40 Days Crash Course for JEE Main Mathematics 9789313199328, 9313199327

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CRASH COURSE PROGRAMME

JEE Main in

40 DAYS

MATHEMATICS

ARIHANT PRAKASHAN (Series), MEERUT

Arihant Prakashan (Series), Meerut All Rights Reserved

© PUBLISHERS

No part of this publication may be re-produced, stored in a retrieval system or distributed in any form or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher. Arihant has obtained all the information in this book from the sources believed to be reliable and true. However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damages or loss suffered there upon. All disputes subject to Meerut (UP) jurisdiction only.

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PREFACE It is a fact that nearly 10 lacs students would be in the race with you in JEE Main, the gateway to some of the prestigious engineering and technology institutions in the country, requires that you take it seriously and head-on. A slight underestimation or wrong guidance will ruin all your prospects. You have to earmark the topics in the syllabus and have to master them in concept-driven-problem-solving ways, considering the thrust of the questions being asked in JEE Main. The book 40 Days JEE Main Mathematics serves the above cited purpose in perfect manner. At whatever level of preparation you are before the exam, this book gives you an accelerated way to master the whole JEE Main Physics Syllabus. It has been conceived keeping in mind the latest trend of questions, and the level of different types of students. The whole syllabus of Physics has been divided into day-wise-learning modules with clear groundings into concepts and sufficient practice with solved and unsolved questions on that day. After every few days you get a Unit Test based upon the topics covered before that day. On last three days you get three full-length Mock Tests, making you ready to face the test. It is not necessary that you start working with this book in 40 days just before the exam. You may start and finish your preparation of JEE Main much in advance before the exam date. This will only keep you in good frame of mind and relaxed, vital for success at this level.

Salient Features Ÿ Concepts discussed clearly and directly without being superfluous. Only the required Ÿ Ÿ Ÿ Ÿ Ÿ

material for JEE Main being described comprehensively to keep the students focussed. Exercises for each day give you the collection of only the Best Questions of the concept, giving you the perfect practice in less time. Each day has two Exercises; Foundation Questions Exercise having Topically Arranged Questions & Progressive Question Exercise having higher Difficulty Level Questions. All types of Objective Questions included in Daily Exercises (Single Option Correct, Assertion & Reason, etc). Along with Daywise Exercises, there above also the Unit Tests & Full Length Mock Tests. At the end, there are all Online Solved Papers of JEE Main 2019; January & April attempts.

We are sure that 40 Days Physics for JEE Main will give you a fast way to prepare for Physics without any other support or guidance.

Publisher

CONTENTS Preparing JEE Main 2020 Mathematics in 40 Days ! Day 1.

Sets, Relations and Functions

Day 2.

Complex Numbers

10-19

Day 3.

Sequences and Series

20-30

Day 4.

Quadratic Equation and Inequalities

31-44

Day 5.

Matrices

45-54

Day 6.

Determinants

55-67

Day 7.

Binomial Theorem and Mathematical Induction

68-77

Day 8.

Permutations and Combinations

78-86

Day 9.

Unit Test 1 (Algebra)

87-94

Day 10. Real Function

1-9

95-103

Day 11. Limits, Continuity and Differentiability

104-116

Day 12. Differentiation

117-126

Day 13. Applications of Derivatives

127-137

Day 14. Maxima and Minima

138-149

Day 15. Indefinite Integrals

150-162

Day 16. Definite Integrals

163-175

Day 17. Area Bounded by the Curves

176-187

Day 18. Differential Equations

188-198

Day 19. Unit Test 2 (Calculus)

199-208

Day 20. Trigonometric Functions and Equations

209-221

Day 21. Properties of Triangle, Height and Distances

222-232

Day 22. Inverse Trigonometric Function

233-242

Day 23. Unit Test 3 (Trigonometry)

243-250

Day 24. Cartesian System of Rectangular Coordinates

251-262

Day 25. Straight Line

263-274

Day 26. The Circle

275-288

Day 27. Parabola

289-300

Day 28. Ellipse

301-313

Day 29. Hyperbola

314-325

Day 30. Unit Test 4 (Coordinate Geometry)

326-335

Day 31. Vector Algebra

336-350

Day 32. Three Dimensional Geometry

351-366

Day 33. Unit Test 5 (Vector & 3D Geometry)

367-374

Day 34. Statistics

375-383

Day 35. Probability

384-394

Day 36. Mathematical Reasoning

395-405

Day 37. Unit Test 6 (Statistics, Probability & Mathematical Reasoning)

406-411

Day 38. Mock Test 1

412-416

Day 39. Mock Test 2

417-423

Day 40. Mock Test 3

424-430

Online JEE Main Solved Papers 2019

1-32

SYLLABUS MATHEMATICS UNIT 1 Sets, Relations and Functions Sets and their representation; Union, intersection and complement of sets and their algebraic properties; Power set; Relation, Types of relations, equivalence relations, functions;. one-one, into and onto functions, composition of functions.

UNIT 2 Complex Numbers and Quadratic Equations Complex numbers as ordered pairs of reals, Representation of complex numbers in the form a+ib and their representation in a plane, Argand diagram, algebra of complex numbers, modulus and argument (or amplitude) of a complex number, square root of a complex number, triangle inequality, Quadratic equations in real and complex number system and their solutions. Relation between roots and co-efficients, nature of roots, formation of quadratic equations with given roots.

UNIT 3 Matrices and Determinants Matrices, algebra of matrices, types of matrices, determinants and matrices of order two and three. Properties of determinants, evaluation of deter-minants, area of triangles using determinants. Adjoint and evaluation of inverse of a square matrix using determinants and elementary transformations, Test of consistency and solution of simultaneous linear equations in two or three variables using determinants and matrices.

UNIT 4 Permutations and Combinations Fundamental principle of counting, permutation as an arrangement and combination as selection, Meaning of P (n,r) and C (n,r), simple applications.

UNIT 5 Mathematical Induction Principle of Mathematical Induction and its simple applications.

UNIT 6 Binomial Theorem and its Simple Applications Binomial theorem for a positive integral index, general term and middle term, properties of Binomial coefficients and simple applications.

UNIT 7 Sequences and Series Arithmetic and Geometric progressions, insertion of arithmetic, geometric means between two given numbers. Relation between AM and GM Sum upto n terms of special series: ∑ n, ∑ n2, ∑ n3. Arithmetico - Geometric progression.

UNIT 8 Limit, Continuity and Differentiability Real valued functions, algebra of functions, polynomials, rational, trigonometric, logarithmic and exponential functions, inverse functions. Graphs of simple functions. Limits, continuity and differentiability. Differentiation of the sum, difference, product and quotient of two functions. Differentiation of trigonometric, inverse trigonometric, logarithmic exponential, composite and implicit functions derivatives of order upto two. Rolle's and Lagrange's Mean Value Theorems. Applications of derivatives: Rate of change of quantities, monotonic - increasing and decreasing functions, Maxima and minima of functions of one variable, tangents and normals. UNIT 9 Integral Calculus Integral as an anti - derivative. Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions. Integration by substitution, by parts and by partial fractions. Integration using trigonometric identities. Evaluation of simple integrals of the type dx , x2 ± a2

dx 2

x ± a

dx , ax 2 + bx + c (px + q) dx ax 2 + bx + c

,

2

,

dx a2 – x2

dx ax 2 + bx + c a 2 ± x 2 dx

,

dx a 2 – x 2

,

(px + q) dx , ax 2 + bx + c

,

x 2 – a 2 dx

and

Integral as limit of a sum. Fundamental Theorem of Calculus. Properties of definite integrals. Evaluation of definite integrals, determining areas of the regions bounded by simple curves in standard form.

UNIT 10 Differential Equations Ordinary differential equations, their order and degree. Formation of differential equations. Solution of differential equations by the method of separation of variables, solution of homogeneous and linear differential equations of the type dy +p (x) y = q(x) dx

UNIT 11 Coordinate Geometry Cartesian system of rectangular coordinates in a plane, distance formula, section formula, locus and its equation, translation of axes, slope of a line, parallel and perpendicular lines, intercepts of a line on the coordinate axes.

Ÿ Straight Lines Various forms of equations of a line, intersection of lines, angles between two lines, conditions for concurrence of three lines, distance of a point from a line, equations of internal and external bisectors of angles between two lines, coordinates of centroid, orthocentre and circumcentre of a triangle, equation of family of lines passing through the point of intersection of two lines.

Ÿ Circles, Conic Sections Standard form of equation of a circle, general form of the equation of a circle, its radius and centre, equation of a circle when the end points of a diameter are given, points of intersection of a line and a circle with the centre at the origin and condition for a line to be tangent to a circle, equation of the tangent. Sections of cones, equations of conic sections (parabola, ellipse and hyperbola) in standard forms, condition for y = mx + c to be a tangent and point (s) of tangency.

UNIT 12 Three Dimensional Geometry Coordinates of a point in space, distance between two points, section formula, direction ratios and direction cosines, angle between two intersecting lines. Skew lines, the shortest distance between them and its equation. Equations of a line and a plane in different forms, intersection of a line and a plane, coplanar lines.

UNIT 13 Vector Algebra Vectors and scalars, addition of vectors, components of a vector in two dimensions and three dimensional space, scalar and vector products, scalar and vector triple product.

UNIT 14 Statistics and Probability Measures of Dispersion Calculation of mean, median, mode of grouped and ungrouped data. Calculation of standard deviation, variance and mean deviation for grouped and ungrouped data. Probability Probability of an event, addition and multiplication theorems of probability, Baye's theorem, probability distribution of a random variate, Bernoulli trials and Binomial distribution.

UNIT 15 Trigonometry Trigonometrical identities and equations. Trigonometrical functions. Inverse trigonometrical functions and their properties. Heights and Distances.

UNIT 16 Mathematical Reasoning Statements, logical operations and implies, implied by, if and only if. Understanding of tautology, contradiction, converse and contra positive.

HOW THIS BOOK IS

USEFUL FOR YOU ? As the name suggest, this is the perfect book for your recapitulation of the whole syllabus, as it provides you a capsule course on the subject covering the syllabi of JEE Main, with the smartest possible tactics as outlined below:

1.

REVISION PLAN The book provides you with a practical and sound revision plan. The chapters of the book have been designed day-wise to guide the students in a planned manner through day-by-day, during those precious 35-40 days. Every day you complete a chapter/a topic, also take an exercise on the chapter. So that you can check & correct your mistakes, answers with hints & solutions also have been provided. By 37th day from the date you start using this book, entire syllabus gets revisited. Again, as per your convenience/preparation strategy, you can also divide the available 30-35 days into two time frames, first time slot of 3 weeks and last slot of 1 & 1/2 week. Utilize first time slot for studies and last one for revising the formulas and important points. Now fill the time slots with subjects/topics and set key milestones. Keep all the formulas, key points on a couple of A4 size sheets as ready-reckner on your table and go over them time and again. If you are done with notes, prepare more detailed inside notes and go over them once again. Study all the 3 subjects every day. Concentrate on the topics that have more weightage in the exam that you are targeting.

2.

MOCK TESTS Once you finish your revision on 37th day, the book provides you with full length mock tests for day 38th, 39th, & 40th, thereby ensures your total & full proof preparation for the final show. The importance of solving previous years' papers and 10-15 mock tests cannot be overemphasized. Identify your weaknesses and strengths. Work towards your strengths i.e., devote more time to your strengths to be 100% sure and confident. In the last time frame of 1 & 1/2 week, don't take-up anything new, just revise what you have studied before. Be examready with quality mock tests in between to implement your winning strategy.

3.

FOCUS TOPICS Based on past years question paper trends, there are few topics in each subject which have more questions in exam than other. So far Mathematics is concerned it may be summed up as below: Calculus, Trigonometry, Algebra, Coordinate Geometry & Vector 3D. More than 80% of questions are normally asked from these topics. However, be prepared to find a completely changed pattern for the exam than noted above as examiners keep trying to weed out 'learn by rot practice'. One should not panic by witnessing a new pattern , rather should be tension free as no one will have any upper hand in the exam.

4.

IMPROVES STRIKE RATE AND ACCURACY The book even helps to improve your strike rate & accuracy. When solving practice tests or mock tests, try to analyze where you are making mistakes-where are you wasting your time; which section you are doing best. Whatever mistakes you make in the first mock test, try to improve that in second. In this way, you can make the optimum use of the book for giving perfection to your preparation. What most students do is that they revise whole of the syllabus but never attempt a mock and thus they always make mistake in main exam and lose the track.

5.

LOG OF LESSONS During your preparations, make a log of Lesson's Learnt. It is specific to each individual as to where the person is being most efficient and least efficient. Three things are important - what is working, what's not working and how would you like to do in your next mock test.

6.

TIME MANAGEMENT Most candidates who don't make it to good medical colleges are not good in one area- Time Management. And, probably here lies the most important value addition that's the book provides in an aspirant's preparation. Once the students go through the content of the book precisely as given/directed, he/she learns the tactics of time management in the exam. Realization and strengthening of what you are good at is very helpful, rather than what one doesn't know. Your greatest motto in the exam should be, how to maximize your scoring with the given level of preparation. You have to get about 200 plus marks out of a total of about 400 marks for admission to a good NIT (though for a good branch one needs to do much better than that). Remember that one would be doomed if s/he tries to score 400 in about 3 hours.

7.

ART OF PROBLEM SOLVING The book also let you to master the art of problem solving. The key to problem solving does not lie in understanding the solution to the problem but to find out what clues in the problem leads you to the right solution. And, that's the reason Hints & Solutions are provided with the exercises after each chapter of the book. Try to find out the reason by analyzing the level of problem & practice similar kind of problems so that you can master the tricks involved. Remember that directly going though the solutions is not going to help you at all.

8.

POSITIVE PERCEPTION The book put forth for its readers a 'Simple and Straightforward' concept of studies, which is the best possible, time-tested perception for 11th hour revision / preparation. The content of the book has been presented in such a lucid way so that you can enjoy what you are reading, keeping a note of your already stressed mind & time span. Cracking JEE Main is not a matter of life and death. Do not allow panic and pressure to create confusion. Do some yoga and prayers. Enjoy this time with studies as it will never come back.

DAY ONE

Sets, Relations and Functions Learning & Revision for the Day u

Sets

u

Law of Algebra of Sets

u

Composition of Relations

u

Venn Diagram

u

Cartesian Product of Sets

u

Functions or Mapping

u

Operations on Sets

u

Relations

u

Composition of Functions

Sets ●

●

●

A set is a well-defined class or collection of the objects. Sets are usually denoted by the symbol A, B, C, ... and its elements are denoted by a, b , c, … etc. If a is an element of a set A, then we write a ∈ A and if not then we write a ∉ A.

Representations of Sets There are two methods of representing a set : ●

●

In roster method, a set is described by listing elements, separated by commas, within curly braces{≠}. e.g. A set of vowels of English alphabet may be described as {a, e, i, o, u}. elements x. In such a case the set is written as { x : P ( x) holds} or { x| P ( x) holds}, which is read as the set of all x such that P( x) holds. e.g. The set P = {0, 1, 4 , 9, 16,...} can be written as P = { x2 | x ∈ Z }.

Types of Sets ●

●

●

PRED MIRROR

In set-builder method, a set is described by a property P ( x), which is possessed by all its

The set which contains no element at all is called the null set (empty set or void set) and it is denoted by the symbol ‘φ ’ or ‘{}’ and if it contains a single element, then it is called singleton set. A set in which the process of counting of elements definitely comes to an end, is called a finite set, otherwise it is an infinite set. Two sets A and B are said to be equal set iff every element of A is an element of B and also every element of B is an element of A. i.e. A = B, if x ∈ A ⇔ x ∈ B.

Your Personal Preparation Indicator u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

2

●

●

●

●

●

●

ONE

40

A ∩ B = { x : x ∈ A and x ∈ B}.

i.e.

Equivalent sets have the same number of elements but not exactly the same elements.

A

A set that contains all sets in a given context is called universal set (U).

B

U

Let A and B be two sets. If every element of A is an element of B, then A is called a subset of B, i.e. A ⊆ B. If A is a subset of B and A ≠ B, then A is a proper subset of B. i.e. A ⊂ B. The null set φ is a subset of every set and every set is a subset of itself i.e. φ ⊂ A and A ⊆ A for every set A. They are called improper subsets of A.

A∩B ●

●

If S is any set, then the set of all the subsets of S is called the power set of S and it is denoted by P(S ). Power set of a given set is always non-empty. If A has n elements, then P( A) has 2 n elements.

If A ∩ B = φ, then A and B are called disjoint sets. Let U be an universal set and A be a set such that A ⊂ U. Then, complement of A with respect to U is denoted by A′ or Ac or A or U − A. It is defined as the set of all those elements of U which are not in A. U

A′

A NOTE

• The set { φ} is not a null set. It is a set containing one element φ.

• Whenever we have to show that two sets A and B are equal show that A ⊆ B and B ⊆ A.

●

• If a set A has m elements, then the number m is called cardinal number of set A and it is denoted by n( A). Thus, n( A) = m.

The difference A − B is the set of all those elements of A which does not belong to B. i.e. A − B = { x : x ∈ A and x ∉ B} and B − A = { x : x ∈ B and x ∉ A}. U

Venn Diagram The combination of rectangles and circles is called Venn Euler diagram or Venn diagram. In Venn diagram, the universal set is represented by a rectangular region and a set is represented by circle on some closed geometrical figure. Where, A is the set and U is the universal set. U

B–A

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

i.e.

A ∆ B = ( A − B) ∪ ( B − A) A

B

Operations on Sets ●

The union of sets A and B is the set of all elements which are in set A or in B or in both A and B. i.e. A ∪ B = { x : x ∈ A or x ∈ B} A

B

U

A∆B

Law of Algebra of Sets If A, B and C are any three sets, then

1. Idempotent Laws (i) A ∪ A = A A∪B

●

The intersection of A and B is the set of all those elements that belong to both A and B.

B

A

A–B ●

A

U

B

A

(ii) A ∩ A = A

2. Identity Laws (i) A ∪ φ = A

(ii) A ∩ U = A

3. Distributive Laws (i) A ∪ (B ∩ C) = ( A ∪ B) ∩ ( A ∪ C) (ii) A ∩ (B ∪ C) = ( A ∩ B) ∪ ( A ∩ C)

U

3

DAY 4. De-Morgan’s Laws (i) (ii) (iii) (iv)

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

5. Associative Laws

Relations ●

●

●

(i) ( A ∪ B) ∪ C = A ∪ (B ∪ C) (ii) A ∩ (B ∩ C) = ( A ∩ B) ∩ C

●

Let A and B be two non-empty sets, then relation R from A to B is a subset of A × B, i.e. R ⊆ A × B. If (a, b ) ∈ R, then we say a is related to b by the relation R and we write it as aRb. Domain of R = {a :(a, b ) ∈ R} and range of R = {b : (a, b ) ∈ R}. If n( A) = p and n(B) = q , then the total number of relations from A to B is 2 pq .

6. Commutative Laws (i) A ∪ B = B ∪ A (iii) A ∆ B = B ∆ A

(ii) A ∩ B = B ∩ A

Types of Relations ●

Let A be any non-empty set and R be a relation on A. Then,

(i) R is said to be reflexive iff ( a, a) ∈ R, ∀ a ∈ A.

Important Results on Operation of Sets 1. A − B = A ∩ B ′

(ii) R is said to be symmetric iff

2. B − A = B ∩ A ′ 3. A − B = A ⇔ A ∩ B = φ 4. ( A − B) ∪ B = A ∪ B 5. ( A − B) ∩ B = φ 6. A ⊆ B ⇔ B ′ ⊆ A ′ 7. ( A − B) ∪ ( B − A) = ( A ∪ B) − ( A ∩ B)

⇒

(iii) R is said to be a transitive iff ( a, b) ∈ R and (b, c) ∈ R ⇒ ( a, c) ∈ R, ∀ a, b, c ∈ A i.e. aRb and bRc ⇒ aRc, ∀ a, b, c ∈ A. ●

8. n ( A ∪ B) = n ( A) + n ( B) − n ( A ∩ B) 9. n ( A ∪ B) = n ( A) + n ( B) ⇔ A and B are disjoint sets. 10. n ( A − B) = n ( A) − n ( A ∩ B) 11. n ( A ∆ B) = n ( A) + n ( B) − 2n ( A ∩ B)

●

The relation I A = {(a, a) : a ∈ A} on A is called the identity relation on A. R is said to be an equivalence relation iff

(i) it is reflexive i.e. ( a, a) ∈ R, ∀ a ∈ A. (ii) it is symmetric i.e. ( a, b) ∈ R ⇒ (b, a) ∈ R, ∀ a, b ∈ A (iii) it is transitive

12. n ( A ∪ B ∪ C ) = n ( A) + n ( B) + n (C ) − n ( A ∩ B) − n ( B ∩ C) − n ( A ∩ C) + n ( A ∩ B ∩ C) 13. n ( A ′ ∪ B ′ ) = n ( A ∩ B) ′ = n (U ) − n ( A ∩ B) 14. n ( A ′ ∩ B ′ ) = n ( A ∪ B) ′ = n (U ) − n ( A ∪ B)

Cartesian Product of Sets Let A and B be any two non-empty sets. Then the cartesian product A × B, is defined as set of all ordered pairs (a, b ) such that a ∈ A and b ∈ B. i.e. A × B = {(a, b ) : a ∈ A and b ∈ B} B × A = {(b , a) : b ∈ B and a ∈ A} and A × A = {(a, b ) : a, b ∈ A}. A × B = φ, if either A or B is an empty set. If n ( A) = p and n (B) = q , then n ( A × B) = n( A) ⋅ n(B) = pq . A × (B ∪ C) = ( A × B) ∪ ( A × C) A × (B ∩ C) = ( A × B) ∩ ( A × C) A × (B − C) = ( A × B) − ( A × C) ( A × B) ∩ (C × D) = ( A ∩ C) × (B ∩ D). ●

●

●

●

●

●

●

●

(a, b ) ∈ R (b , a) ∈ R, ∀ a, b ∈ A

i.e.

( a, b) ∈ R and (b, c) ∈ R

⇒

( a, c) ∈ R, ∀ a, b, c ∈ A

Inverse Relation Let R be a relation from set A to set B, then the inverse of R, denoted by R −1 , is defined by R −1 = {(b , a) : (a, b ) ∈ R}. Clearly, (a, b ) ∈ R ⇔ (b , a) ∈ R −1 . NOTE

• The intersection of two equivalence relations on a set is an equivalence relation on the set. • The union of two equivalence relations on a set is not necessarily an equivalence relation on the set. • If R is an equivalence relation on a set A, then R −1 is also an equivalence relation A.

Composition of Relations Let R and S be two relations from set A to B and B to C respectively, then we can define a relation SoR from A to C such that (a, c) ∈ SoR ⇔ ∃ b ∈ B such that (a, b ) ∈ R and (b , c) ∈ S . This relation is called the composition of R and S.

RoS ≠ SoR

4

ONE

40

Functions or Mapping ●

●

If A and B are two non-empty sets, then a rule f which associates each x ∈ A, to a unique member y ∈ B, is called a function from A to B and it is denoted by f : A → B.

i.e. f : A → B is a many-one function, if it is not a one-one function. ●

f is said to be onto function or surjective function, if each element of B has its pre-image in A. A

The set A is called the domain of f (D f ) and set B is called the codomain of f (C f ).

●

The set consisting of all the f -images of the elements of the domain A, called the range of f (R f ).

NOTE

• A relation will be a function, if no two distinct ordered pairs have the same first element. • Every function is a relation but every relation is not necessarily a function. • The number of functions from a finite set A into finite set B is { n(B )}n ( A).

● ●

● ●

different elements of A have different images in B. A

B b1

f

a2

b2

a3

b3

a4

a2

b2

a3

b3

Find the range of f ( x) and show that range of f ( x) = codomain of f ( x).

f is said to be one-one function or injective function, if

a1

●

Any polynomial function of odd degree is always onto. The number of onto functions that can be defined from a finite set A containing n elements onto a finite set B containing 2 elements = 2 n − 2 . If n ( A) ≥ n (B), then number of onto function is 0. If A has m elements and B has n elements, where m < n, then number of onto functions from A to B is nm − nC1 (n − 1)m + nC2 (n − 2)m − ..., m < n. f is said to be an into function, if there exists atleast one element in B having no pre-image in A. i.e. f : A → B is an into function, if it is not an onto function. A a1

b4

Methods to Check One-One Function Method I

If f ( x) = f ( y ) ⇒ x = y , then f is one-one.

Method II A function is one-one iff no line parallel to X-axis meets the graph of function at more than one point. ●

The number of one-one function that can be defined from a  n( B )

finite set A into finite set B is  0, ●

A a1

f

●

B b1

a2

b2

a3

b3

a4

b4

a5

b5

B

f

b1

a2

b2

a3

b3

a4

b4

a5

b5

f is said to be a bijective function, if it is one-one as well as onto.

NOTE

• If f : A → B is a bijective, then A and B have the same number of elements.

Pn( A ) , if n (B) ≥ n ( A) . otherwise

f is said to be a many-one function, if two or more elements of set A have the same image in B.

b1

Method to Check Onto Function

Different Types of Functions Let f be a function from A to B, i.e. f : A → B. Then,

B f

a1

• If n ( A) = n (B ) = m, then number of bijective map from A to B is m!.

Composition of Functions Let f : A → B and g : B → C are two functions. Then, the composition of f and g, denoted by gof : A → C, is defined as, gof ( x) = g[ f ( x)], ∀ x ∈ A. NOTE

• gof is defined only if f ( x ) is an element of domain of g. • Generally, gof ≠ fog.

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 

1 If Q = x : x = 

1  , where y ∈ N , then y 

(a) 0 ∈Q

(b) 1∈Q

(c) 2∈Q

(d)

2 ∈Q 3

2 If P ( A ) denotes the power set of A and A is the void set, then what is number of elements in P {P {P {P ( A )}}}? (a) 0

(b) 1

(c) 4

(d) 16

3 If X = {4n − 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 j

(a) N

(b) Y-X

(c) X

JEE Mains 2014

(d) Y

4 If A, B and C are three sets such that A ∩ B = A ∩ C and A ∪ B = A ∪ C, then (a) A = C

(b) B = C

(c) A ∩ B = φ (d) A = B

5 Suppose A1, A2,… , A 30 are thirty sets each having

5 elements and B1, B2,… , Bn are n sets each having 30

n

i =1

j =1

3 elements. Let ∪ A i = ∪ Bj = S and each element of S belongs to exactly 10 of Ai ’s and exactly 9 of Bj ’ s. The j NCERT Exemplar value of n is equal to (a) 15 (c) 45

(b) 3 (d) None of these

6 If A and B are two sets and A ∪ B ∪ C = U. Then, {( A − B ) ∪ (B − C ) ∪ (C − A )}′ is equal to (a) A ∪ B ∪ C (c) A ∩ B ∩ C

(b) A ∪ (B ∩ C) (d) A ∩ (B ∪ C)

7 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 n( X ) is equal to (a) 600

(b) 700

(c) 800

(d) 900

8 If n( A ) = 1000, n(B ) = 500, n( A ∩ B ) ≥ 1 and n( A ∪ B ) = P , then (a) 500 ≤ P ≤ 1000 (c) 1000 ≤ P ≤ 1498

(b) 1001 ≤ P ≤ 1498 (d) 1000 ≤ P ≤ 1499

9 If n( A ) = 4, n(B ) = 3, n( A × B × C ) = 24, then n(C ) is equal to (a) 2

(b) 288

(c) 12

(d) 1

10 If R = {( 3 , 3) ,( 6, 6), ( 9, 9), (12 , 12), ( 6 , 12), ( 3 , 9), (3,12), (3, 6)} is a relation on the set A = {3 , 6 , 9 , 12}. The relation is (a) (b) (c) (d)

an equivalence relation reflexive and symmetric reflexive and transitive only reflexive

11 Let R = {( x , y ) : x , y ∈ N and x 2 − 4xy + 3y 2 = 0}, where N is the set of all natural numbers. Then, the relation R is j

JEE Mains 2013

(a) reflexive but neither symmetric nor transitive (b) symmetric and transitive (c) reflexive and symmetric (d) reflexive and transitive

12 If g ( x ) = 1 + x and f {g ( x )} = 3 + 2 x + x , then f ( x ) is equal to (a) 1 + 2 x 2 (c) 1 + x

(b) 2 + x 2 (d) 2 + x

13 Let f ( x ) = ax + b and g ( x ) = cx + d , a ≠ 0, c ≠ 0. Assume a = 1, b = 2 , if ( fog ) ( x ) = ( gof ) ( x ) for all x. What can you say about c and d? (a) c and d both arbitrary (c) c is arbitrary and d = 1

(b) c = 1and d is arbitrary (d) c = 1, d = 1

14 If R is relation from {11, 12 , 13 } to {8 , 10 , 12} defined by y = x − 3. Then, R −1 is (a) {(8 , 11), (10, 13)} (c) {(10, 13), (8 , 11)}

(b) {(11, 18), (13 , 10)} (d) None of these

15 Let R be a relation defined by R = {(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)}, then R −1 OR is

(a) {(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)} (b) {(1, 1), (4, 4), (7, 7), (3, 3)} (c) {(1, 5), (1, 6), (3, 6)} (d) None of the above

16 Let A be a non-empty set of real numbers and f : A → A be such that f (f ( x )) = x , ∀ x ∈ R. Then, f ( x ) is (a) a bijection (c) onto but not one-one

(b) one-one but not onto (d) neither one-one nor onto

17 The function f satisfies the functional equation  x + 59 3f ( x ) + 2f   = 10x + 30 for all real x ≠ 1. The value  x −1  of f (7) is (a) 8

(b) 4

(c) − 8

(d) 11

18 The number of onto mapping from the set A = {1, 2, ...100} to set B = {1, 2} is (a) 2100 − 2

(b) 2100

(c) 2 99 − 2

(d) 2 99

19 Let f : R − {n} → R be a function defined by f ( x ) = where m ≠ n . Then, (a) f is one-one onto (c) f is many-one onto

x −m , x −n

(b) f is one-one into (d) f is many-one into

20 A function f from the set of natural numbers to integers n − 1 , when n is odd  defined by f (n ) =  2 is n  − , when n is even  2 (a) one-one but not onto (c) both one-one and onto

(b) onto but not one-one (d) neither one-one nor onto

6

ONE

40

21 Let f : N → N defined by f ( x ) = x 2 + x + 1 , x ∈ N, then f is (a) one-one onto (c) one-one but not onto

(b) many-one onto (d) None of these

22 Let R be the real line. Consider the following subsets of the plane R × R .

and

S = {( x , y ): y = x + 1and 0 < x < 2} T = {( x , y ): x − y is an integer}

Which one of the following is true? (a) (b) (c) (d)

T is an equivalence relation on R but S is not Neither S nor T is an equivalence relation on R Both S and T are equivalence relations on R S is an equivalence relation on R but T is not

23 Consider the following relations R = {(x , y) | x and y are real numbers and x = wy for some rational number w};  m p  S =  ,  m, n , p and q are integers such that n, q ≠ 0  n q  and qm = pn }. Then, (a) R is an equivalence relation but S is not an equivalence relation (b) neither R nor S is an equivalence relation

(c) S is an equivalence relation but R is not an equivalence relation (d) R and S both are equivalence relations

2 + x , x ≥ 0 , then f (f ( x )) is given by 4 − x , x < 0

24 If f ( x ) = 

 4 + x, (a) f (f (x)) =   6 − x,  4 − x, (c) f (f (x)) =   x,

 4 + x, x ≥ 0 (b) f (f (x)) =  x 0, k ≠ 1 is a real | z − z1| number, then = k represents a circle. | z − z2| For k = 1, it represents perpendicular bisector of the segment joining A(z1 ) and B (z2 ). 8. If end points of diameter of a circle are A(z1 ) and B(z 2) and P(z) be any point on the circle, then equation of circle in diameter form is (z − z1 ) (z − z2 ) + (z − z 2) (z − z1 ) = 0

13

DAY

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 Real part of (a) −

1 2

1 is 1 − cos θ + i sin θ (b)

1 2

(c)

2 A value of θ, for which π (a) 3 13

3

∑ (i

n

11 If z − 1 tanθ / 2 2

−1

(c) sin

1  (d) sin    3

3 4

−1

+ i n + 1 ) is equal to

n =1

(b) i − 1

(a) i

(c) −i

(b) 1

(d) −2

(c) 2

5 If z1 ≠ 0 and z 2 are two complex numbers such that a purely imaginary number, then

(a) 2

6 If f ( z ) =

(b) 5

z2 is z1

2z1 + 3z 2 is equal to 2z1 − 3z 2 j

(c) 3

JEE Mains 2013

(d) 1

7−z , where z = 1 + 2i , then | f ( z )| is equal to 1− z 2

|z | (a) 2 (c) 2| z |

(d) None of these

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

(c) 1

(d) 3 / 4

8 If a complex number z satisfies the equation z + 2 z + 1 + i = 0, then z is equal to (a) 2

(b)

3

(c)

5

j

JEE Mains 2013

(d) 1

9 If α and β are two different complex numbers such that β −α is equal to | α | = 1, | β | = 1, then the expression 1 − αβ 1 2 (c) 2

(b) 1

(a)

z −1 10 If | z | = 1 and ω = (where z ≠ −1), then Re(ω ) is z +1 (b) −

(c)

2 z+1

2

2

1 2

j

JEE Mains 2014

is equal to 5/2 lies in the interval (1, 2) is strictly greater than 5/2 is strictly greater than 3/2 but less than 5/2

(a) 8

(b) 18

(c) 10

(d) 5

14 If z < 3 − 1, then z + 2z cos α is 2

(b) 3 + 1 (d) None of these

(a) less than 2 (c) 3 − 1

15 The number of complex numbers z such that z − 1 = z + 1 = z − i , is (a) 0

(b) 1

(d) ∞

(c) 2

16 Number of solutions of the equation z (a) 1

(b) 2

2

+ 7z = 0 is/are

(c) 4

(d) 6

17 If z z + ( 3 − 4i )z + ( 3 + 4i )z = 0 represent a circle, the area of the circle in square units is (c) 25 π 2

(b) 10π

 π  5

(d) 25 π

 π  5

18 If z = 1 + cos   + i sin  , then {sin (arg( z ))} is equal to (a) (c)

10 − 2 5 4 5+1 4

5 −1 4

(b)

(d) None of these

19 If z is a complex number of unit modulus and argument 1 + z  θ, then arg   equals to 1 + z  (a) −θ

π (b) −θ 2

j

(c) θ

JEE Mains 2013

(d) π − θ

20 Let z and ω are two non-zero complex numbers such that

(d) None of these

(a) 0

(d) 2 +

(c) 2

equal to

7 If 8 iz 3 + 12z 2 − 18z + 27i = 0, then the value of | z | is (b) 2 / 3

5 +1

minimum value of z +

(a) 5 π

(b) | z |

(a) 3 / 2

(b)

12 If z is a complex number such that z ≥ 2, then the

13 If | z1| = 2, | z 2 | = 3 then z1 + z 2 + 5 + 12i is less than or

(d) 0

z −1 4 If is a purely imaginary number ( where, z ≠ −1), then z +1 the value of | z | is (a) −1

3 +1

(a)

(d) 2

2 + 3i sin θ is purely imaginary, is 1 − 2i sin θ

π (b) 6

4 = 2, then the maximum value of | z | is j AIEEE 2009 z

1 z+1

2

(d) None of these

z = ω and arg z + arg ω = π, then z equals (a) ω (c) − ω

(b) ω (d) − ω

21 If z − 1 = 1, then arg ( z ) is equal to 1 arg (z) 2 1 (c) arg (z − 1) 2 (a)

(b)

1 arg (z + 1) 3

(d) None of these

14 40 15

22 Let z = cos θ + i sin θ. Then the value of ∑ Im ( z 2 m −1 ) at,

 1 z

33 If Re   = 3 , then z lies on

m =1

θ = 2°, is 1 (a) sin 2 °

1 (b) 3 sin 2 °

1 (c) 2 sin 2 °

1 (d) 4 sin 2 °

(i )

23 If z = (i )(i ) , where i = −1, then | z | is equal to (b) e − π / 2

(a) 1

(d) e π / 2

(c) 0

π π  1 + i sin + cos  8 8  equals to 24   1 − i sin π + cos π   8 8

(a) a circle (c) a parabola

(d) 1

25 If 1, α 1, α 2 , K , α n − 1 are the nth roots of unity, then ( 2 − α 1 )( 2 − α 2 ) K ( 2 − α n −1 ) is equal to (c) 2 n + 1

(b) 2 n

(a) n

(d) 2 n − 1

26 If ω( ≠ 1) is a cube root of unity and (1 + ω )7 = A + Bω.

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

(c) (−1, 1)

(b) (1, 0)

(d) (0, 1)

27 If α , β ∈C are the distinct roots of the equation x 2 − x + 1 = 0, then α 101 + β107 is equal to (a) −1

(b) 0

j

(c) 1

  r=1 25

28 If x 2 + x + 1 = 0, then ∑  x r +

JEE Mains 2018

(d) 2 2

1  is equal to xr 

(b) 25 ω (d) None of these

(a) 25 (c) 25 ω2

29 Let ω be a complex number such that 2ω + 1 = z, 1 1 1 2 where z = −3. If 1 −ω − 1 ω 2 = 3k, then k is equal to 1 ω2 ω7 j JEE Mains 2017 (a) −z

(c) −1

(b) z

(d) 1

1+ ω ω2 1 + ω2 30 The value − ω − (1 + ω 2 ) (1 + ω) , where ω is cube − 1 − (1 + ω 2 ) 1 + ω root of unity, is equal to (a) 2 ω

(b) 3 ω2

(c) − 3 ω2

(d) 3ω

31 If a , b and c are integers not all equal and ω is a cube root of unity (where, ω ≠ 1), then minimum value of | a + bω + cω 2 | is equal to (a) 0

32 Let ω = e

(b) 1

(c)

3 2

(d)

1 2

iπ /3

, and a, b, c, x , y , z be non-zero complex numbers such that: a + b + c = x ; a + b ω + cω = y ; a + b ω + c ω = z 2

Then the value of (a) 1

x

2

a

2

(b) 2

+ y

2

+ b

2

2

+ z

2

+ c

2

(c) 3

is: (d) 4

z lie on 1− z 2

a line not passing through the origin |z | = 2 the X-axis theY-axis

36 If ω =

Then, ( A, B ) is equal to (a) (11 ,)

(b) a straight line (d) None of these

35 If | z | = 1 and z ≠ ± 1, then all the values of

(c) −1

(b) 0

34 If the imaginary part of ( 2z + 1) / (iz + 1) is −2, then the locus of the point representing z in the complex plane is

8

(a) 2 8

(a) circle with centre onY-axis (b) circle with centre on X-axis not passing through origin (c) circle with centre on X-axis passing through origin (d) None of the above

z z−

i 3

and | ω | = 1, then z lies on

(a) a circle (c) a parabola

(b) an ellipse (d) a straight line

37 If z 1 and z 2 are two complex numbers such that z1 z 2 + = 1, then z 2 z1 (a) (b) (c) (d)

z 1, z 2 are collinear z 1, z 2 and the origin form a right angled triangle z 1, z 2 and the origin form an equilateral triangle None of the above

38 A complex number z is said to be unimodular, if z = 1. Suppose z1 and z 2 are complex numbers such that z1 − 2z 2 is unimodular and z 2 is not unimodular. 2 − z 1z 2 Then, the point z1 lies on a (a) (b) (c) (d)

j

JEE Mains 2015

straight line parallel to X −axis straight line parallel toY −axis circle of radius 2 circle of radius 2

39 If | z 2 − 1 | = | z |2 +1, then z lies on (a) a real axis (c) a circle

(b) an ellipse (d) imaginary axis

40 Let z satisfy z = 1 and z = 1 − z

j

JEE Mains 2013

Statement I z is a real number. Statement II Principal argument of z is π /3. (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

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 For positive integers n1 and n 2 the value of the expression

8 If a complex number z lies in the interior or on the

(1 + i )n1 + (1 + i 3 )n1 + (1 + i 5 )n 2 + (1 + i 7 )n 2 where i = −1, is a real number iff (a) n1 = n2

boundary of a circle of radius 3 and centre at ( − 4 , 0), then the greatest and least value of | z + 1| are

(b) n2 = n2 − 1 (c) n1 = n2 + 1 (d) ∀n1 and n2

(a) 5, 0

z is real, then the point represented by the z −1 complex number z lies

2 If z ≠ 1 and

(a) on the imaginary axis (b) either on the real axis or on a circle passing through the origin (c) on a circle with centre at the origin (d) either on the real axis or on a circle not passing through the origin

2π 2π 3 Let ω be the complex number cos . Then the + i sin 3 3 number of distinct complex numbers z satisfying z +1 ω ω2 1 = 0 is equal to ω z + ω2 ω2

1

(a) 0

(a) 2

(d) 4

(c) x − y < 0

(a) 3eiπ / 4 + 4i (c) (4 + 3i )eiπ / 4

11 If 1, a1, a 2 ... a n −1 are n th roots of unity, then 1 1 1 + + ... + equals to 1 − a1 1 − a 2 1 − a n −1

(c) 3 / 2

(d) None of these

6 Let z = x + iy be a complex number where x and y are

(b) 32

(c) 40

+ + + +

y2 y2 y2 y2

(c)

n n −1

(d) None of these

2

(b) ω or ω /|ω | 2 (d) None of these

  k =1

2kπ 2kπ  + i cos  is 11 11 

(b) −1

(a) 1

(c) − i

(d) i

14 Let z1 and z 2 be roots of the equation z + pz + q = 0, 2

p, q ∈ C. Let A and B represents z1 and z 2 in the complex plane. If ∠AOB = α ≠ 0 and OA = OB; O is the origin, then p 2 / 4q is equal to (a) sin2 (α / 2) (b) tan2 (α / 2) (c) cos2 (α / 2) (d) None of these

15 If 1, ω and ω 2 are the three cube roots of unity α , β, γ are the cube roots of p, q < 0, then for any x , y , z the  x α + y β + z γ expression   is equal to  x β + y γ + z α

then the locus of z is x2 x2 x2 x2

n −1 2

10

(d) 80

7 If α + i β = cot −1( z ), where z = x + iy and α is a constant, (a) (b) (c) (d)

(b)

13 The value of ∑  sin

integers. Then the area of the rectangle whose vertices are the roots of the equation zz 3 + zz 3 = 350 is (a) 48

2n − 1 n

(a) ω or ω (c) ω or ω /|ω | 2

(d) x + y > 0

such that arg ( z − z1 ) / ( z − z 2 ) = π / 4, then z − 7 − 9i is equal to (b) 3 2

(d) 6

(b) (3 − 4i )eiπ / 4 (d) (3 + 4i )eiπ / 4

2

5 Let z1 = 10 + 6i , z 2 = 4 + 6i . If z is any complex number

(a) 18

(c) 5

the North-East (N 45° E) direction. From there, he walks a distance of 4 units towards the North-West (N 45° W) direction to reach a point P. Then the position of P in the Argand plane is

log1/ 2 z − 1 > log1/ 2 z − i is given by (b) x − y > 0

(b) 3

12 For z ,ω ∈ C, if z ω − ω z = z − ω, then z is equal to

4 The locus of z = x + iy which satisfying the inequality (a) x + y < 0

(d) None of these

10 A man walks a distance of 3 units from the origin towards

(a) (c) 2

(c) 6, 0

then the minimum value of 2z − 6 + 5i is

z +ω (b) 1

(b) 6, 1

9 If z is any complex number satisfying z − 3 − 2i ≤ 2,

2

− x cot 2 α − 1 = 0 − 2 x cot α − 1 = 0 − 2 x cot 2 α + 1 = 0 − 2 x cot 2 α − 1 = 0

(b) ω

(a) 1

(c) ω 2

(d) None of these

ANSWERS SESSION 1

1. 11. 21. 31.

SESSION 2

1. (d) 11. (b)

(b) (b) (c) (b)

2. 12. 22. 32.

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

2. (b) 12. (b)

3. 13. 23. 33.

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

3. (b) 13. (c)

4. 14. 24. 34.

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

4. (b) 14. (c)

5. 15. 25. 35.

(d) (b) (d) (d)

5. (b) 15. (c)

6. 16. 26. 36.

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

6. (a)

7. 17. 27. 37.

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

7. (d)

8. 18. 28. 38.

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

8. (c)

9. 19. 29. 39.

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

9. (c)

10. 20. 30. 40.

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

10. (d)

16 40

Hints and Explanations SESSION 1

3

13

∑ (i

n

+ i n +1 ) = (1 + i )

n =1

13

∑i

n

n =1

= (1 + i )

i (1 − i 13 ) 1− i

= i − 1 [Q i 13 = i , i 2 = −1]

4 Let z = x + iy x + iy − 1 ( x − 1) + iy z−1 = = z + 1 x + iy + 1 ( x + 1) + iy ( x + 1) − iy × ( x + 1) − iy ( x − 1)( x + 1) − iy ( x − 1) + iy ( x + 1) − i 2 y 2 ( x + 1)2 − i 2 y 2

= =

x2 − 1 + iy ( x + 1 − x + 1) + y 2 ( x + 1)2 + y 2

i (2 y ) z − 1 ( x2 + y 2 − 1) = + z + 1 ( x + 1)2 + y 2 ( x + 1)2 + y 2 z−1 is purely imaginary. Since, z+1 z − 1 ∴ Re   =0  z + 1 2 2 x + y −1 ⇒ =0 ( x + 1)2 + y 2 ⇒

⇒

x + y −1 = 0 2

2

x2 + y 2 = 1 |z |2 = 1 ⇒ |z |= 1

10 Given,|z | = 1 ⇒

5 Given, z2 is a purely imaginary

zz = 1

Let z = ni. Then, 2z1 + 3z2 2z1 − 3z2

6

=

z2 2 + 3ni z1 = = z2 2 − 3ni 2 − 3⋅ z1 2+ 3⋅

4 + 9n2

=

4 + 9n2

= ∴

=1

z+1

2

=0

[Q zz = 1]

z

z

4 4 |z | ≤  z −  + z  |z |  4 |z | ≤ 2 + |z|

⇒ ⇒

|z |2 − 2 |z | − 4 ≤0 |z |

⇒

Since,|z | > 0 ⇒ |z |2 − 2 |z | − 4 ≤ 0 ⇒ [|z | − ( 5 + 1)] [|z |− (1 − 5)] ≤ 0 ⇒

Now,| f (z )| =

5]

7 Given, 8 iz 3 + 12z 2 − 18z + 27i = 0 4z 2 (2iz + 3) + 9 i (2iz + 3) = 0 (2iz + 3) (4z2 + 9 i ) = 0 2iz + 3 = 0 or 4z2 + 9 i = 0 3 |z | = 2

∴

2zz − 2

2

Re(ω ) = 0. 

7− z and z = 1 + 2i 1 − z2 7 − (1 + 2i ) ∴ f (z ) = 1 − (1 + 2i )2 6 − 2i 6 − 2i = = 1 − (1 − 4 + 4i ) 4 − 4i 6 − 2i 1+ i 6 + 4i + 2 = × = 4(1 − i ) (1 + i ) 4(12 − i 2 ) 8 + 4i 1 = = (2 + i ) 4(2) 2

⇒ ⇒ ⇒

z+1

  z − 4  + 4  11 |z | =     

Given, f (z ) =

4+ 1 5 |z | = = 2 2 2 [Qz = 1 + 2i , given ⇒|z |=

z−1 z−1 + z+1 z+1 (z − 1)(z + 1) + (z − 1)(z + 1)

Now, 2Re(ω ) = ω + ω =

z1

1−

5 ≤ |z | ≤

5+ 1

12 z ≥ 2 is the region on or outside circle whose centre is (0,0) and radius is 2. 1 Minimum z + is distance of z, which 2 1 lie on circle z = 2 from  − , 0 .  2  Y

8 We have, ( x + iy ) + 2 x + iy + 1 + i = 0 [put z = x + iy ] ⇒ ( x + iy ) + 2 ( x + 1)2 + y 2 + i = 0 ⇒

(– 1 2

2

,0

(0, 0)

(2, 0)

X

y+1= 0 x + 2 ( x + 1) + (−1)2 = 0 2

and y = −1 ⇒ x2 = 2[( x + 1)2 + 1] ⇒ x2 = 2 x2 + 4 x + 4 2 ⇒ x + 4 x + 4 = 0 ⇒ ( x + 2)2 = 0 ⇒ x = −2 ∴ z = −2 − i ⇒ z = 4 + 1 = 5 β −α β −α 9 = 1 − αβ β ⋅ β − αβ

β −α β (β − α )

=

Y′

1 ∴ Minimum z + 2 1   = Distance of  − , 0 from (−2, 0)  2  2

=  Q |β | = 1  and |β|2 = ββ = 1  

=

A

(–2, 0)

x + 2 ( x + 1) + y = 0 2

and ⇒

X′

(

1 1 − cos θ + i sin θ 1 = 2sin2 (θ / 2) + 2i sin(θ / 2)cos(θ / 2) 1 1 = 2i sin(θ / 2) [cos(θ / 2) − i sin(θ / 2)] cos(θ / 2) + i sin(θ / 2) 1 1 = = + cot(θ / 2) 2i sin(θ / 2) 2 2i 1 1 = − i ⋅ cot θ / 2 2 2 2 Let z = 2 + 3i sin θ is purely imaginary 1 − 2i sin θ then we have Re (z ) = 0 2 + 3i sin θ Consider, z = 1 − 2i sin θ (2 + 3i sin θ)(1 + 2i sin θ) = (1 − 2i sin θ)(1 + 2i sin θ) (2 − 6sin2 θ) + (4sin θ + 3sin θ )i = 1 + 4sin2 θ Q Re(z ) = 0 2 − 6sin2 θ ∴ =0 1 + 4sin2 θ 1 1 sin2 θ = ⇒ sinθ = ± ⇒ 3 3 −1  1  ⇒ θ = ± sin    3

1 Let z =

⇒ ⇒

1 |β − α| |β − α| = =1 |β| | β − α| |β − α| [Q|z|=|z|]

 −2 + 1  + 0 = 3    2 2

Alternate Method We know,|z1 + z2 |≥||z1 |−|z2|| ∴

z+

1 1 1 ≥ |z| − = |z|− 2 2 2 ≥ z−

1 3 = 2 2

17

DAY ∴

z+

1 3 ≥ 2 2

20 Let z = ω = r and let arg ω = θ

1 3 ∴ Minimum value of z + is ⋅ 2 2

13 Fact: z1 + z2 + ... + z n ∴

≤ z1 + z2 + ... + z n z1 + z2 + (5 + 12i ) ≤ z1 + z2 + 5 + 12i = 2 + 3 + 13 = 18 2

14 Consider z2 + 2z cos α ≤ z + 2 z 2

cos α ≤ z + 2 z < ( 3 − 1)2 + 2( 3 − 1) = 3+ 1−2 3 + 2 3 −2= 2 ∴ z + 2z cos α < 2 2

|z − 1| = |z + 1| Re z = 0 ⇒ x = 0 |z − 1| = |z − i| ⇒ x = y |z + 1| = |z − i| ⇒ y = − x Since, only (0, 0) will satisfy all conditions. ∴ Number of complex number z = 1. 2

16 Given z + 7z = 0 ⇒ z z + 7z = 0 ⇒ z (z + 7) = 0 Case (i) : z = 0, ∴ z = 0 = 0 + i0 Case (ii) : z = −7 ∴ z = −7 + 0i Hence, there is only two solutions. z = 0 and z = −7

17 Given zz + (3 − 4i )z + (3 + 4i )z = 0 Let z = x + iy Then, zz = x2 + y 2 ∴ x2 + y 2 + (3 − 4i )( x + iy ) + (3 + 4i )( x − iy ) = 0 ⇒ x2 + y 2 + 6 x + 8 y = 0 ⇒ ( x + 6 x) + ( y + 8 y ) = 0 2

⇒ ( x + 3)2 + ( y + 4)2 = 32 + 42 ⇒ [ x − (−3)]2 + [ y − (−4)]2 = 52 So, area of circle be π R = 25π [QR = radius = 5] 18 If z = 1 + cos θ + i sin θ, then arg(z ) = θ 2 π/5 π ∴ arg(z ) = = 2 10 ⇒ sin(arg z ) 5−1 π = sin   = sin18° =  10  4 2

19 Given, z = 1 and arg z = θ ∴ z = e iθ and z =

21 Given,|z − 1| = 1 ⇒ z − 1 = e i θ , where arg(z − 1) = θ …(i) ⇒ z = eiθ + 1 ⇒ z = 1 + cos θ + i sin θ [Qe i θ = cos θ + i sin θ] θ θ 2θ = 2 cos + 2i sin ⋅ cos 2 2 2 θ 1 ⇒ arg (z ) = = arg(z − 1) [from Eq. (i)] 2 2

22 Given that z = cos θ + i sin θ = e iθ

15 Let z = x + iy

2

Then, ω = r (cos θ + i sin θ) = re i θ and arg z = π − θ Hence, z = r (cos( π − θ) + i sin( π − θ)) = r (− cos θ + i sin θ) = − r (cos θ − i sin θ) z = −ω

1 z

  1+ z 1+ z  Now, arg    = arg  1 + z 1 + 1   z = arg (z ) = θ

∴

15

∑ lm(z

2 m −1

)=

m=1

15

∑ lm(e

iθ 2 m −1

)

=

∑ lme

i(2 m −1 )θ

m=1

= sin θ + sin 3θ + sin 5θ + ... + sin 29θ 14 ⋅ 2θ   15⋅ 2θ  sin  θ +  sin     2  2  = 2θ sin   2 sin(15θ)sin(15θ) 1 = = [Qθ = 2° ] sin θ 4sin 2° 23 Clearly, i = cos π + i sin π = e iπ /2 2 2 ∴

(i )i = (e iπ /2 )i = e ( i )(i )

Now, (i )

e − π /2

= (i )

i2 ⋅

π

= e − π /2

2

⇒ z = (i )e

− π /2

e − π /2

⇒

|z|=|i | =1 π 24 Let z = cos + i sin π 8 8 1 π π Then, = cos − i sin z 8 8 8  1 + cos π + i sin π  8   8 8  =  1 + z  Now,  −1 π π 1 + z   1 + cos − i sin   8 8 8

8

π π  (1 + z )z   8 =  = z =  cos + i sin    (1 + z )  8 8 π π = cos 8 ⋅ + i sin 8 ⋅ 8 8 [using De-moivre’s theorem] = cos π = −1(Q sin π = 0)

25 Clearly, ( x − 1)( x − α1 )( x − α2 ) …

( x − α n −1 ) = x – 1 Putting x = 2, we get ( 2 − α1 )( 2 − α2 )K( 2 − α n − 1 ) = 2n − 1 n

26 We have, (1 + ω )7 = A + Bω We know that 1 + ω + ω = 0 ∴ 1 + ω = −ω2 ⇒ (− ω2 ) 7 = A + Bω ⇒ − ω14 = A + Bω 2

27 α,β are the roots of x2 − x + 1 = 0 Q Roots of x2 − x + 1 = 0 are − ω, − ω 2 ∴ Let α = −ω and β = − ω 2 ⇒

α101 + β107 = (− ω )101 + (− ω2 )107 = − (ω101 + ω214 ) = − (ω2 + ω ) [Qω3 n + 2 = ω2 and ω3 n+1 = ω] = −(−1) = 1 [1 + ω + ω2 = 0]

28 x2 + x + 1 = 0 ⇒

x = ω , ω2 1 1 So, x + r = ω r + r = − 1 x ω or 2 according as r is not divisible by 3 or divisible by 3. r

∴ Required sum = 17(−1) 2 + 8 ⋅ 2 2 = 49

m=1 15

⇒ − ω2 = A + Bω ⇒ 1 + ω = Α + Βω [Qω14 = ω12 ⋅ ω2 = ω2 ] On comparing both sides, we get A = 1, B = 1

29 Given, z = 2ω + 1 ⇒ω =

−1 + z −1 + 3i ⇒ω = 2 2

[Qz =

−3]

⇒ ω is complex cube root of unity 1 1 1 Now, 1 −ω2 − 1 ω2 = 3k 1 ω2 ω7 ⇒

1 1 1 1 ω ω2 = 3k 1 ω2 ω

Q1 + ω + ω2 = 0   ω7 = ω   Applying R1 → R1 + R2 + R3 , we get 3 1 + ω + ω2 1 + ω + ω2 1 ω ω2 = 3k 2 1 ω ω 3 0 0 ⇒ 1 ω ω2 = 3k 1 ω2 ω ⇒ 3(ω2 − ω 4 ) = 3k ⇒ k = ω2 − ω ⇒ k = −1 − 2ω ⇒ k = −(1 + 2ω ) ⇒ k = − z

30 Using 1 + ω + ω2 = 0, we get 1 + ω ω2 ∆ = 1 + ω2 ω ω2 + ω ω

−ω − ω2 − ω2

Applying C1 → C1 + C2 , 0 ω2 ∆ = 0 ω ω2 + 2ω ω

−ω − ω2 − ω2

= (ω2 + 2ω )(− ω + ω2 ) = − 3ω2

18 40 31 |a + bω + c ω2|2 = (a + bω + c ω2 )

38 Given, z2 is not unimodular i.e. z2 ≠ 1

(a + b ω + c ω ) = (a + bω + c ω2 )(a + bω2 + c ω ) [Q ω = ω2 and ω 2 = ω] 2 2 = a + b + c 2 − ab – bc − ca 1 = [(a − b ) 2 + (b − c ) 2 + (c − a) 2 ] 2 So, it has minimum value 1 for a = b = 1 and c = 2. 2

2

2

z − 2z2 and 1 is unimodular. 2 − z1 z2 ⇒ ⇒

32 Clearly, x + y + z = x x + y y + zz

z1

2

2

2

+ (a + bω + cω )(a + bω + c ω ) 2

2

2

2

= 3( a + b + c ) ⇒

2

2

2

2

x + y

+ z

a + b + c

2

=3

2

 |z| 

 1 z  Q z = |z |2    x ⇒ 2 = 3 ⇒ 3 x2 + 3 y 2 − x = 0 x + y2 So, it is a circle whose centre is on X-axis and passes through the origin. 2z + 1 (2 x + 1) + 2iy 34 = iz + 1 (1 − y ) + ix [(2 x + 1) + 2iy ] ⋅ [(1 − y ) − ix] = (1 − y )2 − i 2 x2 (2 x − y + 1) − (2 x2 + 2 y 2 + x − 2 y )i = 1 + x2 + y 2 − 2 y ∴ Imaginary part − (2 x2 + 2 y 2 + x − 2 y ) = = −2 1 + x2 + y 2 − 2 y

⇒ x + 2 y − 2 = 0, which is a straight line.

35 Clearly,

z z 1 , which = = 1 − z2 zz − z2 z − z

is always imaginary.

36 |ω | = 1 ⇒|z | = z − i

3

It is the perpendicular bisector of the 1 line segment joining (0, 0) to  0,  i.e.  3 the line y =

1 ⋅ 6

37 Given, z1 + z2 = 1 ⇒ z12 + z22 = z1 z2 z2

2

= 4 + z1 2

2

z2

2

− 2z1 z2 − 2z1 z2

2

⇒ ( z2 − 1)( z1 − 4) = 0 Q z2 ∴ z1 Let z1 = Point z1

Either z = z ⇒ real axis or

z1

⇒ z12 + z22 + z32 = z1 z2 + z1 z3 + z2 z3 , where z3 = 0 So, z1 , z2 and the origin form an equilateral triangle.

and

y =±

3 ω = e 2 πi / 3 = imaginary cube root of unity ∴

1 + ω + ω2 = 0 z+1

Now, ∆ =

ω ω2

z = ω ω2

ω z + ω2 1

z z + ω2 1

ω2 1 z+ω

z 1 z+ω

(applying R1 → R1 + R2 + R3 ) 1 1 1 = z ω z + ω2 1 ω2 1 z+ω

1 2

3 2

= z {[(z + ω2 )(z + ω ) − 1] + [ω2 − ω( z + ω)] + [ω − ω2 (z + ω2 )]} = z {z2 + z(ω + ω2 ) + ω3 −1 − ωz − ω2 z} = z3 3 ∴ ∆ = 0 ⇒ z = 0 ⇒ z = 0 is the only solution.

1 3 ± i 2 2 1 3 Now, take, z = + i 2 2  3 / 2 π ∴ θ = tan −1   = 3  1/2  ∴

2

z = z + z ⇒ zz − z − z = 0

i.e. ( x2 + y 2 = 2 x ) represents a circle passing through origin.

40 Let z = x + iy

x2 + y 2 = 1 and 2 x = 1 ⇒ x =

z−1 z −1 zzz − z2 = zzz − z2 2 z (z − z ) − (z − z )(z + z ) = 0 2

Then, |r 2e 2 iθ − 1| = r 2 + 1 ⇒ (r 2 cos 2θ − 1) 2 + (r 2 sin 2θ) 2 = (r 2 + 1) 2 ⇒ r 4 − 2r 2 cos 2θ + 1 = r 4 + 2 r 2 + 1 π ⇒ cos 2θ = − 1 ⇒ θ = 2 π π ⇒ z = r  cos + i sin  = ir  2 2

⇒

⇒ ⇒

⇒ (z − z ) ( z − (z + z )) = 0

≠1 =2 x + iy ⇒ x2 + y 2 = (2)2 lies on a circle of radius 2.

Then, x2 + y 2 = 1 and x + iy = 1 − ( x − iy )

which is purely real ∀ n1 , n2 .

2 2 2 Clearly, z = z

2

39 Let z = re i θ

  33 Given, Re  1  = 3 ⇒ Re  z 2  = 3  z

= 2 − z1 z2

+ 4 z2 − 2z1 z2 − 2z1 z2

= (a + b + c )(a + b + c ) + (a + bω + cω2 )(a + bω + c ω )

2

z1 − 2z2

⇒ (z1 − 2z2 )(z1 − 2z2 ) = (2 − z1 z2 ) 2 (2 − z1 z2 ) [Qzz = z ] ⇒

2

z1 − 2z2 =1 2 − z1 z2

n π n π + ( 2 )n1 cos 1 − i sin i 1   4 4  n π π + ( 2 )n2 cos 2 − i sin n2   4 4  n π n π  + ( 2 )n2 cos 2 + i sin 2  4 4   n1 π  n π n1  = ( 2 ) 2cos + ( 2 )n2 2cos 2    4  4 

z=

4 In the problem, base = 1 / 2∈(0,1)

SESSION 2

∴ z−1 < z− i ⇒ n1

1 Clearly, (1 + i )

+ (1 + i 3 )n1

n1

  π π  +  2  cos  −  + i sin  −     4  4  

n2

π π    2 cos + i sin    4 4   n π n π = ( 2 )n1 cos 1 + i sin 1  4 4  

n2

+

2

⇒ (z − 1)(z − 1) < (z − i )( z + i )

+ (1 + i 5 )n2 + (1 + i 7 )n2 n1 n1 = (1 + i ) + (1 − i ) + (1 + i )n2 + (1 − i )n2 n1 π π   = 2  cos + i sin    4 4     π π  +  2  cos  −  + i sin  −     4  4  

2

z−1 < z− i [Q z

2

= z z]

⇒ (1 + i )z + (1 − i )z > 0 ⇒ (z + z ) + i (z − z ) > 0  z − z  z + z ⇒ >0   + i  2   2   z + z −  z − z > 0 ⇒      2   2i  ⇒ Re(z ) − Im(z ) > 0 ⇒ x − y > 0

5 arg z − (10 + 6i ) = π

 z − (4 + 6i )  4 y−6 π −1 y − 6 − tan −1 = ⇒ tan x − 10 x−4 4 [take z = x + iy ]

19

DAY y−6 y−6 − x − 10 x − 4 = 1 ⇒ ( y − 6)( y − 6) 1+ ( x − 10)( x − 4)

⇒

z − (7 + 9i ) = 3 2

iπ / 4

6 We have, zz = (z + z ) = 350 2

k =1

 5  = 2 Distance of z from  3, −    2    where z lies on circle (i). 9 ∴ min 2z − 6 + 5i = 2PA = 2  − 2 = 5 2 

⇒ x2 + y 2 − 14x − 18y + 112 = 0 ⇒ ( x − 7)2 + ( y − 9)2 = 18 = (3 2)2

2

⇒ 2( x2 + y 2 )( x2 − y 2 ) = 350 ⇒ ( x2 + y 2 )( x2 − y 2 ) = 175 Since x, y ∈ I, the only possible case which gives integral solutions, is …(i) x2 + y 2 = 25 …(ii) x2 − y 2 = 7 From Eqs. (i) and (ii) x2 = 16; y 2 = 9 ⇒ x = ±4; y = ±3 ⇒ Area = 48

10 Clearly, 0 − 3e iπ / 4 = 3 e iπ /2

P(z)

N

( x + y − 1) 2x x2 + y 2 − 2 xcot 2α − 1 = 0 =

∴

2

8 Given,|z + 4 |≤ 3 Now,|z + 1 |= |z + 4 − 3 | ≤ |z + 4| + |3|≤ 3 + 3 = 6 Hence, greatest value of|z + 1 | = 6 Since, least value of the modulus of a complex number is 0. Consider,|z + 1 |= 0 ⇒ z = − 1 Now, |z + 4|= |− 1 + 4 | = 3 ⇒| z + 4 |≤ 3 is satisfied by z = − 1. ∴ Least value of|z + 1| = 0

9 z − 3 − 2i ≤ 2

∴

z − 3e iπ / 4 = 4ie iπ / 4 z = (3 + 4i )e iπ / 4 …(i) z = 1, a1 , a2 ,... an−1 1 1 Let α = , then z = 1 − 1− z α n 1  [by (i)] ∴ 1 −  = 1  α n n ⇒ (α − 1) − α = 0 ⇒ − C1 α n−1 + C2α n−2 + ... + (−1)n = 0 1 1 1 where, α = , ..., 1 − a1 1 − a2 1 − an−1 1 1 1 + + ...+ 1 − a1 1 − a2 1 − an−1

⇒

= 2

C2 n(n − 1) (n − 1) = = 2/ n 2 C1

2

12 z ω − ω z = z − ω

⇒

( x − 3)2 + ( y − 2)2 = 22 = 4

... (i) 2

+ 1)z

z zω =z ⇒ = zω ω ω

B(z2) A(z1) α

⇒

⇒

⇒

∴

... (ii)

⇒ zz ω − ωz ω − z + ω = 0 ⇒ (z ω − 1)(z − ω) = 0 ⇒ z = ω or zω = 1 i.e. zω = 1 2 1 = ω/ ω ⇒ z = ω or z = ω

X

O

⇒

Also, from Eq. (i), zzω − ωω z = z − ω

... (i)

z2 = e iα = cos α + i sinα z1

11 Given z n = 1, where

2

⇒ z lies on or inside the circle

z1 + z2 = − p and z1 z2 = q OA = OB z1 = z2

Y

z +1 z ⇒ = = real 2 ω ω +1

P(3, – 5/2)

k

k =1

where α = e −i 2π /11 = i [α + α2 + α3 + ... + α10 ]

⇒ Q ⇒ E

O

2

A

k =1

10

∑α

14 Given, z2 + pz + q = 0 3eiπ/4

⇒ ( z + 1)ω = ( ω

(3, 2)

=i

11

= −i

7 We have, α + iβ = cot (z)

2

− i 2 kπ

=i

−1

⇒ cot(α + iβ) = x + iy and cot(α − iβ) = x − iy Now, consider cot 2α = cot [(α + iβ) + (α − iβ)] cot(α + iβ ) ⋅ cot(α − iβ ) − 1 = cot(α + iβ ) + cot(α − iβ )

10

= i ∑e

α(1 − α10 ) (α − α11 ) =i 1−α 1−α (α − 1) 11 =i [Qα = cos 2π − sin2π = 1] (1 − α )

z − 3e 4 −3e − iπ / 4 3 = i z − 3e iπ / 4 4

∴

2k π 2k π  + i cos   11 11  10 2kπ 2kπ  = i ∑  cos − i sin  11 11  k =1  10

13 We have, ∑  sin

5 2z − 6 + 5i = 2 z −  3 − i   2 

z1 + z2 = 1 + cos α + i sinα z1 α α α = 2cos  cos + i sin  2 2 2 (z1 + z2 )2 2 α iα = 4cos e z12 2 z = 4cos 2 α / 2 ⋅ 2 z1 α (z1 + z2 )2 = 4cos 2 z1 z2 2 p2 = 4q cos 2 α /2 p2 α = cos 2 4q 2

15 Q p < 0, take

p = − q 3 (q > 0)

∴ p1 /3 = q(−1)1 /3 = − q , − qω, − qω2 Now, take α = −q , β = − qω,γ = − qω2 Then, given expression =

x + yω + zω2 = ω2 xω + yω2 + z

DAY THREE

Sequence and Series Learning & Revision for the Day u

u

u

Definition Arithmetic Progression (AP) Arithmetic Mean (AM)

u

u

u

Geometric Progression (GP) Geometric Mean (GM) Arithmetico-Geometric Progression (AGP)

u

u

Sum of Special Series Summation of Series by the Difference Method

Definition ●

By a sequence we mean a list of numbers, arranged according to some definite rule. or We define a sequence as a function whose domain is the set of natural numbers or some subsets of type {1, 2, 3, ... k}.

●

●

If a1, a2 , a3 ,..., an,.... is a sequence, then the expression a1 + a2 + a3 + ... + an+... is called the series. If the terms of a sequence follow a certain pattern, then it is called a progression.

PRED MIRROR

Arithmetic Progression (AP) ●

Your Personal Preparation Indicator

It is a sequence in which the difference between any two consecutive terms is always same. An AP can be represented as a , a + d, a + 2 d, a + 3 d, … where, a is the first term, d is the common difference.

u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

●

The nth term, t n = a + (n − 1)d

u

●

Common difference d = t n − t n−1

●

●

●

●

The nth term from end, t n = l − (n − 1)d, where l is the last term. n n Sum of first n terms, S n = [2 a + (n − 1)d ] = [a + l ], where l is the last term. 2 2 1 If sum of n terms is S n, then nth term is t n = S n − S n−1 , t n = [t n− k + t n+ k ], where k < n 2

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

21

DAY NOTE

• Three non-zero numbers a, b, c are in GP iff b2 = ac. • If a, b and c are in AP as well as GP, then a = b = c. • If a > 0 and r > 1 or a < 0 and 0 < r < 1, then the GP will be

• Any three numbers in AP can be taken as a − d , a, a + d. • Any four numbers in AP can be taken as a − 3d , a − d , a + d , a + 3d.

an increasing GP.

• Any five numbers in AP can be taken as

• If a > 0 and 0 < r < 1 or a < 0 and r > 1, then the GP will be a

a − 2d , a − d , a, a + d , a + 2d.

decreasing GP.

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

Important Results on GP

An Important Result of AP ●

●

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

●

Arithmetic Mean (AM) ●

●

●

●

If a, A and b are in AP, then A = mean of a and b.

a+b is the arithmetic 2

If a, A1 , A2 , …, An , b are in AP, then A1, A2,..., An are the n arithmetic means between a and b. The n arithmetic means, A1, A2 , ..., An, between a and b are r (b − a) given by the formula, Ar = a + ∀ r = 1, 2, ... n n+1 Sum of n AM’s inserted between a and b is n A i.e.  a + b A1 + A2 + A3 + K + An = n    2 

NOTE

●

●

●

●

●

It is a sequence in which the ratio of any two consecutive terms is always same. A GP can be represented as a, ar , ar 2 , … where, a is the first term and r is the common ratio. The nth term, t n = ar n –1 l The nth term from end, t n′ = n − 1 , where l is the last term. r  1 – r  , r ≠1 a  Sum of first n terms, S n =   1 – r   na, r =1  If| r | < 1, then the sum of infinite GP is S ∞ =

NOTE

●

●

a r a a • Any four numbers in GP can be taken as 3 , , ar, ar 3 . r r a a • Any five numbers in GP can be taken as 2 , , a, ar, ar 2 . r r

• Any three numbers in GP can be taken as , a, ar.

If a, G1, G2,K , Gn, b are in GP, then G1, G2 ,K , Gn are the n geometric means between a and b. The n GM’s, G1, G2 , ..., Gn, inserted between a and b, are r

Product of n GM’s, inserted between a and b, is the nth power of the single GM between a and b, i.e. G1 ⋅ G2 ⋅ ... ⋅ Gn = G n = (ab )n/2 .

NOTE

• If a and b are of opposite signs, then their GM can not exist. • If A and G are respectively the AM and GM between two numbers a and b, then a, b are given by [ A ± ( A + G)( A − G) ] .

• If a1 , a2 , a3 ,... , an are positive numbers, then their GM = ( a1 a2 a3 ... an ) 1 / n .

Arithmetico-Geometric Progression (AGP) ●

a 1−r

If a, G and b are in GP, then G = ab is the geometric mean of a and b.

b n + 1 given by the formula, Gr = a   .  a

( a1 + a2 + a3 + K + an ) n

n

●

●

●

Geometric Progression (GP)

In a finite GP, a1 , a2 , ..., an, 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 term. i.e. a1 an = ak ⋅ an − ( k − 1 ), ∀ k = 1, 2, 3, ..., n − 1.

Geometric Mean (GM)

• The AM of n numbers a1 , a2 , ... , an is given by AM =

If a1 , a2 , a3 ,K , an is a GP of positive terms, then log a1 , log a2 ,K , log an is an AP and vice-versa.

●

A progression in which every term is a product of a term of AP and corresponding term of GP, is known as arithmetico-geometric progression. If the series of AGP be a + (a + d)r + (a + 2 d)r 2 + ... + {a + (n − 1)d}r n−1 + ..., then a dr (1 − r n−1 ) {a + (n − 1) d }r n + − ,r ≠1 1−r 1−r (1 − r )2 a dr = + ,| r | < 1 1 − r (1 − r )2

(i) S n = (ii) S ∞

22

40

Method to find the Sum of n-terms of Arithmetic Geometric Progression

●

Usually, we do not use the above formula to find the sum of n terms. Infact we use the mechanism by which we derived the formula, shown below: Let, S n = a + (a + d) r + (a + 2d) r 2 + K + (a + (n − 1) d) r n − 1 …(i) Step I Multiply each term by r (Common ratio of GP) and obtain a new series ⇒ r S n = ar + (a + d) r 2 + K + (a + (n − 2) d) r

n −1

+ (a + (n − 1) d) r …(ii) n

Step II Subtract the new series from the original series by shifting the terms of new series by one term ⇒ (1 − r ) S n= a + [dr + dr 2 +K + dr n−1 ] − (a + (n − 1) d) r n 1 − rn −1 ⇒ S n(1 − r ) = a + dr   − (a + (n − 1) d) r n  1−r  ⇒

 1 − r n − 1  (a + (n − 1) d) n a Sn = + dr  r  − 1−r 1−r  (1 − r )2 

●

Sum of squares of first n natural numbers, n (n + 1)(2 n + 1) 12 + 22 + K + n2 = Σ n 2 = 6 Sum of cubes of first n natural numbers, 2  n (n + 1)  13 + 23 + 33 + ... + n3 = Σ n3 =   2 (i) Sum of first n even natural numbers 2 + 4 + 6 + K+ 2 n = n (n + 1) (ii) Sum of first n odd natural numbers 1 + 3 + 5 + K + (2 n − 1) = n2

Summation of Series by the Difference Method If nth term of a series cannot be determined by the methods discussed so far. Then, nth term can be determined by the method of difference, if the difference between successive terms of series are either in AP or in GP, as shown below: Let T1 + T2 + T3 + ... be a given infinite series. If T2 − T1 , T3 − T2 ,K are in AP or GP, then Tn can be found by following procedure. Clearly, S n = T1 + T2 + T3 + K + Tn

…(ii)

∴ S n − S n = T1 + (T2 − T1 ) + (T3 − T2 ) + ... + (Tn − Tn−1 ) − Tn

Sum of Special Series ●

…(i)

S n = T1 + T2 + K + Tn−1 + Tn

Again,

Sum of first n natural numbers, n (n + 1) 1 + 2 + ... + n = Σ n = 2

⇒

Tn = T1 + (T2 − T1 ) + (T3 − T2 ) + K + (Tn − Tn−1 )

⇒

Tn = T1 + t 1 + t 2 + t 3 + t n−1 n

where, t 1, t 2 , t 3 ,K are terms of the new series ⇒ S n = ∑ Tr r =1

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE  

7 2

1 If log3 2, log3 ( 2x − 5) and log3  2 x −  are in AP, then x is equal to (a) 2

n −1 a1 an + 1

(b)

1 a1 an + 1

(c)

n+1 a1 an + 1

(d)

n a1 an

+1

5 A man arranges to pay off a debt of ` 3600 by 40 annual (b) 3

(c) 4

(d) 2, 3

2 The number of numbers lying between 100 and 500 that are divisible by 7 but not by 21 is (a) 57

(a)

(b) 19

(c) 38

(d) None of these

3 If 100 times the 100th term of an AP with non-zero common difference equals the 50 times its 50th term, then the 150th term of this AP is (a) − 150 (c) 150

(b) 150 times its 50th term (d) zero

4 If a1, a 2, ..., an + 1 are in AP, then 1 1 1 is + +K+ a1a 2 a 2a 3 anan + 1

instalments which are in AP. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid. The value of the 8th instalment is (a) ` 35 (c) ` 65

(b) ` 50 (d) None of these

6 Let a1, a 2, a 3, ... be an AP, such that a1 + a 2 + ... + a p

a1 + a 2 + a 3 + ... + aq

=

p3 a ; p ≠ q , then 6 is equal to a 21 q3 j

41 (a) 11 11 (c) 41

121 (b) 1681 121 (d) 1861

JEE Mains 2013

23

DAY 19 If | a | < 1 and | b | < 1, then the sum of the series

7 A person is to count 4500 currency notes. Let an denotes the number of notes he counts in the n th minute. If a1 = a 2 = ... = a10 = 150 and a10, a11 ,... are in AP with common difference – 2, then the time taken by him to count all notes, is (a) 24 min

(b) 34 min

8 If log

3

(c) 125 min

(d) 135 min

a + log(3)1/ 3 a + log31/ 4 a + K upto 8th term 2

2

2

(a) ±

3

(b) 2 2

(d) None of these

9 n arithmetic means are inserted between 7 and 49 and (b) 12

(c) 13

(d) 14

(a) 1

(b) 2

(a) g 2

(d) 3

(a)

8 5

(b)

4 3

j

(c) 1

JEE Mains 2016 7 (d) 4

12 Three positive numbers form an increasing GP. If the middle term in this GP is doubled,then new numbers are in AP. Then, the common ratio of the GP is j JEE Mains 2014 (a)

2+

3

(b) 3 +

2

(c) 2 − 3

(d) 2 +

3

13 A GP consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places,then its common ratio is (a) 2

(b) 3

(c) 4

(d) 5

14 The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, …, is

j JEE Mains 2013 7 − 20 (b) [99 − 10 ] 9 7 (d) [99 + 10 − 20 ] 9

7 (a) [179 − 1020 ] 81 7 (c) [179 + 10 − 20 ] 81

15 If x , y and z are distinct prime numbers, then (a) x, y and z may be in AP but not in GP (b) x, y and z may be in GP but not in AP (c) x, y and z can neither be in AP nor in GP (d) None of the above

(c) 64

(b) −10 < x < 0 (c) 0 < x < 10 (d) x > 10

(a) a m

2

(b) 2a m

2

2

(c) 3 a m

2

(b) 6

(d) − 6

(c) 12

23 The sum to 50 terms of the series 2

1 1   1 + 2 1 +  + 3 1 +  + K , is given by   50 50 (a) 2500 (c) 2450

(b) 2550 (d) None of these

24 If (10)9 + 2(11)1(10)8 + 3(11)2(10)7 + ... + 10 (11)9 = k (10)9 , then k is equal to 121 (a) 10

j

441 (b) 100

(c) 100

JEE Mains 2014

(d) 110

25 The sum of the infinity of the series 2 6 10 14 1 + + 2 + 3 + 4 + ... is 3 3 3 3 (a) 3

(b) 4

(c) 6

(d) 2

26 The sum of the series1 + 3 + 5 + K upto 20 terms is 3

3

(a) 319600 (c) 306000

3

(b) 321760 (d) 347500

∑ a 4k + 1 = 416

2 and a 9 + a 43 = 66. If a12 + a 22 + K + a17 = 140 m, then m is j JEE Mains 2018 equal to

(b) 68

(c) 34

(d) 33

28 Let a, b, c ∈ R . if f ( x ) = ax + bx + c be such that 2

2

(d) 4 a m

a + b + c = 3 and f ( x + y ) = f ( x ) + f ( y ) + xy , ∀ x , y ∈ R , 10

square is formed by joining the mid-points of these squares. Then, a third square is formed by joining the mid-points of the second square and so on. Then, sum of the area of the squares which carried upto infinity is 2

(a) 4

(a) 66

(d) 16

18 The length of a side of a square is a metre. A second

2

GM will be

k =0

17 An infinite GP has first term x and sum 5, then x belongs (a) x < − 10

2 3

12

such that (nm + 1) divides (1 + n + n 2 + K + n127 ), is (b) 8

(d) 3 g 2

(c) 2g

27 Let a1, a 2, a 3, ..., a 49 be in AP such that

16 Let n ( > 1) be a positive integer, then the largest integer m (a) 32

(b) −g 2

22 If five GM’s are inserted between 486 and , then fourth

11 If the 2nd, 5th and 9th terms of a non-constant AP are in GP, then the common ratio of this GP is

(b) 20 months (d) 18 months

two numbers a and b, then ( 2p − q )( p − 2 q ) is equal to

x −y is equal to z 2

(c) 1 / 2

service. In each of the subsequent months his saving increases by ` 40 more than the saving of immediately previous month. His total saving from the start of service j AIEEE 2011 will be ` 11040 after

21 If one GM, g and two AM’s, p and q are inserted between

10 If x = 111K1 (20 digits), y = 333K3 (10 digits) and z = 222K 2 (10 digits), then

1 (1 − a) (1 − ab) 1 (d) (1 − a) (1 − b) (1 − ab) (b)

(a) 19 months (c) 21 months

their sum is found to be 364, then n is (a) 11

1 (1 − a) (1 − b) 1 (c) (1 − b) (1 − ab) (a)

20 A man saves ` 200 in each of the first three months of his

= 44, then the value of a is 1 (c) 2

1 + (1 + a ) b + (1 + a + a 2 ) b 2 + (1 + a + a 2 + a 3 ) b 3 + K is

2

then

∑ f (n ) is equal to

n =1

(a) 330

j

(b) 165

(c) 190

JEE Mains 2017

(d) 255

29 The sum of the series ( 2)2 + 2( 4)2 + 3 ( 6)2 + ... upto 10 terms is (a) 11300

j

(b) 11200

(c) 12100

JEE Mains 2013 (d) 12300

24

40

30 Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + K

(a)

If B − 2A = 100λ, then λ is equal to (a) 232

(b) 248

j

(c) 464

7 2

(b)

11 4

(c)

33 The sum of the series1 +

11 2

(d)

j

(c) 142

JEE Mains 2015

18 (a) 11

60 11

1 1 + + ... 1+ 2 1+ 2 + 3

upto 10 terms is

13 13 + 23 13 + 23 + 33 + +... is + 1 1+ 3 1+ 3 + 5 (b) 96

3 5 7 + + + ... upto 11 terms is 12 12 + 22 12 + 22 + 32 j JEE Mains 2013

JEE Mains 2018

(d) 496

31 The sum of first 9 terms of the series

(a) 71

32 The sum

j

22 (b) 13

20 (c) 11

JEE Mains 2013 16 (d) 9

(d) 192

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 The value of12 + 32 + 52 + ... + 252 is (a) 2925

(b) 1469

j

(c) 1728

JEE Mains 2013

(d) 1456

2 If the function f satisfies the relation f ( x + y ) = f ( x ) ⋅ f ( y ) for all natural numbers x , y , f (1) = 2 and

8 If Sn is the sum of first n terms of a GP : {an } and Sn′ is the sum of another GP : {1 / an }, then Sn equals (a)

n

∑ f (a + r ) = 16 ( 2n −1), then the natural number a is (b) 3

3 If the sum of an infinite GP is

(c) 4

(d) 5

7 and sum of the squares 2

147 of its terms is , then the sum of the cubes of its terms 16 is (a)

315 19

(b)

700 39

4 The sum of the infinite series (a)

31 18

(b)

65 32

(c)

985 13

(d)

1029 38

5 55 555 + 2 + + K is 13 13 133 (c)

65 36

(d)

75 36

5 Given sum of the first n terms of an AP is 2n + 3 n 2. Another AP is formed with the same first term and double of the common difference, the sum of n terms of the new AP is (a) n + 4n 2

6 For 0 < θ < z =

∞

(b) 6n 2 − n

(c) n 2 + 4n

(d) 3 n + 2n 2

∞ ∞ π , if x = ∑ cos 2n θ , y = ∑ sin2n θ and 2 n =0 n =0

∑ cos 2n θ sin2n θ, then xyz is equal to

n =0

(a) xz + y

(b) x + y + z (c) yz + x

(d) x + y − z

1 1 1 π4 1 1 1 7 If 4 + 4 + 4 + K to ∞ = , then 4 + 4 + 4 + K to 90 1 2 3 1 3 5 ∞ is equal to (a)

π4 96

(b)

π4 45

(c)

89 π 4 90

(d)

π4 90

(b) a1an Sn′

(c)

a1 ′ Sn an

(d)

an ′ Sn a1

9 If the sum of the first ten terms of the series

r =1

(a) 2

Sn′ a1an

2

2

2

2

16  3  2  1  4 2 m, 1  +  2  +  3  + 4 +  4  + ..., is  5  5  5  5 5 then m is equal to (a) 102

j

(b) 101

(c) 100

JEE Mains 2016

(d) 99

10 If m is the AM of two distinct real numbers I and n (l, n > 1) and G1, G 2 and G 3 are three geometric means between I j and n, then G14 + 2G 24 + G 34 equals JEE Mains 2015 (a) 4l 2mn (c) 4 lmn 2

(b) 4 lm 2n (d) 4l 2m 2n 2

11 The sum of the series ( 2 + 1) + 1 + ( 2 − 1) + K ∞ is (a)

2

(b) 2 + 3 2

(c) 2 − 3 2

(d)

4+ 3 2 2

12 The largest term common to the sequences 1, 11, 21, 31, ... to 100 terms and 31, 36, 41, 46, ... to 100 terms is (a) 531

(b) 471

(c) 281

(d) 521

13 If a, b, c are in GP and x is the AM between a and b, y the AM between b and c, then a c + =1 x y a c (c) + =3 x y (a)

(b)

a c + =2 x y

(d) None of these

14 Suppose a, b and c are in AP and a 2, b 2 and c 2 are in 3 GP. If a < b < c and a + b + c = , then the value of a is 2 (a)

1 2 2

(b)

1 2 3

(c)

1 1 − 2 3

(d)

1 1 − 2 2

25

DAY 15 For any three positive real numbers a, b and c, if

17 Statement I The sum of the series 1 + (1 + 2 + 4) + ( 4 + 6 + 9) + ( 9 + 12 + 16) + K

9 ( 25a + b ) + 25 (c − 3ac ) = 15b ( 3a + c ), then 2

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

2

2

b, c and a are in GP b, c and a are in AP a, b and c are in AP a, b and c are in GP

j

+ ( 361 + 380 + 400) is 8000.

JEE Mains 2017 n

Statement II

k =1

number n.

16 If S1, S 2, S 3,K are the sum of infinite geometric series whose first terms are 1, 2 , 3,K and whose common ratios 1 1 1 , , ,... respectively, then 2 3 4 2 is equal to S12 + S 22 + S 32 + K + S10 (a) 485 (c) 500

∑ [k 3 − (k − 1)3 ] = n 3, for any natural

(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 ture; 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) 495 (d) 505

ANSWERS SESSION 1

SESSION 2

1 (b)

2 (c)

3 (d)

4 (d)

5 (c)

6 (b)

7 (b)

8 (a)

9 (c)

10 (a)

11 (b)

12 (d)

13 (c)

14 (c)

15 (a)

16 (c)

17 (c)

18 (b)

19 (c)

20 (c)

21 (b)

22 (b)

23 (a)

24 (c)

25 (a)

26 (a)

27 (c)

28 (a)

29 (c)

30 (b)

31 (b)

32 (c)

33 (c)

1 (a) 11 (d)

2 (b) 12 (d)

3 (d) 13 (b)

4 (c) 14 (d)

5 (b) 15 (b)

6 (b) 16 (d)

7 (a) 17 (a)

8 (b)

9 (b)

10 (b)

Hints and Explanations SESSION 1 1 Q2 log 3 (2 − 5) = log 3 2 + log 3  2x − 7   2 7  x 2 x ⇒ (2 − 5) = 2  2 −   2 ⇒ t 2 + 25 − 10 t = 2t − 7 x

[put 2x = t ] ⇒

t − 12t + 32 = 0

⇒ ⇒

(t − 8) (t − 4) = 0 2x = 8 or 2x = 4

2

∴ x = 3 or x = 2 At, x = 2, log 3 (2x − 5) is not defined. Hence, x = 3 is the only solution.

2 The numbers between 100 and 500 that are divisible by 7 are 105, 112, 119, 126, …, 490, 497. Let such numbers be n. ∴ t n = an + (n − 1) d ⇒ 497 = 105 + (n − 1) × 7 ⇒ n − 1 = 56 ⇒ n = 57

The numbers between 100 and 500 that are divisible by 21 are 105, 126, 147, …, 483. Let such numbers be m. ∴ 483 = 105 + (m − 1) × 21 ⇒ 18 = m − 1 ⇒ m = 19 ∴ Required number = n − m = 57 − 19 = 38

3 Let a be the first term and d (d ≠ 0) be the common difference of a given AP, then T100 = a + (100 − 1) d = a + 99 d T 50 = a + (50 − 1) d = a + 49 d T150 = a + (150 − 1) d = a + 149 d Now, according to the given condition, 100 × T100 = 50 × T 50 ⇒

100(a + 99 d ) = 50(a + 49 d )

⇒

2(a + 99 d ) = (a + 49 d )

⇒

2a + 198 d = a + 49 d

⇒ ∴

a + 149 d = 0 T150 = 0

4 Let d be the common difference of given AP and let 1 1 1 + +K+ a1 a2 a2 a3 an an

S =

. Then, +1

 d  d d + +K+   a a a a a a 2 3 n n +1   1 2  a2 − a1 + a3 − a2   a1 a2  a2 a3 1  =  a a − d  n +1 n +K+  an an + 1     1 1  1 1   a − a  +  a − a   1 2 2 3 1  =   1 d  1   +K+  −  a an + 1    n     a a  − 1 1 1 1 n +1 1 =  − =   d  a1 an + 1  d  a1 an + 1  1  a + nd − a1  n =  1 = d  a1 an + 1 a a 1 n +1 

S =

1 d

26

40



5 Given, 3600 = 40 [ 2a + (40 − 1) d ] ⇒ ⇒

180 = 2a + 39d

…(i)

After 30 instalments one-third of the debt is unpaid. 3600 Hence, = 1200 is unpaid and 2400 3 is paid. 30 Now, 2400 = {2a + (30 − 1) d } 2 …(ii) ∴ 160 = 2a + 29d On solving Eqs. (i) and (ii), we get a = 51, d = 2 Now, the value of 8th instalment = a + (8 − 1) d = 51 + 7 ⋅ 2 = ` 65

6 Given that,

a1 + a2 + K + a p a1 + a2 + K + a q

=

1 1  +  1 1 log log 3 3 8 S n = log a  2 4 3    + K upto 8th term  log a2 ⇒ [2 + 3 + 4 + K + 9] = 44 log 3

p3 q3

p [ 2a1 + ( p − 1 )d ] p3 2 ⇒ = 3 q q [ 2a2 + (q − 1 )d ] 2 where, d is a common difference of an AP. 2a 1 + ( p − 1)d p2 = 2 ⇒ 2a 2 + (q − 1)d q d 2 2 = p ⇒ 2 d q a2 + (q − 1) 2 On putting p = 11 and q = 41 , we get d a1 + (11 − 1) 2 2 = (11) d (41)2 a2 + (41 − 1) 2 a1 + 5d 121 ⇒ = a2 + 20d 1681 a6 a21

=

121 1681

7 Number of notes that the person counts in 10 min = 10 × 150 = 1500 Since, a10, a11 , a12 ,... are in AP with common difference − 2. Let n be the time taken to count remaining 3000 notes. n Then, [2 × 148 + (n − 1) × −2] = 3000 2 ⇒ n2 − 149 n + 3000 = 0 ⇒

(n − 24)(n − 125) = 0

∴

n = 24 and 125

Then, the total time taken by the person to count all notes = 10 + 24 = 34 min

+

Given, sum of all terms = 5 × sum of terms occupying odd places, i.e. a + ar + ar 2 + ... + a r 2 n −1 = 5 × (a + a r 2 + a r 4 + ... + a r 2 n −2 ) ⇒

[given] ⇒

44 log a2 = 44 log 3 a= ±

∴

3

9 We know that,

A1 + A2 + K + A n = nA, where a+ b A= 2 7 + 49  ∴ 364 =   n  2  364 × 2 ⇒ n= = 13 56 10 Given, x = 1 (999K 9) = 1 (1020 − 1) 9 9 1 1 y = (999K 9) = (1010 − 1) 3 3 2 2 and z = (999K 9) = (1010 − 1) 9 9 ∴

x − y 2 1020 − 1 − (1010 − 1)2 = z 2(1010 − 1) 1010 + 1 − (1010 − 1) = =1 2

a1 + ( p − 1)

⇒

1

2  1 log 3

2 3600 = 20 (2a + 39d )

11 Let a be the first term and d be the common difference. Then, we have a + d , a + 4d , a + 8 d in GP, i.e. (a + 4d )2 = (a + d ) (a + 8d ) ⇒ a2 + 16d 2 + 8ad = a2 + 8ad + ad + 8d 2 ⇒

8d 2 = ad

⇒ 8d = a [Qd ≠ 0] Now, common ratio, a + 4d 8d + 4d 12d 4 r = = = = a+ d 8d + d 9d 3

12 Let a, ar , ar 2 be in GP ( where, r > 1). On multiplying middle term by 2, we get that a, 2ar , ar 2 are in AP.

⇒ ⇒

⇒ ∴

r = 2 + 3 [QAP is increasing] 2

3

13 Let the GP be a, a r , a r , a r , ar 2 n −2 , ar 2 n −1 . where, a, ar 2 , ar 4 , ar 6, ... occupy odd places and ar , ar 3 , ar 5, ar 7 ,... occupy even places.

5(r 2 n − 1) r2n − 1 = r −1 (r − 1)(r + 1) 1=

5 ⇒ r + 1 = 5⇒ r = 4 r+1

14 Let S = 0.7 + 0.77 + 0.777 + … upto 20 terms 7 77 777 = + + + … upto 20 terms 10 102 103  1 + 11 + 111   = 7  10 102 103  + … upto 20 terms   =

 9 + 99 + 999  7  10 100 1000  9 + … upto 20 terms  

1  1   1   7 1 −  + 1 − 2  + 1 − 3    10   10   10   9 + … upto 20 terms   7 = (1 + 1 + … upto 20 terms) 9  =

 1 + 1 + 1   −  10 102 103   + … upto 20 terms    20  1   1    1 −      10  10  7  = 20 −   1 9 1−   10      Qsum of n terms of GP,    a(1 − r n ) , where r < 1  Sn =   1− r

=

20 7 1  1   20 − 1 −      10   9 9  

=

20 7  179 1  1   7 +   = [179 + 10−20]  9 9 9  10   81

⇒ 4 ar = a + ar 2 ⇒ r 2 − 4r + 1 = 0 4 ± 16 − 4 r = = 2± 3 2

a (r 2 n − 1) 5a[(r 2 )n − 1] = r −1 r2 − 1  a(r n − 1) QS n = r − 1   

15 x, y , z are in GP ⇔

y 2 = xz

⇔ x is factor of y. Which is not possible, as y is a prime number. If x = 3, y = 5and z = 7, then they are in AP. Thus, x, y and z may be in AP but not in GP.

27

DAY 16 Clearly, 1+ n+ n + K+ n 2

=

(n

64

127

− 1) (n n−1

64

n128 − 1 = n−1

 a (r n − 1) QS n = r − 1    + 1)

= (1 + n + n2 + K + n 63 ) (n 64 + 1) Thus, the largest value of m for which n m + 1 divides 1 + n + n2 + K + n127 is 64. 5− x 17 Since, S ∞ = x = 5 ⇒ r = 1− r 5

a + a2 + K + an − 1 ) b n − 1

n =1

n =1

⇒

n + 11 n − 522 = 0

⇒

n2 + 29 n − 18 n − 522 = 0

⇒

n (n + 29) − 18(n + 29) = 0

 1 − an  n − 1  b  1− a

∞  1  ∞ n −1 = − ∑ an b n − 1  ∑ b 1 − a n = 1 n =1  ∞ ∞   1 n −1 n −1 = − a ∑ (ab ) ∑ b  1 − a n = 1 n =1  a 1 2 = [1 + b + b + K ∞] − 1− a 1− a

[1 + ab + (ab )2 + K ] 1 1 a 1 = ⋅ − ⋅ 1 − a 1 − b 1 − a 1 − ab [Q|b | < 1 and |ab | = |a||b | < 1] 1 − ab − a (1 − b ) = (1 − a) (1 − b ) (1 − ab ) 1 − ab − a + ab = (1 − a) (1 − b ) (1 − ab ) 1 = (1 − b ) (1 − ab )

20 Let the time taken to save ` 11040 be (n+3) months. For first 3 months he saves ` 200 each month. n In (n+3) months, 3 × 200 + {2(240) 2 + (n − 1) × 40} = 11040

S 50 = 2500.

