersection of Eqs. (i) and (ii) is 2  y − mx  2 x 2 + y2 =   a  c  ⇒

x 2 (c2 − a2m2 ) + 2a2mxy + y2 (c2 − a2 ) = 0 …(iii)

The lines given by Eq. (iii) are at right angles, if Coefficient of x 2 + Coefficient of y2 = 0 ⇒

(c2 − a2m2 ) + (c2 − a2 ) = 0

⇒ 2c = a (1 + m2 ), which is the required condition. 2

[Q m = tanθ]

3 tan θ − tan 3 θ λ= ⇒ 1 − 3 tan 2 θ ⇒ λ = tan 3θ ⇒ tan 3θ = tan φ, where, λ = tan φ or φ = tan −1 λ ⇒

3θ = φ, π + φ, 2π + φ φ π φ 2π φ ⇒ θ= , + , + 3 3 3 3 3 Thus, the given equation represents three lines passing φ π π through the origin and the lines make angles , + 3 3 3 2π φ and + with X-axis. 3 3 Clearly, they are equally inclined with each other. Hence, (a) is the correct answer.

652

Sol. Let the point be P(α , β). Then,

…(iv)

Clearly, lines (iii) and (iv) are parallel and the distance between them is given by a 1 + m2 + a 1 + m2 d1 =   = |2a| 1 + m2   Similarly, lines (iii) and (iv) are parallel and the distance between them is given by a 1 + n2 + a 1 + n2 d2 =   = |2a| 1 + n2   Clearly, d1 = d2 The lines (i), (ii), (iii) and (iv) form a rhombus. Hence, (b) is the correct answer.

⇒

(a) ( x 2 + y 2 )( h 2 + ab ) = λ 2 [( a + b ) 2 − 4h 2 ] (b) ( x 2 + y 2 )( h 2 − ab ) = λ 2 [( a − b ) 2 + 4h 2 ] (c) ( x 2 − y 2 )( h 2 − ab ) = λ 2 [( a − b ) 2 − 4h 2 ] (d) None of the above

Ex 9. A point moves in such a way that the distance between the feet of the perpendiculars drawn from it to the lines ax 2 + 2hxy + by 2 = 0 is a constant 2λ. Then, the locus of the point is

2

Hence, (b) is the correct answer.

Ex 11. All chords of the curve 3x 2 − y 2 − 2x + 4 y = 0 which subtend a right angle at the origin pass through a fixed point. Then, the point is (a) ( −1, 2) (c) (1, − 2)

(b) ( −1, − 2) (d) (1, 2)

Ax + By = 1 be a chord of the curve 3x 2 − y2 − 2x + 4 y = 0. The combined equation of the straight lines joining the origin to the points of intersection of the lines Ax + By = 1 and the curve 3x 2 − y2 − 2x + 4 y = 0 is 3x 2 − y2 − 2 x ( Ax + By) + 4 y( Ax + By) = 0

Sol. Let

⇒ x 2 (3 − 2 A ) + 2xy(2 A − B ) + y2 (4 B − 1) = 0 …(i) Since, the lines represented by Eq. (i) are at right angles. ∴Coefficient of x 2 + Coefficient of y2 = 0 ⇒ 3 − 2A + 4 B − 1 = 0 ⇒ 2A − 4 B = 2 ⇒ A − 2B = 1 ⇒ A = 2B + 1

Ex 12. The lines joining the origin to the other two points of intersection of the curves and ax 2 + 2hxy + by 2 + 2gx = 0 2 2 a ′ x + 2h′ xy + b′ y + 2g ′ x = 0 will be at right angles to one another, if (a) g ( a′ − b′ ) = g ′ ( a + b ) (b) g ( a′ + b′ ) = g ′ ( a − b ) (c) g ( a′ − b′ ) = g ′ ( a − b ) (d) g ( a′ + b′ ) = g ′ ( a + b )

⇒

ax + 2hxy + by = − 2gx and

2

...(i)

a′ x + 2h′ xy + b′ y = − 2g′ x 2

⇒ ⇒

Sol. The given curves are 2

Let the coordinates of mid-point M of AB be (α , β ). Then, h+ a 6h ,β= α= 2 2 …(ii) ⇒ 2α = h + a and β = 3h We have to eliminate a and h from Eqs. (i) and (ii). β β β + 3a From Eq. (ii), h = and 2α = + a = 3 3 3 6α − β 6α = β + 3a ⇒ a = ⇒ 3 Now, from Eq. (i), 2 β2  β 6α – β  4 l2 =  −  + 36 ⋅ 3 3  9

2

...(ii)

The combined equation of the straight lines joining the origin to the points of intersection of Eqs. (i) and (ii) is a homogeneous equation of second degree. To obtain the desired equation we proceed as follows. On multiplying Eq. (i) by g′ and Eq. (ii) by g and subtracting, we get (ag′ − a ′ g )x 2 + 2(hg′ − h′ g )x y + (bg′ − b′ g ) y2 = 0 …(iii) The lines given by Eq. (iii) are at right angles, if g (a′ + b′ ) = g′ (a + b) Hence, (d) is the correct answer.

(2β − 6α )2 (2β − 6α )2 + 36 β 2 + 4β 2 = 9 9 36l 2 = 4β 2 − 24αβ + 36α 2 + 36β 2 + 36α 2 9l 2 = 9α 2 − 6 αβ + 10β 2

Hence, the locus of (α , β ) is 9x 2 − 6xy + 10 y2 = 9l 2 Hence, (b) is the correct answer.

Ex 14. Let (h, k ) be a fixed point, where h > 0, k > 0. A straight line passing through this point cuts the positive direction of the coordinate axes at the points Pand Q. Then, the minimum area of the ∆OPQ, O being the origin, is (a) 4hk sq units (c) 3hk sq units y − k = m(x − h). Y Q

(a) 9x 2 − 6xy − 10 y 2 = 9l 2 (b) 9x 2 − 6xy + 10 y 2 = 9l 2 (c) 9x 2 + 6xy + 10 y 2 = 9l 2

AB 2 = 4 l 2 (h − a)2 + (6h)2 = 4 l 2 Y

X

…(i)

Let S be the area of ∆OPQ, then 1 1 k S = OP × OQ =  h −  (k − mh) 2 m 2 1 (mh − k )(k − mh) = 2 m ⇒ 2mS = hkm − k 2 − h2m2 + khm ⇒

6x y=

P

Let this line cut the X-axis and Y-axis at P and Q. k   Then, P ≡  h − , 0 and Q ≡ (0, k − mh)  m 

Sol. Let the coordinates of A be (a, 0) and those of B be

B (h , 6 h )

M (a , b )

O

A (h, k)

O

(d) 9x 2 + 6xy − 10 y 2 = 9l 2

⇒

(b) 2hk sq units (d) None of these

Sol. Let the equation of any line passing through a(h, k ) be

Ex 13. If a line AB of length 2l moves with the end A always on the X-axis and the end B always on the line y = 6x, then the equation of the locus of the mid-point of AB is

(h, 6h). Given,

12 Straight Line and Pair of Straight Lines

On putting A = 2 B + 1in Ax + By = 1, we get x (2 B + 1) + By = 1 ⇒ (x − 1) + B (2 x + y) = 0 Clearly, it is a line passing through the intersection of the lines x − 1 = 0 and 2x + y = 0 i.e. (1, − 2). Thus, the required point is (1, − 2). Hence, (c) is the correct answer.

X A (a, 0)

h2m2 − 2(hk − S )m + k 2 = 0

Since, m is real. ∴ Discriminant, D≥0 ⇒ 4 (hk − S )2 − 4 h2k 2 ≥ 0 ⇒ S − 2hk ≥ 0 ⇒ S ≥ 2hk Hence, minimum value of S is 2hk sq units. Hence, (b) is the correct answer.

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Ex 15. A straight line passes through a fixed point ( h, k ). Then, the locus of the foot of the perpendicular on it from the origin is (a) x 2 + y 2 − h x − k y = 0 (b) x 2 − y 2 − h x + k y = 0 (c) x 2 − y 2 + h x + k y = 0 (d) x 2 + y 2 + h x + k y = 0 Sol. Let the equation of the straight line be ...(i) x cosα + y sin α − p = 0 Since, the Eq. (i) passes through the point (h, k ), therefore ...(ii) h cosα + k sin α − p = 0 On subtracting Eq. (ii) from Eq. (i), we get ...(iii) (x − h)cosα + ( y − k )sin α = 0 Now, the equation of the straight line which is perpendicular to Eq. (i) and passes through the origin, is ...(iv) x sin α − y cos α = 0 The required locus will be obtained by eliminating α from the Eqs. (iii) and (iv). From Eq. (iv), x sin α = y cos α x y ⇒ = cos α sin α

Hence, the line (i) always passes through a fixed point  a + i − 1 a + 2i − 2 ,Σ Σ .   n n But it is given that this passes through the point  13   , 11 .  2  a + i − 1 13 Σ = ∴ n 2 a + 2i − 2 and Σ = 11 n On solving the above equations, we get a = 2 and n = 10 Hence, (a) is the correct answer.

Ex 17. A variable line is drawn through the origin O. Two points A and B are taken on the line such that OA =1 unit and OB = 2 units. Through points A and B two lines are drawn making equal angle ‘α’ with the line AB. Then, the locus of the point of intersection of the lines is

⇒ ⇒

x (x − h) + y ( y − k ) = 0 x 2 − hx + y2 − ky = 0

9 + tan 2 α 4 2 − tan α 9 (b) x 2 − y 2 = 4 2 + tan α 9 (c) x 2 + y 2 = 2 9 + 2 tan 2 α 2 2 (d) x + y = 4

∴

x 2 + y2 − hx − ky = 0

Sol. Let C be the mid-point of AB.

⇒

x y = = cos α sin α

x 2 + y2 cos α + sin α 2

2

= x 2 + y2

which is the required locus. Hence, (a) is the correct answer.

Ex 16. Abscissae and ordinates of n given points are in AP with first term a and common difference 1 and 2, respectively. If algebraic sum of perpendiculars from these given points on a variable line which always passes through the  13  point  , 11 is zero, then the values of a and  2 n are (a) 2, 10 (c) 3, 9

(a) x 2 + y 2 =

Given that,

AB = 1

⇒

AC = CB =

3 2 1 RC = tanα 2 OR 2 = OC 2 + RC 2 OC =

and ⇒

2

1  3 x 2 + y2 =   + tan 2 α  2 4

⇒ Y

(b) 4, 20 (d) 0, 1

R (x, y) y = mx a B

Sol. Let the n given points be

(a, a), (a + 1, a + 2), (a + 2, a + 4 ), K i.e. [ a + i − 1, a + 2 (i − 1)]; i = 1, 2, 3, ..., n Let the variable line be ...(i) px + qy + r = 0 Since, the algebraic sum of perpendiculars drawn from these n points on the variable line (i) in always zero. p(a + i − 1) + q(a + 2i − 2) + r Σ =0 ∴ p 2 + q2 ⇒ ⇒

654

⇒

Σ[ p(a + i − 1) + q(a + 2i − 2) + r ] = 0 pΣ (a + i − 1) + qΣ (a + 2i − 2) + rn = 0 a+ i −1 a + 2i − 2 pΣ + qΣ + r=0 n n

1 2

a

C

A X′

O

X

Y′

9 + tan 2 α is the required locus of the 4 point of intersection of the lines.

⇒ x 2 + y2 =

Hence, (a) is the correct answer.

(a) y = x (c) xy = 1

(b) y + x = 0 (d) xy + 1 = 0

12 17 (a)  − ,   5 5 6 17 (c)  − ,   5 5

84 13 (b)  − ,   5 5 (d) None of these

Sol. We have, | PA − PB | ≤ AB.

Sol. If A1 , A2 , B1 , B2 are concyclic. ⇒

Ex 20. Consider the point A ≡ (0, 1) and B ≡ (2, 0) and P is a point on the line 4x + 3 y + 9 = 0. Coordinates of the point P such that | PA − PB | is maximum, are

P

Thus, for | PA − PB | to be maximum point A , B and P must be collinear equation of AB is x + 2 y = 2. Solving it with given line,  84 13 we get P ≡  − ,   5 5

m2 − 1 = 0 ⇒ m = 1

A1 ≡ (− c1 , 0), A ≡ (− c2 , 0) B1 ≡ (0, c1 ), B2 ≡ (0, c2 ) B2 y = mx + c 2 B1

A (0, 1)

O B (2, 0)

12 Straight Line and Pair of Straight Lines

Ex 18. Straight lines y = mx + c1 and y = mx + c2 , where m ∈ R + , meet the X-axis at A1 and A2 respectively and Y-axis at B1 and B 2 respectively. It is given that points A1 , A2 , B1 and B 2 are concyclic. Locus of intersection of lines A1 B 2 and A2 B1 is

Hence, (b) is the correct answer. A2

Ex 21. In question number 20, coordinates of the point P such that | PA − PB | is minimum, are

y = mx + c1

A1

x y + =1 c1 c2 x y Equation of A2 B1 is – + =1 c2 c1 Equation of A1 B2 is –

...(i)

3 14 (a)  , −   20 5

3 14 (b)  ,   20 5 

...(ii)

3 14 (c)  − ,   30 5 

9 12 (d)  − , −   20 5

The locus of point of intersection of Eqs. (i) and (ii), is 1 1 1 1 y −  + x  −  =0 ⇒ x + y=0  c2 c1   c2 c1  Hence, (b) is the correct answer.

Ex 19. Consider the point A ≡ (3, 4) and B ≡ (7, 13). If P is a point of the line y = x such that PA + PB is minimum, then coordinates of P are 13 13 (a)  ,   7 7 31 31 (c)  ,   7 7

23 23 (b)  ,   7 7 33 33 (d)  ,   7 7

Sol. Let A1 be the reflection of A in y = x. B (7,13)

A (3, 4)

y=x

P

A1

⇒ A1 = (4 , 3) Now, PA + PB = A1P + PB, which is minimum, if A1 , P and B are collinear. Equation of A1 B is 13 − 3 ( y − 3) = (x − 4 ) 7−4 ⇒ 3 y = 10x − 31 Solving it with y = x , we get  31 31 P≡ ,   7 7 Hence, (c) is the correct answer.

Sol. Minimum value of | PA − PB | is zero. It can be obtained,

if PA = PB. That means P must lie on the right bisector of AB. Equation of right bisector of AB is 1 3 y − = 2 (x − 1) i.e. y = 2x − 2 2 Solving with given line, we get 12  9 P ≡ − , −   20 5 Hence, (d) is the correct answer.

Y Ex 22. In the adjacent B figure, ∆ABC is C right angled at B. If AB = 4 and BC = 3 and side AC slides along the coordinate X axes in such a way X′ O A that always B Y′ remains in the first quadrant, then B always lie on the straight line

(a) y = x (c) 4 y = 3x

(b) 3 y = 3x (d) x + y = 0

Sol. Given, AB = 4 , BC = 3 ⇒ AC = 5 A ≡ (5 cos θ , 0), C ≡ (0, 5 sin θ ) If BC and AB make the angle θ 1 and θ 2 with X-axis, then θ1 = C − θ, θ2 = π − ( A + θ) π  π = π −  − C + θ = + (C − θ ) 2  2

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Sol. AB = BC = CA = 2

B(h, k)

Let ⇒

C

∠BAO = θ A ≡ (2 cosθ , 0), B ≡ (0, 2 sin θ ) Y

X′

θ O

A

C (h, k)

X

B

Y′

Using parametric equation of line for BC, we get h k − 5 sin θ ...(i) = =3 cos (C − θ ) sin (C − θ ) Similarly, using parametric equation of line for AB, we get h k ...(ii) = =4 − sin (C − θ ) cos (C − θ )

X

Y′

π  BC makes an angle  − θ with X-axis and 3 

h k − 2 sin θ = =2 π  π  cos  − θ sin  − θ 3  3 

Y C

and

B X′

A

O

 π  AC makes an angle π −  + θ  with X-axis. 3   If C ≡ (h, k ), then

From Eqs. (i) and (ii), we get h k cos (C − θ ) = = 3 4 ⇒ Locus of B is 3x = 3 y. Hence, (b) is the correct answer.

Ex 23. Adjacent figure represents an equilateral ∆ABC of side length 2 units. Locus of vertex C as the side AB slides along the coordinate axes, is

θ

X′

A

O

X

Y′

⇒

h − 2 cosθ k =2 = π  π  − cos  + θ sin  + θ 3  3  π  π  2 cos  − θ = h, 2 sin  + θ = k 3  3 

⇒

cosθ +

(a) x + y − xy + 1 = 0

⇒

2 cosθ = ( 3k − h), 2 sin θ = (h 3 − k )

(b) x + y + xy 3 = 1

Thus, locus of (h, k ) is

2

2

2

2

(c) x 2 + y 2 = 1 + xy 3

3 sin θ = h, 3 cosθ + sin θ = k

1 = x 2 + y2 − x y 3 or

(d) x 2 + y 2 − xy 3 + 1 = 0

x 2 + y2 = 1 + x y 3

Hence, (c) is the correct answer.

Type 2. More than One Correct Option Ex 24. If the equation a x 2 − 6xy + y 2 + bx + cy + d = 0 represents pair of lines whose slopes are m and m 2 , then value of a is/are (a) a = − 8 (c) a = 27

(b) a = 8 (d) a = − 27

Sol. m and m2 are roots of equation 2

 y  y   −6  + a=0  x  x ⇒ m + m2 = 6 ⇒ m3 = a 3 6 3 ⇒ m + m + 3m (m + m2 ) = 216 ⇒ a + a2 + 3a ⋅ 6 = 216 ⇒ a2 + 19a − 216 = 0 ∴ a = 8, − 27 Hence, (b) and (d) are the correct answers.

656

Ex 25. Two sides of a triangle are the lines ( a + b) x + ( a − b) y − 2ab = 0 and ( a − b) x + ( a + b) y − 2ab = 0. If the triangle is isosceles and the third side passes through point ( b − a, a − b), then the equation of third side can be (a) x + y = 0 (c) x − b + a = 0

(b) x = y + 2 ( b − a ) (d) y − a + b = 0

Sol. We have, (a + b) x + (a − b) y − 2ab = 0 and

(a − b) x + (a + b) y − 2ab = 0

Equation of the angle bisectors are (a + b) x + (a − b) y − 2ab = ± {(a − b) x + (a + b) y − 2ab} ⇒ 2bx − 2by = 0 i.e. x= y

Sol.

∴ Equation of third side is given by (ii) x − y = k satisfying the point (b − a, a − b) ∴ k = 2b − 2a So, the line is x − y = 2 (b − a). (ii) x + y − 2b = k passing through the point (b − a, a − b) ∴ k = − 2b So, the line is x + y = 0. Hence, (a) and (b) are the correct answers.

x y + =1 cuts the coordinate axes at a b y x + = −1 A ( a, 0) and B (0, b) and the line a ′ b′ at A ′ ( −a ′, 0) and B ′ (0, − b ′ ). If the point A, B , A ′ , B ′ are concyclic, then the orthocentre of the ∆ABA ′ is (a) ( 0, 0) (b) ( 0, b ) aa ′  bb ′   (d)  0, (c)  0,    b   a 

12

x y + =1 a b − aa′ = − bb′ i.e. aa′ = bb′ Equation of the altitude through A′ is a y − 0 = (x + a′ ) b If it intersects the altitude x = 0 at aa y= ′ b  aa′ ∴ Orthocentre is  0, .  b

Ex 26. Line

...(i)

Y B (0,b)

X′

O A′(–a′,0)

A (a,0)

Straight Line and Pair of Straight Lines

and 2ax + 2ay − 4 ab = 0 i.e. x + y = 2b

X

B ′ (0,–b ′ ) Y′

which is the same as (0, b′ ) by Eq. (i). Hence, (b) and (c) are the correct answers.

Type 3. Assertion and Reason Directions (Ex. Nos. 27-30) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 27. Statement I If P1 (x1 , y1 ) and P2 (x 2 , y2 ) are the images of point P ( x, y) about lines L1 ≡ ax + by + c = 0 and L2 ≡ bx − ay + c ′ = 0 respectively, then the line joining points P1 and P2 always passes through point of intersection of lines L1 and L2 . Statement II Lines perpendicular.

L1

and

L2

⇒ θ + φ = 90° , which is true as L1 and L2 are perpendicular. ∴ Both the statements are true and Statement II is correct explanation of Statement I. Hence, (a) is the correct answer.

Ex 28. Statement I If ah + bk + c = 0, then the point ( h, k ) lies above the line ax + by + c = 0 where a, b, c are non-zero real numbers. − ( ah + c) Statement II If x = h, y = which is b −( ah + c) less than k (for b > 0) i.e. < k, hence b ( h, k ) lies above the line. Sol. Statement I is false, as Statement II is true from figure O (0, 0) and P (h, k ) lie on same side for ax + by + c = 0

are

Y P (h, k)

Sol. Let the line joining P1 and P2 passes through intersection

ah + bk + c = 0

of L1 and L2, then from the figure, L2 (x2, y2)P2

φ O

φ

θ θ

y=

L1 P1 (x1, y1)

2 (θ + φ ) = 180°

X

O (0, 0)

P (x, y)

– (ah + c) b

x=h

− (ah + c) b Hence, (d) is the correct answer.

⇒

ah + bk + L > 0or k >

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Ex 29. Statement I A homogeneous equation in x and y of degree n represents family of line (not more than n), all intersecting at origin. y Statement II Substituting = m, gives a x polynomial in m of degree n, which can have maximum n real roots of the form y − m1 x = 0, y − m2 x = 0, ...

Ex 30. Statement I Two mutually perpendicular lines intersect with coordinate axes, then the angle which one makes with positive X-axis anti-clockwise is equal to angle that other one makes with positive Y-axis anti-clockwise.

Sol. Statement II is true and correct explanation of

of intersection of pair of lines with coordinate axes are concyclic only when they are perpendicular. Hence, (c) is the correct answer.

Statement I. Hence, (a) is the correct answer.

Statement II All quadrilaterals that can be formed by points of intersection of pair of lines with coordinate axes are concyclic. Sol. Statement II is false as the quadrilateral formed by point

Type 4. Linked Comprehension Based Questions Passage I (Ex. Nos. 31-33) The line 6 x + 8 y = 48 intersects the coordinate axes at A and B, respectively. A line L bisects the area and the perimeter of the ∆OAB , where O is the origin.

B x y

Ex 31. The number of such lines possible is

Q

(a) 1 (b) 2 (c) 3 (d) more than 3

Ex 32. The slope of the line L can be 10 + 5 3 10 10 − 5 6 (b) 10 8+ 3 6 (c) 10 (d) None of the above

(a)

Ex 33. The line L does not intersect the side ... of the ∆OAB. (a) AB (b) OB (c) OA (d) can intersect all the sides Sol. (Ex. Nos. 31-33) Case I Let the line L cuts AO and AB at distances X and Y and Y from A. ∴Area of the triangle with sidesx and y, 3xy = 12 ⇒ x y= 40 10 Also, x + y = 12 [using perimeter bisection]

P

O

A

x, y = 6 ± 2 3

[not possible]

So, there is a unique line possible. Let point P be (α , β ). Using parametric equation AB, 3 β = 6 − (6 + 6 ) 5 4 and α = (6 + 6 ) 5 ∴ Slope of PQ =

β − 6 10 − 5 6 = α −0 10

31. (a) 32. (b) 33. (c) Passage II (Ex. Nos. 34-36) Given the equation of

two sides of a square as 5 x + 12 y − 10 = 0, 5 x + 12 y + 29 = 0. Also, given a point M ( −3, 5) lying on one of its side.

Ex 34. Number of possible squares are (a) 0 (c) 2

(b) 1 (d) 3

Sol. Given, 5x + 12 y − 10 = 0 5x + 12 y + 29 = 0

and C′

A

C

which is not possible. Case II If the line L cuts OB and BA at distances y and x from B, then we have xy = 30 and x + y = 12. ⇒ x = 6 + 6 and y = 6 − 6

658

Case III If the line L cuts the sides OA and OB at distances x and y from O, then x + y = 12 and xy = 24.

M (–3, 5) D′

B

Square can be ABCD or ABC ′D′. Hence, (c) is the correct answer.

D

d=

39

[d = distance between parallel sides]

52 + 122

D′

=3

5x + 12y – 10 = 0

M (–3, 5)

⇒A=9 Hence, (c) is the correct answer.

B

C′

(b) 22 (d) 36

C

5x + 12y + 29 = 0

|61 − λ | 13 ⇒ 61 − λ = ± 39 λ = 100, 22 So, λ = 100, 22, 61 Hence, (d) is the correct answer. d =3=

∴

Ex 36. If the possible equation of the remaining sides is 12x − 5 y + λ = 0, then λ cannot be (a) 61 (c) 100

D

A

Straight Line and Pair of Straight Lines

 5 ⇒ 5 y − 25 = 12x + 36 ⇒ 12x − 5 y + 61 = 0 Now, CD and CD′ ⇒ 12x − 5 y + λ = 0

(a) 3 (b) 6 (c) 9 (d) 12 Sol. Q A = d 2

12

12 Sol. AB ( y − 5) = (x + 3)  

Ex 35. Area of square is

Type 5. Match the Column Ex 37. Match the following : Column I A. a x + by + c = 0 is a variable straight line, where a, b, c are 1st, 4th, 6th terms of an increasing AP. Then, variable straight line always passes through a fixed point

PB =

Similarly, Column II p. ( x + y − 3)2 + ( x − y + 1)2 = 20

B. A variable line L is drawn through q. P(2, 3) to meet lines y − x − 10 = 0 and y − x − 20 = 0 at points A and B respectively. A point Q is taken on L such that PQ 2 = PA ⋅ PB. Locus ofQ is

(0, 3)

C. Vertices of a triangle are (1, 2 ), ( 5 cos θ, 5 sinθ) and ( 5 sinθ, − 5 cos θ). Locus of its orthocentre is

r.

5 2  ,–  3 3

D. A ray of light emerging from the point source placed at (1, 2) is reflected at a point Q on the Y-axis and then passes through the point (6, 9). Coordinates of Q are

s.

and

19 sin θ − cos θ

...(ii)

Q ≡ (h, k ) PQ = r h = 2 + r cos θ , k = 3 + r sin θ     9 19 r2 =      sin θ − cos θ   sin θ − cos θ 

⇒

(r sin θ − r cos θ )2 = 19 × 9

⇒

(k − 3 − h + 2)2 = 171

⇒

( y − x − 1)2 = 171

C. Let A ≡ (1, 2), B ≡ ( 5 cos θ , 5 sin θ ), C ≡ ( 5 sin θ , − 5 cos θ ) Distance of A , B , C from (0, 0) is 5 units. ⇒ (0, 0) is circumcentre C of ∆ABC. ( y − x − 1)2 = 171

Sol. A. b = a + 3d , c = a + 5d

b − a = 3d , c − a = 5d b−a c−a = ⇒ 3 5 ⇒ 5b − 5a = 3c − 3a ⇒ 2a − 5b + 3c = 0 2 5 ⇒ a− b+ c=0 3 3 5 2 So, straight lines passes through  , −  . 3 3 x −2 y−3 B. = =r cos θ sin θ ⇒ x = 2 + r cos θ , y = 3 + r sin θ ⇒ 3 + r sin θ = 2 + r cos θ + 10 ⇒ r (sin θ − cos θ ) = 12 − 3 = 9 9 ...(i) PA = sin θ − cos θ

Let C (h, k ) be centroid of triangle. 3h = 1 +

5 (cos θ + sin θ )

3k = 2 +

5 (sin θ − cos θ )

If O is orthocentre of triangle having coordinates (α , β ). OG : GC = 2 : 1 ⇒ α = 3h, β = 3k α −1 ...(i) cos θ + sin θ = 5 β−2 ...(ii) sin θ − cos θ = 5 α+β−3 sinθ = ⇒ 2 5 α −β + 1 ⇒ cos θ = 2 5 2

2

 α − β + 1  α + β − 3  =1   +  2 5   2 5  ∴ Required locus is (x + y − 3)2 + (x − y + 1)2 = 20.

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Objective Mathematics Vol. 1

12

D. P′ is reflection of P in Y-axis i.e. P′ (−1, 2).

Equation of P′R is

R (6, 9)

Y

y−2= ⇒ ∴

Q (0,k) P (1, 2) O

y−2=x +1 y=x + 3

It meets Y-axis at Q (0, 3). A → r; B → s; C → p; D → q

X

Type 6. Single Integer Answer Type Questions Ex 38. A straight line cuts the X -axis at point A (1, 0) and Y -axis at point B, such that ∠OAB = α (α > π / 4). C is a middle point of AB, if B ′ is a mirror image of point B with respect to line OC and C ′ is a mirror image of point C with respect to line BB ′, then the ratio of the areas of ∆ABB ′ and BB ′C is ______ . Sol. (2) From the figure, OB = tanα Y B

D B′

C

A (1, 0)

O

X

Coordinates of point C 1 0 1 1 0 1 1 kλ λ 1 = 1 2 λ+1 λ+1 λ+1 k (λ − 1) λ −1 1 0 1 λ+1 λ+1

k 1 k = 2 λ+1 0

So, equation of line OC is y = x tanα i.e. ∠COA = α. Now,  λ −1 1 1  λ λ − 1 . − 0  AB = k 2  −  + 1 2  λ + 1 λ + 1 λ + 1 λ + 1  ∴ AB = sec α

...(i)

k  1 λ −1  AC = BC = +   2  λ + 1 (λ + 1)2   2

So,

=

⇒

Area of ∆ABB′ AB × BB′ × sin ∠ABB′ = Area of ∆BB′ C ′ BB′ × BC ′ × sin ∠B′ BC ′ [Q ∠ABB′ = ∠B′ BC ′] AB sec α = = = 2 [from Eqs. (i) and (ii)] BC ′ 1 sec α 2

Ex 39. Sides of a rhombus are parallel to the lines x + y − 1 = 0 and 7x − y − 5 = 0. It is given that diagonal of the rhombus intersect at (1, 3) and one vertex A of the rhombus lies on the line y = 2x. Then, the number of all possible coordinates of the vertex A is ________.

BD = BC sin (180° − 2α ) 3 1 = × sin 2α = × secα × 2 sin α × cosα = sin α 8 3 BD = B′ D′ = sinα  k    0, 2  B′ is a mirror image of B in line OC   

Clearly, ∆BDC and ∆BDC′ are congruent. 1 BC ′ = BC = sec α and ∠ABB′ = ∠B′ BC ′ 2

parallel to the bisectors of the given lines and will pass through (1, 3). Equations of bisectors of the given lines are x + y−1  7x − y − 5 =±   5 2  2 or 2x − 6 y = 0, 6x + 2 y = 5 The equations of diagonal are x − 3 y + 8 = 0 and 3x + y − 6 = 0. Thus, the required vertex will be the point where these lines meet the line y = 2x. Solving these lines, we get possible coordinates as  8 16  6 12  ,  or  ,  . 5 5 5 5 Hence, the number of possible coordinates are 2.

Ex 40. If the lines ax + y + 1 = 0, x + by + 1 = 0 and x + y + c = 0 (a, b and c being distinct and different from 1) are concurrent, then 1 1 1 is equal to ________. + + 1− a 1− b 1− c Sol. (1) Since, the given lines are concurrent

k 2 2λ sec α 2 (λ + 1)2

  k 2λ In ∆BDC, ∠BCD = (180° − 2α ) ∠COA = α    (λ + 1) ⇒

∴

Sol. (2) It is clear that diagonals on the rhombus will be

C′

660

9−2 (x + 1) 6+1

...(ii)

a 1 1 1 b 1 =0 1 1 c Operating C 2 → C 2 − C 1 and C 3 → C 3 − C 1 , we get a 1− a 1− a 1 b−1 0 =0 1

0

c−1

⇒ a(b − 1) (c − 1) − (b − 1) (1 − a) − (c − 1) (1 − a) = 0 a 1 1 ⇒ + + =0 1− a 1− b 1− c a 1 1 1 1 1 1 b 1 =0 ⇒ ⇒ + + =1 1 − a 1− b 1 − c 1 1 c

Target Exercises Type 1. Only One Correct Option (a) has a fixed direction (b) always passes through a fixed point (c) always cuts intercepts on the axes such that their sum is zero (d) forms a triangle with the axes whose area is constant

( b − c ) x + ( c − a ) y + ( a − b ) = 0 and (where, (b − c ) x + (c 3 − a 3 ) y + (a 3 − b 3 ) = 0 a, b and c are not all equal) will represent the same line, if

2. Equations 3

3

(a) b = c (c) c + a = 0

(b) c = a (d) a + b + c = 0

3. A rectangle ABCD , A ≡ ( 0, 0), B ≡ ( 4, 0), C ≡ ( 4, 2) and D ≡ ( 0, 2) undergoes the following transformations successively: (i) f1 ( x, y ) → ( y, x ) (ii) f 2 ( x, y ) → ( x + 3 y, y )  x − y x + y (iii) f 3 ( x, y ) →  , .  2 2  The final figure will be (a) a square (c) a rectangle

(b) a rhombus (d) a parallelogram

4. The points A( 0, − 1) , B( 2, 1) , C( 0, 3) and D( −2, 1) are the vertices of a (a) square (c) parallelogram

(b) rectangle (d) None of these

 1 5. If the points ( −2, 0) ,  −1,  and (cos θ, sin θ ) are  3 collinear, then the number of values of θ ∈[ 0, 2π ] is (a) 0

(b) 1

(c) 2

(d) infinite

6. If a, b, c are in AP and α, β, γ are in GP, then the points ( a, α), ( b, β ) and ( c, γ ) are collinear, if (a) α = β = γ

(b) α = β 2

(c) β = γ 2

(d) None of these

7. The points (α , β ) , ( γ , δ ) , (α , δ ) and ( γ , β ) taken in order, where α, β, γ and δ are different real numbers, are (a) collinear (c) vertices of a rhombus

(b) vertices of a square (d) concyclic

8. The diagonals of a parallelogram PQRS are along the lines x + 3 y = 4 and 6x − 2 y = 7, then PQRS must be a (a) rectangle (c) cyclic quadrilateral

(b) square (d) rhombus

9. A line passes through ( 2, 2) and is perpendicular to the line 3x + y = 3, its y -intercept is (a)

1 3

(b)

2 3

(c) 1

(d)

4 3

10. If one of the diagonals of a square is along the line x = 2 y and one of its vertices is (3, 0), then its sides through this vertex are given by the equations (a) (b) (c) (d)

y − 3x + 9 = 0, 3 y + x − 3 = 0 y + 3x + 9 = 0, 3 y + x − 3 = 0 y − 3x + 9 = 0, 3 y − x + 3 = 0 y − 3x + 3 = 0, 3 y + x + 9 = 0

11. The equation of a line in parametric form is given by (a) (x − x1 )r cos θ = ( y − y1 )r sin θ x − x1 y − y1 (b) = =r cos θ sin θ (c) (x − x1 ) cos θ = ( y − y1 )sin θ = r (d) None of the above

12. A line is drawn perpendicular to the line y = 5x, meeting the coordinate axes at A and B. If the area of ∆OAB is 10 sq units, where O is the origin, then equation of drawn line is (a) x + 5 y = 10 (c) x + 4 y = 10

(b) x − 5 y = 10 (d) x − 4 y = 10

13. The vertices of a ∆ABC are (1, 1), ( 4, − 2) and (5, 5) respectively. The equation of perpendicular dropped from C to the internal bisector of angle A is (a) y − 5 = 0 (c) 2x + 3 y − 7 = 0

Targ e t E x e rc is e s

1. If a, b and c are any three terms of an AP, then the ax + by + c = 0

(b) x − 5 = 0 (d) None of these

14. In an isosceles ∆ABC, the coordinates of the points B and C on the base BC are respectively (2, 1) and (1, 2). 1 If the equation of the line AB is y = x, then the 2 equation of the line AC is (a) 2 y = x + 3 (b) y = 2x 1 (c) y = (x − 1) 2 (d) y = x − 1

15. The equation of the lines through the points (2, 3) and making an intercept of length 2 units between the lines y + 2x = 3 and y + 2x = 5, are (a) x + 3 = 0, 3x + 4 y = 12 (b) y − 2 = 0, 4 x − 3 y = 6 (c) x − 2 = 0, 3x + 4 y = 18 (d) None of the above

661

16. A line through A( − 5, − 4 ) meets the lines x + 3 y + 2 = 0, 2x + y + 4 = 0 and x − y − 5 = 0 at B, C 2 2 2  6   10   15  and D respectively. If   ,  =  +  AD   AC   AB  then the equation of the line is (a) 2x + 3 y + 22 = 0 (c) 3x − 2 y + 3 = 0

(b) 5x − 4 y + 7 = 0 (d) None of these

17. Two lines 2x − 3 y = 1 and x + 2 y + 3 = 0 divide the XY-plane in four compartments which are named as shown in the figure. Consider the locations of the points ( 2, − 1), (3, 2) and ( − 1, − 2), we get

x+

2y

+3

II

III

IV

Ta rg e t E x e rc is e s

(a) (2, − 1) ∈ IV (c) (−2, − 1) ∈ II

O

I

(0, 0) 2x – 3y = 1

(b) (3, 2) ∈ III (d) None of these

(b) (24 , − 3) , (1, 1) (d) (0, 1), (3, 0)

  1  1 19. If A sin α ,  and B  , cos α , − π ≤ α ≤ π, are   2  2 two points on the same side of the line x − y = 0, then α belongs to the interval  π 3π   π π (a)  − ,  ∪  ,  4 4   4 4  π 3π  (c)  ,  4 4 

 π π (b)  − ,   4 4 (d) None of these

20. If the point (1 + cos θ, sin θ ) lies between the region corresponding to the acute angle between the lines 3 y = x and 6 y = x, then (a) θ ∈ R

π (c) θ ∈ R ~ (2n + 1) , n ∈ I 2

(b) θ ∈ R ~ n π , n ∈ I (d) None of these

1 + t 2 + t  21. If P  ,  is any point on a line, then the range  2 2 of values of t for which the point P lies between the parallel lines x + 2 y = 1and 2 x + 4 y = 15, is 4 2 5 2 0 , the distance between (1, 1) and the point of intersection of the lines a x + by + c = 0 and [2013] bx + ay + c = 0 is less than 2 2 , then

(a) a + b − c > 0 (c) a − b + c > 0

(b) a − b + c < 0 (d) a + b − c < 0

3. A straight line L through the point ( 3, − 2) is inclined at an angle 60° to the line 3x + y = 1. If L also intersects the X-axis, then the equation of L is [2011] (a) y +

3x + 2 − 3 3 = 0

(b) y − 3x + 2 + 3 3 = 0 (c) 3 y − x + 3 + 2 3 = 0 (d) 3 y + x − 3 + 2 3 = 0

JEE Main/AIEEE 4. Locus of the image of the point (2, 3) in the line [2015] ( 2x − 3 y + 4 ) + k ( x − 2 y + 3) = 0, k ∈ R, is a

(a) straight line parallel to X -axis (b) straight line parallel to Y -axis (d) circle of radius 3

(b) 3 y = x − 3 (d) 3 y = x − 1

Statement I The ratio PR : RQ equals 2 2 : 5.

5. Let PS be the median of the triangle with vertices P( 2, 2) , Q( 6, − 1) and R( 7, 3). The equation of the line [2014] passing through (1, − 1) and parallel to PS is (b) 2 x + 9 y + 7 = 0 (d) 2 x − 9 y − 11 = 0

6. A ray of light along x + 3 y = 3 gets reflected upon reaching X-axis, the equation of the reflected ray is 670

3

7. The lines L1 : y − x = 0 and L2 : 2x + y = 0 intersect the line L3 : y + 2 = 0 at P and Q, respectively. The bisector of the acute angle between L1 and L2 intersects L3 at R.

(c) circle of radius 2

(a) 4 x − 7 y − 11 = 0 (c) 4 x + 7 y + 3 = 0

(a) y = x +

(c) y = 3x − 3

[2013]

Statement II In any triangle, bisector of an angle divides the triangle into two similar triangles. [2011] (a) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (b) Statement I is true, Statement II is false (c) Statement I is false, Statement II is true (d) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

(b) (0, ∞ )

(c) [1, ∞ )

(d) (−1, ∞ )

x y + = 1passes through the point 5 b (13, 32). The line K is parallel to L and has the x y equation + = 1. Then, the distance between L and c 3 [2010] K is

9. The line L given by

(a)

23 15

(c)

(b) 17

17 15

23 17

(d)

10. The lines and p( p 2 + 1) x − y + q = 0 2 2 2 ( p + 1) x + ( p + 1) y + 2q = 0 are perpendicular to [2009] a common line for (a) exactly one value of p (b) exactly two values of p (c) more than two values of p (d) no value of p

11. The perpendicular bisector of the line segment joining P(1, 4 ) and Q ( k , 3) has y-intercept −4. Then, a possible [2008] value of k is (a) −4 (c) 2

(b) 1 (d) −2

12. If one of the lines my + (1 − m ) xy − mx = 0 is a bisector of the angle between the lines xy = 0, then m is [2007] 2

1 2 (c) ±1 (a) −

2

2

(b) −2 (d) 2

13. A straight line through the point A( 3, 4 ) is such that its intercept between the axes is bisected at A. Its [2006] equation is (a) 3x − 4 y + 7 = 0 (c) 3x + 4 y = 25

(b) 4 x + 3 y = 24 (d) x + y = 7

14. If non-zero numbers a, b and c are in HP, then the x y 1 straight line + + = 0 always passes through a a b c fixed point. That point is [2005] 1  (a) 1, −   2

(b) (1, − 2)

(c) (−1, − 2)

(d) (−1, 2)

15. The equation of the straight line passing through the point ( 4, 3) and making intercepts on the coordinate axes whose sum is −1, is [2004]

12

y x y = − 1 and + = −1 3 −2 1 y x y and = −1 + = −1 3 −2 1 y x y = 1 and + =1 3 −2 1 y x y = 1 and + =1 3 −2 1

16. If the sum of the slopes of the lines given by x 2 − 2cx y − 7 y 2 = 0 is four times their product, then c has the value [2004] (a) 1

(b) −1

(d) −2

(c) 2

17. If one of the lines given by 6x 2 − xy + 4cy 2 = 0 is equals [2004] 3x + 4 y = 0 , then c (a) 1

(b) −1

Straight Line and Pair of Straight Lines

(a) (−1, 1]

x + 2 x (b) − 2 x (c) + 2 x (d) − 2

(a)

(d) −3

(c) 3

18. If x1 , x 2 , x 3 and y1 , y 2 , y 3 are both in GP with the same common ratio, then the points ( x1 , y1 ), ( x 2 , y 2 ) [2003] and ( x 3 , y 3 ) (a) lie on a straight line (c) lie on a circle

(b) lie on an ellipse (d) are vertices of a triangle

19. A square of side a lies above the X-axis and has one vertex at the origin. The side passing through the π  origin makes an angle α  0 < α <  with the positive  4 direction of X-axis. The equation of its diagonal not [2003] passing through the origin is (a) (b) (c) (d)

y (cos α y (cos α y (cos α y (cos α

− sin α ) − x (sin α − cos α ) = a + sin α ) + x (sin α − cosα ) = a + sin α ) + x (sin α + cos α ) = a + sin α ) + x (cos α − sin α ) = a

20. Three straight lines 2x + 11y − 5 = 0, [2002] 24x + 7 y − 20 = 0 and 4x − 3 y − 2 = 0

Targ e t E x e rc is e s

8. The line x + y = | a | and ax − y = 1intersect each other in the first quadrant. Then, the set of all possible values of a in the interval [2011]

(a) form a triangle (b) are only concurrent (c) are concurrent with one line bisecting the angle between the other two (d) None of these

21. A straight line through the point ( 2, 2) intersects the lines 3x + y = 0 and 3x − y = 0 at the points A and B. The equation to the line AB, so that the ∆OAB is equilateral, is [2002] (a) x − 2 = 0 (c) x + y − 4 = 0

(b) y − 2 = 0 (d) None of these

Other Engineering Entrances 22. The value of k such that the lines 2x − 3 y + k = 0 , 3x − 4 y − 13 = 0 and 8 x − 11y − 33 = 0 are concurrent, [BITSAT 2014] is (a) 20 (c) 7

(b) −7 (d) −20

23. The points ( 2 , 5) and ( 5, 1) are the two opposite vertices of a rectangle. If other two vertices are points on the straight line y = 2x + k, then the value of k is [Kerala CEE 2014] (a) 4 (d) −3

(b) 3 (e) 1

(c) −4

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Objective Mathematics Vol. 1

12

24. A straight line perpendicular to the line 2x + y = 3 is passing through (1, 1). Its y-intercept is [Kerala CEE 2014] (a) 1 1 (d) 2

(b) 2 1 (e) 3

(c) 3 (a) x − y = 1 (d) x + y = 1

25. The ratio by which the line 2x + 5 y − 7 = 0 divides the straight line joining the points ( −4, 7) and ( 6, − 5) is [Kerala CEE 2014] (a) 1 : 4 (d) 2 : 3

(b) 1 : 2 (e) 1 : 3

(c) 1 : 1

26. The straight lines x + y = 0, 5x + y = 4 and x + 5 y = 4 [WB JEE 2014] form (a) an isosceles triangle (c) a scalene triangle

(b) an equilateral triangle (d) a right angled triangle

27. The slopes of the lines represented by x 2 + 2hxy + 2 y 2 = 0 are in the ratio 1: 2 , then h equals 1 2 (c) ± 1

3 2 (d) ± 3

Ta rg e t E x e rc is e s

(a) ±

[UP SEE 2014]

[J&K CET 2014] (b) 2x + 3 y = 9 (d) 3x + 2 y = 5

(a) 3x − 2 y = 3 (c) 2x − 3 y = 7

29. A straight line passes through the points ( 5, 0) and ( 0, 3). The length of perpendicular from the point [Karnataka CET 2014] ( 4, 4 ) on the line is 15 34 17 (c) 2

17 2 17 (d) 2 (b)

(a)

30. If p is the length of the perpendicular from the origin to the line, whose intercepts with the coordinate axes 1 1 are and , then the value of p is [Kerala CEE 2014] 3 4 3 4

(d) 12

1 12 1 (e) 5

(b)

(c) 5

(b) 4 a2 (e) a2

(c) 3a2

(a) x − y = 0 (c) bx + ay = 0

(b) 2 5

[UP SEE 2013] (d) 0

(c) 2

35. Let P = ( −1, 0) , O = ( 0, 0) and Q = ( 3, 3 3 ) be three points. Then, the equation of the bisector of the ∠POQ is [BITSAT 2013] (a) y = 3x

(b) 3 y = x

(c) y = − 3x (d) 3 y = − x

36. The point of intersection of line represented by the equation 3x 2 + 8 x y − 3 y 2 + 29x − 3 y + 18 = 0 is  3 5 (a)  ,   2 2

[MP PET 2012]  −3 −5 (b)  ,  (c) (−3, − 5) (d) (3, 5)  2 2

37. If lines (tan 2 θ + cos 2 θ) x 2 − 2 tan θ ⋅ xy + sin 2 θ ⋅ y 2 = 0 make with X -axis angles α and β , then tan α − tan β is equal to [MP PET 2012] (a) 2 (c) tanθ

(b) 4 (d) 2tanθ

38. If two pairs of lines x 2 − 2 mx y − y 2 = 0 and x 2 − 2 nx y − y 2 = 0 are such that one of them represents the bisector of the angles between the [AMU 2012] other, then (a) mn = 1 (c) mn = − 1

(b) m + n = mn (d) m − n = mn

 1  39. If a straight line passes through the points  − , 1 and  2  (1, 2), then its x-intercept is (a) −2 (d) 1

(b) −1 (e) 0

[Kerala CEE 2011] (c) 2

40. The equation of the line passing through ( 0, 0) and intersection 3x − 4 y = 2 and x + 2 y = − 4 is (a) 7x = 6 y (c) 5x = 8 y

(b) 6x = 7 y (d) x = 0

41. The line parallel to the X-axis and passing through the point of intersection of the lines a x + 2by + 3b = 0and bx − 2a y − 3a = 0, where ( a, b ) ≠ ( 0, 0) is [Kerala CEE 2011]

32. If any point P is at the equal distances from points A ( a + b, a − b ) and B ( a − b, a + b ), then locus of a point is [RPET 2014] 672

(a) 2 3

[J&K CET 2011]

31. If p and q are respectively the perpendiculars from the origin upon the straight lines, whose equations are x secθ + y cosec θ = a and x cos θ − y sin θ = a cos 2θ, then 4 p 2 + q 2 is equal to [Kerala CEE 2014] (a) 5a2 (d) 2a2

(b) x − y = 3 (e) x + y = − 1

[Kerala CEE 2013] (c) x + y = 3

34. The equation of second degree x 2 + 2 2xy + 2 y 2 + 4x + 4 2 y + 1 = 0 represents a pair of straight lines. The distance between them is

(b) ±

28. Let A( 2, − 3) and B( −2, 1) be vertices of a ∆ABC. If the centroid of this triangle moves on the line 2x + 3 y = 1, then the locus of the vertex C is the line

(a)

33. The equation of the perpendicular bisector of the line segment joining A( −2, 3) and B( 6, − 5) is

(b) ax + by = 0 (d) x + y = 0

(a) above the X-axis at a distance of 3/ 2 (b) above the X-axis at a distance of 2/ 3 (c) below the X-axis at a distance of 2/ 3 (d) below the X-axis at a distance of 3/ 2 (e) below the X-axis at a distance of 3

[BITSAT 2011] (b) 99x − 27 y + 30 = 0 (d) 21x − 77 y − 100 = 0

43. A variable line passes through a fixed point ( a, b ) and meets the coordinate axes in A and B. The locus of the point of intersection of lines through A , B parallel to coordinate axes is [MP PET 2011] x y + =1 a b x y (c) + = 2 a b

a b + =1 x y x y (d) + = 3 a b

(a)

[Kerala CEE 2010] (a)

(a) x + 4 xy − y = 0

(b) 2x − 4 xy + y = 0

(c) x 2 − 4 xy + y2 = 0

(d) x 2 − 4 xy − y2 = 0

2

2

(a) an equilateral triangle (c) an isosceles triangle

[BITSAT 2010] (b) a right angled triangle (d) a scalene triangle

46. The equation of a straight line which passes through the point ( a cos 3 θ, a sin 3 θ ) and perpendicular to x secθ + y cosec θ = a is [Kerala CEE 2010] x y + = a cosθ a a (b) x cos θ − y sin θ = a cos 2θ (c) x cos θ + y sin θ = a cos 2θ (d) x cos θ + y sin θ − a cos 2θ = 1 (e) x cos θ − y sin θ + a cos 2θ = − 1 (a)

47. If the line px − q y = r intersects the coordinate axes at ( a, 0) and ( 0, b ) , then the value of a + b is equal to  q − p (b) r   pq   p − q (e) r   p + q

[Kerala CEE 2010]  p − q (c) r   pq 

48. The equations of the line through (1, 1) and making angles of 45° with the line x + y = 0 are [WB JEE 2010] (a) x − 1 = 0, x − y = 0 (b) x − y = 0, y − 1 = 0 (c) x + y − 2 = 0, y − 1 = 0 (d) x − 1 = 0, y − 1 = 0

49. If the sum of distances from a point P on two mutually perpendicular straight lines is 1 unit, then the locus of P is [WB JEE 2010] (a) a parabola (b) a circle (c) an ellipse (d) a straight line

(c) (e)

2

45. The equations y = ± 3x, y = 1are the sides of

 q + p (a) r   pq   p + q (d) r   p − q

51. A line has slope m and y-intercept 4. The distance between the origin and the line is equal to

(b)

44. Locus of a point which moves such that its distance from the X -axis is twice its distance from the line [Karnataka CET 2011] x − y = 0 is 2

[Kerala CEE 2010] (a) 3x + 4 y − 3 = 0 (b) 3x + 4 y + 3 = 0 (c) 4 x − 3 y + 3 = 0 (d) 4 x − 3 y − 3 = 0 (e) 4 x − 3 y − 4 = 0

4

4

(b)

1 − m2 4

(d)

m2 + 1 4m m −1

m2 − 1 4m

12 Straight Line and Pair of Straight Lines

(a) 21x + 77 y + 100 = 0 (c) 99x + 27 y + 30 = 0

50. The equation of one of the lines parallel to 4x − 3 y = 5 and at a unit distance form the point ( − 1, − 4 ) is

1 + m2

52. The distance of the point (1, 2) from the line x + y + 5 = 0 measured along the line parallel to 3x − y = 7 is equal to [Kerala CEE 2010] (b) 40 (d) 10 2

(a) 4 10 (c) 40 (e) 2 20

53. A line through the point A( 2, 0) which makes an angle of 30° with the positive direction of X -axis is rotated about A in clockwise direction through an angle of 15°. Then, the equation of the straight line in the new position is [WB JEE 2010] (a) (2 − 3 ) x + y − 4 + 2 3 = 0 (b) (2 − 3 ) x − y − 4 + 2 3 = 0 (c) (2 − 3 ) x − y + 4 + 2 3 = 0

Targ e t E x e rc is e s

42. Equation of the bisector of the acute angle between lines 3x + 4 y + 5 = 0 and 12x − 5 y − 7 = 0 is

(d) (2 − 3 ) x + y + 4 + 2 3 = 0

54. The number of points on the line x + y = 4 which are unit distance apart from the line 2 x + 2 y = 5 is [WB JEE 2010] (a) 0 (c) 2

(b) 1 (d) ∞

55. The slopes of the lines which makes an angle 45° with the line 3x − y = − 5 are [Kerala CEE 2010] (a) (1, − 1) 1  (d)  2, −   2

1  (b)  , − 1 2  1  (e)  −2,   2

 (c) 1, 

1  2

56. The vertices of a triangle are A( 3, 7), B( 3, 4 ) and C( 5, 4 ). The equation of the bisector of the ∠ ABC is (a) y = x + 1 (c) y = 3x − 5 (e) y = − x

[Kerala CEE 2010] (b) y = x − 1 (d) y = x

673

Answers Work Book Exercise 12.1 1. (c)

2. (a)

3. (d)

4. (d)

5. (a)

6. (b)

7. (a)

8. (b)

9. (b)

10. (b)

4. (b)

5. (a)

6. (b)

7. (b)

8. (a)

9. (c)

10. (b)

4. (c)

5. (b)

6. (b)

7. (a)

8. (b)

4. (a)

5. (a)

6. (b)

7. (a)

8. (c)

9. (b)

10. (d)

Work Book Exercise 12.2 1. (d)

2. (a)

11. (b)

12. (d)

3. (d)

Work Book Exercise 12.3 1. (a)

2. (b)

3. (a)

Work Book Exercise 12.4 1. (b)

2. (a)

3. (a)

Ta rg e t E x e rc is e s

Target Exercises 1. (b)

2. (d)

3. (d)

4. (a)

5. (b)

6. (a)

7. (b)

8. (d)

9. (d)

10. (a)

11. (b)

12. (a)

13. (b)

14. (b)

15. (c)

16. (a)

17. (a)

18. (c)

19. (a)

20. (d)

21. (a)

22. (b)

23. (a)

24. (a)

25. (d)

26. (d)

27. (c)

28. (b)

29. (c)

30. (a)

31. (a)

32. (a)

33. (a)

34. (a)

35. (c)

36. (b)

37. (b)

38. (b)

39. (c)

40. (b)

41. (b)

42. (c)

43. (b)

44. (b)

45. (b)

46. (a)

47. (c)

48. (a)

49. (a)

50. (a)

51. (b)

52. (b)

53. (d)

54. (c)

55. (d)

56. (c)

57. (d)

58. (c)

59. (a)

60. (a)

61. (c)

62. (a)

63. (d)

64. (b)

65. (c)

66. (c)

67. (a)

68. (b)

69. (d)

70. (d)

71. (c)

72. (b)

73. (a)

74. (d)

75. (c)

76. (a)

77. (b)

78. (a)

79. (b)

80. (c)

81. (a)

82. (a)

83. (c)

84. (b)

85. (b)

86. (d)

87. (a,b)

88. (a,c)

89. (all)

91. (a,c,d)

92. (all)

93. (b,c,d)

94. (a,b)

95. (d)

96. (a)

97. (d)

98. (c)

99. (b)

100. (a) 110. (a)

90. (b,c)

101. (d)

102. (c)

103. (d)

104. (d)

105. (b)

106. (b)

107. (c)

108. (b)

109. (d)

111. (b)

112. (d)

113. (c)

114. (a)

115. (b)

116. (b)

117. (b)

118. (c)

119. (a)

120. (d)

121. (c)

122. (a)

123. (*)

124. (**)

125. (***)

126. (2)

127. (4)

128. (6)

129. (1)

130. (2)

131. (4) * A → p; B → q; C → s; D → s ** A → p, s; B → p, r; C → p; D → p *** A → p; B → s; C → q; D → r

Entrances Gallery

674

1. (6)

2. (a)

3. (b)

4. (c)

5. (b)

6. (b)

7. (b)

8. (c)

9. (d)

10. (a)

11. (a)

12. (c)

13. (b)

14. (b)

15. (d)

16. (c)

17. (d)

18. (a)

19. (d)

20. (c)

21. (b)

22. (b)

23. (c)

24. (d)

25. (c)

26. (a)

27. (b)

28. (b)

29. (d)

30. (e)

31. (e)

32. (a)

33. (b)

34. (c)

35. (c)

36. (b)

37. (a)

38. (c)

39. (a)

40. (a)

41. (d)

42. (c)

43. (b)

44. (b)

45. (a)

46. (b)

47. (b)

48. (d)

49. (d)

50. (d)

51. (c)

52. (c)

53. (b)

54. (a)

55. (e)

56. (a)

Explanations Target Exercises ∴ a − 2 b + c = 0 and a x + b y + c = 0 Hence, it always passes through the point (1, − 2 ). b3 − c 3 c 3 − a3 a3 − b3 = = b−c c −a a−b ⇒ (b2 + c 2 + bc ) = (c 2 + a2 + ac ) = (a2 + b2 + ab)

2. If

⇒ ⇒

b2 − a2 = − c (b − a) a+ b+c =0

3. On applying Eqs. (i), (ii) and (iii) transformation, we finally get A ≡ (0, 0 ), B ≡ (4, 8), C ≡ (5, 9), 8 9− 8 9−1 8 D ≡ (1, 1), mAB = , mBC = = 1, mCD = = , 4 5− 4 5−1 4 9 8−1 7 mAD = 1, mAC = , mBD = = 5 4−1 3 Hence, final figure will be a parallelogram.

4. All sides are equal and angles are 90° by Pythagoras theorem or m1m2 = − 1. 1 5. 2 ⇒ ⇒

−2

0 1 1 −1 1 =0 3 cos θ sin θ 1

1 2 10. tan (± 45° ) = 1 1 + m⋅ 2 1 ∴ m = 3, − 3 ∴Sides are y − 3 x + 9 = 0 and 3 y + x − 3 = 0. x − x1 y − y1 [by definition] 11. = =r cos θ sin θ x y 12. Let the equation of line be + = 1 a b b − ⋅5 = − 1 ⇒ 5b = a ⇒ a 1 Area of ∆OAB = | ab| 2 1 10 = |5 b2| ⇒ b2 = 4 ⇒ 2 ⇒ b = ± 2, a = ± 10 x y x y The line can be + = 1 or + = −1 10 2 10 2 m−

13. The bisector AD of ∆ABC will divide the opposite side BC in the ratio of arms of the angle i.e. AB : AC or 3 2 : 4 2 or 3 : 4. Hence, the point D on BC by ratio  31  formula is  , 1 and A is (1, 1.)  4 

3 sin θ − cos θ = 2 π π π  sin θ −  = 1 ⇒ θ − =  6 6 2

∴ Slope of AD = 0. Hence, the slope of CL which is perpendicular from C on AD is ∞. Equation of CLis x − 5 = 0. ∴

6. The points A (a, α ), B (b, β) and C (c , γ ) are collinear, if ⇒

Slope of AB = Slope of BC β −α γ −β = b−a c −b

14. Here, ∆ABC is isosceles.

Let (h, k ) be the coordinates of the vertex A of the ∆ABC. Since, AB = AC ⇒ ( 2 − h )2 + (1 − k )2 = (1 − h )2 + (2 − k )2

i.e. (b − a) (γ − β ) − (c − b) (β − α ) = 0 i.e. γ − β = β − α [Q a, b and c are in AP, so b − a = c − b] ⇒ 2β = γ + α [Qα , β and γ are in GP] ⇒ α =β = γ

⇒ h 2 + k 2 − 4h − 2 k + 5 = h 2 + k 2 − 2 h − 4k + 5 ...(i) ⇒ h=k Again, A (h, k ) lies on the line AB whose equation is 1 y= x 2 1 Then, ...(ii) k= h 2 From Eqs. (i) and (ii), we get h = 0, k = 0 Now, equation of the line joining the points A(0, 0 ) and C(1, 2 ) is 2−0 y−0= ( x − 0) ⇒ y = 2 x 1− 0

7. Let the points be A, B, C and D. Clearly, m of AD = 0 = m of BC ∴ AD || BC || X-axis. Also, AD 2 = BC 2 ∴

mof AC = ∞ = mof BD AC || BC || Y-axis. Also, AC 2 = BD 2

Hence, the given points are vertices of a square.

8. Clearly, the lines x + 3 y = 4 and 6 x − 2 y = 7 are at right angles. So, diagonals of the parallelogram PQRS intersect at right angles. Hence, PQRS must be a rhombus.

9. Equation of the line passing through (2, 2 ) and perpendicular to 3 x + y = 3 is ( x − 2) − 3 ( y − 2) = 0 ⇒ x − 3y = − 4 4 ∴ y-intercept of this line is . 3

Targ e t E x e rc is e s

1. As, 2b = a + c

15.

x −2 y − 3 = = r or (r + 2 ) cos θ sin θ ∴ [r sin θ + 3] + 2 (r cos θ + 2 ) = 3 for y + 2 x = 3 and [(r + 2 )sin θ + 3] + 2 [(r + 2 )cos θ + 2 ] = 5 For y + 2 x = 5, On subtracting, we get

675

Ta rg e t E x e rc is e s

Objective Mathematics Vol. 1

12

…(i) 2 sin θ + 4 cos θ = 2 2 cos θ = 1 − sin θ 4 (1 − sin 2 θ ) = (1 − sin θ )2 3 ⇒ sin θ = 1 or − 5 4 From Eq. (i), cos θ = 0, 5 3 ∴ tan θ = ∞ or − 4 ∴ x − 2 = 0 and 3 x + 4 y = 18 x+5 y+4 r r r 16. = = 1 = 2 = 3 cos θ sin θ AB AC AD Since, (r1 cos θ − 5, r1 sin θ − 4) lies on x + 3 y + 2 = 0. 15 r1 = ∴ cos θ + 3 sin θ 10 Similarly, = 2 cos θ + sin θ AC 6 and = cos θ − sin θ AD On putting in the given relation, we get (2 cos θ + 3 sin θ )2 = 0 2 2 tan θ = − ⇒ y + 4 = − ( x + 5) ∴ 3 3 ⇒ 2 x + 3 y + 22 = 0 ∴ ⇒

17. Clearly, (a) is the only solution. 18. (−3 + 8 − 7 ) (9 − 56 − 7 ) ⇒

(−1, − 1) and (3, 7) are required points. 1 1 19. sin α − > 0 and − cos α > 0 2 2 1 1 or sin α − < 0 and − cos α < 0 2 2  π 3π   π π ⇒ α ∈− ,  ∪  ,  4 4  4 4

20. If P lies between the acute region of the given lines, then (6 sin θ − 1 − cos θ ) (3 sin θ − 1 − cos θ ) < 0 θ θ 1  θ 1 72 cos 4  tan −   tan −  < 0 ⇒ 2 2 6  2 3 1  θ 1 ⇒ < tan   <  2 3 6  1 θ  1 nπ + tan −1   < < tan −1   + nπ ⇒  6 2  3  1  1 ⇒ 2 nπ + 2 tan −1   < θ < 2 nπ + 2 tan −1    3  6

21. Let P be on x + 2 y = 1, then 1+

t t   + 2 2 +  =1  2 2

4 2 3 Let P be on 2 x + 3 y = 15, then t  t    2 1 +  + 4 2 +  = 15   2 2 ⇒

⇒

676

∴

t =−

5 2 6 −4 2 5 2 0

⇒

n2 + 3 n − 4 > 0

⇒ ∴

(n + 4) (n − 1) > 0 ⇒ n > 1 n + 4> 0

...(i)

Y R x=0 4x + y – 21 = 0 A (n , n 2 ) S

P

Q y = 0X

Now, A and O lie on the same side of QR. ∴ 4 x + y − 21 = 0 + 0 − 21 < 0 ⇒ A (n, n 2 ) 4 n + n 2 − 21 < 0 ⇒

n 2 + 4n − 21 < 0 ⇒ (n + 7 ) (n − 3) < 0

⇒ 0 PR So, B1 is an obtuse angle bisector and B2 is an acute angle bisector. ∴ θ1 > θ 2 x y 1 3 65. Let the line be + = 1, then + = 1 a b a b A ≡ (a, 0 ), B ≡ (0, b) ⇒ P ≡ (a, b) 1 3 Thus, locus of P is + = 1. x y

66. Let the equation of side BC be y = x + a. ⇒ A ≡ (1 − a, 1), B ≡ (2, 2 + a) Equation of side AD is y − 1 = − { x − (1 − a)} ⇒ D ≡ (−2, 4 − a) Let C ≡ (h, k ) ⇒ h + 1 − a = 2 − 2 ⇒ h = a−1 and k + 1= 2 + a + 4 − a ∴ k=5 Thus, locus of C is y = 5.

67.

x + 4 y − 4 xy + 4 = 1 − x + 2 y 2

2

x y + =1 4 h+4

⇒

2 x − 4y + 3 = 0

…(ii)

Elimination h from Eqs. (i) and (ii), we get x 2 + 2 xy − 16 = 0

69. Family of lines passes through (0, 2 ) pair of given lines S : ( x − 2 y + 3) ( x − y + 1) = 0 ⇒

x 2 + 2 y 2 + 4 x − 3 xy − 5 y + 3 = 0

Chord with middle point (h, k ) is T = S1 3 ( xh + yk ) 5 ⇒ xh + 2 yk + 2 ( x + h ) − − (y + k) 2 2 = h 2 + 2 k 2 + 4h − 4hk − 5 k + 3 It passes through (0, 2 ). 3 5 ∴ 4 k + 2 h − (2 h ) − (2 + k ) + 3 = 1 2 2 ⇒ 8 y + 4 x − 10 − 5 y + 6 = 0 ⇒ 4x + 3y = 4 π 70. We have, ∠B = ∠C = . Let m be the slope of side 4 1 m−2 1= ⇒ m = and m = − 3 AB, then 1 + 2m 3 Thus, equations of equal sides are x − 3 y + 1 = 0 and 3x + y − 7 = 0 Their combined equation is ( x − 3 y + 1) (3 x + y − 7 ) = 0 ⇒ 3 x 2 − 3 y 2 − 8 xy − 4 x + 22 y − 7 = 0 1 ( x − 2) 3 Equation of refracted ray is y = − 3 ( x − 2 ) Combined equation is y  ( x − 2 + y 3)  x − 2 +  =0  3 y ( x − 2 )2 + y 2 + ( x − 2) 4 = 0 ⇒ 3 72. Common point is (3, 4).

71. Equation of incident ray is y = −

Let L1 and L 2 be y − 4 = m( x − 3) and y − 4 = − Now, ⇒ ⇒

Squaring both sides, we get x 2 + 4 y 2 − 4 xy + 4 = 1 + x 2 + 4 y 2 − 2 x + 4 y − 4 xy

680

B(4, 0)

Equation of BN is

B1 (1, 2)P

…(i)

N

12 x + 5 y − 2

θ + θ2 and 63. Angle bisectors will make the angles 1 2  π θ1 + θ 2   with the X-axis. Hence, their equations  + 2 2 

Ta rg e t E x e rc is e s

Equation of AM is

64 x − 8 y = 115 14 x + 112 y = 15 64 8  14 112    ,b=− a =  or  a = , b =    115 115 15 15 

⇒

62.

68. Let M ≡ (0, h ), N ≡ (0, h + 4)

⇒

A1 = A 2 1 (4 m + 3)2 (4 − 3 m)2 1 = × 2 2 m m 2 7 m + 48 m − 7 = 0 48 , m1 m2 = − 1 m1 + m2 = − 7  24 |θ1 + θ 2| = tan −1   7

1 ( x − 3) m

∆ = abc + 2 fgh − af 2 − bg 2 − ch 2 = 0 11 1 31 Here, a = 6, h = , b = − 10, g = , f = , c = λ 2 2 2 ∴ ∆=0 −5415 On comparing, λ= = − 15 361 121 361 Also, h 2 − ab = + 60 = >0 4 4 ∴ λ = − 15 is valid. 74. We have, a x 2 + 2 λ xy + by 2 + 2 k x + 2 ky + 2 k = 0

From Eq. (i),

⇒

⇒ ⇒

Let

Comparing the coefficient of similar term, we get l1l 2 = a, m1m2 = b, n1n2 = c

4k − 4 ⋅ 2 k [(a + b)k − 2 abk ] ≥ 0

⇒

k − 2 (a + b)k + 4ab ≥ 0

l1m2 + l 2 m1 = 2 h, l1n2 + l 2 n1 = 2 g,

2

m1n2 + m2 n1 = 2 f

2

Now, the two lines are equidistant from origin. 0 ⋅ l1 + 0 ⋅ m1 + n1 0 ⋅ l 2 + 0 ⋅ m2 + n2 = ∴ l12 + m12 l 22 + m2

(k − 2 a) (k − 2 b) ≥ 0

⇒ k does not lie between 2a and 2b. ⇒ k ≤ 2 a, k ≥ 2 b

75. x 3 + y 3 = 0 ⇒ ( x + y ) ( x − xy + y ) = 0 x+ y=0 x 2 − xy + y 2 = 0 2

[real line] [imaginary line]

76. Let x = r cos θ and y = r sin θ 3

2

2

2

2

2

+ 3 r cos θ + 1 = 0 r 3 cos θ + 2 r 2 + 3 r cos θ + 1 = 0 Let r1, r2 , r3 be the three roots. 2 r1 + r2 + r3 = − cos θ 3 cos θ r1r2 + r2 r3 + r3 r1 = =3 cos θ 1 r1r2 r3 = − cos θ Q OA, OB and OC are in HP. 2r r r2 = 1 3 ∴ r1 + r3 ⇒

r2 r1 + r2 r3 = 2 r1r3 Y P (r, θ) r

θ

X

O

From Eq. (ii), ⇒ From Eq. (iii), From Eq. (iv), ⇒

3 r1r3 = 3 r1r3 = 1

1 cos θ 2r r r1 + r3 = 1 3 r2 r1 + r3 = − 2 cos θ r2 = −

n12 (l 22 + m22 ) = n22 (l12 + m12 )

⇒

n12 l 22 − n22 l12 = n22 m12 − n12 m22

⇒

…(i)

⇒

f 4 − g 4 = c (bf 2 − ag 2 )

79. y = x should satisfy a x 2 + 2 hxy + by 2 = 0 ⇒

a + b = − 2h

80. If the lines intersect ( x1, 0 ), then a x 2 + h ⋅ 0 + g = 0, ⇒ ∴

…(iv)

4 g 2 [4 g 2 − 4 ac ] = 4f 2 [4 f 2 − 4 bc ]

78. Clearly, coefficient of x 2 + coefficient of y 2 = 0 in option (a).

…(ii) …(iii)

(n1l 2 + n2 l1 )2 [(n1l 2 + n2 l1 )2 − 4 n1n2 l1l 2 ] = (m1n2 + m2 n1 )2 [(m1n2 + m2 n1 )2 − 4 m1m2 n1n2 ]

∴

r cos θ + r cos θ sin θ + 2 r cos θ + 2 r sin θ 3

⇒

Targ e t E x e rc is e s

⇒

3

a x 2 + 2 hxy + by 2 + 2 gx + 2 f y + c

12

≡ (l1 x + m1 y + n1 ) (l 2 x + m2 y + n2 )

For real λ, B 2 − 4 AC ≥ 0

2

−1

a x 2 + 2 hxy + by 2 + 2 gx + 2 f y + c = 0

2 kλ2 − 2 k 2 λ + (a + b)k 2 − 2 abk = 0

⇒

tan θ = 1 or

77. We have,

⇒ ab ⋅ (2 k ) + 2 λk 2 − ak 2 − bk 2 − 2 λ2 k = 0

4

tan 2 θ = 1

∴

On comparing, h = λ, g = k, c = 2 k, f = k ∴ abc + 2 fgh − af 2 − bg 2 − ch 2 = 0 ⇒

1 2 =− cos θ cos θ 2 1 1 2 cos θ = − = cos θ cos θ cos θ 1 cos 2 θ = 2 sec 2 θ = 2

−2 cos θ −

Straight Line and Pair of Straight Lines

73. Given equation will represent a pair of straight lines, if

hx1 + 0 ⋅ b + f = 0 g f x1 = − = − a h af = gh

81. The lines of first pair are 3 x + 4 y = 0 and 4 x − 3 y = 0. The lines of second pair are 3 x + 4 y − 1 = 0 and 4 x − 3 y + 1 = 0. They may enclose a square or a rectangle. But the distance between each pair of parallel lines is 1 same i.e. . Hence, it is a square. 5

82. Since, area of the triangle formed by a x 2 + 2 hxy + by 2 = 0 and lx + my + n = 0, is 81 − 18 4 = am2 − 2 hlm + bl 2 18 ⋅ 1 + 2 ⋅ 9 ⋅ 1⋅ 0 + 1⋅ 0 2 2 81 × 3 27 sq units = = 2 × 18 4 n 2 h 2 − ab

81

681

Objective Mathematics Vol. 1

12

83. Given, 4 x 2 − 9 xy − 9 y 2 = 0 and x = 2 ⇒

4 (2 ) − 9 (2 )y − 9 y = 0

⇒

9 y 2 + 18 y − 16 = 0

⇒

(3 y + 8) (3 y − 2 ) = 0

2

x−

2

88.

=

a a +b 2

8 2 , 3 3 8   2 ∴ Points of intersection are 2, −  and 2,  .   3 3 10 sq units ∴Required area = 3 ⇒

a 2

y−

b 2

=±

a a + b2 a x= + 2 a x= − 2

2

a2 + b2 2

2

y=−

and

b b a ,y= + 2 2 2 b b a ,y= − 2 2 2

Y

(0, b) (a/2, b/2)

84. Equation of a curve which passes through the point of intersection of two curves C1 and C2 , is

(a, 0)

C1 + 3 C2 = 0 ⇒ (λ + 9) x 2 − 9 y 2 + (24 − 2 λ ) xy = 0 ...(i) which represents a pair of straight lines passing through origin. Let y = m1 x and y = m2 x be two lines.

Ta rg e t E x e rc is e s

∴ m1 m2 x 2 − y 2 − (m1 + m2 ) xy = 0

...(ii)

1 −(m1 + m2 ) m1m2 [from Eqs. (i) and (ii)] =− = ⇒ λ+9 9 24 − 2 λ 2 λ − 24 ⇒ m1 + m2 = 9 Since, both lines are equally inclined with X-axis. ∴ m1 + m2 = 0 ⇒ λ = 12

85. x = ∴

5± 2 5 2 5m5 are the point of intersection. ,y= 3 3 Required area of the triangle, thus formed 0 1 5+ 2 5 = 3 2 5−2 5 3

0 1 2 5−5 1 = 5 sq units 3 2 5+5 1 3

86. Making the equation of curve homogeneous with the

fx − g y = 1 and to be λ perpendicular both the lines represented by this homogeneous equation a + b = 0 ⇒ λ + gf − λ − gf = 0 ⇒ 0 = 0 Hence, λ may have any value. help of equation of line

87. Let P (2, − 1) goes 2 units along x + y = 1 upto A and 5 units along x − 2 y = 4 upto B. Slope of PA = − 1 = tan 135° , slope of PB = ∴ ∴

682

1 = tan θ 2

1 2 , cos θ = 5 5 A ≡ ( x1 + r cos 135° , y1 = r sin 135° ) 1 1  = 2 + 2 × − , − 1+ 2 ×   2 2 = (2 − 2 , 2 − 1) B ≡ ( x1 + r cos θ, y1 + r sin θ ) 2 5  , − 1+ = 2 + 5 ×   5 5

sin θ =

= (2 5 + 2, 5 − 1)

∴The required points are  a + b a + b ,  and   2 2 

X

 a − b b − a , .   2 2 

89. The equation of the line in parametric form is

x−3 y−4 = =r cos θ sin θ Any point on this line is (3 + r cos θ, 4 + r sin θ ). It lies on x = 6, if 3 + r cos θ = 6 ⇒ r = 3sec θ ∴ PR = 3sec θ Again, the point lies on y = 8, if 4 + r sin θ = 8 ∴ r = 4 cosec θ or PS = 4 cosec θ 90. Let h be the height of the triangle. Since, the area of the triangle is a2 . 1 ∴ × a × h = a2 2 ⇒ h = 2a Since, the base lies along the line x = a, the vertex lies on the line parallel to the base at a distance 2a from it. So, the required lines are x = a ± 2 a i.e. x = − a or x = 3 a

91. tan 2 (a + 2 )b + a2 = 0 a = 0, tan 2 (a + 2 )b = 0 ⇒

tan 2 b = 0 2

π π ,− 2 2 π  π  (a, b) = (0, 0 ),  0,  ,  0, −   2  2 1 π 1 y − 0 = ( x − 0 ), y − = ( x − 0 ) ∴ 2 2 2 π 1 and y + = ( x − 0) 2 2 ⇒ 2 y = x, 2 y − π = x and 2y + π = x 9m 92. Let the slope ofu = 0 be m,then the slope of v = 0 is . 2 9m m− 7 2 = −7 m = ∴ 9 1 + m ⋅ 9m 2 + 9 m2 2 7 −7 m ⇒ =± 9 2 + 9 m2 ⇒

⇒

b = 0,

2 + 9 m2 = ± 9 m

or

9 m2 + 9 m + 2 = 0

Case II When P lies below AB. The line BP makes angle 105° with positive direction of X -axis. ∴Equation of line BP is x−5 y −4 = = − 2 2 [Q Plies below of (5, 4)] cos 105° sin 105°

81 − 72 18 9± 3 2 1 = = , 18 3 3 −9 ± 3 2 1 or m= =− ,− 18 3 3 ∴ Equations of the lines are (i) 3y = x and 2 y = 3 x (ii) 3 y = 2 x and y = 3 x (iii) x + 3 y = 0 and 3 x + 2 y = 0 (iv) 2 x + 3 y = 0 and 3 x + y = 0 2h 93. Q m1 + m2 = − b a and m1 m2 = b 2 tan θ = 4 cosec 2 θ tan α + tan β = ∴ sin 2 θ tan 2 θ + cos 2 θ Also, tan α tan β = sin 2 θ 2 = sec θ + cot 2 θ m=

9±

⇒ x = 5 − cos 105° (2 2 ) and y = 4 − 2 2 sin 105° ⇒ x = 5 – (1 − 3 ) and y = 4 − (1 + 3 ) ⇒ x = 4 + 3 and y = 3 − 3

95. Statement II is true. Statement I Since, AB may not be equal to AC. ∴ Perpendicular drawn from A to BC may not bisects BC. Hence, Statement I is false.

96. Figure explains the both statements, D (0, 17/6)

B (17/9, 0) A (–19/2,0) C (0, –19/3)

tan α − tan β = 2 tan α 2 + sin 2 θ Also, = tan β 2 − sin 2 θ i.e.

Case I

O 9x+6y–17 = 0

2x+3y+19 = 0

97. Statement II is true and Statement I is false as excentres of the triangle are also equidistant from the sides.

4−2 = 1 = tan 45° 5− 3

94. Clearly, the slope of line, AB =

12 Straight Line and Pair of Straight Lines

9 m2 − 9 m + 2 = 0

98. Statement I is true and Statement II is false as if L1, L 2 and L 3 are parallel, then also ∆ = 0.

When P is above AB.

The line AP makes 105° angle with positive direction of X-axis. ∴Equation of line AP is Y

|3 x + 2 y| |2 x + 3 y + 6| = 13 13 ⇒ x − y − 6 = 0 and 5 x + 5 y + 6 = 0 According to given equations of sides internal angle bisector at C will have negative slope, image of A will lie on BC with respect to both bisectors.

99. Bisector at C,

Targ e t E x e rc is e s

⇒

100. Statement I is true and follows from statement II as the family of lines can a ( x + y − 1) + b ( x − 2 y ) = 0.

P (x, y) B (5, 4)

2√ 2 60° P (x, y)

60° X′

45° 75°

105°

O Y′

x−3 y −2 = =2 2 cos 105° sin 105° ⇒ and ⇒ and

x = 3 + 2 2 cos 105° y = 2 + 2 2 sin 105°  1 − 3 x = 3+ 2 2    2 2   3 + 1 y =2 + 2 2    2 2 

⇒ x = 3 − 3 and y = 3 +

written

as

101. First, let the equation a x + by + c = 0 represents a

60°

(3, 2)A

be

3

105°

X

family of straight lines passing through (a, b) for different values of a, b and c. Then, we have to show that there is a linear relation between a, b and c. Then, we have to prove that the equation a x + by + c = 0 represents a family of lines passing through a fixed point. Let the linear relation be  m l la + mb + nc = 0 ⇒   a +   b + c = 0  n  n ⇒ ax + by + c = 0 always passes through a fixed point  l m  , . n n  3  2 ∴ 3a + 2b + 4 c = 0 ⇒   a +   b + c = 0  4  4 and ax + by + c = 0 represents system of concurrent  3 1 lines passing through  ,  .  4 2 Thus, Statement I is false and Statement II is true.

683

Objective Mathematics Vol. 1

12

102. As ax 2 + 2 hxy + by 2 + 2 gx + 2 f y + c = 0

2

represents the general equation of second degree. But it represents straight line, if ∆ = abc + 2 fgh − af 2 − bg 2 − ch 2 = 0 Thus, Statement II is false. And to the pair of straight lines formed by 2 x − y = 5 and x + 2 y = 3 is given by (2 x − y − 5) ( x + 2 y − 3) = 0 ⇒ 2 x 2 + 3 xy − 2 y 2 − 11x − 7 y + 15 = 0 Thus, Statement I is true. Hence, Statement I is true and Statement II is false.

103. The joint equation of y = x and y = − x is ( x − y ) ( x + y ) = 0, i.e. x 2 − y 2 = 0.

104. The bisectors of the angles between the lines in new

=

 2g    − 4c  l 

Hence, Statement II is false. Also, x 2 + 6 xy + 9 y 2 + 4 x + 12 y − 5 = 0 represents two parallel straight lines. 4+ 5 6 g 2 − ac ∴ Distance between them = 2 =2 = 1+ 9 a ( a + b) 10 Hence, Statement I is true and Statement II is false. 4−1 108. Slope of AO = =3 2 −1  a −  3 = − 1 ⇒ 3a = b  b ⇒

O(2, 4)

px 2 + 2 xy − py 2 = 0 B (1, –2)

Ta rg e t E x e rc is e s

∴ Statement I is false and Statement II is true.

105. Put 2h = − (a + b) in ax 2 + 2 h x y + by 2 = 0 ⇒

a x 2 − (a + b) xy + by 2 = 0

⇒

( x − y ) (a x − by ) = 0

⇒ One of the line bisects the angle between coordinate axes in positive quadrant. Also, put b = − 2 h − a in ax − by, we have a x − by = ax − (−2 h − a) y = a x + (2 h + a) y Hence, ax + (2 h + a) y is a factor of a x 2 + 2 h x y + by 2 = 0.

106. a x 3 + bx 2 + y + c x y 2 + dy 3 = 0 3

2

 y  y  y ⇒ d   +c   + b  + a=0  x  x  x ⇒ dm3 + cm2 + bm + a = 0 m1 m2 m3 = − a / d ⇒ m3 = a /d as two lines are perpendicular, put m3 = a / d ⇒

…(i) …(ii)

a2 + ac + bd + d 2 = 0

be two parallel lines represented by the given equation. ∴ a x 2 + 2 h x y + by 2 + 2 g x + 2 f y + c = (lx + my + n1 )⋅ (lx + my + n2 ) Comparing coefficients of like terms, we get l 2 = a, m2 = b, 2 lm = 2 h, l (n + n2 ) = 2 g, m (n1 + n2 ) = 2 f n1n2 = c Now, the distance between the parallel lines is

109. Consider, −17 a + 4b + c = − 17 a + 4 b + 2 b − a = − 18 a + 6 b = 6 (b − 3 a) = 0 1− 4 3 1 110. Slope of AC (m1 ) = =− =− 1 + 17 18 6 6 1 −2 − 4 Slope of BC (m2 ) = =− =− 1 + 17 18 3 m2 − m1 − 1/ 3 + 1/ 6 tan C = = 0 and a − −1

 6   6   ∑ x i   ∑ yi   i =1   i =1  So, the fixed point must be  ,  .  6   6          But the fixed point is (2, 1), so that 2 + 3 + 7 + h1 + h2 + h3 =2 6 ⇒ h1 + h2 + h3 = 0 ⇒ h1 = 0, h2 = 0, h3 = 0

[as h1, h2 , h3 ≥ 0]

and maximum slope can occur, if it passes through (0, 0 ). 1 i.e. m< 5 1  ⇒ m ∈  − 1,   5 ⇒ a= −1 1 and b= 5 1  ⇒ a +  = − 1+ 5  b =4

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Objective Mathematics Vol. 1

12

Entrances Gallery 1. 2 ≤ d1( p) + d 2 ( p) ≤ 4

4. First of all find the point of intersection of the lines

For P(α , β ), α > β Y y=x

P (α, β) x = √2

X

x = 2 √2

2 2 ≤ 2α ≤ 4 2 ⇒ ⇒ 2 ≤α ≤2 2 ∴ Area of region = (2 2 )2 − ( 2 )2 = 8 − 2 = 6 sq units

2. For point of intersection, (a − b)x1 = (a − b)y1 Point lies on line y = x ⇒ Let point be (r , r ).

Let M be the mid-point of PP′, then AM is perpendicular bisector of PP′ (where, A is the point of intersection of given lines). P (2, 3)

(r − 1)2 + (r − 1)2 < 2 2

Then, ⇒ ⇒

2|(r − 1)| < 2 2 | r − 1| < 2 −1 < r < 3 bx + ay + c = 0 ax + by + c = 0

Ta rg e t E x e rc is e s

…(i)

2 x − 3 y + 4 = 0 and x − 2 y + 3 = 0 (say A). Now, the line (2 x − 3 y + 4) + k( x − 2 y + 3) = 0 is the perpendicular bisector of the line joining points P(2, 3) and image P′ (h, k ). Now, AP = AP′ and simplify. Given line is …(i) (2 x − 3 y + 4) + k( x − 2 y + 3) = 0, k ∈ R This line will pass through the point of intersection of the lines …(ii) 2 x − 3y + 4 = 0 and …(iii) x − 2y + 3 = 0 On solving Eqs. (ii) and (iii), we get x = 1, y = 2 ∴Point of intersection of lines (ii) and (iii) is (1, 2 ).

X'

(1, 2) A

Y P ′ (h, k)

X

O (0, 0) (– r, – r) (–1, –1) Y'

⇒ (−1, 1) lies on the opposite side of origin for both lines. ⇒ −a − b + c < 0 ⇒ a+ b−c > 0 m+ 3 = 3 1− 3 m

3.

m+

⇒ ⇒ ⇒ or ⇒ y

= – √3x+1

M

3 = ± ( 3 − 3 m) 4m = 0 m=0 2m = 2 3 m= 3

∴

AP = AP′

⇒

(2 − 1) + (3 − 2 )2 = (h − 1)2 + (k − 2 )2 2

⇒

2 = h 2 + k 2 − 2 h − 4k + 1 + 4

⇒

2 = h 2 + k 2 − 2 h − 4k + 5

⇒

h 2 + k 2 − 2 h − 4k + 5 = 2

⇒

h 2 + k 2 − 2 h − 4k + 3 = 0

Thus, the required locus is x 2 + y 2 − 2 x − 4y + 3 = 0 which is a equation of circle with radius = 1 + 4 − 3 = 2 Aliter (2 x − 3 y + 4) + k( x − 2 y + 3) = 0 is family of lines passing through (1, 2). By congruency of triangles, we can prove that mirror image (h, k ) and the point (2, 3) will be equidistant from (1, 2). 2x – 3y + 4 = 0 Q

(2, 3)

Y

P m

X′

(1, 2)

(0, 1) X

O 60°(3, –2) Y′

688

∴ ⇒

Equation is y + 2 = 3( x − 3) 3 x − y − (2 + 3 3 ) = 0

R

(h, k)

x – 2y + 3 = 0

∴Locus of (h, k ) is PR = PQ ⇒ (h − 1)2 + (k − 2 )2 = (2 − 1)2 + (3 − 2 )2 or ( x − 1)2 + ( y − 2 )2 = 2 Hence, locus is a circle of radius 2.

13  , 1 , P = (2, 2 ) 2  2 Slope = − 9 y+1 2 Equation will be =− x −1 9 ⇒ ⇒

9y + 9 + 2 x − 2 = 0 2 x + 9y + 7 = 0

6. Take any point B(0, 1) on given

B(0,1) √3y =x–1

line. Equation of AB′ is

10. Lines perpendicular to same line are parallel to each

−1 − 0 A(√3, 0) ( x − 3) 0− 3 B' (0,–1) − 3y = − x + 3 x − 3y = 3 ⇒ 3y = x − 3

other. ∴

y−0=

⇒ ⇒

12

x y + = 1, we get c 3 20 a = − 3 3 c = 5a = − ∴ 4 Hence, the distance between lines −3 −1 | a − 1| 23 20 = = = 1 1 17 17 + 25 400 400 On comparing with

7. Here, L1 : y − x = 0 and L 2 : 2 x + y = 0 and L 3 : y + 2 = 0 shown as

− p( p2 + 1) = p2 + 1 ⇒

p= −1

Hence, there is exactly one value of p. 4− 3 1 11. Since, slope of PQ = = 1− k 1− k

Straight Line and Pair of Straight Lines

5. Q S = 

and slope of AM = (k − 1) A

Y

X'

y=x

O(0, 0)

(–2, –2) P

L3

Q R

–2

Angle bisector

(1, –2) Y'

L2

P (1, 4)

X

y = –2

y = –2x

8. As x + y =| a|and ax − y = 1intersect in I quadrant. Q x and y-coordinates are positive. a| a| − 1 1 + | a| ∴ x= ≥ 0 and y = ≥0 a+1 1+ a ⇒ 1 + a ≥ 0 and a| a| − 1 ≥ 0 ...(i) ⇒ a ≥ − 1 and a| a| ≥ 1 If −1 ≤ a < 0, then [not possible] − a2 ≥ 1

∴

a2 ≥ 1 ⇒ a ≥ 1 a ≥ 1 or a ∈ [1, ∞ )

…(ii)

9. Since, the line L is passing through the point (13, 32). 13 32 32 8 + =1 ⇒ = − ⇒ b = − 20 b 5 b 5 The line K is parallel to the line L, its equation must be x y x y − = a or − =1 5 20 5 a 20 a

∴

Q (k, 3)

k + 1, 7 2 2

∴ Equation of AM is 7   k + 1  y − = (k − 1)  x −    2   2 

| PO| = 4 + 4 = 2 2 |OQ| = 1 + 4 = 5 Since, OR is angle bisector. OP PR PR 2 2 ∴ = ⇒ = OQ RQ RQ 5 ∴ Statement I is true. But it does not divide that ∆ in two similar triangles. ∴ Statement II is false.

If a ≥ 0, then

M

For y-intercept, x = 0, y = − 4 15 k 2 − 1 7  k + 1 ⇒ − 4 − = − (k − 1)  = ⇒  2  2 2 2 2 ⇒ k − 1 = 15 ⇒

k 2 = 16 ⇒ k = ± 4

12. Equations of bisectors of lines xy = 0 are y = ± x.

Targ e t E x e rc is e s

L1

Put y = ± x in my 2 + (1 − m2 )xy − mx 2 = 0, we get mx 2 ± (1 − m2 )x 2 − mx 2 = 0 ⇒

(1 − m2 )x 2 = 0 ⇒ m = ± 1

13. Since, A is the mid-point of line PQ. a+ 0 3= ∴ 2 ⇒ a=6

Y P (0, b) A (3, 4)

0+b 2 X X′ Q O ⇒ b=8 ( a , 0) Hence, the equation of Y′ line is x y + =1 6 8 ⇒ 4 x + 3 y = 24 1 1 1 14. Since, a, b and c are in HP. Then , and are in AP. a b c 2 1 1 1 2 1 = + ⇒ − + =0 ∴ a b c b a c x y 1 Hence, straight line + + = 0 is always passes a b c through a fixed point (1, − 2 ). and

4=

689

Ta rg e t E x e rc is e s

Objective Mathematics Vol. 1

12

Y

15. Let the intercepts on the

coordinate axes be a and b. According to the given b condition, a+ b= −1 ⇒ b = − a − 1 = − (a + 1) X′ O Let equation of line be x y Y′ + = 1. a b x y − =1 ⇒ a a+1

x1 Now,

Taking x1 common from C1 and y1 from C2 , 1 1 1

...(i)

2

and slope of BC =

b(r 2 − r ) b = a(r 2 − r ) a

∴ Slope of AB = Slope of BC But B is a common point. ∴ A, B and C are collinear i.e. the points ( x1, y1 ), ( x2 , y2 ) and ( x3 , y3 ) lie on a straight line.

19. Since, line OA makes an angle α with X-axis and given

OA = a, then coordinates of A are (a cos α , a sin α ). Also, OB ⊥ OA, then OB makes an angle (90° + α ) with then coordinates of are X-axis, B [a cos (90 ° + α ), a sin (90 ° + α )] i.e. (−a sin α , a cos α ).

A (a cos α, a sin α)

3x 4

Equation of the diagonal AB not passing through the origin is a cos α − a sin α ( y − a sin α ) = ( x − a cos α ) − a sin α − a cos α ⇒ (sin α + cos α ) ( y − a sin α ) = (sin α − cos α ) ( x − a cos α ) ⇒ y (sin α + cos α ) + x (cos α − sin α ) = a sin α (sin α + cos α) − a cos α (sin α − cos α ) = a (sin 2 α + sin α cos α − cos α sin α + cos 2 α ) = a

3x 2 9x 2 + 4c ⋅ =0 4 16 2 x (27 + 9 c ) = 0 c =−3

20. The angle bisector for the given two lines

18. Since, x1, x2 , x3 and y1, y2 , y3 are in GP ⋅ Then, x2 = r x1, x3 = r x1 and y2 = r y1, y3 = r y1

690

y3 = br 2

O

 3x   3x  ∴ 6x 2 − x  −  + 4c  −  = 0  4  4

where, r is a common ratio. The points become ( x1, y1 ), (r x1, ry1 ) and (r 2 x1, r 2 y1 ).

and

90° α

will satisfy the equation 6 x 2 − xy + 4 cy 2 = 0

2

y2 = br

[a cos (90° + α), a sin (90° + α)]B

17. Since, one of the two lines is 3 x + 4 y = 0. Then, y = −

⇒

y1 = b ⇒

where, r is a common ratio. The given points will be A(a, b), B(ar , br )and C(ar 2 , br 2 ). b(r − 1) b Now, slope of AB = = a(r − 1) a

a = 1, 2 h = − 2c , b = − 7 2h 2c Now, m1 + m2 = − =− b 7 a and m1m2 = b 1 =− 7 According to the given condition, m1 + m2 = 4 m1 m2 2c 4 − =− ∴ 7 7 ⇒ c =2

⇒

r 1 = x1 y1(0 ) = 0 r2 1

[Q two columns are identical] Hence, these points lie on a straight line. Aliter Let x1 = a ⇒ x2 = ar and x3 = ar 2 and

On comparing the standard equation ax 2 + 2 hxy + by 2 = 0 , we get

6x 2 +

= x1 y1 r r2

X

a

Since, this line passes through a point (4, 3). 4 3 ∴ − =1 a a+1 4a + 4 − 3a ⇒ =1 a (a + 1) ⇒ a + 4 = a2 + a ⇒ a2 = 4 ⇒ a=±2 On putting the values of a in Eq. (i), we get x y − =1 2 3 x y and + =1 −2 1 16. The given pair of lines is x 2 − 2cxy − 7 y 2 = 0.

⇒

x2 x3

x1 y1 1 y1 1 y2 1 = r x1 r y1 1 y3 1 r 2 x1 r 2 y1 1

2

24 x + 7 y − 20 = 0 and 4 x − 3 y − 2 = 0, are 24 x + 7 y − 20 4x − 3y − 2 =± 25 5 Taking positive sign, we get 2 x + 11y − 5 = 0 This equation of line is already given. Therefore, the given three lines are concurrent with one line bisecting the angle between the other two.

−3 −4

∴

k −13 = 0 8 −11 −33

So,

x+ y=0 5x + y = 4 and x + 5y = 4 On solving above equations pairwise, we get  2 2 A(1, − 1), B  ,  , C (−1, 1)  3 3

⇒ 2 (132 − 143) + 3 (− 99 + 104) + k (− 33 + 32 ) = 0 ⇒ − 22 + 15 − k = 0 ⇒ k = −7

23. Now, mid-point of A(2, 5) and B(5, 1) is  2 + 5 5 + 1 , C   2 2 

7  or C  , 3 . 2 

2

2

C

2

2 2   BC =  −1 −  + 1 −    3 3

B (5, 1)

We know that, the diagonals of a rectangle bisect each other. 7  So, the mid-point C  , 3 lies on a given line. 2  7 ∴ 3=2 × + k ⇒ 3=7 + k ⇒ k = − 4 2

24. Any straight line perpendicular to 2 x + y = 3 is 2y − x + c = 0 Since, it passes through (1, 1). ∴ 2(1) − 1 + c = 0 ⇒ c = −1 On putting c = − 1in Eq. (i), we get 2y − x − 1= 0 ⇒ 2y − x = 1 y x ⇒ − =1 1/ 2 1 1 So, the y-intercept is . 2

...(i)

25. Let the given line divide line joining two points in the ratio m1 : m2 . 2x + 5 y − 7 = 0

A(– 4, 7)

m1 P

m2 B(6, –5)

=

`

=

25 1 + 9 9

26 26 = 9 3

Here, we see that AB = BC Hence, given lines form an isosceles triangle. 2h 1 27. Q m + 2 m = − and m × 2 m = 2 2 1 1 ⇒ 3 m = − h and 2 m2 = ⇒ h = − 3m and m = ± 2 2 3  1 ∴ h = −3  ±  ⇒ h = ±  2 2

28. Let the coordinates of C be ( x1, y1 ). The coordinates of centroid of the triangle are  2 − 2 + x1 −3 + 1 + y1  ,     3 3  x1 y1 − 2  i.e.   , 3 3  So, this point will satisfy the given equation 2 x + 3y = 1 2 x1  y1 − 2  + 3 ∴  =1  3  3 ⇒ 2 x1 + 3 y1 = 9 Hence, the locus of the vertex C is the line 2 x + 3y = 9

29. The equation of line passing through points (5, 0) and ( y − 0) =

Since, the point P lies on the line 2 x + 5y − 7 = 0  6 m1 − 4 m2   −5 m1 + 7 m2  ∴ 2  + 5  −7 = 0  m1 + m2   m1 + m2 

2

CA = (1 + 1)2 + (−1 − 1)2 = 4 + 4 = 2 2

and

(0, 3) is

 6 m1 − 4 m2 −5 m1 + 7 m2  ∴ Coordinates of P =  ,   m1 + m2 m1 + m2 

…(i) …(ii) …(iii)

 2 2  AB =  − 1 +  + 1 3  3  1 25 26 26 = + = = 9 9 9 3

Now,

y = 2x + k

A (2, 5)

m1 : m2 = 1 : 1

26. Given lines are

Straight Line and Pair of Straight Lines

22. Given lines are concurrent.

Targ e t E x e rc is e s

with OX and 3 x − y = 0 makes an angle of 60° with OX. So, the required line is y − 2 = 0. 2 3

12

⇒ 12 m1 − 8 m2 − 25 m1 + 35 m2 − 7 m1 − 7 m2 = 0 ⇒ −20 m1 + 20 m2 = 0 m1 1 = ⇒ m1 = m2 ⇒ m2 1

21. Given lines 3 x + y = 0 makes an angle of 120°

⇒

(3 − 0 ) ( x − 5) ⇒ (0 − 5)

y=

3 ( x − 5) −5

−5 y = 3 x − 15 ⇒ 3 x + 5 y − 15 = 0

∴The length of perpendicular from the point (4, 4) to the above line is (3 × 4 + 5 × 4 − 15) 12 + 20 − 15 d = = 9 + 25 32 + 52 =

17 17 = 2 34

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Objective Mathematics Vol. 1

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30. Equation of line in intercept form is x y + =1 a b 1 1 Here, a = and b = 3 4 x y ∴ + =1 1/ 3 1/ 4 ⇒ 3x + 4y − 1 = 0 ∴Perpendicular distance from (0, 0) to the line is |3 (0 ) + 4 (0 ) − 1| =p 32 + 42 1 =p ⇒ 9 + 16 1 p= ⇒ 5

31. Since, the perpendicular distance from (0, 0 ) to the line

Ta rg e t E x e rc is e s

x sec θ + y cosec θ − a = 0 is p. |0 + 0 − a| a ∴ p= = 2 2 1 1 sec θ + cosec θ + cos 2 θ sin 2 θ a sin θ cos θ a sin θ cos θ 2 a sin 2θ = = ⋅ = 2 2 1 2 2 sin θ + cos θ

Y

2π/3 π/3 X′

PA = PB

Given,

⇒ (PA)2 = (PB)2

⇒

{ x − (a + b)} 2 + { y − (a − b)} 2

⇒

x + (a + b) − 2 x (a + b) + y 2 + (a − b)2

= { x − (a − b)} 2 + { y − (a + b)} 2 2

2

− 2 y(a − b) = x 2 + (a − b)2 − 2 x (a − b) + y 2 + (a + b)2 − 2 y (a + b) ⇒ 2 x ( − a − b + a − b) + 2 y ( − a + b + a + b) = 0 ⇒ −2 b x + 2 b y = 0 ⇒ −x + y = 0 ⇒ x−y=0  −2 + 6 3 − 5 33. Mid-point of AB is  ,  i.e. (2, − 1).  2 2  −5 − 3 Slope of the line AB = = −1 6+ 2 So, slope of the required line is 1. ∴ Required equation of the line is given by y + 1 = 1( x − 2) ⇒ y − x + 3= 0 ⇒x − y = 3

2π 2π and the line OM makes an angle with 3 3 positive direction of X-axis. 2π Thus, slope of the line OM = tan =− 3 3 Hence, equation of line OM is y = − 3 x

⇒

3x + y = 0 29 3 36. Here, a = 3, h = 4, b = − 3, g = , f = − , c = 18 2 2 ∴ The point of intersection  bg − fh af − gh  = 2 ,   h − ab h 2 − ab 29  3     −3 × 2  +  2 × 4 = 16 + 9  

692

−3 29    − × 4  3×   2 2 ,  16 + 9  

 −87 12 −9 116  + −  2 , 2 2  =  −3 , −5  = 2  2 2  25 25       2 h 2 tan θ 2 37. Here, m1 + m2 = − = = 2 b sin θ sin θ cos θ m1 m2 =

and

a (tan 2 θ + cos 2 θ ) = b sin 2 θ = sec 2 θ + cot 2 θ

Now, m1 − m2 = (m1 + m2 )2 − 4m1m2 =

4 − 4 (sec 2 θ + cot 2 θ ) sin θ cos 2 θ

=

 1 4 cos 2 θ  − 4 +  2 2 2 sin θ cos θ cos θ sin θ 

=

4 − 4 sin 2 θ − 4 cos 4 θ sin 2 θ cos 2 θ

=

4 − 4 sin 2 θ − 4 (1 − sin 2 θ ) cos 2 θ sin 2 θ cos 2 θ

=

4 − 4 sin 2 θ − 4 cos 2 θ + 4 sin 2 θ cos 2 θ sin 2 θ cos 2 θ

=

4 sin 2 θ cos 2 θ =2 sin 2 θ cos 2 θ

34. We have, x 2 + 2 2 x y + 2 y 2 + 4 x + 4 2 y + 1 = 0 Here, a = 1, b = 2, h = 2 , g = 2, f = 2 2 , c = 1 ∴Required distance 2 ⋅ 2 − 1⋅ 1 g 2 − ac 3 =2 =2 =2 = 2 units 3 1 (1 + 2 ) a ( a + b)

X

O (0, 0)

∴ ∠ POQ is

 a sin 2θ  2 4 p2 + q 2 = 4   + (a cos 2θ )  2 

32. Let coordinate of point P be ( x, y ).

P (–1, 0)

Y′

2

= a2 sin 2 2θ + a2 cos 2 2θ = a2

Q (3, 3Ö3)

M

Also, the perpendicular distance from (0, 0 ) to the line x cos θ − y sin θ − a cos 2 θ = 0 is q. |0 − 0 − a cos 2θ| ∴ q= = a cos 2θ cos 2 θ + sin 2 θ ∴

3 3−0 = 3 3− 0

35. Slope of OQ =

2

2

X-axis i.e. is twice its distance from the line x − y = 0 Y

x −y xy = −m 2 mx 2 + 2 xy − my 2 = 0 2

⇒

44. Let P(h, k ) be the point from which its distance from

2

But it represents by x 2 − 2 nxy − y 2 = 0. m 2 ∴ = ⇒ mn = − 1 1 −2 n 2 −1 1 39. Q ( y − 1) = x +  1 2 1+ 2 2 1 ( y − 1) =  x +  ⇒ 2 x − 3 y = − 4 ⇒ 3 2 x y ⇒ =1 + −2  4    3 So, x-intercept is −2.

40. Intersection point of 3 x − 4 y = 2 and x + 2 y = − 4 is  −6 −7   , .  5 5

−7 −0 7 = ∴Slope of the required line = 5 −6 −0 6 5 ∴Required equation of line is 7 y − 0 = ( x − 0) ⇒ 6y = 7 x 6

x–y = 0 P (h, k) X'

(h, 0)

Y'

| h − k|  2  2 k = 2 (h 2 + k 2 − 2 hk ) k =2

∴ ⇒ ⇒

2 h 2 + k 2 − 4hk = 0

So, the locus of the point P is 2 x 2 + y 2 − 4 xy = 0

45. Equation of OA is y = 3 x. Equation of OB is y = − 3 x and equation of AB is y = 1. Y B

∴Required equation of the line is 3 3 y + = 0 ( x − 0) ⇒ y = − 2 2 So, the line is below the X-axis at a distance of 3/2.

42. Here, a1 a2 + b1 b2 = 3 × 12 + 4 × (−5) = 16 > 0 For acute angle bisector, 3x + 4y + 5 (12 x − 5 y − 7 ) =− 9 + 16 12 2 + (−5)2 3x + 4y + 5 (12 x − 5 y − 7 ) =− 5 13 ⇒ 39 x + 52 y + 65 = − 60 x + 25 y + 35 ⇒ 99 x + 27 y + 30 = 0 43. Let (h, k ) be the point of the intersection, then coordinates of A and B are (h, 0 ) and (0, k ), respectively. Therefore, equation of variables line is k−0 y−0= ( x − h) 0−h k ⇒ y = − ( x − h) h Since, the line passes through (a, b), we get k b a b = (a − h ) ⇒ + =1 h k h a b So, locus is + = 1. x y ⇒

60°

60°

A

60° 60° O

X′

41. The line is parallel to the X-axis i.e. m = 0 Also, point of intersection of the lines ax + 2 by + 3 b = 0 3  and bx − 2 ay − 3 a = 0 is  0, −  .  2

X

12 Straight Line and Pair of Straight Lines

x 2 − 2 mxy − y 2 = 0 is

60°

X

Y′

Clearly, from figure, ∆OAB is an equilateral triangle. − x sec θ a 46. We have,y = + cosec θ cosec θ sec θ ⇒ Slope = − cosec θ ∴ Required equation of line is cosec θ y − a sin 3 θ = ( x − a cos 3 θ ) sec θ cos θ y − a sin 3 θ = ⇒ ( x − a cos 3 θ) sin θ ⇒ y sin θ − a sin 4 θ = x cos θ − a cos 4 θ ⇒

x cos θ − y sin θ + a sin 4 θ − a cos 4 θ = 0

⇒

x cos θ − y sin θ − a cos 2θ = 0

Targ e t E x e rc is e s

38. Bisector of the angles between the lines

47. Points (a, 0 ) and (0, b) will satisfy the equation of line px − qy = r . ⇒ ∴

ap = r , − bq = r r r  q − p a+ b= − =r    pq  p q

48. Let m be the slope of required line. ∴

m+1 m − (−1) =±1 =1 ⇒ 1− m 1 + m (−1)

⇒ ⇒

m + 1 = 1 − m, m + 1 = − 1 + m m = 0, m = ∞

So, equation of line through (1, 1) is y − 1 = 0, x − 1 = 0

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Objective Mathematics Vol. 1

12

49. Locus is | x | + | y | = 1, which separately represents equation of straight lines. 4 × (− 1) − 3 × (− 4) − k 42 + (−3)2

=1

−4 + 12 − k =±1 5 ⇒ 8− k = ± 5 ⇒ k = 3 or k = 13 ∴ Equations of lines are 4 x − 3 y − 3 = 0 and 4 x − 3 y − 13 = 0. ⇒

51. Equation of line is y = mx + 4. ∴

Required distance =

Ta rg e t E x e rc is e s

Hence, no point lies in it.

55. Let m be the required slope. m−3 =1 1+ 3m

∴

m−3 =±1 1+ 3m

⇒ 4 1 + m2

52. Let equation of line parallel to 3 x − y = 7 be 3x − y = λ.

694

2x + 2y = 5 5 x+ y= ⇒ 2 The distance between two parallel lines. 5 4− 3 3 2 2 ∴ d = = = >1 2 2 4 2 2 1 +1 and

50. Required equation can be 4 x − 3 y − k = 0 ∴

54. Given lines are x + y = 4

Since, it passes through (1, 2). ∴ 3−2 = λ ⇒ λ =1 So, line is 3 x − y = 1 The point of intersection of x + y + 5 = 0 and 3 x − y = 1 is (−1, − 4). So, distance between (1, 2 ) and (−1, − 4)

⇒ and

m − 3 = 1+ 3m m − 3 = − 1− 3m m = − 2, m =

⇒

1 2

56. Required line is passing through (3, 4) and having slope 1. A(3, 7)

= {1 − (−1)} 2 + {2 − (−4)} 2 = (2 )2 + (6)2 = 4 + 36 = 40

53. The equation of line in new position is y − 0 = tan 15° ( x − 2 ) y = (2 − 3 ) ( x − 2 ) ⇒ ⇒ (2 − 3 ) x − y − 4 + 2 3 = 0

(3, 4)B

45°

∴ Equation of required line is y − 4 = 1 ( x − 3) ⇒ x − y + 1= 0 ⇒ y=x+1

C(5, 4)

13 Circle Introduction A circle is the locus of a point which moves in a plane such that its distance from a fixed point in plane is always a constant. The fixed point is called the centre and the constant distance is called the radius of the circle.

Standard Equation of a Circle

Chapter Snapshot ●

The equation of a circle whose centre is at (h, k) and radius r, is given by ( x − h) 2 + ( y − k ) 2 = r 2

●

●

P (x, y) ●

r ●

C (h, k)

Proof Let C be the centre of the circle and its coordinates be (h, k). Let the radius of the circle be r and let P ( x, y) be any point on the circumference. Then, CP = r ⇒ CP 2 = r 2 ⇒ ( x − h) 2 + ( y − k ) 2 = r 2

X

Example 1. The equation of the circle whose centre is (2, − 3) and radius 5, is (a) x 2 + y 2 + 4x − 6 y − 12 = 0 (b) x 2 + y 2 − 4x + 6 y − 12 = 0 (c) x 2 + y 2 − 6x + 4 y − 12 = 0 (d) None of the above Sol. (b) The equation of the required circle is ( x − 2 )2 + ( y + 3)2 = 52 ⇒

x2 + y2 − 4 x + 6 y − 12 = 0

Standard Equation of a Circle Circle Passing through Three Points Position of a Point with Respect to a Circle Intersection of a Straight Line and a Circle

●

Equation of Tangent

●

Normal to a Circle

●

Pair of Tangents

●

Director Circle

●

Pole and Polar

●

Diameter of a Circle

●

It is the required equation of the circle having centre at ( h, k ) and radius equal to r. Ø The above equation is known as the central form of the equation of a circle.

Introduction

Angle of Intersection of Two Circles

●

Family of Circles

●

Coaxial System of Circles

●

Limiting Points

Objective Mathematics Vol. 1

13

X

Example 2. The equation of the circle whose centre is (1, 2) and which passes through the point ( 4, 6), is (a) x 2 + y 2 − 8x − 12 y + 20 = 0 (b) x 2 + y 2 − 2x − 4 y − 25 = 0 (c) x 2 + y 2 − 2x − 4 y − 20 = 0 (d) x 2 + y 2 − 8x − 12 y − 20 = 0

The equation of the circle (i) is ( x − h) 2 + ( y − k ) 2 = h 2 + k 2 ⇒ x 2 + y 2 − 2hx − 2ky = 0

iii.

When the circle touches X-axis Let C(h, k) be the centre of the circle. Since, the circle touches the X-axis. ∴ r=k Y

Sol. (c) We have, centre of the circle (1, 2), since point P(4, 6) lies on the circle, therefore radius of circle, r = (4 − 1)2 + (6 − 2 )2 = 5 Now, the equation of the required circle whose centre is (1, 2) and radius 5 is ( x − 1)2 + ( y − 2 )2 = 52 ⇒

C (h, k)

x2 + y2 − 2 x − 4 y − 20 = 0

k O

Some Particular Cases of the Central Form

X

Hence, the equation of the circle is ( x − h) 2 + ( y − r ) 2 = r 2

The equation of a circle with centre at (h, k ) and radius equal to r, is …(i) ( x − h) 2 + ( y − k ) 2 = r 2

i.

M

⇒

iv.

When the centre of the circle coincides with the origin i.e. h = k = 0 Y

x 2 + y 2 − 2hx − 2ry + h 2 = 0

When the circle touches Y-axis Let C ( h, k ) be the centre of the circle. Since, the circle touches the Y-axis. ∴ h=r Y

r X′

O

X

r

M

X

Hence, the equation of the circle is (x − r ) 2 + ( y − k ) 2 = r 2 ⇒ x 2 + y 2 − 2rx − 2ky + k 2 = 0

Then, Eq. (i) reduces to x 2 + y2 = r 2 When the circle passes through the origin Let O be the origin and C (h, k) be the centre of the circle. Draw CM ⊥ OX . In ∆OCM, we have OC 2 = OM 2 + CM 2 r 2 = h2 + k 2

i.e.

C (h, k)

O

Y′

ii.

r

Y

v.

When the circle touches both the axes In this case, h=k =r Hence, the equation of the circle is (x − r ) 2 + ( y − r ) 2 = r 2 ⇒ x 2 + y 2 − 2rx − 2ry + r 2 = 0 Y

C (h, k) r O

h M

X

r O

696

C (r, r)

r

k

X

When the circle passes through the origin and centre lies on X -axis In this case, we have k = 0 and h = r Y

X

Example 4. The equation of the incircle of the triangle formed by the axes and the line 4x + 3 y = 6, is (a) x 2 + y 2 − 6x − 6 y + 9 = 0 (b) 4( x 2 + y 2 − x − y) + 1 = 0 (c) 4( x 2 + y 2 + x + y) + 1 = 0 (d) None of the above

13 Circle

vi.

Sol. (b) If the inradius is r, then the centre is (r, r) and its distance from the line 4 x + 3 y = 6 is r.

C (r, 0)

X′

X

r

O

4r + 3r − 6

So,

42 + 32

7 r − 6 = 5r

⇒

7 r − 6 = ± 5r ⇒ r = 3,

Y′

Hence, the equation of the circle is ( x − r ) 2 + ( y − 0) 2 = r 2

vii.

2

2

2

4x + 3y = 6

When the circle passes through the origin and centre lies on Y-axis In this case, we have h = 0 and k = r Y

C (0,r) X′

1 2

Y

x + y − 2rx = 0

⇒

=r

O

X

O

3/2

X

Clearly, from the figure, r ≠ 3 ∴ The equation of the incircle is 2 2 2  x − 1 +  y − 1 =  1       2   2 2 1 2 2 x + y − x− y+ =0 ⇒ 4 ⇒ 4( x2 + y2 − x − y) + 1 = 0

General Equation of a Circle The equation x2 + y2 + 2gx + 2 fy + c = 0 always

Y′

The equation of the circle is ( x − 0) 2 + ( y − r ) 2 = r 2 ⇒ x 2 + y 2 − 2ry = 0 X

Example 3. The equation of the circle which touches the Y -axis at the origin and passes through (3, 4) is (a) 2 ( x 2 + y 2 ) − 3x = 0 (b) 3( x 2 + y 2 ) − 25x = 0 (c) 4( x 2 + y 2 ) − 25x = 0 (d) None of these Sol. (b) The circle touches Y-axis at origin i.e. centre of circle lies on X-axis. ∴ The equation of circle is x 2 + y2 − 2 r x = 0 The circle passes through (3, 4). ∴ 9 + 16 − 6r = 0 25 r= ⇒ 6 Required equation of circle is 25 x 2 + y2 − 2 × x=0 6 2 2 ⇒ 3 ( x + y ) − 25 x = 0

represents a circle whose centre is ( − g, − f ) and radius is g2 + f

2

− c.

The given equation is …(i) x 2 + y 2 + 2gx + 2 fy + c = 0 2 2 2 2 2 2 ⇒ ( x + 2gx + g ) + ( y + 2 fy + f ) = g + f − c Proof

⇒

(x + g ) 2 + ( y + f ) 2 = ( g 2 + f

2

− c) 2

⇒ {x − ( − g )}2 + { y − ( − f )}2 = { g 2 + f

2

− c}2

This is of the form ( x − h) 2 + ( y − k ) 2 = r 2 , which represents a circle having centre at ( h, k ) and radius equal to r. Hence, the given Eq. (i) represents a circle whose centre is ( − g, − f ) i.e. 1  1   − coefficient of x, − coefficient of y  2  2 and

radius = g 2 + f

2 2

−c

1  1   coefficient of x  +  coefficient of y = 2  2 

2

− constant term

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If g2 + f 2 − c > 0, then the radius of the circle is real and hence the circle is also real. If g2 + f 2 − c = 0, then the radius of the circle is zero. Such a circle is known as point circle. If g2 + f 2 − c < 0, then the radius g2 + f 2 − c of circle is imaginary but the centre is real. Such a circle is called an imaginary circle as it is not possible to draw such a circle. Special features of the general equation x 2 + y 2 + 2 gx + 2 fy + c = 0 of the circle are ➣ It is quadratic in both x and y. 2 2 ➣ Coefficient of x = Coefficient of y In solving problems, it is advisable to keep the coefficient of x 2 and y 2 unity. ➣ There is no term containing xy i.e. the coefficient of xy is zero. ➣ It contains three arbitrary constants viz. g , f and c. On comparing the general equation x 2 + y 2 + 2 gx + 2 fy + c = 0 of a circle with the general equation of second degree ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 We find that it represents a circle, if a = b i.e. coefficient of x 2 = coefficient of y 2 and h = 0 i.e. coefficient of xy = 0.

Example 5. The centre and radius of the circle 3x 2 + 3 y 2 − 8x − 10 y + 3 = 0 are (a) (4, 5), 1 (b) (8, 10), 3  − 4 −5  4 (c)  , , 2  3 3  3  4 5 4 (d)  ,  , 2  3 3 3 Sol. (d) We have, 3 x2 + 3 y2 − 8 x − 10 y + 3 = 0 8 10 y + 1= 0 ⇒ x 2 + y2 − x − 3 3 −4 −5 and f = Here, g= 3 3 and

r= =

g2 + f2 − c 2

2

 − 4  + − 5  − 1      3  3

32 4 2 = 9 3 4 5 4 Hence, centre is  ,  and radius is 2.  3 3 3

Sol. (b) Let the centre of required circle be (h, h). We have, Y

C (h, h)

B (2, 2) O

⇒

698

Example 6. Radius of bigger circle touching the circle x 2 + y 2 − 4x − 4 y + 4 = 0 and both the coordinate axes, is (a) (3 + 2 2 ) (b) 2(3 + 2 2 ) (c) (6 + 2 2 ) (d) 2(6 + 2 2 )

X

D

π , CB = h + 2, BE = h − 2 4 h−2 π 1 = cos = 4 2 h+2

∠COD = ∠CBE =

⇒ ⇒

h=

2( 2 + 1) ( 2 − 1)

= 2(3 + 2 2 ) X

Example 7. If the equation px 2 + (2 − q ) xy + 3 y 2 − 6qx + 30 y + 6q = 0 represents a circle, then the values of p and q are (a) p = 2, q = 3 (b) p = 3, q = 2 (c) p = 1, q = 2 (d) p = 2, q = 1 Sol. (b) Second degree equation ax2 + 2 hxy + by2 + 2 gx + 2 fy + c = 0 represents a circle, iff a = b and h = 0 In equation px2 + (2 − q )xy + 3 y2 − 6qx + 30 y + 6q = 0, it represents circle iff p = 3 and 2 −q = 0 ⇒ q =2 ∴ p = 3 and q = 2

Equation of Circle when End Points of Diameter are Given If A and B are end points of a diameter of a circle whose coordinates are ( x1 , y1 ) and ( x 2 , y2 ), respectively. Then, the equation of circle is ( x − x1 ) ( x − x 2 ) + ( y − y1 ) ( y − y2 ) = 0 Proof Let P ( x, y) be any point on the circle. P (x, y)

=

X

E

A (x1, y1)

O

B (x2, y2)

∴ Slope of PA ⋅ Slope of PB = −1 y − y2 ( y − y1 ) = −1 × ⇒ x − x 2 ( x − x1 ) ⇒ ( x − x1 ) ( x − x 2 ) + ( y − y1 ) ( y − y2 ) = 0

Example 8. If the abscissa and the ordinates of two points A and B are the roots of the equations x 2 + 2ax − b 2 = 0 and y 2 + 2 py − q 2 = 0 respectively. Then, the equation of circle with AB as diameter is (a) x 2 + y 2 + 2ax + 2 py − b 2 − q 2 = 0 (b) x 2 + y 2 − 2ax + 2 py + b 2 + q 2 = 0 (c) x 2 + y 2 − 2ax − 2 py + b 2 + q 2 = 0 (d) x 2 + y 2 + 2ax + 2 py + b 2 + q 2 = 0 Sol. (a) Let ( x1, x2 ) and ( y1, y2 ) be the roots of the equation x2 + 2 ax − b 2 = 0 and

y2 + 2 py − q 2 = 0 respectively.

Then, x1 + x2 = − 2 a, x1 x2 = − b 2 , y1 + y2 = − 2 p and y1 y2 = − q 2 . The equation of circle with A( x1, y1 ) and as diameter is B( x2 , y2 ) ( x − x1 )( x − x2 ) + ( y − y1 )( y − y2 ) = 0 ⇒ x2 + y2 − x( x1 + x2 ) − y ( y1 + y2 ) ⇒

+ x1 x2 + y1 y2 = 0 x2 + y2 + 2 ax + 2 py − b 2 − q 2 = 0

x 2 + y 2 + 2gx + 2 fy + c = 0

x2 + y2 + 2 gx + 2 fy + c = 0

…(i)

It is passing through (1, t,) (t, 1) and (t , t ), then 1 + t 2 + 2 g + 2 ft + c = 0 t 2 + 1 + 2 gt + 2 f + c = 0 and 2t 2 + 2 gt + 2 ft + c = 0

…(ii) …(iii) …(iv)

On solving Eqs. (ii), (iii) and (iv), we get 2 g (t − 1) + 2 f (1 − t ) = 0 ⇒ g − f = 0 and t 2 − 1 + 2 f(t − 1) = 0 (t + 1) f =− =g 2 From Eq. (iv), 2t 2 − t (t + 1) − t (t + 1) + c = 0 ⇒ c = 2t From Eq. (i), x2 + y2 − (t + 1)x − (t + 1)y + 2 t = 0 ⇒

( x2 + y2 − x − y) − t ( x + y − 2 ) = 0

∴ ∴ and Then,

P + tQ = 0 P=0 Q=0 x 2 + y2 − x − y = 0

…(v)

Intercepts on the Axes …(i)

be the circle passing through three non-collinear points P ( x1 , y1 ), Q ( x 2 , y2 ) and R ( x 3 , y3 ). Then, …(ii) x12 + y12 + 2gx1 + 2 fy1 + c = 0 2 2 …(iii) x 2 + y2 + 2gx 2 + 2 fy2 + c = 0 …(iv) x 32 + y32 + 2gx 3 + 2 fy3 + c = 0

The lengths of intercepts made by the circle x 2 + y 2 + 2gx + 2 fy + c = 0 with X and Y -axes are 2 g 2 − c and 2 f

2

−c

Proof Intercepts on X-axis: Suppose the circle x 2 + y 2 + 2gx + 2 fy + c = 0

On solving Eqs. (ii), (iii) and (iv) as simultaneous linear equations in g, f and c, we can obtain the values of g, f and c. Substituting these values in Eq. (i), we obtain the equation of the desired circle.

meets the X -axis at two points A1 and A2 . Since, on X-axis, y = 0. Therefore, x-coordinates (abscissa) of points A1 and A2 are roots of the equation …(i) x 2 + 2 gx + c = 0

Alternately, we eliminate g, f and c from Eqs. (i), (ii), (iii) and (iv) to obtain the equation of the required circle as follows : x 2 + y2 x y 1 2 2 x1 + y1 x1 y1 1 =0 x 22 + y22 x 2 y2 1

Let x1 and x 2 be the abscissa of A1 and A2 , respectively. Then, x1 and x 2 are roots of Eq. (i). ∴ x1 + x 2 = − 2g and x1 x 2 = c Now, intercept on X-axis = A1 A2 = x 2 − x1

x 32 + y32

x3

y3

Y

1

Remember If P is a point and C is the centre of a circle of radius r, then the maximum and minimum distances of P from the circle are CP + r and CP − r respectively. X

13

and …(vi) x+ y−2 = 0 On solving Eqs. (v) and (vi), we get x = 1 and y = 1.

Circle Passing through Three Points Let

Sol. (a) Let the equation of circle be

Circle

X

Example 9. The circle passing through the distinct points (1, t), (t, 1) and ( t, t ) for all values of t, passes through the point (a) (1, 1) (b) ( −1, − 1) (c) (1, − 1) (d) ( −1, 1)

B2

X′

B1 O A1

A2

X

Y′

= ( x 2 + x1 ) 2 − 4x1 x 2 = 4g 2 − 4c =2 g2 − c Similarly, we have intercept on Y-axis = 2 f

2

−c

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If g2 > c, then the roots of the equation x 2 + 2 gx + c = 0 are real and distinct, so the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 meets the X-axis at two real and distinct points and the length of the intercept on X-axis is 2 g2 − c. If g2 = c, then the roots of the equation x 2 + 2 gx + c = 0 are real and equal, so the circle touches X-axis and the intercept on X-axis is zero. If g2 < c , then the roots of the equation x 2 + 2 gx + c = 0 are imaginary, so the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 does not meet X-axis at real points. Similarly, the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 cuts the Y-axis at real and distinct points, touches or does not met at real points according as f 2 >, = or < c.

Example 10. The equation of the circle touching Y -axis at (0, 3) and making intercept of 8 units on the axis (a) x 2 + y 2 − 10x − 6 y − 9 = 0 (b) x 2 + y 2 − 10x − 6 y + 9 = 0 (c) x 2 + y 2 + 10x − 6 y + 9 = 0 (d) x 2 + y 2 + 10x + 6 y + 9 = 0 Sol. (b, c) Let the equation of the circle be x2 + y2 + 2 gx + 2 fy + c = 0 Since, circle touches Y-axis at (0, 3). ∴ 9 + 6f + c = 0 and f 2 = c ⇒

f 2 + 6f + 9 = 0

⇒

(f + 3)2 = 0 f = −3 c = f2 = 9

and ∴

Since, length of the intercept of circle on Y-axis is 8 units. ∴ ⇒

2 g2 − c = 8 g 2 − c = 16

⇒

g 2 = 16 + c

⇒

g 2 = 16 + 9

⇒

g=± 5

The centre of the required circle are (− g , − f ) i.e. (−5, 3) and (5, 3) and radius =

g2 + f2 − c

= 25 + 9 − 9 = 5 Hence, equation of required circle are ( x + 5)2 + ( y − 3)2 = 25 or

X

700

( x − 5)2 + ( y − 3)2 = 25

⇒

x2 + y2 + 10 x − 6 y + 9 = 0

or

x2 + y2 − 10 x − 6 y + 9 = 0

Example 11. A circle touches a given straight line and cuts of a constant length 2d from another straight line perpendicular to the first straight line. The locus of the centre of the circle is (a) y 2 − x 2 = d 2 (b) x 2 + y 2 = d 2 (d) None of these (c) xy = d 2

Sol. (a) Let the two given straight lines be X and Y-axes and the circle be x2 + y2 + 2 gx + 2 fy + c = 0. It touches X-axis at c = g 2 . The circle cuts an intercept of length 2d on Y-axis. ∴

2d = 2 f 2 − c

⇒

d2 = f2 − c

⇒

d2 = f2 − g2

Hence, the locus of (− g , − f ) is d 2 = (− y)2 − (− x)2 ⇒

y2 − x 2 = d 2

Position of a Point with Respect to a Circle A point ( x1 , y1 ) lies outside, on or inside a circle S ≡ x 2 + y 2 + 2gx + 2 fy + c = 0 according as S 1 >, = or < 0, where S 1 = x12 + y12 + 2gx1 + 2 fy1 + c. Proof Let P ( x1 , y1 ) be the given point and let C be the centre of the circle. Then, the coordinates of C are ( −g, − f ). Now, CP = ( x1 + g ) 2 + ( y1 + f ) 2 The point P lies outside, on or inside the circle according as, CP >, = or < radius of the circle i.e. ( x1 + g ) 2 + ( y1 + f 2 ) >, = or < g 2 + f X

2

−c

Example 12. If the point (a, − a ) lies inside the circle x 2 + y 2 − 4x + 2 y − 8 = 0, then a lies in the interval (a) ( −1, 4) (b) ( − ∞, −1) (c) ( 4, ∞ ) (d) [ −1, 4] Sol. (a) Since, the point (a, − a) lies inside the circle x2 + y2 − 4 x + 2 y − 8 = 0. ∴

a2 + a2 − 4a − 2 a − 8 < 0

⇒

2 a2 − 6 a − 8 < 0

⇒

a2 − 3 a − 4 < 0

⇒ ⇒ ⇒

(a − 4) (a + 1) < 0 −1 < a < 4 a ∈ (−1, 4)

Equation of a Circle in Parametric Form (i) Circle with Centre ( 0, 0 ) and Radius r Let P( x, y) be any point on the circle x 2 + y 2 = r 2 and OP make an angle θ with the positive direction of X-axis. Let PM be perpendicular from P on X-axis.Then, ∠MOP = θ.

X

P (x, y) r θ

X′

O

X

M

Y′

Thus, x = r cos θ, y = r sin θ are the required parametric form of the equation x 2 + y 2 = r 2 , where θ is parameter.

and (cos γ, sin γ ) respectively and A ≡ (−1, 0). α 2 β 2 2 AQ = (1 + cos β ) + sin β = 2 cos 2 γ AR = (1 + cos γ )2 + sin2 γ = 2 cos 2 α β γ Hence, AP, AQ and AR are in GP, then cos , cos , cos 2 2 2 are also in GP.

∴

Let P ( x, y) be any point on the circle ( x − a ) 2 + ( y − b) 2 = r 2 Let PL and CM be perpendiculars from P and C, respectively on X-axis and let CN be perpendicular from C on PL. P (x, y) (a , b ) C

X′

O

M

r θ

X

N

L

X

Y′

Let ∠NCP = θ Then, x = OL = OM + ML = OM + CN = a + r cos θ [QCN = CP cos θ = r cos θ] and y = PL = PN + NL = CM + PN = b + r cos θ [Q PN = CP sin θ = r sin θ] Hence, x = a + r cos θ, y = b + r sin θ is the required parametric equation of the circle ( x − a ) 2 + ( y − b) 2 = r 2

13

Sol. (b) Coordinates of P, Q, R are (cos α, sinα ), (cos β, sinβ )

(ii) Circle with Centre ( a , b ) and Radius r

Y

Example 13. α, β and γ are parametric angles of three points P , Q and R respectively on the circle x 2 + y 2 = 1 and A is the point ( −1, 0). If the lengths of the chords AP , AQ and AR are in GP, then β γ α cos , cos and cos are in 2 2 2 (a) AP (b) GP (c) HP (d) None of these

Circle

Y

AP = (1 + cos α )2 + sin2 α = 2 cos

Example 14. Orthocentre of the ∆ABC, where A ≡ ( a cos θ 1 , a sin θ1 ), B ≡ ( a cos θ 2 , a sin θ 2 ) and C ≡ ( a cos θ 3 , a sin θ 3 ) is 2a  2a  (a)  Σ cos θ 1, Σ sin θ 1  3  3 a a  (b)  Σ cos θ 1, Σ sin θ 1  2  2 (c) (a Σ cos θ1 , a Σ sin θ1 ) 3a  3a  (d)  Σ cos θ 1 , Σ sin θ 1  2  2 Sol. (c) A, B and C lie on the circle x2 + y2 = a2 . That means circumcentre of ∆ABC is O ≡ (0,0) and its centroid is a a G ≡  Σ cos θ1, Σ sin θ1  3  3 If orthocentre of triangle is H ( xp , yp ), then ⇒

OG : GH = 1 : 2 xp = a Σ cos θ1, yp = a Σ sin θ1

Work Book Exercise 13.1 1 If 2( x 2 + y 2 ) + 4λx + λ2 = 0 represents a circle of meaningful radius, then the range of real values of λ is a b c d

R ( 0, ∞ ) ( −∞, 0) None of the above

2 If one end of a diameter of the circle

x 2 + y 2 − 4 x − 6 y + 11 = 0 is (3, 4), then find the coordinates of the other end of the diameter.

a b c d

(2, 1) (1, 2 ) (1, 1) None of the above

3 The equation of circle with centre (−a, − b) and radius a2 − b2 is a x 2 + y 2 − 2 ax − 2 by − 2 b 2 = 0 b x 2 + y 2 + 2 ax + by + 2 b 2 = 0 c x 2 + y 2 + 2 ax + 2 by + 2 b 2 = 0 d None of the above

4 Equation of the circle with centre on the Y-axis and passing through the origin and the point (2, 3) is a x 2 + y 2 + 13 y = 0 b 3 x 2 + 3 y 2 + 13 x + 3 = 0 c 6 x 2 + 6 y 2 − 26 y = 0 d x 2 + y 2 + 13 x + 3 = 0

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5 Circle x 2 + y 2 − 2 x − λx − 1 = 0 passes through

a

x 2 + y2 − 2 x + 4 y + 3 = 0

two fixed points, coordinates of the points are

b

x 2 + y2 − 2 x − 4 y + 3 = 0

a ( 0, ± 1) c ( 0, 1) and ( 0, 2 )

c

x 2 + y2 − 2 x + 3 y = 0

d

x 2 + y2 + 2 x + 4 y − 3 = 0

b ( ± 1, 0) d ( 0, − 1) and ( 0, − 2 )

6 Find the equation of the circle which passes

9 If the circle x 2 + y 2 + 4 x + 22 y + c = 0 bisects the

through the points (2, 3) and (4, 5) and the centre lies on the straight line y − 4 x + 3 = 0.

circumference of the circle x 2 + y 2 − 2 x + 8 y − d = 0 (where, c and d are greater than zero). Then, maximum value of cd is

a x 2 + y 2 − 4 x − 10 y + 25 = 0 b x 2 + y 2 − 4 x − 10 y − 25 = 0 c x 2 + y 2 − 4 x + 10 y − 25 = 0 d None of the above

a 25

c 425

d 625

10 From

the point A(0, 3) on the circle x 2 + 4 x + ( y − 3)2 = 0, a chord AB is drawn and extended to a point P, such that AP = 2 AB. The locus of P is

7 The circle x 2 + y 2 − 10 x − 14 y + 24 = 0 cuts an intercept on Y-axis of length a 5 c 1

b 125

b 10 d None of these

and P ≡ (3, 6), Q ≡ (3 cos θ, 3 sin θ) R ≡ (3 sin θ, − 3 cos θ), then the locus of centroid to ∆ PQR is (where, θ is real parameter)

8 Let

a

x 2 + ( y − 3)2 = 0

b

x 2 + 4 x + ( y + 3)2 = 0

c

x 2 + 8 x + ( y − 3)2 = 16

d ( x + 4)2 + ( y − 3)2 = 16

Intersection of a Straight Line and a Circle Let the equation of the circle be …(i) x 2 + y2 = r 2 and the equation of the line be …(ii) y = mx + c Since, the points of intersection of Eqs. (i) and (ii) satisfy each of them. Therefore, to find the points of intersection of Eqs. (i) and (ii), we have to solve them simultaneously. On substituting the value of y from Eq. (ii) in Eq. (i),

iii.

we get x 2 + ( mx + c) 2 = r 2 ⇒ x (1 + m ) + 2mcx + ( c 2 − r 2 ) = 0 2

i.

702

…(iii)

When points of intersection are real and distinct In this case, Eq. (iii) has two distinct roots. Discriminant > 0 ∴ 2 2 ⇒ 4m c − 4(1 + m 2 )( c 2 − r 2 ) > 0 ⇒ 4r 2 (1 + m 2 ) − 4c 2 > 0 c2 ⇒ r 2 (1 + m 2 ) > c 2 ⇒ r 2 > 1 + m2 ⇒

ii.

2

r>

c 2 = r 2 (1 + m 2 ) ⇒ r =

⇒

1 + m2

When the points of intersection are imaginary In this case, Eq. (iii) has imaginary roots. Discriminant < 0 ∴ ⇒ 4m 2 c 2 − 4 (1 + m 2 ) ( c 2 − r 2 ) < 0 ⇒ r 2 (1 + m 2 ) − c 2 < 0 ⇒ r 2 (1 + m 2 ) < c 2 r
8q 2 Sol. (d) The equation of circle is x2 + y2 − px − qy = 0 Y

c 1 + m2

When the points of intersection are coincident In this case, Eq. (iii) has two equal two roots. Discriminant = 0 ∴ ⇒ 4m 2 c 2 − 4(1 + m 2 )( c 2 − r 2 ) = 0

(0, q)

p, q 2 2

x1 + p ,0 2 X′

(0, 0) O (x 1 , – q ) Y′

(p , q )

(p, 0) X x2 + p ,0 2 (x 2 , – q )

and OP = radius = r y=

For real roots, p2 − 8q 2 > 0 ⇒ X

(x2, y2) Q

p2 > 8q 2

+c M

X′

Example 16. L1 be a straight line through the origin and L2 be the straight line x + y =1. If the intercepts made by x 2 + y 2 − x + 3 y = 0 on L1 and L2 are equal, then one possible equation of line L is (a) x + 6 y = 0 (b) x − 7 y = 0 (c) x + 7 y = 0 (d) x − 6 y = 0

P (x1, y1) X

O

Y′

Sol. (c) Centre of circle is  1 , − 3  and its radius is 5 . 2 2 2

∴

PQ = 2PM

⇒

PQ = 2 OP 2 − OM 2

⇒

PQ = 2 r 2 −

⇒

PQ = 2

B

L1 = 0 E

X

A

L2 = 0

(0, 0)

where, L1 : y = mx and L2 : x + y − 1 = 0 Since, length of intercept made by L1 and L2 are equal. ⇒ Distance from centre should also be equal. m 3 1 3 + − −1 2 2 2 2 ⇒ = 2 2 m +1

1 + m2

Sol.(d) Here, OA = 1

5

2

m2 + 6m + 9 = 8m2 + 8

⇒

7 m − 6m − 1 = 0

⇒

(7 m + 1) (m − 1) = 0

P

y = 2x + 1

A Q

2

⇒

1 + m2

Example 17. Length of the chord cut off by y = 2x +1 from the circle x 2 + y 2 = 4 is equal to 13 17 (a) 2 (b) 2 5 5 11 19 (d) 2 (c) 2 5 5

 m + 3  = 2(m2 + 1)    2 

⇒

c2

r 2 (1 + m 2 ) − c 2

D C

⇒

13

Y

mx

Circle

On putting y = − q , we get x2 − px + 2q 2 = 0

O (0, 0)

m = 1, −

1 7

∴ Required straight lines are ⇒

y= x

and

y− x=0

and

1 x 7 x + 7y = 0

y =−

Length of the Intercept Cut off from a Line by a Circle The length of the intercept cut off from the line y = mx + c by the circle x 2 + y 2 = r 2 is 2

Proof

r 2 (1 + m 2 ) − c 2

(1 + m 2 ) Draw OM ⊥ PQ. Join OP.

OM = Length of the perpendicular from O (0, 0) to y = mx + c c = 1 + m2

1 19 = 5 5 19 PQ = 2 PA = 2 5

PA 2 = OP 2 − OA 2 = 4 −

Now, ⇒

Condition of Tangency The line y = mx + c touches the circle x 2 + y 2 = r 2 , if the length of the intercept is zero. i.e. PQ = 0 ⇒ ⇒ ⇒

2

r 2 (1 + m 2 ) − c 2 1 + m2

=0

r 2 (1 + m 2 ) − c 2 = 0 c = ± r 1 + m2

which is the required condition.

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Example 18. The line 3x − 2 y = k meets the circle x 2 + y 2 = 4r 2 at only one point, if k 2 is equal to 52r 2 20r 2 (a) 20r 2 (b) 52r 2 (c) (d) 9 9

X

(a)

Sol.(b) The line 3 x − 2 y = k meets the circle x + y = 4r at 2

2

2

k 3 = ± 2 r 1 +   2 2

⇒

k2 9 = 4r 2  1 +   4 4

2b

2

⇒ k 2 = 52 r 2

a 2 − 4b 2

(b)

a 2 − 4b 2 2b (c) a − 2b

only one point, so it is tangent to the circle. ∴

Example 20. If a > 2 b > 0, then the positive value of m for which y = mx − b 1 + m 2 is a common tangent to and x 2 + y2 = b2 2 2 2 ( x − a ) + y = b is

(d)

2b b a − 2b

Sol. (a) We have, y = mx − b 1 + m2 is a tangent to the circle x2 + y2 = b 2 for all values of m. As it touches the circle

X

Example 19. The condition that the straight line cx − by + b 2 = 0 may touch the circle x 2 + y 2 = ax + by, is (a) abc =1 (b) a = c (c) b = ac (d) None of these

( x − a)2 + y2 = b 2 and the equation of tangent of slope m to the circle ( x − a)2 + y2 = b 2 is y = m( x − a) ± b 1 + m2

Sol. (b) Circle is given as x2 + y2 = ax + by …(i) ⇒ x2 + y2 − ax − by = 0 The line is given as cx − by + b 2 = 0 c …(ii) y =  x + b  ⇒ b  On substituting the value of y from Eq. (ii) in Eq. (i), we get 2 c c x2 +  x + b  − ax − b  x + b  = 0  b  b c2 2 2 2 2 ⇒ x + 2 x + 2cx + b − ax − cx − b = 0 b  c2  …(iii) x2  1 + 2  + x (c − a) = 0 ⇒ b   If it is a perfect square, then c − a = 0 or c = a.

Equation of Tangent i.

Slope form The equation of a tangent of slope m to the circle x 2 + y 2 = r 2 is

∴

y = mx − b 1 + m2

and

y = m( x − a) ± b 1 + m2 are same.

∴ − ma + b 1 + m2 = − b 1 + m2 ⇒

ma = 2 b 1 + m2

⇒ ⇒ ⇒

X

m2 a2 = 4b 2 (1 + m2 ) m (a − 4b 2 ) = 4b 2 2

2

m=

2b a − 4b 2 2

Example 21. A light ray gets reflected from the line x = − 2. If the reflected ray touches the circle x 2 + y 2 = 4 and point of incident is ( −2, − 4), then equation of incident ray is (a) 4 y + 3x + 22 = 0 (b) 3 y + 4x + 20 = 0 (c) 4 y + 2x + 20 = 0 (d) x + y + 6 = 0 Sol. (a) Any tangent of x2 + y2 = 4 is y = mx ± 2 1 + m2 . If it passes through (−2, − 4) , then (2 m − 4)2 = 4 (1 + m2 ). x = –2

y = mx ± r 1 + m 2 The coordinates of the point of contact are   r  ± rm , m   2 2  1+ m 1+ m   Ø

●

and the coordinates of the points of contact are  r   a ± mr , b m  1 + m2 1 + m2  

704

P (–2, – 4)

The equation of tangent of slope m to the circle (x − a)2 + (y − b)2 = r 2 is given by y − b = m(x − a) ± r 1 + m2

●

A

The equation of the tangent of slope m to the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 is

y + f = m (x + g) ± (g2 + f2 − c)(1 + m2)

⇒

4 m2 + 16 − 16 m = 4 + 4 m2 3 m = ∞, m = ⇒ 4 Also, m = ∞ as x = − 2 is reflected ray. Hence, slope of 3 reflected ray is . Thus, equation of incident ray is 4 3 ( y + 4) = − ( x + 2 ) 4 ⇒ 4 y + 3 x + 22 = 0

Sol. (b) Equation of tangent to x2 + y2 = 5 at (1, − 2 ) is

Point form The equation of the tangent at the

x − 2 y = 5. Putting x = 2 y + 5 in second circle, we get (2 y + 5)2 + y2 − 8 (2 y + 5) + 6 y + 20 = 0

point P ( x1 , y1 ) to a circle x2 + y2 + 2gx + 2 fy + c = 0 is

xx1 + yy1 + g ( x + x1 ) + f ( y + y1 ) + c = 0

iii.

Parametric form The equation of the tangent to the circle ( x − a ) 2 + ( y − b) 2 = r 2 at the point ( a + r cos θ, b + r sin θ ) is ( x − a ) cos θ + ( y − b) sin θ = r

Ø Equation of tangent for x + y = r at (x1 , y1) is xx1 + yy1 = r . 2

X

2

2

2

Example 22. The equation of tangent to the at the point circle x 2 + y 2 − 2ax = 0 [ a (1 + cos α ), a sin α ] is (a) x cos α + y sin α = a (b) x cos α + y sin α = 1 + cos α (c) x cos α + y sin α = a (1 + cos α ) (d) None of the above Sol. (c) The equation of tangent of x2 + y2 − 2 ax = 0 at [a(1 + cos α ), a sin α ] is xa(1 + cos α ) + ya sin α − a [ x + a (1 + cos α )] = 0 axcos α + ay sin α − a2 (1 + cos α ) = 0 ⇒ x cos α + y sin α = a (1 + cos α )

X

Example 23. Tangent to circle x 2 + y 2 = 5 at (1, − 2), also touches the circle x 2 + y 2 − 8x + 6 y + 20 = 0. The coordinates of the corresponding point of contact is (a) (2, 0) (b) (3, − 1) (c) ( −1, 3) (d) (1, − 2)

⇒

5 y2 + 10 y + 5 = 0

⇒

( y + 1)2 = 0

⇒ ⇒ Thus, point of contact is (3, − 1).

13 Circle

ii.

y = −1 x = −2 + 5 = 3

Normal to a Circle The normal at any point on a curve is a straight line which is perpendicular to the tangent to the curve at that point. X

Example 24. The normal to the circle x 2 + y 2 − 3x − 6 y − 10 = 0 at the point ( −3, 4) is (a) 2x + 9 y − 30 = 0 (b) 9x − 2 y + 35 = 0 (c) 2x − 9 y + 30 = 0 (d) 2x − 9 y − 30 = 0 Sol. (a) The

equation of normal to the x2 + y2 + 2 gx + 2 fy + c = 0 at any point ( x1, y1 ) is y +f ( x − x1 ) y − y1 = 1 x1 + g

circle

x − x1 y − y1 = x1 + g y1 + f x+ 3 y− 4 = −3 − 3 / 2 4 − 3 2 x + 9 y − 30 = 0

or ⇒ ⇒

Work Book Exercise 13.2 1 The equation of the circle which is touched by

4 If the circle x 2 + y 2 + 4 x + 22 y + c = 0 bisects the

y = x, has its centre on the positive direction of the X-axis and cuts off a chord of length 2 units along the line 3 y − x = 0, is

circumference of the circle x 2 + y 2 − 2 x + 8 y − d = 0, then c + d is equal to

a c

x 2 + y2 − 4 x + 2 = 0 x 2 + y2 − 8 x + 8 = 0

b d

x 2 + y2 − 4 x + 1 = 0 x 2 + y2 − 4 y + 2 = 0

2 The locus of the point of intersection of perpendicular tangents to the circles x 2 + y 2 = a2 and x 2 + y 2 = b2 is a c

x 2 + y2 = a 2 − b 2 x 2 + y 2 = ( a + b )2

a

x 2 + y2 = a 2 + b 2 None of these

3 If the line y = mx does not intersect the circle ( x + 10 )2 + ( y + 10 )2 = 180, then

b 50

c 40

d 56

5 If AB is the intercept of the tangent of the circle x 2 + y 2 = r 2 between the coordinate axes, then the locus of the vertex P of the rectangle OAPB is a c

b d

60

x 2 + y2 = r 2

b

1

d

x2

+

1 y2

= r2

1

x2

+

1 y2

=

1 r2

None of these

6 The area of the triangle formed by the tangent at the point (a, b) to the circle x 2 + y 2 = r 2 and the coordinate axes is

a

m ∈ ( −2, ∞ )

b

1 m ∈  − ∞, −   2

a

r4 2 ab

b

r4 2 ab

c

1 m ∈  −2, −   2

d

None of these

c

r4 ab

d

r4 ab

705

13

two circles a( x 2 + y 2 ) + bx + cy = 0 and A( x + y 2 ) + Bx + Cy = 0 touch each other, then

7 If

2

Objective Mathematics Vol. 1

a c

aC = cA aB = bA

b d

bC = cB aA = bB = cC

c

c

3 ( x + y ) = 4a 2

2

2

b

3( x 2 + y2 ) = a 2

d

4 ( x 2 + y 2 ) = 3a 2

then the parametric 35 x 2 − 24 x − 35 = 0, equations x = α + a cos θ, y = β + a sin θ can represent a family of

(1, 0) and (3,0) and touch the Y-axis intersect at an angle θ, then cos θ is equal to 1 2 1 d − 4

1 2 1 4

x 2 + y 2 = 4a 2

10 If α, β are the roots of the equation

8 The circles which can be drawn, to pass through

a

a

5 7 straight lines passing through  − , −   7 5 5 7  b straight lines passing through  , −  7 5 5 7 7 5  c circles passing through  − ,  and  ,   7 5 5 7 7 5 d concentric circles whose centres are at  , −  5 7

b −

a

9 The locus of the point of intersection of the tangents to the circle x 2 + y 2 = a2 at points whose π parametric angles differ by , is 3

Pair of Tangents Or according as x12 + y12 − a 2 >, = or < 0. Or according as the point P ( x1 , y1 ) lies outside, on or inside the circle x 2 + y 2 = a 2 .

Theorem 1 From a given point, two tangents can be drawn to a circle which are real and distinct, coincident or imaginary according as the given point lies outside, on or inside the circle. Proof Let ( x1 , y1 ) be a given point and the

Theorem 2 The combined equation of the pair of tangents drawn from a point P ( x1, y1 ) to the circle x 2 + y 2 = a 2 is

equation of the circle be x + y = a . Then, the equation of any tangent to the circle is of the form y = mx ± a 1 + m 2 , where m is the slope of the tangent. 2

2

2

( x 2 + y 2 − a 2 ) ( x12 + y12 − a 2 ) = ( xx1 + yy1 − a 2 ) 2 or SS ′ = T 2 where, S = x 2 + y 2 − a 2 , S ′ = x12 + y12 − a 2 and T = xx1 + yy1 − a 2

If this tangent passes through the given point ( x1 , y1 ), we must have y1 = mx1 ± a 1 + m 2 ( y1 − mx1 ) 2 = a 2 (1 + m 2 )

⇒ ⇒

m 2 ( x12 − a 2 ) − 2mx1 y1 + y12 − a 2 = 0 …(i)

This equation being quadratic in m, gives two values of m (real, coincident or imaginary), corresponding to any given values of x1 and y1 . For each value of m, we have a corresponding tangent. T′

X

Example 25. The angle between the pair of  1 tangents from the point 1,  to the circle  2 x 2 + y 2 + 4x + 2 y − 4 = 0, is  4  4 (b) sin −1   (a) cos −1    5  5  3 (c) sin −1   (d) None of these  5 Sol. (b) The equation of the pair of tangents is SS1 = T 2 .

O P (x 1 , y 1 )

2   1 1 ⇒ ( x2 + y2 + 4 x + 2 y − 4) 12 +   + 4 ⋅ 1 + 2 ⋅ − 4   2 2   2

⇒

1 1 =  x ⋅ 1 + y ⋅ + 2( x + 1) +  y +  − 4  2 2   9 2 9 2 ( x + y + 4 x + 2 y − 4) = (2 x + y − 1)2 4 4 3 x2 + 4 xy − 8 x − 4 y + 5 = 0

∴

Required angle = tan−1

T

⇒

706

Hence, in general, two tangents can be drawn from the point ( x1 , y1 ) to a circle. The tangents are real, coincident or imaginary according as the values of m obtained from Eq. (i) are real, coincident or imaginary. Or according as 4x12 y12 − 4( x12 − a 2 )( y12 − a 2 ) ≥ or < 0

2 (2 )2 − 3 × 0 3+ 0

4 4 = tan−1   = sin−1    3  5

Example 26. The number of real tangents that can be drawn from (2, 2) to the circle x 2 + y 2 − 6x − 4 y + 3 = 0 is (a) 0 (b)1 (c) 2 (d) 3

Sol. (a) The equation of a tangent to x2 + y2 = 1 is …(i) x cos α + y sin α = 1 The equation of tangent to x2 + y2 = 3, perpendicular to Eq. (i), is …(ii) x sin α − y cos α = 3 Let the coordinates of P be (h, k). Then, h cos α + k sin α = 1 and h sin α − k cos α = 3 Eliminating (h, k ) from these two equations, we get h2 + k 2 = 4

Sol. (a) The point (2, 2) lies inside the circle. Hence, there cannot be a tangent drawn from (2, 2) to the given circle. X

Example 27. The angle between tangents drawn from point P to the circle 2θ. x 2 + y 2 + 4x − 6 y + 9 sin 2 θ + 13 cos 2 θ = 0 is Locus of point P is (a) x 2 + y 2 + 4x − 6 y − 9 = 0 (b) x 2 + y 2 + 4x − 6 y − 4 = 0

So, locus of (h, k ) is x2 + y2 = 4, which is a circle of radius 2.

Length of the Tangents

(c) x 2 + y 2 + 4x − 6 y + 9 = 0

The length of the tangent from the point P ( x1, y1 ) to the circle x 2 + y 2 + 2gx + 2 fy + c = 0 is equal to

(d) x 2 + y 2 + 4x − 6 y + 4 = 0 A

Sol. (c) Centre is (− 2, 3) and radius is 2 sin θ , CA we have CP = =2 sin θ

θ

x12 + y12 + 2gx1 + 2 fy1 + c = S 1

P (h, k)

T

C B

Thus, (h + 2 )2 + (k − 3)2 = 4

α P α (x 1 , y 1)

Hence, locus is x 2 + y2 + 4 x − 6 y + 9 = 0 X

13 Circle

X

Example 28. PA is tangent to x 2 + y 2 = a 2 and PB is tangent to x 2 + y 2 = b 2 ( b > a ). If π ∠APB = , then locus of point P is 2 (a) x 2 − y 2 = a 2 − b 2

(b) x 2 + y 2 = b 2 − a 2

(c) x 2 + y 2 = a 2 + b 2

(d) x 2 − y 2 = b 2 − a 2

C (–g, –f ) T′

Ø

●

If PT is the length of the tangent from a point P to a given circle, then PT 2is called the power of the point with respect to the given circle.

T

Sol. (c) If ∠APB = π , then OAPB will be a rectangle. 2

B

P (h, k)

P

A

⇒ A

O

B

PT 2 = PA ⋅ PB

All these formulae are applicable when coefficients of x 2 and y 2 is unity. A

⇒

L

OP 2 = OA 2 + OB2

Thus,

h + k =a + b 2

2

2

R

2

O (– g, –f ) R

Thus, required locus is x + y = a + b . 2

2

2

2

D d L

X

Example 29. If the tangent from a point P to the circle x 2 + y 2 = 1 is perpendicular to the tangent from P to the circle x 2 + y 2 = 3, then the locus of P is a (a) circle of radius 2 (b) circle of radius 4 (c) circle of radius 3 (d) None of the above

θ θ

P (x1,y1)

B ●

●

Area of quadrilateral PAOB = 2 ∆POA 1 = 2 ⋅ RL= RL 2 Find AB i.e. length of chord of contact R AB = 2L sinθ, where tanθ = L 2RL = R2 + L2

707

13 Objective Mathematics Vol. 1

●

Area of ∆PAB (triangle formed by pair of tangents and corresponding chord of contact) 1 = AB × PD 2 1 = (2 Lsin θ)(Lcos θ) 2 = L2sin θ cos θ 3

=

●

RL 2

R + L2 Angle 2θ between the pair of tangents 2

tan2θ =

●

X

2 tanθ 2RL = 2 2 2 1 − tan θ L(L − R )

The locus of the point of intersection of two perpendicular tangents to a given conic is known as its director circle. The equation of the director circle of the circle x 2 + y 2 = a 2 is x 2 + y 2 = 2a 2 X

 2RL  2θ = tan −1  2 2 L − R  Equation of the circle circumscribing the ∆PAB. One such circle described on OP as diameter i.e. (x − x1)(x + g) + (y − y1)(y + f ) = 0

h2 + k 2 + 2 gh + 2 fk + α = 0 Length of the tangent from (h, k ) to the second circle =

h2 + k 2 + 2 hg + 2 fk + β

= β −α

Example 31. If OA and OB are tangents from the origin to the circle, x 2 + y 2 + 2gx + 2 f y + c = 0 and C is the centre of circle, then area of the quadrilateral OACB is 1 (b) c ( g 2 + f 2 − c) (a) ( g 2 + f 2 − c) 2 (g 2 + f 2 − c 2 ) (c) c ( g 2 + f 2 − c) (d) c Sol. (b) Since, OA = OB = c (length of the tangent from the origin to the circle) CA = CB =

and

g2 + f2 − c

Example 32. The intersection of tangents y = a sin θ at the points, π differ by is a 2 (a) straight line (c) pair of straight line

locus of the point of to the circle x = a cos θ, whose parametric angles

(b) circle (d) None of these

Sol. (b) Tangents at θ and  θ + π  are

Example 30. The length of the tangent drawn from any point on the circle 2 2 to the circle x + y + 2gx + 2 fy + α = 0 2 2 x + y + 2gx + 2 fy + β = 0, is α (d) (a) β − α (b) α − β (c) αβ β Sol.(a) Let (h, k) be any point on the first circle, then

X

Director Circle



2

x cos θ + y sin θ = a π π and x cos  θ +  + y sin  θ +  = a   2 2 ⇒

x cos θ + y sin θ = a and − x sin θ + y cos θ = a

Squaring on x 2 + y2 = 2 a 2 . X

both

sides

and

adding,

Sol. (b) Clearly, (0, 0) lies on director circle of the given circle. Now, equation of director circle is ( x + g )2 + ( y + f )2 = 2 (g 2 + f 2 − c ) If (0, 0) lies on it, then g 2 + f 2 = 2(g 2 + f 2 − c ) ⇒

g 2 + f 2 = 2c

Chord of Contact The equation of the chord of contact of tangents drawn from a point ( x1 , y1 ) to the circle x 2 + y 2 = a 2 is Y

T (a, b) C

X′

O

P (x1, y1) B

708

Area of the quadrilateral OACB = 2 Area of ∆OAC 1 = 2 ×   OA × CA = c g 2 + f 2 − c 2

get

Example 33. If the angle between tangents drawn to x 2 + y 2 + 2gx + 2 f y + c = 0 from (0, 0) is π , then 2 (a) g 2 + f 2 = 3c (b) g 2 + f 2 = 2c (d) g 2 + f 2 = 4c (c) g 2 + c 2 = 5c

A

(0, 0) O

we

T' (a1, b1) Y′

xx1 + yy1 = a 2 or T = 0

X

X

Since, Eqs. (i) and (ii) passes through (x1 , y1 ). Therefore, αx1 + βy1 = a 2 and α 1 x1 + β1 y1 = a 2 ⇒ (α, β) and (α 1 , β1 ) lies on xx1 + yy1 = a 2 . Hence, the chord of contact is xx1 + yy1 = a i.e. T = 0. X

Y

2

R O X'

X

Example 35. Tangents are drawn to x 2 + y 2 = 1 from any arbitrary point P on the line 2x + y − 4 = 0. The corresponding chord of contact passes through a fixed point whose coordinates are  1  1 1 1  1 1 (c)  ,  (d) 1,  (a)  ,  (b)  , 1  2  4 2 2   2 4 Sol. (c) Let P ≡ (a, 4 − 2 a). Equation of chord of contact is ⇒

x ⋅ a + y ⋅ ( 4 − 2 a) = 1 (4 y − 1) + a ( x − 2 y) = 0

It will always pass through a 1 coordinates are y = and x = 2 y = 4

fixed point whose 1 . 2

90° Q Y'

As shown in diagram, ∠ROQ = π /2 Also, ∠PRO = ∠PQO = π/2. Then, quadrilateral PROQ is square and hence, PR = RO. ⇒ ⇒

…(i) …(ii)

If A and B are the points of intersection of Eqs. (i) and (ii), clearly AB will be the common chord whose equation will be ( x2 + y2 − 12 ) − ( x2 + y2 − 5 x + 3 y − 2 ) = 0 …(iii) ⇒ 5 x − 3 y − 10 = 0 If P is the point, where the tangents at A and B with respect to Eq. (i), meet each other, AB will be the chord of contact of P. Let the coordinates of P be (α, β). Equation of the chord of contact of (α, β ) with respect to Eq. (i) is …(iv) xα + yβ − 12 = 0 As Eqs. (iii) and (iv) represent the same equation, on comparing the coefficients, we get α β −12 = = 5 −3 −10 18 ⇒ α = 6 and β = − 5

X P (h, k)

Sol. (a) The circles are given as and

13

Sol. (c)

Example 34. Tangents are drawn to the circle x 2 + y 2 = 12 at the points where it is met by the circle x 2 + y 2 − 5x + 3 y − 2 = 0, the point of intersection of these tangents is 18    18  (b)  6,  (a)  6, −    5 5  18  (c)  , 6 (d) None of these 5  x2 + y2 = 12 2 2 x + y − 5x + 3y − 2 = 0

Example 36. If the chord of contact of tangents drawn from the point ( h, k ) to the circle x 2 + y 2 = a 2 subtends a right angle at the centre, then h 2 + k 2 is equal to a2 (a) (b) a 2 2 (c) 2a 2 (d) None of these

Circle

Proof Let T (α, β) and T ′ (α 1 , β1 ) be the points of contact of tangents drawn from P ( x1 , y1 ) to x 2 + y 2 = a 2 . The equations of tangents at T (α, β) and T (α 1 , β1 ) to the circle x 2 + y 2 = a 2 are …(i) αx + βy = a 2 …(ii) and α 1 x + β1 y = a 2

X

h2 + k 2 − a2 = a h2 + k 2 = 2 a2

Example 37. If the chord of contact of tangents drawn from a point on the circle x 2 + y 2 = a 2 to the circle x 2 + y 2 = b 2 touches the circle x 2 + y 2 = c 2 , then, a, b and c are in (a) AP (b) GP (c) HP (d) None of the above Sol. (b) Let any point on the circle x2 + y2 = a2 be (acos θ, asinθ). The chord of contact of this point with respect to the circle x2 + y2 = b 2 is axcos θ + aysinθ = b 2 . If this chord touches the circle x2 + y2 = c 2 , then 0 + 0 − b2

⇒ ⇒

=c a2 cos 2 θ + a2 sin2 θ b2 =c a 2 b = ac

Chord Bisected at a Given Point The equation of the chord of the circle x 2 + y 2 = a 2 bisected at the point ( x1 , y1 ) is given by xx1 + yy1 − a 2 = x12 + y12 − a 2 or T = S′ Proof Let AB be the chord of the circle x 2 + y 2 = a 2 which is bisected at P ( x1 , y1 ). Then, OP ⊥ AB .

709

13

We have,

Pole and Polar

y1 − 0 y1 = x1 − 0 x1 − x1 Slope of AB = ∴ y1 So, the equation of AB is x y − y1 = − 1 ( x − x1 ) y1

Objective Mathematics Vol. 1

Slope of OP =

Polar of point with respect to a circle If through a point P ( x1 , y1 ) (inside or outside a circle), there any straight line is drawn to meet the given circle at Q and R, then the locus of the point of intersection of the tangents at Q and R is called the polar of P and P is called the pole of the polar.

Y

Theorem The polar of a point P ( x1, y1 ) with respect to the circle x 2 + y 2 = a 2 is xx1 + yy1 = a 2 or T = 0.

X'

O(0, 0)

X

A

R

B P (x, y1)

P (x1, y1) Y'

⇒ ⇒ ⇒ ⇒ X

yy1 − = − xx1 + x12 xx1 + yy1 = x12 + y12 xx1 + yy1 − a 2 = x12 + y12 − a 2 T = S′ y12

Example 38. Locus of mid-point of chords of x 2 + y 2 + 2gx + 2 fy + c = 0 that pass through the origin, is (a) x 2 + y 2 + 2gx + 2 fy = 0 (b) x 2 + y 2 + gx + fy + c = 0 (c) x 2 + y 2 + gx + fy = 0 (d) 2( x 2 + y 2 + gx + fy) + c = 0

T Q

Proof Suppose a straight line through the given point P ( x1 , y1 ) intersects the given circle at Q and R. Let the tangents to the circle at Q and R meet at T whose coordinates are ( h, k ) (say). Then, QR is the chord of contact of tangents drawn from T ( h, k ) to the circle x 2 + y 2 = a 2 . Therefore, the equation of QR is hx + ky = a 2 . Since, it passes through P ( x1 , y1 ), therefore hx1 + ky1 = a 2 ∴ Locus of T ( h, k ), i.e. polar of P is xx1 + yy1 = a 2

Sol. (c) Let the mid-point of chords of circle be P(h, k ). Equation of this chord is T = S1 i.e. xh + yk + g ( x + h) + f( y + k ) + c = h2 + k 2 + 2 gh + 2 fk + c. It passes through (0, 0), thus h2 + k 2 + gh + fk = 0 i.e. X

required locus is x2 + y2 + g x + fy = 0

Example 39. Tangents PA and PB drawn to x 2 + y 2 = 9 from any arbitrary point P on the line x + y = 25. Locus of mid-point of chord AB is (a) 25( x 2 + y 2 ) = 9( x + y) (b) 25( x 2 + y 2 ) = 3( x + y) (c) 5( x 2 + y 2 ) = 3 ( x + y) (d) None of these

Conjugate points Two points A and B are conjugate points with respect to a given circle, if each lies on the polar of the other with respect to the circle. Conjugate lines If two lines be such that the pole of one line lies on the other, then they are called conjugate lines with respect to the given circle. Ø

●

●

Sol. (a) Let P ≡ (a, 25 − a). Equation of chord AB is T = O i.e.

xa + y(25 − a) = 9. If mid-point of chord AB is C(h, k ), then equation of chord AB is T = S1 i.e. xh + yk = h2 + k 2 .

Comparing the coefficients of like powers, we get 9 a 25 − a = = 2 h k h + k2 a + 25 − a 25 = = h+ k h+ k ⇒

710

⇒

9(h + k ) = 25(h2 + k 2 ) 25( x2 + y2 ) = 9( x + y)

●

The distance of any two points P(x1, y1) and Q (x 2, y 2) from the centre of a circle are proportional to the distance of each from the polar of the other. If the polar of a point P (x1 ,y1) with respect to a circle whose centre is O meets the straight line joining P to the centre of the circle at tthe point Q , then ➣ OP is perpendicular to the polar of P. ➣ P and Q are inverse points with respect to the circle i.e. OP. OQ = (radius)2 The polar of (x1, y1) with respect to the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 is xx1 + yy1 + g (x + x1) + f (y + y1) + c = 0 which is same as the equation of the tangent at (x1, y1), if (x1, y1) lies on the circle.

Example 40. The pole of the line 3x + 5 y + 17 = 0 with respect to the circle x 2 + y 2 + 4x + 6 y + 9 = 0 is (a) (1, 2) (b) (2, 1) (c) (1, − 4) (d) None of these Sol. (a) Given circle is x 2 + y2 + 4 x + 6 y + 9 = 0

…(i)

and given line …(ii) 3 x + 5 y + 17 = 0 Let (λ, β ) be the pole of line with respect to circle. Now, equation of polar of point (α, β ) with respect to the circle is xα + yβ + 2( x + α ) + 3( y + β ) + 9 = 0 …(iii) ⇒ (α + 2 ) x + (β + 3) y + 2α + 3β + 9 = 0 Now, lines (ii) and (iii) are same. α + 2 β + 3 2α + 3β + 9 = = ∴ 3 5 17 Solving these equations, we get α = 1 and β = 2 Hence, required pole is (1, 2 ).

Diameter of a Circle The locus of the middle points of a system of parallel chords of a circle is called a diameter of the circle.

Theorem The equation of the diameter bisecting parallel chord y = mx + c (c is a parameter) of the circle x 2 + y 2 = a 2 is x + my = 0. Proof Let P ( h, k ) be the mid-point of a chord y = mx + c. Then, x-coordinate of points A and B are the roots of the equation x 2 + ( mx + c) 2 = a 2 2 2 …(i) ⇒ x (1 + m ) + 2mcx + c 2 − a 2 = 0 Let ( x1 , y1 ) and ( x 2 , y2 ) be the coordinates of A and B respectively. Then, x1 , x 2 are the roots of Eq. (i). x + x2 −mc 2mc ∴ x1 + x 2 = − = ⇒ 1 2 2 1 + m2 1+ m mc ⇒ h=− 1 + m2 [Q P ( h, k ) is the mid-point of AB] −h …(ii) (1 + m 2 ) = c ⇒ m Since, ( h, k ) lies on y = mx + c. Therefore, …(iii) k = mh + c

⇒ −h = mk So, locus of ( h, k ) is −x = my ⇒ x + my = 0

Common Tangents to Two Circles Let the equations of the two circles be ( x − h1 ) 2 + ( y − k1 ) 2 = a12 and ( x − h2 ) 2 + ( y − k 2 ) 2 = a 22

B Q

P (h, k) y = mx + c

A

…(i) …(ii)

with centres C1 ( h1 , k1 ) and C 2 ( h2 , k 2 ) and radii a1 and a 2 respectively. Then, the following cases of intersection of these two circles may arise : Case I other, when

Circles neither intersect nor touch each C1 C 2 > a1 + a 2

i.e. the distance between the centres is greater than the sum of the radii. In this case, the two circles do not intersect with each other and four common tangents can be drawn to two circles. Two of them are direct common tangents and the other two are transverse common tangents. The point of intersection of direct and transverse common tangents always lie on the line joining the centres C1 and C 2 and divide it internally and externally respectively in the ratio a1 : a 2 i.e. C1T2 a1 [externally] = C 2T2 a 2 C1T1 a1 = C 2T1 a 2

and

[internally]

Transverse common tangents

C1

T1

C2

T2

Direct common tangents

Diameter x + my = 0

13

Eliminating c from Eqs. (ii) and (iii), we get h − (1 + m 2 ) = ( k − mh) m ⇒ −h − hm 2 = mk − m 2 h

Circle

X

Case II when

Circles touch externally, C1C 2 = a1 + a 2

i.e. the distance between the centres is equal to the sum of the radii. In this case, two direct tangents are real and distinct while the transverse tangents are coincident. 711

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Direct common tangents

T1

C1

C2

X

Example 41. How many common tangents can be drawn to the following circles x 2 + y 2 = 6x and x 2 + y 2 + 6x + 2 y + 1 = 0? (a)1 (b) 2 (c) 3

T2

(d) 4

Sol. (d) The coordinates of the centre of the given circles are C1 (3, 0), C 2 (−3, − 1) and corresponding radii are r1 = 3 and r2 = 3 respectively.

Transverse common tangents

C1C 2 = (3 + 3)2 + (0 + 1)2 =

Now,

Case III Circles intersect each other when | a1 − a 2 | < C1C 2 < a1 + a 2 i.e. the distance between the centres is less than the sum of the radii. In this case, the two direct common tangents are real while the transverse tangents are imaginary.

and r1 + r2 = 6 ∴ C1C 2 > r1 + r2 So, four common tangents can be drawn to the given circles. X

C2

Example 42.

The

two

circles

and x + y + 2by + c = 0 x + y + 2ax + c = 0 touching each other, then 1 1 1 (b) a 2 + b 2 = c 2 (a) 2 + 2 = 2 a b c 1 1 1 (d) None of these (c) 2 + 2 = c a b 2

Direct common tangents

C1

37

T2

2

2

2

Sol. (c) We have, x2 + y2 + 2 ax + c = 0 x2 + y2 + 2 by + c = 0

and

Circles touch internally when C1C 2 = a1 − a 2 i.e. the distance between the centres is equal to the difference of the radii. In this case, two tangents are real and coincident while the other two tangents are imaginary. Case IV

C1 (–a, 0)

∴ C2 a2

where

a1

⇒ ⇒ ⇒ ⇒

Circles neither touch nor intersect when C1C 2 < a1 − a 2 i.e. the distance between the centres is less than the difference of the radii. In this case, all four common tangents are imaginary. Case V

X

C2 (0, –b)

C1C 2 = r1 ± r2 ( a2 + b 2 ) = ( a2 − c ) ±

⇒ C1

P

…(i) …(ii)

(b 2 − c )

r1 = C1P = (a2 − c ), r1 = C 2 P = (b 2 − c ) a2 + b 2 = a2 − c + b 2 − c ± 2 (a2 − c )(b 2 − c ) c = ± (a2 − c )(b 2 − c ) c 2 = a2 b 2 − a2c − b 2c + c 2 1 1 1 a2 b 2 = a2c + b 2c ⇒ = + 2 c a2 b

Example 43. The centre of a set of circles, each of radius 3, lies on the circles x 2 + y 2 = 25. The locus of any point in the set is (b) x 2 + y 2 ≤ 25 (a) 4 ≤ x 2 + y 2 ≤ 64 (c) x 2 + y 2 ≥ 25

(d) 3 ≤ x 2 + y 2 ≤ 9

Sol. (a) Let(h, k ) be any point in the set, then equation of circle is ( x − h)2 + ( y − k )2 = 9. C1 C2

But (h, k ) lie on x2 + y2 = 25, then h2 + k 2 = 25 ∴ 2 ≤ Distance between the centres of two circles ≤ 8 ⇒

712

∴

2 ≤ (h2 + k 2 ) ≤ 8 4 ≤ h2 + k 2 ≤ 64 Locus of (h, k ) is 4 ≤ x2 + y2 ≤ 64.

Example 44. The equation of a circle with centre ( 4, 3) touching the circle x 2 + y 2 = 1, is (a) ( x − 4) 2 + ( y − 3) 2 = 15 (b) ( x − 4) 2 + ( y − 3) 2 = 16 (c) ( x − 4) 2 + ( y − 3) 2 = 36 (d) None of the above Sol. (b, c) The given circle is x2 + y2 = 1, which has centre C(0, 0) and radius r1 = 1.

The required circle has centre C 2 (4, 3) and radius r2 . If the circles are touching externally, then r1 + r2 = C1 C 2 ⇒ r2 = 5 − 1 = 4 If the circles are touching internally, then r2 − r1 = C1 C 2 ⇒ r2 = 6 Thus, the required circle are ( x − 4)2 + ( y − 3)2 = 16 ( x − 4)2 + ( y − 3)2 = 36.

or X

Example 45. The common tangents to the circles x 2 + y 2 + 2x = 0 and x 2 + y 2 − 6x = 0 form a triangle which is (a) equilateral (b) isosceles (c) right angled (d) None of these Sol. (a) Clearly, the distance between the centres of the given circles is equal to the sum of their radii. So, two circles touch each other externally.

X

Sol. (a) Common chord S1 − S 2 = 0 passes through the centre of S 2 = 0, where S1 ≡ x2 + y2 + 2 gx + 2 fy + c = 0 and

S 2 ≡ x 2 + y2 + 2 g ′ x + 2 f ′ y + c ′ = 0

Common chord is 2(g − g ′ )x + 2 f ′(f − f ′ ) + (c − c ′ ) = 0 which passes through (− g ′, − f ′ ). 2 g ′(g − g ′ ) + 2 f ′(f − f ′ ) = (c − c ′ )

Angle of Intersection of Two Circles The angle of intersection of two circles is defined as the angle between the tangents or angle between the normals to the two circles at their point of intersection is given by r 2 + r22 − d 2 cos θ = 1 2r1 r2 A′

T3 1 Q 1 T1

3 1

R

1

S′ = 0 P

1

r1

P T2

B′

S=0

C2

C1

13

Example 46. The circle x 2 + y 2 + 2gx + 2 fy + c = 0 bisects the circumference of the circle x 2 + y 2 + 2g ′ x + 2 f ′ y + c′ = 0, if (a) 2g ′ ( g − g ′ ) + 2 f ′ ( f − f ′ ) = c − c′ (b) 2g ′ ( g − g ′ ) − 2 f ′ ( f − f ′ ) = c − c′ (c) g ′ ( g − g ′ ) + f ′ ( f − f ′ ) = c + c′ (d) None of the above

Circle

X

1

C1 B

T4 1

We have, PT1 = PT2 and PT3 = PT4 ⇒ TT 1 3 = T2T4 ⇒ TQ 1 = T2 R ⇒ PT1 + TQ 1 = PT1 + T2 R ⇒ PT1 + TQ 1 = PT2 + T2 R ⇒ PQ = PR = QR = 2 So, ∆PQR is equilateral.

Common Chord The chord joining the points of intersection of two given circles is called their common chord. The equation of the common chord of two circles S 1 ≡ x 2 + y 2 + 2g1 x + 2 f 1 y + c1 = 0 …(i) and S 2 ≡ x 2 + y 2 + 2g 2 x + 2 f 2 y + c2 = 0 …(ii) is 2x ( g1 − g 2 ) + 2 y ( f 1 − f 2 ) + c1 − c2 = 0 i.e. S1 − S 2 = 0

r2

Proof Let and

C2 Q

A

x 2 + y 2 + 2g1 x + 2 f 1 y + c1 = 0 x 2 + y 2 + 2g 2 x + 2 f 2 y + c2 = 0

…(i) …(ii)

be two circles having their centres at C1 ( −g1 , − f 1 ) and C 2 ( −g 2 , − f 2 ), respectively. Let P and Q be the points of intersection of these two circles. Since, the tangents at P are perpendicular to the radii of the corresponding circles and angle between two lines is same as the angle between their perpendiculars. Therefore, the angle of intersection of two circles is equal to the angle between their radii at the point of intersection. Thus, we have, C P 2 + C 2 P 2 − C1C 22 cos ∠C1 PC 2 = 1 2C1 P ⋅ C 2 P 2 r + r22 − d 2 cos θ = 1 ∴ 2r1 r2 where, C1 P = r1 , c2 P = r2 and C1C 2 = d

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Objective Mathematics Vol. 1

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X

Example 47. If θ is the angle of intersection of two circles x 2 + y 2 = a 2 and ( x − c) 2 + y 2 = b 2 , then the length of common chord of two circles is ab 2ab (a) (b) a 2 + b 2 − 2ab cos θ a 2 + b 2 − 2ab cos θ 2ab sin θ (d) None of these (c) 2 a + b 2 − 2ab cos θ Sol. (c) The equation of two circles are and

x 2 + y2 = a 2

...(i)

( x − c )2 + y2 = b 2

…(ii)

If two circles intersect orthogonally, then we say that the circles are orthogonal circles. If circles (i) and (ii) intersect orthogonally, then ∠C1 PC 2 = 90° ⇒ cos ∠C1 PC 2 = 90° ⇒ C1 P 2 + C 2 P 2 − C1C 22 = 0 ⇒ C1 P 2 + C 2 P 2 = C1C 22 2 ⇒ g1 + f 12 − c1 + g 22 + f 22 − c2 = ( − g1 − g 2 ) 2 + ( − f 1 + f 2 ) 2 ⇒ 2( g1 g 2 + f 1 f 2 ) = c1 + c2 This is the condition of orthogonality of two circles. X

The radius of first and second circle are a and b respectively. Let ∠OPM = α and ∠APM = β ∴ ∠OPA = α + β = θ Let PQ = x

+6 = 0 and x 2 + y 2 + 4ay + a = 0 intersect orthogonally, then value of a is equal to 1± 4 6 1± 4 7 (a) (b) 8 8 1 ± 97 1±2 6 (d) (c) 8 8

P a

Ta rg e t E x e rc is e s

O

b β

α M

Example 48. If the circles x 2 + y 2 + 2x + 2ay

Sol.(c) We have, 2 g1g 2 + 2 f1f2 = c1 + c 2 A

⇒

2(1)(0) + 2(a) (2 a) = 6 + a

⇒

Q

4a2 − a − 6 = 0

PM x x and cosβ = = a 2a 2b Now, cos θ = cos(α + β ) = cos α cos β − sin α sin β ∴ sin α sin β = cos α cos β − cos θ ⇒

cosα =

On squaring both sides, we get sin2 α sin2 β = cos 2 α cos 2 β + cos 2 θ − 2 cos θcos α cos β 1 − cos 2 α − cos 2 β + cos 2 α + cos 2 β = cos 2 α cos 2 β + cos 2 α − 2 cos θ cos α cos β ∴

sin θ = cos α + cos β − 2 cos θ cos α cos β 2

2

2

2

2

sin2 θ =

1 + 96 8

(c)

r1 r2

(d)

r12 + r22

2r12 r22 r12 + r22

Sol. (a) Let ∠AB1B2 = θ ⇒ ⇒

AD = r1 sin θ and AD = r2 cos θ 1 1 AD2  2 + 2  = 1 r2   r1 A

Orthogonal Circles

θ

714

B1

B2

P

D

C2

Q

r1

r2

Two circles are said to intersect orthogonally, if their angle of intersection is a right angle.

C1

=

1±

97 8

Example 49. The circles having radii r1 and r2 intersect orthogonally. Length of their common chord is 2r1 r2 2r12 r2 (b) (a) r12 + r22 r12 + r22

2

x x 2x + − ⋅ cos θ 4a2 4b 2 4ab ⇒ 4a2 b 2 sin2 θ = x2 (b 2 + a2 − 2 ab cos θ) 2 ab sin θ ∴ x= 2 a + b 2 − 2 ab cos θ ⇒

X

1±

a=

⇒

⇒

AD =

Thus, length of common chord =

r1r2 r12 + r22 2 r1r2 r12 + r22

Example 50. The centre of circle S lies on 2x − 2 y + 9 = 0 and it cuts orthogonally the circle x 2 + y 2 = 4. Then, the circle passes through two fixed points  1 1 (a) (1, 1) and (3, 3) (b)  − ,  and ( − 4, 4)  2 2 (c) (0, 0) and (5, 5)

(d) None of these

Sol. (b) Let S ≡ x + y + 2 gx + 2 fy + c = 0 2

2

Q It cuts orthogonally. ⇒ c=4 Moreover, − 2 g + 2 f + 9 = 0 [Q(− g , − f ) satisfy the given equation] ∴ S ≡ x2 + y2 + 2 gx + 2 fy + 4 = 0 ⇒

x2 + y2 + (2 f + 9)x + 2 fy + 4 = 0

⇒

( x2 + y2 + 9 x + 4) + 2 f( x + y) = 0

Let P ( h, k ) be a point on the radical axis of Eqs. (i) and (ii). Then, PQ = PR ⇒ PQ 2 = PR 2 2 2 ⇒ h + k + 2g1 h + 2 f 1 k + c1 = h 2 + k 2 + 2g 2 h + 2 f 2 k + c2 ⇒ 2h( g1 − g 2 ) + 2k ( f 1 − f 2 ) + c1 − c2 = 0 Hence, the locus of (h, k) is 2x ( g1 − g 2 ) + 2 y ( f 1 − f 2 ) + c1 − c2 = 0 or S1 − S 2 = 0 This is the equation of the radical axis of circles (i) and (ii). As it is a first degree equation in x and y. So, it represents a straight line. Ø

It is of the form S + λP = 0 and hence passes through the intersection of S = 0 and P = 0 which when solved give  − 1 , 1  , (− 4, 4).    2 2

●

●

●

●

X

Example 51. Locus of the centre of a circle that passes through ( a, b) and cuts the circle x 2 + y 2 = a 2 orthogonally, is (a) 2ax + 2by = a 2 + b 2 (b) 2ax + by = b 2 + 2a 2 (c) ax + 2by = 2b 2 + a 2 (d) 2ax + 2by = b 2 + 2a 2

●

Sol. (d) Let the circle be x2 + y2 + 2 gx + 2 fy + c = 0. It

●

passes through (a, b ), then a2 + b 2 + 2 ag + 2 fb + c = 0 It also cuts x2 + y2 = a2 orthogonally, then 2 g ⋅ 0 + 2 f ⋅ 0 = c − a2 ⇒

⇒ c = a2

X

The equations of radical axis and the common chord of two circles are identical. The radical axis of two circles is always perpendicular to the line joining the centres of the circles. The radical axis of three circles, whose centres are non-collinear, taken in pairs are concurrent. The centre of the circle cutting two given circles orthogonally, lies on their radical axis. or The locus of the centres of the circles cutting two given circles orthogonally, is their radical axis. Radical centre The point of intersection of radical axes of three circles whose centres are non-collinear, taken in pairs is called their radical centre. The circle with centre at the radical centre and radius equal to the length of the tangents from it to one of the circles intersects all the three circles orthogonally.

Example 52. The point from which the lengths of the tangents to the following three circles

x 2 + y 2 + 4x + 6 y − 19 = 0 , x 2 + y 2 − 2x − 2 y − 5 = 0

2 ag + 2 fb + b 2 + 2 a2 = 0

Thus, locus of centre is 2 ax + 2 by = b + 2 a . 2

2

and x 2 + y 2 = 9 are equal, is (a) (2, − 1)

Radical Axis

R C2

(d) (1, − 2)

On solving Eqs. (i) and (ii), we get the point of intersection of R12 and R 23 is (1, 1). X

Q

(c) (1, 1)

∴ The common chord R12 of the first two circles is S1 − S 2 = 0 i.e. 6 x + 8 y − 14 = 0 …(i) ⇒ 3x + 4y = 7 The common chord R 23 of the second and third circles is S2 − S3 = 0 i.e. − 2x − 2y + 4 = 0 …(ii) ⇒ x+ y=2

be two circles. P (h, k)

(b) (2, − 2)

Sol. (c) The required point will be the radical centre.

The radical axis of two circles is the locus of a point which moves in such a way that the lengths of the tangents drawn from it to the two circles are equal. The radical axis of two circles S 1 = 0 and S 2 = 0 is given by S1 − S 2 = 0 Proof Let S 1 ≡ x 2 + y 2 + 2g1 x + 2 f 1 y + c1 = 0 …(i) and S 2 ≡ x 2 + y 2 + 2g 2 x + 2 f 2 y + c2 = 0 …(ii)

C1

13 Circle

X

Example 53. Equation of the circle that cuts the circle x 2 + y 2 = a 2 , ( x − b) 2 + y 2 = a 2 and x 2 + ( y − c) 2 = a 2 orthogonally, is (a) x 2 + y 2 − bx − cy − a 2 = 0 (b) x 2 + y 2 + bx + cy − a 2 = 0 (c) x 2 + y 2 + bx + cy + a 2 = 0 (d) x 2 + y 2 − bx − cy + a 2 = 0

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Sol. (d) S1 ≡ x2 + y2 − a2 = 0

X

S 2 ≡ x2 + y2 − 2 bx + b 2 − a2 = 0 S 3 ≡ x2 + y2 − 2 yc + c 2 − a2 = 0 b c Clearly, the radical centre of these circles is P  ,  . 2 2 b2 c 2 + − a2 . 4 4 Thus, equation of required circle is Length of tangent from P to S1 is 2

2

2 2  x − b  +  y − c  = b + c − a2       2 2 4 4

⇒

Sol. (b) Equation of line AB is

x2 + y2 − bx − cy + a2 = 0

y−2 =

Family of Circles i.

ii.

and family of circles passing through the points of intersection of Eqs. (i) and (ii) x 2 + y2 + 2 x − 3 y + 2 + λ ( x + 2 y − 2 ) = 0 ⇒

∴

X

i.e.

+ λ( x 2 + y 2 + 2g 2 x + 2 f 2 y + c2 ) = 0 716

where, λ( ≠ −1) is an arbitrary real number.

λ=± 7 7( x + 2 y − 2) = 0

Example 55. Radius of the largest circle that can be drawn to pass through the point (0, 1), (0, 6) and touching X-axis,is 5 3 (a) (b) 2 2 7 9 (d) (c) 2 2 Sol.(c) Equation of any circles through (0, 1) and (0, 6) is x2 + ( y − 1)( y − 6) + λx = 0 ⇒

x 2 + y2 − 7 y + 6 + λ x = 0

If it touches X-axis, then x2 + λx + 6 = 0 should have equal roots. ⇒ λ2 = 24 ⇒ λ = ± 24

The equation of a family of circles passing through the intersection of the circles, S 1 ≡ x 2 + y 2 + 2g1 x + 2 f 1 y + c1 = 0 S 1 + λS 2 = 0 2 2 ( x + y + 2g1 x + 2 f 1 y + c1 )

2

There are two such circles possible.

The equation of the family of circles touching the circle S 3 ≡ x 2 + y 2 + 2gx + 2 f y + c = 0 at point P ( x1 , y1 ) is x 2 + y 2 + 2gx + 2 fy + c + λ{xx1 + yy1 + g ( x + x1 )

S 2 ≡ x 2 + y 2 + 2g 2 x + 2 f 2 y + c2 = 0 is

2

 − 2 + λ  +  − 2 λ − 3  − 2 + 2 λ = 10       2  2 

⇒ Hence, required circles are x 2 + y2 + 2 x − 3 y + 2 ±

y1 1 = 0 y2 1

and

…(iii)

⇒ (4 + 4λ + λ2 ) + (4λ2 + 9 − 12 λ) + 8λ − 8 = 40

+ f ( y + y1 ) + c} = 0 or S + λ L =10 where, L =0 is the equation of the tangent to S =0 at ( x1 , y1 ) and λ ∈R.

iv.

x2 + y2 + (2 + λ )x + (2 λ − 3)y + 2 − 2 λ = 0

Eq. (iii) represents a circle of radius 10 units.

The equation of the family of circles passing through the points A ( x1 , y1 ) and B ( x 2 , y2 ) is ( x − x1 ) ( x − x 2 ) + ( y − y1 )( y − y2 ) x y 1

or ( x − x1 )( x − x 2 ) + ( y − y1 )( y − y2 ) + λL = 0 where, L = 0 represents the line passing through A ( x1 , y1 ) and B ( x 2 , y2 ) and λ ∈R.

2 −1 1 ( x + 2) = − ( x + 2) 2 −2 − 0

…(i) ⇒ x + 2y − 2 = 0 Now, equation of circle whose diagonally opposite points are A and B. ( x − 0)( x + 2 ) + ( y − 1)( y − 2 ) = 0 …(ii) ⇒ x 2 + y2 + 2 x − 3 y + 2 = 0

The equation of a family of circles passing through the intersection of a circle x 2 + y 2 + 2gx + 2 fy + c = 0 and line L = lx + my + n = 0 is x 2 + y 2 + 2gx + 2 fy + c + λ( lx + my + n) = 0 or S + λL = 0 where, λ is any real number.

+ λ x1 x2

iii.

Example 54. Let S 1 be a circle passing through A (0, 1), B ( −2, 2) and S 2 be a circle of radius 10 units such that AB is common chord of S 1 and S 2 . Then, the equation of S 2 is (a) x 2 + y 2 + 2x − 3 y + 2 = 0 (b) x 2 + y 2 + 2x − 3 y ± 7 ( x + 2 y − 2) = 0 (c) x 2 + y 2 + 2x − 3 y ± 5 ( x + 2 y − 2) = 0 (d) None of the above

49 7 − 6 = units. Thus, we 4 2 7 can draw two circles but radius of both circles is . 2

Radius of these circles =

X

6+

Example 56. A circle passes through the points A(1, 0), B (5, 0) and touches Y -axis at C (0, h). If ∠ACB is maximum, then (a) h = 5 (b) h = 2 5 (d) h = 2 10 (c) h = 10

(5, 0) is y2 + ( x − 1)( x − 5) + λy = 0 ⇒

x 2 + y2 + λ y − 6 x + 5 = 0

If ∠ACB is maximum, then this circle must touch the Y-axis at (0, h). On putting x = 0 in the equation of circle, we get y2 + λ y + 5 = 0 where y = h is its repeated root. ⇒ h2 = 5 and ⇒

λ = − 2h |h| = 5

Coaxial System of Circles A system of circles is said to be coaxial system of circles, if every pair of the circles in the system has the same radical axis. The following results are direct consequences from the above definition and concepts learnt in earlier sections.

i.

Since, the lines joining the centres of two circles is perpendicular to their radical axis. Therefore, the centres of all circles of a coaxial system lie on a straight line which is perpendicular to the common radical axis.

ii.

Circles passing through two fixed points P and Q form a coaxial system because every pair of circles has the same common chord PQ and therefore, the same radical axis which is perpendicular bisector of PQ.

iii.

iv.

If the equation of a member of a system of coaxial circles is S = 0 and the equation of the common radical axis is L = 0, then the equation representing the coaxial system of circle is S + λL = 0, where λ ∈R In other words, S + λL = 0, λ ∈R represents a family of coaxial circles having L = 0 as the common radical axis. The equation of a coaxial system of circles of which two members are S 1 = 0 and S 2 = 0 is S 1 + λS 2 = 0 or S 1 + λ( S 1 − S 2 ) = 0 or S 2 + λ( S 1 − S 2 ) = 0, where λ ∈R. In other words, if S 1 = 0 and S 2 = 0 are two circles, then S 1 + λS 2 = 0 or S + λ( S 1 − S 2 ) = 0 or S 2 + λ( S 1 − S 2 ) = 0, where λ ∈R represents a family of coaxial circles having S 1 − S 2 = 0 as the common radical axis.

X

Example 57. The equation of a circle which is coaxial with the circles 2x2 + 2 y2 − 2x + 6 y − 3 = 0 and x 2 + y 2 + 4x + 2 y + 1 = 0. It is given that the centre of the circle to be determined lies on the radical axis of the circle (a) 3( x 2 + y 2 ) + 2x + 8 y − 2 = 0

13 Circle

Sol. (a) Equation of any circle passing through (1, 0) and

(b) 4( x 2 + y 2 ) + 6x + 10 y − 1 = 0 (c) 2( x 2 + y 2 ) + 2x + 8 y − 3 = 0 (d) None of the above Sol. (b) Equation of the given circles are 3 =0 2 2 2 S2 = x + y + 4x + 2 y + 1 = 0 S1 = x2 + y2 − x + 3 y −

and

…(i) …(ii)

The radical axis of Eqs. (i) and (ii) is …(iii) S1 − S 2 = 0 ⇒ 10 x − 2 y + 5 = 0 ∴Required circle will have the equation of the form x2 + y2 + 4 x + 2 y + 1 + k(10 x − 2 y + 5) = 0 ⇒

x2 + y2 + 2(2 + 5k )x + 2(1 − k )y + (1 + 5k ) = 0 …(iv)

Its centre is [(− 2 − 5k ), k − 1]. From question, it lies on Eq. (iii). ∴ 10(− 2 − 5k ) − 2(k − 1) + 5 = 0 1 k=− ⇒ 4 On putting the value of k in Eq. (iv), we get 3 5 1 x 2 + y2 + x + y − = 0 2 2 4 ⇒ 4( x2 + y2 ) + 6 x + 10 y − 1 = 0

Limiting Points Limiting points of a coaxial system of circles are the members of the system which are of zero radius. It follows from the above definition that limiting points of a coaxial system of circles are the centres of point circles belonging to the system. In the previous section, we have seen that for a fixed value of c and varying g, the equation x 2 + y 2 + 2gx + c = 0

…(i)

represents a family of coaxial circles having X -axis as the common radical axis. The coordinates of the centre and radius of Eq.(i) are ( − g, 0) and g 2 − c, respectively. For the circle (i) to be a point circle, we must have radius = 0 ⇒ g2 − c =0 ⇒ c = g2 ⇒ g=± c Hence, the limiting points of the coaxial system of circles given by Eq. (i) are ( c, 0) and ( − c, 0). Clearly, these points are real and distinct, if c > 0, coincident for c = 0 and imaginary, if c < 0.

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13 System of Coaxial Circles when Limiting Points are Given

Let ( a, b) and (α, β) be two limiting points of a coaxial system of circles. Then, the corresponding point of circles

are S 1 ≡ ( x − a ) 2 + ( y − b) 2 = 0 S 2 ≡ (x − α ) 2 + ( y − β) 2 = 0

and

So, the coaxial system of circles is given by S 1 + λS 2 = 0, λ ≠ −1 ⇒ {( x − a ) 2 + ( y − b) 2 } + λ{( x − α ) 2 + ( y − β) 2 } = 0, λ ≠ −1 The value of λ is determined from the given condition. X

Example 58. If one circle of a coaxial system is x 2 + y 2 + 2gx + 2 fy + c = 0 and one limiting point is (a, b), then equation of the radical axis will be (a) ( g + a ) x + ( f + b) y + c − a 2 − b 2 = 0 (b) 2( g + a ) x + 2( f + b) y + c − a 2 − b 2 = 0 (c) 2g + 2 fy + c − a 2 − b 2 = 0 (d) None of the above Sol.(b) Given circle is S1 ≡ x2 + y2 + 2 gx + 2 fy + c = 0

…(i)

∴ Equation of the second circle is ( x − a)2 + ( y − b )2 = 0 S1 ≡ x + y − 2 ax − 2 by + a2 + b 2 = 0 2

2

…(ii)

From Eqs. (i) and (ii), equation of radical axis is ⇒ X

S1 − S1 = 0 2(g + a)x + 2(f + b )y + c − a2 − b 2 = 0

Example 59. Two circles, each of radius 5 units touch each other at (1, 2). If the equation of their common tangent is 4x + 3 y − 10 = 0, then equation of one such circle is (a) x 2 + y 2 − 6x + 2 y + 15 = 0 (b) x 2 + y 2 − 10x − 10 y + 25 = 0 (c) x 2 + y 2 + 6x − 2 y − 15 = 0 (d) x 2 + y 2 − 10x − 10 y − 25 = 0 Sol. (b) Equation of circles will be ( x − 1)2 + ( y − 2 )2 + λ(4 x + 3 y − 10) = 0 ⇒

x2 + y2 + 2 x(2 λ − 1) + y(3λ − 4) + 5 − 10λ = 0

Since, its radius is 5 units. ∴ ⇒ Thus, equation of circles are

(2 λ − 1)2 +

1 (3λ − 4)2 − 5 + 10λ = 25 4 λ=±2

x2 + y2 + 6 x + 2 y − 15 = 0 or X

718

x2 + y2 − 10 x − 10 y + 25 = 0

Example 60. The coordinates of the limiting points of a system of coaxial circles determined by the circles S 1 ≡ x 2 + y 2 + 4x + 2 y + 5 = 0 and S 2 ≡ x 2 + y 2 + 2x − 4 y + 7 = 0 are (a) ( − 2, − 1) and (0, − 3) (b) (2, − 1) and (1, − 3) (c) (2, 1) and (0, 3) (d) ( − 2, 1) and (3, 0)

S1 + λ(S1 − S 2 ) = 0, λ ∈ R x2 + y2 + 4 x + 2 y + 5 + λ (2 x − 2 y − 2 ) = 0

⇒ ⇒

x2 + y2 + 2(2 + λ ) x + 2 (1 − λ) y + 5 − 2 λ = 0

…(i)

Circle

13

Sol. (a) The equation of a coaxial of system of circles determined by the given circles is

The coordinates of the centre and radius are [− (2 + λ ), λ − 1] and (2 + λ ) + (1 − λ ) − 5 + 2 λ. 2

R=0 (2 + λ )2 + (1 − λ )2 − 5 + 2 λ = 0

Now, ⇒ ⇒

2 λ2 + 4λ = 0

∴ Hence, the limiting points are (− 2, − 1) and (0, − 3). X

2

λ = 0, − 2

Example 61. The equation of the radical of a coaxial system of circle whose limiting points are (2, − 1) and ( − 3, 2), is (a) 5x + 3 y − 4 = 0 (b) 5x + 3 y + 4 = 0 (c) 5x − 3 y + 4 = 0 (d) 3x − 5 y + 4 = 0 Sol. (c) The equations of circles having (2, − 1) and (− 3, 2 ) as limiting points, are S1 ≡ ( x − 2 )2 + ( y + 1)2 = 0

…(i)

S 2 ≡ ( x + 3) + ( y − 2 ) = 0

…(ii)

2

and

2

Clearly, Eqs. (i) and (ii) are members of the coaxial system of circles whose limiting points are (2, − 1) and (− 3, 2 ). So, the equation of the radical axis is S 2 − S1 = 0 ⇒ 10 x − 6 y + 8 = 0 ⇒ 5x − 3y + 4 = 0

Work Book Exercise 13.3 1 The locus of middle points of the chords of the circle x 2 + y 2 = 49 passing through the point (1, − 1) is a b c d

x2 x2 x2 x2

+ + + +

y2 y2 y2 y2

− − + +

x− x+ x− x+

y=0 y=0 y=0 y=0

2 The point from which the tangents to the circles x 2 + y 2 − 8 x + 40 = 0, 5 x 2 + 5 y 2 − 25 x + 80 = 0, and x 2 + y 2 − 8 x + 16 y + 160 = 0 are equal in length, is 15 a  8,   2  15 b  − 8,   2  15   c  8, −   2  d None of the above

3 For the two circles x 2 + y 2 = 16 and x 2 + y 2 − 2 y = 0, there is/are a b c d

one pair of common tangents two pairs of common tangents three common tangents no common tangents

4 If the chord of contact of tangents from three points A, B, C to the circle x 2 + y 2 = a2 are concurrent, then A, B, C will a b c d

be concyclic be collinear form the vertices of a triangle None of the above

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5 If a line is drawn through a fixed point P(α, β) to cut

8 If the chord of contact of tangents from of a point P

the circle x + y = a at A and B, then PA ⋅ PB is

to a given circle passes through Q, then the circle on PQ as diameter

2

2

2

equal to a b c d

6 If

a b c d

α2 + β2 α 2 + β 2 − a2 a2 α 2 + β 2 + a2

the

two

9 If the chord of contact of tangents from a point P to a1 x 2 + 2 h1 xy + b1 y 2 = c1,

conics

a2 x 2 + 2 h2 xy + b2 y 2 = c 2

intersect

in

four

concyclic points, then a b c d

( a1 ( a1 ( a1 ( a1

− − + +

b1 ) h2 = ( a2 − b 2 ) h1 b1 ) h1 = ( a2 − b 2 ) h2 b1 ) h2 = ( a2 + b 2 ) h1 b1 ) h1 = ( a2 + b 2 ) h2

x 2 + y 2 = a2

touches

( x − c )2 + ( y − d )2 = b2 , then a b c d

720

b 2( p2 + q 2 ) = ( a 2 − cp − dq )2 b 2( p2 + q 2 ) = ( a 2 − cp − dp)2 a 2( p2 + q 2 ) = ( b 2 − cp − dq )2 None of the above

a circle passes through Q. If l1 and l 2 are the lengths of the tangents from P and Q to the circle, then PQ is equal to a c

7 If the polar of a point (p, q) with respect to the circle

cuts the given circle orthogonally touches the given circle externally touches the given circle internally None of the above

the

circle

l1 + l 2 2 l12 + l 22

b d

l1 − l 2 2 l12 − l 22

10 The circle S1 with centre C1(a1, b1 ) and radius r1 touches externally the circle S 2 with centre C2 (a2 , b2 ) and radius r2 . If the tangent at their common point passes through the origin, then a ( a12 + a22 ) + ( b12 + b 22 ) = r12 + r22 b ( a12 − a22 ) + ( b12 − b 22 ) = r12 − r22 c ( a12 − b 22 ) + ( a22 + b 22 ) = r12 + r22 d ( a12 − b12 ) + ( a22 + b 22 ) = r12 + r22

WorkedOut Examples Type 1. Only One Correct Option Ex 1. Let C be any circle with centre (0, 2 ). Then, the number of rational points can be there on C is atmost (A rational point is a point both of whose coordinates are rational numbers) (a) four (b) three (c) two (d) one Sol. The equation of any circle C with centre (0, 2 ) is given

Thus, in both the cases we arrive at a contradiction. This means that our supposition is wrong. Hence, there can be atmost two rational points on circle C. Aliter Let there be three points P (x1 , y1 ), Q (x2 , y2 ) and R (x3 , y3 ) with rational coordinates on circle C having its equation …(i) x 2 + y2 + 2gx + 2 fy + c = 0 Since, P, Q and R lie on circle Eq. (i), therefore x12 + y12 + 2gx1 + 2 fy1 + c = 0 x22 + y22 + 2gx2 + 2 fy2 + c = 0

by (x − 0)2 + ( y − 2 )2 = r2, where r is any positive real number. x 2 + y2 − 2 2 y = r2 − 2 ⇒

…(i)

If possible, let P (x1 , y1 ), Q (x2 , y2 ) and R (x3 , y3 ) be three distinct rational points on circle C, then …(ii) x12 + y12 − 2 2 y1 = r2 − 2 x22 + y22 − 2 2 y2 = r2 − 2 and

x32

+

y32

− 2 2 y3 = r − 2 2

x32 + y32 + 2gx3 + 2 fy3 + c = 0

and

…(iii) …(iv)

We claim atleast two of y1, y2 and y3 are distinct. Here, if y1 = y2 = y3, then P , Q and R lie on a line parallel to X-axis and a line parallel to X-axis does not cross the circle in more than two points. Thus, we have either y1 ≠ y2 or y1 ≠ y3 or y2 ≠ y3. On subtracting Eq. (ii) from Eqs. (iii) and (iv), we get (x22 + y22 ) − (x12 + y12 ) − 2 2 ( y2 − y1 ) = 0 and (x32 + y32 ) − (x12 + y12 ) − 2 2 ( y3 − y1 ) = 0

These are three linear equations in g, f and c with rational coefficients. So, we get rational values of g, f and c but f = 2. Thus, we arrive at a contradiction. Hence, there can be atmost two rational points on circle C. Hence, (c) is the correct answer.

Ex 2. P is a variable point on a circle with centre at C. CA and CB are perpendicular from C to the X and Y-axes, respectively. Then, the locus of the centroid of ∆PAB is (a) a parabola (b) a circle (c) an ellipse (d) None of the above Sol. Let C (a, b) be the centre and r be the radius of the given

⇒

a1 − 2b1 = 0 and a2 − 2b2 = 0

where,

a1 = (x22 + y22 ) − (x12 + y12 ) , b1 = 2( y2 − y1 )

…(v)

a2 = (x32 + y32 ) − (x12 + y12 ) , b2 = 2( y3 − y1 ) Clearly, a1, a2 and b1 , b2 are rational numbers as x1, x2 and x3, y1, y2, y3 are rational numbers. Since, either y1 ≠ y2 or y1 ≠ y3 ∴ Either b1 ≠ 0 or b2 ≠ 0 If b1 ≠ 0, then [from Eq. (v)] a1 − 2b1 = 0 a1 ⇒ = 2 b1 a which is not possible because 1 is a rational number b1 and 2 is an irrational number. If b2 ≠ 0, then a a2 − 2b2 = 0 ⇒ 2 = 2 b2 a which is not possible because 2 is a rational number b2 and 2 is an irrational number.

circle. Let P (α , β) be a variable point on the circle. Then, coordinates of A and B are (a, 0) and (0, b), respectively. Let Q (h, k ) be the centroid of ∆PAB. Then, Y P (α, β)

(0, b) B

X′

C (a, b)

(0, 0) O

A (a, 0)

X

Y′

h= and and

k=

β+b 3

a +α 3

⇒ α = 3h − a β = 3k − b

…(i)

The equation of the circle is (x − a)2 + ( y − b)2 = r2, where r is the radius of the circle. Since, P(α , β ) lies on the circle.

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13

Therefore, (α − a)2 + (β − b)2 = r2 ⇒

=

(3h − 2a)2 + (3k − 2b)2 = r2

[from Eq. (i)]

Hence, the locus of (h, k) is (3x − 2a)2 + (3 y − 2b)2 = r2 2

2

2a 2b  r   x −  +  y −  =    3     3 3

⇒

Clearly, this will be independent of θ, if

2

2

 2a 2b Clearly, it represents a circle with centre  ,  .  3 3 Hence, (b) is the correct answer.

Ex 3. Consider a curve ax 2 + 2hxy + by 2 = 1 and a point P not on the curve. A line drawn from the point P intersects the curve at points Q and R. If the product PQ ⋅ PR is independent of the slope of the line, then the curve is

 a − b 2  +h =0   2  a−b =0 ⇒ 2 and h=0 ⇒ a=b and h=0 On substituting a = b and h = 0 in ax 2 + 2hxy + by2 = 1, we get ax 2 + ay2 = 1 which represents a circle.

(a) a circle (b) a parabola (c) an ellipse (d) a hyperbola

Hence, (a) is the correct answer.

Sol. The equation of a line passing through P (x1 , y1 ) is x − x1 y − y1 = =r cosθ sin θ

Ex 4. If two curves whose equations are ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 and a ′ x 2 + 2h′ xy + b′ y 2 + 2g ′ x + 2 f ′ y + c′ = 0 intersect in four concyclic points, then a − b a′ + b′ = h h′ a + b a′ + b′ (b) = h h′ a − b a′ − b′ (c) = h′ h a − b a′ − b′ (d) = h h′

R

(a)

C Q P (x1 , y 1 )

The coordinates of any point on this line are (x1 + r cosθ , y1 + r sin θ ) If it lies on ax 2 + 2hxy + by2 = 1, then a(x1 + r cosθ )2 + 2h (x1 + r cosθ ) ( y1 + r sin θ ) + b ( y1 + r sin θ )2 = 1 ⇒r2 (a cos2 θ + h sin 2θ + b sin 2 θ ) + 2r(ax1 cosθ + hx1 sin θ + hy1 cosθ + by1 sin θ ) +

ax12

+ 2hx1 y1 +

by12

⇒ PQ ⋅ PR = =

+ 2hx1 y1 + −1 a cos θ + h sin 2θ + b sin 2 θ ax12 2

Sol. The equation of the curve passing through the intersection of given curves is (ax 2 + 2hxy + by2 + 2gx + 2 fy + c) + λ (a′ x 2 + 2h′ xy + b′ y2 + 2g′ x + 2 f ′ y + c′ ) = 0 …(i) ⇒ x 2 (a + λa ′) + y2 (b + λb′ ) + 2xy (h + λh′ ) + 2x (g + λg′ ) + 2 y ( f + λf ′ ) + c + λc′ = 0 This will represent a circle, if Coefficient of x 2 = Coefficient of y2

− 1 = 0 ...(i)

This is a quadratic in r. So, it gives two values of r which represent the distances of two points of intersection of the line and the circle from the point P. Let PQ = r1 and PR = r2, then r1 and r2 are the roots of the Eq. (i). ax12 + 2hx1 y1 + by12 − 1 r1r2 = ∴ a cos2 θ + h sin 2θ + b sin 2 θ

722

ax12 + 2hx1 y1 + by12 − 1 2    a + b   a − b 2   + h  sin (2θ + φ )   +   2    2   

by12

ax12 + 2hx1 y1 + by12 − 1  a + b  a − b  cos 2θ + sin 2θ  +   2   2 

and ⇒ and ⇒ and ⇒ ⇒

coefficient of xy = 0 a + λa′ = b + λb′ h + λh′ = 0 a−b λ=− a′ − b′ h λ=− h′ a−b h = a′ − b′ h′ a − b a ′ − b′ = h h′

Hence, (d) is the correct answer.

(a) PA 2 + PB 2 + PC 2 + PD 2 = 4PA 2 (b) PA 2 + PB 2 + PC 2 + PD 2 = 2PA 2 (c) PA 2 + PB 2 + PC 2 + PD 2 = 3PA 2 (d) PA 2 + PB 2 + PC 2 + PD 2 = 2( PA + PB ) 2 Sol. The equation of a conic passing through the point of intersection of the given curves is ax 2 + 2hxy + by2 − 2gx − 2 fy + c

Sol. The coordinates of the centres and radii of three given circles are as given below : P

C (h, k) C3(–3, –6)

Centre

Radius

Circle (i)

C1( 0, 2 )

r1 = 3

Circle (ii)

C 2( − 6, − 2 )

r2 = 3

Circle (iii)

C 3( − 3, − 6)

r3 = 3

Let C (h, k ) be the centre of the circle passing through the centres of the circles (i), (ii) and (iii). Then, CC 1 = CC 2 = CC 3 ⇒

⇒(a + λl )x + by + 2h(1 − λ )xy − 2x (g + λm) 2

− 2 y ( f + λn) + c + λ (a + l − b)2 + kλ = 0 …(i) This will represent a circle, if a + λl = b and 2h(1 − λ ) = 0 ⇒ λ =1 and a+ l=b On substituting these values in Eq. (i), it reduces to (a + l )x 2 + (a + l ) y2 − 2x (g + m) − 2 y ( f + n) + c + k = 0    f + n c+ k g m + =0 ⇒ x 2 + y2 − 2   x−2  y+  a+ l  a+ l a+ l

CC 12 = CC 22 = CC 32

⇒

(h − 0)2 + (k − 2)2 = (h + 6)2 + (k + 2)2 = (h + 3)2 + (k + 6)2

⇒

− 4 k + 4 = 12h + 4 k + 40

⇒

= 6h + 12k + 45 12h + 8k + 36 = 0

and

6h − 8k − 5 = 0

⇒

3h + 2k + 9 = 0

and

6h − 8k − 5 = 0

⇒

 g + m f + n Clearly, its centre is at P  ,   a+ l a+ l Thus, if A, B, C and D are four points on it. Then, PA 2 = PB 2 = PC 2 = PD 2 = (Radius)2 ⇒

2

2

=

2

−23 12

5 36

2

949

Now,

CP = CC 1 + C 1P

⇒

 5 CP =   36

 949 + 3 

 31 23 Thus, required circle has its centre at  − , −  and  18 12    5 radius = CP =  949 + 3   36 2

2

2

2

2

2

5 31 23     (c)  x +  +  y +  =  3 + 949  18     36 12 (d) None of the above

k=

Hence, its equation is

2

23 31 5     (b)  x +  +  y +  =  3 + 949     12 18 36

−31 18

2

Hence, (a) is the correct answer.

Ex 6. The equation of the circle of minimum radius which contains the three circles …(i) x 2 + y2 − 4 y − 5 = 0 2 2 …(ii) x + y + 12x + 4 y + 31 = 0 2 2 …(iii) and x + y + 6x + 12 y + 36 = 0 is

h=

31 23   CC 1 =  0 +  +  2 +    18 12 

∴

PA 2 + PB 2 + PC 2 + PD 2 = 4 PA 2

5 31 23     (a)  x −  +  y −  =  3 − 949       36 18 12

C1(0, 2)

(–6, –2)C2

+ λ {λx 2 − 2hxy + (a + l − b)2 − 2mx − 2ny + k } = 0 2

13 Circle

Ex 5. If the curves ax 2 + 2hxy + by 2 − 2gx − 2 fy + c = 0 and lx 2 − 2hxy + ( a + l − b) 2 −2mx − 2ny + k = 0 intersect at four concyclic points A, B, C and D  g + m f + n and P is the point  ,  , then  a + l a +l 

2

31 23 5     949 x +  +  y +  = 3 +     18 12  36

2

Hence, (c) is the correct answer. Ø In the above example, radii of all the circles were same. In

case three circles are of distinct radii, then the radius of the required circle will be equal to CC1 + radius of largest circle.

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Ex 7. Three circles touch one another externally. The tangents at their points of contact meet at a point whose distance from a point of contact is 4. Then, the ratio of the product of the radii to the sum of the radii of the circles is (a) 16

(b) 4

(c) 8

(d)

1 16

Sol. LetC 1,C 2 andC 3 be the centres of three circles of radii r1, r2 and r3 respectively, then the lengths of the sides of ∆C 1C 2C 3 are C 1C 3 = r1 + r3, C 2C 3 = r2 + r3 and C 1C 2 = r1 + r2

The coordinates A , B and C are (7, −1), (−2, 8) and (1, 2), respectively. Let the equation of the circumcircle of ∆ABC be …(iv) x 2 + y2 + 2gx + 2 fy + c = 0 It passes through A (7, −1), B (−2, 8) and C (1, 2). Therefore, 50 + 14 g − 2 f + c = 0 …(v) … (vi) 68 − 4 g + 16 f + c = 0 and …(vii) 5 + 2g + 4 f + c = 0 On solving these equations, we get 17 19 g=− , f =− 2 2 c = 50

and

r1

C1

On substituting the values of g, f and c in Eq. (iv), the equation of the required circumcircle is x 2 + y2 − 17x − 19 y + 50 = 0

P r2

r1

C2

Aliter Consider the equation (x + y − 6)(2x + y − 4 ) + λ (2x + y − 4 )(x + 2 y − 5) + µ (x + 2 y − 5)(x + y − 6) = 0 …(i) where, λ and µ are constants.

r2 R

O Q r3

r3

This equation represents a curve passing through the vertices of the triangle formed by the given lines. We have to determine the values of λ and µ, so that the curve given in Eq. (i) is a circle.

C3

Let O be the point of intersection of common tangents in three circles taken in pairs. Then, OP = OQ = OR. Also, OP, OQ and OR are perpendicular to the sides C 1C 2, C 2C 3 and C 3C 1 respectively. Therefore, OP = OQ = OR = r [inradius of ∆C 1C 2C 3] ⇒ r=4 ∆ Area of ∆C 1C 2C 3  ⇒ Qr= =4 s  Semi-perimeter  But area of ∆C 1C 2C 3 = (r1 + r2 + r3 )r1r2r3 ∴

Area of ∆C 1C 2C 3 =4 Semi-perimeter

[Q s = r1 + r2 + r3]

(r1 + r2 + r3 )r1r2r3 =4 r1 + r2 + r3 r1r2r3 = 16 ⇒ r1 + r2 + r3 Hence, (a) is the correct answer. ⇒

Hence, (a) is the correct answer.

and

Ex 9. If Lr = a r x + br y + cr = 0, r =1, 2, 3 be the sides of a ∆ABC, then the equation of its circumcircle is 1 L1 (a) a 2 a 3 − b 2 b 3 a 2 b3 + a 3 b2

Sol. Let the equation of sides AB, BC and CA of ∆ABC are

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(x + 2 y − 5) + (x + 2 y − 5)(x + y − 6) = 0 x 2 + y2 − 17x − 19 y + 50 = 0

This is the equation of the required circle.

(a) x 2 + y 2 − 17x − 19 y + 50 = 0 (b) x 2 + y 2 + 17x − 19 y + 50 = 0 (c) x 2 + y 2 − 19x − 17 y + 50 = 0 (d) None of the above

x + y=6 2x + y = 4 x + 2y = 5

Coefficient of x 2 = Coefficient of y2 and Coefficient of xy = 0 ⇒ 2 + 2λ + µ = 1 + 2λ + 2µ and 3 + 5λ + 3µ = 0 ⇒ µ =1 and 3µ + 5λ + 3 = 0 ⇒ µ =1 −6 and λ= 5 On substituting the values of λ and µ in Eq. (i), we get  6 (x + y − 6) (2x + y − 4 ) −   (2x + y − 4 )  5 ⇒

Ex 8. The equation of the circle circumscribing the triangle formed by the line x + y = 6, 2x + y = 4 and x + 2 y = 5, is

respectively

The curve given in Eq. (i) will be a circle, if

…(i) …(ii) …(iii)

L1 (b) a 2 a 3 − b 2 b 3 a 2 b3 + a 3 b2

1 L2 a 3 a1 − b 3 b1 a 3 b1 + a1 b 3 L2 a 3 a1 − b 3 b1 a 3 b1 + a1 b 3

1 L3 a1 a 2 − b1 b 2 = 0 a1 b 2 + a 2 b1 L3 a1 a 2 − b1 b 2 = 0 a1 b 2 + a 2 b1

a1 a 2 − b1 b 2 1 =0 L3 a1 b 2 + a 2 b1

Sol. The equation of a curve passing through the intersection of the lines L1 = a1x + b1 y + c1 = 0 L2 = a2x + b2 y + c2 = 0 and L3 = a3x + b3 y + c3 = 0 is …(i) λL1L2 + µL2L3 + νL3L1 = 0 ⇒ λ (a1x + b1 y + c1 )(a2x + b2 y + c2 ) + µ(a2x + b2 y + c2 ) (a3x + b3 y + c3 ) + ν(a3 x + b3 y + c3 ) (a1x + b1 y + c1 ) = 0 This equation will represent a circle, if Coefficient of x 2 = Coefficient of y2

⇒ λ (x cosα 1 + y sin α 1 − p1 )(x cosα 2 + y sin α 2 − p2 ) + µ (x cosα 2 + y sin α 2 − p2 )(x cosα 3 + y sin α 3 − p3 ) + ν (x cosα 3 + y sin α 3 − p3 )(x cosα 1 + y sin α 1 − p1 ) =0 This equation will represent a circle, if Coefficient of x 2 = Coefficient of y2 and coefficient of xy = 0 ∴λ (cosα 1 cosα 2 − sin α 1 sin α 2 ) + µ (cosα 2 cosα 3 − sin α 2 sin α 3 ) + ν (cosα 3 cosα 1 − sin α 3 sin α 1 ) = 0 and λ (sin α 1 cosα 2 + cosα 1 sin α 2 ) + µ (sin α 2 cosα 3 + cosα 2 sin α 3 ) + ν (sin α 3 cosα 1 + cosα 3 sin α 1 ) = 0 ⇒ λ cos (α 1 + α 2 ) + µ cos (α 2 + α 3 ) + ν cos (α 3 + α 1 ) = 0 ...(ii) and λ sin (α 1 + α 2 ) + µ sin (α 2 + α 3 ) + ν sin (α 3 + α 1 ) = 0 ...(iii) Eliminating λ , µ, ν from Eqs. (i), (ii) and (iii), we get

and coefficient of xy = 0

L1L2

∴

λ (a1 a2 − b1b2 ) + µ (a2a3 − b2b3 ) + ν (a3a1 − b3b1 ) = 0 …(ii) and λ (a1 b2 + a2 b1 ) + µ (a2b3 + a3b2 ) + ν (a3b1 + a1b3 ) = 0 …(iii) Eliminating λ, µ and v from Eqs. (i), (ii) and (iii), we get L1L2

L2L3

L3L1

a1a2 − b1b2 a2a3 − b2b3 a3a1 − b3b1 = 0 a1b2 + a2b1 a2b3 + a3b2 a3b1 + a1b3 1 1 1 L1 L2 L3 ⇒ a2a3 − b2b3 a3a1 − b3b1 a1a2 − b1b2 = 0 a2b3 + a3b2 a3b1 + a1b3 a1b2 + a2b1

L2L3

L1 L2

L2 L3

This is the equation of the required circle. Hence, (c) is the correct answer.

Ex 11. The general equation of the circle passing through two given points A ( x1 , y1 ) and B ( x 2 , y2 ) may be written as (a) ( x − x1 )( x − x 2 ) + ( y − y1 ) ( y − y 2 ) x y

(b) L3 L2

L1 L3

L2 L1

cos (α1 + α 2 ) cos (α 2 + α 3 ) cos (α 3 + α1 ) = 0 sin (α1 + α 2 ) sin (α 2 + α 3 ) sin (α 3 + α1 ) L1 L2

L2 L3

L3 L1

(c) cos (α1 + α 2 ) cos (α 2 + α 3 ) cos (α 3 + α1 ) = 0 sin (α1 + α 2 ) sin (α 2 + α 3 ) sin (α 3 + α1 )

y1

1 =0

x2

y2

1

(b) ( x − x 2 )( y − y1 ) + ( y − y 2 ) ( x − x1 ) x y

1

+ λ x1

y1

1 =0

x2

y2

1

(c) ( x − x1 )( y − y1 ) + ( x − x 2 ) ( y − y1 ) x y

1

+ λ x1

y1

1 =0

x2

y2

1

(d) None of the above Sol. Let P (h, k ) be any point on the circle passing through points A (x1 , y1 ) and B (x2 , y2 ). Since, the angle in the same segment of a circle is always same. Therefore, ∠APB = θ or π − θ, where θ is a constant. k − y1 Now, , m1 = Slope of AP = h − x1 and

m2 = Slope of BP =

(d) None of the above tanθ = ±

Sol. The equation of three lines are L1 ≡ x cosα i + y sin α i − pi = 0, (i = 1, 2, 3). The equation of any curve passing through the points of intersection of the given straight lines is …(i) λL1L2 + µL2L3 + νL3L1 = 0

1

+ λ x1

L3 L1

(a) cos(α1 − α 2 ) cos(α 2 − α 3 ) cos(α 3 − α1 ) sin (α1 + α 2 ) sin (α 2 + α 3 ) sin (α 3 + α1 )

L3L1

cos (α 1 + α 2 ) cos (α 2 + α 3 ) cos (α 3 + α 1 ) = 0 sin (α 1 + α 2 ) sin (α 2 + α 3 ) sin (α 3 + α 1 )

Hence, (a) is the correct answer.

Ex 10. The equation of the circumcircle of the triangle formed by the lines x cos α i + y sin α i = pi , ( i =1, 2, 3), is

13 Circle

a 2 a 3 + b 2 b 3 a 3 a1 + b 3 b1 1 1 (c) L1 L2 a 2 b 3 + a 3 b 2 a 3 b1 + a1 b 3 (d) None of the above

⇒

k − y2 h − x2

m1 − m2 1 + m1m2

k − y1 k − y2 − h − x1 h − x2 tanθ = ± k − y1 k − y2 1+ × h − x1 h − x2

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⇒ tan θ = ±

Objective Mathematics Vol. 1

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(h − x2 ) (k − y1 ) − (h − x1 ) (k − y2 ) (h − x1 ) (h − x2 ) + (k − y1 ) (k − y2 )

C L3 = 0

P (h, k ) L4 = 0

L2 = 0

θ A

L1 = 0 B

(x1 , y1 ) A

B (x2 , y2 ) π–θ P (h, k)

⇒

(h − x1 )(h − x2 ) + (k − y1 )(k − y2 ) = ± cot θ{(h − x2 )(k − y1 ) − (h − x1 )(k − y2 )}

Hence, the locus of (h, k ) is (x − x1 )(x − x2 ) + ( y − y1 )( y − y2 ) = ± cot θ {(x − x2 )( y − y1 ) − (x − x1 )( y − y2 )} ⇒ (x − x1 )(x − x2 ) + ( y − y1 )( y − y2 ) ± cot θ {( x ( y1 − y2 ) + y(x2 − x1 ) + x1 y2 − x2 y1} = 0 ⇒ (x − x1 )(x − x2 ) + ( y − y1 )( y − y2 ) x y 1 ± cotθ x1 y1 1 = 0 x2 ⇒

(x − x1 )(x − x2 ) + ( y − y1 )( y − y2 ) x + λ x1 x2

y2 1 y 1 y1 1 = 0 y2 1

Hence, (a) is the correct answer.

Ex 12. The equation of the circle circumscribing the quadrilateral formed by the lines in order are 5x + 3 y − 9 = 0, x − 3 y = 0, 2x − y = 0, x + 4 y − 2 = 0 without finding the vertices of the quadrilateral, is 16 2 (a) 2x 2 + 2 y 2 − x + y = 0 9 3 16 2 2 2 (b) 2x + y + x + y = 0 9 3 16 1 2 2 (c) x + y + x + y = 0 9 3 16 1 2 2 (d) x + y − x + y = 0 9 3 Sol. The equation of the sides of the quadrilateral are

and

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D

L1 ≡ 5x + 3 y − 9 = 0 L2 ≡ x − 3 y = 0 L3 ≡ 2x − y = 0 L4 ≡ x + 4 y − 2 = 0

The equation of the curve passing through the vertices of the quadrilateral is = L1L3 + λL2L4 = 0 ⇒ (5x + 3 y − 9)(2x − y) + λ (x − 3 y) (x + 4 y − 2) = 0 …(i) This will represent a circle, if Coefficient of x 2 = Coefficient of y2 and coefficient of xy = 0 ⇒ 10 + λ = − 3 − 12λ ⇒ 1+ λ = 0 ⇒ λ = −1 On substituting the value of λ in Eq. (i), we get (5x + 3 y − 9)(2x − y) − (x − 3 y)(x + 4 y − 2) = 0 16 1 ⇒ x 2 + y2 − x + y = 0 9 3 which is the equation of the required circle. Hence, (d) is the correct answer.

Ex 13. The equation of the circles passing through ( −4, 3) and touching the lines x + y = 2 and x − y = 2, is (a) x 2 + y 2 ± 2(10 + 3 6 )x + ( 55 ± 24 6 ) = 0 (b) x 2 + y 2 ± 2(10 + 6 )x + ( 55 ± 6 ) = 0 (c) x 2 − y 2 ± 2(10 + 3 6 ) y − ( 55 ± 24 6 ) = 0 (d) None of the above

Sol. Let the equation of the required circle be x 2 + y2 + 2gx + 2 fy + c = 0 It passes through (−4 , 3). Therefore, 25 − 8g + 6 f + c = 0

…(i) …(ii)

Since, circle (i) touches the lines x + y − 2 = 0 and x − y − 2 = 0. Therefore, −g − f − 2 −g + f − 2 = 2 2 = g2 + f 2 − c Now, ⇒ ⇒ or ⇒

−g − f − 2 −g + f − 2 = 2 2 − g − f − 2 = ± (− g + f − 2) −g − f − 2 = − g + f − 2 −g − f − 2 = g − f + 2 f = 0 or g = − 2

…(iii)

f =0

When

From Eq. (iii), we have −g + f − 2 = g2 + f 2 − c 2 −g − 2 ⇒ = g2 − c 2 ⇒ ⇒

[Q f = 0 ]

(g + 2)2 = 2(g 2 − c) g 2 − 4 g − 4 − 2c = 0

…(iv)

Putting f = 0 in Eq. (ii), we get 25 − 8g + c = 0 Eliminating c from Eqs. (iv) and (v), we get g 2 − 20g + 46 = 0 ⇒ g = 10 ± 3 6

…(v)

(a) 3 times the radius of either circle (b) 2 times the radius of either circle 1 (c) times the radius of either circle 2 (d) 2 times the radius of either circle circle.

This is the equation of the required circle.

Then, and

b− y a−x −a − y m2 = Slope of PB = b−x m1 = Slope of PA =

g = −2

When

From Eq. (iii), we have −g + f − 2 = g2 + f 2 − c 2 ⇒

f 2 = 2(4 + f 2 − c)

⇒

P (x, y) 45°

[Q g = − 2 ]

f − 2c + 8 = 0 2

…(vi)

On putting g = − 2 in Eq. (ii), we get c = − 6 f − 41 On substituting the value of c in Eq. (vi), we get f 2 + 12 f + 82 + 8 = 0 ⇒

f + 12 f + 90 = 0 2

This equation gives imaginary values of f . Hence, the equations of the required circles are x 2 + y2 ± 2(10 + 3 6 )x + (55 ± 24 6 ) = 0 Hence, (a) is the correct answer.

 1  Ex 14. If  mi ,  are four distinct points on a circle, mi   then (a) m1 m2 m3 m4 = 1 (b) m1 m2 m3 m4 = − 1 1 (c) m1 m2 m3 m4 = 2 1 1 1 1 1 (d) + + + = m1 m2 m3 m4 4  

Sol. Let the points  mi ,

1  , i = 1, 2, 3, 4 lie on the circle mi 

x + y + 2gx + 2 fy + c = 0 1 2f Then, mi2 + 2 + 2gmi + + c = 0, i = 1, 2, 3, 4 mi mi 2

13

Sol. Let P (x , y) be any point on the circumference of the

On substituting the values of g, f and c in Eq. (i), we get x 2 + y2 ± 2(10 ± 3 6 )x + (55 ± 24 6 ) = 0 Case II

Ex 15. Circles are drawn through the points (a, b) and ( b, − a ), such that the chord joining the two points subtends an angle of 45° at any point of the circumference. Then, the distance between the centres is

Circle

Case I

2

⇒ mi4 + 2gmi3 + cmi2 + 2 fmi + 1 = 0, i = 1, 2, 3, 4 ⇒ m1, m2, m3 and m4 are the roots of the equation m4 + 2gm3 + cm2 + 2 fm + 1 = 0 1 ⇒ m1m2m3m4 = = 1 1 Hence, (a) is the correct answer.

(a, b)A 135°

B (b, –a)

P (x, y)

We have, ∠APB = 45° or 135° m1 − m2 ⇒ = tan 45° or tan135° 1 + m1m2 b − y −a − y − a−x b−x ⇒ = ±1 b − y −a − y × 1+ a−x b−x (b − y)(b − x ) + (a + y)(a − x ) = ±1 ⇒ (a − x )(b − x ) − (b − y)(a + y) ⇒

x 2 + y2 = a2 + b2

⇒

{x − (a + b)}2 + { y − (b − a)}2 = a2 + b2

Hence, the equations of the circles are x 2 + y2 = a2 + b2 and {x − (a + b)}2 + y − {(b − a)}2 = a2 + b2 The centres of these circles are O (0, 0) and C (a + b, b − a). ∴ Distance between the centre = (a + b)2 − (a − b)2 = 2 a2 + b2 = 2 (radius of either circle) Hence, (d) is the correct answer.

Ex 16. Let a n , n =1, 2, 3, 4 represent four distinct positive real numbers other than unit such that each pair of the logarithm of a n and the reciprocal of logarithm denotes a point on a circle, whose centre lies on Y-axis. Then, the product of these four numbers is (a) 0

(b) 1

(c) 2

(d) 3

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Sol. Let (0, b) be the centre and r be the radius of the given circle, then its equation is (x − 0)2 + ( y − b)2 = r2

Objective Mathematics Vol. 1

13

⇒

x 2 + y2 − 2 yb + b2 − r2 = 0

…(i)

 1  It is given that the point Pn  log an ,  ; n = 1, 2, 3, 4  log an  lie on the circle given by Eq. (i). Therefore, 1 2b (log an )2 + − + b2 − r 2 = 0 (log an )2 log an where, n = 1, 2, 3, 4 ⇒ loga1, loga2, loga3 and loga4 are the roots of the equation λ4 + (b2 − r2 )λ2 − 2bλ + 1 = 0 ∴ Sum of the roots = 0 log a1 + log a2 + log a3 + log a4 = 0 ⇒ log(a1a2a3a4 ) = 0 ⇒ a1a2a3a4 = 1 Hence, (b) is the correct answer.

Ex 17. If A (a, 0) and B (b, 0) are two point on X-axis and a point C (0, c) is on Y-axis. Then, the locus of a point P such that A, B, P and C are concyclic and the coordinates of a point D on the locus of P such that AB is parallel to CD, are 2

 a + b c 2 + ab    (a)  x −  +y−  2  2c   c4 + a2b2 + b2c2 + c2a2 , ( a + b, c ) = 4c 2 2 2  a + b c 2 + ab    (b)  x −  +y−  2  2c   c4 + a2 b2 + b2c2 + c2a2 , ( a − b, c ) = c2 2 2  a + b c 2 + ab     (c)  x − + − y   2  2c   2 2 2 2 2 c + a b + b c + c2a2 , ( a + c, b ) = 4c 2 (d) None of the above 2

Sol.

Y

Then, the equation of the circle is (x − α )2 + ( y − β )2 = r2

Let the coordinates of A, B and C be (a, 0), (b, 0) and (0, c) respectively.As A , B and C lie on Eq.(i). Therefore, (a − α )2 + β 2 = r2 (b − α )2 + β 2 = r2 and α 2 + (c − β )2 = r2 On solving these equations, we get (a + b) c2 + ab , β= α= 2c 2 c4 + a2b2 + b2c2 + c2a2 2c Let (h, k ) be the coordinates of P. Then, SP = r ⇒ (h − α )2 + (k − β )2 = r2 and

r=

2  a + b c2 + ab  ⇒ h −   + k −  2  2c  

=

2

c4 + a2b2 + b2c2 + c2a2 4 c2

Hence, the locus of P is 2  a + b c2 + ab    + y− x −  2  2c  

2

c4 + a2b2 + b2c2 + c2a2 4 c2 It is given that CD is parallel to AB. So, the equation of CD is y = c. On putting y = c in Eq. (ii), we get =

2

2  a + b c2 + ab c4 + a2b2 + b2c2 + c2a2   = x −  + c −  2  2c  4 c2  2

c4 + a2b2 + b2c2 + c2a2 − (c2 − ab)2  a + b x−  =  2  4 c2 2

2

a + b  a + b  ⇒ x −   =  2    2 ⇒ x=a+ b ∴ The coordinates of D are (a + b, c). Hence, (a) is the correct answer.

Ex 18. The set of value of a for which the point (2a, a + 1) is an interior point of the larger segment of the circle x 2 + y 2 − 2x − 2 y − 8 = 0 made by the chord x − y + 1 = 0, is  5 9  5 (a)  ,  (b)  0,   9 5  9

P

...(i)

 9 (c)  0,   5

 9 (d) 1,   5

Sol. The point (2a, a + 1) will be an interior point of the larger (0, c)C X′

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O A (a, 0) Y′

S (α, β)

segment of the circle x 2 + y2 − 2x − 2 y − 8 = 0

D

B (b, 0)

X

It is given that A, B, P and C are concyclic points. Let S (α , β ) be the centre and r be the radius of the circle.

Since, the point (2a, a + 1) is an interior point and the point (2a, a + 1) and the centre (1, 1) are on the same side of the chord x − y + 1 = 0. ∴ (2a)2 + (a + 1)2 − 2(2a) − 2(a + 1) − 8 < 0 and (2a − a − 1 + 1)(1 − 1 + 1) > 0 ⇒ 5a2 − 4 a − 9 < 0 and a > 0

(5a − 9)(a + 1) < 0 and 9 and −1 < a < 5  9 a ∈  0,   5

⇒ ⇒

a>0 a>0 Now,

Ex 19. Consider the circle x 2 + y 2 = a 2 . Let A (a, 0) and D be given interior point of the circle. If BC is an arbitrary chord of the circle through point D, then the locus of the centroid of ∆ABC is (a) a circle whose radius is less than 2a/ 3 (b) a circle whose radius is greater than 2a/ 3 (c) a circle whose radius is equal to 2a/ 3 (d) None of the above Y

B

X

Y′ 2

α + β 2 < a2

…(i)

Let DB = r. Then, the coordinates of B are (α + r cosθ , β + r sin θ ) It lies on x 2 + y2 = a2 Therefore, (α + r cosθ )2 + (β + r sin θ )2 = a2 ⇒ r2 + 2r(α cosθ + β sin θ ) + α 2 + β 2 − a2 = 0 …(ii) This is quadratic in r. So, it gives two values of r one each for point B and C. Let DB = r1 and DC = r2 Then, r1, r2 are the roots of Eq. (ii). ∴ r1 + r2 = − 2(α cosθ + β sin θ ) and …(iii) r1r2 = α 2 + β 2 − a2 The coordinates of B and C are (r1 cosθ , r1 sin θ ) and (r2 cosθ , r2 sin θ ) respectively. Let C (h, k ) be the centroid of ∆ABC. Then, 3h = a + r1 cosθ + r2 cosθ and 3k = r1 sin θ + r2 sin θ ⇒ (3h − a)2 + (3k )2 = (r1 + r2 )2 Hence, the locus of (h, k ) is (3x − a)2 + (3 y)2 = (r1 + r2 )2 2

Hence, (a) is the correct answer.

Therefore, circumcentre of ∆ABC is at the origin. The coordinates of the centroid C are a a   (cosα + cosβ + cos γ ), (sin α + sin β + sin γ ) , 3  3 Let O1 (x , y) be orthocentre of ∆ABC. Then, O (0, 0), a a  C  (cosα + cosβ + cos γ ), (sin α + sin β + sin γ ) 3  3

The equation of BC is x −α y−β = cosθ sin θ

2

a r + r   2  x −  + y =  1 2  3   3 which represents a circle of radius r given by ⇒

2 2 |α cosθ + β sin θ | ⇒ r ≤ α 2 + β2 3 3 2 [using Eq. (i)] r< a ⇒ 3 Hence, the locus of the centroid of ∆ABC is a circle 2a a  having its centre at  , 0 and radius less than . 3  3

Sol. Clearly, points A, B andC lie on the circle x 2 + y2 = a2.

C

Then,

13

(a) collinear (b) concyclic (c) not concylic (d) None of the above

x2 + y2 = a2 A (a, 0)

O

r2  r2 

Ex 20. The angular points of a triangle are A ( a cos α, a sin α ), B ( a cos β, a sin β) and C ( a cos γ , a sin γ ). The coordinates of the orthocentre of ∆ABC are [ a (cos α + cos β + cos γ ), a (sinα + sin β + sin γ )] If A, B, C and D are four points on a circle, then the orthocentre of the ∆ABC, ∆BCD, ∆CDA and ∆DAB are

Sol. Let D (α , β) be an interior point of circle.

X′

+ 3 + 3

⇒r =

Hence, (c) is the correct answer.

D (α, β)

r r = 1  r r = 1 

Circle

⇒

and O1 (x , y) are collinear and C divides OO1 in the ratio 1 : 2. a 1× x + 2 × 0 (cos α + cos β + cos γ ) = ∴ 3 1+ 2 a 1× y + 2 × 0 and (sin α + sin β + sin γ ) = 3 1+ 2 x = a (cos α + cos β + cos γ ) y = a (sin α + sin β + sin γ ) Thus, the coordinates of the orthocentre are [ a(cosα + cosβ + cos γ ),(a(sin α + sin β + sin γ )] Let D (a cos δ , a sin δ ) be a point on the circle x 2 + y2 = a2. ⇒ and

Then, the coordinates of the orthocentre of ∆BCD, ∆CDA and ∆DAB are O2 [ a(cosβ + cos γ + cosδ ), a(sin β + sin γ + sin δ )] O3 [ a(cos γ + cosδ + cosα ), a(sin γ + sin δ + sin α )] O4 [ a(cosδ + cosα + cosβ ), a(sin δ + sin α + sin β )] Consider the circle {x − a(cosα + cosβ + cos γ + cosδ )}2 + { y − a(sin α + sin β + sin γ + sin δ )}2 = a2 ...(i) We observe that this circle passes through O1, O2, O3 and O4. Hence, the four orthocentre are concyclic. Hence, (b) is the correct answer.

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Ex 21. Three concentric circles of which biggest is x 2 + y 2 = 1, have their radii in AP. If the line y = x +1 cuts all the circles at real and distinct points, then the interval in which the common difference in an AP will lie, is   1  (a) 0, 1 −  2   

The equation of the tangent at O (0, 0) is a b 0x + 0 y − (x + 0) − ( y + 0) = 0 2 2 ⇒ ax + by = 0

 1 (b)  0,   2

B (0, b)

n

 1 1  (d) 0, 1 −   2   2

(c) (1, 1)

X′

O

A (a, 0)

x 2 + y2 = 12

Y′

Clearly, it is centred at O (0, 0) and has radius 1. Let the radii of the other two circles be 1 − r, 1 − 2r, where r > 0. Thus, the equations of the concentric circles are …(i) x 2 + y2 = 1 x 2 + y2 = (1 − r)2 x + y = (1 − 2r) 2

…(ii) 2

…(iii)

Clearly, y = x + 1 cuts the circle (i) at (1, 0) and (0, 1). This line will cut circles (ii) and (iii) at real and distinct points, if  1  < 1 − r and  1  < 1 − 2r  2  2 1 2b 2 (c) a 2 = 2b 2

(b) a 2 < 2b 2 (d) None of these

Sol. The equation of the circle is

2x (x − a) + y(2 y − b) = 0 b x 2 + y2 − ax − y = 0 2

⇒

(c) x 2 + y 2 − mn ( m x + n y ) = 0 (d) None of the above π 2 Therefore, AB is a diameter of the circle. Thus, the equation of the circle is x (x − a) + y( y − b) = 0 ⇒ x 2 + y2 − ax − by = 0

∴ m = Length of the perpendicular from A (a, 0) on Eq. (ii) a2 …(iii) = a 2 + b2

⇒

r
0 4 8  2 a > 2b 2

Let O (0, 0) be the fixed point in the plane of the circles. Then, li2 = Square of the length of the tangent from O (0, 0) to x 2 + y2 + 2gi x + 2 fi y = 0

2

= ci , i = 1, 2, 3, 4

Ex 24. The length of the common chord of the two circles ( x − a ) 2 + ( y − b) 2 = c 2 and ( x − b) 2 + ( y − a ) 2 = c 2 is (a) 4c 2 + 2( a − b ) 2

(b) 4c 2 − ( a − b ) 2

(c) 4c 2 − 2( a − b ) 2

(d) 2c 2 − 2( a − b ) 2

Sol. The equation of two circles are and

S 1 ≡ (x − a)2 + ( y − b)2 = c2

…(i)

S 2 ≡ (x − b)2 + ( y − a)2 = c2

…(ii)

The equation of the common chord of these circle is P

M

Eliminating g, f and c from these four equations, we get g1 f1 −1 − l12 g2 f2 −1 − l22 =0 g3 f3 −1 − l32 g4 f4 −1 − l42 l12 l22 l32 l42

⇒

C2 (b, a)

Q

⇒

S1 − S2 = 0 (x − a)2 − (x − b)2 + ( y − b)2 − ( y − a)2 = 0

⇒ (2x − a − b) (b − a) + (2 y − b − a) (a − b) = 0 ⇒ 2x − a − b − 2 y + b + a = 0 ⇒ x − y=0 Let C 1 and C 2 be centres of circles (i) and (ii) respectively. Then, the coordinates of C 1 and C 2 are (a, b) and (b, a) respectively.  a − b  | a − b| = Now, C 1M =  2  1 + 1 In right angled ∆C 1PM , we have PM = C 1P 2 − C 1M 2 = c2 −

(a − b)2 2

(a − b)2 = 4 c2 − 2(a − b)2 2 Hence, (c) is the correct answer.

∴ PQ = 2PM = 2 c2 −

…(ii)

From Eqs. (i) and (ii), we have 2(ggi + ffi ) = c + li2; i = 1, 2, 3, 4 ⇒ 2gg1 + 2 ff1 − c − l12 = 0 2gg2 + 2 ff2 − c − l22 = 0 2gg3 + 2 ff3 − c − l32 = 0 and 2gg4 + 2 ff4 − c − l42 = 0

c C1 (a, b)

13

Sol. Let x 2 + y2 + 2gi x + 2 fi y + ci = 0 , i = 1, 2, 3, 4 be four

a b2 a  Therefore,  t −  a − = t2 − t  2 8 2  a 2 b2  3 ⇒ t 2 − at +  +  = 0 2 8  2

a2 b 2 ⇒ − >0 ⇒ 4 2 Hence, (a) is the correct answer.

Ex 25. P, Q, R and S are the centres as the four circles each of which is cut by a fixed circle orthogonally. If l12 , l22 , l32 and l42 are the squares of the lengths of tangents to the four circles from a point in their plane, then

Circle

 b Let PQ and PR be two chords drawn from P  a,   2 such that they are bisected by X-axis. Let A (t , 0) be the mid-point of PQ. Then, its equation is a b [QT = S ] tx + 0 y − (x + t ) − ( y + 0) = t 2 − at 2 4 a b a  2 ⇒ t −  x − y = t − t   2 4 2  b This passes through P  a,  .  2

g2 ⇒ l12 g3 g4

f2 f3 f4

g1 g2 g3 g4 1 g1 1 − l22 g3 1 g4 g 1 f1

+ l32 g2

f2

g4

f4

f1 1 f2 1 =0 f3 1 f4 1 f1 1 f3 1 f4 1 g 1 f1 1 1 1 − l42 g2 f2 1 = 0 …(iii) g 3 f3 1 1

The coordinates of the centre of the four circles P ( − 2g1 − 2 f1 ), Q ( − 2g2 − 2 f2 ), R ( − 2g3 − 2 f3 ) and S ( − 2g4 − 2 f4 ). Q Area of ∆ PQR −2g1 −2 f1 1 g 1 f1 1 = −2g2 −2 f2 1 = 2 g2 f2 2 −2g3 −2 f3 1 g 3 f3 g 1 f1 1 ⇒ g2 f2 1 = 2∆PQR g3

f3 1

are

1 1 1

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Similarly, we have g 2 f2 1 g1 g3 f3 1 = 2∆QRS , g3 g 4 f4 1 g4 g1 and

g2 g4

f1 1 f3 1 = 2∆RSP f4 1

f1 1 f2 1 = 2∆PQS f4 1

A′

On substituting these values in Eq. (iii), we get l12∆QRS − l22∆RSP + l32∆SPQ − l42∆PQR = 0 Hence, (c) is the correct answer.

Ex 26. The line Ax + By + C = 0 cuts the circle x 2 + y 2 + ax + by + c = 0 at P and Q. The line A ′ x + B ′ y + C ′ = 0 cuts the circle x 2 + y 2 + a ′ x + b′ y + c′ = 0 at R and S. If P, Q, R and S are concyclic, then a − a ′ b − b ′ c − c′ (a) A B C =0 A′ B′ C′

a + a ′ b + b ′ c + c′ (b) A B C =0 A′ a (c) A

B′ c

B

C =0

B′

C′

Aliter The equation of a family of circles passing through P and Q is x 2 + y2 + ax + by + c + λ ( Ax + By + C ) = 0 …(i) Similarly, the equation of a family of circles passing through R and S is x 2 + y2 + a′ x + b′ y + c′ + µ( A′ x + B′ y + C ′ ) = 0 …(ii) If points P, Q, R and S are concyclic points, then Eqs. (i) and (ii) must represents the same circle for some values of λ and µ. ∴ a + λA = a′ + µA′ b + λB = b′ + µB′ c + λC = c′ + µC ′ …(iii) ⇒ a − a′ + λA − µA′ = 0 …(iv) b − b′ + λB − µB′ = 0 …(v) c − c′ + λC − µC ′ = 0 Eliminating λ and µ from Eqs. (iii), (iv) and (v), we get a − a′ b − b′ c − c′

C′

b

A′ x + B′ y + C ′ = 0 and x (a − a′ ) + y(b − b′ ) + c − c′ = 0 are concurrent. Consequently, we have a − a′ b − b′ c − c′ A B C =0

A A′

A′ B′ C′

B B′

C C′

=0

Hence, (a) is the correct answer.

(d) None of the above Sol. Let the given circles be S 1 = 0 and S 2 = 0. Let P, Q, R and S lie on the circle S 3 = 0. Since, Ax + By + C = 0 cuts the circle S 1 = 0 at P and Q. Therefore, Ax + By + C = 0 is the radical axis of S 1 = 0 and S 3 = 0 Similarly, A′ x + B′ y + C ′ = 0 is the radical axis of S 2 = 0 and S 3 = 0. The radical axis of S 1 = 0 and S 2 = 0 is S 1 − S 2 = 0.

Ex 27. A circle touches the hypotenuse of a right angled triangle at its middle point and passes through the mid-point of the shorter side. If a and b ( a < b) are the length of the sides, then the radius is b a2 + b2 a b (c) a2 + b2 4a

(a)

b a2 − b2 2a

(b)

(d) None of these

x y + = 1 at a b 2 2 a b y   a a  x  P  ,  is  x −  +  y −  + λ  + − 1 = 0. a b    2 2  2 2

Sol. The equation of the circle touching

Y Q C1

R

C2

(– a2 , – b2 )

(– a2 , – b2 )

B (0, b)

(

P a,b 2 2

S

)

P

X′

732

i.e. x (a − a′ ) + y(b − b′ ) + c − c′ = 0 Since, the radical axes of three circles, taken in pairs, are concurrent. Therefore, Ax + By + C = 0

A (a, 0) Q a ,0 2

( )

X

( xa + yb ) = 1 Y′

Ex 29. The limiting points of the coaxial system of circles given by x 2 + y 2 + 2gx + c + λ ( x 2 + y 2 + 2 fy + k ) = 0 subtend a right angle at the origin, if

b2 Therefore, λ= 2 On putting the value of λ in Eq. (i), we get 2 2 a b b2  x y    x −  +  y −  +  + − 1 = 0     2 2 2 a b   b2  by a2 − b2 + =0 ⇒ x 2 + y2 −  a −  x − 2a 2 4 

(c)

2

2

2

2

2

Ex 28. Let P , Q and R be the centres and r1 , r2 , r3 be the corresponding radii of three circles from a system of coaxial circle, then r12 ⋅ QR + r22 ⋅ RP + r32 ⋅ PQ is equal to (a) PQ ⋅ QR ⋅ RP (b) − PQ ⋅ QR ⋅ RP (c) PQ + QR + RP PQ (d) × RP QR

Sol. Let x 2 + y2 + 2gx + c = 0, where g is a variable and c is a constant, be a coaxial system of circle having common radical axis as X-axis. Let x 2 + y2 + 2gi x + c = 0, i = 1, 2, 3 be three members of the given coaxial system of circles. Then, the coordinates of their centres and radii are P (− g1 , 0), Q (− g2 , O ), R (− g3 , 0) r12 = g12 − c, r22 = g22 − c r32 = g32 − c

and

Now, r12 ⋅ QR + r22 ⋅ RP + r32 ⋅ PQ = (g12 − c) (g2 − g3 ) + (g22 − c) + (g3 − g1 ) + (g32 − c) (g1 − g2 ) = − g3 ) + − g1 ) + g32 (g1 − g2 ) − c {(g2 − g3 ) + (g3 − g1 ) + (g1 − g2 )} = − (g1 − g2 ) (g2 − g3 ) (g3 − g1 ) = − PQ ⋅ QR ⋅ RP g12 (g2

2

2

−

+

k f

k f2

2

=2

=2

(b) (d)

c 2

g c

g2

+ +

k f2 k f2

=−2 =2

Sol. The equation representing the coaxial system of circles

a − b 1 b b b (a + b ) − = a −  + 4 2a 16  4  16a2 b ⇒ r= a2 + b2 4a Hence, (c) is the correct answer. r2 =

2

g c

g

Let r be the radius of this circle. Then, 2 2

c

(a) −

13 Circle

a  It passes through Q  , 0 . 2 

g22 (g3

Hence, (b) is the correct aswer.

is x 2 + y2 + 2gx + c + λ (x 2 + y2 + 2 fy + k ) = 0 2g 2fλ c + kλ x 2 + y2 + x+ y+ = 0 ...(i) ⇒ 1+ λ 1+ λ 1+ λ The coordinates of the centre of this circle are  g fλ  , −   1 + λ 1 + λ and radius =

...(ii)

g 2 + f 2λ2 − (c + kλ ) (1 + λ ) (1 + λ )2

For the limiting points, we have Radius = 0 ⇒ g 2 + f 2λ2 − (c + kλ ) (1 + λ ) = 0 ⇒ λ2 ( f 2 − k ) − λ (c + k ) + (g 2 − c) = 0

...(iii)

Let λ 1 and λ 2 be the roots of this equation. Then, c+ k g2 − c and λ 1λ 2 = 2 ...(iv) λ1 + λ2 = 2 f −k f −k Thus, the coordinates of limiting points L1 and L2 are  − f λ1 g , L1  −   1 + λ 1 1 + λ 1 and

 −g − f λ2 , L2    1 + λ 2 1 + λ 2

[from Eq. (iv)]

Now, L1L2 will subtend a right angle at the origin, if Slope of OL1 × Slope of OL2 = − 1 f λ1 f λ2 Then, × = −1 g g ⇒ f 2λ 1λ 2 = − g 2  g2 − c  f2  2 ⇒  = − g2  f − k ⇒

f 2 (g 2 − c) + g 2 ( f 2 − k ) = 0 c k ⇒ + 2 =2 g2 f Hence, (d) is the correct answer.