24 Given, k ⋅ 109 = 109 + 2(11)1 (10)8 + 3(11)2 (10)7 + ... + 10(11)9

2

⇒ ∴ n = 18

11 11 k = 1 + 2  + 3    10   10 

[neglecting n = − 29 ] ∴ Total time = (n + 3) = 21 months

b −a . 3 b − a 2a + b ∴ p = a+ d = a+ = 3 3 b − a a + 2b q =b −d =b − = 3 3 Now, (2 p − q )( p − 2q ) (4a + 2b − a − 2b ) (2a + b − 2a − 4b ) ⋅ 3 3

22 Here, a = 486 and b = 2

2 1  ∴ G 4 = 486  ⋅   3 486  1  = 486    3 ⋅ 243  1  = 486    729 

r n +1

4/6

[Q here, n = 5]

4/6

4/6

= 486 ⋅

1 34

 11  k = 1  11  + 2  11         10   10   10 

=6

23 Let x = 1 + 1 and S 50 be the sum of 50 first 50 terms of the given series. Then, S 50 = 1 + 2 x + 3 x2

...(i)

2

9

11 11 + ... + 9   + 10    10   10 

10

...(ii)

On subtracting Eq.(ii) from Eq.(i), we get 11 11  11  k  1 −  = 1 + +   10  10  10 

2

9

11 11 + ... + 9   − 10    10   10 

10

 11 10  1    − 1 10   10 − 11   10  −10  11  ⇒ k  =    10   10   11 − 1    10 

10 10  11 11  ⇒ − k = 10 10   − 10 − 10     10   10    ∴ k = 100 14 25 Let S = 1 + 2 + 62 + 10 + 4 +K 3 3 33 3 2 3 5 7 =1+ 1 + + 2 + 3 + K  3  3 3 3

2 1 2⋅1 / 3  +  3  1 − 1 / 3 (1 − 1 / 3)2  Qsum of infinite AGP, is   a dr  S∞ = +   1− r (1 − r )2  

=1+

+ ... + 50 x 49 …(i) ⇒

9

 a( r n − 1)  Q in GP,sum of n terms = r − 1 ,    when r > 1 

3

b We know that, G r = a    a

2

11 + ... + 10   10 

(n − 18)(n + 29) = 0

2

+ (1 + a + a2 + a3 ) b 3 + K ∞

∞

⇒

30 + n2 + 11 n = 552

= − ab = − g 2

19 Clearly, 1 + (1 + a)b + (1 + a + a )b

∑

600 + 20 n (n + 11) = 11040

⇒

=

2

=

⇒

So, common difference d is

carried upto infinity a2 a2 = a2 + + + ... 2 4 2 a = = 2a2 m2 1 1− 2

∑ (1 +

− 1 50 50 ⇒ S 50   = − 50 + 50 x − 50 x  50 

in AP.

18 Sum of the area of the squares which

=

n {40(12 + n − 1)} = 11040 2

21 Since, g = ab . Also, a, p , q and b are

For infinite GP,|r | < 1 5− x ⇒ −1 < < 1 ⇒ −10 < − x < 0 5 ∴ 0 < x < 10

∞

⇒ 600 +

x S 50 = x + 2 x2 + K + 49 x 49 + 50 x 50 …(ii)

⇒ (1 − x )S 50 = 1 + x + x2 + x3 + K + x 49 − 50 x 50 [subtracting Eq. (ii) from Eq. (i)] 1 − x 50 ⇒ S 50 (1 − x ) = − 50 x 50 1− x 50  − 1 1 − x ⇒ S 50  − 50 x 50  =  50   − 1    50 

Q x = 1 + 1   50 

=1+

2 3 2 9  2 3 + ⋅ = 1 + ⋅2⋅ = 3 3  2 3 4  3 2

26 13 + 33 + K + 393 = 13 + 23 + 33 + K + 403 – (23 + 43 + 63 + ... + 403 ) 2

40 × 41  3 3 =   − 8(1 + 2   2 + 33 + K + 203 ) = (20 × 41)2 − 8  

20 × 21   2 

= 202 [412 − 2(21)2 ] = 319600

2

28

40

27 Let a1 = a and d = common difference Q a1 + a5 + a9 + L + a49 = 416 ∴ a + (a + 4d ) + (a + 8d ) + … (a + 48d ) = 416 13 (2a + 48d ) = 416 ⇒ 2 ⇒ a + 24d = 32 …(i) Also, we have a9 + a43 = 66 ∴ a + 8d + a + 42d = 66 ⇒ 2a + 50d = 66 ⇒ a + 25d = 33 …(ii) Solving Eqs. (i) and (ii), we get a = 8 and d = 1 Now,

a12

+

a22

+

+L+

a32

= 140m

2 a17

8 + 9 + 10 + … + 24 = 140m 2

2

2

2

⇒ (12 + 22 + 32 + … + 242 ) − (12 + 22 + 32 + … + 72 ) = 140m ⇒ ⇒ ⇒ ⇒ ⇒

24 × 25 × 49 7 × 8 × 15 − = 140m 6 6 3×7× 8× 5 (7 × 5 − 1) = 140m 6 7 × 4 × 5 × 34 = 140m 140 × 34 = 140m m = 34

28 We have, f ( x ) = ax2 + bx + c Now, f ( x + y ) = f ( x ) + f ( y ) + xy Put y = 0 ⇒ f ( x ) = f ( x ) + f (0) + 0 ⇒ f (0) = 0 ⇒ c =0 Again, put y = − x ∴ f (0) = f ( x ) + f (− x ) − x2 ⇒

0 = ax2 + bx + ax2 − bx − x2 1 ⇒ 2ax2 − x2 = 0 ⇒ a = 2 Also, a + b + c = 3 1 5 ⇒ + b + 0 = 3⇒ b = 2 2 x2 + 5x f ( x) = ∴ 2 n2 + 5n 1 2 5 Now, f (n ) = = n + n 2 2 2 10 1 10 2 5 10 ∴ ∑ f (n ) = 2 ∑ n + 2 ∑ n n =1 n =1 n =1

12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + … A = sum of first 20 terms B = sum of first 40 terms ∴ A = 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + … + 2 ⋅ 202 A = (1 + 2 + 3 + … + 202 ) + (22 + 42 2

2

2

+ 62 + … + 202 ) A = (12 + 22 + 32 + … + 202 ) + 4(12 + 22 + 32 + …+102 ) 20 × 21 × 41 4 × 10 × 11 × 21 + 6 6 20 × 21 20 × 21 × 63 A= (41 + 22) = 6 6 Similarly, A=

B = (12 + 22 + 32 + … + 402 ) + 4(12 + 22 +… + 202 ) 40 × 41 × 81 4 × 20 × 21 × 41 + 6 6 40 × 41 × 123 40 × 41 B = (81 + 42) = 6 6 Now, B − 2 A = 100λ 40 × 41 × 123 ∴ 6 2 × 20 × 21 × 63 − = 100λ 6 40 ⇒ (5043 − 1323) = 100λ 6 40 ⇒ × 3720 = 100λ 6 ⇒ 40 × 620 = 100λ 40 × 620 ⇒ λ= = 248 100 B =

31 Write the nth term of the given series and simplify it to get its lowest form.Then, apply, S n = ΣT n . Given series is 13 13 + 23 13 + 23 + 33 + + + K∞ 1 1+ 3 1+ 3+ 5 Let T n be the nth term of the given series. ∴ Tn =

1 10 × 11 × 21 5 10 × 11 ⋅ + × 2 6 2 2 385 275 660 = + = = 330 2 2 2 =

29 Series (2)2 + 2(4)2 + 3(6)2 + K

13 + 23 + 33 + ... + n3 1 + 3 + 5 + K upto n terms 2

 n(n + 1)   2 2  = (n + 1) = 2 4 n (n + 1)2 1 = 4 4 n =1 9

= 4 {1 ⋅ 12 + 2 ⋅ 22 + 3 ⋅ 32 + K } ∴

30 We have,

Now, S 9 = ∑

[(22 + 32 + ... + 102 ) + 12 − 12 ]

Tn = 4n ⋅ n

2

and S n = ΣT n = 4Σ n3 = 4   Now, S 10 = [10 ⋅ (10 + 1)]

2

= (110)2 = 12100

2

n(n + 1)  2

=

1  10(10 + 1)(20 + 1) − 1 4  6 

384 = = 96 4

2n + 1

32 T n =

(12 + 22 + ... + n2 ) 2n + 1 6 = = n (n + 1 ) (2n + 1 ) n (n + 1 ) 6 1 1  = 6 −   n ( n + 1 )

 1 1 1 1 T1 = 6  −  , T2 = 6  −  , …  2 3   1 2 1 1 T11 = 6  −  11 12  1 1  6 × 11 11 ∴S = 6  −  = 12 = 2 1 12  

33 n th term of the series is 1 2 = n (n + 1 ) n (n + 1 ) 2 1  1 ⇒ Tn = 2 −   n n + 1 Tn =

 1 1 1 1 ⇒ T1 = 2  −  , T2 = 2  −  ,  2 3  1 2 1 1 1 1 T3 = 2  −  , ..., T10 = 2  −   3 4  10 11  ∴ S 10 = T1 + T2 + K + T20 1 − 1 + 1 − 1 + 1 − 1   2 2 3 3 4  = 2 1 1 +K+ −    10 11  1 = 2  1 −   11  10 20 = 2⋅ = 11 11

SESSION 2 1 Let S = 12 + 32 + 52 + K + 252 = (12 + 22 + 32 + 42 + K + 252 ) − (22 + 42 + 62 + K + 242 ) = (1 + 2 + 3 + 42 + K + 252 ) 2

2

2

− 22 (1 + 22 + 32 + K + 122 ) 25(25 + 1) (2 × 25 + 1) = 6 12 (12 + 1) (2 × 12 + 1) − 4× 6 25 × 26 × 51 4 × 12 × 13 × 25 = − 6 6 = 25 × 13 × 17 − 4 × 2 × 13 × 25 = 5525 − 2600 = 2925

2 Now, f (2) = f (1 + 1) = f (1) ⋅ f (1) = 22 and f (3) = 23 Similarly, f (n ) = 2n ∴ 16(2n − 1) =

n

∑ f (a +

r =1 a

r )=

n

∑2

a+r

r =1

= 2 (2 + 22 + K + 2n )

29

DAY  2n − 1  = 2a ⋅ 2    2−1

[GP series]

1 = 1 − cos 2 θ sin2 θ z xy − 1 1 =1− = xy xy

and

= 2a + 1 (2n − 1) 2a + 1 = 16 = 24

⇒ ∴

⇒ ∴

a= 3

3 Let GP be a, ar , ar ,K,|r | < 1. 2

According to the question, a 7 a2 147 = , = 1 − r 2 1 − r2 16 On eliminating a, we get 2

147 7 (1 − r 2 ) =   (1 − r ) 2  2 16 ⇒ 3(1 + r ) = 4 (1 − r ) ⇒ r = ∴ Sum of cubes a3 = = 1 − r3

(3)3 1 1 −    7

3

1 ,a = 3 7

1029 = 38

4 Let S = 5 + 552 + 555 …(i) +… 13 13 133 S 5 55 and ...(ii) = + 3 + … 13 132 13 On subtracting Eq. (ii) from Eq. (i), we get 12 5 50 500 +… S = + + 13 13 132 133 10 which is a GP with common ratio . 13 13  5  10   65 ∴ S = × ÷ 1 − =  12  13  13   36  a  QS ∞ = 1 − r   

5 Here, T1 = S1 = 2(1 ) + 3(1 )2 = 5 T2 = S 2 − S 1 = 16 − 5 = 11 [QS 2 = 2(2) + 3(2) = 16] 2

T3 = S 3 − S 2 = 33 − 16 = 17 [Q S 3 = 2(3) + 3 (3)2 = 33 ]

x y = x yz − z x yz = x y + z = x + y + z 7 Let S = 14 + 14 + 14 + K + to ∞ 1 3 5 1 1 1 π4 Since, 4 + 4 + 4 + K to ∞ = 90 1 2 3 1 1 1   ∴  4 + 4 + 4 + K to ∞  1  3 5 1 1 1 π4 +  4 + 4 + 4 + K to ∞  = 2  90 4 6 1 1 1   ⇒  4 + 4 + 4 + K to ∞  1  3 5 1  1 1 1 π4 + 4 + 4 + K + to ∞  = 4  4  90 2 1 2 3 1 π4 π4 ⇒S + ⋅ = 16 90 90  1 1 1 π4  Q 4 + 4 + 4 + K to ∞ =  90  2 3  1 +

15 π 4 π4  1 π4 ⇒ S = = 1 −  = 90  16  16 × 90 96

8 Let an = a r

x=

1 1 − cos 2 θ

=

1 sin2 θ

y =

1

cos 2 θ 1 1 Now, + =1 x y ⇒

and z =

x + y = xy

.

 1  1 −  1         a r   and S n′ = 1 1− r  1  1 Q first term of  a  is a   n   1  and common ratio is  r   1  (r n − 1)    a = ⋅r r n (r − 1) =

1 1 − cos 2 θ sin2 θ

1− r 1 ⋅ 1 − r a⋅ r n − 1 n

1 − rn 1 a (1 − r n ) 1 ⋅ = ⋅ 1 − r an 1− r a an 1 = Sn ⋅ a1 an =

S n = a1 an S n′

[Q sin2 θ + cos 2 θ = 1]

+ ... to 10 terms =

1

(8 + 12 + 16 + 202 + 242 2

52

2

2

+ ... to 10 terms) = =

42 52 42

(2 + 3 + 4 + 5 2

2

2

2

+ ... to 10 terms) (22 + 32 + 42 + 52 + ... + 112 )

52 16 2 = ((1 + 22 + ... + 112 ) − 12 ) 25 16  11 ⋅ (11 + 1) (2 ⋅ 11 + 1) = − 1   25  6 16 16 = (506 − 1) = × 505 25 25 16 16 m= × 505 ⇒ 5 25 ⇒ m = 101

10 Given, m is the AM of l and n. ∴

l + n = 2m

and G1 , G2 , G3 are geometric means between l and n. So, l ,G1 ,G2 ,G3 , n are in GP.

n n = lr 4 ⇒ r =   l

1/4

Now, G14 + 2G24 + G34 = (lr )4 + 2(lr 2 )4 + (lr 3 )4 = l 4 × r 4 (1 + 2r 4 + r 8 ) = l 4 × r 4 ( r 4 + 1)2 2

= l4 ×

nn+ l 2 2   = ln × 4 m = 4 l m n l  l  [Q n + l = 2m]

11 Given series is a geometric series with a = 2 + 1 and r = 2 − 1. ∴ Required sum 2+1 2+1 a = = = 1 − r 1 − ( 2 − 1) 2 − 2 = =

( 2 + 1) (2 +

2)

(2 − 2 ) (2 +

2)

2 2+ 2+ 2+ 4−2

2

=

4+ 3 2 2

12 Clearly, the common terms of the given

the series. Then, we have 2

2

∴ G1 = lr , G2 = lr 2 , G3 = lr 3 ;

9 Let S10 be the sum of first ten terms of

Similarly,

2

n

⇒

,

2

Let r be the common ratio of this GP.

a (1 − r n ) Then, S n = 1− r

Hence, sequence is 5, 11, 17. ∴ a = 5and d = 6 For new AP, A = 5, D = 2 × 6 = 12 n S ′ n = [2 × 5 + (n − 1 )12] ∴ 2 = 6 n2 − n

6 Sum of three infinite GP’s are

n −1

2

8 12 16 24 =   +   +   + 42 +    5  5  5  5

2

3 2 1 S 10 =  1  +  2  +  3   5  5  5 2

2

4 + 42 +  4  + ... to 10 terms  5

sequences are 31, 41, 51, ... Now, 100th term of 1, 11, 21, 31, ... is 1 + 99 × 10 = 991 and 100th term of 31, 36, 41, 46, ... is 31 + 99 × 5 = 526 . Let the largest common term be 526. Then, 526 = 31 + (n − 1) 10 ⇒ (n − 1) 10 = 495

30

40

⇒ n − 1 = 49.5 ⇒ n = 50.5 But n is an integer, n = 50. Hence, the largest common term is 31 + (50 − 1) 10 = 521.

In this case, (a − c )2 = (a + c )2 − 4ac = 0

b 2 = ac

Case II When a + c = 1 and ac = − …(i)

Also, as x is A between a and b a+ b …(ii) x= ∴ 2 b +c Similarly, y = …(iii) 2 a c 2a 2c Now, consider + = + x y a+ b b + c [using Eqs. (ii) and (iii)]  ab + ac + ac + bc  =2  2  ab + ac + b + bc   ab + bc + 2ac  =2   ab + bc + 2ac  [using Eq. (i)]

=2

14 Since, a, b, c are in AP ∴

2b = a + c

…(i)

Also, as a2 , b 2 and c 2 are in GP ∴

b 4 = a2c 2

Q

a+ b + c =

3 2 1 b = ⇒ 2 ⇒ a+ c =1 1 1 and ac = or − 4 4

∴

3b =

Let S=(1)+(1+2+4)+(4+6+9)

⇒ a=c But a ≠ c, as a < c.

13 Since, a, b, c are in GP. ∴

17 Statement I

…(ii)

+(9+12+16)+... +(361+380+400) = (0+0+1)+(1+2+4)+(4+6+9) 1 4

In this case, (a − c )2 = 1 + 1 = 2 ⇒

a−c = ± 2

But a < c, a − c = − 2 On solving a + c = 1 and a − c = − 2, we get 1 1 . a= − 2 2

[using Eq. (i)]

[using Eq. (i)]

Case I When a + c = 1 and ac =

1 4

=

⇒

10

∑ Sr

r =1

2

= 22 + 32 + K + 112 = 12 + 22 + 32 + K + 112 − 1 =

11 × 12 × 23 − 1 = 505 6

+ (r − 1)r + (r )2 ]

 r 3 − (r − 1)3   r =1  r − ( r − 1)  n

∑

[Q(a3 − b 3 ) = (a − b )(a2 + ab + b 2 )] =

1 ⇒ [(15a − 3b )2 + (3b − 5c )2 2 + (5c − 15a)2 ] = 0

1 is common ratio r +1 r Sr = =r +1 1 1− r +1

2

r =1

− (15a)(5c ) − (15a)(3b ) − (3b )(5c ) = 0

⇒ 15a = 3b, 3b = 5c and 5c = 15a ∴ 15a = 3b = 5c a b c = = = λ (say) ⇒ 1 5 3 ⇒ a = λ, b = 5λ, c = 3λ Hence, b, c and a are in AP.

n

∑ [(r − 1)

Sn =

− 75ac − 45ab − 15bc = 0 ⇒ (15a)2 + (3b )2 + (5c )2

first term and

[using Eq. (ii)]

Now, the sum to n terms of the series is

15 We have, 225a2 + 9b2 + 25c 2

16 Here, S r is sum of an infinite GP, r is

3 2

+(9+12+16) +... +(361+380+400) Now, we can clearly observe the elements in each bracket. The general term of the series is T r = (r − 1)2 + (r − 1)r + (r 2 )

n

∑ [r

3

− (r − 1)3 ]

r =1

= (13 − 03 ) + (23 − 13 ) + (33 − 23 ) + ... + [n3 − (n − 1)3 ] Rearranging the terms, we get S n = − 03 + (13 − 13 ) + (23 − 23 ) + (33 − 33 ) + ... + [(n − 1)3 − (n − 1)3 ] + n3 = n3 ⇒ S 20 = 8000 Hence, Statement I is correct. Statement II We have, already proved in the Statement I that Sn =

n

∑ (r

3

− (r − 1)3 ) = n3

r =1

Hence, Statement II is also correct and it is a correct explanation for Statement I.

DAY FOUR

Quadratic Equation and Inequalities Learning & Revision for the Day u

u

u

u

Quadratic Equation

u

Relation between Roots and Coefficients Formation of an Equation

u

Transformation of Equations

u

u

Maximum and Minimum Value of ax 2 + bx + c Sign of Quadratic Expression Position of Roots Inequalities

u

u

Arithmetic-GeometricHarmonic Mean Inequality Logarithm Inequality

Quadratic Equation ●

●

●

An equation of the form ax2 + bx + c = 0, where a ≠ 0, a, b and c, x ∈ R, is called a real quadratic equation. Here a, b and c are called the coefficients of the equation.

PRED MIRROR

The quantity D = b – 4ac is known as the discriminant of the equation ax + bx + c = 0 −b ± D and its roots are given by x = 2a 2

2

An equation of the form az2 + bz + c = 0, where a ≠ 0, a, b and c, z ∈C (complex) is called a −b ± D complex quadratic equation and its roots are given by z = . 2a

Nature of Roots of Quadratic Equation Let a, b , c ∈ R and a ≠ 0, then the equation ax + bx + c = 0 2

(i) has real and distinct roots if and only if D > 0. (ii) has real and equal roots if and only if D = 0. (iii) has complex roots with non-zero imaginary parts if and only if D < 0.

Your Personal Preparation Indicator u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

32

40

Some Important Results (i) If p + iq (where, p, q ∈ R, q ≠ 0) is one root of ax2 + bx + c = 0, then second root will be p − iq (ii) If a, b , c ∈ Q and p + q is an irrational root of ax + bx + c = 0, then other root will be p − q . 2

(iii) If a, b , c ∈Q and D is a perfect square, then ax2 + bx + c = 0 has rational roots. (iv) If a = 1, b , c ∈ I and roots of ax2 + bx + c = 0 are rational numbers, then these roots must be integers. (v) If the roots of ax2 + bx + c = 0 are both positive, then the signs of a and c should be like and opposite to the sign of b. (vi) If the roots of ax2 + bx + c = 0 are both negative, then signs of a, b and c should be like. (vii) If the roots of ax2 + bx + c = 0 are reciprocal to each other, then c = a. (viii) In the equation ax2 + bx + c = 0 (a, b , c ∈ R), if a + b + c = 0, then the roots are 1, c and if a c a − b + c = 0, then the roots are −1 and − . a

Relation between Roots and Coefficients Quadratic Roots If α and β are the roots of quadratic equation ax2 + bx + c = 0; b a ≠ 0, then sum of roots = α + β = − a c and product of roots = αβ = . a And, also ax2 + bx + c = a( x − α )( x − β)

Formation of an Equation Quadratic Equation If the roots of a quadratic equation are α and β, then the equation will be of the form x2 − (α + β) x + αβ = 0.

Cubic Equation If α , β and γ are the roots of the cubic equation, then the equation will be form of x3 − (α + β + γ) x2 + (αβ + βγ + γα ) x − αβγ = 0.

Transformation of Equations Let the given equation be a0 x n + a1 x n −1 + ... + an −1 x + an = 0 …(A) Then, the equation (i) whose roots are k (≠ 0) times roots of the Eq. (A), x is obtained by replacing x by in Eq. (A). k (ii) whose roots are the negatives of the roots of Eq. (A), is obtained by replacing x by − x in Eq. (A). (iii) whose roots are k more than the roots of Eq. (A), is obtained by replacing x by ( x − k ) in Eq. (A). (iv) whose roots are reciprocals of the roots of Eq. (A), is obtained by replacing x by 1/x in Eq. (A) and then multiply both the sides by x n .

Maximum and Minimum Value of ax2 + bx + c (i) When a > 0, then minimum value of ax2 + bx + c is −D 4ac − b 2 −b or at x = 4a 4a 2a (ii) When a < 0, then maximum value of ax2 + bx + c is −D −b 4ac − b 2 or at x = 4a 2a 4a

Cubic Roots If α , β and γ are the roots of cubic equation

b ax3 + bx2 + cx + d = 0; a ≠ 0, then α + β + γ = − a c βγ + γα + αβ = a d and αβγ = − a

Common Roots (Conditions) Suppose that the quadratic equations are ax + bx + c = 0 and a′ x2 + b ′ x + c′ = 0. 2

(i) When one root is common, then the condition is (a′ c − ac′ )2 = (bc′ − b ′ c)(ab ′ − a′ b ). (ii) When both roots are common, then the condition is a b c = = . a′ b ′ c′

Sign of Quadratic Expression Let f ( x) = ax2 + bx + c, where a, b and c ∈ R and a ≠ 0. (i) If a > 0 and D < 0, then f ( x) > 0, ∀x ∈ R. (ii) If a < 0 and D < 0, then f ( x) < 0, ∀x ∈ R. (iii) If a > 0 and D = 0, then f ( x) ≥ 0, ∀x ∈ R. (iv) If a < 0 and D = 0, then f ( x) ≤ 0, ∀x ∈ R. (v) If a > 0, D > 0 and f ( x) = 0 have two real roots α and β, where (α < β), then f ( x) > 0, ∀ x ∈(− ∞, α ) ∪ (β, ∞) and f ( x) < 0, ∀ x ∈(α , β). (vi) If a < 0, D > 0 and f ( x) = 0 have two real roots α and β, where (α < β), then f ( x) < 0, ∀ x ∈ (−∞, α ) ∪ (β, ∞) and f ( x) > 0, ∀ x ∈(α , β).

DAY

33

Position of Roots Let ax + bx + c = 0 has roots α and β. Then, we have the following conditions: 2

(i) with respect to one real number (k ). Situation

Required conditions

α 0, k > − b / 2a

k 0, af (k2 ) > 0, k1 < − b /2a < k2

α < k1 < k2 < β

D > 0, af (k1 ) < 0, af (k2 ) < 0

k1 < α < k2 < β

D > 0, f (k1 ) f (k2 ) < 0

Inequalities Let a and b be two real numbers. If a − b is negative, we say that a is less than b ( a < b ) and if a − b is positive, then a is greater than b (a > b ). This shows the inequalities concept.

Important Results on Inequalities (i) If a > b , then a ± c > b ± c, ∀ c ∈ R.

a (a) for m > 0, am > bm, > m a (b) for m < 0, am < bm, < m

(iii) (a) If a > b > 0 , then •a >b

2

b m b m

•| a| > |b |

(b) If a < b < 0, then • a2 > b 2

The Arithmetic-Geometric-Harmonic Mean of positive real numbers is defined as follows Arithmetic Mean ≥ Geometric Mean ≥ Harmonic Mean (i) If a, b > 0 then

• | a| > |b |

(iv) If a < 0 < b , then

Logarithm Inequality If a is a positive real number other than 1 and ax = m, then x is called the logarithm of m to the base a, written as log a m. In log a m, m should always be positive. (i) If m ≤ 0, then log a m will be meaningless. (ii) log a m exists, if m , a > 0 and a ≠ 1.

Important Results on Logarithm (i) alog a x = x ; a > 0, ≠ 1, x > 0 log b x

=x

log b a

; a , b > 0, ≠ 1, x > 0

(iii) log a a = 1, a > 0, ≠ 1 1 (iv) log a x = ; x , a > 0, ≠ 1 log x a 1 1 • < a b 1 1 • > a b

(a) a2 > b 2 , if| a| > |b | (b) a2 < b 2 , if| a| < |b | (v) If a < x < b and a, b are positive real numbers, then a2 < x2 < b 2 . (vi) If a < x < b and a is negative number and b is positive number, then (a) 0 ≤ x2 < b 2 , if |b | >| a | (b) 0 ≤ x2 < a2 , if | a | >|b | (vii) If ai > b i > 0 , where i = 1, 2, 3, ..., n, then a1 a2 a3 ... an > b1b2b3 ... b n . (viii) If ai > b i , where i = 1, 2, 3, ..., n, then a1 + a2 + a3 + ... + an > b1 + b2 + ... + b n . (ix) If| x | < a, then (a) for a > 0, − a < x < a. (b) for a < 0, x ∈ φ .

a+b 2 ≥ ab ≥ 1 1 2 + a b

(ii) If ai > 0 , where i = 1, 2, 3, K , n, then a 1 + a 2 + ... + an n ≥ (a1 ⋅ a2 ... an )1 / n ≥ 1 1 1 n + + ... + a1 a2 an

(ii) a

(ii) If a > b , then

2

Arithmetic-GeometricHarmonic Mean Inequality

(v) log a x = log a b logb x =

logb x ; a, b > 0, ≠ 1, x > 0 logb a

(vi) For x > 0; a > 0, ≠ 1 1 (a) log an ( x) = log a x n  m (b) log an x m =   log a x  n (vii) 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 (viii) If a > 1 and x > 0, then (a) log a x > p ⇒ x > a p (b) 0 < log a x < p ⇒ 0 < x < a p (ix) If 0 < a < 1 , then (a) log a x > p ⇒ 0 < x < a p (b) 0 < log a x < p ⇒ a p < x < 1

34

40

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 If 3x 2 − 7x − 30 + 2x 2 − 7x − 5 = x + 5, then x is equal to (a) 2

(b) 3

(c) 6

(d) 5

2 The number of solutions for equation x − 5 x + 6 = 0 is 2

(a) 4

(b) 3

(c) 2

(d) 1

2

3 The roots of the equation 2x − 1 − 3 2x − 1 + 2 = 0 are 1 1 1 3 3 1 1 3 (a) − , 0,  (b) − , 0,  (c) − , , 0,1 (d) − , 0,1,  2  2 2  2 2  2 2   2

4 The product of all the values of x satisfying the equation (5 + 2 6)

x2 − 3

(a) 4

+ (5 − 2 6)

x2 − 3

(b) 6

(c) 8

(d) 19

5 The root of the equation 2(1 + i )x − 4( 2 − i )x − 5 − 3i = 0, where i = −1, which has greater modulus, is 5 − 3i (b) 2

3+i (c) 2

3i + 1 (d) 2

6 x 2 + x + 1 + 2k ( x 2 − x − 1) = 0 is perfect square for how (b) 0

(c) 1

(d) 3

7 If the roots of (a + b )x − 2(bc + ad )x + c + d 2 = 0 are 2

2

2

2

equal, then (a)

a c = b d

(b)

a b a b + = 0 (c) = c d d c

(d) a + b = c + d

8 The least value of α for which tan θ and cot θ are roots of the equation x 2 + ax + 1 = 0, is (a) 2

(a) four real roots (b) exactly two real roots (c) either two or four real roots (d) atmost two real roots

15 The rational values of a in ax 2 + bx + 1 = 0 if root, are (a) a = 13, b = − 8 (c) a = 13, b = 8

(b) 1

(c) 1/2

(d) 0

9 If one root of the equation x 2 − λx + 12 = 0 is even prime

16 If 1 − i , is a root of the equation x 2 + ax + b = 0, where (a) 1,−1 (c) 3 ,−3

(b) 16

(c) 24

(d) 32

10 If a + b + c = 0, then the roots of the equation 4ax 2 + 3bx + 2c = 0, where a , b, c ∈ R are (a) real and distinct (c) real and equal

x 2 − 30x + p = 0 is the square of the other, is/are (b) 125 and − 216 (d) Only 216

(a) Only 125 (c) 125 and 215

variable x has real roots, then β lies in the interval (b) (− π, 0)

π π (c)  − ,   2 2

12 If ax 2 + 2bx − 3 c = 0 has no real root and then the range of c is (a) (−1, 1 ) (c) (0, ∞)

(d) (0, π)

3c < a + b, 4

(b) (0, 1) (d) (− ∞, 0)

13 If a , b and c are real numbers in AP, then the roots of ax 2 + bx + c = 0 are real for (a) all a and c c (c) −7 ≥ 4 3 a

x −m x +n are = mx + 1 nx + 1

reciprocal to each other, then (a) n = 0

(b) m = n

(c) m + n = 1

(d) m 2 + n 2 = 1

19 Let α and α 2 be the roots of x 2 + x + 1 = 0, then the equation whose roots are α 31 and α 62 , is (a) x 2 − x + 1 = 0 (c) x 2 + x + 1 = 0

(b) x 2 + x − 1 = 0 (d) x 60 + x 30 + 1 = 0

20 If α and β are the roots of x 2 − a( x − 1) + b = 0, then the value of (a)

1 1 2 is + 2 + α − aα β − aβ a + b

4 a+b

2

(b)

1 a+b

(c) 0

(d) −1

21 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 j least value is AIEEE 2005

(b) imaginary (d) infinite

11 The equation (cos β − 1) x 2 + (cos β)x + sin β = 0 in the (a) (0, 2 π)

(b) 2,− 2 (d) None of these

17 The values of p for which one root of the equation

while x 2 + λx + µ = 0 has equal roots, then µ is equal to (a) 8

1 is a 4+ 3

(b) a = − 13, b = 8 (d) a = − 13, b = − 8

18 If the roots of the quadratic equation

many value of k (a) 2

ac ≠ 0, then the equation P ( x ) ⋅ Q ( x ) = 0 has

a, b ∈ R , then the values of a and b are

= 10 is 2

3 − 5i (a) 2

14 If P ( x ) = ax 2 + bx + c and Q ( x ) = − ax 2 + dx + c = 0 ;

(b) no a and c a (d) +7 ≥2 3 c

(a) 2

(b) 3

(c) 0

(d) 1

22 If α and β be the roots of the equation 2x 2 + 2(a + b )x + a 2 + b 2 = 0, then the equation whose roots are (α + β )2 and (α − β )2 , is (a) x 2 − 2abx − (a 2 − b 2 )2 = 0 (b) x 2 − 4abx − (a 2 − b 2 )2 = 0 (c) x 2 − 4abx + (a 2 − b 2 )2 = 0(d) None of these

23 Let α , β be the roots of x 2 − 2x cos φ + 1 = 0, then the equation whose roots are α n and β n , is (a) x 2 − 2 x cosnφ − 1 = 0 (c) x 2 − 2 x sinnφ + 1 = 0

(b) x 2 − 2 x cosnφ + 1 = 0 (d) x 2 + 2 x sinnφ − 1 = 0

24 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

(d) 8

DAY

35

25 If the ratio of the roots of λx 2 + µx + ν = 0 is equal to the ratio of the roots of x + x + 1 = 0, then λ , µ and ν are in 2

(a) AP (c) HP

(a) 1 : 2 : 3

1 1 1 26 If the roots of the equation + = are equal in x +p x +q r magnitude but opposite in sign, then the product of the roots is (b) − (p 2 + q 2 )

k +1 k+2 and , then (a + b + c )2 is equal to k k +1 (c) b 2 − 4ac

(d) b 2 − 2ac

such that β < α < 0, then the quadratic equation whose roots are α , β , is given by (a) a x + b x + c = 0 (c) a x − b x + c = 0 2

(b) ax − b x + c = 0 2

(d) a x

2

+ b| x + c = 0

(ωα + ω 2β)(ω 2α + ωβ) is equal to α 2 β2 + β α q p

(b) αβ

(c) −

(d) ω

(c) {− 2, 5 }

j

p = 0, 4

(c) 4, 3

(d) −6, − 1

(c) 24

(d) 44

π P Q and tan are the roots of the 2 2 2 equation ax 2 + bx + c = 0, then (b) b = c + a (d) b = c

34 If the equation k ( 6x 2 + 3) + rx + 2x 2 − 1 = 0 and 6 k ( 2x 2 + 1) + px + 4x 2 − 2 = 0 have both roots common, then 2r − p is equal to (a) 2

(b) 1

(c) 0

(b) 0 < c < b 2 (d) | c | > | b | 2

(a) a1 + a2 + … + an (c) n (a1 + a2 + … + an )

(b) 2 (a1 + a2 + a3 + ... + an ) (d) None of these

39 If the roots of the equation bx 2 + cx + a = 0 is imaginary, then for all real values of x, the expression 3b 2 x 2 + 6bcx + 2c 2 is

j

AIEEE 2009

(b) less than 4ab (d) less than −4ab

(d) k

(b) a < − 5 (d) 2 < a < 5

 

1  is non-negative for all x

positive real x, then the minimum value of a must be

(c)

33 In ∆PQR , R = . If tan (a) a = b + c (c) c = a + b

(d) b = a + c 2

( x − a1 )2 + ( x − a 2 )2 +… + ( x − a n )2 assumes its least value at x =

(d) {3, − 5 }

roots of x 2 + 3x + 4 = 0, then (α + γ )(α + δ )(β + γ ) (β + δ ) is equal to (b) 18

2

(a) | c| < | b| 2 (c) | c| < | b| 2

(a) 0

32 If α and β are the roots of x 2 + x + 2 = 0 and γ,δ are the

(a) −18

(c) c = − a

37 If f ( x ) = x + 2bx + 2c and g ( x ) = − x − 2cx + b 2 such

JEE Mains 2013

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). j The correct roots of equation are AIEEE 2011 (b) 6, 1

(b) b = − c

2

41 If the expression ax − 1 +

31 Sachin and Rahul attemped to solve a quadratic

(a) −4, − 3

(a) a = − b

(a) − 5 < a < 2 (c) a > 5

p q

such that | α − β | = 10, then p belongs (b) {− 3, 2 }

ax 2 + bx + c = 0 by 1 is 2x 2 + 8x + 2 = 0, then

40 If x 2 + 2ax + 10 − 3a > 0 for all x ∈ R , then

30 If α and β are roots of the equation x 2 + px + 3. (a) {2, − 5 }

(d) 3 : 1 : 2

36 The equation formed by decreasing each root of

(a) greater than 4ab (c) greater than −4ab

29 If α and β be the roots of x 2 + px + q = 0, then

(a) −

(c) 1 : 3 : 2

38 If a ∈ R and a1, a 2 , a 3 … , a n ∈ R , then

28 If α and β are the roots of the equation ax 2 + bx + c = 0

2

(b) 3 : 2 : 1

AIEEE 2012

that min f ( x ) > max g ( x ), then the relation between b and c is

(d) −pq

27 If the roots of the equation ax 2 + bx + c = 0 of the form

(a) 2b 2 − ac (b) ∑a 2

a, b, c ∈ R , have a common root, then a : b : c is j

(b) GP (d) None of these

(a) −2 (p 2 + q 2 ) − (p 2 + q 2 ) (c) 2

35 If the equations x 2 + 2x + 3 = 0 and ax 2 + bx + c = 0,

1 2

1 4

(d) None of these

42 The number of real solutions of the equation x

 9 2   = − 3 + x − x is 10 (a) 0 (c)2

(b) 1 (d) None of these

43 If α and β be the roots of the quadratic equation ax 2 + bx + c = 0 and k be a real number, then the condition, so that α < k < β is given by (b) ak 2 + bk + c = 0 (d) a 2 k 2 + abk + ac < 0

(a) ac > 0 (c) ac < 0

44 2x − 3 < x + 5 , then x belongs to (a) (−3,5)

(b) (5, 9)

2 (c)  − , 8   3 

45 The least integral value α of x such that satisfies (a) α + 3 α − 4 = 0 (c) α 2 − 7α + 6 = 0 2

2 (d)  −8,   3

x −5 > 0, x 2 + 5x − 14 j

JEE Mains 2013

(b) α − 5 α + 4 = 0 (d) α 2 + 5 α − 6 = 0 2

36

40

46 The solution set of

| x − 2 | −1 ≤ 0 is |x −2| −2

51 If a, b, c are positive real numbers such that a + b + c = 1, then the greatest value of (1 − a ) (1 − b ) (1 − c ), is

(a) [0, 1] ∪ (3, 4)

(b) [0, 1] ∪ [3, 4]

(c) [−1, 1) ∪ (3, 4]

(d) None of these

47 Number of integral solutions of (a) 0

(b) 1

x +2 1 > is x2 +1 2

(c) 2

(d) 3

48 If the product of n positive numbers is 1, then their sum is (a) a positive integer 1 (c) equation to n + n

(b) divisible by n (d) greater than or equal to n

49 The minimum value of P = bcx + cay + abz , when xyz = abc, is (a) 3abc (c) abc

(a)

1 27

(b)

(c)

4 27

(d) 9

52 If a, b, c, d are positive real numbers such that a + b + c + d = 2, then M = (a + b ) (c + d ) satisfies the relation (a) 0 ≤ M ≤ 1 (b) 1 ≤ M ≤ 2

(c) 2 ≤ M ≤ 3

(d) 3 ≤ M ≤ 4

53 log2 ( x − 3x + 18) < 4, then x belongs to 2

(a) (1, 2) (c) (1, 16)

(b) (2, 16) (d) None of these

54 If log0. 3 ( x − 1) > log0. 09 ( x − 1), then x lies in (b) (− ∞, 1) (d) None of these

(a) (1, 2) (c) (2, ∞)

(b) 6abc (d) 4abc

55 What is the solution set of the following inequality?  x + 5 logx   >0 1 − 3x 

50 If a, b and c are distinct three positive real numbers, then  1 1 1 (a + b + c )  + +  is a b c (a) > 1 (c) < 9

8 27

(a) 0 < x < (b) > 9 (d) None of these

(c)

1 3

(b) x ≥ 3

1 < x 0 is (a) 9

(b) 8

(c) 2

(d) 1

9 For a > 0, the roots of the equation logax a + logx a 2 + loga 2 x a 3 = 0 is given by (a) a −4 / 3

(b) a −3 / 4

(c) a1 / 2

(d) a −1

DAY

37 19 If α and β are the roots of the equation ax 2 + bx + c = 0,

10 If a , b and c are in AP and if the equations (b − c ) x + (c − a )x + (a − b ) = 0 and 2 (c + a ) x + (b + c ) x = 0 have a common root, then 2

(a) a 2 , b 2 and c 2 are in AP (c) c 2 , a 2 and b 2 are in AP

2

(b) a 2 , c 2 and b 2 are in AP (d) None of these

11 If the equations x 2 + ax + 12 = 0, x 2 + bx + 15 = 0 and x 2 + (a + b )x + 36 = 0 have a common positive root, then the ordered pair (a , b ) is (a) (− 6, − 7) (c) (− 6, − 8)

(b) (− 7, − 8) (d) (− 8, − 7)

x 2 − 3x + 4 will be x 2 + 3x + 4

(a) 2, 1 (c) 7,

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 j JEE Mains 2014 possible values of a lie in the interval (b) (1, 2) (d) (−∞, − 2) ∪ (2, ∞)

2π 2π , then the quadratic equation 14 If a = cos + i sin 7 7 whose roots are α = a + a 2 + a 4 and β = a 3 + a 5 + a 6 , is (a) x 2 − x + 2 = 0 (c) x 2 + x + 2 = 0

(b) x 2 + 2 x + 2 = 0 (d) x 2 + x − 2 = 0

15 If α and β are roots of 375 x 2 − 25x − 2 = 0 and S n = α n + β n , then lim

n→ ∞

n

∑S

r

is equal to

7 116 29 (c) 358

(b)

1 12

16 If S = {a ∈ N, 1 ≤ a ≤ 100} and [tan x ] − tan x − a = 0 has 2

real roots, where [. ] denotes the greatest integer function, then number of elements in set S equals (b) 5

(c) 6

(d) 9

17 The sum of all real values of x satisfying the equation ( x 2 − 5 x + 5 )x (a) 3

2

+ 4 x − 60

= 1 is

(b) −4

j

(c) 6

JEE Mains 2016

(d) 5

18 If λ is an integer and α , β are the roots of

λ = 0 such that 1 < α < 2 and 2 < β < 3, then 4 the possible values of λ are 4x 2 − 16x +

(a) {60, 64, 68} (c) {49, 50,…, 62, 63}

x 2 − 2kx + k 2 + k − 5 = 0 are less than 5, then k lies in the j AIEEE 2005 interval (b) (−∞, 4)

(c) (6, ∞)

(d) (5, 6)

x 2 − 2mx + m 2 − 1 = 0 are greater than −2 but less than 4 lie in the interval (a) m > 3 (c) 1 < m < 4

(b) −1 < m < 3 (d) −2 < m < 0

22 Let α and β be real and z be a complex number. If

z 2 + αz + β = 0 has two distinct roots on the line j AIEEE 2011 Re ( z ) = 1, then it is necessary that (a) β ∈ (− 1, 0) (b) | β | = 1

23 The equation e (a) (b) (c) (d)

sin x

−e

(c) β ∈ [1, ∞)

− sin x

− 4 = 0 has

(d) β ∈ (0, 1) j

AIEEE 2012

infinite number of real roots no real root exactly one real root exactly four real roots

24 Ifa, b, c, d are positive real numbers such that a+

1 1 1 1 = 4, b + = 1, c + = 4 and d + = 1, then b c d a (b) b = d but a ≠ c (d) cd = 1and ab ≠ 1

(ω and ω2 are complex cube roots of unity)

(d) None of these

(a) 2

− b (1 − x) + c (1 − x)2 = 0 − b (x − 1) + c (x − 1)2 = 0 + b (1 − x) + c (1 − x)2 = 0 + b (x + 1) + c (1 + x)2 = 0

(a) a = c and b = d (c) ab = 1and cd ≠ 1

r =1

(a)

ax 2 ax 2 ax 2 ax 2

21 All the values of m for which both roots of the equation

1 5

(d) None of these

(a) (−1, 0) ∪ (0,1) (c) (−2, − 1)

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

(a) [4,5]

(b) 5,

1 7

β is 1+ β

α and 1+ α

20 If both the roots of the quadratic equation

12 If x is real, then the maximum and minimum value of the expression

then the quadratic equation whose roots are

(b) {61, 62, 63} (d) {62, 65, 68, 71, 75}

25 Let f : R → R be a continuous function defined by f (x ) =

1 e x + 2e − x

1 Statement I f (c ) = , for some c ∈ R . 3 Statement II 0 < f ( x ) ≤

1 2 2

, ∀ x ∈R . j

AIEEE 2010

(a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is true. Statement II is a correct explanation of 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

38

40

ANSWERS SESSION 1

SESSION 2

1. 11. 21. 31. 41. 51.

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

2. 12. 22. 32. 42. 52.

1. (c) 11. (b) 21. (b)

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

3. 13. 23. 33. 43. 53.

2. (d) 12. (c) 22. (c)

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

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

4. 14. 24. 34. 44. 54.

3. (c) 13. (a) 23. (b)

5. 15. 25. 35. 45. 55.

4. (c) 14. (c) 24. (a)

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

6. 16. 26. 36. 46.

5. (a) 15. (b) 25. (b)

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

7. 17. 27. 37. 47.

6. (d) 16. (d)

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

8. 18. 28. 38. 48.

7. (b) 17. (a)

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

8. (b) 18. (c)

9. 19. 29. 39. 49.

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

10. 20. 30. 40. 50.

9. (a) 19. (c)

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

10. (b) 20. (b)

Hints and Explanations SESSION 1

4 Q 5− 2 6 =

1 We have, 3 x2 − 7 x − 30 + 2 x2 − 7 x − 5 = x + 5 ⇒ 3 x2 − 7 x − 30 = ( x + 5) − 2 x2 − 7 x − 5 On squaring both sides, we get

∴

t +

1 5+ 2 6

1 = 10, t 2

where t = (5 + 2 6 )x

−3

⇒ t − 10t + 1 = 0

7 Since, roots are real.

3 x2 − 7 x − 30 = x2 + 25 + 10 x + (2 x2 − 7 x − 5) −2( x + 5) 2 x2 − 7 x − 5

⇒

t = 5± 2 6

or

t = (5 + 2 6 )± 1

⇒ −10 x − 50 = −2( x + 5) 2 x2 − 7 x − 5

From Eqs. (i) and (ii),

x + 5 ≠ 0, 2 x2 − 7 x − 5 = 5 [Q x = −5does not satisfy the given equation] ⇒ 2 x2 − 7 x − 30 = 0 ∴ x=6

2 Given equation is x2 − 5 x + 6 = 0

and ⇒

x2 + 3 x + 2 x + 6 = 0; x < 0 ( x − 3)( x − 2) = 0, x ≥ 0

and

( x + 3)⋅ ( x + 2) = 0, x < 0

∴ x = 3, x = 2 and x = −3, x = −2 There are four solutions of this equation.

3 Given equation is 2

2 x − 1 − 32 x − 1 + 2 = 0 Let 2 x − 1 = t , then t 2 − 3t + 2 = 0 ⇒ (t − 1)(t − 2) = 0 ⇒ t = 1,2 ⇒ 2 x − 1 = 1 and 2 x − 1 = 2 ⇒ 2 x − 1 = ±1 and 2 x − 1 = ±2 3 1 ⇒ x = 1, 0 and x = , − 2 2

∴ {2(bc + ad )}2 = 4(a2 + b 2 )(c 2 + d 2 ) …(ii)

⇒

x = − 2 , 2 , − 2, 2

2(1 + i )x2 − 4(2 − i )x − 5 − 3i = 0

=−

16(2 − i )2 + 8(1 + i )(5 + 3i ) 4(1 + i )

Now,

−1 − i = 2

1 1 + = 4 4

1 2

and

3 − 5i = 2

9 25 + = 4 4

17 2

Also,

1 2

Hence, required root is

2

2

Since, roots are real ∴ a2 − 4 ≥ 0 ⇒ a ≥ 2 Thus, the least value of a is 2.

9 We know that only even prime is 2, ∴ (2)2 − λ(2) + 12 = 0.

i 4− i −1 − i 3 − 5i or or = 1+ i 1+ i 2 2

17 > 2

4a c + 4b d − 8abcd = 0 4(ac − bd )2 = 0 ac = bd a b = d c 2 2

8 Given equation is x2 + ax + 1 = 0

5 The given equation is

⇒x=

+4b 2c 2 + 4b 2d 2

⇒

∴ Required product = 8

4(2 − i ) ±

⇒ 4b 2c 2 + 4a2d 2 + 8abcd = 4a2c 2 + 4a2d 2 ⇒ ⇒ ⇒

x2 − 3 = ± 1 x2 = 2, 4

⇒

When x ≥ 0, x2 − 5x + 6 = 0 and when x < 0, x2 + 5x + 6 = 0 ⇒ x2 − 3 x − 2 x + 6 = 0; x ≥ 0

…(i)

2

If equation is a perfect square then root are equal i.e. (1 − 2k )2 − 4(1 + 2k ) (1 − 2k ) = 0 1 −3 i.e. k = , . 2 10 Hence, total number of values = 2.

3 − 5i . 2

6 Given equation (1 + 2k ) x2 + (1 − 2k )x + (1 − 2k ) = 0

⇒

λ=8

...(i)

Qx2 + λx + µ = 0 has equal roots. [QD = 0] ∴ λ2 − 4µ = 0 ⇒ (8)2 − 4µ = 0 ⇒ µ = 16

10 Here, D = (3b )2 − 4 (4 a)(2c ) = 9b 2 − 32ac = 9 (− a − c )2 − 32ac = 9 a2 − 14 ac + 9c 2  a 2 14 a  = 9c 2    − ⋅ + 1   c 9 c   2   a 7 49   = 9c 2   −  − + 1 > 0   c 9 81   Hence, the roots are real and distinct.

DAY

39

11 For real roots, discriminant, D = b − 4 ac ≥ 0 = cos 2 β − 4(cos β − 1)sin β ≥ 0 = cos 2 β + 4(1 − cos β )sin β ≥ 0 So, sin β should be > 0. [Q cos 2 β ≥ 0,1 − cos β ≥ 0] ⇒ β ∈ (0, π ) 2

12 Here, D = 4b + 12ca < 0 2

⇒ b 2 + 3ca < 0 ⇒ ca < 0 If c > 0, then a < 0 3c Also, < a+ b 4 ⇒

…(i)

3ca > 4 a2 + 4 ab

⇒ b + 3ca > 4 a2 + 4 ab + b 2 …(ii) = (2a + b )2 ≥ 0 From (i) and (ii), c > 0, is not true. ∴ c 0 So, either D1 and D2 are positive or atleast one D’s is positive. Therefore, P ( x )⋅ Q ( x ) = 0 has either two or four real roots.

15 One root =

4− 3 4− 3 1 × = 13 4+ 3 4− 3

4+ 3 13 ∴The quadratic equation is 4+ 3 4 − 3 x2 −  +  x 13   13 ∴ Other root =

4+ 3 4− 3 + ⋅ =0 13 13 or 13 x2 − 8 x + 1 = 0 This equation must be identical with ax2 + bx + 1 = 0; ∴ a = 13 and b = − 8.

16 Sum of roots

−a = (1 − i ) + (1 + i ) ⇒ a = − 2. 1

[since, non-real complex roots occur in conjugate pairs] Product of roots, b = (1 − i )(1 + i ) ⇒ b = 2 1

17 Let roots be α and α2 . Then, α + α2 = 30 and α3 = p ⇒ α2 + α − 30 = 0 ⇒

(α + 6)(α − 5) = 0

∴ ⇒ and ∴

α p p p

= −6, 5 = α3 = (−6)3 = −216 = (5)3 = 125 = 125and −216 x−m x+ n 18 Given, = mx + 1 nx + 1 ⇒ x2 (m − n ) + 2mnx + (m + n ) = 0 1 Roots are α, respectively, then α 1 m+ n α⋅ = α m−n ⇒ m − n = m + n ⇒ n = 0.

19 Since, α,α2 be the roots of the equation x2 + x + 1 = 0 ∴ α + α 2 = −1 ... (i) 3 and α = 1 ... (ii) Now, α 31 + α 62 = α31 (1 + α31 ) 31 ⇒ α + α 62 = α30 ⋅ α(1 + α 30 ⋅ α) ⇒ α31 + α 62 = (α3 )10 ⋅ α{1 + (α3 )10 ⋅ α} ⇒ α31 + α 62 = α(1 + α) [from Eq. (ii)] [from Eq. (i)] ⇒ α31 + α 62 = −1 Again, α31 ⋅ α 62 = α 93 ⇒ α31 ⋅ α 62 = [α3 ]31 = 1 ∴ Required equation is x2 − (α31 + α 62 )x + α31 ⋅ α 62 = 0 ⇒ x2 + x + 1 = 0

20 Since, α and β are the roots of x2 − ax + a + b = 0, then a+ β = a and αβ = a + b ⇒ a2 + αβ = aα ⇒ α2 − aα = − (a + b ) and αβ + β2 = aβ ⇒ β2 − aβ = − (a + b ) 1 1 2 + + ∴ α2 − aα β2 − aβ a+ b 1 1 2 = + + =0 − (a + b ) − (a + b ) (a + b )

21 Let α and β be the roots of equation x2 − (a − 2)x − a − 1 = 0 Then, α + β = a − 2 and αβ = − a − 1 Now, α2 + β2 = (α + β )2 − 2αβ ⇒ α2 + β2 = (a − 2)2 + 2(a + 1) ⇒ α2 + β2 = a2 − 2a + 6 ⇒ α2 + β2 = (a − 1)2 + 5 The value of α2 + β2 will be least, if

a−1 = 0 a=1

⇒

22 Since, α and β are the roots of the equation 2 x2 + 2(a + b )x + a2 + b 2 = 0 ∴

(α + β )2 = (a + b )2 and αβ =

a2 + b 2 2

Now, (α − β )2 = (α + β )2 − 4αβ  a2 + b 2  = (a + b )2 − 4   2 

= − (a − b )2 Now, the required equation whose roots are (α + β )2 and (α − β )2 is x2 − {(α + β )2 + (α − β )2 } x + (α + β )2 (α − β )2 = 0 ⇒ x2 − {(a + b )2 − (a − b )2 } x − (a + b )2 (a − b )2 = 0 2 ⇒ x − 4abx − (a2 − b 2 )2 = 0

23 The given equation is x2 − 2 x cos φ + 1 = 0

4cos 2 φ − 4 2 = cos φ ± i sin φ Let α = cos φ + i sin φ, then β = cos φ − i sin φ ∴ α n + β n = (cos φ + i sin φ)n + (cos φ − i sin φ)n = 2cos nφ and α nβ n = (cos nφ + i sin nφ) ⋅ (cos nφ − i sin nφ) = cos 2 nφ + sin2 nφ = 1 ∴ Required equation is x2 − 2 x cos nφ + 1 = 0 24 Given equation is (5 + 2 )x2 − (4 + 5)x + 8 + 2 5 = 0 Let x1 and x2 are the roots of the equation, then ∴

x=

2cos φ ±

x1 + x2 =

4+ 5 5+ 2

... (i)

8 + 2 5 2(4 + 5) = 5+ 2 5+ 2 ... (ii) = 2( x1 + x2 ) Clearly, harmonic mean 2 x1 x2 4( x1 + x2 ) = = =4 x1 + x2 ( x1 + x2 ) and

x1 x2 =

[from Eq. (ii)]

25 Let α, β and α ′, β ′ be the roots of the given equations, respectively. µ ν ∴ α + β = − , αβ = λ λ and α ′ = ω and β ′ = ω2 α α′ [given] = Q β β′ α ω ⇒ = ⇒ β = αω β ω2 From Eq. (i),

…(i)

40

40

µ 2 ν ,α ω = λ λ µ ν − αω2 = − , α2ω = λ λ [Q− ω2 = 1 + ω] µ2 ν = ⇒ µ 2 = λν λ2 λ

α + αω = − ⇒

⇒

26 Simplified form of given equation is (2x + p + q ) r = ( x + p )( x + q ) ⇒ x + ( p + q − 2r )x − ( p + q ) r + pq = 0 Since, sum of roots = 0 ⇒ − ( p + q − 2r ) = 0 p+q r = ⇒ 2 and product of roots = − ( p + q ) r + pq ( p + q )2 =− + pq 2 1 = − ( p2 + q 2 ) 2 2

27 Clearly, sum of roots, k+1 k+2 b ... (i) + =− k k+1 a and product of roots, k+1 k+2 c × = k k+1 a k+2 c = ⇒ k a 2 c c−a 2a ⇒ = −1 = ⇒k = k a a c−a On putting the value of k in Eq. (i), we get c+a 2c b + =− 2a c+a a ⇒ (c + a)2 + 4ac = −2b(a + c ) ⇒ (a + c )2 + 2b(a + c ) = −4ac ⇒ (a + c )2 + 2b(a + c ) + b 2 = b 2 − 4ac ⇒ (a + b + c )2 = b 2 − 4ac

(Qω + ω2 = −1) = α2 + β2 − αβ 2 2 = (α + β ) − 3αβ = p − 3q α 2 β2 α 3 + β3 Also, + = β α αβ p(3q − p2 ) (α + β )3 − 3αβ(α + β) = = αβ q ∴ The given expression ( p2 − 3q ) q = =− p(3q − p2 ) p q

30 Clearly, α + β = − p and αβ = Also, (α − β )2 = 10 ⇒ (α + β )2 − 4αβ = 10 ⇒ p2 − 3 p = 10 ⇒ ( p + 2)( p − 5) = 0 ∴ p = −2, 5

ax2 + bx + c = 0 and its roots are α and β. Sachin made a mistake in writing down constant term. ∴ 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. α ⋅ β = 6 ⇒ Correct quadratic equation is x2 − (α + β )x + αβ = 0. ⇒ x2 − 7 x + 6 = 0 having roots 1 and 6.

32 Since, α + β = − 1,αβ = 2, γ + δ = − 3, and γδ = 4 ∴ (α + γ )(α + δ )(β + γ )(β + δ ) = (α2 − 3α + 4)(β2 − 3β + 4) = 4 − 3αβ2 + 4β2 − 3α2β + 9αβ − 12β + 4α2 − 12α + 16 = 4 − 3(2)β + 4β2 + 4α2 −3( 2)α + 9( 2) − 12 ( β + α ) + 16

2

= 4 − 6β + 4(α2 + β2 )

∴

Hence, required eqution is b c x2 − x + =0 a a ⇒

a x2 − b x + c = 0

− 6α + 18 + 12 + 16 = 50 + 6 + 4[(α + β )2 − 2αβ] = 56 − 12 = 44

33 Given, R = π ⇒ P + Q = π 2

⇒

⇒

29 Since, α and β are the roots of the equation x2 + px + q = 0 therefore α + β = − p and αβ = q Now, (ωα + ω2β )(ω2α + ωβ ) = α2 + β2 + (ω 4 + ω2 )αβ (Qω3 = 1)

⇒

and (12k + 4)x2 + px + 6 k − 2 = 0 For both common roots, 6k + 2 r 3k − 1 = = 12 k + 4 p 6 k − 2 r 1 ⇒ = ⇒ 2r − p = 0 p 2

35 Given equations are ... (i) x2 + 2 x + 3 = 0 ... (ii) and ax2 + 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.

36 α, β be the roots of ax2 + bx + c = 0

31 Let the quadratic equation be

28 Since, α and β are the roots of the equation ax + bx + c = 0 b c α + β = − and αβ = a a Now, sum of roots = α + β = − α −β (Qβ < α < 0) b b  = − −  = (Qα + β > 0)  a a c and product of roots = α β = a

3p 4

34 Given, (6 k + 2)x2 + rx + 3 k − 1 = 0

2

P Q π + = 2 2 4 Q P tan + tan π 2 2 1 = tan = 4 1 − tan P tan Q 2 2 b − a b 1= = ⇒a+ b =c c c −a 1− a

∴ α + β = − b / a, αβ = c / a Now roots are α − 1, β − 1 Their sum, α + β − 2 = (− b / a) − 2 = − 8 /2 = − 4 Their product, (α − 1) (β − 1) = αβ − (α + β ) + 1 = c /a + b/a + 1 = 1 ∴ b / a = 2 i.e. b = 2a also c + b = 0 ⇒ b = − c . 4b 2 − 8c 2 37 min f ( x ) = − D = −

4a 4 = − (b 2 − 2c 2 ) (upward parabola) 4c 2 + 4b 2 D max g ( x ) = − = 4a 4 = b2 + c 2 (downward parabola) Now, 2c 2 − b 2 > b 2 + c 2 ⇒ c 2 > 2b 2 ⇒ |c| > 2 |b|

38 We have, ( x − a1 )2 + ( x − a2 )2 + … + ( x − an )2 = n x2 − 2 x(a1 + a2 + … + an ) + (a12 + a22 + …+ a2n ) So, it attains its minimum value at 2(a1 + a2 + … + an ) x= 2n a + a2 + … + an  using : x = −b  = 1 n 2a  

39 Given bx2 + cx + a = 0 has imaginary roots. …(i) ⇒ c 2 − 4ab < 0 ⇒ c 2 < 4ab Let f ( x ) = 3b 2 x2 + 6bcx + 2c 2 Here, 3b 2 > 0 So, the given expression has a minimum value. −D ∴ Minimum value = 4a 4(3b 2 )(2c 2 ) − 36b 2c 2 4ac − b 2 = = 4a 4(3b 2 )

DAY

41 =−

12b 2c 2 = − c 2 > −4ab 12b 2 [from Eq. (i)]

40 According to given condition, 4a2 − 4(10 − 3a) < 0 ⇒ a2 + 3a − 10 < 0 ⇒ (a + 5) (a − 2) < 0 ⇒ − 5 < a < 2. 41 We have, ax − 1 + 1 ≥ 0 x ax2 − x + 1 ≥0 ⇒ x 2 ⇒ ax − x + 1 ≤ 0 as x > 0 It will hold if a > 0 and D ≤ 0 1 a > 0 and 1 − 4a ≤ 0 ⇒ a ≥ 4 1 Therefore, the minimum value of a is . 4

42 Let f ( x ) = − 3 + x − x2 . Then, f ( x ) < 0 for all x, because coeff. of x2 < 0 and disc. < 0. Thus, LHS of the given equation is always positive whereas the RHS is always less than zero. Hence, the given equation has no solution.

43 Let f ( x ) = ax2 + bx + c . Then, k lies betweenα andβ, if a f (k ) < 0 ⇒ a (ak 2 + bk + c ) < 0 2 ⇒ a k 2 + abk + ac < 0.

44 We have, 2 x − 3 < x + 5 ⇒

⇒

⇒

⇒ ⇒

2x − 3 − x + 5 < 0   3 − 2 x + x + 5 < 0, x ≤ − 5 3  3 − 2 x − x − 5 < 0, − 5 < x ≤ 2  3  2 x − 3 − x − 5 < 0, x >  2  x > 8, x ≤ −5  2 3  x > − ,− 5< x ≤  3 2  3  x < 8, x >  2 2 3  3   x ∈  − , ∪  , 8  3 2   2  2 x ∈  − , 8  3  x−5 >0 x + 5x − 14

45

2

⇒

( x + 7 )( x − 2) ( x − 5) >0 ( x − 2)2 ( x + 7 )2

⇒ x ∈ (−7, 2) ∪ (5, ∞ ) So, the least integral value α of x is − 6, which satisfy the equation α2 + 5α − 6 = 0

46 Given,

| x − 2| − 1 ≤0 | x − 2| − 2

51 Using GM ≤ AM, we have {(1 − a) (1 − b ) (1 − c )1 /3 } 1− a+ 1−b + 1−c ≤ 3 8 ⇒ (1 − a) (1 − b ) (1 − c ) ≤ 27 8 Hence, the greatest value is . 27

Let | x − 2 |= k Then, given equation, (k − 1) (k − 2) k −1 ≤0 ⇒ ≤0 k −2 (k − 2)2 ⇒ (k − 1) (k − 2) ≤ 0 ⇒ 1 ≤ k ≤ 2 ⇒ 1 ≤ | x − 2|≤ 2

52 Using AM ≥ GM, we have

Case I When 1 ≤ | x − 2| ⇒ ⇒ ⇒

| x − 2|≥ 1 x − 2 ≥ 1 or x − 2 ≤ −1 x ≥ 3 and x ≤ 1

…(i)

Case II When | x − 2|≤ 2 ⇒ −2 ≤ x − 2 ≤ 2 ⇒ −2 + 2 ≤ x ≤ 2 + 2 ⇒ 0≤ x ≤ 4 From (i) and (ii), x ∈ [0, 1] ∪ [3, 4] x+2 1 47 − >0 x2 + 1 2

...(ii)

54. Clearly, x − 1 > 0 ⇒ x > 1

3 + 2 x − x2 >0 2( x2 + 1)

and log 0 . 3 ( x − 1) > log ( 0 . 3 )2 ( x − 1) ⇒ ⇒

such that a1 a2 … an = 1. Using AM ≥ GM, we have a + a2 + … + an ⇒ 1 ≥ (a1 a2 … an )1 / n n ⇒ a1 + a2 + … an ≥ n.

49 Since, AM ≥ GM

bcx + cay + abz ≥ (a2b 2c 2 ⋅ xyz )1 /3 3 ⇒ bcx + cay + abz ≥ 3abc [Qxyz = abc ]

∴

⇒

3  1 + 1 + 1 >    a b c  (abc )1 /3

x ∈ (1, 2)

55 By definition of log x, x > 0 and  x + 5  >0  1 − 3x  ⇒

( x + 5)(1 − 3 x ) >0 (1 − 3 x )2

⇒

( x + 5)(1 − 3 x ) > 0

⇒

( x + 5)(3 x − 1) < 0

⇒

... (i)

− 5 < x < 1/3 x > 0, 0 < x < 1 / 3

As ∴

50 We know that, AM > GM

and

1 log 0 . 3 ( x − 1) 2 log 0 . 3 ( x − 1) > 0 ⇒ x < 2

log 0 . 3 ( x − 1) >

Hence,

48 Let a1 , a2 , … , an be n positive integers

a+ b + c > (abc )1 /3 3 1 1 1 1 /3 + + a b c >  1   abc  3

x2 − 3 x + 18 < 16 (Qlog a b < c ⇔ b < ac , if a > 1) x2 − 3 x + 2 < 0 ( x − 1)( x − 2) < 0 ⇒ x ∈ (1,2)

⇒ ⇒

Since, denominator is positive ∴3 + 2 x − x2 > 0 ⇒ − 1< x < 3 ⇒ x = 0, 1, 2

∴

53 log 2 ( x2 − 3 x + 18) < 4 ⇒

− x2 − 1 + 2 x + 4 ⇒ >0 2( x2 + 1) ⇒

(a + b ) + (c + d ) ≥ {(a + b ) (c + d )}1 /2 2 2 ⇒ ≥ M 1 /2 ⇒ M ≤ 1 2 As a, b, c , d > 0. Therefore, M = (a + b ) (c + d ) > 0. Hence, 0 ≤ M ≤ 1.

x+ 5 < 1⇒ x < − 1 1 − 3x

This does not satisfy 0 < x < 1 / 3. Hence, there is no solution.

SESSION 2 1 We have ... (ii)

From (i) and (ii), we get  a + b + c   1 + 1 + 1      a b c  3 3 > ⋅ (abc )1 /3 (abc )1 /3 1 1 1 ⇒ (a + b + c ) + +  > 9 a b c

2 x −3 + Let ⇒

x ( x − 6) + 6 = 0

x −3 = y x = y+3

∴ 2 y + ( y + 3)( y − 3) + 6 = 0 ⇒ 2 y + y2 − 3 = 0 2 ⇒ y + 2 y −3 = 0 ⇒ ⇒

( y + 3)( y − 1) = 0 y ≠ −3 ⇒ y = 1

42

40

⇒ ⇒ ⇒

y = ±1 x − 3 = ±1 x = 4,2

⇒

5 Given equation can be rewritten as

x = 16, 4

2 We have, 2

x +2

⋅ 33 x / ( x − 1 ) = 9

Taking log on both sides, we get 3x ( x + 2)log 2 + log 3 = 2 log 3 ( x − 1) ⇒ ( x2 + x − 2)log 2 + 3 x log 3 = 2( x − 1)log 3 ⇒ x2 log 2 + (log 2 + log 3)x − 2 log 2 + 2 log 3 = 0 − (log 2 + log 3) ± {(log 2 + log 3)2 − 4 log 2 (− 2 log 2 + 2 log 3)}

∴ x=

2 log 2 − (log 2 + log 3) ±

{(3 log 2)2 − 6 log 2 log 3 + (log 3)2 }

=

2 log 2 =

− (log 2 + log 3) ± (3 log 2 − log 3) 2 log 2

∴ x = − 2, 1 −

log 3 log 2

equation x2 − 6 x − 2 = 0 Qan = α n − β n , n ≥ 1 ⇒ a10 = α10 − β10 a8 = α 8 − β 8 and a9 = α 9 − β 9 Now, consider

[Qα and β are the

roots of α10 − β10 − 2(α 8 − β 8 ) x2 − 6 x − 2 = 0 = 2(α 9 − β 9 ) or x2 = 6 x + 2 = = =

α 8(α2 − 2) − β 8(β2 − 2) ⇒ α2 = 6α + 2 2(α 9 − β 9 ) ⇒ α2 − 2 = 6α and β2 = 6β + 2 α 8 ⋅ 6α − β 8 ⋅ 6β ⇒

2(α 9 − β 9 )

β2 − 2 = 6β]

6α 9 − 6β 9 6 = =3 2(α 9 − β 9 ) 2

4 We have, x2 − 5x + 1 = 0 1 1 = 5 ⇒ x2 + 2 = 52 − 2 = 23 x x 1 and x3 + 3 = 53 − 3 × 5 = 110 x 1 1 Now,  x2 + 2   x3 + 3   x   x  x+

= 23 × 110 = 2530 ⇒ x5 +

Since, p,q and r are in AP. ∴ 2q = p + r 1 1 α+β Also, + =4 ⇒ =4 α β αβ −q 4r ⇒ α + β = 4αβ ⇒ = p p 2q = p + r 2(−4r ) = p + r ⇒ p = −9r −q 4r 4r 4 α+β = = = =− p p −9r 9 r r 1 αβ = = = p −9r −9

Q ⇒ Q and

3 Given, α and β are the roots of the

a10 − 2a8 2a9

3 x2 − (a + c + 2b + 2d )x + ac + 2bd = 0 Now, discriminant, D = (a + c + 2b + 2d )2 − 4⋅ 3(ac + 2bd ) = {(a + 2d ) + (c + 2b )}2 − 12(ac + 2bd ) = {(a + 2d ) − (c + 2b )}2 + 4(a + 2d ) (c + 2b ) − 12(ac + 2bd ) = {(a + 2d ) − (c + 2b )}2 −8ac + 8ab + 8dc − 8bd = {(a + 2d ) − (c + 2b )}2 + 8(c − b )(d − a) Which is + ve, since a < b < c < d . Hence, roots are real and distinct. 6 Since, α and β are roots of px2 + qx + r = 0, p ≠ 0 −q r , αβ = ∴ α+β = p p

1 1 = 2530 −  x +  = 2525  x5 x

Now, consider (α − β )2 = (α + β )2 − 4αβ 16 4 16 + 36 = + = 81 9 81 52 2 ⇒ (α − β ) = 81 2 ⇒ α −β = 13 9

7 α + β = −b / a and αβ = c / a Now, (1 + α + α2 )(1 + β + β2 ) = 1 + (α + β ) + (α2 + β2 ) + αβ + αβ(α + β) + (αβ) 2 = 1 + (α + β ) + (α + β )2 − αβ + αβ(α + β) + (αβ) 2 b b2 c c b c2 = 1 − + 2 − +    −  + 2 a a a  a  a a (a2 + b 2 + c 2 − ab − bc − ca) = a2 2 = [(a − b ) + (b − c )2 + (c − a)2 ] / 2a2 which is positive.

8 Using AM ≥ GM −5

−4

a + a

−3

+ a

−3

+ a

−3

+ a

+ 1 + a8 + a10 ∴ 8 ≥ (a−5 ⋅ a−4 ⋅ a−3 ⋅ a−3 ⋅ a−3 ⋅ 1 ⋅ a8 ⋅ a10 )1 / 8 ⇒ a−5 + a−4 + 3a−3 + 1 + a8 + a10 ≥ 8⋅1 Hence, minimum value is 8.

9 We have, log a a 2 log a a + log a a + log a x log a x 3 log a a + =0 2 log a a + log a x 1 2 3 + + =0 ⇒ 1+ t t 2+ t ⇒

( let log a x = t ) 2t + t 2 + 2t 2 + 6t + 4 + 3t 2 + 3t =0 t (1 + t ) (2 + t )

⇒ ⇒ ⇒ ⇒ ⇒ ∴

6t 2 + 11t + 4 = 0 6t + 8t + 3t + 4 = 0 (2t + 1) (3t + 4) = 0 1 4 t = − ,− 2 3 1 4 log a x = − , − 2 3 x = a−1 /2 , a−4 /3 2

10 Since, x = 1 is a root of first equation. If α is another root of first equation, then a−b α × 1= α = b −c (Product of roots) 2a − 2b 2a − (a + c ) = = =1 2b − 2c a + c − 2c So, the roots of first equation are 1 and 1. Since, the equations have a common root, 1 is also a root of second equation. ⇒ 2(c + a) + b + c = 0 ⇒ 2(2b ) + b + c = 0 [since, a, b and c are in AP] ⇒ c = − 5b Also, a + c = 2b ⇒ a = 2b − c = 2b + 5 b = 7b ∴ a2 = 49b 2 , c 2 = 25 b 2 Hence, a2 , c 2 and b 2 are in AP.

11 On adding first and second equations, we get 2 x2 + (a + b )x + 27 = 0 On subtracting above equation from given third equation, we get x2 − 9 = 0 ⇒ x = 3, − 3 Thus, common positive root is 3. Now, (3)2 + 3 a + 12 = 0 ⇒ a = − 7 and 9 + 3b + 15 = 0 ⇒ b = − 8 Hence, the order pair (a, b ) is (− 7, − 8). 2 12 Let y = x2 − 3 x + 4

x + 3x + 4 ⇒ ( y − 1)x2 + 3( y + 1)x + 4( y − 1) = 0 Q x is real. ∴ D≥0 ⇒ 9( y + 1)2 − 16( y − 1)2 ≥ 0 ⇒ −7 y 2 + 50 y − 7 ≥ 0 ⇒ 7 y 2 − 50 y + 7 ≤ 0 ⇒ ( y − 7)(7 y − 1) ≤ 0

DAY

43

⇒

y ≤ 7 and y ≥

1 7

1 ≤ y≤7 7 Hence, maximum value is 7 and 1 minimum value is . 7

Now, consider n

lim

⇒

13 Here, a ∈ R and equation is

−3{ x − [ x]}2 + 2{ x − [ x]} + a2 = 0 Let t = x − [ x], then −3t 2 + 2t + a2 = 0 1 ± 1 + 3a2 ⇒ t = 3 Q t = x − [ x] = { x} [fractional part] ∴ 0≤ t ≤ 1 1 ± 1 + 3a2 ⇒ 0≤

and

0
0 Now, D≥0 ⇒ 4m2 − 4m2 + 4 ≥ 0 ... (i) ⇒ 4> 0⇒m ∈R

(m + 3) (m + 1) > 0 −∞ < m < −3 and ... (iv) −1 < m < ∞ From (i), (ii), (iii) and (iv), we get m lie between −1 and 3. ⇒

22 Let z = x + iy, given Re(z) = 1 ∴ x = 1 ⇒ z = 1 + iy Since, the complex roots are conjugate to each other. So, z = 1 + iy and 1 − iy are two roots of z2 + α z + β = 0. Q Product of roots = β ⇒ (1 + iy )(1 − iy ) = β ∴ β = 1 + y 2 ≥ 1 ⇒ β ∈ [1, ∞ )

23 Given equation is e sin x − e − sin x = 4 ⇒ e sin x −

1 =4 e sin x

1 =4 t 2 2 ⇒ t − 1 − 4t = 0 ⇒ t − 4t − 1 = 0 4 ± 16 + 4 ⇒ t = 2 t = 2 ± 5 ⇒ e sin x = 2 ± 5 But −1 ≤ sin x ≤ 1 ⇒ e −1 ≤ e sin x ≤ e 1 1 ⇒ e sin x ∈  ,e   e  Also, 0 < e < 2 + 5 Hence, given equation has no solution. Let e sin x = t, then t −

b 0 ⇒ m2 − 8m + 15 > 0 ⇒ (m − 3)(m − 5) > 0 ⇒ −∞ < m < 3 and 5 < m < ∞ ... (iii) and f (−2) > 0 ⇒ 4 + 4m + m2 − 1 > 0 ⇒ m2 + 4m + 3 > 0

24 Using AM > GM, we have 1 a 1 b >2 , b + >2 b b c c 1 c 1 d c + >2 and d + > 2 d d a a  a + 1   b + 1   c + 1   d + 1  > 16         b  c  d  a a+

1 1 1 But,  a +   b +   c +   b  c  d  d + 1  = 4 × 1 × 4 × 1 = 16    a 1 1 1 1 and d = ∴ a = ,b = ,c b c d a 1 1 1 1 ⇒ a = = 2, b = = , c = = 2 b c 2 d 1 1 and d = = a 2 1 1 ⇒ a = 2, b = ,c = 2 and d = 2 2 ⇒ a = c and b = d 1 25 We have, f ( x ) = 2 x e + x e Using AM ≥ GM, we get 2 1 /2 ex + x e ≥  e x ⋅ 2  , as e x > 0 x  2 e  2 x ⇒ e + x ≥2 2 e 1 1 0< ≤ ⇒ 2 ex + x 2 2 e 1 ∴ 0 < f ( x) ≤ ,∀x∈R 2 2 Statement II is true and Statement I is also true as for some ‘c’. 1 ⇒ f (c ) = [for c = 0] 3 1 which lies betwen 0 and . 2 2 So, Statement II is correct explanation of Statement I.

DAY FIVE

Matrices Learning & Revision for the Day u

u

u

Matrix Types of Matrices Equality of Matrices

u

u

u

Algebra of Matrices Transpose of a Matrix Some Special Matrices

u

u

u

Trace of a Matrix Equivalent Matrices Invertible Matrices

Matrix ●

●

●

A matrix is an arrangement of numbers in rows and columns. A matrix having m rows and n columns is called a matrix of order m × n and the number of elements in this matrix will be mn.  a11 a12 a13 ... a1 n  a a22 a23 ... a2 n   A matrix of order m × n is of the form A =  21  ... ... ... ... ...  a   m1 am2 am3 ... amn 

Some important terms related to matrices ●

●

●

The element in the ith row and jth column is denoted by aij. The elements a11 , a22 , a33 , ...... are called diagonal elements. The line along which the diagonal elements lie is called the principal diagonal or simply the diagonal of the matrix.

PRED MIRROR Your Personal Preparation Indicator

Types of Matrices ●

●

●

●

●

If all elements of a matrix are zero, then it is called a null or zero matrix and it is denoted by O. A matrix which has only one row and any number of columns is called a row matrix and if it has only one column and any number of rows, then it is called a column matrix. If in a matrix, the number of rows and columns are equal, then it is called a square matrix. If A = [aij]n × n , then it is known as square matrix of order n. If in a matrix, the number of rows is less/greater than the number of columns, then it is called rectangular matrix. If in a square matrix, all the non-diagonal elements are zero, it is called a diagonal matrix.

u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

46

●

●

●

FIVE

40

If in a square matrix, all non-diagonal elements are zero and diagonal elements are equal, then it is called a scalar matrix. If in a square matrix, all non-diagonal elements are zero and diagonal elements are unity, then it is called an unit (identity) matrix. We denote the identity matrix of order n by I n and when order is clear from context then we simply write it as I. In a square matrix, if aij = 0, ∀ i > j, then it is called an upper triangular matrix and if aij = 0, ∀i < j , then it is called a lower triangular matrix.

NOTE

Transpose of a Matrix Let A be m × n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A and is denoted by A′ or AC or A T . If A be m × n matrix, A′ will be n × m matrix.

Important Results (i) If A and B are two matrices of order m × n, then ( A ± B)′ = A′ ± B ′ (ii) If k is a scalar, then (k A)′ = k A′ (iii) ( A′ )′ = A (iv) ( AB)′ = B′ A′ (v) ( A n )′ = ( A′ )n

• The diagonal elements of diagonal matrix may or may not be zero.

Equality of Matrices Two matrices A and B are said to be equal, if they are of same order and all the corresponding elements are equal.

Algebra of Matrices ●

●

●

If A = [aij]m × n and B = [b ij] m × n be two matrices of same order, then A + B = [aij + b ij]m × n and A − B = [aij − b ij]m × n , where i = 1, 2, ..., m, j = 1, 2, ..., n. If A = [aij] be an m × n matrix and k be any scalar, then, kA = [kaij]m × n . If A = [aij]m × n and B = [b ij]n × p be any two matrices such that number of columns of A is equal to the number of rows of B, then the product matrix AB = [c ij], of order m × p, where c ij =

ik

b kj .

Some Important Properties ●

●

●

●

●

●

●

A + B = B + A (Commutativity of addition) ( A + B) + C = A + (B + C) (Associativity of addition) α ( A + B) = αA + αB, where α is any scalar. (α + β) A = αA + βA, where α and β are any scalars. α (βA) = (αβ) A, where α and β are any scalars. ( AB) C = A (BC) (Associativity of multiplication) AI = A = IA A (B + C) = AB + AC (Distributive property)

NOTE

●

●

●

●

●

●

n

∑a

k =1

●

Some Special Matrices

• A2 = A ⋅ A, A3 = A ⋅ A ⋅ A = A2 ⋅ A1 , K • If the product AB is possible, then it is not necessary that the product BA is also possible. Also, it is not necessary that AB = BA. • The product of two non-zero matrices can be a zero matrix.

●

A square matrix A is called an idempotent matrix, if it satisfies the relation A2 = A. A square matrix A is called nilpotent matrix of order k, if it satisfies the relation A k = O, for some k ∈ N . The least value of k is called the index of the nilpotent matrix A. A square matrix A is called an involutary matrix, if it satisfies the relation A2 = I . A square matrix A is called an orthogonal matrix, if it satisfies the relation AA′ = I or A′ A = I . A square matrix A is called symmetric matrix, if it satisfies the relation A′ = A. A square matrix A is called skew-symmetric matrix, if it satisfies the relation A′ = − A.

NOTE

• If A and B are idempotent matrices, then A + B is idempotent iff AB = − BA.  a1 a2 • If A =  b1 b2   c1 c 2

a3  b3  is orthogonal, then  c 3 

Σ ai2 = Σ bi2 = Σ c i2 = 1 and Σ ai bi = Σ bi c i = Σ ai c i = 0 • If A B are symmetric matrices of the same order, then (i) AB is symmetric if and only if AB = BA . (ii) A ± B , AB + BA are also symmetric matrices. • If A and B are two skew-symmetric matrices, then (i) A ± B, AB − BA are skew-symmetric matrices. (ii) AB + BA is a symmetric matrix. • Every square matrix can be uniquely expressed as the sum of symmetric and skew-symmetric matrices. 1 1 1 1 i.e. A = ( A + A′ ) + ( A − A′ ), where ( A + A′ ) and ( A − A′ ) 2 2 2 2 are symmetric and skew-symmetric respectively.

47

DAY

(iii) Addition of constant multiple of the elements of any row (column) to the corresponding elements of any other row (column), indicated as Ri → Ri + kR j (Ci → Ci + kC j).

Trace of a Matrix The sum of the diagonal elements of a square matrix A is called the trace of A and is denoted by tr( A).

(i) tr( λA) = λ tr( A) (iii) tr( AB) = tr( BA)

(ii) tr( A ) = tr( A ′ )

Invertible Matrices ●

Equivalent Matrices Two matrices A and B are said to be equivalent, if one is obtained from the other by one or more elementary operations and we write A ~ B. Following types operations.

of

operations

are

called

elementary

●

Some Important Results ●

(i) Interchanging any two rows (columns). This transformation is indicated by Ri ↔ R j (Ci ↔ C j) (ii) Multiplication of the elements of any row (column) by a non-zero scalar quantity, indicated as Ri → kRi (Ci → kCi )

A square matrix A of order n is said to be invertible if there exists another square matrix B of order n such that AB = BA = I . The matrix B is called the inverse of matrix A and it is denoted by A −1 .

●

●

●

●

●

Inverse of a square matrix, if it exists, is unique. AA −1 = I = A −1 A If A and B are invertible, then ( AB)−1 = B −1 A −1 ( A − 1 )T = ( A T )− 1 If A is symmetric, then A −1 will also be symmetric matrix. Every orthogonal matrix is invertible.

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 0 1

0 −1 , then which of the following is 0 

1 If A =   and B =  1 1 1 correct?

j

(a) (A + B) ⋅ (A − B) = A + B (c) (A + B) ⋅ (A − B) = I 2

2

NCERT Exemplar

(b) (A + B) ⋅ (A − B) = A 2 − B 2 (d) None of these

2 If p, q , r are 3 real numbers satisfying the matrix 3 4 1  equation, [ p q r ] 3 2 3 = [ 3 0 1 ], then 2p + q − r is   2 0 2 j equal to JEE Mains 2013 (a) − 3

(b) − 1

(c) 4

(d) 2

3 In a upper triangular matrix n × n, minimum number of zeroes is n (n − 1) 2 2n (n − 1) (c) 2

(a)

(b)

n (n + 1) 2

(d) None of these

1 2 a 0 and B =  4 Let A =   ; a , b ∈ N. Then, 3 4 0 b  (a) there exists more than one but finite number of B’s such that AB = BA (b) there exists exactly one B such that AB = BA (c) there exist infinitely many B’s such that AB = BA (d) there cannot exist any B such that AB = BA

2 3  1 −2 3    5 If A =   and B = 4 5, then  −4 2 5  2 1  (a) AB, BA exist and are equal (b) AB, BA exist and are not equal (c) AB exists and BA does not exist (d) AB does not exist and BA exists

6 If ω ≠ 1 is the complex cube root of unity and matrix ω 0  70 H= , then H is equal to 0 ω (a) H

(b) 0

(c) −H

(d) H 2

7 If A and B are 3 × 3 matrices such that AB = A and BA = B, then (a) A 2 = A and B 2 ≠ B (c) A 2 = A and B 2 = B

(b) A 2 ≠ A and B 2 = B (d) A 2 ≠ A and B 2 ≠ B

8 For each real number x such that − 1 < x < 1, let  1 1 − x A( x ) =  −x  1 − x (a) A (z) = (b) A(z) = (c) A (z) = (d) A(z) =

−x  x +y 1− x and z = . Then, 1  1 + xy  1 − x 

A(x) + A(y) A(x) [A(y)]−1 A(x) ⋅ A(y) A(x) − A(y)

48

FIVE

40

cos α

9 If A(α ) =  sin α  

0

(a) A(αβ)

− sin α 0 cos α 0, then A(α ) A(β ) is equal to  0 1 

(b) A(α + β)

(c) A(α − β)

(d) None

10 If A is 3 × 4 matrix and B is a matrix such that A′ B and (b) 3 × 4

(c) 3 × 3

(d) 4 × 4

2 1 2  11 If A = 2 1 −2 is a matrix satisfying the equation   a 2 b  AAT = 9I , where I is 3 × 3 identity matrix, then the ordered j JEE Mains 2015 pair (a,b) is equal to (a) (2, − 1)

(b) (−2, 1)

(c) (2, 1)

0 0 1 

1 0 0

  0 0 0

  0 0 1 

(d) (−2, − 1)

12 If E = 0 0 1  and F = 0 1 0, then E 2 F + F 2E (a) F

(b) E

(c) 0

(d) None

13 If A and B are two invertible matrices and both are symmetric and commute each other, then −1

−1

14 If neither α nor β are multiples of π /2 and the product AB

 x equal to

2

 y 

(a) −1

(b) 1

 cos 2 α sin α cos α  A= cos sin sin2 α  α α   cos 2 β cos β sin β  B= β β cos sin sin2 β  

(a) XY (c) − YX

1 0 0 1 0 0   20 Let A = 0 1 1 , I = 0 1 0 and     0 −2 4 0 0 1  1 2  ( A + cA + dI ) . The values of c and d are A −1 =  6  (a) (− 6, − 11) (c) (− 6, 11)

21 Elements of a matrix A of order 9 × 9 are defined as aij = ωi + j (where ω is cube root of unity), then trace ( A ) of the matrix is (b) 1

(b) multiple of π (d) None of these

1  2  1 −1  4 1  1 − 3  and A −1 = 0 −5   10  1 1 1 1 2 −    α is equal to (b) 5

2 α , then  3 

(d) − 1

(c) 2

symmetric skew-symmetric orthogonal None of the above

24 Let A be a square matrix satisfying A 2 + 5A + 5I = O . The inverse of A + 2I is equal to (a) A − 2I (c) A − 3I

 1

(b) nilpotent (d) orthogonal

 1 (a)  48  (1 / 3)  1 0 (c)   16 1

(b) symmetric (d) orthogonal

(b) A + 3I (d) does not exist

0

0 1

1 0  (b)  3 1 − 1  1  2  3 48   (d) None of these

26 If X is any matrix of order n × p and I is an identity matrix

 a a −1 −2   2 17 If A = a + 1 1 a + 4 is symmetric, then a is   4a 5   −2 2

(c) −1

(d) ω2

48 25 Let A =  . Then A is 1 / 3 1 

cos θ − sin θ  , then cos θ 

(b) 2

(c) ω

22 If A =  2

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

16 If A =   sin θ

(a) A is skew-symmetric (c) idempotent

(b) (6, 11) (d) (6, − 11)

23 If A is skew-symmetric and B = (I − A ) (I + A ) , then B is

3  1 2 15 The matrix  1 2 3 is   −1 −2 −3 (a) idempotent (c) involutary

(b)YX (d) None of these

−1

is null matrix, then α − β is (a) 0 (c) an odd multiple of π/ 2

(d) − 2

(c) 2

X = AB + BA and Y = AB − BA, then ( XY )T is equal to

(a) − 2

of matrices

(a) −2

2

(a) 0

−1

(a) both A B and A B are symmetric (b) neither A −1B nor A −1B −1 are symmetric (c) A −1B is symmetric but A −1B −1 is not symmetric (d) A −1B −1 is symmetric but A −1B is not symmetric

and

2

19 If A and B are symmetric matrices of the same order and

BA′ are both defined, then B is of the type (a) 4 × 3

1

18 If A =  2 1 − 2 and AT A = AAT = I , then xy is

(d) None

of order n × n, then the matrix M = I − X ( X ′ X )−1 X ′ is I. Idempotent matrix II. MX = O (a) Only I is correct (c) Both I and II are correct

(b) Only II is correct (d) None of them is correct

49

DAY 27 Let A and B be two symmetric matrices of order 3. Statement I A (BA) and ( AB ) A are symmetric matrices. Statement II AB is symmetric matrix, if matrix multiplication of A with B is commutative. (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

28 Consider the following relation R on the set of real square matrices of order 3. R = {( A, B ) : A = P −1BP for some invertible matrix P} Statement I R is an equivalence relation. Statement II For any two invertible 3 × 3 matrices M and N,(MN )−1 = N −1M −1. (a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is true; Statement II is correct explanation of Statement I (c) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (d) Statement I is true, Statement II is false

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 2

1 

2 1 If A =   , then I + 2A + 3A + ... ∞ is equal to − 4 − 2 

 4 1 (a)    − 4 0

2 1 2  3  5  5 (b)   (c)  − 8 − 3  (d)  − 3 – 8   − 4 − 1    

 1 2  is  3 4

2 The matrix A that commute with the matrix  1 2 1 (c) A = 3 (a) A =

2b  2a   2a 1  2b (b) A =     2  3a 2a + 3b   3b 2a + 3b  2a   2a + 3b   (d) None of these 3 2 + 3b  a a 

3 The total number of matrices that can be formed using 5 different letters such that no letter is repeated in any matrix, is (a) 5! (c) 2 × (5!)

(b) 2 × 5 5 (d) None of these

4 If A is symmetric and B is a skew-symmetric matrix, then for n ∈ N, which of the following is not correct? (a) A n (b) B n (c) A n (d) B n

is symmetric is symmetric if n is even is symmetric if n is odd only is skew-symmetric if n is odd

2 1

5 4

 5 − 4 Z = . Then, the value of the sum − 6 5 

(a) 6 (c) 12

1 1 I and A + I are orthogonal matrices, then 2 2

(a) A is orthogonal (b) A is skew-symmetric matrix (c) A is symmetric matrix (d) None of the above

 −1 + i 3  2i 7 If A =  + i 3 1   i 2 

−1 − i 3   2i , i = −1 and f ( x ) = x 2 + 2, 1−i 3   2i  then f ( A ) is equal to 5 (a)   1 (c)  0

− i 3   1 0  2   0 1 0 1

 X (YZ )3    + K to ∞ is  8 

(b) 9 (d) None of these

 3 − i 3   1 0 (b)    2   0 1   1 0 (d) (2 + i 3 )    0 1

1 0

n 8 If A =   , then A is equal to 1 1  

(a) 2 n − 1 A − (n − 1) I (c) 2 n − 1 A + (n − 1) I

 cos θ

5 Consider three matrices X =  , Y = 6 5 and 4 1  

 X (YZ )2   XYZ  tr ( X ) + tr   + tr   + tr  2   4 

6 If both A −

(b) nA − (n − 1) I (d) nA + (n − 1) I

sin θ 

n 9 Let A =  . Let A = [bij ]2 × 2 . Define − sin θ cos θ 

 An  lim A n = lim [bij ]2 × 2 . Then lim   is n→ ∞ n→ ∞  n 

n→ ∞

(a) zero matrix  0 1 (c)    −1 0

(b) unit matrix (d) limit does not exist

10 If B is skew-symmetric matrix of order n and A is n × 1 column matrix and AT BA = [ p ], then (a) p < 0 (c) p > 0

(b) p = 0 (d) Nothing can be said

50

FIVE

40

11 If A, B and A + B are idempotent matrices, then AB is

14 If A1, A3 , ..., A2 n − 1 are n skew-symmetric matrices of

equal to (b) − BA

(a) BA

 3  12 If P =  2 − 1  2 is equal to

(d) O

1 2 + 3  4 −1

1  2019 (d)  4 2 + 3

(a)

(c)

1 7 1 7

2 3 −6 −2 3 6

2r − 1

)2 r

−1

will be

(a) symmetric (b) skew-symmetric (c) neither symmetric nor skew-symmetric (d) data not adequate

a b c 

15 Let matrix A = b c a , where a, b, c are real positive   c a b 

numbers with abc = 1. If AT A = I, then a 3 + b 3 + c 3 is

  4 − 2019 3  6057

(a) 3 (c) 2

1   2 − 3

and B = A −1A′, then BB ′ equals

2 − 3  2019 

−3  6  2  −3  6  2 

(b) 4 (d) None of these

16 If A is an 3 × 3 non-singular matrix such that AA′ = A′ A −1

(b)

(d)

1 7 1 7

6 2   3 6 2   −6

2 −3 6 −2 2 2

j

JEE Mains 2014

(b) I + B (d) B −1

(a) (B )′ (c) I

17 A is a 3 × 3 matrix with entries from the set {−1, 0,1} . The

13 Which of the following is an orthogonal matrix? 6 2   3  −6 2   −3

n

∑ (2r − 1) ( A

r =1

1  2  , A = 1 1 and Q = PAP T , then P T Q 2019P  0 1 3   2 

 1 2019 (a)  1  0  4 + 2019 3 (b)  2019  (c)

(c) I

same order, then B =

3  6   − 2  3 −3   3 

probability that A is neither symmetric nor skew-symmetric is (a)

39 − 36 − 33 + 1 39

(b)

39 − 36 − 33 39

(c)

39 − 36 + 1 39

(d)

39 − 33 + 1 39

ANSWERS SESSION 1

1. (d) 11. (d) 21. (a)

2. (a) 12. (b) 22. (b)

3. (a) 13. (a) 23. (c)

4. (c) 14. (c) 24. (b)

5. (b) 15. (b) 25. (c)

6. (a) 16. (d) 26. (c)

7. (c) 17. (b) 27. (b)

8. (c) 18. (c) 28. (c)

9. (b) 19. (c)

10. (b) 20. (c)

SESSION 2

1. (c) 11. (b)

2. (a) 12. (a)

3. (c) 13. (a)

4. (c) 14. (b)

5. (a) 15. (d)

6. (b) 16. (c)

7. (d) 17. (a)

8. (b)

9. (a)

10. (b)

51

DAY

Hints and Explanations SESSION 1

5 Here, A is 2 × 3 matrix and B is 3 × 2

1 Here,

1   0 −1  0 + = 1 1 0  2 1   0 −1  0 − = 1 1 0   0 0 1 0 1  A2 = A ⋅ A = 1 1 1 1   

0 1  0 A−B = 1 

A+ B =

 0 + 1 0 + 1 1 =  0 + 1 1 + 1  1     0 − 1  0 and B 2 = B ⋅ B = ⋅ 1 0  1     −1 0  =  0 −1   =

∴ A2 + B 2 =

0 1 2 1

1 2 −1 0 

1 1  −1 0   0 1 + = 1 2  0 −1 1 1      

1 1  −1 0  2 1 − = A2 − B 2 = 1 2  0 −1 1 3        0 0  0 2 and ( A + B )( A − B ) = 2 1   0 1     0 + 0 0 + 0  0 0 = =  0 + 0 4 + 1   0 5     Clearly, ( A + B )( A − B ) ≠ A2 − B 2 ≠ A + B ≠ I. 2

2

matrix. ∴Both AB and BA exist, and AB is a 2 × 2 matrix, while BA is 3 × 3 matrix. ∴ AB ≠ BA. ω 0  ω2 0  =  0 ω   0 ω2      2 3    0 ω   0 0 ω ω H3 =  = 2   0 ω 3 0 0 ω ω      ω70 0  ω 69 ⋅ ω 0  70 = ∴ H = 70  ω 69 ⋅ ω   0 ω   0 (ω3 )23 ⋅ ω  0 = 3 23 (ω ) ⋅ ω  0  ω 0  = =H [Qω3 = 1]  0 ω   H2 =

3(3 − 1) Here, number of zeroes = 3 = 2 ∴ Minimum number of zeroes n (n − 1) = 2 1 2   a 0  a 2b  4 Clearly, AB =  =     3 4  0 b  3a 4 b   a 0 1 2   a 2 a  and BA = =  0 b  3 4 3b 4 b       If AB = BA, then a = b . Hence, AB = BA is possible for infinitely many values of B’s.