Type 2. More than One Correct Option Ex 30. The coordinates of the point(s) on the line x + y = 5, which is/are equidistant from the lines | x| = | y |, is/are (a) (5, 0) (b) (0, 5) (c) ( −5, 0) (d) ( 0, − 5)

Sol. | x | = | y | ⇒ x + y = 0 and x − y = 0 Bisector of the angles between these are x− y x+ y =± ⇒ y = 0 and x = 0 2 2 By solving y = 0, x + y = 5, we get (5, 0) and x = 0, x + y = 5, we get (0, 5) Hence, (a) and (b) are the correct answers.

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∆PP1P2 =

Ex 31. The tangents drawn from the origin to the circle x 2 + y 2 + 2gx + 2 fy + f 2 = 0 are perpendicular, if (a) g = f (c) g = 2 f

(b) g = − f (d) 2g = f

Sol. Radius of the given circle = | g |. For perpendicular tangents, origin should lie on the director circle, i.e. distance of origin from the centre of the given circle = |g | = 2 ⇒

g 2 + f 2 = 2g 2 ⇒ g = ± f

Hence, (a) and (b) are the correct answers.

Ex 32. If (a, 0) is a point on a diameter of the circle x 2 + y 2 = 4, then x 2 − 4x − a 2 = 0 has (a) exactly one real root in ( −1 ,0] (b) exactly one real root in [ 2, 5] (c) distinct roots greater than − 1 (d) distinct roots less than 5 Sol. Since, (a, 0) is a point on the diameter of the circle x 2 + y2 = 4. 2

Maximum value of a is 4. Let f (x ) = x 2 − 4 x − a2 Clearly, f (−1) = 5 − a2 > 0, f (2) = − (a2 + 4 ) < 0 f (0) = − a2 < 0 and

f (5) = 5 − a2 > 0

Graph of f (x ) will be as shown as 0

2 X 5

–1

Hence, (a), (b), (c) and (d) are the correct answers.

Ex 33. Tangents are drawn to the circle x 2 + y 2 = 50 from a point P lying on the X-axis. These tangents meet the Y -axis at points P1 and P2 . Possible coordinates of P so that area of ∆PP1 P2 is minimum, are (a) (10, 0)

(b) (10 2 , 0)

(c) ( −10, 0)

(d) ( −10 2 , 0)

Sol.

(∆PP1P2 )min = 100 ⇒ θ =

⇒ OP = 10 ⇒ P = (10, 0), (−10, 0) Hence, (a) and (c) are the correct answers.

Ex 34. Consider the circle x 2 + y 2 − 10x − 6 y + 30 = 0. Let O be the centre of the circle and tangent at A ( 7, 3) and B (5, 1) meet at C. Let S = 0 represents family of circles passing through A and B, then (a) area of quadrilateral OACB = 4 (b) the radical axis for the family of circles S = 0 is x + y = 10 (c) the smallest possible circle of the family S = 0 is x 2 + y 2 − 12 x − 4 y + 38 = 0 (d) the coordinates of point C are ( 7, 1)

Sol. Coordinates of O are (5, 3) and radius = 2 Equation of tangent at A (7, 3) is (7x + 3 y − 59x + 7) − 3 ( y + 3) + 30 = 0 i.e. 2x − 15 = 0 ⇒ x = 7 Equation of tangent at B (5, 1) is 5x + y − 5 (x + 5) − 3 ( y + 1) + 30 = 0 i.e. − 2 y + 2 = 0 i.e. y = 1 ∴ Coordinates of C are (7, 1). ∴ Area of OACB = 4 Equation of AB is x − y = 4 (radical axis) Equation of the smallest circle is (x − 7) (x − 5) + ( y − 3) ( y − 1) = 0 i.e. x 2 + y2 − 12x − 4 y + 38 = 0 Hence, (a), (c) and (d) are the correct answers.

Ex 35. The range of values of a such that the angle θ between the pair of tangents drawn from ( a, 0) π to the circle x 2 + y 2 = 1 satisfies < θ < π, 2 lies in (a) (1, 2) (c) ( − 2 , − 1)

(b) (1, 2 ) (d) ( − 2 , − 1) ∪ (1, 2 )

(a2 − 1) y2 − x 2 + 2ax − a2 = 0 If θ is the angle between the tangents, then

P1

tan θ =

θ P

O

X

2 h2 − ab 2 − (a2 − 1) (−1) 2 (a2 − 1) = = a+ b a2 − 2 a2 − 2

If θ lies in II quadrant, than tanθ < 0 2 (a2 − 1)

∴ P2

⇒

a2 − 2 a2 − 1 > 0

and

1 and | a| < 2 a ∈ (− 2 , − 1) ∪ (1, 2 )

(a) 20 (c) −30

(b) 22 (d) −28

Sol. Since, the given line touches the given circle, then length of the perpendicular from the centre (2, 4) of the circle from the line 3x − 4 y − k = 0 is equal to the radius 4 + 16 + 5 = 5 of the circle. 3×2−4 ×4 −k ⇒ =±5 9 + 16 ⇒ k = 15 or − 35 Now, equation of the tangent at (a, b) to the given circle is xa + yb − 2 (x + a) − 4 ( y + b) − 5 = 0 ⇒ (a − 2) x + (b − 4 ) y − (2a + 4 b + 5) = 0 If it represents the given line 3x − 4 y − k = 0 a − 2 b − 4 2a + 4 b + 5 [say]…(i) Then, = = =λ −4 k 3 Then, a = 3λ + 2, b = 4 − 4 λ and 2a + 4 b + 5 = kλ ⇒ 2 (3λ + 2) + 4 (4 − 4 λ ) + 5 = 15λ [if k = 15] ⇒ λ =1 ⇒ a = 5, b = 0 and k + a + b = 20 Again, if k = − 35, from Eq. (i), 25 − 10λ = − 35λ ⇒ λ = − 1⇒ a = − 1 b = 8and k + a + b = − 35 − 1 + 8 = − 28 Hence, (a) and (d) are the correct answers.

Ex 37. If the circle x 2 + y 2 + 2gx + 2 fy + c = 0 cuts each of the circles x 2 + y 2 − 4 = 0 x 2 + y 2 − 6x − 8 y + 10 = 0 and x 2 + y 2 + 2x − 4 y − 2 = 0 at the extremities of a diameter, then (a) c = − 4 (b) g + f = c − 1 (c) g 2 + f 2 − c = 17 (d) gf = 6

Sol. Let S 1 = x 2 + y2 + 2gx + 2 fy + c = 0 S 2 = x 2 + y2 − 4 = 0 S 3 = x 2 + y2 − 6x − 8 y + 10 = 0 S 4 = x 2 + y2 + 2x − 4 y − 2 = 0

13 Circle

Ex 36. If the line 3x − 4 y − k = 0 touches the circle x 2 + y 2 − 4x − 8 y − 5 = 0 at ( a, b), then k + a + b can be equal to

Equation of the common chords S 1 − S 2 = 0, S 1 − S 3 = 0, S 1 − S 4 = 0. Satisfy these equations by the coordinates of the centre of corresponding circles. Hence, (a), (b), (c) and (d) are the correct answers.

Ex 38. If the area of the quadrilateral formed by the tangents from the origin to the circle x 2 + y 2 + 6x − 10 y + c = 0 and the radii corresponding to the points of contact is 15, then a value of c is (a) 9 (b) 4 (c) 5 (d) 25 Sol. Area of the quadrilateral = c × 9 + 25 − c = 15 ∴ c = 9, 25 Hence, (a) and (d) are the correct answers

Ex 39. The set of real values of a for which atleast one tangent to the parabola y 2 = 4ax becomes normal to the circle x 2 + y 2 − 2ax − 4ay + 3a 2 = 0, is (a) [1, 2] (b) [ 2 , 3] (c) R (d) φ (null set) Sol. Any tangent of parabola will be of the form ty = at 2 at the point (at 2 , 2at ). If this is normal to the circle, then this will passes through centre of the circle which is (a, 2a). 2at = a + at 2 ⇒

t 2 − 2t + 1 = 0

⇒ t = 1, 1 for any value of a. So, the condition satisfies for all real values of a. Hence, (a), (b) and (c) are the correct answers.

Type 3. Assertion and Reason Directions (Ex. Nos. 40-49) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d) out of which only one is correct. The correct choices are

Ex 40. Statement I The circle of smallest radius passing through two given points A and B must 1 be of radius AB. 2

(a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

Statement II A straight line is a shortest distance between two points.

(b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

Sol. Let C 1 be a circle which passes through A, B and C

(c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

whose diameter is AB and C 2 be another circle which passes through A and B, then centres of C 1 and C 2 must lie on perpendicular bisector of AB. Indeed, centre of

735

C 1 is mid-point M of AB and centre of any other circle lies somewhere else on bisector.

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13

M1

Statement II Two orthogonal circles intersect to generate a common chord which subtends supplementary angles at their centres.

M A

B C2 C1

M 1 A > AM [hypotenuse of right angled ∆AMM 1 ] Radius of C 2 > 1/ 2 AB ⇒ ⇒ C 1 is the circle whose radius is least. Thus, Statement I is true but does not actually follow from Statement II which is certainly true. Hence, (b) is the correct answer.

Then,

Ex 41. Statement I The equation of chord of the circle x 2 + y 2 − 6x + 10 y − 9 = 0 which is bisected at ( −2, 4) must be x + y − 2 = 0. Statement II In notations, the equation of the chord of the circle S = 0 bisected at ( x1 , y1 ) must be T = S 1 . Sol. The Statement II is well known result but if applied to the data given in Statement I will yield 5x − 9 y + 46 = 0 ⇒ Statement I is false, Statement II is true. Hence, (d) is the correct answer.

Ex 42. Statement I If two circles x 2 + y 2 + 2g ′x + 2 f y′ = 0 and x 2 + y 2 + 2g ′x + 2 f y′ = 0 touch each other, then f g′ = fg ′. Statement II Two circles touch each other, if line joining their centres is perpendicular to all possible common tangents. Sol. The Statement II is false, since line joining centres may not be parallel to common tangents. The assertion can be proved easily be using distance between centres = sum of radii. Hence, (c) is the correct answer.

Ex 43. Statement I The equations of the straight lines joining origin to the points of intersection of x 2 + y 2 − 4x − 2 y = 4 and x 2 + y 2 − 2x − 4 y − 4 = 0 is ( y − x ) 2 = 0. Statement II y + x = 0 is a common chord of x 2 + y 2 − 4x − 2 y = 4 and x 2 + y 2 − 2x − 4 y − 4 = 0. Sol. Statement I is true because common chord itself passes 736

Ex 44. Statement I Two orthogonal circles intersect to generate a common chord which subtends complementary angles at their circumferences.

through origin. Statement II is false (common chord is x − y = 0). Hence, (c) is the correct answer.

Sol. Common chord of two orthogonal circles subtends supplementary angles at the centres and so complementary angles on the circumferences of the two circles. ∴ Both the statements are correct and Statement II is correct explanation for Statement I. Hence, (a) is the correct answer.

Ex 45. Statement I For two non-intersecting circles, direct common tangents subtends a right angle at either of point of intersection of circles with line segment joining the centres of circles. Statement II If distance between the centres is more than sum of radii, then circles are non-intersecting. Sol. Draw circle C with direct common tangent as diameter, then both the points of intersection of line joining the centres lie in the interior of the circle C. Therefore, the angles are obtuse angles. So, Statement I is false and Statement II is true. Hence, (d) is the correct answer.

Ex 46. Tangents are drawn from the point (17, 7) to the circle x 2 + y 2 = 169. Statement I The tangents are mutually perpendicular. Statement II The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is x 2 + y 2 = 338. Sol. Since, the tangents are perpendicular. ⇒ Locus of perpendicular tangents to circle x 2 + y2 = 169 is a director circle having equation x 2 + y2 = 338 Hence, (a) is the correct answer.

Ex 47. Statement I The equation x 2 + y 2 − 4x + 8 y − 5 = 0 represents a circle. Statement II The general equation of degree two ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 represents a circle, if a = b and h = 0 circle will be real, if g 2 + f 2 − c > 0. Sol. Statement I is true and Statement II is true. Also, Statement II is the correct explanation of Statement I. Hence, (a) is the correct answer.

Sol. Centre of the circle = (2, 1)

Ex 49. Statement I The chord of contact of tangent from three points A, B and C to the circle x 2 + y 2 = a 2 are concurrent, then A, B and C will be collinear. Statement II A, B and C always lie on the normal to the circle x 2 + y 2 = a 2 . Sol. Equations of chord of contact from A (x1 , y1 ) are xx1 + yy1 − a2 = 0 xx2 + yy2 − a2 = 0 xx3 + yy3 − a2 = 0

r = 25 = 5 units

and

Distance of (10, 7) from (2, 1) is 10 units. Thus, required distances are 5 and 15 units, respectively. Hence, (b) is the correct answer.

13 Circle

Ex 48. Statement I The least and greatest distances of the point P (10, 7) from the circle x 2 + y 2 − 4x − 2 y − 20 = 0 are 5 and 15 units respectively. Statement II A point ( x1 , y1 ) lies outside a circle S = x 2 + y 2 + 2gx + 2 fy + c = 0, if S 1 > 0, where S 1 = x12 + y12 + 2gx1 + 2 fy1 + c.

i.e. A , B and C are collinear. Hence, (c) is the correct answer.

Type 4. Linked Comprehension Based Questions Passage I (Ex. Nos. 50-54) Consider the two circles C1 : x 2 + y 2 = r12 and C 2 : x 2 + y 2 = r22 (r2 < r1). Let A be a fixed point on the circle C1, say A (r1, 0) and B be a variable point on the circle C 2 . The line BA meets the circle C 2 again at C.

Y

C B

X′

2

A

O

X

Ex 50. The maximum value of BC is (a) 4r12

(b) 4r22

(c) 4r22 − 4r12

(d) None of these

x 2+y 2 r 2 = 2 x2 + y2 = r12

Y′

Ex 51. The minimum value of BC 2 is (a) 4r12

(b) 4r22

(c) 4r22 − 4r12

(d) None of these

Ex 52. The set of values of OA 2 + OB 2 + BC 2 is (a) [ 5r22 − 3r12 , 5r22 + r12 ]

∴ Coordinates of any point on this line are, (r1 + r cos θ , r sin θ ) If it lies on x 2 + y2 = r22. Then, we have ⇒

(r1 + r cos θ )2 + (r sin θ )2 = r22 r2 + 2rr1 cos θ + r12 − r22 = 0

…(i)

Let AB = rB and AC = rC . Then, rB and rC are the roots of Eq. (i). ∴ rB + rC = − 2r1 cos θ and rB rC = r12 − r22

(b) [ 4r22 − 4r12 , − 4r12 ] (c) [ 4r12 , 4r22 ]

( BC )2 = (rC − rB2 ) = (rC + rB2 ) − 4 rB rC

(d) [ 5r22 − 3r12 , 5r22 + 3r12 ]

= 4 r12 cos2 θ − 4 r12 + 4 r22

Ex 53. The locus of the mid-point of AB, O being the origin is 2

2

r r2 r r2   (a)  x − 1  + y 2 = 2 (b)  x − 1  + y 2 = 2   2 2 2 4 2

2

r  r2 r  r2   (c)  x − 2  + y 2 = 1 (d)  x − 2  + y 2 = 1   2 2 2 4

which is maximum, when cos2 θ = 1. ∴

( BC )2max = 4 r22

Again, ( BC )2 is minimum when cos2 θ = 0. ∴

( BC )2min = 4 r22 − 4 r12

Now, OA 2 + OB 2 + BC 2 ⇒ r12 + r22 + 4 r12 cos2 θ − 4 r12 + 4 r22 ⇒ 5r22 − 3r12 + 4 r12 cos2 θ

Ex 54. The locus of the mid-point of AB, when BC 2 is maximum, is (a) x 2 + y 2 = r22

(b) x 2 + y 2 = ( r1 + r2 ) 2

(c) x 2 + y 2 = ( r1 − r2 ) 2 (d) None of these Sol. (Ex. Nos. 50-54) Let the equation of AB be x − r1 y − 0 = =r cosθ sin θ

Now, 0 ≤ cos2 θ ≤ 1 ⇒ 0 ≤ 4 r12 cos2 θ ≤ 4 r12 ⇒ 5 r22 − 3r12 ≤ 5r22 − 3r12 + 4 r12 + cos2 α ≤ 5r22 + r12 ⇒ OA 2 + OB 2 + BC 2 ∈ [ 5r22 − 3r12 , 5r22 + r12 ] Let (h, k ) be the mid-point of AB and let (α , β ) be the coordinates of B. Then, α + r1 β = h and =k 2 2 ⇒ α = 2h − r1 and β = 2k where, (α , β ) lies on x 2 + y2 = r22

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Objective Mathematics Vol. 1

13

∴

α 2 + β 2 = r22

⇒

(2h − r1 ) + (2k )2 = r22

Ex 56. The equation of the circle which passes through the origin and belongs to the coaxial system of which the limiting points are (1, 2) and (4, 3), is given by

2

∴ Locus of mid-point of AB is 2 r r2  2  x − 1 + y = 2  2 4 Again, the locus of mid-point of AB when BC is maximum is a fixed point M on X-axis shown as Y

(c) 3( x 2 + y 2 ) − x − 7 y = 0 (d) None of the above Sol. The coaxial circle is (x − 1)2 + ( y − 2)2+ λ {(x − 4 )2 + ( y − 3)2} = 0

X′

C

B

X

O

BC is maximum Locus of mid-point Y′

50. (b) 51. (c)

52. (a) 53. (b) 54. (d)

coaxial system of circles are the members of the system which are of zero radius. Let (a, b) and (α, β) be two limiting points of a coaxial system of circles. Then, the corresponding point of circles are S1 ≡ ( x − a) 2 + ( y − b) 2 = 0 and S 2 ≡ ( x − α) 2 + ( y − β ) 2 = 0 So, the coaxial system of circles is given by S 1 + λS 2 = 0 , λ ≠ − 1 or {( x − a) 2 + ( y − b) 2 } + λ {( x − α) 2 + ( y − β) 2 } = 0 λ ≠ −1 The value of λ is determined from the given condition.

Ex 55. The coordinates of the limiting points of a system of coaxial circles determined by the circles S 1 ≡ x 2 + y 2 + 4x + 2 y + 5 = 0 and S 2 ≡ x 2 + y 2 + 2x + 4 y + 7 = 0 are (a) (2, 1) and (0, 3) (b) ( −2, − 1) and ( 0, − 3) (c) ( −2, 1) and ( 0, − 3) (d) None of the above

R = (2 + λ ) + (1 − λ ) − 5 + 2λ 2

Now, R=0 ⇒ (2 + λ )2 + (1 − λ )2 − 5 + 2λ = 0 ⇒

(a) x − 3 y + 4 = 0 (c) 5x − 3 y + 4 = 0

(b) x + 3 y − 4 = 0 (d) None of these

Sol. Here, S 1 ≡ (x − 2)2 + ( y + 1)2 = 00 S 2 ≡ (x + 3)2 + ( y − 2)2 = 0 ∴ ⇒ or

Radical axis is S 2 − S 1 = 0. 10x − 6 y + 8 = 0 5x − 3 y + 4 = 0 Hence, (c) is the correct answer.

Passage III (Ex. Nos. 58-60) Let y = f ( x ) and

y = g( x ) be the pair of curves such that the tangents at point with equal abscissae intersect on Y-axis, the normals drawn at points with equal abscissae intersect on X-axis. One curve passes through (1, 1) and the other passes through (2, 3).

Ex 58. The curve f (x ) is given by 2 −x x 2 (c) 2 − x x

(a)

(a) x −

Coordinates of centre are (− (2 + λ ), 1 − λ )

and

Ex 57. The equation of the radical axis of a coaxial system of circles whose limiting points are (2, − 1) and ( −3, 2) is given by

(b) 2x 2 −

1 x

(d) None of these

Ex 59. The curve g (x ) is given by

Sol. Using, S 1 + λ (S 1 − S 2 ) = 0 2

...(i)

which will pass through origin. ⇒ 1 + 4 + λ (16 + 9) = 0 1 ⇒ λ=− 5 1 Putting λ = − in Eq. (i), we get 5 2(x 2 + y2 ) − x − 7 y = 0 Hence, (a) is the correct answer.

Passage II (Ex. Nos. 55-57) Limiting points of a

2λ2 + 4 λ = 0

⇒ λ = 0, − 2 ∴ Points are (−2, − 1) and (0, − 3).

738

(a) 2 ( x 2 + y 2 ) − x − 7 y = 0 (b) x 2 + y 2 − x − 7 y = 0

Hence, (b) is the correct answer.

1 x

(c) x 2 −

(b) x + 1 x2

2 x

(d) None of these

Ex 60. The number of positive integral solutions for f ( x ) = g ( x ) is (a) 4 (b) 5 (c) 6 (d) None of the above

required curves. The equations of the tangents to these two curves at points with equal abscissae x are Y − f (x ) = f ′ (x ) ( X − x ) and Y − g (x ) = g′ (x ) ( X − x ) These two lines intersect on Y-axis. ∴ Y − f (x ) = − xf ′ (x ) and Y − g (x ) = − xg′ (x ) ⇒ Y = f (x ) − xf ′ (x ) = g (x ) − xg′ (x ) ⇒ f (x ) − g (x ) = x{ f ′ (x ) − g′ (x )} d ⇒ f (x ) − g (x ) = x { f (x ) − g (x )} dx d{ f (x ) − g (x )} dx = ⇒ f (x ) − g (x ) x On integrating both sides we get log{ f (x ) − g (x )} = log x + log c ...(i) f (x ) − g (x ) = cx The equation of the normals at points with equal abscissae x to the two curves are 1 Y − f (x ) = − ( X − x) f ′ (x ) 1 and Y − g (x ) = − ( X − x) g′ (x ) These intersect on X-axis. ∴ and ⇒ and ⇒ ⇒

1 ( X − x) f ′ (x ) 1 0 − g (x ) = − ( X − x) g′ (x )

...(ii)

59. (b)

...(i)

Put (0, 0) in Eq. (i) to get λ. ∴

λ=−

1 2

The equation of circle is x 2 + y2 + 4 y = 0. Hence, (d) is the correct answer.

Ex 62. If origin is a limiting point of a coaxial system one of whose member is x 2 + y 2 − 2αx − 2βy + c = 0, then the other limiting point is  cα cβ   (a)  2 ,− 2 2 α + β2  α + β

x 2 + y2 − 2αx − 2βy + c + λ (x 2 + y2 ) = 0 2α 2β c x 2 + y2 − x− y+ =0 λ+1 λ+1 λ+1 Radius of this circle = 0

...(iii)

⇒

...(iv)

60. (d)

(b) x 2 + y 2 + 4x = 0 (d) x 2 + y 2 + 4 y = 0

Sol. (x − 1)2 + ( y + 1)2 + λ [(x − 2)2 + y2 ] = 0

2

Since, Eq. (iii) passes through (1, 1) and Eq. (iv) passes through (2, 3). C C 2 = C + 1 and 6 = 1 − 2C ∴ 2C C ⇒ C = − 2 and C 1 = − 8 2 2 and g (x ) = + x ∴ f (x ) = − x + x x For number of positive integral solutions, f (x ) = g (x ) 2 2 x+ = +x ⇒ x x ∴ x = 0, no solution.

58. (a)

(a) x 2 + y 2 − 4x = 0 (c) x 2 + y 2 − 4 y = 0

Sol. Family of circle,

⇒

and

Ex 61. The equation of the circle which belong to the coaxial system of circles for which the limiting points are (1, − 1), (2, 0) and which passes through the origin, is

 cβ cα   (d)  − 2 ,− 2 2 α + β2   α +β

{ f (x )}2 − {g (x )}2 = C 1

1 C  g (x ) =  1 + Cx  2  Cx

13

 cβ cα   (c)  2 ,− 2 2 α + β2  α + β

X = x + f (x ) f ′ (x ) X = x + g (x )g′ (x ) x + f (x ) f ′ (x ) = x + g (x )g′ (x ) f (x ) f ′ (x ) = g (x )g′ (x )

{ f (x ) + g (x )} Cx = C 1 C ⇒ f (x ) + g (x ) = 1 Cx On solving Eqs. (i) and (ii), we get 1 C  f (x ) = Cx + 1  2 x

said to be coaxial when every pair of the circles has the same radical axis.

 cα cβ   (b)  2 , 2 2 α + β α + β 2 

0 − f (x ) =

On integrating, we get { f (x )}2 = {g (x )}2 + C 1 ⇒

Passage IV (Ex. Nos. 61-63) A system of circles is

Circle

Sol. (Ex. Nos. 58-60) Let y = f (x ) and y = g (x ) be the

Put

2

 α   β  c =0   +  −  λ + 1  λ + 1 λ+1 α 2 + β 2 − c(λ + 1) = 0 α 2 + β2 λ + 1= c  α cβ  β   cα , ,    =  λ + 1 λ + 1  α 2 + β 2 α 2 + β 2 

Hence, (b) is the correct answer.

Ex 63. The equation of the radical axis of the system of coaxial circles x 2 + y 2 − 2ax + 2by + c + 2λ( ax − by + 1) = 0 is (a) ax − by + 1 = 0 (c) 2 ( ax + by ) + 1 = 0

(b) bx + ay − 1 = 0 (d) 2 ( bx − ay ) + 1 = 0

Sol. S + λL = 0 , L = 0 represents the radical axis ∴ ax − by + 1 = 0 Hence, (a) is the correct answer.

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Objective Mathematics Vol. 1

13 Type 5. Match the Columns Ex 64. Match the statement of Column I with values of Column II. Column I

Column II

A.

P, Q lie on the X-axis and R, S lie on the Y-axis. Maximum number of parabolas that can be drawn through these points is

p.

3

B.

From a point on the radical axis of circles S1 ≡ x 2 + y 2 + 4 x − 6 y − 12 = 0, S 2 ≡ x 2 + y 2 − 6 y − 16 = 0, tangents are drawn to S 2. Chord of contact passes through( − 25, α 2 ), then[α ] can be

q.

−3

If(α, β ) is the foot of normal drawn from the point (6, 2) on the parabola y 2 = 4 x, then roots of equation x 2 = β (α − 4) is/are

r.

A variable circle whose centre lies on cuts rectangular y 2 − 26 = 0 hyperbola at xy = 16  4 , then 4 t , , i 1 , 2 , 3 , 4 =  i  ti  1 1 1 1 can be + + + t1 t 2 t 3 t 4

s.

C.

D.

For r = 2, the circle and the branch of the hyperbola intersect at two points. For r = 1, there is no point of intersection. If m is the slope of the common tangent, then r2 + 3 m2 − 3 = r2 (1 + m2 ) ⇒ m2 = 1 − r2 Hence, no common tangents for r > 1and two common tangents for r < 1. A → p; B → r; C → r; D → p

Ex 66. Match the statements of Column I with values of Column II. Column I 2

−2

A. The radical axis of two circles

p. is the square of the distance between their centres

B. The common tangent to two intersecting circles of equal radii

q. is perpendicular to the line joining the centres

C. The common chord of two intersecting circles

r. is parallel to the line joining the centre

D. The sum of the squares of the radii of two circles intersecting orthogonally

s. is bisected by the line joining the centres

Sol. Let the equations of two circles be x 2 + y2 + 2g1x + 2 f1 y + c1 = 0

Sol. A. (ax + by + c) (a′ x + b′ y + c′ ) + λxy = 0

represents parabola for maximum two values of λ.

B. Equation of radical axis is x = − 1. Let P ≡ (−1, β ) Equation of chord of contact is (x + 3 y + 16) − β ( y − 3) = 0 ⇒ α2 = 3

and

So that radical axis is perpendicular to the line joining the centre.

41 1 1 1 0 ± 6 D.  + + +  = =±3 4  t1 t2 t3 t4  2

B.

A → r; B → s; C → s, r; D → p, q

Ex 65. A (−2, 0) and B (2, 0) are the two fixed points and P is a point such that PA − PB = 2. Let S be the circle x 2 + y 2 = r 2 , then match the following : Column I

Column II

A.

If r = 2, then the number of points P satisfying p. PA − PB = 2 and lying on x 2 + y 2 = r 2 is

2

B.

If r = 1, then the number of points satisfying q. PA − PB = 2 and lying on x 2 + y 2 = r 2 is

4

C. D.

If r = 2, then number of common tangents is 1 If r = , then number of common tangents is 2

r.

0

s.

1

740

Common tangent to the intersecting circles of equal radii will be at the same distance from the centres of the two circles and hence will be parallel to the line joining the centres.

C.

Since, the line joining the centres to the middle point of the common chord is perpendicular to the chord, the line joining the centres bisects the chord. D. Since, ∠C 1PC 2 = 90° [see in the figure] T1

T2 P

Sol. Locus of point P satisfying PA − PB = 2 is a branch of y2 the hyperbola x − = 1. 3

x 2 + y2 + 2g2x + 2 f2 y + c2 = 0

A. Equation of the radical axis is 2 (g1 − g2 ) x + 2 ( f1 + f2 ) y + c1 − c2 = 0 g − g2 Slope of the radical axis = − 1 f1 − f2 and slope of the line joining the centres f − f2 . (− g1 , − f1 ) and (− g2 , − f2 ) is 1 g1 − g2

⇒ [α ] = 1 or −2 C. Equation of normal at (α , β ) is 2 ( y − β ) + β (x − α ) = 0 ⇒β (α − 4 ) = 4

2

Column II

C1

C2

(C 1C 2 )2 = (C 1P )2 + (C 2P ) A → q; B → r; C → s; D → p

13

Type 6. Single Integer Answer Type Questions

Sol. (1) Let (h, k ) be the point on the curve y = tan −1 x. Image of (h, k ) in y = x is (k , h) which is the centre of a π circle of radius . 2 2

π  h = tan  h −  = − cot h  2 ⇒ − h = cot h which gives only one solution for h ∈ (0, π ). So, there is only one point on the curve.

Circle

Ex 67. Find the number of points on y = tan −1 x, ∀ x (0, π ) whose image in y = x is π unit and the centre of the circle with radius 2 2 π which is at a minimum distance of unit 2 2 from the circle.

Ex 68. P (a, b) is a point in the first quadrant, circles are drawn through P touching the coordinate axes, such that the length of common chord of these circles is maximum, if possible values of . a / b is k1 and k 2 , then k1 + k 2 is equal to Sol. (6) Let r be the radius of the circle. Its equation is x 2 + y2 − 2r (x + y) + r2 = 0

C (k, h)

Since, it passes through P (a, b). a2 + b2 − 2r (a + b) + r2 = 0

y=x

M

Solving y = tan–1x P (h, k)

Given, and ⇒

π 2 2 π CM = 2 2 π  π  CP =  2=  2 2 2 PM =

[shortest distance] [radius of circle]

π Now, CP = (h − k ) + (k − h) = 2 π 2 |h − k | = ⇒ 2 π |h − k | = ⇒ 2 2

2

y = cot x

Y

y = –x π/2 π

O

⇒ ⇒

h−k =±

π 2

k =h±

π 2

X

Since, (h, k ) lies on y = tan −1 x π k =h− ⇒ 2 π −1 Now, h − = tan h 2 Since, 0

13 13 or λ < − 8 8

(d) None of these

53. The range of values of m for which the line y = mx + 2 cuts the circle x 2 + y 2 = 1 at distinct or coincident points, is (a) ( −∞ , − 3 ] ∪ [ 3 , ∞ )

(b) [ − 3 , 3 ]

(c) [ 3 , ∞ ]

(d) None of these

54. A foot of the normal from the point (4, 3) to a circle is (2, 1) and a diameter of circle has the equation 2x − y = 2. Then, the equation of the circle is (a) x 2 + y2 + 2x − 1 = 0 (c) x 2 + y2 − 2 y − 1 = 0

(b) x 2 + y2 − 2x − 1 = 0 (d) None of these

55. Lines are drawn through the point P( −2, − 3) to meet the circle x 2 + y 2 − 2x − 10 y + 1 = 0. The length of the line segment PA, A being the point on the circle, where the line meets the circle at coincident points, is (a) 16 units (c) 48 units

(b) 4 3 units (d) None of these

56. A ray of light incident at the point ( −2, − 1) gets reflected from the tangent at ( 0, − 1) to the circle x 2 + y 2 = 1. The reflected ray touches the circle. The equation of the line along which the incident ray moved, is 746

(a) 4 x − 3 y + 11 = 0 (c) 3x + 4 y + 11 = 0

58. If the circumcircle of the triangle formed by the line ax + by + c = 0, bx + cy + a = 0, cx + ay + b = 0passes through the origin, then

b2

52. The range of λ for which the circles x 2 + y 2 = 4 and x 2 + y 2 − 4λx + 9 = 0 have two common tangents, is  13 13 (a) λ ∈  − ,   8 8 13 (c) 1 < λ < 8

(a) a pair of straight lines (b) an ellipse (c) a circle of radius a2 − r2 (d) an ellipse of major axis a and minor axis r

b2 c2 a2 b2 c2 a2 (a) ab bc ca = bc ca ab ac ab bc ac ab bc

51. There are two circles whose equations are x 2 + y 2 = 9 and x 2 + y 2 − 8x − 6 y + n 2 = 0, n ∈ I . If the two circles have exactly two common tangents, then the number of possible values of n is (a) 2 (c) 9

57. The point P moves in the plane of a regular hexagon such that the sum of the squares of its distances from the vertices of the hexagon is 6a 2 . If the radius of the circumcircle of the hexagon is r (< a ), then the locus of P is

(b) 4 x + 3 y + 11 = 0 (d) None of these

c2

a2

b2

c2

a2

(b) ab bc ca = − bc ca ab ac ab bc ac ab bc b2

c2

a2

b2

c2

a2

(c) − ab bc ca = bc ca ab ac bc ab ac ab bc (d) None of the above

59. The equation of the sides of a quadrilateral are given by Lr = a r x + br y + c r = 0; r = 1, 2, 3, 4. If the quadrilateral is cyclic, then a1a3 − b1b3 a b + a3b1 =− 1 3 a2a4 − b2b4 a2b4 + a4b2 a1a3 + b1b3 a1b3 − a3b1 (b) = a2a4 + b2b4 a2b4 − a4b2 a a − b1b3 a1b3 + a3b1 (c) 1 3 = a2a4 − b2b4 a2b4 + a4b2 (d) None of the above

(a)

60. A point moves in such a manner that the sum of the squares of its distances from the vertices of a triangle is constant. Then, the locus of the point is (a) a hyperbola (c) an ellipse

(b) a parabola (d) a circle

61. A line L1 intersects X and Y-axes at P and Q respectively and a line L2 , perpendicular to L1 cuts X and Y-axes at R and S , respectively. Then, the locus of point of intersection of the lines PS and QR is (a) 2x 2 + 2 y2 − ax − by = 0

(b) x 2 + y2 − ax − by = 0

(c) x − y − ax − by = 0

(d) None of these

2

2

62. A circle of constant radius r passes through the origin O and cuts the axes at A and B. Then, the locus of the perpendicular from O to AB is (a) (x 2 + y2 ) = 4 r2x 2 y2

(b) (x 2 − y2 )3 = 4 r2x 2 y2

(c) (x 2 + y2 )3 = r2x 2 y2

(d) (x 2 + y2 )3 = 4 r2x 2 y2

63. O is a fixed point and a point R moves along a fixed line L not passing through O. If S is a point on OR such that OR ⋅ OS = λ 2 , then the locus of S is (a) an ellipse (b) a parabola (c) a circle

(d) a polygon

value of k is equal to (a) 2

(b) 1

(c) 3

(d) − 2

65. Consider the two circles C1 : x 2 + y 2 = r12 and C 2 : x 2 + y 2 = r22 ( r2 < r1 ). Let A be a fixed point on the circle C1 , say A ( r1 , 0) and B be a variable point on the circle C 2 . Then, line BA meets the circle C 2 again at C. Then, the set of values of OB 2 + OA 2 + BC 2 is (a) [ 5r22 − 3r12 , 5r22 + r12 ] (c) [ 5r22 − 3r12 , r22 + 5r12 ]

(b) [ 3r22 − 5r12 , 5r22 + r12 ] (d) None of these

66. Two rods of length a and b slide along the coordinate axes, which are rectangular, in such a way that their ends are always concyclic, then the locus of the centre of the circle passing through these ends, is the curve (a) 4 (x 2 + y2 ) = a2 + b2 (c) 4 (x 2 − y2 ) = a2 − b2

(b) (x 2 − y2 ) = a2 − b2 (d) None of these

67. Through the point of intersection P, which has integral coordinates, of the circles x 2 + y 2 = 1 and x 2 + y 2 + 2x + 4 y + 1= 0, a common chord APB is drawn terminating on the two circles such that chords AP and BP of the given circles subtend equal angles at the centre. Then, the equation of this chord is (a) y = x + 1 (c) x 2 + y2 = − 1

(b) y = x − 1 1 1 (d) + = 0 x y

68. The circle x 2 + y 2 − 6x − 10 y + λ = 0 does not touch or intersect the coordinate axes and the point (1, 4) is inside the circle. Then, the set of values of λ is (a) (20, 25) (c) (5, 9)

(b) (25, 30) (d) (25, 29)

69. If 4l 2 − 5m2 + 6l + 1 = 0 and the line lx + my + 1 = 0 touches a fixed circle. Then, the equation of circle is (a) (x + 3)2 + ( y − 0)2 = 5 (c) (x − 3)2 + ( y − 0)2 = 0

(b) (x − 3)2 + ( y − 0)2 = 5 (d) None of these

70. If al 2 − bm2 + 2dl + 1 = 0, where a, b, d are fixed real numbers such that a + b = d 2 and lx + my + 1 = 0 touches a fixed circle. Then, the equation of fixed circle is (a) x 2 + y2 − dx + a = 0 (c) x 2 + y2 − 2dx + a = 5

(b) 2x 2 + y2 − 2dx + 2a = 0 (d) x 2 + y2 − 2dx + a = 0

71. The range of parameter a for which the variable line y = 2x + a lies between the circle x 2 + y 2 − 2x − 2 y + 1 = 0 and x 2 + y 2 − 16x − 2 y + 61 = 0 without intersecting or touching either circle, is (a) (−15 + 2 5 , − 5 − 1) (c) (−15 − 2 5 , − 5 + 1)

(b) (15 + 2 5 , 5 − 1) (d) (−15 + 2 5 , 5 − 1)

 a2 − b2  2 2 (a) x 2 + y2 − 2(x + y)  + (a + b ) = 0  a+ b  (b) x 2 + y2 − 2(x + y) + (a2 + b2 ) = 0  a2 + b2  2 2 (c) x 2 + y2 − 2(x + y)   + (a + b ) = 0  a+ b  (d) x 2 + y2 − 2(x + y) + (a2 − b2 ) = 0

13 Circle

72. If P ( a, b ) and Q ( b, a ) are two points such that a ≠ b, then the equation of the circle touching OP and OQ and Q, where Q is the origin, is

73. A circle touches the line y = x at a point P such that OP = 4 2 units where O is the origin. The circle contains the point (−10, 2) in its interior and the length of its chord on the line x + y = 0 is 6 2 units. Then, the equation of the circle is 1 + x2 1 (b) 2 + x (c) x 2 + (d) x 2 + (a)

1 + 18x − 2 y + 32 = 0 y2 1 + 18x − 16 y + 32 = 0 y2 y2 + 18x − 16 y + 32 = 0 y2 + 18x − 2 y + 32 = 0

74. A tangent drawn from the point (4, 0) to the circle x 2 + y 2 = 8 touches it at a point A in the first quadrant. Then, the coordinates of another point B on the circle such that AB = 4, is (a) (2, − 2) and (2, 2) (c) (− 2, − 2) and (2, − 2)

(b) (2, − 2) and (− 2, 2) (d) (− 2, 2) and (− 2, − 2)

75. The locus of the point of intersection of tangents to the circle x 2 + y 2 = a 2 , which are inclined at an angle α with each other, is (a) (x 2 + (b) (x 2 + (c) (x 2 + (d) (x 2 +

y2 − 2a2 )tan 2 α = 4 a2 (x 2 + y2 − a2 ) y2 − a2 )tan 2 α = 4 a2 (x 2 + y2 − 2a2 ) y2 − 2a2 )2 tan 2 α = 4 a2 (x 2 + y2 − 2a2 ) y2 − 2a2 )2 tan 2 α = 4 a2 (x 2 + y2 − a2 )

Targ e t E x e rc is e s

64. The circle x 2 + y 2 − 4x − 4 y + 4 = 0 is inscribed in a triangle which has two of its sides along the coordinate axes. If the locus of the circumcentre of the triangle is x + y − xy + k x 2 + y 2 = 0, then the

76. If four points P, Q, R and S in the plane are taken and the square of the length of the tangent from P to the circle on QR as diameter is denoted by {P , QR}, then (a) {P , RS } − {P , QS } + {Q , PR} − {Q , RS } = 0 (b) {P , RS } + {P , QS } + {Q , PR} + {Q , RS } = 0 (c) {P , RS } − {P , QS } + {Q , PR} − {Q , RS } = 1 (d) {P , RS } − {Q , RS } − {Q , PR} − {P , QS } = 0

77. If the chord of contact of tangents from the point (α, β) to the circle x 2 + y 2 = r12 is a tangent to the circle ( x − a ) 2 + ( y − b ) 2 = r22 , then (a) r22 (α 2 − β 2 ) = (r12 − aα − bβ )2 (b) r12 (α 2 − β 2 ) = (r22 − aα − bβ )2 (c) r22 (α 2 + β 2 ) = (r12 − aα − bβ )2 (d) r12 (α 2 + β 2 ) = (r22 − aα − bβ )2

78. The locus of the middle points of the chords of the circle x 2 + y 2 = a 2 which subtend a right angle at the centre, is (a) (x 2 + y2 ) − a2 = 0 (c) 2(x 2 − y2 ) + a2 = 0

(b) 2(x 2 + y2 ) − 2a2 = 0 (d) 2(x 2 + y2 ) − a2 = 0

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79. The locus of the middle of chords of the circle x 2 + y 2 = a 2 which subtend a right angle at (c, 0), is (a) 2(x 2 + y2 ) − 2cx + c2 − a2 = 0 (b) (x 2 + y2 ) − 2cx + c2 − a2 = 0 (c) 2(x 2 + y2 ) − cx + c2 − a2 = 0 (d) 2(x 2 + y2 ) − 2cx + 2c2 − 2a2 = 0

80. The locus of the mid-points of the chords of the circle x 2 + y 2 = 4 such that the segment intercepted by the chord on the curve x 2 = 2( x + y ) subtends a right angle at the origin, is (a) x 2 + y2 + 2 x + 2 y = 0 (b) x 2 + y2 − 2 x − 2 y = 0 (c) x 2 + y2 − x − y = 0 (d) None of the above

81. Through a fixed point ( x1 , y1 ), secants are drawn to the circle x 2 + y 2 = a 2 . Then, the locus of mid-points of the secants intercepted by the given circle is

Ta rg e t E x e rc is e s

(a) x 2 + (b) x 2 + (c) x 2 + (d) x 2 −

y2 = xx1 + yy1 y2 = xx1 − yy1 y2 = yy1 − xx1 y2 = yy1 + xx1

82. The intervals of values of a for which the line y + x = 0 bisects two chords drawn from a point  1 + 2a 1 − 2a  to the circle ,   2   2 2x 2 + 2 y 2 − (1 + 2a )x − (1− 2a ) y = 0, is (a) (2, ∞) (b) (− ∞ , − 2) ∪ (2, ∞ ) (c) (− ∞ , 2) (d) None of the above

83. The polar of a given point with respect to any one of the circles x 2 + y 2 − 2kx + c 2 = 0, (k being a variable) always passes through a fixed point whatever be the value of k. Then, the point of intersection is  x 2 − c2  (a)  − x1 , 1  y1    x 2 + c2  (c)  − x1 , 1  y1  

 x 2 − c2  (b)  x1 , 1  y1   (d) None of these

84. Let C1 and C 2 be two circles with C 2 lying inside C1 . If a circle C lying inside C1 touches C1 internally and C 2 externally. Then, the locus of the centre of C is (a) a circle (c) a hyperbola

(b) a parabola (d) an ellipse

85. Two circles, each of radius 5 units, touch each other at (1, 2). If the equation of their common tangent is 4x + 3 y = 10, then the equations of the circles are

748

(a) (x + 5)2 − ( y + 5)2 = 52, (x + 3)2 − ( y + 1)2 = 52 (b) (x + 5)2 + ( y + 5)2 = 52, (x + 3)2 − ( y + 1)2 = 52 (c) (x − 5)2 + ( y − 5)2 = 52, (x + 3)2 + ( y + 1)2 = 52 (d) None of the above

86. Two circles have centres (a, 0), (− a, 0) and radii b, c respectively such that a > b > c. Then, the points of contact of the common tangents to the two circles lie on (a) x 2 + y2 = a2 ± bc (c) 2x 2 + y2 = a2 ± bc

(b) x 2 − y2 = a2 ± bc (d) None of these

87. The equation of the circle which touches the straight lines x + y = 2, x − y = 2 and also touches the circle x 2 + y 2 = 1, is (a) (x − 2 )2 + y2 = ( 2 + 1)2 (b) (x − 2 )2 + y2 = ( 2 − 1)2 (c) (x − 2 )2 − y2 = ( 2 − 1)2 (d) (x + 2 )2 + y2 = ( 2 − 1)2

88. The equation to the circle cutting orthogonally the three circles x 2 + y 2 − 2x + 3 y − 7 = 0, x 2 + y 2 + 5x − 5 y + 9 = 0 and x 2 + y 2 + 7x − 9 y + 29 = 0, is (a) x 2 + (b) x 2 + (c) x 2 + (d) x 2 +

y2 − 18x − 16 y − 4 = 0 y2 − 16x − 18 y − 16 = 0 y2 + 16x + 18 y − 4 = 0 y2 − 16x − 18 y − 4 = 0

89. The equation of a circle which is coaxial with the circles and x 2 + y 2 + 4x + 2 y + 1 = 0 3 2 2 x + y − x + 3 y − = 0 and having its centre on the 2 radical axis of these circles, is (a) 4 x 2 + 4 y2 + 6x + 10 y − 1 = 0 (b) 4 x 2 + y2 + 6x + 10 y − 1 = 0 (c) x 2 + y2 + 6x + 10 y − 1 = 0 (d) 4 x 2 + 4 y2 − 6x − 10 y − 1 = 0

90. As λ varies, the circles x 2 + y 2 + 2ax + 2by +2λ ( ax − by ) = 0 form a coaxial system. Then, the equation of the circle which are orthogonal of the circles of the above system, is x y (a) x2 + y2 + c  +  + c = 0  a b c x y (b) x2 + y2 +  −  + c = 0 2  a b x y (c) x2 + y2 + c  −  + c = 0  a b c x y (d) x2 + y2 +  +  + c = 0 2  a b

91. The polar of any point with respect to a system of coaxial circles pass through (a) a fixed point (b) a variable point (c) radical axis (d) None of the above

92. If P, Q and R are the centres of three circles from a coaxial system of vectors and l1 , l2 , l3 are the lengths of the tangents to them from a fixed point. Then, l12 ⋅ QR + l22 ⋅ RP + l32 ⋅ PQ is equal to (a) 1 (c) 0

(b) 2 (d) None of these

(b) x + y − 8 = 0 (d) None of these

(b) x 2 + y2 = 16 (d) x 2 + y2 = 25

95. If the circle x 2 + y 2 + 2a1 x + c = 0 lies completely inside the circle x 2 + y 2 + 2a 2 x + c = 0, then (a) a1a2 > 0, c < 0 (c) a1a2 < 0, c < 0

(b) a1a2 > 0, c > 0 (d) a1a2 < 0, c > 0

(b) | c | = 2 5 (c) | c | = 10 (d) | c | = 2 10

97. A circle touches the lines y =

x

, y = x 3 and has 3 unit radius. If the centre of this circle lies in the first quadrant, then possible equation of this circle is (a) x 2 + (b) x 2 + (c) x 2 + (d) x 2 +

y2 − 2x ( 3 + 1) − 2 y ( 3 + 1) + y2 − 2x (1 + 3 ) − 2 y (1 + 3 ) + y2 − 2x (1 + 3 ) − 2 y (1 + 3 ) + y2 − 2x (1 + 3 ) − 2 y (1 + 3 ) +

8+ 5+ 7+ 6+

4 4 4 4

3=0 3=0 3=0 3=0

98. The equation of the circle which passes through the point (2a, 0) and whose radical axis with respect to a the circle x 2 + y 2 = a 2 is the line x = , is 2 (a) x 2 − y2 + 2ax = 0 (c) x 2 + y2 + 2ax = 0

(b) x 2 + y2 − 2ax = 0 (d) None of these

99. The equation of the system of coaxial circles that are tangent at ( 2, 4) to the locus of the point of intersection of two mutually perpendicular tangents to the circle x 2 + y 2 = 9, is (a) x 2 + (b) x 2 + (c) x 2 + (d) x 2 +

(d) x 2 + y2 − x − y + 9 = 0

102. The equation of the radical axis of the circle 2x 2 + 2 y 2 + 14x − 18 y + 15 = 0 and 4x 2 + 4 y 2 − 3x − y + 5 = 0, is (a) 31x + 35 y − 25 = 0 (c) 35x + 31 y − 25 = 0

(b) 31x − 35 y + 25 = 0 (d) 35x − 31 y + 25 = 0

103. The equation of the circle whose diameter is the common chord of the circles ( x − a ) 2 + y 2 = a 2 and x 2 + ( y − b ) 2 = b 2 , is (a) (x 2 + y2 )(a2 + b2 ) = ab(bx + ay) (b) (x 2 + y2 )(a2 + b2 ) = 2ab(bx + ay)

96. If the angle between tangents drawn to x 2 + y 2 − 2x − 4 y + 1 = 0 at the points, where it is π cut by the line y = 2x + c, is . Then, 2 (a) | c | = 5

(b) x 2 + 2 y2 − 6x − 6 y + 9 = 0 (c) x 2 + y2 − 6x − 6 y + 9 = 0

94. Tangents are drawn to x 2 + y 2 = 1from any arbitrary point P on the circle C1 : x 2 + y 2 − 4 = 0. These tangent meets the circle C1 again in A and B. Locus of point of intersection of tangents drawn to C1 at A and B is (a) x 2 + y2 = 10 (c) x 2 + y2 = 5

(a) 2x 2 + 2 y2 − 6x − 6 y + 9 = 0

13

y2 + 9 − λ ( 2x + 4 y − 18) = 0 y2 + 9 + λ ( 2x + 4 y − 18) = 0 y2 + λ ( 2x + 4 y − 18) = 0 y2 − 9 + λ ( 2x + 4 y − 18) = 0

100. The equation of the circle passing through the origin and cutting the circles x 2 + y 2 − 4x + 6 y + 10 = 0 and x 2 + y 2 + 12 y + 6 = 0 at right angles, is (a) 2(x 2 + y2 ) − 7x + 2 y = 0 (b) (x 2 + y2 ) − 7x + 2 y = 0 (c) 2(x 2 + y2 ) + 7x − 2 y = 0 (d) (x 2 + y2 ) + 7x − 2 y = 0

(c) (2x 2 + 2 y2 )(a2 + b2 ) = ab(bx + ay) (d) x 2 + y2 + a2 + b2 − bx − ay = 0

104. The equation of the circle which touches the line x − y = 0 at the origin and bisects the circumference of the circle x 2 + y 2 + 2 y − 3 = 0, is (a) x 2 + y2 − 5x − 5 y = 0

(b) 2x 2 + 2 y2 − 5x + 5 y = 0

(c) 2x 2 + y2 − 5x + 5 y = 0

(d) x 2 + y2 − 5x + 5 y = 0

105. The equation of the circle which passes through the origin and belong to the coaxial system of which the limiting points are (1, 2) and (4, 3), is (a) x 2 + y2 − x − 7 y = 0 (b) 3(x 2 + y2 ) − x − 7 y = 0 2 2 (c) 2(x + y ) − x − 7 y = 0 (d) None of these

106. The equation of the circle through the intersection of the circles x 2 + y 2 − 8x − 2 y + 7 = 0 and x 2 + y 2 − 4x + 10 y + 8 = 0 and that passes through the point (− 1, − 2), is

Targ e t E x e rc is e s

(a) 3 y + 4 x − 25 = 0 (c) 3x + 4 y − 31 = 0

101. The equation of the circle which cuts orthogonally the circle x 2 + y 2 − 6x + 4 y − 3 = 0, passes through (3, 0) and touches Y-axis, is

Circle

93. One possible equation of the chord of x 2 + y 2 = 100 2π that passes through (1, 7) and subtends an angle at 3 the origin, is

(a) x 2 + 9 y2 − 40x + 78 y + 71 = 0 (b) 9x 2 + y2 − 40x + 78 y + 71 = 0 (c) x 2 + y2 − 40x + 78 y + 71 = 0 (d) 9x 2 + 9 y2 − 40x + 78 y + 71 = 0

107. The lengths of the tangents from any point of a fixed circle of coaxial system to another fixed circles of the system are in the ratio (a)

g2 − g1 (b) g3 − g1

g2 + g1 (c) g3 + g1

g3 − g1 (d) g2 + g1

g3 − g2 g2 − g1

108. The radius of the smallest circle which touches the straight line 3x − y = 6 at (1, − 3) and also touches the line y = x, is (a) 1

(b) 1.6

(c) 1.49

(d)

1 2

109. If the circle x 2 + y 2 + 2gx + 2 fy + c = 0 is touched by the line y = x at point P such that OP = 6 2 units, where O is the origin, then the value of c is (a) 74

(b) 62

(c) 64

(d) 72

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Type 2. More than One Correct Option 110. The equation(s) of circle passing through (3, − 6) and touching both the axes, is/are (a) x 2 + (b) x 2 + (c) x 2 + (d) x 2 +

y2 − 6x + 6 y + 9 = 0 y2 + 6x − 6 y + 9 = 0 y2 + 30x − 30 y + 225 = 0 y2 − 30x + 30 y + 225 = 0

111. The equation of the tangents drawn from the origin to the circle x 2 + y 2 − 2rx − 2hy + h 2 = 0, is/are (a) x = 0 (c) (h2 − r2 )x − 2rhy = 0

(b) y = 0 (d) (h2 − r2 )x + 2rhy = 0

112. Equation of a circle with centre (4, 3) touching the circle x 2 + y 2 = 1 is (a) x 2 + (b) x 2 + (c) x 2 + (d) x 2 +

y2 − 8x − 6 y − 9 = 0 y2 − 8x − 6 y + 11 = 0 y2 − 8x − 6 y − 11 = 0 y2 − 8x − 6 y + 9 = 0

Ta rg e t E x e rc is e s

113. The equation(s) of a tangent to the circle x 2 + y 2 = 25 passing through (−2, 11), is/are (a) 4 x + 3 y = 25 (c) 24 x − 7 y + 125 = 0

(b) 3x + 4 y = 38 (d) 7x + 24 y = 230

114. The centre(s) of the circle(s) passing through the points (0, 0), (1, 0) and touching the circle x 2 + y 2 = 9 is/are  3 1 (a)  ,   2 2 1  (c)  , 21/ 2 2 

 1 3 (b)  ,   2 2 1  (d)  , − 21/ 2 2 

115. Let x, y be real variable satisfying the equation of circle x 2 + y 2 + 8x − 10 y − 40 = 0. If a = max{( x + 2) 2 + ( y − 3) 2 } and b = min{( x + 2) 2 + ( y − 3) 2 }, then (a) a + b = 18 (b) a + b = 4 2 (c) a − b = 4 2 (d) a ⋅ b = 73

116. Coordinates of the centre of a circle, whose radius is 2 units and which touches the line pair x 2 − y 2 − 2x + 1 = 0, are (a) (4, 0) (b) (1 + 2 2 , 0) (c) (4, 1) (d) (1, 2 2 )

117. Point

M

moved

on

the

circle

( x − 4 )2 + ( y − 8 )2 = 20. Then, it broke away from it

and moving along a tangent to the circle, cuts the X-axis at the point ( −2, 0). The coordinates of a point on the circle at which the moving point broke away, are  3 46 (a)  − ,   5 5 (c) (6, 4)

 2 44  (b)  − ,   5 5 (d) (3, 5)

118. If PQR is the triangle formed by the common

tangents to the circles x 2 + y 2 + 6x = 0 and x 2 + y 2 − 2x = 0, then (a) centroid of ∆PQR is (1, 0) (b) incentre of ∆PQR is (1, 0) (c) circumcentre of ∆PQR is (1, 0) (d) orthocentre of ∆PQR is (2, 0)

Type 3. Assertion and Reason Directions (Q. Nos. 119-127) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

750

119. Statement I The number of circles that pass through the points (1, − 7) and (− 5, 1) and of radius 4, is two. Statement II The centre of any circle that pass through the points A and B lies on the perpendicular bisector of AB. 120. Statement I If the chord of contact of tangent from three points A, B, C to the circle x 2 + y 2 = a 2 are concurrent, then A, B, C will be collinear.

Statement II Lines ( a1 x + b1 y + c1 ) + k ( a 2 x + b2 y + c 2 ) = 0 always pass through a fixed point for k ∈ R. 121. Statement I Circles x 2 + y 2 = 144 and x 2 + y 2 − 6x − 8 y = 0 do not have any common tangent. Statement II If two circles are concentric, then they do not have common tangents. 122. Statement I The least and greatest distances of the point 7) from the circle P(10, x 2 + y 2 − 4x − 2 y − 20 = 0 are 5 and 15 units, respectively. Statement II A point ( x1 , y1 ) lies outside a circle S ≡ x 2 + y 2 + 2gx + 2 fy + c = 0, if S 1 > 0, where S 1 ≡ x12 + y12 + 2g x1 + 2 fy1 + c. 123. Statement I Number of circles passing through (1, 2), (4, 8) and (0, 0) is one. Statement II Every triangle has one circumcircle.

125. Statement I The equation x + y − 2x − 2ay − 8 = 0 represents for different values of a. A system of circle passing through two fixed points lying on X-axis. Statement II If S = 0 is a circle and L = 0 is a straight line, then S + lL = 0 represents the family of circles passing through the point of intersection of circle and straight line, where l is an arbitrary parameter. 2

2

126. Statement I The circles x 2 + y 2 + 2 px + r = 0, x 2 + y 2 + 2qy + r = 0 touch, if

1 p

2

+

1

1 = . r q

13 Circle

124. Statement I Let S 1 : x 2 + y 2 − 10x − 12 y − 39 = 0, S 2 : x 2 + y 2 − 2x − 4 y + 1 = 0 and S 3 : 2x 2 + 2 y 2 − 20x − 24 y + 78 = 0. The radical centre of these circles taken pairwise is (− 2 , − 3). Statement II Point of intersection of three radical axes of three circles taken in pairs is known as radical centre.

2

Statement II Two circles with centres C1 , C 2 and radii r1 , r2 touch each other, if r1 ± r2 = C1C 2 . 127. Statement I The equation of chord of the circle x 2 + y 2 − 6x + 10 y − 9 = 0 which is bisected at ( −2, 4 ), must be x + y − 2 = 0. Statement II In notations, the equation of the chord of the circle S = 0bisected at (x1 , y1 ) must be T = S 1 .

Passage I (Q. Nos. 128 -130) Each side of a square

Passage III (Q. Nos. 134-136) In the problems of

has length 4 units and its centre is at (3, 4). If one of the diagonals is parallel to the line y = x .

coordinate geometry, use of pure geometry may be helpful rather than developing the equations especially in the case of circles. Simple facts like angles in the same segments of a circle are equal and the angle at the centre of a circle is double of the angle at the circumference standing on the same arc.

128. Which of the following is not the vertex of the square? (a) (1, 6)

(b) (5, 2)

(c) (1, 2)

(d) (4, 6)

129. The radius of the circle inscribed in the triangle formed by any three vertices is (a) 2 2 ( 2 + 1) (b) 2 2 ( 2 − 1) (c) 2( 2 + 1) (d) None of the above

130. The radius of the circle inscribed in the triangle formed by any two vertices of square and the centre is (a) 2( 2 − 1) (c) 2 ( 2 − 1)

(b) 2( 2 + 1) (d) None of these

Passage II (Q. Nos. 131-133) Tangents PA and PB are drawn to the circle ( x − 4) 2 + ( y − 5) 2 = 4 from the point P on the curve y = sin x , where A and B lie on the circle. Consider the function y = f ( x ) represented by the locus of the centre of the circumcircle of ∆PAB, then answer the following questions: 131. Range of y = f ( x ) is (a) [− 2, 1]

(b) [−1, 4]

(c) [0, 2]

(d) [2, 3]

132. Period of y = f ( x ) is (a) 2π (c) π

(b) 3π (d) None of these

133. Which of the following is true? (a) f (x ) = 4 has real roots (b) f (x ) = 1has real roots

π   π (c) Range of y = f −1 (x ) is − + 2, + 2   4 4 (d) None of the above

134. A(2, 0), B(0, 2) and C are the vertices of a triangle inscribed in the circle x 2 + y 2 = 4. The bisector of ∠C belongs to either of two families of concurrent lines, whose points of concurrency are (a) ( 2, 2) (b) (2, 2) (c) ( 2, − 2) (d) (− 2, 2)

Targ e t E x e rc is e s

Type 4. Linked Comprehension Based Questions

135. A(2, 0) and B(0, 2) are two points on the circle x 2 + y 2 − 2x − 2 y = 0. C is any point on the circle and l is the incentre of ∆ABC. If Cl intersects the circle at O, then the circumcentre of ∆AlB can be (a) (3, 3) (c) (1, 1)

(b) (2, 2) (d) (2, − 2)

136. A and B are two points on a circle and P is any point on arc AB. If the bisectors of ∠PBA and ∠PAB intersect at O, then the locus of the point O is (a) minor arc of a circle, if AB is a minor arc (b) major arc of a circle, if AB is a major arc (c) always on a major arc of a circle (d) always on a minor arc of a circle

Passage IV (Q. Nos. 137-139) A variable circle of

radius 2 units roots outside the circle x 2 + y 2 + 4 x = 0. If C and C1 denote the centres of respectively circles.

137. Locus of centre of variable circle i.e. c is (a) x 2 + y2 + 4 y = 0 (b) x 2 + y2 + 4 x − 12 = 0 (c) x 2 + y2 + 4 y − 12 = 0 (d) None of the above

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138. If the line joining c and c1 makes an angle of 60° with X-axis, then equation of common tangents of the circle can be (a) 3x − y ± 2 = 0, 3 y + x − 2 = 0 (b) 3x + y ± 2 = 0, 3 y + x − 2 = 0 (c) 3x − y ± 2 = 0, 3 y − x − 2 = 0 (d) None of the above

equation is (a) x − y +

6=0

(b) x − y = 2 3 (c) x − y = 3

139. The area of the region by the common tangents and the line x + 3 y + 2 = 0, is (a) 2 sq units (b) 4 sq units (c) 8 sq units (d) 16 sq units

Passage V (Q. Nos. 140-142) Let α chord of a circle

be that chord of the circle which subtends an angle α at the centre.

140. If x + y = 1is α chord of x 2 + y 2 = 1, then α is equal to (a) π /6 (c) π /6

π chord of x 2 + y 2 = 4 is 1, then its 3

141. If slope of

(b) π /6 (d) x + y = 1is not a chord

(d) x − y +

3=0

2π chord of x 2 + y 2 + 2x + 4 y + 1 = 0 3 from the centre, is

142. Distance of (a) 1 unit (b) 2 units (c) 2 units 1 unit (d) 2

Type 5. Match the Columns

Ta rg e t E x e rc is e s

143. Match the statements of Column I with values of Column II. Column I

Column II

A. Number of circles touching given three p. non-concurrent lines is

1

B. Number of circles touching y = x at (2, 2) q. and also touching line x + 2 y − 4 = 0, is

2

C. Number of circles touching lines x ± y = 2 and passing through the point (4, 3) is

r.

4

D. Number of circles intersecting given three circles orthogonally

s.

Infinite

144. Let x + y + 2gx + 2 fy + c = 0 be an equation of circle. 2

2

Column I

Column II g0

C. If circle lie on the left of Y-axis, then

r.

g2 − c < 0

D. If circle touches positive X-axis and does not intersect Y-axis, then

s.

c>0

145. Match the statements of Column I with values of Column II.

752

Column I

Column II

A. If the length of the common chord of two p. circles of radii 3 and 4 units which intersect k orthogonally is , then k is equal to 5

1

q. B. If the circumference of the circle x 2 + y 2 + 4 x + 12 y + p = 0 is bisected by the circle x 2 + y 2 − 2 x + 8 y − q = 0, then p + q is equal to

24

Column I

Column II

C. Number of distinct chords of the circle r. 2 x( x − 2 ) + y(2 y − 1) = 0; chords are 1 passing through the point  2 ,  and are  2 bisects on X-axis, is

32

D. One of the diameters of the circle s. circumscribing the rectangle ABCD is 4 y = x + 7 . If A and B are the points (−3, 4) and (5, 4) respectively, then the area of the rectangle is

36

146. Match the statements of Column I with values of Column II. Column I

Column II

A. If one of the diameter of the circle x 2 + y 2 − 2 x − 6 y + 6 = 0 is a chord to the circle with centre (2, 1), then the radius of the circle is equal to

p.

2 −

3

B. If the straight line y = mx touches or lies outside the circle x 2 + y 2 − 20 y + 90 = 0, then| m| can be equal to

q.

2 +

3

C. AB is a chord of the circle x 2 + y 2 = 25 and the tangents at A and B intersect at C. If (2, 2 3) is the mid-point of AB, then the area of quadrilateral OACB is equal to (O being origin)

r.

2

D. A circleC of radius unity touches both the coordinate axes and lies in the first quadrant. If the circle C1, which also touches both the axes and lies in the first quadrant, intersect C orthogonally, then the radius C1 can be equal to

s.

3

t.

75/4

148. Line segments AC and BD are diameters of circle of radius one. If ∠BDC = 60°, then the length of line segment AB is _________ . 149. Let the lines ( y − 2) = m1 ( x − 5) and ( y + 4 ) = m2 ( x − 3) intersect at right angles at P, where m1 and m2 are parameters. If locus of P is x 2 + y 2 + g x + fy + 7 = 0, then the value of | f + g | is _________ . 150. Consider the family of circles x 2 + y 2 − 2x − 2λy −8 = 0 passing through two fixed points A and B. Then, the distance between the points A and B is___ . 151. A circle touches the hypotenuse of a right angled triangle at its middle point and passes through the middle point of shorter side. If 3 and 4 units are the

length of the sides and r is the radius of the circle, then find the value of 3r. 152. Let 2x 2 + y 2 − 3xy = 0 be the equations of a pair of tangents drawn from the origin O to a circle of radius 3 units with centre in the first quadrant. If A is one of the points of contact, and the length of OA is 3( 3 + 2λ ), then find the numerical quantity λ. 153. If line L touch a circle C1 of diameter 6. If the centre of C1 , lies in the first quadrant, then the radius of the circle C 2 which is concentric with C1 and cuts intercepts of length 8 on these lines is λ, then find the numerical quantity λ. 154. Suppose the line y = 2 x + a does not intersect or touch the circles x 2 + y 2 − 2 x − 2 y + 1 = 0 and x 2 + y 2 − 16 x − 2 y + 61 = 0. If the circles lie on opposite side of the line, then a must lie in the interval (2 5 − 5k, − 5 − 1). Find the numerical quantity k.

Targ e t E x e rc is e s

147. The length of a common internal tangent to two circles is 7 and a common external tangent is 11. If the product of the radii of the two circles is p, then the value of p/ 2 is _________ .

Entrances Gallery IIT JEE/JEE Advanced 1. A circle S passes through the point (0, 1) and is orthogonal to the circles ( x − 1) 2 + y 2 = 16 and [2014] x 2 + y 2 = 1. Then, (a) radius of S is 8 (c) centre of S is (−7, 1)

(b) radius of S is 7 (d) centre of S is (−8, 1)

1 2. Let complex numbers α and lie on circles α ( x − x 0 ) 2 + ( y − y 0 ) 2 = r 2 and ( x − x0 ) 2 + ( y − y0 ) 2 = 4r 2 ,

respectively.

If

z 0 = x 0 + iy 0 satisfies the equation 2| z 0 |2 = r 2 + 2 , then |α | is equal to 1 (a) 2

1 (b) 2

[2013] 1 (c) 7

1 (d) 3

3. Circle(s) touching X-axis at a distance 3 from the origin and having an intercept of length 2 7 on Y-axis is/are [2013] (a) x 2 + (b) x 2 + (c) x 2 + (d) x 2 +

y2 − 6x + 8 y + 9 = 0 y2 − 6x + 7 y + 9 = 0 y2 − 6x − 8 y + 9 = 0 y2 − 6x − 7 y + 9 = 0

4. The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line [2012] 4x − 5 y = 20 to the circle x 2 + y 2 = 9 is

Circle

13

Type 6. Single Integer Answer Type Questions

(a) 20(x 2 + (b) 20(x 2 + (c) 36(x 2 + (d) 36(x 2 +

y2 ) − 36x + 45 y = 0 y2 ) + 36x − 45 y = 0 y2 ) − 20x + 45 y = 0 y2 ) + 20x − 45 y = 0

Directions (Q.Nos. 5-6) A tangent PT is drawn to the

circle x 2 + y 2 = 4 at the point P( 3, 1). A straight line L, perpendicular to PT is a tangent to the circle [ 2012] ( x − 3) 2 + y 2 = 1.

5. A possible equation of L is (a) x − 3 y = 1 (c) x − 3 y = − 1

(b) x + (d) x +

3y =1 3y = 5

6. A common tangent of the two circles is (a) x = 4 (c) x + 3 y = 4

(b) y = 2 (d) x + 2 2 y = 6

7. The circle passing through the point (−1, 0) and touching Y-axis at (0, 2) also passes through the point [2011]  3  (a)  − , 0  2 

 5  (b)  − , 2  2 

 3 (c)  − ,  2

5  2

(d) (−4 , 0)

8. Two parallel chords of a circle of radius 2 are at a distance 3 + 1 apart. If the chords subtend at the π 2π centre, angles of and , (k > 0), then the value of k k [ k ], where [k] denotes the largest integer ≤ k, is ________. [2010]

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JEE Main/AIEEE 9. The number of common tangents to the circles x 2 + y 2 − 4x − 6 y − 12 = 0 and x 2 + y 2 + 6x + 18 y + 26 [2015] = 0 is (a) 1

(b) 2

(c) 3

(d) 4

[2007]

10. Let C be the circle with centre at (1, 1) and radius 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius [2014] of T is equal to 3 2

(a) (c)

1 2

(a) (−5, 2)

(b) (2, −5)

(c) (5, −2)

10 3 6 (c) 5

3 5 5 (d) 3

Ta rg e t E x e rc is e s

(c) | a | = 2c

(d) 2| a | = c

14. The equation of the circle passing through the point (1, 0) and (0, 1) and having the smallest radius, is [2011] (a) x + y + x + y − 2 = 0 2

2

(b) x 2 + y2 − 2x − 2 y + 1 = 0 (c) x 2 + y2 − x − y = 0 (d) x 2 + y2 + 2x + 2 y − 7 = 0

15. The circle x 2 + y 2 = 4x + 8 y + 5 intersects the line [2010] 3x − 4 y = m at two distinct points, if (a) −85 < m < − 35 (c) 15 < m < 65

(b) −35 < m < 15 (d) 35 < m < 85

16. If P and Q are the points of intersection of the circles x + y + 3x + 7 y + 2 p − 5 = 0 2

2

and

x + y + 2x + 2 y − p = 0, then there is a circle [2009] passing through P, Q and (1, 1) for 2

2

2

(a) all values of p (b) all except one value of p (c) all except two values of p (d) exactly one value of p

17. The point diametrically opposite to the point P(1, 0) [2008] on the circle x 2 + y 2 + 2x + 4 y − 3 = 0 is (a) (3, 4) (c) (−3, 4)

(b) (3, −4) (d) (−3, −4)

(c) x 2 + y2 − 2x + 2 y − 47 = 0

20. Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid-points of the chords of the circle C that subtend an angle of 2π at its centre, is [2006] 3 (a) x 2 + y2 = 1

13. The two circles x 2 + y 2 = ax and x 2 + y 2 = c 2 [2011] ( c > 0) touch each other, if (b) a = 2c

(b) x 2 + y2 − 2x + 2 y − 62 = 0 (d) x 2 + y2 + 2x − 2 y − 47 = 0

(d) (−2 , 5)

(b)

(a) | a | = c

1 (b) k ≥ 2 1 (d) k ≤ 2

(a) x 2 + y2 + 2x − 2 y − 62 = 0

12. The length of the diameter of the circle which touches the X-axis at the point (1, 0) and passes through the point (2, 3), is [2012] (a)

1 (a) 0 < k < 2 1 1 (c) − ≤ k ≤ 2 2

19. If the lines 3x − 4 y − 7 = 0 and 2x − 3 y − 5 = 0 are two diameters of a circle of area 49 π sq units, then equation of the circle is [2006]

3 2 1 (d) 4 (b)

11. The circle passing through (1, −2) and touching the X-axis at (3, 0) also passes through the point [2013]

754

18. Consider a family of circles which are passing through the point (−1, 1) and are tangent to X-axis. If ( h, k ) are the coordinates of the centre of the circles, then the set of values of k is given by the interval

(c) x 2 + y2 =

9 4

27 4 3 2 2 (d) x + y = 2 (b) x 2 + y2 =

21. If the circles x 2 + y 2 + 2 ax + c y + a = 0 and x 2 + y 2 − 3a x + d y − 1 = 0 intersect at two distinct points P and Q , then the line 5x + by − a = 0 passes through P and Q for [2005] (a) exactly two values of a (b) infinitely many values of a (c) no value of a (d) exactly one value of a

22. If a circle passes through the point (a, b) and cuts the circle x 2 + y 2 = p 2 orthogonally, then the equation [2005] of the locus of its centre is (a) 2ax + 2by − (a2 + b2 + p2 ) = 0 (b) x 2 + y2 − 2ax − 3by + (a2 − b2 − p2 ) = 0 (c) 2ax + 2by − (a2 − b2 + p2 ) = 0 (d) x 2 + y2 − 3ax − 4 by + (a2 + b2 − p2 ) = 0

23. If a circle passes through the point ( a, b ) and cuts the circle x 2 + y 2 = 4 orthogonally, then the locus of its centre is [2004] (a) 2ax + 2by + (a2 + b2 + 4 ) = 0 (b) 2ax + 2by − (a2 + b2 + 4 ) = 0 (c) 2ax − 2by + (a2 + b2 + 4 ) = 0 (d) 2ax − 2by − (a2 + b2 + 4 ) = 0

24. A variable circle passes through the fixed point A ( p, q ) and touches X-axis. The locus of the other [2004] end of the diameter through A is (a) (x − p)2 = 4 qy (c) ( y − p)2 = 4 qx

(b) (x − q)2 = 4 py (d) ( y − q)2 = 4 px

(a) x 2 + y2 − 2x + 2 y − 23 = 0 (b) x 2 + y2 − 2x − 2 y − 23 = 0

(a) 2 < r < 8 (b) r < 2

(c) r = 2

(d) r > 2

13

28. The greatest distance of the point P(10, 7) from the circle x 2 + y 2 − 4x − 2 y − 20 = 0 is [ 2002]

(c) x + y + 2x + 2 y − 23 = 0 2

27. If the two circles ( x − 1) 2 + ( y − 3) 2 = r 2 and x 2 + y 2 − 8x + 2 y + 8 = 0 intersect in two distinct [2003] points, then

Circle

25. If the lines 2x + 3 y + 1 = 0 and 3x − y − 4 = 0 lie along diameters of a circle of circumference 10 π , then the equation of the circle is [2004]

2

(a) 10 units (c) 5 units

(d) x 2 + y2 + 2x − 2 y − 23 = 0

26. The intercept on the line y = x by the circle x 2 + y 2 − 2x = 0 is AB. Equation of the circle on AB as a diameter is [2004] (a) x 2 + y2 − x − y = 0

(b) x 2 + y2 − x + y = 0

(c) x 2 + y2 + x + y = 0

(d) x 2 + y2 + x − y = 0

(b) 15 units (d) None of these

29. The equation of the tangent to the circle x 2 + y 2 + 4x − 4 y + 4 = 0 which make equal intercepts on the positive coordinate axes, is [2002] (b) x + y = 2 2 (d) x + y = 8

(a) x + y = 2 (c) x + y = 4

Other Engineering Entrances 30. Equation of circle with centre (−a, −b) and radius 2

(a) x + (b) x 2 + (c) x 2 + (d) x 2 + 2

[Karnataka CET 2014]

y + 2ax + 2by + 2b = 0 y2 − 2ax − 2by − 2b2 = 0 y2 − 2ax − 2by + 2b2 = 0 y2 − 2ax + 2by + 2a2 = 0 2

2

145 11 cm (b) cm (a) 4 2

(b) (x + 2)2 + ( y + 2)2 = 8 (d) (x + 3)2 + ( y + 3)2 = 8

32. Compute the shortest distance between the circle x 2 + y 2 − 10x − 14 y − 151 = 0 and the point (−7, 2). (a) 0

(b) 1

(c) 2

[GGSIPU 2014] (d) 4

33. If one end of a diameter of the circle x 2 + y 2 − 4x − 6 y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter. [RPET 2014] (b) (1, 2) (d) None of these

(a) (2, 1) (c) (1, 1)

34. The measure of the chord intercepted by circle x 2 + y 2 = 9 and the line x − y + 2 = 0is [AMU 2014] (a) 28

(b) 2 5

(c) 7

(d) 5

35. A circle with centre at (2, 4) is such that the line x + y + 2 = 0 cuts a chord of length 6. The radius of [EAMCET 2014] the circle is (a) 41 cm (c) 21 cm

[EAMCET 2014] 135 (c) 135 cm (d) cm 4

38. The shortest distance between the circles ( x − 1) 2 + ( y + 2) 2 = 1 and ( x + 2) 2 + ( y − 2) 2 = 4 is

31. A circle of radius 8 is passing through origin and the point (4, 0). If the centre lies on the line y = x, [Kerala CEE 2014] then the equation of circle is (a) (x − 2)2 + ( y − 2)2 = 8 (c) (x − 3)2 + ( y − 3)2 = 8 (e) (x − 4 )2 + ( y − 4 )2 = 8

37. The length of the common chord of the two circles x 2 + y 2 − 4 y = 0 and x 2 + y 2 − 8 x − 4 y + 11 = 0 is

(b) 11 cm (d) 31 cm

36. The condition for the lines lx + my + n = 0 and l1 x + m1 y + n1 = 0 to be conjugate with respect to the circle x 2 + y 2 = r 2 , is [EAMCET 2014] (a) r2 (ll1 + mm1 ) = nn1 (b) r2 (ll1 − mm1 ) = nn1 2 (c) r (ll1 + mm1 ) + nn1 = 0 (d) r2 (lm1 + l1m) = nn1

[Kerala CEE 2014] (a) 1 (d) 4

(b) 2 (e) 5

(c) 3

39. The centre of the circle, whose radius is 5 and which touches the circle x 2 + y 2 − 2x − 4 y − 20 = 0 at (5, 5) is [Kerala CEE 2014] (a) (10, 5) (d) (8, 9)

(b) (5, 8) (e) (9, 8)

(c) (5, 10)

40. A circle passes through the points (0, 0) and (0, 1) and also touches the circle x 2 + y 2 = 16. The radius of the circle is [Kerala CEE 2014] (a) 1 (d) 4

(b) 2 (e) 5

Targ e t E x e rc is e s

a − b is 2

(c) 3

41. If the circle x 2 + y 2 + 2g x + 2 fy + c = 0 cuts the three circles x 2 + y 2 − 5 = 0, x 2 + y 2 − 8 x − 6 y + 10 = 0 and x 2 + y 2 − 4x + 2 y − 2 = 0 at the extremities of [WB JEE 2014] their diameters, then (a) c = − 5 (c) g + 2 f = c + 2

147 25 (d) 4 f = 3g

(b) fg =

42. If the squares of the tangents from a point P to the circles x 2 + y 2 = a, x 2 + y 2 = b 2 and x 2 + y 2 = c 2 [AMU 2014] are in AP, then (a) a, b, c are in AP (c) a2, b2, c2 are in AP

(b) a, b, c are in GP (d) a2, b2, c2 are in GP

43. The point at which the circles x 2 + y 2 − 4x − 4 y + 7 = 0 and x 2 + y 2 − 12x − 10 y + 45 = 0 touch each other, is [EAMCET 2014]  13 14  (a)  ,   5 5  14 13 (c)  ,   5 5

 2 5 (b)  ,   5 6  12 (d)  , 2 +  5

21  5 

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44. The locus of the centre of the circle, which cuts the circle x 2 + y 2 − 20x + 4 = 0 orthogonally and touches the line x = 2, is [EAMCET 2014] (a) x 2 = 16 y (c) y2 = 16x

(b) y2 = 4 x (d) x 2 = 4 y

(a) x + y = 2 (c) x 2 + y2 = 2

45. If the equation of the tangent to the circle x 2 + y 2 − 2x + 6 y − 6 = 0 parallel to 3x − 4 y + 7 = 0 is 3x − 4 y + k = 0, then the values of k are [Kerala CEE 2013] (a) 5, −35 (c) 7, −32 (e) None of these

(b) −5, 35 (d) −7, 32

46. Circles are drawn through the point (2, 0) to cut intercept of length 5 units on X -axis. If their centres lies in the first quadrant, then their equation is [UP SEE 2013]

Ta rg e t E x e rc is e s

(a) x 2 + y2 − 9x + 2 f y + 14 = 0 (b) 3x 2 + 3 y2 + 27x − 2 f y + 42 = 0 (c) x 2 + y2 − 9x − 2 f y + 14 = 0 (d) x 2 + y2 − 2 fx − 9 y + 14 = 0

2

2

1 4  (b) x 2 −  y +  =  3 3 (d) None of these

48. The least and the greatest distances of the point (10, 7) from the circle x 2 + y 2 − 4x − 2 y − 20 = 0 are (a) 10, 5

`(b) 15, 20

[Karnataka CET 2012] (c) 12, 16 (d) 5, 15

49. Equation of unit circle concentric with circle x 2 + y 2 + 8 x + 4 y − 8 = 0 is [OJEE 2012] (a) x 2 + y2 + 8 x + 4 y + 19 = 0 (b) x 2 + y2 − 8 x + 4 y + 19 = 0 (c) x 2 + y2 − 8 x − 4 y + 19 = 0 (d) None of the above

is

a

common

tangent

to

x 2 + y 2 = b 2 and ( x − a ) 2 + y 2 = b 2 is [Manipal 2012] (a)

2b

a 2 − 4 b2 2b (c) a − 2b

(b) (d)

a 2 − 4 b2 2b b a − 2b

51. The equation of the two tangents from (−5, −4) to the circle x 2 + y 2 + 4x + 6 y + 8 = 0 are [Karnataka CET 2012]

756

(a) intersecting in two points (b) non-intersecting (c) touching externally (d) touching internally

54. If the circles x 2 + y 2 + 2x + 2 ky + 6 = 0 and x 2 + y 2 + 2k y + k = 0 intersect orthogonally, then k [WB JEE 2012] is equal to (a) 2 or − (c) 2 or

3 2

(b) −2 or −

3 2

(a) 20 (d) 16

(d) −2 or

3 2

3 2

(a) x + 2 y + 13 = 0, 2x − y + 6 = 0 (b) 2x + y + 13 = 0, x − 2 y = 6 (c) 3x + 2 y + 23 = 0, 2x − 3 y + 4 = 0 (d) x − 7 y = 23, 6x + 13 y = 4

(b) 12 (e) 22

(c) 10

56. If the straight line y = mx lies outside the circle x 2 + y 2 − 20 y + 90 = 0 , then the value of m will [WB JEE 2011; Guj CET 2011] satisfy (b) | m | < 3 (d) | m | > 3

57. The intercept on the line y = x by the circle x 2 + y 2 − 2x = 0 is AB. Equation of the circle with [WB JEE 2011] AB as diameter is (a) x 2 + y2 = 1 (b) x (x − 1) + y ( y − 1) = 0 (c) x 2 + y2 = 2 (d) (x − 1)(x − 2) + ( y − 1)( y − 2) = 0

58. The length (in units) of tangent from point (5, 1) to the circle x 2 + y 2 + 6x − 4 y − 3 = 0 is [BITSAT 2011] (a) 81

50. If a > 2b > 0, then positive value of m for which y = mx − b 1 + m2

53. The circles and x 2 + y 2 − 6x − 8 y = 0 2 2 [MP PET 2012] x + y − 6x + 8 = 0 are

(a) m < 3 (c) m > 3

2

1 4  (c) x 2 +  y −  =−  3 3

(b) x 2 + y2 = 1 (d) x + y = 1

55. The sum of the minimum distance and the maximum distance from the point (4, −3) to the circle [Kerala CEE 2011] x 2 + y 2 + 4x − 10 y − 7 = 0 is

47. Two vertices of an equilateral triangle are ( −1, 0) and (1, 0) and its circumcircle is [OJEE 2013] 1 4  (a) x 2 +  y −  =  3 3

52. The locus of the mid-points of the chord of the circle x 2 + y 2 = 4 which subtends a right angle at the origin is [Manipal 2012]

(b) 29

(c) 7

(d) 21

59. Circle a x 2 + ay 2 + 2g x + 2 f y + c = 0 touches X -axis, if [MP PET 2011] (a) f 2 > ac

(b) g 2 > ac

(c) f 2 = bc

(d) g 2 = ac

60. The locus of the point of intersection of the tangents at the extremeties of a chord of the circle which touches the circle x2 + y2 = a2, x 2 + y 2 − 2ax = 0 passes through the point, is [AMU 2011] a  (a)  , 0 2  (c) (a, 0)

 a (b)  0,   2 (d) (0, 0)

61. If the lines joining the origin to the intersection of the line y = mx + 2 and the circle x 2 + y 2 = 1 are at right angles, then [AMU 2011] (a) m = 3 (c) m = 1

(b) m = ± 7 (d) m = 5

(b) 2

(c) 3

[Karnataka CET 2011] (d) 1

63. The centre of a circle which cuts x 2 + y 2 + 6x − 1 = 0, x2 + y2 − 3y + 2 = 0 2 2 x + y + x + y − 3 = 0 orthogonally is

and

67. The straight line x + y − 1 = 0 meets the circle x 2 + y 2 − 6x − 8 y = 0 at A and B. Then, the equation of the circle of which AB is a diameter, is [WB JEE 2010] (a) x 2 + y2 − 2 y − 6 = 0 (b) x 2 + y2 + 2 y − 6 = 0 (c) 2(x 2 + y2 ) + 2 y − 6 = 0 (d) 3(x 2 + y2 ) + 2 y − 6 = 0

[Karnataka CET 2011]  1 9 (b)  − , −   7 7  1 9 (d)  − ,   7 7

68. The equation of normal of x 2 + y 2 − 2x + 4 y − 5 = 0 [WB JEE 2010] at (2, 1) is

64. The line segment joining the points (4, 7) and (−2, −1) is a diameter of a circle. If the circle intersects X -axis at A and B, then AB is equal to [BITSAT 2010]

69. The equation of the chord of the circle x 2 + y 2 = 81, which is bisected at the point (−2, 3), is

 1 9 (a)  ,   7 7  1 9 (c)  , −   7 7

(a) 4 (c) 6

(b) 5 (d) 8

65. The distance of the mid-point of line joining two points (4, 0) and (0, 4) from the centre of the circle x 2 + y 2 = 16 is [VITEEE 2010] (a) 2

(b) 2 2

(c) 3 2

(d) 2 3

66. The equation of family of circles with centre at (h, k) [Kerala CEE 2010] touching X -axis, is given by (a) x 2 + (b) x 2 + (c) x 2 + (d) x 2 + (e) x 2 +

y2 − 2hx + h2 = 0 y2 − 2hx − 2ky + h2 = 0 y2 − 2hx − 2ky − h2 = 0 y2 − 2hx − 2ky = 0 y2 + 2hx + 2ky = 0

13 Circle

(a) 4

of

(a) y = 3x − 5 (c) y = 3x + 4

(b) 2 y = 3x − 4 (d) y = x + 1

[Kerala CEE 2010] (a) 3x − y = 13 (b) 3x − 4 y = 13 (c) 2x − 3 y = 13 (d) 3x − 3 y = 13 (e) 2x − 3 y = − 13

70. If the two circles ( x + 7) 2 + ( y − 3) 2 = 36 and ( x − 5) 2 + ( y + 2) 2 = 49 touch each other externally, [Kerala CEE 2010] then the point of contact is  −19 19 (a)  ,   13 13  17 9  (c)  ,   13 13  19 19 (e)  ,   13 13

 −19 9  (b)  ,   13 13  −17 9  (d)  ,   13 13

Targ e t E x e rc is e s

62. The total number of common tangents x 2 + y 2 − 6x − 8 y + 9 = 0 and x 2 + y 2 = 1 is

757

Answers Work Book Exercise 13.1 1. (a)

2. (b)

3. (c)

4. (c)

5. (a)

6. (a)

7. (b)

8. (b)

9. (d)

10. (d)

4. (b)

5. (b)

6. (b)

7. (b)

8. (b)

9. (c)

10. (d)

4. (b)

5. (b)

6. (a)

7. (a)

8. (a)

9. (c)

10. (b)

10. (a)

Work Book Exercise 13.2 1. (a)

2. (b)

3. (c)

Work Book Exercise 13.3 1. (b)

2. (c)

3. (d)

Ta rg e t E x e rc is e s

Target Exercises 1. (d)

2. (b)

3. (d)

4. (c)

5. (d)

6. (a)

7. (b)

8. (a)

9. (a)

11. (d)

12. (c)

13. (a)

14. (d)

15. (c)

16. (a)

17. (d)

18. (a)

19. (c)

20. (d)

21. (d)

22. (d)

23. (a)

24. (b)

25. (c)

26. (b)

27. (b)

28. (d)

29. (b)

30. (b)

31. (c)

32. (c)

33. (c)

34. (c)

35. (c)

36. (a)

37. (c)

38. (d)

39. (b)

40. (b)

41. (b)

42. (c)

43. (c)

44. (b)

45. (b)

46. (a)

47. (a)

48. (b)

49. (c)

50. (b)

51. (c)

52. (b)

53. (a)

54. (b)

55. (b)

56. (b)

57. (c)

58. (a)

59. (c)

60. (d)

61. (b)

62. (d)

63. (c)

64. (b)

65. (a)

66. (c)

67. (a)

68. (d)

69. (b)

70. (d)

71. (a)

72. (c)

73. (d)

74. (b)

75. (d)

76. (a)

77. (c)

78. (d)

79. (a)

80. (b)

81. (a)

82. (b)

83. (a)

84. (d)

85. (c)

86. (a)

87. (b)

88. (d)

89. (a)

90. (d)

91. (a)

92. (c)

93. (a)

94. (b)

95. (b)

96. (c)

97. (c)

98. (b)

99. (d)

101. (c)

102. (b)

103. (b)

104. (d)

105. (c)

106. (d)

107. (a)

108. (c)

109. (d)

100. (a) 110. (a,d)

111. (a,c) 112. (c,d)

113. (a,c) 114. (c,d) 115. (a,c,d)

116. (b,d)

117. (b,c)

118. (a,b,c) 119. (d)

120. (a)

121. (b)

122. (b)

123. (d)

124. (d)

125. (a)

126. (a)

127. (d)

128. (d)

129. (b)

130. (a)

131. (d)

132. (c)

133. (c)

134. (a)

135. (b)

136. (d)

137. (b)

138. (a)

139. (c)

140. (b)

141. (a)

142. (a)

143. (*)

144. (**)

145. (***)

146. (****)

147. (9)

148. (1)

149. (6)

150. (6)

151. (5)

152. (5)

153. (5)

154. (3)

4. (a)

5. (a)

6. (d)

7. (d)

8. (3)

9. (c)

10. (d)

* A → r; B → q; C → q; D → p ** A → p, r, s ; B → r, s; C → q, s; D → p, s *** A → q; B → s; C → q; D → r **** A → s; B → s; C → t; D → p, q

Entrances Gallery 1. (b,c)

758

2. (c)

3. (a,c)

11. (c)

12. (a)

13. (a)

14. (c)

15. (b)

16. (c)

17. (d)

18. (b)

19. (c)

20. (c)

21. (c)

22. (a)

23. (b)

24. (a)

25. (a)

26. (a)

27. (a)

28. (b)

29. (b)

30. (a)

31. (a)

32. (c)

33. (b)

34. (a)

35. (a)

36. (a)

37. (d)

38. (b)

39. (e)

40. (b)

41. (d)

42. (c)

43. (c)

44. (c)

45. (a)

46. (a)

47. (a)

48. (d)

49. (a)

50. (a)

51. (a)

52. (c)

53. (d)

54. (a)

55. (a)

56. (b)

57. (b)

58. (c)

59. (d)

60. (a)

61. (b)

62. (c)

63. (d)

64. (d)

65. (b)

66. (b)

67. (a)

68. (a)

69. (e)

70. (b)

Explanations Target Exercises So, point ([(p + 1)], [ p] ) lies outside the circle x2 + y2 − 2 x − 7 = 0 ∴ ([ p + 1]2 + [ p]2 − 2([ p + 1] ) − 7 > 0 ⇒ ([ p] + 1)2 + [ p]2 − 2([ p] + 1) − 7 > 0 ⇒ 2[ p]2 − 8 > 0 ∴ [ p]2 > 4 From Eqs. (i) and (ii), we get 4 < [ p]2 < 8, which is impossible.

15 2

15 5 =− 2 2 2. Let φ( x, y ) ≡ x 2 + y 2 + 2 gx + 2 f y + c = 0 ⇒

( x + 1)2 + ( y + 2 )2 = 12 + 2 2 −

∴

φ(0, λ ) = 0 + λ2 + 0 + 2 f λ + c = 0

have equal roots. Then,

2+2=−

∴ and and

and ∴

and 2 ⋅ 2 =

c 1

f = −2 c =4 φ(λ , 0 ) ≡ λ2 + 0 + 2 gλ + 0 + c = 0

∴ Here, ⇒

2f 1

λ + 2 gλ + c = 0 c =4

∴ For no value of p, the point will be within the region.

6. A( x, 3) and B(3, 5) are the end points of diameter. Centre, C = (2, y ) Now, C is the mid-point of AB. ∴ x = 1 and y = 4

2

7. If ( x, y ) is the point, then by ratio formula,

4 , 5. 5 4 −29 + 5 = − 2g ⇒ g = 5 10  29  Centre ≡ ( − g, − f ) =  , 2  10  λ2 + 2 gλ + 4 = 0 have roots

and ∴

8. The circles g, f and c passes through (2, 0). ∴

centre different. If centre is (α, β), then α − 3 β − 2 −2(3 + 2 − 19) = = 1 1 12 + 12

…(i)

∴ ⇒

[from Eq. (i)] 4 (g 2 + 4g + 4) = 25 (2 g + 9)(2 g − 1) = 0 −9 1 ∴ g= , 2 2 Since, centre (−g, −f) lies in Ist quadrant, therefore we choose 9 9 g = − , so that − g = (+ ve). 2 2 [from Eq. (i)] ∴ c = 14

Hence, required circle is ( x − 17 )2 + ( y − 16)2 = 1.   λ2 (1 − λ )2 − 5 ≤ 5  + 4 4   ⇒ λ2 + (1 − λ )2 − 20 ≤ 100

9. Let coordinates of A and B be (α, β) and (γ, δ),

2 λ2 − 2 λ − 119 ≤ 0

respectively.

1 − 239 1 + 239 ∴ ≤λ≤ 2 2 ⇒ −7.2 ≤ λ ≤ 8.2 (approx) ∴ λ = − 7, −6, −5, …, 7, 8

Then, and

α + γ = − 2a αγ = − b2 β + δ = − 2 p, β δ = − q 2

Equation of circle with AB as diameter is ( x − α )( x − γ ) + ( y − β )( y − δ ) = 0 ⇒ x 2 + y 2 − x(α + γ ) − y(β + δ ) + γ + βδ = 0 ⇒ x 2 + y 2 + 2 a x + 2 py − b2 − q 2 = 0

([ p + 1], [ p]) lies inside the circle x 2 + y 2 − 2 x − 15 = 0.

5. Since, the point ∴

4 + 4g + c = 0

Intercept on X-axis is 2 g 2 − c = 5.

⇒ α − 3 = β − 2 = 14 ∴ α = 17, β = 16

⇒

x = 4 cos θ + 3 y = 4sin θ ( x − 3)2 + y 2 = 16

which is the equation of a circle.

3. The image of the circle has same radius but different

4.

…(ii)

Targ e t E x e rc is e s

1. x 2 + y 2 + 2 x + 4 y = −

[ p + 1]2 + [ p]2 − 2[ p + 1] − 15 < 0

∴

Radius = (a2 + p2 + b2 + q 2 )

10. (1, 0) O √8

⇒ ⇒

P(5, 7) 3 P2

4

O(0, 0) P1 2 B N A d1

([ p] + 1)2 + [ p]2 − 2([ p] + 1) − 15 < 0 2[ p]2 − 16 < 0 ⇒ [ p]2 < 8

Since, circles are concentric.

…(i)

D

d2 M C

Let equation of line be y = x + c or

y − x =c

…(i)

759

Objective Mathematics Vol. 1

13

Perpendicular from (0, 0) on line (i) is  c  22 −    2

2

2 − c 32 −    2 

2

In ∆AON, and in∆CPM,

−c c . = 2 2

OA2 = 10 (2 − 4m)2 ⇒ = 10 1 + m2 ⇒ 4 − 16 m + 16 m2 = 10(1 + m)2 ⇒ 3 m2 − 8 m − 3 = 0 1 ⇒ m = 3 or − 3 Hence, the required equation is y = 3 x or x + 3y = 0 So,

= AN = CM

AN = CM c2 (2 − c )2 ⇒ 4− = 9− 2 2 −3 ⇒ c= 2 3 Therefore, equation of the line is y = x − 2 or 2x − 2y − 3 = 0 Given,

13. S(−3, − 2 ) = 9 + 4 − 6 − 6 + 1= 2 > 0 So, (−3, − 2) lies outside of S and S′ (−3, − 2 ) = 9 + 4 − 12 − 6 + 2 = − 3 < 0 This implies, (−3, − 2 ) lies inside of S′.

14. Since, y = | x| + c and x 2 + y 2 − 8| x| − 9 = 0 both are

11. ∆ABC and ∆DBA are similar.

symmetrical about Y-axis for x > 0, y = x + c . Equation of tangent to circle x 2 + y 2 − 8 x − 9 = 0

B θ

D

π/2 –

x

Parallel to y = x + c is y = ( x − 4) + 5 1 + 1 y = x + (5 2 − 4) ⇒ For no solution, c > 5 2 − 4. c ∈ (5 2 − 4, ∞ ) ∴

y θ

A

lx = y l 2 + x 2

Ta rg e t E x e rc is e s

∴ ⇒ ⇒ ⇒

C

I

2

2

2

n(n − 1) (α x )2 + K 1⋅ 2 = 1 + 8 x + 24 x 2 + ...

1 + n(α x ) +

⇒

l 2( x 2 − y 2 ) = x 2 y 2 xy AB × AD l= = 2 2 x −y AB 2 − AD 2

12. Let the equation of the chord OA of the circle x 2 + y 2 − 2 x + 4y = 0

C

…(ii) A

(1 + m2 )x 2 − (2 − 4m)x = 0

⇒

x=0 2 − 4m 1 + m2

θ P (2, 4)

x=

and Y

A C

B

X O

Hence, the points of intersection are  2 − 4m m(2 − 4 m) (0, 0) and A  ,   1 + m2 1 + m2  2

⇒

B

…(i)

be y = mx On solving Eqs. (i) and (ii), we get x 2 + m2 x 2 − 2 x + 4mx = 0 ⇒

760

15. Given, (1 + α x )n = 1 + 8 x + 24 x 2 + K

l x = y (l + x ) 2 2

 2 − 4m (2 − 4m)2 OA2 =   (1 + m2 ) = 2 1+ m  1 + m2

Since, OAB is an isosceles right angled triangle. 1 OA2 = AB 2 ∴ 2 where, AB is a diameter of the given circle.

On equating the coefficients of x and x 2 , we get na(nα − α ) = 24 nα = 8, 1⋅ 2 8 (8 − α ) or = 24 2 ⇒ 8−α = 6 ∴ α = 2 and n = 4 x −2 y − 4 Equation of the line is = = r , then point cos θ sin θ (2 + r cos θ, 4 + r sin θ ) lies on the circle x 2 + y 2 = 4. ⇒

(2 + r cos θ )2 + (4 + r sin θ )2 = 4

or

r 2 + 4r (cos θ + 2 sin θ ) + 16 = 0

∴

PA ⋅ PB = r1r2 16 = = 16 1

Aliter PA ⋅ PB = (PC )2 = 2 2 + 44 − 4 = 16

4

21.

B b L D

N

4

α

C M

In ∆MLN,

a−b a+ b  a − b α = sin −    a + b

sin α =

D

Thus, Y-axis is their common chord. So, ∆ABC is equilateral. 2 3 2 Hence, area of ABCD is (4) i.e. 8 3 sq units. 4

∴ Angle between AB and AD  a − b = 2α = 2 sin −1    a + b

17. From the figure, we have y = 2x + 5

22. Centre is (−a, − b) and radius is a2 + b2 . Thus, equation of incircle is ( x + a)2 + ( y + b)2 =

⇒

x1 = − 6 and

PQ = 2 r co t θ

23. Here,

y1 + 6 × 2 = −1 x1 + 8

RS = 2 r tan θ 2r = PQ ⋅ RS

⇒

y1 = − 7

18. For smallest circle, OA will become common normal. ∴

OA = 5 ⇒ AB = 4 3 So, equation of line OA is y = x. 4 Putting this value of y in x 2 + y 2 = 1, we get

24. Clearly, (0, 0) lies on director circle of the given circle. Now, equation of director circle is ( x + g )2 + ( y + f )2 = 2(g 2 + f 2 − c ) If (0, 0) lies on it, then g 2 + f 2 = 2(g 2 + f 2 − c ) ⇒

4 9x 2 =1 ⇒ x = ± 5 16  4 3 B≡ ,   5 5 Thus, equation of required circle is 24 18 x2 + y2 − x− y+ 5= 0 5 5 x2 +

g2 + f 2 = 2c

25. Centre of circle is (3, − 1) . Equation of any line passing through(3, − 1)and perpendicular to2 x − 5 y + 18 = 0 is 5 ( y + 1) = − ( x − 3) 2 ⇒ 2 y + 5 x − 13 = 0 On solving it with the given chord, we get x = 1, y = 4. Thus, required mid-point is (1, 4).

19. Here, 5 x − 2 y + 6 = 0 meets Y-axis at (0, 3). ∴ Q(0, 3)

26. If (h, k ) is the centroid of the ∆PAB, then P

and

C (–3, –3)

PQ = 0 + 9 + 0 + 18 − 2 = 5 units

27. Q C1C2 = | r1 − r2|

20. Qc = f 2 g2 + f 2 − c  

2

r r   ⇒ h −  + k −    3 3 Hence, locus of (h, k ) is 2 2 2 r r    r x −  + y −  =      3 3 3

Here, length of tangent = S1

∴ (g − 3)2 + (f − 3)2 =  9 + 9 − 14 +  ⇒ g 2 + f 2 − 6g − 6f + 18 = (2 + g )2 ⇒ f 2 − 10 g − 6f + 14 = 0

k 2

x 2 + y 2 + 6x + 6 y – 2 = 0

Hence, locus of centre (g, f ) is y 2 − 10 x − 6 y + 14 = 0

r (1 + cos θ ) 3 r (1 − sin θ ) = 3 2  r =   3

h=

Q(0,3)

⇒

a2 + b2 . 4

⇒ 4 x 2 + 4 y 2 + 8 ax + 8 by + 3 a2 + 3 b2 = 0

(–8, –6)

y1 = 2 x1 + 5 and

A

E

B (2,0)

(x1, y1)

α

Targ e t E x e rc is e s

A (–2,0)

13

C

Circle

16. Circles with centres (2, 0) and (−2, 0) each with radius 4.

2

circles touch internally. Hence, the number of common tangents is one.

28. Tangents are clearly x = − 1, y = − 1. Their combined equation is ( x + 1)( y + 1) = 0 ⇒ x + y + xy + 1 = 0

761

Objective Mathematics Vol. 1

13

29. Diameter of circle = Distance of the point (1, 0) from

3x − 6y + 7 = 0

3 (1) − 6 (0 ) + 7 32 + (−6)2 ∴

2 10 = 5 units = 45 3

Radius of circle =

1 2  5 5 = unit    2 3 3

30. Equation of normal is y − 1 = 2( x + 1) ⇒ 2x − y + 3 = 0 x = − 3 and y = − 3, solving tangent and normals.

31. Q C1C2 = r1 + r2 = 4 So, circles touch each other externally. ∴ Common tangent is ( x 2 + y 2 − 8 x + 12 ) − ( x 2 + y 2 − 4) = 0 ⇒

−8 x + 16 = 0 ⇒

x =2

37. Let C be centre. We have, CD = 4 and AD = 3 ⇒ CA2 = 32 + 42 ⇒

CP = CA = 5 h−4 k−4 If C is (h, k), then = =±5 1 1 − 2 2 5 5 ,k = 4+ ⇒ h = 4− 2 2 5 5 ,k = 4− or h = 4+ 2 2

38. A1B1 = 4 + 4 = 2 2 AB = 2 2 − 2 = 2( 2 − 1) ⇒ Thus, equation of required circle is x 2 + y 2 = ( 2 − 1)2 = 3−2 2

32. ( x − 1)( y − 2 ) = 0 ⇒ x − 1 = 0 and y − 2 = 0 3 (1) + 4 (2 ) − 6 5 ∴ Radius of circle = = = 1unit 5 9 + 16 Now, equation of the circle is ( x − 1)2 + ( y − 2 )2 = 1 ⇒

Ta rg e t E x e rc is e s

On putting x = 0 in the equation of circles, we get y2 + b = 0 It should have equal roots, hence b = 0. a(1) + b(0 ) − 2 34. Here, =2 a2 + b2

Also, normal to the circle will pass through the centre (0, 2). a − 2 = 2 a2 + 1 ⇒ ⇒

(a − 2 ) = 4 (a + 1)

⇒

3 a2 + 4a = 0 ⇒ a = 0, −

2

4 3

35. Equation of AB is T = 0 i.e. 3 x + y ⋅ 0 = 4 4 4 x = , OD = 3 3 ⇒ AD 2 = OA2 − OD 2 16 20 ⇒ AD 2 = 4 − = 9 9 2 5 4 5 ⇒ ⇒ AB = AD = 3 3 1 4 5  4 10 5 sq units ⋅ 3 −  = ∴ Area of ∆PAB = ⋅ 2 3  3 9 ⇒

36. Equation of AB is x ⋅ 4 + y ⋅ 4 = 9 ⇒ Also,

762

OD =

9 4 2

AD 2 = OA2 − OD 2 = 9 −

∴ ⇒

π 4

CB = h + 2 BE = h − 2 π h −2 1 = cos = h+2 4 2 2( 2 + 1) = 2(3 + 2 2 ) units h= ( 2 − 1)

40. C1C2 = (0 − 1)2 + (4 − 2 )2 = 1 + 4 = 5 = r2 − r1 Thus, two circles touch each other internally.

41. C1C2 = (4 − 2 )2 + (3 − 4)2

a − 2 = 2 a2 + b2

2

We have, ∠COD = ∠CBE = and

x 2 + y 2 − 2 x − 4y + 4 = 0

33. Equation of common tangent is S1 − S 2 = 0 i.e. x = 0

or

39. Let the centre of required circle be (h,h).

81 207 = 32 32

207 ⇒ AB = 2 ⋅ AD = 2 ⋅ 32 1 9 207 9 207 sq units = ∴ Area of triangle = ⋅ 2 4 2 2 2 32

= 4 + 1 = 5 = r1 − r2 This show that these circles touch each other internally.

42. Radical axes of circles are indeed the altitudes of the triangle. Thus, radical centres of S1 = 0, S 2 = 0, S 3 = 0 will be the orthocentre of the triangle.

43. S1 = T at (h, k ) ⇒

hx + ky = h 2 + k 2

So, (1, − 2) is mid-point of chord whose equation is 1⋅ x + (−2 )y = 12 + (−2 )2 ⇒

x − 2y − 5 = 0

44. The sides are x + y − 4 = 0, x − 1 = 0,y − 2 = 0. So, the triangle is right angled at (1, 2). The hypotenuse is x + y − 4 = 0, whose ends are (1, 3) and (2, 2). This,  1 + 2 3 + 2 circumcentre =  ,   2 2  1 1 and circumradius = (1 − 2 )2 + (3 − 2 )2 = 2 2 Hence, the equation of circumcircle is x 2 + y 2 − 3 x − 5 y + 8 = 0.

45. On solving y = x, x 2 + y 2 − 2 x = 0, we get ( x, y ) = (0, 0 ), (1, 1) So, the ends of the diameter are (0, 0), (1, 1). Hence, the equation of the required circle is ( x − 0 )( x − 1) + ( y − 0 )( y − 1) = 0 or x2 + y2 = x + y

2

 13  x 2 +   = 4 or  4λ 

PC = 12 2 + 52 = 13 ∴ and

PA = 13 − 5 5 PB = 13 + 5 5 GM = PA ⋅ PB = (13 − 5 5)(13 + 5 5 ) = 2 11

3 47. Here, centre of circle =  , 0 4  3 and radius of circle = 4 2

121 137 + 1= 16 4 137 − 3 137 3 and PA = − = >2 4 4 4 Hence, there are two tangents. =

48. Let (a, b) should be an internal or end point of the line segment joining A, B. So, the x-coordinate a will vary from 2 to 3 i.e. from the x-coordinate of A to the x-coordinate of B. Similarly, b ∈ (−3, 2 )

a + 1 > 0 and a2 + (a + 1)2 − 25 < 0 −1 < a < 3

50. The given point is an interior point 2

2

r  r    ∴  −5 +  +  −3 +  − 16 < 0   2 2 ⇒ r 2 − 8 2 r + 18 < 0 ⇒ 4 2 − 14 < r < 4 2 + 14 So, the point is on the major segment. This, the centre and the point are on the same side of the line x + y = 2. r r −5 + − 3+ d ∴

3 + 25 − n 2 > 42 + 33

⇒

25 − n 2 > 2 ⇒ 25 − n 2 > 4

∴

n 2 < 21 ⇒ − 21 < n < 21

Therefore, common values of n should satisfy − 21 < n < 21 But n ∈ I. So, n = − 4, −3,…,4.

13

53. The length of the perpendicular from the centre (0,0) to 2 1 + m2

.

The radius of the circle = 1. For the line to cut the circle at distinct or coincident points, 2 ≤1 1 + m2 ⇒

1 + m2 ≥ 4

⇒

m2 ≥ 3

54. The line joining (4, 3) and (2, 1) is also along a diameter. So, the centre is the intersection of the diameters. 3−1 = ( x − 4) 2 x − y = 2 and y − 3 = 4−2 Solving these, we get centre = (1, 0 ) Also, radius = Distances between (1, 0) and (2, 1) = 2 units

55. PA = Length of the tangent from P to the circle

49. y = 25 − x 2 , y = 0 bound the semi-circle above the X-axis. ∴ ⇒

2

It should have two real and distinct values. 2 13 13  13  or λ > So, 4 −   > 0 ⇒ λ < −  4λ  8 8 the line is

3  PC =  + 2 + (0 − 1)2 4 

∴

 13  x = 4−    4λ 

Circle

52. On solving the equation,

radius = 52 + 7 2 + 51 = 5 5

and

= (−2 )2 + (−3)2 − 2(−2 ) − 10(−3) + 1 = 48 = 4 3 units

56. Any line through P(−2, − 1) is y + 1 = m( x + 2 ). It touches the circle, if

2m − 1 1 + m2

⇒

= 1.

m = 0,

4 3

Targ e t E x e rc is e s

46. The centre C of the circle is (5, 7)

∴ The equation of PB is

4( x + 2) 3 ⇒ 4x − 3y + 5 = 0 A point on PB is (−5, − 5). Its image by the line y = − 1is P(−5, 3). So, the equation of the incident ray is 3+ 1 y − 3= ( x + 5) −5 + 2 y + 1=

⇒

4 x + 3 y + 11 = 0

vertices are A(r , 0 ), B(r cos 60 ° , r sin 60 ° ), C(− r cos 60 ° , r sin 60 ° ), D(− r , 0 ), E(− r cos 60 ° ,− r sin 60 ° ) and F(r cos 60 ° , r sin 60 ° ). If P = ( x, y ), then ΣPA2 = 6 a2

57. The

⇒ ⇒

x 2 + y 2 + r 2 = a2 x 2 + y 2 = ( a2 − r 2 )2

58. Let ABC be the triangle formed by the lines

a x + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0. The equation of a second degree curve passing through the vertices A, B and C is λ (ax + by + c )(bx + cy + a) + µ(bx + cy + a) (cx + ay + b) + γ(cx + ay + b)(ax + by + c ) = 0 …(i)

763

and

This will represent a circle, if Coefficient of x 2 = Coefficient of y 2 and coefficient of xy = 0 …(ii) ⇒ λ (ab − bc ) + µ(bc − ca) + γ(ca − ab) = 0 and λ (b2 + ac ) + µ(c 2 + ab) + γ(a2 + bc ) = 0 …(iii)

⇒ ⇒

A

⇒

y+

0

cx + a

a=

B

C

ax +by + c = 0

The circle given in Eq.(i) will pass through the origin (0, 0), if …(iv) λ ac + µ ab + γ bc = 0 Eliminating λ,µ and γ from Eqs. (ii), (iii) and (iv), we get b + ac c + ab a + bc 2

2

ab − bc ac

bc − ca ca − ab = 0 ab bc

Ta rg e t E x e rc is e s

b2

c2

ab − bc ac

⇒

2

bc − ca ca − ab = 0 ab bc [applying R1 → R1 − R3 ]

⇒

ac

b2 a2 ca + − bc ac ab bc b2

⇒

⇒

c2

a2

c2 a2 − ca − ab = 0 ab b2

bc

c2

ab bc ca − bc ac ab bc ac

a2

ca ab = 0 ab bc

b2 c 2 ab bc ac

b2 c 2 a2 a2 ca = bc ca ab ab bc ac ab bc

59. The equation of a second degree curve passing through the vertices of the quadrilateral ABCD is D

C L3 = 0

µ L4 = 0 A

L1 = 0

L2 = 0

λ

B

...(i) λL1L3 + µL2 L4 = 0 or λ(a1 x + b1 y + c1 )(a3 x + b3 y + c 3 ) + µ(a2 x + b2 y + c 2 )(a4 x + b4 y + c 4 ) = 0 ...(ii) This will represent a circle, if Coefficient of x 2 = Coefficient of y 2

764

⇒ ⇒

(a1b2 − a2 b1 )(a3 a4 + b3 b4 ) = − (a1a2 + b1b2 )(a3 b4 − a4 b3 ) (a1a3 − b1b3 )(a2 b4 + a4 b2 ) = (a2 a4 − b2 b4 )(a1b3 + a3 b1 ) a1a3 − b1b3 a b + a3 b1 ⇒ = 1 3 a2 a4 − b2 b4 a2 b4 + a4 b2

60. Let A( x1, y1 ), B( x2 , y2 ), C( x3 , y3 ) and D( x4 , y4 ) be the

a2

b2 c 2 ab bc

λ (a1b3 + b1a3 ) + µ(a2 b4 + b2 a4 ) = 0  a a − b2 b4   a b + a4 b2  λ = − 2 4  = − 2 4   a1a3 − b1b3   a1b3 + a3 b1  µ a2 a4 − b2 b4 a2 b4 + a4 b2 = a1a3 − b1b3 a1b3 + a3 b1 a1a3 − b1b3 a b + a3 b1 = 1 3 a2 a4 − b2 b4 a2 b4 + a4 b2

Aliter The quadrilateral ABCD will be a cyclic quadrilateral, if θ + φ + π ⇒ θ = π − φ ⇒ tan θ = − tan φ a a a a − 2 + 1 − 4 + 3 b2 b1 b4 b3 ⇒ =− a a a a 1+ 2 × 1 1+ 4 × 3 b2 b1 b4 b3 a1b2 − a2 b1 a b −a b =− 3 4 4 3 ⇒ a1a2 + b1b2 a3 a4 + b3 b4

y+ +c

b=

0

bx

Objective Mathematics Vol. 1

13

and coefficient of xy = 0 ⇒ λ (a1a3 − b1b3 ) + µ(a2 a4 − b2 b4 ) = 0

vertices of ∆ABC. Again, let P(h, k ) be a variable point such that [constant] PA2 + PB 2 + PC 2 = λ2 ⇒

( x1 − h )2 + ( y1 − k )2 + ( x2 − h )2 + ( y2 − k )2

⇒

3 h + 3k − 2 h( x1 + x2 + x3 ) − 2 k ( y1 + y2 + y3 )

+ ( x3 − h )2 + ( y3 − k )2 = λ2 2

2

+ x12 + x22 + x32 + y12 + y22 + y32 − λ2 = 0 Hence, the locus of (h, k) is 3 x 2 + 3 y 2 − 2( x1 + x2 + x3 )x − 2 ( y1 + y2 + y3 )y + x12 + x22 + x32 + y12 + y22 + y32 − λ2 = 0  y + y2 + y3   x + x2 + x3  ⇒ x2 + y2 − 2  1 y  x −2  1     3 3 1 2 + ( x1 + x22 + x32 + y12 + y22 + y32 − λ2 ) = 0 3 which represents a circle.

61. Let the equation of the line L1 be x y ...(i) + =1 a b It cuts the coordinate axes at P(a, 0 ) and Q(0, b). The x y equation of the line L2 is − = λ, where λ is b a parameter. x y ⇒ − =1 bλ aλ This cuts the coordinate axes at R(bλ , 0 ) and S (0, − aλ ). Equation of lines PS and QR are x y x y − = 1 and + =1 a aλ bλ b Let T (h, k ) be the point of intersection of PS and QR. h k h k Then, − = 1 and + =1 a aλ bλ b h k h k and ⇒ −1= = 1− a aλ bλ b

⇒ ⇒

Let S (h, k ) be a variable point on OR such that OS = r1. Then, h = r1 cos θ, k = r1 sin θ Y

Y

R (r cos q, r sin q) S (h, k)

r

Q (0, b)

q

X′

L2

13 Circle

h−a 1 b−k 1 and = = k λ h λ h−a b−k = k h h 2 + k 2 − ah − bk = 0

⇒

X

O L = ax + by + c = 0

P (a, 0)

X′

O

R

S

X

L1 s

Y′

It is given that OR ⋅ OS = λ2

Hence, the locus of T (h, k ) is x 2 + y 2 − ax − by = 0 which represents a circle passing through the origin.

62. Let the coordinates of A and B be (a, 0) and (0, b), respectively. Then, the equation of the circle passing through O(0, 0 ), A(a, 0 ) and B(0, b) is ...(i) x 2 + y 2 − ax − by = 0 It is given that the radius of circle (i) is r. a2 b2 ∴ + r2 = 4 4 2 2 2 …(ii) ⇒ a + b = 4r x y The equation of sides AB is + = 1 a b Let P(h, k )be the foot of the perpendicular from O to AB. Then, k −0 b−0 × = −1 h−0 0−a h k [say] = =λ ⇒ b a k h and b = a= ⇒ λ λ x y Since, (h, k ) lies on + = 1. a b h k  h k + = 1 ⇒ λ +  = 1 ∴  k h a b hk ⇒ λ= 2 h + k2

⇒

rr1 = λ2

⇒

h=

λ2 cos θ r

and

…(i)

λ4 (cos 2 θ + sin 2 θ ) r2 λ4 ...(ii) h2 + k 2 = 2 r Since, R(r cos θ, r sin θ ) lies on the line ax + by + c = 0. ∴ (a cos θ + b sin θ ) r + c = 0  ahr bkr  [from Eq. (i)] ⇒  2 + 2 r + c = 0 λ λ   ah + bk  2 …(iii) ⇒  r + c = 0  λ2  h2 + k 2 =

⇒

On eliminating r from Eqs. (ii) and (iii), we get λ2 (ah + bk ) 2 +c =0 h + k2 λ2 (ah + bk ) = 0 c Thus, the locus of (h, k ) is λ2 x2 + y2 + (ax + by ) = 0 c which represents a circle. ⇒

h2 + k 2 +

64. Let OAB be the triangle in which the circle x 2 + y 2 − 4 x − 4 y + 4 = 0 is inscribed. Let equation of AB be

x y + = 1. a b

Y

h2 + k 2 h2 + k 2 and b = h k On substituting the values of a and b in Eq. (ii), we get (h 2 + k 2 )2 (h 2 + k 2 )2 + = 4r 2 h2 k2 ⇒ (h 2 + k 2 )3 = 4h 2 k 2 r 2

∴

λ2 r λ2 sin θ k= r

⇒ r1 =

Targ e t E x e rc is e s

Y′

a=

B (0, b)

C

Hence, the locus of (h, k ) is ( x 2 + y 2 )3 = 4 x 2 y 2 r 2

x y a + b =1

2 X′

(a, 0)A

O

X

63. Let the fixed point O be at the origin (0, 0) and the fixed line L be ax + b y + c = 0 such that c ≠ 0. Again, let R be a variable point on line Lsuch thatOR = r and ∠XOR = θ. Then, the coordinates of R are (r cos θ, r sin θ).

Y′

Since, AB touches the circle x 2 + y 2 − 4x − 4y + 4 = 0

765

Objective Mathematics Vol. 1

13

Let AB = rB and AC = rC . Then, rB and rC are the roots of Eq. (i). Therefore, rB + rC = − 2 r1 cos θ and rB rC = r12 − r22

Therefore, 2 2 + −1 a b =2 1 1 + a2 b2  2 2  + − 1 a b  =2 1 1 + a2 b2

∴

⇒

= 4r12 cos 2 θ − 4r12 + 4r22 Now, OA2 + OB 2 + BC 2 = r12 + r22 + 4r12 cos 2 θ − 4r12 + 4r22

2 a + 2 b − ab + 2 a + b = 0 2

BC 2 = (rC − rB )2 = (rC + rB )2 − 4rB rC

[QO (0, 0 )andC (2, 2 )lies on the same 2 2 side of AB,therefore + − 1 < 0] a b (2 b + 2 a − ab) − =2 a2 + b2

⇒ ⇒

∴

2

...(i)

= 5r22 − 3r12 + 4r12 cos 2 θ Now, 0 ≤ cos θ ≤ 1 2

⇒

0 ≤ 4r12 cos 2 θ ≤ 4r12

⇒

5 r22

− 3 r12 ≤ 5r22 − 3r12 + 4r12 cos 2 θ ≤ 5r22 − 3r12 + 4r12

⇒

5r22 − 3r12 ≤ OA2 + OB 2 + BC 2 ≤ 5r22 + r12

Let P (h, k )be the circumcentre of ∆OAB.

Ta rg e t E x e rc is e s

Since, ∆OAB is a right angled triangle. So, its circumcentre is the mid-point of AB. a h= ∴ 2 b and k= 2 ⇒ a = 2 h and b = 2 k

⇒ OA + OB + BC 2

h + k − hk +

∈[5 r22

− 3r12 ,

[from Eq.(ii)] 5 r22

+

r12 ]

slide along OX and OY respectively and x 2 + y 2 + 2 gx + 2 fy + c = 0 ...(ii)

be the circle passing through A1, A2 , B1 and B2 . Then, A1 A2 = Intercept on X-axis ⇒ A1 A2 = 2 g 2 − c ⇒

4k + 4k − 4hk + 2 4h 2 + 4k 2 = 0 ⇒

2

66. Let A1 A2 and B1B2 be two rods of lengths a and b which

Form Eqs. (i) and (ii), we get

a = 2 g2 − c

...(i)

and B1B2 = Intercept on Y-axis

h +k =0 2

2

2

Y

So, the locus of P (h, k ) is x + y − xy + x 2 + y 2 = 0

B2

But the locus of the circumcentre is given to be x + y − xy + k x 2 + y 2 = 0

C (h, k)

b

Thus, the value of k is 1. X′

65. Let the equation of line AB be x − r1 y − 0 = =r cos θ sin θ

B1 O

a A1

A2

X

Y′

Y

C

X

O

A (r1,0)

Hence, the locus of the centre of the circle is a2 − b2 = 4 ( x 2 − y 2 )

The coordinates of any point on this line are (r1 + r cos θ, r sin θ ) If it lies on x + y = 2

r22 , 2

67. On solving the equations x 2 + y 2 = 1 and

then

(r1 + r cos θ ) + (r sin θ )2 = r22

766

⇒

r 2 + 2 rr1 cos θ + r12 − r 22 = 0

b=2 f −c

Thus, the locus of (− g, − f ) is a2 − b2 = 4 ( x 2 − y 2 )

Y′

2

⇒

...(ii)

2

On eliminating c from Eqs. (i) and (ii), we get a2 − b2 = 4 (g 2 − f 2 )

x2 + y2 =r12

x2 + y2 =r22

B1B2 = 2 f 2 − c

Here, c is the variable which is to be eliminated to find the locus of the centre (− g, − f ) of the circle x 2 + y 2 + 2 gx + 2 fy + c = 0

B X′

⇒

…(i)

x 2 + y 2 + 2 x + 4y + 1 = 0

We find that the two circles intersect at P (−1, 0 ) and 4 3 Q  , − . 5 5

Y

(–1, 0) P

X'

α

B

A α

O

X β Q 3 ,– 5 2

(

π –α C 2 (–1,–2)

( )

It is given that the point (1, 4) lies inside the circle x 2 + y 2 − 6 x − 10 y + λ = 0 Therefore,

)

17 − 6 − 40 + λ < 0 λ < 29

⇒

…(ii)

From Eqs. (i) and (ii), we get λ ∈(25, 29)

69. Let the line lx + my + 1 = 0 touch the circle

Y'

( x − a)2 + ( y − b)2 = r 2

Let y − 0 = m ( x + 1) or y = m ( x + 1) be a chord AB passing through P meeting the two circles at Aand B. It is given that chords AP and BP make the same angle 2α at the centres O (0, 0 ) and C (−1, − 2 ) respectively of the given circles.

⇒

We have,

⇒ l 2 (a2 − r 2 ) + m2 (b2 − r 2 ) + 2 ablm + 2 al

Slope of OP = 0 slope of AP = m

and

⇒ Also, where, We have,

m = cot α 2 slope of CP = m′ m′ = 0

+ 2 bm + 1 = 0

a2 − r 2 = 4

…(i)

Slope of BP = m and angle between CP and BP is π   − α . 2  2 m− π  m ′ Therefore, tan  − α  = 2 2  1+ m × m′ mm′ − 2 1 = = m′ + 2 m m 1 ...(ii) cot α = ⇒ m From Eqs. (i) and (ii), we get 1 m= m ⇒ m| m| = 1 ⇒ m2 = 1 ⇒

la + mb + 1  2 = r 2  l +m  (la + mb + 1)2 = r 2 (l 2 + m2 )

On comparing this with the equation 4l 2 − 5m2 + 6l + 1 = 0, we have

π  m = tan  − α  2 

⇒

Then,

m±1

Clearly, for m = − 1, we obtain chord PAB when Aand B lie on the same side of P. So, we take m = 1.

b2 − r 2 = − 5 2 ab = 0 2a = 6 2b = 0 a = 3, b = 0 r2 = 5

and ⇒ and

Clearly, above equations are consistent. Thus, the line touches the fixed circle ( x − 3)2 + ( y − 0 )2 = 5

70. Let the equation of the circle be x 2 + y 2 + 2 gx + 2 fy + c = 0 If lx + my + 1 = 0 touches this circle, then − lg − mf + 1 l 2 + m2

= g2 + f 2 − c

⇒ (f 2 − c )l 2 + (g 2 − c ) m2 + 2 l (g − mfg ) + 2 mf − 1 = 0 ⇒ (c − f 2 )l 2 + (c − g 2 ) m2 + 2 l (mfg − g ) − 2 mf + 1 = 0 On comparing this equation with al 2 − bm2 + 2dl + 1 = 0, we get c − f 2 = a, c − g 2 = − b ⇒ and

mfg − g = d and f = 0 c = a, g = − d c −d2 = −b

⇒

a −d2 = −b

On putting m = 1in y = m( x + 1,) we obtain y=x+1

⇒ a + b = d , which is the required condition.

which is the equation of the chord APB.

Hence, the fixed circle is x 2 + y 2 − 2dx + a = 0.

68. We know that, the circle x + y + 2 gx + 2 fy + c = 0 does not intersect or touch with the coordinate axes, if g 2 − c < 0 and f 2 − c < 0 2

Targ e t E x e rc is e s

( π2 –α)

13 Circle

Therefore, the circle x 2 + y 2 − 6 x − 10 y + λ = 0 will not intersect or touch the coordinate axes, if 9 − λ < 0 and 25 − λ < 0 ⇒ λ > 9 and λ > 25 …(i) ⇒ λ > 25

Clearly, point P has integral coordinates.

2

71. The equations of the given circles are x2 + y2 − 2 x − 2 y + 1 = 0

…(i)

x + y − 16 x − 2 y + 61 = 0

…(ii)

2

and

2

2

767

Objective Mathematics Vol. 1

13

The coordinates of the centres and radii of these two circles are C1(1, 1), r1 = 1 and C2 (8, 1), r2 = 2 respectively. For the line y = 2 x + a not to touch or intersect circle (i), we must have 1 + a > 1  5 [Q length of perpendicular from centre C1 > radius r1] Y

C1 (1, 1) X′

On solving Eqs. (i) and (ii), we obtain the coordinates of C as  a2 + b2 a2 + b2  ,   a+ b   a+ b Now, CP = Length of perpendicular from C on bx − ay = 0  a2 + b2   a2 + b2  b  −a   a+ b   a+ b  = b2 + (− a)2

C2 (8, 1) X

O

Similarly, the equation of a line through Q and perpendicular to OQ is −b y −a= ( x − b) a ⇒ bx + ay = a2 + b2

y = 2k + a

=

Y′

| a + 1| > 5 a ∈ (−∞, − 5 − 1) ∪ ( 5 − 1, ∞ )

Ta rg e t E x e rc is e s

⇒ ⇒

...(iii)

Similarly, for the line y = 2 x + a not to touch or intersect circle (ii), we must have 15 + a > 2  5  |15 + a| > 2 5 ⇒ …(iv) a ∈ (−∞, − 15 − 2 5, − 15 + 2 5 ) ⇒ The line y = 2 x + a will be between the circles, if their centres C1 and C2 are on the opposite side of it.

Thus, the equation of the required circle is 2 2   x 2 + b2  a2 + b2  + − x y −     a+ b  a+ b    2

 a2 + b2   2 2  a b + ⇒ x 2 + y 2 − 2( x + y )   + (a2 + b2 ) = 0  a+ b  b− a = a+ b

73. The parametric form of OP is

x−0 y−0 = cos π / 4 sin π / 4

Y a y= x b (b, a)Q

| b − a| 2 a + b2 a+ b

Y b y= x a

B(x2,y2) M 3√2

C P(a, b)

y=x

O X′

(h,k)C

X

(x1,y1) A X

X′

4√2 P (–4,–4)

Y′

∴ ⇒ ⇒

(2 − 1 + a) (16 − 1 + a) < 0 (a + 1)(a + 15) < 0 a ∈ (−15, − 1)

Y′

Since, OP = 4 2 units ...(v)

From Eqs. (iii), (iv) and (v), we get a ∈ (−15 + 2 5, − 5 − 1)

72. The equations of OP and OQ are y =

768

b a x and y = x, a b

So, the coordinates of P are given by x−0 y−0 = = −4 2 cos π / 4 sin π / 4 ⇒ x = − 4, y = − 4 Thus, the coordinates of P are (−4 ,−4).

respectively.

Let C (h, k) be the centre of the circle and r be its radius.

Clearly, centre C of the required circle is the point of intersection of CP and CQ.

Now,

The equation of a line passing through P and perpendicular to OP is a y − b = − ( x − a) b ⇒ ax + by = a2 + b2

⇒ ⇒ ⇒ Also, ⇒

CP ⊥ OP k+4 × 1= −1 h+4 k + 4= − h −4 h + k = −8 CP 2 = (h + 4)2 + (k + 4)2 (h + 4)2 + (k + 4)2 = r 2

...(i) ...(ii)

AC = AM + CM 2

2

⇒ 2

⇒

 h + k r 2 = (3 2 )2 +    2 

⇒

 −8  r = 18 +    2

⇒

2

⇒

x 2 + y2 = 0 y2 = − x 2 x22 + y22 = 8 2 x22 = 8 ⇒ x22 = 4 ⇒

∴

...(iii)

Thus, the equations of the circles are ( x + 9)2 + ( y − 1)2 = (5 2 )2

y2 = − x 2 ⇒ y2 = m 2

75. Let P(h , k )be the point of intersection of tangents drawn from P to the circle x 2 + y 2 = a2 The equation of any tangent to the circle x 2 + y 2 = a2 is y = mx + a 1 + m2 Y

( x − 1)2 + ( y + 9)2 = (5 2 )2

or

x + y + 18 x − 2 y + 32 = 0

and

x 2 + y 2 − 2 x + 18 y + 32 = 0

2

x2 = ± 2

Hence, the possible coordinates of B are (2, − 2 ) and (−2, 2 ).

From Eqs. (i) and (iv), we get h = − 9, k = 1 or h = 1, k = − 9

and

2

Q

Clearly, (−10, 2 ) lies in the interior of x 2 + y 2 + 18 x − 2 y + 32 = 0

a

X′

Hence, the equation of the required circle is x 2 + y 2 + 18 x − 2 y + 32 = 0

O

P(h, k) X

R

x2 + y2 = a2

74. Let A( x1, y1 )be a point on the circle x 2 + y 2 = 8 such that the tangent at A passes through P(4, 0 ).

Y′

Y

If it passes through P(h, k ) then k = mh + a 1 + m2

B (–2, 2) X′

4

A(x1,y1)

O

L 4

P(4, 0)

⇒

(k − mh )2 = a2 (1 + m2 )

⇒ m2 (h 2 − a2 ) − 2 mhk + k 2 − a2 = 0 X

B(2, –2)

Let m1 and m2 be the roots of this equation. Then, 2 hk m1 + m2 = 2 h − a2 and

m1m2 =

Y′

The equation of the tangent at A( x1, y1 ) to the circle x 2 + y 2 = 8 is xx1 + yy1 = 8 It passes through, P (4, 0 ). ∴ 4 x1 = 8 ⇒

x1 = 2

Since, ( x1, y1 ) lies on x + y = 8. 2

x12 ⇒ ⇒ ⇒ ⇒

13

2

+

4+

y12 y12 y12

=8 =8 =4

y1 = ± 2 y1 = 2 [Q A( x1, y1 ) lies in first quadrant]

Targ e t E x e rc is e s

⇒

⇒

r 2 = 18 + 32 r = 5 2 units CP = r h−k h−k =r ⇒ =5 2 2 2 h − k = 10 ⇒ h − k = ± 10

8 − 4 ( x 2 + y2 ) = 8 [Q B( x2 , y2 ) lies on x 2 + y 2 = 8, x22 + y22 = 8 ]

⇒ ⇒ Now,

2

2

⇒ ⇒ Also,

x22 + y22 − 4 ( x2 + y2 ) = 8

Circle

In ∆ACM, we have

k 2 − a2 h 2 − a2

Clearly, m1 and m2 represent slopes of two tangents drawn from P which are inclined at an angle α with each other. m − m2 tan α = 1 1 + m1m2 ⇒

(1 + m1m2 ) tan α = (m1 + m2 )2 − 4m1m2 2

 4h 2 k 2  k 2 − a2    k 2 − a2  2 tan α ⇒ 1 + 2 = − 4  2    2 2 2 h − a2   h − a2     (h − a ) ⇒ (h 2 − k 2 − 2 a2 )2 tan 2 α = 4a2 (k 2 + k 2 − a2 ) Hence, the locus of P(h, k ) is ( x 2 + y 2 − 2 a2 )2 tan 2 α = 4a2 ( x 2 + y 2 − a2 )

76. Let P( x1, y1 ), Q( x2 , y2 ), R( x3 , y3 ) and

Thus, the coodinates of A are (2, 2).

S ( x4 , y4 ) be four points in a plane. Then,

Let B ( x2 , y2 ) be a point on the given circle such that AB = 4. Then, AB = 4 ⇒ ( x2 − 2 )2 + ( y2 − 2 )2 = 42

the equation of a circle with RS as diameter is ( x − x 3 ) ( x − x 4 ) + ( y − y3 ) ( y − y4 ) = 0 ∴ { P, RS } = ( x1 − x3 ) ( x1 − x4 ) + ( y1 − y3 ) ( y1 − y4 )

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Objective Mathematics Vol. 1

13

Similarly, we have { P, QS } = ( x1 − x2 ) ( x1 − x4 ) + ( y1 − y2 ) ( y1 − y4 ) {Q, PR} = ( x2 − x1 ) ( x2 − x3 ) + ( y2 − y1 ) ( y2 − y3 ) and {Q, RS } = ( x2 − x3 ) ( x2 − x4 ) + ( y2 − y3 ) ( y2 − y4 ) Hence, { P, RS } − { P, QS } + {Q, PR} − {Q, RS } = 0

77. The equation of the chord of contact of tangent drawn from a point {α , β} to the circle x 2 + y 2 = r12 is αx + βy = r12 .

∴

a2 + β 2

⇒

r22

(α + β

) = (r12

⇒

a2 = h 2 + k 2 + (h − c )2 + k 2 [from Eq. (i)]

⇒

2(h + k 2 ) − 2 hc + c 2 − a2 = 0 2

Hence, the locus of P(h, k ) is 2( x 2 + y 2 ) − 2cx + c 2 − a2 = 0

Then, the equation of the chord AB is hx + ky = h 2 + k 2

r22 (α 2 + β 2 ) = (aα + bβ − r12 )2 2

a2 = h 2 + k 2 + PB 2

x 2 + y 2 = 4.

= r22

⇒

2

⇒

80. Let P (h, k ) be the mid-point of a chord AB of the circle

This line touches the circle ( x − a)2 + ( y − b)2 = r22 aα + bβ − r12

Also, in ∆OPB, we have OB 2 = OP 2 + PB 2

Y

− aα − bβ )

2

B

78. Let (h, k ) be the mid-point of a chord AB of the circle

R

x + y = a . Then, the equation of AB is 2

2

2

hx + ky − a2 = h 2 + k 2 − a2 ⇒

hx + ky = h + k 2

[QT = S ]

2

90° Q

X′

...(i)

x2= 2 (x + y)

P (h, k)

A

X

O

x2 + y2 = 4

Ta rg e t E x e rc is e s

O π/2 A

Y′

Let QR be the segment intercepted by the chord AB on the curve x 2 = 2( x + y ) such that ∠QOR = 90 °.

B

P (h,k)

The combined equation of OQ and OR is  hx + ky  x 2 = 2( x + y )  2   h + k2

The combined equation of OA and OB is 2  hx + ky  x 2 + y 2 = a2  2   h + k2 ⇒ (h 2 + k 2 )2 ( x 2 + y 2 ) − a2 (hx + ky )2 = 0

⇒

x 2 (h 2 + k 2 ) = ( x + y )(hx + ky )

which represents a pair of perpendicular lines.

Since, OA and OB will be perpendicular, if Coefficient of x 2 + Coefficient of y 2 = 0

∴Coefficient of x 2 + Coefficient of y 2 = 0

⇒(h 2 + k 2 )2 − a2 h 2 + (h 2 + k 2 )2 − a2 k 2 = 0

⇒

⇒

2(h + k ) − a (h + k ) = 0

⇒

2(h + k ) − a = 0

Hence, the locus of P(h, k ) is x2 + y2 − 2 x − 2 y = 0

2

2 2

2

2

2

2

2

2

locus of (h, k ) is 2( x 2 + y 2 ) − a2 = 0

So,

79. Let P(h, k ) be the mid-point of a chord AB of the circle x 2 + y 2 = a2 which subtends a right angle at the point C(c , 0 ). Since, mid-point of the hypotenuse of a right triangle is equidistant from the vertices. ∴

h2 + k 2 − 2h − 2k = 0

81. Let P(h, k )be the mid-point of chord BC intercepted by the circle x 2 + y 2 = a2 on the secant drawn through A( x1, y1 ). Then, the equation of chord BC is hx + ky = h 2 + k 2 Y

PA = PB = PC x2 + y 2 =a 2

Y

X′ A X′

(h, k) P

C (c, 0) O

Y′

⇒

C

X

B

770

A (x1,y1) B

PA = PB = (h − c )2 + k 2

X

O P (h,k)

Y′

Since, it passes through A( x1, y1 ). ∴ hx1 + ky1 = h 2 + k 2 Hence, the locus of (h, k) is x 2 + y 2 = xx1 + yy1

…(i)

Y

Let P (h, k ) be the centre of circle C and r be its radius. Since, circle C touches circle C1 internally and the circle C2 externally. ∴ OP = r1 − r and PQ = r + r2 h 2 + k 2 = r1 − r ⇒ (h − α )2 + (k − β )2 = r + r2

and ⇒

h 2 + k 2 + (h − α )2 + (k − β )2 = r + r2

R

y+x=0

[on eliminating r]

X′

C α, β 2 2

(

Thus, the locus of (h, k ) is

X

)

x 2 + y 2 + ( x − α )2 + ( y − β )2 = r1 + r2 Q

Clearly, it represents an ellipse having its foci at O (0, 0 ) and Q (α , β ) and the length of the major axis is r1 + r2 .

S(t, – t) P (a, b) y + x = 0

Y′

13 Circle

1 + 2a 1 − 2a and β = . 2 2 Then, the equation of the circle is 2 x 2 + 2 y 2 − 2αx − 2βy = 0 ⇒ x 2 + y 2 − αx − βy = 0

82. Let α =

85. We have,

The coordinates of the given point are (α , β). Clearly, it lies on the circle (i). C1

Then, its equation is α β t x − t y − ( x + t ) − ( y − t ) = 2t 2 − αt + βt 2 2 α β 4t 2 − αt + βt   ⇒ x t −  − y t −  =   2 2 2 ⇒ 2αt − α 2 + 2βt − β 2 = 4 t 2 − αt + βt 2 ⇒ 4 t − 3 α t + 3 β t + (α 2 + β 2 ) = 0 ⇒ 4 t 2 − 3(α − β )t + (α 2 + β 2 ) = 0

polar of ( x1, y1 ) with respect to the circles x 2 + y 2 − 2 kx + c 2 = 0 is xx1 + yy1 − k( x + x1 ) + c 2 = 0 ⇒ ( xx1 + yy1 + c 2 ) − k( x + x1 ) = 0 This is a straight line passing through a fixed point which is the point of intersection of the lines xx1 + yy1 + c 2 = 0 and x + x1 = 0  x2 − c 2  Hence, the point of intersection is  − x1, 1 . y1  

and

x ( x − α )2 + ( y

...(i) ...(ii)

C P (h,k)

C1 O(0,0) Q(a, b)

C2

Slope of C1C2 =

...(i)

Since, C1 and C2 are points on Eq. (i) at a distance of 5 units from P. So, the coordinates of C1 and C2 are given by x −1 y −2 = =±5 4/ 5 3/ 5 ⇒ x = 1 ± 4 and y = 2 ± 3 Thus, the coordinates of C1 and C2 are (5, 5) and (−3, − 1), respectively. Hence, the equations of the two circles are ( x − 5)2 + ( y − 5)2 = 52 and ( x + 3)2 + ( y + 1)2 = 52

86. The equations of the given circles are

84. Let the equations of the circles C1 and C2 be + y = r12 − β )2 = r22

4 3

3 4 If C1C2 makes an angle θ with X-axis, then 4 3 and sin θ = cos θ = 5 5 So, the equation of C1C2 in parametric form is x −1 y −2 = 4/ 5 3/ 5

∴

83. Let (x1, y1) be the given point. Then, the equation of the

2

5 C P (1,2) 2

Slope of the tangent = −

Since, t is real. Therefore, roots of the above equation are real. Hence, 9 (α − β )2 − 16 (α 2 + β 2 ) = 0 ⇒ 9 × 2 a2 − 8 (1 + 2 a2 ) > 0 ⇒ 2 a2 − 8 > 0 ⇒ a2 − 4 > 0 ⇒ a ∈ (−∞, − 2 ) ∪ (2, ∞ )

2

5

Targ e t E x e rc is e s

Let the chord PQ be bisected at S (t , − t ).

( x − a)2 + y 2 = b2 ⇒ x + y − 2 ax + a2 − b2 = 0 and ( x + a)2 + y 2 = c 2 2 2 ⇒ x + y + 2 ax + a2 − c 2 = 0 2

2

...(i) ...(ii)

Let P (h, k )be a point of contact of common tangents to these circles such that it lies on Eq. (i). Then, ...(iii) h 2 + k 2 − 2 ah + a2 − b2 = 0 ∴Equation of the tangent at P (h, k ) to x 2 + y 2 − 2 ax + a2 − b2 = 0 is hx + ky − a ( x + h ) + a2 − b2 = 0 ⇒ x (h − a) + ky − ah + a2 − b2 = 0

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Objective Mathematics Vol. 1

13

This will touch circle (ii), if − a (h − a) + k × 0 − ah + a2 − b2 (h − a)2 + k 2

On solving these equations, we get g = − 8, f = − 9 and c = − 4

=c

On putting the values of g, f and c in Eq. (i), we obtain the required circle as x 2 + y 2 − 16 x − 18 y − 4 = 0.

−2 ah + 2 a2 − b2 =c b

⇒ ⇒

89. The given circles are

2 ah − (2 a2 − b2 ) = ± bc

⇒

2 ah − (a2 − b2 ) = a2 ± bc

⇒

h + k = a ± bc 2

Since,

2

and

2

2 ah − (a2 − b2 ) = h 2 + k 2 [from Eq. (iii)]

Hence, P (h, k ) lies on x + y = a ± bc . 2

2

2

87. Clearly, lines x + y = 2 and x − y = 2 intersect at (2, 0 ). So, the centre of the required circle lies on the bisector of the angle containing the origin.Let r be the radius of the required circle. Then, the the coordinates of the centre are C1(1 + r , 0 ) or C2 [−(1 + r ), 0 ]. As its touches x + y = 2.

Ta rg e t E x e rc is e s

x+y=2 (0 x2 + y2 =1 ,2) (1, 0) C2[–(1 + r), 0] C1 (2, 0) X O (–1, 0) (1 + r, 0) x–y=2 (0, –2)

⇒ ⇒ or ⇒ or

...(iii)

The equation of the family of circles coaxial with the circles (i) and (ii) is S1 + λ (S1 − S 2 ) = 0 ⇒ ( x 2 + y 2 + 4 x + 2 y + 1) + λ (10 x − 2 y + 5) = 0 ...(iv) ⇒ x 2 + y 2 + 2 x (2 x + 5λ ) + 2 (1 − λ )y + 1 + 5λ = 0

coaxial circles having common radical axis L = 0. So, the family of circles x 2 + y 2 + 2 ax + 2 by + 2 λ (ax − by ) = 0 represents a

|± (1 + r ) + 0 − 2| =r 2 ± (1 + r ) − 2 = ± 2 r 1+ r − 2 = − 2 r −1 − r − 2 = − 2 r 1 r= 2 +1 3 r= 2 −1

family of coaxial circles having common radical axis ax − by = 0. Let ...(i) x 2 + y 2 + 2 gx + 2 fy + c = 0 be the circle orthogonal to the above system of coaxial circles.

88. Let the required circle be x 2 + y 2 + 2 gx + 2 fy + c = 0 Since, it is orthogonal to three given circles. 3   ∴ 2  − g + f  = c − 7 ⇒ −2 g + 3f = c − 7  2  5  5 ⇒ 2  g − f  = c + 9 ⇒ 5g − 5f = c + 9 2 2  9  7 ⇒ 2  g − f  = c + 29 ⇒ 7 g − 9f = c + 29 2 2 

Thus circle will have its centre on the radical axis of the given circles i.e. on Eq. (iii), if −10 (2 + 5λ ) + 2 (1 − λ ) + 5 = 0 1 ⇒ λ=− 4 1 On putting λ = − in Eq. (iv), we get 4 4 x 2 + 4 y 2 + 6 x + 10 y − 1 = 0

90. We know that, S + λL = 0, λ ∈ R represents a family of

⇒ r = 2 −1 or r = 3 ( 2 + 1) Thus, coordinates of the centre and radius of the required circle are ( 2 , 0 ) and ( 2 − 1) respectively. Hence, the equation of the required circle is ( x − 2 )2 + y 2 = ( 2 − 1)2

772

...(ii)

as the equation of the required circle.

Y′

∴

The radical axis of these two circles is S1 − S 2 = 0 5 5x − y + = 0 ⇒ 2 ⇒ 10 x − 2 y + 5 = 0

...(i)

This coordinates of the centre are [−(2 + 5λ ), − (1 − λ )]

Y

X′

S1 ≡ x 2 + y 2 + 4 x + 2 y + 1 = 0 3 S 2 ≡ x 2 + y 2 − x + 3y − = 0 2

...(i)

Then, 2 { ga (l + λ ) + fb (1 − λ )} = c + 0 [applying the condition of orthogonality] ⇒ 2 (ag + bf ) − c + 2 λ (ag − bf ) = 0 for all λ ∈ R This is true for all real values of λ. ∴ 2 (ag − bf ) = 0 and 2 (ag + bf ) − c = 0 c ⇒ ag = bf and ag + bf = 2 c ag = bf = ⇒ 4 c c and f = ⇒ g= 4a 4b On substituting the values of g, f andc in Eq. (i), we get c  x y x2 + y2 +  +  + c = 0 2  a b This is the equation of the circle orthogonal to the given coaxial system of circles.

Let P (α , β ) be any point in the same plane. Then, the polar of P, (α , β ) with respect to Eq. (i) is αx + βy + g( x + α ) + c = 0 ⇒ (αx + βy + c ) + g ( x + α ) = 0 This represents a family of straight lines passing through the point of intersection of the lines αx + βy + c = 0 and x+α=0 On solving these two equations, we get α2 − c x = − α, y = β

OD =

⇒ ⇒

m + 49 − 14m = 25 + 25 m

Then, the coordinates of their centres are P(− g1, 0 ), Q(− g 2 , 0 ) and R(− g 3 , 0 ). Let A (α , β ) be any point and let l1, l 2 , l 3 be the lengths of the tangents drawn from A to the three circles. Then, Ii2 = α 2 + β 2 + 2 g iα + c , i = 1, 2, 3 We have, PQ = g1 − g 2 , QR = g 2 − g 3 and RP = g 3 − g1 ∴ I12 ⋅ QR + I22 ⋅ RP + I32 ⋅ PQ = I12 (g 2 − g 3 ) + I22 (g 3 − g1 ) + I32 (g1 − g 2 ) = (α 2 + β 2 + c ) (g 2 − g 3 + g 3 − g1 − g 2 ) =0

+ 2α { g1(g 2 − g 3 ) + g 2 (g 3 − g1 ) + g 3 (g1 − g 2 )}

93. Let D be the mid-point of chord AB, drawn through the

2

24m2 + 14m − 24 = 0 3 4 m = ,− ⇒ 4 3 ⇒ Equation of chord AB is 4 y − 3 x − 25 = 0 ⇒ 3 y + 4 x − 25 = 0

94. We have, OD = 1, AO = 2 If ∠AOD = θ ⇒ cos θ =

1 2

A P C1 O D Q

B

θ=

⇒

π 3

π 3 = 2 OA = 4 Thus, locus of Q is x 2 + y 2 = 16. OQ = OA ⋅sec

⇒

95. Equation of radical axis of the given circles is x = 0 . If one circle lies completely inside the other, then the centre of both circles should lie on the same side of radical axis and radical axis should not intersect the circles. ⇒ (− a1 ) (− a2 ) > 0 ⇒ a1a2 > 0 and y2 + c = 0 should have imaginary roots ⇒ c > 0 π 96. Since, ∠APB = 2 P

A

point P(1, 7 ). We have,

π ∠DOA = 3

⇒

13

=5

⇒

92. Let x 2 + y 2 + 2 gx + c = 0, where g is a variable and c is

be three members of the given coaxial system of circles

1 + m2

2

Hence, the polar of P(α , β ) with respect to Eq.(i) passes  α2 − c through the fixed point  − α , . β   a constant, be a coaxial system of circles having X-axis as the common radical axis. Let x 2 + y 2 + 2 g i x + c = 0, i = 1, 2, 3

| m − 7|

Circle

...(i) where g ∈ R and c is a constant, be a family of coaxial circles having Y-axis as the common radical axis.

Targ e t E x e rc is e s

91. Let x 2 + y 2 + 2 gx + c = 0

D

π OD = OA ⋅ cos = 5 3

C B

B D

(1, 7) A

O (0, 0)

Let equation of chord AB be ( y − 7 ) = m ( x − 1) ⇒ y − mx + m − 7 = 0

⇒ ⇒

π 2 π ∠DCB = 4 ∠ACB =

π 4

⇒

CD = CB ⋅cos

⇒

2 = 2 2 |c| = 10 =

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Objective Mathematics Vol. 1

13

97. Angle between lines is 60 ° − 30 ° = 30 °. Thus, equation of their acute angle bisector is y = tan (30 ° + 15° )⋅ x i.e. y = x. Let C ≡ (h, h ), then | h − h 3| =1 2 ⇒ ( 3 − 1) h = 2 Y

y = x√3

O

⇒

f =

101. Let the equation of the required circle be x 2 + y 2 + 2 gx + 2 fy + c = 0 It passes through (3, 0) and touches Y-axis. ∴ 9 + 0 + 6 g + 0f + c = 0 ⇒ 9 + 6g + c = 0 and c = f2

X

h=

∴ ⇒

⇒ x 2 + y 2 − 2 x ( 3 + 1) − 2 y ( 3 + 1) + 7 + 4 3 = 0 a 98. The equation of the family of coaxial circles having x = 2 as radical axis with respect to x 2 + y 2 = a2 , is a  …(i) x 2 + y 2 − a2 + λ  x −  = 0  2 If it passes through (2 a, 0 ), then a  4a2 − a2 + λ 2 a −  = 0  2 3λa 3a2 + =0 ⇒ 2 ⇒ λ = − 2a On putting λ = − 2a in Eq. (i), we get x 2 + y 2 − 2 ax = 0

and

 9 + f 2 g=−   6 

On substituting these values in Eq. (iv), we get 9 + f 2 + 4f = f 2 − 3 ⇒ f = − 3 ∴

c = f2  9 + f 2 g=−   6 

⇒ c =9 ⇒ g=−3

15 =0 2 3 1 5 and S2 ≡ x2 + y2 − x − y + = 0 4 4 4 The radical axis of these two circles is given by S1 − S 2 = 0 31x 35 y 25 ⇒ − + =0 3 4 4 S1 ≡ x 2 + y 2 + 7 x − 9 y +

The equation of the tangent to this circle at ( 2 , 4) is 2 x + 4 y = 18 Hence, the required equation of the system of coaxial circle is x 2 + y 2 − 9 + λ ( 2 x + 4 y − 18) = 0

100. Let the required circle be ...(i)

It passes through the origin. So, (0, 0) satisfies Eq. (i) consequently, we have c =0

orthogonally. Therefore, 2 (−2 g + 3f ) = c + 10 and 2 (0 g + 6f ) = c + 6

c = f2

102. The equation of the circles are

x 2 + y 2 = 18

x 2 + y 2 + 12 y + 6 = 0

...(iv)

This is the equation of required circle.

x 2 + y 2 = 9 is its director circle given by

and

2 (−3 g + 2 f ) = c + (−3) −6 g + 4f = c − 3

On substituting the values of g, f andc in Eq. (i), we get x 2 + y 2 − 6x − 6y + 9 = 0

99. The locus of the point of perpendicular tangents to

The Eq. (i) intersects the circles x 2 + y 2 − 4 x + 6 y + 10 = 0

...(ii) ...(iii)

From Eqs. (ii) and (iii), we get

and

This is the equation of the required circle.

x 2 + y 2 + 2 gx + 2 fy + c = 0

...(i)

It is given that Eq. (i) is orthogonal to x 2 + y 2 − 6x + 4y − 3 = 0

2 2 ( 3 + 1) = = ( 3 + 1) 2 ( 3 − 1)

Thus, equation of circle is ( x − ( 3 + 1))2 + ( y − ( 3 + 1))2 = 1

Ta rg e t E x e rc is e s

−2 g + 3f = 5 and

This is the equation of the required circle. x y= √3 C

774

1 2 7 1 and f = g=− ⇒ 4 2 On substituting the values of g, f and c in Eq. (i), we get 7 x2 + y2 − x + y = 0 2 ⇒ 2 ( x2 + y2 ) − 7 x + 2 y = 0 ⇒

⇒

...(i) ..(ii)

31x − 35 y + 25 = 0

103. The common chord of the given circles is ax − by = 0

...(i)

The equation of the family of circles passing through the intersection of the given circles is ( x − a)2 + y 2 − a2 + λ { ax − by} = 0 ⇒

x 2 + y 2 + ax (λ − 2 ) − λ by = 0

...(ii)

The coordinates of the centre of this circle are  − a ( λ − 2 ) λ b ,    2 2 If the common chord i.e. Eq. (i) is a diameter of circle (ii), then λ b2 a2 − (λ − 2 ) − =0 2 2

⇒

λ=

107. Let x 2 + y 2 + 2 g x + c = 0; i = 1, 2, 3 2

2a a2 + b2

be three fixed circles from a system of coaxial circles having Y-axis as the common radical axis.

On substituting the value of λ in Eq.(ii), we obtain the equation of the required circle which is ( x 2 + y 2 ) (a2 + b2 ) = 2 ab (bx + ay )

104. The equation of the family of circles touching the line x − y = 0 at the origin is

Let P(α , β ) be any point on the circle x 2 + y 2 + 2 g1 x + c = 0 Then,

α 2 + β 2 + 2 g1α + c = 0

⇒

x + y + λ ( x − y) = 0 2

= α 2 + β 2 + 2g 2α + c = (2 (g 2 − g1 )α

...(i)

This circle bisects the circumference of the circle ...(ii) x2 + y2 + 2 y − 3 = 0

The equation of the common chord of Eqs. (i) and (ii) is ...(iii) λx − (λ + 2 )y + 3 = 0 This will be a diameter of Eq.(ii). If the centre (0, − 1) of circle (ii) satisfies the Eq. (iii). ∴ λ + 2 + 3= 0 ⇒ λ=−5 On putting λ = − 5 in Eq. (i), we get x 2 + y 2 − 5x + 5y = 0

= α 2 + β 2 + 2g 3α + c = 2 (g 3 − g1 )α

[using Eq. (ii)]

g 2 − g1 = Constant g 3 − g1

108. The equation of the family of circles touching the straight line 3 x − y = 6 at (1, − 3) is ( x − 1)2 + ( y + 3)2 + λ (3 x − y − 6) = 0

...(i)

The coordinates of the centre of the circle given in Eq. (i). Then, 2

( x − 1)2 + ( y − 2 )2 = 0 ( x − 4)2 + ( y − 3)2 = 0 ...(i)

This will pass through the origin, if 1 + 4 + λ (16 + 9) = 0 1 ⇒ λ=− 5 1 On putting λ = − in Eq. (i), we get 5 2 ( x2 + y2 ) − x − 7 y = 0 This is the equation of the required circle.

106. The equation of any circle passing through the points of intersection of the circles S1 = 0 and S 2 = 0 is S1 + λS 2 = 0 Hence, the equation of the required circle is ( x 2 + y 2 − 8 x − 2 y + 7 ) + λ (x 2 + y 2 − 4 x + 10 y + 8) = 0 x 2 (1 + λ ) + y 2 (1 + λ ) − 4 x (2 + λ ) − 2 y (1 − 5λ ) + 7 + 8λ = 0 ...(i) This circle passes through the point (−1, − 2 ). ∴ (1 + λ ) + 4 (1 + λ ) + 4 (2 + λ) + 4 (1 − 5λ ) + 7 + 8λ = 0 ⇒ 24 − 3λ = 0 ⇒ λ=8 On putting λ = 8 in Eq. (i), we get the required circle 9 x 2 + 9 y 2 − 40 x + 78 y + 71 = 0

2

5  λ − 6  2 − 3λ  r=  | λ|  − 10 + 6λ =  +  2   2  2

are

⇒

l1 = l2

∴

105. Equations of circles with limiting points (1, 2) and (4, 3)

Any circle coaxial with these circles is ( x − 1)2 + ( y − 2 )2 + λ {( x − 4)2 + ( y − 3)2 } = 0

[using Eq. (i)]

and l 2 = Length of the tangent from P (a, b) to x 2 + y 2 + 2 g3 x + c = 0

Therefore, the common chord of Eqs. (i) and (ii) is a diameter of circle (ii).

and

...(i)

Now, l1 = Length of the tangent from P(α , β ) to x 2 + y 2 + 2 g2 x + c = 0

( x − 0 )2 + ( y − 0 )2 + λ ( x − y ) = 0 2

13 Circle

a2 (λ − 2 ) + λb2 = 0

If the circle in Eq. (i) touches y = x, then 2 − 3λ λ − 6 − 2 2 = 5 | λ| 2 1+ 1 ⇒ ⇒ ⇒ ⇒ ⇒ We have,

4 − 2λ 5 = | λ| 2 2 |4 − 2 λ| = 5 | λ| 4 − 2λ = ± 5 λ 4 λ= 2− 5 4 λ= 2+ 5 r=

Targ e t E x e rc is e s

⇒

5 | λ| 4

Clearly, r will be least, if λ is least. The smaller of the two 4 . values of λ is 2+ 5 For this value of λ, we have 5 4 r= × = 10 2 − 4 10 2 2+ 5 = 10 × 1414 . − 4 × 3162 . = 149 .

109. The equation of the line y = x in parametric form is x y = cos π / 4 sin π / 4

775

Since,OP = 6 2. So, the coordinates of P are given by Y

Objective Mathematics Vol. 1

13

y = x cot 2α x (1 − tan 2 α ) y= 2 tan α  r2  x 1 − 2  h    r y= Qin ∆ODC, tan  h   r 2   h

y=x C (– g,– f ) P (6, 6) 6√2 45°

X′

X

O

⇒ (h 2 − r 2 )x − 2 rhy = 0

Y′

112. Let radius of required circle be r.

x y = =6 2 cos π / 4 sin π / 4 ⇒

∴ Now, to touch the line pair Distance between centres = Radius of I circle ± Radius of II circle 5= r ± 1 Q r = 5± 1 = 6, 4 ∴ Equation of circles are ( x − 4)2 + ( y − 3)2 = 62

x=y=6

So, the coordinates of P are (6, 6). The equation of the circle touching y = x at (6,6) is ( x − 6)2 + ( y − 6)2 + λ ( x − y ) = 0 x 2 + y 2 + (λ − 12 ) − y(λ + 12 ) + 72 = 0 This will be same as x 2 + y 2 + 2 gx + 2 fy + c = 0, if λ − 12 = 2 g, (λ + 12 ) = − 2 f

110. Equation of circle is ( x − c )2 + ( y + c )2 = c 2

( x − 4)2 + ( y − 3)2 = 42

and

and c = 72

Hence, the required value of c is 72.

Ta rg e t E x e rc is e s

 π y = x tan  − 2α   2

and

...(i)

⇒

x 2 + y 2 − 8 x − 6 y − 11 = 0

and

x 2 + y 2 − 8x − 6y + 9 = 0

113. The equation of tangent in terms of slope of x 2 + y 2 = 25 is

It passes through (3, − 6). ⇒ (3 − c )2 + (c − 6)2 = c 2 ⇒ c 2 − 18 c + 45 = 0 (c − 15) (c − 3) = 0

y = mx ± 5 (1 + m2 )

...(i)

Eq.(i) pass through (−2, 11,) then 11 = − 2 m ± 5 (1 + m2 )

Y

On squaring both sides, we get 21m2 − 44m − 96 = 0 X′

X

O c c (c,–c)

(3,–6) Y′

∴ c = 3, 15 From Eq. (i), we get equation of circles are x 2 + y 2 − 6x + 6y + 9 = 0 and

⇒ ∴

From Eq. (i), we get required tangents as 24 x − 7 y ± 125 = 0 and 4 x + 3 y = ± 25 Hence, tangents are 24 x − 7 y + 125 = 0 and 4 x + 3 y = 25

114. Consider family of circles through (0, 0) and (1, 0) x( x − 1) + y 2 + λy = 0 touches x 2 + y 2 = 9.

x + y − 30 x + 30 y + 225 = 0 2

(7 m − 24) (3m + 4) = 0 m = − 4/ 3, 24 / 7

2

Y

111. The given equation is ( x − r )2 + ( y − h )2 = r 2 Tangents are x = 0 Y X′

D

r

A

O (0, 0)

(3, 0)

X

(1, 0)

C (r, h) r

h α O

776

α

E π/2– 2α

Y′

X

∴

Common chord = − x + λy + 9 = 0

...(i)

9 1+ λ

2

⇒

=3

λ2 = 8

Let the centre be (α, 0), then its distance from α −1 x + y − 1 = 0 is = 2 (radius) 2 i.e. α = 1± 2 2 ∴ Centre may be (1 + 2 2 ,0 ) , (1 − 2 2, 0 ) 1+ β − 1 Now, let the centre be (1, β), then =2 2 β = ±2 2 ⇒

⇒ λ=±2 2 Circle x( x − 1) + y 2 + 2 2 y = 0 1  ∴Centre  , − 2 −1/ 2  2 

∴Centre may be (1, 2 2 ), (1, − 2 2 ).

117. x 2 + y 2 − 8 x − 16 y + 60 = 0

…(i)

Equation of chord of contact from (−2, 0) is −2 x − 4( x − 2 ) − 8 y + 60 = 0 3 x + 4 y − 34 = 0 2 34 3 x 34 − 3x  −    x2 +   + 60 = 0  − 8 x − 16    4  4 

P (1/2 λ) O

13 Circle

∴ Perpendicular from (0, 0) on Eq. (i) is equal to 3.

(1,0)

...(ii)

16 x 2 + 1156 − 204 x + 9 x 2 − 128 x − 2176 5 x 2 − 28 x − 12 = 0 OP =

⇒

3 2

x = 6, −

1 9 1 3 + λ2 = + λ2 = ⇒ 4 4 4 2 λ2 = 2 ⇒ λ = ± 2

⇒ ⇒

( x − 6) (5 x + 2 ) = 0

115. x 2 + y 2 + 8 x − 10 y − 40 = 0

 2 44 ∴Points are (6, 4) and  − ,  .  5 5

118. Centre of the first circle is (−3, 0 ) and the radius is 3 and radius of the second circle is (1, 0) and the radius is 1 unit. Since, the distance between the centres is equal to the sum of the radii, the two circles touch each other externally at the origin, the common tangent at the origin is Y-axis.

Centre of the circle is (−4, 5). Its radius = 9 units

(–4, 5) 9

(–2, 3)

2 5

Q P

(1,0)

(–3,0)

Targ e t E x e rc is e s

Aliter

+ 192 x + 960 = 0

R

Distance of the centre (−4, 5) from the point (−2, 3) is 4+ 4 =2 2 ∴ and ∴

a=2 2 + 9 b = −2 2 + 9 a + b = 18 a−b=4 2 a ⋅ b = 81 − 8 = 73

116. Line pair is ( x − 1)2 − y 2 = 0

X

Y′

i.e.

x + y − 1= 0 x − y − 1= 0

and

−6 cm + c 2 = 9 2cm + c 2 = 1 cm = − 1 and c 2 = 3

1 3 ⇒ Equation of the common tangents are 1 1 y=− x + 3, y = x − 3, x = 0 3 3 1 Since, the lines y = − x + 3 and make angle of 60° 3 with x = 0. ⇒

x=1

O

⇒ ⇒

Y

X′

Let y = mx + c be a direct common tangent to the two circles, then −3m + c m+c = ± 3 and = ±1 2 1+ m 1 + m2

c = ± 3 and

m=±

The ∆PQR formed by these tangents is equilateral, so that the centroid, circumcentre and orthocentre of the triangle coincide with its incentre (1, 0), the centre of the circle of smaller radius inscribed in the ∆PQR.

777

Objective Mathematics Vol. 1

13

119. Statement II is true as the centre is equidistant from A and B, hence lies on the perpendicular bisector of AB. Statement I is false as the distance between the given points is 10 and hence any circle through A and B has radius more than or equal to 5 and hence there is no circle of radius 4 through A and B is possible.

120. We know that, chords of contact of given circle generated by any point on given line passes through the fixed point, as they form family of straight lines, hence both the statements are true and Statement II is the correct explanation of Statement I.

121. For circle x 2 + y 2 = 144, centre C1(0, 0) and radius

r1 = 12. For circle x 2 + y 2 − 6 x − 8 y = 0, centre C2 = (3, 4) and radius r2 = 5. Now, C1C2 = 5 and r1 − r2 = 7, thus C1C2 < r1 − r2 , hence one circle is completely lying inside other without touching it, hence there is no common tangent. Therefore, Statement I is true. Therefore, both the statements are true but Statement II is not correct explanation of Statement I.

Ta rg e t E x e rc is e s

122. Centre of the circle C(2, 1) and radius r = 5. Distance of P(10, 7 ) from C(2, 1) is 10 units, hence required distances are 5, 15, respectively. Therefore, Statement I is true. Statement II is true but not the correct explanation of Statement I, as the information is not sufficient to get distance said in Statement I.

123. Given points are collinear, therefore circle is not possible. Hence, Statement I is false, however Statement II is true.

124. Since, S1 = 0 and S 3 = 0 has no radical axis. ∴Radical centre does not exist.

Now, equation of one of diagonals is x−3 y−4 = = r = ±2 2 1 1 2 2 ∴ x − 3= y − 4=± 2 ⇒ x = 5, 1 and y = 6, 2 So, two of the vertices are (1, 2) and (5, 6). The other diagonal is parallel to the line y = − x, so that its equation is x−3 y−4 = r = ±2 2 = 1 1 − 2 2 Hence, the two vertices on this diagonal are (1, 6) and (5, 2). G

D

r F

E

C

B

(a)

⇒ In Fig.(a), ⇒ ⇒

E

p +q = 2

⇒ ⇒

2

2

(b)

B

In Fig. (b), EG + GF = EF ⇒

2 r + r =2 r=

2 = 2( 2 − 1) units 2 +1

Solutions (Q. Nos. 131-133) Centre of the given circle is C(4, 5) . Points P, A, C and B are concyclic such that PC is diameter of the circle. Hence, centre D of the circumcircle of ∆ABC is mid-point of PC, then we have A C(4,5)

D

B

p −r + q −r 2

2

p + q = p − r + q 2 − r + 2 ( p2 − r ) (q 2 − r ) 1 1 1 = 2 + 2 r p q 2

C

AB = 4 units, AC = 4 2 units AE = 2 2 units EF + FA = AE 2r + r = 2 2 2 2 = 2 2 ( 2 − 1) units r= 2 +1

125. x 2 + y 2 − 2 x − 2 ay − 8 = 0 ( x 2 + y 2 − 2 x − 8) − 2 a( y ) = 0 S + IL = 0 solving the two equations x2 + y2 − 2 x − 8 = 0 y=0 x 2 − 4x + 2 x − 8 = 0 x( x − 4) + 2( x − 4) = 0 x = 4, x = − 2 So, (4, 0) and ( −2, 0) are the points of intersection which lie on X-axis.

A r

⇒

⇒

D G

126. Two circles touch each other C1C2 = r1 ± r2 2

127. The Statement II is well known result but if applied to the data in Statement I will yield 5 x − 9 y + 46 = 0 ⇒ Statement I is false, Statement II is true. Solutions (Q. Nos.128-130)

778

A

It is given that one of the diagonals of the square is parallel to the line y = x . Also, the length of the diagonal of the square is 4 2 unit.

P (t, sin t)

t +4 sin t − 5 and k = 2 2 sin(2 h − 4) + 5 Eliminating t, we have k = 2 sin(2 x − 4) + 5 or y= 2 sin −1(2 x − 5) + 4 −1 ⇒ f ( x) = 2 sin(2 x − 4) + 5 Thus, range of y = is [2, 3] and period is π. 2 h=

and

⇒ sin(2 x − 4) = 3, which has no real solutions. For f( x) = 1

⇒ ⇒

⇒ sin(2 x − 4) = − 3, which has no real solutions.

y = 3 x ± 2 and

AB with

the

3y + x = 2 1 139. Here, Area = 4  × 2 × 2 = 8 sq units 2  Y

circle

(0, 2)

Hence, the points are ( 2 , 2 ) or (− 2 , − 2 ).

135. Circumcentre lies on the point of intersection of

X′

O

perpendicular bisector of AB with the circle. Hence, circumcentre is (0, 0) or (2, 2). A + B 136. ∠AOB = π −    2  π ⇒ ∠AOB is fixed and is rather than . 2 ⇒ O lies always on a minor arc of a circle.

137.

4 3

X

(2, 2)

Y′

140.

Y B(0, 1)

Y

X′

X A(1, 0) Y′

Here, x + y = 1 subtends an angle

60° O (–2, 0)

X′

X

π at centre. 2

141. Slope of chord = 1

π . 3 ∴ Distance from the origin is 3 units. Since, the chord is

Y′

CC1 = r + r1 (h + 2 ) + k 2 = 2 + 2 = 4

⇒

2

∴Locus of c is x 2 + y 2 + 4 x − 12 = 0.

138. If (h, t) denotes the centre of variable circle, then 1 =0 2 k = 0 + 4 sin 60 ° = 2 3 ∴Common tangent at the point of contact is S1 − S 2 = 0 i.e. x 2 + y 2 + 4 x − ( x 2 ( y − 2 3 )2 − 4) = 0 h = − 2 + 4 ⋅ cos 60 ° = − 2 + 4

Let the equation of the chord be x− y+ k =0 k ∴ = 3 i.e. k = ± 6 2

142. Radius of the circle = 2 units Distance from the origin = 2 cos

π =1 3

143. A. A

4x + 4 3y − 8 = 0 x + 3y − 2 = 0

i.e.

Targ e t E x e rc is e s

of

y = (−1 / 3 )x + 2

⇒

134. The points of concurrency lie on the intersection of bisectors

m− 3=0 m − 3 = 2 1 + m2

Equation of tangent is

sin −1 (2 x − 5) + 4 is But range of y = 2 π  π  − + 2, + 2  4  4 perpendicular x 2 + y 2 = 4.

13

m − 3 − 1 + m2 = ± 1 + m2

Circle

f( x) = 4

Also,

B

Equation of tangent to given circle is

C

y = m( x + 2 ) ± 2 1 + m2 mx − y + 2 m ± 2 1 + m2 =0 m × 0 − 2 3 + 2 m ± 2 1 + m2 1 + m2

B.

y=x

=2 (2, 2)

x + 2y – 4 = 0

m − 3 ± 1 + m2 = ± 1 + m2 m − 3 + 1 + m2 = ± 1 + m2

⇒ i.e. m = 3, −

1 3

779

Objective Mathematics Vol. 1

13

C.

1  Since, it passes through the point  2 ,  .  2 1 ∴ 2 2α − 2 ( 2 + α ) − = 2α 2 − 2 2α 4 8 α 2 − 12 2α + 9 = 0 ⇒

x+y=2 (4, 3) O

(2 2α − 3)2 = 0

⇒

α=

i.e. x–y=2

∴ Number of chords is 1.

D. Obviously one (see theory).

144. A. (−g, − f ) lies in first quadrant, then g < 0 and f < 0; also X and Y-axes must not cut the circle. Solving circle and X-axis, we have x 2 + 2 gx + c = 0, which must have imaginary roots, then g 2 − c < 0, then c must be positive.

D. Mid-point of AB = (1, 4) ∴Equation perpendicular bisector of AB is x = 1. A diameter is 4 y = x + 7 ⇒ Centre of the circle is (1, 2). ⇒ Sides of the rectangle are 8 and 4. ∴

f2 −c < 0

Also,

B. If circle lies above X-axis, then x + 2 gx + c = 0 must have imaginary roots, then g 2 − c < 0 and c > 0. 2

Area = 32 sq units

146. A. CD = 5 and AC = 2

C. (−g, −f) lies in third or fourth quadrant, then g > 0. Also, Y-axis must not cut the circle.

A

D

Ta rg e t E x e rc is e s

Solving circle and Y-axis, we have y 2 + 2 gy + c = 0, which must have imaginary roots, then f 2 − c < 0, then c must be positive.

C

D. x 2 + 2 gx + c = 0 must have equal roots, then g 2 = c , hence c > 0 Also ,

B

−g>0 ⇒ g O, we must have, h < k < 2 h h = 3( 5 + 2) ⇒ and k = 3( 5 + 2 2 )

B

∠A = 60 ° = ∠D AC = 2 ∠ABC = 90 ° x = 1unit

⇒

3r = 5

Now,

152. Let (h, k) be the centre of circle with radius 3 units.

D

A

A

[given]

3 (h, k)

149. Clearly, locus of point of intersection of lines is ( x − 5). ( x − 3) + ( y − 2 )( y + 4) = 0 x 2 + y 2 − 8x + 2 y + 7 = 0

⇒

| f + g| = |2 + (−8)| = 6

Hence,

… (i)

⇒ k =1 Put k = 1in Eq. (i), we get 6h = 1 1 h= ⇒ 6 ∴

7

148.

13

⇒

Circle

A

150. x 2 + y 2 − 2 x − 2 λy − 8 = 0 ⇒ ( x 2 + y 2 − 2 x − 8) − 2 λy = 0, which is of the form of S + λL = 0 All the circles passing through the point of intersection of circle x 2 + y 2 − 2 x − 8 = 0 and y = 0. Solving, we get x 2 − 2 x + 8 = 0 ⇒ ( x − 4)( x + 2 ) = 0 ⇒ A ≡ (4, 0 ) and B ≡ (−2, 0 ) ⇒ Distance between A and B is 6 units.

Targ e t E x e rc is e s

147.

O

If A is point of contact, then OA2 = (h 2 + k 2 ) = 9 ⇒

OA2 = 9 (3 + 10 )2

⇒ ⇒

OA = 3 (3 + 10 ) λ=5

153. Radius of C1 = 3.If a tangent to C1 cuts and intercept of length 8 onC2 , thenC1 and C2 being concentric. Radius of C1on the point of contact = Perpendicular drawn from center of C2 on its chord. 8

151. Let centre be C(h, k )

L

Y

3

B (0, 4)

X′

O Y′

C1

P (3/2, 2)

r C

r

A (3, 0) X Q (3/2, 0)

C2

⇒ Chord is bisected at the point of contact of C1. ∴

Radius ofC2 = 42 + 32 = 5

[see the figure]

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Objective Mathematics Vol. 1

13

154. The circles are and

|1 − 2 − a| >1 5 |1 − 16 − a| >2 5 | 1 + a| > 5,| 15 + a| > 2 5

⇒

( x − 1) + ( y − 1) = 1 ( x − 8)2 + ( y − 1)2 = 4 2

2

If circles lie on opposite sides, then centre of circles must be on opposite sides. ⇒ (1 − 2 − a) (1 − 16 − a) < 0 … (i) ⇒ a ∈ (− 15, − 1) Again, if line does not cut any circle, then distance from centre > radius

On taking the intersection of intervals, i.e. Eqs. (i) and (ii), we get a ∈ (2 5 − 15, − 15 − 1) ⇒ k=3

Entrances Gallery 1. Given circle

x 2 + y 2 − 2 x − 15 = 0 x2 + y2 − 1 = 0 Radical axis x + 7 = 0 Centre of circle lies on Eq.(i). Let the centre be (−7, k ). Let equation be x 2 + y 2 + 14 x − 2 ky + c = 0 Orthogonality gives −14 = c − 15 ⇒ (0, 1) → 1 − 2 k + 1 = 0 ⇒

c =1 k =1

Ta rg e t E x e rc is e s

Aliter Given circles x 2 + y 2 − 2 x − 15 = 0 x2 + y2 − 1 = 0

Circle passes through (0, 1). ⇒ 1 + 2f + c = 0 Applying condition of orthogonality, −2 g = c − 15, 0 = c − 1 ⇒ c = 1, g = 7, f = − 1 r = 49 + 1 − 1 = 7 units; centre (−7, 1)

2. OB =|α| 1 1 = |α| |α|

| z0|2 + |α|2 − r 2 2| z0||α|

cos θ = In ∆OCD,

A θ

…(i)

…(ii)

Comparing Eqs. (i) and (ii), we get h2 + k 2 h k = = 9 α 4α − 4 5 20 h α= ⇒ 4h − 5k h(4h − 5k ) h 2 + k 2 = 20 h 9 20(h 2 + k 2 ) = 9 (4h − 5k )

⇒ 20( x 2 + y 2 ) − 36 x + 45 y = 0 Y x2 + y 2 = 9

2r

X′

r

O

hx + ky = h 2 + k 2  4  Let any point on line be α , α − 4 .  5  Equation of the chord of contact is 4  ⇒ α x +  α − 4 y = 9 5 

C (1/α) B (α)

1 7

( x − 3)2 + y 2 + λ ( y ) = 0 ⇒ x + y 2 − 6 x + λy + 9 = 0 Now, (Radius )2 = 7 + 9 = 16 λ2 9+ − 9 = 16 ⇒ 4 ⇒ λ2 = 64 ⇒ λ = ± 8 2 ∴Equation is x + y 2 − 6 x ± 8 y + 9 = 0.

⇒

1 − 4r 2 |α|2 1 2| z0| |α|

Y

|α| =

3. Equation of circle can be written as

Now,

| z0|2 + cos θ =

⇒

4. Equation of the chord bisected at P(h, k ) is

Let equation of circle x 2 + y 2 + 2 gx + 2 fy + c = 0

In ∆OBD,

1 − 4r 2 |α|2 1 2| z0| |α|

| z0|2 +

…(i)

…(ii)

= 49 + 1 − 1 = 7 units

OC =

| z0|2 + |α|2 − r 2 = 2| z0||α|

2

Hence, radius = 7 2 + k 2 − c

O

x–y =1 5 5

X P (h, k)

D z0

(a, 45 a – 4 )

X Y′

782

…(ii)

1 1 ( x − 3) ± 1 1 + 3 3 3 y = x − 3 ± (2 ) 3y = x − 1 ⇒ 3 y = x − 5. y=

6. Point of intersection of direct common tangent is (6, 0)

x2 + y 2 = 4 2 y2 =1 (x – 3 ) +

R(6, 0)

So, let the equation of common tangent be y − 0 = m( x − 6) as it touches x 2 + y 2 = 4 ⇒

0 − 0 + 6m 1 + m2

=2

9m2 = 1 + m2 1 m=± 2 2 So, equation of common tangent 1 1 ( x − 6), y = − ( x − 6) and also x = 2 y= 2 2 2 2

the circles with respect to each other. (i) If circles touch externally ⇒ C1C2 = r1 + r2 , 3 common tangents (ii) If circles touch internally ⇒ C1C2 = r2 + r1, 1 common tangent (iii) If circles do not touch each other, 4 common tangents Given equations of circles are …(i) x 2 + y 2 − 4 x − 6 y − 12 = 0 x 2 + y 2 + 6 x + 18 y + 26 = 0

…(ii)

Centre of circle (i) is C1(2, 3) and radius = 4 + 9 + 12 [say] = 5 = r1 Centre of circle (ii) is C2 (−3, − 9) and radius = 9 + 81 − 26 [say] = 8 = r2 Now,

C1C2 = (2 + 3)2 + (3 + 9)2

⇒

C1C2 = 52 + 12 2 C1C2 = 25 + 144 = 13 r1 + r2 = 5 + 8 = 13 C1C2 = r1 + r2

⇒ ∴ Also,

Thus, both circles touch each other externally. Hence, there are three common tangents.

10. According to the figure, (1 + y )2 = (1 − y )2 + 1 1 y= 4

⇒

7. Circle touching Y-axis at (0, 2) is

[Q y > 0 ]

(1, 1)

( x − 0 )2 + ( y − 2 )2 + λ x = 0

1+y

passes through (−1, 0 ). ∴ 1+ 4 − λ = 0 ⇒ λ = 5 x 2 + y 2 + 5x − 4y + 4 = 0 Put y = 0 ⇒ x = − 1, −4 ∴Circle passes through (−4, 0 ). π π 8. 2 cos + 2 cos = 3 + 1 2k k π π 3+1 cos + cos = 2k k 2 θ 3+1 π Let = 0, cos θ + cos = 2 2 k θ θ + 3 1 ⇒ 2 cos 2 − 1 + cos = 2 2 2 θ 3 +3 Now, cos = t, 2 t 2 + t − =0 2 2 −1 ± 1 + 4 (3 + 3 ) −1 ± (2 3 + 1) ∴ t = = 4 4 −2 − 2 3 3 = , 4 2 θ 3 Q t ∈ [−1, 1], cos = 2 2 θ π = ⇒ k=3 2 6

13 Circle

9. Number of common tangents depend on the position of

3x + y = 4 Slope of line perpendicular to above tangent is 1/ 3. So, equation of tangents with slope 1/ 3 to ( x − 3)2 + y 2 = 1 will be

Targ e t E x e rc is e s

5. Equation of tangent at P( 3, 1) is

(0, y)

1–y 1

11. Let the equation of circle be ( x − 3)2 + ( y − 0 )2 + λy = 0 As it passes through (1, − 2 ) ∴ (1 − 3)2 + (−2 )2 + λ (−2 ) = 0 ⇒ 4 + 4 − 2λ = 0 ⇒ λ = 4 Y

A(3, 0) X X'

O (1, –2)P

Y'

∴ Equation of circle is ( x − 3)2 + y 2 + 4 y = 0 By hit and trial method, we see that point (5, − 2 ) satisfies equation of circle.

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13

12. Given (i) A circle which touches X-axis at the point (1, 0).

(ii) If circles touch externally, then Y

(ii) The circle also passes through the point (2, 3). To find The length of the diameter of the circle. Y X′

C

a/2 a/2

X

(h, k) (2, 3)

k

Y′

k X

O

∴ Let us assume that the coordinates of the centre of the circle are C(h, k ) and its radius be r. Now, since the circle touches X-axis at (1, 0), hence its radius should be equal to ordinate of centre. ⇒ r =k Hence, the equation of the circle is

(1 − h )2 + (0 − k )2 = k 2

…(i)

(2 − h ) + (3 − k ) = k

…(ii)

2

2

From Eq. (i), we get h = 1

5 3

Hence, the diameter of the circle is 2 k =

r 2 = 4 + 16 + 5 = 25

The line will intersect the circle at two distinct points, if the distance of (2, 4) from 3 x − 4 y = m is less than radius of the circle. |6 − 16 − m| ∴ 0 The circles should touch internally and | a| = c

15. Since, the coordinates of the centre of the circle are (2, 4).

( x − h )2 + ( y − k )2 = k 2

Ta rg e t E x e rc is e s

C (0, 0)

10 . 3

13. x 2 + y 2 − ax = 0 and x 2 + y 2 = c 2 touch each other. Y

Equation of the required circle is S + λS ′ = 0 As it passes through (1, 1), the value of − (7 + 2 p) λ= ( 6 − p2 ) Here, λ is not defined at p = ± 6 Hence, it is true for all except two values of p.

17. Given equation can be rewritten as ( x + 1)2 + ( y + 2 )2 = (2 2 )2 Let required point be Q(α , β ).

C a/2 X′

O

a/2

C

X

Then, mid-point of P(1, 0 ) and Q(α , β ) is the centre of the circle α+1 β+0 i.e. = − 1 and = −2 2 2 ⇒ α = − 3 and β = − 4 Hence, required point is (−3, − 4).

18. Equation of circle which touches X-axis and coordinates Y′

784

(i) If circles touch internally, then a a c− = 2 2 a a ⇒ c− = 2 2 ⇒ c = a, c > 0 ∴ | a| = c

of centre are (h, k ), is ( x − h )2 + ( y − k )2 = k 2 Since, it is passing through (−1, 1,) then (−1 − h )2 + (1 − k )2 = k 2 ⇒ h2 + 2h − 2k + 2 = 0 For real circles, D ≥ 0 ⇒ (2 )2 − 4 (−2 k + 2 ) ≥ 0 1 ⇒ k≥ 2

3x − 4y − 7 = 0 2 x − 3y − 5 = 0

On solving Eqs.(i) and (ii), we get x = 1 and y = − 1

x 2 + y 2 + 2 gx + 2 fy + c = 0 It cuts the circle x 2 + y 2 = p2 orthogonally. ∴ ⇒

2 g( 0 ) + 2 f ( 0 ) = c − p 2 c = p2

∴ The intersection of two diameters is the centre of circle, i.e. (1, − 1).

Also, it passes through (a, b). ∴ a2 + b2 + 2 ga + 2 fb + p2 = 0

Let r be the radius of circle, then Area of circle = 49 π sq units ⇒ πr 2 = 49 π

So, locus of (− g, − f ) is a2 + b2 − 2 ax − 2 by + p2 = 0

⇒ r = 7 units ∴ Equation of required circle is ( x − 1)2 + ( y + 1)2 = 49 ⇒

x − 2 x + 1 + y + 2 y + 1 = 49

⇒

x 2 + y 2 − 2 x + 2 y − 47 = 0

2

⇒

2 ax + 2 by − (a2 + b2 + p2 ) = 0

23. Let the equation of circle be x 2 + y 2 + 2 gx + 2 fy + c = 0. It cuts the circle x 2 + y 2 = 4 orthogonally, if

2

20. Let the coordinates of a point P be (h, k ) which is mid-point of the chord AB. Now,

OP = (h − 0 )2 + (k − 0 )2

∴ ⇒

2 g1g 2 + 2 f1f2 = c1 + c 2 2g ⋅ 0 + 2f ⋅ 0 = c − 4 c =4

Equation of circle is x 2 + y 2 + 2 gx + 2 fy + 4 = 0 Since, it passes through the point (a, b). ∴ a2 + b2 + 2 ag + 2 fb + 4 = 0

= h2 + k 2

Locus of centre (− g, − f ) will be a2 + b2 − 2 xa − 2 yb + 4 = 0 ⇒

O (0, 0) 3 A

π/3

π 3

P (h, k)

B

π OP = 3 OA h2 + k 2 1 = ⇒ 2 3 9 2 2 ⇒ h +k = . 4 Hence, the required locus is 9 x2 + y2 = 4

In ∆AOP,

cos

21. Let equation of circles be S1 ≡ x 2 + y 2 + 2 ax + cy + a = 0 and

13

S 2 ≡ x 2 + y 2 − 3 ax + dy − 1 = 0

Chord through intersection points P and Q of the given circles is S1 − S 2 = 0. ∴ ( x 2 + y 2 + 2 ax + cy + a) −( x 2 + y 2 − 3 ax + dy − 1) = 0 ⇒ 5 ax + (c − d )y + a + 1 = 0 On comparing it with 5 x + by − a = 0, we get 5a c − d a + 1 = = 5 b −a ⇒ a(− a) = a + 1 ⇒ a2 + a + 1 = 0 which gives no real value of a. Hence, the line will pass through P and Q for no value of a.

2 ax + 2 by − (a2 + b2 + 4) = 0

Aliter Let the centre of required circle be (− g, − f ). This circle cuts the circle x 2 + y 2 = 4 orthogonally. The centre and radius of circle x 2 + y 2 = 4

are (0, 0) and 2,

respectively. ∴ g 2 + f 2 = 4 + (a + g )2 + (b + f )2 ⇒ ⇒

Targ e t E x e rc is e s

and

22. Let the equation of circle be …(i) …(ii)

Circle

19. The given equations of diameters are

g 2 + f 2 = 4 + a2 + g 2 + 2 ag +2 bf + b2 + f 2 4 + a2 + b2 + 2 ag + 2 bf = 0

Hence, the locus of centre is 2 ax + 2 by − (a2 + b2 + 4) = 0.

24. Since, the coordinates of one end of a diameter of a circle A are ( p, q ) and let the coordinates of other end B be ( x1, y1 ). Equation of circle in diameter form is ( x − p)( x − x1 ) + ( y − q )( y − y1 ) = 0 ⇒ x 2 − ( p + x1 )x + px1 + y 2 − (q + y1 )y + qy1 = 0 ⇒ x 2 − ( p + x1 )x + y 2 −( y1 + q )y + px1 + qy1 = 0 Since, this circle touches X-axis. i.e. y=0 ∴ x 2 − ( p + x1 )x + px1 + q y1 = 0 Also, the discriminant of above equation will be ( p + x1 )2 = 4( px1 + qy1 ) ⇒

p2 + x12 + 2 px1 = 4 px1 + 4qy1

⇒

x12 − 2 px1 + p2 = 4qy1

Hence, locus of point B is ( x − p)2 = 4qy.

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25. Given lines 2 x + 3 y + 1 = 0 and 3 x − y − 4 = 0 are the diameters of circle.

whose centre is (−2, 2 ) and radius = 2 units.

The intersection of two lines is the centre of circle [given] (1, − 1). Circumference of circle = 10π ⇒ 2 πr = 10 π ⇒ r =5 ∴Equation of circle having centre (1, − 1) and radius 5 is

Let the equation of required tangent be x+ y=a The perpendicular distance from centre to the circle is equal to radius of circle. −2 + 2 − a ∴ =2 2 a=2 2 ⇒ Hence, equation of the tangent is x + y =2 2

( x − 1)2 + ( y + 1)2 = 52 ⇒

x 2 + 1 − 2 x + y 2 + 2 y + 1 = 25

⇒

x 2 + y 2 − 2 x + 2 y − 23 = 0

26. Given equation of line is y = x and equation of circle is x2 + y2 − 2 x = 0

…(i) …(ii)

From Eqs. (i) and (ii), we get x2 + x2 − 2 x = 0 ⇒ 2 x( x − 1) = 0 ⇒ x = 0, y = 1 On putting the value of x in Eq. (i) respectively, we get y = 0, x = 1. Let coordinates of A be (0, 0 ) and coordinates of B be (1, 1).

Ta rg e t E x e rc is e s

29. Given equation of circle is x 2 + y 2 + 4 x − 4 y + 4 = 0

∴Equation of circle when AB as a diameter, is ( x − 0 )( x − 1) + ( y − 0 )( y − 1) = 0 ⇒ x2 − x + y2 − y = 0 ⇒

x2 + y2 − x − y = 0

27. The equation of first circle is ( x − 1)2 + ( y − 3)2 = r 2 whose centre is C1(1, 3) and radius, r1 = r . and equation of second circle is x 2 + y 2 − 8x + 2 y + 8 = 0 whose centre is C2 (4, − 1) and radius, r2 = 42 + 12 − 8 = 17 − 8 = 3 units If two circles intersect at two distinct points, then r1 − r2 < C1C2 < r1 + r2 ⇒ r − 3 < (4 − 1)2 + (−1 − 3)2 < r + 3 ⇒ ⇒ ⇒ ⇒ ⇒

r − 3 < 9 + 16 < r + 3 r − 3< 5< r + 3 r − 3 < 5 and 5 < r + 3 r < 8 and 2 < r 2 0

45. We have equation of circle x + y − 2 x + 6y − 6 = 0 2

2

So, the point P (10, 7 ) lies outside the circle.

⇒

g = − 1, f = 3, c = − 6 Radius (r ) = 1 + 9 + 6 = 4 and centre = (1, − 3) Now, distance of tangent 3 x − 4 y + k = 0 from the centre is |3 + 12 + k| k + 15 ⇒ 4= 4= ±5 25 ⇒ ± 20 = k + 15 ∴ k = 5, − 35

46. The circle x 2 + y 2 + 2 gx + 2 fy + c = 0 passes through (2, 0). ∴

4 + 4g + c = 0

...(i)

Its intercept on X-axis is 2 × g 2 − c = 5

[given]

25 4g − 25 ⇒ c= 4 4 4g 2 − 25 [using Eq. (i)] 4 + 4g + =0 4 2 16 + 16 g + 4g − 25 = 0

⇒

2

g2 − c =

∴ ⇒ ⇒

4 g 2 + 16 g − 9 = 0 −9 1 g= , 2 2

⇒

Hence, x 2 + y 2 − 9 x + 2 fy + 14 = 0 is the required equation.

47. The third vertex of the equilateral triangle must lie on Y-axis. Let it be C(0, b). Now, AC = BC = AB b2 = 4 − 1

⇒

b= 3

Also, centre of given circle is C (2, 1) and r =5 ∴ PC = (10 − 2 )2 + (7 − 1)2 = 10 units [Q B is a point on circle and line PC] BC = 5 ∴ Greatest distance = 10 + 5 = 15 units Least distance = 10 − 5 = 5 units and

49. Q g = 4, f = 2, c = − 8 ∴ Centre = (−4, − 2 ) = Centre of concentric circle Also, radius of unit circle = 1unit ∴ Required equation of circle is ( x + 4)2 + ( y + 2 )2 = 1 ⇒

50. Given, y = mx − b 1 + m2 touches both the circles. ∴Distance from centre = Radius of both the circles | ma − 0 − b 1 + m2| ⇒ ⇒

|− b 1 + m2| m2 + 1

=b

| ma − b 1 + m2| = |− b 1 + m2| m2 a2 − 2 abm 1 + m2 + b2 (1 + m2 )

⇒ ma − 2 b 1 + m2 = 0 ⇒ ⇒

m2 a2 = 4b2 (1 + m2 ) 2b m= a2 − 4b2

...(i)

If it is a tangent, then perpendicular from centre (−2, − 3) is equal to radius = (2 )2 + (3)2 − 8 ⇒

Y′

= b and

= b2 (1 + m2 )

C (0, b)

A (–1, 0)

x 2 + y 2 + 8 x + 4 y + 19 = 0

y + 4 = m ( x + 5) mx − y + (5 m − 4) = 0

Y

X′

P (10, 7)

C

51. Any line through the point (−5, − 4) is

1 + b2 = 1 + b2 = 2

∴

(2, 1) B

m2 + 1

9 So, g = − , since centre lies in first quadrant, also 2 c = 14.

∴

13

Targ e t E x e rc is e s

⇒

As triangle is an equilateral, so circumcentre is centroid of ∆ABC. Now, centroid = Circumcentre 1  b  =  0,  =  0,   3  3

Circle

So, the perpendicular distance from centre to the circle is equal to radius of the circle. |− g − 2| = g2 + f 2 − c ∴ 1

B (1, 0)

X

⇒ ⇒

= 4+ 9− 8 = 5 m(−2 ) − (−3) + (5m − 4) = 5 m2 + 1 − 2 m + 3 + 5 m − 4 = 5 m2 + 1 (3 m − 1)2 = 5 (m2 + 1)

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Objective Mathematics Vol. 1

13

⇒

9 m2 + 1 − 6m = 5m2 + 5

⇒

4 m2 − 6 m − 4 = 0

⇒

2 m2 − 3 m − 2 = 0

1 ,2 2 Hence, required equations are 2 x − y + 6 = 0 and x + 2 y + 13 = 0 ⇒

m=−

52. As we have to find the locus of mid-point of chord and we know perpendicular bisect the chord. A 45°

O C

(h, k)

2

2 B

56. The intersection of line and circle is x 2 + m2 x 2 − 20 mx + 90 = 0 x 2 (1 + m)2 − 20 mx + 90 = 0 D |C1C2| One circle lies inside the other and both the circles are touching internally.

= 49 = 7 units

59. Since, ax 2 + ay 2 + 2 gx + 2 fy + c = 0 touches X-axis, i.e. y = 0 satisfy the equation ∴ ax 2 + 2 gx + c = 0 ⇒ (2 g )2 = 4 × ac ⇒ 4g 2 = 4ac ⇒ g 2 = ac

60. We have, x 2 + y 2 = a2 i.e. centre = (0, 0 ) and radius = a Also, we have x 2 + y 2 − 2 ax = 0 Centre = (a, 0 ) and radius = a So, the circles are intersecting each other. Y

⇒

2(g1g 2 + f1f2 ) = C1 + C2 2 (1⋅ 0 + k ⋅ k ) = 6 + k 2k 2 − k − 6 = 0 2 k 2 − 4k + 3k − 6 = 0 2 k (k − 2 ) + 3 (k − 2 ) = 0 (2 k + 3) (k − 2 ) = 0 3 or 2 k=− 2

55. Centre of the circle x 2 + y 2 + 4 x − 10 y − 7 = 0 is (−2, 5) and radius (r ) = 4 + 25 + 7 = 6 units Distance of point (4, − 3) from centre (− 2, 5) is (− 2 − 4)2 + (5 + 3)2 = 100 = 10 units Also, we check whether the point lies inside the circle or not. ∴ (4)2 + (−3)2 + 4 ⋅ 4 − 10 ⋅ (−3) − 7 = 80 > 0

790

So, minimum distance = 10 − 6 = 4 units and maximum distance = 10 + 6 = 16 units ∴Sum of the minimum and maximum distances = 4 + 16 = 20 units

X′

(x1,y1)

54. Since, circles intersect orthogonally. ∴ ⇒ ⇒ ⇒ ⇒ ⇒

[since, x is a real]

(0,0)

(a,0) (2a, 0)

X

Y′

Locus of the tangent to the extremeties of the chord passes through ( x1, y1 ) to circle x 2 + y 2 = a2 is ⇒

T =0 xx1 + yy1 − a2 = 0

...(i)

Also, above equation is tangent to circle x 2 + y 2 − 2 ax = 0 Equation of tangent line is x = 2a ⇒ 1⋅ x + 0 ⋅ y − 2 a = 0 On comparing Eqs. (i) and (ii), we get x1 y1 a2 = = 1 0 2a a x1 = , y1 = 0 ⇒ 2

a  ∴Line will passes through the fixed point  , 0 . 2 

...(ii)

and

Centre = (0, 0 ) and radius = 1

−1  13  −9  + 2 =   3 7 7

 −1 9 So, the centre of the required circle is  ,  .  7 7

B

64. Equation of circle is

O (0, 0)

( x − 4) ( x + 2 ) + ( y − 7 ) ( y + 1) = 0 ⇒ x2 − 2 x − 8 + y2 + y − 7 y − 7 = 0 ⇒ x 2 + y 2 − 2 x − 6 y − 15 = 0

1 A

Here, g = − 1, c = − 15 ⇒ AB = 2 g 2 − c ⇒ Since, lines are intersecting at right angles. i.e. AB is the diameter of another circle which passes through (0, 0 ) and have radius 2 1 = = unit 2 2 ∴ Distance of line mx − y + 2 = 0 from (0, 0) is 1 | m ⋅ 0 − 0 + 2| = 2 m2 + 1 ⇒

(m2 + 1) = 2 ⋅ 2 ⋅ 2

⇒

m2 = 7

∴

AB = 2 1 + 15

AB = 8 units

65. Mid-point of (4, 0) and (0, 4) is (2, 2). ∴ Required distance = (2 − 0 )2 + (2 − 0 )2 = 8 = 2 2 units

66. Required equation is ( x − h )2 + ( y − k )2 = k 2 ⇒

x 2 + y 2 − 2 hx − 2 ky + h 2 = 0 x 2 + y 2 − 6 x − 8 y + λ ( x + y − 1) = 0

⇒

x 2 + y 2 − (6 − λ ) x − (8 − λ ) y − λ = 0 λ λ  whose centre is  3 − , 4 −   2 2 and which lies on the line x + y − 1 = 0. λ λ ⇒ 3 − + 4 − − 1= 0 2 2 ⇒ λ=6 Hence, required equation is x 2 + y 2 − 6x − 8y + 6x + 6y − 6 = 0

62. C1 = (3, 4) and C2 = (0, 0 ), ∴

∴

67. Required equation of circle is

m = ± 7 unit r1 = 32 + 42 − 9 = 4

13

and

r2 = 1

|C1C2| = | (0 − 3)2 + (0 − 4)2| = 5

r1 + r2 = 5 |C1C2| = r1 + r2 Hence, there are three common tangents.

So,

63. Let equation of required circle be x 2 + y 2 + 2 gx + 2 fy + c = 0 ∴ 2 (g ⋅ 3 + f ⋅ 0 ) = c − 1 ⇒ 6g = c − 1 3  Also, 2 g ⋅ 0 − f ⋅  = c + 2  2 ⇒ −3f = c + 2 1  1 and 2 g ⋅ + f ⋅  = c − 3  2 2

⇒ ...(i)

...(ii)

…(iii) ⇒ g + f =c − 3 From Eqs. (i) and (ii), we get …(iv) 6 g + 3f = − 3 Also, g + f = 6g + 1− 3 [using value of c from Eqs. (i) and (iii)] ⇒ 5g − f = 2 …(v) ⇒ 15 g − 3f = 6 From Eqs. (iv) and (v), we get 21g = 3 1 6 13 and c = + 1 = ⇒ g= 7 7 7

x2 + y2 − 2 y − 6 = 0

68. Centre of circle is (1, − 2 ).

Targ e t E x e rc is e s

√2

f =

Circle

61. We have, x 2 + y 2 = 1

∴ Required equation of normal Equation of straight line passing through (1, − 2 ) and (2, 1) −2 − 1 i.e. y+2= ( x − 1) 1− 2 ⇒ y + 2 = 3x − 3 ⇒ 3x − y − 5 = 0

69. Required equation of chord is ⇒ ⇒

T = S1 −2 x + 3 y − 81 = 4 + 9 − 81 2 x − 3 y = − 13

70. Here, C1(−7, 3), r1 = 6 and C2 (5, − 2 ), r2 = 7 ∴ Required point of contact  r x + r2 x1 r1 y2 + r2 y1  =1 2 ,   r1 + r2 r1 + r2   6 × 5 + 7 × (−7 ) 6 × (−2 ) + 7 × 3  19 9  = ,  = − ,     13 13 6+ 7 6+ 7

791

14 Parabola Conic Sections A conic section, as the name implies, is a section cut off from a circular (not necessary a right circular) cone by a plane in various ways. The shape of the section depends upon the position of the cutting plane. Consider a double napped right circular cone of semi-vertical angle α and let it be cut by a plane inclined at an angle θ to the axis of the cone. We will get different sections (curves) as follows : Case I If the plane passes through the vertex V The curve of intersection is a pair of straight lines passing through the vertex which are (i) real and distinct, for θ < α. (ii) coincident, for θ = α i.e. the plane touches the cone. (iii) imaginary, for θ > α.

Chapter Snapshot ●

●

●

●

●

●

●

●

α

●

●

V

Pair of lines ●

●

●

●

Case II If the plane does not pass through the vertex V The curve of intersection is called π (i) a circle, if θ = . 2 (ii) a parabola, for θ = α i.e. if the plane is parallel to the generator PQ. π  (iii) an ellipse, for θ > α  θ ≠  i.e. if the plane cuts both the generating  2 lines PQ and RS. (iv) a hyperbola, for θ < α i.e. if the plane cuts both the cones.

Conic Sections Parabola Other Standard Forms of Parabola Position of a Point with Respect to a Parabola Intersection of a Line and a Parabola Equation of Tangent Angle of Intersection of Two Parabolas Equation of Normal to Parabola Number of Normals and Conormal Points Combined Equation of Pair of Tangents Director Circle Diameter of a Parabola Pole and Polar of a Parabola Lengths of Tangent, Subtangent, Normal and Subnormal

α

α

(θ = π2 )

V

P

Q

Q

P

θ

Ellipse

) π2 > θ > α)

Circle

If the focus, directrix and eccentricity of a conic are given, we can find its equation as given below. Let S (α, β) be the focus, ax + by + c = 0 be the directrix and e be the eccentricity of a conic. Let P ( h, k ) be any point on the conic such that PM is the perpendicular from P on the directrix. Then, by definition, we have

S

S

R

R

P (h, k)

M

P α

α

Parabola (θ = α)

14

Z

ax + by + c = 0

P

General Equation of a Conic

R

Parabola

R

Hyperbola (θ < α)

S (α, β)

Z′

S

Q

S

Q

Z

Directrix

S (Focus)

Z′

The fixed point is called the focus of the conic section and the fixed line is known as its directrix. The constant ratio e is called the eccentricity of the conic section. If e =1, the conic is called a parabola. If e 1, the conic is called a hyperbola.

2

 ax + by + c   (x − α ) 2 + ( y − β) 2 = e 2   2 2  a + b 

2

2

2  ah

This is the equation of the required conic. This equation when simplified can be written in the form ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 which is the general equation of second degree. X

P

M

+ bk + c  a 2 + b 2 

 ⇒ (h − α ) + (k − β) = e   So, the locus of ( h, k ) is 2

Analytical Definition of Conic Section A conic section or a conic is the locus of a point which moves in such a way that its distance from a fixed point always bears a constant ratio to its distance from a fixed line, all being in the same plane. Thus, if S is a fixed point in the plane of the paper and ZZ′ is a fixed line in the same plane, then the locus of a point P which moves in the same plane in such a way that SP [say] = Constant = e PM is called a conic section or simply a conic.

⇒ SP 2 = e 2 ⋅ PM 2

SP = e ⋅ PM

Example 1. The equation of a conic section whose focus is at ( −1, 0), directrix is the line 1 , is 4x − 3 y + 2 = 0 and eccentricity is 2 (a) 34x 2 + 24xy + 41 y 2 + 84x + 12 y + 46 = 0 (b) 41x 2 + 24xy + 34 y 2 + 84x + 12 y + 46 = 0 (c) 34x 2 + 24xy + 41 y 2 + 12x + 84 y + 46 = 0 (d) 34x 2 + 24xy + 41 y 2 + 84x + 12 y + 81 = 0 Sol. (a) Let P(h, k ) be a point on the conic. Then, by definition

⇒ ⇒

SP = e ⋅ PM 1 4h − 3k + 2  (h + 1)2 + (k − 0)2 =   2 42 + (−3)2   1 2 2 2 (h + 1) + k = (4h − 3k + 2 ) 50

793

14

So, the locus of (h, k ) is 1 (4 x − 3 y + 2 )2 50 50{( x + 1)2 + y2 } = (4 x − 3 y + 2 )2

Objective Mathematics Vol. 1

( x + 1)2 + y2 =

⇒

⇒ 34 x2 + 41y2 + 24 xy + 84 x + 12 y + 46 = 0 Z

ax2 + 2 hxy + by2 + 2 gx + 2 fy + c = 0, we get a = 0, b = 0, h = 1, c = 17, g = 2, f = −3 ∴ ∆ = a bc + 2 f g h − a f 2 − bg 2 − ch2 S (–1, 0)

and

Z′

Hence, the equation of the conic is 34 x2 + 24 xy + 41y2 + 84 x + 12 y + 46 = 0

Recognition of Conic Section The general equation of second degree viz. ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 represents (i) a pair of straight lines, if ∆ = abc + 2 fgh − af

2

− bg 2 − ch 2 = 0

(ii) a pair of parallel (or coincident) straight lines, if ∆ = 0 and h 2 = ab (iii) a pair of perpendicular straight lines, if ∆ = 0 and a + b = 0 (iv) a point, if ∆ = 0 and h 2 < ab The general equation of the second degree ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0, will represent (i) a circle, if ∆ ≠ 0, a = b and h = 0. (ii) a parabola, if ∆ ≠ 0 and h 2 = ab. (iii) an ellipse, if ∆ ≠ 0 and h 2 < ab. (iv) a hyperbola, if ∆ ≠ 0 and h 2 > ab. (v) a rectangular hyperbola, if ∆ ≠ 0 and h 2 > ab and a + b = 0. X

Example 2. The name of the conic represented by the equation x 2 + y 2 − 2xy + 20x + 10 = 0 is (a) a hyperbola (b) an ellipse (c) a parabola (d) a circle Sol. (c) On comparing the given equation with the equation ax2 + 2 hxy + by2 + 2 gx + 2 fy + c = 0, we have a = 1, b = 1, h = − 1, c = 10, g = 10 and f = 0 ∴ abc + 2 fgh − af 2 − bg 2 − ch2 = 10 − 100 − 10 = − 100 ≠ 0 and h2 − ab = (−1)2 − 1 × 1 = 0 ⇒

h2 = ab

Thus, we have ∆ ≠ 0 and

h2 = ab

So, the given equation represents a parabola.

794

Example 3. The name of the conic represented by the equation 2xy + 4x − 6 y + 17 = 0 is (a) an ellipse (b) a circle (c) a parabola (d) a rectangular hyperbola Sol. (d) On comparing the given equation with

P (h, k)

4 x – 3y + 2 = 0

M

X

= − 12 − 17 = − 29 ≠ 0 h2 − ab = 1 − 0 > 0

Thus, we have ∆ ≠ 0 and h2 > ab, a + b = 0 So, the given equation represents a rectangular hyperbola.

Some Useful Terms Used in Conic Sections (i) Axis The straight line passing through the focus and perpendicular to the directrix is called the axis of the conic section. (ii) Vertex The point(s) of intersection of the conic section and the axis is (are) called the vertex (vertices) of the conic section. (iii) Centre The point which bisects every chord of the conic passing through it, is called the centre of the conic section. (iv) Latusrectum The latusrectum of a conic is the chord passing through the focus and perpendicular to the axis. (v) Focal chord Any chord passing through the focus is called a focal chord of the conic section. (vi) Double ordinate A chord perpendicular to the axis of a conic is known as its double ordinate. (vii) If S ≡ ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 is the equation of a conic, then the coordinates of its centre are  hf − bg hg − af    ,  ab − h 2 ab − h 2  In order to obtain the coordinates of the centre, we may solve the equations ∂S =0 ∂x ∂S and =0 ∂y ⇒ and

2ax + 2hy + 2g = 0 2hx + 2by + 2 f = 0

Example 4. The centre of the conic 14x 2 − 4xy + 11 y 2 − 44x − 58 y + 71 = 0 is (a) (3, 2) (b) (2, 4) (c) (2, 3) (d) (4, 2)

By definition of parabola, Z

x–y+5=0

∂S = 0 and ∂x

Then,

∂S =0 ∂y

⇒ 28 x − 4 y − 44 = 0 and 22 y − 4 x − 58 = 0 ⇒ 7 x − y − 11 = 0 and 2 x − 11y + 29 = 0 On solving these two equations, we get x = 2 and y = 3 Hence, the coordinates of the centre of given conic is (2, 3).

Parabola A parabola is the locus of a point which moves in a plane such that its distance from a fixed point in the plane is always equal to its distance from a fixed straight line in the same plane. It is evident from this definition of a conic, that a parabola is a conic section with eccentricity e =1.

P (x, y)

M

Sol. (c) Let S ≡ 14 x2 − 4 xy + 11y2 − 44 x − 58 y + 71 = 0 be the given conic.

14

Sol. (a) Let P( x, y) be any point on the parabola.

Parabola

X

S(3, – 4)

Z′

∴

SP = PM  x − y + 5 ( x − 3) + ( y + 4)2 =   2 2  1 +1  2

On squaring both sides, we get ( x − y + 5)2 ( x − 3)2 + ( y + 4)2 = ( 2 )2 2 2 ⇒ 2( x + y − 6 x + 8 y + 25) = x2 + y2 − 2 xy + 10 x − 10 y + 25 ⇒ x + y + 2 xy − 22 x + 26 y + 25 = 0 which is the required equation of parabola. 2

2

Standard Equation of Parabola Consider the focus of the parabola as S ( a, 0) and directrix is x + a = 0 and axis as X-axis.

Z

x=–a Y P

M

Directrix

M

S (Focus)

X′

Z

N

A

X

Example 5. The equation of parabola whose focus is (3, − 4) and directrix is x − y + 5 = 0, is (a) x 2 + y 2 + 2xy − 22x + 26 y + 25 = 0 (b) x 2 + y 2 + 2xy + 22x − 26 y + 25 = 0 (c) x 2 + y 2 + 2xy + 22x + 26 y − 25 = 0 (d) None of the above

X

S (a,0) y2 = 4ax

Z′

The fixed point is called the focus and the fixed straight line is called the directrix of the parabola. The line through the focus and perpendicular to the directrix is the axis of the parabola. The point on the axis mid-way between the focus and directrix is called the vertex of the parabola. Let S be the focus, ZZ′ be the directrix and P be any point on the parabola. Then, by definition SP = PM where, PM is the length of the perpendicular from P on the directrix ZZ′.

P(x, y)

Y′

Now, according to the definition of the parabola, for any point on the parabola, we must have SP = PM ⇒ ⇒ ⇒ ⇒

( x − a ) 2 + ( y − 0) 2 = PN + NM = x + a (x − a ) 2 + y 2 = (a + x ) 2 y 2 = (a + x ) 2 − (x − a ) 2 y 2 = 4ax

Vertex = (0, 0) Tangent at vertex, x = 0 Equation of latusrectum, x = a Extremities of latusrectum are ( a, 2a ), ( a, − 2a ) Length of latusrectum = 4a Focal distance, SP = PM = x + a Parametric forms are x = at 2 and y = 2at, where t is a parameter.

795

Objective Mathematics Vol. 1

14

Some Terms Related to Parabola (i) Axis The axis of a parabola is a straight line through the focus and perpendicular to the directrix. (ii) Vertex The point of intersection of the parabola and its axis is called its vertex. (iii) Double ordinate A straight line that is drawn perpendicular to the axis and terminates at both ends of the parabola, is a double ordinate of the parabola. (iv) Latusrectum The double ordinate passes through the focus is called latusrectum of parabola.

Coordinates of vertex Coordinates of focus Equation of the directrix Equation of the axis Length of the latusrectum Focal distance of a point P( x, y) Ø

●

●

(v) Chord A chord of a parabola is the line segment joining any two points on the parabola.

●

●

(vi) Focal chord A chord of a parabola passing through the focus is called focal chord.

●

(vii) Focal distance The focal distance of any point P ( x, y) on the parabola is its distance from the focus.

Other Standard Forms of Parabola y 2 = – 4ax

●

X

Y

P

M

L

X′

x=a

L′

x 2 = − 4ay

(0, 0)

(0, 0)

(0, 0)

(0, 0)

( a, 0)

( −a, 0)

( 0, a )

( 0, − a )

x= −a

x=a

y= −a

y=a

y=0

y=0

x=0

x=0

4a

4a

4a

4a

x+ a

a− x

y+ a

a− y

If the vertex of the parabola is at the point A(h, k) and its latusrectum is of length 4a, then its equation is (y − k)2 = 4 a(x − h) , if its axis is parallel to OX i.e. parabola opens rightward. (y − k)2 = − 4 a(x − h) , if its axis is parallel to OX′ i.e. parabola opens leftward. (x − h)2 = 4 a(y − k) , if its axis is parallel to OY i.e. parabola opens upward. (x − h)2 = − 4 a(y − k) , if its axis is parallel to OY ′ i.e. parabola opens downward. The general equation of a parabola whose axis is parallel to X-axis is x = ay 2 + by + c and the general equation of a parabola whose axis is parallel to Y-axis is y = ax 2 + bx + c .

Example 6. If the equation of parabola is x 2 = − 8 y, then (a) focus is (0, − 2) (b) directrix is y = 2 (c) length of latusrectum is 8 (d) All of the above

∴The focus is on Y-axis in the negative direction and parabola opens downward. On comparing the given equation with standard form, we get a = 2.

Y′ Y

Therefore, the coordinates of the focus are (0, − 2 ), then the equation of the directrix is y = 2 and the latusrectum is 4a i.e. 8.

x2= 4ay

S(0, a) P L¢

Hence, all of these are correct.

L

X′

X

A y = –a

M Y′ Y

M

y=a A

X′ L

X L′

S(0, –a) Y′

796

x 2 = 4ay

a is positive.

S(–a, 0)

P

y 2 = − 4ax

Sol. (d) The given equation is of the form x2 = − 4ay, where

X

A

y 2 = 4ax

x2= – 4ay

X

Example 7. The focus of the parabola y 2 − x − 2 y + 2 = 0, is 5  1  (b)  , 0 (a)  , 1 4  4  (c) (1, 1) (d) None of these Sol. (a) The equation of parabola can be rewritten as y2 − 2 y + 1 = x − 1 ⇒ ( y − 1)2 = x − 1 On putting x − 1 = X and y − 1 = Y , then the equation becomes 1 Y 2 = X ⇒ Y 2 = 4⋅ ⋅ X 4 5 1 =  , 1 ∴ Focus =  , 0  4 ( X , Y )  4 

X

Example 8. The vertex of the parabola x 2 + 2 y = 8x − 7 is  7 (a)  4,   2  9 (b)  4,   2 9  (c)  , 4 2  (d) (1, 0) Sol. (b) We have, x2 − 8 x = − 2 y − 7 9 ( x − 4)2 = − 2  y −   2 9 Clearly, the vertex is at  4,  .  2 ⇒

X

Sol. (b) On shifting the origin at (−1, 1,) we have …(i)

The coordinates of the end points of latusrectum are ( X = 1, Y = 2 ) and ( X = 1, Y = − 2 ). From Eq. (i), the coordinates of the end points of the latusrectum are (0, 3) and (0, − 1). X

(a)

4 ( 3 + 2) 3

(b)

4 (2 − 2 ) 3

(c)

4 3 2

(d)

2 ( 3 + 2) 3

y−0 x− 3 …(i) = =r 3 /2 1/ 2 On solving Eq.(i) with the parabola y2 = x + 2, we get r 3r 2 = + 4 2 ⇒ 3r 2 − 2 r − (4 3 + 8) = 0

L

V

S

V′

4( 3 + 2 ) 3

Parametric Equations of a Parabola The simplest and the best form of representing the coordinates of a point on the parabola y 2 = 4ax is ( at 2 , 2at ) because these coordinates satisfy the equation y 2 = 4ax for all values of t. The equations x = at 2 and y = 2at taken together are called the parametric equations of the parabola y 2 = 4ax, t being the parameter. Ø

●

Parabola

y 2 = 4 ax y 2 = − 4 ax x 2 = 4 ay x 2 = − 4 ay

Parametric (at 2 , 2 at) (− at 2 , 2 at) (2 at , at 2) (2 at , − at 2) coordinates Parametric equations

Sol. (b) Given, L, L′ are the ends of the latusrectum, S is mid-point of LL′ and VSV′ is the perpendicular bisector of 1 LL′, where VS =   LL′ = V ′S.  4

3+2

PA ⋅ PB = |r1 r2 | =

∴

Example 10. If the two ends of the latusrectrum are given, then the maximum number of parabolas that can be drawn, is (a) 1 (b) 2 (c) 0 (d) infinite

Clearly, two parabolas are possible.

14

Sol. (a) Given y − 3 x + 3 = 0 can be rewritten as

Example 9. The coordinates of an end point of the latusrectum of the parabola ( y − 1) 2 = 4( x + 1) are (a) (0, − 3) (b) (0, − 1) (c) (0, 1) (d) (1, 3) x = X − 1  y = Y + 1 From Eq. (i), the given parabola becomes Y 2 = 4X

Example 11. If the line y − 3 x + 3 = 0 cuts the parabola y 2 = x + 2 at A and B, then PA ⋅ PB is equal to [where, P = ( 3, 0)]

Parabola

X

●

X

x = at 2, y = 2 at

x = − at 2, y = 2 at

x = 2 at, y = at 2

x = 2 at, y = − at 2

The parametric equations of (y − k)2 = 4 a(x − h) are x = h + at 2 and y = k + 2 at.

Example 12. The point on y 2 = 4ax nearest to the focus has its abscissa (a) x = − a (b) x = a 3 (c) x = 2 (d) x = 0 Sol. (d) The focus is S (a, 0). If P(at 2 , 2 at ) is any point on the parabola, then SP 2 = a2 (t 2 − 1)2 + 4a2t 2 = a2 (t 2 + 1)2

L′

This is least for t = 0. Hence, the abscissa of P is x = 0 for SP to be least.

797

14

Work Book Exercise 14.1 5 If the parabola y 2 = 4ax passes through (3, 2),

Objective Mathematics Vol. 1

1 The name of the curve described parametrically by the equations x = t + t + 1, y = t − t + 1, is 2

2

a a circle c a hyperbola

b d

then the length of its latusrectum is

an ellipse a parabola

a c

2 The axis of parabola 9 y 2 − 16 x − 12 y − 57 = 0 is a 3y = 2 c 2x = 3

b d

x + 3y = 3 y=3

x = −1 3 x=− 2

b d

a (9, 6) c (25, − 10)

x=1 3 2

a

b d

y−a=0 y+ a=0

c

Position of a Point with Respect to a Parabola The point ( x1 , y1 ) lies outside, on or inside the parabola y 2 = 4ax according as y12 − 4ax1 >, = or < 0.

⇒

y12

>, = or
, = or < 4ax 2

Example 13. The ends of a line segment are P (1, 3) and Q (1, 1). R is a point on the line segment PQ such that PR : QR = 1 : λ. If R is an interior point of a parabola y 2 = 4ax, then

1+ λ 

Example 14. The number of points with non-negative integral coordinates that lie in the interior of the region common to the circle x 2 + y 2 = 16 and the parabola y 2 = 4x, is (a) 8 (b) 10 (c) 16 (d) None of these Sol. (a) (λ, µ ) is interior to both the curve, if λ 2 + µ 2 − 16 < 0 and

y 2 = 4ax

 4, 13     2   15 , 4   2 

b

2

X

y22

(25, 10) None of these

 1 + 3λ  It is an interior point of y2 − 4 x = 0, if   − 4 < 0.  1+ λ  3 Therefore, − < λ 0 ⇒ 0< λ < 1 ∴ λ ∈(0, 1)

Now, point ( x 2 , y2 ) will be outside, on or inside the parabola y 2 = 4ax according as PM >, = or < QM PM 2 >, = or < QM 2

 13 , 4   2   4, 15     2 



y22 = 4ax 2

⇒

b d

  Sol. (a) The coordinates of R are  1, 1 + 3λ  .

Let P ( x1 , y1 ) be a point in the plane. From P, draw PM ⊥ AX , meeting the parabola y 2 = 4ax at Q. Let the coordinates of Q be ( x 2 , y2 ). Since, Q lies on the parabola y 2 = 4ax. ∴

4

are (7, 5) and (7, 3), then possible coordinates of its vertex are

x = − 2 at , y = − at 2 , t ∈ R, is x−a=0 x+ a=0

d

7 The end points of the latusrectum of a parabola

x=

4 The equation of the directrix of the parabola a c

4 3

y 2 = 4 x and above its axis, is 10 units. Its coordinates are

y 2 + 4 y + 4 x + 2 = 0 is c

b

6 The focal distance of a point on the parabola

3 The equation of the directrix of the parabola a

2 3 1 3

Now,

µ 2 − 4λ < 0

µ µ 2 − 4λ < 0 ⇒ λ >   2

2

If µ = 0, λ = 1, 2, 3; if µ = 1, λ = 1, 2, 3; if µ =2, λ = 2, 3 and if µ = 3, λ = 3, 4. Also, λ 2 + µ 2 − 16 < 0 ⇒

λ 2 < 16 − µ 2

If µ = 0, λ = 1, 2, 3; if µ = 1, λ = 1, 2, 3; if µ = 2, λ = 2, 3 and if µ = 3, λ = 1, 2. Hence, (1, 0), (2, 0), (3, 0), (1, 1), (2, 1), (3, 1), (2, 2), (3, 2) are the possible points.

(v) The length of the focal chord which makes an angle θ with positive direction of X-axis is 4a cosec 2θ.

(i) Parametric equation of a chord Let P ( at12 , 2at1 ) and Q ( at 22 , 2at 2 ) be any two points on the parabola y 2 = 4ax. Then, the equation of the chord PQ is 2at 2 − 2at1 y − 2at1 = ( x − at12 ) 2 2 at 2 − at1 ⇒ ⇒

(vi) Circle described on the focal length as diameter touches the tangent at vertex. 2, 2 (a t P

Y

2 ( x − at12 ) t 2 + t1 y ( t1 + t 2 ) = 2x + 2at1 t 2

14 Parabola

Equation of a Chord

at)

y − 2at1 =

(ii) The equation of the chord joining points t1 and t 2 on the parabola y 2 = 4ax is y ( t1 + t 2 ) = 2x + 2at1 t 2 If the chord PQ is a focal chord of the parabola, then ( a, 0) must satisfy the above equation. 0 = 2a + 2at1 t 2 ⇒ t1 t 2 = − 1 It follows from this that t is the parameter for one end of a focal chord of the parabola y 2 = 4ax, then the parameter for the other end is −1/ t and the coordinates of the end points of a focal chord PQ of the parabola y 2 = 4ax can be 2a  a taken as P ( at 2 , 2at ) and Q  2 , −  . t t  (iii) Length of a focal chord Let P ( at 2 , 2at ) be one end of a focal chord PQ of the parabola y 2 = 4ax. Then, the coordinates of the other end  a − 2a  Q are  2 , . t t 

X′

(vii) Circle described on the focal chord as diameter touches directrix. Y 2,

P(at

X′ (–a, 0)

S O

X

(a, 0)

Q

a , – 2a t2 t

Y′ X

Example 15. If (2, − 8) is at an end of a focal chord of the parabola y 2 = 32x, then the coordinate of the other end of chord is (a) (8, − 2) (b) (16, 32) (c) (32, 32) (d) None of the above Sol. (c) We have, y2 = 32 x Here, a=8 Then, any point on this parabola is (8t 2 , 16 t ). On comparing this with point (2, − 8), we get t = −

1 2

a −2 a  Now, the other end of focal chord is  2 , . t t  Hence, the coordinates of the other end of the focal chord is (32, 32).

⇒ PQ ≥ 4a

i.e. Length of latusrectum = 2 (Harmonic mean of focal segment)

2at)

x = –a

2

Thus, the length of the smallest focal chord of the parabola is 4a, which is the length of its latusrectum. Hence, the latusrectum of a parabola is the smallest focal chord. (iv) If l1 and l2 are length of focal segments, then 4l l 4a = 1 2 l1 + l2

X

S(a, 0)

Y′

The length of the focal chord of the parabola 2  1 y 2 = 4ax is a  t +  , where t is the parameter  t for one end of the chord. We know that,   ≥ 2, ∀ t ≠ 0  t + 1 t   1 ∴ a  t +  ≥ 4a  t

O

X

Example 16. Circles are drawn with diameter being any focal chord of the parabola y 2 − 4x − y − 4 = 0 will always touch a fixed line. The equation of line is (a)16x + 17 = 0 (b)16x + 33 = 0 (c) 4 y + 17 = 0 (d) 4 y + 33 = 0

799

14

Sol. (b) We have, y2 − 4 x − y − 4 = 0

Objective Mathematics Vol. 1

Intersection of a Line and a Parabola

17 1 = 4x + 4 4 2  y − 1  = 4  x + 17        16  2

y2 − y +

⇒ ⇒

Y

Circle drawn with diameter being any focal chord of the parabola always touches the directrix of the parabola. 17 Thus, circle will touch the line at x + = −1 16 i.e. 16 x + 33 = 0 X

Example 17. The circles on focal radii of a parabola as diameter touch (a) the tangent at the vertex (b) the axis (c) the directrix (d) None of the above Sol. (a) The circle on SP as diameter is

O

P (at 12 , 2at1)

X S (a, 0)

D>0 Y′

Let the parabola be y 2 = 4ax

…(i)

and the given line be y = mx + c Eliminating y from Eqs. (i) and (ii), we get

…(ii)

( mx + c) 2 = 4ax or

m 2 x 2 + 2x ( mc − 2a ) + c 2 = 0

It meets Y-axis when x = 0. ⇒ y2 − 2 at 1 y + a2 t 12 = 0

a >, = or < mc

( y − at 1 )2 = 0

Example 18. The length of the chord of the parabola x 2 = 4ay passing through the vertex and having slope tan α, is (a) 4a cosec α cot α (b) 4a tan α secα (c) 4a cos α cot α (d) 4a sin α tan α Sol. (b) Let A (0, 0) be the vertex and AP be a chord of x2 = 4ay such that slope of AP is tan α.

The line y = mx + c touches the parabola y 2 = 4ax, a if c = and the coordinates of the point of contact are m  a 2a   2 , . m m The conditions of tangency of a line and a parabola in different forms are listed below for ready reference:

at 2 t = 2 at 2 t tan α = 2

Slope of AP =

Now,

t = 2 tan α AP = {(2 at − 0) + (at − 0) } 2

2

2

= at (4 + t 2 ) = 2 a tan α (4 + 4 tan α ) 2

= 4a tan α sec α

800

Condition of tangency

Parabola

Line

Point of contact

y 2 = 4ax

y = mx + c

 a , 2a  2  m m

c=

y 2 = − 4ax

y = mx + c

 − a , − 2a    m2 m

c=−

x 2 = 4ay

x = my + c

 2a, a     m m2 

c=

x 2 = − 4ay

x = my + c

 − 2a, − a     m m2 

c=−

Again, let the coordinates of P be (2 at , at 2 ). Then,

⇒

…(iv)

Condition of Tangency

Clearly, Y-axis meets the circle in two coincident point. Hence, the circle touches the tangent at the vertex.

⇒

…(iii)

4a 2 − 4amc >, = or < 0

or

X

X

D < 0D = 0

or

⇒

S (a, 0)

O

This equation, being quadratic in x, gives two values of x and shows that every straight line will cut the parabola in two points may be real, coincident or imaginary, according as discriminant of Eq. (iii) is greater, equal or less than 0. i.e. 4( mc − 2a ) 2 − 4m 2 c 2 >, = or < 0

( x − at 12 )( x − a) + ( y − 2 at 1 )y = 0 Y

X′

( y − k )2 = 4a( x − h)

y = mx + c

y = mx + c ( y − k )2 = − 4a( x − h)

a m a m

a m a m

 h + a , k + 2a    m m2

mh + c

 h − a , k − 2a    m m2

mh + c

=k+

a m

=k−

a m

( x − h)

2

Line

Point of contact

x = my + c

 h + 2a, k + a     m m2 

= 4a( y − k )

x = my + c

( x − h)2 = − 4a( y − k )

X

(i) Point form The equation of the tangent to the parabola y 2 = 4ax at a point ( x1 , y1 ) is given by

Condition of tangency

 h − 2a, k − a     m m2 

mk + c

yy1 = 2a ( x + x1 )

a =h+ m

Ø

●

mk + c =h−

a m

●

Example 19. If the straight line y = mx + c meets the parabola y 2 = 8x in real and distinct points, then 2 2 2 2 (b) c > (c) | c | < (d) | c | > (a) c < m m m m

y = mx + c (mx + c )2 = 8 x

and ⇒

m x + 2 mxc + c − 8 x = 0

⇒

m2 x2 + (2 mc − 8)x + c 2 = 0

2 2

⇒ X

y = 4 ax

yy1 = 2 a(x + x1)

y = − 4 ax

yy1 = − 2 a(x + x1)

x = 4 ay

xx1 = 2 a(y + y1)

x 2 = − 4 ay

xx1 = − 2 a(y + y1)

2

(ii) Parametric form The equation of the tangent to the parabola y 2 = 4ax at ( at 2 , 2at ) is given by ty = x + at 2

2

Ø The parametric equations of tangents to all standard forms of

parabola are as given below :

For real and distinct points, (2 mc − 8)2 − 4m2c 2 > 0 ⇒

Equation of tangent

2

Sol. (a) Given, equation of parabola is y = 8 x and equation y2 = 8 x

Equation of parabola 2

2

of line is y = mx + c. From

The equation of the tangent at (x1 , y1) to any second degree curve can also be obtained by replacing x 2 by xx1, y 2 by yy1, x x + x1 y + y1 xy + x y , y by and xy by 1 1 and without by 2 2 2 changing the constant (if any) in the equation of the curve. The equation of tangents to all standard forms of parabola at point (x1 , y1) are given below for ready reference.

4m2c 2 − 32 mc + 64 − 4m2c 2 > 0 2 32 mc < 64 ⇒ c < m

Example 20. If the straight line x cos α + y sin α = p touches the parabola y 2 = 4ax, then the point of contact is (a) a cos α, a sin α (b) − p sec α, 2 p cosec α (d) None of these (c) a cot 2α, − 2a cot α Sol. (b) The given line is xcos α + ysin α = p or y = − xcot α + p cosec α Since, the given line touches the parabola. a or a = cm c= ∴ m ⇒ ( p cosec α )(− cot α ) = a and the point of contact a 2a =  2 ,  m m a −2 a  , =    cot 2 α cot α  = − p sec α, 2 p cosec α

Equation of Tangent In this section, we will obtain the equation of tangents to various forms of the parabola in terms of slope of the tangent or in terms of the coordinates of the point of contact of the tangent line and the parabola or in terms of value of the parameter of the point of contact.

14 Parabola

Parabola

Equation of parabola

Point of contact

Equation of the tangent

y 2 = 4 ax

(at 2 , 2 at)

t y = x + at 2

y 2 = − 4 ax

(−at 2 , 2 at)

x 2 = 4 ay

(2 at , at 2)

x 2 = − 4 ay

(2 at , − at 2)

t y = − x + at 2 t x = y + at 2 t x = − y + at 2

(iii) Slope form The equation of the tangent of slope m to the parabola y 2 = 4ax is y = mx + Ø

●

●

a  a 2a  at  , . m  m2 m 

The equation of a tangent of slope m to the parabola a (y − k)2 = 4 a(x − h) is given by y − k = m(x − h) + . The m 2 a a  coordinates of the point of contact are h + 2 , k +  .  m m It should be noted that to obtained tangent and the coordinates of the point of contact of tangent to x 2 = 4 ay, we can interchange x and y in the equation of the tangent to y 2 = 4 ax and also in the coordinates of the point of contact. Similarly, relationship exists between x 2 = − 4 ay and y 2 = − 4 ax.

(iv) Point of intersection of tangents Let tangent at P ( at12 , 2at ) and Q ( at 22 , 2at 2 ) intersect at R. Then, their point of intersection is R [ at1 t 2 , a ( t1 + t 2 )] i.e. (GM of abscissa, AM of ordinate).

801

X

Objective Mathematics Vol. 1

14

Example 21. If the line y = 3x + c touches the parabola y 2 = 12x at point P, then the equation of the tangent at point Q, where PQ is a focal chord, is (a) x + 3 y + 27 = 0 (b) 3x − y − 27 = 0 (c) x + 3 y − 27 = 0 (d) None of the above

X

Sol. (a) Line y = 3 x + c touches y2 = 12 x, then we must

Sol. (c) The equation of any tangent to y2 = 4 ( x + 1) is y = m ( x + 1) +

have a 3 i.e. c = = 1 c= m 3 a 2a  1   The point of contact P is  2 ,  =  , 2  m m 3 

X

2 m′ It is given that Eqs. (i) and (ii) are perpendicular. 1 On putting m′ = − in Eq. (ii), we get m 1 y =  −  ( x + 2 ) − 2 m  m y = m′ ( x + 2 ) +

Example 22. The equation of tangent to the parabola y 2 = 8x having slope 2, is (a) y = 2x + 1 (b) y = 2x + 4 (c) y = 2x + 3 (d) None of these a . m For the given parabola y2 = 8 x, we have a = 2 and m = 2 2 ∴ Equation of tangent is y = 2 x + 2 ⇒ y = 2x + 1 is y = mx +

Sol. (d) Let P( x1, y1 ) be the point of contact of the two

⇒

802

x1 y1 = 4a2

Hence, the locus of ( x1, y1 ) is xy = 4a2 , which is a hyperbola.

…(ii)

…(iii)

X

Example 25. From an external point P, tangents are drawn to the parabola y 2 = 4ax, then the equation to the locus of P when these tangents makes angles θ 1 and θ 2 with the axis, such that tan θ 1 + tan θ 2 is constant ( = b), is x (a) y = (b) y = bx b (c) y = b 2 x (d) None of these Sol. (b) Let the coordinates of P (h, k ) and the given

Example 23. If two parabolas y 2 = 4a (x − λ 1 ) and x 2 = 4a ( y − λ 2 ) always touch each other, where λ 1 and λ 2 being variable parameters. Then, their points of contact lie on a (a) straight line (b) circle (c) parabola (d) hyperbola parabolas.Tangents at P to the two parabolas are yy1 = 2 a( x + x1 ) − 4aλ1 ⇒ 2 ax − yy1 = 2 a(2 λ1 − x1 ) and xx1 = 2 a( y + y1 ) − 4aλ 2 ⇒ xx1 − 2 ay = 2 a( y1 − 2 λ 2 ) Clearly, Eqs. (i) and (ii) represent the same line. y 2a ∴ = 1 x1 2 a

…(i)

On subtracting Eq. (iii) from Eq. (i), we get 1 1 0 =  m +  x + 3  m +    m m Q m + 1 ≠ 0 ⇒ x+ 3=0  m 

x + 3 y + 27 = 0

Sol. (a) Equation of the tangent to y2 = 4ax having slope m

X

1 m

and tangent to y2 = 8 ( x + 2 ) is

On comparing this point to (at 2 , 2 at ), we get 1 2 at = 2 or t = 3 1 Hence, point P has parameter , then point Q has 3 parameter −3. Now, tangent at point Q is (−3)y = x + 3 (−3)2 or

Example 24. The locus of the point of intersection of tangents to the parabola y 2 = 4( x + 1) and y 2 = 8( x + 2) which are perpendicular to each other, is (a) x + 7 = 0 (b) x − y = 4 (c) x + 3 = 0 (d) y − x =12

…(i) …(ii)

equation of the parabola be y2 = 4ax. Any tangent on the a parabola is given by y = mx + . m If this passes through (h, k ), then a k = mh + m …(i) ⇒ m2 h − mk + a = 0 which is a quadratic in m. k Let its roots be m1 and m2 , then m1 + m2 = and h a m1 m2 = . Now, if the two tangents through P make h angles θ1 and θ2 with X-axis and m1 = tan θ1 and m2 = tan θ2 . k …(ii) tan θ1 + tan θ2 = ∴ h a and …(iii) m1 m2 = h a …(iv) tan θ1 tan θ2 = ⇒ h By hypothesis, tan θ1 + tan θ2 = b From Eq. (ii), we get k = b ⇒ k = bh h Generalising, the locus of (h, k ) is y = bx.

X

In this section, we will be discussing some useful properties of tangents to a parabola. (i) The tangent at any point on a parabola bisects the angle between the focal distance of the point and the perpendicular on the directrix from the point. (It is known as reflection property of parabola). (ii) The tangent at the extremities of a focal chord of a parabola intersect at right angles on the directrix. (iii) The portion of the tangent to a parabola cut off between the directrix and the curve subtends a right angle at the focus. (iv) The perpendicular drawn from the focus on any tangent to a parabola intersect it at the point where it cuts the tangent at the vertex. (v) If SZ is perpendicular to the tangent at a point P of a parabola, then Z lies on the tangent at the vertex and SZ 2 = AS⋅ SP, where A is the vertex of the parabola. Y

14

Sol. (b) Let the coordinates of P and Q be (at 12 , 2 at 1 ) and

(at 22 , 2 at 2 ) respectively. Then, y1 = 2 at 1 and y2 = 2 at 2 . The coordinates of the point of intersection of the tangents at P and Q are [at 1 t 2 a (t 1 + t 2 )]. ∴ y3 = a (t 1 + t 2 ) y + y2 y3 = 1 ⇒ 2 So, y1, y3 , y2 are in AP.

X

P

Example 28. If parabola of latusrectum l, touches a fixed equal parabola, the axes of the two curves being parallel, then the locus of the vertex of the moving curve is (a) a parabola of latusrectum 2 l (b) a parabola of latusrectum l (c) an ellipse whose major axis is 2 l (d) an ellipse whose minor axis is 2 l Sol. (a) Let (h, k ) be the coordinates of the vertex of the moving parabola and its equation be ( y − k )2 + l ( x − h) = 0

Z X′

Example 27. If y1 , y2 are the ordinates of two points P and Q on the parabola and y3 is the ordinate of the point of intersection of tangents at P and Q, then (a) y1 , y2 , y3 are in AP (b) y1 , y3 , y2 are in AP (c) y1 , y2 , y3 are in GP (d) y1 , y3 , y2 are in GP

Parabola

Some Useful Results on Tangents

A

S

The equation of the fixed curve y2 − l x = 0; then the tangents at the point (lt 2 , lt ) to the two curves are

X

−2 yt + x + lt 2 = 0 l x + l 2 t 2 + 2 k 2 − 2 l h − 2 k lt = 0 2 Since, they are coincident. k − lt l / 2 l 2 t 2 + 2 k 2 − 2 l h − 2 lt ∴ = = 2t 1 2 lt 2 k from which and k 2 = l h = k lt t = 2l and

Y′

(vi) The orthocentre of any triangle formed by three tangents to a parabola lies on the directrix. (vii) The circumcircle formed by the intersection points of tangents at any three points on a parabola passes through the focus of the parabola. (viii) The tangent at any point of a parabola is equally inclined to the focal distance of the point and the axis of the parabola. (ix) The length of the subtangent at any point on a parabola is equal to twice the abscissae of the point. X

Example 26. A ray of light moving parallel to the X -axis gets reflected from a parabolic mirror whose equation is ( y − 2) 2 = 4( x + 1). After reflection, the ray must pass through the point (a) (0, 2) (b) (2, 0) (c) (0,− 2 ) (d) (−1, 2) Sol. (a) The equation of the axis of the parabola is y − 2 = 0, which is parallel to the X-axis. So, a ray parallel to the axis of the parabola.We know that any ray parallel to the axis of a parabola passes through the focus after reflection. Here, (0, 2) is the focus.

y(lt − k ) +

Eliminating t, we get k 2 = 2 l h. Hence, the locus of the vertex is y2 = 2 lx, which is a parabola having latusrectum as 2 l.

Common Tangents In this section, we shall discuss some problems on finding the common tangents to two conics. X

Example 29. If two equal parabolas having the same vertex and their axes are at right angles. Then, the length of their common tangent is 3 (a) 3a (b) 3 2a (c) (d) 2a a 2 Sol. (b) Let the two equal parabolas be y2 = 4ax and x2 = 4ay. The equation of any tangent to y2 a at the point y = mx + m

= 4ax is  a , 2a .  2  m m

…(i)

803

14

y = mx +

Now,

a m a

y − m m2 It also touches the parabola x2 = 4ay. −a a Then, = 2 1/ m m 3 ⇒ m = −1

Objective Mathematics Vol. 1

⇒

x=

⇒

⇒

So, the coordinates of the point of contact are P(a, − 2 a). The common tangent of Eq. (i) touches x2 = 4ay at Q x = my + a touches   a  m  2a 2 ,     ≡ Q (2 am, am ) a a 2  1 / m 1 / m2  2    x = 4ay at  ,   2  m m    ≡ Q (− 2 a, a) Hence, length of the common tangent PQ = (a + 2 a)2 + (− 2 a − a)2 = 3 2a

Example 30 If two lines are drawn at right angles, one being a tangent to y 2 = 4ax and the other to x 2 = 4by. Then, the locus of their point of intersection is the curve (a) ( ax + by) ( x 2 + y 2 ) + ( bx − ay) 2 = 0 (b) ( ax + by) 2 ( x 2 + y 2 ) + ( bx − ay) = 0 (c) ( bx + ay) ( x 2 + y 2 ) + ( ax − by) 2 = 0 (d) None of the above …(i)

m be a tangent to y2 = 4ax x = m1 y +

and

b m1

…(ii)

be a tangent to x2 = 4by. Lines (i) and (ii) will be perpendicular, if 1 m× = −1 m1 ⇒ m1 = − m Let (h, k ) be the point of intersection of Eqs. (i) and (ii). a Then, k = mh + m b and h = m1k + m1 a k = mh + ⇒ m b and h = − mk − m ⇒ m2 h − mk + a = 0 and

804

The angle θ between the tangents at point of intersection of two parabolas C1 and C 2 is defined as the angle of intersection of two parabolas and is given by m − m2 tan θ = 1 1 + m1 m2 where, m1 and m2 are slopes of tangent at point of intersection of two parabolas C1 and C 2 , respectively. π If θ = , then two parabolas are said to be intersect 2 orthogonally and condition for the same is m1 m2 = − 1 If θ = 0 or π, then the two parabolas touch each other and the condition for same is m1 = m2 . X

Example 31. If two parabolas y 2 = 4ax and π x 2 = 4 by intersect at an angle at a point other 3 1/ 3 1/ 3 a  b than the origin, then the value of   +    b a is 3 (a) 3 (b) 2 3 (c) − (d) None of these 2 Sol. (b) By solving y2 = 4ax and x2 = 4by, we get the points of intersection are (0, 0) and (4a1/ 3 b1/ 3 , 4a2 / 3 b1/ 3 ). The tangent to y2 = 4ax at (4a1/ 3 b1/ 3 , 4a2 / 3 b1/ 3 ) is y ⋅ 4a2 / 3 b1/ 3 = 2 a( x + 4a1/ 3 b 2 / 3 )

1 m2 m = = − kb − ah ak − bh h2 + k 2 m=

kb + ah and bh − ak

[by cross-multiplication] ak − bh h + k 2

2

…(i)

Similarly, x2 = 4by is x4a1/ 3 b1/ 3 = 2 a( y + 4a2 / 3 b1/ 3 ) …(ii) Slope of tangents are

a1/ 3

and

2 b1/ 3

a1/ 3 π tan = 3 ⇒ ⇒

 a    b

1/ 3

b +    a

1/ 3

=

1/ 3

2 a1/ 3 b1/ 3 −

, respectively.

2 a1/ 3

b1/ 3 a 2 a1/ 3 1+ . 1/ 3 1/ 3 b 2b 3  a1/ 3 b1/ 3  3 =  2/ 3  2 a + b2 / 3 

m2 k + mh + b = 0

∴

⇒

⇒ ( x2 + y2 ) (ax + by) + (bx − ay)2 = 0

Angle of Intersection of Two Parabolas

m= −1

Sol. (a) Let y = mx + a

(h2 + k 2 ) (kb + ah) = − (bh − ak )2

Hence, the locus of (h, k ) is ( x2 + y2 ) (ax + by) = − (bx − ay)2

Qusing c = a   m 

a 2a The common tangent touches y2 = 4ax at  2 ,  . m m

X

kb + ah ak − bh = bh − ak h2 + k 2

⇒

2b

1/ 3

3 2

4 If a focal chord of y 2 = 16 x is a tangent to

1 A tangent to the parabola y 2 = 4ax is inclined at

( x − 6)2 + y 2 = 2, then the values of the slope of this chord will be

π / 3 with the axis of the parabola. The point of contact is  a −2 a  a  ,  3 3 c ( 3a, 2 3a )

a { − 1, 1} c { − 2, 1 / 2}

b ( 3a, − 2 3a )

parabola y 2 = 8 x orthogonally at the point P. The equation of the tangent to the parabola at P can be

which goes through the point (4, 10), is a x + 4y + 1 = 0 b 9x + 4y + 4 = 0 c x − 4 y + 36 = 0 d 9x + 4y + 9 = 0

a c

7 The angle between the tangents drawn from the point (1, 4) to the parabola y 2 = 4 x is a

(iii) Slope form The equation of normal to the parabola y 2 = 4ax in terms of its slope m, is given by y = mx − 2am − am 3 at the point (am 2 , − 2am). The equation of normals to various standard forms of the parabola in terms of the slope of the normal are as given below: Slope of the normal

Point of contact (Foot of the normal)

3

m

(am ,− 2 am)

y 2 = − 4 ax y = mx + 2 am + am3

m

(− am2 , 2 am)

1/ m

(− 2 am, am2)

x 2 = 4 ay

Equation of the normal y = mx − 2 am − am

x = my − 2 am − am3

2

π/6

b

π/4

c

x 2 = − 4 ay x = my + 2 am + am3

In this section, we will discuss equation of normal to a parabola in different forms. (i) Point form The equation of normal to the parabola y 2 = 4ax at a point ( x1 , y1 ) is given by y y − y1 = − 1 ( x − x1 ) 2a (ii) Parametric form The equation of normal to the parabola y 2 = 4ax at point ( at 2 , 2at ) is given by y + tx = 2at + at 3

y = 4 ax

b

c tan−1 (1/2 )

Equation of Normal to Parabola

2

1 tan−1 (1/2 ) 2 π d 4

a tan−1 ( 3 / 4)

AP GP HP None of the above

Equation of the parabola

b 2x + y − 2 = 0 d 2x − y + 1= 0

P(16, 16) of the parabola y 2 = 16 x. If S is the focus of the parabola, then ∠TPS can be equal to

parabola y 2 = 4ax with coordinates (at i2 , 2 at i ); i = 1, 2, 3, where t1, t 2 and t 3 are in AP. If AA′, BB′ and CC′ are focal chords and coordinates of A′, B′, C′ are (at i′ 2, 2 at i′ ); i = 1, 2, 3, then t1′ , t 2′ and t 3′ are in

●

x− y−4=0 x+ y−4=0

6 From a point T, a tangent is drawn at the point

3 Let A, B and C be three points taken on the

Ø

b { − 2, 2} d {2, − 1 / 2}

5 The circle x 2 + y 2 − 2 x − 6 y + 2 = 0 intersects the

d None of these

2 The equation of tangent to the parabola y 2 = 9 x

a b c d

Parabola

14

Work Book Exercise 14.2

●

X

π/3

1/ m

d

π /2

(2 am , − am2)

The equation of normals to the parabolas reducible to one of the standard forms are given below for ready reference : Parabola

Equation of the normal

(y − k)2 = 4 a (x − h)

y − k = m(x − h) − 2 am − am3

(y − k)2 = − 4 a (x − h)

y − k = m(x − h) + 2 am + am3

(x − h)2 = 4 a (y − k)

x − h = m(y − k) − 2 am − am3

(x − h)2 = − 4 a (y − k)

x − h = m(y − k) + 2 am + am3

Example 32. The equation of the normals at the ends of the latusrectum of the parabola y 2 = 4ax are given by (a) x 2 − y 2 − 6ax + 9a 2 = 0 (b) x 2 − y 2 − 6ax − 6ay + 9a 2 = 0 (c) x 2 − y 2 + 6ay + 9a 2 = 0 (d) None of the above Sol. (a) The coordinates of the ends of the latusrectum of the parabola y2 = 4ax are (a, 2 a) and (a, − 2 a) respectively. The equation of the normal at (a, 2 a) to y2 = 4ax is 2a ( x − a) y − 2a = − 2a ⇒ x + y − 3a = 0

…(i)

805

Objective Mathematics Vol. 1

14

Similarly, the equation of the normal (a, − 2 a) is x − y − 3a = 0 On combining Eqs. (i) and (ii), we get x2 − y2 − 6ax + 9a2 = 0 X

…(ii)

Example 33. If the tangents and normals at the extremities of a focal chord of a parabola intersect at ( x1 , y1 ) and ( x 2 , y2 ) respectively, then (b) x1 = y2 (a) x1 = y2 (c) y1 = y2 (d) x 2 = y1 Sol. (c) Let P (at 12 , 2 at 1 ) and Q (at 22 2 at 2 ) be the extremities of a focal chord of the parabola y2 = 4ax. The tangents at P and Q intersect at [at 1 t 2 , a (t 1 + t 2 )]. ∴ x1 = at 1 t 2 and y1 = a(t 1 + t 2 ) ⇒ x1 = − a and y1 = a(t 1 + t 2 ) [QPQ is a focal chord. ∴t 1 t 2 = − 1] The normals at P and Q intersect at [2 a + a (t 12 + t 22 + t 1t 2 ), − at 1 t 2 (t 1 + t 2 )] ∴

x2 = 2 a + a (t 12 + t 22 + t 1 t 2 )

and ⇒

y2 = − at 1 t 2 (t 1 + t 2 ) x2 = 2 a + a (t 12 + t 22 − 1) = a + a (t 12 + t 22 )

and Clearly,

y2 = a(t 1 + t 2 ) y1 = y2

(v) If the normals at two points P and Q of a parabola y 2 = 4ax intersect at a third point R on the curve, then the product of the ordinates of P and Q is 8a 2 . (vi) If the normal chord at a point P ( at 2 , 2at ) to the parabola y 2 = 4ax subtends a right angle at the vertex of the parabola, then t 2 = 2. (vii) The normal chord of a parabola at a point whose ordinate is equal to the abscissae, subtends a right angle at the focus. (viii) The normal at any point of a parabola is equally inclined to the focal distance of the point and the axis of the parabola. X

Example 34. The normal at the point P ( ap 2 , 2ap) meets the parabola at Q ( aq 2 , 2aq ) such that the lines joining the origin to P and Q are at right angle. Then, (a) p 2 = 2 (b) q 2 = 2 (c) p = 2q (d) q = 2 p Sol. (a) Since, the normal at (ap2 , 2 ap) to y2 = 4ax meets the parabola at (aq 2 , 2 aq ).

q = − p−

parabola y 2 = 4ax at ( at 22 , 2at 2 ), then 2 t 2 = − t1 − t1

⇒ ⇒ ⇒ X

Y

Q (h,k)

X

y 2= 4ax

(ii) The tangent at one extremity of the focal chord of a parabola is parallel to the normal at the other extremity. (iii) The normals at points P ( at12 , 2at1 ) and Q ( at 22 , 2at 2 ) to the parabola y 2 = 4ax intersect at the point [2a + a ( t12 + t 22 + t1 t 2 ), − at1 t 2 ( t1 + t 2 )].