ω 0   0 ω  

B =I

⇒

A=I

⇒

A2 = A

cos α

9 A(α ) A(β ) =  sin α   0

∴

and A (z ) = 1−

=

=

1 + xy 1 + xy − x

1 + xy (1 − x ) (1 −

…(i) …(ii)

1 (x + y) 1 + xy

( x + y )  1 −  1 + xy   (x + y)  1 −  xy 1 +   ( x + y )  1 −  1 + xy    − y − ( x + y ) 1  xy 1 +   ( x + y )  1 −  1 + xy    y ) − ( x + y ) 1  xy 1 +  

1 = (1 − x ) (1 − y ) − ( x + y )  1 + xy …(iii)  − ( x + y ) 1 + xy    Now, consider

− sin α cos α 0

cos β ×  sin β   0

0 0  1

− sin β cos β 0

0 0  1

cos(α + β ) − sin (α + β ) 0 =  sin(α + β ) cos(α + β ) 0   0 0 1  = A(α + β ) Now, for A ′ B to be defined, order of B should be 3 × m and for BA′ to be defined, order of B should be n × 4. Thus, for both A ′ B and BA′ to be defined, order of B should be 3 × 4.

8 We have,

1  1 − x 1 − x  − x 1  1  1 −y A( y ) = 1 − y  − y 1 

1 ⋅ (1 − x ) (1 − y ) − ( x + y )  1 + xy …(iv)  − ( x + y ) 1 + xy   

10 Clearly, order of A′ is 4 × 3.

Hence, A2 = A and B 2 = B

A ( x) =

−y 1 

From Eqs. (iii) and (iv), we get A(z ) = A( x ) ⋅ A( y ).

⇒ B2 = B

Similarly, BA = B

3 We know that, a square matrix A = [aij] is said to be an upper triangular matrix if aij = 0, ∀ i > j. Consider, an upper triangular matrix 1 2 3  A =  0 5 6    0 0 7 3 × 3

=

7 Since, AB = A ∴

1 ⋅ (1 − x ) (1 − y )  1 − x  1 − x 1  − y  

6 Clearly,

2 [3 p + 3q + 2r , 4 p + 2q + 0, p + 3q + 2r ] = [3 0 1] ⇒ 3 p + 3q + 2r = 3, 4 p + 2q = 0, p + 3q + 2r = 1 ⇒ p = 1,q = −2, r = 3 ∴ 2 p + q − r = 2 − 2 − 3 = −3

A ( x ) ⋅ A( y ) =

1

2

2

11 Given, A =  2 1 −2  a 2 1 2 A T = 2 1  2 −2

 b

a 2  b  1 2 2 Now, AA T =  2 1 −2   a 2 b 

⇒

1 2 a 2 1 2   2 −2 b 

9 0 a + 4 + 2b   = 0 9 2a + 2 − 2b    2 2  a + 4 + 2b 2a + 2 − 2b a + 4 + b 

It is given that, AA T = 9I

9 0 a + 4 + 2b  0 9 2a + 2 − 2b    2 2  a + 4 + 2b 2a + 2 − 2b a + 4 + b  

⇒

 9 0 0 =  0 9 0    0 0 9 On comparing, we get a + 4 + 2b = 0 ⇒ a + 2b = −4 2a + 2 − 2b = 0

…(i)

52

FIVE

40

⇒ a − b = −1 …(ii) and …(iii) a2 + 4 + b 2 = 9 On solving Eqs. (i) and (ii), we get a = −2,b = −1 This satisfies Eq. (iii) also. Hence, (a,b ) ≡ (−2,−1)

12 F is unit matrix ⇒ F 2 = F

= A −1 B −1

= A −1 (BA ) A −1 ⇒ BA −1 = A −1 B] ⇒ A −1 B is symmetric. Now, consider ( A −1 B −1 )T = ((BA )−1 )T

A

T −1

= (A )

[Q AB = BA]

) = (A T −1

(B )

−1 T

) (B

= A

−1

−1 T

B

)

−1

⇒ A −1 B −1 is also symmetric. 2   14 AB =  cos α cos α2sin α  cos α sin α sin α





 cos 2 β cos β sin β  × sin2 β  cos β sin β =

21 Clearly, tr ( A ) = a11 + a22 + a33 + a44

∴ X = AB + BA will be a symmetric matrix and Y = AB − BA will be a skew-symmetric matrix. Thus, we get X T = X and Y T = − Y Now, consider ( XY )T = Y T X T = (− Y )( X ) = − YX ⇒ (6 A −1 )A = ( A2 + cA + dI ) A

= (( AB )−1 )

cos α cos β cos(α − β)  sin α cos β cos(α − β )  cos α sin β cos(α − β ) sin α sin β cos(α − β ) 

 0 0 =  0 0   ⇒ cos(α − β ) = 0 ⇒ α − β = (2n + 1) π / 2 3  1 2 15 Let A =  1 2 3     − 1 −2 − 3

 0 0 0 Then, A2 =  0 0 0    0 0 0 Hence, A is nilpotent matrix of index 2. cos θ sin θ  16 A ′ =  ≠ A or − A.   − sin θ cos θ cos θ − sin θ A A′ =  sin θ cos θ   

[Q Post multiply both sides by A] ⇒ 6( A −1 A ) = A3 + cA2 + dIA

+ a55 + a66 + a77 + a88 + a99 = ω2 + ω 4 + ω 6 + ω 8 + ω10 + ω12 + ω14 + ω16 + ω18 = (ω + ω + 1) + (ω + ω + 1) 2

+ (ω2 + ω + 1) [Qω3 n = 1, n ∈ N ] = 0+ 0+ 0

2

⇒

A + cA + dA − 6I = O 2

22 Clearly, AA −1 = I Now, if R1 of A is multiplied by C3 of A −1 , we get 2 − α + 3 = 0 ⇒ α = 5

23 Consider,

= (I − A )−1 (I + A ) (I − A ) (I + A )−1 = (I − A )−1 (I − A ) (I + A ) (I + A )−1

…(i)

1 Here, A2 = A ⋅ A =  0  0 1 0 0 1   0 −2

0 0 1 1 ×  −2 4 0 1 0 0 1 =  0 − 1 5     4  0 −10 14 0 0 1 and A3 = A2 ⋅ A =  0 −1 5  ×    0 −10 14 0 0 1 0 0 1  0 1 1  =  0 −11 19       0 −2 4  0 −38 46 Now, from Eq. (i), we get 0 0 0 0 1 1  0 −11 19  + c  0 −1 5       0 −38 46  0 −10 14 1 + d 0  0 1 0 0  0 − 6  0 1 0 =  0     0 0 1  0

[Q1 + ω + ω2 = 0]

=0

[Q A −1 A = I and IA = A ] 3

2

BB T = (I − A )−1 (I + A ) (I + A )T [(I − A )−1 ] T

⇒ 6I = A + cA + dA 3

   6

 0 0 0 =  0 0 0    0 0 0 ⇒ 1 + c + d − 6 = 0; − 11 − c + d − 6 = 0 ⇒ c + d = 5; − c + d = 17 On solving, we get c = − 6, d = 11. These value also satisfy other equations.

a = 2.

20 Clearly, 6 A −1 = A2 + cA + dI

T

= (B

⇒ a2 − a − 2 = 0, a2 − 4a + 4 = 0

19 Since, A and B are symmetric matrices

[Q A T = A and B T = B ]

−1 T

2

orthogonal to the other rows. ⇒ R1 ⋅ R3 = 0 ⇒ x + 4 + 2y = 0 Also, R2 ⋅ R3 = 0 ⇒ 2x + 2 − 2y = 0 On solving, we get x = − 2, y = − 1 ∴ xy = 2

= B T ( A T )−1 = B A −1

−1

⇒ a − 1 = a + 1, a + 4 = 4a 2

18 Since, A is orthogonal, each row is

13 Consider, ( A −1 B )T = B T ( A −1 )T

[Q AB = BA ⇒ A ( AB )A

0 1 + c + d − 6 ⇒ 0 − 11 − c + d − 6  0 − 38 − 10c − 2d  0 19 + 5c + d 46 + 14c + 4 d −

17 A is symmetric

⇒

+ F 2 E = E2 + E 0 1  0 0 1 0 1 ×  0 0 1    0 0  0 0 0  0 0 0 =  0 0 0    0 0 0 ∴ E 2 + E = E. E2 F 0 Also, E 2 =  0  0

and

−1

 cos θ sin θ  1 0 = =I  − sin θ cos θ  0 1      ∴ A is orthogonal.

0 0 1 1  −2 4 0 0 0 0  0 0

= I⋅I = I Hence, B is an orthogonal matrix.

24 We have, A2 + 5A + 5I = O ⇒

A2 + 5A + 6I = I

⇒ ⇒

( A + 2I ) ( A + 3I ) = I A + 2I and A + 3I are inverse of

each other.    25 If A =  , then A2 =  2a 1  a 1     1 0

1

0

 1 0  1 0 , …, A n = 3a 1   na 1      Here, a = 1 / 3,  1 0 ∴ A 48 = 16 1    A3 =

26 We have, M = I − X ( X ′ X )−1 X ′ = I − X ( X −1 ( X ′ )−1 )X ′ [Q( AB )−1 = B −1 A −1 ] = I − ( XX −1 ) (( X ′ )−1 X ′ ) =I −I ×I =I −I =O

[by associative property] [Q AA −1 = I = A −1 A] [Q I 2 = I ]

53

DAY Clearly, M 2 = O = M So, M is an idempotent matrix. Also, MX = O .

27 Given, A = A and B = B T

T

Statement I [ A (BA )] T = (BA )T ⋅ A T T

T

= (A B )A

T

= (AB) A = A (BA) So, A(BA ) is symmetric matrix. Similarly, ( AB ) A is symmetric matrix. Hence, Statement I is true. Also, Statement II is true but not a correct explanation of Statement I.

28 Given, R = {( A, B ) : A = P −1 BP for some invertible matrix P} For Statement I (i) Reflexive ARA ⇒ A = P −1 AP which is true only, if P = I . Thus, A = P −1 AP for some invertible matrix P. So, R is Reflexive. (ii) Symmetric ARB ⇒ A = P −1 BP ⇒ PAP −1 = P (P −1 BP ) P −1 ⇒ PAP −1 = ( PP −1 ) B( PP −1 ) B = PAP −1

∴

Now, let Q = P Then,

−1

B = Q −1 AQ ⇒ BRA

⇒ R is symmetric. (iii) Transitive ARB and BRC ⇒ A = P −1 BP

= (P −1Q −1 )C (QP ) = (QP )−1 C (QP ) So, ARC. ⇒ R is transitive So, R is an equivalence relation. For Statement II It is always true that (MN )−1 = N −1 M −1 Hence, both statements are true but second is not the correct explanation of first.

SESSION 2   1 Clearly, A2 =     − 4 − 2  − 4 − 2 1

2

 a b   1 2  1 2  a b    =       c d   3 4  3 4  c d   a + 3b 2a + 4 b  ⇒    c + 3d 2c + 4 d  b + 2d   a + 2c =   3a + 4c 3b + 4d  On equating the corresponding elements, we get …(i) a + 3b = a + 2c ⇒ 3b = 2c …(ii) 2a + 4b = b + 2d ⇒ 2a + 3b = 2d …(iii) c + 3d = 3a + 4c ⇒ a + c = d …(iv) 2c + 4d = 3b + 4d ⇒ 3b = 2c Thus, A can be taken as b  1  2a  a 2b   3b 3  = a + b  2  3b 2a + 3b    2 2 

3 Clearly, matrix having five elements is

of order 5 × 1 or 1 × 5. ∴Total number of such matrices = 2 × 5!.

4 ( A n )′ = ( A AL A )′ = ( A ′ A ′L A ′ ) = ( A ′ )n = A n for all n

1

 0 0 =O  0 0   ∴ I + 2 A + 3 A2 + ... = I + 2 A =

2  5 2 1 0  4 = + =  0 1   − 8 − 4  − 8 − 3      

1 1 ⇒  A ′ − I   A − I  = I  2   2  1 ′ and  A + I   2  ⇒

From Eq. (i), we get 1 1 1 A ′ A − IA ′ − IA + I = I 2 2 4 1 1 1 ⇒ A ′ A − A ′− A + I = I 2 2 4

…(ii)

…(iii)

Similarly, from Eq. (ii), we get 1 1 1 A ′ A + A ′ + A + I = I …(iv) 2 2 4 On subtracting Eq. (iii) from Eq. (iv), we get A + A′ = O A′ = − A

or

Hence, A is a skew-symmetric matrix. ω2  ω i  =ω 1   − ω − 1 −ω i    i  1 − ω2 0  A2 = − ω2  0 1 − ω2  

 ω  7 We have, A =  i 2 − ω  i ∴

∴ A is symmetric for all n ∈ N .

Q

 − ω2 + ω 4  0 = 2 4 0 − + ω ω    − ω2 + ω  0 = 0 − ω2 + ω   2 f ( x) = x + 2 [given]

∴

f ( A ) = A2 + 2I

Also, B is skew-symmetric ⇒ B ′ = − B. ∴ (B n )′ = (B B L B )′ = (B ′ B ′ L B ′ ) = (B ′ )n

 − ω2 + ω  2 0 0 =  +  0 2 2 − + 0 ω ω     1 0 2 = (− ω + ω + 2)  0 1   1 0   = (3 + 2ω )  0 1   1 0 = (2 + i 3 )  0 1  

5 4  5 − 4 1 0 5 Here, YZ =  = 6 5  − 6 5   0 1    X (YZ )2  XYZ  ∴ tr ( X ) + tr    + tr   2  4    X (YZ )3  + tr   +K 8   X X = tr ( X ) + tr   + tr   + K 2  4 1 1 = tr ( X ) + tr ( X ) + tr ( X ) + K 2 4 1 1  = tr ( X ) 1 + + 2 + K   2 2 1 = tr ( X ) 1 1− 2 = 2 tr ( X ) = 2 (2 + 1) = 6 6 Since, both A − 1 I and A + 1 I are 2 2 orthogonal, therefore, we have  A − 1 I ′  A − 1 I = I      2   2 

…(i)

 A + 1 I = I    2 

 A′ + 1 I   A + 1 I  = I      2   2 

n

⇒ B n is symmetric if n is even and is skew-symmetric if n is odd.

A = P −1 (Q −1CQ ) P

2

a b  be a matrix that c d   1 2 commute with   . Then,  3 4

= (− B )n = (−1)n B n .

and B = Q −1CQ ⇒

2 Let A = 

    8 A2 =  =     1 1  1 1  2 1  1 0 1 0

1 0

1 0 1 0 1 0 = 2 1  1 1  3 1       .................... ....................  1 0 An =  n 1   A3 =

0   n 0  n − 1 −  n n  0 n − 1    = nA − (n − 1) I =

54

FIVE

40

 cos θ

sin θ  cos θ

sin θ

  9 A2 =     − sin θ cos θ  − sin θ cos θ

13 We know that a matrix

 cos 2θ sin 2θ  =  − sin 2θ cos 2θ    cos 3θ sin 3θ  3 Similarly, A = etc  − sin 3θ cos 3θ    cos nθ sin nθ  ∴ An =  − sin nθ cos nθ   b11 b12  = b   21 b22  b ij A n  0 0 as lim = =0  0 0 n→∞ n n  

Now, lim

n→ ∞

10 A T B A = [ p] ⇒ ( A T BA)T = [ p] T = [ p] ⇒

A T B T A = A T (−B ) A = [ p]

⇒

[− p] = [ p] ⇒ p = 0.

matrix ∴ A2 = A; B 2 = B and ( A + B )2 = A + B Now, consider ( A + B )2 = A + B A2 + B 2 + AB + BA = A + B

⇒ ⇒

A + B + AB + BA = A + B AB = − BA Q 2019 = (PAP T ) (PAP T ) ...(PAP T ) = PA2019P T

∴ P Q Now,

P = P ⋅ PA

2019

P ⋅P = A

2019

T

2019

1 1 1 1 1 2 A = =  0 1  0 1  0 1      2

A3 = ⇒

T

A2019 =

1 2 1 1 1 3 =  0 1  0 1  0 1      1 2019 0 1  

B = A1 + 3 A33 + K + (2n − 1) A 22 nn −− 11

Now, B T = ( A1 + 3 A33 + K + (2n − 1) A22nn −− 11 )T = A1T + (3 A33 )T + K + ((2n − 1) A22nn −− 11 )T = A1T + 3( A3T )3 + K + (2n − 1) ( A 2T n − 1 )2 n − 1 =− A−

12 P is orthogonal matrix as P P = I T

T

and Σaib i = Σb ic i = Σc i ai = 0 Now, from the given options, only  6 2 −3 1 2 3 6  satisfies these conditions.  7 3 − 6 2    6 2 −3 1 Hence, 2 3 6  is an orthogonal  7  3 −6 2  matrix.

14 We have,

11 Since, A, B and A + B are idempotent

⇒

 a1 a2 a3  A = b1 b2 b3  will be orthogonal if   c 1 c 2 c 3  AA′ = I, which implies Σa2i = Σb 2i = Σc 2i = 1

3 A33

− K − (2n −

1) A22nn −− 11

[Q A1 , A3 , K , A 2n − 1 are skewsymmetric matrices ∴( A i )T = − A i ∀ i = 1, 3, 5, ... 2n − 1] = − [ A + 3 A33 + K + (2n − 1)A22nn −− 11 ] =−B Hence, B is a skew-symmetric matrix. a b c 

a b c 

 c

 c

15 A T A = b c a × b c a  a b

 a b

 a2 + b 2 + c 2 ab + bc + ca =  ab + bc + ca b 2 + c 2 + a2   ab + bc + ca ab + bc + ca  ac + ab + bc  ab + bc + ca   a2 + b 2 + c 2  A T A = I ⇒ a2 + b 2 + c 2 = 1 and ab + bc + ca = 0 Since a, b, c > 0, ∴ ab + bc + ca ≠ 0 and hence no real value of a3 + b 3 + c 3 exists.

16 AA′ = A′ A, B = A −1 A′. BB ′ = ( A −1 A′ ) ( A −1 ⋅ A′ )′ = ( A −1 A′ ) [( A′ )′ ( A −1 )′] = ( A −1 A′ ) [ A( A′ )−1 ] [Q( A −1 )′ = ( A′ )−1 ] = A

−1

−1

( A′ A) ( A′ )

= A −1 ( AA′ ) ( A′ )−1

[Q A′ A = AA′]

= ( A −1 A) [ A′( A′ )−1 ] =I⋅I =I

17 Total number of matrices = 39. A is

symmetric, then aij = a ji . Now, 6 places (3 diagonal, 3 non-diagonal), can be filled from any of −1, 0, 1 in 36 ways. A is skew-symmetric, then diagonal entries are ‘o’ and a12 , a13 , a23 can be filled from any of −1, 0, 1 in 33 ways. Zero matrix is

common. ∴Favourable matrices are 39 − 36 − 33 + 1. Hence, required probability 39 − 36 − 33 + 1 = 39

DAY SIX

Determinants Learning & Revision for the Day u

Determinants

u

Properties of Determinants

u

Cyclic Determinants

u

Area of Triangle by using Determinants

u

Minors and Cofactors

u

Adjoint of a Matrix

u

u

Inverse of a Matrix Solution of System of Linear Equations in Two and Three Variables

Determinants Every square matrix A can be associated with a number or an expression which is called its determinant and it is denoted by det (A) or |A| or ∆ . a11 a12 ... a1 n  a11 a12 ... a1 n   a a22 ... a2 n a a22 ... a2 n  , then det ( A) = 21 If A =  21  M M M M  M M M M  a a ... a a a ... a  n1 n2 nn  n1 n2 nn ●

a b a b  If A =  , then | A| = = ad − bc  c d c d 

●

a b c a b c  If A =  p q r  , then | A| = p q r   u v w u v w  =a

PRED MIRROR Your Personal Preparation Indicator

q r p r p q [expanding along R1 ] −b +c v w u w u v

There are six ways of expanding a determinant of order 3 corresponding to each of three rows (R1 , R2 , R3 ) and three columns (C1 , C2 , C3 ). • Rule to put + or − sign in the expansion of determinant of order 3.

• A square matrix A is said to be singular, if | A| = 0 and non-singular, if | A| ≠ 0.

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

= a (qw − vr ) − b ( pw − ur ) + c ( pv − uq )

NOTE

u

+ − + − + − + − +

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

56

SIX

40

Properties of Determinants (i) If each element of a row (column) is zero, then ∆ = 0. (ii) If two rows (columns) are proportional, then ∆ = 0. (iii) | A T| = | A|, where A T is a transpose of a matrix. (iv) If any two rows (columns) are interchanged, then ∆ becomes −∆. (v) If each element of a row (column) of a determinant is multiplied by a constant k, then the value of the new determinant is k times the value of the original determinant (vi) det (kA) = k n det( A), if A is of order n × n. (vii) If each element of a row (column) of a determinant is written as the sum of two or more terms, then the determinant can be written as the sum of two or more determinants i.e. a1 + a2 b c a1 b c a2 b c p1 + p2 q r = p1 q r + p2 q r u1 v w u1 + u2 v w u2 v w (viii) If a scalar multiple of any row (column) is added to another row (column), then ∆ is unchanged a b c a b c i.e. p q r = p + ka q + kb r + kc , which is u v w u v w obtained by the operation R2 → R2 + kR1

Product of Determinants a1 If| A| = a2 a3

b1 b2 b3

c1 α1 c2 and| B| = α 2 c3 α3

β1 β2 β3

γ1 γ2 , then γ3

a1α 1 + b1β1 + c1 γ1 a1α 2 + | A| × | B| = a2α 1 + b2β1 + c2 γ1 a2α 2 + a3α 1 + b3β1 + c3 γ1 a3α 2 + a1α 3 + b1β3 a2α 3 + b2β3 a3α 3 + b3β3

b1β2 + c1 γ2 b2β2 + c2 γ2 b3β2 + c3 γ2 + c1 γ3 + c2 γ3 =| AB| + c3 γ3

[multiplying row by row] We can multiply rows by columns or columns by rows or columns by columns NOTE

• | AB | = | A| |B | = |BA| = | AT B | =| ABT | = | AT BT | • | An | = | A| n , n ∈ Z +

Cyclic Determinants In a cyclic determinant, the elements of row (or column) are arranged in a systematic order and the value of a determinant is also in systematic order. 1 (i) 1 1

x y z

x2 y2 = ( x − y) ( y − z) (z − x) z2

1 (ii) 1 1

x y z

1 (iii) 1 1

x2 y2 z2

(iv)

x3 y3 = ( x − y) ( y − z) (z − x)( x + y + z) z3 x3 y3 = ( x − y) ( y − z) (z − x)( xy + yz + zx) z3

a b c b c a = − (a + b + c)(a 2 + b 2 + c2 – ab – bc – ca) c a b = − (a3 + b 3 + c3 − 3 abc) abc a a2 abc = b b 2 c c2 abc

a bc (v) b ca c ab

a3 b 3 = abc (a − b )(b − c)(c − a) c3

Area of Triangle by using Determinants If A( x1 , y1 ), B( x2 , y2 ) and C( x3 , y3 ) are vertices of ∆ABC, then Area of ∆ABC =

1 2

x1 x2 x3

y1 y2 y3

1 1 1

1 [ x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )] 2 x1 y1 1 If these three points are collinear, then x2 y2 1 = 0 x3 y3 1 and vice-versa. =

Minors and Cofactors The minor M ij of the element aij is the determinant obtained by deleting the ith row and jth column of ∆. a11 a12 a13 If ∆ = a21 a22 a23 , then M11

a31 a = 22 a32

a32 a33 a23 a21 , M12 = a33 a31

a23 etc. a33

The cofactor Cij of the element aij is (−1)i + j M ij. a11 a12 a13 a a23 a If ∆ = a21 a22 a23 , then C11 = 22 , C12 = − 21 a32 a33 a31 a31 a32 a33

a23 etc. a33

The sum of product of the elements of any row (or column) with their corresponding cofactors is equal to the value of determinant. i.e.

∆ = a 11 C 11 + a 12 C 12 + a 13 C 13 = a 21 C21 + a 22 C 22 + a 23 C 23

= a 31 C 31 + a 32 C 32 + a 33 C 33 But if elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero.

57

DAY Adjoint of a Matrix

1. Matrix Method

If A = aij

In this method we first write the above system of equations in matrix form as shown below.  a1 b1 c1   x  d1   a b c   y  = d  or AX = B 2 2  2     2 a b c  3 d3  3 3  z

[ ]

nxn

, then adjoint of A, denoted by adj ( A), is defined

as [Cij]Tnxn , where Cij is the cofactor of aij. a11 If A = a21  a31 NOTE

a12 a22 a32

a13  C11 a23 , then adj ( A) = C21   a33  C31

d a b , then adj ( A) =  • If A =   c

 − c

d 

c12 C22 C32

C13  C23   C33 

T

− b . a 

Properties of Adjoint of a Matrix Let A be a square matrix of order n, then

 a1 where, A =  a2   a3

●

●

(iii) adj (AB) = (adj B) (adj A) ●

(iv) adj ( A T ) = (adj A)T

c1 c2 c3

 x d1   , X =  y  and B = d      2  z d3 

Case I When system of equations is non-homogeneous (i.e. when B ≠ 0).

(i) (adj A) A = A (adj A) = | A| ⋅ I n (ii) |adj A | = | A|n − 1 , if| A| ≠ 0

b1 b2 b3

(v) adj (adj A) = | A|n – 2 A, if| A| ≠ 0

If| A| ≠ 0, then the system of equations is consistent and has a unique solution given by X = A −1 B. If| A| = 0 and (adj A), B ≠ 0, then the system of equations is inconsistent and has no solution. If| A| = 0 and (adj A) ⋅ B = O, then the system of equations may be either consistent or inconsistent according as the system have infinitely many solutions or no solution.

2

(vi) |adj (adj A)| = | A |( n – 1 ) , if| A| ≠ 0

Inverse of a Matrix Let A be any non-singular (i.e.| A| ≠ 0) square matrix, then inverse of A can be obtained by following two ways.

1. Using determinants In this,

A −1 =

1 adj ( A) | A|

Case II When system of equations is homogeneous (i.e. when B = 0). ●

●

2. Cramer’s Rule Method In this method we first determine

2. Using Elementary operations In this, first write A = IA (for applying row operations) or A = AI (for applying column operations) and then reduce A of LHS to I, by applying elementary operations simultaneously on A of LHS and I of RHS. If it reduces to I = PA or I = AP, then P = A −1 .

Properties of Inverse of a Matrix (i) A square matrix is invertible if and only if it is non-singular. (ii) If A = diag (λ 1 , λ 2 ,..., λ n ),

Solution of System of Linear Equations in Two and Three Variables Let system of linear equations in three variables be a1 x + b1 y + c1 z = d1 , a2 x + b2 y + c2 z = d2 and a3 x + b3 y + c3 z = d3 . Now, we have two methods to solve these equations.

a1 D = a2 a3

b1 b2 b3

c1 d1 c2 , D1 = d2 c3 d3

b1 b2 b3

a1 D2 = a2 a3

d1 d2 d3

c1 a1 c2 and D3 = a2 c3 a3

c1 c2 , c3 b1 b2 b3

d1 d2 d3

Case I When system of equations is non-homogeneous ●

●

then A −1 = diag (λ−11 , λ−12 ,..., λ−1n ) provided λ i ≠ 0 ∀, i = 1, 2, … n.

If| A| ≠ 0, then system of equations has only trivial solution, namely x = 0, y = 0 and z = 0. If| A| = 0, then system of equations has non-trivial solution, which will be infinite in numbers.

●

If D ≠ 0, then it is consistent with unique solution given by D D D x = 1 , y = 2 ,z = 3. D D D If D = 0 and atleast one of D1 , D2 and D3 is non-zero, then it is inconsistent (no solution). If D = D1 = D2 = D3 = 0, then it may be consistent or inconsistent according as the system have infinitely many solutions or no solution.

Case II When system of equations is homogeneous If D ≠ 0, then x = y = z = 0 is the only solution, i.e. the trivial solution. If D = 0, then it has infinitely many solutions. Above methods can be used, in a similar way, for the solution of system of linear equations in two variables. ●

●

SIX

40

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 If x = cy + bz , y = az + cx and z = bx + ay , where x , y and z are not all zero, then a 2 + b 2 + c 2 is equal to (a) 1 + 2abc (c) 1 + abc

(b) 1 − 2abc (d) abc −1

2 Consider the set A of all determinants of order 3 with entries 0 or 1 only. Let B be the subset of A consisting of all determinants with value 1 and C be the subset of A consisting of all determinants with value –1. Then, (a) C is empty (b) B and C have the same number of elements (c) A = B ∪ C (d) B has twice as many elements as C

1

3 If x , y and z are positive, then logy x logz x

logx y logx z logy z is 1 logz y 1

equal to (a) 0

(c) −1

(b) 1

(d) None of these

4 If a , b and c are cube roots of unity, then ea eb ec (a) 0

e2a e 2b e2c

e3a −1 e 3b − 1 is equal to e3c −1 (c) e 2

(b) e

(d) e 3

x −1 x + 3 −2x x − 4 , where p, q , r , s x + 4 3x

and t are constants, then t is equal to (a) 0

(b) 1

(c) 2

(d) –1

1 x x +1 6 If f ( x ) = 2x x ( x − 1) ( x + 1)x , 3x ( x − 1) x ( x − 1)( x − 2) ( x + 1)x ( x −1) then f ( 50) is equal to (a) 0

(b) 50

(d) −50

(c) 1

7 If α , β and γ are the roots of the equation x + px + q = 0 , 3

α β γ then the value of the determinant β γ α is γ α β (a) 0

(b) −2

(c) 2

(d) 4

8 If ω is a cube root of unity, then a root of the following x − ω − ω2 ω ω2 (a) x = 0 (c) x = ω

a bω cω 2

bω 2 c aω

aω bω2 is c

(a) a 3 + b 3 + c 2 − 3 abc (c) 0

ω ω2 x − ω −1 1 = 0 is 1 x − 1 − ω2

(b) x = − 1 (d) None of these

(b) a 2 b − b 2 c (d) a 2 + b 2 + c 2

x − 4 2x 2x 10 If 2x x − 4 2x = ( A + Bx )( x − A )2 , then the 2x 2x x −4 ordered pair ( A, B ) is equal to (a) (−4,−5) (c) (−4,5)

j

JEE Mains 2018

(b) (−4, 3) (d) (4,5)

11 If x, y, z are non-zero real numbers and 1+ x 1 1 1 + y 1 + 2y 1 = 0, 1 + z 1 + z 1 + 3z then x −1 + y −1 + z −1 is equal to (b) −1 (d) −6

(a) 0 (c) −3

5 If px 4 + qx . 3 + rx 2 + sx + t x 2 + 3x = x +1 x −3

9 If ω is an imaginary cube root of unity, then the value of

b+c c+a a+b a b c 12 If c + a a + b b + c = k b c a , then k is equal to a+b b+c c+a c a b (a) 0

(b) 1

(c) 2

(d) 3

13 Let a, b and c be such that (b + c ) ≠ 0. If

a a + 1 a −1 a +1 b +1 c −1 −b b + 1 b − 1 + a −1 b −1 c + 1 = 0, ( −1)n + 2 a (−1)n +1b ( −1)n c c c −1 c + 1

then the value of n is (a) zero (c) an odd integer

(b) an even integer (d) an integer

14 If one of the roots of the equation 7 6 x 2 − 13 2 x 2 − 13 2 = 0 is x = 2, then sum of all 2 x − 13 3 7 other five roots is (a) – 2 (c) 2 5

(b) 0 (d) 15

15 If a, b and c are sides of a scalene triangle, then the value a b c of b c a is c a b (a) non-negative (c) positive

JEE Mains 2013 (b) negative (d) non-positive j

59

DAY

x + ky + 3z = 0, 3x + ky − 2z = 0 2x + 4y − 3z = 0 xz has a non-zero solution ( x , y , z ), then 2 is equal to y and

j

(a) −10 (c) −30

JEE Mains 2018

(b) 10 (d) 30

17 If A = [aij ]nxn and aij = (i 2 + j 2 − ij )( j − i ), n is odd, then which of the following is not the value of tr ( A ). (a) 0 (c) 2 | A |

(b) | A | (d) None of these

18 If the coordinates of the vertices of an equilateral triangle with sides of length a are ( x1, y1 ),( x 2 , y 2 ) and ( x 3 , y 3 ), 2 x1 y1 1 then x 2 y 2 1 is equal to y3 1

x3 (a)

a4 4

j

(b)

3a2 4

(c)

5a4 4

NCERT Exemplar 3a4 (d) 4

19 The points A (a, b + c ), B (b, c + a ) and C (c, a + b ) are (a) (b) (c) (d)

vertices of an isoscele triangle vertices of an equilateral triangle collinear None of the above

(a) x1 ∆ (c) (x1 + y1)∆

2

21 If A =   −4

z1 B z 2 , then 2 B3 z3

C2 equals to C3

(b) x1 x 3 ∆ (d) None of these

−3  , then adj ( 3A 2 + 12A ) is equal to 1  j JEE Mains 2017

72 − 84  (a)   − 63 51  51 84  (c)   63 72 

51 63  (b)   84 72  72 − 63  (d)   − 84 51 

22 Which of the following is/are incorrect? (i) Adjoint of a symmetric matrix is symmetric (ii) Adjoint of a unit matrix is a unit matrix (iii) A(adj A ) = (adj A ) =| A| I (iv) Adjoint of a diagonal matrix is a diagonal matrix . (a) (i) (c) (iii) and (iv)

(b) (ii) (d) None of these

1 α 3 23 If P = 1 3 3 is the adjoint of a 3 × 3 matrix A and   2 4 4 j JEE Mains 2013 | A| = 4, then α is equal to (a) 4

(b) 11

(c) 5

to

j

(a) − 1

(b) 5

(c) 4

JEE Mains 2016 (d) 13

a b c   q −b y    25 If A = x y z , B = −p a −x  and if A is invertible,     −c z  p q r  r then which of the following is not true? (a) | A | = −| B | and | adj A| ≠| adj B| (b) | A | = − | B | and | adj A | = | adj B | (c) A is invertible iff B is invertible (b) None of the above

26 If A an 3 × 3 non-singular matrix such that AA′ = A′ A and B = A −1A′, then BB′ is equal to (a) I + B

(b) I

j

(c) B −1

JEE Mains 2014 (d) (B −1)′

−1

 1

27 If  tan θ

− tan θ   1 tan θ  a −b  , then =   1  − tan θ 1  b a 

(a) a = 1, b = 1 (b) a = sin 2θ, b = cos 2θ (c) a = cos 2θ, b = sin2θ (d) None of the above

1 0 0

elements x1, y1, z1,.... of the determinant y1 y2 y3

24 If A =  3

28 If A = 2 1 0 and u1 , u 2 are column matrices such

20 If A1 , B1,C1 , ......... are respectively the co-factors of the x1 ∆ = x2 x3

− b and A adj A = AAT , then 5a + b is equal 2 

5a

16 If the system of linear equations

(d) 0

  3 2 1  1  0 that Au1 = 0 and Au 2 = 1 , then u1 + u 2 is equal to     0 0  −1 (a)  1    0

 −1 (b)  1    −1

 −1 (c)  −1    0

 1 (d)  −1    −1

29 The number of values of k, for which the system of equations (k + 1)x + 8y = 4k and k x + (k + 3)y = 3k − 1 has no solution, is (a) infinite (c) 2

(b) 1 (d) 3

j

JEE Mains 2013

30 The system of equations αx + y + z = α − 1, x + αy + z = α − 1 and x + y + αz = α − 1 has no solution, if α is (a) 1 (c) either –2 or 1

(b) not –2 (d) –2

31 If the system of linear equations

j

JEE Mains 2013

x1 + 2x 2 + 3x 3 = 6, x1 + 3x 2 + 5x 3 = 9 2x1 + 5x 2 + ax 3 = b is consistent and has infinite number of solutions, then (a) (b) (c) (d)

a = 8, b can be any real number b = 15, a can be any real number a ∈ R − {8 } and b ∈ R − {15 } a = 8, b = 15

60

SIX

40

32 If the trivial solution is the only solution of the system of equations x − ky + z = 0, kx + 3y − kz = 0 and 3x + y − z = 0 Then, the set of all values of k is (a) { 2 , − 3 }

(b) R − { 2 , − 3 } (c) R − { 2 } (d) R − { − 3 }

33 Let A, other than I or − I, be a 2 × 2 real matrix such that

A 2 = I , I being the unit matrix. Let tr ( A ) be the sum of j JEE Mains 2013 diagonal elements of A. Statement I tr ( A ) = 0 Statement II det ( A ) = − 1 (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

34 The set of all values of λ for which the system of linear equations 2x1 − 2x 2 + x 3 = λx1, 2x1 − 3x 2 + 2x 3 = λx 2 and – x1 + 2x 2 = λx 3 has a non-trivial solution. (a) (b) (c) (d)

is an empty set is a singleton set contains two elements contains more than two elements

j

JEE Mains 2015

35 Statement I Determinant of a skew-symmetric matrix of order 3 is zero. Statement II For any matrix A , det( AT ) = det( A ) and det( − A ) = − det( A ). Where, det ( A ) denotes the determinant of matrix A. j JEE Mains 2013 Then, (a) (b) (c) (d)

Statement I is true and Statement II is false Both statements are true Both statements are false Statement I is false and Statement II is true

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 6 If the equations a ( y + z ) = x , b( z + x ) = y , c( x + y ) = z

1 If a 2 + b 2 + c 2 = −2 and 1 + a x (1 + b )x (1 + c )x f ( x ) = (1 + a 2 )x 1 + b 2 x (1 + c 2 )x , (1 + a 2 )x (1 + b 2 )x 1 + c 2 x 2

2

2

have non-trivial solution, then to (a) 1

(b) 3

(c) 0

(d) 1

2 If A is a square matrix of order 3 such that | A| = 2, then −1 −1

| (adj A ) | is (a) 1

(b) 2

(c) 4

(d) 8

3 The equations (k − 1)x + ( 3k + 1)y + 2kz = 0, (k − 1)x + ( 4k − 2)y + (k + 3)z = 0 and 2x + ( 3k + 1)y + 3(k − 1)z = 0 gives non-trivial solution for some values of k, then the ratio x : y : z when k has the smallest of these values. (a) 3 : 2 :1

(b) 3 : 3 : 2

(c) 1 : 3 : 1

(d) 1:1:1

4 If x = 1 + 2 + 4 +.... upto k terms, y = 1 + 3 + 9+.... upto k terms and c = 1 + 5 + 25+ ... upto k terms. Then, x ∆= 3 2k (a) (20)k

2y 3 3k

4z 3 equals to 5k (b) 5 k

(c) 0

1 + 2x x 3 (b) 4

(d) −2

(c) – 1

system of equations in x , y and z. x2 y2 z2 x2 y2 z2 x2 y2 z2 + 2 − 2 = 1, 2 − 2 + 2 = 1 and − 2 + 2 + 2 = 1 2 a b c a b c a b c has (a) no solution (b) unique solution (c) infinitely many solutions (d) finitely many solutions

8 If S is the set of distinct values of b for which the following system of linear equations x + y + z = 1, x + ay + z = 1 and ax + by + z = 0 j has no solution, then S is JEE Mains 2017 (a) an infinite set (b) a finite set containing two or more elements (c) singleton set (d) a empty set

9 If M is a 3 × 3 matrix, where M T M = I and

5 Product of roots of equation 2 − x 1 (a) 2

(b) 2

7 Let a, b and c be positive real numbers. The following

then f ( x ) is a polynomial of degree. (a) 2

1 1 1 is equal + + 1+ a 1+ b 1+ c

4 (c) 3

(d) 2 k + 3 k + 5 k

1 1− x 2+ x 3+ x = 0 1+ x 1− x 2 1 (d) 4

det(M) = 1, then the value of det (M − I ) is (a) –1 (b) 1 (c) 0 (d) None of these

10 If a1, a 2 , . . ., a n , . . . are in GP, then the determinant log a n ∆ = log a n + 3 log a n + 6 (a) 2

(b) 4

log a n + 1 log a n + 4 log a n + 7 (c) 0

log a n log a n log a n

+ 2 + 5

is equal to

+ 8

(d) 1

61

DAY 11 Let A be a square matrix of order 2 with | A| ≠ 0 such that | A + | A| adj (A)| = 0, then the value of | A−| A| adj (A)| is (a) 1 (c) 3

(b) 2 (d) 4

14 Area of triangle whose vertices are (a,a 2 ),(b,b 2 ),(c,c 2 ) is 1 , and the area of triangle whose vertices are 2 (p, p 2 ),(q,q 2 ) and (r,r 2 ) is 4, then the value of

12 Let P and Q be 3 × 3 matrices P ≠ Q. If P 3 = Q 3 and

(1 + ap )2 (1 + aq )2 (1 + ar )2

P 2Q = Q 2P, then the determinant of (P 2 + Q 2 ) is (a) − 2 (c) 0

(b) 1 (d) − 1

3 1 + f (1) 1 + f ( 2) 13 If α, β ≠ 0, f (n ) = α + β and 1 + f (1) 1 + f ( 2) 1 + f ( 3) 1 + f ( 2) 1 + f ( 3) 1 + f ( 4) n

(b) 4

j

1 (b) αβ

(c) 1

JEE Mains 2014

(d) − 1

(1 + cp )2 (1 + cq )2 is (1 + cr )2 (c) 8

(d) 16

l

m n r and if (l − m)2 + (p − q )2 = 9, 1 1 1

n

= K (1 − α )2 (1 − β)2 (α − β)2 , then K is equal to (a) αβ

(a) 2

(1 + bp )2 (1 + bq )2 (1 + br )2

15 Let det A = p q

(m − n )2 + (q − r )2 = 16, (n − l )2 + (r − p)2 = 25, then the value of (det A)2 equals to (a) 36

(b) 100

(c) 144

(b) 169

ANSWERS SESSION 1

SESSION 2

1 (b)

2 (b)

3 (a)

4 (a)

5 (a)

6 (a)

7 (a)

8 (a)

9 (c)

10 (c)

11 (c)

12 (c)

13 (c)

14 (a)

15 (b)

16 (b)

17 (d)

18 (d)

19 (c)

20 (a)

21 (b)

22 (d)

23 (b)

24 (b)

25 (a)

26 (b)

27 (c)

28 (d)

29 (d)

30 (d)

31 (d)

32 (b)

33 (b)

34 (c)

35 (a)

1 (a) 11 (d)

2 (c) 12 (c)

3 (d) 13 (c)

4 (c) 14 (d)

5 (a) 15 (c)

6 (b)

7 (d)

8 (c)

9 (c)

10 (c)

62

SIX

40

Hints and Explanations SESSION 1 1 Given equation can be rewritteen as x − cy − bz = 0 cx − y + az = 0 bx + ay − z = 0 Since, x, y and z are not all zero ∴The above system have non-trivial solution. 1 −c −b ∴ c −1 a = 0 b a −1 ⇒ 1(1 − a2 ) + c (−c − ab ) − b( ac + b ) = 0 ⇒ 1 − a2 − c 2 − abc − abc − b2 = 0 ⇒ a2 + b 2 + c 2 = 1 − 2abc

1 e a e 2a 1 e a e 2a b 2b = 1 e e − 1 e b e 2b c 2c 1 e e 1 ec e2c

− a b aω2 = ω ⋅ ω − b c bω2 − c a cω2

= 0 [Q a + b + c = 0 ⇒ e a + b + c = 1 ]

a b a = − ω5 b c b = 0 c a c

5 Put x = 0 in the given equation, we get t =

log x log x log x = log y log x log z

log y log x log y log y log y log z

log z log x log z log y log z log z

By taking common factors from the rows, we get 1 ∆ = log x ⋅ log y ⋅ log z log x log y log x log y log x log y

C2 , ( x + 1) from C3 and ( x − 1) from R3 , we get 1 1 1 f ( x ) = x ( x2 − 1) 2 x x − 1 x 3x x − 2 x

e 3a e a e 2a 3b e − e b e 2b 3c e ec e2c

1 0 0 = x ( x − 1) 2 x − ( x + 1) 1 3 x − 2( x + 1) 2 2

1 1 1

= e a ⋅ eb ⋅ ec 1 e a e 2a e a 1 e 2a b 2b + e b 1 e 2b 1 e e c 2c 1 e e ec 1 e2c

C3 → C3 − C2 , C → C − C  2 1   2

=0 ∴

f (50) = 0

7 Clearly, α + β + γ = 0 On applying C1 → C1 + C2 + C3 in the given determinant, we get α +β+ γ β γ ∴∆ = α +β+ γ γ α α +β+ γ α β

⇒ ∴

x x

ω x + ω2

ω2 1

x

1

x+ω

ω x + ω2 1

2x x−4 2x

2x  2x   x − 4

= ( A + Bx )( x − A ) 2 On applying C1 → C1 + C2 + C3 , we get  5x − 4  5x − 4   5x − 4

2x x−4 2x

2x  2x   x − 4

= ( A + Bx )( x − A )2 On taking common (5x − 4) from C1 , we get 2x  1 2 x (5x − 4)1 x − 4 2 x    x − 4 1 2 x = ( A + Bx )( x − A )2 On applying R2 → R2 − R1 and R3 → R3 − R1 , we get 2x 2x 1 ∴ (5x − 4) 0 − x − 4 0  0 0 − x − 

   4

=0

[Q1 + ω + ω2 = 0] ω2 =0 1 x+ω

On comparing, we get A = − 4 and B = 5 1+ x 1 1 11 Let ∆ = 1 + y 1 + 2 y 1 =0 1 + z 1 + z 1 + 3z On applying C1 → C1 − C3 and C2 → C2 − C3 , we get ∆ =

x=0

a(1 + ω ) bω2 aω 9 ∆ = b(ω + ω2 ) c bω2 c (ω2 + 1) aω c − aω2 bω2 = −b c − cω aω

x − 4  2x   2x

On expanding along C1 , we get (5x − 4)( x + 4)2 = ( A + Bx )( x − A ) 2

8 On applying C1 → C1 + C2 + C3 , we get

1 x 1 1

10 Given,

= ( A + Bx )( x − A )2

0 β γ = 0 γ α =0 0 α β

log z log z log z

Now, by taking common factor from the 1 1 1 columns, we get 1 1 1 = 0 1 1 1 e a e 2a 4 Let ∆ = e b e 2b ec e2c

0 −1 3 1 0 − 4 = − 12 + 12 = 0 −3 4 0

6 On taking common factors x from

2 If we interchange any two rows of a determinant in the set B, its value becomes −1 and hence it is in C. Likewise, for every determinant in C, there is corresponding determinant in B. So , B and C have the same number of elements. 1 log x y log x z 3 Let ∆ = log y x 1 log y z log z x log z y 1

2

[QC1 → C1 + C3 ] aω bω2 c

x 0 1 y 2y 1 −2z −2z 1 + 3z

On expanding along R1 , we get ∆ = x[2 y (1 + 3z ) + 2z] +1[−2 yz + 4 yz] = 0 ⇒ 2[ xy + 3 xyz + xz] + 2 yz = 0 ⇒

xy + yz + zx + 3 xyz = 0 1 1 1 + + = −3 ⇒ x y z

63

DAY b + c c + a a+ b

12 Let ∆ = c + a a + b b + c

⇒ ( x2 − 4)

a+ b b + c c + a 2 (a + b + c ) c + a a + b = 2 (a + b + c ) a + b b + c 2 (a + b + c ) b + c c + a [using C 1 → C 1 + C 2 + C 3 ] a+ b + c c + a a+ b =2 a+ b + c a+ b b + c a+ b + c b + c c + a [taking common 2 from C1 ] a + b + c −b −c = 2 a + b + c −c − a a + b + c − a −b [using C 2 a =2 b c

→ b c a

C2 − C1 , C3 → C3 − C1 ] c a b

[using C1 → C1 + C2 + C3 ] On comparing, k = 2 a a+ 1 a−1 13 − b b + 1 b − 1 + (−1)n c c −1 c + 1

a a+ 1 a−1 = −b b + 1 b − 1 c c −1 c + 1 a+ 1 a−1 a + (−1)n b + 1 b − 1 −b c −1 c + 1 c [Q| A ′|= | A|] a a+ 1 a−1 = −b b + 1 b − 1 c c −1 c + 1 a−1 b −1 c +1

a a+ 1 a−1 = [1 + (− 1)n +2 ] − b b + 1 b − 1 c c −1 c + 1 This is equal to zero only, if n + 2 is an odd i.e. n is an odd integer. 7 14 2 x2 − 13

6 x2 − 13 3

1 1 x2 − 13 2 = 0 3 7

Now, on applying C2 → C2 − C1 and C3 → C3 − C1 , we get 1 0 0 ( x2 − 4) 2 x2 − 15 0 =0 x2 − 13 16 − x2 20 − x2 On expanding along R1 , we get ( x2 − 4)( x2 − 15)( x2 − 20) = 0 Thus, other five roots are −2, ± 15, ± 2 5 Hence, sum of other five roots is −2. a b c 15 Let ∆ = b c a c a b = = = =

a(bc − a2 ) − b (b2 − ac ) + c ( ab − c 2 ) abc − a3 − b 3 + abc + abc − c 3 − (a3 + b 3 + c 3 − 3 abc ) − (a + b + c ) (a2 + b 2 + c 2 − ab − bc − ca) 1 = − (a + b + c ) 2 {(a − b ) 2 + (b − c ) 2 + (c − a) 2 } < 0, where a ≠ b ≠ c

18 If ( x1 , y 1 ),( x2 , y 2 ) and ( x3 , y 3 ) are the vertices of x1 y 1 1 = x2 y 2 2 x3 y 3

x2 − 13 2 =0 7

x + ky + 3z = 0; 3 x + ky − 2z = 0; 2 x + 4 y − 3z = 0 System of equation has non-zero solution, if  1 k 3  3 k −2 = 0    2 4 −3 ⇒ (−3k + 8) − k (−9 + 4) + 3(12 − 2k ) = 0 ⇒ −3k + 8 + 9k − 4k + 36 − 6k = 0 ⇒ −4k + 44 = 0 ⇒ k = 11 Let z = λ , then we get …(i) x + 11 y + 3λ = 0 …(ii) 3 x + 11 y − 2λ = 0 and 2 x + 4 y − 3λ = 0 …(iii) On solving Eqs. (i) and (ii), we get 5λ −λ ,y = ,z = λ x= 2 2 xz 5λ2 ⇒ = = 10 2 2 y λ 2 ×  −   2

17 We have, ∴

aij = (i 2 + j2 − ij )( j − i ) a ji = (i 2 + j2 − ij )(i − j ) = −(i 2 + j2 − ij )( j − i ) = −aij

a triangle, then 1 1 1

Area …(i)

Also, we know that, if a is the length of an equilateral triangle, then Area =

3 2 a 4

…(ii)

From Eqs. (i) and (ii), we get

⇒

x1 3 2 1 a = x2 4 2 x3

y1 y2 y3

1 1 1

x1 3 2 a = x2 2 x3

y1 y2 y3

1 1 1

On squaring both sides, we get 2 x1 y 1 1 3 4 a = x2 y 2 1 4 x3 y 3 1

19 Clearly, area of a b +c 1 b c + a 2 c a+ b a 1 = (a + b + c ) b 2 c

(∆ABC ) =

16 We have,

a+ 1 b + 1 c −1 a−1 b −1 c + 1 a −b c

a+ 1 a + (−1)n + 1 b + 1 − b c −1 c

1 2 x2 − 13

1 1 1 1 1 1 1 =0 1 1 [QC2 → C2 + C1 ]

x1 z1 = x1 z3 − x3 z1 x3 z3 x y1 C2 = − 1 = −( x1 y 3 − x3 y 1 ) x3 y 3 x z1 B3 = − 1 = −( x1 z2 − x2 z1 ) x2 z2 x y1 and C3 = 1 = x1 y 2 − y 1 x2 x2 y 2 B C2 ∴ 2 B3 C 3 x1 z3 − x3 z1 −( x1 y 3 − y 1 x3 ) = −( x1 z2 − x2 z1 ) x1 y 2 − x2 y 1

20 Clearly, B2 =

=

x1 z3 − x1 z2 + x2 z1

− x1 y 3 x1 y 2 − x2 y 1

− x3 z1 y 1 x3 − x1 z2 + x2 z1 x1 y 2 − x2 y 1 − x1 y 3 x z x z − x1 y 3 = 1 3 + 1 3 − x1 z2 x1 y 2 x2 z1 − x2 y 1 − x3 z1 y 1 x3 − x3 z1 y 1 x3 + + − x1 z2 x1 y 2 x2 z1 − x2 y 1 +

On applying R1 → R1 + R2 + R3 , we get

⇒ A is a skew-symmetric matrix of odd order.

= x12 (z3 y 2 − z2 y 3 ) − x1 x2 (z3 y 1 − z1 y 3 )

x2 − 4 2 x2 − 13

∴

= x1 [ x1 (z3 y 2 − z2 y 3 ) − x2 (z3 y 1 − z1 y 3 )

x2 − 4 x2 − 13 3

x2 − 4 2 =0 7

tr ( A ) = 0 and| A|= 0

− x1 x3 (z1 y 2 − z2 y 1 ) + x2 x3 (z1 y 1 − z1 y 1 )

= x1 ∆

+ x3 (z2 y 1 − z1 y 2 )]

64

SIX

40

21 We have, A =  − 4 2

− 3 1 

 a −b  =  b a 

2b 10   + 3b = 13  15

∴ A2 = A ⋅ A  2 − 3  2 − 3 =   − 4 1  − 4 1   4 + 12 − 6 − 3  16 − 9 = =    − 8 − 4 12 + 1   − 12 13   16 − 9 Now, 3 A2 + 12 A = 3    − 12 13   2 − 3 + 12   − 4 1   48 − 27  24 − 36 = +    − 36 39   − 48 12   72 − 63 =   − 84 51   51 63 ∴adj (3 A2 + 12 A ) =    84 72 

20b + 45b = 13 15 65b = 13 ⇒ b = 3 ⇒ 15 Now, substituting the value of b in Eq. (iii), we get 5a = 2

⇒

Hence, 5a + b = 2 + 3 = 5 q −b y 25 Clearly,|B|= − p a − x r −c z

c

1 α 3    2 4 4

Find the determinant of P and apply the formula |adj A|=| A|n −1 5a − b  24 Given, A =  2  3 and A adj A = AA T Clearly, A(adj A ) = A I2 [Q if A is square matrix of order n, then A(adj A ) = (adj A ) ⋅ A = A I n ] 5a − b = I2 3 2 = (10a + 3b ) I2

a x =−b y c z

cos 2θ − sin 2θ  a −b  ⇒ =    sin 2θ cos 2θ  b a 

1   0 1 + Au1 + Au2 =  0 + 1  =  0 +       0  0  0 +

z

(taken (–1) common from C2 ) p a x = − q b y (QR1 ↔ R2 ) r c z

23 Given, adj A = P = 1 3 3 

p q r

⇒

Since, A is non-singular matrix, i.e. | A | ≠ 0, on multiplying both sides by A −1 , we get 1  A −1 A(u1 + u2 ) = A −1 1     0 −1

(QC1 ↔ C2 and then C2 ↔ C3 ) = −| A T | = −| A| Thus,| A|= −|B| Hence,| A| ≠ 0 iff|B|≠ 0 ∴ A is invertible iff B is invertible Also,|adj A|= | A| = |B| = |adj B | 2

2

1 0 0 1  ⇒ u1 + u2 = 2 1 0 × 1      3 2 1   0 1 0 0 Now, |A | = 2 1 0 3 2 1

25a2 + b 2 15a − 2b  = 13   15a − 2b

...(ii)

Also we have, AA T = A T A and B = A −1 A T

T

∴ A −1

Now, BB ′ = ( A −1 A T ) ( A −1 A T )T = A −1 A T A ( A −1 )T −1

−1 T

[Q( A ′ )′ = A]

= A AA ( A ) [Q AA = A T A ]

Q A(adj A ) = AA T 0 10a + 3b  ∴ 0 10a + 3b   25a2 + b 2 15a − 2b  = 13   15a − 2b [using Eqs. (i) and (ii)] ⇒ 15a − 2b = 0 2b ...(iii) ⇒ a= 15 and 10a + 3b = 13 ...(iv) On substituting the value of ‘a’ from Eq. (iii) in Eq. (iv), we get

T

T

= IA T ( A −1 )T = A T ( A −1 )T = ( A −1 A )T [Q( AB )T = B T A T ] = IT = I

27 We have,

− tan θ  1 tan θ  1  tan θ   − tan θ 1 1    

−1

 a −b  =  b a 

− tan θ 1  1 ⋅ ⇒ 1  1 + tan2 θ  tan θ − tan θ  a −b   1 =  tan θ 1  b a   ⇒

1 − tan2 θ −2 tan θ  1 2 1 + tan θ  2 tan θ 1 − tan2 θ

…(i)

= 1 (1 − 0) − 0 + 0 = 1 ∴ adj of A

⇒| A|≠ 0.

...(i)

0 1  0

1  A(u1 + u2 ) = 1     0

26 Given, A is non-singular matrix

1 0 = (10a + 3b )    0 1 10 a 3 b 0 +   = 0 10a + 3b    5a − b   5a 3 and AA T =  2   − b 2 3

 a −b  b a   

⇒ a = cos 2θ and b = sin 2θ

(taken (–1) common from R2 ) q b y = (+ ) p a x r

−2 tan θ  1 + tan2 θ  = 1 − tan2 θ  2  1 + tan θ 

28 On adding Au1 and Au2 , we get

q −b y = − p −a x r −c z

22 All the given statements are true.

 1 − tan2 θ  1 + tan2 θ ⇒  2 tan θ  1 + tan2 θ 

0 0  1 −2 1   1 =  0 1 −2 =  −2 1 0      0 0 1   1 −2 1  0 0  1 =  −2 1 0 [Q| A | = 1]    1 −2 1 

From Eq. (i), 0 1 u1 + u2 = −2 1   1 −2

0 1  0 × 1     1   0

 1 + 0 + 0  1  ⇒ u1 + u2 =  −2 + 1 + 0 =  −1      1 − 2 + 0  −1

29 Given equations can be written in

matrix form as AX = B 8  k + 1  x where, A =  , X =  y  k k + 3     4 k  and B =   3 k − 1  For no solution,| A |= 0 and (adj A) B ≠ 0

65

DAY Now, | A |=

k+1 8 =0 k k+3

⇒

(k + 1) (k + 3) − 8 k = 0

⇒

k2 + 4 k + 3 − 8 k = 0

⇒

k2 − 4 k + 3 = 0

⇒

(k − 1) (k − 3) = 0

∴

⇒ ∴

But α = 1 makes given three equations same. So, the system of equations has infinite solution. Hence, answer is α = − 2 for which the system of equations has no solution.

31 For consistency| A | = 0 and

k = 1, k = 3

 k + 3 −8  Also, adj A =    − k k + 1   k + 3 −8  4 k ∴(adj A) B =   3 k − 1 − k k + 1    (k + 3) (4 k ) − 8 (3 k − 1) = 2   −4 k + (k + 1) (3 k − 1)   4 k 2 − 12k + 8 =  2  − k + 2k − 1 

Put k = 3, then (adj A) 36 − 36 + 8  8  B =  =   ≠ 0, true.  −9 + 6 − 1   −4 Hence, the required value of k is 3. Alternate Method Condition for the system of equations has no solution is 4k a1 b c k+1 8 = 1 ≠ 1 ⇒ = ≠ a2 b2 c2 k k + 3 3k − 1

kx + 3 y − kz = 0 and 3 x + y − z = 0 has trivial solution. 1 −k 1 k 3 −k ≠0 ∴ 3 1 −1

2− λ 2 –1

∴

35 Determinant of skew-symmetric matrix of odd order is zero and of even order is perfect square. Also, det( A T ) = det( A ) det(− A ) = (−1)n det( A )

and

Hence, Statement II is false.

SESSION 2 1 Given that,

⇒ 1 (− 3 + k ) + k (− k + 3k ) + 1 (k − 9) ≠ 0 ⇒ k − 3 + 2 k2 + k − 9 ≠ 0

1 + a2 x (1 + b 2 )x (1 + c 2 )x f ( x ) = (1 + a2 )x 1 + b 2 x (1 + c 2 )x (1 + a2 )x (1 + b 2 )x 1 + c 2 x

⇒ ⇒ ⇒

On applying C1 → C1 + C2 + C3 , we get 1 + a2 x + x + b 2 x + x + c 2 x f ( x ) = x + a2 x + 1 + b 2 x + x + c 2 x x + a2 x + x + b 2 x + 1 + c 2 x

2 k 2 + 2 k − 12 ≠ 0 k2 + k − 6 ≠ 0 (k + 3) (k − 2) ≠ 0 k ≠ 2, − 3 k ∈ R − {2, − 3}

2

[false] [true]

30 The system of given equations has no solution, if 1 1 α 1 =0 1 α

On applying C1 → C1 + C2 + C3 , we get α+2 1 1 α +2 α 1 =0 α +2 1 α On applying R2 → R2 − R1 , R3 → R3 − R1 ,

−2 1 −(3 + λ ) 2 = 0 2 −λ

⇒ (2 − λ )(3λ + λ2 − 4) + 2(−2λ + 2) +1(4 − 3 − λ ) = 0 ⇒ (2 − λ )(λ2 + 3λ − 4) + 4(1 − λ ) + (1 − λ ) = 0 ⇒ (2 − λ )( λ + 4)( λ − 1) + 5 (1 − λ ) = 0 ⇒ (λ − 1)[(2 − λ )(λ + 4) − 5] = 0 ⇒ (λ − 1)(λ2 + 2λ − 3) = 0 ⇒ (λ − 1)[(λ − 1)(λ + 3)] = 0 ⇒ (λ − 1)2 (λ + 3) = 0 ⇒ λ = 1,1,−3

(1 + b 2 )x 1 + b2 x (1 + b 2 )x

a b 33 Let A =   such that A ≠ I , − I c  d

∴ k =3 Hence, only one value of k exists.

1 1 1 (α + 2) 0 α − 1 0 =0 0 0 α −1

⇒ 1(3 a − 25) − 2(a − 10) + 3(5 − 6) = 0 ⇒ a= 8 On solving, (adj A ) B = O , we get b = 15

∴

Take

α 1 1

(adj A ) B = O 1 2 3 1 3 5 =0 2 5 a

32 Since, x − ky + z = 0,

Put k = 1, then  4 − 12 + 8  0 (adj A) B =   =  , not true.  −1 + 2 − 1   0

k+1 8 = k k+3 ⇒ k 2 + 4 k + 3 = 8k ⇒ k2 − 4 k + 3 = 0 ⇒ (k − 1) (k − 3) = 0 ∴ k = 1, 3 8 4⋅1 If k = 1, then ≠ , 1+ 3 2 8 4⋅3 , and k = 3, then ≠ 6 9−1

(α + 2)(α − 1)2 = 0 α = 1, − 2

a b  1 0 and   =  0 1 c d      a2 + bc ⇒  ac + cd

ab + bd  1 0 = bc + d 2   0 1 

⇒ b(a + d ) = 0,c (a + d ) = 0 and a2 + bc = 1,bc + d 2 = 1 ⇒ a = 1,d = −1,b = c = 0 or a = −1,d = 1,b = c = 0 1 0  A=  , then  0 − 1 det( A ) = −1 and tr ( A ) = 1 − 1 = 0

34 Given system of linear equations are 2 x1 − 2 x2 + x3 = λx1 …(i) ⇒ (2 − λ) x1 − 2 x2 + x3 = 0 2 x1 − 3 x2 + 2 x3 = λx2 …(ii) ⇒ 2 x1 − (3 + λ )x2 + 2 x3 = 0 and − x1 + 2 x2 = λx3 …(iii) ⇒ − x1 + 2 x2 − λx3 = 0 Since, the system has non-trivial solution.

(1 + c 2 )x (1 + c 2 )x 1 + c2 x

1 + (a2 + b 2 + c 2 + 2)x (1 + b 2 )x = 1 + (a2 + b 2 + c 2 + 2)x (1 + b 2 )x 1 + (a2 + b 2 + c 2 + 2)x (1 + b 2 )x

1 = 1 1

(1 + b 2 )x (1 + c 2 )x 1 + b 2 x (1 + c 2 )x (1 + b 2 )x 1 + c 2 x

(1 + c 2 )x (1 + c 2 )x 1 + c2 x

[Qa2 + b 2 + c 2 = − 2, given] On applying R1 → R1 − R3 , R2 → R2 − R3 , we get 0 0 x −1 = 0 1− x x −1 1 (1 + b 2 )x 1 + c 2 x =

0 1− x

x −1 = ( x − 1) 2 x −1

Hence, f ( x ) is of degree 2.

66

SIX

40

−1

2 Clearly, adj A −1 = | A −1 |2 = 1 2 | A|

and |(adj A −1 )−1 |=

6 Here, b c

1 |(adj A −1 )|

On applying, C2 → C2 − C1 and C3 → C3 − C1 , we get −1 a+ 1 a+ 1 b −(b + 1) 0 =0 c 0 − (1 + c )

= | A|2 = 22 = 4

3 For non - trivial solution, k − 1 3k + 1 2k k − 1 4k − 2 k + 3 = 0 2 3k + 1 3(k − 1) ⇒

On applying R1 →

−x + y = 0

...(i)

− x − 2 y + 3z = 0

...(ii)

2 x + y − 3z = 0

(iii)

From (i), we get x = y = λ (say), Then, from Eq. (ii), we get − λ − 2λ + 3z = 0 ⇒

3z = 3λ

⇒

z=λ

∴

x: y : z = 1:1:1

k 4 Clearly, x = 2k − 1, y = 3 − 1

2 5k − 1 and z = 4  Q sum to n terms of a GP is  given by

a(r n − 1) r − 1 

2k − 1 3k − 1 5k − 1 Now, ∆ = 3 3 3 2k 3k 5k 2k 3k = 3 3 2k 3k

R1 R2 , R2 → a+ 1 b+1

R3 , we get c+1 1 − 1 1 a+ 1 b −1 0 = 0 b+1 c 0 −1 c+1 1 b c ⇒ − + + =0 a+ 1 b + 1 c + 1 1 1 c + 1− + 1− =0 ∴− a+ 1 b+1 c+1 1 1 1 ⇒ + + =2 a+ 1 b + 1 c + 1

For a = 1 ∆ = ∆1 = ∆2 = ∆3 = 0 For b = 1 only x + y + z = 1, x + y + z = 1 and x+ y + z=0 i.e. no solution (QRHS is not equal) Hence, for no solution, b = 1 only.

9 Clearly, det(M − I ) = det(M − I ) ⋅ det(M ) [Qdet(M ) = 1] = det(M − I )det(M T )

and R3 →

k = 0 or 3

Now, when k = 0, then the equation becomes

and

a a −1 b = 0 c −1

5k 1 1 3 −3 1 1 5k 2k 3k

1 1 5k

= 0−3× 0= 0 1 + 2x 1 1− x 5 Let p( x ) = 2 − x 2 + x 3 + x = 0 x 1 + x 1 − x2 Clearly, product of roots Constant term = Coefficient of x 4 Here, constant term is given by 1 1 1 P(0) = 2 2 3 = −1 0 1 1 and coefficient of x 4 is −2 1 ∴ Product of roots is . 2

2 2 y2 7 Let x2 = X , 2 = Y , z2 = Z a b c Then, X + Y − Z = 1, X − Y + Z = 1, −X + Y + Z = 1 Now, coefficient matrix is  1 1 −1 A =  1 −1 1     −1 1 1 

Here,| A|≠ 0 ∴ It has unique solution, viz., X = 1, Y = 1 and Z = 1 Hence , x = ± a; y = ± b and z = ± c . 1 1 1 8 Here, ∆ = 1 a 1 a b 1 = 1 (a − b ) − 1 (1 − a) + 1 (b − a2 ) = − (a − 1)2

1 1 1 ∆1 = 1 a 1 0 b 1 = 1 (a − b ) − 1 (1) + 1 (b ) = (a − 1) 1 1 1 ∆2 = 1 1 1 a 0 1 = 1 (1) − 1 (1 − a) + 1 (0 − a) = 0 1 1 1 and ∆ 3 = 1 a 1 a b 0 = 1 ( − b ) − 1 (− a) + 1 (b − a2 ) = − a(a − 1)

[Q det( A T ) = det( A )] = det(MM T − M T ) = det(I − M T )

[Q MM T = I ]

= − det(M − I ) T

= − det(M T − I )T = − det(M − I ) ⇒ 2det(M − I ) = 0 ⇒ det(M − I ) = 0

10 Since, a1 , a2 , . . . , an, . . . are in GP, then, log an , log an + 1 , log an + 2 , ..., log an + 8 , ...are in AP. Given that, log an log an + 1 ∆ = log an + 3 log an + 4 log an + 6 log an + 7 a ∴ ∆ = a + 3d a + 6d

a+ d a + 4d a + 7d

log an +2 log an + 5 log an + 8 a + 2d a + 5d a + 8d

where a and d are the first term and common difference of AP. On applying R2 → 2R2 , we get a a+ d a + 2d 1 ∆ = 2a + 6d 2a + 8d 2a + 10d 2 a + 6d a + 7d a + 8d On applying R2 a 1 ∆ = a 2 a + 6d

→ R2 − a+ d a+ d a + 7d

R3 , we get a + 2d a + 2d = 0 a + 8d

 11 Let A =  . Then, c d  a b

| A|= ad − bc = k (say)  d −b  and adj ( A ) =    −c a  Now,| A +| A| adj ( A )|= 0 a + kd (1 − k )b =0 ⇒ (1 − k )c d + ka ⇒ ⇒

(a + kd )(d + ka) − (1 − k )2 bc ad + a2 k + k d 2 + k 2 ad −(1 + k 2 − 2k )bc ⇒ (ad − bc ) + ( ad − bc )k 2 + k( a2 + d 2 + 2bc ) = 0

67

DAY ⇒ (ad − bc ) + (ad − bc )k 2 + k (a2 + d 2 ) +2(ad − k )) = 0 [Qbc = ad − k] ⇒ (ad − bc ) + (ad − bc )k 2 + k(a + d )2 − 2k 2 = 0 3 ⇒ (k + k − 2k 2 ) + k(a + d )2 = 0 ⇒ k[(k − 1)2 + (a + d )2 ] = 0 ⇒ k = 1 and a + d = 0 Now,|A−|A|adj A| a − kd (1 + k )b a−d 2b = = (1 + k )c d − ak 2c d−a = −(a − d ) − 4bc = −((a + d )2 − 4ad ) − 4bc = 4ad − 4bc = 4k = 4 2

12 On subtracting the given equation, we get P 3 − P 2Q = Q 3 − Q 2 P ⇒ P 2 (P − Q ) = Q 2 (Q − P ) ⇒ (P − Q ) (P 2 + Q 2 ) = 0 Now, if |P 2 + Q 2 | ≠ 0 (P 2 + Q 2 ) is invertible . On post multiply both sides by (P 2 + Q 2 )−1 , we get P − Q = 0, which is a contradiction. Hence,|P 2 + Q 2|= 0 3 1 + f (1) 1 + f (2) 13 Let ∆ = 1 + f (1) 1 + f (2) 1 + f (3) 1 + f (2) 1 + f (3) 1 + f (4)

...(i)

3 1+ α+ β ⇒ ∆ = 1 + α + β 1 + α 2 + β2 1 + α 2 + β2 1 + α 3 + β3 1 + α 2 + β2 1 + α 3 + β3 1 + α4 + β4 1⋅1 + 1⋅1 + 1⋅1 1⋅1 + 1⋅α + 1⋅β = 1⋅1 + α ⋅1 + β ⋅1 1⋅1 + α ⋅α + α ⋅β 1 ⋅ 1 + 1 ⋅ α 2 + 1 ⋅ β2 1 ⋅ 1 + α 2 ⋅ α + β2 ⋅ β 1 ⋅ 1 + 1 ⋅ α 2 + 1 ⋅ β2 1 ⋅ 1 + α ⋅ α 2 + β ⋅ β2 1 ⋅ 1 + α 2 ⋅ α 2 + β2 ⋅ β2 1 1 1 1 1 1 1 1 1 = 1 α β 1 α β = 1 α β 1 α 2 β2 1 α 2 β2 1 α 2 β2

2

On expanding, we get ∆ = (1 − α )2 (1 − β)2 (α − β2 ) Hence, K (1 − α )2 (1 − β)2 (α − β)2 = (1 − α )2 (1 − β)2 (α − β)2 ∴ K =1 (1 + ap )2 (1 + bp )2 (1 + cp )2 14 Let ∆ = (1 + aq )2 (1 + bq )2 (1 + cq )2 (1 + ar )2 (1 + br )2 (1 + cr )2 1 + 2ap + a p 1 + 2bp + b p = 1 + 2aq + a2q 2 1 + 2bq + b 2q 2 1 + 2ar + a2 r 2 1 + 2br + b 2 r 2 1 + 2cp + c 2 p2 1 + 2cq + c 2q 2 1 + 2cr + c 2 r 2 2

2

2

2

1 p p2 1 2a a2 2 = 1 q q × 1 2b b 2 1 r r2 1 2c c 2  1 p p2 1 a a2    = 2 1 q q 2 × 1 b b 2   1 r r2 1 c c2    = 2(2A1 × 2A2 ) = 2(8 × 1) = 16

15 According to given conditions we get a right angled triangle whose vertices are (n, r ),(m,q ) and (l, p) . A(n, r)

4

B(m, q)

5

C(l, p)

3

l m n Also, we have,|A|= p q r 1 1 1 2

l p 1 2 ⇒|A| = m q 1 = [2ar(∆ABC )] n r 1 2

2

1 = 2 × × 3 × 4 = 144 2  

DAY SEVEN

Binomial Theorem and Mathematical Induction Learning & Revision for the Day u

u

Binomial Theorem Binomial Theorem for Positive Index

u

u

Properties of Binomial Coefficient Applications of Binomial Theorem

u

u

Binomial Theorem for Negative/Rational Index Principle of Mathematical Induction

Binomial Theorem Binomial theorem describes the algebraic expansion of powers of a binomial. According to this theorem, it is possible to expand ( x + y)n into a sum involving terms of the form axb yc , where the exponents b and c are non-negative integers with b + c = n. The coefficient a of each term is  n a specific positive integer depending on n and b, is known as the binomial coefficient   . b 

Binomial Theorem for Positive Index An algebraic expression consisting of two terms with (+ ) ve or (−)ve sign between them, is called binomial expression. If n is any positive integer, then ( x + a)n = nC0 x n + nC1 x n − 1 a + L + n Cnan =

Your Personal Preparation Indicator

n

∑ nCr ⋅ x n − r ar , where x and a are real (complex) numbers.

r=0

(i) (ii) (iii) (iv)

PRED MIRROR

The coefficient of terms equidistant from the beginning and the end, are equal. n ( x − a)n = nC0 x n − nC1 x n − 1 a + L + (− 1)n Cnan (1 + x)n = nC0 + nC1 x + nC2 x2 + L + nCn x n Total number of terms in the expansion ( x + a)n is (n + 1). (n + 1) (n + 2) (v) If n is a positive integer, then the number of terms in ( x + y + z)n is . 2

u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

69

DAY (vi) The number of terms in the expansion of

Properties of Binomial Coefficients

n + 2 , if n is even  n n ( x + a) + ( x − a) =  2 n+1  , if n is odd  2 (vii) The number of terms in the expansion of

In the binomial expansion of (1 + x)n, (1 + x)n = nC0 + nC1 ⋅ x + nC2 ⋅ x2 + ... + nCr ⋅ x r + ... nCn ⋅ x n, where, nC0, nC1,...,n Cn are the coefficients of various powers of x are called binomial coefficients and it is also written as

 n , if n is even  n n ( x + a) − ( x − a) =  2 n+1  , if n is odd  2

 n  n  n C0, C1, ..., Cn or   ,   , … ,    0  1  n

General Term and Middle Term (i) Let (r + 1)th term be the general term in the expansion of ( x + a)n. Tr + 1 = nCr x n − r ar

n

●

n

(iii) The middle term in the expansion of (a + x)n. n  (a) Case I If n is even, then  + 1 th term is middle 2  term. (n + 1) (b) Case II If n is odd, then th term and 2 (n + 3) th terms are middle terms. 2 (iv) ( p + 1)th term from end = (n − p + 1)th term from beginning. (v) For making a term independent of x we put r = n in general term of ( x + a)n, so we get nCnan, that is independent of x. NOTE

If the coefficients of rth, ( r + 1)th, ( r + 2)th term of ( 1 + x ) are in AP, then n2 − ( 4 r + 1) n + 4 r 2 = 2

Tr + 1 Tr

=

Cr ⋅ x n−r +1 = ⋅x r −1 r Cr − 1 ⋅ x n

r

n

Let numerically, Tr + 1 be the greatest term in the above Tr + 1 expansion. Then, Tr + 1 ≥ Tr or ≥ 1. Tr n−r +1 (n + 1) …(i) ∴ | x | ≥ 1 or r ≤ | x| r (1 + | x |) (i) Now, substituting values of n and x in Eq. (i), we get r ≤ m + f or r ≤ m, where m is a positive integer and f is a fraction such that 0 < f < 1. (ii) When r ≤ m + f , Tm + 1 is the greatest term, when r ≤ m, Tm and Tm +1 are the greatest terms and both are equal. (iii) The coefficients of the middle terms in the expansion of (a + x)n are called greatest coefficients.

n+ 1

•

Cr

n−1

•

Cr n−r +1 = r Cr −1

n n

n+ 1 Cr Cr + 1 = r +1 n+1 n

●

C0 + C1 + C2 + ...+ Cn = 2 n

●

C0 + C2 + C4 + ... = C1 + C3 + C5 +... = 2 n − 1

●

C0 − C1 + C2 − C3 + ... + (− 1)n ⋅ Cn = 0

●

C20 + C12 + C22 + ... + C2n = 2 nCn =

●

(−1)n/2 ⋅n Cn/2 , if n is even C20 − C12 + C22 − C32 + .... =  0, if n is odd 

●

C0 ⋅ Cr + C1 ⋅ Cr + 1 + ... + Cn − r ⋅ Cn

Cr −1

= 2 nCn− r =

(2 n)! (n !)2

(2 n)! (n − r )!(n + r )!

●

C1 − 2C2 + 3C3 − .... = 0

●

C0 + 2C1 + 3C2 + ... + (n + 1) ⋅ Cn = (n + 2) 2 n − 1

●

C0 − C2 + C4 − C6 +K = 2 n ⋅ cos

●

If Tr and Tr + 1 be the rth and (r + 1)th terms in the expansion of (1 + x)n, then

Cr + n Cr −1 =

Cr1 = n Cr2 ⇒ r1 = r2 or r1 + r2 = n

n

r ⋅ nCr = n ⋅

n

Greatest Term

●

●

(ii) If expansion is ( x − a)n, then the general term is

( − 1) r ⋅ n C r x n − r a r .

Cr = nCn− r

●

nπ 4 nπ n C1 − C3 + C5 − C7 + ... = 2 ⋅ sin 4

Applications of Binomial Theorem 1. R-f Factor Relation Here, we are going to discuss problems involving ( A + B)n = I + f , where I and n are positive integers 0 ≤ f ≤ 1,| A − B2| = k and| A − B| < 1.

2. Divisibility Problem In the expansion, (1 + α )n. We can conclude that, (1 + α )n − 1 is divisible by α, i.e. it is a multiple of α.

3. Differentiability Problem Sometimes to generalise the result we use the differentiation. (1 + x)n = nC0 + nC1 x + nC2 x 2 + … + nCn x n

70

40

On differentiating w.r.t. x, we get n(1 + x)n−1 = 0 + n C1 + 2 ⋅ x ⋅ nC2 + … + n ⋅ nCn ⋅ x n−1

Principle of Mathematical Induction

Put x = 1, we get, n 2 n−1 = nC1 + 2 nC2 + … + n nCn

In algebra, there are certain results that are formulated in terms of n, where n is a positive integer. Such results can be proved by a specific technique, which is known as the principle of mathematical induction.

Binomial Theorem for Negative/Rational Index Let n be a rational number and x be a real number such that n(n − 1) 2 n(n − 1)(n − 2) 3 x < 1, then (1 + x)n = 1 + nx + x + x + ... 2! 3! ●

●

If n is a positive integer, then (1 + x)n contains (n + 1) terms i.e. a finite number of terms. When n is any negative integer or rational number, then expansion of (1 + x)n contains infinitely many terms. When n is a positive integer, then expansion of (1 + x)n is valid for all values of x. If n is any negative integer or rational number, then expansion of (1 + x)n is valid for the values of x satisfying the condition| x| < 1. (i) (1 + x)−1 = 1 − x + x2 − x3 + ... (ii) (1 − x)−1 = 1 + x + x2 + x3 + K (iii) (1 + x)−2 = 1 − 2 x + 3 x2 − 4 x3 + ... (iv) (1 − x)−2 = 1 + 2 x + 3 x2 + 4 x3 + ...

First Principle of Mathematical Induction It consists of the following three steps Step I Actual verification of the proposition for the starting value of i. Step II Assuming the proposition to be true for k, k ≥ i and proving that it is true for the value (k + 1) which is next higher integer. Step III To combine the above two steps. Let p(n) be a statement involving the natural number n such that (i) p(1) is true i.e. p(n) is true for n = 1. (ii) p(m + 1) is true, whenever p(m) is true i.e. p(m) is true ⇒ p(m + 1) is true. Then, p(n) is true for all natural numbers n. Product of r consecutive integers is divisible by r !

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 If (1 + ax )n = 1 + 8x + 24x 3 + ..., then the values of a and n are (a) 2, 4

(b) 2, 3

(c) 3, 6

(d) 1, 2

2 The coefficient of x n in the expansion of (1 + x )2n and (1 + x )2n −1 are in the ratio (a) 1 : 2 (c) 3 : 1

j

NCERT Exemplar

3 The value of (1. 002)12 upto fourth place of decimal is (b) 1.0245 (d) 1.0254

n

(b) (d)

C4 C4 + n C2 + n C2

n n

C4 + n C2 C4 + n C2 + n C1 ⋅ n C2

1  + x sin x  x 

5 If the middle term of 

10

is equal to 7

the value of x is (a) 2nπ +

π 6

(c) nπ + (− 1)n

j

(b) nπ + π 6

(a) e 2

(b) e

(c) e / 2

(d) 2e

7 If the number of terms in the expansion of 4  2 1 − + 2  , x ≠ 0 is 28, then the sum of the coefficients  x x  j of all the terms in this expansion, is JEE Mains 2016 (b) 2187

(c) 243

(d) 729

8 In the binomial expansion of (a − b ) , n ≥ 5, the sum of 5th n

is n

9

 3 ln x  , x > 0 is equal to 729, then x can be  j JEE Mains 2013

(a) 64

4 The coefficient of x 4 in the expansion of (1 + x + x 2 + x 3 )n (a) (c)

 3 + 3  84

n

(b) 1 : 3 (d) 2 : 1

(a) 1.0242 (c) 1.0004

6 If the 7th term in the binomial expansion of

π 6

(d) nπ + (− 1)

7 , then 8

NCERT Exemplar

π 3

and 6th terms is zero, then 5 n−4 n −5 (c) 6 (a)

a is equal to b 6 n −5 n−4 (d) 5 (b)

9 In the expansion of the following expression 1 + (1 + x ) + (1 + x )2 + ... + (1 + x )n , the coefficient of x 4( 0 ≤ k ≤ n ) is (a) (c)

n +1 n

Ck + 1

Cn −k −1

(b) n Ck (d) None of these

71

DAY 10 The coefficient of t 24 in the expansion of

21 If the sum of the coefficients in the expansion of

(1 + t 2 )12(1 + t 12 )(1 + t 24 ) is (a)

12

C6 + 2

(b)

12

11 The coefficient of x

C5

53

(c)

12

12

(d)

C6

( x − 2y + 3z )n is 128, then the greatest coefficient in the expansion of (1 + x )n is

C7

(a) 35

in the following expansion

∑ 100Cm ( x − 3)100−m ⋅ 2m is (a)

C47

(b)

100

C53

(c) −

100

(d) −

100

C53

C100

(a) n = 2r (c) n = 2r + 1

12 If p is a real number and if the middle term in the 8

p  expansion of  + 2 is 1120, then the value of p is 2 

(b) 517

23 If an = ∑

r =0

2  , is x

(c) 569

(d) 581

(c) 12

(a)

AIEEE 2012

(b) an odd positive integer (c) an even positive integer (d) a rational number other than positive integers



16 If the (r + 1) th term in the expansion of  3 

17 If x

2k

(b) 10

(c) 8

1  occurs in the expansion of  x + 2   x 

(a) n − 2k is a multiple of 2 (c) k = 0

21

15

 20 1 (b)    6  81

 20 1 (a)    7  27 1  20 (c)   9  9

JEE Mains 2013

(d) 1 : 32

1   3

(b) 2 21 − 210

1 (b) 20C10 2

(a) (n + 2)2n −1 (c) (n + 1)2n −1

is

(d) None of these

(c) 2 20 − 2 9

JEE Mains 2017

(d) 2 20 − 210

(c) 0

j

(d)

AIEEE 2007 20

(c) 7th

(d) 6th

C10

29 If n > ( 8 + 3 7 )10, n ∈ N, then the least value of n is (a) (b) (c) (d)

(8 + 3 (8 + 3 (8 + 3 (8 + 3

7 )10 − (8 − 3 7 )10 + (8 − 3 7 )10 − (8 − 3 7 )10 − (8 − 3

7 )10 7 )10 7 )10 + 1 7 )10 − 1

(a) 3

(b) 19

31 If A = 1000

1000

(c) 64

and B = (1001)

999

(d) 29

, then

(b) A = B (d) None of these

32 If n − 1Cr = (k 2 − 3) ⋅ nCr + 1, then k belongs to (a) (− ∞, − 2 ] (b) [2 , ∞)

(c) [ − 3 , 3 ] (d) ( 3 , 2 ]

33 The remainder left out when 82n − ( 62)2n + 1 is divided by 9, is

(b) 3th

C55

10

(b) (n + 1)2n (d) (n + 2)2n

(a) A > B (c) A < B

1 is 5

(a) 5th

21

30 49n + 16n − 1 is divisible by

20

20 The largest term in the expansion of ( 3 + 2x )50, where x =

10

C0 + 2C1 + 3C2 + ...+ (n + 1)Cn will be

(b) n − 2k is a multiple of 3 (d) None of these

 

65

(d)

C10

28 If (1 + x )n = C0 + C1x + C2x 2 +...+ Cn X n , then the value of

, then

19 The greatest term in the expansion of 3 1 +

30

C0 − 20 C1 + 20 C2 − 20 C3 + ...+ 20C10 is

(a) − C10

n −3

(c) 1 : 4

(c)

C10

27 The sum of the series 20

2  independent of x in the expansion of  x 2 +  , is  x (b) 7 : 64

60

26 The value of ( C1 − C1) + ( C2 − C2 ) 21

20

18 The ratio of the coefficient of x 15 to the term

(a) 7 : 16

(b)

C11

j

(d) 6

j

30

(a) 2 21 − 211

has the same power of a and b, then the value of r is (a) 9

(d) 0

(c) n

+ ( 21C3 −10 C3 ) + ( 21C4 −10 C4 ) + ... + ( 21C10 −10 C10 ) is

b   3 a

a + b

(b) −1

(a) 1

 30  30  30  30  30  30    −     + ... +     is equal to  0  10  1   11  20  30

15 If n is a positive integer, then ( 3 + 1)2n − ( 3 − 1)2n is (a) an irrational number

1 + rx

∑ ( −1)r (n Cr ) 1 + nx is equal to

25 

(d) 24 j

(b) nan (d) None of these

r =0

and x 2 are 3 and −6 respectively, them m is (b) 9

n 1 r , then ∑ n C is equal to n Cr r r =0

n

24

14 If in the expansion of (1 + x )m (1 − x )n , the coefficient of x (a) 6

(d) None of these

(b) n = 3r (d) None of these

(a) (n − 1)an 1 (c) nan 2

6

 

13 The constant term in the expansion of 1 + x + (a) 479

n

j NCERT Exemplar (b) ±1 (d) None of these

(a) ±3 (c) ± 2

(c) 10

( 3r )th and (r + 2)th powers of x in the expansion of (1 + x )2n are equal, then

m =0

100

(b) 20

22. If for positive integers r > 1, n > 2, the coefficient of the

100

(a) 0

(b) 2

(c) 7

(d) 8

72

40

34 If x is positive, the first negative term in the expansion of (1 + x )

27 / 5

is

j

(a) 7th term

(b) 5th term

(c) 8th term

37 For each n ∈ N, 23n − 1 is divisible by

AIEEE 2003

(a) 8 (c) 32

(d) 6th term

35 Let P (n ) : n 2 + n + 1 ( n ∈ N ) is an even integer. Therefore,

(b) 16 (d) None of these

38 Let S (k ) = 1 + 3 + 5+...+( 2k − 1) = 3 + k 2. Then, which of the following is true?

P (n ) is true (a) for n > 1

(b) for all n

(c) for n > 2 (d) None of these j

(a) (n + 1) ! − 2 (c) (n + 1) ! − 1

AIEEE 2004

(a) S(1) is correct (b) S (k) ⇒ S (k + 1) (c) S (k) ⇒ / S (k + 1) (d) Principle of mathematical induction can be used to prove the formula

36 For all n ∈ N, 1 × 1! + 2 × 2 ! + 3 × 3! + ...+ n × n ! is equal to

j

NCERT Exemplar

(b) (n + 1)! (d) (n + 1)! − 3

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 The coefficient of x 2m + 1 in the expansion of

7 If a and d are two complex numbers, then the sum to (n + 1) terms of the following series aC0 − (a + d )C1 + (a + 2d )C2 − ... + ... is

1 E = , x < 1 is 2 (1 + x )(1 + x )(1 + x 4 )(1 + x 8 )...(1 + x 2m ) (a) 3

(b) 2

(c) 1

a 2n (c) 0

(d) 0

(a)

C C C 2 C1 − 2 + 3 – ... + ( − 1)n − 1 n is equal to 2 3 n 1 + 2 1 (c) 1 + + 2 (a) 1 −

(−1)n − 1 1 −... + 3 n 1 1 + ... + 3 n −1

n

1 1 1 + + ... + 2 3 n

(b) 1 +

(d) None of these 10

 

3 If the coefficient of x 5 in ax 2 +

1  is a times and equal bx  10

1   to the coefficient of x −5 in ax − 2 2 , then the value  b x  of ab is (a) (b)−3

(b) − (b)6

(c) (b)−1

(d) None of these

4 The sum of coefficients of integral powers of x in the binomial expansion of (1 − 2 x )50, is 1 50 (a) (3 + 1 ) 2 1 50 (c) (3 − 1 ) 2

j

JEE Mains 2015

(a) 4

(b) 120

(c) 210

(b)

(a) 3n (c) 3n + 2n

(b) 2n (d) 3n − 2n

9 The sum of the series 1

n

3r

7r



15r

∑ ( −1)r nCr  2r + 22r + 23r + 24r + ... + m terms is

r =0

2mn − 1 2 (2n − 1) 2mn + 1 (c) n 2 +1 (a)

(b)

mn

2mn − 1 2n − 1

(d) None of these

10 The value of x, for which the 6th term in the expansion of 9 x −1 + 7

+

7

  is 84, is equal to (1/ 5 )log2 (3 x −1 + 1)  2 1

(b) 3

(c) 2

(d) 5

 

JEE Mains 2013

(d) 310

C02 + C12 + C22 + C32 + ... + Cn2 is equal to n! n !n ! (2n)! (c) n!

p =1 m = p

 n  m     is equal to m  p 

11 If the last term in the binomial expansion of  21/ 3 −

6 If (1 + x )n = C0 + C1x + C2x 2 + ... + Cn x n , then (a)

∑ ∑

(a) 4

5 The term independent of x in expansion of j

(d) None of these

n

 log2 2 

1 50 (b) (3 ) 2 1 50 (d) (2 + 1 ) 2

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

8

(b) na

(2n)! n !n !

(d) None of these

 1  is  5/ 3  3 

1   2

n

log3 8

, then the 5th term from the beginning is

(a) 210 (c) 105

(b) 420 (d) None of these

12 The sum of the coefficients of all odd degree terms in the expansion of ( x + x 3 − 1)5 + ( x − x 3 − 1)5,( x > 1) is (a) −1

(b) 0

(c) 1

j

JEE Mains 2018

(d) 2

73

DAY 13 The greatest value of the term independent of x, as α sin α   varies over R, in the expansion of  x cos α +   x  (a)

20

(b)

C10

20

(c)

C15

20

C19

15 If the ratio of the fifth term from the beginning to the fifth n

20

1   term from the end in the expansion of  4 2 + 4  is  3 6 : 1, then

is

(d) None of these

Statement I The value of n is 10.

14 Statement I For each natural number

n− 4

n,(n + 1)7 − n 7 − 1 is divisible by 7. Statement II For each natural number n, n 7 − n is divisible j AIEEE 2011 by 7.

2

Statement II

4

2⋅3

(a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is true, Statement II is correct explanation of Statement I. (c) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (d) Statement I is true, Statement II is false

⋅ 3 −1

= 6

4 +n

j

NCERT Exemplar

4

(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

ANSWERS SESSION 1

1 (a)

SESSION 2

2 (d)

3 (a)

4 (d)

5 (c)

6 (b)

7 (d)

8 (d)

9 (a)

10 (a)

11 (c)

12 (c)

13 (d)

14 (c)

15 (a)

16 (a)

17 (b)

18 (d)

19 (a)

20 (c)

21 (a)

22 (c)

23 (c)

24 (d)

25 (c)

26 (d)

27 (b)

28 (a)

29 (b)

30 (c)

31 (a)

32 (d)

33 (b)

34 (c)

35 (d)

36 (c)

37 (d)

38 (b)

1 (c) 11 (a)

2 (b) 12 (d)

3 (b) 13 (d)

4 (a) 14 (b)

5 (c) 15 (c)

6 (b)

7 (c)

8 (d)

9 (a)

10 (c)

Hints and Explanations 1 Given that,(1 + ax )n = 1 + 8 x + 24 x2 + ... ⇒ 1+

n n(n − 1) 2 2 ax + a x + ... 1 1⋅ 2

rewritten as (1 + 0. 002) ⇒ (1. 002)12 = 1 + 12 C1 ( 0. 002) 12

= 1 + 8 x + 24 x2 + ... On comparing the coefficients of x, x2 , we get na = 8,

n(n − 1) 2 a = 24 1⋅ 2

⇒ na(n − 1)a = 48 ⇒ 8(8 − a) = 48 ⇒ 8 − a = 6 ⇒ a=2 ⇒ n= 4

2 Coefficient of x in (1 + x ) = 2n

n

and coefficient of x n in (1 + x )2 n −1 =2 n −1 C n ∴ Required ratio =

2n

Cn

2 n −1

Cn

2n

3 We have, (1. 002)12 or it can be

Cn

(2n )! n ! n ! = 2 :1 = (2n − 1)! n !(n − 1)!

+ 12 C2 (0. 002)2 + 12 C3 (0. 002)3 + ... We want the answer upto 4 decimal places and as such we have left further expansion. ∴(1. 002)12 = 1 + 12( 0. 002) 12⋅ 11 12⋅ 11⋅ 10 + (0. 002)2 + (0. 002)3 + ... 1⋅ 2 1⋅ 2⋅ 3 = 1 + 0. 024 + 2. 64 × 10−4 +1.76 × 10−6 + ... = 1. 0242

4 (1 + x + x2 + x3 )n = {(1 + x )n (1 + x2 )n } = (1 + nC1 x + nC2 x2 + nC3 x3 + nC 4 x 4 +… nC n x n ) (1 + nC1 x2 + nC2 x 4 + K + nC n x2 n ) Therefore, the coefficient of x 4 = nC2 + nC2 nC1 + nC 4 = nC 4 + nC2 + nC1 nC2

5  1 + x sin x 

10

x  Here, n = 10 [even] 10 ⇒ Middle term =  + 1 th = 6th  2  10− 5

1 C 5   ( x sin x )5  x 7 63 ⇒ 252(sin x )5 = 7 = 8 8 1 1 5 (sin x ) = ⇒ sin x = ⇒ 32 2 ⇒ sin x = sin π/6 π ∴ x = nπ + (− 1)n 6 T6 =

10

3

6 T7 = 9C 6  3 3  ( 3 ln x )6 = 729   84

⇒ ⇒

84 × 33 × 33 × (ln x )6 = 729 84 = (ln x )6 = 1 x=e

74

40

⇒

7 Clearly number of terms in the expansion of n  1 − 2 + 4  is (n + 2)(n + 1) or n +2 C .  2 2   x x 2 1 1 [assuming and 2 distinct] x x (n + 2)(n + 1) ∴ = 28 2 ⇒ (n + 2)(n + 1) = 56 = (6 + 1)(6 + 2) ⇒ n=6 Hence, sum of coefficients = (1 − 2 + 4)6 = 36 = 729

⇒

=

n +1

(1 + x ) −1 = x −1 [(1 + x )n +1 − 1] (1 + x ) − 1

∴ The coefficient of x k in S. = The coefficient of x k +1 in [(1 + x )n +1 − 1] =

n +1

C k +1

10 We have, (1 + t 2 )12 (1 + t 12 )(1 + t 24 ) = (1 + 12 C1t 2 + 12 C2t + ...+ 12 C 6t 12 + ...+ 12 C12t 24 + ... )(1 + t 12 + t 24 + t 36 ) ∴Coefficient of t 24 in (1 + t 2 )12 (1 + t 12 )(1 + t 24 ) =

12

C6 +

12

C12 + 1 =

12

C6 + 2

∑

100− m

C m ( x − 3)

100

∴ Constant term  6  2  6 = 1 +     21 +   2 1      4

14 (1 + x )m (1 − x )n   m(m − 1)x = 1 + mx + + ... 2 !   1 − nx + n(n − 1) x2 −... 2!   2

= 1 + (m − n )x  n2 − n (m2 − m ) 2 + − mn +  x + ... 2  2  Given, m − n = 3 ⇒ n = m − 3 n2 − n m2 − m and − mn + = −6 2 2 (m − 3)(m − 4) − m(m − 3) ⇒ 2 m2 − m + = −6 2 ⇒ m2 − 7m + 12 − 2m2 + 6m + m2 − m + 12 = 0 ⇒ −2m + 24 = 0 ⇒ m = 12

15 ( 3 + 1) = 2n

+

2n

⋅ 2 can be written

as [( x − 3) + 2]100 = ( x − 1)100 = (1 − x )100 ∴Coefficient of x 53 in

100

C 53 = −100C 53

p + 2 2 

2n

C 0( 3 )

2 n −2

C2 ( 3 )

+

2n

+ ...+

2 n −1

C1 ( 3 )

2n

2 n −2 n

C2 n ( 3 )

( 3 − 1)2 n =2 n C 0( 3 )2 n (−1)0 + 2 n C1 ( 3 )2 n −1 (−1)1 + 2 n C2 ( 3 )2 n −2 (−1)2 + ... + 2 n C2 n ( 3 )2 n −2 n (−1)2 n Adding both the binomial expansions above, we get ( 3 + 1)2 n − ( 3 − 1)2 n = 2[2 n C1 ( 3 )2 n −1 + ... + 2 n C2 n −1 ( 3 )2 n −(2 n −1 )]

8

n = 8 [even] 8 ⇒ Middle term =  + 1 th term 2  = 5 th term T 5 = 8C 4 ( p / 2)8− 4 (24 )

which is most certainly an irrational number because of odd powers of 3 in each of the terms.

16 ∴General term is Tr

Here,

⇒

2n

+ 2 n C3 ( 3 )2 n −3 + 2 n C 5( 3 )2 n − 5

m= 0

12 Given expression is 

 4 2  2 +  2  6  6 3    2  6  3 

= 1 + 60 + 360 + 160 = 581

m

(1 − x )100 = (−1)53

x+ 1     x2 

2 6  6  6 2 2 +    x +  + ... +    x +  x x  2   6 

11 The given sigma expansion 100

17 The general term in the expansion of

p=±2

6 6 13  1 + x + 2  = 1 +    x + 2   x x  1 

8 Since, in a binomial expansion of (a − b )n , n ≥ 5, the sum of 5th and 6th terms is equal to zero. ∴ n C 4 an − 4 (−b )4 + n C 5an − 5(−b )5 = 0 n! ⇒ an − 4 ⋅ b 4 (n − 4)! 4! n! − an − 5b 5 = 0 (n − 5)! 5! n! a b ⇒ an − 5 ⋅ b 4  −  = 0  n − 4 5 (n − 5)! 4! a n−4 ⇒ = b 5 9 The given expression is 1 + (1 + x ) + (1 + x )2 + ...+ (1 + x )n being in GP. Let, S = 1 + (1 + x ) + (1 + x )2 + ...+ (1 + x )n

p 4 = 16

8 × 7 × 6 × 5 p4 × × 24 = 1120 4 × 3 × 2 × 1 24

+1

 a  = 21C r  3   b =

21

Cr

r 7− a 2

⋅

21 − r

  

b   3 a

r

2r 7 − b3 2

Q Power of a = Power of b [given] ⇒ ∴

7−

n −3

is given by

1 T r +1 = n −3 C r ( x )n −3 − r  2  x  = n −3 C r x n −3 −3 r

r

As x2 k occurs in the expansion of n −3

 x + 1  , we must have    x2  n − 3 − 3r = 2k for some non-negative integer r. ⇒ 3(1 + r ) = n − 2k ⇒ n − 2k is a multiple of 3.

18 T r + 1 = 15C r ( x2 )15 − r ⋅  2 

r

 x

=

15

=

15

30 − 2 r

Cr x

⋅ 2 ⋅ x−r

30 − 3 r

Cr ⋅ x

r

⋅ 2r

…(i)

For coefficient of x , put 30 − 3 r = 15 15

⇒ 3 r = 15 ⇒ r = 5 ∴ Coefficient of x15 =

C 5 ⋅ 25

15

For coefficient of independent of x i.e. x 0 put 30 − 3r = 0 ⇒ r = 10 ∴ Coefficient of x 0 = By condition ⇒

C10 ⋅ 210

15

Coefficient of x15

Coefficient of x 0 15 C 5 ⋅ 25 C ⋅ 25 = 15 = 15 10 10 = 1 : 32 10 C10 ⋅ 2 C10 ⋅ 2 15

19 Greatest term in the expansion of (1 + x )n is T r

+1

 (n + 1)x  where, r =    1+ x  1 Here, n = 20, x = 3  21  ∴ r =   3 + 1 = [10.5 ( 3 − 1)] = (7.69) ≈ 7 Hence, greatest term is 7  20  1   20 1 3   .  =   7   3  7  27

20 Q

2x  (3 + 2 x )50 = 350  1 +   3

Here, and But T r +1 Tr

50

2x T r +1 = 350 50C r    3

r

2x T r = 350 50C r −1    3 1 x = (given) 5 50 C 2 1 ≥ 1 ⇒ 50 r ⋅ ≥ 1 C r −1 3 5

r 2 7 = r − 2 3 2

∴

r =9

⇒ 102 − 2r ≥ 15r ⇒ r ≤ 6

r −1

75

DAY 21 Sum of the coefficients in the expansion of ( x − 2 y + 3z )n is (1 − 2 + 3)n = 2n (put x = y = z = 1) ∴ 2n = 128 ⇒ n = 7 Therefore, the greatest coefficient in the expansion of (1 + x )7 is 7 C3 or 7 C 4 because both are equal to 35.

22 In the expansion of (1 + x )2 n , the general term =2 n C k x k , 0 ≤ k ≤ 2n As given for r > 1, n > 2, 2n C3 r =2 n C r +2 ⇒ Either 3r = r + 2 or 3r = 2n − (r + 2) (Q n C x = nC y ⇒ x + y = n or x = y ) ⇒ r = 1 or n = 2r + 1 We take the relation only n = 2r + 1

23 Let b =

n

∑

r

n r=0 C r n

= n∑

=

1

r=0

n

Cr

= nan −

n

n

∑

r=0

n−r

(1 + x )20 =20 C 0 + 20 C1 x + ... + 20C10 x10 + ... + 20 C20 x20 On putting x = −1 in the above expansion, we get 0 = 20C 0 − 20C1 + ... − 20C a + 20C10

(Q C r = C n − r ) n

n− r

n an 2

20

⇒

20

1  =   (0 + 0) = 0  1 + nx 

25 Let

C10 = 2(20C 0 − 20C1 + ... + 20C10 ) 1 ⇒ 20C 0 − 20C1 + ... + 20C10 = 20C10 2 20

28 Since, x(1 + x )n = xC 0 + C1 x2

On differentiating w.r.t. x, we get (1 + x )n + nx(1 + x )n −1 = C 0 + 2C1 x + 3C2 x2 + ... + (n + 1)C n x n Put x = 1, we get C 0 + 2C1 + 3C2 + ... + (n + 1)C n = 2n + n2n −1 = 2n −1 (n + 2) ∴ (8 + 3 7 )10 + (8 − 3 7 )10 is an integer, hence this is the value of n. 49 + 16n − 1 = (1 + 48) + 16n − 1 n

n

= 1 + C1 (48) + C2 (48) + ... n

 30  30  30  30 +     − ... +     2 12     20  30

2

n

+ n C n (48)n + 16n − 1 = ( 48n + 16n ) + C2 ( 48)2 + n C3 ( 48)3 + n

C 0 . C10 − C1 . C11 30

30

... + n C n (48)n

+ 30C2 . 30C12 − ... + 30C20 . 30C30 = Coefficient of x

in (1 + x )30(1 − x )30

= Coefficient of x

in (1 − x )

= Coefficient of x

in

20

20 20

20

30 We have,

 30  30  30  30 A =    −    0   10  1   11

30

20

29 Let f = (8 − 3 7 )10, here 0 < f < 1

[Q n C 0− nC1 + nC2 − nC3 + ...(−1)n nC n = 0]

2 30

= 64n + 8 [ C2 ⋅ 6 + C3 ⋅ 6 ⋅ 8 2 n

r=0

= (−1)10 30C10 (for coefficient of x20, let r = 10) =30 C10

2

3

n

+ n C 4 ⋅ 64 82 + ... + n C n ⋅ 6n ⋅ 8n −2 ]

Hence, 49n + 16n − 1 is divisible by 64. n

31 Since,  1 + 1  < 3 for ∀n ∈ N 

30

r 30 2 r ∑ (−1) C r ( x )

r+1 n ⇒ 0 < k2 − 3 ≤ 1 Qn ≥ r ⇒ r + 1 ≤ 1and n, r > 0  n  ⇒ 3 < k2 ≤ 4 Hence, k ∈ [−2, − 3] ∪ ( 3,2) ⇒

k2 − 3 =

33 82 n − (62)2 n +1 = (1 + 63)n − (63 − 1)2 n +1 = (1 + 63)n + (1 − 63)2 n +1 = [1 + C1 ⋅ 63 + n C2 ⋅ ( 63)2 + ... + ( 63)n ] n

+ [1 −2 n −1 C1 ⋅ 63 + (2 n +1 ) C2 ⋅ (63)2 − ... n

+ (63)n −1 −(2 n +1 ) C1 +

20

20

Now, =

n

(1001)999 (1000)1000

=

1  1001    1001  1000 

1  1  1 +  1001  1000 

1000


1)

(1001)999 < (1000)1000

26 (21 C1 −10 C1 ) + (21 C2 −10 C2 )

1000

1 ⋅3 < 1 1001

(2 n +1 )

C2 (63) − ...+ (−1)(63)(2 n )]

Hence, remainder is 2.

34 Since, (r + 1)th term in the expansion of (1 + x )27 / 5 27  27 27 − 1 ...  − r + 1    5  r 5 5 = x r! Now, this term will be negative, if the last factor in numerator is the only one negative factor. 27 32 − r + 1< 0 ⇒ 1

3

5

⇒ Divisible by 7. So, both statements are true and Statement II is correct explanation of Statement I.

3

6

20

1

10

n

6

= 7λ + 7{k 6 + 3k 5 + ... + k} [Using Eq. (i)]

3

= 3−( 5/3 )log3 2 = 2−5

(−1)

5

is sin α   x 

r

= 20C r x20−2 r (cos α )20− r (sin α )r

  1 =  9x −1 + 7 + x −1 (3 + 1)1 / 5   + 7)

7

= 2( x + 10 x ( x − 1) + 5x( x3 − 1)2 )

 +1 )  

7−5

log3 8

By mathematical induction For n = 1, P(1) = 0, which is divisible by 7. For n = k P (k ) = k 7 − k

+ 7 C7 ) − (k + 1) = (k − k ) + 7{k + 3k + ... + k}

2n /2

+ 5C 4 x( x3 − 1 )4 )

7

x −1

(−1)n

n

5

7

x −1

2

=

n

= 2( 5C 0 x 5 + 5C2 x3 ( x3 − 1 )2

10 We have,

7

 1   5/3  3 

(x +

n

1

1 n /2

14 Let P (n ) = (n )7 − n

Let P (k ) be divisible by 7. ... (i) ∴ k 7 − k = 7λ, for some λ ∈ N For n = k + 1, P (k + 1) = (k + 1)7 − (k + 1) = (7 C 0k 7 + 7 C1 k 6 + 7 C2 k 5 + ...+ 7 C 6 ⋅ k

2

We have

+ ...

1 3 7 =  1 −  +  1 −  +  1 −  + ...    2 4 8 upto m terms 1 1 1 = n + 2 n + 3 n ... upto m terms 2 2 2 m 1  1   1−  n  n  mn   2  2  = 2 −1 = mn 1 − 1  2 (2n − 1)   n  2 

∴T 6 = C 5( 9

x = 1,2

Thus,

r=0

7

( y − 3)( y − 9) = 0 y = 3, 9 3x = 3, 9

(put y = 3x )

Also, we have

r

+

⇒ ⇒ ⇒

= nC n (−1)n

n

9x −1 +7

y 2 − 12 y + 27 = 0

 1  T n +1 = nC n (21 /3 )n − n  −   2

∑ (−1)

 log2 2 

⇒



n    n 1 1 n    ⋅ p = 2   1 +  − 1  2  p 2  

n

32x − 12(3x ) + 27 = 0

n

 n − p    t 

∑

 n n −  2  p

n

∑

= 2n

n− p

20

⇒

 n − p    m − p

  n   ∑  p m = p

∑

Thus, the greatest possible value of β is 10 1 C10   .  2

 3x  + 1 4  3 

11 Last term of  21 /3 − 1  is

 n  n − p       p  m − p

n

32x +7= 9

⇒

∴Given series can be rewritten as

∑ ∑

⇒

5

For this term to be independent of x, we get 20 − 2r = 0 ⇒ r = 10 Let β = Term indeoendent of x

So, from the given condition, we have Fifth term from the beginning 6 = Fifth term from the end 1 n−4

n

⇒

C4 ⋅2

n

−1

4

2

n−8

⋅ 3 −1

−n + 4

C 4. ⋅ 2 ⋅ 3

n−4

⇒

4

⋅3

=

4

 4 −n  −1 −    4 

=

6 1

6

n−8

⇒

2

= 20C10(cos α )10(sin α)10

⇒

= 20C10(cos α sin α)10

⇒

(2 × 3) 4 = (2 ⋅ 3)1 /2 n−8 = 1/2 4 n = 2 + 8 ∴ n = 10

sin 2α  = C10    2  20

10

4

3

4

= 61 /2

n−8

⇒

DAY EIGHT

Permutations and Combinations Learning & Revision for the Day u

u

u

Fundamental Principle of Counting Factorial Notation Permutations

u

u u

Circular Permutations Combinations Applications of Permutations and Combinations

u

u

Prime Factors Division of Objects into Groups

Fundamental Principle of Counting The fundamental principle of counting is a way to figure out the total number of ways in which different events can occur. If a certain work A can be done in m ways and another work B in n ways, then (i) the number of ways of doing the work C, which is done only when either A or B is done, is m + n. (addition principle) (ii) the number of ways of doing the work C, which is done only when both A and B are done, is mn. (multiplication principle)

Factorial Notation The product of first n natural numbers is denoted by n! and read as ‘factorial n’. Thus, n ! = n (n − 1) (n − 2) … 3 ⋅ 2 ⋅1

PRED MIRROR

Properties of Factorial Notation

Your Personal Preparation Indicator

1. 0 ! = 1 ! = 1 2. Factorials of negative integers and fractions are not defined. 3. n ! = n (n − 1)! = n (n − 1) (n − 2)! n! 4. = n (n − 1) (n − 2)… (r + 1) r!

Permutations ●

●

Permutation means arrangement of things. The number of permutations of n different things taken r at a time is n Pr . n! n ,0 ≤ r ≤ n Pr = n (n − 1) (n − 2)...(n − r + 1) = (n − r )!

u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

79

DAY Properties of n Pr

(i) n P0 = 1, n P1 = n , n Pn = n ! (ii)

(iii) Pr = n n

(v)

n −1

n −1

n

Pr + r ⋅n Pr − 1 =

n +1

Pr

(iv) Pr = (n − r + 1) Pr − 1 n

Pr − 1

Pr = (n − r )

n −1

n

Pr − 1

Important Results on Permutations (i) Number of permutations of n different things taken r at a time when a particular thing is to be always included in each arrangement is r ⋅ n −1 Pr −1 . (ii) Number of permutations of n different things taken r at a time, when a particular thing is never taken in each arrangement is n −1 Pr . (iii) The number of permutations of n different things taken r at a time, allowing repetitions is nr . (iv) The permutations of n things of which p are identical of one sort, q are identical of second sort, r are identical of n! third sort, is , where p + q + r = n. p!q ! r ! (v) Arrangements (a) The number of ways in which m different things and n different things (m + 1 ≥ n) can be arranged in a row, so that no two things of second kind come together is m ! ( m +1 )Pn . (b) The number of ways in which m different things and n different things (m ≥ n) can be arranged in a row so that all the second type of things come together is (m + 1)! n !. (vi) Dearrangement The number of dearrangements (No object goes to its scheduled place) of n objects, is 1 1 1 1 n! − + − ... (− 1)n . 2 ! 3 ! 4 ! n ! (vii) Sum of Digits (a) Sum of numbers formed by taking all the given n digits (excluding 0) is (sum of all the n digits) × (n − 1)! × (111 ... n times). (b) Sum of the numbers formed by taking all the given n digits (including 0) is (sum of all the n digits) × [(n − 1)! × (111 ... n times) − (n − 2) × {111 ...(n − 1) times}].

Circular Permutations ●

●

●

●

Number of circular permutations of n different things, taken r at a time, when clockwise and anti-clockwise orders are n P not different, is r . 2r

●

If different objects are arranged along a closed curve, then permutation is known as circular permutation. The number of circular permutations of n different things taken all at a time is (n − 1)!. If clockwise and anti-clockwise orders are taken as different. If clockwise and anti-clockwise circular permutations are (n − 1)! . considered to be same, then it is 2 Number of circular permutations of n different things, taken r at a time, when clockwise and anti-clockwise n P orders are taken as different, is r . r

Important Results on Circular Permutations (i) The number of ways in which m different things and n different things (where, m ≥ n) can be arranged in a circle, so that no two things of second kind come together is (m − 1)! m Pn . (ii) The number of ways in which m different things and n different things can be arranged in a circle, so that all the second type of things come together is m ! n !. (iii) The number of ways in which m different things and n different things (where, m ≥ n) can be arranged in the form of garland, so that no two things of second kind come together is (m − 1)! m Pn / 2. (iv) The number of ways in which m different things and n different things can be arranged in the form of garland, so that all the second type of things come together is m ! n !/ 2 .

Combinations Combination means selection of things. The number of combinations of n different things taken r at a time is  n n Cr or   .  r n

n  n n (n − 1) (n − 2)...(n − r + 1) n! P Cr =   = = = r r r! r !(n − r )! r!

Properties of n Cr (i)

n

Cr = nCn − r

(iii)

n

Cr + Cr − 1 = n

(ii) n Cx = nC y ⇒ x = y or x + y = n n +1

Cr

Important Results on Combinations (i) The number of combinations of n different things, taken r at a time, when p particular things always occur is n− p Cr − p. (ii) The number of combinations of n different things, taken r at a time, when p particular things never occur is n − pCr . (iii) The number of selections of zero or more things out of n different things is n C0 + nC1 + … + nCn = 2 n . (iv) The number of selections of one or more things out of n different things is n C1 + nC2 + … + nCn = 2 n − 1. (v) The number of selections of zero or more things out of n identical things = n + 1. (vi) The number of selections of one or more things out of n identical things = n.

80

40

(vii) The number of selections of one or more things from p + q + r things, 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. (viii) The number of selections of one or more things from p identical things of one kind, q identical things of second kind, r identical things of third kind and n different things, is ( p + 1) (q + 1) (r + 1) 2 n − 1 (ix) If there are m items of one kind, n items of another kind and so on. Then, the number of ways of choosing r items out of these items = coefficient of x r in (1 + x + x2 + K + x m ) (1 + x + x2 + K + x n )K (x) If there are m items of one kind, n items of another kind and so on. Then, the number of ways of choosing r items out of these items such that atleast one item of each kind is included in every selection = coefficient of x r in ( x + x2 + K + x m )( x + x2 + K + x n )K

(ii) Sum of divisors of nis

(iii) If p is a prime such that pr divides n! but pr +1 does not divide n ! .  n  n  Then, r =   +  2  + K  p  p 

Division of Objects into Groups Objects are Different (i) The number of ways of dividing n different objects into 3 groups of size p, q and r ( p + q + r = n) is n! (a) ; p, q and r are unequal. p!q ! r ! (b)

Applications of Permutations and Combinations Functional Applications If the set A has m elements and B has n elements, then (i) the number of functions from A to B is nm . (ii) the number of one-one functions from A to B is n Pm , m ≤ n. (iii) the number of onto functions from A to B is  n  n nm −   (n − 1)m +   (n − 2)m − ..., m ≤ n. 1 2 (iv) the number of bijections from A to B is n!, if m = n.

Geometrical Applications (i) Number of triangles formed from n points, when no three points are collinear, is n C3 . (ii) Out of n non-concurrent and non-parallel straight lines, the points of intersection are n C2 . (iii) The number of diagonals in a polygon of nsides is n C2 − n. (iv) The number of total triangles formed by the n points on a plane of which m are collinear, is n C3 − mC3 . (v) The number of total different straight lines, formed by the n points on a plane, of which m are collinear, is n C2 − mC2 + 1 .

Prime Factors Let n = p1α 1 ⋅ p2α 2 ⋅ p3α 3 K prα r , where pi , i = 1, 2, K , r are distinct primes and α i , i = 1, 2 , K , r are positive integers. (i) Number of divisor of n is (α 1 + 1) (α 2 + 1) (α 3 + 1)K (α r + 1).

( p1α 1 + 1 − 1) ( p2α 2 + 1 − 1) ( prα r + 1 − 1) ... ( p1 – 1) ( p2 − 1) ( pr − 1)

n! ;q = r p ! 2 !(q !)2

(c)

n! ; p =q = r 3 !( p !)3

(ii) The number of ways in which n different things can be distributed into r different groups, if empty groups are allowed, is r n . (iii) The number of ways in which n different things can be distributed into r different groups, if empty groups are not allowed, is  r  r r n −   (r − 1)n +   (r − 2)n − … + (−1)r − 1 rCr − 1 ⋅ 1 1 2

Objects are Identical (i) The number of ways of dividing n identical objects among r persons such that each one may get atmost n objects, is  n + r − 1   . In other words, the total number of ways of  r −1  dividing n identical objects into r groups, if blank groups are allowed, is n + r − 1 Cr − 1 . (ii) The total number of ways of dividing n identical objects among r persons, each one of whom, receives atleast one item, is n − 1 Cr − 1 . In other words, the number of ways in which n identical things can be divided into r groups such that blank groups are not allowed, is n − 1 Cr − 1 . (iii) Number of non-negative integral solutions of the equation x1 + x2 + … + x r = n is equivalent to number of ways of distributing n identical objects into r groups if empty groups are allowed, which is n + r −1 Cr −1 . (iv) Number of positive integral solutions of the equation x1 + x2 + …+ x r = n is equivalent to the number of ways of distributing n identical objects into r groups such that no group empty, which is n −1 Cr −1 . (v) Number of integral solutions of the equation x1 + x2 + … + x r = n, where a ≤ x i ≤ b , ∀ i = ,1, 2,… , r , is given by coefficient of x n in ( x a + x a + 1 + x a + 2 + …+ xb )r .

81

DAY DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 The value of 2n [1 ⋅ 3 ⋅ 5 K ( 2 n − 3) ( 2 n − 1)] is (2n)! (a) n!

(2 n)! (b) 2n

n! (c) (2 n)!

11 The total number of permutations of n ( > 1) different things

(d) None of these

2 The value of1⋅1! + 2 ⋅ 2! + 3 ⋅ 3! + … + n ⋅ n !, is (a) (n + 1) ! (c) (n + 1) ! − 1

(b) (n + 1)! + 1 (d) None of these

from the digits of the number 223355888 by rearrangement of the digits so that the odd digits occupy even places? (b) 36

(c) 60

(d) 180

4 A library has a copies of one book, b copies of each of two books, c copies of each of three books and single copies of d books. The total number of ways in which these books can be arranged, is (a + b + c + d)! a! b ! c! (a + 2b + 3c + d) (c) a! b ! c!

(a)

(b)

(a + 2b + 3c + d)! a !(b !)2 (c !)3

(d) None of these

(c) 7!

(d)

4

C4 × 3 C3

6 In how many ways the letters of the word ‘ARRANGE’ can be arranged without altering the relative positions of vowels and consonants? (a) 36 (c) 62

(b) 26 (d) None of these

7 If the letters of the word ‘SACHIN’ are arranged in all possible ways and these words are written in dictionary order, then the word ‘SACHIN’ appears at serial number (a) 600

(b) 601

(c) 602

(d) 603

8 The number of integers greater than 6000 that can be formed, using the digits 3, 5, 6, 7 and 8 without j repetition, is JEE Mains 2015 (a) 216

(b) 192

(c) 120

(d) 72

9 The number of 5-digits telephone numbers having atleast one of their digits repeated, is (a) 90000

(b) 100000

(c) 30240

j

NCERT Exemplar

(d) 69760

10 If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in dictionary, then the position of the word j SMALL is JEE Mains 2016 (a) 46th (c) 52nd

Pr

−1

a

=

(b)

nr − 1 n −1

(d) None of these

n Pr + 1 Pr , then = b c

n

(a) b 2 = a (b + c) (c) ab = a 2 + bc

(b) c 2 = a (b + c) (d) bc = a 3 + b 2

13 The range of the function f ( x ) = (a) {1, 2, 3 } (c) {1, 2, 3, 4}

7 − x

Px − 3 is

j

AIEEE 2004

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

14 Find the number of different words that can be formed from the letters of the word TRIANGLE, so that no vowels j NCERT Exemplar are together. (b) 14500

(c) 14400

(d) 14402

15 In a class of 10 students, there are 3 girls. The number of

letters of the word ARTICLE, so that vowels occupy the j even place is NCERT Exemplar (b) 144

n

12 If

(a) 14000

5 The number of words which can be formed out of the

(a) 1440

n (n n − 1) n −1 n (n r − 1) (c) n −1 (a)

3 How many different nine-digit numbers can be formed

(a) 16

taken not more than r at a time, when each thing may be repeated any number of times is

(b) 59th (d) 58th

ways they can be arranged in a row, so that no 2 girls are consecutive is k ⋅ 8!, where k is equal to (a) 12

(b) 24

(c) 36

(d) 42

16 The sum of all the 4-digit numbers that can be formed using the digits 1, 2, 3, 4 and 5 without repetition of the digits is (a) 399960

(b) 288860

(c) 301250

(d) 420210

17 If eleven members of a committee sit at a round table so that the President and Secretary always sit together, then the number of arrangements is (a) 10! × 2 (c) 9! × 2

(b) 10! (d) None of these

18 Let f : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} that are onto and f ( x ) ≠ x , is equal to (a) 9 (c) 16

(b) 44 (d) None of these

19 There are 4 balls of different colours and 4 boxes of same colours as those of the balls. The number of ways in which the balls, one in each box, could be placed such that a ball does not go to box of its own colour is (a) 8

(b) 9

(c) 7

(d) 1

20 How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in which no two S are j adjacent? AIEEE 2008 (a) 7 ⋅ 6 C4 ⋅ 8 C4 (c) 6 ⋅ 7 ⋅ 8 C4

(b) 8 ⋅ 6 C4 ⋅ 7 C4 (d) 6 ⋅ 8 ⋅ 7 C4

82

40

21 The number of ways in which seven persons can be arranged at a round table, if two particular persons may not sit together is (a) 480 (c) 80

at a round table, if no two women are to sit together, is j AIEEE 2003 given by

23 The value of

(c) 5 ! × 4 !

47

C4 +

5

∑

52 − r

(d) 7 ! × 5 !

(a) (c)

24 If n C3 + nC4 >

(a) 312 − 1

C3 , then (b) n > 7 (d) None of these

(b) 126 (d) None of these

balls. The number of ways of drawing 3 balls from the box, if atleast one black ball is included, is (c) 56

(d) 64

27 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 select atleast one book is 63, then n is equal to (a) 3

(b) 4

(c) 6

(d) 5

28 Let A and B two sets containing 2 elements and 4 elements, respectively. The number of subsets of A × B j having 3 or more elements is JEE Mains 2013 (a) 256

(b) 220

(c) 219

(d) 211

29 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

30 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 done, is j

(a) 3

(b) 36

(c) 66

(b) 312

(c) (12)3 − 1

(d) (12)3

(a) N > 190 (c) 100 < N ≤ 140

(b) N ≤ 100 (d) 140 < N ≤ 190

35 Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If j JEE Mains 2013 Tn +1 − Tn = 10, then the value of n is

26 A box contains 2 white balls, 3 black balls and 4 red

(b) 42

(d) 36

collinear. If N is the number of triangles formed by joining j AIEEE 2011 these points, then

committee from four men and six women so that the committee includes atleast two men and exactly twice as j NCERT Exemplar many women as men is

(a) 36

(c) 32

34 There are 10 points in a plane, out of these 6 are

25 The number of ways in which we can choose a

(a) 94 (c) 128

(b) 24

horses, cows and calves (not less than 12 each) ready to be shipped. In how many ways, can the ship load be made?

n +1

(a) n > 6 (c) n < 6

(d) 0

33 In a steamer, there are stalls for 12 animals and there are

(b) 52 C5 (d) None of these

C6 52 C4

(c) (m + 1) λ

all possible ways. If the number of times two particular soldiers A and B are together on guard is thrice the number of times three particular soldiers C, D, E are together on guard, then n is equal to (a) 18

C3 is equal to

r =1 47

(b) mλ

32 A guard of 12 men is formed from a group of n soldiers in

22 The number of ways in which 6 men and 5 women can sit

(b) 30

A = {a1, a 2 , a 3 , ..., a n } is λ times the number of m elements subsets containing a 4 , then n is (a) (m − 1) λ

(b) 120 (d) None of these

(a) 6! × 5 !

31 If the total number of m elements subsets of the set

AIEEE 2010

(d) 108

(a) 7 (c) 10

(b) 5 (d) 8

36 On the sides AB, BC, CA of a ∆ABC, 3, 4, 5 distinct points (excluding vertices A , B , C) are respectively chosen. The number of triangles that can be constructed using these j chosen points as vertices are JEE Mains 2013 (a) 210 (c) 215

(b) 205 (d) 220

37 The number of divisors of the number of 38808 (excluding 1 and the number itself) is (a) 70 (c) 71

(b) 72 (d) None of these

38 If a, b, c, d , e are prime integers, then the number of divisors of ab 2c 2de excluding 1 as a factor is (a) 94

(b) 72

(c) 36

(d) 71

39 The number of ways of distributing 8 identical balls in 3 distinct boxes, so that no box is empty, is (a) 5 (c) 3 8

8 (b)    3 (d) 21

40 If 4 dice are rolled, then the number of ways of getting the sum 10 is (a) 56 (c) 72

(b) 64 (d) 80

83

DAY DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 From 6 different novels and 3 different dictionaries, 4 novels

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

and 1 dictionary are to be selected and arranged in a row on a shelf, so that the dictionary is always in the middle. j JEE Mains 2018 The number of such arrangements is

j

(a) 485

(a) atleast 1000 (b) less than 500 (c) atleast 500 but less than 750 (d) atleast 750 but less than 1000

(a) 27378

(b) 163 × 15 2 × 14 (d) 162 × 15 × 14

(c)

52 ! 17 !

(a) 36

(a) 36

 a11 a12    : aij ∈ {0, 1, 2 }, a11 = a22 . Then, the number a a    21 22 

(a) 27

(b) 24

(c) 10

(a)

(b) 629

30

(b)

C7

(d) 879

21

C8

(c)

21

(d)

C7

30

C8

number 1010 1111 1313 is (a) 750

(d) 7200

(c) 630

13 The number of divisors of the form 4n + 1, n ≥ 0 of the

(d) 20

(c) 7600

(d) None of these

30 marks to 8 questions, giving not less than 2 marks to j any question, is JEE Mains 2013

(b) 840

(c) 924

(d) 1024

14 Out of 8 sailors on a boat, 3 can work only one particular side and 2 only the other side. Then, number of ways in which the sailors can be arranged on the boat is

to contain atleast 3 men and 2 women. The number of ways this can be done, if two particular women refuse to serve on the same group, is (b) 7800

(c) 120

12 The number of ways in which an examiner can assign

JEE Mains 2013

6 A group of 6 is chosen from 10 men and 7 women so as

(a) 8000

(d) None of these

colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls j AIEEE 2012 is

(d) 309

j

(b) 90

(a) 880

5 Let S = 

of non-singular matrices in the set S, is

(c) 0

11 Assuming the balls to be identical except for difference in

(d) None of these

(c) 308

(b) 18

xyz = 24 is

possible orders and these words are written out as in a dictionary, then the rank of the word MOTHER is (b) 261

(d) None of these

by four different even digits is

4 If the letters of the word MOTHER are written in all

(a) 240

(c) 27399

10 The total number of integral solutions ( x , y , z ) such that

players so that three players have 17 cards each and the fourth players just one card, is (b) 52!

(b) 27405

9 The number of numbers divisible by 3 that can be formed

3 The number of ways of dividing 52 cards amongst four

52 ! (17 !)3

(d) 484

from 1 to 30 so as to exclude every selection of four consecutive numbers is

swimming and riding. How many prize lists could be made, if there were altogether 6 prizes for different values, one for running, 2 for swimming and 3 for riding?

(a)

(c) 469

8 The number of ways in which we can select four numbers

2 Sixteen men compete with one another in running,

(a) 16 × 15 × 14 (c) 163 × 15 × 14 2

(b) 468

JEE Mains 2017

(a) 2718

(b) 1728

(c) 7218

(d) None of these

15 In a cricket match between two teams X and Y , the team X requires 10 runs to win in the last 3 balls. If the possible runs that can be made from a ball be 0, 1, 2, 3, 4, 5 and 6. The number of sequence of runs made by the batsman is

7 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

(a) 12

(b) 18

(c) 21

(d) 36

ANSWERS (a) (c) (a) (b)

SESSION 1

1. 11. 21. 31.

SESSION 2

1. (a) 11. (d)

2. 12. 22. 32.

(c) (a) (a) (c)

2. (b) 12. (c)

3. 13. 23. 33.

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

3. (a) 13. (c)

4. 14. 24. 34.

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

4. (d) 14. (b)

5. 15. 25. 35.

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

5. (d) 15. (d)

6. 16. 26. 36.

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

6. (b)

7. 17. 27. 37.

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

7. (a)

8. 18. 28. 38.

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

8. (a)

9. 19. 29. 39.

(d) (b) (b) (d)

9. (a)

10. 20. 30. 40.

(d) (a) (d) (d)

10. (c)

84

40

Hints and Explanations SESSION 1

Four digit number can start from 6,7 or 8.

1 Clearly, [1 ⋅ 3 ⋅ 5 ...( 2n − 3)(2 n − 1 )]2 =

1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5⋅ 6...(2 n − 1 ) (2 n ) 2 2 ⋅ 4 ⋅ 6... 2 n

=

(2 n ) ! 2n (2 n )! = 2 (1 ⋅ 2 ⋅ 3... n ) n!

n

n

2 Clearly, given expression =

n

∑

r ⋅r!

r =1

=

n

∑ r =1

((r + 1) − 1) r ! =

n

∑

6, 7 or 8

n

((r + 1)! − r !)

r =1

= (2! − 1!) + (3!− 2!) + …+ ((n + 1)! − n !) = (n + 1)! − 1

3 In a nine digits number, there are four even places for the four odd digits 3, 3, 5, 5. 4! 5! ⋅ ∴ Required number of ways = 2! 2! 2! 3! = 60

4 Total number of books = a + 2b + 3c + d ∴ The total number of arrangements (a + 2b + 3c + d )! = a!(b !) 2 (c !) 3

5 In a word ARTICLE, vowels are A, E, I and consonants are C, L, R, T. In a seven letter word, there are three even places in which three vowels are placed in 3! ways. In rest of the four places, four consonants are placed in 4! ways. ∴ Required number of ways = 3! × 4! = 6 × 24 = 144

6 Clearly, the consonants in their 4! positions can be arranged in = 12 2! ways and the vowels in their positions 3! can be arranged in = 3 ways 2! ∴ Total number of arrangements = 12 × 3 = 36

7 The letters of given word are A, C, H, I, N, S. Now, the number of words starting with A = 5! the number of words starting with C = 5! the number of words starting withH = 5! the number of words starting with I = 5! and the number of words starting with N = 5!. Then, next word is SACHIN. So, the required serial number is = (5⋅ 5!) + 1 = 601.

8 The integer greater than 6000 may be of 4 digits or 5 digits. So,here two cases arise. Case I When number is of 4 digit.

3

4

3

2

Thus, total number of 4-digit number, which are greater than 6000 = 3 × 4 × 3 × 2 = 72 Case II When number is of 5 digit. Total number of 5-digit number, which are greater than 6000 = 5! = 120 ∴ Total number of integers = 72 + 120 = 192

9 Using the digits 0, 1, 2, ..., 9 the number of five digits telephone numbers which can be formed is 10 5 (since, repetition is allowed). The number of five digits telephone numbers, which have none of the digits repeated = 10 P5 = 30240 ∴ The required number of telephone number = 105 − 30240 = 69760

10 Clearly, number of words start with 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, position of word ‘‘SMALL’’ is 58th. A=

11 When we arrange things one at a time, the number of possible permutations is n. When we arrange them two at a time the number of possible permutations are n × n = n2 and so on. Thus, the total number of permutations are n (n r − 1) n + n2 + K + n r = [Q n > 1] n−1 n

12 Q

Pr − 1 a

=

n

Pr = b

n

Pr + 1 c

b =n−r +1 a c From last two terms =n−r b b c Hence, = + 1 ⇒ b 2 = a (b + c ) a b From first two terms

13 Given that, f ( x ) =

7−x

14 In a word TRIANGLE, vowels are (A, E, I) and consonants are (G, L, N, R, T). First, we fix the 5 consonants in alternate position in 5! ways. _G_L_N_R_T_ In rest of the six blank position, three vowels can be arranged in 6 P3 ways. ∴ Total number of different words = 5! × 6 P3 = 120 × 6 × 5 × 4 = 14400

15 The 7 boys can be arranged in row in 7! ways. There will be 6 gaps between them and one place before them and one place after them. The 3 girls can be arranged in a row in 8 P3 = 8 ⋅ 7 ⋅ 6 ways ∴ Required number of ways = 7! × 8 ⋅ 7 ⋅ 6 = 42 ⋅ 8!

16 Required sum = (Sum of all the n-digits) × n −1 Pr −1 × (111 ... r times) = (1 + 2 + 3 + 4 + 5) 4 P3 × (1111) = 15 × 24 × 1111 = 399960

17 Since, out of eleven members two members sit together, the number of arrangements = 9! × 2 (Q two members can be sit in two ways.) 18 Total number of functions = Number of dearrangement of 5 objects 1 1 1 1 = 5!  − + −  = 44  2! 3! 4! 5!

19 Use the number of dearrangements 1 1 1 1 i.e. n ! 1 − + − + ... + (− 1)n  n !  1! 2! 3! Here, n = 4 So, the required number of ways 1 1 1 = 4!  − +  = 12 − 4 + 1 = 9  2! 3 ! 4!

20 Given word is MISSISSIPPI. Here, I = 4 times, S = 4 times, P = 2 times, M = 1 time _M_I_I_I_I_P_P_ ∴ Required number of words 7 × 6! 7! = 8C 4 × = 8C 4 × = 7 ⋅ 8C 4 ⋅ 6C 4 4!2! 4!2!

21 Clearly, remaining 5 persons can be seated in 4! ways. Now, on five cross marked places person can sit in 5 P2 ways. P4

Px − 3 . The above

function is defined, if 7 − x ≥ 0 and x − 3 ≥ 0 and 7 − x ≥ x − 3. ⇒ x ≤ 7, x ≥ 3 and x ≤ 5 ∴ D f = {3, 4, 5} Now, f (3) = 4 P0 = 1 f (4) = 3 P1 = 3 and f (5) = 2 P2 = 2 ∴ R f = {1, 2, 3}

P3 P5

P2 P1 So, number of arrangements 5! = 4! × = 24 × 20 = 480 ways 3!

85

DAY 22 First, we fix the position of men, number of ways in which men can sit = 5!. Now, the number of ways in which women can sit = 6 P5

29 Clearly, the candidate is unsuccessful, if he fails in 9 or 8 or 7 or 6 or 5 papers. ∴ Numbers of ways to be unsuccessful = 9C 9 + 9C 8 + 9C7 + 9C 6 + 9C 5 = 9C 0 + 9C1 + 9C2 + 9C3 + 9C 4 1 = ( 9 C 0 + 9C1 + K + 9C 9 ) 2 1 = (29 ) = 28 = 256 2

M M

M

M

M

5

C4 +

∑

52 − r

r =1

= =

24

n

51 52

C3 + C4

50

+ C3 +

C3 + nC 4 >

C3 +

50 49

n +1

C4 +

47

C3 +

51

C3 +

49 48

C3 C3 +

48

C3 + ( C3 + 47

47

47

C3

C4)

C3

C4 > C3 ( nC r + n C r + 1 = n +1 C r +1 ) n +1 n−2 C ⇒ n +1 4 > 1 ⇒ > 1 ⇒n> 6 C3 4 ⇒

n +1

C3 =

choose a committee = Choose two men and four women + Choose three men and six women = 4C2 × 6C 4 + 4C3 × 6C 6 = 6 × 15 + 4 × 1 = 90 + 4 = 94

26 The number of ways of drawing 1 black and 2 non-black balls is 3 C1 ⋅ 6C2 = 3 ⋅ 15 = 45 The number of ways of drawing 2 black and 1 non-black ball is 3 C2 ⋅ 6C1 = 3 ⋅ 6 = 18 The number of ways of drawing 3 black balls is 3 C3 = 1 ∴ Number of ways = 45 + 18 + 1 = 64

27 He can select 1, 2, ... or n books. The number of ways to select atleast one book is 2n + 1 C1 + 2 n + 1 C2 + K + 2 n + 1 C n 1 = (2 n + 1 C1 + 2 n + 1 C2 + K + 2 n + 1 C n 2 + 2 n + 1 C n + 1 + K + 2 n + 1 C2 n) 1 2n + 1 2n + 1 C 0 − 2 n + 1 C2 n + 1 ) = (2 − 2 [given] = 22n − 1 = 63 ⇒ 22 n = 64 = 26 ⇒ n = 3

28 Given, n( A ) = 2, n(B ) = 4. ∴ n (A × B) = 8 The number of subsets of A × B having 3 or more elements = 8C 3 + 8C 4 +…+ 8C 8 = ( 8 C 0 + 8C1 + 8C2 + 8C3 +…+ 8C 8 ) − ( 8 C 0 + 8C1 + 8C2 ) 8 8 8 = 2 − C 0 − C1 − 8C2 = 256 − 1 − 8 − 28 = 219 [Q 2n = nC 0 + nC1 +…+ nC n ]

3 distinct red balls

9 distinct blue balls

constructed using these chosen points as vertices = 12C3 − 3C3 − 4C3 − 5C3 Here, we subtract those cases in which points are collinear = 220 − 1 − 4 − 10 = 220 − 15 = 205

Urn A

Urn B

38 The number of divisors of ab2c 2de = (1 + 1 ) (2 + 1 ) (2 + 1 ) (1 + 1 ) (1 + 1 ) − 1 = 2 ⋅ 3 ⋅ 3 ⋅ 2 ⋅ 2 − 1 = 71

39 Required number of ways is equal to the

The number of ways in which 2 balls from urn A and 2 balls from urn B can be selected = 3C2 × 9C2 = 3 × 36 = 108

31 Total number of m elements subsets of A = nC m

n +1

25 The number of ways in which we can

36 Required number of triangles that can be

∴ Number of divisors = 4 × 3 × 3 × 2 − 2 = 72 − 2 = 70

∴ Total number of ways = 5! × 6 P5 = 5! × 6! 47

⇒

n (n − 1) × 3 = 10 3! 2 n − n − 20 = 0 ⇒ n = 5

37 Since, 38808 = 23 × 32 × 72 × 111

30

M

23

⇒

…(i)

and number of m elements subsets of A each containing element a4 = n − 1 C m − 1 According to the question, = λ ⋅ n −1 C m −1 n n −1 ⇒ ⋅ C m −1 = λ ⋅ n −1C m −1 m n ⇒ λ= ⇒ n = mλ m

32 Number of times A and B are together  n − 2 on guard is  .  10 

number of positive integer solutions of the equation x + y + z = 8 which equal  8 − 1  7 to   =   = 21  3 − 1   2

40 Coefficient of x10 in ( x + x2 + ...+ x 6 )4 = Coefficient of x 6 in (1 + x + ...+ x 5 )4 = (1 − x 6 )4 (1 − x ) − 4 in (1 − 4 x 6 + ... )    4 1 +  1  x + ...    9 6 Hence, coefficient of x is   − 4 = 80.  6

SESSION 2 1 Given, 6 different novels and 3 different

stall can be filled in 3 ways and so on. ∴ Number of ways of loading steamer = 3C1 × 3C1 × K × 3C1 (12 times) = 3 × 3 × K × 3 (12 times) = 312

dictionaries. Number of ways of selecting 4 novels 6! from 6 novels is 6 C 4 = = 15 2! 4! Number of ways of selecting 1 dictionary from 3 dictionaries is 3! 3 C1 = =3 1!2! Number of arrangement of 4 novels and 1 dictionary where dictionary is always in the middle, is 4! Required number of arrangement 15 × 3 × 4! = 45 × 24 = 1080

34 If out of n points, m are collinear, then

2 Number of ways of giving one prize for

Number of times C , D and E are together  n − 3 on guard is  .  9  According to the question,  n − 2  n − 3   = 3   10   9  ⇒

n − 2 = 30 ⇒ n = 32

33 First stall can be filled in 3 ways, second

Number of triangles = C3 − C3 ∴ Required number of triangles = 10C3 − 6C3 = 120 − 20 ⇒ N = 100 n

35 T n = nC3 , hence T n+1 = Now, ⇒ ⇒

⇒

n +1

m

C3

T n +1 − T n = 10 n +1 C3 − nC3 = 10 [given] (n + 1) n (n − 1) 3! n (n − 1) (n − 2) − = 10 3! n (n − 1) (n + 1 − n + 2) = 10 3!

running = 16 Number of ways of giving two prizes for swimming = 16 × 15 Number of ways of giving three prizes for riding = 16 × 15 × 14 ∴ Required ways of giving prizes = 16 × 16 × 15 × 16 × 15 × 14 = 163 × 152 × 14

3 For the first player, cards can be distributed in the 52 C17 ways. Now, out of 35 cards left 17 cards can be distributed for second player in 52 C17 ways.

86

40

Similarly, for third player in 18 C17 ways. One card for the last player can be distributed in 1 C1 way. Therefore, the required number of ways for the proper distribution. = 52C17 × 35C17 × 18C17 × 1 C1 52! 35! 18! 52! = × × × 1! = 35! 17! 18!17! 17!1! (17!) 3

4 The number of words starting from

E = 5! = 120 The number of words starting from H = 5! = 120 The number of words starting from ME = 4! = 24 The number of words starting from MH = 4! = 24 The number of words starting from MOE = 3! = 6 The number of words starting from MOH = 3! = 6 The number of words starting from MOR = 3! = 6 The number of words starting from MOTE = 2! = 2 The number of words starting from MOTHER = 1! = 1 Hence, rank of the word MOTHER = 2 (120) + 2 (24) + 3 (6) + 2 + 1 = 309

5 A matrix whose determinant is non-zero is called a non-singular matrix. Here, we have   a11 a 12   S =   ; aij ∈ {0, 1, 2}, a11 = a22  a a   22  21  Clearly, n(s ) = 27 [Qfor a11 = a22 , we have 3 choices, for a12 , we have 3 choices and for a21 , we have 3 choices] a11 a12 Now, =0 a21 a22 ⇒ a11 ⋅ a22 − a12 ⋅ a21 = 0 ⇒ (a11 ) 2 = a12 a21 = 0 [Q a11 = a22 ] ⇒ a12 a21 = 0, 1, 4 [Q a11 ∈ {0, 1, 2}] Consider a12 a21 = 0, this is possible in 5 cases a12 a21 = 1, this is possible in only 1 case a12 a21 = 4, this is possible in only 1 case Thus, number of singular matrices in S are 7. Hence, number of non-singular matrices in S are 27 − 7 = 20.

6 Let the men be M1 , M2 , ..., M10 and women be W1 , W2 , ..., W7 . Let W1 and W2 do not want to be on the same group. The six members group can contain 4 men and 2 women or 3 men and 3 women. The number of ways of forming 4M , 2W group is 10 C 4 ( 5C2 + 2C1 ⋅ 5C1 ) = 4200

where, 5C2 is the number of ways without W1 and W2 and 5C1 is the number of ways with W1 and without W2 or with W2 and without W1 . The number of ways of forming 3M , 3W 5 group is 10 C3 ( 5C3 + 2C1 C2 ) = 3600 where, 5C3 is the number of ways without W1 and W2 and 5C2 is the number of ways with W1 or W2 but not both. ∴ Number of ways = 4200 + 3600 = 7800

7 Given, X has 7 friends, 4 of them are ladies and 3 are men while Y has 7 friends, 3 of them are ladies and 4 are men. ∴ Total number of required ways = 3C3 × 4C 0 × 4C 0 × 3C3 + 3C2 × 4C1 × 4C1 × 3C2 + 3C1 × 4C2 × 4C2 × 3C1 + 3C 0 × 4C3 × 4C3 × 3C 0 = 1 + 144 + 324 + 16 = 485

8 The number of ways of selecting four numbers from 1 to 30 without any restriction is 30 C 4 . The number of ways of selecting four consecutive numbers [i.e. (1, 2, 3, 4), (2, 3, 4, 5), ..., (27, 28, 29, 30)] is 27. Hence, the number of ways of selecting four integers which excludes selection of consecutive four numbers is 30 × 29 × 28 × 27 30 C 4 − 27 = − 27 24 = 27378

9 Possible even digits are 2,4,6,8,0. Case I Number has digits 4,6,8,0. (Here, sum of digits is 18, divisible by 3)

Number of arrangements = 3 × 3! [Ist place can be filled using 4, 6, 8] = 3 × 6 = 18 Case II Number has digits 2, 4, 6, 0 (Here, sum of digits is 12, divisible by 3)

[Qall white balls are mutually identical] Number of ways to choose zero or more from green balls = (9 + 1) [Qall green balls are mutually identical] Number of ways to choose zero or more from black balls = (7 + 1) [Qall black balls are mutually identical] Hence, number of ways to choose zero or more balls of any colour = (10 + 1) (9 + 1) (7 + 1) Also, number of ways to choose zero balls from the total = 1 Hence, the number of ways to choose atleast one ball [irrespective of any colour] = (10 + 1) (9 + 1) (7 + 1) − 1 = 880 − 1 = 879

12 Let x1 , x2 , ..., x 8 denote the marks assign to 8 questions. ∴ x1 + x2 + ...+ x 8 = 30 Also, x1 , x2 , ...., x 8 ≥ 2 Let, u1 = x1 − 2, u2 = x2 − 2 ... u 8 = x8 − 2 Then, (u1 + 2 + u2 + 2 + ...+ u 8 + 2) = 30 ⇒

u1 + u2 + ...+ u 8 = 14

∴ Total number of solutions =

14 + 8 − 1

C 8 −1 =

21

c7

13 210 510 1111 1313 has a divisor of the form 2α ⋅ 5β ⋅ 11γ ⋅ 13δ , where α = 0, 1, 2,...,10; β = 0, 1, 2,...,10; γ = 0, 1, 2,...,11; δ = 0, 1, 2,...,13 It is of the form 4 n + 1, if α = 0 ; β = 0, 1, 2,...,10; γ = 0, 2, 4,...,10; δ = 0, 1, 2,...,13. ∴ Number of divisors = 11 × 6 × 14 = 924

14 Let the particular side on which 3 1st place cannot be filled by 0. Number of arrangements = 3 × 3! = 18 ∴ Number of numbers = 18 + 18 = 36

10 24 = 2 ⋅ 3 ⋅ 4, 2 ⋅ 2 ⋅ 6, 1 ⋅ 6 ⋅ 4, 1 ⋅ 3 ⋅ 8,

1 ⋅ 2 ⋅ 12, 1 ⋅ 1 ⋅ 24 [as product of three positive integers] ∴ The total number of positive integral solutions of xyz = 24 is 3! 3! equal to3! + + 3! + 3! + 3! + i.e. 30. 2! 2! Any two of the numbers in each factorisation may be negative. So, the number of ways to associate negative sign in each case is 3 C2 i.e. 3. ∴ Total number of integral solutions = 30 + 3 × 30 = 120

11 The number of ways to choose zero or more from white balls = (10 + 1)

particular sailors can work be named A and on the other side by B on which 2 particular sailors can work. Thus, we are left with 3 sailors only. Selection of one sailor for side A = 3C1 = 3 and, then we are left with 2 sailors for the other side. Now, on each side, 4 sailors can be arranged in 4! ways. ∴ Total number of arrangements = 3 × 24 × 24 = 1728

15 Required number is the coefficient of x10 in (1 + x + x2 + ... + x 6 ) 3 = (1 − x7 )3 (1 − x )−3 = (1 − 3 x7 + ... )    3  4 2 1 +  1 x +  2  x + K   Hence, coefficient of x10 is  12  5   − 3   = 36.  10  3

87

DAY

DAY NINE

Unit Test 1 (Algebra) cos x 1 If f ( x ) = 2 sin x tan x

x x2 x

1 f ′ (x ) is equal to 2x , then lim x→ 0 x 1

(b) −1

(a) 1

(d) − 2

(c) 2

2 If log0. 5 ( x − 1) < log0. 25 ( x − 1), then x lies in the interval (a) (2 , ∞)

(b) (3 , ∞)

(c) (− ∞, 0)

(d) (0, 3)

3 Sum of n terms of series 12 + 16 + 24 + 40 + … will be (a) (b) (c) (d)

2 (2 n 2 (2 n 3 (2 n 4 (2 n

− 1) + − 1) + − 1) + − 1) +

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 of a 2 x 2 + 2bx + 2c = 0 has a root γ, that always satisfies (b) γ = β (d) α < γ < β

1 an even number 6 of arithmetic means are inserted. If the sum of these means exceeds their number by unity, then the number of means are

5 Between two numbers whose sum is 2

(b) 10 (d) None of these

1 0 6 If A =  , then which of the following is not true? 1 1  (a) lim

n→ ∞

1 − n  0 0 A =  n2  −1 0

 1 0 (c) A − n =  , ∀ n ∈ N  −n 1

(a) 11

(b) 12

(c) 13

(d) 14

given by

4 Let a, b and c ∈ R and a ≠ 0. If α is a root of

(a) 12 (c) 8

Physics and Chemistry, 37 passed Mathematics, 24 Physics and 43 Chemistry. Atmost 19 passed Mathematics and Physics, atmost 29 passed Mathematics and Chemistry and atmost 20 passed Physics and Chemistry. The largest possible number that could have passed all three examinations is

8 The inequality | z − 4 | < | z − 2 | represents the region

8n 6n 8n 8n

(a) γ = α (c) γ = (α + β) / 2

7 From 50 students taking examinations in Mathematics,

(b) lim

n→ ∞

1 − n  0 0 A =  n  −1 0

(d) None of these

(a) Re(z) > 0

(b) Re(z) < 0

(c) Re(z) > 3

(d) None of these

9 If 1, ω and ω 2 be the three cube roots of unity, then (1 + ω )(1 + ω 2 )(1 + ω 4 ) K 2n factors is equal to (b) −1 (d) None of these

(a) 1 (c) 0

10 If a < 0, then the positive root of the equation x 2 − 2a | x − a | − 3 a 2 = 0 is (a) a (− 1 − 6) (c) a (1 − 6)

(b) a (1 − (d) a (1 +

2) 2)

11 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, ω2

(c) ω, ω

(d) None of these

2

12 Let z1, z 2 and z 3 be three points on | z | = 1. If θ1, θ 2 and θ 3 are the arguments of z1, z 2 and z 3 respectively, then cos (θ1 − θ 2 ) + cos (θ 2 − θ 3 ) + cos (θ 3 − θ1 ) 3 2 −3 (c) ≤ 2 (a) ≥

(b) ≥ −

3 2

(d) None of these

88

40

13 If the roots of the equation

(a + b )x + 2 (bc + ad )x + (c + d ) = 0 are real, then a 2 , bd and c 2 are in 2

2

2

(a) AP (c) HP

2

2

(b) GP (d) None of these

14 150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped the second day, four more workers dropped the third day and so on. It takes 8 more days to finish the work now. Then, the number of days in which the work was completed is (a) 29 days

(b) 24 days

(c) 25 days

(d) 26 days

15 Let R be a relation defined by R = {( x , x ) : x is a prime 3

number < 10 }, then which of the following is true?

(a) R = {(1, 1), (2 , 8), (3 , 27), (4 , 64), (5 , 125),(6, 216), (7, 343), (8, 512), (9, 729)} (b) R = {(2 , 8), (3 , 27), (5 , 125), (7, 343)} (d) None of the above

16 If b > a , then the equation ( x − a )( x − b ) − 1 = 0 , has

(c)

(b) 2

1 2

18 If f ( x , n ) =

(d) None of these n

∑

r =1

r logx   , then the value of x satisfying the x

equation f ( x , 11) = f ( x , 12) is (a) 10 (c) 12

(b) 11 (d) None of these

19 The three numbers a , b and c between 2 and 18 are such that their sum is 25, the numbers 2 ,a and b are consecutive terms of an AP and the numbers b, c and18 are consecutive terms of a GP. The three numbers are (a) 3, 8, 14 (c) 5, 8, 12

(b) 2, 9, 14 (d) None of these

20 If X is the set of all complex numbers z such that | z | = 1, then the relation R defined on X by 2π , is | arg z1 − arg z 2 | = 3 (a) reflexive (c) transitive

(b) symmetric (d) anti-symmetric

21 If α and β are the roots of the equation ax − 2 bx + c = 0, 2

then α 3β 3 + α 2β 3 + α 3β 2 is equal to c2 (a) 3 (c − 2 b) a bc 2 (c) 3 a

(b) 2 ≤ a ≤ 3

(c) 3 ≤ a ≤ 4

(d) a > 4

23 The integer k for which the inequality x 2 − 2 ( 4 k − 1)x + 15 k 2 − 2k − 7 > 0 is valid for any x, is (a) 2 (c) 4

(b) 3 (d) None of these

24 The maximum sum of the series 20 + 19

1 2 + 18 + 18 + K is 3 3

(a) 310 (c) 320

(b) 290 (d) None of these

25 The number of common terms to the two sequences 17, 21, 25, ..., 417 and 16, 21, 26, ..., 466 is (a) 21

(b) 19

(c) 20

(d) 91

26 If the sum of the first three terms of a GP is 21 and the

(a) 3 , 4 (c) 3 , 2

(b) 2 , 4 (d) None of these

1 2 3 + + + ..., is 1 + 12 + 14 1 + 22 + 24 1 + 32 + 34

17 The value of x satisfying log2 ( 3x − 2) = log1/ 2 x is 1 3

(a) a < 2

27 The sum to n terms of the series

both the roots in [a , b] both the roots in (−∞, a ] both the roots in (b, ∞) one root in (−∞, a) and other in (b, ∞)

(a) −

real and less than 3, then

sum of the next three terms is 168, then the first term and the common ratio is

(c) R = {(2 , 8), (4 , 64), (6, 216), (8, 512)}

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

22 If the roots of the equation x 2 − 2ax + a 2 + a − 3 = 0 are

c2 (b) 3 (c + 2 b) a (d) None of these

n2 + n 2 (n 2 + n + 1) n2 + n (c) 2 (n 2 − n + 1) (a)

(b)

n2 − n 2 (n 2 + n + 1)

(d) None of these

28 If C is a skew-symmetric matrix of order n and X is n × 1 column matrix, then X ′ C X is a (a) scalar matrix (c) null matrix

(b) unit matrix (d) None of these

29 Which of the following is correct? (a) Skew-symmetric matrix of an even order is always singular (b) Skew-symmetric matrix of an odd order is non-singular (c) Skew-symmetric matrix of an odd order is singular (d) None of the above

a1

b1

c1

30 If a 2 b2 c 2 = 5 , then the value of a3

b3

c3

b2c3 − b3c2 a3c2 − a2c3 a2b3 − a3b2 ∆ = b3c1 − b1c3 a1c3 − a3c1 a3b1 − a1b3 is b1c2 − b2c1 a2c1 − a1c2 a1b2 − a2b1 (a) 5

(b) 25

(c) 125

(d) 0

31 The number of seven letter words that can be formed by using the letters of the word ‘SUCCESS’ so that the two C are together but no two S are together, is (a) 24 (c) 54

(b) 36 (d) None of these

89

DAY 32 The greatest integer less than or equal to ( 2 + 1)6 is (a) 196 (c) 198

(b) 197 (d) 199

33 If A and B are square matrices such that B = − A −1BA , then (a) (b) (c) (d)

AB + BA = O (A + B)2 = A 2 − B 2 (A + B)2 = A 2 + 2 AB + B 2 (A + B)2 = A + B

xp + y 34 The determinant yp + z 0 (a) x, y, z are in AP (c) x, y, z are in HP

x y xp + y

y = 0 , if z yp + z

(b) x, y, z are in GP (d) xy, yz, zx are in AP

35 The value of k, for which the system of equations

x + ky + 3z = 0, 3x + ky − 2z = 0 and 2x + 3y – 4z = 0 possess a non-trivial solution over the set of rationals, is 33 2 (d) None of these

33 2 (c) 11 (a) −

(b)

36 The number of groups that can be made from 5 different green balls, 4 different blue balls and 3 different red balls, if atleast 1 green and 1 blue ball is to be included is (a) 3700 (c) 4340

(b) 3720 (d) None of these

37 If n is an integer greater than 1, then

 

2   3x 2  ascending power of x, is

42 The 8th term of  3x +

228096 x3 328179 (c) x9 (a)

(b) 0

(c) a 2

(d) 2 n

38 If α , β and γ are the roots of the equation x ( x + e ) = e ( x + 1). Then, the value of the determinant 1+ α 1 1 1 1+ β 1 is 1 1 1+ γ 2

(b) 1

(c) 0

(d) 1 +

1 1 1 + + α β γ

39 For all natural number n > 1, 24 n − 15 n − 1 is divisible by (a) 225 (c) 325

(b) 125 (d) None of these

40 If x is so small that its two and higher power can be

neglected and if (1 − 2x )−1/ 2 (1 − 4x )−5 / 2 = 1 + kx, then k is equal to (a) − 2 (c) 10

(b) 1 (d) 11

2x + 1 4 41 If x = − 5 is a root of 2 2x 7 6

8 2 = 0 , then the 2x

other two roots are (a) 3, 3.5

(b) 1, 3.5

(c) 3 , 6

(d) 2 , 6

(b)

228096 x9

(d) None of these

(a) 55 × 3 9 (c) 55 × 3 6

(b) 46 × 3 9 (d) None of these

44 In an examination a candidate has to pass in each of the papers to be successful. If the total number of ways to fail is 63, how many papers are there in the examination? (a) 6

(b) 8

(c) 10

(d) 12

45 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 contains exactly two elements is (a) 9 ⋅ n C2 (c) 2 ⋅ n Cn

(b) 3 n − n C2 (d) None of these

46 If the sets A and B are defined as 1 ,0 ≠ x ∈ R } x B = {( x , y ): y = − x , x ∈ R }, then A = {( x , y ) : y =

and

(a) A ∩ B = A (c) A ∩ B = φ

(b) A ∩ B = B (d) None of these

47 There are 16 points in a plane no three of which are in a straight line except 8 which are all in a straight line. The number of triangles can be formed by joining them equals (a) 1120

(a) − 1

when expanded in

43 The greatest term in the expansion of ( 3 − 5x )11 when 1 x = , is 5

a − nC1(a − 1) + nC2 (a − 2) + ...+ ( − 1)n (a − n ) is equal to (a) a

12

(b) 560

(c) 552

(d) 504

48 The value of the natural numbers n such that inequality 2n > 2n + 1 is valid, is (a) for n ≥ 3

(b) for n < 3

(c) for mn

(d) for any n

 2π   2π  49 Let w = cos   + i sin   and α = w + w 2 + w 4 and  7  7 β = w 3 + w 5 + w 6 , then α + β is equal to (a) 0

(b) −1

(c) −2

(d) 1

f (x ) 50 Let H( x ) = , where f ( x ) = 1 − 2 sin2 x and g( x ) g ( x ) = cos 2x , ∀ f : R → [ −1,1] and g : R → [ −1, 1]. Domain and range of H( x ) are respectively (a) R and {1} (b) R and {0, 1}

π (c) R ~ {(2n + 1) } and {1},n ∈ I 4 π (d) R ~ (2n + 1)  and {0, 1},n ∈ I 2 

90

40

55 The general term in the expansion of (a + x )n is

51 Let Tn denote the number of triangles which can be

n

formed using the vertices of a regular polygon of n sides. If Tn +1 − Tn = 21, then n equals (a) 4 (c) 7

Statement I The third term in the expansion of m 2 x + 1  does not contain x . The value of x for    x 2

(b) 6 (d) None of these

52 If r is a real number such that | r | < 1 and if a = 5 (1 − r ),

which that term equal to the second term in the expansion of (1 + x 3 ) 30 is 2.

then (a) 0 < a < 5 (c) 0 < a < 10

Cr a n − r x r .

(b) − 5 < a < 5 (d) 0 ≤ a < 10

Statement II (a + x ) n =

Directions (Q. Nos. 53-57) Each of these questions contains two statements : Statement I (Assertion) and Statement II (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below. (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 p is a prime, then number of divisors of p n is p n + 1 − 1.

n

Cr a n − r x r .

r = 0

56 Sets A and B have four and eight elements, respectively. Statement I The minimum number of elements in A ∪ B is 8. Statement II A ∩ B = 5 a b c 57 Let a ≠ 0, p ≠ 0 and ∆ = 0 p q p q 0

Statement I If the equations ax 2 + bx + c = 0 and px + q = 0 have a common root, then ∆ = 0. Statement II If ∆ = 0, then the equations ax 2 + bx + c = 0 and px + q = 0 have a common root.

53 Statement I The number of natural numbers which divide 102009 but not 102008 is 4019.

n

∑

58 Assume X ,Y , Z ,W and P are matrices of order

2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively.

Now, consider the following statements

 1 0 2 0 54 Suppose A =  and B =    . Let X be a 2 × 2 0 −1 0 2 matrices such that X ′ AX = B.

I. PY + WY will be defined for k = 3 and p = n. II. The order of the matrix 7X − 5Z is n × 2 (if p = n). Choose the correct option.

Statement I X is non-singular and det (X) = ± 2 . Statement II X is a singular matrix.

(a) Only I is true (c) Both I and II are true

(b) Only II is true (d) Neither I nor II is true

ANSWERS 1. 11. 21. 31. 41. 51.

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

2. 12. 22. 32. 42. 52.

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

3. 13. 23. 33. 43. 53.

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

4. 14. 24. 34. 44. 54.

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

5. 15. 25. 35. 45. 55.

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

6. 16. 26. 36. 46. 56.

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

7. 17. 27. 37. 47. 57.

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

8. 18. 28. 38. 48. 58.

(c) (c) (c) (c) (a) (a)

9. 19. 29. 39. 49.

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

10. 20. 30. 40. 50.

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

91

DAY

Hints and Explanations 1 Applying R 1 → R 1 − R 3 cos x − tan x f ( x) =

2 sin x tan x

0 x2

2x

0

x

1

= (cos x − tan x)( x − 2 x ) = − x 2 (cos x − tan x) ∴ f ′ ( x) = − 2 x(cos x − tan x) − x 2 (− sin x − sec2 x) f ′ ( x) = lim [−2 (cos x − tan x) ∴ lim x→ 0 x→ 0 x + lim x(sin x + sec2 x)] 2

2

x→ 0

= −2 ×1= −2

2 Given, log 0.5( x − 1) < log 0.25( x − 1) ⇒

⇒ ∴

log 0.5( x − 1) < log( 0.5 )2 ( x − 1) 1 log 0. 5 ( x − 1) < log 0.5( x − 1) 2 log 0.5 ( x − 1) < 0 ⇒ x − 1 > 1 x >2

3 Let,

S n = 12 + 16 + 24 + … + Tn

⇒

Again, S n = 12 + 16 + K + Tn 0 = (12 + 4 + 8 + 16 + K upto n terms) −Tn 4 (2 n − 1 − 1) ∴ Tn = 12 + = 2n + 1 + 8 2 −1 On putting n = 1, 2, 3,K , we get T1 = 22 + 8, T2 = 23 + 8, T3 = 2 4 + 8 … ∴ S n = T1 + T2 + K + Tn = (22 + 23 + … upto n terms) + (8 + 8 + … upto n terms) 22 (2 n − 1) = + 8n 2 −1 = 4 (2 n − 1) + 8 n

4 Let f ( x) = a2 x2 + 2bx + 2c Q We have, a2α 2 + bα + c = 0 and a2β2 − bβ − c = 0 ∴ f (α ) = a2α 2 + 2bα + 2c = bα + c = − a2α 2 2 2 f (β) = a β + 2bβ + 2c = 3 (bβ + c ) = 3 a2β2 But 0 < α < β ⇒ α , β are real number. ∴ f (α ) < 0, f (β) > 0 Hence, α < γ < β.

5 Let 2n arithmetic means be A 1 , A 2 ,K , A 2 n between a and b . a+b Then, A 1 + A 2 +K + A 2 n = × 2n 2 13 / 6 13 n = × 2n = 2 6

Given, A 1 + A 2 + K + A 2 n = 2 n + 13 n 2n + 1 = ⇒ 6 ⇒ 12 n + 6 = 13 n ∴ n=6 Hence, the number of means = 2 × 6 = 12 1 0  6 A −1 =   −  1 1  1 0  1 0  1 ⇒ ( A −1 )2 =  =    −1 1   −1 1   −2

1

10 If x ≥ a, then x2 − 2 a( x − a) − 3 a2 = 0 ⇒ x 2 − 2 ax − a2 = 0 ∴ x=

0 1 

 0 0 =   −1 0  n→ ∞

1 / n2 0  1 −n A = lim  1  2 2 n → ∞ − n  n 1 / n  0 0  =  0 0 

7 Given, n (M ∪ P ∪ C) = 50, n (M ) = 37, n (P) = 24 , n (C) = 43 n (M ∩ P) ≤ 19, n (M ∩ C) ≤ 29, n (P ∩ C) ≤ 20 ∴ n (M ∪ P ∪ C) = n (M ) + n (P) + n (C) − n ( M ∩ P ) − n ( M ∩ C) − n (P ∩ C) + n (M ∩ P ∩ C) ⇒ 50 = 37 + 24 + 43 − n(M ∩ P) − n (M ∩ C) − n (P ∩ C) + n (M ∩ P ∩ C) ⇒ n (M ∩ P ∩ C) = n (M ∩ P ) + n (M ∩ C) + n (P ∩ C) − 54 ∴ n (M ∩ P ∩ C) ≤ 19 + 29 + 20 − 54 = 14

8 |( x − 4) + iy|2 < |( x − 2) + iy|2 ⇒ ⇒ ∴

2a ±

4 a2 + 4 a2 2

= a(1 ± 2 )

Since, x ≥ a ∴ x = a(1 + 2 ), it is impossible

 1 0 Similarly, ( A −1 )n =   −n 1 0  1 / n 1 Now, lim A − n = lim   n→ ∞ n n→ ∞  − 1 1 / n

and lim

= (1 + ω + ω2 + ω3 )n = (0 + ω3 )n = ω3 n = 1

[let z = x + iy] ( x − 4)2 + y2 < ( x − 2)2 + y2 − 4 x < − 12 ⇒ x > 3 Re(z) > 3

9 (1 + ω)(1 + ω2 )(1 + ω 4 )(1 + ω 8) … to 2n factors = (1 + ω)(1 + ω2 )(1 + ω)(1 + ω2 ) … to 2n factors = [(1 + ω)(1 + ω) … to n factors] [(1 + ω2 )(1 + ω2 ) … to n factors] = (1 + ω)n (1 + ω2 )n

because a < 0 ∴ x = a (1 − 2 ) If x < a, then x2 + 2 a ( x − a) − 3 a2 = 0 ⇒ x 2 + 2 ax − 5 a2 = 0 ∴ x = (− 1 ± 6 ) a [impossible x < a and a < 0]

11 z3 + 2 z2 + 2 z + 1 = 0 ⇒ (z + 1)(z2 + z + 1) = 0 ⇒ z = – 1, ω, ω2 But z = − 1 does not satisfy the second equation. Hence, common roots are ω and ω2 .

12 We have,| z1| = | z2| = | z3| = 1 Now,| z1 + z2 + z3| ≥ 0 ⇒ | z1|2 + | z2|2 + | z3|2 + 2 Re (z1 z2 + z2 z3 + z3 z1 ) ≥ 0 ⇒ 3 + 2 [cos(θ1 − θ2 ) + cos(θ2 − θ3 ) + cos(θ3 − θ1 )] ≥ 0 ⇒ cos(θ1 − θ2 ) + cos(θ2 − θ3 ) 3 + cos(θ3 − θ1 ) ≥ − 2

13 Here, D ≥ 0 ∴4 (bc + ad )2 − 4 (a2 + b 2 )(c 2 + d 2 ) ≥ 0 ⇒ b 2c 2 + a2d 2 + 2 abcd − a2c 2 − a2d 2 − b 2c 2 − b 2d 2 ≥ 0 ⇒ (ac − bd )2 ≤ 0 ⇒ ac − bd = 0 [since, square of any expression cannot be negative] ∴ b 2d 2 = a2c 2 Hence, a2 , bd and c 2 are in GP.

14 Here, a = 150 and d = − 4 n [2 × 150 + (n − 1) (− 4)] 2 = n (152 − 2 n) Had the workers not dropped, then the work would have finished is (n − 8) days with 150 workers working on each day. Sn =

92

40

∴ ⇒ ⇒ ∴

n(152 − 2 n) = 150 (n − 8) n2 − n − 600 = 0 (n − 25) (n + 24) = 0 n = 25 [since, n cannot be negative]

15 Given, x is a prime < 10 ∴ x = {2 , 3 , 5 , 7} Now, from R = {( x, x 3 ) : x = 2, 3, 5 , 7} = {(2 , 8), (3, 27), (5 , 125), (7, 343)}

16 Let f ( x) = ( x − a)( x − b ) − 1 We observe that the coefficient of x in f ( x) is positive and f (a) = f (b ) = − 1. Thus, the graph of f ( x) is as shown in figure given below

2

Y y = f(x) (a, 0)

X'

(b, 0)

O (a, –1)

X

(b, –1)

Y'

It is evident from the graph that one of the roots of f ( x) = 0 lies in (− ∞, a) and the other root lies in (b , ∞). log2 (3 x − 2) = − log 2 x

⇒ log2 (3 x − 2) = log2 x −1 ⇒ 3 x − 2 = x −1 ⇒ 3 x2 − 2 x − 1 = 0 ⇒ (3 x + 1) ( x − 1) = 0 1 ⇒ x = 1 or x = − 3 ∴ x =1 [since, negative of x cannot satisfy the given equation]

18 f ( x, n) =

n

∑

r =1

=

n

∑

r =1

(log x r − log x x)

(log x r − 1) = log x (1⋅ 2 ... n) − n

= log x n ! − n Given, f ( x, 11) = f ( x, 12) ⇒ log x (11 !) − 11 = log x (12 !) − 12  12 ! ⇒ log x   = 1 ⇒ log x (12) = 1  11 ! ∴

x = 12

19 Given, a + b + c = 25

25 First series has common difference

20 Q | z| = 1 ⇒ z = cos θ + i sin θ ∴ arg (z) = θ Let, arg (z1 ) = θ1 and arg (z2 ) = θ2

Then, z1 Rz2 ⇒|arg z1 − arg z2 | = ⇒

z1 Rz2 ⇒|θ1 − θ2 | =

2π 3

2π 3

2π ⇒ |θ2 − θ1 | = 3 ⇒ z2 Rz1 Hence, it is symmetric. 2b c 21 Here, α + β = and αβ = a a Now, (αβ)3 + α 2β2 ( β + α ) 3 c2  2 b  c =  + 2    a a  a =

c 2 (c + 2 b ) a3

22 According to the question,

17 Given, log2 (3 x − 2) = log1 /2 x ⇒

∴ Maximum sum, 31   2  S31 = 2 × 20 + (31 − 1)  −     3  2  31 = (40 − 20) = 310 2

From Eqs. (i) and (ii), we get 3b = 48 − 2 c From Eq. (iii), we get c 2 = 6 (48 − 2 c ) = 288 − 12 c ⇒ c 2 + 12 c − 288 = 0 ⇒ (c + 24)(c − 12) = 0 ⇒ c = 12 as c ≠ – 24 ∴ b = 8 and a = 5

…(i)

Since, 2, a, b are in AP, therefore …(ii) 2a = 2 + b Since, b , c , 18 are in GP, therefore …(iii) c 2 = 18 b

D ≥ 0 and f (3) > 0 ⇒ 4 a2 − 4 (a2 + a − 3) ≥ 0 and 32 − 2 a (3) + a2 + a − 3 > 0 ⇒ − a + 3 ≥ 0 and a2 − 5 a + 6 > 0 ⇒ a ≤ 3 and a < 2 or a > 3 ∴ a 0 ∴ D 0) (c) | x ± y | ≤ | x | + | y | (d) | x ± y | ≥ || x | − | y ||

X

97

DAY (ii) Signum Function The function f ( x) x  1, if | x| , if x ≠ 0  or  | x| = − 1, if = sgn ( x) =  x 0, if x = 0  0, if 

Y 4

x > 0 x < 0 x = 0

3 2 1

called signum function.

X¢

Y y=1 X¢

–1 0 1 –1 –2

–4 –3 –2

f (x) = (x)

Y¢

Its domain ∈ R and range ∈ I.

Y¢

Its domain is R and range is {− 1, 0, 1}. (iii) Greatest Integer Function The symbol [ x] indicates the integral part of x which is nearest and smaller than to x. It is also known as floor of x. x ∀ x ∈ I The function f ( x) = [ x] =  is called n, n ≤ x < n + 1, n ∈ I greatest integer function. Y

(vi) Transcendental function The function which is not algebraic is called transcendental function. (vii) Exponential Function The function f ( x) = ax , a > 0, a ≠ 1, is called an exponential function. Y

Y 0 0) and a ≠ 1 is called logarithmic function.

–2

Y

Y

Y¢

a>1

0 0  f {g ( x )} is equal to (a) x

Statement II sin−1 x and cos−1 x is defined in | x | ≤ 1and tan−1 x is defined for all x. (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

f ( x ) = 2 cos

14 If f 2 ( x )⋅ f  (a) 1/ x

18. If domain of f ( x ) and g ( x ) are D1 and D2 respectively,

(b) 1

(c) f (x)

(d) g (x)

1 1 ( x − π ) + 4 sin ( x − π ) is 3π. 3 3

Statement II IfT is the period of f ( x ), then the period of T . f (ax + b) is |a | (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

16. The function f ( x ) = log ( x + x 2 + 1), is (a) an even function (b) an odd function (c) a periodic function (d) neither an even nor an odd function

17. Statement I f ( x ) = | x − 2 | + | x − 3 | + | x − 5 | is an odd function for all values of x lie between 3 and 5. Statement II For odd function f ( − x ) = − f ( x ) (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

20. If the range of f ( x ) is collection of all outputs f ( x ) corresponding to each real number in the domain, then

 1  Statement I The range of log  is ( − ∞, ∞). 2 1 + x  Statement II When 0 < x ≤ 1, log x ∈ ( − ∞, 0]. (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

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE x −1 , where {}⋅ denotes the x − 2 {x } fractional part of x, is

1 Domain of f ( x ) =

(a) (− ∞, 0) ∪ (0, 2] (c) (− ∞, ∞) ~ [0, 2)

(b) [1, 2) (d) (− ∞, 0) ∪ (0, 1] ∪ [2, ∞)

2 Range of f ( x ) = [| sin x | + | cos x | ], where [⋅] denotes the greatest integer function, is (a) {0}

(b) {0, 1}

(c) {1}

(d) None of these

3 If [ x 2 ] + x − a = 0 has a solution, where a ∈ N and a ≤ 20, then total number of different values of a can be (a) 2

(b) 3

(c) 4

(d) 6

4 Total number of solutions of [ x ] = x + 2 {x }, where [⋅] 2

and {}⋅ denotes the greatest integer function and fractional part respectively, is equal to (a) 2 (c) 6

(b) 4 (d) None of these

101

DAY 5 If f ( x ) = sin x + cos x , g ( x ) = x 2 − 1, then g { f ( x )} is invertible in the domain π (a)  0,   2  π π (c)  − ,   2 2 

8 If f ( x ) and g ( x ) are two functions such that

f ( x ) = [ x ] + [ − x ] and g ( x ) = {x } ∀x ∈ R and h( x ) = f (g ( x )); then which of the following is incorrect ? ([⋅] denotes greatest integer function and {}⋅ denotes fractional part function).

π π (b)  − ,   4 4  (d) [0, π]

(a) f (x) and h (x) are inertial functions (b) f (x) = g (x) has no solution (c) f (x) + h (x) > 0 has no solution (b) f (x) − h (x) is a periodic function

6 Let f ( x ) = x 10 + a ⋅ x 8 + b ⋅ x 6 + cx 4 + dx 2 be a polynomial function with real corfficient. If f (1) = 1 and f ( 2) = −5, then the minimum number of distinct real zeroes of f ( x ) is (a) 5 (c) 7

9 The period of the function f ( x ) = [ 6x + 7] + cos πx − 6x ,

(b) 6 (d) 8

where [⋅] denotes the greatest integer function, is

−1

−1

(c) 2

(d) None of these

10 The number of real solutions of the equation

f og ( 23) equals (a) 2 (c) 4

(b) 2 π

(a) 3

7 If f : R → R , f ( x ) = x 3 + 3, and g : R → R , g ( x ) = 2x + 1, then

log0. 5 | x | = 2| x | is.

(b) 3 (d) 5

(a) 1

(b) 2

(c) 0

(d) None of these

ANSWERS SESSION 1

SESSION 2

1 (d)

2 (a)

3 (b)

4 (d)

5 (b)

6 (a)

7 (c)

8 (d)

9 (c)

10 (b)

11 (d)

12 (d)

13 (b)

14 (d)

15 (b)

16 (b)

17 (b)

18 (a)

19 (d)

20 (d)

1 (d)

2 (c)

3 (c)

4 (b)

5 (b)

6 (a)

7 (a)

8 (b)

9 (c)

10 (b)

Hints and Explanations SESSION 1 1 Set A represents the set of points lying on the graph of an exponential function and set B represents the set of points lying on the graph of the polynomial. Take e2 x = x 2 , then the two curves does not intersect. Hence, there is no point common between them.

2 For f ( x) to be defined, π sin (2 x) + ≥0 6 π π ⇒ − ≤ sin −1 (2 x) ≤ 2 6  π  π sin  −  ≤ 2 x ≤ sin   ⇒  6  2 1 1 ⇒ − ≤ x≤ 4 2  1 1 ⇒ x∈ − ,  4 2  1 3 y= | x| − x −1

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

3 + log10 ( x3 − x) 4 − x2 For domain of f ( x), x3 − x > 0 ⇒ x( x − 1)( x + 1) > 0

4 Given, f ( x) =

–

+ –1

– 0

+ 1

⇒ x ∈ (−1, 0) ∪ (1, ∞) and 4 − x2 ≠ 0 ⇒ x≠±2 ⇒ x ∈ (−∞, − 2) ∪ (−2 , 2) ∪ (2 , ∞) So, common region is (−1, 0) ∪ (1, 2) ∪ (2 , ∞).

6 We have, f ( x + 1) + f ( x + 3) = 2 ...(i) On replacing x by x + 2, we get ...(ii) f ( x + 3) + f ( x + 5) = 2 On subtracting Eq. (ii) from Eq. (i), we get f ( x + 1) − f ( x + 5) = 0 ⇒ f ( x + 1) = f ( x + 5) Now, on replacing x by x − 1, we get f ( x) = f ( x + 4) Hence, f is periodic with period 4.

 3 sin x − sin 3 x  4

6 f ( x) =

3 cos x + cos 3 x   4 ∴ Period of f ( x) = LCM of period of +

{sin x, cos x, sin 3 x, cos 3 x} LCM of {2 π , 2 π } = = 2π HCF of {1, 3}

7 Let y = x2 + 1 ⇒ ∴ ∴

x = ± y −1 f

−1

( x) = ±

x −1

−1

f (17) = ± 17 − 1 = ± 4

and f −1 (−3) = ± −3 − 1 = ± −4 ∉ R ∴

−1

f (−3) = φ

8 Let y = ( x + 1)2 for x ≥ − 1 ⇒

± y = x +1⇒

⇒ ⇒

y ≥ 0, x + 1 ≥ 0 x = y −1

⇒

f −1 ( y) =

⇒

−1

f ( x) =

y= x+1

y −1 x – 1, x ≥ 0

TEN

102 x 1 + x2 1  1 x ∴ f  =  x 1 1+ 2 x x = = f ( x) 1 + x2  1 ∴ f   = f (2)  2  1 or f   = f (3)  3 and so on. So, f ( x) is many-one function. Again, let y = f ( x) x ⇒ y= 1 + x2 ⇒ y + x2 y = x ⇒ yx2 − x + y = 0 As, x ∈R ∴ (− 1)2 − 4 ( y)( y) ≥ 0 ⇒ 1 − 4 y2 ≥ 0 − 1 1 y∈ , ⇒  2 2  − 1 1 ∴Range = Codomain = ,  2 2 

9 We have, f ( x) =

So, f ( x) is surjective. Hence, f ( x) is surjective but not injective. ex − e−x 10 Given, y = x +2 e + e−x ⇒ ⇒ ⇒

e2 x − 1 +2 y = 2x e +1 1− y y –1 e2 x = = y −3 3 − y  y − 1 1 x = loge   3 − y 2

 y − 1 ⇒ f −1 ( y) = loge   3 − y ⇒

 x − 1 f −1 ( x) = loge   3 − x

13 Q[ x] denotes the integral part of x. 50  1 each +  2 100  term will be one. Hence, the sum of given series will be 50.

Hence, after term

1 − x 3 …(i)  = x 1 + x 1− x On replacing x by , we get 1+ x

14 f 2 ( x) ⋅ f 

1 − x 1 − x f2   f ( x) =   1 + x 1 + x

3

From Eqs. (i) and (ii), 3 1 + x f 3 ( x) = x 6   1 − x 1 + x f ( x) = x 2   1 − x −4 f (− 2) = ⇒ [ f (− 2)] = − 2 3 | [ f (− 2)]| = 2

⇒ ⇒ ⇒

15 Q g ( x) = 1 + x − [ x] [put x = n ∈ Z ] ∴

g ( x) = 1 + x − x = 1

and g( x) = 1 + n + k − n = 1 + k [put x = n + k ] [where, n ∈ Z , 0 < k < 1 ]  −1, g( x) < 0  Now, f { g( x)} =  0, g( x) = 0  1, g ( x) > 0  Clearly, g ( x) > 0 , ∀ x So, f { g( x)} = 1 , ∀ x

16 Given that, f ( x) = log ( x + Now,

1 /2

11 We have, g( x) = 2 f ( x) + 5 Now, on replacing x by g −1 ( x), we get g(g −1 ( x)) = 2 f (g −1 ( x)) + 5 ⇒ x = 2 f (g −1 ( x)) + 5 x −5 ⇒ f (g −1 ( x)) = 2 −1 −1  x − 5  ⇒ g ( x) = f    2 

12 f : (2 , 3) → (0, 1) and f ( x) = x − [ x] ∴ f ( x) = y = x − 2 ⇒ x = y + 2 ⇒ f −1 ( x) = x + 2

x2 + 1 )

f (− x) = log (− x +

∴ f ( x) + f (− x) = log ( x + 1 /2

…(ii)

x2 + 1 ) x + 1) 2

+ log (− x + x 2 + 1 ) = log (1) = 0 Hence, f ( x) is an odd function.  −3 x + 10 , ∀ x ≤ 2  − x + 6, ∀ 2 < x ≤ 3 17 Here, f ( x) =   x, ∀ 3 < x ≤ 5 3 x − 10, ∀ x >5 ∴ ⇒

f ( x) = x, ∀ 3 < x < 5 f (− x ) = − x = − f ( x )

18 Since, sin −1 x is defined in [−1, 1] , cos −1 x is defined in [−1, 1] and tan −1 x is defined in R. Hence, f ( x) is defined in [−1, 1]. 1 19 Period of 2 cos ( x − π ) and 3 1 2π 2π or 6 π , 6 π 4 sin ( x − π ) are , 3 1/3 1/3 ∴ Period of their sum = 6π

1 is (0, 1) and 1 + x2 domain R

20 Range of

∴

 1  log   ∈ (− ∞, 0]  1 + x2 

SESSION 2 1 We have,

x −1 ≥ 0, here two x − 2 { x}

cases arise Case I x ≥ 1 and x > 2 { x} ⇒ x ≥2 ∴ x ∈ [2, ∞). Case II x ≤ 1 and x < 2 { x} ⇒ x < 1 and x ≠ 0. ∴ x ∈ ( − ∞, 0) ∪ (0, 1). Finally, x = 1 is also a point of the domain.

2 y = | sin x | + | cos x | ⇒ y2 = 1 + | sin 2 x | ⇒ 1 ≤ y2 ≤ 2 ⇒ y ∈[1, 2 ] ∴

f ( x) = [ y] = 1, ∀ x ∈ R

3 Since, [ x2 ] + x − a = 0 ∴ x has to be an integer. ⇒ a = x 2 + x = x ( x + 1) Thus, a can be 2, 6, 12, 20.

4 [ x]2 = x + 2{ x} ⇒ [ x] 2 = [ x] + 3 { x} [ x] 2 − [ x] ⇒ { x} = 3 ⇒ 0≤

[ x] 2 − [ x] 0 x→ a

u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

[quotient rule]

(vi) If lim f ( x) = + ∞ or − ∞, then lim x→ a

Your Personal Preparation Indicator

x→ a

x→ a

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

105

DAY (ix) If f ( x) ≤ g( x), ∀x, then lim f ( x) ≤ lim g( x) x→ a

g (x )

(x) lim[ f ( x)] x→ a

= {lim f ( x)}

x→ a

lim g ( x )

x →a

x→ 0

x→ a

continuous at lim g( x) = m. x→ a

(xii) Sandwich Theorem If f ( x) ≤ g( x) ≤ h( x)∀x ∈ (α , β) − {a} and lim f ( x) = lim h( x) = l , then lim g( x) = l where a ∈(α , β) x→ a

log a (1 + x) = log a e; a > 0, ≠ 1 x loge (1 + x) loge (1 − x) (ii) In particular, lim = 1 and lim = −1 x→ 0 x→ 0 x x (i) lim

x→ a

(xi) lim f {g( x)} = f {lim g( x)} = f (m) provided f is x→ a

3. Logarithmic Limits

x→ a

x→ a

Important Results on Limit Some important results on limits are given below

1. Algebraic Limits x n − an = nan −1 , n ∈ Q, a > 0 x→ a x −a 1 (ii) lim n = 0, n ∈ N x→ ∞ x (iii) If m, n are positive integers and a0, b 0 are non-zero real numbers, then a x m + a x m −1 + ... + am −1 x + am lim 0 n 1 n −1 x→ ∞ b x + b x + ... + b n −1 x + b n 0 1  a0 if m = n b  00 if m n, a0 b 0 > 0  −∞ if m > n, a0 b 0 < 0 (i) lim

(1 + x)n − 1 =n x→ 0 x

(1 + x)m − 1 m = x → 0 (1 + x)n − 1 n

4. Expotential Limits ax − 1 = loge a, a > 0 x ex −1 e λx − 1 (ii) In particular, lim = 1 and lim =λ x→ 0 x→ 0 x x 0, 0 ≤ a 1  Does not exist, a < 0 (i) lim x→ 0

5. 1∞ Form Limits (i) If lim f ( x) = lim g( x) = 0, then x→ a

x→ a

lim{1 + f ( x)}1 / g ( x ) = e

lim

x → a

f (x )

g (x )

x→ a

(ii) If lim f ( x) = 1 and lim g( x) = ∞, then x→ a

x→ a lim { f ( x ) − 1 } g ( x )

lim{ f ( x)}g ( x ) = e x → a x→ a

In General Cases x

(i) lim(1 + x)1 / x = e x→ 0

(v) lim

(iv) lim

(iii) lim(1 + λx)1 / x = e λ x→ 0

2. Trigonometric Limits sin x x = 1 = lim x → 0 sin x x tan x x lim = 1 = lim x→ 0 x → 0 tan x x sin −1 x x lim = 1 = lim x→ 0 x → 0 sin −1 x x tan −1 x x lim = 1 = lim x→ 0 x → 0 x tan −1 x sin x ° π sin x cos x (vi) lim lim = = lim =0 x→ 0 x→ ∞ x→ ∞ x 180 x x lim sin x or lim cos x oscillates between −1 to 1.

a  (v) lim (1 + ax) b / x = lim 1 +  x→ 0 x→ ∞  x

1  (ii) lim 1 +  = e x→ ∞  x x λ  (iv) lim 1 +  = e λ x→ ∞  x bx

= e ab

(i) lim x→ 0

(ii) (iii) (iv) (v) (vii)

x→ ∞

x→ ∞

sin P mx  m  =  x → 0 sin P nx  n

P

(viii) lim (ix) lim x→ 0

tan P mx  m  =  tan P nx  n 

P

1 − cos m x m2 cos ax − cos bx a2 – b 2 = 2 ; lim = x → 0 1 − cos n x n x → 0 cos cx − cos dx c2 – d2

(x) lim

cos mx − cos nx n2 − m2 = x→ 0 x2 2

(xi) lim

Methods To Evaluate Limits To find lim f ( x), we substitute x = a in the function. x→ a

If f (a) is finite, then lim f ( x) = f (a). x→ a

0 ∞ If f (a) leads to one of the following form ; ; ∞ − ∞; 0 × ∞; 1 ∞ , 0 0 ∞ and ∞ 0 (called indeterminate forms), then lim f ( x) can be x→ a evaluated by using following methods (i) Factorization Method This method is particularly used when on substituting the value of x, the expression take the form 0/0. (ii) Rationalization Method This method is particularly used when either the numerator or the denominator or both involved square roots and on substituting the value of x, 0 ∞ the expression take the form , . 0 ∞ NOTE To evaluate x → ∞ type limits write the given expression in the

form N /D and then divide both N and D by highest power of x occurring in both N and D to get a meaningful form.

TEN

106

L’Hospital’s Rule

Continuity of a Function in an Interval

If f ( x) and g( x) be two functions of x such that

A function f ( x) is said to be continuous in (a, b ) if it is continuous at every point of the interval (a, b ). A function f ( x) is said to be continuous in [a, b ], if f ( x) is continuous in (a, b ). Also, in addition f ( x) is continuous at x = a from right and continuous at x = b from left.

(i) lim f ( x) = lim g( x) = 0. 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 x = a, then f ( x) f ′ ( x) provided that g(a) ≠ 0. lim = lim x → a g ( x) x → a g ′ ( x)

Results on Continuous Functions (i) Sum, difference product and quotient of two continuous functions are always a continuous f ( x) function. However, r ( x) = is continuous at x = a g ( x) only if g(a) ≠ 0. (ii) Every polynomial is continuous at each point of real line. (iii) Every rational function is continuous at each point where its denominator is different from zero. (iv) Logarithmic functions, exponential functions, trigonometric functions, inverse circular functions and modulus function are continuous in their domain. (v) [ x] is discontinuous when x is an integer. (vi) If g( x) is continuous at x = a and f is continuous at g(a), then fog is continuous at x = a. (vii) f ( x) is a continuous function defined on [a, b ] such that f (a) and f (b ) are of opposite signs, then there is atleast one value of x for which f ( x) vanishes, i.e. f (a) > 0, f (b ) < 0 ⇒ ∃ c ∈ (a, b ) such that f (c) = 0.

Above rule is also applicable, if lim f ( x) = ∞ and lim g( x) = ∞. x→ a

x→ a

If f ′ ( x), g′ ( x) satisfy all the conditions embeded in L’Hospital’s rule, then we can repeat the application of this rule on f ′ ( x) f ′ ′ ( x) f ′ ( x) to get lim . = lim x → a g ′ ( x) x → a g ′ ′ ( x) g ′ ( x) Sometimes, following expansions are useful in evaluating limits. ●

●

●

●

●

●

x2 x3 x 4 x 5 + − + + ... + (−1 < x ≤ 1) 2 3 4 5 x2 x3 x 4 x 5 log(1 − x) = − x − − − − − ...(−1 < x < 1) 2 3 4 5 2 3 4 x x x x ex = 1 + + + + + ... 1 ! 2 ! 3 ! 4! 2 x ax = 1 + x (loge a) + (loge a)2 + ... 2! x3 x 5 x7 sin x = x − + − + ... 3! 5! 7! x2 x 4 x 6 cos x = 1 − + − − ... 2 ! 4! 6 ! log(1 + x) = x −

Continuity If the graph of a function has no break (or gap), then it is continuous. A function which is not continuous is called a 1 discontinuous function. e.g. x2 and e x are continuous while x and [ x], where [ ⋅ ] denotes the greatest integer function, are discontinuous.