806

(iv) If the normals at points P ( at 12, 2at ) and Q ( at 22 , 2at 2 ) on the parabola y 2 = 4ax meet on the parabola, then t1 t 2 = 2.

at 2 = 2 at

⇒ t =2

So, the coordinates of P are (4a, 4a). Let PQ be a normal chord at P to the parabola. Let the coordinates of Q be (at 12 , 2 at 1 ). 2 2 Then, t1 = t − = − 2 − = − 3 2 t So, the coordinates of Q are (9a, − 6a). 4a −6a ∴Slope of SP × Slope of SQ = × = −1 3a 8a Thus, the normal chord makes a right angle at the focus.

A

Y′

[from Eq. (i)]

Sol. (a) Let P(at 2 , 2 at ) be a point on the parabola y2 = 4ax.

P (at 2, 2at) G

pq = − 4 2  p − p −  = − 4  p p2 = 2

Example 35. The normal chord of a parabola y 2 = 4ax at a point whose ordinate is equal to abscissa, subtends a right angle at the (a) focus (b) vertex (c) end of the latusrectum (d) None of these Then,

X′

…(i)

Since, OP ⊥ OQ. 2 ap − 0 2 aq − 0 = −1 × ap2 − 0 aq 2 − 0

Some Useful Results on Normals to a Parabola In this section, we will be discussing some useful results on normals to a parabola. These results are very useful to solve problems of parabola. (i) If the normal at the point P ( at12 , 2at1 ) meets the

2 p

X

Example 36. P is the point t on the parabola y 2 = 4ax and PQ is a focal chord. PT is the tangent at P and QN is the normal at Q. If the angle between PT and QN is α and the distance between PT and QN is d, then (a) 0 < α < 90° (b) α = 90°

 1 (d) d = a  1 + t 2 +  2 t 1+ t2 

are called conormal points. The conormal points are also called the feet of the normals.

   

A

Sol. (d) Since, PQ is the focal chord. a 2a So, coordinates of P and Q are (at , 2 at ) and  2 , − , t t  respectively. ∴ Equation of the tangent at P is t y = x + at 2 . 1 Slope of PT = ∴ t Equation of the normal at Q is xt 2 − yt 3 − a − 2 at 2 = 0 1 Slope of QN = t QPT and QN are parallel. ∴ α=0 Given, distance betwene PT and QN is d.    − at 4 − a − 2 at 2  1 . So, d =   = a  1+ t2 +    t 2 1 + t 2  (t 4 + t 6 ) 

X′

2

X

Example 37. If the normal to the parabola y 2 = 4ax at the point ( at 2 , 2at ) cuts the parabola again at ( aT 2 , 2aT ), then (a) − 2 ≤ T ≤ 2 (b) T ∈ ( − ∞, − 8) ∪ (8, ∞ ) (c) T 2 < 8 (d) T 2 ≥ 8 Sol. (d) Equation of the normal of the parabola y2 = 4ax at the point (at 2 , 2 at ) is y + tx = 2 at + at 3

…(i)

Since, Eq. (i) cuts the parabola again at (aT 2 , 2 aT ). ∴ ⇒

2 aT + taT = 2 at + at 2

⇒

3 2

2 = − t (T + t )

[Qt ≠ T]

t 2 + tT + 2 = 0

Since, t is a real. ∴ ⇒

y 2 = 4ax P

O

X B C

Y′

In figure A, B , C are conormal points with respect to point P.

Some Useful Results on Conormal Points Let P ( h, k ) be a point and y 2 = 4ax be a parabola. The equation of any normal to the parabola y 2 = 4ax is y = mx − 2am − am 3 If it passes through the point P ( h, k ), then k = mh − 2am − am 3 ⇒

am 3 + m(2a − h) + k = 0

…(i)

This is cubic equation in m. So, it gives three values of m, say m1 , m2 and m3 . Corresponding to each value of m, there is a normal passing through the point P ( h, k ). Let A, B , C be the feet of the normals. Then, AP, BP and CP are three normals passing through point P. Let m1 , m2 and m3 respectively be their slopes. Then, their equations are y = m1 x − 2am1 − am13 y = m2 x − 2am2 − am23 y = m3 x − 2am3 − am33

2 a(T − t ) = − at (T − t ) 2

⇒

14

Y

Parabola

(c) d = 0

T 2 − 4 ⋅ 1⋅ 2 ≥ 0 T2 ≥ 8

Number of Normals and Conormal Points In this section, we will see that in general, three normals can be drawn from a point to a parabola. We shall also study about the relations between their slopes and conditions, so that three normals are distinct. (i) Three normals can be drawn from a point to a parabola. (ii) Conormal points The point on the parabola at which the normals pass through a common point

The coordinates of A, B and C are ( am12 , − 2am1 ), ( am22 , − 2am2 ) and ( am32 , − 2am3 ) respectively. Since, m1 , m2 , m3 are the roots of Eq. (i). …(ii) m1 + m2 + m3 = 0 2a − h …(iii) m1 m2 + m2 m3 + m3 m1 = a −k and …(iv) m1 m2 m3 = a We have the following results related to conormal points and the slopes of the normals at conormal points : (a) The algebraic sum of the slopes of the normals at conormals points is zero. (b) The sum of the ordinates of the conormal points is zero. (c) The centroid of the triangle formed by the conormal points on a parabola lies on its axis.

807

X

Objective Mathematics Vol. 1

14

Example 38. If the normals drawn from any point to the parabola y 2 = 4ax cut the line x = 2a in points whose ordinates are in arithmetic progression, then tangents of the angles which the normals makes with the axis, are (a) in AP (b) in GP (c) in HP (d) None of these

Example 40. The normal chord at a point t on the parabola y 2 = 4ax subtends a right angle at the vertex. Then, t 2 is equal to (a) 4 (b) 2 (c) 1 (d) 3 Sol. (b) The equation of any normal to y2 = 4ax at (at 2 , 2 at ) is

Sol. (b) Let the equation of the parabola be y2 = 4ax. Then, the feet of the normals passing throught a point are (am12 , − 2 am1 ), (am22 , − 2 am2 ) and (a m32 , − 2 am3 ). y = m1 x − 2 am1 − am13 .

⇒

On solving with the line x = 2 a, we get the point of intersection as (2 a, − am13 ). Similarly, on solving with the other normals, the ordinates of the points of intersection with x = 2 a are − am13 , − am23 and − am33 . These ordinates are in AP. So, − am13 − am33 = − 2 am23 But

m13 + m33 = 2 m23

…(i)

as

Σ m1 = 0 m13 + m23 + m33 = 3m1m2 m3

…(ii) …(iii)

From Eqs. (i) and (iii), we get 3m22 = 3m1m3 ⇒

m22 = m1m3

Hence, m1, m2 and m3 are in GP. X

Example 39. The number of distinct normals that can be drawn from ( −2, 1) to the parabola y 2 − 4x − 2 y − 3 = 0, is (a) 1 (b) 2 (c) 3 (d) 0 Sol. (a) Here, y2 − 2 y + 1 = 4( x + 1) ⇒

( y − 1)2 = 4( x + 1)

So, the axis is y − 1 = 0. Also, (− 2, 1) lies on the axes and it is exterior to the parabola because 12 − 4(−2 ) − 2(1) − 3 = 4 > 0 Hence, only one normal is possible.

Combined Equation of Pair of Tangents The combined equation of the pair of tangents drawn from an external point ( x1 , y1 ) to the parabola y 2 = 4ax is ( y 2 − 4ax ) ( y12 − 4ax1 ) = { yy1 − 2a ( x + x1 )}2 or

SS 1 = T 2

where, S ≡ y 2 − 4ax, S 1 ≡ y12 − 4ax1 and

T ≡ yy1 − 2a ( x + x1 )

y + t x = 2 at + at 3

…(i)

The combined equation of the lines joining the vertex (origin) to the points of intersection of the parabola and Eq. (i) is  y + tx  y2 = 4ax   3  2 at + at 

Equation of normal at (am12 , − 2 am1 ) is

808

X

(2t + t 3 ) y2 = 4 x( y + tx)

If Eq. (i) makes a right angle at the vertex, then Coefficient of x2 + Coefficient of y2 = 0 4 t − 2t − t 3 = 0 ⇒

t2 = 2

Director Circle The locus of the point of intersection of perpendicular tangents to a conic is known as its director circle. The director circle of a parabola is its directrix.

Chord of Contact The chord joining the points of contact of two tangents drawn from an external point P to a parabola is known as the chord of contact of tangents drawn from P. (i) The chord of contact of tangents drawn from a point P ( x1 , y1 ) to the parabola y 2 = 4ax is yy1 = 2a ( x + x1 ) (ii) The length of the chord of contact of tangents drawn from ( x1 , y1 ) to the parabola y 2 = 4ax is 1 ( y1 2 − 4ax1 ) ( y12 + 4a 2 ) a (iii) The area of the triangle formed by the tangents drawn from ( x1 , y1 ) to y 2 = 4ax and their chord ( y12 − 4ax1 ) 3/ 2 . 2a (iv) The locus of the point of intersection of tangents to the parabola y 2 = 4ax which meet an angle α is ( x + a ) 2 tan 2 α = y 2 − 4ax of contact is

(v) The angle between the pair of tangents drawn from ( x1 , y1 ) to the parabola y 2 = 4ax is  s1   tan −1   x1 + a 

Example 41. If the chord of contact of tangents from a point P to the parabola y 2 = 4ax, touches the parabola x 2 = 4by, then the locus of P is a/an (a) circle (b) parabola (c) ellipse (d) hyperbola

Theorem The equation of chord of the parabola y 2 = 4ax which is bisected at ( x1 , y1 ) is yy1 − 2a ( x + x1 ) = y12 − 4ax1

Sol. (d) Let P (h, k ) be a point. Then, the chord of contact of

This touches the parabola x = 4by. So, it should be of the form b …(ii) x = my + m Eq. (i) can be rewritten as k x =   y − h  2a

where, S ′ =

…(i)

2

meets the parabola y 2 = 4 ax + b at Q and R , then the mid-point of QR is also (x1 , y1).

X

Sol. (a) Let the point of intersection of tangents be ( x1, y1 ), then the equation of the chord of contact of tangents drawn from P to the parabola y2 = 4ax is yy1 = 2 a ( x + x1 ). Clearly, lx + my + n = 0 is also the chord of contact of tangents. Therefore, yy1 = 2 a ( x + x1 ) and lx + my + n = 0 are same line. 2 a − y1 2 a x1 = = ∴ l m n n x1 = ⇒ l −2 am and y1 = l n −2 am  Hence, required point is  , . l l 

− 4ax1 and T = yy1 − 2a ( x + x1 ). 2

…(iii)

Example 42. Tangents are drawn at the points, where the line lx + my + n = 0 intersected by the parabola y 2 = 4ax. The coordinates of the point of intersection of tangents are  n −2am  (a)  ,  l l   l −2am  (b)  ,  n n   n −2al  (c)  ,  m m  (d) None of the above

y12

Ø If the tangent at the point P (x1 , y1) to the parabola y = 4 ax

Since, Eqs. (ii) and (iii) represent the same line. k m= ∴ 2a b and =−h m Eliminating m from these two equations, we get 2ab = − hk Hence, the locus of P (h, k ) is xy = − 2 ab, which is a hyperbola. X

T = S′

or

tangents from P to y2 = 4ax is ky = 2 a ( x + h)

14

Equation of the Chord Bisected at a Given Point

Parabola

X

Example 43. The mid-point of the chord intercepted on the line 4x − 3 y + 4 = 0 by the parabola y 2 = 8x is (a) (5/ 4, 3) (b) (2, 4) (c) (5/ 2, 14/ 3) (d) (5, 8) Sol. (a) Let (α,β) be the mid-points of chord of the parabola y2 = 8 x.

∴ Equation of chord whose mid-point is (α,β ) is yβ − 4 ( x + α ) = β 2 − 8α 4 x − yβ + β 2 − 4α = 0

or

Clearly, 4 x − 3 y + 4 and 4 x − yβ + β 2 − 4α = 0 are same line. ∴

4 3 4 = = 4 β β 2 − 4α

⇒ β=3 and α = 5/ 4 Hence, required coordinates is (5/ 4, 3). X

Example 44. The locus of the mid-points of chords of the parabola y 2 = 4ax which passes through the focus is (a) y 2 + 2ax + 2a 2 = 0 (b) y 2 − ax + 2a 2 = 0 (c) y 2 − 2ax + 2a 2 = 0 (d) y 2 − 2ax + a 2 = 0 Sol. (c) Let (h, k ) be the coordinates of the mid-points of one of the chord which passes through the focus (a, 0). Then, the equation of the chord, where mid-point is (h, k ), is ky − 2 a ( x + h) = k 2 − 4ah This line passes through the focus (a, 0). ∴ − 2 a2 + 2 ah − k 2 = 0 ⇒

k 2 − 2 ah + 2 a2 = 0

Hence, the locus of (h, k ) is y2 − 2 ax + 2 a2 = 0

809

Objective Mathematics Vol. 1

14 Diameter of a Parabola Diameter of a parabola is the locus of the mid-point of a series of its parallel chords.

Equation of Diameter of a Parabola The equation of the diameter bisecting chords of 2a slope m of the parabola y 2 = 4ax is y = . m X

Example 45. The locus of the middle points of parallel chords of a parabola x 2 = 4ay is a (a) straight line parallel to X -axis (b) straight line parallel to Y -axis (c) circle (d) straight line parallel to a bisector of the angles between the axes Sol. (b) The locus of the middle points of parallel chords is a line parallel to the axes of the parabola.

Pole and Polar of a Parabola If a line through the point P, cuts the parabola at points A and B, then the locus of the point of intersection of tangents at A and B is a straight line. This line is called the polar of point P with respect to the parabola. If line l is the polar of point P with respect to a parabola, then point P is called the pole of line l with respect to the parabola. l

A P (x1 , y 1 )

(i) The equation of polar of point P ( x1 , y1 ) with respect to the parabola y 2 = 4ax is T = 0 i.e. yy1 − 2a ( x + x1 ) = 0 (ii) Properties of pole and polar (a) If polar of a point P with respect to a parabola passes through point Q, then polar of Q with respect to the parabola will pass through P. Such points P and Q are called conjugate points. (b) If the pole of line l1 lies on line l2 , then the pole of line l2 lies on line l1 . Such lines l1 and l2 are called the conjugate lines.

810

Example 46. The locus of the poles of tangents to the parabola y 2 = 4ax with respect to the parabola y 2 = 4bx, is (a) a circle (b) a parabola (c) a straight line (d) an ellipse Sol. (b) The equation of any tangent to y2 = 4ax is y = mx + a / m Let (h, k ) be the pole of Eq. (i) with respect to the parabola y2 = 4bx. Then, the equation of polar is ky = 2 b ( x + h) or 2 bx − ky + 2 bh = 0

…(i)

…(ii)

Clearly, Eqs. (i) and (ii) represent the same straight line. 2 b − k 2 bh = = ∴ −1 a/m m 2b k= ⇒ m a and h= 2 m 2b m= ⇒ k a 2 and m = h 2 2b  a  ⇒   =  k  h 4b 2 h 2 ⇒ k = a 4b 2 x 2 , which is a parabola. Hence, the locus of (h, k ) is y = a

Lengths of Tangent, Subtangent, Normal and Subnormal

B

Q (a, b)

X

Subtangent and subnormal Let the tangent and normal at any point P ( x1 , y1 ) on a parabola y 2 = 4ax meet the axis in T and G, respectively. Then, PT is called the length of the tangent at P, PG is called the length of the normal at P, NT is called the subtangent at P and NG is called the subnormal at P. Y (x1, y1) P yy1 = 2a (x+x1) X′

T

A

S(a ,0) N G (x1+ 2a, 0) (x1, 0)

y2= 4ax Y′

X

2

 dx  = y1 1 +    dy  ( x ,

1 y1 )

(ii) The length of the normal = PG = PN sec ψ 2

 dx  = y1 1 +    dy  ( x ,

1 y1 )

(iii) The length of the subtangent = NT = PN cot ψ  dx  = y1    dy  ( x , y ) 1

2a =m y1

Example 47. The subtangent, ordinate and subnormal to the parabola y 2 = 4ax at a point, different from the origin, are in (a) AP (b) GP (c) HP (d) None of the above

1

1

14

Sol. (b) The equation of parabola is y2 = 4ax. ⇒

2y

⇒

dy = 4a dx dy 2 a = dx y

At ( x1, y1 ), we have Subtangent =

1

(iv) The length of the subnormal = NG = PN tan ψ  dy  = y1    dx  ( x , y ) where, tan ψ =

X

Parabola

Subtangent = NT = Twice the abscissa of P and subnormal = NG = 2a = Semi-latusrectum It can be easily seen, with the help of the right angled ∆NPT and ∆PNG, that (i) The length of the tangent = PT = PN cosec ψ

=

y1 y1 = (dy /dx) 2 a / y1 y12 2a

Subnormal = y1 ⋅

dy = 2a dx

y12 , ordinate = y1 2a and subnormal = 2a are in GP.

Clearly, subtangent =

Work Book Exercise 14.3 1 The chord of contact of the pair of tangents drawn from each point on the line 2 x + y = 4 to the parabola y 2 = − 4 x pass through the point a (2, −1) c ( − 1 / 2, − 1 / 4)

b (1/2, 1/4) d ( − 2, 1)

2 The inclination of the normal with positive direction of X-axis, drawn at another end of a normal to the parabola y 2 = 4 x is always a 60° c more than 60°

b less than 60° d less than 45°

3 Two parabolas with the same axis, focus of each being exterior to the other and the latusrectum being 4a and 4b. The locus of middle points of the intercepts between the parabolas made on the lines parallel to the common axis is a/an a straight line, if a > b c parabola, ∀ a, b

b parabola, if a ≠ b d ellipse, if b > a

4 The set of real values of a for which atleast one tangent to y 2 = 4ax becomes normal to the circle x 2 + y 2 − 2 ax − 4ay + 3a2 = 0, is a [1, 2] c R

b [ 2 , 3] d None of these

5 If P, Q and R are three points on parabola y 2 = 4 x and normal at P and R meet at Q, then the locus of mid-point of the chord PR is a parabola whose vertex is at a (2, 0) c ( − 2, 0)

b ( 0, − 2 ) d None of these

6 Tangent and normal at any point P of the parabola y 2 = 4ax (a > 0 ) meet the X-axis at T and N, respectively. If the lengths of subtangent and subnormal at this point are equal, then the area of ∆PTN is given by a b c d

4a 2 6 2 a2 4 2 a2 None of the above

7 If ( x1, y1 ), ( x2 , y2 ) and ( x3 , y3 ) are the feet of the three normals drawn from a point to the parabola y 2 = 4ax, then x1 − x2 x − x3 x − x1 is equal to + 2 + 3 y3 y1 y2 a 4a c a

b 2a d 0

811

WorkedOut Examples Type 1. Only One Correct Option Ex 1. The condition for all the three normals drawn from a given point ( h, k ) to the parabola y 2 = 4ax to be real and distinct, is (a) 4 ( h − 2a ) 3 > 27ak 2 (b) ( h − 2a ) 3 > 27ak 2 (c) 4 ( h − 2a ) < 27ak 3

Sol. Equation of the normal to the parabola y2 = 4 ax at any point (at 2 , 2at ) is ...(i) y = − tx + 2at + at If this normal passes through the point (h, k ), then …(ii) k + th = 2at + at 3 3

f (t ) = at 3 + (2a − h)t − k f ′ (t ) = 3at 2 + (2a − h) = 0 for critical points

h − 2a = ± α [say] ⇒ h > 2a 3a All the roots of Eq. (ii) are real, if f (α ) f (− α ) < 0 ⇒[ aα 3 + (2a − h) α − k ][ − aα 3 − (2a − h) α − k ] < 0 t=±

⇒

[ aα 3 + (2a − h) α ]2 − k 2 > 0

⇒

α 2[ aα 2 + (2a − h)]2 − k 2 > 0 h − 2a   h − 2a  2 a  + 2a − h − k > 0 3a   3a  

Ex 2. The locus of centres of a family of circles passing through the vertex of the parabola y 2 = 4ax and cutting the parabola orthogonally at the other point of intersection, is (a) y 2 ( 2 y 2 + x 2 − 12ax ) = ax ( 3x − 4a ) 2 (b) 2 y 2 ( 2 y 2 + x 2 − 12ax ) = ax ( 3x + 4a ) 2 (c) 2 y 2 ( 2 y 2 + x 2 − 12ax ) = ax ( 3x − 4a ) 2 (d) None of the above

...(vi)

Eliminating t from Eqs. (v) and (vi), we have t2 t = 2 −4 ak − hk h (4 a − 3h) − 4 k 1 = − k 2 − a (4 a − 3h) ⇒ 2k 2 (2k 2 + h2 − 12ah) = ah (3h − 4 a)2

Ex 3. Normals are drawn from the point P with slopes m1 ⋅ m2 = α and it is a part of the parabola itself, then α is equal to (a) 1

(b) 2

(c) 3

(d) −2

Sol. Let the point P be (h, k ) . ∴ ⇒

k = mh − 2m − m3 m + m (2 − h) + k = 0 3

m1m2m3 = − k k m3 = − α

⇒ 3

k  k ⇒  −  − (2 − h) + k = 0  α α ⇒ ...(i)

Tangent at P will be normal to the circle as circle and parabola intersect orthogonally at P.  at 2  , at and slope AP is a chord, whose mid-point is   2 

812

...(iv)

On multiplying Eq. (iii) by t and subtracting from Eq. (iv), we get ...(v) t 2k + t (4 a − 3h) − 4 k = 0

⇒

Sol. Let P (at 2 , 2at ) be any point on y2 = 4 ax.

is (2/t).

2th + 4 k = at 3 + 4 at

Hence, (c) is the correct answer.

⇒ 4 (h − 2a)3 > 27 ak 2 Hence, (a) is the correct answer.

Vertex A = (0, 0) The equation of tangent at P is ty = x + at 2

⇒

Hence, the required locus is 2 y2 (2 y2 + x 2 − 12ax ) = ax (3x − 4 a)2

2

⇒

tx + 2 y =

Also, from Eq. (iii), at 2 − tk + h = 0

So,

i.e.

at 3 ...(ii) + 2at 2 Eqs. (i) and (ii) both pass through (h, k ), which is the centre of the circle. ...(iii) ∴ tk = h + at 2 ⇒

2

(d) ( h − 2a ) 3 < 27ak 2

Let

∴ Line passing through the mid-point of AP and perpendicular to AP will pass through the centre of the circle. t at 2  ∴ y − at = −  x −  2 2

k 2 = α 2h − 2α 2 + α 3

Thus, the locus of (h, k ) is ⇒ y2 = α 2x − 2α 2 + α 3 On comparing it with y2 = 4 x, we get α2 = 4 and ⇒

− 2α + α 3 = 0 2

α=2

t2

Y t1

(a) 2x 2 − 2x + 2 y 2 − 2 y + 1 = 0 (b) x 2 − 2x + 2 y 2 − 2 y + 1 = 0 (c) 2x 2 − 2x + 2 y 2 + 2 y + 2 = 0 (d) 2x 2 + 2x − 2 y 2 − 2 y − 2 = 0

m1 X′

X

m3

Ex 5. A movable parabola touches the X and Y-axes at (1, 0) and (0, 1). Then, the locus of the focus of the parabola is

m2

14 Parabola

Aliter Since, locus of P is a part of the parabola.

Sol. Since, X -axis and Y-axis are two perpendicular tangents Y′

to the parabola and both meet at the origin, the directrix passes through the origin.

P (y, k)

Y

So, normals at any two points t1 and t2 meet at P. ⇒ t1t2 = 2 ⇒ (− m1 ) (− m2 ) = 2 ⇒ α=2 Hence, (b) is the correct answer.

Ex 4. Consider two concentric circles C1 : x 2 + y 2 − 1 = 0 and C 2 : x 2 + y 2 − 4 = 0. A parabola is drawn through the points, where C1 meets the X -axis and having arbitrary tangent of C 2 as its directrix. Then, the locus of the focus of drawn parabola is 4 2 x − y2 = 3 3 4 (c) x 2 + y 2 = 3 3

3 2 x − y2 = 3 4 3 (d) x 2 + y 2 = 3 4

(a)

(b)

Sol. Clearly, the parabola should pass through (1, 0) and (−1, 0). Let directrix of this parabola be x cosθ + y sin θ = 2 Y P (2 cos θ, 2 sin θ)

X′

X C2 S

X′ N

(h,k)

Y′

If S (h, k ) is the focus of this parabola, then distance of (± 1, 0) from S and from the directrix should be same. ...(i) ⇒ (h − 1)2 + k 2 = (cosθ − 2)2 and ...(ii) (h + 1)2 + k 2 = (cosθ + 2)2 On subtracting Eq. (i) from Eq. (ii), we get h 4 h = 8 cosθ ⇒ cosθ = 2 On adding Eqs. (i) and (ii), we get 2 (h2 + k 2 + 1) = 2 (cos2 θ + 4 ) h2 ⇒ h2 + k 2 + 1 = 4 + 4 3 2 h + k2 = 3 ⇒ 4 Hence, locus of focus is 3 2 x + y2 = 3 4 Hence, (d) is the correct answer.

X

A(1,0) M Y′ y = mx

Let y = mx be the directrix and (h, k ) be the focus. FA = AM ⇒

(h − 1)2 + k 2 =

m

…(i)

1 + m2

FB = BN

and ⇒

h2 + (k − 1)2 =

1

…(ii)

1 + m2

From Eqs. (i) and (ii), we get (h − 1)2 + h2 + k 2 + (k − 1)2 = 1 ⇒

C1 O

) B(0,1 F(h,k)

2x 2 − 2x + 2 y2 − 2 y + 1 = 0

which is the required locus. Hence, (a) is the correct answer.

Ex 6. The locus of vertices of the equilateral ∆ABC such that its sides touch the parabola y 2 = 4ax, is (a) y 2 = ( x + a ) 2 + 4ax (b) y 2 = 3 ( x + a ) 2 + 4ax (c) y 2 = 3 ( x + a ) 2 − ax (d) None of these

Sol. Any point on the parabola y2 = 4 ax is (at 2 , 2at ) and point of intersection [ at1t2 , a (t1 + t2 )] .

of

tangents

at

t1 , t2

is

Y A B

t1

t3

X′

X C t2 Y′

⇒ A ≡ [ at1t3 , a (t1 + t3 )], B ≡ [ at1t2 , a (t1 + t2 )] and C ≡ [ at2t3 , a (t2 + t3 )]

813

Objective Mathematics Vol. 1

14

Given, ∆ABC is an equilateral triangle. tan 60° = 3 =

⇒

mAB − mAC 1 − mAB ⋅ mAC

1 1 − t1 t3 = 1 1 1+ ⋅ t1 t3

⇒

3 (1 + t1t3 )2 = (t1 − t3 )2

⇒

3 (1 + t1t3 ) = (t1 + t3 ) − 4 t1t3 2

2

If A is (h, k ), then h = at1t3 , k = a(t1 + t3 ) 2 2 h 4h   k 3 1 +  =   − ⇒   a a a So, the required locus is y2 = 3 (x + a)2 + 4 ax.

So, coordinates of Q and N are (4 a + at 2 , 0) and (at 2 , 0), respectively. ∴ Length of projection = 4 a + at 2 − at 2 = 4 a Hence, (d) is the correct answer.

Ex 9. If on a given base, triangle is described such that the sum of tangents of the base angles is constant ( k ), then the locus of third vertex is (a) a hyperbola (c) a circle

(b) an ellipse (d) a parabola

Sol. Let AB be the given base. Then,

tan A + tan B = k

Hence, (b) is the correct answer. Y

Ex 7. A line L passing through the focus of the parabola y 2 = 4 ( x − 1) intersects the parabola in two distinct points. If m is the slope of the line L, then (a) − 1< m < 1 (c) m ∈ R

(b) m < − 1or m > 1 (d) None of these

Sol. Let y = Y , x − 1 = X . Then, the equation is Y 2 = 4 X . So, the focus = (1, 0) X , Y = (2, 0) Any line through the focus is y = m (x − 2). Solving this with y2 = 4 (x − 1), we get m2 (x − 2)2 = 4 (x − 1) ⇒ m2x 2 − 4 (m2 + 1)x + 4 (m2 + 1) = 0 If m ≠ 0, then D = 16 (m2 + 1)2 − 16m2 (m2 + 1) = 16 (m2 + 1) > 0, for all m But, if m = 0, then x does not have two real distinct values. So, m ∈ R, except m = 0. Hence, (d) is the correct answer.

Ex 8. Vertex A of a parabola y 2 = 4ax is joined to any point P on it and line PQ is drawn at right angle to AP to meet the axis at Q. Then, the projection of PQ on the axis is always equal to (a) 3a (c) 3a

(b) 2a (d) 4a

Sol. Let P ≡ (at 2 , 2at ) Y

X′

C (a, b)

X

A

B (a, 0) Y′

β β , tan B = α α−a β β Now, − =k α α−a ⇒ − aβ = k (α 2 − aα ) a x 2 = − y + ax ⇒ k which is the equation of a parabola. Hence, (d) is the correct answer. ⇒

tan A =

Ex 10. From a point A (t ) on the parabola y 2 = 4ax, a focal chord and a tangent are drawn. Two circles are drawn in which one circle is drawn taking focal chord AB as diameter and other is drawn by taking intercept of tangent between point A and point P on the directrix as diameter. Then, the common chord of the circles is (a) line joining focus and P (b) line joining focus and A (c) tangent to the parabola at point A (d) None of the above Sol. Circle S 2 , taking focal chord AB as diameter will touch

P

y2 =

4ax

directrix at point P and circle S 1, taking AP as diameter will pass through focus S (since, AP subtends angle 90° at focus of parabola). Y

X′

A

N

Q

S1

X

A

y 2 = 4ax

S2

Y′

814

Equation of the line PQ is t y − 2at = − (x − at 2 ) 2 On putting y = 0, we get x = 4 a + at 2

P X′

S(a, 0) B

x = –a Y′

X

Hence, (c) is the correct answer.

Ex 11. A tangent is drawn to the parabola y 2 = 4 ax at P such that it cuts Y-axis at Q. A line perpendicular to this tangent is drawn through Q which cuts the axis of the parabola at R. If the rectangle PQRS is completed, then the locus of S is

Sol. Let θ be the angle in which the line FP makes with positive direction of X-axis, then 2t tanθ = 2 t −1 sin θ =

t2 − 1 2t and cosθ = 2 1+ t 1 + t2

(a) y 4 = ( x + a ) [ a ( x − a ) 2 − y 2 ( x 2 − 2a )]

 t2 − 1 2t  B ≡ 1 + [1 + t 2 + f (t )] ,[1 + t 2 + f (t )]  2 + + t2  t 1 1  ∴ B lies on y − x = 2. t2 − 1 2t 2 +2 = 1 + [1 + t 2 + f (t )] ⇒[1 + t + f (t )] 2 1 + t2 1+ t

(b) y 4 = ( x − a ) [ a ( x − a ) 2 + y 2 ( x 2 − 2a )]

⇒

 2t − t 2 + 1 f (t )  = 3 + t 2 − 1 − 2t  1 + t2 

⇒

f (t ) =

(c) y 4 = ( x − a ) [ a ( x − a ) 2 − y 2 ( x 2 + 2a )] (d) None of the above Sol. From the property of parabola, R is focus.

14 Parabola

So, common chord of given circles is line AP (which is intercept of tangent at point A between point A and directrix.

(t 2 − 2t + 2) (1 + t 2 ) − t 2 + 2t + 1

Hence, (a) is the correct answer.

Y

Ex 13. Two equal parabolas have the same focus and their axes are at right angles, a normal to one is perpendicular to a normal to the other. Then, the locus of the point of intersection of these normal is

P (at 2, 2at)

Q

S (h, k) X′

R (a, 0)

O

X

(a) an ellipse (c) another parabola

(b) a circle (d) a hyperbola

Sol. Taking the common focus as origin and axis of the

Y′

and ⇒

2at − k k − 0 × = −1 at 2 − h h − a k 1 = h−a t h−a t= k

parabola as axis of the coordinate, the equation of the parabola may be written as ...(i) y2 = 4 a (x + a) x 2 = 4 a ( y + a)

and

...(ii)

where, 4a is the length of latusrectum of each parabola. Y

So, required locus is   h − a 2     h − a 2a  k  − k  k = (a − h) a  k  − h     ⇒ y4 = (x − a) [ a (x − a)2 − y2 (x + 2a)]

X′

O

X

Y′

Hence, (c) is the correct answer.

Ex 12. A point P (t 2 , 2t ) lies on the parabola y 2 = 4x, where FP is produced to B, where F is focus. If PB = f ( t ) and point B always lies on the line y − x = 2, then f ( t ) is equal to (a) (b) (c)

( t 2 − 2t + 2) (1 + t 2 ) − t 2 + 2t + 1 ( t 2 − 2t + 2) (1 + t 2 ) t 2 + 2t + 1 ( t 2 − 2t + 2) (1 − t 2 ) − t 2 + 2t + 1

(d) None of the above

Equation of any normal to parabola (i) is y = m (x + a) − 2am − am3

...(iii)

Equation of any normal to parabola (ii) is ...(iv) x = m′ ( y + a) − 2am′ − am′ 3 But the two normals are at right angles, then 1 m× = −1 m′ ⇒ m′ = − m On substituting m′ = − m in Eq. (iv), we get x = − m ( y + a) + 2am + am3 ...(v) The locus of the point of intersection of the normals can be obtained by the elimination of m from Eqs. (iii) and (v). So, on adding Eqs. (iii) and (v) after taking the terms to RHS in both cases, we get 0 = mx − y − x − my

815

Objective Mathematics Vol. 1

14

⇒

x + y = m (x − y) ⇒ m =

x+ y x− y

On putting the value of m in Eq. (iii), we get 3  x + y x+ y x+ y y= −a (x + a) − 2a   x − y x− y x− y ⇒ y (x − y)3 = (x + y) [(x + a) (x − y)2 − 2a (x − y)2 − a (x + y)2 ] = (x + y) [(x − y)2 (x + a − 2a) − a (x + y)2 ] = (x + y) [ x (x − y)2 − 2a (x 2 + y2 )] ⇒ 2a (x 2 + y2 ) (x + y) = (x + y) x (x − y)2 − y (x − y)3 ⇒ 2a (x 2 + y2 ) (x + y) = (x − y)2 [ x 2 + xy − xy + y2 ] ⇒ 2a (x + y) = (x − y)2 which is the required locus of a parabola. Hence, (c) is the correct answer.

 4 at  So, the coordinates of T are  0, .  3  4 at or AT = 3 2 Clearly, AT = × PN 3 Hence, (d) is the correct answer.

Ex 15. The sides of a triangle touch y 2 − 4ax = 0 and two of its angular points are on y 2 − 4b ( x + c) = 0 . Then, the locus of the third angular point is equal to 2

 2b  (a) y 2 =  + 1 ( ax + 4bc ) a  2

Ex 14. PN is ordinate of the parabola, a straight line is drawn parallel to the axis which bisect PN and meets the curve at Q, if NQ meets the tangent at the vertex at a point T, then AT is equal to (where, A is the vertex) (a)

PN 3 2

2 2 3 (b) PN (c) NP 3 2

2 (d) PN 3

 2b  (b) y 2 = 4  − 1 ( ax + 4bc ) a   2b  (c) y 2 = 4  + 1 ( ax + 4bc ) a  (d) None of the above Sol. Let the three tangents be yt1 − x − at12 = 0 yt2 − x −

Sol. Let the equation of the parabola be y2 = 4 ax

...(i)

Tangent at the vertex is Y-axis or x=0

...(ii)

Y

P (at 2, 2at)

T Q X′

M N

A

X

O Y′

Again, let P be any point (at 2 , 2at ) on the parabola. So, PN = 2at and coordinates of N are (at 2 , 0).  1 If M is the mid-point of PN , then NM =   PN = at.  2 As MQ is parallel to X-axis, its equation is ...(iii) y = at On putting the value of y from Eq. (iii) in Eq. (i), we get 1 a2t 2 = 4 ax ⇒ x = at 2 4 1  So, the coordinates of Q are  at 2 , at . 4  Equation of NQ will be

at − 0 (x − at 2 ) 1 2 2 at − at 4 −4 ...(iv) ⇒ y= (x − at 2 ) 3t If NQ meets the tangent at the vertex i.e. x = 0 at T , then on putting x = 0 in Eq. (iv), we get 4 at y= 3 ( y − 0) =

816

and

yt3 − x −

at22 at32

...(i)

=0

...(ii)

=0

...(iii)

Then, the three angular points are [ at2t3 , a (t2 + t3 )], [ at3t1 , a (t3 + t1 )], [ at1t2 , a (t1 + t2 )] Let the last two be on the second parabola, then ...(iv) a2 (t3 + t1 )2 − 4 bat3t1 − 4 bc = 0 and

a2 (t1 + t2 )2 − 4 bat1t2 − 4 bc = 0

...(v)

On subtracting Eq. (v) from Eq. (iv), we get ...(vi) a (t2 + t3 ) = (4 b − 2a) t1 From Eqs. (v) and (vi), a4 (t2 + t3 )2 − 4 bc ...(vii) a2t2t3 = a2t12 − 4 bc = 4 (2b − a)2 But for the third angular point, we have x = at2t3 and ...(viii) y = a (t2 + t3 ) Hence, the required locus of the parabola is a2 y2 − 4 bc ax = 4 (2b − a)2 From Eqs. (vii) and (viii), we get 2  2b  − 1 (ax + 4 bc) y2 = 4   a  Hence, (b) is the correct answer.

Ex 16. The parabola is y 2 = λx 25 [( x − 3) 2 + ( y + 2) 2 ] = (3x − 4 y − 2) 2 equal, if λ is equal to (a) 1 (b) 2 (c) 3 (d) 6

and are

their latusrectum are equal. Length of the latusrectum of y2 = λx is λ.

AS = Focal distance of A AS = 1 + t 2

⇒

∴ Equation of the second parabola is 25 {(x − 3)2 + ( y + 2)2} = (3x − 4 y − 2)2 |3x − 4 y − 2| ⇒ (x − 3)2 + ( y + 2)2 = 32 + 4 2

Y

3 × 3 − 4 × (−2) − 2 32 + (−4 )2

Ex 17. Set of values of h for which the number of distinct common normals of ( x − 2) 2 = 4 ( y − 3) and x 2 + y 2 − 2x − hy − c = 0 (c > 0) is 3, is (a) ( 2, ∞ ) (c) ( 2, 4 )

(b) ( 4, ∞ ) (d) (10, ∞ )

Sol. The equation of any normal to the parabola (x − 2)2 = 4 ( y − 3) is x − 2 = m ( y − 3) − 2m − m3 If it passes through (1, h / 2), then h  1 − 2 = m  − 3 − 2m − m3 2  ⇒

2m3 + m (10 − h) − 2 = 0

This equation will give three distinct values of m, if f ′ (m) = 0 has two distinct roots, where f (m) = 2m3 + m (10 − h) − 2 f ′ (m) = 6m2 + (10 − h) f ′ (m) = 0 h − 10 m=± ⇒ 6 The values of m are real and distinct, if h > 10 i.e. h ∈ (10, ∞ ). Hence, (d) is the correct answer.

Now, ⇒

Ex 18. Area of the trapezium whose vertices lie on the parabola y 2 = 4x and its diagonals pass 25 through (1, 0) and having length units each, 4 is 75 sq units 4 625 (b) sq units 16 25 (c) sq units 4 25 sq units (d) 8 (a)

X′

X

S(1, 0)

O D (t 2, – 2t )

=6

Thus, the two parabolas are equal, if λ = 6. Hence, (d) is the correct answer.

B

A (t 2, 2 t )

Clearly, it represents a parabola having focus at (3, − 2) and equation of the directrix as 3x − 4 y − 2 = 0 ∴ Length of latusrectum = 2 (Distance between focus and directrix) =2

14

Sol. We have,

Parabola

Sol. Let us recall that two parabolas are equal, if the length of

C Y′

∴

CS = AC − AS =

25 − (1 + t 2 ) 4

If PQ is a focal chord of parabola y2 = 4 ax , then 1 1 1 + = SP SQ a 1 1 1 + = ∴ AS CS a 1 1 + =1 ⇒ 2 25 1+ t − (1 + t 2 ) 4 1 4 ⇒ + =1 1 + t 2 25 − 4 (1 + t 2 ) ⇒

25 (1 + t 2 ) − 4 (1 + t 2 )2 = 25

⇒ 4 (1 + t 2 )2 − 25 (1 + t 2 ) + 25 = 0 ⇒

{(1 + t 2 ) − 5} {4 (1 + t 2 ) − 5} = 0 5 1 ⇒ 1 + t 2 = 5, ⇒ t2 = 4 , 4 4 1 t = ± 2, ± ⇒ 2 Thus, the coordinates of A , B , C and D are A ≡ (1/ 4 , 1), B ≡ (4 , 4 ), C ≡ (4 , − 4 ) and D ≡ (1/ 4 , − 1) ∴ AD = 2, BC = 8 15 and distance between AD and BC = 4 1 15 ∴Area of trapezium ABCD = (2 + 8) × 2 4 75 sq units = 4 Hence, (a) is the correct answer.

Ex 19. If (h, k ) is a point on the axis of the parabola 2( x − 1) 2 + 2 ( y − 1) 2 = ( x + y + 2) 2 from where three distinct normals may be drawn, then (a) h > 2

(b) h < 4

(c) h > 8

(d) h = 8

Sol. We have, 2 (x − 1)2 + 2 ( y − 1)2 = (x + y + 2)2 ⇒

(x − 1)2 + ( y − 1)2 =

x + y+ 2 1+ 1

Clearly, it represents a parabola having its focus at (1, 1) and directrix is x + y + 2 = 0.

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Objective Mathematics Vol. 1

14

The equation of the axis is y − 1 = 1 (x − 1) i.e. y = x. Semi-latusrectum = Length of perpendicular from (1, 1) on the directrix 1+ 1+ 2 = =2 2 1+ 1 The coordinates of the vertex are (0, 0). So, the equation of the axis in parametric form is x−0 y−0 ...(i) = cos π / 4 sin π / 4

The coordinates of a point on Eq. (i) at a distance 2 2 from the vertex are given by x y =2 2 = cos π / 4 sin π / 4 ⇒ x = 2, y = 2 Hence, (a) is the correct answer.

Type 2. More than One Correct Option Ex 20. Consider a circle with its centre lying on the focus of the parabola y 2 = 2 px such that it touches the directrix of the parabola. Then, a point of intersection of the circle and the parabola is p  (a)  , p 2 

 p (b)  ,− p  2

 p  (c)  − , p  2 

  p (d)  − , − p   2

  Sol. The focus of the parabola y2 = 2 px is at  , 0 and its p 2  p directrix is x = − . Since, the circle touches the directix. 2 p  ∴ Radius = Distance between the point  , 0 and the 2  p line x = − = p 2 So, the equation of the circle is 2 p  2 2  x −  + ( y − 0) = p  2 3 p2 x 2 + y2 − px − =0 ⇒ 4 On sloving this equation with y2 = 2 px, we obtian

3 p2 = 0 ⇒ (2x − p) (2x + 3 p) = 0 4 p 3p or x = − x= ⇒ 2 2 p On putting x = in y2 = 2 px, we obtain y = ± p 2 3p For x = − , y is imaginary. Hence, the circle and the 2  p  p parabola intersect at  , p and  , − p .  2  2 x 2 + px −

Hence, (a) and (b) are the correct answers.

Ex 21. A variable circle is described to pass through the point (1, 0) and tangent to the curve y = tan (tan −1 x ). The locus of the centre of the circle is a parabola whose

818

We know that three distinct normals can be drawn from a point (h, 0) on the axis of the parabola y2 = 4 ax, if h > 2a ( = Semi-latusrectum).

(a) length of the latusrectum is 2 2 (b) axis of symmetry has the equation x + y = 1 (c) vertex has the coordinates (3/4, 1/4) (d) None of the above

Sol. Let (h, k ) be the centre, then

h−k = (h − 1)2 + k 2 2

∴

(h − k )2 = 2 (h2 + k 2 − 2h + 1)

⇒ h2 + k 2 + 2hk − 4 h + 2 = 0 ⇒

(h + k )2 = 4 h − 2

⇒

(h + k )2 = 2h + 2k + 2h − 2k − 2

1  ⇒ (h + k )2 − 2(h + k ) + 1 = 2  h − k −   2 ∴ Locus is

(x + y − 1)2 = 2 (x − y − 1/ 2) x − y − 1 2 2   x + y − 1 2   =  2  2  2  ∴ Axis of symmetry is x + y = 1 Length of latusrectum = 2 Vertex is the point of intersection of x + y = 1 and x − y = 1/ 2 i.e. (3/4, 1/4) Hence, (b) and (c) are the correct answers.

Ex 22. The locus of the mid-point of the focal radii of a variable point moving on the parabola y 2 = 4ax is a parabola whose (a) latusrectum is half the latusrectum of the original parabola a  (b) vertex is  , 0 2  (c) directrix is Y-axis (d) focus has the coordinates ( a, 0) Sol. Any point on the parabola is (at 2 , 2at ).  a + at 2  , at . ∴ Mid-point of (a , 0) and (at 2 , 2at ) is   2  a + at 2 2  y2  y2  y 2x = a 1 + 2  = a + Q y = at ⇒ t = ⇒  a a  a   ⇒ 2ax = a2 + y2 ⇒ y2 = 2a (x − a / 2) a  Its a parabola with vertex at  , 0 , latusrectum = 2a 2  Directrix is x − a/ 2 = a/ 2 i.e. (a, 0) ⇒ x=a Hence, (a), (b), (c) and (d) are the correct answers.

∴ Locus is given byx =

(a) k = − 5 (b) k = 0 (c) k = 3 (d) k takes any real value

14

Sol. x 2 − x + y = 0 ⇒

x 2 − x + 4 − kx = 0

⇒

x − (1 + k )x + 4 = 0

∴

(1 + k )2 − 16 = 0

2

⇒ k + 1 = ± 4 ⇒ k = 3, − 5 Hence, (a) and (c) are the correct answers.

Parabola

Ex 23. The straight line kx + y = 4 touches the parabola y = x − x 2 , if

Type 3. Assertion and Reason Directions (Ex. Nos. 24-27) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 24. Statement I Circumcircle of a triangle formed by the lines x = 0, x + y + 1 = 0 and x − y + 1 = 0 also passes through the point (1, 0). Statement II Circumcircle of a triangle formed by three tangents of parabola passes through its focus. Sol. Statement I Consider a parabola y = 4 ax 2

Let

∠SAC = α

⇒

1 t1 + t2 − 1 t2 t1t2 − 1 tanα = = t1 1  t1 + t2  1+   t2  t1t2 − 1

Similarly, tanβ =

Ex 25. Statement I Length of focal chord of a parabola y 2 = 8x making an angle of 60° with X-axis is 32. Statement II Length of focal chord of a parabola y 2 = 4ax making an angle α with X-axis is 4a cosec 2 α. Sol. Let AB be a focal chord. ∴ Slope of AB =

A (at 2, 2at) X′

( y − 2) = 0

X

B

⇒ Length of

P

α S (a, 0) a , –2a t2 t

O

Y′

Let P (t1 ), Q (t2 ) and R (t3 ) be three points on it. A

2t = tanα t −1 2

Y

2

Y

⇒ α = β or α + β = π

Hence, (a) is the correct answer.

∴ x + y + 1 = 0is a tangent to the parabola Again, y2 = 4 ( y − 1) ⇒ y2 − 4 y + 4 = 0 So, x − y + 1 = 0 is a tangent to the parabola and (1, 0) is the focus. ∴ Statement I is true. Consider a parabola y2 = 4 ax.

∠SBC = β

So, A , B , C and S are concyclic.

Clearly, x = 0 is tangent to the parabola at (0, 0). y2 = 4 (− y − 1) ⇒ ( y + 2)2 = 0

⇒

1 t1

and

α 1 α = ⇒ t = cot 2 t 2 2 1  AB = a  t +  = 4 a cosec2 α  t

tan

So, Statement II is true but Statement I is false. Hence, (d) is the correct answer.

Q

B X′

X

S (a, 0)

C R

Y′

Tangents are drawn at these points which intersect A ≡ [ at1t2 , a (t1 + t2 )] B ≡ [ at1t3 , a(t1 + t3 )] C ≡ [ at2t3 , a (t2 + t3 )]

Ex 26. Statement I Area of triangle formed by pair of tangents drawn from a point (12, 8) to the parabola y 2 = 4x and their corresponding chord of contact is 32 sq units. Statement II If from a point P ( x1 , y1 ), tangents are drawn to a parabola y 2 = 4ax, then area of triangle formed by these tangents and their corresponding chord of contact is ( y12 − 4ax1 ) 3/ 2 sq units. 4 | a| 819

Sol. Statement II

Area of triangle formed by these tangents and their corresponding chord of contact is ( y12 − 4 ax1 )3/ 2 . 2 | a|

Objective Mathematics Vol. 1

14

∴ Statement II is false. x1 = 12, y1 = 8

Statement I

( y12 − 4 x1 )3/ 2 2 (64 − 48)3/ 2 = 2 = 32 Statement I is true. Hence, (c) is the correct answer. Area =

Ex 27. Statement I The perpendicular bisector of the line segment joining the points( −a, 2at ) and ( a, 0) is tangent to the parabola y 2 = 4ax, where t ∈ R . Statement II Number of parabolas with a given point as vertex and length of latusrectum equal to 4, is 2. Sol. Image of (a, 0) with respect to tangent yt = x + at 2 is (−a, 2at ). ∴ Perpendicular bisector of (a, 0) and (−a, 2at ) is the tangent line yt = x + at 2 to the parabola. ∴ Statement I is true. Statement II Infinitely many parabolas are possible. ∴ Statement II is false. Hence, (c) is the correct answer.

Type 4. Linked Comprehension Based Questions Ex 30. The value of lim [[x − 5] + [4 − x ]], where [ ]

Passage I (Ex. Nos. 28-30) If a =

lim

x → π /2

tan x − sin [tan (tan x )] 1 and r = λ tan x + cos 2 (tan x )

where, λ is the least integral value for which three real normals can be drawn to the parabola y 2 = 8 x from ( λ, 0).

Ex 28. Sum of infinite GP whose first term is a and common ratio is r, is 2 3 5 (c) 4

1 5 4 (d) 3

(a)

(b)

Sol. Q λ = 4, so r = ∴

1 4 π− 2

⇒

a = lim

⇒

a=1

∴

tan x − sin [tan −1 (tan x )] tan x + cos2 (tan x )

a = lim x→

1 − cos x =1 cos2 (tan x ) 1+ tan x

π− x→ 2

Sum =

1

=

4 3

1 4 Hence, (d) is the correct answer. 1−

Ex 29. Radius of the circle whose centre is (a, λ ) which touches the line 3x + 4 y = 4, is (a) 5 (b) 3 (c) 4 (d) 2 Sol. The point (a, λ) is (1, 4).

820

x → aλ

−1

Its distance from the line 3x + 4 y − 4 = 0 is 3 + 16 − 4 =3 5 Hence, (b) is the correct answer.

denotes greatest integer function, is (a) −2 (b) −1 (c) 2 (d) Does not exist Sol. lim+ [[ x − 5 ] + [ 4 − x ]] = − 2 x→ 4

When x → 4 + , then [ x − 5 ] = − 1 and [ 4 − x ] = − 1 lim [[ x − 5 ] + [ 4 − x ]] = − 2

x → 4−

When x → 4 − then [ x − 5 ] = − 2 and [ 4 − x ] = 0 Hence, (a) is the correct answer.

Passage II (Ex. Nos. 31-33) The straight line y + tx = 2at + at 3 is a normal to the parabola y 2 = 4ax for all real values of t and foot of the normal is given by (at 2, 2at ) .

Ex 31. Number of real normals that can be drawn from the point (4, 0) to the parabola y 2 = 16x, is (a) 1 (c) 3

(b) 2 (d) None of these

Sol. y + tx = 2at + at 3 Since, a = 4 and normal passes through (4, 0). ∴ 0 + 4 t = 8t + 4 t 3 ⇒ 4 t 3 + 4 t = 0 ⇒ 4 t (t 2 + 1) = 0 ⇒ Only one real value of t i.e. t = 0 So, number of real normals is 1. Hence, (a) is the correct answer.

Ex 32. If the ordinates of points P and Q on the parabola y 2 = 12x are in the ratio 1 : 2, then the locus of the point of intersection of normals to the parabola at P and Q, is (a) 343 y 2 = 12 ( x − 6) 3 (b) 343 y 2 = 12 ( x + 6) 3 (c) 343 y 2 = − 12 ( x − 6) 3 (d) None of the above

∴ ⇒

k + th = 2at + at 3 3 at + (2a − h) t − k = 0 k ⇒ t1t2t3 = , t1 + t2 + t3 = 0 a 2at1 1 Also, = 2at2 2 ⇒ 2t1 = t2 , t3 = − 3t1 k (t1 )(2t1 )(−3t1 ) = ∴ a k 3 ⇒ t1 = − 6a Since, t1 satisfies Eq. (i). ∴ (2a − h)3 t1 = (k − at13 )

…(i)

3

⇒ ∴ ⇒

k k  = k +  6a  6 343 2 3 y − (6 − x ) = 12 343 y2 = − 12 (6 − x )3

− (2a − h)3

= 12 (x − 6)3 Hence, (a) is the correct answer.

Ex 33. A point through which three normals of parabola y 2 = 4ax are passing, two of which are making angles α and β with X-axis, where tan α tan β = 2, lies on the curve (a) y ( y 2 − 2ax ) = 0 (b) y ( y 2 + 2ax ) = 0 (c) y ( y 2 − 4ax ) = 0 (d) y ( y 2 − ax ) = 0 k a ⇒ t1t2 = (− t1 ) (− t2 ) = tan α tan β = 2 k ⇒ t3 = 2a Since, t3 satisfy the equation at 2 + (2a − h)t − k = 0

Sol. t1t2t3 =

k  k a  3  + (2a − h) − k = 0 2a  8a  3

⇒ ⇒

y [ y2 + 4 a (2a − x ) − 8a2 ] = 0 y ( y2 − 4 ax ) = 0

Hence, (c) is the correct answer.

Passage III (Ex. Nos. 34-36) y = f ( x ) is parabola of

the form y = x 2 + ax + 1, its tangent at the point of intersection of Y-axis and parabola also touches the circle x 2 + y 2 = r 2 . It is known that no point of the parabola is below X-axis.

Ex 34. The radius of circle when a attains its maximum value, is 1

(a)

(b)

1

10

(c) 1

(d) 5

5

14 Parabola

Sol. Let (h, k ) be the point of intersection of the normals.

Sol. Since, no point of the parabola is below X-axis. ∴

a2 − 4 ≤ 0

[QD < 0]

So, maximum value of a is 2. Now, equation of the parabola, when a = 2 is y = x 2 + 2x + 1 It intersects Y-axis at (0, 1). Equation of the tangent at (0, 1) is y = 2x + 1. Since, y = 2x + 1touches the circle x 2 + y2 = r2. 1 r= ∴ 5 Hence, (b) is the correct answer.

Ex 35. The slope of the tangent when radius of the circle is maximum, is (a) 0 (c) −1

(b) 1 (d) not defined

Sol. Equation of the tangent at (0, 1) to the parabola y+1 a = (x + 0) + 1 2 2 ax − y + 1 = 0 1 r= 2 a +1

y = x 2 + ax + 1 is i.e. ∴

Radius is maximum, when a = 0. ∴ Equation of the tangent is y = 1 and slope of the tangent is 0. Hence, (a) is the correct answer.

Ex 36. The minimum area bounded by the tangent and the coordinate axes is (a)

1 4

(b)

1 3

(c)

1 2

(d) 1

Sol. Equation of tangent is y = ax + 1 whose intercepts are 1 − and 1. a ∴ Area of the triangle bounded by tangent and the axes 1 1 = − ⋅1 2 a 1 = 2 | a| It is minimum, when a = 2. 1 ∴ Minimum area = 4 Hence, (a) is the correct answer.

821

Objective Mathematics Vol. 1

14 Type 5. Match the Columns Ex 37. Match the statements of Column I with values of Column II. Column I

Ex 38. Match the statements of Column I with values of Column II. Column I

Column II

Column II

A. If the parabola y 2 = 4 x and the circle having its centre at (6, 5) intersects at right angle, at the point ( a, a ), then one value of a is

p.

13

q.

8

A. Area of a triangle formed by the tangents drawn p. from a point ( −2, 2 ) to the parabola y 2 = 4 ( x + y) and their corresponding chord of contact is

8

B. Length of the latusrectum of the conic q. 25 {( x − 2 )2 + ( y − 3)2} = ( 3 x + 4 y − 6)2 is

4 3

C. If focal distance of a point on the parabola r. y = x 2 − 4 is 25/4 and points are of the form ( ± a, b ), then the value of a + b is

4

B. If the angle between the tangents drawn to ( y − 2 )2 = 4 ( x + 3) at the points, where it is 4π intersected by the line 3 x − y + 8 = 0 is , p then p has the value

r.

10 5

D. Length of side of an equilateral triangle inscribed s. in a parabola y 2 − 2 x − 2 y − 3 = 0 whose one angular point is vertex of the parabola is

24/5

C. If the line x − 1 = 0 is the directrix of the parabola y 2 − kx + 8 = 0, then one of the value of k is D. Length of the normal chord of the parabola y 2 = 8 x at the point, where abscissa and ordinate are equal, is

s.

4

Sol. A. Point of contact of tangent drawn from (−2, 2) on y2 = 4 (x + y) are (0, 4 ) and (0, 0).

Sol. A. Since, point (a, a) lies on y2 = 4 x.

Area = 4 ∴ B. The conic is a parabola having focus (2, 3) and directrix 3x + 4 y − 6 = 0. ∴ Latusrectum = 2 × (Perpendicular distance of focus from the directrix)  6 + 12 − 6 24 =2  =   5 5 1 2 2 C. y + 4 = x ⇒ x = 4 ⋅ ( y + 4 ) 4 25 Focal distance = 4 −17 ∴ Distance from directrix y = 4 Now, ordinate of points of the parabola whose 25 focal distance is 4 −17 25 = + =2 4 4 So, points are (± 6 , 2). ⇒

a+ b=8

D. Length of side = 8 3 a = 8 3 ⋅ A → r; B → s; C → p; D → q

∴ a2 = 4 a ⇒ a = 0, 4

⇒ a=4

B. The line 3x − y + 8 = 0 passes through the focus (−2, 2), so the tangents at the end points on the π chord is . 2 4π π = ⇒ p=8 ∴ p 2 y2 = k (x − 8 / k ) 8 k Equation of directrix is x − = − 4 k 8 k ⇒ x= − k 4 But it is given that directrix is x = 1. 8 k ⇒ − = 1 ⇒ k 2 + 4 k − 32 = 0 k 4 ∴ k = 4, − 8 D. End points of the normal chord will be (8, 8) and   −5 2  −5  2   , 2 ⋅ 2    2 2   So, length of the chord will be 10 5. A → s; B → q; C → s; D → r C.

1 =4 3 2

Type 6. Integer Answer Type Questions Ex 39.The sides of ∆ABC are tangents to the parabola y 2 = 4ax. Let D, E , F be the point of contact of sides BC , CA and AB, respectively. If lines AD, BE and CF are concurrent at focus of the parabola and ∠ABC of ∆ABC is π , then k equals . k Sol. (3) As AD passes through focus. E(t3)

C [at2t3,a(t1 + t3)] A

822

Focus D(t2) B

F(t1) .

2t1t2t3 − 2t2 = (t1 + t3 ) t22 − (t1 + t3 ) t23 − (t1 + t2 + t3 ) t22 − 3t2 + 2t1t2t3

⇒ ⇒

+ (t1 + t2 + t3 ) = 0 ⇒ t1 , t2 , t3 are the roots of the equation t 3 − (t1 + t2 + t3 ) t 2 − 3t + 2t1t2t3 + (t1 + t2 + t3 ) = 0. t1t2t3 = − [ 2t1t2t3 + (t1 + t2 + t3 )] (t + t + t ) ⇒ t − (t1 + t2 + t3 ) t 2 − 3t + 1 2 3 = 0 3 3  3t − t t1 + t2 + t3 = ⇒  3  1 − 3t 2  1 1 1 As, , , are the slopes of the sides taking t1 t2 t3 1 1 1 = tan θ 1 , = tan θ 2 and = tan θ 3 t1 t2 t3 ⇒

...(i)

2

Since, (−1, 0) lies on the directrix of the parabola, hence tangents to the parabola are these lines y = x + 1and y = − x − 1, also the tangents to the given circle. Here, ∆PAB is possible in two ways, one shown by using PAB and other by using PA1 B1. Length of PA = PA1 = 1, also B (1, 2) and B1 (1, − 2) as B and B1 lies on the latusrectum of the parabola. Here, ∆PAB and ∆PA1 B1 are right angled triangles. 1 1 So, area of ∆PAB = × PA × PB = × 1 × 2 2 2 2 = 2 sq units 1 Also, area of ∆PA1 B1 = × PA1 × PB 2 1 = ×1× 2 2 2 = 2sq units

14 Parabola

a (t1 + t3 ) 2at2 = a (t22 − 1) a (t1t3 − 1)

∴

∴ Area of ∆PAB = 2 sq units

3 cot θ − cot θ = cot α 1 − 3 cot 2 θ t +t +t where, 1 2 3 = cotα and the roots of the 3 equations are θ 1 , θ 2 , θ 3. ⇒ cot 3θ = cot α nπ + α θ= ⇒ 3 2π + α α π+α , θ3 = θ1 = , θ2 = ⇒ 3 3 3 π ∴ θ2 − θ1 = θ3 − θ2 = 3 3

⇒

Ex 40. If a point P on the axis of the parabola y 2 = 4x is taken such that the point is at shortest distance from the circle x 2 + y 2 + 2x − 2 2 y + 2 = 0 tangent are drawn to the circle and the parabola, then the area of ∆PAB is a, where A and B are the points of contact on two different lines on circle and parabola respectively, find a. Sol. (2) Axis of the parabola y2 = 4 x is y = 0 and the point

Ex 41. 64, 27 are the lengths of any two perpendicular chords of parabola passing through the vertex of the parabola. If l is the length of latusrectum of the parabola, then write the value of 5l/16 is________ . Sol. (9) Let the equation of parabola be y2 = lx or x 2 = ly

8 (0, 0)

θ 27 P (27 sin θ, –27 cos θ)

By substituting P , Q in y2 = lx and eliminating θ, we get 144 l= ⇒ 5l = 144 5 5l 144 = =9 ∴ 16 16

Ex 42. If tangent to the parabola y 2 = 4x is normal to 1 , then the numerical x 2 = 4by and | b | < k . quantity of k should be

which is at shortest distance from the circle. x 2 + y2 + 2x − 2 2 y + 2 = 0 is (−1, 0) Y

Sol. (8) Any tangent to y2 = 4 x is y = mx +

(–1, √2)

A1

X′

X P (1,0)

Y′

B1(1, –2)

b . m′ 2 1 b If these are same lines, then m = m′ and = 2b + 2 , m m′ 2bm2 − m + b = 0 normal to x 2 = 4 by is y = m′ x + 2b +

B (1, 2)

A

1 and any m

This will give real roots, if 1 − 8b2 > 0 1 1 ⇒ b2 < ⇒ | b| < 8 8 ∴ k =8

823

Target Exercises Type 1. Only One Correct Option 1. Equation of parabola having the extremities of its latusrectum as (3, 4) and (4, 3) is 2

2

7 7  x + y + 6   (a)  x −  +  y −  =       2 2 2 2

2

2

2

2

(a) x 2 + y2 = 4 (c) x 2 + y2 = 80

7 7 (x + y − 8)2   (b)  x −  +  y −  =   2 2 2 (x + y − 4 ) 7 7   (c)  x −  +  y −  =     2 2 2 (d) None of the above

Ta rg e t E x e rc is e s

(b) 2 2a

(c) 4a

(d) 2a

3. Double ordinate AB of the parabola y 2 = 4ax π subtends an angle at the vertex of the parabola, 2 then tangents drawn to parabola at A and B will intersect at (a) (−4 a, 0) (c) (−3a, 0)

(b) (−2a, 0) (d) None of these

4. AB is a focal chord of x 2 − 2x + y − 2 = 0, whose focus is S. If AS = l1 , then BS is equal to 4 l1 4 l1 − 1 2l1 (c) 4 l1 − 1

(a)

(b)

l1 4 l1 − 1

(d) None of these

5. OA and OB are two mutually perpendicular chords of y 2 = 4ax, O being the origin. Line AB will always passes through the point (a) (2a, 0)

(b) (6a, 0)

(c) (8a, 0)

(d) (4 a, 0)

6. The length of latusrectum of the parabola, whose focus is ( a sin 2θ, a cos 2θ) and directrix is the line y = a, is (a) |4 a cos θ | (c) |4 a cos 2 θ | 2

(b) |4 a sin θ | (d) | a cos 2 θ | 2

7. The point ( a, 2a ) is an interior point of the region bounded by the parabola y 2 = 16x and the double ordinate through the focus. Then, a belongs to the open interval

824

(a) a < 4 (b) 0 < a < 4 (c) 0 < a < 2 (d) a > 4

(b) x 2 + y2 = 20 (d) None of these

9. AB is a chord of the parabola y 2 = 4ax with vertex A. BC is drawn perpendicular to AB meeting the axis at C. The projection of BC on the axis of the parabola is

2

2. The length of the latusrectum of the parabola 2{( x − a ) 2 + ( y − a ) 2 } = ( x + y ) 2 is (a) 2a

8. The vertex of parabola y 2 = 8 x is at the centre of a circle and parabola cuts the circle at ends of its latusrectum. Then, equation of the circle is

(a) a (c) 4a

(b) 2a (d) 8a

.

10. The circle x 2 + y 2 + 2λ x = 0, λ ∈ R , touches the parabola y 2 = 4x externally. Then, (a) λ > 0 (c) λ > 1

(b) λ < 0 (d) None of these

11. Vertex of the parabola whose parametric equation is x = t 2 − t + 1, y = t 2 + t + 1, t ∈ R, is (a) (1, 1)  1 1 (c)  ,   2 2

(b) (2, 2) (d) (3 ,3)

12. The circle x 2 + y 2 + 2gx + 2 fy + c = 0 cuts the parabola x 2 = 4ay at point A i ( x i , y i ), i = 1, 2, 3, 4, then (a) Σy1 = 0 (b) Σy1 = − 4 ( f + 2a) (c) Σx1 = − 4 (g + 2a) (d) Σx1 = − 2(g + 2a)

13. A circle is drawn to pass through the extremities of latusrectum of the parabola y 2 = 8 x. It is given that this circle also touches the directrix of the parabola. Radius of this circle is equal to (a) 4 (c) 3

(b) 21 (d) 26

14. A circle having its centre at (2, 3) is cut orthogonally by the parabola y 2 = 4x. The possible intersection point of these curves can be (a) (1, 2) or (3, 2 3 ) (b) (9, 6) or (2, 2 2 ) (c) (1, 2) or (4, 4) (d) None of the above

15. If the length of a focal chord of the parabola y 2 = 4ax at a distance b from the vertex is c, then (a) 2a2 = bc (c) ac = b2

(b) a3 = b2c (d) b2c = 4 a3

π 6

π 4 π (d) 2

(b)

(c) 0

17. The equation of a parabola is y 2 = 4x. If P (1, 3) and Q(1, 1) are two points in the XY -plane. Then, for the parabola, (a) P and Q are exterior points (b) P is an interior point while Q is an exterior point (c) P and Q are interior points (d) P is an exterior point while Q is an interior point

18. A double ordinate parabola y 2 = 8 px is of length 16p. The angle subtended by it at the vertex of the parabola is π 4 π (c) 3

(a)

(b)

π 2

(d) None of these

19. The parabola y 2 = kx makes an intercept of length 4 on the line x − 2 y = 1. Then, k is equal to 105 − 5 10 5 + 105 (c) 10

(b)

(a)

5 − 105 10

(d) None of these

(b) 4

(c) 2

(d) 5

21. The condition that the straight line lx + my + n = 0 touches the parabola x 2 = 4ay is (a) bn = am2 (c) ln = am2

(b) al 2 − mn = 0 (d) am = ln2

22. The point ( −2m, m + 1) is an interior point of the smaller region bounded by the circle x 2 + y 2 = 4 and parabola y 2 = 4x. Then, m belongs to the interval (a) −5 − 2 6 < m < 1 (b) 0 < m < 4 3 (c) −1 < m < 5 (d) −1 < m < − 5 + 2 6

(a) y = ax (c) 9 y2 = ax

π 3

(b)

π 4

(c)

π 2

2π 3

(d)

27. A water jet from a fountain reaches its maximum height of 4 m at a distance 0.5 m from the vertical passing through the point O of water outlet. The height of the jet above the horizontal OX at a distance of 0.75 m from the point O, is (a) 5 m (c) 3 m

(b) 6 m (d) 7 m

28. If the normals at three points P , Q and R of the parabola y 2 = 4ax meet in a point O and S is its focus, then | SP | ⋅ | SQ | ⋅ | SR | is equal to (b) a (SO )3 (d) None of these

 29. If the 4th term in the expansion of  px +  5 is and three normals to the parabola 2 drawn through a point ( q, 0), then (a) q = p (c) q < p

n

1  ,n ∈N x y 2 = x are

(b) q > p (d) pq = 1

30. Tangent at the vertex divides the distance between directrix and latusrectum in the ratio (a) 1 : 1 (b) 1 : 2 (c) depends on directrix and focus (d) None of the above

B = ( at 22 , 2at 2 ) and AC : AB = 1: 3, then

(b) 9 y = 4 ax (d) y2 = 9ax 2

24. If the tangents at P and Q on a parabola meet in T , then SP , ST and SQ are in (a) AP (b) GP (c) HP (d) None of the above

(a)

31. The chord AB of the parabola y 2 = 4ax cuts the axis of the parabola at C. If A = ( at 12 , 2at 1 ),

23. The locus of the points of trisection of the double ordinates of parabola y 2 = 4ax, is 2

26. If the tangent at P on y 2 = 4ax meets the tangent at the vertex in Q and S is the focus of the parabola, then ∠SQP is equal to

(a) a2 (c) a (SO )2

20. Radius of the largest circle which passes through the focus of the parabola y 2 = 4x and contained in it, is (a) 8

(a) a parabola whose axis is horizontal (b) a parabola whose axis is vertical (c) integer point on the parabola y = x 2 (d) None of the above

14 Parabola

(a)

25. Let us define a region R in XY -plane as a set of points ( x, y ) satisfying [ x 2 ] = [ y], where [ x] denotes greatest integer ≤ x, then the region R defines

Targ e t E x e rc is e s

16. The angle of intersection between the curves x 2 = 4( y + 1) and x 2 = − 4( y + 1) is

(a) t2 = 2t1 (c) t1 + 2t2 = 0

(b) t2 + 2t1 = 0 (d) None of these

32. AB is a chord of the parabola y 2 = 4ax. If its equation is y = mx + c and it subtends a right angle at the vertex of the parabola, then (a) c = 4 am (c) c = − 4 am

(b) a = 4 mc (d) a + 4 mc = 0

33. Locus of trisection point of any double ordinate of y 2 = 4ax is (a) 3 y 2 = 4 ax (c) 9 y 2 = 4 ax

(b) y 2 = 6ax (d) None of these

825

Objective Mathematics Vol. 1

14

34. Circle drawn having its diameter equal to focal distance of any point lying on the parabola x 2 − 4x + 6 y + 10 = 0 , will touch a fixed line whose equation is (a) y = 2 (c) x + y = 2

(b) y = − 1 (d) x − y = 2

35. An equilateral ∆SAB is inscribed in the parabola y 2 = 4ax having its focus at S. If chord AB lies towards the left of S , then side length of this triangle is (a) 2a (2 − 3 ) (b) 4 a (2 − 3 ) (c) a(2 − 3 ) (d) 8a(2 − 3 )

36. Normals AO , AA1 , AA 2 are drawn to parabola y 2 = 8x from the point A ( h, 0) . If ∆OA1 A 2 is equilateral, then possible values of h is

Ta rg e t E x e rc is e s

(a) 26 (c) 28

(b) 24 (d) None of these

37. The locus of centre of a circle that passes through P ( a, b ) and touch the line y = mx + c (it is given that b ≠ ma + c ), is (a) a straight line (c) a parabola

(b) a circle (d) a hyperbola

and the circle y 2 = 4x ( x − 6) + y = r will have no common tangent, if

38. The

parabola

2

2

2

(a) r > 20 (c) r > 18

(b) r < 20 (d) r ∈ ( 20 , 28 )

39. Parabola y 2 = 4x and the circle having its centre at (6, 5) intersect at right angle. Possible point of intersection of these curves can be (a) (9, 6) (c) (1, 2)

(b) (2, 8 ) (d) (3, 2 3 )

40. y = 2x + c′, c being variable, is a chord of the parabola y 2 = 4x, meeting the parabola at A and B. Locus of point dividing the segment AB internally in the ratio 1: 1, is (a) y = 1 (c) y = 2

(b) x = 1 (d) x = 2

41. Radius of the circle that passes through origin and touches the parabola y 2 = 4ax at the point ( a, 2a ), is 5 a 2 (c) 5 2a

(a)

(b) 2 2a (d) 3 2a

42. A line L passing through the focus of the parabola ( y − 2) 2 = 4( x + 1) intersects the parabola in two distinct points. If m is the slope of the line L , then m can take values in the interval 826

(a) (− 1, 1) (c) (−∞ , 0) ∪ (0, ∞ )

(b) (−∞ , − 1) ∪ (1, ∞ ) (d) (−∞ , ∞ )

43. A variable parabola having length of its latusrectum equal to 4a and having its axis parallel to X -axis drawn in such a way that it always touches the parabola y 2 = 4ax. Locus of the vertex of the variable parabola is (a) y2 = 8ax (c) y2 = 6ax

(b) y2 + 8ax = 0 (d) y2 + 6ax = 0

44. Set of values of m for which a chord of slope m of the circle x 2 + y 2 = 4 touches the parabola y 2 = 4x, is   2 − 1 (a)  −∞ , − ∪    2    (b) (−∞ , 1) ∪ (1, ∞ ) (c) (− 1, 1) (d) R

2 − 1  ,∞  2 

45. If two tangents drawn from the point (α, β) to the parabola y 2 = 4x is such that the slope of one tangent is double of the other, then 2 2 α 9 (c) 2α = 9β 2

2 (b) α = β 2 9 (d) None of these

(a) β =

46. The tangents from the origin to the parabola y 2 + 4 = 4x are inclined at π 6 π (c) 3

π 4 π (d) 2 (b)

(a)

47. The minimum distance between the parabolas y 2 − 4x − 8 y + 40 = 0 and x 2 − 8x − 4 y + 40 = 0 is (a) 0

(b) 3

(c) 2 2

(d) 2

48. The equation of common tangent to the equal parabolas y 2 = 4ax and x 2 = 4ay, is (a) x + y + a = 0 (c) x − y = a

(b) x + y = a (d) None of these

49. The length of common chord of the parabola 2 y 2 = 3( x + 1) and the circle x 2 + y 2 + 2x = 0, is (a) 3 3 (c) 2

(b) 2 3 (d) None of these

50. The locus of the middle points of chords of a parabola which subtend a right angle at the vertex of the parabola, is (a) a circle (b) an ellipse (c) a parabola (d) None of the above

51. The locus of a point from which tangents to a parabola are at right angles, is (a) a straight line (b) a pair of straight lines (c) a circle (d) a parabola

(b) a circle (d) an ellipse

53. The locus of the middle points of chords of the parabola y 2 = 8x drawn through the vertex is a parabola whose (a) focus is (2, 0) (c) focus is (0, 2)

(b) latusrectum is 8 (d) latusrectum is 4

54. Circle drawn with diameter being any focal chord of the parabola y 2 − 4x − y − 4 = 0 will always touch a fixed line, whose equation is (a) 16 + 33x = 0 (c) 13x + 32 = 0

(b) 32x + 13 = 0 (d) 16x + 33 = 0

55. Tangents are drawn to y 2 = 4ax at point, where the line lx + my + n = 0 meets this parabola. Intersection point of these tangents is  n 2am (a)  ,  l l   n −2am (c)  ,  l l 

 n am (b)  ,  l l   n − am (d)  ,  l l 

y2 + 2ax + ay − 6a2 = 0 y2 + 2ax − ay + 6a2 = 0 y2 − 2ax + ay + 6a2 = 0 y2 − 2ax − ay + 6a2 = 0

(a) m1 + m2 = 0 (c) m1m2 = − 1

(b) m1m2 = 1 (d) m1 + m2 = 1

58. Tangents drawn to parabola y = 4ax at the points A and B intersect at C. If S is the focus of the parabola, then SA , SC and SB forms 2

(a) an AP (c) an HP

(b) a GP (d) None of these

59. If the line ax + by + c = 0 is a tangent to the parabola y 2 − 4 y − 8x + 32 = 0, then (a) 4 b = a(7a + 2c + 4 b) (c) 4 b2 = a(7a + 2c + b) 2

(b) 4 b = a(7a + c − 4 b) (d) 4 b2 = a(7a + 2c − b) 2

60. If the tangent drawn from a point to the parabola y 2 = 4ax are normals to the parabola x 2 = 4by, then (a) a2 ≥ b2 (c) a2 ≥ 8b2

(b) a2 ≥ 4 b2 (d) 8a2 ≥ b2

(b) 6a

(c) 8a

(d) 2a

(b) a2 ( y2 + 4 x )x 2 = 16 (d) a2 ( y2 − 4 x )x 2 = 16

64. Locus of the mid-point of any focal chord of y 2 = 4ax is (a) y2 = a(x − 2a) (c) y2 = 2a(x − a)

(b) y2 = 2a(x − 2a) (d) None of these

65. Chord AB of the parabola y 2 = 4ax subtends a right angle at the origin. Point of intersection of tangents drawn to parabola at A and B lie on the line (b) y + 2a = 0 (d) y + 4 a = 0

66. A normal is drawn to the parabola y 2 = 4ax at a point P other than the vertex. If it cuts the parabola at point Q, then the least value of OQ, where O is the vertex of the parabola, is (b) 2 6a (d) None of these

67. Tangents PA and PB are drawn to x 2 = 4ay. If mPA and mPB are the slope of these tangents and 2 2 + mPB = 4, then locus of P is mPA (a) y2 = 2x (x − a) (c) y2 = 2x (2x − a)

(b) y2 = x (x − a) (d) y2 = 2x (x − 2a)

68. If a line is drawn from A( −2, 0) to intersect the curve y 2 = 4x in P and Q in the first quadrant such that 1 1 1 + < , then slope of the line is always AP AQ 4 (a) > 3 (c) > 2

1 3 1 (d) > 3 (b)
− 2}

61. If mutually perpendicular tangents TA and TB are drawn to y 2 = 4ax, then minimum length of AB is equal to (a) 4a

(a) a2 ( y2 + 2x )x 2 = 8 (c) a2 ( y2 − 4 x )x 2 = 8

(a) 4 6a (c) 3 6a

57. If straight lines ( y − b ) = m1 ( x + a ) and ( y − b ) = m2 ( x + a )are the tangents of y 2 = 4ax, then

(b) y2 = x 2 − a2 + 6ax (d) y2 = x 2 − a2 − 6ax

63. PA and PB are the tangents drawn to y 2 = 4ax from point P. These tangents meet Y -axis at the points A1 and B1 , respectively. If the area of ∆PA1 B1 is 2 sq units, then locus of P is

(a) x + 2a = 0 (c) x + 4 a = 0

56. Locus of mid-point of chords of the parabola y 2 = 4ax that pass through the point ( 3a, a ), is (a) (b) (c) (d)

(a) y2 = x 2 + a2 + 6ax (c) y2 = x 2 + a2 − 6ax

14 Parabola

(a) a straight line (c) a parabola

62. PA and PB are tangents drawn to y 2 = 4ax from π arbitrary point P. If the angle between tangents is , 4 then locus of point P is

Targ e t E x e rc is e s

52. P is a point. Two tangents are drawn from it to the parabola y 2 = 4 x, such that the slope of one tangent is three times the slope of the other. The locus of P is

(b) {(k , 0)| k > − 2} (d) None of these

70. If three distinct and real normals can be drawn to y 2 = 8x from the point ( a, 0) , then (a) a > 2 (c) a ∈ (2, 4 )

(b) a > 4 (d) None of these

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71. Length of the shortest normal chord of the parabola y 2 = 4x is (a) a 27 units (c) 2a 27 units

(b) 3a 3 units (d) None of these

72. Tangent and normal drawn to parabola at A ( at 2 , 2at ), t ≡ 0 meet X-axis at points B and D, respectively. If the rectangle ABCD is completed, then the locus of C is (a) y = 2a (c) x = 2a

(b) y + 2a = 0 (d) x + 2a = 0

73. Normals PO , PA and PB, O being the origin, are π drawn to y 2 = 4x from P ( h, 0). If ∠AOB = , then 2 area of quadrilateral OAPB is equal to (a) 12 sq units (b) 24 sq units (c) 6 sq units (d) 18 sq units

Ta rg e t E x e rc is e s

(a) 1 + m2 (b) 2 + m2 (c) (1 + m2 )3/ 2 (d) (2 + m2 )3/ 2

77. AB is a double ordinate of the parabola y 2 = 4ax. Tangents drawn to parabola at A and B meets Y -axis at A1 and B1 , respectively. If the area of trapezium AA1 B1 B is equal to 12a 2 , then angle subtended by A1 B1 at the focus of the parabola is equal to (a) 2 tan −1 (3) (b) tan −1 (3) (c) 2 tan −1 (2) (d) tan −1 (2)

78. Maximum number of common normals of y 2 = 4ax and x 2 = 4by can be equal to

74. Locus of the mid-point of any normal chords of y 2 = 4a x is  4 a2 y2  (a) x = a  2 − 2 + 2  2a   y 2  4a y2  (b) x = a  2 + 2 + 2  2a   y 2  4a y2  (c) x = a  2 − 2 − 2  2a   y

(b) 4

(c) 18

(d) 5

(c) 6

(d) 5

79. If two different tangents of y 2 = 4x are the normals to x 2 = 4by , then 1 2 2 1 (c) | b | > 2

75. The radius of the circle whose centre is ( −4, 0) and which cuts the parabola y 2 = 8x at A and B such that its common chord AB subtends a right angle at the vertex of the parabola, is equal to (b) 3

(a) 3

(a) | b | >

 4 a2 y2  (d) x = a  2 + 2 − 2  2a   y

(a) 4

76. If the line y = mx + c is a tangent to y 2 = 4mx, then distance of this tangent from the parallel normal is

1 2 2 1 (d) | b | < 2

(b) | b |
b (b) parabola, if a ≠ b (c) parabola for all a, b (d) ellipse, if b > a

Type 2. More than One Correct Option 81. The coordinates of an end point of the latusrectum of the parabola ( y − 1) 2 = 2( x + 2) are (a) (−2, 1)

(b) (−3/ 2, 1)

(c) (−3/ 2, 2) (d) (−3/ 2, 0)

82. Which of the following is/are true about the parabola y 2 = 4ax ( a > 0)? (a) If t1 , t2 are end points of a focal chord, then t1t2 = − 1 (b) Tangent at the end of a focal chord cuts at right angle at directrix (c) Distance of any point on the parabola from directrix is equal to the sum of a abscissa of the point (d) End points of latusrectum are (a, 2a), (− a, 2a)

83. The equation of the tangent of the parabola y 2 = 9x which goes through the point (4, 10), is 828

(a) x + 4 y + 1 = 0 (c) x − 4 y + 36 = 0

(b) 9x + 4 y + 4 = 0 (d) 9x − 4 y + 4 = 0

84. The line(s) tangent to the curve y = x 2 − x is/are (a) x − y = 0 (c) x − y = 1

(b) x + y = 0 (d) x + y = 1

85. A straight line touches the circle x 2 + y 2 = 2a 2 and the parabola y 2 = 8ax. The equation of the line is (a) (b) (c) (d)

y = x + 2a y = − x − 2a y = x − 2a y=x −a

π at 2 the vertex of the parabola, then its slope is equal to

86. If a normal chord of y 2 = 4ax subtends an angle (a) 2 (c) 1

(b) − 2 (d) − 1

questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

87. Statement I

89. Statement I Two parabolas y 2 = 4ax and x 2 = 4ay have common tangent x + y + a = 0. Statement II x + y + a = 0 is common tangent to the parabolas y 2 = 4ax and x 2 = 4ay and point of contacts lie on their end points of latusrectum. 90. Statement I A is a point on the parabola y 2 = 4ax. The normal at A cuts the parabola again at point B. If AB subtends a right angle at the vertex of the parabola, then slope of AB is 1/ 2. Statement II If normal at ( at 12 , 2at 1 ) cuts again 2 the parabola at ( at 22 , 2at 22 ), then t 2 = t 1 − . t1

Slope of tangents drawn from (4, 10) 1 9 to parabola y = 9x are , . 4 4

91. Statement I Through ( h, h + 1) there cannot be more than one normal to the parabola y 2 = 4x, if h < 2.

Every parabola is symmetric about

Statement II The point ( h, h + 1) lies outside the parabola for all h ≠ 1.

2

Statement II its directrix.

88. Statement I Equation ( 5x − 5) 2 + ( 5 y + 10) 2 = ( 3x + 4 y + 5) 2 is a parabola. Statement II If distance of the point from the given line and from the given point (not lying on the given line) is equal, then locus of variable point is parabola.

92. Statement I If normal at the ends of double ordinate x = 4 of parabola y 2 = 4x meet the curve again at P and P ′ respectively, then PP ′ = 12 units. If normal at t 1 of y 2 = 4ax meet the 2 parabola again at t 2 , then t 2 = − . t2 Statement II

Type 4. Linked Comprehension Based Questions Passage I (Q. Nos. 93-95) parabola y = ax 2 + c.

y = x is tangent to the

(a)

1 8

(b) −

1 2

(c)

1 2

(d) 1

1 2

(b)

1 3

(c)

1 4

(d)

1 6

(b) (4, 4)

(c) (6, 6)

(d) (3, 3)

Passage II (Q. Nos. 96-100) Let S be the focus considered as pole (origin) and take SA as the initial line, a vertical line ZM be considered as directrix. Let P be any point on the conic and its coordinates be (r, θ), so that SP = r and ∠ZSP = θ and further SL = l (semi-latusrectum). L

P

M

l e (d) le

(b)

l = e cos θ r l cos θ (c) = r e

(a)

r = e cos θ l r e (d) = l cos θ

(b)

99. The polar equation of the parabola must be l  θ sin 2    2 2 l 2  θ (c) r = cos    2 2

(a) r =

l  θ cosec2    2 2 l 2  θ (d) r = sec    2 2 (b) r =

100. If the axis SZ of the conic makes an angle α with the initial line SA, then the equation of conic will be

r θ S

l e2 (c) le2

98. The equation of directrix of the conic corresponding to the focus which has been chosen as the pole, is

95. If c = 2 , then point of contact is (a) (2, 2)

(b) r = l (1 − e cos θ) l (d) r = (1 − e cos θ )

97. If e is the eccentricity of the conic, then SZ is equal to (a)

94. If (1, 1) is point of contact, then a is (a)

96. The polar equation of conic is (a) r = l (1 + e cos θ) l (c) r = 1 + e cos θ

93. If a = 2 , then the value of c is

14

Targ e t E x e rc is e s

Directions (Q. Nos. 87-92) For the following

Parabola

Type 3. Assertion and Reason

N

Z

A

(a) r = l[1 + e cos (θ + α)] l (c) r = 1 + e cos (θ + α)

(b) r = l[1 − e cos (θ − α)] l (d) r = 1 + e cos (θ − α)

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Passage III (Q. Nos. 101-103) Normally, the various proposition you study, i.e. equation of tangent, normal, chord, focal chord, formula for focal distance etc, are derived for the parabola y 2 = 4ax . However, all the results with slight transformation are valid for any parabola. Suppose we represent the equation of parabola y 2 − 4ax = 0 by S ( x , y , a) = 0 and any equation derived for this parabola by P ( x , y , a) = 0. Now, if the given parabola is y 2 = − 4ax , i.e. y 2 + 4ax = 0, we can write it S ( x , y , − a) = 0, so that corresponding equation of P will be P ( x , y , − a) = 0. Similarly, for x 2 = 4ay , it can be written as S ( y , x , a) = 0 and corresponding transformation is P ( y , x , a) i.e. interchange x and y . Transformation for x 2 = − 4ay , it is P ( y , x , − a) = 0. For example, The equation of tangent to y 2 = 4ax is a y = mx + , so corresponding tangent to x 2 = 4ay is m 2a a a and the point of contact will be  , 2 . x = my + m m  m The equation of y = mx − 2am − am 3 .

normal

y 2 = 4ax

to

is

The equation of normal to x 2 = − 4ay is x = my + 2am + am 3 . Further, if the coordinates of vertex are not (0, 0) but (h, k ), then we replaced x by x − h and y by y − k in addition to above transformation.

101. The focal distance of the point ( x, y ) on the parabola x 2 − 8x + 16 y = 0, is (a) | y − 4 | (c) | y − 2|

(b) | y − 5| (d) | x − 4 |

102. The points on the axis of the parabola x 2 + 2x + 4 y + 13 = 0 from where three distinct normals can drawn, are given by (a) (− 1, k ), k ∈ (− 1, 2) (c) (− 1, k ), k ∈ (− ∞ , − 5)

(b) (− 1, − 6) (d) (− 1, 1)

103. The line x cos α + y sin α = p touches the parabola x 2 + 4a ( y + a ) = 0, if (a) a = p sec α (b) a cos 2α = p sin α (c) a2 cos α + p2 sin α = 0 (d) a tan α = p sec α

Ta rg e t E x e rc is e s

Type 5. Match the Columns 104. Match the statements of Column I with statements of Column II. Column I

Column II

A. If a parabola represented p. Equation of directrix is 4 x − 3 y + 12 = 0 by 25( x 2 + y 2 )

Column I If AB subtends 90° at point (0, 0), r. then

t2 =

D.

If AB is inclined at 45° to the axis s. of parabola, then

t 2 = t1 −

= ( 4 x + 3 y − 12 )2, then B. If a parabola represented q. Equation of axis of parabola is 4 x + 3 y = 0 by 25( x 2 + y 2 )

= ( 3 x − 4 y + 12 )2, then

The shortest normal chord of the parabola x 2 = 16 y is of length

p.

1 12

B.

The y-coordinate of the point on the circle ( x − 6)2 + y 2 = 1 which is at a minimum distance from the parabola x 2 = 8 y, is

q.

24 3

C.

A variable chord PQ of the parabola y = 4 x 2 subtends a right angle at the vertex O. The locus of the centroid of ∆OPQ is a parabola whose latusrectum is of length

r.

1

D.

PQ is a variable focal chord of the parabola y 2 = 4 x. The normals at P and Q meet at R. Then, the locus of R is a parabola whose latusrectum is of length

s.

1 2

105. The parabola y = 4ax has a chord AB joining points A ( at 12 , 2at 1 ) and B ( at 22 , 2at 2 ). Match the statements of Column I with values of Column II. 2

Column I

Column II

A.

If AB is normal chord, then

p.

B.

If AB is a focal chord, then

q.

t1 = − t 2 + 2 −4 t2 = t1

Column II

A.

s. Equation of axis of parabola is 3 x + 4 y = 0 t. Equation of directrix is 3 x − 4 y + 12 = 0

2 t1

106. Match the statements of Column I with values of Column II. Column I

C. If a parabola represented r. Equation of directrix is 4 x + 3 y = 12 by 25( x 2 + y 2 )

−1 t1

C.

= ( 4 x − 3 y + 12 )2, then

830

Column II

107. If circle ( x − 6) 2 + y 2 = r 2 and parabola y 2 = 4x have maximum number of common chord, then least integral value of r is _____ .

110. Normal at a point P ( a, − 2a ) intersects the parabola y 2 = 2x at point Q. If the tangent at P and Q meet a point R, then the area of ∆PQR is ka 2 (1 + m2 ) 3

108. The minimum number of normals that can be drawn through any point in the cartesian plane to the parabola y 2 = 4ax is _____ . 109. If the locus of centres of a family of circles passing through the vertex of the parabola y 2 = 4a and cutting the parabola orthogonally at the other point of intersection is 2 y 2 ( 2 y 2 + x 2 − 12ax ) = ax( kx − 4a ) 2 , then find the value of k.

m3

Parabola

14

Type 6. Single Integer Answer Type Questions

.

Find the numerical value of k. 111. Tangents are drawn from the point ( − 2, − 1) to the parabola y 2 = 4ax. If α is the angle between these tangents, then tan α equals___ .

Entrances Gallery 1. The common tangents to the circle x 2 + y 2 = 2 and the parabola y 2 = 8x touch the circle at the points P, Q and the parabola at the points R, S. Then, the area [2014] of the quadrilateral PQRS is (a) 3 sq units (c) 9 sq units

(b) 6 sq units (d) 15 sq units

Passage I (Q. Nos. 2-3) Let a, r, s, t be non-zero real numbers. Let P (at 2, 2at ), Q, R (ar 2, 2ar ) and S (as 2, 2as ) be distinct points on the parabola y 2 = 4ax . Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (2a, 0). [2014] 2. The value of r is (a) −

1 t

(b)

t2 + 1 t

(c)

1 t

(d)

t2 − 1 t

3. If st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is (t 2 + 1)2 2t 3 a(t 2 + 1)2 (c) t3

a(t 2 + 1)2 2t 3 2 a(t + 2)2 (d) t3

(a)

(b)

the parabola y = 4ax . The tangents to the parabola at P and Q meet at a point lying on the line y = 2 x + a, a > 0. 2

[2013]

4. Length of chord PQ is (b) 5a

(c) 2a

(d) 3a

5. If chord PQ subtends an angle θ at the vertex of y 2 = 4ax, then tan θ is equal to (a)

2 7 3

(b)

−2 7 3

(c)

2 5 3

7. Consider the parabola y 2 = 8x. Let ∆ 1 be the area of the triangle formed by the end points of its 1  latusrectum and the point P  , 2 on the parabola 2  and ∆ 2 be the area of the triangle formed by drawing tangents at P and at the end points of the latusrectum. ∆ [2011] Then, 1 is _____ ∆2 8. Let ( x, y ) be any point on the parabola y 2 = 4x. If P is the point that divides the line segment from (0, 0) to ( x, y ) in the ratio 1 : 3. Then, the locus of P is [2011] (a) x 2 = y (c) y2 = x

(b) y2 = 2x (d) x 2 = 2 y

9. Let L be a normal to the parabola y 2 = 4x. If L passes [2011] through the point (9, 6), then L is given by

Passage II (Q. Nos. 4-5) Let PQ be a focal chord of

(a) 7a

6. Let S be the focus of the parabola y 2 = 8 x and PQ be the common chord of the circle x 2 + y 2 − 2x − 4 y = 0 and the given parabola. The [2012] area of ∆PQS is ________ .

Targ e t E x e rc is e s

JEE Advanced/IITJEE

(d)

−2 5 3

(a) (b) (c) (d)

y−x + 3=0 y + 3x − 33 = 0 y + x − 15 = 0 y − 2x + 12 = 0

10. Let A and B be two distinct points on the parabola y 2 = 4x. If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of [2010] the line joining A and B can be (a) − (c)

2 r

1 r

(b)

1 r

(d) −

2 r

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11. The slope of the line touching both the parabolas y 2 = 4x and x 2 = − 32 y is [2014] (a)

1 2

(b)

3 2

(c)

1 8

(d)

2 3

12. Given A circle, 2x 2 + 2 y 2 = 5 and a parabola, y 2 = 4 5x. Statement I An equation of a common tangent to these curves is y = x + 5.

5 ( m ≠ 0) is the m common tangent, then m satisfies m4 − 3m2 + 2 = 0.

Statement II If the line y = mx +

[2013] (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

JEE Main/AIEEE 13. Let O be the vertex and Q be any point on the parabola x 2 = 8 y. If the point P divides the line segment OQ internally in the ratio 1: 3, then the locus [2015] of P is (a) x 2 = y

(b) y2 = x

(c) y2 = 2x

(d) x 2 = 2 y

14. The shortest distance between line y − x = 1 and curve x = y 2 is [2011]

Ta rg e t E x e rc is e s

3 2 (a) 8

(b)

8 3 2

4 (c) 3

3 (d) 4

15. If two tangents drawn from a point P to the parabola y 2 = 4x are at right angles, then the locus of P is (a) x = 1

(b) 2x + 1 = 0 (c) x = − 1

[2010] (d) 2x − 1 = 0

16. A parabola has the origin as its focus and the line x = 2as the directrix. Then, the vertex of the parabola is at [2008] (a) (2, 0) (c) (1, 0)

(b) (0, 2) (d) (0, 1)

17. The equation of a tangent to the parabola y 2 = 8x is y = x + 2. The point on this line from which the other tangent to the parabola is perpendicular to the given [2007] tangent, is (a) (−1, 1) (c) (2, 4)

(b) (0, 2) (d) (− 2, 0)

18. The locus of the vertices of the family of parabolas a3x2 a2x [2006] y= + − 2a is 3 2 3 4 64 (c) xy = 105

35 16 105 (d) xy = 64

(a) xy =

(b) xy =

19. Let P be the point (1, 0) and Q be the point on [2005] y 2 = 8x. The locus of mid-point of PQ is (a) x 2 − 4 y + 2 = 0 (c) y2 + 4 x + 2 = 0

(b) x 2 + 4 y + 2 = 0 (d) y2 − 4 x + 2 = 0

20. A circle touches X-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the [2005] centre of the circle is (a) a parabola (c) a circle

(b) a hyperbola (d) an ellipse

21. If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y 2 = 4ax and x 2 = 4ay, then [2004] (a) d 2 + (b) d 2 + (c) d 2 + (d) d 2 +

(2b + 3c)2 = 0 (3b + 2c)2 = 0 (2b − 3c)2 = 0 (3b − 2c)2 = 0

22. If the normal at the point ( bt 12 , 2bt 1 ) on a parabola meets the parabola again in the point ( bt 22 , 2bt 2 ), then [2003] 2 t1 2 (b) t2 = − t1 + t1 2 (c) t2 = t1 − t1 2 (d) t2 = t1 + t1 (a) t2 = − t1 −

23. The equation of the directrix of the parabola y 2 + 4 y + 4x + 2 = 0 is [2002] (a) x = − 1

(b) x = 1

(c) x = −

3 2

(d) x =

3 2

Other Engineering Entrances 24.The value of λ for which the curve ( 7x + 5) 2 + ( 7 y + 3) 2 = λ 2 ( 4x + 3 y − 24 ) 2 represents a parabola, is [WB JEE 2014] 832

(a) ±

6 5

(b) ±

7 5

(c) ±

1 5

(d) ±

2 5

25. If the equation of parabola is x 2 = − 9 y, then equation of directrix and length of latusrectum are [RPET 2014] 9 ,8 4 9 (c) y = − , 9 4

(a) y = −

(b) y =

9 ,9 4

(d) None of these

4 ( 3 + 2) 3 2( 3 + 2) (d) 3

(b)

(a) −

[BITSAT 2014] (b) a hyperbola (d) None of these

28. The normals at three points P, Q and R of the parabola y 2 = 4ax meet at ( h, k ). The centroid of [VITEEE 2014] ∆PQR lies on (a) x = 0

(b) y = 0

(c) x = − a

(d) y = a

29. AB is a chord of the parabola y 2 = 4ax with vertex A, BC is drawn perpendicular to AB meeting the axis at C. The projection of BC on the axis of the parabola is (a) a

(b) 2a

(c) 4a

[Manipal 2014] (d) 8a

30. The slope of the straight line joining the centre of the circle x 2 + y 2 − 8 x + 2 y = 0 and the vertex of the [Kerala CEE 2014] parabola y = x 2 − 4x + 10is −5 −7 (b) 2 2 (e) None of these

(a)

(c)

−3 2

(d)

7 2

31. If y = 4x + 3 is parallel to a tangent to the parabola y 2 = 12x , then its distance from the normal parallel [WB JEE 2014] to the given line is 213 17 211 (c) 17

(a)

219 17 210 (d) 17 (b)

(b) (1, 81) (d) (− 9, − 24)

33. If a normal chord at a point on the parabola y 2 = 4ax subtends a right angle at the vertex, then t equals [EAMCET 2014] (b) 2 (d) 3

34. The slopes of the focal chords of the parabola y 2 = 32x, which are tangents to the circle [EAMCET 2014] x 2 + y 2 = 4, are 1 −1 , 2 2 −1 1 (c) , 15 15

(a)

1 −1 , 3 3 2 −2 (d) , 15 15

(b)

(a) (x + 1)2 = 4 ( y − 1) (c) (x + 1)2 = − 4 ( y + 1)

(b) (x − 1)2 = 4 ( y + 1) (d) (x − 1)2 = − 4 ( y − 1)

37. The angle between the tangents drawn from the point (1, 4) to the parabolas y 2 = 4x and x 2 = 4 y, is [AMU 2013] (b) π /6 (d) π /3

(a) 0 (c) π /4

38. The line 2x + y + k = 0 is a normal to the parabola [AMU 2013] y 2 = − 8x, if k is equal to (a) − 24

(b) 12

(d) − 12

(c) 24

39. The equation y + 4x + 4 y + k = 0 represents a parabola whose latusrectum is [WB JEE 2012] 2

(a) 1

(b) 2

(c) 3

(d) 4

40. Let P and Q be the points on the parabola y 2 = 4x, so that the line segment PQ subtends right angle at the vertex. If PQ intersects the axis of the parabola at R, then the distance of the vertex from R is (a) 1

(b) 2

[WB JEE 2012] (d) 6

(c) 4

41. If the straight lines y = ± x intersect the parabola y 2 = 8 x in points P and Q, then length of PQ is (b) 4 2

[MP PET 2011] (d) 16

(c) 8

42. The sum of the reciprocals of focal distances of a focal chord PQ of y 2 = 4ax is [Karnataka CET 2011]

[WB JEE 2014]

(a) 1 (c) 2

36. The image of the parabola y 2 = 4x about the line [Karnataka CET 2013] x − y + 1 = 0 is

(a) 4

32. The point on the parabola y 2 = 64x which is nearest to the line 4x + 3 y + 35 = 0, has coordinates (a) (9, − 24) (c) (4, −16)

(b)

(c) 1

27. If a variable chord PQ of the parabola y 2 = 4ax subtends a right angle at the vertex, then the locus of the points of intersection of the normal at P and Q is (a) a parabola (c) a circle

1 2 (d) None of these

1 2

14 Parabola

4 (2 − 3 ) 3 4 3 (c) 3

(a)

35. Let A and B be two distinct points on the parabola y 2 = 4x. If the axis of the parabola touches a circle of radius 2 having AB as its diameter, then the slope of [AMU 2014] the line joining A and B can be

Targ e t E x e rc is e s

26. If the line y − 3x + 3 = 0 cuts the parabola y 2 = x + 2 at A and B, then PA ⋅ PB is equal to [where, [Manipal 2014] P = ( 3 , 0)]

(a)

1 a

(b) a

(c) 2a

(d)

1 2a

43. For the parabola y 2 + 8x − 12 y + 20 = 0, [UP SEE 2011] (a) vertex is (2, 6) (c) latusrectum is 4

(b) focus is (0, 6) (d) axis Y = 6

44. The distance between the vertex of the parabola y = x 2 − 4x + 3 and the centre of the circle [Kerala CEE 2011] x 2 = 9 − ( y − 3) 2 , is (a) 2 3 units (c) 2 2 units (e) 2 5 units

(b) 3 2 units (d) 2 units

45. The parabola y 2 = 4x 2 2 x + y − 6x + 1 = 0 will

and

(a) intersect at exactly one point (b) touch each other at two distinct points (c) touch each other at exactly one point (d) intersect at two distinct points

the

circle

[AMU 2011]

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Objective Mathematics Vol. 1

14

46. The equation of tangent of y 2 = 12 x and making an π [Guj CET 2011] angle with X -axis is 3 (a) ± y − 3x + 3 = 0 (c) ± y − 3x − 3 = 0

(b) ± y + (d) ± y +

t1 =1 t2 (d) t1 + t2 = − 1

(a) t1t2 = 1

(b)

(c) t1t2 = − 1

48. One of the points on the parabola y 2 = 12x with focal distance 12, is [Kerala CEE 2010]

3x + 3 = 0 3x − 3 = 0

(a) (3, 6) (c) (7, 2 21) (e) (1, 12)

47. If t 1 and t 2 are the parameters of the end points of a focal chord for the parabola y 2 = 4ax, then which one is correct? [VITEEE 2010]

(b) (9, 6 3) (d) (8, 4 6)

Answers Work Book Exercise 14.1 1. (d)

2. (a)

3. (d)

4. (b)

5. (b)

6. (a)

7. (a)

4. (a)

5. (d)

6. (c)

7. (c)

4. (c)

5. (c)

6. (a)

7. (d)

Work Book Exercise 14.2

Ta rg e t E x e rc is e s

1. (a)

2. (d)

3. (c)

Work Book Exercise 14.3 1. (d)

2. (c)

3. (b)

Target Exercises 1. (b)

2. (b)

3. (a)

4. (b)

5. (d)

6. (b)

7. (b)

8. (b)

9. (c)

10. (a)

11. (a)

12. (b)

13. (a)

14. (c)

15. (d)

16. (c)

17. (d)

18. (b)

19. (a)

20. (b)

21. (b)

22. (d)

23. (b)

24. (b)

25. (d)

26. (c)

27. (c)

28. (c)

29. (b)

30. (a)

31. (b)

32. (c)

33. (c)

34. (b)

35. (b)

36. (c)

37. (c)

38. (b)

39. (a)

40. (a)

41. (a)

42. (c)

43. (a)

44. (a)

45. (b)

46. (d)

47. (d)

48. (a)

49. (a)

50. (c)

51. (a)

52. (c)

53. (d)

54. (d)

55. (c)

56. (d)

57. (c)

58. (b)

59. (a)

60. (c)

61. (a)

62. (a)

63. (d)

64. (c)

65. (c)

66. (a)

67. (c)

68. (a)

69. (d)

70. (b)

71. (c)

72. (c)

73. (b)

74. (b)

75. (a)

76. (c)

77. (c)

78. (d)

79. (b)

80. (b)

81. (c,d)

82. (a,b,c)

83. (c,d)

84. (b,c)

85. (a,b)

86. (a,b)

87. (c)

88. (d)

89. (a)

90. (b)

91. (a)

92. (c)

93. (a)

94. (a)

95. (b)

96. (c)

97. (b)

98. (a)

99. (d)

100. (d)

101. (b)

102. (b,c)

103. (b)

104. (*)

105. (**)

106. (***)

107. (5)

108. (1)

109. (3)

110. (4)

9. (a,b,d)

111. (3) * A → r; B → p,s; C → t, q ** A → s; B → r; C → q; D → p *** A → q; B → s; C → p; D → r

Entrances Gallery

834

1. (d)

2. (d)

3. (b)

4. (b)

5. (d)

6. (4)

7. (2)

8. (c)

10. (c,d)

11. (a)

12. (a)

13. (d)

14. (a)

15. (c)

16. (c)

17. (d)

18. (d)

19. (d)

20. (a)

21. (a)

22. (a)

23. (d)

24. (b)

25. (b)

26. (b)

27. (a)

28. (b)

29. (c)

30. (b)

31. (b)

32. (a)

33. (b)

34. (c)

35. (c)

36. (a)

37. (a)

38. (c)

39. (d)

40. (c)

41. (d)

42. (a)

43. (a,b,c)

44. (e)

45. (d)

46. (c)

47. (c)

48. (b)

Explanations Target Exercises 7 7 2 2

1 2 . Corresponding value of a is ( 1 + 1) = 4 4 Let the equation of its directrix be y + x + λ = 0. |3 + 4 + λ | 2 ⇒ = 2⋅ 4 2 ⇒ λ = − 6, −8 Thus, equation of parabola is 2 2 7 7 ( x + y − 6)2   , x −  + y −  =   2 2 2

The ends of latusrectum are (2, 4) and (2, − 4). ∴ Centre of circle is (0, 0 ) and radius of circle = 2 2 + 42 = 20 ∴ Equation of circle is x2 + y2 = r 2 ⇒

x 2 + y 2 = 20

9. In ∆BCD, we have tan θ =

y CD

B

Y

(x, y)

2

( x + y − 8)2 7 7   x −  + y −  =   2 2 2

90

2

8. Vertex = (0, 0 )

°–θ

1. Focus is  ,  and its axis is the line y = x.  

X′

A

θ

X

C

D

2. We have, 2{( x − a)2 + ( y − a)2 } = ( x + y )2

⇒

1 | x + y| 2 x + y ( x − a)2 + ( y − a)2 =   2  ( x − a)2 + ( y − a)2 =

Clearly, the equation represents a parabola having its focus at (a, a) and directrix x + y = 0. ∴Latusrectum = 2 (Distance between focus and directrix a+ a =2 = 2 2a 1+ 1

3. Let A ≡ (at 2 , 2 at ), B ≡ (at 2 , − 2 at ), mOA = and

mOB = −

2 t

Y′

⇒

CD = y cot θ y In ∆BAD, we have tan(90° − θ ) = x y2 CD = y cot θ = ⇒ x 4ax ⇒ CD = = 4a x

2  −2  ⋅   = −1 ⇒ t 2 = 4 t  t  Thus, tangents will intersect at (−4 a, 0 ).

X′

6. Distance of focus from directrix is| a cos 2θ − a|. Thus, length of latusrectum is| 4a sin 2 θ |.

7. (a, 2 a) is an interior point of y − 16 x = 0, if 2

(2 a)2 − 16 a < 0

⇒ a2 − 4 a < 0

V (0, 0) and (a, 2 a) are on the same side of x − 4 = 0. So, a− 4< 0 ⇒ a< 4 Now, a2 − 4 a < 0 ⇒ 0 < a < 4

X

(–1, 0) (0, 0) Y′

4. x 2 − 2 x + y − 2 = 0 ⇒ x 2 − 2 x + 1 = 3 − y

Thus, t1t 2 = − 4. Equation of line AB is y(t1 + t 2 ) = 2( x + at1t 2 ) i.e. y(t1 + t 2 ) = 2( x − 4 a), which passes through a fix point (4a, 0 ) .

[Q y 2 = 4 ax]

y 2 = 4x

Y

Thus,

5. Let A ≡ (at12 , 2 at1 ) and B ≡ (at 22 , 2 at 2 ).

[from Eq. (i)]

10. The graph shows that λ > 0.

2 t

⇒ ( x − 1)2 = − ( y − 3) Length of its latusrectum is 1 unit. 1 Since, AS, and BS are in HP. 2 l1 1 2 ⋅ AS ⋅ BS ∴ = ⇒ BS = (4l1 − 1) 2 AS + BS

…(i)

Targ e t E x e rc is e s

⇒

11. x = t 2 − t + 1, y = t 2 + t + 1 ⇒

x + y = 2(t 2 + 1) and

y − x = 2t

x+ y = 1+ 2 ( y − x )2 = 2( x ( y − x )2 = 2( x

⇒ ⇒ ⇒

 y − x    2 

2

+ y) − 4 + y − 2)

∴ Vertex will be the point, where lines y − x = 0 and x + y − 2 = 0 meet at the point (1, 1.) x2 in the equation of circle, we get 4a fx 2 x4 x2 + + 2 gx + +c =0 2 2a 16 a

12. On putting y =

⇒ x 4 + x 2 (16 a2 + 8 a f ) + 32 a2 gx + 16 a2c = 0 ⇒ Σx1 = 0, Σx1 x2 = 8 a(f + 2 a), Σx1 x2 x3 = − 32 a2 g and x1 x2 x3 x4 = 16 a2c 1 1 ⇒ Σy1 = Σ x12 = [(Σ x1 )2 − 2 Σ x1 x2 ] 4a 4a = − 4 (f + 2 a)

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Objective Mathematics Vol. 1

14

13. Extremities of the latusrectum are (2, 4) and (2, − 4). Since, any circle drawn with any focal chord as its diameter touches the directrix. Thus, equation of required circle is ( x − 2 )2 + ( y − 4)( y + 4) = 0 i.e. x 2 + y 2 − 4 x − 12 = 0, its radius is 4 + 12 = 4.

14. Any tangent of parabola is yt = x + t 2 . If it passes through (2, 3), then t 2 − 3 t + 2 = 0 ⇒ t = 1, 2. That point can be (1, 2 ) or (4, 4).

15. Let P(at12 , 2 at1 ) and Q(at 22 , 2 at 2 ) be the end points of a focal chord of the parabola y 2 = 4 ax. Then, PQ = a (t 2 − t1 )2 The equation of PQ is [Qt1 t 2 = − 1] ( t1 + t 2 ) y = 2 x − 2 a It is given that PQ = c and it is at a distance b from the vertex.   −2 a ∴ a ( t 2 − t1 )2 = c and  = b 4  (t1 + t 2 ) + 4  2a c and ⇒ = b [Q t1 t 2 = − 1] ( t 2 − t1 )2 = (t 2 − t1 ) a

19. On solving x − 2 y = 1and y 2 = k x, we get ∴ ∴

y 2 = k(1 + 2 y ) ⇒ y 2 − 2 ky − k = 0 y1 + y2 = 2 k, y1 ⋅ y2 = − k 16 = ( x1 − x2 )2 + ( y1 − y2 )2 2

 y2 y2  =  1 − 2  + ( y1 − y2 )2 k  k ( y + y )2  = ( y1 − y2 )2 ⋅  1 2 2 + 1 k    4k 2  2 = {( y1 + y2 ) − 4 y1 y2 }  2 + 1 k   = 5 { 4 k 2 + 4 k} = 20 k 2 + 20 k 105 − 5 ∴ 5 k 2 + 5k − 4 = 0 ⇒ k = 10 20. Equation of circle is ( x − r − 1)2 + y 2 = r 2 Y

X′

(1, 0)

(r +1, 0)

X

2

c  2 a ⇒ 4 a3 = b2c   =  b a 16. The point of intersection between the curves x 2 = 4 ( y + 1) and x 2 = − 4 ( y + 1) is (0, − 1). The slopes of first and second curve at the point (0, − 1) are respectively 2x m1 = =0 4 −2 x and m2 = =0 4 m − m2 ∴ tan θ = 1 = 0 ⇒ θ = 0° 1 + m1m2

Ta rg e t E x e rc is e s

⇒

Y¢

⇒ ( x − r − 1)2 + 4 x = r 2 ⇒ x 2 + { 4 − 2(r + 1)} x + (2 r + 1) = 0 ∴ D=0 ⇒ 4 (1 − r )2 − 4 (2 r + 1) = 0 ⇒ r =4 21. Q lx + my + n = 0 (lx + n ) ⇒ y=− m and parabola x 2 = 4 ay

X′

3) P (1, , 1) Q (1

y 2 = 4x (a, 2a)

O

S

X

(Focus)

S(1, 3) = 3 − 4 ⋅ 1 > 0 2

So, P(1, 3) is an exterior point. S(1, 1) = 12 − 4 ⋅ 1 < 0

⇒ Q ∴ ⇒ ⇒

 (lx + n ) x 2 = 4a −  m   mx 2 + 4 alx + 4 an = 0 B 2 = 4 AC 2 2 16 a l = 4 m(4 an ) al 2 = mn 2 al − mn = 0

parabola, then (−2 m)2 + (m + 1)2 − 4 < 0 and (m + 1)2 − 4 (−2 m) < 0 Y

Thus, Q(1, 1) is an interior point.

18. Clearly, A = (8 p, 8 p) and B = (8 p, − 8 p) A

X′

X

O

y 2 = 8px 8p

V 8p B

836

…(ii)

22. It is clear that point (−2 m, m + 1) lies inside the circle and

Y′

V = (0, 0 ) m of AV = 1, m of BV = − 1 Therefore, AV ⊥ BV .

Also,

…(i)

From Eqs. (i) and (ii), we get

17. Here, S ≡ y 2 − 4 x = 0 Y

[Q y 2 = 4 x]

x2 + y2= 4

y2= 4x Y′

∴ 5 m2 + 2 m − 3 < 0 and m2 + 10 m + 1 < 0 ⇒ (m + 1)(5 m − 3) < 0 and (m + 5)2 − 24 < 0 3 ⇒ −1 < m < 5 and −5 − 2 6 < m < −5 + 2 6 Hence, −1 < m < −5 + 2 6

26. Let P ≡ (at 2 , 2 at )

L(m, l )

∴Tangent at P is

trisected points. B divides LL′ in 1 : 2.

B

Then, coordinate of X′ A  1⋅ m + 2 ⋅ m − l + 2 l  B ,   1+ 2 3 

X

m C

L′(m, –l ⇒ B(m, l / 3) ) Y′ Let x1 = m, y1 = l / 3 ∴ (m, l ) = ( x1, 3 y1 ) But (m, l ) lie on parabola. (3 y1 )2 = 4 ax1 ⇒ 9 y12 = 4 ax1 Locus is 9 y 2 = 4 ax ∴

24. Let parabola y 2 = 4 ax P ≡ (at12 , 2 at1 ), Q ≡ (at 22 , 2 at 2 ) T ≡ [(at1 t 2 , a (t1 + t 2 ))] Y

P (at 12, 2at1) S (a, 0)

M A

X′

Tangent at vertex is x = 0 On solving Eqs. (i) and (ii), we get Q ≡ (0, at ) and S ≡ (a, 0 ) 2 at − at 1 = = m1 Q Slope of QP is at 2 − 0 t 0 − at and slope of QS is = − t = m2 a−0

…(ii)

14

[say] [say]

m1m2 = − 1 ∠SQP = π /2

Q ∴

27. The path of the water jet is a parabola. Let its equation be y = ax 2 + bx + c Y E(0.5, 4) 4

X

3

T

x+a=0

…(i)

P (0.75, 3)

4

2 N

Q (at 12, 2at2)

Y′

SP = PM = a + at12 SQ = QN = a + at 22 and ST = (a − at1t 2 )2 + [0 − a(t1 + t 2 )]2 = a (1 − t1 t 2 )2 + (t1 + t 2 )2 = a (1 + t12 t 22 + t12 + t 22 ) = a (1 + t12 )(1 + t 22 ) = a(1 + t12 )⋅ a(1 + t 22 ) = (SP )(SQ ) So, SP, ST and SQ are in GP.

1

Q(1, 0) X

(0, 0) O

It should pass through (0, 0 ), (0.5, 4), (1, 0 ). ⇒ c = 0, a = − 16, b = 16 ⇒ y = − 16 x 2 + 16 x If x = 0.75, then y = 3.

28. Equation of normal in terms of slope is y = mx − 2 am − am3 .

25. Q[ x 2 ] = [ y ]

Y

If 0 ≤ y < 1, then [ y ] = 0. ∴ [ x 2 ] = 0 ⇒ 0 ≤ x 2 < 1 ⇒ x ∈ (−1, 1) If 1 ≤ y < 2, then [ y ] = 1 ∴ [x2 ] = 1 ⇒ 1 ≤ x2 < 2 ⇒ x ∈ (− 2 , − 1) ∪ [1, 2 ] If 2 ≤ y < 3, then [ y ] = 2 ⇒ [x2 ] = 2 ⇒ 2 ≤ x2 < 3 x ∈ (− 3, − 2 ] ∪ [ 2 , 3 ) ∴ ............... The graph of the region will not only contain of the parabola y = x 2 but [ x 2 ] = [ y ] contain points within the rectangles of side 1, 2 ; 1, 2 − 1; 1, 3 − 2 , etc. 3

2 1

–√3 –√2 –1

1 √2√3

P

M

X′

O

S

A

Targ e t E x e rc is e s

Let Then,

ty = x + at 2

Parabola

Y

23. Let B and C be the

X

Q

N

R

L Y′

Let O ≡ (h, k ) Then, am3 − (h − 2 a) m + k = 0 or m1 + m2 + m3 = 0 (h − 2 a) ⇒ m1m2 + m2 m3 + m3 m1 = − a k and m1m2 m3 = − a Let P ≡ (am12 , − 2 am1 ) , Q ≡ (am22 , − 2 am2 ) R ≡ (am32 , − 2 am3 ) and S ≡ (a, 0 ) ∴|SP |⋅|SQ |⋅|SR | = | PM |⋅|QN |⋅| RL| = | a + am12 || a + am22 || a + am32 | = a3|(1 + m12 )(1 + m22 )(1 + m32 )|

Hence, a, b, c are incorrects.

= a3|1 + Σm12 + Σm12 m22 + m12 m22 m32 |

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Objective Mathematics Vol. 1

14

= a3|1 + (Σm1 )2 − 2 Σm1m2 + (Σm1m2 )2

33. Let AB be a double ordinate, where A ≡ (at 2 , 2 at ),

− 2 m1m2 m3 Σm1 + (m1m2 m3 ) | 2

= a3 1 + (0 )2 +

2(h − 2 a) (h − 2 a)2 k2 + −0+ 2 2 a a a

= a| k 2 + 2 a(h − 2 a) + a2 + (h − 2 a)2 | = a| k 2 + (h − 2 a + a)2 | = a| k 2 + (h − a)2 | = a|(k − 0 )2 + (h − a)2 | = a (SO )2

34. x 2 − 4 x + 6 y + 10 = 0 ⇒ x 2 − 4 x + 4 = − 6 − 6 y

3

29. Given,

5 n  1 = C3 ( px )n − 3   = nC3 ⋅ pn − 3 x n − 6  x 2

…(i)

Ta rg e t E x e rc is e s

Since, LHS of Eq. (i) is independent of x. ∴ n − 6= 0 ⇒ n = 6 From Eq. (i), we get 5 6 = C3 p3 = 20 p3 2 3 1  1 p3 =   ⇒ p = ⇒  2 2 Given, parabola is y 2 = x. …(ii) 1 Here, 4a = 1 ⇒ a = 4 Since, three normals are drawn from point (q, 0 ). 1 1  or q > p ∴ q > 2 a or q > Qp=  2  2

30. We know that, distance from vertex to the parabola is equal to the focus and directrix. Y S X′

O

x = –a

X

(a, 0)

 2 at 2 + at 22 4 at1 + 2 at 2  31. Here, C ≡  1 ,  3 3   Since, it lies on y = 0. t21, 2a

t 1)

1 V

X

35. Let A ≡ (at12 , 2 at1 ), B ≡ (at 22 , 2 at 2 )  5π  We have, mAS = tan    6

⇒

2 at1 1 =− 2 3 at1 − a

Y A

X′

π 6 X

O

S (a, 0) B Y′

⇒ t12 + 2 3 t1 − 1 = 0 t1 = − 3 ± 2 ⇒ Clearly, t1 = − 3 − 2 is rejected. Thus, t1 = (2 − 3 ) Hence, AB = 4 at1 = 4 a(2 − 3 ) π 6

⇒

2 1 = t1 3

t1 = 2 3 ⇒ Y Equation of normal at A1 is y = − t1 x + 4 t1 + 2 t13 X′ O ⇒ h = 4 + 2 t12 = 4 + 2 ⋅ 12 = 28 Y′

2

A (h, 0)

X

A2

from P and the line y = mx + c will be equal. Thus, locus of D will be a parabola.

4 at1 + 2 at 2 3 4at 2 + 2 at 2 4t1 + 2t 2 2t1 + t 2

through (6, 0 ), then −6 t + 2 t + t 3 = 0

=0 =0 =0 =0

Y A

X′

C (6, 0)

O

32. The equation of the pair of lines VA and VB, where V is the vertex = (0, 0 ) is y2 = 4 a x ⋅

838

A1

38. Any normal of parabola is y = − t x + 2 t + t 3 . If it pass B (at2,2at2 )

⇒ ⇒ ⇒

Circle drawn on focal distance as diameter always touches the tangent drawn to parabola at vertex. Thus, circle will touch the line y + 1 = 0.

37. Let O(h, k ) be the centre of circle. Clearly, distance of O C 2

∴

( x − 2 )2 = − 6 ( y + 1)

Clearly, ∠A1OA =

Thus, the tangent at the vertex divides in the ratio 1 : 1.

X′

⇒

36. Let A1 ≡ (2 t12 , 4 t1 ), A 2 ≡ (2 t12 , − 4 t1 )

Y′

A (a

B ≡ (at 2 , − 2 at ). If P(h, k ) is its trisection point, then h 3k 3 h = 2 at 2 + at 2 , 3k = 4 at − 2 at ⇒ t 2 = , t = a 2a Thus, locus is 9k 2 h = ⇒ 9 y 2 = 4 ax 4 a2 a

y − mx c

⇒ cy 2 − 4 a xy + 4 amx 2 = 0 Since, lines are perpendicular to each other. ∴Coefficient of x 2 + Coefficient of y 2 = 0 ⇒ c + 4 am = 0 ⇒ c = − 4 am

Y′

⇒ t = 0, t 2 = 4 Thus, A ≡ (4, 4) For no common tangent, AC > 4 + 16 > r r < 20 ⇒

X

Equation of AB is y(t1 + t 2 ) = 2( x + t1 t 2 ) A

P (h, k)

⇒

 ⇒  m − 

Y

O

⇒

⇒ X

B Y′

It must be same as y = 2 x + c . t1 + t 2 2 2 t1 t 2 ⇒ = = 1 2 c Let P (h, k ) be such that AP : PB = 1 : 1, then 2 h = t12 + t 22 , 2 k = 2 t1 + 2 t 2 ⇒ t1 + t 2 = k = 1 ⇒ k =1 Thus, locus of P is y = 1.

41. Equation of tangent of parabola at (a, 2 a) is y ⋅ 2 a = 2 a( x + a) ⇒ y − x −a=0 Equation of circle touching the parabola at (a , 2 a) ( x − a)2 + ( y − 2 a)2 + λ ( y − x − a) = 0 Since, it passes through (0, 0). ∴ a2 + 4 a2 + λ (− a) = 0 ⇒ λ = 5a Thus, required circle is x 2 + y 2 − 7 a x + ay = 0 and radius 49 2 a2 5 a + a = = 4 4 2

42. We know that, every line passing through the focus of a parabola intersects the parabola in two distinct points except lines parallel to the axis. The equation ( y − 2 )2 = 4 ( x + 1) represents a parabola with vertex (−1, 2 ) and axis parallel to X-axis. So, the line of slope m will cut the parabola in two distinct points, if m ≠ 0 i.e. m ∈ (−∞, 0 ) ∪ (0, ∞ ).

14

  1  < 2 2 m m + 1 4m4 + 4m2 − 1 > 0  2 2 − 1  2 1 + 2  m −  m +  >0 2  2  

⇒

40. Let A ≡ (t12 , 2 t1 ) and B ≡ (t 22 , 2 t 2 )

X′

i.e.

Parabola

Equation of tangent at this point is yt = x + t 2 . Since, it must pass through (6, 5). ∴ t 2 − 5t + 6 = 0 ⇒ t = 2, 3 Hence, possible points are (4, 4) and (9, 6).

 2 2 − 1 m −  >0 2   2 − 1  m+ 2    m ∈  −∞, − 

2 − 1 >0 2  2 − 1  ∪ 2  

2 − 1  , ∞ 2 

45. Any tangent to the parabola y 2 = 4 x is y = mx + It passes through (α , β ), if β = mα + ⇒

1 . m

1 m

α m2 − β m + 1 = 0

It will have roots m1 and 2 m1, if β 1 m1 + 2 m1 = and m1 ⋅ 2 m1 = α α 2  β  1 ⇒ 2⋅  =  3α  α ⇒ ∴

Targ e t E x e rc is e s

39. Let the possible point be (t 2 , 2 t ).

2 β2 1 = 2 α 9α 2 α = β2 9

46. Equation of parabola is y 2 + 4 = 4 x. ⇒ y 2 = 4 ( x − 1) Any tangent to parabola is y = m( x − 1) +

1 . m

It passes through the origin, if 0 = −m+ ⇒

1 m

m = 1, − 1

So, the two tangents are inclined at

π . 2

47. The equations of the parabolas are and

y 2 − 4 x − 8 y + 40 = 0 x 2 − 8 x − 4 y + 40 = 0

…(i) …(ii)

44. The equation of tangent of slope m to the parabola

We observe that, if we interchange x and y in Eq. (i), we obtain Eq. (ii). So, the two parabola are symmetric about y = x. If the two parabolas intersect on y = x, then the minimum distance between them is zero. On solving y = m and y 2 − 4 x − 8 y + 4 = 0, we get x 2 − 12 x + 40 = 0, which has imaginary roots.

1 . m This will be a chord of the circle x 2 + y 2 = 4, if length of the perpendicular from the centre (0, 0 ) is less than the radius.

So, the parabolas do not intersect. Consequently, distance between them is not zero. Minimum distance between the two parabolas is the distance between tangents to the two parabolas which are parallel to y = x.

43. Let the vertex be P(h, k ), then its equation is ( y − k )2 = − 4 a( x − h ). It will touch y 2 = 4 ax, if ( y − k )2 = − y 2 + 4 a h has equal roots. ⇒ k 2 = 8 ah 2 Thus, locus is y − 8 ax = 0. y 2 = 4 x is y = mx +

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Objective Mathematics Vol. 1

14

On differentiating Eq. (ii) w.r.t. x, we get dy 2x − 8 − 4 =0 dx dy x − 4 ⇒ = dx 2 If the tangent is parallel to y = x, then x−4 =1 2 ⇒ x=6 On putting x = 6 in Eq. (ii), we get y = 7 Thus, the coordinates of a point on parabola (ii) are P(6, 7 ). The corresponding point on parabola (i) is Q(7, 6). ∴ Required distance, PQ = 1 + 1 = 2 a 48. Any tangent to y 2 = 4 ax is y = mx + . m a  It touches x 2 = 4 ay, if x 2 = 4 a mx +  , i.e.  m 2 4a x 2 − 4 amx − = 0 has equal roots. m So, m2 + 1 = 0, i.e. m = − 1 Hence, the common tangent is y = − x − a.

Ta rg e t E x e rc is e s

49. On solving 2 y 2 = 3 ( x + 1) and x 2 + y 2 + 2 x = 0, we get x2 + ⇒ ⇒ ⇒

3 ( x + 1) + 2x = 0 2 7x 3 x2 + + =0 2 2 2 x2 + 7 x + 3 = 0 (2 x + 1)( x + 3) = 0

parabola.

51. The locus is the directrix, which is a line. 52. Let P = (α, β). Any tangent to the parabola is y = mx + Since, it passes through (α , β ). ∴

β = mα +

⇒ m α − βm + 1 = 0 Its roots are m1, 3m1. β 1 So, m1 + 3 m1 = , m1 ⋅ 3 m1 = α α 2  β  1 ∴ 3⋅   =  4α  α ⇒ 3 β 2 = 16 α 2

1 m

a . m

[Q a = 1]

2t2 + 0 4t + 0 ,β = 2 2 2  β Eliminating t, we get α =   .  2 2 So, the locus is y = 4 x. α=

2

− 3 3  −1  1  =  −  +  − −   = 3 2   2  2  2 

54. y 2 − 4 x − y − 4 = 0 ⇒ ⇒

1 17 = 4x + 4 4 2 1 17     y −  = 4 x +    2 16

y2 − y +

Circle drawn with diameter being any focal chord of the parabola always touches the directrix of the parabola. 17 Thus, circle will touch the line x + = −1 16 i.e. 16 x + 33 = 0

55. Let the tangents intersect at P(h, k ), then lx + my + n = 0

at 2 + at 22 2 at1 + 2 at 2 ,β = 50. Here, α = 1 = a(t 2 + t 2 ) 2 2 (2at 12 , 2at1)

2 x y2 = 2 + 8 which is a a a

53. If the middle point of a chord is (α, β), then

1 x = − ,−3 2 But x = − 3 makes y imaginary, because 2 y 2 = 3 ( x + 1). 1 So, x=− 2 3 ⇒ y=± 2 ∴The length of the chord = The distance between  1 3  1 3 − ,  and  − , −  2   2 2   2

Y

So, the locus of (α , β ) is

Thus, the locus is 3 y 2 = 16 x, which is a parabola.

⇒

2

2 at1 − 0 2 at 2 − 0 = −1 × 2 at12 − 0 at 2 − 0 ⇒ t1 t 2 = − 4 2α Now, = t12 + t 22 = ( t1 + t 2 )2 − 2t1 t 2 a 2  β =   − 2 ⋅ (−4)  a Also,

will be the chord of contact of P i.e. lx + my + n = 0 and yk − 2 ax − 2 ah = 0 will represent the same line. Thus, k −2 a −2 ah = = m l n n 2 am ⇒ h= ,k =− l l

y2 = 4ax

56. Let the mid-point of chord be P(h, k ), then its equation is X′

(a, b)

(0, 0)

X

T = S1 i.e. yk − 2 a( x + h ) = k 2 − 4 ah. It must pass through(3 a, a), hence ak − 2 a(3 a + h ) = k 2 − 4 ah. Thus, locus of P is y 2 − 2 a x − ay + 6 a2 = 0.

57. Clearly, both the lines pass through(−a, b), which a point Y′

840

lying on the directrix of the parabola. Thus, m1m2 = − 1. Because tangents drawn from any point on the directrix are always mutually perpendicular.

64. Let the mid-point be P(h, k ). Equation of this chord is

Thus, and

SC = (at1 t 2 − a) + a (t1 + t 2 )

⇒

SC = a (t1 t 2 − 1)2 + (t1 + t 2 )2

2

2

It must pass through (a, 0 ). ⇒ −2 a(a + h ) = k 2 − 4 ah

2

Thus, required locus is y 2 = 2 ax − 2 a2 .

= a t12 + t 22 + 1 + t12 t 22 = a (1 + t12 )(1 + t 22 ) Clearly, SC 2 = SA ⋅ SB

59. Line

will

touch the parabola, 8 (− by − c ) has equal roots. y − 4 y + 32 = a ⇒ 4b2 = 7 a2 + 2 a(c + 2 b)

if

2

60. The equation of a tangent to y = 4 ax is 2

a y a …(i) − ⇒ x= m m m2 This will be a normal to the parabola x 2 = 4 by, if it is of the form y 2b b …(ii) − − x= m m m3 2b a b − 2 =− − ∴ m m3 m ⇒ 2 bm2 − am + b = 0 Since, m is a real. ∴ a2 − 8 b2 ≤ 0 ⇒ a2 ≥ 8 b2 y = mx +

61. Chord of contact of mutually perpendicular tangents is always a focal chord. Thus, minimum length of AB is 4 a.

62. Let P ≡ (h, k ). Tangent in terms of m is y = mx +

a m

⇒ m2 x − my + a = 0 If it passes through (h, k ), then m2 h − mk + a = 0. π Angle between these tangents is . 4 π | m1 − m2 | Thus, tan = 4 |1 + m1m2 | ⇒ (m1 − m2 )2 = (1 + m1m2 )2 ⇒ (m1 + m2 )2 − 4m1m2 = (1 + m1m2 )2 2

65. Let A ≡ (at12 , 2 at1 ), B ≡ (at 22 , 2 at 2 ). Then, 2 2 and m OB = t1 t2 Since, m OA ⋅ m OB = − 1 ⇒ t1 t 2 = − 4 So, tangents will intersect at P[at1 t 2 , a (t1 + t 2 )]. Thus, locus of P is x = − 4 a or x + 4 a = 0. m OA =

66. Let P(at 2 , 2 at ) be a point on the parabola y 2 = 4 ax such that it cuts the parabola again at Q(at12 , 2 at1 ). Then, 2 t1 = − t − t 2  |t1| = t + ⇒ t    2 [using AM ≥ GM] ⇒ |t1| ≥ 2 t ×  t   |t1| ≥ 2 2

⇒ Now,

OQ = (at12 − 0 )2 + (2 at1 − 0 )2

⇒

OQ = at1 t12 + 4

⇒ ⇒

OQ ≥ 2 a 2 8 + 4 OQ ≥ 4 6 a

67. Let the coordinates of point P be (h, k ). Equation of tangent of x 2 = 4 ay, in terms of m, is y a + 2. x= m m If it passes through P, then m2 h − km − a = 0 k a ⇒ mPA + mPB = , mPA ⋅ mPB = − h h 2 2 Now, 4 = mPA + mPB = (mPA + mPB )2 − 2 mPA ⋅ mPB k 2 2a 4= 2 + ⇒ h h Locus is y 2 = 4 x 2 − 2 a x. ∴

68. Let P(−2 + r cos θ, r sin θ ) and Q lies on parabola Y

2

4a  k a − = 1 +  h  h h2 Thus, the locus is y 2 = 4 ax + (a + x )2 . ⇒

14 Parabola

T = S1, i.e. yk − 2 a( x + h ) = k 2 − 4 ah.

C ≡ [at1t 2 , a(t1 + t 2 )] SA = a + at12, SB = a + at 22

Targ e t E x e rc is e s

58. If A ≡ (at12 , 2 at1 ), B ≡ (at 22 , 2 at 2 ), then

Q P

63. Let A ≡ (t12 , 2 t1 ), B ≡ (t 22 , 2 t 2 ), then P ≡ (t1 t 2 , t1 + t 2 ). Also, equations of PA and PB are y t1 = x + at12 , yt 2 = x + at 22 Thus, A1 ≡ (0, at1 ), B1 ≡ (0, at 2 ) 1 Now, area of ∆PA1 B1 = ⋅| A1B1 ||t1 t 2 | 2 1 2 = a|t1 − t 2||t1 t 2 | 2 ⇒ a2 (t1 − t 2 )2 (t1 t 2 )2 = 16 ⇒ a2 [(t1 + t 2 )2 − 4 t1t 2 ](t1 t 2 )2 = 16 Thus, locus of P is a2 ( y 2 − 4 x )x 2 = 16.

X′

θ A(–2, 0) O

X

Y′

[given]

⇒ ⇒ Now,

r sin θ − 4(−2 + r cos θ ) = 0 4 cos θ 8 , r1r2 = r1 + r2 = sin 2 θ sin 2 θ 1 1 r +r cos θ + = 1 2 = AP AQ r1r2 2 2

2

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Objective Mathematics Vol. 1

14

1 1 1 + < AP AQ 4 1 ⇒ cos θ < ⇒ tan θ > 3 2 [Qcos θ is decreasing and tan θ is increasing in (0, π /2 )] m> 3 ⇒

Given,

69. The equation of the normal at (−2 + t 2 , 2 t ) is y − 2t =

2t ⋅(x + 2 − t 2 ) 2 ⇒ t x + y = 2t − 2t + t3 ⇒ t x + y =t3 It passes through (k, 0), if tk = t 3 . ⇒ t (t 2 − k) = 0

Y

X′

y − 2t = −

⇒

P

≡ (at 22 , 2 at 2 ).

B 2 We have, t 2 = − t1 − t1 Now, AB 2 = [a2 (t12 − t 22 )]2 + 4 a2 (t1 − t 2 )2 = a2 (t1 − t 2 )2 {(t1 + t 2 )2 + 4}

2 2 mOA = , mOB = − t t ⇒ t2 = 4 ⇒ t =2 Equation of normal AP is y = − 2 x + 4 + 8 ⇒ P ≡ (6, 0 ). Thus, area of quadrilateral OAPB 1 = (OP )( AB) 2 1 = ⋅ 6 ⋅ 8 = 24 sq units 2

74. Let AB be a normal chord, where If its mid-point is P(h, k ), then 2 h = a(t12 + t 22 ) = [(t1 + t1 )2 − 2 t1t 2 ] 2 k = 2 a(t1 + t 2 )

and We have,

t 2 = − t1 −

2

  2  4 = a2 t1 + t 2 +   2 + 4  t1   t1 

⇒

16 a (1 + d ( AB ) = 4 dt1 t1 4 2 2  3 1 + ⋅ 2 t1 ] − (1 + t12 )3 ⋅ 4 t13  t [ ( t ) 1 = 16 a2   t18   2

=

t12 )3

a2 ⋅ 32(1 + t12 )2 2 (t1 − 2 ) t15 t1 = 2 is indeed the point of minima of AB 2 . 4a (1 + 2 )3 / 2 = 2 a 27 units ABmin = ∴ 2

72. Equations of tangent and normal at A are yt = x + at 2 and y = − t x + 2 at + at 3 . (at 2, 2at) A X′

B O

D

X

C Y′

⇒

t1 + t 2 = −

2

=

Y

X

A ≡ (at12 , 2 at1 ) and B ≡ (at 22 , 2 at 2 ).

71. Let AB be a normal chord, where A

O B

If it passes through (a, 0), then am − 4m − 2 m3 = 0. ⇒ m(a − 4 − 2 m2 ) = 0 a−4 ⇒ m = 0, m2 = 2 For three distinct normals, a− 4= 0 ⇒ a> 4 ≡ (at12 , 2 at 2 ),

A π 2

Y′

70. Equation of normal in terms of m is y = mx − 4 m − 2 m3 .

Ta rg e t E x e rc is e s

73. Let A ≡ (t 2 , 2 t ) and B ≡ (t 2 , 2 t )

−1 ⋅(x + 2 − t 2 )  dy     dx  ( −2 + t 2, 2 t )

It has three real values of t, if k > 0. So, the set of points (k, 0 ) is the such that k > 0.

842

If ABCD is a rectangle, then mid-points of BD and AC will be coincident. ⇒ h = 2 a , k + 2 at = 0 k ⇒ h = 2a , t = − 2a Thus, locus is x = 2 a.

B ≡ (− at 2 , 0 ), D ≡ (2 a + at 2 , 0 )

⇒

t1 = −

2 t1

2 t1

and t1t 2 = − t12 − 2

2a and k

 2 h = at12 + 2 + 2   t1 

 4a2 y2  Thus, required locus is x = a  2 + 2 + 2  . 2a   y

75. Let r be the radius of the circle. Then, its equation is ( x + 4)2 + y 2 = r 2 . This cuts the parabola y 2 = 8 x at points A( x2 , y2 ) and B( x2 , y2 ). The abscissae of A and B are the roots of the equation ( x + 4)2 + 8 x = r 2 2 ⇒ x + 16 x + 16 − r 2 = 0 …(i) ∴ x1 x2 = 16 − r 2 2 The ordinates of A and B are given by y1 = 8 x1 and y22 = 8 x2 , respectively. y1 = ± 2 2 x1 and y2 = ± 2 2 x2 ∴ Since, AB subtends a right angle at the vertex of the parabola. y1 y2 ∴ × = −1 x1 x2 ⇒ x1 x2 + y1 y2 = 0

x1 x2 + 8 x1 x2 = 0 x1 x2 = 0 16 − r 2 = 0 r =4

76. Equation of normal is bx a2 − a b |c + 2 m2 + m4|

y=− ∴Required distance =

y 2 = 4 a( x − l ) and y = − 4 b( x + m). Further, let y = h be a line parallel to the common axis. This intersects the two parabolas at  h2   −h 2  A + I, h and B  − m, h respectively. 4 4 a b    

80. Let the two parabolas be [from Eq. (i)]

14

Let P(α , β ) be the mid-point of AB. h 2  1 1 Then, 2α =  −  + I − m and β = h 4  a b

1 + m2

= (1 + m2 )3 / 2

2

Parabola

⇒ ⇒ ⇒ ⇒

[Q c = 1]

77. Let A ≡ (at12 , 2 at1 ), B ≡ (at12 , − 2 at1 ).

⇒

2α =

β 2  1 1  −  +I−m 4  a b Y

Equation of tangents at A and B are yt1 = x + at12 and yt1 = x + at12 , respectively. A1 ≡ (0, at1 ), B1 ≡ (0, − at1 )

B

P (a , b )

Y

X′

(–m,0)

(t,0)

y=h A

X

A A1 y 2 = –4b(x + m)

O B1

S

C

y 2 = 4a(x – l )

X

B

Y′

∴ Area of trapezium A A 1 B1B 1 = ( AB + A 1 B1 )⋅ OC 2 1 ⇒ 24 a2 = ⋅ (4 at1 + 2 at1 ) (at12 ) 2 ⇒ t13 = 8 ⇒ t1 = 2 ⇒ A1 ≡ (0, 2 a) If ∠OSA1 = θ, then 2a tan θ = = 2 ⇒ θ = tan−1(2 ) a Thus, required angle is 2 tan −1(2 ).

Y′

Thus, the locus of (α , β ) is y 2  1 1 2x =  −  +I−m 4  a b Clearly, it represents a parabola, if a ≠ b. If a = b, then we have 2 x = I − m, which represents a line.

Targ e t E x e rc is e s

X′

81. End point of latusrectum x = a, y = ± 2 a Y P t 1

X′

O (0, 0)

X

78. Normals to y 2 = 4 ax and x 2 = 4 by in terms of m are y = mx − 2 am − am3 b and y = mx + 2 b + 3 m For a common normal, b 2 b + 2 − 2 am + am3 = 0 m ⇒ am5 + 2 am3 + 2 bm2 + b = 0 Thus, there can be atmost 5 common normals.

79. Tangent to y = 4 x in terms of m is 2

1 m Normal to x 2 = 4by in terms of m is b y = mx + 2 b + 2 m 1 b If these are same lines, then = 2b + 2 m m ⇒ 2 bm2 − m + b = 0 For two different tangents, 1 1 − 8 b2 > 0 ⇒| b| < 8 y = mx +

Q

t2

Y′

t 2 = − t1 −

2 t1

82. By properties of parabola, options (a), (b) and (c) are always correct.

83. Equation of any tangent to y 2 = 9 x is of the form 9 2 t . 4 Since, it passes through (4, 10). 9 ∴ 10t = 4 + t 2 4 ⇒ 9 t 2 − 40 t + 16 = 0 4 ⇒ t = 4, 9 ∴ Equation of tangent can be x − 4 y + 36 = 0 or 9x − 4y + 4 = 0 yt = x +

843

Objective Mathematics Vol. 1

14

84. Given curve is y = x 2 − x.

…(i)

When x − y = 0 or y = x Then, from Eq. (i), x 2 − 2 x = 0, its D ≠ 0 When x + y = 0 or y = − x Again, from Eq. (i), x 2 = 0 , its D = 0 When x − y = 1or y = x − 1, then from Eq. (i), x 2 − 2 x + 1 = 0, its D = 0 When x + y = 1or y = 1 − x, then from Eq. (i), x 2 − 1 = 0, its D ≠ 0 Hence, only x + y = 0 and x − y = 1are the tangents to given parabola. 2a 85. The equation of tangent to y 2 = 8 ax is y = mx + . m Y

O

X′

(0, 0) r

X

2a

P

Ta rg e t E x e rc is e s

(√2a, 0)

y = mx + m

Y′

2a is tangent to x 2 + y 2 = 2 a2 . m OP = r 2a 0−0+ m = 2a 1 + m2 2 = 1 + m2 m2 m4 + m2 − 2 = 0 2 (m + 2 )(m2 − 1) = 0 m = ±1

Since, y = mx + ∴ ⇒ ⇒ or ⇒ ∴

⇒ Equation of tangent is y = x + 2 a and y = − x − 2 a

86. The equation of tangent to parabola is 2a …(i) m Length of a perpendicular from (0, 0) to Eq. (i) = 2a (Radius) y = mx +

89. Clearly, Statement I is true. Also, the point of contact is given by (a, − 2 a) and (− 2a, − a). Hence, common tangent can be found by joining these two points also. Hence, Statement II is also correct and is the correct explanation of Statement I.

90. If t1and t 2 are the parameters of a and b, then Also,

9 4m Satisfying (4, 10) with the above equation, we get 1 9 m= , 4 4 So, Statement I is true and Statement II is false.

88. Statement II is true as it is the definition of parabola. From Statement I, we have

|3 x + 4 y + 5| 5 which is not parabola as point (1, − 2 ) lies on the line 3 x + 4 y + 5 = 0 . Hence, Statement I is false. ( x − 1)2 + ( y + 2 )2 =

844

…(i) …(ii)

On solving Eqs. (i) and (ii), we get t1 = 2 2 Also, mAB = = − t1 = − 2 t 2 + t1

91. Any normal at t to the parabola y 2 = 4 x is y + t x = 2t + t3 Since, it passes through (h, h + 1.) ∴ h + 1 + th = 2 t + t 3 ⇒ t 3 − (h − 2 )t − h − 1 = f (t ) ∴ f ′ (t ) = 3 t 2 − (h − 2 ) = 3 t 2 + (2 − h ) > 0

[say] [Q h < 2]

So, f (t ) = 0 will have only one real root. Also, (h + 1)2 − 4h = (h − 1)2 > 0, h ≠ 1 Hence, (h, h + 1) lies outside the parabola.

92. End points of double ordinate x = 4 of parabola y 2 = 4 x are (4, ± 4) ⇒ t1 = ± 2 2 ⇒ t 2 = − t1 − = ± 3 t1 ⇒ P(9, 6) and P′ (9, − 6) ∴ PP′ = 12 units

93. y = ax 2 + c dy = 2a x = 1 dx So, point of contact of the tangent is  1 1  + c  ,  2a 4 a 

∴

Since, it lies on y = x. ∴

1 4a 1 for a = 2 c= 8 c=

Thus,

1 2

94. If (1, 1) is point of contact, then a = .

87. Let the equation of tangent from (4, 10) be y = mx +

t1 t 2 = − 4  2 t1  − t1 −  = − 4  t1 

 1 1  , + 2 .  2a 4 a 

95. If c = 2, then point of contact is 

Since, it lies on the line y = x. 1 1 ∴ = +2 2a 4 a 1 i.e. a= 8 So, point of contact is (4, 4).

96. QSP = e ⋅ PM = e ⋅ NZ = e (SZ − SN ) = e ⋅ SZ − e ⋅ SN = l − e ⋅ (r cos θ ) or r = l − er cos θ

∴ ⇒

2

…(i)

97. Qe ⋅ SZ = Semi-latusrectum SL = l

[given]

l SZ = e

∴

98. Q r cos θ = SZ =

l e

l = e cos θ r

∴

99. Put e = 1in Eq. (i), then the equation of parabola is l  θ = 1 + cos θ = 2 cos 2    2 r l 2  θ r = sec    2 2

∴

100. Since, the axis SZ of the conic makes an angle α with the initial line SA, so the equation of the conic will be l = 1 + e cos (θ − α ) r l r= ∴ 1 + e cos(θ − α ) M

P

Z

r θ α A

S

101. ( x − 4)2 = − 16 ( y − 1) ∴ Focal distance of the parabola ( x − 4)2 = − 16 ( y − 1) is y − 1− 4 = y − 5

102. Equation of the normal is x + 1 = m ( y + 3) + 2 m + m3 . It passes through (−1, k ) on the axis. If m(m2 + k + 5) = 0 ∴ m=0 and m = ± −1(k + 5), if k < −5 Hence, (−1, − 6) and (−1, k ), k ∈ (− ∞, − 5).

103. Equation of tangent is mx − m2 y = m2 a − a and given equation is x cos α + y sin α = p On comparing Eqs. (i) and (ii), we get a cos 2 α = p sin α

…(i) …(ii)

104. A. Q 25 ( x 2 + y 2 ) = (4 x + 3 y − 12 )2 ⇒ 25 x 2 + 25 y 2 = 16 x 2 + 9 y 2 + 24 xy − 96 x − 72 y + 144 ⇒ 9 x 2 + 16 y 2 − 24 xy = − 96 x − 72 y + 144 ⇒

 4 x + 3 y − 6  3x − 4y 25   ×5  = − 24     5  5 2 24  4 x + 3 y − 6  3x − 4y ⇒    =−    5  5  5 3x − 4y 4x + 3y − 6 Let = Y and =X 5 5 24 Y2 = − X ∴ 5 2 On comparing with Y = − 4 ρX, we get 24 6 4ρ = ⇒ ρ= 5 5 ∴ Equation of directrix is X −ρ = 0 4x + 3y − 6 6 i.e. − = 0 ⇒ 4 x + 3 y = 12 5 5 Axis of the parabola is Y=0 3x − 4y i.e. =0 5 ⇒ 3x − 4y = 0 B. ∴ 25 ( x 2 + y 2 ) = (4 x − 3 y + 12 )2 or 25 x 2 + 25 y 2 = 16 x 2 + 9 y 2 − 24 xy + 96 x − 72 y + 144 ⇒ 9 x 2 + 16 y 2 + 24 xy = 96 x − 72 y + 144 ⇒ (3 x + 4 y )2 = 24 (4 x − 3 y + 6) 2  4 x − 3 y + 6  3x + 4y ⇒ 25 ×   ×5  = 24      5 5 2 24  4 x − 3 y + 6  3x + 4y ⇒    =    5  5  5 3x + 4y 4x − 3y + 6 Let = Y and =X 5 5 24 Y2 = X ∴ 5 On comparing with Y 2 = 4 ρX, we get 24 6 4ρ = ⇒ ρ= 5 5 ∴ Equation of direction is X + ρ = 0 4x − 3y + 6 6 i.e. + =0 5 5 ⇒ 4 x − 3 y + 12 = 0 and axis of the parabola is Y = 0 3x + 4y i.e. =0 5 or 3x + 4y = 0 C. Q 25 ( x 2 + y 2 ) = (3 x − 4 y + 12 )2 ⇒ 25 x 2 + 25 y 2 = 9 x 2 + 16 y 2 − 24 xy + 72 x − 96 y + 144 ⇒ 16 x 2 + 9 y 2 + 24 xy = 72 x − 96 y + 144 ⇒ (4 x + 3 y )2 = 24 (3 x − 4 y + 6) ⇒

(3 x − 4 y )2 = − 24 (4 x + 3 y − 6)

14 Parabola

r (1 + e cos θ ) = l l = 1 + e cos θ r l r= 1 + e cos θ

Targ e t E x e rc is e s

⇒

2

⇒

 3 x − 4 y + 6  4x + 3y 25 ×   ×5  = 24     5 5 2

⇒ Let ∴

24  3 x − 4 y + 6  4x + 3y    =    5  5  5 4x + 3y 3x − 4y + 6 = Y and =X 5 5 24 Y2 = X 5

845

Objective Mathematics Vol. 1

14

On comparing with Y 2 = 4 ρX, we get 24 4ρ = 5 6 ⇒ ρ= 5 ∴ Equation of directix is X+ρ=0 3x − 4y + 6 6 i.e. + =0 5 5 ⇒ 3 x − 4 y + 12 = 0 and axis of the parabola is Y = 0 i.e. 4x + 3y = 0  2 105. A. If AB is a normal chord, then t 2 = t1 +   t1  B. If AB is a focal chord, then t1 t 2 = − 1. C. If AB subtend 90° at (0,0), then t1 t 2 = − 4 . 2 a (t1 − t 2 ) 2 D. tan 45° = = a (t1 − t 2 ) (t1 + t 2 ) t1 + t 2 2 ∴ 1= t1 + t 2 ⇒ t1 + t 2 = 2

C. P( t , 4 t 2 ) and Q (t12 , 4 t12 ) ∴ ∠POQ = 90 ° ⇒ 16 t t1 = −1 Centroid of ∆OPQ is  t + t1 4 2  (x, y) =  , ( t + t12 )  3  3 3 1 ∴ (3 x )2 − y = (t + t1 )2 − (t 2 + t12 ) = 2t t1 = − 4 8 ⇒ 72 x 2 − 6 y = − 1 1  1 x2 = ⇒ y −  12  6 which is a parabola whose latusrectum is of length 1 . 12 D. If P is t1, Q is t 2 , then t1 t 2 = − 1 Equation of the normal at P and Q are y + xt1 = 2t1 + t13 and y + x t 2 = 2 t 2 + t 23

Ta rg e t E x e rc is e s

106. A. Any point on x 2 = 16 y is p(8 t , 4 t 2 ). The equation of the normal at P is x + yt = 8 t + 4 t 3 This normal meets the parabola at Q whose 2 parameter is t1, where t1 = − t − . t Now, PQ 2 = (8 t − 8 t1 )2 + (4 t 2 − 4 t12 )2 = (2 t − 2 t1 )2 {16 + (2 t + 2 t1 )2 } 2

4  16  256 (1 + t 2 )3  =  4 t +  16 + 2  =  t  t  t4 −4 / 3 2/ 3 3 = 256 {t +t } 1 1  = 256  t 2 / 3 + t 2 / 3 + t −4 / 3  2 2 

3

3

∴

 1 [using AM ≥ GM] ≥ 256  3 3  4  256 × 27 = = 64 × 9 × 3 4 PQ ≥ 24 3

B. The pair of points A, B such that AB is minimum, is such that AB passes through the centre of the circle also, AB is normal at A to the parabola. A can be taken as (4 t , 2 t 2 ) and the normal at A is x + y t = 4t + 2 t 3. Y

B

C (6, 0)

X

Y′

846

∴

− y 3 − (2 − x ) y − y = 0

So, locus of ( x, y ) is y 2 + 2 − x + 1 = 0 which is a parabola whose latusrectum is 1.

107. For maximum number of common chord, circle and parabola must intersect in four distinct points. Let us first find the value r when circle and parabola touch each other. For that solving the given curves, we have ( x − 6)2 + 4 x = r 2 or x 2 − 8 x + 36 − r 2 = 0 curves touch, if discriminant D = 64 − 4(36 − r 2 ) = 0 or r 2 = 20. Hence, least integral value of r for which the curves intersect, is 5.

108. The normal to the parabola y 2 = 4ax is y = mx − 2 am − am3 . If it passes through a point (h, k), then k = 3m − 2 am − am3 This equation is cubic in m, it has three roots. Atleast one of them must be real. Hence, the minimum number of normals is one.

109. Let (at 2 , 2 at )be any point on y 2 = ax.Then, vertex A(0,0). ∴Equation of tangent at P is ty = x + at 2

…(i)

Tangent at P will be normal to the circle, AP is a chord   at 2 2 whose mid-point is  , at  and slopes is . t   2

A X′

If the normals intersect at ( x, y ), then t1 and t 2 are the two roots of the cubic equation in t. …(i) t 3 + (2 − x )t − y = 0 If t 3 is the third root, then t1 t 2 t 3 = y ⇒ t 3 = − y, then t1t 2 = 1 Since, (− y ) satisfies Eq. (i).

As this has to pass through (6, 0), this gives t = 1, equation of ABC is x + y = 6. This meets the circle ( x − 6)2 + y 2 = 1 at B whose y-coordinate is given 1 by 2 y 2 = 1 or y = . 2

∴ Equation of the line passing through mid-point of AP and perpendicular to AP is 1 at 2  y − at = −  x −  2 2  at 3 …(ii) ⇒ tx + 2 y = + 2 at 2 Eqs. (i) and (ii) both pass through ( x1, y1 ), which is the centre of the circle.

…(iii) …(iv)

On multiplying Eq. (ii) by t and subtracting from Eq. (iv), we get …(v) t 2 y1 + t (4 a − 3 x1 ) − 4 y1 = 0 Also, from Eq. (iii), at 2 − ty1 + x1 = 0 On eliminating t from Eqs. (v) and (vi), we get t2 1 = x1(4a − 3 x1 ) − 4 y12 − 4ay1 − x1 y1 1 = 2 − y1 − a(4a − 3 x1 ) On simplifying, we get 2 y12 (2 y12 − 12 ax1 ) = ax1(3 x1 − 4a)2 ⇒ k=3

110. Q P = (am2 , − 2 am) Q = (am12 , − 2 am1 ) 2 m1 = − m − ∴ m Tangents at P and Q are − my = x + am2 and − m1 y = x + am12 and

…(i)

Then, area of ∆PQR 1 = PQ ⋅ RM 2 2 2 2 2 1 ( y1 + 4a ) ( y1 − 4 ax1 ) ( y1 − 4ax1 ) = ⋅ 2 a y12 + 4a2

14 Parabola

ty1 = x1 + at 2 2 tx1 + 4 y = at 3 + 4at

( y12 − 4 ax1 )3 / 2 2a But x1 = − 2 a − am2 2a y1 = m 3/ 2   4a2  2 + 8 a2 + 4 a2 m2   m ∴Area of ∆PQR = 2a 4 a2 (1 + m2 )3 = m3 ⇒ k=4 =

111. Any tangent to y 2 = 4 x is 1 m If is drawn from (−2, − 1,) then y = mx +

…(ii) …(iii)

Point of intersection of Eqs. (ii) and (iii) 2a [from Eq. (i)] y = − a(m + m1 ) = m 2 and x = − 2 a − am 2a  R ≡ −2 a − am2 , ∴  m  Let R be ( x1, y1 ), then PQ is the chord of contact of parabola w.r.t. R, is yy1 = 2 a( x + x1 ) Length of perpendicular from R to PQ is y 2 − 4 ax1 RM = 1 y12 + 4 a2

−1 = − 2 m + ⇒

1 m

2 m2 − m − 1 = 0

If m = m1, m2 , then m1 + m2 =

Targ e t E x e rc is e s

∴ and

1 2

1 2 (m1 + m2 )2 − 4 m1m2 m − m2 = ∴ tan α = 1 1 + m1m2 1 + m1m2 and

m1m2 = −

1 +2 = 4 =3 1 1− 2

Entrances Gallery ty = x + at 2 x ⇒ y = + at t x 2a a Normal at S, y + = + 3 t t t

S (2,4)

1.

3. Tangent at P,

P (–1,1)

On solving Eqs. (i) and (ii), we get 2 y = at +

Q (–1, –1) R (2, –4)

 1 Area of quadrilateral PQRS =  (1 + 4) × 3 × 2  2 = 15 sq units

∴

⇒

2a − 2 ar t a − ar 2 t2   1  +r  t t =−  t2 −2  1 − r2  t2

2a a + 3 t t

a (t 2 + 1)2 2t 3

a 2a 4. Let P (at 2 , 2 at ), Q  2 , −  as PQ is focal chord. t  Point of intersection of tangents at P and Q is 1    R − a, a t −    t 

−

⇒ r=

y=

…(ii)

t

2. Slope (PK ) = Slope (QR) 2 at − 0 = at 2 − 2 a

⇒

…(i)

t2 −1 t

Since, point of intersection lies on y = 2 x + a. 1  a t −  = − 2 a + a ∴  t 1 t − = −1 ⇒ t

847

Objective Mathematics Vol. 1

14

2

10. Given, A = (t12 , 2 t1 ) and B = (t 22 , 2 t 2 )

1  t +  = 5  t

⇒

t 2 + t 22  Centre =  1 , ( t1 + t 2 )  2  t1 + t 2 = ± r 2 (t − t ) 2 2 =± m = 21 22 = t1 + t 2 r t1 − t 2

2

1  ∴ Length of focal chord = a t +  = 5 a  t

5. Angle made by chord PQ at vertex (0, 0 ) is given by  2  + 2t 2 t tan θ =   =  1− 4   

1   + t  t −3

=

−2 5 3

Q So,

11. Equation of tangent at A (t 2 , 2 t ) yt = x + t 2 is tangent to x2 + 32 y = 0 at B. Y

6. The parabola is x = 2t 2 and y = 4 t .

y 2 = 4x

Solving it with the circle, we get 4 t 4 + 16 t 2 − 4 t 2 − 16 t = 0 ⇒ t 4 + 3t 2 − 4t = 0 ⇒ t = 0, 1 So, the points P and Q are (0, 0) and (2, 4), which are also diametrically opposite points on the circle. The focus is S ≡ (2, 0 ). 1 Area of ∆PQS = × 2 × 4 = 4sq units ∴ 2

7. Q y 2 = 8 x = 4 ⋅ 2 ⋅ x L (2,4)

Ta rg e t E x e rc is e s

A P (1/2, 2)

X (2,0)

C

M (2,– 4)

Y′

∆LPM =2 ∆ABC ∆1 =2 ∆2

∴ Thus,

X′

X

O

B

x2= –32y

Y′

x  x 2 + 32  + t  = 0 t  32 2 x + x + 32 t = 0 ⇒ t 2 ∴ b − 4ac = 0 2  32  ⇒   − 4 (32 t ) = 0  t   32  ⇒ 32  2 − 4t  = 0 t  ⇒ t3 = 8 ⇒ t =2 1 1 Hence, slope of tangent is i.e. . t 2 ⇒

Y

B (0,2) X′ O

A(t 2, 2t)

12. Equation of circle can be rewritten as x2 + y2 =

8. Q y 2 = 4 x

5 2

Centre = (0, 0) and radius = Y

5 2

Let common tangent be 1 X′

:3

2 P (x, y) = y , y 4 X

Y′

and P will lie on it. ∴ (4k )2 = 4 × 4h ⇒ k 2 = h ⇒ y 2 = x [replacing h by x and k by y]

9.

848

y 2 = 4x Equation of normal is y = mx − 2 m − m3 . Since, it passes through (9, 6). ∴ m3 − 7 m + 6 = 0 ⇒ m = 1, 2, − 3 ∴Normals are y − x + 3 = 0; y + 3 x − 33 = 0 and y − 2 x + 12 = 0

y = mx + ⇒

m2 x − my +

5=0

5 m

The perpendicular from centre to the tangent is equal to radius of the circle. 5 5 m ∴ = 2 1 + m2 ⇒

m 1 + m2 = 2

⇒

m2 (1 + m2 ) = 2

⇒

m4 + m2 − 2 = 0

⇒

(m + 2 ) (m2 − 1) = 0

⇒

2

m=±1 [Q m2 + 2 ≠ 0, as m ∈ R]

∴y = ± ( x + 5 ), both statements are correct as m = ± 1 satisfies the given equation of Statement II.

15. We know that, the locus of point P from which two

x 2 = 8y Let any point Q on the parabola (i) is (4t , 2t 2 ).

…(i)

Let P(h, k ) be the point which divides the line segment joining (0,0) and (4t , 2t 2 ) in the ratio 1 : 3.

perpendicular tangents are drawn to the parabola, is the directrix of the parabola. Hence, the required locus is x = − 1.

16. From the figure, it is clear that vertex of the parabola at (1, 0).

Y

Y

P 1 : (h ,

k)

X′ Q (4t,2t2)

3

(1, 0) (0, 0)

X

(0, 0) O

X (2, 0)

Y′ X′

14 Parabola

13. Given, equation of parabola is

x=2

17. Since, perpendicular tangents intersect on the directrix, so point must lie on the directrix x = − 2. Hence, the required point is (−2, 0 ).

18. The given equation of parabola is

1 × 4t + 3 × 0 4 ⇒ h =t 1× 2t 2 + 3 × 0 and k= 4 t2 ⇒ k= 2 1 k = h2 [Q t = h ] ⇒ 2 ⇒ 2 k = h 2 ⇒ 2 y = x 2 , which is required locus.

∴

14. To find the shortest distance between y − x = 1 and x = y 2 is along the common normal. ∴ Tangent at P is parallel to y=x+1 Y 1

y=1+x Q P (t 2, t)

X′

–1

X

O

y2 = x Y′

∴ Slope of tangent at P (t 2 , t ) is dy  1  1 =  = dx  2 y  (t 2, t ) 2 t ⇒ ⇒  1 1 ∴ P ,   4 2 1 1 − +1 3 4 2 = ∴ Shortest distance = | PQ| = 1+ 1 4 2 3 2 8

3 9 9  a3  2 x+ − x +  2 3  2a 16 a 16 a2 

⇒

y + 2a =

⇒

3 9 a3  a3 y + 2a = ×  − x + 2 3 3  4 a 16 a

2

⇒

3 a a3  3 = x +  16 3  4a

2

35 a a3  3  y +  = x +   16  3  4a

2

y + 2a +

 3 35 a Thus, the vertices of parabola is  − ,− .  4a 16  3 Let h=− 4a 35 a and k=− 16 105 Now, hk = 64 ∴ Locus of vertices of a parabola is 105 xy = 64

19. Since, the coordinates of P are (1, 0). ...(ii)

1 = 1 [as Eqs. (i) and (ii) are parallel] 2t 1 t = 2

=

⇒

⇒ ...(i)

a3 x 2 a2 x + − 2a 3 2 3 3  a  2 y + 2a = x x + 3  2a  y=

h=

Targ e t E x e rc is e s

Y′

Let any point Q on y 2 = 8 x be (2 t 2 , 4 t ). Again, let mid-point of PQ be (h, k ), then 2t2 + 1 h= 2 ...(i) ⇒ 2h = 2 t 2 + 1 4t + 0 and k= 2 k ...(ii) ⇒ t = 2 On putting the value of t from Eq. (ii) in Eq. (i), we get 2k 2 2h = +1 4 ⇒ 4h = k 2 + 2 Hence, locus of (h, k ) is y 2 − 4 x + 2 = 0.

849

the centre at (0, 3) and radius 2, so C1C2 = r1 + r2 2 ∴ h + (k − 3)2 = (| k| + 2 )2 2 ⇒ h + k 2 + 9 − 6k = k 2 + 4 + 4| k|

Objective Mathematics Vol. 1

14

20. Since, circle touches X-axis and also touches circle with

Y

(0,3) C1

2 (h,k) C2 k

X′

X

O

25. Given, equation of parabola is x 2 = − 9 y, which is of the

Y′

∴ Locus of centre of circle is x 2 = − 5 + 6 y + 4| y| ⇒ x 2 = 10 y − 5 which represents a parabola.

[Q y > 0 ]

Ta rg e t E x e rc is e s

21. Given, equation of parabolas are y 2 = 4 ax and x 2 = 4 ay The point of intersection of parabolas are A (0, 0 ) and B (4a, 4a). Also, given line 2 bx + 3 cy + 4 d = 0 passes through the points A and B, respectively. ...(i) ∴ d =0 and 2 b ⋅ 4 a + 3 c ⋅ 4 a + 4d = 0 ⇒ 2 ab + 3 ac + d = 0 ⇒ a (2 b + 3 c ) = 0 [Q d = 0 ] ...(ii) ⇒ 2b + 3c = 0 On squaring and adding Eqs. (i) and (ii), we get d 2 + (2 b + 3 c )2 = 0

22.

Equation of the normal at point(bt12 , 2 bt1 )on parabola is y = − t1 x + 2 bt1 + bt13 It also passes through (bt 22 , 2 bt 2 ), then 2 bt 2 = − t1 ⋅ bt 22 + 2 bt1 + bt13 ⇒ 2 t 2 − 2 t1 = − t1(t 22 − t12 ) ⇒

2( t 2 − t1 ) = − t1(t 2 + t1 ) (t 2 − t1 )

⇒

On comparing with a x 2 + 2 hxy + by 2 + 2 g x + 2 f y + c = 0, we get a = 49 − 16λ2 , b = 49 − 9λ2 , h = − 12 λ2 Condition for parabola is ab = h 2 . ∴ (49 − 16λ2 ) (49 − 9λ2 ) = (−12 λ2 )2 ⇒ 2401 − 441λ2 − 784λ2 + 144λ4 − 144λ4 = 0 ⇒ 2401 − 1225λ2 = 0 2401 ⇒ λ2 = 1225 49 ⇒ λ=± 35 7 ∴ λ=± 5

2 = − t1(t 2 + t1 ) ⇒ t 2 = − t1 −

2 t1

form x 2 = − 4 ay i.e. focus lies on the negative direction of Y-axis. Here, 4a = 9 ⇒ a = 9/ 4 ∴ Equation of directrix is y = a ⇒ y = 9/ 4 9 and length of latusrectum = 4a = 4 × = 9 4

26. Given, P = ( 3, 0 ) ∴ Equation of line AB is

A 60°

r 3 r x= 3+ , y= 2 2  r r 3 So, point  3 + ,  lies on y 2 = x + 2. 2 2  

⇒

850

(7 x + 5)2 + (7 y + 3)2 = λ2 (4 x + 3 y − 24)2 ⇒ 49 x 2 + 25 + 70 x + 49 y 2 + 9 + 42 y = λ2 (16 x 2 + 9 y 2 + 576 + 24 xy − 144 y − 192 x) ⇒ (49 − 16λ2 ) x 2 + (49 − 9λ2 ) y 2 + (70 + 192 λ2 ) x + (42 + 144λ2 ) y − 24λ2 xy + (25 − 576λ2 ) = 0

3r2 r = 3+ +2 4 2

∴

3r2 r − − (2 + 3 ) = 0 4 2 Let the roots be r1 and r2 , then the product

⇒

r1 × r2 = PA ⋅ PB =

2

24. Given, curve is

[say]

B

23. Given, equation of parabola can be rewritten as 1  ( y + 2) = − 4  x −   2 1 Let y + 2 = Y and x − = X 2 ∴ Y 2 = − 4X Here, a=1 ∴Equation of directrix is X = a 1 3 x − =1 ⇒ x = ⇒ 2 2

x− 3 y−0 = =r cos 60 ° sin 60 °

−(2 + 3 ) 4 (2 + 3 ) = 3 3 4

27. Let coordinates of P and Q be (at12 , 2 at1 ) and (at 22 , 2 at 2 ), respectively. Since, PQ subtends a right angle at the vertex (0, 0 ). Then, ...(i) t1 t 2 = − 4 If (h, k ) is the point of intersecting of normals at P and Q, then ...(ii) h = 2 a + a ( t12 + t 22 + t1 t 2 ) and

k = − at1 t 2 ( t1 + t 2 )

...(iii)

In order to find the locus of (h, k ), we have to eliminate t1 and t 2 between Eqs. (i) and (iii). …(iv) k = 4 a (t1 + t 2 ) and h − 2 a = a [(t1 + t 2 )2 − t1 t 2 ]

[from Eqs. (i) and (iv)]

k2 16 a Hence, the required locus is y 2 = 16 a ( x − 6 a). ⇒

h − 6a =

28. We know that, the sum of ordinates of foot of normals drawn from a point to the parabola y 2 = 4ax is always zero. Now, normals at three points P, Q and R of parabola y 2 = 4ax meet at (h, k ). So, the normals from (h, k ) to y 2 = 4ax meet the parabola at P, Q and R. Thus y-coordinates y1, y2 and y3 of these points P, Q and R will be zero. Now, y-coordinate of the centroid of ∆PQR y + y2 + y3 0 = 1 = =0 3 3 Hence, centroid lies on y = 0.

29. Given, equation of parabola is y 2 = 4ax. Let the coordinates of B be (at 2 , 2 at ), then the slope of 2 AB = . t Y

B (at 2, 2at )

∴ Slope of normal curve = Slope of line y − =4 ⇒ 6 ⇒ y = − 24 From Eq. (i), we get (−24)2 = 12 x ⇒ 24 × 24 = 12 x ⇒ x = 48 Thus, normal point on a curve is (48, − 24). ∴Distance from (48, − 24) to the line 4 x − y + 3 = 0 4 × 48 + 24 + 3 219 = = 17 42 + 12

32. Given, equation of parabola is y 2 = 64 x.

...(i) The point at which the tangent ot the curve is parallel to the line is the nearest point on the curve. On differentiating both sides of Eq. (i), we get dy 2y = 64 dx dy 32 ⇒ = dx y Also, slope of the given line is −4/ 3. 4 32 − = ∴ 3 y ⇒ y = − 24 From Eq. (i), (−24)2 = 64 x ⇒ x = 9 Hence, the required point is (9, − 24).

X′

(0, 0) A

33. The perpendicular of the normal to the parabola D

C

y 2 = 4ax at P is

X

y + t x = 2 at + at 3 Y Y′

Since, ∴

P (at 2, 2at)

BC ⊥ AB −t Slope of BC = 2

X′

t ( x − at 2 ). 2 Since, this line meets X-axis at point C. So, put y = 0, we get x = 4 a + at 2 So, the distance CD = 4 a + at 2 − at 2 = 4 a The equation of BC is y − 2 at = −

X

O

Q Y′

Suppose, it meets the parabola at Q. If O is the vertex of the parabola, then the combined equation of OP and OQ is a homogeneous equation of second degree.  y+tx  ∴ y 2 = 4 ax    2 at + at 3 

30. Since, the centre of given circle x 2 + y 2 − 8 x + 2 y = 0 is C1(4, − 1). Given, equation of parabola is y = x 2 − 4 x + 10 ⇒ y = x 2 − 4x + 4 + 6 ⇒ y − 6 = ( x − 2 )2 Vertex, V = (2, 6) ∴ 6+ 1 7 7 and slope of C1V = = =− 2 − 4 −2 2

⇒ ⇒

31. Given equation of parabola is y 2 = 12 x

14 Parabola

  k2 + 4 h − 2a = a  2  16 a

Targ e t E x e rc is e s

⇒

...(i)

On differentiating both sides w.r.t. x, we get dy dy 6 2y = 12 ⇒ = dx dx y Since, the normal to the curve is parallel to the line y = 4 x + 3.

y 2 (2 at + at 3 ) = 4 ax ( y + t x ) 4 at x 2 + 4 axy − (2 at + at 3 ) y 2 = 0

Since, OP and OQ are the right angles, so Coefficient of x 2 + Coefficient of y 2 = 0 ∴ 4at − 2 at − at 3 = 0 ⇒ t2 =2 ⇒ t = 2

34. Given equation of circle is x 2 + y 2 = (2 )2 ∴ Equation of tangent to the circle is y = mx ± 2 1 + m2

[Q y = mx ± r 1 + m2 ]

Also, equation of parabola is y = 32 x. 2

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14

So, focus of parabola is (8, 0). Since, the line passes through focus (8, 0). ∴

0 = 8m ± 2 1 + m

⇒

− 4m = ± 1 + m2

⇒

16 m2 = 1 + m2

⇒

15 m2 = 1 ⇒ m2 =

1 15

1 15

m=±

∴

2

35. Given, equation of parabola is y 2 = 4 x. Let two points on the parabola be A (at12 , 2 at1 ) and B (at 22 , 2 at 2 ). Here, a=1 ∴ A (t12 , 2t1 ) and B (t 22 , 2 t 2 ) Since, A and B are the end points of a diameter.  t 2 + t 22  Mid-point of AB =  1 ∴ , t1 + t 2   2  Also, radius of circle is r. ∴ Equation of circle is 2  t 2 + t 22   + [ y − (t1 + t 2 )]2 = (2 )2 x − 1 2  

Ta rg e t E x e rc is e s

2

⇒

⇒

Here, g =

 t 2 + t 22   t 2 + t 22  x2 +  1   − 2x  1  2   2  + y 2 + ( t1 + t 2 )2 − 2 y ( t1 + t 2 ) = 4 ( t 2 + t 22 ) − 2 y ( t1 + t 2 ) x2 + y2 − 2 x 1 2 2  t 2 + t 22  +1  + ( t1 + t 2 )2 − 4 = 0  2  t12 + t 22 2

2

 t 2 + t 22  and c =  1  + ( t1 + t 2 )2 − 4  2 

Since, the circle touch X-axis. ∴ g2 = c 2

⇒

2

 t 2 + t 22   t12 + t 22   + ( t1 + t 2 )2 − 4  =1   2   2 

⇒ ⇒

( t1 + t 2 )2 = 4 t1 + t 2 = 2 2 (t − t ) Thus, slope of AB = 2 2 21 ( t 2 − t1 ) 2 ( t 2 − t1 ) = ( t 2 − t1 ) ( t 2 + t1 ) 2 2 = = =1 ( t 2 + t1 ) 2

36. Let P (h, k )be any point on y 2 = 4 x and its image about

852

the line x − y + 1 = 0 be Q (α , β ). Since, mid-point of P and Q lies on the line. h+α k+β ⇒ − + 1= 0 2 2 ⇒ h + α − k −β + 2 = 0 ...(i) ⇒ h − k = −α + β −2 β−k Also, × 1= − 1 [Q m1m2 = − 1] α−h ⇒ β − k = −α + h ...(ii) ⇒ h + k =α + β

From Eqs. (i) and (ii), h = β − 1, k = α + 1 Since, k 2 = 4h ⇒ (α + 1)2 = 4 (β − 1) So, required curve is ( x + 1)2 = 4 ( y − 1).

37. Given curve is y 2 = 4 x

...(i) ...(ii)

x 2 = 4y dy dy 2 From Eq. (i), 2 y =4 ⇒ = dx dx y 2 1 dy  = = = m1 ⇒ dx  4 2 ( 1, 4 ) and

Now, from Eq. (ii), dy dx dy  dx 

2x = 4 ⇒

( 1, 4 )

⇒ =

dy x = dx 2

1 = m2 2

Q m1 = m2 So, tangents are parallel.

38. We know that, the equation of normal to parabola y 2 = − 4 ax is y = mx + 2 am + am3 8 Here, 4a = 8 ⇒ a = = 2 4 and equation of normal is 2x + y + k = 0 ⇒ y = − 2x − k On comparing Eqs. (i) and (ii), we get m = − 2 and − k = 2 am + am3 ⇒ − k = 2(2 ) (−2 ) + 2(−2 )3 ⇒ − k = − 8 − 16 ∴ k = 24

...(i)

...(ii)

39. y 2 + 4 x + 4 y + k = 0 ⇒ y 2 + 2 × 2 y + 4 − 4 + 4x + k = 0 ⇒ ( y + 2 )2 = − 4 x − k + 4 −4 +  ⇒ ( y + 2 )2 = − 4  x +  4 ∴Latusrectum = 4 units

k  

40. We have, equation of parabola is y 2 = 4 x. Let the coordinates of P and Q be ( t 2 , 2 t ) and ( t12 , 2 t1 ) respectively. R will be the mid-point of PQ. Let the coordinate of R be ( x, 0 ). t 2 + t12 and t + t1 = 0 Then, x = 2 ( t + t1 )2 − 2 t1 t ⇒ x= 2 ...(i) ⇒ x = − t1 t Now, in ∆POQ, we have PQ 2 = OP 2 + OQ 2 ⇒ ( t12 − t 2 )2 + (2 t1 − 2 t )2 = ( t12 )2 + 2 ( t1 )2 + ( t 2 )2 + ( 2 t )2 4 4 2 2 2 2 ⇒ t1 + t − 2 t1 t + 4 t1 + 4 t − 8 t1 t = t14 + 4 t12 + t 4 + 4 t 2 ⇒ −2 t12 t 2 = 8 t1 t ⇒ t1 t = − 4 From Eq. (i), x = 4

∴

y = ± 8y

[Q x = ± y ]

= 4 + 16 = 2 5 units

(8 ,

8)

Y

∴ Required distance = (2 − 0 )2 + (−1 − 3)2

P

y = –x

y2 =

8x

45. The parabola y 2 = 4 x and the circle x 2 + y 2 − 6 x + 1 = 0 O

X′ (0,0)

X Q (8,–8)

y=x Y′

⇒ y m 8 y = 0 ⇒ y ( y ± 8) = 0 ⇒ y = 0, ± 8 ∴ P (8, 8) and Q(8, − 8) Length of PQ = (16)2 + (0 )2 = 16 ∴ 2

42. Required sum =

1 1 2 1 + = = 2a 2a 2a a

43. Given, equation of parabola is y 2 + 8 x − 12 y + 20 = 0. ⇒ ( y − 6)2 + 8 x − 16 = 0 ⇒ ( y − 6)2 = 16 − 8 x ⇒ ( y − 6)2 = − 4 (2 x − 4) Let Y = y − 6 and X = 2 x − 4 Also, Y = 0 and X = 0 ∴ y − 6= 0 ⇒ y = 6 and 2x − 4 = 0 ⇒ x = 2 ∴ Vertex is (2, 6). and focus is (0, 6). Now, latusrectum = 4 × 1 = 4 and X -axis = 4

44. We have, x 2 − 4 x + 3 = y ⇒ ( x − 2 ) = ( y + 1) Let X = x − 2 and Y = y + 1 ⇒ x − 2 = 0 and y + 1 = 0 ⇒ x = 2 and y = − 1 So, vertex of the parabola is (2,− 1). Also, ( y − 3)2 + ( x − 0 )2 = 32 2

Parabola

14

Thus, centre of the circle is (0, 3). 2

will intersect, if x 2 + 4 x − 6 x + 1 = 0. ⇒

( x 2 − 2 x + 1) = 0

⇒

( x − 1)2 = 0

⇒ x = 1, 1 ∴ y= 4=±2 So, the intersecting points are (1, − 2 ) and (1, 2 ).

46. We have, y 2 = 12 x ∴Equation of tangent in slope form is a y = mx + m 3 y = 3x + 3 y = 3 ( x + 1) ⇒ ⇒ ± y − 3x − 3 = 0

47. Since, equation of the chord of the parabola y 2 = 4 ax joining the points ( at12 , 2 at1 ) and ( at 22 , 2 at 2 ) is y (t1 + t 2 ) = 2 x + 2 at1 t 2

...(i)

Also, line (i) joining t1, t 2 on the parabola and passes through (a, 0 ) i.e. 2 a + 2 at1 t 2 = 0 ⇒ t1 t 2 = − 1

48. Let the point be (3 t 2 , 6 t ). Focal distance = 3 t 2 + 3 ∴ ⇒ 3 t 2 + 3 = 12 ⇒ 3t 2 = 9 ⇒ t2 = 3 ⇒ t = 3 Hence, the required point is (9, 6 3 ).

Targ e t E x e rc is e s

41. We have, y 2 = 8 x

853

15 Ellipse Introduction An ellipse is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from a fixed straight line (called directrix) is always constant which is always less than unity.

Chapter Snapshot ●

●

Z P

●

M Directrix

●

S (Focus)

●

Z′

●

The constant ratio is generally denoted by e and is known as the eccentricity of the ellipse. If S is the focus, ZZ ′ is the directrix and P is any point on the ellipse, then by definition SP = e ⇒ SP = e ⋅ PM PM or P1

P2

Focus F1

P3

Focus F2

P1 F1 + P1 F2 = P2 F1 + P2 F2 = P3 F1 + P3 F2 An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. The two fixed points are called the foci (plural of focus) of the ellipse. The constant which is the sum of the distances of a point on the ellipse from the two fixed points is always greater than the distance between the two fixed points. The mid-point of the line segment joining the foci is called the centre of the ellipse. The line segment through the foci of the ellipse is called the major axis and the line segment through the centre and perpendicular to the major axis is called the minor axis. The end points of the major axis are called the vertices of the ellipse.

Introduction Position of a Point with Respect to an Ellipse Equation of the Chord Intersection of a Line and an Ellipse Tangent Combined Equation of the Pair of Tangents

●

Director Circle

●

Normal

●

Number of Normals and Conormal Points

●

Pole and Polar

●

Conjugate Lines

●

Diameter

Sol. (a) We have,

Let P ( x, y) be any point on the ellipse. Join SP and draw PM ⊥ ZK . Then, by definition, we have SP = ePM ⇒ SP 2 = e 2 PM 2 SP 2 = e 2 ( NK ) 2

⇒

 a ⇒ ( x − ae) 2 + ( y − 0) 2 = e 2  − x   e ⇒ ⇒

a

2

+

y2 a (1 − e ) 2

2

x2

⇒

+

y2

X

2

=1

Y

(0, b)B 8 P (h,k)

X

Example 4. The equation to the ellipse, whose focus is the point ( −1, 1), whose directrix is the straight line x − y + 3 = 0 and whose eccentricity is 1 is 2 (a) 7x 2 + 2xy + 7 y 2 + 10x − 10 y + 7 = 0 (b) x 2 + 2xy + 10x − 10 y + 3 = 0 (c) 3x 2 + xy + 10x − 10 y + 3 = 0 (d) None of the above and PN be the perpendicular from P on directrix, then by definition of an ellipse PS 2 = e 2 PN2 , hence

X

O

A (a, 0)

( x + 1)2 + ( y − 1)2 =

Y′

Given,

OA 2 + OB2 = 12 2

a2 + b 2 = 144 PB 8 Also, = =2 PA 4 2a b and k = ⇒ h= 3 3 3h and b = 3k a= 2 9h2 From Eq. (i), we get + 9k 2 = 144 4 x2 y2 ⇒ + =1 64 16 which is equation of an ellipse. i.e.

2

X

2

Sol. (a) Let P( x, y) be any point on the ellipse, S be its focus

4 X′

2

 x  +  y  = 2(sin2 t + cos 2 t ) = 2      5 2 2 x y2 + =1 ⇒ 8 50 which is equation of an ellipse.

∴

Example 1. A ladder 12 units long slides in a vertical plane with its ends in contact with a vertical wall and a horizontal floor along X -axis. Then, the locus of a point on ladder 4 units from its foot is (a) circle (b) parabola (c) straight line (d) ellipse Sol. (d)

x = cos t + sin t 2 y = cos t − sin t 5

and

= 1, where b 2 = a 2 (1 − e 2 )

15

Example 3. The curve represented by x = 2(cos t + sin t ), y = 5 (cos t − sin t ) is (a) a circle (b) a parabola (c) an ellipse (d) a hyperbola Sol. (c) Here,

a 2 b2 which is the standard equation of the ellipse. X

…(i)

Eq. (i) represents an ellipse 10 − a > 0 and 4 − a > 0 ∴ 10 > a and a < 4 ⇒ a < 10 and a < 4, then a < 4

x 2 (1 − e 2 ) + y 2 = a 2 (1 − e 2 ) x2

x2 y2 + =1 10 − a 4 − a

Ellipse

Standard Form of the Ellipse

1  x − y + 3    4 2

2

( x − y + 3)2 8 [as focus is (− 1, 1) and directrix is ( x − y + 3 = 0)] 8( x2 + y2 + 2 x − 2 y + 2 ) = x2 + y2 + 9 − 2 xy =

…(i) ⇒

+ 6x − 6y ⇒

7 x2 + 2 xy + 7 y2 + 10 x − 10 y + 7 = 0

Some Basic Terms Related to an Ellipse Z′

Y

Q 2

y x Example 2. The equation + =1 10 − a 4 − a represents an ellipse, if (a) a < 4 (b) a > 4 (c) 4 < a < 10 (d) a >10

B (0, b) L

K′ A ′ (–a, 0)

C

Q′ x = –a/e

P (x, y)

(ae, 0) S

S ′(–ae, 0)

X′

Z

X N

B′ L′A (a, 0) (0,–b) Y′

M K

x = a/e

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Objective Mathematics Vol. 1

15

i.

ii.

Vertices The points A and A′, where the curve meets the line joining the foci S and S ′, are called the vertices of the ellipse. The coordinates of A and A′ are ( a, 0) and ( − a, 0), respectively. Major and minor axes The distance AA ′ = 2a and BB ′ = 2b are called the major and minor axes of the ellipse, respectively. Since, e b ⇒

AA ′ > BB ′

iii.

Foci The points S ( ae, 0) and S ′ ( − ae, 0) are the foci of the ellipse.

iv.

Directrices

v.

vi.

Centre Since, the centre of a conic section is a point which bisects every chord passing x 2 y2 through it. In case of the ellipse 2 + 2 = 1 a b every chord is bisected at C (0, 0). Therefore, C is the centre of the ellipse and C is the mid-point of AA′. x2 y2 Eccentricity of the ellipse 2 + 2 = 1, a > b a b x 2 y2 For the ellipse 2 + 2 = 1, we have a b 2 2 b = a (1 − e 2 )

⇒ ⇒ ⇒

e2 = 1 − e2 = 1 −

ix.

x. X

a2 4b 2

Diameter A chord of the ellipse passing through the centre is called the diameter of the ellipse all such chords are bisected by the centre. Focal chord A chord of the ellipse passing through its focus is called a focal chord.

Example 5. The length of the latusrectum of an ellipse is one-third of the major axis. Its eccentricity would be 2 2 1 1 (d) (a) (b) (c) 3 3 3 2 2 Sol. (b) Given, 2 ⋅ b = 1 (2 a)

4a 2 2

 Minor axis  e = 1−    Major axis 

⇒

a 3 6 b 2 = 2 a2

⇒

3 a (1 − e ) = a

⇒ 2

This formula gives the eccentricity of the ellipse.

856

b2 a Hence, length of the latusrectum 2b 2 = 2a (1 − e 2 ) = 2( SL) = a SL = SL′ =

Thus,

b2

 2b  e2 = 1 −    2a 

Latusrectum A double ordinate passing through the focus is called the latusrectum and LS is called the semi-latusrectum. Q, S ′ Q ′ is also a latusrectum. The coordinates of L are x 2 y2 ( ae, SL). As L lies on the ellipse 2 + 2 = 1 a b the coordinates of L will satisfy the equation ofthe ellipse, therefore ( ae) 2 ( SL) 2 + = 1 ⇒ ( SL) 2 = b 2 (1 − e 2 ) a2 b2  b2  ⇒ ( SL) 2 = b 2  2  a   2 b2  2 2 2 Q b = a (1 − e ) ⇒ 1 − e = 2  a  

ZK and Z ′ K ′ are two directrices a of the ellipse and their equations are x = and e a x = − , respectively. e

⇒

vii.

viii.

Ordinate and double ordinate Let P be a point on the ellipse and let PN be perpendicular to the major axis AA′ such that, PN produced meets the ellipse at P ′. Then, PN is called the ordinate of P and PNP ′ the double ordinate of P.

⇒ X

2

⇒ 3 b 2 = a2

⇒ 3 − 3e 2 = 1 2 3e 2 = 2 ⇒ e 2 = 3 2 e= 3 2

2

Example 6. An ellipse has its centre at (1, − 1) and semi-major axis is equal to 8 and which passes through the point (1, 3). Then, the equation of the ellipse is ( x − 1) 2 ( y + 1) 2 ( x − 1) 2 ( y + 1) 2 (a) + = 1 (b) + =1 64 16 64 32 ( x − 1) 2 ( y + 1) 2 ( x + 1) 2 ( y − 1) 2 (c) + = 1 (d) + =1 16 64 64 16

( y + 1) + = 1, where a = semi-major a2 b2 axis and b = semi-minor axis. If a = 8, then equation of an ellipse is ( x − 1)2 ( y + 1)2 + =1 64 b2 This passes through point (1, 3). 16 0+ 2 =1 ∴ b ⇒ b 2 = 16 written as

Now,

2

Hence, equation of ellipse is ( x − 1)2 ( y + 1)2 + =1 64 16

X

Example 8. With a given point and line as focus and directrix, a series of ellipses are described. The locus of the extremities of their minor axes is (a) an ellipse (b) a parabola (c) a hyperbola (d) None of these Sol. (b) Let S be the given focus and ZM be the given line. SZ = CZ − CS =

Then,

a − ae e M

Y

B

Focal Distances of a Point on the Ellipse Y

Z P

M

N

X′ K′

S′

O S(ae, 0)

x = –a/e

X

Y′

K

x = a/e

Let P ( x, y) be any point on the ellipse

x2 a2

+

y2 b2

= 1.

Then by definition, we have SP = e ⋅ PM and S ′ P = e ⋅ PM ′ ⇒ SP = e( NK ) and S ′ P = e( NK ′ ) ⇒ SP = e(CK − CN ) and S ′ P = e(CK ′ + CN )   a a ⇒ SP = e  − x  and S ′ P = e  + x    e e ⇒ SP = a − ex and S ′ P = a + ex Thus, the focal distances of a point P ( x, y) on the x 2 y2 ellipse 2 + 2 = 1 are a − ex and a + ex. a b Also, SP + S ′ P = a − ex + a + ex = 2a = Major axis = Constant Hence, the sum of the focal distances of a point on the ellipse is constant and is equal to the length of the

Sol. (a) Clearly, the race course is an ellipse with flag-posts as its foci. If a and b are the semi-major and semi-minor axes of the ellipse, then 2 a = 10 and 2 ae = 8. 4 ⇒ a = 5, e = 5

[say]

Let ( x, y) be the coordinates of B with respect to these axes. y2 b 2 Then, x = SC = ae and y = CB = b, hence = =k x ae Therefore, y2 = kx, is the required locus and is clearly a parabola.

Other Forms of the Ellipse x2

+

y2

= 1, if a > b or a 2 b2 a 2 > b 2 (denominator of x 2 is greater than that of y 2 ), then the major and minor axes lie along X and Y-axes respectively. But, if a < b or a 2 < b 2 (denominator of x 2 is less than that of y 2 ), then the major axis of the ellipse lies along the Y-axis and is of length 2b and the minor axis along the X-axis and is of length 2a. The equation of the ellipse

Y y = b/e K B (0,b)

major axis of the ellipse.

Example 7. A man running round a race course notes that the sum of the distances of two flag-posts from him is always 10 m and the distance between the flag-posts is 8 m . The area of the path he encloses in square metres is (a)15π (b)12π (c)18π (d) 8π

x = a/e

b2 a = (1 − e 2 ) = =k ae e 2 2 2 b = a (1 − e )

As

Y′

X

X

S

C M′

Ellipse

( x − 1)

2

15

b 2 = a2 (1 − e 2 ) 16 b 2 = 25  1 −  = 9 ⇒ b = 3 ⇒  25  So, area of the ellipse = πab = 15 π

Sol. (a) Equation of the ellipse with centre at (1, − 1) can be

Z

S (0, be) X′

A (a, 0)

A′ (– a, 0)

X

S′ (0, –be) B′ (0,– b) K′ y = –b/e Y′

Z′

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Objective Mathematics Vol. 1

15

The coordinates of foci S and S ′ are (0, be) and (0, − be), respectively. The equations of the directrices b ZK and Z ′ K ′ are y = ± and eccentricity e is given by e the formula e = 1−

a 2 = b 2 (1 − e 2 ) or Various result related to x2 2

+

y2 2

x2 a

2

+

y2 b

2

b2

= 1, ( a > b) and

a

+

y2 b

2

= 1, a > b

x2 a

+

2

y2 b2

= 1, a < b

Coordinates of the centre

(0, 0)

(0, 0)

Coordinates of the vertices

( a, 0) and ( − a, 0)

(0, b) and ( 0, − b )

Coordinates of foci

( ae, 0) and ( − ae, 0) ( 0, be ) and ( 0, − be )

2( x2 − 2 x) + 3( y2 − 4 y) + 13 = 0 2( x − 1)2 + 3( y − 2 )2 = 1 ( x − 1)2

⇒

 1     2

+

2

a=

1 2

a2 − b 2 =

Length of minor axis

2b

2a

Equation of the major axis

y=0

x=0

1  Then, foci is  1 ± , 2 .  6 

Equation of the minor axis

x=0

y=0

Eccentricity

Length of latusrectum Focal distances of a point ( x, y)

e = 1−

b

2

a

2

b b and y = − e e

e = 1−

2 b2 a

2 a2 b

a ± ex

b ± ey

a

2

b

2

ae =

Equation of such an ellipse is ( x − h) 2 ( y − k ) 2 + = 1, a > b a2 b2 Foci : ( h ± ae, k ) Vertex : ( h ± a, k ) a Directrix : x=h± e If a < b, then Foci : ( h, k ± be) Vertex : ( h, k ± b) b Directrix : y=k ± e

1 3

b=

 1   1    −   3  2

2

1 1 = 6 6

Equation of an Ellipse Referred to Two Perpendicular Lines Consider the ellipse

x2 a2

+

y2 b2

= 1 as shown in figure.

Y

B (0, b)

X′

Equation of an Ellipse whose Axes are Parallel to Coordinate Axes and Centre is ( h, k )

=1

2

ae =

or ⇒

y=

2

a> b

Now, ∴ Foci is (h ± ae, k )

2b

a a and x = − e e

 1     3

and

2a

x=

( y − 2 )2

Here, h = 1, k = 2

Length of major axis

Equation of the directrices

858

(d) None of these

Sol. (c) The given equation can be rewritten as ⇒

a b ready reference:

2

Example 9. The foci of the ellipse 2x 2 + 3 y 2 − 4x − 12 y + 13 = 0 is   2  1  (b)  2, 1 ± (a)  2,     3 6  1  (c) 1 ± , 2  6 

a2

= 1, ( a < b) are given in the following table for

x2

X

A ′(–a, 0)

P (x, y)

N

A (a, 0) O

M

X

B′(0,–b)

Y′

Let P ( x, y) be any point on the ellipse. Then, PM = y and PN = x. x 2 y2 + =1 ∴ a 2 b2 PN 2 PM 2 ⇒ + 2 =1 a2 b It follows from this, that if perpendicular distances p1 and p2 of a moving point P ( x, y) from two mutually perpendicular coplanar straight lines L1 ≡ a1 x + b1 y + c1 = 0, L2 ≡ b1 x − a1 y + c2 = 0 respectively satisfy the equation p12 p22 + =1 a 2 b2

2

+

a2

b x −a y+c  1 2  1  2 2 b1 + a1  

The equation of the ellipse is

2

b2

p12

=1 ⇒

Then, the point P describes an ellipse in the plane of the given lines such that

i.

⇒

The centre of the ellipse is the point of intersection of the lines L1 = 0 and L2 = 0

⇒

L1º ≡ a1x + b1y + c1= 0 B P (x, y)

b A′

p1

a

p2

a

L2º ≡ b1x – a1y + c2= 0 X A

b B′

ii.

The major axis lies along L2 = 0 and the minor axis lies along L1 = 0, if a > b. If a < b, then the major axis lies along L1 = 0 and the minor axis lies along L2 = 0.

iii.

The length of the major and minor axes are 2a and 2b, if a > b.

(c) 21x 2 − 6xy + 29 y 2 + 6x − 58 y − 151 = 0 (d) None of the above Sol. (c) Let P( x, y) be any point on the ellipse and p1, p2 be the lengths of perpendiculars drawn from P on the major and minor axes of the ellipse. Then, p1 =

x − 3y + 3 1+ 9

p2 =

3x + y − 1 9+1

Let 2a and 2b be the lengths of major and minor axes of the ellipse. 2 a = 6 and 2 b = 2 6 a = 3 and

b=

6

x 2 y2 + =1 9 4 and C be the circle x 2 + y 2 = 9. If P and Q are the points (1, 2) and (2, 1), respectively. Then, (a) Q lies inside C but outside E (b) Q lies outside both C and E (c) P lies inside both C and E (d) P lies inside C but outside E

Example 11. Let E be the ellipse

x2 y2 + = 1. 9 4 For x = 1, y = 2, the value of the expression x2 y2 + − 1 is positive 9 4 and for x = 2, y = 1, the value of the above expression is negative. Therefore, P lies outside E and Q lies inside E. The value of the expression x2 + y2 − 9 is negative for both the points P and Q. Therefore, P and Q both lie inside C. ∴P lies inside C but outside E.

(b)15x 2 − 3xy + 9 y 2 + 6x − 48 y + 151 = 0

⇒

( x − 3 y + 3)2 (3 x + y − 1)2 + =1 60 90 3( x − 3 y + 3)2 + 2(3 x + y − 1)2 = 180

Sol. (d) The given ellipse is

Example 10. The equation to the ellipse whose axes are of lengths 6 and 2 6 and their equations are x − 3 y + 3 = 0 and 3x + y − 1 = 0 respectively, is (a)19x 2 − 9xy + 6 y 2 + 3x − 68 y − 156 = 0

We have,

15

The point P ( x1 , y1 ) lies outside, on or inside the x 2 y2 x 2 y2 ellipse 2 + 2 = 1 according 12 + 12 − 1 is > 0, =, < 0. a b a b

Y′

and

=1 b2 a2 2 2  x − 3y + 3   3 x + y − 1     1+ 9  9+1     +   =1 6 3        

Position of a Point with Respect to an Ellipse

X

X

p22

⇒ 21x2 − 6 xy + 29 y2 + 6 x − 58 y − 151 = 0

Y

X′

+

Ellipse

⇒

a x +b y+c  1 1  1  2 2 a1 + b1  

X

Example 12. The number of real tangents that can be drawn to the ellipse 3x 2 + 5 y 2 = 32 passing through (3, 5) is (a) 0 (b) 1 (c) 2 (d) infinite Sol. (c) Since, 3(3)2 + 5(5)2 − 32 = 120 > 0. So, the given point lies outside the ellipse. Hence, two real tangents can be drawn from this point to the ellipse.

859

Work Book Exercise 15.1

Objective Mathematics Vol. 1

15

1 If the eccentricity of the ellipse

x2 y2 + 2 =1 a +1 a +2

3 The length of major axis of the ellipse

2

(5 x − 10 )2 + (5 y + 15)2 =

1 , then the length of latusrectum is is 6

a

5 a 6 10 b 6 8 c 6 d None of the above

b

20 3

c

20 7

d 4

x2 y2 + =1 25 16 and P is any point on it, then range of value of SP ⋅ S ′ P is

4 If S and S′ are the foci of the ellipse

a 9 ≤ f(θ) ≤ 16 c 16 ≤ f(θ) ≤ 25

2 If S1, S 2 are foci of an ellipse of major axis of

b 9 ≤ f(θ) ≤ 25 d 1 ≤ f(θ) ≤ 16

5 The equation of the ellipse, whose focus is (1, − 1),

length 10 units and P is any point on the ellipse such that perimeter of ∆PS1S 2 is 15. Then, eccentricity of the ellipse is a b c d

10

(3 x − 4 y + 7 )2 is 4

the directrix is the line x − y = 3 and eccentricity is 1/ 2, is a b c d

0.5 0.25 0.28 0.75

7 x 2 + 2 xy + 7 y 2 − 10 x + 10 y + 7 = 0 7 x 2 + 2 xy + 7 y 2 + 7 = 0 7 x 2 + 2 xy + 7 y 2 + 10 x − 10 y − 7 = 0 None of the above

Parametric Coordinates and Equations i.

ii.

Auxiliary circle The circle described on the major axis of an ellipse as diameter is called an auxiliary circle of the ellipse. x 2 y2 If 2 + 2 = 1 is an ellipse, then its auxiliary a b circle is x 2 + y 2 = a 2 .

iii.

Parametric coordinates of a point on an ellipse Let P ( x, y) be a point on the ellipse x 2 y2 + = 1 and Q be the corresponding point a 2 b2 on the auxiliary circle x 2 + y 2 = a 2 . Let the eccentric angle of P be φ. Then, ∠ACQ = φ. Now, x = CM ⇒ x = CQ cos φ = a cos φ [QCQ = radius of x 2 + y 2 = a 2 ]

Eccentric angle of a point Let P be any x 2 y2 point on the ellipse 2 + 2 = 1. Draw PM a b perpendicular from P on the major axis of the ellipse and produce MP to the auxiliary circle in Q. Join CQ. The ∠ACQ = φ is called the eccentric angle of the point P on the ellipse. Y

⇒

Q

X′

φ C

A′

B′ x2 +

y2 =

a2

M

A

x2 y2 = 1 + a2 b2

Y′

Ø The ∠ACP is not the eccentric angle of point P.

860

+

y2

⇒ y = b sin φ Thus, the coordinates of point P having eccentric angle φ can be written as ( a cos φ, b sin φ ) and are known as the parametric coordinates.

P (x, y)

B

x2

= 1. Therefore, a 2 b2 a 2 cos 2 φ y 2 + 2 =1 a2 b 2 2 2 y = b (1 − cos φ ) = b 2 sin 2 φ

Since, P ( x, y) lies on

X

iv.

Parametric equations The equations x = a cos φ, y = b sin φ taken together are called the parametric equations of the ellipse x 2 y2 + = 1, where φ is the parameter. a 2 b2

Equation of the tangent at point P(acos φ, b sin φ) is  x  cos φ +  y  sin φ = 1 …(ii)      b  a

Example 13. The sum of the squares of the reciprocals of two perpendicular diameter of an ellipse is 1 1 1  1 1 1  (b)  2 + 2  (a)  2 + 2  4 a 2 a b  b  1 1 (d) None of these (c) 2 + 2 a b

which meets the auxiliary circle at points A and B. Y A

2 2 Sol. (a) Let the equation of the ellipse be x2 + y 2 = 1.

a On putting x = r cosθ and y = r sin θ We have, the polar equation as r 2 cos 2 θ r 2 sin2 θ + =1 a2 b2

b

Auxiliary P (φ) circle B 90° O

X

∴Equation of pair of lines OA and OB is obtained by making homogeneous, Eq. (i) with the help of Eq. (ii) as 2 x y x2 + y2 = a2  cos φ + sin φ  a b  a2 2 xyasin φcos φ  +  1 − 2 sin2 φ y2 = 0 ⇒ (1 − cos 2 φ)x2 − b b  

Y

(r cos θ, r sin θ) r X′

15 Ellipse

X

But ∠AOB = 90° ∴Coefficient of x2 + Coefficient of y2 = 0

X

 a2  ⇒ 1 − cos 2 φ + 1 −  2  sin2 φ = 0 b  ⇒ Y′

⇒

cos 2 θ

+

sin2 θ

=

⇒

1

a2 b2 r2 where, r is the radius vector and if r and r1 be the perpendicular radii vectors respectively, then the diameters will be 2r and 2 r1 and the corresponding angles will be θ and (90° + θ). 1 1 So, 2 + 2 4r 4r1

⇒ ⇒ ⇒

X

1   cos 2 θ sin2 θ   cos 2 (90° + θ) sin2 (90° + θ)     +  =   + + 4   a2 b2 b 2   a2   2 2 2 2 1  cos θ sin θ sin θ cos θ  =  2 + + +  4 a b2 a2 b2  1 1 1 =  2 + 2 4  a b  which is a constant quantity. X

Example 14. The tangent at a point x 2 y2 P ( a cos φ, b sin φ ) of an ellipse 2 + 2 = 1, meets a b its auxiliary circle in two points, the chord joining which subtends a right angle at the centre, then the eccentricity of the ellipse is (a) (1 + sin 2 φ ) −1 (b) (1 + sin 2 φ) −1/ 2 (c) (1 + sin 2 φ ) −3/ 2 (d) (1 + sin 2 φ ) −2 Sol. (b) Equation of the auxiliary circle is x 2 + y2 = a 2

...(i)

 a2  sin2 φ  1 − 2  + 1 = 0 b   (a2 − b 2 ) sin2 φ = b 2 a2e 2 sin2 φ = a2 (1 − e 2 ) (1 + sin2 φ) e 2 = 1 1 e= = (1 + sin2 φ)−1/ 2 2 1 + sin φ

Example 15. The area of the triangle inscribed in an ellipse is α − β γ −α β − γ  2ab sin   , where α, β,  sin   sin   2   2   2  γ are the eccentric angles of the vertices, then the condition that the area of a triangle inscribed in an ellipse may be maximum, is 2π 4π (a) α, α − ,α + 3 3 2π 4π (b) α, α + ,α + 3 3 2π π (c) 2α, α + , α + 3 3 π 2π (d) 2α, α − , α + 3 3 Sol. (b) Let P, Q and R be the angular points of the triangle and let α, β and γ be their eccentric angles, respectively. The vertices of the ∆PQR are (acos α, b sinα ), (acos β, b sin β ) and (acos γ, b sin γ ). Therefore, the area of ∆PQR β − γ  γ − α  sin  α − β  = 2 ab sin   sin      2   2   2 

861

15 Objective Mathematics Vol. 1

Let p, q and r be the corresponding points on the auxiliary circle. Then, area of β − γ  γ − α  sin  α − β  ∆pqr = 2 a2 sin   sin      2   2   2  On putting b = a, ar (∆PQR ) b = ar (∆pqr ) a b ⇒ ar (∆PQR ) = ar (∆pqr ) a Hence, ∆PQR will be maximum when ∆pqr is maximum. But ∆pqr is maximum when it is an equilateral triangle and in that case 2π α −β =β − γ = γ −α = 3 Hence, when a triangle inscribed in an ellipse is maximum the eccentric angles of its angular points are 2π 4π α, α + ,α + . 3 3 X

Also, − sin(α + β + γ ) =

⇒

(a2 + 3b 2 )2 2

x2 +

(b 2 + 3a2 )2

a which is the required locus.

b2

y2 = (a2 − b 2 )2

Equation of the Chord Let P ( a cos θ, b sin θ ), Q ( a cos φ, b sin φ ) be any two x 2 y2 points of the ellipse 2 + 2 = 1. Then, the equation of a b the chord joining these two points is b sin φ − b sin θ y − b sin θ = ( x − a cos θ ) a cos φ − a cos θ φ +θ cos    2  −b ⋅ ( x − a cos θ ) ⇒ y − b sin θ = a θ +φ sin    2  ⇒

1 θ +φ ( y − b sin θ )sin    2  b =−

⇒

1 θ +φ cos   ( x − a cos θ )  2  a

x θ +φ θ +φ y cos    + sin   2    2 a b θ +φ θ +φ = cos θ cos   + sin θ sin    2   2 

Sol. (d) Let it be remembered at the outset that the centroid

⇒

of an equilateral triangle is the same as the circumcentre of the circumscribing circle.

862

b(a2 − b 2 )

On squaring and adding, we get (a2 + 3b 2 )2 2 (b 2 + 3a2 )2 2 y x + 2 2 1= 2 2 2 2 b (a − b 2 )2 a (a − b )

Example 16. The locus of the centroid of an equilateral triangle inscribed in the ellipse x 2 y2 + = 1 is a 2 b2 ( a 2 + 3b 2 ) 2 2 ( b 2 + 3a 2 ) 2 (a) x + y = (a 2 + b 2 ) 2 2 2 a b ( a 2 − 3b 2 ) 2 2 ( b 2 − 3a 2 ) 2 (b) x + y = (a 2 + b 2 ) 2 a2 b2 ( a 2 + 3b 2 ) 2 2 ( b 2 + 3a 2 ) 2 (c) x − y = (a 2 − b 2 ) 2 2 2 a b ( a 2 + 3b 2 ) 2 2 ( b 2 + 3a 2 ) 2 2 (d) x + y = (a 2 − b 2 ) 2 2 2 a b

Let α, β and γ be the vertices of the equilateral triangle, so that the centroid is given by 1 x = (acos α + acos β + acos γ ) 3 1 = a(cos α + cos β + cos γ ) 3 1 and y = (b sinα + b sinβ + b sin γ ) 3 1 = b(sinα + sinβ + sin γ ) 3 ∴ cos α + cos β + cos γ = 3x / a 3y and sin α + sin β + sin γ = b a2 − b 2 But x = [cos α + cos β + cos γ + cos(α + β + γ )] 4a 2 2 b −a and y = [sin α + sin β + sin γ − sin(α + β + γ )] 4b a2 − b 2  3 x Hence, x = + cos(α + β + γ )  4a  a   4a x(a2 + 3b 2 ) 3 ⇒ −  = cos(α + β + γ ) = x  2 2 a a(a2 − b 2 ) a − b

y(b 2 + 3 a2 )

x θ −φ θ +φ θ +φ y cos    = cos   + sin   2   2   2  b a

Thus, the equation of the chord joining two points having eccentric angles θ and φ on the ellipse x 2 y2 + = 1 is a 2 b2 x θ −φ θ +φ θ +φ y cos    = cos   + sin   2      2 2 a b X

Example 17. If α and β are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is cos α + cos β sin α − sin β (a) (b) cos (α − β) sin (α − β) cos α − cos β sin α + sin β (d) (c) cos (α − β) sin (α + β) Sol. (d) The equation of a chord joining points having eccentric angles α and β, is given by α + β x y  α + β  = cos cos   + sin      2  a b 2

α − β    2 

X

This is a quadratic equation in x. So, it gives two values of x and corresponding to each value of x, there is a point of intersection of ellipse (i) and the line (ii). In order that the line (ii) may be tangent to ellipse (i), the two points of intersection may be coincident. Therefore, the discriminant of Eq. (iii) must be zero. i.e. 4a 4 m 2 c 2 − 4a 2 ( a 2 m 2 + b 2 )( c 2 − b 2 ) = 0 ⇒

Sol. (d) Let the eccentric angles of the vertices P, Q, R of a ∆PQR be θ1, θ2 , θ3 . Then, the equations of PQ and PR are respectively x y  θ + θ2   θ1 + θ2   θ1 − θ2  cos  1  + sin   = cos    2  b  2   2  a x y  θ + θ3   θ1 + θ3   θ1 − θ3  and cos  1  + sin   = cos    2  b  2   2  a If PQ and PR are parallel to given straight lines, then we have [say] θ1 + θ2 = Constant = 2α and θ1 + θ3 = Constant = 2β Hence, ...(i) θ2 − θ3 = 2(α − β ) Now, the equation of QR is x y  θ + θ3   θ2 + θ3   θ2 − θ3  cos  2  + sin   = cos   ...(ii)  2  b  2   2  a

a

2

+

y2 b

2

=1

…(i)

and a line …(ii) y = mx + c The abscissae of the points of intersection of the line (ii) and the ellipse (i) are the roots of the equation x 2 ( mx + c) 2 + =1 a2 b2 ⇒ ( a 2 m 2 + b 2 ) x 2 + 2a 2 mcx + a 2 ( c 2 − b 2 ) = 0 …(iii)

c2 < a 2 m2 + b2

⇒

Similarly, the line will not intersect the ellipse (i), if the Eq. (iii) has imaginary roots i.e. Discriminant < 0 ⇒ c2 > a 2 m2 + b2 X

Example 19. The number of values of c such that the straight line y = 4x + c touches the curve x2 + y 2 = 1 is 4 (a) 0 (b) 1 (c) 2 (d) infinite Sol. (c) We know that the line y = mx + c touches the curve x2 2

+

y2

a b2 Here, ∴

= 1 iff c 2 = a2 m2 + b 2 a2 = 4, b 2 = 1, m = 4 c 2 = 64 + 1

⇒ X

Intersection of a Line and an Ellipse

c = ± a 2m2 + b2

Thus, the line y = mx + c touches the ellipse x 2 y2 + 2 = 1, if c 2 = a 2 m 2 + b 2 2 a b The line will intersect the ellipse, if Eq. (iii) has real and distinct roots. This is possible when its discriminant is greater than zero. i.e. 4a 4 m 2 c 2 − 4a 2 ( a 2 m 2 + b 2 )( c 2 − b 2 ) > 0

[from Eq. (i)] = cos(α − β ) which shows that the Eq. (ii), for different values of θ2 + θ3 is tangent to the ellipse x2 y2 + 2 = cos 2 (α − β ) 2 a b which is the required envelope of QR.

Consider the ellipse x2

c2 = a 2 m2 + b2

⇒

Example 18. If a triangle is inscribed in an ellipse and two of its sides are parallel to given straight lines, then envelope of the third side is (a) a circle (b) a parabola (c) a hyperbola (d) an ellipse

15 Ellipse

If it passes through (ae, 0), then α + β α − β e cos   = cos    2   2  α − β cos    2  e= ⇒ α + β cos    2  α + β α − β 2 sin   cos    2   2  ⇒ e= α + β α + β 2 sin    cos   2   2  sin α + sin β ⇒ e= sin (α + β )

c=±

65

x2

+

y2

= 1 and the a 2 b2 straight line y = mx + c intersect in real points only, if (a) a 2 m 2 < c 2 − b 2

Example 20. The ellipse

(b) a 2 m 2 > c 2 − b 2 (c) a 2 m 2 ≥ c 2 − b 2 (d) c ≥ b 2 2 Sol. (c) The ellipse x2 + y 2 = 1 and the straight line

a b y = mx + c intersect in real points, then the quadratic x2 (mx + c )2 equation 2 + = 1 has real roots. Therefore, a b2 Discriminant ≥ 0 ⇒ c 2 ≤ a2 m2 + b 2

863

Objective Mathematics Vol. 1

15 Tangent

iii.

Tangents at the extremities of latusrectum passes through the corresponding foot of directrix on major axis.

iv.

Length of tangent between the point of contact and the point, where it meets the directrix subtends right angle at the corresponding focus.

In this section, we shall derive the equations of tangents to an ellipse in different forms viz. slope form, point form and parametric form. These equations will be used to solve various problems related to tangents to an ellipse. (i) Slope form The equations of tangents of slope x 2 y2 m to ellipse 2 + 2 = 1 are y = mx ± a 2 m 2 + b 2 a b and the coordinates of the points of contact are  a 2m b2 ± ,K  a 2m2 + b2 a 2m2 + b2 

X

   

(ii) Point form The equation of the tangent to the x 2 y2 ellipse 2 + 2 = 1 at the point ( x1 , y1 ) is a b xx1 yy1 + 2 =1 a2 b

Sol. (d) The equations of tangents at two points having eccentric angles θ1 and θ2 are x y cos θ1 + sinθ1 = 1 a b x y cos θ2 + sinθ2 = 1 a b The point of intersection of Eqs. (i) and (ii) is   θ + θ2   θ1 + θ2    a cos  1  b sin      2  2  ,   θ1 − θ2   θ1 − θ2    cos    cos   2    2 

(iii) Parametric form The equation of the tangent x 2 y2 to the ellipse 2 + 2 = 1 at the point a b ( a cos θ, b sin θ ) is y x cos θ + sin θ =1 a b (iv) Point of intersection of tangents The equations of the tangents to the ellipse at points P ( a cos θ 1 , b sin θ 1 ) and Q ( a cos θ 2 , b sin θ 2 ) are y x cos θ 1 + sin θ 1 = 1 a b y x and cos θ 2 + sin θ 2 = 1 a b and these two intersect at the point   θ1 + θ 2    θ1 + θ 2   a cos    b sin   2   2   ,   θ1 − θ 2   θ1 − θ 2    cos   cos     2   2   

Important Properties Related to Tangent i. ii.

864

Locus of foot of perpendiculars from foci upon any tangent is an auxiliary circle. Product of perpendiculars from foci upon any x 2 y2 tangent of the ellipse 2 + 2 = 1 is b 2 . a b

Example 21. The locus of the point of intersection of tangents to an ellipse at two points, sum of whose eccentric angles is constant, is a/an (a) parabola (b) circle (c) ellipse (d) straight line

…(i) …(ii)

It is given that (θ1 + θ2 ) = k = constant. Therefore, if ( x1, y1 ) is the point of intersection of Eqs. (i) and (ii), then a cos k ′ Q k ′ = k  x1 =   θ1 − θ2  2  cos    2  b sin k ′ and y1 =  θ − θ2  cos  1   2  x1 a b = cot k ′, y1 =  tan k ′ x1 ⇒  a y1 b b  ⇒ ( x1, y1 ) lies on the straight line y =  tan k ′ x.  a X

Example 22. The points on the ellipse such that the tangent at each of them makes equal angles with the axes, is   a2 b2  (a)  ± ,± 2 2 2 2   a b a b + +     b2 a2  (b)  ± ,± 2 2 2 2   a +b a +b     1 1  (c)  ± ,± 2 2 2 2   a +b a +b   (d) None of the above

But the tangent equation is given by 3x + 4y = 7 Identifying Eqs. (i) and (ii), we get 1 cos θ sin θ = = 3 2 7

x y at P is cos φ + sin φ = 1. a b

Y

X

O

X

Y′

− cos φ ⋅ b = ± tan 45° = ± 1 sin φ ⋅ a a cot φ = ± b a cos φ = ± 2 a + b2 b sinφ = ± a2 + b 2

Slope of tangent = ⇒ ⇒ and

∴Coordinates of required points are  a2 b2 ,± ± 2 2 2  a + b a + b2 X

x2

Y L (1/√3 cos θ, 1/2 sin θ)

X

y = mx ±

a2 m2 + b 2

...(i)

Since, 3 x − 2 y − 20 = 0 i.e. 2 y = 3 x − 20 3 i.e. y = x − 10 2 which is tangent to the ellipse. Therefore, comparing with Eq. (i), 3 m= 2 and a2 m2 + b 2 = 100 9 ⇒ a2 ⋅ + b 2 = 100 4 ⇒ 9a2 + 4b 2 = 400

…(ii)

1 10 Similarly, since x + 6 y − 20 = 0, i.e. y = − x + 6 3 which is tangent to the ellipse. Therefore, comparing with Eq. (i), 1 m= − 6 100 and a2 m2 + b 2 = 9 a2 100 2 ⇒ + b = 36 9 ⇒ a2 + 36 b 2 = 400

...(iii)

Now, solving Eqs. (ii) and (iii), we get a2 = 40

M

and Y′

The equation of the tangent at L is 3 x cos θ + 2 y sin θ = 1

+

y2

=1 a b2 We know that the general equation of the tangent to the ellipse is 2

2 2 Sol. (a) The equation of the ellipse is x + y = 1 1/ 3 1/ 4 1 1 ,b= i.e. a = 3 2

O

Example 24. The equation of the ellipse whose axes are coincident with the coordinate axes and which touches the straight line 3x − 2 y − 20 = 0 and x + 6 y − 20 = 0, is x 2 y2 (a) + =1 5 8 x 2 y2 (b) + = 10 40 10 x 2 y2 (c) + =1 40 10 x 2 y2 (d) + =1 10 40 Sol. (c) Let the equation of the ellipse be

 . 

Example 23. If the line 3x + 4 y = 7 touches the ellipse 3x 2 + 4 y 2 = 1, then the point of contact is  1 1  (a)  ,   7 7  1 1  (b)  ,−   3 3  1 1  (c)  ,−   7 7 (d) None of the above

X′

15

The point of contact = (a cos θ, b sin θ) 1 1  , =    7 7

(a cos φ, b sin φ) P X′

…(ii)

Ellipse

Sol. (a) Let P be (a cos φ, b sin φ), then equation of tangent

…(i)

b 2 = 10

∴The required equation of the ellipse is x2 y2 + =1 40 10

865

15

Work Book Exercise 15.2

Objective Mathematics Vol. 1

1 A tangent to the ellipse 4 x 2 + 9 y 2 = 36 is cut by

5 The slope of a common tangent to the ellipse

the tangent at the extremities of the major axis at T and T′. The circle on TT′ as diameter passes through the point

x2 y2 + = 1 and a concentric circle of radius r is a2 b2

a (−

d (3, 2)

a

x2 y2 2 If PSQ is a focal chord of the ellipse 2 + 2 = 1, a b a > b, then the harmonic mean of SP and SQ is

c

a

5, 0)

b2 a

b ( 5, 1)

a2 b

b

c ( 0, 0)

2 b2 a

c

d

2 a2 b

a

π 6

b

π 8

c

d

r2 − b2 a −r 2

r2 − b2

d

a2 − r 2

a2 − r 2 r2 − b2

π 4

perpendicular from a focus and the centre of an ellipse with semi-major axis of length a, respectively on a tangent to the ellipse and r denotes the focal distance of the point, then ap = rp′ ap = rp′ + 1

a c

rp = ap′ ap′ + rp = 1

b d

x2 y2 + =1 a2 b2 (a > b) and the circle x 2 + y 2 = a2 at the points,

7 Tangents are drawn to the ellipse

4 If F1 and F2 are the foot of the perpendiculars from

where a common ordinate cuts them on the same side of the X-axis. Then, the greatest acute angle between these tangents is given by

x2 y2 the foci S1 and S 2 of an ellipse + = 1 on the 5 3 tangent at any point P on the ellipse, then

 a − b tan−1    2 ab   2 ab   tan−1    a − b

a

(S1F1 ) (S 2 F2 ) is equal to a 2 c 4

2

b

6 If p and p′ denote the lengths of the

x2 3 Tangent is drawn to the ellipse + y 2 = 1 at 27  π (3 3 cos θ,sin θ ), where θ ∈  0,  , then the value  2 of θ such that sum of intercepts on axes made by the tangent is minimum, is π 3

 r2 − b2   tan−1   a2 − r 2   

b 3 d 5

c

 a + b tan−1    2 ab   2 ab   tan−1    a + b

b d

Combined Equation of the Pair of Tangents The combined equation of the pair of tangents x 2 y2 drawn from a point ( x1 , y1 ) to the ellipse 2 + 2 = 1 is a b 2   xx1 yy1   x12 y12  x 2 y2     + − 1  2 + 2 − 1 =  2 + 2 − 1  b b  a  a 2 b2  a Y

If the angle between these lines is θ, then tan θ =

2 h2 − ab a+ b

⇒ tan θ = ⇒

, where a = 9, h = − 12, b = − 4

2 144 + 36 9−4 tan θ =

12 5

2 180 12 5 = 5 5 −1 12 ⇒ θ = tan 5 ⇒ tan θ =

B

Chord of Contact of Tangents

Q X′

A

A′

The equation of the chord of contact of tangents x 2 y2 drawn from a point ( x1 , y1 ) to the ellipse 2 + 2 = 1 is a b xx1 yy1 + 2 = 1. a2 b

X P (x1, y1)

B′ T (h,k) Y′ X

Example 25. The angle between the pair of tangents from the point (1, 2) to the ellipse 3x 2 + 2 y 2 = 5 5 π π 12 (a) tan −1 (b) (c) (d) tan −1 3 4 2 5 Sol. (d) The combined equation of the pair of tangents

866

(3 x + 2 y − 5)(3 + 8 − 5) = (3 x + 4 y − 5) ⇒ 9 x2 − 24 xy − 4 y2 + 30 x + 40 y − 55 = 0 2

2

B Q (a1, b1) X′

A′

O

A

P (x1, y1) R (a2, b2)

B′

drawn from (1, 2) to the ellipse 3 x2 + 2 y2 = 5 is 2

Y

[QSS1 = T ] 2

Y′

X

Sol. (a) Let A( x1, x1 − 5) be a point on x − y − 5 = 0, then chord of contact of x2 + 4 y2 = 4 w.r.t. A is ⇒

xx1 + 4 y( x1 − 5) = 4 ( x + 4 y)x1 − (20 y + 4) = 0

It is passes through a fixed point. ∴ x + 4y = 0 and [QP + λQ = 0] 20 y + 4 = 0 1 y=− ⇒ 5 4 and x= 5 4 1 The coordinates of fixed point are  , −  . 5 5

Example 27. Tangent are drawn from any x 2 y2 point on the ellipse 2 + 2 = 1 to the circle a b x 2 + y 2 = r 2 , then the chord of contact is a tangent to the ellipse, is (a) a 2 x 2 − b 2 y 2 = r 2 (b) a 2 x 2 + b 2 y 2 = r 4 (c) a 4 x 4 + b 4 y 4 = r 2 (d) a 4 x 4 + b 4 y 4 = r 4 Sol. (b) The coordinates of any point on the ellipse are

Example 28. The locus of the point of the chord of contact of tangents from which the ellipse x 2 y2 + = 1 touches the circle, is a 2 b2 x 2 y2 1 (a) 4 + 4 = 2 a b c 2 2 y x 1 (b) 2 + 2 = 2 a b c x 2 y2 1 (c) 4 + 4 = 4 a b c (d) None of the above

15

Sol. (a) Let the point be (h, k ) from which two tangents have been drawn to the given ellipse. Then, the equation of the chord of contact is xh yk …(i) + 2 =1 a2 b If the chord of contact of Eq. (i) touches the circle x2 + y2 = c 2 , then the length of the perpendicular from the origin on Eq. (i) is the radius of the circle. 1 Hence, =c h2 k2 + 4 a4 b 1 h2 k2 + 4 = 2 ⇒ 4 a b c 1 x2 y2 Hence, the locus of (h, k ) is 4 + 4 = 2 . a b c

Director Circle The locus of the point of intersection of perpendicular tangents to an ellipse is known as its director circle.

Equation of the Director Circle x2

+

y2

= 1 be the ellipse and ( h, k ) be a point a 2 b2 on its director circle. Let

(acos θ, b sinθ).

Y

The equation of the chord of contact to the circle x2 + y2 = r 2 is xa cos θ + yb sin θ = r 2

...(i)

B (a , b )

If Eq. (i) is tangent to the ellipse a2 x2 + b 2 y2 = r 4 , the

90°

2

 r 2 − xa cos θ   = r 4 quadratic equation in x viz. a x +  sin θ   must have two equal roots i.e. the equation a2 x2 − 2 xar 2 cos θ + r 4 cos 2 θ = 0

P (h, k)

2 2

A′ (–a, 0)

C

b)

X′

,–

X

X

Ellipse

Example 26. Tangent are drawn from the points on the line x − y − 5 = 0 to x 2 + 4 y 2 = 4, then all the chords of contact passes through a fixed point, whose coordinates are  4 1 (a)  , −   5 5  4 1 (b)  ,   5 5  4 1 (c)  − ,   5 5 (d) None of the above

must have two equal roots. The condition for this is 4 a2 r 4 cos 2 θ = 4 a2 r 4 cos 2 θ which is clearly satisfied.

A (a, 0)

X

B′ (a

X

Y′

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Objective Mathematics Vol. 1

15

The equation of any tangent to the ellipse y2 + = 1 is a 2 b2

i.

x2

y = mx ± a 2 m 2 + b 2 If it passes through ( h, k ), then k = mh ± a 2 m 2 + b 2 ⇒

ii.

( k − mh) 2 = a 2 m 2 + b 2

⇒

m 2 ( h 2 − a 2 ) − 2mhk + ( k 2 − b 2 ) = 0

This is quadratic equation in m. So, it gives two values of m, say m1 and m2 . Corresponding to each to these values of m, there is a tangent to the ellipse passing through ( h, k ). If the two tangents are perpendicular, then m1 m2 = − 1 k 2 − b2 ⇒ = −1 h2 − a 2 ⇒

x2

x 2 + y2 = a 2 + b2

iii.

Remark If follows from the definition of the director circle that the tangents drawn from any point on the director circle of a given ellipse to the ellipse are always at right angle.

Example 29. If two tangents drawn to the ellipse x 2 + 4 y 2 = 4 intersect perpendicularly at P, then the locus of P is (a) x 2 + y 2 = 4 (b) x 2 + y 2 = 5 (c) x 2 + y 2 = 8 (d) x 2 + y 2 = 2

then c2 =

x2 + y2 = 1 4 By definition of director circle, P lies on the director circle formed by the ellipse x2 y2 + =1 4 1 Hence, locus of P is x2 + y2 = 4 + 1 [Q x2 + y2 = a2 + b 2 ] x 2 + y2 = 5

Normal 868

The equations of the normals of x 2 y2 slope m to the ellipse 2 + 2 = 1 are given by a b 2 2 m (a − b ) at the points y = mx ± a 2 + b2m2   a2 b2m . ± ,±  2 2 2 2 2 2  a +b m a +b m   Slope form

Ø If the line y = mx + c is a normal to the ellipse

In this section, we shall discuss forms of the equation of the normal to an ellipse which will be used to solve various types of problems on ellipse.

m 2 (a2 − b 2)2

a2 + b 2m2 line of the ellipse.

Sol. (b) Equation of ellipse is rewritten as

⇒

+

y2

= 1 at two points (a cos θ1 , b sin θ1) and a2 b 2 (a cos θ 2 , b sin θ 2) are (λ, µ ), where  θ + θ2  cos  1   2  (a2 − b 2) cos θ1 cos θ 2 λ= a  θ − θ2  cos  1   2   θ + θ2  sin  1  2 2  2  (a − b ) and µ = − sin θ1 sin θ 2 b  θ − θ2  cos  1   2 

The equation of the director circle is

X

Parametric form The equation of the normal x 2 y2 to the ellipse 2 + 2 = 1 at ( a cos θ, b sin θ) a b is ax sec θ − by cosec θ = a 2 − b 2 .

Ø The point of intersection of normals to the ellipse

h2 + k 2 = a 2 + b2

Clearly, it is a circle concentric to the ellipse and radius is equal to a 2 + b 2 .

The equation of the normal at x 2 y2 ( x1 , y1 ) to the ellipse 2 + 2 = 1 is a b 2 2 a x b y − = a 2 − b2 x1 y1 Point form

X

x2 a2

+

y2 b2

= 1,

is the condition of normality of the

Example 30. If the normal at an end of a latusrectum of an ellipse passes through one extremity of the minor axis, then the eccentricity of the ellipse is given by (a) e 4 + e 2 − 1 = 0 (b) e 2 + e − 5 = 0 (c) e 3 = 5 (d) None of the above 2 2 Sol. (a) Let x2 + y 2 = 1 be the ellipse. The coordinates of

a

b

 b2  . an end of the latusrectum is  ae, a  

Ø

a2 x b2 y − 2 = a2 − b 2 ae b /a ax − ay = a2 − b 2 ⇒ e It passes through one extremity of the minor axis whose coordinates are (0, − b ). ∴ ab = a2 − b 2 ⇒ ⇒

●

●

a2 b 2 = (a2 − b 2 )2 a × a (1 − e 2 ) = (a2e 2 )2 2

2

⇒

1 − e2 = e4

⇒ X

●

e + e2 − 1 = 0 4

Example 31. If Q is the point on the auxiliary circle corresponding to a point P on an ellipse. Then, the normals at P and Q meet on (a) a fixed circle (b) an ellipse (c) a hyperbola (d) None of these

●

X

Sol. (a) The coordinates of Q are (a cos φ, a sin φ) and the coordinates of P are (a cos φ, b sin φ). The equation of the normal at P is ax sec φ − by cosec φ = a2 − b 2 The equation of the normal at Q is ax sec φ − ay cosec φ = 0 On solving Eqs. (i) and (ii), we get ay cosec φ − by cosec φ = a2 − b 2 ⇒

…(i)

which is the required circle.

Some Useful Properties of an Ellipse In this section, we shall state and prove some useful results on an ellipse as theorems given below : Y X

T

P (x1, y1) M

N′ A′ (–a, 0)

B (0, b)

N A G S (ae, 0) (a, 0)

S′ C (–ae, 0)

X

B′ (0, –b)

Y′

The tangent and normal at any point of an ellipse bisect the external and internal angles between the focal radii to the point.

If SM and S ′ M′ are perpendiculars from the foci upon the tangent at any point of the ellipse, then SM ⋅ S ′ M′ = b 2 and M, M′ lie on the auxiliary circle. x2 2

+

y2

=1 a b2 meets the major axis inTand minor axis inT ′, then CN ⋅ CT = a2, CN ′⋅ CT ′= b 2, where N and N′ are the foot of the perpendiculars from P on the respective axes. If the tangent at any point P on the ellipse

If SM and S ′ M′ are perpendiculars from the foci S and S′, respectively upon a tangent to the ellipse, then CM and CM′ are parallel to S ′ P and SP, respectively. The common chords of an ellipse and a circle are equally inclined to the axes of the ellipse.

Example 32. If a number of ellipses is described having the same major axis but a variable minor axis, such that each one of the tangents at the ends of their latusrectum passes through the fixed point, then the point of concurrency is (a) (1, 1) (b) (0, 1) (c) (1, 0) (d) (0, 2) a b Since, major axis is fixed. ∴a is variable. Hence, e is variable. Now, the ends of the latusrectum  b2  . are  ± ae, ± a   Hence, the tangents at these points are b2 ± y⋅ xx1 yy ± x ⋅ ae a =1 + + 21 = 1 ⇒ b2 a2 a2 b xe y ± ± = 1 ⇒ ± xe ± y = a ⇒ a a ⇒ ± ex ± y − a = 0 Hence, there are 4 tangents. Since, e is variable. ∴Each of the tangents passes through the point (0, 1) which is fixed.

y cosec φ (a − b ) = a2 − b 2

T′

15

It follows from the above property that if an incident light ray passing through the focus (S) strikes the concave side of the ellipse, then the reflected ray will passes through the other focus (S′).

2 2 Sol. (b) Let the equation of the ellipse be x2 + y 2 = 1.

…(ii)

…(iii) ∴ y = (a + b ) sin φ From Eq. (ii), x sec φ = (a + b ) …(iv) ∴ x = (a + b )cos φ On squaring Eqs. (iii) and (iv) and adding, we get x2 + y2 = (a + b )2

X′

●

Ellipse

 b2   is The equation of the normal at  ae, a  

Example 33. If the normal at the point P (θ ) to x 2 y2 the ellipse + = 1 intersects it again at the 14 5 point Q(2θ), then cos θ is equal to 2 1 3 2 (a) (b) − (d) (c) 3 2 2 3 2 2 Sol. (b) We have, x + y = 1

14 5 The equation of normal at P( 14 cos θ, 5 sin θ) is ( 14 sec θ)x − ( 5 cosec θ) y = 9 This meets the ellipse (i) at Q ( 14 cos 2θ, 5 sin 2θ). ∴ 14 sec θ cos 2θ − 5 cosec θ sin 2 θ = 9 ⇒ 18 cos 2 θ − 9 cos θ − 14 = 0 ⇒ (6 cos θ − 7 ) (3 cos θ + 2 ) = 0 ⇒ cos θ = − 2 / 3 or cos θ ≠ 7 / 6

…(i)

869

Objective Mathematics Vol. 1

15

(a) b 2 b2 (c) 2

Number of Normals and Conormal Points In the previous section, we have learnt that at a point on a given ellipse, exactly one normal can be drawn. If there is a point not lying on a given ellipse, infinitely many lines can be drawn which intersect the ellipse. Out of these lines, there are atmost four lines which are normals to the ellipse at the points, where they cut the ellipse. Such points on the ellipse are called conormal points. In this section, we shall learnt about the conormal points and various relations between their eccentric angles.

i. ii.

iii.

iv.

v.

x2

870

+

y2

= 1 such a 2 b2 that sin (θ 1 + θ 2 ) + sin (θ 2 + θ 3 ) + sin (θ 3 + θ 1 ) = 0, then the normals at these points are concurrent.

X

Y F

If the normals at four points P ( x1 , y1 ), Q ( x 2 , y2 ), R ( x 3 , y3 ) and S ( x 4 , y4 ) on the x 2 y2 ellipse 2 + 2 = 1 are concurrent, then a b 1 1 1 1 + +  =4 ( x1 + x 2 + x 3 + x 4 )  +  x1 x 2 x 3 x 4 

Example 34. If CF is the perpendicular from x 2 y2 the centre C of the ellipse 2 + 2 = 1 on the a b tangent at any point P and G is the point when the normal at P meets the major axis, then CF ⋅ PG is

P (a cos φ, b sin φ)

B X′

C

A′

G

X

A

B′ Y′

∴ CF =

1  cos φ sin φ    + 2 b 2   a 2

2

=

ab (a sin φ + b 2 cos 2 φ) 2

2

Equation of normal at P is axsec φ − by cosec φ = a2 − b 2 ,

Properties of eccentric angles of conormal points The sum of the eccentric angles of the x 2 y2 conormal points on the ellipse 2 + 2 = 1 is a b an odd multiple of π.

three points on the ellipse

vi.

a b x y Tangent at P is cos φ + sin φ = 1 a b

Conormal points The points on the ellipse at which four normals to the ellipse passes through a given point are called conormal points.

If θ 1 , θ 2 and θ 3 are the eccentric angles of

(d) None of these

2 2 Sol. (a) Equation of an ellipse is x2 + y 2 = 1.

The four normals can be drawn from a point to an ellipse.

If θ 1 , θ 2 , θ 3 and θ 4 are eccentric angles of four points on the ellipse the normals at which are concurrent, then (a) Σ cos (θ 1 + θ 2 ) = 0 (b) Σ sin (θ 1 + θ 2 ) = 0 (c) (θ 1 + θ 2 ) + sin (θ 2 + θ 3 ) + sin(θ 3 + θ 1 ) = 0

(b) 2b 2

then

  ( a2 − b 2 ) cos φ, 0 G =  a   2

PG = =

  ( a2 − b 2 )  acos φ − cos φ + (b sin φ − 0)2 a     b4  cos 2 φ + b 2 sin2 φ 2  a

b (a2 sin2 φ + b 2 cos 2 φ) a ⇒ CF ⋅ PG = b 2 =

X

Example 35. If the normals at the points A, B , C and D of an ellipse meet in a point O, then SA ⋅ SB ⋅ SC ⋅ SD is equal to (where, S is one of the foci) b2 (a) 2 ⋅ SO 2 e b3 (b) 3 ⋅ SO 3 e 2b 2 (c) 2 ⋅ SO 2 e b2 (d) 2 ⋅ 2SO 2 3e Sol. (a) Let S be the point (− ae, 0), we have SA = a + ex1 etc. Therefore, SA ⋅ SB ⋅ SC ⋅ SD = (a + ex1 )(a + ex2 )(a + ex3 )(a + ex4 ) = a4 + a3eΣ x1 + a2e 2 Σ x1 x2 + ae 3 Σ x1 x2 x3 + e 4 x1 x2 x3 x4 =

b2

{(h + ae )2 + k} e2 On substitution and simplification, b2 = 2 ⋅ SO 2 e

(a)

The equation of the chord of the ellipse x 2 y2 + = 1 bisected at the point ( x1 , y1 ) is given by a 2 b2 x12 y12 xx1 yy1 + − 1 = + − 1 or T = S ′ a 2 b2 a2 b2 X

(c)

α2 a4 α2 a

4

x2 − x2 +

β2 b4 β2 b

4

y2 = 1

(d) None of these

a2

x4 +

β2

(b)

b2

15

y4 = 1

Sol. (c) The equation of any tangent to the curve (ellipse) x2 α

2

+

y2 β2

= 1 is y = mx +

α 2 m2 + β 2 .

...(i)

Example 36. The locus of the mid-points of a x 2 y2 focal chord of the ellipse 2 + 2 = 1 is a b 2 2 2 y y 2 ex x ex x (b) 2 − 2 = (a) 2 + 2 = a a a b a b (c) x 2 + y 2 = a 2 + b 2 (d) None of these

Let ( x1, y1 ) be the pole of Eq. (i) with respect to the ellipse x2 y2 + 2 =1 2 a b xx yy Then, its polar viz. 21 + 21 = 1 must be identical with a b a2 m b2 Eq. (i), therefore = − α 2 m2 + β 2 =− x1 y1

Sol. (a) Let the mid-point of the focal chord of the given

∴

ellipse be (h, k ). Then, its equation is hx ky h2 k2 [using T = S1] + 2 = 2 + 2 2 a b a b Since, this passes through (ae, 0). hae h2 k2 he h2 k2 = 2 + 2 ⇒ = 2 + 2 ∴ 2 a a a b a b ex x2 y2 Locus of (h, k ) is ∴ = + 2. a a2 b

Pole and Polar Let P be a point inside or outside an ellipse. Then, the locus of the point of intersection of tangents to the ellipse at the point, where secants drawn through P intersect the ellipse, is called the polar of point P with respect to the ellipse and the point P is called the pole of the polar. The polar of a point ( x1 , y1 ) with respect to the xx yy x 2 y2 ellipse 2 + 2 = 1 is 21 + 21 = 1. a b a b Y

R

A′ Pole (–a, 0)

Q

y12

+ β2 =

b4

y12

⇒

y12

Hence, the locus of ( x1, y1 ) is

α 2 x12 a α2

X

touch

α2

+

β2

+

x2 +

β 2 y12 b4 β2 b4

=1

y2 = 1.

x2

−

β 2 y2

=1 α2 4a 2 (c) α 2 x 2 + β 2 y 2 = 1 (d) None of the above (b)

2

B′ (0, –b)

Example 37. Chord of the ellipse y2

a

4

4

Example 38. If the polar with respect to y2 x2 y 2 = 4x touches the ellipse 2 + 2 = 1, then locus α β of its pole is y2 x2 (a) 2 − =1  4a 2α 2  α    β2 

2

 2 ah  = α 2  2 a  + β 2      k  k 

So,

[using c 2 = a2 m2 + b 2 ]

Y′

x2

x1 y1

ky = 2 a ( x + h) 2a 2 ah y= x+ ⇒ k k x2 y2 This line touches the ellipse 2 + 2 = 1. α β

Q′

P R′ (x 1 , y 1 )

a

4

⋅

polar is

A(a, 0)

Polar

b 4 x12

a2

b4

α 2 m2 + β 2 =

⇒ α2

b2

Sol. (a) Let P (h, k ) be the pole. Then, the equation of the

b) B (0 ,

(h, k)T X′

m= −

Also,

X

X

α2

y2 = 1

Ellipse

Chord Bisected at a Given Point

x2 a2

+

y2 b2

=1

= 1, then the locus of their pole is

⇒

4a h = 4a α + k 2β 2 2 2

2 2

So, locus of (h, k ) is 4a2 x2 = 4a2α 2 + β 2 y2 . ⇒

x2 α

2

−

y2  4a2α 2    2   β 

=1

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Objective Mathematics Vol. 1

15

Conjugate Points Two points are said to be conjugate points with respect to an ellipse, if each lies on the polar of the other. In other words, two points P and Q are conjugate points with respect to an ellipse, if the polar of P passes through Q and the polar of Q passes through P.

Conjugate Lines Two lines are said to be conjugate lines with respect to an ellipse, if each passes through the pole of the polar. X

Example 39. The locus of the pole of any tangent to the auxiliary circle w.r.t. the ellipse x 2 y2 + = 1 is a 2 b2 (a) b 4 x 2 + a 4 y 2 = a 4 b 2 (b) b 4 x 2 + a 4 y 2 = 2a 2 b 4 (c) b 2 x 4 + a 4 y 2 = a 4 b 2 (d) b 4 x 2 + a 4 y 2 = a 2 b 4 Sol. (d) The equation of any tangent to the auxiliary circle x2 + y2 = a2 is …(i) xcos θ + ysinθ = a Let (α, β ) be the pole of Eq. (i) with respect to the ellipse x2 y2 xα yβ + 2 = 1, then its polar viz. 2 + 2 = 1 must be 2 a b a b identical with Eq. (i). a2 cos θ b 2 sinθ Therefore, = =a α β r ∴ cos θ = a a and sin θ = β ⋅ 2 b On squaring and adding, we get a2 α2 + β2 ⋅ 4 = 1 2 a b ⇒ b 4α 2 + a4β 2 = a2 b 4 .

Then, the pole of (α, β ) with respect to the ellipse viz. xα yβ …(ii) + 2 =1 a2 b must represent the same straight line as Eq. (i). On comparing Eqs. (i) and (ii), we get b 3 cosec φ a3 sec φ =− = a2 − b 2 β α ∴ cos φ =

a3 α( a − b ) 2

2

and sin φ = −

b3 β(a − b 2 ) 2

On squaring and adding, we have a6 b6 + 2 2 1= 2 2 2 2 α (a − b ) β ( a − b 2 )2 a6

⇒

α2

+

b6 β2

= (a2 − b 2 )2

Hence, the locus of (α, β ) is

a6 x

2

. +

b6 y2

= (a2 − b 2 )2 .

Diameter The locus of the mid-point of a system of parallel chords of an ellipse is called a diameter whose equation b2 of diameter is y = − 2 x. a m

Conjugate Diameters Two diameters of an ellipse are said to be conjugate diameters, if each bisect the chords parallel to the other. Y

(–a cos θ, b sin θ) D

X′

B

P (a cos θ, b sin θ)

A

A′

X

C D′

P′ B′

Hence, the locus of (α, β ) is b 4 x2 + a4 y2 = a2 b 4 . X

872

Example 40. The locus of the poles of normal x 2 y2 chords of the ellipse 2 + 2 = 1 is a b 6 6 a b (a) 2 − 2 = ( a 2 + b 2 ) 2 x y 6 a b6 (b) 2 + 2 = ( a 2 − b 2 ) 2 x y 2 a b2 (c) 6 + 6 = ( a 6 + b 6 ) 2 x y (d) None of the above Sol. (b) Let (α, β ) be the pole of the normal ax sec φ − by cosec φ = a − b 2

2

…(i)

Y′

Let y = m1 x and y = m2 x be conjugate diameters x 2 y2 with respect to the ellipse 2 + 2 = 1. Then, y = m2 x a b bisects the system of chords parallel to y = m1 x. b2 So, its equation is y = − 2 x. …(i) a m1 Clearly, Eq. (i) and y = m2 x represent the same line. −b 2 ∴ m2 = 2 a m1 ⇒

m1 m2 =

−b 2 a2

X

Ø In an ellipse, the major axis bisects all chords parallel to the

minor axis and vice-versa, therefore major and minor axes of an ellipse are conjugate diameters of the ellipse but they do not satisfy the condition m1m2 = − b 2 / a2 and are the only perpendicular conjugate diameters.

Example 42. If the point of intersection of the y2 x 2 y2 x2 ellipses 2 + 2 = 1 and 2 + 2 = 1 are at the α β a b extremities of the conjugate diameters of the former, then a 2 b2 α 2 β2 (a) 2 + 2 = 2 (b) 2 + 2 = 2 α β a b 2 2 a b (c) 2 − 2 = 2 (d) None of these α β

15 Ellipse

Thus, two straight lines y = m1 x and y = m2 x are x 2 y2 conjugate diameters of the ellipse 2 + 2 = 1, if a b −b 2 m1 m2 = 2 a

Sol. (a) The points of intersection of the two ellipses will be X

Example 41. The eccentricity of an ellipse whose pair of conjugate diameter are y = x and 3 y = − 2x, is 2 1 (b) (a) 3 3 1 (c) (d) None of these 3

given by x2

2 2 Sol. (c) Let the ellipse be x2 + y 2 = 1, since y = x

a b and 3 y = − 2 x is a pair of conjugate diameters, therefore b2 m1m2 = − 2 a b2 2  ⇒ (1)  −  = − 2  3 a ⇒ ⇒ ⇒

2 a2 = 3 a2 (1 − e 2 ) ⇒ 2 = 3 (1 − e 2 ) 1 1 ⇒ e= e2 = 3 3

ii.

The sum of the squares of any two conjugate semi-diameters of an ellipse is constant and equal to the sum of the squares of the semi-axes of the ellipse i.e. CP 2 + CD 2 = a 2 + b 2 .

iii.

The product of the focal distances of a point on an ellipse is equal to the square of the semi-diameter which is conjugate to the diameter through the point.

iv.

The tangents at the ends of a pair of conjugate diameters of an ellipse form a parallelogram.

v.

The area of the parallelogram formed by the tangents at the ends of conjugate diameters of an ellipse is constant and is equal to the product of the axes.