Differentiability The function f ( x) is differentiable at a point P iff there exists a unique tangent at point P. In other words, f ( x) is differentiable at a point P iff the curve does not have P as a corner point, i.e. the function is not differentiable at those points where graph of the function has holes or sharp edges. Let us consider the function f ( x) = | x − 1|. It is not differentiable at x = 1. Since, f ( x) has sharp edge at x = 1. f (x) = – x + 1 Lf ' (x) = – 1

Continuity of a Function at a Point

X'

A function f ( x) is said to be continuous at a point x = a of its domain if and only if it satisfies the following conditions (i) f (a) exists, where (‘a’ lies in the domain of f (ii) lim f ( x) exist, i.e. lim− f ( x) = lim+ f ( x) x→ a

or

x→ a

x→ a

LHL = RHL

f (x) = x – 1 Rf ' (x) = 1

Y

O

1

2

3

X

Y'

(graph of f ( x) describe differentiability)

Differentiability of a Function at a Point A function f is said to be differentiable at x = c, if left hand

(iii) lim f ( x) = f (a), f ( x) is said to be x→ a

left continuous at x = a, if lim− f ( x) = f (a) x→ a

right continuous at x = a, if lim+ f ( x) = f (a) x→ a

and right hand derivatives at c exist and are equal. ●

Right hand derivative of f ( x) at x = a denoted by f ′ (a + 0) or f ′ (a + ) is lim h→ 0

f (a + h) − f (a) ⋅ h

107

DAY ●

Left hand derivative of f ( x) at x = a denoted by f ′ (a − 0) or f ′ (a− ) is lim h→ 0

●

●

f (a − h) − f (a) . −h

Thus, f is said to be differentiable at x = a, if f ′ (a + 0) = f ′ (a − 0) = finite. The common limit is called the derivative of f ( x) at x = a f ( x) − f (a) denoted by f ′ (a). i.e. f ′ (a) = lim . x→ a x−a

Results on Differentiability (i) Every polynomial, constant and exponential function is differentiable at each x ∈ R. (ii) The logarithmic, trigonometric and inverse trigonometric function are differentiable in their domain. (iii) The sum, difference, product and quotient of two differentiable functions is differentiable. (iv) Every differentiable function is continuous but converse may or may not be true.

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 lim | x | [cos x ] , where [. ] is the greatest integer function, is x→ 0

(a) 1 (c) Does not exist

9 lim

x→ 0

(b) 0 (d) None of these

(a) 4

2 Let f : R → [ 0, ∞ ) be such that lim f ( x ) exists and x→ 5

(b) 0

(c) 1

(a) 0

(d) 2

2 x  m = (where [⋅] denotes greatest integer x  5  n function), then m + n (where m, n are relatively prime) is (c) 5

(c) (1, − 1)

(b)

3 5

4 7

(d) −

(a) 0

lim

x→ 3

(a) − (2)3 / 2

(c) log2

x  2

(a) 1

x  4

(d) e

x  8

 x  is  2n 

x (c) sin x

2 3

(c)

(c) 0

(d) 1

3 2

(d) None of these

lim

x → π /2

(d) 3

j

JEE Mains 2013

j

JEE Mains 2017

1 (b) − 2 (d) 0

(c) 2

15

(c) (2)− 3 / 2

1  −1  x + 1  π  is tan   −  2x + 1 4  x 

(a) 1

(d) None of these

 f ( x + T ) + 2f ( x + 2T )+ ...+nf ( x + nT )  lim n   is equal to n→ ∞  f ( x + T ) + 4f ( x + 4T )+ ...+ n 2f ( x + n 2T ) (b)

(d) π

tan x f (x ) , then lim is equal to x x→ 0 x 2 1

(b) (2)1 / 2 x→ 0

8 If f is periodic with period T and f ( x ) > 0∀x ∈ R , then

(a) 2

JEE Mains 2014

1 − cos( x 2 − 10x + 21) ( x − 3)

14 The value of lim

4

7 The value of lim cos   cos   cos   … cos  n→ ∞ sinx (b) x

(c) − π

(b) −1

20 7



1 (b) 2

cos x x2 1

j

1 n

13 The limit of the following is

(c) −



1 2

(d) n +

(c) n

(b) 1

(a) 3

6 lim  x + x + x − x  is equal to x→ ∞

n+1 n

sin( π cos 2 x ) is equal to x2

π (a) 2

(d) (0, 1)

3 ⋅ 2 n +1 − 4 ⋅ 5 n +1 5 lim is equal to n→ ∞ 5 ⋅ 2n + 7 ⋅ 5n (a) 0

x→ 0

(b)

sin x 12 If f ( x ) = x 3 2x

x2 +1  lim  − α x − β = 0 are respectively x→ ∞  x + 1  (b) (−1, 1)

11 lim

(d) 6

4 The value of the constant α and β such that

(a) (1, 1)

(d)

x→ 0

x→ ∞

(b) 7

(c) 2

10 If lim

3 If lim

(a) 2

(b) 3

[(a − n ) nx − tan x ] sin nx = 0, where n is non-zero x2 real number, then a is equal to

[f ( x )]2 − 9 lim = 0. Then, lim f ( x ) is equal to x→ 5 x→ 5 | x − 5| (a) 3

(1 − cos 2x )( 3 + cos x ) is equal to x tan 4x j JEE Mains 2015, 13

cot x − cos x equals ( π − 2 x )3

1 (a) 24 1 (c) 8

1 (b) 16 1 (d) 4

TEN

108 16 If m and n are positive integers, then (cos x )1/ m − (cos x )1/ n equals to lim x→ 0 x2 (a) m − n

17 lim

x→ ∞

(b)

1 1 − n m

(c)

m −n 2mn

(d) None of these

cot −1( x + 1 − x ) is equal to x   −1  2 x + 1 sec       x −1 

(a) 1

27 Let f ( x ) = 1 + | x − 2 | and g ( x ) = 1 − | x | , then the set of all points, where fog is discontinuous, is (a) {0, 2 }

π (c) 2

(b) 0

(d) Does not exists

ln(cos 2x ) sin2 2x and , q = lim 2 x→ 0 x → 0 x (1 − e x ) 3x x −x . Then p, q , r satisfy r = lim x →1 ln x

18 Let p = lim

(a) p < q < r

(b) q < r < p

(c) p < r < q

(d) q < p < r

19 Let p = lim (1 + tan2 x )1/ 2 x , then log p is equal to x → 0+

(a) 2

j

1 (c) 2

(b) 1

 3x − 4  20 The value of lim   x → ∞  3x + 2 (a) e −1 / 3

 

21 If lim 1 + x→ ∞

x +1

(b) e −2 / 3

a b +  x x2

JEE Mains 2016 1 (d) 4

3

is (c) e −1

(d) e −2

= e 2 , then the values of a and b are (b) a = 1, b ∈ R (d) a = 1, b = 2

22 If f ′ ( 2) = 6 and f ′ (1) = 4 , then lim

h→ 0

(b) –3/2

(c) 3/2

(d) Does not exist

23 Let f (a ) = g (a ) = k and their nth derivatives f n (a ), g n (a ) exist and are not equal for some n. Further, if f (a )g ( x ) − f (a ) − g (a )f ( x ) + g (a ) lim = 4, x→ a g( x ) − f ( x ) then the value of k is (a) 4

(b) 2

(c) 1

(d) 0

24 If f : R → R be such that f (1) = 3 and f ′ (1) = 6. Then,  f (1 + x ) lim   x→ 0  f (1)  (a) 1

1/ x

(b) e1 / 2

(c) e 2

(d) e 3

 sin x , x is rational 25 Let f ( x ) =  , then set of points, 2 − sin x , x is irrational 2

where f ( x ) is continuous, is (b) a null set (d) set of all rational numbers

26 Let f :[a, b ] → R be any function which is such that f ( x ) is rational for irrational x and f ( x ) is irrational for rational x. Then, in [a, b ]

(d) an empty set

3π 7π 2π 5π 5π 7π (b)  ,  (c)  ,  (d)  ,   4 4   3 3   4 3 

29 If f ( x ) is differentiable at x = 1 and lim

h→ 0

1 f (1 + h ) = 5, then h

f ′ (1) is equal to (a) 6

(b) 5

30 If f ( x ) = 3x

(c) 4

(d) 3

− 7x + 5x − 21x + 3x − 7, then the f (1 − h ) − f (1) value of lim is h→ 0 h3 + 3h (a)

10

53 3

8

(b)

6

22 3

3

2

(c) 13

(d)

22 13

31 Let f ( 2) = 4 and f ′ ( 2) = 4. Then, x f ( 2) − 2f ( x ) is given by x −2

(a) 2

(b) –2

(c) – 4

(d) 3

32 If f is a real-valued differentiable function satisfying | f ( x ) − f ( y )| ≤ ( x − y )2 ; x , y ∈ R and f ( 0) = 0, then f (1) is equal to (a) 1

(b) 2

(c) 0

(d) –1

log(a + x ) − log a log x − 1 33 If lim + k lim = 1, then x→ 0 x → 0 x x −e 1 (a) k = e  1 −   a (c) k = e (2 − a)

(b) k = e (1 + a) (d) Equality is not possible

34 The left hand derivative of f ( x ) = [ x ] sin ( πx ) at x = k , k is an integer, is (a) (− 1)k (k − 1) π (c) (− 1)k kπ

is equal to

π (a) (2n + 1) ,n ∈ I  2   (c) {nπ,n ∈ I }

π 3π (a)  ,  2 4 

lim

equal to

(c) {0}

JEE Mains 2013

points of discontinuity of f {g ( x )} in ( 0 , 2π ) is

x→ 2

f ( 2h + 2 + h 2 ) − f ( 2) is f (h − h 2 + 1) − f (1)

(b) {0, 1, 2 }

j

− 1 , x < 0  28 If f ( x ) =  0 , x = 0 and g ( x ) = sin x + cos x , then the 1 , x > 0 

2x

(a) a ∈ R, b ∈ R (c) a ∈ R, b = 2

(a) 3

(a) f is discontinuous everywhere (b) f is continuous only at x = 0 (c) f is continuous for all irrational x and discontinuous for all rational x (d) f is continuous for all rational x and discontinuous for all irrational x

(b) (− 1)k − 1 (k − 1) π (d) (− 1)k − 1 kπ

 ex, x ≤0 , then |1 − x | , x > 0

35 If f ( x ) =  (a) (b) (c) (d)

f (x) is differentiable at x = 0 f (x) is continuous at x = 0, 1 f (x) is differentiable at x = 1 None of the above

 

36 If f ( x ) = xe 

 1 1 − +  x | x |

0

, x ≠ 0 , then f ( x ) is ,x = 0

109

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

continuous as well as differentiable for all x continuous for all x but not differentiable at x = 0 neither differentiable nor continuous at x = 0 discontinuous everywhere

37 The set of points, where f ( x ) = (a) (− ∞, − 1) ∪ (−1, ∞) (c) (0, ∞)

x is differentiable, is 1+ |x|

(b) (− ∞, ∞) (d) (− ∞, 0) ∪ (0, ∞)

k x + 1 , 0 ≤ x ≤ 3 is differentiable,  mx + 2 , 3 < x ≤ 5

39 If the function g ( x ) = 

then the value of k + m is (c)

10 5

(b) 3

(c) 5

(d) 4

(a) f is everywhere differentiable (b) f is everywhere continuous but not differentiable at x = nπ,n ∈Z (c) f is everywhere continuous but not differentiable at π x = (2n + 1) ,n ∈ Z 2 (d) None of the above

(a) g (x) is discontinuous at x = π (b) g (x) is continuous for x ∈ [0, ∞] (c) g (x) is differentiable at x = π (d) g (x) is differentiable for x ∈ [0, ∞]

16 5

the greatest integer function. Then, the total number of points, where f ( x ) is non-differentiable, is

42 If f ( x ) = | sin x | , then

min{f (t ): 0 ≤ t ≤ x } , x ∈ [ 0, π ] then g( x ) =  (sin x ) − 1 , x >π 

(b)

41 If f ( x ) = [sin x ] + [cos x ], x ∈ [ 0, 2π ], where [. ] denotes

(a) 2

38 Let f ( x ) = cos x and

(a) 2

(a) f (x) is continuous and differentiable (b) f (x) is continuous but not differentiable (c) f is not continuous but differentiable (d) f is neither continuous nor differentiable

(d) 4

sin(cos −1 x ) + cos(sin−1 x ), x ≤ 0 −1 −1 sin(cos x ) − cos(sin x ), x > 0

40 If f ( x ) = 

then at x = 0

43 Statement I f ( x ) = | log x | is differentiable at x = 1 . Statement II Both log x and – log x are differentiable at x = 1. (a) Statement I is false, Statement II is true (b) Statemnt I is true, Statement II is true; Statement II is a correct explanation of Statement I (c) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (d) Statement I is true, Statement II is false

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 For each t ∈ R , let [t ] be the greatest integer less than or 1   2  15  equal to t. Then, lim+ x  + +…+  x  x→ 0   x   x 

JEE Mains 2018 (b) is equal to 15 (d) does not exist (in R) j

(a) is equal to 0 (c) is equal to 120 x

 x 2 + 5x + 3  is equal to  x2 + x + 2 

2 lim  x→ ∞

(a) e 4

(b) e 2

(c) e 3

(d) e

1 − cos (ax 2 + bx + c ) is equal to x→ α ( x − α )2 lim

1 a2 (α − β)2 (b) − (α − β)2 (c) 0 2 2

(d)

4 lim sin[ π n + 1] is equal to n→ ∞

(b) 0 (d) None of these

5 If x > 0 and g is a bounded function, then lim

n→ ∞

f ( x ) ⋅ e nx + g ( x ) is equal to e nx + 1

 a x + bx + c x     3

2/x

6 The value of lim  x→ 0 (a) (abc)3 (c) (abc)1 / 3

( where a , b , c > 0) is

(b) abc (d) None of these

 3x − x 3  1 − x 2  and g ( x ) = cos −1   , then 2   1 − 3x  1 + x 2  f ( x ) − f (a ) 1 lim , where 0 < a < , is equal to x → a g ( x ) − g (a ) 2 (a)

2

(a) ∞ (c) Does not exist

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

7 If f ( x ) = cot −1 

3 If α and β are the distinct roots of ax 2 + bx + c = 0, then

(a)

(a) 0 (c) g (x)

a2 (α − β)2 2

3 3 3 (b) (c) 2 (1 + a 2 ) 2 (1 + x 2 ) 2

(d) −

3 2

8 If f : R → R is a function defined by

 2x − 1 f ( x ) = [ x ] cos   π, where [ x ] denotes the greatest  2  integer function, then f is (a) (b) (c) (d)

continuous for every real x discontinuous only at x = 0 discontinuous only at non-zero integral values of x continuous only at x = 0

TEN

110

18 The function f ( x ) is discontinuous only at x = 0 such that

9 Let f be a composite function of x defined by

f 2 ( x ) = 1 ∀x ∈ R. The total number of such function is

1 1 f (u ) = 2 , u( x ) = ⋅ u +u−2 x −1 Then, the number of points x, where f is discontinuous, is JEE Mains 2013

j

(a) 4

(b) 3

(c) 2

(d) 1

10 If f : R → R be a positive increasing function with lim

x→ ∞

f ( 3x ) f ( 2x ) is equal to = 1 . Then, lim x → ∞ f (x ) f (x )

(a) 1

(b) 2/3

(c) 3/2

(d) 3

 π 3π 

11 Let f ( x ) = max {tan x , sin x , cos x }, where x ∈ − , .  2 2  Then, the number of points of non-differentiability is (a) 1

(b) 3

(c) 0

(d) 2

12 Let S = {t ∈ R : f ( x ) = | x − π|⋅(e| x | − 1)} sin| x | is not differentiable at t}. Then, the set S is equal to (a) φ (an empty set) (c) { π }

 π x +   3

n

(b) {0} (d) {0, π }

j

JEE Mains 2018

n→ ∞

 π x n −1 +    3

n −1

, where n is an even integer,

Then which of the following is incorrect? π π (a) If f:  , ∞  →  , ∞  , then f is both one-one and onto  3   3  (b) f (x) = f (− x) has infinitely many solutions (c) f (x) is one-one for all x ∈ R (d) None of these n −1

x . Then, 14 Let f ( x ) = lim ∑ n→ ∞ k = 0 (kx + 1){(k + 1)x + 1} (a) f is continuous but not differentiable at x = 0 (b) f is differentiable at x = 0 (c) f is neither continuous nor differentiable at x = 0 (d) None of the above

15 If x1, x 2 , x 3 ,..., x 4 are the roots of x n + ax + b = 0, then the value of ( x1 − x 2 )( x1 − x 3 )...( x1 − x n ) is equal to (b) nx1n − 1 + a (d) nx1n

(a) nx1 + b (c) nx1n − 1

   16 If *lim x ln  x→ ∞    

1 x 0 1

−b 1 x 0

 c   −1  = −4, where a, b, c are real  a / x  

numbers, then (a) a = 1,b ∈ R , c = −1 (c) a = 1,b = 1,c ∈ R

 

17 If lim 1 + x +  x→ 0

(a) e

1/ x

f (x ) x 

(b) e 2

(b) a ∈ R,b = 2,c = 4 (d) a ∈ R,b = 1,c = 4 1/ x

 f (x ) = e 3 , then lim 1 + x→ 0  x   (c) e 3

(b) 3 (d) None of these

19 Let f ( x ) = x | x | and g ( x ) = sin x Statement I gof is differentiable at x = 0 and its derivative is continuous at that point. Statement II gof is twice differentiable at x = 0. (a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I (c) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (d) Statement I is true, Statement II is false

20 Statement I The function f ( x ) = ( 3x 1 x = and 2

− 1) | 4x 2 − 12x + 5 | cos πx is differentiable at 5 . 2

π = 0, ∀ n ∈ I . 2 (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 cos(2n + 1)

n

13 If f ( x ) = lim

(a) 2 (c) 6

is equal to

(d) None of these

21 Define f ( x ) as the product of two real functions f1( x ) = x , x ∈ IR

 sin 1 , if x ≠ 0 and f2 ( x ) =  as follows x  0 , if x = 0  f ( x ) ⋅ f2 ( x ), if x ≠ 0 f (x ) =  1 0 , if x = 0  Statement I f ( x ) is continuous on IR. Statement II f1( x ) and f2 ( x ) are continuous on IR. (a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is true; Statement II is correct explanation of Statement I. (c) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (d) Statement I is true, Statement II is false

22 Consider the function f ( x ) = | x − 2 | + | x − 5 | , x ∈ R .

Statement I f ′ ( 4) = 0 Statement II f is continuous in [ 2, 5 ] and differentiable in ( 2, 5) and f ( 2) = f ( 5 ). (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 correct explanation for Statement I (c) Statement I is true; Statement II is false (d) Statement I is false; Statement II is true

111

DAY

ANSWERS SESSION 1

SESSION 2

1 (a)

2 (a)

3 (b)

4 (c)

5 (d)

6 (b)

7 (b)

8 (c)

9 (c)

10 (d)

11 (d)

12 (d)

13 (a)

14 (b)

15 (b)

16 (c)

17 (a)

18 (d)

19 (c)

20 (b)

21 (b)

22 (a)

23 (a)

24 (c)

25 (c)

26 (a)

27 (d)

28 (b)

29 (b)

30 (a)

31 (c)

32 (c)

33 (a)

34 (a)

35 (b)

36 (b)

37 (b)

38 (b)

39 (a)

40 (d)

41 (c)

42 (b)

43 (a)

1 (c) 11 (b) 21 (d)

2 (a) 12 (d) 22 (b)

3 (d) 13 (c)

4 (b) 14 (c)

5 (b) 15 (b)

6 (d) 16 (d)

7 (d) 17 (b)

8 (a) 18 (c)

9 (b) 19 (b)

10 (a) 20 (a)

Hints and Explanations SESSION 1

n

1 RHL = lim ( x )0 = 1

= lim

x → 0+

n→ ∞

2 6 ⋅   − 20  5 n

0 − 20 20 =− 0+ 7 7

=

LHL = lim− (− x )0 = lim− 1 = 1

2 5⋅   + 7  5

RHL = LHL lim | x |[cos x ] = 1

  6 xlim x+ x+ x − x →∞

x→ 0

Q ∴

x→ 0

x→ 0

2 Given, lim f ( x ) exists and x→ 5 lim

[ f ( x )]2 − 9 | x − 5|

x→ 5



= lim

x→ ∞

x+

x − x

x+

x+

x +

x+

x

x→ ∞

x+

x +

x+

2

⇒ ∴

8 Clearly, f ( x + T ) = f ( x + 2T ) = ...

x+

= lim

x→ 5

 lim [ f ( x )] = 9  x→ 5 

n→ ∞



=0

⇒ lim[ f ( x )]2 − 9 = 0

sin x x / 2n ⋅ x sin( x / 2n ) sin x = x = lim

= lim

x→ ∞

lim f ( x ) = 3, − 3

x

x

1 + x −1 / 2 1+

x

−1

+ x

−3 / 2

+1

=

1 2

= f ( x + nT ) = f ( x )  f ( x + T ) + 2 f ( x + 2T ) + ...    + nf ( x + nT )  ∴lim n   f ( x + T ) + 4 f ( x + 4T )  n→ ∞   2 2 + ...+ n f ( x + x T )  nf ( x )(1 + 2 + 3+ ...+ n ) = lim n→ ∞ f ( x )(1 + 22 + 32 + ...+ n2 ) n(n + 1) n     3 2 = lim = n→ ∞ n( n + 1)(2n + 1) 2 6

x→ 5

But f : R → [0, ∞ ) ∴ Range of f ( x ) ≥ 0 ⇒ lim f ( x ) = 3

7 We know that,

cos A ⋅ cos 2 A ⋅ cos 4 A K cos 2n −1 A=

x→ 5

2  x 2x x  3 Clearly, lim = lim  −    x→ ∞ x  5    x→ ∞ x  5  5   2 2 = −0= 5 5 ∴ m+ n =7  x2 + 1  4 xlim − α x − β = 0  →∞ x + 1   ⇒ lim

x→ ∞

x (1 − α ) − x (α + β ) + 1 − β =0 x+1

∴ ⇒

2

1 − α = 0, α + β = 0 α = 1, β = − 1

x Take A = n , 2 x x then cos  n  ⋅ cos  n − 1  ... 2  2 

n→ ∞

3 ⋅ 2n +1 − 4 ⋅ 5n +1 6 ⋅ 2n − 20 ⋅ 5n = lim n n n → ∞ 5⋅ 2 + 7 ⋅ 5 5⋅ 2n + 7 ⋅ 5n

9 We have, lim x→ 0

2sin 2 x(3 + cos x ) tan 4 x x× × 4x 4x 2 (3 + cos x ) 2sin x = lim × lim x→ 0 x→ 0 x2 4 x→ 0

sin x x 2 sin  n  2  x x ∴ limcos   cos   ... n→ ∞  2  4 =

×

n

x cos  n − 1 2 = lim

n→ ∞

sin x x 2n sin  n  2 

(1 − cos 2 x )(3 + cos x ) x tan 4 x

= lim

x x cos   cos    4  2

5 Clearly, lim

sin 2n A 2n sin A

1 tan 4 x lim x→ 0 4x

4 ×1 = 2 4 Q lim sin θ = 1 and lim tan θ = 1  x→ 0  x → 2 θ  θ

= 2×  ⋅ cos  x    n  2 

10 Since, tan x  sin nx lim (a − n ) n − ⋅ =0 x→ 0  x  x 

TEN

112 ⇒ [(a − n ) n − 1] n = 0 ⇒ (a − n ) n = 1 a= n+

sin θ = π Qlim = 1  θ → 0 θ 

12 Q f ( x ) = x( x − 1)sin x − ( x3 − 2 x2 ) cos x − x3 tan x = x sin x − x cos x − x tan x + 2 x2 cos x − x sin x f ( x) ∴ lim 2 = lim x→ 0 x x→ 0  sin x − x cos x − x tan x    sin x   + 2cos x −  x  2

3

1 − cos( x − 10 x + 21) 2

=

= =

=

( x − 3) ( x − 3) ( x − 7) 2 sin 2 lim x→3 ( x − 3) ( x − 3) ( x − 7) 2 sin ( x − 7) 2 lim ⋅ x→3 ( x − 3) ⋅ ( x − 7) 2 2 lim ( x − 7) x→3 ( x − 3) ( x − 7) sin 1 2 ⋅ lim × x→3 ( x − 3) ( x − 7) 2 2 − (2)3 /2

 x+1 π  −  x 4  2x + 1 x + 1     −1 = lim  tan −1   − tan (1) x→ 0  2x + 1    x + 1 − 1 1  2x + 1  = lim ⋅ tan −1  x→ 0 x x+1 1 +   2x + 1   x + 1 − 2x − 1  1 = lim ⋅ tan −1   x→ 0 x  2x + 1 + x + 1  

1 −1 14 lim  tan  x→ 0

 −x 1 = lim ⋅ tan −1  x→ 0 x  4x +

  2

 form 0   0 

[using L’ Hospital rule] 1

= lim x→ 0

1+

=

=

=

x2 (4 x + 2)2

=

=

x→ 0

cot x − cos x ( π − 2 x )3 1 cos x(1 − sin x ) lim ⋅ 3 x → π /2 8 π sin x  − x  2  π π     cos  − h  1 − sin  − h  2   2   1 lim ⋅ 3 h→ 0 8 π π π sin  − h   − + h  2  2  2 sin h (1 − cos h ) 1 lim 8 h→ 0 cos h ⋅ h3 h sin h  2sin2   1 2 lim 8 h → 0 cos h ⋅ h3 h sin h ⋅ sin2    2 1 lim 4 h→ 0 h3 cos h 2

 sin h   sin h   1 2 ⋅ 1 ⋅ 1 lim    4 h → 0  h   h  cos h 4  2  1 1 1 = × = 4 4 16 =

1/m

(cos x ) 16 lim x→ 0

− (cos x )1 / n x2

1

(cos x )m = lim x→ 0 x2

−

−1

⋅ lim x→ 0

1

 1 − 2sin2 x  m    2 = lim 2 x→ 0 x 1 1 = lim − 2 −  x→ 0  m n

−

1 (cos x )1 / n

1 n

x→1

= lim x→1

= lim x→1

x (1 − x ) 1 + x − 1  ln   ⋅ ( x − 1)  x −1  x (1 − x ) 1 + ( x − 1)  ln   ⋅ ( x − 1)(1 +  x −1 

x 2 = m−n x2 2mn

sin2

= lim( x + 1 − x→ ∞

x→ ∞

1 x+1+

1 2 Hence, q < p < r . 1

19 Given, p = lim (1 + tan2 x→ 0

lim

= ex→0

tan +

x

2x

1

lim

= e2 x→0

 tan x     x 

+

1

= e2 1

1 2

∴ log p = log e 2 =  3x − 4   3x + 2

20 xlim  →∞

x +1 3

 6  = lim  1 −  x→ ∞ 3x + 2 

x +1 3

x +1 3

3x + 2 3x + 2    6  −6   = lim  1 −  x→ ∞   3x + 2    3x + 2

x→ ∞

=0

x

 2+ 1    2x + 1  x  =∞ lim    = lim x→ ∞  x − 1  x→ ∞ 1 − 1 / x     cot ( x + 1 − 2x + 1 sec    x −1 

x) x

=

−1

=

x +1 3

= e −2 / 3

1 + a + b  21 Now, xlim  2 →∞  

=e

−1

cot (0) sec −1 (∞ ) π /2 =1 π /2

3x ln(1 + cos 2 x − 1) cos 2 x − 1 = lim ⋅ x→ 0 (cos 2 x − 1) 3 x2

2x

x

a b = lim  1 + + 2  x→ ∞  x x 

x

x→ ∞

⋅

−2 ( x + 1 )

= lim e

x

∴ lim

2

 −2( x + 1) − 2  = Q xlim →∞ 3x + 2 3  

and

−1

x ) 2x

+

(1∞ form) 2

x) x

x)

=−

17 Clearly, = lim

x−x ln(1 + x − 1)

−6

−1

ln(1 + cos 2 x − 1) 18 Clearly, p = lim 2 x→ 0   4x + 2 − 4x   × −   2   (4 x + 2)  

and r = lim

sin2 2 x 4 x2 ⋅ = −4 2 4x x (1 − e x )

 3x + 2 − 6 = lim   x→ ∞  3x + 2 

1 n

2 3

q = lim

15 xlim → π /2

3

=2−1=1

13 lim x→3

=−

x→ 0

1 n sin( π cos 2 x ) sin( π − π sin2 x ) 11 lim = lim x→ 0 x→ 0 x2 x2 2 sin( π sin x ) = lim x→ 0 x2 [Q sin( π − θ) = sin θ ] sin( π sin 2 x )  sin2 x  = lim × ( π )  2 x→ 0 π sin x  x2  ∴

(+2) x2 + (4 x + 2)2 (+2) −2 −1 =− = = 0 + (0 + 2)2 4 2 = lim −

  2x    

b   x2  b  +  x x2 

a

x a

+

a b  lim 2 x  +   x x2 

x →∞

Q lim (1 + x )1 / x = e  = e 2 a  x→ ∞  a b But lim  1 + + 2  x→ ∞  x x  ⇒ ⇒ and

e 2a = e 2 a=1 b ∈R

2x

= e2

113

DAY 22 lim h→ 0

23 lim x→ a

f (2h + 2 + h2 ) − f (2) f (h − h2 + 1) − f (1) f ′ (2h + 2 + h2 )(2 + 2h ) = lim h → 0 f ′ ( h − h2 + 1)(1 − 2h ) f ′ (2) × 2 = f ′ (1) × 1 6×2 = =3 4×1 f (a) g ( x ) − f ( a) − g ( a) f ( x ) + g ( a) =4 g( x) − f ( x)

Applying L’Hospital rule, we get f (a) g ′ ( x ) − g (a)f ′ ( x ) lim =4 x→ a g ′( x ) − f ′( x ) k g ′( x ) − k f ′( x ) ⇒ lim =4 x→ a g ′( x ) − f ′( x ) ∴

k =4  f (1 + x )   f (1) 

1/x

24 Let y = 

⇒ log y =

1 [log f (1 + x ) − log f (1)] x

0 < x < 3 π /4 x = 3 π / 4 , 7 π /4  − 1 , 3 π / 4 < x < 7 π /4   1,

28 f { g ( x )} =  0,

or 7 π /4 < x < 2 Clearly, [ f {g ( x )}] is not continuous at 3 π 7π x= , . 4 4 f (1 + h ) − f (1) 29 f ′(1) = lim h→ 0 h f (1 + h ) f (1) = lim − lim h→ 0 h→ 0 h h f (1 + h ) f (1) Since, lim must = 5, so lim h→ 0 h→ 0 h h f (1) be finite as f ′(1) exists and lim can h→ 0 h be finite only, if f (1) = 0 and f (1) lim = 0. h→ 0 h f (1 + h ) ∴ f ′ (1) = lim =5 h→ 0 h f (1 − h ) − f (1) 30 lim h→ 0 h3 + 3h = lim

  1 ⇒ lim log y = lim  f ′ (1 + x ) x→ 0 x→ 0  f (1 + x )  ⇒ ∴

lim log y = x→ 0

h→ 0

−1 53 = f ′ (1) ⋅   =  3  3

f ′ (1) 6 = f (1) 3

x f (2) − 2 f ( x ) x−2 x f (2) − 2 f (2) + 2 f (2) − 2 f ( x ) = lim x→2 x−2 f (2)( x − 2) − 2 { f ( x ) − f (2)} = lim x→2 x−2 f ( x ) − f (2) = f (2) − 2 lim x→2 x−2

31 lim x→2

lim y = e 2 x→ 0

25 Clearly, f ( x ) is continuous only when sin2 x = − sin2 x ⇒ 2sin2 x = 0 ⇒ x = nπ

26 We have,

rational, if x ∉ Q in[a,b] f ( x ) =  irrational , if x ∈ Q in[a,b]  Let C ∈ [a,b] and c ∈ Q . Then, f (c ) = irrational and lim f ( x ) = lim f (c + h ) = rational or x→c h→ 0 irrational Thus, f is discontinuous everywhere.

1 + x, x < 0 27 g ( x ) =  1 − x,

1 + | x − 1|, ∴ f {g ( x )} =  1  + |− x − 1|,

x< 0 x≥ 0

x, x < 0 1, x ≥ 0 x< 0 x≥ 0

It is a polynomial function, so it is continuous in everywhere except at x = 0. Now, LHL = lim 2 − x = 2, x→ 0

RHL = lim 2 + x = 2 x→ 0

 f ( x ) − f (a) Q f ′ ( x ) = lim x→ a x − a   = f (2) − 2 f ′ (2) = 4−2× 4= −4

32 Q ∴

x≥ 0

1+ 1− =  1 + x + 2 − x, =  2 + x,

f (1 − h ) − f (1) −1 ⋅ 2 −h h +3

Also, f (0) = 2 + 0 = 2 Hence, it is continuous everywhere.

⇒

| f ( x ) − f ( y )| ≤ ( x − y ) | f ( x ) − f ( y )| lim ≤ lim | x − y | x→ y x→ y |x − y| 2

| f ′ ( y )| ≤ 0 ⇒

f ′( y ) = 0

34 If x is just less than k, then [ x] = k − 1 ∴

f ( x ) = (k − 1) sin π x

LHD of f ( x ) = lim x→ k

(k − 1) sin πx − k sin πk x−k (k − 1) sin πx , = lim x→ k x−k where x = k − h (k − 1)sin π (k − h ) = lim h→ 0 −h = (k − 1)(−1)k π x x≤ 0  e , 35 f ( x ) = 1 − x, 0 < x ≤ 1  x − 1, x > 1 

f (0 + h ) − f (0) h 1− h−1 = lim = −1 h→ 0 h f (0 − h ) − f (0) Lf ′ (0) = lim h→ 0 −h e−h − 1 = lim =1 h→ 0 −h So, it is not differentiable at x = 0. Rf ′ (0) = lim h→ 0

Similarly, it is not differentiable at x = 1 but it is continuous at x = 0 and 1. h 36 RHL = lim (0 + h ) e −2 / h = lim 2 / h = 0 h→ 0 h→ 0 e LHL = lim (0 − h ) e

1 1  − −   h h

=0

h→ 0

Hence, f ( x ) is continuous at x = 0. Now, Rf ′ ( x ) = lim

(0 + h ) e

h→ 0

= lim e − 2 / h h→ 0

1 1  − +   h h

1 1  − −  h

(0 − h ) e  h and Lf ′ ( x ) = lim h→ 0 −h h→ 0

∴ Lf ′ ( x ) ≠ Rf ′ ( x ) Hence, f ( x ) is not differentiable at x = 0. g( x) 37 Since, f ( x ) = x = [say] 1 + | x| h( x )

⇒

f ( y ) = Constant

⇒

f( y) = 0

(− ∞, 0) ∪ (0, ∞ ).

⇒

f (1) = 0

33 Let f ( x ) = log x 1 x Therefore, given function = f ′ (a) + k f ′ (e ) = 1 1 k ⇒ + =1 a e a − 1 ⇒ k = e    a  ⇒

f ′( x ) =

−0

= lim e − 0 = 1

It is clear that g ( x ) and h( x ) are differentiable on (− ∞, ∞ ) and

[Q f (0) = 0, given]

−0

h =∞

Now, x −0 f ( x ) − f (0) 1 + | x| lim = lim =1 x→ 0 x→ 0 x−0 x Hence, f ( x ) is differentiable on (− ∞, ∞ ). x ∈ [0, π] x> π sin x − 1, Also, g ( π − ) = g ( π ) = g ( π + ) = −1 and g ′( π − ) ≠ g ′( π + ) ∴ g is continuous at x = π but not differentiable at x = π

cos x , 38 Clearly, g ( x ) = 

TEN

114 39 Since, g ( x ) is differentiable ⇒ g ( x ) must be continuous.  k x + 1, 0 ≤ x ≤ 3 g ( x) =   mx + 2, 3 < x ≤ 5 At, x = 3, RHL = 3 m + 2 and at x = 3, LHL = 2k ∴

2k = 3 m + 2

Also, ∴

k  , 0≤ x < 3 g ′( x ) = 2 x + 1 m , 3< x ≤ 5  L(g ′ (3)) =

k 4

and R {g ′ (3)} = m ⇒

k =m 4

k = 4m

i.e.

On solving Eqs (i) and (ii), we get 8 2 k = ,m = 5 5 ⇒

− log x, x < 1 43 f ( x ) =   log x, x ≥ 1

SESSION 2 1 We have,

1 15   1 lim x    +   + …+     x     x   x 

x→ 0 +

We know, [ x] = x − { x} 1 = 1 − 1 ∴    x  x  x  n n n Similarly,   = −    x  x  x 

+

π , π, 2 π and [cos x] is 2 non-differentiable at π 3π x = 0, , and 2 π. 2 2 Thus, f ( x ) is definitely 3π non-differentiable at x = π, , 0. 2 Also, π π f   = 1, f  − 0 = 0,  2 2  f (2 π ) = 1, f (2 π − 0) = − 1 Thus, f ( x ) is also non-differentiable at π and 2π. x= 2 x=

42 Let u( x ) = sin x

Q u( x ) = sin x is a continuous function and v ( x ) = | x| is a continuous function. ∴ f ( x ) = vou ( x ) is also continuous everywhere but v ( x ) is not differentiable at x = 0 ⇒ f ( x ) is not differentiable where sin x = 0 ⇒ x = nπ , n ∈ Z Hence, f ( x ) is continuous everywhere but not differentiable at x = nπ, n ∈ Z .

a2 (α − β )2 2

 1 4 nlim sin  nπ  1 + 2  →∞ 

function. n→ ∞

1  4x + 1  = lim   1 + 2  x→ ∞  x + x + 2   lim

x →∞

=e

2 + x x2

 ax + b x + c x  y = lim   x→ 0 3  

2/ x

 ax + b x + c x  2 log   x→ 0 x 3  

⇒ log y = lim

log(ax + b x + c x ) − log 3 x Apply L’Hospital’s rule, ax log a + b x log b + c x log c ax + b x + c x = 2 lim x→ 0 1 log y = log (abc ) 2 /3 ⇒ y = (abc ) 2 /3 x→ 0

x

( 4x + 1 ) x2 + x + 2

( 4 x +1 )x

 x2 + x +2   

1

= e4 x  1 + 1  = e    Q lim  x→ 0  x  

1 − cos (ax + bx + c ) ( x − α )2 2

3 Now, xlim →α

6 Let

= 2 lim

x

 1  4 +  x

1 +

f ( x ) ⋅ e nx + g ( x ) e nx + 1

   f ( x) g( x)  = lim  + nx  n→ ∞ e + 1  1 +  1  nx e    f ( x) Finite = + = f ( x) 1+ 0 ∞

 15     x 

= 120 − 0 = 120 n   Q 0 ≤   < 1, therefore    x   n n  0 ≤ x   < x ⇒ lim x   = 0     + x→ 0    x  x

  2

  

5 Given, x > 0 and g is a bounded

2 2 15 − + … − x  x x

 4x + 1 = lim  1 + 2 x→ ∞ x + x+ 

n 

 1  1 = lim (−1)n sin π  − +K ∞ 3 n→ ∞  2n 8 n 

Then, lim

 x2 + 5x + 3  2 Now, xlim  2  →∞  x + x+2

1 /2

    1 1 = limsin  nπ  1 + − + K∞  2 4 n→ ∞ 2 n 8 n    π π   = lim sin  nπ + − + K∞ n→ ∞ 2n 8 n3  

1 + 2 + 3+ ...+15) − x = lim( x→ 0 + 1 2  15     +   + ... +     x  x  x 

v( x) = |x | ∴ f ( x ) = vou ( x ) = v (u( x )) = v (sin x ) =|sin x|

=

Q lim sin x = 1  x → 0 x 

a2 ( x − β )2 2

=0

∴Given limit 1 1 = lim+ x  −   x→ 0  x  x

2  40 Clearly, f ( x ) = 2 1 − x , x ≤ 0

41 [sin x] is non-differentiable at

x→ α

 − 1 / x, x < 1 f ′( x ) =  1 , x>1  x ∴ f ′ (1− ) = − 1 and f ′ (1+ ) = 1 Hence, f ( x ) is not differentiable.

k+ m=2

0, x> 0  ∴ f ( x ) is discontinuous and hence non-differentiable at x = 0.

= lim

 ax2 + bx + c  2sin2   2   = lim 2 x→ α (x − α) a 2sin2  ( x − α )( x − β ) 2  = lim 2 x→ α  a  ( x − α )2 ( x − β )2    2 2

 a  ( x − β )2    2

 3 x − x3  2   1 − 3x 

7 f ( x ) = cot −1 

 1 − x2  and g ( x ) = cos −1  2 1 + x  On putting x = tan θ in both equations, we get  3 tan θ − tan3 θ  f (θ) = cot −1   2  1 − 3 tan θ  ⇒ f (θ) = cot −1 (tan 3 θ) π ⇒ f (θ) = cot −1 cot − 3 θ    2 π = − 3θ 2 ∴ f ′ (θ) = − 3  1 − tan2 θ  and g(θ) = cos −1  2   1 + tan θ  = cos −1 (cos 2θ) = 2θ

… (i)

115

DAY ∴ g ′ (θ) = 2 Now,  f ( x ) − f (a)  f ( x ) − f (a) lim   = lim   x → a g ( x ) − g ( a) x→ a x−a     1 ×  g ( x ) − g (a) lim   x→ a x−a   1 1 3 = f ′ (a) ⋅ = –3× = – g ′ (a) 2 2

Hence, the composite function y = f ( x ) is discontinuous at three points, 1 x = , x = 1 and x = 2 . 2

10 Since, f ( x ) is a positive increasing function. ∴ 0 < f ( x ) < f (2 x ) < f (3 x ) ⇒ 0< 1
π  Clearly, lim− f ′( x ) = 0 = lim+ f ′( x ) x→ 0

x→ π −

13 We have, f ( x ) = lim n→ ∞

f (2 x ) f (3 x ) < f ( x) f ( x)

lim 1 ≤ lim

lim f ′( x ) = 0

and

x→ 0

n  π  x  1 +     3x   

π 1 +    3x 

n −1

π x n +    3 π x n −1 +    3

= x, if x >

n

n −1

π 3

  3x  n     + 1   π  π  π Also, f ( x ) = lim = , n→ ∞ 3 3    3 x  n −1 + 1   π    if x < π / 3

π π Note that f   =  3 3  x, if x ≥ π  3 f ( x) =  π , if π x < 3  3 From the given options, it is clear that option (c ) is incorrect. x 14 Let ak +1 = (kx + 1){(k + 1)x + 1} ∴

1 1  =  −  kx + 1 ( k + 1 )x + 1   n −1 1 1  ∴ f ( x ) = lim ∑  −  n→ ∞ (k + 1)x + 1  k = 0  kx + 1 1  = lim 1 − n→ ∞  nx + 1   1 , if x ≠ 0 =   0 , if x = 0

Clearly, f ( x ) is neither continuous nor differentiable at x = 0.

15 Clearly, x n + xa + b = ( x − x1 ) ⇒

( x − x2 )...( x − x n ) x n + xa + b = ( x − x2 ) x − x1

x n + ax + b ⇒ lim x→ x 1 x − x1

( x − x3 )...( x − x n )

= ( x1 − x2 )( x1 − x3 )...( x1 − x n ) nx n −1 + a x→ x 1 1 = ( x1 − x2 )( x1 − x3 )...( x1 − x n ) [using L’Hospital rule]

⇒ lim

⇒ nx1n −1 + a = ( x1 − x2 ) ( x1 − x3 )...( x1 − x n ).

TEN

116    16 Let L = lim x ln  x→ ∞    

1 x 0

 c   −1   a  x 

−b 1 x 0

1

 1,  (v) f ( x) =  1,  −1 

Clearly, for limit to be exist, b = 1. a c Thus, L = lim x ln  1 + 3 −  x→ ∞  x x a c = lim x  3 −  x→ ∞ x x   x2 x3 x 4 Q ln(1 + x ) = x − + − + ...   2 3 4   = −c Q L = −4 ∴ c =4 Hence, a ∈ R,b = 1 and c = 4. 

f( x )  1 lim 1 + x + −1  x x

⇒ e x →0 

lim

⇒

e

= e3

= e3

f( x )  1 lim  1 + −1   x x

1/x

f ( x ) x 

= e x →0  lim

f( x )

= e x →0 x = e 2 2

18 We have, f 2 ( x) = 1 ∀x ∈ R ∴ f can take values +1 or −1 Since f is discontinuous only at x = 0 ∴ f may be one of the followings

 1, x ≤ 0 (i) f ( x) =  −1, x > 0  1, x < 0 (ii) f ( x) =  −1, x ≥ 0 −1, x ≤ 0 (iii) f ( x) =   1, x > 0 −1, x < 0 (iv) f ( x) =   1, x ≥ 0

1 Here, lim f2 ( x ) = lim sin   x→ 0 x→ 0  x which does not exist. So, f2 ( x) is not continuous at x = 0. Hence, Statement II is false.

19 f ( x) = x |x | and g( x) = sin x 2  − sin x , x < 0 ∴ gof ( x ) = sin ( x | x |) =  2  sin x , x ≥ 0 2  − 2 x cos x , x < 0 (gof )′ ( x ) =  2  2 x cos x , x ≥ 0 Clearly, L ( gof )′ (0) = 0 = R ( gof )′(0) So, gof is differentiable at x = 0 and also its derivative is continuous at x = 0. Now,

22 ∴ f ( x) = |x − 2 | + |x − 5|  (2 − x ) + (5 − x ),  = ( x − 2) + (5 − x ), ( x − 2) + ( x − 5),  7 − 2 x,  =  3, 2 x − 7, 

2 2 2  − 2 cos x + 4 x sin x ,  x< 0  ( gof )′ ′ ( x ) =  2 2 2 cos x − 4 x sin x2 ,   x≥ 0

∴

Now, lim 1 + x→ 0  

x =0

L ( gof )′′(0) ≠ R ( gof )′′(0)

x 5

Y y = 2x – 7

y = 7 – 2x

Hence, gof ( x) is not twice differentiable at x = 0. Therefore, Statement I is true, Statement II is false.

20 Statement I is correct as though

|4x2 − 12x + 5|is non-differentiable at 1 5 x = and but cos πx = 0 those points. 2 2 1 5 So, f ′   and f ′   exists.  2  2

 1  21 Here, f ( x ) =  x ⋅ sin  x  , ,  0

x≠0 x=0

To check continuity at x = 0, 1 LHL = lim (− h ) sin  −   = 0 h→ 0  h  1 RHL = lim  h sin    = 0 h→ 0  h  f (0) = 0 So, f ( x) is continuous at x = 0.

x 5

Now, we can draw the graph of f very easily.

∴ L ( gof )′ ′(0) = − 2 and R ( gof )′′(0) = 2

f( x )   1+  x2 

= e3 f ( x) lim 2 = 2 x→ 0 x

x →0

⇒

1/x

f ( x ) x 

Hence, Statement I is correct. sin  1  , x ≠ 0    f2 ( x ) =   x  0 , x=0

 −1, x > 0  (vi) f ( x ) =  −1, x < 0  1, x = 0 

a c = lim x ln  3 + b −  x→ ∞ x x

1 + x + 17 We have, lim x→ 0

x >0 x 0 dx x d x (a ) = ax log a, for a > 0 dx

●

●

●

●

●

●

●

d dx d dx

x n = nx n −1 n  1  n  = − n +1 x  x

d (cos x) = − sin x dx d (sec x) = sec x tan x dx d (cosec x) = − cosec x cot x dx d x (e ) = e x dx d 1 (log a x) = , for x > 0, a > 0, a ≠ 1 dx x log a

PRED MIRROR Your Personal Preparation Indicator u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

TEN

118

●

●

●

●

●

●

(x) If u = f ( x) and v = g( x), then the differentiation of u  du    du  dx  with respect to v is = . dv  dv     dx  Differentiation of a function  w. r. t another function   

d 1 (sin −1 x) = , for −1 < x < 1 dx 1 − x2 d 1 (cos −1 x) = − , for −1 < x < 1 dx 1 − x2 d 1 (sec −1 x) = , for| x| > 1 dx | x| x 2 − 1 d 1 , for| x| > 1 (cosec −1 x) = − dx | x| x 2 − 1 d 1 (tan −1 x) = , for x ∈ R dx 1 + x2 d 1 (cot −1 x) = − , for x ∈ R dx 1 + x2

Differentiation Using Substitution In order to find differential coefficients of complicated expressions, some substitution are very helpful, which are listed below

S. No.

Methods of Differentiation (i) If y = f ( x) ± g( x), then

dy d = { f ( x) ± g( x)} dx dx

= f ′ ( x) ± g ′ ( x) (ii) If y = c ⋅ f ( x), where c is any constant, then dy d = (c ⋅ f ( x)) = c ⋅ f ′( x). [Scalar multiple rule] dx dx dy d (iii) If y = f ( x) ⋅ g( x), then = { f ( x) ⋅ g( x)} dx dx = f ( x) ⋅ g ′ ( x) + g ( x) ⋅ f ′ ( x) [Product rule] f ( x) dy d  f ( x)  g( x) ⋅ f ′ ( x) − f ( x) ⋅ g′ ( x) (iv) If y = , then =  = g ( x) dx dx  g( x)  {g( x)}2 g ( x) ≠ 0 (v) If y = f (u) and u = g( x), then dy df du = ⋅ = f ′(u) ⋅ g′( x) dx du dx

2

(vii) If given function cannot be expressed in the form y = f ( x) but can be expressed in the form f ( x, y) = 0, then to find derivatives of each term of f ( x, y) = 0 w.r.t x. [differentiation of implicit function] (viii) If y is the product or the quotient of a number of complicated functions or if it is of the form ( f ( x))g ( x ), then the derivative of y can be found by first taking log on both sides and then differentiating it. [logarithmic differentiation rule] dy When y = ( f ( x))g ( x ), then = ( f ( x))g ( x ) dx  g ( x)   f ( x) ⋅ f ′( x) + log f ( x) ⋅ g′( x)   (ix) If x = φ (t ) and y = Ψ (t ), where t is parameter, then dy dy/dt [Parametric differentiation rule] = dx dx/dt

x = a sin θ or a cos θ

2

(i)

a –x

(ii)

x 2 – a2

x = a sec θ or a cosec θ

(iii)

x 2 + a2

x = a tan θ or a cot θ

(iv)

a+ x or a− x

(v)

a2 + x2 or a2 − x2

a– x a+ x

x = a cos 2θ

a2 – x 2 a2 + x 2

x2 = a2 cos 2θ

(vi)

x a+ x

x = a tan2 θ

(vii)

( x – a)( x – b )

x = a sec2 θ – b tan2 θ

(viii)

ax – x 2

x = a sin2 θ

(ix)

x a– x

x = a sin2 θ

(x)

( x – a) (b – x)

x = a cos2 θ + b sin2 θ

[Chain rule]

This rule can be extended as follows. If y = f (u),u = g(v) dy df du dv and v = h ( x), then = ⋅ ⋅ , dx du dv dx d (vi) ( f {g( x)}) = f ′ (g( x)) ⋅ g′( x) dx

Substitution

Function

Usually this is done in case of inverse trigonometric functions.

Second Order Derivative d  dy    is called the second order derivative dx  dx  d2 y of y w.r.t x. It is denoted by 2 or f ′′( x) or y′′ or y2 . dx

If y = f ( x), then

Differentiation of a Determinant dp p q r dx dy If y = u v w , then = u dx l m n l p du + dx l

q dv dx m

dq dx v m

dr dx w n

r p dw + u dx dl n dx

q v dm dx

r w dn dx

119

DAY DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE  3π   is equal to  4

11 If y = (1 − x ) (1 + x 2 ) (1 + x 4 )...(1 + x 2 n ), then

1 If f ( x ) = | cos x |, then f ′  1 (a) 2

(b) 2

j

NCERT Exemplar

1 (c) 2

(d) 2 2

2 If f ( x ) = | x − 1 | and g ( x ) = f [f {f ( x )}], then for x > 2, g ′ ( x ) is equal to (a) − 1, if 2 ≤ x < 3 (c) 1, if x > 2

(d) (−1)n − 1 (n − 1)!

(b) (−1)(n − 1)! (c) n! − 1

4 If f ( x ) = x , then the value of n

f (1) − (a) 2 n

f ′ (1) f ′′ (1) f ′ ′′ (1) ( −1)n f n (1) is + − + ...+ 1! 2! 3! n! (b) 0

(c) 2 n − 1

(d) None of these

x −x d −1 5 If f ( x ) = 2 , where x ≠ 0, − 2, then [f ( x )] x + 2x dx 2

(whenever it is defined) is equal to (a)

−1 (1 − x)2

(b)

3 (1 − x)2

(c)

j

1 (1 − x)2

JEE Mains 2013 −3 (d) (1 − x)2

6 If f ( x ) = 2+| x |−| x − 1|−| x + 1| , then

 1  1  3  5 f ′  −  + f ′   + f ′   + f ′   is equal to  2  2  2  2 (b) −1

(a) 1

(c) 0

(d) None of these

(c) 1

(d)

dy 10 If sin y = x sin(a + y ), then is equal to dx (a)

sina sin2 (a + y)

(c) sina sin2 (a + y)

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

sin (a + y) sina sin2 (a − y) (d) sina 2

(b)

x (1 − x)2

(d) −2

AP GP Arithmetic-Geometric progression None of the above

14 If y = f ( x ) is an odd differentiable function defined on ( − ∞, ∞ ) such that f ′ ( 3) = − 2, then f ′ ( − 3) is equal to (a) 4

(c) −2

(b) 2

(d) 0

15 If f and g are differentiable function satisfying g ′ (a ) = 2, g (a ) = b and fog = I (identity function). Then, f ′ (b ) is equal to 1 2

(b) 2

(c)

2 3

(d) None of these

16 If y is an implicit function of x defined by (b) 1

x+ y xy

(b) xy

(c)

(b) −n 2 y

(a) n 2 y

19 If x =

3 2

(d) − log 2

(c) log 2

dy +n is equal to , then dx x y

(d)

y x

d 2y dy is equal to +x 2 dx dx

(c) −y

(d) 2 x 2 y

2t 1−t dy and y = is equal to , then 2 2 1+ t 1+ t dx 2

2t t2 + 1 2t (c) 1− t2 (a)

1 2

(d)

(c) 0

18 If y = ( x + 1 + x 2 )n , then (1 + x 2 )

 π f ′   is equal to  4

(b)

x (1 + x 2 )

f (1) = f ( −1) and a, b, c are in AP, then f ′ (a ), f ′ (b ) and f ′ (c ) are in.

(a)

9 If f ( x ) = cos x ⋅ cos 2x ⋅ cos 4x ⋅ cos 8x ⋅ cos 16x , then

2

(b) −4

(a) 4

17 If x m y n = ( x + y ) m

 π 8 If f ( x ) = | cos x − sin x |, then f ′   is equal to  2

(a)

(c)

13 Let f ( x ) be a polynomial function of second degree. If

(a) − 1

1 1 , where x ≠ 0 (b) for | x| > 1 | x| x 1 (c) − for | x| > 1 x 1 1 (d) for x > 0 and − for x < 0 x x (a)

(b) −1

1 (1+ x)2

x 2 x − 2x x cot y − 1 = 0. Then, y ′ (1) is equal to

7 If f ( x ) = loge | x |, then f ′( x ) equals

(a) 1

(b)

12 If f : ( −1, 1) → R be a differentiable function with f ( 0) = 1

(a) (d) −2

(c) 2

(a) −1

and f ′ ( 0) = 1. Let g ( x ) = [f ( 2f ( x ) + 2 )]2 . The, g′ ( 0) is equal to

(b) 1, if 2 ≤ x < 3 (d) None of these

3 The derivative of y = (1 − x )( 2 − x )K (n − x ) at x = 1 is (a) 0

is equal to

dy at x = 0 dx

(b)

2t t2 − 1

(d) None of these

π 2

 

20 For a > 0, t ∈  0,  , let x = a sin−1 t and y = a cos −1 t . 2

 dy  Then, 1 +   equals  dx  2

x (a) 2 y

2

y (b) 2 x

j

x + y (c) y2 2

2

JEE Mains 2013 x2 + y2 (d) x2

TEN

120  x + 1  x − 1 dy −1  + sin   , then dx is equal to  x − 1  x + 1

21 If y = sec−1 

1 x +1 (d) None of these

(a) 0

(a) 2 (c) 0

(b)

(c) 1

(a)

9 1 + 9x 3

(b)

3x x 1 − 9x 3

23 If y = sec(tan−1 x ), then

j

(c)

3x 1 − 9x 3

JEE Mains 2017 3 (d) 1 + 9x 3

dy at x = 1 is equal to dx j

JEE Mains 2013

1 (a) 2

1 (b) 2

24 If y = tan−1

x

(b) −

1− x4

1− x2

(d) 2

  , then dy is equal to  dx 

 1+ x 2 − 1− x 2   1+ x 2 + 1− x 2 

x

(a)

(c) 1

(c)

2x 1− x4

(d) None of these

dy 25 If 1 − x + 1 − y = a ( x − y ), then is equal to dx 2

2

1− x2 1− y2

(a)

(b)

1− y2 1− x2

(c)

x2 − 1 1− y2

26 If y = sin−1( x 1 − x + x 1 − x 2 ), then −2 x

(a) (c)

a cos 1+ a

(a) − (c)

2 x − x2 1

+

1− x2

27 If y =

1

+

1− x2 1 −1

2 x − x2

x

cos −1 x

and z = a cos

1 1 + a cos 1

(1 + a cos

−1

−1

(b)

−1

−

1 2 x − x2

(d) None of these

−1

x

, then

(b)

x

y2 − 1 1− x2

(d)

dy is equal to dx

1− x2

dy is equal to dz 1

1 + a cos

−1

)

28 Let g ( x ) be the inverse of f ( x ) such that f ′( x ) = 2

then

d (g ( x )) is equal to dx 2

1 1 + (g (x))5 (c) 5 (g (x))4 (1 + (g (x))5 (a)

29

 d2y  (a) −  2   dx 

−1

 dy     dx 

g ′(x) 1 + (g (x))5 (d) 1 + (g (x))5

 d 2 y  dy −3 (c) −  2     dx   dx 

 d 2 y  dy −2 (b)  2     dx   dx   d2y  (d)  2   dx 

−1

(a) f (0)g (0) (b) 0 (c) can’t be determined (d) None of the above

32 If

f ′( x ) f ( x ) = 0, where f ( x ) is continuously f ′′( x ) f ′( x )

differentiable function with f ′( x ) ≠ 0 and satisfies f ( 0) = 1 f ( x ) −1 and f ′( 0) = 2, then lim is x→ 0 x (a) 1 1 (c) 2

(b) 2 (d) 0

3x 33 If f ( x ) = 6 P Then,

2

cos x −1 P2

sin x 0 , where P is a constant. P3

d2 { f ( x )} at x = 0 is equal to dx 2

(a) P (c) P + P 3

(b) P + P 2 (d) independent of P

34 Which of the following statements is/are true? Statement I If y = (log x )log x , then dy  1 log (log x )  = (log x )log x + .  x  dx x Statement II If y = cos(a cos x + b sin x ) for some constants a and b, then y ′ = (a sin x − b cos x ) sin(a cos x + b sin x ) Only I is true Only II is true Both I and II are true Neither I nor II is true

35 Statement I If u = f (tan x ), v = g (sec x ) and f ′ (1) = 2, 1  du  . g′ ( 2 ) = 4 , then   =  dv  x = π / 4 2

Statement II If u = f ( x ), v = g ( x ), then the derivative of f du du /dx with respect to g is = . dv dv /dx j

−3

1 , 1+ x 5

(b)

d 2x is equal to dy 2

that of y = g ( x ) is symmetrical about the origin and if d 2h( x ) at x = 0 is h( x ) = f ( x ) ⋅ g ( x ), then dx 2

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

x

(d) None of these

x 2

(b) 1 (d) −1

31 If graph of y = f ( x ) is symmetrical about the Y-axis and

 6x x   1 22 For x ∈  0,  , if the derivative of tan−1   is  4 1 − 9x 3  x ⋅ g ( x ), then g ( x ) equals

 log(e / x 2 ) d 2y −1  3 + 2 log x  is , then + tan     1 − 6 log x  dx 2  log(ex 2 ) 

30 If y = tan−1 

AIEEE 2011

(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

121

DAY DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 For x ∈R ,f ( x ) = | log 2 − sin x | and g ( x ) = f (f ( x )), then (a) (b) (c) (d)

g is not differentiable at x = 0 g′ (0) = cos(log 2) g′ (0) = − cos (log 2) g is differentiable at x = 0 and g′ (0) = − sin(log 2) (b) y ⋅ ∑ k cot kx

k =1

n

(c) y ⋅ ∑ k ⋅ tankx

(d)

k =1

∑ cotkx

2 7

(b)

1 2

(c) 2

(d)

7 2

 1 + (where, a > 0), then 

4 If f ( x ) = (cos x + i sin x ) ⋅ (cos 2x + i sin 2x )

(cos 3x + i sin 3x ) ... (cos nx + i sin nx ) and f (1) = 1, then f ′ ′ (1) is equal to n (n + 1) 2 2 n (n + 1)   (c) −   2 

n (n + 1)  (b)    2

(a)

5 If x 2 + y 2 = ae

2

is equal to 2 −π / 2 e a 2 (c) − e − π / 2 a

2 π /2 e a (d) None of these (b) −

(a)

6 If y = | sin x || x| , then the value of (a) (c)

2

−π 6

6 2

π − 6

6

dy π at x = − is dx 6 π 6

[6log 2 −

3 π]

2 (b) [6log 2 + 6

[6log 2 +

3 π]

(d) None of these

3 π]

1 ( 5) 2 x + 1 2

and g ( x ) = 5x + 4x loge 5 is (b) (0, 1)

(c) (∞, 0)

(d) (0, ∞)

8 Let f ′ ′ ( x ) = − f ( x ), where f ( x ) is a continuous double differentiable function and g ( x ) = f ′ ( x ). 2

 x  x If F ( x ) = f    + g      2    2  equal to (a) 0 (c) 10

(b) a 2

(c)

2 3a

(d) g (x )

1  f ′ n +   3 then 1   f n +   3

2

and F ( 5) = 5, then F (10) is

2a 3

and g ( x + 1) = x + g ( x ) ∀ x ∈ R. If n ∈ I + ,  1 f′    3 is equal to −  1 f   3  1 1 1  (b) 3  1 + + + ... +  3 5 2 n − 1  (d) 1

1 1 1 (a) 3  1 + + + K +   2 3 n (c) n 1− x

13 If the function f ( x ) = − 4e

2

+1+ x +

x2 x3 and + 2 3

 −7  g ( x ) = f −1( x ), then the value of g′   is equal to  6 1 5 6 (c) 7

1 5 6 (d) − 7 (b) −

(a)

7 The solution set of f ′ ( x ) > g ′ ( x ), where f ( x ) = (a) (1, ∞)

a 3

12 Let f ( x ) = elne , a>0 assuming y > 0, then y′ ′ ( 0)

3

2  dy   2     dx   at π is given by d 2y 6 2 dx

(a)

(d) None of these

y tan−1   x

JEE Mains 2013

(b) sin2 x cos (cos x) (d) cos2 x sin(sin x)

11 If x = a cos t cos 2 t and y = a sin t cos 2 t

k =1

3 If 3f ( x ) − 2f (1 /x ) = x , then f ′ ( 2) is equal to (a)

j

(a) cos2 x cos (sin x) (c) sin2 x sin(cos x)

k =1

n

−1 81 −1 (d) 3 (b)

g ( x ) is equal to

n

(a) ∑ k ⋅ tankx

−1 9 −1 (c) 27 (a)

10 If f ( x ) = sin (sin x ) and f ′ ′ ( x ) + tan x f ′ ( x ) + g ( x ) = 0, then

2 If y = sin x ⋅ sin 2x ⋅ sin 3x ... sin nx , then y ′ is n

9 If f ( 2) = 4, f ′ ( 2) = 3, f ′′ ( 2) = 1, then (f −1 )′′ (4) is equal to

14. If f ( x ) = ( x − 1)100 ( x − 2)2 ( 99 ) ( x − 3)3 ( 98 ) ...( x − 100)100 , then the value of (a) 5050 (c) 3030

f ′ (101) is f (101)

(b) 2575 (d) 1250

15 The derivative of the function represented parametrically as x = 2t − | t | , y = t 3 + t 2 | t | at t = 0 is (b) 5 (d) 25

(a) −1 (c) 1

(b) 0 (d) does not exist.

TEN

122

ANSWERS SESSION 1

SESSION 2

1 (a)

2 (a)

3 (b)

4 (b)

5 (b)

6 (d)

7 (b)

8 (a)

9 (a)

10 (b)

11 (a)

12 (b)

13 (a)

14 (c)

15 (a)

16 (a)

17 (d)

18 (a)

19 (b)

20 (d)

21 (a)

22 (a)

23 (a)

24 (a)

25 (b)

26 (c)

27 (c)

28 (c)

29 (c)

30 (c)

31 (b)

32 (b)

33 (d)

34 (c)

35 (a)

1 (b) 11 (d)

2 (b) 12 (c)

3 (b) 13 (a)

4 (c) 14 (a)

5 (c) 15 (b)

6 (a)

7 (d)

8 (b)

9 (c)

10 (d)

Hints and Explanations SESSION 1 1 When π < x < π, cos x < 0, so that 2 |cos x |= − cos x,

f ′ (1) f ′ ′ (1) f ′ ′ ′ (1) + − 1! 2! 3! (− 1)n f n (1) + ... + n! n n n = C 0 − C 1 + C 2 − nC 3 + ... + (−1)n nC n = (1 − 1)n = 0

∴

f (1) −

i.e. f ( x ) = − cos x, f ′( x ) = sin x 3π   3π  = 1 Hence, f ′  = sin    4  4 2

2 We have, f ( x ) = | x − 1|

[Q x > 2]

f [ f ( x )] = f ( x − 1) = |( x − 1) − 1 | =| x − 2| g ( x ) = f [ f { f ( x )}] = f ( x − 2) =|( x − 2) − 1 |= | x − 3 | x − 3, if x ≥ 3 =   − x + 3, if 2 ≤ x < 3

5 Let ⇒ ⇒ ∴

1, if x ≥ 3 ∴ g ′ ( x ) =   −1, if 2 ≤ x < 3

3

dy = − [(2 − x )(3 − x )K (n − x ) + dx (1 − x )(3 − x )K(n − x ) + K + (1 − x )(2 − x )K (n − 1 − x )] dy ⇒   = − [(n − 1)! + 0 + K + 0]  dx  x = 1

x2 − x x2 + 2 x 2y + 1 x= ;x≠ 0 −y + 1

y =

f −1 ( x ) =

(− x + 1 ) ⋅ 2 − (2 x + 1 ) (−1 ) d −1 [ f ( x )] = dx (− x + 1 )2 3 = (− x + 1 )2

6 We have, f ( x ) = 2+| x|−| x − 1|−| x + 1| ∴

2 − 2 − f ( x) =  2 + 2 +

= (−1)(n − 1)!

⇒

f (1) = 1 = nC 0 f ′ (1) n n = = C1 1! 1! f ′ ′ (1) n (n − 1) n = = C2 2! 2! f ′ ′ ′ (1) n (n − 1) (n − 2) n = = C3 3! 3! M

M

M M f n (1) n ! n = = Cn n! n!

x + ( x − 1) + ( x + 1), x + ( x − 1) − ( x + 1), x + ( x − 1) − ( x + 1), x − ( x − 1) − ( x + 1),  if x < −1  if − 1 ≤ x < 0   if 0 ≤ x < 1  if x ≥ 1

4 We have, f ( x ) = x n ⇒

2x + 1 −x + 1

if x < −1  x + 2,  − x, if − 1 ≤ x < 0 = if 0 ≤ x < 1  x, 2 − x, if x ≥ 1  1,  −1, ⇒ f ′( x ) =   1,  −1,

if x < −1 if −1 ≤ x < 0 if 0 ≤ x < 1 if x≥1

1 1 3 5 Hence, f ′ −  + f ′  + f ′  + f ′   2  2  2  2 = (−1) + 1 + (−1) + (−1) = −2

7 We have, f ( x ) = log e | x| ∴

 log(− x )  − log(− x ) f ( x) =   − log x  log x

, x < −1 , −1 < x < 0 , 0< x < 1 , x >1

   − f ′( x ) =  −   

1 , x< −1 x 1 , − 1< x < 0 x ⇒ 1 , 0< x < 1 x 1 , x>1 x 1 Clearly, f ′( x ) = for| x|> 1 x π 8 When 0 < x < , cos x > sin x 4 ∴ cos x − sin x > 0 π Also, when < x < π,cos x < sin x 4 ∴ cos x − sin x < 0 ∴ |cos x − sin x|= − (cos x − sin x ), when π < x< π 4 ⇒ f ′ ( x ) = sin x + cos x π π π ⇒ f ′   = sin + cos = 1 + 0 = 1  2 2 2 2sin x ⋅ cos x ⋅ cos 2 x ⋅ cos 4 x ⋅ cos 8 x ⋅ cos 16 x 9 f ( x) = 2sin x sin 2 x cos 2 x cos 4 x ⋅ cos 8 x ⋅ cos 16 x = 2 sin x sin 32 x = 5 2 sin x

123

DAY 1 ⋅ 32 32cos 32 x ⋅ sin x − cos x ⋅ sin 32 x sin2 x 1 1 32 × − ×0 π  2 2 ⇒ f ′  = = 2 2  4  1  32  2 

∴ f ′( x ) =

10 ∴ ⇒

sin y = x sin( a + y ) x=

sin y sin(a + y )

On differentiating w.r.t. y, we get dx sin(a + y )cos y − sin y cos( a + y ) = dy sin2 (a + y ) dx sin a = ⇒ dy sin2 (a + y ) ⇒

dy sin2 (a + y ) = dx sin a

11 Given, y = (1 − x ) (1 + x2 ) (1 + x 4 )...(1 + x2 n ) (1 − x ) (1 + x2 )...(1 + x2 n ) or y = (1 + x ) 1 − ( x )4 n = (1 + x ) (1 + x ) ⋅ (0 − 4n ⋅ x 4 n − 1 )

⇒ ⇒

− (1 − x 4 n ) ⋅ 1 (1 + x )2

∴ ∴

 dy  = −1    dx  x = 0

12 We have, f :(−1,1) → R f (0) = −1, f ′(0) = 1 g ( x ) = [ f (2 f ( x ) + 2)]2 ⇒ g ′( x ) = 2[ f (2 f ( x ) + 2)] × f ′(2 f ( x ) + 2) × 2 f ′( x ) ⇒ g ′(0) = 2[ f {2 f (0) + 2}] × f ′{2 f (0) + 2} × 2 f ′(0) = 2[ f (0)] × f ′( 0) × 2 f ′( 0) = 2 × (−1) × 1 × 2 × 1 = −4

⇒ ⇒ ⇒

14 Since, f ( x ) is odd. ∴

f (− x ) = − f ( x )

f ′ ( g ( x )) g ′ ( x ) = 1 for all x 1 1 f ′(g (a)) = = g ′ (a) 2 1 [Q g (a) = b] f ′( b ) = 2

16 x2 x − 2 x x cot y − 1 = 0

…(i)

Now, x = 1, 1 − 2 cot y − 1 = 0 ⇒ cot y = 0 π ⇒ y = 2 On differentiating Eq. (i) w.r.t. x, we get dy 2 x2 x (1 + log x ) − 2 [ x x (− cosec2 y ) dx + cot y x x (1 + log x )] = 0 π At  1,  , 2 (1 + log 1)  2   dy   − 2 1 (− 1)   + 0 = 0    1 , π  dx     2  dy ⇒ 2 + 2   =0  dx  1, π  

∴

= ⇒ (1 + x2 )

⇒

dy  m + n n  m m + n −  = −  dx  x + y y x x+ y

⇒

dy  my + ny − nx − ny    dx  y( x + y ) 

2t 1 − t2 ,y = 2 1+ t 1 + t2

Put t = tan θ 2 tan θ = sin 2θ and 1 + tan2 θ

1 − tan2 θ = cos 2θ 1 + tan2 θ dy dy / dθ − 2sin 2θ ∴ = = = − tan 2θ dx dx / dθ 2cos 2θ − 2 tan θ = 1 − tan2 θ − 2t 2t = = 2 1 − t2 t –1 y =

∴

1 sin

2 a

−1

t

 −1  asin t ×  

dy 1 = cos −1 dt 2 a asin

dy =− dx

acos

=−

t −1

−1

a

cos

a

sin

−1

−1

 −1  acos t ×  

  1 − t  1

2

  1 − t 2  −1

 acos t   × 1 sin −1 t a  −1

t t t t

−1

2

 x + 1

mx + my − mx − nx x( x + y ) dy y = dx x

dx

  1 + x  x

dy = n ( x + 1 + x2 )n dx  d2 y dy  x  ⇒ ( 1 + x2 ) + 2 2  dx  1 + x  dx  x = n2 ( x + 1 + x2 )n − 1  1 +  1 + x2 

 x − 1

−1 21 Q y = sec −1   + sin  x + 1   x − 1  

18 d ( y ) = n ( x + 1 + x2 )n − 1  1 +  

d y dy + x = n2 y dx dx2

dy y2 acos t ∴ 1 +   = 1 + =1+ 2 −1 sin t  dx  x a x2 + y 2 = x2

=

∴

1 + x2 )n

2

∴ x=

17 Given that, x m y n = ( x + y )m + n

⇒

1 + x2

d y dy + x dx dx2

19 We have, x =

and

Taking log on both sides, we get m log x + n log y = (m + n )log( x + y ) On differentiating w.r.t. x, we get dy  m n dy (m + n )  + = 1 +  x y dx ( x + y )  dx 

1 + x2 )n

2

(1 + x2 )

dt

2 

n2 ( x +

= n2 ( x +

20 Q dx =

2 

 dy  = −1    dx  1, π  

13 Let f ( x ) = Ax2 + Bx + C ∴ f (1) = A + B + C and f (−1) = A − B + C Q f (1) = f (−1) [given] ⇒ A+ B+C = A−B+C ⇒ 2B = 0 ⇒ B = 0 ∴ f ( x ) = Ax2 + C ⇒ f ′( x ) = 2 Ax ∴ f ′(a) = 2 Aa f ′(b ) = 2 Ab and f ′(c ) = 2 Ac Also, a, b, c are in AP. So, 2 Aa, 2 Ab and 2 Ac are in AP. Hence, f ′(a), f ′(b )and f ′(c )are also in AP.

d2 y dy (1 + x2 ) + ⋅x 2 dx dx ⇒ 1 + x2

15 Since, fog = I ⇒ fog ( x ) = x for all x

2

dy = dx

f ′ (− x )(−1) = − f ′ ( x ) f ′(− x ) = f ′( x ∴ ) f ′ (−3) = f ′ (3) = −2

2

⇒ ( 1 + x2 )

   

 x − 1 x − 1 π −1  = cos −1   + sin  x + 1  = 2 x + 1     dy π = 0 Qsin −1 x + cos −1 x =  ⇒  dx 2    22 Let y = tan −1  6 x x1   1 − 9x 

 2 ⋅ (3 x3 /2 )  = tan −1  3 /2 2   1 − (3 x )  −1 3 /2 = 2 tan (3 x ) Q2 tan −1 x = tan −1 2 x   1 − x2  dy 1 3 1 /2 ∴ = 2⋅ ⋅ 3 × ( x) dx 2 1 + (3 x3 /2 )2

TEN

124 9 ⋅ x 1 + 9 x3 9 ∴g ( x ) = 1 + 9 x3 =

27 y = ⇒

−1

23 Given, y = sec (tan x ) Let tan −1 x = θ ⇒ x = tanθ ∴

y = secθ = 1 + x2

On differentiating both sides w.r.t. x, we get dy 1 = ⋅ 2x dx 2 1 + x2

⇒

 1 + x2 − 1 − x2  24 Given, y = tan  . 2 2  1 + x + 1 − x  −1

Put x = cos 2θ 2

 cos θ − sin θ  y = tan −1    cos θ + sin θ   1 − tan θ  = tan    1 + tan θ  −1

π   = tan −1 tan  − θ  4   π π 1 = −θ= − cos −1 x2 4 4 2 dy x x ∴ = 0+ = dx 1 − x4 1 − x4

25 On putting x = sinθ and y = sin φ , we get Given equation becomes cos θ + cos φ = a(sin θ − sin φ) θ + φ  θ − φ ⇒ 2cos   cos    2   2 

⇒ ⇒ ⇒ ⇒

θ + φ  θ − φ  = a 2cos   sin    2   2   θ−φ = cot −1 a 2 θ − φ = 2cot −1 a sin −1 x − sin −1 y = 2cot −1 a dy 1 1 − =0 1 − x2 1 − y 2 dx dy = dx

∴

1 − y2 1 − x2

26 On putting x = sin A and x = sin B y = sin −1 (sin A 1 − sin2 B + sin B 1 − sin2 A ) −1

= sin (sin A cos B + sin B cos A ) = sin −1 [sin( A + B )] = A + B = sin −1 x + sin −1 x dy 1 1 ⇒ = + 2 dx 1− x 2 x − x2

−1

x

cos −1 x

1+ a

, z = acos

−1

x

32 Since,

z y = 1+ z dy (1 + z )1 − z (1) = dz (1 + z ) 2 1 = (1 + z ) 2 1 = cos −1 x 2 (1 + a )

28 Since g ( x ) is the inverse of f ( x ) ∴ ⇒

At x = 1 dy 1 = dx 2

∴

⇒

acos

⇒

g ′′( x ) = 5 (g ( x ))4 ⋅ g ′ ( x ). = 5 (g ( x ))4 (1 + (g ( x ))5 ) dy

⇒

−1

 dx 

dy d2 x = −    dx  dy 2

−2

d y dx ⋅ dx2 dy

⇒

( f ′( x ))2 − f ′′( x ) ⋅ f ( x ) =0 ( f ′( x ))2

⇒

d  f ( x)  =0 dx  f ′( x )

⇒

f ( x) = c , (constant) f ′( x )

⇒

f ′( x ) =2 f ( x)

⇒

d (log f ( x )) = 2 dx

⇒

log( f ( x )) = 2 x + k

On putting x = 0, we get 0 = k

−3

⇒ log( f ( x )) = 2 x ⇒

 log(e / x2 ) 30 Given, y = tan   2  log(ex )  −1

x→ 0

 log e − log x2  y = tan −1  2   log e + log x   3 + 2 log x  + tan −1    1 − 6 log x   1 − 2 log x  = tan −1    1 + 2 log x   3 + 2 log + tan −1   1 − 6 log

f ( x) = e2 x

Now, lim

 3 + 2 log x  + tan −1    1 − 6 log x 

−1

( f ′( x ))2 − f ′′( x ) ⋅ f ( x ) = 0

2

 d 2 y  dy = −  2     dx   dx 

∴

∴

On putting x = 0, we get 1 =c 2 f ( x) 1 ⇒ = f ′( x ) 2

f (g ( x )) = x f ′(g ( x )) ⋅ g ′( x ) = 1 1 g ′( x ) = = 1 + (g ( x ))5 f ′(g ( x ))

dy 29 Since, dx =  

f ′( x ) f ( x ) =0 f ′′( x ) f ′( x )

x  x

−1

= tan (1) − tan (2 log x ) + tan −1 (3) + tan −1 (2 log x ) −1 = tan (1) + tan −1 (3) dy d2 y Now, = 0 and 2 = 0 dx dx

31 Since, y = f ( x ) is symmetrical about the Y-axis ∴ f ( x ) is an even function. Also, as y = g ( x ) is symmetrical about the origin ∴ g ( x ) is an odd function. Thus, h( x ) = f ( x ) ⋅ g ( x ) is an odd function. or h( x ) = − h(− x ) Now, h ′( x ) = h ′ (− x ) and h "( x ) = − h ′′(− x ) ⇒ h ′′(0) = − h ′′(0) ⇒ h ′′(0) = 0

f ( x) − 1 e2 x − 1 = lim ⋅ 2 = 2. x → 0 x 2x

33 f ′′( x ) d2 d2 d2 (3 x2 ) (cos x ) (sin x ) 2 2 dx dx dx2 = 6 −1 0 P P2 P3 6 = 6 P

− cos x − sin x −1 0 P2 P3

6 ∴ f ′′(0) = 6 P

−1 0 −1 0 = 0, which is P2 P3

independent of P.

34 I. Let y = (log x ) log x On taking log both sides, we get log y = log ( log x ) log x ⇒ log y = log x log [log x] [Q log m n = n log m] On differentiating both sides w.r.t. x, we get 1 dy d = (log x ) {log (log x )} y dx dx d + log (log x ) log x dx 1 1 1 = (log x ) + log (log x ) log x x x 1 = {1 + log (log x )} x

125

DAY ⇒

dy y = {1 + log (log x )} dx x (log x ) log x = {1 + log (log x )} x log(log x ) 1 = (log x ) log x  +  x  x

II. Let y = cos (a cos x + b sin x ). On differentiating w.r.t. x, we get d {cos(acos x + b sin x} dx = − sin(a cos x + b sin x ) d (acos x + b sin x ) dx = − sin(acos x + b sin x ) [− asin x + b cos x] = (asin x − b cos x ) sin (acos x + b sin x )

35 Given, u = f (tan x ) du = f ′(tan x )sec2 x dx and v = g (sec x ) dv ⇒ = g ′(sec x )sec x tan x dx du (du / dx ) f ′(tan x ) 1 ∴ = = ⋅ dv (dv / dx ) g ′(sec x ) sin x f ′(1) du = ⋅ 2 ∴    dv  x = π 4 g ′( 2 ) 2 1 = ⋅ 2= 4 2 ⇒

SESSION 2 1 We have, f ( x ) = |log 2 − sin x| and g ( x ) = f ( f ( x )), x ∈ R Note that, for x→ 0, log 2 > sin x ∴ ⇒

f ( x ) = log 2 − sin x g ( x ) = log 2 − sin( f ( x )) = log 2 − sin(log 2 − sin x ) Clearly, g ( x ) is differentiable at x = 0 as sin x is differentiable. Now, g ′ ( x ) = − cos (log 2 − sin x ) ( − cos x ) = cos x.cos(log 2 − sin x ) ⇒ g ′ (0) = 1 ⋅ cos (log 2)

2 We have, ∴

y = sin x ⋅ sin 2 x ⋅ sin 3 x⋅....sin nx y ′ = cos x ⋅ sin 2 x ⋅ sin 3 x...⋅ sin nx + sin x ⋅ (2cos 2 x ) sin 3x...sin nx + sin x ⋅ sin 2 x(3cos 3 x )...sin nx + ...+ sin x sin 2 x sin 3 x...(cos nx )

(by product rule) ⇒ y ′ = cot x ⋅ y + 2 ⋅ cot 2 x ⋅ y +3 ⋅ cot 3 x ⋅ y + ...+ n ⋅ cot nx ⋅ y ⇒ y ′ = y [cot x + 2cot 2 x +3cot 3 x + ...+ n cot nx] ⇒ y′ = y ⋅

n

∑ k cot kx k =1

3 3 f ( x ) − 2 f (1 / x ) = x

…(i)

Let 1 / x = y , then 3 f (1 / y ) − 2 f ( y ) = 1 /y ⇒ −2 f ( y ) + 3 f (1 / y ) = 1 /y ⇒ −2 f ( x ) + 3 f (1 / x ) = 1 /x

…(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 2 and adding, we get 2 5 f ( x) = 3 x + x 1 2 ⇒ f ( x ) =  3 x +  5 x ⇒ ⇒

1 3 − 5 1 f ′(2) =  3 − 5

f ′( x ) =

2  x2  2 1 = 4 2

When x = 0, we get from Eq. (i), y ′ = −1 2 −2 − π /2 ⇒ y ′′ (0) = = e a − ae π /2

6 Given, y = |sin x || x | In the neighbourhood of π − , | x |and|sin x | both are negative 6 i.e. y = (− sin x )( − x ) On taking log both sides, we get log y = (− x ) ⋅ log (− sin x ) On differentiating both sides, we get 1 dy 1  ⋅ = (− x )   ⋅ (− cos x )  − sin x  y dx + log(− sin x ) ⋅ (−1) = − x cot x − log (− sin x ) = − [ x cot x + log ( − sin x )]

4 f ( x ) = (cos x + i sin x ) (cos 2 x + i sin 2 x )(cos 3 x + i sin 3 x ) … (cos nx + i sin nx ) = cos( x + 2 x + 3 x + ...+ nx ) + i sin ( x + 2 x + 3 x + ...+ nx ) n (n + 1) n (n + 1) = cos x + i sin x 2 2 n (n + 1) ⇒ f ′( x ) =    2  − sin n (n + 1) x + i cos n (n + 1) x  2 2   2

n (n + 1) f ′′ ( x ) = −    2 n ( n + 1 ) n (n + 1)  cos x + i sin x   2 2 = − 

2

n (n + 1) ⋅ f ( x)  2

 ∴ f ′ ′ (1) = −  = − 

2

n (n + 1) f (1)  2

On differentiating both sides w.r.t. x, we get 1 xy ′− y 1 2 x + 2 yy ′ × 2 = × 2 2 x + y2 x2 y 1 +    x …(i)

Again, on differentiating both sides w.r.t. x, we get 1 + ( y ′ )2 + yy ′′ = xy ′ ′+ y ′− y ′ ⇒

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

−

x=−

π 6

π

(2) 6 [6 log 2 − 3 π] = 6

7 Since, f ′( x ) > g ′( x ) 1 ⇒   52 x  2

+1

log e 5 × 2 >

5x log e 5 + 4 log e 5 52 x ⋅ 5 > 5 x + 4 5⋅ 52 x – 5 x – 4 > 0 x (5 – 1) (5⋅ 5 x + 4) > 0 5x > 1 x> 0 d 8 Given, { f ′( x )} = − f ( x ) dx ⇒ g ′( x ) = − f ( x ) ⇒ ⇒ ⇒ ∴ ⇒

=  f 

On taking log both sides of the given equation, we get y 1 log ( x2 + y 2 ) = log a + tan −1    x 2

⇒

dy ∴    dx  at

[Q g ( x ) = f ′ ( x ), given]

n (n + 1)  2

x + yy ′ = xy ′− y

dy = − y [ x cot x + log (− sin x )] dx

Also, given F ( x )

2

5 When x = 0, y > 0 ⇒ y = ae π /2

⇒

⇒

2

2

 x   + g  x         2   2

⇒ F ′( x ) = 2  f 

 x f ′  x ⋅ 1      2  2 2

x x 1 + 2  g    ⋅ g ′   ⋅ = 0  2 2   2 Hence, f ( x ) is constant. Therefore, F(10) = 5 .

9 Let y = f ( x ), then x = f −1 ( y ). Now, Q ∴

d2 x = ( f −1 )′′ ( y ) d y2 dx  dy  =  dy  dx 

−1

d2 x d d y  =   dy  d x  d y2 =

d dx

 dy     dx 

−1

−1

⋅

dx dy

TEN

126 dy = −    dx 

−2

⋅

d 2 y dx ⋅ dx2 dy

−d 2 y 2 = dx 3  dy     dx 

3 /2

2   dy   ∴ 1 +     dx   

2   dy   ⇒ 1 +     dx   

⇒ f ′( x) = cos x⋅ cos (sin x) ⇒ f ′ ′( x) = − sin x⋅cos (sin x) − cos 2 x⋅sin(sin x) Now, g ( x) = − [ f ′ ′( x) + f ′( x)⋅ tan x] = sin x ⋅cos (sin x) + cos 2 x ⋅sin(sin x) − tan x⋅cos x⋅cos (sin x) = sin x ⋅ cos (sin x) + cos2 x ⋅ sin(sin x) − sin x ⋅ cos (sin x) = cos 2 x ⋅ sin (sin x)

π 3 cos 3

12 Clearly, f ( x) = e g (x ) Now, as g ( x + 1) = x + g ( x) ∴ e g (x + 1 ) = e x + g (x ) = e x ⋅ e g (x ) ⇒ f ( x + 1) = e x f ( x) On taking log both sides, we get ln f ( x + 1) = ln (e x ⋅ f ( x)) 1 . f ′ ( x + 1) ⇒ f ( x + 1) =1+

−asin 3 t cos 2 t

 dy sin t ⋅ sin 2t  and = a cos t cos 2t −  dt cos 2t   acos 3 t cos 2 t

dy dy / dt = = − cot3t dx dx / dt 2 d y dt = 3cosec 2 3t ⋅ ⇒ dx dx2 −3cosec2 3t ⋅ cos 2t = asin 3t 3  = −   cosec3 3t ⋅ cos 2t  a

⇒

1 . f ′( x ) f ( x)

f ′ ( x + 1) f ′ ( x ) − =1 f ( x + 1) f ( x)

⇒

11 We have,

 cos t ⋅ sin 2t  dx = a  − sin t cos 2t −  dt cos 2t  

=

1 1 f '  1 +  f '    3  3 =1 − 1 1 f  1 +  f    3  3 f ′  2 +   f 2 +  f ′  n +   f n + 

1 f ′  1 +   3 − 1  f 1 +   3

1 f ′  (n − 1) +   3 − 1  f  (n − 1) +   3

1  3 =1 1  3 1  3 =1 1  3

on adding columnwise, we get 1 1 f ′  n +  f ′     3 3 − =n 1 1 f  n +  f     3 3

⇒

2a 3

f (g ( x)) = x ⇒ f ′(g ( x))⋅ g ′( x) = 1 1 g ′( x ) = f ′ (g ( x )) −7 g ′   =  6

1   7  f ′g−    6  1 =  −1  − 7   f ′ f    6    Q f (1) = − 4 + 1 + 1 + 1 + 1 = − 7  2 3 6 7  ∴ f −1   = 1  6  ⇒

d2 y π at t = is 6 dx2 a

10 f ( x) = sin (sin x)

∴

3 /2

−1 = 27

x =2

=

∴

−3 = (1 + cot2 3 t )3 /2   cosec3 3t cos 2t  a

Since, y = 4 when x = 2 d2 y dx2 x = 2 ∴ ( f −1 )′′ (4) = − 3  dy     dx 

=

13 Since, g ( x) = f −1 ( x)

d2 y dx2

=

1 5

1−x   2  − 1  + 1 + x + x2   Q f ′ ( x ) = − 4e   2  

14 We have, f ( x ) = 100 ( x − i )i (101 − i ) 11 i =1

100

⇒ log f ( x) = Σ i (101 − i ) log ( x − i ) i =1

100 1 1 ⋅ f '( x ) = ∑ i (101 − i )⋅ f ( x) x−i i =1 f ′ (101) 100 (101 − i ) = Σ i ⇒ f (101) i = 1 (101 − i ) 100 100(101) = Σ i= = 5050 i =1 2

15 Given, x = 2t − |t|and y = t 3 + t 2 |t| Clearly,x = t , y = 2t 3 when t ≥ 0 and x = 3t, y = 0 when t < 0 On eliminating the parameter t, we get 2 x3 , when x ≥ 0 y =  0, when x < 0 dy  6 x2 , when x > 0 = dx  0, when x < 0 Q (LHD) at x = 0 = (RHD) at x = 0 = 0 ∴ Its derivative at x = 0 (i.e. at t = 0 is 0) Now,

DAY THIRTEEN

Applications of Derivatives Learning & Revision for the Day u

u

Derivatives as the Rate of Change Increasing and Decreasing Function

u

u

Tangent and Normal to a Curve Angle of Intersection of Two Curves

u

u

Rolle’s Theorem Lagrange’s Mean Value Theorem

Derivatives as the Rate of Change dy is nothing but the rate of change of y, relative to x. If a variable quantity y is some dx function of time t i.e. y = f (t ), then small change in time ∆t have a corresponding change ∆y ∆y in y. Thus, the average rate of change = ∆t When limit ∆t → 0 is applied, the rate of change becomes instantaneous and we get the rate of change with respect to at the instant t , i.e. ∆y dy lim = ∆t → 0 ∆ t dt dy is positive if y increases as t increase and it is negative if y decrease as t increase. dt

PRED MIRROR Your Personal Preparation Indicator

Increasing and Decreasing Function ●

●

●

●

●

A function f is said to be an increasing function in ]a, b [, if x1 < x2 ⇒ f ( x1 ) ≤ f ( x2 ), ∀ x1 , x2 ∈] a, b [.

u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

A function f is said to be a decreasing function in ] a, b [, if x1 < x2 ⇒ f ( x1 ) ≥ f ( x2 ), ∀ x1 , x2 ∈] a, b [. f ( x) is known as increasing, if f ′ ( x) ≥ 0 and decreasing, if f ′ ( x) ≤ 0. f ( x) is known as strictly increasing, if f ′ ( x) > 0 and strictly decreasing, if f ′ ( x) < 0. Let f ( x) be a function that is continuous in [a, b ] and differentiable in (a, b ). Then, (i) f ( x) is an increasing function in [a, b ], if f ′( x) > 0 in (a, b ). (ii) f ( x) is strictly increasing function in (a, b ), if f ′( x) > 0 in (a, b ).

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

128 (iii) f ( x) is a decreasing function in [a, b ], if f ′( x) < 0 in (a, b ). (iv) f ( x) is a strictly decreasing function in [a, b ], if f ′( x) < 0 in (a, b ).

Monotonic Function A function f is said to be monotonic in an interval, if it is either increasing or decreasing in that interval.

Results on Monotonic Function (i) If f ( x) is a strictly increasing function on an interval [a, b ], then f −1 exists and it is also a strictly increasing function. (ii) If f ( x) is strictly increasing function on an interval [a, b ] such that it is continuous, then f −1 is continuous on [ f (a), f (b )]. (iii) If f ( x) is continuous on [a, b ] such that f ′ (c) ≥ 0 [ f ′ (c) > 0] for each c ∈(a, b ), then f ( x) is monotonically increasing on [a, b ]. (iv) If f ( x) is continuous on [a, b ] such that f ′ (c) ≤ 0 ( f ′ (c) < 0) for each c ∈[a, b ], then f ( x) is monotonically decreasing function on [a, b ]. (v) Monotonic function have atmost one root.

(a) The slope of the tangent is  dy  = tan ψ   dx  ( x , y ) 1

Y

Tangent

Normal

P (x1, y1)

A

C

B

X

graph desenber tangents and normal

1

 dy  (b) Equation of tangent is y − y1 =    dx  ( x

( x − x1 ) 1,

PA = y1 cosec ψ = y1

y1 )

1,

2

y1 )

Angle of Intersection of Two Curves The angle of intersection of two curves is defined to be the angle between their tangents, to the two curves at their point of intersection. The angle between the tangents of the two curves y = f1 ( x) and y = f2 ( x) is given by  dy   dy  −    dx I ( x 1 , y1 )  dx  II ( x 1 , y1 ) m1 − m2 or tan φ =  dy   dy  1 + m1 m2 1+   dx  I ( x , y )  dx  II ( x , y ) 1

1

1

1

If the angle of intersection of two curves is a right angle, the two curves are said to intersect orthogonally and the curves are called orthogonal curves. π  dy   dy  If φ = , m1 m2 = − 1 ⇒     = − 1  dx  I  dx  II 2

y1 )

• Two curves touch each other, if m1 = m2 .

Rolle’s Theorem Let f be a real-valued function defined in the closed interval [a, b ], such that (i) f ( x) is continuous in the closed interval [a, b ]. (ii) f ( x) is differentiable in the open interval (a, b ). (iii) f (a) = f (b ), then there is some point c in the open interval (a, b ), such that f ′ (c) = 0. Geometrically, Y

Y 2

 dy    dx ( x 1 , y1 )

(d) Length of subtangent AC = y1 cot ψ =

1,

 dy  (d) Length of subnormal is BC = y1 tan ψ = y1    dx  ( x

(c) Length of tangent is   dy   1 +      dx  ( x 1 , y1 

( x − x1 )

  dy   (c) Length of normal is PB = y1 sec ψ = y1 1 +       dx  ( x 1 , y1 )

NOTE

ψ

ψ O

1  dy    dx ( x

Orthogonal Curves

Tangent and Normal to a Curve (i) If a tangent is drawn to the curve y = f ( x) at a point P( x1 , y1 ) and this tangent makes an angle with positive ψ X -direction, then

(b) Equation of normal is y − y1 = −

f(c) O

y1 (dy / dx)( x 1 , y1 )

(ii) The normal to a curve at a point P( x1 , y1 ) is a line perpendicular to tangent at P and passing through P, then 1 (a) The slope of the normal is − (dy / dx)( x 1 , y1 )

a

c

b

X

O a

b

X

graph of a differentiable function, satisfying the hypothesis of Rolle’s theorem. There is atleast one point c between a and b, such that the tangent to the graph at (c, f (c)) is parallel to the X -axis.

Algebraic Interpretation of Rolle’s Theorem Between any two roots of a polynomial f ( x), there is always a root of its derivative f ′ ( x).

129

Lagrange’s Mean Value Theorem Let f be a real function, continuous on the closed interval [a, b ] and differentiable in the open interval (a, b ). Then, there is atleast one point c in the open interval (a, b ), such that Y (b, f(b)) (a, f(a)) O a

(c, f(c)) c

f (b ) − f (a) . b −a

Geometrically For any chord of the curve y = f ( x), there is a point on the graph, where the tangent is parallel to this chord. Remarks In the particular case, where f (a) = f (b ). f (b ) − f (a) becomes zero. The expression b −a Thus, when f (a) = f (b ), f ′ (c) = 0 for some c in (a, b ). Thus, Rolle’s theorem becomes a particular case of the mean value theorem.

X

b

f ′ (c) =

graph of a continuous functions explain Lagrange’s mean value theorem.