=

x2 2

+

y2 …(i)

But by the question, the two lines are conjugate. Hence, 1 1 − 2 2 b2 b2 α a =− 2 m1m2 = − 2 ⇒ 1 1 a a − b2 β 2

2 a2 = 3b 2

The eccentric angles of the ends of a pair of conjugate diameter of an ellipse differ by a right angle.

b

2

Thus, if m1 and m2 are the gradients of the two lines, then  1 1 1 1 m1m2 =  2 − 2  ÷  2 − 2  a α  b β 

1−

⇒

Properties of Conjugate Diameters i.

y2

+

α β2  1 1 1 1 ⇒ x2  2 − 2  + y2  2 − 2  = 0  a b α β  which being a homogeneous equation of second degree represents a pair of straight lines passing through the origin. Let y = mx be one such straight line. Then, putting y = mx in Eq. (i), we get  1 1 1 1 m2  2 − 2  +  2 − 2  = 0 b β  a α  a

2

a

∴ X

2

α2

+

a2 α b2 2

β2

= − 1+

b2 β2

=2

Example 43. If λ and µ are the angles subtended by the major axis to an ellipse at the extremities of a pair of conjugate diameters, then cot 2 λ + cot 2 µ is equal to (a 2 − b 2 ) 2 (a 2 + b 2 ) 2 (b) (a) 4a 2 b 2 4a 2 b 2 2 2 2 (a + b ) (a 2 − b 2 ) 2 (d) (c) a 2b2 a 2b2 Sol. (a) D′

P(θ) λ A(a, 0)

(–a, 0)A′ µ P′

D π +θ 2

Let PP′ and DD′ be the conjugate diameters, so that P ≡ (a cos θ, b sin θ) and D ≡ (− a sin θ, b cos θ). Given, ∠A′ PA = λ and ∠A′ DA = µ.

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Objective Mathematics Vol. 1

15

Now, slopes AP and A′ P are b sin θ a cos θ − a

and

b sin θ a cos θ + a

b sin θ b sin θ − a cos θ − a a cos θ + a ∴ tan λ = b sin θ b sin θ 1+ ⋅ a cos θ − a a cos θ + a − 2 ab sin θ 2 ab = 2 = b sin2 θ − a2 sin2 θ (a2 − b 2 ) sin θ

= Thus, we get

(a2 − b 2 ) sin θ 2 ab (a2 − b 2 ) cos θ and cot µ = 2 ab 2 Hence, cot λ + cot 2 µ cot λ =

=

From this, it follows that tan µ =

2 ab

=

π (a − b ) sin  + θ  2 2

2 ab (a2 − b 2 ) cos θ

(a2 − b 2 )2 4a2 b 2 2 (a − b 2 )2

(sin2 θ + cos 2 θ)

4a2 b 2 which is constant.

2

Work Book Exercise 15.3 1 Chords of an ellipse are drawn through the

perpendicular tangents to

a a circle c an ellipse

x2 y2 + 2 = 1 is a +λ b +λ

b d

a parabola a hyperbola

2. The area of the rectangle formed by the perpendiculars from the centre of the ellipse to the tangent and normal at the point, whose eccentric angle is π /4, is a

 a2 − b 2   ab  2 2 a + b 

c

1  a2 − b 2   ab  ab  a 2 + b 2 

a2 b

c −

2

a4 b4

2

a c

x 2 + y2 = a 2 + λ x 2 + y2 = a 2 + b 2 + λ

b d

x 2 + y2 = b 2 + λ x 2 + y2 = a 2 + b 2

5 The locus of the mid-points of the chords

b

2 x + 3 y + 1 = 0 of the ellipse x 2 + 4 y 2 = 1, 1 being parameter, is

d

1  a2 + b 2   ab  ab  a 2 − b 2 

a c

points ( x1, y1 ) and ( x2 , y2 ) to the ellipse x2 y2 xx + 2 = 1 are at right angles, then 1 2 is 2 y1 y2 a b equal to a

x2 y2 + 2 = 1 and 2 a b

 a2 + b 2   ab  2 2 a − b 

3 If the chords of contact of tangent from two

874

4 The locus of the point of intersection of

positive end the minor axis. Then, their mid-point lies on

b

−

d

−

b2 a2 b4 a4

8x − 3y = 0 3x − 8y = 0

b d

8x + 3y = 0 3x + 8y = 0

x2 y2 + 2 = 1, the length 2 a b perpendiculars from the centre upon all chords which join ends of perpendicular diameters, is

6 In the ellipse

a

ab (a2 + b 2 )

b

(a2 + b 2 )

c ab d None of the above

WorkedOut Examples Type 1. Only One Correct Option Ex 1. The coordinates of all the points P on the x 2 y2 ellipse 2 + 2 = 1, for which the area of the a b ∆PON is maximum, where O denotes the origin and N , the foot of the perpendicular from O to the tangent at P, is   a2 b2  (a)  ± ,±  2 2 2 2 a +b a +b  

Thus, for area to be minimum, b tanθ = a a , sinθ = ± ∴ cosθ = ± a2 + b 2

b a + b2 2

 . a2 + b2 

 a2 ,± ∴ Points are  ±  a2 + b 2 

b2

Hence, (a) is the correct answer.

Ex 2. If two points are taken on the minor axis of an ellipse at the same distance from the centre as the foci, then the sum of the squares of the perpendicular distances from these points on any tangent to the ellipse is

  a2 b2  (b)  ± ,±  2 2 2 2 a b a b − −    2a 2 2b 2  (c)  ± ,±  a2 + b2 a 2 + b 2  

(a) 2a 2 (c) 4a 2

 2a 2 2b 2  (d)  ,  a2 + b2 a 2 + b 2  

(b) 3a 2 (d) None of these

Sol. Let the equation of the ellipse be x2 y2 + 2 =1 2 a b Y

Sol. Equation of tangent at P is

P1

x cosθ y sin θ + =1 a b

S1

Y

P2 X′ N P(a cos θ, b sin θ)

X′

S

S 1′

X

O

X

O

S′

Y′

L

Then, the distance of a focus from the centre = ae b2 = a2 − b2 a2 So, the two points on the minor axis are S 1 (0, a2 − b2 ) and S ′1 (0, − a2 − b2 ) = a 1−

Y′

ON =

1

ab

=

cos θ sin θ b cos θ + a2 sin 2 θ + 2 2 a b Equation of the normal at P is ax secθ − by cosec θ = a2 − b2 OL =

2

2

a2 − b2 a2 sec2 θ + b2cosec2 θ

2

=

2

Now, any tangent to the ellipse is y = mx +

where, m is a parameter. The sum of the squares of the perpendiculars on this tangent from the two points S 1 and S 1′

(a2 − b2 )sin θ ⋅ cos θ a2 sin 2 θ + b2 cos2 θ

where, L is the foot of perpendicular from O on the normal. ⇒ NP = OL 1 Area of ∆PON = × ON × OL 2 (a2 − b2 ) ab tan θ (a2 − b2 ) ab = 2 2 = 2 2 a tan θ + b a tan θ + b2 cot θ which is minimum maximum.

when

a2 tan θ + b2 cot θ

a2m2 + b2

is

 a2 − b2 − a2m2 + b2   =   1 + m2  

2

 − a2 − b2 − a2m2 + b2   +   1 + m2   =

2

2(a2 − b2 + a2m2 + b2 ) = 2a2 1 + m2

Hence, (a) is the correct answer.

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Objective Mathematics Vol. 1

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Ex 3. If y = mx −

(a 2 − b 2 )m

is normal to the

a 2 + b2m2 y2 x ellipse + = 1 for all values of m a 2 b2 belonging to (a) (0, 1) (b) (0, ∞) (c) R (d) None of these 2

Sol. The equation of the normal to the given ellipse at the point P (a cos θ , b sin θ ) is ax sec θ − by cosec θ − a2 = b2 (a2 − b2 )  a y =  tan θ x − sin θ  b b bm a tanθ = m, so that sinθ = 2 b a + b2m2

⇒ Let

Now, Eq. (i) becomes

…(i)

a2 + b2m2

a tan θ ∈ R b Hence, (c) is the correct answer.

x2

x

2

+

( ar + bs ) 2 y (r + s )

Sol. Given that,

2

= 1 (d)

2

a2 x

2

−

+

y2

= 1, a 2 b2 0 < b < a. Let the line parallel to Y-axis passing through P meet the circle x 2 + y 2 = a 2 at the point Q such that P and Q are on the same side of X-axis. For two positive real numbers r and s, then the locus of the point R on PQ such that PR : RQ = r : s as P varies over the ellipse, is y 2 (r + s )2 y 2 (r + s )2 x2 x2 (b) (a) 2 − = =1 1 + ( ar + bs ) 2 a a 2 ( ar + bs ) 2

Ex 4. Let P be a point on the ellipse

a2

( ar + bs ) 2 y (r + s ) 2

PR r = RQ s

P O

(a cos θ, b sin θ) X

Y′

⇒

876

x2 y2 (r + s)2 =1 + a2 (ar + bs)2

Hence, (b) is the correct answer.

Ex 5. In an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse to the point of contact meet on

x2 y2 + 2 = 1 and O be the centre. 2 a b xx1 yy Tangent at P (x1 , y1 ) is 2 + 21 − 1 = 0 a b b2x1 whose slope = − 2 a y1

2

=1

Focus is S (ae, 0). Equation of the line perpendicular to tangent at S is a2 y …(i) y = 2 1 (x − ae) b x1 y Equation of OP is y = 1 x …(ii) x1 Eqs. (i) and (ii) intersect. y1 a2 y x = 2 1 (x − ae) ∴ x1 b x1 ⇒

x (a2 − b2 ) = a3e

⇒

x ⋅ a2e2 = a3e a x= e

which is the corresponding directrix. Q (a cos θ, a sin θ) R (a cos θ, α)

⇒

Hence, locus of R is

⇒ Y

X′

h2 k 2 (r + s)2 =1 + a2 (ar + bs)2

Sol. Let the ellipse be

∴ m ∈ R , as m =

(c)

⇒

(ar + bs)sin θ r+ s h k (r + s) cos θ = , sin θ = a ar + bs k =α =

(a) tangent at vertex (b) corresponding directrix (c) tangent at (0, b) (d) None of the above

(a2 − b2 )m

y = mx −

and

α − b sin θ r = a sin θ − α s (ar + bs)sin θ α= r+ s

Let the coordinates of R be (h, k ). h = a cos θ

Hence, (b) is the correct answer.

Ex 6. A parabola is drawn whose focus is one of the x 2 y2 foci of the ellipse 2 + 2 = 1 (where, a > b) a b and whose directrix passes through the other focus and perpendicular to the major axes of the ellipse. Then, the eccentricity of the ellipse for which the latusrectum of the ellipse and the parabola are same, is (a) 2 − 1 (b) 2 2 + 1 (c) 2 + 1 (d) 2 2 − 1

2

∴

2

sec φ = 1 + tan 2 φ

x y + =1 a 2 b2 Equation of the parabola with focus S (ae, 0) and directrix x + ae = 0 is y2 = 4 aex 2b2 and a

Now, length of latusrectum of the ellipse is

= 1+

x 2 + y2 x2 sec φ cosec φ = tan φ

and

=

Y

= O S′ (–ae, 0)

ax

Y′

For the two latusrectums to be equal, 2b2 = 4 ae a 2a2 (1 − e2 ) ⇒ = 4 ae a ⇒ 1 − e2 = 2e 8

x 2 + y2 x 2 + y2 − by = a2 − b2 x y (a − b) x 2 + y2 = a2 − b2

⇒

x 2 + y2 = a + b

⇒

x 2 + y2 = (a + b)2

Ex 8. If the focal distance of an end of the minor axis of any ellipse (its axes as X and Y-axes respectively) is k and the distance between the foci is 2h, then its equation is (a)

= −1±

2

(b)

Hence, (a) is the correct answer.

Ex 7. Any ordinate NP of an ellipse meets the auxiliary circle at Q. Then, the locus of the intersection of the normal at P and O is the circle (a) x 2 + y 2 = a 2 (b) x + y = ( a + b ) 2

2

(c) (d)

x2 k2 x2 k2 x2 k2 x2 k2

+ + − +

y2 h2

=1 y2

k 2 − h2 y2 k 2 − h2 y2 k 2 + h2

=1 =1 =1

Sol. Let the equation of ellipse be

2

(c) x 2 + y 2 = b 2 (d) None of the above x2 y2 + 2 = 1, 2 a b therefore the equation of the auxiliary circle will be x 2 + y2 = a2.

Sol. Suppose the equation of the ellipse is

Let the coordinates of any point P be (a cos φ − b sin φ), then coordinates of Q will be (a cos φ , a sin φ). Normal at P is …(i) ax secφ − by cosec φ = a2 − b2 and normal at Q will be y = x tan φ y tan φ = ⇒ x

x 2 + y2 y

Hence, (b) is the correct answer.

e2 + 2e − 1 = 0

−2± Therefore, e= 2 So, e = 2 − 1, as 0 < e < 1.

x 2 + y2 x2 y/ x

On putting the values in Eq. (i), we get

X S (ae, 0)

⇒

⇒

15

=

that of the parabola is 4ae.

X′

y2 x2

Ellipse

Sol. Equation of the ellipse is

…(ii)

x2 y2 + 2 = 1 and e be the 2 a b

eccentricity of ellipse. Therefore, 2h = 2ae ...(i) ae = h Focal distance of one end of minor axis say (0, b) is h. Therefore, a + e(0) = k ...(ii) a=k From Eqs. (i) and (ii), b2 = a2 (1 − e2 ) ⇒

b2 = a2 − a2e2 = k 2 − h2

Therefore, equation of ellipse is x2 y2 + =1 k 2 k 2 − h2 Hence, (b) is the correct answer.

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Objective Mathematics Vol. 1

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Ex 9. If the normal at point P on the ellipse x 2 y2 + = 1 meets the axes in R and S a 2 b2 respectively, then PR : RS is equal to (a) a : b

(b) a 2 : b 2

(c) b 2 : a 2 − b 2

(d) b : a

Sol. Let P (a cosθ, b sin θ ) Equation of normal is ax sec θ − by cosec θ = a2 − b2

3k …(ii) b On squaring and adding Eqs. (i) and (ii), we get h2 9k 2 + 2 =1 a2 b Therefore, locus of R is x 2 9 y2 + 2 =1 a2 b Hence, (a) is the correct answer. ⇒

Ex 11. At a point P on the ellipse

Point of intersection with X-axis,   a2 − b 2 cos θ, 0 R≡   a

2

⇒

(b)

(b) (c)

PR : RS = b : (a − b )

(d)

Ex 10. Locus of the point which divides double x 2 y2 ordinate of the ellipse 2 + 2 = 1 in the ratio a b 1 : 2 internally, is (a)

(a)

2

Hence, (c) is the correct answer.

x2 a2 x2 2

a 9x 2

+

9y2

+

b2 9y2 2

b 9y2

b2

= 1, tangent

x2 a

2

x

2

a

2

x

2

a

2

x2 a

2

+ − + −

y2 b

2

y

2

b

2

y

2

b

2

y2 b

2

2 2

a b 1

=1 − = =

2 2

a b

1 2 2

a b 1

2 2

a b

The distance of the tangent from the origin is ab

p=

1 9

1

=1 +

x − a cos θ y − b sin θ = a sin θ − b cos θ

⇒

+ 2 =1 a2 b (d) None of the above (c)

y2

Sol. Equation of the tangent at P is

=1 =

a2

+

1 p from the point P, where p is distance of the tangent from the origin, then the locus of the point Q is

 a2 − b2     cos θ  + b2 sin 2 θ RP 2 = a cos θ −  a     b2 2 = 2 (b cos2 θ + a2 sin 2 θ ) a (a2 − b2 )2 2 2 and RS = (b cos2 θ + a2 sin 2 θ ) a2b2 Hence, PR 2 : RS 2 = b4 : (a2 − b2 )2 2

x2

PQ is drawn. If the point Q is at a distance

 a2 − b2     sin θ  and with Y-axis, S ≡ 0, −   b    Therefore,

2

sinθ =

b2 cos2 θ + a2 sin 2 θ b2 cos2 θ + a2 sin 2 θ ab

1 = p

Y

Q

1/p P (a cos θ, b sin θ)

Sol. Let P(a cosθ , b sin θ), Q (a cosθ , − b sin θ ) PR : RQ = 1 : 2

X′

X

O

Y P (a cos θ, b sin θ) R (h, k) X′

X

Y′

Q (a cos θ, – b sin θ) Y′

Therefore, ⇒ and

878

h = a cosθ h cosθ = a b k = sinθ 3

...(i)

Now, the coordinates of the point Q are given as follows x − a cos θ y − b sin θ − a sin θ b cos θ = 2 2 2 2 2 b cos θ + a sin θ b cos2 θ + a2 sin 2 θ =

1 = p

b2 cos2 θ + a2 sin 2 θ ab

x = a cos θ −

and 2

2

1  y  x ⇒   +   = 1 + 2 2 is the required locus.  b  a ab Hence, (a) is the correct answer.

Ex 12. A triangle is drawn such that it is right angled x 2 y2 at the centre of the ellipse 2 + 2 = 1 ( a > b) a b and its other two vertices lie on the ellipse with eccentric angles Then, α, β. + α β   1 − e 2 cos 2    2  is equal to 2 α + β cos    2  (a) (b) (c) (d)

a2 a +b 2

y = mx +

y2

15

a2m2 + b2

...(i)

S ′ R′ RS is a trapezium and its area 1 ∆ 1 = (SR + S ′ R′ ) × SD 2 Y B′

X′

A′ (–a, 0)

R′

S′ (–ae, 0)

P

R

S (ae, 0)

C

B

A (a, 0)

X

a2 + b2 a2 a2

Y′

Equation of the line S ′ R ′ is 1 y = − (x + ae) m ⇒ x + my + ae = 0

a2 − b2 a2 − b2 a2

extremities as α, β is x y  α − β  α + β  α + β cos    = cos   + sin   2   2   2  b a α+β Let =∆ 2 α −β and = δ, so that 2 x y cos ∆ + sin ∆ = cos δ a b As the triangle is right angled, homogenizing the equation of the curve, we get 2

x2 y2  x cos ∆ y sin ∆  + 2 − +  =0 2 b cos δ   a cos δ a b  1 cos2 ∆   1 sin 2 ∆   =0    2− 2 + − a cos2 δ   b2 b2 cos2 δ  a [Qcoefficient of x 2 + coefficient of y2 = 0] ⇒

+

Sol. Equation of tangent at P is

2

Sol. Equation of the chord with eccentric angles of the

⇒

x2

= 1 ( a > b) with foci a 2 b2 at S and S ′ and vertices at A and A′. A tangent is drawn at any point P on the ellipse and let R, R ′, B , B ′ respectively be the foot of the perpendiculars drawn from S , S ′ , A, A ′ on the tangent at P. Then, the ratio of the areas of the quadrilaterals S ′ R ′ RS and A ′ B ′ BA is (a) e : 2 (b) e : 3 (c) e :1 (d) e : 4

Ex 13. Given an ellipse

Ellipse

a sin θ ab b cos θ y = b sin θ + ab

⇒

Therefore,

SD = SR =

and

S ′ R′ =

⇒

cos2 δ( a2 + b2 ) = a2 (1 − e2 cos2 ∆ )

⇒

 α + β 1 − e2 cos2    2  a2 + b 2 =  α − β a2 cos2    2 

Hence, (b) is the correct answer.

ae + ae 1 + m2 aem +

a2m2 + b2 1 + m2

− aem +

a2m2 + b2

1 + m2

1 (S ′ R′ + SR ) × SD 2  a2m2 + b2   ⇒ ∆ 1 = 2ae   1 + m2    1 Area of A′ B ′ BA is ∆ 2 = ( A′ B′ + AB ) × AE 2 1 Equation of A′ B′ is y = − (x + a) m …(iii) ⇒ x + my + a = 0 ∆1 =

and

AB =

and

A′ B′ =

am +

⇒

a2m2 + b2 1 + m2

b2 (cos2 δ − cos2 ∆) + a2 (cos2 δ − sin 2 ∆ ) = 0

⇒ cos2 δ( a2 + b2 ) − b2 cos2 ∆ − a2 + a2 cos2 ∆ = 0

…(ii)

− am +

a2m2 + b2 1 + m2

1 ( A′ B′ + AB ) × AE 2  a2m2 + b2   ∆ 2 = 2a   1 + m2    ∆2 =

Hence, ∆ 1 : ∆ 2 = e : 1

879

Objective Mathematics Vol. 1

15

Aliter Taking point P at the end of minor axis. ⇒ ∆ 1 = 2ae × b = 2abe and ∆ 2 = 2a × b = 2ab ∆1 e = ∴ ∆2 1 Y R′

B′ X′

A′

R

B

S

S′

A

points such that its subtends a right angle at the origin, then the locus of the point of tangency, is (given eccentricity of both the ellipses are same) (a) 2x 2 + 2 y 2 = 2a 2 (b) 2x 2 + y 2 = a 2 2 2 2 (c) x + 2 y = a (d) x 2 + y 2 = 2a 2 Sol. Equation of first ellipse is x2 y2 + 2 =1 2 a b

X

Y A

Y′

R B

Hence, (c) is the correct answer. X′

Ex 14. The locus of chords of contact of perpendicular tangents to the ellipse

x y + 2 = 1 touch another fixed 2 a b

ellipse is

(a) (b) (c) (d)

x2 a2 x2 a2 x2 a4 x2 a2

+ + + −

y2 b2 y2 b2 y2 b4 y2 b2

X

O

Q

2

2

Y′

= = = =

1

(a 2 + b 2 ) 2

Since, the circle passes through the foci and the extremities of the minor axis. ∴ 2ae = 2b ⇒ a2e2 = b2 ⇒ e2 = 1 − e2 1 = Eccentricities of both the ellipses. ⇒e= 2 Now, equation of the auxiliary circle will be …(ii) x 2 + y2 = b2

( 3a 2 − b 2 )

Let the second ellipse be

( 2a 2 + b 2 ) 2 (a 2 − b 2 ) 1

Sol. We know that locus of the point of intersection of perpendicular tangents of the given ellipse is x 2 + y2 = a2 + b2 Any point on this circle can be taken as P ≡ ( a2 + b2 cos θ, a2 + b2 sin θ) The equation of the chord of contact of tangents from y x P is 2 a2 + b2 cos θ + 2 a2 + b2 sin θ = 1 b a Let this line be a tangent to the fixed ellipse x2 y2 + =1 A2 B2 x y cos θ + sin θ = 1, A B 2 b2 a where ,B= A= a2 + b2 a2 + b2 1 x2 y2 + 4 = 2 4 (a + b2 ) a b which is an ellipse.

x2

+

y2

= 1, ( a > b). a 2 b2 A circle is drawn which passes through the foci and the extremities of the minor axis of this ellipse and let this circle be the auxiliary circle for a second ellipse whose vertices are the foci of the given ellipse. If any tangent to the second ellipse meets its auxiliary circle at two

Ex 15. Given an ellipse

x2 y2 + = 1 whose major A2 B2

axis = 2ae Also, B 2 = A 2 1 − e2 ⇒ a2 2 2

B 2 = a2e2 1 − e2

1 2 a2 and A 2 = a2e2 ⇒ A 2 = 2 Therefore, equation of the second ellipse is x2 y2 + =1 a2 / 2 a2 / 2 2 a2 ...(iii) x 2 + 2 y2 = 2 Equation of the tangent at R (h, k ) to Eq. (iii) is a2 ...(iv) xh + 2 yk = 2 From Eqs. (ii) and (iv) 2 2 x 2 + y2 − b2  2 (xh + 2 yk  = 0  a  ⇒

B2 =

i.e. e =

4 b2 2 2 [ x h + 2 y2k 2 + 2 2khxy ] = 0 a4 Since, AB subtends a right angle at the origin. 4 b2h2 4 b2 ∴ 1− + 1 − 4 2k 2 = 0 4 a a   a2 a4 a2 2 2 ⇒ h + 2k = 2 ⇒ h2 + 2k 2 = ⋅ 2 Q 2 = 2 2 b 2b   ⇒ h2 + 2k 2 = a2 Hence, the locus of (h, k ) is x 2 + 2 y2 = a2. Hence, (c) is the correct answer. ⇒ x 2 + y2 −

Hence, (c) is the correct answer.

880

…(i)

 x2 y2  (a) ( a − b )  2 − 2  b  a 2

2

2 2

If it passes through (α, β).

 x2 y2  = ( a 2 + b 2 ) 2 ( x 2 + y 2 )  2 + 2  b  a

2

22

x y (b) ( a 2 + b 2 ) 2  2 + 2  b  a 2

 x2 y2  = ( a 2 − b 2 ) 2 ( x 2 + y 2 )  2 − 2  b  a  x2 y2  (c) ( a − b )  2 − 2  b  a 2

15

∴ Eq. (i) can be written as bm bm ax ⋅ − by = (a2 − b2 ) ⋅ 2 2 a b m + a2 bm ⇒ bmx − by = (a2 − b2 ) ⋅ 2 2 b m + a2

Ellipse

Ex 16. The locus of the intersection of two normal to x 2 y2 are the ellipse + = 1 which a 2 b2 perpendicular to each other, is

2

2

2

 x2 y2  = ( a 2 + b 2 ) 2 ( x 2 − y 2 )  2 + 2  b  a (d) None of the above

2

Sol. Equation of normal to the ellipse is ax sec θ − by cosec θ = a2 − b2 ⇒ ax tan θ − by = (a2 − b2 ) sin θ a Its slope = tanθ = m (say) b bm tanθ = ⇒ a

…(i)

∴ b2 α 2m4 − 2b2αβm3 + {a2α 2 + b2β 2 − (a2 − b2 )2} − 2a2 αβm + a2 β 2 = 0 Let m1 , m2, m3 and m4 be four roots of this equation. 2β …(i) m1 + m2 + m3 + m4 = α m1m2 + m1m3 + m1m4 + m2m3 + m2m4 + m3m4 a2α 2 + b2β 2 − (a2 − b2 )2 …(ii) = b2α 2 2a2β m1m2m3 + m1m3m4 + m2m3m4 + m1m2m4 = 2 …(iii) bα a2β 2 …(iv) m1m2m3m4 = 2 2 bα Also, …(v) m1m2 = − 1 Eliminating m1 , m2 , m3 and m4 from Eqs. (i), (ii), (iii), (iv) and (v), we get (a2 − b2 )2 (b2α 2 − a2β 2 )2 = (a2 + b2 )2 (α 2 + β 2 ) (b2α 2 + a2β 2 )2 ∴ Required locus is 2

 x2  x 2 y2  y2  (a2 − b2 )2  2 − 2  = (a2 + b2 )2 (x 2 + y2 )  2 + 2  b  b  a a

2

Hence, (a) is the correct answer.

Type 2. More than One Correct Option Ex 17. If the chord through the point whose eccentric x 2 y2 angles are θ and φ on the ellipse, 2 + 2 = 1 a b passes through the focus, then the value of tan (θ / 2) tan (φ / 2) is (a)

e+1 e−1

(b)

e−1 e+1

(c)

1+ e 1− e

(d)

Sol. The equation of the line joining θ and φ.

1− e 1+ e

x y  θ − φ  θ + φ  θ + φ cos    = cos   + sin   2   2   2  b a

If it passes through the point (ae, 0). ae  θ − φ  θ + φ cos    = cos   2    2 a  θ − φ cos    2  e=  θ + φ cos    2   θ + φ  θ − φ cos    + cos   2   2  e+1 =  θ − φ  θ + φ e−1 cos   − cos    2   2 

2 cos θ / 2 cos φ / 2 2 sin φ / 2 sin θ / 2 θ φ e−1 tan tan = 2 2 e+1 =

⇒

If it passes through the point (−ae, 0), then φ θ e+1 tan tan = 2 2 e−1 Hence, (a) and (b) are the correct answers.

Ex 18. Identify the statements which are true. (a) The equation of the director circle of the ellipse, 5x 2 + 9 y 2 = 45 is x 2 + y 2 = 14 (b) The sum of the focal distance of the point y2 x2 (0, 6) on the ellipse + = 1 is10 25 36 (c) The point of intersection of any tangent to a parabola and the perpendicular to it from the focus lies on the tangent at the vertex (d) The line through focus and ( at 12 , 2at 1 ) on y 2 = 4ax , meelts it again in the point ( at 22 , 2at 2 ), if t 1 t 2 = − 1 881

Objective Mathematics Vol. 1

15

x2 y2 + = 1 is x 2 + y2 = 14 9 5 (b) Using focal property PS + PS ′ = 2 for a < b i.e. | PS + PS ′ | = 12 (c) Property (d) Slope of focal chord joining (at12 , 2at1 ) and (at22 , 2at2 ) is equal to slope joining (at12 , 2at ) and (a, 0). 2 2at1 i.e. = ⇒ t1t2 = − 1 t1 + t2 a(t12 − 1)

Sol. (a) Director circle of

Hence, (a), (c) and (d) are the correct answers.

Ex 19. The parametric angle θ, where − π ≤ θ ≤ π, of x 2 y2 the point on the ellipse 2 + 2 = 1 at which a b the tangent drawn cuts the intercept of minimum length on the coordinate axes, is/are (a) tan −1

b a

(c) π − tan −1

(b) − tan −1 b a

b a

(d) π + tan −1

b a

x cos θ y sin θ + =1 Sol. Equation of tangent at θ is a b x y ⇒ + =1 a sec θ b cosec θ ∴

i.e. ∴

d 2 = a2 sec2 θ + b2 cosec2 θ

⇒

d (d 2 ) sin θ cosθ = 2a2 − 2b2 3 = 0 dθ cos3 θ sin θ

tan 4 θ =

b2 a2

⇒

tan θ = ±

b a

b b , − tan −1 a a b b θ = π − tan −1 , − π + tan −1 a a Hence, (a), (b) and (c) are the correct answers. θ = tan −1

Ex 20. Extremities of the latusrectum of the ellipse x 2 y2 + = 1, ( a > b) having a given major axis a 2 b2 2a, lies on (a) x 2 = a( a − y ) (c) y 2 = a( a + x ) Sol. h = ± ae, k = ±

(b) x 2 = a ( a + y ) (d) y 2 = a ( a − x )

b2 a

  h2  h2  k = ± a (1 − e2 ) = ± a 1 − 2  = ±  a −  a a    On taking +ve sign, we get h2 h2 =a−k k =a− ⇒ a a ⇒ h2 = a(a − k ) On taking −ve sign, we get

Intercept, d = a2 sec2 θ + b2 cosec2 θ

∴

a2 sin θ b2 cosθ = cos3 θ sin 3 θ

h2 a h2 = a(a + k ) k = −a+

⇒

Hence, (a) and (b) are the correct answers.

Type 3. Assertion and Reason Directions (Ex. Nos. 21-25) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 21. Statement I Locus of centre of a variable circle touching two circles ( x − 1) 2 + ( y − 2) 2 = 25 and ( x + 2) 2 + ( y − 1) 2 = 16 is an ellipse.

882

Statement II If a circle S 1 = 0 lies completely inside the circle S 2 = 0, then locus of centre of variable circle S = 0 which touches both the circles is an ellipse.

Sol. LetC 1,C 2 be the centres and R1, R2 be the radii of the two circles. Let S 1 = 0 lies completely inside in the circle S 2 = 0. Let C and r be the centre and radius of the variable circle. Then, CC 2 = R2 − r and C 1C = R1 + r [constant] ∴ C 1C + C 2 C = R1 + R2 ∴ Locus of C is an ellipse. ∴ S 2 is true. Statement I is false (two circles are intersecting). Hence, (d) is the correct answer.

3 3  Ex 22. Statement I If P  , 1 is a point on the  2  2 2 ellipse 4x + 9 y = 36. Circle drawn AP as diameter touches another circle x 2 + y 2 = 9, where A ≡ ( − 5, 0). Statement II Circle drawn with focal radius as diameter touches the auxiliary circle.

Ex 23. Statement I Foot of the perpendicular drawn from foci of an ellipse 4x 2 + y 2 = 16 on the line 2 3x + y = 8 lie on the circle x 2 + y 2 = 16. Statement II If perpendicular are drawn from foci of an ellipse to its any tangent, then foot of these perpendicular lie on director circle of the ellipse. Sol. 4 x 2 + (8 − 2 3x )2 = 16 i.e. x − 2 3x + 3 = 0 i.e. (x − 3 )2 = 0 2

∴ 2 3x + y = 8 is a tangent to the ellipse, the auxiliary circle is x 2 + y2 = 16. ∴ Statement I is true and Statement II is false. Hence, (c) is the correct answer.

Ex 24. Statement I In a ∆ABC, if base BC is fixed and perimeter of the triangle is constant, then vertex A moves on an ellipse.

Statement II If sum of distance of a point P from two fixed points is constant, then locus of P is a real ellipse. Sol. Statement II is false (locus of P may be a line segment

15 Ellipse

x2 y2 + =1 9 4 ∴ Auxiliary circle is x 2 + y2 = 9 and (− 5 , 0) and ( 5 ,0) are foci. ∴Statement I is true. Statement II is also true. Hence, (a) is the correct answer.

Sol. The ellipse is

also). Statement I is true. Hence, (c) is the correct answer.

Ex 25. Statement I Let tangent at a point P on the ellipse, which is not an extremity of major axis, meets a directrix at T. If circle drawn on PT as diameter cuts the directrix at Q, then PQ = ePS , where S is the focus corresponding to the direcrix. Statement II Let tangent at a point P on a ellipse, which is not an extremity of major axis, meets the directrices at T ′ and T. Then, PT subtends a right angle at the focus corresponding the directrix at which T lies. Sol. Statement I is false. Since, ∠PQT = π /2 ∴ PQ perpendicular to the directrix. ∴ By definition, PS = ePQ ∴ Statement I is false. Hence, (d) is the correct answer.

Type 4. Linked Comprehension Based Questions Passage I (Ex. Nos. 26-28) An ellipse E has its centre C (1, 3) focus at S (6, 3) and passing through the point P(4, 7).

Sol. (Ex. Nos. 26-28) (x − 1)2 ( y − 3)2 + =1 45 20 PS + PS ′ = 2a PS = 64 + 16 = 4 5

Equation of ellipse is

Ex 26. The product of the lengths of the perpendicular segments from the foci on tangent at point P is (a) 20 (b) 45 (c) 40 (d) Cannot be determined

PS′ = 4 + 16 = 2 5 2a = 6 5 ⇒ a = 3 5 ⇒

Ex 27. The point of intersection of the lines joining each focus to the foot of the perpendicular from the other focus upon the tangent at point P, is 5  (a)  , 5 3  8  (c)  , 3 3 

(a) e′ = (c) e′ =

P

3 10 3 5

(b) e′ = 10 (d) e′ = 3

5 3

S G (6, 3)

S′ C (– 4, 3) (1, 3)

4  (b)  , 3 3   10  (d)  , 5 3 

Ex 28. If the normal at a variable on the ellipse (E) meets its axes in Q and R, then the locus of the mid-point of QR is a conic with an eccentricity ( e′ ), then

ae = 5 e×3 5 =5

∴ Q ⇒ ∴

e=

5 3

b2 a2 2 5 4 b =1− = 9 9 a2 4 2 b = ×9×5 9 b=2 5 e2 = 1 −

883

Objective Mathematics Vol. 1

15

26. Q Product of the length of the perpendicular segments

b2 = a2 − a2e2 = a2 − 36

…(iii)

from the foci on tangent at P(4 , 7) is b2 = 20

From Eq. (ii), we get 25b2 = 36 + 12a + a2

Hence, (a) is the correct answer.

⇒25(a2 − 36) = 36 + a2 + 12a

perpendicular from the other focus upon the tangent at P(4 ,7) meet the normal PG and bisects it. ∴ Required point is mid-point of PG. Now, equation of normal at P(4, 7) is given by 3x − y − 5 = 0 8  ∴ G  , 3 3   10  ∴Required point is mid-point of PG i.e.  , 5 .  3 

28. Q Locus of mid-point of QR is another ellipse having the same eccentricity as the ellipse (e). 5 ⇒ e′ = e = 3 Hence, (b) is the correct answer.

Passage II (Ex. Nos. 29-31) Consider an ellipse (E) x2 y2 + = 1 ,centerd at point Q and having AB and CD a2 b 2 as its major and minor axes respectively, if S1 is one of the focus of the ellipse radius of incircle of ∆OCS1, be 1 unit and OS1 = 6 units, then

+ + + +

y2 y2 y2 y2

∴ ⇒

29. Area of ellipse = πab 65π sq units 4 Hence, (a) is the correct answer. =

30. Q Perimeter of ∆OCS 1 =6+ a+ b 13 15 =6+ + 2 2 = 15 units Hence, (c) is the correct answer.

31. Q S : x 2 + y2 = a2 + b2 97 = r2 2 ∴ Equation of director circle of S is x 2 + y2 = 2r2 ⇒ S : x 2 + y2 =

equation ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 a h g represents an ellipse, if h b f ≠ 0 and h 2 < ab. g

(b) 10 units (d) 25 units

= ( 48.5) = 97 = 97 = 48.5

f

c

Intersection of major axis and minor axis gives centre of ellipse.

Ex 32. There are exactly n integral values of λ for which equation x 2 + λxy + y 2 = 1 represents an ellipse, then n must be (a) 0 (b) 1 (c) 2

(d) 3

Sol. Using conditions for ellipse,

Sol. (Ex. Nos. 29-31) OS 1 = ae = 6, OC = b Also, CS 1 = a 1 Q Area of ∆OCS 1 = (OS 1 ) × (OC ) = 3b 2 Q Semi-perimeter of ∆OCS 1 1 1 = (OS 1 + OC + CS 1 ) = (6 + a + b) 2 2 ∴Inradius of ∆OCS 1 = 1 3b ⇒ = 1 ⇒ 5b = 6 + a 1 (6 + a + b) 2

x 2 + y2 = 97

Passage III (Ex. Nos. 32-34) Second degree

Ex 31. If S is the director circle of ellipse (E), then the equation of director circle of S is (a) x 2 (b) x 2 (c) x 2 (d) x 2

2a − a − 78 = 0 13 a= ,−6 2 13 a= 2 5 b= 2

Hence, (c) is the correct answer.

64π (b) sq units 5 (d) 65π sq units

Ex 30. Perimeter of ∆OCS is (a) 20 units (c) 15 units

⇒

⇒

Ex 29. The area of ellipse (E) is 65π sq units (a) 4 (c) 64π sq units

[from Eq. (iii)]

2

27. Q Lines joining each focus to the foot of the

Hence, (d) is the correct answer.

884

Also,

[let]

− 2 < λ < 2 ⇒ λ = − 1, 0, 1 Hence, (d) is the correct answer.

Ex 33. Length of the longest chord of the ellipse x 2 + y 2 + xy = 1 is …(i)

…(ii)

1

(a) 2

(b)

(c) 2 2

(d) 1

2 Sol. Consider, f = x 2 + xy + y2 − 1 = 0 Hence, (c) is the correct answer.

(a)

1

(b)

2

3 2

(c) 2

2 3

(d)

1 3

∂f ∂f centre of ellipse comes out to be =0= Sol. Solving ∂x ∂y (0, 0). Longest chord of ellipse is major axis. While chord perpendicular to it and through centre of ellipse is minor axis, so it is the smallest chord. Let P be a point

on ellipse such that OP = r and ∠POX = θ, then as P lies on ellipse P (r cosθ , r sin θ ) r2 cos2 θ + r2 sin θ cosθ + r2 sin 2 θ − 1 = 0 1 1 ⇒ r2 = = 1 + sin θ cosθ 1 + 1/ 2 sin 2θ

15 Ellipse

Ex 34. Length of the chord perpendicular to longest chord as in above question and passing through centre of ellipse is

∴

r is smallest, if sin 2θ is greatest. 2 2 i.e. ⇒ r= sin 2θ = 1 ∴ r2 = 3 3 2 Now, length of required chord = 2 3 Hence, (c) is the correct answer.

Type 5. Match the Columns Ex 35. Match the statements of Column I with values of Column II. Column I

Column II

p. x2 y2 + = 1 having 27 48 4 slope − cuts the X and Y-axes at the points A 3 and B, respectively. If O is the origin, then area of ∆OAB is equal to

A. A tangent to the ellipse

B. Product of the perpendiculars drawn from the q. points ( ± 3, 0) to the line y = mx − 25m2 + 16 is

C. 2a = 5 + 10 = 15, a = ⇒

36

10 2

∴

s. x2 y2 + = 1, 25 16 which passes through S ≡ ( 3, 0) and PS = 2, then length of chord PQ is

16

⇒ Hence,

Sol. A. Let ( 27 cosθ, 48 sin θ) be a point on the ellipse. Thus, equation of tangent of their points is 27 cosθx 48 sin θy + =1 27 48 − cos θ 48 Slope = × ∴ sin θ 27 4 = − cotθ 3 4 =− 3 i.e. cotθ = 1 π i.e. θ= 4 x y ∴ Equation of the tangent is + =1 54 96 1 Hence, area of triangle = ⋅ 3 6 × 4 6 = 36 2 B. Product of perpendicular = =

3m − 25m2 + 16 −3m − 25m2 + 16 1 + m2 25m2 + 16 − 9m2 = 16 1 + m2

1 + m2

x = a/e

S (3,0)

10

D. If PQ is focal chord of ellipse

2ae = (6 − 3)2 + (8 − 4 )2 = 5 1 e= 3 (x, y)

r.

C. An ellipse passing through the origin has its foci ( 3, 4) and ( 6, 8), then length of its minor axis is

15 2

225  1 1 −  = 50 4  9 b=5 2 2b = 10 2

b2 =

D. a = 5, b = 4, e =

3 5

∴ ae = 3 Q SA = 2 Also, SP = 2 Since, P coincides with A and Q coincides with A′. ∴ PQ = 2a = 10 A → p; B → s; C → q; D → r

Ex 36. Match the statements of Column I with values of Column II. Column I

Column II

A. A stick of length 10 m slides on coordinate axes, p. then locus of a point dividing this stick reckoning from X-axis in the ratio 6 : 4 is a curve whose eccentricity is e, then 9 e is equal to

6

B. AA′ is major axis of an ellipse q. 3 x 2 + 2 y 2 + 6 x − 4 y − 1 = 0 and P is a variable point on it, then the greatest area of ∆APA′ is

2 7

C. Distance between foci of the curve represented by the equation x = 1 + 4cosθ, y = 2 + 3sinθ is

r.

128 3

x2 y2 + = 1 at 16 7

s.

3 5

D. Tangents are drawn to the ellipse

end points of latusrectum. The area of equilateral, so formed is

885

15

Sol. A. x = 10cos θ − 6 cos θ = 4 cos θ

Ex 37. Match the statements of Column I with values of Column II.

y = 6sinθ

2

2

Objective Mathematics Vol. 1

 y  x ∴ Locus is   +   = 1  6  4 2

Column I

x y + =1 16 36

⇒

36 − 16 36 20 5 = = 36 3 9e = 3 5

A.

If the mid-point of a chord of the ellipse p. x2 y2 + = 1 is ( 0, 3), then length of the chord is 16 25 4k , then k is 5

6

B.

If the line y = x + λ touches the ellipse q. 9 x 2 + 16 y 2 = 144, then the sum of values of λ is

8

C.

If the distance between a focus and corresponding directrix of an ellipse be 8 and 1 the eccentricity is , then length of the minor 2 k , where k is axis is 3

r.

0

D.

Sum of the distances of a point on the ellipse x2 y2 + = 1 from the foci, is 9 16

s.

16

e=

∴

∴

B. 3(x 2 + 2x + 1) + 2( y2 − 2 y + 1) = 3 + 2 + 1 (x + 1)2 ( y − 1)2 + =1 2 3 3−2 1 = 3 3 a = 3, b = 2 1 Area = ⋅ 2 3 ⋅ 2 = 6 2 e=

∴ C.

(x − 1)2 ( y − 2)2 + =1 16 9

Sol. A. Equation of the chord whose mid-point is (0, 3), is 3y 9 −1= − 1 i.e. y = 3 25 25 x2 y2 If it intersects the ellipse + =1 16 25 16 9 x2 At = =1− 16 25 25 16 x=± ⇒ 5 32 ∴ Length of the chord = 5 4 k 32 Thus, = 5 5 ∴ k =8 B. If the line y = x + λ touches the ellipse 9x 2 + 16 y2 = 144, then

7 7 a = 4e = = 16 4 ae = 7

Q ∴

Hence, distance between the foci is 2 7. D.

x2 y2 + =1 16 7 =

9 3 = 16 4

One end of latusrecltum is  b2   7   ae,  =  3,  a   4  ∴ Equation of the tangent is 3x 7 y + ⋅ =1 16 4 7

λ2 = 16(1)2 + 9

Y

⇒ a C. − ae = 8 e X′

O

X

Y′

3x y + =1 ⇒ 16 4  16  It meets X -axis at  , 0 and Y-axis at (0, 4 ).  3  16 128 ×4= ∴ Area of rhombus = 2 × 3 3 A → s; B → p; C → q; D → r

886

Column II

2

λ=±5

a =8 2 16 ⇒ a= 3 Now, b2 = a2 (1 − e2 ) 256  1  64 = 1 −  = 9  4 3 16 k = ∴ Length of minor axis = 2b = 3 3 Hence, k = 16 D. Sum of the distances of a point on the ellipse from the foci 2a = 8 A → q; B → r; C → s; D → q ⇒

2a −

Ex 38. A straight line PQ touches the ellipse x 2 y2 and the circle + =1 16 9 x 2 + y 2 = r 2 (3 < r < 4). RS is a focal chord of the ellipse which is parallel to PQ and meets the circle at points R and S. Then, the length of RS is ______. Sol. (6) y = mx + a2m2 + b2 is tangent to the ellipse 2

Sol. (4) 5x 2 + 9 y2 = 45 2 a = 3, b = 5, e = , one end of latusrectum in first 3  5 quadrant =  2,   3 Y E2

(i)

E1

X′

⇒

0− 0 +

±r =

am +b 2

1+ m

m=

2

[Q b < r < a ]

Q

R

(b, 0)

(0, –c)

P

X

S′ S

E2

(0, 0)

X′

(ii) (–a, 0)

2

r 2 − b2 a2 − r 2

(0, 0) O

(0, b) (0, c)

2

Y

X′

X

2

x y + =1 a 2 b2 This tangent also touches the circle x 2 + y2 = r2 ⇒

Ellipse

15

Type 6. Single Integer Answer Type Questions

E1

(a, 0)

X

Y′

Equation of tangent, 2x + 3 y = 9 9  It meets axes at  , 0 and (0, 3). 2  Area of the quadrilateral 1 9 = 4 ⋅ ⋅ ⋅ 3 = 27 2 2 Let the equation of the ellipse E2 and E1 be

x2 y2 + 2 =1 2 a b

x2 y2 + 2 =1 2 b c Fig. (i) and (ii) give the same area which is required area of E2 = πab and area of E1 = πbc Required area = πb(a − c) Now, b2 = a2 (1 − e2 ) and

Y′

RS passes through (ae, 0) and parallel to PQ. Equation of RS is y − 0 = m(x − ae) mx − y − ame = 0 Let T be foot of the perpendicular dropped from origin on RS. Then, RT 2 = OR 2 − OT 2 ⇒ ⇒ ∴ ⇒

RT 2 = r2 − RT = b RS = 2b RS = 6

and

c2 = a2 (1 − e2 )2

⇒

c = a(1 − e2 )

⇒

a2m2e2 1 + m2

a − c = ae2

∴Required area = πabe2 = 9 × [Qπab =

[Q b = 3, given]

Ex 39. Let E1 and E 2 be two ellipse. The area of the ellipse E 2 is one-third the area of the quadrilateral formed by the tangents at the ends of the latusrectum of the ellipse E1 ( E1 ; 5x 2 + 9 y 2 = 45). The eccentricities of E1 and E 2 are equal. E1 is inscribed in E 2 in such a way that both E1 and E 2 touch each other at one end of their common major axis. If the length of the major axis of E1 is equal to the length of the minor axis of E 2 , then find the area of the ellipse E 2 outside the ellipse E1 .

4 = 4 sq units 9

1 × area of quadrilateral] 3

Ex 40. P1 , P2 , …, Pi ,…, Pn are the points on the x 2 y2 ellipse + = 1 and Q1 , Q2 , …, Qi , …Qn , 16 9 are the corresponding points on the auxiliary circle of the ellipse. It the line joining C and Qi meets the normal at Pi w.r.t. the given ellipse at n

K i and

∑ (CK i ) = 175,

then find the value

i =1

n of . 5

887

Objective Mathematics Vol. 1

15

Sol. (5) Equation of normal at Pi is through K i

4CK i − 3CK i = 7 CK i = 7 n

i=1

⇒ ⇒

7n = 175 n = 25 n 25 = =5 5 5

∴

Ex 41. If the normals at the four points (x1 , y1 ), ( x 2 , y2 ), ( x 3 , y3 ) and ( x 4 , y4 ) on the ellipse x 2 y2 + = 1 are concurrent, then find the value a 2 b2 1 1 1 1 of ( x1 + x 2 + x 3 + x 4 ) ×  + + + .  x1 x 2 x 3 x 4 

Ex 42. The equation of the common tangent, Ist quadrant to the circle x 2 + y 2 = 16 and the x 2 y2 ellipse + = 1, makes intercept between 25 4 the coordinate axes. Then, find the value of k 14 . for which length of this intercept is k Sol. (3) Let the common tangent to circle x 2 + y2 = 16 and ellipse

Sol. (4) Let points of concurrent be (h, k ). Equation of normal at (x′ , y′ ) is

x − x ′ yi − y ′ = x′ y′ a2 b2

It passes through (h, k ), then y′ 2{a2 (h − x′ ) + b2x′ }2 = b4k 2x′ 2 But

b2 x′ 2 y′ 2 + 2 = 1or y′ 2 = 2 (a2 − x′ 2 ) 2 a a b

2

25m2 + 4 m2 + 1

…(i)

⇒

−2a4h(a2 − b2 ) x′ + a6h2 = 0 Above the equation being of fourth degree in x′, therefore roots of the above equation are x1, x2, x3, x4 , then 2ha2 (a2 − b2 ) 2ha2 = 2 (x1 + x2 + x3 + x4 ) = − 2 2 2 − (a − b ) (a − b2 ) …(iii)

=4

…(ii)

Q length of perpendicular   from (0, 0) to Eq. (i) = 4   

…(ii)

Arranging above as a fourth degree equation in x′, we get − (a2 − b2 )2 x′ 4 + 2ha2 (a2 − b2 )x′ 3 + x′ 2 (...)

888

…(i)

As it is tangent circle x + y = 16 we should have

b 2 (a − x′ 2 ){a2h + (b2 − a2 )x′ }2 = b4k 2x′ 2 a2 b4k 2x′ 2

1 Σx1x2x3 1 1 1 + +  =  +  x1 x2 x3 x4  (x1 ⋅ x2 ⋅ x3 ⋅ x4 )

25m2 + 4 2

2

⇒

x2 y2 + = 1 be 25 4 y = mx +

Value of y′ 2 from Eq. (ii), putting in Eq. (i), we get

…(iv)

On multiplying Eqs. (iii) and (iv), we get 1 1 1 1 (x1 + x2 + x3 + x4 ) ×  + + +  =4  x1 x2 x3 x4 

∑ 7 = 175

∴

2a4h(a2 − b2 ) 2(a2 − b2 ) − (a2 − b2 )2 = = 6 2 ah a2h − (a2 − b2 )2

4x 3y − = 7, it passes cosθ sin θ

⇒

25m2 + 4 = 16m2 + 16 9m2 = 12

2 p 3 [leaving +ve sign, to consider tangent in Ist quadrant] ∴ Equation of common tangent is 4 7 2 2 y=− x + 25 ⋅ + 4 ⇒ y = − x+4 3 3 3 3 ⇒

m=−

This tangent meets the axes at A(2 7 , 0) and  7 B  0, 4  . 3  ∴ Length of intercepted portion of tangent between axes, 2

 7 14 ⇒ k =3 AB = (2 7 )2 +  4  = 3  3

Target Exercises Type 1. Only One Correct Option x2

y2

+

= 1 will

r − r − 6 r − 6r + 5 ellipse, if r lies in the interval 2

(a) (− ∞ , 2)

2

(b) (3, ∞ )

represents

(c) (5, ∞ )

2. Length of latusrectum of the ellipse 2a2 , if a < b b 2 2b (c) , if a < b a

(b)

(a)

the

(d) (1, ∞ )

x2 a2

+

y2 b2

= 1, is

2a2 , if a > b b

(d) None of these

x2

+

y2 = 1 is inscribed in a 13 − 5a

a2 − 7 square of side length 2a, then a is equal to

13  (b) (− ∞ , − 7 ) ∪  7 ,   5

6 (a) 5

 13  (c) (− ∞ , − 7 ) ∪  , 7 (d) No such a exists  5 

5. A parabola is drawn with focus is at one of the foci of y2 x2 the ellipse 2 + 2 = 1 (where, a > b) and directrix a b passing through the other focus and perpendicular to the major axes of the ellipse. If latusrectum of the ellipse and the parabola are same, then the eccentricity of the ellipse is 1 2 (b) 2 2 − 2 (c) 2 − 1 (d) None of the above (a) 1 −

x2

(a) PS + PS ′ = 2α , if α > β (b) PS + PS ′ = 2β, if α > β α β 1+ e (c) tan tan = 2 2 1− e a2 − b2 b2

(a − a2 − b2 )

+

y2

+

 a2 + b2  (a) 4    7   a2 + b 2  (c) 12   5 

(d) None of these

y2

 a2 + b 2  (b) 4    3   a2 + b 2  (d) 8   5 

9. A circle has the same centre as an ellipse and passes through the foci F1 and F2 of the ellipse, such that the two curves intersect in 4 points. Let P be any one of their points of intersection. If the major axis of the ellipse is 17 and the area of the ∆PF1 F2 is 30, then the distance between the foci is (a) 13 (b) 10 (c) 11 (d) None of the above

10. If α and β are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is cosα + cosβ cos (α − β ) cosα − cosβ (c) cos (α − β )

= 1, a2 b2 whose foci are S and S ′. Let ∠PSS ′ = α and ∠PS ′ S = β, then which of the following is/are true?

α β tan = 2 2

x2

+ β  2  − β  2 

= 1 is inscribed in a rectangle a2 b2 whose length to breadth ratio is 2 : 1, then the area of the rectangle is

(a)

6. If P is any point lying on the ellipse

(d) tan

e−1 α β = tan tan e+1 2 2

8. If the ellipse

(a) the arithmetic mean of the segments of its focal chord (b) the geometric mean of the segments of its focal chord (c) the harmonic mean of the segments of its focal chord (d) None of the above

α cos   (b) e = α cos  

sin α − sin β (a) e = sin (α − β ) (c)

3. The semi-latusrectum of an ellipse is

4. If the ellipse

7. If ( a cos α , b sin α ), ( a cos β , b sin β ) are the end points of a focal chord, then which of the following is correct?

Targ e t E x e rc is e s

1.

sin α − sin β sin (α − β ) sin α + sin β (d) sin (α + β ) (b)

11. Three points A, B and C are taken on the ellipse y2 x2 + = 1 with eccentric angles θ, θ + α and a2 b2 θ + 2α, then the (a) area of ∆ABC is independent of θ (b) area of ∆ABC is independent of α 3 (c) maximum value of area is ab 4 3 3 (d) maximum value of area is ab 4

889

Objective Mathematics Vol. 1

15

12. The equation of the line passing through the centre and bisecting the chord 7x + y − 1 = 0 of the ellipse y2 x2 + = 1 is 1 7 (a) x = y (c) x = 2 y

(b) 2x = y (d) x + y = 0

x2

 a − b (a) tan −1    2 ab   2ab  (c) tan −1    a − b

+

y2

= 1( a > b ) a2 b2 and the circle x 2 + y 2 = a 2 at the points, where a common ordinate cuts them (on the same side of the X-axis). Then, the greatest acute angle between these tangents is given by

13. Tangents are drawn to the ellipse

 a + b (b) tan −1    2 ab   2ab  (d) tan −1    a + b

Ta rg e t E x e rc is e s

the

value

of

i=1

1 5 3 (c) 5

2 5 4 (d) 5 (b)

(c) 1

(d) 4

16. Let (α , β) be a point from which two perpendicular tangents can be drawn to the ellipse 4x 2 + 5 y 2 = 20. If F = 4α + 3β, then (a) − 15 ≤ F ≤ 15 (c) − 5 ≤ F ≤ 20

(b) F ≥ 0 (d) F ≤ − 5 5 or F ≥ 5 5

17. The point at shortest distance from the line x + y = 7 and lying on an ellipse x 2 + 2 y 2 = 6, has coordinates (a) ( 2 , 2 ) (c) (2, 1)

(b) (0, 3 ) 1  (d)  5 ,   2

18. If equation of the ellipse is 2x 2 + 3 y 2 − 8x + 6 y + 5 = 0, then which of the following is true?

890

(b) ± 6

(c) ± 8

(d) ± 132

y2 x2 + = 1. Then, angle subtended by QR 4 9

at origin is  2 6 (a) tan −1    65   8 2 (c) tan −1    65 

(a) x − 2 y + 4 = 0 (c) x − 2 y − 2 = 0

y2 x2 + = 1 from 50 20 which pair of perpendicular tangents are drawn to the y2 x2 ellipse + = 1 is 16 9 (b) 2

(a) ± 4

 4 6 (b) tan −1    65  (d) None of these

22. Equation of the tangent to the ellipse

15. Number of points on the ellipse

(a) 0

y2 x2 + = 1, then c will be equal to 8 4

y2 x2 + =1 4 3 perpendicular to the line 2x + y + 7 = 0, is

eccentricity is (a)

20. If the straight line y = 4x + c is a tangent to the ellipse

the ellipse

10

then

(a) (x 2 + y2 − a2 − b2 )2 = 4 (a2x 2 + b2 y2 − 1) (b) (x 2 + y2 − a2 − b2 )2 = 4 (b2x 2 + a2 y2 − 1) (c) (x 2 + y2 − a2 − b2 )2 = (b2x 2 + a2 y2 − 1) (d) None of the above

21. From point P( 8, 27) tangents PQ and PR are drawn to

14. Let Pi and Pi′ be the foot of the perpendiculars drawn from foci S, S ′ on a tangent Ti to an ellipse whose length of semi-major axis is 20, if

∑ ( SPi )( S ′ Pi′ ) = 2560,

19. Locus of all such points from where the tangents y2 x2 drawn to the ellipse 2 + 2 = 1 are always inclined a b at 45°, is

(a) Equation of director circle is x 2 + y2 − 4 x + 2 y = 10 (b) Director circle will pass through (4 , − 1) (c) Equation of auxiliary circle is x 2 + y2 − 4 x + 2 y + 2 = 0 (d) None of the above

(b) x − 2 y + 5 = 0 (d) x − 2 y − 5 = 0

y2 x2 + =1 9 16 which are parallel to the line x + y + 1 = 0, is

23. Equation of the tangents to the ellipse (a) x + y + 5 = 0 (c) 2x + y − 5 = 0

(b) x + y + 6 = 0 (d) x + y − 6 = 0

24. If from a point P tangents PQ and PR are drawn to the x2 + y 2 = 1, so that equation of QR is 2 x + 3 y = 1, then coordinates of P is ellipse

(a) (2, − 3) (b) (2, 3) (c) (− 2, 3) (d) None of the above

25. The angle between pair of tangents drawn to the ellipse 3x 2 + 2 y 2 = 5 from the points (1, 2), is  12 (a) tan −1    5  12  (c) tan −1    5

 (b) tan −1    (d) tan −1  

6  5 12   5 

26. If tangents drawn to the ellipse through point (1, 2 3 ) to the ellipse x 2 / 9 + y 2 / b 2 = 1 are at right angle, then value of b is (a) 1 (c) 2

(b) 4 (d) None of these

a (b) x = − , if a < b e b (d) y = − , if a > b e

28. Locus of the point of intersection of the tangents at the end points of the focal chord of an ellipse y2 x2 + = 1, ( b < a ) is a2 b2 (a) x = ± (c) x = ±

a2 a −b ab 2

(b) y = ±

2

b2 a − b2 2

(d) None of these

a2 − b2

y2 x2 29. If pair of tangents drawn to the ellipse + =1 16 9 from a point P, so that angle between the tangents are at right angle, then possible coordinates of the point P is/are (a) (3, 4) (c) (2 5 , 5 )

(b) (5, 0) (d) All of these

30. The locus of the point of intersection of tangents to an ellipse at two points, sum of whose eccentric angles is constant, is a/an (a) parabola (c) ellipse

(b) circle (d) straight line

x2

2

+

y2

= 1 is always touching the a b2 parabola y 2 = 4ax, then locus of the pole is

31. If polar of

(a) a3 y2 = b4x (c) a4 y = − b4x

32. If polar of x2 b2

+

y2 a2

(b) b3 y = a4 y (d) None of these

x2 a2

+

y2 b2

= 1 is always touching the curve

= 1, then locus of pole is

(a) parabola

(b) ellipse

x2

(c) hyperbola (d) circle

y2

+ = 1 is always touching the circle a2 b2 x 2 + y 2 = c 2 , then locus of pole is

33. If polar of

(a) c2 (a4 y2 + b4x 2 ) = a4b4 (b) c2 (a4x 2 + b4 y2 ) = a4b4 (c) c2 (a4 y2 + b4x 2 − a3b3 ) = a4b4 (d) None of the above

34. If polar of y 2 = 4ax is always touching the ellipse x2 a

2

+

y2 b

2

= 1, then locus of the pole is

(a) 4 a2x 2 − b2 y2 = 4 a4 (c) 4 a2x 2 + b2 y2 = 4 a4

(b) a2x 2 − b2 y2 = a4 (d) None of these

+

y2

15 Ellipse

a (a) x = , if a > b e b (c) y = , if a > b e

x2

= 1 at P and Q a2 b2 and the parabola y 2 = 4d ( x + a ) at R and S . The line segment PQ subtend a right angle at the centre of the ellipse. Then, the locus of the point of intersection of tangents to the parabola at R and S is

35. A line intersects the ellipse

1 1 (a) y2 − d 2 = d 2 (x − 2a)2  2 + 2  a b  1 1 (b) y2 − 2d 2 = 2d 2 (x − 2a)2  2 + 2  a b  1 1 (c) y2 + d 2 = d 2 (x + 2a)2  2 + 2  a b  1 1 (d) y2 + 4 d 2 = 4 d 2 (x + 2a)2  2 + 2  a b 

36. The equation of the largest circle with centre (1, 0) that can be inscribed in the ellipse x 2 + 4 y 2 = 16, is 11 3 11 2 2 (b) (x − 1) − ( y − 0) = 3 11 2 2 (c) (x + 1) − ( y + 0) = 3 11 (d) (x + 1)2 + ( y + 0)2 = 3 (a) (x − 1)2 + ( y − 0)2 =

37. Locus of the mid-points of the chord of ellipse y2 x2 + = 1, so that chord is always touching the a2 b2 circle x 2 + y 2 = c 2 ( c < a, c < b ) is (a) (b2x 2 + a2 y2 )2 = c2 (b4x 2 + a4 y2 ) (b) (a2x 2 + b2 y2 )2 = c2 (a4x 2 + b4 y2 ) (c) (b2x 2 + a2 y2 )2 = c2 (b2x 4 + a2 y4 ) (d) None of the above

Targ e t E x e rc is e s

27. Locus of the point of intersection of tangents at the end points of focal chord is

 b2  at = 1 ae ,   is a  a2 b2 passing through ( 0, − 2b ), then value of eccentricity is

38. If normal to ellipse

x2

+

y2

(a) 2 − 1

(b) 2( 2 − 1)

(c) 2( 2 − 1)

(d) None of these

39. The area of the parallelogram formed by the tangents at the ends of conjugate diameters of an ellipse is (a) constant and is equal to the product of the axes (b) cannot be constant (c) constant and is equal to the two lines (d) None of the above

40. An ellipse of semi-axes a, b slides between two perpendicular lines, then the locus of its foci is (the two lines being taken as the axes of coordinates) (a) (x 2 + (b) (x 3 + (c) (x 2 + (d) (x 2 −

y2 )(x 2 y2 + b6 ) = 4 a2x 2 y4 y3 )(x 3 y3 + b6 ) = 4 a3x 3 y6 y2 )(x 2 y2 + b4 ) = 4 a2x 2 y2 y2 )(x 2 y2 − b4 ) = 4 a2x 2 y2

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41. If ω is one of the angles between the normals to the y2 x2 ellipse 2 + 2 = 1 at the points whose eccentric a b π 2cos ω angles are θ and + θ, then is 2 sin 2θ (a)

e2 1 − e2

(b)

e2

(c)

1 + e2

e2 1 − e2

(d)

e2 1 + e2

42. If the normals at P ( x1 , y1 ), Q ( x 2 , y 2 ) and R ( x 3 , y 3 ) to the ellipse are concurrent, then x2 (a) x1 x3 x2 (c) x3 x1

y2 x2 y2

x1 (b) x2

y1 x1 y1 = − 1 y3 x3 y3 y2 x2 y2 y3 x3 y3 = 1

x3

y2 x2 y2 = 0 y3 x3 y3

(d) None of these

Ta rg e t E x e rc is e s

(b) a circle (d) a parabola

48. The condition on a and b for which two distinct y2 x2 chords of the ellipse 2 + 2 = 1 passing through 2a 2b ( a, − b ) are bisected by the line x + y = b, is (a) a2 + 6ab ≤ 7b2 (c) a2 + ab ≤ 7b2

(b) a2 + 6ab ≥ 7b2 (d) a2 + ab ≥ 7b2

x2 y2 1 + 4 = 2 4 b a c x2 y2 1 (c) 4 + 4 = 2 a b c

(a)

43. If the tangent drawn at point ( t 2 , 2t ) on the parabola y 2 = 4x is same as the normal drawn at point ( 5 cos θ, 2sin θ ) on the ellipse 4x 2 + 5 y 2 = 20. Then, the values of t and θ are 1  1 (a) θ = cos−1  −  and t = −  5 5 1 −1  1  (b) θ = cos   and t =  5 5 2 2 −1  (c) θ = cos  −  and t = −  5 5 (d) None of the above

x2

+

y2

=1 a2 b2 meet in the point ( h, k ). Then, the mean position of the four points is

44. The normals at four points on the ellipse

+

y2

= 1, four normals cannot be a2 b2 drawn through a point ( h, 0) unless (e being the eccentricity of the ellipse) (a) | h| < 2ae2 (b) | h| > 2ae2 (c) | h| < ae2

x2

(a) a2l1l2 + b2m1m2 = 1 (c) a2l1l2 − b2m1m2 = − 1

+

(d) | h| > ae2

y2

= 1 at the ends of the a2 b2 chords l1 x + m1 y = 1 and l2 x + m2 y = 1 are concurrent, then

46. If the normals to

x 2 y2 1 − = a4 b4 c2

(d) None of these

50. If the product of the perpendiculars from the foci of y2 x2 the ellipse 2 + 2 = 1 upon the polar of P is always a b c 2 , then locus of P is the ellipse

51. The locus of the poles of the chords of the ellipse subtending a right angle at its centre is 1 1 x4 y4 + 2 = 4+ 4 2 a b a b 1 1 x2 y2 (c) 4 + 4 = 2 + 2 a b a b

(a)

(b)

1 1 x 2 y2 − = + a4 b4 a2 b2

(d) None of these

52. The locus of the mid-points of the chords of the y2 x2 ellipse 2 + 2 = 1 whose poles are on the auxiliary a b circle on the tangents at the extremities of which intersect on the auxiliary circle, is

  ah bk (c)  , 2 2 2 2   2(a − b ) 2(a − b )  a2h b2k  (d)  , 2 2 2 2   2(a − b ) 2(a − b )

x2

(b)

(a) b4 (c2 + a2 − b2 ) x 2 + c2a4 y2 = a4b4 (b) b2 (c2 + a2 + b2 ) x 2 + c2a4 y2 = a4b4 (c) b3 (c3 + a3 − b2 ) x 2 + c2a4 y2 = a2b2 (d) None of the above

  a2h b2k (a)  , 2 2 2 2   2(a + b ) 2(a + b )   a3h b3k (b)  , 2 2 2 2   2(a + b ) 2(a + b )

892

(a) an ellipse (c) a hyperbola

49. The perpendicular from the centre of an ellipse y2 x2 + = 1 to polar of a point with respect to the a2 b2 ellipse is constant equal to c. Then, the locus of the point is the ellipse

y1 x1 y1

y1 x1 y1

45. For the ellipse

47. The locus of the middle points of chords of an ellipse which pass through a fixed point is

(b) a2l1l2 = b2m1m2 = − 1 (d) None of these

 x2 y2  (a) x 2 + y2 =  2 + 2  b  a

2

 x2 y2  (b) x 2 − y2 =  2 + 2  b  a 2

2

 x2 1  x2 y2  y2  (c) x + y = 2  2 + 2  (d) x 2 + y2 = a2  2 + 2  b  a a b  a 2

2

2

53. If CP and CD are conjugate semi-diameters of the y2 x2 ellipse 2 + 2 = 1. Then, the locus of the mid-point a b of PD is x2 y2 + 2 =2 2 a b x 2 y2 1 (c) 2 − 2 = 2 a b

(a)

(b)

x2 y2 1 + 2 = 2 2 a b

(d) None of these

a2x 2 b2 y2 + 4 =1 α4 β a2x 2 b2 y2 (c) 4 + 4 = − 1 α β

(a)

a2x 2 b2 y2 − 4 =1 α4 β a2x 2 b2 y2 (d) 4 − 4 = − 1 α β (b)

55. The locus of pole of tangents to the ellipse y2 x2 + 2 = 1with respect to the parabola y 2 = 4ax, is 2 a b (a) b2 y2 = 4 a2 (x 2 + a2 ) (c) 2b2 y2 = a2 (x 2 − a2 )

(b) b2 y2 = a2 (x 2 − a2 ) (d) b2 y2 = 4 a2 (x 2 − a2 )

56. The locus of the poles of normal chords of the ellipse y2 x2 + = 1 , is a2 b2 a6 b6 + = (a2 + b2 )2 x2 y2 a6 b6 (c) 2 + 2 = (a2 + b2 ) x y

(a)

a6 b6 − = (a2 − b2 )2 x 2 y2 a6 b6 (d) 2 + 2 = (a2 − b2 )2 x y (b)

57. If CP and CD are conjugate semi-diameters of the ellipse, then PD touches the ellipse 2

2

1 x y − 2 = 2 2 a b x2 y2 1 (c) 2 + 2 = 2 a b

(a)

2

2

1 x y + 2 =− 2 2 a b x2 y2 (d) 2 + 2 = − 1 a b (b)

(a) cot 2 α + cot 2 β = Constant (b) cot 2 α + cot 2 β = Variable (c) cot 2 α − cot 2 β = Constant (d) None of the above

59. If PCP′ and DCD′ form a pair of conjugate diameters y2 x2 of the ellipse 2 + 2 = 1 and R is any point on the a b circle x 2 + y 2 = c 2 , then (a) PR 2 − DR 2 − P′ R 2 − D′ R 2 = 2(a2 − b2 − 2c2 ) (b) PR 2 + DR 2 + P′ R 2 + D′ R 2 = 2(a2 − b2 − 2c2 ) (c) PR 2 + DR 2 + P′ R 2 + D′ R 2 = (a2 + b2 + 2c2 ) (d) PR 2 + DR 2 + P′ R 2 + D′ R 2 = 2(a2 + b2 + 2c2 )

60. PCP′ and DCD′ are conjugate diameters of an ellipse y2 x2 + 2 = 1 and θ is the eccentric angle of P. Then, 2 a b eccentric angle of the point where the circle PP ′ D again cuts the ellipse π 2 π (b) 2 π (c) 2 π (d) 2 (a)

+θ + 3θ −θ − 3θ

Type 2. More than One Correct Option 61. The equation of a directrix of the ellipse  x2   y2    +   = 1, is  16   25  25 3 (b) x = 3 (c) x = − 3 25 (d) y = − 3

(a) y =

straight lines bx − ay = 0 and bx + ay = 0 in the points Q and R respectively, then (a) CQ ⋅ CR = a2 + b2 (c) CQ ⋅ CR = a2b2

(b) CQ ⋅ CR = a2 − b2 (d) CQ ⋅ CR = a2 / b2

64. If the tangent at the point P(θ ) to the ellipse 16x 2 + 11y 2 = 256 is also a tangent to the circle x 2 + y 2 − 2x = 15, then θ is equal to 2π 3 5π (c) 3

(a)

62. The equation 3x 2 + 4 y 2 − 18x + 16 y + 43 = c (a) cannot represent a real pair of straight lines for any value of c (b) represents an ellipse, if c > 0 (c) no locus, if c < 0 (d) a point, if c = 0

63. C is the centre of hyperbola

15 Ellipse

58. If α and β are the angles subtended by the major axis of an ellipse at the extremities of a pair of conjugate diameters, then

Targ e t E x e rc is e s

54. The locus of poles of tangents to the ellipse y2 x2 + = 1 with respect to concentric ellipse a2 b2 y2 x2 + = 1, is α2 β2

x2

−

y2

= 1. The

a2 b2 tangents at any point P on this hyperbola meets the

4π 3 π (d) 3 (b)

65. Let F1 and F2 be two foci of the ellipse and PT and PN be the tangent and the normal respectively to the ellipse at point P. Then, (a) PN bisects ∠F1PF2 (b) PT bisects ∠F1PF2 (c) PT bisects angle (180° − ∠F1PF2 ) (d) None of the above

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Objective Mathematics Vol. 1

15

Type 3. Assertion and Reason Directions (Q. Nos. 66-71) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

66. Statement I The major and minor axes of the ellipse 5x 2 + 9 y 2 − 54 y + 36 = 0 are 6 and 10, respectively. Statement II The equation 5x 2 + 9 y 2 − 54 y + 36 = 0 can be expressed as 5x 2 + 9( y − 3) 2 = 45.

Ta rg e t E x e rc is e s

67. Statement I

The condition on a and b for which y2 x2 two distinct chords of the ellipse + =1 2a 2 2b 2 passing through ( a, − b ) are bisected by the line x + y = b is a 2 + 6ab − 7b 2 ≥ 0 Statement II Equation of chord of the ellipse y2 x2 + = 1 whose mid-point is ( x1 , y1 ), is T = S 1 . a2 b2

Statement II Circle described on the focal distance as diameter of the ellipse 4x 2 + 9 y 2 = 36 touch the auxiliary circle x 2 + y 2 = 9 internally. 69. Statement I If the tangents from the point ( λ , 3) to y2 x2 the ellipse + = 1 are at right angles, then λ is 9 4 equal to ± 2 . Statement II The locus of the point of the intersection of two perpendicular tangents to the y2 x2 ellipse 2 + 2 = 1, is x 2 + y 2 = a 2 + b 2 . a b 70. Statement I The latusrectum of an ellipse is shortest focal chord. Statement II The semi-latusrectum of an ellipse is the harmonic mean of the segment of a focal chord. The circle x 2 + y 2 = 4 is auxiliary y2 x2 circle of ellipse + 2 = 1, ( b < 2). 4 b

71. Statement I

Statement II A given circle is auxiliary circle of exactly one ellipse.

Type 4. Linked Comprehension Based Questions x2 y2 + =1 a2 b 2 is such that it has the least area but contains the circle ( x − 1) 2 + y 2 = 1.

Passage I (Q. Nos. 72-74)

The ellipse

72. The eccentricity of the ellipse is (a)

2 3

1 (c) 2

(b)

1 3

(d) None of these

73. Equation of auxiliary circle of ellipse is (a) x 2 + y2 = 65 (c) x 2 + y2 = 45

(b) x 2 + y2 = 5 (d) None of these

74. Length of latusrectum of the ellipse is (a) 2 units (c) 3 units

(b) 1 unit (d) 2.5 units

Passage II (Q. Nos. 75-77) On a level plain, the

894

68. Statement I Circle x 2 + y 2 = 9 and the circle ( x − 5 )( 2x − 3) + y( 2 y − 2) = 0 touch each other internally.

crack of the rifle and the thud of the ball striking the target are heared at the same instant. If V1 = 330 m/s and V 2 = 990 m/s are the velocities of sound and bullet respectively and P is the position of hearer, A is the target and B is the firing point.

75. The locus of the hearer is (a) parabola (c) hyperbola

(b) ellipse (d) None of these

76. If the distance between AB is 600 m and at some instant the hearer is 200 m from the target A, then how far is he from B? (a) 200 m (c) 400 m

(b) 300 m (d) 500 m

77. If Y-axis is perpendicular bisector of AB at origin and 100 m = 1 unit in cartesian plane, then equation of the locus of hearer is (a) 8x 2 − y2 = 8 x2 y2 (c) + =1 4 3

(b) x 2 − 8 y2 = 8 (d) None of these

Passage III (Q. Nos. 78-80) Consider the standard equation of an ellipse whose focus and corresponding  16  feet of directrix are ( 7, 0) and  , 0 and a circle with  17  equation x 2 + y 2 = r 2 . If in the first quadrant, the common tangent to a circle of this family and the above ellipse meets the coordinate axes at A and B.

(a) 16x + 9 y = 144 (c) 16x 2 + y2 = 144 2

2

(b) 9x + 16 y = 144 (d) x 2 + 9 y2 = 144 2

2

79. Let P be a variable point on the ellipse with foci at S and S ′. If ∆ is the area of ∆PSS ′, then the maximum value of ∆ is

15

(b) 2 7 sq units (d) 4 7 sq units

80. If mid-point of A and B is ( x1 , y1 ) and slope of common tangent is m, then (a) 2mx1 + y1 = 0 (c) my1 + x1 = 0

(b) 2my1 + x1 = 0 (d) mx1 + y1 = 0

Ellipse

(a) 7 sq units (c) 3 7 sq units

78. The equation of the ellipse is

Type 5. Match the Columns 81. Match the statements of Column I with values of Column II. Column II

A. An ellipse passing through the origin has its foci (3, 4) and (6, 8), then length of its minor axis is

p.

8

B. If PQ is focal chord of ellipse x2 y2 + = 1 which passes through 25 16 S ≡ ( 3, 0) and PS = 2, then length of chord PQ is

q.

10 2

C. If the line y = x + k touches the ellipse 9 x 2 + 16 y 2 = 144, then the difference of

r.

10

Column II

C. Tangents are drawn from the points on the line x − y − 5 = 0 to x 2 + 4 y 2 = 4. Then, all the chords of contact pass through a fixed point, whose abscissa is

r.

8

D. The sum of the distances of any point on the ellipse 3 x 2 + 4 y 2 = 12 from its directrix is

s.

1 3

83. Match the statements of Column I with values of Column II. Column I

values of k is s.

D. Sum of distances of a point on the x2 y2 ellipse + = 1 from the foci 9 16

12

82. Match the statements of Column I with values of Column II. Column I

A.

A point on the curve x + 3 y = 9 where the tangent is parallel to the line y − x = 0, is

p.

( − 1, 2 )

B.

A focus of the curve 25( x + 1)2 + 9( y + 2 )2 = 225 is at

q.

 9, − 8   5 5

C.

A point on the ellipse

x2 y2 + = 1, 9 4 where the normal is parallel to the line 2 x + y = 1, is

r.  3 3 3 ,  − 2 2  

D.

Tangents are drawn from points on the line x − y + 2 = 0 to the ellipse x 2 + 2 y 2 = 2 , then all the chords of contact pass through the point

s.

 − 1, 1     2

t.

( − 1, − 6)

Column II

A. If P is a point on the ellipse x2 y2 + = 1 whose foci are S and 16 20 S ′, then PS + PS ′ is

p.

B. The eccentricity of the ellipse 2 x 2 + 3 y 2 − 4 x − 12 y + 13 = 0 is

q.

4 5

Column II

4 5

2

2

Targ e t E x e rc is e s

Column I

Column I

Type 6. Single Integer Answer Type Questions (x − 4 )2 y2 + = 1, 4 9 then the difference between the largest and smallest y2 x2 is ________ . value of the expression + 4 9

84. If x, y ∈ R, satisfying the equation

x2

is ________ .

+

y2

= 1 ( a > b ), if a2 b2 the extremities of the latusrectum of the ellipse having positive ordinate lies on the parabola x 2 = − 2( y − 2),

85. The value of a for the ellipse

86. An ellipse has it centre at (1, − 1) and semi-major axis = 8and which passes through the point (1, 3). If l is the length of its latusrectum, then find l/ 4. 87. If (5, 12) and (24, 7) are the foci of the conic passing through the origin, then the eccentricity of conic is 386 k / 24, then find the value of k. 88. If sum of the squares of the perpendicular on any y2 x2 tangent to the ellipse + = 1 from the ( 67) 2 ( 33) 2 points on the minor axis, each at a distance 10 34 from the centre is 4489 l, then find the value of l.

895

Entrances Gallery JEE Advanced/IIT JEE 1. The locus of the foot of perpendicular drawn from the centre of the ellipse x 2 + 3 y 2 = 6 on any tangent to [2014] it, is (a) (x − (b) (x 2 − (c) (x 2 + (d) (x 2 + 2

y ) = 6x + 2 y y2 )2 = 6x 2 − 2 y2 y2 )2 = 6x 2 + 2 y2 y2 )2 = 6x 2 − 2 y2 2 2

2

2

(0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E 2 is [2012] (a)

2 2

(b)

3 2

(c)

1 2

point P( 3, 4) to the ellipse

x2 y2 + = 1, touching the 9 4

ellipse at point A and B.

Ta rg e t E x e rc is e s

3 4

Passage (Q. Nos. 5-7) Tangents are drawn from the

2. A vertical line passing through the point (h, 0) y2 x2 intersects the ellipse + = 1 at the points P 4 3 and Q. Let the tangents to the ellipse at P and Q meet at the point R. If ∆( h ) = area of the ∆PQR, ∆ 1 = max ∆ ( h ) and ∆ 2 = min ∆ ( h ), then 1/ 2≤ h ≤ 1 1/ 2 ≤ h ≤ 1 8 [2013] ∆ 1 − 8∆ 2 = ______ . 5 3. The equation of the circle passing through the foci of y2 x2 the ellipse + = 1 and having centre at (0, 3) is 16 9 [2013] (a) x 2 + y2 − 6 y − 7 = 0 (c) x 2 + y2 − 6 y − 5 = 0

(d)

(b) x 2 + y2 − 6 y + 7 = 0 (d) x 2 + y2 − 6 y + 5 = 0

y2 x2 + = 1 is inscribed in a 9 4 rectangle R whose sides are parallel to the coordinate axes. Another ellipse E 2 passing through the point

4. The ellipse E1 :

[2010]

5. The coordinates of A and B are (a) (3, 0) and (0, 2)  8 2 161  9 8 (b)  − ,  and  − ,   5 5  5 15   8 2 161 (c)  − ,  and (0, 2)  5 15  9 8 (d) (3, 0) and  − ,   5 5

6. The orthocentre of the ∆PAB is 8 (a)  5,   7

 7 25 (b)  ,  5 8

11 8 (c)  ,   5 5

8 7 (d)  ,   25 5

7. The equation of the locus of the point whose distances from the point P and the line AB are equal, is (a) 9x 2 + y2 − 6xy − 54 x − 62 y + 241 = 0 (b) x 2 + 9 y2 + 6xy − 54 x + 62 y − 241 = 0 (c) 9x 2 + 9 y2 − 6xy − 54 x − 62 y − 241 = 0 (d) x 2 + y2 − 2xy + 27x + 31 y − 120 = 0

AIEEE 8. An ellipse is drawn by taking a diameter of the circle ( x − 1) 2 + y 2 = 1 as its semi-minor axis and a diameter of the circle x 2 + ( y − 2) 2 = 4 is semi-major axis. If the centre of the ellipse is at the origin and its axis are the coordinate axes, then the [2012] equation of the ellipse is (a) 4 x 2 + y2 = 4 (c) 4 x 2 + y2 = 8

(b) x 2 + 4 y2 = 8 (d) x 2 + 4 y2 = 16

9. Statement I An equation of a common tangent to the parabola y 2 = 16 3x and the ellipse 2x 2 + y 2 = 4 is y = 2x + 2 3. 4 3 , ( m ≠ 0) is a m common tangent to the parabola y 2 = 16 3x and the

Statement II

896

If the line y = mx +

ellipse 2x 2 + y 2 = 4, then m satisfies m4 + 2m2 = 24.

[2012] (a) Statement I is incorrect, Statement II is correct (b) Statement I is correct, Statement II is correct; Statement II is a correct explanation for Statement I (c) Statement I is correct, Statement II is correct; Statement II is not a correct explanation for Statement I (d) Statement I is correct, Statement II is incorrect

10. Equation of the ellipse whose axes are the axes of coordinates and which passes through the point [2011] ( − 3, 1) and has eccentricity 2/ 5, is (a) 5x 2 + 3 y2 − 48 = 0 (c) 5x 2 + 3 y2 − 32 = 0

(b) 3x 2 + 5 y2 − 15 = 0 (d) 3x 2 + 5 y2 − 32 = 0

11. The ellipse x 2 + 4 y 2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which is turn in inscribed in another ellipse that passes through the point (4, 0). Then, the equation of the ellipse is [2009]

12. A focus of an ellipse is at the origin. The directrix is the line x = 4 and the eccentricity is 1/ 2 , then the [2008] length of semi-major axis is (a)

5 3

(b)

8 3

(c)

2 3

(d)

4 3

13. In an ellipse, the distances between its foci is 6 and minor axis is 8. Then, its eccentricity is [2006] 1 (a) 2 1 (c) 5

4 (b) 5 3 (d) 5

(b) ab sq unit

15. An ellipse has OB as semi-minor axis, F and F ′ its foci and the ∠FBF ′ is a right angle. Then, the [2005] eccentricity of the ellipse is (a)

1 3

(b)

1 4

(c)

17. The radius of the circle passing through the foci of y2 x2 the ellipse + = 1and having its centre at (0, 3), 16 9 is [2002]

1 2

(b) 3 units 7 (d) units 2

(c) 12 units

18. The equation of the ellipse whose foci are ( ± 2, 0) and [2002] eccentricity is 1/2, is x2 y2 + =1 12 16 2 2 x y (c) + =1 16 8

(a)

(d) 2ab sq unit

(d)

(b)

x2 y2 + =1 16 12

(d) None of these

1 2

Other Engineering Entrances 19. The parametric form 4( x + 1) 2 + ( y − 1) 2 = 4 is

of

the

ellipse

 −15 63  (c)  ,  4   4

[Kerala CEE 2014] 8 5 (c) 10 (e) None of these

5 8 (d) 12

(b)

21. An ellipse passing through ( 4 2 , 2 6 ) has foci at ( − 4, 0) and ( 4, 0). Then, its eccentricity is [EAMCET 2014] (a) 2 1 (c) 2

 −15 63  (b)  ,  2   4  −15 63  (d)  ,  2   2

y2 x2 + = 1. Then, 144 25 the radius of the circle with centre (0, 2) and passing through the foci of the ellipse is

23. Let the equation of an ellipse be

20. A point P on an ellipse is at a distance 6 units from a focus. If the eccentricity of the ellipse is 3/ 5, then the distance of P from the corresponding directrix is (a)

(a) (− 15, 63 )

[Kerala CEE 2014]

(a) x = cos θ − 1, y = 2 sinθ − 1 (b) x = 2 cos θ − 1, y = sinθ + 1 (c) x = cos θ − 1, y = 2 sin θ + 1 (d) x = cos θ + 1, y = 2 sin θ + 1 (e) x = cos θ + 1, y = 2 sin θ − 1

15

(a) 3x 2 + 4 y2 = 1 (b) 3x 2 + 4 y2 = 12 (c) 4 x 2 + 3 y2 = 12 (d) 4 x 2 + 3 y2 = 1

(a) 4 units

14. Area of the greatest rectangle that can be inscribed in y2 x2 [2005] the ellipse 2 + 2 = 1is a b a (a) sq unit b (c) ab sq unit

16. The eccentricity of an ellipse with its centre at the 1 origin, is . If one of the directrices is x = 4, then the 2 equation of the ellipse is [2004]

Ellipse

(b) 4 x 2 + 48 y2 = 48 (d) x 2 + 16 y2 = 16

Targ e t E x e rc is e s

(a) x 2 + 12 y2 = 16 (c) 4 x 2 + 64 y2 = 48

1 (b) 2 1 (d) 3

22. On the ellipse 9x 2 + 25 y 2 = 225, find out the point of the distance from which the focus F1 is four times the distance to the other focus F2 . [GGSIPU 2014]

(a) 9

(b) 7

(c) 11

[WB JEE 2014] (d) 5

24. The locus of the points of intersection of the tangents at the extremities of the chords of the ellipse which touches the ellipse x2 + 2y2 = 6 x 2 + 4 y 2 = 4, is [BITSAT 2014] (a) x 2 + y2 = 4 (c) x 2 + y2 = 9

(b) x 2 + y2 = 6 (d) None of these

25. The minimum area of triangle formed by any tangent y2 x2 to the ellipse 2 + 2 = 1with the coordinate axes, is a b [VITEEE 2014] (a) a2 + b2 (c) ab

(a + b)2 2 (a − b)2 (d) 2

(b)

897

Objective Mathematics Vol. 1

15

26. If tangents are drawn from any point on the circle y2 x2 + = 1, then the x 2 + y 2 = 25 to the ellipse 16 9 [EAMCET 2014] angle between the tangents is 2π 3 π (c) 3

π 4 π (d) 2

(a)

1 3

(b)

1 3

(c)

2 3

(d)

x2

y2

+

2 2 3

Ta rg e t E x e rc is e s

1 (c) 2

[BITSAT 2012] 3 (d) 2

x2 29. The line x = 2 y intersects the ellipse + y 2 = 1 at 4 the points P and Q. The equation of the circle with PQ as diameter is [WB JEE 2012] (a) x 2 + y2 =

1 2

(c) x 2 + y2 = 2

(b) x 2 + y2 = 1 (d) x 2 + y2 =

5 2

x2 + y 2 = 1 subtend a right 9 angle at a point P. Then, the locus of P is

30. If the foci of the ellipse

(a) x 2 + y2 = 1 (c) x 2 + y2 = 4

[WB JEE 2012] (b) x 2 + y2 = 2 (d) x 2 + y2 = 8

31. If the centre, one of the foci and semi-major axis of an ellipse are (0, 0), (0, 3) and 5, then its equation is [BITSAT 2011]

898

x2 y2 + =1 25 16

(d) None of these

32 units 5 5 (d) unit 32 (b)

y2 x2 + =1 25 16 and C is the centre of the ellipse, then the sum of maximum and minimum values of CP is

33. If a point P ( x, y )moves along the ellipse

[Kerala CEE 2011]

= 1 and B

a2 b2 is end of the minor axis. If ∆STB is an equilateral triangle, then eccentricity of the ellipse is 1 (b) 3

5 unit 16 16 units (c) 5 (a)

28. S and T are the foci of the ellipse

1 (a) 4

(b)

32. The length of the latusrectum of the ellipse [WB JEE 2011] 16x 2 + 25 y 2 = 400 is

(b)

27. If the length of the major axis of the ellipse y2 x2 + = 1 is three times the length of minor axis, a2 b2 then its eccentricity is [BITSAT 2012] (a)

x2 y2 + =1 16 25 x2 y2 (c) + =1 9 25

(a)

(a) 25 (d) 5

(b) 9 (e) 16

(c) 4

17 times 8 the length of the minor axis, then the eccentricity of the ellipse is [Kerala CEE 2011]

34. If the length of the major axis of an ellipse is

8 17 9 (c) 17 13 (e) 17 (a)

15 17 2 2 (d) 17 (b)

35. Area of a triangle formed by tangent and normal to y2 x2  a b the curve 2 + 2 = 1at P  ,  with the X -axis  2 2 a b [Karnataka CET 2011] is (a) 4ab ab a2 + b2 (b) 4 ab a2 − b2 (c) 4 b(a2 + b2 ) (d) 4a

Answers Work Book Exercise 15.1 1. (b)

2. (a)

3. (b)

4. (c)

5. (a)

4. (b)

5. (b)

6. (a)

4. (c)

5. (c)

6. (a)

Work Book Exercise 15.2 1. (c)

2. (a)

3. (b)

7. (a)

Work Book Exercise 15.3 1. (a)

2. (a)

3. (c)

1. (c)

2. (a)

3. (c)

4. (d)

5. (c)

6. (d)

7. (c)

8. (d)

9. (a)

10. (d)

11. (d)

12. (a)

13. (a)

14. (c)

15. (d)

16. (a)

17. (c)

18. (c)

19. (d)

20. (d)

21. (d)

22. (c)

23. (a)

24. (b)

25. (c)

26. (c)

27. (a)

28. (a)

29. (d)

30. (d)

31. (d)

32. (b)

33. (a)

34. (a)

35. (d)

36. (a)

37. (a)

38. (c)

39. (a)

40. (c)

41. (a)

42. (b)

43. (a)

44. (d)

45. (c)

46. (b)

47. (a)

48. (b)

49. (c)

50. (a)

51. (c)

52. (d)

53. (b)

54. (a)

55. (d)

56. (d)

57. (c)

58. (a)

59. (d)

60. (d)

61. (a,d)

62. (all)

63. (a,c)

64. (c,d)

65. (a,c)

66. (d)

67. (a)

68. (a)

69. (a)

70. (a)

71. (c)

72. (a)

73. (c)

74. (b)

75. (c)

76. (c)

77. (a)

78. (b)

79. (c)

80. (d)

81. (*)

82. (**)

83. (***)

84. (8)

85. (2)

86. (1)

87. (2)

88. (2)

* A → q ; B → r; C → r; D → p ** A → q ; B → s; C → p; D → r *** A → r; B → p, t; C → q ; D → s

Entrances Gallery 1. (c)

2. (9)

3. (a)

4. (c)

5. (d)

6. (c)

7. (a)

8. (d)

9. (c)

10. (d)

11. (a)

12. (b)

13. (d)

14. (d)

15. (d)

16. (b)

17. (a)

18. (b)

19. (c)

20. (c)

21. (b)

22. (c)

23. (c)

24. (c)

25. (c)

26. (d)

27. (d)

28. (c)

29. (d)

30. (d)

31. (a)

32. (b)

33. (b)

34. (b)

35. (d)

Targ e t E x e rc is e s

Target Exercises

899

Explanations Target Exercises 1. r 2 − r − 6 > 0 and r 2 − 6 r + 5 > 0 ⇒ and ⇒ ⇒

=

(r − 3) (r + 2 ) > 0 (r − 1) (r − 5) > 0 { r < − 2 or r > 3} ∩ { r < 1 or r > 5} r < − 2 or r > 5

2 b2 and when a 2 a2 . a < b, then length of the latusrectum is b

2. If a > b, then length of the latusrectum is

3. Let PQ be a focal chord of the ellipse

x2 y2 + 2 = 1 with 2 a b

foci S (ae , 0 ) and S ′ (− ae , 0 ). Y B(0, b)

Ta rg e t E x e rc is e s

X′

(–a, 0)A′

S′ (–ae, 0)

O

P

θ A (a, 0) S (ae, 0)

X

Q B′(0, –b) Y′

The equation of chrod PQ is x − ae y − 0 = cos θ sin θ The coordinates of any point on this line are given by x − ae y − 0 = =r cos θ sin θ ⇒ x = ae + r cos θ and y = r sin θ This point will lie on the ellipse, if (ae + r cos θ)2 (r sin θ)2 + =1 a2 b2  cos 2 θ sin 2 θ   2 rae cos θ  2 + ⇒ r2  +  + e −1= 0  b2   a2  a2  cos 2 θ sin 2 θ   2e cos θ  2 ⇒ r2  +  +  r + (e − 1) = 0  a b2    a2 Let r1, r2 be the roots of this equation. Then, − 2e cos θ r1 + r2 = cos 2 θ sin 2 θ  a 2 +  b2   a e2 − 1 and r1r2 = 2 cos θ sin 2 θ  +   2 b2   a Let SP = r1 and SQ = r2 . Clearly, r1 and r2 are of opposite signs. Now, ( r1 + r2 )2 = (r1 − r2 )2

900

= (r1 + r2 ) − 4r1r2 2

4 e 2 cos 2 θ cos 2 θ sin 2 θ  a2  +  2 b2   a

2

−

4 (e 2 − 1) cos 2 θ sin 2 θ  +   2 b2   a

2  2 sin 2 θ  2  2 2  cos θ + a  e cos θ + (1 − e )  2 b2    a  =4 2 cos 2 θ sin 2 θ  a2  +  2 b2   a   2 a2 2  2 2  2 e cos θ + (1 − e ) cos θ + 2 sin θ  b   =4 2 2 2 cos θ sin θ  a2  +  2 b2   a 4 [e 2 cos 2 θ + (1 − e )2 cos 2 θ + sin 2 θ ] = 2 cos 2 θ sin 2 θ  a2  +  2 b2   a 4 = 2 2 cos θ sin 2 θ  + a2   2 b2   a 2 r1 + r2 = ∴ cos 2 θ sin 2 θ  + a  2 b2   a cos 2 θ sin 2 θ  a +  2 b2  a2 2 r1 r2 2 (1 − e )  ⇒ = ⋅ 2 cos 2 θ sin 2 θ  r1 + r2 +   2 2 b   a = a (1 − e 2 ) a2 (1 − e 2 ) b2 = = = Semi-latusrectum a a 2 SP ⋅ SQ ∴HM of SP and SQ = SP + SQ 2 r1 r2 = r1 + r2

=

b2 = Semi-latusrectum a

4. Since, sides of the square are tangent and perpendicular to each other, so two vertices lie on director circle. Y A

X′

B

D

C

∴

X

O

Y′

x + y = (a − 7 ) + (13 − 5a) = a2 [ 2a is side of the square] 2

2

2

(a2 − 7 ) + (13 − 5 a) = a2 6 ⇒ a= 5 But for an ellipse to exist, a2 − 7 > 0 and 13 − 5 a > 0 ⇒ a ∈ (− ∞, − 7 ) 6 ∴ a≠ 5 Hence, no such a exists. x2 y2 + 2 = 1. 2 a b Equation of the parabola with focus S (ae , 0 ) and directrix x + ae = 0 is y 2 = 4aex.

5. Equation of the ellipse is

Y

X′

O (–ae, 0)S′

ae 7. a cos α a cos β ⇒

b sin β 1  α − β  α − β 2 cos    sin   2   2   α − β  α + β 2 cos    sin   2   2   α − β cos    2  e=  α + β cos    2   α + β  α − β cos    − cos   2   2  −1 =  α + β +1  α − β cos   + cos    2   2  −1 α β = tan tan +1 2 2

sin (α − β ) e= = sin α − sin β

⇒

⇒

e e

⇒

e e

X

S (ae, 0)

15

0 1 b sin α 1 = 0

Ellipse

⇒

8. Since, mutually perpendicular tangents can be drawn from vertices of rectangle. So, all the vertices of rectangle should lie on director circle x 2 + y 2 = a2 + b2 .

Y′

2 b2 and that a

of the parabola is 4ae. For the two latusrectum to be equal, we get 2 b2 = 4ae a 2 a2 (1 − e 2 ) = 4ae ⇒ a ⇒ 1 − e 2 = 2e 2 ⇒ e + 2e − 1 = 0 2± 8 Therefore, e = − = − 1± 2 2 Hence, e = 2 −1

D

=

b2

(a − a − b ) 2

2

E

2l B

x2 y2 + = 1 and circle x 2 + y 2 = a2e 2 a2 b2 Radius of circle = ae Point of intersection of circle and ellipse is a a  2e 2 − 1, (1 − e 2 ) . e  e Now, area of ∆PF1F2 a a 2e 2 − 1 (1 − e 2 ) 1 e 1 e 1 a = 0 1 = (1 − e 2 )(2 ae ) = 30 ae 2 2 e − ae 0 1

9. Let ellipse

2a 2b = 2 b, if a < b = 2 a, if a > b = e ⋅ e e PS + PS ′ + SS ′ = 2 a + 2 ae = 2 a (1 + e ) ∴ s = a (1 + e ) α β (s − c ) (s − a) (s − a) (s − b) Now, tan tan = ⋅ 2 2 s ( s − b) s (s − c ) s−a e = = s 1+ e 2 a − b2  a2 − b2  = × 1− 2 2   a a × b /a  

a2 − b2

A l

Let breadth = 2l and length = 4l, then Area of rectangle = 2 l × 4l = 8l 2 8 = (a2 + b2 ) 5 [Q in ∆OAE, a2 + b2 = l 2 + 4l 2 ⇒ a2 + b2 = 5 l 2 ]

e ⋅ ZZ′ = e ⋅

=

a2 + b2 O

6. PS + PS ′ = ePZ + ePZ′ = e (PZ + PZ′ )

 Q e =  a2 − b2  a − a2 − b2  ×a   a b2  

C

Targ e t E x e rc is e s

Now, length of latusrectum of the ellipse is

a2 − b2   a 

⇒ ⇒

⇒

a2 (1 − e 2 ) = 30 a2e 2 = a2 − 30 2  17  =   − 30 2 169 = 4 2 ae = 13

[given]

10. The equation of a chord joining points having eccentric angles α and β is given by x y  α − β  α − β  α + β cos    = cos   + sin   2   2   2  b a

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Objective Mathematics Vol. 1

15

b  cot α 1 −   a−b a = = b a tan α + b cot α 1 + cot 2 α a

If it passes through (ae , 0 ), then  α − β  α + β e cos    = cos   2   2   α − β cos    2  ⇒ e=  α + β cos    2   α − β  α + β 2 sin    cos   2   2  ⇒ e=  α + β  α + β 2 sin    cos   2   2  sin α + sin β e= ⇒ sin (α + β )

=

Now, the greatest value of the above expression is a−b when a tan α = b cot α 2 ab  a − b ⇒ θmax = tan − 1    2 ab  10

14. Q

11. A ≡ (a cos θ, b sin θ )

Ta rg e t E x e rc is e s

a−b ( a tan α − b cot α )2 + 2 ab

∑ (SPi ) (S ′ Pi′ ) = 2560 i =1

B ≡ { a cos (θ + α ), b sin (θ + α )} C ≡ { a cos (θ + 2α ), b sin (θ + 2α )} ∆ ≡ Area of ∆ABC a cos θ b sin θ 1 1 = 1 a cos (θ + α ) b sin (θ + α ) 2 1 a cos (θ + 2α ) b sin (θ + 2α ) α = 2 ab sin 2   sin α  2 ∆(α ) = ab sin α (1 − cos α) ab = (2 sin α − sin 2α ) 2 ∆′ (α ) = 0 1 ⇒ cos α = 1or cos α = − 2 cos α = 1gives triangle = 0 3 3 1 cos α = − gives maximum value of triangle = ab 4 2

12. Let (h, k ) be the mid-point of the chord 7 x + y − 1 = 0 hx ky h 2 k 2 + = + 1 7 1 7 and 7x + y = 1 represents same straight line. h k = ⇒ h=k ⇒ 7 7 Equation of the line joining (0, 0 ) and (h, k ) is y−x=0 ⇒

13. Tangent to the ellipse at P(a cos α, b sin α) is x y cos α + sin α = 1 a b Y Q P X′

θ X

…(i) …(ii)

∴ ⇒ ⇒ ⇒ ⇒ ⇒

10 b2 = 2560 b2 = 256 b = 16 256 = 400 (1 − e 2 ) 16 1− e2 = 25 3 e= 5

x2 y2 + = 1, equation of director circle is 16 9 x 2 + y 2 = 25. The director circle will cut the ellipse x2 y2 + = 1 at 4 points. 50 20 Hence, number of points = 4.

15. For the ellipse

16. (α, β) lies on the director circle of the ellipse i.e. on x 2 + y 2 = 9. So, we can assume α = 3 cos θ, β = 3 sin θ ∴ F = 12 cos θ + 9 sin θ − 3 (4 cos θ + 3 sin θ ) ⇒ − 15 ≤ F ≤ 15

17. The tangent at the point of shortest distance from the line x + y = 7 parallel to the given line. Any point on the given ellipse is ( 6 cos θ, 3 sin θ). Equation of the tangent is x cos θ y sin θ + =1 6 3 which is parallel to x + y = 7. cos θ sin θ = ⇒ 6 3 1 cos θ sin θ ⇒ = = 1 3 2 ∴The required point is (2, 1.) ( x − 2 )2 ( y + 1)2 + =1 3 2 ∴Equation of the director circle is ( x − 2 )2 + ( y + 1)2 = 3 + 2 …(i) ⇒ x 2 + y 2 − 4x + 2 y = 0

18. 2 x 2 + 3 y 2 − 8 x + 6 y + 5 = 0 ⇒ Y′

902

Tangent to the circle at Q (a cos α , a sin α ) is cos αx + sin αy = a Now, angle between tangents is θ, then b − cot α − (− cot α ) a tan θ =  b  1 +  − cot α  (− cot α )  a 

Q(4, − 2 ) is satisfying Eq. (i), therefore (4, − 2) is lying on the director circle. and equation of circle is ( x − 2 )2 + ( y + 1)2 = 3 2 ⇒ x + y 2 − 4x + 2 y + 2 = 0

2

x y + = 1are inclined at an angle 45° be a2 b2 (h, k ), then equation of the pair of tangents is 2   hx ky   x2  h2 k 2 y2   2 + 2 − 1  2 + 2 − 1 =  2 + 2 − 1  b b b  a  a a to the ellipse

1 k2 − a2 b2 a2 1 h2 Coefficient of y 2 , B = 2 2 − 2 a b b hk and coefficient of 2 xy, H = − 2 2 a b 1 h2k 2 h2k 2 k2 h2 2 4 4 − 4 4 + 2 4+ 4 2 − 2 2 a b a b a b a b a b ∴ tan 45° = h 2 + k 2 − a2 − b2 a2 b2 ∴Required locus is ( x 2 + y 2 − a2 − b2 )2 = 4(b2 x 2 + a2 y 2 − a2 b2 ) Coefficient of x 2 , A =

20. c 2 = a2 m2 + b2 , where a2 = 8, b2 = 4, m = 4 ⇒ ⇒ ⇒ ∴

c 2 = 8 ⋅ 42 + 4 c 2 = 128 + 4 c 2 = 132 c = ± 132

x y …(i) + 27 ⋅ = 1 ⇒ 2 x + 3 y = 1 4 9 Now, equation of the pair of the lines passing through origin and points Q, R is given by  x2 y2  +  = (2 x + 3 y )2  9  4 ⇒ 9 x + 4 y = 36 (4 x + 12 xy + 9 y ) 2 ⇒ 135 x + 432 xy + 320 y 2 = 0  2 2162 − 135 ⋅ 320   ∴Required angle = tan − 1    455   2

2

 8 2916 − 2700   8 216   48 6  = tan −1   = tan −1   = tan −1   455    455   455 

22. Since, tangent is perpendicular to the line 2x + y + 7 = 0 Let equation of the tangents be x − 2 y + c = 0 Q c 2 = a2 m2 + b2 1 c 2 = 4⋅ + 3 = 4 ⇒ c = ± 2 4 ∴ Equation of the required tangents are x − 2 y + 2 = 0 and x − 2 y − 2 = 0. ∴

c =a m +b

⇒ c = 9 + 16 = 25 c =±5 ∴Equation of the required tangents are x + y ± 5 = 0 2

2

2

2

24. Let coordinates of P be (h, k ), then equation of QR is hx + ky = 1 2 but is given as x + 3y = 1 Q Eqs. (i) and (ii) are identical.

∴ ⇒

2 h 2 − ab a+ b 2 144 + 36  12  tan θ = ⇒ θ = tan − 1    5 5 tan θ =

26. Since, tangents drawn from (1, 2 3 ) to the ellipse x2 y2 + = 1 are right angles, therefore (1, 2 3 ) will lie on a2 b2 the director circle of the ellipse i.e. (1, 2 3 ) will lie on x 2 + y 2 = 9 + b2 ∴ 13 = 9 + b2 ⇒ b2 = 4 ⇒ b = 2

28. Since, locus of the point of intersection of the tangents at the end points of a focal chord is directix. a a2 ∴ Required locus is x = ± = ± e a2 − b2

29. Since, all the points (3, 4), (5, 0 )and(2 5, 5 )are lying on the director circle x 2 + y 2 = 25. Therefore, all the choices are true.

30. The point of intersection of tangents is   θ + θ2    θ1 + θ 2    b sin  1  a cos   2   2   ,  θ1 − θ 2   θ1 − θ 2    cos co s       2   2    It is given that (θ1 + θ 2 ) = k = constant. Since, given θ1 + θ 2 = constant. Therefore, a cos k b sin k and y1 = x1 =  θ1 − θ 2   θ − θ2  cos  cos  1    2   2  x a b  ⇒ 1 = cot k, y1 =  cot k x1 a  y1 b  b ⇒ ( x1, y1 ) lies on the straight line y =  cot k x.  a

31. Let coordinate of the pole be (h, k ), then equation of the

23. Let equation of the tangent be x + y + c = 0 2

(3 x 2 + 2 y 2 − 5) [3 (1)2 + 2(2 )2 − 5] = [3 × (1) + 2 y(2 ) − 5]2 ⇒ 9 x 2 − 4 y 2 − 24 xy + 40 y + 30 x − 30 = 0 ∴ a = 9, b = − 4, h = − 12

point of focal chord is directrix. a x = ,when a > b ∴ e′

8⋅

2

25. SS1 = T 2

27. The locus of point of intersection of tangents at end

21. Equation of QR is

2

15

h k = =1 2 3 ⇒ h = 2 and k = 3 Coordinates of P are (2, 3).

∴

Targ e t E x e rc is e s

2

Ellipse

19. Let coordinate of the point from where tangents drawn

…(i) …(ii)

x2 y2 hx ky + 2 = 1 is 2 + 2 = 1. 2 a b a b b2 b2 h ⇒ y=− 2 x+ k a k Since, line (i) is touching the parabola y 2 = 4ax. b2 a ⋅ a2 ∴ = k k − b2 ⋅ h ∴Required locus is a3 y 2 = − b4 x. polar

…(i)

903

Objective Mathematics Vol. 1

15

32. Let coordinates of the pole be (h, k ), then equation of the 2

polar of

2

x y + 2 = 1 is 2 a b hx ky + =1 ⇒ a2 b2

b2 h b2 + k a2 k 2 x y2 Since, line (i) is touching the ellipse 2 + 2 = 1. b a 4 2 b4 2 2 b h =b ⋅ 4 2 +a ∴ a k k2 ∴ Required locus is b6 x 2 + a6 y 2 = a4 b4 y=−

…(i)

36. The largest circle inscribed in the ellipse x 2 + 4 y 2 = 16

33. Let coordinates of pole be (h, k ), then equation of the x2 y2 + = 1 is a2 b2

b2 b2 h hx ky + 2 =1 ⇒ y = − 2 x + 2 k a b a k Since, line (i) is touching the circlex 2 + y 2 = c 2 b4 h 2  b4 2  ∴ = 1 + + c   a4 k 2  k2 

Hence, the locus of (h, k ) is

1  1 y 2 + 4 d 2 = 4 d 2 ( x + 2 a)2  2 + 2  a b 

which is equation of the ellipse.

polar of the ellipse

1 4d 2 1 k2 − + 2 − =0 2 2 2 2 a 4 d (h + 2 a) b 4 d (h + 2 a)2 1  1 ⇒ k 2 + 4 d 2 = 4 d 2 (h + 2 a)2  2 + 2  a b  ⇒

will touch the ellipse at some point. So, let r be the radius of the largest circle centred atC (1, 0 )and inscribed in the ellipse x 2 + 4 y 2 = 16. Suppose it touches the ellipse at P(4 cos θ, 2 sin θ). Then, the equation of the tangent to the ellipse at P is 4 x cos θ + 8 y sin θ = 16 …(i) ⇒ x cos θ + 2 y sin θ = 4 Y

… (i)

P (4 cos θ, 2 sin θ) X

X′

O C(1, 0)

∴Required locus is c 2 (b4 x 2 + a4 y 2 ) = a4 b4 .

x cos θ + 2y sin θ = 4

Ta rg e t E x e rc is e s

34. Let coordinates of the pole be (h, k ), then equation of the polar of y 2 = 4ax is ky = 2 a ( x + h ) 2a 2 ah y= x+ ⇒ k k

…(i)

35. Let T (h, k )be the point of intersection of tangents to the parabola at R and S. Then, the equation of the chord of contact of tangents is ky = 2d ( x + h ) + 4ad ⇒ ky − 2dx = 2d (h + 2 a) ky − 2ds ⇒ =1 2d (h + 2 a) The combined equations of OP and OQ is 2 x2 y 2  ky − 2dx  + − =0   a2 b2 2d (h + 2 a) Since, ∠POQ = 90 °. Therefore, Coefficient of x 2 + Coefficient of y 2 = 0 4d (x+a)

904

h,

k)

° Q S

O

x2 y2 + =1 B′(0, –b) a2 b2

Y′

1 32 11 + = 9 9 3 Hence, the equation of the circle is 11 ( x − 1)2 + ( y − 0 )2 = 3

37. Let mid-point of the chord be (h, k ), then equation of the

∴ A (a, 0)

X

2

=

hx ky 2 h2 + 2 − 1= 2 2 a b a b2 h y=− 2⋅ x+ a k

k2 −1 b2 2 h k 2  b2  2 + 2 b  k a +

Since, line (i) is touching the circle x 2 + y 2 = c 2

B (0, b)

90

T(

A ′(–a, 0)

2   2 2 4  − 0 =  − 1 + 2 × 3  3  

⇒

P X′

Clearly, CP is perpendicular to Eq. (i). Therefore, 2 sin θ − 0 cos θ ×− = −1 4 cos θ − 1 2 sin θ ⇒ − cos θ = − 4 cos θ + 1 1 cos θ = ⇒ 3 ∴ r = CP = (4 cos θ − 1)2 + (2 sin θ − 0 )2

chord is

Y y2 =

4y 2 = 16

Y′

x2 y2 Since, line (i) is touching the ellipse 2 + 2 = 1. a b 2 4a2 h 2 2 2 4a Therefore, =a ⋅ 2 +b k k2 ∴Required locus is 4a2 x 2 = 4a4 + b2 y 2 ⇒ 4a2 x 2 − b2 y 2 = 4a4

R

x2 +

2 2

h k  2 + 2 b  a 2

…(i) ...(ii)

2

 b h b = c 2 1 + 4 2  2 a k  k  4

4

∴Required locus is (b2 x 2 + a2 y 2 )2 = c 2 (b4 x 2 + a4 y 2 ). 

38. Equation of the normal at  ae ,

b2   is a

 x1 y1 ( x − x1 ) − 2 ( y − y1 ) = 0 a b2

…(i)

∴(0, − 2 b) is lying on Eq. (i), we get e  b − e + b 2 +  = 0 ⇒ b(2 a + b) = a2 a  a 2 ab = a2 − b2 ⇒ 2 ab = a2e 2 ⇒ 4a2 (1 − e 2 ) = a2e 4 − 4 ± 32 ⇒ e4 + 4e2 − 4 = 0 ⇒ e2 = 2 ∴ e = 2 ( 2 − 1)

39. Area of parallelogram T1 T2 T3 T4

⇒

⇒ (α 2 + β 2 )b4 + α 2β 2 (α 2 + β 2 ) = 4α 2β 2 (b2 + a2 − b2 ) ⇒

T2

Hence, the locus of (α, β) is ( x 2 + y 2 ) ( x 2 y 2 + b4 ) = 4a2 x 2 y 2 Similarly, by eliminating α and β, we can show that the locus of S ′ (h, k ) is ( x 2 + y 2 ) ( x 2 y 2 + b4 ) = 4a2 x 2 y 2 . x2 y2 + 2 =1 2 a b π at the points whose eccentric angles areθ and + θ are 2 ax sec θ − by cosec θ = a2 − b2 and − ax cosec θ − by sec θ = a2 − b2 respectively. Since, ω is the angle between these two normals.

B (–a sin θ, b cos θ) D

P (a cos θ, b sin θ) A

C

X

∴

T1 D′

⇒

B' T4

⇒

Y′

0 0 1 1 = 8× a cos θ b sin θ 1 2 − a sin θ b cos θ 1

⇒

= 4 (ab cos 2 θ + ab sin 2 θ ) = 4ab = 2 a × 2 b = Product of the axes of the ellipse

⇒ ⇒

40. Let S(α, β) and S ′ (h, k ) be the foci of the ellipse. We know that the product of the perpendiculars from the foci of an ellipse to any tangent is equal to the square of semi-minor axes. Therefore, kβ = b2 and hα = b2 Y

P (0, β) C Q (0, k)

S(α, β)

S'(h, k)

X′

X

M (h, 0) L(α, 0) Y'

Now, ⇒

SS ′ = 2 ae (α − h )2 + ( β − k )2 = 4a2e 2 2

⇒ ⇒ ⇒

(α 2 + β 2 ) (α 2β 2 + b4 ) = 4a2α 2β 2

(α 2 − b2 )2 (β 2 − b2 )2 + = 4a2e 2 α2 β2 β 2 (α 2 − b2 )2 + α 2 (β 2 − b2 )2 = 4a2α 2β 2e 2

2 ab (sin 2θ ) (b2 − a2 ) 2 ab tan ω = 2 (a − b2 ) sin 2θ

tan ω =

tan ω =

2 a2 1 − e 2 a2e 2 sin 2θ e2

2 cot ω = sin 2θ 1− e2

42. The equations of the normals at P( x1, y1 ), Q ( x2 , y2 ) and R ( x3 , y3 ) to the ellipse are a2 x b2 y − = a2 − b2 x1 y1 a2 x b2 y − = a2 − b2 x2 y2 a2 x b2 y and − = a2 − b2 x3 y3 These three lines will be concurrent, if a2 − b2 a2 − b2 x1 y1 a2 − b2 a2 − b2 = 0 x2 y2 a2 − b2 a2 − b2 x3 y3

2

  b2  b2  α −  + β −  = 4a2e 2 α β  

a a tan θ + cot θ b b tan ω = a2 1− 2 b ab (tan θ + cot θ ) tan ω = b2 − a2

Targ e t E x e rc is e s

Y

T3 A′ P′

15

41. The equations of the normals to the ellipse

= 4 (Area of parallelogram CPT2 D ) = 4 (2 × Area of ∆CPD ) = 8 (Area of ∆CPD )

X′

(α 2 + β 2 )b4 + β 2α 4 + α 2β 4 − 4α 2β 2 b2  a2 − b2  = 4a2α 2β 2    a2 

Ellipse

e  1 b2  ( x − ae ) −  y −  = 0 a a a

⇒

⇒

1 x1 1 − a2 b2 (a2 − b2 ) x2 1 x3

1 y1 1 y2 1 y3

1 1 =0 1

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Objective Mathematics Vol. 1

15 ⇒

1 x1 1 x2 1 x3

1 y1 1 y2 1 y3

∴

1 1 =0 ⇒ 1

y1

x1

x1 y1

y2 y3

x2 x3

x 2 y2 = 0 x 3 y3

y3

x 3 y3

43. The equation of the tangent at (t 2 , 2 t ) to the parabola

Ta rg e t E x e rc is e s

y 2 = 4 x is 2 ty = 2( x + t 2 ) ⇒ ty = x + t 2 …(i) ⇒ x − ty + t 2 = 0 The equation of the normal at point ( 5 cos θ, 2 sin θ ) the ellipse 5 x 2 + 5 y 2 = 20 is ( 5 sec θ )x − (2 cosec θ )y = 5 − 4 …(ii) ( 5 sec θ )x − (2 cosec θ )y − 1 = 0 ⇒ It is given that Eqs. (i) and (ii) represent the same line. Therefore, 5 sec θ − 2 cosec θ − 1 = = 2 1 −1 t 2 cosec θ 1 and t = − t = ⇒ 2 cosec θ 5 sec θ 2 1 ⇒ cot θ and t = − sin θ t = 2 5 2 1 cot θ = − sin θ ⇒ 2 5 4 cos θ = − 5 sin 2 θ ⇒ ⇒ 4 cos θ = − 5 (1 − cos 2 θ ) 5 cos 2 θ − 4 cos θ − 5 = 0 ⇒ 2 ⇒ 5 cos θ − 5 cos θ + cos θ − 5 = 0 ⇒ 5 cos θ(cos θ − 5 ) + (cos θ − 5 ) = 0 (cos θ − 5 ) ( 5 cos θ + 1) = 0 ⇒ 1 [Qcos θ ≠ − 5] ⇒ cos θ = − 5  1 θ = cos − 1  − ⇒   5 1 1 On putting cos θ = − in t = − sin θ, we get 2 5 1 1 1 t =− 1− = − 2 5 5 1 1 −1  Hence, θ = cos  −  and t = −  5 5

44. The equation of the normal at point( x, y )on the ellipse is

906

a2 X b2 Y − = a2 − b2 x y If it passes through (h, k ), then a2 h b2 k − = a2 − b2 x y ⇒ a2 hy − b2 kx = (a2 − b2 )xy b2 kx ⇒ y= 2 a h − (a2 − b2 )x x2 y2 Since, ( x, y ) lies on the ellipse 2 + 2 = 1. a b

x2 b4 k 2 x 2 [from Eq. (i)] + 2 2 =1 2 a b { a h − (a2 − b2 )x} 2 ⇒ x 2 { a2 h − (a2 − b2 )x} 2 + a2 b2 k 2 x 2 = a2 { a2 h − (a2 − b2 )x} 2 2 2 2 4 2 2 2 ⇒ (a − b ) x − 2 a (a − b )hx 3 + a2 (a2 h 2 − a2 + b2 + b2 k 2 )x 2 + 2 a4 (a2 − b2 )hx − a6 h 2 = 0 …(ii) ⇒

[on multiplying R1 by x1 y1, R2 by x2 y2 and R3 by x3 y3 ] x1 y1 x1 y1 ⇒ x 2 y2 x 2 y2 = 0 x3

x2 y2 + 2 =1 2 a b

…(i)

This is a fourth degree equation in x. So, it given four values of x. Corresponding to each value of x, there is a point on the ellipse such that the normal to the ellipse at that point passes through (h, k ). So, let P ( x1, y1 ), Q ( x2 , y2 ), R ( x3 , y3 ) and S ( x4 , y4 ) be four points on the ellipse such that the normals at these points pass thought (h, k ). Clearly, x1, x2 , x3 and x4 are the roots of the Eq. (ii). 2 a2 h ∴ x1 + x2 + x3 + x4 = 2 ( a − b2 ) x1 + x2 + x3 + x4 a2 h ⇒ = 4 2 (a2 − b2 ) Similarly, we have y1 + y2 + y3 + y4 b2 k = 4 2 (a2 − b2 ) Hence, the mean position of the four points is  a2 h b2 k  ,  . 2 2  2 (a − b ) 2 (a2 − b2 )

45. We know that from a point (h, k ), four normals real or x2 y2 + 2 = 1 such 2 a b that the eccentric angles of the foot of the normals are given by θ θ bk tan 4 + 2(ah + a2e 2 ) tan 3 + 2(ah − a2e 2 ) 2 2 θ tan − bk = 0 2 ⇒ bkt 4 + 2(ah + a2e 2 ) t 3 + 2(ah − a2e 2 ) t − bk = 0 …(i) θ where, t = tan 2 In the given case, we have k = 0. Therefore, Eq. (i) reduces to …(ii) 2 (ah + a2e 2 ) t 3 + 2(ah − a2e 2 ) t = 0 imaginary can be drawn to the ellipse

Thus, the degree of the Eq. (i) reduce by one which corresponds to one root t → ∞. The corresponding value of the eccentric angle of the foot of the normal is given by θ tan = ∞ ⇒ θ = π 2 This means that one normal from (h, 0 ) coincides with the negative direction of X-axis which is a part of the major axis. From Eq. (ii), we have 2(ah + a2e 2 ) t 3 + 2(ah − a2e 2 ) t = 0 ah − a2e 2 ⇒ t = 0 and t 2 = ah + a2e 2 For t = 0, we have θ tan = 0 ⇒ θ = 0 2

are real and distinct. This is possible only when ah − a2e 2 ⇒ >0 − ah + a2e 2 h − ae 2 ⇒ a, a = 4, b = 5, e = 3/ 5 b c 25 y= 3 25 y=− 3

Equation of directrix is y = ± B′ (0, – b) Y′

∴

⇒

tan α =

tan α =

m1 − m2 1 + m1m2 −

b θ b θ cot − tan a 2 a 2 b2 1− 2 a

i.e.

62. ( 3 x + 3 3 )2 + (2 y + 4)2 = c Comparing this with ax 2 + 2 hxy + by 2 + 2 gx + 2 fy + c = 0, we get h 2 < ab So, no locus for c < 0 Ellipse for c > 0 and point for c = 0.

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Objective Mathematics Vol. 1

15

⇒ t 2 (a2 + b2 ) − ab(3 a + b)t + 2 a2 b2 = 0 Qt is real. ∴ B 2 − 4 AC ≥ 0 ⇒ a2 b2 (3 a + b)2 − 4(a2 + b2 ) 2 a2 b2 ≥ 0 ⇒ 9a2 + 6 ab + b2 − 8 a2 − 8 b2 ≥ 0 ∴ a2 + 6 ab − 7 b2 ≥ 0

63. Let P (a sec θ, b tan θ )

x sec θ y tan θ − =1 a b x y It meets bx − ay = 0 i.e. − = 0 at Q. a b   a b ∴ Q ,   sec θ − tan θ sec θ − tan θ  x y It meets bx + ay = 0 i.e. + = 0 at R. a b   a −b ∴R ,   sec θ + tan θ sec θ + tan θ  Tangent at P is

∴ CQ ⋅ CR =

68. Ellipse is

64. Given equation of ellipse is 16 x 2 + 11y 2 = 256. 16   Equation of tangent at  4 cos θ, sin θ is   11  16  16 x (4 cos θ) + 11y  sin θ = 256  11  2 2 2 It touches ( x − 1) + y = 4 , if 16 cos θ + 11 sin θ

Ta rg e t E x e rc is e s

2

⇒ ⇒

2

Equation of the circle as the diameter, joining the point  3 2  ,   and focus ( 5, 0 ) is  2 2 ( x − 5 ) ( 2 x − 3) + y ( 2 y − 2 ) = 0

69. (λ, 3) should satisfy the equation x 2 + y 2 = 13 ∴

λ=±2

70. Let PSP′ be any focal chord. We know the property described in Statement II is true. SP + SP′ 2 ⇒ Latusrectum ≤ PP′ ⇒ Statement I is true.

=4

∴ Semi-latusrectum = HM of SP and SP′ ≤

(cos θ − 4)2 = 16 cos 2 θ + 11 sin 2 θ 4 cos θ + 8 cos θ − 5 = 0 1 cos θ = ⇒ 2 π 5π ∴ θ= , 3 3 PF1 NF1 65. Q = PF2 NF2 2

71. A circle concentric with ellipse and having major axis as diameter is called auxiliary circle x2 + y2 = 4 is auxiliary circle of infinite ellipse x2 y2 + 2 = 1, b < 2 4 b

Y

72. Solving both equations, we have B

X′

5 3  3 2  A point on ellipse  , .  2 2 Focus ≡ ( 5, 0 ), e =

a2 + b2 a2 + b2 ⋅ = a2 + b2 (sec θ − tan θ ) (sec θ + tan θ )

4 cos θ − 16

x2 y2 + =1 9 4

F1

A′ F2 C

x 2 1 − ( x − 1)2 + =1 a2 b2

P

A

X

Y

T

N B′ Y′

X′

X (1, 0)

∴ PN bisects ∠F1PF2 . Q Bisectors are perpendicular to each other. ∴ PT bisects the angle (180 ° − ∠F1PF2 ). 5 x 2 + 9 y 2 − 54 y + 36 = 0 ⇒ 5 x 2 + 9 ( y − 3)2 = 45 x 2 ( y − 3)2 ⇒ + =1 ( 5 )2 32 Length of major axis = 2 × 3 = 6 ∴ and length of minor axis = 2 × 5 = 2 5

Y′

66. Q

⇒ ⇒

67. Let (t , b − t ) be a point on the line x + y = b, then

910

equation chord whose mid-point (t , b − t ), is (b − t )y (b − t )2 tx t2 + − 1= + −1 2 2 2 2a 2b 2a 2 b2 (a, − b) lies on Eq. (i), then (b − t )2 ta b(b − t ) t2 − = + 2 2 2 2a 2b 2a 2 b2

…(i)

b x + a [1 − ( x − 1)2 ] = a2 b2 (b2 − a2 )x 2 + 2 a2 x − a2 b2 = 0 2 2

2

For least area, circle must touch the ellipse. ∴Discriminant of Eq. (i) is zero. ⇒ 4a4 + 4a2 b2 (b2 − a2 ) = 0 ⇒ a2 + b2 (b2 − a2 ) = 0 ⇒ a2 + b2 (− a2e 2 ) = 0 ⇒ 1 − b2e 2 = 0 1 b= ⇒ e 1 b2 Also, = 2 a2 = 2 1− e e (1 − e 2 )

…(i)

1  Qb=  e 

1 e 1− e

78. Q ae = 7 ,

2

Let S be the area of the ellipse. Then, π π S = π ab = = e2 1− e2 e4 − e6 Area is minimum, if f (e ) = e 4 − e 6 is maximum. When f ′ (e ) = 4 e 3 − 6 e 5 = 0 e = 2/ 3 ⇒ which is point of maxima for f (e ). 2 S is least when e = 3

73. Clearly, b =

1 3 = 2 2/ 3

and

a=

1 3 = 2 2 2 1− 3 3

∴Equation of ellipse is 2 x 2 + 6 y 2 = 9. Equation of auxiliary circle of ellipse is x 2 + y 2 = 45.

74. Length of latusrectum of ellipse is 9 2⋅ 2 b2 = 4 =1 9 a 2 x2 + y2 9 = + 330 990

75.

∴ ⇒

a2 = 16 a=4 7 Then, e= 4 and b2 = a2 (1 − e 2 ) 7  = 16 1 −   16 = 16 − 7 =9 x2 y2 ∴ Ellipse is + =1 16 9

79. Let P (4 cos φ, 3 sin φ ) be the point on the ellipse x2 y2 + = 1 and foci are S ≡ ( 7 , 0 ) and S′ ≡ (− 7 , 0 ). 16 9 1 ∆ = × S ′ S × PM ∴ 2 1 = (2 7 ) × 3 sin φ 2 = 3 7 sin φ sq units ∴Maximum value of ∆ is 3 7 sq units.

80. Q Mid-point of A and B is ( x1, y1 ). ∴ A ≡ (2 x1, 0 ) and B ≡ (0, 2 y1 ) Q Slope of AB is m. 2 y1 − 0 =m ∴ 0 − 2 x1 ⇒ y1 + mx1 = 0 ⇒ mx1 + y1 = 0

( x − a)2 + y 2 330 P (x, y)

B (0, 0)

x 2 + y 2 − ( x − a)2 + y 2 =

a 9

which is hyperbola. x 600 200 76. = + 330 990 330

2a = OP + OQ = 5 + 10 = 15 15 a= ⇒ 2 Also, distance between foci,

P

⇒ x

⇒ 600

∴

77.

Y

2 ae = (6 − 3)2 1 e= 3 225  2 b = 1 − 4 

+ (8 − 4)2 = 5

1  = 50 9

b=5 2 2 b = 10 2 1 1 2a B. We know that, + = 2 SP SQ b 1 1 10 ⇒ + = 2 SQ 16 ⇒ SQ = 8 ⇒ PQ = 10 ⇒ ⇒

A

x = 400 m ( x + 3)2 + y 2 6 = + 330 990

[given]

81. A. Points are O(0, 0 ), P(3, 4) and Q(6, 8).

A (a, 0)

B

15

a 16 = e 7

Ellipse

a=

Targ e t E x e rc is e s

⇒

( x − 3)2 + y 2 330 P

C. If the line y = x + k touches the ellipse B (–3, 0)

⇒ ⇒

O

A (3, 0)

X

( x + 3)2 + y 2 = 2 ( x − 3)2 + y 2 8x 2 − y 2 = 8

9 x 2 + 16 y 2 = 144, then k 2 = 16 (1)2 + 9 ⇒ k=±5 D. Sum of the distances of a point on the ellipse from the foci = 2 a = 8.

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Objective Mathematics Vol. 1

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82. A. Here, a2 = 16, b2 = 20 ⇒ a2 < b2

Ta rg e t E x e rc is e s 912

⇒ 9 + 3 = k2

2

2

x y + =1 9 3 ⇒ k=± 2

83. A. Let x − y + k = 0 be the tangent to a2 l 2 + b2 m2 = n 2

Point of contact  − a2 l − b2 m  − 9 × 1 − 3 (− 1) , , =  =  n   ± 12 ± 12   n  = m 

9 ,m 12

3   3 3 3 ,m   = m  2 2 12  

( x + 1)2 ( y + 2 )2 + =1 9 25 ⇒ a2 = 9, b2 = 25 and centre = (− 1, − 2 ) Foci = (− 1, − 2 ± 4) = (− 1, 2 ) and (− 1, − 6) ax by C. − = a2 − b2 cos θ sin θ 3x 2y …(i) − =5 cos θ sin θ …(ii) 2x + y = k Comparing Eqs. (i) and (ii), we get 3 5 −2 = = 2 cos θ sin θ k 9 2 4 2 ⇒cos 2 θ + sin 2 θ = 1 ⇒ k + k =1 100 25 ⇒ k=±2 3 4 If k = 2, then cos θ = and sin θ = − 5 5 8 9 Point = (a cos θ, b sin θ ) =  , −  5 5  9 8 Other point =  − ,   5 5 x2 y2 D. Pole of the line x − y + 2 = 0 w.r.t. + =1 2 1  − a2 l − b2   − 2 × 1 − 1 × (−1)  1 , = ,  =  =  − 1,      2 2 2 n n  

B.

x − 4 = 2 cos θ x = 2 cos θ + 4 y = 3 sin θ

84. Let

PS + PS ′ = 2 b = 2 20 = 4 5. ∴ B. Given ellipse is 2 x 2 + 3 y 2 − 4 x − 12 y + 13 = 0 ( x − 1)2 ( y − 2 )2 1 1 ⇒ + = 1. Here, a2 = , b2 = 1 1 2 3 2 3 1 1 e = 1− × 2 = ∴ 3 3 1 4 C.  , −  Any point on the line x − y − 5 = 0 will be of 5 5 the form ( t , t − 5). Chord of contact of this point w.r.t. curve x 2 + 4 y 2 = 4 is tx + 4 (t − 5)y − 4 = 0 ⇒ (− 20 y − 4) + t ( x + 4 y ) = 0 which is a family of straight lines, each members of this family passes through point of intersection of straight lines − 20 y − 4 = 0 and x + 4 y = 0. 1 x2 y2 D. + = 1 ⇒ a = 2, b = 3 and e = 4 3 2 2a 4 Sum of distances = = =8 e 1/ 2

⇒ and

Y

X′

O

1

(6,0)

X

(2,0) Y′

x2 y2 + 4 9 (2 cos θ + 4)2 = + sin 2 θ 4 4 cos 2 θ + 16 + 16 cos θ + 4 sin 2 θ = 4 20 + 16 cos θ = = 5 + 4 cos θ 4 Hence, Emax − Emin = (9 − 1) = 8

Now, E =



85.  ± ae , 

b2   are extremities of the latusrectum having a

positive ordinates. ⇒ But

 b2  a2e 2 = − 2  − 2 a   b2 = a2 (1 − e 2 )

…(i) …(ii)

From Eqs. (i) and (ii), we get a2e 2 − 2 ae 2 + 2 a − 4 = 0 ⇒ ae 2 (a − 2 ) + 2(a − 2 ) = 0 ∴ (ae 2 + 2 ) (a − 2 ) = 0 Hence, a=2

86. Equation of the ellipse with centre at(1, − 1)can be written ( x − 1)2 ( y + 1)2 + =1 a2 b2 where, a = semi-major axis and b = semi-minor axis. ( x − 1)2 ( y + 1)2 If a = 8, then ellipse is + = 1. 64 b2 This passes through (1, 3). 16 ∴ 0 + 2 = 1 ⇒ b2 = 16 b ( x − 1)2 ( y + 1)2 Hence, equation of ellipse is + = 1. 64 16 as

87. Let S ≡ (5, 12 ) and S′ ≡ (24, 7 ) Q Conic passing through origin. Let P = (0, 0 ) ∴ S ′ P − SP = 12 and S ′ P + SP = 38 Distance between foci = 2ae 386 = 2 ae If conic is ellipse, then S ′ P + SP = 38 = 2 a 386 ∴Eccentricity, e = 38 and if conic is hyperbola, then S ′ P − PS = 12 = 2 a 386 Eccentricity, e = ⇒ k =2 ∴ 12

=

2

 10 34   10 34  sin θ − 1 +  sin θ + 1  33 33     cos θ sin θ + (67 )2 (33)2 2

=

2

15

2 [100 × 34 sin 2 θ + (33)2 ] (33)2 cos 2 θ + (67 )2 sin 2 θ (67 )2 2 × (67 )2 [(67 )2 − (33)2 sin 2 θ + (33)2 ] (33)2 cos 2 θ + (67 )2 sin 2 θ

Ellipse

x y cos θ + sin θ = 1. 67 33 Sum of the squares of the lengths of the perpendicular from (0, ± 10 34 ) on this tangent

88. Any tangent to the ellipse is

= 2 × (67 )2

2

= 8978 l =2

⇒

.

Entrances Gallery 1. Let the foot of perpendicular be P(h, k ). Equation of tangent with slope m and passing through P(h, k ) is h y = mx ± 6m2 + 2, where m = − k 6 h2 h2 + k 2 +2 = ⇒ k k2 ⇒ 6 h 2 + 2 k 2 = (h 2 + k 2 )2

3. Given equation of ellipse is Y

r X′

So, required locus is 6 x + 2 y = ( x + y ) . 2

2

2 2

X

x2 y2 + =1 4 3 ⇒

Y′

3 y= 2 Y

4 − h at x = h 2

P h,√3 √4 – h2 2 T (h, 0)

X R (x1,0)

Q h,–√3 √4 – h2 2

Let R( x1, 0 ) PQ is chord of contact. xx1 So, =1 4 4 ⇒ x= x1 which is equation of PQ,x = h. 4 So, =h x1 4 x1 = ⇒ h 1 ∆(h ) = Area of ∆PQR = × PQ × RT 2 1 2 3 3 = × 4 − h 2 × ( x1 − h ) = (4 − h 2 )3 / 2 2 2 2h − 3 (4 + 2 h 2 ) 4 − h2 ∆′ (h ) = 2h2 which is always decreasing. 1 45 5 So, at h = ∆1 =Maximum of ∆(h ) = 8 2 9 ∆ 2 = Minimum of ∆(h ) = at h = 1 2 8 45 5 9 8 So, ∆1 − 8∆ 2 = × − 8 ⋅ = 45 − 36 = 9 8 2 5 5

Targ e t E x e rc is e s

2.

2

x2 y2 + = 1. 16 9

9 7 = 16 4  7  , 0 = (± 7 , 0 ) ∴Foci is (± ae , 0 ) =  ± 4 × 4   Here, a = 4, b = 3, e = 1 −

∴Radius of the circle, r = (ae )2 + b2 = 7 + 9 = 16 = 4 Now, equation of circle is ( x − 0 )2 + ( y − 3)2 = 16 ⇒

x 2 + y 2 − 6y − 7 = 0

4. Equation of ellipse is( y + 2 ) ( y − 2 ) + λ( x + 3) ( x − 3) = 0 If passes through (0, 4), then λ = Equation of ellipse is ∴

4 3

x2 y2 + =1 12 16 1 e= 2 Y

(0, 4) (–3, 2) x = –3

y=2 x=3 X

X′ (–3, –2)

y = –2

(3, –2)

Y′

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Objective Mathematics Vol. 1

15

Aliter x2 y2 Let the ellipse be 2 + 2 = 1 as it is passing through a b (0, 4) and (3, 2 ). 9 4 So, b2 = 16 and + =1 2 16 a 2 ⇒ a = 12 2 So, 12 = 16 (1 − e ) ⇒ e = 1/ 2

5. Equation of chord of contact x y ⋅3 + ⋅4 =1 ⇒ 9 4

x + y =1 3 P (3, 4)

Y A

D

X′

O

X

B

Y′

Ta rg e t E x e rc is e s

⇒

x = 3 (1 − y ) x2 y2 Solving with ellipse + = 1, we get 9 4 2 y (1 − y )2 + =1 4 2 2 ⇒ 4 ( y + 1 − 2 y) + y = 4 ⇒ 5y 2 − 8y = 0 8 ⇒ y = 0 and y = 5 8 −9  ⇒ x = 3 and 3 1 −  ⇒ x = 3 and  5 5  − 9 8 , . ∴Points are (3, 0) and   5 5

6. Slope of AD must be 0.

8 9  = 0x +  ⇒  5 5 Thus, y-coordinate of D is 8 / 5. ⇒

y−

y=

8 5

8 Hence, y-coordinate of the orthocentre must be . 5

9. Statement I Standard equation of tangent with slope m to the parabola y 2 = 16 3 x is 4 3 …(i) y = mx + m Standard equation of tangent with slope m to the ellipse x2 y2 + = 1 is 2 4 …(ii) y = mx ± 2 m2 + 4 If a line L is a common tangent to both parabola and ellipse, then L should be tangent to parabola i.e. its equation should be like Eq. (i). L should be tangent to ellipse i.e. its equation should be like Eq. (ii). i.e. L must be like both of the Eqs. (i) and (ii). Hence, on comparing Eqs. (i) and (ii), we get 4 3 = ± 2 m2 + 4 m On squaring both sides, we get m2 (2 m2 + 4) = 48 ⇒ m4 + 2 m2 − 24 = 0 ⇒ (m2 + 6) (m2 − 4) = 0 [Q m2 ≠ − 6] ⇒ m2 = 4 ⇒ m=±2 On substituting m = ± 2 in the Eq. (i), we get the required equation of the common tangents as y = 2x + 2 3 and y = − 2x − 2 3 Hence, Statement I is correct. Statement II We have already seen that, if the line 4 3 is a common tangent to the parabola y = mx + m x2 y2 y 2 = 16 3 x and the ellipse + = 1, then it satisfies 2 4 4 2 the equation m + 2 m − 24 = 0. Hence, Statement II is also correct but it is not able to explain the Statement I.

10. Given,

x2 y2 + 2 =1 2 a b

which passes through P(− 3, 1) and e =

7. Since, locus is the parabola.

Y

3x 4y x + =1 ⇒ + y =1 3 9 4 x + 3y − 3 = 0 ( x + 3 y − 3)2 2 ( x − 3) + ( y − 4)2 = 10 10 x 2 + 90 − 60 x + 10 y 2 + 160 − 80 y = x 2 + 9 y 2 + 9 + 6 xy − 6 x − 18 y 9 x 2 + y 2 − 6 xy − 54 x − 62 y + 241 = 0

Equation of AB is ⇒ ∴ ⇒ ⇒

(–3, 1) P X′

8. Diameter of circle ( x − 1)2 + y 2 = 1 is 2 units and that of

914

circle x 2 + ( y − 2 )2 = 4 is 4 units. Semi-minor axis of ellipse, b = 2 units and semi-major axis of ellipse, a = 4 units. Hence, equation of the ellipse is x2 y2 x2 y2 + = 1 ⇒ + =1 16 4 a2 b2 ⇒ x 2 + 4 y 2 = 16

X

O

Y′

∴ ⇒ and

9 1 b2 + 2 = 1 and e 2 = 1 − 2 2 a b a 9 5 + =1 a2 3 a2 2 b2 = 1− 2 5 a

2 5

14. Let the coordinates of the vertices of rectangle ABCD be A(a cos θ, b sin θ), B (− a cos θ, b sin θ), C (− a cos θ, − b sin θ ) and D (a cos θ, − b sin θ ), then Length of rectangle, AB = 2 a cos θ and breadth of rectangle, AD = 2 b sin θ Area of rectangle, A = AB × AD ∴ = 2 a cos θ × 2 b sin θ …(i) ⇒ A = 2 ab sin 2θ Y

x2 y2 11. Let the equation of the required ellipse be + 2 = 1. 16 b But the ellipse passes through the point (2, 1.) x2 y2 + =1 1 4

Y

(–a cos θ, b sin θ) B

X′

A (a cos θ, b sin θ) X

O

A (2,1) (0, 1) X′

D (a cos θ, – b sin θ)

C (–a cos θ, –b sin θ)

X (2, 0)

15 Ellipse

27 + 5 b2 3 = = 1 and 2 a2 5 3a 32 32 and b2 = ⇒ a2 = 3 5 ∴ Equation of ellipse is 3x 2 5y 2 + =1 32 32 2 2 ⇒ 3 x + 5 y = 32 ⇒

(4, 0)

Y′

4 1 1 1 3 ⇒ + 2 = 1 ⇒ 2 = ⇒ b2 = 3 4 b 4 b x 2 3y 2 Hence, equation is + =1 16 4 ⇒ x 2 + 12 y 2 = 16

12. Since, ∴

1 a − ae = 4 and e = 2 e a 2a − = 4 2 Y

X′

X

Y′ x = 4

⇒

8 3a =4 ⇒ a= 3 2

2 ae = 6 and 2 b = 8 ⇒ ae = 3 and b = 4 ae 3 = ⇒ b 4 2 b 16 e 2 = ⇒ 9 a2 2 b We know that, = 1− e2 a2 16 e 2 = 1− e2 ⇒ 9  16 + 9 2 ⇒   e =1  9   25 2 ⇒   e =1  9 9 3 e2 = ⇒ e= ⇒ 25 5

Targ e t E x e rc is e s

Y′

On differentiating Eq. (i) w.r.t. θ, we get dA = 2 × 2 ab cos 2θ dθ dA For maxima or minima, put =0 dθ ⇒ 4ab cos 2θ = 0 π π 2θ = ⇒ θ= ⇒ 2 4 d 2A Now, = − 8ab sin 2θ dθ 2 π  d 2 A At θ = ,  2  < 0 4  dθ  π ∴ Area is maximum at θ = . 4 Maximum area of rectangle = 2ab sq units [from Eq. (i)] Aliter From Eq. (i), Area of rectangle, A = 2 ab sin 2 θ ∴ A is maximum, when sin 2 θ = 1 ⇒ Maximum area of rectangle = 2 ab sq units

15. Since, ∠FBF′ = 90 °, then ∠OBF′ = 45° and

13. Given that,

∠BF′ O = 45°

Y B (0, b) 45° X′

O F F′ (–ae, 0) (ae, 0)

A (a, 0)

X

Y′

⇒ and ⇒

ae = b [Q ∆BOF′ is an isosceles triangle] b2 a2e 2 e2 = 1− 2 ⇒ e2 = 1− 2 a a e2 = 1− e2

915

Objective Mathematics Vol. 1

15

⇒

2e 2 = 1 ⇒ e =

1 2 [Qe cannot be negative]

Aliter Since, F and F′ are foci of an ellipse, whose coordinates are (ae , 0 ) and (− ae , 0 ) respectively and coordinates of B are (0, b). b Slope of BF = ∴ − ae b and Slope of BF′ = ae Q ∠FBF′ = 90 ° b  b − ⋅   = − 1 ⇒ b2 = a2e 2 ∴ ae  ae  b2 a2e 2 ∴ e2 = 1− 2 = 1− 2 a a 1 2 ⇒ 2e = 1 ⇒ e = 2

16. Since, equation of directrix is x = 4, then major axis of an ellipse is along X-axis. a 1 = 4 ⇒ a= 4× e 2 ⇒ a=2 Now, b2 = a2 (1 − e 2 ) 1 3  ∴ b2 = 4  1 −  = 4 ×  4 4 ⇒ b2 = 3 x2 y2 Hence, equation of ellipse is + =1 4 3 2 2 ⇒ 3 x + 4 y = 12

Ta rg e t E x e rc is e s

∴

17. The equation of an ellipse is x2 y2 + =1 16 9 Here, a = 4, b = 3 b2 a2 9 7 = 1− = 16 4 Foci of an ellipse are (± ae , 0 ) i.e. (± 7 , 0 ). ∴Radius of required circle Eccentricity, e = 1 −

= ( 7 − 0 )2 + (0 − 3)2 = 7 + 9 = 16 = 4units 1 18. Given that, e = , ae = 2 ⇒ a = 4 2 ∴ b2 = a2 (1 − e 2 ) 1  = 16 1 −  = 12  4 Hence, the equation of ellipse are x2 y2 + =1 16 12

19. Given equation of ellipse can be rewritten as

916

( x + 1)2 ( y − 1)2 + =1 1 4 ∴Parametric equations of ellipse are x + 1 = cos θ and y − 1 = 2 sin θ ⇒ x = cos θ − 1 and y = 2 sin θ + 1

1  Qe = 2  

3 5

20. Given, PS = 6 and e = PS =e PM

Q

⇒

6 3 = PM 5

Y P X′

M

S

X

Y′

∴

PM =

6× 5 = 10 3

21. The y-coordinate of foci is zero. ∴Major axis is on X-axis. ∴ ae = 4 x2 y2 Let equation of ellipse be 2 + 2 = 1 a b (4 2 )2 (2 6 )2 ∴ + 2 =1 a2 a − 16 [Q b2 = a2 (1 − e 2 ) = a2 − 16] 32 24 + 2 =1 ⇒ 2 a a − 16 ⇒ 32 a2 − 512 + 24a2 = a2 (a2 − 16) ⇒ 56a2 − 512 = a4 − 16 a2 ⇒ a4 − 72 a2 + 512 = 0 4 ⇒ a − 64a2 − 8 a2 + 512 = 0 ⇒ a2 (a2 − 64) − 8 (a2 − 64) = 0 ⇒ (a2 − 8) (a2 − 64) = 0 ⇒ a2 = 64 ⇒ a = 8 [Q a2 = 8 is not possible] Q ae = 4 ⇒ 8× e = 4 1 ∴ e= 2

22. Equation of ellipse is 9 x 2 + 25 y 2 = 225 or

x2 y2 + =1 25 9

Here, a = 5, b = 3 Now, PF1 = 4 PF2 ∴ PF1 + PF2 = 5PF2 , 2 a = 5PF2 ⇒ 10 = 5PF2 ⇒ 2 = PF2 Now, c 2 = a2 − b2 = 25 − 9 = 16 ∴ F1 ≡ (4, 0 ), F2 ≡ (− 4, 0 ) Let the coordinates of P be (5 cos θ, 3 sin θ ). ∴ PF2 = 2 2 ⇒ (5 cos θ + 4) + (3 sin θ )2 = 4 ⇒ 16 cos 2 θ + 40 cos θ + 21 = 0 ⇒ ⇒ or

(4 cos θ + 3) (4 cos θ + 7 ) = 0 3 4 7 cos θ = − 4 cos θ = −

[rejected]

7 4

25. Equation of tangent at (a cos θ, b sin θ ) to the ellipse is x y cos θ + sin θ = 1 a b Y

4

Q ( a co s

√7 θ

X′

θ, b

sin

O

15 Ellipse

sin θ =

and

θ) P

X

3

 15 3 7   15 63  P ≡ − ,  ≡ − ,  4   4 4   4

23. Given equation of ellipse is

Here, and

x2 y2 + =1 144 25 a2 = 144 b2 = 25

b2 Now, e = 1− 2 a 25 = 1− 144 119 = = 144 ∴Foci of an ellipse = (± ae , 0 )  =  ± 12 × 

Y′

  a  b  Coordinates of P and Q are  , 0 and  0, ,  sin θ   cos θ  respectively. 1 a b ab Now, area of ∆OPQ = × = 2 cos θ sin θ sin 2θ ∴

Minimum area = ab

26. Given circle is a director circle to the given ellipse. ∴Angle between tangent is θ = 119 12 119  , 0 12 

27. Q2 a = 3 ⋅ 2 ⋅ b ⇒ a = 3 b ∴ l=

(3b)2 − b2 a2 − b2 = = (3 b)2 a2

28. In ∆BOT,

Y B 60

Since, the circle with centre (0, 2 ) and passing through foci (± 119, 0 ) of the ellipse.

°

Radius of circle = ( 119 − 0 )2 + (0 − 2 )2

X′

= 119 + 2 = 121 = 11

60° 60° O S T (–ae, 0) (ae, 0)

24. The given equation of second ellipse can be rewritten as

Since, Eqs. (i) and (iii) represent the same line. h/ 6 k/3 = =1 cos θ sin θ 2 ⇒ h = 3 cos θ and k = 3 sin θ Hence, locus is x 2 + y 2 = 9.

X (a, 0)

Y′

x2 y2 + =1 4 1 Equation of tangent to this ellipse is x …(i) cos θ + y sin θ = 1 2 Equation of the first ellipse can be rewritten as x2 y2 …(ii) + =1 6 3 Let Eq. (i) meets the first ellipse at P and Q and the tangents at P and Q to the first ellipse intersected at (h, k ), then Eq. (i) is the chord of contact of (h, k ), with respect to the ellipse (ii) and thus its equation is hx ky …(iii) + =1 6 3

8 b2 2 2 = 3 9 b2

b = tan 60 ° ⇒ b = ae 3 ae

= (± 119, 0 )

∴

π . 2

Targ e t E x e rc is e s

∴

b a2e 2 3 = 1 − a2 a2 e 2 = 1 − 3e 2 ⇒ 4e 2 = 1 1 1 e =± ⇒ e= 2 2 [since, e cannot be negative] e = 1−

∴ ⇒

2

x = 2y x2 and + y2 = 1 4 On putting x = 2 y in Eq. (ii), we get (2 y )2 + y2 = 1 4 4y 2 ⇒ + y2 = 1 4 ⇒ 2 y2 = 1 1 y=± ⇒ 2 From Eq. (i), x = ± 2 1 1   P  2, ∴   and Q  − 2 , −   2 2

29. We have,

…(i) …(ii)

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Objective Mathematics Vol. 1

15

∴Equation of circle is 1  1  (x − 2 )(x + 2 ) +  y −  y +  =0  2  2 1 ⇒ x2 − 2 + y2 − = 0 2 ⇒ x 2 + y 2 = 5/ 2

30. Given,

x2 y2 + =1 9 1

1 2 2 = 9 3 Two foci are (± ae , 0 ) i.e. (± 2 2 , 0 ). Let P(h, k ) be any point on the ellipse. k−0 k−0 ∴ × = − 1 [from given condition] h −2 2 h + 2 2 ⇒ h2 − 8 = − k 2 2 ⇒ x + y2 = 8 e = 1−

⇒

31. We have, b = 5 and b > a c 2 = b2 − a2 9 = 25 − a2 a2 = 16

Since, ⇒ ⇒

Y

Ta rg e t E x e rc is e s

(0, 5)

(0, 3) c O (0, 0)

X′

X

Y′

∴Required equation of ellipse is x2 y2 + =1 16 25

32. We have, 16 x 2 + 25 y 2 = 400 16 x 2 25 y 2 + =1 400 400 x2 y2 ⇒ + =1 25 16 2 b2 ∴Length of latusrectum = a 2 × 16 = 5 32 units = 5 ⇒

33. Here, a = 5, b = 4 ∴ Required sum = a + b = 9

34. According to the question, 2a =

918

17 17 b ⋅ 2b ⇒ a = 8 8

b2 = a2 (1 − e 2 ) 289 2 b2 = b (1 − e 2 ) ⇒ 64 64 ⇒ 1− e2 = 289 225 ⇒ e2 = 289 15 ∴ e= 17  a b 35. Equation of tangent at P  ,  is  2 2 x⋅a y⋅b + =1 2 2a 2 b2 x y ⇒ + =1 2a 2b ⇒ xb + ya = 2 ab  a b Equation of normal at P  ,  is  2 2 a2 ⋅ x b2 ⋅ y − = a2 − b2 a b 2 2 ⇒ 2 ax − 2 by = a2 − b2 ⇒ 2 (ax − by ) = a2 − b2 Also, equation of X-axis is y = 0. Now, from Eqs. (i) and (iii), xb = 2 ab x = 2a ⇒ ∴Point of intersection is ( 2 a, 0 ). Again, from Eqs. (ii) and (iii), we get 2 ax = a2 − b2 a2 − b2 x= ⇒ 2a 2  a − b2  ∴Point of intersection is  , 0 .   2a Q

…(i)

…(ii) …(iii)

On solving Eqs. (i) and (ii), we get b a and x = y= 2 2  a b , ∴ Point of intersection is  .  2 2 2a 0 1 1 a2 − b2 ∴Area of triangle = 0 1 2 2a a b 1 2 2  a2 − b2 b   1  b ⋅ =  2a  −   + 1  2 2 2a 2    b 2 1 2 ab = − + (a − b2 ) 2a 2 2  =

− b (a2 + b2 ) b(a2 + b2 ) = 4a 4a

16 Hyperbola Introduction

The constant ratio is generally denoted by e and is known as the eccentricity of the hyperbola. If S is the focus, ZZ′ is the directrix and P is any point on the hyperbola as shown in the adjacent figure then by definition, we have SP = e ⇒ SP = e PM PM

Z M Directrix

A hyperbola is the locus of a point in a plane, which moves in the plane in such a way that the ratio of its distance from a fixed point (called focus) to its distance from a fixed line (called directrix) is always constant which is always greater than unity.

Chapter Snapshot P

S (Focus)

●

Introduction

●

Conjugate Hyperbola

●

Z′ ●

Position of a Point with Respect to a Hyperbola Intersection of a Line and a Hyperbola

●

Tangent to a Hyperbola

●

Director Circle

Or The hyperbola is the set of all points in a plane, the difference of whose distance from two fixed points in the plane is a constant.

●

Normals to a Hyperbola

The term ‘difference’ that means the distance to the farther point minus the distance to the closer point. The two fixed points are called the foci of the hyperbola. The mid-point of the line segment joining the foci is called the centre of the hyperbola. The line through the foci is called the transverse axis and the line through the centre and perpendicular to the transverse axis is called the conjugate axis. The points at which the hyperbola intersects the transverse axis are called the vertices of the hyperbola.

●

Equations of Chord

●

Pole and Polar

●

Conjugate points

●

Conjugate Lines

●

Diameter

●

Asymptotes

●

Rectangular Hyperbola

Conjugate axis P2 P3

●

P1 F1 Focus

Centre

F2 Focus

●

Transverse axis

Vertex Vertex

P1 F2 − P1 F1 = P2 F2 − P2 F1 = P3 F1 − P3 F2 We denote the distance between the two foci by 2c, the distance between the two vertices (the length of the transverse axis) by 2a and we define the quantity b as b = c 2 − a 2 .

●

Equation of the Pair of Tangents

Tangent to a Rectangular Hyperbola Normals to a Rectangular Hyperbola

Objective Mathematics Vol. 1

16

Standard Equation of Hyperbola

⇒

SP 2 = e 2 PM 2

In this section, we shall derive the equation of the hyperbola in the standard form for which we have to choose its focus and directrix in a specific way as given below: Let S be the focus, ZK be the directrix and e be the eccentricity of the hyperbola whose equation is required. Draw SK perpendicular from S on the directrix ZK and divide SK internally and externally at A and A ′ (on SK produced) respectively in the ratio e :1.

⇒

SP 2 = e 2 KN 2

⇒

SP 2 = e 2 (CN − CK ) 2

Z′

Y

M′

Z

S′ (–ae, 0)

P (x, y)

M

A′ K′ (–a, 0)

x= –

C

a e

K A (a, 0)

⇒

x 2 ( e 2 − 1) − y 2 = a 2 ( e 2 − 1) x2

⇒

a2

N S (ae, 0)

−

y2 a 2 ( e 2 − 1) x2 a2

−

where,

2

=1

y2

=1 b2 b 2 = a 2 ( e 2 − 1)

This is the equation of the hyperbola in the standard form. X

a x =e

Y′

920

a  ( x − ae) 2 + y 2 = e 2  x −   e

⇒ L

X′

⇒

Then, …(i) SA = e AK and …(ii) SA ′ = e A ′K Since, A and A ′ are such points that their distances from the focus bear constant ratio e ( >1) to their respective distances from the directrix. Therefore, these points lie on the hyperbola. Let AA ′ = 2a and C be the middle point of AA ′. Then, CA = CA ′ = a On adding Eqs. (i) and (ii), we get SA + SA ′ = e ( AK + A ′K ) ⇒ CS − CA + CS + CA ′ = e (CA ′ − CK + CA ′ + CK ) ⇒ 2 CS = 2ae ⇒ CS = ae On subtracting Eq. (i) from Eq. (ii), we get SA ′ − SA = e ( A ′ K − AK ) ⇒ (CS + CA ′ ) − (CS − CA ) = e (CA ′ + CK − CA + CK ) ⇒ AA ′ = 2e (CK ) ⇒ 2a = 2e (CK ) a CK = ⇒ e Let C be the origin, CSX be the X -axis and a straight line CY through C perpendicular to CX as the Y-axis. Let P ( x, y) be any point on the hyperbola and PM, PN be the perpendiculars from P on KZ and KX . By definition, we have SP = e PM

Or Let ( ± c, 0) be the foci of a hyperbola. Then, its centre is (0, 0). Let P ( x, y) be any point on the hyperbola. According to the definition of hyperbola, PF1 − PF2 = 2a (2a < 2c, i.e. c > a) Y P X′

X F1(–c,0)

O

F2(c,0)

Y′

⇒

( x + c) 2 + y 2 − ( x − c) 2 + y 2 = 2a

i.e.

( x + c) 2 + y 2 = 2a + ( x − c) 2 + y 2

On squaring both sides, we get ( x + c) 2 + y 2 = 4a 2 + 4a ( x − c) 2 + y 2 + ( x − c) 2 + y 2 On simplifying, we get cx − a = ( x − c) 2 + y 2 a On squaring again and further simplifying, we get y2 x2 − =1 a 2 c2 − a 2 x 2 y2 [Q c 2 − a 2 = b 2 ] i.e. − =1 a 2 b2 Hence, any point on the hyperbola satisfies the equation x 2 y2 − =1 a 2 b2

Example 1. Two circles are given such that they neither intersect nor touch. The locus of centre of variable circle which touches both the circles externally, is (a) circle (b) parabola (c) ellipse (d) hyperbola

Some Basic Terms Associated to Hyperbola In this section, we shall define various terms x 2 y2 associated to the hyperbola 2 − 2 = 1 as given below: a b Z′

Y

16 Hyperbola

X

Z

Sol. (d) In the figure, circles with solid line have centres C1 and C 2 and radii r1 and r2 , respectively.

r2

r1

C1

r1

T

P (x, y)

M L

X′

C2

r2 r

M′

S′ (–ae, 0)

C

A′ K′ (–a, 0)

r T′

x= –

C

N S (ae, 0)

K A (a, 0)

X

a x = e L'

a e

Y′

Let the circle with centre C and radius r is a variable circle which touches both circles externally. Now, CC1 = r + r1 and CC 2 = r + r2 Hence, [constant] CC1 − CC 2 = r1 − r2 Then, by definition, the locus of C is hyperbola, whose foci are C1 and C 2 . X

Example 2. If base of triangle and ratio of tangents of half of base angles are given, then the locus of opposite vertex is (a) hyperbola (b) ellipse (c) parabola (d) circle Sol. (a) In ∆ABC, base BC = a

[given]

A

c

b

B

B 2 =k ⇒ C tan 2 tan

Also,

⇒ ⇒ ⇒ ⇒ ⇒

C

a

( s − c ) ( s − a) s (s − b) ( s − a) ( s − b ) s (s − c )

s−c =k s−b (s − c ) − (s − b) k − 1 = (s − c ) + (s − b) k + 1 b−c k −1 = 2 s − (b + c ) k + 1 (k − 1) b−c = a (k + 1) AC − AB = Constant

So, locus of vertex of ∆ABC is a hyperbola.

=k

Vertices The points A and A′, where the curve meets the line joining the foci S and S ′, are called the vertices of the hyperbola. The coordinates of A and A′ are ( a, 0) and ( −a, 0), respectively. Transverse and conjugate axes The straight line joining the vertices A and A′ is called the transverse axis of the hyperbola. Its length AA′ is generally taken to be 2a. The straight line through the centre which is perpendicular to the transverse axis does not meet the hyperbola in real points. But if B, B ′ are the points on this line such that CB = CB ′ = b, the line BB ′ is called the conjugate axis such that BB ′ = 2b. Foci The points S ( ae, 0) and S ′ ( −ae, 0) are the foci of the hyperbola. Directrices ZK and Z ′ K ′ are two directrices of the a a hyperbola and their equations are x = and x = − , e e respectively. Centre The middle point C of AA′ bisects every chord of the hyperbola passing through it and is called the centre of the hyperbola. x 2 y2 Eccentricity For the hyperbola 2 − 2 = 1, we a b have b 2 = a 2 ( e 2 − 1) ⇒ ⇒ ⇒

e2 = e = 1+

b2 a2

a 2 + b2 a2

=1 +

⇒ e = 1+ e = 1+

b2 a2

(2b) 2 (2a ) 2

(Conjugate axis) 2 (Transverse axis) 2

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16

Its shape is shown in the figure.

Latusrectum LSL′ is the latusrectum and LS is called the semi-latusrectum. TS ′ T ′ is also a latusrectum.

Y

The coordinates of L are ( ae, SL). As L lies on the x 2 y2 hyperbola 2 − 2 = 1 and the coordinates of L will a b satisfy the equation of the hyperbola. Therefore, ( ae) 2 ( SL) 2 − =1 a2 b2 ⇒ ( SL) 2 = b 2 ( e 2 − 1)

X′

C

K′

b2 a Hence, length of the latusrectum = 2 ( SL) 2b 2 = = 2a ( e 2 − 1) a Focal distances of a point Let P ( x, y) be any point on the hyperbola x 2 y2 − =1 a 2 b2 Then, by definition, we have SP = e ⋅ PM and S ′ P = e ⋅ PM ′ ⇒ SP = e ⋅ NK = e(CN − CK ) a  = e  x −  = ex − a  e S ′ P = e ⋅ PM ′ = e( NK ′ ) = e(CN + CK ′ ) a  = e  x +  = ex + a  e S ′ P − SP = ( ex + a ) − ( ex − a ) = 2a = Transverse axis

Conjugate Hyperbola The hyperbola whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola, is called the conjugate hyperbola of the given hyperbola.

a2

−

y2 b2

= 1 is −

x2 a2

+

Z' y= –b/e

The eccentricity of the conjugate hyperbola is given by a 2 = b 2 ( e 2 − 1) and length of the latusrectum is

y2 b2

=1

2a 2 . b

Various results related to the hyperbola and its conjugate −

x2

Hyperbola

+

x2 a2

−

y2 b2

=1

y2

= 1 are given in the a 2 b2 following table for ready reference: Conjugate hyperbola

Hyperbola x

Terms

2

a2

Coordinates of the centre

−

y

2

b2

=1

−

x2 a2

+

y2 b2

=1

(0, 0)

(0, 0)

Coordinates of the vertices

( a, 0) and ( −a, 0)

( 0, b ) and ( 0, − b )

Coordinates of foci

( ± ae, 0)

( 0, ± be )

Length of transverse axis

2a

2b

Length of conjugate axis

2b

2a

a x=± e

Equation of the directrices Eccentricity

e =

a2 + b 2 a

2

or b 2 = a 2(e 2 − 1)

y=± e =

b e

b 2 + a2

b2 or a 2 = b 2(e 2 − 1)

Length of latusrectum

2 b2 a

2 a2 b

Equation of transverse axis

y=0

x=0

Equation of the conjugate axis

x=0

y=0

Ø

●

The hyperbola conjugate to the hyperbola x2

X

Y′

SL = SL′ =

⇒

922

A(a, 0)

B′(0, –b)

b2 SL = a

⇒

y = b/e Z

K

S′ (0,–be)

 2 b2  2 2 2 ( ) Q b = a e − 1 ⇒ e − 1 =   a 2  

∴

B (0,b) A′(_ a, 0)

 b2  ( SL) 2 = b 2  2  a 

⇒

and

S (0, be)

●

If e1 and e 2 are the eccentricities of a hyperbola and its conjugate respectively, then 1 1 + =1 e12 e 22 The foci of a hyperbola and its conjugate are concyclic and form the vertices of a square.

X

Equation of such hyperbola is ( x − h) 2 ( y − k ) 2 − =1 a2 b2 Various results ( x − h )2 ( y − k )2 − =1 a2 b2

( y − k )2 b2

−

( x − h) 2

Hyperbola Terms

a2

related

and

to

hyperbola

its

Coordinates of the vertices

( h ± a, k )

( h, k ± b )

Coordinates of foci

( h ± ae, k )

( h, k ± be )

Length of transverse axis

2a

2b

Length of conjugate axis

2b

2a

x=h± e =

a e

a2 + b 2 a

2

y=k± e =

b2

2 b2 a

2 a2 b

Equation of transverse axis

y=k

x=h

Equation of conjugate axis

x=h

y=k

X

b e

a2 + b 2

Length of latusrectum

Example 3. The distance between the foci of a hyperbola is 16 and its eccentricity is 2. Its equation is (a) x 2 − y 2 = 32 (b) 9x 2 − 4 y 2 = 36 (c) 2x 2 − 3 y 2 = 7 (d) None of the above 2 2 Sol. (a) Let the equation of the hyperbola be x2 − y 2 = 1.

a b The coordinates of the foci are (ae, 0) and (−ae, 0). [Qe = 2 ] ∴ 2 ae = 16 ⇒ 2 a × 2 = 16 a=4 2 ⇒ Also, b 2 = (ae )2 − a2 = 64 − 32 = 32 Thus,

a2 = 32

and

b 2 = 32

Hence, the equation of hyperbola is x2 − y2 = 32.

…(i)

x2 y2 1 The foci of the hyperbola − = 144 81 25 x2 y2 12 i.e. − = 1 are given by  ± e ′, 0 , where  5  144 81 25 25 81 144 2 = (e ′ − 1) 25 25 ⇒ 81 = 144(e ′2 − 1) 81 e′2 − 1 = ⇒ 144 81 ⇒ e′2 = 1 + 144 225 = 144 15 5 e′ = = ⇒ 12 4 12 Given, e′ 5e = 5 12 5 5e = × =3 ⇒ 5 4 3 e= ⇒ 5 From Eq. (i),  32  b 2 = 25  1 − 2   5  25 × 16 = 25 = 16

Conjugate hyperbola Hyperbola ( y − k ) 2 ( x − h) 2 ( x − h) 2 ( y − k ) 2 − =1 − =1 b2 a2 a2 b2 ( h, k )

Eccentricity

25 b b 2 = 25(1 − e 2 )

= 1 are given in the following table:

( h, k )

Equation of directrices

2 2 Sol. (c) The foci of the ellipse x + y 2 = 1are(± 5e, 0), where

conjugate

Coordinates of the centre

16

Example 4. If the foci of the ellipse y2 1 x 2 y2 x2 + 2 = 1 and the hyperbola − = 25 b 144 81 25 coincide, then the value of b 2 is (a) 18 (b) −16 (c) 16 (d) −18

Hyperbola

Equation of Hyperbola whose Axes are Parallel to Coordinates Axes and Centre is ( h, k )

Parametric Coordinates Let P ( x, y) be any point on the hyperbola y2 x − = 1. Draw PL perpendicular from P on OX and a 2 b2 then a tangent LM from L to the circle described on A ′ A as diameter. Then, x = CL = CM sec θ = a sec θ 2

On putting x = a sec θ in

x2

−

y2

= 1, we obtain a 2 b2 y = b tan θ. Thus, the coordinates of any point on the x 2 y2 hyperbola 2 − 2 = 1 are ( a sec θ, b tan θ), where θ is a a b parameter such that 0 ≤ θ ≤ 2π.

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Objective Mathematics Vol. 1

16

These coordinates are known as the parametric coordinates. The parameter θ is also called as the eccentric angle of point P on the hyperbola. Y

M a X′

A′

C

P (x, y)

θ AL

X

x2 - y 2 =1 a2 b2 Y′

The equation x = a sec θ and y = b tan θ are known as x 2 y2 the parametric equation of the hyperbola 2 − 2 = 1. a b Ø

●

●

The circle x 2 + y 2 = a2 is known as the auxiliary circle of the hyperbola.

 eθ + e−θ   eθ − e−θ   and y = b   are also 2 2    

The equations x = a 

known as the parametric equations of the hyperbola x2 y2 − = 1. a2 b 2 X

Example 5. If equation of hyperbola is x 2 y2 − = 1, then equation of auxiliary circle is 25 16 (b) x 2 + y 2 = 9 (a) x 2 + y 2 = 41 (c) x 2 + y 2 = 25 (d) x 2 + y 2 = 16 Sol. (c) Given equation of hyperbola is x2 y2 − =1 25 16 Here, foci is lie on X-axis. ∴ a2 = 25 and b 2 = 16 Hence, equation of auxiliary circle is x2 + y2 = 25

X

Example 6. Any point on the hyperbola ( x + 1) 2 ( y − 2) 2 − = 1 is of the form 16 4 (a) ( 4 sec θ, 2 tan θ) (b) ( 4 sec θ −1, 2 tan θ + 2) (c) ( 4 sec θ − 1, 2 tan θ − 2) (d) ( 4 sec θ − 4, 2 tan θ − 2) Sol. (b) Given equation of hyperbola is

924

( x + 1)2 ( y − 2 )2 − =1 16 4 Here, a = 4 and b = 2 Now, x + 1 = 4 sec θ and y − 2 = 2 tan θ ⇒ x = 4 sec θ − 1 and y = 2 tan θ + 2 Hence, the required point is (4sec θ − 1, 2 tan θ + 2 ).

Position of a Point with Respect to a Hyperbola The point ( x1 , y1 ) lies outside, on or inside the x 2 y2 hyperbola 2 − 2 = 1 according as a b x12 y12 − − 1 0. a 2 b2 X

Example 7. The position of the point (5, − 4) relative to the hyperbola 9x 2 − y 2 − 1 = 0, is (a) outside (b) on the hyperbola (c) inside (d) None of these Sol. (c) The equation of hyperbola is 9 x2 − y2 − 1 = 0 At point (5, − 4), we have 9(5)2 − (−4)2 − 1 = 225 − 16 − 1 = 208 ⇒ 208 > 0 So, the point (5, − 4) lies inside the hyperbola 9 x 2 − y2 = 1

Intersection of a Line and a Hyperbola Consider the hyperbola

x2 a2

−

y2 b2

=1

…(i)

and the line y = mx + c …(ii) The coordinates of the points of intersection of Eqs. (i) and (ii) satisfy both the equations and therefore can be obtained by solving Eqs. (i) and (ii) as simultaneous equations. Substituting the value of y from Eq. (ii) in Eq. (i), we get x 2 ( mx + c) 2 − =1 a2 b2 ⇒ x 2 ( a 2 m 2 − b 2 ) + 2a 2 mcx + a 2 ( c 2 + b 2 ) = 0 …(iii) This equation being a quadratic in x, gives two values x which may be real and distinct, coincident or imaginary. Thus, a line intersects a hyperbola in two points which may be real and distinct, coincident or imaginary. The line will intersect the hyperbola in two distinct points, if the roots of the equation are real and distinct and the condition for the same is 4a 4 m 2 c 2 − 4a 2 ( c 2 + b 2 )( a 2 m 2 − b 2 ) > 0 ⇒ c2 > a 2 m2 − b2 The line will touch the hyperbola, if the roots of Eq. (iii) are equal. ∴ 4a 4 m 2 c 2 − 4a 2 ( c 2 + b 2 )( a 2 m 2 − b 2 ) = 0 ⇒ c2 = a 2 m2 − b2 ⇒

c = ± a 2m2 − b2

On substituting c = ± a m − b in Eq. (iii), we get 2

2

2

( a 2 m 2 − b 2 ) x 2 ± 2a 2 m a 2 m 2 − b 2 x + a 4 m 2 = 0

Equation of a Hyperbola Referred to Two Perpendicular Lines Consider the hyperbola

x=±

a 2m2 − b2

Sol. (b) The given line is x cos α + y sin α = p or y = − x cot α + p cosec α On comparing with y = mx + c, we get m = − cotα and c = pcosec α Now, the given line touches the hyperbola x2 y2 − 2 = 1 iff c 2 = a2 m2 − b 2 2 a b ⇔ p2 cosec 2 α = a2 (− cot α )2 − b 2 ⇔

= 1 as shown in the

N

Example 8. The line x cos α + y sin α = p x 2 y2 touches the hyperbola 2 − 2 = 1, if a b 2 2 2 2 a cos α − b sin α is equal to (c) − p 2 (d) 2 p 2 (a) p (b) p 2

p cosec α = 2

b2

P (x, y)

a 2m

Thus, the line y = mx + c touches the hyperbola x 2 y2 − = 1, if c 2 = a 2 m 2 − b 2 and the coordinates of a 2 b2 the point of contact are   a 2m b2 ± . ,±  2 2 2 2 2 2  a m −b a m −b  

2

y2

Y

On putting the values of x and c in Eq. (ii), we get b2 y=± a 2m2 − b2

X

a2

−

figure given below.

⇒ ( a 2 m 2 − b 2 x ± a 2 m) 2 = 0 ⇒

x2

16 Hyperbola

This is the condition for the line y = mx + c to touch x 2 y2 the hyperbola 2 − 2 = 1. a b

a2 cos 2 α − b 2 sin2 α sin2 α

X'

O

M

X

Y'

Let P ( x, y) be any point on the hyperbola. Then, PM = y and PN = x. PN 2 PM 2 ∴ − 2 =1 a2 b It follows from this that if perpendicular distance p1 and p2 of a moving point P ( x, y) from two mutually perpendicular coplanar straight lines L2 ≡ b1 x − a1 y + c2 = 0, L1 ≡ a1 x + b1 y + c1 = 0, respectively, satisfy the equation p12 p22 − =1 a 2 b2 a x +b y+c  1 1  1   2 2 a1 + b1  

2

b x −a y+c  1 2  1   2 2 b1 + a1  

2

=1 − b2 a2 Then, the locus of point P denotes a hyperbola in the plane of the given lines such that (i) The centre of the hyperbola is the point of intersection of the lines L1 = 0 and L2 = 0 . i.e.

P (x, y)

L1 ≡ a1x + b1y +c1=0

p2

p1

L2 ≡ b1x – a1y +c2=0

⇔ a2 cos 2 α − b 2 sin2 α = p2 X

Example 9. If the line y = 3x + λ touches the hyperbola 9x 2 − 5 y 2 = 45, then λ is equal to (b) ±6 (c) ±3 (d) ± 4 (a) ±3 6

a

Sol. (b) The equation of the hyperbola is 9 x2 − 5 y2 = 45. x2 y2 − =1 5 9 x2 y2 This is of the form 2 − 2 = 1, where a2 = 5 and b 2 = 9. a b The line y = 3 x + λ touches the given hyperbola, if λ2 = 5 (3)2 − 9 [using c 2 = a2 m2 − b 2 ] ⇒

⇒

λ2 = 45 − 9

⇒

λ = 36 ⇒ λ = ± 6 2

(ii) The transverse axis lies along L2 = 0 and the conjugate axis lies along L1 = 0. (iii) The length of the transverse and conjugate axes are 2a and 2b.

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Objective Mathematics Vol. 1

16

X

2y − x − 3 2x + y − 1 = X and = Y , the given 5 5 equation can be written as On putting

Example 10. The eccentricity of the conic 4(2 y − x − 3) 2 − 9(2x + y − 1) 2 = 80 is 3 13 (d) 3 (b) (c) 13 (a) 13 3

4 X 2 − 9Y 2 = 16

Sol. (b) Here, 2 y − x − 3 = 0 and 2 x + y − 1 = 0 are perpendicular to each other. Therefore, the equation of the conic can be written as 2

X2 Y2 =1 − 2 4  4    3

⇒

The eccentricity is given by

2

 2 x + y − 1 2 y − x − 3  4×5  = 80  −9×5  2 2 + 12   2 2 + 12  4 

⇒

2

e = 1+

2

2 x + y − 1 2 y − x − 3 = 16 −9  5 5  

 4    3

2

4

13 = 3

Work Book Exercise 16.1 4 If

1 The eccentricity of the conjugate hyperbola of the

the eccentricity of the hyperbola x 2 − y 2 sec 2 α = 5 is 3 times the eccentricity of

hyperbola x 2 − 3 y 2 = 1 is a

2

b

2 3

c 4

d

the ellipse x 2 sec 2 α + y 2 = 25, then the value of α is

4 5

a

2 The equation of the transverse axis of the hyperbola ( x − 3)2 + ( y + 1)2 = (4 x + 3 y )2 is a c

x + 3y = 0 3 x − 4 y = 13

b d

b

π 4

c

π 3

π 2

d

5 If ax + by = 1is tangent to the hyperbola

4x + 3y = 9 4x + 3y = 0

x2 y2 − = 1, then a2 − b2 is equal to a2 b2

3 If the distance between the foci and the distance

a

between the two directrices of the hyperbola x2 y2 − = 1 are in the ratio 3 : 2, then b : a is a2 b2 a 1: 2 c 1: 2

π 6

1 a2 e 2

b a2 e 2 c

b2 e 2

d None of the above

b 3: 2 d 2 :1

Tangent to a Hyperbola (θ1 − θ 2) (θ − θ 2)   b sin 1   a cos 2 2  R , + + ( ) θ θ ( θ θ ) 2 2   cos 1 cos 1   2 2

(i) Slope form The equations of tangent of slope m x 2 y2 to the hyperbola 2 − 2 = 1, is given by a b y = mx ± a 2 m 2 − b 2 The coordinates of the points of contact are   a 2m b2  ± ,±  2 2 2 2 2 2  a m −b a m −b   (ii) Point form The equation of the tangent to the xx yy x 2 y2 hyperbola 2 − 2 = 1 at ( x1 , y1 ) is 21 − 21 = 1. a b a b (iii) Parametric form The equation of the tangent x 2 y2 to the hyperbola 2 − 2 = 1 at ( a sec θ, b tan θ) is a b y x sec θ − tan θ =1. a b Ø The tangent at the points P(a sec θ1 , b tan θ1) and

926

Q(a sec θ 2 , b tan θ 2) intersect at the point

X

Example 11. S is the focus of the hyperbola y2 x2 − = 1 and M is the foot of the a2 b2 perpendicular drawn from S on a tangent to the hyperbola. Then, the locus of M is (b) x 2 + y 2 = 2a 2 (a) x 2 + y 2 = a 2 2 2 2 (d) x 2 − y 2 = a 2 (c) x − y = 2a Sol. (a) Let the coordinates of M be (α, β ). We know that the equation of any tangent to the hyperbola y = mx ±

x2

a

2

−

y2

b2

= 1 is

a2 m2 − b 2

Since, M (α, β ) lies on it. ∴

β = mα ±

a2 m2 − b 2

Now, SM perpendicular to the given tangent.

…(i)

β − mα = ±

From Eq. (i),

a2 m2 − b 2

From Eq. (ii), (mβ + α ) = ae On squaring both sides and then adding, we get ( β − mα )2 + (mβ + α )2 = (a2 m2 − b 2 ) + a2e 2 ⇒

(β 2 + m2α 2 − 2 mαβ ) + (m2β 2 + α 2 + 2 mαβ ) = [a2 m2 − a2 (e 2 − 1)] + a2e 2

⇒

(β + α ) + m (α + β ) = a2 m2 − a2e 2 + a2 + a2e 2 2

2

2

2

2

⇒

(α 2 + β 2 )(1 + m2 ) = a2 (m2 + 1)

⇒

α 2 + β 2 = a2

Hence, the locus of M(α, β ) is x2 + y2 = a2 . X

Example 12. The equation of the tangent to the hyperbola 4x 2 − 9 y 2 = 1, which is parallel to the line 4 y = 5x + 27, is 24 y − 30x = k , where k is equal to (a) ± 161 (c) ± 170

(b) ± 160 (d) None of these

Sol. (a) The given equation can be written as

So that,

x2 y2 − =1 1 1 4 9 1 1 and b 2 = a2 = 4 9

We know that, y = mx + a m − b is always a tangent x2 y2 to the hyperbola 2 − 2 = 1, ∀ m. a b 2

In our case, y = mx ±

2

2

m2 1 − is a tangent to the given 4 9

hyperbola. But given equation of tangent is 24 y = 30 x + k 30 k 5 k y= x+ = x+ ⇒ 24 24 4 24 is a tangent to the given hyperbola. Identifying the two equations, we get 5 m2 1 k2 m = and − = 2 4 4 9 24 k2 25 1 ⇒ − = 64 9 576 k2 161 = ⇒ 64 × 9 576 161 × 576 ⇒ = 161 k2 = 64 × 9 ∴

k=±

161

X

16

Example 13. The equations of the common x 2 y2 tangents to the two hyperbolas 2 − 2 = 1 and a b y2 x 2 − = 1 are a 2 b2

Hyperbola

1 m Also, S = (ae, 0) 1 Equation of SM is y − 0 = − ( x − ae ) ∴ m 1 y = − ( x − ae ) ⇒ m But M(α, β ) lies on it. 1 …(ii) β = − (α − ae ) ∴ m To get the locus of M, we have to eliminate m from Eqs.(i) and (ii). ∴ Slope of the segment, SM = −

(a) y = ± x ± a 2 + b 2 (b) y = ± x ± a 2 − b 2 (c) y = ± x ± a 3 + b 3 (d) None of the above Sol. (b) Let y = mx + c be the common tangent of the given hyperbolas. Since, y = mx + c is a tangent to the hyperbola x2 y2 − 2 =1 2 a b ∴ c 2 = a2 m2 − b 2

…(i)

Also, since y = mx + c is a tangent to the hyperbola y2 x2 − 2 = 1. 2 a b (mx + c )2 x2 − 2 =1 ∴ 2 a b ⇒ b 2 (mx + c )2 − a2 x2 = a2 b 2 ⇒

b 2 (m2 x2 + 2 mcx + c 2 ) − a2 x2 − a2 b 2 = 0

⇒ (b m2 − a2 )x2 + 2 b 2 mcx + b 2c 2 − a2 b 2 = 0 2

For tangency, this quadratic equation will have two roots equal. The condition for this 4b 4 m2c 2 = 4(b 2 m2 − a2 )b 2 (c 2 − a2 ) ⇒

b 2 m2c 2 = (b 2 m2 − a2 )(c 2 − a2 ) = b 2 m2c 2 − a2 b 2 m2 − a2c 2 + a4

⇒

a c = a4 − a2 b 2 m2

⇒

c 2 = a2 − b 2 m2

2 2

…(ii)

From Eqs. (i) and (ii), we get a2 m2 − b 2 = a2 − b 2 m2 ⇒

m2 (a2 + b 2 ) − (a2 + b 2 ) = 0

⇒

(m2 − 1)(a2 + b 2 ) = 0

⇒

m2 − 1 = 0

⇒ ∴

m= ±1 c = ± a2 − b 2

Hence, the equations of the common tangents are y = ± x±

a2 − b 2

Number of Tangents Drawn from a Point to a Hyperbola In this section, we shall prove that two tangents can be drawn from a point to a hyperbola. Proof Let P ( h, k ) be a point in the plane of the hyperbola x 2 y2 − =1 a 2 b2

927

The equation of any tangent to the hyperbola is y = mx ± a m − b 2

Objective Mathematics Vol. 1

16

2

X

2

Y

X′

C

X

A (a, 0)

(–a, 0)A′

x2

P (h, k)

a2

−

Sol. (b) Given, point P ≡ (4, 3) y2 b2

x2 y2 − −1 16 9 16 9 S1 ≡ − − 1 = − 1< 0 16 9

Hyperbola, S ≡

=1

Y′

If it passes through P ( h, k ). Then,

⇒ Point P(4, 3) lies outside the hyperbola.

k = mh + a 2 m 2 − b 2 ⇒

Hence, two tangents can be drawn from the point P(4, 3).

( k − mh) 2 = a 2 m 2 − b 2

⇒ m 2 ( h 2 − a 2 ) − 2khm + k 2 + b 2 = 0

…(i)

which is a quadratic equation in m. So, it gives two values of m. Corresponding to each value of m there is a tangent to the hyperbola passing through the point P ( h, k ). Thus, two tangents drawn from P are real and distinct, coincident or imaginary according as the roots of the Eq. (i) are real and distinct, coincident or imaginary. Now, the following cases arise: Case I Two tangents are real and distinct, if 4k 2 h 2 − 4( h 2 − a 2 )( k 2 + b 2 ) > 0 [on putting D > 0] 2 2 2 2 2 2 ⇒ −h b + a k + a b > 0 h2

⇒

−

Two tangents are coincident, if 4k 2 h 2 − 4( h 2 − a 2 )( k 2 + b 2 ) = 0 [on putting D = 0] ⇒ −h b + k a + a b = 0 2 2

⇒ ⇒

2

2

2 2

h b −k a =a b 2 2

2

h2

−

2

2 2

k2

=1 a 2 b2 ⇒ P ( h, k ) lies on the hyperbola

Case III Two tangents are imaginary, if 4k 2 h 2 − 4( h 2 − a 2 )( k 2 + b 2 ) < 0 [on putting D < 0] 2 2 2 2 2 2 ⇒ −h b + k a + a b < 0 ⇒ ⇒

The locus of the point of intersection of perpendicular tangents to a hyperbola is known as its director circle. Let m1 and m2 be the slopes of two tangents drawn x 2 y2 from a point P ( h, k ) to the hyperbola 2 − 2 = 1. a b Then, m1 and m2 are the roots of equation m 2 ( h 2 − a 2 ) − 2khm + k 2 + b 2 = 0.

h2b2 − k 2a 2 > a 2b2 h

2

k

>1 a b2 ⇒ P ( h, k ) lies inside the hyperbola. 2

−

2

m1 m2 =

Therefore,

k 2 + b2 h2 − a 2

If the tangents are perpendicular, then k 2 + b2 = −1 m1 m2 = − 1 ⇒ h2 − a 2 ⇒ h2 + k 2 = a 2 − b2

k2

0, b > 0 (b) a > 0, b < 0 (c) a < 0, b > 0 (d) a < 0, b < 0 Sol. (b, c) Given equation of the curve is xy = 1. On differentiating w.r.t. x, we get dy −1 = dx x2 If the line ax + by + c = 0 is a normal to the curve xy = 1, a then its slope − is equal to the slope of the normal. b a a a 0 and b < 0 or a < 0 and b > 0.

(ii) If θ 1 , θ 2 , θ 3 and θ 4 are eccentric angles of four x 2 y2 points on the hyperbola 2 − 2 = 1, the normals a b at which are concurrent, then (a) Σ cos (θ 1 + θ 2 ) = 0 (b) Σ sin (θ 1 + θ 2 ) = 0 (c) sin (θ 1 + θ 2 ) + sin (θ 2 + θ 3 ) + sin (θ 3 + θ 1 ) = 0 (iii) If θ 1 , θ 2 and θ 3 are the eccentric angles of three x 2 y2 points on the hyperbola 2 − 2 = 1 such that a b sin (θ 1 + θ 2 ) + sin (θ 2 + θ 3 ) + sin (θ 3 + θ 1 ) = 0 Then, the normals at these points are concurrent.

Work Book Exercise 16.2 1 The tangents to the hyperbola x 2 − y 2 = 3 are

parallel to the straight line 2 x + y + 8 = 0 at the following points

a

b (2, −1)

(2, 1)

c (−2, 1)

d (−2, −1)

2 The equation of a tangent to the hyperbola

16 x − 25 y − 96 x + 100 y − 356 = 0, which makes π an angle with the transverse axis, is 4 2

a c

2

y= x+2 y= x+ 3

b d

y= x−5 x= y+2

3 The point of intersection of two tangents to the x2 y2 hyperbola 2 − 2 = 1, the product of whose a b slopes is c 2 , lies on the curve a b c d

930

y2 y2 y2 y2

− + + −

b 2 = c 2( x 2 + a 2 = c 2( x 2 − b 2 = c 2( x 2 − a 2 = c 2( x 2 +

a2 ) b2) a2 ) b2)

4 The focus of the middle points of the portions of x2 y2 − =1 a2 b2 intercepted between the coordinate axes, is

the

tangents

to

the

hyperbola

a 4 x 2 y2 = a 2 y2 − b 2 x 2 b x 2 y2 = 2 a 2 y2 + b 2 x 2 c x 2 y 2 = 2( a 2 y 2 − b 2 x 2 )2 d None of the above

5 If radii of director circle of

x2 y2 + 2 = 1 and 2 a b

x2 y2 − = 1 are 2r and r respectively and ee and e h a2 b2 are the eccentricities of the ellipse and hyperbola respectively, then a b c d

2e h2 − ee2 = 6 ee2 − 4e h2 = 6 4e h2 − ee2 = 6 None of the above

Sol. (d) The equations of chord of contact of tangents from

The combined equation of the pair of tangents drawn from a point P(x1 , y1 ), lying outside the hyperbola x 2 y2 x 2 y2 to the hyperbola − = 1 − = 1 is a 2 b2 a 2 b2 2   xx1 yy1  x 2 y2   x12 y12     − − 1  2 − 2 − 1 =  2 − 2 − 1  b b  a 2 b2  a  a SS ′ = T 2 x 2 y2 x 2 y2 S = 2 − 2 − 1, S ′ = 12 − 12 − 1 a b a b xx1 yy1 T = 2 − 2 −1 a b

or where, and

Equations of Chord i.

Chord joining two points Let P ( a sec θ 1 , b tan θ 1 ) and Q ( a sec θ 2 , b tan θ 2 ) x 2 y2 be two points on the hyperbola 2 − 2 = 1. a b Then, the equation of the chord PQ is b tan θ 2 − b tan θ 1 y − b tan θ 1 = ( x − a sec θ 1) a sec θ 2 − a sec θ 1 x a

( x1, y1 ) and ( x2 , y2 ) to the given hyperbola are xx1 yy1 − 2 =1 a2 b xx2 yy2 and − =1 a2 b2 Since, lines are perpendicular to each other. b 2 x1 b 2 x2 × = −1 ∴ a2 y1 a2 y2 x1 x2 − a4 ⇒ = 4 y1 y2 b

Equation of the Chord Bisected at a Given Point The equation of the chord of the hyperbola y2 − = 1, bisected at the point (x1 , y1 ) is a 2 b2 xx1 yy1 x12 y12 − − 1 = − −1 a2 b2 a 2 b2 T = S 1 , where T and S 1 have their usual meanings. x2

X

y  θ1 + θ 2   θ1 + θ 2   θ1 − θ 2    = cos   − sin   2   2  b  2 

⇒ cos 

ii.

X

Chord of contact If the tangents drawn from x 2 y2 a point P(x1 , y1 ) to the hyperbola 2 − 2 = 1 a b touch the hyperbola at Q and R, then the equation of the chord of contact QR is xx1 yy1 − 2 = 1. a2 b

mid-point is (h, k), is T = S1 i.e. xh − yk = h2 − k 2 ⇒

⇒ ⇒

y=

h3 − hk 2 = − ak 2 h3 = (h − a) k 2

Hence, the locus of (h, k) is x3 = ( x − a)y2 .

drawn from (2, −1) to the hyperbola 16 x2 − 9 y2 = 144 is

Example 20. The chords of contact of tangents from two points ( x1 , y1 ) and ( x 2 , y2 ) to the y2 x2 hyperbola 2 − 2 = 1 are at right angles, then a b x1 x 2 is y1 y2 −a 2 −b 2 −b 4 −a 4 (a) 2 (b) 2 (c) 4 (d) 4 b a a b

yk = xh − h2 + k 2

h2 − k 2 h . x− k k We know that the line y = mx + c will touch the parabola, a if c = . m h2 − k 2 ka ∴ − = k h ⇒ h(h2 − k 2 ) = − ak 2 ⇒

Sol. (c) The equation of the chord of contact of tangents

X

Example 21. If chord of the hyperbola x 2 − y 2 = a 2 touch the parabola y 2 = 4ax, then the locus of the mid-points of these chords is (b) x 2 = ( x − a ) y 2 (a) x 3 = ( x − a ) y 2 (c) x 3 = ( x + a ) y 2 (d) x 3 = ( x − a ) y Sol. (a) The equation of the chord of the hyperbola, whose

Example 19. The equation of chord of contact of tangents drawn from a point (2, − 1) to the hyperbola 16x 2 − 9 y 2 = 144, is (a) 9x + 32 y = 144 (b) 32x − 9 y = 144 (c) 32x + 9 y = 144 (d) None of these

32 x + 9 y = 144.

Hyperbola

16

Equation of the Pair of Tangents

X

Example 22. Chords of the circle x 2 + y 2 = a 2 x 2 y2 touch the hyperbola 2 − 2 = 1. Then, their a b mid-points lie on the curve (a) ( x 2 + y 2 ) 2 = a 2 x 2 − b 2 y 2 (b) ( x 2 + y 2 ) = a 2 x 2 − b 2 y 2 (c) ( x 2 − y 2 ) 2 = a 2 x 2 − b 2 y 2 (d) ( x 2 + y 2 ) = a 2 x 2 + b 2 y 2

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Objective Mathematics Vol. 1

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Sol. (a) The equation of the chord of the circle x2 + y2 = a2 whose mid-point is (α, β), is x α + yβ = α 2 + β 2

…(i)

which is obtained by using the formula S1 = T. Again, the equation of the tangent to the hyperbola xx yy x2 y2 − 2 = 1 at the point ( x1, y1 ) is 21 − 21 = 1. 2 a b a b Now, putting x1 = asec θ and y1 = b tan θ, the equation of the tangent to the hyperbola at the point (a sec θ, b tan θ) is x ⋅ asec θ y ⋅ b tan θ − =1 a2 b2 x y …(ii) ⇒ sec θ − tan θ = 1 a b Since, Eqs. (i) and (ii) are the same, therefore αa βb α2 + β2 = = sec θ − tan θ 1 aα βb and tan θ = − 2 ⇒ sec θ = 2 α + β2 α + β2 sec 2 θ − tan2 θ =

⇒

a α −β b 2 2

⇒

2 2

(α + β ) 2

⇒

2 2

a2α 2 (α + β ) 2

2 2

−

β 2 b2 (α + β 2 )2 2

⇒

Y (h,k) T

Q

C A (a, 0) P (x1, y1)

X

Y′

932

x

2

−

y

2

= 1 be the hyperbola and P ( x1 , y1 ) be a

a b point such that the secant drawn through P intersects the hyperbola at Q and R. Let the tangents at Q and R intersect at T ( h, k ). Clearly, QR is the chord of contact of x 2 y2 tangents drawn from T to the hyperbola 2 − 2 = 1. a b So, the equation QR is hx ky − =1 a 2 b2

x2 + y2 = a2e 2

…(ii)

Let P(α, β ) be the pole of the chord of the hyperbola (i). Then, polar of P w.r.t. Eq. (i) is x α yβ …(iii) − 2 =1 a2 b If it is tangent to the circle (ii), then 1 = ae 2 α β2 + a4 b4 ⇒

R

Let

y2

…(i) =1 a b2 are G (− ae, 0) and H(ae, 0). So, the equation of the circle on GH as diameter is ( x + ae )( x − ae ) + ( y − 0)( y − 0) = 0 ⇒ x2 − a2e 2 + y2 = 0

Let P be a point inside or outside a hyperbola. Then, the locus of the point of intersection of two tangents to the hyperbola at the points, where secants drawn through P intersect the hyperbola, is called the polar of point P with respect to the hyperbola and the point P is called the pole of the polar.

2

−

2

Pole and Polar

2

Example 23. A series of chords of the x 2 y2 hyperbola 2 − 2 = 1 are tangents to the circle a b described on the straight line joining the foci of the hyperbola as diameter. Then, the locus of their poles with respect to the hyperbola is x 2 y2 x 2 y2 1 (b) 2 + 2 = 2 (a) 2 + 2 = a 2 + b 2 a b a b a + b2 x 2 y2 1 (d) None of these (c) 4 + 4 = 2 a b a + b2 x2

Hence, the locus of the mid-point is ( x2 + y2 )2 = a2 x2 − b 2 y2

(–a, 0) A′

X

Sol. (c) The foci of the hyperbola

=1

(α 2 + β 2 )2 = a2α 2 − β 2 b 2

X′

Clearly, QR passes through P ( x1 , y1 ). Therefore, hx1 ky1 − 2 =1 a2 b Hence, the locus T ( h, k ) i.e. the polar of P ( x1 , y1 ) is xx1 yy1 − 2 =1 a2 b

α2 a

4

+

β2 b

4

=

The locus of (α, β) is x2 a

4

1 2 2

ae +

y2 b4

=

=

1 a + b2 2

[Qb 2 = a2 (e 2 − 1)]

1 a2 + b 2

Conjugate Points Two points are said to be conjugate points with respect to a hyperbola, if each lies on the polar of the other. If P ( x1 , y1 ) and Q ( x 2 , y2 ) are conjugate points with x 2 y2 respect to the hyperbola 2 − 2 = 1, then P ( x1 , y1 ) lies a b xx yy on the polar of Q ( x 2 , y2 ) i.e. on 22 − 22 = 1. Therefore, a b x1 x 2 y1 y2 − 2 =1 a2 b

Two lines are said to be conjugate lines with respect x 2 y2 to a hyperbola 2 − 2 = 1, if each passes through the a b pole of the other. X

Example 24. The locus of the poles of the x 2 y2 chords of the hyperbola 2 − 2 = 1 which subtend a b a right angle at its centre, is x 2 y2 1 1 x 2 y2 1 1 (a) 4 + 4 = 2 − 2 (b) 4 − 4 = 2 + 2 a b a b a b a b x 2 y2 1 1 x 2 y2 1 1 (c) 2 + 2 = 2 − 2 (d) 2 + 2 = 2 + 2 a b a b a b a b Sol. (a) Let P(h, k) be the pole of a chord of the hyperbola x2 a

2

−

y2 b2

= 1. Then, the equation of the polar is

hx ky …(i) − =1 a2 b 2 The combined equation of the lines joining the origin to x2 y2 the points of intersection of the hyperbola 2 − 2 = 1 a b and the line (i) is 2 x2 y2 hx ky − 2 =  2 − 2  2 a a b b 

(i) Equation of a diameter The equation of a diameter bisecting a system of parallel chords of b2 x 2 y2 slope m of the hyperbola 2 − 2 = 1 is y = 2 x. a m a b (ii) Conjugate diameters Two diameters of a hyperbola are said to be conjugate diameters, if each bisects the chords parallel to the other. (iii) In a pair of conjugate diameters of a hyperbola, only one meets the hyperbola in real points. (iv) Let P(a sec θ, b tan θ) be a point on the hyperbola x 2 y2 − = 1 such that CP and CD are conjugate a 2 b2 diameters of the hyperbola. Then, coordinates of D are (a tan θ, bsec θ). (v) If a pair of conjugate diameters meet the hyperbola and its conjugate in P and D respectively, then CP 2 − CD 2 = a 2 − b 2 X

Since, the lines given by the above equation are at right angle. Therefore, Coefficient of x2 + Coefficient of y2 = 0 1 1 h2 k2 − 4 − 2 − 4 =0 ⇒ 2 a a b b 1 1 h2 k2 ⇒ + 4 = 2 − 2 4 a b a b ∴ The locus of P (h, k ) is 1 1 x2 y2 + 4 = 2 − 2 4 a b a b

Diameter Y

Q

so that a2 = 3 and b 2 = 7.

(–a, 0) A′

C

y = mx+c X

A (a, 0)

2

3

7

The slope of the line 7 x + y − 2 = 0 is −7. Let the slope of its diameter be m1. b2 7 7 1 Then, ⇒ m1 = =− mm1 = 2 = 3 3(−7 ) 3 a Hence, the required equation of the diameter is 1 y = m1 x = − x ⇒ 3 y + x = 0 3

Example 26. If the line lx + my + n = 0 meets x 2 y2 the hyperbola 2 − 2 = 1 at the extremities of a a b pair of conjugate diameters, then (b) a 2 l 2 − b 2 m 2 = 1 (a) a 2 l 2 − b 2 m 2 = 0 (c) a 2 l 2 − b 2 m 2 = 2 (d) a 2 l 2 − b 2 m 2 = 3 Sol. (a) Let CP and CD be a pair of conjugate diameters of

P (h,k) X′

Example 25. The equation of the diameter which bisects the chord 7x + y − 2 = 0 of the x 2 y2 hyperbola − = 1, is 3 7 (a) y − 3x = 0 (b) 3 y − x = 0 (c) 3 y + x = 0 (d) None of these 2 2 Sol. (c) The equation of the hyperbola is x − y = 1,

X

The locus of the mid-points of a system of parallel chords of a hyperbola is called a diameter.

16 Hyperbola

Conjugate Lines

y2

x – =1 a2 b2 R

Y′

The point, where a diameter intersects the hyperbola, is known as the vertex of the diameter.

x2

and ⇒ and

2

−

y2

= 1. Then, the coordinates of P a b2 and D are (asec θ, b tan θ) and (a tan θ, b sec θ) respectively. It is given that the line lx + my + n = 0 meets the hyperbola at P and D. Therefore, the hyperbola

al sec θ + bm tan θ + n = al tan θ + bmsec θ + n = al sec θ + bm tan θ = al tan θ + bmsec θ =

0 0 −n −n

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Objective Mathematics Vol. 1

16

⇒

(al sec θ + bm tan θ)2 = n2

and

(al tanθ + bmsec θ) = n 2

●

2

⇒ a l (sec θ − tan2 θ) + b 2 m2 (tan2 θ − sec 2 θ) = 0 [on subtracting] ⇒ a2 l 2 − b 2 m2 = 0 2 2

X

2

 x 2 y 2   x2 y 2   x2 y 2   x2 y 2   2 − 2 − 1 −  2 − 2  =  2 − 2  −  2 − 2 + 1 b b b  a b  a a   a

Example 27. If the equation

● ●

lx 2 + 2mxy + ny 2 = 0 represents a pair of conjugate x2

−

y2

(a) la 2 + nb 2 = 0

= 1, then a 2 b2 (b) la 2 = nb 2

(c) 2la 2 = nb 2

(d) None of these

diameters of the hyperbola

X

Sol. (b) Let y = m1 x and y = m2 x be the lines represented by by lx2 + 2 mxy + ny2 = 0.

l …(i) n But y = m1 x and y = m2 x are conjugate diameters of the hyperbola. b2 …(ii) m1m2 = 2 ∴ a From Eqs. (i) and (ii), we get l b2 ⇒ la2 = nb 2 = n a2 m1m2 =

Then,

Y Asymptotes x2 – y 2 =1 a2 b2

B X′

A′

A

C

X

2 y2 – x2 + 2 = 1 a b

The equations of two asymptotes of the hyperbola 2 y2 x − = 1 are a 2 b2 b x y y = ± x or ± =0 a a b Ø

●

●

934

●

Example 28. The equation of the hyperbola whose asymptotes are the straight lines x + 2 y + 3 = 0 and 3x + 4 y + 5 = 0, which passes through (1, − 1), is (a) x 2 − 8xy + 8 y 2 + 14x + 22 y − 7 = 0 (b) 3x 2 + 10xy + 8 y 2 + 14x + 22 y + 7 = 0 (c) x 2 + 10xy + 8 y 2 + 14x + 22 y + 7 = 0 (d) None of the above that of the joint equations of the asymptotes by a constant, therefore the equation of the hyperbola will be of the form …(i) ( x + 2 y + 3)(3 x + 4 y + 5) + k = 0 Since, the hyperbola passes through the point (1, − 1). ∴ (1 − 2 + 3)(3 − 4 + 5) + k = 0 ⇒ 2 (4) + k = 0 ∴ k = −8 From Eq. (i), ( x + 2 y + 3)(3 x + 4 y + 5) − 8 = 0 ⇒ 3 x2 + 10 xy + 8 y2 + 14 x + 22 y + 7 = 0

X

Example 29. The equation of the hyperbola conjugate to the hyperbola 2x 2 + 3xy − 2 y 2 − 5x + 5 y + 2 = 0 is (a) 2x 2 + 3xy − 2 y 2 − 5x + 5 y − 8 = 0 (b) x 2 + 3xy − 2 y 2 − 5x + 5 y − 8 = 0 (c) x 2 + 3xy − 2 y 2 − 5x + 5 y − 6 = 0 (d) None of the above Sol. (a) The required equation is obtained by the equation.

B′ Y′

The asymptotes pass through the centre of the hyperbola. The bisectors of the angle between the asymptotes are the coordinates axes.

Sol. (b) Since, the equation of the hyperbola differs from

Asymptotes An asymptote to a curve is a straight line at a finite distance from the origin, to which the tangent to a curve tends as the point of contact goes to infinity. In other words, asymptote to a curve touches the curve at infinity.

The equation of the pair of asymptotes differ the hyperbola and the conjugate hyperbola by the same constant only i.e. Hyperbola − Asymptotes = Asymptotes − Conjugate hyperbola or

The combined equation of the asymptotes to the hyperbola x2 y2 x2 y2 − 2 = 1 is 2 − 2 = 0. 2 a b a b When b = a i.e. the asymptotes of rectangular hyperbola x 2 − y 2 = a2 are y = ± x, which are at right angle. A hyperbola and its conjugate hyperbola have the same asymptotes.

Hyperbola + Conjugate hyperbola = 2(Asymptotes) Now, let the equation of the asymptotes be 2 x2 + 3 xy − 2 y2 − 5 x + 5 y + λ = 0. This will represent a pair of straight lines, if abc + 2 fgh − af 2 − bg 2 − ch2 = 0 ⇒

5 − 5  3  5 2 (− 2 ) λ + 2       − 2  2 2  2  2 2

2

2

5 3 − (− 2 )  −  − λ   = 0 2  2 75 50 50 9 −4λ − − + − λ=0 ⇒ 4 4 4 4 25 75 50 50  75 λ = −  + − ⇒  =−  4 4 4 4 4 ⇒ λ = −3 ∴ The equation of the conjugate hyperbola is 2(Asymptotes) − Hyperbola i.e. 2 x2 + 3 xy − 2 y2 − 5 x + 5 y − 8 = 0

Example 30. If e is the eccentricity of the x 2 y2 hyperbola 2 − 2 = 1 and θ is the angle between a b θ the asymptotes, then cos   is equal to  2 1 1 2 (a) (c) e (d) (b) − e e e Sol. (a) The equations of the asymptotes of the hyperbola x2 2

−

y2 2

= 1 are y = ±

b b b x i.e. y = x and y = − x. a a a

a b If θ is the angle between the two asymptotes, then b b + 2b 2 ab a2 × 2 = 2 tan θ = a a = 2 b b a a −b a − b2 1− ⋅ a a a2 − b 2 From this, we get cos θ = 2 a + b2 ⇒

a2 − b 2

2 θ

2 cos   − 1 = 2 2 a + b2

⇒

a2 − b 2 θ 2 cos 2   = 1 + 2 2 a + b2

⇒

θ 2 a2 2 cos 2   = 2  2  a + b2

⇒

θ a2 cos 2   = 2  2  a + b2

∴ ⇒

a 2 − b2 = 0

⇒ a=b ⇒ 2a = 2b Thus, the transverse and conjugate axes of a rectangular hyperbola are equal and the equation of the hyperbola is x 2 − y 2 = a 2 .

Ø Since, the transverse and conjugate axes of a rectangular

hyperbola are equal. So, its eccentricity e is given by e = 1+

b2 a2 = 1 + = 2 a2 a2

Equation of the Rectangular Hyperbola Referred to its Asymptotes as the Axes of Coordinates Referred to the transverse and conjugate axes along the axes of coordinates, the equation of the rectangular hyperbola is …(i) x 2 − y2 = a 2 Y

xy = c2

a + b =ae 2

2

2 2

θ 1 a2 cos 2   = 2 2 = 2 2 a e e θ 1 cos   = 2 e

X′

A hyperbola whose asymptotes are at right angles to each other is called a rectangular hyperbola. The equation of the asymptotes of the hyperbola b x 2 y2 − 2 = 1 are given by y = ± x. 2 a a b The angle θ between these two asymptotes is given b  b 2b − −  a  a 2ab tan θ = = a 2 = 2 b  b b a − b2 1+ −  1− a  a a2 If the asymptotes are at right angle, then π π θ= ⇒ tan θ = tan 2 2 π 2ab ⇒ = tan 2 a 2 − b2

X

O

Rectangular Hyperbola

by

16

The equation of the asymptotes of the rectangular hyperbola are y = ± x i.e. y = x and y = − x. Clearly, each of these two asymptotes is inclined at 45° to the transverse axis.

b 2 = a2 (e 2 − 1) = a2e 2 − a2

But ⇒

⇒

Hyperbola

X

Y′

The asymptotes of Eq. (i) are y = x and y = − x. Each of these two asymptotes are inclined at an angle of 45° with the transverse axis. So, if we rotate the coordinate π axes through an angle of − keeping the origin fixed, 4 then the axes coincide with the asymptotes of the hyperbola, we have  π  π X +Y x = X cos  −  − Y sin  −  =  4  4 2 and

 π  π Y − X y = X sin  −  + Y cos  −  =  4  4 2

On substituting the values of x and yin Eq. (i), we get 2

2

 X +Y  Y − X    −  = a2   2  2  ⇒

XY =

a2 2

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16

Then, the equation of PQ is c c − t1 t 2 c ( x − ct 1 ) y− = t 1 ct 1 − ct 2

a2 2 Thus, the equation of the hyperbola referred to its asymptotes as the coordinate axes is a2 xy = c 2 , where c 2 = 2 The equation of a rectangular hyperbola having coordinate axes as its asymptotes is xy = c 2 . If the asymptotes of a rectangular hyperbola are x = α, y = β, then its equation is ( x − α )( y − β) = c 2 ⇒

⇒ X

XY = c 2 , where c 2 =

=

⇒ t 1 t 2 y − ct 2 = − x + ct 1 ⇒ t 1 t 2 y + x = c (t 1 + t 2 ) This is parallel to y + m1 x = 0, if t1t 2 1 = 1 m1 ⇒

xy − αy − βx + λ = 0

Example 31. A straight line has its extremities on two fixed straight lines (say a and b) and cuts off from them a triangle of constant area. Then, locus of the middle point of the line is (a) xy = c 2 (b) xy + c 2 = 0 (c) x 2 y 2 = c (d) None of the above X

Hence, the locus of (α, β) is xy = c 2 .

(b) m1 m2 xy = c 2 ( m1 + m2 ) (c) 2m1 m2 xy = c 2 ( m1 + m2 ) 2 (d) 4m1 m2 xy = c 2 ( m1 − m2 ) 2 Sol. (a) Let PQR be a triangle inscribed in the hyperbola    c c c xy = c 2 and P ≡  ct 1,  , Q ≡  ct 2 ,  and R ≡  ct 3 ,  . t1  t2  t3    

936

[from Eq. (i)]

= c(m1 + m2 ) t 3 m1 x + m2 t 32 y = c (m1 + m2 ) t 3

Example 33. The asymptotes of the hyperbola xy = hx + ky are (a) x = k , y = h (b) x = h, y = k (c) x = h, y = h (d) x = k , y = k Sol. (a) The equation of asymptotes are …(i) xy − hx − ky + λ = 0 where, λ is a constant and Eq. (i) represents a pair of straight line. Here, A = 0, B = 0, C = λ, 2 H = 1, 2G = − h and 2F = − k Then, ABC + 2 FGH − AF 2 − BG 2 − CH2 = 0 k h 1 1 ⇒ 0 + 2  −   −    − 0 − 0 − λ ⋅ = 0  2  2 2 4 hk λ = ⇒ 4 4 ⇒ λ = hk On putting λ = hk in Eq. (i), we get xy − hx − ky + hk = 0 ⇒ ( x − k )( y − h) = 0 Hence, asymptotes are x = k and y = h.

If (α, β) are the coordinates of the middle points of AB, then a b and β = …(ii) α= 2 2 On eliminating a and b from Eqs. (i) and (ii), we get αβ = c 2

Example 32. A triangle is inscribed in xy = c 2 and two of its sides are parallel to y + m1 x = 0 and y + m2 x = 0. Then, the third side envelopes the hyperbola is (a) 4m1 m2 xy = c 2 ( m1 + m2 ) 2

…(i)

Now, the equation of QR is t 2 t 3 y + x = c(t 2 + t 3 ) ⇒ m1 t 2 t 3 y + m1 x = c(m1 t 2 + m1 t 3 ) ⇒ m2 t 32 y + m1 x = c(m2 t 3 + m1 t 3 )

The envelope of which for different values of t 3 is 4m1m2 xy = c 2 (m1 + m2 )2

This line cuts the coordinate axes at the point A(a, 0) and B(0, b ). 1 Therefore, area of ∆AOB = ab = Constant 2 [say = 4c 2 ] ...(i) ⇒ ab = Constant

X

m1 t 1 t 2 = 1

Similarly, PR is parallel to y + m2 x = 0, if m2 t 1 t 3 = 1 We have, m1 t 1 t 2 = m2 t 1 t 3 ⇒ m1 t 2 = m2 t 3

⇒

Sol. (a) Let the given straight line be axes of coordinates and let the equation of the variable line be x y + =1 a b

c (t 2 − t 1 ) 1 ( x − ct 1 ) ( x − ct 1 ) = − t 1t 2 (c ) (t 1 − t 2 ) t1t 2

X

Example 34. Consider the set of hyperbola xy = k , x ∈ R . Let e1 be the eccentricity when k = 4 and e2 be the eccentricity when k = 9, then e1 − e2 is equal to (a) 0 (b)1 (c) 2 (d) 3 Sol. (a) We know that the eccentricity of xy = k, ∀ k ∈ R is 2. ∴ Also, Hence,

e1 = 2 e2 = 2 e1 − e 2 = 0

i.`

Point form

The equation of tangent at (x1 ,

y1 ) to the parabola xy = c is y x xy1 + yx1 = 2c 2 or + =2 x1 y1

i.

Point form The equation of the normal at ( x1 , y1 ) to the hyperbola xy = c 2 is xx1 − yy1 = x12 − y12

ii.

Parametric form The equation of the normal  c at  ct,  to the hyperbola xy = c 2 is  t

2

ii.

Parametric form The equation of tangent at  c 2  ct,  to the hyperbola xy = c is  t x + yt = 2c t

xt 3 − yt − ct 4 + c = 0 Ø

●



 c c  to the rectangular hyperbola t2  t1    2 ct t 2c  xy = c2 intersect at  1 2 , .  t1 + t 2 t1 + t 2 

Example 35. A variable straight line of slope 4 intersects the hyperbola xy =1 at two points. Then, the locus of the point which divides the line segment between these points in the ratio of 1 : 2, is (a) x 2 + y 2 = 10xy (b)16x 2 − y 2 = 10xy (c)16x 2 + y 2 = 10xy − 18 (d) x 2 + y 2 = xy Sol. (c) Let R (α,β) be the point which divides the line segment PQ in the ratio of 1 : 2. The equation of any line passing through R(α, β) is x−α y−β [say] = =r cos θ sinθ where, tan θ = 4 4 1 and cosθ = sinθ = ∴ 17 17 ∴ x = α + r cos θ and y = β + r sinθ Since, (x, y) lies on xy = 1. ∴ (α + r cos θ)( β + r sinθ) = 1 ⇒ αβ + r (α sinθ + β cos θ) + r 2 sinθcos θ = 1 This is a quadratic equation in r and hence it will have two roots of r. Let r1 and r2 be the two roots. (α sinθ + β cos θ) …(i) ∴ r1 + r2 = − sinθcos θ αβ − 1 and ...(ii) r1r2 = sinθcos θ r1 1 Also, = ⇒ 2 r1 = r2 r2 2 On putting r2 = 2 r1 in Eqs. (i) and (ii), we get −(α sinθ + β cos θ) αβ − 1 and 2 r12 = 3r1 = sinθcos θ sinθcos θ 4 Now, eliminating r1 and using sinθ = and 17 1 , we get (4α + β )2 = 18(αβ − 1). cosθ = 17 Hence, the locus of (α, β) is (4 x + y)2 = 18 xy − 18 i.e.

16 x + y = 10 xy − 18. 2

2

c The equation of the normal at  ct ,  is a fourth degree 

t

equation in t. So, in general, four normals can be drawn from a point to the hyperbola xy = c2.

Ø Tangent at P  ct1 ,  and Q  ct 2 ,

X

16

Normals to a Rectangular Hyperbola

Hyperbola

Tangent to a Rectangular Hyperbola

●

●

●

The equation of the polar of any point P(x1 , y1) with respect to xy = c2 is xy1 + yx1 = 2 c2. The equation of the chord of the hyperbola xy = c2 whose middle point is (x1 , y1) , is xy1 + yx1 = 2 x1 y1 orT = S ′ where, Tand S′ have their usual meanings. The equation of the chord of the contact of tangents drawn from a point (x1 , y1) to the rectangular hyperbola xy = c2 is xy1 + yx1 = 2 c2

X

Example 36. The locus of the foot of the perpendicular from the centre upon any normal to the hyperbola x 2 − y 2 = a 2 is (a) ( x 2 + y 2 ) 2 ( y 2 − x 2 ) = a 2 x 2 y 2 (b) ( x 2 − y 2 ) 2 ( y 2 − 2x 2 ) = 4a 2 x 2 y 2 (c) ( x 2 + y 2 ) 2 ( y 2 − x 2 ) = 4a 2 x 2 y 2 (d) None of the above Sol. (c) The equation of normal line at an arbitrary point P (asec φ, a tan φ) is axsin φ + ay = (a2 + a2 )tan φ = 2 a2 tan φ …(i) ⇒ xsin φ + y = 2 a tan φ Again, the equation of the perpendicular from the centre i.e. the origin Eq. (i) is …(ii) x − sin φ⋅ y = 0 The required locus is obtained by eliminating φ from Eqs. (i) and (ii). From Eq. (i), ( xsin φ + y)2 = 4a2 tan2 φ ⇒

( x sin φ + y)2 = 4a2

sin2 φ

…(iii)

1 − sin2 φ

x y Therefore, putting the value of sin φ in Eq. (iii), we get x2 2 x   y2 2  x ⋅ + y = 4a ⋅  y  x2 1− 2 y But from Eq. (ii), sinφ =

⇒ ⇒

( x2 + y2 )2 y

2

= 4a2 ⋅

x2

⋅

y2

=

4a2 x2

y − x y2 − x 2 2 2 ( x + y ) ( y − x ) = 4a2 x2 y2 2

y

2

2 2

2

2

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Objective Mathematics Vol. 1

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X

Example 37. The locus of the middle points of normal chords of the rectangular hyperbola x 2 − y 2 = a 2 is (a) ( y 2 − x 2 ) 2 = 4a 2 x 2 y 2 (b) ( y 2 − x 2 ) 3 = 4a 2 x 2 y 2 (c) ( y 2 + x 2 ) 3 = 4a 2 x 2 y 2 (d) ( y 2 − x 2 ) 3 = a 2 x 2 y 2 Sol. (b) The equation of the normal to the hyperbola x2 a

2

−

y2 b2

x − x1 y − y1 = x1 − y1

= 1 at the point (x1, y1) is

a2

b2

So, the equation of the normal to the hyperbola x2 − y2 = a2 at the point (asec θ, a tan θ) is x − asec θ y − a tan θ x y −a=− + a = ⇒ sec θ − tan θ sec θ tan θ x y …(i) ⇒ + = 2a sec θ tan θ Again, the equation to the chord of the given hyperbola whose middle point is (α, β), is …(ii) x α − yβ = α 2 − β 2 which is obtained from the formula S1 = T. Since, Eqs. (i) and (ii) represent the same line, therefore on comparing Eqs. (i) and (ii), we get α2 − β2 α sec θ = − β tan θ = 2a α2 − β2 α2 − β2 and tan θ = − ⇒ sec θ = 2aα 2aβ ⇒

sec 2θ − tan2 θ =

(α 2 − β 2 )2 4a α

−

(α 2 − β 2 )2

4a2 β 2 (α − β )  1 1 1=  2 − 2 α β 4a2   4a2α 2β 2 = (α 2 − β 2 )2 (β 2 − α 2 ) 2 2

2

⇒ ⇒

2 2

⇒

(β 2 − α 2 )3 = 4a2α 2β 2 ∴ The locus of (α, β) is ( y2 − x2 )3 = 4a2 x2 y2 X

Example 38. The locus of poles of normal chords of the rectangular hyperbola xy = c 2 is the curve (a) ( x 2 − y 2 ) 2 + 4c 2 xy = 0 (b) ( x 2 − y 2 ) + c 2 xy = 0 (c) ( x 2 + y 2 ) 2 + 4c 2 xy = 0 (d) None of these Sol. (a) The equation of the normal to the curve xy = c 2 is c i.e. xt 3 − ty = ct 4 − c …(i) t Let (α, β) be the pole of Eq. (i) w.r.t. the given curve. Also, the equation of the polar of (α, β) w.r.t. xy = c 2 xt 2 − y = ct 3 −

(i.e. 2 xy = 2c 2 ) is

xβ + yα = 2c 2

...(ii)

Since, Eqs. (i) and (ii) are identical, therefore 2c 2 β α = = 4 3 − t ct − c t β β From the first two, 2 = − α ⇒ t 2 = − α t From the second and third, we have −2ct −2ctα 2 −2ct −2c 2t = 2 = 2 = 4 α= 4 ct − c t − 1 β β − α2 −1 2 α 2ctα 2 2 1= 2 ⇒ ⇒ α − β = 2cαt α − β2 β Qt 2 = − β  ⇒ (α 2 − β 2 )2 = 4c 2α 2t 2 = 4c 2α 2  −   α  α  ⇒ (α 2 − β 2 )2 = − 4c 2αβ ⇒ (α 2 − β 2 )2 + 4c 2αβ = 0 So, the locus of (α, β) is ( x2 − y2 )2 + 4c 2 xy = 0

Work Book Exercise 16.3 1 If PN is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus of the mid-point of PN is a c

a circle an ellipse

b a parabola d a hyperbola

2 If angles subtended by any chord of a rectangular hyperbola at the centre is a and angle between the tangents at ends of chord is β, then a α = 2β c α +β= π

b β = 2α

π d α +β= 2

et + e − t et − e − t ,y= is a point 2 2 2 2 on the hyperbola x − y = 1. Then, the area bounded by this hyperbola and the lines joining its centre to the points corresponding to t1 and − t1, is

3 For any real t, x =

938

a b c d

t1t 2 t −1 t 2 t t −1 t1

4 The combined equation of the asymptotes of the hyperbola 2 x 2 + 5 xy + 2 y 2 + 4 x + 5 y = 0 is a b c d

2 x 2 + 5 xy + 2 y 2 + 4 x + 5 y + 2 = 0 2 x 2 + 5 xy + 2 y 2 + 4 x + 5 y − 2 = 0 2 x 2 + 5 xy + 2 y 2 = 0 None of the above

5 The equation of the line passing through the centre of a rectangular hyperbola is x − y − 1 = 0. If one of its asymptotes is 3 x − 4 y − 6 = 0, then the equation of the other asymptote is a c

4 x + 3 y + 17 = 0 3 x − 2 y + 15 = 0

b d

4x − 3y + 8 = 0 None of these

6 The asymptotes of a hyperbola are parallel to 2 x + 3 y = 0 and 3 x + 2 y = 0. Its centre is (4, 2) and hyperbola passes through (5, 3). Then, its equation is a x 2 + xy + y 2 − 74 x − 76 y + 199 = 0 b 6 x 2 + 13 xy + 6 y 2 − 74 x − 76 y + 199 = 0 c 6 x 2 + 13 xy + y 2 − 74 x − 76 y + 199 = 0 d None of the above

WorkedOut Examples Type 1. Only One Correct Option Ex 1. A series of hyperbolas is drawn having a common transverse axis of length 2a. Then, the locus of a point P on each hyperbola, such that its distance from the transverse axis is equal to its distance from an asymptote, is (a) ( x 2 − y 2 ) 2 = 4x 2 ( x 2 − a 2 ) (b) ( x 2 − y 2 ) 2 = x 2 ( x 2 − a 2 ) (c) ( x 2 − y 2 ) = 4x 2 ( x 2 − a 2 ) (d) None of the above Sol. The equations of series of a hyperbola is x 2 y2 − = 1, where, λ is a parameter. a2 λ2 x y The asymptotes of this hyperbola is = ± . λ a Suppose (x′, y′ ) is a point P on the hyperbola which is equidistant from the transverse axis and an asymptote. x′ y′ − x′ 2 y′ 2 a λ Then, − 2 = 1 and y′ = 2 1 1 λ a + 2 2 a λ y′ 2 x′ 2 i.e. = 2 −1 λ2 a x ′2 y′ 2 2x′ y′ 1 1   and y′ 2  2 + 2  = 2 + 2 − a aλ λ λ  a The second relation gives on simplification, 4 x ′ 2 y′ 2 a2 = 4 x ′ 2 (x ′ 2 − a2 ) ( y′ 2 − x ′ 2 )2 = λ2 [by the first relation] So, the locus of P is ( y2 − x 2 )2 = 4 x 2 (x 2 − a2 ) Hence, (a) is the correct answer.

Ex 2. A variable straight line of slope 4 intersects the hyperbola xy =1 at two points. Then, the locus of the point which divides the line segment between these points in the ratio 1 : 2, is (a) 16 x 2 + y 2 + 8 xy = 2 (b) x 2 + y 2 + 10 xy = 2 (c) 16 x 2 + y 2 + 10 xy = 2 (d) 16 x 2 + y 2 + 10 xy = 0 Sol. Let the line be y = 4 x + c. It meets the curve xy = 1at ⇒

x (4 x + c) = 1 4 x 2 + cx = 1

⇒

x1 + x2 = −

Also, ⇒ ⇒

y ( y − c) = 4 y2 − cy − 4 = 0 y1 + y2 = c

c 4

Let the point which divides the line segement in the ratio 1 : 2 be (h, k ). x1 + 2x2 ⇒ =h 3 c x2 = 3h + ⇒ 4 c ⇒ x1 = − − 3h 2 y1 + 2 y2 Also, =k 3 ⇒ y2 = 3k − c ⇒ y1 = − 3k + 2c Now, (h, k ) lies on the line y = 4 x + c ⇒ k = 4h + c ⇒ c = k − 4h k k ⇒ x1 = − + 2h − 3h = − h − 2 2 and y1 = − 3k + 2k − 8h = − k − 8h k  ⇒  h +  (k + 8h) = 1  2 ⇒ ⇒

k2 + 4 hk = 1 2 2 2 16h + k + 10 hk = 2

hk + 8h2 +

∴Locus of (h, k ) is 16x 2 + y2 + 10xy = 2. Hence, (c) is the correct answer.

Aliter Let P (h, k ) divide the chord of the hyperbola xy = 1 in the ratio 1 : 2. The equation of the line of slope 4 is x −h y−k = =r 1 4 17 17 4r  r  Any other point on this line is  h + ,k + .  17 17  Since, this point lies on xy = 1. r   4r   h +  k +  =1  17   17  ⇒ 4 r2 + r(4 h + k ) 17 + 17(hk − 1) = 0

…(i)

Also, the point (h, k ) divides the chord in the ratio 1 : 2. The roots of Eq. (i) are r1 and −2r1, so that − (4 h + k ) 17 −2r1 + r1 = 4 17(hk − 1) and −2r1 ⋅ r1 = 4 (4 h + k ) 17 ⇒ r1 = 4 −17 (hk − 1) 2 and r1 = 8 17 17 2 (hk − 1) = 0, so that locus (h, k ) is ∴ (4 h + k ) + 16 8 16x 2 + y2 + 10xy = 2. Hence, (c) is the correct answer.

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Ex 3. If A, B and P are three points on the hyperbola xy = c 2 such that AB subtends a right angle at P. Then, AB is parallel to (a) tangent at P (c) axis of hyperbola

(b) normal at P (d) None of these

Sol. Let coordinates of A, B and P be  c  c  ct1 ,  ,  ct2 ,  t1   t2  

 c  ct3 ,  t3   1 Slope of PA = − t1t3 1 Slope of PB = − t2t3 ⇒ t1t2t32 = − 1 1 Slope of AB = − t1t2 1 Slope of normal at P is t32 i.e. − t1t2 which is same as slope of AB. ⇒ AB is parallel to normal at P. Hence, (b) is the correct answer. and

Q Coefficient of x 2 + Coefficient of y2 = 0 For all values of λ , it will represent a rectangular hyperbola. Hence, (c) is the correct answer.

Ex 6. A line parallel to the Y-axis intersects the x 2 y2 hyperbola − = 1 and its conjugate a 2 b2 hyperbola at P and Q respectively. Then, the normals at P and Q to the respective curves meet on (a) Y -axis (c) asymptote

Ex 4. If two points P and Q on the hyperbola x 2 y2 − = 1, whose centre C is such that CP is a 2 b2 1 1 perpendicular to CQ, a < b, then + 2 CP CQ 2 is equal to (a) − (c) −

1 a 1

2

a2

+ −

1 b 1

2

b2

1

(b) (d)

a 1

2

a2

+ −

1 b2 1 b2

Sol. Let CP = r1 be inclined to transverse axis at an angle θ, so that P is (r1 cosθ , r1 sin θ) and P lies on the hyperbola.  cos2 θ sin 2 θ   =1 ⇒ r12  2 − b2   a

 sin 2 θ cos2 θ   =1 Replacing θ by 90° + θ, then r22  2 − b2   a 1 1 cos2 θ sin 2 θ sin 2 θ cos2 θ + 2= − + − 2 r1 r2 a2 b2 a2 b2 1 1 1 1  1  1 + = cos2 θ  2 − 2  + sin 2 θ  2 − 2  ⇒ a a r12 r22 b  b  1 1 1 1 + 2= 2− 2 ⇒ 2 r1 r2 a b 1 1 1 1 ⇒ + = 2− 2 2 2 CP CQ a b ⇒

Hence, (d) is the correct answer.

Ex 5. The locus of all conics passing through the intersection of rectangular hyperbola is a/an (a) parabola (c) hyperbola 940

Let two rectangular hyperbolas be S 1 ≡ ax 2 + 2hxy − ay2 + 2gx + 2 fy + c = 0 S 2 ≡ a′ x 2 + 2h′ xy − a′ y2 + 2g′ x + 2 f ′ y + c′ = 0 ∴Equation of any curve passing through intersection of S 1 and S 2 is S 1 + λ S 2 = 0, where λ is real parameter. (a + λa′ )x 2 + 2(h + λh′ )xy − (a + λa′ ) y2 + 2x (g + λg ′ ) + 2 y ( f + λf ′ ) + (c + λc′ ) = 0

(b) ellipse (d) None of these

Sol. We know that in the equation of rectangular hyperbola, sum of the coefficients of x 2 and y2 should be zero.

(b) X -axis (d) None of these

Sol. Let the line parallel to Y-axis be x = α, then points are  b   b  α 2 − a2  and Q ≡ α , α 2 + a2  . P ≡ α ,  a   a  Equation of normals are b − a2 b y− α 2 − a2 = 2 ⋅ α 2 − a2 (x − α ) a b aα b − a2 b and y − α 2 + a2 = 2 ⋅ α 2 + a2 (x − α ) a b aα Now, putting y = 0 in both the normal, we get   b2 x =  2 + 1 α  a So, they intersect on X-axis. Hence, (b) is the correct answer.

Ex 7. The foot of perpendicular drawn from the focus to any tangent of the hyperbola lies on (a) x 2 + y 2 = a 2 (c) x 2 + y 2 = 3a 2

(b) x 2 + y 2 = 2a 2 (d) None of these

x 2 y2 − = 1. a2 b 2 Equation of any tangent is

Sol. Let hyperbola be

y = mx ±

a2m2 − b2

…(i)

A line perpendicular to it and passing through the focus is …(ii) my + x = ae Let (h, k ) be their point of intersection. On squaring and adding Eqs. (i) and (ii), we get (k 2 + h2 )(1 + m2 ) = a2m2 − b2 + a2e2 = a2 (m2 + 1) ⇒

h + k = a2 2

2

⇒ The locus of (h, k ) is x 2 + y2 = a2, which is the auxiliary circle. Hence, (a) is the correct answer.

(a) orthocentre of ∆ABC (b) centroid of ∆ABC (c) incentre of ∆ABC (d) None of the above

of the circle passing through points A, B, C and D be

⇒

⇒ t1, t2, t3 and t4 are the roots of the equation ...(i)

  c Coordinates of D are  , ct1t2t3 . t t t   123

The orthocentre of ∆ABC is   c ,– ct1t2t3  H ≡ −   t1t2t3 ⇒ The reflection of H with respect to the origin is point D. Hence, (a) is the correct answer.

Ex 9. Equation of a rectangular hyperbola whose asymptotes are x = 3 and y = 5 and passing through (7, 8), is (a) xy − 3 y + 5x + 3 = 0 (b) xy + 3 y + 4x + 3 = 0 (c) xy − 3 y + 5x − 3 = 0 (d) xy − 3 y − 5x + 3 = 0

Sol. The equation of rectangular hyperbola is (x − 3)( y − 5) + λ = 0 which passes through (7, 8). ⇒ 4 ⋅ 3 + λ = 0 ⇒ λ = − 12 ∴ xy − 5x − 3 y + 15 − 12 = 0 ⇒ xy − 3 y − 5x + 3 = 0

Ex 10. From a point (h, k ), normals are drawn to the x 2 y2 ellipse 2 + 2 = 1. Then, the feet of these a b normals lie on a hyperbola, whose asymptotes are (a) ( a − b ) x − a h = 0 and 2

2

(a 2 + b 2 ) y + b 2 k = 0 (b) ( a 2 − b 2 ) x − a 2 h = 0 and (a 2 − b 2 ) y + b 2 k = 0 (c) ( a 2 + b 2 ) x + a 2 h = 0 and (a 2 + b 2 ) y + b 2 k = 0 (d) ( a 2 + b 2 ) x + a 2 h = 0 and (a 2 − b 2 ) y + b 2 k = 0

2

which is equation of hyperbola. Its asymptotes are given by xy (b2 − a2 ) + a2hy − b2kx + λ = 0 ∴

∆=0  − b2k   a2h  b2 − a2  (b2 − a2 )2  −λ   =0 ⇒ 0 + 2 4  2  2  2 

⇒

λ=

− (b2a2kh) b2 − a2

Asymptotes are given by b2a2kh =0 b2 − a 2 ⇒ (a2 − b2 )2 xy − a2 (a2 − b2 )hy + b2 (a2 − b2 ) kx (b2 − a2 )xy + a2hy − b2kx −

− a2b2kh = 0 ⇒ [(a − b ) x − a h ][(a − b ) y + b k ] = 0 2

2

2

2

2

2

Hence, (b) is the correct answer.

Ex 11. The angle between the rectangular hyperbolas ( y − mx )( my + x ) = a 2 and ( m 2 − 1)( y 2 − x 2 ) + 4mxy = b 2 is π 2 π (c) 4

π 3 π (d) 6

(a)

(b)

Sol. For first hyperbola,  dy   dy  ( y − mx )  m − m = 0 + 1 + (my + x )   dx   dx 

Hence, (d) is the correct answer.

2

a2 (h − x ) b2 (k − y) = x y

a2hy − xya2 = b2kx − b2 yx 2

Again, let points be A (t1 ), B (t2 ), C (t3 ) and D (t4 ).

⇒

Let us consider a curve

16

⇒ xy (b − a ) + a2hy − b2kx = 0

x 2 + y2 + 2gx + 2 fy + k = 0

⇒

(x4 , y4 ). Equation of normals is a2 (x − xi ) b2 ( y − yi ) , where i = 1, 2, 3, 4 = xi yi These normals pass through (h, k ), so a2 (h − xi ) b2 (k − yi ) = xi yi

Sol. Let the equation of the hyperbola be xy = c2 and equation

c2t 4 + 2gct 3 + kt 2 + 2 fct + c2 = 0 1 t1t2t3t4 = 1 ⇒ t4 = t1t2t3

Sol. Let normal points be (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) and

Hyperbola

Ex 8. ABCD is a cyclic quadrilateral in which the three vertices A, B and C lie on a rectangular hyperbola. Then, the reflection of the fourth vertex D with respect to the origin is the

dy (my + x + my − m2x ) + y − mx − m2 y − mx = 0 dx dy − y + m2 y + 2mx ⇒ = m1 = dx 2my + x − m2x ⇒

For second hyperbola,  dy   dy  + y = 0 (m2 − 1)  2 y − 2x + 4 m  x  dx   dx  dy 2 [ 2 y(m − 1) + 4 mx ] = − 4 my + 2x (m2 − 1) ⇒ dx dy −2my + m2x − x ⇒ = m2 = dx m2 y − y + 2mx m1m2 = − 1 π ∴ Angle between the hyperbolas = 2 Hence, (a) is the correct answer.

Q

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Objective Mathematics Vol. 1

16

Ex 12. From the centre C of hyperbola

x2 2

−

y2 2

= 1,

a b perpendicular CN is drawn on any tangent to it at the point P ( a sec θ, b tan θ ) in the first quadrant. Then, the value of θ, so that area of ∆CPN is maximum, is a b a b (b) sin −1 (c) cos −1 (d) tan −1 (a) sin −1 b a b a

Sol. Equation of tangent is

Ex 14. Portion of the asymptotes of hyperbola x 2 y2 − = 1 (between centre and the tangent at a 2 b2 vertex) in the first quadrant is cut by the line y + λ ( x − a ) = 0 (λ is a parameter), then

b sec θ x − a tan θ y − ab = 0 ab ∴ CN = 2 2 b sec θ + a2 tan 2 θ Equation of normal at P is ax cosθ + by cot θ = a2 + b2 ∴

CM =

⇒ y1 + y2 + y3 + y4 = k From Eqs. (ii) and (iii), h2 t12 + t22 + t32 + t42 = 4 h2 Given that = k ⇒ Locus of (h, k ) is x 2 = 4 y 4 which is a parabola. Hence, (a) is the correct answer.

(a) λ ∈ R (c) λ ∈ ( − ∞ , 0)

a 2 + b2

(b) λ ∈ ( 0, ∞ ) (d) None of these

Sol. The line y + λ (x − a) = 0 will intersect the portion of

a2 cos2 θ + b2 cot 2 θ

1 A = CM × CN 2 ab (a2 + b2 ) = a2b2 + b4cosec2 θ + a4 sin 2 θ

the asymptote in the first qudrant only if its slope is negative. Y

A is maximum when b4cosec2 θ + a4 sin 2 θ is minimum. Now, b4cosec2 θ + a4 sin 2 θ ≥ 2a2b2 ab (a2 + b2 ) a2 + b2 = 2ab 2 −1 b where, θ = sin a Hence, (b) is the correct answer.

X′

Amax =

Ex 13. A point P moves such that sum of the slopes of the normals drawn from it to the hyperbola xy = 4 is equal to the sum of ordinates of feet of normals. Then, the locus of P is (a) a parabola (c) an ellipse

(b) a hyperbola (d) a circle 

2 t

Sol. Any point on the hyperbola xy = 4 is  2t ,  .   2 2  Now, normal at  2t ,  is y − = t 2 (x − 2t )  t t [its slope is t 2] If the normal passes through P (h, k ), then 2 k − = t 2 (h − 2t ) t …(i) ⇒ 2t 4 − ht 3 + tk − 2 = 0 Roots of Eq. (i) give parameters of foot of normals passing through (h, k ). Let roots be t1 , t2 , t3 and t4, then h ...(ii) t1 + t2 + t3 + t4 = 2 ...(iii) t1t2 + t1t3 + t1t4 + t2t3 + t2t4 + t3t4 = 0 k ...(iv) t1t2t3 + t1t2t4 + t1t3t4 + t2t3t4 = − 2 and ...(v) t1t2t3t4 = − 1

942

On dividing Eq. (iv) by Eq. (v), we get 1 1 1 1 k + + + = t1 t2 t3 t4 2

O

(a, 0)

X

Y′

Hence −λ0 ∴ λ ∈ (0, ∞ ) Hence, (b) is the correct answer.

Ex 15. If the lines x = h and y = k are conjugate with respect to the hyperbola xy = c 2 , then the locus of ( h, k ) is (a) xy = 2c 2 (c) 2xy = c 2

(b) xy = c 2 (d) xy = 4c 2

Sol. Let (x1 , y1 ) be the pole of the line x − h = 0 with respect

to the hyperbola xy = c2. Then, the equation of the polar is xy1 + yx1 = 2c2.

Thus, the equations x − h = 0 and xy1 + yx1 = 2c2 represent the same line. y1 x1 2c2 ∴ = = 1 0 h 2c2 ⇒ x1 = 0 and y1 = h Since, the lines x − h = 0 and y − k = 0 are conjugate lines. Therefore, the pole of the line x − h = 0 lies on the line y − k = 0. ∴ y1 − k = 0 2c2 − k = 0 ⇒ hk = 2c2 ⇒ h So, the locus of (h, k ) is xy = 2c2. Hence, (a) is the correct answer.

Ex 16. The equation 16x − 3 y − 32x + 12 y − 44 = 0 represents a hyperbola 2

2

(a) the length of whose transverse axis is 4 3 (b) the length of whose conjugate axis is 4 (c) whose centre is (1, 2) 19 (d) whose eccentricity is 3 Sol. 16 (x 2 − 2x + 1) − 3 ( y2 − 4 y + 4 ) = 48 (x − 1)2 ( y − 2)2 − =1 3 16 So, length of transverse axis is 3 and length of conjugate axis is 4. 19 Centre = (1, 2) and e = 3 Hence, (b), (c) and (d) are the correct answers. ⇒

Ex 19. If P is a point on a hyperbola, then (a) locus of excentre of the circle described opposite to ∠P for ∆PSS ′ (S , S ′ are foci), is triangle at vertex (b) locus of excentre of the circle described opposite to ∠S ′, is hyperbola (c) locus of excentre of the circle described opposite to ∠P for ∆PSS ′ (S , S ′ are foci), is hyperbola (d) locus of excentre of the circle described opposite to ∠S ′, is tangent at vertex Sol. Let (h, k ) be excentre, then h=

2

Sol. Q

1 (b) 2e

e2 − 1

(c)

e

(d) 0

2

x y − =1 a2 b2 a θ = cos = 2 2 a + b2

a 2 2

=

ae

θ

θ

1 e

β

e −1 θ β = sin = 2 2 e Hence, (a) and (c) are the correct answers. 2

cos

Ex 18. If the circle x 2 + y 2 = a 2 intersect the hyperbola xy = c 2 in four points P ( x1 , y1 ), Q ( x 2 , y2 ), R ( x 3 , y3 ) and S ( x 4 , y4 ), then (a) x1 + x 2 + x 3 + x 4 = 0 (b) y1 + y 2 + y 3 + y 4 = 0 (c) y1 y 2 y 3 y 4 = c 4 (d) x1 x 2 x 3 x 4 = c 4 c4 = a2 x2 x 4 − a2x 2 + c4 = 0 Σxi = 0; Σyi = 0 x1 x2 x3 x4 = c4 x2 +

⇒ ⇒ ⇒

y1 y2 y3 y4 = c4

Hence, (a), (b), (c) and (d) are the correct answers.

[for a, sec θ > 0]

P (a cos θ, b tan θ)

S′ H (ae, 0)

S (ae, 0)

Similarly, x = a ⇒ Locus is x 2 = a2

[for a, sec θ < 0]

Again, let (h, k ) be excentre opposite to ∠S ′. 2a2e sec θ + a2e2 sec θ + a2e + a2e2 sec θ − a2e ∴ h= 2a + 2ae 2aeb tan θ ⇒ h = ae sec θ, k = 2a + 2ae ⇒ Locus is the hyperbola. Hence, (a) and (b) are the correct answers.

Ex 20. If foci of

x2 a2

−

y2 b2

= 1 coincide with the foci of

x 2 y2 + = 1 and eccentricity of the hyperbola 25 9 is 2, then (a) a 2 + b 2 = 16 (b) there is no director circle to the hyperbola (c) centre of the director circle is (0, 0) (d) length of latusrectum of the hyperbola is 12 Sol. For an ellipse, and

Sol. On solving xy = c2 and x 2 + y2 = a2, we get

a (ae sec θ + a) − ae (ae sec θ − a) − 2 ae (a sec θ ) 2ae (sec θ − 1) h=−a ⇒ x=−a

Ex 17. If e is the eccentricity of the hyperbola x 2 y2 − = 1 and θ is angle between the a 2 b2 asymptotes, then cos (θ / 2 ) is equal to 1 (a) e

Hyperbola

16

Type 2. More than One Correct Option

∴

a=5 e=

25 − 9 4 = 5 25

ae = 4

∴ The foci are (−4, 0) and (4, 0). For the hyperbola, ae = 4 and e = 2 ∴ a=2 ⇒ b2 = 4 (4 − 1) = 12 ⇒

b = 12

Hence, (a), (b) and (d) are the correct answers.

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Objective Mathematics Vol. 1

16 Type 3. Assertion and Reason Directions

(Ex.Nos. 21-25) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Statement II If difference of distances of a point P from the two fixed points is constant and less than the distance between the fixed points, then locus of P is a hyperbola. Sol. Let P be the position of the gun and Q be the position of the target. Again, let u be the velocity of sound, v be the velocity of bullet and R be the position of the man, then we have P

Q

Ex 21. Let a, b, α ∈ R − {0}, where a, b are constants and α is a parameter. All the members of the family x 2 y2 1 of hyperbolas 2 + 2 = 2 have the same a b α pair of asymptotes.

Statement I

Statement II Change in α, does not change the slopes of the asymptotes of a member of the x 2 y2 1 family 2 + 2 = 2 . a b α Sol. Both statements are true and Statement II is true reason for Statement I, as for any member, semi-transverse and a b semi-conjugate axes are and respectively and hence α α b asymptotes are always y = ± x. a Hence, (a) is the correct answer.

x 2 y2 and + =1 25 16 2 2 12x − 4 y = 27 intersect each other at right angle. Statement II Whenever confocal conics intersect, they intersect each other orthogonally.

Ex 22. Statement I Ellipse

3 ,a=5 5 ∴ Foci are (± 3, 0). x2 y2 For hyperbola, − =1 27/12 27/ 4

Sol. Here, e =

12 + 4 3 = 2, a = 4 2 Now, foci are (± 3, 0). ∴The two conics are confocal. Hence, (a) is the correct answer. ⇒

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e=

Ex 23. Statement I A bullet is fired and hit a target. If an observer in the same plane heard two sounds the crack of the rifle and the thud of the bullet striking the target at the same instant, then locus of the observer is hyperbola, where velocity of sound is smaller than velocity of bullet.

R

PR QR PQ = + u u v PR QR PQ i.e. − = u u v u u i.e. PR − QR = ⋅ PQ = Constant and ⋅ PQ < PQ v v ∴Locus of R is a hyperbola. Hence, (a) is the correct answer.

Ex 24. Statement I With respect to a hyperbola x 2 y2 − = 1, if perpendiculars are drawn from 9 16 a point (5, 0) on the lines 3 y ± 4x = 0, then their feet lie on circle x 2 + y 2 = 16. Statement II If from any foci of a hyperbola, perpendicular are drawn on the asymptotes of the hyperbola, then their feet lie on auxiliary circle. x 2 y2 − = 1 and 9 16 3 y ± 4 x = 0 are asymptotes, then the auxiliary circle is x 2 + y2 = 9.

Sol. (5, 0) is a focus of the hyperbola

∴ The feet lie on x 2 + y2 = 9. So, Statement I is false and Statement II is true. Hence, (d) is the correct answer.

Ex 25. Statement I If eccentricity of a hyperbola is 2, then eccentricity of its conjugate hyperbola 2 . is 3 Statement II If e and e′ are the eccentricities of a hyperbola and its conjugate hyperbola, 1 1 then 2 + 2 = 1. e e′ Sol. Statement II is true. 2

Q

1  3 +   =1 22  2 

∴Statement I is also true. Hence, (a) is the correct answer.

Passage I (Ex. Nos. 26-28) If P is a variable point and and F2 are two fixed points such that F1 | PF1 – PF2 | = 2a. Then, the locus of the point P is a hyperbola, with points F1 and F2 as the two foci (F1F2 > 2a). If

x2 y2 − 2 = 1 is a hyperbola, then its 2 a b

conjugate hyperbola is

x2 y2 − 2 = − 1. Let P ( x , y ) be a 2 a b

variable point such that | ( x − 1) 2 + ( y − 2) 2 − ( x − 5) 2 + ( y − 5) 2 | = 3.

Ex 26. If the locus of the point P represents a hyperbola of eccentricity e, then the eccentricity e′ of the corresponding conjugate hyperbola is 5 3 5 (c) 4

(b)

(a)

(d)

4 3 3

Sol. Distance between the foci (1, 2) and (5, 5) is 5. 2ae = 5 5 ∴ e= 3 1 1 Now, + =1 e2 e′ 2 5 e′ = ⇒ 4 Hence, (c) is the correct answer. Q

[Q 2a = 3 ]

3 4 Hence, (b) is the correct answer.

∴Angle of rotation, θ = tan −1

x2 y2 − 2 = 1, the normal at P meets the transverse axis 2 a b AA′ in G and the conjugate axis BB ′ in g and CF is perpendicular to the normal from the centre.

Ex 29. If PF ⋅ PG = k CB 2 , then k is equal to (a) 2

(b) 1

2

(c) 1/2

(d) 4

ab b2sec2 φ + a2tan 2 φ 2

b (b2sec2 φ + a2 tan 2 φ ) a2 ab PF = ⇒ PG ⋅ PF = b2 = CB 2 ∴ k = 1 a ⋅ PG b Hence, (b) is the correct answer. PG 2 =

7 55 (a) ( x − 3) 2 +  y −  =  2 4

Ex 30. PF ⋅ Pg equals (a) CA 2

2

7 25 (b) ( x − 3) +  y −  =  2 4 2

(b) CF 2

(c) CB 2

(d) CA ⋅ CB

2

a 2 2 (b sec φ + a2 tan 2 φ ) b2 ab ⇒ PF ⋅ Pg = a2 = CA 2 PF = ∴ b Pg a Hence, (a) is the correct answer.

Sol. Pg 2 =

2

7 7 (c) ( x − 3) 2 +  y −  =  2 4 (d) None of the above Sol. Director circle (x − h)2 + ( y − k )2 = a2 − b2,  1 + 5 2 + 5  7 where (h, k) is centre, is  ,  ≡  3,   2 2   2   3  5 b2 = a2 (e2 − 1) =     − 1 = 4  2  3   2

 (x − 3)2 +  y − 

Sol. Slope of transverse axis is 3/ 4 .

Sol. Q PF =

Ex 27. Locus of intersection of two perpendicular tangents to the given hyperbola is

⇒

16

Passage II (Ex. Nos. 29-31) For the hyperbola

7

 Director circle, (x − 3)2 +  y − 

 7 Ex 28. If origin is shifted to point  3,  and the axes  2 are rotated through an angle θ in clockwise sense, so that equation of given hyperbola x 2 y2 changes to the standard form 2 − 2 = 1, then a b θ is 4 3 (b) tan −1   (a) tan −1    3  4 −1  5 −1  3 (c) tan   (d) tan    3  5

Hyperbola

Type 4. Linked Comprehension Based Questions

2

2

7 9  = −4  2 4 2

7 −7  = 2 4

which does not represent any real point. Hence, (d) is the correct answer.

Ex 31. Locus of middle point of G and g is a hyperbola of eccentricity (a)

1 e −1 2

(b)

e e −1 2

(c) 2 e 2 − 1 (d)

x2 y2 − 4 4 =1 2 4 ae ae 4 4 b2 2 4 4 4 ae ae + e 4 4 b2 = e1 = 2 4 ae e2 − 1 4 Hence, (b) is the correct answer.

e 2

Sol. Locus of middle point is

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Objective Mathematics Vol. 1

16

Passage III (Ex. Nos. 32-34) If a circle with centre C(α, β) intersects a rectangular hyperbola with centre L(h, k ) at four points P ( x 1, y 1), Q( x 2, y 2 ), R ( x 3, y 3 ) and S ( x 4, y 4 ), then the mean of the four points P, Q, R, S is the mean of the points C and L. In other words, the mid-point of CL coincides with the mean point of P, Q, R, S analytically i.e. x1 + x 2 + x 3 + x 4 α + h = 4 2 y1 + y 2 + y 3 + y 4 β + k and = 4 2

Ex 32. Five points are selected on a circle of radius a. The centres of the rectangular hyperbola, each passing through four of these points, all lie on a circle of radius (a) a

(b) 2a

(c)

a

(d)

2

a 2

Sol. Let the circle be x 2 + y2 = a2 and centre of rectangular hyperbola be (h, k ) . Again, let given points on circle (a cosθ i , a sin θ i ), i = 1, 2, 3, 4 , 5, so that

be

Sol. Let centre of circle and hyperbola be (α , β) and (h, k ) and points be A (x1 , y1 ), B (x2 , y2 ), C (x3 , y3 ) and D (x4 , y4 ). Then, h + α x1 + x2 + x3 + x4 …(i) = 2 4 k + β y1 + y2 + y3 + y4 …(ii) and = 2 4 As any chord passing through centre of hyperbola is bisected at the centre. ∴AB is bisected at (h, k ). x1 + x2 …(iii) ⇒ =h 2 y1 + y2 …(iv) and =k 2 From Eqs. (i) and (iii), x + x2 + x3 + x4 x + x4 h+α = 1 ⇒ α= 3 2 2 y3 + y4 From Eqs. (ii) and (iv), β = 2 ⇒ (α , β ) is mid-point of CD. ⇒ (α , β ) lies on CD. ⇒ Centre of circle lies on CD. Hence, (b) is the correct answer.

4

∑ a cos θ i 4

⇒

Ex 34. If the normals drawn at four concyclic points on a rectangular hyperbola xy = c 2 meet at point (α, β), then the centre of the circle has the coordinates

0+ h = 3

i=1 4

∑ a cos θ i − a cos θ 5 = 2h

i=1 4

Similarly,

(a) (α , β ) α β (c)  ,   2 2

∑ a sin θ i − a sin θ 5 = 2k

i=1

As the five points are given,

4

4

i=1

i=1

∑ a sin θ i and ∑ a cos θ i

are known. Let us assume their values of be µ and λ , respectively. ∴ λ − a cos θ 5 = 2h and µ − a sin θ 5 = 2k ⇒ 2h − λ = − a cos θ 5 and 2k − µ = − a sin θ 5 ⇒ (2h − λ )2 + (2h − µ )2 = a2 2

2

λ µ  a   ⇒ h −  + k −  =    2   2 2

2

a Centre (h, k ) lies on circle of radius . 2 Hence, (d) is the correct answer.

Ex 33. A, B , C and D are the points of intersection of a circle and a rectangular hyperbola which have different centres. If AB passes through the centre of the hyperbola, then CD passes through

946

(a) centre of the hyperbola (b) centre of the circle (c) mid-point of the centres of circle and hyperbola (d) none of the points mentioned in the three options

(b) ( 2α , 2β ) α β (d)  ,   4 4

Sol. Let the four concyclic points at which normals to rectangular hyperbola are concurrent be A (x1 , y1 ), B (x2 , y2 ),C (x3 , y3 ) and D (x4 , y4 ) and centre of circle be (h, k ). x1 + x2 + x3 + x4 h + 0 = ∴ 4 2 y1 + y2 + y3 + y4 k + 0 and = 4 2 …(i) ⇒ x1 + x2 + x3 + x4 = 2h and …(ii) y1 + y2 + y3 + y4 = 2k  c Normal to rectangular hyperbola xy = c2 at  ct ,  is  t ct 4 − xt 3 + yt − c = 0 As all normal pass through (α , β ). ⇒ x1 + x2 + x3 + x4 = c (t1 + t2 + t3 + t4 ) α …(iii) = c  =α  c 1 1 1 1 and y1 + y2 + y3 + y4 = c  + + +   t1 t2 t3 t4   −β / c =c  =β  − c / c From Eqs. (i) and (iii), we get 2h = α From Eqs. (ii) and (iv), we get 2k = β  α β ∴ (h, k ) =  ,   2 2 Hence, (c) is the correct answer.

… (iv)

Ex 35. Match the statements of Column I with values of Column II. Column I

Column II

A. The eccentricity of the conic represented by x 2 − y 2 − 4 x + 4 y + 16 = 0 is

p.

7

2 2 B. If the foci of the ellipse x + y = 1 and 16 b2 x2 y2 1 the hyperbola coincides, − = 144 81 25 2 then the value of b is

q.

0

C. The product of lengths of perpendiculars from any point of the hyperbola x 2 − y 2 = 8 to its asymptotes is

r.

2

D. The number of point(s) outside the x2 y2 hyperbola − = 1, where two 25 36 perpendicular tangents can be drawn to the hyperbola, is/are

s.

4

(x 2 − 4 x ) − ( y2 − 4 y) = − 16

⇒ (x 2 − 4 x + 4 ) − ( y2 − 4 y + 4 ) = − 16 ⇒

(x − 2)2 − ( y − 2)2 = − 16

(x − 2)2 ( y − 2)2 − = −1 42 42 Shifting the origin at (2, 2), we obtain X 2 Y2 − = − 1, where x = X + 2, y = Y + 2 42 42 which is a rectangular hyperbola, whose eccentricity is always 2. B. For the given ellipse, a2 = 16 ⇒

e = 1−

Therefore,

b2 a2

16 − b2 b2 = 16 16 So, the foci of the ellipse are ( ± ae, 0).  ± 16 − b2 , 0 i.e.   e = 1−

⇒

2

 12  9 For the hyperbola, a =   , b2 =    5  5

2

2

The eccentricity e is given by b2 a2 81 15 5 = 1+ = = 144 12 4

e = 1+

2

∴ ⇒

⇒ x 2 + y2 = − 9 which is not possible. ∴Number of points is zero. A → r; B → p; C → s; D → q

Ex 36. Match the statements of Column I with values of Column II.

Sol. A. We have, x 2 − y2 − 4 x + 4 y + 16 = 0 ⇒

x2 y2 D. − =1 25 36 Equation of director circle is x 2 + y2 = a2 − b2

Hyperbola

16

Type 5. Match the Columns

 12 5  16 − b2 =  ×  = 9  5 4 b2 = 7

C. x 2 − y2 = 8 when rotate d = 45° to positive X -axes becomes xy = 4 has asymptotes X and Y-axes. ∴ p1 p2 = 4

Column I

Column II

A. The area of the triangle that a tangent at a point p. x2 y2 of the hyperbola − = 1 makes with its 16 9 asymptotes, is

12

B. If the line y = 3 x + λ touches the curve 9 x 2 − 5 y 2 = 45, then| λ | is

q.

6

C. If the chord x cos α + y sin α = p of the x2 y2 hyperbola − = 1 subtends a right angle 16 18 at the centre, then the diameter of the circle, concentric with the hyperbola to which the given chord is a tangent, is

r.

24

D. If λ is the length of the latusrectum of the hyperbola 16 x 2 − 9 y 2 + 32 x + 36 y − 164 = 0, then 3λ is equal to

s.

32

Sol. A. Equation of tangent at (a, 0) is x = a b x a Area = a ⋅ b = 4 × 3 = 12 ∴ B. y = 3x + λ touches 9x 2 − 5 y2 = 45 ∴ 9x 2 − 5(3x + λ )2 = 45 i.e. − 36x 2 − 30λx − 5λ2 − 45 = 0 i.e. 36x 2 + 30λx + 5λ2 + 45 = 0 has equal roots. ∴ 900λ2 − 720λ2 − 180 × 36 = 0 i.e. λ2 = 36 ⇒ λ=±6 ∴ |λ|= 6 C. 18x 2 − 16 y2 − 288 = 0 y=

2

 x cos α + y sin α  i.e. 9x 2 − 8 y2 − 144   =0 p   Since, these lines are perpendicular to each other. ∴ 9 p2 − 8 p2 − 144 (cos2 α + sin 2 α ) = 0 ⇒ p2 = 144 = ± 12 ∴ Radius of the circle = 12 ∴ Diameter of the circle = 24 D. 16x 2 + 32 + 16 − 9( y2 − 4 y + 4 ) − 144 = 0 i.e. 16(x + 1)2 − 9( y − 2)2 = 144 (x + 1)2 ( y − 2)2 i.e. − =1 9 16 16 32 Length of latusrectum = 2 × = 3 3 ∴ 3λ = 32 A → p; B → q; C → r; D → s

947

Ex 37. A variable of line of slope 4 intersects the hyperbola xy =1 at two points. If the locus of the point which divides the line segment between these two points in the ratio 1 : 2 is λx 2 + y 2 + µxy − γ = 0, then the value of λ − (µ + γ ) is equal to_______ .

(i) x1 + x2 + x3 + x4 = ct1 + ct2 + ct3 + ct4 = 2 = l (ii) Σ x1x2 = c2 Σ t1t2 = − 20 ∴Σ x12 = (Σ x1 )2 − 2 Σ x1x2 = 44 = m (iii) Σ y1 = Σ

Σ y1 y2 = c2 Σ

Sol. (4) P (t1 , 1 / t1 ), Q (t2 , 1 / t2 ) ∴

Y

xy =1 X

Ex 39. If the product of the perpendicular distances x 2 y2 from any point on the hyperbola 2 − 2 = 1 of a b eccentricity e = 3 from its asymptotes is equal to 6, then the length of the transverse axis of the hyperbola is ________.

Q (t2)

Y′

1 Slope of PQ = − =4 t1t2 1 …(i) ∴ t1t2 = − 4 t + 2t2 …(ii) α= 1 3 1 2 + t1 t2 …(iii) and β= 3 On solving Eqs. (i), (ii) and (iii), we get the required locus is 16x 2 + y2 + 10xy − 2 = 0 ⇒

Σ y12 = (Σ y1 )2 − 2 Σ y1 y2 = 56 = n

Sol. (6) p1 p2 =

a2b2 a2 ⋅ a2 (e2 − 1) = =6 a2 + b 2 a2e2

p

X′

(a cos θ, b tan θ) p2

λ − (µ + γ ) = 16 − (10 + 2) = 4

Ex 38. Let the circle (x − 1) 2 + ( y − 2) 2 = 25 cuts a rectangular hyperbola with transverse axis along y = x at four points A, B , C and D having coordinates ( x i , yi ), i =1, 2, 3, 4 respectively, O being the centre of the hyperbola. If l = x1 + x 2 + x 3 + x 4 , m = x12 + x 22 + x 32 + x 42 and n = y12 + y22 + y32 + y42 , then find the value

Hence, 2a = 6

Ex 40. If e1

and e2 are the eccentricities of the x 2 y2 hyperbola − = 1 and its conjugate a 2 b2 hyperbola, while e1−2 + e2−2 = λ, then the value of λ is ________ .

Sol. (1) The eccentricity e1 of the given hyperbola is obtained

of n − m − 4l.

from b2 = a2 (e12 − 1)

Sol. (4) The circle is x 2 + y2 − 2x − 4 y − 20 = 0 and let the hyperbola be xy = c .  c If  ct ,  is the point  t 2

c2t 2 + ⇒

of

intersection,

4c c2 − 2ct − − 20 = 0 2 t t c2t 4 − 2ct 3 − 20t 2 − 4 ct + c2 = 0

If t1, t2, t3 and t4 are its roots, then 2 4 20 Σ t1 = ; Σ t1t2 = − 2 , Σ t1t2t3 = and t1t2t3t4 = 1 c c c

948

1 = 20 t1t2

⇒ n − m − 4l = 4

P (t1) R (α, β)

c Σt t t =c 123 t1 t1t2t3t4

1

Objective Mathematics Vol. 1

16 Type 6. Single Integer Answer Type Questions

then

…(i)

The eccentricity e2 of the conjugate hyperbola is given by …(ii) a2 = b2 (e22 − 1) On multiplying Eqs. (i) and (ii), we get 1 = (e12 − 1)(e22 − 1) ⇒ ⇒ ⇒

0 = e12e22 − e12 − e22 e1−2

+

e2−2

=1

λ =1

Target Exercises Type 1. Only One Correct Option (x − 1)2 ( y − 5)2 − =1 16 9 2 2 (x − 1) ( y − 5) (c) + =1 16 9

(a)

(b)

x 2 y2 − =1 16 9

(d) None of these

(b) 4 x 2 − 3 y2 + 12 = 0 (d) None of these

3. The eccentricity of the hyperbola −

x2 a2

+

y2 b2

= 1 is

given by a2 + b2 a2 2 b − a2 a2

(a) e = (c) e =

a2 − b2 a2 2 a + b2 b2

(b) e = (d) e =

(c) −

7 3

(b) 7 3

7 2

(d) −

7 2

5. The vertices of the 2 2 9x − 16 y − 36x + 96 y − 252 = 0 are (a) (6, 3) and (− 6, 3) (c) (− 6, 3) and (− 6, − 3)

(b) x 2 − 3 y2 = a2 (d) None of these

7. If the coordinates of a point are a tan (θ + α ) and b tan (θ + β ), where θ is a variable, then locus of the point is (a) a hyperbola (c) an ellipse

(b) a rectangular hyperbola (d) None of these

8. Let A and B be two fixed points and P be another point in the plane, moves in such a way that k1 PA + k 2 PB = k 3 , where k1 , k 2 and k 3 are real constants. Then, which one of the following is not the locus of P? (a) A circle, if k 1 = 0 and k 2 , k 3 > 0 (b) A circle, if k 1 > 0, k 2 < 0 and k 3 = 0 (c) An ellipse, if k 1 = k 2 > 0 and k 3 > 0 (d) A hyperbola, if k 2 = − 1 and k 1 , k 3 > 0

5+1 2

(d) 6

(b)

11 + 1 2

(c)

13 + 1 2 3

(d)

13 − 1 2 3

11. Let LL′ be the latusrectum through the focus of the y2 x2 hyperbola 2 − 2 = 1 and A′ be the farther vertex. a b If ∆A ′ LL′ is equilateral, then the eccentricity of the hyperbola is (b) 3 + 1

(c)

3+1 2

(d)

3+1 3

12. If PQ is a double ordinate of the hyperbola y2 x2 − = 1 such that OPQ is an equilateral triangle, a2 b2 O being the centre of the hyperbola. Then, the eccentricity e of the hyperbola satisfies

hyperbola

6. The equation of the hyperbola of given transverse axis whose vertex bisects the distance between the centre and the focus, is given by (a) 3x 2 − y2 = 3a2 (c) x 2 − y2 = 3a2

(a)

(a) 1 < e < 2 3

(b) (6, 3) and (− 2, 3) (d) None of these

(c) 5

10. If the latusrectum of a hyperbola forms an equilateral triangle with the vertex at the centre of the hyperbola, then eccentricity of the hyperbola is

(a) 3

4. The eccentricity of the hyperbola 3x 2 − 4 y 2 = − 12 is (a)

(b) 3

(a) 2

2. The equation of the hyperbola with vertices (3, 0), ( −3, 0) and semi-latusrectum 4, is given by (a) 4 x 2 − 3 y2 + 36 = 0 (c) 4 x 2 − 3 y2 − 36 = 0

9. If the latusrectum of a hyperbola through one focus subtends 60° angle at the other focus, then its eccentricity e is

(c) e =

3 2

3 3 2 (d) e > 3 (b) e =

Targ e t E x e rc is e s

1. The equation of the hyperbola whose foci are (6, 5), ( − 4, 5) and eccentricity is 5/4, is

13. Equation of tangent to the hyperbola 2x 2 − 3 y 2 = 6 which is parallel to the line y = 3x + 4, is (a) y = 3x + 5 (b) y = 3x − 5 (c) y = 3x + 5 and y = 3x − 5(d) None of these

14. The line x cos α + y sin α = p touches the hyperbola y2 x2 − = 1 , if a2 b2 (a) a2 cos2 α (b) a2 cos2 α (c) a2 cos2 α (d) a2 cos2 α

− b2 sin 2 α = p2 − b2 sin 2 α = p + b2 sin 2 α = p2 + b2 sin 2 α = p

15. The angle between lines joining the origin to the points of intersection of the line 3 x + y = 2 and the curve y 2 − x 2 = 4, is  (a) tan −1    (c) tan −1  

2  3 3  2

π 6 π (d) 2 (b)

949

Objective Mathematics Vol. 1

16

16. A common tangent to 9x 2 − 16 y 2 = 144 and x 2 + y 2 = 9, is 3x 15 + 7 7 3 (c) y = 2 x + 15 7 7 (a) y =

(b) y = 3

2 15 x+ 7 7

(d) None of these

17. The equation of a tangent parallel to y = x drawn to

y2

x2

− = 1, N is the a2 b2 foot of the perpendicular from P on the transverse axis. The tangent to the hyperbola at P meets the transverse axis at T. If O is the centre of the hyperbola, then OT ⋅ ON is equal to

Ta rg e t E x e rc is e s

(b) a2

(c) b2

(d)

b2 a2

19. The line lx + my + n = 0 will be a normal to the hyperbola b 2 x 2 − a 2 y 2 = a 2 b 2 , if a2 b2 (a2 + b2 )2 + 2= 2 l m n2 2 2 2 a b (a + b2 )2 (c) 2 − 2 = n l m

(a)

(b)

a2 b2 (a2 − b2 )2 − = l 2 m2 n2

21. Equation of the chord of the hyperbola 25x 2 − 16 y 2 = 400 which is bisected at the point ( 6, 2), is (a) 16x − 75 y = 418 (c) 25x − 4 y = 400

(b) 75x − 16 y = 418 (d) None of these

22. The diameter of 16x 2 − 9 y 2 = 144 which is conjugate to x = 2 y, is 16x 9 16 y (c) x = 9

(a) y =

32x 9 32 y (d) x = 9

(b) y =

x2

950

−

y2

= 1 meets the a2 b2 transverse and conjugate axes in M and N and the lines MP and NP are drawn at right angle to the axes. The locus of P is

23. A normal to the hyperbola

(a) the parabola y2 = 4 a (x + b) (b) the circle x 2 + y2 = ab (c) the ellipse b2x 2 + a2 y2 = a2 + b2 (d) the hyperbola a2x 2 − b2 y2 = (a2 + b2 )2

y2

26. The equation of a hyperbola, conjugate to the hyperbola x 2 + 3xy + 2 y 2 + 2x + 3 y + 1 = 0, is (a) x 2 + (b) x 2 + (c) x 2 + (d) x 2 +

3xy + 3xy + 3xy + 3xy +

2 y2 + 2 y2 + 2 y2 + 2 y2 +

2x + 2x + 2x + 2x +

3y + 1= 0 3y + 2 = 0 3y + 3 = 0 3y + 4 = 0

27. The eccentricity of the rectangular hyperbola is (a) 2 (c) 0

(b) 2 (d) None of these

28. The angle between the asymptotes of

(b) 3 y − 4 x + 4 = 0 (d) 3x − 4 y = 2

−

(a) a2 sec θ (b) a2 (e4 sec2 θ + 1) (c) a2 (e4 sec2 θ − 1) (d) None of the above

(d) None of these

20. The locus of the middle points of chords of hyperbola 3x 2 − 2 y 2 + 4x − 6 y = 0 parallel to y = 2x, is (a) 3x − 4 y = 4 (c) 4 x − 4 y = 3

x2

= 1meets a2 b2 the transverse axis at G, then AG ⋅ A ′ G is (where, A and A′ are the vertices of the hyperbola)

(b) x − y + 2 = 0 (d) x + y − 2 = 0

18. P is a point on the hyperbola

(a) e2

(a) 9x 2 − 8 y2 + 18x − 9 = 0 (b) 9x 2 − 8 y2 − 18x + 9 = 0 (c) 9x 2 − 8 y2 − 18x − 9 = 0 (d) 9x 2 − 8 y2 + 18x + 9 = 0

25. If the normal at θ on the hyperbola

y2 x2 − = 1, is 3 2 (a) x − y + 1 = 0 (c) x + y − 1 = 0

24. If x = 9 is the chord of contact of the hyperbola x 2 − y 2 = 9, then the equation of the corresponding pair of tangent is

 b (a) 2 tan −1    a  a (c) 2 tan −1    b

x2 a2

−

y2 b2

= 1 is

 b (b) tan −1    a  a (d) tan −1    b

29. The equation of the asymptotes of the hyperbola 2x 2 + 5xy + 2 y 2 − 11x − 7 y − 4 = 0, is (a) 2x 2 + 5xy + 2 y2 − 11x − 7 y − 5 = 0 (b) 2x 2 + 4 xy + 2 y2 − 7x − 11 y + 5 = 0 (c) 2x 2 + 5xy + 2 y2 − 11x − 7 y + 5 = 0 (d) None of the above

30. The asymptotes of a hyperbola having centre at the point (1, 2) are parallel to the lines 2x + 3 y = 0 and 3x + 2 y = 0. If the hyperbola passes through the point (5, 3), then its equation is (a) (2x + 3 y − 8) (3x + 2 y − 7) = 154 (b) (2x + 3 y − 8) (3x + 2 y + 7) = 152 (c) (2x + 3 y + 8) (3x + 2 y − 7) = 152 (d) None of the above

31. The equation of the hyperbola whose asymptotes are the straight lines 3x − 4 y + 7 = 0 and 4x + 3 y + 1 = 0 and which passes through the origin, is (a) 12x 2 − 7xy − 12 y2 + 31x + 17 y = 0 (b) 12x 2 − 7xy − 12 y2 + 31x = 0 (c) x 2 − 7xy − 12 y2 + 31x + 17 y = 0 (d) None of the above

(b) at 90° (d) None of these

33. The number of triangles can be inscribed in the rectangular hyperbola xy = c 2 whose all sides touch the parabola y 2 = 4ax, is (a) infinite

(b) 4

(c) 3

(d) 2

34. If ( x − 1) ( y − 2) = 5 and ( x − 1) + ( y + 2) 2 = r 2 intersect at four points A,B, C, D and centroid of ∆ABC lies on line y = 3x − 4, then locus of D is 2

(a) y = 3x (c) 3 y = x + 1

(b) x 2 + y2 + 3x + 1 = 0 (d) y = 3x + 1

35. If foci of hyperbola lie on y = x and one of the asymptotes is y = 2x, then equation of the hyperbola, given that it passes through (3, 4), is 5 (b) 2x 2 − 2 y2 + 5xy + 5 = 0 xy + 5 = 0 2 (c) 2x 2 + 2 y2 − 5xy + 10 = 0 (d) None of these

(a) x 2 − y2 −

36. The exhaustive set of values of α 2 such that there exists a tangent to the ellipse x 2 + α 2 y 2 = α 2 such that the portion of the tangent intercepted by the hyperbola α 2 x 2 − y 2 = 1subtends a right angle at the centre of the curves, is  5+1  (a)  , 2  2  (b) (1, 2 ]  5−1  (c)  ,1   2  5+1   5 + 1 (d)  , 1 ∪ 1,  2     2

(b) a hyperbola (d) a circle

(a) 2x 2 + 5xy + 2 y2 + 4 x + 5 y + 2 = 0 (b) 2x 2 + 5xy + 2 y2 + 4 x + 5 y − 2 = 0 (c) 2x 2 + 5xy + 2 y2 = 0 (d) None of the above

(a) α ∈ (− 2, 0) (c) α ∈ (− ∞ , − 2)

(b) α ∈ (− 3, 0) (d) α ∈ (− ∞ , − 3)

42. If tangents OQ and OR are drawn to variable circles having radius r and the centre lying on the rectangular hyperbola xy = 1, then locus of circumcentre of ∆OQR is (O being the origin) 1 4 (d) None of these

(a) xy = 4

(b) xy =

(c) xy = 1

43. Four points are such that the line joining any two points is perpendicular to the line joining other two points. If three points out of these lie on a rectangular hyperbola, then the fourth point will lie on (b) conjugate hyperbola (d) one of the asymptotes

39. From any point on the hyperbola tangents are drawn to the hyperbola

a2 x2

− −

y2 b2 y2

= 1, = 2.

a2 b2 Then, area cut-off by the chord of contact on the asymptotes is equal to (b) ab sq unit (d) 4ab sq unit

(a) 2

(b)

2 3

(c) 2

(d) 3

45. If a hyperbola passes through (2, 3) and has asymptotes 3x − 4 y + 5 = 0 and 12x + 5 y − 40 = 0, then the equation of its transverse axis is (a) 77x − 21 y − 265 = 0 (c) 21x − 77 y − 265 = 0

(b) 21x − 77 y + 265 = 0 (d) 21x + 77 y − 265 = 0

46. If P ( x1 , y1 ), Q ( x 2 , y 2 ), R ( x 3 , y 3 ) and S ( x 4 , y 4 ) are four concyclic points on the rectangular hyperbola xy = c 2 , then coordinates of orthocentre of ∆PQR are (a) (x4 , − y4 ) (b) (x4 , y4 )

x2

16

41. If two tangets can be drawn to the different branches y2 x2 of hyperbola − = 1from the paint (α , α 2 ), then 1 4

44. From a point P (1, 2), two tangents are drawn to a hyperbola H in which one tangent is drawn to each arm of the hyperbola. If the equations of asymptotes of hyperbola H are 3 x − y + 5 = 0 and 3 x + y − 1 = 0, then eccentricity of H is

38. The combined equation of the asymptotes of the hyperbola 2x 2 + 5xy + 2 y 2 + 4x + 5 y = 0 is

(a) a/2 sq unit (c) 2ab sq unit

y3

(a) a fixed circle concentric with the hyperbola (b) a fixed ellipse concentric with the hyperbola (c) a fixed hyperbola concentric with the hyperbola (d) a fixed parabola having vertex at (0, 0)

(a) the same hyperbola (c) one of the directrix

37. The locus of the point which is such that the chord of contact of tangents drawn from it to the ellipse y2 x2 + = 1 forms a triangle of constant area with a2 b2 the coordinate axes, is (a) a straight line (c) an ellipse

−

Hyperbola

(a) at 45° (c) Nothing can be said

x2

=1 a2 b2 ( b > a ) subtends a right angle at the centre of the hyperbola, if this chord touches

40. A variable chord of the hyperbola

Targ e t E x e rc is e s

32. Two concentric hyperbolas, whose axes meet at angle of 45°, cut

(c) (− x4 , − y4 ) (d) (− x4 , y4 )

47. A rectangular hyperbola whose centre is C, is cut by any circle of radius r in four points P, Q, R and S. Then, CP 2 + CQ 2 + CR 2 + CS 2 is equal to (a) r2

(b) 2r2

(c) 3r2

(d) 4 r2

48. PQ and RS are two perpendicular chords of the rectangular hyperbola xy = c 2 . If C is the centre of the rectangular hyperbola, then the product of the slopes of CP, CQ, CR and CS is equal to (a) −1

(b) 1

(c) 0

(d) None of these

951

Objective Mathematics Vol. 1

16

49. If a circle cuts the rectangular hyperbola xy = 1in the then points ( x r , yr ); r = 1, 2, 3, 4, x1 x 2 x 3 x 4 = y1 y 2 y 3 y 4 is equal to (a) 1

(b) 2

(c) 3

(d) 4

50. The product of the perpendicular from any point on y2 x2 the hyperbola 2 − 2 = 1 to its asymptotes, is equal a b to ab a +b 2a2b2 (c) 2 a + b2

(b)

(a)

a2b2 a + b2 2

(d) None of these

51. If H ( x, y ) = 0 represents the equation of a hyperbola and A( x, y ) = 0, C ( x, y ) = 0 are the equation of its asymptotes and the conjugate hyperbola respectively, then for any point (α , β ) in the plane; H (α , β ), A (α , β ) and C(α , β ) are in

Ta rg e t E x e rc is e s

(a) AP (c) HP

(b) GP (d) None of these

52. If four points are taken on a rectangular hyperbola such that the chord joining any two is perpendicular to the chord joining the other two and α , β , γ , δ are the inclinations to either asymptote of the straight line joining these points to the centre, then (a) tan α (b) tan α (c) tan α (d) tan α

tan β tan γ tan δ = 1 tan β tan γ tan δ = 2 tan β tan γ tan δ = − 1 tan β tan γ tan δ = – 2

53. Let P be a point on the hyperbola x 2 − y 2 = a 2 , where a is a parameter such that P is nearest to the line y = 2x. Then, the locus of P is (a) x ± y = 0 (c) 2x ± y = 0

(b) x ± 2 y = 0 (d) 2x ± 3 y = 0

54. The coordinates of the point of intersection of two tangents to a rectangular hyperbola referred to its asymptote as axes, are (a) geometric means between the coordinates of the point of contact (b) arithmetic means between the coordinates of the point of contact (c) harmonic means between the coordinates of the point of contact (d) None of the above

55. From any point P ( h, k ), four normals can be drawn to the rectangular hyperbola xy = c 2 such that sum of the ordinates of the feet of the normals is equal to (a) 3k (c) k 2

(b) 2k (d) k

56. From any point P ( h, k ), four normals can be drawn to the rectangular hyperbola xy = c 2 such that product of the abscissae of the feet of the normals = product of the ordinates of the feet of the normals is equal to (a) c4 (c) 2c4

(b) −c4 (d) 2c2

57. A point P moves in such a way that the sum of the slopes of the normals drawn from it to the hyperbola xy = 4 is equal to the sum of the ordinates of feet of the normals. The locus of P is a parabola x 2 = 4 y. Then, the least distance of this parabola from the circle x 2 − y 2 − 24x + 128 = 0 is (a) 4 5 4 (b) 5 (c) − 4 5 (d) None of the above

Type 2. More than One Correct Option 58. The equation 16x 2 − 3 y 2 − 32x − 12 y − 44 = 0 represents hyperbola with (a) length of transverse axis = 2 3 (b) length of conjugate axis = 8 (c) centre at (1, − 2) (d) eccentricity = 19

59. Which of the following equation(s) in parametric form can represent a hyperbola, where t is a parameter? a 1 b  1  t +  and y =  t −  t t 2 2 tx y x ty (b) − + t = 0 and + − 1 = 0 a b a b (c) x = et + e− t and y = et − e− t (d) x 2 − 6 = 2 cos t and y2 + 2 = 4 cos2 t / 2 (a) x =

y2

− = 1, let n be the number of a2 b2 points on the plane through which perpendicular

60. For hyperbola 952

x2

tangents are drawn. Which of the following conditions satisfied the given hyperbola? (a) If n = 1, then e = 2 (b) If n > 1, then 0 < e < 2 (c) If n = 0, then e > 2 (d) None of the above

61. A straight line touches the rectangular hyperbola 9x 2 − 9 y 2 = 8 and the parabola y 2 = 32x. The equation of the line is (a) 9x + 3 y − 8 = 0 (c) 9x + 3 y + 8 = 0

(b) 9x − 3 y + 8 = 0 (d) 9x − 3 y − 8 = 0

62. The coordinates of a point common to a directrix and y2 x2 an asymptote of the hyperbola − = 1 are 25 16  25 20  (a)  ,   41 41  25 20 (c)  ,   3 3

 25 20  (b)  − ,   41 41  25 20  (d)  ,   3 3

Directions (Q.Nos. 63-67) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

63. Statement I The equation x 2 + 2 y 2 + λxy + 2x + 3 y + 1 = 0 can never represent a hyperbola. Statement II The general equation of second degree represents a hyperbola, if h 2 > ab.

between −

x2

+

16

66. Statement I The equation of the director circle to the hyperbola 4x 2 − 3 y 2 = 12 is x 2 + y 2 = 1. Statement II Director circle is the locus of the point of intersection of perpendicular tangents to a hyperbola. 67. Statement I

If a point ( x1 , y1 ) lies in the shaded y2 x region 2 − 2 = 1, shown in the figure, then a b x12 y12 − < 0. a2 b2 2

Y

The slope of the common tangent

the

hyperbola

x2 a2

−

y2 b2

=1

and

X'

A F1(–c, 0)

F2(c, 0)

O

X

y2

= 1may be 1 or −1. b2 a2 Statement II The locus of the point of intersection x y x y 1 of lines − = m and + = is a hyperbola a b a b m (where, m is a variable and ab ≠ 0).

Y'

Statement II hyperbola

65. Statement I Let ( 2, 2 ) be any point on hyperbola x 2 − y 2 = 2, then the product of distances of foci from P is equal to 6.

x

2

a2

−

If P ( x1 , y1 ) lies outside the y2 b2

= 1, then

x12 a2

−

y12 b2

< 1.

Targ e t E x e rc is e s

64. Statement I

Statement II If S and S ′ are the foci, C is the centre and P is any point on a hyperbola x 2 − y 2 = a 2 , then SP ⋅ S ′ P = CP 2 .

Hyperbola

Type 3. Assertion and Reason

Type 4. Linked Comprehension Based Questions Passage I (Q. Nos. 68-70) For the hyperbola ( 3 x − 4 y − 12) 2 (4 x + 3 y − 12) 2 − =1 100 225

68. The centre is  84 12  (a)  ,   25 25  − 84 12  (c)  ,   25 25

 84 − 12 (b)  ,   25 25   84 − 12 (d)  − ,   25 25 

69. The length of latusrectum of hyperbola is (a) 40 units (b) 45 units (c) 9/2 units (d) 8/9 unit

x 2 − ( y − 1) 2 = 1 has one tangent line with positive slope that passes through the origin. The point of tangency being (a, b).

a 71. The value of sin −1   is  b 5π 12 π (c) 3

(a)

π 6 π (d) 4

(b)

72. Length of latusrectum of the conic is (a) 1 (c) 2

70. If a straight line cuts the hyperbola in P and Q and its asymptotes in R and S, then PR is equal to (a) QS (c) 3QS

Passage II (Q. Nos. 71-73) The graph of the conic

(b) 2QS (d) 4QS

(b) 2 (d) None of these

73. Eccentricity of the conic is 4 3 (c) 2

(a)

(b) 3 (d) None of these

953

Objective Mathematics Vol. 1

16

Passage III (Q. Nos. 74-76) The vertices of ∆ABC lie on a rectangular hyperbola such that the orthocentre of the triangle is ( 3, 2) and the asymptotes of the rectangular hyperbola are parallel to the coordinate axes. The two perpendicular tangents of the hyperbola intersect at the point (1, 1).

74. The equation of the asymptotes is (a) xy − 1 = x − y (c) 2xy = x + y

(b) xy + 1 = x + y (d) None of these

(a) xy = 2x + y − 2 (b) 2xy = x + 2 y + 5 (c) xy = x + y + 1 (d) None of the above

76. Number of real tangents that can be drawn from the point (1, 1) to the rectangular hyperbola, is (a) 4 (c) 3

(b) 0 (d) 2

Type 5. Match the Columns 77. Let the foci of the hyperbola

x2 A

vertices of the ellipse

x

2

a2

+

y

2

b2

2

−

y2 B

2

= 1 be the

Column II

Column I A.

b is equal to B

p.

1

B.

e H + e E is always greater than

q.

2

C. If angle between the asymptotes of 2π hyperbola is , then 4e E is equal to 3

r.

3

s.

4

1 and ( x, y) is point of 2 intersection of ellipse and the 9 x2 hyperbola, then 2 is 2y

D. If e E2 =

78. Match the statements of Column I with values of Column II.

= 1and the foci of the

ellipse be the vertices of the hyperbola. Let the eccentricities of the ellipse and hyperbola be e E and e H respectively, then match the following:

Ta rg e t E x e rc is e s

75. Equation of the rectangular hyperbola is

Column I

Column II

A. Value of c for which p. 3 x 2 − 5 xy − 2 y 2 + 5 x + 11y + c = 0 are the asymptotes of the hyperbola 3 x 2 − 5 xy − 2 y 2 + 5 x + 11y − 8 = 0, is

3

B. If locus of a point, whose chord of contact with respect to the circle x 2 + y 2 = 4 is a tangent to the hyperbola xy = 1, is xy = c 2, then the value of c 2 is

q.

–4

C. If equation of a hyperbola whose conjugate axis is 5 and distance between its foci is 13, is ax 2 − by 2 = c, where a and b are coprime natural ab is numbers, then the value of c

r.

−12

D. If the vertex of a hyperbola bisects the distance between its centre and the corresponding focus, then ratio of square of its conjugate axis to the square of its transverse axis is

s.

4

Type 6. Single Integer Answer Type Questions 79. If the equation of the chord of the hyperbola 25x 2 − 16 y 2 = 400 which is bisected at (5, 3), is 125x − 48 y = λ + 480, then the value of λ is_______ . 80. If the line 25x + 12 y − 45 = 0 meets the hyperbola 5λ 25x 2 − 9 y 2 = 255 only at point  5, −  , then the  3 value of λ is _______ . 81. If a variable line has its intercepts on the coordinate axes e, e ′, where e / 2 , e ′ /2 are the eccentricities of a hyperbola and its conjugate hyperbola, then the line always touches the circle x 2 + y 2 = r 2 , where r is equal to_______ .

954

82. If the chord x cos α + y sin α = p of the hyperbola y2 x2 − = 1 subtends a right angle at the centre and 16 18 the diameter of the circle, concentric with the hyperbola, to which the given chord is a tangent, is d , then the value of d / 4 is_______ . 83. If L is the length of latusrectum of hyperbola for which x = 3and y = 2are the equations of asymptotes and which passes through the point ( 4, 6), then the value of L/ 2 is_______ .

Entrances Gallery JEE Advanced/IIT JEE

1  9 (a)  ,  2 2 2 (c) (3 3 , − 2 2 )

1  9 (b)  − ,−   2 2 2 (d) (−3 3 , 2 2 )

x2

5 2 (c) 2

− = 1 be a2 b2 reciprocal to that of the ellipse x 2 + 4 y 2 = 4. If the hyperbola passes through a focus of the ellipse, then

2. Let the eccentricity of the hyperbola

[2011] x 2 y2 − =1 3 2 (b) a focus of the hyperbola is (2, 0) 5 (c) the eccentricity of the hyperbola is 3 (d) the equation of the hyperbola is x 2 − 3 y2 = 3 (a) the equation of the hyperbola is

−

y2

3 2 (d) 3

(a)

y2

x2

= 1. a2 b2 If the normal at the point P intersects the X -axis at (9, 0), then the eccentricity of the hyperbola is [2011]

3. Let P( 6, 3) be a point on the hyperbola

(b)

Passage (Q. Nos. 4-5) The circle x 2 + y 2 − 8 x = 0 and hyperbola

x2 y2 − = 1 intersect at the points A and B. [2010] 9 4

4. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is (a) 2x − 5 y − 20 = 0 (c) 3x − 4 y + 8 = 0

(b) 2x − 5 y + 4 = 0 (d) 4 x − 3 y + 4 = 0

5. Equation of the circle with AB as its diameter, is (a) x 2 + y2 − 12x + 24 = 0 (c) x 2 + y2 + 24 x − 12 = 0

(b) x 2 + y2 + 12x + 24 = 0 (d) x 2 + y2 − 24 x − 12 = 0

(a) a hyperbola (c) a circle

(b) a parabola (d) an ellipse

AIEEE 6. The equation of the hyperbola, whose foci are ( − 2, 0) and ( 2, 0) and eccentricity is 2, is given by [2011] (a) − 3x 2 + y2 = 3 (c) 3x 2 − y2 = 3

(b) x 2 − 3 y2 = 3 (d) − x 2 + 3 y2 = 3

x

2

y

= 1, which of the cos α sin 2 α following remains constant when α varies? [2007]

7. For the hyperbola

(a) Eccentricity (c) Abscissae of vertices

2

−

2

(b) Directrix (d) Abscissae of foci

8. The locus of a point P(α , β ) moving under the condition that the line y = α x + β is a tangent to the y2 x2 hyperbola 2 − 2 = 1, is [2005] a b

9. If the foci of the ellipse

y2 x2 + 2 = 1and the hyperbola 16 b

y2 x2 1 coincide. Then, the value of b 2 is − = 144 81 25 [2003] (a) 1

(b) 5

(c) 7

Targ e t E x e rc is e s

x2 y2 − = 1, 9 4 parallel to the straight line 2x − y = 1. The points of contacts of the tangents on the hyperbola are [2012]

1. Tangents are drawn to the hyperbola

(d) 9

10. The equation of the chord joining two points ( x1 , y1 ) and ( x 2 , y 2 ) on the rectangular hyperbola xy = c 2 is x y + =1 x1 + x2 y1 + y2 x y (c) + =1 y1 + y2 x1 + x2 (a)

[2002] x y + =1 x1 − x2 y1 − y2 x y (d) + =1 y1 − y2 x1 − x2 (b)

Other Engineering Entrances 11. If the length of the latusrectum and the length of transverse axis of a hyperbola are 4 3 and 2 3 respectively, then the equation of the hyperbola is x 2 y2 (a) − =1 3 4 x 2 y2 (c) − =1 6 9 x 2 y2 (e) − =1 3 6

[Kerala CEE 2014] x 2 y2 (b) − =1 3 9 x 2 y2 (d) − =1 6 3

x2

y2

5 = 1 is 4 a2 b2 and 2x + 3 y − 6 = 0 is a focal chord of the hyperbola, then the length of transverse axis is equal to

12. If the eccentricity of the hyperbola

−

[Kerala CEE 2014] 12 (a) 5 24 (d) 5

(b) 6 (e)

24 (c) 7

12 7

955

Objective Mathematics Vol. 1

16

13. The length of the transverse axis of a hyperbola is 2cos α. The foci of the hyperbola are the same as that of the ellipse 9x 2 + 16 y 2 = 144. The equation of the hyperbola is [Kerala CEE 2014] x2 y2 − =1 cos2 α 7 − cos2 α x2 y2 (b) − =1 2 cos α 7 + cos2 α x2 y2 (c) − =1 1 + cos2 α 7 − cos2 α x2 y2 (d) − =1 2 1 + cos α 7 + cos2 α x2 y2 (e) − =1 2 cos α 5 − cos2 α

(a)

Ta rg e t E x e rc is e s

(b) 4 (e) 7

x2 y2 − =1 144 9 2 2 x y (c) − =1 144 25

(c) 5

x2 y2 − =1 169 25 2 2 x y (d) − =1 25 9 (b)

[GGSIPU 2014] (b) parabola (d) parallel straight lines

17. If the line lx + my − n = 0 will be a normal to the a 2 b 2 (a 2 + b 2 )2 , where k is hyperbola, then 2 − 2 = k l m equal to [VITEEE 2014] (b) n2 (d) None of these

18. If the chords of the hyperbola x − y = a touch the parabola y 2 = 4ax. Then, the locus of the middle points of these chords is [Manipal 2014] 2

(a) y2 = (x − a)x 3 (c) x 2 (x − a) = x 3

2

2

(b) y2 (x − a) = x 3 (d) None of these

19. The equation of the common tangent with positive slope to the parabola y 2 = 8 3 x and the hyperbola [WB JEE 2014] 4x 2 − y 2 = 4 is

956

(a) (b) (c) (d)

y= y= y= y=

6x + 6x − 3x + 3x −

2 2 2 2

2 (b) ± 3 2 2 (d) ± 3

22. The eccentricity of the hyperbola with latusrectum 12 and semi-conjugate axis 2 3, is [AMU 2011] (a) 3

(b)

3 2

(c) 2 3

(d) 2

23. The equation to the common tangents to the two y2 y2 x2 x2 hyperbolas 2 − 2 = 1 and 2 − 2 = 1 are a b a b [MP PET 2011] (a) y = ± x ±

(b2 − a2 )

(c) y = ± x ± (a − b ) 2

2

(b) y = ± x ±

(a2 − b2 )

(d) y = ± x ±

(a2 + b2 )

24. The equation of the common tangent to the parabola y 2 = 8x and rectangular hyperbola xy = − 1is (a) x − y + 2 = 0 (c) 2x − y + 1 = 0

16. The curve 5x + 12xy − 22x − 12 y − 19 = 0 is

(a) n (c) n3

(d) None of these

[MP PET 2012]

2

(a) ellipse (c) hyperbola

40 (b) 3

4 2 (a) ± 3 8 (c) ± 9

15. A hyperbola passing through a focus of the ellipse y2 x2 + = 1. Its transverse and conjugate axes 169 25 coincide respectively with the major and minor axes of the ellipse. The product of eccentricities is 1. Then, the equation of the hyperbola is [EAMCET 2014] (a)

[MP PET 2012] 20 (a) 3 50 (c) 3

21. The value of m, for which the line y = mx + 2 is a tangent to the hyperbola 4x 2 − 9 y 2 = 36, are

14. The number of points ( a, b ), where a and b are positive integers, lying on the hyperbola [Kerala CEE 2014] x 2 − y 2 = 512 , is (a) 3 (d) 6

20. A hyperbola passes through (3, 3) and the length of its conjugate axis is 8. The length of latusrectum is

[AMU 2011] (b) 9x − 3 y + 2 = 0 (d) x + 2 y − 1 = 0

25. The equation of a hyperbola, whose asymptotes are 3x ± 5 y = 0 and vertices are ( ± 5, 0), is (a) 3x 2 − 5 y2 = 25

[Karnataka CET 2011] (b) 5x 2 − 3 y2 = 225

(c) 25x 2 − 9 y2 = 225

(d) 9x 2 − 25 y2 = 225

26. Equation of auxiliary circle of

y2 x2 − = 1, is 16 25

(a) x 2 + y2 = 16

[Guj CET 2011] (b) x 2 + y2 = 25

(c) x 2 + y2 = 9

(d) x 2 + y2 = 41

27. The distance between the foci of the conic [Kerala CEE 2010] 7x 2 − 9 y 2 = 63 is equal to (a) 8 (d) 7

(b) 4 (e) 12

(c) 3

28. For different values of α, the locus of the point of intersection of the two straight lines 3x − y − 4 3 α = 0 and 3 αx + αy − 4 3 = 0 is [WB JEE 2010] (a) a hyperbola with eccentricity 2 2 (b) an ellipse with eccentricity 3 (c) a hyperbola with eccentricity (d) an ellipse with eccentricity

3 4

19 16

Answers Work Book Exercise 16.1 1. (a)

2. (c)

3. (a)

4. (b)

5. (a)

4. (a)

5. (c)

4. (a)

5. (a)

Work Book Exercise 16.2 1. (b,c)

2. (a)

3. (c)

Work Book Exercise 16.3 1. (d)

2. (c)

3. (d)

6. (b)

1. (a)

2. (c)

3. (d)

4. (a)

5. (b)

6. (a)

7. (a)

8. (d)

9. (b)

10. (c)

11. (d)

12. (d)

13. (c)

14. (a)

15. (c)

16. (b)

17. (a)

18. (b)

19. (d)

20. (a)

21. (b)

22. (b)

23. (d)

24. (b)

25. (c)

26. (b)

27. (b)

28. (a)

29. (c)

30. (a)

31. (a)

32. (b)

33. (a)

34. (a)

35. (c)

36. (a)

37. (b)

38. (a)

39. (d)

40. (a)

41. (c)

42. (b)

43. (a)

44. (b)

45. (d)

46. (b)

47. (d)

48. (b)

49. (a)

50. (b)

51. (a)

52. (a)

53. (b)

54. (c)

55. (d)

56. (b)

57. (a)

58. (a,b,c)

59. (all)

60. (a,b,c)

61. (a,c)

62. (a,b)

63. (d)

64. (b)

65. (a)

66. (d)

67. (d)

68. (b)

69. (b)

70. (a)

71. (d)

72. (c)

73. (d)

74. (b)

75. (c)

76. (d)

77. (*)

78. (**)

79. (1)

80. (4)

81. (2)

82. (6)

83. (4)

* A → p; B → q; C → p; D → s ** A → r; B → s; C → s; D → p

Entrances Gallery 3. (b)

4. (b)

5. (a)

6. (c)

7. (d)

8. (a)

9. (c)

10. (a)

11. (e)

1. (a,b)

12. (d)

2. (b,d)

13. (a)

14. (b)

15. (c)

16. (c)

17. (b)

18. (b)

19. (a)

20. (b)

21. (d)

22. (d)

23. (b)

24. (a)

25. (d)

26. (b)

27. (a)

28. (a)

Targ e t E x e rc is e s

Target Exercises

957

Explanations Target Exercises 1. Centre of the hyperbola is the mid-point of the line joining the two foci, therefore the coordinates of the centre are (1, 5). Now, distance between the foci = 10 ⇒ 2 ae = 10 ⇒ ae = 5 ⇒ a = 4 Now, b 2 = a 2 (e 2 − 1) ⇒ b = 3 Hence, the equation of the hyperbola is ( x − 1)2 ( y − 5)2 − =1 16 9

2. We have, a = 3 and

b2 = 4 ⇒ b2 = 12 a

Hence, the equation of the hyperbola is ⇒

x2 y2 − =1 9 12

4 x 2 − 3 y 2 = 36

8. If k1 = 0, then k1PA + k2 PB = k3

3. Equation of hyperbola is 2

2

Ta rg e t E x e rc is e s

y x − 2 = 1 ⇒ a2 = b2 (e 2 − 1) ⇒ e = 2 b a

a +b b2 2

2

x2 y2 4. The given equation can be written as − + = 1. The 4 3 eccentricity of this hyperbola is given by e = 1+

4 7 a2 = 1+ = 2 3 3 b

5. We have, 9 ( x 2 − 4 x + 4) − 16 ( y 2 − 6 y + 9) = 144 ( x − 2 )2 ( y − 3)2 − =1 42 32 X2 Y 2 Shifting the origin at (2, 3), we have 2 − 2 = 1 4 3 where, x = X + 2, y = Y + 3. Then, coordinates of the vertices are X = ± 4, Y = 3 i.e. x = 6, y = 3 and x = − 2, y = 3. ⇒

6. Let the equation to the hyperbola be x2 y2 ...(i) − =1 a2 b2 Hence, the transverse axis is 2a. If C is the centre, S is the focus and A is the vertex to the hyperbola, then CS = ae [distance between the centre and the focus] ae or e = 2 ∴ a= 2 2 2 2 Again, b = a (e − 1) = a2 (4 − 1) = 3 a2 On substituting the value of b2 in Eq. (i), we get x2 y2 − 2 =1 2 3a a 2 ⇒ 3 x − y 2 = 3 a2

7. Given that, x = a tan(θ + α ) and y = b tan(θ + β )  x or …(i) tan −1   = θ + α  a  y …(ii) and tan −1   = θ + β  b To get the required locus, we have to eliminate θ from

958

Eqs. (i) and (ii).

On subtracting Eq. (ii) from Eq. (i), we get  x  y tan −1   − tan −1   = α − β  a  b x y   − −1  a b  =α −β tan  ⇒ x y 1 + ⋅   a b x y − a b = tan (α − β ) ⇒ x y 1+ ⋅ a b On simplifying, we get the required locus as xy + ab = (bx − ay ) cot (α − β ) which is a hyperbola. ⇒

PB =

k3 >0 k2

⇒ P describes a circle with B as centre and radius =

k3 k2

If k3 = 0, then k1 PA + k2 PB = 0 PA k = 2 =k>0 ⇒ PB − k1 ⇒ P describes a circle with P1P2 as its diameter, P1 and P2 being the points which divide AB internally and externally in the ratio k : 1. If k1 = k2 > 0 and k3 > 0, then k PA + PB = 3 = k > 0 k1 ⇒ P describes an ellipse with A and B as its foci.

9. Let LSL′ be a latusrectum through the focus S (ae , 0 ) of x2 y2 − = 1. It subtends angle 60° at the a2 b2 other focusS ′ (− ae , 0 ). the hyperbola

Y 2 L ae, b a

X′

30° S′ 30° (–ae, 0)

O

S (ae, 0) 2 L′ ae,− b a

Y′

We have, ∠ LS ′ L′ = 60 ° ∴ ∠ LS ′ S = 30 ° In ∆LS ′ S , we have LS 1 b2 / a tan 30° = ⇒ = S′ S 2 ae 3 b2 1 ⇒ = 3 2 a2e 1 e2 − 1 = ⇒ 2e 3 ⇒ 3 e 2 − 2e − 3 = 0 (e − 3 ) ( 2e + 1) = 0 ⇒ ⇒ e= 3

X

x2 y2 hyperbola − = 1. It is given that ∆CLL′ is an a2 b2 equilateral triangle. Therefore, ∠LCS = 30 ° In ∆CSL, we have SL tan 30° = CS Y

Y′

⇒ ⇒ ⇒

2

1 + 13 2 3

∠LA′ S = 30 °

LS In ∆A′ LS , we have tan ∠LA′ S = A′ S b2 / a ⇒ tan 30 ° = a + ae 1 b2 = 2 ⇒ 3 a (1 + e ) 1 e2 − 1 ⇒ = 3 1+ e 1 ⇒ =e −1 3 1 3+1 ⇒ e = 1+ = 3 3

12. (OP )2 = (OQ )2 = (PQ )2

⇒ ⇒ ⇒ ⇒

a2 a2 2 (b + l 2 ) + l 2 = 2 (b2 + l 2 ) + l 2 = 4l 2 2 b b a2 b2 2 >0 l = (3 b2 − a2 ) 3 b2 − a2 > 0 4 3b2 > a2 ⇒ 3 a2 (e 2 − 1) > a2 ⇒ e 2 > 3 2 e> 3

13. Here, a2 = 3, b2 = 2, m = 3 and the equation of tangent to hyperbola is y = mx ± (a2 m2 − b2 ) ⇒

2 h 2 − ab 2 3− 0 = tan −1 a+ b 4+ 0

Here, a = 4, h = 3, b = 0

 3 θ = tan −1    2 

∴

y = mx + c be a common tangent to 9 x 2 − 16 y 2 = 144 and x 2 + y 2 = 9. Since,y = mx + c is a tangent to 9 x 2 − 16 y 2 = 144. Therefore, …(i) c 2 = a2 m2 − b2 ⇒ c 2 = 16 m2 − 9 Now, y = mx + c is a tangent to x 2 + y 2 = 9, therefore c …(ii) = 3 ⇒ c 2 = 9 (1 + m2 ) 2 m +1

16. Let

11. It is given that ∆A′ LL′ is an equilateral triangle. Therefore,

⇒

4 x 2 + 2 3 xy + 0 ⋅ y 2 = 0 θ = tan −1

X

L′ ae, – b a x2 y2 – 2= 1 2 b a

3e2 − e − 3 = 0 ⇒ e =

⇒

16

If θ is the angle between these lines,then

2 L ae, b a S (ae, 0)

1 b2 / a = ae 3 b2 1 = 2 3 ae 1 e2 − 1 = e 3

⇒

straight line 3 x + y = 2, combined equation of lines joining the points of intersection of the curve y 2 − x 2 = 4 and the given line to the origin is given by 2  3x + y y2 − x2 = 4   2  

y = 3x ± 5

14. The equation of line is x cos α + y sin α = p ⇒ y = − x cot α + p cosec α Now, c 2 = a2 m2 − b2 ⇒ p2 cosec 2α = a2cot 2 α − b2 2 2 ⇒ a cos α − b2 sin 2 α = p2

From Eqs. (i) and (ii), we get 16 m2 − 9 = 9 + 9 m2

⇒ m=± 3

2 7

On putting the value of m in Eq. (ii), we get c = ± Hence, y = 3

15 7

2 15 is a common tangent. x+ 7 7

x2 y2 − = 1, then 3 2 c2 = 3 − 2 = 1 ⇒ c = ± 1 So, the required tangents are y = x ± 1.

17. Let y = x + c be a tangent to

Targ e t E x e rc is e s

X′

30° A′(–a,0) C(0,0) A S′ (–ae, 0) (a,0)

15. Making y 2 − x 2 = 4 homogeneous with the help of the

Hyperbola

10. Let LSL′ be a latusrectum and C be the centre of the

18. Let P ( x1 , y1 ) be a point on the hyperbola. Then, the coordinates of N are( x1, 0 ). The equation of the tangent at xx yy ( x1, y1 ) is 21 − 21 = 1. a b  a2  This meets X-axis at T  , 0 .  x1  a2 ∴ OT ⋅ ON = × x1 = a2 x1

19. The equation of the normal at (a sec φ, b tan φ ) to the x2 y2 − = 1 is a2 b2 …(i) ax sin φ + by = (a2 + b2 ) tan φ and the equation of the line is …(ii) lx + my + n = 0 Now, Eqs. (i) and (ii) represent the same line, therefore a sin φ b (a2 + b2 ) tan φ (a2 + b2 )sin φ = = = l m −n − n cos φ bl (a2 + b2 ) l and cos φ = sin φ = ∴ am − na 2 2 (a2 + b2 )2 a b ⇒ − = l 2 m2 n2 hyperbola

959

Objective Mathematics Vol. 1

16

20. ByT = S1, the equation of chord whose mid-point is (α, β)

21. Equation of hyperbola is 25 x 2 − 16 y 2 = 400 x2 y2 − =1 16 25

⇒

Chord is bisected at (6, 2 ). 6 x 2 y 62 2 2 ∴ − = − 16 25 16 25 ⇒ 75 x − 16 y = 418

22. Diameters y = m1 x and y = m2 x are conjugate diameters b2 x2 y2 − 2 = 1, if m1m2 = 2 . 2 a a b 1 2 2 Here, a = 9, b = 16 and m1 = 2 b2 1 16 (m2 ) = m1m2 = 2 ⇒ ∴ 2 9 a 32 ⇒ m2 = 9 32 x . Thus, the required diameter is y = 9

Ta rg e t E x e rc is e s

a = b ⇒ b2 = a2 ⇒ a2 (e 2 − 1) = a2 ⇒ e = 2

28. Asymptotes of the given hyperbola are y = ±  b Therefore, angle between them is 2 tan −1   .  a

x2 y2 − = 1 at the point a2 b2

Let coordinates of P be (h, k ), then (a2 + b2 ) (a2 + b2 ) h= sec φ and k = tan φ a b …(i) ⇒ ah = (a2 + b2 )sec φ …(ii) and bk = (a2 + b2 ) tan φ On squaring and subtracting Eq. (ii) from Eq. (i), we get a2 h 2 − b2 k 2 = (a2 + b2 )2 Then, locus of P is a2 x 2 − b2 y 2 = (a2 + b2 )2 Then, equations of tangent at these points are and The 3x − 2 2 y − 3 = 0 3 x + 2 2 y − 3 = 0. combined equations of these two is 9 x 2 − 8 y 2 − 18 x + 9 = 0.

25. The equation of the normal at (a sec θ , b tan θ) to the given hyperbola is ax cos θ + by cot θ = (a2 + b2 ) This meets the transverse axis i.e. X-axis at G. So, the coordinates of G are  a2 + b2  sec θ, 0  a   The coordinates of the vertices A and A′ are A (a, 0 ) and A′ (− a, 0 ), respectively.    a +b a +b ∴ AG ⋅ A′ G =  − a + sec θ  a + sec θ a a   

960

2 x 2 + 5 xy + 2 y 2 − 11x − 7 y − 4 = 0

…(i)

Let asymptotes of the hyperbola (i) be 2 x 2 + 5 xy + 2 y 2 − 11x − 7 y + c = 0

…(ii)

Since, Eq. (ii) represents a pair of lines. 2 5/ 2 −11/ 2 −7 / 2 = 0 ∴ 5/ 2 2 −11/ 2 −7 / 2 c 11  35 7 55 25     ⇒− + 11 +  −7 +  + c  4 −  = 0 −      2 4 2 4 4 9 99 189 90 ⇒ c =− + = ⇒ c =5 4 8 8 8 ∴ Equation of the required asymptotes is 2 x 2 + 5 xy + 2 y 2 − 11x − 7 y + 5 = 0

30. Let the asymptotes be

24. If x = 9 meets the hyperbola at (9, 6 2 ) and (9, − 6 2 ).

2

bx . a

29. Equation of hyperbola is

Q(a sec φ, b tan φ ) is ax cos φ + by cot φ = a2 + b2

2

∴Equation of asymptotes is x 2 + 3 xy + 2 y 2 + 2 x + 3 y + λ = 0 ∴ ∆ = 0, abc + 2 fgh − af 2 − bg 2 − ch 2 = 0, then λ =1 ∴ A ( x, y ) ≡ x 2 + 3 xy + 2 y 2 + 2 x + 3 y + 1 = 0 ∴ C ( x, y ) ≡ x 2 + 3 xy + 2 y 2 + 2 x + 3 y + 2 = 0 [from Eq. (i)]

27. In case of rectangular hyperbola,

of the hyperbola

23. Normal to the hyperbola

26. We know that, Equation of hyperbola + Equation of conjugate hyperbola = 2 (Equation of asymptotes) ∴Equation of conjugate hyperbola = 2 Equation of asymptotes − Equation of hyperbola ...(i) C ( x, y ) = 2 A ( x, y ) − H ( x, y ) Q H ( x, y ) ≡ x 2 + 3 xy + 2 y 2 + 2 x + 3 y = 0

is 3 xα − 2 yβ + 2 ( x + α ) − 2 ( y + β ) = 0 ⇒ x (3 α + 2 ) − y(2β + 3) = 2 as it is parallel to y = 2 x ∴ 3α − 4β = 4 Required locus is 3 x − 4 y = 4. ∴

2

2

= (− a + ae 2 sec θ ) (a + ae 2 sec θ) = a2 (e 4 sec 2 θ − 1)

2 x + 3 y + c1 = 0 and

3x + 2 y + c 2 = 0

Since, the asymptotes passes through the centre (1, 2 ) of the hyperbola. ∴ 2 + 6 + c1 = 0 and 3 (1) + 2 (2 ) + c 2 = 0 ⇒ c1 = − 8 and c 2 = − 7 Thus, the equations of the asymptotes are 2 x + 3y − 8 = 0 and 3x + 2 y − 7 = 0 Let the equation of the hyperbola be (2 x + 3 y − 8)(3 x + 2 y − 7 ) + λ = 0 It passes through (5, 3). ⇒ (2 x + 3 y − 8) ⋅ (3 x + 2 y − 7 ) = 154 which is the equation of the required hyperbola.

31. The combined equation of the asymptotes is

(3 x − 4 y + 7 ) (4 x + 3 y + 1) = 0 So, the combined equation of the hyperbola is (3 x − 4 y + 7 ) (4 x + 3 y + 1) + λ = 0 It passes through the origin. ∴ 7 × 1+ λ = 0 ⇒ λ = −7

…(i)

∑ xi

x 2 − y 2 = a2

…(i)

As the asymptotes of this are the axes of the other and vice-versa, hence the equation of the other hyperbola may be written as ...(ii) xy = c 2 Let Eqs. (i) and (ii) meet at some point whose coordinates are (a sec α , a tan α ). Then, the tangent at the point(a sec α , a tan α )to Eq.(i) is x − y sin α = a cos α …(iii) and the tangent at the point (a sec α , a tan α ) to Eq. (ii) is  2c 2  …(iv) y + x sin α =   cos α  a  Clearly, the slopes of the tangents given by Eqs. (iv) and 1 (iii) are respectively − sin α and , so their product is sin α  1  − sin α ⋅   = −1  sin α  Hence, the tangents are at right angle.   c  c  c A ct1,  , B ct 2 ,  and C ct 3 ,    t1   t2  t3 

be

three

points on the hyperbola xy = c 2 . Then, the equation of the side AB is x + yt1 t 2 = c (t1 + t 2 ) t + t  x ⇒ y= − + c  1 2  t1 t 2  t1 t 2 This will touch the parabola y 2 = 4ax, if t + t  a c  1 2 =  t1 t 2   1  −   t1 t 2 

[Qc = a/ m]

⇒ (at12 )t 22 + ct 2 + ct 2 = 0 This is a quadratic equation in t 2 . So, it gives two values of t 2 corresponding a given value of t1. Thus,  c corresponding to one vertex A ct1,  , there is another  t1   c position of vertex i.e. B ct 2 ,  .  t2  Similarly,corresponding to each position of B, there are two positions of C. Thus, corresponding to a given position of vertex A, there can be four triangles whose sides touch the parabola y 2 = 4ax. But A can attain infinitely many positions. Hence, there are infinitely many triangles inscribed in xy = c 2 such that its sides touch the parabola y 2 = 4ax.

34. If ( xi, yi ) is the point of intersection of given curves, then 4

∑

i =1

4

4

xi =

1+ 1 and 2

∑

i =1

4

=0

=

16

35. Foci of hyperbola lie on y = x. So, the major axis is y = x. Major axis of hyperbola bisects the asymptote. ⇒ Equation of hyperbola is x = 2 y ⇒ Equation of hyperbola is ( y − 2 x ) ( x − 2 y ) + k = 0 Given that, it passes through (3, 4). ⇒ k = 10 Hence, required equation is 2 x 2 + 2 y 2 − 5 xy + 10 = 0

36. Equation of tangent at point P(α cos θ, sin θ) is x y …(i) cos θ + sin θ = 1 α 1 Let it cut the hyperbola at points P and Q. Homogenizing the hyperbola α 2 x 2 − y 2 = 1with the help of the above the equation, we get 2 x  α 2 x 2 − y 2 =  cos θ + y sin θ α  This is a pair of straight lines OP and OQ. π Given, ∠POQ = 2 ⇒ Coefficient of x 2 + Coefficient of y 2 = 0 cos 2 θ − 1 − sin 2 θ = 0 α2 − ⇒ α2 cos 2 θ ⇒ − 1 − 1 + cos 2 θ = 0 α2 − α2 α 2 (2 − α 2 ) ⇒ cos 2 θ = α2 − 1 Now, 0 ≤ cos 2 θ ≤ 1 α 2 (2 − α 2 ) ⇒ ≤1 0≤ α2 − 1 On solving, we get  5+1  α2 ∈ ,2   2 

37. The chord of contact of tangents from ( x1, y1 ) is xx1 yy1 + =1 a2 b2  a2   b2  It meets the axes at the points  , 0 and  0, . y1   x1   1 a2 b2 [constant] Area of the triangle is ⋅ ⋅ =k 2 x1 y1 ⇒

yi

i =1

Targ e t E x e rc is e s

which is the equation of the required hyperbola.

33. Let

y 4 − x4 i =1 and =− 4 3 3 3 4 3  3   ∑ x i ∑ yi   i =1 i =1  Centroid  ,  lies on the line y = 3 x − 4. 3   3     − y4 3 ( 4 − x 4 ) Hence, = −4 3 3 ⇒ y4 = 3 x 4 Therefore, the locus of D is y = 3 x. Now,

32. Let the equation to the rectangular hyperbola be

3

∑ yi

Hyperbola

3

On putting the value of λ in Eq. (i), we get (3 x − 4 y + 7 ) (4 x + 3 y + 1) − 7 = 0 ⇒ 12 x 2 − 7 xy − 12 y 2 + 31x + 17 y = 0

x1 y1 =

a2 b2 = c 2 , where c is a constant. 2k

⇒ xy = c 2 is the required locus.

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Objective Mathematics Vol. 1

16

38. Let the equation of asymptotes be 2 x + 5 xy + 2 y + 4 x + 5 y + λ = 0 2

2

…(i)

This equation represents a pair of straight lines. Therefore, abc + 2 fgh − a f 2 − bg 2 − ch 2 = 0 25 25 We have, 4λ + 25 − − 8− λ =0 2 4 9λ 9 ⇒ − + =0 4 2 ⇒ λ =2 Putting the value of λ in Eq. (i), we get 2 x 2 + 5 xy + 2 y 2 + 4 x + 5 y + 2 = 0

41. Given that,

x2 y2 − =1 1 4 Y

X′

X O

Ta rg e t E x e rc is e s

which is equation of the asymptote.

Y′

x2 y2 39. Let P ( x1, y1 ) be a point on the hyperbola 2 − 2 = 1. a b The chord of contact of tangents from P to the hyperbola is given by xx1 yy1 …(i) − 2 =1 a2 b The equation of the asymptotes are x y − =0 a b x y and + =0 a b The points of intersection of Eq. (i) with the two asymptotes are given by 2a 2b , y = x1 = x1 y1 1 x1 y1 − − a b a b 2a 2b , y = x2 = x1 y1 2 x1 y1 + + a b a b 1 Area of the triangle = ( x1 y2 − x2 y1 ) 2     1  4ab × 2  = − 2 = 4absq units 2 2  x1 y1  −    a2 b2 

40. Let the variable chord be …(i) x cos α + y sin α = p Let this chord intersect the hyperbola at A and B.Then, the combined equation of OA and OB is given by 2 x 2 y 2  x cos α + y sin α  − =    p a2 b2  ⇒

962

 1 cos α   1 sin α  x2  2 −  − y2  2 +  2 p  p2  a b 2 sin α cos α − xy = 0 p 2

2

This chord subtends a right angle at centre. Therefore, Coefficient of x 2 + Coefficient of y 2 = 0 1 cos 2 α 1 sin 2 α ⇒ − − 2 − =0 2 2 a p b p2 1 1 1 ⇒ − = a2 b2 p2 a2 b2 ⇒ p2 = 2 b − a2 Hence, p is a constant i.e. it touches the fixed circle.

Since, (α , α 2 ) lie on the parabola y = x 2 , then (α , α 2 ) must lie between the asymptotes of hyperbola x2 y2 − = 1 in 1st and 2nd quadrants. 1 4 So, the asymptotes are y = ± 2 x ∴ 2α < α 2 ⇒ α < 0 or α > 2 and − 2α < α 2 α < − 2 or α > 0 ∴ α ∈ (− ∞, − 2 ) or (2, ∞ ) 1 42. LetS,say t ,  be a point on the rectangular hyperbola.  t Y

R

S (t, 1/t) Q

X′

O

X

Y′

Now, circumcircle of ∆OQR also passes through S. Therefore, circumcentre is the mid-point of OS. t 1 Hence, and y = x= 2 2t 1 So, the locus of the circumcentre is xy = . 4

43. The points are such that one of the points is the orthocentre of the triangle formed by other three points. When the vertices of a triangle lie on a rectangular hyperbola the orthocentre also lies on the same hyperbola.

44. Since, c1 c 2 (a1a2 + b1b2 ) < 0, therefore origin lies in acute angle and P (1, 2 ) lies in obtuse angle. π Acute angle between the asymptotes is . 3 θ π 2 Hence, e = sec = sec = 2 6 3

45. Transverse axis is the equation of the angle bisector passing containing point (2, 3), which is given by 3 x − 4 y + 5 12 x + 5 y − 40 = 5 13 ⇒ 21x + 77 y = 265

Slope of QR is −

c , for r = 1, 2, 3, 4 xr

c2 . x2 x3

∴Equation of line passing through A and perpendicular to QR is …(i) x1 x2 x3 x − c 2 x1 y = x12 x2 x3 − c 4 Similarly, equation of line through Q and perpendicular to PR is x1 x2 x3 x − c 2 x2 y = x1 x22 x3 − c 4 …(ii) From Eqs. (i) and (ii), we get xx x y = − 1 22 3 c c4 ∴ x= x1 x2 x3 Thus, orthocentres is again lying on xy = c 2 i.e. the fourth point ( x4 , y4 ).

47. Let equation of the rectangular hyperbola be xy = c 2

…(i)

and equation of circle be …(ii) x2 + y2 = r 2 2 c in Eq. (ii), we get Put y = x c4 x2 + 2 = r 2 x …(iii) ⇒ x4 − r 2 x2 + c 4 = 0 Now, CP 2 + CQ 2 + CR 2 + CS 2 = x12 + y12 + x22 + y22 + x32 + y32 + x42 + y42 2

2

 4   4  =  ∑ xi − 2 Σx1 x2 +  ∑ yi − 2 Σy1 y2     i =1  i =1  = 2r 2 + 2r 2 = 4 r 2

[from Eq. (iii)]

48. Let t1, t 2 , t 3 and t 4 be the parameters of the points P, Q, R and S, respectively. Then, the coordinates of P,Q, R   c  c c  c and S are ct1,  , ct 2 ,  , ct 3 ,  and ct 4 ,  ,   t1   t4  t2   t3  respectively. c c c c − − t 2 t1 t4 t3 Now, PQ ⊥ RS ⇒ × = −1 ct 2 − ct1 ct 4 − ct 3  1   1  ⇒ −  × −  =1  t1 t 2   t 3 t 4 

This is a fourth degree equation in x giving four values of x say x1, x2 , x3 and x4 . ∴ x1 x2 x3 x4 = 1 Corresponding to every value of x, there is a value of y given by xy = 1. 1 ∴ =1 y1 y2 y3 y4 = x1 x2 x3 x4

50. Let (a sec θ, b tan θ ) be the any point on the hyperbola x2 y2 − 2 =1 a2 b The equations of the asymptotes of the given hyperbola are x y x y − = 0 and + =0 a b a b Now, p1 = length of the perpendicular from x y (a sec θ, b tan θ) on + = 0 a b sec θ − tan θ = 1 1 + 2 2 a b and p2 = length of the perpendicular from x y (a sec θ, b tan θ ) on + = 0 a b sec θ + tan θ = 1 1 + 2 2 a b a2 b2 sec 2 θ − tan 2 θ ∴ p1 p2 = = 2 1 1 a + b2 + 2 2 a b

51. Let the equation of the hyperbola be

x2 y2 − − 1= 0 a2 b2 x2 y2 Then, A( x, y ) ≡ 2 − 2 = 0 a b is the equation of asymptotes x2 y2 and C( x, y ) ≡ 2 − 2 + 1 = 0 a b is the equation of its conjugate hyperbola. Since, H(α , β ) + C(α , β ) = 2 A(α , β ) ∴ Hyperbola, asymptotes and conjugate hyperbola are in AP. H( x, y ) ≡

52. Let xy = c 2 be the equation of the rectangular hyperbola referred to its asymptotes as the coordinate axes and    c c  c c P ct1,  , Q ct 2 ,  , R ct 3 ,  and S ct 4 ,  be four    t1  t2   t3  t4  points on its such that PQ is perpendicular to RS.

…(i) ⇒ t1 t 2 t 3 t 4 = 1 ∴ Product of the slopes of CP, CQ, CR and CS 1 1 1 1 1 = 2 × 2 × 2 × 2 = 2 2 2 2 = 1 [from Eq. (i)] t1 t 2 t 3 t 4 t1 t 2 t 3 t 4

Y

1  1 x + 2 + 2 gx + 2 f   + k = 0  x x 4 3 2 x + 2 gx + kx + 2 fx + 1 = 0 2

⇒

xy = c 2

B

49. Let the equation of the circle be x 2 + y 2 + 2 gx + 2 fy + k = 0 The equation of the hyperbola is xy = c 2 . Eliminating y from these two equations, we get

16 Hyperbola

yr =

∴

2

Targ e t E x e rc is e s

46. Since, points ( xr , yr ) is lying on xy = c 2 for r = 1, 2, 3, 4

β

X'

R ct3,

α O

Q ct2, c t2

S ct4, c t4

P ct1, c t1

X

c t3

Y′

963

Objective Mathematics Vol. 1

16

Since, OP makes angle α with OX. Therefore, c 1 t1 tan α = = ct1 t12 1 1 1 Similarly, tan β = 2 , tan γ = 2 and tan δ = 2 t2 t3 t4 1 ∴ tan α tan β tan γ tan δ = 2 2 2 2 t1 t 2 t 3 t 4 Now, PQ is perpendicular to RS. c c c c − − t 2 t1 t4 t3 ⇒ × =1 ct 2 − ct1 ct 4 − ct 3 1 −1 − × = −1 ⇒ t1 t 2 t 3 t 4 1 ⇒ = −1 t1 t 2 t 3 t 4 ⇒ t1 t 2 t 3 t 4 = − 1 From Eqs. (i) and (ii), we get tan α tan β tan γ tan δ = 1

 2ct1 t 2 2c  These two intersect at R  , .  t1 + t 2 t1 + t 2  Now, HM of ct1 and ct 2 =

…(i)

Ta rg e t E x e rc is e s

c c 2 ⋅ c c 2c t1 t 2 and HM of and = = c c t1 + t 2 t1 t2 + t1 t 2 Hence, the coordinates of R are the harmonic means of the coordinates of the points of contact P and Q.

55. The equation of the normal to the hyperbola xy = c 2 at  c any point ct ,  is xt 3 − yt − ct 4 + c = 0.  t

…(ii)

53. Let P(a sec θ, a tan θ) be a point on the hyperbola x 2 − y 2 = a2 . The distance of P from the line y = 2 x is given by 2 a sec θ − a tan θ S = 4+ 1 a S = (2 sec θ − tan θ ) ⇒ 5 dS a ⇒ = (2 sec θ tan θ − sec 2 θ ) dθ 5 dS Put = 0, then dθ 2 sec θ tan θ − sec 2 θ = 0 ⇒ 2 tan θ = sec θ ⇒ 4 tan 2 θ = 1 + tan 2 θ 1 tan θ = ± ⇒ 3 d 2S a 3 Now, = (2 sec θ + 2 sec θ tan 2 θ 2 dθ 2 − 2 sec 2 θ tan θ ) 2 1 π d S Clearly, i.e. for θ = ± > 0 for tan θ = ± 2 6 3 dθ Thus, the coordinates of P are given by 2a h= 3 a and k=± 3 ⇒ h = ± 2k ⇒ h m 2k = 0 Hence, the locus of (h, k ) is x ± 2 y = 0.

54. The equation of the hyperbola referred to its asymptotes as the coordinate axes is xy = c 2 .

964

2 c t1 ⋅ c t 2 2ct1 t 2 = c t1 + c t 2 t1 + t 2

 c The equation of the tangents to it at P ct1,  and  t1   x c Q ct 2 ,  are + yt1 = 2c  t2  t1 x and + yt 2 = 2c t2

If it passes through P (h, k ), then ht 3 − kt − ct 4 + c = 0 ⇒ ct 4 − ht 3 + kt − c = 0

…(i)

This being a fourth degree equation in t, gives four values of t, say t1, t 2 , t 3 and t 4 such that corresponding to each value of t, there is a normal passing through P(h, k ).   c  c  c c Let A ct1,  , B ct 2 ,  , C ct 3 ,  and D ct 4 ,    t1   t2   t3  t4  be four points on the hyperbola such that normals at A, B, C and D passes through P. From Eq. (i), we have h t1 + t 2 + t 3 + t 4 = c k and t1 t 2 t 3 + t1 t 2 t 4 + t 3 t 4 t1 + t 2 t 3 t 4 = − c c Also, t1 t 2 t 3 = − = − 1 c  t 2 t 3 t 4 + t1 t 3 t 4 + t1 t 2 t 4 + t1 t 2 t 3  c c c c ∴ + + + =c   t1 t 2 t 3 t 4 t1 t 2 t 3 t 4   c (− k / c ) = =k −1 Thus, sum of the ordinates A, B, C and D is k.

56. The equation of the normal to the hyperbola xy = c 2 at  c any point ct ,  is xt 3 − yt − ct 4 + c = 0.  t If it passes through P (h, k ), then ht 3 − kt − ct 4 + c = 0 ⇒

ct 4 − ht 3 + kt − c = 0

…(i)

This being a fourth degree equation in t, gives four values of t, say t1, t 2 , t 3 and t 4 such that corresponding to each value oft, there is a normal passing through P(h, k ).     c c c c Let A ct1,  , B ct 2 ,  , C ct 3 ,  and D ct 4 ,  be     t1  t2  t3  t4  four points on the hyperbola such that normals at A, B, C and D passes through P. Product of the abscissae of A, B, C and D = ct1 × ct 2 × ct 3 × ct 4 = c 4 t1 t 2 t 3 t 4 = −c4

[Q t1 t 2 t 3 t 4 = − 1]

Product of the ordinates of A, B, C and D c c c c = × × × t1 t 2 t 3 t 4 c4 = = −c4 t1 t 2 t 3 t 4

(d) We have, x − 6 = 2 cos t and y 2 + 2 = 4 cos 2 t / 2 = 2 + 2 cos t ⇒ y 2 = 2 cos t

60. Locus of point of intersection of perpendicular tangents is director circle x 2 + y 2 = a2 − b2 . b2 Now, e2 = 1+ 2 a If a2 > b2 , then there are infinite (or more than 1) points on the circle. ⇒ e2 < 2 ⇒ e < 2 If a2 < b2 , there does not exist any point on the plane. ⇒ e2 > 2 ⇒ e > 2 2 2 If a = b , there is exactly one point (centre of the hyperbola). e= 2 ⇒

61. The equation of tangent in terms of slope of y 2 = 32 x is 8 …(i) m which is also tangent of the hyperbola 9 x 2 − 9 y 2 = 8 8 i.e. x2 − y2 = 9 2 8 2 8 8 m2 1  8 Then,   = m − ⇒ = − 2  m 9 9 9 9 m ⇒ 72 = m4 − m2 ⇒ m4 − m2 − 72 = 0 ⇒ (m2 − 9)(m2 + 8) = 0 ⇒ m2 = 9 [Q m2 + 8 ≠ 0 ] ⇒ m=±3 From Eq. (i), we get 8 y = ± 3x ± ⇒ 3y = ± 9x ± 8 3 ⇒ ±9 x − 3 y ± 8 = 0 ⇒ 9 x − 3 y + 8 = 0, 9 x − 3 y − 8 = 0 ⇒ −9 x − 3 y + 8 = 0, −9 x − 3 y − 8 = 0 or 9 x − 3 y + 8 = 0, 9 x − 3 y − 8 = 0 ⇒ 9x + 3y − 8 = 0 and 9x + 3y + 8 = 0 y = mx +

4

AC = 4 5

Hence, the least distance = 4 5

58.

( x − 1)2 ( y − 2 )2 − =1 3 16 Transverse axis = 2 3 units Conjugate axis = 8 units Centre = (1, − 2 ) Eccentricity =

59. Verify each option in

19 3

x2 y2 x2 y2 and − = 1 − = − 1and a2 b2 a2 b2

xy = c 2 2x 1 =t + a t 2y 1 and =t − b t On solving Eqs. (i) and (ii), we get x y x y 1 + = t and − = a b a b t

(a) We have,

x2 y2 − =1 a2 b2 tx y (b) We have, − +t =0 a b x ty and + − 1= 0 a b On solving Eqs. (i) and (ii), we get

…(i) …(ii)

62.

∴

x 1− t2 = a 1+ t2

and

y 2t = b 1+ t2

…(i) …(ii)

Targ e t E x e rc is e s

AC 2 = 80 ⇒

…(ii)

x2 − y2 = 6

Z = 4 (2 − 6) + 2 = 80 ⇒

…(i)

From Eqs. (i) and (ii), we get

t =2

d 2Z Clearly, > 0 , for t = 2 dt 2 Thus, Z is minimum when t = 2. The minimum value of Z is given by

x2 − y2 = 4 2

16 Hyperbola

⇒

distance of any point on the parabola from the centre of the circle. Let A (2t , t 2 ) be any point on the parabola x 2 = 4 y and C(12, 0 ) be the centre of the circle. Then, AC 2 = (2 t − 12 )2 + ( t 2 − 0 )2 Let Z = AC 2 Then, Z = 4 ( t − 6)2 + t 4 dZ ∴ = 8 ( t − 60 + 4 t 3 ) dt d 2Z and = 12 + 12 t 2 dt 2 For maximum and minimum values of Z, we must have dZ =0 dt 3 ⇒ 8 ( t − 6) + 4 t = 0 ⇒ t 3 + 2 t − 12 = 0

2

…(i) …(ii)

On solving Eqs. (i) and (ii), we get x + y = 2e t and x − y = 2e − t

57. The distance of the parabola from the circle means the

⇒ ( t − 2 )( t 2 + 2 t + 6) = 0 ⇒

x = e t + e −t y = e t − e −t

(c) We have, and

x2 y2 − =1 25 16 Asymptotes are 4 x + 5 y = 0 ⇒ 4x − 5y = 0 25 Directrices are x = ± 41 and common points are  25 −20   25 20  , ,  ,  ,  41 41  41 41   −25 20   −25 −20  , ,  ,  .  41 41  41 41

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16

63. Statement I is false, because this will represent

70.

Y

hyperbola, if h 2 > ab.

R

λ2 >2 4

⇒

P O

| λ| > 2 2 ⇒ Statement II, being a standard result, is true.

X'

X

64. If y = mx + c be the common tangent, then and

Q

c 2 = a2 m2 − b2

…(i)

c 2 = − b2 m2 + a2

…(ii)

On eliminating c 2 , we get m2 = 1 ⇒ m = ± 1 For Statement II, On eliminating m, we get x2 y2 − =1 a2 b2 which is a hyperbola.

S Y'

From the figure, obviously PR = QS

71. On differentiating the curve, we have 2 x − 2( y − 1)

dy =0 dx

a b b dy  = =  mOP =  dx  b−1 a  a a, b

⇒ ⇒

a2 = b2 − b

…(i)

Y P (a, b)

65. First of all, we will verify Statement II. Let P(a sec θ, a tan θ) be any point on x 2 − y 2 = a2 , then

Ta rg e t E x e rc is e s

SP ⋅ S′ P = (ea sec θ − a)(ea sec θ + a) = e 2 a2 sec 2 θ − a2 = 2 a2 sec 2 θ − a2 and

X'

CP 2 = a2 sec 2 θ + a2 tan 2 θ = 2 a2 sec 2 θ − a2

⇒ SP ⋅ S′ P = CP 2 ∴ Statement II is true. π If we put a = 2, θ = , then Statement I is verified. 4 4 x 2 − 3 y 2 = 12 x2 y2 ⇒ − =1 3 4 Then, director circle is x 2 + y 2 = 3 − 4 = − 1, which does not exist for existence a > b.

66. Now,

For the points (2, 2), (4, 1) and (6, 2/3), t1 = 1, t 2 = 2 and t 3 = 3, respectively. 1 For the point (1/4, 16), t 4 = 8 3 Now, t1 t 2 t 3 t 4 = ≠ 1 4 Hence, Statement I is false.

68. By solving 3 x − 4 y − 12 = 0 4 x + 3 y − 12 = 0, we get centres 3 x − 4 y − 12 = 0 and 4 x + 3 y − 12 = 0  84 12  On solving Eqs. (i) and (ii), we get  , −  .  25 25 and

69. Length of laustrectum =

2b a

2

2 (225) = = 45 10

(1, 1)

(–√2, 1)

O (√2, 0)

X

Y'

= a2 sec 2 θ + a2 sec 2 θ − a2

67. Statement II is true.

966

(−1, 1)

[Q e = 2 ]

…(i) …(ii)

Also, (a, b) satisfy the curve a2 − (b − 1)2 = 1 2 ⇒ a − (b2 − 2 b + 1) = 1 a2 − b2 + 2 b = 2 ∴ −b + 2 b = 2 ⇒ b=2 a= 2 ∴ π −1  a  sin   = ∴  b 4

[Q a2 = b2 − b] [Q a ≠ − 2 ]

2 b2 a = 2a = Distance between the vertices = 2

72. Length of latusrectum =

73. Curve is a rectangular hyperbola. ⇒

e= 2

74. Perpendicular tangents intersect at the centre of rectangular hyperbola. Hence, centre of hyperbola is (1, 1) and equation of asymptotes are x − 1 = 0 and y − 1 = 0.

75. Let the equation of hyperbola be xy − x − y + 1 + λ = 0. It passes through (3, 2), hence λ = − 2. So, the equation of hyperbola is xy = x + y + 1.

76. From the centre of hyperbola, we can draw two real tangents to the rectangular hyperbola.

77. We have, A = ae E and a = Ae H ⇒

eE eH = 1

⇒

eE + eH > 2 B 2 = A2 (e H2 − 1) = a2 (1 − e E2 ) = b2

⇒

b =1 B

Also, the angle between the asymptotes is B 2π 2 tan −1 = 3 A 1 B b Also, = 3 ⇒ = 3 ⇒ e E2 = 4 A ae E 2 2 2 2 x y x y On solving 2 + 2 = 1 and 2 2 − 2 = 1, we get a b a eE b

Hyperbola

Equation of chord is T = S1. On simplifying, we get 125 x − 48 y = 481 ∴ λ =1

80. On eliminating y between the line and hyperbola, we get

2 a2 e E2 =4 b (1 − e E2 )

2

 45 − 25 x  25 x 2 − 9   = 225  12 

2

3 x 2 − 5 xy − 2 y 2 + 5 x + 11y + c = 0 asymptotes. ∴ It represents a pair of a straight lines. 2 11  5  −5  11 ∴ 3 (−2 )c + 2 ⋅     − 3   2 2  2  2 

78. A. Since,

2

are

⇒ hx 2 − 4 x + k = 0 ⇒ hk = 4 ∴ Locus of (h, k ) is xy = 4 ⇒ c2 = 4

On simplifying, we get x 2 − 10 x + 25 = 0 ⇒

x=5

Hence,

y=−

2

 5  5 − (−2 )   − c  −  = 0  2  2 275 363 25 25 ⇒ −6 c − − + − c =0 4 4 2 4 ⇒ − 24 c − 275 − 363 + 50 − 25 c = 0 ⇒ 49 c = − 588 ⇒ c = − 12 B. Let the point be (h, k). The equation of the chord of contact is hx + ky = 4. Since, hx + ky = 4 is tangent to xy = 1.  4 − hx  x  = 1 has two equal roots.  k 

C. Equation of the hyperbola is

16

x2 y2 − = 1. 16 25

20 5× 4 =− 3 3

⇒

λ=4 e e′ 81. Since, and are eccentricities of a hyperbola and its 2 2 conjugate. 4 4 + =1 ∴ e 2 e′ 2 e 2e ′ 2 i.e. 4= 2 e + e′ 2 Line passing through the points (e , 0 ) and (0, e′ ). x2 y2 − =1 16 18 9 x 2 − 8 y 2 − 144 = 0

82. Equation of hyperbola is or

Homogenization of this equation using x cos α + y sin α =1 p x2 y2 − =1 c/a c/b

c 5 13 c a+ b and = = ⋅ b 2 2 a b c = 36 ∴ a ∴ The hyperbola is 25 x 2 − 144 y 2 = 900

∴

∴

a = 25, b = 144, c = 900 ab ∴ =4 c x2 y2 D. Let the hyperbola be 2 − 2 = 1. a b Then, 2a = ae i.e. e = 2 b2 ∴ = e2 − 1= 3 a2 (2 b)2 =3 ∴ (2 a)2

2

 x cos α + y sin α  We have, 9 x 2 − 8 y 2 − 144   =0   p Since, these lines are perpendicular to each other. ∴ 9 p2 − 8 p2 − 144 (cos 2 α + sin 2 α ) = 0 ⇒

Targ e t E x e rc is e s

x2 =

79. The equation of hyperbola is

p2 = 144

or p = ± 12 ∴ Radius of the circle = 12 ∴ Diameter of the circle = 24

83. Equation of hyperbola ( x − 3)( y − 2 ) = c 2 xy − 2 x − 3 y + 6 = c 2

or

It passes through (4, 6), then 4 × 6 − 2 × 4 − 3 × 6 + 6 = c2 ⇒ ∴

c =2 Latusrectum = 2 2c =2 2 ×2 = 4 2

Entrances Gallery 1. Slope of tangent = 2 The tangents are y = 2 x ± i.e. ⇒

2x − y = ± 4 2 y − =1 2 2 4 2 x

x y + =1 2 2 4 2 xx yy On comparing it with 1 − 1 = 1, we get 9 4 and

9× 4− 4

−

9 −1 1   9 Points of contact as  , . ,  and  −  2 2 2 2 2 2

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Objective Mathematics Vol. 1

16

Aliter  tan θ   sec θ  Equation of tangent at P(θ ) is   x−  y =1  3   2  2 sec θ Slope = ⇒ =2 3 tan θ 1 ⇒ sin θ = 3 9 1 1   9 Points are  ,− ,  and  − ⇒ .  2 2 2 2 2 2

2. Ellipse is

Ta rg e t E x e rc is e s

where,

Y

X'

2

( y − 3) =

− a2 ( x − 6) 2 b2

⇒

a2 =1 2 b2

⇒

e=

3 2

x2 y2 − = 1 is 9 4 y = mx + 9 m2 − 4, m > 0

4m +

⇒ ⇒ ⇒ ∴

(2, 0)

9 m2 − 4

1 + m2

=4

⇒

495 m4 + 104 m2 − 400 = 0 4 2 or m = ⇒ m2 = 5 5 2 4 x+ ∴The tangent is y = 5 5 ⇒ 2 x − 5y + 4 = 0

5. A point on hyperbola is (3 sec θ, 2 tan θ ). It lies on the circle, so 9 sec 2 θ + 4 tan 2 θ − 24 sec θ = 0 ⇒ 13 sec 2 θ − 24 sec θ − 4 = 0 2 sec θ = 2 − ⇒ 13 ∴ sec θ = 2 ⇒ tan θ = 3 ∴ The point of intersection are A(6, 2 3 ) and B(6, − 2 3 ). ∴ The circle with AB as diameter is ( x − 6)2 + y 2 = (2 3 )2 2 ⇒ x + y 2 − 12 x + 24 = 0

a=1 a2e 2 = a2 + b2 4 = 1 + b2 b2 = 3

Thus, equation of hyperbola is x2 y2 − =1 1 3 2 2 ⇒ 3x − y = 3

7. The given equation of hyperbola is x2 y2 − =1 2 cos α sin 2 α 2 2 Here, a = cos α and b2 = sin 2 α Now,

3. Equation of normal is

4. A tangent to

X

O

(–2, 0)

Y'

3 ⇒ 1 = 2 (1 − e ) ⇒ e = 2 2 . ∴ Eccentricity of the hyperbola is 3 4  b2 = a2  − 1 ⇒ 3 b2 = a2 ⇒ 3  Foci of the ellipse are ( 3, 0 ) and (− 3, 0 ). Hyperbola passes through ( 3, 0 ). 3 ⇒ = 1 ⇒ a2 = 3 and b2 = 1 a2 ∴Equation of hyperbola is x 2 − 3 y 2 = 3. Focus of hyperbola is 2   (ae , 0 ) ≡  3 × , 0 ≡ (2, 0 )  3  2

It is tangent to x 2 + y 2 − 8 x = 0

968

x2 y2 − =1 a2 b2 2 ae = 4 and e = 2

x2 y2 + 2 =1 2 2 1 2

∴

6. Let equation of hyperbola be

e = 1+

b2 a2

sin 2 α = 1 + tan 2 α cos 2 α = sec α Coordinates of foci are (± ae , 0 ) i.e. (±1, 0 ). Hence, abscissae of foci remain constant, when α varies. = 1+

8. Since, the line y = αx + β is tangent to the hyperbola

∴

x2 y2 − =1 a2 b2 β 2 = a2α 2 − b2

So, locus of (α , β ) is y 2 = a2 x 2 − b2 ⇒ a2 x 2 − y 2 − b2 = 0 Since, this equation represents a hyperbola, so locus of a point P(α , β ) is a hyperbola.

9. Given equation of hyperbola can be rewritten as x2  12     5

2

−

y2  9    5

2

=1

b′ 2 a′ 2 9 25 5 ⇒ e′ 2 = 1 + = ⇒ e′ = 16 16 4 The foci of a hyperbola are  12 5  (± a′ e ′, 0 ) =  ± × , 0 = (±3, 0 )  5 4 

∴ Eccentricity is e ′ 2 = 1 +

Given equation of ellipse is x2 y2 + 2 =1 16 b

10. The mid-point of the chord is

 x1 + x2 y1 + y2  ,    2 2  The equation of the chord T = S1  x + x2   y1 + y2   x + x2   y + y2  ∴ x 1     =2  1 + y 1  2   2   2   2  ⇒ x( y1 + y2 ) + y ( x1 + x2 ) = ( x1 + x2 ) ( y1 + y2 ) x y ⇒ + =1 x1 + x2 y1 + y2

11. Given length of latusrectum is 4 3. 2 b2 …(i) =4 3 a and length of transverse axis is 2 3. 2a = 2 3 a= 3 ⇒ On putting a = 3 in Eq. (i), we get 2 b2 =4 3 3 ⇒ 2 b2 = 12 ⇒ b2 = 6 x2 y2 ∴ Equation of hyperbola is 2 − 2 = 1 a b x2 y2 ⇒ − =1 3 6 5 12. Given, e = 4   5 Foci of a hyperbola is (± ae , 0 ) i.e.  ± a, 0 .   4 Since, the focal chord is 2 x + 3 y − 6 = 0 …(i)  5  ∴Points  ± a, 0 satisfy Eq. (i).  4  5   ⇒ 2 ×  ± a + 3 (0 ) − 6 = 0  4  5 ⇒ ± a=6 2 12 ⇒ a=± 5 12  12  take a = ∴Length of transverse axis = 2 a = 2 ⋅ 5  5  24 = 5 ∴

13. Let equation of hyperbola be x2 y2 − =1 a12 b12 Given, 2 a1 = 2 cos α

⇒ a1 = cos α x2 y2 Also, given equation of ellipse is + = 1. 16 9 Here, e = 1 −

b2 9 7 = 1− = 2 16 4 a

16 Hyperbola

According to the given condition, Foci of hyperbola (e1 ) = Foci of ellipse (e ) ⇒ ± a1 e1 = ± ae 7 ⇒ cos α ⋅ e1 = 4 ⋅ 4 cos α ⋅ e1 = 7 ⇒ b12 = 7 cos 2 α ⇒ cos 2 α + b12 = 7 ⇒ b12 = 7 − cos 2 α ∴The equation of hyperbola is x2 y2 − =1 cos 2 α 7 − cos 2 α ⇒

cos α 1 +

14. Given equation is x 2 − y 2 = 512 ∴ a2 − b2 = 512 ⇒ ( a + b) ( a − b) = 2 9 ⇒ (a + b, a − b) = (2 8 , 2 ), (2 7, 2 2 ), (2 6 , 2 3 ), (2 5 , 2 4 ) So, the number of order pairs is 4.

15. Let the equation of hyperbola be x2 y2 − =1 a2 b2 x2 y2 Given equation of ellipse is + 2 =1 2 (13) (5) Here, a = 13, b = 5

…(i)

b2 25 144 12 = 1− = = 169 169 13 a2 12   ∴Foci = (± ae , 0 ) =  ± 13 × , 0 = (± 12, 0 )  13  Since, Eq. (i) passes through (± 12, 0 ). 144 0 ∴ − 2 =1 a2 b ⇒ a2 = 144 ⇒ a = ± 12 e = 1−

∴

Targ e t E x e rc is e s

Foci of an ellipse are (± ae, 0 ) = (± 4 e, 0 ). But given focus of ellipse and hyperbola coincide, then 3 4e = 3 ⇒ e = 4 Also, b2 = a2 (1 − e 2 ) 9  = 16 1 −  = 16 − 9 = 7  16

Now, eccentricity of hyperbola, e′ = 1 +

b2 b2 = 1+ 2 144 a

According to the question, ee′ = 1 ⇒ ⇒

⇒

12 b2 × 1+ =1 13 144

13 b2 b2 169 = ⇒ 1+ = 144 12 144 144 25 b2 169 b2 = −1 ⇒ = 144 144 144 144 1+

⇒

b2 = 25 x2 y2 ∴ Equation of hyperbola is − = 1. 144 25

16. The given curve is ...(i) 5 x 2 + 12 xy − 22 x − 12 y − 19 = 0 Comparing the curve (i) with general form ax 2 + by 2 + 2 hxy + 2 gx + 2 fy + c = 0, we get a = 5, h = 6, b = 0, g = − 11, f = −6, c = − 19 ∴ ∆ = abc + 2 fgh − af 2 − bg 2 − ch 2 = 5 × 0 × (−19) + 2 × (−6) (−11) (6) − 5 × 36 − 0 − (−19) (36)

969

Objective Mathematics Vol. 1

16

⇒ ∆ = 792 − 180 + 19 × 36 ≠ 0 Q ∆≠0 Now, ab = 5 × 0 = 0, h 2 = 36 ⇒ ab − h 2 = 0 − 36 = − 36 Q ab − h 2 < 0 So, the curve is a hyperbola. x2 y2 17. The equation of any normal to 2 − 2 = 1 is a b ax cos φ + by cot φ = a2 + b2 …(i) ⇒ ax cos φ + by cot φ − (a2 + b2 ) = 0 The straight line lx + my − n = 0 will be normal to the x2 y2 hyperbola 2 − 2 = 1, then Eq. (i) and lx + my − n = 0 a b represent the same line. a cos φ b cot φ a2 + b2 = = l m n na sec φ = ⇒ l (a2 + b2 ) nb and tan φ = m (a2 + b2 ) Q sec 2 φ − tan 2 φ = 1 n 2 a2 n 2 b2 − 2 2 =1 2 2 2 l (a + b ) m (a + b2 )2 a2 b2 (a2 + b2 )2 ⇒ − = l 2 m2 n2 But given equation of normal is a2 b2 (a2 + b2 )2 − = k l 2 m2 ⇒ k = n2

Ta rg e t E x e rc is e s

∴

h (h − k ) x− k k This will touch the parabola y 2 = 4ax, if 2

2

⇒

970

Since, hyperbola passes through (3, 3). 9 9 − 2 =1 ∴ 2 a b ⇒ 9 (b2 − a2 ) = a2 b2 ⇒

9 (16 − a2 ) = a2 ⋅ 16

⇒ ⇒

144 − 9 a2 = 16a2 25 a2 = 144

⇒

 12  a2 =    5

⇒ m4 − 6 m2 + 2 m2 − 12 = 0

2

⇒ a=

12 5

x2 y2 − =1 9 4

4x 2 9y 2 − =1 ⇒ 36 36

Also,

y = mx ±

a2 m2 − b2

⇒

y = mx ±

9 m2 − 4

…(i)

which is the equation of tangent. On comparing with y = mx + 2, we get

∴

…(i)

…(ii)

2

⇒

20. Q 2 b = 8 ⇒ b = 4

9 m2 − 4 = 4 ⇒ m2 =

19. Equation of tangent in slope form parabola y = 8 3 x is

⇒

y = 6x + 2

∴

On squaring both sides, we get

2

 2 3   = m2 − 4  m  12 = m2 − 4 m2 m4 − 4 m2 − 12 = 0

and m2 + 2 ≠ 0 m2 = 6 m=± 6

i.e. m = 6, as m is positive slope. 2 3 From Eq. (i), y = 6x + 6

2

y = mx + c a where, c= m 2 3 c= m Also, tangent to the hyperbola 4 x 2 − y 2 = 4 x2 y2 or − = 1 is c 2 = a2 m2 − b2 1 4 ⇒ c 2 = 1⋅ m2 − 4

m2 − 6 = 0

9 m2 − 4 = 2

h − k a ⇒ ak 2 = h 3 + k 2 h −  = h k   k ∴ Locus of the mid-point is x 3 = y 2 ( x − a). 2

⇒ ⇒ ⇒

21. We have,

mid-point as (h, k ) is given by y=

(m2 + 2 ) (m2 − 6) = 0

2 b2 a 16 × 5 40 =2 × = 12 3

18. Equation of chord of hyperbola x 2 − y 2 = a2 with ⇒

⇒

∴ Length of latusrectum =

2

xh − yk = h 2 − k 2

⇒ m2 (m2 − 6) + 2 (m2 − 6) = 0

[from Eq. (ii)]

22. Since, ∴ ∴

m=±

8 9

2 2 3

2 b2 = 12 and b = 2 3 a b2 a= =2 6 l=

a2 + b2 = a2

4 + 12 =2 4

23. Since, y = mx + c is tangent to the hyperbola x2 y2 …(i) − = 1, c 2 = a2 m2 − b2 a2 b2 Again, since y = mx + c is tangent to the hyperbola y2 x2 …(ii) − = 1, c 2 = a2 − b2 m2 a2 b2 On solving Eqs. (i) and (ii), we get m = ± 1 and c = ± a2 − b2 Hence, equation of common tangents y=± x±

a2 − b2

25. Vertices of hyperbola = (± 5, 0 ) ⇒ a = 5 Also, asymptotes of hyperbola x2 y2 x y − = 1 is ± = 0 a b a2 b2 b y=± x ⇒ a 3 Here, y = ± x ⇒ b = 3 and a = 5 5

Hyperbola

26. Q x 2 + y 2 = b2 ⇒ x 2 + y 2 = 25

27. Given, equation of hyperbola is a  Qc =  m 

x2 y2 − =1 9 7 Distance between foci = 2 ae = 2 a2 + b2 =2 9+ 7 = 8

28. Equation of two straight lines are 3x − y = 4 3 α

and

3x + y =

4 3 α

On solving above equations, we get x2 y2 3 x 2 − y 2 = 48 ⇒ − =1 16 48 which is a hyperbola. 16 + 48  Whose eccentricity, e = Q e = 16  = 4 =2

a2 + b2   a2 

Targ e t E x e rc is e s

1  xy = − 1 is t , −  .  t 1  ∴Equation of tangent at t , −  is  t 1 xt − y = − 2 t ⇒ y = xt 2 + 2 t which is also a tangent to the parabola. 2 2t = 2 ∴ t ⇒ t3 =1 ⇒ t =1 ∴ y=x+2 ⇒ x−y+2=0

16

∴ Required equation of hyperbola is x2 y2 − =1 25 9 2 2 ⇒ 9 x − 25 y = 225

24. Any point on the rectangular hyperbola

971

17 Introduction to Three Dimensional [3D] Geometry Coordinate Axes and Coordinate Planes in Three Dimensional Space Three mutually perpendicular lines in space define three mutually perpendicular planes which in turn divide the space into eight parts known as octants and the lines are known as the coordinate axes. Let X ′OX , Y ′OY and Z ′OZ be three mutually perpendicular lines intersecting at O. O is the origin and the lines X ′OX , Y ′OY and ZOZ′ are called X, Y and Z-axes, respectively. These three lines are also called the rectangular axes of coordinates. The planes containing these three lines in pairs, determine three mutually perpendicular planes XOY , YOZ and ZOX . y=0

Z X'

Y'

x=0 Origin O (0, 0 0)

z=0

Y X

Z'

The three planes divide space into eight cells called octants. The following table show the signs of coordinates of points in various octants. Octant coordinates

Octant

x

y

z

OXYZ

I

+

+

+

OX ′ Y Z

II

−

+

+

OX ′ Y ′ Z

III

−

−

+

Chapter Snapshot ●

●

●

Coordinate Axes and Coordinate Planes in Three Dimensional Space Coordinates of a Point in Space Distance between Two Points

●

Section Formulae

●

Centroid of a Triangle

Octant

x

y

z

OXY ′ Z

IV

+

−

+

OXYZ ′

V

+

+

−

OX ′ YZ ′

VI

−

+

−

OX ′ Y ′ Z ′

VII

−

−

−

OXY ′ Z ′

VIII

+

−

−

X

Example 2. Which of the following points lies in the fifth octant? (a) (1, 2, 3) (b) (4, −2, 3) (c) (4, − 2, − 5) (d) (4, 2, − 5) Sol. (d)

Coordinates of a Point in Space An arbitrary point P in three dimensional space is assigned coordinates ( x 0 , y0 , z 0 ) provided that (1) The plane through P parallel to the YZ-plane intersects the X-axis at ( x 0 , 0, 0). (2) The plane through P parallel to the XZ-plane intersects the Z-axis at (0, y0 , 0). (3) The plane through P parallel to the XY -plane intersects the Z-axis at (0, 0, z 0 ). The space coordinates ( x 0 , y0 , z 0 ) are called the cartesian coordinates of P or simple the rectangular coordinates of P. The cartesian coordinates ( x, y, z ) of a point P in a space are the numbers at which the planes through P are perpendicular to the axes. The coordinates of a point on X-axis are ( x, 0, 0), on Y-axis are (0, y, 0) and on Z-axis are (0, 0, z ). The standard equations of XY -plane, YZ-plane and ZX -plane are z = 0, x = 0 and y = 0, respectively.

Points ( 4, − 2, 3) ( 4, − 2, − 5)

(0, y, z)

(0, 0, z)

The distance between two points P ( x1 , y1 , z1 ) and Q ( x 2 , y2 , z 2 ) is given by Z P(x1, y1, z1) Q(x2, y2, z2) Y O X

PQ = ( x 2 − x1 ) 2 + ( y2 − y1 ) 2 + ( z 2 − z1 ) 2 Ø

(x, y, 0)

●

X (x, 0, z)

respectively, then position vector of point P is x$i + y$j + z k$ or simply ( x, y, z ).

Example 1. A plane is parallel to XZ-plane, so it is perpendicular to (a) X- axis (b) Y-axis (c) Z- axis (d) None of the above Sol. (b) If any plane is parallel to XZ-plane, then it is perpendicular to Y-axis.

●

The distance of a point P(x , y , z) from Y-axis is x2 + z2 .

●

Z

X

The distance of a point P(x , y , z) from X-axis is y2 + z2 .

●

(x, 0, 0)

The distance of a point P(x , y , z) from the origin is x2 + y2 + z2 .

●

●

Consider a point P in space whose position is given by triad ( x, y, z ), where x, y, z are perpendicular distances from YZ, ZX and XY-planes, respectively. If we assume $i, $j, k$ unit vectors along OX , OY , OZ

(v) ( z-coordinate is negative)

Distance between Two Points

P (x, y, z)

O

(iv) (y-coordinate is negative) (viii) (y and z-coordinates are negative)

( 4, 2, − 5)

Y (0, y, 0)

Octant (i) (all the coordinates are positive)

(1, 2, 3)

17 Introduction to Three Dimensional [3D] Geometry

Octant coordinates

The distance of a point P(x , y , z) from Z-axis is x2 + y2 . If P(x , y , z) is a point in a space, then distances from YZ , ZX and XY-planes are respectively x, y and z. A parallelopiped is formed by planes drawn through the points (x1 , y1 , z1) and (x 2 , y 2 , z 2) parallel to the coordinate planes. The length of edges are x 2 − x1 , y 2 − y1 , z 2 − z1 and length of diagonals is (x2 − x1)2 + (y2 − y 1)2 + (z2 − z 1)2 .

X

Example 3. The point on Y-axis, which is at a distance of 10 units from the point (1, 2, 3) is (a) (0, 3, 0) (b) (0, 2, 0) (c) (0, 1, 0) (d) None of the above Sol. (b) Since the point P is on Y-axis, therefore P(0, y, 0). The point (1, 2, 3) is at a distance of 10 units from (0, y, 0). ∴ (1 − 0)2 + (2 − y)2 + (3 − 0)2 = 10 ⇒ ⇒

1 + (2 − y)2 + 9 = 10 (2 − y)2 = 0

⇒ y−2 = 0 ⇒ y=2 Hence, the required point is (0, 2, 0).

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Objective Mathematics Vol. 1

17

Example 4. The equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, −1) is (a) x − 3 z = 0 (b) x − 2 z = 0 (c) x − 4 z = 0 (d) x + 2 z = 0 Sol. (b) Let the given points be A

m P (x1, y1, z1)

Ø

( x − 1)2 + ( y − 2 )2 + ( z − 3)2

A

B

= ( x − 3) + ( y − 2 ) + ( z + 1) 2

2

If P and Q are such that R divides the join of P and Q externally in the ratio m : n. Then, the coordinates of R are  mx 2 − nx1 my2 − ny1 mz 2 − nz1  , ,   m−n m−n   m−n

P

and B. Let P( x, y, z) be any point equidistant from A and B ∴ PA = PB i.e. distance between P and A = distance between P and B ⇒

Section Formula for External Division

X

= ( x − 3)2 + ( y − 2 )2 + ( z + 1)2 ⇒ x2 + 1 − 2 x + y2 + 4 − 4 y + z2 + 9 − 6 z = x2 + 9 − 6 x + y2 + 4 − 4 y + z2 + 1 + 2 z

If A(x1 , y1 , z1) and B(x 2 , y 2 , z 2) are two points, then the  x + x 2 y1 + y 2 z1 + z 2  mid-point of AB is  1 , , .  2 2 2 

where the line AB meets the XY-plane divides it in the ratio k : 1. ∴The coordinates of P are  −3k + 1 4k + 2 −5k + 3  , ,   k+1   k+1 k+1

Example 5. If a parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7) parallel to the coordinate plane, then the length of edges and diagonal of parallelopiped are (a) (2, 3, 5), 7 units (b) (3, 6, 2), 7 units (c) (5, 9, 7), 7 units (d) None of the above Sol. (b) Length of edges of the parallelopiped are | x2 − x1|,| y2 − y1|,| z2 − z1| ∴Length of edges are |5 − 2|,|9 − 3|,|7 − 5| ⇒ (3, 6, 2 ) Length of diagonal = ( x2 − x1 )2 + ( y2 − y1 )2 + ( z2 − z1 )2 = (3)2 + (6)2 + (2 )2

Example 6. The ratio in which the line joining (1, 2, 3) and (−3, 4, −5) is divided by XY -plane, is (a) 5 : 3 (b) 3 : 5 (c) 2 : 3 (d) None of these Sol. (b) Let A ≡ (1, 2, 3) and B ≡ (−3, 4, − 5). Let the point P

⇒ 4x − 8z = 0 ⇒ x − 2z = 0 Which is the required equation.

Since, the point P lies upon XY-plane. ∴ z-coordinates of P must be zero. −5k + 3 3 i.e. = 0 ⇒ −5k + 3 = 0 ⇒ k = k+1 5 3 Hence, required ratio is : 1 i.e. 3 : 5 internally. 5 X

Example 7. A point R with x-coordinate 4 lies on the line segment joining the points P (2, − 3, 4) and Q(8, 0, 10). Find the coordinates of the point R. (a) ( 4, − 2, − 6) (b) ( 4, 2, 6) (c) ( 4, − 2, 6) (d) None of these Sol. (c) Let the point R( x, y, z) divides PQ in the ratio k : 1. k

= 9 + 36 + 4 = 49 = 7 units

Section Formula for Internal Division Let P ( x1 , y1 , z1 ) and Q ( x 2 , y2 , z 2 ) be two points. Let R be a point on the line segment joining P and Q, which divides it internally in the ratio m : n. Then, the coordinates of R are  mx 2 + nx1 my2 + ny1 mz 2 + nz1  , ,   m+n m+n   m+n

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R

P (2, –3, 4)

1 Q R (4, y, z) (8, 0, 10)

k × 8 + 1× 2 =4 k+1 8k + 2 =4 k+1

⇒ x-coordinate of R =

Section Formulae

m P (x1, y1, z1)

R (x2, y2, z2)

2

⇒ ( x − 1)2 + ( y − 2 )2 + ( z − 3)2

X

●

n Q

n Q (x2, y2, z2)

⇒ ⇒ ⇒ ⇒ ⇒

[given]

8k + 2 = 4k + 4 8k − 4k = 4 − 2 4k = 2 1 k= 2

⇒ k : 1 = 1: 2 Hence, the point R divides PQ internally in the ratio 1 : 2. Therefore, y-coordinate of R  1 × 0 + 2 × (−3) −6 = −2 =  = 1+ 2 3  

Hence, coordinates of R are (4, − 2, 6).

Centroid of a Triangle (i) The coordinates of the centroid of the ∆ABC, whose vertices are A ( x1 , y1 , z1 ), B ( x 2 , y2 , z 2 ) and C ( x 3 , y3 , z 3 ) are  x1 + x 2 + x 3 y1 + y2 + y3 z1 + z 2 + z 3  , ,  .   3 3 3 (ii) If ( x1 , y1 , z1 ), ( x 2 , y2 , z 2 ), ( x 3 , y3 , z 3 ) and ( x 4 , y4 , z 4 ) are the vertices of a tetrahedron, then its centroid G is given by   x1 + x 2 + x 3 + x 4 y1 + y2 + y3 + y4 , ,   4 4 .  z1 + z 2 + z 3 + z 4       4 X

Example 8. If the origin is the centroid of ∆PQR with vertices P(2a, 2, 6), Q ( −4, 3b, − 10) and R (8, 14, 2c), then the values of a, b and c are, respectively 16 16 (a) −2, − , 3 (b) 2 , , −2 3 3 16 (d) None of these (c) −2, − , 2 3 Sol. (c) Centroid of the ∆PQR is

 x1 + x2 + x3 , y1 + y2 + y3 , z1 + z2 + z3      3 3 3 2 a − 4 + 8 2 + 3b + 14 6 − 10 + 2c   i.e.  , ,    3 3 3

X

Example 9. A(3, 2, 0), B (5, 3, 2), C (−9, 6, − 3) are three points forming a triangle. If AD the bisector of ∠BAC, meets BC in D, then coordinates of D are  19 57 17   19 57 17  (a)  − , ,  (b)  , − ,   8 16 16  8 16 16   19 57 17  (c)  , , −   8 16 16 

17 Introduction to Three Dimensional [3D] Geometry

P(2a, 2, 6) 2a + 4 = 0 a = −2 2 + 3b + 14 Also, =0 3 ⇒ 3b + 16 = 0 Q R 16 b = − (– 4, 3b, –10) ⇒ (8, 14, 2c) 3 6 − 10 + 2c and =0 3 ⇒ 2c − 4 = 0 ⇒ c = 2 −16 ,c =2 ∴ a = − 2, b = 3

⇒ ⇒

 1 × 10 + 2 × 4  and z-coordinate of R =   1+ 2   10 + 8 18 =6 = = 3 3

 19 57 17  (d)  , ,   8 16 16 

Sol. (d) Since, AD is the bisector of ∠BAC. ∴

BD AB = DC AC

Now,

AB = (5 − 3)2 + (3 − 2 )2 + (2 − 0)2 = 4 + 1+ 4 = 9=3 AC = (−9 − 3)2 + (6 − 2 )2 + (−3 − 0)2

and

= 13 3 BD = DC 13 ⇒ D divides BC in the ratio 3 : 13. ∴ The coordinates of D are  3(−9) + 13(5) 3(6) + 13(3) 3(−3) + 13(2 ) , ,   3 + 13 3 + 13 3 + 13   ∴

19 57 17  ,  =  ,  8 16 16 

Origin is the centroid i.e. the coordinates of the centroid 2a − 4 + 8 are (0, 0, 0), then =0 3

Work Book Exercise 1 The distance of the point P(a, b, c ) from the X- axis is a

b +c

2

c

a + b

2

2

2

b

a +c

d

a2 + b 2 + c 2

2

2

2 Ratio in which the xy-plane divides the join of (1, 2, 3) and (4, 2, 1) is a b c d

A(1, 2, 3), B(−1, − 2, − 1) and C(2, 3, 2 ). Then, the fourth vertex D is a (3,5,5) c (1, 2, 1)

b ( 4, 7, 6) d None of these

5 If the point ( x, y, 1 − x 2 − y 2 ) is at a unit distance from origin, then x 2 + y 2 is equal to a 0

3 : 1, internally 3 : 1, externally 1 : 2 ,internally 2 : 1, externally

b 1

c 2

d 3

6 Let A(2, 2, − 3), B(5, 6, 9) and C(2, 7, 9) be the

3 The mid-point of the sides of a triangle are (1, 5, − 1), (0, 4, − 2 ) and ( 2, 3, 4 ). Then, the centroid of the triangle is a (3, 4, 5) c (1, 2, 3)

4 Three vertices of a parallelogram ABCD are

b ( −1, 6, − 7 ) d (1, 4, 1 / 3)

vertices of a triangle. The internal bisector of ∠A meets BC at the point D. Then, the coordinates of D are a (7, 13, 9)

 7 13 9  b  , ,  2 2 2

 7 13  c  , , 9  2 2

d None of these

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WorkedOut Examples Type 1. Only One Correct Option Ex 1. L is the foot of the perpendicular drawn from a point P (3, 4, 5) on the XY -plane. The coordinates of point L are (a) (3, 0, 0) (c) (3, 0, 5)

(b) (0, 4, 5) (d) None of these

Sol. Since, in XY-plane, the z-coordinate is zero. Hence, the coordinates of the foot of the point L is (3,4 ,0). Hence, (d) is the correct answer.

Ex 2. Three vertices of a parallelogram ABCD are A(3, − 1, 2), B (1, 2, − 4) and C ( −1, 1, 2). Find the coordinates of the fourth vertex. (a) ( −1, − 2, 8) (c) (1, − 2, − 8)

Ex 3. What is the length of foot of perpendicular drawn from the point P (3, 4, 5) on Y-axis? (a) 41 units (b) 34 units (c) 5 units (d) None of the above Sol. When we draw perpendicular from the point P(3,4 ,5) on Y-axis, the x and z-coordinates will be zero and y-coordinate will be 4 i.e. coordinate on Y-axis will be Q (0, 4 , 0). ∴Distance between P and Q = PQ = (3 − 0)2 + (4 − 4 )2 + (5 − 0)2

(b) (1, − 2, 8) (d) (1, 2, 8)

= 32 + 02 + 52 = 9 + 25

Sol. Let ABCD be a parallelogram and coordinates of point D be (α , β , γ ) and the diagonals AC and BD intersect at P. In a parallelogram, diagonals bisect each other. ∴Coordinates of mid-point of BD = Coordinates of mid-point of AC  α + 1 β + 2 γ − 4   3 − 1 −1 + 1 2 + 2 ⇒ , , , ,   =  2 2 2   2 2 2  4  α + 1 β + 2 γ − 4  2 ⇒ , ,  =  , 0,   2 2 2  2 2  α + 1 β + 2 γ − 4 , , ⇒  = (1, 0, 2)  2 2 2  D(α, β, γ)

= 34 units Hence, (b) is the correct answer.

Ex 4. If the origin is the centroid of a ∆ABC having vertices A ( a, 1, 3), B ( − 2, b, − 5) and C ( 4, 7, c), then (a) a = − 2 (c) c = − 2 Sol. For centroid of ∆ABC,

C(–1, 1, 2)

and B(1, 2, – 4)

On comparing the corresponding coordinates, we get α+1 β+2 γ−4 = 1, = 0, =2 2 2 2 ⇒ α + 1 = 2, β + 2 = 0, γ − 4 = 4 ⇒ α = 1, β = − 2, γ = 8 ∴Coordinates of point D are (1, − 2, 8). Hence, (b) is the correct answer.

a−2+ 4 a+ 2 = 3 3 1+ b + 7 b + 8 y= = 3 3 3−5+ c c−2 z= = 3 3

x=

P A(3, –1, 2)

(b) b = 8 (d) None of these

But given centroid is (0, 0, 0). a+ 2 ∴ =0 ⇒ a=−2 3 b+ 8 =0 ⇒ b=−8 3 c−2 =0 ⇒ c=2 3 Hence, (a) is the correct answer.

Type 2. More than One Correct Option Ex 5. Which of the following statements is/are correct?

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(a) Lines X ′OX , Y ′OY and Z ′OZ are called rectangular axes (b) Each plane divide the space in two parts (c) Three coordinate planes together divide the space into eight regions (parts) called octant (d) None of the above

Sol. Let X ′OX , Y′OY and Z′OZ be three mutually perpendicular lines that pass through a point O such that Y' X ′OX and Y′OY lies in the plane of the paper and line Z′OZ is perpendicular to the plane of paper. These three

Z

X'

Y O X Z'

Each plane divide the space in two parts and the three coordinate planes together divide the space into eight regions (parts) called octants, namely (i) OXYZ (ii) OX ′ Y Z (iii) OXY′Z (iv) OXYZ′ (v) OXY′ Z′ (vi) OX ′ YZ′ (vii) OX ′ Y′ Z (viii) OX ′ Y′ Z′ Hence, (a), (b) and (c) are the correct answers.

Ex 6. Which of the following statements is/are incorrect? (a) The coordinates of the origin O are (0, 0, 0) (b) The coordinates of any point on the X -axis will be as ( 0, y, z ) (c) The coordinates of any point in the YZ-plane will be ( x, 0, 0) (d) None of the above Sol. The coordinates of the origin O are (0, 0, 0). The coordinates of any point on the X -axis will be (x , 0, 0) and the coordinates of any point in the YZ-plane will be (0, y, z). Hence, (a), (b) and (c) are the correct answers.

Type 3. Assertion and Reason Directions (Ex. Nos. 7-8) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices line (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 7. Statement I The coordinates of the point which divides the line segment joining the points (1, − 2, 3) and (3, 4, − 5) in the ratio 2 : 3  9 2 1 internally, are  , , −  .  5 5 5 Statement II The coordinates of the point which divides the line segment joining the points (1, − 2, 3) and (3, 4, − 5) in the ratio 2 : 3  19 5 −1 externally, are  , ,  . 2 2 2

Statement II Let P (x , y, z) be the point which divides line segment joining A(1, − 2, 3) and B(3, 4 , − 5) externally in the ratio 2 : 3. 2(3) − 3(1) Then, x= = −3 2−3 2(4 ) − 3(−2) y= = − 14 2−3 2(−5) − 3(3) z= = 19 2−3 Therefore, the required point is (−3, − 14 , 19). Hence, (c) is the correct answer.

Ex 8. Statement I The points P (−2, 3, 5), Q(1, 2, 3) and R ( 7, 0, − 1) are collinear. Statement II If PQ + QR = PR, then P, Q and R will be collinear. Sol. We know that points are said to be collinear, if they lie on a line. Now,

 9 2 −1 Thus, the required point is  , ,  . 5 5 5

PQ = (1 + 2)2 + (2 − 3)2 + (3 − 5)2 = 9 + 1+ 4 = 14 QR = (7 − 1)2 + (0 − 2)2 + (−1 − 3)2

Sol. Statement I Let P (x , y, z) be the point which divides line segment joining A(1, − 2, 3) and B(3, 4 , − 5) internally in the ratio 2 : 3. Then, 2(3) + 3(1) 9 x= = 2+ 3 5 2(4 ) + 3(−2) 2 y= = 2+ 3 5 2(−5) + 3(3) 1 z= =− 2+ 3 5

17 Introduction to Three Dimensional [3D] Geometry

lines are called rectangular axes (lines X ′OX , Y′OY and Z′ OZ are called X ,Y and Z-axes). We call this coordinate system a three dimensional space or simply space. The three axes taken together in pairs determine XY, YZ and ZX -planes i.e. three coordinate planes.

= 36 + 4 + 16 = 56 = 2 14 and

PR = (7 + 2)2 + (0 − 3)2 + (−1 − 5)2 = 81 + 9 + 36 = 126 = 3 14

Thus, PQ + QR = PR Hence, P, Q and R are collinear. Hence, (a) is the correct answer.

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Objective Mathematics Vol. 1

17 Type 4. Linked Comprehension Based Questions Z

Directions (Ex. Nos.9-11) Consider the figure given alongside:

(a) Only I is incorrect (b) Only II is incorrect (c) Both I and II are incorrect (d) Both I and II are correct

X'

In this figure, it is shown that Y' three planes are intersecting at a point O, such that these X three planes are mutually perpendicular to each other.

Ex 9. The lines X ′OX , Y ′OY constituting the

Y O

Sol. (Ex. Nos. 9-11) These problems are based on coordinate axes and coordinate planes in three dimensional space. Z

Z'

and Z ′OZ

X'

are

Y'

Y O

(a) rectangular non-perpendicular system (b) rectangular coordinate system (c) rectangular parallel system (d) None of the above

X Z'

9.

Ex 10. Which of the following statement is correct? (a) The plane XOY is called the ZX-plane (b) The plane YOZ is called the XY-plane (c) The plane ZOX is called the YZ-plane (d) The XY , YZ and ZX -planes are called the three coordinate planes

Ex 11. Which of the following statement(s) is/are incorrect? I. If we take the XOY plane as the plane of the paper, then the line Z ′ OZ is perpendicular to the plane XOY . II. If the plane of the paper is considered as horizontal, then the line Z ′ OZ will be horizontal.

Considering the three planes intersecting at a point O such that these three planes are mutually perpendicular to each other (see figure). These three planes intersect along the lines X ′OX , Y′OY and Z′OZ , called the X, Y and Z-axes respectively. We see that these lines are mutually perpendicular to each other. These lines constitute the rectangular coordinates system. Hence, (b) is the correct answer.

10. The planes XOY, YOZ and ZOX called respectively the XY, YZ and ZX-planes, are known as the three coordinate planes. Hence, (d) is the correct answer.

11. If we take the XOY plane as the plane of the paper, then line Z′OZ is perpendicular to the plane XOY. If the plane of the paper is considered as horizontal, then the line Z′OZ s will be vertical. Hence, (b) is the correct answer.

Type 5. Match the Column Ex 12. Match each term given under the Column I to its correct answer given under Column II and choose the correct option from the codes given below: Column I

Column II

A. If the centroid of the triangle is origin and two of its vertices are ( 3, − 5, 7 ) and ( −1, 7, − 6), then the third vertex is

p. Parallelogram

B. The points A( 3, − 1, − 1), B( 5, − 4, 0), C(2, 3, − 2 ) and D( 0, 6, − 3) are the vertices of a

q. An isosceles angled triangle

C. Points A(1, − 1, 3), B(2, − 4, 5) and C( 5, − 13, 11) are

r. ( −2, − 2, − 1)

D. Points A(2, 4, 3), B( 4, 1, 9) and C(10, − 1, 6) are the vertices of

s. Collinear

right

Sol. A. Let A(3, − 5, 7), B(−1, 7, − 6), C (x , y, z) be the 978

vertices of a ∆ABC with centroid (0, 0, 0).

Therefore,

 3 − 1 + x −5 + 7 + y 7 − 6 + z  (0, 0, 0) =  , ,    3 3 3 x+2 y+ 2 z+1 This implies = 0, = 0, =0 3 3 3 Hence, x = − 2, y = − 2 and z = − 1. B. Mid-point of diagonal AC −3  3 + 2 −1 + 3 −1 − 2  5 , , =  =  , 1,   2 2 2  2 2 Mid-point of diagonal BD − 3  5 + 0 − 4 + 6 0 − 3  5 , , =   =  , 1,  2 2 2  2 2  Diagonals of parallelogram bisect each other. C. | AB | = (2 − 1)2 + (−4 + 1)2 + (5 − 3)2 = 14 | BC | = (5 − 2)2 + (−13 + 4 )2 + (11 − 5)2 = 3 14 | AC | = (5 − 1)2 + (−13 + 1)2 + (11 − 3)2 = 4 14 Now, | AB | + | BC | = | AC |. Hence, points A, B and C are collinear.

AB = 4 + 9 + 36 = 7 BC = 36 + 4 + 9 = 7 CA = 64 + 25 + 9 =7 2

Now, AB 2 + BC 2 = AC 2. Hence, ABC is an isosceles right angled triangle. A → r; B → p; C → s; D → q

Type 6. Single Integer Answer Type Questions Ex 13. If x 2 + y 2 = 1, then the point P (x, y, 1 − x 2 − y 2 ) is at a distance of A unit from the origin. Here, A refers to___________. Sol. (1)The coordinates of point P are (x , y, 1 − x 2 − y2 ). Then,

OP = (x − 0)2 + ( y − 0)2 + ( 1 − x 2 − y2 − 0)2 = x 2 + y2 + 1 − x 2 − y2 = 1 = 1

17 Introduction to Three Dimensional [3D] Geometry

D. Here,

Ex 14. The perpendicular distance of the point P (6, 7, 8) from XY-plane is__________. Sol. (8) Let L be the foot of perpendicular drawn from the point P(6, 7, 8) to the XY-plane, then the distance of this foot L from P is z-coordinate of P i.e. 8 units.

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Target Exercises Type 1. Only One Correct Option 1. The coordinates of the point where the line segment joining A( 5, 1, 6) and B( 3, 4, 1) crosses the YZ-plane are  17 13 (a)  0, ,   2 2 13  17 (c)  0, , −   2 2

17 13  (b)  0,− ,   2 2 17 13  (d)  0, − , −   2 2

2. The equation of the set of point P, the sum of whose distances from A( 4, 0, 0) and B( −4, 0, 0) is equal to 10, is (a) 3x 2 + 25 y2 + 16z2 − 225 = 0

Ta rg e t E x e rc is e s

(b) 9x 2 + 25 y2 + 25z2 − 225 = 0 (c) 5x 2 + 16 y2 + 18z2 − 225 = 0 (d) None of the above

3. The points ( 0, 7, 10), ( −1, 6, 6) and ( −4, 9, 6) form (a) a right angled isosceles triangle (b) a scalene triangle (c) a right angled triangle (d) an equilateral triangle

4. Let A ( 2, − 1, 4 ) and B( 0, 2, − 3) be two points and C be a point on AB produced such that 2AC = 3AB, then the coordinates of C are 1 5 (a)  , , − 2 4

5  4

(c) (6, − 7, 18)

5. If the coordinates of the vertices of a ∆ABC are A( −1, 3, 2), B( 2, 3, 5) and C( 3, 5, − 2), then ∠A is equal to (a) 45° (c) 90°

(b) 60° (d) 30°

6. The coordinates of the points which trisect the line segments joining the points P( 4, 2, − 6) and Q(10, − 16, 6), are (a) (6, − 4 , − 2) and (8, 10, − 2) (b) (6, − 4 , − 2) and (8, − 10, 2) (c) (−6, 4 , 2) and (−8, 10, 2) (d) None of the above

7. The mid-points of the sides of a triangle are ( 5, 7, 11), ( 0, 8, 5) and ( 2, 3, − 1), then the vertices are (a) (7, 2, 5), (3, 12, 17), (−3, 4 , − 7) (b) (7, 2, 5), (3, 12, 17), (3, 4 , 7) (c) (7, 2, 5), (−3, 11, 11), (3, 4 , 8) (d) None of the above

8. If A and B are the points ( 3, 4, 5) and ( −1, 3, − 7) respectively, then the equation of set of point P such that PA 2 + PB 2 = k 2 , where k is constant, is (a) 2(x 2 + y2 + z2 ) − 4 x − 14 y + 4 z + 109 − k 2 = 0 (b) 3(x 2 + y2 + z2 ) − 2x − 14 y + 4 z − 109 + k 2 = 0

13  1 7 (b)  − , , −   2 4 4

(c) 2(x 2 − y2 + z2 ) + 2x − 14 y + 8z + 109 − k 2 = 0

(d) None of these

(d) None of the above

Type 2. More than One Correct Option 9. Which of the following statements(s) is/are incorrect? (a) The plane x = 1represents the plane parallel to XZ-plane at a distance 1 unit above YZ-plane (b) The equation of plane y = 3 represents the plane parallel to YZ-plane (c) The equation of the plane z = 3 represents the plane parallel to XY-plane at a distance 3 units above the XY-plane (d) All of the above

980

10. Which of the following statements(s) is/are correct? (a) (0, 7, − 10), (1, 6, − 6) and (4 , 9, − 6) are the vertices of an isosceles triangle (b) (0, 7, 10), (−1, 6, 6) and (−4 , 9, 6) are the vertices of a right angled triangle (c) (−1, 2, 1), (1, − 2, 5), (4 , − 7, 8) and (2, − 3, 4 ) are the vertices of a parallelogram (d) None of the above

Directions (Q. Nos. 11-12) In the following questions,

11. Statement I If three vertices of a parallelogram are A(1, 2, 3), B( −1, − 2, − 1) and C( 2, 3, 2), then the coordinates of fourth vertex are ( 4, 6, 7).  x − 1 y − 2 z − 1 Statement II Mid-point of BD is  , , ,  2 2 2  where coordinates of D are ( x, y, z ).

each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

12. Statement I The XY -plane divides the line joining the points ( −1, 3, 4 ) and ( 2, − 5, 6) externally in the ratio 2 : 3. Statement II For a point in XY -plane, its z-coordinate should be zero.

(b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is true (d) Statement I is false, Statement II is true

Type 4. Linked Comprehension Based Questions 14. PA, AN and NQ are respectively

Directions (Q. Nos. 13-16) Consider the figure given below.

(a) y2 − y1, z2 − z1, x2 − x1 (c) z2 − z1, y2 − y1, x2 − x1

Z Q

(a) (x2 − x1 )2 + ( y2 − y1 )2 + (z2 − z1 )2 (b) (x2 − x1 )2 − ( y2 − y1 )2 − (z2 − z1 )2

Y

O X

(c) (x1 + x2 )2 + ( y1 + y2 )2 + (z1 + z2 )2

In the figure shown above, P ( x 1, y 1, z1) and Q( x 2, y 2, z 2 ) are two points referring to a system of rectangular axes OX, OY and OZ. Through the points P and Q, planes are drawn parallel to the coordinate planes forming a rectangular parallelopiped with one diagonal PQ.

(d) None of the above

16. The distance OQ between the origin O and any point Q ( x2 , y2 , z 2 ) is

13. PQ 2 is equal to (a) PA + AN (c) PA 2 + NQ 2

2

(b) AN + NQ (d) None of these 2

2

(a) x22 + y22 + z22

(b) x22 + y22 + z22

(c) (x2 + y2 + z2 )2

(d) None of these

Targ e t E x e rc is e s

N

2

(b) y2 − y1, x2 − x1, z2 − z1 (d) None of these

15. The distance PQ between two points P ( x1 , y1 , z1 ) and Q ( x 2 , y 2 , z 2 ) is A

P

Introduction to Three Dimensional [3D] Geometry

17

Type 3. Assertion and Reason

Type 5. Match the Column 17. Match the terms of Column I with the terms of Column II and choose the correct option from the codes given below. Column I (Points)

Column II (Name of the octants)

A. (1, 2, 3)

p.

I–XOYZ

B. ( 4, − 2, 3) C. ( 4, − 2, − 5)

q.

II–X ′ OYZ

r.

III–X ′ OY ′ Z

D. ( 4, 2, − 5) E. ( −4, 2, − 5)

s.

IV–XOY ′ Z

t.

F. ( −4, 2, 5) G. ( −3, − 1, 6)

u. v.

V– XOYZ ′ VI–X ′ OYZ ′ VII–X ′ OY ′ Z ′

H. (2, − 4, − 7 )

w.

VIII–XOY ′ Z ′

Type 6. Single Integer Answer Type Questions 18. The perimeter of triangle with vertices (1, 0, 0), (0, 1, 0) and (0, 0, 1) is k 2, where the value of k is _______. 19. The perpendicular distance of the point ( 6, 5, 8) from Y-axis is_________. 20. The distance between the points (1, 4, 5) and ( 2, 2, 3) is___________.

981

Entrances Gallery JEE Main 1. The distance of the point (1, 0, 2) from the point of x− 2 y+1 z − 2 and the intersection of the line = = 3 4 12 plane x − y + z = 16 is [2015] (b) 8

(a) 2 14

(c) 3 21

(d) 13

2. The equation of the plane containing the line 2x − 5 y + z = 3, x + y + 4 z = 5 and parallel to the [2015] plane x + 3 y + 6z = 1is (a) 2x + 6 y + 12z = 13 (c) x + 3 y + 6z = 7

(b) x + 3 y + 6z = − 7 (d) 2x + 6 y + 12z = − 13

Other Engineering Entrances 3. The distance between the X -axis and the point [Kerala CEE 2014] ( 3, 12, 5) is (a) 3 (d) 12

(b) 13 (e) 5

(c) 14

4. If the line joining A(1, 3, 4 ) and B is divided by the point ( −2, 3, 5) in the ratio 1: 3, then B is [EAMCET 2014] (b) (−11, 3, − 8) (d) (13, 6, − 13)

Ta rg e t E x e rc is e s

(a) (−11, 3, 8) (c) (−8, 12, 20)

5. The ratio in which ZX -plane divides the line segment AB joining the points A( 4, 2, 3) and B( −2, 4, 5) is equal to [J&K CET 2011] (a) 1 : 2 , internally (c) −2 : 1

(b) 1 : 2 , externally (d) None of these

6. The shortest distance of the point (1, 2, 3) from X, Y and Z-axes, respectively are [MP PET 2011] (a) 1, 2, 3 (b) 5, 13, 10 (c) 10, 13, 5 (d) 13, 10, 5

7. Let A(1, − 1, 2) and B( 2, 3, − 1) be two points. If a point P divides AB internally in the ratio 2 : 3, then the [Kerala CEE 2010] position vector of P is 1 $ $ $ (i + j + k ) 5 1 $ $ $ (c) (i + j + k ) 3 (e) None of these

1 $ (i + 6$j + k$ ) 3 1 (d) (7i$ + 3$j + 4 k$ ) 5

(a)

(b)

Answers Work Book Exercise 1. (a)

2. (b)

3. (d)

4. (b)

5. (b)

6. (c)

Target Exercises 1. (c)

2. (b)

3. (a)

4. (d)

5. (c)

6. (b)

7. (a)

11. (d)

12. (a)

13. (d)

14. (b)

15. (a)

16. (a)

17. (*)

6. (d)

7. (d)

*A → p ; B → s ; C → w ; D → t ; E → u; F → q ; G → r; H → w

Entrances Gallery 1. (d)

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2. (c)

3. (b)

4. (a)

5. (b)

8. (a) 18. (3 2)

9. (a,b) 19. (10)

10. (a,b,c) 20. (3)

Explanations Target Exercises 1. Suppose YZ-plane divides the line segment joining A and B in the ratio λ : 1. The coordinates of the point C of  3λ + 5 4λ + 1 λ + 6 division are  , , .  λ + 1 λ + 1 λ + 1

2. Let the point be P( x, y, z ). Then, it is given PA + PB = 10 ( x − 4)2 + ( y − 0 )2 + ( z − 0 )2 + ( x + 4)2 + ( y − 0 )2 + ( z − 0 )2 = 10 ( x − 4)2 + y 2 + z 2 = 10 − ( x + 4)2 + y 2 + z 2

Squaring on both sides, we get ( x − 4)2 + y 2 + z 2 = 100 + ( x + 4)2 + y 2 + z 2 − 20 ( x + 4) + y + z 2

⇒

2

2

− 20 ( x + 4) + y + z 2

∴ Coordinates of C are

5. Q Vertices of ∆ABC are A(−1, 3, 2 ), B(2, 3, 5) and C(3, 5, − 2 ). ⇒

AB = 9 + 0 + 9 = 18 CA = 16 + 4 + 16 = 6

and Q

BC = 1 + 4 + 49 = 54 AB + CA2 = BC 2 2

∆ABC is right angled triangle at A. ∴ ∠A = 90 °

2

2

the line in the ratio 1 : 2. 1

⇒ − 8 x − 8 x − 100 = − 20 ( x + 4)2 + y 2 + z 2 ⇒

− 16 x − 100 = − 20 ( x + 4)2 + y 2 + z 2

⇒

4 x + 25 = 5 ( x + 4)2 + y 2 + z 2

[dividing both sides by −4] Again, squaring on both sides, we get (4 x + 25)2 = 25 [( x + 4)2 + y 2 + z 2 ] ⇒ 16 x 2 + 625 + 200 x = 25 [( x + 4)2 + y 2 + z 2 ] ⇒ 16 x 2 + 625 + 200 x = 25 [ x 2 + 16 + 8 x + y 2 + z 2 ]

P (4, 2, –6)

⇒

9 x + 25 y + 25 z − 225 = 0

Again, let the point R2 divides PQ internally in the ratio 2 : 1. 2

three points. Here,

PQ = 1 + 1 + 16 = 3 2 QR = 9 + 9 + 0 = 3 2 PR = 16 + 4 + 16 = 6

Now,

P (4, 2, –6)

2

3. Let P(0, 7, 10 ), Q(− 1, 6, 6) and R(− 4, 9, 6) be the given

PQ 2 + QR 2 = (3 2 )2 + (3 2 )2 = 18 + 18 = 36 = (PR )2

Q (10, –16, 6)

 1 × 10 + 2 × 4 1 × (−16) + 2 × 2 1 × 6 + 2 × (−6)  , , ⇒ R1 =     1+ 2 1+ 2 1+ 2  10 + 8 −16 + 4 6 − 12   18 −12 −6 = , , ,   = ,  3 3 3   3 3 3 = (6, − 4, − 2 )

+ 25 y 2 + 25 z 2 Which is the required equation.

2 R1

⇒ 16 x 2 + 625 + 200 x = 25 x 2 + 400 + 200 x 2

C

6. Let the points R1 and R2 trisects the line PQ i.e. R1 divides

x 2 + 16 − 8 x = 100 + x 2 + 16 + 8 x

2

B(0, 2, –3)

Targ e t E x e rc is e s

⇒

A(2, –1, 4)

 3 × 0 − 2 × 1 3 × 2 − 1 × (−1) 3 × (−3) − 1 × 4 , ,     3−1 3−1 3−1 7 −13  or  −1, ,   2 2 

This point lies on YZ-plane. 5 3λ + 5 ∴ =0 ⇒ λ=− 3 λ+1  17 13 Hence, the coordinates of C are  0, , − .  2 2 ⇒

⇒ C divides AB externally in the ratio 3 : 1.

1 R2

Q (10, –16, 6)

 2 × 10 + 1 × 4 2 × (−16) + 1 × 2 2 × 6 + 1 × (−6) ⇒ R2 =  , ,    2+1 2+1 1+ 2 20 4 32 2 12 6 24 30 6 − + − + −     = , , ,   = ,  3 3 3   3 3 3 = (8, − 10, 2 ) Hence, required points are (6, − 4, − 2 ) and (8, − 10, 2 ).

7. Let the vertices of a triangle are A( x1, y1, z1 ), B( x2 , y2 , z2 ) and C( x3 , y3 , z3 ). A(x1,y1,z1)

Therefore, ∆PQR is a right angled triangle at Q. Also, PQ = QR. Hence, ∆PQR is a right angled isosceles triangle.

(0,8,5) E

D(5,7,11)

4. We have, 2 AC = 3 AB ⇒ ⇒ ⇒

B(x2,y2,z2)

2 AC = 3 ( AC − BC ) AC = 3BC AC 3 = BC 1

F(2, 3, – 1)

C(x3,y3,z3)

Since, D, E and F are the mid-points of AC, AB and BC.  x1 + x2 y1 + y2 z1 + z2  , ,   = (0, 8, 5)  2 2 2 

∴

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Objective Mathematics Vol. 1

17

⇒ x1 + x2 = 0, y1 + y2 = 16, z1 + z2 = 10  x 2 + x 3 y 2 + y 3 z2 + z3  , ,   = (2, 3, − 1)  2 2 2 

…(i)

⇒ x2 + x3 = 4, y2 + y3 = 6, z2 + z3 = − 2  x1 + x3 y1 + y3 z1 + z3  and , ,   = (5, 7, 11)  2 2 2 

…(ii)

⇒ x1 + x3 = 10, y1 + y3 = 14, z1 + z3 = 22

…(iii)

On adding Eqs. (i), (ii) and (iii), we get 2( x1 + x2 + x3 ) = 14, 2( y1 + y2 + y3 ) = 36 2( z1 + z2 + z3 ) = 30 ⇒ x1 + x2 + x3 = 7, y1 + y2 + y3 = 18, z1 + z2 + z3 = 15

Z

z=3 Y

10. (a) Let A(0, 7, − 10 ), B(1, 6, − 6) and C(4, 9, − 6) be the …(iv)

8. Let the point P be ( x, y, z ). Given, (PA)2 + (PB)2 = k 2 ⇒ ( x − 3)2 + ( y − 4)2 + ( z − 5)2 + ( x + 1)2 + ( y − 3)2

Ta rg e t E x e rc is e s

+ ( z + 7 )2 = k 2 ⇒ x 2 + 9 − 6 x + y 2 + 16 − 8 y + z 2 + 25 − 10 z + x 2 + 1 + 2 x + y + 9 − 6 y + z + 49 + 14 z = k

vertices of a triangle. Then, side AB = Distance between points A and B = (0 − 1)2 + (7 − 6)2 + (−10 + 6)2 = 1 + 1 + 16 = 18 = 3 2 units and side BC = Distance between points B and C

Hence, vertices of a triangle are (7, 2, 5), (3, 12, 17 ) and (− 3, 4, − 7 ).

2

O

X

On solving Eqs. (i), (ii), (iii) and (iv), we get x3 = 7, x1 = 3, x2 = − 3 y3 = 2, y1 = 12, y2 = 4 and z3 = 5, z1 = 17, z2 = − 7

2

(c) The equation of the plane z = 0 represents the XY-plane and z = 3 represents the plane parallel to XY-plane at a distance 3 units above XY-plane.

= (1 − 4)2 + (6 − 9)2 + (−6 + 6)2 = 9 + 9 + 0 = 18 = 3 2 units Clearly, AB = BC Hence, triangle is an isosceles triangle. (b) Let A(0, 7, 10 ), B(−1, 6, 6) and C(−4, 9, 6) be the vertices of a triangle. Then, side AB = (0 + 1)2 + (7 − 6)2 + (10 − 6)2

2

⇒

2 x + 2 y 2 + 2 z 2 − 4 x − 14 y + 4 z + 109 − k 2 = 0

⇒

2( x 2 + y 2 + z 2 ) − 4 x − 14 y + 4 z + 109 − k 2 = 0

= 1 + 1 + 16 = 18 = 3 2 units

2

side BC = (−1 + 4)2 + (6 − 9)2 + (6 − 6)2

which is the required equation.

= 9+ 9+ 0 = 18 = 3 2 units

9. (a) The equation of the plane x = 0 represents the YZ-plane and equation of the plane x = 1represents the plane parallel to YZ-plane at a distance 1 unit above YZ-plane. A plane parallel to YZ-plane at a distance 1 unit above YZ-plane is shown below Z

and

side CA = (−4 − 0 )2 + (9 − 7 )2 + (6 − 10 )2 = 16 + 4 + 16

= 36 = 6 units Now, AB 2 + BC 2 = (3 2 )2 + (3 2 )2 = 62 = CA2 ∴ ∆ABC is a right angled triangle at B. (c) Let A(−1, 2, 1,) B(1, − 2, 5), C(4, − 7, 8) and D(2, − 3, 4) be the vertices of a quadrilateral ABCD.

C

D

D

x=1 Y

(1,0,0) B(1,1,0)

A

X

P

(b) The equation of the plane y = 0 represents the XZ-plane and the equation of the plane y = 3 represents the plane parallel to XZ-plane at a distance 3 units above XZ-plane. Z

A B (1,0,0) X

984

C

Y O C

3 y= D

A

B

 −1 + 4 2 − 7 1 + 8 Then, mid-point of AC =  , ,   2 2 2   3 −5 9 ,  = ,  2 2 2  1 + 2 − 2 − 3 5 + 4 and mid-point of BD =  , ,   2 2 2   3 − 5 9 ,  = ,  2 2 2 So, mid-points of both the diagonals are same (i.e. they bisect each other). Hence, ABCD is a parallelogram.

PQ 2 = PA2 + AQ 2

Also, ∆ANQ is right angled triangle with ∠ANQ as right angle.

C(2,3,2)

(x, y, z)D

…(i)

Therefore, AQ 2 = AN 2 + NQ 2 (1,2,3)A

From Eqs. (i) and (ii), we have

 x − 1 y − 2 z − 1  3 5 5 , , ∴ Mid-point of BD is   i.e.  , ,  .  2   2 2 2 2 2 ∴

and

x −1 3 = ⇒ 2 2 y −2 5 = ⇒ 2 2 z −1 5 = ⇒ 2 2

x=4

PQ 2 = ( x2 − x1 )2 + ( y2 − y1 )2 + ( z2 − z1 )2 ∴ PQ = ( x2 − x1 )2 + ( y2 − y1 )2 + ( z2 − z1 )2

z=6

This gives us the distance between two points ( x1, y1, z1 ) and ( x2 , y2 , z2 ).

points in the ratio λ : 1. The coordinates of the points of 2 λ − 1 −5λ + 3 6λ + 4  . Since, the point division are  , , λ + 1  λ+1 λ+1 ∴ ⇒

6λ + 4 =0 λ+1 −2 λ= 3

●

●

16. In particular, if x1 = y1 = z1 = 0, i.e. point P is origin O, then OQ =

The ratio in which XY-plane divides the line segment = − z1 : z 2 The ratio in which YZ-plane divides the line segment = − x1 : x 2 The ratio in which ZX-plane divides the line segment = − y1 : y 2

Solutions (Q. Nos. 13-16)

x22 + y22 + z22 , which gives the distance

between the origin O and any point Q( x2 , y2 , z2 ).

17.

Point A.

(1, 2, 3)

B.

( 4, − 2, 3)

C.

Aliter The ratio in which XY-plane divides the line segment 2 4 = − z1 : z2 = − = − 6 3 ●

14. PA = y2 − y1, AN = x2 − x1 and NQ = z2 − z1

y =7

12. Suppose XY-plane divides the line joining the given

lies on the XY-plane

PQ 2 = PA2 + AN 2 + NQ 2

15. Q PQ 2 = PA2 + AN 2 + NQ 2

The coordinates of D are (4, 7, 6).

Ø

...(ii)

B(–1,–2,–1)

17 Introduction to Three Dimensional [3D] Geometry

13. Now, since ∠PAQ is a right angle, it follows that in ∆PAQ,

other. Now, let the coordinates of D are ( x, y, z ).

Octant

Name

p. (all the coordinates are + ve)

XOYZ

s. ( y-coordinate is −ve)

XOY ′ Z

( 4, − 2, − 5) w. (y and z-coordinates are −ve) ( 4, 2, − 5)

E.

( − 4, 2, − 5) u. (x and z-coordinates are −ve )

X ′ OYZ ′

F.

( − 4 , 2, 5)

q. (x-coordinate is −ve)

X ′ OYZ

r. (x and y-coordinates are −ve)

X ′ OY ′ Z

H. (2, − 4 , − 7 ) w. (y and z-coordinates are −ve)

XOY ′ Z ′

G. ( − 3 , − 1, 6)

t. (z-coordinate is −ve )

XOY ′ Z ′

D.

XOYZ ′

Targ e t E x e rc is e s

11. We know that, in a parallelogram, diagonals bisect each

18. Let A = (1, 0 , 0 ), B = (0, 1, 0 ) and C = (0,0,1). Now, AB = (0 − 1)2 + (1 − 0 )2 + 0 2 = 2 units BC = 0 2 + (0 − 1)2 + (1 − 0 )2 = 2 units and CA = (1 − 0 )2 + 0 2 + (0 − 1)2 = 2 units

Z

∴ Perimeter of triangle = AB + BC + CA = 2+ 2+ 2

Q

= 3 2 units P

90° 90° A

19. Perpendicular distance of the point(6, 5, 8 )from Y-axis

N O

Y

= 62 + 82 = 10 units

20. Distance between given points = (2 − 1)2 + (2 − 4)2 + (3 − 5)2

X

= 1 + 4 + 4 = 3 units

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Objective Mathematics Vol. 1

17

Entrances Gallery 1. Given equation of line is

x −2 y + 1 z −2 [say]…(i) = = =λ 3 4 12 and equation of plane is …(ii) x − y + z = 16 Any point on the line (i) is (3λ + 2, 4λ − 1, 12 λ + 2) Let this point be point of intersection of the line and plane. ∴ (3λ + 2 ) − (4λ − 1) + (12 λ + 2 ) = 16 ⇒ 11λ + 5 = 16 ⇒ 11λ = 11 ⇒ λ =1 ∴ Point of intersection is (5, 3, 14). Now, distance between the points (1, 0, 2) and (5, 3, 14) = (5 − 1)2 + (3 − 0 )2 + (14 − 2 )2 = 16 + 9 + 144 = 169 = 13 units

Ta rg e t E x e rc is e s

2. Let

equation of plane containing the lines 2 x − 5 y + z = 3 and x + y + 4 z = 5 be (2 x − 5 y + z − 3) + λ ( x + y + 4 z − 5) = 0 ⇒ (2 + λ )x + (λ − 5)y + (4λ + 1)z − 3 − 5λ = 0 …(i) This plane is parallel to the plane x + 3 y + 6 z = 1. 2 + λ λ − 5 4λ + 1 = = ∴ 1 3 6 On taking first two equalities, we get 6 + 3λ = λ − 5 ⇒ 2 λ = − 11 11 ⇒ λ=− 2 On taking last two equalities, we get 6λ − 30 = 3 + 12 λ ⇒ −6λ = 33 11 λ=− ⇒ 2 So, the equation of required plane is 11 11   −11   44  − 5 y +  − + 1 z − 3 + 5 × = 0 2 −  x +    2   2  2 2 ⇒ ⇒

986

−

7 21 42 49 =0 x− y− z+ 2 2 2 2 x + 3y + 6z − 7 = 0

3. Distance between X-axis and point (3,12, 5) = Distance between PQ Y P (3, 12, 5)

O

Q (3, 0, 0)

X

Z

= (3 − 3)2 + (12 − 0 )2 + (5 − 0 )2 = 0 2 + 12 2 + 52 = 144 + 25 = 169 = 13 units

4. Using internal ratio formula,  1 × x + 3 × 1 1 × y + 3 × 3 1× z + 3 × 4  (−2, 3, 5) =  , ,    1+ 3 1+ 3 1+ 3  x + 3 y + 9 z + 12  ⇒ (− 2, 3, 5 ) =  , ,   4 4 4  ⇒ x + 3 = − 8 , y + 9 = 12, z + 12 = 20 ⇒ x = − 11,y = 3,z = 8 Hence, required point = B (−11, 3, 8)

5. The ratio in which ZX- plane divides the line segment = − y1 : y2 = − (2 ) : 4 = − 2 : 4 = − 1 : 2 Hence, it divides 1 : 2, externally.

6. Shortest distance of the point (a, b, c ) from X, Y and Z-axes respectively are b2 + c 2 ,

a2 + c 2 ,

a2 + b2

∴ Shortest distance of the point (1, 2, 3) from the X, Y and Z-axes, respectively are 13, 10, 5.

7. Let the coordinates of P be (α, β, γ ). 2 ⋅ 2 + 3 ⋅1 7 = 2+3 5 2 ⋅ 3 + 3 ⋅ (−1) 3 β= = 2+3 5

α=

A (1,–1,2)

and ∴

2

P

3

(α, β, γ)

2 ⋅ (−1) + 3 ⋅ 2 4 = 2+3 5 1 $ OP = (7 i + 3$j + 4 k$ ) 5 γ=

B (2,3,–1)

18 Introduction to Limits and Derivatives Chapter Snapshot

Limits

●

If f ( x ) is a function of x such that, if x approaches to a constant value a, then the value of f ( x ) also approaches to another constant k, then constant k is known as limit of f ( x ) at x = a. It is written as lim f ( x ) = k or A real number l is called the limit of the function f , if for every ε > 0, however small, there exists δ > 0 such that | f ( x ) − l | < ε, whenever 0 < | x − a| < δ and we write it as x→ a

lim f ( x ) = l

●

●

●

●

●

●

●

x→ a

Left Hand Limit (LHL)

●

A function f is said to approach l as x approaches a from the left, if corresponding to an arbitrary positive number ε, there exists a positive number δ such that | f ( x ) − l | < ε, whenever a − δ < x < a It is written as lim f ( x ) = l or f ( a − 0) = l

●

●

x → a−

The working rule for finding the left hand limit is put a − h for x in f ( x ), where h is positive and very small and make h approach zero. i.e. f ( a − 0) = lim f ( a − h) h→ 0

Right Hand Limit (RHL) A function f is said to approach l as x approaches a from right, if corresponding to an arbitrary positive number ε, there exists a positive number δ such that | f ( x ) − l | < ε, whenever a < x < a + δ. It is written as lim f ( x ) = l or f ( a + 0) = l x→ a

+

●

●

●

●

●

●

●

Limits Existence of Limit Algebra of Limits Evaluation of Limits by Using L' Hospital’s Rule Evaluation of Algebraic Limits Evaluation of Trigonometric Limits Evaluation of Exponential and Logarithmic Limits Evaluation of Exponential Limits of the Form 1∞ Sandwich Theorem for Evaluating Limits Some Useful Expansions Use of Newton-Leibnitz’s Formula in Evaluating the Limits Derivative Geometrical Meaning of a Derivative Derivative from First Principle Differentiation of Some Important Functions Algebra of Derivative of Functions Chain Rule Logarithmic Differentiation

Objective Mathematics Vol. 1

18

The working rule for finding the right hand limit is put a + h for x in f ( x ), where h is positive and very small and make h approach zero. i.e. f ( a + 0) = lim f ( a + h) h→ 0

X

x 2 − 1, 0 < x < 2 , then Example 1. If f (x ) =  2x + 3, 2 ≤ x < 3 lim f ( x ) and lim f ( x ) are x → 2−

x → 2+

(a) 3 ,7

(c) −3, 7

(b) 7, 3

x→ a

X

(d) 3, − 7

h→ 0

x→1

lim f( x) = lim f(2 + h)

x→ 2+

h→ 0

x → 1+

= lim [2 ( 2 + h) + 3] = 4 + 3 = 7

Example 2. The left hand and right hand limits of the following function 5x − 4, 0 < x ≤ 1 f (x ) =  3 4x − 3x, 1 < x < 2 at x =1, are (a) 1, 1 (b) −1, 1

(c)1, – 1

(d) − 1, − 1

Sol. (a) LHL of f( x) at x = 1 = lim f( x) = lim f(1 − h) x → 1−

h→ 0

= lim {5 (1 − h) − 4} = lim 1 − 5h = 1 h→ 0

h→ 0

RHL of f( x) at x = 1 = lim f( x) = lim f(1 + h) x → 1+

h→ 0

= lim {4 (1 + h)3 − 3 (1 + h)} h→ 0

= 4 (1)3 − 3 (1) = 1

Example 3. Left hand limit and right hand x at x = 0, are limit of the function | x| + x 2 (a) 1, 0 (b) 0, − 1 (c)1, − 1 (d) −1, 1 Sol. (d) lim − f( x) = lim 0 − h2 h→ 0

x→ 0

and

lim f( x) = lim

x→ 0+

h→ 0

x→1

= 3 (1 + 1) = 6 lim f( x) = lim 3( x + 1) = lim [3(1 + h + 1)] = 6

∴

h→ 0

X

x→ 1

Sol. lim− f( x) = lim− 3( x + 1) = lim [3(1 − h + 1)]

h→ 0

X

Example 4. Check whether lim f (x ) exists or 2x + 3, x ≤ 0 not, where f ( x ) =  . Also, find the 3( x + 1), x > 0 limit.

= lim [( 2 − h)2 − 1] = 4 − 1 = 3 and

x→ a

If however, either both of these limits do not exist or both these limits exist but are not equal in value, then lim f ( x ) does not exist.

Sol. (a) lim − f( x) = lim f( 2 − h) x→ 2

lim f ( x ) = l

Then,

h+ h h

h + h2 1 = lim =1 h→ 0 1+ h h→ 0

Ø LHL may or may not be equal to RHL.

x → 1+

h→ 0

lim f( x) = lim f( x) = 6

x → 1−

x → 1+

Hence, lim f( x) exists and is equal to 6. x→1

X

 1 Example 5. If f (x ) = sin   , then at x = 0  x (a) only LHL exists (b) only RHL exists (c) both LHL and RHL do not exist (d) LHL and RHL exist and are equal Sol. (c) lim − f( x) = lim f(0 − h) = lim sin 1 x→ 0

h→ 0

h→ 0

−h

1 h = − (An oscillating number which oscillatesbetween −1 and 1) So, lim f ( x) does not exist. = − lim sin h→ 0

x → 0−

lim f( x) = lim sin

and

x→ 0+

h→ 0

1 h

= (An oscillating number which oscillates between − 1 and 1) So, lim f( x) does not exist. x→ 0+

Algebra of Limits Let lim f ( x ) = l and lim g ( x ) = m. If l and m exist, x→ a

x→ a

then (i) lim ( f ± g ) ( x ) = lim f ( x ) ± lim g ( x ) = l ± m x→ a

x→ a

x→ a

(ii) lim ( fg ) ( x ) = lim f ( x ) lim g ( x ) = lm x→ a

x→ a

x→ a

lim f ( x )

f l x→ a (iii) lim   ( x ) = = , provided m ≠ 0 x → a g  lim g ( x ) m

Existence of Limit

x→ a

If both right hand limit and left hand limit exist and are equal, then their common value, evidently will be the limit of f as x → a i.e. If lim f ( x ) = lim f ( x ) = l 988

x → a+

x → a−

(iv) lim ( kf ) ( x ) = lim k f ( x ) = k lim f ( x ) = kl, x→ a

x→ a

where k is constant. (v) lim | f ( x )| = lim f ( x ) = | l | x→ a

x→ a

x→ a

x→ a

x→ a

= lm

  (vii) lim fog ( x ) = f  lim g ( x )  = f ( m), only if f is  x→ a x→ a continuous at g ( x ) = m and lim g ( x ) = m. x→ a In particular,   (a) lim log f ( x ) = log  lim f ( x )  = log l  x→ a x→ a lim f ( x )

(b) lim e f ( x ) = e x → a

= el

x→ a

(viii) If lim f ( x ) = + ∞ or − ∞, then lim x→ a

x→ a

1 =0 f (x )

(ix) If f ( x ) ≤ g ( x ) for every x in the neighbourhood of a, then lim f ( x ) ≤ lim g ( x ). x→ a

Generalisation f ′ (x ) assumes the indeterminate form 0/0 If lim x → a g ′ (x ) and f ′ ( x ), g ′ ( x ) satisfy all the conditions embodied in L’Hospital’s rule, then we can repeat the application of f ′ (x ) this rule on to get g ′ (x ) f (x ) f ′ ′ (x ) lim = lim x → a g (x ) x → a g ′ ′ (x ) Sometimes, it may be necessary to repeat this process a number of times till our goal of evaluating limit is achieved. X

x→ 0

x→ c

following cases: (a) Both lim f ( x ) and lim g ( x ) exist. x→ c

(b) lim f ( x ) exists and lim g ( x ) does not x→ c

x→ c

exist. (c) Both lim f ( x ) and lim g ( x ) do not exist. x→ c

ii.

x→ c

If lim [ f ( x ) + g ( x )] exists, then we can have x→ c

the following cases: (a) If lim f ( x ) exists, then lim g ( x ) must x→ c

x→ c

exist. (b) Both lim f ( x ) and lim g (x ) do not exist. x→ 1

x→ c

X

x→ a

x→ a

(ii) both are continuous at x = a (iii) both are differentiable at x = a (iv) f ′ ( x ) and g ′ ( x ) are continuous at the point f (x ) f ′ (x ) , provided that = lim x = a, then lim x → a g (x ) x → a g ′ (x ) g ′ ( a ) ≠ 0. Ø The above rule is also applicable, if

lim g (x) = ∞.

lim f (x) = ∞ and

x→ a

tan − 1 x − sin − 1 x x3

x→ 0

1 2 (c)1

is equal to

1 2 (d) −1

(b) −

(a)

−1 −1 Sol. (b) lim tan x −3 sin x x→ 0

= lim

x→ 0

= lim

 0 form  0 

x

1 x→ 0

If f ( x ) and g ( x ) are two functions of x such that (i) lim f ( x ) = lim g ( x ) = 0

is equal to

 0 form  0  x + tan x − xsin x + cos x + cos x = lim x→ 0 2 x + sec 2 x [using L’ Hospital’s rule] 0 + 1+ 1 = =2 0+1

Example 7. lim

= lim

Evaluation of Limits by Using L’ Hospital’s Rule

x 2 + tan x (b) 0 (d) 2

x→ 0

x + sin x Sol. (d) lim xcos 2

If lim f ( x ) g ( x ) exists, then we can have the

x→ c

x cos x + sin x

(a) −1 (c) 1

x→ a

Points to be Remembered

i.

Example 6. lim

18 Introduction to Limits and Derivatives

lim g ( x )

(vi) lim [ f ( x )]g ( x ) = lim f ( x ) x → a

1 + x2

1

−

1 − x2

3 x2

⇒ = lim

x→ 0

1 − x2 − (1 + x2 ) 3 x2 (1 + x2 ) 1 − x2

(1 − x2 ) − (1 + x2 )2 3 x2 (1 + x2 ) 1 − x2 [ 1 − x2 + (1 + x2 )] − 3 − x2

x→ 0

3 x2 (1 + x2 ) 1 − x2 [ 1 − x2 + (1 + x2 )] 3 1 =− =− 6 2

Evaluation of Algebraic Limits Let f ( x ) be an algebraic function (polynomial or rational) and a be any real number, then lim f ( x ) is x→ a known as an algebraic limit. The limit of algebraic functions can be find by the following methods:

x→ a

989

Objective Mathematics Vol. 1

18

Direct Substitution Method

x→

If by direct substitution of the point in the given expression we get a finite number, then the number obtained is the limit of the given expression. x −1 1 −1 e.g. =0 lim = 2 x → 1 ( x + 2) (1 + 2) 2

x + 3 2x − 8 2

2

= lim x→

= lim x→

=

x

X

 x + 3 Example 8. The value of lim   is x → 0  x +1 (a) 0 (b) 3 (c) 1 (d) None of these x

x→ 0

 x + 1

0

 0 + 1

Factorisation method is used when we have rational f (x ) function i.e. function of the form . g (x )

X

 2 1  . Example 9. Evaluate lim  + 2 x → 1 1 − x x − 1 (a) 1 (b) 2 (c)1/ 2 (d) 3 / 2 



= lim

2 − (1 + x)

1 − x2 1 1 = lim = x→1 1+ x 2 x→1

X

x→1

1− x 1 − x2

x→

(a) −

8 5

7 (b) 5 8 (c) 5 (d) None of the above

2

(b) 3

x 2 + 3 2x − 8

is

.

x −1

x→ 1

(c) 2

(d) 4

Sol. (b) lim x − x x −1

x→1

 x − x ( x + 1) ( x2 + x ) × × = lim   x → 1 ( x − 1) ( x + 1) ( x2 + x )  x( x3 − 1) ( x + 1) ( x4 − x) ( x + 1) = lim = lim 1 x → 1 ( x − 1) ( x2 + x → x) ( x − 1) ( x2 + x ) 2

= lim

x ( x − 1) ( x2 + x + 1) ( x + 1) ( x − 1) ( x2 +

x→1

= lim

x)

x( x + x + 1) ( x + 1) 2

(x + 2

x→1

x)

Example 12. lim

=

1 × (12 + 1 + 1) ( 1 + 1) (12 +

x→ 0 3

( 8xz − 4x 2 + 3 8xz ) 4

equal to z (a) 11/ 3 2 1 (b) 23/ 3 ⋅z 2

Sol. (b) lim

x→ 0 3 (

x 3 z2 − ( z − x)2 8 xz − 4 x2 + = lim

x→ 0 3 (

3

8 xz − 4 x2 +

2z

[2 3 8 z}4

3

8 xz )4

x4 / 3 3 2 z − x

x → 0 x4 / 3 ( 3 3

8 xz )4

x 3 2 xz − x2

= lim =

1)

x 3 z 2 − ( z − x) 2

(c) 2 21/ 3 ⋅ z (d) None of the above x4 − 4

Example 10. The value of lim

x2 − x

2

X

 0 form  0  = lim

( 2 + 2 ) (2 + 2 ) 8 2 8 = = ( 2 + 4 2) 5 2 5

Example 11. Evaluate lim (a) 1

f (x ) . If by putting x = a, the x → a g (x ) f (x ) 0 ∞ rational function takes the form , etc., then 0 ∞ g (x ) ( x − a ) is a factor of both f ( x ) and g ( x ). In such a case we factorise the numerator and denominator and then cancel out the common factor ( x − a ). After cancelling out the common factor ( x − a ), we again put x = a in the given expression and see whether, we get a meaningful number or not. This process is repeated till we get a meaningful number. Consider, lim

Sol. (c) lim  2 2 − 1  x→ 1 1 − x 1 − x

2

( x + 2 ) ( x2 + 2 ) (x + 4 2 )

 0 form  0 

Rationalisation method is used when we have 1 1   radical signs  like , , etc.,  in numerator or   2 3 denominator or in both of an rational expression. After rationalisation, the terms are factorised which on cancellation gives the required result.

Factorisation Method

X

2

( x + 2 ) ( x − 2 ) ( x2 + 2 ) (x − 2 )(x + 4 2 )

Rationalisation Method

    Sol. (c) lim  x + 3  =  0 + 3  = 30 = 1

990

x4 − 4

Sol. (c) lim

=

8z − 4x + 1 2 23 / 3 ⋅ z

3

8 z )4

is

=3

lim

x→ a

⇒

∴

X

X

xn − an = lim ( x n − 1 + x n − 2 a x→ a x−a + x n − 3 a + ... + a n − 1 )

Example 15. The values of constants a and b  1  x 2 +1 so that lim  − ax − b = , are x→ ∞  x +1  2 (a) a = 1, b = −

3 2

3 2 (d) a = 2, b = − 1

(b) a = − 1, b =

(c) a = 0, b = 0

[using expansion] xn − an n− 1 n −1 lim =a +a + an −1 x→ a x−a + ... + a n − 1 xn − an lim = na n − 1 x→ a x−a





Sol. (a) We have, lim  x + 1 − ax − b = 1 x→ ∞  x + 1  2 ⇒ ⇒

(1 − x ) n − 1 Example 13. Evaluate lim . x→ 0 x (a) n (b) − n (c) 0 (d) 1

⇒

lim

x→ ∞

lim

x→ ∞

2

( x2 + 1) − (ax + b )( x + 1) 1 = x+1 2

x2 (1 − a) − (a + b ) x − b + 1 1 = 2 x+1 1 − a = 0 and

∴

a = 1 and

a+ b=− b=−

Sol. (b) We have, lim (1 − x) − 1 n

x→ 0

x

(1 − x)n − 1 = − lim x → 0 (1 − x) − 1 y n − 1n , where y = 1 − x = − lim y→1 y − 1 As x → 0, y → 1 = − n(1)n − 1 = − n

3 2

If m and n are positive integers and a 0 , b0 are non-zero real numbers, then a 0 x m + a1 x m − 1 + ...+ a m − 1 x + a m lim x → ∞ b x n + b x n − 1 + ...+ b 0 1 n − 1 x + bn  a0  b , if m = n  0 = 0, if m < n ∞, if m > n, a b > 0 0 0  −∞, if m > n, a 0 b0 < 0

Step II

Evaluation of Trigonometric Limits

 x3 x2  . Example 14. Evaluate lim  2 − x → ∞  3x − 4 3x + 2  1 2 3 2 (b) (c) (d) (a) 3 9 4 3

In order to evaluate the said type of limits, we use the following standard results: sin θ tan θ (ii) lim (i) lim =1 =1 θ→ 0 θ θ→ 0 θ sin x ° π (iii) lim (iv) lim cos θ = 1 = θ→ a θ→ 0 x 180 sin (θ − a ) tan (θ − a ) (vi) lim (v) lim =1 =1 θ→ a θ→ a θ−a θ−a

If k is the highest power of x in numerator and denominator both, then divide each term in numerator and denominator by x k . 1 Step III Use the result lim n = 0, where n > 0. x→ ∞ x

X

1 2

Important Result

Method of Evaluating Algebraic Limits When x → ∞ To evaluate this type of limits, we follow the following procedure: Step I Write down the given expression in form of f (x ) a rational function i.e. , if it is not so. g (x )



3

2

= lim

x→ ∞

= lim

(vii) lim

x3 (3 x + 2 ) − x2 (3 x2 − 4) (3 x2 − 4) (3 x + 2 ) 2 x3 + 4 x2

9 x3 + 6 x2 − 12 x − 8 4 2+ 2 x = lim = 6 12 8 x→ ∞ 9 9+ − 2 − 3 x x x x→ ∞

sin − 1 x =1 x→ 0 x



x  Sol. (b) lim  2x − x → ∞  3x − 4 3 x + 2 

18 Introduction to Limits and Derivatives

Evaluation of Limits Using Standard Results

X

tan − 1 x =1 x→ 0 x

(viii) lim

Example 16. The value of lim

x→ 0

1 2 1 (c) 4

(a)

1 4 1 (d) − 2

sin 2 2x sin 2 4x

is

(b) −

991

18

2 Sol. (c) lim sin2 2 x x→ 0

X

sin 4 x

Objective Mathematics Vol. 1

= lim

2

sin 2 x

x→ 0

(2 x)2

×

2

( 4 x)

sin2 4 x

×

1 4

2

X

Example 17.

lim

2

x tan 2x − 2x tan x

x→ 0

(a) 2 1 (c) 2

(1 − cos 2x ) 2 (b) − 2 1 (d) − 2

log 9 log 8

Sol. (a) lim

x→ 0

(1 − cos 2 x) 2 tan x − 2 x tan x x 1 − tan2 x = lim x→ 0 (2 sin2 x)2   1 −1 2 x tan x  2 − tan 1 x   = lim x→ 0 4sin4 x  1 − 1 + tan2 x  2 x tan x   2  1 − tan x  = lim 4 x→ 0 4sin x x tan3 x = lim x→ 0 2 sin4 x(1− tan2 x)

X

=

=

2(1)4 (1 − 0)

=

Evaluation of Exponential and Logarithmic Limits To evaluate the exponential and logarithmic limits, we use the following results: a x −1 (i) lim = log e a, a > 0 x→ 0 x log (1 + x ) (ii) lim =1 x→ 0 x 992

ex − 1 =1 x→ 0 x

(iii) lim

log 3 log 2

is

(d)

log 8 log 9

32 x − 1 × 2x = lim 32x x − 1 x→ 0 2 − 1 × 3x 3x x 2  3 − 1  lim  2 x→ 0  2 x  = ×  2 3 x − 1 3  lim  x→ 0   3x  2  log 3  log 32 log 9 =  = = 3  log 2  log 2 3 log 8

32 x − 1 23 x

 x 2 

log[1 + ( x − 1)] π π sin  − x  2 2  log[1 + ( x − 1)] ⋅ ( x − 1) x−1 = lim x→1 π π sin  − x  2 2   π π ⋅ −   π − π  2 2x   x 2 2  ( x − 1) ( x − 1) = lim = lim x→1 π π  x → 1  2 x − 1 − 1  − x  π  x 2 2   2  ( x − 1) 2 2 = = lim π x → 1 2 x − 1 − 1 πlog 2 = lim

x→1

3

1 2

(c)

2 3x − 1

Sol. (c) lim sec  πx  log x = lim log x x→1 x→1 2  π cos  

1 lim 2 x→ 0  sin x  4 2   (1 − tan x)  x  1 (1)3

log 2 log 3

3 2x − 1

π Example 19. The value of lim sec  x  log x is 2  x→1 2 (b) π log 2 (a) log 2 2 (c) (d) None of these π log 2

3   tan x  ⋅ x3     x   1  = lim  4 x 2 x→ 0 2  sin x (1 − tan x)    

 tan x     x 

(b)

is equal to

x Sol. (c) lim x tan 2 x − 2 x tan 2 x→ 0

x→ 0

(a)

1 sin2 x   4x  lim    × lim  x → 0  sin 4 x  4 x→ 0  2 x  1 Q lim sin x = 1 = × 1× 1  x → 0 x  4 1 = 4 =

Example 18. The value of lim

Evaluation of Exponential Limits of the Form 1∞ To evaluate the exponential limits of the form 1∞ , we use the following results: (i) If lim f ( x ) = lim g ( x ) = 0, then x→ a

x→ a

lim [1 + f ( x )]

1/ g ( x )

x→ a

=e

lim

f (x)

x → a g (x)

(ii) If lim f ( x ) =1 and lim g ( x ) = ∞, then x→ a

lim [ f ( x )]

x→ a

x→ a

g (x)

= lim [1 + ( f ( x ) − 1)]g ( x ) x→ a

lim ( f ( x ) − 1) g ( x )

= ex → a

(i) lim (1 + x )

1/ x

=e

x→ 0

(iii) lim (1 + λx )

1/ x

x→ 0

X

(ii)

= eλ

 lim 1 + x→ ∞  

(iv) lim 1 + x→ ∞ 

1  x =e x 

Sol. (i) lim (log 3 3 x) log x 3

x→ 0

(b) e − 1 (d) None of these

(a) 1 (c) e Sol. (c) lim(1 + x)cosec x

λ  x λ  =e x

Example 20. Evaluate the following limits tan ( πx / 2a )  log x 3 a (ii) lim 2 −  (i) lim  log 3 3x x→1 x → a x

Example 22. lim(1 + x ) cosec x is equal to

x→ 0

x

lim

x

1  x→ 0 sin x 1  sin x   = lim (1 + x) x  =  lim(1 + x) x  = e1 = e x→ 0 x→ 0        

Sandwich Theorem for Evaluating Limits

x→1

=

lim (log 3 3 + log 3 x)

=

1 log 3 x

log x 3

y = h(x)

x→1

lim (1 + log 3 x)

=e

x→1 1

y = g(x)

1 lim log 3 x × log 3 x x→ 1

y = f(x)

=e =e (ii)

tan ( πx / 2 a )  a    = lim 1 +  1 −     x→ a  x    a πx πx  x − a  lim  1 −  tan lim   tan x→ a  x  x→ a  x 2a 2a =e =e

lim  2 − a  x x→ a 

tan( πx / 2 a )

x − a πx where l = lim   tan x→ a  2a x  x − a πx l = lim   tan x → a 2a x  x − a  π πx  = lim    cot  − x → a  2 2a   x x − a π = lim   cot (a − x) x → a x  2a ( a − x) 1 = − lim × π x→ a tan (a − x) x 2a π ( a − x) 2a 2 × = − lim 2 a =− π x→ a π x π tan (a − x) 2a l

=e ,

Now,

X

Example 21. (a)

 x + 6 lim   x → ∞  x +1

x+4

is equal to

1

e5 (b) e 5 (c) 5e 5 (d) None of the above   Sol. (b) lim  x + 6  x→ ∞

x+ 4

 x + 1

= e5

x=β

x=a

If f ( x ) ≤ g ( x ) ≤ h ( x ), ∀ x ∈ (α, β) − {a} and lim f ( x ) = L = lim h( x ), then lim g ( x ) = L, where x→ a x→ a x→ a a ∈(α, β) Remarks In the Sandwich theorem, we assume that f ( x ) ≤ g ( x ) ≤ h( x ) for all x near a, ‘except possibly at a’. This means that it is not required that when x = a, we have the inequality for the functions i.e. it is not required that f ( a ) ≤ h( a ). The reason is that we are dealing with limits as x approaches a. So, we have x that is moving closer and closer to a. As long as f ( x ) ≤ g ( x ) ≤ h( x ) is true for all these x, we can be sure the limit, i.e. the point where the function values are heading must behave as the Sandwich theorem indicates.In particular, unless we are given extra information about the functions and their values at a, the Sandwich theorem does not allow us to make conclusions about functions values at a. So, none of the following claims can be guaranteed by the assumptions in the Sandwich theorem. 1. f ( a ) = g ( a ) = h( a ) [well, not even f ( a ) ≤ h( a )] 2. g ( a ) = lim f ( x ) = lim h( x ) = L x→ a

3. lim g ( x ) = g ( a ) 1 + 6   x = lim  1 x→ ∞  1+   x x

e 6  1   e  1

x=α

x→ a

1 + 6     1+  x   = lim x x→ ∞ 1 + 1   1 +     x =

18 Introduction to Limits and Derivatives

X

In particular cases

6  x 1 x

x→ a

x+ 4

4

x  1 + λ  = e λ    Q xlim  → ∞ x  

X

Example 23. Evaluate lim f (x ), x→ 1

where, 2x ≤ f ( x ) ≤ x ( x + 1), ∀ x that are near to 1 but not equal to 1. (a) 2 (b) 3 (c) 4 (d) 1 2

2

Sol. (a) lim 2 x2 = 2(1)2 = 2 x→1

and

lim x( x2 + 1) = 1 (12 + 1) = 2

x→1

It follows from the Sandwich theorem that lim f ( x) = 2 x→1

993

Objective Mathematics Vol. 1

18 Some Useful Expansions i. ii.

iii. iv. v. vi. vii. viii. ix. x. X

x2 x3 + + ... 2! 3! x2 x3 a x =1+ x log a + (log a ) 2 + (log a ) 3 2! 3! +..., a > 0 2 3 4 x x x log (1 + x ) = x − + − + ..., − 1 < x ≤ 1 2 3 4 x2 x3 x4 log (1 − x ) = − x − − − −… , −1 ≤ x < 1 2 3 4 n( n − 1) 2 (1 + x ) n = 1 + nx + x + ... , − 1 < x < 1 2! x3 x5 sin x = x − + − ... 3! 5! x2 x4 cos x = 1 − + − ... 2! 4! x 3 2x 5 tan x = x + + + ... 3 15 1 x3 1 3 x5 1 3 5 x7 sin − 1 x = x + ⋅ + ⋅ ⋅ + ⋅ ⋅ ⋅ + ... 2 3 24 5 2 4 6 7 1 1 tan − 1 x = x − x 3 + x 5 − ... 3 5 ex = 1 + x +

Example 24. lim

sin x + log(1 − x )

x→ 0

(a) 0 (c) −

1 2

x2 (b) 1/2

I (x ) = ∫

Ψ (x)

φ( x )

f ( t ) dt

d {I ( x )} dx  d  d = f {ψ ( x )} ψ( x ) − f {φ ( x )} φ ( x )   dx   dx

Newton-Leibnitz’s formula states that

X

Example 26. If f (x ) = ∫ is equal to (a) 0 (c) 2

= lim

f ( x) =

lim f ′( x) = lim

⇒

x→ ∞

is equal to

− = lim

x→ 0

X

2

x→ 0

(a) 1 1 (c) 2

x2 (b) 0

3 2 (c) −1

x→ 0

= lim

x→ 0

cos t 2 dt

x sin x

is

(d) None of these

Sol. (b) lim ∫0

cos t 2dt

 0 form  0 

x→ 0

xsin x 2 xcos x4 = lim x→ 0 xcos x + sin x

[using L’ Hospital’s rule]

2 cos x4 − 8 x4 sin x4 [using L’ Hospital’s rule] x→ 0 2 cos x − xsin x 2−0 = =1 2−0 = lim

x

is equal to

x   x2 x3 (1 + x + x2 ) −  1 + x + + + … 2! 3!   x2

sin x =0 x

(b) 1 x2

3

(d) None of these

2 x Sol. (c) We have, lim 1 + x + 2x − e

x→ ∞

dt

x→ 0

x x 1 1 – ... − x3  +  −  3! 3  1 2 4 =− 2 x2

1 + x + x 2 − ex

x→ ∞

∫ Example 27. The value of lim 0

4

Example 25. lim

dt, then lim f ′ ( x )

x2

x2 [applying expansions of sin x and log (1 − x)] 2

x sint

∫1

t sin x f ′ ( x) = x

∴

    x x x x  x − – ... − + − ... +  − x − ! ! 3 5 2 3    

x→ 0

t

t

(b) 1 (d) None of these

Sol. (a) Given,

x2 5

x sin

1

sin x + log(1 − x) 3

994

Let us consider the definite integral

(a)

x→ 0

1 − 1  = 1    2 2

Use of Newton-Leibnitz’s Formula in Evaluating the Limits

(d) None of these

Sol. (c) lim

X

3 4  1 − 1  x2 − x − x − …    2 ! 3! 4 = lim x→ 0 x2 x x2 1 − −… ⇒= = lim  1 −  − x→ 0  2 ! 3! 4!

X

∫ cos t dt is The value of lim 0

Example 28.

x→ 0 x (b) −1 (d) None of these

(a) 1 (c) ∞ x

Sol.

∫ (d) lim 0 x→ 0

cos t dt x

cos x 1 = cos 0° = 1

= lim

x→ 0

 0 form  0  [using L’ Hospital’s rule]

1 If [ ] denotes the greatest integer function n ∈ N,  n sin x  is equal to x→ 0   x 

10 The value of 1⋅

then lim a n

c n−1

b 0

d n+1

(cos x − 1) (cos x − e x ) x→ 0 xn is a finite non-zero number, is

2 The integer n for which lim a 1 c 3

b 2 d 4

e 1/ x − 1 is equal to x→ 0 e 1/ x + 1

lim

b 1 d Does not exist

x→ ∞ 4

x + 2 x 3 − cx 2 + 3 x − d } is finite, then the value

r =1

r =1

∑ r + ... + n ⋅ 1 is

n4

1 24 1 b 12 1 c 6 d None of the above a

[ x ] + [ x 2 ] + [ x 3 ] + ... + [ x 2 n + 1 ] + n + 1 , n ∈N 1 + [x2 ] + [x] + 2 x

11 lim− x→ 0

is equal to b n d 0

12 The value of n→ ∞

3 5 2 any real number

 x  a 5 lim   + sin x → ∞  x + 1 

...log( n k a ∞ c k x

1   is equal to x 

b e1 − a

a ea − 1

c e

d 0

b 0 d None of these

1 3

7 If f ( x ) = f ( x + 1) +

5   and f ( x ) > 0,∀ x ∈ R, f ( x + 2 )

then lim f ( x ) is x→ ∞

2 a 5 c ∞

5 b 2 d None of these

( x + 1)10 + ( x + 2 )10 + L + ( x + 100 )10 is x→ ∞ x 10 + 10 10 equal to

8 lim

a 0 c 10

b 1 d 100

9 If φ( x ) = lim

x→ ∞

x 2 n f ( x ) + g( x ) , then 1 + x 2n

a φ( x ) = g ( x ), ∀ x ∈ R b φ( x ) = f( x ), ∀ x ∈ R g ( x ), if − 1 < x < 1 c φ( x ) =   f( x ), if | x | ≥ 1  if | x| < 1  g ( x ),  d φ( x ) =  f( x ), if | x | > 1  f( x ) + g ( x ) , if | x | = 1   2

(n k ) is

b n d None of these

 12  3 52 7 + + + + ... is 3 2 3 2 − x + x − x + x 1 1 1 1  

{ x} is equal to tan{ x}

a 1 c −1

− 1)

13 lim  x→ ∞

6 If { x} denotes the fractional part of x, then x→ 0

r =1

lim|log( n − 1) n ⋅ log n (n + 1) ⋅ log( n + 1)(n + 2 )

of a is

lim

n− 2

∑ r + 3⋅

a n+1 c 1

4 If lim { x 4 + ax 3 + 3 x 2 + bx + 2 −

a b c d

n− 1

∑r + 2⋅

n→ ∞

3 lim

a 0 c −1

n

Introduction to Limits and Derivatives

18

Work Book Exercise 18.1

equal to 5 6 10 b − 3 5 c 6 d None of the above

a −

{(a − n ) nx − tan x} sin nx = 0, where n is a x2 non-zero real number, then a is equal to

14 If lim

x→ 0

n+1 n 1 d n+ n

a 0

b

c n

15 The value of

 x  x  x  x lim cos   cos   cos   … cos  n  is 2   8  4  2

n→ 0

a 1 sin x b x x c sin x d None of the above sec 2 x

16 lim

π x→ 4

a c

∫2

f ( t ) dt x2 −

8 f(2 ) π 2  1 f  π 2

π2 16

is equal to

b

2 f(2 ) π

d 4f(2 )

995

Objective Mathematics Vol. 1

18 Derivative Suppose f is a real valued function, the function f ( x + h) − f ( x ) defined by lim , wherever the limit h→ 0 h exists, is defined to be the derivative of f at x and is denoted by f ′ ( x ). This definition of derivative is also called the first principle of derivative. f ( x + h) − f ( x ) Hence, f ′ ( x ) = lim h→ 0 h Clearly, the domain of definition of f ′ ( x ) is wherever the above limit exists. Sometimes, f ′ ( x ) is dy d denoted by [ f ( x )] or if y = f ( x ), it is denoted by . dx dx This is referred to as derivative of f ( x ) or y with respect to x. It is also denoted byD[ f ( x )] . Further, the derivative of f at x = a is also denoted by  df     dx  a

or

 df     dx  x = a

Geometrical Meaning of a Derivative

Y

Q

Ψ

φ

Ψ

O T

L

P

R

M

N

Let f ( x ) be a function finitely differentiable at every point on the real number line. Then, its derivative which is given by the first principle, is f ( x + h) − f ( x ) f/ ′ ( x ) = lim h→ 0 h

Example 29. The value of lim

h→ 0

f ( x + h) − f ( x ) , h

x +1 where f ( x ) = , is x −1 −2 (a) ( x − 1) 2 −1 (b) ( x − 1) 2 1 (c) (1 − x ) 2 3 (d) (1 − x ) 2 x−1 f ( x + h) − f( x) Now, lim h→ 0 h x+ h+1 x+1 − x+ h−1 x−1 = lim h→ 0 h ( x + h + 1) ( x − 1) − ( x + 1) ( x + h − 1) = lim h→ 0 ( x + h − 1) ( x − 1) h

X

( x2 − x + hx − h + x − 1) − ( x2 + hx − x + x + h − 1) ( x + h − 1) ( x − 1) h x2 + hx − h − 1 − x2 − hx − h + 1 = lim h→ 0 ( x + h − 1) ( x − 1) h − 2h −2 = lim = h → 0 ( x + h − 1) ( x − 1) h ( x + 0 − 1) ( x − 1)

= lim

h→ 0

=

996

Derivative from First Principle

Sol. (a) We have, f( x) = x + 1

In right angled ∆QPL, QL NQ − NL tan φ = = PL MN NQ − MP = ON − OM f ( x 0 + h) − f ( x 0 ) ( x 0 + h) − x 0 f ( x 0 + h) − f ( x 0 ) = h

Thus, the limit turns out to be equal to the slope of the tangent. Hence, f ′ ( x 0 ) = tan ψ

X

Let P {x 0 , f ( x 0 )} and Q{x 0 + h, f ( x 0 + h)} be two points very near to each other on the curve y = f ( x ). Draw PM and QN perpendiculars from P and Q on X-axis, PL perpendicular from P on QN . Let the chord QP produced meet the X-axis at R and ∠QPL = ∠QRN = φ.

y = f(x)

When h → 0, the point Q moving along the curve tends to P i.e. Q → P , then the chord PQ approaches the tangent line PT at the point P and then φ → Ψ. Now, applying lim in Eq. (i), we get h→ 0 f ( x 0 + h) − f ( x 0 ) lim tan φ = lim h→ 0 h→ 0 h f ( x 0 + h) − f ( x 0 ) tan Ψ = lim h→ 0 h

...(i)

=

−2 ( x − 1)2

Example 30. The value of lim

h→ 0

where f ( x ) = sin ( x +1), is (a) sin ( x +1) (b) cos ( x +1) (c) − sin ( x + 1) (d) − cos ( x + 1)

f ( x + h) − f ( x ) , h

Sol. (b) We have, f( x) = sin ( x + 1)

f( x + h) − f( x) h sin( x + h + 1) − sin ( x + 1) = lim h→ 0 h x + h + 1 + x + 1   x + h + 1 − x − 1 2 cos   sin       2 2 = lim h→ 0 h 2x + h + 2 h 2 cos  sin   2 2 = lim h h→ 0 2× 2 2x + 0 + 2 = cos   × 1 = cos( x + 1)   2

Now, lim

h→ 0

Differentiation of Some Important Functions (i) (ii) (iii) (iv) (v)

d dx d dx d dx d dx d dx

( c) = 0, c is independent of x ( x n ) = nx n −1 (sin x ) = cos x (cos x ) = − sin x

(vi)

(viii) (ix) (x) (xi) (xii)

d dx d dx d dx d dx d dx

(e x ) = e x 1 (log x ) = , ( x > 0) x 1 (log a x ) = x log e a

18

Example 31. The derivative of the function 2x 2 + 6x + 1 at x = 3 is (a) 18 (b) 17 (c) 16 (d) 12 Sol. (a) We have, f( x) = 2 x2 + 6 x + 1

π  2

(cosec x ) = − cosec x cot x, {x ≠ nπ} ( a x ) = a x log e a

Let f and g be two functions such that their derivatives are defined in a common domain. Then, (i) Derivative of sum of two functions is sum of the derivatives of the functions. d d d i.e. f (x ) + [ f ( x ) + g ( x )] = g (x ) dx dx dx (ii) Derivative of difference of two functions is the difference of the derivatives of two functions. d d d i.e. f (x ) − g (x ) [ f ( x ) − g ( x )] = dx dx dx (iii) Derivative of product of two functions is given by the following product rule. d d d g (x ) [ f ( x ) ⋅ g ( x )] = g ( x ) f (x ) + f (x ) dx dx dx (iv) Derivative of scalar multiplication of a function is the scalar multiple of derivative of the function. d d i.e. [ kf ( x )] = k [ f ( x )] dx dx where, k is any constant. (v) Derivative of quotient of two functions is given by the following quotient rule whenever the denominator is non-zero. d d g (x ) ⋅ f (x ) − f (x ) ⋅ g (x ) d  f (x )  dx dx = dx  g ( x )  [ g ( x )]2 X

π  (tan x ) = sec 2 x, x ≠ n π +  2 

d (cot x ) = − cosec 2 x, {x ≠ nπ} dx d  (vii) (sec x ) = sec x tan x, x ≠ nπ + dx 

Algebra of Derivative of Functions

Introduction to Limits and Derivatives

X

On differentiating f( x) w.r.t. x, we get d [2 x2 + 6 x + 1] f ′ ( x) = dx d d d (2 x2 ) + (6 x) + (1) = dx dx dx d d ( x) + 0 = 2 ( x2 ) + 6 dx dx Q d (constant) = 0 and d {a f( x)} = a d f( x)  dx  dx dx d n  n − 1 x = nx Q = 2 ⋅2 x + 6⋅1  dx  = 4x + 6 f ′ ( x) = 4 x + 6 ∴ At x = 3, f′(3) = 4 ⋅ 3 + 6 = 18

997

X

Objective Mathematics Vol. 1

18

Chain Rule

Example 32. For the function x 100 x 99 x2 f (x ) = + + ... + + x + 1, 100 99 2 f ′ (1) = mf ′ (0), where m is equal to (a) 50 (b) 0 (c) 100 (d) 200

If y = [ u {v ( x )}], then

100 99 2 Sol. (c) Given, f( x) = x + x + ... + x + x + 1

100 99 2 100 x99 99 x98 2x f ′ ( x) = + + ... + + 1 + 0 ...(i) ⇒ 100 99 2 On putting x = 1, we get f′(1) = (1)99 + (1)98 + ... + 1 + 1 14444244443

known as chain rule. If y = u [ v {w( x )}], then dy du [ v {w( x )}] dv {w ( x )} dw ( x ) = × × dx dv {w( x )} dw ( x ) dx X

100 times

= 1 + 1 + 1 + ... + 1 + 1 144424443 100 times

⇒ f′(1) = 100 Again, f′(0) = 0 + 0 + ... + 0 + 1 ⇒ f′(0) = 1 From Eqs. (ii) and (iii), we get f ′(1) = 100 f ′(0) Hence, m = 100 X

On differentiating w.r.t. x, we get dy d d = ( x2 + 1) (cos x) + (cos x) ( x2 + 1) dx dx dx [by product rule] = ( x2 + 1) (− sin x) + cos x(2 x) = − x2 sin x − sin x + 2 xcos x

1 1+ x is Example 34. The derivative of 1 1− x −2 −1 2 3 (b) (c) (d) (a) 2 2 2 (1 + x ) (1 − x ) 2 (1 − x ) (1 − x ) 1 x ⇒ y= x+1 Sol. (b) Let y = 1 x−1 1− x On differentiating w.r.t. x, we get d d ( x − 1) ( x + 1) − ( x + 1) ( x − 1) dy dx dx = dx ( x − 1)2 ( x − 1) (1 + 0) − ( x + 1) (1 − 0) = ( x − 1)2 x − 1− x − 1 −2 = = 2 ( x − 1) (1 − x)2 1+

998

dy = dx

...(iii)

Sol. (a) Let y = ( x2 + 1)cos x

X

Example 35. If y = log tan x , then the value of dy π at x = , is given by 4 dx 1 (a) ∞ (b) 1 (c) 0 (c) 2 Sol. (b) We have,

...(ii)

Example 33. The derivative of (x 2 + 1) cos x is given by (a) − x 2 sin x − sin x + 2x cos x (b) x 2 sin x − sin x + 2x cos x (c) x 2 sin x + sin x + 2x cos x (d) x 2 sin x − sin x − 2x cos x

dy du {v ( x )} d = × v ( x ) is dx d {v ( x )} dx

At x =

X

sec 2 x 1 1 ⋅ ⋅ sec 2 x = 2 tan x tan x 2 tan x

π dy sec 2 π / 4 2 , = = =1 4 dx 2 tan π / 4 2

 2x − 1  Example 36. If y = f  2  and  x + 1 dy is equal to f ′ ( x ) = sin x 2 , then dx 2  2x − 1   2 + 2x + x 2     (a) sin  2  x + 1  ( x 2 + 1) 2  2

 2x − 1   2 + 2x − 2x 2     (b) sin  2  x + 1  ( x 2 + 1) 2  2

 2x − 1   2 + 2x − x 2     (c) sin  2  x + 1  ( x 2 + 1) 2  (c) None of the above   Sol. (b) We have, y = f  22x − 1  x + 1

⇒

 2 x − 1 dy  = f ′  2 dx  x + 1

 ( x2 + 1) 2 − (2 x − 1) 2 x    ( x2 + 1)2   2

 2 x − 1  2 + 2 x − 2 x2    = sin  2  2 2  x + 1  ( x + 1)  2   2 x − 1  2 x − 1   Q f ′( x) = sin x2 ⇒ f ′     = sin  x2 + 1  x2 + 1        

Logarithmic Differentiation So far, we have discussed derivatives of the functions of the form [ f ( x )] n , n f ( x ) and n x , where f ( x ) is a function of x and n is a constant. In this section, we will be mainly discussing derivatives of the functions of the form [ f ( x )] g ( x ) , where f ( x ) and g ( x ) are

Taking logarithm on both sides, we have log y = g ( x ) log { f ( x )} On differentiating w.r.t. x, we get dg ( x ) 1 dy 1 df ( x ) ⋅ = g (x ) ⋅ ⋅ + log { f ( x )} ⋅ y dx f ( x ) dx dx  g ( x ) df ( x ) dy dg ( x )  = y ⋅ + log { f ( x )} ⋅ dx dx   f ( x ) dx  g ( x ) df ( x ) dy dg ( x )  ⇒ = [ f ( x )]g ( x )  ⋅ + log{ f ( x )} ⋅ dx dx   f ( x ) dx ∴

Logarithmic Differentiation (Objective Approach)

or when we have to differentiate the function of the form (Variable) variable , take log on both sides and differentiate. X

 a + x Example 37. If f (x ) =    b+ x f ′ (0) is equal to a+b  a a 2 − b2   a    (a)  2 log + b ab   b    a b2 − a 2   (b)  2 log + b ab  

a    b

a + b + 2x

, then

a+b

18 Introduction to Limits and Derivatives

functions of x. To find the derivatives of this type of functions, we proceed as follows: Let y = [ f ( x )]g ( x )

a+b  a a 2 + b2   a    (c)  2 log + b ab   b  

(d) None of the above Sol. (b) We have,

φ (x)

If y = { f ( x )} , then dy = Derivative of { f ( x )}φ ( x ) w.r.t. x taking φ( x ) as dx a constant + Derivative of { f ( x )}φ ( x ) w.r.t. x taking f ( x ) as a constant dy df ( x ) = φ ( x ) ⋅ { f ( x )}φ ( x ) − 1 ⋅ ⇒ dx dx dφ ( x ) φ (x) + { f ( x )} ⋅ log f ( x ) ⋅ dx

log f( x) = (a + b + 2 x) [log (a + x) − log (b + x)] On differentiating both sides w.r.t. x, we get f ′( x) = 2 [log (a + x) − log (b + x)] f ( x)  1 1  + ( a + b + 2 x)  −   a + x b + x 1 1   ⇒ f ′(0) = f(0) 2 (log a − log b ) + (a + b )  −    a b   a+ b  a a b 2 − a2   2 log +  =    b b ab  

Work Book Exercise 18.2 1 The derivative of ( x 2 sin x + cos 2 x ) is a

x cos x + 2 x sin x − sin2 x

b

x 2 cos x + 2 x sin x − 2 sin2 x

c

x cos x + x sin x − 2 sin2 x

d

x 2 cos x + 2 x sin x + 2 sin2 x

6 If f ( x ) =

2

a

2

2 If y = x + a 1

b

1 2

1 2

c

d 0

c

3 sin2 2 x cos 2 x 5 3 sin2 x cos 2 2 x 4

4 If y =

3 sin2 2 x cos 2 x 4 3 d sin2 2 x cos 2 x 2 b

dy sin ( x + 9) at x = 0, is , then dx cos x

a cos 9 c 0

b sin 9 d 1

b−

π 6

c

1 6

c 1

d 0

sin x + cos x dy , then at x = 0, is sin x − cos x dx

a −2 1 c 2

b 0 d None of these

1 e d None of these

a e c

b

1

9 If f ′( x ) = 2 x 2 − 1 and y = f ( x 2 ), then

log (log x )

d

π 6

, then

y log y (2 log log x + 1) x log x 2 y log y c (log log x + 1) x log x

a

dy at x = 1, is dx

b 1 d None of these

10 If y = x(log x)

 π  is  2 

π 6

4 5

a 2 c −2

5 If f ( x ) = 1 + cos 2 ( x 2 ), then f ′  a

b

8 If f ( x ) = log x (log x ), then f ′( x ) at x = e is

3 The derivative of sin 3 x cos 3 x is a

5 4

7 If y =

1 dy , then at x = 1, is dx x

x−4 , then f′ (1) is equal to 2 x

dy is equal to dx b

x log x (2 log log x + 1) y log y

d None of these

999

WorkedOut Examples Type 1. Only One Correct Option  2x  1  is sin −1  x→ 0 x 1 + x 2  (a) 2 (b) ∞ (c) Does not exist (d) None of these

Ex 1. The value of lim

Sol. Using sin −1

 2x  = 2 tan −1 x, we have  2  ( ) x 1 +  

Given limit = lim

x→ 0

= 2 lim

x→ 0

2 tan −1 x x

tan −1 x = 2 (1) = 2 x

  tan −1 x = 1 Q xlim x  →0 

Ex 3. lim sin || | x| − 2| − 3| is equal to x→1

(a) sin 2 (c) 0

Sol. sin ||| x | − 2| − 3| is continuous function. So, its limit will same as that of its value. sin |||1| − 2 | − 3| = sin |1 − 3| = sin 2 Hence, (a) is the correct answer.

   1/ 2 1/ 2 − 1  a + x   Ex 4. lim   1/ 4 x→ a   a − x 1/ 4  

Hence, (a) is the correct answer.

 1  x tan  2  + 3 | x| 2 + 7  πx 

−

5

Ex 2. lim

| x| 3 + 7 | x| + 8

x→ − ∞

(a) −

x→ − ∞

 1  x tan  2  + 3 | x | 2 + 7  πx  |x| + 7 |x| + 8 3

 1  x 5 tan  2  + 3x 2 + 7  πx  − x3 − 7 x + 8 [Q x < 0 ⇒ | x | = − x ]

7  1  3 x 2 tan  2  + + 3  πx  x x = lim 7 8 x→ − ∞ −1− 2 + 3 x x [dividing numerator and denominator by x 3 ]

= lim

x→ − ∞

 1  tan  2  1  (πx )  + 3 + 7 ⋅ 1 x x3 π (πx 2 ) 7 8 −1− 2 + 3 x x

1 ⋅1+ 0 + 0 1 π =− = π −1− 0 + 0 tan θ 1   Q x → − ∞ ⇒ 2 → 0 and lim =1   θ→ 0 θ πx

1000

 3/ 4  −a 

− 2

log 4 a

−1

  

8

(b) a 3/ 4 (d) None of these

Sol. Simplifying the expression in brackets by setting

5

= lim

x 3/ 4 − a 1/ 4 x 1/ 2 + a 1/ 2 x 1/ 4

is equal to (a) a (c) a 2

(b) 0 (c) ∞ (d) Does not exist

x→ − ∞

2 ( ax )1/ 4

is equal to

1 π

Sol. We have, lim

(b) sin 1 (d) Does not exist

Hence, (a) is the correct answer.

a1 / 4 = b and x1 / 4 = y, the function whose limit is required can be written as  2 2   b + y      b − y  

−1

−1  1  log 4 b 4 2by  2  −2 − 3  y − by2 + b2 y − b3    

−1    2by  b− y  b − − =  2   2 2 2 ( y − b) ( y + b )   b + y 

 1  =   b −

 y 

−1

8

  − b = y8 = x 2 

So, the required limit as x → a is a2 . Hence, (c) is the correct answer.

Ex 5. The value of lim

x→ 0

1 − cos x 2 1 − cos x

is

1 2 (b) 2 (c) 2 (d) None of the above (a)

Sol. The required limit = lim

x→ 0

(1 − cos x 2 ) (1 + cos x 2 ) (1 − cos x ) 1 + cos x 2

8

8

= lim

1

 sin x / 2  2 x2  

x→ 0

2

1 + cos x 2

=

1 = 2 1 2× × 2 4

Hence, (c) is the correct answer.

x −2 + x − 2

Ex 6. lim

x→ 2

|x + π| is equal to x→ − π sin x (a) −1 (b) 1 (c) π (d) Does not exist

Ex 9. lim

Sol. lim

is equal to

x2 − 4

1 2 (c) 2

x→ − π

(b) 1

(a)

(d) None of these

1 x − 2  +  x+2 x 2 − 4   1 x−2 = + lim ⋅ 2 x→ 2 x + 2  

Sol. lim  x→ 2

=

1 + lim 2 x→ 2

1 x+

2

⋅

So, limit does not exist.

1 (x + 2) (x − 2)

Hence, (d) is the correct answer.

 log (1 + x ) x − 1 Ex 10. lim  e 2 +  is equal to x→ 0  x  x

x−2 1 = x+2 2

Hence, (a) is the correct answer.

(a)

 1 1 Ex 7. lim  −  is equal to h→ 0 h⋅3 8 + h 2h   (a)

1 2

Sol. lim

h→ 0

(b) −

4 2

(c) −

(d) −

1 48

2−38+ h

h→ 0

8 − (8 + h) 2h ⋅ 3 8 + h {82 / 3 + 81 / 3 ⋅ 8 + h)1 / 3 + (8 + h)2 / 3 }

1 48 Hence, (d) is the correct answer.

x − 2a + x − 2a

Ex 8. The value of lim

x→ 2a

x 2 − 4a 2 1 (b) 2 a

1

a a (c) 2

x→ 0

is

x→ 0

 x − 4a 

1 2 a

x − 2a  x 2 − 4 a2 

+ lim

x→ 2 a

x − 2a ( x+

2a ) (x + 2a)

1 2 a Hence, (b) is the correct answer. =

2 (a) 5 3 (c) 2 Sol. lim

x→ 0

=

1 2 a

+0

1 − cos 3 x is x sin x cos x 3 (b) 5 3 (d) 4

1 − cos3 x x sin x cos x

= lim

  1 x − 2a = lim  +  x→ 2 a − + + ( ) ( ) 2 2 2 x a x a x a   x − 2a 1 = + lim 2 a x→ 2 a ( x − 2a ) ( x + 2a ) (x + 2a) =

x2 1 2 1 3   2  x − x + x − ... + x − x   1 2 3 = lim = 2 x→ 0 2 x Hence, (a) is the correct answer.

Ex 11. The value of lim

(d) 2 a

 x − 2a + Sol. lim  2 2 x→ 2 a

loge (1 + x ) + x 2 − x

x→ 0

=−

(a)

1 2

(c) 1 (d) None of the above Sol. lim

2h ⋅ 3 8 + h

= lim

1 2

(b) −

16 3

|x + π | |t| [put x + π = t] = lim t → 0 sin (t − π ) sin x |t| = lim t → 0 − sin t t RHL = lim = − 1 [Q | t | = + t , t > 0 ] t → 0 + − sin t −t LHL = lim [Q | t | = − t , t < 0 ] =1 t → 0 − − sin t

18 Introduction to Limits and Derivatives

 sin x 2   2   x 

(1 − cos x ) (1 + cos x + cos2 x ) x sin x cos x

 x 2 sin 2    2 (1 + cos x + cos2 x ) × = lim x→ 0  x  x cos x x ⋅ 2 sin   cos    2  2  x sin    2 1 + cos x + cos2 x = lim × x→ 0  x  x cos   cos x 2   2  2 1 3 ×3= 2 2 Hence, (c) is the correct answer.

=

1001

Objective Mathematics Vol. 1

18

Ex 12. The value of   x +1 −1  x   lim x  tan −1    is  − tan  x→ ∞   x + 2   x + 2 1 1 (a) 1 (b) − 1 (c) (d) − 2 2 

 x + 1



x 

−1 Sol. lim x tan −1   − tan   x→ ∞    x + 2  + 2 x 

x  x+1  −   2 2 x + x +  = lim x tan −1  x→ ∞ 1 + x + 1 ⋅ x     x + 2 x + 2   x+2  = lim x tan −1  2 x→ ∞  2x + 5x + 4   −1   x+2  tan   2   2x + 5x + 4   x (x + 2) = lim   × 2 x+2 x→ ∞  2x + 5x + 4    2x 2 + 5x + 4   1 1 =1× = 2 2 Hence, (c) is the correct answer.

Ex 13. f (x ) = 3x 10 − 7x 8 + 5x 6 − 21x 3 + 3x 2 − 7, then f (1 − h) − f (1) the value of lim is h→ 0 h 3 + 3h 50 3 (c) 13

22 3 (d) None of these

(a)

Sol. lim

(b)

f (1 − h) − f (1)

h→ 0

h3 + 3h

= lim

h→ 0

f (1 − h) − f (1) − 1 ⋅ 2 −h h +3

 1 53 = f ′ (1)  −  =  3 3 Hence, (d) is the correct answer.

Ex 14. If lim

4 + sin 2x + A sin x + B cos x

exists, then x2 the values of A and B are (a) − 2 and − 4 (b) − 4 and − 2 (c) − 3 and − 2 (d) None of these x→ 0

Sol. Since, the given limit exists and denominator

approaches zero as x → 0. Numerator must approach zero as x → 0 which is possible, if 4 + B = 0 which is obtained by substituting zero in numerator in place of x. ∴ B=−4 4 + sin 2x + A sin x − 4 cos x 0  Now, lim form  0  x→ 0 x2 2 cos 2x + A cos x + 4 sin x = lim x→ 0 2x [using L’ Hospital’s rule] Again, D r → 0, while N r = 2 + A

1002

2 cos 2x − 2 cos x + 4 sin x 0  form  0  x→ 0 2x − 4 sin 2x + 2 sin x + 4 cos x 4 = lim = =2 x→ 0 2 2 ∴ A = − 2 and B = − 4 Hence, (a) is the correct answer. Now, lim

Hence, for above limit to exist A + 2 = 0 ⇒ A=−2

Ex 15. Let f (x ) = x 2 − 1, 0 < x < 2 and 2x + 3, 2 ≤ x < 3. The quadratic equation whose roots are lim f ( x ) and lim f ( x ), is x→ 2− 0

x→ 2+ 0

(a) x 2 − 6x + 9 = 0

(b) x 2 − 10x + 21 = 0 (c) x 2 − 14x + 49 = 0 (d) None of the above Sol. lim

x→ 2 − 0

f (x ) = lim

x→ 2 − 0

f (x ) = lim

lim

x→ 2 + 0

x→ 2 + 0

(x 2 − 1) = 22 − 1 = 3

(2x + 3) = 2 × 2 + 3 = 7

So, required quadratic equation is x 2 − 10x + 21 = 0.

Hence, (b) is the correct answer.

Ex 16. lim

a n + bn

a n − bn (a) −1 (c) 0 n→ ∞

Sol. lim

n→ ∞

 b 1+    a

n

 b 1−    a

n

, where a > b >1, is equal to (b) 1 (d) None of these = 1, because 0 < b / a < 1

implies (b / a)n → 0 as n → ∞ Hence, (b) is the correct answer.

Ex 17. The limiting value of (cos x )1/ sin x as x → 0 is (a) 1 (c) 0

(b) e (d) None of these

Sol. Put cos x = 1 + y as ; x → 0 ; y → 0 lim (cos x )1 / sin

x→ 0

lim

= ex→ 0

x

= lim [(1 + y)1 / y ]y / sin

x

y→ 0

cos x − 1 sin x

x  lim  − tan   2

= e x→ 0

= e0 = 1

Hence, (a) is the correct answer.

 x 2 − 1 Ex 18. The value of lim  2  x → ∞  x + 1

is

(b) e −1 (d) e −3

(a) 1 (c) e −2  x 2 + 1 − 2  Sol. lim  2 x→ ∞  x + 1 

x2

x2

 −2   = lim 1 + 2 x→ ∞  x + 1 x 2 +1 − 2 x 2 ⋅

x2

 − 2  − 2 x 2 +1 = e− 2 = lim 1 + 2  x→ ∞ 1 x +   Hence, (c) is the correct answer.

Then,

(a) 0 (c) g ( x )

lim f ( x )

x→ 2

(a) exists only when k = 1 (b) exists for every real k (c) exists for every real k, except k = 1 (d) Does not exist x→ 2

∴

lim

n→ ∞

= lim k (− 2 − x ) = − 4 k x→ 2 +

lim f (x ) = lim

x→ 2 −

x→ 2 −

k (x 2 − 4 ) 2−x

x→ 2 −

Hence, (b) is the correct answer.

Ex 20. In a circle of radius r, an isosceles ∆ABC is inscribed with AB = AC . If ∆ABC has perimeter p = 2 [ 2hr − h 2 + 2hr ] and area A = h 2hr − h 2 , where h is the altitude from A A to BC, then lim 3 is h→ 0+ p

(c)

(b)

1 64 r

Sol. lim+ h→ 0

A p3

h 2hr − h2

h→ 0 +

h⋅ h

= lim h→ 0

+

= lim h→ 0

=

8 [ 2hr − h2 +

2hr ] 3 2r ]

3

2r − h

+

8 [ 2r − h +

2r 8 (2 2r )3

=

2r ]

Ex 21. Let f (x ) = x (− 1)[1/ x ] , x ≠ 0, where [x ] denotes the greatest integer less than or equal to x, then lim f ( x ) x→ 0

(b) is equal to 2 (d) is equal to − 1

 1

Sol. Since,   is an integer, so (−1)[1 / x ] is either1 or −1. x So, | f (x )| ≤ x and thus, lim f (x ) = 0. Hence, (c) is the correct answer.

Let f (x ) = a0 x n + a1 x n − 1 + K + an − 1 x + an , a0 ≠ 0

or

f (x ) = x n + 1

Then, ⇒

f (2) > 1 f (x ) = x n + 1 lim f (x ) = lim (x n + 1) = 2

x→ 1

x→ 1

Hence, (a) is the correct answer.

Ex 24. The value of lim[sin −1 x ] is

Hence, (b) is the correct answer.

x→ 0

(b) 1 (d) None of these

 1  1 Sol. f (x ) ⋅ f   = f (x ) + f   , f (2) > 1  x  x

x→1

3

1 128r

(a) does not exist (c) is equal to 0

x→1

(a) 2 (c) − 1

∴

2r − h

8 h ⋅ h [ 2r − h +

Ex 23. If f (x ) is a polynomial satisfying  1  1 f ( x ) ⋅ f   = f ( x ) + f   and f (2) > 1, then  x  x lim f ( x ) is

Then, by comparing the coefficients of like powers, we get an = 1, a0 = 1, a1 = a2 = K = an − 1 = 0 ∴ f (x ) = − x n + 1

1 128 r

(d) None of these = lim

f (x ) enx + g (x ) enx + 1 f (x ) g (x ) + nx = lim n→ ∞  1 +1 e 1 +  nx  e  f (x ) finite = f (x ) = + 1+ 0 ∞

18

Hence, (b) is the correct answer.

= lim k (− 2 − x ) = − 4 k

(a) 128 r

(b) f ( x ) (d) None of these

Sol. Given, x > 0 and g (x ) is bounded function.

k (x 2 − 4 ) 2−x

Sol. lim+ f (x ) = lim+ x→ 2

Ex 22. If x > 0 and g is a bounded function, then f ( x ) e nx + g ( x ) is lim n→ ∞ e nx +1

Introduction to Limits and Derivatives

x 2 , when x is an integer  Ex 19. If f (x ) =  k (x 2 − 4) , otherwise   2−x

(a) Does not exist

(b) 1 π (d) 2

(c) 0 Sol. lim[sin −1 x ] = lim [ t ] = 1 x→ 1

t → π /2

[put sin −1 x = t]

Hence, (b) is the correct answer.

Ex 25. The value of lim [tan −1 x ] is x→ − ∞

(a) − 2 (b) − 1 (c) ∞ (d) None of the above Sol. lim [tan −1 x ] = x→ − ∞

lim [ t ] = − 2

t → −π /2

[put tan −1 x = t]

Hence, (a) is the correct answer.

1003

Objective Mathematics Vol. 1

18

Ex 26. The value of lim [sin sin −1 x ] is x→1

n

Ex 32. lim Π

−

(a) 0 π (c) 2

(b) Does not exist (d) None of these

Sol. lim− [sin sin −1 x ] = lim− [ x ] = 0 x→ 1

x→ 1

−

here, x → 1 means x → 1 , since    −1 sin sin x is not defined for x > 1 Hence, (a) is the correct answer.

n→ ∞ r =3

r3 − 8 r3 + 8

is equal to

2 7 (c) 1

7 2 (d) Does not exist

(a)

(b)

 33 − 8   4 3 − 8   n3 − 8      3  K ∞  33 + 8  4 3 + 8  n + 8

Sol. lim  n→

 3 − 2 32 + 4 + 2(3)  4 − 2 4 2 + 4 + 2(4 )   ⋅ = lim  ⋅ n → ∞  3 + 2 32 + 4 − 2(3)   4 + 2 4 2 + 4 − 2(4 )   n − 2 n2 + 4 + 2n  … ⋅  n + 2 n2 + 4 − 2n 

Ex 27. The value of lim [sin −1sin x ] is x→ π/2

π 2 (d) None of these

(a) 1

(b)

(c) 0

3−2 4 −2 5−2 n − 2 = lim  ⋅ ⋅ K  n → ∞ 3 + 2 4 + 2 5 + 2 n + 2  32 + 4 + 2(3) 4 2 + 4 + 2(4 ) n2 + 4 + 2n  2  K 2 ⋅ 2 n + 4 − 2n   3 + 4 − 2(3) 4 + 4 − 2(4 )

Sol. lim [sin −1 sin x ] = 1 x → π /2

Hence, (a) is the correct answer.

 sin x  Ex 28. The value of lim   is + x→ 0  x  (a) 1 (c) Does not exist

(b) 0 (d) None of these

 sin x  = lim [ t ] = 0 Sol. lim+   t→1− x→ 0  x 

sin x   put =t  x 

Hence, (b) is the correct answer.

 sin x  Ex 29. The value of lim   is − x→ 0  x  (a) 1 (c) Does not exist

(b) 0 (d) None of these x   put sin = t   x

 sin x  = lim [ t ] = 0 Sol. lim−  x→ 0  x   t→1− Hence, (b) is the correct answer.

 sin x  Ex 30. The value of lim  is x→ 0  x   (a) 0 (b) 1 (c) Does not exist (d) None of these  sin x  = lim [ t ] = 0 Sol. lim  x→ 0  x   t→1−

x   put sin = t   x

Ex 33. lim | x|[cos x ] is equal to x→ 0

(a) 1 (b) 0 (c) Does not exist (d) None of the above Sol. lim + (x )0 = 1, x→ 0

lim (− x )0 = lim 1 = 1

x → 0−

4 + 3a n , n ≥ 1 and if a n has 3 + 2a n a limit l as n → ∞, then (a) l = − 2 (b) l = 2 (c) l = 2 (d) None of the above

Ex 34. If a1 = 1 and a n + 1 =

2

Ex 31. lim

a2 =

is equal to

x2 −9 (a) ∞ (b) 0 (c) Does not exist (d) None of the above

x→ 3

Sol. lim− x→ 3

[ x ]2 − 9 x2 − 9

= ∞ and lim

x→ 3 +

∴

x2 − 9

So, limit does not exist.

1004

(4 + 3) 7 = >1 (3 + 2) 5

∴ Let

[ x ]2 − 9

Hence, (c) is the correct answer.

=0

x → 0−

Hence, (a) is the correct answer.

Sol. a1 = 1

Hence, (a) is the correct answer.

[x ] − 9

 1⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7...  19 ⋅ 28 ⋅ 39 ⋅ 52 ⋅ 63...  =    5 ⋅ 6 ⋅ 7 ⋅ 8 ...   7 ⋅ 12 ⋅ 19 ⋅ 28 ⋅ 39 ⋅ 52 ... 1⋅ 2 ⋅ 3 ⋅ 4 2 = = 7 ⋅ 12 7 Hence, (a) is the correct answer.

= =

an + 2

a2 > a1 an + 1 > an 4 + 3an + 1 4 + 3an − an + 1 = − 3 + 2an + 1 3 + 2an

(4 + 3an + 1 )(3 + 2an ) − (4 + 3an )(3 + 2an + 1 ) (3 + 2an + 1 )(3 + 2an ) an + 1 − an (3 + 2an + 1 )(3 + 2an )

>0

[Q an + 1 > an ]

n→ ∞

∴

n→ ∞

l=

4 + 3l 3 + 2l

⇒

3l + 2l 2 = 4 + 3l

⇒ l2 = 2 l= 2

⇒

[Ql > 0]

Hence, (b) is the correct answer.

Ex 35. If [ ] denotes the greatest integer function, then [ x ] + [2x ] + K + [ nx ] is lim n→ ∞ n2 (a) 0

(b) x

(c)

x 2

(d)

x2 2

Sol. nx − 1 < [ nx ] ≤ nx

Putting n = 1, 2, 3, ..., n and adding them, we get

xΣn − n < Σ[ nx ] ≤ xΣn Σn 1 Σ[ nx ] Σn ...(i) ≤ x⋅ 2 x⋅ 2 − < ∴ 2 n n n n Σn 1 x  Σn 1  Now, lim x ⋅ 2 −  = x ⋅ lim 2 − lim = n → ∞ n→ ∞ n n→ ∞ n 2 n n xΣn x and lim = n→ ∞ n2 2 As the two limits are equal. By Sandwich theorem, we have Σ[ nx ] x lim = n → ∞ n2 2 Hence, (c) is the correct answer. x

∫ |t − 1| dt is equal to Ex 36. lim 1 x→ 1

sin( x − 1)

(a) 0 (c) − 1 1+h

Sol. lim

∫1

(b) 1 (d) None of these | t − 1| dt

sin (1 + h − 1) d 1+h ∫1 | t − 1| dt dh [by Leibnitz’s rule] = lim d h→ 0 (sin h) dh d (1 + h) |1 + h − 1|⋅ |h | dh = lim = lim =0 h→ 0 h → 0 cos h cos h

h→ 0

Hence, (a) is the correct answer.

Ex 37. The value of 2 ( x ) 1/ 2 + 3 ( x ) 1/ 3 + 4 ( x ) 1/ 4 + K + n ( x ) 1/ n lim x → ∞ (2x − 3) 1/ 2 + (2x − 3) 1/ 3 + K + (2x − 3) 1/ n is (a) 2 1 (c) 3

(b) 2 (d) 0

18

Sol. Given, lim

2x1 / 2 + 3x1 / 3 + K + nx1 / n

+ (2x − 3)1 / 3 + K + (2x − 3)1 / n  1   1   1  2  1/ 2  + 3  1/ 3  + K + n  1/ n  h  h  h  = lim 1 h→ 0  1  (2 − 3h)1 / 2 + 1 / 3 (2 − 3h)1 / 3  h1 / 2  h   1  + K + 1 / n (2 − 3h)1 / n    h 1 [on putting x = , as x → ∞, h → 0] h

x→ ∞

(2x − 3)

1/ 2

1 1  −   

1 1  −   

1 1  −   

2 + 3h 2 3 + 4 h 2 4 + K + nh 2 n = lim 1 1 h→ 0    −      (2 − 3h)1 / 2 + h 2 3 (2 − 3h)1 / 3   1 1 1   −     + K + h 2 n (2 − 3h) n  2+ 0+ 0+ 0+K = 2 = 1/ 2 2 + 0+ 0+K

Introduction to Limits and Derivatives

Now, an + 2 > an + 1 So, the sequence of the values of an is increasing and since a1 = 1, an > 0 ∀ n. Let lim an = l = lim an + 1

Hence, (a) is the correct answer.

Ex 38. The derivative of x 2/ 3 by using first principle is 2 −1 / 3 x 3 2 (c) x1/ 3 3

(a)

(b) −

2 −1 / 3 x 3

(d) None of these

Sol. Let f (x ) = x 2 / 3 . We know by first principle, f (x + h) − f (x ) h (x + h)2 / 3 − x 2 / 3 ∴ f ′ (x ) = lim h→ 0 h  2 / 3 2 −1 / 3  h + K − x 2 / 3 + x x   2 3 = lim = 1/ 3 h→ 0 h 3x Hence, (a) is the correct answer. f ′ (x ) = lim

h→ 0

Ex 39. The derivative of cos (x 2 + 1) by using first principle is (a) 2x sin( x 2 + 1)

(b) − 2x sin( x 2 + 1)

(c) − 2x sin( x 2 − 1)

(d) None of these

Sol. Let f (x ) = cos(x 2 + 1) We know by first principle, f (x + h) − f (x ) f ′ (x ) = lim h→ 0 h cos[(x + h)2 + 1] − cos (x 2 + 1) ⇒ f ′ (x ) = lim h→ 0 h [Q f (x ) = cos (x 2 + 1)]  (x + h)2 + 1 + x 2 + 1 1 = lim − 2 sin   h→ 0 h  2    2 2  (x + h) + 1 − (x + 1)  × sin   2   C +D C − D  Q cosC − cos D = − 2 sin sin  2 2 

1005

Objective Mathematics Vol. 1

18

= lim

− 2 sin

h→ 0

− 2 sin = lim

h→ 0

(x + h)2 + x 2 + 2 (x + h)2 − x 2 sin 2 2 h (x + h)2 + x 2 + 2  (x + h)2 − x 2  sin  2 2   (x + h)2 − x 2 h× 2  (x + h)2 − x 2  ×  2  

(x + h)2 + x 2 + 2 (x + h + x )(x + h − x ) −2 sin × 2 2 = lim h→ 0 h  (x + h)2 + x 2 + 2  h − 2 sin   (2x + h) 2 2   = lim h→ 0 h  2x 2 + 2 1  × 2x × = − 2 sin  2 2  

Ex 40. The derivative of an odd function is always (a) an even function (c) Does not exist

(b) an odd function (d) None of these

Sol. Let f be an odd function. Then, f (− x ) = − f (x ) ∴ f ′ (− x ) ⋅ (− 1) = − f ′ (x ) ⇒ f ′ (− x ) = f ′ (x ) ∴ f ′ (x ) is an even function. Hence, (a) is the correct answer.

Ex 41. If f (x ) = x m , m being a non-negative integer, then the value of m for which f ′ (α + β) = f ′ (α ) + f ′ (β), for all α, β > 0, is (a) 1 (c) − 2

(b) 2 (d) None of these

Sol. We have, f ′ (α + β) = f ′ (α ) + f ′ (β)

(α + β )m − 1 = α m − 1 + β m − 1

⇒

Since, for m > 2, the above equality is not valid. ∴We must have, m = 2. Also, for m = 0, f ′ (x ) = 0 for all x. So, the equality is trivially true. Hence, (b) is the correct answer.

= − 2x sin(x 2 + 1) Hence, (b) is the correct answer.

Type 2. More than One Correct Option π x +1 2 Ex 42. If lim 4x  − tan −1  = y + 4 y + 5, x→ ∞ x + 2 4 then y can be equal to (b) − 1

(a) 1

(c) − 4

(d) − 3

−1 

1  tan    2x + 3 1 × =2 Sol. lim 4 x x→ ∞  1  2x + 3    2x + 3 ⇒ ⇒

y2 + 4 y + 5 = 2 y = − 1, − 3

Hence, (b) and (d) are the correct answers.

Ex 43. lim

x→ 0

1 − cos ( x 2 ) x 3 ( 4 x − 1)

is equal to

1  e2  ln   = ln 2 2  4

Hence, (b) and (d) are the correct answers.

Ex 44. If f (x ) = e[cot x ] , where [ y] represents the greatest integer less than or equal to y, then (a) (c)

lim

x → π / 2+

lim

x → π /2

−

f (x ) = 1

(b)

1 e

(d)

f (x ) =

1 e

lim

f (x ) =

lim

f (x ) = 1

x → π / 2+ x → π / 2−

Sol. f (x ) = e [cot x ], as cot x is negative in the II quadrant and cot

π = 0. 2 [cot x ] = − 1

1 (a) ln 2 2 (c) ln 4 2

f (x ) = e−1 =

1 e

(b) ln 2

∴

1  e2  (d) 1 − ln   2  4

π− , cot x is positive (being in I quadrant) and As x → 2 hence [cot x ] = 0 ∴ lim f (x ) = e0 = 1

 x2   x2  2 sin 2      2  2 1 x ⋅ = ln 4 = ln 2 ⋅ x Sol. lim x→ 0 2 (4 − 1) x 4 2

1006

1−

Also,

lim

x → π /2

+

x → π / 2−

Hence, (b) and (d) are the correct answers.

Directions (Ex. Nos. 45-47) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Ex 46. Statement I lim sec

x  =0   x 

−1  sin

x→ 0

 sin x  Statement II lim   =1 x→ 0  x   π  sin x sin x in  0,  , 

2

x

 sin x  −1  sin x  ∴ sec−1   is not defined and lim sec    x   x  x→ 0 does not exist.

Ex 45. Let a n = 299 . 2... 1 39, n ∈ N

18 Introduction to Limits and Derivatives

Type 3. Assertion and Reason

Hence, (d) is the correct answer.

n times

  Statement I  lim a n  = lim [ a n ], where [ ] n → ∞  n → ∞ denotes the greatest integer function.

Ex 47. Statement I lim

 sin  π sin 2  x2

x→ 0

Statement II lim a n = 3

x  2

=π

sin x =1 x→ 0 x

n→ ∞

Statement II lim

Sol. lim 2.9 = 3, as there is no real number between 2. 9 n→ ∞

and 3. So, 2. 9 = 3   ∴ lim an = [ 3 ] = 3, while [ an ] = 2, ∀ n ∈ N and n → ∞    lim [ an ] = 2 ≠ lim an n → ∞  n→ ∞ Hence, (d) is the correct answer.

x x   sin  π sin 2  π  sin 2    2 2 π × = and not π Sol. lim 2 x x→ 0 4  x π sin 2   ×4 2  2 Statement II is evidently true. Hence, (d) is the correct answer.

Type 4. Linked Comprehension Based Questions 1

Passage (Ex. Nos. 48-49) If lim f ( x ) = 1 and x→a

lim [f (x ) − 1] × g (x )

lim g( x ) = ∞, then lim {f ( x )} g (x ) = e x → a

.

x→a

x→a

 sin x  Ex 48. lim   x→ 0  x 

sin x x − sin x

−

is equal to

(b) e (c) e 1 (d) e

(b) −

L=e

 sin x  sin x − 1 lim   x − sin x x→ 0  x

1 e Hence, (a) is the correct answer. = e−1 =

=e

1 2

x − 1 + cos x = lim 1 − Sol. lim x→ 0 x→ 0 x

sin x sin x x = lim =∞ Sol. lim sin x x → 0 x − sin x x→ 0 1− x ∴

1

(a) e 2

1 e (d) − e

1 e (c) e (a)

 x − 1 + cos x  x Ex 49. lim   is equal to  x→ 0  x

2 sin 2 x

x 2 =1

 1  x − 1 + cos x lim  − 1 ×  x x x→ 0 

− sin x x x→ 0 lim

L=e

 2x  − 2 sin  2 lim   2 x→ 0  x   

=e

= e−2 / 4 = e−1 / 2

Hence, (b) is the correct answer.

1007

Objective Mathematics Vol. 1

18 Type 5. Match the Columns Ex 50. Match the limits of the functions given in Column I with their corresponding values given in Column II. Column II

Column I A.

sin ( π − x ) lim x → π π ( π − x)

B.

cos x lim x→ 0 π − x

C.

cos 2 x − 1 lim x → 0 cos x − 1

D.

lim

x→ 0

Sol. A. Given, lim

x→ π

p.

4

q.

1 π

r.

a+1 b

sin(π − x ) π (π − x )

a + cos 0 a + 1 = b×1 b A → q; B → q; C → p; D → r

Column II p. − 12 x −5 + 36 x −10

B. x 5 ( 3 − 6 x −9 )

q.

−2 ( x + 1)2

C. x −4 ( 3 − 4 x −5 ) D.

−

x( 3 x − 2 ) ( 3 x − 1)2

r. − 15 x −4 − 6 x −3 s. 15 x 4 + 24 x −5

2 x2 − x + 1 3x − 1

t. 24 x 3 − 2 x2

Sol. A. Let y = x −3 (5 + 3x )

[multiplying and dividing by x 2 and then multiplying by 4 / 4 in the numerator] x2 4× 2 sin x 4 = lim × x x → 0 x2 sin 2 2 2  x  2    sin x  = lim   ×  2  × 4=1×1× 4 = 4 x→ 0  x   sin x   2 ax + x cos x D. Given, lim x→ 0 b sin x ax x cos x + x [dividing each term by x] = lim x b sin x x→ 0 x a + cos x = lim x → 0  sin x  b   x 

1008

Column I A. x −3 ( 5 + 3 x )

ax + x cos x b sin x

Let π − x = h, as x → π, then h → 0 1 sin h sin(π − x ) sin h = lim = lim × ∴ lim x → π π (π − x ) h → 0 πh h→ 0 π h 1 1  sin h  = ×1= Q lim =1  π π  h → 0 h cos x B. Given, lim x→ 0π − x cos 0 1 Put the limit directly, we get = π−0 π cos 2x − 1 1 − cos 2x C. Given, lim = lim x → 0 cos x − 1 x → 0 1 − cos x 2 sin 2 x = lim x x→ 0 2 sin 2 2 x  2 Q 1 − cos 2x = 2 sin x and 1 − cos x 2 sin 2  2  cos 2x − 1 = lim x → 0 cos x − 1

=

Ex 51. Match the functions given in Column I with their corresponding derivatives given in Column II.

sin x   Q lim =1  x → 0 x 

y = 5x −3 + 3x −2

⇒

On differentiating y w.r.t. x, we get dy = 5(− 3)x − 3 − 1 + 3(− 2)x − 2 − 1 dx = − 15x −4 − 6x −3 B. Let y = x 5 (3 − 6x −9 ) = 3x 5 − 6x 5 − 9 = 3x 5 − 6x −4 On differentiating y w.r.t. x, we get dy = 3 × 5x 5 − 1 − 6(− 4 )x − 4 − 1 dx = 15x 4 + 24 x −5 C. Let y = x −4 (3 − 4 x −5 ) ⇒

y = 3x −4 − 4 x − 4 − 5

⇒

y = 3x −4 − 4 x −9

On differentiating y w.r.t. x, we get dy = 3(−4 )x −4 −1 − 4 (−9)x −9 −1 = − 12x −5 + 36x −10 dx 2 x2 D. Let y = − x + 1 3x − 1 On differentiating y w.r.t. x, we get d d (x + 1) (2) − 2 (x + 1) dy dx dx = dx (x + 1)2 d d (3x − 1) (x 2 ) − x 2 (3x − 1) dx dx − (3x − 1)2 ⇒

dy (x + 1) × 0 − 2 × 1 (3x − 1)(2x ) − x 2 (3) − = dx (x + 1)2 (3x − 1)2 =−

2 (x + 1)2

−

(6x 2 − 2x − 3x 2 ) (3x − 1)2

2 dy (3x 2 − 2x ) =− − 2 dx (x + 1) (3x − 1)2 dy −2 x (3x − 2) − = ⇒ 2 dx (x + 1) (3x − 1)2

⇒

A → r; B → s; C → p; D → q

Ex 52. The value of lim

cot −1 ( x − a log a x )

x→ ∞

sec

−1

x

( a log x a )

( a > 1)

is _________.  log cot −1  aa 

x  x  ; lim  loga x  → 0 Sol. (1) Let I = lim   x→ ∞  ax  x → ∞  x a  sec−1    loga x  and

[using L’ Hospital’s rule]

x→ ∞

∴

Ex 54. lim [1 + (arc cos x )1 − x ]2 has the value _____. x → 1−

Sol. (1) I = lim (cos−1 x )1 − x

ln I = lim (1 − x )ln(cos−1 x )

θ→ 0

ln θ 1 (1 − cosθ ) (1 − cosθ )2 0 sin θ ⋅ θ

= − lim

cot −1 ( x + 1 − x )  2x + 1 sec −1    x − 1 

x

θ→

is equal to _______.

  

(1 − cosθ )2 =0 sin θ 0 θ2 θ

= − lim θ→

∴

cot −1 ( x + 1 − x ) x→ ∞  2x + 1 x  sec−1     x −1    ⇒ lim ( x + 1 − x ) = 0

I =1

Ex 55. If f (x ) = x cos x, then the value of f ′ (2π ) is _______. Sol. (1) ⇒

x→ ∞

x

cot −1 (0) =

x = cosθ

Put θ→ 0

Sol. (1) Let I = lim

⇒

[ 00 from ]

x→1

But lim (1 − cosθ )ln θ = lim

π I = 2 =1 π 2

Ex 53. lim

⇒

x→1

 ax  →∞   loga x 

∴

π sec (∞ ) = 2 I =1 −1

Introduction to Limits and Derivatives

18

Type 6. Single Integer Answer Type Questions

 2x + 1 π ⇒ lim   =∞ x → ∞ x − 1  2

∴

f (x ) = x cos x f ′ (x ) = x (− sin x ) + cos x ⋅ 1 = − x sin x + cos x f ′ (2π ) = − 2π sin 2π + cos 2π = 0+ 1 =1

1009

Target Exercises Type 1. Only One Correct Option x

1. lim

x→ 0

4− x − x

  x−3 9. lim   is equal to x→ 3  x − 2 − 4 − x 

is equal to

(a) 0 (c) − 1

(b) 1 (d) Does not exist

2 2  1 1   2. lim  3x +  −  2x −   is equal to x→ 0   x x  

(a) 5 (c) 10

(b) 2 (d) 0

3. The value of lim

x 7 − 2x 5 + 1

x→1

Ta rg e t E x e rc is e s

(a) 0 (c) − 1 3

4. lim

x−2

x→ 2

3

1 (b) − 12 1 (d) − 6

5. lim

x2 − 2 3 x + 1 ( x − 1)

x→1

2

1 9 1 (c) 3

(b)

x − 2a + x − 2a

x → 2a

(a)

x 2 − 4a 2

1 a

1 (c) − a

x−2

x→ 2

1

8. lim

x→ −1

(a) 1 (c) 0

1010

1 2 a 1 2 a

x+1 6x + 3 + 3x 2

2− 2+ x

11. lim

x→ 2 3

(c)

is equal to

2−3 4−x

3 2

(b) −

3 2

(d) − a

(c) a

4/ 3

(d) −

3/ 4

3 2

4/ 3

2

3/ 4

3

12. lim (1 + x ) (1 + x ) (1 + x ) ... (1 + x 2n ), | x| < 1 , is 2

4

n→ ∞

equal to 1 1− x (d) x − 1

(b)

n→ ∞

1 (a) 3

is equal to

is equal to (b) − 1 (d) None of these

(b) −

1 3

(c)

2 3

(d) −

 n2 + 1 + n   is equal to 14. lim  n → ∞ 4 3 4  + − n n n   (a) 0 (c) − 1

15. lim 1 (b) 3 (d) 3

8 3 (c) 8 3

(b) − a

, is equal to

13. The value of lim [ 3 n 2 − n 3 + n] is

(d) −

1+ 2 + x − 3

7. lim (a)

1 6

is equal to (b)

a+x− a−x

(a)

(d) None of these

6. lim

a 2 − ax + x 2 − a 2 + ax + x 2

1 x −1 (c) 1 − x

is equal to

(a)

x→ 0

(a)

is equal to

1 (a) 12 1 (c) 6

10. lim f ( x ), where

(a) a

(b) 1 (d) None of these

10 − x − 2

(d) None of these

f (x ) =

is

x 3 − 3x 2 + 2

(b) − 1

(a) 1 1 (c) 2

x→∞

(b) 1 (d) ∞

x − sin x x + cos 2 x

is equal to

(a) − 1 (c) 0

16. lim (sin x→∞

(b) 1 (d) None of these

x + 1 − sin

(a) 1 (b) − 1 (c) 0 (d) None of the above

x ) is equal to

2 3

h→ 0

(a)

h

cos x 2 x

(b) sin

x→0

x 2 sin 3x

6 5

(b)

27. lim

is equal to (c)

x

(1 − cos 2x ) sin 5x

18. lim (a)

x

1 (d) cos x 2 sin x

(c)

10 3

(d) −

3 10

(b)

1 2

1 3

(c)

(d) 3

 π  π  2  3 sin  + h − cos  + h  6  6  20. lim  is equal to h→ 0 3h( 3 cos h − sin h ) (a)

4 3

(b) −

4 3

(c)

x→∞

(a) 0 (c) − 1

22. lim

n→ ∞

2 3

(d)

3 4

1 + n sin 2 nx

is equal to

(a) − 1 (c) 3

2a π

(b) −

30. lim

 x + 1, x > 0 24. If f ( x ) =   2 − x, x ≤ 0

x→

π 4

(a) 5 2 (c) 2

32. lim

x→∞

sin 2x + a sin x 3

( 3x − 1) ( 2x + 5) is equal to ( x − 3) ( 3x + 7) (b) 2 (d) None of these

1 + x 4 − (1 + x 2 ) x2

2x 3 − 4x + 7 3x 3 + 5x 2 − 4

2 3

(c) −

(c) 2

(d) − 2

(b) − 1 (d) None of these

is finite, then the value of a and (c) 2, 1

(d) 2, − 1

(a)

1 2

36. lim

x→e

(a)

1 e

37. lim

x →1

(a)

1 2

is

is equal to (b)

7 4

3 2

(d) None of these

e x − (1 + x )

x→0

(b) − 2, − 1

4a π

(b) 0 (d) None of these

35. The value of lim

x the limit are given by

(a) − 2, 1

(b) 1 (d) None of these

(a) − 1 (c) 2

(a)

25. The value of  2n n+1 n n ( − 1) n  lim  2 cos −  is n → ∞ 2n − 1 2n − 1 1 − 2n n 2 + 1  

x→0

(d) −

(b) 3 2 (d) None of these

1 2 (c) 0

x→∞

Then, lim g[ f ( x )] is

26. If lim

4a π

31. lim cos [ π n 2 + n ], n ∈ I is equal to

34. lim

x 0 is equal to (c) − 1

(b) 0

2x − x 2 x x − 22

log 2 − 1 log 2 + 1

69. If lim

log 2 + 1 log 2 − 1

x n − sin x n

x→0

x − sin n x

(d) ∞

(c) 1

(d) − 1

equal to (b) 2 (d) None of these

 x  2 , where [⋅] denotes the greatest integer 70. lim π ln (sin x ) x→ 2

function, is equal to (a) Does not exist (c) 0

71. lim lim (1 + cos m→ ∞ n→ ∞

(a) − 2 (c) 0

(c) 0

2 π

(d) 1

27x − 9x − 3x + 1 2 − 1 + cos x

x→0

is

(b) 8 2 (log 3)2

2

(d) None of these

π − cos −1 x x+1

is equal to

1 2π 1 (c) 2 (a)

(b)

1 π

(d) None of these

77. The value of lim [ 3 ( n + 1) 2 − 3 ( n − 1) 2 ] is n→∞

(b) −1 (d) None of these

(a) 1 (c) 0

α is a repeated root of ax 2 + bx + c = 0, then tan ( ax 2 + bx + c ) is lim x→α (x − α )2 (b) b

x

79. lim

x→0

(a)

tan

−1

1 2

n ! πx ) is equal to (b) 1 (d) None of these

(c) c

(d) 0

(c) 0

(d) 1

is equal to 2x (b) ∞

cos [ x], x ≥ 0 80. Let f ( x ) =  , then the value of a, so that  x + a, x < 0 lim f ( x ) exists, where [ x] denotes the greatest integer h→ 0

function ≤ x, is equal to (a) 0

(b) − 1

(c) 2

(d) 1

81. If a = min {x 2 + 4x + 5, x ∈ R}and b = lim n

(b) 1 (d) −1 2m

(d) −

is equal to

75. The value of lim

(c) 2 2 (log 3)

π 2

(c) −

(b) 2

(a) a

is non-zero finite, then n may be

(a) 1 (c) 3

2 π

78. If

is equal to (b)

1 2

x→ −1

(b) − 1 (d) None of these

(log x ) 2

(a)

x

76. lim

, m, n ∈ N , n > m is equal to

(a) 2 (c) 0 x→∞

1+ x − 1

x→0

2

(a) 1

67. lim

(b)

74. lim

sin x − (sin x ) sin x 65. lim equals π x→ 1 − sin x + ln sin x

x→0

π 2

(a) 4 2 (log 3)2

(a) l exists but m does not (c) both l and m exist

66. lim

(a)

n

(b) e− 1 (d) None of these

(b) 1 (d) sin 1

 πx 73. lim (1 − x ) tan   is equal to x →1  2

1/ 2

x  is equal to n

(a) e1 (c) 1

64. Let lim

(a) 0 (c) Does not exist

18 Introduction to Limits and Derivatives

(b) − e

(a) 1

sin ([ x − 3]) 72. lim  , where [ ] represents greatest x→0  [ x − 3]  integer function, is equal to

Targ e t E x e rc is e s

x

1  61. The value of lim 1 + n  , n > 1is x→∞  x 

then the value of

1 − cos 2θ

θ→ 0

∑ a r ⋅ b n − r is

θ2

,

r=0

2n + 1 − 1 (a) 4 ⋅ 2n n+ 1 −1 2 (c) 3 ⋅ 2n

(b) 2n + 1 − 1 (d) None of these

1013

Objective Mathematics Vol. 1

18

82. lim

x→0

log (1 + x + x 2 ) + log (1 − x + x 2 ) is equal to sec x − cos x (b) − 1

(a) 1

(d) ∞

(c) 0

 x + 1 π is 83. The value of lim x tan − 1 − x→∞ x + 2 4   (a)

1 2

(b) −

1 2

84. If f ( x ) = x − [ x] , where [ x] denotes the greatest { f ( x )}2n − 1 integer ≤ x and g ( x ) = lim , then g ( x ) n → ∞ { f ( x )}2 n + 1 is equal to (a) 0 (c) − 1

(b) 1 (d) None of these

 1 2 n  85. lim  + + ... +  is equal to 2 n→ ∞ 1− n2 1− n 1− n2   1 (b) − 2

(a) 0 1 (c) 2

(d) None of these

Ta rg e t E x e rc is e s

86. lim (1 + cos πx ) cot 2 πx is equal to (b) − 1

(a) 1 1 (c) 2

87. lim

x →1 y→ 0

(d) None of these

y

3

x − y −1 3

2

as ( x, y ) → (1, 0) along the line

(b) ∞ (d) None of these

(a) 1 (c) 0

88. lim (tan x cot β )1/ (x − β) is equal to x →β

1 sin β cosβ 1 (c) − sin β cos β (a)

89. lim

x→∞

xn e

x

(b) sin β cos β

(a) no value of n (c) only negative values of n

(b) all values of n (d) only positive values of n

90. If f ( a ) = 2, f ′ ( a ) = 1and g ( a ) = − 1, g ′ ( a ) = 2, then g (x ) f (a ) − g (a ) f (x ) is equal to lim x→a x−a (a) 3

(b) 5

(c) − 3

(d) 0

g ( x ) − g (1) 91. If g ( x ) = − 25 − x 2 , then lim x →1 x−1 is equal to 3 24 1 (c) − 24

(a)

1014

x →1

(a) 0 (c) 2

(b)

1 24

(d) None of these

(b) 1 (d) None of these

(b) 1 (d) None of these

2x 2 − 4 f ( x ) is x→2 x−2

94. If f ( 2) = 2 and f ′ ( 2) = 1, then lim equal to (b) − 4 (d) − 2

(a) 4 (c) 2

x sin ( x − [ x]) , where [ ] denotes the greatest x →1 x−1 integer function, is equal to

95. lim

(b) −1 (d) Does not exist

96. lim [ x + x + x − x ] is equal to x→0

(a) 0

1 2 (d) e4 (b)

97. lim log tan x (sin x ) is equal to x → 0+

(b) − 1 (d) None of these

(a) 1 (c) 0 x→∞

= 0 (n integer), for

is equal

f (1) g ( x ) − f ( x ) g (1) − f (1) + g (1) is equal to g (x ) − f (x )

lim

98. lim f ( x ), where

(d) None of these

3− x

93. If f ( x ) and g ( x ) are differentiable functions and f (1) = g (1) = 2 , then

(c) log 2

y = x − 1is given by

f (x )

to

(a) 1 (c) ∞

x →1

3−

x→9

(a) 0 (c) − 1 (d) − 1

(c) 1

92. If f ( 9) = 9 and f ′ ( 9) = 1, then lim

(a) 1 (c) −1

2x − 3 2x 2 + 5x , is equal to < f (x ) < x x2 (b) 2 (d) − 2

 sin ( 2k − 3) x ,x< 0  4x  99. If f ( x ) =  k + 1, x=0  tan ( 3k − 4 ) x ,x> 0  2x and lim f ( x ) exists, then the value of k is given by x→0

5 (a) 4

(b) −

5 4

4 5

(d) −

4 5

(c)

100. lim ( − 1)[ x ] , where [ x] denotes the greatest integer x→n

less than or equal to x, is equal to (a) (− 1)n (c) 0

(b) (− 1)n − 1 (d) Does not exist

n ∈I

where

and

 x 2 + 1, x ≠ 2 , then lim g [ f ( x )] is equal to g (x ) =  x→0 x=2  3, (a) 1 (c) 3

(b) 0 (d) Does not exist

 tan [ x] , [ x] ≠ 0  102. If f ( x ) =  [ x] , where [ x] denotes the  0, [ x] = 0 greatest integer less than or equal to x, then lim f ( x ) x→0 equals (b) − 1 (d) Does not exist

(a) 1 (c) 0

 tan −1 ([ x] + x )  , [ x] ≠ 0, where denotes 103. If f ( x ) =  [ x] − 2x [ x]  [ x] = 0 0, the greatest integer less than or equal to x, then lim f ( x ) is equal to x→0

(a) − π (c) 4

1 2

(d) Does not exist

(a) 1 (c) e − 1

(a) 1 (c) −

(b) 1 2

(a) 1 (c)

(b)

1 3

(d) None of these

2 sin x − sin 2x

1/ cos2 x

dx, x ≠ 0, then lim f ′ ( x ) is x→ 0

x3 (b) ∞

(c) −1

(b) − 1

3 + 3+ x4 3 (c) 3x 2 + 4 + 3 − x (a) 3x 2 +

(d) 1

(c) −

2 7

(d) 0

+ ... + n

1/ cos2 x

3 x2 3 x2

(b) 3x 2 −

3 3 + 3− 2 4 x x

(d) None of these

115. The derivative of ( 3x + 5) (1 + tan x ) is (a) 3x sec2 x − 6 sec2 x − 3 + 3 tan x (b) 3x sec2 x − 5 sec2 x − 3 − 3 tan x (c) 3x sec2 x + 5 sec2 x + 3 + 3 tan x (d) None of the above

116. The derivative of +2

1 2

3

107. The value of x

(d) None of these

1  114. The derivative of  x +  is  x

(b) n (d) 0

2

1 2

18

111. If the rth term t r of a series is given by n r , then tr = 4 lim ∑ t r is n→ ∞ r + r2 + 1 r =1

x→3

11/ x + 21/ x + 31/ x + ... + n1/ x  106. The value of lim   is x→ ∞ n  

2

] cos x is

(b) n n (n + 1) (d) 2

  1 1 1 1 108. lim  is equal to + + + ... + n→ ∞ 1⋅ 2 2⋅ 3 3⋅ 4 n ⋅ ( n + 1) (a) 1 (c) 0

  1 1 1 1 110. lim  is equal + + +...+ n→ ∞ 1⋅ 3 3 ⋅ 5 5 ⋅ 7 ( 2n − 1) ( 2n + 1) to

(a) − 2

nx

(c) ∞

(d) None of these

2( 2x + 1) 2 x−3 113. If f ( x ) = and h( x ) = − 2 , , g (x ) = x+4 x−3 x + x − 12 then lim [ f ( x ) + g ( x ) + h( x )] is

(b) 0 (d) e + 1

(a) n! (c) (n − 1) !

is equal to

(b) − 1

(a) 1 1 (c) 3

(a) 0

 1 e1 / n e 2 / n e (n − 1)/ n  105. The value of lim  + + + ... +  is n→∞ n n n n  

(a) 0

n3

equal to

(a) 1 (b) 0 (c) ∞ (d) None of the above

π x→ 2

1⋅ 2 + 2⋅ 3 + 3⋅ 4 + K + n ⋅ ( n + 1)

n→ ∞

112. If f ( x ) = ∫

(b) 1

1  x sin , x ≠ 0 104. If f ( x ) =  , then lim f ( x ) equals x x→0  0, x=0

lim [11/ cos

109. lim

Introduction to Limits and Derivatives

sin x, x ≠ nπ , f (x ) =  x = nπ  2,

Targ e t E x e rc is e s

101. If

(b) − 1 (d) None of these

x 5 − cos x is sin x

5x 4 sin x − 1 − x 5 cos x sin 2 x 4 5x sin x − 1 − x 3 cos x (c) sin 2 x

(a)

(b)

5x 4 sin x + 1 − x 5 cos x sin 2 x

(d) None of these

117. The derivative of ( ax 2 + cot x ) ( p + q cos x ) is (a) (ax 2 + cot x ) (− q sin x ) + ( p + q cos x )(2ax − cosec2 x ) (b) (ax 2 + cot x ) (q sin x ) + ( p + q cos x )(2ax − cosec2 x ) (c) (ax 2 + cot x ) (− q sin x ) + ( p − q cos x )(2ax + cosec2 x ) (d) None of the above

1015

Objective Mathematics Vol. 1

18

118. If f ( x ) = ( ax + b )sin x + ( cx + d )cos x, then the values of a, b, c and d such that f ′ ( x ) = x cos x for all x are

121. If f ( x ) = log |2x|, x ≠ 0, then f ′ ( x ) is equal to (a)

(a) b = c = 0, a = d = 1 (b) b = d = 0, a = c = 1 (c) c = d = 0, a = b = 1 (d) None of the above

(b) −

f ( x ) be a polynomial function satisfying  1  1 f ( x ) ⋅ f   = f ( x ) + f   . If f ( 4 ) = 65 and I 1 , I 2 , I 3  x  x are in GP, then f ′ ( I 1 ), f ′ ( I 2 ), f ′ ( I 3 ) are in

1 |x| (d) None of the above

φ′ ( x ) is equal to (a) 1 (c) −1

dy  1 at 120. If 8 f ( x ) + 6 f   = x + 5 and y = x 2 f ( x ), then  x dx x = − 1, is equal to

(c) −

122. If f ( x ) = | x − 3| and φ ( x ) = ( fof )( x ), when x > 10, then

(b) GP (d) None of these

(a) 0

(b) 1 14

1 x

(c)

119. Let

(a) AP (c) HP

1 x

(b) 0 (d) None of these

123. If f ( x ) is a polynomial of degree n(> 2) and f ( x ) = f ( k − x ), where k is a fixed real number, then degree of f ′ ( x ) is

1 14

(a) n (b) n − 1 (c) n − 2 (d) None of the above

(d) None of these

Ta rg e t E x e rc is e s

Type 2. More than One Correct Option  sin x 124. lim m , where m ∈ I and [ ] denotes x→ 0  x  greatest integer function, is equal to

(c) a =

(a) m, if m ≤ 0 (b) m − 1, if m > 0 (c) m − 1, if m < 0 (d) m, if m > 0

(b) a =

(a) a = 3, b = 0

the

3 ,b=4 2

3 ,b=1 2

(d) a = 2, b = 3

126. If f ( x ) = 5, then which of the following is/are true? (a) (b) (c) (d)

125. If lim (1 + ax + bx 2 ) 2/ x = e 3 , then x→ 0

f ′ (1) and f ′ (−1) are negative f ′ (1) and f ′ (−1) are positive f ′ (1) and f ′ (−1) are equal f ′ (1) = 0 = f ′ (−1)

Type 3. Assertion and Reason Directions (Q. Nos. 127-128) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d), out of which only one is correct. The choices are (a) Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

 12 22 32 x2  127. Statement I lim  3 + 3 + 3 + ... + 3  x→∞ x x x x  = lim

12

x→∞

Statement II

x3 lim [ f1 ( x ) + f 2 ( x ) + ... + f n ( x )]

x→a

x→∞

22 x3

+ ... + lim

x→∞

x→a

x→a

The derivative of the function f ( x ) = 3x at x = 2 is 3. d d Statement II [ af ( x )] = a ⋅ [ f ( x )] , where a is a constant. dx dx

128. Statement I

x2 x3

=0

= lim f1 ( x ) + lim f 2 ( x ) +...+ lim f n ( x ), where n ∈ N . x→a

1016

+ lim

Passage I (Q. Nos. 129-130) Let f ( x ) = x and

Passage

1 , x ≠ 0 define a function h( x ) as follows f (x ) f ( x ) − g( x ) h( x ) = , x ≠ ±1 f ( x ) + g( x )

Ai =

g( x ) = −

(d) None of these

(a)

8 9

(c) −

(b) 8 9

131-133)

+ x → am

the

value

Let

of

(b) always − 1 (c) (−1)n − m + 1(d) (− 1)n − m

132. If 1< m < n, m ∈ N , then R = lim ( A1 A 2 ... A n ) is (a) always 1

130. The value of h′ ( 2) is

Nos.