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 If the volume of a sphere is increasing at a constant rate, then the rate at which its radius is increasing, is (a) (b) (c) (d)

a constant proportional to the radius inversely proportional to the radius inversely proportional to the surface area

(a) 10

(c) 8 cm/s

(d) 12 cm/s

3 An object is moving in the clockwise direction around the unit circle x 2 + y 2 = 1. As it passes through the point  1 3  ,  , its y-coordinate is decreasing at the rate of 3 2 2  units per second. The rate at which the x-coordinate changes at this point is (in unit per second) (a) 2

(b) 3 3

(c)

3

(d) 2 3

4 The position of a point in time ‘ t ’ is given by x = a + bt − ct 2 , y = at + bt 2 . Its acceleration at time ‘ t ’ is (a) b − c

(b) b + c

(c) 2 b − 2 c

(d) 2 b 2 + c 2

5 Water is dripping out from a conical funnel of

π semi-vertical angle at the uniform rate of 2 cm 2 /s in the 4 surface area, through a tiny hole at the vertex of the bottom. When the slant height of cone is 4 cm, the rate of decrease of the slant height of water, is j NCERT Exemplar 2 cm/s 4π 1 (c) cm/s π 2 (a)

(b)

(b)

10

(c) 100

JEE Mains 2013

(d) 10 10

7 Oil is leaking at the rate of 16 cm 3 /s from a vertically kept

x = 10 + 6 t , x = 3 + t 2 . The speed with which they are reaching from each other at the time of encounter is (x is in centimetre and t is in seconds) (b) 20 cm/s

35 cc/min. The rate of increase in the surface area (in cm 2 /min) of the balloon when its diameter is 14 cm, is j

2 Moving along the X -axis there are two points with

(a) 16 cm/s

6 A spherical balloon is being inflated at the rate of

1 cm/s 4π

(d) None of these

cylindrical drum containing oil. If the radius of the drum is 7 cm and its height is 60 cm. Then, the rate at which the level of the oil is changing when oil level is 18 cm, is (a)

−16 49π

(b)

−16 48 π

(c)

16 49π

(d)

−16 47 π

8 Two men A and B start with velocities v at the same time from the junction of two roads inclined at 45° to each other. If they travel by different roads, the rate at which j NCERT Exemplar they are being separated. (a)

2−

(c)

2 ⋅v

(b)

2+

2 ⋅v

2 − 1⋅ v

(d)

2+

2 ⋅v

9 The interval in which the function f ( x ) = x 1 / x is increasing, is (a) (−∞, e) (c) (−∞, ∞)

(b) (e, ∞) (d) None of these

10 The function f ( x ) = (a) (b) (c) (d)

x is 1+ | x |

strictly increasing strictly decreasing neither increasing nor decreasing not differential at x = 0

11 The length of the longest interval, in which the function 3 sin x − 4 sin3 x is increasing, is (a)

π 3

(b)

π 2

(c)

3π 2

(d) π

130 π 2

12 An angle θ, 0 < θ < , which increases twice as fast as its sine, is

j

π (a) 2

3π (b) 2

π (c) 4

NCERT Exemplar π (d) 3

13 The value of x for which the polynomial 2 x 3 − 9x 2 + 12 x + 4 is a decreasing function of x, is (a) −1 < x < 1 (c) x > 3

14 If f ( x ) = (a) (b) (c) (d)

(b) 0 < x < 2 (d) 1 < x < 2

(a) a + b = 0 (b) a > 0

point having abscissa 2 is (a) 5 2 (a) m ≤1

(b) 3 3 (b) m > − 1

(b) a = φ (d) a = −2 ± 2 3

is parallel to the X-axis, is

4 , that x2

AIEEE 2010

(d) y = 3

a −b 1 + ab a+b (c) 1 − ab

3

(d) 3 2

x = 3 t 2 + 1, y = t 3 − 1, at x = 1 is

22 Coordinates of a point on the curve y = x log x at which the normal is parallel to the line 2 x − 2 y = 3, are

(d) m >1

x

loga − logb 1 + loga logb loga + logb (d) 1 − loga logb (b)

at right angles, then the value of b is (a) 6 (c) 4

(d) − 2

(c) m >1

32 If the curves y 2 = 6x , 9x 2 + by 2 = 16 intersect each other

21 The slope of the tangent to the curve

(b) (e,e) (d) (e − 2 , − 2 e − 2 )

(c)

31 If the curves y = a and y = b intersects at angle α , then x

(a)

j

x at the 1− x2

30 If m is the slope of a tangent to the curve e = 1 + x 2 , then

20 The equation of the tangent to the curve y = x +

(a) (0, 0) (c) (e 2 , 2 e 2 )

(c) a < 0, b < 0 (d) a = − 2b

29 The length of subnormal to the curve y =

tan α is equal to

(c) ∞

(d) (−6, − 7)

y

ax the curve y = , then 1+ x

(a) 0

(c) (6, − 7)

(d) λ > 4

19 Line joining the points (0, 3) and (5, −2) is a tangent to

1 (b) 2

(b) (−6, 7)

then

y = a , is

(c) y = 2

(b) square of the ordinate (d) None of these

28 If the line ax + by + c = 0 is normal to curve xy + 5 = 0,

(b) 2a (d) None of these

(b) y = 1

(d) 4 , 2

2

1 3 , b = − and c = 3 2 4 1 3 (b) a = , b = − and c = − 3 2 4 1 1 (c) a = , b = − and c = 3 2 4 (d) None of the above

(a) (6, 7)

18 The sum of intercepts on coordinate axes made by

(a) y = 0

(c) −4 ,14

(b) 4 ,14

circle x 2 + y 2 + 16x + 12y + c = 0 at

λ sin x + 6 cos x is monotonic increasing, if 2 sin x + 3 cos x

(a) a = 1 ± 3 (c) a = −1 ± 3

(a) 4 ,− 14

27 The tangent at (1, 7) to the curve x 2 = y − 6 touches the

f (x) is decreasing in (−∞, 1) f (x) is decreasing in (−∞, 2) f (x) is increasing in (−∞, 1) f (x) is increasing in (−∞, 2)

(a) a (c) 2 a

of the form ax − 5y + b = 0 , then a and b are

(a) square of the abscissa (c) constant

16 If f ( x ) = − 2x 3 + 21x 2 − 60x + 41, then

tangent to the curve x +

24 If the normal to the curve y 2 = 5x − 1 at the point (1,−2) is

at any point of a curve is

(d) None of these

(c) λ < 4

(d) (0, 0)

26 The product of the lengths of subtangent and subnormal

π π (b)  ,   4 2 

(b) λ < 1

(c) (2, 0)

(a) a = −

decreasing in 0 ≤ x ≤ 2 π, is

(a) λ > 1

1 (b)  − , 0  2 

P( −2, 0) and cuts the Y -axis at a point Q, where its gradient is 3. Then,

an increasing function a decreasing function both increasing and decreasing function None of the above

17 Function f ( x ) =

1 (a)  , 0 2 

25 The curve y = ax + bx + cx + 5 touches the X -axis at

15 If f ( x ) = sin x − cos x , the interval in which function is

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

y = e 2 x , meets X -axis at the point

3

1 − log(1 + x ), x > 0, then f is x +1

5π 3π (a)  ,  6 4  3π 5π (c)  ,  2 2 

23 The tangent drawn at the point (0, 1) on the curve

7 2 9 (d) 2 (b)

33 Angle between the tangents to the curve y = x 2 − 5x + 6 at the points (2, 0) and (3, 0) is π 2 π (c) 4

(a)

π 6 π (d) 3

(b)

131 x2 y2 + = 1 and y 3 = 16x intersect at right α 4 j JEE Mains 2013 angles, then the value of α is

34 If the curves

(a) 2

(b)

4 3

(c)

1 2

(d)

3 4

2

and f ( x ) is continuous in [1, 2], then ∫ f ′ ( x ) dx is equal to 1

(b) 0

(c) 1

(d) 2

36 If the function f ( x ) = x 3 − 6x 2 + ax + b satisfies Rolle’s

 2 3 + 1 theorem in the interval [1, 3] and f ′   = 0, then  3  (a) a = −11

(b) a = − 6

(c) a = 6

theorem holds for the function f ( x ) = loge x on the interval j AIEEE 2007 [1, 3] is

(d) a = 11

5

then ∫ f ( x ) dx is equal to

39 The abscissa of the points of the curve y = x 3 in the interval [ −2, 2], where the slope of the tangents can be obtained by mean value theorem for the interval [ −2, 2], are (a) ±

2 3

(b) +

(c) ±

3 2

(d) 0

3

40 In the mean value theorem, f (b ) − f (a ) = (b − a )f ′ (c ), if

3

(b) −1

(b)

(c) log3 e

37 If f ( x ) satisfies the conditions for Rolle’s theorem in [3, 5],

(a) 2

1 loge 3 2 (d) loge 3

(a) 2 log3 e

35 f ( x ) satisfies the conditions of Rolle’s theorem in [1, 2] (a) 3

38 A value of C for which the conclusion of mean value

(d) −

(c) 0

4 3

a = 4 , b = 9 and f ( x ) = x , then the value of c is (a) 8.00

(b) 5.25

(c) 4.00

(d) 6.25

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 A kite is moving horizontally at a height of 151.5 m. If the speed of kite is 10 m/s, how fast is the string being let out, when the kite is 250 m away from the boy who is flying the j NCERT Exemplar kite? The height of boy is 1.5 m. (a) 8 m/s

(b) 12 m/s

(c) 16 m/s

(d) 19 m/s

2 The normal to the curve y ( x − 2) ( x − 3) = x + 6 at the point, where the curve intersects the Y -axis passes through the point 1 1 (a)  − , −   2 2 1 1 (c)  , −  2 3

1 (b)  , 2 1 (d)  , 2

1  2 1  3

3 The values of a for which the function (a + 2)x 3 − 3 ax 2 + 9 ax − 1 = 0 decreases monotonically throughout for all real x, are (a) a < − 2 (c) −3 < a < 0

(b) a > − 2 (d) − ∞ < a ≤ − 3

x x and g ( x ) = , where 0 < x ≤ 1, then in sin x tan x this interval

4 If f ( x ) = (a) (b) (c) (d)

both f (x) and g (x) are increasing functions both f (x) and g (x) are decreasing functions f (x) is an increasing function g (x) is an increasing function x

5 f ( x ) = ∫ | log2 [log3 {log4 (cos t + a )}]| dt . If f ( x ) 0

is increasing for all real values of x, then (a) a ∈ (−11 ,) (c) a ∈ (1, ∞)

(b) a ∈ (15 , ) (d) a ∈ (5 , ∞)

6 If f ( x ) satisfy all the conditions of mean value theorem in [ 0 , 2 ]. If f ( 0) = 0 and | f ′ ( x )| ≤ (a) f (x) < 2 (c) f (x) = 2 x

1 for all x in [ 0, 2 ], then 2

(b) | f (x)| ≤ 1 (d) f (x) = 3 for atleast one x in [0, 2]

π 2 and g ( x ) = f (sin x ) + f (cos x ), then g ( x ) is decreasing in 

7 If f ′ (sin x ) < 0 and f ′ ′ (sin x ) > 0 , ∀x ∈  0 ,    π π (a)  ,   4 2

π (b)  0,   4

π (c)  0,   2

π π (d)  ,   6 2

8 If f ( x ) = ( x − p )( x −q )( x −r ), where p < q < r , are real numbers, then application of Rolle’s theorem on f leads to (a) (p + q + r )(pq + qr + rp) = 3 (b) (p + q + r )2 = 3 (pq + qr + rp) (c) (p + q + r )2 > 3 (pq + qr + rp) (d) (p + q + r )2 < 3 (pq + qr + rp)

9 If f ( x ) is a monotonic polynomial of 2m −1 degree, where m ∈ N, then the equation [f ( x ) + f ( 3x ) + f ( 5x ) + ...+ f ( 2m − 1)x ] = 2m − 1 has (a) atleast one real root (c) exactly one real root

(b) 2m roots (d) (2m + 1) roots

10 A spherical balloon is filled with 4500 π cu m of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72π cu m/min, then the rate (in m/min) at which the radius of the balloon decreases 49 min after j AIEEE 2012 the leakage began is (a)

9 7

(b)

7 9

(c)

2 9

(d)

9 2

132 11 The normal to the curve x 2 + 2xy − 3y 2 = 0 at (1, 1)

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

j JEE Mains 2015 (a) does not meet the curve again (b) meets the curve again in the second quadrant (c) meets the curve again in the third quadrant (d) meets the curve again in the fourth quadrant

15 The angle of intersection of curves, y = [| sin x | + | cos x | ] and x 2 + y 2 = 5, where [⋅] denotes greatest integral function is

12 If f and g are differentiable functions in ( 0, 1) satisfying f ( 0) = 2 = g (1), g ( 0) = 0 and f (1) = 6, then for some j JEE Mains 2014 c ∈] 0, 1[ (a) 2f ′ (c) = g ′ (c) (c) f ′ (c) = g ′ (c)

(a)

(b) 2f ′ (c) = 3g ′ (c) (d) f ′ (c) = 2g ′ (c)

1  1 − x, x<  2 2 (a) f (x) =  (b) f (x) = 2 1 1  − x  , x ≥  2  2

increasing function of x. Then, 0

∫

b 0

(d) None of these

applicable to

(b) f ′ (1) = 1 (d) 2f (0) = 1 + f ′ (0)

14 Let a + b = 4, a < 2 and g ( x ) be a monotonically a

1 (b) tan−1    2

16 In [0, 1], Lagrange’s mean value theorem is not

by the line y = x at the point where x = 1, then

f ( x ) = ∫ g ( x ) dx +

π 4

(c) tan−1 (2)

13 If y = f ( x ) is the equation of a parabola which is touched (a) 2f ′ (0) = 3f ′ (1) (c) f (0) + f ′ (1) + f ′ ′ (1) = 2

increases with increase in (b − a) decreases with increase in (b − a) increases with decreases in (b − a) None of the above

(c) f (x) = x | x |

g ( x ) dx

 sin x , x ≠ 0   x  1, x=0

(d) f (x) = | x |

ANSWERS SESSION 1

SESSION 2

1 (d)

2 (c)

3 (b)

4 (d)

5 (a)

6 (a)

7 (c)

8 (a)

9 (a)

10 (a)

11 (a)

12 (d)

13 (d)

14 (b)

15 (d)

16 (b)

17 (d)

18 (a)

19 (b)

20 (d)

21 (a)

22 (d)

23 (b)

24 (a)

25 (a)

26 (b)

27 (d)

28 (c)

29 (d)

30 (a)

31 (b)

32 (d)

33 (a)

34 (b)

35 (b)

36 (d)

37 (d)

38 (a)

39 (a)

40 (d)

1 (a) 11 (d)

2 (b) 12 (d)

3 (d) 13 (b)

4 (c) 14 (a)

5 (d) 15 (c)

6 (b) 16 (a)

7 (b)

8 (c)

9 (a)

10 (c)

Hints and Explanations SESSION 1 1 Given that, dV = k (say) dt 4 3 dV dR = 4 πR 2 Q V = πR ⇒ 3 dt dt dR k = ⇒ dt 4 πR 2 Rate of increasing radius is inversely proportional to its surface area. 2 They will encounter, if 10 + 6t = 3 + t 2 2 ⇒ t − 6t − 7 = 0 ⇒ t = 7 At t = 7 s, moving in a first point d v 1 = (10 + 6t ) = 6 cm/s dt At t = 7 s, moving in a second point d v 2 = (3 + t 2 ) = 2t = 2 × 7 = 14 cm/s dt ∴ Resultant velocity = v 2 − v 1 = 14 − 6 = 8 cm/s

3 The equation of given circle is x + y =1 On differentiating w.r.t. t, we get dy dx 2x + 2y =0 dt dt dy dx ⇒ x + y =0 dt dt 2

2

1 3 But we have, x = , y = 2 2 dy 1 dx 3 = − 3, then + (− 3) = 0 dt 2 dt 2 dx =3 3 ⇒ dt and

4 Given point is x = a + bt − ct 2 Acceleration in x direction and point d2 x is y = at + bt 2 = 2 = − 2c dt

and point is y = at + bt 2 acceleration in y direction d2 y = 2 = 2b dt ∴ Resultant acceleration 2

=

2

2 d y  d x  2 + 2   dt   dt 

2

= (−2c )2 + (2b )2 = 2 b 2 + c 2

5 If S represents the surface area, then dS = 2 cm2 / s dt r

h

l π/4

133 S = πrl = πl ⋅ sin

π π 2 l = l 4 2

Therefore, dS 2 π dl dl = l⋅ = 2 πl ⋅ dt dt dt 2 dl 2 = dt 2π ⋅ 4 1 2 = = cm / s 2 2π 4π

when l = 4 cm,

From ∆LOM, OL2 + OM 2 − LM 2 cos 45° = 2 ⋅ OL ⋅ OM x2 + x2 − y 2 2 x2 − y 2 1 = = ⇒ 2 x2 2 2⋅ x ⋅ x 2 2 2 ⇒ 2x = 2x − y ⇒

(2 − 2 ) x2 = y 2

∴

6 V = 4 πr 3 ⇒ dV = 4 π3r 2 dr 3

dt 3 dt dr dr 5 ⇒ = dt dt 28 π Surface area of balloon, S = 4 πr 2 dS dr ∴ = 8πr dt dt 5 = 8π ×7× = 10 cm2 / min 28 π ⇒ 35 = 4 π (7)2

7 Let h be height of oil level at any instant t and V be the volume of oil in cylindrical drum. Given, h = 60 cm, r = 7 cm dV and = − 16 cm3 /s dt

H h

So, height of oil is decreasing at the rate 16 of cm/s. 49π

8 Let L and M be the positions of two men A and B at any time t. Let OL = x and LM = y Then, OM = x dy dx Given, = v and we have to find dt dt M

y x

O A

x

L

For decreasing, f ′ ( x ) < 0 π  π 3π <  x −  < 2  4 2 (within 0 ≤ x ≤ 2 π) π π  π π 3π π < x − +  < + ⇒ + 2 4  4 4 2 4 3π 7π ⇒ < x< 4 4

9 Given, f ( x ) = x1 / x 1 (1 − log x ) x1 / x x2 f ′ ( x ) > 0, if 1 − log x > 0 ⇒ log x < 1 ⇒ x < e ∴ f ( x ) is increasing in the interval (− ∞, e ). x 10 Given, f ( x) = 1 + | x| | x| (1 + | x |) ⋅ 1 − x ⋅ x ∴ f ′( x ) = (1 + | x |)2 1 = > 0, ∀ x ∈ R (1 + | x |)2 f ′( x ) =

Since, sin x is increasing in the interval − π , π  ⋅  2 2  π π π π ≤ 3x ≤ ⇒ − ≤ x≤ 2 2 6 6 Thus, length of interval π  π π = − −  = 6  6 3

∴

−

12 Given, 2 d (sinθ) = dθ dt dt dθ dθ = ⇒ 2 × cos θ dt dt ⇒ 2cos θ = 1 ⇒ cos θ =

16 Given, …(i) f ( x ) = − 2 x 3 + 21 x2 − 60 x + 41 On differentiating Eq. (i) w.r.t. x, we get f ′ ( x ) = − 6 x2 + 42 x − 60 = − 6 ( x2 − 7 x + 10 ) = − 6 ( x − 2) ( x − 5) If x < 2, f ′ ( x ) < 0 i.e. f ( x ) is decreasing. 17 Q f ( x ) = λ sin x + 6cos x …(i) 2sin x + 3cos x On differentiating w.r.t. x, we get  (2sin x + 3cos x )   (λ cos x − 6sin x )   − (λ sin x + 6cos x )    (2cos x − 3sin x )  f ′( x ) = (2sin x + 3cos x )2 The function is monotonic increasing, if f ′ ( x ) > 0 ⇒

⇒ f ′ ( x ) = 6 x2 − 18 x + 12 f ′ ( x ) < 0 for function to be decreasing ⇒ 6( x2 − 3 x + 2) < 0 ⇒ ( x2 − 2 x − x + 2) < 0 ( x − 2)( x − 1) < 0 ⇒ 1 < x < 2

1 − log(1 + x ) x+1 On differentiating w.r.t. x, we get

3λ (sin2 x + cos 2 x ) − 12 (sin2 x + cos 2 x )> 0

⇒ 3λ − 12 > 0

1 π ⇒θ = 2 3

13 Let f ( x ) = 2 x3 − 9 x2 + 12 x + 4

⇒

45°

On differentiating w.r.t. x, we get f ′ ( x ) = cos x + sin x 1 1 = 2  cos x + sin x   2  2 π π = 2  cos cos x + sin sin x    4 4 π   = 2 cos  x −   4   

Hence, they are being separated from each other at the rate 2 − 2 v.

14 Given curve is f ( x ) = B

15 Q f ( x ) = sin x − cos x

11 Let f ( x ) = 3 sin x − 4 sin3 x = sin 3 x

dV dh Q V = πr 2 h ⇒ = πr 2 dt dt (since, r is constant all the time) dh dh 16 ⇒ =− ⇒ −16 = π(7)2 dt dt 49π dh 16 ⇒   =−  dt  at h =18 49π

⇒ f ′ ( x ) = − ve, when x > 0 ∴ f ( x ) is a decreasing function.

On differentiating w.r.t. t, we get dy dx = 2− 2 dt dt Q dx = v  = 2 − 2v    dt 

⇒

1 1 − ( x + 1)2 1 + x

1 1  ⇒ f ′( x ) = −  +  x + 1 ( x + 1)2 

y = 2 − 2x

⇒ f ( x ) is strictly increasing.

r

f ′( x ) = −

⇒

18

x +

(Q sin2 x + cos 2 x = 1)

λ> 4 y =

a ; (i )

1 2

x

1

+

2

y dy =− ∴ dx x Hence, tangent at ( x, y ) is y Y − y =− ( X − x) x ⇒ X y + Y x = xy ( x + ⇒

X y +Y x =

dy =0 y dx

y)

xy a (using Eq. (i))

134 ⇒

X + a x

Y =1 a y

dy dy = 2e 2 x ⇒   = 2e 0 = 2 dx  dx  ( 0, 1 )

Clearly, its intercepts on the axes are a ⋅ x and a ⋅ y . Sum of intercepts = =

a( x + a⋅ a = a

and (5, −2) is y = 3 − x. If this line is ax tangent to y = , then ( x + 1) (3 − x ) ( x + 1) = ax should have equal roots. Thus, (a − 2)2 + 12 = 0 ⇒ no value of a ⇒ a ∈ φ. 20 We have, y = x + 42 x On differentiating w.r.t. x, we get dy 8 = 1− 3 dx x Since, the tangent is parallel to X -axis, therefore dy = 0 ⇒ x3 = 8 dx ∴ x = 2 and y = 3

21 Given curve is x = 3t 2 + 1, y = t 3 − 1 For x = 1, 3t 2 + 1 = 1 ⇒ t = 0 dy dx ∴ = 6t , = 3t 2 dt dt  dy  dy  dt  3t 2 t Now, = = = dx  dx  6t 2  dt  0  dy  = =0    dx  ( t = 0 ) 2

22 Given curve is y = x log x On differentiating w.r.t. x, we get dy = 1 + log x dx The slope of the normal −1 1 =− = (dy /dx ) 1 + log x The slope of the given line 2 x − 2 y = 3 is 1. Since, these lines are parallel. −1 ∴ =1 1 + log x ⇒ ⇒ and

y − 1 = 2( x − 0) ⇒ y = 2 x + 1

y)

19 Equation of line joining the points (0, 3)

∴

Equation of tangent at (0, 1) with slope 2 is

log x = − 2 x = e −2 y = − 2e − 2

∴ Coordinates of the point are (e − 2 , − 2e − 2 ).

23 Given curve is y = e 2 x On differentiating w.r.t. x, we get

This tangent meets X -axis. ∴

y = 0 ⇒ 0 = 2x + 1 ⇒ x = −

1 2

∴ Coordinates of the point on X -axis is  − 1 , 0 .    2 

24 We have, y 2 = 5x − 1 At (1, − 2),

…(i)

dy  5  −5 = =  dx  2 y  (1 , −2 ) 4

∴ Equation of normal at the point (1, − 2) is −5 [ y − (− 2)]   + x − 1 = 0  4 ∴ 4 x − 5y − 14 = 0 As the normal is of the form ax − 5y + b = 0

…(ii)

On comparing this with Eq. (ii), we get a = 4 and b = − 14

25 Since, we have the curve y = ax3 + bx2 + cx + 5 touches X -axis at P(−2, 0), then X -axis is the tangent at (−2, 0). The curve meets Y-axis in (0, 5). dy = 3ax2 + 2bx + c dx dy (given) ⇒   = 0 + 0 + c = 3  dx  0, 5 ⇒

c =3

…(i)

dy and   =0  dx  ( −2, 0 ) [from Eq. (i)] ⇒ 12a − 4b + c = 0 …(ii) ⇒ 12a − 4b + 3 = 0 and ( −2, 0) lies on the curve, then 0 = − 8a + 4b − 2c + 5 (Q c = 3) ⇒ 0 = − 8a + 4b − 1 …(iii) ⇒ 8a − 4b + 1 = 0 From Eqs. (ii) and (iii), we get 1 3 a = − ,b = − 2 4 26 Length of subtangent = y dx dy dy and length of subnormal = y dx ∴

Product = y 2

⇒ Required product is the square of the ordinate.

27 The tangent to the parabola x2 = y − 6 at (1, 7) is 1 x(1) = ( y + 7) − 6 ⇒ y = 2 x + 5 2

which is also a tangent to the given circle. i.e. x2 + (2 x + 5)2 + 16 x + 12 (2 x + 5) + c = 0 ⇒ (5x2 + 60 x + 85 + c = 0) must have equal roots. Let the roots be α = β. 60 α +β=− ∴ 5 ⇒ α =−6 ∴ x = − 6 and y = 2x + 5 = − 7

28 Given curve, is xy = − 5 < 0 dy dy −y + y = 0⇒ = >0 dx dx x [as xy = − 5 < 0] −1 x Slope of normal = = 0 b b ⇒ a > 0, b > 0 or a < 0, b < 0 29 Given, y = x 2 1− x ⇒

At

x

x = 2, y = − 2, therefore

point is ( 2, − 2 ). dy 1 + x2 = dx (1 − x2 )2

∴ dy ⇒    dx  (

= 2, − 2 )

1+ 2 =3 (1 − 2)2

∴ length of subnormal at ( 2, − 2 ) = |(− 2 )(3)| = 3 2

30 We have, e y = 1 + x2 ⇒ ⇒ ⇒

ey

dy =2x dx dy 2x = dx 1 + x2 2x , 1 + x2 2| x| 2| x| |m|= = ≤1 |1 + x2 | 1 + | x|2 m=

Q

| x|2 + 1 − 2| x|≥ 0 ≥

⇒

(| x|− 1 )2 ≥ 0

⇒ ⇒

| x|2 + 1 ≥ 2| x| 2 | x| 1≥ 1 + | x|2

135 Now, apply the condition of perpendicularity of two curves, i.e. m1 m2 = − 1 4 and get α = with the help of equation 3 of curves.

∴

⇒ 3  2 + 

Also, ⇒ ⇒

3 y1

9 x + by = 16 dy 18 x + 2by =0 dx dy −9 x = dx by 2

−9 x1 Slope of tangent at ( x1 , y 1 ) is m2 = by 1 Since, these are intersection at right angle. 27 x1 ∴ m1 m2 = − 1 ⇒ =1 by 12 27 x1 [Q y 12 = 6 x1 ] ⇒ =1 6bx1 9 ⇒ b = 2

33 Q ∴ Now, and Now,

⇒

1 Let AB be the position of boy who is flying the kite and C be the position of the kite at any time t. C y

2

a = 11

37 Since, f ( x ) satisfies all the conditions of Rolle’s theorem in [3, 5]. Let f ( x ) = ( x − 3)( x − 5) = x2 − 8 x + 15

∫

Now,

5 3

f ( x )dx =

∫

5 3

( x2 − 8 x + 15)dx

5

=

50 4 − 18 = − 3 3

38 Using mean value theorem, f (3) − f (1) 3−1

⇒

1 log e 3 − log e 1 = c 2

dy m2 =   = 6 − 5= 1  dx  (3, 0 )

∴

c =

m1 m2 = − 1 × 1 = − 1

2 2 34 Slope of the curve, x + y = 1 is

α

4

f (b ) − f (a)  Q f ′ (c ) =  b − a 

2 = 2 log 3 e log e 3

39 Given that, equation of curve y = x = f ( x) So, f (2) = 8 and f (− 2) = − 8 f (2) − f (– 2) Now, f ′( x ) = 3 x2 ⇒ f ′( x ) = 2 − (− 2) 3

⇒

Now, slope of the curve, y = 16 x is 16 . m2 = 3 y2

1.5 m

3

∴

8 − (− 8) = 3 x2 4 x= ±

B

D

x

Let BD = x and AC = y , then AE = x Given, AB = 1.5 m, CD = 151.5 m ∴

CE = 150 m dx Given, = 10 m/s dt dy when dt

y = 250 m

125 =  − 100 + 75 − (9 − 36 + 45)  3 

f ′ (c ) =

E

A

Here, we have to find

3 8 x2 x  − + 15x 2  3 3

=

1 2

x f (b ) − f (a) 3 − 2 1 Also, f ′ (c ) = = = b −a 9− 4 5 1 1 25 = ⇒C = = 6.25 ∴ 5 4 2 c

1  1    − 12  2 +  + a= 0  3 3

y = x − 5x + 6 dy = 2x − 5 dx dy m1 =   = 4 − 5= −1  dx  (2 , 0 )

−4 x . αy

9 = 3; f ′ ( x ) =

f (b ) =

⇒ 12 + 1 + 4 3 − 24 − 4 3 + a = 0

2

π Hence, angle between the tangents is . 2

m1 =

4=2

1 4  1   ⇒ 3  4 + +  − 12  2 +  + a= 0 3   3 3

Slope of tangent at ( x1 , y 1 ) is m1 = 2

2

On differentiating w.r.t. x, we get f ′ ( x ) = 3 x2 − 12 x + a By the definition of Rolle’s theorem 1  f ′ (c ) = 0 ⇒ f ′  2 +  =0  3

dy =6 dx dy 3 = dx y

∴ f (a) =

SESSION 2

f (2) = f (1)

36 Q f ( x )= x − 6 x + ax + b

32 We have, y 2 = 6 x

⇒

f ′ ( x )dx = [ f ( x )]21 = f (2) − f (1) = 0

3

m1 − m2 1 + m1 m2 log a − log b = 1 + log a log b

2y

2 1

[Q f ( x ) satisfies the conditions of Rolle’s theorem]

∴ tan α =

⇒

∫

x

151.5 m

Slope of tangent of second curve, dy m2 = = b x log b dx dy ⇒ m2 =   = log b  dx  ( 0 , 1 )

35

f ( x) =

40

150 m

curves is (0, 1). Now, slope of tangent of first curve, dy m1 = = ax log a dx dy ⇒   = m1 = log a  dx  ( 0 , 1 )

1.5 m

31 Clearly, the point of intersection of

2 3

Now, from ∆CAE, y 2 = x2 + 1502 On differentiating, we get dy dx 2y = 2x dt dt ⇒

dy x dx x = ⋅ = ⋅ 10 dt y dt y

In ∆ACE, x = 2502 − 1502

…(i) (Q y = 250)

= 200 m ∴ From Eq. (i), we get dy 200 = × 10 = 8 m/s dt 250

2 Given curve is y ( x − 2)( x − 3) = x + 6 …(i) Put x = 0 in Eq. (i), we get y(− 2) (− 3) = 6 ⇒ y = 1 So, point of intersection is (0, 1). x+ 6 Now, y = ( x − 2)( x − 3) 1 ( x − 2)( x − 3) − ( x + 6) dy ( x − 3 + x − 2) ⇒ = dx ( x − 2)2 ( x − 3)2

136 dy 6 + 30 36 ⇒   = = =1  dx  ( 0, 1 ) 4 × 9 36 ∴ Equation of normal at (0, 1) is given by −1 y −1= ( x − 0) ⇒ x + y − 1 = 0 1 1 1 which passes through the point  ,  .  2 2

3 Let f ( x ) = (a + 2) x3 − 3 ax2 + 9 ax − 1 decreases monotonically for all x ∈ R, then f ′ ( x ) ≤ 0 for all x ∈ R ⇒ 3(a + 2) x2 − 6 ax + 9 a ≤ 0 for all x ∈ R ⇒ (a + 2) x2 − 2 ax + 3 a ≤ 0 for all x ∈ R ⇒ a + 2 < 0 and discriminant ≤ 0 ⇒ a < − 2, − 8 a2 − 24a ≤ 0 ⇒ a < − 2 and a (a + 3) ≥ 0 ⇒ a < − 2, a ≤ − 3 or a ≥ 0 ⇒ a ≤ − 3 ∴ − ∞ < a≤ − 3 sin x − x cos x 4 Now, f ′( x ) = sin2 x cos x (tan x − x ) = sin2 x ∴ f ′ ( x ) > 0 for 0 < x ≤ 1 So, f ( x ) is an increasing function. tan x − x sec2 x Now, g ′ ( x ) = tan2 x sin x cos x − x sin 2 x − 2 x = = sin2 x 2sin2 x d Again, (sin 2 x − 2 x ) = 2 cos 2 x − 2 dx = 2 (cos 2 x − 1) < 0 So, sin 2 x − 2 x is decreasing. sin 2 x − 2 x < 0 ⇒ ∴ g ′ ( x) < 0 So, g ( x ) is decreasing.

5 f ′( x ) = | log 2 [log 3 {log 4 (cos x + a)}] | Clearly, f ( x ) is increasing for all values of x, if log 2 [log 3 {log 4 (cos x + a)}] is defined for all values of x. ⇒ log 3 [log 4 (cos x + a)] > 0, ∀ x ∈ R log 4 (cos x + a) > 1, ∀ x ∈ R ⇒ cos x + a > 4, ∀ x ∈ R ⇒ ∴ a> 5 f (2) − f (0) 6 Since, = f ′( x ) 2− 0 f (2) − 0 df ( x ) f (2) ⇒ = f ′( x ) ⇒ = 2 dx 2 f (2) ⇒ f ( x) = x+C 2 Q f (0) = 0 ⇒ C = 0 f (2) …(i) ∴ f ( x) = x 2 f (2) 1 1 Also, | f ′ ( x )|≤ ⇒ …(ii) ≤ 2 2 2

From Eq. (i), | f ( x )| f (2) f (2) 1 = x = | x| ≤ | x| 2 2 2 [from Eq. (ii)] In interval [0, 2], for maximum x, 1 | f ( x )| ≤ ⋅ 2 ⇒ | f ( x )| ≤ 1 [Q x = 2] 2

7 g ′( x ) = f ′(sin x ) ⋅ cos x − f ′(cos x ) ⋅ sin x ⇒ g ′ ′ ( x ) = − f ′ (sin x ) ⋅ sin x + cos2 x f ′ ′ (sin x ) + f ′ ′ (cos x ) ⋅ sin2 x − f ′ (cos x ) ⋅ cos x π > 0, ∀ x ∈  0,   2 π ⇒ g ′( x ) is increasing in  0,  .  2 π Also, g ′   = 0  4 π π ⇒ g ′ ( x ) > 0, ∀ x ∈  ,   4 2 π and g ′ ( x ) < 0, ∀ x ∈  0,   4 π Thus, g ( x ) is decreasing in  0,  .  4

8 We have, f ( x ) = ( x − p )( x − q )( x − r ) ⇒ f ( p ) = 0 = f (q ) = f (r ) ⇒ p,q and r are three distinct real roots of f ( x ) = 0 So, by Roolle’s theorem, f '( x ) has one real root in the interval ( p,q ) and other in the interval (q, r ). Thus, f '( x ) has two distinct real roots. Now, f ( x ) = ( x − p )( x − q )( x − r ) ⇒ f ( x ) = x3 − x2 ( p + q + r ) + x( pq + qr + rp ) − pqr ⇒ f '( x ) = 3 x2 − 2( p + q + r )x + ( pq + qr + rp ) As f '( x ) has distinct real roots ∴ 4( p + q + r )2 − 12( pq + qr + rp ) > 0 ⇒ ( p + q + r )2 > 3( pq + qr + rp )

9 Given that, f ( x ) is monotonic. ⇒ f ′( x ) = 0 or f ′( x ) > 0, ∀ x ∈ R ⇒ f ′( px ) < 0 or f ′( px ) > 0, ∀ x ∈ R So, f ′( px ) is also monotonic. Hence, f ( x ) + f (3 x ) + ... + f [(2m − 1)x] is a monotonic. Polynomial of odd degree (2m − 1), so it will attain all real values only once.

10 Since, the balloon is spherical in shape, hence the volume of the balloon is 4 V = πr 3. 3 On differentiating both the sides w.r.t. t, we get dV 4 dr  = π  3 r 2 ×  dt 3  dt  dr dV / dt …(i) = ⇒ dt 4 πr 2

dr at the rate t = 49 min, dt dV we require the radius (r ) at that dt dV stage. = − 72 π m3 / min dt Also, amount of volume lost in 49 min = 72 π × 49 m3 ∴ Final volume at the end of 49 min = (4500 π − 3528 π ) m3 = 972 π m3 If r is the radius at the end of 49 min, 4 then πr 3 = 972 π ⇒ r 3 = 729 3 ⇒ r =9 Radius of the balloon at the end of 49 min = 9 m From Eq. (i),  dV     dt  t = 49 dr dV / dt dr   = ⇒  = 2  dt  t = 49 4π (r 2 ) t = 49 dt 4 πr Now, to find

72 π 2  dr  = = m/min    dt  t = 49 4 π(9 2 ) 9

11 Given equation of curve is x2 + 2 xy − 3 y 2 = 0 On differentiating w.r.t. x, we get 2 x + 2 xy ′ + 2 y − 6 yy ′ = 0 x+ y ⇒ y′ = 3y − x dy At x = 1, y = 1, y ′ = 1 i.e.   =1  dx  (1 , 1 ) Equation of normal at (1, 1) is 1 y − 1 = − ( x − 1) ⇒y − 1 = − ( x − 1) 1 ⇒ x+ y =2 On solving Eqs. (i) and (ii) simultaneously, we get x2 + 2 x (2 − x ) − 3 (2 − x )2 = 0 ⇒ x2 + 4 x − 2 x2 − 3(4 + x2 − 4 x ) = 0 ⇒ − x2 + 4 x − 12 − 3 x2 + 12 x = 0 ⇒ − 4 x2 + 16 x − 12 = 0 4 x2 − 16 x + 12 = 0 ⇒ x2 − 4 x + 3 = 0 ⇒ ( x − 1) ( x − 3) = 0 ⇒ ⇒ x = 1, 3 Now, when x = 1, then y = 1 and when x = 3, then y = − 1 ∴ P = (1, 1) and Q = (3, − 1) Hence, normal meets the curve again at (3, − 1) in fourth quadrant. Alternate Method Given, x2 + 2 xy − 3 y 2 = 0 ⇒ ( x − y ) ( x + 3y ) = 0 ⇒ x − y = 0 or x + 3 y = 0 Equation of normal at (1, 1) y − 1 = − 1 ( x − 1) ⇒ x + y − 2 = 0 It intersects x + 3 y = 0 at (3, − 1) and hence normal meets the curve in fourth quadrant.

137 x+y=2 x + 3y = 0

y=x

Y

14 a + b = 4 ⇒ b = 4 − a and b − a = 4 − 2a = t Now, ∫

a 0

g( x) dx +

0

dx +

(1, 1) X′

∫

b

X O (3, –1) Y′

0

∫

0

g( x)

g ( x ) dx = I (a)

dI (a) = g (a) − g (4 − a) da As a < 2 and g( x) is increasing. ⇒ 4 − a > a ⇒ g(a) − g(4 − a) < 0

Now,

and

dI (a) dI (a) dt dI (a) = = − 2⋅ d (a) dt da dt

dI (a) >0 dt Thus, I(a) is an increasing function of t. Hence, the given expression increasing with (b − a). ⇒

15 We know that, 1 ≤ |sin x | + |cos x |≤ 2 √5 P

O

[equation of parabola] As it touches y = x at x = 1. ∴ y = a+ b + c and y =1 ⇒ a+ b + c =1 dy Now, = 2ax + b dx dy = 2a + b ⇒ 2a + b = 1 ⇒    dx  at x =1

–√5

y=1 x

O

x 2+y 2=5 –√5 ⇒

±2 Q(−2, 1) + y 2 = 5 w.r.t. x, ⇒

dy x =− dx y

 dy  = −2    dx  (2, 1 )

dI (a) 0 (very small quantity) Local minimum D X X′ In such a case f (a) is said to be the maximum value of f ( x) at x = a. A function f ( x) is said to attain a minimum at B x = a, if there exists a neighbourhood (a − δ, a + δ) Least value/ minimum absolute Y′ such that f ( x) > f (a), ∀x ∈(a − δ, a + δ), x ≠ a. Graph of a continuous function explained local maxima (minima) and absolute maxima (minima). In such a case f (a) is said to be the minimum value of f ( x) at x = a. The points at which a function attains either the maximum or the minimum values are known as the extreme points or turning points and both minimum and maximum values of f ( x) are called extreme values. The turning points A and C are called local maximum and points B and D are called local minimum.

PRED MIRROR

Critical Point ●

●

A point c in the domain of a function f at which either f ′ (c) = 0 or f is not differentiable is called a critical point of f. Note that, if f is continuous at point c and f ′ (c) = 0, then there exists h > 0 such that f is differentiable in the interval (c − h, c + h). The converse of above theorem need not be true, that is a point at which the derivative vanishes need not be a point of local maxima or local minima.

Your Personal Preparation Indicator u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

Method to Find Local Maxima or Local Minima First Derivative Test Let f be a function defined on an open interval I and f be continuous at a critical point c in I. Then, (i) If f ′ ( x) changes sign from positive to negative as x increases through c, i.e. if f ′ ( x) > 0 at every point sufficiently close to and to the left of c and f ′ ( x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima.

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

139

DAY (ii) If f ′ ( x) changes sign from negative to positive as x increases through point c, i.e. if f ′ ( x) < 0 at every point sufficiently close to and to the left of c and f ′ ( x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima. (iii) If f ′ ( x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflection. Y

Point of local maxima

f′ (

0

Point of non-differentiability and point at local maxima

f′(c1)=0

0

x) >0

f′ (

c1

O

●

●

●

●

Point of non-differentiability but it is a point of local minima

f ′(c2)=0 Point of local minima X′

●

x)
f ′(

Important Results

c2

c3

●

X

c4

If n is even and f n (a) < 0 ⇒ x = a is a point of local maximum. If n is even and f n (a) > 0 ⇒ x = a is a point of local minimum. If n is odd ⇒ x = a is a point of neither local maximum nor a point of local minimum. ax + b The function f ( x) = has no local maximum or cx + d minimum regardless of values of a, b , c and d. The function f (θ) = sin m θ ⋅ cos n θ attains maximum values at  m θ = tan −1   .  n If AB is diameter of circle and C is any point on the circumference, then area of the ∆ ABC will be maximum, if triangle is isosceles.

Y′

Graph of f around c explained following points. (iv) If c is a point of local maxima of f, then f (c) is a local maximum value of f . Similarly, if c is a point of local minima of f, then f (c) is a local minimum value of f .

Second or Higher Order Derivative Test (ii) Find f ′ ′ ( x) and at x = a1 , (a) if f ′ ′ (a1 ) is positive, then f ( x) is minimum at x = a1 . (b) if f ′ ′ (a1 ) is negative, then f ( x) is maximum at x = a1 . (iii) (a) If at x = a1 , f ′ ′ (a1 ) = 0, then find f ′ ′ ′ ( x). If f ′ ′ ′ (a1 ) ≠ 0 , then f ( x) is neither maximum nor minimum at x = a. (b) If f ′ ′ ′ (a1 ) = 0, then find f iv ( x). iv

(c) If f ( x) is positive (minimum value) and f ( x) is negative (maximum value). (iv) If at x = a1 , f (a1 ) = 0 , then find f ( x) and proceed similarly. iv

v

Point of Inflection At point of inflection (i) It is not necessary that 1st derivative is zero. (ii) 2nd derivative must be zero or 2nd derivative changes sign in the neighbourhood of point of inflection.

nth Derivative Test Let f be a differentiable function on an interval I and a be an interior point of I such that (i) f ′ (a) = f ′′(a) = f ′′′(a) = ... = f n −1 (a) = 0 and (ii) f n (a) exists and is non-zero.

●

●

(i) Find f ′ ( x) and equate it to zero. Solve f ′ ( x) = 0 let its roots be x = a1 , a2 ,...

iv

Concept of Global Maximum/Minimum Let y = f ( x) be a given function with domain D and [a, b ] ⊆ D, then global maximum/minimum of f ( x) in [a, b ] is basically the greatest / least value of f ( x) in [a, b ]. Global maxima/minima in [a, b ] would always occur at critical points of f ( x) within [a, b ] or at end points of the interval.

Global Maximum/Minimum in [a, b] In order to find the global maximum and minimum of f ( x) in [a, b ]. Step I

Find out all critical points of f ( x) in [a, b ] [i.e. all points at which f ′ ( x) = 0] points are c1 , c2 ,..., c n .

Step II

and let these

Find the value of f (c1 ), f (c2 ) ,..., f (c n ) and also at the end points of domain i.e. f (a) and f (b ).

Step III

Find M1 → Global maxima or greatest value and M2 → Global minima or least value. where, M1 = max { f (a), f (c1 ), f (c2 ),..., f (c n ), f (b )} and M2 = min { f (a), f (c1 ), f (c2 ),..., f (c n ), f (b )}

Some Important Results on Maxima and Minima (i) Maxima and minima occur alternatively i.e. between two maxima there is one minimum and vice-versa. (ii) If f ( x) → ∞ as x → a or b and f ′ ( x) = 0 only for one value of x (say c) between a and b, then f (c) is necessarily the minimum and the least value. (iii) If f ( x) → − ∞ as x → a or b, then f (c) is necessarily the maximum and greatest value. (iv) The stationary points are the points of the domain, where f ′ ( x) = 0.

140

40

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 If f is defined as f ( x ) = x +

1 , then which of following is x

true?

j

NCERT Exemplar

(a) Local maximum value of f (x) is − 2 (b) Local minimum value of f (x) is 2 (c) Local maximum value of f (x) is less than local minimum value of f (x) (d) All the above are true

2 If the sum of two numbers is 3, then the maximum value of the product of the first and the square of second is j

(a) 4

(b) 1

(c) 3

NCERT Exemplar

(d) 0

3 If y = a log x + bx + x has its extremum value at x = 1 2

and x = 2, then (a, b ) is equal to 1 (a)  1,   2

1 (b)  , 2  2 

−1 (c)  2 ,   2

−2 −1 (d)  ,   3 6

4 The function f ( x ) = a cos x + b tan x + x has extreme π values at x = 0 and x = , then 6 2 ,b = −1 3 2 (c) a = − , b = 1 3

2 ,b = −1 3 2 (d) a = , b = 1 3

(a) a = −

(b) a =

5 The minimum radius vector of the curve

(b) 5

(c) 7

(d) None of these

6 The function f ( x ) = 4x 3 − 18x 2 + 27x − 7 has (a) one local maxima

j

NCERT Exemplar

(b) one local minima (c) one local maxima and two local minima (d) neither maxima nor minima

7 The function f ( x ) = (a) (b) (c) (d)

k − 2x , if x ≤ − 1 . 2x + 3, if x > − 1

If f has a local minimum at x = −1, then a possible value j of k is AIEEE 2010 (d) −1

(b) 28

(c) 38

(d)

1 2

11 If f ( x ) = x 2 + 2bx + 2c 2 and g ( x ) = − x 2 − 2cx + b 2 such that minimum f ( x ) > maximum g ( x ), then the relation between b and c is (a) 0 < c < b 2 (c) | c | > | b | 2

(b) | c | < | b | 2 (d) No real values of b and c

12 Let f ( x ) be a polynomial of degree four having extreme

 f (x ) values at x = 1 and x = 2. If lim 1 + 2 = 3, then f ( 2) is x→ 0  x   j JEE Mains 2015 equal to (a) −8 (c) 0

(b) −4 (d) 4

13 If a differential function f ( x ) has a relative minimum at x = 0, then the function φ( x ) = f ( x ) + ax + b has a relative minimum at x = 0 for (b) all b, if a = 0 (d) all a > 0

(a) all a and all b (c) all b > 0

14 The denominator of a fraction is greater than 16 of the square of numerator, then least value of fraction is (b) − 1/ 8 (d) 1/16

b x

15 The function f ( x ) = ax + , b, x > 0 takes the least value at x equal to (a) b

(b)

a

(c)

(d)

b

b a

(d) 18

j

AIEEE 2011

(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

17 The absolute maximum and minimum values of the function f given by f ( x ) = cos 2 x + sin x , x ∈[ 0, π ] j

9 The minimum value of 9x + 4y , where xy = 16 is (a) 48

(c) 2

Statement I x = 0 is point of minima of f . Statement II f ′ (0) = 0.

8 Let f : R → R be defined by f ( x ) = 

(b) 0

(b) 1

 tan x , x ≠0 x  1 , x = 0

no point of local minima no point of local maxima exactly one point of local minima exactly one point of local maxima

(a) 1

(a) 3

16 Let f be a function defined by f ( x ) = 

x2 − 2 has x2 − 4

1 (c) − 2

attains its maximum and minimum at p and q respectively such that p 2 = q , then a is equal to

(a) −1/4 (c) 1/12

4 9 + = 1 is of length x2 y2 (a) 1

10 If the function f ( x ) = 2x 3 − 9 ax 2 + 12 a 2 x + 1, where a > 0

(a) 2.25 and 2 (c) 1.75 and 1.5

NCERT Exemplar

(b) 1.25 and 1 (d) None of these

141

DAY 18 The maximum value of f ( x ) = (a) −

1 4

(b) −

1 3

x on [ −1,1] is 4+ x + x2

(c)

1 6

(d)

1 5

(a) e

1 1 (b) log e e

2

(c) e log e (d) None of these

20 If x is real, the maximum value of

3x 2 + 9x + 17 is 3x 2 + 9x + 7 AIEEE 2007 1 (d) 4 j

(a) 41

17 (c) 7

(b) 1

f ( x ) = sec x + log cos 2 x , 0 < x < 2π are respectively (a) (b) (c) (d)

NCERT Exemplar

(1, − 1) and {2 (1 − log 2), 2 (1 + log 2)} (1, − 1) and {2 (1 − log 2), 2 (1 − log 2)} (1, − 1) and (2, − 3) None of the above

(a) π

(b) 2 π

(c) 3 π

(b) (0, 0)

j

NCERT Exemplar π (d) 2

(c) (1, 0)

3

(d) (0, 1)

(a)

h 2 2

(b)

JEE Main 2013 h2 (d) 4

j

2

h 2

(c)

h

2

2

25 A straight line is drawn through the point P ( 3, 4) meeting the positive direction of coordinate axes at the points A and B. If O is the origin, then minimum area of ∆OAB is equal to (a) 12 sq units (c) 24 sq units

(b) 6 sq units (d) 48 sq units

roots, where p > 0 and q > 0. Then, which one of the following holds?

2

− (a − 2)x − a + 1 = 0 assume

(b) 1

(c) 3

(d) 0

30 The minimum intercepts made by the axes on the tangent to the ellipse

x2 y2 + = 1 is 16 9

(b) 7

(c) 1

(d) None of these

sphere is maximum, if (a) h =

4R 3

(b) h =

R 3

(c) h =

2R (d) None of these 3

32 The volume of the largest cone that can be inscribed in a sphere of radius R is

(b) (c) (d)

j

NCERT

3 of the volume of the sphere 8 8 of the volume of the sphere 27 2 of the volume of the sphere 7 None of the above

33 Area of the greatest rectangle that can be inscribed in x2 y2 the ellipse 2 + 2 = 1 is a b (a)

ab

(b)

a b

(c) 2ab

minimum value of the sum at x equal to (a) 2

(b) 1

(c) –1

(d) ab

p 3 p 3

27 If A ( x1, y1 ), B ( x 2 , y 2 ) and C ( x 3 , y 3 ) are the vertices of a ∆ ABC. A parallelogram AFDE is drawn with D, E and F on the line segment BC, CA and AB, respectively. Then, maximum area of such parallelogram is

j

AIEEE 2003

(d) –2

35 The greatest value of f ( x ) = ( x + 1)1/ 3 − ( x − 1)1/ 3 on [0, 1] is

p p and − 3 3

p and maxima at − 3 p (c) The cubic has minima at − and maxima at 3 p p (d) The cubic has minima at both and − 3 3 (b) The cubic has minima at

(d) 46, −6

34 The real number x when added to its inverse gives the

26 Suppose the cubic x 3 − px + q has three distinct real

(a) The cubic has maxima at both

roots of the equation x the least value is

(a)

24 The maximum area of a right angled triangle with hypotenuse h is

(c) 40, −6

(b) 46, 6

29 The value of a , so that the sum of the squares of the

(a) 25

23 The point of inflection for the curve y = x 5 / 2 is (a) (1, 1)

(a) 36, 3

31 The curved surface of the cone inscribed in a given

22 The difference between greatest and least values of the  π π function f ( x ) = sin 2x − x , on − , is  2 2 

that x = 3 t 2 − 18 t + 7 and y = 2 t 3 − 15 t 2 + 24 t + 10, ∀ x ∈ [ 0, 6]. Then, the maximum and minimum values of y = f ( x ) are

(a) 2

21 The maximum and minimum values of j

1 ( area of ∆ABC) 4 1 ( area of ∆ABC) (d) 8

(b)

28 If y = f ( x ) is a parametrically defined expression such

19 In interval [1, e ], the greatest value of x 2 log x is 2

1 ( area of ∆ABC) 2 1 (c) ( area of ∆ABC) 6 (a)

(a) 1

(b) 2

(c) 3

j

AIEEE 2002

(d) 1/3

36 The coordinate of a point on the parabola y 2 = 8x whose distance from the circle x 2 + ( y + 6) 2 = 1 is minimum, is (a) (2,− 4)

(b) (2,4)

(c) (18,−12)

(d) (8, 8)

37 The volume of the largest cylinder that can be inscribed in a sphere of radius r cm is (a)

4 πr 3 3

(b)

4 πr 3 3 3

(c)

4 πr 3 2 3

(d)

4 πr 3 5 2

38 Maximum slope of the curve y = − x 3 + 3x 2 + 9x − 27 is (a) 0

(b) 12

(c) 16

(d) 32

TEN

142

39 If ab = 2 a + 3b, a > 0, b > 0, then the minimum value of ab is

40 The perimeter of a sector is p. The area of the sector is maximum, when its radius is

(a) 12 1 (c) 4

(b) 24

(a)

p

(b)

(d) None of these

1 p

(c)

p 2

(d)

p 4

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 The minimum radius vector of the curve

a2 b2 + = 1 is of x2 y2

` 75 while at 40 km/h, it is ` 65. Then, the most economical speed (in km/h) of the bus is j JEE Mains 2013 (a) 45

length (a) a − b (c) 2a + b

(b) a + b (d) None of these

2 f ( x ) = x 2 − 4 | x | and min {f (t ) : − 6 ≤ t ≤ x }, x ∈ [ − 6, 0] , then g ( x ) has g( x ) =  max {f (t ) : 0 < t ≤ x }, x ∈ ( 0, 6] (a) exactly one point of local minima (b) exactly one point of local maxima (c) no point to local maxima but exactly one point of local minima (d) neither a point of local maxima nor minima

4x − x 3 + log (a 2 − 3 a + 3), x − 18 ,

3 f (x ) = 

0≤x 0 dx2 d2 y and (at x = − 1) < 0 dx2 Hence, local maximum value of y is at x = − 1 and the local maximum value = − 2. Local minimum value of y is at x = 1 and local minimum value = 2. Therefore, local maximum value − 2 is less than local minimum value 2.

2 Let two numbers be x and (3 − x). Then, product P = x(3 − x )2 dP = −2 x(3 − x ) + (3 − x )2 dx dP d 2P = (3 − x )(3 − 3 x ) and 2 = 6 x − 12 dx dx dP For maxima or minima, put =0 dx ⇒ (3 − x )(3 − 3 x ) = 0 ⇒x = 3, 1

At x = 3, d 2P = 18 − 12 = 6 > 0 [minima] dx2 At x = 1, d 2P = −6 < 0 dx2 So, P is maximum at x = 1. ∴ Maximum value of P = 1 (3 − 1)2 = 4 dy a = + 2bx + 1 dx x

3 Q ⇒

 dy  = a + 2b + 1 = 0    dx  x =1

⇒ a = − 2b − 1 dy a and   = + 4b + 1 = 0  dx  x =2 2 − 2b − 1 + 4b + 1 = 0 2 1 −1 ⇒ – b + 4b + = 0 ⇒ 3b = 2 2 −2 1 −1 and a = − 1 = ⇒ b = 6 3 3 ⇒

4 f ′( x ) = − a sin x + b sec2 x + 1 Now,

π f ′ (0) = 0 and f ′   = 0  6

⇒ b + 1 = 0 and −

a 4b + + 1= 0 2 3

⇒

b = − 1, a = −

2 3

5 The given curve is 42 + 92 = 1 x

y

Put x = r cos θ, y = r sin θ, we get r 2 = (2 sec θ)2 + (3 cosec θ)2 So, r 2 will have minimum value ( 2 + 3)2 . or r have minimum value equal to 5.

6 f ( x )= 4 x 3 − 18 x2 + 27 x − 7 f ′ ( x ) = 12 x2 − 36 x + 27 = 3(4 x2 − 12 x + 9) = 3(2 x − 3)2 3 f ′ ( x ) = 0 ⇒ x = (critical point) 2 3 Since, f ′ ( x ) > 0 for all x < and for all 2 3 x> 2 3 Hence, x = is a point of inflection i.e., 2 neither a point of maxima nor a point of minima. 3 x = is the only critical point and f 2 has neither maxima nor minima.

7 For y =

x2 − 2 − 4x dy ⇒ = x2 − 4 dx ( x2 − 4)2

TEN

144

dy > 0, for x < 0 dx dy and < 0, for x > 0 dx Thus, x = 0 is the point of local maxima 1 for y. Now, ( y ) x = 0 = (positive). Thus, 2 x = 0 is also the point of local x2 − 2 maximum for y = 2 . x −4 ⇒

8 If f ( x ) has a local minimum at x = −1, then lim f ( x ) = lim− f ( x )

x→ −1 +

⇒ ⇒

x→ −1

lim+ 2 x + 3 = lim− 1 < −2 x

x→ −1

x→ −1

−2 + 3 = k + 2 ⇒ k = −1 Y f(x)=2x+3

f(x)=1 0, therefore a = 2 11 Minimum of f ( x ) = − D 4a − ( 4b 2 − 8 c 2 ) = 4 = 2c 2 − b 2

and maximum of g ( x ) = −

( 4 c 2 + 4b 2 ) 4 (− 1)

= b2 + c 2 Since, min f ( x ) > max g ( x ) ⇒ 2c 2 − b 2 > b 2 + c 2 ⇒ c 2 > 2b 2 ⇒ |c |> 2 |b |

12 Central Idea Any function have extreme values (maximum or minimum) at its critical points, where f ′( x ) = 0. Since, the function have extreme values at x = 1 and x = 2. ∴ f ′( x ) = 0 at x = 1 and x = 2 ⇒ f ′(1) = 0 and f ′(2) = 0 Also it is given that f ( x) f ( x) lim 1 + 2  = 3 ⇒ 1 + lim 2 = 3 x→ 0  x→ 0 x x   f ( x) ⇒ lim 2 = 2 x→ 0 x ⇒ f ( x ) will be of the form ax 4 + bx3 + 2 x2 [Q f ( x ) is of four degree polynomial] Let f ( x ) = ax 4 + bx3 + 2 x2 ⇒ f ′( x ) = 4ax3 + 3bx2 + 4 x ⇒ f ′(1) = 4a + 3b + 4 = 0 ...(i) and f ′(2) = 32a + 12b + 8 = 0 ...(ii) ⇒ 8a + 3b + 2 = 0 On solving Eqs. (i) and (ii), we get 1 a = ,b = − 2 2 x4 ∴ f ( x) = − 2 x3 + 2 x2 2 ⇒ f (2) = 8 − 16 + 8 = 0

13 φ′( x ) = f ′( x ) + a Q φ′ (0) = 0 ⇒ f ′ (0) + a = 0 ⇒ a= 0 [Q f ′ (0) = 0] Also, φ′ (0) > 0 [Q f ′ ′ (0) > 0] ⇒ φ( x ) has relative minimum at x = 0 for all b, if a = 0

14 Let the number be x, then f (x ) =

x x2 + 16

On differentiating w.r.t. x, we get ( x2 + 16) ⋅ 1 − x (2 x ) f ′( x ) = ( x2 + 16)2 2 2 x + 16 − 2 x 16 − x2 …(i) = = 2 2 2 ( x + 16) ( x + 16)2 Put f ′ ( x ) = 0 for maxima or minima f ′ ( x ) = 0 ⇒ 16 − x2 = 0 ⇒ x = 4, − 4 Again, on differentiating w.r.t. x, we get ( x2 + 16)2 (−2 x ) − (16 − x2 ) f ′ ′( x ) = x = 4, f ′ ′ ( x ) < 0

2( x2 + 16) 2 x At ( x2 + 16)4

∴ f ( x ) is maximum at x = 4. and at x = − 4, f ′ ′ ( x ) > 0, f ( x ) is minimum. ∴ Least value of −4 1 f ( x) = =− 16 + 16 8

15 Given, f ( x ) = ax + b

x On differentiating w.r.t. x, we get b f ′( x ) = a − 2 x For maxima or minima, put f ′ ( x ) = 0 b ⇒ x= a Again, differentiating w.r.t. x, we get 2b f ′ ′( x ) = 3 x b At x = , f ′ ′ ( x ) = positive a b . ⇒ f ( x ) is minimum at x = a b . ∴ f ( x ) has the least value at x = a

 tan x 16 f ( x ) =  x , x ≠ 0

x=0 1, tan x As > 1, ∀ x ≠ 0 x ∴ f (0 + h ) > f (0) and f (0 − h ) > f (0) At x = 0, f ( x ) attains minima. f (h ) − f (0) Now, f ′ (0) = lim h→ 0 h tan h −1 tan h − h = lim h = lim h→ 0 h→ 0 h h2 [using L’ Hospital’s rule] sec 2 h − 1 = lim [Q tan2 θ = sec 2 θ − 1] h→ 0 2h tan2 h 1 = lim ⋅h= ⋅0 = 0 h→ 0 2h2 2 Therefore, Statement II is true. Hence, both statements are true but Statement II is not the correct explanation of Statement I.

17 Given, f ( x ) = cos2 x + sin x, x ∈ [0, π] Now, f ′ ( x ) = 2 cos x (− sin x ) + cos x = − 2sin x cos x + cos x For maximum or minimum put f ′( x ) = 0 ⇒ −2sin x cos x + cos x = 0 ⇒ cos x (− 2sin x + 1 ) = 0 ⇒ ⇒

cos x = 0 or sin x = x=

π π , 6 2

1 2

For absolute maximum and absolute minimum, we have to evaluate

145

DAY π π f (0), f   , f   , f ( π )  6  2 At x = 0, f (0) = cos 2 0 + sin 0 = 12 + 0 = 1 π π π π At x = , f   = cos 2   + sin  6 6  6 6 2

At x =

π , 2

 3 1 5 = .  + = = 125 2 4  2 

π π π f   = cos 2   + sin = 02 + 1 = 1  2  2 2 At x = π, f ( π ) = cos 2 π + sin π = (−1)2 + 0 = 1 Hence, the absolute maximum value of π f is 1.25 occurring at x = and the 6 absolute minimum value of f is 1 π occurring at x = 0, and π. 2 Note If close interval is given, to determine global maximum (minimum), check the value at all critical points as well as end points of a given interval. x 18 Q f ( x ) = 4 + x + x2 On differentiating w.r.t. x, we get 4 + x + x2 − x ( 1 + 2 x ) f ′( x ) = (4 + x + x2 )2 For maximum, put f ′ ( x ) = 0 4 − x2 ⇒ = 0 ⇒ x = 2, − 2 (4 + x + x2 )2 Both the values of x are not in the interval [−1, 1] . −1 −1 = ∴ f ( −1 ) = 4−1+ 1 4 f (1 ) =

1 1 = (maximum) 4+ 1+1 6

19 Given, f ( x ) = x2 log x On differentiating w.r.t. x, we get f ′ ( x ) = (2 log x + 1 ) x For a maximum, put f ′ ( x ) = 0 ⇒ (2 log x + 1 ) x = 0 ⇒ x = e −1 /2 , 0 Q 0 < e −1 / 2 < 1 None of these critical points lies in the interval [1, e ] . So, we only compute the value of f ( x ) at the end points 1 and e. We have, f (1 ) = 0, f (e ) = e 2 Hence, greatest value of f ( x ) = e 2 10 20 Let f (x ) = 1 + 7  2 3  x + 3 x +   3

10 2  3 1 3   x +  +    2 12   So, the maximum value of f ( x ) at 3 x = − is 2 =1+

f  − 

3 =1+ 2

10 = 1 + 40 = 41 1  3    12  21 Given, f ( x ) = sec x + log cos2 x ⇒ f ( x ) = sec x + 2 log(cos x ) Therefore, f ′ ( x ) = sec x tan x − 2 tan x = tan x (sec x − 2) f ′( x ) = 0 ⇒

tan x = 0 or sec x = 2 ⇒ cos x =

1 2

Therefore, possible values of x are π 5π . x = 0, x = π and x = or x = 3 3 2 Again,f ′ ′ ( x ) = sec x (sec x − 2) + tan x (sec x tan x ) = sec3 x + sec x tan2 x − 2 sec2 x = sec x (sec2 x + tan2 x − 2 sec x ) ⇒ f ′ ′ (0 ) = 1 (1 + 0 − 2) = − 1 < 0 Therefore, x = 0 is a point of maxima. f ′ ′ ( π ) = − 1 (1 + 0 + 2) = − 3 < 0 Therefore, x = π is a point of maxima. π f ′ ′   = 2 (4 + 3 − 4) = 6 > 0  3 π Therefore, x = is a point of minima. 3 5π  f ′ ′   = 2 (4 + 3 − 4) = 6 > 0  3  5π Therefore, x = is a point of minima. 3 Maximum value of y at x = 0 is 1 + 0 = 1. Maximum value of y at x = π is − 1 + 0 = − 1. π Minimum value of y at x = is 3 1 2 + 2 log = 2 (1 − log 2). 2 5π Minimum value of y at x = is 3 1 2 + 2 log = 2 (1 − log 2). 2 22 Given, f ( x ) = sin 2 x − x ⇒ f ′ ( x ) = 2 cos 2 x − 1 1 2 π π π π or ⇒ 2x = − or ⇒ x = − 3 3 6 6 π π π Now, f  −  = sin(− π ) + =  2 2 2 Put f ′ ( x ) = 0 ⇒ cos 2 x =

π 2π  π 3 π f  −  = sin  − =− +  +  6  6 6 2 6 π 2π  π 3 π f   = sin  = −  −  6  6 6 2 6

π π π and f   = sin( π) − = −  2 2 2 π π Clearly, is the greatest value and − 2 2 is the least. π π Therefore, difference = + = π 2 2 23 Given, y = x 5/2 d 2 y 15 1 /2 dy 5 x ∴ = x 3 /2 , 2 = dx 4 dx 2 2 dy d y At x = 0, =0 = 0, dx dx2 3 d y and 3 is not defined, dx when x = 0, y = 0 ∴(0, 0) is a point of inflection. 24 Area of triangle, ∆ = 1 x h2 − x2 2 x ( −2 x )  d∆ 1  2 2 =  h − x + =0 dx 2  2 h2 − x2   ⇒

x=

h 2 C h

A

⇒ ∴

x

 √h2 – x2

B

d ∆ h < 0 at x = dx2 2 2

∆ =

1 h h2 h2 × h2 − = 2 2 2 4

25 Let the equation of drawn line be y x + = 1, where a > 3, a b b > 4, as the line passes through (3, 4) and meets the positive direction of coordinate axes. 4a 3 4 We have, + = 1 ⇒ b = a b (a − 3) Now, area of ∆AOB, 1 2a2 ∆ = ab = 2 (a − 3) d∆ 2a (a − 6) = da (a − 3)2 Clearly, a= 6 is the point of minima for ∆. 2 × 36 Thus, ∆ min = = 24 sq units 3

26 Let f ( x ) = x3 − px + q Maxima

−√p/3 Minima √p/3

TEN

146 f ′ ( x ) = 3 x2 − p f ′( x ) = 0 p p x= ,− 3 3 f ′′ ( x ) = 6 x p p , f ′′ ( x ) = 6 >0 3 3 [minima]

Then, Put ⇒ Now, At x =

and at x = −

p f ′′ ( x ) < 0 3

[maxima]

27 We have, AF || DE and AE || FD A y F

29 Let α and β be the roots of the equation

E

x2 − (a − 2) x − a + 1 = 0

x B

D

C

Now, in ∆ABC and ∆EDC, ∠DEC = ∠BAC , ∠ACB is common. ⇒

∆ABC ≅ ∆EDC b − y x c Now, = ⇒ x = (b − y ) b b c Now, S = Area of parallelogram AFDE = 2 (area of ∆AEF ) 1 ⇒ S = 2  xy sin A  2  c = (b − y )y sin A b dS  c =  sin A  (b − 2 y ) b  dy Sign scheme of +

dS , dy b/2

d2 y 0 dx ⇒ t = 4 is a point of local minima. d2 y dy At (t = 3), and 2 are not defined dx dx and change its sign. d2 y is unknown in the vicinity of t = 3, dx2 thus t = 3 is a point of neither maxima nor minima. Finally, maximum and minimum values of expression y = f ( x ) are 46 and −6, respectively. At (t = 1),

So, z has minima at a = 1. So, α2 + β2 has least value for a = 1. This is because we have only one stationary value at which we have minima. Hence, a = 1.

30 Any tangent to the ellipse is

–

Hence, S is maximum when y =

Then, α + β = a − 2, αβ = − a + 1 Let z = α 2 + β2 = (α + β )2 − 2αβ = (a − 2)2 + 2 (a − 1) = a2 − 2a + 2 dz ⇒ = 2a − 2 da dz Put = 0 , then da ⇒ a=1 d 2z ∴ = 2> 0 da2

b ⋅ 2

c b b   × sin A b  2 2 1 1 1 =  bc sin A  = (area of ∆ABC )  2 22

∴ S max =

28 We have, dy = 6 t 2 − 30 t + 24 = 6 (t − 1) (t − 4) dt dx and = 6 t − 18 = 6 (t − 3) dt dy (t − 1) (t − 4) Thus, = dx (t − 3) which indicates that t = 1, 3 and 4 are the critical points of y = f ( x ). d2 y d  dy  dt Now, =  ⋅ dx2 dt  dx  dx t 2 − 6 t + 11 1 = × (t − 3)2 6 (t − 3)

y x cos t + sin t = 1, where the point of 4 3 contact is (4cos t , 3sin t ) y x or + = 1, 4 sec t 3 cosec t It means the axes Q (4sec t , 0) and R (0, 3 cosec t ). ∴The distance of the line segment QR is QR2 = D = 16 sec2 t + 9 cosec2 t So, the minimum value of D is (4 + 3)2 or QR = 7.

31 Let S be the curved surface area of a cone. C

h O

R B

A

r

OA = AC − OC = h − R In ∆OAB, R2 = r 2 + (h − R )2 ⇒ r = 2Rh − h2 ∴

S = πrl = π ( 2Rh − h2 )( h2 + r 2 ) = ( π 2Rh − h2 )( 2Rh )

Let S 2 = P ∴ P = π2 2R ( 2 Rh2 − h3 ) Since, S is maximum, if P is maximum, then dP = 2 π2 R (4Rh − 3h2 ) = 0 dh 4R ∴ h = 0, 3 dP Again, on differentiating , we get dh 2 d P = 2 π2 R ( 4 R − 6 h ) dh2 4R d 2P < 0 at h = dh2 3

32 Let OC = x, CQ = r Now, OA = R [given] Height of the cone = h = x + R ∴ Volume of the cone 1 = V = π r 2h 3 A

R O x P

…(i)

h

r

Q

C

Also, in right angled ∆OCQ, OC 2 + CQ 2 = OQ 2 ⇒ x2 + r 2 = R2 …(ii) ⇒ r 2 = R2 − x2 From Eqs. (i) and (ii), 1 V = π (R2 − x2 )( x + R ) …(iii) 3 [Q h = x + R ] On differentiating Eq. (iii) w.r.t. x, we get dV 1 = π [(R2 − x2 ) − 2 x( x + R )] dx 3 dV π ⇒ = (R2 − x2 − 2 x2 − 2 xR ) dx 3 dV π = (R2 − 2 xR − 3 x2 ) ⇒ dx 3 dV π ⇒ = (R − 3 x )(R + x ) …(iv) dx 3 dV For maxima, put =0 dx π (R − 3 x )(R + x ) = 0 ⇒ 3 R R or x = − R ⇒ x = x= ⇒ 3 3 [since, x cannot be negative]

147

DAY On differentiating Eq. (iv) w.r.t. x, we get d V π = [(−3)(R + x ) + (R − 3 x )] dx2 3 π π = (−2R − 6 x ) = − (2 R + 6 x ) 3 3 6R  R d 2V −π  At x = , =  2R +  3 dx2 3  3  4π =− R 0, hence minimum. dt 2 So, point is (2, − 4).

37 We know that, volume of cylinder, V = πR 2 h

B′

Y (–a cos θ, b sin θ) (a cos θ, b sin θ) A B X′

O

D C (–a cos θ, –b sin θ) (a cos θ, –b sin θ) Y′

35 Given that, f ( x ) = ( x + 1)1 /3 − ( x − 1)1 /3 On differentiating w.r.t. x, we get  1 1 1 f ′( x ) =  − 3  ( x + 1)2 /3 ( x − 1)2 /3  =

( x − 1)2 /3 − ( x + 1)2 /3 3( x2 − 1)2 /3

Clearly, f ′( x ) does not exist at x = ± 1. Now, put f ′ ( x ) = 0, then ( x − 1)2 /3 = ( x + 1)2 /3 ⇒ x = 0 At x = 0 f ( x ) = (0 + 1)1 /3 − (0 − 1)1 /3 = 2 Hence, the greatest value of f ( x ) is 2.

36 Q y 2 = 8 x. But y 2 = 4ax ⇒ 4a= 8

⇒ a= 2

r

h/2 B

C

A R 2

h In ∆OCA, r 2 =   + R2  2 ⇒

R2 = r 2 −

On differentiating w.r.t. x, we get g ′( x ) = − 6 x + 6 For maxima or minima put g ′ ( x ) = 0 ⇒ x=1 Now, g ′ ′ ( x ) = − 6 < 0 and hence, at x = 1, g ( x ) (slope) will have maximum value. ∴[g ( 1 )] max = − 3 × 1 + 6( 1 ) + 9 = 12

39 Given, ab = 2 a + 3b ⇒

⇒ (a − 3) b = 2 a 2a b = a−3

Now, let z = ab =

2 a2 a−3

40 Q Perimeter of a

h2 4

sector = p

 h2  V = π  r2 − ∴ h 4  π V = πr 2 h − h3 …(i) ⇒ 4 On differentiating Eq. (i) both sides w.r.t. h, we get dV 3 πh2 = πr 2 − dh 4 d2 V −3 πh = ⇒ dh2 2 For maximum or minimum value of V , dV 3 πh2 = 0 ⇒ πr 2 − =0 dh 4 2 4r 2 ⇒ h= r ⇒ h2 = 3 3  d2 V  Now,  2  = − 3 πr < 0  dh  h = 2 r 3

Thus, V is maximum when h =

2r , then 3

2

h2 1  2r  2 2 = r2 −   = r 4 4 3 3 4 πr 3 2 Max V = πR h = 3 3 R2 = r 2 −

38 Let f ( x ) = − x 3 + 3 x2 + 9 x − 27 The slope of this curve f ′ ( x ) = − 3 x2 + 6 x + 9

g ( x ) = f ′ ( x ) = − 3 x2 + 6 x + 9

On differentiating w.r.t. x, we get dz 2 [(a − 3) 2 a − a2 ] 2 [a2 − 6a] = = (a − 3)2 da (a − 3)2 dz For a minimum, put =0 da ⇒ a2 − 6a = 0 ⇒ a = 0, 6 d 2z At a = 6, 2 = positive da When a = 6, b = 4 ∴ (ab )min = 6 × 4 = 24

A′

h/2

X

Area of rectangle ABCD = (2a cos θ) (2b sin θ) = 2ab sin 2θ Hence, area of greatest rectangle is equal to 2ab when sin 2θ = 1. 1 34 Let f ( x) = x + x 1 f ′( x ) = 1 − 2 x For maxima and minima, put f ′ ( x ) = 0 1 ⇒ 1 − 2 = 0⇒ x = ± 1 x 2 Now, f ′ ′ ( x ) = 3 x At x = 1, f ′ ′ ( x ) = + ve [minima] and at x = − 1, f ′ ′ ( x ) = −ve [maxima] Thus, f ( x ) attains minimum value at x = 1.

C′

Let

Let AOB be the sector with radius r. If angle of the sector be θ radians, then area of sector, 1 A = r 2θ 2

A r O θ

s

r

and length of arc, s = r θ ⇒ θ =

B

…(i) s r

∴Perimeter of the sector p = r + s + r = 2 r + s …(ii) s On substituting θ = in Eq. (i), we get r 2A 1 s 1 A =  r 2    = rs ⇒ s = 2   r  2 r Now, on substituting the value of s in Eq. (ii), we get 2A 2 p = 2 r +   ⇒ 2 A = pr − 2 r  r  On differentiating w.r.t. r, we get dA 2 = p − 4r dr For the maximum area, put dA =0 dr ⇒ p − 4r = 0 p ⇒ r = 4

TEN

148

SESSION 2 1 Let radius vector is r. ∴ r 2 = x2 + y 2 a2 y 2 ⇒ r2 = 2 + y2 y − b2

 a2  b2 Q 2 + 2 = 1 y  x 

For minimum value of r, −2 yb 2 a2 d (r 2 ) =0 ⇒ + 2y = 0 dy ( y 2 − b 2 )2 y 2 = b( a + b ) x2 = a(a + b ) r 2 = (a + b )2 ⇒ r = a + b

⇒ ∴ ⇒

Clearly, f ′( x ) changes its sign at x = − 1 from positive to negative and so f ( x ) has local maxima at x = − 1. Also, f ′(0) does not exist but f ′ (0− ) < 0 and f ′ (0+ ) > 0. It can only be inferred that f ( x ) has a possibility of a minimum at x = 0. Hence, it has one local maxima at x = − 1 and one local minima at x = 0 So, total number of local maxima and local minima is 2.

6 Total length = 2r + r θ = 20

y = g ( x ) , clearly g ( x ) has neither a point of local maxima nor a point of local minima.

r

y = x2 – 4|x| –2

0

2

–6 –4

4

6

3 Clearly, f ( x ) in increasing just before x = 3 and decreasing after x = 3. For x = 3 to be the point of local maxima. f (3) ≥ f (3 − 0) ⇒ − 15 ≥ 12 − 27 + log (a2 − 3 a + 3) ⇒ 0 < a2 − 3 a + 3 ≤ 1 ⇒ 1 ≤ a ≤ 2

4 (Slope) f ′( x ) = e x cos x + sin xe x f ′ ′ ( x ) = 2e x {sin ( x + π/4) + cos ( x + π/4)} = 2e x ⋅ sin ( x + π/ 2) For maximum slope, put f ′ ′ ( x ) = 0 ⇒ sin ( x + π/2 ) = 0 ⇒ cos x = 0 ∴ x = π / 2, 3 π / 2 f ′ ′ ′ ( x ) = 2e x cos ( x + π/ 2) f ′ ′ ′ ( π/2 ) = 2e x ⋅ cos π = − ve Maximum slope is at x = π /2. 2 3 (2 + x ) , − 3 < x ≤ − 1 5 f ′( x ) =  2 −1 /3 x , − 1< x < 2  3

Y A X′

(–2, 0) (–1, 0) O

X

∴

Y′

⇒

Y 2 y = |x2 – 2| y= x 3 1

A = 10r − r 2 dA = 10 − 2r dr

For maxima or minima, put ⇒

10 − 2r = 0 ⇒ r = 5

dA = 0. dr

1 2  20 − 2 (5) (5)   2 5 1 = × 25 × 2 = 25sq m 2 7 Let c = av + b …(i) v When v = 30 km/ h, then c = ` 75 b …(ii) 75 = 30 a + ∴ 30 When v = 40 km/h, then c = ` 65 b …(iii) ∴ 65 = 40 a + 40 On solving Eqs. (ii) and (iii), we get 1 and b = 1800 a= 2 On differentiating w.r.t. v in Eq. (i), dc b = a− 2 dv v For maximum or minimum c,3 dc b =0 ⇒ v =± dv a ⇒

(–3, 0)

1 × 1800 = 2 × 30 2 c = 60

rθ

⇒

A max =

d 2c 2b = 3 dv 2 v

at v =

b a +b = 2 ab a b

c =2

r

20 − 2r ⇒ θ= r Now, area of flower-bed, 1 A = r 2θ 2 20 − 2r  1 ⇒ A = r 2    2  r

∴

= e x 2 sin ( x + π/4)

c =a

 | x2 − 2 |, − 1 ≤ x < 3  8 f ( x ) =  x , 3 ≤ x < 2 3  3 3 − x, 2 3 ≤ x ≤ 4

θ

2 Bold line represents the graph of

b the speed is most a economical. ∴Most economical speed is So, at v =

b dx , >0 a dv 2

X′

–1

√2 √3

O

4 2√3

X

y=3 – x Y′

From the above graph, Maximum occurs at x = 0 and minimum at x = 4. 3 2 | x + x + 3 x + sin x |  x≠0 9 f ( x ) =   3 + sin  1   ,     x   0, x=0 Let g ( x ) = x3 + x2 + 3 x + sin x g ′ ( x ) = 3 x2 + 2 x + 3 + cos x 2x = 3  x2 + + 1 + cos x   3  1 8 = 3   x +  +  + cos x > 0 3 9  1  and 2 < 3 + sin   < 4  x 2

Hence, minimum value of f ( x ) is 0 at x = 0. Hence, number of points = 1

10 According to given information, we have Perimeter of square + Perimeter of circle = 2 units ⇒ 4 x + 2 πr = 2 1 − 2x ...(i) ⇒ r = π Now, let A be the sum of the areas of the square and the circle. Then, A = x2 + πr 2 (1 − 2 x ) 2 = x2 + π π2 ( 1 − 2 x )2 ⇒ A( x ) = x2 + π

149

DAY Now, for minimum value of A( x ), dA =0 dx 2 (1 − 2 x ) ⇒ 2x + ⋅ (− 2) = 0 π 2 − 4x ⇒ x= π ⇒ πx + 4 x = 2 2 ...(ii) ⇒ x= π+ 4 Now, from Eq. (i), we get 2 1 − 2⋅ π+ 4 r = π π + 4− 4 1 ...(iii) = = π(π + 4) π+ 4 From Eqs. (ii) and (iii), we get x = 2r

11 We have, f ( x ) = x2 +

1 1 and g ( x ) = x − x2 x

f ( x) g( x)

⇒ h( x ) =

⇒

a7 =

=

49 is the greatest term. 543

13 f ( x ) = x 3 − 3(7 − a)x2 − 3(9 − a2 )x + 2 f ′ ( x ) = 3 x2 − 6(7 − a)x − 3(9 − a2 ) For real root D ≥ 0, ⇒ 49 + a2 − 14a + 9 − a2 ≥ 0 58 ⇒a≤ 14 For local minimum f ′ ′ ( x ) = 6 x − 6(7 − a) > 0 ⇒ 7− x

 x − 1 + 2 1   2  x x ∴ h( x ) = = 1 1 x− x− x x 1 2  ⇒ h( x ) =  x −  + 1  x x− x 1 1 2 x − > 0, x −  + ∈ [2 2, ∞ ) 1  x x x− x 1 x − < 0, x 2  x − 1 + ∈ (−∞, 2 2]   1  x x− x ∴ Local minimum value is 2 2. x2 +

When

Thus constradictory, i.e., for real roots 58 and for negative point of local a≤ 14 minimum a > 7. No possible values of a.

12 Consider the function x2 ( x + 200) (400 − x 3 ) f ′( x ) = x 3 =0 ( x + 200)2 3

when x = (400)1 /3 , (Q x ≠ 0) x = (400)1 /3 − h ⇒ f ′ ( x ) > 0 x = (400)1 /3 + h ⇒ f ′ ( x ) < 0

(− 1, 0 ). The equation of circle centre at Q with variable radius r is R

Since, 7 < (400) < 8, either a7 or a8 is the greatest term of the sequence. 49 543

and

a8 =

8 89

P

[on solving Eqs. (i) and (ii) A = Area of ∆QSR 1 = × QS × RT 2 1 r = r  . 4 − r 2   2 2

1 /3

a7 =

S

…(i) ( x + 1 )2 + y 2 = r 2 This circle meets the line segment QP at S, where QS = r It meets the circle x2 + y 2 = 1 at …(ii)  r2 − 2 r  R , 4 − r2  2  2 

∴ f ( x ) has maxima at x = (400)1 /3

Q

T O

∴

4 4 − r2

− (8r − 3 r 3 )

(−4r ) 4 − r2

16 (4 − r ) 2

r =

8 d2 A , then 2 < 0 3 dr

Hence, A is maximum when r =

has x must be negative ⇒ 7 − a< 0 ⇒ a> 7

Q

8r − 3 r 3

dA = 0, when r (8 − 3 r 2 ) = 0 giving dr 8 r = 3 4 4 − r 2 (8 − 9r 2 ) d2 A ⇒ 2 = dr

14 From the given figure coordinate of Q is 2

f (x ) =

49 8 > 543 89

and

[since, RT is the y-coordinate of R] 1 = [r 2 4 − r 2 ] 4 r 2 (− r )  dA 1  = 2r 4 − r 2 +  dr 4 4 − r 2   {2r (4 − r 2 ) − r 3 } = 4 4 − r2

8 . 3

Then, maximum area 8 8 4 3 sq unit = 4− = 4×3 3 9

15 Given,

P ( x ) = x 4 + a x3 + bx2 + cx + d ⇒ P ′ ( x ) = 4 x3 + 3 ax2 + 2 bx + c Since, x = 0 is a solution for P ′ ( x ) = 0, then c =0 …(i) ∴ P ( x ) = x 4 + ax3 + bx2 + d Also, we have P (− 1) < P (1) ⇒ 1− a+ b + d 0 Since, P ′ ( x ) = 0, only when x = 0 and P ( x ) is differentiable in (− 1, 1), we should have the maximum and minimum at the points x = − 1, 0 and 1. Also, we have P (− 1) < P (1) ∴ Maximum of P ( x ) = Max {P (0), P (1)} and minimum of P ( x ) = Min {P (− 1), P (0)} In the interval [0, 1], P ′ ( x ) = 4 x3 + 3ax2 + 2 bx = x (4 x2 + 3ax + 2 b ) Since, P ′( x ) has only one root x = 0, then 4 x2 + 3 ax + 2 b = 0 has no real roots. ∴ ⇒

(3a)2 − 32 b < 0 9 a2 0 Thus, we have a > 0 and b > 0. ∴ P ′ ( x ) = 4 x3 + 3 ax2 + 2 bx > 0, ∀ x ∈ (0, 1) Hence, P ( x ) is increasing in [0, 1]. ∴ Maximum of P ( x ) = P (1) Similarly, P ( x ) is decreasing in [−1 , 0]. Therefore, minimum P ( x ) does not occur at x = − 1.

DAY FIFTEEN

Indefinite Integrals Learning & Revision for the Day u

Integral as an Anti-derivative

u

Fundamental Integration Formulae

u

Methods of Integration

Integral as an Anti-derivative A function φ ( x) is called a primitive or anti-derivative of a function f ( x), if φ′ ( x) = f ( x). If f1 ( x) and f2 ( x) are two anti-derivatives of f ( x), then f1 ( x) and f2 ( x) differ by a constant. The collection of all its anti-derivatives is called indefinite integral of f ( x) and is denoted by ∫ f ( x) dx. d {φ ( x) + C} = f ( x) ⇒ ∫ f ( x) dx = φ ( x) + C dx where, φ ( x) is an anti-derivative of f ( x), f ( x) is the integrand and C is an arbitrary constant known as the constant of integration. Anti-derivative of odd function is always even and of even function is always odd. Thus,

Properties of Indefinite Integrals ●

●

●

∫ { f ( x) ± g( x)} dx = ∫ f ( x) dx ± ∫ g( x) dx ∫ k ⋅ f ( x) dx = k ⋅ ∫ f ( x) dx, where k is any non-zero real number. ∫ [k f ( x) + k f ( x) +...+ k f ( x)] dx = k ∫ f ( x) dx + k ∫ f ( x) dx + ... + k ∫ f ( x) dx, 1 1

2 2

1

n n

1

2

2

n

PRED MIRROR Your Personal Preparation Indicator

n

where k1 , k2 ,... k n are non-zero real numbers.

Fundamental Integration Formulae

u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

There are some important fundamental formulae, which are given below

1. Algebraic Formulae (i)

∫

xn + 1 x ndx = + C, n ≠ − 1 n+1

(ii)∫ (ax + b )n dx =

1 (ax + b )n ⋅ a n+1

+1

+ C, n ≠ − 1

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

151

DAY

∫

1 dx = log| x | + C x 1 1 dx = (log| ax + b |) + C ax + b a

(v) (vi)

(iii) (iv)

∫

(ix)

∫

1 1 a+ x dx = log +C 2 2 a −x 2a a− x

(iii)

x

∫

1 1 x−a dx = log +C x2 − a2 2a x+a

∫a

(iv)

∫a

(bx + c )

∫a

2

1

∫

x − a2 2

x2 + a2

(xi)

∫

a2 − x2

(xiv) ∫

∫ ∫

(xvii) ∫

dx =

1 a (bx + c ) ⋅ + C, a > 0 and a ≠ 1 b loge a

Methods of Integration Following methods are used for integration

The method of reducing a given integral into one of the standard integrals, by a proper substitution, is called method of substitution.

 x dx = sin −1   + C  a −1

−1 x −a

2

dx =

1  x cosec −1   + C  a a

1 1  x a2 − x2 dx = x a2 − x2 + a2 sin −1   + C  a 2 2 1 1 x2 − a2 dx = x x2 − a2 − a2 log| x + x2 − a2 | + C 2 2 1 1 x2 + a2 dx = x x2 + a2 + a2 log| x + x2 + a2 | + C 2 2

∫ sin x dx = − cos x + C (ii) ∫ cos x dx = sin x + C (iii) ∫ tan x dx = − log|cos x | + C = log|sec x | + C (iv) ∫ cot x dx = log|sin x | + C = − log| cosec x | + C (i)

x π (v) ∫ sec xdx = log| sec x + tan x |+ C = log tan  +  + C  4 2 x (vi) ∫ cosec x dx = log| cosec x − cot x | + C = log tan + C 2

∫ sec x dx = tan x + C (viii) ∫ cosec x dx = − cot x + C (ix) ∫ sec x ⋅ tan x dx = sec x + C (x) ∫ cosec x ⋅ cot x dx = − cosec x + C 2

2

+C

ax + C, a > 0 and a ≠ 1 loge a

To evaluate an integral of the form

∫ f {g( x)} ⋅ g′ ( x) dx,

we

substitute g( x) = t , so that g′ ( x)dx = dt and given integral reduces to ∫ f (t ) dt . [ f ( x )]n +1 +C n+ 1

• ∫ [ f ( x )]n ⋅ f ′( x ) =

NOTE

• If ∫ f ( x ) dx = φ( x ), then ∫ f ( ax + b) dx =

1 φ( ax + b) + C a

(i) To evaluate integrals of the form

∫ ax ∫

2

dx dx or ∫ or 2 + bx + c ax + bx + c

ax 2 + bx + c dx

b c  We write, ax2 + bx + c = a  x2 + x +   a a

2. Trigonometric Formulae

(vii)

+ b)

x2 + a2 | + C

1  x dx = sec −1   + C 2 2  a a x −a

x

dx =

1 ( ax ⋅e a

dx = log| x +

1

2

dx =

1. Integration by Substitutions

 x dx = cos   + C 2 2  a a −x x

+C

x2 − a2 | + C

−1

∫

(xiii) ∫

(xvi)

1

( ax + b )

x

dx = log| x +

1

∫

(xv)

x

∫e

(x)

(xii)

∫ e dx = e

(i) (ii)

1 1  x dx = tan −1   + C  a + x2 a −1 1  x (viii) ∫ 2 dx = cot −1   + C  a a + x2 a

(vii)

3. Exponential Formulae

2

b c b2  = a x +  + −  2 a a 4c This process reduces the integral to one of following forms dX dX dX , =∫ 2 , or ∫ 2 X − A2 ∫ X 2 + A2 A − X2 dX dX dX ∫ A 2 − X 2 , ∫ x2 − A 2 , ∫ X 2 + A 2 or

∫

A2 − X 2 dX , ∫

X 2 − A2 dX , ∫

A2 + X 2 dX

(ii) To evaluate integrals of the form (px + q) dx or 2 + bx + c

∫ ax or

∫ (px + q)

∫

(px + q) ax 2 + bx + c

dx

ax 2 + bx + c dx

We put px + q = A {differentiation of (ax2 + bx + c)} + B, where A and B can be found by comparing the coefficients of like powers of x on the two sides.

152

2. Integration using Trigonometric Identities

●

In this method, we have to evaluate integrals of the form ●

∫ sin mx ⋅ cos nx dx or ∫ sin mx ⋅ sinnx dx or ∫ cos mx ⋅ cos nx dx or ∫ cos mx ⋅ sinnx dx

To evaluate integrals of the form 1 1 ∫ a sin x + b cos x dx or ∫ a + b sin x dx 1 1 or ∫ dx or ∫ dx a + b cos x a sin x + b cos x + c x 2 (i) Put sin x = and cos x = x x 1 + tan2 1 + tan2 2 2 x x x (ii) Replace 1 + tan2 by sec2 and put tan = t . 2 2 2 2 tan

In this method, we use the following trigonometrical identities (i) 2 sin A ⋅ cos B = sin ( A + B) + sin ( A − B) (ii) 2 cos A ⋅ sin B = sin ( A + B) − sin ( A − B) (iii) 2 cos A ⋅ cos B = cos ( A + B) + cos ( A − B) (iv) 2 sin A ⋅ sin B = cos ( A − B) − cos ( A + B)

●

(v) 2 sin A ⋅ cos A = sin 2 A  1 + cos 2 A  (vi) cos2 A =     2  1 − cos 2 A  (vii) sin2 A =     2 (ix) sin2 A + cos2 A = 1

To evaluate integrals of the form ∫ sin p x cos q x dx

Where p, q ∈ Q, we use the following rules : (i) If p is odd, then put cos x = t (ii) If q is odd, then put sin x = t (iii) If both p, q are odd, then put either sin x = t or cos x = t (iv) If both p, q are even, then use trigonometric identities only.  p + q − 2 (v) If p, q are rational numbers and   is a negative   2 integer, then put cot x = t or tan x = t as required.

To evaluate integrals of the form ∫ dx

dx or a + b cos 2 x

dx , x a sin x + b cos 2 x dx dx ∫ ( asin x + bcos x) 2 or ∫ a + bsin2 x + c cos 2 x

∫ a + bsin

2

or ∫

a sin x + b cos x dx, c sin x + d cos x d we write a sin x + b cos x = A ( c sin x + d cos x) dx +B( c sin x + d cos x)

To evaluate integral of form∫

a sin x + b cos x + c dx. p sin x + q cos x + r d We write a sin x + b cos x + c = A (p sin x + q cos x + r) dx + B (p sin x + q cos x + r) + C To evaluate integral of the form∫

3. Integration of Different Types of Functions

●

1 − tan2

Where A and B can be found by equating the coefficient of sin x and cos x on both sides.

(viii) cos2 A − sin2 A = cos 2 A

●

x 2

2

(i) Divide both the numerator and denominator by cos2 x. (ii) Replace sec2 x by 1 + tan2 x in the denominator, if any. (iii) Put tan x = t , so that sec2 x dx = dt

Where A, B and C can be found by equating the coefficient of sin x, cos x and the constant term. ●

To evaluate integrals of the form ∫ or ∫

x2 − 1 dx x 4 + kx 2 + 1

x2 + 1 dx x 4 + kx 2 + 1

We divide the numerator and denominator by x2 and make 2 1  perfect square in denominator as  x ±  and then put  x 1 1 x + = t or x – = t as required. x x ●

Substitution for Some Irrational Integrand (i)

a− x a+ x , , x = a cos 2θ a+ x a− x

(ii)

x a+ x , , x (a + x), a+ x x

1 , x = a tan2 θ x (a + x)

or x = a cot2 θ (iii)

x a− x , , x (a − x), a− x x or x = a cos2 θ

1 , x = a sin2 θ x (a − x)

153

DAY x , x−a

(iv)

x−a , x ( x − a) , x

dx

∫ ( x − α)(β − x) , ∫

(v)

∫ (vi) (vii)

∫

(ix)

 x − α   dx β − x

( x − α )(β − x) dx, put x = α cos2 θ + β sin2 θ

2

dx + bx + c)

px + q

dx ( px + q ) ax2 + bx + c

∫ ( px

dx 2

+ q ) (ax + b ) 2

, put px + q = t 2 , put px + q =

first put x =

1 t

1 t

and then a + bt 2 = z2

4. Integration by Parts (i) If u and v are two functions of x, then  du  ∫ uI vII dx = u ∫ v dx − ∫  dx ⋅ ∫ v dx dx We use the following preference in order to select the first function I Inverse function → L Logarithmic function → A Algebraic function → T Trigonometric function → E Exponential function → (ii) If one of the function is not directly integrable, then we take it as the first function. (iii) If both the functions are directly integrable, then the first function is chosen in such a way that its derivative vanishes easily or the function obtained in integral sign is easily integrable. (iv) If only one which is not directly integrable, function is there e.g. ∫ log x dx, then 1 (unity) is taken as second function.

Some more Special Integrals Based on Integration by Parts (i)

∫e

x

(ii)

∫e

ax

{ f ( x) + f ′ ( x)}dx = f ( x)e x + C sin (bx + c) dx =

e ax a + b2 {a sin (bx + c) − b cos (bx + c)} + k 2

(iii)

e ax a + b2 {a cos (bx + c) + b sin (bx + c)} + k Here, c and k are integration constant.