131. If then 1< m < n, m ∈ N , L = lim ( A1 A 2 ... A n ) is (a) always 1

4 x3 (b) 2 (x − 1)2

(Q.

x − ai , i = 1, 2, ..., n and a1 < a2 < a3 < ... < an . x − ai

− x → am

129. The derivative of h( x ) is equal to 4x (a) 2 (x − 1)2 − 4x (c) (1 − x 2 )2

II

the

(b) always − 1 (c) (−1)m + 1

value

of

Introduction to Limits and Derivatives

18

Type 4. Linked Comprehension Based Questions

(d) (− 1)n − m

133. If 1< m < n, m ∈ N , then lim ( A1 A 2 ... A n ) is equal x → am

32 9

to

(d) None of these

(a) −1 (c) Does not exist

(b) −1 (d) 1 or −1

134. Match the limits of the functions given in Column I with their corresponding values given in Column II. Column I A.

lim

x→ 0

( x + 1)5 − 1 x

B. If

1   L = lim  x − x 2 loge  1 +  , x→ ∞  x   then the value of 8L is  z1/ 3 − 1 C. lim  1/ 6  z→1  z − 1

D.

1 1 lim  1 − 2   1 − 2  n → ∞ 2  3   1 − 1  ...  1 − 1       42   n2 

Column I

Column II

Column II p.

4

A. cosec x

p. 5 cos x + 6 sin x

q.

5

B. 3 cot x + 5 cosec x

q. − 3 cosec 2 x − 5 cosec x cot x

C. 5 sin x − 6 cos x + 7

r. 2 sec 2 x − 7 sec x tan x

r.

2

s.

1 2

D. 2 tan x − 7 sec x

s. − cot x cosec x

t.

Targ e t E x e rc is e s

Type 5. Match the Columns

t. 6 sin x + 5 cos x

3/2

135. Match the functions given in Column I with their corresponding derivatives given in Column II.

Type 6. Single Integer Answer Type Questions  x + 2, x ≤ − 1 . If lim f ( x ) exists, then 136. Let f ( x ) =  2  cx , x > − 1 x → − 1 the value of c is _________ . sin x    1  is _______. 137. For x > 0, lim  (sin x )1/ x +    x→ 0   x  

138. If lim

1− cos 2x ⋅ 3 cos 3x ⋅ 4 cos 4x ... n cos nx

has x2 the value equal to 10, then the value of n equals____. x→ 0

139. If

 sin (1 + [ x]) ,  f (x ) =  [ x]  0,

for [ x] ≠ 0 , for [ x] = 0

where

[ x]

denotes the greatest integer not exceeding x, then lim f ( x ) is ________.

x → 0−

  x2 2  ∫0 sec t dt  140. The value of lim   is ________. x→ 0  x sin x    1017

Entrances Gallery JEE Advanced/IIT JEE 1. The largest value of the non-negative integer a for

4. Let α ( a ) and β ( a ) be the roots of the equation

 − ax + sin ( x − 1) + a  1 − lim   x → 1  x + sin ( x − 1) − 1 

( 3 1 + a − 1) x 2 + ( 1 + a − 1) x + ( 6 1 + a − 1) = 0,

1−x

which

x

1 = 4

_______ .

is

a→ 0+

[2012] 5 (a) − and 1 2 7 (c) − and 2 2

(1a + 2a +...+ n a ) ( n + 1) a − 1 [( na + 1) + ( na + 2) +...+ ( na + n )]

h→ ∞

= then a is equal to (a) 5

(b) 7

(c)

− 15 2

1 , 60

[2013] − 17 (d) 2

x + x + 1  3. If lim  − ax − b = 4, then x→ ∞  x+1 

Ta rg e t E x e rc is e s

(a) a = 1, b = 4 (c) a = 2, b = − 3

1 (b) − and − 1 2 9 (d) − and 3 2

5. The value of lim

1

x→ 0

x

3

x

∫0

t log (1 + t )

1 (b) 12

(a) 0

t4 + 4

dt is

1 (c) 24

x→ 0

θ ∈ ( − π , π ] , then the value of θ is

[2012]

(b) a = 1, b = − 4 (d) a = 2, b = 3

[2011] 1 (d) 64

lim [1 + x ln (1 + b 2 )]1/ x = 2b sin 2 θ, b > 0

6. If

2

π (a) ± 4

π (b) ± 3

π (c) ± 6

and

[2011] π (d) ± 2

JEE Main/AIEEE 7. lim

x→ 0

(1 − cos 2x )( 3 + cos x ) is equal to x tan 4x

(a) 4

[2015]

(b) 3 1 (d) 2

(c) 2

k x + 1 , 0 ≤ x ≤ 3 is 8. If the function g ( x ) =   mx + 2 , 3 < x ≤ 5 [2015] differentiable, then the value of k + m is 16 (b) 5

(a) 2

10 (c) 3

(d) 4

then f ( 2) is equal to (a) −8

(b) −4

sin ( π cos x )

[2015] (c) 0

x→ 0

(a)

π 2

x2

[2014]

x→ 5

[ f ( x )] − 9 2

lim

| x − 5|

= 0. Then, lim f ( x ) equals x→ 5

(a) 3 (c) 1

(b) 1

(c) − π

(d) π

1 (b) 2

(c) 1

[2013] (d) 2

[2011]

(b) 0 (d) 2

14. Let α and β be the distinct roots of ax 2 + bx + c = 0, then lim

1 − cos ( ax 2 + bx + c )

x→ α

(x − α )2

1 (α − β )2 2

 a b 15. If lim 1 + + 2  x→∞  x x  are (a) a ∈ R , b ∈ R (c) a ∈ R , b = 2

is equal to

(b) −

(c) 0

(1 − cos 2x ) ( 3 + cos x ) is equal to 11. lim x→ 0 x tan 4x 1 (a) − 4

(d) Does not exist

13. Let f : R → [ 0, ∞ ) be such that lim f ( x ) exists and

(a)

is equal to

[2011]

(b) − 2

(a) 2 1 (c) 2

(d) 4

2

10. lim

 1 − {cos 2 ( x − 2)}  is equal to 12. lim   x→ 2  x−2 

x→ 5

9. Let f ( x ) be a polynomial of degree four having f ( x )  extreme values at x = 1and x = 2. If lim 1 + 2  = 3, x→ 0  x 

1018

a→ 0+

[2014]

2. For a ∈ R (the set of all real numbers), a ≠ − 1, lim

where a > − 1. Then, lim α ( a ) and lim β ( a ) are

(d)

[2005]

a2 (α − β )2 2

a2 (α − β )2 2

2x

= e 2 , then the values of a and b [2004] (b) a = 1, b ∈ R (d) a = 1, b = 2

 x     (1 − sin x )  2   x  3    ( π − 2x )  2 

1 − cos 2x

18. lim is equal to

x→ 0

[2003]

is equal to

[2002]

2x (b) − 1 (d) Does not exist

(a) 2 (c) zero x

1 8 1 (c) 32

 x 2 + 5x + 3 19. lim  2  is equal to x→ ∞  x + x + 2 

(b) 0

(a)

(d) ∞

log ( 3 + x ) − log ( 3 − x ) 17. If lim = k , then value of k is x→ 0 [2003] x (b) −1/3 (d) −2/3

(a) 0 (c) 2/3

(c) e3

(b) e2

(a) e4

[2002] (d) e

x

 x − 3 20. For x ∈ R , lim   is equal to x→ ∞  x + 2 (c) e−5

(b) e−1

(a) e

[2002] (d) e5

18 Introduction to Limits and Derivatives

 1 − tan 16. lim  π  x→ 2 1 + tan 

Other Engineering Entrances x→ 0

(1 + x ) 8 − 1 (1 + x ) 2 − 1

is equal to

(a) 8 (c) 4

22. lim

x→1

(a)

[BITSAT 2014]

28. lim

xn − 1

n m

3 5 −5 (d) 3

(a)

is equal to (b)

[BITSAT 2014]

m n

(c)

2

2m n

2

(d) 2

23. lim (1cosec x + 2cosec x + ... + n cosec x ) sin

2

x

x→ 0

(a) 1

(b) 1/ n

x ( e 5x − 1)

x→ 0

(b) 6 (d) 2

xm − 1

log (1 + 3x 2 )

2n m

[WB JEE 2014] (a) 0

(b) 2

x→1

lim f ( x ) is equal to

f (x ) − 2 x2 − 1

(a) 1 (c) 0 x→ 3

5 (a) 8 1 (d) 27

(b)

x 5 − 35 x −3 8

5 64

8

is

= π, then

and g ( x ) = ( x 2 + 2x + 3) f ( x ), f ( 0) = 5  f ( x ) − 5 lim   = 4, then g ′ ( 0) is equal to x→ 0  x  [J&K CET 2014]

x→ 3

(a) 0

33.

f ( x ) = ( x 5 −1) ( x 3 + 1), g ( x ) = ( x 2 − 1) ( x 2 − x + 1) x→1

[Kerala CEE 2014] (b) 1 (e) 5

(c) 3

(b) 18

32. Evaluate lim

x → tan − 1 3

(a) 1

34. lim

x→ ∞

(a) 0 (c) 2

(c) 20

3− x 4 + x − 1 + 2x (b) 7 2

lim

and h( x ) such that f ( x ) = g ( x ) h( x ), then lim h( x ) is (a) 0 (d) 4

(d) None of these

31. If

5 216

equal to

(b) a = 1

(a) 22

(e) None of these

x 4 , a > 0 and L is finite, [AMU 2014]

(a) a = 2 1 (c) a = 3

[Kerala CEE 2014] (c)

x4

(d) 1

2

then

(b) 2 (d) 3

26. The value of lim

(c) 2

a − a2 − x2 −

x→ 0

[Karnataka CET 2014]

x→1

27. If

30. If L = lim

(b) abc (d) None of these

25. If the function f ( x ) satisfies lim

−3 5

(e) 1

2/ x

(a) (abc) (c) (abc)1/ 3

(c)

is equal to

 ax + bx + cx  24. The value of the lim   , ( a, b, c > 0) x→ 0  3  is [Manipal 2014] 3

5 3

[Kerala CEE 2014]

29. If [ x] denotes the greatest integer less than or equal to [ n 2] is equal to x for any real number x. Then, lim n→∞ n

[Manipal 2014] (d) 0

(c) n

(b)

is equal to

Targ e t E x e rc is e s

21. lim

(c) 4 7

tan 2 x − 2 tan x − 3 tan 2 x − 4 tan x + 3 (b) 2

 1 x 4 sin   + x 2  x 1+ |x3 |

.

(d) 25 [J&K CET 2014] (d) 2 7

is equal to [BITSAT 2014]

(c) 0

equals

(d) 3

[BITSAT 2014] (b) −1 (d) 1

1019

Objective Mathematics Vol. 1

18

35. lim

x→ 0

e sin x − 1 is equal to x

[Manipal 2014]

(a) 0 (c) 1

(b) e (d) Does not exist

36. If lim

2a sin x − sin 2x

x→ 0 tan x the value of a is 3

(a) 2

sin

37. lim

x→ 2

( x − 2)

x2 − 4

(a) 0

[WB JEE 2014] (d) − 1

(c) 0

is equal to

(b) 2

(c)

[J&K CET 2014] 1 4

(d) 1

38. The graph of the function y = f ( x ) has a unique tangent at the point ( a, 0) through which the graph passes.  log e {1 + 6 f ( x )} Then, lim   is x→ a 3 f (x )  

Ta rg e t E x e rc is e s

(a) 0 (c) 2

1 + 2 + ... + n 3n + 5

n→ ∞

(a)

[UP SEE 2013]

(b) 1 (d) None of these

39. The value of lim 1 3

40. lim

h→ 0

(b)

2

1 5

(c)

is equal to

1 6

(d) 6

(a) a cos a + a sin a

(b) a cos a + 2a sin a (d) None of these [Kerala CEE 2013]

(b) 1

1/ x

is equal to

2

e

x→ 0

1020

(b) log 2

πx − 1 1+ x − 1

is

(a) Does not exist (c) 1

[Manipal 2012]

(b) e1/ 2 (d) e3

x2 x − x 43. lim is equal to x → 0 1 − cos x

44. lim

greatest integer less than or equal to x, then lim f ( x ) x→ 0 [Manipal 2012] is equal to (a) 1 (b) 0 (c) − 1 (d) None of the above

x2

(a) 1/5 (c) 1/3

1 log 2 2

(d)

1 2

[WB JEE 2012] (b) loge (π 2 ) (d) lies between 10 and 11

is [MP PET 2012]

(b) 1/4 (d) 1/2

sin x cos x tan x 49. If f ( x ) =

x3 2x

x2 1

, then lim

x

x→ 0

x

f (x ) x2

is

[Karnataka CET 2012] (c) 2 (d) 1

(b) 3

 a + 2x − 3x  50. lim   is equal to [Karnataka CET 2012] x→ a  3a + x − 2 x  2 3 3 3 (c) 2

2 3 2 (d) 3 3

(b)

 13 + 23 + 33 + ... + k 3  51. lim   is equal to k→∞ k4  [Kerala CEE 2011] (a) 0 (d) ∞

[Karnataka CET 2012] (c)

x cos x − log e (1 + x )

x→ 0

(a)

(d) e2

42. If f : R → R is such that f (1) = 3 and f ′ (1) = 6. Then,

(a) 2 log 2

 sin [ x] , [ x] ≠ 0  47. If f ( x ) =  [ x] , where [ x] denotes the  0, [ x] = 0

(a) 0

1 e (c) e (e) None of these

(a) 1 (c)

(a) sin 2 (b) cos 2 (c) 1 (d) 2 cos 2 + sin 2

equal to

cosec x

(a)

 f (1 + x ) lim   x→ 0  f (1) 

 ( 2 + x ) sin ( 2 + x ) − 2 sin 2 46. lim   is equal to x→ 0  x [BITSAT 2012]

2

is equal to

1 e2 1 (d) e

1 2e

[OJEE 2013]

(c) 2a2 cos a + a sin a

 1 + tan x 41. lim   x → 0  1 + sin x 

(c)

[WB JEE 2012] (b)

48. The value of lim

( a + h ) 2 sin ( a + h ) − a 2 sin a is equal to h [Manipal 2013]

2

n→ ∞

( n !)1/ n is n

(a) 1

exists and is equal to 1, then

(b) 1 −1

45. The value of lim

(b) 2 (e) 1/4

 4  52. lim 1 +  x→ ∞  x − 1 (a) e4 (c) e3

(c) 1/3

x+3

is equal to

[Guj CET 2011]

(b) e2 (d) e

53. If f ′ ( x ) = f ( x ), f ( 0) = 1, then lim

x→ 0

f (x ) − 1 equals x [Guj CET 2011]

(a) 0 (c) − 1

(b) 1 (d) 2

is equal to

(a) 25/2 (c) 1

[GGSIPU 2011]

(a) 0 (d) −1

(b) 12 (d) 1/4

55. The value of lim x→

π 2

sin (cos x ) cos x is sin x − cosec x

(a) ∞ (c) 0

[UP SEE 2011]

(b) 1 (d) − 1

56. If lim

x→ 0

ae x − b cos x + ce − x = 2, then x sin x

[AMU 2011]

(a) 35 (b) − 35 (c) 28 (d) − 28

1/ x 2

(b) e

(c)

sin | x| is equal to x

62. The value of lim

x→ 0

is

[WB JEE 2010]

1 e

(d)

1 e2

[WB JEE 2010; BITSAT 2010]

sin 2 x + cos x − 1 x2

is

[WB JEE 2010; BITSAT 2010] (a) 1 1 (b) 2 1 2

2

(b) −

1 2

(c) 0

(e) None of these

18

(b) 0 (d) Does not exist

(d) 0

 x x  58. lim  2 −  is equal to [Kerala CEE 2010] x → ∞  3x − 4 3x + 2

2 9

x→ 0

(c) − 3

(d)

61. lim

[Kerala CEE 2010]

(c) 2

(a) 1 (c) positive infinity

xf ( 5) − 5 f ( x ) is 57. If f ( 5) = 7 and f ′ ( 5) = 7, then lim x→ 5 x−5 equal to [WB JEE 2010]

1 4

(b) 1 (e) − 2

 1 + 5x 2  60. The value of lim   x → 0  1 + 3x 2  (a) e2

(a) a = 1, b = 2 and c = 1 (b) a = 1, b = 1 and c = 2 (c) a = 2, b = 1 and c = 1 (d) a = b = c = 1

(a) −

  x  is equal to 59. lim  x → 0  1+ x − 1− x 

Introduction to Limits and Derivatives

1/ x

63. The value of lim

x→ 0

1 2 1 (c) 6

(a)

1 − cos (1 − cos x ) x4 1 4 1 (d) 8 (b)

is

[WB JEE 2010]

Targ e t E x e rc is e s

 16x + 9x  54. lim   x→ 0  2 

1021

Answers Work Book Exercise 18.1 1. (c)

2. (c)

3. (d)

4. (c)

5. (b)

6. (d)

11. (d)

12. (c)

13. (b)

14. (d)

15. (b)

16. (a)

4. (a)

5. (b)

6. (a)

7. (b)

8. (d)

9. (d)

10. (a)

7. (a)

8. (b)

9. (a)

10. (a)

Work Book Exercise 18.2 1. (b)

2. (d)

3. (b)

Ta rg e t E x e rc is e s

Target Exercises

1022

1. (d)

2. (c)

3. (b)

4. (b)

5. (a)

6. (b)

7. (a)

8. (a)

9. (a)

10. (d)

11. (b)

12. (b)

13. (a)

14. (d)

15. (b)

16. (c)

17. (a)

18. (c)

19. (c)

20. (a)

21. (b)

22. (d)

23. (a)

24. (b)

25. (c)

26. (b)

27. (d)

28. (c)

29. (a)

30. (a)

31. (a)

32. (b)

33. (b)

34. (a)

35. (a)

36. (a)

37. (c)

38. (c)

39. (a)

40. (c)

41. (a)

42. (a)

43. (c)

44. (c)

45. (c)

46. (d)

47. (b)

48. (c)

49. (b)

50. (b)

51. (a)

52. (c)

53. (b)

54. (d)

55. (a)

56. (b)

57. (a)

58. (b)

59. (a)

60. (a)

61. (a)

62. (b)

63. (c)

64. (b)

65. (b)

66. (c)

67. (b)

68. (a)

69. (a)

70. (c)

71. (d)

72. (c)

73. (b)

74. (a)

75. (b)

76. (a)

77. (c)

78. (a)

79. (a)

80. (d)

81. (b)

82. (a)

83. (b)

84. (c)

85. (b)

86. (c)

87. (c)

88. (d)

89. (b)

90. (b)

91. (b)

92. (b)

93. (c)

94. (a)

95. (d)

96. (b)

97. (a)

98. (b)

99. (a)

100. (d) 110. (b)

101. (a)

102. (d)

103. (d)

104. (b)

105. (c)

106. (a)

107. (b)

108. (a)

109. (c)

111. (b)

112. (d)

113. (c)

114. (b)

115. (c)

116. (b)

117. (a)

118. (a)

119. (b)

120. (c)

121. (a)

122. (a)

123. (b)

124. (a,b)

125. (b,c)

126. (c,d)

127. (d)

128. (a)

129. (c)

130. (c)

131. (c)

132. (d)

133. (c)

134. (*)

135. (**)

136. (1)

137. (1)

138. (6)

139. (0)

140. (1)

* A → q; B → p; C → r; D → s ** A → s; B → q; C → p; D → r

Entrances Gallery 3. (b)

4. (b)

5. (b)

6. (d)

7. (c)

8. (a)

9. (c)

10. (d)

11. (d)

1. (2)

12. (d)

2. (b,d)

13. (a)

14. (d)

15. (b)

16. (c)

17. (c)

18. (d)

19. (a)

20. (c)

21. (c)

22. (b)

23. (c)

24. (d)

25. (b)

26. (c)

27. (e)

28. (a)

29. (c)

30. (a)

31. (a)

32. (d)

33. (b)

34. (d)

35. (d)

36. (b)

37. (c)

38. (c)

39. (c)

40. (b)

41. (b)

42. (c)

43. (a)

44. (b)

45. (d)

46. (d)

47. (d)

48. (d)

49. (d)

50. (d)

51. (e)

52. (a)

53. (b)

54. (b)

55. (d)

56. (a)

57. (d)

58. (d)

59. (b)

60. (a)

61. (d)

62. (b)

63. (d)

Explanations Target Exercises 4−

x→ 0

= lim

x −

h→ 0

x

h 4− h − h

1+ 2 + x − 3

7. lim

x −2

x→ 2

0 = =0 2 The function is not defined for negative values of x, so LHL does not exist. x does not exist. lim ∴ x→ 0 4− x − x

= lim

x→ 2

= lim

x→ 2

= lim

x→ 2

 1 1 2. lim  3 x +  − 2 x −    x→ 0  x x 2

2





1 1  1 1   = lim  3 x + − 2 x +   3 x + + 2 x −   x → 0  x x  x x  2   = lim  x +  5 x  x → 0  x  2 = lim (5 x + 10 ) = 10

=

x→ −1

x→1

3

4. lim

x→ 2

10 − x − 2 (10 − x )1/ 3 − 81/ 3 = lim x→ 2 x −2 x −2 1 (10 − x ) − 8 ⋅ = lim x → 2 (10 − x )2 / 3 + 82 / 3 + 81/ 3 (10 − x )1/ 3 x − 2  a3 − b3  − = using a b   a2 + b2 + ab   1 −1 =− = 2/ 3 12 8 + 82 / 3 + 82 / 3

( x − 2 ) [ 1 + 2 + x + 3 ][ 2 + x + 2 ] 1 [ 1+ 2 + x +

6x + 3 + 3x ( x + 1) [ 6x 2 + 3 − 3x ] = lim x → −1 3 (1 − x 2 ) 1 1 = lim [ 6x 2 + 3 − 3x ] = (3 + 3) = 1 x→−1 3 (1 − x ) 3⋅ 2



 x−3   x −2 − 4− x ( x − 3) [ x − 2 + 4 − x ] = lim x→ 3 ( x − 2 ) − (4 − x )

9. lim  x→ 3

= lim

x→ 3

= lim

x→ 3

( x − 3) [ x − 2 + 2( x − 3) x −2 + 2

4− x

x→ 0

x → 2a

x − 2a

= lim

x → 2a

= lim

x 2 − 4 a2 x − 2a

x → 2a

1 2 a

x 2 − 4a2

+ lim

= lim − x→ 0

= − a⋅

x → 2a

2 ax × 2x

a+ x + a − ax + x + 2

2

a+ x + a − ax + x + 2

2

a− x a2 + ax + x 2 a− x a2 + ax + x 2

2 a =− a 2a

2 − 2+ x

2−x 2 2 / 3 + 21/ 3 ⋅ (4 − x )1/ 3 + (4 − x )2 / 3 ⋅ x→ 2 2 + 2 + x 2 − (4 − x ) −1 = lim ⋅ [2 2 / 3 + 21/ 3 ⋅ (4 − x )1/ 3 + (4 − x )2 / 3 ] x→ 2 2 + 2 + x = lim

x + 2a + lim

=

×

2 − 3 4− x 4 − (2 + x ) 1 × = lim 1/ 3 x→ 2 2 − (4 − x )1/ 3 2 + 2 + x

x − 2a

x → 2a

1+ 1 =1 2

x→ 2 3

x − 2a

x → 2a

=

(a − ax + x 2 ) − (a2 + ax + x 2 ) x→ 0 (a + x ) − (a − x )

11. lim + lim

4− 3

2

= lim

x − 2a

x − 4 a2

3−2 + 2

=

a+ x − a− x

x→ 0

x 2 − 23 x + 1 x→1 ( x − 1)2 y2 − 2 y + 1 [putting 3 x = y;as x → 1, y → 1] = lim y → 1 ( y 3 − 1)2 1 1 ( y − 1)2 = lim 2 = = lim 2 y → 1 ( y − 1) ( y 2 + y + 1)2 y → 1 ( y + y + 1)2 9 2

4 − x]

a2 − ax + x 2 − a2 + ax + x 2

10. lim f ( x ) = lim

3

x − 2a +

3 ][ 2 + x + 2 ]

1 1 = 3) ( 2 + 2 + 2) 8 3

( 1+ 2 +

5. lim

6. lim

3]

x −2

2

x→ 0

x − 2 x5 + 1 x 5 ( x 2 − 1) − ( x 5 − 1) = lim x 3 − 3 x 2 + 2 x → 1 x 2 ( x − 1) − 2 ( x 2 − 1) x 5 ( x + 1) − ( x 4 + x 3 + x 2 + x + 1) = lim x→1 x 2 − 2 ( x + 1) 1 (1 + 1) − (1 + 1 + 1 + 1 + 1) − 3 = = =1 −3 1 − 2 (1 + 1)

( x − 2) [ 1 + 2 + x +

x+1

8. lim

7

3. lim

1+ 2 + x − 3

Targ e t E x e rc is e s

x

1. RHL = lim +

( x − 2 a) ⋅ x + 2a

1 x − 2a x + 2a

1 1 1 x − 2a = +0= ⋅ 2 a x + 2a x + 2a 2 a

=−

1 2/ 3 − 3⋅ 22/ 3 −3 ⋅ (2 + 22/ 3 + 22/ 3 ) = = 4/ 3 4 4 2

1023

Objective Mathematics Vol. 1

18

12. lim (1 + x ) (1 + x 2 ) (1 + x 4 ) ... (1 + x 2 n ) (1 − x ) (1 + x ) (1 + x 2 ) (1 + x 4 ) ... (1 + x 2 n ) n→ ∞ 1− x (1 − x 2 ) (1 + x 2 ) (1 + x 4 ) ... (1 + x 2 n ) = lim n→ ∞ 1− x M M M 1 − x 4n 1 , for x < 1 = lim = n→ ∞ 1− x 1− x = lim

 1 13. lim [ 3 n 2 − n 3 + n ] = lim n  − 1 +  n→ ∞

= lim n ⋅ n→ ∞

 1   − 1 + 1 n 

1   − 1 n 

2/ 3

1/ 3

n

1  + 1 −  − 1 n 

 + 11/ 3  

1/ 3

 a3 + b3   using a + b = 2 a − ab + b2   1

= lim

n→ ∞

1   − 1 n  1 1 = = 1+ 1+ 1 3

Ta rg e t E x e rc is e s



n→ ∞

2/ 3

1  + 1 −  − 1 n 

1/ 3

 1 1  n  1+ 2 +   n2 + 1 + n  n n  14. lim   = lim 4 n→ ∞ 4 3 n→ ∞  1 1   n + n − n  n 3 / 4 4 1 + 2 −  n n   1 1    1+ 2 + n  = ∞⋅1= ∞ n = lim n1/ 4  n→ ∞ 1 4 1+ 1 − 1   n  n2

15. lim

x→ ∞

sin x 1− 0 x = =1 1+ 0 cos 2 x 1+ x   sin x  1 = lim y sin   = 0  Q xlim y→0 →∞  y x   × (an oscillating quantity between − 1and 1) = 0    2  Similarly, lim cos x = 0 x→ ∞  x 

16. lim (sin x + 1 − sin x ) x→ ∞

x→ ∞

= lim 2 cos x→ ∞

x + 1+ 2 x + 1+ 2

x + 1− x 2 x 1 ⋅ sin 2 ( x + 1 + x) x

⋅ sin

= 2 × (an oscillating quantity between − 1and 1) × sin 0 = 0

17. lim

h→ 0

1024

sin

x + h − sin x h cos x + h

= lim

h→ 0

2

x+h 1

 π  π  2  3 sin  + h − cos  + h  6  6   20. lim h→ 0 3h ( 3 cos h − sin h )  1   3 sin h   3  cos h + 2 2   2   3   1 −  cos h − sin h  2     2 = lim  h→ 0 3h ( 3 cos h − sin h ) 2 [2 sin h ] = lim h→ 0 3h ( 3 cos h − sin h ) sin h 4⋅ 4 4 h = lim = = h→ 0 3 ( 3 − 0) 3 3 ( 3 cos h − sin h )

21. lim

x→ ∞

0  form   0 −0 =

cos x 2 x

sin x x = 1+ 0 = 1 cos x 1− 0 1− x cos x   sin x Q → 0, → 0 as x → ∞  x x 

x + sin x = lim x − cos x x → ∞

1−

x − sin x = lim x + cos 2 x x → ∞

= lim 2 cos

(1 − cos 2 x ) sin 5 x 2 sin 2 x sin 5 x = lim ⋅ 2 x→ 0 x→ 0 sin 3x x sin 3 x x2 sin 5 x 2 5 10  sin x  = lim 2   ⋅ 5x ⋅ = sin 3 x 3 x→ 0  x  3 3x π  sin  − x 3   0 19. lim form   0 x → π / 3 2 cos x − 1 π  − cos  − x 3  [using L ’Hospital’s rule] = lim x→ π /3 − 2 sin x 1 cos 0 = = π 3 2 sin 3

18. lim

n→ ∞

22. Case I x ≠ m π

1+

[mis an integer] 1 1 lim = =0 n → ∞ 1 + n sin 2 nx ∞ Case II x = mπ [m is an integer] 1 1 lim = =1 n → ∞ 1 + n sin 2 nx 1 log (tan 2 x ) 23. lim + log tan x (tan 2 x ) = lim + log (tan x ) x→ 0 x→ 0  log m  using log n m = log n    1 2 ( 2 sec 2 x ) tan 2 x [using L’ Hospital’s rule] = lim 1 x→ 0+ (sec 2 x ) tan x 2 sin x cos x 2 sin 2 x 0  = lim = lim form  0  + sin 2 x cos 2 x + sin 4 x x→ 0 x→ 0 4 cos 2 x [using L’ Hospital’s rule] = lim x → 0 + 4 cos 4 x cos 2 h = lim =1 h → 0 cos 4h

− 2x

24. As x → 0 − ⇒ f ( x ) → f (0 − ) = 2 + = lim

x→ a

Also, as x → 0 + ⇒ f ( x ) → f (0 + ) = 1+ ⇒ lim g[f ( x )] = g(1+ ) = − 3 x→ 0+

=

Hence, lim g[f ( x )] exists and is equal to −3 .

 cos x 2 cos 3 x cos 5 x  = lim  4 − +  x → 0  sin x sin 4 x sin 4 x  cos x (1 − cos 2 x )2 x→ 0 sin 4 x = lim cos x = 1 = lim

x→ 0

sin 2 x + a sin x  0 form x3   0 2 cos 2 x + a cos x [using L’ Hospital’s rule] = lim x→ 0 3x 2 We require 2 cos 2 x + a cos x = 0 for x = 0 as denominator is zero. ∴ a= −2 2 cos 2 x − 2 cos x  0 Hence, k = lim form   0 x→ 0 3x 2 − 4 sin 2 x + 2 sin x 0  = lim form  0  x→ 0 6x − 8 cos 2 x + 2 cos x − 8 + 2 = lim = = −1 x→ 0 6 6 sin (a + 3h ) − 3 sin (a + 2 h ) + 3 sin (a + h ) − sin a 27. lim h→ 0 h3 0  form  0  3 cos (a + 3 h ) − 6 cos (a + 2 h ) + 3 cos (a + h ) = lim h→ 0 3 h2 0  form  0  − 9 sin (a + 3 h ) + 12 sin (a + 2 h ) − 3 sin(a + h ) = lim h→ 0 6h 0  form  0  − 27 cos(a + 3h ) + 24 cos(a + 2 h ) − 3 cos(a + h ) = lim h→ 0 6 − 27 cos a + 24 cos a − 3 cos a = = − cos a 6

30. lim

x→ π /4

= lim

x→ a

a− x a+ x

a2 − x 2 π tan 2

a− x a+ x

[0 ⋅ ∞ form] 0  form   0

4 2 − (cos x + sin x )5 1 − sin 2 x = lim

x→ π /4

π  4 2 − 4 2 cos 5  x −   4 1 − sin 2 x

4 2 (1 − cos 5 y ) y→0 1 − cos 2 y π π  putting x − = y; as x → , y →  4 4 4 2 (1 − cos 5 y ) = lim y → 0 2 (1 − cos 2 y )  2 2 (1 − cos y )(1 + cos y   2 3 4 + cos y + cos y + cos y ) = lim  y→0 (1 − cos y ) (1 + cos y ) = lim

x→ 0

π 2

cot 4 x 



26. Let k = lim

a2 − x 2 cot

a− x π 2a × × a + x 2 2 (a + x ) a2 − x 2

29. lim cosec 3 x cot x − 2 cot 3 x cosec x + x→ 0 sec x  

  1  n  1+  1 ( − ) 1 1 2 n − = lim ⋅ cos  ⋅   n→ ∞ n  2 − 1   1 − 2 1 + 1   2 − 1 2 2   n   n  n n   1 1 (either 1 or −1 ) 2 =0 =0× × cos + × 2  2 2 1

x→ a

π − sec 2 2

4a π

x→ 0

 2n n+1 n n(− 1)n  25. lim  2 − ⋅ 2 cos n → ∞ 2n − 1 2 n − 1 1 − 2 n n + 1    2  1 + 1/ n   1 ⋅ cos    1 n  2 − 1 / n   2 − 2 n  = lim  n→ ∞  1  1 (− 1)n ⋅ ⋅  −   1 − 2 1 + 1  n    n2     n

28. lim

18

2 a2 − x 2

Introduction to Limits and Derivatives

lim g[f ( x )] = g(2 ) = − 3

x → 0−

=

0

 

2 2⋅5 =5 2 2

 1 31. lim cos [π n 2 + n ] = lim cos nπ 1 +    n→ ∞ n→ ∞ 

n

   1 1 = lim cos nπ 1 + − + ...  2 n→ ∞ n 2   8 n    π π = lim cos  nπ + − + ... n→ ∞   2 8n

1/ 2 

Targ e t E x e rc is e s

⇒

+

 

  π = − lim sin  nπ − + ... n→ ∞   8n  π  = − lim (− 1)n − 1 sin  − .... n→ ∞  8n   π  Q 8 n − ... → 0, as n → ∞    1 5      3 −  2 +   (3 x − 1) (2 x + 5) x  x 32. lim = lim x → ∞ ( x − 3) (3 x + 7 ) x→ ∞  3  7 1 −   3 +     x x (3 − 0 ) (2 + 0 ) 6 = = =2 (1 − 0 ) (3 + 0 ) 3 =0

33. lim

x→ ∞

 1 1 + x 4 − (1 + x 2 )  1  = lim  4 + 1 −  2 + 1  = 0   x→ ∞ x2 x x 

1025

Objective Mathematics Vol. 1

18

4 7 2− 2 + 3 2 x 3 − 4x + 7 x x 34. lim = lim 5 4 x → ∞ 3x 3 + 5x 2 − 4 x→ ∞ 3+ − 3 x x 2−0+0 2 = = 3+ 0 − 0 3 e x − (1 + x ) 0  form  0  x→ 0 x2 x x e −1 1 e −1 1 1 = lim = lim = × 1= x→ 0 2x 2 x→ 0 x 2 2

35. lim

log x − 1 x −e (1/ x ) = lim x→e 1 1 = e loge x 37. lim x→1 x − 1 1/ x = lim =1 x→1 1

0  form  0 

36. lim

x→e

[using L’ Hospital’s rule]

0  form   0 [using L’ Hospital’s rule]

2

e x − cos x x→ 0 x2

0  form  0 

38. lim

2

Ta rg e t E x e rc is e s

e x + e − x + 2 cos x − 4 x→ 0 x4 x e − e − x − 2 sin x = lim x→ 0 4x 3 x −x e + e − 2 cos x = lim x→ 0 12 x 2 x − x e − e + 2 sin x = lim x→ 0 24 x e x + e − x + 2 cos x = lim x→ 0 24 4 1 = = 24 6

e x ⋅ 2 x + sin x [using L’ Hospital’s rule] x→ 0 2x 1 sin x  1 1 3  2 = e 0 + (1) = 1 + = = lim e x +  x→ 0  x 2 2 2 2   x→ 0

x→ 0

1 (− sin x ) cos x = lim x→ 0 1 = − lim tan x = 0

[using L’ Hospital’s rule]

x→ 0

e x − e−x − 2x 40. lim x→ 0 x − sin x e x + e− x − 2 = lim x→ 0 1 − cos x e x − e −x = lim x→ 0 sin x e x + e−x = lim =2 x → 0 cos x

0  form   0 [using L’ Hospital’s rule] 0  form   0

1− x 3x  1 − x

 1+   2 + 3x 

46. lim  x→ ∞

x→ ∞

1026

3x + 1 − 5x + 1 3⋅ 3x − 5⋅ 5x = lim x x x→ ∞ 3 −5 3x − 5x x  3 3⋅   − 5  5 −5 = lim = =5 x x→ ∞ −1  3   −1  5

 

0 / −1

 1  − 1   x

= 10 = 1

n 1   1  1−     n n  2 1 2  = 1 −  1 47. ∑ r =     2 1  r =1 2 1 −    2 which tends to 1 as n → ∞ but infact always remains less than 1.  n 1 Thus, lim ∑ r  = 0 n→ ∞  r =1 2 

48. Given, f ( x ) = x 2 − π 2 ∴

lim

x → −π

x−3   x + 6 − 3   ( x − 3) ( x + 6 + 3) = lim log a  x→ 3 ( x − 3)  

42. lim

1  1  −  x

 x 1  x + 3 = lim   x→ ∞ 2  + 3  x  3 =   3



x→ 3

form

 

x→ 0

41. lim log a x→ 3

= lim log a ( x + 6 + 3) = log a 6

form

loge sin x ∞  form = lim x→ 0 cot x  ∞  cot x [using L’ Hospital’s rule] = lim x → 0 − cosec 2 x = lim (− cos x ⋅ sin x ) = 0 x→ 0  4 sin  x  3   4 x 45. lim 3 sin  x  = lim ⋅ 4 = 1⋅ 4 = 4  3  x→∞  4  x→∞  x 3 

 0 form   0

log cos x x

   form  form

44. lim loge (sin x )tan X = lim tan x ⋅ loge sin x [0 ⋅ ∞ form]

= lim

39. lim

0  0 0  0 0  0 0  0

43. lim

x2 − π 2 (− π + h )2 − π 2 = lim sin (sin x ) h → 0 sin [sin (− π + h )] −2 hπ + h 2 = lim h → 0 − sin (sin h ) h − 2π = 2π = lim h → 0 − sin (sin h ) sin h × sin h h 1/ x

49.

 π  lim tan  + x  4  x→ 0  lim

= e x→ 0

1/ x

1 + tan x  = lim  x→ 0  1 − tan x  

 1  1 + tan x − 1  x  1 − tan x 

 1 + tan x 1 Q 1 − tan x → 1 and x → ∞, as x →  =e

lim

2 tan x

x→ 0 x ( 1 − tan x )

=e

1  tan x  lim 2   x  1 − tan x

x→ 0 

= e2

 0 

1

lim

[( 1 + ax 2 + bx + c ) − 1]

x → α ( x − α)

=e lim g( x ) [f ( x ) − 1]   g( x ) x→ a f x e using lim [ ( )] provided  =  x→ a   f ( x ) → 1and g ( x ) → ∞, as x → a   =e

lim

x→ α

( ax 2 + bx + c ) ( x − α)

lim

x→ α

a ( x − α ) ( x − β) ( x − α)

=e [Qα, β are roots of ax 2 + bx + c = 0]

= e a( α − β) sin {sgn ( x )}  sin 1 =0  = lim +    sgn ( x )  x → 0  1  sin {(sgn x )}  sin (−1) and lim  = lim  − − sgn ( x )  x → 0  −1  x→ 0  = lim [sin 1] = 0 x → 0−

Hence, the given limit is 0. a

lim

a

52. lim (cos x + a sin bx ) x = e x → 0 x

(cos x + a sin bx − 1)

x→ 0

lim φ ( x ) [f( x ) − 1]   [f ( x )]φ ( x ) =e x → a , using xlim →a   where f ( x ) → 1 and φ ( x ) → ∞, as x → a a ( − sin x + ab cos bx ) 1 x→ 0 lim

 2 x2 +  2 x2 +

53. lim  x→ ∞

= lim

x→ ∞

3  5

5   1 +   2 x2 

 1 +   = lim  x→ ∞  1 +  

2

8x + 3

3   1 +   2 x2 

=e

c + dx

 c + dx   lim  x → ∞  a + bx 

 x 2 − 2 x + 2   x 2 − 4 x + 1

c + dx

a + bx  a + bx  1  = lim 1 +   x → ∞  a + bx   

=e

c   +d  lim  x  x→ ∞  a  + b  x 

= ed / b

x

56. lim  x→ ∞

     1 2x + 1   = lim 1 + 2  = lim 1 + 2 x→ ∞  x→ ∞  x − 4 x + 1 x − 4 x + 1    2x + 1 

x

x

51. lim +  x→ 0

=e

 1  55. lim 1 +  x→ ∞  a + bx 

= lim

x→ ∞

8x

2

8 x2

3   2 x2 

= ea

2

3   1 +   2 x2  5   1 +   2 x2 

8 x2 + 3

8 x2 + 3

3

3   1 +   2 x2  × 3 5    1 +  2 x2 

12 2 x2  3 

  2 20 2x  5  5    2 x2  

lim

=e

x → ∞ x2 − 4 x + 1

=e

sin 2 x 2 +2+ x x 54. The given limit is lim x→ ∞  sin 2 x  sin x e 2 +  x  0+2+0 = 1 (2 + 0 ) × (a value between and e ) e   Q lim sin x ∈ (−1, 1)  x → ∞  Hence, limit does not exist.

= lim   x → ∞  4

 7   3 1 + 1  12 = lim   x ⋅ 1 +  x → ∞  4 x2 −   x2

lim

3

 1    2+ x    4 1  lim  1 − +  x x→ ∞  x2 

x3  3 1 + x

 3x 2 + 57. lim  2  x → ∞  4 x − 1

 7  x → ∞  1 + = 0 ⋅  12   e 

3   1 +   e 12 1 2 x2  × = × = e− 8 3 e 20 1 5   1 +   2 x2 

     

= e2 x3

x3 1  1 +x

b

x ( 2 x + 1) x2 − 4 x + 1  x2 − 4 x + 1 2x + 1 

     1  = lim 1 + 2 x → ∞  x − 4 x + 1   2x + 1    x ( 2 x + 1)

18 Introduction to Limits and Derivatives

x→ α

   1 4

 2 1 1 + x +  3 .  x2 − 1  4

x

4 x3 1 1 + x ⋅ 2 4x −1 x − 4 2

    

Targ e t E x e rc is e s

50. lim (1 + ax 2 + bx + c )1/( x − α)

4 1  1   4 − 2 x  x 

= 0 ⋅ (e 7/ 12 )1 = 0

58. Let y = lim x x x→ 0

⇒ log y = lim x log x = lim x→ 0

x→ 0

log x 1/ x

∞  form  ∞ 

1/ x = − lim x = 0 x→ 0 − 1/ x 2 0 y =e =1 = lim

x→ 0

⇒

59. lim (cos x + sin x→ 0

lim

x→ 0

=e = e1 = e

1 x)x

=e

lim

1

x→ 0 x

( − sin x + cos x ) 1

(cos x + sin x − 1)

[using L’ Hospital’s rule] b   lim n  1 + tan − 1   n

n

60. lim 1 + tan  = e n → ∞   b n

n→ ∞

=e

lim n tan

n→ ∞

b n

=e

lim

n→ ∞

tan b/ n ×b b/ n

= eb

1027

Objective Mathematics Vol. 1

18

xn  x  1 1 61. lim 1 + n  = lim 1 + n   x → ∞  x→ ∞  x  x     e 0 = 1, if n > 1 lim x1 − n  = e x→ ∞ = e ∞ = ∞, if n < 1  1 e = e , if n = 1

 x 4 + x 2 + x + 1 62. lim   x→ −1  x2 − x + 1 

2 x − x2 x→ 2 x x − 2 2 2 x log 2 − 2 x x → 2 x x (1 + log x ) 4 log 2 − 4 = 4 (1 + log 2 ) log 2 − 1 = log 2 + 1

= lim

( x + 1) 2

69. Clearly, for n = 1, lim

 x 4 + x 2 + x + 1 = lim   x→ −1  x2 − x + 1  

x

x→ 0

 x + 1 2 sin 2    2 

70. Q

( x + 1) 2

  x + 1  sin    2  1   2   x + 1       2  

 2 =   3



Ta rg e t E x e rc is e s

m→ ∞ n → ∞

2

m→ ∞ n → ∞

 0, if 0 < x < 1 [x]  64. 2 =  1 , if − 1 < x < 0 x  x 2

sin ([ x − 3])

sin 4  = −1  4  sin [ x − 3] RHL = lim   + x → 0  [ x − 3]  =

Q ∴

x→ ∞

x→1

m

∞  form   ∞

1 x = lim 2 log x = lim n −1 x→ ∞ x→ ∞ nx nx n 2 = lim 2 n = 0 x→ ∞ n x 2 log x ⋅

1028

∞  form  ∞ 

3π   Q π < 4<  2 

sin (− 3) =   −3  sin 3   π  = =0 Q < 3< π  3   2  LHL ≠ RHL sin ([ x − 3]) does not exist. lim x→ 0   [ x − 3]  πx  [0 ⋅ ∞ form]  2 (1 − x ) = lim x→1  πx  cot   2 −1 [using L’ Hospital’s rule] = lim x→1  πx  π − cosec 2   2 2 2 π x  2  = lim sin 2   = 2 π π x→1

73. lim (1 − x ) tan 

66. lim

(log x )2 xn

sin (− 4)

72. LHL = lim −  = [ x − 3]   − 4  x→ 0 

⇒ l does not exist. [ x ]2 0, if 0 < x < 1 = x2 0, if − 1 < x < 0 ⇒ m exists and is equal to 0. π 65. Let sin x = h, then as x → , h → 1 2 h − hh ∴ lim h → 1 1 − h + In h 1 − h h − h h In h [using L’Hospital’s rule] = lim 1 h→1 − 1+ h − h h − 2 h h In h − h h − 1 − h h ( In h )2 − 1 − 1 = lim = =2 h→1 −1 − 1/ h2  sin x n  n − m  x  sin x n = ⋅ lim   ⋅x  x → 0  xn  x → 0 (sin x )m  sin x  1, if n = m  = 0, if n > m ∞, if n < m 

lim (1 + 12 m )

m→ ∞ n → ∞

= lim lim (1 + 1) = 2

=e0 = 1

2

67. lim

=0

cos n ! πx = 1, x ∈ rational lim lim (1 + cos 2 m n ! πx ) = lim

2

x  sin ( x / 2 n )  − 2 lim  n  ⋅ x/ 2n  4n2 n→ ∞  x

In (sin x )

m→ ∞ n → ∞

2

n → ∞ 4n

lim

x→ π / 2

So, if cos n ! πx < 1, when x ∈irrational lim lim (1 + cos 2 m n ! πx ) = (1 + 0 ) = 1and if

lim − 2 n ⋅ sin    2n 

− 2 × lim

x  2 

71. We know that, cos θ ≤ 1∀ θ.

= e n→ ∞

=e

∴

x − sin x =1 x − sin x

π  =0  4 

1/ 2

2 x 

=e

π 0 or

x>0

2 x < 0 or x < 0 1 x>0 , x>0  = x 1 xm x  sin x   sin x  ⇒ m =m = m ⇒ lim m  x→ 0  x  x   If m > 0, then for values of x sufficiently close to 0, we have 1 sin x 1− < 0 for i = 1, 2, … , m. x − ai Ai = = − 1for i = m + 1, … , n ∴ − ( x − ai ) x − ai and Ai = = 1for i = 1, 2, … , m x − ai Now, lim ( A1 A2 … An ) = (− 1)n − m + 1 − x → am

and

lim ( A1 A2 … An ) = (− 1)n − m

+ x → am

Hence, lim ( A1 A2 … An ) does not exist.

( ax + bx 2 )

= e3 ⇒ ex→ 0x Since, limit value e 2 a does not involve b. ∴ b can have any value. 3 Thus, a = , b ∈ R 2 ⇒

22 32 x2  + 3 + L + 3 3 x→ ∞  x x x x  x ( x + 1) (2 x + 1) 1 = lim = x→ ∞ 3 6x 3 d d  d  128. f ′ ( x ) = (3 x ) = 3 Q [af ( x )] = a f ( x)  dx dx  dx ⇒ f′ (2 ) = 3 Hence, both statements are true and Statement II is the correct explanation for Statement I.  1 x − −   x x2 + 1 129. h ( x ) = = 2  1 x −1 x + −   x ( x 2 − 1) ⋅ 2 x − ( x 2 + 1) ⋅ 2 x ⇒ h′ ( x ) = ( x 2 − 1)2 3 2 x − 2 x − 2 x3 − 2 x = ( x 2 − 1)2 −4 x −4 x = = 2 ( x − 1)2 (1 − x 2 )2 3

Introduction to Limits and Derivatives

 log 2 x, log (−2 x ),  1 × 2,  f′ ( x) =  2 x 1  × − 2,  −2 x 1 f′ ( x) = , x ≠ 0 x

121. f ( x ) = log| 2 x | = 

 12

127. lim 

Targ e t E x e rc is e s

From Eqs. (i) and (ii), we have 1  6  f( x) =  8 x − + 10  28  x 1 ∴ y = x 2f ( x ) = (8 x 3 − 6 x + 10 x 2 ) 28 dy 1 ⇒ = (24 x 2 + 20 x − 6) dx 28 1 1 dy  = (24 − 20 − 6) = − ∴ dx  28 14 x = −1

f′ (1) = 0 and

 d  Q (constant ) = 0  dx  f′ (− 1) = 0

x → am

( x + 1)5 − 1 x→ 0 x Let x + 1 = h When x → 0, then h → 1 h5 − 1 h 5 − 15 = lim ∴Given limit = lim h→1 h − 1 h→1 h − 1 5 −1 = 5 (1) = 5 × 14 = 5 B. Let x = 1/ y 1    ∴Given limit = lim  x − x 2 loge 1 +    x→ ∞  x 

134. A. lim

0  form  0 

1033

Objective Mathematics Vol. 1

18

 1 loge (1 + y ) = lim  −  y→ 0  y  y2

= 0 + lim e

n

∏ n→ ∞

= lim

n→ ∞

r =2 n

1 −1 (1/ 3) (1)3 1 −1 (1/ 6) (1)6

=

6 × 1= 2 3

=e

 r − 1  r  r =2

n

∏

r =2

 r + 1    r 

 1 2 3 n − 1  3 4 5 n + 1 = lim  ⋅ ⋅ …    ⋅ ⋅ … n→ ∞  2 3 4 n  2 3 4 n  1 n+1 1 = lim ⋅ = n→ ∞ n 2 2 1 135. A. Let y = cosec x = sin x On differentiating y w.r.t. x, we get d d sin x (1) − (1) (sin x ) sin x × 0 − 1 × cos x dy dx dx = = 2 dx sin 2 x sin x 0 − cos x − cos x 1 = = × sin x sin x sin 2 x dy ⇒ = − cot x cosec x dx B. Let y = 3 cot x + 5 cosec x On differentiating y w.r.t. x, we get dy = − 3 cosec 2 x − 5 cosec x cot x dx C. Let y = 5 sin x − 6 cos x + 7 On differentiating y w.r.t. x, we get dy = 5 cos x − 6 (− sin x ) + 0 dx = 5 cos x + 6 sin x D. Let y = 2 tan x − 7 sec x On differentiating y w.r.t. x, we get dy = 2 sec 2 x − 7 sec x tan x dx  x + 2, x ≤ − 1 136. Given, f ( x ) =  2 cx , x > − 1 At x = − 1, limit exists. RHL = LHL ∴ ⇒ lim f ( x ) = lim f ( x ) x → − 1+

⇒ ⇒ ⇒ ∴

x → − 1−

lim f (− 1 + h ) = lim f (− 1 − h )

h→ 0

h→ 0

lim c (− 1 + h )2 = lim (− 1 − h + 2 )

h→ 0

=e =e

138. Let l = lim

h→ 0

c (− 1 + 0 )2 = 1 − 0 c =1

sin x

log ( 1/ x )

lim

sin x

Q lim (sin x )1/ x → 0    x→ 0 as (decimal)∞ → 0   

x → 0 cosec x

 1  x −   x2  lim − x → 0 cosec x cot x sin x tan x x→ 0 x lim

[applying L’ Hospital’s rule]

= e0 =1

1 − cos 2 x

3

4

cos 3 x

cos 4 x … n cos nx

2

x→ 0

x   n D  ∏ (cos rx )1/ r   r = 2 = − lim x→ 0 2x [using L’Hospital’s rule]

 r 2 − 1  2   r 

∏ 

 1 log    x

x→ 0

[using L’Hospital’s rule] y 1 1 = lim = lim = y → 0 ( y + 1)2 y y → 0 2( y + 1) 2 z1/ 3 − 1 z1/ 3 − 1 z −1 C. Given, lim 1/ 6 = lim × 1/ 6 z →1 z z → 1 z−1 z −1 −1

D. Given limit = lim

Ta rg e t E x e rc is e s

1

1   1−   y − loge (1 + y ) + 1 y = lim   = lim  2 y→ 0   y→ 0  2 y  y    

=

1034

137. Given limit = lim (sin x )1/ x + lim   x→ 0 x → 0  x

y=

Let

n

∏ (cos rx )1/ r

r =2 n

∑

1   log (cos rx ) r 

⇒

log y =

⇒

n 1 dy = − ∑ tan (rx ) y dx r =2

⇒

− Dy = y ∑ tan (rx )

r =2

n

r =2

⇒

 n  D  ∏ (cos rx )1/ r  = − y r = 2  y⋅

⇒

⇒ ⇒

n

∑

tan (rx )

r =2

n

∑

tan (rx )

1 [2 + 3 + 4 + L + n ] 2x 2 2 1  n (n + 1)  n + n − 2 = −1 =  2  2 4 2 n + n −2 2 = 10 ⇒ n + n − 42 = 0 4 (n + 7 ) (n − 6) = 0 ⇒ n = 6

L = lim

x→ 0

r =2

=

sin (1 + [ x ]) , for [ x ] ≠ 0  [x] 0, for [ x ] = 0  sin (1 + [ x ]) sin (1 − 1) lim f ( x ) = lim = =0 [x] −1 x → 0− x → 0−

139. Given, f ( x ) =  ⇒

  x2 2  ∫ sec t dt  140. Given, lim  0  x→ 0 x sin x     d x2 2 ∫0 sec t dt sec 2 x 2 ⋅ 2 x = lim dx = lim d x→ 0 sin x + x cos x x→ 0 ( x sin x ) dx 2 x ⋅ sec 2 x 2 = lim x → 0  sin x  x + cos x   x sin x  2 ×1 2  = Q lim =1 = =1  x → 0 x  1+ 1 2

⇒

− ax + sin ( x − 1) + a  1 −  x + sin ( x − 1) − 1     lim  x→1  

 sin ( x − 1) − a  ( x − 1)   sin ( x − 1) + 1 ( x − 1) 

(1 +

x

=

1 4

x)

=

1 4

2

1  1 − a  =   2  4 a = 0 or a = 2 a=2

⇒ ⇒ ⇒

1

2. Required limit =

   x → ∞ 

3. Given, lim

⇒ ⇒

1

a

∫ 0 (a +

dx

⇒ ∴

 x2 + x + 1  − ax − b = 4  x+1 

x 2 + x + 1 − ax 2 − ax − bx − b =4 x→ ∞ ( x + 1) (1 − a) x 2 + (1 − a − b)x + (1 − b) lim =4 x→ ∞ ( x + 1) 1 − a = 0 and 1 − a − b = 4 ⇒ b = − 4, a = 1 lim

4. Let 1 + a = y ⇒ ⇒

( y1/ 3 − 1) x 2 + ( y1/ 2 − 1)x + y1/ 6 − 1 = 0  y1/ 3 − 1 2  y1/ 2 − 1 y1/ 6 − 1 =0   x +  x+ y −1  y −1   y −1 

Now, taking lim on both sides, we get y →1

1 2 1 1 x + x+ =0 3 2 6 2 x 2 + 3x + 1 = 0

⇒ ∴

x = − 1, −

1 2

t log (1 + t ) dt t4 + 4 5. Given limit = lim x→ 0 x3 x log (1 + x ) x4 + 4 [using L’ Hospital’s rule] = lim x→ 0 3x 2 log (1 + x ) 1 1 1 1 = lim ⋅ 4 = ⋅ = x→ 0 3x x + 4 3 4 12 x

⇒ ⇒ ∴

2

)

= 2 b sin 2 θ 1 + b2 2b 1 + b2 2 sin θ = 1 as ≥1 2b θ = ± π /2

sin 2 θ =

g( x ) must be continuous. k x + 1 , 0 ≤ x ≤ 3 g( x ) =   mx + 2 , 3 < x ≤ 5

At x = 3, RHL = 3m + 2 and at x = 3, LHL = 2k ∴ 2 k = 3m + 2  k , 0≤x 2 (ii) (1 + i) is a real number or it is a complex number. (iii) If 7 > 3, then 6 < 14 (iv) 4 > 2 iff 0 < ( 4 − 2) Sol. (i) Let p : 3 > 5 q : 1> 2 Then, p ∧ q : 3 > 5 and 1 > 2. Here, pis false andq is false, therefore p ∧ q is false. Hence, the given statement is false and so its truth value is F. (ii) Let p : (1 + i ) is a real number. q : (1 + i ) is a complex number. Then, ( p ∨ q ) : (1 + i ) is a real number or it is a complex number. Here, pis false andq is true and therefore p ∨ q is true. Hence, the given statement is true and so its truth value is T. (iii) Let p : 7 > 3 q : 6 < 14 ∴ p ⇒ q : If 7 > 3, then 6 < 14 Here, p is true and q is true and therefore p ⇒ q is true. Hence, the given statement is true and its truth value is T. (iv) Let p : 4 > 2 and q : 0 < (4 − 2 ) The given statement is p ⇔ q . Here, both p and q are true and therefore p ⇔ q is true. Hence, the given statement is true and its truth value is T.

Quantifiers and Quantified Statements Quantifiers Other phrases which are frequently found in mathematical statements are ‘there exist’ ( ∃) and ‘for all’ ( ∀). These two phrases are called quantifiers. Depending upon the context, the phrase ‘there exist’ can also be replaced by the equivalent phrases ‘there is’ or ‘there is atleast one’ or ‘it is possible to find’ or ‘some’. Consider the statement p :There exists a rectangle whose all sides are equal. This means that there is atleast one rectangle whose all sides are equal.

Quantified Statement An open statement with a quantifier becomes a statement, called a quantified statement.

Example 8. Use quantifiers to convert each of the following open sentences defined on N , into a true statement. (i) x + 5 = 8

T

F

T

F

F

F

F

F

T

T

F

F

T

T

T

F

F

T

F

F

F

T

T

T

T

F

F

F

F

(ii) x 2 > 0

F

T

F

T

Sol. (i) ∃ x ∈ N such that x + 5 = 8 is a true statement, since

T

F

T

F

T T

i. ii.

~ {∀ x ∈ A: p( x ) is true} = {∃ x ∈ A: ~ p( x ) is true} ~ {∃ x ∈ A: p( x ) is true} = {∀ x ∈ A : ~ p( x ) is true}

Example 9. Write the negation of (i) For all positive integers n, n 2 + 41n + 41 is a prime number. (ii) There is a real number which is not a complex number. Sol. The required negation is (i) There is atleast one positive integer n for which n2 + 41n + 41 is not prime. (ii) Every real number is a complex number.

Logical Equivalence Logically Equivalence Statement Two compounds S 1 ( p, q, r, ... ) and S 2 ( p, q, r, ... ) are said to be logically equivalent or simply equivalent, if they have the same truth values for all logically possibilities. If statements S 1 ( p, q, r, K ) and S 2 ( p, q, r, ... ) are logically equivalent, then we write S 1 ( p, q, r, ... ) ≡ S 2 ( p, q, r, ... ) It follows from the above definition that two statements S 1 and S 2 are logically equivalent, if they have identical truth tables i.e. the entries in the last column of the truth tables are same. ● ●

X

F

r

(ii) x2 > 0, ∀ x ∈ N is a true statement, since the square of every natural number is positive.

Ø

S1

q

x = 3 ∈ N satisfies x + 5 = 8

X

q ⇔r

p

Negation of a Quantified Statement

p ⇒ q ≡ (~ p ∨ q) p ⇔ q ≡ (~ p ∨ q) ∧ (p ∨ ~ q)

Example 10. Prove that the statements S 1 : p ∧ ( q ⇔ r ) and S 2 : p ∧ ((~ q ∨ r ) ∧ (~ r ∨ q )) are equivalent.

19

Sol. Since, S1 : p ∧ (q ⇔ r ), hence

Mathematical Reasoning

X

Also, S 2 : p ∧ [(~ q ∨ r ) ∧ (~ r ∨ q )], we have ~ r ~q ∨ r ~ r ∨ q

p

q

r

~q

F

F

F

T

T

F

F

T

T

F

T

F

F

F

T

T

T

F

F

T

F

T

T

T

T

T

S2

T

T

F

F

T

F

F

T

F

T

F

F

F

T

T

F

T

T

T

T

T

T

T

F

T

F

F

F

F

T

F

T

F

F

F

T

T

T

Since, the last columns representing the truth values of statements S 1 and S 2 have same truth value pattern corresponding to a respective substatement truth pattern. Hence, S 1 and S 2 are equivalent.

Negation of a Compound Statement We have learnt about negation of a simple statement. Writing the negation of compound statements having conjunction, disjunctions, implication, equivalence, etc., is not very simple. So, let us discuss the negation of compound statement.

Negation of Conjunction If p and q are ~ ( p ∧ q ) ≡ (~ p∨ ~ q ).

two

statements,

then

statements,

then

statements,

then

Negation of Disjunction If p and q are ~ ( p ∨ q ) ≡ (~ p∧ ~ q ).

two

Negation of Implication If p and q are ~ ( p ⇒ q ) ≡ ( p∧ ~ q ).

two

Negation of Biconditional Statement or Equivalence If p and q are two statements, then ~ ( p ⇔ q ) ≡ ( p∧ ~ q ) ∨ ( q ∧ ~ p) 1045

X

Objective Mathematics Vol. 1

19

Example 11. Write down the negation of (i) 3 + 6 > 8 and 2 + 3 < 6 (ii) Ramesh is cruel or he is strict. (iii) If ∆ABC is isosceles, then the base angles A and B are equal. (iv) | a | < 2 if and only if (a > − 2 and a < 2). Sol. (i) Let p : 3 + 6 > 8 q :2 + 3< 6 ∴ p ∧ q : 3 + 6 > 8 and 2 + 3 < 6 Q ~( p ∧ q ) ≡ ~ p ∨ ~ q So, negation of the given statement is 3 + 6 ≤ 8 or 2 + 3 ≥ 6

Inverse of Conditional Statement If p and q are two statements, then the inverse of ‘if p then q’ is ‘if ~ p then ~ q. e.g. Consider the statement ‘if a number is a multiple of 9, then it is a multiple of 3’. It is an implication with ‘if − then’ having antecedent ( p) and consequent ( q ) as given below : p : A number is multiple of 9. q : A number is multiple of 3. The given statement is ‘if p then q’. Its inverse is ‘if ~ p then ~ q’ i.e. if a number is not a multiple of 9, then it is not a multiple of 3.

Contrapositive of Conditional Statements

(ii) Let p : Ramesh is cruel. q : He is strict. ∴ p ∨ q : Ramesh is cruel or he is strict. Q~( p ∨ q ) ≡ ~ p ∧ ~ q ∴ Negation of the given statement is, Ramesh is not cruel and he is not strict. That is, Ramesh is neither cruel nor strict. (iii) Let p : ∆ABC is isosceles. q : The base angles A and B are equal. Q ~( p ⇒ q ) ≡ ( p ∧ ~ q ) ∴ The negation of the given statement is given by ∆ABC is isosceles and the base angles A and B are not equal.

If p and q are two statements, then contrapositive of the implication ‘if p then q’ is ‘if ~ q then ~ p’ i.e. let p → q be conditional statement, then contrapositive of this conditional statement is ~ q → ~ p. e.g. Let the conditional statement be ‘if there is a strike of buses, then I will hire taxi’, then its contrapositive statement is as follows: ‘I will not hire taxi, if there is no strike of buses’. Ø The contrapositive of a conditional statement have identical truth

values as p → q have, then the truth table is given by the following way:

(iv) Let p :|a| < 2 q : a > − 2 and a < 2 The given statement is p ⇔ q . of the given statement is ∴ Negation ( p ∧ ~ q ) ∨ (q ∧ ~ p) given by Either |a| < 2 and (a ≤ − 2 or a ≥ 2) or (a > − 2 and a < 2) and|a| ≥ 2. X

Example 12. Which of the following logically equivalent to ~ (~ p ⇒ q )? (a) p ∧ q (b) p ∧ ~ q (c) ~ p ∧ q (d) ~ p ∧ ~ q

is

X

~(~ p ⇒ q ) ≡ ~ p ∧ ~ q

Converse, Inverse and Contrapositive of an Implication

Ø If p → q is true, then its converse q → p may or may not be true

1046

e.g. Consider the following conditional statement ‘if a triangle is equilateral, then it is equiangular’. The statement is true and converse statement ‘if a triangle is equiangular, then it is equilateral’ is also true. But if we consider another conditional statement, ‘if a number is real number, then it is complex number’. It is true but its converse i.e. ‘if a number is complex number, then it is real number’ is not true.

III

IV

V

VI

p

q

p→ q

~p

~q

~q→~p

T

T

T

F

F

T

T

F

F

F

T

F

F

T

T

T

F

T

F

F

T

T

T

T

Example 13. The inverse of the proposition ( p ∧ ~ q ) ⇒ r is (a) ~ r ⇒ ~ p ∨ q (b) ~ p ∨ q ⇒ ~ r (c) r ⇒ p ∧ ~ q (d) None of these Sol. (b)QInverse of p ⇒ q is ~ p ⇒ ~ q ∴ Inverse of ( p ∧ ~ q ) ⇒ r is ~( p ∧ ~ q ) ⇒ ~ r i.e. (~ p ∨ q ) ⇒ ~ r

Converse of Conditional Statement If p and q are two statements, then converse of the implication is ‘if p then q’ is ‘if q then p’.

II

On comparing III and VI column, it is verified that p→ q ≡~q→ ~p

Sol. (d) Since, ~( p ⇒ q ) ≡ p ∧ ~ q ∴

I

X

Example 14. The contrapositive of ( p ∨ q ) ⇒ r is (a) r ⇒ ( p ∨ q ) (b) ~ r ⇒ ( p ∨ q ) (c) ~ r ⇒ ~ p ∧ ~ q (d) p ⇒ ( q ∨ r ) Sol. (c)QContrapositive of p ⇒ q is ~ q ⇒ ~ p ∴ Contrapositive of ( p ∨ q ) ⇒ r is ~ r ⇒ ~( p ∨ q ) i.e. ~ r ⇒ (~ p ∧ ~ q )

Let p, q, r, ... be statements, then any statement involving p, q, r, ... and the logical connectives ∧, ∨, ~, ⇒, ⇔ is called a statement pattern or a Well Formed Formula (WFF). e.g. (i) p ∨ q (ii) p ⇒ q (iii) [( p ∧ q ) ∨ r ] ⇒ ( s∧ ~ s) (iv) ( p ⇒ q ) ⇔ (~ q ⇒ ~ p) etc., are statement patterns. A statement is also a statement pattern. Thus, we can define statement pattern as follows:

A compound statement with the repetitive use of the logical connectives is called a statement pattern or a well formed formula.

A statement pattern is called a tautology, if it is always true, whatever may be the truth values of constitute statements. A tautology is called a theorem or a logically valid statement pattern. A tautology contains only T in the last column of its truth table.

A statement pattern is called a contradiction, if it is always false, whatever may the truth values of its constitute statements. In the last column of the truth table of contradiction, there is always F.

X

The negation of a tautology is a contradiction and vice-versa. If a statement is neither tautology nor contradiction, then the statement is called contigency.

Example 15. The statement p ∨ ~ ( p ∧ q ) is a (a) tautology (b) contradiction (c) contigency (d) None of these Sol. (a) p

q

p∧q

~( p ∧ q )

T

T

T

F

p ∨ ~( p ∧ q ) T

T

F

F

T

T

F

T

F

T

T

F

F

F

T

T

Hence, all values of a given statement are true, so it is a tautology. X

q ⇔r T F F T T F F T

p ∧ (q ⇔ r ) T F F T F F F F

19

Idempotent Laws For any statement p, we have (i) p ∧ p ≡ p (ii) p ∨ p ≡ p

For any two statements p and q, we have (i) p ∧ q ≡ q ∧ p (ii) p ∨ q ≡ q ∨ p

Associative Laws For any three statements p, q and r, we have (i) ( p ∧ q ) ∧ r ≡ p ∧ ( q ∧ r ) (ii) ( p ∨ q ) ∨ r ≡ p ∨ ( q ∨ r )

Distributive Laws For any three statements p, q and r,we have (i) p ∨ ( q ∧ r ) ≡ ( p ∨ q ) ∧ ( p ∨ r ) (ii) p ∧ ( q ∨ r ) ≡ ( p ∧ q ) ∨ ( p ∧ r )

Contradiction

●

r F T F T F T F T

Algebra of Statements

Tautology

●

q F F T T F F T T

Commutative Laws

Statement Pattern

Ø

p T T T T F F F F

Mathematical Reasoning

Sol. (c)

Tautologies and Contradictions

Example 16. The statement p ∧ (q ⇔ r ) is a (a) tautology (b) contradiction (c) contigency (d) None of these

Involution Laws For any statement p, we have ~ (~ p) ≡ p

De-Morgan’s Laws If p and q are any two statements, then (i) ~ ( p ∧ q ) = ~ p ∨ ~ q (ii) ~ ( p ∨ q ) =~ p ∧ ~ q

Contrapositive Law For any two statements p and q, we have p ⇒ q ≡ (~ q ) ⇒ (~ p)

Identity Law For any statement p, we have (i) p ∧ T = p (ii) p ∨ F = p (iii) p ∨ T = T where, T is a true statement and F is a false statements. X

Example 17. ( p ∧ ~ q ) ∧ (~ p ∧ q ) is (a) a tautology (b) a contradiction (c) both a tautology and a contradiction (d) neither a tautology nor a contradiction Sol. (b) ( p ∧ ~ q ) ∧ (~ p ∧ q ) = ( p ∧ ~ p) ∧ (~ q ∧ q ) = f ∧ f

=f [by using associative laws and commutative laws] ∴ ( p ∧ ~ q ) ∧ (~ p ∧ q ) is a contradiction.

1047

Objective Mathematics Vol. 1

19

X

Example 18. Simplify ( p ∨ q ) ∧ ( p ∨ ~ q ). (a) p (b) T (c) F (d) q Sol. (a) ( p ∨ q ) ∧ ( p ∨ ~ q ) ≡ p ∨ (q ∧ ~ q ) ≡ p∨F = p

[distributive law] [complement law] [F is identity for ∨]

Duality Two compound statements S 1 and S 2 are said to be duals of each other, if one can be obtained from the other by replacing ∧ by ∨ and ∨ by ∧. Remarks (i) The connectives ∧ and ∨ are also called duals of each other. (ii) If a compound statement contains the special variable t (tautology) or c (contradiction), then to obtain its dual, we replace t by c and c by t in addition to replacing ∧ by ∨ and ∨ by ∧.

(iii) Let S ( p, q ) be a compound statement containing two substatements and S * ( p, q ) be its dual. Then, (a) ~ S ( p, q ) ≡ S * (~ p, ~ q ) (b) ~ S * ( p, q ) ≡ S (~ p, ~ q ) (iv) The above result can be extended to the compound statements having finite number of substatements. Thus, if S ( p1 , p2 , ..., pn ) is a compound statement containing n substatements p1 , p2 , K, pn and S * ( p1 , p2 , ..., pn ) is its dual. Then, (a) ~ S ( p1 , p2 , ..., pn ) ≡ S * (~ p1 , ~ p2 , ..., ~ pn ) (b) ~ S * ( p1 , p2 , ..., pn ) ≡ S (~ p1 , ~ p2 , ..., ~ pn ) X

Example 19. The dual of the statement ( p ∨ q ) ∧ ~ ( r ∨ s) is (a) ( p ∨ q ) ∨ ~ ( r ∨ s) (b) ( p ∧ q ) ∨ ~ ( r ∧ s) (c) ( p ∧ q ) ∨ (~ r ∨ ~ s) (d) ( p ∧ q ) ∨ (~ r ∧ ~ s) Sol. (b,c) Let S : ( p ∨ q ) ∧~(r ∨ s ), then dual of S is S * : ( p ∧ q ) ∨ ~(r ∧ s ) ≡ ( p ∧ q ) ∨ (~ r ∨ ~ s )

Work Book Exercise 1 Which of the following is an open statement? a x is a natural number b Give me a glass of water c Wish you best of luck d Good morning to all

conjunction? Ram and Shyam are friends Both Ram and Shyam are tall Both Ram and Shyam are enemies None of the above

3 Negation of ‘2 + 3 = 5 and 8 < 10’ is a 2 + 3 ≠ 5 and 8 < 10 c 2 + 3 ≠ 5 or 8 0.

be

20. The given conditional statement is p → ~ q. The contrapositive of this statement is

p q ( p ∧ q ) ~ ( p ∧ q ) q ⇔ p ~ (q ⇔ p)

~( p ∧ q ) ∨ ~ (q ⇔ p)

T

T

T

F

T

F

F

T

F

F

T

F

T

T

F

T

F

T

F

T

T

F

F

F

T

T

F

T

It is clear that, the given statement is neither a tautology nor a contradiction.

~(~ q ) → ~ p ≡ q → ~ p

21. ~( p ∨ q ) ∨ (~ p ∧ q ) ≡ (~ p ∧ ~ q ) ∨ (~ p ∧ q ) ≡ ~ p ∧ (~ q ∨ q ) ≡ ~ p

22. S ( p, q, r ) = (~ p) ∨ [~(q ∧ r )] = (~ p) ∨ [~ q ∨ ~ r ] ⇒

S (~ p, ~ q, ~ r ) = p ∨ (q ∨ r )

23. ~( p ∨ q ) ≡ ~ p ∧ ~ q Hence, 7 is greater than 4 and Paris is not in France.

1057

20 Statistics Measures of Central Tendency An average value generally lies in the central part of the distribution and therefore, such values are called the measures of central tendency. The following are the nine measures of central tendency: (i) Arithmetic mean (ii) Weighted arithmetic mean (iii) Combined mean (iv) Geometric mean (v) Harmonic mean (vi) Relation among AM, GM and HM (vii) Median (viii) Quartiles, deciles and percentiles (ix) Mode

Chapter Snapshot ●

●

Measures of Dispersion

●

Skewness

●

●

Arithmetic Mean

X

●

Arithmetic mean of ungrouped or individual data If x1 , x 2 , x 3 , … , x n are n observations, then mean by (a) Direct method x + x 2 +…+x n 1 n x= 1 = ∑ xi n n i =1 (b) Shortcut method 1 n x = A + ∑ di n i =1 where, A = Assumed mean and d i = x i − A

i.

Example 1. The AM of n numbers of a series is X . If the sum of first ( n −1) terms is k, then the nth number is (a) X − k (b) nX − k (c) X − nk (d) nX − nk Sol. (b) Let the n numbers be x1, x2 , …, xn . Then, X= ⇒ ⇒ ⇒

X=

1 n ∑ xi ni =1 x1 + x2 + … + xn − 1 + xn

k + xn n xn = nX − k X=

n [Q x1 + x2 + …+ xn − 1 = k]

Measures of Central Tendency

●

●

●

Some Results to be Remembered Correlation Analysis Characteristics of Correlation Coefficient Regression Analysis Properties of Regression Coefficients Properties of Lines of Regression

Example 2. The mean of data 55, 155, 255, 355, 455 and 555 is (a) 310 (b) 305 (c) 300 (d) None of these

X

Example 4. Calculate the arithmetic mean of the following frequency data. x

12.58

13.58

14.58

15.58

16.58

f

4

3

2

4

2

Sol. (b) Let A = 255, then

Sol. Let us take assumed mean (A) = 14.58

d1 = − 200, d 2 = − 100, d 3 = 0, d 4 = 100, d 5 = 200 and d 6 = 300 6

∴Mean ( x ) = A +

ii.

∑di

i =1

6

 −200 − 100 + 0 + 100   + 200 + 300  = 255 + 6 300 = 255 + 6 = 255 + 50 = 305

Arithmetic mean of grouped data If x1 , x 2 , x 3 , … , x n are n observations whose corresponding frequencies are f 1 , f 2 , f 3 , … , f n , then mean by (a) Direct method

∑

i =1

f i xi

n

∑ fi

x

f

d = x− A

fd

12.58

4

−2

−8

13.58

3

−1

−3

14.58

2

0

0

15.58

4

1

4

16.58

2

2

4

Total

Σf = 15

Σfi d i Σfi 3 x = 14.58 − 15 x = 14.38

∴ ⇒ X

Marks Number of students

(b) Shortcut method n

∑ f i di

x =A+

n

∑ fi

(a) 20

where, A = Assumed mean and d i = x i − A Ø If the grouped data represents the frequencies included between a

X

1 1 1 1 1 Example 3. If the values 1, , , , , … , 2 3 4 5 n occur at frequencies 1, 2, 3, 4, 5, … , n, in a distribution, then the mean is (a)1 (b) n 1 2 (d) (c) n n +1 Sol. (d) Mean = =

1⋅ 1 +

1 1 1 1 1 ⋅2 + ⋅ 3 + ⋅ 4 + ⋅ 5 + … + ⋅ n n 5 4 3 2 1+ 2 + 3 + … + n

1+ 1+ 1+ 1+ … + 1 n 2 = = n(n + 1) n(n + 1) n + 1 2 2

0-10 10-20 20-30 30-40 40-50 50-60 12

18

(b) 28

27

20

(c) 2800

17

6

(d)100

Sol. (b)

i =1

specific class interval. So, we actually do not know how many times a particular observation lying in between that class limit. So, assume that the given frequency is the frequency of mid-point of that particular class interval. This mid-point is known as class mark. Thus, class mark of an interval (Lower limit of interval + Upper limit of interval) = 2

[Qassumed mean (A) = 14.58]

Example 5. The arithmetic mean of the marks from the following table is

i =1

i =1

Σfd = − 3

x= A+

Q

n

x f + x 2 f 2 +… + x n f n = x= 1 1 f 1 + f 2 +… + f n

20 Statistics

X

Marks

Class marks ( x )

f

fx

0-10

5

12

60

10-20

15

18

270

20-30

25

27

675

30-40

35

20

700

40-50

45

17

765

50-60

55

Total

∴

iii.

x=

6

330

Σf = 100

Σfx = 2800

Σfx 2800 = = 28 Σf 100

(c) Step deviation method When the class intervals in a grouped data are given and equal in length, then the arithmetic mean can be obtained by using the formula  Σf u  x = A+ i i h  Σf i  where, A = Assumed mean x −A ui = i h f i = Frequency h = Width of class interval

1059

X

Objective Mathematics Vol. 1

20

Sol. (a) The required mean

Example 6. Find the mean for the data in following table. Income per day

Number of persons

0-100

4

100-200

8

200-300

9

300-400

10

400-500

7

500-600

5

600-700

4

700-800

3

(a) 350 (c) 358

fi

Mid value ( xi )

=

xi − A , h

A = 350, h = 100

fu i i

4

50

−3

−12

8

150

−2

−16

200-300

9

250

−1

−9 00

300-400

10

350

00

400-500

7

450

01

07

500-600

5

550

02

10

600-700

4

650

03

12

700-800

3

750

04

Total

Σfi = 50

12 Σfi u i = 4

Σfi u i 4 × h = 350 + × 100 Σfi 50

= 350 + 8 = 358

Weighted Arithmetic Mean If w1 , w2 , w3 , … , wn are the weights assigned to the values x1 , x 2 , x 3 , … , x n respectively, then the weighted arithmetic mean is defined as w x + w2 x 2 +… + wn x n Weighted AM = 1 1 w1 + w2 +… + wn n

∑ wi x i

1060

n

=

n

∑r⋅ r

n −1

Cr

r =1

n

∑

Cr

n

−1

= n

n∑

Cr

r =1

n −1

r =1 n

∑

Cr

n

−1

Cr

r =1

n⋅2n − 1 2n − 1

i =1

where, x12 = Combined mean of the two data sets x1 = Mean of the first data set x 2 = Mean of the second data set n1 = Size of the first data set n2 = Size of the second data set X

Example 8. The mean height of 25 male workers in a factory is 61 cm and the mean height of 35 female workers in the same factory is 58 cm. The combined mean height of 60 workers in the factory is (a) 59. 25 (b) 59. 5 (c) 59.75 (d) 58.75 Sol. (a) Qn1 = 25, n2 = 35, x1 = 61, x2 = 58 ∴

Geometric Mean i.

n

∑ wi

Example 7. The weighted mean of first n natural numbers whose weights are equal to the number of selections out of n natural numbers of corresponding numbers, is 3n( n + 1) n ⋅ 2n − 1 (a) n (b) 2(2n + 1) 2 −1 ( n + 1)(2n + 1) n( n +1) (c) (d) 6 2

n1 x1 + n2 x2 n1 + n2 (25)(61) + (35)(58) = 25 + 35

x=

= 59.25

i =1

X

∑

n

Cr

If we are given the AM of two data sets and their sizes, then the combined AM of two data sets can be obtained by the formula n x + n2 x 2 x12 = 1 1 n1 + n2

0-100

=

r =0 n

n

Combined Mean

100-200

Mean ( x ) = A +

∑r⋅ r =1

Sol. (c)

Class

1⋅n C1 + 2 ⋅n C 2 + 3 ⋅n C 3 + … + n ⋅n C n n C1 + nC 2 + … + nC n n

=

(b) 359 (d) None of these ui =

x=

Geometric mean for ungrouped data If x1 , x 2 , x 3 , … , x n are n observations, none of them being zero, then their geometric mean given by GM = ( x1 ⋅ x 2 ⋅ x 3 … x n )1/ n  log x1 + log x 2 +L+ log x n  or GM = antilog     n

ii.

Geometric mean for grouped data Geometric mean of n observations x1 , x 2 , … , x n whose corresponding frequencies are f 1 , f 2 , f 3 , … , f n , is given by

f ⋅ x22…

1 fn N xn )

, where N =

n

∑ fi

X

i =1

 n   ∑ f i log x i  i =1  or GM = antilog     N

Example 10. The harmonic mean of 4, 8 and 16 is (a) 6.4 (b) 6.7 (c) 6.85 (d) 7.8 Sol. (c) HM of 4, 8 and 16 =

Ø In case of continuous frequency distribution, the values of the

variate x are taken to be the values corresponding to the mid-points of the class intervals.

iii.

X

Combined geometric mean If G1 , G2 , G3 , … , Gk are GM of k series of sizes n1 , n2 , … , nk , then GM of combined series is given by n log G1 + n2 log G2 +… + nk log Gk log G = 1 n1 + n2 +… + nk =

1 N

k

∑ ni log Gi ,

Sol. (b) We know that,

i =1

Harmonic mean =

Sol. (d) GM = (3 ⋅ 32 … 3n )1/ n 1+ 2 + … + n

=3

n n( n + 1)

=3

n+1

=3

2n

2

Σf = 1 + 2 + 3 + … + n =

n n {2 × 2 + n − 1} = (3 + n) 2 2 n(n + 1) 2 ∴ Harmonic mean = n(3 + n) 2 n(n + 1) × 2 n+1 = = n(3 + n) × 2 3 + n =

Harmonic Mean

ii.

Σf f Σ    x

n(n + 1) 2 f n 1 2 3 and Σ   = + + + …+  x  1/2 2 / 3 3/ 4 n/(n + 1) 3×2 4× 3 n(n + 1) =2 + + + …+ 2 3 n = 2 + 3 + 4 + … + n + (n + 1) which is an arithmetic progression with a = 2 and d = 1. By the formula of sum of n terms of an AP, f n Σ   =  {2 a + (n − 1) d }  x   2 

∴

Example 9. The geometric mean of the numbers 3, 3 2 , 3 3 , … , 3 n is (a) 3 2/ n (b) 3 ( n − 1)/ 2 n/2 (c) 3 (d) 3 ( n + 1)/ 2

i.

3 48 = = 6.85 1 1 1 7 + + 4 8 16

Example 11. Find the harmonic mean of 1 2 3 n occurring with frequencies , , , …, 2 3 4 n +1 1, 2, 3, … , n, respectively. n −1 n +1 (a) (b) 3−n 3+n n +1 (d) None of these (c) 3−n

where N = n1 + n2 + ... + nk X

20 Statistics

GM =

f ( x1 1

Harmonic mean for ungrouped data The harmonic mean of n observations x1 , x 2 , … , x n is defined as n HM = 1 1 1 + +… + x1 x 2 xn n or HM = n 1 ∑x i =1 i Harmonic mean for grouped data Harmonic mean of n observations x1 , x 2 , x 3 , … , x n which occur with frequencies f 1 , f 2 , … , f n respectively, is given by n

HM =

∑ fi

i =1 n

f 

∑  x i  i =1

i

Relation among AM, GM and HM The arithmetic mean (AM), geometric mean (GM) and harmonic mean (HM) for a given set of observations are related as under AM ≥ GM ≥ HM Equality sign hold only when all the observations are equal. X

Example 12. For three numbers a, b and c, product of the average of the numbers a 2 , b 2 , c 2 1 1 1 and 2 , 2 , 2 cannot be less than a b c (a)1 (b) 3 (c) 9 (d) None of these

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Objective Mathematics Vol. 1

20

h = Width of the median class cf = Cumulative frequency of the class preceding the median class

Sol. (a) Since, AM ≥ GM, therefore we have a2 + b 2 + c 2 …(i) ≥ (a2 b 2c 2 )1/ 3 3 1 1 1 + 2 + 2 1/ 3 2 1 1 1 a b c and …(ii) ≥  2 2 2  a b c  3 On multiplying Eqs. (i) and (ii), we get 1 1   1  a2 + b 2 + c 2   a2 + b 2 + c 2    ≥1   3 3       1 1 1 2 2 2 ∴Product of the averages of a , b , c and 2 , 2 , 2 a b c cannot be less than 1.

Ø Cumulative frequency The cumulative frequency of a value is its

frequency plus the frequencies of all smaller values. e.g.

X

Median i.

Median of an individual data Let n be the number of observations, then (a) Arrange the data in ascending or descending order. (b) If n is odd, then 1 Median = Value of the ( n + 1) th 2 observation If n is even, then n   n Median = Mean of the   th and  + 1 th 2   2 observations i.e. Median  1  n  n  =    th item +  + 1 th item  2  2  2  

iii.

1062

Median of a continuous frequency data (a) Prepare the cumulative frequency table. (b) Find the median class i.e. the class in N which the   th observation lies. 2 (c) The median value is given by the formula N   − cf 2 Median = l + ×h f where, l = Lower limit of the median class N = Total frequency f = Frequency of the median class

cf

1

4

4

2

6

10 (4 + 6)

3

4

14 (4 + 6 + 4)

Example 13. If a variable takes the discrete 7 5 1 1 values α + 4, α − , α − , α − 3, α − 2, α + , α − , 2 2 2 2 α + 5 (α > 0), then the median is 5 1 5 (a) α − (b) α − (c) α − 2 (d) α + 4 2 4 7 5 1 1 α − , α − 3, α − , α − 2, α − , α + , α + 4, α + 5 2 2 2 2 1 Median = [Value zof 4th item + Value of 5th item] 2 [Qn = 8, which is even] 1 α −2 + α − 2 Median = ∴ 2 5 2α − 2 =α − 5 = 2 4

■

Median of a discrete frequency data (a) Arrange the values of the variate in ascending or descending order. (b) Prepare a cumulative frequency table. N (c) Find , where N = Σ f i and see the 2 cumulative frequency just greater than or N equal to , then the corresponding value 2 of x is median.

f

Sol. (a) Arrange the data in ascending order as

■

ii.

x

X

X

Example 14. The value of median for the data Income (in `)

1000

1100

1200

1300

1400

1500

Number of persons

14

26

21

18

28

15

is (a)1300

(b)1200

(c)1250

(d)1150

Sol. (c) Income ( x )

Frequency ( f )

cf

1000

14

14

1100

26

40

1200

21

61

1300

18

79

1400

28

107

1500

15

122

Here, Q Q

N = 122 n 1 n  Median = th term +  + 1 th term  2  2  2 1 Median = (61th + 62 th) terms 2 1 = (1200 + 1300) = 1250 2

Example 15. From the data given, the median of the average deposit balance of saving for the branch during March 1982 is Average deposit balance (in `)

Number of deposit

Less than 100

26

100-200

68

200-300

145

300-400

242

400-500

188

500-600

65

600-700

16

(a) 356 (c) 56. 2

where, l is lower limit of the required class, i is width of class interval, f is frequency of the class and cf is sum of all frequencies just above the class of quartile/decile/percentile. X

Cumulative frequency (cf )

Less than 100

26

26

100-200

68

94

200-300

145

239

300-400

242

481

400-500

188

669

500-600

65

734

600-700

16

750

N 750 Here, = = 375 2 2 The frequency 375 lies in the class 300- 400. So, median class is 300-400.  N − cf     Median = l + 2 ×h ∴ f 375 − 239 = 300 + × 100 242 = 300 + 56.2 = 356.2

Quartiles, Deciles and Percentiles i.

20

 Nr  i Pr = l +  − cf  ; r = 1, 2, 3, …, 99  100  f

(b) 300 (d) 356. 2

f

For grouped data Arrange data in ascending order and prepare cumulative frequency.  Nr  i Qr = l +  − cf  ; r = 1, 2, 3  4  f  i  Nr Dr = l +  − cf  ; r = 1, 2, 3, … , 9  f  10

Sol. (d) Average deposit balance (in `)

ii.

Statistics

X

For ungrouped and discrete frequency data Quartiles are also a kind of positional averages which divide the complete distribution into four equal parts. ( N + 1) r Qr = ; r = 1, 2, 3 4 Deciles divide the frequency distribution into 10 equal parts. ( N + 1) r Dr = ; r = 1, 2, 3, … , 9 10 Percentiles divide frequency distribution into 100 equal parts. ( N + 1) r Pr = ; r = 1, 2, 3, … , 99 100

Example 16. The profits earned by 100 companies during 2012-13 are given below: Profits (in ` lakh)

Number of companies

Profits (in ` lakh)

Number of companies

20-30

4

60-70

15

30-40

8

70-80

10

40-50

18

80-90

8

50-60

30

90-100

7

Calculate Q1 , median, D4 and P80 and interpret the values of cf. Sol. Calculation of Q1, Q2 , D4 and P80 Profits (in ` lakh)

f

cf

20-30

4

4

30-40

8

12

40-50

18

30

50-60

30

60

60-70

15

75

70-80

10

85

80-90

8

93

90-100

7

100

100 N th observation = = 25th observation 4 4 So, Q1 lies in the class 40-50. N / 4 − cf ×i Q1 = l + f 25 − 12 = 40 + × 10 18 = 40 + 7.22 = 47.22 25% of the companies earn an annual profit of ` 47.22 lakh or less. 2N Median or Q2 = Size of th observation 4 200 = = 50th observation 4 So, Q2 lies in the class 50-60. 2N / 4 − c f ∴ Q2 = l + ×i f 50 − 30 = 50 + × 10 = 50 + 6.67 = 56.67 30 Q1 = Size of

1063

Objective Mathematics Vol. 1

20

50% of the companies earn an annual profit of ` 56.67 lakh or less. 4N th observation = 40th observation D4 = Size of 10 So, D4 lies in the class 50-60. 4N / 10 − cf 40 − 30 ∴ × i = 50 + D4 = l + × 10 f 30 = 50 + 3.33 = 53.33

X

Example 17. The mode of the following distribution is (a) 46 (b) 666 . (c) 4667 (d) None of these .

Thus, 40% of the companies earn an annual profit of ` 53.33 lakh or less. 80N th observation P80 = Size of 100 80 × 100 = = 80th observation 100 So, P80 lies in the class 70-80. 80N /100 − c f 80 − 75 × i = 70 + P80 = l + × 10 f 10 = 70 + 5 = 75 This means that 80% of the companies earn an annual profit of ` 75 lakh or less and 20% of the companies earn an annual profit of more than ` 75 lakh.

ii.

iii.

0-10

5

10-20

8

20-30

7

30-40

12

40-50

28

50-60

20

60-70

10

70-80

10

40-50 is the modal class. ∴

Mode = 40 +

10 (28 − 12 ) (2 × 28 − 12 − 20)

= 40 + 6.666 = 46.67 (approx.)

Mode of individual data In the case of individual series, the value which is repeated maximum number of times is the mode of the series. e.g. Mode of 6, 10, 7, 5, 9, 3, 7, 5 is 7. Mode of discrete data In the case of discrete frequency distribution, mode is the value of the variate corresponding to the maximum frequency. Mode of continuous data (a) Find the modal class i.e. the class which has maximum frequency. The modal class can be determined either by inspection or with the help of grouping table. (b) The mode is given by the formula fm − fm−1 Mode = l + ×i 2fm − fm−1 − fm+1 where, l = Lower limit of the modal class i = Width of the modal class f m − 1 = Frequency of the class preceding the modal class f m = Frequency of the modal class f m + 1 = Frequency of the class succeeding modal class In case, the modal value lies in a class other than the one containing maximum frequency, we take the help of the following formula: fm+1 Mode = l + × i, fm−1 + fm+1

1064

Frequency

Sol. (c) Here, maximum frequency is 28. Thus, the class

Mode i.

Class interval

where, symbols have usual meaning.

Measures of Dispersion The dispersion of a statistical distribution is the measure of the deviations of its values about the average value. It gives an idea of scatteredness of different values from the mean. The following measures of dispersion are commonly used: (i) Range (ii) Quartile deviation (iii) Mean deviation (iv) Standard deviation

i.

Range It is the difference between the greatest and the smallest observations of the distribution. If L is the largest and S is the smallest observation in a distribution, then its Range = L − S L−S Also, coefficient of range = L+S e.g. Range of observations 3, 5, 9, 8, 7, 6, 5, 7, 7 4, 2 is 9 − 2 = 7 and their coefficient is . 11

ii.

Quartile deviation Quartile deviation or semi-interquartile range is given by 1 QD = (Q3 − Q1 ) 2 (a) Interquartile range Interquartile range is Q3 − Q1 . (b) Coefficient of quartile deviation Q − Q1 = 3 Q3 + Q1

Example 18. The quartile deviation of daily wages (in `) of 11 persons given below, is 140, 145, 130, 165, 160, 125, 150, 170, 175, 120, 180 (a) 22. 5 (b) 25 (c) 20 (d) 21. 2 Sol. (c) The given data in ascending order of magnitude is 120, 125, 130, 140, 145, 150, 160, 165, 170, 175, 180 n + 1 11 + 1 12 Here, Q1 =  = = 3rd term  th term =  4  4 4 = 130 3 (n + 1) 3 × (11 + 1) th = = 9th term = 170 Q3 = 4 4 Q − Q1 QD = 3 ∴ 2 170 − 130 = 2 40 = = 20 2

iii.

Mean deviation (a) For ungrouped data The mean deviation from M (mean or median or mode) is given by Σ| x i − M | MD = n (b) For grouped data The mean deviation from M (mean or median or mode) is given by n

∑ f i | xi − M |

MD =

i =1

n

∑ fi

i =1

(c) Coefficient of MD Mean deviation = Corresponding average (mean or median or mode) X

Example 19. The mean deviation from the mean of the AP a, a + d, a + 2d, …, a + 2nd is n( n + 1) d (a) n( n +1) d (b) 2n + 1 n( n +1) d n( n − 1) d (d) (c) 2n 2n + 1 Sol. (b) The mean of the series a, a + d , a + 2d , …, a + 2 nd is 1 [a + a + d + a + 2d + … + a + 2 nd ] x= 2n + 1 1 2 n + 1 = (a + a + 2 nd ) = a + nd  2n + 1 2  Therefore, mean deviation from mean 2n 2n 1 1 |(a + rd ) − (a + nd )| = |r − n|d = ∑ ∑ 2n + 1r = 0 2n + 1 r = 0 =

n(n + 1) 1 d ⋅ 2d (1 + 2 + … + n) = 2n + 1 2n + 1

iv.

Standard Deviation (SD) The standard deviation of a statistical data is defined as the positive square root of the squared deviations of observations from the AM of the series under consideration and it is generally denoted by σ read as sigma. The square of SD is called the variance and is denoted by σ 2 . (a) Computation of standard deviation Standard ● For ungrouped data deviation for ungrouped set of observations is given by (i) Direct method

20 Statistics

X

n

∑ (x i − x ) 2

SD (σ ) =

i =1

n

Σx 2  Σx  = −   n  n

2

(ii) Shortcut method Σ d 2  Σd  σ= −   n  n

2

where, A is assumed mean and d=x− A ● For grouped data SD for frequency distribution is given by (i) Direct method n

∑ f i (x i − x ) 2

SD (σ ) =

i =1

n

∑ fi

i =1

=

1 N

n

∑

i =1

f i x i2

 Σf x  − i i   N 

2

where, f i is the frequency of x i (1 ≤ i ≤ n). (ii) Shortcut method σ=

Σfd 2  Σfd  −   N  N

2

where, N = Σf and d = x − A (iii) Step deviation method σ=h

Σ fu 2  Σ fu  −   N  N

2

x−A and symbols have h their usual meaning.

where, u =

1065

20 Objective Mathematics Vol. 1

(b) Combined standard deviation Let A1 and A2 be two series having n1 and n2 observations respectively. Let their AM be x1 and x 2 and standard deviations be σ 1 and σ 2 . Then, the combined standard deviation σ or σ 12 of A1 and A2 is given by σ 12 or σ =

n1σ 12 + n2σ 22 + n1 d12 + n2 d 22 n1 + n2 n1 (σ 12

+

n2 (σ 22

and σ 2 = 7

d 22 )

Ø When the values of the variable are given in the form of class interval,

then their respective mid-points are taken as the values of the variable. X

Example 20. Find the standard deviation for the following data. 95-105 105-115 115-125 125-135 135-145

Height (in cm)

19

No. of children

23

36

70

52

Sol. Class Midinterval value 95-105

f

100

u =

x − 120 10

fu 2

19

−2

−38

76

105-115 110

23

−1

−23

23

36

0

0

0

125-135 130

70

1

70

70

135-145 140

52

2

104

208

Σf = 200

Σfu = 113 Σfu 2 = 377

We know that, σ=

∴Combined SD =

n1σ12 + n2 σ 22 n n ( x − x2 )2 + 1 2 1 n1 + n2 (n1 + n2 ) 2

=

60 × 64 + 80 × 49 60 × 80 (650 − 660)2 + 60 + 80 (60 + 80)2

=

3840 + 3920 (4800 × 100) + 140 (140)2

7760 480000 776 4800 + = + 140 19600 14 196 = 55.42 + 24.49 = 79.91 = 8. 94

=

X

Example 22. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below: Subject

Mathematics

Physics

Chemistry

Mean

42

32

40.9

Standard deviation

12

15

20

Which of these three subjects shows the highest variability in marks and which shows the lowest? (a) Chemistry, Mathematics (b) Mathematics, Physics (c) Chemistry, Physics (d) None of the above Sol. (a) Here, n = 50

fu

115-125 120

Total

Example 21. The mean life of a sample of 60 bulbs was 650 h and the standard deviation was 8 h. A second sample of 80 bulbs has a mean life of 660 h and standard deviation 7 h. Find the overall standard deviation. (a) 8. 97 (b) 8. 98 (c) 8. 94 (d) None of these Sol. (c) Given, n1 = 60, x1 = 650, σ1 = 8, n2 = 80, x2 = 660,

+ + = n1 + n2 where, d1 = ( x1 − x12 ), d 2 = ( x 2 − x12) and n x + n2 x 2 is the combined mean x12 = 1 1 n1 + n2 d12 )

n σ 2 + n2σ 2 n1 n2 ( x1 − x 2 ) 2 or σ 12 = 1 1 + n1 + n2 ( n1 + n2 ) 2 (c) Coefficient of variation (CV) or coefficient of standard deviation To compare two or more series for variability, the relative measure called coefficient of standard deviation or coefficient of variation. This measure is define as σ CV = × 100 x

2  1  Σfu  h2 2    Σfu −   N     N

2  377  113   σ = 100  − . − (0.565)2 }   = 100 {1885  200     200

. − 0.319225) = 100 (1565775 . ) ⇒ σ = 100(1885 = 12.51 So, SD of height is 12.51 cm.

1066

X

σ × 100 x 12 = × 100 42 2 = × 100 7 200 …(i) = = 28.57 7 15 σ For Physics, CV = × 100 = × 100 32 x 1500 …(ii) = = 46.87 32 σ For Chemistry, CV = × 100 x 2000 20 = × 100 = = 48.89 …(iii) 40.9 40.9 From Eqs. (i), (ii) and (iii), we have CV of Chemistry > CV of Physics > CV of Mathematics ∴ Chemistry shows the highest variability and Mathematics shows the least variability. For Mathematics,CV =

We study skewness to have an idea about the shape of the curve which we can draw with the help of the given data. The term skewness refers to lack of symmetry. We can define skewness of a distribution as the tendency of a distribution to depart from symmetry. (i) In a symmetrical distribution, we have Mean = Median = Mode

●

M = Mo = M d

Symmetrical distribution ●

(ii) When the distribution is not symmetrical, it is called asymmetrical or skewed. In a skewed distribution, Mean ≠ Median ≠ Mode (a) Positively skewed distribution In such a distribution, we have Mean > Median > Mode

M

X

Bowley’s coefficients of skewness (Q − Median) − (Median − Q1) Sk = 3 (Q3 − Median) + (Median − Q1) Q + Q1 − 2 Median = 3 Q3 − Q1 where, Q1 is the first quartile and Q3 is the third quartile. Thus, S k = + 1, if median = Q1 and S k = − 1, if Q3 = median. Bowley’s coefficient of skewness lies between −1and +1. Kelly’s coefficient of skewness P + P − 2P50 [based on percentiles] S k = 90 10 P90 − P10 D + D1 − 2D5 [based on deciles] = 9 D9 − D1 It should be noted that three different formulae of calculating skewness are based on different assumptions and hence the answer obtained from the same question by different method may differ.

Example 23. In a moderately skewed distribution, the values of mean and median are 5 and 6, respectively. The value of mode for such distribution is (a) 8 (b)11 (c)16 (d) None of these Sol. (a) For moderately skewed distribution, we have

M d Mo

Mode = 3 Median − 2 Mean = 18 − 10 = 8

Positively skewed distribution

(b) Negatively skewed distribution distribution, we have Mean < Median < Mode

M

In such a X

Md Mo

Ø

●

Absolute measures of skewness ã S = Mean − Median k ã S = Mean − Mode k ã S = Q + Q − 2Q or S = Q + Q − 2 (Median). k 3 1 2 k 3 1 Relative measures of skewness The following are three important relative measures of skewness, which can be used to compare the distributions with regard to symmetry. ã Karl Pearson’s coefficient of skewness Mean − Mode Sk = Standard deviation If mode is not defined, then using the relation, Mode = 3 Median − 2 Mean for a moderately skewed distribution, we get 3 (Mean − Median) Sk = Standard deviation It follows that S k = 0, if Mean = Mode = Median Skewness is positive, if Mean > Mode or Mean > Median and negative, if Mean < Mode or Mean < Median Pearson’s coefficient of skewness lies between −3 and 3.

Example 24. Karl Pearson’s coefficient of skewness of a distribution is 0. 32. Its SD is 6.5 and mean 396 . . Then, the median of the distribution is given by (a) 28.61 (b) 38.90 (c) 29.13 (d) 28.31 Sol. (b) We know that, S k = M − Mo ,

Negatively skewed distribution ●

20 Statistics

●

Skewness

σ

where M = Mean, Mo = Mode, σ = SD 39.6 − Mo i.e. 0.32 = 6.5 ⇒ Mo = 37.52 and also we know that, Mo = 3 Median − 2 Mean ⇒ 37.52 = 3 (Median) − 2(39.6) Median = 38.90 (approx.) ∴ X

Example 25. For a symmetrical distribution Q1 = 25 and Q3 = 45, then median is (a) 28 (b) 35 (c) 30 (d) 40 Sol. (b) For a symmetrical distribution, coefficient of skewness is zero. ∴ ⇒

70 − 2 M e =0 20 M e = 35

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Objective Mathematics Vol. 1

20 Some Results to be Remembered

2 (b) Semi-interquartile range = σ 3 = Quartile deviation

(i) In a statistical data, the sum of the deviations of individual values from AM is always zero i.e.

2 (c) Probable error of standard deviation = σ 3 = Semi-interquartile range 5 (d) Quartile deviation = MD 6 From these relationships, we have 4 SD = 5 MD = 6 QD

n

∑ (x i − x ) = 0

i =1

(ii) In a statistical data, the sum of squares of the deviations individual values from AM is least i.e. n

∑ f i (x i − x ) 2 is least.

i =1

X

(iii) If each of the n given observations is increased/decreased by a, then their mean is also increased/decreased by a. (iv) If mean of some observations is x and if each observation is multiplied or divide by k ≠ 0, then the new mean, so obtained will be equal to kx or x , respectively. k (v) Median is the point of intersection of less than and more than ogive. (vi) SD remains unchanged, if each variate value is raised or reduced by a constant scalar quantity i.e. SD is independent of change of origin. (vii) SD is not independent of change of scale, if y = ax + b, then σ y = | a | σ x . (viii) Standard deviation of n natural numbers 1  σ =  ( n 2 − 1)   12 

Sol. (c) Let X be the mean of x1, x2 ,…, xn , then the mean of x1 x2 x3 x x , , , …, n is and the mean of α α α α α x x x observations 1 + 10, 2 + 10, …, n + 10 is given by α α α x + 10 α x . + 10 i.e. α α observations

X

Example 27. An ogive is used to determine (a) mean (b) median (c) mode (d) harmonic mean Sol. (b) The intersection point of less than and more than

1/ 2

(ix) Standard deviation shows the limits of variability by which the individual observation in a distribution will vary from the mean. For a symmetrical distribution with mean X , the following area relationship holds good: X ± σ covers 68.27% observations. X ± 2σ covers 95.44% observations. X ± 3σ covers 99.73% observations. These limits are illustrated by the following curve known as normal curve:

Example 26. The arithmetic mean of a set of observations is X . If each observation is divided by α and then is increased by 10, then the mean of the new series is X +10 X +10α X (c) (d) α X +10 (a) (b) α α α

ogive is Median.

O Median

X

68.27%

Example 28. If the standard deviation of the observations −5, − 4, − 3, − 2, − 1, 0, 1, 2, 3, 4, 5 is 10. Then, standard deviation of observations 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 will be (b) 10 + 10 (a) 10 + 20 (d) None of the above (c) 10 Sol. (c) The new observations are obtained by adding 20 to each term. Hence, σ does not change.

X

95.44% 99.73% – X –3σ X–2σ X – σ

1068

X

X + σ X + 2σ X + 3σ

(x) Empirical relationships If the data is moderately non-symmetrical, then the following empirical relationships hold: 4 (a) Mean deviation = σ 5

Example 29. For a series, the value of mean deviation is 15. The most likely value of its quartile deviation is (a)12.5 (b)11.6 (c)13 (d) 9.7 Sol. (a) Since, MD = 4 σ, QD = 2 σ ⇒

5 3 5 5 MD 6 = ⇒ QD = (MD) = ⋅ 15 = 12.5 6 6 QD 5

20

1 Which of the following is most unstable average? a Mode c Mean

b Median d Geometric mean

2 Which of the following is affected most by

13 Coefficient of quartile deviation is c

(M→ Mean, Me → Median, Mo → Mode)

extreme observations? b Median d HM

4 The most stable measure of central tendency is b mode d None of these

5 Which of following is correct relation? a HM > GM > AM c AM > GM > HM

b HM > AM > GM d GM > HM > AM

6 Which of the following is correct relation for a symmetrical distribution? a b c d

AM − Mo = 3 (AM − M d ) AM − Mo = 2 (AM − Md ) Md = 2 AM − 3 Md AM + Mo = 3 (AM − Md )

7 One of the methods of determining mode is a b c d

mode mode mode mode

= 2 median − 3 mean = 2 median + 3 mean = 3 median − 2 mean = 3 median + 3 mean

8 Coefficient of variation is given by σ a   × 100  X X c σ

σ b X  X d   × 100 σ

9 The Karl Pearson’s coefficient of skewness is a b c d

always positive always negative both positive and negative None of the above

10 Bowley’s coefficient of skewness is given by Q 3 − Q1 + 2 Me Q 3 + Q1 Q 3 − Q1 − 2 M c Q 3 − Q1 a

Q 3 + Q1 − 2 Md Q 3 − Q1 Q 3 + Q1 − 2 M d Q 3 + Q1 b

11 The value of coefficient of skewness in Bowley’s formula lies between a −3 and 3 c −∞ and ∞

Q 3 − Q1 Q 3 + Q1

a M > Mo c M = Me

3 Which of the following is affected least with

a median c mean

b

14 A negative coefficient of skewness implies

Median Mode Harmonic mean Arithmetic mean

a Mode c GM

1 (Q 3 − Q1 ) 2 σ d X

a Q 3 − Q1

extreme observations? a b c d

Statistics

Work Book Exercise 20.1

b −1 and 1 d −∞ and 0

12 The value of coefficient of skewness in Karl

b M = Mo d M < Mo

15 In a positively skewed distribution, which is true? a M < Me < Mo c Mo > M > Me

b Me > M > Mo d M > Me > Mo

16 In a frequency curve of scores, it is found that the mode is greater than mean. Then, which is correct about frequency curve? a b c d

Positive skewed Negative skewed Symmetric None of the above

17 The correct empirical relation between the measures of dispersion is 3 SD 4 5 c MD = SD 4

4 SD 3 4 d MD = SD 5

a MD =

b MD =

18 If m is mean of distribution, then Σ( xi − m) is equal to a mean deviation c 0

b standard deviation d −1

19 The semi-interquartile range = λσ, then λ is a

2 5

b

2 3

c

1 5

d

20 The root mean square deviation is least when deviations are measured from a mean c median

b mode d origin

21 In a symmetrical frequency distribution, lower and upper quartiles are equidistant from a mean deviation c standard deviation

b median d None of these

22 SD : MD : QD is equal to a 4:5:6 c 1:2:3

b 15 : 12 : 10 d None of these

23 Which of following statement is true for a given distribution? a b c d

Mean deviation > Standard deviation Mean deviation < Standard deviation Mean deviation = Standard deviation They are not related

24 If r is the range and σ is the SD of a set of

Pearson’s formula lies between

n observations, then correct relation is

a −3 and 3 c −∞ and ∞

a σ≤r c σ =r

b −1 and 1 d 0 and ∞

4 5

b σ≥r d None of these

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Objective Mathematics Vol. 1

20 Correlation Analysis If two quantities vary in such a way that fluctuation in one are accompanied by fluctuation in other, these quantities are said to be correlated. The statistical tool by which the relationship between two or more than two variables studied is called correlation. The measures of correlation called the coefficient of correlation and denoted by r. (i) Covariance The covariance between two variables x and y with n pairs of observations ( x1 , y1 ), ( x 2 , y2 ), … , ( x n , yn ) is defined as Σ ( x i − x )( yi − y ) Cov ( x, y) = n  Σx i yi  = − x y  n  Σx i Σy where, and y = i x= n n (ii) Karl Pearson’s correlation coefficient Σ ( x i − x )( yi − y ) Cov ( x, y) r= = Var ( x ) ⋅ Var ( y) Σ ( x i − x ) 2 ⋅ Σ ( yi − y) 2 =

nΣxy − ( Σx )( Σy) {nΣx 2 − ( Σx ) 2 }{nΣy 2 − ( Σy) 2 }

(iii) Coefficient of rank correlation This formula is applied to the problems in which data cannot be measured quantitatively but qualitative assessment is possible such as beauty, honesty etc. In this case, the best individual is given rank number 1, next rank 2 and so on. The coefficient of rank correlation is given by the formula R =1 −

6 Σd i2 n( n 2 − 1)

where, d i is the difference of corresponding rank and n is the number of pairs of observations. Ø There should be formula for equal rank or tie in ranks. X

1070

Example 30. A computer while calculating rxy from 25 pairs of observations obtained the following constant: n = 25, Σx =125, Σx 2 = 650, Σy =100, Σy 2 = 460, Σxy = 508 A recheck showed that he had copied down two pairs (6, 14), (8, 6) while the correct values were (8, 12), (6, 8). The correct value of the correlation coefficient is 2 3 (b) (a) 3 4 1 (c) − (d)1 2

Sol. (a) Correct value of Σ x = 125 − 6 − 8 + 8 + 6 = 125 Correct value of Σ x2 = 650 − 36 − 64 + 64 + 36 = 650 Correct value of Σ y = 100 − 14 − 6 + 12 + 8 = 100 Correct value of Σ y2 = 460 − 196 − 36 + 144 + 64 = 436 Correct value of Σ xy = 508 − 84 − 48 + 96 + 48 = 520 ∴Correct correlation coefficient nΣ xy − (Σ x)(Σy) = 2 {nΣ x − (Σ x)2 }{nΣy2 − (Σy)2 } 2 = 3 X

Example 31. The rank correlation coefficient between marks secured in Mathematics and Physics by a certain number of students in a class is 08 . and the sum of the squares of differences in ranks is 33, then number of students in the class is (a) 30 (b)11 (c)10 (d)15 2 Sol. (c) r = 1 − 6Σ2 d i = 1 − 6 ×2 33

n(n − 1)

n(n − 1)

⇒

n(n − 1) = 990

⇒

n(n2 − 1) = 10 (102 − 1)

⇒

n = 10

2

Characteristics of Correlation Coefficient (i) −1 ≤ r ≤ 1 (ii) If r = −1, then there is perfect negative correlation between x and y i.e. corresponding to an increase (or decrease) in one variable, there is a proportional decrease (or increase) in the other variable. (iii) If r =1, then there is perfect positive correlation between x and y i.e. corresponding to an increase (or decrease) in one variable, there is proportional increase (or decrease) in the other variable. (iv) If r = 0, then x and y are not correlated i.e. the changes in one variable are not followed by changes in the other. (v) If 0 < r < 1, then there is a positive correlation between x and y i.e. an increase (or decrease) in one variable corresponds to an increase (or decrease) in the other. (vi) If −1 < r < 0, then there is negative correlation between x and y i.e. an increase (or decrease) in one variable corresponds to decrease (or increase) in the other.

X

Example 32. If r is correlation coefficient, then correct relation is (a) r >1 (b)1 ≤ r (c) | r | ≤1 (d) | r | ≥1

i.

Line of regression of y on x The line of regression of y on x gives the best estimate of the value of y given value of x and is given by y − Y = b yx ( x − X ); b yx = r ⋅ σ y / σ x

ii.

Line of regression of x on y The line of regression of x on y gives the best estimate of the value of x for given value of y is given by σ x − X = bxy ( y − Y ); bxy = r ⋅ x σy

iii.

Coefficient of regression of y on x The coefficient of regression of y on x denoted by b yx and is given by σ y Cov( x, y) r1 = b yx = r ⋅ = σx σ 2x

Sol. (c) By definition of correlation coefficients, we have

=

−1 ≤ r ≤ 1 ⇒ |r|≤ 1 X

Example 33. For 12 pairs of observations on x and y, the r calculated is 0674 . . The standard error is (a) 0.2563 (b) 0.1575 (c) 0.4384 (d) None of the above Sol. (b) Using the formula, 1 − r2 SE = n

iv.

X

1 − 0.4542 3.4641 0.5458 = 0.1575 = 3.4641 =

X

⇒ ⇒ ⇒ ∴

PE = (0.6745) SE , we have (0.6745)(1 − r 2 ) PE = n (0.6745)[1 − (0.917 )2 ] 0.034 = n (0.6745)[1 − (0.917 )2 ] n= 0.034 n = 3156 . n = 10 (approx.)

nΣ x i yi − Σ x i Σyi nΣ x i2 − ( Σ x i ) 2

Coefficient of regression of x on y The coefficient of regression of x on y denoted by bxy and is given by σ Cov( x, y) nΣ x i yi − Σx i Σyi r2 = bxy = x = = rσ y σ 2y nΣyi2 − ( Σyi ) 2

Example 35. If x and y are respectively heights of father and son and M x = 6825 . inch, M y = 68.5 inch, σ x = 2 .49, σ y = 2.23 and r = 0.47, then which of following is height of son when height of father is 67.5 inch? (a) 6819 (b) 6881 (c) 6566 (d) 7502 . . . . Sol. (a) Let us first find the line of regression of y on x ,

Example 34. In a question on correlation, the value of r was 0. 917 and its probable error was 0034 . . The number of pairs of observations was (a) 8.5 (b)12.02 (c)10 (d)10.627 Sol. (c) Using the formula,

20

Regression Analysis

Statistics

(vii) The value r 2 is called coefficient of determination. If r = 0.5, then r 2 = 0. 25, which means only 25% of variations are explained and remaining 75% are unexplained. They may due to external causes. 1− r2 . (viii) Standard Error (SE) is defined as SE = n (ix) Probable Error (PE) is defined as PE = (06745 . ) SE. It is used to test reliability of particular value of r. Infact ( r ± PE ) gives limits with in which the coefficient of correlation sample must always lie when the sample is drawn from the universe.

which is given by y − Y = r⋅

σy σx

(x − X)

y − 68.5 = (0.47 )

⇒

(2.23) ( x − 68.25) (2.49)

When x = 67.5, then y − 68.5 = 0.42(67.5 − 68.25) ∴ y = 6819 . X

Example 36. If Σ x = 30, Σy = 42, Σ xy =199, Σ x 2 = 184, Σy 2 = 318 and n = 6, then regression coefficient bxy is (a) −0. 36 (b) −0.46 (c) 0. 26 (d) 0. 38 Sol. (b) We know that, b xy =

Σx Σy  Σ xy −    n  (Σy)2 Σy − n 2

=

30 × 42  199 −    6  (42 )2 318 − 6

=

−11 = − 0.46 24

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Objective Mathematics Vol. 1

of Regression 20 Properties Coefficients i.

ii.

iii.

Both regression coefficients have the same sign i.e. either both are positive or both are negative.

Properties of Lines of Regression The two lines of regression have the following properties:

i.

The sign of correlation coefficient is same as that of regression coefficient i.e. r > 0, if bxy > 0 and b yx > 0 and r < 0, if bxy < 0 and b yx < 0. The coefficients of correlation is the geometric mean between the two regression coefficients r = ± b yx × bxy

ii.

v. vi. X

iv.

The angle θ between two lines of regression is given by 1 − r 2  σ xσ y  tan θ =   | r|  σ 2x + σ 2y =

Regression coefficients are independent of change of origin but not scale. AM of the regression coefficients is greater than the correlation coefficient.

v.

Example 37. If the correlation coefficient between two variables x and y is 0.4 and regression coefficient of x on y is 0. 2 , then regression coefficient of y on x is (a) 0.4 (b) ±0.8 (c) 0.8 (d) 0.6

vi.

Sol. (c) Here, r = 0.4, b xy = 0.2 r 2 = b xy × byx ⇒ ∴ X

016 . = 0.2 × byx byx = 0.8

Example 38. If ax + by + c = 0 is line of regression of y on x and a1 x + b1 y + c1 = 0 is line of regression of x on y, then which of the following is correct? (b) | ab1 | = | a1 b | (a) | ab1 | ≥ | a1 b | (c) ab1 + ba1 = 0 (d) | ab1 | ≤ | a1 b | Sol. (d) Line of regression of y on x is ax + by + c = 0 or ∴

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c a y =  −  x −  b b a  r1 =  −   b

1 . bxy

Slope of the line of regression of x on y =

Both the regression coefficients cannot be numerically greater than one.

We know, that

Slope of the line of regression of y on x = b yx .

iii.

The sign to be taken outside the square root is that of the regression coefficients.

iv.

The two lines of regression pass through the point ( x , y ).

X

b yx bxy − 1 b yx + bxy

=

m1 − m2 1 + m1 m2

where, m1 = Slope of line of regression y on x and m2 = Slope of line of regression x on y π If r = 0, then tan θ = ∞ ⇒ θ = i.e. if two 2 variables are uncorrelated, the lines of regression are perpendicular to each other. If r = ±1, then tan θ = 0 ⇒ θ = 0 or π i.e. in the case of perfect correlation ( r = ±1) the two lines of regression coincide.

Example 39. For two variables x and y with the same mean, the two regression lines are x + 2 y − 5 = 0; 2x + 3 y − 8 = 0 and variance ( x ) =12. Then, σ y is (a) 4 (b) 5 (c) 2 (d) None of these Sol. (c) Slopes of regression lines are − 1 and − 2 . ∴ ∴

2 1 3 and b xy = − 2 2 1 3 3 r 2 =  −   −  = (< 1)  2  2 4

⇒ Also,

3

byx = −

r=− byx = r ⋅

3 [Qbyx , b xy are less than zero] 2 σy σx

Line of regression of x on y is b c x=− 1y− 1 a1 a1 b1 ∴ r2 = − a1

and

σx =2 3

Q

σ 2x = 12

∴

−

Then,| r1 r2 | ≤ 1 or| ab1| ≤ | a1b|.

⇒

1 3 σy =− ⋅ 2 2 2 3 σy = 2

Example 40. If b yx =1.6 and bxy = 0.4 and the angle between two regression lines is θ, then tan θ is (a) 0.18 (b) 0.16 (c) 0.3 (d) 0.24

1 σy r σx 5 1 = × 2 = = 2.5 2 0.8 σ m2 = r ⋅ y σx = 0.8 × 2 = 16 . m1 − m2 tan θ = 1 + m1m2 0.9 . = = 018 5 m1 =

∴

and

Sol. (a) Clearly, r = 16 . × 0.4 = 0.8 byx = r ⋅

Now,

σy

⇒

σx

=

σy

∴

σx

. 16 =2 0.8

20 Statistics

X

Work Book Exercise 20.2 1 If r is coefficient of correlation between x and y, then r = 0 means a b c d

y = 0.99 x + 1, then means of x and y series are

x and y are perfectly related x and y are negatively related x and y are partially related x and y are not related

a b c d

2 The regression coefficients are independent of a b c d

change of origin and scale change of scale only change of origin only None of the above b 64%

c 20%

d 36%

4 The coefficient of correlation is, which of following mean between two regression coefficients a HM

b GM

c AM

66.5, 65.5 65.2, 65.55 65.4, 65.68 64.5, 64.8

8 The two lines of regression are parallel to axes, if r is

3 If r = 0.8, then explained data would be a 80%

7 If two lines of regression are x = 0.85 y + 9.48 and

d 2 × AM

5 Let d i represent difference in ranks of rth item. If

a b c d

1 0 −1 0.5

9 If lines of regression of y on x and x on y are respectively y = kx + 4, x = 4 y + 5, then which is true for k? a 0≤ k≤ 4 b 0 ≤ k ≤ 1/ 4

d i = 0, ∀ i, then correct relation is

c k > 1/ 4

a b c d

d None of the above

r=0 r = −1 r=1 None of the above

6 If two regression coefficients are 0.8 and 0. 2, then r is a b c d

−0.4 −0.5 0.5 0.4

10 Let x1, x2 and x3 be uncorrelated variables each having the same standard deviation. The coefficient of correlation between ( x1 + x2 ) and ( x2 + x3 ) is a 1/4 b 1/3 c 1/2 d 1

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WorkedOut Examples Type 1. Only One Correct Option Ex 1. The geometric mean G of the product of n series of data with geometric means G1 , G2 ,..., Gn respectively, then (a) G = G1 , G2 , K , Gn (b) G = G1 ⋅ G2 ⋅ K ⋅ Gn (c) G = − G1 , G2 , K , Gn

(d) None of the above Sol. Let x1 , x2 , ..., xn be the variates corresponding to nsets of data, each having the same number of observations say K and x be their product. Then, x = x1 ⋅ x2 ... xn i.e. log x = log x1 + log x2 + K + log xn Σ log x Σ log x1 Σ log x2 Σ log xn or = + +K+ K K K K or logG = logG1 + logG2 + K + logGn ⇒ G = G1 ⋅ G2 K Gn Hence, (b) is the correct answer.

Ex 2. A motor car when travelling from rest travels the first twentieth of a mile at 6 m/h the next three twentieths of the mile at 8, 12, 24 m/h respectively. But its average speed over the first one-fifth of a mile is not 12.5 m/h, then the correct average is (a) 9.6 m/h (b) 19.6 m/h (c) 6.6 m/h (d) None of the above Sol. Let A and H be the arithmetic and harmonic means of speeds. So, and

6 + 8 + 12 + 24 50 = = 12.5 m/h 4 4 4 H = 1 1 1 1 + + + 6 8 12 24 4 × 24 48 = = = 9.6 m/h 10 5 A=

It is obvious from the data that the average speed would not be arithmetic mean but it is the harmonic mean, because Total distance travelled Average speed = Total time of journey 1 3 + 20 20 = 1 1 1 1 1 1 1 1 × + × + × + × 20 6 20 8 20 12 20 24 480 = = 9.6 = HM 5 × 10

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Hence, (a) is the correct answer.

Ex 3. The average rate of (i) motion in the case of a person who rides the first mile at 10 m/h, the next mile at 8 m/h and the third mile at 6 m/h (ii) increase in population which is the first decade has increased 20% in the next 25% and in the third 44% are (where, log 20 =1.3010, , log 25 =1.3979, log 2802 . = 14475 . ) log 44 = 16435 . (a) 7.66, 1.7 (c) 7.66, 28.02

(b) 7.5, 2.9 (d) None of these

Sol. (i) In this case, the harmonic mean is the suitable average. The average rate of motion is, Total distance 3 H = = 1 1 1 Total time taken + + 10 8 6 3 × 120 = = 7.66 m/h 47 (ii) Here, GM is the appropriate average. ∴ G = (20 × 25 × 44 )1/ 3 1 logG = [log 20 + log 25 + log 44 ] ⇒ 3 1 = [1.3010 + 1.3979 + 1.6435 ] 3 1 = × 4.3425 = 14475 . 3 ∴ G = 28.02 = The average increase in population. Hence, (c) is the correct answer.

Ex 4. A man motors from A to B. In motoring a distance uphill he gets a mileage of only 10 miles per gallon of gasoline. On the return trip, he makes 15 miles per gallon, then the harmonic mean of his mileage (verify that this is the proper average to be used, here assuming that the distance from A to B is 60 miles) is (a) 12 (c) 10 Sol. The harmonic mean,

(b) 11 (d) None of these

2 2 × 30 = = 12 miles per gallon 1 1 5 + 10 15 Let us verify it. The distance between A and B is 60 miles. 60 He has consumed = 6 gallons of gasoline in going 10 60 uphill and = 4 gallons in returning i.e. 10 gallons in 15 all. The total distance covered = (60 + 60) [both ways] = 120 miles H =

Ex 5. The median of the following items 25, 15, 23, 40, 27, 25, 23, 25 and 20 is (a) 27 (c) 25

(b) 40 (d) 23

Ex 6. If a variate takes values a, ar, ar 2 , ..., ar n − 1 , then which of the relation between means hold? A+H (b) =G 2 (d) A = G = H

(c) A < G < H Sol. A =

a(1 + r + r2 + K + rn − 1 ) a  1 − rn  =   n n  1− r  G = (a ⋅ ar ⋅ ar2... arn − 1 )1/ n 1 + 2 + K + (n − 1)

⇒ ∴

n− 1

n = a⋅ r = ar 2 1 1  1 1 1  = 1 + + 2 + K + n − 1  H na  r r r  1 1− n 1 r = ⋅ na 1 − 1 r na(1 − r)rn − 1 H = 1 − rn 2 n− 1 AH = a r = G2

Hence, (a) is the correct answer.

Ex 7. An aeroplane flies around a square, the sides of which measure 100 miles each. The aeroplane covers at a speed of 100 m/h the first side, at 200 m/h the second side, at 300 m/h the third side and 400 m/h the fourth side. The average speed of aeroplane around the square is (a) 190 m/h (c) 192 m/h

(b) 195 m/h (d) 200 m/h

Sol. Using the weighted HM formula, 1 1 = H N H =

or

1 N

n

fi x i=1 i 1

∑ n

f ∑ xii i=1

∴ Average speed 400 = 192 m/h 1 1 1   1 100  + + +   100 200 300 400 Hence, (c) is the correct answer. =

a σ c c2 (d) 2 σ a (b)

Var (ax + b) = a2 Var (x )

15, 20, 23, 23, 25, 25, 25, 27, 40  9 + 1 ∴Median = Size of   th item = 5th item = 25  2  Hence, (c) is the correct answer.

(a) AH = G

c σ a b (c) σ c

(a)

20

Sol. Make use of the results

Sol. First arrange in ascending order as

2

Ex 8. If SD of x is σ, then SD of the variable ax + b , where a, b and c are constants, is u= c

Statistics

120 = 12 gallons per mile 10 = HM Thus, harmonic mean is the proper average here. Hence, (a) is the correct answer.

∴ Average consumption =

2 a2 2  ax + b  a We have, Var   = 2 Var (x ) = 2 σ c  c  c ax + b a   SD  ⇒ = σ  c  c Hence, (b) is the correct answer.

Ex 9. The first of two samples has 100 items with mean 15 and SD 3. If the whole group has 250 items with mean 15.6 and SD = 13.44, the SD of the second group is (a) 5 (b) 4 (c) 6 (d) 3.52 n1 (σ 12 + d12 ) + n2 (σ 22 + d22 ) n1 + n2 where, d1 = m1 − a, d2 = m2 − a, a being the mean of the whole group. Let m2 = mean of the second group 100 × 15 + 150 × m2 ⇒ m2 = 16 ∴ 15.6 = 250 (100 × 9 + 150 × σ 2 ) + 100 × (0.6)2    + 150 × (0.4 )2  Thus, 13.44 =  250 ⇒ σ=4 Hence, (b) is the correct answer.

Sol. Use σ 2 =

Ex 10. The mean and SD of 63 children in an arithmetic test are respectively 27.6 and 7.1. They are added to a new group of 26 who had less training and whose mean is 19.2 and SD 6.2. The values of the combined group of the mean and SD are (a) 25.1, 7.8 (b) 2.3, 0.8 (c) 1.5, 0.9 (d) None of the above Sol. Mean and SD (σ) of the combined group are m=

63 × 27.6 + 26 × 19.2 = 251 . 63 + 26

63(71 . )2 + 26(6.2)2 + 63(27.6 − 251 . )2    + 26(19.2 − 251 . )2  and σ 2 =  89 ⇒ σ = 7.8 (approx. ) Hence, (a) is the correct answer.

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Objective Mathematics Vol. 1

20

Ex 11. Coefficient of skewness for the values median =188 . , Q1 = 146 . , Q3 = 25.2 is (a) 0.2 (c) 0.7

(b) 0.5 (d) None of these

Sol. Coefficient of skewness =

Q3 + Q1 − 2(Median) 25.2 + 14.6 − 2 × 18.8 = = 0.20 Q3 − Q1 25.2 − 14.6

Hence, (a) is the correct answer.

Ex 12. An incomplete frequency distribution is given below: Variate

Frequency

10-20

12

20-30

30

30-40

x

40-50

65

50-60

45

60-70

25

70-80

18

Total

229

If median value is 46, then the missing frequency is (a) 33.5 (b) 35 (c) 34 (d) 26 Sol. Median = 46, which lies in 40-50 class N   − cf  2  Median = l + h  f

⋅ {(x1 − x1 ) +

σ1 (x2 − x2 )}] = 0 σ2

 σ ⇒ Σ (x1 − x1 )2 + a ⋅ 1 (x2 − x2 )2 + (x1 − x1 ) (x2 − x2 ) σ2   σ1   σ + a = 0  2  ⇒

σ 21 + a

 σ1 2  σ1 ⋅ σ2 +  + a r σ 1σ 2 = 0  σ2  σ2

⇒ aσ 1σ 2 (1 + r) = − σ 12 (1 + r) σ a = − 1, since here r ≠ − 1 ⇒ σ2 Hence, (a) is the correct answer.

Ex 14. Correlation coefficient r of two variables x and y is positive, when (a) σ x + y > σ x − y (b) σ x − y > σ x + y (c) σ x + y = 0 (d) None of the above Sol. σ 2x + y = Var(x + y) = Var (x ) + Var ( y) + 2Cov (x , y) or

σ 2x +

Similarly,

σ 2x − y = σ x2 + σ 2y − 2r σ xσ y

y

= σ x2 + σ 2y + 2r σ xσ y

On subtracting, we get σ x2 + y − σ x2 − y = 4 r σ xσ y Since, σ x and σ y are always positive, so r is positive, if σ 2x +

where, f = Frequency of median class cf = Cumulative frequency of the class preceding the median class  229  − (x + 42)  2  46 = 40 + 10 ∴ 65 where, x = Frequency of class 30-40 ⇒ x = 33.5 = 34 [Q x is an integer ] Hence, (c) is the correct answer.

Ex 13. x1 and x 2 are two variates with variances σ 12 and σ 22 respectively and r is the correlation between them. The value of a such that σ x1 + ax 2 and x1 + 1 x 2 are uncorrelated, is σ2 σ (a) − 1 σ2

σ (b) r⋅ 1 σ2

(c) σ 1σ 2

(d) None of these

Sol. Let u = x1 + ax2 and v = x1 +

σ1 ⋅ x2 σ2

Therefore, u = x1 + ax2 and v = x1 +

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Therefore, Σ(u − u )(v − v ) = 0 or Σ[{(x1 − x1 ) + a(x2 − x2 )}

σ1 ⋅ x2 σ2

If u and v are uncorrelated, then Cov (u, v ) = 0

y

− σ 2x − y > 0

or σx + y > σx − y Hence, (a) is the correct answer.

Ex 15. Two variates x and y have zero means, the same variance and zero correlation. Then, u = x cos α + y sin α, v = x sin α − y cos α have correlation (a) − 1 (c) 0

(b) 1 (d) None of these

Sol. Given, x = 0, y = 0, Var (x ) = Var ( y) = σ 2 and 1 zero correlation = Cov(x , y) = 0 or Σ(xy) = 0 n Thus, Var (u) = cos2 α Var (x ) + sin 2 α Var ( y) = σ 2 and similarly Var(v ) = σ 2 Also,

u = x cosα + y sin α = 0

v = x sin α − y cos α = 0 1 1 Therefore, Cov (u, v ) = Σ (u − u )(v − v ) = Σ (uv ) n n 1 = Σ{(x cos α + y sin α )(x sin α − y cos α )} n 1 = Σ{x 2 sin α cosα − y2 sin α cosα n + xy (sin 2 α − cos2 α )}

1 + (sin α − cos α ) Σ (xy) n = sin α cosα Var (x ) − sin α cosα Var ( y) + 0  1  Q Σ(xy) = 0  n  2 Thus, ruv = 0 [Q Var (x ) = Var ( y) = σ ] 2

2

Hence, (c) is the correct answer.

Ex 16. The correlation between x and a − x is (a) − 1

(b) 1

(c)

1 2

(d) 0

Sol. Let u = a − x and therefore

Var (u) = Var (a − x ) = (− 1)2 Var (x )

[say] = Var(x ) = σ 2 1 Now, Cov (x , a − x ) = Cov (x , u) = Σ{(x − x )(u − u )} n 1 2 [Q u − u = − (x − x )] = − Σ(x − x ) n 2 = − Var(x ) = − σ ∴ r(x , u) =

Cov (x , u) = Var (x ) Var (u)

− σ2 σ2 ⋅ σ2

= −1

Hence, (a) is the correct answer.

Ex 17. The two variates x and y are uncorrelated and have standard deviations σ x and σ y respectively, the correlation coefficient between x + y and x − y is (a)

(c)

σ xσ y σ 2x + σ 2y σ 2x

− σ 2y

σ 2x + σ 2y

1 1 1  (b)  + 2  σ x σ y 

(d) None of these

u = x + y and v = x − y Therefore, u = x + y and v = x − y 1 ∴ Cov (u, v ) = Σ{(u − u )(v − v )} n 1 = Σ[{(x − x ) + ( y − y)} ⋅ {(x − x ) − ( y − y)}] n 1 = Σ{(x − x )2 − ( y − y)2} = σ 2x − σ 2y n 1 1 Var(u) = Σ (u − u )2 = Σ{(x − x ) + ( y − y)} 2 n n = σ 2x + σ 2y

Sol. Let

Q x and y are uncorrelated,   1 so Σ(x − x )( y − y) = 0    n Similarly, Var(v ) = σ 2x + σ 2y Thus,

r(u, v ) =

Cov (u, v ) = σ u σv

Hence, (c) is the correct answer.

σ 2x σ 2x

− +

σ 2y σ 2y

Ex 18. x and y are two correlated variables with the same standard deviation and the correlation coefficient r, the correlation coefficient between x and x + y is (a)

r 2

(b)

1− r 2

1+ r (d) 0 2

(c)

20 Statistics

1 1 = sin α cosα Σ (x )2 − sin α cosα Σ ( y)2 n n

Sol. Let u = x + y and σ = SD of x or y ∴

σ 2u = σ 2x + σ 2y + 2 Cov (x , y)

= σ 2 + σ 2 + 2r σ 2 = 2σ 2 (1 + r) 1 ∴Cov(u, x ) = Σ{(u − u )(x − x )} n 1 = Σ{[(x − x ) + ( y − y)](x − x )} n 1 = Σ{(x − x )2 + (x − x )( y − y)} = σ 2 (1 + r) n r+1 σ 2 (1 + r) Thus, r(x , x + y) = = 2 σ ⋅ σ 2(1 + r) Hence, (c) is the correct answer.

Ex 19. x is the arithmetic mean of n independent variates x1 , x 2 , x 3 ,..., x n each of standard deviation σ, then variance ( x ) is σ2 n ( n + 1) σ 2 (c) 3

(a)

Sol. We have, x = ∴ Var (x ) =

(b)

nσ 2 2

(d) None of these

1 n ∑ xi ni =1  1 n Var (xi ) + 2 ∑ Cov (xi , x j ) 2 ∑ n i = 1  i≠ j

1 σ2 [ nσ 2 ] = 2 n n Q Var (xi ) = σ 2 for every i and Cov (xi , x j ) = 0   as variates are independent   Hence, (a) is the correct answer. =

Ex 20. If rxy between two variates x and y is 0.6, Cov ( x, y) = 48, . σ 2x = 9, then σ y is 8 9 8 (c) 3

(a)

(b)

5 8

(d) None of these

4.8 Cov (x , y) , we have 0.6 = 3σ y σ xσ y 8 ⇒ σy = 3 Hence, (c) is the correct answer.

Sol. Using r =

Ex 21. If two random variables have the regression lines 3x + 2 y − 26 = 0 and 6x + y − 31 = 0, then correlation coefficient rxy is (a) − 0.5 (c) 1/2

(b) 1/2 (d) None of these

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Objective Mathematics Vol. 1

20

3 and − 6. 2 3 1 and bxy = − Since, r2 < 1, byx = − 2 6 1 ∴ r2 = byx ⋅ bxy = ⇒ r = − 0.5 4 [as both byx and bxy are negative] Hence, (a) is the correct answer.

Sol. Slopes of regression lines are −

Ex 22. rxy < 0, according as (b) σ x + y < σ x − y (d) None of these

Sol. Since, σ 2x + y = σ 2x + σ 2y + 2r σ xσ y =

σ x2

+

σ 2y

...(i)

− 2r σ xσ y

From Eqs. (i) and (ii), 4 r σ xσ y = σ 2x + As σ x σ y is always positive. ∴ r = rxy < 0, if σ 2x + y < σ 2x − y or σ x +

...(ii) y

y

− σ 2x − y < σx − y

Hence, (b) is the correct answer.

Ex 23. The correlation between two variables x and y is given to be r. The values of x and y change such that Cov( x, y) remains unaltered and Var ( x ) and Var( y) are changed to four times their original values. The new correlation coefficient is changed to (a) 4r

(b) 16r

Sol. We have, r =

(c)

r 16

(d)

r 4

Cov(x , y) σ xσ y

Changed correlation coefficient, 1 Cov(x , y) r Cov (x , y) = = r1 = 2 2 4 σ xσ y 4 4σ ⋅ 4σ x

y

Hence, (d) is the correct answer.

Ex 24. If z = ax + by and r is the correlation coefficient between x and y, then σ 2z is equal to (a) 2ab r σ x σ y (b) a 2σ x2 + b 2σ 2y − 2abr σ x σ y (c) a

2

σ x2

+b

2

σ 2y

+ 2ab r σ x σ y

(d) None of the above Sol. Q z = ax + by ∴ z = ax + by 1 1 Σ (z − z )2 = Σ{a(x − x ) + b( y − y)} 2 n n 1 21 2 21 = a Σ (x − x ) + b Σ ( y − y)2 +2ab Σ(x − x )( y− y) n n n = a2σ 2x + b2σ 2y + 2ab r σ x σ y

Now, σ 2z =

Hence, (c) is the correct answer.

Ex 25. If both the regression coefficients b yx and bxy are positive, then 1 1 2 + > b yx bxy r 1 1 r (c) + < b yx bxy 2

(a)

1078

So, considering the regression coefficients byx and bxy, we have 2(b yx ⋅ b xy ) 2(bxy ⋅ byx ) or r > [ b yx ⋅ bxy ]1/ 2 > byx + bxy bxy + byx ⇒

2 1 1 < + r byx bxy

⇒

1 1 2 + > byx bxy r

Hence, (a) is the correct answer.

(a) σ x + y > σ x − y (c) σ x + y > 0 σ x2 − y

Sol. Since, GM > HM

(b)

1 1 2 + < b yx bxy r

(d) None of these

Ex 26. If the slopes of the line of regression of y on x and of x on y are 30° and 60° respectively, then r ( x, y) is (a) − 1

(b) 1

1

(d) 3

3

Sol. We have, byx = tan 30° = ∴

(c)

byx ⋅ bxy =

1 1 and = tan 60° = 3 bxy 3 1 3

1 1 , as byx and bxy are positive. ⇒ r=± 3 3 Hence, (c) is the correct answer.

⇒ r2 =

Ex 27. If the two lines of regression are 3x + y = 15 and x + 4 y = 3, then value of x when y = 3 is (a) − 4 (c) 4

(b) − 8 (d) None of these

Sol. Here, we need the line of regression of x on y for

estimating the value of x corresponding to y = 3. The 1 slope of x + 4 y = 3 is − and that of 3x + y = 15 is − 3. 4 1 1 1 So, r2 = byx ⋅ bxy = − × − = which is less than 4 3 12 unity.

Thus, 3x + y = 15 is the line of regression of x on y. When y = 3, then 3x + 3 = 15 ⇒ x = 4 Hence, (c) is the correct answer.

Ex 28. Linear relation between the variables is given by the equation ax + by + c = 0 such that ab > 0. Then, r ( x, y) is given by (a) 1 (b) − 1 (c) 0 (d) any number lying between −1and 1 Sol. ax + by + c = 0 ⇒

ax + by + c = 0

⇒

x−x =−

b ( y − y) a

1 Σ (x − x )( y − y) n b 1 b = − ⋅ Σ( y − y)2 = − Var ( y) a n a 2 2 1 1 b b Var (x ) = Σ (x − x )2 = 2 ⋅ Σ ( y − y)2 = 2 Var ( y) n a a n b − Var ( y) Cov (x , y) = a =− 1, if ab> 0 ∴ r(x , y)= b (Var x )(Var y) Var ( y) a Hence, (b) is the correct answer. ∴

Cov (x , y) =

Target Exercises Type 1. Only One Correct Option (nX − x1 + a) n (c) {(n − 1) X − x1 + a}

( X − x1 + a) n (d) None of these (b)

(a)

(a) 48

2. A batsman in his 16th innings makes a score of 70 runs and thereby increases his average by 2 runs. If he had never been not out, then his average after 16th innings is (a) 36

(b) 38

(c) 40

(d) 42

(b) 50 kg (d) 58 kg

(a) 52

5. An additional observation 15 is included in a series of 11 observations and its mean remains unaffected. The mean of series was (a) 12 (c) 20

(b) 4

(c) 6

(d) 1

7. A frequency distribution has arithmetic mean 7.85, then missing term is Daily wages (i inn `) Number of workers

(a) 10.5 (c) 15.5

5 10

6 f

7 13

10 8

12 5

15 4

(b) 10.05 (d) 15

8. The mean of 20 observations is 15. On checking it was found that the two observations were wrongly copied as 3 and 6. The correct values are 8 and 4, then correct mean will be given by (a) 15.15 (c) 14.74

(b) 14.69 (d) 15.25

(d) 41.5

(c) 46

(d) 54

11. The mean of series 1, 2, 4, 8, ...., 2n is 2n + 1 − 1 n+1 2n − 1 (d) n+1

2n− 1 n 2n + 1 (c) n

(b)

12. The algebraic sum of the deviations of 20 observations measured from 30 is 2. The mean of observations is (b) 30.1

(c) 30.5

(d) 29.6

13. A class of 30 boys and 15 girls is given a test in Mathematics. The average marks obtained by boys is 15 and by girls is 6. The average of whole class is (b) 12 (d) None of these

14. If the arithmetic mean of 9 observations is 100 and that of 6 is 80. Then, the combined mean of all the 15 observations will be (a) 100

(b) 15 (d) None of these

6. In a monthly test marks obtained in Mathematics by 15 students of a class are 0, 0, 2, 2, 3, 3, 3, 5, 5, 5, 5, 6, 6, 7, 8. The arithmetic mean of marks is (a) 8

(b) 48

(a) 10.5 (c) 4.5

(b) 169 (d) 169.18

(c) 50

10. The average weight of a class of 14 students is 42 kg. If the teacher is included then average weight increases by 400 g, then weight of teacher is

(a) 28.5

4. The mean income of a group of 50 persons was calculated as ` 169. Later, it was discovered that one figure was wrongly taken as 134 instead of correct value 143. The correct mean (in `) should be (a) 168 (c) 168.92

(b) 85.5

(a)

3. The average weight of students in a class of 35 students is 40 kg. If the weight of the teacher be 1 included the average rises by kg, then weight of the 2 teacher is (a) 40.5 kg (c) 41 kg

9. The mean of 100 items is 49. It was later discovered that three items taken as 40, 20, 50 were actually 60, 70, 80, respectively. The correct mean is

Targ e t E x e rc is e s

1. The average of n numbers x1 , x 2 , x 3 , ... , x n is X . If x1 is replaced by a, then new average is

(b) 80

(c) 90

(d) 92

15. Combined mean of two series x1 and x 2 , when Σx1 = 210, Σx 2 = 150 , n1 = 50, n 2 = 100 is (a) 150

(b) 160

(c) 170

(d) 180

16. The mean income of a group of workers is X and that of another group is Y . If the number of workers in the second group is 10 times the number of workers in the first group, then mean income of combined group is X + 10Y 3 X +Y (c) Y

(a)

X + 10Y 11 X + 10Y (d) 9 (b)

17. The mean age of a combined group of men and women is 25 yr. If mean age of men is 26 and that of women is 21, then percentage of men and women in the group are (a) 60, 40 (c) 20, 80

(b) 80, 20 (d) 30, 70

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18. A set of 7 observations has mean 10 and another set of 3 observations has mean 5. The mean of combined set is (a) 15

(b) 10

(c) 8.5

(d) 7.5

19. If x1 and x 2 are means of two distribution such that x1 < x 2 and x is the mean of the joint distribution, then x + x2 (a) x = 1 2 (c) x2 < x1

(b) x > x2 (d) x1 < x < x2

20. If G1 and G2 are geometric means of two series and G is GM of the ratio of corresponding series, then G is G1 G2 (c) log G1 + log G2 (a)

(b) log G1 − log G2 (d) None of these

21. The geometric mean of 5, 8, 10, 15, 20, 25, 30 is (b) 10 (9)1/ 7 (d) None of these

(a) 16.9 (c) 18

22. The geometric mean of the series 1, 2, 4, ..., 2n , is (a) 2n + (1/ 2)

(b) 2(n + 1) / 2

(c) 2n − (1/ 2)

(d) 2n/ 2

23. The geometric mean of observations 2, 4, 16 and 32 is

Ta rg e t E x e rc is e s

(a) 6

(b) 7

(c) 8

(d) 9

24. If G1 and G2 are geometric means of two series of sizes n1 and n 2 and G is geometric mean of combined series, then log G is (a) log G1 + log G2 (c)

log G1 − log G2 n1 − n2

n log G1 + n2 logG2 (b) 1 n1 + n2

(d) None of these

25. On thirteen consecutive days, the number of persons booked for violating speed limit of 40 km/h were as follows: 58, 61, 68, 57, 62, 50, 55, 62, 53, 54, 51, 59, 52 The median number of speed violations per day is (a) 61

(b) 52

(c) 55

(d) 57

26. The test marks in Statistics for a class are 20, 24, 27, 38, 18, 42, 35, 21, 44, 18, 31, 36, 41, 26, 29. The median score of the class is (a) 8

(b) 21

(c) 29

(d) 31

27. The median of a series is 10. Two additional observations 7 and 20 are added to series. The median of new series is (a) 9 (c) 7

(b) 20 (d) 10

28. Which of the following is correct about the series 24, 38, 55, 69, 89? (a) Median = Mode (c) Mean = Median

(b) Mean = Mode (d) None of these

29. The median value of observations 83, 54, 78, 64, 90, 59, 67, 72, 70, 73 is 1080

(a) 70 (c) 72

(b) 71 (d) 73

30. In a test of Mathematics, 15 students got following marks 16, 25, 35, 20, 0, 12, 30, 22, 5, 8, 4, 38, 32, 2, 1. The median score of these 15 students is (a) 7

(b) 22

(c) 20

(d) 16

31. The mean and median of 100 items are 50 and 52, respectively. The value of largest item is 100. It was later found that it is 110 and not 100. The true mean and median are (a) 50.10, 51.5 (c) 50, 51.5

(b) 50.10, 52 (d) None of these

32. Median is based on the (a) highest 50% of items (c) highest 25% of items

(b) lowest 25% of items (d) middle 50% of items

33. The 70th percentile is equal to (a) 7th decile (c) 6th decile

(b) Q3 (d) None of these

34. Which of the following is correct for data –1, 0, 1, 2, 3, 5, 5, 6, 8, 10, 11? (a) Mean = Mode = Median (c) Mean = Mode

(b) Mean = 5 (d) Mode = Median

35. The mode of a set of observations 7, 12, 8, 5, 6, 4, 9, 10, 8, 9, 7, 9, 6, 5, 9 is (a) 7

(b) 8

(c) 9

(d) 12

36. Mode of data 3, 2, 5, 2, 3, 5, 6, 6, 5, 3, 5, 2, 5 is (a) 6

(b) 4

(c) 5

(d) 3

37. For the data 2, 2, 3, 3, 3, 4, 4, 5, 6, 8, 8, 8, 8, the mean, median and mode are (a) 64/13, 5, 3 (c) 5, 4, 3

(b) 64/13, 14, 3 (d) 64/13, 4, 8

38. The quartile deviation of daily wages of 7 persons which are ` 12, 7, 10, 15, 17, 17, 25, is (a) 14.5

(b) 13.5

(c) 7

(d) 3.5

(c) 3σ / 2

(d) σ / 3

39. Semi-interquartile range is (a) 4σ / 5

(b) 2σ / 3

40. If 25% of items are less than 10 and 25% are more than 40, then coefficient of quartile deviation is (a) 4/5

(b) 5/3

(c) 1/3

(d) 3/5

41. Consider any set of observations x1 , x 2 , x 3 , ... , x101 ; it being given that x1 < x 2 < x 3 < .... < x100 < x101 , then the mean deviation of this set of observations about a point k is minimum, when k equals (a) x1 x + x2 + ... + x101 (c) 1 101

(b) x51 (d) x50

42. The sum of absolute deviations about median is (a) least (c) greatest

(b) zero (d) None of these

43. For a series, the information available is n = 10, Σx = 60 and Σx 2 = 1000. The standard deviation is (a) 8

(b) 64

(c) 24

(d) 128

(b) 6 / 5

(c) 4 / 5

(d) 3 / 5

(a) 25, 20

45. The variance of 6, 8, 10, 12, 14 is (a) 1

(b) 8

(c) 12

(d) 16

46. The mean and standard deviation of 1, 2, 3, 4, 5, 6 are (a)

7 35 , 2 12

7 (b) , 3 2

(c) 3, 3

(d) 3, 35/12

(b) 18, 22

(a) 0.0003

(b) 0.0002

(c) 0.0004

(d) 0.0012

48. The standard deviation of certain data is 4. Then, their variance is (a) 1

(b) 16

(c) 8

(d) 2

2 3 22 (c) 3

(a) 2

(c) −2

(b) 4

(d) −4

50. The mean and SD of the marks of 200 candidates were found to be 40 and 15, respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and SD are respectively (a) 14.98, 39.95 (c) 39.95, 224.5

(b) 39.95, 14.98 (d) None of these

51. Suppose values taken by a variable x are such that a ≤ x i ≤ b, where x i denotes the value of x in the ith case for i = 1, 2, 3, …, n. Then, (a) a ≤ Var (x ) ≤ b a2 (c) ≤ Var (x ) b

(b) a2 ≤ Var (x ) ≤ b2 (d) (b − a)2 ≥ Var (x )

1 n 52. Let r be the range and S = ∑ ( xi − x )2 be the n − 1i=1 SD of a set of observations x1 , x 2 , ... , x n , then 2

(a) S ≤ r

n n−1

(b) S = r

(c) S ≥ r

n n−1

(d) None of these

n n−1

53. The best statistical measure used for comparing two series is (a) mean deviation (c) coefficient of variation

(b) range (d) interquartile range

54. The sum of square of deviations for 10 observations taken from mean 50 is 250. The coefficients of variation is (a) 50

(b) 10

(c) 30

(d) 40

55. For a given distribution of marks, mean is 35.16 and its standard deviation is 19.76. The coefficient of variation is (a) 56.2 (c) 1.779

(b) 6 (d) 3

58. One set containing five members has mean 8, variance 18 and the second set containing three members has mean 8 and variance 24. The variance of combined set of numbers is (a) 24 (c) 22.25

49. The standard deviation of − 1, − 2, − 3, − 4, − 5, − 6, − 7 is

(b) 0.562 (d) 177.935

(d) 16, 24

57. The standard deviations of two sets containing 10 and 20 members are 2 and 3, respectively measured from their common mean 5. The SD for the whole set of 30 members is (a)

47. Given 5 sample measurement of a part of machine as 6.33, 6.37, 6.36, 6.32 and 6.37. The sample variance is

(c) 22, 18

20 Statistics

(a) − 3 / 5

56. The coefficient of variation of two series are 70 and 90 and their standard deviation are 17.5 and 18, respectively. The mean of two series are

(b) 20.25 (d) None of these

59. The mode of a moderately symmetrical series is 18 and mean is 24. The median is (a) 18

(b) 24

(c) 22

(d) 21

60. The value of mean, median and mode coincides when distribution shows (a) positive skewness (c) negative skewness

(b) symmetrical distribution (d) All of these

61. If in a moderately skewed distribution, the values of mode and mean are 6λ and 9λ respectively, then the value of median is (a) 8λ

(b) 6λ

(c) 7λ

(d) 5λ

62. Karl Pearson’s coefficient of skewness is (M = Mean, M e = Median, M o = Mode) (M − M e ) σ (M − M o ) (c) σ

(a)

(b)

Targ e t E x e rc is e s

44. The sum of 10 items is 12 and sum of their squares is 18, then standard deviation will be

(M e − M o ) σ

(d) None of these

63. If mean and mode of a frequency distribution have same value, then coefficient of skewness is (a) zero (c) greater than zero

(b) less than zero (d) None of these

64. For a given series, the mean, SD and mode are respectively 622.08, 3.06 and 625. The coefficient of skewness is (a) − 0.95 (c) − 0.84

(b) 0.95 (d) 0.84

65. For a symmetrical distribution, the coefficient of skewness is (a) 1

(b) 0

(c) − 1

(d) 3

66. Mean, mode and standard deviation of a frequency distribution are 41, 45 and 8 respectively, Pearson’s coefficient of skewness is (a) 0.5 (c) 1.0

(b) 1.5 (d) − 0.5

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67. The arithmetic mean of 10 observations is 12.45. If each reading is increased by 5, then resulting mean is increased by (a) 5

(b) 29

(c) 0.5

(d) 50

68. The mean of 18 observations is 7 and if in each observation 5 is added, then the new mean will be (a) 2 (c) 7

(b) 12 (d) None of these

69. In a test of Statistics, marks were awarded out of 40. The average of 15 students was 38. Later, it was decided to give marks out of 50. The new average marks will be (a) 40

(b) 47.5

(c) 95

(d) 41.5

70. The standard deviation of the wages of 85 employees is ` 15.40. After one year, each of them is given an increment of ` 25. The standard deviation of new wages (in `) is (a) 15.40

(b) 40.40

(c) 20.40

(d) 10.40

Ta rg e t E x e rc is e s

71. The standard deviation for the set of numbers 1, 4, 5, 7, 8 is 2.45 nearly. If 10 is added to each number, then new standard deviation is (a) 24.45

(b) 12.45

(c) 2.45

(b) 22, 6

(c) 18, 6

(d) 22, 4

73. Let σ be standard deviation of n observations. Each observation is multiplied by a constant c. The standard deviation of resulting numbers is (a) σ (c) σ c

(b) cσ (d) None of these

74. The variance of 15 observations was found to be 8. If each observation is multiplied by 4, the new variance of the series is (a) 32

(b) 24

(c) 8

(d) 128

75. The variates x and u are related by hu = x – a, then correct relation between σ x and σ u is (a) σ x = hσ u (c) σ x = a + h σ u

(b) σ u = h σ x (d) σ u = a + h σ x

76. The marks of some students were listed out of 75. The SD of marks was found to be 9. Subsequently, the marks were raised to a maximum of 100 and variance of new marks was calculated. The new variance is (a) 144 (c) 81

(b) 122 (d) None of these

77. The mean deviation of any series is 15, then the value of quartile deviation is (a) 10.5

1082

(b) 11.5

(c) 12.5

(d) 13.5

78. The coefficient of correlation between x and y is 1, then coefficient of correlation between x and x + y is (a) 1

(b) − 1

(c) − 0.5

(a) 4

(b) 8

(c) 16

(d) 64

80. If a linear relation aX + bY + c = 0 exists between the variables X and Y and ab < 0, then coefficient of correlation between X and Y , is (a) 1 (b) − 1 (c) 0 (d) a real number between − 1and 1

81. If σ x = 3, σ y = 4 and σ x + y = 7, then r ( x, y ) is (b) − 1 (d) 0.75

(a) 1 (c) 0

82. If variance of X , Y , X − Y are respectively σ 2x , σ 2y , σ x2 − y , then the coefficient of correlation r between two variables X and Y is given by (a) (c)

σ x2 + σ 2y − σ 2x + y

(b)

2σ x ⋅ σ y σ 2x

+

σ 2y

σ 2x + σ 2y − σ x2 − y 2σ x ⋅ σ y

(d) None of these

2σ x ⋅ σ y

(d) 0.245

72. The median and SD of a distribution are 20 and 4, respectively. If each item is increased by 2, then new median and SD will be (a) 20, 6

79. The coefficient of correlation between x and y is 0.5 and their covariance is 16 and SD of x is 4, then SD of y is

(d) 0.5

83. If x, y are independent variables, then Cov ( x, y ) is (a) σ xσ y (c) σ y / σ x

(b) σ x / σ y (d) 0

84. If x = X − M x , y = Y − M y and Σxy = 40, Σx 2 = 80, Σy 2 = 20. Then, coefficient of correlation between X and Y is (a) 0 (c) − 1

(b) 0.5 (d) 1

85. The coefficient of correlation between x and y is x

1

2

3

4

5

y

3

7

12

19

28

(a) − 1 (c) 0.82

(b) − 0.82 (d) 0.988

86. If Cov ( x, y ) = − 16.5, Var ( x ) = 2.89, Var ( y ) = 100, then the coefficient of correlation is (a) − 0.97 (c) 0.97

(b) − 0.87 (d) None of these

87. If x and y are deviations from arithmetic mean, r = 0.8, Σxy = 60, σ y = 2.5 and Σx 2 = 90, then the number of items in the series, is (a) 10 (b) 15 (c) 18 (d) None of the above

88. The correlation coefficient 0.5 between x and y means (a) 50% of data are explained (b) 25% of variations are explained (c) out of total variations of y only 25% is due to x and rest due to other factors (d) out of total variation of y only 50% is due to x and rest due to other factors

(b) (x , y) (d) (r1 , r2 )

90. Out of two lines of regression given by x + 2 y = 4 and 2x − 3 y − 5 = 0, the regression of x on y is (a) x + 2 y = 4 (b) 2x + 3 y − 5 = 0 (c) they are not lines of regression (d) None of the above

(a) x = 0, y = 0 (b) x = 0, y = constant (c) x = constant, y = 0 (d) x = constant, y = constant

(a) − 0.82

(b) 0.82

(b) 3, 5 (d) 3, − 5

93. The two lines of regression are 8 x − 10 y = 66 and 40 x − 18 y = 214 and variance of x is 9. The standard deviation of y series is (b) 6 (d) 3

(c) − 0.72

(d) 0.76

96. If r1 and r2 are regression coefficients of y on x and x on y respectively, then (a) r1 + r2 = 2r (c) r1 + r2 ≥ 2r

92. The two lines of regression are 13x − 10 y + 11 = 0 and 2x − y − 1 = 0, then mean of x and y series are

(a) 8 (c) 4

1− a b α (d) 1−α

(b)

95. The two lines of regression are x + 2 y − 5 = 0 and x + 3 y − 8 = 0. The coefficient of correlation between x and y is

91. If x and y are independent variables, then two lines of regression are

(a) 3, −5 (c) −3, 10

b β or 1− a 1−α β (c) 1−β

(a)

20

(b) r1 + r2 < 2r (d) None of these

97. If two regression lines are coincident, then which is correct? (a) r = 0 (c) r = 1 / 2

(b) r = − 1 / 2 (d) r = ± 1

98. Angle between two lines of regression is  r − 1 / r2  (a) tan − 1  1   1 + r2 / r1   r − 1 / r1  (c) tan − 1  2   1 + r1 / r2 

 r r − 1 (b) tan − 1  1 2   r1 + r2  r − r2  −1  1 (d) tan    1 + r1r2 

Targ e t E x e rc is e s

(a) (0, 0) (c) (σ x , σ y )

94. Lines of regression of y on x and x on y are respectively y = ax + b and x = αy + β. If mean of x and y series is same, then its value is

Statistics

89. The point that lies on both the lines of regression for a bivariate distribution, is

Entrances Gallery JEE Advanced/IIT JEE 1. The minimum value of the sum of real numbers a − 5 , a − 4 , 3a − 3 , 1, a 8 and a10 with a > 0is

.

[2011]

JEE Main/AIEEE 2. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data is [2015] (a) 16.8

(b) 16.0

(c) 15.8

(d) 14.0

3. The variance of first 50 even natural numbers is 833 (a) 4

(b) 833

(c) 437

[2014] 437 (d) 4

4. All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even [2014] after the grace marks were given? (a) Mean (c) Mode

(b) Median (d) Variance

5. Let x1 , x 2 , ... , x n be n observations, x be their arithmetic mean and σ 2 be the variance. Statement I Variance of 2x1 , 2x 2 , ... , 2x n is 4σ 2 . Statement II Arithmetic mean of 2x1 , 2x 2 , ... , 2x n is 4x.

[2012] (a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (c) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (d) Statement I is true, Statement II is false

6. If the mean deviations about the median of the numbers a, 2a, ... , 50 a is 50, then a equals [2011] (a) 3 (b) 4 (c) 5 (d) 2

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Objective Mathematics Vol. 1

20

7. A scientist is weighting each of 30 fish. Their mean weight worked out is 30 g and a standard deviation of 2 g. Later, it was found that the measuring scale was misaligned and always under reported every fish weight by 2g. The correct mean and standard deviation (in gram) of fishes are respectively [2011] (a) 28, 4 (c) 32, 4

(b) 32, 2 (d) 28, 2

8. For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is [2010] (a)

5 2

11 2 13 (d) 2

(b)

(c) 6

9. If the mean deviation of number 1, 1 + d , 1 + 2d , ... , 1 + 100d from their mean is 255, then d is equal to [2009]

Ta rg e t E x e rc is e s

(a) 10.0 (c) 10.1

(b) 20.0 (d) 20.2

Statement II The sum of first n natural numbers is n ( n + 1) and the sum of squares of first n natural 2 n ( n + 1) ( 2n + 1) [2009] . numbers is 6 (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

11. The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then, which one of the following gives possible values of a and b? [2008] (b) a = 0, b = 7 (d) a = 1, b = 6

12. The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys is the class is [2007] (a) 40% (c) 80%

(b) 20% (d) 60%

13. Suppose a population A has 100 observations 101, 102, ..., 200 and another population B has 100 observations 151, 152, ..., 250. If V A and VB represent the variances of the two populations respectively, then VA [2006] is VB 1084

(a)

9 4

(b)

4 9

(c)

2 3

Σ x i2 = 400 and Σ x i = 80. Then, a possible value of n [2005] among the following is (a) 12 (b) 9 (c) 18 (d) 15

15. If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately [2005] (a) 24.0 (b) 25.5 (c) 20.5 (d) 22.0

16. Consider the following statements: I. Mode can be computed from histogram. II. Median is not independent of change of scale. III. Variance is independent of change of origin and scale. [2004] Which of these is/are correct?

`10. Statement I The variance of first n even natural n2 − 1 numbers is . 4

(a) a = 3, b = 4 (c) a = 5, b = 2

14. Let x1 , x 2 , ... , x n be n observations such that

(d) 1

(a) Only I (b) Only II (c) I and II (d) I, II and III

17. In a series of 2n observations, half of them equal a and remaining half equal − a. If the standard deviation of [2004] the observations is 2 , then a equals (a)

1 n

(b) 2

(c) 2

(d)

2 n

18. The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set [2003] (a) is increased by 2 (b) is decreased by 2 (c) is two times the original median (d) remains the same as that of the original set

19. In an experiment with 15 observations on x, the following results were available Σ x 2 = 2830, Σ x = 170. One observation that was 20 was found to be wrong and was replaced by the correct value 30. Then, the [2003] corrected variance is (a) 78.00 (b) 188.66 (c) 177.33 (d) 8.33

20. In a class of 100 students, there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls? [2002] (a) 73 (b) 65 (c) 68 (d) 74

[BITSAT 2014] n (b) x + 2

(a) x + n (c) x +

n+1 2

(a) 30 (c) 25 (e) 20

(d) None of these

22. In a moderately asymmetrical distribution, the mean and median are 36 and 34, respectively. Find out the value of emprical mode. [J&K CET 2014] (a) 30 (c) 42

(b) 32 (d) 22

23. The variance of first n natural numbers is [Manipal 2014] (n + 1) (n + 5) (b) 12 (n2 − 1) (d) 12

n (n + 1) (a) 2 (n + 1) (n − 5) (c) 12

24. If the coefficient of variation and standard deviation are 60 and 21 respectively, then arithmetic mean of distribution is [Karnataka CET 2014] (a) 60

(b) 30

(c) 35

9

25. If

∑ ( xi − 5) = 9

(d) 21

9

and

i=1

∑ ( xi − 5)2 = 45,

then the

i=1

standard deviation of the 9 items x1 , x 2 , ... , x 9 is [Kerala CEE 2014] (a) 9 (d) 2

(b) 4 (e) 1

(c) 3

[Kerala CEE 2014] (b) 9 (e) 16

(c) 12

27. The mean of four observations is 3. If the sum of the squares of these observations is 48, then their standard deviation is [EAMCET 2014] (a) 7

(b) 2

(c) 3

(d) 5

28. The sum of deviations of the variates from the [GGSIPU 2014] arithmetic mean is always (a) + 1 (c) − 1

(b) 0 (d) real number

29. The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2 and 6, [UP SEE 2013] then the other two are (a) 2 and 9 (c) 4 and 7

(b) 3 and 8 (d) 5 and 6

(b) 27 (d) 23

31. If the median of

x x x x , x, , , ( x > 0) is 8, then the 5 4 2 3

value of x is

[Kerala CEE 2011]

(a) 24 (c) 8 (e) 40

(b) 32 (d) 16

32. If x and y are deviations from arithmetic mean, r = 0.5, Σ xy = 120, Σ x 2 = 90, σ y = 8, then n is equal to [MP PET 2011] (a) 100 (c) 15

(b) 10 (d) 50

33. If the standard deviation of 3, 8, 6,10, 12, 9, 11, 10, 12, 7 is 2.71, then the standard deviation of 30, 80, 60, 100, 120, 90, 110, 100, 120, 70 is [Kerala CEE 2011] (a) 2.71 (c) (2.71) 10 (e) 0.271

(b) 27.1 (d) (2.71) 2

34. Coefficient of variation of two distributions are 50% and 60% and their arithmetic means are 30 and 25, respectively. Difference of their standard deviation is (a) 1

26. The standard deviation of 9, 16, 23, 30, 37, 44, 51 is (a) 7 (d) 14

30. The AM of 9 terms is 15. If one more term is added to this series, then the AM becomes 16. The value of the added term is [Kerala CEE 2011]

(b) 1.5

(c) 2.5

[GGSIPU 2011] (d) 0

35. Mean deviation of 6, 8, 12, 15, 10, 9 through mean is (a) 10 (c) 2.5

[GGSIPU 2011] (b) 2.33 (d) None of these

Targ e t E x e rc is e s

21. The mean of n terms is x. If the first term is increased by 1, second by 2 and so on, then the new mean is

Statistics

20

Other Engineering Entrances

36. If the value observed are 1, 2, 3, ..., n each with frequency 1 and n is even, then the mean deviation from mean equals [J&K CET 2011] (a) n (c)

n 4

(b)

n 2

(d) None of these

37. The mean and variance of n observations x1 , x 2 , x 3 , ... , x n are 5 and 0, respectively. If n

∑ xi2 = 400, then the value of n is equal to

i=1

(a) 80 (c) 20 (e) 4

[Kerala CEE 2010]

(b) 25 (d) 16

1085

Answers Work Book Exercise 20.1 1. (a)

2. (d)

3. (b)

4. (a)

5. (c)

6. (a)

7. (c)

8. (a)

9. (c)

10. (b)

11. (b)

12. (a)

13. (c)

14. (d)

15. (d)

16. (b)

17. (d)

18. (c)

19. (b)

20. (c)

21. (b)

22. (b)

23. (b)

24. (d)

5. (c)

6. (d)

7. (b)

8. (b)

9. (b)

10. (c)

Work Book Exercise 20.2 1. (d)

2. (c)

3. (b)

4. (b)

Target Exercises 1. (a)

2. (c)

3. (d)

4. (d)

5. (b)

6. (b)

7. (d)

8. (a)

9. (c)

10. (b)

11. (b)

12. (b)

13. (b)

14. (d)

15. (c)

16. (b)

17. (b)

18. (c)

19. (d)

20. (a)

21. (b)

22. (b)

23. (c)

24. (b)

25. (d)

26. (c)

27. (d)

28. (c)

29. (b)

30. (d)

31. (b)

32. (d)

33. (a)

34. (d)

35. (c)

36. (c)

37. (d)

38. (d)

39. (b)

40. (d)

41. (b)

42. (a)

43. (a)

44. (d)

45. (b)

46. (a)

47. (c)

48. (b)

49. (a)

50. (b)

51. (d)

52. (a)

53. (c)

54. (b)

55. (a)

56. (a)

57. (c)

58. (b)

59. (c)

60. (b)

61. (a)

62. (c)

63. (a)

64. (a)

65. (b)

66. (d)

67. (a)

68. (b)

69. (b)

70. (a)

71. (c)

72. (d)

73. (b)

74. (d)

75. (a)

76. (a)

77. (c)

78. (a)

79. (b)

80. (a)

81. (a)

82. (b)

83. (d)

84. (d)

85. (d)

86. (a)

87. (a)

88. (b)

89. (b)

90. (c)

91. (d)

92. (b)

93. (c)

94. (a)

95. (a)

96. (c)

97. (d)

98. (b)

Ta rg e t E x e rc is e s

Entrances Gallery

1086

1. (8)

2. (d)

3. (b)

4. (d)

5. (d)

6. (b)

7. (b)

8. (b)

9. (c)

10. (d)

11. (a)

12. (c)

13. (d)

14. (c)

15. (a)

16. (c)

17. (c)

18. (d)

19. (a)

20. (b)

21. (c)

22. (a)

23. (d)

24. (c)

25. (d)

26. (d)

27. (c)

28. (b)

29. (c)

30. (c)

31. (a)

32. (b)

33. (b)

34. (d)

35. (b)

36. (c)

37. (d)

Explanations Target Exercises Σ x n n X − x1 + a New, mean = n X=

∴ ⇒

2. Old average is x,then new average is ( x + 2 ).

11. M =

According to the question, 15 x + 70 = 16 ( x + 2 ) ⇒ 15 x + 70 = 16 x + 32 ⇒ x = 38 New average = 40 ∴

=

20

⇒

i =1

1 ⇒ 20

20

20

i =1

i =1

∑ xi − ∑ (30 ) = 2

20

20 2 ∑ xi − 20 (30 ) = 20 i =1

∴

x = 0.1 + 30 = 30.1 30 × 15 + 15 × 6 13. M = = 12 30 + 15

14. M =

9 × 100 + 6 × 80 = 92 15

15. Combined mean =

n1 x1 + n2 x2 50 × 210 + 100 × 150 = n1 + n2 50 + 100

= 170 x1 + x2 + ... + xn y + y2 + ... + ym 16. Given, X = ,Y = 1 n m Further, m = 10 n Hence, mean of combined group is nX + mY nX + 10 nY X + 10 Y M= = = n+m n + 10 n 11

6. The frequency distribution is Marks (x)

0

2

3

5

6

7

8

Number of students (f)

2

2

3

4

2

1

1

Σ fx 60 = =4 Σ f 15

50 + 6 f + 91 + 80 + 60 + 60 40 + f

314 + 7.85 f = 341 + 6 f f = 14.59 ≈ 15 300 − (3 + 6) + (8 + 4) 8. Correct mean = = 1515 . 20 Σx 9. Given, 49 = 100 Now, Σ x = 4900 Infact, Σ x should have been more. The correct Σx is new corrected Σx = 4900 + 20 + 50 + 30 = 5000 5000 Correct mean = ∴ 100 = 50 ⇒

[Q (n + 1) terms present]

2n + 1 − 1 n+1

∑ ( xi − 30 ) = 2

1 35 × 40 + w 40 + = 2 35 + 1 1 36 × 40 + 36 × = 35 × 40 + w ⇒ 2 ⇒ w = 58 ∴Weight of the teacher = 58 kg Σx 4. Given, 169 = 50 Correct value is 143. Clearly, sum is short by 9. 169 × 50 + 9 M= ∴ 50 9 = 169 + = 16918 . 50 Σ x 5. M = 11 Σ x + 15 11M + 15 Also, M= = 12 12 ∴ M = 15

7. Mean = 7.85 =

1 + 2 + 4 + ... + 2 n n+1

12. Given that,

3. Let weight of the teacher be w kg, then

Arithmetic mean =

Σ x = 42 14 Σ x+w 42.400 = 15 w = 48kg

10. Given,

17. Q

25 =

Targ e t E x e rc is e s

1. Given,

26 x1 + 21x2 x1 + x2

⇒ x1 = 4 x2 ∴ x1 = 80% , x2 = 20% 7(10 ) + 3 (5) 18. M = = 8.5 10 n x + n2 x2 19. Q x = 1 1 n1 + n2 ⇒ ⇒

n1 x1 + n2 x2 − x1 n1 + n2 n ( x − x1 ) x − x1 = 2 2 n1 + n2 x − x1 =

x − x1 > 0 x > x1 n ( x − x2 ) x − x2 = 1 1 n1 + n2

[Q x2 > x1] …(i)

x − x2 < 0 ⇒ x < x2 ⇒ From Eqs. (i) and (ii), x1 < x < x2

[Q x2 > x1 ] ...(ii)

⇒ ⇒ and

1087

Objective Mathematics Vol. 1

20

20. Given, G1 = ( x1 x2 ... xn )1/ n

34. Since, the series is in ascending order, hence median is

G2 = ( y1 y2 ... yn )1/ n x x x New series is 1 , 2 , … , n y1 y2 yn

and

∴

x x x  G =  1 ⋅ 2 ... n   y1 y2 yn  =

measure of

Mode is also, clearly 5. 1/ n

( x1 x2 ... xn )1/ n G1 = ( y1 y2 ... yn )1/ n G2

21. GM = (5 ⋅ 8 ⋅ 10 ⋅ 15 ⋅ 20 ⋅ 25 ⋅ 30 )1/ 7 = 10 (9)1/ 7 22. GM = (1⋅ 2 ⋅ 4 ... 2 n )1/ n = (2 0 ⋅ 21 ⋅ 2 2 ... 2 n )1 / n n+1 = [2( 0 + 1 + 2 + … + n ) ]1/ n = 2 2

23. GM = (2 ⋅ 4 ⋅ 16 ⋅ 32 )1/ 4 = 8 24. By definition, log G =

n1 log G1 + n2 log G2 n1 + n2

∴ Median = Mode −1 + 0 + 1 + 2 + 3 + 5 + 5 + 6 + 8 + 10 + 11 Mean = 11 50 = = 4.54 (approx.) 11

35. Mode = Measure of most repeated item = 9 36. Mode = Measure of most repeated item = 5 2 + 2 + 3 + 3 + … + 8 + 8 64 = 13 13 13 + 1 Median = Measure of th item  2 

37. AM =

= Measure of 7th item = 4 Mode = Measure of most repeated item = 8

25. Let us first write the ascending series as shown below: 50, 51, 52, 53, 54, 55, 57, 58, 59, 61, 62, 62, 68 (13 + 1) th i.e. 7th item is median 2 value, which is 57.

Thus, the measure of

Ta rg e t E x e rc is e s

26. Arranging in ascending order, we get 18, 18, 20, 21, 24, 26, 27, 29, 31, 35, 36, 38, 41, 42, 44 (15 + 1) The measure of th = 8th item gives median. 2 Median = 29 ∴

27. Median being positional average will not change. (24 + 38 + 55 + 69 + 89) = 55 5 (5 + 1) Median is measure of = 3 rd item = 55 2

28. Mean =

∴ Mean = Median 5th + 6th 29. Me = Measure of term = 71 2

30. Arranging in ascending order, we get 0, 1, 2, 4, 5, 8, 12, 16, 20, 22, 25, 30, 32, 35, 38  15 + 1 Median = Value of   th term  2  =Value of 8th term = 16

31. Given, n = 100, Mean = 50, Median = 52 Σ x = 50 n ∴ Σ x = 5000 5000 − 100 + 110 Corrected mean = ∴ 100 = 50.10 Median remains same. M=

32. Middle 50% of items. 7n 33. D7 = 10 70 n 7 n P70 = = 100 10 ∴ 70th percentile = 7th decile

1088

(11 + 1) th = 6 th item = 5 2

38. We see that,  7 + 1 Q1 = Size of   th item = 2nd item  4   7 + 1  Q3 = Size of   th item = 6th item  4   1 ∴ Quartile deviation = (Q3 − Q1 ) = 3.5 2 2 3

39. By definition, semi-interquartile range is σ. 40. Coefficient of quartile deviation =

Q3 − Q1 40 − 10 3 = = Q3 + Q1 40 + 10 5

41. Mean deviation is minimum when it is considered about the item, equidistant from the beginning and the end i.e.  101 + 1 the median. In this case, median is   th term  2  i.e. 51st item i.e. x51.

42. We know that, median divides the series into two equal parts (i.e. it is positional average) and hence absolute deviations will be least when measured from median. Σx 43. Clearly, M = =6 n and Σ ( x − M )2 = Σ x 2 − 2 M Σx + Σ M 2 ∴ ⇒

= 1000 − 12 × 60 + 10 × 36 = 640 Σ ( x − M )2 640 σ = = = 64 10 n σ=8 2

44. Given, x1 + x2 + ... + x10 = 12 and

2 x12 + x22 + … + x10 = 18 2

∴

⇒

1 18  12  1  Σ x 2 −  Σ x = −  n  n 10  10  9 36 9 = − = 5 25 25 3 SD = 5 σ2 =

2

1976 . × 100 = 56.2 3516 . 17.5 56. Clearly, 70 = × 100 ⇒ x1 = 25 x1 18 and 90 = × 100 x2

( x) =

55. CV =

∴

n −1 35 = 12 12 6.33 + 6.37 + 6.36 + 6.32 + 6.37 47. M = = 6.35 5 Σ( x − m)2 0.0022 = = 0.00044 5 n

σ2 =

61. Mode = 3 Median − 2 Mean ⇒ 6λ = 3 Median − 18λ Median = 8λ ∴

= 200[152 + 40 2 ] = 365000 Correct Σx = 365000 − 2500 + 1600 2

= 364100 364100 Corrected σ = − (39.95)2 200 = 1820.5 − 1596

65. For symmetrical distribution, Mean = Median = Mode, it is zero. 41 − 45 66. Coefficient of skewness = = − 0.5 8 5 × 10 67. Mean increased = =5 10

= 224.5 = 14.98

51. Since, SD ≤ Range = b − a ∴ Var ( x ) ≤ (b − a)2 or (b − a)2 ≥ Var( x )

52. We have, r = max| xi − x |and S2 = j i≠ j

68. New mean = 7 + 5 = 12 n

1 ( xi − x )2 n − 1 i∑ =1 2

x + x2 + ... + xn   Now, ( xi − x )2 =  xi − 1    n 1 = 2 [( xi − x1 ) + ( xi − x2 ) + ... + ( xi − xi− 1 ) n [(n − 1) r ]2 n2 [Q| xi − x |j ≤ r ]

+ ( xi − xi+ 1 ) + ... + ( xi − xn )]2 ≤ n

∑ ( xi − x )2 ≤ nr 2

i =1

⇒ ⇒

n

1 nr 2 ( xi − x )2 ≤ ∑ (n − 1) n − 1i =1 S2 ≤

2

nr n ⇒ S ≤r n −1 (n − 1)

53. Coefficient of variation. Σ ( x − x )2 250 54. σ = = 10 n ⇒ σ=5 σ 5 Now, CV = × 100 = × 100 = 10 x 50 2

M − Mo σ M − Mo 63. By definition, Karl Pearson’s coefficient = =0 σ 622.08 − 625 64. Coefficient of skewness = = − 0.95 3.06

62. By definition, Karl Pearson’s coefficient =

Corrected x = 7990 / 200 = 39.95 Incorrect Σx 2 = n[σ 2 + x 2 ]

⇒

⇒ Me = 22

Mean = Median = Mode

50. Corrected Σx = 40 × 200 − 50 + 40 = 7990

( xi − x )2 ≤ r 2

5 ⋅ 18 + 3 ⋅ 24 15 + (8 − 8) = 20.25 5+ 3 64

60. For symmetrical distribution,

− (1 + 2 + ... + 7 ) =−4 7 Σ xi2 − ( x )2 = 4 ∴ σ2 = n ∴ σ =2

⇒

22 n2 (σ 22 + d 22 ) = 3 n1 + n2

⇒ 24 − 18 = 3 (24 − Me )

49. AM =

∴

58. σ 2 =

+

d12 ) +

59. M − Mo = 3(M − Me )

48. σ 2 = 42 = 16

∴

57. σ =

x2 = 20 n1 (σ12

Targ e t E x e rc is e s

SD =

2

20 Statistics

6 + 8 + 10 + 12 + 14 = 10 5 2 Σ( x − M ) Q σ2 = =8 n 2 ∴Variance = σ = 8 1+ 2 + 3 + 4 + 5 + 6 7 46. Mean = = 6 2

45. Mean

5x 4 marks out of 50. Thus, each observation will be 5 multiplied by . 4 5 Hence, mean is also multiplied by , giving mean 4 5 = 38 × = 47.5 4

69. Let a student gets x marks out of 40. Then, he gets

70. Since, SD is independent of change of origin. Therefore, the standard deviation remains same i.e. 15.40.

71. The standard deviation remains same = 2.45 72. Median = 20 + 2 = 22, SD = 4 73. Resulting standard deviation = cσ 74. New variance = 8 ⋅ (4)2 = 128 [Q Var (kx ) = k 2 Var ( x )] x a − h h σ ⇒ σu = x h ⇒ σ x = hσu

75. Here,u =

[each x is divided by h] σcx = c σ x  Q  and σ c + x = σx  

Ø SD does not change by adding or subtracting a fixed value

from each variate value.

1089

Objective Mathematics Vol. 1

20

76. Given, σ = 9

⇒

Let a student obtains x marks out of 75, then his marks 4x 4 out of 100 are . Each observation is multiplied by . 3 3 4 2 New, σ = × 9 =12, therefore new variance = σ = 144 3 4 5 75   77. SD = (15) = Q MD = SD   5 4 4 2 2  75 Quartile deviation = (SD ) =   = 12.5 3 3  4

∴

78. r = 1

Increase/decrease in brings ⇒ x increase/decrease in y. Further, x + y also increases or decreases. Hence, r between x and x + y is 1. σy =

Cov ( x, y ) =8 r σx

Ta rg e t E x e rc is e s

80. Given, …(i) aX + bY + c = 0 …(ii) ∴ aX + bY + c = 0 From Eqs. (i) and (ii), we get a( X − X ) + b (Y − Y ) = 0 b …(iii) X − X = − (Y − Y ) ⇒ a 1 Cov ( X, Y ) = Σ( xi − X ) ( yi − Y ) n b 1  Σ( yi − Y )2 =−  a  n b …(iv) = − Var(Y ) a 1 Also, Var ( X ) = Σ( xi − X )2 n b2  1  Σ( yi − Y )2 = 2  a  n 2 b …(v) = 2 Var (Y ) a b b − Var(Y ) − a ∴ r= = a b   b2  2 Var(Y ) { Var(Y )} a  a Q ab < 0 ⇒ a and b are  − b/ a = =1   − b/ a  of opposite sign 

81. Let E = σ 2x + y = =

1 n 1 n

n

1 Σ( yi − y )2 n 2 n ( xi − x )( yi − y ) n i∑ =1

E = σ 2x + σ 2y + 2 Cov ( x, y )

We have, σ 2x + y = σ 2x + σ y2 + 2 Cov ( x, y )

1090

⇒

σ 2x − y = σ 2x + σ y2 − 2 rσ x σ y

∴

83. r =

r=

σ 2x + σ 2y − σ 2x − y 2σ x σ y

Cov ( x, y ) , since independent, therefore r = 0 σ xσ y

⇒

84. r =

Σ xy Σ x2 Σ y2

=

Cov ( x, y ) = 0 40 =1 80 × 20

85. Here, Σ xy = 269, Σ x 2 = 55, Σ y 2 = 1347 ∴

r=

269 = 0.988 55 × 1347

86. Here, σ x = 2.89, σ y = 100 ∴

r=

− 16.5 = − 0.97 17 . (10 )

87. We know that, 2

and

∴ ⇒

Cov ( x, y ) r2 =    σ x σy  Σ xy 60 Cov ( x, y ) = = n n Σ x 2 90 2 = σx = n n 3600 / n 2 0.64 = (90 / n ) (6.25) n = 10

88. Here, r 2 = 0.25. Hence, only 25% of variations are explained.

90. If we solve first for x and second for y, then product of

+ ⇒

Z =X −Y Z − Z = X − X − (Y − Y ) (Z − Z )2 = ( X − X )2 + (Y − Y )2 − 2 ( X − X ) (Y − Y ) ⇒ 1 1 1 ⇒ Σ(Z − Z )2 = Σ( X − X )2 + Σ( Y − Y )2 n n n 2 − Σ( X − X )(Y − Y ) n ⇒ σ 2z = σ 2x + σ 2y − 2 Cov ( X, Y )

regression. Hence, ( x, y ) lies on both.

yi − ( x + y )]2

∑ ( xi − x )2 +

i =1

∴

89. We know, ( x, y ) is the point of intersection of two lines of

∑ [ xi +

i =1 n

82. Let Z = X − Y

⇒

79. Given, r = 0.5, Cov ( x, y ) = 16, σ x = 4 ∴

Cov ( x, y ) = 12 Cov ( x, y ) r ( x, y ) = σ x σy 12 = =1 (3) (4)

49 = 9 + 16 + 2 Cov ( x, y )

regression coefficient is negative, which is not possible. Similarly, reversing the process we again find that product of regression coefficient is negative, which is not possible. Therefore, they are not the lines of regression.

91. x − x = 0 and y − y = 0 92. Solving 13 x − 10 y + 11 = 0 and 2 x − y − 1 = 0, we get x=3

and

y=5

4 5

From 40 x − 18 y − 214 = 0, we get r2 = 36 r = 100

96. Given, r1 = 9 20

⇒ r = 0.6

2

But

94. Let mean of x and y series be x and y. Lines of regression passes through mean ( x, y ), hence y = ax + b, x = αy + β b β Given that, x = y ⇒ x = = 1− a 1− α

95. x = − 2 y + 5 ⇒ and ∴

r1r2 = r

r1 = − 2

1 8 1 y=− x+ ⇒ r2 = − 3 3 3 2 2 r = ⇒ r = − 0.82 3

σx

20

rσx σy

, r2 =

2

⇒ r is GM of r1 and r2 . Now,

12 σy = =4 3.0

∴

∴

r σy

∴

r1 + r2 2 r +r AM ≥ GM ⇒ 1 2 ≥ r 2 r1 + r2 ≥ 2 r AM =

Statistics

93. From 8 x − 10 y − 66 = 0, we get r1 =

97. r = ± 1 98. The slope of line of regression of y on x is m1 = r1. Similarly, for x on y, m2 =

1 r2

1 m − m2 r2 r1r2 − 1 tan θ = 1 = = 1 + m1m2 1 + r1 r2 + r1 r2 r1 −

∴

Entrances Gallery a− 5 + a− 4 + a− 3 + a− 3 + a− 3 + a8 + a10 + 1 ≥1 8

Minimum value = 8 x + x2 + x3 + ... + x16 2. (d) Given, 1 = 16 16 16

∑ xi = 16 × 16

⇒

i=1

18

= ∑ yi = (16 × 16 − 16) + (3 + 4 + 5) i =1

18

i =1

18

=

252 = 14 18

3. Clearly, x = Q

r =1

50

∴Statement I is correct. Finally, Statement I is true and Statement II is false.

6. Median of a, 2 a, 3 a, 4 a, ..., 50 a 25 a + 26 a = (25.5) a 2 MD (about median) =

50

∑ 2r

From here, only we can tell that option (d) is the correct answer. However, here we will formally check the validity of Statement I. Variance of 2 x1, 2 x2 , 2 x3 , ..., 2 xn = Variance (2 x ) = 2 2 Variance ( x ) = 4σ 2

= 252 Number of observations = 18 New mean =

n observations x1, x2 , x3 , ..., xn . Now, AM of 2 x1, 2 x2 , 2 x3 , ..., 2 xn 2 x + 2 x2 + 2 x3 + ... + 2 xn = 1 n  x1 + x2 + x3 + ...+ xn  =2   = 2x   n Hence, Statement II is false.

Sum of new observations

∑ yi

5. Given, x is the AM and σ 2 is the variance of

Targ e t E x e rc is e s

1.

= 51

50

 Σ x2  σ2 =  i  − x 2  n 

=

50

σ = 2

∑ 4r

⇒

2

r =1

50 = 833

− (51)

2

4. If initially marks were xi, then σ12 =

Σ( xi − x )2 n

Now, each is increased by 10 Σ[( xi + 10 ) − ( x + 10 )]2 ∴ = σ12 σ 22 = n So, variance will not change whereas mean, median and mode will increase by 10.

⇒ ∴

∑| xi − Median| i =1

n 1 50 = {2| a| ⋅ (0.5 + 15 . + 2.5 + ... + 24.5)} 50 25 2500 = 2| a| ⋅ (25) 2 | a| = 4

7. Correct mean = Old mean + 2 = 30 + 2 = 32 As, standard deviation is independent of change of origin. ∴ It remains same. ⇒ Standard deviation = 2

8. We have, σ 2x = 4 and σ 2y = 5 Also,

x = 2 and

y=4

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Objective Mathematics Vol. 1

20

Now, ⇒ Also, ⇒

Σ xi Σ yi = 2 and =4 5 5 Σ xi = 10 and Σyi = 20 1  1 σ 2x =  Σ xi2  − ( x )2 = (Σxi2 ) − 4  5 5 Σxi2 = 40

Similarly,

Σyi2

= 105

Therefore, variance of observations 101, 102,..,200 is same as variance of 151, 152,..., 250. V ∴ VA = VB ⇒ A = 1 VB

14. Given that, Σ xi2 = 400 and Σ xi = 80, since σ 2 ≥ 0 2

2

1  x + y (Σ xi2 + Σyi2 ) −    2  10 1 145 − 90 55 11 = = = (40 + 105) − 9 = 10 10 10 2 101 (a + l ) Sum of quantities 9. x = = 2 n 101 1 = [1 + 1 + 100 d ] = 1 + 50 d 2 1 ∴ MD (from mean) = Σ| xi − x| = 255 n 1 = [50 d + 49 d + 48 d + K + d + 0 + d 101 + K + 50 d ] 2d  50 × 51 255 × 101 ⇒ d = = = 10.1 101  2  50 × 51 Now, σ z2 =

10. Statement II is true. Statement I Sum of n even natural numbers = n(n + 1) n(n + 1) Mean ( x ) = =n+1 n 1  Variance = Σ( xi )2 − ( x )2  n  1 2 = [2 + 42 + K + (2 n )2 ] − (n + 1)2 n 1 2 2 = 2 [1 + 2 2 + K + n 2 ] − (n + 1)2 n 4 n(n + 1)(2 n + 1) = ⋅ − (n + 1)2 n 6 (n + 1)[2(2 n + 1) − 3 (n + 1)] = 3 (n + 1)(n − 1) n 2 − 1 = = 3 3 ∴ Statement I is false.

11. According to the given condition,

[(6 − a)2 + (6 − b)2 + (6 − 8)2 + (6 − 5)2 + (6 − 10 )2 ] 5 ⇒ 34 = (6 − a)2 + (6 − b)2 + 4 + 1 + 16 ⇒ (6 − a)2 + (6 − b)2 = 13 = 9 + 4 ⇒ (6 − a)2 + (6 − b)2 = 32 + 2 2 ⇒ a = 3, b = 4 6.80 =

12. Let the number of boys and girls be x and y.

1092

13. Since, variance is independent of change of origin.

∴ 52 x + 42 y = 50( x + y ) ⇒ 52 x + 42 y = 50 x + 50 y ⇒ 2 x = 8y ⇒ x = 4y ∴Total number of students in the class = x + y = 4y + y = 5y ∴Required percentage of boys  4y  = × 100 = 80%  5y 

Σ xi2  Σ xi  −  ≥0  n  n 400 6400 − 2 ≥0 n n n ≥ 16

⇒ ⇒ ∴

15. Given that, mean = 21 and median = 22 Using the relation, Mode = 3Median − 2 Mean Mode = 3 (22 ) − 2(21) = 66 − 42 = 24 ∴

16. It is true that mode can be computed from histogram and median is not independent of change of scale. But variance is independent of change of origin and not of scale.

17. In the 2n observations, half of them equal to a and remaining half equal to − a. Then, the mean of total 2n observations is equal to zero. Σ( x − x )2 SD = ∴ N ⇒ ⇒ ⇒ ⇒

Σ x2 2n Σ x2 4= 2n 2 na2 4= 2n 2 a = 4 ⇒ | a| = 2 2=

18. Since, the four largest observations either be the first four observation or last four observations, therefore median of new set remains the same as that of the original set.

19. Given, N = 15, Σ x 2 = 2830, Σ x = 170 Since, one observation 20 was replaced by 30, then Σ x 2 = 2830 − 400 + 900 = 3330 Σ x = 170 − 20 + 30 = 180 2 Σ x2  Σ x −  ∴ Variance, σ 2 =  N N and

2

3330 − 12 × 180 3330  180  −  =  15  15 15 3330 − 2160 1170 = = = 78.0 15 15 =

20. Since, total number of students = 100 and number of boys = 70 ∴ Number of girls = (100 − 70 ) = 30 Now, the total marks of 100 students = 100 × 72 = 7200 and total marks of 70 boys = 70 × 75 = 5250 Total marks of 30 girls = 7200 − 5250 = 1950 1950 Average marks of 30 girls = ∴ = 65 30

+ (30 − 30 )2 + (37 − 30 )2

Now, mean ( x ) =

22. Given, mean = 36, median = 34 We know that, Mode = 3 Median − 2 Mean ∴ Mode = 3 × 34 − 2 × 36 = 102 − 72 = 30

23. Variance of first n natural numbers is given by 2

n(n + 1)(2 n + 1)  n(n + 1) = −  6n  2n  2 n + 1 (n + 1) = (n + 1) − 4   6

(− 21)2 + (− 14)2 + (− 7 )2 + (0 )2 + (7 )2 7 441 + 196 + 49 + 49 + 196 + 441 = 7 1372 = 7 = 196 = 14

27. Let four observations be x1, x2 , x3 and x4 . Given, mean ( x ) = 3 and Σ xi2 = 48 ∴

mean will be x=

2

and standard deviation, σ = 21 σ CV = × 100 Q x 21 21 60 = × 100 ⇒ x = × 100 ∴ x 60 ∴ x = 35

25. We have, Σ xi − 45 = 9 ⇒ Σ xi = 54

⇒ ⇒

Σ( xi − 5)2 = 45 Σ( xi2 − 2 xi ⋅ 5 + 25) = 45 × 54 × 5 + 25 × 9 = 45 = 45 + 540 − 225 = 360

σ=

360  54 −   9 9

= 40 − 36 = 2 9 + 16 + 23 + 30 + 37 + 44 + 51 7 210 = = 30 7 Σ( xi − x )2 ∴ Standard deviation = 7 x=

∑ xi

i =1

…(i)

n

Now, sum of deviations of the variates from the AM n

n

n

i =1 n

i =1 n

i =1 n

i =1

i =1

i =1

= ∑ ( xi − x ) = ∑ xi − ∑ x = ∑ xi − nx = ∑ xi − ∑ xi

[from Eq. (i)]

=0

29. Let two unknown items be x and y, then 1+ 2 + 6 + x + y =4 5 ⇒ x + y = 11 and variance = 5.2 12 + 2 2 + 62 + x 2 + y 2 ⇒ − ( x )2 = 5.2 5 ⇒ 41 + x 2 + y 2 = 5 (5.2 + 42 ) Mean = 4 ⇒

⇒

...(i)

41 + x 2 + y 2 = 106

∴

x 2 + y 2 = 65

...(ii)

On solving Eqs. (i) and (ii), we get x = 4, y = 7 or x = 7, y = 4

2

26. Now, mean of given observation is,

48 Σ xi2 − ( x )2 = − (3)2 n 4 = 12 − 9 = 3

n

24. Given, coefficient of variation, CV = 60

Σ( xi2 ) − 2 Σ xi2

SD =

28. Let x1, x2 , ..., xn be n variates. Then, their arithmetic

(n + 1) [4 n + 2 − 3 n − 3] 12 (n + 1) n2 − 1 = × (n − 1) = 12 12

⇒

+ (14)2 + (21)2

=

=

and

+ (44 − 30 )2 + (51 − 30 )2 7

=

Targ e t E x e rc is e s

Σ n2  Σ n variance = −   n n

Statistics

x1 + x2 + K + xn n When first term is increased by 1, second by 2 and so on, then the observation will be ( x1 + 1), ( x1 + 2 ), ( x1 + 3),..., ( xn + n ). Then, new mean ( x + 1) + ( x2 + 2 ) + K + ( xn + n ) x= 1 n ( x1 + x2 + K + xn ) + (1 + 2 + 3 + K + n ) = n x1 + x2 + K + xn 1 + 2 + 3 + K + n = + n n n(n + 1) n(n + 1)  =x+ Q1+ 2 + 3 + K + n =  2  2n n+1 =x+ 2

20

(9 − 30 )2 + (16 − 30 )2 + (23 − 30 )2

21. Let the observations be x1, x2 , x3 , ..., xn .

n

30. Q AM =

∑ xi i =1

n

9

⇒ 15 =

∑ xi i =1

9

9

∑ xi = 135

⇒

i =1 10

Also, ∴

16 =

∑ xi i =1

10

⇒

10

∑ xi = 160 i =1

Added term = 160 − 135 = 25

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20

x x x x 5 4 3 2 Median = 8

x =8 3 x = 24

⇒ ⇒

32. σ 2x =

Σ x 2 90 = n n

Σ xy 120 = n n σ 2y = 82 = 64

Cov ( x, y ) =

∴

r2 =

⇒

0.25 =

⇒

[Cov( x, y )]2 σ 2x σ y2

(120 )2 / n 2 90 ⋅ 64 n (120 )2 n= 0.25 × 90 × 64 14400 = = 10 1440

Ta rg e t E x e rc is e s

33. Standard deviation = 10 × 2.71 = 27.1

1094

34. Since, CV1 = ⇒ Similarly, ∴

6 + 8 + 12 + 15 + 10 + 9 = 10 6 Σ| xi − x| ∴Mean deviation mean = n |6 − 10| + |8 − 10| + |12 − 10| + |15 − 10|  + |10 − 10| + |9 − 10|  = 6 4 + 2 + 2 + 5 + 0 + 1 14 = = = 2.33 6 6

35. Here, mean =

31. In , , , , x

σ1 × 100 x 30 100 25 σ 2 = 60 × 100 σ1 = 50 ×

σ1 − σ 2 = 0

36. We have, 1, 2, 3, ..., n

n(n + 1) n + 1 = 2n 2 ∴ Mean deviation from mean 1 = Σ|( xi − x )| n 1  n + 1 n + 1 n + 1    =  1 −   + ... +  n −  + 2 −   2  n  2  2   1 1− n 3− n n−3 n − 1 =  + + ... + + n 2 2 2 2  1 n − 1 n − 3 n − 3 n − 1 = + + ... + + n  2 2 2 2  1 = [(n − 1) + (n − 3) + ... + 1] [Q n is even] n 2 1 n n  n = (n − 1 + 1) = =   n 4  4n 4 ∴

x=

37. We have, x = 5 and variance = ∴ ⇒

1 Σ xi2 − ( x )2 n

1 ⋅ 400 − 25 n 400 n= = 16 25 0=

21 Fundamentals of Probability Introduction Probability is a measure of uncertainty of various phenomenon and to measure it. There are three approaches: (i) Statistical approach (ii) Classical approach (iii) Axiomatic approach In statistical approach, probability is find on the basis of observations and collected data, which is not used in general and to understand classical and axiomatic approach, we must know about some definitions.

Some Basic Definitions Deterministic Experiment Those experiments which when repeated under identical conditions produce the same result or outcome are known as deterministic experiments. When experiments in science or engineering are repeated under identical conditions, we get almost the same result every time.

Probabilistic or Random Experiment If an experiment, when repeated under identical conditions, do not produce the same outcome every time but the outcome in a trial is one of the several possible outcomes, then such an experiment is known as a probabilistic experiment or a random experiment.

Outcomes A possible result of a random experiment is called outcome. e.g. If the experiment consists of a tossing a coin twice, then some of the outcomes are HH, HT, etc.

Sample Space The set of all possible outcomes of a random experiment is called the sample space associated with it and it is generally denoted by S. e.g. The sample space for the experiment of tossing a die is given by S = {1, 2, 3, 4, 5, 6 }

Chapter Snapshot ●

Introduction

●

Some Basic Definitions

●

Event

●

Important Events

●

Algebra of Events

●

Probability

●

Geometrical Probability

●

Addition Theorem of Probability

●

Independent Events

●

Booley’s Inequality

X

Objective Mathematics Vol. 1

21

Sol. (d) The sample space for this experiment is {RR, RB, BR, BB}, where R denotes the red ball and B denotes the black ball.

Event An event is associated with a subset of sample space. Events can be classified into various types on the basis of the elements they have.

Types of Events (i) Impossible and sure events The empty set φ is called an impossible event. e.g. Getting two heads in tossing a unbiased coin one time. The event which is certain to occur is said be the sure event. e.g. In tossing a die, any number will come is a sure event. (ii) Simple event If an event E has only one sample point of a sample space, it is called a simple (or elementary) event. e.g. In the experiment of tossing a coin twicely getting both head i.e. E = {H, H }. (iii) Compound event If an event has more than one sample point, it is called a compound event. e.g. In the experiment of tossing a coin thrice the events E : Exactly one head appeared F : Atleast one head appeared G : Atmost one head appeared, etc., are all compound events. As E = {HTT , THT , TTH } F = {HTT , THT , TTH , HHT , THH , HTH , HHH } G = {HTT , THT , TTH , TTT } Each of the above subsets contain more than one sample point. Hence, they are all compound events. X

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Sol. (b) A = Getting the sum of the number on the dice is 2

Example 1. A bag contains 4 identical red balls and 3 identical black balls. The experiment consists of drawing one ball, then putting it into the bag and again drawing a ball. Then, the possible outcomes of this experiment is (a) {RR, BB} (b) {RR, BB, RR} (c) {BB, R} (d) None of these

Example 2. A pair of dice is rolled. Describe the following events associated with this random experiment A = Getting the sum of the number on the dice is 2. B = Getting the sum of the number is multiple of 11. C = Getting the sum of the number is multiple of 12. Which of these events are compound events? (a) A (b) B (c) C (d) All of these

= {1, 1} B = Getting the sum of the number is multiple of 11 = {(5, 6), (6, 5)} C = Getting the sum of the number is multiple of 12 = (6, 6) ∴ B is the compound event.

Important Events Equally Likely Events Events are said to be equally likely when we have no reason to believe that one is more likely to occur than the other. e.g. When an unbiased die is thrown, all the six faces 1, 2, 3, 4, 5, 6 are equally likely to come up.

Exhaustive Events A set of events is said to be exhaustive, if one of them must necessarily happen every time the experiment is performed. e.g. When a die is thrown, events 1, 2, 3, 4, 5, 6 form an exhaustive set of events. Ø We can say that the total number of elementary events of a

random experiment is called the exhaustive number of cases.

Mutually Exclusive Events Two or more events are said to be mutually exclusive, one of them occurs, others cannot occur. Thus, two or more events are said to be mutually exclusive, if no two of them can occur together. Hence, A1 , A2 ,A3 ,…,An are mutually exclusive if and only if Ai ∩ A j = φ, ∀ i ≠ j. e.g. When a die is thrown, the sample space is S = {1, 2, 3, 4, 5, 6}. Let A be an event of occurrence of number greater than 4, i.e. {5, 6}. B is an event of occurrence of an odd number, i.e. {1, 3, 5}. C is an event of occurrence of an even number, i.e. {2, 4, 6}. Here, events B and C are mutually exclusive but the events A and B or A and C are not mutually exclusive. X

Example 3. Two dice are thrown. The events A, B and C are as follow : A : Getting an even number on the first die. B : Getting an odd number on the first die. C : Getting the sum of the numbers on the dice ≤ 5. Which of the following statement is false? (a) A and B are mutually exclusive (b) A and B are mutually exclusive and exhaustive (c) A = B ′ (d) A and C are mutually exclusive

(i) Complementary event For every event A, there corresponds another event A′ or A C or A called the complementary event to A. It is also called the event not A. e.g. Let S = {HHH , HHT , HTH , THH , HTT , THT , TTH , TTT } be a sample space and A = {HTH , HHT , THH } be a subset of S, then complementary of A is defined as A = {HHH , HTT , THT , TTH , TTT }

outcomes, S = 6 × 6 = 36 which are as follow: →

↓ 1 2 3 4 5 6

1

2

3

4

5

6

1, 1 2, 1 3, 1 4, 1 5, 1 6, 1

1, 2 2, 2 3, 2 4, 2 5, 2 6, 2

1, 3 2, 3 3, 3 4, 3 5, 3 6, 3

1, 4 2, 4 3, 4 4, 4 5, 4 6, 4

1, 5 2, 5 3, 5 4, 5 5, 5 6, 5

1, 6 2, 6 3, 6 4, 6 5, 6 6, 6

A = Getting an even number on the first die = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} B = Getting an odd number on the first die = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)} Here, B′ = Event getting an odd number on the first die. (a) True Q A = Getting an even number on the first die B = Getting an odd number on the first die ⇒ A∩B= φ ∴ A and B are mutually exclusive events. (b) True ∴ A ∪ B = S i.e. exhaustive. Also, A ∩ B = φ (c) True Q B = Getting an odd number on the first die ⇒ B′ = Getting an even number on first die = A A = B′ (d) False Q A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠ φ, so A and C are not mutually exclusive.

Algebra of Events

(ii) Event A or B It is the set of all those elements, which are either in A or B or in both. And it is denoted by the symbol A ∪ B . e.g. Let A = {1, 2, 3} and B = {3, 4}, then A ∪ B = {1, 2, 3, 4} (iii) Events A and B It is the set of all those elements, which are common in A and B. And it is denoted by the symbol A ∩ B . e.g. Let A = {2, 3, 5} and B = {1, 2, 3, 4, 5}, then A ∩ B = {2, 3, 5} (iv) Event A but not B It is the set of all those elements, which are in A but not in B. And it is denoted by the symbol A − B or A ∩ B ′. e.g. Let A = {1, 2, 3, 4} and B = {2, 4, 6} , then A − B = {1, 3} X

Example 4. Let S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}, then E ′ is (a) {2, 4} (b) {3, 6} (c) {1, 2, 4} (d) {2, 4, 6} Sol. (d) We have,

Let A, B and C be events associated with an experiment, whose sample space is S.

21 Fundamentals of Probability

Sol. (d) If two dice are thrown, then total number of possible

and

S = {1, 2, 3, 4, 5, 6} E = {1, 3, 5} E ′ = S − E = {2, 4, 6}

Work Book Exercise 21.1 1 Two coins are tossed once, then the sample space is a {HH, HT, TH, TT} c {HT, TH, HH}

b {TT, HH} d None of these

2 A coin is tossed twice, if the second throw result in a tail, a die is thrown. Then, the total number of the outcomes of this experiment is a 11

b 13

c 14

d 16

3 Event can be classified into various types on the basis of the a b c d

experiment sample space elements None of the above

4 An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events.

A : The sum is greater than 8. B : 2 occurs on either die. C : The sum is atleast 7 and a multiple of 3. Which pair of these events is mutually exclusive? a A and B c Both A and B

b B and C d None of these

5 A pair of dice is rolled, if the outcomes is a doublet, a coin is tossed. Then, the total number of outcomes for this experiment is a 40 c 42

b 41 d 43

6 An experiment consists of rolling a die until a 2 appears. Then, number of elements of the sample space correspond to the event that the 2 appears not later than the Kth roll of the die is a

5K 4

b 5K − 1

c

5K − 1 4

d

5K − 1 5

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Objective Mathematics Vol. 1

21 Probability Classical Approach of Probability If there are n elementary equally likely events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A is denoted by P ( A ) and m is defined as the ratio . n m Thus, P ( A) = n Clearly, 0 ≤ m ≤ n. Therefore, m 0 ≤ ≤1 n ⇒ 0 ≤ P ( A) ≤ 1

Let S be the sample space and A be an event, such that n ( S ) = n and n ( A ) = m. If each outcomes is equally likely, then it follows that m Number of outcomes favourable to A P ( A) = = n Total number of possible outcomes Ø

●

●

●

Axiomatic Approach of Probability Let S be the sample space of a random experiment. The probability P is a real valued function whose domain is the power set of S and range is the interval [0, 1] satisfying the following axioms: (i) For any event E, P ( E ) ≥ 0 (ii) P ( S ) =1 (iii) If E and F are mutually exclusive events, then P (E ∪ F ) = P (E ) + P (F ) It follows from axion (iii) that P (φ ) = 0 Let S be a sample space containing outcomes α 1 , α 2 , K , α n i.e. S = {α 1 , α 2 , … , α n }. It follows from the axiomatic definition of probability that (i) 0 ≤ P (α i ) ≤ 1, for each α i ∈ S (ii) P (α 1 ) + P (α 2 ) + K + P (α n ) = 1 (iii) P ( A ) = ΣP (α i ), for any event A containing elementary events wi .

Probabilities of Equally Likely Outcomes Let a sample space of an experiment be S = {w1 , w2 , K, wn } and suppose that all the outcomes are equally likely to occur i.e. the chance of occurrence of each simple event must be the same. i.e. P ( wi ) = p, for all wi ∈ S , where 0 ≤ p ≤ 1

●

X

If the outcomes are equally likely, then classical approach and axiomatic approach are equivalent. If P(A) = 1 , then A is called certain event and A is called an impossible event, if P(A) = 0. The number of elementary event which will ensure the non-occurrence of A i.e. which ensure the occurrence of A is (n − m). Therefore, n−m m = 1 − = 1 − P(A) ⇒ P(A) + P(A) = 1 P(A) = n n The odds in favour of occurrence of the event A are defined by m :(n − m) i.e. P(A) : P(A) and the odds against the occurrenuce of A are defined by n − m : m i.e. P(A) : P(A).

Example 5. There are four men and six women in the city council. If one council member is selected for a committee at random, then how likely is it that it is a woman? 4 3 1 1 (a) (b) (c) (d) 5 5 7 5 Sol. (b) Let E be the event that woman is selected. ∴ n(E ) = 6 Number of sample space n(S ) = Total number of persons in the council = 4 + 6 = 10 n(E ) 6 3 ∴ P= = = n(S ) 10 5

X

Example 6. Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Now, the probability of getting atleast one tail is 1 2 (a) (b) 4 4 3 (c) (d) None of these 4 Sol. (c) The outcomes of the experiment can be represented in following diagrammatic manner called the tree diagram. (H,H) Head (H)

n

Since,

(H,T)

∑ P (wi ) = 1 i =1

i.e. p + p + p +… + p (n times) =1 1 ⇒ np =1 i.e. p = n 1098

Tail (T)

(T,1) (T,2) (T,3) (T,4) (T,5) (T,6)

1/4

(H,H)

1/4

(H,T)

Sol. (d) Out of 20 numbers, six numbers can be chosen in 20

∴Total number of outcomes =

X

(T,1) 1/12 (T,2)

1/2

1/12

Tail (T )

1/12 1/12

(T,3)

1/12 (T,5)

∴

(T,6)

X

X

Example 7. What is the probability that a number selected from the numbers 1, 2, 3, 4,…,25, is prime number, when each of the given numbers is equally likely to be selected? 12 11 10 9 (b) (c) (d) (a) 25 25 25 25

∴Probability of an event Favourable number of outcomes = Total number of outcomes 9 Hence, required probability = 25 X

Example 8. In a lottery, a person chooses six different natural numbers at random from 1 to 20 and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize, then the probability of winning the prize is 6 1 6 1 (b) (c) (d) (a) 38760 20 20 38760

9! and n(E ) = 3! 4! 3! 2 ! (3!) (4!) (3!) (2 !) 1 = P(E ) = 9! 210

Example 10. In a convex hexagon, two diagonals are drawn at random. The probability that the diagonals intersect at an interior point of the hexagon, is 5 7 (a) (b) 12 12 2 (d) None of these (c) 5 Sol. (a) n(S ) = Total number of selections of two diagonals 9×8 = 36 2 n(E ) = Number of selections of two diagonals which intersect at an interior point = Total number of selections of four vertices = 6C 4 = 15 15 5 ∴ The required probability = = 36 12 = 9C 2 =

Sol. (d) Let S be the sample space associated with the given experiment and A be the event ‘selecting a prime number’. Then, S = {1, 2, 3, 4, 5,…,25} A = {2, 3, 5, 7, 11, 13, 17, 19, 23} Total number of outcomes = 25 ∴ Favourable number of outcomes = 9

Example 9. 4 five rupee coins, 3 two rupee coins and 2 one rupee coins are stacked together in a column at random. The probability that the coins of the same denomination are consecutive, is 13 1 (b) (a) 9! 210 1 (c) (d) None of these 35

21

Sol. (b) n(S ) =

(T,4)

Now, let F be an event of getting atleast one tail. Then, F = {(H, T ), (T, 1), (T, 2 ), (T, 3), (T, 4), (T, 5), (T, 6)} ∴P(F )|P(H, T ) + P(T, 1)| P(T, 2 ) + P(T, 3) + P(T, 4) + P(T, 5) + P(T, 6) 1 1 1 1 1 1 1 = + + + + + + 4 12 12 12 12 12 12 1 1 3 = + = 4 2 4

C 6 = 38760

20

It is given that a person wins the prize, if six selected numbers match with the six numbers already fixed ∴ Favourable number of outcomes = 1 1 1 Hence, required probability = 20 = C 6 38760

Head (H ) 1/2

C 6 ways.

Fundamentals of Probability

The sample space of the experiment may be described as S = {(H, H), (H, T ), (T, 1), (T, 2 ), (T, 3), (T, 4), (T, 5), (T, 6)} where, (H, H) denotes that both the tosses result into head and (T, i ) denotes the first toss result into a tail and the number i appeared on the die for i = 1, 2, 3, 4, 5, 6. Thus, the probabilities assigned to the 8 elementary events (H, H), (H, H), (T, 2 ), (T, 3), (T, 4),(T, 5), (T, 6) are 1 1 1 1 1 1 1 1 , , , , , , , , respectively which is clear 4 4 12 12 12 12 12 12 from the

X

Example 11. Three numbers are chosen at random from numbers 1 to 30. The probability that the minimum of the chosen numbers is 9 and maximum is 25, is 1 1 (b) (a) 406 812 3 (c) (d) None of these 812 Sol. (c) Out of first 30 natural numbers, three natural numbers can be chosen in 30 C 3 ways. If the minimum and maximum of the numbers chosen are 9 and 25 respectively, then we have to a select a number from the remaining 15 numbers i.e. 10, 11, …, 24. ∴Favourable number of elementary events = 15C1 Hence, required probability =

15

C1

30

C3

=

3 812

1099

X

Objective Mathematics Vol. 1

21

Example 12. Three different numbers are selected at random from the set A = {1, 2, … , 10}. The probability that the product of two of the numbers is equal to third, is 3 1 (a) (b) 4 40 1 39 (c) (d) 8 40

X

Sol. (b) Since, a person can alight at anyone of n floors. Therefore, the number of ways in which m passengers can alight at n floors is n × n × n × … × n = nm . 14442444 3

Sol. (b) Out of 10 numbers, three numbers can be chosen in 10 C 3 ways.

m times

∴Total number of elementary events =

10

The number of ways in which all passengers can alight at different floors is n C m × m! = n Pm . n P Hence, required probability = mm n

C3

The product of two numbers, out of the three chosen numbers, will be equal to the third number, if the numbers are chosen in one of the following ways: (2, 3, 6), (2, 4, 8), (2, 5, 10) ∴Favourable number of elementary events = 3 3 1 Hence, required probability = 10 = C 3 40 X

X

Example 13. The probability that out of 10 persons, all born in June, atleast two have the same birthday, is 30 30 C10 C10 (b) (a) 10 30! (30) (c)

30

10

−

30

C10

(30)10

choose a number from 1 to 25 is 25, so the total number of ways of choosing number is 25 × 25 = 625. There are 25 ways in which the numbers chosen by both players is the same.Therefore, the probability they will win a prize in a 25 1 single trial, is = . 625 25

(d) None of these

days of June month as his (her) birthday. Therefore, Number of ways in which 10 persons can have birthdays in the month of June = 30 × 30 × 30 × … × 30 (10 times) = 3010 ∴Required probability = 1 − Probability that no two persons have the same birthday 30 C10 3010 − 30C10 = 1− = 3010 3010

1100

Example 14. The probability that when 12 balls are distributed among three boxes, the first will contain three balls, is 12 C3 × 29 29 (b) (a) 12 312 3 12 12 C 3 × 212 C3 (d) (c) 12 3 12 3 Sol. (b) Since, each ball can be put into anyone of the three boxes. So, the total number of ways in which 12 balls can be put into three boxes is 312 . Out of 12 balls, 3 balls can be chosen in 12 C 3 ways. Now, remaining 9 balls can be put in the remaining 2 boxes in 2 9 ways. So, the total number of ways in which 3 balls are put in the first box and the remaining in other two boxes is 12 C 3 × 2 9 . 12 C3 × 2 9 Hence, required probability = 312

Example 16. A and B play a game, where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial, is 1 24 (b) (a) 25 25 2 (c) (d) None of these 25 Sol. (b) The number of ways in which either player can

Sol. (c) Since, each person can have anyone of the thirty

X

Example 15. An elevator starts with m passengers and stops at n floor ( m ≤ n). The probability that no two passengers alight at the same floor, is n n n n P P C C (a) nm (b) mm (c) nm (d) mm m n m n

Hence, the probability that they will not win a prize in a 1 24 single trial = 1 − = . 25 25 X

Example 17. Two integers x and y are chosen with replacement out of the set {0, 1, 2, 3, … , 10}. Then, the probability that | x − y | > 5, is 81 30 25 20 (a) (b) (c) (d) 121 121 121 121 Sol. (b) Since, x and y each can take values from 0 to 10. So, the total number of ways of selecting x and y is 11 × 11 = 121. Now, | x − y|> 5 ⇒ x − y < −5 or x − y > 5 There are 30 pairs of values of x and y satisfying these two inequalities. So, favourable number of ways = 30 30 Hence, required probability = 121

X

Example 18. Three persons A, B and C are to speak at a function along with five others. If they all speak in random order, then the probability that A speaks before B and B speaks before C, is 3 1 (a) (b) 8 6 3 (c) (d) None of these 5

speak is P8 = 8!. The number of ways in which A, B and C 8

= Coefficient of x12 in

8

(1 − 3 x6 + 3 x12 − x18 )( 2 C 0 + 3C1 x + 4C 2 x2 + … )

can be arranged in the specified speaking order is C 3 . There are 5! ways in which the other five can speak. So, favourable number of ways is 8 C 3 × 5!. Hence, required probability = X

8

=

X

X

24

C14

19

(c)

C2 × 50

30

C2

(d) None of these

C5

Sol. (b) Here, n(S ) = 50C 5 , n(E ) = 30C 2 × 19C 2 P(E ) =

∴

X

30

C2 × 50

19

C2

Sol. (d) n(S ) = 6 × 6 × 6 n(E ) = The number of solutions of x + y + z = 15 where, 1 ≤ x ≤ 6, 1 ≤ y ≤ 6, 1 ≤ z ≤ 6 = Coefficient of x15 in ( x + x2 + … + x6 )3 = Coefficient of x

4

= Coefficient of x8 in (1 − x6 )4 ⋅ ( 3 C 0 + 4C1 x + 5C 2 x2 + … ) = 11C 8 − 4 ⋅ 5C 2 = 125 ∴

X

P(E ) =

125 6×6×6×6

Example 23. Three six faced dice are thrown together. The probability that the sum of the numbers appearing on the dice is k (9 ≤ k ≤ 14), is 21k − k 2 − 83 (a) 216 k 2 − 3k + 2 (b) 432 21k − k 2 − 83 (c) 432 (d) None of the above Sol. (a) Favourable number of elementary events = Coefficient of xk − 3 in (1 − x6 )3 (1 − x)−3

C5

Example 21. Three dice are thrown simultaneously. The probability of getting a sum of 15 is 1 5 (b) (a) 72 36 5 (c) (d) None of these 72

12

(d) None of these

 1 − x6   = Coefficient of x8 in   1− x 

15 = 92

Example 20. x1 , x 2 , x 3 ,…, x 50 are fifty real numbers such that x r < x r + 1 for r = 1, 2, 3, … , 49. Five numbers out of these are picked up at random. The probability that the five numbers have x 20 as the middle number, is 20 30 C 2 × 30C 2 C 2 × 19C 2 (a) (b) 50 50 C5 C5

2

n(E ) = The number of integral solutions of x1 + x2 + x3 + x4 = 12, where 1 ≤ x1 ≤ 6,…, 1 ≤ x4 ≤ 6 = Coefficient of x12 in ( x + x2 + … + x6 )4

Excluding the neighbouring places there are 22 places in which 14 cars can be parked in 22 C14 ways. Hence, required probability =

1  5   36  6 

Sol. (a) n(S ) = 6 × 6 × 6 × 6

includes the owner’s car also and hence 14 other cars are parked. There are 24 places (excluding places at the two end) out of which 14 places can be chosen in 24 C14 ways.

C14

P(E ) =

Example 22. The probability of getting a sum of 12 in four throws of an ordinary dice is 3 4 1  5  5 (b)   (a)    6 6  6 (c)

Sol. (c) It is given that 15 places are occupied. This

22

10 6×6×6 5 = 108

∴

C 3 × 5! 1 = 8! 6

Example 19. A car is parked by an owner amongst 25 cars in a row, not at either end. On his return he finds that exactly 15 places are still occupied. The probability that both the neighbouring places are empty, is 91 15 (a) (b) 276 184 15 (d) None of these (c) 92

C12 − 3 × 8C 6 + 3 × 2C 0 = 10

14

21 Fundamentals of Probability

= Coefficient of x12 in (1 − 3 ⋅ x6 + 3 ⋅ x12 − x18 ) ⋅ (1 − x)−3

Sol. (b) The total number of ways in which 8 persons can

in (1 + x + … + x )

5 3

= Coefficient of xk − 3 in (1 − 3C1 x6 + … )(1 − x)−3 [Q9 ≤ k ≤ 14 ∴6 ≤ k − 3 ≤ 11] = Coefficient of xk − 3 in (1 − x)−3 − 3C1 [Coefficient of xk − 9 in (1 − x)−3 ] =

k − 3 + 3 −1

=

k −1

C 3 − 1 − 3C1 ×

k − 9 + 3 −1

C3 − 1

k −7

C2 − 3 ⋅ C2 (k − 1)(k − 2 ) (k − 7 )(k − 8) = − 3⋅ = 21k − k 2 − 83 2 2 The total number of elementary events associated to the random experiment of throwing three dice is 6 × 6 × 6 = 63 . Hence, the probability of the required event 21k − k 2 − 83 = 63

1101

Objective Mathematics Vol. 1

21 Geometrical Probability

Sol. (a) Let AB be a straight line segment of length a and let P and Q be two points on it such that AP = x and AQ = y.

We observe that the definition of the probability of occurrence of an event fails, if the total number outcomes (elementary events) of a trial in a random experiment is infinite. For example, if it is asked to find the probability that a point selected at random in a given region will lie in a specified part of that region, then the definition of the probability, which is discussed earlier is unable to answer this. In such cases, the definition of probability is modified and extended to what is called probability or probability in continuum. In such cases, the general expression for the probability p of occurrence of an event is given by Measure of the specified part of the region p= Measure of the whole region

Then, we have to find the probability for|PQ|> c i.e. | x − y| > c. Clearly, 0 ≤ x ≤ a and 0 ≤ y ≤ a We observe that there are two variables x and y satisfying 0 ≤ x ≤ a and 0 ≤ y ≤ a. In the cartesian plane the inequalities 0 ≤ x ≤ a and 0 ≤ y ≤ a represent the region enclosed by the square having x = 0, y = 0, x = a and y = a as its sides. Clearly, area enclosed by this square is a2 . Now,| x − y|> c ⇒ x − y > c and −( x − y) > c ⇒ x − y > c and y − x > c In order to find P (| x − y|> c ), we require the area of the region enclosed by x − y > c, y − x > c, 0 ≤ x ≤ a and 0 ≤ y ≤ a. Clearly, shaded region in figure represents the area enclosed. P

where, ‘measure’ means length, area or volume of the region, if we are dealing with one, two or three dimensional space respectively. X

Example 24. If a ∈[−20, 0] then the probability that the equation 16x 2 + 8( a + 5) x − 7a −5 = 0 has imaginary roots, is 17 13 7 3 (a) (b) (c) (d) 20 20 20 20

a

We have, Area of the shaded region = Area APQ + Area BRS 1 1 = (a − c )2 + (a − c )2 = (a − c )2 2 2 Y

a–c

B

imaginary roots, then Discriminant < 0 ⇒ 64(a + 5)2 + 64(7 a + 5) < 0

a –c

a + 17 a + 30 < 0 ⇒ (a + 15)(a + 2 ) < 0

x–y=c x=a a –c

c

−2

Hence, the required probability =

c Q

R

−15 < a < −2

S

x=0

2

⇒

y–x=c c C

y=a

Sol. (b) If the equation 16 x + 8(a + 5)x − 7 a − 5 = 0 has

X

B y

2

⇒

Q

x

A

∫−15 dx = 13 0 20 ∫−20 dx

Example 25. Two points are taken at random on the given straight line segment of length a. The probability for the distance between them to exceed a given length c, where 0 < c < a, is 2 2 c2 a2 c   a (d)1 − 2 (a) 1 −  (b) 1 −  (c)1 − 2  a  c a c

O

c

Py=0A a–c

X

Hence, the required probability Area enclosed by | x − y|> c, 0 ≤ x ≤ a and 0 ≤ y ≤ a = Area enclosed by 0 ≤ x ≤ a, 0 ≤ y ≤ a =

Area of the shaded region (a − c )2  c = =  1 −   Area of the square a a2

2

Work Book Exercise 21.2 1 In a single cast with two dice, the odds against drawing 7 is a 1:6

b 1 : 12

c 5:1

d 1:3

2 If the letters of the word ATTEMPT are written down at random, then the chance that all T’s are consecutive, is 1 42 1 c 7

a

1102

b

6 7

d None of these

3 Let x = 33n . The index n is given a positive integral value at random. The probability that the value of x will have 3 at the unit’s place, is 1 4 1 b 2 1 c 3 d None of the above a

7 Three six-faced dice are thrown together. The

6 persons. The probability that atleast one of them will receive none, is 14

6 143 137 c 143 a

b

probability that the sum of the numbers appearing on the dice is k(3 ≤ k ≤ 8), is

C4

k2 432 ( k − 1)( k − 2 ) c 432

C5

d None of these

5 A point is selected at random from the interior of

c

3 4 1 4

b

6 is thrown n times and the list of n numbers showing up is noted. Then, the probability that among the numbers 1, 2, 3, 4, 5, 6 only three numbers appear in this list, is

1 2

d None of these

6 Three dice are thrown. The probability of getting 2 5 1 c 4

b

b

9 20

●

c

●

●

or

ii.

Ø

●

●

− ∑ P ( Ai ∩ A j ) + i 8 7 11 (c) ≤ P ( A ) + P ( B ) ≤ 8 8 (d) None of the above (a) P ( A ) + P ( B )
0, then (a) occurrence of E ⇒ occurrence of F (b) occurrence of F ⇒ occurrence of E (c) non-occurrence of E ⇒ non-occurrence of F (d) None of the above implications holds

Sol. Consider the following events in a single throw of a die.

(d) None of these 6

C Sol. The probability of 4 being the minimum number = 10 2 C3 [because after selecting 4, any two can be selected from 5, 6, 7, 8, 9, 10] 6 C The probability of 8 being maximum number = 10 2 C3

E = Getting a multiple 3 F = Getting an even number 1 1 We have, P (E ) = and P (F ) = 3 2

Clearly,

P (E ) ≤ P (F )

and

P (E ∩ F ) =

But none of the options (a), (b) and (c) are true. Hence, (d) is the correct answer.

1 > 0. 6

1109

Objective Mathematics Vol. 1

21

Ex 21. There are four machines and it is known that exactly two of them are faulty. They are tested, one-by-one, in a random order till both the faulty machines are identified. Then, the probability that only two tests are needed, is (a)

1 3

(b)

1 6

(c)

1 2

(d)

1 4

Sol. The total number of ways in which two machines can be chosen out of four machines is 4C 2 = 6. If only two tests are required to identify faulty machines, then in first two tests faulty machines are identified. This can be done in one way only. ∴Favourable number of ways = 1 1 Now, the required probability = 6 Hence, (b) is the correct answer.

Ex 22. The numbers 1, 2, 3, ..., n are arranged in a random order. The probability that the digits 1, 2, 3, ...., k ( k < n) appears as neighbours in that order, is 1 (a) n! ( n − k )! (c) n!

Ex 24. If four dice are thrown together, then the probability that the sum of the numbers appearing on them is 13, is (a)

(b)

11 216

 1 − x6  = Coefficient of x 9 in    1− x

( n − k )! (a) n! n− k +1 (b) n Ck n−k (c) n Ck k! (d) n!

= Coefficient of x 9 in (1 − 4 x 6 + ... ) (1 − x )− 4 = Coefficient of x 9 in (1 − x )− 4 − 4 × Coefficient of x 3 in (1 − x )− 4 =

9 + 4− 1

C4 − 1 − 4 ×

3+ 4− 1

C4 − 1

[Qcoefficient of x in (1 − x )− r = n

n + r− 1

C r − 1]

12 × 11 × 10 6×5×4 −4⋅ 3× 2×1 3× 2×1

140 35 = 4 324 6 Hence, (c) is the correct answer.

So, the required probability =

Ex 25. Six faces of a die are marked with the numbers 1, −1, 0, − 2, 2 and 3. The die is thrown thrice. The probability that the sum of the numbers thrown is six, is (a)

1 72

(b)

1 12

(c)

5 108

(d)

1 36

Sol. Total number of elementary events = 63 Favourable number of elementary events = Coefficient of x6 in (x + x − 1 + x 0+ x − 2 + x 2 + x 3 )3  1 + x + x2 + x3 + x4 + x5 = Coefficient of x6 in   x2  

3

= Coefficient of x 12 in (1 + x + x 2 + x 3 + x 4 + x 5 )3  1 − x6  = Coefficient of x 12 in    1− x 

3

= Coefficient of x 12 in (1 − x6 )3 (1 − x )− 3 = Coefficient of x 12 in (1 − 3C 1x6 + 3C 2x 12 ...) (1 − x )− 3 = Coefficient of x 12 in (1 − x )− 3 − 3C 1 ⋅ Coefficient of

(x6 in 1 − x )− 3 + 3C 2 ⋅ Coefficient of x 0 in (1 − x )− 3

Sol. The numbers 1, 2, 3, ..., n can be arranged in a row in n! ways. The total number of ways in which the digits 1, 2, 3, ..., k (k < n) occur together is k !(n − k + 1)!. So, the required probability k ! (n − k + 1)! n − k + 1 = = n n! Ck Hence, (b) is the correct answer.

4

= Coefficient of x 9 in (1 − x6 )4 (1 − x )− 4

= 220 − 80 = 140

Ex 23. The numbers 1, 2, 3, ..., n are arranged in a random order. The probability that the digits 1, 2, 3, ..., k ( n > k ) appear as neighbours, is

11 432

= Coefficient of x 9 in (1 + x + x 2 + ... + x 5 )4

(d) None of these

So, the total number of ways in the digits 1, 2, 3, ..., k appear as neighbours in the same order is (n − k + 1)!. (n − k + 1)! So, the required probability = n! Hence, (d) is the correct answer.

(d)

random experiment of throwing 4 dice is 6 × 6 × 6 × 6 = 64 Favourable number of elementary events = Coefficient of x 13 in (x + x 2 + x 3 + ... + x6 )4

= 12C 3 − 4 × 6C 3 =

n !. Considering digits 1, 2, 3, 4, ..., k as one digit, we have (n − k + 1) digits which can be arranged in (n − k + 1)! ways.

35 324

(c)

Sol. Total number of elementary events associated to the

k! (b) n!

Sol. The number of ways of arranging n numbers in a row is

1110

5 216

=

12 + 3 − 1

C 3 − 1 − 3C 1 ⋅6 + 3 − 1 C 3 − 1 + 3C 2

= 14C 2 − 3C 1 ⋅

6 + 3− 1

C 3 − 1 + 3C 2

= 14C 2 − 3C 1 × 8C 2 + 3C 2 = 91 − 84 + 3 = 10 10 5 So, the required probability = 3 = 108 6 Hence, (c) is the correct answer.

7 16 9 (c) 16

101 201 10 (d) 16

(a)

(b)

Ex 28. Let X be a set containing n elements. If two subsets A and B of X are picked at random, then the probability that A and B have the same number of elements, is 2n

(a) 2

2n n !

product = xy = 10000,

where

3 × 10000 4 xy ≥ 7500 x (200 − x ) ≥ 7500 x 2 − 200x + 7500 ≤ 0 (x − 50) (x − 150) ≤ 0 50 ≤ x ≤ 150 xy ≥

Now, ⇒ ⇒ ⇒ ⇒ ⇒

So, the favourable number of ways = 101 Total number of ways = 201 101 Now, required probability = 201 Hence, (b) is the correct answer.

Ex 27. Let A be a set containing n elements. A subset P of the set A is chosen at random. The set A is reconstructed by replacing the elements of P and another subset Q of A is chosen at random. The probability that P ∩ Q contains exactly m ( m < n) elements, is (a)

3n − m n

(c)

4n C m ⋅ 3n − m 4n

n

(b)

C m ⋅ 3m 4n

(d) None of these

Sol. We know that, the number of subsets of a set containing the number of ways of n elements is 2n. Therefore, n n choosing P and Q is 2 C 1 × 2 C 1 = 2n × 2n = 4 n . Out of n elements, m elements can be chosen in nC m ways. If P ∩ Q contains exactly m elements, then from the remaining n − m elements either an element belongs to P or Q but not both P and Q. Suppose P contains r elements from the remaining (n − m) elements. Then, Q may contain any number of elements from the remaining (n − m) − r elements. Therefore, P and Q can be chosen in n − mC r 2(n − m) − r . But r can vary from 0 to (n − m). So, P and Q can be chosen in general in  n − m n m n m r − − − n n m n − ( )  C m = (1 + 2) ∑ Cr 2 Cm    r=0 = nC m ⋅ 3n − m Cm ⋅ 3 4n Hence, (c) is the correct answer. So, the required probability =

n

n− m

(d)

1 2n

Cn

3n 4n

Sol. Required probability n

non-negative quantities is fixed, the product will be maximum when they are equal. So, the greatest x = y = 100

(b)

1⋅ 3⋅ 5 ... ( 2n + 1)

(c)

Sol. Let x and y be the two quantities. When the sum of two

Cn 2n

21 Fundamentals of Probability

Ex 26. Given that the sum of two non-negative quantities is 200, the probability that their 3 product is not less than times their greatest 4 product value, is

=

∑ nC r × nC r

r=0

2 ×2 n

n

=

C 02 + C 12 + C 22 + ... + C n2 4n

{1 ⋅ 3 ⋅ 5 ... (2n − 1)} 2nC n = 2n 2n ⋅ n! 2 Hence, (a) is the correct answer. =

Ex 29. A and B are two events. Odds against A are 2 :1 and odds in favour of A ∪ B are 3 :1. If x ≤ P ( B ) ≤ y , then the ordered pair ( x, y) is  5 3 (a)  ,   12 4   1 3 (c)  ,   3 4

 2 3 (b)  ,   3 4

(d) None of these

1 3 and P ( A ∪ B ) = 3 4 ∴ P ( A ∪ B) = P ( A) + P ( B) − P ( A ∩ B) 3 1 ⇒ = + P ( B) − P ( A ∩ B) 4 3 5 ⇒ = P ( B) − P ( A ∩ B) 12 5 5 …(i) ⇒ P ( B) = + P ( A ∩ B) ⇒ P ( B) ≥ 12 12 5 Again, P ( B) = + P ( A ∩ B) 12 5 ⇒ P ( B) ≤ + P ( A ) [QP ( A ∩ B ) ≤ P ( A )] 12 5 1 3 …(ii) ⇒ P( B) ≤ + = 2 3 4 5 3 From Eqs. (i) and (ii), we get ≤ P ( B) ≤ 12 4 5 3 So, and y = x= 12 4 Hence, (a) is the correct answer.

Sol. We have, P ( A ) =

Ex 30. Two numbers b and c are chosen at random with replacement from the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9. The probability that x 2 + bx + c > 0, ∀ x ∈ R is (a)

10 27

(b)

31 81

(c)

32 81

(d)

11 27

Sol. Since, b and c both can assume values from 1 to 9. So, total number of ways of choosing b and c is 9 × 9 = 81 Now, x 2 + bx + c > 0, ∀ x ∈ R ⇒ D < 0 i.e.

b2 − 4 c < 0

1111

The following table shows the possible values of b and c for which b2 − 4 c < 0

Objective Mathematics Vol. 1

21

c

b

Total

1

1

1

2

1, 2

2

3

1, 2, 3

3

4

1, 2, 3

3

c

b

Total

5

1, 2, 3, 4

4

6

1, 2, 3, 4

4

7

1, 2, 3, 4, 5

5

8

1, 2, 3, 4, 5

5

9

1, 2, 3, 4, 5

5

∴ Favourable number of ways = 32 32 So, the required probability is = . 81 Hence, (c) is the correct answer.

Type 2. More than One Correct Options Ex 31. If A and B are two events such that P ( A ) = 3 / 4 and P ( B ) = 5 / 8, then (a) P ( A ∪ B ) ≥ 3 / 4 (b) P ( A ′ ∩ B ) ≤ 1/ 4 (c) 3 / 8 ≤ P ( A ∩ B ) ≤ 5 / 8 (d) None of the above

Sol. Q AC ⊆ A ∪ B ⇒

P( A) ≤ P( A ∪ B) 3 ⇒ P( A ∪ B) ≥ 4 Also, P ( A ∩ B ) = P ( A ) + P ( B ) − P ( A ∪ B ) ≥ P( A) + P( B) − 1 3 5 3 = + − 1= 4 8 8 Now, A ∩ B ⊆ B 5 ⇒ P( A ∩ B) ≤ P( B) = 8 3 5 …(i) ≤ P( A ∩ B) ≤ ∴ 8 8 Now, consider P ( A′ ∩ B ) = P ( B ) − P ( A ∩ B ) …(ii) From Eq. (i), we have −5 −3 ≤ −P ( A ∩ B ) ≤ 8 8 5 3 ⇒ P( B) − ≤ P( B) − P( A ∩ B) ≤ P( B) 8 8 1 [from Eq. (ii)] 0 ≤ P ( A′ ∩ B ) ≤ ⇒ 4 Hence, (a), (b) and (c) are correct answers.

Ex 32. The probability that a 50 yr old man will be alive at 60, is 0.83 and the probability that a 45 yr old woman will be alive at 55, is 0.87. Then,

(a) the probability that both will be alive is 0.7221 (b) atleast one of them will alive is 0.9779 (c) atleast one of them will alive is 0.8230 (d) the probability that both will be alive is 0.6320 Sol. Required probability = Probability that both will be alive for 10 yr = 0.83 × 0.87 = 0.7221

The probability that atleast one of them will be alive 1 − P (none of them remains alive 10 yr) ∴ = 1 − (1 − 0.83)(1 − 0.87) = 1 − (017 . × 013 . ) = 0.9779 Hence, (a) and (b) are correct answers.

Ex 33. The chance of an event happening is the square of the chance of a second event but the odds against the first are the cube of the odds against the second. The chances of the events are (a) P1 = 1/ 9 (c) P2 = 1/ 3

(b) P1 = 1/ 16 (d) P2 = 1/ 4

Sol. Let P1 and P2 be the chances of happening of the first and second events, respectively. Then, according to the given conditions 3 1 − P1  1 − P2  = P1 = P22 and   P2  P1 ∴

1 − P22  1 − P2  =   P2  P22

3

⇒

P2 (1 + P2 ) = (1 − P2 )2 1 ⇒ 3P2 = 1 ⇒ P2 = 3 1 P1 = 9 Hence, (a) and (c) are correct answers.

Type 3. Assertion and Reason

1112

Directions (Ex. Nos. 34-37) In the following examples, each example contains Statement I (Assertion) and Statement II (Reason). Each example has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are. (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Sol. Here, n(S ) = 1, length of the interval [ 0, 5 ] = 5 and n(E ) = length of the interval ≤ [ 0, 5 ] in which P belongs such that the given equation has real roots. 1 Now, x 2 + Px + (P + 2) = 0 will have real roots, if 4 1 P 2 − 4 ⋅ 1⋅ (P + 2) ≥ 0 ⇒ P 2 − P − 2 ≥ 0 4 ⇒ (P + 1) (P − 2) ≥ 0 ⇒ P ≤ − 1or P ≥ 2 But P ∈[ 0, 5 ] So, E = [ 2, 5 ] ∴ n (E ) = Length of the interval [ 2, 5 ] = 3 3 ∴ Now, required probability = 5 Hence, (a) is the correct answer.

1 + 4P 1 − P 1 − 2P Ex 35. Statement I If are , and 4 4 4 probabilities of three pairwise mutually exclusive events, then the possible values of P  1 1 belong to the set  − , .  4 2 Statement II If three events are pairwise mutually exclusive and exhaustive, then sum of their probabilities is equal to 1. Sol. We have,

1 + 4P 1− P 0≤ ≤ 1, 0 ≤ ≤1 4 4 1 − 2P and 0≤ ≤1 4 1 3 3 1 ⇒ − ≤ P ≤ , − 3 ≤ P ≤ 1, − ≤ P ≤ 4 4 2 2 Again, the events are pairwise mutually exclusive, so 1 + 4 P 1 − P 1 − 2P 0≤ + + ≤1 4 4 4 ⇒ − 3≤ P ≤1 Taking intersection of all four intervals of P, we get 1 1 − ≤P≤ 4 2 Hence, (b) is the correct answer.

Ex 36. Statement I Two non-negative integers are chosen at random. The probability that the sum of their squares is divisible by 5 is 9/25. Statement II If at the unit place in any number is zero, that number is only divisible by 5. Sol. Let the two non-negative integers be x and y. Then, x = 5a + α and y = 5b + β, where 0≤β ≤ 4 Now, x 2 + y2 = (5a + α)2 + (5b + β)2

0 ≤ α ≤ 4,

= 25 (a2 + b2 ) + 10 (aα + bβ ) + α 2 + β 2

21 Fundamentals of Probability

Ex 34. Statement I If P is chosen at random in the closed interval [0, 5], then the probability that 1 the equation x 2 + Px + ( P + 2) = 0 has real 4 3 root is . 5 Statement II If discriminant ≥ 0, then roots of the quadratic equation are always real.

∴ x 2 + y2 is divisible by 5 if and only if 5 divides α 2 + β 2. The total number of ways of choosing α and β = 5 × 5 = 25 Further, α 2 + β 2 will be divisible by 5, if (α, β) ∈ {(0, 0), (1, 2), (1, 3), (2, 1), (2, 4 ), (3, 1), ` (3, 4 ), (4 , 2), (4 , 3)} ∴Favourable number of ways of choosing α and β = 9 9 Now, required probability = 25 Hence, (c) is the correct answer.

Ex 37. Statement I Out of 5 tickets consecutively numbered, three are drawn at random, the chance that the numbers on them are in an AP 2 is . 15 Statement II Out of (2n + 1) tickets consecutively numbered, three are drawn at random, the chance that the numbers on them 3n are in an AP is 2 . 4n − 1 Sol. 2n + 1 = 5 ⇒

2n = 4

3n 4 n2 − 1 6 2 = = 15 5 For a, b and c in an AP, a + c = 2b ⇒ a + c is even, so a and c are both even or odd. Now, a and c can be chosen in nC 2 + n + 1C 2 = n2 ways ⇒

n = 2; P (E ) =

n2 × 3 × 2 × 1 C 3 (2n + 1) 2n (2n − 1) 3n = 2 4n − 1 Hence, (d) is the correct answer.

∴ P (E ) =

n2

(2n + 1)

=

1113

Objective Mathematics Vol. 1

21 Type 4. Linked Comprehension Based Questions Passage (Ex. Nos. 38-40) In an objective paper, there are two sections of 10 questions each. For ‘section 1’, each question has five options and only one option is correct and ‘section 2’ has four options with multiple answers and marks for a question in this section is awarded only if he ticks all correct answers. Marks for each question in ‘section 1’ is 1 and in ‘section 2’ is 3. (There is no negative marking).

Ex 38. If a candidate attempts only two questions by guessing, one from ‘section 1’ and one from ‘section 2’, then the probability that he scores in both questions is (a) 74 / 75

(b) 1/ 25

(c) 1/ 15

(d) 1/ 75

Sol. Let P1 be the probability of being an answer correct from section 1. Then, P1 = 1 / 5. Let P2 be the probability of being an answer correct from section 2. Then, P2 = 1 / 15. So, the required probability is 1 / 5 × 1 / 15 = 1 / 75. Hence, (d) is the correct answer.

Ex 39. If a candidate in total attempts four questions all by guessing, then the probability of scoring 10 marks is

(a) 1/ 15 (1/ 15) 3 (c) 1/ 15 (14 / 15)

(b) 4 / 5 (1/ 15) 3 3

(d) None of these

Sol. Scoring 10 marks from four questions can be done in 3 + 3 + 3 + 1 = 10 ways, so as to answer three questions from section 2 and 1 question from section 1 correctly. So, the required probability 3 3 10 C 10C 1 1  1  4  1 = 203   =   5  15 5  15 C4 Hence, (b) is the correct answer.

Ex 40. The probability of getting a score less than 40 by answering all the questions by guessing in this paper is (a) (1/ 75)10

(b) 1− (1/ 75)10

(c) ( 74 / 75)10

(d) None of these

Sol. To get 40 marks, he has to answer all questions correctly and its probability is (1 / 5)10 (1 / 15)10. So, the probability of getting a score less than 40 10 10 10  1  1   1 =1−     =1−    5  15  75 Hence, (b) is the correct answer.

Type 5. Match the Columns Ex 41. ‘Three coins are tossed’. Then, match the terms of Column I with the terms of Column II and choose the correct option from the codes given below. Column I

Column II

A.

Two events which are mutually exclusive.

p. A, B and C, where A is the event ‘three heads show’. B is the event ‘two heads show’ and C is the event ‘three tails show’.

B.

Two events which are mutually exclusive and exhaustive.

q. A and B, where A is the event ‘three heads show’ and B is event ‘atleast two heads show’.

C.

Two events which are not mutually exclusive.

r. A and B, where A is the event atleast one head and B is the event getting three tails.

D.

Two events which are mutually exclusive but not exhaustive.

s. A and B, where A is the event ‘three heads show’ and B is the event ‘three tails show’.

E.

Three events which are mutually exclusive but not exhaustive.

Sol. If three coins are tossed, then possible number of

1114

outcomes = 23 = 8 Sample space, S = {HHH , HHT , HTH , HTT , THH , THT , TTH , TTT } A. Let A be the event ‘three heads show’. ⇒ A = {HHH }

and B be the event ‘three tails show’. ⇒ B = {TTT } ⇒ A ∩ B = φ Hence, A and B are mutually exclusive events. B. Let A be the event ‘atleast one head’. ⇒ A = {HHH , HHT , HTH , HTT ,THH ,THT , TTH } and B be the event getting three tails. ⇒ B = {TTT } ⇒ A ∩ B = φ and A ∪ B = S [two events are exhaustive, if A ∪ B = S ] Hence, A and B are mutually exclusive and exhaustive. C. Let A be the event ‘three heads show’ and B be the event ‘atleast two heads show’. ⇒ A = {HHH } and B = {HHT , HTH , THH , HHH } ⇒ A ∩ B = {HHH } ≠ φ Hence, A and B are not mutually exclusive. D. Same as in C , A andB are not mutually exclusive. Also, A ∪ B ≠ S . E. Let A be the event ‘three heads show’, B be the event ‘two heads show’ and C be the event ‘three tails show’. ⇒ A = {HHH } ⇒ B = {HHT , HTH , THH } and C = {TTT } ⇒ A ∩ B ∩ C = φ and A ∪ B ∪ C ≠ S ⇒ A , B and C are mutually exclusive but not exhaustive. A → s; B → r : C → q; D → q; E → p

Column I

Column I

Column II

A.

If E1 and E 2 are two mutually exclusive events, then

p. E1 ∩ E 2 = E1

B.

If E1 and E 2 are the mutually exclusive and exhaustive events, then

q. ( E1 − E 2 ) ∪ ( E1 ∩ E 2 ) = E1

C.

If E1 and E 2 have common outcomes, then

r. E1 ∩ E 2 = φ, E1 ∪ E 2 = S

D.

If E1 and E 2 are two events such that E1 ⊂ E 2, then

s. E1 ∩ E 2 = φ

A.

Column II

Mutually exclusive events

p. A and B, A and C, B and C, C and D, A, B and C

B.

Simple events

q. B, D

C.

Compound events

r. A and C

Sol. When three coins are tossed, then there are 23 = 8

21 Fundamentals of Probability

match the terms of Column I with the terms of Column II and choose the correct option from the codes given below:

Ex 42. Match the terms of Column I with the terms of Column II and choose the correct option from the codes given below.

possible outcomes. i.e. S = {HHH , HHT , HTH , HTT , THH , THT , TTH , TTT } A = Three heads show = {HHH } B = Two heads and one tail show = {HHT , HTH , THH } C = Three tails show = {TTT } D = A head shows on the first toss = {HHH , HHT , HTH , HTT } A. Here, A∩ B=φ A ∩C =φ B ∩C =φ C ∩D=φ A ∩ B ∩C = φ Here, A and B, A and C, B and C, C and D and A, B and C are mutually exclusive. B. Since, events A, C have only one sample point. Here, A and C are simple events. C. Since, events B , D have two or more sample points. Here, B and D are compound events. A → p; B → r : C → q

Sol. A. If E1 and E2 are two mutually exclusive events, then E1 ∩ E2 = φ

B. If E1 and E2 are the mutually exclusive and exhaustive events, then E1 ∩ E2 = φ and E1 ∪ E2 = S where, S is the sample space for the events E1 and E2 C. If E1 and E2 have common outcomes, then (E1 − E2 ) ∪ (E1 ∩ E2 ) = E1 D. If E1 and E2 are two events such that E1 ⊂ E2 and E1 ∩ E2 = E1 A → s; B → r; C → q; D → p

Ex 43. Three coins are tossed once. Let A denotes the event ‘three heads show’, B denotes the event ‘two heads and one tail show’, C denotes the event ‘three tails show’ and D denotes the event ‘a head shows on the first coin’. Then,

Type 6. Single Integer Answer Type Questions Ex 44. Out of 21 tickets consecutively numbered, three are drawn at random. Find the probability that the numbers on them are in an AP, is a / b, then(14a − b) is_________. Sol. (7) Any three tickets out of 21 tickets can be chosen in 21

C 3 ways. For the favourable choice, if the chosen numbers are a, b and c, a < b < c, then we should have a+ c = b. Obviously, either both a and care even or both 2 are odd and then b is fixed. Hence, for the favourable choice, we have to choose two numbers from 1 to 21, which are either both even or both odd. This can be done in 11C 2 + 10C 2 ways. Hence, the required probability 11 C 2 + 10C 2 10 a = = = 21 133 b C3 ⇒ 14 a − b = 140 − 133 = 7

Ex 45. A has 3 shares in a lottery containing 3 prizes and 9 blanks. B has 2 shares in a lottery containing 2 prizes and 6 blanks and their 100a + 10b + c , then chances of success is 100 p + 10q + r ( a − p) is_________ .

Sol. (2) Let E1 be the event of success of A and let E2 be the event of success of B. Since, A has 3 shares in a lottery containing 3 prizes and 9 blanks, A will draw 3 tickets out of 12 tickets (containing 3 prizes and 9 blanks). A will get success, if he draws atleast one prize out of 3 draws. 9 C 21 ⇒ P (E1′ ) = 12 3 = C 3 55 21 34 ⇒ P (E1 ) = 1 − = 55 55 6 C 15 Again, P (E2′ ) = 8 2 = C 2 28 15 Now, P (E2 ) = 1 − 28 13 = 28 P (E1 ) 34 28 = × ∴ P (E2 ) 55 13 952 = 715 ⇒ P (E1 ) : P (E2 ) = 952 : 715 9 × 100 + 5 × 10 + 2 = 7 × 100 + 1 × 10 + 5 ⇒ a − p = 9 − 7= 2

1115

Target Exercises Type 1. Only One Correct Option 1. From a group of 2 boys and 3 girls, two children are selected. Then, the total number of the outcomes of this experiment is (a) 11

(b) 13

(c) 10

(d) 16

2. Two dice are thrown simultaneously. Then, the set of event ‘a total of atleast 10’ is (a) {(6, 4), (4, 6), (5, 5), (6, 5), (5, 6), (6, 6)} (b) {(5, 5), (6, 4), (4, 6)} (c) {(6, 5), (5, 6), (6, 6)} (d) {(4, 4), (4, 5), (6, 5), (5, 5)}

Ta rg e t E x e rc is e s

(a) A and B are mutually exclusive (b) A and BC are independent (c) AC and BC are independent (d) AC and B are independent

4. The probability of the occurrence of an event A is 0.5 and that of occurrence of another event B is 0.3. If A and B are mutually exclusive events, then the probability that none of A and B will occur, is (b) 0.5 (d) None of these

5. Three mangoes and three apples are kept in a box. If two fruits are selected at random from the box, then the probability that the selection will contain one mango and one apple, is (a) 3/5 (c) 1/36

(b) 5/6 (d) None of these

6. Two friends A and B have equal number of daughters. There are three cinema tickets which are to be distributed among the daughters of A is 1/20. The number of daughters each of them have, is (a) 4

(b) 5

(c) 6

(d) 3

7. If 6n tickets numbered 0, 1, 2, ..., ( 6n − 1) are placed in a bag and three are drawn at random, then the probability that the sum of numbers on the ticket is 6n, is (a)

3n C3

6n

(b)

3n2 C3

6n

(c)

1 3n ⋅ 2 6nC 3

(d)

1 3n2 ⋅ 2 6nC 3

8. A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. Two cards are drawn at random from this pack without replacement. The probability that atleast one of them will be an ace, is 1116

(a) 1/5

(b) 9/20

(c) 1/6

(a) 1/169

(d) 1/9

(b) 1/201

(c) 1/2652

(d) 4/663

10. A bag contains 5 brown socks and 4 white socks. A man selects two socks from the bag without replacement. The probability that the selected socks will be of the same colour, is (a) 5/108

3. If A and B are independent events such that 0 < P ( A ) < 1 and 0 < P ( B ) < 1, then which one is not correct?

(a) 0.6 (c) 0.7

9. Two cards are drawn successively with replacement from a well, shuffled deck of 52 cards, the probability that both of these will be aces, is

(b) 1/6

(c) 5/18

(d) 4/9

11. Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently, equals (a) 1/2

(b) 7/15

(c) 2/15

(d) 1/2

12. Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first, then the probability of all 5 persons leaving at different floors is 7

(a)

P5 75

(b)

75 P5

7

(c)

6

6 P5

5

(d)

P5 55

13. The chance of throwing a total of 3 or 5 or 11 with two dice is (a) 5/36

(b) 1/9

(c) 2/9

(d) 19/36

14. The probability that a teacher will give an unannounced test during any class meeting is 1/5. If a student is absent twice, then the probability that the student will miss atleast one test, is (a) 4/5 (c) 7/75

(b) 2/5 (d) 9/25

15. Given two events are A and B. If odds against A are as 2 : 1 and those in favour of A ∪ B are as 3 : 1, then 1 3 ≤ P( B) ≤ 2 4 5 3 (b) ≤ P( B) ≤ 12 4 1 3 (c) ≤ P ( B ) ≤ 4 5 (d) None of the above (a)

16. A six faced die is so biased that it is twice as likely to show an even number as an odd number when thrown. It is thrown twice. The probability that the sum of two numbers thrown is even, is (a) 1/12

(b) 1/6

(c) 1/3

(d) 5/9

18. The probability that a company executive will travel by train is 2/3 and that he will travel by plane is 1/5. The probability of his travelling by train or plane is (a) 2/15 (c) 15/13

(b) 13/15 (d) 15/2

19. The probability that Krishna will be alive 10 yr hence, is 7/15 and the probability that Hari will be alive after 10 yr, is 7/10. The probability that both Krishna and Hari will be alive 10 yr hence, is (a) 21/150 (c) 49/150

(b) 24/150 (d) 56/150

20. If P ( A ) = 0.65 and P ( B ) = 0.80, then P ( A ∩ B ) lies in the interval (a) [0.30, 0.80] (c) [0.4, 0.70]

(b) [0.35, 0.75] (d) [0.45, 0.65]

21. A and B are two independent events. The probability that both A and B occur, is 1/6 and the probability that none of them occurs, is 1/3. The minimum value of probability of occurrence of A, is (a) 1/2 (b) 1/3 (c) 1/4 (d) None of the above

22. A circle is inscribed in an ellipse. If P is the probability that a point within the ellipse chosen at random lies outside the circle, then the eccentricity of the ellipse is (a) 1 − P (b) 1 − P 2 (c) 1 − (1 − P )2 (d) (1 + P )2 − 1

23. A and B are two independent events. The probability that both A and B occur is 1/6 and the probability that neither of them occurs is 1/3. Then, the probability of the two events are respectively (a) 1/2, 1/3 (b) 1/5, 1/6 (c) 1/2, 1/6 (d) 2/3, 1/4

24. Let A and B be two events such that P ( A ) = 0.3 and P ( A ∪ B ) = 0.8. If A and B are independent events, then P ( B ) is equal to (a) 5/7 (b) 2/3 (c) 1 (d) None of the above

1133 1296 1131 (c) 1296

(a)

1137 1296 1000 (d) 1296 (b)

26. The probability that a man will live 10 more years, is 1/4 and the probability that his wife will live 10 more years, is 1/3. Then, the probability that none of them will be alive after 10 yr, is (a) 5/12 (c) 7/12

21 Fundamentals of Probability

(a) 1/84 (b) 1/21 (c) 5/84 (d) None of the above

25. Two dice are thrown together first and secondly three dice are thrown together. Then, the probability that the total in the first throw is 4 or more and at the same time the total in the second thrown is 6 or more, is

(b) 1/2 (d) 11/12

27. A fair coin is tossed. If the result is a head, a pair of fair dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well-shuffled pack of 11 cards numbered 2, 3, 4, ..., 12 is picked up and the number on the card is noted. The probability that the noted number is 7 or 8, is (a) 193/792 (b) 195/792 (c) 191/792 (d) None of the above

28. A pair of dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7, is (a) 2/5 (b) 1/5 (c) 3/5 (d) None of the above

29. A sample space consists of three mutually independent and equally likely events. The probability of happening of each one of them is equal to

Targ e t E x e rc is e s

17. In a box, there are 2 red, 3 black and 4 white balls. Out of these, three balls are drawn together. The probability of these being of same colour is

(a) 0 (b) 1/3 (c) 1 (d) None of the above

30. Let E and F be two independent events. The probability that both E and F happens is 1/12 and the probability that neither E nor F happens is 1/ 2. Then, (a) P (E ) = 1 / 3, P (F ) = 1 / 4 or P (E ) = 1 / 4 , P (F ) = 1 / 3 (b) P (E ) = 1 / 2, P (F ) = 1 / 6 (c) P (E ) = 1 / 6, P (F ) = 1 / 2 (d) P (E ) = 1 / 4 , P (F ) = 1 / 3

31. If the odds against a certain event are 5 : 2 and the odds in favour of other event, independent of the former, are 6 : 5. Then, the probability that atleast one of the events will happen is 52 77 25 (b) 77 32 (c) 77 (d) None of the above (a)

1117

Objective Mathematics Vol. 1

21

Type 2. More than One Correct Option 32. If A and B are two events, then the probability that exactly one of them occurs is given by (a) P ( A ) + P ( B ) − 2P ( A ∩ B ) (b) P ( A ∩ B ) + P ( A ∩ B ) (c) P ( A ∪ B ) − P ( A ∩ B ) (d) P ( A ) + P ( B ) − 2P ( A ∩ B )

33. If A , B and C are three events, then which of the following is/are correct? (a) P (Exactly two of A , B and C occur) ≤ P ( A ∩ B ) + P ( B ∩ C ) + P (C ∩ A ) (b) P ( A ∪ B ∪ C ) ≤ P ( A ) + P ( B ) + P (C ) (c) P (Exactly one of A , B and C occur) ≤ P ( A ) + P ( B ) + P (C ) − P ( B ∩ C ) − P (C ∩ A ) − P ( A ∩ B ) (d) P (A and atleast one of B and C occur) ≤ P( A ∩ B) + P ( A ∩ C )

1 and 2 2 P ( B ) = , then which of the following is correct? 3

34. If A and B are two events such that P ( A ) =

2 3 1 (b) P ( A ∩ B ) ≤ 3 1 1 (c) ≤ P ( A ∩ B ) ≤ 6 2 1 1 (d) ≤ P ( A ∩ B ) ≤ 6 2 (a) P ( A ∪ B ) ≥

35. If A and B are two independent events such that P ( A ∩ B ) = 2 / 15and P ( A ∩ B ) = 1/ 6, then P ( B )is (a)

4 5

(b)

1 6

(c)

1 5

(d)

5 6

Type 3. Assertion and Reason

Ta rg e t E x e rc is e s

Directions (Q. Nos. 36-39) In the following questions, each question contains Statement I (Assertion) and Statement II (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are (a) Statement I is true, Statement II is true: Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

36. A fair die is rolled once. Statement I The probability of getting a composite number is 1/ 3. Statement II There are three possibilities for the obtained number (i) the number is a prime number (ii) the number is a composite number (iii) the number is 1 and hence probability of getting a prime number = 1/ 3.

37. Let A and B be two events such that P ( A ∪ B ) ≥ and

1 3 ≤ P (A ∩ B)≤ . 8 8

Statement I Statement II

3 4

7 8 11 P ( A ) + P (B ) ≤ 8

P ( A ) + P (B ) ≥

38. Let A and B be two independent events. Statement I If P ( A ) = 0.3 and P ( A ∪ B ) = 0.8, then 2 P ( B ) is . 7 Statement II P ( E ) = 1 − P ( E ), where E is any event. 39. Statement I If A = {2, 4, 6} and B = {1, 2, 3}, where A and B are the events of numbers occurring on a die, then P ( A ) + P ( B ) = 1. Statement II If A1 , A 2 , A 3 , K, A n are all mutually exclusive events, then P ( A1 ) + P ( A2 ) + K + P ( An ) = 1.

Type 4. Linked Comprehension Based Questions Passage I (Q. Nos. 40-42) Two dice are thrown. The events A, B and C are described as follows : A : Getting an even number on the first die. B : Getting an odd number on the first die. C : Getting atmost 5 as sum of the numbers on two dice.

40. Then, the set of the events A and B is (a) φ (b) {1, 2} (c) {2, 4} (d) {1, 3}

1118

41. The total number of outcomes in the event ‘B or C’ is (a) 20 (b) 22 (c) 21 (d) 23

42. The set of the event ‘B and C’ is (a) {(1, 1), (1, 2), (1, 3), (1, 4)} (b) {(1, 2), (1, 3), (1, 4), (1, 5)} (c) {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)} (d) None of the above

43. If the balls are different and any number of balls can go to any urns, then P ( A ) is equal to M! nM n! (c) M Pn

n! Mn 1 (d) n M

(b)

(a)

44. If the balls are identical and any number of balls can go to any urns, then P ( A ) equals (a) (c)

1 Mn

1 CM − 1 1 (d) M + n − 1 PM − 1

(b) 1

M+ n− 1

Cn − 1

M+ n− 1

45. If the balls are identical but atmost one ball can be put in any box, then P ( A ) is equal to (a)

1 Pn

M

(b)

n! CM

n

(c)

n! Cn

M

(d)

1 Cn

21 Fundamentals of Probability

Passage II (Q. Nos. 43-45) Consider the experiment of distribution of balls among urns. Suppose we are given M urns, numbered 1 to M , among which we are to distribute n balls (n < M ). Let P ( A) denotes the probability that each of the urns numbered 1 to n will contain exactly one ball. Then, answer the following questions.

M

Type 5. Match the Columns Column I

47. Match the statements of Column I with values of Column II.

Column II

Column I

Column II

A. There are 7 seats in a row. Three persons p. take seats at random. The probability that the middle seat is always occupied and no two persons are consecutive, is

1 2

B. Three numbers are chosen at random q. without replacement from the set A = { x |1 ≤ x ≤ 10, x ∈ N} The probability that the minimum of the chosen numbers is 3 and maximum is 7, is

11 40

C. From a group of 10 persons consisting of r. 5 lawyers, 3 doctors and 2 engineers, four persons are selected at random. The probability that the selection contains atleast one of each category is

1 40

D. Three numbers are chosen at random s. without replacement from 1, 2, 3, ..., 10. The probability that the minimum of the chosen numbers is 4 or their maximum is 8, is

4 35

A.

7 white balls and 3 black balls are placed in a row at random. The probability that no two black balls are adjacent, is

p.

5 11

B.

4 gentlemen and 4 ladies take seats at random round a table. The probability that they are sitting alternately, is

q.

1 16

C.

10 different books and 2 different pens are given to 3 boys so that each gets equal number of things. The probability that the same boy does not receive both the pens, is

r.

7 15

D.

A fair coin is tossed repeatedly. The probability of getting a result in the fifth toss different from those obtained in the first four tosses is

s.

1 35

Targ e t E x e rc is e s

46. Match the statements of Column I with values of Column II.

Type 6. Single Integer Answer Type Questions 48. A four digit number is formed using the digits 0, 1, 2, 3, 4 without repetition. If the probability that it is a divisible by 4, is , then b − 2a is _______. b 49. In throwing of a die, let A be the event ‘an odd number turns up’, B be the event ‘a number divisible by 3 turns up’ and C be the event ‘a number ≤ 4 turns up’. If the probability that exactly two of A , B and C occur a is , then a + b is ______ . b

50. A bag contains a large number of white and black marbles in equal proportions. Two samples of 5 marbles are selected (with replacement) at random. If the probability that the first sample contains exactly 1 black marble and the second sample contains 10a + b exactly 3 black marbles, is , then a + b is___ . 512 51. If two switches S 1 and S 2 have respectively 90% and 80% chances of working. Find the probabilities that if each of the following circuits will work = a / b, then b − a is ______ .

1119

Entrances Gallery JEE Advanced/IIT JEE 1. Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is atleast one more than the number of girls ahead of her, is [2014] (a)

1 2

(b)

1 3

(c)

2 3

(d)

3 4

Passage (Q. Nos. 2-3) Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5 and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let x i be the number on the card drawn from the ith box, i = 1, 2, 3. [2014]

equations (α − 2β) p = αβ and (β − 3γ) p = 2 βγ . All the given probabilities are assumed to lie in the interval (0, 1). Then,

6. Four fair dice D1 , D2 , D3 and D4 , each having six faces numbered 1, 2, 3, 4, 5 and 6, are rolled simultaneously. The probability that D4 shows a number appearing on one of D1 , D2 and D3 , is [2012] (a)

2. The probability that x1 + x 2 + x 3 is odd, is (a)

29 105

(b)

53 105

(c)

57 105

(d)

Ta rg e t E x e rc is e s

3. The probability that x1 , x 2 and x 3 arithmetico-progression, is (a)

9 105

(b)

10 105

(c)

11 105

1 2

are (d)

in

an

7 105

4. Four persons independently solve a certain problem 1 3 1 1 correctly with probabilities , , , . Then, the 2 4 4 8 probability that the problem is solved correctly by [2013] atleast one of them, is 235 256 3 (c) 256

21 256 253 (d) 256

(a)

(b)

5. Of the three independent events E1 , E 2 and E 3 , the probability that only E1 occurs is α, only E 2 occurs is β and only E 3 occurs is γ. Let the probability p that none of events E1 , E 2 or E 3 occurs satisfy the

Probability of occurrence of E1 is ______ . Probability of occurrence of E 3 [2013]

91 216

(b)

108 216

(c)

125 216

(d)

127 216

7. Let E and F be two independent events. The 11 probability that exactly one of them occurs is and 25 2 the probability of none of them occurring is . If 25 P(T) denotes the probability of occurrence of the [2011] event T, then 4 3 , P (F ) = 5 5 2 1 (c) P (E ) = , P (F ) = 5 5 (a) P (E ) =

1 , P (F ) = 5 3 (d) P (E ) = , P (F ) = 5 (b) P (E ) =

2 5 4 5

8. Let ω be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1 , r2 and r3 are the numbers obtained on the die, then the probability that [2010] ω r1 + ω r2 + ω r3 = 0, is 1 18 2 (c) 9

(a)

1 9 1 (d) 36

(b)

JEE Main/AIEEE 9. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes [2015] contains exactly 3 balls, is (a)

55  2   3  3

11

 2 (b) 55   3

10

 1 (c) 220   3

12

 1 (d) 22   3

11

1 10. If A and B are two events such that P ( A ∪ B ) = , 6 1 1 P ( A ∩ B ) = and P ( A ) = , where A stands for the 4 4 complement of the event A. Then, the events A and B [2014] are (a) (b) (c) (d)

1120

mutually exclusive and independent equally likely but not independent independent but not equally likely independent and equally likely

11. Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ......, 20}. Statement I The probability that the chosen numbers when arranged in some order will form an 1 AP, is . 85 Statement II If the four chosen numbers form an AP, then the set of all possible values of common difference is {±1, ± 2, ± 3, ± 4, ± 5,}. [2010] (a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

(b) 3/5

(c) 0

(d) 1

13. Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house, is [2005] (a)

7 9

(b)

8 9

(c)

1 9

14. The probability that A speaks truth, is

(d)

2 9

4 while this 5

3 . The probability that they 4 contradict each other when asked to speak on a fact,is probability for B is

3 (a) 20

1 (b) 5

7 (c) 20

4 (d) [2004] 5

15. Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse, is [2003] (a)

4 5

(b)

3 5

(c)

1 5

(d)

2 5

 1 1  3 , 2   1 13  (c) ,  3 3  (a)

(b)

 1 2  3 , 3 

(d) [ 0, 1]

17. A and B play a game, where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial, is [2002] 1 25 2 (c) 25

(b)

(a)

24 25

(d) None of these

18. A problem in Mathematics is given to three students A , B , C and their respective probability of solving the 1 1 1 problem is , and . Probability that the problem is 2 3 4 solved, is [2002] (a)

3 4

(b)

1 2

(c)

2 3

(d)

1 3

Other Engineering Entrances 19. Five persons A , B , C , D and E are in queue of a shop. The probability that A and E are always together, is 1 (a) 4

2 (b) 3

2 (c) 5

[BITSAT 2014] 3 (d) 5

20. A four digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd, is [VITEEE 2014] (a) zero

(b)

1 (c) 4

1 3

21. If the integers m and n are chosen at random from 1 to 100, then the probability that a number of the form [VITEEE 2014] 7n + 7m is divisible by 5, equals (a)

1 4

(b)

1 2

(c)

1 8

(d)

1 3

22. One mapping (function) is selected at random from all the mappings of the set A = {1, 2, 3,... , n} into itself. The probability that the mapping selected is one-one, is [Manipal 2014] n! nn − 1 n! (c) n 2n

(b)

(a)

n! nn

1 9

(d) None of these

(b)

1 18

(c)

1 36

(d)

5 36 13 (d) 18

(a)

1 12

1 6 3 (e) 18 (b)

(c)

5 18

25. A fair six faced die is rolled 12 times. The probability that each face turns up twice is equal to [WB JEE 2014] 12! 6! 6! 612

(b)

212 2 612 6

(c)

12! 26 612

(d)

12! 62 612

26. A poker hand consists of 5 cards drawn at random from a well-shuffled pack of 52 cards. Then, the probability that a poker hand consists of a pair and a triple of equal face values (for example, 2 sevens and 3 kings or 2 aces and 3 queens, etc.), is [WB JEE 2014] (a)

6 4165

(b)

23 4165

(c)

1797 4165

(d)

1 4165

27. Ram is visiting a friend. Ram knows that his friend has 2 children and 1 of them is a boy. Assuming that a child is equally likely to be a boy or a girl, then the probability that the other child is a girl, is 1 (a) 2

23. Two dice are thrown simultaneously. The probability of obtaining a total score of 5 is [Karnataka CET 2014] (a)

24. If two dice are thrown simultaneously, then the probability that the sum of the numbers which come upon the dice to be more than 5, is [Kerala CEE 2014]

(a)

(d) None of these

21 Fundamentals of Probability

[2008] (a) 2/5

16. Events A , B and C are mutually exclusive events such 3x + 1 1− x 1 − 2x that P ( A ) = , P (B ) = and P (C ) = . 3 4 2 The set of possible values of x are in the interval [2003]

Targ e t E x e rc is e s

12. A die is thrown. Let A be the event that the number obtained is greater than 3 and B be the event that the number obtained is less than 5. Then, P ( A ∪ B ) is

1 (b) 3

2 (c) 3

[WB JEE 2014] 7 (d) 10

28. The probability that a leap year selected at random contains 53 Sunday, is [AMU 2014] (a)

7 366

(b)

28 183

(c)

1 7

(d)

2 7

1121

Objective Mathematics Vol. 1

21

29. Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral, equals [AMU 2014] 1 (a) 2

1 (b) 2

1 (c) 10

1 (d) 20

30. A six faced unbiased die is thrown twice and the sum of the numbers appearing on the upper face is observed to be 7. The probability that the number 3 [EAMCET 2014] has appeared atleast once, is (a)

1 5

(b)

1 2

(c)

1 3

(d)

1 4

31. A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel, is [GGSIPU 2014] (a)

8 11

(b)

4 11

(c)

2 11

(d)

3 11

32. If three squares are chosen on a chess board, then the chance that they should be in a diagonal line, is

Ta rg e t E x e rc is e s

7 (a) 744

5 (b) 744

7 (c) 544

[GGSIPU 2014] 11 (d) 744

33. If two cards are drawn simultaneously from the same set, then probability that atleast one of them will be the ace of hearts, is [GGSIPU 2014] (a)

1 13

(b)

1 26

(c)

1 52

(d)

3 13

34. In a class, there are 10 boys and 8 girls. When 3 students are selected at random, then the probability that 2 girls and 1 boy are selected, is [GGSIPU 2014] (a)

35 102

(b)

15 102

(c)

55 102

(d)

25 102

35. The probability of simultaneous occurrence of atleast one of two events A and B is p. If the probability that exactly one of A , B occurs is q, then P ( A ′ ) + P ( B ′ ) is equal to [BITSAT 2014] (a) 2 − 2 p + q (c) 3 − 3 p + q

(b) 2 + 2 p − q (d) 2 − 4 p + q

36. If M and N are any two events, then the probability that exactly one of them occurs, is (for an event set A, the complement is A C ) [GGSIPU 2014] (a) P (M ) + P (N ) − 2P (M ∩ N ) (b) P (M ) + P (N ) − P (M ∪ N ) (c) P (M C ) + P (N C ) − 2P (M C ∪ N C ) (d) P (M ∪ N C ) + P (M C ∪ N )

37. Let A and B be two events such that P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ) P ( B ). If 0 < P ( A ) < 1 and 0 < P ( B ) < 1, then P ( A ∪ B )′ is [Kerala CEE 2014] equal to

1122

(a) 1 − P ( A ) (b) 1 − P ( A′ ) (c) 1 − P ( A ) P ( B ) (d) [1 − P ( A )] P ( B′ ) (e) 1

38. If the events A and B are independent, if P ( A ′ ) =

2 3

2 and P ( B ′ ) = , then P ( A ∩ B ) is equal to 7 [Karnataka CET 2014] 4 21 1 (c) 21

5 21 3 (d) 21

(a)

(b)

39. If P ( A ∪ B ) = 0.83, P ( A ) = 0.3 and P ( B ) = 0.6, then the events will be [RPET 2014] (a) dependent (c) cannot say anything

(b) independent (d) None of these

40. If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9, is [UP SEE 2013] 2n 5n 4n (c) n 5 (a)

(b)

4 n − 2n 5n

(d) None of these

41. An urn contains 8 red and 5 white balls. Three balls are drawn at random. Then, the probability that balls of both colours are drawn, is [WB JEE 2012] (a)

40 143

(b)

70 143

(c)

3 13

(d)

10 13

42. Two decks of playing cards are well-shuffled and 26 cards are randomly distributed to a player. Then, the probability that the player gets all distinct cards, is [WB JEE 2012] C 26 (b) 2 × 104 C 26 52 C 26 (d) 2 × 104 26 C 26

52

C (a) 104 26 C 26 (c) 213 ×

52

52

C 26 C 26

104

43. Probability of product of a perfect square when 2 dice are thrown together, is [OJEE 2012] 2 9 5 (c) 18

1 9 (d) None of these

(b)

(a)

44. A bag contains 3 white and 5 black balls. One ball is drawn at random. Then, the probability that it is white, [BITSAT 2011] is (a)

1 8

(b)

3 8

(c)

5 8

(d)

7 8

45. A die is rolled three times. The probability of getting a larger number than the previous number each time, is (a)

5 72

(b)

5 54

(c)

13 216

[MP PET 2011] 1 (d) 18

46. 4 boys and 2 girls occupy seats in a row at random. Then, the probability that two girls occupy seats side by side, is [WB JEE 2011] (a)

1 2

(b)

1 4

(c)

1 3

(d)

1 6

(a)

1 52

(b)

15 52

(c)

[UP SEE 2011] 19 (d) 52

18 52

48. A fair coin is tossed 100 times the probability of getting tails an odd number of time is [AMU 2011] (a)

1 2

(b)

1 4

(c) 0

(d) 1

49. The probability that a leap year will have only 52 Sunday, is [GGSIPU 2011] (a)

4 7

(b)

5 7

(c)

6 7

(d)

1 7

1 7!

(b)

8 35

(c)

19 35

(d)

[UP SEE 2011] (a) P ( A ∩ BC ) = P ( B ) − P ( A ) (b)P ( A ∩ BC ) = P ( A ) − P ( B ) (c) P ( B ) ≤ P ( A ) (d) P ( A ) ≤ P ( B )

52. If both to male child and birth to female child are equal probable, then what is the probability that atleast one of the three children born to a couple is male? [UP SEE 2010] (a)

50. A bag has four pairs of balls of four distinct colours. If four balls are picked at random (without replacement), then the probability that there is atleast one pair among [J&K CET 2011] them have the same colour, is (a)

51. If A and B are two non-empty sets with B C denoting the complement of set B such that B ⊂ A, then which of the following probability statements hold correct?

27 35

4 5

(b)

7 8

(c)

8 7

(d)

1 7

53. Two dice are tossed once. The probability of getting even number at the first die or a total of 8, is [WB JEE 2010] 1 (a) 36 11 (c) 36

3 (b) 36 5 (d) 9

Targ e t E x e rc is e s

Answers Work Book Exercise 21.1 1. (a)

2. (c)

3. (c)

4. (c)

5. (c)

6. (c)

4. (c)

5. (c)

6. (d)

7. (c)

4. (a)

5. (a)

6. (a)

7. (b)

Work Book Exercise 21.2 1. (c)

2. (c)

3. (a)

8. (a)

Work Book Exercise 21.3 1. (c)

2. (b)

3. (d)

21 Fundamentals of Probability

47. A card is drawn from a well-shuffled pack of playing cards. The probability that it is a heart of a king, is

Target Exercises 1. (c)

2. (a)

3. (a)

4. (d)

5. (a)

6. (d)

7. (b)

8. (b)

9. (a)

10. (d)

11. (b)

12. (a)

13. (c)

14. (d)

15. (b)

16. (d)

17. (c)

18. (b)

19. (c)

20. (d)

21. (b)

22. (c)

23. (a)

24. (a)

25. (a)

26. (b)

27. (a)

28. (a)

29. (c)

30. (a)

31. (a)

32. (c,d)

33. (b,c,d)

34. (a,c,d)

35. (a,b)

36. (c)

37. (b)

38. (b)

39. (c)

40. (a)

42. (c)

43. (b)

44. (b)

45. (d)

46. (*)

47. (**)

48. (6)

49. (7)

50. (7)

41. (b) 51. (1)

.

* A → r; B → s; C → p; D → q ** A → s; B → r; C → p; D → q

Entrances Gallery 1. (a)

2. (b)

3. (c)

4. (a)

5. (6)

6. (a)

8. (c)

9. (a)

11. (c)

12. (d)

13. (c)

14. (c)

15. (d)

16. (a)

17. (b)

7. (a,d)

18. (a)

19. (c)

10. (c) 20. (d)

21. (a)

22. (b)

23. (a)

24. (d)

25. (c)

26. (a)

27. (c)

28. (d)

29. (c)

30. (c)

31. (b)

32. (a)

33. (b)

34. (a)

35. (a)

36. (a)

37. (d)

38. (b)

39. (a)

40. (a)

41. (d)

42. (d)

43. (a)

44. (b)

45. (b)

46. (c)

47. (a)

48. (a)

49. (b)

50. (d)

51. (c)

52. (b)

53. (d)

1123

Explanations Target Exercises 1. Let the two boys be taken as B1 and B2 and the three girls be taken as G1, G2 and G3 . Clearly, there are 5 children, out of which two children can be chosen in 5 C2 ways. So, there are 5 C2 = 10 elementary events associated to this experiments and are given by B1 B2 , B1G1, B1G2 , B1G3 , B2G1, B2G2 , B2G3 , G1G2 , G1G3 and G2G3 .

2. The sample space associated with the given random experiment is given by S = {(1, 1), (1, 2 ), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2 ), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} Let event E = a total of atleast 10 E = {(6, 4), (4, 6), (5, 5), (6, 6), (5, 6), (6, 5)}

Ta rg e t E x e rc is e s

3. A and B are not mutually exclusive. 4. Since, the events A and B are mutually exclusive, these events are not independent. Hence, the required probability cannot be found.

5. Out of 6 fruits, 2 fruits can be selected in 6 C2 ways. Also, 1 apple and 1 mango can be selected in 3 × 3 ways. 9 3 Hence, the required probability = 6 = C2 5

6. Let each of the friend have x daughters. Then, the probability that all the tickets go to the daughters of A is x C3 . 2x C3 x ( x − 2) 1 C3 1 = ⇒ = ∴ 2x 4 (2 x − 1) 20 C3 20 ⇒ ⇒

5 x − 10 = 2 x − 1 x=3

7. The total number of ways of selecting 3 tickets out of 6n 6n

numbers is C3 . The sum of the numbers on three tickets drawn will be 6 n in each of the following cases : Lowest number

Total number of ways

0

(0, 1, 6n − 1) (0, 2, 6n − 2 ) ...( 0, 3n − 1, 3n + 1)

( 3n − 1)

1

(1, 2, 6n − 3) (1, 3, 6n − 4) ... (1, 3n − 1, 3n + 1)

( 3n − 2 )

2

(2, 3, 6n − 5) (2, 4, 6n − 6) ...(2, 3n − 2 , 3n)

( 3n − 4)

3

(3, 4, 6n − 7 ) (3, 5, 6n − 8) ...(3, 3n − 2 , 3n − 1)

( 3n − 5)

M

1124

Ways of selection

M

M

(2 n − 2 ) (2n – 2, 2n – 1, 2n + 3), (2n – 2, 2n, 2n + 2),

2

(2 n − 1) (2n – 1, 2n, 2n + 1)

1

Total number of ways of getting 6n as the sum = (3 n − 1) + (3 n − 2 )(3 n − 3) + (3 n − 5) + K + 2 + 1 = { 3 n − 1 + 3 n − 2} + { 3 n − 4 + 3 n − 5} + K + { 5 + 4} + {2 + 1} = (6 n − 3) + (6 n − 9) + K + 9 + 3 = 3 n 2 3 n2 ∴ Required probability = 6 n C3

8. P (none of the two cards is an ace) = ∴P (atleast one ace) =

11 20

9 20

9. P (first card is an ace and the second card is an ace) =

1 169

10. From the bag, two socks can be drawn in 9 C2 ways. Also, two socks of the same colour can be drawn in 5 C2 + 4C2 ways. 5

Hence, the required probability is

C2 + 4C2 4 = . 9 9 C2

11. Total number of ways in which all the balls can be placed is

10 ! = 120. 7 ! 3!

If no two black balls are placed adjacently, then the possible number of ways is 56. 56 7 Hence, the required chance is = . 120 15

12. There are seven floors besides the ground floor. ∴Number of possible outcomes = 7 5 Number of favourable cases = 7P5 7 P ∴Required probability = 55 7

13. 3 can be thrown as (1, 2), (2, 1). 5 can be thrown as (1, 4), (4, 1), (2, 3), (3, 2). 11 can be thrown as (5, 6), (6, 5). 8 2 ∴ Required probability = = 36 9 1 4 4 1 8 14. Probability that one test is held = × + × = 5 5 5 5 25 1 ∴Probability that the test is held on both days = 25 ∴Probability that the student misses atleast one test 8 1 9 = + = 25 25 25 1 3 15. P( A) = , P( A ∪ B) = 3 4 Now, P( A ∪ B) = P( A) + P(B) − P( A ∩ B) ≤ P( A) + P(B) 3 1 5 ⇒ ≤ + P(B) ⇒ ≤ P(B) 4 3 12 3 Also, B ⊆ A ∪ B ⇒ P(B) ≤ P( A ∪ B) = 4 5 3 ∴ ≤ P(B) ≤ 12 4

17. Required probability =

3

C3 + 4C3 5 = 9 84 C3

18. Let E1 denotes the event of travelling by train and E2 denotes the event of travelling by plane. 2 1 , P(E2 ) = 3 5 13 ∴ P(E1 ∪ E2 ) = P(E1 ) + P(E2 ) = 15 Then, P(E1 ) =

19. The probability that both Krishna and Hari will be alive after 10 yr, is

49 . 150

20. P( A ∩ B) ≤ min { P( A), P(B)} = min {0.65, 0.80} = 0.65 ∴ P( A ∩ B) ≤ 0.65 Also, P( A ∩ B) = P( A) + P(B) − P( A ∪ B) ≥ 0.65 + 0.80 − 1 = 0.45 ∴ 0.45 ≤ P( A ∩ B) ≤ 0.65

21. Let P( A) = a and P(B) = b, then we have ab = 1 (1 − a)(1 − b) = . 3

Solving

simultaneously, we have a =

these

1 and 6

equations

1 1 or . Minimum value of 2 3

1 these is . 3

24. Let P(B) = b. We have, P( A ∪ B) = P( A) + P(B) − P( A)⋅ P(B), if A and B are independent events. 5 Substituting the given values, we have P(B) = 7

25. Consider the events. A = Getting a total of 4 or more in the throw of two dice B = Getting a total of 6 or more in the simultaneous throw of three dice ...(i) ∴Required probability = P( A ∩ B) = P( A)⋅ P(B) For P( A) Total number of elementary events = 62 Favourable events = 33 33 11 P( A) = 2 = ∴ 12 6 For P(B) Total number of elementary events = 63 Favourable number of elementary events = 63 − {Sum of the coefficients of x 3 , x 4 and x 5 in ( x + x 2 + K + x 6 )3 } 3 = 6 − (1 + 3 + 6) = 206 206 103 ∴ P(B) = = 216 108 11 103 1133 × = ∴Required probability = 12 108 1296

26. Let the name of the man be A and that of his wife be B. Then, P (A will not be alive after 10 yr and B will not be 3 2 1 alive after 10 yr) = × = 4 3 2

27. Probability of getting a total of 7 or 8 on a pair of dice, is 11/36.

2

x y + 2 = 1, the inscribed circle has 2 a b the equation x 2 + y 2 = b2 , P = probability that the point π b2 of the ellipse lies outside the circle = 1 − πab a 1− e2 b ⇒ P = 1− = 1− = 1− 1− e2 a a where, e is the eccentricity of the ellipse. ⇒ 1 − e 2 = (1 − P )2

22. Let the ellipse be

e = 1 − (1 − P )2

⇒

1 3 1 P( A)P(B) = ⇒ 6 1 2 and P( A ∪ B) = ⇒ P( A ∪ B) = 3 3 2 ∴ P( A) + P(B) − P( A ∩ B) = 3 5 ⇒ P( A) + P(B) = 6 On solving Eqs. (i) and (ii), we get 1 1 P( A) = , P(B) = 2 3 1 1 or P( A) = , P(B) = 3 2

1 11 × . 2 36 Also, P (tail shows up and number 7 or 8 is noted) 1 2 = × . 2 11 193 Hence, the required probability = 792 1 28. The probability of getting a total of 5 on a pair of dice, is 9 1 13 and that of getting a total of 7, is . Also, P( 5 7 ) = 6 18 because there are 10 cases in favour of getting a sum of 5 or 7.5 can come before 7 in the following cases : 5 or( 5 7 )5 or( 5 7 )( 5 7 )5 or ... Hence, the required probability = 2 / 5 P (head shows up and number 7 or 8 is noted) =

2

1 6

23. P( A ∩ B) = , P( A ∩ B ) =

...(i)

21 Fundamentals of Probability

Then, probability for appearance of even number is 2 p. 1 ∴ p + 2p = 1 ⇒ p = 3 Sum of two numbers is even means either both numbers are even or odd. 5 ∴Required probability = 9

Targ e t E x e rc is e s

16. Let probability for appearance of odd number be p.

29. Required probability = 1 1 1 and P(E ∩ F ) = 12 2 1 P(E )⋅ P(F ) = 12 1 {1 − P(E )}{1 − P(F )} = 2 P(E ) = x and P(F ) = y 1 1 and (1 − x )(1 − y ) = xy = 12 2 1 1 1 1    x = and y =  or  x = and y =    3 4 4 3

30. P(E ∩ F ) = ⇒ ...(ii)

and Let ⇒ ∴

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Objective Mathematics Vol. 1

21

31. P(first event does not happen) =

5 5 = 5+ 2 7

P(second event does not happen) = ∴P(both the event fail to happen) =

5 5 = 5 + 6 11

5 5 25 × = 7 11 77

Therefore, the probability that atleast one of the events will happen = 1 − P (none of two happens) 25 52 = 1− = 77 77

32. We have, P (exactly one of A, B occurs)

Ta rg e t E x e rc is e s

= P[( A ∩ B ) ∪ ( A ∩ B)] = P( A ∩ B ) + P( A ∩ B) = P( A) − P( A ∩ B) + P(B) − P( A ∩ B) = P( A) + P(B) − 2 P( A ∩ B) = P ( A ∪ B) − P( A ∩ B) Also, P (exactly one of A, B occurs) = P( A ∪ B) − P( A ∩ B) = {1 − P( A ∪ B)} − {1 − P( A ∩ B)} = {1 − P( A ∩ B )} − {1 − P( A ∪ B )} = P( A ∪ B ) − P( A ∩ B ) = P( A ) + P(B) − 2 P( A ∩ B )

1126

33. We have, P(exactly two of A, B, C occur) = P( A ∩ B ∩ C ) + P( A ∩ B ∩ C ) + P( A ∩ B ∩ C ) = P( A ∩ B) − P( A ∩ B ∩ C ) + P( A ∩ C ) − P( A ∩ B ∩ C ) + P(B ∩ C ) − P( A ∩ B ∩ C ) = P( A ∩ B) + P(B ∩ C ) + P( A ∩ C ) − 3P( A ∩ B ∩ C ) ≤ P( A ∩ B) + P(B ∩ C ) + P( A ∩ C ) Also, P( A ∪ B ∪ C ) = P( A ∪ B) + P(C ) − P{( A ∪ B) ∩ C} ≤ P( A ∪ B) + P(C ) ≤ P( A) + P(B) + P(C ) [Q P( A ∪ B) ≤ P( A) + P(B)] Now, P (exactly one of A, B, C occurs) = P( A ∩ B ∩ C ) + P( A ∩ B ∩ C ) + P( A ∩ B ∩ C ) = P( A ∩ B ∪ C ) + P( A ∪ B ∩ C ) + P(B ∩ A ∪ C ) = P( A) − P{ A ∩ (B ∪ C ) ∪ ( A ∩ C )} + P(C ) − P{(C ∩ A) ∪ (C ∩ B)} + P(B) − P{(B ∩ A) ∪ (B ∩ C )} = P( A) + P(B) + P(C ) − 2 P( A ∩ B) − 2 P(B ∩ C ) − 2 P( A ∩ C ) + 3P( A ∩ B ∩ C ) = [P( A) + P(B) + P(C ) − P( A ∩ B) − P(B ∩ C ) − P( A ∩ C )] − [P( A ∩ B) + P(B ∩ C ) + P( A ∩ C ) − 3P ( A ∩ B ∩ C )] = P( A) + P(B) + P(C ) − P( A ∩ B) − P(B ∩ C ) − P( A ∩ C ) − P (exactly two of A, B, C occur) ≤ P( A) + P(B) + P(C ) − P( A ∩ B) − P(B ∩ C ) − P( A ∩ C ) Finally, P (A and atleast one of B, C occurs) = P[ A ∩ (B ∪ C )] = P[( A ∩ B) ∪ ( A ∩ C )] = P( A ∩ B) + P( A ∩ C ) − P[( A ∩ B) ∩ ( A ∩ C )] = P( A ∩ B) + P( A ∩ C ) − P( A ∩ B ∩ C ) ≤ P( A ∩ B) + P( A ∩ C ) 2 34. We have, P( A ∪ B) ≥ max{ P( A), P(B)} = 3 2 and P( A ∩ B) ≤ min{ P( A), P(B)} = 3

Now, ⇒ ∴ Also, and Hence, ⇒

P( A ∩ B) = P( A) + P(B) − P( A ∪ B) 1 P( A ∩ B) = P( A) + P(B) − 1 = 6 1 1 ≤ P( A ∩ B) ≤ 6 2 1 1 1 P( A ∩ B ) = P( A) − P( A ∩ B) ≤ − = 2 6 3 2 1 1 P( A ∩ B) = P(B) − P( A ∩ B) ≤ − = 3 6 2 2 1 2 1 − ≤ P( A ∩ B) ≤ − 3 2 3 6 1 1 ≤ P( A ∩ B) ≤ 6 2

35. Let P( A) = x and P(B) = y. Since, A and B are independent events. Therefore, 2 15 2 P( A ) P(B) = ⇒ 15 2 ⇒ {1 − P( A)} P(B) = 15 2 (1 − x )y = ⇒ 15 2 y − xy = ⇒ 15 1 and P( A ∩ B ) = 6 1 ⇒ P( A)P(B ) = 6 1 x(1 − y ) = ⇒ 6 1 x − xy = ⇒ 6 On subtracting Eq. (i) from Eq. (ii), we get 1 1 x−y= + y ⇒ x= 30 30 Putting this value of x in Eq. (i), we get 2  1  y−y + y =  30  15 ⇒ 30 y − y − 30 y 2 = 4 ⇒ 30 y 2 − 29 y + 4 = 0 ⇒ (6 y − 1)(5 y − 4) = 0 1 4 or y = y= ⇒ 6 5 1 4 or P(B) = ⇒ P(B) = 6 5 P( A ∩ B) =

...(i)

...(ii)

36. Statement I is true, as there are six equally likely possibilities of which only two are favourable (4 and 6). 2 1 = 6 3 Statement II is not true, as three possibilities are not equally likely. Hence, P (obtained number is composite) =

37. P( A ∪ B) = P( A) + P(B) − P( A ∩ B) ∴ ⇒

1 ≥ P( A) + P(B) − P( A ∩ B) ≥ P( A) + P(B) −

1 3 ≥ 8 4

3 4

1  Q minimum value of P( A ∩ B) =  8 

38. P( A ∪ B ) = 1 − ( A ∪ B ) = 1 − ( A ∩ B) = 1 − P( A ) P(B) 0.8 = 1 − 0.7 × P(B) 2 P(B) = 7

⇒

39. P( A) + P(B) = 1is true, as A and B are mutually exclusive and exhaustive events but Statement II is false, as it is not given that the events are exhaustive.

40. A = Getting an even number on the first die = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} B = Getting an odd number on first die = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} A and B = A ∩ B = φ

41. Event B given above and C = Getting atmost 5 as sum of the numbers on the two dice C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)} B or C = B ∪ C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} ∴ n(B ∪ C ) = 22

42. The set of event B as above and the set of event C as above. Now, set of events B and C = B ∩ C = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}

43. Total number of ways = M n Number of favourable cases = n !

44. Total number of ways =

M + n −1

CM − 1

Number of favourable cases = 1

45. Total number of ways = MCn Number of favourable cases = 1 B

B

B

B

B

B

B

46. A. | W | W | W | ... | W | W | n(S ) =

10 ! (7 !)(3 !)

8! because there are 8 places for (3 !)(5 !) 3 black balls. 8! (3 !)(5 !) (8 !)(7 !) ∴ = P(E ) = 10 ! (10 !)(5 !) (7 !)(3 !) 7⋅6 7 = = 10 ⋅ 9 15 n(E ) = 8C3 =

B. n(S ) = 7 !, n(E ) = (3 !) × (4 !) [Q after making 4 gentlemen sit in 3! ways, 4 ladies can sit in 4!ways in between the gentlemen] (3 !) × (4 !) 6 1 P(E ) = = = ∴ 7! 7 × 6 × 5 35 C. n(S ) = 12C4 × 8C4 × 4C4 n(E ) = n(S ) − number of ways in which one boy gets both the pens = n(S ) − 10C2 × 8C4 × 4C4 × (3 !) 10 C2 × 8C4 × 4C4 × (3 !) ∴ P(E ) = 1 − 12 C4 × 8C4 × 4C4 6 5 = 1− = 11 11 D. Required probability = P(EEEEE ) + P(EEEEE ) = { P(E )} 4 ⋅ P(E ) + { P(E )} 4 ⋅ P(E ) 1 = 16 7 ⋅ 6⋅ 5 7 47. A. n(S ) = C3 × 3 ! = ⋅ 6 = 210 6 n(E ) = 2C1 × 2C1 × 1C1 × 3 ! 3

3

3

21 Fundamentals of Probability

P( A) + P(B) ≥

3

On the left and the other has to sit at any one of the two marked seats on the right. n(E ) 2 × 2 × 6 4 P(E ) = = = ∴ n(S ) 210 35 B. n(S ) = 10C3 and n(E ) = 3C1 because on selecting 3, 7 and we have to select one from 4, 5 and 6 3 C 1 P(E ) = 10 1 = ∴ C3 40 C. n(S ) = 10C4 = 210 n(E ) = 5C2 × 3C1 × 2C1 + 5C1 × 3C2 × 2C1 + 5C1 × 3C1 × 2C2 = 105 105 1 ∴ P(E ) = = 210 2 D. The probability of 4 being the minimum number 6 C = 10 2 C3 [after selecting 4, any two can be selected from 5, 6, 7, 8, 9, 10] The probability of 4 being the minimum number and 7 C 8 being the maximum number = 10 3 C3 ∴Required probability = P( A ∪ B) = P( A) + P(B) − P( A ∩ B) 7 6 C C 3 11 = 10 2 + 10 3 − 10 = C3 C3 C3 40

Targ e t E x e rc is e s

1 3 7 + = 8 4 8 3 As the maximum value of P( A ∩ B) = , we get 8 3 1 ≥ P( A) + P(B) − 8 3 11 P( A) + P(B) ≤ 1 + = ⇒ 8 8 ⇒

48. Total 4-digit numbers formed 4 × 4 × 3 × 2 = 96 Each of these 96 numbers are equally likely and mutually exclusive of each other. Now, a number is divisible by 4, if last two digits of the number is divisible by 4.

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Objective Mathematics Vol. 1

21

Hence, we can have 0

4

1

2

2

0

→ first two places can be filled in 3 × 2 = 6 ways → first two places can be filled in 2 × 2 = 4 ways 6 ways →

2

4

→

4 ways

3

2

→

4 ways

4

0

→

6 ways

Total number of ways 30 ways Favourable outcomes 30 5 a ∴Probability = = = = Total outcomes 96 16 b ⇒ b − 2a = 6

Ta rg e t E x e rc is e s

49. Event A = {1, 3, 5}, event B = {3, 6} and event C = {1, 2, 3, 4} ∴ A ∩ B = { 3}, B ∩ C = { 3}, A ∩ C = {1, 3} and A ∩ B ∩ C = { 3} Thus, P (exactly two of A, B and C occur) = P( A ∩ B) + P(B ∩ C ) + P(C ∩ A) − 3P( A ∩ B ∩ C ) 1 1 2 1 = + + − 3× 6 6 6 6 1 a = = 6 b ⇒ a+ b=7

50. Let the number of marbles be 2n, where n is large. Required probability = lim

n→ ∞

n × nC4 n C3 × nC2 × 2n 2n C5 C5

n→ ∞

51. Consider the following events: A = Switch S1 works, B = Switch S 2 works We have, 90 9 P( A) = = 100 10 80 8 and P(B) = = 100 10 The circuit will work, if the current flows in the circuit. This is possible only when atleast one of the two switches S1, S 2 works. Therefore, required probability = P( A ∪ B) = 1 − P( A )P(B ) [Q A and B are independent events] 9  8  = 1 − 1 −  1 −   10   10  1 2 = 1− × 10 10 49 a = = ⇒ b− a=1 50 b

Entrances Gallery 1. Either a girl will start the sequence or will be at second position and will not acquire the last position as well. (3 C1 + 2C1 ) 1 ∴ Required probability = = 5 2 C2

2. Case I 1 odd, 2 even Total number of ways = 2 × 2 × 3 + 1 × 3 × 3 + 1 × 2 × 4 = 29 Case II All 3 odd Number of ways = 2 × 3 × 4 = 24 Favourable ways = 53 53 Required probability = ∴ 3× 5×7 53 = 105

3. Here, 2 x2 = x1 + x3 ⇒ x1 + x3 = even Hence, number of favourable ways = 2C1 ⋅4 C1 + 1C1 ⋅3 C1 = 11 ∴ Required probability =

2

C1 × 4C1 + 1C1 × 3C1 11 = 3 5 7 105 C1 × C1 × C1

4. P(atleast one of them solves correctly)

1128

n × n(n − 1)(n − 2 )(n − 3) n(n − 1)(n − 2 ) × 3! 4! n(n − 1) (5)2 ((2 n − 5)!)2 × × 2! (2 n !)2 4 3 2 n (n − 1) (n − 2 ) (n − 3)((2 n − 5)!)2 × 5 × 5 × 4 × 3 ! = lim n→ ∞ 3 !2 !(2 n !)2 4 3 50 ⋅ n (n − 1) (n − 2 )2 (n − 3) = lim n → ∞ (2 n(2 n − 1)(2 n − 2 )(2 n − 3) (2 n − 4))2 50 25 2 × 10 + 5 = = = 1024 512 512 ∴ a+ b=7 = lim

= 1 − P(none of them solves correctly)  1 1 3 7  235 = 1−  × × ×  =  2 4 4 8 256

5. Let P(E1 ) = x, P(E2 ) = y and P(E3 ) = z (1 − x )(1 − y )(1 − z ) = p x(1 − y )(1 − z ) = α (1 − x )y(1 − z ) = β and (1 − x )(1 − y )( z ) = γ α 1− x p = ⇒ x= ∴ α+ p x α γ Similarly, z= γ+ p p α γ+ p 1+ P(E1 ) α + p γ γ = = = ∴ p γ α+ p P(E3 ) 1+ γ+ p α α αβ 2βγ 5 αγ Also, given = p= ⇒ β= α − 2β β − 3γ α + 4γ   5 αγ   α ⋅ 5 αγ Substituting back α − 2   p =  α + 4γ   α + 4γ  ⇒ αp − 6 pγ = 5 αγ p 6p − =5 ⇒ γ α p 6p ⇒ = 5+ γ α p  6 1 +   α P(E1 ) ∴ = =6 p P(E3 ) 1+ α Then, ⇒

…(i) …(ii) …(iii) …(iv)

P(E ∪ F ) − P(E ∩ F ) =

12. Q A = { 4, 5, 6} and B = {1, 2, 3, 4} ...(i)

∴ Now, ⇒

13. All the three persons have three options to apply for a ...(ii)

house. ∴ Total number of cases = 33 Now, favourable cases = 3 (either all has applied for house 1 or 2 or 3) 3 1 Required probability = 3 = ∴ 9 3

14. Given probabilities of speaking truth are

8. r1, r2 , r3 ∈ {1, 2, 3, 4, 5, 6} r1, r2 and r3 are of the form 3k, 3k + 1, 3k + 2. 3 ! × 2C1 × 2C1 × 2C1 6 × 8 2 ∴ Required probability = = = 6× 6× 6 216 9

9. There seems to be ambiguity in this question. It should be mentioned that boxes are different and one particular box has 3 balls. According to the question, 12 ! × 3 ! 3 C1 × 12C3 2 9 − 3C212C3 9C3 + 3! 3! 6! 3! 312 12 C3 × 2 9 Then, number of ways = 312 11 55  2  =   3  3 1 10. P( A ∪ B) = 6

⇒

5 3 1 1 P(B) = − + = 6 4 4 3

P( A) =

Statement I

3 4

15. The probability that Mr. A selected the loosing horse =

4 3 3 × = 5 4 5

The probability that Mr. A selected the winning horse 3 2 = 1− = 5 5

5 6

and ∴ ⇒

⇒ Common difference is 1.

Total number of cases = 17 Common difference is 2. Total number of cases = 14 Common difference is 3. Total number of cases = 11 Common difference is 4. Total number of cases = 8 Common difference is 5. Total number of cases = 5 Common difference is 6.

P(B) =

and their corresponding probabilities of not speaking truth are 1 1 and P (B ) = P (A) = 5 4 The probability that they contradict each other = P( A) × P(B) + P( A ) × P(B) 4 1 1 3 = × + × 5 4 5 4 1 3 7 = + = 5 20 20

3 1 1 ⇒ P( A ∩ B) = P( A)⋅ P(B) ⇒ × = 4 3 4

11. n (S ) = 20C4

4 and 5

16. Q 0 ≤ P ( A) ≤ 1, 0 ≤ P (B) ≤ 1, 0 ≤ P (C ) ≤ 1

5 3 ⇒ P( A ∪ B) = , P( A) = 6 4 ⇒ P( A ∪ B) = P( A) + P(B) − P( A ∩ B) =

( A ∩ B) = { 4} P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B ) 3 4 1 6 P ( A ∪ B) = + − = = 1 6 6 6 6

Fundamentals of Probability

7. Let P(E ) = e and P(F ) = f 11 25 11 e + f − 2ef = ⇒ 25 2 and P(E ∩ F ) = 25 2 ⇒ (1 − e )(1 − f ) = 25 2 ⇒ 1 − e − f + ef = 25 From Eqs. (i) and (ii), we get 12 7 and e + f = ef = 25 5 On solving, we get 4 3 3 4 or e = , f = e = ,f = 5 5 5 5

21

Total number of cases = 2 Hence, the required probability 17 + 14 + 11 + 8 + 5 + 2 1 = = 20 85 C4

125 91 6 ⋅ 53 = 1− = 216 216 64

Targ e t E x e rc is e s

6. Required probability = 1 −

⇒ and

0 ≤ P ( A) + P (B) + P (C ) ≤ 1 3x + 1 0≤ ≤1 3 1 2 …(i) − ≤x≤ 3 3 1− x 0≤ ≤1 4 …(ii) − 3≤ x ≤1 1− 2x 0≤ ≤1 2 1 1 …(iii) − ≤x≤ 2 2 3x + 1 1 − x 1 − 2 x 0≤ + + ≤1 3 4 2

⇒

0 ≤ 13 − 3 x ≤ 12 1 13 ≤x≤ ⇒ 3 3 From Eqs. (i), (ii), (iii) and (iv), we get 1 1 ≤x≤ 3 2

…(iv)

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Objective Mathematics Vol. 1

21

17. The total number of ways in which numbers can be

two dice, n (S ) = 36

The number of ways in which either players can choose same numbers = 25 25 1 = ∴ Probability that they win a prize = 625 25

Let

Thus, the probability that they will not win a prize in a 1 24 single trial = 1 − = 25 25

18. Since, the probabilities of solving the problem by A, B 1 1 1 and C are , and , respectively. 2 3 4

E = Event of getting score of 5 = {(1, 4), (4, 1), (2, 3), (3, 2)} n (E ) = 4 n (E ) 4 1 P (E ) = = = n (S ) 36 9

⇒ ∴

24. Total number of outcomes in sample space, n (S ) = 6 × 6 = 36 Let E = Event of getting maximum sum of numbers on two dice is 5

∴ Probability that the problem is not solved 1  1   = P ( A ) P(B ) P(C ) = 1 −  1 −  1 −    2 3 

= {(1, 1), (1, 2), (2, 1), (1, 3), (3,1), (2, 2), (1, 4), (4, 1), (2, 3), (3, 2)}

1  4

n (E ) = 10 ∴ Probability that the maximum sum of numbers on two dice is 5 n(E ) 10 5 = = = n(S ) 36 18

1 2 3 1 = × × = 2 3 4 4 Hence, the probability that the problem is solved 1 3 = 1− = 4 4

∴ Probability (the sum of numbers on two dice is more than 5 = 1 − Probability of the maximum sum of numbers on two dice is 5 5 13 = 1− = 18 18

19. Q Total number of ways = 5!

Ta rg e t E x e rc is e s

23. Total number of elements in sample space of throwing

choosed = 25 × 25 = 625

and favourable number of ways = 2 ⋅ 4! 2 ⋅ 4! 2 = ∴Required probability = 5! 5

25. Required probability

20. Given numbers are 1, 2, 3 and 4. Possibilities for unit’s place digit (either 1 or 3) = 2 Possibilities for ten’s place digit = 3 Possibilities for hundred’s place digit = 2

 1 = 12C2 × 10C2 × 8 C2 × 6C2 × 4C2 × 2C2 ×    6 12 ! 10 ! 8! 6! = × × × 10 ! × 2 ! 8 ! × 2 ! 6 ! × 2 ! 4 ! × 2 !

Possibilities for thousand’s place digit = 1

×

∴ Number of favourable outcomes = 2 × 3 × 2 × 1 = 12 Number of numbers formed by 1, 2, 3 and 4 (without repetition) = 4! 12 1 = ∴ Required probability = 4× 3×2 2

7 m + 7 n is divisible by 5, if (i) mis of the form 4 k + 1and n is of the form 4 k − 1or

=

12 ! 2 6 × 612

13

27. Total sample space = { B1B2 , B1G1, G1B2 } Favourable ways = { B1G1, G1B2 } 2 ∴Required probability = 3

(ii) m is of the form 4k + 2 and n is of the form 4 k or

28. A leap year has 366 days in which 52 weeks and 2 days

(iii) mis of the form 4 k − 1and n is of the form 4 k + 1or

are extra. i.e. (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday). So, probability that a leap year contains 53 Sunday =2/7

(iv) m is of the form 4 k and n is of the form 4 k + 2 Thus, number of (m, n ) = 4 × 25 × 25

favourable

Hence, the required probability =

ordered

pairs

4 × 25 × 25 1 = 4 (100 )2

22. Total number of mappings from a set A into itself is n n . 1130

12

C1 × 4C2 × 12C1 × 4C3 52 C5 13 × 6 × 12 × 4 = 52 × 51 × 50 × 49 × 48 5× 4× 3×2 ×1 13 × 6 × 12 × 4 6 = = 52 × 51 × 10 × 49 × 2 4165

26. Required probability =

21. Let l = 7 n + 7 m, then we observe that 71, 7 2 , 7 3 and 7 4

ends in 7, 9, 3 and 1, respectively. Thus, 7 i ends in 7, 9, 3 or 1 according as i is of the form 4k + 1, 4k + 2, 4k − 1 or 4k, respectively. If S is the sample space, then n (S ) = (100 )2

4! 2!  1 × ×  2 ! × 2 ! 2 ! × 0 !  6

12

And the total number of one to one mapping is n !. n! ∴ Required probability = n n

29. In a regular hexagon, two equilateral triangles are possible. ∴Required probability =

6

2 2 2 1 = = = × × 6 5 4 20 10 C3 3×2

8

i.e. S = {(1, 6), (6, 1), (2, 5), (5, 2 ), (3, 4), (4, 3)} ∴ n(S ) = 6 Let E = Event of getting 3 atleast once n(E ) = 2 n(E ) 2 1 = = ∴Required probability = n(S ) 6 3

31. Total number of letters in the word PROBABILITY = 11 Total number of vowels in the word PROBABILITY is 4, i.e. O, A, I, I. Hence, probability that the single letter to be selected 4 at random is vowel = 11

32. We can choose three squares in a diagonal line parallel to BD in the ∆ABD. Hence, three squares in ∆ABD and a diagonal line can be chosen in 3 C3 + 4C3 + 5C3 + 6C3 + 7C3 + 8C3 ways A

35. Since, P(exactly one of A, B occurs) = q (given), we get P ( A ∪ B) − P ( A ∩ B) = q ⇒ p − P ( A ∩ B) = q ⇒ P ( A ∩ B) = p − q ⇒ 1 − P ( A′ ∪ B′ ) = p − q ⇒ P ( A′ ∪ B′ ) = 1 − p + q ⇒ P( A′ ) + P(B′ ) − P ( A′ ∩ B′ ) = 1 + q − p ⇒ P ( A′ ) + P(B′ ) = (1 − p + q ) + [1 − P ( A ∪ B)] = (1 − p + q ) + (1 − p) = 2 − 2 p + q

21 Fundamentals of Probability

8×7 × 10   C2 × 10C1 n! 2 = = Q nCr = 18 18 × 17 × 16  r ! (n − r )! C3 3×2 4 × 7 × 10 × 6 70 = = 18 × 17 × 16 3 × 17 × 4 35 35 = = 6 × 17 102

30. Sum of numbers appearing on the dice is 7.

36. We have two events M and N. N

M

B M–N

N–M

Clearly, P(exactly one of them occurs) = P(M − N ) + P (N − M ) = P(M ) − P(M ∩ N ) + P(N ) − P(M ∩ N ) = P(M ) + P(N ) − 2 P(M ∩ N )

37. Given, P( A ∪ B) = P ( A) + P(B) − P( A) P(B) D

C

Similarly, in ∆BCD, the squares can be chosen parallel to BD in an equal number of ways. Hence, the total number of ways in which three squares can be chosen in a diagonal line parallel to BD is 2 (3 C3 + 4C3 + 5C3 + 6C3 + 7C3 ) + 8C3 Since, BD is common to both the triangles. Similarly, squares can be chosen in a diagonal line parallel to AC and hence the total number of favourable ways = 4 (3 C3 + 4C3 + 5C3 + 6C3 + 7C3 ) + 2 ⋅8 C3 = 392 Hence, the required probability 392 392 × 6 7 = 64 = = C3 64 ⋅ 63 ⋅ 62 744

33. Total number of cards = 52 Total number of ace of hearts = 1 Q Two cards are drawn simultaneously, such that atleast one of them is ace of heart. ∴Required probability 1  n! C × 51C1 1 × 51 × 2  n = 152 = Q Cr =   × r n − r ! ( )! 52 51 C2   2 1 = = 52 26

34. Total number of boys = 10 Total number of girls = 8 Number of students have to be selected at random = 3 If 2 girls and 1 boy are selected, then the required probability

∴

P( A ∪ B)′ = 1 − P ( A ∪ B) = 1 − [P( A) + P(B) − P( A) P(B)] = 1[1 − P( A)] − P(B) [1 − P( A)] = [1 − P( A)][1 − P(B)] = [1 − P( A)] P(B′ ) 2 2 38. Given, P( A′ ) = and P(B ′ ) = 3 7 Since, events A and B are independent. ∴P( A ∩ B) = P ( A)⋅ P(B) = [1 − P( A′ )][1 − P(B ′ )] 2  2 1 5 5  = 1 −  1 −  = × =    3 7  3 7 21

Targ e t E x e rc is e s

M∩N

39. Given, P ( A ∪ B) = 0.83, P( A) = 0.3 and P (B) = 0.6 Q P ( A ∪ B) = P ( A) + P(B) − P ( A ∩ B) ∴ 0.83 = 0.3 + 0.6 − P( A ∩ B) ⇒ P( A ∩ B) = 0.9 − 0.83 = 0.07 Now, P( A)⋅ P(B) = 0.3 × 0.6 = 0.18 ∴ P( A)⋅ P(B) ≠ P ( A ∩ B) Hence, events are dependent.

40. In any number, the last digit can be 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. Therefore, last digit of each number can be chosen in 10 ways. Thus, exhaustive number of ways = 10 n If the last digit is 1, 3, 7 or 9, none of the numbers can be even or end in 0 to 5. Thus, we have a choice of 4 digits, viz. 1,3, 7 or 9 with each of n numbers should end. So, favourable number of ways = 4n Hence,

required probability =

4n  2 =  n  5 10

n

1131

Objective Mathematics Vol. 1

21

41. 8 Red 5 White Urn Total number of ways of selecting 3 balls = 13C3 and number of ways of selecting 3 balls including both colours = 8C2 × 5C1 + 8C1 × 5C2

Hence, probability that balls of both colours are drawn 8 C2 × 5C1 + 8C1 × 5C2 = 13 C3 140 + 80 = 286 220 110 10 = = = 286 143 13

42. Since, there are 52 distinct cards in decks and in two decks each distinct card is 2 in number. Therefore, 2 decks i.e. 104 cards will also contain only 52 distinct cards. ∴Probability that the player gets all distinct cards 52 C26 × (2 C1 )26 52 C26 × 2 26 = = 104 104 C26 C26

Ta rg e t E x e rc is e s

43. Total number of outcomes in sample space, i.e. n(S ) = 6 × 6 = 36 Let E be the event of getting the product perfect square. ∴ E = {(1, 1), (2, 2 ), (4, 1), (1, 4), (3, 3), (4, 4), (5, 5), (6, 6)} ∴ n(E ) = 8 n(E ) 8 2 Hence, the required probability = = = n(S ) 36 9

44. 3 white 5 black Bag

Total number of balls = 8 Number of white balls = 3 3 Required probability = ∴ 8 45. Total number of outcomes = 216 and number of favourable outcomes is equal to 20%, which are given below (1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (1, 5, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6) 20 5 Therefore, the required probability = = 216 54

46. Girls sit side by side, it means both girls are seated together in 2 ! ways. Let E = 4 boys and 2 girls occupy seats side by side n(E ) = 5 !2 ! 5 !2 ! 2 1 Required probability = = = 6! 6 3

1132

47. Total number of cards = 52 Number of king of a heart = 1 1 Required probability = ∴ 52

48. Total number of cases are 2100 . The number of favourable ways = 100C1 + 100C3 + ... + 100C99 = 2100 − 1 = 2 99 2 99 1 Hence, required probability = 100 = 2 2

49. A leap year 366 days in which 52 weeks and 2 days are extra. i.e. (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday). So, probability that a leap year contains only 5 52 Sunday = 7

50. Total number of ways of selecting 4 balls at random = 8C4 Now, the number of ways of selecting 4 balls such that none of the them have same colour = 2C1 × 2C1 × 2C1 × 2C1 Thus, the probability that none of them have same 2 C × 2C1 × 2C1 × 2C1 colour = 1 8 C4 16 × 4 × 3 × 2 × 1 = 8×7 × 6× 5 8 = 35 8 27 Hence, required probability = 1 − = 35 35

51. Q ∴ and

B⊂ A P( A ∩ B ) = P( A) − P(B) P(B) ≤ P( A) C

52. Let S = { BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG} and E = { BBB, BBG, BGB, GBB, GGB, GBG, BGG} n(E ) 7 ∴ P(E ) = = n(S ) 8

53. Let A = Event of getting even number on first die = {(2, 1), ..., (2, 6) (4, 1), ..., (4, 6) (6, 1), ... ,(6, 6)} and B = Event of getting a sum of 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2 )} A ∩ B = {(2, 6), (4, 4), (6, 2 )} ∴ n( A) = 18, n(B) = 5 and n( A ∩ B) = 3 ⇒ n( A ∪ B) = 18 + 5 − 3 = 20 n( A ∪ B) 20 5 P ( A ∪ B) = = = ∴ n(S ) 36 9

SOLVED PAPER 2015 JEE Advanced Paper

1

Section 1 (Maximum Marks : 16) l

l

l

l

This section contains four questions. The answer to each question is a single digit integer ranging from 0 to 9 both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme + 4 If the bubble corresponding to the answer is darkened. 0 In all other cases.

1. Let the curve C be the mirror image of the parabola y 2 = 4x with respect to the line x + y + 4 = 0. If A and B are the points of intersection of C with the line y = − 5, then the distance between A and B is 2. Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. m Then, the value of is n

3. If the normals of the parabola y 2 = 4x drawn at the end points of its latusrectum are tangents to the circle ( x − 3) 2 + ( y + 2) 2 = r 2 , then the value of r 2 is 4. The number of distinct solutions of the equation 5 cos2 2x + cos 4 x + sin 4 x + cos 6 x + sin 6 x = 2 in the 4 interval [ 0, 2π ] is

Section 2 (Maximum Marks : 4) l

l

l

l

This section contains one question. Each question has four options (a), (b), (c) and (d). One or more than one of these four options are correct. For each question, darken the bubble(s) corresponding to all the correct options in the ORS. Marking scheme +4 If only the bubble(s) corresponding to all the correct options are darkened. 0 If none of the bubbles is darkened. −2 In all the other cases.

5. Let P and Q be distinct points on the parabola y 2 = 2x such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of ∆OPQ is 3 2, then which of the following is/are the coordinates of P? (a) (4 , 2 2 )

(b) (9 , 3 2 )

1 1  (c)  ,  4 2

(d) (1, 2 )

Paper

2

Section 1 (Maximum Marks : 16) l

l

l

l

This section contains four questions. The answer to each question is a single digit integer ranging from 0 to 9 both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme +4 If the bubble corresponding to the answer is darkened. 0 in all other cases.

1. Suppose that all the terms of an arithmetic progression are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this AP is 2. The coefficient of x 9 in the (1 + x ) (1 + x 2 ) (1 + x 3 ) ... (1 + x100 ) is

expansion

of

x2 y2 + = 1 are( f1 , 0) 9 5 and ( f 2 , 0), where f1 > 0 and f 2 < 0. Let P1 and P2 be two parabolas with a common vertex at ( 0, 0) with foci at( f1 , 0) and (2 f 2 , 0), respectively. Let T1 be a tangent to P1 which passes through (2 f 2 , 0) and T2 be a tangent to P2 which

passes through( f1 , 0). If m1 is the slope of T1 and m2 is the  1  slope of T2 , then the value of  2 + m22  is  m1   kπ   kπ  4. For any integer k, let α k = cos   + i sin   , where  7  7 12

∑ |α k + 1 − α k |

3. Suppose that the foci of the ellipse

i = −1. The value of the expression

k =1 3

∑ |α 4k − 1 − α 4k − 2 |

k =1

is

Section 2 (Maximum Marks : 16) l

l

l

l

This section contains four questions. Each question has four options (a), (b), (c) and (d). One or more than one of these four options are correct. For each question, darken the bubble(s) corresponding to all the correct options in the ORS. Marking scheme + 4 If only the bubble(s) corresponding to all the correct options are darkened. 0 If none of the bubbles is darkened. − 2 In all other cases.

5. Let S be the set of all non-zero real numbers α such that the quadratic equation αx 2 − x + α = 0 has two distinct real roots x1 and x 2 satisfying the inequality x1 – x 2 < 1. Which of the following interval(s) is/are a subset of S? 1 1  1    1  1 (a)  – , – , 0 (c)  0, ,  (d)   (b)  –      2   5 5 5 5

1  2

 6  4 6. If α = 3sin −1   and β = 3 cos −1   , where the inverse  11  9 trigonometric functions take only the principal values, then the correct option(s) is/are (a) cos β > 0 (b) sin β < 0 (c) cos (α + β ) > 0 (d) cos α < 0

7. Let E1 and E 2 be two ellipses whose centres are at the origin. The major axes of E1 and E 2 lie along the X-axis and Y-axis, respectively. Let S be the circle x 2 + ( y − 1) 2 = 2. The straight line x + y = 3 touches the curves S, E1 and E 2 at P, Q and R, respectively.

2 2 . If e1 and e 2 are the 3 eccentricities of E1 and E 2 respectively, then the correct expression(s) is/are

Suppose that PQ = PR =

43 40 5 (c) | e12 – e22 | = 8 (a) e12 + e22 =

7 2 10 3 (d) e1 e2 = 4 (b) e1 e2 =

8. Consider the hyperbola H: x 2 − y 2 = 1 and a circle S with centre N ( x 2 , 0). Suppose that H and S touch each other at a point P( x1 , y1 ) with x1 > 1 and y1 > 0. The common tangent to H and S at P intersects the X -axis at point M. If ( l , m ) is the centroid of ∆PMN, then the correct expression(s) is/are (a)

dl 1 = 1 − 2 for x1 > 1 dx1 3x1

(c)

dl 1 = 1 + 2 for x1 > 1 dx1 3x1

dm x1 for x1 > 1 = dx1 3( x 2 − 1) 1 dm 1 (d) = for y1 > 0 dy1 3 (b)

Answer with Explanations Paper 1 1. (4) Let P (t 2 , 2t ) be a point on the curve y2 = 4 x, whose image is Q(x , y) on x + y + 4 = 0, then −2(t 2 + 2t + 4 ) x − t2 y − 2t = = 1 1 12 + 12 ⇒

i.e.

y−2 y+ 2 2 2 = − and = x −1 x −1 2 2

⇒

x + y = 3 and x − y = 3

which is tangent to (x − 3)2 + ( y + 2)2 = r2. Y

x = − 2t − 4 and y = − t 2 − 4

y 2 = 4x

Y

L(1, 2)

–4

X′

O P (t 2, 2t)

Normal

X

F

X′

O

y 2 = 4x

Normal

Q –4 y = –5

x+y+4=0 Y′

L′(1, –2)

Mirror image

Now, the straight line y = − 5 meets the mirror image. ∴

Y′

− t − 4 = −5 2

⇒

∴ Length of perpendicular from centre = Radius |3 − 2 − 3| =r ⇒ 12 + 12

t =1 2

⇒ t=±1 Thus, points of intersection of A and B are (− 6, − 5) and (−2, − 5).

∴

B1

B2

B3

B4

5 4

⇒

B5

Out of 5 girls, 4 girls are together and 1 girl is separate. Now, to select 2 positions out of 6 positions between boys …(i) = 6C 2 4 girls are to be selected out of 5 = 5C 4

…(ii)

Now, 2 groups of girls can be arranged in 2!ways. …(iii) Also, the group of 4 girls and 5 boys is arranged in 4 ! × 5! ways . …(iv) Now, total number of ways = 6C 2 × 5C 4 × 2! × 4 ! × 5! [from Eqs. (i), (ii), (iii) and (iv)] ∴

m = 6C 2 × 5C 4 × 2! × 4 ! × 5!

and

n = 5! × 6!

⇒

m 6C 2 × 5C 4 × 2! × 4 ! × 5! = 6! × 5 ! n =

15 × 5 × 2 × 4 ! =5 6 × 5 × 4!

3. (2) End points of latusrectum are (a, ± 2a) i.e. (1, ± 2). Equation of normal at (x1 , y1) is y − y1 y =− 1 x − x1 2a

r2 = 2

4. (8) Here, cos2 2x + (cos 4x + sin 4 x ) + (cos 6x + sin 6 x ) = 2

∴ Distance, AB = (−2 + 6)2 + (−5 + 5)2 = 4

2. (5) Here,

X

(1, 0)

5 cot 2x + [(cos2 x + sin 2 x )2 − 2 sin 2 x cos2 x ] 4

+ (cos2 x + sin 2 x )[(cos2 x + sin 2 x )2 − 3 sin 2 x cos2 x ] = 2 5 cos2 2x + (1 − 2 sin 2 x cos2 x ) + (1 − 3 cos2 x sin 2 x ) = 2 ⇒ 4 5 ⇒ cos2 2x − 5 sin 2 x cos2 x = 0 4 5 5 ⇒ cos2 2x − sin 2 2x = 0 4 4 5 5 5 cos2 2x − + cos2 2x = 0 ⇒ 4 4 4 5 5 ⇒ cos2 2x = 2 4 1 2 cos 2x = ⇒ 2 cos2 2x = 1 ⇒ 2 ⇒ 1 + cos 4 x = 1 ⇒ cos 4 x = 0, as 0 ≤ x ≤ 2π  π 3π 5π 7π 9π 11π 13π 15π  ∴ 4x =  , , , , , , ,  2 2 2  2 2 2 2 2 as ⇒

0 ≤ 4 x ≤ 8π π 3 π 5π 7π 9π 11π 13π 15π   x= , , , , , , ,  8 8 8  8 8 8 8 8

Hence, the total number of solutions is 8.

1138

JEE Advanced Solved Paper 2015

5. (a, d) Since, ∠ POQ = 90° P 2 t1

Y

X′

ar (∆OPQ ) = 3 2

Q t 12 ,

y 2 = 2x

∴

0 0 1 1 2 t1 / 2 t1 1 = ± 3 2 2 2 t2 / 2 t2 1

⇒

1  t12t2 t1t22    =±3 2 − 2 2 2 

X

O (0, 0)

1 ( −4 t1 + 4 t2) = ± 3 2 4 4 t1 + = 3 2 t1

⇒ ⇒ t 22 ,

Y′

Q 2 t2

t1 − 0 t2 − 0 = −1 ⋅ 2 t12 t2 −0 −0 2 2

⇒

⇒

⇒

t12 − 3 2t 1 + 4 = 0

⇒

(t1 − 2 2 ) (t1 − 2 ) = 0

⇒

t1t2 = − 4

…(i)

t1 = 2 or

∴

P (1, 2 ) or

[Q t1 > 0 for P]

2 2

P (4 , 2 2 )

Paper 2 1. (9) Given,

Tangent to P1 passes through (2 f2 , 0) i.e. (−4 , 0). 2 2 ⇒ 0 = −4 m1 + T1 : y = m1x + ∴ m1 m1

S7 6 = and 130 < t7 < 140 S11 11 7 [ 2a + 6d ] 7 (2a + 6d ) 6 2 =6 = ⇒ 11 11 (2a + 10d ) [ 2a + 10d ] 2

⇒ ⇒ Also, ⇒ ⇒

⇒

a = 9d 130 < t7 < 140 130 < a + 6d < 140 130 < 9d + 6d < 140

⇒

…(i)

[from Eq. (i)]

∴

0 = 2m2 −

⇒

m22 = 2 1

∴

m12

(1 + x )(1 + x 2 )(1 + x 3 ) K(1 + x100 ) = Terms having x 9 = [199 ⋅ x 9 , 198 ⋅ x ⋅ x 8 , 198 ⋅ x 2 ⋅ x 7 , 198 ⋅ x 3 ⋅ x 6 , 198 ⋅ x 4 ⋅ x 5 , 197 ⋅ x ⋅ x 2 ⋅ x 6 , 197 ⋅ x ⋅ x 3 ⋅ x 5 ,197 ⋅ x 2 ⋅ x 3 ⋅ x 4]

∴ α k are vertices of regular polygon having 14 sides. Let the side length of regular polygon be a. ∴ α k + 1 − α k = length of a side of the regular polygon = a and α 4k −1 − α 4k − 2 = length of a side of the regular polygon …(ii) =a

T2 Y

12

T1

∑ α k + 1 −α k

∴

(– 4, 0) F1′ (–2, 0) O F1(2, 0)

Y′

k =1 3

=

∑ α 4k −1 −α 4k − 2

X

12 (a) =4 3 (a)

k =1

y 2 = 8x

y2 = –16x

+ m22 = 2 + 2 = 4

…(i)

∴Coefficient of x 9 = 8

F2

…(ii)

kπ   kπ   2kπ   2kπ   + i sin   = cos   + i sin    7  7  14   14 

2. (8) Coefficient of x in the expansion of

X′

4 m2

4. (4) Given, α k = cos 

d=9 9

3. (4)

…(i)

Also, tangent to P2 passes through ( f1, 0) i.e. (2, 0). (−4 ) T2 : y = m2x + ⇒ m2 ⇒

130 < 15d < 140 26 28 [since, d is a natural number] 0 +

–

∴

+

–1/√5

Since, PR = PQ =

1/√5

1  1   α ∈ −∞ , − , ∞  ∪  5  5 

∴

⇒

1− 4α > 0 or

…(ii)

6 4 6 1 6. (b, c, d ) Here, α = 3sin −1   and β = 3cos−1   as >  11

sin

 9

6 π −1  1   > sin   =  11  2 6

−1 

6 π  >  11 2

−1 

∴

α = 3 sin

Now,

 4 β = 3 cos−1    9

∴

Now, equation of tangent at Q on ellipse E1 is x⋅ 5 y⋅ 4 + =1 a2 ⋅ 3 b2 ⋅ 3 On comparing with x + y = 3, we get

e12 = 1 −

∴

∴ 7. (a, b) Here, E1: x

2

c2

+

2

y

d2

x2 a

2

+

y2 b2

e22 = 1 −

= 1, (c < d ) and

Also,

S : x 2 + ( y − 1)2 = 2

e12 ⋅ e22 =

a2 2

b

= 1−

1 7 = 8 8

1 7 43 + = 5 8 40 1 7 27 e12 − e22 = − = 5 8 40

8. (a, b, d) Y

,y

R

M

X′

P

(y – +

2

(x P

2 1) 2 =

Y

(–1, 0)

(0, 1)

(1, 0)

X N(x2, 0)

Q

S

:x

…(ii)

7 7 ⇒ e1 e2 = 40 2 10

as tangent to E1 , E2 and S is x + y = 3.

X′

…(i)

e12 + e22 =

and

= 1, (a > b)

4 1 = 5 5

a2 = 1, b2 = 8

Now,

cos (α + β ) > 0

= 1−

a2

On comparing with x + y = 3, we get

∴ 3π . 2

b2

Also, equation of tangent at R on ellipse E2 is x ⋅1 y⋅ 8 + =1 a2 ⋅ 3 b2 ⋅ 3

cosβ < 0 and sinβ < 0

Now, α + β is slightly greater than

E2 :

⇒ cosα < 0

 4 β = 3 cos−1   > π  9

∴

2

 1 8 R= ,   3 3

a2 = 5 and b2 = 4

4 1  1 π  4 < ⇒ cos−1   > cos−1   =  2 3  9 9 2

As

11

1 8  and  x = , y =   3 3

 5 4 Q =  ,  and  3 3

∴

From Eqs. (i) and (ii), we get  1 −1   1 1 α ∈ − ,  ∪  ,   2 5   5 2

⇒

5 4  x = , y =   3 3

⇒

 1 1 α ∈ − ,   2 2

2

2 2 . Thus, by parametric form, 3

x −1 y− 2 2 2 = =± 3 −1/ 2 1/ 2

…(i)

D >0

Also,

P = (1, 2)

)

⇒

2

1

or

Let the point of contact of tangent be (x1 , y1 ) to S. ∴ x ⋅ x1 + y⋅ y1 − ( y + y1 ) + 1 = 2 or x x1 + y y1 − y = (1 + y1 ), same as x + y = 3. x1 y − 1 1 + y1 = 1 = ⇒ 1 1 3 i.e. x1 = 1 and y1 = 2

1

⇒

1139

O

x2 – y2 = 1

X E1

x+y=3

Y′

Equation of family of circles touching hyperbola at (x1 , y1 ) is E2 Y′

(x − x1 )2 + ( y − y1 )2 + λ( x x1 − y y1 −1)= 0 Now, its centre is (x2 , 0).

1140 ∴ ⇒ and ∴

JEE Advanced Solved Paper 2015  − (λx1 − 2x1 ) – (−2 y1 − λy1 )  , = (x2 , 0)   2 2 2 y1 + λy1 = 0 ⇒ λ = − 2 2x1 − λx1 = 2x2 ⇒ x2 = 2x1 P (x1 , x12 − 1) and N (x2 , 0) = (2x1 , 0)

1  As tangent intersect X-axis at M  ,0 . x  Centroid of ∆PMN = (l , m) 1     3x1 + 0 0 y + + x 1 1  = (l , m)  , ⇒ 3 3       ⇒

1 3x1 + x1 l= 3

On differentiating w.r.t. x1, we get 1 3− 2 dl x1 = dx1 3 ⇒

dl 1 = 1 − 2 , for x1 > 1 and m = dx1 3x1

x12 − 1 3

On differentiating w.r.t. x1, we get dm x1 2x1 , for x1 > 1 = = dx1 2 × 3 x 2 − 1 3 x 2 − 1 1 1 Also,

m=

y1 3

On differentiating w.r.t. y1, we get dm 1 = , for y1 > 0 dy1 3

JEE Main 2016 1 1. If f ( x ) + 2 f æç ö÷ = 3x, è xø

10. Two sides of a rhombus are along the lines, x - y + 1 = 0 and 7 x - y - 5 = 0. If its diagonals intersect at (- 1, - 2), then which one of the following is a vertex of this rhombus?

x ¹ 0 and S = {x Î R : f ( x ) = f ( - x )}; then S (a) is an empty set (b) contains exactly one element (c) contains exactly two elements (d) contains more than two elements

(a) (- 3, - 9)

2 + 3i sin q is purely imaginary, is 2. A value of q for which 1 - 2i sin q (a)

p 3

(b)

æ 3ö (c) sin-1 ç ÷ è 4 ø

p 6

1 (d) sin-1 æç ö÷ è 3ø

(b) - 4

(c) 6

(b) 59th

(c) 52nd

(d) 58th

æ 2 4ö 5. If the number of terms in the expansion of ç1 - + 2 ÷ , è x x ø x ¹ 0, is 28, then the sum of the coefficients of all the terms in this expansion, is (b) 2187

(c) 243

(d) 729

6. If the 2nd, 5th and 9th terms of a non-constant AP are in GP, then the common ratio of this GP is (a)

8 5

(b)

4 3

(c) 1

(d)

7 4

2

2

2

æ1 3 ö + æ 2 2 ö + æ 3 1 ö + 4 2 + æ 4 4 ö + K, ç ÷ ç ÷ ç ÷ ç ÷ è 5ø è 5ø è 5ø è 5ø then m is equal to (a) 102

(b) 101

8. Let p = lim (1 + tan 2 x ® 0+

(a) 2

(c) 100

æ ( n + 1)( n + 2) K 3n ö 9. lim ç ÷ n ®¥ è ø n 2n (a)

18 e4

(b)

27 e2

is

16 m, 5

(d) 99

x )1/ 2x , then log p is equal to

(b) 1

(c)

1 2

(d)

1 4

1/ n

is equal to (c)

9 e2

12. if one of the diameters of the circle, given by the equation, x 2 + y 2 - 4x + 6 y - 12 = 0, is a chord of a circle S , whose centre is at ( -3, 2), then the radius of S is (b) 5 3

(c) 5

(d) 10 2

13. Let P be the point on the parabola, y = 8x, which is at a minimum distance from the centre C of the circle, x 2 + ( y + 6) 2 = 1. Then, the equation of the circle, passing through C and having its centre at P is (b) x2 + y2 - x + 4 y - 12 = 0 x (c) x2 + y2 - + 2 y - 24 = 0 4 (d) x2 + y2 - 4 x + 9 y + 18 = 0

14. The eccentricity of the hyperbola whose length of the latusrectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is (a)

4 3

(b)

4 3

(c)

2 3

(d) 3

15. The distance of the point (1, - 5, 9) from the plane x - y + z = 5 measured along the line x = y = z is

7. If the sum of the first ten terms of the series 2

10 7 (d) æç - , - ö÷ è 3 3ø

(a) x2 + y2 - 4 x + 8 y + 12 = 0 n

(a) 64

1 8 (c) æç , - ö÷ è3 3ø

11. The centres of those circles which touch the circle, x 2 + y 2 - 8x - 8 y - 4 = 0, externally and also touch the X -axis, lie on

(a) 5 2

(d) 5

4. If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary, then the position of the word SMALL is (a) 46th

(b) (- 3, - 8)

(a) a circle (b) an ellipse which is not a circle (c) a hyperbola (d) a parabola

3. The sum of all real values of x satisfying the equation 2 ( x 2 - 5x + 5) x + 4x - 60 = 1 is (a) 3

(Solved)

(d) 3log 3 - 2

(a) 3 10

(b) 10 3

(c)

10 3

(d)

20 3

16. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true? (a) 3a2 - 26a + 55 = 0 2

(c) 3a - 34a + 91 = 0

(b) 3a2 - 32 a + 84 = 0 (d) 3a2 - 23a + 44 = 0

17. Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E 2 is the event that die B shows up two and E 3 is the event that the sum of numbers on both dice is odd, then which of the following statements is not true? (a) E1 and E2 are independent (b) E2 and E3 are independent (c) E1 and E3 are independent (d) E1, E2 and E3 are independent

2

JEE Main and Advanced Solved Papers 2016

18. If 0 £ x < 2p, then the number of real values of x, which satisfy the equation cos x + cos 2x + cos 3x + cos 4x = 0, is (a) 3 (c) 7

(b) 5 (d) 9

19. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 min from A in the same

direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then, the time taken (in minutes) by him, from B to reach the pillar, is (a) 6

(b) 10

1. A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include atmost one boy, the number of ways of selecting the team is (Single Option Correct) (a) 380

(b) 320

(c) 260

(d) 95

p p < q < - . Suppose a 1 and b 1 are the roots of the 6 12 equation x 2 - 2x secq + 1 = 0 , and a 2 and b 2 are the roots

2. Let -

2

of the equation x + 2x tan q - 1 = 0. If a 1 > b 1 and a 2 > b 2 , then a 1 + b 2 equals (a) 2(sec q - tan q) (c) -2 tanq

(Single Option Correct)

(b) 2sec q (d) 0

p 3. Let S = ìí x Î ( - p , p ): x =/ 0, ± üý. The sum of all distinct 2þ î solutions of the equation 3 sec x + cosec x + 2(tan x - cot x ) = 0 in the set S is equal to (Single Option Correct) 7p (a) 9

2p (b) 9

(c) 0

(d)

5p 9

4. Let RS be the diameter of the circle x 2 + y 2 = 1, where S is the point (1, 0) . Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then, the locus of E passes through the point(s) (One or More Than One Option Correct) 1 1 ö (a) æç , ÷ è3 3ø

1 1 (b) æç , ö÷ è4 2ø

1 1 ö 1ö æ1 (c) æç , ÷ (d) ç , - ÷ è4 2ø è3 3ø

5. The circle C1 : x 2 + y 2 = 3 with centre at O intersects the parabola x 2 = 2 y at the point P in the first quadrant. Let the tangent to the circle C1 at P touches other two circles

(d) 5

20. The Boolean expression ( p Ù ~ q ) Ú q Ú (~ p Ù q ) is equivalent to (a) ~ p Ù q (c) p Ú q

JEE Advanced 2016 Paper I

(c) 20

(b) p Ù q (d) p Ú ~ q

(Solved)

C 2 and C 3 at R 2 and R 3 , respectively. Suppose C 2 and C 3 have equal radii 2 3 and centres Q2 and Q3 , respectively. If Q2 and Q3 lie on the Y-axis, then (One or More Than One Option Correct) (a) Q2Q3 = 12

(b) R 2 R 3 = 4 6

(c) area of the DOR 2 R 3 is 6 2 (d) area of the DPQ2Q3 is 4 2

6. Let m be the smallest positive integer such that the in the expansion of coefficient of x2 2 3 (1 + x ) + (1 + x ) + K + (1 + x ) 49 + (1 + mx ) 50 is ( 3n + 1) 51

C 3 for some positive integer n. Then, the value of n is (Single Option Correct)

-1 + 3i , where i = -1, and r, s Î {1, 2, 3}. Let 2 é ( - z ) r z 2s ù P = ê 2s ú and I be the identity matrix of order 2. zr û ë z

7. Let z =

Then, the total number of ordered pairs ( r, s ) for which P 2 = - I is (Single Digit Integer) x 2 sin (bx ) 8. Let a , b Î R be such that lim = 1. Then, x ® 0 ax - sin x (Single Digit Integer) 6 (a + b ) equals

Paper II 13

1. The value of

1 is equal k p ö æ p kp ö ( 1 ) p k = 1 sin æ ÷ sin ç + ÷ ç + è4 6 ø è4 6ø

å

to (a) 3 -

(Single Option Correct) 3

(b) 2(3 -

3)

(c) 2( 3 - 1)

(d) 2(2 +

3)

2. Let bi > 1 for i = 1, 2, ... , 101. Suppose log e b1 , log e b2 , ... , log e b101 are in AP with the common difference log e 2 . Suppose a1 , a 2 , ... , a101 are in AP, such that a1 = b1 and If and a 51 = b51 . t = b1 + b2 + ... + b51 s = a1 + a 2 + ... + a 51 , then (Single Option Correct)

(a) s > t and a101 > b101 (c) s < t and a101 > b101 p/ 2

x 2 cos x

-p/ 2

1+ ex

3. The value of ò (a)

p2 -2 4

(b) s > t and a101 < b101 (d) s < t and a101 < b101

(b)

p2 +2 4

(c) the X-axis for a ¹ 0, b = 0 (d) the Y-axis for a = 0, b ¹ 0

Paragraph 1

dx is equal to (Single Option Correct) (c) p 2 - e - p / 2 (d) p 2 + e p / 2 x

é nö æ æ ê n n ( x + n ) ç x + ÷ ... ç x + è 2ø è 4. Let f ( x ) = lim ê n ® ¥ê æ n2 ö æ 2 2 2 2 ÷ ... ç x ê n ! (x + n ) ç x + 4ø è è ë

ùn nö ú ÷ nø ú , n2 ö ú + 2÷ú n øû

for all x = 0. Then (One or More Than One Option Correct) 1 1 2 (a) f æç ö÷ ³ f(1) (b) f æç ö÷ £ f æç ö÷ (c) f¢ (2 ) £ 0 è2 ø è 3ø è 3ø

(d)

f ¢(3) f ¢(2 ) ³ f(3) f(2 )

5. Let P be the point on the parabola y 2 = 4x, which is at the shortest distance from the centre S of the circle x 2 + y 2 - 4x - 16 y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then, (a) SP = 2 5

(One or More Than One Option Correct)

(b) SQ : QP = ( 5 + 1) : 2 (c) the x-intercept of the normal to the parabola at P is 6 1 (d) the slope of the tangent to the circle at Q is 2

6. Let

a, b Î R

and

a 2 + b 2 ¹ 0.

Suppose

ü ì 1 , t Î R , t ¹ 0ý, where i = - 1. If S = í z ÎC : z = a + i bt þ î z = x + iy and z Î S , then ( x, y ) lies on (One or More Than One Option Correct) 1 1 and centre æç , 0ö÷ for a > 0, b ¹ 0 (a) the circle with radius è2 a ø 2a 1 1 ö and centre æç (b) the circle with radius , 0÷ for a < 0, è 2a ø 2a b¹0

Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1 winning, 1 1 1 drawing and losing a game against T2 are , and , 2 6 3 respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2 , respectively, after two games. 7. P ( X > Y ) is (a)

1 4

æ 1ö ...(i) 1. (c) We have, f (x ) + 2 f ç ÷ = 3x, x ¹ 0 è xø 1 On replacing x by in the above equation, we get x 3 æ 1ö f ç ÷ + 2 f (x ) = è xø x æ 1ö 3 ...(ii) Þ 2 f (x ) + f ç ÷ = è xø x On multiplying Eq. (ii) by 2 and subtracting Eq. (i) from Eq. (ii), we get æ 1ö 6 4 f (x ) + 2 f ç ÷ = è xø x

5 12

(c)

1 2

(d)

7 12

(b)

1 3

(c)

13 36

(d)

1 2

8. P ( X = Y ) is (a)

11 36

Paragraph 2 Let F1 ( x1 , 0) and F2 ( x 2 , 0), for x1 < 0 and x 2 > 0, be the y2 x2 foci of the ellipse + = 1. Suppose a parabola having 9 8 vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. 9. The orthocentre of DF1 MN is 9 (a) æç - , 0ö÷ è 10 ø

2 (b) æç , 0ö÷ è3 ø

9 (c) æç , 0ö÷ è 10 ø

2 (d) æç , 6 ö÷ è3 ø

10. If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the X-axis at Q, then the ratio of area of MF1 NF2 is (a) 3 : 4

SOLUTIONS

JEE Main

(b)

DMQR to area of the quadrilateral

(b) 4 : 5

(c) 5 : 8

(d) 2 : 3

æ 1ö f (x ) + 2 f ç ÷ = 3x è xø 6 3 f (x ) = - 3x x 2 f (x ) = - x Þ x Now, consider f (x ) = f (- x ) 2 2 4 - x = - + x Þ = 2x Þ 2x 2 = 4 Þ x x x Þ

x2 = 2 Þ x = ±

2

Hence, S contains exactly two elements.

4

JEE Main and Advanced Solved Papers 2016

2. (d) Let z =

2 + 3i sin q is purely imaginary. Then, we have 1 - 2i sin q

Re(z) = 0 2 + 3i sin q 1 - 2i sin q (2 + 3i sin q) (1 + 2i sin q) = (1 - 2i sin q) (1 + 2i sin q)

z=

Now, consider

=

2 + 4 i sin q + 3i sin q + 6i 2 sin 2 q 12 - (2i sin q) 2

=

2 + 7i sin q - 6 sin 2 q 1 + 4 sin 2 q

=

2 - 6 sin 2 q 7 sin q +i 1 + 4 sin 2 q 1 + 4 sin 2 q

Re(z) = 0 2 - 6 sin 2 q \ =0 1 + 4 sin 2 q Q

Þ Þ

When x = 3, x 2 + 4 x - 60 = 9 + 12 - 60 = - 39, which is not an even integer. Thus, in this case, we get x = 2. Hence, the sum of all real values of x = - 10 + 6 + 4 + 1 + 2 = 3 4. (d) Clearly, number of words start with A =

4! = 12 2!

Number of words start with L = 4 ! = 24 4! Number of words start with M = = 12 2! 3! Number of words start with SA = = 3 2! Number of words start with SL = 3! = 6 Note that, next word will be “SMALL”. Hence, the position of word “SMALL” is 58th. 5. (d) Clearly, number of terms in the expansion of n

1 2 = 6 sin q Þ sin q = 3 1 1 ö -1 æ -1 æ 1 ö sin q = ± Þ q = sin ç ± ÷ = ± sin ç ÷ è è 3ø 3 3ø 2

2

3. (a) Given, (x 2 - 5x + 5)x

2

+ 4x - 60

=1

Clearly, this is possible when I. x 2 + 4 x - 60 = 0 and x 2 - 5x + 5 ¹ 0 or II. x 2 - 5x + 5 = 1 or III. x 2 - 5x + 5 = - 1 and x 2 + 4 x - 60 = Even integer. 2

Case I

When x + 4 x - 60 = 0

Þ

x 2 + 10x - 6x - 60 = 0

Þ x (x + 10) - 6(x + 10) = 0 Þ (x + 10) (x - 6) = 0 Þ x = - 10 or x = 6 Note that, for these two values of x, x 2 - 5x + 5 ¹ 0 Case II When

x 2 - 5x + 5 = 1

Þ

x 2 - 5x + 4 = 0

Þ

x2 - 4 x - x + 4 = 0

Þ Þ

x (x - 4 ) - 1 (x - 4 ) = 0 (x - 4 ) (x - 1) = 0

Þ

x = 4 or x = 1

Case III When

x 2 - 5x + 5 = - 1

Þ

x 2 - 5x + 6 = 0

Þ

x 2 - 2x - 3x + 6 = 0

Þ Þ Þ

x (x - 2) - 3 (x - 2) = 0 (x - 2) (x - 3) = 0 x = 2 or x = 3

Now, when x = 2, x 2 + 4 x - 60 = 4 + 8 - 60 = - 48, which is an even integer.

(n + 2) (n + 1) 2 4ö æ or ç1 - + 2 ÷ is è x x ø 2

n+ 2

C 2.

[assuming (n + 2) (n + 1) = 28 2

\

1 1 and 2 distinct] x x

Þ (n + 2) (n + 1) = 56 = (6 + 1) (6 + 2) Þ n = 6 Hence, sum of coefficients = (1 - 2 + 4 )6 = 36 = 729 1 1 and 2 are functions of same variables, therefore x x number of dissimilar terms will be 2n + 1, i.e. odd, which is not possible. Hence, it contains error.

Note As

6. (b) Let a be the first term and d be the common difference. Then, we have a + d, a + 4 d, a + 8 d in GP, i.e. (a + 4 d ) 2 = (a + d ) (a + 8 d ) Þ

a 2 + 16 d 2 + 8ad = a2 + 8ad + ad + 8 d 2 8 d 2 = ad

Þ Þ

[Q d ¹ 0] 8d = a a + 4 d 8 d + 4 d 12 d 4 Now, common ratio, r= = = = a+ d 8d + d 9d 3

7. (b) Let S 10 be the sum of first ten terms of the series. Then, we have 2 2 2 2 æ 3ö æ 2ö æ 1ö æ 4ö S 10 = ç1 ÷ + ç 2 ÷ + ç 3 ÷ + 4 2 + ç 4 ÷ + ... to 10 terms è 5ø è 5ø è 5ø è 5ø 2

2

2

2

æ 8ö æ 24 ö æ 12 ö æ 16 ö = ç ÷ + ç ÷ + ç ÷ + 4 2 + ç ÷ + ... to 10 terms è 5ø è 5ø è 5ø è 5ø 1 2 2 2 2 = 2 (8 + 12 + 16 + 20 + 24 2 + ... to 10 terms) 5 42 2 = 2 (2 + 32 + 4 2 + 52 + ... to 10 terms) 5 42 2 = 2 (2 + 32 + 4 2 + 52 + ... + 112 ) 5 16 2 = ((1 + 22 + ... + 112 ) - 12 ) 25 16 æ 11 × (11 + 1) (2 × 11 + 1) ö = - 1÷ ç ø 25 è 6

5

JEE Main and Advanced Solved Papers 2016 = Þ

16 16 ´ 505 (506 - 1) = 25 25 16 16 m= ´ 505 Þ m = 101 5 25

8. (c) Given,

p = lim (1 + tan

2

x)

x ® 0+

tan 2 x 2x x ® 0+ lim

=e \

log p = log

1 e2

=e

1 2x

-1=

Then,

(1 form) 2

=

Þ

x + 1 = - 2 and y = - 4 - 2

Þ

x = - 3 and y = - 6

Hence, coordinates of C = (- 3, - 6)

¥

æ tan x ö 1 lim ç ÷ 2 x ® 0 + çè x ÷ø

1 e2

1 = 2 1

Note that, vertices B and D will satisfy x - y + 1 = 0 and 7x - y - 5 = 0, respectively. Since, option (c) satisfies 7x - y - 5 = 0, therefore coordinate of æ 1 - 8ö vertex D is ç , ÷. è3 3 ø 11. (d) Given equation of circle is x 2 + y2 - 8x - 8 y - 4 = 0, whose centre is C (4 , 4 ) and radius

æ ( n + 1) × ( n + 2 ) K (3 n) ö n 9. (b) Let l = lim ç ÷ ø n®¥ è n2 n

= 4 2 + 4 2 + 4 = 36 = 6

1

æ (n + 1) × (n + 2) ... (n + 2n) ö n = lim ç ÷ ø n® ¥ è n2n æ æ n + 1ö æ n + 2 ö æn+ = lim ç ç ÷ç ÷Kç è n n ® ¥ èè n ø è n ø

1 2n ö ö n

÷÷ øø

Taking log on both sides, we get 1é 1ö æ 2ö æ 2n öü ù ìæ log l = lim ê log í ç1 + ÷ ç1 + ÷ ... ç1 + ÷ý ú è ø è ø è n® ¥n n n n øþ û î ë 1 Þ log l = lim n® ¥n 1ö 2ö 2n ö ù é æ æ æ ê log çè1 + n ÷ø + log çè1 + n ÷ø + ... + log çè1 + n ÷ø ú û ë 2 1 2n r æ ö Þ log l = lim å log ç1 + ÷ Þ log l = ò log (1 + x ) dx 0 è n® ¥n nø r=1 2

ù é 1 Þ log l = ê log (1 + x ) × x - ò × x dx ú 1+ x û0 ë 2 x 1 1 + dx Þ log l = [log (1 + x ) × x ]20 - ò 0 1+ x 2æ 1 ö Þ log l = 2 × log 3 - ò ç1 ÷ dx 0 è 1+ xø

Þ log l = 2 × log 3 - [ x - log 1 + x ]20 Þ log l = 2 × log 3 - [ 2 - log 3 ] Þ log l = 3 × log 3 - 2 Þ log l = log 27 - 2 27 \ l = elog 27 - 2 = 27 × e- 2 = 2 e

Let the centre of required circle be C 1 (x , y). Now, as it touch the X -axis, therefore its radius = y . Also, it touch the circle x 2 + y2 - 8x - 8 y - 4 = 0, therefore CC 1 = 6 + y (x - 4 )2 + ( y - 4 )2 = 6 + y

Þ

x 2 + 16 - 8x + y2 + 16 - 8 y

Þ

= 36 + y2 + 12 y 2

x - 8x - 8 y + 32 = 36 + 12 y

Þ

x 2 - 8x - 8 y - 4 = 12 y

Þ

Case I If y > 0, then we have x 2 - 8x - 8 y - 4 = 12 y Þ x 2 - 8x - 20 y - 4 = 0 Þ

x 2 - 8x - 4 = 20 y

Þ

(x - 4 )2 - 20 = 20 y (x - 4 ) 2 = 20 ( y + 1), which is a parabola.

Þ

Case II If y< 0, then we have x 2 - 8x - 8 y - 4 = - 12 y Þ x - 8x - 8 y - 4 + 12 y = 0 Þ x 2 - 8x + 4 y - 4 = 0 Þ x 2 - 8x - 4 = - 4 y Þ (x - 4 ) 2 = 20 - 4 y 2

Þ (x - 4 )2 = - 4 ( y - 5), which is again a parabola. 12. (b) Given equation of a circle is x 2 + y2 - 4 x + 6 y - 12 = 0, whose centre is (2, - 3) and radius = 2 2 + (- 3) 2 + 12 = 4 + 9 + 12 = 5 Now, according to given information, we have the following figure. S

C (x, y) A (–3, 2)

(–1, –2)

7x –

y–5

=0

10. (c) As the given lines x - y + 1 = 0 and 7 x - y - 5 = 0 are not parallel, therefore they represent the adjacent sides of the rhombus. On solving x - y + 1 = 0 and 7x - y - 5 = 0, we get x = 1 and y = 2. Thus, one of the vertex is A(1, 2). D

x+1 y+ 2 and - 2 = 2 2

A x–y+1=0 (1, 2)

O (2,–3)

B

Let the coordinate of point C be (x , y).

B

C

x 2 + y2 - 4 x + 6 y - 12 = 0

16. (b) We know that, if x1, x2 , ..., xn are n observations, then their standard deviation is given by

Clearly, AO ^ BC , as O is mid-point of the chord. Now, in DAOB, we have

1 2 æ S xi ö S xi - ç ÷ è n ø n

OA = (- 3 - 2) 2 + (2 + 3) 2 = 25 + 25 = 50 = 5 2

We have, (3.5)2 =

and

OB = 5

\

AB = OA 2 + OB 2 = 50 + 25 = 75 = 5 3

Þ

13. (a) Centre of circle x 2 + ( y + 6)2 = 1 is C (0, - 6). Þ

2

Let the coordinates of point P be (2t , 4 t ). Now, let Þ

D = CP = (2t 2 )2 + (4 t + 6)2 4

Þ

2

D = 4 t + 16 t + 36 + 48 t

Þ

Squaring on both sides Þ D 2 (t ) = 4 t 4 + 16 t 2 + 48 t + 36 Let

Þ

F (t ) = 4 t 4 + 16 t 2 + 48 t + 36

CP = 22 + (- 4 + 6)2 = 4 + 4 = 2 2

14. (c) We have, 2b2 = 8 and 2b = ae Þ b2 = 4 a and 2b = ae a Consider, 2b = ae Þ 4 b2 = a2e2

Þ

4 a2 (e2 - 1) = a2e2 4 e2 - 4 = e2 2 3 e2 = 4 Þ e = 3

[Q a ¹ 0] [Q e > 0]

15. (b) Equation of line passing through the point (1, - 5, 9) and parallel to x = y = z is x -1 y+ 5 z - 9 (say) = = =l 1 1 1 Thus, any point on this line is of the form (l + 1, l - 5, l + 9). Now, if P(l + 1, l - 5, l + 9) is the point of intersection of line and plane, then (l + 1) - (l - 5) + l + 9 = 5 Þ l + 15 = 5 Þ l = - 10 \Coordinates of point P are (- 9, - 15, - 1). Hence, required distance = (1 + 9)2 + (- 5 + 15)2 + (9 + 1)2 = 102 + 102 + 102 = 10 3

49 4 + 9 + a2 + 121 æ 16 + a ö -ç = ÷ è 4 ø 4 4

2

2

49 134 + a2 256 + a2 + 32a = 4 4 16 49 4 a2 + 536 - 256 - a2 - 32a = 4 16 49 ´ 4 = 3a2 - 32a + 280 3a2 - 32a + 84 = 0

E2 = {(1, 2),(2, 2),(3, 2),(4 , 2),(5, 2),(6, 2)} E3 = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)} 6 1 P (E1 ) = = , Þ 36 6 6 1 P (E2 ) = = 36 6 18 1 and P (E3 ) = = 36 2 Now, P (E1 Ç E2 ) = P (getting 4 on die A and 2 on die B) 1 = = P (E1 ) × P (E2 ) 36 (getting 2 on die B and sum of numbers P (E2 Ç E3 ) = P on both dice is odd) 3 = = P (E2 ) × P (E3 ) 36 (getting 4 on die A and sum of numbers P (E1 Ç E3 ) = P on both dice is odd) 3 = = P (E1 ) × P (E3 ) 36 and P (E1 Ç E2 Ç E3 ) = P [getting 4 on die A, 2 on die B and sum of numbers is odd] = P (impossible event) = 0 Hence, E1, E2 and E3 are not independent. and

Hence, the required equation of circle is (x - 2)2 + ( y + 4 )2 = (2 2 )2 2 Þ x + 4 - 4 x + y2 + 16 + 8 y = 8 Þ x 2 + y2 - 4 x + 8 y + 12 = 0

Þ Þ

(22 + 32 + a2 + 112 ) æ 2 + 3 + a + 11ö -ç ÷ è ø 4 4

17. (d) Clearly, E1 = {(4 ,1),(4 , 2),(4 , 3),(4 , 4 ),(4 , 5),(4 , 6)}

For minimum, F ¢ (t ) = 0 Þ 16 t 3 + 32t + 48 = 0 Þ t 3 + 2t + 3 = 0 Þ (t + 1) (t 2 - t + 3) = 0 Þ t = -1 Thus, coordinate of point P are (2, - 4 ). Now ,

2

18. (c) Given equation is cos x + cos 2x + cos 3x + cos 4 x = 0 Þ (cos x + cos 3x ) + (cos 2x + cos 4 x ) = 0 Þ 2 cos 2x cos x + 2 cos 3x cos x = 0 Þ 2 cos x (cos 2x + cos 3x ) = 0 5x xö æ 2 cos x ç 2 cos cos ÷ = 0 Þ è 2 2ø 5x x Þ cos x × cos × cos = 0 2 2 Þ cos x = 0

7

JEE Main and Advanced Solved Papers 2016 or Now, Þ

Þ Þ and Þ Þ Hence,

5x x = 0 or cos = 0 2 2 cos x = 0 p 3p [Q 0 £ x < 2p] x= , 2 2 5x cos =0 2 5x p 3p 5p 7p 9p 11p , ..., = , , , 2 2 2 2 2 2 2 p 3p 7p 9p [Q 0 £ x < 2p] x= , , p, , 5 5 5 5 x cos = 0 2 x p 3p 5p = , , , ... 2 2 2 2 [Q 0 £ x < 2p] x=p p 3p p 3p 7p 9p x= , , p, , , , 2 2 5 5 5 5

cos

19. (d) According to given information, we have the following figure D

Pillar h

A

30º x

B

60º y

From Eqs. (i) and (ii), x+ y = 3 y 3 Þ x +y = 3y Þ x - 2y = 0 x y= Þ 2 Q Speed is uniform. \ Distance y will be cover in 5 min. Q Distance x covered in 10 min. x \Distance will be cover in 5 min. 2 20. (c) Consider, ( p Ù ~ q) Ú q Ú (~ p Ù q) º [( p Ù ~ q) Ú q ] Ú (~ p Ù q) º [( p Ú q) Ù (~ q Ú q)] Ú (~ p Ù q) º [( p Ú q) Ù t ] Ú (~ p Ù q) º ( p Ú q) Ú (~ p Ù q) º ( p Ú q Ú ~ p) Ù ( p Ú q Ú q) º (q Ú t ) Ù ( p Ú q) º t Ù ( p Ú q) º pÚ q

Paper I 1. (a) We have, 6 girls and 4 boys. To select 4 members (atmost one boy) i.e. (1 boy and 3 girls) or (4 girls) = 6 C 3 ×4 C 1 + 6C 4 …(i) Now, selection of captain from 4 members = 4C 1

…(ii)

\ Number of ways to select 4 members (including the selection of a captain, from these 4 members) = ( 6C 3 ×4 C 1 + 6C 4 ) 4C 1 = (20 ´ 4 + 15) ´ 4 = 380 2. (c) Here, x 2 - 2x secq + 1 = 0 has roots a 1 and b 1. 2 sec q ± 4 sec2 q - 4 2 sec q ± 2 |tan q | = 2 2´1 æ p pö Since, q Î ç - ,- ÷ , è 6 12 ø 2 sec q m 2 tan q i.e. q Î IV quadrant = 2 [as a 1 > b 1] \ a 1 = sec q - tan q and b 1 = sec q + tan q and x 2 + 2x tan q - 1 = 0 has roots a 2 and b 2 .

\

a1 , b1 =

i.e. \ and Thus,

C

Now, from DACD and DBCD, we have h h and tan 60° = tan 30° = x+ y y x+ y h= Þ 3 and h= 3 y

JEE Advanced

4 tan 2 q + 4 2 a 2 = - tan q + sec q b 2 = - tan q - sec q a 1 + b 2 = -2 tan q

a2 , b2 =

-2 tan q ±

[as a 2 > b 2]

3. (c) Given, 3 sec x + cosec x + 2(tan x - cot x ) = 0,

...(i)

Þ

(- p < x < p ) - {0, ± p / 2} 3 sin x + cos x + 2 (sin 2 x - cos2 x ) = 0 3 sin x + cos x - 2 cos 2x = 0

Þ ...(ii)

Multiplying and dividing by a2 + b2 , i.e. 3 + 1 = 2 , we get ö æ 3 1 2ç sin x + cos x ÷ - 2 cos 2x = 0 2 ø è 2 Þ Þ \ Þ

p pö æ ç cos x × cos + sin x × sin ÷ - cos 2x = 0 è 3 3ø pö æ cos ç x - ÷ = cos 2x è 3ø é since, cos q = cos a ù pö æ 2x = 2np ± ç x - ÷ ê Þ q = 2 np ± a ú è 3ø ë û p p 2x = 2np + x - or 2x = 2np - x + 3 3

Þ

x = 2np -

p p or 3x = 2np + 3 3

Þ

x = 2np -

p 2np p or x = + 3 3 9

-p p -5p 7p or x = , , 3 9 9 9 - p p 5p 7p Now, sum of all distinct solutions = + + =0 3 9 9 9

\

x=

8

JEE Main and Advanced Solved Papers 2016

4. (a,c) Given, RS is the diameter of x2 + y2 = 1. Here, equation of the tangent at P(cos q, sin q) is x cos q + y sin q = 1.

5. (a,b,c) Given, C 1 : x 2 + y2 = 3 intersects the parabola x 2 = 2 y. 3

P

P (cos θ, sin θ) − 3

Q

O

E (h, k) R (–1,0)

O

3 − 3

S(1,0) x cos θ + y sin θ = 1

On solving x 2 + y2 = 3 and x 2 = 2 y, we get y2 + 2 y = 3 Þ

Intersecting with x = 1, 1 - cos q y= sin q \

æ 1 - cos q ö Q ç1, ÷ sin q ø è

\ Equation of the line through Q parallel to RS is q 2 sin 2 q 1 - cos q 2 y= = = tan q q sin q 2 2 sin cos 2 2 sin q Normal at P : y = ×x cos q Þ y = x tan q Let their point of intersection be (h, k ). q Then, k = tan and k = h tan q 2 q ö æ ÷ ç 2 tan 2h × k 2 k =hç \ ÷ Þ k= q 1- k2 ç 1 - tan 2 ÷ è 2ø 2 Þ k (1 - k ) = 2hk 2

\ Locus for point E : 2x = (1 - y ) 1 When x = , then 3 2 1 - y2 = 3 2 2 Þ y =13 1 Þ y=± 3 1 1 æ ö 2 \ç ,± ÷ satisfy 2x = 1 - y . è3 3ø 1 When x = , then 4 2 1 1 - y2 = Þ y2 = 1 4 2 1 Þ y=± 2 1ö æ1 \ ç , ± ÷ does not satisfy 1 - y2 = 2x. è4 2ø

y2 + 2 y - 3 = 0

Þ \

( y + 3)( y - 1) = 0 y = 1, - 3 [neglecting y = - 3, as - 3 £ y £ 3]

\

y=1 Þx = ±

Þ

2

P( 2 , 1) Î I quadrant

Equation of tangent at P( 2 , 1) to C 1 : x 2 + y2 = 3 is 2x + 1× y = 3

…(i)

…(i)

Now, let the centres of C 2 and C 3 be Q2 and Q3, and tangent at P touches C 2 and C 3 at R2 and R3 shown as below Y

Q2

…(ii)

R2

C2

P( 2,1)

O C1

X

Q3 C3

…(iii)

2x + y = 3

Let Q2 be (0, k ) and radius is 2 3. |0 + k - 3| \ = 2 3 Þ |k - 3 | = 6 2+1 Þ k = 9, - 3 \ Q2 (0, 9) and Q3 (0, - 3) Hence, Q2Q3 = 12 \ Option (a) is correct. Also, R2R3 is common internal tangent to C 2 and C 3, and r2 = r3 = 2 3 \

R2R3 = d 2 - (r1 + r2 )2 = 122 - (4 3 )2 = 144 - 48 = 96 = 4 6

9

JEE Main and Advanced Solved Papers 2016 r = 1, then w2 + w4s = - 1

If

Q2 (0,9)

which is only possible, when s = 1. As, w2 + w4 = - 1

R2

\ r = 1, s = 1 Again, if r = 3, then w6 + w4s = - 1

C2

w4s = -2 (never possible)

Þ

d

\ r¹3 Þ (r, s) = (1, 1) is the only solution. Hence, the total number of ordered pairs is 1.

R3

Q3 (0, −3)

x 2 sin (bx ) =1 x ® 0 ax - sin x

8. (7) Here, lim C3

\ Option (b) is correct. Q Length of perpendicular from O(0, 0) to R2R3 is equal to radius of C 1 = 3. 1 \ Area of DOR2R3 = ´ R2R3 ´ 3 2 1 = ´4 6 ´ 3=6 2 2 \ Option (c) is correct. 2 1 Also, area of DPQ2Q3 = Q2Q3 ´ 2 = ´ 12 = 6 2 2 2 \ Option (d) is incorrect. 2

6. (5) Coefficient of x in the expansion of {(1 + x )2 + (1 + x )3 + K + (1 + x )49 + (1 + mx )50} Þ 2C 2 + 3C 2 + 4C 2 + K +

49

50

C 2 × m2 = (3n + 1) ×51 C 3

C2 + 50

Þ r

[Q C r +

C3 r+ 1

+

50

C 2m2 = (3n + 1) ×51 C 3

C r + K+ nC r =

n+ 1

C r + 1]

51 ´ 50 ´ 49 50 ´ 49 ´ 48 50 ´ 49 + ´ m2 = (3n + 1) 3´ 2´1 3´ 2´1 2

Þ

æ ö (bx )3 (bx )5 + - K÷ x 2 ç bx ! ! 3 5 è ø lim =1 x®0 æ ö x3 x5 + - K÷ ax - ç x 3! 5! è ø

Þ

=1 x3 x5 + -K 3! 5! Limit exists only, when a - 1 = 0 Þ a =1 3 2 æ ö x b b 5x 4 + - K÷ x 3 çb 3! 5! è ø lim =1 \ 2 x®0 æ ö x 1 3 x ç - K÷ è 3! 5! ø Þ

lim

x®0

\ m = 16 and n = 5 Hence, the value of n is 5. -1+ i 3 =w 2 é (- w)r w2s ù P = ê 2s ú wr û ë w é (- w)r w2s ù é (- w)r P 2 = ê 2s úê wr û ë w2s ë w

7. (1) Here, Q

z=

Given, \

1 k 1 p ( ) p ü æ p kp ö ì k = 1 sin ÷ ý sin ç + í + è4 4 6 6 ø þ î Converting into differences, by multiplying and dividing by é æ p k p ö ì p (k - 1)p ü ù æpö sin ê ç + ÷ -í + ý ú , i.e. sin ç ÷ . è 6ø è ø 4 6 4 6 î þû ë

\

é w2r + w4s wr + 2s [(- 1)r + 1]ù = ê r + 2s ú r w4s + w2r [(- 1) + 1] ëw û P2 = - I

w + w4s = - 1 and wr + 2s [(- 1)r + 1] = 0 2r

Since, and

r Î{1, 2, 3} (- 1)r + 1 = 0 Þ r = {1, 3}

Also,

w2r + w4s = - 1

…(ii)

13

1. (c) Here,

13

w2s ù ú wr û

…(i)

Paper II

\ Minimum value of m2 for which (51n + 1) is integer (perfect square) for n = 5. \ m2 = 51 ´ 5 + 1 Þ m2 = 256

(a - 1)x +

Þ 6b = 1 From Eqs. (i) and (ii), we get 6(a + b ) = 6 a + 6b = 6 + 1 = 7

m2 = 51n + 1

Þ

æ ö b 3x 2 b 5x 4 x 3 çb + - K÷ 3! 5! è ø

å k=1

å

pö ìp püù éæ p sin ê ç + k ÷ - í + (k - 1) ý ú è ø 4 6 4 6þû î ë pü p öü p ì ìp æp sin í sin í + (k - 1) ý sin ç + k ÷ý è4 6þ 6 øþ 6 î î4 pü ù é æ p kp ö ìp ê sin çè 4 + 6 ÷ø cos í 4 + (k - 1) 6 ý ú î þ ú ê pü ê æ p kp öú ìp + (k - 1) ý cos ç + ÷ 13 ê - sin í è4 6þ 6 øúû î4 ë = 2å pö pü æp ìp k=1 sin í + (k - 1) ý sin ç + k ÷ è4 6þ 6ø î4

10

JEE Main and Advanced Solved Papers 2016 13 é ìp = 2 å ê cot í + (k - 1) î4 k=1 ë

éì = 2 êí cot ëî

æpö ç ÷ - cot è4ø

pü p öù æp ý - cot ç + k ÷ ú è4 6þ 6 øû

3. (a) Let I = ò

p p öü ì æp = 2 í cot - cot ç + 13 ÷ý è4 4 6 øþ î é é æ 29 p ö ù = 2 ê1 - cot ç ÷ = 2 ê1 - cot è 12 ø úû ë ë 5p ù é = 2 ê1 - cot 12 úû ë = 2 (1 - 2 + 3 )

5p ö ù æ ç 2p + ÷ è 12 ø úû 5p é ù êëQ cot 12 = (2 - 3 )úû

…(i)

or and

\

a1 + 50 D = 250 a1

[Q a1 = b1 ] …(ii)

t = b1 + b2 + K + b51 (251 - 1) t = b1 2-1

…(iii)

s = a1 + a2 + K + a51 51 = (2a1 + 50 D ) 2 t = a1 (251 - 1)

…(iv) [Q a1 = b1 ]

t = 251 a1 - a1 < 251 a1 51 s = [ a1 + (a1 + 50 D )] 2 51 = [ a1 + 250 a1 ] 2 51 51 50 = a1 + 2 a1 2 2 s > 251 a1

…(v) [from Eq. (ii)]

From Eqs. (v) and (vi), we get s > t Also, a101 = a1 + 100 D and b101 = 2100 b1 \

æ 250 a1 - a1 ö a101 = a1 + 100 ç ÷ 50 è ø

and

b101 = 2100 a1

Þ

a101 = a1 + 251 a1 - 2a1 = 251 a1 - a1

Þ

51

a101 < 2 a1 and b101 > 2

p/ 2

-p / 2

x 2 cos (- x ) dx 1 + e- x

51

…(ii)

On adding Eqs. (i) and (ii), we get p/ 2 2 é 1 1 ù x cos x ê + 2I = ò ú dx x -p/ 2 1 + e- x û ë1 + e p/ 2

x 2 cos x × (1) dx

é ù a a êQ ò f (x ) dx = 2 ò f (x ) dx , when f (- x ) = f (x )ú -a 0 êë úû Þ

2I = 2ò

p/ 2 2

0

x cos x dx

Þ \

é p2 ù 2I = 2ê - 2ú ë 4 û 2 p I = -2 4

4. (b,c) Here,

Also, a1 , a2 , ..., a101 are in AP. Given, a1 = b1 and a51 = b51 Þ a1 + 50 D = 250 b1

\

…(i)

Using integration by parts, we get 2 2I = 2 [ x 2 (sin x ) - (2x ) (- cos x ) + (2) (- sin x )] p/ 0

\ b1 = 20 b1 , b2 = 21 b1 , b3 = 22 b1,…, b101 = 2100 b1

and

I =ò

-p/ 2

2. (b) If log b1 , log b2 , ..., log b101 are in AP, with common difference loge 2, then b1 , b2 , ..., b101 are in GP, with common ratio 2.

Þ

Þ

=ò

= 2 ( 3 - 1)

Now,

x 2 cos x dx 1 + ex

éQ b f (x ) dx = b f (a + b - x ) dx ù òa ûú ëê òa

æ p p öü ç + ÷ý è4 6 øþ

ì æp pö æ p 2p öü + í cot ç + ÷ - cot ç + ÷ý è ø è4 4 6 6 øþ î pö p öü ù ì æp æp + K + í cot ç + 12 ÷ - cot ç + 13 ÷ý ú è ø è 4 6 4 6 øþ û î

Þ

p/ 2

- p/ 2

a1 Þ b101 > a101

…(vi)

é ê nn (x + n) æç x + è f (x ) = lim ê n® ¥ ê æ ê n! (x 2 + n2 ) ç x 2 + è êë Taking log on both sides, we get

x n nö æ nö ù ÷ ¼ çx + ÷ ú nø ú 2ø è ,x>0 2ö æ n n2 ö ú ÷ K ç x2 + 2 ÷ ú 4ø n ø úû è x

é ùn n n ê nn (x + n) æç x + ö÷ K æç x + ö÷ ú è è nø ú 2ø loge { f (x )} = lim log ê 2ö ê n ®¥ æ æ n n2 ö ú ê n!(x 2 + n2 ) ç x 2 + ÷ K ç x 2 + 2 ÷ ú 4ø n ø úû è è êë n é ù 1 ö æ P çx + ÷ ê ú ø r=1è x / r n ú = lim × log ê n® ¥n ê n æ 2 ú 1 ö n çx + ÷ P (r / n) ú ê rP 2 (r / n) ø r = 1 ë =1è û é ù n ê ú x+ 1 n ê ú r = x lim log å 2ö êæ ú n® ¥ n n r r=1 ê ç x2 + 2 ÷ ú r ø n ûú êë è ö æ r ç ×x + 1 ÷ 1 n n ÷ = x lim å log çç r2 n® ¥ n ÷ 2 r=1 ç 2 × x + 1÷ ø èn Converting summation into definite integration, we get 1 æ xt + 1 ö loge{ f (x )} = x ò log ç 2 2 ÷ dt 0 è x t + 1ø

11

JEE Main and Advanced Solved Papers 2016 tx = z xdt = dz

Put, Þ

Þ x

æ 1 + z ö dz x

\

loge { f (x )} = x

Þ

æ 1+ z ö loge{ f (x )} = ò log ç ÷ dz 0 è 1 + z2 ø

ò0 log çè 1 + z2 ÷ø

\

x

Using Newton-Leibnitz formula, we get æ 1+ x ö 1 × f ¢ (x ) = log ç ÷ f (x ) è 1 + x2 ø

x=1

\ At x = 1, function attains maximum. Since, f (x ) increases on (0, 1). \ f (1) > f (1 / 2) \option (a) is incorrect. f (1 / 3) < f (2 / 3) \option (b) is correct. Also, f ¢ (x ) < 0, when x > 1 Þ f ¢ (2) < 0 \option (c) is correct. æ 1+ x ö f ¢ (x ) Also, = log ç ÷ f (x ) è 1 + x2 ø

\

1 = 5 -1

5+1 4

a ¹ 0, b ¹ 0 a \ x= 2 a + b2t 2 - bt and y= 2 a + b2t 2 y - bt ay = Þ t= Þ x a bx a , we get On putting x = 2 a + b2t 2

= log (2 / 3) < 0 f ¢ (3) f ¢ (2) < f (3) f (2)

æ a2 y2 ö x ç a2 + b 2 × 2 2 ÷ = a bx ø è a2 (x 2 + y2 ) = ax x or x 2 + y2 - = 0 a 2 1ö 1 æ 2 or çx - ÷ + y = 2 è 2a ø 4a \Option (a) is correct. For a ¹ 0 and b = 0, 1 1 x + iy = Þ x = , y = 0 a a Þ z lies on X-axis. \ Option (c) is correct. For a = 0 and b ¹ 0, 1 x + iy = ibt 1 Þ x = 0, y = bt Þ z lies on Y-axis. \ Option (d) is correct. Þ

5. (a,c,d) Tangent to y2 = 4 x at (t2 , 2 t) is

(t2, 2t) X

O

a - ibt 1 ´ a + ibt a - ibt a - ibt x + iy = 2 a + b2t 2

Let

\option (d) is incorrect.

y2=4x

y(2 t ) = 2(x + t 2 ) Þ

SQ = QP

6. (a,c,d) Here, x + iy =

f ¢ (3) f ¢ (2) æ4ö æ 3ö = log ç ÷ - log ç ÷ è ø è 5ø f (3) f (2) 10

S(2, 8) Q P

Thus,

\option (c) is correct. 1 1 Slope of tangent = = t 2 \option (d) is correct.

–

Y

PQ = SP - SQ = 2 5 - 2

\option (b) is incorrect. Now, x-intercept of normal is x = 2 + 22 = 6

\ f ¢ (1) = 0 Now, sign scheme of f ¢(x ) is shown below

Þ

\ … (i)

f ¢ (1) = log (1) = 0 f (1)

\

SP = (4 - 2)2 + (4 - 8)2 = 2 5

\option (a) is correct. Also, SQ = 2

Here, at x = 1,

+

t = 2, i.e. P(4 , 4 ) [since, shortest distance between two curves lie along their common normal and the common normal will pass through the centre of circle]

yt = x + t 2

Equation of normal at P (t 2, 2t) is y + tx = 2t + t 3 Since, normal at P passes through centre of circle S (2, 8). \ 8 + 2 t = 2 t + t3

…(i)

… (i)

12

JEE Main and Advanced Solved Papers 2016

7. (b) Here, P( X > Y ) = P(T1win) P(T1 win ) + P(T1 win) P(draw) + P( draw) P(T1 win) æ 1 1ö æ 1 1ö æ 1 1ö =ç ´ ÷+ç ´ ÷+ç ´ ÷ è 2 2ø è 2 6ø è 6 2ø 5 = 12 8. (c) P [ X = Y ] = P (draw) × P (draw) + P (T1 win) × P (T2 win) + P (T2 win) × P (T1 win) = (1 / 6 ´ 1 / 6) + (1 / 2 ´ 1 / 3) + (1 / 3 ´ 1 / 2) = 13 / 36 x2 y2 + =1 9 8 has foci (± ae, 0) where, a2e2 = a2 - b2

9. (a) Here,

Þ i.e.

…(i)

a2e2 = 9 - 8 Þ ae = ± 1 F1, F2 = (± 1, 0)

Y

5 y- 6 5 …(iii) = Þ ( y - 6) = (x - 3 / 2) x - 3/ 2 2 6 2 6 and equation of altitude through F1 is y = 0 …(iv) æ 9 ö On solving Eqs. (iii) and (iv), we get ç - , 0÷ as orthocentre. è 10 ø 10. (c) Equation of tangent at M (3 / 2, 6 ) to

x2 y2 + = 1 is 9 8

3 x y × + 6× =1 2 9 8 which intersect X-axis at (6, 0). Also, equation of tangent at N (3 / 2, - 6 ) is 3 x y × - 6 =1 2 9 8 Eqs. (i) and (ii) intersect on X-axis at R(6, 0) . - 6 Also, normal at M (3 / 2, 6 ) is y - 6 = 2 On solving with y = 0, we get Q (7 / 2, 0)

…(i)

…(ii) …(iii) 3ö æ çx - ÷ è 2ø …(iv)

Y

M (3/2, √ 6 )

X′

F1 (–1,0)

O

M (3/2, √ 6 )

X

F2 (1,0)

R(6, 0)

X′ F1

O

y 2=4x

F2 Q(7/2,0) N

N (3/2,–√6 ) Y′

Equation of parabola having vertex O(0, 0) and F2 (1, 0)(as, x2 > 0) …(ii) y2 = 4 x x2 y2 + = 1 and y2 = 4 x, we get 9 8 x = 3 / 2 and y = ± 6 Equation of altitude through M on NF1 is

On solving

Y′

1æ 7ö ç6 - ÷ 2è 2ø

5 6 sq units 4 1 and area of quadrilateral MF1NF2 = 2 ´ {1 - (- 1)} 6 2 = 2 6 sq units Area of DMQR 5 = \ Area of quadrilateral MF1NF2 8 \ Area of DMQR =

6=

X

Solved Papers 2017

1

JEE Main & Adv./BITSAT/KERALA CEE/KCET AP & TS EAMCET/VIT/MHT CET

SOLVED PAPERS 2017 JEE Main 1. The statement ( p ® q) ® [(~ p ® q) ® q] is (a) a tautology (c) equivalent to p ® ~ q

(b) equivalent to ~ p ® q (d) a fallacy

2. If 5 (tan2 x - cos2 x) = 2 cos 2 x + 9, then the value of cos 4x is (a) -

3 5

(b)

1 3

(c)

2 9

(d) -

7 9

3. For three events A, B and C, if P (exactly one of A or B occurs) = P(exactly one of B or C occurs) = P (exactly 1 one of C or A occurs) = and P (all the three events 4 1 , then the probability that occur simultaneously) = 16 atleast one of the events occurs, is (a)

7 32

(b)

7 16

(c)

7 64

(d)

3 16

4. Let w be a complex number such that 2w + 1 = z, where 1 1 1 2 z = - 3. If 1 - w - 1 w2 = 3 k , then k is equal to 1 (a) - z

w2

(b) z

(d) 1

5. Let k be an integer such that the triangle with vertices ( k , - 3k ), (5, k ) and ( - k , 2) has area 28 sq units. Then, the orthocentre of this triangle is at the point 1 (a) æç2, - ö÷ è 2ø

(b) æç1, è

3ö ÷ 4ø

3 (c) æç1, - ö÷ è 4ø

(a) 2 y - x = 2 (c) 4 x + 2 y = 7

1 (d) æç2, ö÷ è 2ø

(a) (3 2 , 2 3 ) (c) ( 3, 2 )

10. lim

x ® p/ 2

(a)

6 (a) 7

1 (b) 4

2 (c) 9

4 (d) 9

7. For any three positive real numbers a, b and c, if 9 (25a2 + b2 ) + 25 (c2 - 3ac) = 15b (3a + c), then (a) b, c and a are in GP (c) a, b and c are in AP

(b) b, c and a are in AP (d) a, b and c are in GP

(b) (2 2 , 3 3 ) (d) (- 2 , - 3 )

cot x - cos x ( p - 2 x)3

1 24

(b)

equals

1 16

(c)

1 8

(d)

1 4

11. If two different numbers are taken from the set {0, 1, 2, 3, …, 10}, then the probability that their sum as well as absolute difference are both multiple of 4, is 6 55

(b)

12 55

(c)

14 45

(d)

7 55

12. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then, the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is (a) 485

6. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2 AB. If ÐBPC = b, then tan b is equal to

(b) 4 x - 2 y = 1 (d) x + 2 y = 4

9. If a hyperbola passes through the point P( 2, 3) and has foci at ( ± 2, 0), then the tangent to this hyperbola at P also passes through the point

(a)

w7 (c) - 1

8. The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directrices is x = - 4, then the æ 3ö equation of the normal to it at ç1, ÷ is è 2ø

(b) 468

(c) 469

(d) 484

13. The value of (21C 1 - 10C 1) + (21C2 - 10C2 ) 21 10 21 + ( C3 - C3 ) + ( C 4 - 10C 4) + ... + (21C 10 - 10C 10) is (a) 2 21 - 211

(b) 2 21 - 210

(c) 2 20 - 2 9

(d) 2 20 - 210

14. Let a, b, c Î R. If f ( x) = ax2 + bx + c be such that a + b + c = 3 and f ( x + y ) = f ( x) + f ( y ) + xy , 10

" x, y Î R, then

å f ( n) is equal to n=1

(a) 330

(b) 165

(c) 190

(d) 255

2

Solved Papers 2017

15. The radius of a circle having minimum area, which touches the curve y = 4 - x2 and the lines y =| x |, is (b) 2 ( 2 - 1) (d) 4 ( 2 + 1)

(a) 2 ( 2 + 1) (c) 4 ( 2 - 1)

16. For a positive integer n, if the quadratic equation, x( x + 1) + ( x + 1) ( x + 2) + ... + ( x + n - 1) ( x + n) = 10 n has two consecutive integral solutions, then n is equal to (a) 12 (c) 10

(b) 9 (d) 11

JEE Advanced Matching Type Questions

Paper 1 1. Let a, b, x and y be real numbers such that a - b = 1 and y ¹ 0. If the complex number æ az + b ö z = x + iy satisfies Im ç ÷ = y , then which è z+1 ø of the following is(are) possible value(s) of x? (One or More Than One Correct Option)

(a) 1 - 1 + y2 (c) 1 +

(b) - 1 - 1 - y2

1 + y2

(d) - 1 +

1 - y2

2. If 2 x - y + 1 = 0 is a tangent to the hyperbola x2 y2 = 1 then which of the following 2 16 a CANNOT be sides of a right angled triangle?

Directions (Q.Nos. 7-9) by appropriately matching the information given in the three columns of the following table. Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively. Column-1

(I) x2 + y2 = a2

Column-2

Column-3

(i) my = m2 x + a

(P) æ a , 2 a ö ç 2 ÷ èm m ø

(II) x2 + a2 y2 = a2 (ii) y = mx + a m2 + 1 (Q) æ - ma ç , ç m2 + 1 è (III) y2 = 4ax

ö ÷ m + 1 ÷ø a

2

(iii) y = mx +

a2 m2 - 1 (R) æç - a2 m , ç a2 m2 + 1 è

ö ÷ a2 m2 + 1 ÷ø

(IV) x2 - a2 y2 = a2 (iv) y = mx +

a2 m2 + 1 (S) æç - a2 m , ç a2 m2 - 1 è

ö ÷ a2 m2 - 1 ÷ø

1

(One or More Than One Correct Option)

(a) a, 4, 1

(b) 2 a, 4, 1

(c) a, 4, 2

(d) 2 a, 8, 1

3. If a chord, which is not a tangent, of the parabola y 2 = 16 x has the equation 2x + y = p, and mid-point ( h, k ), then which of the following is(are) possible value(s) of p, h and k ? (One or More Than One Correct Option)

(a) p = - 1, h = 1, k = - 3 (c) p = - 2, h = 2, k = - 4

(b) p = 2, h = 3, k = - 4 (d) p = 5, h = 4, k = - 3

4. The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side? (Single Digit Integer)

5. For how many values of p, the circle x2 + y 2 + 2 x + 4 y - p = 0 and the coordinate axes have exactly three common points? (Single Digit Integer)

6. Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let x be the number of such words where no letter is repeated; and let y be the number of such words where exactly one letter is repeated twice and no other letter is y repeated. Then, = 9x (Single Digit Integer)

-1

7. For a = 2 , if a tangent is drawn to a suitable conic (Column 1) at the point of contact ( -1, 1), then which of the following options is the only CORRECT combination for obtaining its equation? (a) (I) (ii) (Q) (c) (III) (i) (P)

(b) (I) (i) (P) (d) (II) (ii) (Q)

1ö æ 8. The tangent to a suitable conic (Column 1) at ç 3, ÷ is found è 2ø to be 3 x + 2 y = 4, then which of the following options is the only CORRECT combination? (a) (IV) (iv) (S) (c) (IV) (iii) (S)

(b) (II) (iv) (R) (d) (II) (iii) (R)

9. If a tangent to a suitable conic (Column 1) is found to be y = x + 8 and its point of contact is (8, 16), then which of the following options is the only CORRECT combination? (a) (III) (i) (P) (c) (II) (iv) (R)

(b) (I) (ii) (Q) (d) (III) (ii) (Q)

Solved Papers 2017

3

Paper 2 1. Let S = {1, 2, 3, ¼ ¼ , 9}. For k = 1, 2 , ¼ ¼ 5,let N k be the number of subsets of S, each containing five elements out of which exactly k are odd. Then (Single Option Correct) N 1 + N2 + N3 + N 4 + N 5 = (a) 210

(b) 252

(c) 126

(d) 125

2. Let a and b be non zero real numbers such that 2(cos b - cos a ) + cos a cos b = 1. Then which of the following is/are true? (One or More Than One Correct Option)

a b (a) 3 tanæç ö÷ - tanæç ö÷ = 0 è2 ø è2 ø a b (c) tanæç ö÷ + 3 tanæç ö÷ = 0 è2 ø è2 ø

3. Let f ( x) =

a b (b) tanæç ö÷ - 3 tanæç ö÷ = 0 è2 ø è2 ø a b (d) 3 tanæç ö÷ + tanæç ö÷ = 0 è2 ø è2 ø

(b) lim x ® 1- f( x) does not exist (d) lim x ® 1+ f( x) does not exist

Paragraph Based Questions (Q. Nos. 4-5) Let p , q be integers and let a , b be the roots of the equation, x 2 - x - 1 = 0 where a ¹ b. For n = 0, 1, 2, ¼¼ , let a n = pa n + qb n . FACT : If a and b are rational numbers and a + b 5 = 0, then a = 0 = b. 4. a12 = (a) a11 + 2 a10 (c) a11 - a10

æ 1 ö 1 - x (1 + |1 - x|) cos ç ÷ è1 - x ø |1 - x|

for x ¹ 1. Then

(a) lim x ® 1+ f( x) = 0 (c) lim x ® 1- f( x) = 0

(b) 2 a11 + a10 (d) a11 + a10

5. If a4 = 28, then p + 2q = (a) 14

(b) 7

(c) 21

(d) 12

(One or More Than One Correct Option)

BITSAT 1. The coefficient of x 5 in the (1 + x)21 + (1 + x)22 + ... + (1 + x)30 is (a)

51

(c)

31

expansion

of

(b) 9 C 5

C5 21

(d)

C6 - C6

30

C5 +

(b) 1 : 3

(c) - 1 : 3

C5

(d) None of these

3. If p and p¢ denote the lengths of the perpendicular from a focus and the centre of an ellipse with semi-major axis of length a, respectively, on a tangent to the ellipse and r denotes the focal distance of the point, then (a) ap = rp¢ (c) ap = rp¢ + 1

år.

n

r =1

(a) 5 (2 n - 9) (c) 9 (n - 4)

n

Cr

Cr - 1

(c) 40

(d) 53

6. For the equation 3 x2 + px + 3 = 0, p > 0, if one of the roots is square of the other, then p is equal to 1 (a) 2

(b) 1

(d) None of these

8. If cos ( x - y ), cos x and cos ( x + y ) are in HP, then cos x sec ( y / 2) is equal to (a) ±

(b) ± 1 / 2

2

(c) ± 2

(d) None of these

9. The number of times the digit 5 will be written when listing the integers from 1 to 1000, is (a) 271

(b) 272

(a) A = B

(b) 10 n (d) None of these

(b) - 12

2 ac + 1 2a + c + a

(c) 300

(d) None of these

(b) A = X

(c) B = X

(d) A È B = X

is equal to

5. The numbers 32 sin 2 a - 1, 14 and 34 - 2 sin 2 a form first three terms of an (a)P., its fifth term is (a) - 25

(b)

10. Let A and B be two sets such that A Ç X = B Ç X = f and A È X = B È X for same set X. Then,

(b) rp = ap¢ (d) ap¢ + rp = 1 10

4. The value of

2 ac + 1 2c + abc + 1 2 ac + 1 (c) 2c + ab + a

(a) 20

2. If z = a + ib satisfies arg ( z - 1) = arg ( z + 3i), then ( a - 1) : b = (a) 2 : 1

7. If a = log 2 3, b = log 2 5 and c = log 7 2, then log 140 63 in terms of a, b, c is

(c) 3

2 (d) 3

11. The general solution of sin x - 3 sin 2 x + sin 3 x = cos x - 3 cos 2 x + cos 3 x is (a) np +

p 8

(b)

np p np p + (c) (-1)n + 2 8 2 8

(d) 2 np + cos -1

3 2

12. Two equal sides of an isosceles triangle are 7 x - y + 3 = 0 and x + y - 3 = 0 and its third side passes through the point (1 , - 10). Find the equation of the third side (a) x - 3 y = - 31 (c) x + 3 y = 31

(b) x - 3 y = 31 (d) x + 3 y = - 31

4

Solved Papers 2017

13. If two distinct chords drawn from the point ( p, q) on the circle x2 + y 2 = px + q y (where pq ¹ 0 ) are bisected by the X-axis, then 2

(a) p = q

2

2

(b) p = 8q

2

2

(c) p < 8q

2

2

(d) p > 8q

2

(b) 0 (d) None of these

15. If the curve y = ax and y = b x intersect at angle a, then tan a is equal to a-b 1 + ab

(b)

log a - log b a+ b (c) 1 + log a log b 1 - ab

log a + log b (d) 1 - log a log b

æ 1 + n2 ö 16. If lim ç an ÷ = b, where ais finite number, then x ®¥ è 1+ n ø (a) a = 2

(b) a = 0

(c) b = 1

12 49

(b)

6 49

(c)

9 49

(d)

2

2

18. If equation (10 x - 5) + (10 y - 4) represents a hyperbola, then (a) - 2 < l < 2 (c) l < - 2 or l > 2

(a) 0 (c) 1.3

15 49

2

2

= l (3 x + 4 y - 1)

(b) 1 (d) 2.5

21. A coin is tossed 7 times. Each time a man calls hea(d) The probability that he wins the toss atleast 4 occasions is (a)

1 4

(d) b = - 1

17. If the papers of 4 students can be checked by anyone of the 7 teachers, then the probability that all the 4 papers are checked by exactly 2 teachers, is equal to (a)

(b) as2 s2 (d) 2 a

20. Coefficient of variation of two distributions are 50 and 60 and their arithmetic means are 30 and 25, respectively. Difference of their standard deviation is

n®¥

(a)

(a) s2 (c) a2s2

14. lim sin [p n2 + 1] is equal to (a) ¥ (c) does not exist

19. If the variance of the observations x1, x2 , ......., xn is s2 , then the variance of ax1, ax2 , ......, axn , a ¹ 0 is

22. The value of

(b)

5 8

(c)

(d)

1 6

2 2+ 4 2+ 4+6 + + + .... is 1! 2! 3!

(a) e (c) 3e

(b) 2e (d) None of these

23. If z1 , z2 and z3 represent the vertices of an equilateral triangle such that| z1| = | z2| = | z3|, then (a) z1 + z2 = z3 1 z3

(b) z1 + z2 + z3 = 0 (d) z1 - z2 = z3 - z2

(c) z1 z2 =

(b) l > 2 (d) 0 < l < 2

1 2

KERALA CEE 1. If ( x, y ) is equidistant from ( a + b, b - a) and ( a - b, a + b), then (a) x + y = 0 (c) ax - by = 0 (e) ax + by = 0

6. The equation of the line passing through ( -3, 5) and perpendicular to the line through the points (1, 0) and ( - 4, 1) is

(b) bx - ay = 0 (d) bx + ay = 0

2. If the points (1, 0), (0, 1) and ( x, 8) are collinear, then the value of x is equal to (a) 5 (c) 6 (e) - 7

(b) - 6 (d) 7

(b) 5 x - y + 20 = 0 (d) 5 x + y + 20 = 0

7. The coefficient of (1 + x2 ) 5 (1 + x) 4 is (a) 30

(b) 60

x5

in

the

(c) 40

expansion (d) 10

of

(e) 45

4

3. If f (9) = f ¢ (9) = 0, then lim

x ®9

(a) 0

(b) f(0)

f ( x) - 3 x -3

(c) f¢(3)

8. The coefficient of x in the expansion of(1 - 2 x) 5 is equal to

is equal to

(d) f(9)

(a) 40

(e) 1

4. The value of cos ( p / 4 + x) + cos ( p / 4 - x) is 2

(a) 2 sin x (e) 2 cos x

(b) 2 sin x

2

(c) 2 cos x

(b) 3/5

(c) 29/2

(b) 320 2

(c) - 320

and

(d) 33/2 (e) 35/2

(d) - 32

(e) 80

2

9. The equation 5 x + y + y = 8 represents (a) an ellipse (c) a hyperbola (e) a straight line

(d) 3cos x

5. Area of the triangle with vertices ( - 2, 2), (1, 5) (6, - 1) is (a) 15

(a) 5 x + y + 10 = 0 (c) 5 x - y - 10 = 0 (e) 5 y - x - 10 = 0

(b) a parabola (d) a circle

10. The centre of the ellipse 4 x2 + y 2 - 8 x + 4 y - 8 = 0 is (a) (0, 2) (e) (1, - 2)

(b) (2, - 1)

(c) (2, 1)

(d) (1, 2)

5

Solved Papers 2017 11. If f ( x) = 2 x + (a) 0

4 2x

, then f ¢ (2) is equal to

(b) - 1

(c) 1

(d) 2

(e) - 2

12. The equation of the hyperbola with vertices (0, ± 15) and foci (0, ± 20) is x2 y2 (a) =1 175 225 2 2 y x (c) =1 225 125 2 2 y x (e) =1 225 175

x2 y2 (b) =1 625 125 2 2 y x (d) =1 65 65

(b) (- 1, 2 )

(c) (2, - 1)

(d) (3, - 1)

(b) 6 ´ 50 ´ 102 (d) 6 ´ 25 ´ 102 2

(b) 1 / 2 2

(c) 1/2

(d) 1/4

(b) p

(c) p / 4

(d) p /2

(e) 0

17. The sum S = 1 / 9 ! + 1 / 3! 7! + 1 / 5! 5! + 1 / 7! 3! + 1 / 9 ! is equal to 10

(a) 2 / 8! (e) 2 5 / 8!

9

(b) 2 / 10!

7

(c) 2 / 10!

6

(d) 2 / 10!

(b) 0

(b) 2/5

(c) 3 / 2

(d) 1/2

(e) 1/ 2

(c) 7/5

(d) 3/5

(e) 1

20. The difference between the maximum and minimum x value of the function f ( x) = ò (t 2 + t + 1) dt on [2, 3] is 0

(a) 39/6

(b) 49/6

(c) 59/6

(d) 69/6

(e) 79/6

21. If a and b are the non-zero distinct roots of x2 + ax + b = 0, then the minimum value of x2 + ax + b is (a) 2/3

(b) 9/4

2 (e) ± 13

(b) - 3, 5

(c) 3,- 5

(d) 4, 5

(b) 8.3

(c) 188.6

(d) 177.3 (e) 78

26. The area of the triangular region whose sides are y = 2 x + 1, y = 3 x + 1 and x = 4 is (a) 5

(b) 6

(c) 7

(d) 8

(e) 9

of r is (a) 9

(b) 3

(c) 4

(d) 5

(e) 6

28. Let S = {1, 2, 3, .....,10}. The number of subsets of S containing only odd numbers is (a) 15

(b) 31

(c) 63

(d) 7

(e) 5

29. The area of the parallelogram with vertices (0, 0), (7, 2), (5, 9) and (12, 11) is (b) 54

(c) 51

(d) 52

(e) 53

(d) 3

(e) 5

(d) - 4

(e) - 5

30. The value of| 4 + 2 3| -| 4 - 2 3| is

19. The distance of the point (3, - 5) from the line 3 x - 4 y - 26 = 0 is (a) 3/7

(d) ±

25. In an experiment with 15 observations on x, the following results were available Sx2 = 2830, Sx = 170 On observation that was 20, was found to be wrong and was replaced by the correct value 30. Then, the corrected variance is

(a) 50

18. sin 765° is equal to (a) 1

62

27. If nC r - 1 = 36, nC r = 84 and nC r + 1 = 126, then the value

16. If x Î[0, p / 2], y Î[0, p / 2] and sin x + cos y = 2, then the value of x + y is equal to (a) 2p

(c) ±

65

24. Let A (6, - 1) B (1, 3) and C ( x, 8) be three points such that AB = BC . The values of x are

(a) 9.3

3ö 1 ( x - 1)2 æ 15. The eccentricity of the ellipse + çy + ÷ = è 2 4ø 16 is (a) 1 / 2 (e) 1/4 2

(b) ±

23. The set {( x, y ) :| x| + | y | = 1} in the xy-plane represents

(a) 3, 5 (e) - 3, - 5

14. For all rest numbers x and y, it is known as the real valued function f satisfies f ( x) + f ( y ) = f ( x + y ). If f (1) = 7, thenS r100 = 1 f ( r) is equal to (a) 7 ´ 51 ´ 102 (c) 7 ´ 50 ´ 102 (e) 7 ´ 50 ´ 101

(a) 0

(a) a square (b) A circle (c) An ellipse (d) A rectangle which is not a square (e) A rhombus which is not a square

13. The vertex of the parabola y 2 - 4 y - x + 3 = 0 is (a) (- 1, 3) (e) (1, 2 )

22. If the straight line y = 4 x + c touches the ellipse x2 + y 2 = 1, then c is equal to 4

(c) - 9 / 4

(d) - 2 / 3 (e) 1

(a) 1

(b) 2

31. The value of 8 (a) - 1

2 /3

(c) 4

- 16

(b) - 2

1/ 4

-9

1/ 2

is

(c) - 3 2

32. Let x = 2 be a root of y = 4 x - 14 x + q = 0. Then y is equal to (a) ( x - 2 ) (4 x - 6) (c) ( x - 2 ) (- 4 x - 6) (e) ( x - 2 ) (4 x + 3)

(b) ( x - 2 ) (4 x + 6) (d) ( x - 2 ) (- 4 x + 6)

33. If x1 and x2 are the roots of 3 x2 - 2 x - 6 = 0, then x21 + x22 is equal to (a)

50 9

(b)

40 9

(c)

30 9

(d)

20 9

(e)

10 9

6

Solved Papers 2017

34. Let x1 and x2 be the roots of the equations x2 - px - 3 = 0. If x21 + x22 = 10, then the value of p is equal to (a) - 4 or 4 (e) 0

(b) - 3 or 3

(c) - 2 or 2

(a) 9

(d) - 1or 1

(a) 1/3

(b) 1

(c) 3

1

-

2

x + 4x + 4

(d) - 1

4 4

3

2

+

x + 4x + 4x

(e) - 3

4 3

x + 2 x2

,

æ1ö then f ç ÷ is equal to è2ø (a) 1

(b) 2

(c) -1

(d) 3

(e) 4

37. If x and y are the roots of the equation x2 + bx + 1 = 0, 1 1 then the value of is + x+b y +b (a) 1/b

(b) b

(c) 1 / 2 b

(d) 2b

(b) - 2

(c) - 3

(d) - 1

(e) 0

39. The root of ax2 + x + 1 = 0, where a ¹ 0, are in the ratio 1 : 1. Then a is equal to (a) 1/4

(b) 1/2

(c) 3/4

(d) 1

(e) 0

40. If z2 + z + 1 = 0 where z is a complex number, then the 2

2

2

1ö 1ö 1ö æ æ æ value of ç z + ÷ + ç z2 + 2 ÷ + ç z3 + 3 ÷ equal è è è zø z ø z ø (a) 4

(b) 5

(c) 6

æpö 41. If z = cos ç ÷ - i sin è3ø (a) 0

(b) 1

(c) - 1

(b) - 1

43. If| z| = 5 and w = (a) 0

q 2 (e) 2 tanq (a) cot

(e) 8

(d)

p 2

(e) p

(d) 20

(e) 1

(d) 1/2 (e) -1 / 2

(c) 25

(c) i cot

(d) 1

q 2

(d) i tan

(c) - 77

(d) - 67

(e) - 57

47. The arithmetic mean of two numbers x and y is 3 and geometric mean is 1. Then x2 + y 2 is equal to (a) 30

(b) 31

(c) 32

(d) 33

(e) 34

48. The solution of 32 x - 1 = 811 - x is (a) 2/3 (c) 7/6 (e) 1/3

(b) 1/6 (d) 5/6

1 49. The sixth term in the sequence is 3, 1, , ... is 3 (a) 1/27

(b) 1/9

(c) 1/81

(d) 1/17

(e) 1/7

50. Three numbers are in arithmetic progression.Their sum is 21 and the product of the first number and the third number is 45. Then the product of these three number is (a) 315

(b) 90

(c) 180

(d) 270

(e) 450

51. If a + 1, 2a + 1, 4a - 1 are in arithmetic progression, then the value of a is (a) 1

(b) 2

(c) 3

(d) 4

(e) 5

52. Two numbers x and y have arithmetic mean 9 and geometric mean 4. Then, x and y are the roots of (a) x2 - 18 x - 16 = 0

(b) x2 - 18 x + 16 = 0

(c) x2 + 18 x - 16 = 0

(d) x2 + 18 x + 16 = 0

2

(e) x - 17 x + 16 = 0

53. Three unbiased coins are tossed. The probability of getting atleast 2 tails is

(a) 1/36 (c) 1

(b) - 87

(b) 1/4

(c) 1/2

(d) 1/3

(e) 2/3

54. A single letter is selected from the word TRICKS. The probability that it is either T or R is

is equal to

(e) - 1

1+ a is equal to 1- a

(b) tanq

(a) - 97

(a) 3/4

72

z-5 , then the Re(w ) is equal to z+5

(b) 1/25

44. If a = e iq , then

(d) 7

æpö 2 ç ÷, then z - z + 1 is equal to è3ø

æ æ p öö æpö ç 1 + cos ç ÷ + i sin ç ÷ ÷ è 12 ø è 12 ø ÷ 42. ç æpö÷ æpö ç ç 1 + cos çè ÷ø - i sin çè ÷ø ÷ è 12 12 ø (a) 0

(c) 21

(e) 1

38. The equations x 5 + ax + 1 = 0 and x 6 + ax2 + 1 = 0 have a common root. Then a is equal to (a) - 4

(b) 10

46. The 30th term of the arithmetic progression 10, 7, 4 is

35. If the product of roots of the equation mx2 + 6 x + (2m - 1) = 0 is - 1, then the value of m is

36. If f ( x) =

45. Three numbers x, y and z are in arithmetic progression. If x + y + z = - 3 and xyz = 8, then x2 + y 2 + z2 is equal to

q 2

(b) 1/4

(c) 1/2

(d) 2/3

(e) 1/3

55. From 4 red balls, 2 white balls and 4 black balls, four balls are selected. The probability of getting 2 red balls is (a) 7/21

(b) 8/21

(c) 9/21

(d) 10/21 (e) 11/21

56. In a class, 60% of the students know lesson I, 40% know lesson II and 20% know I and II. A student is selected of random. The probability that the student does not know lesson I and lesson II is (a) 0 (d) 1/5

(b) 4/5 (e) 2/5

(c) 3/5

Solved Papers 2017 57. Two distinct numbers x and y are chosen from 1, 2, 3, 4, 5. The probability that the arithmetic mean of x and y is an inter is (a) 0

(b) 1/5

(c) 3/5

(d) 2/5

(e) 4/5

58. The number of 3 ´ 3 matrices with entries - 1 or + 1 is (a) 2

-4

(b) 2

5

(c) 2

6

(d) 2

7

(e) 2

9

59. Let S be the set of all 2 ´ 2 symmetric matrices whose entries are either zero or one. A matrix X is chosen from S. The probability that the determinant of X is not zero is (a) 1/3 (e) 2/9

(b) 1/2

(c) 3/4

(d) 1/4

(b) 6!

(c) 7!

(b) 9

(c) 10

(e) 12

62. The sum of odd integers from 1 to 2001 is (a) 10012

(b) 10002

(c) 1002 2

(d) 10032 (e) 9992

63. Two balls are selected from two black and two red balls. The probability that the two balls will have no black balls is (a) 1/7

(b) 1/5

(c) 1/4

(d) 1/3

(e) 1/6

(b) 1 + 0i

(c) 0 + i

(d) 1 + 2 i (e) 1 + 3i

65. The mean for the data 6, 7, 10, 12, 13, 4, 8, 12 is (a) 9

(b) 8

(c) 7

(d) 6

(e) 5

66. The set of all real numbers satisfying the inequality x - 2 < 1 is

(d) (- ¥, - 3)

(c) x Î(2, ¥)

(d) x Î(1, ¥)

68. The mode of the data 8, 11, 9, 8, 11, 9, 7, 8, 7, 3, 2 is (a) 11

(b) 9

(c) 8

(d) 3

(e) 7

69. If the mean of six numbers is 41, then the sum of these numbers is (b) 236

(c) 226

(d) 216

(e) 206

70. In a flight 50 people speak Hindi, 20 speak English and 10 speak both English and Hindi. The number of people who speak atleast one of the two languages is (b) 50

(c) 20

(d) 80

(e) 60

x+1 , then the value of f ( f ( x)) is equal to x-1

(a) x

(b) 0

(c) - x

(d) 1

(e) 2

72. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even? (a) 3/4

73. lim

x ®0

(b) 1/4

(c) 1/2

(d) 2/3

(e) 1/16

2+ x - 2- x is equal to x

1 2 1 (e) 2 2

(a)

64. If z = i 9 + i 19, then z is equal to (a) 0 + 0i

(a) x Î(- 3, ¥) (b) x Î(3, ¥) (e) x Î(- ¥, 3)

71. If f ( x) =

(d) 11

(c) [- 3, ¥)

| x - 3| >, then x-3

(a) 40

(d) 8!

61. The number of diagonals in a hexagon is (a) 8

67. If

(b) [3, ¥)

(a) 246

60. The number of words that can be formed by using all the letters of the word PROBLEM only one is (a) 5! (e) 9!

(a) (3, ¥) (e) (- ¥, 3)

(b) 2

(c) 0

(d) Does not exist

æp qö æp qö 74. tan ç + ÷ + tan ç - ÷ is equal to è 4 2ø è 4 2ø (a) sec q

(b) 2sec q

(c) sec q / 2

(d) sinq (e) cos q

Karnataka CET 1. The total number of terms in the expansion of ( x + a) 47 - ( x - a) 47 after simplification is (a) 96

(b) 48

(c) 47

(d) 24

2. The contrapositive statement of the statement “If x is prime number, then x is odd” is (a) If x is not odd, then x is not a prime number (b) If x is a prime number, then x is not odd (c) If x is not a prime number, then x is not odd (d) If x is not a prime number, then x is odd

3. A box has 100 pens of which 10 are defective. The probability that out of a sample of 5 pens drawn one by one with replacement and atmost one is defective, is

7

9 10 5 4 9 1 9 (c)æç ö÷ + æç ö÷ è 10 ø 2 è 10 ø

(a)

1 2 1 (d) 2

(b)

æ9ö ç ÷ è 10 ø

4

æ9ö ç ÷ è 10 ø

5

4. If n C 12 = nC 8, then n is equal to (a) 12

(b) 20

(c) 26

(d) 6

(c) n2

(d) n(n - 2 )

5. 3 + 5 + 7 + ... to n terms is (a) n(n + 2 )

(b) (n + 1)2

6. If A and B are finites sets and A Ì B, then (a) n( A È B) = n(B) (c) n( A Ç B) = n(B)

(b) n( A Ç B) = f (d) n( A È B) = n( A)

8

Solved Papers 2017 1 - cos 4q is q ® 0 1 - cos 6q

14. If| x - 1| £ 1, then

7. The value of lim (a) 9/4

(b) 9/3

(a) x Î[1, 3]

(c) 4/9

(d) 3/4

8. The area of triangle with vertices (K, 0), (4, 0), (0, 2) is 4 sq units, then value of K is (a) 8

æ1 + i ö 9. If ç ÷ è1 - i ø m is

(b) 0 or 8

(c) 0

(a)

2 +1 (a) 2 2

(c) 3

3 -1 (b) 2 2

2t 1 + t2

(a) 0

(d) 1

(b) 7

, tan y +

2t 1 - t2

(b) 2

3 (d) 4

dy , then is equal to dx

(c) 1

(d) - 1

(c) 5

13. If y = log(log x), then - (1 + log x)

d2 y dx2

(d)

( xlog x)2

2 5 4

(c)

x2 y2 + = 1 is 36 16

2 5 6

2 13 (d) 4

17. The range of the function f ( x) = 9 - x2 is (b) (0, 3)

(c) [0, 3]

(d) [0, 3)

18. Reflexion of the point (a , b, g ) in XY -plane is (a) (0, 0, g ) (c) (a, b, – g )

(b) (– a,–b, g ) (d) (a, b, 0)

19. If coefficient of variation is 60 and standard deviation is 24, then Arithmetic mean is (a) 40

(b) 7/20

(c) 20/7

(d) 1/40

(a) P ( A ¢ Ç B¢ ) = 1 (1 - P( A)) (1 - P(B)) (b) P( A) + P(B) = 1 (c) P( A) = P(B) (d) A and B are mutually exclusive

is equal to (b)

x2 log x (1 + log x)

(b)

(d) 8

20. Two events A and B will be independent if

(b) 83 x (d) 8 x3

(a) 8x (c) (8 x)1/ 3

2 13 6

(a) (0, 3]

3 (c) 2

12. If f ( x) = 8 x3 , g ( x) = x 1/ 3 , then fog ( x) is

(c)

(a) 6

= 1, then the least positive integral value of

(b) 4

(d) x Î(1, 3)

15. The perpendicular distance of the point P(6, 7, 8) from XY -plane is

16. The eccentricity of the ellipse

10. The value of cos2 45° - sin2 15° is

(a)

(c) x Î[1, 3)

(d) 0 or - 8

m

(a) 2

11. If sin x =

(b) x Î(- 1, 3)

- (1 + log x)

21. Equation of line passing through the point (1, 2) and perpendicular to the line y = 3 x - 1

( xlog x)2 (1 + log x)

(a) x + 3 y = 0 (c) x + 3 y + 7 = 0

x2 log x

(b) x + 3 y - 7 = 0 (d) x - 3 y = 0

Andhra Pradesh EAMCET 1. In D ABC, if b cos q = c - a, (where q is an acute angle), then (c - a) tan q = (a) 2 ca cos (c) 2ca cos

B 2

(b) 2 ca sin

B 2

(d) 2casin

B 2

(a)

B 2

2. If the pair of lines x2 - 16 pxy - y 2 = 0 and x2 - 16qxy - y 2 = 0 are such that each pair bisects the angle between the other pair, then pq = (a)

-1 64

(b)

1 64

(c)

-1 8

(d)

1 8

3. If three numbers are drawn at random successively without replacement from a set S = {1, 2,...10}, then the probability that the minimum of the chosen numbers is 3 or their maximum is 7. (a)

11 40

(b)

5 40

(c)

3 40

4. The coefficient of x 5 in the 22 (1 + x)21 + (1 + x) + ... + (1 + x)30 is

(d)

1 40

31

C6 -

21

C 6 (b)

51

C5

(c) 9 C 5

expansion (d)

30

C5 +

of

20

C5

5. Let A = { - 4, - 2, - 1, 0, 3, 5} and f : A ® IR be defined by x>3 ì3 x - 1 for ï 2 f ( x) = í x + 1 for - 3 £ x £ 3 ï2 x - 3 for x< -3 î Then the range of f is (a) {- 11, 5, 2, 1 ,10, 14} (c) {- 11, 5, 2, 1, 8, 14}

(b) {- 11, - 2, 2, 1, 8, 14} (d) {- 11, - 7, - 5, 1, 10, 14}

6. The incentre of the triangle formed by the straight lines y = 3 x, y = - 3 x and y = 3 is (a) (0, 2) (c) (2, 0)

(b) (1, 2) (d) (2, 1)

9

Solved Papers 2017 7. If a circle with radius 2.5 units passes through the points (2, 3) and (5, 7), then its centre is (a) (1.5, 2)

(b) (7, 10)

(c) (3, 4)

16. If a, b, g are the lengths of the tangents from the vertices of a triangle to its incircle. Then

(d) (3.5, 5)

1 (abg ) r2 1 (c) a + b + g = (abg ) r

8. The circumcentre of the triangle formed by the points (1, 2, 3), (3, - 1, 5) , (4, 0, - 3) is (a) (1, 1, 1)

(b) (2, 2, 2)

(c) (3, 3, 3)

7 -1 (d) æç , , 1ö÷ è2 2 ø

9. The lines y = 2 x + 76 and 2 y + x = 8 touch the x2 y2 ellipse + = 1. If the point of intersection of these 16 12 two lines lie on a circle. whose centre coincides with the centre of that ellipse, then the equation of that circle is (a) x2 + y2 = 28 2

2

(c) x + y = 12

17. The angle between the tangents drawn from the point (1, 2) to the ellipse 3 x2 + 2 y 2 = 5 is æ 12 5 ö (a) tan-1 ç ÷ è 5 ø p (c) 4

2

2

(d) x + y = (4 +

8)

(b) xy + x - 2 y - 2 = 0 (d) xy - x + 2 y - 2 = 0

8 11. The harmonic mean of two numbers is - and their 5 geometric mean is 2. The quadratic equation whose roots are twice those numbers is (a) x2 + 5 x + 4 = 0

(b) x2 + 10 x + 16 = 0

(c) x2 - 10 x + 16 = 0

(d) x2 - 5 x + 4 = 0

24 5

(b)

26 5

(c)

23 5

(d)

14. If a is a non-real a (1 + a ) (1 + a 2 + a 4) =

29 5

(a) 1 (c) - 1

(b) 2 (d) 3

root

of

x7 = 1,

then

(b) 2 (d) - 2

15. The point to which the origin is to be shifted to remove the first degree terms from the equation 2 x2 + 4 xy - 6 y 2 + 2 x + 8 y + 1 = 0 is 7 3 (a) æç , ö÷ è 8 8ø -7 3 (c) æç , ö÷ è 8 8ø

-7 -3 (b) æç , ö÷ è 8 8ø 7 -3 (d) æç , ö÷ è8 8 ø

a

y2 b2

= 1,

(a) (c)

m2 l2 l2 m2

(a2 + b 2 )2

(b) (l 2 + m2 )(a2 + b 2 )2

(a2 + b 2 )2

(d) l 2 m2 (a2 + b 2 )2

19. The equation of the circle whose diameter is the common chord of the circles x2 + y 2 + 2 x + 2 y + 1 = 0 and x2 + y 2 + 4 x + 6 y + 4 = 0 is (a) 10 x2 + 10 y2 + 14 x + 8 y + 1 = 0 (b) 3 x2 + 3 y2 - 3 x + 6 y - 8 = 0 (d) x2 + y2 - x + 2 y + 4 = 0

13. If the roots of the equation x3 - 7 x2 + 14 x - 8 = 0 are in geometric progression, then the difference between the largest and the smallest roots is (a) 4 1 (c) 2

-

2

(c) 2 x2 + 2 y2 - 2 x + 4 y + 1 = 0

12. If z is a complex number with | z| ³ 5. Then the least 2 value of z + is z (a)

x2

then a2 m2 - b2 l2 =

10. The equation of the pair of lines through the point (2, 1) and perpendicular to the pair of lines 4 xy + 2 x + 6 y + 3 = 0 is (a) xy - x - 2 y + 2 = 0 (c) xy + x + 2 y - 6 = 0

æ 12 5 ö (b) tan-1 ç ÷ è 13 ø p (d) 4

18. If lx + my = 1 is a normal to the hyperbola

(b) x2 + y2 = 16

2

1 1 1 + + = r (abg ) a b g 2 (d) a2 + b 2 + g 2 = (abg ) r

(b)

(a) a + b + g =

20.

x-1 x-3 holds, for all x in the interval < 3x + 4 3x - 2 - 5ö (b) æç - ¥, ÷ è 4 ø 3 -5 (d)æç - ¥, ö÷ È æç , - ¥ö÷ ø è 4 ø è4

-4 2 (a) æç , ö÷ è 3 3ø 3 (c) æç , ¥ö÷ è3 ø

21. There are 10 intermediate stations on a railway line between two particular stations. The number of ways that a train can be made to stop at 3 of these intermediate stations so that no two of these halting stations are consecutive is. (a) 56

(b) 20

(c) 126

22. The figure formed by 6 x2 + 13 xy + 6 y 2 = 0 and

the

(d) 120

pairs

of

lines

6 x2 + 13 xy + 6 y 2 + 10 x + 10 y + 4 = 0, is a (a) Square (c) Rhombus

(b) Parallelogram (d) Rectangle

23. If the point of intersection of the tangents drawn at the points where the line 5 x + y + 1 = 0 cuts the circle x2 + y 2 - 2 x - 6 y - 8 = 0 is ( a, b), then 5a + b = (a) 3

(b) - 44

(c) - 1

(d) 4

10

Solved Papers 2017

24. The tangents to the parabola y 2 = 4ax from an external point P make angles q 1 and q 2 with the axis of the parabola. Such that tan q 1 + tan q 2 = b where b is constant. Then P lies on (a) y = x + b

(b) y + x = b (c) y =

x b

(d) y = bx

11 (b) æç 4, ö÷, (- 1, - 1) è 4ø

-1 4

(b)

1 2

(c) 1 2

2

2

27. An angle between the curves x = 3 y and x + y = 4 is (a) tan-1

5 3

5 3

(b) tan-1

(c) tan-1

2 3

(d)

(b) 43200

(c) 86400

(d) 151200

(b) p

(c) q

(d)

n-1

n > 1, S r( r + 1 - w)( r + 1 - w2 ) = r =1

31.

37

(b)

39

(c)

C4

38

(d)

C4

42

C4

3n 2(3n + 2 )

(b)

3n 3n + 2

(d) (2, 0, 3)

1 1 1 + + + ... is 2 × 5 5 × 8 8 × 11

(c)

n 2(3n + 2 )

(d)

n 3n + 2

38. The equations of the parabola whose axis is parallel to the X-axis and which passes through the points ( - 2, 1),(1, 2)( - 1, 3) is (a) 18 y2 - 12 x - 21y - 21 = 0 (b) 5 y2 + 2 x - 21y + 20 = 0

39. In if q is D ABC b cos (C + q ) + c cos ( B - q ) = (b) acos q

any

(c) a tanq

angle,

then

(d) asinq

40. The number of solutions of the trigonometric equation 1 + cos x , cos 5 x = sin2 x in [0, 2p] is (a) 8

(b) 12

(c) 10

(d) 6

(a) (r - p)2 + (r - q )2

(b) (1 + p)2 + (1 + q )2

2

(d) (r - p)2 + (q - 1)2

(a)

xi

6

10

14

18

24

28

30

fi

2

4

7

12

8

4

3

2p 3

2

(b)

43. If log

1 3

(c) 43.4

p 2

(c)

p 6

(d) p

ì| z|2 - | z| + 1ü ý > - 2, then z lies inside í î 2 + | z| þ

(a) a triangle (b) 34.3

(c) (2, 3, 0)

37. The sum of first n terms of

32. The variance of the following data is

(a) 33.4

(b) (3, 0, 2)

42. The angle between the two circles, each passing through the centre of the other is

r =1

C4

(d) 2

A1 A2 x + A3 , then( A1, A2 , A3 ) = + x x2 + 1

(c) (r + p) + (q + 1)

5

41

=

(c) 1

41. If a, b, g are the roots of x3 + px2 + qx + r = 0, then the value of (1 + a 2 )(1 + b2 ) (1 + g 2 ) is

n(n + 1) (2 n + 1) (b) 6 n (n + 1) (2 n + 1) (d) 4

C 4 + S ( 42 - r)C r =

(a)

x3 + x

(a) acot q

p p+ q

30. If w is a complex cube root of unity, then for any

n2 (n + 1)2 (a) 4 n(n - 1) 2 (c) (n + 3n + 4) 4

(d) 5

(c) 15 y2 + 12 x - 11y + 20 = 0 (d) 25 y2 - 2 x - 65 y + 36 = 0

29. If tan a and tan b are the roots of the equation then the value of x2 + px + q = 0, sin2 (a + b) + p cos (a + b) sin (a + b) + q cos2 (a + b) is (a) p + q

5 x2 + 2

(a)

p 3

28. The number of different ways of preparing a garland using 6 distinct white roses and 5 distinct red roses such that no two red roses come together is (a) 21600

(b) 4

(a) (0, 2, 3)

(d) 2

(c) 2

35. If 2kx + 3 y - 1 = 0, 2 x + y + 5 = 0 are conjugate lines with respect to the circle x2 + y 2 - 2 x - 4 y - 4 = 0, then k =

36. If

(1 - cos 2 x) (3 + cos x) 26. lim = x ®0 x tan 4 x (a)

(b) 3

(a) 3

(d) (7, 5), (-1, - 1)

(b) b ( x - p) = a( y - q ) (d) p( x - a) = q ( y - b )

34. If a ¹ 0 and the mean deviation of the observations { ka }, for k = 1, 2, ……, 50 about its median is 50, then |a| = (a) 4

25. The points on the straight line 3 x - 4 y + 1 = 0 which are at a distance of 5 units from the point (3, 2) are -5 7 (a) æç - 2, - ö÷, æç - 3 , ö÷ è 2 ø 4ø è 1ö æ 5ö æ (c) ç1, ÷, ç2, ÷ è 2 ø è 4ø

(a) a( x - p) = b ( y - q ) (c) a( x - p) = b ( y - q )

(b) an ellipse (c) a circle

(d) a square

(d) 44.3

33. p, x1, x2 ..., xn and q, y 1, y2 ,..., y n are two arithmetic progressions with common differences a and b respectively. If a and b are the arithmetic means of x1, x2 .... xn and y 1, y2 ,..., y n respectively. Then the locus of P (a , b) is

44. A circle having centre at the origin passes through the three vertices of an equilateral triangle the length of its median being 9 units. Then the equation of that circle is (a) x2 + y2 = 9

(b) x2 + y2 = 18

(c) x2 + y2 = 36

(d) x2 + y2 = 81

Solved Papers 2017 45. 1 + cos 10° + cos 20° + cos 30° = (a) 4sin10° sin20° sin 30° (c) 4cos 10° cos 20° cos 30°

1×3×5×7 1×3 1×3×5 + ... to infinite terms, + + 3 × 6 3 × 6 × 9 3 × 6 × 9 × 12 then 9 x2 + 24 x =

47. If x =

(b) 4cos 5° cos 10° cos 15° (d) 4sin 5° sin10° sin15°

46. The set of all values of a such that both the points (1, 2) and (3, 4) lie on the same side of the line 3 x - 5 y + a = 0 (a) { x ÎIR : x > 11} (c) { x ÎIR : x < 7}

11

(b){ x ÎIR : x > 11} È { x ÎIR : x < 7} (d) f

(a) 31 (b) 11 (c) 41 (d) 21

Telangana State EAMCET 1. If tan 20° = l, then (a)

1 + l2 2l

(b)

tan 160° - tan 110° = 1 + (tan 160° ) (tan 110° )

1 + l2 l

(c)

1 - l2 l

(d)

1 - l2 2l

2. Consider the circle x2 + y 2 - 6 x + 4 y = 12. The equations of a tangent of this circle that is parallel to the line 4 x + 3 y + 5 = 0 is (a) 4 x + 3 y + 10 = 0 (c) 4 x + 3 y + 9 = 0

(b) 4 x + 3 y - 9 = 0 (d) 4 x + 3 y - 31 = 0

3. The mean deviation from the mean 10 of the data 6, 7, 10, 12, 13, a, 12, 16 is (a) 3.5 (c) 3

(b) 3.25 (d) 3.75

37 44 21 (c) 44

(a)

31 44 41 (d) 44

1 2

(b)

(b) 2 x - 4 y + 7 = 0 (d) 18 x - 12 y + 1 = 0

6. If cosec q - cot q = 2017, then quadrant in which q lies is (c) III

(d) II

7. If A = (5, 3), B = (3, - 2) and a point P is such that the area of the triangle PAB is 9, then the locus of P represents (a) a circle (c) a pair of parallel lines

(b) a pair of coincident lines (d) a pair of perpendicular lines

8. A straight line makes an intercept on the Y -axis twice as long as that on X-axis and is at unit distance from the origin. Then the line is represented by the equations (a) 2 x + 3 y = ± (c) x + y = ± 2

5

(d)

1 3

II. the directrix is y + 3 = 0 Which of the following is correct? (a) Both I and II are correct (c) Both I and II are false

x2 + 5

=

2

( x + 1) ( x - 2)

(a) - 1

(c) x - iy =

(b) IV

3 2

I. the vertex is ( - 2, - 3)

(b)

(b) I is true, II is false (d) I is false, II is true

A Bx + C , then A + B + C = + x - 2 x2 + 1

2 5

(a) x + iy = 1 - i

5. The radical centre of the circles x2 + y 2 - 2 x - 4 y - 1 = 0 x2 + y 2 - 4 x - 6 y + 5 = 0, and x2 + y 2 - 6 x - 2 y = 0 = 0 lies on the line

(a) I

(c)

(c)

-3 5

(d) 0

12. If the conjugate of ( x + iy ) (1 - 2 i) is (1 + i), then

(b)

(a) x + y - 5 = 0 (c) 4 x - 6 y + 5 = 0

1 2

10. For the parabola y 2 + 6 y - 2 x = - 5

11. If

4. A bag contains 5 red balls, 3 black balls and 4 white balls. Three balls are drawn at random. The probability that they are not of same colour is (a)

9. Let S and S ¢ be the foci of an ellipse and B be one end of its minor axis. If SBS ¢ is a isosceles right angled triangle then the eccentricity of the ellipse is

(b) x + y = ± 2 (d) 2 x + y = ± 5

1- i 1 + 2i

1- i 1 - 2i 1- i (d) x - iy = 1+ i

(b) x + iy =

13. The sides of a triangle are in the ratio 1 : 3 : 2 . Then the angles are in the ratio (a) 1 : 2 : 3 (c) 1 : 4 : 5

(b) 1 : 2 : 4 (d) 1 : 3 : 5

14. The sum of the complex roots of the equations ( x - 1)3 + 64 = 0 is (a) 6

(b) 3

(c) 6i

(d) 3 i

2z + 1 is - 2, then the locus of iz + 1 the point representing z in the complex plane is

15. If the imaginary part of

(a) a circle (c) a straight line

(b) a parabola (d) an ellipse

16. If the perpendicular distance between the point (1, 1) to the line 3 x + 4 y + c = 0 is 7, then the possible values of c are (a) - 35, 42 (c) 42, - 28

(b) 35, 28 (d) 28, - 42

12

Solved Papers 2017

17. If

x2

y2

+

2

a

(a) -

b b

2

= 1, then

4

a 2 y3

(b)

b

d2 y 2

28. The angle between the curves x2 = 8 y and xy = 8 is

=

dx

2

ay2

(c)

-b

3

(d)

a 2 y3

b

3

a 2 y2

1 2

(b)

1 3

(c)

1 4

(b) 14

(c) 16

(d) 20

20. Let f ( x) be a quadratic expression such that f (0) + f (1) = 0. If f ( - 2) = 0, then -2 2 (a) f æç ö÷ = 0 (b) f æç ö÷ = 0 è 5ø è 5ø

-3 (c) f æç ö÷ = 0 è 5 ø

3 (d) f æç ö÷ = 0 è 5ø

21. If the line x + y + k = 0 is a normal to the hyperbola x2 y2 = 1 then k = 9 4 (a) ±

5 13

(b) ±

13 5

(c) ±

22. The

product of all 8 1 x - 8 x + 9 - + 2 = 0 is x x

13 5

the

(d) ±

real

5 13

roots

of

2

(a) 2

(b) 1

(c) 3

(d) 7

23. A village has 10 players. A team of 6 players is to be formed. 5 members are chosen out of these 10 players and then the captain is chosen from the remaining players. Then total number of ways of choosing such teams is (a) 1260

(b) 210

(c) (10 C 6 )5!

(d) (10 C 5 ) 6

24. The equation of the straight line passing through the point of intersection of 5 x - 6 y - 1, 3 x + 2 y + 5 = 0 and perpendicular to the line 3 x - 5 y + 11 = 0 is (a) 5 x + 3 y + 18 = 0 (c) 5 x + 3 y + 8 = 0

(b) - 5 x - 3 y + 18 = 0 (d) 5 x + 3 y - 8 = 0

25. An integer is chosen from {2k / -9 £ k £ 10}. The probability that it is divisible by both 4 and 6 is (a)

1 10

(b)

1 20

(c)

1 4

(d)

3 20

26. a and b are the roots of x2 + 2 x + c = 0. If a 3 + b3 = 4, then the value of c is (a) - 2

(b) 3

(c) 2

(d) 4

27. Using the letters of the word TRICK, a five letter word with distinct letters is formed such that C is in the middle. In how many ways this is possible? (a) 6

(b) 120

(c) tan-1 (-

-1 (d) tan-1 æç ö÷ è 3ø

3)

(a) (5, 1, 0) (c) (5, 3, 1)

(d) 0

19. If the coefficients of (2r + 1) th term and ( r + 1) th term in the expansion of (1 + x) 42 are equal then r can be (a) 12

(b) tan-1(- 3)

29. A parallelogram has vertices A ( 4, 4, - 1), B (5, 6, - 1), C (6, 5, 1) and D ( x, y , z). Then the vertex D is

æ 1 2 ö 18. lim ç 2 - 4 ÷= y ®1 è y - 1 y - 1ø (a)

-1 (a) tan-1 æç ö÷ è3ø

(c) 24

(d) 72

(b) (- 5, 0, 1) (d) (5, 1, 3)

30. If 2 x2 - 10 xy + 2ly 2 + 5 x - 16 y - 3 = 0 represents a pair of straight lines, then point of intersection of those lines is (a) (2, - 3) -7 ö (c) æç - 10, ÷ è 2 ø

(b) (5, - 16) -3 (d) æç - 10, ö÷ è 2 ø

31. In order to eliminate the first degree terms from the equation 4 x2 + 8 xy + 10 y 2 - 8 x - 44 y + 14 = 0 the point to which the origin has to be shifted is (a) (- 2, 3) (c) (1, - 3)

(b) (2, - 3) (d) (-1 , 3)

32. Two circles of equal radius a cut orthogonally. If their centres are (2, 3) and (5, 6) then radical axis of these circles passes through the point (a) (3a, 5a) 5a (c) æç a, ö÷ è 3ø

(b) (2a, a) (d) (a, a)

33. If tan q 1 = k cot q 2 , then

cos(q 1 + q 2 ) = cos(q 1 - q 2 )

1+ k 1- k k+1 (c) k -1

1- k 1+ k k -1 (d) k+1 (b)

(a)

34. In the expansion of (1 + x) n , the coefficients of pth and ( p + 1)th terms are respectively p and q then p + q = (a) n + 3 (c) n

(b) n + 2 (d) n + 1 n

35. For any integer n ³ 1, S K ( K + 2) = K =1

n (n + 1) (n + 2 ) (a) 6 n (n + 1) (2 n + 1) (c) 6

n (n + 1) (2 n + 7 ) 6 n (n - 1) (2 n + 8) (d) 6 (b)

36. The foci of the ellipse 25 x2 + 4 y 2 + 100 x - 4 y + 100 = 0 are æ 5 ± 21 ö (a) ç , - 2÷ è 10 ø

æ 5 ± 21 ö (b) ç - 2, ÷ 10 ø è

æ 2 ± 21 ö (c) ç , - 2÷ è 10 ø

æ 2 ± 21 ö (d) ç - 2, ÷ 10 ø è

Solved Papers 2017 é æpö ê 1 + cos çè 12 ÷ø + i sin 37. ê ê 1 + cos æç p ö÷ - i sin è 12 ø êë (a) 0

æ p öù ç ÷ú è 12 ø ú æ p öú ç ÷ è 12 ø úû

(b) - 1

72

between the pair ax2 - 13 xy - 7 y 2 + x + 23 y - 6 = 0 is =

(c) 1

(d)

5 ö (a) cos -1 æç ÷ è 218 ø 5 ö (c) cos -1 æç ÷ è 173 ø

1 2

38. If the range of the function f ( x) = - 3 x - 3 is {3, - 6, - 9, - 18}, then which of the following elements is not in the domain of f ? (a) - 1

(b) - 2

(c) 1

(d) 2

(b) 3

(c)

3 2

(b) ab + bc + ca (d) abc

æpö æpö 41. If A ç ÷, B ç ÷ are the points on the circle represented è 3ø è6 ø in parametric from with centre (0, 0) and radius 12 then the length of the chord AB is (a) 6( 6 - 2 ) (b) 6( 6 -

1 ö (b) cos -1 æç ÷ è 10 ø 1 (d) cos -1 æç ö÷ è 5ø

43. The number of solutions of cos 2q = sin q in (0, 2p) is (a) 4 (c) 2

(a) 84

(d) 9

40. If a and b are the roots of the equation ax2 + bx + c = 0 1-a 1-b and the equation having roots and is a b px2 + qx + r = 0, then r = (a) a + 2 b (c) a + b + c

lines

(b) 3 (d) 5

44. The lengths of the sides of a triangle are 13, 14 and 15. If R and r respectively denote circumradius and inradius of that triangle, then 8R + r =

39. In DABC, if a = 1, b = 2, ÐC = 60° then 4D2 + c2 = (a) 6

of

13

3 ) (c) 2 ( 3 - 1) (d) 6( 3 - 1)

42. If the pair of straight lines xy - x - y + 1 = 0 and the line x + ay - 3 = 0 are concurrent, then the acute angle

(b)

65 8

(c) 4

(d) 69

45. If A and B are variances of the 1st ‘n’ even numbers and 1st ‘n’ odd numbers respectively then (a) A = B (c) A < B

(b) A > B (d) A = B - 1

46. If the line x - y = - 4K is a tangent to the parabola y 2 = 8 x at P, then the perpendicular distance of normal at P from ( K , 2K ) is (a)

5 2 2

(b)

7 2 2

(c)

9 2 2

(d)

1 2 2

47. If A and B are events having probabilities, P ( A) = 0.6, P ( B) = 0.4 and P ( A Ç B) = 0, then probability that neither A nor B occurs is (a)

1 4

(b) 1

(c)

1 2

(d) 0

VIT 1. The principal amplitude of (sin 40° + i cos 40° ) 5 is (a) 70°

(b) - 110°

(c) 110°

(d) - 70°

2. Let w be the complex number cos

2p 2p + i sin . 3 3

Then, the number of distinct complex number z satisfying w2 1

z+1 w w z + w2 w2

1

(a) 1

= 0 is equal to

(c) 2

x-1 3. If f ( x) = , then f (2 x) is x+1 f ( x) + f ( x) + f ( x) + (c) f ( x) + (a)

1 3 3 1

3f( x) + 1 f ( x) + 3 f ( x) + 3 (d) 3f( x) + 1

(b)

(a) {(2, 4), (3, 4)} (c) {(2, 4), (3, 4), (4, 4)}

(d) 3

(b) {(4, 2), (4, 3)} (d) {(2, 2), (3, 3), (4, 4) (5, 5)}

5. Let A and B be two non-empty sets having n elements in common. Then, the number of elements common to A ´ B and B ´ A is (a) 2n (c) n2

z+w (b) 0

4. If A = { x : x2 - 5 x + 6 = 0}, B = {2, 4}, C = { 4, 5}, then A ´ ( B Ç C ) is

(b) n (d) None of these

6. The foci of a hyperbola are ( - 5, 18) and (10, 20) and it touches the Y -axis. The length of its transverse axis is (a) 100

(b) 89 /2

(c) 89

(d) 50

7. The locus of mid-point of the line segment joining the locus to a moving point on the parabola y 2 = 4ax is another parabola with directrix (a) x = - a (c) x = 0

(b) x = a (d) x = a / 2

14

Solved Papers 2017

8. The equation of the ellipse with its centre at (1, 2), one locus at (6, 2) and passing through (4, 6) is 2

2

( x - 1) ( y - 2) + =1 45 20 ( x + 1)2 ( y + 2 )2 (c) + =1 45 20

(a)

2

(b)

(a) e 4 - e 2 + 1 = 0

(b) e 2 - e + 1 = 0

2

(d) e 4 + e 2 - 1 = 0

(c) e + e + 1 = 0

2

( x - 1) ( y - 2) + =1 20 45

10. lim (tan x cot a )

1 x -a

x ®a

(d) None of the above

9. If the normal at one end of the latusrectum of an ellipse x2 y2 + = 1 passes through the one end of the minor a2 b2 axis, then

is equal to

(a) 2 cosec 2 a 1 (b) sin2 a 2 (c) - 2 cosec 2 a (d) None of these

MHT CET 1. The number of principal solutions of tan 2q = 1 is (a) one (c) three

(b) two (d) four

2. If z1 and z2 are z-coordinates of the points of trisection of the segment joining the points A(2, 1, 4), B ( -1, 3 , 6), then z1 + z2 = (a) 1 (c) 5

(b) 4 (d) 10

3. The statement pattern (~ p ^ q) is logically equivalent to (a) ( p Ú q ) Ú ~ p (c) ( p Ù q ) ® p

(b) ( p Ú q ) Ù ~ p (d) ( p Ú q ) ® p

4. O(0, 0), A(1, 2), B(3, 4) are the vertices of D OAB. The joint equation of the altitude and median drawn from O is (a) x2 + 7 xy - y2 = 0 2

2

(b) x + 7 xy + y = 0

(a) 1

(b)

(d) 3 x2 + xy - 2 y2 = 0

5. In D ABC, if sin2 A + sin2 B = sin2 C and l( AB) = 10, then the maximum value of the area of D ABC is (b) 10 2 (d) 25 2

3 2

(c) 3

(d) 4

7. Which of the following statement pattern is a tautology? (a) p Ú (q ® p) (c) (q ® p) Ú (~ p « q )

(b) ~ q ® ~ p (d) p Ù ~ p

8. If lines represented by equation px2 - qy 2 = 0 are distinct, then (a) pq > 0

(b) pq < 0

(c) pq = 0

(d) p + q = 0

9. If slopes of lines represented by kx2 + 5 xy + y 2 = 0 differ by 1, then k = (a) 2

(b) 3

(c) 6

(d) 8

10. If c denotes the contradiction, then dual of the compound statement ~ p Ù (q Ú c) is (a) ~ p Ú (q Ù t ) (c) p Ú (~q Ú t )

(c) 3 x2 - xy - 2 y2 = 0

(a) 50 (c) 25

6. A boy tosses fair coin 3 times. If he gets ` 2 X for X heads, then his expected gain equals to ` ……

(b) ~ p Ù (q Ú t ) (d) ~ p Ú (q Ù c )

11. Probability that a person will develop immunity after vaccinations is 0.8. If 8 people are given the vaccine, then probability that all develop immunity is = (a) (02 . )8 (c) 1

(b) (0.8)8 (d) 8 C 6 (02 . )6 (0.8)2

15

Solved Papers 2017

Answer with Explanations JEE Main 1. (a) p

~p q px®ºq ~p ~p ® q ®y qº)(® q

T T F F

T F T F

T F T T

F F T T

T T T F

T F T T

x® y

T T T T

2. (d) Given, 5 (tan2 x - cos2 x ) = 2cos 2x + 9 Þ

æ 1 - cos 2x 1 + cos 2x ö 5ç ÷ 2 è 1 + cos 2x ø

= 2cos 2x + 9 Put cos2x = y, we have æ1 - y 1 + y ö 5ç ÷ = 2y + 9 2 ø è1 + y Þ 5 (2 - 2 y - 1 - y2 - 2 y ) Þ Þ Þ \ Þ

= 2(1 + y )(2 y + 9 ) 9 y2 + 42 y + 13 = 0 (3 y + 1)(3 y + 13) = 0 1 13 y=- ,3 3 1 13 cos 2x = - , 3 3 1 éQcos2x ¹ - 13 ù cos2x = êë 3 3 úû

= P ( A) + P ( B ) + P (C ) P ( A Ç B) - P ( B Ç C ) - P (C Ç A) + P ( A Ç B Ç C ) 1ù 3 1 7 é = + = Q P(A Ç B Ç C ) = 16 úû 8 16 16 êë

[Q z = - 3] -3 - 1 + 3i Here, , w= 2 - 1 - 3i and w3 n = 1 w2 = 2 1 1 1 2 Now, 1 - w - 1 w2 = 3k w2 w7 1 Þ

= P(A È B ) - P(A Ç B ) = P ( A) + P ( B ) - 2P ( A Ç B ) According to the question, 1 …(i) P ( A) + P ( B ) - 2P ( A Ç B ) = 4 1 …(ii) P ( B ) + P (C ) - 2P ( B Ç C ) = 4 1 and P (C ) + P ( A) - 2P (C Ç A) = …(iii) 4 On adding Eqs. (i), (ii) and (iii), we get 2 [ P ( A) + P ( B ) + P (C ) - P ( A Ç B ) 3 - P ( B Ç C ) - P (C Ç A)] = 4 Þ P ( A) + P ( B ) + P (C ) - P ( A Ç B ) 3 - P ( B Ç C ) - P (C Ç A) = 8 \ P (atleast one event occurs) = P(A È B È C )

1 1 1 w 1 w2

w

3 0 Þ 1 w 1 w2 Þ Þ \

1+ w+ w w2

3( w - w ) = 3k ( w2 - w) = k æ - 1 - 3i ö æ - 1 + 3i ö k=ç ÷-ç ÷ 2 2 è ø è ø = - 3i = - z - 3k 1 1 = ± 28 1

k 2

[Qk Î I ] Þ k =2 Thus, the coordinates of vertices of triangle are A(2, - 6 ), B(5, 2) and C( - 2, 2). Y C (–2, 2)

h/2

β α A

P 2h

7. (c) 225 a2 + 9 b2 + 25 c2 - 75 ac - 45ab - 15bc = 0 (15a)2 + (3b )2 + (5c )2 - (15a) (5c ) - (15a)(3b ) - (3b )(5c ) = 0 1 [(15a - 3b )2 + (3b - 5c )2 2 + (5c - 15a)2 ] = 0

Now, b2 = a2 (1 - e2 ) 2 é 1 ù 1 = (2)2 ê1 - æç ö÷ ú = 4 æç1 - ö÷ = 3 è2 ø ú è ø 4 êë û b= 3 Þ \ Equation of the ellipse is x2 y2 x2 y2 + = 1 Þ + =1 4 3 (2)2 ( 3 )2 3 Now, the equation of normal at æç1, ö÷ is è 2ø 4x 3y =4 -3 Þ 1 (3 / 2)

9. (b) Let the equation of hyperbola be x2 y2 = 1. a2 b2 ae = 2 Þ a2 e2 = 4

\

D

B (5, 2) (2, 1/2)

Þ

E X′

C

8. (b)\ a = 2

4

k 1 5 2 -k

B

h

Þ

5. (d) \

…(ii)

h/2

= 3k

w

0 w2 = 3k w

2

3x + 8 y - 10 = 0 1 x = 2 and y = . 2

6. (c)

Þ

2

3 1+ w+ w w 1 w2 1

Þ

Now, in DABP, AB h 1 tan ( a + b ) = = = AP 2h 2

1 w2 = 3k

2

1 2 7 = 2 æç - ö÷ - 1 = - 1 = è 3ø 9 9

3. (b) We have, P (exactly one of A or B occurs)

2w + 1 =

On applying R1 ® R1 + R2 + R3 , we get

Now cos 4 x = 2cos2 2x - 1 2

Þ

4. (a) Given, 2 w + 1 = z Þ

…(i) Þ x =2 Equation of altitude from vertex C is -1 y-2= [ x - ( - 2)] é2 - ( - 6 ) ù ê ú ë 5-2 û

2

X

O

a2 + b2 = 4

\

A (2, –6) Y′

Now, equation of altitude from vertex A is -1 y - (- 6 ) = ( x - 2) æ 2-2 ö ç ÷ è - 2 - 5ø

Þ

b2 = 4 - a2

2

x y =1 a2 4 - a2

Since, ( 2, 3 ) lie on hyperbola. 2 3 =1 \ a2 4 - a2 Þ

a4 - 9 a2 + 8 = 0

Þ( a4 - 8)( a4 - 1) = 0

16

Solved Papers 2017 a4 = 8, a4 = 1 Þ a = 1

Þ

13. (d) ( 21C1 - 10C1 ) + (21C2 - 10C2 ) +

Now, equation of hyperbola is x2 y2 = 1. 1 3 \ Equation of tangent at ( 2, 3 ) is given by 3y =1 2x 3

10. (b) lim x ® p /2

=

lim

cot x - cos x ( p - 2 x )3

(21C3 - 10C3 ) + ... + (21C10 - 10C10 )

Þ

= ( 21C1 +

Þ

1 = (21C1 + 2 1 = (21C1 + 2

21

C2 + ... + 21C10 ) - (10 C1 + 10C2 + ... + 10C10 )

21

21

or

21

21

But

C2 + ... + C2 + ... +

1 = (221 - 2) - (210 - 1) = 220 - 210 2

1 cos x(1 - sin x ) × 3 p sin x æç - x ö÷ è2 ø

2

11. (a) Total number of ways of selecting 2 different numbers from {0, 1, 2, ..., 10} = 11C2 = 55 Let two numbers selected be x and y. Then, …(i) x + y = 4m and …(ii) x - y = 4n Þ 2x = 4( m + n ) and 2 y = 4( m - n ) Þ x = 2( m + n ) and y = 2( m - n ) \ x and y both are even numbers.

Now, f ( x + y ) = f ( x ) + f ( y ) + xy Put y=0 Þ f ( x ) = f ( x ) + f (0 ) + 0 Þ f(0 ) = 0 Þ c = 0 Again, put y = - x \ f (0 ) = f ( x ) + f ( - x ) - x2 Þ

0 = ax2 + bx + ax2 - bx - x2

2ax2 - x2 = 0 1 a= Þ 2 Also, a+ b+ c=3 1 + b+ 0 =3 Þ 2 5 b= Þ 2 x2 + 5x f(x ) = \ 2 n2 + 5n Now, f ( n ) = 2 1 2 5 = n + n 2 2 10 1 10 2 5 \ å f( n) = 2 å n + 2 n =1 n =1 Þ

0 2 4 6 8 10

4, 8 6, 10 0, 8 2, 10 0, 4 2, 6 6 \ Required probability = 55

\ Þ

|a - b| = 1 ( a - b )2 = 1

Again,

( a - b )2 = ( a + b )2 - 4 ab

Þ

ån n =1

–r

X′ 3

+ C1 ´ C2 ´ C2 ´ C1 + 3 C0 ´ 4 C3 ´ 4 C3 ´ 3C0

O

)

C1 ´ 4 C1 ´ 3C2

3 = 3n2 - 4 n2 + 124 n = 121 2

10

,4

r

2 æ n2 - 31 ö - 3n ö 1 = æç ÷ ÷ - 4ç è 3 ø è 3 ø

Þ

(0

r

3x2 + 3nx + n2 - 31 = 0

Let a and b be the roots.

Þ \

X y=4 – x2

Y′

Since, circle touches the line y = x in first quadrant.

n = ± 11 n = 11

[Qn > 0]

JEE Advanced Paper 1 1. (b,d) =

y=|x|

= 3C3 ´ 4 C0 ´ 4 C0 ´ 3C3 + 3C2 ´

= 1 + 144 + 324 + 16 = 485

[(1 + 3 + 5 + ... + (2n - 1)] x + [(1 × 2 + 2 × 3 + ... + ( n - 1)n] = 10 n n( n2 - 1) - 10 n = 0 Þ nx2 + n2 x + 3 n2 - 1 Þ - 10 = 0 x2 + nx + 3

Then, then coordinates of centre = (0, 4 - r ).

12. (a)\Total number of required ways

4

x( x + 1) + ( x + 1)( x + 2) + ... + ( x + n - 1) ( x + n ) = 10 n Þ ( x2 + x2 + ... + x2 ) +

Since, a and b are consecutive.

15. (c) Let the radius of circle with least area be r.

Y

4 = 4 ( 2 - 1) 2 +1

16. (d) Given quadratic equation is

Þ

1 10 ´ 11 ´ 21 5 10 ´ 11 = × + ´ 2 6 2 2 385 275 660 = + = = 330 2 2 2

y

4

r=

\

14. (a) We have, f ( x ) = ax + bx + c

æ sin h ö ÷ sin h ö ç 1 1 1 æ × = lim ç ÷ ç h2 ÷ × 0 h ® è ø h 4 ÷ cos h 4 ç è 2 ø 1 1 1 = ´ = 4 4 16

3

éQ 4 < 0 ù êë 1 - 2 úû

2

p p é ù cos æç - h ö÷ ê1 - sin æç - h ö÷ ú è2 øë è2 øû 1 = lim × 3 h ®08 p p p sin æç - h ö÷ æç - + h ö÷ ø è2 ø è2 2 sin h (1 - cos h ) 1 = lim 8 h ® 0 cos h × h3 h sin h æç2sin2 ö÷ è ø 1 2 = lim 8 h ® 0 cos h × h3

4

C21 - 1) - (210 - 1)

x ® p /2 8

x

C20 ) - (210 - 1)

0 - (4 - r ) =r 2 r -4 = ± r 2 4 r= 2+1 4 1- 2 4 r¹ 1- 2

\

az + b ax + b + aiy = z+1 ( x + 1) + iy ( ax + b + aiy )(( x + 1) - iy ) ( x + 1)2 + y2

æ az + b ö \ Im ç ÷ è z+1 ø - ( ax + b ) y + ay( x + 1) = ( x + 1)2 + y2 ( a - b) y = y Þ ( x + 1)2 + y2 Q

a- b=1

Solved Papers 2017 \

( x + 1)2 + y2 = 1 x = -1 ±

\

1 - y2

2. (a,b,c) Tangent º 2 x - y + 1 = 0 x2 y2 =1 2 16 a It point º ( asec q, 4 tan q), x sec q y tan q tangent º =1 a 4 sec q = - 2a and tan q = - 4 17 Þ 4 a2 - 16 = 1 Þ a = 2 Hyperbola º

9. (a) Either option (a), (b) or (c) is correct. Satisfying the point (8, 16 ) in (III), we get \ The conic is y2 = 32 x = 16(2 x ) Now, equation of tangent at (8, 16) is y × 16 = 16( x + 8 ) Þ y = x + 8, which is the given in equation. So, option (a) is correct (16-18) f ( x ) = x + ln x - x ln x f(1) = 1 > 0 f ( e2 ) = e2 + 2 - 2e2 = 2 - e2 < 0

3. (b) 4. (6) Let the sides are a - d, a and a + d. Then, a( a - d) = 48 and

a2 - 2ad + d2 + a2 = a2 + 2ad + d2

Þ

a2 = 4 ad Þ a = 4 d

Thus, a = 8, d = 2 Hence, a- d=6

5. (2) The circle and coordinate axes can have 3 common points, if it passes through origin. [ p = 0] If circle is cutting one axis and touching other axis. Only possibility is of touching X-axis and cutting Y-axis. [ p = - 1]

6. (5) x = 10!

7. (a) Either option (a) or (c) is correct. But a = 2 and ( -1, 1) satisfy x2 + y2 = a2 . 8. (b) Either option (b) or (c) is correct. 1ö æ Satisfying the point ç 3 , ÷ in (II), è 2ø we get a2 = 4. 2

2

\ The conic is x + 4 y = 4 1 Now, equation of tangent at ( 3 , ) 2 is 3 x + 2 y = 4

Now, lim f ( x ) x ® 1-

æ 1 ö 1 - x(1 + 1 - x ) cos ç ÷ 1-x è1 - x ø

= lim

x ® 1-

æ 1 ö = lim (1 - x )cos ç ÷=0 x ® 1è1 - x ø and

lim f ( x )

x ® 1+

æ 1 ö 1 - x(1 - 1 + x ) cos ç ÷ x -1 è1 - x ø

= lim x ®1

+

1 ö = lim - ( x + 1) × cos æç ÷, which è x + 1ø x ® 1+ does not exist.

4. (d) a2 = a + 1 Þ b2 = b + 1 an = pan + qb n

Þ f ( x ) = 0 for some x Î(1, e2 )

= p( an - 1 + an - 2 ) + q(b n - 1 + b n - 2 )

\ I is correct

= an - 1 + an - 2 \ a12 = a11 + a10

1 1 - ln x - 1 = - ln x x x f ¢( x ) > 0 for (0, 1) f ¢( x ) < 0 for ( e, ¥) \ P and Q are correct, II is correct, III is incorrect. -1 1 f ¢¢( x ) = 2 x x f ¢¢( x ) < 0 for (0, ¥) \ S, is correct, R is incorrect. IV is incorrect. lim f ( x ) = - ¥ f ¢( x ) = 1 +

x ®¥

10 ! y = 10C1 ´ 9C8 ´ 2! 10 ! = 10 ´ 9 ´ 2 y 10 = =5 9x 2 Solutions. (7-9) Using standard equations I-ii-Q; II-iv-R; III-i-P; IV-iii-S Comparing with equations in given questions, we get

17

lim f ¢( x ) = - ¥ x ®¥

5. (d) a =

1+ 5 1- 5 , b= 2 2

a4 = a3 + a2 = 2a2 + a1 = 3a1 + 2a0 28 = p(3a + 2) + q(3b + 2) æ3 5 ö 3 28 = ( p + q )æç + 2ö÷ + ( p - q )ç ÷ è2 ø è 2 ø \

p - q =0 7 = 28 2 Þ p + q =8 Þ p = q =4 \ p + 2q = 12 and

( p + q) ´

lim f ¢¢( x ) = 0 x ®¥

BITSAT

\ ii, iii, iv are correct.

21

1.(c) (1 + x ) + (1 + x )22 + .... + (1 + x )30

Paper 2 5

4

1. (c) N i = C k ´ C5 - k 2. (b,c) We have, 4(cos b - cos a) + 2cos acos b = 2 Þ 1 - cos a + cos b - cos acos b = 3 + 3cos a - 3cos b - 3cos acos b Þ

(1 - cos a)(1 + cos b )

= 3(1 + cos a)(1 - cos b ) (1 - cos a) 3(1 - cos b ) Þ = (1 + cos a) 1 + cos b a b tan2 = 3 tan2 Þ 2 2 a b \ tan ± 3 tan = 0 2 2

3. (c,d) f ( x) =

æ 1 ö 1 - x( 1 + | 1 - x|) cos ç ÷ è1 - x ø |1 - x|

= (1 + x )21 [1 + (1 + x )1 + .... + (1 + x )9 ] é (1 + x )10 - 1 ù = (1 + x )21 ê ú ë (1 + x ) - 1 û 1 = [(1 + x )31 - (1 + x )21] x \ Coefficient of x5 in the given expression = Coefficient of x5 in 1 [(1 + x )31 - (1 + x )21] x = Coefficient of x6 in [(1 + x )31 - (1 + x )21] = 31C6 - 21C6

2. (b) arg ( z - 1) = arg ( z + 3 i ) Þarg (( a - 1) + ib ) = arg ( a + ( b + 3) i ) æ b ö -1 æ b + 3 ö Þ tan -1 ç ÷ = tan ç ÷ è a ø è a - 1ø Þ

b+3 b = a-1 a

18

Solved Papers 2017

3. (a) Let the ellipse be

x2 y2 + 2 =1 2 a b

x cos q y sin q + =1 a b

and let

…(i)

be a tangent to it at point ( a cos q, b sin q). Then, p = length of the perpendicular from S( ae, 0 ) on Eq. (i)

6. (c) Let a, a2 be the roots of \ Product of the roots, a × a2 = 1 Þ a3 = 1 a = 1, w, w2

Þ

p 3 p 1+1=3 p = -6

a + a2 = -

Again, Þ

Þ

e cos q - 1

p=

cos2 q sin2 q + a2 b2 p¢ = length of the perpendicular from O(0, 0 ) on Eq. (i)

Þ

1

p¢ =

cos2 q sin2 q + a2 b2

4. (a) We have, n

10 r =1

=

r -1 n

r =1

å{( n + 1) - r} r =1

n

Þ

n

Cr

å r . nC

r =1

=

10

C Þ å r . n r = 10 ( n + 1) C r -1 r =1 10

7. (d) log 140 63 = log 22

å( n - r + 1)

C Þ år . n r = Cr - 1 r =1 10

år r =1

= 10 ( n + 1) - 55

5. (d) Since, 32 sin 2 a -1, 14 and 34 - 2 sin 2 a are in AP. Therefore, 2 ´ 14 = 32 sin 2 a - 1 + 34 - 2 sin 2 a a 34 , where a = 32 sin 2 a + 3 a Þ a2 - 84 a + 243 = 0 28 =

( a - 81) ( a - 3) = 0 a = 81, a = 3 32 sin 2 a = 34

or

32 sin 2 a = 3

Þ

(3 ´ 3 ´ 7 )

log2 (3 ´ 3 ´ 7 ) log2 (22 ´ 5 ´ 7 )

cos ( x - y ), cos x, cos ( x + y ) are in HP. 2 cos ( x - y ) cos ( x + y ) Then, cos x = cos ( x - y ) + cos ( x + y )

r -1

Þ Þ Þ

´5 ´7

8. (a) Given that, 10

= 10 n - 45 = 5 (2n - 9 )

Þ

Þ But p > 0 \ a = 1 is not possible. If a = w, then a + a2 = w + w2 = - 1 p \ -1= 3 Þ p =3 Again, if a = w2 , then

10

Cr

10

(if a = 1)

a + a2 = w2 + w4 = w2 + w = - 1 p -1= \ 3 Þ p =3

and r = ae cos q - a Clearly, rp ¢ = ap

å r . nC

10. (a) Given, that, A È X = B È X

3x2 + px + 3

2 sin 2a = 1 [Q2sin 2 a ¹ 4] 1 sin2a = Þ 2 Þ 2a = 30° [Qsin 30° = 1 / 2] Thus, the first three terms of the AP are 1, 14, 27. Hence, its fifth term a5 = a1 + (5 - 1) d = 1 + 4 ´ 13 = 1 + 52 = 53

Þ

cos x =

2 (cos2 x - sin2 y ) 2 cos x cos y

Þ cos2 x cos y = cos2 x - sin2 y 2

Þ

cos x (1 - cos y )

Þ

= (1 - cos y ) (1 + cos y ) cos2 x = 1 + cos y

Þ

cos2 x = 2cos2 y / 2

Þ

cos2 x sec2 ( y / 2) = 2

Þ

cos x sec ( y / 2) = ±

2

9. (c) Any number between 1 to 999 is of the form abc when 0 £ a, b, c £ 9. Let us first count the number in which 5 occurs exactly once. Since, 5 can occur at one place in 1 ´3 C1 ´ 9 ´ 9 = 243 ways, next 5 can occur in exactly two places in 3 C2 ´ 9 = 27. Lastly, 5 can occur in all three digits in only one way. Hence, the number of times 5 occurs = 1 ´ 243 + 27 ´ 2 + 1 ´ 3 = 243 + 54 + 3 = 300

Þ A Ç (A È X ) = A Ç (B È X ) Þ ( A Ç A) È ( A Ç X ) = (A Ç B) È (A Ç X ) Þ A È f = ( A Ç B ) È f [Q A Ç X = f] …(i) Þ A = AÇ B Again, consider A È X = B È X Þ B Ç (A È X ) = B Ç (B È X ) Þ ( B Ç A) È ( B Ç X ) = (B Ç B)È (B Ç X ) Þ ( B Ç A) È f = B È f [ B Ç X = f] …(ii) Þ AÇB=B Thus from Eq. (i) and (ii), we get A=B

11. (b) We have, sin x - 3sin 2x + sin 3x = cos x - 3cos 2x + cos 3x Þ sin x + sin 3x - 3sin 2x = cos x + cos 3x - 3cos 2x Þ 2sin 2x cos x - 3sin 2x 2cos 2x cos x + 3cos 2x = 0 Þ sin 2x (2 cos x - 3) - cos 2x (2 cos x - 3) = 0 Þ (sin 2x - cos 2x ) (2 cos x - 3) = 0 [Qcos x ¹ 3 / 2] Þ sin 2x = cos 2x p æ ö Þ 2x = 2xp ± ç - 2x ÷ è2 ø np p [Qneglect - ve sign] Þ x= + 2 8

12. (b) The third side is parallel to a bisector of the angle between equal sides. The bisectors are 7 x - y + 3 = ± 5 ( x + y - 3) Þ 2x - 6 y + 18 = 0 or 12x + 4 y - 12 = 0 Þ x - 3y + 9 = 0 or 3x + y - 3 = 0 Let the third side be x - 3 y = k or 3x + y = L It passes through (1, - 10 ). k = 31, L = - 7 Hence, required lines are x - 3 y = 31, 3x + y = - 7

13. (d) Let (t, m) be the other end of the chord drawn from the point ( p, q ) on the circle x2 + y2 = px + q y t + p m + qö Their mid-point is æç , ÷ è 2 2 ø Since, mid-point lies on X-axis i.e. y=0 …(i) m+ q =0

Solved Papers 2017 Also, (t, m ) lies on the circle. \ t2 + m2 - pt - qm = 0

21. (c)\ Required probability …(ii)

From Eqs. (i) and (ii), we get t2 - pt + 2q2 = 0 Which is quadratic in t such that, Discriminant > 0 Þ p2 - 8q2 > 0 Þ p2 > 8q2 1/ 2 1 ö ïü ïì æ 14. (b) lim sin í np ç1 + 2 ÷ ý è n®¥ n ø þï îï

1 1 ü ì = lim sin ínp æç1 + 2 - 4 + ....ö÷ ý n® ¥ øþ 2n 8n î è p p = lim sin ìínp + + ....üý n® ¥ 2n 8n3 î þ 1 1 æ n = lim ( -1) sin p ç + ....ö÷ = 0 n® ¥ è 2n 8n3 ø

15. (b) Clearly, the point of intersection of curves is (0, 1). Now, slope of tangent of first curve dy = ax log a m1 = dx dy = m1 = log a Þ æç ö÷ è dx ø (0, 1) Slope of tangent of second curve, dy = b x log b m2 = dx dy = log b Þ m2 = æç ö÷ è dx ø (0, 1)

16. (c) lim n®¥

an (1 + n) - (1 + n2 ) 1+ n

( a - 1) n2 + an - 1 = ¥, n® ¥ n+1

= lim

If a - 1 ¹ 0 limit does not exist and if a - 1 = 0, then an - 1 =a=b lim n® ¥ n + 1 Þ

a= b=1

17. (b) Total ways in which papers can be checked is equal to 7 4 . Now, two teachers who have to check all the papers can be selected in 7 C2 ways and papers can be checked by them is (2 4 - 2 ) favourable ways. 18. (c) 1 19. (c) Variance = S( x - x )2 s 2 n 1 New variance = å( ax - ax )2 n 1 = a2 å( x - x )2 = a2 s2 n

20. (a) Þ

C =

s ´ 100 x

s 50 = 1 ´ 100 30

4

5

3

2

1 1 1 1 = 7C4 æç ö÷ æç ö÷ + 7C5 æç ö÷ æç ö÷ è2 ø è2 ø è2 ø è2 ø 7 6 1 1 1 + 7C6 æç ö÷ æç ö÷ + 7C7 æç ö÷ è2 ø è2 ø è2 ø

n( n + 1) n - 1 + 2 22. (c) Here, Tn = = n! ( n - 1)! 1 2 = + ( n - 2)! ( n - 1)! ¥ ¥ 1 \ S = å Tn = å ( 2)! n n =1 n =1 ¥ 1 +2 å ( n 1)! n =1 = e + 2e = 3e

23. (b) Since,|z1| = |z2| = |z3| \ 0 is the circumcentre of an equilateral DABC. x1 + x2 + x3 y + y2 + y3 \ =0 = 1 3 3 x1 + x2 + x3 y1 + y2 + y3 ö æ + iç Þ ÷ è ø 3 3 =0

= ( x - ( a - b ))2 + ( y - ( a + b ))2

2. (e) f ( x) - 3 x -3

0 [ form] 0

f ¢( x ) xf ¢( x ) 9 f ¢ ( a) 2 f(x ) = lim = lim = 1 x ®9 x ®9 f(x ) f ( a) 2 x 3´0 = =0 3

4. (e) -2 5. (d) Required area =

6. (b) ( - 4,1) =

1 2

4

C3 x3 + 4 C4 x4

= 1 + 4 x + 6 x2 + 4 x3 + x4 \ Coefficient of x5 in the product of (1 + x2 )5 (1 + x )4 = (5x2 ) . (4 x3 ) + (10 x4 ) . (4 x ) = 20 x5 + 40 x5 = 60 x5

8. (e) Here, Tr + 1 = 5C r ( - 2 x )r = 5C r ( - 2)r x r For coefficient of x4 , power of x = 4 \

r =4

9. (a) Given equation can be rewriten ( y + 1 / 2 )2 x2 as =1 + æ 33 ö æ 33 ö ç ÷ ç ÷ è8 ø è 40 ø 10. (e) (4 x2 - 8 x ) + ( y2 + 4 y ) - 8 = 0 Þ

4 ( x - 1)2 + ( y + 2)2 = 16

Þ

( x - 1)2 ( y + 2)2 + =1 4 16

11. (c)

1. (b) ( x - ( a + b ))2 + ( y - ( b - a))2

x ®9

= 1 + 5x2 + 10 x4 + 10 x6 + 5x8 + x10 (1 + x )4 = 4 C0 x0 + 4 C1x1 + 4 C2 x2 +

\ Centre = (1, - 2)

KERALA CEE

3. (a) lim

19

1

2 5

1 1

6

-1 1

1-0 -1 = -4 - 1 5

\ Slope of line perpendicular to the -1 above line = =5 æ- 1 ö ç ÷ è 5ø \ Equation of required line is given by y - 5 = 5 ( x - ( - 3))

7. (b) We have, (1 + x2 )5 = 5C0 ( x2 )0 + 5C1( x2 )1 + 5C2 ( x2 )2 + 5C3 ( x2 )3 + 5C4 ( x2 )4 + 5C5 ( x2 )5

12. (e) Since, both foci and vertices lies on Y-axis, then equation of hyperbola will be y2 x2 - 2 =1 b2 a Now, vertices = (0, ± b ) \ b = 15 Again, foci = (0, ± be ) \ be = 20 20 4 e= = Þ 15 3 a2 4 é a2 ù Þ 1+ 2 = êQ e = 1 + 2 ú 3 êë b b úû

13. (b) We have, Þ

( y - 2)2 = ( x + 1)

14. (e) We have, f ( x + y) = f ( x ) + f ( y) Put, x = y = 1 in Eq. (i),

…(i)

we get f (2) = f (1) + f (1) = 2f (1) = 2 ´ 7 Again, put x = 1, y = 2 in Eq. (i), we get f (3) = f (1) + f (2) = 7 + 2 ´ 7 = 3 ´ 7 \ f( n) = n ´ 7 Now, 100

S f ( r ) = f (1) + f (2) + f (3) + ...+ f (100 ) r =1

= 7 + 2 ´ 7 + 3 ´ 7 + K + 100 ´ 7

20

Solved Papers 2017 Þ 2a + b = 0 Product of roots a ´ b = b Þ b( a - 1) = 0 Þ a = 1, b ¹ 0 From Eq. (i), 2a + b = 0 Þ 2(1) + b = 0 b = -2 2 æ 12 + 4 ´ 2 ö a æ Now, ç x + ö÷ - ç ÷ è 2ø è 4 ø

15. (a) We have, 2

2

( x - 1) ( y + 3 / 4) + =1 1/8 1 / 16 1 1 a2 = and b2 = 8 16 1 1 and b = a= 2 2 4

Þ \ Þ

16. (d) We have, sin x + cos y = 2

a 9 = æç x + ö÷ è 2ø 4 9 \ Minimum value = 4

22. (b) We have,

17. (b) Let

=

1 1 1 1 1 + + + + 9 ! 3! 7 ! 5! 5! 7 ! 3! 9 ! 1 10 !

é 10 ! 10 ! 10 ! + + ê 3! 7 ! 5! 5! ë 9! 10 ! 10 ! ù + 7 ! 3! 9 ! úû

+ =

1 [ 10 !

10

C1 +

10

C3 +

10

C5

+

10

C7 +

10

2 1 10 - 1 (2 ) = 10 ! 10 !

18. (e)

… (i) y = 4x + c x2 2 and … (ii) + y =1 4 Put value of y from Eqs. (i) into (ii), we get x2 + (4 x + c )2 = 1 4 Þ x2 + 4 (4 x + c )2 = 4

19. (d)

2

2

2

Þ

x + 4 (16 x + 8cx + c ) = 4

Þ

x2 + 64 x2 + 32cx + 4 c2 = 4

Þ C9 ]

9

=

2

2

y = 2x + 1, y = 3x + 1, x = 4 Intersecting points of above lines are (0, 1), (4, 9), (4, 13) x1 y1 1 1 x1 y2 1 \ Area of triangle = 2 x3 y3 1 0 1 1 1 = 4 9 1 2 4 13 1

x 2

20. (c) Given f ( x ) = ò (t + t + 1)dt 0

f ¢( x ) = ( x2 + x + 1) ´ 1 - 0 = x2 + x + 1 For x Î[2, 3) f ¢( x ) > 0 \ Minimum is at x = 2 and maximum is at x = 3. Now, minimum value 2 2 ù é t3 t2 + tú = ò (t2 + t + 1)dt = ê + 0 2 û0 ë3 3 2

ò0 (t

+ t + 1)dt 3

ù é t3 t2 =ê + + tú 2 û0 ë3

21. (c) For district now zero roots D > 0 a2 - 4 b > 0 x2 + ax + b 2 æ a a2 ö = æç x + ö÷ + ç b - ÷ è 2ø 4 ø è

Þ Now,

2

a 4bö = æç x + ö÷ - æç a2 ÷ è 2ø è 4 ø We know, sum of roots a + b = - a

1 [0 (9 - 13) - 1 (4 - 4 ) + 1 (52 - 36 )] 2 1 = ´ 16 = 8 2 =

n

27. (b)\

Þ

Þ

28. (b)

Clearly, ABCD is a square.

24. (b) 25. (e)\ Correct Sxi = Incorrect Sx i - 20 + 30 = 170 + 10 = 180 Sx i (correct) \ Correct mean = x = N 180 = 12 = 15 Also, correct Sx2i = Incorrect Sx2i - (20 )2 + (30 )2 = 2830 - 400 + 900 = 2830 + 500 = 3330 \ Correct variance Sx2i (correct) = - ( x )2 N 3330 = - (12)2 = 222 - 144 = 78 15

=

Cr

36 84

n! ( n - r + 1)! ( r - 1) ! 3 = n! 7 ( n - r )! r ! n

65x + 32cx + 4( c - 1) = 0

|x| + | y| = 1 x, y Î I quadrant ì x + y = 1, ï - x + y = 1, x, y Î II quadrant ï =í ï- x - y = 1, x, y Î III quadrant ïî x - y = 1, x, y Î IV quadrant

Cr - 1 n

Again,

Since, given line is a tangent to the ellipse. \ Discriminant = 0

23. (a) We have,

and maximum value =

26. (d) We have,

2

Since, x Î[0, p / 2] and y Î[0, p / 2] \ sin x = 1 and cos y = 1 p x = and y = 0 Þ 2 p p \x + y = +0 = 2 2

S=

…(i)

n

Cr 84 = C r + 1 126

n! ( n - r )! r ! 2 = n! 3 ( n - r - 1)! ( r + 1)!

0 0 1 1 29. (e)\ Area of DABC = 7 2 1 2 5 9 1 1 53 sq unit = × 1(63 - 10 ) = 2 2 53 sq unit 2 \ Area of parallelogram ABCD = Area of DABC + Area of DACD 53 53 = + = 53 sq unit 2 2 Similarly Area of =

30. (b) We have, = | 3 + 1 + 2 3| - | 3 + 1 - 2 3| = | ( 3 )2 + (1)2 + 2 × 3 × 1| -| ( 3 )2 + (1)2 - 2 × 3 × 1| = | ( 3 + 1)2| - | ( 3 ) - 1)2| = | 3 + 1| - | 3 - 1| = ( 3 + 1) - ( 3 - 1) = 2

31. (a) 32. (a) We have y = 4 x2 - 14 x + q = 0 Since, x = 2 is the root \ 4 (2)2 - 14(2) + q = 0

Solved Papers 2017 Þ \

72

q = 12 y = 4 x2 - 14 x + 12

33. (b)

34. (c)

37. (b) Here, x + y = - b and xy = 1

43. (a) Let

y+ b+ x+ b 1 1 Now, + = x+ b y + b) ( x + b) ( y + b)

or

x6 + ax2 + x = 0

x2 + y2 = 25 z - 5 x + iy - 5 Now w = = z + 5 x + iy + 5 ( x - 5) + iy = ( x + 5) + iy

2

x + ax + 1 = 0

and

44. (c) We have,

2

6

2

( x + ax + x ) - ( x + ax + 1) = 0 Þ

x =1

\ x = 1 is the common root. (1)5 + a(1) + 1 = 0

\ Þ

a= -2

39. (a) 2

40. (c) We have, z + z + 1 = 0 z = w or w2

\

Let z = w, then 2

2

2

2

2

1 1 1 = æç w + ö÷ + æç w2 + 2 ö÷ + æç w3 + 3 ö÷ è è è wø w ø w ø = ( w + w2 )2 + ( w2 + w)2 + ( w3 + 1)2

41. (a) We have z = cos

p p - i sin 3 3

1 i 3 1 - 3i = 2 2 2 é - 1 + 3i ù =-ê ú 2 ë û é -1 + 3 i ù = -w êQ w = ú 2 ë û

=

Now, z2 - z + 1 = ( - w)2 - ( - w) + 1 = w2 + w + 1 =0

42. (c) Let

Now, we have x + y+ z = -3 \ a - r + a + a + r = -3Þ a = -1 Again, xyz = 8 \ ( a - r )( a) ( a + r ) = 8 Þ a( a2 - r2 ) = 8 Þ

- 1 (1 - r2 ) = 8

Þ

r2 = 9

Þ r=±3 \ x, y, z are - 4, - 1, 2 or 2, - 1, - 4 \ x2 + y2 + z2 = ( - 4 )2 + ( - 1)2 + (2)2 = 16 + 1 + 4 = 21

46. (c)

47. (e) 2x - 1

[Q1 + w + w2 = 0]

p p ö æ + i sin ÷ ç 1 + cos 12 12 z=ç ÷ ç 1 + cos p - i sin p ÷ è 12 12 ø

iq

45. (c) Let x = a - r , y = a, z = a + r

æ z + 1 ö + æ z2 + 1 ö + æ z3 + 1 ö ç ÷ ç ÷ ç ÷ è è è zø z2 ø z3 ø 2

… (i)

= cos q + i sin q 1 + a 1 + (cos q + i sin q) Now, = 1 - a 1 - (cos q + i sin q) (1 + cos q) + i sin q = (1 - cos q) - i sin q q q q 2cos2 + i2sin cos 2 2 2 = q q q 2sin2 - i2sin cos 2 2 2 q q q 2cos éêcos + i sin ùú 2 2 2 ë û = qé q qù 2sin êsin - i cos ú 2 ë 2 2û

\ Common root is given by 6

a=e

72

48. (d) 3 Þ \ Þ Þ

= (34 )1 - x 32 x

-1

= 34

53. (c)

Let E be the event of selecting T or R \ E = {T, R} \ n( E ) = 2

x2 + y2 = 5

Þ

38. (b) We have, x + ax + 1 = 0

52. (b)

54. (e) Number of ways of selecting one letter from the word TRICKS. n(S ) = 6C1 = 6

x2 + y2

|z| =

Þ

5

6

51. (b)

z = x + iy

\

( x + y ) + 2b xy + b ( x + y ) + b2

x6 + ax2 + 1 = 0

\ a + d + a + a - d = 21 and, ( a - d ) ( a + d ) = 45

72

35. (a)

and

50. (a) Let the numbers be a - d, a, a + d

æ cos p + i sin p ö ÷ ç 24 24 ÷ =ç p p ç cos - i sin ÷ è 24 24 ø

36. (e)

=

49. (c)

æ 2cos2 p + 2i sin p cos p ö ç ÷ 24 24 24 ÷ =ç p p p 2 ç 2cos - 2i sin cos ÷ è 24 24 24 ø

= 4 x - 8x - 6 x + 12 = 4 x ( x - 2) - 6 ( x - 2) = ( x - 2) (4 x - 6 )

21

-4x

2x - 1 = 4 - 4 x 6x = 5 5 x= 6

55. (c)\Total balls = 4 + 2 + 4 = 10 Number of ways of selecting 4 balls from 10 balls, n( S ) = 10C4 Let E = Event getting 2 red balls \ n( E ) = 4 C2 ´ 6C2

56. (d) Now, according to the question, P ( E1 ) = 0.60, P ( E2 ) = 0.40, P ( E1 Ç E2 ) = 0.20 \ Required probability = P ( E ¢1 Ç E ¢2 ) = P ( E1 È E2 )¢ = 1 - P ( E1 È E2 ) = 1 - [ P ( E1 ) + P ( E2 ) - P ( E1 Ç E2 )]

57. (d) Here, n( S) = 5C2 = 10 and \

E = {(1, 3), (1,5), (2,4 ), (3,5)} n( E ) = 4

58. (e) In 3 ´ 3 matrix, total number of elements = 3 ´ 3 = 9 \ Total number of 3 ´ 3 matrices with enteries either - 1 or 1 = 29

59. (b) S = {2 ´ 2 symmetric matrices whose entries are either zero or one} ìé1 0 ù é1 0 ù é0 0 ù é0 0 ù = íê ú, ê ú, ê ú, ê ú, îë0 1 û ë0 0 û ë0 1 û ë0 0 û é0 1 ù é0 1ù é1 1 ù é1 1ù ü ê1 0 ú, ê1 1ú, ê1 0 ú, ê1 1ú ý ë û ë û ë û ë ûþ \ n( S ) = 8 Let x = {matrix whose determinant is non-zero} ìé1 0 ù é0 1 ù é0 1ù é1 1 ù ü = íê úý ú, ê ú, ê ú, ê îë0 1 û ë1 0 û ë1 1û ë1 0 û þ \ n( x ) = 4 n( x ) 4 1 \ P(x ) = = = n( s ) 8 2

60. (c) 61. (b) Number of diagonals in a hexagon =

6(6 - 3) =9 2

22

Solved Papers 2017

62. (a) 65. (a) 68. (c)

63. (e) 66. (e) 69. (a)

64. (a) 67. (b) 70. (e)

4. (b)

x+1 +1 æ x + 1ö x - 1 f ( f ( x )) = f ç ÷= èx - 1ø x + 1 - 1 x -1 x + 1 + x - 1 2x = = =x x+1-x+1 2

= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3) (6, 4), (6, 5), (6, 6)} n( E ) = 27

73. (a) lim x ®0

= lim

\ A Ç B = A Þ n ( A Ç B ) = n( A) … (i) Again, we know that n ( A È B ) = n( A) + n( B ) - n( A Ç B ) Þ n( A È B ) = n( A) + n( B ) - n( A) [from Eq. (i)] Þ n ( A È B ) = n( B ) 2

7. (c) lim

´

2+ x +

2-x

2+ x +

2-x

2

1 4 4 sin 2q ö × = = lim æç ÷ × 2 q ® 0 è 2q ø æ sin 3q ö 9 9 ÷ ç è 3q ø

KCET 1. (d) Total number of terms in ( x + a)n - ( x - a)n ìn + 1 ï =í 2 n ï î 2

m

Þ

æ 2t ö x = sin -1 ç ÷ 2 è1 + t ø

Þ

x = 2 tan -1 t

Þ

\ The contrapositive of the statement “If x is prime number, then x is odd” is “If x is not odd, then x is not a prime number”.

… (i)

2t 1 - t2

y = 2 tan -1 t

… (ii)

p = probability that a bulb is defective 10 1 = = 100 10 9 q =1- p = \ 10 \ Required probability = P ( x = 0 ) + P ( x = 1) 4

1 9 1 9 = 5C0 æç ö÷ æç ö÷ + 5C1 æç ö÷ æç ö÷ è 10 ø è 10 ø è 10 ø è 10 ø

dy =1 dx

12. (a) 13. (b) We have, y = log(log x ) \

3. (c) We have

y = 9 - x2

Þ

y2 = 9 - x2

Þ

x2 = 9 - y2 x = 9 - y2

dy 1 1 = . dx log x x 1 = ( x log x )-1 = x log x

Again differentiating w.r.t. x, we get d2 y 1 = - ( x log x )- 1 - 1 éê1 × log x + x × ùú xû dx2 ë = - ( x log x )-2 (log x + 1) - (1 + log x ) = ( x log x )2

(3 - y ) (3 + y ) ³ 0

y Î [ - 3, 3]

But,

f ( x ) = 9 - x2

\

f ( x ) Î[0, 3]

\ Reflexion of ( a, b, g ) in XY-plane is ( a, b,- g ).

y=x \

16. (c)

18. (c) Reflexion of any point ( x, y, z ) in XY-plane is ( x, y, - z ).

From Eqs. (i) and (ii), we get

1

|x - 2| £ 1

19. (a) We know that s C × V = ´ 100 x 24 ´ 100 x 24 ´ 100 Þ x= 60 Þ x = 40 \ Arithmetic means = 40 Þ

if n is even

5

Note Correction in the question

\

2t Þ y = tan 1 - t2

if n is odd

x Î[1, 3]

Þ

=1

-1

2. (a) If p and q are two statements, then the contrapositive of the implication “if p, then q” is “if ~q then ~p”.

0

1£ x £3

Since, f ( x ) ³ 0 \ 9 - y2 ³ 0

11. (c) We have,

and tan y =

Þ \

Þ

8. (b)

74. (b)

-1+ 2£ x -2+ 2£1+ 2

Let

= cos(45° + 15° ) cos (45° - 15° )

x

Þ

17. (c) We have, f ( x ) = 9 - x2

10. (d) cos2 45° - sin2 15°

2+ x - 2-x

-1£ x -2£1

15. (d)

1 - cos 4 q 2 sin 2 q = lim - cos 6 q q ® 0 2 sin2 3 q

q ®01

é1 + i 1 + i ù 9. (b) ê ´ ú ë1 - i 1 + iû

2+ x - 2-x x

x ®0

Þ

sin2 2q 1 4 q2 = lim × . 2 2 2 q ® 0 4q sin 3q 9 q 9 q2

72. (a) Let E = outcomes in which product of two number is even

\

14. (a) We have, | x - 2| £ 1

6. (a) We have, A Ì B

x+1 71. (a) We have, f ( x ) = x -1 \

5. (a)

60 =

20. (a) Here, P ( A Ç B ) = P ( A ) × P ( B ) and P ( A¢ Ç B ¢ ) = P ( A¢ ) × P ( B ¢ ) = (1 - P ( A)) (1 - P ( B )) Again, if A and B are mutually exclusive, then P ( A Ç B ) = 0.

21. (b) Given line is y = 3x - 1 \ Slope of the given line = 3 \ Slope of a line perpendicular to the given 1 line = 3 \ Equation of required line will be -1 ( x - 1) y-2= 3 Þ 3y - 6 = - x + 1 Þ x + 3y - 7 = 0

Solved Papers 2017

1. (b) We have, cos q =

c-a b

b2 - ( c - a)2

Þ( c - a)tan q =

b

2

=

2

= 21C5 +

30

C5 + K +

21

C5 +

22

C5

= (22 C6 +

b2 - ( c - a)2

23

= ( C6 + 30

= ( C6 +

2

= 2ac - 2ac cos B [Using cosine rule] B = 2ac 1 - cos B = 2 ac 2sin2 2 B = 2 ac sin 2

22 23

C5 + ... +

30

C5 + ... +

30

30

30

C5 ) - 21C6

C5 ) - 21C6

x2 - 16 qxy - y2 = 0

…(ii)

C5 ) - C6 = 31C6 - 21C6

6. (a) The triangle formed by the given lines is shown in the adjacent figure.

y=3

C(√3, 3)

B(√3, 3)

The, equations of bisectors of these line are i.e.

4 px + xy - 4 py = 0 4 qx2 + xy - 4 qy2 = 0

…(iii)

y=–√3x

… (iv)

According to the given condition Eqs. (i) and (iv), and Eqs. (ii) and (iii) must be coincident -16 p -1 1 = = -4 q 4q 1

\ Þ

Þ Þ

-1 1 = - 64 pq Þ pq = 64

3. (a) Total number of possible outcomes = 10C3 Let A be the event that minimum of chosen number is 3 and B be the event that maximum of chosen number is 7. Then, n( A ) = Number of ways of choosing remaining two numbers from the set {4, 5, 6, 7, 8, 9, 10} = 7C2 = 21

Similarly, n( B ) = Number of ways of choosing remaining two numbers from the set

Clearly, the triangle ABC is an isosceles triangle.

Q D is mid-point of BC \ OD is median to the base BC. Thus, incentre lie on Y-axis.

and n ( A Ç B ) = Number of ways of choosing remaining one number from the set {4, 5, 6} = 3C1 = 3

OA = OB = OC OA2 = OB2

Þ ( x - 1)2 + ( y - 2)2 + ( z - 3)2 = ( x - 3)2 + ( y + 1)2 + ( z - 5)2 …(i)

Similarly OB = OC ( x - 3)2 + ( y + 1)2 + ( z - 5)2

{1, 2, 3, 4, 5, 6} = C2 = 15

= ( x - 4 )2 + ( y - 0 )2 + ( z + 3)2 Þ

x + y - 8z + 5 = 0

Similarly OA = OC ( x - 1)2 + ( y - 2)2 + ( z - 3)2

1 2

y - 1 = 0 and x - 2 = 0

\ Combining equation of lines are Þ

( y - 1) ( x - 2) = 0 xy - x - 2 y + 2 = 0

11. (b) Let the numbers are a and b. Then, we have 2ab 8 = - and ab = 2 a+ b 5 2´4 8 (Qab = 4) =a+ b 5

2 2 ³ |z| z z æ -1 ö 2 = |z| = |z| + 2 ç ÷ è|z| ø |z|

12. (c) Consider, z +

³5-

2 23 = 5 5 2 23 is , z 5

13. (d) Let the roots of given equations a are , a, ar . Then, we get r

8. (d) Let O ( x, y, z ) be the circumcentre of DABC

6

y=-

Thus, the least value of z +

7. (d)

Þ 4 x - 6 y + 4z - 21 = 0

-3 2

Equations of line passing through (2, 1) and perpendicular to the line 23 1 andy = are x= 2 2

\ The incentre lie on the median to the base.

\ Now

2x (2 y + 1) + 3 (2 y + 1) = 0 (2x + 3) (2 y + 1) = 0

and

Þ

and - 8qx2 - 2xy + 8qy2 = 0, i.e.

9. (a)

A(0, 0)

- 8 px2 - 2xy + 8 py2 = 0, 2

7 1 , y= - ,z =1 2 2

C5 ) - C6

21

D

… (i)

2

x=

y=√3x

x2 - 16 pxy - y2 = 0

…(iii)

Solving Eqs. (i), (ii) and (iii), we get

Equation of lines are x =

21

x >3 ì 3 x - 1 for ï 5. (a) f ( x ) = í x2 + 1 for - 3 £ x £ 3 ï2 x - 3 for x < -3 î

2. (a) Given equations of pair of straight lines are

6 x - 4 y - 12z - 11 = 0

10. (a) We have 4 xy + 2 x + 6 y + 3 = 0

C5

+ ... +

- ( c + a - b ) + 2ac

and

22

( c - a)

b2 - c2 - a2 + 2ac

=

(1 + x )21 + (1 + x )22 + K + (1 + x )30 = (21C6 +

b2 - ( c - a)2

tan q =

Now,

Þ

4. (a) Coefficient of x5 in the expansion of

Þ sin q = 1 - cos2 q ( c - a)2 = 1= b2

= ( x - 4 )2 + ( y - 0 )2 + ( z + 3)2

Thus, required probability 21 + 15 - 3 33 11 = = = 120 120 40

AP EAMCET

23

…(ii)

a + a + ar = 7 r a a × a + a × ar + ar × = 14 r r a and × a × ar = 8 r From Eq. (iii), we get a=2 Now, from Eq. (i), we get 2 + 2r = 5 Þ 2 + 2r 2 = 5r r Þ 2r 2 - 5r + 2 = 0 Þ

2r 2 - 4 r - r + 2 = 0

Þ 2 r( r - 2 ) - 1 ( r - 2 ) = 0 Þ ( r - 2) (2r - 1) = 0 1 Þ r = 2 or r = 2

… (i) … (ii) …(iii)

24

Solved Papers 2017 Thus, the roots are 1, 2, 4. Hence, required difference =4 -1=3

14. (c) Given, a is non-real root of x7 = 1

= a (1 + a2 + a4 + a + a3 + a5 )

Also,

= a + a3 + a5 + a2 + a4 + a6 3

4

5

7

a- a a-1 = 1-a 1-a

[Qa ¹ 1] [Q a7 = 1]

= -1

15. (c) Required point = Point of intersection of given lines

γ r

γ

β r

Þ

γ

β

C

S = a+ b + g

Area of DABC =

S1 - S2 = 0 Þ - 2x - 4 y - 3 = 0 i.e. 2x + 4 y + 3 = 0

The equation of required circle is s1 + l ( s2 - s1 ) = 0 ( x2 + y2 + 2 x + 2 y + 1) + l (2 x + 4 y + 3 ) = 0 2 y(2l + 1) + 3l + 1 = 0

A

B

19. (a) Then, the equation of common chord of these two circles is

Since, 2 x + 4 y + 3 = 0 is a diameter of this circle, therefore centre ( - (1 + l ), - (2 l + 1)) lies on it.

16. (a)

r

s( s - a)( s - b ) ( s - c )

= ( a + b + g ) ( a)(b ) ( g )

We know that

So, 2 ( - (1 + l )) + 4( - (2l + 1)) + 3 = 0 Þ - 2 - 2l - 8l - 4 + 3 = 0 -3 Þ 10 l = - 3 Þ l = 10 Thus, the required circle is 3ö æ 6 + 1ö x2 + y2 + 2x æç1 ÷ ÷ + 2 y çè 10 ø è 10 ø 9 + 1=0 10

20. (a) We have, Þ

r = D/s Q Þ Þ

r2 =

( a + b + g ) ( abg ) ( a + b + g )2

Þ

2

SS1 = T2 Þ (3x2 + 2 y2 - 5) (3(1)2 - 2(2)2 - 5) = (3x (1) + 2 y (2) - 5)2 9 x2 - 4 y2 - 24 xy + 30 x + 40 y - 55 = 0

On comparing this equation with 2

Also tan q =

2 h - ab a+ b

3x+2y+2=0

A 2x+3y+2=0 B

Þ

23. (b) Chord of contact of two tangents drawn from the point P ( a, b ) to the given circle is xa + yb - ( x + a) - 3 ( y + b ) - 8 = 0 Þ

Þ Þ

( a - 1) x + ( b - 3) y ( a + 3b + 8) = 0

Clearly, this line coincides with 5 x + y + 1 = 0. a - 1 b - 3 - ( a + 3b + 8) = = 5 1 1 Þ a - 1 = - 5 ( a + 3b + 8) and b - 3 = - a - 3b - 8 Þ 6 a + 15b = - 39 and a + 4 b = - 5 … (i) Þ 2a + 5b = - 13 and … (ii) a + 4b = - 5

\

24. (d) t 1)

P(at1t2, a(at22,

2at2))

2

Slope of line PA i.e. tan q1 =

2 at1 - at1 - at2 at12 - at1t2

=

a(t1 - t2 ) 1 = at1 (t1 - t2 ) t1

(3 x + 4 x - 9 x - 12 ) 0 have a common line, then a = (a)

(b) 2

(a) y2 = 8( x + 3)

x2 - ( a + 2) xy + y 2 + a( x + y - 1) = 0, a ¹ - 2, then the value of q is

2

(a) 1

38. If two tangents to the parabola y 2 = 8 x meet the tangent at its vertex in M and N such that MN = 4, then the locus of the point of intersection of those two tangents is

31. A pair of straight lines is passing through the point (1, 1). One of the lines makes an angle q with the positive direction of X-axis and the other makes the same angle with the positive direction of Y -axis. If the equation of the pair of straight lines is

æ 2 ö 1 (a) sin-1 ç ÷ 2 èa + 2ø æ 2 ö 1 (c) tan-1 ç ÷ 2 èa + 2ø

35. If the circles given by S º x2 + y 2 - 14 x + 6 y + 33 = 0 and S ¢ º x2 + y 2 - a2 = 0( a Î N ) have 4 common tangents, then the possible number of circles S ¢ = 0 is

(a)

x2 b

2

+

y2 a

2

=1

(c) b 2 x2 + a2 y2 = 4

(d) ( 6, 1)

the

circle

(d) 4a = c

x2

y2

= 1 makes a2 b2 intercepts on both the axes. The locus of the middle point of the portion of the tangent between the coordinate axes is

41. A variable tangent to the ellipse

(b) (d)

a2 2

x a2 x2

+ +

b2 y2 b2 y2

+

=1 =4

42. If the eccentricity of a conic satisfies the equation 2 x3 + 10 x - 13 = 0, then that conic is (a) a circle (c) an ellipse

(b) a parabola (d) a hyperbola

Engineering Entrances Solved Papers 2018 43. Assertion (A) If ( -1, 3, 2) and (5, 3, 2) are respectively the orthocentre and circumcentre of a triangle, then (3, 3, 2) is its centroi(d) Reason (R) Centroid of the triangle divides the line segment joining the orthocentre and the circumcentre in the ratio 1 : 2. Which one of the following is true?