∫e

ax

cos(bx + c)dx =

2

5. Integration by Partial Fractions To evaluate the integral of the form

dx , put ax + b = t 2 ( px + q ) ax + b

∫ (ax

(viii) ∫

1 , x = a sec2 θ x ( x − a)

P( x )

∫ Q( x) dx, where P( x), Q( x)

are polynomial in x with degree of P( x) < degree of Q( x) and Q( x) ≠ 0, we use the method of partial fraction. The partial fractions depend on the nature of the factors of Q( x). (i) According to nature of factors of Q( x), corresponding form of partial fraction is given below: If Q ( x) = ( x − a1 ) ( x − a2 ) ( x − a3 ) .... ( x − an ), then we assume that A2 A3 P ( x) A1 An = + + + ... + , Q( x) ( x − a1 ) ( x − a 2 ) ( x − a3 ) ( x − an ) where the constants A1 , A2 , K , An can be determined by equating the coefficients of like power of x or by substituting x = a1 , a2 , K , an . (ii) If Q ( x) = ( x − a)k ( x − a1 ) ( x − a2 ) K ( x − ar ), then we assume that P( x ) A1 A2 Ak = + + ... + Q( x) ( x − a) ( x − a)2 ( x − a)k B1 B2 Br + + + ... + ( x − a1 ) ( x − a2 ) ( x − ar ) where the constants A1 , A2 ,..., Ak , B1 , B2 ..., Br can be obtained by equating the coefficients of like power of x. (iii) If some of the factors in Q( x) are quadratic and non-repeating, corresponding to each quadratic factor ax2 + bx + c (non-factorisable), we assume the partial Ax + B fraction of the type 2 , where A and B are ax + bx + c constants to be determined by comparing coefficients of like powers of x. (iv) If some of the factors in Q( x) are quadratic and repeating, for every quadratic repeating factor of the type (ax2 + bx + c)k where ax2 + bx + c cannot be further factorise, we assume A2 k − 1 x + A2 k A1 x + A2 A3 x + A4 + + ... + 2 2 2 ax + bx + c (ax + bx + c) (ax2 + bx + c)k If degree of P ( x) > degree of Q( x), then we first divide P( x) by P( x ) P ( x) is expressed in the form of T ( x) + 1 , where Q( x) Q( x) P ( x) is a proper rational function T ( x) is a polynomial in x and 1 Q( x) (i.e. degree of P1 ( x) < degree of Q( x))

Q( x) so that

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 If ∫

2

dx x6 to = p ( x ), then ∫ x + x 7 dx is equal x + x7 j JEE Mains 2013

(a) log | x | − p (x) + C

(b) log | x | + p (x) + C

(c) x − p (x) + C

(d) x + p (x) + C

∫

x 3 −1 dx is equal to ( x 4 + 1)( x + 1)

8 If ∫ x + x 2 + 5 dx = P {x + x 2 + 5} 3 / 2

(a)

3

10

∫ ( x + 1)( x + 2) ( x + 3) dx is equal to 7

4 The integral ∫

dx 3

11

12

equals

 x + 1 4 (a)   +C  x4 

j

(c) − (x + 1) + C

(b) (x 4 + 1) 4 + C

4

(a) (sinα,cosα) (c) (− sinα,cosα)

sin8 x − cos 8 x ∫ 1 − 2 sin2 x cos 2 x dx is equal to

14

(b) (cosα,sinα ) (d) (− cosα,sinα)

(b) cos x (d) cot x 2

 (log x − 1)  dx is eqal to 2  

∫ 1 + (log x )

x +C (log x)2 + 1 x (c) 2 +C x +1

∫

1 sin 2 x + C 2 (d) − sin2x + C (b) −

1 sin 2 x + C 2

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

1− x 2

cos 6x + 6 cos 4x + 15 cos 2x + 10   dx = f ( x ) + C, 2 x + 5 cos x cos 3x + cos x cos 5x  then f (10) is equal to 

(b) 10

∫x

(a) (c)

15

5

x 2 −1 3

2x 4 − 2x 2 + 1

dx 2

xe x +C 1+ x 2 log x (d) +C (log x)2 + 1

(d) 2 cos10

x10 +C 2 (x + x 3 + 1)2 − x10 (d) +C 5 2 (x + x 3 + 1)2 (b)

5

dx is equal to

2x 4 − 2x 2 + 1 +C x2 4 2 2x − 2x + 1 +C 2x

∫ (1 + x

(c) 2 sin10

2x 12 + 5x 9 dx is equal to ( x 5 + x 3 + 1)3 j JEE Mains 2016

) p 2 + q 2 (tan−1 x )2

(b) (d)

2x 4 − 2x 2 + 1 +C x 4 2 2x − 2x + 1 +C 2x 2

is equal to

1 log[q tan−1 x + p 2 + q 2 (tan−1 x)2 ] + C q (b) log[q tan−1 x + p 2 + q 2 (tan−1 x)2 ] + C 2 (c) (p 2 + q 2 tan−1 x)3 / 2 + C 3q

(a) (b)

dx ;x ∈( 0,1) equals

∫ 10 cos

(a)

f (x ) dx = log log sin x + C, then f ( x ) is equal to log sin x

(a) sinx (c) logsinx

(a)

1 x 3π 1 x 3π log tan  − log tan  +  + C (d)  +C 2 2 2 8  2 8 

− x5 +C (x + x 3 + 1)2 x5 (c) +C 5 2 (x + x 3 + 1)2

 x 4 + 1 4 (d) −   +C  x4 

sin x dx = Ax + B log sin ( x − α ) + C, sin( x − α ) then the value of ( A , B ) is

7

(c)

13 The integral ∫

5 If ∫

6 If ∫

1 x π 1 x log tan  −  + C (b) log cot   + C 2 8  2 2 2

(a) 20

JEE Mains 2015

1

1

1

(d) −5

(a) 21/ 6 x + C (b) 21/12 x + C (c) 21/ 3 x + C (d) None of these

x ( x 4 + 1) 4

4

(c) −3

(a)

(c)

2

1

(b) −4

dx is equal to cos x − sin x

8

(a)

4

∫

+ C, then the value of 3PQ is

(a) sin2x + C

(x + 2) (x + 2) − +C 10 8 (x + 1)2 (x + 2)8 (x + 3)2 (b) − − +C 2 8 2 10 (x + 2) (c) +C 10 2 (x + 1) (x + 2)8 (x + 3)2 (d) + + +C 2 8 2 10

x x2 + 5

(a) −1

9

1 1 log(1 + x 4) + log(1 + x 3) + C 4 3 1 1 (b) log(1 + x 4) − log(1 + x 3) + C 4 3 1 (c) log(1 + x 4) − log(1 + x) + C 4 1 (d) log(1 + x 4) + log(1 + x) + C 4

Q

+

(d) None of the above

155

DAY 16 In the integral ∫

cos 8x + 1 dx = A cos 8x + k , where cot 2x − tan 2x

24

k is an arbitrary constant, then A is equal to j

(a) −

17

∫

∫

(b)

1 16

(c)

1 8

(sin θ + cos θ ) dθ is equal to sin 2 θ

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

18

1 16

j

log | cosθ − sinθ + sin 2 θ | log | sinθ − cosθ + sin 2 θ| sin−1 (sinθ − cosθ) + C sin−1 (sinθ + cosθ) + C

equal to 1  g (x)  (b)   2  f (x)  2

(c)

1   g (x)    log  + C 2   f (x)  

25

20 If ∫

21

∫ cos

−

x sin

−

11 7

1 (c)  , − 1 5 

(b)

4

22

∫

1 (d)  − , 0  5 

4 tan7 x + C 7 3

28 If ∫

+C

+C +C

∫ g ( x ) e dx is equal to x

(b) f (x) + g (x) + C (d) f (x) − g (x) + C

29

j

∫ tan

−1

(b) cos x

(c) 2 tan−1 

tan (x / 2)   +C  2

(d) None of these

(d) cot x j

NCERT Exemplar

(b) x tan −1 x − x + C (d) x − (x + 1) tan −1 x + C

30 If I n = ∫ (log x )n dx , then I n + n I n − 1 is equal to (a) x (log x)n

(b) (x log x)n

(a) g −1(x) (c) xf −1(x) − g −1(x)

32

(c) (log x)n −1

(d) n (log x)n

∫

23 If the integral

∫ tan x − 2 dx = x + a ln | sin x − 2 cos x | + k ,

(b) xf −1(x) − g (f −1(x)) (d) f −1(x)

( x + 3) e x dx is equal to ( x + 4) 2

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

(b) − 2 (d) 2

(c) tanx

xdx is equal to

(a) (x + 1) tan−1 x − x + C (c) x − x tan−1 x + C

NCERT Exemplar

tan (x / 2) + 1 −1  tan (x / 2) + 1  + C (b) tan   +C    2 2

then a is equal to

JEE Mains 2013

31 If ∫ f ( x ) dx = g ( x ), then ∫ f −1( x ) dx is equal to

(a) 2 tan−1 

5 tan x

j

1 − 6 cos 2 x f (x ) dx = + C, then f ( x ) is equal to 6 2 sin x cos x (sin x )6

(a) sinx

(d) log| cos 7 x | + C

dx is equal to 2 + sin x + cos x

(a) − 1 (c) 1

3 −1  x  sin   a 2

2/ 3

(d)

3/ 2

1 3 [x Ψ (x 3)] − ∫ x 2 Ψ (x 3)dx + C 3 1 (b) [x 3 Ψ (x 3)] − 3 ∫ x 3 Ψ (x 3)dx + C 3 1 (c) [x 3 Ψ (x 3) − ∫ x 2 Ψ (x 3)dx] + C 3 1 (d) [x 3 Ψ (x 3)] − ∫ x 3 Ψ (x 3)dx + C 3

4

− −7 tan 7 x + C 4

2 −1  x  sin   a 3

3/ 2

(b)

(a)

4

(c)

3 −1  x  sin   a 2

+C

27. If ∫ f ( x ) dx = Ψ( x ), then ∫ x 5f ( x 3 ) dx is equal to

x dx is equal to

(a) log | sin7 x | + C

 x 2 − x + 1 1 log  2  +C 2  x + x + 1

3/ 2

(a) f (x) ⋅ g (x) + C (c) e x cos x + C

2

(b) 2 (d) None of these 3 7

(d)

3

then ∫ f ( x ) cos x dx +

tan x 2 dx = x − tan−1 1 + tan x + tan2 x A  2 tan x + 1   + C, then the value of A is   A

(a) 1 (c) 3

 x 2 − x − 1 1 log  2  +C 2  x + x + 1

26 If an anti-derivative of f ( x ) is e x and that of g ( x ) is cos x,

2

 g (x)  (d) log  +C  f (x) 

(b)

x dx is equal to a −x3

∫

(c)

I 4 + I 6 = a tan5 x + bx 5 + C, where C is a constant of integration, then the ordered pair (a, b ) is equal to 1 (b)  , 0 5 

 x 2 + x + 1 1 log  2  +C 2  x − x + 1

 x 2 − x + 1 (c) log  2  +C  x + x + 1

19 Let I n = ∫ tann x dx (n > 1). If

1 (a)  − , 1  5 

x 2 −1 dx is equal to + x2 +1

x (a) sin−1  a

 f ( x ) ⋅ g ' ( x ) − f ' ( x ) ⋅ g ( x )   ((log g ( x ) − log f ( x )) dx is   f ( x ) ⋅ g( x )

 g (x)  (a) log  +C  f (x) 

4

(a)

JEE Mains 2013 1 (d) − 8

JEE Mains 2017

∫x

33 If ∫

ex +C (x + 4) 2 ex (d) +C x+ 3 (b)

−1 x 2 − x + 1 cot −1 x e dx = A ( x ) e cot x + C, then A ( x ) is 2 x +1

equal to (a) − x

j

(b) x

(c)

1− x

JEE Mains 2013

(d)

1+ x

156 34 If g ( x ) is a differentiable function satisfying

35

d { g ( x )} = g ( x ) and g( 0) = 1, then dx  2 − sin 2x  ∫ g ( x ) 1 − cos 2x  dx is equal to (a) g (x) cot x + C g (x) (c) +C 1 − cos 2 x

(b) − g (x) cot x + C

∫x

3

dx 3 is equal to ( x n + 1)

(a)

3  xn  ln  +C n  x n + 1

(b)

1  xn  ln  +C n  x n + 1

(c)

3  x n + 1 ln  +C n  xn 

 x n +1  (d) 3n ln n  + C  x 

(d) None of these

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 If ∫ f ( x ) dx = f ( x ), then ∫ {f ( x )} 2 dx is equal to (a)

1 {f (x)}2 2

(b) {f (x)}3

0 2 If f ( x ) = sin x − x 2 2 − cos x

(c)

x 2 − sin x 0 2x − 1

{f (x)}3 3

(d) {f (x)}2

cos x − 2 1 − 2x , 0

6

x − x 2 sin x + sin 2 x + C 3 x3 (b) − x 2 sin x − cos 2 x + C 3 x3 (c) − x 2 cos x − cos 2 x + C 3 (d) None of the above (a)

∫e

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

 cosθ + sinθ  (b) (cosθ + sinθ) 2 log   +C  cosθ − sinθ 

1 − cos 2a x dx is equal to 1 + sin 2a x

7

1 π − e 2ax cos  + ax  + C 4  a 1 2ax π  − e cot  + ax  + C 4  2a 1 2ax π  − e cos  + ax  + C 4  2a 1 2ax π  − e cosec  + ax  + C 4  a

∫x (a) log (c)

1 log | sec (x 2 + 1) | + C 2

 cosθ − sinθ  (cosθ − sin θ) 2 log   +C 2  cosθ + sinθ 

(d)

1 π 1 sin 2 θ logtan  + θ − log sec 2 θ + C 4  2 2

∫ (x

x2 dx is equal to sin x + cos x ) 2

sin x + cos x +C x sin x + cos x x sin x − cos x (b) +C x sin x + cos x (c)

sin x − x cos x +C x sin x + cos x

(d) None of the above

2 sin ( x + 1) − sin 2 ( x + 1) dx is equal to 2 sin ( x 2 + 1) + sin 2 ( x 2 + 1) 1 sec (x 2 + 1) + C 2

(c)

(a)

4 If x 2 ≠ nπ − 1, n ∈ N. Then, the value of 2

 cos θ + sin θ 

∫ cos 2 θ log  cos θ − sin θ  dθ is equal to  cosθ + sinθ  (a) (cosθ − sinθ) 2 log   +C  cosθ − sinθ 

3

3

2

 x 2 + 1 (b) log sec   +C  2  (d) None of these

8 If f ( x ) = ∫

x 2 dx (1 + x )(1 + 1 + x 2 ) 2

5

sin2 x cos 2 x dx x + cos x sin2 x + sin3 x cos 2 x + cos 5 x )2

is equal to

3

j

JEE Mains 2018

and f ( 0) = 0, then the

value of f (1) is (a) log (1 +

2)

(c) log (1 +

2) +

5 The integral

∫ (sin

(b)

(where C is a constant of integration)

then∫ f ( x ) dx is equal to

2 ax

−1 +C 3 (1 + tan3 x) −1 (d) +C 1 + cot 3 x

1 +C 3 (1 + tan3 x) 1 (c) +C 1 + cot 3 x

(a)

9 If I = ∫

(b) log (1 + π 2

(a) 2/3 (c) 1/3

( x + 1) ( x − 1) 2

π 4

(d) None of these

dx 3

2) −

4

=k

3

1+ x + C, then k is equal to 1− x

(b) 3/2 (d) 1/2

157

DAY 10

dx

∫ (sin x + 2)(sin x − 1) is equal to

 

1  x   2 tan +     2 2 −1 2 2 (a) tan  − +C x 3   3  tan − 1 3 3     2 2 (b) +C  tan x + 1     2  2 tan x − 1  2 1 2 −1  2 (c) − + tan   +C 3  tan x − 1 3 3 3         2  2 tan x − 1   2 2 2 2 (d) − + tan−1   +C 3  tan x − 1 3 3 3         2

11

∫

(a) (x − 1) e

13

1 x2  6  2 + 3 x 2 

3/ 2

(b)

1 x2  6  2 + 3 x 2 

7/ 2

(c)

x+

x

x+

+C

(b) xe

1 x

x+

(d) − xe

+C 99

JEE Mains 2014

1 x

+C

x+

1 x

+C

x )dx is equal to cos(100 x) (sin x)100 +C 100 100 cos(100 x)(cos x) (d) +C 100

sin(100x) (sin x) +C 100 100 cos(100 x)(cos x) (c) +C 100

(b)

(a)

∫

dx is equal to j

100

14

1 x

1

∫ (sin(101 x )⋅ sin

−1 x ( 2018)2 x ( 2018)sin ( 2018 ) dx is equal to 1 − ( 2018)2 x

(a) (log2018 e)2 (2018)sin

x dx is equal to (2 + 3x 2 )5 / 2 3/ 2

x+

(c) (x + 1) e

2

1 x2  (a)  5  2 + 3 x 2 

1 x

12 The integral ∫ 1 + x −  e

−1

( 2018 ) x

+C

2

x + sin −1 ( 2018 ) x

+C

2

x − sin −1 ( 2018 ) x

+C

(b) (log2018 e) (2018) (c) (log2018 e) (2018) −1

x

(2018)sin ( 2018) (d) +C (log2018 e)2

+C

15

+C

∫ (∫ e

x

2 1  log x + − 2  ) dx is equal to  x x 

(a) e x logx + C1x + C2

+C

(c)

(d) None of the above

logx + C1x + C2 x

(b) logx +

1 + C1x + C2 x

(d) None of these

ANSWERS SESSION 1

SESSION 2

1 (a)

2 (c)

3 (a)

4 (d)

5 (b)

6 (d)

7 (a)

8 (d)

9 (d)

10 (b)

11 (a)

12 (a)

13 (b)

14 (d)

15 (a)

16 (a)

17 (c)

18 (c)

19 (b)

20 (c)

21 (c)

22 (a)

23 (d)

24 (d)

25 (b)

26 (c)

27 (a)

28 (c)

29 (a)

30 (a)

31 (b)

32 (c)

33 (b)

34 (b)

35 (a)

1 (a) 11 (b)

2 (d) 12 (b)

3 (b) 13 (a)

4 (b) 14 (a)

5 (b) 15 (a)

6 (d)

7 (c)

8 (b)

9 (b)

10 (a)

158

Hints and Explanations 1 Let I =

x6

∫x+

7

x

dx =

x6

∫ x (1 +

6

x )

(1 + x 6 ) − 1 =∫ dx x (1 + x 6 ) dx dx ⇒ I =∫ −∫ x x + x7

∫ (x

= cot x

x −1 dx + 1)( x + 1) 3

4

7

∫

=

∫

x ( x + 1) − ( x + 1) dx ( x 4 + 1)( x + 1)

=

∫  x

=

1 log( x 4 + 1) − log( x + 1) + C 4

3



=

x3 1  − dx + 1 x + 1 

4

= =

9

7

t t ( x + 2) ( x + 2) = − +C= − +C 10 8 10 8 dx dx 4 ∫ =∫ 3 3 2 4 4 1 4 x ( x + 1) x 5  1 + 4   x  8

(log x ) + 1 − 2 log x dx [(log x ) 2 + 1] 2

d 

10

8

x

∫ dx  (log x )

∫

⇒

1/4

 x4 + 1 + C = −   x4 

Put x − α = t ⇒ dx = dt sin (t + α ) ∴I =∫ dt sin t cos t

∫ sin α ⋅ sin t

dt

= cos α ⋅ t + sin α log sin t + C = x cos α + sin α log {sin( x − α )} + C ∴

A = cos α, B = sin α

∫

x2 + 5 dx

x2 + 5 = t − x

⇒ ⇒ ⇒

x2 + 5 = t 2 + x2 − 2 xt 5 = t 2 − 2 xt 2 xt = t 2 − 5 1 5 x =  t −  2 t 1 5 dx =  1 + 2  2 t 

⇒

2

I = ∫t

1 /2

=

∫

=

∫

=

∫

1/4

+C

3

6

3

 [ x 2 + ( 2 − x2 )2 − 2 x 2 − x 2 ]   6   2   3

x + 2− x2 1/6 3

2 3

21 / 6 3 1 − x 2

=

∫

(sin 4 x + cos 4 x ) (sin 4 x − cos 4 x ) dx 2 x + cos 2 x )2 − 2sin2 x cos 2 x

dx

= 2x + C Clearly, f (10) = 20 2 x12 + 5x 9 dx 5 + x3 + 1) 3

13 Let I =

∫ (x

∫ (1 +

∫ (sin

dx

⋅ cos x + 10 ⋅ (2cos 2 x ) dx = 2∫ dx 10cos 2 x + 5cos x ⋅ cos 3 x + cos x cos 5x

=

I =

1 − x2

2cos 5x ⋅ cos x + 10 ⋅ cos 3 x

1 x 3π log tan  +  +C 2 8  2 sin x − cos x dx 2 x cos 2 x

2− x2 − x

dx

(cos 6 x + cos 4 x ) + 5 (cos 4 x + cos 2 x ) + 10(cos 2 x + 1) =∫ dx 10cos 2 x + 5cos x cos 3 x + cos x cos 5x

=

∫ 1 − 2 sin

dx

  cos 6 x + 6cos 4 x   x + cos + 15 2 10  dx 12 Let I = ∫  2  10cos x + 5cos x cos 3 x    + cos x cos 5x  

∫x

I =

3

(2 − x 2 ) − x 2

=

10 Let

1− x2

3

1 π x π log tan  + +  + C 4 2 2 8

8

1 − x2

dx

x + 2− x2

=

8

1 − x2

= 21 / 6 ∫ dx = 21 / 6 x + C

1 5 ⋅  1 + 2  dt 2 t 

1 1 /2 (t + 5t −3 / 2 )dt 2∫ 12 10  1 3 /2 5 =  t 3 /2 − +C +C= t − 23 3 t t Clearly, 3PQ = −5 dx 9 Let I = 1 ∫ 2  1 cos x − 1 sin x     2  2 1  x + π  dx = sec ∫  4  2

( x + 2 − x2 )( 1 − x 2 − x2 )

3

3

=

5 Let I = ∫ sin x dx sin( x − α )

= ∫ cos α dt +

x+

⇒

Now,

Hence, the integral becomes − t 3dt ∫ t 3 = − ∫ dt = − t + C 1 = −  1 + 4   x 

 dx + 1 

x + 5=t

and

1 = t4 x4 4 − 5 dx = 4t 3dt x dx = − t 3dt x5

2

Put x +

Put 1 + ⇒

=

x +C (log x ) 2 + 1

8 Let I =

= ∫ (t − 1)⋅ t dx = ∫ (t − t )dx 7

∫

∫

sin2 x +C 2

 1  x + 2 − x 2  6 (2 − 2 x 2 − x 2 )  2 

3

2

∫ − cos 2x dx

=−

1 (log x )2 + 1 − 2 x  log x ⋅   x =∫ dx [(log x ) 2 + 1]2

Put x + 2 = t ⇒ x = t − 2 and dx = dt ∴ I = ∫ (t − 1)t 7 (t + 1)dx 10

 (log x − 1)  ∫ 1 + (log x )2  dx

4

3 Let I = ∫ ( x + 1)( x + 2) 7 ( x + 3)dx

2

=

11 Let I =

2

x3 + x 4 − x 4 − 1 dx ( x 4 + 1)( x + 1)

=

= ∫ (sin2 x − cos 2 x )dx

  f '( x ) Q ∫ f ( x ) dx = log( f ( x )) + C   

= log| x| − p( x ) + C

2 Let I =

= ∫ (sin 4 x − cos 4 x )dx

f ( x) = log(log sin x ) + C log sin x d ∴ f ( x) = (log sin x ) dx

6 We have, ∫

dx

15

2 x12 + 5x 9 dx (1 + x − 2 + x − 5 ) 3

2 x − 3 + 5x − 6 dx x − 2 + x − 5)3

Now, put 1 + x − 2 + x − 5 = t ⇒ (− 2 x − 3 − 5x − 6 ) dx = dt ⇒ (2 x − 3 + 5x − 6 ) dx = − dt dt I = − ∫ 3 = − ∫ t − 3 dt ∴ t

159

DAY =− =

14 Let I =

t −3 +1 1 +C= 2 +C −3+ 1 2t

x10 +C 2( x + x3 + 1) 2 5

x2 − 1

∫x

dx 2x4 − 2x 2 + 1 x2 − 1 =∫ dx 2 1 x5 2 − 2 + 4 x x 1 1 − x3 x 5 =∫ dx 2 1 2− 2 + 4 x x 2 1 Now, putting 2 − 2 + 4 = t, we get x x 1 1 4 3 − 5  dx = dt x x  1 dt ∴I = ∫ 4 t 3

1 = ⋅2 t + C = 4

g( x) =t f ( x) f ( x )⋅ g '( x ) − g ( x )⋅ f '( x ) ⇒ dx = dt ( f ( x )) 2 f ( x )⋅ g '( x ) − g ( x ) ⋅ f '( x ) ⇒ f ( x )⋅ g ( x ) g( x) ⋅ ⋅ dx = dt f ( x)

Put

 f ( x )⋅ g '( x ) − g ( x )⋅ f '( x ) dt ⇒   dx = t f ( x )⋅ g ( x )   Now, I =

1

∫ t ⋅ log t dt =

2

=

 g ( x )  1  log   + C 2  f ( x ) 

19 We have, I n = ∫ tan n x dx

∫ tan = ∫ tan = ∫ tan

∴ I n + I n +2 =

2x4 − 2x 2 + 1 +C 2x 2

=

15 Put q tan −1 x = t q 1 dt dx = dx = dt ⇒ 1 + x2 q 1 + x2 dt 1 = log [ t + p2 + t 2 ] ∴∫ q p2 + t 2 q ⇒

=

1 log [q tan −1 x q +

16 LHS =

∫

p + q (tan 2

2

−1

x) ] + C 2

2

2cos 4 x dx cos 2 2 x − sin2 2 x cos 2 x sin 2 x

2cos 2 4 x × cos 2 x sin 2 x dx cos 4 x 1 = ∫ cos 4 x × sin 4 xdx = ∫ sin 8 xdx 2 −1 cos 8 x = + k 2 8 −1 Hence, we get A = 16 sin θ + cos θ 17 Let I = ∫ dθ 1 − (1 − 2sin θ cos θ ) =

∫

= Put

sin θ + cos θ

∫

1 − (sin θ − cos θ) 2

dθ

sin θ − cos θ = t

⇒ (cos θ + sin θ) dθ = dt dt ∴ I =∫ = sin −1 ( t ) + C 1 − t2 −1

= sin (sin θ – cos θ) + C  f ( x ) ⋅ g '( x ) − f '( x )⋅ g ( x )  f ( x )⋅ g ( x )  

18 Let I = ∫ 

 g ( x ) log   dx  f ( x )

(log t )2 +C 2

∫ tan

n +2

n

x dx +

n

x(1 + tan x ) dx

n

x sec2 x dx

x dx

2

n +1

tan x +C n+1

Put n = 4, we get I 4 + I 6 =

tan 5 x +C 5

1 and b = 0 5 tan x 20 Let I = ∫ 1 + tan x + tan2 x ∴

a=

sin x cos x dx = ∫ dx sin2 x sin x 1+ + cos 2 x cos x sin 2 x =∫ dx 2 + sin 2 x dx = ∫ dx − 2 ∫ 2 + sin 2 x = x−2∫

sec2 x dx 2 2 sec x + 2 tan x

Let tan x = t ⇒ sec2 x dx = dt 2 dt = x− ∫ 2 2 t +t +1 dt = x−∫ 2 2 t + 1 +  3       2  2  ⇒I = x−

2 tan −1 3

 2 tan x + 1    +C   3

Hence, we get A = 3.

21 Here, m + n =

− 3  −11  +   = −2  7  7

I = ∫ cos −3 /7 x(sin( − 2 + 3 /7 ) x )dx = ∫ cos −3 /7 x sin −2 x sin3 /7 xdx

=

cosec2 x dx = 3 /7 x    sin3 / 7 x 

∫  cos

cosec2 x dx 3 /7 x

∫ cot

Put cot x = t ⇒ − cosec2 x dx = dt dt −7 4 / 7 t +C ∴ I = − ∫ 3 /7 = t 4 7 = − tan − 4 / 7 x + C 4 dx 22 Let I = ∫ 2 + sin x + cos x dx ⇒ I =∫ x x 2 tan 1 − tan2 2 + 2 2+ x x 1 + tan2 1 + tan2 2 2 x sec2 dx 2 =∫ x x x 2 + 2 tan2 + 2 tan + 1 − tan2 2 2 2 x sec2 dx 2 I =∫ x x tan2 + 2 tan + 3 2 2 x 1 x Put tan = t ⇒ sec2 dx = dt 2 2 2 2dt ∴I = ∫ 2 t + 2t + 3 dt 2dt = 2∫ 2 = 2∫ t + 2t + 1 + 2 (t + 1)2 + ( 2 ) 2 = 2⋅

1  t + 1 tan −1   +C  2  2

 tan x + 1    2 ⇒ I = 2 tan   +C 2     5 tan x 23 Given, ∫ dx tan x − 2 −1

= x + a ln|sin x − 2 cos x | + k Now, let us assume that 5 tan x I =∫ dx tan x − 2

…(i)

On multiplying by cos x in numerator and denominator, we get 5 sin x I =∫ dx sin x − 2 cos x Let 5 sin x = A (sin x − 2 cos x ) + B (cos x + 2 sin x ) ⇒ 0 ⋅ cos x + 5 sin x = ( A + 2B ) sin x + (B − 2 A ) cos x On comparing the coefficients of sin x and cos x, we get A + 2B = 5 and B − 2 A = 0 ⇒ A = 1 and B = 2 ⇒ 5 sin x = (sin x − 2 cos x ) + 2 (cos x + 2 sin x )

160

⇒I =

5 sin x

∫ sin x − 2 cos x dx

(sin x − 2cos x ) + 2 (cos x + 2 sin x ) dx (sin x − 2 cos x ) d (sin x − 2 cos x ) ⇒ I = ∫ 1 dx + 2 ∫ (sin x − 2 cos x ) =

∫

I = x + 2 log|(sin x − 2 cos x )| + k …(ii) where, k is the constant of integration. On comparing the value of I in Eqs. (i) and (ii), we get a = 2

24

∫

1 − 1     x2 

dx 2 1 x + − 1    x 1 1 Put x + = t ⇒  1 − 2  dx = dt  x x  ∴

 t − 1 dt 1 ∫ t 2 − 1 = 2 log  t + 1  + C  x2 − x + 1  1 = log  2  +C 2  x + x + 1

25 Let I = ∫ 3 x 3 dx a −x Put x = a(sin θ) 2 / 3 2 ⇒ dx = a(sin θ)−1 / 3 cos θ dθ 3 2 a1 / 2 (sin θ) 1 / 3 ⋅ 3 a(sin θ)−1 /3 ⋅ cos θ ∴ I =∫ dθ a3 − a3 sin2 θ 2 a3 / 2 ⋅ cos θ 2 = ∫ 3 /2 dθ = ∫ dθ 3 a cos θ 3 3 /2 2 2 −1  x  = θ + C = sin   + C  a 3 3 26

∫ =

f ( x ) cos x dx +

x ∫ g ( x ) e dx

−

29 Let I = ∫ tan −1 x (1) dx = tan −1 x ∫ 1dx

d −  ∫ (tan −1  dx x⋅x−∫

= tan −1 = x tan

−1

e (sin x − cos x ) + C 2

27 Given, ∫ f ( x ) dx = Ψ( x ) I =

∫

x )∫ (1) dx dx 

1 1 × . xdx 1+ x 2 x

1 x x− ∫ dx 2 (1 + x ) x x−

x 5 f ( x3 ) dx

x3 = t dt 2 …(i) ⇒ x dx = 3 1 1 ∴ I = ∫ t f ( t ) dt = [t Ψ (t ) − ∫ Ψ (t ) dt ] 3 3 1 = [ x3 Ψ ( x3 ) − 3 ∫ x2 Ψ( x3 ) dx] + C 3 [from Eq. (i)] 1 3 3 2 3 = x Ψ ( x ) − ∫ x Ψ ( x ) dx + C 3

= ∫ 1 ⋅ e cot = xe cot

x2

∫

⇒

= ( x + 1)tan −1 x −

x+C

30 I n = ∫ ( log x ) dx = x (log x ) n

n

− n ∫ ( log x ) n –1 ⋅

1 ⋅ x dx x

31 Consider, ∫ f −1 ( x )dx = ∫ f −1 ( x )⋅1dx d −1 ( f ( x ))⋅ xdx dx −1 Now, let f ( x ) = t , then d −1 dt ( f ( x )) = dx dx d −1 ( f ( x ))⋅ dx = dt ⇒ dx ∴ ∫ f −1 ( x )dx = x ⋅ f −1 ( x ) − ∫ f (t )dt

32 Let I = ∫ e  x + 3 2  dx  ( x + 4)  x

 x + 4 − 1 = ∫ex  dx  ( x + 4) 2  1  1  = ∫ex −  dx  x + 4 ( x + 4) 2 

x

x

dx −

∫x

2

− ∫ x ⋅ e cot

e cot

−1

x

−1

−1 x e cot x dx +1

x

 1  −  dx 2  1+ x 

dx + C = xe cot

−1

x

+C

g ′( x ) dx = ∫ 1 dx g( x)

log e {g ( x )} = x + log C1 g ( x ) = C1 e x

Now,

g(0) = 1 ⇒ C1 = 1

∴

g( x) = e x  2 − sin 2 x 

∫ g ( x )  1 − cos 2 x  dx = ∫ e (cosec x − cot x ) dx 2

x

= − e cot x + C = − g ( x ) cot x + C x

3 x2 dx dx3 dx = ∫ 3 n dx n + 1) x ( x + 1) x n −1 dx dx = 3∫ = 3∫ n n n x( x + 1) x ( x + 1)

∫ x (x 3

On putting x n = t , we get 3 dt 3 1 1  I = ∫ = − dt n t (t + 1) n ∫  t t + 1  3 = [log t − log(t + 1)] + C n 3 t  = log  +C  t + 1 n

= f −1 ( x )⋅ x − ∫

[Qf −1 ( x ) = t ⇒ x = f (t )] = x ⋅ f −1 ( x ) − g (t ) = x ⋅ f −1 ( x ) − g ( f −1 ( x ))

−1

−1

⇒

35 Let I =

= x tan −1 x − t + tan −1 t + C = x tan −1 x − x + tan −1 x + C



∴ A( x ) = x 34 We have, d {g ( x )} = g ( x ) dx ⇒ g ′( x ) = g ( x )

2

t dt 1 + t2

−1

x

∫1+

−

1 t2 2t dt 2 ∫ (1 + t 2 ) ⋅ t

x−∫

= x tan −1

 x2 + 1

x cot x 33 LHS = ∫  2 − 2 dx e  x + 1 x + 1

∴

Put x = t 2 ⇒ dx = 2t ⋅ dt ∴ I = x tan −1

1 +C x+ 4 x [Q∫ e ( f ( x ) + f '( x )dx = e x ⋅ f ( x ) + C ]

= ex ⋅

⇒

∴ I n + n I n − 1 = x (log x ) n x

ex = (2cos x ) + C 2 x = e cos x + C

Put

∫ sin

x

e (cos x + sin x ) 2

Let

1 − 6cos 2 x dx 6 x cos 2 x 1 dx =∫ dx − 6∫ sin 6 x cos2 x sin 6 x (say) = I1 − I2 sec2 x Here, I1 = ∫ dx sin 6 x 1 1 =∫ ⋅ sec2 xdx = ⋅ tan x sin 6 x sin 6 x (−6) −∫ ⋅ cos x tan xdx sin7 x tan x tan x = + I2 ⇒ I1 − I2 = +C sin 6 x sin 6 x tan x Thus, I = +C sin 6 x Hence, f ( x ) = tan x

28 Let I =

=

 xn  3 log  n +C n  x + 1

SESSION 2 1 We have,

f ( x)

d { f ( x )} = f ( x ) dx 1 d [ f ( x ) ] = dx f ( x)

⇒ ⇒ ⇒ ⇒ ⇒ ∴

∫ f ( x ) dx =

∫

log { f ( x )} = x + log C f ( x ) = Ce x { f ( x )}2 = C 2e 2 x { f ( x )}2 dx = ∫ C 2e 2 xdx =

C 2e 2 x 1 = { f ( x )}2 2 2

161

DAY 2 We have, 0 f ( x ) = sin x − x2 2 − cos x

x − sin x cos x − 2 0 1 − 2x 2x − 1 0 2

0 sin x − x2 2 − cos x ⇒ f ( x ) = x − sin x 0 2x − 1 2

cos x − 2

1 − 2x

0

[interchanging rows and columns] ⇒ f ( x ) = (− 1)3 0 x2 − sin x cos x − 2 2 sin x − x 0 1 − 2x 2 − cos x 2x − 1 0

∫x

=

∫ x tan 

∴

∫ tan 

⇒I =

 x2 + 1   dx 2 

 x2 + 1   x2 + 1   d  2   2 

 x2 + 1  = log sec   +C  2 

5 We have, I =

⇒ I =−

1 2t π 1 e cot  + t  + ∫ e 2 t 4  2a a π 1 cot  + t  dt − ∫ e 2 t 4  a π cot  + t  dt + C 4 

1 2t π e cot  + t  + C 4  2a 1 2 ax π  ∴ I =− e cot  + ax  + C 4  2a

=

⋅ cos ( x 2 + 1)

⇒

tan2 x sec2 x dx tan3 x )2

∫ (1 +

=

1 π sin 2θ log tan  + θ 4  2 1 − log sec 2θ + C 2 x2

7 Let I =

∫ ( x sin x + cos x )

=

∫ ( x sin x + cos x )

∴

2

x cos x

2

dx ⋅

x dx cos x

Q d ( x sin x + cos x ) = x cos x    dx −1 x I = . ( x sin x + cos x ) cos x 1 + ∫ ( x sin x + cos x )

tan2 θ sec 2 θ

∫ sec θ(1 + sec θ)dθ 2

1 − cos 2 θ

∫ cos θ (1 + cos θ) dθ

f (0) = log (0 +

⇒ ∴

− tan −1 (0) + C

f (1) = log (1 +

9 Let I =

∴

∴

2) –

π +0 4

dx

∫

2

 x + 1   1 − x 1+ x 2 =t ⇒ dx = dt 1− x (1 − x ) 2 1 dt 3 I = ∫ 2 /3 = [ t 1 /3 ] + C 2 t 2  3 1+ x =  + C 2 1− x  (1 − x )2

10

1 + 0)

C=0

Put

 π + θ  2 1 π ⇒ ∫ sec 2θ dθ = log tan  + θ 4  2 d π  2sec 2θ = log tan  + θ 4  dθ 1 π   ∴I = sin 2θ log tan  + θ − ∫ tan2θ dθ 4  2

1 + x2 )

= log (sec θ + tan θ) – θ + C = log ( x + 1 + x2 ) − tan – 1 x + C

∫ sec θ dθ = log tan  4

and

x )(1 + 2

= ∫ sec θ dθ – ∫ dθ

∫ (sin

 cos θ + sin θ  π  log   = log tan  + θ 4   cos θ − sin θ 

2 sin ( x 2 + 1) − 2 sin ( x 2 + 1) ⋅ cos ( x2 + 1) dx 2 sin ( x2 + 1) + 2 sin ( x 2 + 1)

=

6 Since,

2 sin ( x 2 + 1) − sin 2 ( x 2 + 1) x dx 2 sin ( x 2 + 1) + sin 2( x 2 + 1)

∫x

dx

3

4 We have,

=

∴ f ( x) =

Put tan x = t ⇒ 3 tan2 x sec2 xdx = dt 1 dt −1 +C ∴I = ∫ ⇒I = 3 (1 + t )2 3(1 + t ) −1 ⇒ I = +C 3(1 + tan3 x )

x 2 dx

∫ (1 +

Put x = tanθ ⇒ dx = sec2 θ dθ = (1 + x2 ) dθ

dx

sin2 x cos 2 x dx 3 x + cos 3 x )2 2 2 sin x cos x =∫ dx cos 6 x(1 + tan3 x )2

⇒ I =−

∫

5

2

sin2 x cos 2 x 3 {sin x(sin2 x + cos2 x ) +

=

⇒ I =

1 π − ∫ e 2 t cot  + t  dt 4  a

sin x ⋅ cos x x + cos 3 x ⋅ sin2 x 2

cos 3 x(sin2 x + cos2 x )}2

1 2 t 1 − cos 2t e dt , [where, ax = t ] a∫ 1 + sin 2t

 1 cosec2  π + t  − cot  π + t   dt      4  4  2 1 π e 2 t cosec2  + t  dt ⇒ I = II 4  2a ∫ I

∫

=

1 − cos 2ax dx 1 + sin 2ax

1 2t e a∫ π π 1 − 2 sin  + t  ⋅ cos  + t  4  4  dt  2  π 2 sin  + t  4  1 ⇒ I = ∫ e2 t a

∫ (sin

8 f ( x) =

+ sin3 x ⋅ cos 2 x + cos 5 x )2

[taking (−1) common from each column] ⇒ f ( x) = − f ( x) ⇒ f ( x) = 0 ∴ ∫ f ( x ) dx = ∫ 0 dx = C

3 Let I = ∫ e 2 ax

cos x − x (− sin x ) dx cos 2 x −x = + tan x + C ( x sin x + cos x )cos x − x + x sin2 x + sin x ⋅ cos x = +C ( x sin x + cos x ) ⋅ cos x sin x − x cos x = +C x sin x + cos x

1 − cos ( x2 + 1) dx 1 + cos ( x2 + 1)

=

k =

3

3 2

dx

∫ (sin x + 2) (sin x − 1) 1 dx 1 dx − 3 ∫ (sin x − 1) 3 ∫ (sin x + 2) 1 dx = ∫ x 3    2 tan  2 − 1    1 + tan2 x   2 1 dx − ∫ x 3    2 tan  2 + 2   1 + tan2 x    2 x Put tan = t 2 =

⇒

1 2 x sec dx = dt 2 2

∴

1 2dt 1 2dt − 3 ∫ 2t − 1 − t 2 3 ∫ 2t + 2t 2 + 2

162 dt dt 2 1 − 2 2 3 ∫ (t − 1)2 3 ∫   3 1  t +  +   2  2  t + 1    2 1 1 2 2 = − tan −1 +C 3 (t − 1) 3 3 3 2 2 1 2 = − tan −1 x 3   3 3  tan − 1   2  tan x + 1     2 2 +C 3 2 2 = x 3  tan − 1   2 =−

−

11

2 tan −1 3 3

x 1    2 tan +    2 2   +C 3    

x2 ∫ (2 + 3 x2 )5/2 dx

dx = −

2t dt x ( t − 3)2

= ∫e

 1 dt 1 1 x = +C=   2 ∫ t 4 6t 3 6  2 + 3 x2  2

x+

1

x−

1  x+x dx e x

1

dx +

x

= ∫e

x+

1

= ∫e

x+

1



dx + xe

x

x+

1 x

x+

1

− ∫e

x

x+

=

x

dx

1 x

+C

13 Let I = ∫ (sin(101 x)⋅sin 99 x)dx = ∫ sin(100 x + x)sin

99

=

xdx

⋅ sin x)sin = ∫ sin100x⋅ (cos x ⋅ sin 99 x)dx

99

xdx

II

+ ∫ cos(100x)⋅ sin

100

xdx

sin(100 x )⋅ sin100 x 100 − ∫ sin100 x ⋅ cos(100x)dx

+ ∫ cos(100x)⋅ sin100 xdx

3 /2

+C

sin(100 x ) ⋅ sin100 x = +C 100

1 − (2018) 2 x

⋅ (2018) sin

−1

(2018 )x

(2018) t +C ln2 (2018)

= (log 2018 e ) 2 ⋅ (2018) sin

= ∫ (sin(100x)⋅ cos x + cos(100x)

I

(2018)x

∫

⋅(2018)x l n(2018)dx = dt 1 (2018) t dt ∴I = ln (2018) ∫ 1 (2018)t = ⋅ +C ln (2018) ln 2018

= ∫ e x +1 / xdx + xe x +1 / x − ∫ e x + 1 / xdx x+

x −1 (2018) 2 x ⋅(2018) sin (2018) dx 1 − (2018) 2 x

Put sin −1 (2018)x = t 1 ⇒ 1 − (2018x ) 2

1

1 x+   1  x+x  dx = e x  Q ∫  1 − x2  e  

= xe

∫

1

1  x+x dx e x2  1 x+ d − ∫ ( x ) e x dx dx

∫ x  1 −

dx + x e

x

14 Let I =

1

=

2

x2  − 2t  dt ∴∫ ⋅  dt = − 2∫ 4 4t ( tx )5  x( t 2 − 3)2  =−



∫  1 +

 sin100 x = sin(100 x )⋅ − ∫ cos(100 x )  100  sin100 x  ⋅100 ⋅ dx + ∫ cos(100x)⋅ sin100 xdx  100 

On substituting 2 + 3x2 = t 2 x2 2 x2 = 2 ⇒ ( t − 3) ∴

12

−1

(2018 ) x

+C

15 Let I = ∫ (∫ e x (log x + 2 − 12 )dx)dx

x x 1 1 1 = ∫ (∫ e x  log x + + − 2  dx )dx  x x x 

 log x + 1  dx + e x  1 − 1   dx   ∫  x x2    x 1 = ∫  e x logx + e x + C1  dx   x =



∫  ∫ e

x

[Q∫ e x ( f ( x) + f '( x))dx = e x ⋅ f ( x) + C] 1 = ∫ e x  logx +  dx + ∫ C1 dx  x = e x log x + C1 x + C2

DAY SIXTEEN

Definite Integrals Learning & Revision for the Day u

Concept of Definite Integrals

u

Walli’s Formula

u

Leibnitz Theorem

u

Inequalities in Definite Integrals

u

Definite Integration as the Limit of a sum

Concept of Definite Integrals Let φ ( x) be an anti-derivative of a function f ( x) defined on [a, b ] i.e. definite integral of f ( x) over [a, b ] is denoted by

∫

b a

∫

b a

d [φ ( x)] = f ( x). Then, dx

f ( x) dx and is defined as [φ (b ) − φ (a)] i.e.

f ( x) dx = φ (b ) − φ (a). The numbers a and b are called the limits of integration, where a is

called lower limit and b is upper limit. NOTE

• Every definite integral has a unique value.

PRED MIRROR

• The above definition is nothing but the statement of second fundamental theorem of integral calculus.

Your Personal Preparation Indicator

Geometrical Interpretation of Definite Integral b

In general,

∫ f ( x)dx represents an algebraic sum of areas of the region bounded by the curve a

y = f ( x), the X -axis, and the ordinates x = a and x = b as show in the following figure. Y′

X′

Y

y=f(x)

a

b

X

u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

164 (xi) Some important integrals, which can be obtained with the help of above properties.

Evaluation of Definite Integrals by Substitution When the variable in a definite integral is changed due to substitution, then the limits of the integral will accordingly be changed. b

b

Step II Find the limits of integration in new system of variable. Here, the lower limit is g(a), the upper limit is g(b ) and the integral is now ∫

g (b ) g ( a)

a

Properties of Definite Integrals ∫ f ( x)dx = 0 β

∫

(iii)

∫

β

∫

β

(iv)

α

β

f ( x) dx = ∫ f (t ) dt

α

β

f ( x) dx = ∫

c1 α

(v) (a) ∫ f ( x) dx = α

(b)

f ( x) dx + ∫

c2

f ( x) dx + ... +

c1

∫

β cn

f ( x) dx

α −α

∫

α 0

∫

β α

f ( x) dx =

f (α + β − x) dx

∫

α 0

f (α − x) dx

f ( x) dx

2 α f ( x) dx, if f (− x) = f ( x)  ∫0  i.e. f ( x) is an even function = if f (− x) = – f ( x) 0,  i.e. f ( x) is an odd function  (vii)

2α

α

α

0

0

0

∫ f ( x)dx = ∫ f ( x)dx + ∫ f (2α − x)dx

(viii)

∫

(ix)

∫

a

x2

∫

xn

f ( x)dx + ... +

x1

∫

b

f ( x)dx +

xn−1

∫ f ( x)dx.

xn

Leibnitz Theorem

Walli’s Formula

α

f ( x) dx = − ∫ f ( x) dx

α

β

∫

x1

f ( x)dx = ∫ f ( x)dx +

α

where, α < c1 < c2 < ... < c n < β

(vi)

π log 2. 8

d  ψ (x ) d d   f (t ) dt  =  ψ ( x) f (ψ ( x))−  {φ ( x)} { f (φ( x)}  dt dx ∫ φ( x )   dx

α

(ii)

log(1 + tan x)dx =

If function φ( x) and ψ ( x) are defined on [α , β] and differentiable on [ α , β ] and f (t ) is continuous on [ψ (α ), φ (β)], then

α

(i)

∫

f (t ) dt .

StepIII Evaluate the integral, so obtained by usual method.

0

 1

(xii) If a function f ( x) is discontinuous at points x1 , x2 ,..., x n in (a, b ), then we can define sub-intervals (a, x1 ),( x1 , x2 ),...,( x n −1 , x n ), ( x n , b ) such that f ( x) is continuous in each of these sub-intervals and

Substitute g( x) = t so that g ′ ( x) dx = dt

Step I

0 π /4

π

0

f [ g ( x )] ⋅ g ′ ( x ) dx, we use the following steps

a

π /2

∫

(b)

For example, to evaluate definite integral of the form

∫

π /2

∫ log sin xdx = ∫ log cos xdx = 2 log  2 .3

(a)

2α 0

β α

2 α f ( x) dx, if f (2α − x) = f ( x) f ( x) dx =  ∫ 0 if f (2α − x) = − f ( x)  0, 1

f ( x) dx = (β − α )∫ f [(β − α) x + α]dx 0

(x) If f ( x) is a periodic function with period T, then

(a) (b) (c)

∫ ∫ ∫

α + nT

α βT αT

T

f ( x) dx = n ∫ f ( x) dx, n ∈ I 0

α + nT

∫

π /2 0

sin n x dx =

∫

π /2 0

cos n x dx

(n − 1) (n − 3)(n − 5) K 5 ⋅ 3 ⋅ 1 π × , n = 2 m (even)  n(n − 2)(n − 4)K 6 ⋅ 4 ⋅ 2 2  = (n − 1) (n − 3) (n − 5) K 6 ⋅ 4 ⋅ 2 , n = 2 m + 1 (odd)  n(n − 2) (n − 4) K 5 ⋅ 3 ⋅ 1  where, n is positive integer. π /2

∫ sin 0

m

x ⋅ cos n x dx

(m − 1)(m − 3)...(2 or 1).(n − 1)(n − 3)...(2 or 1) π ,  (m + n)(m + n − 2)...(2 or 1) 2  when both m and n are even positive integers  =  (m − 1)(m − 3)...(2 or 1) ⋅ (n − 1)(n − 3)...(2 or 1) ,  (m + n)(m + n − 2)...(2 or 1)   when either m or n or both are odd  positive integers

Inequalities in Definite Integrals β

β

(i) If f ( x) ≥ g( x) on [α , β], then ∫ f ( x) dx ≥ ∫ g( x) dx α

α

T

f ( x) dx = ( β – α) ∫ f ( x) dx, α, β ∈ I

β + nT

This is a special type of integral formula whose limits from 0 to π /2 and integral is either integral power of cos x or sin x or cos x sin x.

0

f ( x) dx =

∫

β α

f ( x) dx, n ∈ I

β

(ii) If f ( x) ≥ 0 in the interval [α , β], then ∫ f ( x) dx ≥ 0 α

(iii) If f ( x), g( x) and h( x) are continuous on [a, b ] such that b

b

b

a

a

a

g( x) ≤ f ( x) ≤ h( x), then ∫ g( x) dx ≤ ∫ f ( x)dx ≤ ∫ h( x)dx

165

DAY (iv) If f is continuous on [α , β] and l ≤ f ( x) ≤ M , ∀ β

x ∈[α , β], then l (β − α ) ≤ ∫ f ( x) dx ≤ M (β − α ) α

(v) If f is continuous on [α , β],

∫

then

β α

The converse is also true, i.e. if we have an infinite series of the above form, it can be expressed as definite integral.

α

Some Particular Cases n

f ( x) dx ≤ k (β − α )

(i) lim

Definite Integration as the Limit of a Sum

(ii) lim

∫

then

α

as n → ∞

β

f ( x) dx ≤ ∫ | f ( x)| dx

(vi) If f is continuous on [α , β] and| f ( x)| ≤ k , ∀ x ∈ [α , β], β

b–a →0 n

where, h =

n→ ∞

∑

r =1 pn

n→ ∞

∑

r =1

r =0 h r β = lim = p n→ ∞ h

interval [a, b ], then ∫ f ( x) dx = lim h ∑ f (a + rh) n→ ∞

(Qr = 1)

n→ ∞

n –1

a

β 1  r f  = f ( x) dx n  n ∫ α

where, α = lim

Let f ( x) be a continuous function defined on the closed b

n −1

1 1  r 1  r f   or lim ∑ f   = ∫ f ( x) dx n→ ∞ 0   n  n n n r =1

and

r=0

(Qr = pn)

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1

∫

3π /4 π /4

dx is equal to 1 + cos x

(a) − 2

(b) 2

7 JEE Mains 2017

j

2 If f ( x ) is continuous function, then (a) (b) (c) (d)

3

∫

π /4 0

(a)

4

∫

1 0

∫ ∫ ∫ ∫

2 −2 5 −3 5

2 0

2 f (x) dx =

f (x) dx =

−3 5

∫

f (x) dx =

f (x) dx =

−3

[ tan x + 2π

∫ ∫

4 −4 6 −2

∫

6

f (x − 1) dx

9

f (x − 1) dx

π 2

(c)

5 If I1( n ) = ∫

(b) π /4 π 2 0

π 2

(d) 2 π

(c) π / 2

(d) π π 2

sin( 2n − 1) sin nx dx and I 2 ( n ) = ∫ dx , 0 sin x sin2 x 2

n ∈ N, then

6

(a) I 2 ( n + 1) − I 2 ( n ) = I1 ( n )

(b) I 2 ( n + 1) − I 2 ( n ) = I1 ( n + 1)

(c) I 2 ( n + 1) + I1( n ) = I 2 ( n )

(d) I 2 ( n + 1) + I1 ( n + 1) = I 2 ( n )

∫

1 −1

{x 2 + x − 3} dx, where {x } denotes the fractional part

of x, is equal to 1 (1 + 3 5 ) 3 1 (c) (3 5 − 1) 3 (a)

1 (1 + 3 5 ) 6 1 (d) (3 5 − 1) 6 (b)

| 2 t | dt = f ( x ), then for any x ≥ 0 , f ( x ) is equal to

∫

π /2 −π / 2

(b) 4 − x 2 1 (d) (4 − x 2 ) 4

cos x dx is equal to 1 + ex

(a) 1 (c) −1

 1+ x sin  2 tan−1  dx is equal to 1− x  

(a) π / 6

x −2

(b) 2 + 2 (d) − 2 − 3 + 5

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

f (x − 1) dx

cot x ] dx is equal to (b)

[ x 2 ] dx is equal to

8 If ∫

[f (x) − f (− x)] dx 10

2 0

(a) 2 − 2 (c) 2 − 1

(d) − 1

(c) 4

∫

(b) 0 (d) None of these

10 Let a, b and c be non-zero real numbers such that

∫

3 0

3

( 3 ax 2 + 2bx + c ) dx = ∫ ( 3 ax 2 + 2 bx + c ) dx , then 1

(a) a + b + c = 3 (c) a + b + c = 0

11 The value of ∫

a 1

(b) a + b + c = 1 (d) a + b + c = 2

[ x ] f ′ ( x ) dx , a > 1, where [ x ] denotes the

greatest integer not exceeding x, is (a) (b) (c) (d)

[a]f (a ) − {f (1) + f (2) + . . . + f ([a])} [a]f ([a]) − {f (1) + f (2) + . . . + f (a )} af ([a]) − {f (1) + f (2) + . . . + f (a )} af (a ) − {f (1) + f (2) + . . . + f ([a])}

12 The correct evaluation of ∫ (a) 2 + 2 (c) − 2 + 2

π /2 0

π     sin  x −  dx is  4  

(b) 2 − (d) 0

2

166

13

∫

  2x  sin   dx, where [ ⋅ ] denotes the greatest   π 

3π /2 0

integer function is equal to π (sin1 + cos1) 2 π (c) (sin1 − cos1) 2

n→ ∞

(b)

of ∫

f ( x ) = f (a − x ) and g ( x ) + g (a − x ) = a, then

∫

(a)

x [ x ] dx is

5 4

(b) 0

15 The value of ∫ (a)

16

∫

π /2

(d)

π log 2 2

(b)

17 The value of ∫

e e

(a) 3/2

2

−1

n

k =1

1 0

NCERT Exemplar

f (x) dx (b)

1 0

∫

(c) 3

satisfies ∫

n

2 0

f (x) dx (c)

∫

n 0

(d) n

f (x) dx

f ( x ) dx =

2

∫

1 0

f (x) dx

28

n , ∀ n ∈ I, then ∫ f (| x | ) dx is −3 2

1

7π / 3 7π / 4

(b) 4

2 1

tan2 x dx is

2

2 x dx and

π 2

(b)

π

−π

JEE Mains 2015

(d) 6

π π4 (c)   − 1(d)  4 32

[f ( x ) + f ( − x )] [g ( x ) − g ( − x )] dx is (b) π

(a) 0

j

JEE Mains 2013

(b)

32

∫

π −π

(a)

π 2

(d) None of these 3π 0

π

f (cos 2 x ) dx and Q = ∫ f ( cos 2 x ) dx , then 0

(a) P − Q = 0 (c) P − 3 Q = 0

NCERT Exemplar

1 log( 2 + 1) 2 (d) None of these

(d) None of these

n 1   is + ... + (n + 3) 3 ( 2n + 3) 2n 3 

π 3 π (c) 4

31 If P = ∫ j

(c) 1

 n n 30 The value of lim  + n → ∞ (n + 1) 2n + 1 (n + 2) 2 ( 2n + 2) 

(a)

3 (b) log 2 2 (d) log 2

(b)

(c) 1

[( x + π )3 + cos 2 ( x + 3π )] dx is equal to

+

sin 2x dx is equal to sin x +cos x

1 log( 2 + 1) 2 (c) − log( 2 + 1) (a)

∫

(b) I 3 = I 4 (d) I 2 > I1

(c) 2 log 2 0

2

3

(a) log 2 2

∫

log x 2 dx is equal to 2 log x + log( 36 − 12 x + x 2 ) 4

2 x dx , then

21 The value of ∫

π /2

−π /2 −3π /2

π 2

function, then the value of the integral

3

0

(a) I 3 > I 4 (c) I1 > I 2

22

∫

(d) None of these 2

(d) −

29 If f : R → R and g : R → R are one to one, real valued

35 (b) 2

0

2

(c) − 1

(b) 1

 π4  π (a)   + 2  32 

5

20 If I1 = ∫ 2 x dx ,I 2 = ∫ 2 x dx ,I 3 = ∫ 1

π 2

(a) 2

19 (a) 2 17 (c) 2

I4 = ∫

(c) 4 π

JEE Mains 2018, 13 π (d) 4

j

equal to

1

j

f (k − 1 + x ) dx is

19 f ( x ) is a continuous function for all real values of x and n +1

sin x dx is 1 + 2x

π (b) 2

27 The integral ∫

(d) 5

−π / 2

[cot x ] dx , where [] denotes the greatest integer

(a)

loge x dx is x

∫

0

function, is equal to

(d) 2 ( 2 − 1)

(c) 2

π 0

NCERT Exemplar

2

π /2

(a) π

∫

j

a a (b) ∫ f (x)d x 2 0 a (d) a ∫ f (x)d x

0

26 j

(b) 5/2

18 The value of ∑

∫

3 4

(d) π log2

(c) log2

(b) 2 ( 2 + 1)

(a) 2 2

f ( x ) ⋅ g ( x ) d x is equal to

25 The value of ∫

1 − sin 2x is equal to

0

(a)

3 2

8 log (1 + x ) dx is 1+ x2

1 0

π log 2 8

(c)

a 0

a (a) 2 a (c) ∫ f (x)d x

2

0

(b) 4 − 2 ln 2 (d) 2 − 2 ln 2

24 If f and g are continuous functions in [0, 1] satisfying

14 If [. ] denotes the greatest integer function, then the value 1.5

1 1 2 3n  + + ...+   is n n + 1 n + 2 4n 

(a) 5 − 2 ln 2 (c) 3 − 2 ln 2

π (sin1 − sin 2) 2 π (d) ( sin1 + sin 2) 2

(a)

23 The value of lim

(b) P − 2 Q = 0 (d) P − 5 Q = 0

2x (1 + sin x ) dx is equal to 1 + cos 2 x π2 4

(b) π 2

(c) 0

(d)

π 2

167

DAY 33 If f ( x ) is differentiable and ∫

t2 0

x f ( x ) dx =

 4 f   is equal to  25 (a)

2 5

(b) −

34 If f ( x ) =

1 x2

∫

x 4

5 2

(c) 1

(b)

32 9

∫

x2 0

dx

2

1+ x2

1

and I 2 = ∫

2 1

(b) I 2 > I1

(b) y

(d) I1 > 2 I 2

1+ y

j

JEE Mains 2013

2

38 Let f : R → R be a differentiable function having f ( 2) = 6,  1 f ′ ( 2) =   . Then, lim x→ 2  48 (a) 18

∫

4t3 dt is equal to x −2

f (x ) 6

(b) 12

(c) 36

4

4

3

3

3

39 lim

n→ ∞

(a)

1 30

40 If I = ∫

(b) 0

π /2 0

dx 1 + sin3 x

(c)

1 4

(d)

π 2 2

(b) I >

(c) I
2 3 2 (d) I < and J > 2 3 (b) I >

Statement II f ( x ) is continuous in [0, 2].

(b) 2 / 9 (d) 2 / 3

(a) I1 > I 2

(c)

32 3

is

(a) 0 (c) 1 / 3

2 and J < 2 3 2 (c) I < and J < 2 3 (a) I >

42 Statement I

sin t dt x3

x→ 0

37 If x = ∫

5 2

(d) None of these

35 The value of lim

36 If I1 = ∫

(d)

[ 4 t 2 − 2f ′ ( t )] dt , then f ′ ( 4) is equal to

(a) 32 (c)

2 5 t , then 5

(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 1

1

0

0

45 Statement I ∫ e − x cos 2 x dx < ∫ e − x cos 2 x dx b

b

a

a

2

Statement II ∫ f ( x ) dx < ∫ g ( x ) dx , ∀ f ( x ) < g ( x ) (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

168

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE  1 x

1 If f ( x ) is a function satisfying f   + x 2f ( x ) = 0 for all cosec θ

non-zero x, then ∫ (a) sinθ + cosec θ (c) cosec2 θ

(b) sin2 θ (d) None of these

4 3 3 e sin x d ; x > 0. If ∫ e sin x dx = f (k ) − f (1), then f (x ) = 1 x dx x the possible value of k, is

2 If

(a) 15

(b) 16

(c) 63

(d) 64

2

3 If g (1) = g ( 2), then ∫ [ f {g ( x )}]−1 f ′ {g ( x )} g ′ ( x ) dx is 1

equal to (a) 1

(b) 2

(c) 0

(d) None of these

π 4 For 0 ≤ x ≤ , the value of 2 sin2 x cos 2 x −1 sin ( t ) dt + ∫ cos −1( t ) dt is ∫ 0

(a)

0

π 4

(b) 0

(c) 1

 1 x

x

5 If F ( x ) = f ( x ) + f   , where f ( x ) = ∫ 1

6

(b) 0

∫ The expression ∫

n

(c) 1

JEE Mains 2013 π (d) − 4

log t dt . Then, F (e ) 1+ t (d) 2

, where [ x ] and {x } are integral {x } dx

0

and fractional part of x and n ∈ N, is equal to (a)

1 n −1

(b)

1 n

(c) n

(d) n − 1

f (a ) ex , I1 = ∫ x g [ x (1 − x )] dx and x f (− a ) 1+ e f (a ) I I2 = ∫ g [ x (1 − x )] dx , then the value of 2 is f (− a ) I1

7 If f ( x ) =

(a) 2

(b) – 3 x

8 The value of lim

x→ ∞

(a) 1 / 3

9 The integral ∫ 0

(c) 4 3 − 4

{∫ e dt } 0 x

∫

0

2

2

e 2 t dt

(b) 2 / 3 π

(a) π − 4

(c) –1 t

x f (sin x ) dx is equal to π 0

∫

(c) π

π π /2 f (sin x ) dx 2 ∫0 π (d) π ∫ f (cos x ) dx

f (sin x ) dx

π /2 0

11 If f ( x ) = ∫

(b)

f (cos x ) dx

x2 + 1 x2

0

e − t dt , then f ( x ) increases in 2

(a) (2, 2) (c) (0, ∞)

(b) no value of x (d) (− ∞, 0)

 (n + 1)(n + 2) K 3n     n2n

1/ n

12 lim  n→ ∞

18 (a) 4 e 9 (c) 2 e

is equal to j

(d) 1

2

is (c) 1

(d) None of these

x x 1 + 4 sin2 − 4 sin dx is equal to 2 2 j JEE Mains 2014 2π −4−4 3 3 π (d) 4 3 − 4 − 3 (b)

JEE Mains 2016

27 (b) 2 e (d) 3 log 3 − 2

f (x ) = ∫

x 5π /4

( 3 sin u + 4 cos u ) du on the interval

 5π 4π  ,  4 3 

is 3 2 3 1 (c) + 2 2 3 +

(a)

(b) − 2 3 +

3 1 + 2 2

(d) None of these

x

[ x ] dx

0 n

π 0

13 The least value of the function j

is equal to (a) 1/2

∫

(a) π ∫

f ( x ) dx is equal to

sin θ

10

14 If g ( x ) = ∫ f (t ) dt , where f is such that, 0

1 ≤ f ( t ) ≤ 1, for 2

1 t ∈[ 0, 1] and 0 ≤ f ( t ) ≤ , for t ∈[1, 2]. Then, g( 2) satisfies 2 the inequality 3 1 ≤ g (z) < 2 2 3 5 (c) < g(2) ≤ 2 2 (a) −

(b)

1 3 ≤ g(2) ≤ 2 2

(d) 2 < g(2) < 4

15 Let n ≥ 1,n ∈ z . The real number a ∈( 0, 1) that minimizes 1

the integral ∫ | x n − a n | dx is 0

1 (a) 2

(b) 2

(c) 1

(d)

1 3

16 The value of 1/ n

  π  2π   nπ    3π  lim tan   tan   tan  K tan    n→ ∞  2n   2n   2n    2n   (a) 1 (c) 3

(b) 2 (d) Not defined

is

169

DAY 17 If f ( x ) =

x −1 2 , f (x ) = f (x ), ..., f k x +1

+1

(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

(x ) = f {f k (x )},

k = 1, 2, 3,K and g(x ) = f 1998 (x ), then ∫

1 1 /e

g ( x ) dx is equal

to (a) 0

(c) −1

(b) 1

(d) e

20 Consider sin6 x and cos 6 x is a periodic function with π.

18 If f (x ) is a function satisfying f ′ (x ) = f (x ) with f (0) = 1 and g(x ) is a function that satisfies

Statement I

1

f (x ) + g(x ) = x . Then, the value of ∫ f (x )g(x ) dx, is 2

π  , 8

0

2

e 5 − 2 2 e2 3 (c) e − − 2 2 (a) e −

2

e 3 − 2 2 e2 5 (d) e + + 2 2 (b) e +

1 dx dx = 1 + x n ∫ 0 (1 − x n )1/ n

b

b

a

a

0

( sin6 x + cos 6 x ) dx lie in the interval

π . 2

(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

∞ 0

π /2

Statement II sin6 x + cos 6 x is periodic with period π / 2.

19 If n > 1, then Statement I ∫

∫

Statement II ∫ f ( x ) dx = ∫ f (a + b − x ) dx

ANSWERS SESSION 1

SESSION 2

1 (b)

2 (d)

3 (c)

4 (b)

5 (b)

6 (b)

7 (d)

8 (a)

9 (a)

10 (c)

11 (a)

12 (b)

13 (d)

14 (d)

15 (d)

16 (d)

17 (b)

18 (c)

19 (b)

20 (c)

21 (b)

22 (a)

23 (c)

24 (b)

25 (d)

26 (d)

27 (c)

28 (b)

29 (a)

30 (a)

31 (c)

32 (b)

33 (a)

34 (c)

35 (d)

36 (b)

37 (a)

38 (a)

39 (d)

40 (b)

41 (c)

42 (d)

43 (d)

44 (d)

45 (a)

1 (d) 11 (d)

2 (d) 12 (b)

3 (c) 13 (b)

4 (a) 14 (b)

5 (a) 15 (a)

6 (d) 16 (a)

7 (a) 17 (c)

8 (d) 18 (c)

9 (d) 19 (b)

10 (c) 20 (b)

170

Hints and Explanations SESSION 1

=

∫

3π / 4

(cosec2 x − cosec x cot x )dx

π/4

π /2

=

∫

0

=

∫

0

=

∫

π /2

π /2 0

[sin nx − sin (n − 1) x] dx sin2 x 2

2

sin( 2n − 1) x ⋅ sin x dx sin2 x

∴ I =

∫

0

sin (2n − 1) x dx = I1( n) sin x

=

∫

0

=

∫

0

= [− cot x + cosec x]3ππ/ /44

⇒ I2 ( n + 1 ) − I2 ( n ) = I1 ( n + 1 )

= [(1 +

∫

2 ) − (− 1 +

2 )] = 2

6

1 −1

5

Therefore, ∫

π /4

∫

0

=

∫

0

−3

[ tan x +

π /4

= 2

∫ =∫

f ( x ) dx =

6

6

sin x cos x

∫

0

dx dx

7

∫

2

[ x2 ] dx =

0

2

∫

+

1

∫

1

1

∫

=

[ x + x] dx

3

∫

2

= [ x]1 2 + [2 x]

3 2

∫

⇒

2 dx +

0

(3 ax2 + 2bx + c ) dx

1

(3 ax2 + 2bx + c ) dx

∫

2 3

8

∫

1

3 1

∴

a

∫

+

2 3

3 dx

3 2

π 4

∫

2 1

f ′ ( x ) dx

2f ′ ( x ) dx + . . . +

∫

a

[a] f ′ ( x ) dx

[ a]

= [ f ( x )]21 + 2 [ f ( x )]32 + . . . + [a][ f ( x )][aa ] = f (2) − f (1) + 2 f (3) − 2 f (2) + . . . + [a] f (a) − [a] f ([a]) = [a] f (a) − { f (1) + f (2) + . . . + f ([ a])}

[ x2 ] dx

  sin  x − π    dx  4   π /4 π =−∫ sin  x −  dx 0  4

∫

12 Let I =

π /2

0

+

∫

π /2 π/4

π sin  x −  dx  4

π/4

∫

0

|2t|dt +

−2

∫

x 0

π /2

9 Let I =

∫

=

∫

−π / 2 0 − π /2

2t dt x

13

π /2 2x 2x sin   dx = ∫ sin   dx 0  π   π  π 3π / 2 2x 2x + ∫ sin   dx + ∫ sin   dx π /2 π  π   π 

∫

3 π /2

0

cos x dx 1 + ex

= 0 + sin 1 ∫

cos x dx 1 + ex

=

+

∫

0

cos x dx …(i) 1 + ex

π

π /2

dx + sin 2 ∫

3π / 2

π

dx

π (sin 1 + sin 2) 2 1. 5

π /2

π /2

π  π    = cos  x −  − cos  x −      4   0 4   π / 4 1  1  = 1− − − 1 = 2 − 2  2  2

|2t | dt = f ( x )

t 2  t 2  +2 2 2   −2  0  x2  = − 2 [0 − 2] + 2 − 0 = 4 + x2 2   

1 − x dx

(3 ax2 + 2bx + c ) dx

(3 ax2 + 2bx + c ) dx

11 Since, ∫ 1 [ x] f ′( x ) dx =

= −2

2

=

−2

3 1

(3 ax2 + 2bx + c ) dx = 0

0

+ [3 x]2 3

0

 1 + x dx ∴ ∫ sin 2 tan −1 0 1 − x  

1 0

x

=

1

1

∫

∫

1

[ x2 ] dx

∫

=1

 3 ax3  2bx2  3 + 2 + cx  = 0  0 a+ b + c = 0

⇒

1 dx

1

3

∫ 1

∫ =

[ x2 + x] dx

π /2 0

cos x dx = [sin x]

+

5 − 1  2 

2

π /2

(1 + e x ) cos x dx (1 + e x )

(3 ax2 + 2bx + c ) dx

= 2−1+ 2 3−2 2+ 6−3 3

 π θ  = sin 2 tan –1  tan  −     2 2      π θ  = sin 2  −    2 2   = sin ( π − θ) = sin θ = 1 − cos 2 θ = 1 − x2

1 1 =  x 1 − x2  + [sin −1 x]  2  0 2

0

3

π /2

π / 2 cos x e x cos x dx + ∫ dx 0 1 + ex 1 + ex

2

2

( 5 − 1 )/ 2

∫ 2

∫

0 dx +

0

3 0

π /2

= 5− 3 − 2

θ   = sin 2 tan −1  cot    2  

0

∫

[ x2 ] dx +

+

 1 + x 4 ∫ 0 sin 2 tan −1 dx 1 − x   Put x = cos θ, then  1 + cos θ  sin 2 tan −1 1 − cos θ  

∫

∫ 1

5 −1

[ x2 ] dx

0

∫

⇒

[ x2 + x] dx

+

π 2

10

=

5−1 1 = (1 + 3 5) 2 6

2 + 3

=

1

=

−1

− −

Put sin x − cos x = t ⇒ (cos x + sin x ) dx = dt 0 dt ∴ I= 2∫ ⇒ I = 2 [sin −1 t ] −01 −1 1 − t2

1

0

∫

 2 = + 1 − 0 − 1 1 − 3 

1 − (sin x − cos x ) 2

= 2 [0 − (− π / 2)] =

2

f ( x − 1) dx

sin x + cos x

π /4

−1

1

x x  = +  − 2  −1  3 3

cot x ] dx

sin x + cos x

{ x + x} dx 2

( x2 + x − [ x2 + x]) dx

−1

f (t − 1) dt

−2 −2

∫

=

Let x = t −1 ∴ dx = dt When x tends to − 3 and 5, then t tends to − 2 , 6.

1

1

∫

{ x + x − 3} dx = 2

2 Since, f is continuous function.

3 Let I =

On putting x = − x in 1st integral, we get x 0 π / 2 e cos x cos x ∫ − π /2 1 + e x dx = ∫ 0 1 + e x dx

5 I2 ( n ) − I2 ( n − 1 )

dx 1 Let I = ∫π / 4 1 + cos x 3 π / 4 1 − cos x =∫ dx π/4 1 − cos 2 x 3 π / 4 1 − cos x dx =∫ π/4 sin2 x 3π / 4

14 Here, ∫ 0 I =

∫

1 0

x [ x2 ] dx

x ⋅ 0 dx +

∫

2 1

x ⋅ 1 dx +

∫

1. 5 2

x ⋅ 2dx

171

DAY 0

2

1 {2 − 1} + {(1.5) 2 − 2} 2 1 1 1 3 = + 2. 25 − 2 = + = 2 2 4 4 1 8 log (1 + x ) 15 Let I = ∫ 0 dx (1 + x2 ) =

When x = 0 ⇒ tan θ = 0 ∴ θ=0 When x = 1 = tanθ ⇒ θ =

∫

∴I =

0

= 8∫

∫

⇒

∫

π/4

π/4

I = 8∫

0

⇒

π 4

19

log (1 + tan θ) d θ

…(i)

π /2

0

+∫

∫

π /4

0

(cos x − sin x ) dx

17

∫

e

log e x dx = x

−1

1

=

∫

e

=

∫

−1

−1

0

π /2 π /4

(sin x − cos x ) dx

log e x dx ∫ e −1 x e 2 log x e + ∫ dx 1 x 2 e log x − log e x e dx + ∫ dx 1 x x

− z dz +

3

= 2∫

1

∫

2

2 0

z dz

[put log e x = z ⇒ (1 / x ) dx = dz)

f ( x ) dx + ...+ ∫

1

∫

f ( x ) dx +

0

3

f (|x|) dx +

−3

∫ ∫

5

n n −1

f ( x ) dx

∫

5 3

f (|x|)dx

f ( x ) dx + ∫ f ( x ) dx  2 

1

∫ 4 f ( x ) dx 

∫

I3 =

x3

2

2 x dx

3

2 x dx,

2 1

3

1

2

x2

dx

tan2 x dx

= − [log sec x]

2 +1 2−1

2π 7π / 3 2π

2π 7π / 4

|tan x |dx tan x dx

+ [log sec x] 72 ππ /3

7π  = −  log sec 2 π − log sec  4  7π +  log sec − log sec 2 π    3 π  = −  log 1 − log sec 4   π +  log sec − log 1  3  1 = log 2 + log 2 = log 2 + log 2 2 3 = log 2 2

 ( 2 + 1)2  1 log   1 2  

=

2 = log ( 2 + 1 )] 2 1 ∴ l= log ( 2 + 1) 2

= lim

n→ ∞

∫

0

=

∫

0

3n

∑ r =1

+

3n  2 + ...+  n+ 2 n + 3n 1  r    n n + r

x dx 1+ x

3

=

 1  1 −  dx + x 1 

3

= [ x − ln (1 + x )] 30 = 3 − ln 4 = 3 − 2 ln 2

∫ =∫ =∫

24 Ql = 7π / 3

∫ tan x dx + ∫

|tan x |dx +

7π / 4

1 log 2

dx 2

I2 < I1 and I 4 > I3

21 I =

=

n n + 1

2 x dx x2

1 [log ( 2 + 1) − log ( 2 − 1 )] 2

1  1 23 nlim  →∞

2

2 x dx

3

and

=

1 0

2 x > 2 x for x > 1

∫

∴

2 1

∫

and I 4 = Since, 2

1 0

∫

=

5

 1 2  9 16  35 = 2  0 + +  +  + = 2 2  2 2 2 

20 Given that, I1 =

π π  − log sec  −  + tan  −     4  4   

3

2

∫

π /2

1  π π   log sec  x −  + tan  x −        2 4 4   0  1  π π = log  sec + tan   2  4 4

f ( x ) dx

3 2

4 +  ∫ f ( x ) dx +  3

I2 =

π sin2  − x  2  =∫ dx 0 π π    sin  − x  + cos  − x  2  2  π /2 cos 2 x l =∫ dx 0 sin x + cos x 1 π /2 dx Thus, 2l = ∫ 2 0 cos  x − π     4 1 π /2 π  = ∫ sec  x − 4  dx 2 0 =

f (| x |) dx =

−3

and

= [sin x + cos x] π0 / 4 + [− cos x − sin x] ππ // 24 π π   =  sin + cos  − (sin 0 + cos 0)    4 4 π π π π   +  − cos − sin  −  − cos − sin    2 2  4 4    1 1 1     1 =  + − 1 + −1 −  − −   2     2 2 2     2  = − 1 + [−1 + 2]  2  = ( 2 − 1) + (−1 + 2 ) = 2( 2 − 1) e

∫

f ( x ) dx

0

(sin x − cos x ) dx

+∫

2

n

2

π [Qcos x − sin x > 0 when x ∈  0,  and  4 cos x − sin x < 0 when x ∈ ( π / 4, π / 2)] =

f ( x ) dx

k −1

= 2  ∫ f ( x )dx +  0

(cos x − sin x ) 2 dx

π /4

5

k

1

1 − sin2 xdx

π /2

∫

f ( x ) dx

k −1

sin2 x dx sin x + cos x

π /2

22 We have, l = ∫0 π /2

f ( t ) dt , where

f ( x ) dx +

0

∫

=

log (1 + tan θ) d θ

1

∫

=

= π log 2 0 π /4

∑ ∫ k =1

Q π / 4 log (1 + tan θ)dθ = π log 2 ∫ 8   0

∫ =∫

k k −1

k

∫

I = n

∴

π = 8 ⋅  ⋅ log 2 2 

16 Let l =

I =

f (k − 1 + x ) dx

0

t = k −1+ x ⇒ dt = dx

8 log (1 + tan θ) ⋅ sec2 θd θ 1 + tan2 θ

0

1

18 Let I =

Put x = tan θ ⇒ dx = sec2 θd θ

π/4

2

 z2   z2  1 5 + = + 2= ∴I = −  2 2 2   −1  0 2

 x2  + [ x2 ] 1 .25 2  1

= 0+

a

0 a

f ( x )⋅ g ( x ) dx f (a − x ) g ( a − x )dx

0 a

f ( x ){ a − g ( x )} dx

0

a

a

0 a

0

= a ∫ f ( x )d x− ∫ f ( x ) ⋅ g ( x )dx = a ∫ f ( x)d x − I 0

a a ∴ l = ∫ f ( x ) dx 2 0

25 Let I =

⇒ I =

∫ ∫

π /2 − π /2

π /2 − π /2

sin2 x dx 1 + 2x

−π π sin2  + − x  dx  2  2 −π

1+ 22

Q b f ( x ) dx =  ∫a

∫

b a

+

π 2

−x

dx

f (a + b − x ) dx  

172

⇒ I =

∫

⇒

π /2 − π /2

sin2 x dx 1 + 2− x

⇒ I =

2x ⋅ sin2 x ∫ − π /2 1 + 2x dx

⇒ 2I =

π /2

I =

π /2

⇒ 2I =

∫

− π /2

=

∫

0

π /2

∫

sin2 x dx =

⇒

π

I =

∫

0

I =

∫

0

π /2 0

=

∫

2sin2 x dx

…(i)

…(ii)

∫ ∫

=

π

[cot x] dx +

0 π

∫

π 0

[− cot x] dx

(− 1) dx

0

= =

∫

4

2

∫

4

∫

4

2

2

log xdx [log x + log(6 − x )] 4 log(6 − x ) ⇒ l =∫ dx 2 log (6 − x ) + log x 4

2

4

∫

⇒

2l =

∫

⇒

2l = 2 ⇒ l = 1

∫

− π /2 −3 π / 2

2 4

2

r n  1 +   n

r =1

∫

1

dx

0

(1 + x ) 2 x + x2

=

∫

1

dx

0

(1 + x ) (1 + x )2 − 1

∫ =∫

3π 0

1 0

π

33 Using Newton-Leibnitz’s formula, we get d t 2 { f (t 2 )}  (t )2  − 0  dt  d ⋅ f (0)  (0) = 2t 4  dt  ⇒ t 2 { f (t 2 )} 2t = 2t 4 ⇒ f (t 2 ) = t 4 2  put t = ± 2  f   = ± ⇒  25 5 5  4 2  ∴ f  =  25 5 [neglecting negative sign]

34 We have, 1 x [4t 2 − 2 f ′(t )] dt x2 ∫ 4 1 ∴ f ′ ( x ) = 2 [4 x2 − 2 f ′ ( x )] x 2 x − 3 ∫ [4 t 2 − 2 f ′(t )] dt x 4 1 ⇒ f ′ (4) = [64 − 2 f ′ (4)] − 0 16 32 ∴ f ′ (4) = 9

f (cos 2 x ) dx and f (cos 2 x ) dx

...(ii)

Also, P = 3∫ f (cos 2 x )dx = 3Q

π

0

∴

P − 3Q = 0 π 2 x(1 + sin x ) 32 Let I = ∫ − π dx 1 + cos 2 x

∫

35

2x dx 1 + cos 2 x

π −π

+

dx = [ x]24

[( x + π )3 + cos 2 x] dx

⇒ I = 0+ 4∫

π 0

∫

π −π

⇒ I = 4∫

x sin x dx 1 + cos 2 x

0

x2 0

sin t dt

 form 0   0  sin x ⋅ 2 x = lim x→ 0 3 x2 sin x 2 = lim 3 x→ 0 x 2 2 = ⋅1 = 3 3 x3

x→ 0

…(i)

 2 x sin x is an even function  1 + cos 2 x  π

∫ lim

2 x sin x dx 1 + cos 2 x

 2x Q 1 + cos 2 x is an odd function  and

1 dt 1 + t2

= π2

π 3

Q

=

−1 1

f ( x) =

...(i)

0

∫

π π = 2 π[tan −1 t ]1−1 = 2 π  +   4 4 

r  r 2 +  n n

S =

31 P =

3  π 3 π  ∫ −3π /2   − 2 − 2 − x + π   π 3π  + cos 2  − − − x  dx  2   2 Q b f ( x ) dx = b f ( a + b − x ) dx  ∫a  ∫a  −π / 2

Put cos x = t

1

= sec −1 2 − sec −1 1 =

log x + log (6 − x ) dx log x + log(6 − x )

2l =

⇒ I =

n

= lim ∑ n→ ∞

0

n (n + r ) r (2n + r )

S = [sec (1 + x )]

Q b f ( x ) dx = b f ( a + b − x ) dx  ∫a  ∫a  On adding Eqs.(i) and (ii),we get

28 Let I =

n

30 S = nlim →∞∑

−1

2 log x dx 2[log x + log (6 − x )]

∫

φ ( x ) dx = 0

sin x dx 1 + cos 2 x

⇒ − sin x dx = dt

(Qφ( x ) is an odd function)

log x 2 dx log x2 + log(36 − 12 x + x2 )

2 log x dx 2 log x + log(6 − x ) 2

⇒ l =

π

−π

0

π

∫

∴ I = − 2π

r =1

  − 1 if x ∉ Z  Q[ x] + [− x] =  0, if x ∈ Z     = [− x] π0 = − π π ∴I =− 2

27 l =

⇒I = 2π

∴ φ (− x ) = [ f (− x ) + f ( x )] [g (− x ) − g ( x )] = − φ ( x )

∫

x sin x dx 1 + cos 2 x

[from Eq. (i)]

29 Let φ ( x ) = [ f ( x ) + f (− x )] [g ( x ) − g (− x )]

∴

π 0

sin x dx − I 1 + cos 2 x

π

∫

⇒ I = 4π

π 2

I =

On adding Eqs. (i) and (ii), we get 2I =

(1 + cos 2 x ) dx

π sin (− π ) = − +  2 2

∴

[− cot x] dx

0

−4 ∫

3 π sin (−3 π )  −  − +  = π  2  2

[cot ( π − x )] dx

π

−3 π / 2

π 2

=

[cot x] dx

π

− π /2

π sin x dx 1 + cos 2 x

π 0

2cos 2 x dx

− π /2

π /2

26 Let

− π /2

⇒ I = 4∫

sin 2 x  = x + 2  −3 π /2 

(1 − cos 2 x ) dx

sin 2 x  ⇒ 2I =  x − 2  0  π ∴ I = 4

[ − ( x + π ) 3 + cos 2 x] dx

−3 π / 2

∫

=

2x ⋅ sin2 x ∫ − π /2 1 + 2x dx

−3 π / 2

∫

π /2

I =

− π /2

∫

( π − x ) sin ( π − x ) dx 1 + cos 2 ( π − x )

36 We have, (1 + x2 ) > x2 , ∀ x 1 + x2 > x, ∀ x ∈ (1, 2)

⇒

1

⇒ ∴ ⇒

1 + x2

∫


x2 ⇒

∫

− u (2cos u sin udu )

x

dx ≠ 200λ

sin x

=

x

x π /2

cos 2z sin 2z  = − z⋅ +  2 4  0

is 2 π.

sin x

b

π /2

∫

b

a

0

t = cos 2 u in 2nd integral, we get dt = 2sin z cos z and dt = − 2cos u sin udu

Hence, Statement I is false but ∫ f ( x )dx

3

∫

[ f {g ( x )}] −1 f ′ {g ( x )} {g ′ ( x )} dx

4 Put t = sin2 z in 1st integral and

∴I =

If f ( x ) ≥ g ( x ), then ∫ f ( x ) dx ≥

1 + sin x

2 1

Put f {g ( x )} = z ⇒ f ′ {g ( x )} g ′ ( x ) dx = dz When x = 1, then z = f {g (1)} When x = 2, then z = f {g (2)} f { g(2 )} 1 ∴ I =∫ dz = [log z] ff {{ gg((21 )})} f { g(1 )} z ⇒ I = log f {g (2)} − log f {g (1)} = 0 [Q g (2) = g (1)]

dx

2l = [ x] ππ //36 π 1 π π l =  −  = 2  3 6  12

⇒

3

∫

π /3

∫

3 Let I = ...(i)

π /6

1

x dx < ∫ dx, 0 x x because in x ∈ (0, 1), x > sin x 1 2 ⇒ I 0, 2 then 2 x (1 − e 2 x ⇒ ⇒

⋅ 2 x − e −(x

4

1

0

 1 − 2sin x  dx − π  1 − 2sin x  dx    ∫ π   2 2 3

= 4 3 − 4−

+ 1 )2

 12 Let l = nlim  →∞ 

2

…(ii)

f (a) + f (− a) = 1 Let

0 0 

 1 − 2sin x  dx = π |1 − 2sin x | dx   ∫0  2 2

π 0

⇒I =

On adding Eqs. (i) and (ii), we get ⇒

dt

0

x→ ∞

2I =

e −a 1 = 1 + e −a 1 + e a

and f (− a) =

∫

e 2x t2

On adding Eqs. (i) and (ii), we get

ea f (a) = 1 + ea

∴

 form 0   0 

2

{x } dx

0

7 Given that, f ( x ) =

2 I1 =

x 2 2  2 ∫ e t dt  (e x ) 0  

x

∫

2 dx + K

+

2

e2 t d t

x→ ∞

= e −(x

2

x − a, x≥ a to break given | x − a|=   −( x − a) x < a integral in two parts and then integrate separately.

n

0

x 0

2

9 Use the formula,

6 We have, ∫ 0 [ x] dx

∫

∫

log t log t dt − ∫ dt 1 (1 t + t) e

 (log t )2  =   2 1 1 = [(log e)2 − (log 1)2 ] 2 1 = 2

=

x 2  ∫ e t dt   0 

f ′( x ) = e − ( x

2

+ 1 )2

d ⋅  ( x 2 + 1)  dx 

− e −(x

2 2

)

d ⋅  ( x ) 2   dx 

⇒ log l = 2 ⋅ log 3 −

∫

2 0

∫

2 0

x+ 1−1 dx 1+ x

 1  1 −  dx 1 + x 

⇒

log l = 2 ⋅ log 3 − [ x − log 1 + x ] 20

⇒

log l = 2 ⋅ log 3 − [2 − log 3]

⇒

log l = 3 ⋅ log 3 − 2

⇒

log l = log 27 − 2

⇒

l = e log 27 − 2 = 27 ⋅ e − 2 =

27 e2

13 We have, f ′( x ) = 3 sin x + 4 cos x 5π 4 π  Since, in  , , f ′ ( x ) < 0, so assume  4 3  4π the least value at the point x = . 3 Thus, 4π  f =  3 

∫

4 π /3 5π / 4

(3 sin u + 4cos u ) du

175

DAY = [– 3cos u + 4 sin u] 45 ππ //34



3 1 −2 3+ 2 2

=

14 We have, g ( x ) = ⇒

x

∫

∫

g (2) =

⇒ g (2) =

∫

1 0

f ( t )dt +

n→ ∞

2 0

f ( t ) dt

∫

2 1

f ( t ) dt

We know that, m ≤ f ( x ) ≤ M for x ∈ [a, b] ⇒ m (b − a) ≤

b

∫

a

f ( x ) dx ≤ M (b − a)

1 ≤ f ( t ) ≤ 1, for t ∈[0, 1] 2 1 and 0 ≤ f ( t ) ≤ , for t ∈ [1, 2] 2 1 1 ⇒ (1 − 0) ≤ ∫ f ( t ) dt ≤ 1 (1 − 0) 0 2 2 1 and 0 (2 − 1) ≤ ∫ f ( t ) dt ≤ (2 − 1) 1 2 1 2 1 1 ⇒ ≤ ∫ f ( t ) dt ≤ 1 and 0 ≤ ∫ f ( t ) dt ≤ 0 1 2 2 1 2 1 3 ≤ f ( t ) dt + ∫ f ( t ) dt ≤ ⇒ 1 2 ∫0 2 1 3 ≤ g(2) ≤ ∴ 2 2 ∴

15 Let f (a) = =

∫

a 0

∫

1 0

(a − x ) dx + n

n

⇒

a

∫

a

( x n − an ) dx

   x x = an x − + − an ⋅  n + 1  0  n + 1  n +1

x

2an +1 1 − an + n+1 n+1 n  1 = 2an +1  − an +  n + 1  n+1

∴ ⇒ ⇒

2

= 2an +1 −

n −1

⇒ f ′(a) = n (2a − 1) a Thus, only critical point in (0, 1) is a = 1/2 1 Also, f ′(a) < 0 for a ∈  0,   2 1 and f ′(a) > 0 for a ∈  , 1 . 2  ∴

1 f (a) is minimum for a = . 2

2

∴

g( x) = −

∫

1

⇒

∫

1 1 /e

∫

1

 − 1  dx    x

= − [log e x] 11 /e 1 g ( x ) dx = −  log e 1 − log e  e   = − [0 + 1] = − 1 f (0) = 1 f ( x) = e x

0

0

x

…(ii)

f ( x )g ( x ) dx

e x ( x2 − e x ) dx [from Eqs. (i) and (ii)]

=

∫

1

x2e

0

x

dx −

∫

1 0

e 2 x dx

1 2x 1 [e ] 0 2 1 = [ x2e x − 2 xe x + 2e x ]10 − (e 2 − 1) 2 1 1 = [( x2 − 2 x + 2)e x ]10 − e 2 + 2 2 = [(1 − 2 + 2)e 1 − (0 − 0 + 2)e 0 ] 1 1 − e2 + 2 2 e2 1 e2 3 =e −2− + =e − − 2 2 2 2 = [x e 2

x

−

∫ 2 xe

dx]10 −

x

nx n − 1 dx = 2 tan θ sec2 θ dθ ∞ 0

dx 2 = 1 + xn n

∫

2 n

∫

=

18 Given, f ′( x ) = f ( x ) and Let

∫

1

1 x

2

1 /e

=

∴∫

1 x

g ( x ) dx =

1 /e

∫

Now,

⇒

− 1  + 1

499 times

⇒

g ( x ) = x2 − e 1

19 In LHS, put x n = tan2 θ

1 f ( x ) = f [ f ( x )] = f  −   x −1 = = x 1 − x g ( x ) = f 1998 ( x ) = f 2of 1996 ( x ) g ( x ) = f 2 [ f 1996 ( x )] g ( x ) = f 2 ( x ) [Q f 1996 ( x ) = {( f 4of 4of 4o K f 4 )} ( x ) = x] 14243 4

  a

 1  an +1 − an − + an +1 n + 1  n+1  

x−1 x+1

x f 2 ( x ) = f [ f ( x )] = f  x x−1 −1 x+1 = =− x−1 +1 x+1

⇒

⇒

 an +1  = an +1 − +  n + 1  

r =1

17 We have, f ( x ) =

1

n +1

∑

P =1

n

1

n

2 π /2 ( ln tan x + lncot x )dx π ∫0 2 π /2 = ln 1 dx = 0 π ∫0 ln P = 0

| x − a |dx n

⇒

2 ln P =

∴

f ( x ) + g ( x ) = x2

Also,

1 rπ ln tan    2n  n 1 πx  = ∫ ln tan   dx 0  2  2 π /2 …(i) ln tan x dx ⇒ ln P = π ∫0 2 π /2 and ln P = …(ii) ln cot x dx π ∫0 On adding Eqs. (i) and (ii), we get

∴ ln P = lim

f ( t ) dt

0

1/n



n

rπ 16 Let P = nlim tan     rΠ →∞ =1  2n   

…(i)

π /2 0

π /2 0

tan1 − 2 + 2 / n θ d θ

tan (2 / n) −1 θ d θ

In RHS, put x n = sin2 θ nx n − 1 dx = 2 sin θ cos θ dθ dx 2 π /2 1 ∴∫ = ∫ 0 (1 – x n ) 1 / n n 0 cos 2/n θ ⇒

1

2

sin n

−1

θ cos θ dθ =

2 n

∫

π /2 0

tan (2 / n) − 1 θ dθ

20 sin 6 x + cos 6 x = (sin2 x ) 3 + (cos 2 x ) 3 = (sin2 x + cos2 x ) 3 − 3 sin2 x cos 2 x (sin2 x + cos 2 x ) 2 2 = 1 − 3 sin x cos x 3 Q period = π  = 1 − sin2 2 x  4 2  So, the least and greatest value of 1 sin 6 x + cos 6 x are and 1. 4 π 1   Hence,  − 0 × 2  4 π /2 π < ∫ (sin 6 x + cos 6 x )dx <  − 0 × 1 0 2  π /2 π π 6 6 ∴ < ∫ (sin x + cos x ) dx < 0 8 2

DAY SEVENTEEN

Area Bounded by the Curves Learning & Revision for the Day u

Curve Area

u

Area between a Curve and Lines

u

Area between Two Curves

Curve Area The space occupied by a continuous curve, which is bounded under the certain conditions, is called curve area or the area of bounded by the curve.

Area between a Curve and Lines 1. The area of region shown in the following figure, bounded by the curve y = f ( x) β

β

α

α

defined on [α , β], X-axis and the lines x = α and x = β is given by ∫ y dx or ∫ f ( x) dx. Y

y = f (x) R Q y

X′

O

S P x=α dx x=β

X

Y′

2. If the curve y = f ( x) lies below X-axis, then area of region bounded by the curve y = f ( x), X-axis and the lines x = α and x = β will be negative as shown in the following figure. So, we consider the area as

β

∫α

y dx

or

Y

β

∫α

PRED MIRROR Your Personal Preparation Indicator

f ( x) dx u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

X′

x=β

x=α dx O

y y = f (x) Y′

X

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

177

DAY 3. Area of region shown in the following figure, bounded by the curve x = f ( y), Y-axis and the lines y = c and y = d is given by

d

d

∫ c f ( y) dy or ∫ c

Area between Two Curves 1. (i) Area of region shown in the following figure, bounded between the curves y = f ( x), y = g( x), where f ( x) ≤ g( x), and the lines x = a, x = b (a < b ) is given by

x dy

Y

Y

y=d

S

R x

dy y=c X′

y=

x = f (y)

y = f (x)

Q

P

x=a

O

X

O

) g (x

X

x=b

b

Area = ∫ [ g( x) − f ( x)] dx a

Y′

(ii) If f ( x) ≥ g( x) in [a, c] and f ( x) ≤ g( x) in [c, b ] where a < c < b , then Area

If the position of the curve under consideration is on the left side of Y-axis, then area is given by

c

= ∫ [ f ( x) − g( x)] dx +

d

∫c

f ( y) dy .