(b) (A) and (R) are true, but (R) is not the correct explanation to (A) (c) (A) is true, (R) is false (d) (A) is false, (R) is true

log(1 + x3 ) 3| x| - x - lim = x ®-¥ | x| - 2 x x ®0 sin3 x

44. lim

(a) 1

(a) (A) and (R) are true and (R) is the correct explanation to (A)

11

(b)

1 3

(c)

4 3

(d) 0

VIT 1. The value of {sin(log i i)}3 + {cos(log i i)}3 , is (a) 1

(b) - 1

2. If f ( x) = 27 x3 -

1 x3

(c) 2

(d) 2 i

and a, b are roots of 3 x -

(a) f(a) = f(b) (c) f(b) = - 10

1 = 2, then x

(b) f(a) = 10 (d) None of these

3. Let A and B be two sets. Then, ( A È B) ¢ È ( A ¢ Ç B) is equal to (a) A¢ (c) B¢

(b) A (d) None of these

(b) 16

(c) 8

(b) ~( p Ù r ) ® ~(r Úq ) (d) ( p Ù r ) Ú(r Úq )

6. The eccentricity of the hyperbola with latus rectum 12 and semi-conjugate axis 2 3, is (a) 2

(b) 3

(c)

3 2

(d) 2 3

7. If the line x - 1 = 0 is the directrix of the parabola y 2 - kx + 8 = 0, then one of the value of k is (a)

1 8

(b) 8

(c) 4

(d)

1 4

8. The number of values of C such that the straight line x2 + y 2 = 1, is y = 4 x + c touches the curve 4 (a) 0

(b) 1

(c) 2

9. If any tangent to the ellipse

(d) ¥

x2

+

y2

= 1 intercepts

a2 b2 equal lengths l on the axes, then l is equal to

(a) a2 + b 2 2

2 2

(c) (a + b )

(b) a2 + b 2 (d) None of these

(b) 9

(c) 3

(a) (- 3, 0)

(b) (- 4, 1)

(c) (- 4, 0)

(d) (- 3,1)

(a)|a| = 2

(d) 13

(b)|a| = 1 1 (d)|a| < 2

(c)|a| 1

(x +

(6, 7)

Sum of coefficients of all odd degree terms is 2 (1 - 10 + 5 + 5) = 2

6. (c) We have, a1, a2 , a3 , … a49 are in AP. 12

å a4 k + 1 = 416 and a9 + a43 = 66 k =0

Let a1 = a and d = common difference Q a1 + a5 + a9 + L + a49 = 416 \ a + ( a + 4 d ) + ( a + 8d ) + ¼( a + 48d ) = 416 13 (2a + 48d ) = 416 Þ 2 …(i) Þ a + 24 d = 32 Also , a9 + a43 = 66 \ a + 8d + a + 42d = 66 Þ 2a + 50 d = 66 …(ii) Þ a + 25d = 33 Solving Eqs. (i) and (ii), we get a = 8 and d = 1 2 Now, a12 + a22 + a32 + L + a17 = 140 m 82 + 92 + 102 + ¼ + 242 = 140 m 2

3

(Q w = 1)

= - ( -1) [Q1 + w + w2 = 0]

Þ (1 + 22 + 32 + ¼ + 242 ) - (12 + 22 + 32 + ¼ + 72 ) = 140 m 24 ´ 25 ´ 49 7 ´ 8 ´ 15 Þ = 140 m 6 6

=1

4. (a) Given 6 different novels and 3 different dictionaries. Number of ways of selecting 4 novels from 6 novels is 6! 6 C4 = = 15 2!4 ! Number of ways of selecting 1 dictionary is from 3 dictionaries is

Þ Þ Þ

3´7 ´8´5 (7 ´ 5 - 1) = 140 m 6 7 ´ 4 ´ 5 ´ 34 = 140 m 140 ´ 34 = 140 m Þ m = 34

7. (b) We have, 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 + ¼ A = sum of first 20 terms B = sum of first 40 terms

13

40 ´ 41 ´ 81 4 ´ 20 ´ 21 ´ 41 + 6 6 40 ´ 41 B= (81 + 42) 6 40 ´ 41 ´ 123 = 6 Now, B - 2A = 100 l 40 ´ 41 ´ 123 2 ´ 20 ´ 21 ´ 63 \ 6 6 = 100 l 40 (5043 - 1323) = 100 l Þ 6 40 ´ 3720 = 100 l Þ 6 B=

Þ Þ

40 ´ 620 = 100 l 40 ´ 620 l= = 248 100

8. (c) Key idea Use property of greatest integer function [ x] = x - {x}. We have, 2 15 ö æ 1 lim x ç é ùú + éê ùú + ¼+ éê ùú ÷ x ®0 + è ê ëx û ëx û ë x ûø We know, [ x] = x é1 ù = 1 \ êë x úû x

- { x} ì1 ü í ý îx þ nù n ìnü é Similarly, ê ú = - í ý ëx û x îx þ

\ Given limit 1 2 2 æ1 = lim x ç - ìí üý + - ìí üý + x ®0 + è x îx þ x îx þ …

15 ì15 üö - í ý÷ x î x þø

= lim (1 + 2 + 3+ ...+15) - x x ®0 +

æì 1 ü ì 2 ü ì15 üö ç í ý + í ý + ... + í ý÷ èî x þ î x þ î x þø = 120 - 0 = 120

14

Engineering Entrances Solved Papers 2018 ù é ìnü ú êQ0 £ íî x ýþ < 1, therefore ú ê nü nü ú ê ì ì xí ý = 0ú ê0 £ x íî x ýþ < x Þ x lim ®0 + îx þ û ë

9. (b) We have, sin2 x × cos2 x I =ò 5 (sin x + cos3 x × sin2 x

dx

+ sin3 x × cos2 x + cos5 x )2 =

sin2 x cos2 x ò {sin3 x(sin2 x + cos2 x )

Þ I =

p /2

ò0

sin x cos x ò (sin3 x + cos3 x )2 dx

=

ò cos6 x(1 + tan3 x )2

sin2 x cos2 x

tan2 x sec2 x

ò (1 + tan3 x )2

2I =

Þ

2I = [ x]0p / 2

2

dx

2

b

b

a

a

p /2

sin2 x

ò- p /2 1 + 2x

dx

p p sin2 æç - + - x ö÷ è 2 ø 2 dx p p

p /2

ò

-

-p /2

1+2

+

2

-x

2

b ù é b êQ ò f ( x )dx = ò f ( a + b - x )dx ú úû êë a a p /2

sin2 x ò 1 + 2- x dx -p /2 p /2

2x sin2 x dx 2x + 1 -p / 2

ò

p /2

Þ 2I =

æ 2x + 1 ö sin2 x ç x ÷ dx è2 + 1 ø -p / 2

ò

p /2 2

Þ 2I =

ò sin

x dx

-p / 2 p /2

Þ 2I = 2 ò sin2 x dx 0

[Qsin2 x is an even function] Þ I =

p /2

ò0

p 4 R(α, β)

Q(0, β)

14. (b) Equation of tangent and normal to

(2, 3)

Equation of line PQ is

= ò f ( x )dx = ò f ( a + b - x )dx

Þ I =

Þ I =

dx

10. (d) Key idea Use property

Þ I =

dx

the curve y2 = 16 x at (16, 16) is x - 2 y + 16 = 0 and 2x + y - 48 = 0, respectively.

P(α, 0)

Þ 3 tan x sec xdx = dt dt 1 I = ò \ 3 (1 + t )2 -1 +C I = Þ 3(1 + t ) -1 Þ I = +C 3(1 + tan3 x )

Þ I =

ò0

11. (c) dx

Put tan3 x = t

Let I =

p /2

Þ

Since, line 2x - y + 5 = 0 also touches the circle. ½2( -8) - ( -6 ) + 5½ 100 - c = ½ \ ½ ½ ½ 22 + 12 ½ ½ ½-16 + 6 + 5½ ½ 100 - c = ½ Þ 5 ½ ½ 100 - c = |- 5| Þ Þ 100 - c = 5 Þ c = 95

2

=

=

r = 82 + 62 - c = 100 - c

a ù é a êQ ò f ( x )dx = ò f ( a - x )dx ú úû êë 0 0

+ cos3 x(sin2 x + cos2 x )}2 2

Centre ( -8, - 6 )

cos2 xdx

sin2 xdx

Since this line is passes through fixed point (2, 3). 2 3 \ + =1 a b \ Locus of R is 2b + 3a = ab i.e. 2 y + 3x = xy Þ 3x + 2 y = xy

12. (c) Key idea Otrhocentre, centroid and circumcentre are collinear and centroid divide orthocentre and circumcentre in 2 : 1 ratio. We have orthocentre and centroid of a triangle be A( -3, 5) and B(3, 3) respectively and C circumcentre. A(–3, 5)

B(3,3) 2

6) ,1 16 ( P 2x + q y– 48

Y

x y + =1 a b

C 2

Clearly, AB = (3 + 3) + (3 - 5)

= 36 + 4 = 2 10 We know that, AB : BC = 2 : 1 BC = 10 Þ Now, AC = AB + BC = 2 10 + 10 = 3 10 Since, AC is a diameter of circle. 5 3 10 AC \ r= Þr= =3 2 2 2

13. (d) Key idea Equation of tangent to the curve x2 = 4 ay at ( x1, y1 ) is y + y1 ö xx1 = 4 aæç ÷ è 2 ø Tangent to the curve x2 = y - 6 at (1, 7 ) is y+7 x= -6 2 …(i) Þ 2x - y + 5 = 0 Equation of circle is x2 + y2 + 16 x + 12 y + c = 0

0

X′

1 y+ x–2

6=

=

0

A(–16, 0)

C(4, 0)

B (24, 0)

X

Y′ A = ( -16, 0 ) ; B = (24, 0 ) C is the centre of circle passing through PAB i.e. C = (4, 0 ) Slope of 16 - 0 16 4 = = = m1 PC = 16 - 4 12 3

Slope of PB =

16 - 0 16 = = - 2 = m2 16 - 24 -8

½ m - m2 ½ tan q = ½ 1 ½ ½1 + m1m2½ ½ ½ 4 ½ 3 +2 ½ Þ tan q = ½ Þ tan q = 2 4 ½ ½1 - æç ö÷(2)½ ½ è3 ø ½

15. (a) Tangents are drawn to the hyperbola 4 x2 - y2 = 36 at the point P and Q. Tangent intersects at point T(0, 3) Y T (0, 3)

X

O (–3√5, –12)Q

S(0, –12)

P(3√5, –12)

Clearly, P Q is chord of contact.

15

Engineering Entrances Solved Papers 2018 \ Equation of PQ is -3 y = 36 Þ y = - 12 Solving the curve 4 x2 - y2 = 36 and y = - 12 , we get x = ±3 5 1 Area of DPQT = ´ PQ ´ ST 2 1 = (6 5 ´ 15) = 45 5 2

16. (c) Key idea Standard deviation is remain unchanged, if observations are added or subtracted by a fixed number We have, 9

9

2

å(x1 - 5) = 9 and å(x1 - 5) i =1 9

2 å(x1 - 5)

SD =

= 45

i =1

i =1

9

9

æ ö ç å(x1 - 5) ÷ çi =1 ÷ -ç ÷ 9 ç ÷ ç ÷ è ø

30°

Q

90°

30° M

R

Let height of tower TM be h. TM h In DPMT, tan45° = Þ1= PM PM Þ PM = h h ; QM = 3h In DTQM, tan30° = QM In DPMQ, PM2 + QM2 = PQ2 2

2

4 h = (200 )

Thus, the points A( z1 ), B( z2 ), C( z3 ) and D( z ) taken in order would be concyclic if [( z - z1 )( z2 - z3 ) / ( z2 - z1 )( z - z3 )] purely real. Hence, it is a circle. C(z3)

D(z)

B(z2)

Þ h = 100 m

distributive law. We have, ~( p Ú q ) Ú (~ p Ù q ) º (~ p Ù ~ q ) Ú (~ p Ù q ) [QBy De-Morgan’s law ~( p Ú q ) = (~ p Ù ~ q )] º ~ p Ù (~ q Ú q ) [By distributive law] º ~ p Ù t [~ q Ú q = t] º~ p

cos( x + y )cos( x - y ) = cos2 x - sin2 y and cos 3x = 4 cos3 x - 3cos x We have, p p 1ö æ 8cos x çcos æç + x ö÷ cos æç - x ö÷ - ÷ è6 ø 2ø ø è è6

A(z1)

\ (a), (b), (d) are false statement. Hence, option (a), (b), (d) are correct answer.

2. (b, c, d) We have, In D PQR Ð PQR = 30° PQ = 10 3 QR = 10 P

JEE Advanced 1. (a, b, d) =1

p 1 Þ 8cos x æçcos2 - sin2 x - ö÷ = 1 è 6 2ø

Þ

æ z - z1 ö æ z2 - z3 ö Þ ç ÷ is purely real ÷ç è z - z3 ø è z2 - z1 ø

19. (a) Key idea Use De-Morgan’s and

17. (b) Key idea Apply the identity

3 1 8cos x æç - sin2 x - ö÷ = 1 è4 2ø 3 1 8cos x æç - - 1 + cos2 x ö÷ = 1 è4 2 ø æ -3 + 4 cos2 x ö 8cos x ç ÷ =1 4 ø è 3

200 m

T

Þ

Þ SD = 5 - 1 = 4 = 2

Þ

So, given expression is multiple of 2p. (d) We have, æ ( z - z1 ) ( z2 - z3 ) ö arg ç ÷=p è ( z - z3 ) ( z2 - z1 ) ø

h2 + ( 3h )2 = (200 )2

2

Þ

45°

200 m

2

45 æ 9 ö -ç ÷ Þ SD = è9 ø 9

Þ

P

18. (a)

2(4 cos x - 3cos x ) = 1 1 Þ 2cos 3x = 1 Þ cos3x = 2 p 5p 7 p [0 £ 3x £ 3p] Þ 3x = , , 3 3 3 p 5p 7p Þ x= , , 9 9 9 p 5p 7 p 13p Sum = + + = Þ 9 9 9 9 13p kp = 9 13 Hence, k = 9

(a) Let z = - 1 - i and arg(z) = q ½ im( z )½ ½ - 1½ ½= 1 ½= ½ Now, tan q = ½ ½ Re( z )½ ½ -1½ p q= Þ 4 Since, x < 0, y < 0 p 3p arg ( z ) = - æç p - ö÷ = \ è 4ø 4 (b) We have, f (t ) = arg( -1 + it ) arg ì p - tan -1 t, t ³ 0 ( -1 + it ) = í -1 î- ( p + tan t ), t < 0 This function is discontinuous at t = 0. (c) We have, æz ö arg ç 1 ÷ - arg ( z1 ) + arg ( z2 ) è z2 ø æz ö Now, arg ç 1 ÷ = arg( z1 ) - arg ( z2 ) è z2 ø æz ö \ arg ç 1 ÷ - arg( z1 ) + arg ( z2 ) + 2np è z2 ø = arg( z1 ) - arg( z2 ) + 2np - arg( z1 ) + arg( z2 ) = 2np

10 3 30° Q

10

R

By cosine rule cos30° =

PQ2 + QR2 - PR2 2PQ × QR

Þ

3 300 + 100 - PR2 = 2 200 3

Þ

300 = 300 + 100 - PR2

Þ PR = 10 Since, PR = QR = 10 \ ÐQPR = 30° and ÐQRP = 120° 1 Area of D PQR = PQ × QR × sin 30° 2 1 1 = ´ 10 3 ´ 10 ´ 2 2 = 25 3 Radius of incircle of Area of D PQR D PQR = Semi- perimetre of D PQR

16

Engineering Entrances Solved Papers 2018 i.e. r =

1/n

25 3 D = s 10 3 + 10 + 10 2 25 3 = 5( 3 + 2)

1 öæ 2 öæ 3ö ù éæ ê çè1 + n ÷ø çè1 + n ÷ø çè1 + n ÷ø ú Þ L = lim ê ú n ®¥ ú ê... æ1 + n ö ÷ ç úû êë è nø

Þ r = 5 3 (2 - 3) = 10 3 - 15 and radius of circumcircle abc 10 3 ´ 10 ´ 10 = 10 (R ) = = 4 ´ 25 3 4D \ Area of circumcircle of D PQR = pR2 = 100 p

3. (8) ((log2 9 ) )

´ ( 7)

2 log (log 9 ) 2 2

= (log2 9 )

= (log2

22 log 9 ) (log 2 9 )

n® ¥

n

æ

å log çè1 +

r =1

[Q alog a b = b] = 22 ´ 2 = 8

4. (625) A number is divisible by 4 if last 2 digit number is divisible by 4. \ Last two digit number divisible by 4 from (1, 2, 3, 4, 5) are 12, 24, 32, 44, 52 \ The number of 5 digit number which are divisible by 4, from the digit (1, 2, 3, 4, 5) and digit is repeated is 5 ´ 5 ´ 5 ´ (5 ´ 1) = 625

5. (3798) Here, X = {1, 6, 11, ¼, 10086} [Q an = a + ( n - 1)d] and Y = {9, 16, 23, ¼, 14128} X Ç Y = {16, 51, 86,¼} t n of X Ç Y is less than or equal to 10086 \ t n = 16 + ( n - 1) 35 £ 10086 Þ n £ 2887 . \ n = 288 Q n( X È Y ) = n( X ) + n(Y ) - n( X Ç Y ) \ n ( X È Y ) = 2018 + 2018 - 288 = 3748

log L = log 2 -

1 [( n + 1) ( n + 2) ¼ ( n + n )]1/ n n and lim yn = L n® ¥ 1 Þ L = lim [( n + 1) ( n + 2) ( n + 3) n® ¥ n ¼ ( n + n )]1/ n

1é

d

ù

ò0 êë dx (log(1 + x ) ò dx úû dx

1æ x

+1

1

ö

ò0 çè x + 1 - x + 1 ÷ø dx

3 a cos x + 2b sin x = c …(i) \ 3 acos a + 2b sin a = c and …(ii) 3 a cos b + 2b sin b = c On subtracting Eq. (ii) from Eq. (i), we get 3a (cos a - cos b ) + 2b(sin a - sin b ) = 0

a + b öö æ æa - bö = 0 + 2b ç2cos æç ÷ ÷ sin ç ÷ è 2 øø è 2 ø è

9. (d) We have, x2 + y2 = 4 Let P(2cos q, 2sin q) be a point on a circle. \ Tangent at P is 2cos q x + 2sin q y = 4 Þ x cos q + y sin q = 2 Y N P(2 cos θ, 2 sin θ) X′

M

O

Þ

yn =

1 2b = 3 3a b = 0.5 a

éQ a + b = p , given ù êë úû 3 b 1 Þ = a 2

X

x2 + y2 = 4

a + bö 2b ÷= 2 ø 3a

p 2b tan æç ö÷ = è6 ø 3a

Þ

Point E3 (0, 4), F3 (4, 0) and G3 (2, 2) satisfies the line x + y = 4.

a + bö æa + bö 3 a sin æç ÷ = 2b cos ç ÷ è 2 ø è 2 ø tan æç è

x

Equation of tangent at E1( - 3, 1) is - 3x + y = 4 and at E2 ( 3, 1) is 3x + y = 4 Intersection point of tangent at E1 and E2 is (0, 4 ) \ Coordinates of E3 is (0, 4 ) Similarly, equation of tangent at F1(1, - 3 ) and F2 (1, 3 ) are x - 3 y = 4 and x + 3 y = 4, respectively and intersection point is (4, 0), i.e., F3 (4, 0) and equation of tangent at G1(0, 2) and G2 (2, 0 ) are 2 y = 4 and 2x = 4, respectively and intersection point is (2, 2) i.e., G3 (2, 2).

a + b öö æ æa - bö 3 a ç - 2sin æç ÷ ÷ sin ç ÷ è 2 øø è 2 ø è

Þ

G2(2, 0)

Y′

7. (0.5) We have, a ,b are the roots of

Þ

O

F1(1, – 3)

Þ log L = log 2 - 1 + log 2 - 0 4 Þ log L = log 4 - log e = log e 4ù 4 é Þ L= Þ [ L] = ê ú = 1 e ëeû

6. (1) We have,

(–2,0)

E2( 3,1)

(0, –2)

Þ log L = log 2 - [ x]10 + [log( x + 1)]10

Þ

Po(1,1)

1

Þ

Þ

x′

F2(1, 3)

rö ÷ nø

Þ log L = ( x × log (1 + x ))10

log 7 2

G1 (0,2)

E1(– 3,1)

[by using integration by parts] 1 x dx Þ log L = [ x log(1 + x )]10 - ò 01 + x ù aú û

Y E3(0,4)

1 + x ) dx ò0 1II ´ log( I

-

1 × log 7 4 ´ 72

´7

1 n

Þ log L = lim

1 log 4 7

é 1 = log b êQ ë log a b

æ1 + 1 ö ç ÷ è nø

2 n + log æç1 + ö÷ ¼ log æç1 + ö÷ è è nø nø

Þ log L =

Hence, option (b), (c) and (d) are correct answer. 1 2 log 2 (log 2 9 )

1é Þ log L = lim ê log n® ¥ n ë

8. (a)

Y′

2 \ The coordinates at M æç , 0 ö÷ and è cos q ø 2 ö N æç0, ÷ è sin q ø Let ( h, k ) is mid-point of MN 1 1 and k = h= cos q sin q 1 1 and sin q = Þ cos q = h k

\

17

Engineering Entrances Solved Papers 2018 1 1 + 2 h2 k h2 + k2 1= 2 2 h ×k h2 + k2 = h2 k2

Since, chord is passing through (1, 1). \ Locus of mid-point of chord ( h, k ) is h + k - 1 - k = h2 + k2 - 2k

\ Mid-point of MN lie on the curve x2 + y2 = x2 y2

Now, after checking options, (a) and (d) are correct.

Þ Þ

cos2 q + sin2 q =

S5 and five seats R1, R2 , R3 , R4 and R5 \ Total number of arrangement of sitting five students is 5! = 120 Here, S1 gets previously alloted seat R1 \ S2 , S3 , S4 and S5 not get previously seats. Total number of way S2 , S3 , S4 and S5 not get previously seats is 1 1 1 1 - + ö÷ 4 ! æç1 - + è 1! 2! 3! 4 ! ø 1 1 1ö æ = 24 ç1 - 1 + - + ÷ è 2 6 24 ø 12 - 4 + 1 ö = 24 æç ÷ =9 è ø 24 \

Required probability =

9 3 = 120 40

11. (c) Here, n(T1 Ç T2 Ç T3 Ç T4 ) = Total - n (T1 È T2 È T3 È T4 ) Þ n (T1 Ç T2 Ç T3 Ç T4 ) = 5! - [4 C1 4 ! 2! - (3 C1 3! 2! + 3C1 3! 2! 2!) + (2 C1 2! 2! + 4 C1 2 × 2!) - 2] Þ n(T1 Ç T2 Ç T3 Ç T4 ) = 120 - [192 - (36 + 72) + (8 + 16 ) - 2] = 120 - [192 - 108 + 24 - 2] = 14 14 7 Required probability = = \ 120 60

12. (a, d) It is given that T is tangents to S1 at P and S2 at Q and S1 and S2 touch externally at M.

13. (a, c) We have,

Y y2=4x

Q

x′

√2

Y′

Þ

It is true

\ L may be an empty set. It is false. (c) If elements of set L represents line, then this line and given circle intersect at maximum two point. Hence, it is true. (d) In this case locus of z is a line, so L has infinite elements. Hence, it is true.

0 + 0 + 1/ m 1 = 2 1 + m2 m4 + m2 - 2 = 0

14. (a, c, d) We have,

(b) If|s| = |t|, then rt - rs may or may not be zero. So, z may have no solutions.

Let the equation of common tangent of parabola and circle is 1 y = mx + m 1 Since, radius of circle = 2 \

15. (646) We have,

Þ m=±1

X = (10 C1 )2 + 2(10 C2 )2 + 3(10 C3 )2 + ...

\ Equation of common tangents are y = x + 1 and y = - x - 1 Intesection point of common tangent at Q ( -1, 0 ) x2 y2 \ Equation of ellipse + =1 1 1/2 where, a2 = 1, b2 = 1 / 2

+ 10 (10 C10 )2 10

Þ

s2

Q(2,–5)

C r )2

10

Þ

X =

år

10

C r 10C r

r =1 10

Þ

X =

10 9 Cr - 1 r

år ´

b2 1 1 = 1- = 2 2 a2 and length of latusrectum 2b2 2 (1 / 2) = = =1 a 1

T

\ MN = NP = NQ \ Locus of M is a circle having PQ as its diameter of circle \ Equation of circle ( x - 2) ( x + 2) + ( y + 5)( y - 7) = 0 Þ x2 + y2 - 2 y - 39 = 0 Hence, E1 : x2 + y2 - 2 y - 39 = 0, x ¹ ± 2 Locus of mid-point of chord ( h, k ) of the circle E1 is xh + yk - ( y + k ) - 39 = h2 + k2 - 2k - 39 Þ xh + yk - y - k = h2 + k2 - 2k

y

x′ (–1,0)

O

10

Þ

\ Area of shaded region 1 1 1 - x2 dx =2ò 1/ 2 2

å

X = 10

10

Þ

å

X = 10

9

Cr - 1

10

Cr

Cr - 1

10

C10 - r

C r - 1 ùú û

n -1

9

r =1

√2

x= 1 0, –1  √2 y′ √2

Cr

r =1

0, –1 

1 ,0 √2

10

éQ nC = n r êë r

( e) = 1 -



P(–2,7) N

10

å r(

r =1



M

X =

r =1

Now, eccentricity s1

2

(a) For unique solutions of z |s|2 - |t|2 ¹ 0 Þ |s| ¹ |t| x

 1 ,0

O

1 sin -1 x ùú 2 û1/

…(i) sz + tz + r = 0 On taking conjugate …(ii) sz + tz + r = 0 On solving Eqs. (i) and (ii), we get rt - rs z= 2 |s| - |t|2

1 2 and Equations of parabola y2 = 4 x Equations of circle x2 + y2 =

x2 + y2 = 1/2 1/√2

1 - x2 +

p 1 p ù é = 2 ê æç0 + ö÷ - æç + ö÷ ú 4 ø è4 8 øû ëè p 2 p 1 = 2 æç - ö÷ = è8 4 ø 4 2

Locus is E2 : x2 + y2 - x - 2 y + 1 = 0

10. (a) Here, five students S1, S2 , S3 , S4 and

1

x = 2 éê ë2

Þ h2 + k2 - 2k - h + 1 = 0



Þ

[Q nC r = nC n - r ] (1,0)

x

Þ

X = 10 ´ [Q

19

n -1

C9

C r - 1nC n - r =

2n - 1

C n - 1]

Now, 19 10 ´ 19C9 19 C9 C9 1 = = X = 1430 1430 143 11 ´ 13 19 ´ 17 ´ 16 = = 19 ´ 34 = 646 8

18

Engineering Entrances Solved Papers 2018

16. (c) Given 6 boys M1, M2 , M3 , M4 , M5 , M6 and 5 girls G1, G2 , G3 , G4 , G5 (i) a1 ® Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls. i..e, 6 C3 ´ 5C2 = 20 ´ 10 = 200 \

a1 = 200

(ii) a2 ® Total number of ways selecting at least 2 member and having equal number of boys and girls i.e., 6 C15C1 + 6C25C2 + 6C35C3 6

5

5

6

Now, DLNM is an equilateral triangle whose sides is 2b [Q DLON ~ = DMOL; \ ÐNLO = ÐNMO = 60°] 3 Area of DLMN = (2b )2 \ 4 Þ 4 3 = 3b2 Þ b = 2 1 Also, area of DLMN = a(2b ) = ab 2 4 3 = a(2) Þ a = 2 3 Þ (P) Length of conjugate axis = 2b = 2(2) = 4 (Q) Eccentricity ( e ) = 1 +

+ C4 C4 + C5 C5 = 30 + 150 + 200 + 75 + 6 = 461 Þ a2 = 461 (iii) a3 ® Total number of ways of selecting 5 members in which at least 2 of them girls i.e., 5 C26C3 + 5C36C2 + 5C4 6C1 + 5C56C0 = 200 + 150 + 30 + 1 = 381 a3 = 381 (iv) a4 ® Total number of ways of selecting 4 members in which at least two girls such that M1 and G1 are not included together. G1 is included ® 4 C1 × 5C2 + 4 C2 × 5C1 + 4 C3 = 40 + 30 + 4 = 74 M1 is included ® 4 C2 × 5C1 + 4 C3 = 30 + 4 =34 G1 and M1 both are not included 4 C4 + 4 C3 × 5C1 + 4 C2 × 5C2 1 + 20 + 60 = 81 \ Total number = 74 + 34 + 81 = 189 a4 = 189 Now, P ® 4; Q ® 6; R ® 5; S ® 2 Hence, option (c) is correct. Equation of hyperbola x2 y2 - 2 =1 2 a b Y L (0,–b)

2

2

y x – 2 =1 b a2

b 60° O b

4 4 2 = = 12 2 3 3

(R) Distance between the foci 2 = 2ae = 2 ´ 2 3 ´ =8 3 (S) The length of latusrectum 2b2 2(4 ) 4 = = a 2 3 3 P ® 4; Q ® 3; R ® 1; S ® 2 Hence, option (b) is correct. =

N(a,0)

It is given, Ð LNM = 60° and Area of DLMN = 4 3

n® ¥

r =1 r

n® ¥

= cos 0° + i sin 0° = cos 2rp + i sin 2rp = e i 2 rp =e

1 n

3. (b) Let x = 1

; r = 0, 1, 2, ¼

n( n + 1) ( n + 2) 2

åt

= lim

7

x

lim

20

i (2 rp ) / n

( n - 1) n ( n + 1)( n + 2) = Sn -1 8

n

\

r

1 25840 T8 = 3 × 20C7 æç ö÷ = è 3ø 9

Þ

=

Now, t n = Sn - Sn - 1 =

= Cn = 1

Þ r £ 7.692 Þ r = 7 \ The greatest term is

1/ n

åt r r =1

n

1 and Tr = 3 ×20 C r - 1 æç ö÷ è 3ø T 20 - r + 1 æ 1 ö Now, r + 1 = ç ÷ è 3ø Tr r

M (0,–b) Y′

n( n + 1) ( n + 2)( n + 3) = Sn 8

n -1

\

r -1

x

=

(say)

1ù 1ö é ÷ = ê(2 + x ) + ú xø xû ë 1 n n æ1 ö n = C0 (2 + x ) × ç ÷ + ¼ + C n × n èx ø x

2. (c) Tr + 1

and 2(log 2b - log 3c ) = log a - log 2b + log 3c - log a Þ b2 = ac and 2b = 3c 4a 2a and b = Þ c= 9 3 5a 10 a a+ b= > c, b + c = >a Q 3 9 13a and c + a = > b. 9 \ a, b, c are the sides of a triangle. Also, a is the greatest side b2 + c2 - a2 29 = 0

Þ ( x - 1) ( x - 2) > 0 Þ x Î( - ¥, 1) È (2, ¥) Again, ( x + 1)2 < 7 x - 3 Þ

æ z - 1ö Re ç ÷ =1 èz + iø

x2 + 2x + 1 < 7 x - 3

Þ

x2 - 5x + 4 < 0

\ x( x - 1) + y( y + 1) = x2 + ( y + 1)2

Þ Þ

( x - 1) ( x - 4 ) < 0 x Î(1, 4 )

Since,

Þ x2 - x + y2 + y = x2 + y2 + 2 y + 1

21

\

x =3

[Qx is integer]

22

Engineering Entrances Solved Papers 2018 \ Required sum = 3054 + 3504 + 5034 + 5304 + 3450 + 3540 + 4350 + 4530 + 5340 + 5430 = 43536

8. (a) We have, 2

2

f ( x ) = x + 2bx + 2c

= ( x + b )2 + 2c2 - b2 \ Minimum value of f ( x ) = 2c2 - b2 Again,

12. (b) 6 Boys can be seated in a row in

g( x ) = - x2 - 2cx + b2 = - [ x2 + 2cx - b2 ] = - [( x + c )2 - b2 - c2 ] = - ( x + c )2 + b2 + c2

\ Maximum value of g( x ) = b2 + c2 Now, according to the question. 2c2 - b2 > b2 + c2 Þ c2 > 2b2

9. (a) Given, a, b and c are the roots of equation x3 + qx + r = 0, \

a+ b+ c=0 ab + bc + ca = q abc = - r a+ b+ c=0 ( a + b + c )2 = 0

and Q Þ

Þ a2 + b2 + c2 + 2ab + 2bc + 2ca = 0 2

2

2

Þ a + b + c + 2( ab + bc + ca) = 0 a2 + b2 + c2 = - 2q

Þ

2

2

…(i) 2

Now, ( a - b ) + ( b - c ) + ( c - a) = a2 + b2 - 2ab + b2 + c2 2

2

- 2bc + c + a - 2ca = 2a2 + 2b2 + 2c2 - 2( ab + bc + ca) 2

2

2

= 2( a + b + c ) - 2( ab + bc + ca) = 2( - 2q ) - 2( q ) = - 4 q - 2q = - 6 q.

[by Eq. (i)]

10. (a) Let the two roots of equation is m, - m. Now, sum of three roots = 2 p Hence, third root will be 2 p. Now, m ´ ( - m ) + m ´ (2 p ) + ( - m ) ´ 2 p = 3q Þ - m2 = 3q Now, m ´ ( - m ) ´ 2 p = 4 r Þ 3q ´ 2 p = 4 r 3 pq ´ 2 3 pq Þ Þ r= r= 4 2

11. (b) Four digit numbers which are even and formed with the digits 0, 3, 5, 4 without recepetion are 3054, 3504, 5034, 5304, 3450, 3540, 4350, 4530, 5340, 5430

6 P6 ways = 6 !. Now, in the 7 gaps 6 girls can be arranged in 7 P6 ways. \ x = 6 ! ´ 7 P6 = 6! ´ 7! 6 Boys can be seated in a circle in (6 - 1)! ways = 5! Now, in the 6 gaps 6 girls can be arranged in 6 P6 ways. \ y = 5! ´ 6 P6 = 5! ´ 6 ! Now,x : y = 6 ! ´ 7 ! : 5! ´ 6 ! Þ

x : y = 7 !:5!

Þ x : y = 7 ´ 6 ´ 5! : 5! Þ x : y = 42 : 1 13. (c) There are 7 letters in the word ‘SARANAM’ of which there are 3 A and all other are distinct. The five letter words may consist of (i) all different letters (using S, A, R, N, M) \ Required words = 5! = 120 (ii) 2 alike and other different (using 2 A and 3 other) \ Required words 5! 120 = 4 C3 ´ = 4 ´ = 240 2! 2 (iii) 3 alike and other different (using 3A and 2 other) 5! \ Required words = 4 C2 ´ 3! 4 ´ 3 120 = ´ = 120 2´1 6 \ Total words = 120 + 240 + 120 = 480

14. (c) We have, (4 5 +

5

4 )100 =

100

å

100

C r (4 5 )100 - r (5 4 )r

r =0 100

=

å

100 - r 100

4

Cr 5

r

45

r =0 100

=

åTr + 1 r =0 100 - r

Where,

Tr + 1 =

100

Cr 5

4

r

45

Clearly, Tr + 1 will be an integer if 100 - r and 4 r are integers. This is possible when 5 100 - r is a multiple of 4 and r is a multiple of 5

Þ100 - r = 0, 4, 8, 12, K, 96, 100 and r = 0, 5, 10, ..., 100 Þ r = 0, 4, 8, 12, K, 100 and r = 0, 5, 10, 100 Þ r = 0, 20, 40, 60, 80, 100 Hence, there are 6 rational terms.

15. (d) We have, (2a - 3b )19 19

3b ö = 219 × a19 æç1 ÷ è 2a ø

We know that, the r th term of greatest term of é ( n + 1)| x |ù (1 + x )n = ê ú ë 1 + | x| û n = 19, x = -

Here,

é ê (20 ) \ r=ê ê 1+ êë

3b 2a 3b 2a

3b 2a

ù ú é 20 ù ´ 4ú ú =ê ú ë1 + 4 û úû éQ b = 2 , a = 1 ù êë 3 4 úû

r = 16 \ greatest term of (2a - 3b )19 is 16

=

19

3b C16 ´ 219 × a19 æç ö÷ è 2a ø

=

19

1 C3 ´ 219 ´ æç ö÷ è4 ø

19

=

19

=

19

19

C3 ´ 2

´ (4 )16

[Q 19C16 = 1 ´ 38 ´ 2 32 2

19

C3 ]

C 3 ´ 213

16. (d) It is given that, p p cos æç x - ö÷, cos x, cos æç x + ö÷ are in H.P. è è 3ø 3ø p p 2cos æç x - ö÷ cos æç x + ö÷ è è ø 3 3ø Þ cos x = p pö æ æ cos ç x - ÷ + cos ç x + ö÷ è è 3ø 3ø æ 2 2 pö 2çcos x - sin ÷ è 3ø Þ cos x = p 2cos x cos 3 p p 2 2 Þ cos x cos = cos x - sin2 3 3 p p Þ cos2 x æç1 - cos ö÷ = sin2 è 3ø 3 pö æ p 2 æ Þ cos x ç1 - cos ÷ = ç1 - cos2 ö÷ è 3ø è 3ø pö 2 æ Þ cos x ç1 - cos ÷ è 3ø p p = æç1 - cos ö÷ æç1 + cos ö÷ è è ø 3 3ø

Engineering Entrances Solved Papers 2018 Þ cos2 x = 1 + cos Þ cos2 x = 2cos2

p 3

Since, AM is the median, M is the mid-point of line BC. …(i) Þ BM = CM = 2BC

p 6 2

æ 3ö Þ cos2 x = 2 ´ ç ÷ è 2 ø 3 3 Þ cos2 x = 2 ´ Þ cos2 x = 4 2 3 . Þ cos x = 2

17. (c) cos3 10 + cos3 110 + cos3 130° We know that, cos3 x + cos3 (120 - x ) + cos3 (120 + x ) 3 = cos 3x 4 Here,x = 10 Now, cos3 10 + cos3 110° + cos3 130° = cos 10 3

3

+ cos (120° - 10° ) cos (120° + 10° ) 3 3 = æç ö÷ cos(3 ´ 10° ) = cos 30° è4 ø 4 3 3 3 3 ´ = 4 2 8

18. (b) Given, Þ Þ Þ Þ Þ Þ

…(iii) Þ ÐAMC = 90° - ÐC Again at point M. ÐBMP + ÐPMA + ÐAMC = 180° Þ ÐBMP + 90° + 90° - ÐC = 180° … (iv) Þ ÐBMP = ÐC But in DBPM,

sin 5x = cos 2x p sin 5x = sin æç - 2x ö÷ è2 ø p 5x = np + ( - 1)n æç - 2x ö÷ è2 ø n p n 5 x = np + ( - 1 ) - ( - 1) 2x 2 p 5x + ( - 1)n (2x ) = {2n + ( - 1)n } 2 p x(5 + ( - 1)n 2) = (2n + ( - 1)n ) 2 p æ 2n + ( - 1)n ö x= ç ÷ 2 è 5 + 2( - 1)n ø

p Þ x =an × 2 2 n + (- 1)n So, an = 5 + 2(- 1)n

è

D ì 1 1 ü cD + ý= í s î s - a s - b þ s( s - a) ( s - b ) 2cD = ( a + b + c ) ( s - a) ( s - b ) s(s - c ) 2c = ( a + b + c ) ( s - a) ( s - b ) c 2 c cot 2 = a+ b+c =

21. (d) A α

B

M

P

ÐBPM + ÐC + ÐB = 180° … (v) [From Eqs. (i) and (iv)] …(vi) ÐBPM = ÐA Now, it is evident that ABC and MPB are equivalent triangles. BM BP PM 1 …(vii) = = = Þ BC AB AC 2 Also, ÐBAC = ÐBAM + ÐMAC Þ ÐBAM = ÐBAC - ÐMAC Þ ÐBAM = ÐA - 90° In DAPM, ÐPAM + ÐAPM + ÐAMP = 180° Þ ÐA - 90° + ÐAPM + 90° = 180° Þ ÐAPM = 180° - A AM tan( ÐAPM ) = PM AM Þ tan (180° - A) = PM AM - tan A = Þ PM …(viii) Þ AM = - PM tan A AM Now, in DACM, tanC = AC …(ix) Þ AM = AC tan C [From Eqs. (viii) and (ix),] AM = AC tan C = - PM tan A

Þ

AC = 2MP

Hence, PM tan A + 2PM tan C = 0 Þ PM(tan A + 2 tan C ) = 0 A

C

ABC is the triangle while AM is the median AC and AM are perpendicular. ÐMAC = 90°

α

ÐBPM + ÐBMP + ÐPBM = 180°

Þ PM tan A + AC tan C = 0 MP 1 = Þ AC 2

19. (c)

Þ

A B + tan ö÷ 2 2ø D D = + s( s - a) s( s - b )

20. (b) æç tan

Þ

3

=

Draw a line perpendicular to AM (through M) let it intersect the line AB at P. In DABC, …(ii) ÐA + ÐB + ÐC = 180° In DAMC, ÐAMC + ÐCAM + ÐMCA = 180° Þ ÐAMC + 90° + ÐC = 180°

23

Þ Þ Þ

tan A + 2 tan C = 0 tan A = - 2 tan C tan A = -2 tan C

r

β

r O

γ

β

B

γ

r C

From figure ar( DABC ) = ar( DAOB ) + ar( DBOC ) + ar( DCOA) 1 1 1 S = cr + ar + br Þ 2 2 2 [where ar( DABC ) = S] 1 Þ S = r( a + b + c ) 2 1 Þ S = r(2s ) 2 [where s is semi perimeter] S2 S Þ S = rs Þ r = Þ r2 = 2 s s s( s - a) ( s - b ) ( s - c ) 2 r = Þ s2 a× b × g 2 Þ r = a+ b + g [Q2s = 2a + 2b + 2g]

22. (d) We have, P

N

… (x)

C

Q

M

O R

M is the mid-point of QR. C is the mid-point of PM. Draw MO such that MO is parallel to CN. QN is the mid-point of PO 1 \ CN = MO 2

24

Engineering Entrances Solved Papers 2018 25. (b) Here, total number of possible out

M is the mid-point of QR \ MO is parallel to QN 1 MO = QN Q 2 1 1 1 CN = æç QN ö÷ CN = QN \ ø 2 è2 4 QN \ = 4. CN

Now, we can choose any three numbers out of 6 numbers and we can place them in ascending order and we can place them in ascending order in only one way. So, favourable out come = 6C3 = 20 20 5 Now, required probability = = 216 54

23. (a) We have, n = 8, x = 25 and s = 5 Sx i n Sx i = nx = 8 ´ 25 = 200 Þ Þ In corrected Sx i = 200 and s = 5 Þ s2 = 25 1 Þ Sx2i - (mean)2 = 25 n Sx2i - 625 = 25 Þ 8 Þ Sx2i = 5200 x=

Q

26. (d) Given, points are A(2, 3), B(3, - 6 ), C(5, - 7 ). Let point P be ( x, y ), then according to condition PA2 + PB2 = 2PC2 Þ ( x - 2)2 + ( y - 3)2 + ( x - 3)2 + ( y + 6 )2 = 2[( x - 5)2 + ( y + 7 )2 ] Þ x 2 + 4 - 4 x + y2 + 9 - 6 y + x 2 + 9 - 6 x + y2 + 36 + 12 y

Corrected Sx2i = 5200 + 225 + 625

= 2[ x2 + 25 - 10 x + y2 + 14 y + 49]

= 6050 and corrected mean = 200 + 15 + 25 = 240

Þ 2x2 + 2 y2 - 10 x + 6 y + 58

2

\ Corrected variance =

6050 æ 240 ö -ç ÷ è 10 ø 10

= 605 - (24 )2

fi

Cumulative (d i ) = | x i - 15| frequency

fi |d i |

4

4

9

36

9

5

9

6

30

3

3

12

12

36

12

2

14

3

6

15

5

19

0

0

13

4

23

2

8

21

4

27

6

24

3

30

7

N = Sfi = 30

21 S| fd i i| = 161

N = 15 2 The cumulative frequency just greater N than is 19 and corresponding value 2 of x is 15. Therefore, median = 15. Clearly, Sfi | x i - 15| = Sfi di = 161 and N = 30 161 \ Mean deviation = = 5 × 40 (approx) 30 Clearly,

N = 30

1. 10 (2) - 22( - 5) = 20 + 110 = 130

Þ

x y + =1 a b The line meets the coordinate axes at A( a, 0 ) and B(0, b ) respectively. The coordinate of the point which divides the line joining A( a, 0 ) and B(0, b ) in the ratio 3 : 2 are æ3 ´ 0 + 2 ´ a 3 ´ b + 2 ´ 0 ö , ç ÷ 3+2 3+2 è ø 2a 3b Þ æç , ö÷ è5 5 ø If is given that, the point (2, - 1) divides the AB in the ratio 3 : 2. 2a 3b = 2 and = -1 \ 5 5 -5 Þ a = 5 and b = 3 Hence, the required equation of the required line is x y + =1 -5 5 3 x 3y = 1 Þ x - 3y = 5 Þ 5 5 Þ x - 3y - 5 = 0

29. (b) The equation of a line passing 10 ( - 2) - 22(5) = - 20 - 110 = - 130

6

22

= 2x2 + 2 y2 - 20 x + 28 y + 148 Þ 10 x - 22 y = 90 By checking options, 2.

= 605 - 576 = 29 24. (*) xi

28. (d) Let the equation of the line be

come = 63 = 216

3. 10 (13) - 22(10 ) = 130 - 220 = - 90 4. 10 ( - 13) - 22( - 10 ) = - 130 + 220 = 90 So, point ( - 13, - 10 ) lies on the locus of P.

through the intersection of 2x + y - 4 = 0 and x - 3 y + 5 = 0 is (2x + y - 4 ) + l( x - 3 y + 5) = 0 …(i) Þ x(2 + l ) + y(1 - 3l ) + 5l - 4 = 0 This is at a distance of 5 units from the origin.

[Qq = 135° and P ( x, y ) = (2, - 6 )] - 1ö æ 1 ö x = 2 æç ÷+6ç ÷ è 2ø è 2ø - 1ö 1 and y = 2 æç ö÷ - 6 æç ÷ è 2ø è 2ø 4 8 and y = Þ x= 2 2 Þ x = 2 2 and y = 4 2. Þ

\ Coordinates of P in the original system are (2 2, 4 2 ).

= 5

(2 + l )2 + (1 - 3l )2

27. (c) If ( x, y ) is old coordinates and ( X , Y ) are new coordinates, when axes are rotated through an angle of q, then x = X cos q - Y sin q and y = X sin q + Y cos q \ x = 2cos 135° - ( - 6 )sin 135° and y = 2sin 135° + ( - 6 )cos 135°

5l - 4

\

Þ

(5l - 4 )2 =5 4 + l + 4 l + 1 + 9 l2 - 6 l

Þ

(5l - 4 )2 =5 10 l2 - 2l + 5

2

Þ 25l2 + 16 - 40 l = 50 l2 - 10 l + 25 Þ 25l2 + 30 l + 9 = 0 3 5 3 Putting the value of l = - in Eq. (i), 5 we get 3 Þ (2x + y - 4 ) - ( x - 3 y + 5) = 0 5 Þ 10 x + 5 y - 20 - 3x + 9 y - 15 = 0 Þ 7 x + 14 y - 35 = 0 Þ x + 2y - 5 = 0 By solving, we get

l=-

25

Engineering Entrances Solved Papers 2018 30. (b) Let AD and BE are altitudes of the triangle. \ Equation of AD is given by -1 ( x + 2) Slope of BC -1 y-3= ( x + 2) æ0 + 1 ö ç ÷ è4 - 2 ø y-3= Þ

We know that, than 60° So, q = 60°

y - 3 = - 2x - 4 2x + y + 1 = 0 … (i)

\ Equation of BE is given by -1 y+1= ( x - 2) Slope of AC Þ y+1=

-1 æ0 - 3 ö ç ÷ è4 + 2 ø

( x - 2)

y + 1 = 2( x - 2) y + 1 = 2x - 4 y = 2x - 5

Þ

\ q = 60° Þ a = 60° Hence, line from a equilateral triangle.

… (ii)

\ Equation of line joining (0, - 1) and æ 4 , 2 ö is ç ÷ è3 3ø 2 +3 ( x - 1) y+3= 3 4 -1 3 Þ y + 3 = 11( x - 1) y + 3 = 11x - 11 11x - y - 14 = 0.

31. (c) We have, 23x2 - 48xy + 3 y2 = 0 Þ

3 y2 - 48xy + 23x2 = 0

Here, and

M1 + M2 = 16 23 M1M2 = 3

\ m1 - m2 = ( m1 + m2 )2 - 4( m1m2 ) 23 92 = 256 3 3 768 - 92 26 = = 3 3 = (16 )2 - 4 ´

x12 + y12 + 2gx1 + 2fy1 + c Let P ( h, k ) \ According to the question,

Þ h2 + k2 - 2h + 4 k - 20 = 4 h2 + 4 k2 - 8h - 32k + 4 Þ3h2 + 3k2 - 6 h - 36 k + 24 = 0 Þ h2 + k2 - 2h - 12k + 8 = 0 \ Locus of point P is x2 + y2 - 2x - 12 y + 8 = 0.

35. (c) Since, circle touches both

Þ

2

k2 x2 - k2 xy + k2 y2 + 3kx2 + 6 kxy + 3kxy + 6 ky2 - 2x - 8xy - 8 y2 = 0 2

Þ x 2 ( k2 + 3 k - 2 ) - ( k2 - 9 k + 8 ) xy + ( k2 + 6 k - 8) y2 = 0 Since, ÐAOB = 90° \ k2 + 3 k - 2 + k2 + 6 k - 8 = 0 2k2 + 9 k - 10 = 0

33. (c) As we know that, If the lines ax2 + 2hxy + by2 = 0 be two sides of a parallelogram and the line lx + my = 1 be one of its diagonals, then other diagonal is .....(i) y( bl - hm ) = x( am - hl ) 3 Here, a = 2, b = - 2, h = 2 l = - 3, m = - 1 Putting all values in Eq. (i), we get 9 3 \ y æç6 + ö÷ = x æç - 2 + ö÷ è è 2ø 2ø 15 5 Þ y æç ö÷ = x æç ö÷ è2 ø è2ø 15 y = 5x 3y = x Þ x - 3y = 0

3h - 4 h - 12

=h

(3)2 + ( - 4 )2

x + 2yö æx + 2yö = 0 + 3 y æç ÷ - 2ç ÷ è k ø è k ø

Þ

2 1

Þ h2 + k2 - 2h + 4 k - 20 = 4( h2 + k2 - 2h - 8k + 1)

\

and x + 2y =1 x + 2y = k Þ k By homogeneous of Eq. (i), we get x + 2yö x2 - xy + y2 + 3x æç ÷ è k ø

Þ

=

h2 + k2 - 2h - 8k + 1

x2 - xy + y2 + 3x + 3 y - 2 = 0 …(i)

\ On solving Eqs. (i) and (ii), we get orthocentre (1, - 3). Also, centroid of - 2 + 2 + 4 3 - 1 + 0ö , DABC = æç ÷ è ø 3 3 4 2ö æ =ç , ÷ è3 3ø

Þ

drawn from ( x1, y1 ) to the circle x2 + y2 + 2gx + 2fy + c = 0 is

coordinates axes, then centre will be ( h, h ) and radius = h

32. (c) We have,

Since, orthocentre is the intersecting point of altitudes.

Þ

34. (d) We know that, length of tangent

h2 + k2 - 2h + 4 k - 20

2 Slope of line 2x + 3 y + 4 = 0 is 3 13 Angle between line of slope 8 + 3 2 and - is 3 13 2 + 8+ 3 3 tan a = = 3 13 ö æ 2 ö 1 + æç8 + ÷ç ÷ è 3 ø è3 ø

y - 3 = - 2( x + 2) Þ

13 13 and m2 = 8 3 3 26 m1 - m2 3 = = 3 tan q = 23 1 + m1m2 1+ 3

\ m1 = 8 +

- h - 12 =h 5

Þ - h - 12 = ± 5h Þ - 12 = ± 5h + h Þ - 12 = 6 h or - 12 = - 4 h Þ h = - 2 or 3 Þ h =3 [Q h > 0] \ Equation of circle will be ( x - 3)2 + ( y - 3)2 = 32 Þ x2 - 6 x + 9 + y2 - 6 y + 9 = 9 Þ

x2 + y2 - 6 x - 6 y + 9 = 0

36. (b) Let ( h, k ) be the pole of the line 9 x + y - 28 = 0 with respect to the 3 5 7 circle x2 + y2 - x + y - = 0. 2 2 2 Then, the equation of polar is 3 5 7 ( x + h) + ( y + k ) - = 0 4 4 2 3ö 5ö 3 æ æ Þ x çh - ÷ + y çk + ÷ - h è è 4ø 4ø 4 5 7 + k - =0 4 2 Þ x(4 h - 3) + y(4 k + 5) - 3h + 5k - 14 = 0 This equation and 9 x + y - 28 = 0 represent the same line. 4h - 3 4k + 5 = \ 9 1 - 3h + 5k - 14 = l (say) = - 28 hx + ky -

26

Engineering Entrances Solved Papers 2018 3 + 9l l -5 ,k = , 4 4 - 3h + 5k - 14 = - 28l 3 + 9l ö æ l - 5 ö - 14 Þ - 3 æç ÷ + 5ç ÷ è 4 ø è 4 ø

Þ a2 + b2 + 2ga + 2fb + c 3 = a2 + b2 + a + 4 b + c = 0 2 3ö æ a ç2g - ÷ + b(2f - 4 ) = 0 è 2ø

= - 28l Þ - 9 - 27 l + 5l - 25 - 56 = - 112l Þ - 22l - 90 = - 112l Þ 90 l = 90 Þ l = 1 Hence, the pole of the given line is (3, - 1).

this is the locus of radical axis. 3 So, x æç2g - ö÷ + y(2f - 4 ) = 0 is radical è 2ø axis of given circles.

Þ

h=

37. (d) The direct common tangents to two circles meet on the line of centres and divide it externally in the ratio of the radii centres of the two circles are ( - 11, 2) and (11, - 2) and their radii are 15 and 5. \ Point of intersection æ 11 ´ 15 - ( -11) ´ 5 -2 ´ 15 - 2 ´ 5 ö =ç , ÷ 15 - 5 15 - 5 è ø 165 + 55 - 30 - 10 ö , = æç ÷ = (22, - 4 ). è 10 ø 10

38. (b) We know that, angle between two circles is given by r2 + r22 - d2 , where r1 and r2 are cos q = 1 2r1r2 radius and d is distance between centres. ( 2 )2 + (1)2 (A) cos q =

- [ (2 - 2)2 + (1 - 0 )2 ]2 2´ 2 ´1

[\ r1 = 2, r2 = 1, c1 = (2, 0 ), c2 = (2, 1)] 2+ 1-1 1 = = 2 2 2 \ q = 45° or 135° (3)2 + ( 17 )2 (B) cos q =

- [ (3 - 2)2 + (3 + 2)2 ]2 2 ´ 3 ´ 17

[Qr1 = 3, r2 = 17 , c1 = (3, 3), c2 = (2, - 2)] 9 + 17 - 26 =0 = 6 17 q = 90° (C) cos q =

(5 )2 + (3 )2 - [ ( -2 + 2 )2 + (7 - 0 )2 ]2 2´5´3

[Qr1 = 5, r2 = 3, c1 = ( - 2, 7 ), c2 = ( - 2, 0 )] 25 + 9 - 49 - 15 - 1 = = = 30 30 2 So, q = 120° or 60°.

39. (a) Let point P ( a, b ) S1( a, b ) = S2 ( a, b )

This touched the x2 + y2 + 2x + 2 y + 1 = 0 So,

radius = 12 + 12 - 1 = 1

and centre = ( - 1, - 1) So, radius of circle = distance between centre and touching point. æ 3 - 2g ö + (2f - 4 ) ç ÷ è2 ø 1= 2 æ 3 - 2g ö + (2f - 4 )2 ç ÷ è2 ø Taking square both sides, 3 2 æç - 2g ö÷ (2f - 4 ) = 0 è2 ø 3 So, - 2g = 0 or 2 2f - 4 = 0 3 2g = or 2f = 4 2 3 g = or f = 2 Þ 4

40. (c) Given, equation of line is y = 6 x + 1 and equation of parabola is y2 = 24 x. The locus of the point of intersection of perpendicular tangent to a parabola is its directrix. So, required point is the point of intersection of y = 6 x + 1 and directrix x = - 6.

Let P ( x, y ) be any point on the parabola and M be the foot of the perpendicular drawn from P on the directrix. \ PS = PM As, PS2 = PM2 2

y - 3ö As, ( x - 1)2 + ( y + 1)2 = æç ÷ è 1 ø Þ ( x - 1)2 + ( y + 1)2 = ( y - 3)2

By checking option (a), 1 Þ (3 - 1)2 = 8 æç1 - ö÷ è 2ø 1 (2)2 = 8 ´ Þ 2 Þ 4 =4 1 Hence, point æç3, ö÷ lies on the parabola è 2ø ( x - 1)2 = 8(1 - y ).

42. (c) Let the equation of ellipse is x2 y2 + 2 =1 2 a b

focus S(1, - 1) and vertex A(1, 1) is -1-1 =0 m= 1-1 Let Q( h, k ) be the point of intersection of the axis AS with the directrix. The A(1, 1) will be the mid-point of QS. h+1 k -1 = 1 and =1 \ 2 2 Þ

h = 1 and k = 3

\Q is the point (1, 3) So, the directrix passes through the point (1, 3) and has the gradient 0. The equation of the directrix is y-3=0

e=

Given,

… (i)

2 5

Þ

a2 - b2 2 = 5 a2

Þ

a2 - b2 2 = 5 a2

Þ

5a2 - 5b2 = 2a2

Þ

3a2 = 5b2

5b2 3 Now, ellipse passes through ( - 3, 1) a2 =

Þ

Þ

Hence, its coordinates are ( - 6, - 35).

41. (a) The gradient of the line joining the

( x - 1)2 = 8(1 - y )

Þ

( - 3)2 (1)2 + 2 =1 a2 b

Þ 9 b2 + a2 = a2 b2 32b2 5b4 5b2 5b4 = = Þ 3 3 3 3 32 2 Þ b = 5 From Eq. (i), we get x2 y2 Þ + =1 32 32 3 5 Þ 3x2 + 5 y2 = 32 Þ 9 b2 +

Þ

3x2 + 5 y2 - 32 = 0.

43. (b) Let P(6 cos q, 3 3 sin q) be any point on the ellipse

x2 y2 + = 1. The 36 27

27

Engineering Entrances Solved Papers 2018 equation of the tangent at P(6 cos q, 3 3 sin q) is x y cos q + sin q = 1. 6 3 3 Þ3 3x cos q + 6 y sin q - 18 3 = 0 …(i) The product of the lengths of the perpendiculars from (3, 0) and ( - 3, 0 ) an Eq. (i) is given by P =

3 ´ 3 3 cos q - 18 3 27 cos2 q + 36 sin2 q

- 3 + 2x 5 + 2 y 2 + 2z ö , , Þ æç ÷ = (3, 3, 4 ) è 3 3 3 ø

n

1 n ® ¥ n3

å[ r

1 = lim 3 n® ¥ n

å( r

= lim

27 cos2 q + 36 sin2 q 36 ´ 27 - 9 ´ 27 cos2 q = 36 sin2 q + 27 cos2 q

r =1 n

x]

2

x - {r2 x})

= lim

x n n + 1 (2n + 1) ö = lim æç ´ ´ ´ ÷ n ® ¥è 6 n n n ø

9 ´ 27(4 - cos2 q) = 27. 9(4 - cos2 q)

{r2 n} 3 r =1 n n

44. (b) Equation to the asymptotes are given as … (i) 3x + 4 y - 2 = 0 and 2x + y + 1 = 0 … (ii) Eqs.(i) and (ii) may be given by (3x + 4 y - 2) (2x + y + 1) = 0 … (iii) As, the equation to the hyperbola will differ from Eq. (iii) only by a constant, it may be given by (3x + 4 y - 2) (2x + y + 1) = l …(iv) (where l is a constant) (1, 1) lies on the curve given by Eq. (iv), we have (3 + 4 - 2) (2 + 1 + 1) = l Þ (5) (4 ) = l Þ l = 20 Hence, the equation of the hyperbola will be (3x + 4 y - 2) (2x + y + 1) = 20 Þ 6 x2 + 3xy + 3x + 8xy + 4 y2 + 4 y - 4 x - 2 y - 2 = 20 Þ 6 x + 4 y2 + 11xy - x + 2 y - 22 = 0 2

2

- lim n® ¥

=

TS EAMCET Þ Þ

Þ f

C,(3, 3,4)

O(-3, 5-2)

æ - 3 + 2x 5 + 2 y 2 + 2z ö , , \ ç ÷ = (3, 3, 4 ) 1+2 1+2ø è 1+2

= y (Let)

x =2 -1

(x ) = 2

\

f

2 B

= | - 2 - i |= 5 |Z2 - Z3| = |( - 1 + 6 i ) - ( - 2 + 8i )| = | 1 - 2i | = 5 |Z3 - Z4| = |( - 2 + 8i ) - ( - 4 + 7 i )| = | 2 + i |= 5 |Z4 - Z1| = |( - 4 + 7 i ) - ( - 3 + 5i )| = | - 1 + 2i | = 5 |Z1 - Z3| = |( - 3 + 5i ) - ( - 2 + 8i )| = | - 1 - 3i |= 10 and|Z2 - Z4| = |( - 1 + 6 i ) - ( - 4 + 7 i )| = | 3 - i |= 10 QLength of sides are same and equal to 5 and length of diagonals are same and equal to 10 . \ Points in the argand plane represents a square.

5. (a)Q 3 + i = - 2wi,

(2 y - 1)2 - 1 log2 x = 4

3 - i = 2w2 i

\

where, w is complex cube root of unity. \ ( 3 + i )n + ( 3 - i )n

4

= ( - 2wi )n + (2w2 i )n

æ (2 x - 1)2 - 1 ö ç ÷ ç ÷ 4 è ø

n =8

when 8

( - 2wi ) + (2w2 i )8 = 28 [ w8 + w16 ]

(2 ´ 3 - 1)2 -1 -1

(3) = 2

= 26 = 64

4

= 28 [ w2 + w]

a is greatest divisors of n( n2 - 1)

1

6. (b) 2 cis

2

= value of n( n - 1) for

[Qw3 = 1]

= - 28 = - 256 [Qw + w2 = -1]

2. (b) For all n Î N

n =2 = 2 [22 - 1] = 6

Þ

7p 7p 7p = 2 æçcos + i sin ö÷ = z 5 è 5 5 ø 5

z = 25 (cos 7 p + i sin 7 p)

2

and b is greatest divisors of 2n( n + 2 ) = value of 2n( n2 + 2 ) for n = 1

1:2 S(x, y, z)

1 + 4 log2 x

2 1 + 4 log2 x = (2 y - 1)2

Þ

Þ 6 x + 11xy + 4 y - x + 2 y - 22 = 0. orthocentre, G is centroid and S is circumcentre, then SG:G0 =1:2 Let circumcentre be ( x, y, z ).

1+

(2 y - 1)2 - 1

2

45. (a) We know that, in a triangle, if O is

å

x x ´1 ´2 -0 = 6 3

1. (c) Q f ( x ) =

Real axis

4. (d)Q|Z1 - Z2| = |( - 3 + 5i ) - ( - 1 + 6 i )|

x ´ n( n + 1)(2n + 1) n {r2 n} -å 3 n® ¥ n3 ´ 6 r =1 n

9 ´ 27(4 - cos2 q) = 36 - 9 cos2 q

π/3 π/3

O

æ n 2 n 2 ö çç x å r - å{r n}÷÷ ø è r =1 r =1

1 n ® ¥ n3

9 ´ 27(4 - cos q) 36(1 - cos2 q) + 27 cos2 q

2

Area of sectorial area OAB is 2p 4 p units = 2´ = 3 3

2

r =1

= lim

2

=

A

- 3 + 2x 5 + 2y 2 + 2z = 3, =4 = 3, Þ 3 3 3 Þ x =6 , y =2, z =5 \ S(6, 2, 5) 1 46. (b) lim 3 {[12 x] + [22 x] + [32 x] n® ¥ n + ¼+ [ n2 x]}

3 ´ 3 3 cos q + 18 3

=

Imaginary axis

So,

= 2 ´ 1(1 + 2 ) = 6 ab = 6 ´ 6 = 36

3. (c) Diagram of| Z| £ 2 and -

p p £ amp Z £ , is 3 3

= 25 ( - 1) = - 32 ì - 2 x,

7. (a) Q| x - 1| + |x + 1| ïí - 2, ï 2 x, î For, x < - 1 - 2x < 4 Þ x > - 2 Þ x Î( - 2, - 1)

x 1

...(i)

28

Engineering Entrances Solved Papers 2018 For, x Î[ - 1, 1] -2 < 4 Þ x Î[ - 1, 1] and for x > 1, 2x < 4 Þ x 0 Þ

a Î( - ¥, - 2 ) È (0, ¥)

and D = 1 - 4 a ³ 0 1 Þ a Î æç - ¥, ö÷ è 4ø 1 1 and - > a Þ a < 2 2 1ö æ Þ a Îç - ¥, - ÷ è 2ø

...(i)

...(ii)

...(iii)

From Eqs. (i), (ii) and (iii), a< - 2

9. (b) Let

x =y 1-x

So,

1 5 y+ = y 2

So, a + b = - 3 and ab = 1 So, a + b + ab = - 2 11. (d) Number of possible 5 digit numbers is 120 Number of possible 4 digit numbers is 120 Number of possible 3 digit numbers start with 7 is 12 Number of possible 3 digit numbers start with 5 is 12 Then, according to order 3 digit numbers are 375, 372, 371, 357, 352, 351, 327 So, rank of number 327 is 271. 12. (b)Q10800 = 24 3352

= 1 y =2, 2 4 1 x= , 5 5 p>q

So, Q

4 1 p= , q = 5 5 Now, for the equation p ( p + q )x4 - pqx2 + = 0 q pq Sa = 0, Sab = p+ q p and abgd = q( p + q )

\

pq So, (Sa) - Sab + abgd = 0 + p +q p + q( p + q ) 2

p æ 1ö çq + ÷ p +q è qø 4 1 = 5 æç + 5ö÷ 4 1 è5 ø + 5 5 4 26 104 = ´ = 5 5 25

=

13

Cp

a

bp

1 ( bx ) p

C q ( - 1)q

13 - q

a

bq

q

x13 - 3 q

for x - 5 , q = 6 So, coefficient of x - 5 in the expansion 13 a7 1 of æç ax - 2 ö÷ is 13 C6 6 . è b bx ø According to the question, a6 a7 13 C7 7 = 13C6 6 Þ ab = 1 b b 14. (b)QSum of all binomial coefficients is 2n . and 200 < 2n < 400 Þ n = 8 QGeneral term 1 Tr + 1 = nC r æç 3 / 4 ö÷ èx ø

n-r

Then, A(1) A(4 ) + A(2 ) A(5 ) = (sin a + cos a) (sin4 a + cos4 a) + (sin2 a + cos2 a) (sin5 a + cos5 a) = (sin a + cos a) [(sin2 a + cos2 a)2

+ (sin5 a + cos5 a) = (sin a + cos a) (sin6 a + cos6 a)

= (sin a + cos a) (sin6 a + cos6 a)

Similarly, the general term in the 13 1 expansion of æç ax - 2 ö÷ è bx ø

13

15. (b)QA( n ) = sin n a + cos n a

+ (sin3 a cos2 a + sin5 a) + (sin2 acos3 a + cos5 a)

x26 - 3 p

1 ö Tq + 1 = 13C q ( ax )13 - q æç ÷ è bx2 ø

æ 5ö ar ç x 4 ÷ ç ÷ è ø

r

56 a3 = 448 Þ a3 = 8 Þ a =2

+ sin2 a cos2 a(sin a + cos a) + (sin5 a + cos5 a)

For x5 , p = 7 So, coefficient of x5 in the expansion of 13 6 æ ax2 + 1 ö is 13 C a [Q p =7] ç ÷ 7 è bx ø b7

=

Þ

= (sin a + cos a) [sin6 a + cos6 a + sin2 a cos2 a]

13

13 - p

For independent of ‘x’ 3 [Qn = 8] - n + 2r = 0 4 Þ 2r = 6 Þ r =3 According to the question, n C r ar =448 Þ 8 C3 a3 = 448

- 2 sin2 a cos2 a] + (sin5 a + cos5 a)

1 ö of æç ax2 + ÷ è bx ø

p

3 3 5 n+ r + r 4 4 4

= (sin a + cos a) [(sin2 a + cos2 a)3

13. (a) The general term in the expansion

T( p+1) = 13C p ( ax2 )13 -

-

- 2 sin2 a cos2 a] + (sin5 a + cos5 a)

So, number of all even divisors ‘a’ = 4 ´ 4 ´ 3 = 48 and number of all odd divisors ‘b’ = 4 ´ 3 = 12 So, 2a + 3b = (2 ´ 48) + (3 ´ 12 ) = 96 + 36 = 132

Þ 2 y2 - 5 y + 2 = 0 Þ

= nC r ar x

2

= (sin a + cos a)(sin6 a + cos6 a) + (sin2 a + cos2 a)(sin3 a + cos3 a) = A(1) A(6 ) + A(2 ) A(3 ) sin 9 q sin 3 q sin q 16. (c) Given, + + cos 27 q cos 9 q cos 3 q =k (tan 27 q - tan q) Q LHS sin 9 q cos 9 q + sin 3 q cos 27 q = cos 27 q cos 9 q sin q + cos 3 q sin 18 q + sin 30 q - sin 24 q = 2 cos 9 q cos 27 q sin q + cos 3 q [sin 21 q + sin 15 q + sin 33 q + sin 27 q - sin 27 q - sin 21 q + 4 sin q cos 9 q cos 27 q] 4 cos 3 q cos 9 q cos 27 q sin 15 q + sin 33 q + 4 sin q cos 9 q cos 27 q = 4 cos 3 q cos 9 q co s 27 q =

Engineering Entrances Solved Papers 2018 2 sin 24 q cos 9 q + 4 sin q cos 9 q cos 27 q = 4 cos 3 q cos 9 q cos 27 q sin 24 q + 2 sin q cos 27 q = 2 cos 3 q cos 27 q sin(27 q - 3 q) sin q = + 2 cos 3 q cos 27 q cos 3 q é sin 27 q sin 3 q ù sin q ê ú+ cos 27 q cos 3 q û cos 3 q ë 1 sin 27 q 1 æ sin 3 q - 2 sin q ö = - ç ÷ 2 cos 27 q 2 è cos 3 q ø 1 = tan 27 q 2 1 3 sin q - 4 sin3 q - 2 sin q 2 cos q(4 cos2 q - 3) =

1 2

sin q(1 - 4 sin2 q) ù 1é ú ê tan 27 q 2ë cos q(1 - 4 sin2 q) û 1 = [tan 27 q - tan q] 2 1 So, k = 2 =

åcos

2n

q=

n =0

=

1 sin2 q

y=

åsin

¥ 2n

n =0

=

19. (d)Qcos h2 x - sin h2 x = 1 Þ

1 1 - cos2 q

ìQ0 < q < p ü ý í 2þ î 1 q= 1 - sin2 q

1 cos2 q ¥

z=

åcos

2n

q sin2 n q

n =0

1 1 - cos2 q sin2 q 1 \ z= 1 1xy =

Þ Q

x=

cos h2 x - cos2 q = 1

[Qsin h x = cos q] cos2 q= cos h2 x - 1 5 14 = -1= 9 9 5 4 So, sin2 q = 1 - cos2 q = 1 - = 9 9 p Q -p < q < 2 2 sin q = \ 3 A- B a -b C 20. (c)Qtan cot = a+ b 2 2 a -b A+B = tan 2 a+b a -b 1 \ = a+b 4 5-b 1 Þ = 5+ b 4 Þ

¥

17. (a)Qx =

p ...(i) 2 or sin x(2cos x - 1) = cos x - 1 Þ 2sin x cos x + 1 = sin x + cos x Þ 4 sin2 x cos2 x + 1 + 4 sin x cos x = 1 + 2 sin x cos x [on squaring both sides] Þ 2 sin x cos x(2sin x cos x + 1) = 0 Either sin x = 0 or cos x = 0 or sin 2x = - 1 Q x Î(0, p) \ sin x ¹ 0 So, sin 2x = - 1 3p 3p ...(ii) Þ x= 2x = Þ 2 4 From Eqs. (i) and (ii), p 3p x = or 2 4 Only two solutions possible in x Î(0, p). Þ

...(i) xyz - z = xy 1 1 + = sin2 q + cos2 q = 1 x y

Þ x + y = xy From Eqs. (i) and (ii), ( x + y )z - z = xy Þ xz + yz = xy + z

...(ii)

Þ Þ

20 - 4 b = 5 + b 5b = 15 Þ b = 3

\

a2 - b2 = 25 - 9 = 16 = 4

21. (c) According to sine law,

18. (c) Given equation

Þ

sin x - sin 2x + sin 3x = 2 cos2 x - 2 cos x Þ 2 sin 2x cos x - sin 2x = 2 cos x(cos x - 1) Þ 2 sin x cos x(2 cos x - 1) = 2 cos x(cos x - 1) Either, cos x = 0

Þ

sin A sin B = a+ 1 a - 1 2 sin B cos B sin B = [Q A = 2B] a+ 1 a-1 a+ 1 cos B = 2( a - 1)

Þ

( a + 1)2 + a2 - ( a - 1)2 a+ 1 = 2 a(a + 1) 2( a - 1)

Þ

a+ 1 4 a + a2 = 2 a(a + 1) 2( a - 1)

29

(4 + a) ( a - 1) = ( a + 1)2

Þ

a2 + 3 a - 4 = a2 + 2 a + 1

Þ Þ

a=5 p 2 p B +C= \ 2 A B C and r = 4 R sin sin sin 2 2 2 p B 1 r B sin sin æç - ö÷ =4 Þ è4 2 2 2ø R Bé 1 B Bù 5 1 sin ú Þ = 2 2 sin ê cos 2 2 ë 2 2 2û 2 B B 5 2 B ...(i) = sin cos - sin Þ 4 2 2 2 A B C and r1 = 4 R sin cos cos 2 2 2 p Bö 1 r1 B æ cos cos ç - ÷ =4 ´ Þ è4 2 2 2ø R Bé 1 B B 1 l sin ù Þ = 2 2 cos ê cos + 2 úû 2 2 2 ë 2 2 B B B l ...(ii) Þ = cos2 + cos sin 4 2 2 2 From Eqs. (i) and (ii), l 5 - = 1 Þ l- 5 = 4 Þ l = 9 4 4 So, the roots of quadratic equation x2 - 4 x + 3 = 0 are 1, 3 x + x2 + x3 + K + x n 23.(c) (a)Qx = 1 n

22. (c)Q

A=

Þ nx = x1 + x2 + x3 + K + x n Þ ( x1 - x ) + ( x2 - x ) + K + (xn - x ) = 0 n

Þ

å( x i - x ) = 0 i =1

(b) QVariance ( s2 ) =

1 n

n

2

å( x i - x )

=

i =1

Mean of the squares of the deviations from mean. (c) QMean deviation of observations is mean of the absolute deviations from any measure of central tendency. (d) QCoefficient of variation is used to find the homogeneity of given two series. \ (a) ® (iii), (b) ® (v), (c) ® (iv), (d) ® (ii)

24. (d)QThe variance is remains unchanged if we add or subtract a positive number to each observation, but if any positive number ‘k’ is multiplied to each observation, then variance becomes k2 times. So, variance of the new data is 62 ´ 7 = 36 ´ 7 = 252.

30

Engineering Entrances Solved Papers 2018

25. (a) Since number of getting prime numbers {2, 3, 5} is 3 therefore number of getting prime numbers on both the dice is 3 ´ 3. And getting a head and a tail on two coins = 1 ´ 1 ´ 2 And total number of possibilities = (6 ´ 6 ) ´ (2 ´ 2) So, required probability 3 ´ 3´ 1 ´ 1 ´ 2 1 = = 16 ´ 6 ´ 2 ´ 2 8

26. (d) Let D( h, k ) 3 4 1 1 Now, DABC = | - 3 6 1 | 2 -5 1 1 1 = | [3(6 - 1) + 4( -5 + 3) + 1( -3 + 30 )]| 2 1 = | [15 - 8 + 27]| = 17 2 3 4 1 1 and DADC = | h k 1 | 2 -5 1 1 1 = | [3( k - 1) + 4( - 5 - h ) + 1( h + 5k )]| 2 1 = | (3k - 3 - 20 - 4 h + h + 5k )| 2 1 = | 8k - 3h - 23| 2 Q DABC = DADC Þ 34 = | 3h - 8k + 23| By taking locus of point ( h, k ), | 3x - 8 y + 23| = 34 Þ3x - 8 y - 11 = 0 or 3x - 8 y + 57 = 0 or (3x - 8 y - 11) (3x - 8 y + 57 ) = 0 3 3 -1 27. (d)Q = cos a + 2sin a 2 2 3+3 and = 2 cos a - sin a 2 2 æ 3 + 1ö æp pö 5 cos a = 5ç \ ÷ = 5 cos ç - ÷ è4 6 ø è 2 2 ø p cos a = cos 12

p Þ Þ a= 12 ap + bq + c , 28. (b) Since, L( p, q ) = a2 + b2 " p, q Î R 2 1 1 2 Now, L æç , ö÷ + L æç , ö÷ + L(2, 2 ) = 0 è3 3 ø è3 3 ø 2 1 1 2 Þ a + b + c+ a + b 3 3 3 3 + c + 2a + 2b + c = 0 (given) Þ 3a + 3b + 3c = 0 ...(i) Þ a+ b+c=0 So, line ax + by + c = 0 passes through the point (1, 1).

29. (d)

y

Þ

sin2 q =

Þ

q=

B (0, 4) 4

O

5

3

(3, 0) A

æ 2 ö 1 sin - 1 ç ÷ 2 è a + 2ø

32. (a) Since, 6 x2 + xy - y2 = 0

x

\ Incentre of DOAB is æ (5 ´ 0 ) + (4 ´ 3) + (3 ´ 0 ) , ç 5+4 +3 ç ç (5 ´ 0 ) + (4 ´ 0 ) + (3 ´ 4 ) ç 5+4 +3 è

2 a+2

Þ 6 x2 + 3xy - 2xy - y2 = 0 Þ

ö ÷ ÷ ÷ ÷ ø

12 12 = æç , ö÷ = (1, 1) è 12 12 ø

30. (b) Let the equation of required line is ...(i) x + 2y + c = 0 According to the question, 2|C - 3| | 8 - C | = 5 5 Þ 2(C - 3) = 8 - C Þ 3C = 14 14 C= Þ 3 So, equation will be 14 ...(ii) x + 2y + =0 3 Let the normal form is ...(iii) x cos a + y sin a = p From Eqs. (ii) and (iii), cos a sin a p = = 14 -1 -2 3 Þ a = p + tan - 1 2 14 14 and p = Þ p= 3 5 45 So, required equation is 14 , x cos a + y sin a = 45 a = p + tan - 1 2

31. (a) According question equation of lines ( y - 1) - tan q( x - 1) = 0 and ( y - 1) - cot q( x - 1) = 0 So, combined equation is ( y - 1)2 - ( x - 1)( y - 1) [tan q + cot q] + ( x - 1)2 = 0 2 Þ x2 xy + y2 sin 2 q æ 2 ö + ç - 2÷ ( x + y - 1) = 0 è sin 2 q ø On comparing with the equation x2 - ( a + 2 )xy + y2 + a( x + y - 1) = 0 2 Þ a+2= sin 2 q

(3x - y ) (2x + y ) = 0 y = 3 or - 2 Þ x Now, the second equation 2

For Þ Þ For

Þ

æ y ö + aæ y ö - 3 = 0 ç ÷ ç ÷ èx ø èx ø y =3 x 32 + 3a - 3 = 0 a= -2 y = -2 x ( - 2 )2 + a( - 2 ) - 3 = 0

[Qa >0]

4 - 2a - 3 = 0 1 a= 2

Þ

33. (b) The chord y - mx - 1 = 0 is tangent to the circle S1 º x2 + y2 - 4 x + 1 = 0 or So,

( x - 2)2 + ( y - 0 )2 = 3 | 2m + 1| = 3 1 + m2

Þ

4 m2 + 1 + 4 m = 3 + 3m2

Þ

( m + 2)2 = 6

Þ m = - 2± 6 Now, equation of chord is L º y + (2 ± 6 )x - 1 = 0, Let possible point is ( x1, y1 ) for which the chord L = 0 is chord of contact of circle S º x2 + y2 - 1 = 0. So, xx1 + yy1 - 1 = 0 and y + (2 ± 6 )x - 1 = 0 represent same line, so -1 x1 y = 1 = -1 2± 6 1 Þ

( x1, y1 ) º (2 ± 6 , 1)

34. (b) The given circle is x2 + y2 - 6 x - 2 y + 1 = 0 Þ

( x - 3)2 + ( y - 1)2 = 9

...(i)

and the line y + c = 0 is tangent to the circle at point ( a,4 ), so ...(ii) 4 + c=0 Þ c= -4 and slope of line joining the point of contact and centre is infinite, so

Engineering Entrances Solved Papers 2018 a-3 = 0 Þ a =3 4 -1 So,

QEquation of tangent at the vertex of parabola is x = 0, so let M(0, y1 ) and N (0, y2 ), so we will get y1 and y2 by putting x = 0 in Eq. (i).

ac = - 12

35. (b) For four common tangent, the

2

distance between centres of two circles C1C2 should be more than the sum of radii ( r1 + r2 ). \ 49 + 9 > 49 + 9 - 33 + a Þ 58 > 25 + a Þ a < 58 - 5 Q a Î N so, a ={1, 2} \ Two circles are possible only. x2 + y2 + 2gx + 2fy + c = 0

...(i)

Since, circle (i) passes through point (1, 0 ), ...(ii) 1 + 2g + c = 0 QCircle (i) cutting the circles x2 + y2 - 2x + 4 y + 1 = 0 and x2 + y2 + 6 x - 2 y + 1 = 0, orthogonally, so ...(iii) 2g( - 1) + 2f (2) = c + 1 and 2g(3) + 2f ( - 1) = c + 1 ...(iv) From Eqs. (ii), (iii) and (iv) f = 0 and g = 0 So, c =- 1 So, equation of required circle is x2 + y2 = 1 \ Centre is (0,0 ).

37. (a) Let the equation of required circle is x2 + y2 + 2gx + 2fy + c = 0

Þ Þ

...(i)

Since, circle (i) cuts the circles x2 + y2 - 2x + 6 y = 0, x2 + y2 - 4 x - 2 y + 6 = 0 and x2 + y2 - 12x + 2 y + 3 = 0 orthogonally, then ...(ii) - 2g + 6 f = c ...(iii) - 4 g - 2f = c + 6 and ...(iv) - 12g + 2f = c + 3 From Eqs. (ii), (iii) and (iv), we get 3 9 g = 0, f = - and c = 4 2 So, equation of required circle is 3 9 x2 + y2 - y - = 0 2 2 Now, equation of tangent at point(0, 3) is 9 3 3 y - ( y + 3) - = 0 Þ y = 3 2 4

38. (c) Let point of intersection of required tangents is ( h, k ), then equation of pair of tangents drawn through the point ( h, k ) to the parabola y2 = 8x is ( y2 - 8x )( k2 - 8h ) = [ yk - 4( x + h )]2 ...(i)

y2 k2 - 8hy2 = y2 k2 + 16 h2 - 8hky 2

y - ky + 2h = 0 having roots y1

and y2 . \

2 é b ù e2 = 1 - e2 êQ e = 1 - æç ö÷ ú èaø ú ê û ë

e4 + e2 = 1

Þ

2

y ( k - 8h ) = ( yk - 4 h )

Þ

36. (d) Let the equation of circle is

2

Þ

31

k2 - 8 h

| y1 - y2 | =

= 4 (given)

1

k2 - 8h = 16

Þ k2 = 8( h + 2 )

By taking locus of point ( h, k ) we are getting y2 = 8( x + 2 ).

39. (c) Since, equation of normal to parabola y2 = x, having slope ‘m ’ is 1 1 y = mx - 2 æç ö÷ m - m3 è4 ø 4

...(i)

QNormal drawn through the point( c, 0 ). \

m3 + 2m - 4 mc = 0

Þ

m[ m2 + (2 - 4 c )] = 0

41. (d) Equation of tangent at point ( x1, y1 ) x2 y2 + 2 = 1 is 2 a b xx1 yy1 ...(i) + 2 =1 a2 b Intersecting point of tangent with the 2 2 axes is Aæç a , 0 ö÷ and B æç0, b ö÷. è x1 ø è y1 ø Now, let middle point of point A and B is P ( h, k ), then a2 b2 and k = h= 2x1 2 y1 on the ellipse

a2 2h b2 and y1 = 2k

Þ

x1 =

Q ( x1, y1 ) on the ellipse

x2 y2 + 2 = 1, 2 a b

Either m = 0 {Qnormal is X-axis also} or m2 + (2 - 4 c ) = 0

so

QRemaining two are perpendicular to each other. 3 \ 2 - 4c = - 1 Þ 4c = 3 Þ c = 4

On taking locus of point P ( h, k ), we are a2 b2 getting 2 + 2 = 4. x y

42. (d) Let a function f ( x )= 2x3 + 10 x - 13

40. (c) Given, equation of ellipse is x2 y2 + = 1, let ( a > b ) the an end point a2 b2 of a lat us rectum be æ b2 ö ç ae, ÷, then the equation of the aø è normal at this end point is y Normal A

(–a, 0) (–ae, 0)

x - ae = ae a2

(ae, b2/a) (a, 0)

(0, –b)

y-

x

(ae, 0)

b2 a,

b2 ab2 It will pass through the end of the minor axis (0, - b ), if - a2 = - ab - b2 2

Þ

1=

a2 b2 + =4 h2 k2

2

b æbö b b + ç ÷ Þ 1 - æç ö÷ = èaø a èaø a

Q

f ¢( x ) = 6 x2 + 10 > 0, " x Î R

\ f ( x ) is strictly increasing function. Now,f(1) = 2 + 10 - 13 = - 1 < 0 and f(2 ) = 16 + 20 - 13 = 23 > 0 So, the one and only one root e Î(1, 2 ), where e is the eccentricity of the conic. Q 1 < e < 2, \ Conic is a hyperbola.

43. (c) Except the equilateral triangle, the centroid, orthocentre and circumcentre are collinear and centroid divides the line segment joining the orthocentre and the circumcentre in the ratio 2 : 1. So, if ( - 1, 3, 2 ) and (5, 3, 2 ) are respectively the orthocentre and circumcentre of triangle, then coordinate of centroid is æ ( - 1 ´ 1) + (5 ´ 2 ) , (3 ´ 1) + (3 ´ 2 ) ,ö ÷ ç 1+2 1+2 ÷ ç ÷ ç (2 ´ 1) + (2 ´ 2 ) ÷ ç 1+2 ø è = (3,3, 2 ) So, (A) is true and (R) is false.

32

Engineering Entrances Solved Papers 2018 3| x | - x log(1 + x3 ) - lim ® x 0 2x sin3 x - 3x - x lim ® - ¥ - x - 2x

44. (b) lim

x ® - ¥| x| -

= x

log(1 + x3 ) ´ x3 x3 , - lim 3 x ®0 æ sin x ö ´ x3 ÷ ç è x ø [Q|x | = - x, x < 0] 3+1 1 4 - 1 = - 1= = 3 3 1+2

Þ R = {( - 2, 2 ), (0,0 ), (1,1), (2,2 )} Clearly, R has four elements.So, the number of elements in power set of R is 24 = 16.

5. (a) We know that, ~( p ® q ) @ p Ù ~ q \ ~(( p Ù r ) ® ( r Ú q )) @ ( p Ù r ) Ù (~( r Ú q )) @ ( p Ù r ) Ù (~r Ù~ q )

6. (a) We have, 2b2 = 12 and b = 2 3 a e= 1+

\

VIT log z = log| z | + i arg ( z ) log i i = i log i = i[log|i| + i arg ( i )]

p ù é é = êsin æç - ö÷ ú + êcos è 2 øû ë ë

æ- p öù ç ÷ è 2 ø úû

= ( - 1)3 + (0 )3 = - 1

2. (a) We have, f ( x ) = 27 x3 -

1 x3 3

1 1 = æç3x - ö÷ + 9 æç3x - ö÷ è è xø xø Since, a and b are the roots of 3x \

3a -

1 =2 x

1 1 = 2 and 3 b - = 2 a b 3

1 1 Now, f( a) = æç3a – ö÷ + 9 æç3a - ö÷ è è aø aø = 23 + 9 ´ 2 = 26 Similarly, we have f(b) = 26

3. (a) We have, ( A È B )¢ È( A¢ Ç B ) = ( A¢ Ç B ¢ ) È ( A¢ Ç B ) = ( A¢ È A¢ ) Ç ( A¢ÈB ) Ç ( B ¢ È A¢ ) Ç( B ¢ È B ) = A¢ Ç ( A¢ È B ) Ç ( B ¢ È A¢ ) Ç U = A¢ Ç ( A¢ È B ) Ç ( B ¢ È A¢ ) = A¢ Ç {( A Ç B ¢ )¢ È( A Ç B )¢} = A¢ Ç {( A Ç B ¢ ) Ç( A Ç B )}¢ = A¢ Ç A¢ = A¢

4. (b) We have,

and

binomial distribution. Then, np + npq =15 and ( np )2 + ( npq )2 = 117

Þ

A = {x :|x| < 3, x ÎZ} = {- 2, - 1, 0, 1, 2} R = {( x, y ): y =|x|, x ¹ - 1}

Þ

The equation of the directrix of this parabola is 8 k x- =k 4 8 k x= Þ k 4 But the equation of the directrix is given as x - 1 = 0 8 k - = 1 Þ k2 + 4 k - 32 = 0 \ k 4 Þ k = - 8, 4

8. (c) We know that the line y = mx + c touches the curve

x2 y2 + = 1 iff a2 b2

c2 = a2 m2 + b2 Here, a2 = 4, b2 = 1, m = 4 2

2

\

c =4 ´4 + 1

Þ Þ

c2 = 65 c = ± 65

9. (b) The equation of any tangent to the given ellipse is x y cos q + sin q = 1 a b This line meets the coordinate axes at æ a ö æ b ö , 0 ÷ and Q ç0, Pç ÷ è cos q ø è sin q ø \

a b =l= cos q sin q

a2 + b2

10. (b) Let n and p be parameters of the

y - kx + 8 = 0

3

a2 b2 + 2 l2 l

l=

7. (c) The equation of the parabola is

\{sin(log i i )}3 + {cos(log i i )}3

cos2 q + sin2 q =

Þ

\

8 y2 = k æç x - ö÷ è kø 8 ( y - 0 )2 = k æç x - ö÷ è kø

a b and sin q= l l

l2 = a2 + b2

Þ

Þ

cos q=

Þ

2

p p = i éê log1 + i × ùú = 2û 2 ë 3

b2 a2

Þ

(2 3 )2 = 1+ 3=2 (2 )2

= 1+

1. (b) We know that \

Þ a=2

Þ

Þ

n2 p2 (1 + q2 ) 117 = 2 ( np + npq )2 15 1 + q2 117 = (1 + q )2 225 1 + q2 13 = 1 + q2 + 2q 25 6 q2 - 13 q + 6 = 0

Þ (2 q - 3)(3 q - 2) = 0 3 2 q = or q = Þ 2 3 3 When q = , then 2 3 -1 p = 1 - q = 1 - = , which is 2 2 2 neglected and when q = , then 3 2 1 p =1- = 3 3 2 1 q = and p = Þ 3 3 \ np + npq = 15 1 2 n ´ + n ´ = 15 Þ 3 9 Þ n =27 \ Mean, np = 9

11. (c) For a real chord of contact of tangents drawn from (a, a) to the circle x2 + y2 + 2gx + 4 y + 2 = 0, the point (a, a) must lie outside the circle. \ a2 + a2 + 2ga + 4 a + 2 > 0 Þ

a2 + ga + (2a + 1) > 0

Þ

a2 + ( g + 2 )a + 1 > 0

Since, a ÎR, then ( g + 2 )2 - 4 > 0 Þ Þ

g2 + 4 g > 0

Þ

-4 < g 0 x+y=2

P(a, a) X′

X

O

Þ -5< x 1ú êQa + ar + ar + ¼ + ar r 1 ë û

y = - 8t 1 Given, t = 2 and

2

\

1 x = - 4 æç ö÷ è2 ø

and

1 y = - 8 æç ö÷ è2 ø

x = -1 and y= -4 Hence, cartesian coordinates are ( -1, - 4 ).

2. (c) We have, x2 - 4 ,x ¹2 x -2 ( x + 2) ( x - 2) ,x ¹2 f(x ) = x -2 f(x ) =

f ( x ) = ( x + 2), x ¹ 2 \ Range of f ( x ) = R - {4}

3. (c) Given, 5x + y - 1 = 0 coincides with one of the lines given by 5x2 + xy - kx - 2 y + 2 = 0 On putting the value of y = - 5x + 1 in the above equation, we get

10 (10 n - 1) - n 9 Put n = 10 10 10 (10 - 1) - 10 Þ S10 = 9 10 10 (10 - 1 - 9 ) = 9 10 10 = (10 - 10 ) 9 100 9 ` = (10 - 1) 9 Þ Sn =

6. (b) Given, Þ Þ Þ Þ

A+ B+C =p A+ B = p -C cot ( A + B ) = cot ( p - C ) cot A cot B - 1 = - cot C cot A + cot B

cot A cot B - 1 = - cot A cot C - cot B cot C Þ cot A cot B + cot B cot C + cot C cot A = 1

34

Engineering Entrances Solved Papers 2018

7. (d) We have,

3 1 cos q + sin q = 2 2 Þ 2sin q + 2 3 cos q = 3 cos q + sin q Þ sin q = - 3 cos q Þ tan q = - 3

8. (d) Given P (4, 5, x ), Q (3, y, 4 ) and R (5, 8, 0 ) are collinear. ¾®

\

¾®

¾®

¾®

13. (c) We know that point of intersection of lines represented by ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is

...(ii)

2

8h2 = 9 ab

Þ

10. (a) Since, the line bisects the angle between coordinate axes. \

q = 135° Y

¾®

X′

¾®

O

θ = 135° X

and QR = OR - OQ = (5i$ + 8$j ) - (3i$ + yj$ + 4 k$ ) = 2i$ + (8 - y )$j - 4 k$ $ \ - i + ( y - 5)$j + (4 - x ) k$ = l[2i$ + (8 - y )$j - 4 k$ ] On equating the component of vector both sides, we get -1 =l 2 y-5 =l 8- y 4-x =l -4 On putting the value of l, we get 1 ( y - 5) = (8 - y ) æç - ö÷ è 2ø and 4 - x = - 4 æç è Þ and Þ and \

1ö ÷ 2ø

2 y - 10 = y - 8 4 - x =2 y=2 x =2 x + y=2+ 2=4

9. (a) We have, ax2 + 2hxy + by2 = 0 Let slope of one line is m \ Slope of another line is 2 m. We know that, 2h m1 + m2 = b

Y′

Now, equation of the line passing through the point ( -3, 1) and making angle q = 135° with positive direction of X-axis, is y - 1 = m( x + 3) Þ y - 1 = tan 135° ( x + 3) [Qm = tan q and q = 135° ] Þ y - 1 = - 1 ( x + 3) Þ y - 1 = -x - 3 Þ x + y+ 2=0 Hence, option (a) is correct.

11. (b) We have, Getting above 95% marks is necessary condition for Hema to get the admission in good college. Here, p : Hema get the admission in good college q : Hema gets above 95% marks We know symbolic form of q is necessary condition for p is p ® q Negation of ( p ® q ) is ( p Ù ~ q ) or (~ q Ù p ) \ Negation of the above statement is Hema does not get above 95% marks and she gets admission in good college.

12. (a) cos 1 °× cos 2° × cos 3° .... cos 90° cos 91°.... cos 179° = 0 [Qcos 90° = 0]

æ hf - bg hg - af ö , ÷. ç 2 2 è ab - h ab - h ø

...(i)

- 2h ö a 2 æç ÷ = è 3b ø b

¾®

PQ = OQ - OP = (3i$ + yj$ + 4 k$ ) - (4 i$ + 5$j + xk$ ) = - i$ + ( y - 5)$j + (4 - x )k$

a b

2h \ m + 2m = b - 2h 3m = Þ b a and m (2m ) = b a 2 Þ 2m = b On eliminating m, we get

PQ = l QR ¾®

Þ

m1m2 =

and

p p 2sin æç q + ö÷ = cos æç q - ö÷ è è 3ø 6ø p p Þ 2 éêsin qcos + cos q sin ùú 3 3û ë p p = cos q cos + sin q sin 6 6 é sin q 3 cos q ù + Þ 2ê ú 2 û ë 2

\ Point of intersection of lines represented by 1 3 x2 - y2 + x + 3 y - 2 = 0 is æç - , ö÷ è 2 2ø where, a = 1, b = - 1, c = - 2, h = 0, 1 g= 2 3 and f = 2

14. (c) A die is rolled \ Sample space = { 1, 2, 3, 4, 5, 6 } Given X denotes the number of positive divisors of the outcomes \ X = { 1, 2, 3, 4, 5, 6 }

15. (c) Given a, b, c are in AP \

2b = a + c C A Now, a cos2 + c cos2 2 2 1 + cos C ö 1 + cos A ö æ æ Þ aç ÷+ c ç ÷ è ø è ø 2 2 x éQ2 cos2 = 1 + cos x ù êë úû 2 1 = [ a + c + a cos C + c cos A] 2 3b 1 = [2b + b] = 2 2 [Q b = a cos C + c cos A]

16. (b) We have, sin x + sin 3x + sin 5x = 0 sin x + sin 5x + sin 3x = 0 2 sin 3x cos 2x + sin 3x = 0 sin 3x (2 cos 2x + 1) = 0 sin 3x = 0 1 2p or cos 2x = - = cos 2 3 Þ 3x = n p 2p or 2 n = 2 np ± 3 np x= Þ 3 p or x = np ± 3 p 3p ù é But, it is given that x Î ê , ú ë2 2 û 2p 4p x= , p, \ 3 3 \ Number of solutions is 3. Þ Þ Þ Þ

Engineering Entrances Solved Papers 2018 17. (b) Given statement is ‘‘If the weather is fine then my friends will come and we go for a picnic’’. Here, p = the weather is fine q = my friends will come r = we go for a picnic Symbolic form is p ® ( q Ù r ) contrapositive of p ® q is ~ q Ù ~ p \ Contrapositive of p ® ( q Ù r ) ~( q Ù r ) ~ p Þ (~ q Ú ~ r ) ® ~ p \ Contrapositive of the given statement is, if my friends will not come or we do not go for a picnic then weather will not fine.

18. (a) We have, X = { 4 n - 3n - 1 : n Î N } = {0, 9, 54, ......} and Y = {9 ( n - 1) : n Î N } = {0,9, 18, 27, 36, 45, 54, ......} \ X ÇY = X

19. (b) Let us prepare the following truth table

p

q

~p

~pÙq

p Ù (~ p Ù q )

T

T

F

F

F

Since, (2,3) satisfies the curve y2 = ax3 + b.

T

F

F

F

F

\

F

T

T

T

F

F

F

T

F

F

Þ b = 9 - 16 = - 7 \ 7 a + 2b = 7(2) + 2 ( - 7 ) = 0

Clearly, last column of the truth table contains F only. So, it is a contradiction.

20. (a) We have,

(3)2 = 2(2)3 + b

21. (b) Given sides of rectangle are x = ± a and y = ± b \ Centre of circle = (0, 0 ) and radius of circle =

y2 = ax3 + b On differentiating both sides w.r.t. x, we get dy = 3ax2 2y dx dy 3ax2 = Þ 2y dx

a2 + b2

Y y=b r X′

x=–a

O a

3a(2) æ dy ö = = 2a ç ÷ è dx ø (2, 3 ) 2(3)

Now, slope of line y = 4 x - 5 is 4 \ Slope of curve is 2a = 4 Þ a = 2 Now, y2 = ax3 + b

b x= a y=–b

2

Þ

35

Y′

\ Equation of circle x2 + y2 = ( a2 + b2 )

X

Engineering Entrance Questions 2019-20 JEE Main & Adv., BITSAT, AP & TS EAMCET, MHTCET, WBJEE Sets, Fundamentals of Relations and Function 1. The greatest positive integer k, for which 49k + 1 is a factor of the sum 49125 + 49124 + ...... + 49 2 + 49 + 1, is JEE Main 2020

(a) 32 (c) 65

(b) 63 (d) 60

2. Let f : R → R be such that for all x ∈ R (21 + x + 21 − x ), f ( x ) and (3 x + 3 −x ) are in A.P, then the minimum value of f ( x ) is JEE Main 2020 (a) 4 (c) 2

(b) 3 (d) 0

3. Let f : (1, 3 ) → R be a function defined x[ x ] , where [ x ] denotes the by f ( x ) = 1 + x2 greatest integer ≤ x. Then the range of JEE Main 2020 f is 2 3 3 (a)  ,  ∪  ,  5 5   4 2 4 (b)  ,   5 5 

3 4 (c)  ,  5 5  2 1 3 (d)  ,  ∪  ,  5 2 5

4  5

(a) (15 )! × 6 ! (b) ( −1, 1) × { 0 } 1 1 (c) R −  − ,   2 2  (d) R − [ −1, 1]

6. Two newspapers A and B are published in a city. It is known that 25% of the city population reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A and B look into advertisements. Then, the percentage of the population who look into advertisements is JEE Main 2019

(a) 13.5 (b) 13

(c) 12.8

7. If x ∈ R, then the range of

4 5 

is

4. If A = { x ∈ R :| x | < 2 } and B = { x ∈ R :| x − 2 | ≥ 3 }; then JEE Main 2019

(a) (b) (c) (d)

5. The number of functions f from {1, 2, 3, ...,20 } onto {1, 2, 3,...,20 } such that f (k ) is a multiple of 3, whenever k is a JEE Main 2019 multiple of 4, is

B − A = R − ( −2 , 5 ) A − B = [ −1, 2 ) A ∪ B = R − (2, 5) A ∩ B = ( −2 , − 1)

(d) 13.9

x x 2 − 5x + 9

AP EAMCET 2019

−1 (b) −∞ ,  ∪ (1, ∞ )  11  1  (d) −1,  11 

1 (a)  − , 1  11  −1 (c)  , 1  11 

8. If | x | denotes the greatest integer function, then the domain of the function

f (x ) =

x − |x | , is log( x 2 − x ) TS EAMCET 2019

(a) (1, ∞ ) (b) (1, ∞ ) − Z 1 − 5 1 + 5  (c) R −  . 2   2 1 − 5 5 + 1 (d)  . 2   2

9. If [ x ] denotes the greatest integer ≤ x, then the domain of the function 4 − x2 is f (x ) = [x ] + 2 TS EAMCET 2019 (a) ( −∞ , − 2] ∪ [ −1, 2 ) (b) ( −∞ , − 2 ) ∪ [ −1, 2] (c) ( − ∞ , − 2 ) ∪ ( −1, 2 ) (d) ( −∞ , − 1] ∪ [1, 2]

10. If A = { x | x ∈ N , x is a pirme number less than 12} and B = { x | x ∈ N , x is a factor 10}, then A ∩ B = ……… MHTCET 2019

(a) { 2 } (c) { 2 , 5 , 10 }

(b) { 2 , 5 } (d) {1, 2 , 5 , 10 }

11. Let P and T be the subsets of k ,y -plane defined by P = {( x, y ) : x > 0, y > 0 and x 2 + y 2 = 1} T = {( x, y ) : x > 0, y > 0 and x 8 + y 8 < 1} Then, P ∩ T is

(a) the void set φ (c) T

WB JEE 2019

(b) P (d) P − T

C

Sequence and Series 1. Five numbers are in A.P., whose sum is 25 and product is 2520. If one of these 1 five numbers is − , then the greatest 2 number amongst them is JEE Main 2020

(a) 7 (c) 27

(b) 16 21 (d) 2

2. Let a1, a 2, a 3, ... be a G.P. such that a1 < 0, a1 + a 2 = 4 and a 3 + a 4 = 16. If 9

∑ ai = 4λ, then λ is equal to

i =1

(a) − 171 511 (c) 3

JEE Main 2020

(b) 171 (d) − 513

1

1

1

1

3. The product 2 4 ⋅ 416 ⋅ 8 48 ⋅ 16128 ⋅.... to ∞ is equal to JEE Main 2020 (a)

1 24

(b) 2

(c)

1 22

(d) 1

a 4. Let a1, a 2, ...., a10 be a GP. If 3 = 25, a1 a9 then equals a5 JEE Main 2019 3

(a) 5 (c) 4 (5 2 )

2

(b) 2 (5 ) (d) 5 4

5. If three distinct numbers a, b and c are in GP and the equations ax 2 + 2bx + c = 0 and dx 2 + 2ex + f = 0 have a common

root, then which one of the following statements is correct? JEE Main 2019 (a) d , e and f are in GP d e f (b) , and are in AP a b c (c) d , e and f are in AP d e f (d) , and are in GP a b c 10

6. Let

∑ f (a + k ) = 16(210 − 1), where the

k =1

function f satisfies f ( x + y ) = f ( x ) f (y ) for all natural numbers x, y and f (1 ) = 2. Then, the natural number ‘a’ is JEE Main 2019

(a) 2 (c) 3

(b) 4 (d) 16

2

Engineering Entrance Questions 2019-20

7. The sum of first n terms of the series 3 21 117 + + + … is APEAMECET 2019 5 25 125 (a) n + (c) n +

2n + 1 2 − 3 × 5n 3

(b) n −

2n + 1 2 − 3 × 5n 3

2n + 1 2 + n 3 3 ×5

(d) n −

2n + 1 2 + n 3 3 ×5

8. If α ∈ R, n ∈ N and n + 2 (n −1 ) + 3 (n − 2 ) + ... + (n − 1 )2 + n.1 = αn (n + 1 ) (n + 2 ), then TS EAMCET 2019 α= (a)

1 2

(b)

1 3

(c)

1 5

(d)

9. For a GP, if Sn = 1 9 7 (c) 9

2 9 4 (d) 9

(b)

(a)

1 6

4n − 3n , then t 2 = …… 3n MHT CET 2019

Complex Numbers  z −1 1. If Re   = 1, where z = x + iy ,  2z + i  then the point ( x, y ) lies on EE a Main 2020 2 straight line whose slope is − . 3 1 3 circle whose centre is at  − , −  .  2 2 3 straight line whose slope is . 2 5 circle whose diameter is . 2

(a) (b) (c) (d)

2. If the equation,x 2 + bx + 45 = 0 (b ∈R) has conjugate complex roots and they satisfy | z + 1| = 2 10, then JEE Main 2020 (a) b (c) b

2

2

+ b = 72 − b = 30

(b) b

2

+ b = 12

(d) b

2

− b = 42

10

(b)

7 2

(c)

15 4

π 4

(b)

π 6

(c) 0

(b) 85

3 6. If z = + 2 (1 + iz + z 5

(c) – 85

(d)

(d) – 91

i (i = −1 ), then 2 + iz 8 ) 9 is equal to

π 3

BIT SAT 2019

(a) 3 (c) 5

(b) ( −1 + 2i ) (d) 0

(a) 1 (c) −1

(a) 2 x 2 + 2 y 2 + 5 x + 5 y − 12 = 0

9

(b) 2 x 2 − 3 xy + y 2 + 5 x + y − 12 = 0 (c) 2 x 2 + 3 xy + y 2 + 5 x + y + 12 = 0 (d) 2 x 2 + 2 y 2 − 11x + 7 y − 12 = 0

7. Let S be the set of all complex numbers

z satisfying | z − 2 + i | ≥ 5. If the

1 | z0 − 1 |

10. If P is a complex number whose

modulus is one, then the equation 4  1 + iz   = P has   1 − iz  TS EAMCET 2019

is the maximum of the set  1  : z ∈ S , then  | z − 1 |  the principal argument of π 4

(b)

3π 4

(c) −

(a) real and equal roots (b) real and distinct roots (c) two real and two complex roots (d) all complex roots

4 − z0 − z0 is z 0 − z 0 + 2i

JEE Advance 2019

(a)

π 2

(d)

π 2

8. The value of ‘λ’ for which the loci arg π z = and | z − 2 3 i | = λ on the argand 6 plane touch each other is

11. The general value of the real angle θ,

which satisfies the equation, (cosθ + i sin θ ) (cos2θ + i sin 2θ ) ... (cosnθ + i sin nθ ) = 1 is given by, (assuming k is an integer) WB JEE 2019

(a) 5 2

(b) 10

(c) 10 2 (d) 5

2. The number of real roots of the equation, e 4 x + e 3x − 4e 2x + e x + 1 = 0 is JEE Main 2020

(a) 3

(b) 4

(c) 1

(d) 2

3. let a, b ∈ R, a ≠ 0 be such that the equation,ax 2 − 2bx + 5 = 0 has a repeated rootα, which is also a root of the equation,

x 2 − 2bx − 10 = 0. If β is the other root of is equal to this equation, then α 2 + β 2JEE Main 2020 (a) 26

(b) 24

(c) 28

(d) 25

4. The value of λ such that sum of the squares of the roots of the quadratic equation, x 2 + (3 − λ ) x + 2 = λ has the JEE Main 2019 least value is (a)

4 9

(b) 1

(c)

15 8

(d) 2

5. The sum of the solutions of the equation | x − 2 | + x ( x − 4 ) + 2 = 0 ( x > 0 ) is JEE Main 2019 equal to (a) 9

(b) 12

(c) 4

(d) 10

2kπ n+ 2 4kπ (c) n+1

4kπ n(n + 1) 6kπ (d) n(n + 1) (b)

(a)

Inequalities and Quadratic Equation 1. Let α and β be two real roots of the equation (k + 1 ) tan 2 x − 2 ⋅ λ tan x = (1 − k ), where k( ≠ − 1 ) and λ are real numbers. If tan 2(α + β ) = 50, then a JEE Main 2020 value of λ is

(b) 4 (d) 6

9. If z = x + iy , x, y ∈ R, ( x, y ) ≠ ( 0, − 4 )  2z − 3  π and Arg   = , then the locus of  z + 4i  4 AP EAM CET 2019 z is

JEE Main 2019

(d) 2 3

4. Let z 0 be a root of the quadratic equation, x 2 + x + 1 = 0, If z = 3 + 6iz 081 − 3iz 093, then arg z is equal to JEE Main 2019 (a)

(a) 91

complex number z 0 is such that

3. Let z be a complex number such that 5 z −i = 1 and | z | = . Then the value of z + 2i 2 JEE Main 2020 | z + 3i | is (a)

3

1  x + iy  5. Let  −2 − i  = (i = −1 ), where   3 27 x and y are real numbers, then y − x equals JEE Main 2019

6. If α and β are the roots of the quadratic equation, x 2 + x sin θ − 2 sin θ = 0, α 12 + β12  π is θ ∈  0,  , then −12  2 (α + β −12 )(α − β ) 24 equal to JEE Main 2019 (a)

212 (sin θ + 8 )12

(b)

(c)

212 (sin θ − 4 )12

(d)

26 (sin θ + 8 )12 212 (sin θ − 8 ) 6

7. All the pairs ( x, y ) that satisfy the

inequality 2

sin 2 x − 2 sin x + 5

1

⋅ 4

also satisfy the equation

sin 2 y

≤1

JEE Main 2019

3

Engineering Entrance Questions 2019-20 (a) 2 |sin x | = 3 sin y (b) sin x = |sin y| (c) sin x = 2 sin y (d) 2 sin x = sin y

8. Let p and q be roots of the equation x 2 − 2 x + A = 0 and let r and s be the roots of the equation x 2 − 18 x + B = 0, If p < q < r < s are in AP, then A and B are TS EAMCET 2019

(a) − 3 , − 77 (c) − 3, 77

(a) a > 0, c > 0 (c) a < 0, c > 0

(b) 3, − 77 (d) 3, 77

9. Let a, b, c be real numbers such that a + b + c < 0 and the quadratic equation ax 2 + bx + c = 0 has imaginary roots. Then, WB JEE 2019

10. The value of lim

x → 0+

(a)

[q] p

(b) 0

(b) a > 0, c < 0 (d) a < 0, c < 0

x q  is p  x  WB JEE 2019 (d) ∞

(c) 1

Permutation and Combination 1. If the number of five digit numbers with distinct digits and 2 at the 10th place is 336 k, then k is equal to JEE Main 2020 (a) 8 (c) 4

(b) 7 (d) 6

2. The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is JEE Main 2020 (a) 1256 (c) 1356

(b) 1465 (d) 1365

3. All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is JEE Main 2020 (a) 180 (c) 160

(b) 175 (d) 162

4. The number of four-digit numbers strictly greater than 4321 that can be formed using the digits 0, 1, 2, 3, 4, 5 (repetition of digits is allowed) is

5. Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total number of beams is JEE Main 2019 (a) 180

(b) 210

(b) 310 (d) 288

(d) 190

BIT SAT 2019

(a) 719

(b) 265

(c) 454

(d) 720

7. The number of five digit numbers that are divisible by 6 which can be formed by choosing digits from {0, 1, 2, 3, 4, 5}, when repetition is allowed, is AP EAMCET 2019

JEE Main 2019

(a) 306 (c) 360

(c) 170

6. A person writes letter to six friends and addresses the corresponding envelopes. Let x be the numbers of ways so that at least two of the letters are in wrong envelopes and y be the numbers of ways so that all the letters are in wrong envelopes. Then x − y =

(a) 648 (c) 1296

(b) 540 (d) 1080

JEE Main 2019

(a) (b) (c) (d)

Both P (3 ) and P (5 ) are true P (3 ) is false but P (5 ) is true Both P (3 ) and P (5 ) are false P (5 ) is false but P (3 ) is true

APEAMCET 2019

(b) 12

(c) 10

9. A student is asked to answer 10 out of

13 questions is an examination such that he must answer atleast four questions from the first five questions. The number of choices available to him is TS EAMCET 2019 (a) 140

(b) 176

(d) 5

3. For all n ∈ N , if 1 2 + 2 2 + 3 2 + ...+ n 2 > x, then x is equal to TS EAMCET 2019

(c) 196

(d) 280

10. There are 7 greeting cards, each of a different colour and 7 envelopes of same 7 colours as that of the cards. The number of ways in which the cards can be put in envelopes, so that exactly 4 of the cards go into envelopes of WB JEE 2019 respective colour is, (b) 2. 7 C 3 (d) 3 !7 C 3 4C 3

4

2. If 15k divides 47! but 15k + 1 does not divide it, then k is euqal to (a) 15

TS EAMCET 2019

(b) 3 5 × 70 (d) 3 4 × 60

(a) 630 (c) 3 6 × 70

(a) 7 C 3

Mathematical Induction 1. Consider the statement : ‘‘ P (n ) : n 2 − n + 41 is prime.’’ Then, which one of the following is true?

8. The number of 4 letter permutations formed with English alphabet such that the number of distinct vowels is equal to the number of distinct consonants, when repetition is allowed, is

(c) 3 ! C 4

(a)

n3 3

(b)

n3 2

(c) n3

(d)

n4 4

4. 7 2n + 16n − 1 (n ∈ N ) is divisible by WB JEE 2019

(a) 65 (c) 61

(b) 63 (d) 64

Binomial Theorem 1. If the sum of the coefficients of all even powers of x in the product (1 + x + x 2 + … + x 2n ) (1 − x + x 2 − x 3 + … + x 2n ) is 61, then n is equal to …… . JEE Main 2020

2. If the fractional part of the number 2 403 k is , then k is equal to 15 15 JEE Main 2019 (a) 14 (c) 4

(b) 6 (d) 8

3. If the third term in the binomial expansion of (1 + x log 2 x ) 5 equals 2560, then a possible value of x is JEE Main 2019 1 (a) 4 2 (b) 4

1 (c) 8

(a) 2 26 (d) 2 2

4. The positive value of λ for which the coefficient of x 2 in the expression 10 λ  x 2  x + 2  is 720, is  x  JEE Main 2019 (a) 3 (c) 2 2

(b) 5 (d) 4

5. The sum of the sereis 2 ⋅ 20C 0 + 5 ⋅10 C1 + 8. 20 C 2 + 11 ⋅ 20C 3+ ...... + 62.20 C 20 is equal to JEE Main 2019 (b) 2 25

(c) 2 23

(d) 2 24

6. If the coefficient of x 3 and x 4 in the expansion of (1 + ax + bx 2 )(1 − 2 x )18 in power of x are both zero, then (a, b ) is BITSAT 2019

251  (a)  16 ,   3  272  (c)  14 ,   3 

251  (b)  14 ,   13  272  (d)  16 ,   3 

4

Engineering Entrance Questions 2019-20

Trigonometric Functions and Equations 1. The value of  π  3π   π  3π  cos3   ⋅ cos  + sin 3   ⋅ sin    8  8  8  8 JEE Main 2020 is (a)

1 4

2. If x =

(b)

1 2 2

1 2

(c)

(d)

1 2

∞

∞

∑ (−1)n tan2n θ and y = ∑ cos2n θ, n=0

n=0

π for 0 < θ < , then 4 (a) y (1 + x ) = 1 (c) x (1 + y ) = 1

(a)

3π 8

(b)

5π 4

(c)

π 2

(d) π

5. Let α and β be the roots of the quadratic equation x 2 sin θ − x(sin θ cosθ + 1 ) + cosθ = 0 ( 0 < θ < 45 º ) and α < β. Then, ∞  ( − 1 )n  ∑ αn + βn  is equal to n=0 JEE Main 2019

(b) y (1 − x ) = 1 (d) x (1 − y ) = 1

 π π 3. For any θ ∈  ,  , the expression  4 2 3 (sin θ − cosθ ) 4 + 6 (sin θ + cosθ ) 2 + 4 sin 6 θ equals JEE Main 2019

(c) 13 − 4 cos θ + 6 sin θ cos θ 2

2

2

(d) 13 − 4 cos θ 6

 π 4. The sum of all values of θ ∈  0,   2 3 2 4 satisfying sin 2θ + cos 2θ = is 4

JEE Main 2019

JEE Main 2019

(b) 5

4 and let 5

5 π , where 0 ≤ α, β ≤ , 4 13 then tan2α = BITSAT 2019

sin(α – β ) =

a.

20 7

b.

25 16

c.

56 33

d.

19 2

AP EAMCET 2019

6. The number of solutions of the equation  5π 5π  is 1 + sin 4 x = cos2 3 x, x ∈ − ,  2 2  (a) 3

8. Let cos(α + β ) =

(a) 0 3 (c) 2

(a) 13 − 4 cos 4 θ + 2 sin 2 θ cos 2 θ (b) 13 − 4 cos 2 θ + 6 cos 4 θ

(b) 2 (d) infinitely many

9. cos2 5 ° − cos2 15 ° − sin 2 15 ° + sin 2 35 ° + cos15 °sin 15 ° − cos5 °sin 35 ° =

1 1 − 1 − cos θ 1 + sin θ 1 1 (b) + 1 − cos θ 1 + sin θ 1 1 (c) − 1 + cos θ 1 − sin θ 1 1 (d) + 1 + cos θ 1 − sin θ (a)

JEE Main 2020

(a) 0 (c) 1

(c) 7

(d) 4

7. Number of solution of the equation | cos x| = 2 [ x ] are (where | x|, [ x ] are modulus and greatest integer function respectively). BITSAT 2019

(b) 1 (d) 2

10. If A and B are acute angles satisfying 3 cos2 A + 2 cos2 B = 4 and 3 sin A 2 cos B , then A + 2 B = = sin B cos A TS EAMCET 2019 (a) 30° (c) 60°

(b) 45° (d) 90°

11. In ∆ABC, with the usual notations, if A  B 3   tan   tan  = then a + b = ……  2  2 4 MHT CET 2019 (a) 4c (c) 7c

(b) 2c (d)

Properties of Triangles, Heights and Distances, Cartesian System of Rectangular Coordinates 3  1. Let A(1, 0 ), B(6, 2 ) and C  , 6 be the 2  vertices of a triangle ABC. If P is a point inside the triangle ABC such that the triangles APC, APB and BPC have equal areas, then the length of the line segment PQ, where Q is the point  7 7  − , −  , is ……… .  6 3 JEE Main 2020 2. Let C be the centroid of the triangle with vertices (3,−1 ), (1, 3) and (2, 4). Let P be the point of intersection of the lines x + 3y − 1 = 0 and 3 x − y + 1 = 0. Then the line passing through the points C and P also passes through the JEE Main 2020 point (a) ( −9 , − 7 ) (c) (7, 6)

(b) ( −9 , − 6 ) (d) (9, 7)

3. Consider a triangular plot ABC with sides AB = 7 m, BC = 5 m and CA = 6 m. A vertical lamp-post at the mid-point D of AC subtends an angle 30° at B. The height (in m) of the lamp-post is JEE Main 2019

(a)

2 21 3

(c) 7 3

(b) 2 21 (d)

3 21 2

4. A point P moves on the line 2 x − 3y + 4 = 0. If Q (1, 4 ) and R(3, − 2 ) are fixed points, then the locus of the centroid of ∆PQR is a line JEE Main 2019 2 3 (b) with slope 3 2 (c) parallel toY-axis (d) parallel to X-axis (a) with slope

5. If the line 3 x + 4y − 24 = 0 intersects the x - axis at the point A and the Y -axis at the point B then the incentre of the triangle OAB. Where O is the origin is JEE Main 2019

(a) (4, 3) (c) (4, 4)

(b) (3, 4) (d) (2, 2)

6. Two vertices of a triangle are (0, 2) and (4, 3). If its orthocentre is at the origin, then its third vertex lies in which JEE Main 2019 quadrant? (a) Fourth (c) Second

(b) Third (d) First

7. In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x 2 − c 2 = y , where c is the length of the third side of the triangle, then the circumradius of the triangle is (a)

c 3

(b)

c 3

(c)

3 y 2

(d)

y 3

8. Let O ( 0, 0 ) and A( 0, 1 ) be two fixed points, then the locus of a point P such that the perimeter of ∆AOP is 4, is (a) 8 x 2 − 9 y 2 + 9 y = 18

JEE Main 2019

(b) 9 x 2 − 8 y 2 + 8 y = 16 (c) 9 x 2 + 8 y 2 − 8 y = 16 (d) 8 x 2 + 9 y 2 − 9 y = 18

9. A rectangle is inscribed in a circle with a diameter lying along the line 3y = x + 7. If the two adjacent vertices of the rectangle are (–8, 5) and (6, 5), then the area of the rectangle (in sq JEE Main 2019 units) is (a) 72

(b) 84

(c) 98

(d) 56

5

Engineering Entrance Questions 2019-20 10. Two poles standing on a horizontal ground are of heights 5 m and 10 m, respectively. The line joining their tops makes an angle of 15º with the ground. Then, the distance (in m) between the JEE Main 2019 poles, is (a) 5 ( 3 + 1)

5 (2 + 3 ) 2 (d) 5 ( 2 + 3 )

(b)

(c) 10 ( 3 − 1)

9 10

(b)

10 11

(c)

11 12

(d)

12 13

12. In any triangle ABC, if a : b : c = 2 : 3 : 4, TS EAMCET 2019 then R : r = (a) 8 : 3

(b) 16 : 9

(c) 5 : 16

(d) 16 : 5

13. If A( −2, 1 ), B( 0, − 2 ), C (1, 2 ) are the vertices of a triangle ABC, then the perpendicular distance from its circumcentre to the side BC is TS EAMCET 2019

7 13 3 17 (a) (b) 22 22

5 10 (c) 11

d)

2026 22

2 2

a b c 4∆ 2  1 1 1  (d) 4  2 + 2 + 2  p 2 p3   p1 (b)

(a) p1 p 2 p 3

b +c c +a a +b , = = 9 10 11 cos A + cos B then = APEAMCET 2019 cosC

m+n 2 (c) m + n

3 (m + n) 4 (d)2(m + n)

(a)

(b)

17. In ∆ABC, if tan A + tan B + tan C = 6 and tan A. tan B = 2 then tanC =……… MHT CET 2019

TS EAMCET 2019 2

11. In triangle ABC, if

(a)

14. If p1, p 2, p 3 are the altitudes of a triangle ABC from the vertices A, B, C respectively, then with the usual 1 1 1 1 notation, 2 + 2 + 2 + 2 = r1 r2 r3 r

a 2b 2c 2 (c) ∆2

15. The distance between the circumcentre and the centroid of the triangle formed by the vertices (1, 2 ), (3, − 1 ) and ( 4, 0 ) is TS EAMCET 2019

1 45 2 7 2 (c) 15

(a)

(b) 4 (d)

9 2 5

16. A line meets the coordinate axes at A and B. If the perpendicular distances from A and B to the tangent drawn at the origin to the circumcircle of ∆OAB are m and n respectively, then the diameter of that circle is TS EAMCET 2019

(a) 3

(b) 4

(c) 1

(d) 2

18. The three sides of right angled triangle are in GP (geometric progression). If the two acute angles be α and β, then WB JEE 2019 tanα and tanβ are 5+1 5 −1 and 2 2 5+1 5 −1 (b) and 2 2 1 (c) 5 and 5 5 2 (d) and 2 5

(a)

19. The angles of a triangle are in the ratio 2 : 3 : 7 and the radius of the circumscribed circle is 10 cm. The length of the smallest side is WB JEE 2019

(a) 2 cm (b) 5 cm

(c) 7 cm

(d) 10 cm

Straight Line and Pair of Straight Lines 1. The locus of the mid-points of the perpendiculars drawn from points on the line, x = 2y to the line x = y is JEE Main 2020

(a) (b) (c) (d)

5x − 7 y 2x − 3y 3x − 2y 7 x − 5y

=0 =0 =0 =0

+ 6x 3 − 4 = 0 − 12 x 3 + 4 = 0 − 6x 3 + 4 = 0 − 12 x 3 − 4 = 0

(a) Each line passes through the origin (b) The lines are concurrent at the point  3 , 1    4 2 (c) The lines are all parallel (d) The lines are not concurrent

5. The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then, the sum of perpendicular distances from A and B on the tangent to the JEE Main 2019 circle at the origin is (a) 2 5

(b)

5 4

(c) 4 5

(d)

5 2

6. Slope of a line passing through P (2, 3) and intersecting the line, x + y = 7 at a distance of 4 units from P, is JEE Main 2019

3. Let two points be A(1, − 1 ) and B( 0, 2 ). If a point P ( x′ ,y ′ ) be such that the area of ∆PAB = 5 sq. units and it lies on the line, 3 x + y − 4 λ = 0, then a value of λ is JEE Main 2020

(a) 1 (c) 3

8. If the pairs of straight lines represented by 3 x 2 + 2hxy − 3y 2 = 0 and

JEE Main 2019

2. For a > 0, let the curves C1 : y 2 = ax and C 2 : x 2 = ay intersect at origin O and a point P. Let the line x = b ( 0 < b < a ) intersect the chord OP and the x-axis at points Q and R, respectively. If the line x = b bisects the area bounded by the curves, C1 and C 2, and the area of 1 ∆OQR = , then ‘a’ satisfies the 2 JEE Main 2020 equation (a) x 6 (b) x 6 (c) x 6 (d) x 6

4. Consider the set of all lines px + qy + r = 0 such that 3 p + 2q + 4 r = 0. Which one of the following statements is true?

(b) − 3 (d) 4

(a)

2

(b)

3

(c) 2 2

(d) 3 3

7. The larger of two angles made with the X -axis of a straight line drawn through (1, 2) so that it intersectce the line x + y = 4 at a point distance 6 / 3 from BIT SAT 2019 the point (1, 2) is (a) 60° (c) 105°

(b) 75° (d) None of these

3 x 2 + 2hxy − 3y 2 + 2 x − 4y + c = 0 form a square, then (h, c ) = TS EAMCET 2019

(a) ( 4 , − 1) (c) ( −4 , 1)

(b) ( −1, 4 ) (d) (1, − 4 )

9. If the angle between the pair of lines x 2 + 2 2 xy + ky 2 = 0, k > 0 is 45°, then the area (in square units) of the triangle formed by the pair of bisectors of angles between the given lines and the line x + 2y + 1 = 0 is TS EAMCET 2019 1 3 2 (c) 3 (a)

(b) 1 (d) 2

10. The joint equation of pair of straight lines passing through origin and having  1  slopes (1 + 2 ) and   is …… 1 + 2 MHT CET 2019

(a) x 2 − 2 2 xy + y 2 = 0 (b) x 2 − 2 2 xy − y 2 = 0 (c) x 2 + 2 xy − y 2 = 0 (d) x 2 + 2 xy + y 2 = 0

6

Engineering Entrance Questions 2019-20

11. A variable line passes through a fixed point ( x1, y1 ) and meets the axes at A and B. If the rectangle OAPB be completed, the locus of P is, WBJEE 2019 (O being the origin of the system of axes). (a) ( y − y1 ) 2 = 4 ( x − x1 )

(b)

x1 y + 1 =1 x y

(c) x 2 + y 2 = x12 + y12

(d)

x2 y2 + 2 =1 2 2 x1 y1

Circle 1. Let the tangents drawn from the origin to the circle, x 2 + y 2 − 8 x − 4y + 16 = 0 touch it at the points A and B. The ( AB ) 2 is equal to JEE Main 2020 (a)

56 5

(b)

52 5

(c)

64 5

(d)

32 5

2. A circle touches the Y -axis at the point (0, 4) and passes through the point (2, 0). Which of the following lines is not a tangent to this circle? JEE Main 2020 (a) 4 x − 3 y + 17 = 0 (b) 3 x + 4 y − 6 = 0 (c) 4 x + 3 y − 8 = 0 (d) 3 x − 4 y − 24 = 0

3. Three circles of radii a, b, c(a < b < c ) touch each other externally. If they have X -axis as a common tangent, then JEE Main 2019

(a) a , b , c are in AP 1 1 1 (b) = + a b c a , b , c are in AP 1 1 1 (d) = + b a c

(c)

4. If the circles x 2 + y 2 − 16 x − 20y + 164 = r 2 and ( x − 4 ) 2 + (y − 7 ) 2 = 36 intersect at two distinct points, then JEE Main 2019 (a) 0 < r < 1 (c) 1 < r < 11

(b) r > 11 (d) r = 11

5. A square is inscribed in the circle x 2 + y 2 − 6 x + 8y − 103 = 0 with its sides parallel to the coordinate axes. Then, the distance of the vertex of this square which is nearest to the origin is JEE Main 2019

(a) 6 (c) 41

(b) 13 (d) 137

6. The area (in sq units) of the smaller of the two circles that touch the parabola, y 2 = 4 x at the point (1, 2) and the X -axis JEE Main 2019 is (a) 8 π(3 − 2 2 ) (c) 8 π( 2 − 2 )

(b) 4 π(3 + 2 ) (d) 4 π( 2 − 2 )

7. A line y = mx + 1 intersects the circle ( x − 3 ) 2 + (y + 2 ) 2 = 25 at the points P and Q. If the mid-point of the line 3 segment PQ has x-coordinate − , then 5 which one of the following options is JEE Advance 2019 correct? (a) 6 ≤ m < 8 (c) 4 ≤ m < 6

(b) − 3 ≤ m < − 1 (d) 2 ≤ m < 4

8. The equation of the circle whose radius is 3 and which touches internally the circle x 2 + y 2 − 4 x − 6y − 12 = 0 at the point ( −1, − 1 ) is AP EAMCET 2019

(a) 5 x 2 + 5 y 2 + 9 x − 6 y − 7 = 0 (b) 5 x 2 + 5 y 2 − 8 x − 14 y − 32 = 0 (c) 5 x 2 + 5 y 2 − 6 x + 8 y − 8 = 0 (d) 5 x 2 + 5 y 2 + 6 x − 8 y − 12 = 0

9. If the number of common tangents to the pair of circles x 2 + y 2 − 2 x + 4y − 4 = 0 and x 2 + y 2 + 4 x − 4y + α = 0 is 4, then the least integral value of α is TS EAMCET 2019

(a) 4

(b) 5

(c) 6

(d) 7

10. The intercept on the line y = x by the circle x 2 + y 2 − 2 x = 0 is AB. The equation of the circle with AB as a diameter is ……… MHT CET 2019 (a) x 2 + y 2 + x + y = 0 (b) x 2 + y 2 − x − y = 0 (c) x 2 + y 2 − 3 x + y = 0 (d) x 2 + y 2 + 3 x − y = 0

11. A variable circle passes through the fixed point A( p, q ) and touches X -axis. The locus of the other end of the WB JEE 2019 diameter through A is (a) ( x − p ) 2 = 4qy

(b) ( x − q ) 2 = 4 py

(c) ( y − p ) = 4qx

(d) ( y − q ) 2 = 4 px

2

Parabola 1. If y = mx + 4 is a tangent to both the parabolas, y 2 = 4 x and x 2 = 2by , then b JEE Main 2020 is equal to (a) −32

(b) −128 (c) −64

(d) 128

2. If one end of a focal chord AB of the  1 parabola y 2 = 8 x is at A  , − 2 , then  2 the equation of the tangent to it at B is JEE Main 2020

(a) x − 2 y + 8 = 0 (b) x + 2 y + 8 = 0 (c) 2 x + y − 24 = 0 (d) 2 x − y − 24 = 0

3. If the parabolas y 2 = 4b( x − c ) and y 2 = 8ax have a common normal, then which one of the following is a valid choice for the ordered triad (a, b, c ) ? JEE Main 2019

1 (a)  , 2 , 0  2  (c) (1, 1, 3)

(b) (1, 1, 0) 1 (d)  , 2 , 3  2 

4. Equation of a common tangent to the parabola y 2 = 4 x and the hyperbola JEE Main 2019 xy = 2 is (a) x + 2 y + 4 = 0 (b) x − 2 y + 4 = 0 (c) 4 x + 2 y + 1 = 0 (d) x + y + 1 = 0

5. If the area of the triangle whose one vertex is at the vertex of the parabola, y 2 + 4( x − a 2 ) = 0 and the other two vertices are the points of intersection of the parabola and Y -axis, is 250 sq units, then a value of ‘a’ is JEE Main 2019 (a) 5 5 (c) 5 ( 21 / 3 )

(b) 5 (d) (10 ) 2 / 3

6. Suppose a parabola y = ax 2 + bx + c has two x intercepts, one positive and one negative, and its vertex is (2, − 2 ), then which of the following is true? JEE Main 2019

(a) ab > 0 (c) ac > 0

(b) bc > 0 (d) a + b + c > 0

7. Variable straight lines y = mx + c make intercepts on the curve y 2 − 4ax = 0 which subtend a right angle at the origin. Then the point of concurrence of these lines y = mx + c is APEAMCET 2019 (a) ( 4a , 0 ) (c) ( −4a , 0 )

(b) ( 2a , 0 ) (d) ( −2a , 0 )

8. If the normal at one end of the latusrectum of the parabola y 2 = 16 x meets the X -axis at the point P, then the length of the chord passing through P and perpendicular to the normal is TS EAMCET 2019

(a) 48 (b) 32 (c) 24 (d) 20

2 2 2 2

7

Engineering Entrance Questions 2019-20

Ellipse & Hyperbola 1. If 3 x + 4y = 12 2 is a tangent to the x2 y 2 ellipse 2 + = 1 for some a ∈ R, then 9 a the distance between the foci of the ellipse is JEE Main 2020 (a) 2 7

(b) 4

(c) 2 2

(d) 2 5

2. Let the line y = mx and the ellipse 2 x 2 + y 2 = 1 intersect at a point P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes 1   at  − , 0 and ( 0, β), then β is equal  3 2  to JEE Main 2020 2 2 2 (a) (b) 3 3

2 (c) 3

2 (d) 3

3. If a hyperbola passes through the point P(10, 16 ) and it has vertices at ( ± 6, 0 ), then the equation of the normal to it at JEE Main 2020 P is (a) 3 x + 4 y = 94 (c) 2 x + 5 y = 100

(b) x + 2 y = 42 (d) x + 3 y = 58

π 4. Let 0 < θ < . If the eccentricity of the 2 x2 y2 hyperbola − = 1 is greater cos2 θ sin 2 θ than 2, then the length of its latus rectum lies in the interval JEE Main 2019 3 (a) (1, ] 2 3 (c) ( , 2] 2

(b) (3, ∞) (d) (2, 3]

(b) ( 4 3 , 2 2 ) (d) ( 4 3 , 2 3 )

7. If tangents are drawn to the ellipse x 2 + 2y 2 = 2 at all points on the ellipse other than its four vertices, then the mid-points of the tangents intercepted between the coordinate axes lie on the curve JEE Main 2019 x2 y2 + =1 4 2 x2 y2 (c) + =1 2 4 (a)

1 1 + =1 4x 2 2y 2 1 1 (d) + =1 2x 2 4y 2

(b)

8. In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at ( 0, 5 3 ), then the length of its JEE Main 2019 latusrectum is (a) 5 (c) 8

(b) 10 (d) 6

9. If the tangent to the parabola y 2 = x at a point (α , β ), (β > 0 ) is also a tangent to the ellipse, x 2 + 2y 2 = 1, then α is equal to JEE Main 2019 (a) 2 + 1 (c) 2 2 + 1

(b) 2 − 1 (d) 2 2 − 1

10. If a directrix of a hyperbola centred at the origin and passing through the point ( 4, − 2 3 ) is 5 x = 4 5 and its JEE Main 2019 eccentricity is e, then (a) 4e − 12e − 27 = 0 4

  y x 5. Let S = ( x, y ) ∈ R 2 : − = 1 , + r − r 1 1   where r ≠ ± 1. Then, S represents 2

(a) ( 4 2 , 2 3 ) (c) ( 4 2 , 2 2 )

2

JEE Main 2019

(a) a hyperbola whose eccentricity is 2 , when 0 < r < 1. 1−r (b) a hyperbola whose eccentricity is 2 , when 0 < r < 1. r +1 (c) an ellipse whose eccentricity is 2 , when r > 1. r +1 (d) an ellipse whose eccentricity is

2

(b) 4e 4 − 24e 2 + 27 = 0 (c) 4e 4 + 8e 2 − 35 = 0 (d) 4e 4 − 24e 2 + 35 = 0

11. The tangent and normal to the ellipse 3 x 2 + 5y 2 = 32 at the point P(2, 2 ) meets the X-axis at Q and R, respectively. Then, the area (in sq units) of the ∆PQR JEE Main 2019 is (a)

16 3

(b)

14 3

(c)

34 15

(d)

68 15

12. The locus of the foot of perpendicular drawn from the centre of the ellipse x 2 + 3y 2 = 6 on any tangent to it is 1 , r +1

when r > 1.

6. Let the length of the latusrectum of an ellipse with its major axis along X-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it? JEE Main 2019

BIT SAT 2019

(a) ( x 2 − y 2 ) 2 = 6 x 2 + 2 y 2 (b) ( x 2 − y 2 ) 2 = 6 x 2 − 2 y 2 (c) ( x 2 + y 2 ) 2 = 6 x 2 + 2 y 2 (d) ( x 2 + y 2 ) 2 = 6 x 2 − 2 y 2

13. If the line joining the points A(α ) and x2 y 2 B(β ) on the ellipse + = 1 is a focal 25 9

chord, then one possible value of α β ⋅ cot is AP EAMCET 2019 2 2

cot

(a) −3

(b) 3

(c) −9

(d) 9

14. The major and minor axes of an ellipse are along the X -axis and Y -axis respectively. If its latusrectum is of length 4 and the distance between the foci is 4 2, then the equation of that ellipse is AP EAMCET 2019 (a) 2 x 2 + y 2 = 16 (c)

x2 y2 + =1 2 3

(b) x 2 + 2 y 2 = 16 (d)

x2 y2 + =1 3 2

15. If x + y + n = 0, n > 0 is a normal to the ellipse x 2 + 3y 2 = 3 and x + my + 3 = 0, m < 0 is a tangent to the ellipse x 2 + 5y 2 = 5, then the point of intersection of these two lines satisfy the equation TS EAMCET 2019 x2 y2 − =1 64 25 2 (c) x 2 = y + 1 3 (a)

(b) x − 5 y + 5 = 0 (d) y 2 = − 25 x + 3

16. The normal drawn at the point π π   9 cos , 7 sin  to the ellipse  4 4 x2 x2 + = 1 intersects its major axis at 9 7 TS EAMCET 2019 the point  2 (a)  0 ,  7    2 (c)  0 , −  7 

 2  (b)  − , 0 9    2  (d)  , 0  9 

17. The equation of the directrices of the hyperbola 3 x 2 − 3y 2 − 18 x + 12y + 2 = 0 is WB JEE 2019

(a) x = 3 ± (c) x = 6 ±

13 6 13 3

6 13 3 (d) x = 6 ± 13 (b) x = 3 ±

18. If the lengths of the transverse axis and the latusrectum of a hyperbola are 6 8 and respectively, then the equation of 3 MHT CET 2019 the hyperbola is …… (a) 4 x 2 − 9 y 2 = 72 (b) 4 x 2 − 9 y 2 = 36 (c) 9 x 2 − 4 y 2 = 72 (d) 9 x 2 − 4 y 2 = 36

19. S and T are the foci of an ellipse and B is the end point of the minor axis. If STB is equilateral triangle, the eccentricity of the ellipse is WB JEE 2019 (a)

1 4

(b)

1 3

(c)

1 2

(d)

2 3

8

Engineering Entrance Questions 2019-20

Introduction to Three Dimensional (3D) Geometry 1. A(2, 3, 5 ), B(α , 3, 3 ) and C(7, 5, β ) are the vertices of a triangle. If the median through A is equally inclined with the α co-ordinate axes, then cos−1   =  β AP EAMCET 2019

(a) cos (c)

−1     9

−1 

π 3

π (b) 2

2. A(3, 2, − 1 ), B( 4, 1, 1 ), C(6, 2, 5 ) are three points. If D, E , F are three points which divide BC, CA, AB respectively in the same ratio 2 : 1, then the centroid of TS EAMCET 2019 ∆DEF is 13 5 5 (a)  , ,   3 3 3

(b) (13, 5, 5)

(c) (4, 2, 1)

11 4 1 (d)  , ,   3 3 3

2 (d) cos −1   5

3. If (1, 0, 3), (2, 1, 5), (−2, 3, 6) are the mid-points of the sides of a triangle, then the centroid of the triangle is TS EAMCET 2019

1 4 14 (a)  , , −  3 3 3 1 4 14   (c)  , − ,  3 3 3

1 4 14 (b)  , ,  3 3 3  1 4 14 (d)  − , ,   3 3 3

4. If P (6, 10, 10 ), Q(1, 0, − 5 ), R(6, − 10, λ ) are vertices of a triangle right angled at Q, then value of λ is…… TS EAMCET 2019 (a) 0

(b) 1

(c) 3

(d) 2

Introduction to Limits and Derivatives 3 x + 3 3 − x − 12 is equal to ……… . x→ 2 3 − x / 2 − 31 − x

1. lim

JEE Main 2020

2. Let y = y ( x ) be a function of x

x→ 0

(b) 2 (d) 4

(a) 4 2 (c) 2 2

1 − x2 = k − x 1 − y 2 1  1 where k is a constant and y   = − .  2 4 dy 1 at x = , is equal to Then dx JEE Main 2020 2

5. If f (1 ) = 1, f ′ (1 ) = 3, then the derivative of f ( f ( f ( x ))) + ( f ( x )) 2 at x = 1 is

5 2 2 (c) 5

6. The value of 4 sin 2 x cos x − cos x + sin x is lim 3π sin x + cos x x→

satisfying y

5 2 5 (d) − 4

(b) −

(a)

3. lim

y→ 0

y4

JEE Main 2019

(a) exists and equals

(a) 12 (c) 15

1

(d) exists and equals

1

β = lim

x→ 0

2 2 1

(c)

to x is …… 2 (a) x log x 1 (c) x log x 2

(b) 0 (d) None of these

x ⋅ 2x − x and 1 − cos x x ⋅ 2x − x

1 + x2 − 1 − x2

10. , then

AP EAMCET 2019

2 2 ( 2 + 1)

8 3

64 (b) 27 7 (d) 3

9. Derivative of loge 2 (log x ) with respect

BIT SAT 2019

x→ 0

(b) does not exist (c) exists and equals

(a) 0

4

7. If α = lim

4 2

TS EAMCET 2019

(b) 9 (d) 33

(a) − 1 (c) 1

(b) 2α = β (d) α = 3 β

8. If [ x ] is the greatest integer function,  [ x ]3  x  3  = then lim  −  3   x→ 2 +  3

JEE Main 2019

equal to

1 + 1 +y4 − 2

(a) α = β (c) α = 2β

sin 2 x equals 2 − 1 + cos x JEE Main 2019

4. lim

AP EAMCET 2019

1 x log x 2 (d) log x (b)

lim (e x + x )1/x

x → 0+

WB JEE 2019

(a) Does not exist finitely (b) is 1 (c) is e 2 (d) is 2

Mathematical Reasoning 1. The logical statement ( p ⇒ q ) ∧ (q ⇒ ~ p ) is equivalent to JEE Main 2020

(a) ~ p (c) p

(b) q (d) −q

2. Which one of the following is a JEE Main 2020 tautology? (a) ( P ∧ ( P → Q )) → Q (b) P ∧ ( P ∨ Q ) (c) P ∨ ( P ∧Q ) (d) Q → ( P ∧ ( P → Q )

3. The logical statement [~ (~ p ∨ q ) ∨ ( p ∧ r )] ∧ (~ q ∧ r ) JEE Main 2019 is equivalent to (a) ~ p ∨ r (c) ( p ∧ r ) ∧ ~ q

(b) ( p ∧ ~ q ) ∨ r (d) (~ p ∧ ~ q ) ∧ r

4. Consider the following three statements: P : 5 is a prime number. Q : 7 is a factor of 192. R : LCM of 5 and 7 is 35.

Then, the truth value of which one of the following statements is true ? JEE Main 2019

(a) ( P ∧ Q ) ∨ (~ R ) (b) P ∨ (~ Q ∧ R ) (c) (~ P ) ∨ (Q ∧ R ) (d) (~ P ) ∧ (~ Q ∧ R )

5. If q is false and p ∧ q ←→ r is true, then which one of the following statements JEE Main 2019 is a tautology? (a) p ∨r (c) ( p ∨r ) → ( p ∧r )

(b) ( p ∧r ) → ( p ∨r ) (d) p ∧r

9

Engineering Entrance Questions 2019-20 6. Contrapositive of the statement “If two numbers are not equal, then their squares are not equal” is JEE Main 2019 (a) If the squares of two numbers are not equal, then the numbers are not equal. (b) If the squares of two numbers are equal, then the numbers are equal. (c) If the squares of two numbers are not equal, then the numbers are equal. (d) If the squares of two numbers are equal, then the numbers are not equal.

7. The statement pattern ( p ∧ q ) ∧ [~ r ∨ ( p ∧ q )] ∨ (~ p ∧ q ) is MHT SET 2019 equivalent to ……… (a) r (c) p ∧ q

(b) q (d) p

8. If p and q are true and r and s are false statements, then which of the following is true? MHT CET 2019 (a) (q ∧ r ) ∨ (~ p ∧ s ) (b) (~ p → q ) ↔ (r ∧ s )

(c) ( p → q ) ∨ (r ↔ s ) (d) ( p ∧ ~ r ) ∧ (~ q ∨ s )

9. Let a : ~ ( p ∧ ~ r ) ∨ (~ q ∨ s ) and b : ( p ∨ s ) ↔ (q ∧ r ). If the truth values of p and q are true and that of r and s are false, then the truth values of a and MHT CET 2019 b are respectively…… (a) F , F (c) T , F

(b) T , T (d) F , T

(a) 3.0 (c) 2.5

(b) 2.8 (d) 3.2

Statistics 1. The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by p and then reduced by q,where p ≠ 0 and q ≠ 0. If the new mean and new s.d. become half of their original values, then q is equal JEE Main 2020 to (a) 10

(b) − 10

(c) − 5

(d) − 20

2. Let the observations xi (1 ≤ i ≤ 10 ) 10

satisfy the equations, ∑ ( xi − 5 ) = 10 i =1

10

and ∑ ( xi − 5 ) = 40. If µ and λ are the 2

i =1

mean and the variance of the observations, x1 − 3, x 2 − 3,....., x10 − 3, then the ordered pair (µ, λ ) is equal to JEE Main 2020

(a) (6, 3) (b) (3, 6) (c) (3, 3) (d) (6, 6)

3. 5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm,

joined them. The variance (in cm2) of the height of these six students is JEE Main 2019

(a) 16

(b) 22

(c) 20

(d) 18

4. In a group of data, there are n observations, x, x 2, ...., xn . If n

n

i =1

i =1

Σ ( xi + 1 ) 2 = 9n and Σ ( xi − 1 ) 2 = 5n, the

standard deviation of the data is JEE Main 2019

(a) 2 (c) 5

(b) 7 (d) 5

5. The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is JEE Main 2019

(a) 4 : 9 (b) 6 : 7

(c) 10 : 3 (d) 5 : 8

6. If for some x ∈ R, the frequency distribution of the marks obtained by 20 students in a test is Then, the mean of the marks is

7. In a data with 15 number of observations x1, x 2, x 3,…, x15, 15

15

i =1

i =1

∑ xi2 = 3600 and ∑ xi = 175. If the value of one observation 20 was found wrong and was replaced by its correct value 40, then the corrected variance of TS EAMCET 2019 that data is (a) 151

(b) 149

(c) 145

(d) 144

8. For a group of 100 observations, the arithmetic mean and standard deviation are 8 and 10.5 respectively. The mean and standard deviation of 50 items selected from these 100 observations are 10 and 2 respectively. Then the standard deviation of the remaining 50 TS EAMCET 2019 observation is (a) 2 (c) 3.5

(b) 3 (d) 4

JEE Main 2019

Fundamentals of Probability 1. Let A and B be two events such that the probability that exactly one of them 2 occurs is and the probability that A 5 1 or B occurs is , then the probability of 2 both of them occur together is JEE Main 2020

(a) 0.10 (b) 0.20

(c) 0.01

(d) 0.02

2. An unbiased coin is tossed. If the outcome is a head, then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail, then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, …, 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is JEE Main 2019

15 72 19 (c) 72

13 36 19 (d) 36

5. A bag contains 6 white and 4 black balls. Two balls are drawn at random. The probability that they are of the same colour is …… MHT CET 2019

(b)

(a)

3. If there of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is JEE Main 2019 (a)

1 10

(b)

1 5

(c)

3 10

(d)

3 20

4. If two unbiased dice are rolled simultaneously until a sum of the number appeared on these dice is either 7 or 11, then the probability that 7 comes before 11, is TS EAMCET 2019 3 8 5 (c) 6

(a)

3 4 2 (d) 9

(b)

5 7 7 (c) 15

(a)

1 7 1 (d) 15

(b)

6. A problem in mathematics is given to 4 students whose chances of solving 1 1 1 1 individually are , , and . The 2 3 4 5 probability that the problem will be solved at least by one student is WB JEE 2019

2 3 4 (c) 5

(a)

3 5 3 (d) 4 (b)

Solutions with Explanation Sets, Fundamentals of Relations and Function 1. (b) The sum of series 49125 + 49124 + … + 49 2 + 49 + 1 = 49 0 + 491 + 49 2 + … + 49124 + 49125 =

1( 49126 − 1) ( 49 63 ) 2 − 1 2 = 49 − 1 48

( 49 63 − 1)( 49 63 + 1) 48 Q49 63 + 1 is a factor of the above sum. =

∴The greatest positive integer k, for which 49k + 1 is a factor of the sum 49126 + 49125 + … + 49 2 + 49 + 1 is 63. 2. (b) It is given that 21 + x + 21 − x , f ( x ) and3 x + 3 − x are in A.P. So, 2 f ( x ) = 2 ( 2 x + 2 − x ) + 3 x + 3 − x  3x + 3−x  ⇒ f (x ) = (2 x + 2 −x ) +   2   According to AM-GM, −x x a +a ≥ (a x a − x )1 / 2 2 ⇒a x + a − x ≥ 2, at x = 0 2 ∴ Minimum value of f ( x ) = 2 +    2 = 2 + 1=3 Hence, option (b) is correct. 3. (d) The given function f : (1, 3 ) → R, x[ x] defined by f ( x ) = 1+ x2 x (1) , x ∈ (1, 2 ) 1+ x2 = x (2) , x ∈ [ 2, 3) 1+ x2

Q

x , x ∈ (1, 2 ) 1+ x2 f (x ) = 2x , x ∈ [ 2, 3) 1+ x2

is a decreasing function, so 2 (3 ) 3 lim f (3 − h ) → = h→ 0 1+ 9 5 4 f (2) = 5 2 and Similarly, f ( x → 2 − ) tends to 5 1 + f ( x → 1 ) tends to . 2 So range of the given function ‘f ’ is  2 , 1 ∪ 3 , 4     5 2   5 5  Hence, option (d) is correct.

4. (a) Given sets A = { x ∈ R : | x | < 2 }, and B = { x ∈ R : | x − 2 | ≥ 3 } then, A = { x ∈ R : − 2 < x < 2 }and B = { x ∈ R : ( x − 2 ) ∈ ( −∞ , − 3] ∪ [3 , ∞ )} {x ∈ R : x ∈ ( −∞ , − 1] ∪ [5 , ∞ )} ∴B − A = R − ( −2 , 5 ) Hence, option (a) is correct . 5. (a) According to given information, we have if k ∈ {4, 8, 12, 16, 20} Then, f (k ) ∈ {3, 6, 9, 12, 15, 18} [QCodomain ( f ) = {1, 2, 3, …, 20}] Now, we need to assign the value of f (k ) for k ∈{4, 8, 12, 16, 20} this can be done in 6 C 5 ⋅ 5 ! ways = 6 ⋅ 5 ! = 6 ! and remaining 15 element can be associated by 15! ways. ∴ Total number of onto functions = 15 6 = 15 ! 6 ! 6. (d) Let the population of city is 100. Then, n( A) = 25, n( B ) = 20 and n( A ∩ B ) = 8 A

B

U

12

n(U)=100

1 ⇒ y ∈  − , 1  11  Hence, option (c) is correct. 8. (c) We have, x − [ x] f (x ) = log( x 2 − x ) It is defined log( x 2 − x ) > 0 [Qx − [ x ] = [ x ] ≥ 0 , ∀ x ∈ R] ⇒ x 2 − x > 1⇒ x 2 − x − 1> 0 1± 5 Q x 2 − x −1 = 0 ⇒ x = 2 1 − 5 1 + 5  , ∴ Domain of f ( x ) = R −  2   2 9. (b) We have, 4−x2 f ( x ) is defined if f (x ) = [ x] + 2 4−x2 x2 −4 ≤0 ≥ 0⇒ [ x] + 2 [ x] + 2 ⇒ ( x + 2 ) ( x − 2 ) ([ x ] + 2 ) ≤ 0 By wavy curve method –2

17

8

Venn diagram

So, n( A ∩ B ) = 17 and n( A ∩ B ) = 12 According to the question, Percentage of the population who look into advertisement is 30 40 = × n( A ∩ B )  +  × n( A ∩ B )   100   100  50 +  × n( A ∩ B )   100  = 5.1 + 4.8 + 4 = 13.9 x 7. (c) Let 2 =y x − 5x + 9

–

+

–

–1

2

∴ x ∈ ( −∞ , − 2 ) ∪ [ −1, 2] 10. (b) We have, A = { x | x ∈ N , x | is a prime number less than 12} = { 2 , 3 , 5 , 7 , 11} and B = { x | x ∈ N , x is a factor of 10 } = {1, 2 , 5 , 10 } ∴ A ∩ B = {2, 5} 11. (b) We have, Y

X′

x8+y 8=1

O

⇒ yx 2 − (5 y + 1) x + 9 y = 0

X x2+y 2=1

Q x ∈R ⇒ D ≥ 0 ⇒ (5 y + 1) 2 − 36 y 2 ≥ 0 ⇒ 25 y 2 + 10 y + 1 − 36 y 2 ≥ 0

+

Y′

⇒ 11y 2 − 11y + y − 1 ≤ 0

From the graph, it is clear that there is common region which satisfies x 2 + y 2 = 1 and x 8 + y 8 < 1

⇒ (11y + 1)( y − 1) ≤ 0

Q

⇒ 11y 2 − 10 y − 1 ≤ 0

P ∩T = P

11

Engineering Entrance Questions 2019-20

Sequence and Series 1 (b) Let five numbers, which are in A.P. is a − 2d , a − d , a, a + d , a + 2d . According to given information, 5a = 25 ⇒a = 5 and a (a 2 − d 2 )(a 2 − 4d 2 ) = 2520 ⇒

5 ( 25 − d 2 )( 25 − 4d 2 ) = 2520

⇒

(d 2 − 25 )( 4d 2 − 25 ) = 504

⇒

4d 4 − 125d 2 + 625 = 504

⇒

4d − 125d

⇒ ⇒

4

4d − 4d 4

2

2

− 121d

 

+ 121 = 0 2

4d (d

⇒

2

+ 121 = 0 2

ar 2 (1 + r ) = 16

…(ii)

From Eqs. (i) and (ii), we get r 2 = 4 ⇒ r = ±2

9

∑ ai

 

1 /4   

4 (d) Let r be the common ratio of given GP, then we have the following sequence a1 , a 2 = a1r , a 3 = a1r 2 , ... , a10 = a1r 9 Now, a 3 = 25 a1 ⇒ r 2 = 25 Consider,

a 9 a1 r 8 = = r 4 = ( 25 ) 2 = 5 4 a 5 a1 r 4

5 (b) Given, three distinct numbers a , b and c are in GP. …(i) ∴ b 2 = ac and the given quadratic equations ax 2 + 2bx + c = 0

…(ii)

dx 2 + 2ex + f = 0

…(iii)

For quadratic Eq. (ii), the discriminant D = ( 2b ) 2 − 4ac = 4 (b 2 − ac ) = 0

[from Eq. (i)]

⇒ Quadratic Eq. (ii) have equal roots, and it b is equal to x = − , and it is given that a quadratic Eqs. (ii) and (iii) have a common root, so b d  −   a

…(iii)

From Eqs. (i) and (iii), we get 4 a = > 0 (rejected), if r = 2 3 and a = − 4 < 0, if r = − 2 Now,

1 /4

2  1 − 1 / 2  = 2  1 / 2  = 21 / 2

− 1) − 121(d − 1) = 0 121 11 ⇒ d = ± 1, ± ⇒ d 2 = 1, 4 2 If d = 1, then terms are 3, 4, 5, 6, 7, and if d = − 1, then terms are 7, 6, 5, 4, 3, and 21 11 1 if d = , then terms are −6, − , 5, , 16. 2 2 2 11 When a = 5 and d = ± , then one of these 2 1 five numbers of A.P. is − and the greatest 2 number amongst them is 16. 2 (a) Let first term and common ratio of given G.P. a1 , a 2 , a 3 , … are a1 = a < 0 and ‘r’ respectively. Now, a1 + a 2 = 4 (given) ⇒ a + ar = 4 ⇒ a (1 + r ) = 4 …(i) (given) a 3 + a 4 = 16 ⇒ ar 2 + ar 3 = 16 2

 1 + 1 + 1 + 1 + K to ∞ 

 = 2  4 8 16 32 1 1 1 1 Q + + + + K to ∞ is a G.P. of 4 8 16 32 1 infinite terms having first term and 4 1 common ratio , So, the product is 2

⇒ ⇒

2

b + 2e  −  + f = 0  a

d (ac ) − 2eab + a f = 0 [Qb

i =1

−4 (( −2 ) 9 − 1) 4 = − (512 + 1) −2 − 1 3 4 (given) = − (513 ) = 4λ 3 ⇒ λ = − 171. 3 (c) Given product =

2

= ac ]

1 24

1 ⋅ 416

1 ⋅ 8 48

1 ⋅ 16128 ⋅… to

1 24

2 ⋅ 216

3 ⋅ 2 48

4 ⋅ 2128 ⋅… to

∞

∞

[Qa ≠ 0] dc − 2eb + af = 0 e d f  [Qb 2 = ac ] 2  = + ⇒ b  a c d e f so, , , are in AP. a b c 6 (c) Given, f ( x + y ) = f ( x ) ⋅ f ( y ) Let f ( x ) = λx Q f (1) = 2 ∴ λ=2

[where λ > 0] (given)

10  10  Σ f (a + k ) = Σ λa + k = λa  Σ λk  k =1  k =1  a 1 2 3 10 = 2 [ 2 + 2 + 2 + ...... +2 ] 10

So,

k =1

Hence, option (a) is correct. 8 (d) We have, Tr = r [n − (r − 1)] = r (n − r + 1) = nr − r 2 + r ∴ Sn = =

n

n

r =1

r =1

∑ Tr = ∑ (nr − r n

∑ [(n + 1)r − r

2

2

+ r)

]

n

n

r =1

r =1

= (n + 1) ∑ r − ∑ r 2

⇒

= a1 + a 2 + a 3 + ...... + a 9

7 (a) Given, series 3 21 117 + + + …+ upto n terms 5 25 125 8  4 2 =  1 −  +  1 −  +  1 −  + …+  125  25   5  upto n terms 2 3 2 2 2 = n −  +   +   + 5 5 5 … + upto n terms] n 2  2    1−    5   5  =n− 2 1− 5 n 2  2  1 −    5  5 =n−  3 /5 2n + 1 2 =n+ − 3 × 5n 3

r =1

db 2 − 2eba + a 2 f = 0 2

⇒ ⇒

 2 ( 210 − 1)  = 2a    2 −1  2a + 1 ( 210 − 1) = 16 ( 210 − 1) (given) 2a + 1 = 16 = 2 4 ⇒a + 1 = 4 ⇒a = 3

(n + 1)n(n + 1) n(n + 1)( 2n + 1) = − 2 6 n(n + 1)(n + 2 ) = 6 1 ∴ α= 6 4n − 3n 9 (d) We have, S n = 3n 4 −3 1 For n = 1, S 1 = t1 = = 3 3 2 2 4 −3 7 For n = 2 , S 2 = t1 + t 2 = = 9 32 7 7 1 7 −3 4 ∴ = t 2 = − t1 = − = 9 9 3 9 9

12

Engineering Entrance Questions 2019-20

Complex Numbers 1 (d) For , ( x − 1) + iy z −1 = 2z + i 2 x + ( 2 y + 1)i [( x − 1) + iy][ 2 x − i ( 2 y + 1)] = ( 2 x ) 2 + ( 2 y + 1) 2 (on rationalization)  z − 1  2 x ( x − 1) + y ( 2 y + 1) ∴Re  = 4 x 2 + ( 2 y + 1) 2  2z + i  Now, it is given that  z −1 Re   =1  2z + i  ⇒

⇒ 2 x 2 + 2 y 2 + 2 x + 3 y + 1 = 0, is a circle 1 3 whose centre is  − , −  and radius is  2 4 5 1 9 1 5 , so diameter is . + − = 4 4 16 2 2 2 (c) It is given that roots of quadratic equation x 2 + bx + 45 = 0, b ∈ R has conjugate complex roots, now let roots are α + iβ and α − iβ . …(i) ∴Sum of roots = 2α = − b and product of roots = α 2 + β 2 = 45 …(ii)

⇒

(α + 1) 2 + β 2 = 40

⇒

α + 2α + 1 + β = 40 2

⇒

x2

=

x 2 + (y + 3) 2

=

1 6 +  − + 3   2 

 = 1+ 

24 + 25 25 = 4 4 49 7 = = 4 2 Hence, option (b) is correct. 4 (a) Given, x 2 + x + 1 = 0 x =

ω2 =

−1− 3 i 2

[Qα = − 3]

Hence, option (c) is correct. 3 (b) It is given for complex number z, z −i … (i) =1 z + 2i 5 … (ii) 2 From Eq. (i), | z − i | = | z + 2i | ⇒ Locus of z is a straight line and it is perpendicular bisector of line joining points 1 ( 0 , − 2 ) and (0, 1), so locus of z is y = − , 2 where z = x + iy.

9

9

9 1 π π 3i   = +  =  cos + i sin   2  3 3 2

7 (c) The complex number z satisfying | z − 2 + i| ≥ 5, which represents the region outside the circle (including the circumference) having centre ( 2 , − 1) and radius 5 units.

−1+

3i

2

and

Y

are the cube roots of unity z0 (x,y)

and ω, ω 2 ≠ 1)

(1,0)

X¢

¾ O Ö5

3

x + iy  1 =  − 2 − i   27 3  3 –1 =  (6 + i )  3  x + iy 1 =− ( 216 + 108i + 18i 2 + i 3 ) ⇒ 27 27 1 (198 + 107 i ) =− 27 On equating real and imaginary part, we get x = − 198 and y = − 107 3  1 6 (c) Given, z = +  i  2 2

5 (a) We have,

π

i π π + i sin = e 6 6 6 so, (1 + iz + z 5 + iz 8 ) 9

X (2,-1)

(Qω3n = (ω 2 ) 3n = 1)

= 3 + 3i = 3 (1 + i ) If ‘θ’ is the argument of z, then Im(z ) [Qz is in the first quadrant] tan θ = Re (z ) 3 π = = 1 ⇒ θ= 3 4

= cos

 cos 11 π + i sin 11 π      6 6  

= cos 3 π + i sin 3 π [Qfor any natural number ‘n’ (cos θ + i sin θ)n = cos(nθ) + i sin(nθ)] = − 1

− 1 ± 3i 2

⇒ z 0 = ω, ω 2 [where ω =

9

 1 i 3 3 1 3 i = 1 − + − + i+ −  2 2 2 2 2 2 

2

6+

⇒

π  i  Qi = e 2   

 cos 2 π + i sin 2 π     3 3 

5π 5π +  cos + i sin  +  6 6 

(Qx 2 = 6 and y = − 1 / 2) =

9

2π 5π 11 π   i i i = 1 + e 3 + e 6 + e 6     

Now, | z + 3i | = | x + (y + 3) i |

= 3 + 6i − 3i

and b 2 − b = 36 − 6 = 30

|z | =

1 4 =6

5 = 2 2 5  =    2 25 = 4

Now consider z = 3 + 6i z 081 − 3i z 093

From relation (ii), on putting the value of α 2 + β 2 , we get

and

2

x2 +

2

45 + 2α + 1 = 40 ⇒ 2α = − 6 ⇒α = − 3 From relation (i), we get −b = 2 ( −3 ) ⇒ b =6 ∴ b 2 + b = 36 + 6 = 42

− 1 x 2 +    2 

2

⇒

QRoots α ± iβ satisfy | z + 1| = 2 10, so

+ y

2

x

2 x ( x − 1) + y ( 2 y + 1) =1 4 x 2 + ( 2 y + 1) 2

⇒ 2x 2 − 2x + 2y 2 + y 2 = 4x + 4y 2 + 4y + 1

5π 8π  π  i i i =  1 + ie 6 + e 6 + ie 6     

From Eq. (ii),

Y¢

1 is maximum. Now, for z 0 ∈ S | z 0 − 1| When | z 0 − 1| is minimum. And for this it is required that z 0 ∈ S , such that z 0 is collinear with the points ( 2 , − 1) and (1, 0 ) and lies on the circumference of the circle | z − 2 + i| = 5. So let z 0 = x + iy, and from the figure 0 < x < 1 and y > 0. 4 − x − iy − x + iy 4 −z0 −z0 So, = z 0 − z 0 + 2i x + iy − x + iy + 2i 2−x 2(2 − x ) = −i  =   y + 1 2i ( y + 1) 2−x is a positive real number, so Q y+1 4 −z0 −z0 is purely negative imaginary z 0 − z 0 + 2i number.  4 −z0 −z0  π ⇒ arg   =− 2  z 0 − z 0 + 2i 

9

13

Engineering Entrance Questions 2019-20 z = x + iy

8 (a) Let

∴

⇒

y arg z = tan −1   x π arg (z ) = 6 π y tan –1 = x 6 π 1 y = tan = 6 3 x x = 3y

Also given, | z − 2 3i| = λ | x + iy − 2 3 i| = λ x 2 + ( y − 2 3 ) 2 = λ2 This is the equation of circle whose centre is (0, 2 3) and r = λ. Now, x – 3 y = 0 touches the circle ∴

∴

λ=

0 − 3 (2 3 ) 1+ 3

λ=

6 =3 2

9 (a) For z = x + iy, x , y ∈ R, (x , y ) ≠ (0, − 4) 2z − 3 ( 2 x − 3 ) + 2iy x − i ( y + 4 ) × = z + 4i x + i (y + 4) x − i (y + 4) =

( 2 x − 3 x + 2 y + 8 y ) + i (12 + 3 y − 8 x ) x 2 + (y + 4) 2 2

2

 2z − 3  arg    z + 4i 

So,

 π  12 + 3 y − 8 x = tan −1  2  = 2  2x − 3x + 2y + 8y  4 (given) 12 + 3 y − 8 x ⇒ 1 = 2x 2 − 3x + 2y 2 + 8y ⇒

2 x 2 + 2 y 2 + 5 x + 5 y − 12 = 0

Hence, option (a) is correct. 10 (d) We have, 4

 1 + iz    =P  1 − iz  4

⇒

λ =3

i − z    =P i + z 

4

⇒ ⇒

z − i   = | P| z + i | z − i| 4 = | z + i| 4

[Q| P| = 1]

⇒ | z − i| = | z + i| ∴z lies on perpendicular bisector of i and −i. ∴z lies on y-axis. ∴z has all complex roots. 11 (b) We have, (cos θ + i sin θ) (cos 2 θ + i sin 2 θ) ... (cos nθ + i sin nθ) = 1 ⇒ e iθ ⋅ e i ( 2 θ ) ⋅ e i ( 3 θ ) ⋅ ... e i (nθ ) = 1 i n (n + 1 ) θ

⇒

=1

2

e

n(n + 1)  n(n + 1)  θ + i sin  cos  ⇒     2 2  θ = 1 + 0i n(n + 1)  cos  θ = 1 ⇒   2 n(n + 1) ⇒ θ = 2kπ 2 4k ⇒ θ= π n(n + 1)

Inequalities and Quadratic Equation 1 (b) Given equation (k + 1) tan 2 x − 2 λ tan x = 1 − k, w there k ≠ −1 and λ ∈ R having roots α and β, so tan α + tan β =

2λ k+1

and tan α ⋅ tan β =

k −1 k+1

2λ tan α + tan β k+1 Qtan(α + β ) = = 1 − tan α tan β 1 − k − 1 k+1 2λ λ = 2 2

⇒

tan(α + β ) =

Q

tan (α + β ) = 50

⇒

⇒

e x = y > 0, y 4 + y 3 − 4y 2 + y + 1 = 0  2 1   1 y + 2  + y +  − 4 = 0 y  y   (on dividing by y 2 )

⇒

⇒

t 2 + 3t − 2t − 6 = 0

⇒

t (t + 3 ) − 2 (t + 3 ) = 0

⇒

(t − 2 ) (t + 3 ) = 0

⇒ ∴ ⇒ ⇒

λ 2 = 100

Let

 1 y +   y

2

 1 + y +  − 6 = 0  y

1 = t , then y t2 + t − 6 = 0

2

⇒ λ = ± 10 2 (c) Given equation is e 4 x + e 3 x − 4e 2 x + e x + 1 = 0 so

Again, let y +

t = − 3, 2 1 y + = − 3 or 2 y 1 = − 3 or 2 ex 1 ex + x = 2 e Qe x > 0 ⇒ e x + 1 ≠ − 3 ∀x ∈ R    ex ex +

⇒

(e x − 1) 2 = 0

⇒

ex = 1

⇒ x =0 ∴Number of real roots of given equation is 1. Hence, option (c) is correct. 3 (d) It is given that for a , b ∈ R, a ≠ 0, the quadratic equation ax 2 − 2bx + 5 = 0 has a repeated root α. b …(i) So, α = and D = 0 a ⇒ 4b 2 − 20a = 0

⇒

…(ii) b 2 = 5a 5 and …(iii) α2 = a and α, β are roots of another quadratic equation x 2 − 2bx − 10 = 0. So, α + β = 2b, αβ = − 10 and α 2 − 2bx − 10 = 0

⇒ ⇒ ⇒

b2 b2 b −2 − 10 = 0 Qα =  2  a a a 5a 2 (5a ) − − 10 = 0 (Qb 2 = 5a ) a a2 5 − 10 − 10 = 0 a

⇒

20a = 5 1 a= 4 5 2 b = 4

⇒ So,

5 b2 α = 2 = 4 = 20 ∴ 1 a 16 Since, αβ = − 10 ⇒ α 2β 2 = 100 2

⇒ ∴

β 2 =5 α 2 + β 2 = 20 + 5 = 25

Hence, option (d) is correct.

14

Engineering Entrance Questions 2019-20

4 (d) Given quadratic equation is x 2 + (3 − λ ) x + 2 = λ

6 (a) Given quadratic equation is π x 2 + x sin θ − 2 sin θ = 0, θ ∈  0 ,   2 and its roots are α and β. So, sum of roots = α + β = − sin θ and product of roots = αβ = − 2sin θ …(i) ⇒ αβ = 2(α + β ) Now, the given expression is α12 + β12 −12 (α + β −12 )(α − β) 24

x 2 + (3 − λ ) x + ( 2 − λ ) = 0 … (i) Let Eq. (i) has roots α and β, then α + β = λ − 3 and αβ = 2 − λ b [QForax 2 + bx + c = 0, sum of roots = − a c and product of roots = ] a Now, α 2 + β 2 = (α + β ) 2 − 2αβ = (λ − 3) 2 − 2(2 − λ )

12

  αβ   αβ = =  2 2  (α + β ) − 4αβ   (α − β ) 

= λ2 − 6 λ + 9 − 4 + 2 λ = λ 2 − 4 λ + 5 = ( λ 2 − 4 λ + 4 ) +1

12

  2 (α + β) =  [from Eq. (i)] 2 8 ( α β ) ( α β) + − +  

= (λ − 2)2 + 1 Clearly, a 2 + β 2 will be least when λ = 2.

  2 =   (α + β ) − 8 

5 (d) Given equation is | x − 2| + x ( x − 4) + 2 = 0 ⇒

⇒

⇒

  2 =   − sinθ − 8 

12

212 = (sin θ + 8 )12

⇒sin x = |sin y |

[from the options]

8 (c) x 2 − 2 x + A = 0 ⇒

x =1±

∴

p = 1 − 1 − A, q = 1 +

1− A 1− A

Similarly, r = 9 − 81 − B , s = 9 + Q ⇒

81 − B

p , q , r , s are in AP. q − p = s −r 2 1 − A = 2 81 − B

⇒ and ⇒

B = 80 + A q − p =r −q 3 1 − A = 8 − 81 − B

⇒ A = − 3, B = 77 9 (d) Let f ( x ) = ax 2 + bx + c ⇒ f (1) = a + b + c < 0 Again, f ( x ) has imaginary zeros. So, a < 0. Also, f ( 0 ) = c . Since f ( x ) is downward parabola. So, c < 0. Y

7 (b) Given, inequality is

y + 2y − y − 2 = 0 2

⇒ y ( y + 2 ) − 1( y + 2 ) = 0 ⇒ ( y + 2 )( y − 1) = 0 ⇒ y = 1, − 2 [Qy = | x − 2| ≥ 0] ∴ y =1 ⇒

12

[Q α + β = − sin θ]

(| x − 2 | ) 2 + | x − 2 | − 2 = 0

Let | x − 2 | = y, then above equation reduced to y2 + y − 2 = 0

12

⇒sin x = 1 and sin 2 y = 1

| x − 2| = 1 x = 3 or 1

⇒ x = 9 or 1 ∴ Sum of roots = 9 + 1 = 10

2

sin 2 x − 2 sin x + 5

⋅

1

≤1

sin 2 y

4 ⇒ ⇒

2

(sin x − 1 ) 2 + 4

⋅ 2 −2 sin

2

y

X′

X

≤1

f (x)

(sin x − 1) 2 + 4 ≤ 2 sin 2 y Y′

QRange of (sin x − 1) 2 + 4 is [ 2 , 2 2 ] and range of 2 sin 2 y is [ 0 , 2].

10 (a) Given, lim

x → 0+

∴The above inequality holds, if

= lim

(sin x − 1) 2 + 4 = 2 = 2 sin 2 y

x → 0+

x q  p  x 

[q] x  q  q   [q] − 0=  −   = p p  x  x  p

Permutation and Combination 1 (a) It is given that the 10th place of 5 digit number with distinct digits is ‘2’, then the ten thousand place, thousand place, hundred place and unit place we can fill in 8, 8, 7 and 6 ways only. 2

8 8 7 1 6

So, required number of five digit numbers is (given) 8 × 8 × 7 × 6 = 336 k 8 ×8 ×7 × 6 k= ⇒ 336 8 ×8 ×7 ⇒ =8 = 56 Hence, option (a) is correct. 2 (c) Clearly, the two digit number which leaves remainder 2 when divided by 7 is of the form N = 7 k + 2 [by Division Algorithm]

For, k = 2, N = 16 k = 3, N = 23 M M k = 13, N = 93 ∴ 12 such numbers are possible and these numbers forms an AP. 12 Now, S = [16 + 93] = 654 2 QS = n (a + l )   n    2 Similarly, the two digit number which leaves remainder 5 when divided by 7 is of the form N = 7k + 5 For k = 1, N = 12 k = 2, N = 19 M k = 13, N = 96 ∴ 13 such numbers are possible and these numbers also forms an AP.

S′ =

Now,

13 [12 + 96] = 702 2

QS = n (a + l )   n    2 Total sum = S + S ′ = 654 + 702 = 1356 3 (a) Given digits are 1, 1, 2, 2, 2, 2, 3, 4, 4. According to the question, odd numbers 1, 1, 3 should occur at even places only. 3y=x+7 (–8, β)

(6, β)

(–8, 5)

(6, 5)

∴ The number of ways to arrange odd 3! numbers at even places are 4 C 3 × 2!

15

Engineering Entrance Questions 2019-20 and the number of ways to arrange 6! . remaining even numbers are 4!2! So, total number of 9-digit numbers, that can be formed using the given digits are 3! 6! 4 = 4 × 3 × 15 = 180 C3 × × 2! 4!2! 4 (b) Following are the cases in which the 4-digit numbers strictly greater than 4321 can be formed using digits 0, 1, 2, 3, 4, 5 (repetition of digits is allowed) Case-I 4

3

2 4 numbers

2/3/4/5

Case-II 4

3 3/4/5 0/1/2/3/4/5 3×6=18 numbers 3 ways

6 ways

Case-III 4 4/5

0/1/2/3/4/5 2×6×6=72 numbers

2 ways

6 ways

Case-IV 5

6×6×6=216 numbers 0/1/2/3/4/5 6 ways

So, required total numbers = 4 + 18 + 72 + 216 = 310

5 (c) It is given that, there are 20 pillars of the same height have been erected along the boundary of a circular stadium. Now, the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then total number of beams = number of diagonals of 20-sided polygon. Q 20 C 2 is selection of any two vertices of 20-sided polygon which included the sides as well. So, required number of total beams = 20C 2 − 20 [Qthe number of diagonals in a n-sided closed polygon = nC 2 − n] 20 × 19 = − 20 = 190 − 20 = 170 2 6 (c) Since only one letters can not be in wrong envelope so that at least two letters are in wrong envelope means all the letters are not in right envelopes. ∴ x = 6 ! − 1 = 720 − 1 = 719 Y = number of ways, so that all the letters are in wrong envelops 1 1 1 1 1 1 = 6 !  1 − + − + − +   1! 2 ! 3 ! 4 ! 5 ! 6 !  = 360 − 120 + 30 − 6 + 1 = 265 ∴ x − y = 719 − 265 = 454 7 (d) If number is divisible by 6, then it should be an even number and divisible by 3 also, so sum of the digits should be divisible by 3. Now last digit can be fill by 0 or 2 or 4. First place can be filled by 5 choices (except 0) second and third can be filled by 6 choices.

Now fourth place is filled such that it must be of form 3k or 3 k + 1 or 3k + 2 . So it can be fill by 2 choices (0 or 3) as whole number must be divisible by 3 so, total numbers = 5 × 6 × 6 × 2 × 3 = 1080 Hence, option (d) is correct. 8 (c) The permutation of 4 letters such that vowel and consonants are equal 4! 21 C 1 × 5C 1 × + 21C 2 × 5C 2 × 4 ! 2! 2! = 51030 = 729 × 70 = 3 6 × 70 9 (c) Case I 4 questions from first 5 questions and 6 questions from remaining 8 questions 5 ∴ C 4 × 8C 6 = 5 × 28 = 140 Case II Choose 5 questions from first 5 questions and next 5 questions from remaining 8 questions 5 Q C 5 × 8C 5 = 1 × 56 = 56 Total number = 140 + 56 = 196 10 (b) Required number of ways = 7C 4 × D (3 ) 1 1 1   = 7C 3 × 3 ! 1 − + −    1! 2 ! 3 !    Q nC r = nC n − r , and    D (n) = n!1 − 1 + 1 − 1 + K       1! 2 ! 3 ! 7 7 = C3 × 2 = 2( C3 )

Mathematical Induction 1 (a) Given statement is “P (n) : n 2 − n + 41 is prime”. Clearly P(3) : 3 2 − 3 + 41 = 9 − 3 + 41 = 47 which is a prime number. and P (5 ) : 5 2 − 5 + 41 = 25 − 5 + 41 = 61, which is also a prime number. ∴ Both P (3 ) and P (5 ) are true. 2 (c) Since, 15 = 3 × 5, so to find the exponent of 15 in 47! It is enough to find the exponent of 5 in 47 !. So, exponent of 5 in 47 !.

47 47 =   +  2 + …,  5   5  {where, [ x ] is greatest integer of x]. =9 + 1 So, required value of k is 10. Hence, option (c) is correct. 3 (a) Given, 1 2 + 2 2 + 3 2 +…… + n 2 > x n(n + 1) ( 2n + 1) >x ⇒ 6

n(n + 1) ( 2n + 1) n(n)( 2n) > 6 6 2n3 n3 x = ∴ ⇒x = 6 3 4 (d) We have, 7 2n + 16n − 1 Q

Put n = 1, we get 7 2 + 16 − 1 = 49 + 15 = 64 which is divisible by 64.

Binomial Theorem 1 (30) Let (1 + x + x 2 + … + x 2n )(1 − x + x 2 − x 3 + … + x 2n ) = a 0 + a1 x + a 2 x

2

+ … + a 4n x 4n

On putting x = 1, we get (1 + 1 + 1 + … + 1)(1 − 1 + 1 − 1 + … + 1) ( 2n + 1) terms ( 2n + 1) terms

= a 0 + a1 + a 2 + … + a 4n ⇒ ( 2n + 1) = a 0 + a1 + a 2 + … + a 4n …(i) Similarly, on putting x = − 1, we get (1 − 1 + 1 − 1 + … + 1)(1 + 1 + 1 + … + 1) ( 2n + 1) terms ( 2n + 1) terms = a 0 − a1 + a 2 − a 3 + … + a 4n ⇒ ( 2n + 1)

…(ii) = a 0 − a1 + a 2 − a 3 … + a 4n On adding Eqs. (i) and (ii), we get 2 (a 0 + a 2 + a 4 + … + a 4n ) = 2 ( 2n + 1) ⇒a 0 + a 2 + a 4 + … + a 4n = 2n + 1 ⇒Sum of coefficients of all even powers of x in the product (1 + x + x 2 + … + x 2n )

16

Engineering Entrance Questions 2019-20

= 2n + 1 = 61 ⇒ 2n = 60 ⇒ n = 30 2 (d) 2 403 = 8 ⋅ 2 400 = 8 (16 )100 Note that, when 16 is divided by 15, gives remainder 1. ∴ When (16 )100 is divided by 15, gives remainder 1100 = 1 100

and when 8 (16 ) is divided by 15, gives remainder 8.  2 403  8 ∴ = .   15  15 (where {⋅} is the fractional part function) ⇒ k =8 3 (b) The (r + 1)th term in the expansion of (a + x )n is given by Tr + 1 = nC r a n − r x r ∴3 rd term in the expansion of (1 + x log 2 x ) 5 is 5 C 2 (1) 5 − 2 ( x log 2 x ) 2 ⇒ 5 C 2 (1) 5 − 2 ( x log 2 x ) 2 = 2560 (given) ⇒ x ( 2 log 2 x ) = 256 ⇒

2 (log 2 x )(log 2 x ) = 8 (Qlog 2 256 = log 2 2 8 = 8)

(log 2 x ) = 4 2

⇒ log 2 x = ± 2 ⇒ x = 4 or x = 2 −2 =

1 4

4 (d) The general term in the expansion of binomial expression (a + b )n is Tr + 1 = n C r a n −r b r , so the general term in the expansion of 10 λ binomial expression x 2  x + 2  is  x  r  λ  Tr + 1 = x 2  10 C r ( x )10 −r  2       x  =10 C r λr x

2 +

10 −r − 2r 2

Now, for the coefficient of x 2 , 10 − r put 2 + − 2r = 2 2 ⇒ 10 − r = 4r ⇒ r = 2 So, the coefficient of x 2 is 10 C 2 λ 2 = 720 [given] 10 ! 2 λ = 720 2! 8!

⇒

20

∑ (3r + 2 ) ⋅

20

[Q General term of the sequence 2, 5, 8, …, which forms an AP, is 2 + (n − 1)3 = 3n − 1, where n = 1, 2 , 3 ... and it can be written as 3n + 2 , where n = 0 , 1, 2 , 3] 20

∑r

r =0

20

20

Cr

r =0

Q nC = n n − 1C r r r  20

20

= 3 × 20 ∑ 19 C r

−1

r =1

+ 2∑

20

20

Cr + 2 ∑

20

Cr

r =0

 

−1 

Cr

r =0

19

20

r =0

r =0

= 60 ∑ 19C r + 2 ∑

20

Cr

 20 19 Q ∑ C r  r = 1

−1

=

19

∑

19

r =0

 Cr  

  n = ( 60 × 219 ) + ( 2 × 2 20 ) Q ∑ nC r = 2n    r = 0 = (15 × 2 21 ) + 2 21 = 16 × 2 21 = 2 25 6 (d) We have, (1 + ax + bx 2 )(1 − 2 x )18 18

18

C 1 2 x + 18 C 2 ( 2 x 2 )

C 3 ( 2 x ) 3 + 18 C 4 ( 2 x ) 4 ...... )

Coefficient of x 3 is − 18C 3 ( 2 ) 3 + a ⋅18 C 2 ( 2 ) 2 − b 18C 1 ( 2 ) and coefficient of x 4 is 18 C 4 ( 2 ) 4 −18 C 3 ( 2 ) 3 a +

Cr

r =0

= 3⋅

+2∑

−1

(1 + ax + bx 2 )(1 −

⇒ λ=±4 5 (b) Given series is 2 ⋅ 20C 0 + 5 ⋅ 20C 1 + 8 ⋅ 20C 2 + … + 62 ⋅ 20C 20 =

20

20

20 = 3 ∑ r   19C r r  r =1

18

C 2 (2) 2 b

Coefficient of x 3 and x 4 are zero. ∴ 18 C 3 ( 2 ) 3 + 18 C 2 ( 2 ) 2 (a ) −18 C 1 ( 2 )b = 0 32 …(ii) and 80 − a + b = 0 3 Sovling Eqs. (i) and (ii), we get 272 a = 16 , b = 3

Trigonometric Functions and Equations 1. (b) Given trigonometric expression

π 3π π 3π cos   cos   + sin 3   sin 8 8  8  8 π π π π  3  π 3  π cos   cos  −  + sin   sin  −  8 2 8 2 8 8 Q π − π = 3 π  8   2 8 π π   3  π 3  π = cos   sin   + sin   cos   8 8 8 8 3

π π = sin   cos   8 8 π π   cos 2   + sin 2    8  8   π π 1 = × 2 sin   cos   8 8 2 1 π 1 = sin = 2 4 2 2 Hence, option (b) is correct.

2. (b) It is given that ∞

x = Σ ( −1)n tan 2n θ n =`0

= 1 − tan 2 θ + tan 4 θ − tan 6 θ + .... + upto ∞ 1 π {Q θ ∈  0 ,  =  4 1 + tan 2 θ ⇒

x = cos θ 2

…(i)

∞

y = Σ cos 2n θ n =0 2

= 1 + cos θ + cos 4 θ + cos 6 θ + K =

1 π {Q θ ∈  0 ,   4 1 − cos 2 θ

+ upto ∞

1 ⇒ cos 2 θ ∈  0 ,  }  2 1 1 = ⇒sin 2 θ = y sin 2 θ On adding Eqs. (i) and (ii), we get 1 1 = x + ⇒ y (1 − x ) = 1 y Hence, option (b) is correct.

= 3 (sin θ − cos θ) 4 + 6 (sin θ + cos θ) 2 + 4 sin 6 θ = 3 ((sin θ − cos θ) 2 ) 2

tan 2 θ ∈ ( 0 , 1) }

and

3. (d) Given expression

…(ii)

+6 (sin θ + cos θ) 2 + 4 (sin 2 θ) 3 = 3 (1 − sin 2 θ) 2 + 6 (1 + sin 2 θ) + 4 (1 − cos 2 θ) 3 = 3 (1 2 + sin 2 2 θ − 2 sin 2 θ) + 6 (1 + sin 2 θ) + 4 (1 − cos 6 θ − 3 cos 2 θ + 3 cos 4 θ) 2 = 3 + 3 sin 2 θ − 6 sin 2 θ + 6 + 6 sin 2 θ + 4 − 4 cos 6 θ − 12 cos 2 θ + 12 cos 4 θ 2 6 = 13 + 3 sin 2 θ − 4 cos θ − 12 cos 2 θ + 12 cos 4 θ = 13 + 3 ( 2 sin θ cos θ) 2 − 4 cos 6 θ − 12 cos 2 θ(1 − cos 2 θ) = 13 − 4 cos 6 θ

17

Engineering Entrance Questions 2019-20 3 4

4. (c) Given, sin 2 2 θ + cos 4 2 θ = ⇒

3 4 (Qsin 2 x = 1 − cos 2 x )

(1 − cos 2 2 θ) + cos 4 2 θ =

⇒

4 cos 4 2 θ − 4 cos 2 2 θ + 1 = 0

⇒

( 2 cos 2 2 θ − 1) 2 = 0

⇒

cos 2 θ = ±

1 2

π If θ ∈  0 ,  , then 2 θ ∈ ( 0 , π )  2 1 cos 2 θ = ± ∴ 2 π 3π π 3π ⇒ 2θ = , ⇒ θ= , 4 4 8 8 π 3π π Sum of values of θ = + = 8 8 2

5. (b) Given, x 2 sin θ − x sin θ cos θ − x + cos θ = 0, where 0 < θ < 45 ° ⇒ x sin θ( x − cos θ) − 1( x − cos θ) = 0 ⇒ ( x − cos θ) ( x sin θ − 1) = 0 ⇒ α = cos θ and β = cosecθ 2 < cosecθ < ∞ ⇒ cos θ < cos ecθ) Now, consider, ∞  ∞ ( −1)n  n n ∑  α + βn  = ∑ α + n=0 n=0 1 1 =  1 1−α 1 − −   β 1 1 = + 1 − cos θ 1 + sin θ =

1 + 1−α

∞

(−1)n n n=0 β

∑

+

1 1 1+ β

  1 Q = sin θ   β

6.(b) Given equation is 1 + sin 4 x = cos 2 (3 x ) Since, range of (1 + sin 4 x ) = [1, 2] and range of cos (3 x ) = [ 0 , 1] 2

So, the given equation holds if 1 + sin 4 x = 1 = cos 2 (3 x ) ⇒

sin 4 x = 0 and cos 2 3 x = 1

5π 5π Since, x ∈  − ,  2 2  ∴ x = − 2 π, − π, 0, π, 2 π. Thus, there are five different values of x is possible.

7.(a) Given, | cos x | = 2[ x ]

∴

10. (d) We have, 3 cos 2 A + 2 cos 2 B = 4 ⇒ 3 (1 − sin 2 A) + 2 (1 − sin 2 B ) = 4

0 ≤ | cos x | ≤ 1 0 ≤ 2 [ x] ≤ 1 1 0 ≤ [ x] ≤ 2 x =0

∴

cos 15 ° + cos 5 ° sin 25 ° 2 + sin 50 ° sin 20 ° cos 15 ° 1 =− + [ 2 cos 45 ° cos 15 ° ] 2 2 cos 15 ° cos 15 ° =− + =0 2 2 =−

⇒ ⇒

5 − 3 sin 2 A − 2 sin 2 B = 4 3 sin 2 A + 2 sin 2 B = 1

⇒

| cos x | = 0 π x :n π + 2 For any value of n ∈ N , [n] ≠ 0 Hence, the equation has no solution. π π 8. (c) 0 ≤ α, β ≤ ; α + β ≤ 4 2 cos (α + β ) = 4 / 5 ⇒ tan (α + β ) = 3 / 4 sin (α − β ) = 5 / 13 ⇒α − β ∈ θ1 and tan (α − β ) = 5 / 12 tan 2α = tan (α + β + α − β ) tan(α + β ) + tan (α − β ) = 1 − tan (α + β ) tan (α − β ) 3 / 4 + 5 / 12 56 = = 5 33 1−3/4 × 12

9.(a) cos 2 5 ° − cos 2 15 ° − sin 2 15 ° + sin 2 35 ° + cos 15 ° sin 15 ° − cos 5 ° sin 35 ° = cos 5 ° (cos 5 ° − sin 35 ° )) − cos 15 ° (cos 15 ° − sin 15 ° + sin 2 35 ° − sin 2 15 ° = cos 5 ° ( 2 sin 30 ° sin 25 ° ) − cos 15 ° ( 2 sin 45 ° sin 30 ° ) + sin 50 ° sin 20 °

3 sin 2 A = cos 2 B 3 sin A 2 cos B Again, = sin B cos A ⇒ 3 sin A cos A = 2 sin B cos B ⇒ 3 sin 2 A = 2 sin 2 B Now, cos( A + 2 B ) = cos A cos 2 B − sin A sin 2 B = 3 cos A sin 2 A − 3 sin 2 A cos A = 0

∴A + 2 B = 90 °

11. (c) We have, In ∆ABC  tan A     2 ⇒ ⇒

⇒

⇒ ⇒ ⇒

 tan B  = 3    2 4

(s − b ) (s − c ) s (s − a )

(s − a ) (s − c ) 3 = 4 s (s − b )

(s − b ) (s − c ) (s − a ) (s − c ) 3 = s (s − a ) ⋅ s (s − b ) 4 a+b+c −c 3 (s − c ) 3 2 = = ⇒ a+b+c 4 s 4 2 a + b −c 3 = a+b+c 4 4a + 4b − 4c = 3a + 3b + 3c a + b = 7c

Properties of Triangles, Heights and Distances, Cartesian System of Rectangular Coordinates 1 (05) Given vertices of triangle ABC are 3 A(1, 0 ), B( 6 , 2 ) and C  , 6  and the 2  centroid of triangle G divides the triangle such that area of triangles AGB, BGC and CGA are equal, so point P is the centroid of triangle ‘G’. ∴Coordinate of 1 + 6 + 3 / 2 0 + 2 + 6  17 8  P  ,  = P ,   6 3   3 3 7 1 and given point Q  − , −  .  6 3

So length of the line segment PQ =

 17 + 7     6 6 2

2

8 1 +  +  3 3

9 +   3

2

2

=

 24     6

=

4 2 + 3 2 = 16 + 9

=

25 = 5

2 (b) The centroid of the triangle having vertices (3 , − 1), (1, 3) and (2, 4) is

 3 + 1 + 2 − 1 + 3 + 4 , C  = ( 2, 2)   3 3 And equation of line passes through point of intersection P of lines x + 3 y − 1 = 0 and 3 x − y + 1 = 0 is ( x + 3 y − 1) + λ (3 x − y + 1) = 0 … (i) QLine (i) passes through point C ( 2 , 2 ), so ( 2 + 6 − 1) + λ ( 6 − 2 + 1) = 0 7 ⇒ 7 + 5λ = 0 ⇒ λ = − 5

18

Engineering Entrance Questions 2019-20

∴ Equation of line passes through points C and P is 7 ( x + 3 y − 1) − (3 x − y + 1) = 0 5 ⇒5 x + 15 y − 5 − 21x + 7 y − 7 = 0 ⇒ 16 x − 22 y + 12 = 0 ⇒ 8 x − 11y + 6 = 0 … (ii) From the options, point ( − 9 , − 6 ) satisfy the line. Hence, option (b) is correct. 3 (a) According to given information, we have the following figure. E

c =7

A

D

b=6

30° B

a=5

C

1 Clearly, length of BD = 2a 2 + 2c 2 − b 2 , 2 (using Appollonius theorem) where, c = AB = 7 , a = BC = 5 and b = CA = 6 ∴ 1 2 × 25 + 2 × 49 − 36 BD = 2 1 1 112 = 4 7 = 2 7 = 2 2 Now, let ED = h be the height of the lamp post. E

h B

30°

D

Then, in ∆BDE , tan30° = ⇒ ⇒

h BD

h 1 = 3 2 7 2 7 2 21 h= = 3 3

2 x1 + 4 + 2 3 [from Eq. (i)] ⇒ k= 3 2x + 4 + 6 ⇒ 3k = 1 3 …(iii) ⇒ 9k − 10 = 2 x1 Now, from Eqs. (ii) and (iii), we get 2 (3h − 4 ) = 9k − 10 ⇒ 6h − 8 = 9k − 10 ⇒ 6h − 9k + 2 = 0 Now, replace h by x and k by y. ⇒ 6 x − 9 y + 2 = 0, which is the required 2 locus and slope of this line is 3 Q slope of ax + by + c = 0 is − a   b 

7 (b) We know that, a b c = = = 2 R and given that, sin A sin B sin C a + b = x , ab = y and x 2 − c 2 = y A

5 (d) Given equation of line is 3 x + 4 y − 24 = 0 For intersection with X-axis put y = 0 ⇒ x =8 For intersection with Y-axis, put x = 0 ⇒ 4 y − 24 = 0 ⇒ y = 6 ∴ A(8 , 0 ) and B ( 0 , 6 )

c

O

b

R B

C

a

B(0,6)

∴

(a + b ) 2 − c 2 = ab

a2 + b2 −c2 − ab 1 = =− 2ab 2ab 2 1 ∴cos C = − ⇒C = 120 ° 2 c Now, = 2R sin C 1 c c 2 ⇒ R= = 2 sin (120 ° ) 2 3 c R= ∴ 3 ⇒

O

A(8,0)

Let AB = c = 8 2 + 6 2 = 10 OB = a = 6 and OA = b = 8 Also, let incentre is (h k ), then 6 × 8 + 8 × 0 + 10 × 0 48 = =2 = 24 6 + 8 + 10 and =

6 × 0 + 8 × 6 + 10 × 0 48 = =2 24 6 + 8 + 10

∴ incentre is (2, 2) 6 (c) Let ABC be a given triangle with vertices B( 0 , 2 ), C ( 4 , 3 ) and let third vertex be A(a , b ) A (a, b)

4 (a) Let the coordinates of point P be ( x1 , y1 ) QP lies on the line 2 x − 3 y + 4 = 0 ∴ 2 x1 − 3 y1 + 4 = 0 2x + 4 …(i) y1 = 1 ⇒ 3 Now, let the centroid of ∆PQR be G (h , k ), then x + 1+ 3 h= 1 3 …(ii) ⇒ x1 = 3h − 4 y1 + 4 − 2 and k= 3

b −2 3−0 = −1 × a−0 4−0 …(ii) ⇒ 4a + 3b = 6 From Eqs. (i) and (ii), we get −b + 3b = 6 ⇒ 2b = 6 ⇒ b = 3 3 [from Eq. (i)] and a=− 4 So, the third vertex 3 (a , b ) ≡  − , 3  , which lies in II quadrant.  4  ⇒

8 (c) Given vertices of ∆AOP are O ( 0 , 0 ) and A( 0 , 1) Let the coordinates of point P are ( x , y ). Clearly, perimeter = OA + AP + OP = 4 (given) ⇒

( 0 − 0 ) 2 + ( 0 − 1) 2

+

( 0 − x ) 2 + (1 − y ) 2 +

⇒1 +

E F (0,0)

x2 + y2 = 4

x 2 + ( y − 1) 2 +

x2 + y2 = 4

⇒ x 2 + y 2 − 2y + 1 +

x2 + y2 =3

⇒ x 2 + y 2 − 2y + 1 = 3 − (0, 2) B

D

C (4,3)

Also, let D , E and F are the foot of perpendiculars drawn from A, B and C respectively. Then, b −0 3−2 = −1 × AD⊥ BC ⇒ a−0 4−0 …(i) ⇒ b + 4a = 0 and CF⊥ AB

⇒ ⇒ ⇒

x2 + y2

1 − 2y = 9 − 6 x 2 + y 2 6 x 2 + y 2 = 2y + 8

9(x 2 + y 2 ) = (y + 4) 2 + [squaring both sides] ⇒ 9 x 2 + 9 y 2 = y 2 + 8 y + 16 Thus, the locus of point P ( x , y ) is 9 x 2 + 8 y 2 − 8 y = 16

19

Engineering Entrance Questions 2019-20 9 (b) Given points are ( −8 , 5 ) and (6, 5) in which y-coordinate is same, i.e. these points lie on horizontal line y = 5. 3y=x+7

(s − a ) 2 (s − b ) 2 (s − c ) 2 s 2 + + + 2 ∆2 ∆2 ∆ ∆2 1 2 2 = 2 [ 4 s − 2 s (a + b + c ) + (a + b 2 + c 2 )] ∆ 4∆2  1 1 1  = 2  2 + 2 + 2 ∆  p1 p 2 p3 

5 27 + 40 48 = 11 = 45 / 60 12 Hence, option (c) is correct.

=

12 (d) We have, (6, β)

(–8, β)

Let ∴ (6, 5)

(–8, 5)

∴ Let (−8, β) and ( 6 , β ) are the coordinates of the other vertices of rectangle as shown in the figure. Since, the mid-point of line joining points ( −8 , 5 ) and ( 6 , β ) lies on the line 3 y = x + 7 . 5 + β  −8 + 6 ∴ 3  +7  =  2  2 ⇒ 15 + 3 β = − 2 + 14 ⇒ 3 β = − 3 ⇒β = − 1 Now, area of rectangle = | −8 − 6 | × | β − 5 | = 14 × 6 = 84

⇒

a :b :c = 2:3: 4 a = 2k , b = 3k , c = 4k abc ∆ and r = R= 4∆ s R s ⋅ abc = r 4∆ 2 R s ⋅ ( 2k )(3k )( 4k ) = 4 s ( s − 2k )( s − 3k )( s − 4k ) r

 1 1 1  =4 2 + 2 + 2 p 2 p3   p1 15 (*) We have, A(1, 2 ), B (3 , −1), C ( 4 , 0 ) Centroid of ∆ABC is A(1, 2)

R 6 ⋅ 23 ⋅ k 3 R 16 = = ⇒ ⇒ 5k ⋅ 3k ⋅ k 5 r r ∴ R : r = 16 : 5 13 (b) Let (h , k ) be circumcentre of ∆ABC and A( −2 , 1), B( 0 , − 2 ), C (1, 2 ) are the vertices of ∆ABC. A (–2, 1)

10 (d) Given heights of two poles are 5 m and 10 m. 5m

15°

O (h, k) B (0, –2)

E

C (1, 2)

d 5m 15°

d

C

D

F

i.e. from figure AC = 10 m, DE = 5 m ∴ AB = AC − DE = 10 − 5 = 5 m Let d be the distance between two poles. Clearly, ∆ABE ~ ∆ACF [by AA - similarity criterion] ∴ ∠AEB = 15 ° In ∆ABE, we have 3 −1 5 AB = tan15° = ⇒ 3 +1 d BE 5 ( 3 + 1) ( 3 − 1)

⇒

d=

⇒

d =5

3 +1 3 +1 × 3 −1 3 +1

2 × 5( 3 + 2) = 5(2 + 3 ) m 2 b +c c +a a+b 11 (c) Let = = =k 9 10 11 ⇒b + c = 9k , c + a = 10k and a + b = 11k and a + b + c = 15k ∴a = 6k , b = 5k and c = 4k cos A + cos B Q cos C b2 + c2 −a2 a2 + c2 −b2 + 2bc 2ac = a2 + b2 −c2 2ab =

∴

OA = OB = OC OA = OB (h + 2 ) 2 + (k − 1) 2 = h 2 + (k + 2 ) 2 ⇒

4h − 6k + 1 = 0 OB = OC h 2 + (k + 2 ) 2 = (h − 1) 2 + (k − 2 ) 2

B(3, –1)

C(4, 0)

 1 + 3 + 4 , 2 − 1 + 0  =  8 , 1       3 3 3 3 Let (h , k ) be circumcentre of ∆ABC ∴ OA = OB = OC , OA = OB (h − 1) 2 + (k − 2 ) 2 = (h − 3 ) 2 + (k + 1) 2

A B 10 m

O(h, k)

…(i)

…(ii) = 2h + 8k − 1 = 0 On solving Eqs. (i) and (ii), we get 1 3 h = − ,k = 22 22 Equation of BC 2+ 2 y+ 2= (x ) 1− 0 4x − y − 2 = 0 Distance from circumcentre to line BC 4  −1  − 3 − 2     22 d =  22  2 2 4 +1     3 17 51 d= = 22 17 22 14 (d) We have, p1 , p 2 , p 3 are the altitudes of a triangle ABC from the vertices A, B , C respectively. 1 1 1 ∆ = ap1 = bp 2 = cp 3 ∴ 2 2 2 ∆ ∆ ∆ ∆ r1 = , r2 = , r3 = ,r = s −a s −b s −c s 1 1 1 1 ∴ 2 + 2 + 2 + 2 r1 r 2 r3 r

⇒ h 2 − 2h + 1 + k 2 − 4k + 4 = h 2 − 6h + 9 + k 2 + 2k + 1 ⇒ 4h − 6k = 5 …(i) Similarly, OA = OC Q (h − 1) 2 + (k − 2 ) 2 = (h − 4 ) 2 + k 2 ⇒ 6h − 4k = 11 …(ii) Solving Eqs. (i) and (ii), we get 23 7 h= , k= 10 10 Distance between circumcentre and centroid of ∆ABC is 2

 23 − 8  +  7 − 1       10 3   10 3 

2

=

11 2 30

16 (c) Equation of circle passing through

B (0, b) n Q

m P

and (a , 0 ) ( 0 , b ) x 2 + y 2 − ax − by = 0

A (a, 0)

(0,

0)

is

Equation of tangent at origin, − ax − by = 0 ⇒ ax + by = 0

20

Engineering Entrance Questions 2019-20 a2

PA = a

2

a

2

a2

m=

∴

and

+b

2

b

BQ = n =

… (i)

+ b2

and …(ii) tan A. tan B = 2 From Eqs. (i) and (ii) we get, tanC = 3 18 (b) Let ∆ABC be a right angled triangle at B. Let ∠A and ∠C be α and β A

… (ii)

ar2

a

a4 + b4 m2 + n2 = 2 a + b2 mn =

β

a b a2 + b2

(a 2 + b 2 ) 2 =a2 + b2 a2 + b2 ∴Diametre of circle (m + n) 2 =

= a2 + b2 =

ar

B

2

(m + n) 2 = m + n

17 (a) We have, tan A + tan B + tan C = 6 …(i) ⇒ tan A tan B tan C = 6

tanα =

=

α

2

a2 + b2

2

BC ar = ⇒r = AB a AB a 1 and tanβ = = = BC ar r

∴

C

Since, sides are in GP so sides are a, ar, ar 2 Now, AC 2 = AB 2 + BC 2 ⇒

(ar 2 ) 2 = a 2 + (a ⋅ r ) 2

⇒

a 2r 4 = a 2 + a 2r 2

⇒

r4 − r 2 − 1 = 0

⇒

r =

5 +1 2

1 5 +1 2

=

5 +1 2

5 −1 2

19 (d) Given, that angles of a triangles are 2x, 3x and 7 x. Since, 2 x + 3 x + 7 x = 180 ⇒ 12 x = 180 º ⇒ x = 15º So, angles are 30º, 45º and 105º a Now, = 2R sin 30 º a = 2 × 10 [Q R = 10 cm] ⇒ 1 2 ⇒ 2a = 20 ⇒ a = 10 cm

Straight Line and Pair of Straight Lines 1 (a) Let a point A( x1 , y1 ) on the line x = 2 y and the middle point of the perpendicular drawn from point A( x1 , y1 ) to the line x = y is P (h , k ). X

⇒ 5h = 7 k On taking the locus of point P (h , k ), we get 5 x − 7 y = 0.

C 2 : x 2 = ay, (a > 0 ), intersect each other A'

P (h, k) 2y=x A (x1, y1)

X′

X

at origin ‘O’ and a point P (a , a ). Q O ( 0 , 0 ), Q and P (a , a ) are collinear. So, Q (b , b ).

2 a 1 2 a 1 2a 2 a 2 = − − + − 3 3 3 3a 3 3a 4 a a2 2 ⇒ = + ⇒ 4a a = a 3 + 2 3 3 3a ⇒ 16a 3 = a 6 + 4 + 4a 3

2

x =ay x2=ax P (a, a)

Y′

Q (b, b) X′

O

X

R (b, 0)

(0 0]

x2 − y2 xy = 0 2 x2 − y2 = 0

and 3 x 2 + 2hxy − 3 y 2 + 2 x − 4 y + c = 0 …(ii) both represents pair of perpendicular lines. Now, the point of intersection the lines given by ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0  hf − bg gh − af  is  ,  , So for the pair of 2 2  ab − h ab − h  lines (ii), the point of intersection is  h ( −2 ) − ( −3 ) (1) h − 3 ( −2 )  A ,  − 9 − h2 − 9 − h2  

[By Eq. (iv)]

9 (a) We have, x 2 + 2 2 xy + ky 2

Area of triangle formed by line x 2 − y 2 = 0 and x + 2 y + 1 = 0 ∆=

| 1 0 − (1) ⋅ (1) | | 1( 2 ) 2 − ( 0 ) − (1) 2 |

∆=

| 0 + 1| | 4 − 1|

=

1 3

10 (a) Required equation of pair of straight lines passing through origin and having slopes, m1 = (1 + 2 ) and 1 = 2 −1 m2 = 1+ 2 ⇒

[ y − (1 +

2 ) x ] [ y − ( 2 − 1) x ] = 0

⇒ y − 2 2 xy + x 2 = 0 2

x y + =1 a b Since, the line passing through a fixed point ( x1 , y1 ).

11 (b) Let the equation of line be

B (0, b)

And the line (iv) passes through the mid-point ‘M’of line joining points 1 1 1 2 A , −  and O ( 0 , 0 ). So, M =  , −  . 5  10 5 5

5

2

7 (b) Let A = (1, 2 ) equation of line is x −1 y − 2 6 = =r cos θ sin θ 3

2 4 + +c =0 10 5 ⇒ c = −1 So, (h , c ) = ( 4 , − 1) Hence, option (a) is correct.

∴

⇒3h = 12 ⇒ h = 4  8 −3 2 10   1 So, pointA =  ,− = ,−  5  9 + 16 9 + 16   5

8

3

2

8 (a) Given, equations of pair of straight lines 3 x 2 + 2hxy − 3 y 2 = 0 …(i)

(1, 0) x+2y=1

1 2 ⋅ 0 + 1    2

|2 + 3 − 7| 2 = = 2 1+ 1

O

(a, b) P

(a, 0) A

x1 y + 1 =1 a b Since, OAPB is a rectangle, therefore the coordinate of P will be (a , b ). Hence, locus of P is x1 y + 1 =1 x y

∴

22

Engineering Entrance Questions 2019-20

Circle 1 (c) The equation of chord of contact AB to circle x 2 + y 2 − 8 x − 4 y + 16 = 0 w.r.t point origin ( 0 , 0 ) is T = 0 A 2 C (4, 2)

M

F (0, 0)

2 B 2

2

x +y –8x–4y+16=0

⇒ 0 x + 0 y − 4 ( x + 0 ) − 2 ( y + 0 ) + 16 = 0 …(i) ⇒ 2x + y − 8 = 0 In right angled triangle AMC, |2(4) + 2 − 8 | 2 = CM = 5 2 2 + 12

⇒ ⇒ and

4 y = − 3 x + 31 ± 25 3 x + 4 y − 56 = 0 3x + 4y − 6 = 0 4 Similarly, when m = − , equation of 3 tangent is 20 25  4 y = − x +  4 + ±   3 3  3

AM 2 = CA 2 − CM 2

⇒

2 4 16 = 2 2 −   = 4 − =  5 5 5 4 AM = 5 2 2 × 4 64 AB 2 = ( 2 AM ) 2 =   =  5  5

and ⇒ 4 x + 3 y − 57 = 0 4x + 3y − 7 = 0 ∴The line 4 x + 3 y − 8 = 0 is not a tangent to the circle. Hence, option (c) is correct. 3 (b) According to given information, we have the following figure.

2 (c) Since, equation of a circle which touches the Y-axis at point (0, 4) is x 2 + ( y − 4 ) 2 + λx = 0 … (i)

∴ Centre = (3 , − 4 ) and radius = 8 2 Now, according to given information, we have the following figure. Y C

D

B b

F

45°

AD =

(a + b ) 2 − (b − a ) 2

= 2 ab

⇒

∴In ∆BCF , BF =

Similarly, AC = a + c and CE = c − a

y − 4 = m( x − 5 ) ± 5 1 + m 2 y = mx + ( 4 − 5m ± 5 1 + m 2 )

… (iii) From the option, on taking slope of line, we 4 have when m = , equation of tangent by Eq. 3 (iii), we get  16  20 4 y = x + 4 − ± 5 1+  9  3 3  ⇒ 3 y = 4 x + (12 − 20 ± 25 ) ⇒ 4 x − 3 y − 8 ± 25 = 0 ⇒ 4 x − 3 y + 17 = 0 and 4 x − 3 y − 33 = 0 Similarly, when m = − 3 / 4, equation of tangent is  9  15 3 y = − x + 4 + ± 5 1+  16 4 4  

(a + c ) 2 − (c − a ) 2

= 2 ac Similarly, BC = b + c and CF = c − b (b + c ) 2 − (c − b ) 2 = 2 bc

Now, equation of tangent to the circle (ii) having slope ‘m’ is Q ∴ ⇒

AD + AE = BF 2 ab + 2 ac = 2 bc 1 1 1 + = c b a

For the coordinates of A and C. x −3 y + 4 Consider, = =±8 2 1 1 2 2 [using distance (parametric) form of line, x − x1 y − y1 = = r] cos θ sin θ ⇒ x = 3 ± 8, y = − 4 ± 8 ∴ A( − 5 , − 12 ) and C (11, 4 ) Similarly, for the coordinates of B and D, consider x −3 y + 4 = = ±8 2 1 1 − 2 2 [in this case, θ = 135 °] ⇒ x = 3 m 8, y = − 4 ± 8 ∴ B (11, − 12 ) and D ( − 5 , 4 ) Now,

4 (c) Circle I is x 2 + y 2 − 16 x − 20 y + 164 = r 2 ⇒

⇒ C 2 ( 4 , 7 ) is the centre of 2nd circle and r 2 = 6 is its radius. Two circles intersect if | r1 − r 2 | < C 1C 2 < r1 + r 2 (8 − 4 ) 2 + (10 − 7 ) 2 < r + 6

⇒ |r − 6| 4 ∴Least integral value of α = 5 10 (b) We have equation of line y = x and equation of circle x 2 + y 2 − 2 x = 0

O (0, 0)

X

(1, 0)

x2+y2–2x=0 Y′

Now, intersecting points of given line and circle, x 2 + x 2 − 2x = 0 ⇒ 2 x ( x − 1) = 0 ⇒ x = 0 , 1 when x = 0 then y = 0 and when x = 1 then y =1 ∴Coordinates of end points of diameter AB are (0, 0) and (1, 1). ∴Required equation of circle with diameter AB ( x − 0 )( x − 1) + ( y − 0 )( y − 1) = 0 ⇒ x2 − 2 + y2 − y = 0 ⇒

x2 + y2 − x − y = 0

11 (a) In a circle, AB is a diameter where the coordinate of A is ( p , q ) and let the coordinate of B ( x1 , y1 ). Equation of circle in diameter form is ( x − p ) ( x − x1 ) + ( y − q ) ( y − y1 ) = 0 ⇒ x 2 − ( p + x1 ) x + px1 + y 2 − ( y1 + q ) y + qy1 = 0 ⇒ x 2 − ( p + x1 ) x + y 2 − ( y1 + q ) y + px1 + qy1 = 0 Since, this circle touches X-axis ∴ y=0 ⇒ x 2 − ( p + x1 ) x + px1 + qy1 = 0 Also, the discriminant of above equation will be equal to zero because circle touches X-axis. ∴ ( p + x1 ) 2 = 4 ( px1 + qy1 ) ⇒ p 2 + x12 + 2 px1 = 4 px1 + 4qy1 ⇒ x12 − 2 px1 + p 2 = 4qy1 ⇒

( x1 − p ) 2 = 4qy1

Therefore, the locus of point B is ( x − p ) 2 = 4qy

Parabola 1 (b) As we know, equation of tangent to the parabola y 2 = 4 x , having slope ‘m’ is 1 …(i) y = mx + m On comparing the Eq. (i) with the equation of given tangent y = mx + 4, we get 1 1 = 4 ⇒m = 4 m

1 x + 4, 4 2 now it is tangent to the parabola x = 2by, so on solving the equation of parabola 1 x 2 = 2by and tangent y = x + 4, we must 4 get only a common point, so 1 x 2 = 2b  x + 4  4 

∴ Equation of the tangent is y =

⇒ 2 x 2 − bx − 16b = 0 is a quadratic equation having one solution. So, D = 0 ⇒ b 2 + 4 ( 2 )(16b ) = 0 [Qb cannot be zero] ⇒ b = − 128 2 (a) Equation of given parabola y 2 = 8 x and 1 one end of a focal chord AB is A , − 2  2 

24

Engineering Entrance Questions 2019-20

As, we know, if one end of a focal chord of parabola y 2 = 4ax is (at 2 , 2at ), then other end will be  a , −2a  , so other end point  2  t t 

1 x  mx +  = 2.  m ⇒

⇒ 8y = 4(x + 8) ⇒ x − 2y + 8 = 0 3 (c) Normal to parabola y 2 = 4ax is given by y = mx − 2am − am3

m 2 x 2 + x − 2m = 0 D = 0 ⇒1 = 4 (m 2 ) ( − 2m) 1 m3 =  −   2

3

1 2 ∴Required equation of tangent is x y = − − 2 ⇒ 2y = − x − 4 2 ⇒ x + 2y + 4 = 0 5 (b) Vertex of parabola y 2 = − 4 ( x − a 2 ) is (a 2 , 0 ). ⇒

Now, equation of tangent of parabola y 2 = 8 x at point B(8 , 8 ) is T = 0

1 in xy = 2, m

we get

⇒

    − 2 4  = B (8 , 8 ) , B 2  −1  1     −  2    2

⇒

m=−

y = 4b ( x − c ) is

For point of intersection with Y-axis, put x = 0 in the given equation of parabola. This gives, y 2 = 4a 2

y = m( x − c ) − 2bm − bm3

⇒

∴ Normal to parabola 2

[replacinga byb and x by x − c ] = mx − ( 2b + c )m − bm3

… (i)

and normal to parabola y 2 = 8 ax is y = mx − 4am − 2am3

…(ii)

⇒

m(( 2a − b )m 2 + ( 4a − 2b − c )) = 0

⇒

m=0 2b + c − 4a m = 2a − b 2

c = −2 2a − b c As, m 2 > 0, therefore >2 2a − b Note that if m = 0, then all options satisfy (Qy = 0 is a common normal) and if common normal is other than the axis, then only option (c) satisfies.   c 3 Q for option (c), 2a − b = 2 − 1 = 3 > 2    a 4 (a) We know that, y = mx + is the m equation of tangent to the parabola y 2 = 4ax . 1 ∴ y = mx + is a tangent to the parabola m [Qa = 1] y 2 = 4x. Let, this tangent is also a tangent to the hyperbola xy = 2

Clearly from graph c>0 b − > 0 ⇒ −b> 0 a ⇒ b 0 7 (a) On homogenisation of the curve y 2 − 4ax = 0 by line y = mx + c , we are getting combined equation of straight lines which subtend a right angle at the origin, so B y 2 =4 ax

O

A

y = ± 2a

Thus, the point of intersection are ( 0 , 2a ) and ( 0 , − 2a ). Y

[replacing a by 2a] For common normal, we should have mx − 4am − 2am3 = mx − ( 2b + c )m − bm3 [using Eqs. (i) and (ii)] 4am + 2am3 = ( 2b + c )m + bm3

or

Now, on substituting y = mx +

B (0, 2a) O

A (a2, 0)

X

y − mx  y 2 − 4ax   =0  c  …(i) ⇒ c + 4am = 0 On putting the value of ‘c’ in the line, we get y = m( x − 4a ), represent family of line passes through ( 4a , 0 ). Hence, option (a) is correct. 8 (b) We have, y 2 = 16 x Coordinate of latusrectum (4, 8) Equation of normal at (4, 8) is y = − x + 12

C(0, –2a)

From the given condition, we have Area of ∆ABC = 250 1 ( BC )(OA) = 250 ∴ 2 1 [QArea = × base × height] 2 1 2 ⇒ ( 4a )a = 250 2 ∴ a =5 6 (b) Given, y = ax 2 + bx + c has two x-intercepts one is positive and one is negative and vertex is ( 2 , − 2 ).The graph of y = ax 2 + bx + c is

y=mx+c

Q (36, 24) 4, 8 (12, 0)

P (4, –8)

It cuts x-axis at (12, 0) Equation of chord passing through (12, 0) and perpendicular of normal is y = x − 12 Put the value of y in y 2 = 16 x , we get (12 − x ) 2 = 16 x 144 − 24 x + x 2 = 16 x x 2 − 40 x + 144 = 0

(0, 0) (2, –2)

( x − 36 ) ( x − 4 ) = 0 x = 4 , 36 ∴ y = 4 − 12 = − 8 and y = 36 − 12 = 24 ∴Length of chord =

(36 − 4 ) 2 + ( 24 + 8 ) 2

= 32 2 + 32 2 = 32 2

25

Engineering Entrance Questions 2019-20

Ellipse & Hyperbola 1. (a) Since line 3 x + 4 y = 12 2 is a tangent x2 y2 + = 1 for some a ∈ R, 2 9 a then the equation of tangent to ellipse having slope ‘m’ is

to the ellipse

y = mx ±

a 2m 2 + 9

…(i)

QSlope of line (i), m 3 4 On putting the value of m in Eq. (i), we get 3 9 2 y=− x ± a +9 4 16 = slope of line3 x + 4 y = 12 2 ⇒ m = −

⇒ 3x + 4y = ± 3 a

2

+ 16

…(ii)

Since tangent 3 x + 4 y = 12 2 represented by Eq. (ii) for some a ∈ R, therefore 12 2 = ± 3 a 2 + 16 ⇒

4 2 = ± a 2 + 16

⇒ 32 = a 2 + 16 [on squaring both sides] ⇒ a 2 = 16 ⇒ a = ± 4 ∴ Eccentricity 9 7 7 b2 e = 1− 2 = 1− = = 16 16 4 a ∴ Distance between the foci = | 2ae | 7 =2×4× =2 7 4 2. (b) Since equation of normal to the ellipse at 1 P meets the co-ordinate axes at  − , 0   3 2  and ( 0 , β ) is x y …(i) + =1 1 β − 3 2 Now, let point P ( x1 , y1 ), so equation of x2 y2 normal to ellipse + = 1 at point P is 1 1 2 x y 1 − = −1 ⇒ 2 x1 y1 2 x y 1 …(ii) + = ⇒ −2 x 1 y1 2 QEqs. (i) and (ii) represents same line y 2 x1 2 − = 1 = ∴ 1 β 1 − 3 2 1 and y1 = 2 β x1 = ⇒ 3 2 QPoint P ( x1 , y1 ) lies on ellipse, so  1  2 2  + 4β = 1 9 × 2 ⇒

β2 =

2 2 ⇒β = ± 3 9

Q Point P ( x1 , y1 ) is in first quadrant so y1 = 2 β > 0 2 ∴ β= 3 Hence, option (b) is correct.

3. (c) Since, the vertices of hyperbola on

X-axis at ( ±6 , 0 ), so equation of hyperbola x2 y2 we can assume as 2 − 2 = 1 and | a | = 6 a b and hyperbola passes through point P (10 , 16 ) so (10 ) 2 (16 ) 2 − =1 2 6 b2 (16 ) 2 100 − 36 64 ⇒ = = 36 36 b2 36 × 256 2 = 144 b = ⇒ 64 x2 y2 So, equation of hyperbola is − = 0, 36 144 and the equation of normal to hyperbola at point P is 36 x 144 y + = 36 + 144 10 16 18 x + 9 y = 180 ⇒ 5 ⇒ 2 x + 5 y = 100 Hence, option (c) is correct.

4. (b)∴ For the given hyperbola,

⇒

sin 2 θ e = 1+ >2 cos 2 θ 1 + tan 2 θ > 4

⇒

tan θ ∈ ( − ∞ , − 3 ) ∪ ( 3 , ∞ )

π But θ ∈  0 ,  ⇒ tan θ ∈ ( 3 , ∞ )  2 ⇒

π π θ ∈  ,  3 2

Now, length of latusrectum sin 2 θ =2 = 2sin θ tan θ cos θ Since, both sinθ and tanθ are increasing π π functions in  ,  3 2 ∴ Least value of latus rectum is π π 3 π > 2 sin ⋅ tan = 2 ⋅ ⋅ 3 = 3 at θ =   2 3 3 3 and greatest value of latusrectum is < ∞ Hence, latusrectum length ∈ (3 , ∞ )   y2 x2 5.(c) Given, S = ( x , y ) ∈ R 2 : − = 1 + r − r 1 1     y2 x2 = ( x , y ) ∈ R 2 : + = 1 r r 1 1 + −  

y2 x2 + = 1, represents a 1+ r r −1 vertical ellipse. [Qfor r > 1, r − 1 < r + 1 and r − 1 > 0] r −1 Now, eccentricity (e ) = 1 − r +1 (r + 1) − (r − 1) = r +1 2 = r +1

For r > 1,

6.(b) Let the equation of ellipse be x2 y2 + 2 =1 2 a b Then, according the problem, we have 2b 2 = 8 and 2ae = 2b a b b b   = 4 and = e ⇒ a  a ⇒

b (e ) = 4

1 …(i) e Also, we know that b 2 = a 2 (1 − e 2 ) 1 …(ii) ⇒ 2e 2 = 1 ⇒e = 2 From Eqs. (i) and (ii), we get. b =4 2 ⇒

b = 4.

b2 32 = = 64 1−e2 1− 1 2 x2 y2 + =1 ∴Equation of ellipse be 64 32 Now, check all the options. Only ( 4 3 , 2 2 ), satisfy the above equation. Now,

a2 =

7.(d) 8.(a) One of the focus of ellipse

x2 y2 + =1 a2 b2

is on Y-axis ( 0 , 5 3 ) ∴

be = 5 3

…(i)

[where e is eccentricity of ellipse] According to the question, 2b − 2a = 10 …(ii) ⇒ b −a =5 On squaring Eq. (i) both sides, we get b 2e 2 = 75  a2 b 2  1 − 2  = 75 b  

 2 a2 Qe = 1 − 2  b   …(iii) [from Eq. (ii)] ⇒ b + a = 15 On solving Eqs. (ii) and (iii), we get b = 10 and a = 5 2 × 25 2a 2 So, length of latusrectum is = =5 b 10

⇒

26

Engineering Entrance Questions 2019-20

9. (a) Since the point (α , β ) is on the parabola y = x , so α = β 2

2

Since, line (ii) is also a tangent of the ellipse x 2 + 2y 2 = 1 ∴

β    2

 1  = (1) 2    2β 

2

+

1 2

[Qcondition of tangency of line y = mx + c x2 y2 to ellipse 2 + 2 = 1 is c 2 = a 2m 2 + b 2 , a b β 1 1 , a = 1, b = and c =  here m = 2  2β 2 ⇒

β 1 1 = + 4 2 4β 2

⇒

β2 =

⇒

β2 = 1 +

Q

α = β2 = 1 +

2

2±

4+ 4 2 2 [Qβ 2 > 0] 2

10. (d) Let the equation of hyperbola is x2 y2 …(i) − 2 =1 2 a b Since, equation of given directrix is5 x = 4 5 so

a 5   = 4 5 e 

a [Qequation of directrix is x = ] e 4 a …(ii) ⇒ = 5 e and hyperbola (i) passes through point (4, − 2 3 ) 16 12 so, …(iii) − =1 a2 b2 The eccentricity e = 1 + ⇒

b2 =

…(i)

Now, equation of tangent at point (α , β ) to the parabola y 2 = x , is T = 0 1 ⇒ yβ = ( x + α ) 2 [Qequation of the tangent to the parabola y 2 = 4ax at a point ( x1 , y1 ) is given by yy1 = 2a ( x + x1 )] [from Eq. (i)] ⇒ 2 yβ = x + β 2 x β …(ii) y= + ⇒ 2β 2

2

⇒

From Eqs. (ii) and (iv), we get 16 4 16 2 e − e =b2 5 5 From Eqs. (ii) and (iii), we get 12 16 5 12 − =1 ⇒ 2 − 2 =1 16 2 b 2 e b e 5

…(vi)

From Eqs. (v) and (vi), we get  12e 2  16e 4 − 16e 2 = 5   2 5 − e  ⇒

4e 4 − 24e 2 + 35 = 0

11. (d) Equation of given ellipse is 3 x 2 + 5 y 2 = 32

…(i)

Now, the slope of tangent and normal at point P ( 2 , 2 ) to the ellipse (i) are respectively dy dx and mN = − mT = dx ( 2 , 2 ) dy ( 2 , 2 ) On differentiating ellipse (i), w.r.t. x, we get dy 3x dy =− 6 x + 10 y =0 ⇒ dx 5y dx So,

mT = −

and mN =

5y 3y

3x 5y

=− (2, 2)

= (2, 2)

3 5

5 3

Now, equation of tangent and normal to the given ellipse (i) at point P ( 2 , 2 ) are 3 (y − 2) = − (x − 2) 5 5 and ( y − 2 ) = ( x − 2 ) respectively. 3 It is given that point of intersection of tangent and normal are Q and R at X-axis respectively. 16 4 So, Q  , 0  and R  , 0  3 5   1 ∴ Area of ∆PQR = (QR ) × height 2 1 68 68 sq units = × × 2= 2 15 15

12. (c) Let the foot of perpendicular be (h , k ) . Equation of tangent with slope m passing (h , k ) is y = mx ± 6m 2 + 2 , where m = − h / k ⇒

h2 + k2 6h 2 + 2 = 2 k k 6h 2 + 2k 2 = (h 2 + k 2 ) 2

So, required locus is 6x 2 + 2y 2 = (x 2 + y 2 ) 2 .

b2 a2

a 2e 2 − a 2 = b 2

12e 2 5 −e2

…(iv) …(v)

QChord (i) is the focal chord so, it will pass through focus (4, 0) α+β α −β 4 cos = cos 5 2 2 α β α β ⇒ 4  cos cos − sin sin   2 2 2 2 α β α β = 5  cos cos + sin sin   2 2 2 2 α β α β    ⇒ 4  cot cot − 1 = 5  cot cot +    2 2 2 2 α β ⇒ cot cot = − 9 2 2 Hence, option (c) is correct.

14. (b) Let the equation of ellipse x2 y2 …(i) + 2 = 1, (a > b ) 2 a b QLength of latusrectum of ellipse (i) is 2b 2 …(ii) = 4 (given) a and distance between the foci is 2ae = 4 2 (given) ⇒a 1 −

points A(α ) and B(β ) on the ellipse x2 y2 + = 1 is 25 9 α+β y α+β α −β x cos + sin = cos 5 2 3 2 2

…(i)

b2 =2 2 a2

  b2 Qe = 1 − 2 , (b > a )  a   2 So, b = 8

{Qa > 0}

∴ Equation of required ellipse is x2 y2 + = 1 ⇒ x 2 + 2 y 2 = 16 16 8 Hence, option (b) is correct.

15. (b) We have, x + y + n = 0 is a normal to the ellipse x 2 + 3y 2 = 3 (3 − 1) n=± ∴ 3+1 n=±1 Hence, n > 0 ∴ n=1 Equation of normal = x + y + 1 = 0 … (i) x + my + 3 = 0 is tangent of ellipse x 2 + 5y 2 = 5 3 5 =± +1 m m2 m = ± 2 ,m < 0 ∴m = − 2 Equation of tangent = x − 2 y + 3 = 0 … (ii) On solving Eqs. (i) and (ii), we get 5 2 x = − ,y = 3 3 Which satisfies the equation x − 5y + 5 = 0

∴

13. (c) Since equation of chord joining the

1 

27

Engineering Entrance Questions 2019-20 16. (d) Equation of normal drawn of the point  9 cos π , 7 sin π  to    4 4 2 2 x y + =1 9 7 9x 7y is − =9 −7 π π 9 cos 7 sin 4 4 9 x − 7y = 2

here, a = b =

ellipse ∴

x2 y2 + 2 =1 2 a b B (0, b)

[Qa = b ]

2

∴ Equation of the directrix are 13 x =3 ± 6

18. (b) We have, Length of transverse axis 2a = 6 ⇒a = 3 2b 2 8 and length of latusrectum, = a 3 8 2 ⇒ 2b = × 3 = 8 ⇒b 2 = 4 3 ∴Required equation of hyperbola is x2 y2 x2 y2 − =1 − = 1 ⇒ 9 4 a2 b2 ⇒ 4 x 2 − 9 y 2 = 36

17. (a) Given equation is 3x

19. (c) Equation of the ellipse is

2

b a2 = 1+ 1

e = 1+

=

 2  It cuts major axis at  , 0  9  2

13 3

− 3 y − 18 x + 12 y + 2 = 0 2

It can be written as (x − 3) 2 (y − 2) 2 − =1 2 2  13   13       3   3 

T

S

Foci are S (ae , 0 ), T ( − ae , 0 ). B ( 0 , b ) is the end of the minor axis. STB is an equilateral triangle. SB = ST ⇒ SB 2 = ST 2 ⇒

a 2e 2 + b 2 = 4a 2e 2

⇒

1 − e 2 = 3e 2

1 2 1 Eccentricity of the ellipse, e = 2

⇒

4e 2 = 1

⇒

e=

Introduction to Three Dimensional (3D) Geometry 1 (a) Given, points A( 2 , 3 , 5 ), B(α , 3 , 3 ) and C (7 , 5 , β ) 3 + β α+7 , 4, ∴Mid-point of BC is D    2 2  QDirection ratios of line joining points 3 + β α+7 A( 2 , 3 , 5 ) and D  , 4,   2 2   α + 3 , 1, β − 7 .    2 2 

is

Q The line segment AD is equally inclined with the co-ordinate axes, so α+3 β −7 =1= 2 2 ⇒ α = − 1 and β = 9 α 1 ∴cos −1   = cos −1  −   9 β Hence, option (a) is correct. 2 (a) A(3, 2, –1) 1 E

F O

B (4, 1, 1) 2 D 1

2

C(6, 2, 5)

2 × 6 + 4 ×1 2 × 2 + 1×1 Here, D =  , ,  3 3 2 × 5 + 1 × 1   3 16 5 11 =  , ,  3 3 3 12 6 3 Similarly, E =  , ,   3 3 3 11 4 1 ⇒ F =  , ,   3 3 3 Let centroid of ∆DEF is O ( x , y , z ). 16 12 11 + + 3 3 = 13 ∴ x = 3 3 3 5 6 4 + + 5 y=3 3 3 = 3 3 11 1 + 1+ 3 =5 and z = 3 3 3 13 5 5 Hence, coordinates are  , ,  .  3 3 3 3 (b) We have, (1, 0, 3), (2, 1, 5), (−2, 3, 6) is mid point of the sides of triangle. Centroid of ∆ABC is also the centroid of ∆DEF

A

F

B

E

D

C

∴Centroid 1 + 2 − 2 0 + 1 + 3 3 + 5 + 6 =  , ,    3 3 3 1 4 14   = , ,  3 3 3  4 (a) P (6, 10, 10)

R (6, –10, λ)

Q (1, 0, –5)

In ∆PQR is right angled, at θ ∴ PR 2 = PQ 2 + RQ 2 ⇒ ( 6 − 6 ) 2 + ( −10 − 10 ) 2 + ( λ − 10 ) 2 + [(1 − 6 ) 2 + ( 0 − 10 ) 2 + ( −5 − 10 ) 2 ] ⇒ λ − 20 λ + 500 = 350 + 150 + 10 λ + λ 2 ⇒ −20 λ = 10 λ ⇒ 30 λ = 0 ⇒ λ = 0 2

28

Engineering Entrance Questions 2019-20

Introduction to Limits and Derivatives 1 (36) lim

x→ 2

3 x + 3 3 − x − 12 3 − x / 2 − 31 − x

 0 form   0 

Put x = 2 + h as x → 2 ⇒ h→0 3 2 + h + 31 − h − 12 = lim h h→ 0

3

−1 −

− 3 −1 − h

2

−h

9 ⋅ 3 + 3 ⋅ 3 − 12 1 −h / 2 − 3 −h ) (3 3 9 (3 (3h − 1) + (3 −h − 1))

= lim

h

1− y2 dy =− dx 1− x2

⇒

1 1 Q y   = −  2 4

∴

dy dx

x =

h→ 0

= lim

h→ 0

[Qlim 3h = 3 0 = 1 h→ 0

ah − 1 = log e a ] h→ 0 h

and lim  3 − 1 = 9  = 36  1/ 2 

2 (b) Given functional relation is

y 1 − x2 = k − x 1 − y 2 On differentiating both sides w.r.t. x, we get dy −2 yx + 1− x2 dx 2 1− x2 dy dx − 1 − y 2 =0+ 2 1− y2 2 xy

xy dy  2  1− x − dx  1− y2 

  

 yx  = − 1− y2  2  1 − x  ⇒

2 2 dy  1 − x 1 − y − xy    dx  1− y2  

=

xy − 1 − x 2 1 − y 2 1− x2

1 1 −  −   4

=−

1 1 −    2 =−

3 −h (3h / 2 − 1)

  3h − 1    3 −h − 1   h +   ( −h )   3  − h h      = lim 9 ⋅ 3h    h→ 0  h / 2 − 1 h  h/ 2      2 3     3h − 1   3 −h − 1    −     3   h   −h   = lim 9 ⋅ 3h    h→ 0 1  3h / 2 − 1      2 h /2       3 log e 3 − log e 3  = 9 ×1   1 log e 3     2

⇒

1 2

(given) 2

2

16 − 1 16 =− 4 −1 4

dy = f ′ ( f ( f ( x ))) ⋅ f ′ ( f ( x )) ⋅ f ′ ( x ) dx + 2 f ( x ) f ′ ( x ) [by chain rule] So, dy dx

∴

= f ′ ( f ( f (1))) ⋅ f ′ ( f (1)) ⋅ f ′ (1) + 2 f (1) f ′ (1) at x = 1

dy dx

= f ′ ( f (1)) ⋅ f ′ (1) ⋅ (3 ) + 2 (1)(3 ) x =1

[Qf (1) = 1 and f ′ (1) = 3] = f ′ (1) ⋅ (3 ) ⋅ (3 ) + 6 = (3 × 9 ) + 6 = 27 + 6 = 33

15 5 =− 2 3 2 6 (a) Let

3 (a) Clearly, 1+

lim

1+ y

y→ 0

y

4

−

3π x → 4

2

4

(1 +

= lim

y→ 0

L = lim

y4 ( 1 +

1 + y4 ) − 2 1 + y4 +

2)

[Q(a + b ) (a − b ) = a 2 − b 2 ] = lim

y→ 0

1 + y4 − 1 y4 ( 1 +

1 + y4 +

2) ×

1 + y4 + 1 1 + y4 + 1

4 sin 2 x cos x − cos x + sin x sin x + cos x

L = lim 3π 2 sin x (sin 2x x→ +4 cos 2 x + 2 sin x cos x − 1) − cos x + sin x sin x + cos x 2 sin x (sin x + cos x ) 2 − 2 sin x − cos x + sin x sin x + cos x (sin x + cos x ) ( 2 sin x (sin x + cos ( x − 1) L = lim 3π sin x + cos x x → 4

L = lim 2 sin x (sin x + cos x ) − 1 = − 1 [again, rationalising the numerator] 3π x → y4 4 = lim y→ 0 y 4 ( 1 + 1 + y 4 + 2 ) ( 1 + y 4 + 1) x ⋅ 2x − x 1 7 (c) α = lim 4 (by cancelling y and then by = x → 0 1 − cos x 2 2 ×2 x ( 2 x − 1) x ( 2 x − 1) direct substitution). α = lim = lim x x → 0 1 − cos x x → 0 1 2 sin 2 . = 2 4 2 2x − 1 2 sin x 4 (a) Given limit is lim 1 2x − 1 1 x = lim ⋅ = lim x → 0 2 − 1 + cos x x x → 0 x → 0 2 2 x 2 sin 2 sin  0 form  2 2  0  x x2 [Qdivided by x] sin 2 x = lim x   x→0 2 − 2 cos   2   Q1 + cos x = 2 cos 2 x    x   2 −1 1  1 2  =  lim ×  2 2 x→0 x sin 2 x 1     = lim sin x   1 x→0 x  2  ×  lim  2  1 − cos   x  x → 0   4 2   2   2   sin x 16 = lim = =4 2 sin x   x→0 x =1 Q lim 2 × 2 sin 2   2 2  4  x → 0 x  x 2   a −1 5 (d) Let y = f ( f ( f ( x ))) + ( f ( x )) = log a  Q xlim → 0 x   On differentiating both sides w.r.t. x, we get

29

Engineering Entrance Questions 2019-20 1 1 1 × log 2 × = log 2 × 4 1 2 2 1× 4 …(i) α = 2. log 2 x ⋅ 2x − x Given, β = lim x → 0 1+ x2 − 1− x2

β = log 2 ∴ 2 β = 2 ⋅ log 2 From Eqs. (i) and (ii), we get α = 2β Hence, option (c) is correct.

=

β = lim

x → 0

x ( 2 − 1) x

1+ x2 − 1− x2

Rationalise the denominator x ( 2 x − 1) = lim x → 0 1+ x2 − 1− x2 × = lim

x → 0

1+ x2 +

1− x2

1+ x

1− x

2

+

1 2 −1 × × ( 1+ x2 + 2 x 1 = × log 2 × 2 2 x

2

…(ii)

8 (c) We have,  [ x ]3  x  3   − lim   3   x→ 2 +   3  [ 2 + h]3  2 + h  3   = lim  −  3   h→ 0  3   =

10 (c) Let

x → 0+

log(e x + x ) x → 0 x 1 ⋅ex + 1 (e x + x ) ⇒ log L = lim x → 0+ 1

3

y = log e 2 (log x )

1 ⇒ y = log e (log x ) 2 Q log ( x ) = 1 log x   a  an   n

1− x2)

L = lim (e x + x )1 / x

⇒ log L = lim

(2 ) 2  2  8 8 = −0 = −  3  3 3 3

9 (c) Let

On differentiating both sides w.r.t. x, we get dy 1 1 d (log x ) = dx 2 log x dx 1 1 1 = . = log x 2 x x log x 2

+

[using L’ Hospital rule] ex + 1 ⇒ log L = lim x x → 0+ e + x ⇒ log L = 2 ⇒

L =e2

Mathematical Reasoning 1 (a) From the truth table p ⇒ ~p q ⇒~p q

(p ⇒ q ) ∧ (q ⇒ ~ p )

p

q

T

T

T

F

F

F

T

F

F

F

T

F

F

T

T

T

T

T

F

F

T

T

T

T

4 (b) Since, the statements P : 5 is a prime number, is true statement. Q : 7 is a factor of 192, is false statement and R : LCM of 5 and 7 is 35, is true statement. So, truth value of P is T , Q is F , R is T Now let us check all the options. P Q R ~ P ~Q ~ R P ∧Q Q ∧ R ~Q ∧ R

2 (a) From the truth table P ∧ (P → Q ) → Q

Q → ( P ∧ ( P → Q ))

T

T

T

T

T

F

T

T

∴P ∧ ( P → Q ) → Q is a tautology. Hence, option (a) is correct. 3 (c) Clearly, [~ (~ p ∨ q ) ∨ ( p ∧ r )] ∧ (~ q ∧ r ) ≡ [( p ∧ ~q ) ∨ ( p ∧ r )] ∧ (~ q ∧ r ) (Q~ (~ p ∨ q ) ≡ ~ (~ p ) ∧ ~ q ≡ p ∧ ~ q by De Morgan’s law) ≡ [p ∧ (~ q ∨ r )] ∧ (~q ∧ r )] (distributive law) (associative ≡ p ∧ [(~ q ∨ r ) ∧ (~ q ∧ r )] law) ≡ p ∧ [(~ q ∧ r ) ∧ (~q ∨ r )] (commutative law) ≡ p ∧ [{(~ q ∧ r ) ∧ (~ q )} ∧ {(~ q ∧ r ) ∧ r ] (distributive law)

T

F T

F

T

F

F

F

T

(P ∧ Q ) ∨ (~ R)

P ∨ (~ Q ∧ R)

(~ P) ∨ (Q ∧ R)

(~ P) ∧ (~ Q ∧ R)

F

T

F

F

Clearly, the truth value of P ∨ (~ Q ∧ R ) is T . 5 (b) Given, ( p ∧ q ) ↔ r is true. This is possible under two cases Case I When both p ∧ q and r are true, which is not possible because q is false. Case II When both ( p ∧ q ) and r are false. ⇒ p ≡ T or F ; q ≡ F , r ≡ F (b) ( p ∧ r ) → ( p ∨ r ) is F→ (T or F ) , which always result in T. 6 (b) We know that, contrapositive of p → q is ~q → ~p Therefore, the contrapositive of the given statement is

‘‘If the squares of two numbers are equal, then the numbers are equal’’. 7 (b) We have, ( p ∧ q ) ∧ [~ r ∨ ( p ∧q )] ∨ (~ p ∧ q ) ≡ ( p ∧q ) ∨ (~ p ∧q ) ≡ (p ∨ ~ p ) ∧ q ≡ T ∧q = q 8 (c) We have statements p , q → T and r , s → F Option (c) ( p → q ) ∨ (r ↔ s ) ≡ (~ p v q ) ∨ ((~ r ∨ s ) ∧ (r ∨ ~ s )) (Q p → q ≡ ~ p ∨ q ) ≡ ( F ∨ T ) ∨ ((T ∨ F ) ∧ ( F ∨ T )) ≡ T ∨ (T ∧ T ) ≡ T ∨ T ≡ T 9 (a) ey Idea se p → q ≡ ~ p v q and p ↔ q ≡ (~ p v q ) ∧ ( p v ~ q ) Given, p , q → T and r , s → F ∴ a : ~ ( p ∧ ~ r ) ∨ (~ q ∨ s ) ≡ ~ (T ∧ T ) ∨ ( F ∨ F ) ≡ ~ (T ) ∨ ( F ) ≡ F ∨ F ≡ F and b : ( p ∨ s ) ↔ (q ∧ r ) ≡ (~ ( p ∨ s ) ∨ (q ∧ r )) ∧ (( p ∨ s ) ∨ ~ (q ∧ r )) Qp ↔ q ≡ (~ p ∨ q ) ∧ ( p ∨ ~ q ) ≡ (~ (T ∨ F ) ∨ (T ∧ F )) ∧ ((T ∨ F ) ∨ ~ (T ∧ F )) ≡ ( F ∨ F ) ∧ (T ∨ T ) ≡ F ∧T ≡ F

30

Engineering Entrance Questions 2019-20

Statistics 1. (d) Let the 10 observations are xi (for i = 1, 2, 3, …, 10) and mean and standard deviation is x = 20 and σ1 = 2, respectively. Now, it is given that each of these 10 observations is multiplied by ‘p’ and then reduced by q , where p ≠ 0 and q ≠ 0, so then new mean 20 = p (x ) − q 2 …(i) ⇒ 10 = 20 p − q 2 and new standard deviation = | p| σ1 1 1 | p| = ⇒ 2 1 p=± ⇒ 2 1 Now, at p = 2 (from Eq. (i)) q=0 1 at p=− 2 q = − 20 (from Eq. (i)) Hence, option (d) is correct.

2. (c) For the observations xi (1 ≤ i ≤ 10 ), we have 10

∑ ( xi

− 5 ) = 10 10

∑ xi

five students. Then, we have 5

Mean, Σ xi = 750 variance 5

Σ x i2

i =1

⇒

− (150 ) 2 = 18

5

Σ xi2 = 112590 6

∑ xi

∑ ( xi

= 60

Σ xi + 156

750 + 156 [using Eq. (i)] = 6 6 ⇒ xnew = 151 and new variance =

i =1

5

=

Σ xi2 + (156 ) 2

i =1

2

∑ xi

2

= 390

… (ii)

i =1

10

∑ ( xi

10

∑ xi

− 3)

i =1

=

− 30

10

=

∑ xi

2

∑ ( xi

− 3) 2

i =1

10

− (µ )

i =1

10

=1 n ∴mean ( x ) = 1 n

2

n

=

∑ ( xi

− 1)

∑ ( xi

i =1

2

i =1

n

=

5n = 5 n

5. (a) Let 1, 3, 8, x and y be the five

10

− 6 ∑ xi + 10 (3 2 )

∑ xi

i =1

Now, standard deviation =

observations. −9

Then, mean x =

Σx i n

…(iii)

Putting y = xt in Eq. (ii), we get x 2 (1 + t 2 ) = 97

… (iv)

…(ii)

Dividing Eqs. (iii) by (iv), we get x 2 (1 + t ) 2 169 = x 2 (1 + t 2 ) 97 4 or t= 9 some x ∈ R, of marks obtained by 20 students is

Marks

2

3 2x − 5 x

2

2

5

7

− 3x

x

QNumber of students = 20 = Σ f i ⇒ ( x + 1) 2 + ( 2 x − 5 ) + ( x 2 − 3 x ) + x = 20 ( x + 4 )( x − 3 ) = 0 ⇒ x = 3 [as x > 0] Σ f i xi Now, mean ( x ) = Σ fi 2 ( x + 1) 2 + 3 ( 2 x − 5 ) + 5 ( x 2 − 3 x ) + 7 x = 20 2 ( 4 ) 2 + 3 (1) + 5 ( 0 ) + 7 (3 ) 32 + 3 + 21 = = 20 20 56 = = 2 .8 20 Hence, option (b) is correct.

⇒

i =1

i =1

10 10 60 − 30 30 = = =3 10 10

and variance λ =

n

n

⇒

x 2 + y 2 = 97 y Let =t x ⇒ y = xt Putting y = xt in Eq. (i), we get x (1 + t ) = 13 ⇒ x 2 (1 + t ) 2 = 169

Frequency ( x + 1)

⇒ ∑ xi = n

i =1

[from Eq. (i)]

i =1

...(ii)

i =1 n

10

10

− 1) 2 = 5n

n

⇒ ∑ 4 xi = 4n

⇒

6. (b) The given frequency distribution, for ...(i)

⇒ ∑ {( xi + 1) 2 − ( xi − 1) 2 } = 4n

− 10 ( 60 ) + 250 = 40,

Now, mean µ =

Eq.

On subtracting Eq. (ii) from Eq. (i) is, we get

i =1

⇒

[using

i =1

− 5 ) 2 = 40

i =1

∑ xi

⇒

n

∑ ( xi

and

10

10

− (151) 2

i =1

… (i)

⇒ ∑ xi2 − 10 ∑ xi + 10 (5 2 ) = 40 i =1

6

5

2 ∑ (xi + 1) = 9n

i =1 10

i =1

n

i =1

and

Σ xi

4. (d) We have,

− 50 = 10 10

10

…(ii)

i =1

Now, new mean =

x =

( y 2 − 10 y + 25 ) = 46

5

⇒

i =1

⇒

…(i)

i =1

and

1+ 3 + 8 + x + y = 5 (given) 5 …(i) ⇒ x + y = 13 Σ( xi . − x ) 2 2 and variance = σ = n  (1 − 5 ) 2 + (3 − 5 ) 2 + (8 − 5 ) 2    + ( x − 5 ) 2 + ( y − 5 ) 2   =  = 9. 2 5 (given) ⇒16 + 4 + 9 + ( x 2 − 10 x + 25 ) +

⇒

3. (c) Let x1 , x 2 , x 3 , x 4 , x 5 be the heights of

6 (ii)] = 22821 − 22801 = 20

i =1

⇒

390 − 6 ( 60 ) + 90 − 90 30 = =3 10 10 So, the ordered pair (µ , λ ) = (3 , 3 ) Hence, option (c) is correct. =

n

− x)2

7.(a) Given, 15

∑ xi

i =1

2

= 3600,

15

∑ xi

i =1

When 20 is replace by 40, then

= 175

31

Engineering Entrance Questions 2019-20 15

∑ xi

i =1 15

∑ xi

= 175 − 20 + 40 = 195 2

= 3600 − ( 20 ) 2 + ( 40 ) 2 = 4800

i =1

∴Corrected variance =

Σxi2  Σxi  −   n  n

2

2

4800  195  −   15  15 = 320 − 169 = 151 =

8.(b) Let σ1 and x1 are the standard deviation

and mean of 50 observation and σ 2 and x 2 are standard deviation one mean of another 50 observation respectively x1 = 10 , σ1 = 2 , n1 = 50 x = Mean of 100 observation = 8 σ = SD of 100 observation = 10.5 50 x1 + 50 x 2 x = 50 + 50 ⇒

800 = 500 + 50 x 2 x =6

σ2 = +

n1 ( σ1 ) 2 + n 2 ( σ 2 ) 2 n1 + n 2

n1n 2 ( x1 − x 2 ) 2 (n1 + n 2 ) 2

10.5 =

50 ( 4 ) + 50 ( σ 2 ) 2 2500 (10 − 6 ) 2 + 100 (100 ) 2

⇒ 1050 = 600 + 50 ( σ 2 ) 2 1050 − 600 =9 ⇒ σ22 = 50 ⇒ σ2 =3

Fundamentals of Probability 1 (a) For two events A and B it is given that probability of occurrence of exactly one of 2 them is . 5 2 So, P ( A) + P ( B ) − 2 P ( A ∩ B ) = …(i) 5 1 and probability that A or B occurs is , 2 1 so P (A ∪ B) = 2 1 …(ii) ⇒ P ( A) + P ( B ) − P ( A ∩ B ) = 2 From Eqs. (i) and (ii), we get 1 2 P (A ∩ B) = − 2 5 5−4 1 = = 10 10 Probability of both of them occur together 1 = = 0.10 10 Hence, option (a) is correct. 2 (c) Clearly, 1 P ( H ) = Probability of getting head = 2 1 and P (T ) = Probability of getting tail = 2 Now, let E 1 be the event of getting a sum 7 or 8, when a pair of dice is rolled. ⇒ P ( E 1 ) = Probability of getting 7 or 8 when 11 a pair of dice is thrown = 36

Also, let P ( E 2 ) = Probability of getting 7 or 8 when a card is picked from cards numbered 2 1, 2, ...., 9 = 9 ∴Probability that the noted number is 7 or 8 = P ( H ∩ E 1 ) + P (T ∩ E 2 ) [Q ( H ∩ E 1 ) and (T ∩ E 2 ) are mutually exclusive] = P ( H ) ⋅ P ( E 1 ) + P (T ) ⋅ P ( E 2 ) [Q { H , E 1 } and {T , E 2 } both are sets of independent events] 1 11 1 2 19 = × + × = 2 36 2 9 72 3 (a) Since, there is a regular hexagon, then the number of ways of choosing three vertices is 6 C 3 . And, there is only two ways i.e. choosing vertices of a regular hexagon alternate, here A1 , A3 , A5 or A 2 , A4 , A6 will result in an equilateral triangle. A6

A1 A2

A5

A3 A4

Required probability 2 ×3 × 2 ×3 × 2 2 2 1 = = 6 = = 6! 6 × 5 × 4 × 3 × 2 × 1 10 C3 3 !3 ! 4 (b) Let A be the events sum appeared on two unbiased dice is 7 and B be the event

sum appeared on two unbiased dice is either 7 or 11. 6 6 2 2 ∴P ( A) = , P ( B ) = + = 36 36 36 9 ∴Required probability = P ( A) + P ( BA) + P ( BBA) + P ( BBBA) + … 2

1 7 1 7  1 + × +   × + … 6 9 6 9 6   1 1  1 9 3 =  = × = 6 1 − 7  6 2 4  9 =

5 (c) Total number of balls = 10 Required

probability

=

6

C 2 + 4C 2 10 C2

15 + 6 21 7 = = 45 45 15 1 1 6 (c) Let P ( A) = , P ( B ) = 2 3 1 1 P (C ) = , P ( D ) = 4 5 Now, P ( A ∪ B ∪ C ∪ D ) = 1 − P (A ∩ B ∩ C ∩ D ) = 1 − P ( A ) P ( B ) P (C ) P ( D ) 1 1 1 = 1 −  1 −   1 −   1 −   1 −  4  3  2  =

1 4 1 2 3 4 = 1 −         = 1 − =  2  3   4  5  5 5

1  5