4. If the curve crosses the X-axis number of times, the area of region shown in the following figure, enclosed between the curve y = f ( x), X-axis and the lines x = α and x = β is given by

a

b

∫ c [g( x) − f ( x)] dx

2. Area of region shown in the following figure, bounded by the curves y = f ( x), y = g( x), X -axis and lines x = a, x = b is given by c

Area = ∫ f ( x) dx + a

Y

b

∫ c g( x) dx

Y y=

y = g(x)

) f(x

α1

O x =α

α1

∫α

f ( x) dx +

α2

∫ α1

α2 x = β

f ( x) dx +

β

∫ α2

X

X′

f ( x) dx

x=a

O

x=c

X

x=b

Y′

Area and the shape of some important curves S.No.

Curves

(i)

x2 y2 x y f ( x, y) : 2 + 2 ≤ 1, + ≥ 1 a b a b

Point of intersection

Area of shaded region Y B (0, b)

A (a , 0), B(0 , b ) ⇒ or

Area = ab

x2 y2 x y + 2 ≤1≤ + 2 a b a b

(π – 2) sq units 4

(–a, 0) A′ X′

A (a, 0) X

(0, 0)

Y′ Y A

(ii)

Parabola y2 = 4ax and its latusrectum x = a

A (a , 2a), B(a , − 2a)

Area =

8 2 a sq units 3

X′

O

Y′

(a

,2

a)

y2=4ax

(a, 0)

B (a, –2a)

X

178

Y A

(iii)

 4a 4a f ( x, y) : y2 = 4ax and y = mx O(0, 0), A  2 ,   m m

Area =

2

8a sq units 3 m3

X′

X

O (0, 0)

Y′ x2=4ay

(iv)

f ( x, y) : x2 = 4ay y = 4 bx 2

O(0, 0), 2 /3

A (4a

b

1 /3

1 /3 2 /3

, 4a b

)

Area =

16 (ab ) sq units 3

X′

(4a2/3 b1/3,4a1/3 b2/3) A

Y

X

(0, 0) O y2=4bx Y′

Y

(v)

A (b – a, 2√ab)

Area bounded by y2 = 4a ( x + a) y = 4 b (b − x) 2

and

A(b − a , 2 ab ) Area =

B(b − a , − 2 ab )

8 ab (a + b ) sq units 3

X′

A′ (b, 0)

(0, 0)

X

B′(–a, 0) O

B (b – a, – 2√ab) Y′

(vi)

Common area bounded by the ellipses 2

2

x y 1 and + = b 2 a 2 a 2b 2 x2 y2 1 + 2 = 2 2, 0 < a < b 2 a b a b

Y

x=y= x=y=−

1 a + b2 2

1 a +b 2

2

Area = Area of region PQRS = 4 × Area of OLQM 4  a = tan −1   sq units b ab

P

m Q L

R

S

X′ O (0, 0)

X

Y′

x=b

Y

(vii)

If α , β > 0, α > β the area between the hyperbola xy = p2 , the X-axis and the ordinates x = α, x = β

—

α Area = p2 log    β

X′

x=a

X

(0, 0) O

Y′

179

DAY

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 Let f : [ − 1, 2] → [ 0, ∞ ) be a continuous function such that f ( x ) = f (1 − x ) for all x ∈ [ − 1, 2]. Let R1 = ∫

2 −1

x f ( x ) dx, and

R 2 be the area of region bounded by y = f ( x ),x = − 1, x = 2 and the X -axis. Then (a) R1 = 2R 2

(b) R1 = 3R 2

(c) 2R1 = R 2

(d) 3R1 = R 2

2 Let A1 be the area of the parabola y 2 = 4ax lying between the vertex and the latus rectum and A2 be the area between the latus rectum and the double ordinate x = 2a. Then A1 / A2 is (a) (2 2 − 1) /7 (c) (2 2 + 1)

(b) (2 2 + 1) /7 (d) (2 2 − 1)

x 2 + y 2 = 9 by x = 1 is −1

(b) 9 sec

(3) −

8

(b) 1 : 1

(c) 1 : 2

(d) 2 : 1

lines x = 1, x = b is ( b 2 + 1 − 2 ) for all b > 1, then f ( x ) (b) x + 1

(c) x / x + 1 (d) None of these

8  curve y = 1 + 2  , X -axis and the ordinates x = 2, 4 into  x  two equal parts, then a is equal to (b) 2 2

(c) 2

(d)

2

7 Let the straight line x = b divide the area enclosed by

y = (1 − x )2, y = 0 and x = 0 into two parts R1 ( 0 ≤ x ≤ b ) and R 2 (b ≤ x ≤ 1) such that R1 − R 2 = 1/ 4. Then b equals (a) 3/4

(b) 1/2

(c) 1/3

(d) 1/4

8 The curve y = a x + bx passes through the point (1, 2) and the area enclosed by the curve, the X -axis and the line x = 4 is 8 square units. Then a − b is equal to (a) 2

(b) − 1

(c) − 2

(d) 4

9 If y = f ( x ) makes positive intercepts of 2 and 1 unit on x and y-coordinates axes and encloses an area of 2 3 sq unit with the axes, then ∫ xf ′ ( x ) dx , is 0 4 (a)

3 2

(c)

7 3

(d)

14 9

11 The area bounded by the curve y = ln ( x ) and the lines y = 0, y = ln ( 3) and x = 0 is equal to

j

JEE Mains 2013

(b) 3 ln (3) −2 (c) 3 ln (3)+2 (d) 2

(a) 3

12 The area of the region bounded by the curve ay 2 = x 3, the Y -axis and the lines y = a and y = 2a, is 3 2 a (2 ⋅ 2 2 / 3 − 1) sq unit 5 3 (c) a 2 (2 2 / 3 + 1) sq unit 5 (a)

(b)

NCERT Exemplar

2 a (2 2 / 3 − 1) sq unit 5

(d) None of these

(a) 4.5 sq units (c) 3.5 sq units

(b) 6.3 sq units (d) None of these

14 The area between the curve y = 4 − | x | and X -axis is (a) 16 sq units (c) 12 sq units

(b) 20 sq units (d) 18 sq units

2

6 If the ordinate x = a divides the area bounded by the

(a) 8

14 3

15 The area under the curve y = | cos x − sin x | , 0 ≤ x ≤

equals to (a) x + 1

(b)

y = | x − 3 | and y = 2, is

5 The area bounded by the curve y = f ( x ), X -axis and the

2

12 9

13 The area of the region bounded by

and y = cos 2 x between x = 0, π / 3 and X -axis, is 2 :1

(a)

(d) None of these

4 The ratio of the area bounded by the curves y = cos x (a)

bounded by the parabola y = 9x 2 and the lines x = 0, j JEE Mains 2013 y = 1 and y = 4, is

j

3 The area of smaller segment cut off from the circle 1 (a) (9 sec−1 3 − 8 ) 2 (c) 8 − 9 sec−1 (3)

10 The area of the region (in sq units), in the first quadrant,

(b) 1

(c)

5 4

(d) −

3 4

and above X-axis is (b) 2 2 − 2

(a) 2 2

j

(c) 2 2 + 2

π 2

JEE Mains 2013

(d) 0

16 The area of the region enclosed by the curves y = x , x = e, y =

1 and the positive X-axis is x

(a) 1 sq unit (b)

3 5 1 sq units (c) sq units (d) sq unit 2 2 2

17 The area bounded by y = sin x , X -axis and the lines x = π is

(a) 2 sq units (c) 4 sq units

(b) 3 sq units (d) None of these

18 For 0 ≤ x ≤ π, the area bounded by y = x and y = x + sin x , is (a) 2 (c) 2 π

(b) 4 (d) 4 π

19 The area (in sq unit) of the region described by {( x , y ): y 2 ≤ 2x and y ≥ 4x − 1} is 7 (a) 32 15 (c) 64

5 (b) 34 9 (d) 32

j

JEE Mains 2015

180 20 The area bounded by the curves y 2 = 4x and x 2 = 4y, is (a) 0

32 (b) 3

16 (c) 3

8 (d) 3

21 If the area bounded by y = ax 2 and x = ay 2, a > 0 is 1, then a is equal to

1 3

(d) None of these

y 22 The area bounded between the parabolas x = and 4 x 2 = 9 y and the straight line y = 2, is (b)

10 2 3

(c)

20 2 3

(d) 10 2

23 The area enclosed the curves y = x 3 and y = x is 5 (a) sq units 3 5 (c) sq unit 12

5 (b) sq units 4 12 (d) sq units 5

x 2 + y 2 ≤ 4x, x ≥ 0, y ≥ 0} is 4 3 4 2 (c) π − 3

j

JEE Mains 2016

(b) π −

(b)

π 4 − 2 3

(c)

π 2 − 2 3

j

JEE Mains 2014 π 2 (d) + 2 3

3π + 2 9π − 2 2π − 9 (d) 9π + 2 (b)

27 The area bounded by the curves y = 2 − | 2 − x | and | x | y = 3 is

(a)(5 − 4 ln 2)/ 3 (c)(4 − 3 ln 3)/ 2

y = x , 2y − x + 3 = 0, X-axis and lying in the first j JEE Mains 2013 quadrant is (a) 9

(b) 36 27 (d) 4

(c) 18

31 The area bounded by the curves y = x 2 and (b) (3 π − 2)/ 3 (d) None of these

32 The area bounded by y = tan x, y = cot x, X -axis in 0≤x ≤

π is 2

(b) log2 (d) None of these

y = loge x and y = 2x is equal to

two parts. Then, the ratio of the areas of these parts is 3π − 2 10 π + 2 6π − 3 (c) 11π − 5

30 The area (in sq units) bounded by the curves

33 Area of the region bounded by curves x = 1/2, x = 2,

26 The parabola y 2 = 2x divides the circle x 2 + y 2 = 8 in (a)

(b) (π − 2)/ 2 (d) None of these

(a) 3log2 (c) 2log2

25 The area of the region described by A = {(x, y ): x 2 + y 2 ≤ 1and y 2 ≤ 1 − x} is

(a) (π − 2)/ 4 (c) (π + 2)/ 2

(a) (3 π + 2)/ 3 (c) (3 π − 2)/ 6

8 3 π 2 2 (d) − 2 3

(a) π −

π 4 + 2 3

(b) 2 sq units (d) 4 sq units

y = 2/(1 + x 2 ) is

24 The area (in sq units) of the region {(x, y ) : y 2 ≥ 2x and

(a)

(a) 1 sq unit (c) 2 2 sq units

x 2 + y 2 = 1 and the lines | y | = x + 1 is

2

(a) 20 2

y = − | x | + 1, is

29 The area of the smaller region bounded by the circle 1 (b) 3

(a) 1 (c)

28 The area bounded by the curves y = | x | − 1 and

(b) (2 − ln 3)/ 2 (d) None of these

(4 − 2 )/log 2 + b − c log 2. Then, b + c equals (b) − 1 (d) None of these

(a) 2 (c) 4

34 The area bounded by the curve y = (x + 1)2, y = (x − 1)2 and the line y = 1 sq unit 6 1 (c) sq unit 4 (a)

1 is 4 2 sq unit 3 1 (d) sq unit 3 (b)

35 The area of bounded region by the curve y = loge x and y = (loge x )2, is (a) 3 − e 1 (c) (3 − e) 2

(b) e − 3 1 (d) (e − 3) 2

181

DAY

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 Let f ( x ) = x 2, g ( x ) = cos x and h( x ) = f (g ( x )). Then area bounded by the curve y = h( x ), and X -axis between x = x1 and x = x 2, where x1, x 2 are roots of the equation 18x 2 − 9πx + π 2 = 0, is (a) π/12

(b) π/ 6

(c) π/ 3

(d) None of these

9 The area enclosed between the curves y = loge ( x + e ), x = log(1/ y ) and the X -axis is equal to (a) 2e

(b)

1 4

(c)

1 3

(d)

f (1 + x ) = 3. The area x bounded by the curve y = f ( x ), y-axis and the line y = 3 is equal to x→0

1 2

(a) e

3 A point P ( x , y ) moves in such a way that [| x | ] + [| y | ] = 1, [⋅] = G.I.F. Area of the region representing all possible positions of the point P is equal to (a) 8

(b) 4

(c) 16

(d) None of these

(c) 12 π

6 Area of the region bounded by the parabola ( y − 2)2

= x − 1, the tangent to it at the point with the ordinate 3, and the X -axis is given by (b) 9 sq units (d) None of these

(a)

(a) 2

5 (b) 2

59 (c) 12

(d) 2 π 2

(b) 3

(c) 6

(d) 9

maximum, then b is equal to (b) − 1

(a) 1

(c) b = ± 1 (d) None of these

15 Let An be the area bounded by the curve y = (tan x )n and the lines x = 0 and x = π / 4. Then, for n > 2 1 1 < An < 2n 2n − 2 1 1 (c) < An < 2n + 2 2n − 2 (a)

x } is

JEE Mains 2017

(d)

(c) π 2

14 If area bounded by the curves y = x − bx and by = x 2 is

8 The area (in sq units) of the region

7 (a) 3

π2 2

2

(b) π / (3 3 + π) (d) None of these

j

(b)

f ( x ) − f ( x /7) = x /7 ∀x ∈ R . The area bounded by the curve y = f ( x ) and the coordinate axes is

tangent and normal at (1, 3 ) and X -axis in A1 and A2. A Then 1 equals to A2

{( x , y ) : x ≥ 0, x + y ≤ 3, x 2 ≤ 4y and y ≤ 1 +

π2 4

13 Let f ( x ) be continuous function such that f ( 0) = 1,

7 Let the circle x 2 + y 2 = 4 divide the area bounded by

(a) π / (3 3 − π) (c) π / (3 − π 3 )

(b) 2 x , x ≥ 1 (d) None of these

y = cos −1 (sin x ) − sin−1 (cos x ) and the lines y = 0, 3π x = , x = 2π is 2

3π 5π π π sq units (b) sq units (c) sq units (d) sq unit 8 8 2 8

(a)9/2 sq units (c)18 sq units

(d) None of these

12 The area bounded by the curve

x 2 + y 2 ≤ 1 } is (a)

(a) 2 2 x , x ≥ 1 (c) 2 x , x ≥ 1

(d) 6π

5 The area of the region R = {( x , y ) : | x | ≤ | y | and

(c) 3e

curve y = f ( x ), which cuts the last two lines above the 2 first line for all a ≥ 1, is equal to [( 2a )3/ 2 − 3a + 3 − 2 2 ]. 3 Then f ( x ) =

4 The area bounded by the curve xy = 4 ( 2 − x )and Y-axis is (b) 4 π

(b) 2e

11 The area bounded by the lines y = 2, x = 1, x = a and the

2

(a) 2 π

(d) None of these

f ( x / y ) = f ( x ) − f ( y ) and lim

[ −1, 1] and X-axis, is (in sq unit) 1 5

(c) 2/e

10 Let f ( x ) be a real valued function satisfying the relation

2 The area bounded by f ( x ) = min(| x | ,| x − 1| ,| x + 1| ) in (a)

(b) 2

3 2

1 1 < An < 2n + 1 2n − 1 1 1 (d) < An < 2n + 2 2n

(b)

ANSWERS SESSION 1

SESSION 2

1 11 21 31

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

1 (a) 11 (a)

2 12 22 32

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

2 (d) 12 (a)

3 13 23 33

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

3 (a) 13 (b)

4 14 24 34

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

4 (b) 14 (c)

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

6 (b) 16 (b) 26 (b)

7 (b) 17 (c) 27 (c)

8 (d) 18 (a) 28 (b)

9 (d) 19 (d) 29 (b)

10 (d) 20 (c) 30 (a)

5 (c) 15 (c)

6 (b)

7 (a)

8 (b)

9 (b)

10 (c)

5 15 25 35

182

Hints and Explanations SESSION 1

5

1 Given, R1 =

2

∫− 1

2

∫− 1 f ( x ) dx and f (1 −

and R2 =

Consider, R1 = = =

x f ( x ) dx, x) = f ( x)

2

∫− 1 x f ( x ) dx

2

∫− 1 (1 −

b

∫1

Now, on differentiating both sides w.r.t. b, we get 2b f (b ) = ∀b > 1 ⇒ 2 b2 + 1 ⇒

x ) f (1 − x ) dx

f ( x ) = x / x2 + 1.

∫− 1 (1 −

x ) f ( x ) dx

=∫

= R2 − R1 ∴ R2 = 2R1

4 2

2 A 1 = 2 ∫ 2 a x dx 0

A

2 = 4 a × [ x3 /2 ] a0 = 8a2 / 3 3 a

Now, ∫ x f ′ ( x ) dx = [ xf ( x )]20 −

3 4 3 x f ′ ( x ) dx = 2 × 0 − 0 × 1 − 4 3 =− 4

⇒

∫0

2

(2, 3) C

10 Required area =

B (4, 3/2)

y=9x2

y

4

∫1

3

4

dy =

3

3 Required area, A = 2 ∫ 1

O

M

D x=a

=

∫2

8 8   1 + 2  dx ⇒  x –  = 2  x 2 x  

⇒ a = ± 2 2⇒ a =2 2

(3, 0)

(1, 0) (0, 0)

X′

a

a

7 We have, R1 − R2 =

X′

y=1

Area of region, ACDM

9 − x2 dx

Y

y=4

X

N

b

1  y 3 /2  3  3 / 2  1

Y

2

2 3 /2 2 a 8a [ x ]a = (2 2 − 1) 3 3 (2 2 + 1) 1 = ∴ A 1 / A2 = 7 2 2−1

∫ 0 f ( x ) dx

∫ 0 x f ′( x ) dx = [2 f (2) − 0 f (0)] −

4

 1 + 8  dx =  x − 8  = 4    x 2 x2  

2

2

⇒

2 a x dx

=4 a×

∫0 (1 − x ) 1 − ∫ (1 − b

2

2 2 14 = (43 /2 − 13 /2 ) = × 7 = 9 9 9

dx x )2 dx =

X Y′

[Q a > 0]

1 4

X

11 Required area =

log 3

∫0

xdy =

log 3

∫0

e ydy

3 = [e y] log = [e log 3 − e 0] = 3 − 1 = 2 0

12 We have,

x=1

R1

Y

Y′

C y=2a

2a

N

3

 x 9 − x2 + 9 sin −1 x   3 1

1 = 2⋅ 2

 π = 9 −  2

1   = 9 cos −1   − 8    3 −1 = [ 9 sec (3) − 8]

4 Here, A 1 =

π /3

∫0

= [sin A2 =

R2

1  8 − 9 sin −1    3  

 π 1  =  9  − sin −1    −   2 3  

and

[given]

2

Y

a

A2 = 2 ∫

(2, 0) and (0, 1). ∴ 0 = f (2) and 1 = f (0) 2 3 Also, ∫ f ( x ) dx = 0 4 0

6 Area of region AMNB

2

2a

9 Clearly, y = f ( x ) passes through

f ( x ) dx = b 2 + 1 − 2

π /3

∫0

=

3 4

…(i)

(a x + bx ) dx = 8

16a + 8b = 8 3 On solving Eqs. (i) and (ii), we get a = 3 and b = −1 ∴ a−b = 4 ⇒

X Y′

1 2 (1 − b )3 1 1 − = ⇒ (1 − b )3 = 8 3 3 4 1 ∴ b = 2

∫0

M

By=a

X′ O

(1,0)

⇒

4

cos 2 x dx

sin 2 x  =  2  0 ∴ A1 : A2 = 2:1

y=(1 – x)2

At x = 1, y = 2, we get 2 = a + b

3 = 2

π /3

O (b,0)

8 y = a x + bx ( x ≥ 0).

cos x dx x] π0 / 3

 8 

a

Required area = Area BMNC = 2 a 1 /3

=

∫a

=

3 a3 5

=

1 5 5  3 3 3  3 a a (2) − 1   5  

a

1

…(ii)

y 2 /3 dy =

2a

∫a

1 3 a3

5

xdy

[ y 5/3 ]2aa

5 5   (2a)3 − a3     

2  3  = a2  2⋅ 23 − 1 sq unit   5  

183

DAY e1 1 1 (1 × 1) + ∫ dx = + [log | x |] e1 1 2 x 2 1 1 3 = + [log| x|] e1 = + 1 = sq units 2 2 2

13 The curve y = | x − 3| meets the line

=

y = 2, when x = 1 and x = 5

Y

y=|x – 3|

Hence, required area  y + 1 y2  −  dy 4 2  2

1

=

∫−1 

=

 1  y2 1 + y − ( y 3 )1−1 /2  4 2  −1 /2 6

=

1  1 1   1 1  1    + 1 −  −   − 1 +    8 2  6  4  2 8

1

17 We have,

y=2 X¢

sin x, if x ≥ 0 y = sin x =  −  sin x, if x < 0 and| x|= π ⇒ x = ± π

X

01 3 5

1 3 3  1  9   + −   4 2 8  6  8  1 15 3 9 = × − = 4 8 16 32

Y

Y¢ The area of the shaded region 1 = × 2 × 4 = 4 sq units 2

1 X¢

–p

x=–p Y A (0, 4)

20 For the point of intersection of

Y¢

Y

x=p

y2=4x

x2=4y

π

Now, required area = 2∫ sin xdx 2

y=4–x C (0, 0) (4, 0)

(4, 4)

= 2[− cos x] π0

x 4+

X

p

y 2 = 4 x and x2 = 4 y

4 + x, x < 0 y =   4 − x, x > 0

y=

y=|sin x|

O

14 y = 4 − | x|represents two curves as,

B X¢ (– 4, 0)

=

= −2[cos π − cos 0] = −2[−1 − 1] = 4 sq units

X

X¢

(0, 0)D

X

(4, 0)

18 Given, curves y = x and y = x + sin x, which intersect at (0, 0) and ( π, π ). ∴ Area, A =

Y¢ The area of the shaded portion 1 = × 8 × 4 = 16 sq units 2

=

=

|cos x − sin x|dx

π /4

∫0

π

xdx

∫ π / 4 (sin x − cos x )dx

+

= [sin x + cos x] π0 / 4 + [− cos x − sin x] ππ //24 = ( 2 − 1) − (1 − 2 ) = 2 2 − 2

2

parabola y 2 = 2 x

…(i)

∴ Area bounded between curves 4  x2  = ∫  4x −  dx 0 4  4

and y ≥ 4 x − 1 represents a region to the left of the line y = 4 x − 1 …(ii) The point of intersection of the curves (i) and (ii) is given by

  x3   x3 /2 = 2 ⋅ − 3 12     0 2 (4 )3 4 3 /2 = ⋅ (4 ) − 3 12 32 16 16 = − = 3 3 3

(4 x − 1)2 = 2 x ⇒ 16 x2 + 1 − 8 x = 2 x

16 Given, y = x, x = e and y = 1 , x ≥ 0

x Since, y = x and x ≥ 0 ⇒ y ≥ 0 ∴ Area to be calculated in 1st quadrant shown in figure

Y

⇒ 16 x2 − 10 x + 1 = 0 1 1 ⇒ x= , 2 8 So, the points where these curves 1 1 1 intersect are  ,1 and  ,−  2   8 2 Y

x in y 2 = 4 x 4

 x2  ⇒   = 4 x ⇒ x 4 = 43 x ⇒ x = 0, 4  4

y 2 ≤ 2 x represents a region inside the

π /2

2

Substitute y =

{( x, y ): y 2 ≤ 2 x and y ≥ 4 x − 1}

(cos x − sin x )dx

Y¢

sin x dx = [− cos x] π0

∫0

19 Given region is

π /2

∫0

π

= − cos π + cos 0 = − (−1) + 1 = 2

15 Required area =

π

∫ 0 ( x + sin x ) dx − ∫ 0

21 The intersection point of two curves is  1 , 1 .    a a

y=4x–1

Y

y=x y2=2x

1

)

A

1 (1,

1/2 X¢

X¢

B

O D (1, 0) Y¢

C (e, 0) x=e

y=1/x

X

∴Required area = Area of ∆ODA + Area of DABCD

(1/2,1)

–1 –1/2 1/4 1/2 1 1, – 1 –1/2 8 2 –1 Y¢

y=ax2 y2=x/a

X¢

X

O

X

Y¢ ∴ Area, A =

1/a

∫0

 x  − ax2  dx  a  

184

1 2 3 /2 1 / a a 3 1 / a ⋅ [x ] 0 − [x ] 0 3 a 3 1 1 2 a = ⇒a= 3 3

⇒

Required area =

1=

⇒

22

Y

X¢

2

= 2∫ =

0

2

 y  3 y − 2 

  3 /2  5 y  dy = 5 y     3  2     2 

∫0 Ycircle − ∫0 Yparabola dx

=

πr − 4 2

2

∫0

2 2

x 8 x  + 2 8 − x2 + sin −1 2 2 2 2 2  16 π   = + 2 2 π −  2 + 4 ×    3 4   4 =  + 2π  sq units 3  = 8π sq units 4 Now, A2 = 8 π − A1 = 6 π − 3 A1 and the required ratio, is A2

(–1, 0)

(1,0)

X

4 + 2π 2 + 3π 3 = = 4 9π − 2 6π − 3

Y¢

y=0

1 1 2 πr + 2∫ (1 − y 2 )dy 0 2 1

 y3  1 π 4 = π(1)2 + 2 y −  = + 2 3 0 2 3 

x are given by x = 0

26

Y

2=

y 2 x + B

Y y= x X¢

X

8

C

Y

y=–3/x

y 2 =2x A (2Ö2, 0)

O

: x 0 | x| 

x 27 y = 2 − |2 − x| = 

4

y=x

3

X 2 1 X¢

∴A =

1

1

∫0

( x3 −

 x 4 2 x3 / 2  x ) dx =  − 3  0  4

1 2 5 sq unit =  −  =  4 3  12

24 We have, x2 + y 2 ≤ 4 x and y 2 ≥ 2 x To find point of intersection, substitute y 2 = 2 x in x2 + y 2 = 4 x x + y = 4x ⇒ x + 2x = 4x 2

⇒ ⇒ ⇒

2

2

x2 − 2 x = 0

(2,2)

(2,0)

Y¢

(4,0)

The point B is a point of intersection (lying in the first quadrant) of the given parabola and the circle, whose coordinates can be obtained by solving the two equations y 2 = 2 x and ⇒ x2 + 2 x = 8 ⇒ ( x − 2) ( x + 4) = 0

x = 0 or x = 2 y = 0 or y = 2

O

Let the area of the smaller part be A 1 and that of the bigger A1 part be A 2 . We have to find . A2

x2 + y 2 = 8.

⇒ x( x − 2) = 0

Y

X¢

O

Y¢

Y¢

X

 

2

Y

X¢

and x = 1 .

8 − x2 dx

Clearly, area of the circle = π (2 2 )2

y=2

23 Clearly, the intersection of two curves

2 2

∫2

2 ⇒ A 1 = 2  2 ⋅ ⋅ x3 /2  3   0

2( x )1 /2 dx

Required area =

X¢

 2 2 xdx +  ∫ 0

A = {( x, y ): x2 + y 2 ≤ 1and y 2 ≤ 1 − x}

  dy 

20 2 10 3 /2 (2 − 0) = 3 3

y = x3 and y =

2

=

25 Given,

X

0

⇒ A1 = 2

2

y=1 x2 9 y=2

Required area = 2 ∫

− y 2 ) dx

3 π×4 2  = − 2  x2  4 3   0 2 2 3 /2 8 = π− (2 − 0) = π − 3 3

y=4x2

Y¢

2

2

∫0 ( y 1

⇒ x = 2, − 4 x = − 4 is not possible as both the points of intersection have the same positive x-coordinate. Thus, C ≡ (2, 0). Now, A 1 = 2 [Area (OBCO ) + Area (CBAC )] 2 2  2  = 2 ∫ y 1 dx + ∫ y 2 dx , 2  0  where, y 1 and y 2 are respectively the values of y from the equations of the parabola and that of the circle.

1Ö3 2

3

4

y=4–x

Y¢

⇒

y=3/x X

x = 3/ x ⇒ x = 3

and 4 − x = 3 / x ⇒ x = 3 ( x > 2). ∴ Required area 2 3 3 3 = ∫  x −  dx + ∫  4 − x −  dx 3  2  x x = (4 − 3 log 3)/2

28 The region is clearly square with

vertices at the points (1, 0), (0, 1), (−1, 0) and (0, − 1).

Y (0,1) y=|x| –1 X¢

X (–1,0)

y=–|x|+1

(1,0) (0,–1)

Y¢ So, its area = 2 × 2 = 2 sq units

185

DAY

Y

35 Required area, A =

(0,–1) 0

e

∫ 1 [ log

x − (log x ) 2 ] dx

Y

[ 1 − x − ( x + 1)] dx

y=(logex)2

2

−1

1

or Area = 2 [area of sector AOBA − ∆AOB ] π 1  ( π − 2)  =2 − =  4 2  2

X¢

O

p/4

X

p/2

(e, 1)

=

...(i)

x

and 2 y − x + 3 = 0

y=logex

Y¢

X¢

Now, required area

30 Given curves are y =

y=cot x

= 2∫

Y

X

O

–y=x+1

 ( x − 1)3 x  1 1 1 = 2 −  = 2 − − +  24 8 3  3 4  0 1 = sq unit 3

 π ,1 .   4 

y=cot x

A (–1,0)

1 /2

32 Clearly, the two curves will intersect at

B (0,1)

y=x+1

∴ Required area 1 /2 1 = 2∫ ( x − 1)2 −  dx 0  4  

1

 2 x3  3π − 2 =  4 tan −1 x − = π − 2/3 = 3 3  0 

29 Due to symmetry, required area

π /4

∫0

π /2

∫π / 4 cot x dx

tan xdx +

Y¢

= [log sec x] π0 / 4 + [log sin x] ππ // 24

...(ii)

X

(1, 0)

O

A=

= [log 2 − 0] + [0 + log 2]

e

∫ 1 log

x dx −

e

∫ 1 ( log

x ) 2 dx

= [ x log x − x] e1 − [ x (log x )2

= 2 log 2 = log 2 sq units

−2( x log x − x )] e1

33 Required area

= [e − e − (− 1)] − [e (1) 2

y=2x

− 2e + 2e − (2)]

= 1 − (e − 2) = 3 − e

SESSION 2 1

On solving Eqs. (i) and (ii) , we get

½ 1

2 x − ( x )2 + 3 = 0 ⇒

( x )2 − 2 x − 3 = 0

⇒

=

( x − 3)( x + 1 ) = 0

∴ y =3 3

3

∫0 ( x2

= ∫ {( 2 y + 3 ) − y } dy 0

3

= 9+ 9− 9= 9 0

31 Curves x = y and y = 2 / (1 + x ) are 2

2

⇒

4− 2 5 3 − log 2 + log 2 2 2

=

4− 2 + b − c log 2 log 2

2

x dx = π / 12

2 Graph of

f ( x ) = min (| x |, | x − 1 |, | x + 1 |) is

Y 1/2

b = 3 / 2, c = 5/ 2 ⇒ b + c = 4

X¢

–1

–1/2

O

1/2

1

Y¢

34

Required area 1 1 1 = 2 ×  × 1 ×  = sq unit  2 2  2

y=(x–1)2

y=(x+1)2

symmetrical about Y-axis.

π /3

∫π / 6 cos

∴ Required area =

− log e x ) dx

= − x1 ) dy

2

 y3  =  y2 + 3 y − 3  

x

18 x2 − 9 πx + π2 = 0 ⇒ (6 x − π ) (3 x − π ) = 0 ⇒ x1 = π / 6, x2 = π / 3

2

[Q x = − 1 is not possible] ∴ Required area =

2

∫1 /2 (2

1 h( x ) = f (cos x ) = cos2 x

2

  2x = − x log x + x  log 2  1 /2

x = 3,

⇒

y=logex

Y

Y

3

Y 2

1 y= 4 X¢ O 1

1

 2 Area = 2 ∫  − x2  dx 0 1 + x2  

X

O –1 –1/2 1/2 1

Y¢

X

–2

O

–2

[| x |] + [| y |] = 1

2

X

X

186 ⇒

[| x |] = 1, [| y |] = 0

or

[| x |] = 0, [| y |] = 1

⇒

1 ≤ | x | < 2, 0 ≤ | y | < 1

6 ( y − 2) 2 = x − 1

…(i)

8 On solving x2 = 4 y and x + y = 3

Curve (i) is a parabola with vertex at the point A(1, 2), axis y − 2 = 0 i.e. y = 2 and concavity towards positive X -axis. For the point, say P, at which ordinate y = 3 and x = 2 Equation of tangent at P (2, 3) is dy y −3=   ( x − 2)  dx (2, 3 )

or 0 ≤ | x | < 1 or 1 ≤ | y | < 2 ⇒ x ∈ (− 2, − 1] ∪ [1, 2), y ∈ (− 1, 1) or x ∈ (− 1, 1), y ∈ (− 2, − 1] ∪ [1, 2) ∴ Required area = 8 × 1 × 1 = 8

4 In the equation of curve xy 2 = 4(2 − x ) , the degree of y is even. Therefore, the curve is symmetrical about X-axis and lies in 0 < x ≤ 2. The area bounded by the curve and the

1 or y − 3 = ( x − 2) i.e. 2 x − 2y + 4 = 0

⇒

⇒ ( x + 6) ( x − 2) = 0 ⇒ x = 2, y = 1 Solving y = 1 + x and y = 3 − x, we get 1 +

Y-axis is 2∫ y dx

x = 3 − x ⇒ x = 1, y = 2

4y=x2

Y (0,3) (1,2)

(2,1) X¢

) 2,3

0

P(

2 2 − x 2− x dx = 4∫ dx 0 x x Put x = 2sin2 θ ⇒ dx = 4sin θ ⋅ cos θ dθ

O1

2 (3,0)

2

= 2∫ 2 0

∴Required area = 4∫ = 8∫ = 8∫

π /2 0 π /2 0

π /2

Y¢

2 − 2sin2 θ

O

∴ Area =

2cos 2 θ dθ

3

∫0

1

(1 + cos 2θ)dθ

7

=

= 4∫

0

O p 3

( y 1 − y 2 ) dx ( 1 − x − x ) dx 1/ 2

1 1 x2  = 4  x 1 − x2 + sin −1 x − 2 2  0 2 Y P(x1, y1)

(–1/Ö2, 1/Ö2) X¢

Q (x 2, y 2)

(–1/Ö2, –1/Ö2)

2

Normal B(1, Ö3) A2 A1 X A C(4, 0)

9

5 2 y = log e ( x + e ) y =e

…(i)

−x

…(ii)

Required area =

∞

0

∫1 − e log( x + e ) dx + ∫0

e − x dx

Let area of portion OAB = A 1 and area of portion ABC = A 2 The equation of tangent at (1, 3 ) is x+

3y = 4

X

(1/Ö2, –1/Ö2)

Y¢

–e (1– e)

[Q xx1 + yy 1 = a2 is the tangent for

O

X

= [ x log ( x + e )]10−e 0 x −∫ dx + [− e − x ] ∞0 1 −e x + e

Now, the area of the 1 ∆OBC = × OB × BC 2 1 = × 2× 2 3 = 2 3 2 The area of portion OAB i.e. A 1 =

= 0 + 1 − [ x − e log ( x + e )]10 − e r 2θ 2

4 ⋅ π /3 2 π . = 2 3 Now,A2 = ∆OBC − OAB = 2 3 − 2 π / 3 A1 2 π /3 = A2 6 3 − 2π 3 2π π . = = 6 3 − 2π 3 3 − π =

1 1 π 1  1 =4 × + × −  2 2 2 2 4 4  π = sq units 2

1

the circle x2 + y 2 = a2 at ( x1 , y 1 )]

y=x

(1/Ö2, 1/Ö2)

O

x2 dx 4

Y

2

y=–x

2 0

2

Y

2

2

1

x ) dx + ∫ (3 − x )dx − ∫

1 2 3  x + 2 x3 /2  + 3 x − x  −  x       3 2 1  12  0 0  2 1  8  =  1 +  + (6 − 2) −  3 −  −     3   2    12 

Tangent

region = 4 (Area of the shaded region in first quadrant)

1/ 2

∫0 (1 +

[ y 2 − 4 y + 5 − (2 y − 4)] dy

5 Required area = Area of the shaded

1/ 2

Required Area

3

π

0

X

1

 y3  = − 3 y2 + 9 y   3 0 = 9 − 27 + 27 − 0 = 9.

2 θ 2 = 8 θ + sin 2  0  π  =8 + 0 − 0 = 4π   2

= 4∫

X x+ y=

3

(1,2)

2sin2 θ ⋅ 4 sin θ ⋅ cos θ dθ

0

y=1+Öx

…(ii)

Y

2

x2 + x=3 4 x2 + 4 x − 12 = 0

we get,

=1+ 1=2

10 f ( x / y ) = f ( x ) − f ( y )

…(i)

x = y = 1 ⇒ f (1) = 0 f ( x + h) − f ( x) f ′ ( x ) = lim h→ 0 h x + h  f   x  [using Eq. (i)] = lim h→ 0 h f (1 + h / x ) 1 3 = lim ⋅ = h→ 0 h/ x x x

187

DAY π cos −1 (sin x ) = cos −1 cos  2 π + − x    2 π = 2π + − x 2 and sin −1 (− cos x)

⇒ f ( x) = 3 ln x + c f (1) = 0 ⇒ c = 0. ∴ y = f ( x) = 3 log e x

3

O

∴ y = 2π + π /2 − x + 3π /2 − x = 4π − 2x 3

∫− ∞ e

y/3

dy

13 f ( x) − f ( x /7) = x /7

Y

∴ An = =

Adding, we get

y=2

x 1 1  1 + + K+ n − 1   7 7 7 x 1 = 1 − n  6 7 

f ( x ) − f ( x / 7n ) = a

X

Y′ a

2

3 /2

− 3a + 3 − 2 2]

1

On differentiating both sides w.r.t. a, we get 2 3 f (a) − 2 =  (2a)1 /2 ⋅ 2 − 3  3  2 ⇒ f (a) − 2 = 2 2a − 2 ⇒

f (a) = 2 2a

⇒

f ( x) = 2 2x, x ≥ 1

12 3 π ≤ x ≤ 2 π ⇒ − 2 π ≤ − x ≤ − 3 π 2 2

∆′(b) =

Taking limit as n → ∞, we get x x f ( x ) − f (0) = ⇒ f ( x ) = 1 + 6 6 [Q f (0) = 1] 0  x Required area = ∫  1 +  = 3 −6  6

⇒ ⇒

y = x − bx and by = x

∆(b ) =

b /b + 1

∫0

+

–

 x2  − x + bx2  dx  b   b /(b2 + 1 )

  b 2 + 1  x3 x2  =  −   2   b  3 0

π /4

∫0

π /4

∫0

tan n x dx

tan n − 2 x(sec2 x − 1) dx π /4

For n > 2, 0 < x < π / 4 ⇒ 0 < tan x < 1 ⇒ tan n − 2 x > tan n x

2

Solving these, we get x = 0, b /(1 + b 2 ) ∴

–

–1 0 1

π /4  tan n − 1 x  = −∫ tan n − 2 x dx  0 n − 1  0 1 An = − An − 2 ⇒ n−1 1 ⇒ An + An − 2 = n−1

⇒

14 Given curves are 2

+

15 In (0, π / 4), tan x > 0

f ( x /7) − f ( x /72 ) = x /72 M M M M M M f ( x /7n − 1 ) − f ( x /7n ) = x /7n .

∫ [ f ( x) − 2]dx = 3 [(2a)

(b 2 + 1)3

⇒ ∆(b ) is max. for b = 1, − 1.

f ( x /72 ) − f ( x /73 ) = x /73

1

=

− 2b 2 (b 2 + 1) × 2b 1 ⋅ 6 (b 2 + 1)4 2b (1 − b ) (1 + b )

π2 Required area = ∫ (4 π − 2 x )dx = 3 π /2 4

11 According to given condition, we have

X′

∆ ′ (b ) =

2π

= 3[e y/3 ]3− ∞ = 3e

y=f(x)

1 b2 2 6 (b + 1)2 2b (b 2 + 1)2

3π 3π = sin −1 sin  − x  = − x  2  2

1

Required area =

=

A n − 2 > An 1 2 An < n−1 1 An < 2n − 2 + An =

Also, A n

+2

⇒

2A n >

∴

1 n+1

1 n+1

1 1 < An < 2n + 2 2n − 2

DAY EIGHTEEN

Differential Equations Learning & Revision for the Day u

Solution of Differential Equation

u

Solutions of Differential Equations of First Order and First Degree

u

Linear Differential Equation

u

Inspection Method

An equation involving independent variables, a dependent variable and the derivatives of dependent variable w.r.t. independent variables, is called differential equation. A differential equation which contains only one independent variable, is called an ordinary differential equation.

Order and Degree of a Differential Equation ●

●

The order of a differential equation is the order of the highest derivative occurring in the differential equation. The degree of differential equation is the degree of the highest order derivative, when differential coefficients are made free from radicals and fractions.

Solution of Differential Equation A solution of the differential equation is a relation between the dependent and independent variables of the equation not containing the derivatives, but satisfying the given differential equation.

General Solution A general solution of a differential equation is a relation between the variables (not involving the derivatives) which contains the same number of the arbitrary constants as the order of the differential equation.

Particular Solution Particular solution of the differential equation is obtained from the general solution by assigning particular values to the arbitrary constant in the general solution.

PRED MIRROR Your Personal Preparation Indicator u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

189

DAY Solutions of Differential Equations of the First Order and

First Degree

NOTE

• Sometimes given equation becomes a linear differential dx + Rx = S, where R and S are dy function of y (or constants). R dy • The integrating factor in this case is IF = e ∫ equation of the form

and the solution is given by x ⋅ (IF ) = ∫ ( S × IF ) dy + C.

A differential equation of first order and first degree may be of the following types:

Differential Equation with Variable Separable ‘Variable separable method’ is used to solve such an equation in which variable can be separated completely, e.g. the term f ( x) with dx and the term g( y) with dy i.e.

g( y)dy = f ( x)dx

Reducible to Variable Separable Form dy = f (ax + by + c) can be dx reduced to variable separable form by the substitution ax + by + c = z. dz dy dz  dz  1 − a = f (z) ⇒ = a + bf (z) ⇒ a+b = ⇒   dx  b dx dx dx Differential equations of the form

Homogeneous Differential Equation A function f ( x, y) is said to be a homogeneous function of degree n if f (λx, λy) = λn f ( x, y) for some non-zero constant λ. ●

Let f ( x, y) and g( x, y) be two homogeneous functions of same degree, then a differential equation expressible in the form dy f ( x, y) = dx g( x, y) is called a homogeneous differential equation.

●

To solve a homogeneous differential equation of the type dy f ( x, y)  y = = h  , we put y = vx and to solve a  x dx g( x, y) homogeneous differential equation of type  x dx f ( x, y) = = h  , we put x = vy.  y dy g( x, y)

Linear Differential Equation A differential equation of the form

dy + Py = Q, dx

where, P and Q are the functions of x (or constants), is called a linear differential equation. To solve such an equation, first find integrating Factor, P dx IF = e ∫ . Then, the solution of the differential equation is given by y (IF) = ∫ Q(IF) dx + C.

Bernoulli’s Differential Equation (Reducible to Linear Form) Let the differential equation be of the form dy dy + P y = Q yn ⇒ y−n + P y−n dx dx dy dz Here, put y − n + 1 = z ⇒(− n + 1) y − n = dx dx dz ⇒ + (1 − n) Pz = (1 − n) Q dx

+1

=Q

which is a linear differential equation in z. The solution of this equation is given by ze ∫

(1 − n )P dx

( 1 − n ) Pdx  = ∫ ( 1 − n ) ⋅ Q ⋅ e ∫ dx + C  

Inspection Method If we can write the differential equation in the form f { f1 ( x, y)} d{ f1 ( x, y)} + φ { f2 ( x, y)} d{ f2 ( x, y)} + ... = 0, then each term can be easily integrated separately. For this use the following results. (i) d ( x ± y) = dx ± dy  x  ydx − x dy (iii) d   =  y y2

(ii) d ( xy) = x dy + ydx

 x  y dx − x dy (iv) d  tan −1  =  y x 2 + y2 x dy + y dx (v) d [log( xy)] = xy   x   y dx − x dy (vi) d log    =  y  xy  1  x dx + y dy log ( x2 + y2 ) = 2  x 2 + y2  1  x dy + y dx (viii) d  −  =  xy  x 2 y2 (vii) d

 e x  ye x dx − e x dy (ix) d   =  y y2 (x) d( x2 + y2 ) =

x dx + y dy x 2 + y2

1 x + y  x dy − y dx (xi) d  log  = 2 x − y x 2 − y2

190

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 The order of the differential equation whose general solution is given by y = c1 e 2 x + c 2 + c 3 e x + c 4 sin( x + c 5 ), is (a) 5

(b) 4

(c) 3

Then, y( −3) is equal to

(d) 2

(a) 3 (c) 1

d y  d y dy  +3 = x 2 log 2 , is 2   dx dx   dx  2

2

(a) 1 (c) 3

10 If

(b) 2 (d) None of these

(b) 2 (d) 0

dy = y + 3 > 0 and y( 0) = 2, then y(log 2) is equal to dx

(a) 5

representing the family of curves y = 2c( x + c ), where c is a positive parameter, are respectively equal to 2

(b) 2, 3 (d) 1, 2

4 The differential equation of all circles in the first quadrant which touch the coordinate axes is of order (a) 1 (c) 3

A + B, r

where A and B are arbitrary constants, is d 2 v 1 dv + =0 dr 2 r dr 2 d v 2 dv (c) + =0 dr 2 r dr (a)

(b)

d 2 v 2 dv − =0 dr 2 r dr

(d) None of these

x dy − y dx = 0, is x2 + y2

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

variable radii and a fixed centre at (0, 1) variable radii and a fixed centre at (0, − 1) fixed radius 1 and variable centre along the X-axis fixed radius 1 and variables centre along the Y-axis

dy ax + 3 = dx 2y + f

represents a circle, then the value of a is (b) − 2

(a) 2

(d) − 4

(c) 3

dy dy 13 Solution of the differential equation y − x is = y2 + dx dx (a) Cy = (1 − x)(1 − y) (c) Cy = (1 + x)(1 − y)

(b) Cx = (1 + x)(1 − y) (d) Cx = (1 − x)(1 + y)

 dy   = ax + by , is  dx 

ae −by + be ax + C = 0 aeby + be ax + C = 0 aeby + be − ax + C = 0 None of the above

15 The solution of the differential equation dy = e x − y + x 2e − y is dx x3 +C 3 y x 3 (c) e = e + x + C (a) e y = e x +

(d) None of the above

7 The differential equation of the family of parabolas with focus at the origin and the X -axis as axis, is 2

determines family

of circles with,

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

 x2 + y2 + C (a) y = x tan   2   2 2  x + y + C (b) x = y tan   2   C − x2 − y2  (c) y = x tan   2  

dy dy (a) y   + 4 x = 4y  dx  dx 2 dy dy (c) y   + y = 2 xy  dx  dx

1− y dy = dx y

14 Solution of the equation ln 

6 The solution of the differential equation x dx + y dy +

(d) 7 2

12 If the solution of the differential equation

(b) 2 (d) None of these

5 The differential equation of the family of curves v =

(c) − 2

(b) 13

11 The differential equation

3 Order and degree of the differential equation,

(a) 1, 3 (c) 2, 4

(b) 3 y12 y 2 − (1 + y12 )y 3 = 0 (d) 3 y1 y 22 − (1 + y12 )y 3 = 0

9 If x dy = y (dx + ydy ), y (1) = 1 and y ( x ) > 0.

2 The degree of the differential equation 2

(a) 3 y1 y 2 − (1 + y12 )y 3 = 0 (c) 3 y1 y 22 + (1 + y12 )y 3 = 0

2

dy dy (b) y   = 2 x −y  dx  dx 2 dy dy (d) y   + 2 xy + y=0  dx  dx

8 The differential equation whose solution is x 2 + y 2 + 2ax + 2by + c = 0, where a, b, c are arbitrary constants, is

16 If ( 2 + sin x )

(b) e y = e x + 2 x + C (d) y = e x + C

dy  π + ( y + 1) cos x = 0 and y( 0) = 1, then y    2 dx

is equal to 1 (a) − 3

j

4 (b) 3

1 (c) 3

JEE Mains 2017

(d) −2 / 3

 

7 2

17 If a curve passes through the point  2,  and has slope 1  1 − 2  at any point ( x , y ) on it, then the ordinate of the  x 

191

DAY point on the curve, whose abscissa is − 2, is j

3 (a) − 2

3 (b) 2

5 (c) 2

JEE Mains 2013 5 (d) − 2

dy 18 Solution of the equation ( x + y )2 = 4, y ( 0) = 0 is dx x + y (a) y = 2 tan−1    2  x + y (c) y = 4 tan−1    2 

19 The solution of

x + y (b) y = 4 tan−1    4  (d) None of these

dy + 1 = e x + y is dx

(a) e − ( x + y ) + x + C = 0 (c) e x + y + x + C = 0

(b) e − ( x + y ) − x + C = 0 (d) e x + y − x + C = 0

20 Solution of the equation xdy = ( y + xf ( y / x ) / f ′( y / x )) dx is (a) f (y / x) = C x , C ∈R

(b) f (y / x) = x + C, C > 0

(c) f (y / x) = C x , C > 0

(d) None of these

dy 21 If x = y (log y − log x + 1), then the solution of the dx equation is x (a) log   = Cy y y (c) x log   = Cy x

y (b) log   = Cx x

22 The general solution of y 2 dx + ( x 2 − xy + y 2 ) dy = 0, is x (a) tan−1   + log y + C = 0 y x (b) 2 tan−1   + log x + C = 0 y

(b) y − 2 x = c 2 x 2 / y (d) None of these

4 5

(b)

2e 2 − 1 / 2

(d)

e2 + 1/ 2

(c)

2 5

IIT 2012

π π (b) y ′  =  4  18 π 4π π2 (d) y ′  = +  3 3 3 3

29 The solution of the equation dy + y tan x = x m cos x , is dx (m + 1) y = x m + 1 cos x + C (m + 1) cos x my = (x m + C ) cos x y = (x m + 1 + C ) cos x None of the above

30 Let the population of rabbits surviving at a time t be

t 2

dP (t ) 1 = P (t ) − 200. dt 2 j

(b) 300 − 200 e (d) 400 − 300 e

JEE Mains 2014 −

t 2

−

t 2

(d)

j

( xy 5 + 2y ) dx − xdy = 0, is (a) 9x + 4 x 9 y 4 = 9y 4 C (c) x 8 (9 + 4 y 4 ) = 10y 4 C 8

4 5

JEE Mains 2013

(b) 9x 8 − 4 x 9 y 4 − 9y 4 C = 0 (d) None of these

32 Let Y = y ( x ) be the solution of the differential equation

dy  π + y cos x = 4x , x ∈( 0, π ). If y   = 0, then  2 dx j JEE Mains 2018 y( π / 6) is equal to sin x

(a) −

26 The equation of the curve passing through the origin and satisfying the differential equation dy (1 + x 2 ) + 2xy = 4x 2 is dx

j

2

31 The solution of differential equation

satisfies the differential equation y (1 + xy )dx = xdy , then  1 f  −  is equal to  2 j JEE Mains 2016 (b) −

π π (a) y   =  4 8 2 π π2 (c) y   =  3 9 2

t

25 If a curve y = f ( x ) passes through the point (1,−1) and

2 5

y ′ − y tan x = 2x sec x and y( 0) = 0, then

(c) 600 − 500 e 2

24 If ( x 2 + y 2 )dy = xydx and y ( x 0 ) = e, y (1) = 1, then x 0 is

(a) −

28 If y ( x ) satisfies the differential equation

(a) 400 − 300 e

is

e2 − 1 / 2

1 sin 2 x + C 2 1 (b) y (1 + x 3 ) = Cx + sin 2 x 2 1 (c) y (1 + x 3 ) = Cx − sin 2 x 2 x 1 (d) y (1 + x 3 ) = − sin 2 x + C 2 4 (a) y (1 + x 3 ) = x +

If P( 0) = 100, then P (t ) is equal to

23 Solution of the differential equation x 2dy + y ( x + y )dx = 0

(c)

dy 3x 2 sin2 x , is + y = 3 dx 1 + x 1+ x3

governed by the differential equation

(c) log(y + x 2 + y 2 ) + log y + C = 0 x (d) sin−1   + log y + C = 0 y

(a) e 3

(b) 3 (1 + x 2 )y = 2 x 3 (d) 3 (1 + x 2 ) y = 4 x 3

27 The solution of the differential equation

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

x (d) y log   = Cx y

(a) y + 2 x = c 2 x 2 y (c) y + 2 x = c 2 x 2 / y

(a) (1 + x 2 )y = x 3 (c) (1 + x 2 )y = 3 x 3

8 8 π 2 (b) − π 2 9 3 9

(c) −

4 2 π 9

(d)

4 π2 9 3

33 Statement I The slope of the tangent at any point P on a parabola, whose axis is the axis of x and vertex is at the origin, is inversely proportional to the ordinate of the point P. Statement II The system of parabolas y 2 = 4ax satisfies a differential equation of degree 1 and order 1. j

JEE Mains 2013

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

192 (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

34 Statement I The elimination of four arbitrary constants in y = (c1 + c 2 + c 3 ec 4 )x results into a differential equation dy of the first order x = y. dx

Statement II Elimination of n independent arbitrary constants results in a differential equation of the nth order. (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

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 The differential equation which represents the family of curves y = c1ec 2 x where c1 and c 2 are arbitrary constant, is (a) y ′′ = y ′y (c) yy ′′ = (y ′)2

(b) yy ′′ = y ′ (d) y ′ = y 2

2 The population p ( t ) at time t of a certain mouse species dp(t ) = 0.5 (t ) − 450. If dt p( 0) = 850, then the time at which the population becomes zero is satisfies the differential equation

(a) 2 log18 1 (c) log18 2

(b) log9 (d) log 18

 π  6

3 A curve passes through the point 1,  . Let the slope of y y + sec   , x > 0. x x Then the equation of the curve is j JEE Advanced 2013 the curve at each point (x, y) be 1 y (a) sin  = log x + x 2 y (b) cosec  = log x + 2 x 2y  (c) sec  = log x + 2  x  2y  1 (d) cos  = log x +  x  2

1

1 ey (d) 1 + − y e

6 The equation of the curve y = f ( x ) passing through the origin which satisfies the differential equation dy = sin(10x + 6y ) is dx 1  −1 5 tan 4 x − 5 x  tan 3 4 − 3 tan 4 x  1  −1 5 tan 4 x (b) y = tan + 5 x  3 4 − 3 tan 4 x  1  −1 5 tan 4 x  (c) y = tan − 5x  3 4 + 3 tan 4 x  (d) None of the above (a) y =

7 Let I be the purchase value of an equipment andV (t ) be

(a) I −

df ( x ) and g ( x ) is a given non-constant f ′( x ) denotes dx differentiable function on R with g ( 0) = g ( 2) = 0. Then the value of y( 2) is (b) 4 (d) 0

5 Consider the differential equation

1  y dx +  x −  dy = 0. If y(1) = 1, then x is given by  y 2

1 2 ey (b) 4 − − y e

the value after it has been used for t yr. The value V (t ) depreciates at a rate given by differential equation dV (t ) = − k (T − t ), where k > 0 is a constant and T is the dt total life in years of the equipment. Then, the scrap value V (T ) of the equipment is

4 Let y ′( x ) + y ( x )g ′( x ) = g ( x )g ′( x ), y( 0) = 0, x ∈ R , where

(a) 2 (c) 5

1 1 ey (a) 1 − + y e 1 y 1 (c) 3 − + e y e

(c) e − kT

kT 2 2

k (T − t )2 2 1 (d) T 2 − k (b) I −

8 The degree of the differential equation satisfying the relation 1 + x 2 + 1 + y 2 = λ ( x 1 + y 2 − y 1 + x 2 ), is (a) 1 (c) 3

(b) 2 (d) None of these

9 If length of tangent at any point on the curve y = f ( x ) intercepted between the point of contact and X -axis is of length 1, the equation of the curve is

193

DAY (a)

1 − y 2 + ln (1 − 1 − y 2 ) / y = ± x + C

14 Let f be a real-valued differentiable function on R (the set of all real numbers) such that f (1) = 1 . If the y-intercept of the tangent at any point P ( x , y ) on the curve y = f ( x ) is equal to the cube of the abscissa of P, then the value of f ( −3) is equal to

(b) 1 − y 2 − ln (1 − 1 − y 2 ) / y = ± x + C (c)

1 − y 2 + ln (1 + 1 − y 2 ) / y = ± x + C

(d) None of the above

(a) 3

π cos x dy = y (sin x − y ) dx , where, 0 < x < , is 2

f (1) = 1 and lim t→x

(b) y sec x = tan x + C (d) tan x = (sec x + C ) y

1 2x 2 + 3x 3 1 2 (c) − + 2 2x x

(b) ellipse and circles (d) None of these

(b) − (d)

1 4x 2 + 3x 3

1 x

16 Let a solution y = y ( x ) of the differential equation x x 2 − 1 dy − y y 2 − 1 dx = 0 satisfy y( 2) =

12 A curve passes through (2, 0) and the slope of tangent at a point P ( x , y ) is equal to (( x + 1) + y − 3) / ( x + 1). Then j IIT 2004 equation of the curve is 2

2 . 3

π  Statement I y ( x ) = sec  sec−1 x −  .  6

(b) y = x 2 − 2 x (d) None of these

Statement II y ( x ) is given by

13 The solution of the differential equation x3 x5 + + ... dx − dy 3! 5! is = 2 4 x x dx + dy 1+ + + ... 2! 4! x+

(a) 2 ye 2 x = Ce 2 x + 1 (c) ye 2 x = Ce 2 x + 2

t 2f ( x ) − x 2f (t ) = 1 for each x > 0. Then f ( x ) t −x

(a)

(1 − x 2 )y ′ + xy = ax are

(a) y = x 2 + 2 x (c) y = 2 x 2 − x

(d) 1

is

11 The curves satisfying the differential equation (a) ellipse and parabola (c) ellipse and hyperbola

(c) 9

15 Let f ( x ) be differentiable in the interval ( 0, π ) such that

10 The solution of differential equation

(a) sec x = (tan x + C ) y (c) y tan x = sec x + C

(b) 6

1 2 3 1 = − 1− 2 . y x x

(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) 2 ye 2 x = Ce 2 x − 1 (d) None of these

ANSWERS SESSION 1

1. 11. 21. 31.

SESSION 2

1. (c) 11. (c)

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

2. 12. 22. 32.

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

2. (a) 12. (b)

3. 13. 23. 33.

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

3. (a) 13. (b)

(a) (a) (a) (a)

5. (c) 15. (a) 25. (d)

6. (c) 16. (c) 26. (d)

7. (b) 17. (a) 27. (d)

8. (d) 18. (a) 28. (a)

9. (a) 19. (a) 29. (a)

10. (d) 20. (c) 30. (a)

4. (d) 14. (c)

5. (d) 15. (a)

6. (a) 16. (c)

7. (a)

8. (a)

9. (a)

10. (a)

4. 14. 24. 34.

194

Hints and Explanations SESSION 1 1 Given,

y = c 1 ⋅ e 2 x + c 2 + c 3 e x + c 4 sin( x + c 5 ) = c 1 ⋅ e c 2 e 2x + c 3 e x + c 4 (sin x cos c 5 + cos x sin c 5 )

= Ae 2 x + c 3 e x + B sin x + D cos x Here, A = c 1 e c 2 , B = c 4 cos c 5 and D = c 4 sin c 5 Since, equation consists four arbitrary constants. So, the order of differential equation is 4.

On integrating, we get y 1 2 C ( x + y 2 ) + tan −1   =  x 2 2 C − x2 − y 2 −1  y  ⇒ = tan    x 2 ∴

3 Given, y 2 = 2c ( x + c )

7 Equation of family of parabolas with focus at (0, 0) and axis as X -axis is y 2 = 4 a( x − a) …(i) On differentiating Eq. (i) w.r.t. x, we get 2 yy 1 = 4 a

On differentiating both side w.r.t. x, we get dy dy 2y = 2c ⇒ c = y dx dx On putting in Eq. (i), 3 /2 dy dy y 2 = 2 xy + 2 y 3 /2    dx  dx 3

dy dy  ⇒ 8 y   =  y 2 − 2 xy   dx   dx 

2

3

Which is the differential equation of order one and degree 3.

4 Clearly, the equation of family of circle which touch both the axes is ( x − a)2 + ( y − a)2 = a2 , where a is a parameter. Since, there is only one parameter, therefore order of differential equation representing this family is 1. 5 We have, v = A + B , where A and B are r parameters On differentiating twice w.r.t. r, we get dv −A ... (i) = 2 ⇒ dr r d 2v 2 A and ... (ii) = 3 dr 2 r Now, on substituting the value of A from Eq. (i) in Eq. (ii), we get d 2v 2 dv  −2 dv = 3  − r 2  = dr 2 r  dr  r dr d 2 v 2 dv + = 0, which is the ⇒ dr 2 r dr required differential equation. xdy − y dx 6 We have, x dx + y dy + =0 x2 + y 2 ⇒

yy 1   y 2 = 2 yy 1  x −   2 

∴ ...(i)

y 1 d ( x2 + y 2 ) + d  tan −1  = 0  2 x

⇒

 C − x2 − y 2  y = x tan   2  

2 Since, the equation is not a polynomial in all differential coefficients, so its degree is not defined.

11 Clearly,

⇒ ⇒

y = 2 x y 1 − y y 12 y y 12 = 2 x y 1 − y

8 x2 + y 2 + 2ax + 2by + c = 0

...(i)

On differentiating Eq. (i) three times, we get 2 x + 2 yy 1 + 2a + 2by 1 = 0 ⇒ x + yy 1 + a + by 1 = 0 ...(ii) ...(iii) 1 + y 12 + yy 2 + by 2 = 0 ...(iv) 3 y 1 y 2 + yy 3 + by 3 = 0 On eliminating b from Eqs. (iii) and (iv), we get 3 y 1 y 22 − y 3 − y 12 y 3 = 0

9 We have, x dy = y (dx + y dy ), y > 0 ∴

x dy − y dx = dy y2

⇒

x =− y +C y

Now, ∴

y(1) = 1 ⇒ C = 2 x + y =2 y

For x = − 3, − 3 + y = 2 y ⇒ y2 − 2 y − 3 = 0 ⇒ ( y + 1) ( y − 3) = 0 ∴ y =3 [Q y > 0] dy 10 Here, = y + 3 > 0 and y(0) = 2 dx dy ⇒ ∫ y + 3 = ∫dx 2

⇒ log e | y + 3 | = x + C But y(0) = 2 ∴ log e |2 + 3 | = 0 + C ⇒ C = log e 5 ⇒ log e | y + 3 | = x + log e 5 When x = log e 2, then ⇒ log e | y + 3 | = log e 2 + log e 5 = log e 10 ∴ y + 3 = 10 ⇒ y = 7

y 1 − y2

dy = ∫ dx

− 1 − y2 = x + C

⇒ ( x + C )2 + y 2 = 1 Hence, the centre is (− C , 0) and radius is 1. dy ax + 3 12 We have, = dx 2 y + f ⇒ (ax + 3) dx = (2 y + f ) dy On integrating, we obtain x2 a⋅ + 3 x = y 2 + fy + C 2 a ⇒ − x2 + y 2 − 3 x + fy + C = 0 2 a This will represent a circle, if − = 1 2 [Qcoefficient of x2 = coefficient of y 2 ] ⇒ a = −2 dy 13 ( x + 1) = y − y 2 dx dy dx ⇒ = y (1 − y ) x + 1 1 1  dx ⇒  +  dy = y 1 − y x+1  ⇒ log y − log(1 − y ) = log( x + 1) + log C y ⇒ = C ( x + 1) 1− y ⇒

[integrating]

∫

( x + 1)(1 − y ) = Cy .

dy = e ax .e by dx ⇒ e −by dy = e axdx 1 1 − e −by = e ax + C ⇒ b a ⇒ be ax + ae −by + C ′ = 0

14 We have,

[C ′ = abC ]

15 We have, dy = e x − y + x2e − y = e − y ( x2 + e x ) dx On separating the variables, we get e y dy = ( x2 + e x )dx On integrating both sides, we get x3 ey = + ex + C 3

16 We have, dy + ( y + 1)cos x = 0 dx dy cos x =− dx y+1 2 + sin x dy cos x = − dx y + 1 ∫ 2 + sin x

(2 + sin x ) ⇒ ⇒

∫

⇒ ln( y + 1) = − ln(2 + sin x ) + ln C ⇒ ( y + 1)(2 + sin x ) = C Now, y(0) = 1

195

DAY ⇒ (2)(2 + 0) = C ⇒ C =4 Thus, ( y + 1) (2 + sin x ) = 4 π π Now, at x = , ( y + 1) 2 + sin  = 4  2 2

1 ⇒ dy =  1 − 2  dx  x  1 [integrating] ⇒ y = x+ +C x Since, the curve passing through the 7 point  2,  .  2 7 1 i.e. =2+ + C ⇒ C =1 2 2 1 y = x+ +1 …(i) ∴ x Now, at x = − 2 1 Ordinate, y = − 2 − + 1 = − 3 / 2 2 2 dy 18 We have, ( x + y ) =4 dx dy dv On putting x + y = v and 1 + , = dx dx we get dv v 2  − 1 = 4 ⇒  dx  2 dv v = v 2 + 4. ⇒ dx v2 + 4 − 4 ∴ dv = dx v2 + 4 ⇒ v − 2 tan −1 (v / 2) = x + C  x+ y = x+ C ⇒ x + y − 2 tan −1    2  ∴

y(0) = 0 ⇒ C = 0. x+ y y = 2 tan −1    2 

dy 19 We have, + 1 = e x + y dx On putting x + y = z and 1 +

dy dz = , dx dx

we get dz = ez dx ⇒ e − zdz = dx On integrating both sides, we get e −z = x+C −1 ⇒ x + e −( x + y ) + C = 0 dy y f ( y / x) 20 = + dx x f ′( y / x ) dy dv On putting y / x = v and =v+ x , dx dx we get ∴

⇒ log f (v ) = log x + log C ,C > 0 ⇒ f ( y / x ) = C x ,C > 0. dy  y   y  =   log   + 1   x dx  x    Put y = tx, dy dt and =t + x dx dx dt Then, we get t + x = t log t + t dx dt dx ⇒ = t log t x ⇒ ∴

log log t = log x + log C y log   = Cx  x

22 Given, dx + dy

x2 − xy + y 2 =0 y2 2

⇒ Put and

 x  x dx +   −   + 1= 0 dy  y  y x v = or x = v y y dx dv =v + y dy dy

Then, we get v + y ⇒

1 + log v = − log x + C 2v 2 2 y x ⇒ 2 + C = log + log x = log y . 2y x

⇒−

21 Given,

( y + 1)(2 + 1) = 4 4 1 y = −1 = 3 3 dy 1 17 Given, =1− 2 dx x

Now,

f ′(v ) dx dv = f (v ) x

dv + v2 − v + 1 = 0 dy

dy dv + =0 v2 + 1 y

Now, ⇒ and ⇒ ⇒

25 We have, y (1 + xy ) dx = xdy dy y (1 + xy ) y ⇒ = = + y2 dx x x dy dv On putting y = vx and =v+ x , dx dx we get dv v+ x = v + v2 x 2 dx dv ⇒ = xdx v2 dv ⇒ ∫ v 2 = ∫ xdx 1 x2 ⇒ − = +C v 2 x x2 ⇒ − = +C 2 y Put (1, −1 ), then

∴

dy y( x + y ) + =0 dx x2 On putting y = vx, we get dv v+ x + v (1 + v ) = 0 dx dx dv or + =0 x v (v + 2) dx 1  1 1  or +  −  dv = 0 x 2  v v + 2

∴

On integrating, we get 1 log x + [log v − log(v + 2)] + log C = 0 2 or log(v + 2) = log x2 vC 2 or y + 2 x = C 2 x2 y . dy xy 24 Given, = 2 dx x + y2 dy dv Put y = vx ⇒ = v + x , we get dx dx xdv v ⇒ v+ = dx 1 + v2 dv v v3 ⇒ x = −v = − 2 dx 1 + v 1 + v2 2 1+ v dx dv = − ⇒ v3 x

1 2 x2 1 = + 2 2 1 4 − ,y = 2 5 4 = 5

C =

⇒ tan −1 (v ) + log y + C = 0 [integrating] −1  x  ∴ tan   + log y + C = 0  y

23 We have

x = 1, y = 1 C = −1 / 2 x = x0, y = e x 02 1 − =1 2e 2 2 x 02 = 3e 2 .

−

x y

Now, put x = 1 f  −   2

dy + 2 xy = 4 x 2 dx dy 2x 4x 2 + y = 2 dx 1 + x 1 + x2

26 Given, (1 + x 2 ) ⇒

∫

2x

dx

∴ IF = e 1 + x = e log (1 + x ) = 1 + x 2 and the solution is y ⋅ (1 + x 2 ) = ∫ 4 x 2 dx + C 2

y (1 + x 2 ) =

2

4 x3 +C 3

Now, x = 0, y = 0 ⇒ C =0 4 x3 y (1 + x 2 ) = ∴ 3 ⇒ 3 y (1 + x 2 ) = 4 x3

27 We have,

dy 3 x2 sin2 x + y = 3 dx 1 + x 1 + x3

Since, it is a linear equation with 3 x2 P = 1 + x3 ∴IF = e ∫

P dx

= e log (1 + x

3

)

= 1 + x3

196 and the solution is sin2 x y (1 + x3 ) = ∫ (1 + x3 ) dx 1 + x3 1 − cos 2 x =∫ dx 2 1 sin 2 x +C ∴ y (1 + x3 ) = x − 2 4 dy 28 Given, − y tan x = 2 x sec x, y (0) = 0 dx − tan xdx IF = e ∫ = e − log sec x ∴

IF = cos x cos x ⋅ y = ∫ 2 x sec x ⋅ cos xdx

⇒ cos x ⋅ y = x2 + C ⇒ y (0) = 0 ⇒ C = 0 ∴ y = x2 sec x and y ' = 2 x ⋅ sec x + x2 sec x ⋅ tan x π π2 π2 y   = ⋅ 2=  4  16 8 2 π π π2  y ′  = ⋅ 2 + ⋅ 2  4 2 16 π π2 2 π2 y   = ⋅2 =  3 9 9 π π π2  y ′  = 2 ⋅ 2 + ⋅2⋅ 3  3 3 9 =

4 π 2 π2 3 + 3 9

29 This is the linear equation of the form dy + Py = Q . dx where, P = tan x and Q = x m cos x Now, integrating factor (IF) P dx tan x dx = e∫ = e∫ = e log s ec x = sec x and the solution is given by, P dx P dx = Q ⋅e ∫ dx + C y ⋅e ∫ ⇒ y ⋅ sec x = ⇒

y sec x =

∫ ∫x

m

⋅ cos x ⋅ sec xdx + C

xm +1 +C m+1

∴ (m + 1)y = x m + 1 cos x + C (m + 1) cos x

30 Given differential equation is dP 1 − P (t ) = − 200, which is a linear dt 2 differential equation. −1 Here, P ( t ) = and Q ( t ) = − 200 2 1  t − ∫ −   dt Now, IF = e  2  = e 2 and the solution is P ( t ). IF = ∫ Q (t ) IF dt + K P ( t )⋅ e P ( t )⋅ e

−

−

t 2 t 2

−

= − ∫ 200 e = 400 e

−

t 2

dt + K

t 2

+ K t

⇒ P (t ) = 400 + Ke 2

If P(0) = 100, then K = − 300 t

⇒ P (t ) = 400 − 300 e 2

31 We have, ( xy 5 + 2 y )dx = xdy dy − 2 y = xy 5 dx dy 2 y ⇒ − = y5 dx x dy 2 y −4 ⇒ y −5 − =1 dx x Put, y −4 = t dt −5 dy ⇒ −4 y = dx dx dy −1 dt ⇒ y −5 = dx 4 dx From Eqs. (i) and (ii), we get 1 dt 2t − − =1 4 dx x 8t dt ⇒ + = −4 dx x ⇒

x

8

Now, IF = e ∫ x

dx

…(i)

34 Let c 1 + c 2 + c 3 e c = A

= e 8 log x = x 8

x8 4 ⋅ x9 =− +C 4 y 9

⇒ 9 x 8 + 4 x 9 ⋅ y 4 = 9 y 4C dy 32 We have, sin x + y cos x = 4 x dx ⇒ sin xdy + y cos xdx = 4 xdx ⇒ d ( y sin x ) = 4 xdx On integrating both sides, we get y sin x = 2 x2 + C π Since, it is passes through  ,0 2  π2 + C ⇒ C = − π2 / 2 2 ⇒ y sin x = 2 x2 − π2 / 2 π2 ⇒ y = 2 x2cosec x − cosec x 2  π2  π π2 ⇒ y( π / 6) = 2  cosec − cosec π/6 6 2  36 

∴

0=

 π2  π2 = 2  2 − ⋅2 2  36  =−

[constant]

4

…(ii)

and the solution is t ⋅ x 8 = ∫ (−4)x 8dx + C ⇒

Statement II y 2 = 4 ax …(i) dy ⇒ 2y = 4a dx y dy [from Eq. (i)] ⇒ a= ⋅ 2 dx y dy ⇒ y2 = 4 x ⋅ ⋅ 2 dx dy ⇒ y 2 = 2 xy dx dy ⇒ y = 2x ⋅ , dx which has order = 1 and degree = 1

8 π2 9

33 Statement I Let the equation of parabola whose axis is the axis of x and vertex at the origin is y 2 = 4 ax dy ⇒ 2y = 4a dx dy 2a ⇒ = dx y dy 1 ⇒ ∝ dx y [where, a → parameter]

Then, y = Ax dy = A dx dy ⇒ y = x dx dy ∴ x = y dx ⇒

SESSION 2 1 Given, y = c 1 e c ⇒

2

x

y ′ = c 1 c 2e c 2 x ⇒ c 2 =

y′ y

 y ′ and y ′′ = c 1 c 22e c 2 x ⇒ y ′′ = y .    y

2

⇒ yy ′′ = ( y ′) 2

2 Given, dp ( t ) = 0.5 p (t ) − 450 dt 2dp ( t ) ⇒ = dt p ( t ) − 900 2dp ( t ) ⇒ ∫ = dt p ( t ) − 900 ∫ p ′ (t ) =

⇒ 2 log | p (t ) − 900 | = t + C To find the value of C, let’s substitute t = 0 and p(0) = 850 ⇒ 2 log | p (0) − 900 | = 0 + C ⇒ C = 2 log |850 − 900 | ⇒ C = 2 log 50 Now, 2 log | p (t ) − 900 | = t + 2 log 50 Now, put p (t ) = 0, then 2 log |0 − 900 | = t + 2 log 50 900 ⇒ t = 2 log = 2 log 18 50

3 Given slope at ( x, y ) is dy y = + sec( y / x ) dx x y Let = t ⇒ y = xt x dy dt and =t+ x dx dx dt Now, t + x = t + sec(t ) dx

197

DAY 1

∫ cos t dt = ∫ x dx sin t = ln x + C sin( y / x ) = ln x + C QThis curve passes through (1, π / 6) 1 sin( π / 6) = ln(1) + C ⇒ C = 2 y 1 Thus, sin   = ln x +  x 2

4 We have, y ′( x ) + y ( x )g ′( x ) = g ( x )g ′( x ) Linear differential equation with integrating factor e g( x ) ⇒ y ( x ) ⋅ e g( x ) = ∫ g ( x ) ⋅ g ′( x ) ⋅ e g( x )dx ⇒ y ( x ) ⋅ e g( x ) = e g( x )(g ( x ) − 1) + C Since, y(0) = 0 and g(0) = 0, therefore C =1 ⇒ y ( x ) = (g ( x ) − 1) + e − g( x ) ⇒ y (2) = (g (2) − 1) + e − g(2 ) = 0, as g(2) = 0. 5 Here, dx + 12 ⋅ x = 13 dy y y [linear differential equation in x]

∫y

1

IF = e

Clearly,

2

dy

=e

−

1 y

Now, complete solution is, x ⋅e

−

1

−

1

x ⋅e

⇒

=

y

1

∫

− 1 ⋅ e y dy y3

∫

− 1 1 ⋅ ⋅ e y dy y y2

=

y

1

Put

1 − =t ⇒ y

⇒

xe

⇒

xe

⇒

xe

⇒

xe

−

1

−

1

y

y

−1

−

y

=

y

−1

1 dy = dt y2

∫ − t ⋅ e dt t

= − {t ⋅ e t − ∫ 1 ⋅ e tdt } + C = − te t + e t + C

1

1

=

1

− 1 −y ⋅e +e y +C y −1

+ e + C [Q y (1) = 1] 1 C =− e 1 1 1 − − 1 −y 1 y xe = ⋅e +e y − y e

⇒

e

⇒ ⇒

=e

−1

1

x=

∴

1 1 + 1 − ⋅e y y e

6 On putting 10 x + 6 y = t and dy dt = , we get dx dx  2 tan t / 2  dt = 6sin t + 10 = 6  + 10 2 dx  1 + tan t / 2 

10 + 6

⇒

2

dz [z = tan(t / 2)] = dx 5z2 + 6z + 5 1 dz ⇒ = ∫ dx 2 5∫  3 2  z +  + (4 / 5)  5 1 5z + 3 tan −1 = x+C ⇒ 4 4

= ∫ (cosecθ − sin θ)dθ

⇒

sec t /2 dt =dx 10 tan2 t /2 + 12 tant / 2 + 10

⇒ 5tan t / 2 + 3 = 4 tan 4( x + C ) ⇒ 5tan(5x + 3 y ) + 3 = 4 tan 4( x + C ) 1 Now, x = 0, y = 0 ⇒ C = tan −1 (3 / 4) 4 Hence, equation of the curve is 25tan 4 x 5tan(5x + 3 y ) = 4 − 3 tan 4 x 1 5tan 4 x ⇒ y =  tan −1 − 5x . 3 4 − 3 tan 4 x 

⇒

dt d {V (t )} = − k (T − t ) dt

∫

T 0

d {V (t )} =

∫

T 0

…(i)

− k (T − t )dt T

(t − T )2  ⇒ V (T ) − V (0) = k   2  0 k 2 ⇒ V (T ) − I = [(T − T ) − (0 − T )2 ] 2 ∴

[Qwhen t = 0, then V (t ) = I] k V (T ) = I − T 2 2

8 On putting x = tan A, and y = tan B in the given relation, we get cos A + cos B = λ (sin A − sin B ) A − B 1 ⇒ tan   =  2  λ 1 ⇒ tan −1 x − tan −1 y = 2 tan −1    λ On differentiating w.r.t. x, we get dy 1 1 − ⋅ =0 1 + x2 1 + y 2 dx dy 1 + y 2 = dx 1 + x2

⇒

Clearly, it is a differential equation of degree 1.

9 y 1 + (dx / dy ) 2 = 1  ⇒ y2 1 +   dy ∴ =± dx ⇒∫

1− y y

2

⇒ C ± x = log

 dx     =1  dy   y 2

1 − y2 dy = ± x + C.

On putting y = sinθ and dy = cos θ dθ, we get cos θ ⋅ cos θ dθ C± x=∫ sin θ

1 − 1 − y2 y

+ 1 − y2 .

cos x dy = y sin x dx − y 2 dx 1 dy 1 ⇒ − tan x = − sec x y 2 dx y 1 1 dy dz Put − =z ⇒ 2 = y y dx dx dz ∴ + (tan x ) z = − sec x dx This is a linear differential equation. tan x dx Now, IF = e ∫ = e log sec x = sec x Hence, the solution is z ⋅ (sec x ) = ∫ − sec x ⋅ sec x dx + C1

10 Given,

⇒

7 Given, d {V (t )} = − k (T − t ) ∴

⇒ C ± x = log cosecθ − cot θ + cos θ

−

1 sec x = − tan x + C1 y

∴ sec x = y (tan x + C ) where C = − C1 . dy 11 We have, (1 − x2 ) + xy = ax dx dy x ax [L. D. E.] ⇒ + y = dx 1 − x2 1 − x2 1

IF = e ∫ =e 2 1  , if − 1 < x < 1 2  =  1− x 1 , if x < −1or x > 1.  2  x −1 If −1 < x < 1, then solution is 1 x y⋅ = a∫ dx + C (1 − x2 )3 /2 1 − x2 a 1 y⋅ = +C 1 − x2 1 − x2 ( x / 1 − x2 )dx

⇒

−

log 1 − x2

y = a + C 1 − x2

⇒ ( y − a) 2 = C 2 (1 − x2 ) ⇒ C 2 x2 + ( y − a)2 = C 2 which is an ellipse. If x < −1 or x > 1, solution is 1 x y⋅ = C + a∫ dx 2 2 x −1 (1 − x ) x2 − 1 x a dx = C + = C − a∫ 2 ( x − 1)3 /2 x2 − 1 ⇒

y = a + C x2 − 1

⇒ ( y − a)2 = C 2 x2 − c 2 ⇒ C 2 x2 − ( y − a)2 = C 2 which represents a hyperbola.

12 Given,

dy ( x + 1) 2 + y − 3 y −3 = = ( x + 1) + dx x+1 x+1 dy dY Putting x + 1 = X , y − 3 = Y , = , dx dX the equation becomes dY Y = X+ dX X

198 dY 1 − ⋅Y = X dX X

or IF = e ∫

( −1 / X )dX

[L. D. E.]

= e − log X = X −1 =

1 X

∴ The solution is 1 1 Y ⋅   = C + ∫ X ⋅   dx = C + X X X ( y − 3) or =C + x+1 ( x + 1) x = 2, y = 0 0−3 ⇒ = C + 2+ 1 2+ 1 ⇒ C = −4 ∴ The equation of the curve is y −3 = x − 3 or y = x2 − 2x. x+1 Now,

3

5

x x x+ + +K dx − dy 3 ! 5! 13 We have, = x2 x4 dx + dy 1+ + +K 2! 4! On applying componendo and dividendo, we get     x3 x5 x2 x4 + + K +  1 + + + K x+ 3! 5! 2! 4!         x3 x4 x2 x4 + + K −  1 + + + K x+ 3! 4! 2! 4!     =

⇒

(dx − dy ) + (dx + dy ) (dx − dy ) − (dx − dy )

  x2 x3 + + K 1 + x + 2! 3!     x2 x3 − 1 − x + − + K 2! 3!  

=

dx 2dx =− dy −2dy

⇒

ex dx =− −e − x dy

⇒ dy = e −2xdx On integrating both sides, we get e −2 x y = + C1 −2 ⇒ 2 ye 2 x = −1 + 2C1 e 2x ⇒ 2ye 2x = Ce 2X − 1, where C = 2C1 dy 14 Equation of tangent is Y − y = ( X − x) dx dy Y-intercept of tangent is y − x dx dy From given condition, y − x = x3 , dx dy 1 ...(i) ⇒ − ⋅ y = − x2 dx x 1 − dx 1 Now, IF = e ∫ x = e − loge x = x and solution is 1 x2 ⋅ y = − ∫ xdx = − +C x 2 − x3 ⇒ f ( x) = + Cx 2 1 3 Now, f (1) = 1 ⇒ C = 1 + = 2 2 − x3 3 + x ∴ f ( x) = 2 2 27 9 − =9 ⇒ f (−3) = 2 2 t 2 f ( x ) − x2 f ( x ) + x2 f ( x ) − x2 f (t ) =1 t−x f ( x ) − f (t )   lim (t + x )f ( x ) + x2   =1 t→ x   t − x   

dy − 2 xy 1 dx = − 4, x4 x where y = f ( x) d  y 1   =− 4 dx  x2  x 1 ⇒ x −2 y = 3 + C 3x 1 2 ⇒ x −2 y = 3 + 3x 3 1 2 x2 . ⇒ y = + 3x 3 ⇒

x2

16 Given, ⇒

y y2 − 1 dy = dx x x2 − 1

∫y

dy y2 − 1

=

∫x

dx x2 − 1

sec −1 y = sec −1 x + C 2 Now, x = 2, y = 3 π π π ⇒ = + C ⇒C = − 6 3 6 π ∴ y = sec  sec −1 x −   6 ⇒

⇒

⇒

 1 1 3 = cos  cos −1 − cos −1  y x 2     3 1 = cos cos −1  + y  2x 

15 lim t→ x

2xf ( x) + x2 (− f ′( x)) = 1

[Q f (1) = 1]

1−

1 x2

⋅ 1− ∴

1 3 1 1 = + 1− 2 y 2x x 2

3  4  

199

DAY

DAY NINTEEN

Unit Test 2 (Calculus)

1 Let f : ( 2, 3) → ( 0, 1) be defined by f ( x ) = x − [ x ], then

(b) increasing for every value of x (c) decreasing for every value of x (d) None of the above

−1

f ( x ) is equal to (a) x − 2

2

∫

(b) x + 1

dx sin x − cos x +

2

(c) x − 1

1 x π (b) tan  +  + C 2 8 2 1 x π (d) − cot  +  + C 2 8 2

(c) e −1

(b) −

(a) − 1

3 4

(c)

4 3

(d) 1

1  (b)  1 −  sq units  2 π 1 (d)  + − 1 sq units 4 2  2

(c) 2

(a)

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

x x2 x2 x2

+ y + y2 − y2 − y2 2

(a) 1 −

x3 is 3

1 1 (a) increasing for x > and decreasing for x < 4 4

(c)

1 60

(d)

3 20

(b) 0 (d) not defined

2 3

a = 0, b =

(b)

8 7

1 1 cos 2x − cos 3x is 2 3 (c)

9 4

(d)

f (b ) − f (a ) = f ′ (c ), if b −a

3 8

1 and f ( x ) = x ( x − 1)( x − 2), then value of c is 2 15 6

(b) 1 +

15

(c) 1 −

21 6

(d) 1 +

21

14 If the sides and angles of a plane triangle vary in such a way that its circumradius remains constant. Then, da db dc is equal to (where, da, db and dc + + cos A cos B cos C are small increments in the sides a, b and c, respectively).

+ xy − x + y = C − xy + x + y = C + 2 xy − x + y = C − 2 xy + x − y = C

8 The function f ( x ) = x 4 −

1 40

13 In the mean value theorem

( 2x − y + 1) dx + ( 2y − x + 1) dy = 0 is 2

(b)

the function f ( x ) = cos x +

(d) None of these

7 The general solution of the differential equation

xn n!

12 The difference between the greatest and least values of

 2 − sin θ  6 The value of ∫ log   d θ is −π /2  2 + sin θ  (b) 1

1 20

(a) −2 (c) 2

π /2

(a) 0

(d) y − 1 −

| x − 1| + | x − 3| at the point x = 2 is

5 The area bounded by y = sin−1 x , x = 1/ 2 and X-axis is 1 (a)  + 1 sq units  2  π (c) sq units 4 2

xn n!

11 The differential coefficient of the function

4 If the normal to the curve y = f ( x ) at the point ( 3, 4)

makes an angle 3 π / 4 with the positive X -axis, then f ′ ( 3) is equal to

(c) y −

x 1/ 4 − x 1/ 5 is x →1 x 3 −1

(a) (d) 1

xn n!

10 The value of lim

is equal to

(b) e 2

x2 x3 xn dy is equal to + + ... + , then 2! 3! n! dx (b) y +

(a) y

n ( n − 1)

3 lim  n→ ∞

(a) e

9 If y = 1 + x +

is equal to

1 x π (a) − tan  +  + C 2 8 2 1 x π (c) cot  +  + C 2 8 2

 n 2 − n + 1   n 2 − n − 1

(d) x + 2

(a) 1

15

∫

1.5 0

(b) 2

(c) 0

(d) 3

2

[ x ] dx , where [.] denotes the greatest integer

function, is equal to (a) 2 +

2

(b) 2 −

2

(c) −2 +

2

(d) −2 −

2

200

16 The integrating factor of the differential equation dy = y tan x − y 2 sec x , is dx (a) tanx

(b) sec x

(a 2 + b 2 ) (u 2 + v 2 ) is equal to

(c) − sec x

(d) cot x

17 The area bounded by the straight lines x = 0 , x = 2 and the curve y = 2x , y = 2x − x 2 is 4 1 − 3 log 2 4 (c) −1 log 2

3 4 + log 2 3 3 4 (d) − log 2 3

(a)

(b)

19 If I1 = ∫ I2 = ∫

k 1− k

k 1− k

(c)

20 If I1 = ∫

and I 2 = ∫

sec 2 z

3 2 (c) 1

(b)

1 sq unit 6

I1 is equal to I2

1 2 (d) None of these

(b)

21 If f and g are continuous functions on [ 0, π ] satisfying f ( x ) + f ( π − x ) = g ( x ) + g ( π − x ) = 1, then π ∫ [f ( x ) + g ( x )] dx is equal to (b) 2 π

(c)

(a) a = 1, b = 2 (c) a = 2, any b

(d)

3π 2

 2x − 3 ⋅ [ x ], x ≥ 1   πx  sin   , x < 1   2 

x→

(b)

f ′ (x) (b) 2y + 1

f ′ (x) (c) 1− 2 y

π 2

is

(c)

π 4

(d)

π 8

π 

dy is equal to dx

x, then h is increasing whenever f is increasing h is increasing whenever f is decreasing h is decreasing whenever f is increasing Nothing can be said in general

(b) 4

f ′ (x) (d) 4 + 2y

(c) 5

(d) 6

 sin x  π   , then f ′  3  is equal to 1 + cos x    

32 If f ( x ) = tan−1  (a)

24 If h( x ) = f ( x ) − (f ( x ))2 + (f ( x ))3 for every real numbers (a) (b) (c) (d)

t dt

sin( 2x − π )



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

continuous at x = 2 differentiable at x = 1 continuous but not differentiable at x = 1 None of the above

f ′ (x) (a) 2y − 1



x π 2

31 Let f ( x ) =  2 cos  x +   , 0 < x ≤ 2π (where, [.]  4  

(a) 3

(where, [x] denotes the greatest integer ≤ x ) is

23 If y = f ( x ) + f ( x ) + f ( x ) + ... ∞ , then

2

∫

denotes the greatest integer ≤ x ). The number of points of discontinuity of f ( x ) are

22 The function f ( x ) = 

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

(b) a = 2, b = 4 (d) any a, b = 4

30 The value of limπ (a) ∞

π 2

(b) h is differentiable for all x (d) None of these

e 2 x − 1 , x 1

f {x ( 3 − x )} dx , where f is a continuous

(a)

[g (b)]2 − [g (a)]2 2

28 If h ( x ) = min{x , x 2 }, for every real number of x. Then,

x f {x ( 3 − x )} dx

function and z is any real number, then

(d)

27 If y be the function which passes through (1 , 2) having

(c)

(b) 2 I1 = I 2 (d) None of these

2 − tan2 z

[f (b)] − [f (a)] 2

(a) 6 sq units

sin {x (1 − x )} dx , then

sec 2 z

(b) g (b) − g (a) 2

slope ( 2x + 1). Then, the area bounded between the curve and X-axis is

x sin {x (1 − x )} dx and

2 − tan2 z

b d [f ( x )] = g ( x ) for a ≤ x ≤ b, then ∫ f ( x ) ⋅ g ( x ) dx is a dx equal to

26 If

2

(b) 1 sq unit (d) None of these

(a) I1 = 2 I 2 (c) I1 = I 2

(b) (a 2 + b 2 ) e 2 ax (d) (a 2 − b 2 ) e 2 ax

(a) 2e ax (c) e 2 ax

(a) f (b) − f (a)

18 The area of the region bounded by y = | x − 1| and y = 1 is (a) 2 sq units (c) 1 / 2 sq unit

25 If u = ∫ e ax cos bx dx and v = ∫ e ax sin bx dx , then

(b)

1 2

(d) None of these

33 Let f ( x ) be a polynomial function of the second degree. If f (1) = f ( −1) and a1, a 2 , a 3 are in AP, then f ′ (a1 ), f ′ (a 2 ) and f ′ (a 3 ) are in (a) AP (c) HP

(b) GP (d) None of these

34 If y = cos −1(cos x ), then y ′ ( x ) is equal to (a) (b) (c) (d)

1 for all x −1 for all x 1in 2nd and 3rd quadrants −1in 3rd and 4th quadrants

201

DAY 35 In ( −4 , 4) the function f ( x ) = ∫ (a) no extreme (c) two extreme

x −10

(t 4 − 4) e −4t dt has

44 I1 = ∫ sin−1x dx and I 2 = ∫ sin−1 1 − x 2 dx , then (a) I1 = I 2

(b) one extreme (d) four extreme

(c) I1 + I 2 =

36 If f ( x ) = kx 3 − 9x 2 + 9x + 3 is monotonically increasing in each interval, then (a) k < 3 (c) k > 3

(b) k ≤ 3 (d) None of these

2  x + 1 −1 −1 x + tan tan   dx is equal to 2 ∫ −1  x +1 x  3

(a) π (c) 3 π

(b) 2 π (d) None of these

199 + 299 + 399 + ... + n 99 is equal to n→ ∞ n100

38 lim

1 (b) 100

39 If ∫ ( sin 2x + cos 2x ) dx =

1 (c) 99

1 (d) 101

1 sin( 2x − c ) + a, then the 2

value of a and c is

π and a = k 4 π π (b) c = − and a = 4 2 π (c) c = and a is an arbitrary constant 2 (d) None of the above (a) c =

40

cos 2x

∫ (cos x + sin x )

2

dx is equal to

(a) log cos x + sin x + C (c) log(cos x + sin x) + C

1 2 1 (c) k = − 4 (a) k = −

42

(b) log(cos x − sin x) + C 1 (d) − +C cos x + sin x

cos 4x + 1 dx = k cos 4x + C, then cot x − tan x

41 If ∫

∫

π /4 0

πx 2

1 8 (d) None of these (b) k = −

dx

∫ sin( x − a ) sin( x − b ) is equal to (a)

sin(x − a) 1 log +C sin(a − b) sin(x − b)

(b)

sin(x − a) −1 log +C sin(a − b) sin(x − b)

(c) log sin(x − a) ⋅ sin(x − b) + C sin(x − a) (d) log +C sin(x − b)

( x 2 + 1) 43 ∫ e dx is equal to ( x + 1) 2 x

 x − 1 x (a)  e + C  x + 1

 x + 1 (b) e x   +C  x − 1

(c) e x (x + 1) (x − 1) + C

(d) None of these

π 2I1

(d) I1 + I 2 =

π 2

dx is equal to cos 4 x − cos 2 x sin2 x + sin4 x

π 2 π (c) 3 (a)

37 The integral

9 (a) 100

45

(b) I 2 =

(b)

π 4

(d) None of these

46 If f ( x ) = | x − 1| + | x − 3 | + | 5 − x | , ∀ x ∈ R . If f ( x ) is increases, then x belongs to (a) (1, ∞)

47 The value of ∫

(b) (3, ∞)

(c) (5, ∞)

(d) (1, 3)

dx is 1/ 2 (1 + x ) − (1 + x )1/ 3

(a) 2 λ1 / 2 + 3 λ1 / 3 + 6λ1 / 6 + 6 ln | λ1 / 6 − 1| + C (b) 2 λ1 / 2 − 3 λ1 / 3 + 6λ1 / 6 + 6 ln | λ1 / 6 − 1| + C (c) 2 λ1 / 2 + 3 λ1 / 3 − 6λ1 / 6 + 6 ln | λ1 / 6 − 1| + C (d) 2 λ1 / 2 + 3 λ1 / 3 + 6λ1 / 6 − 6 ln | λ1 / 6 − 1| + C (where, λ = 1 + x)

48 The value of ∫ (a) − (b)

x dx is x −1

x1/ 4 − 1 4 3/4 x + 4 x1 / 4 + 2 ln 1 / 4 +C 3 x +1

x1/ 4 − 1 4 3/4 x + 4 x1 / 4 + 2 ln 1 / 4 +C 3 x +1

(c) − (d)

4

x1/ 4 − 1 4 3/4 x − 4 x1 / 4 + 2 ln 1 / 4 +C 3 x +1

x1/ 4 − 1 4 3/4 x − 4 x1 / 4 + 2 ln 1 / 4 +C 3 x +1

49 The value of n for which the function  (( 5)x − 1)3 , x ≠0   x x2 f ( x ) =  sin   ⋅ log 1 +  n  3  3 15 (log 5) , x =0 may be continuous at x = 0, is (a) 5

(b) 3

(c) 4

(d) 2

50 The longest distance of the point (4, 0) from the curve 2x (1 − x ) = y 2 is equal to (a) 3 units (c) 5 units

(b) 4.5 units (d) None of these

51 The normal to the curve 5x 5 − 10x 3 + x + 2y + 4 = 0 at P ( 0, − 2) meets the curve again at two points at which equation of tangents to the curve is equal to (a) y = 3 x + 2 (c) 3 y + 2 x + 7 = 0

(b) y = 2 (x − 1) (d) None of these

202

52

∫

2 −1

{|x − 1| + [ x ]} dx , where [ x ] is greatest integer is equal

58 Statement I The tangents to curve y = 7x 3 + 11 at the points, where x = 2 and x = − 2 are parallel. Statement II The slope of the tangents at the points, where x = 2 and x = − 2, are equal.

to (a) 8

(b) 9

(c)

5 2

(d) 4

59 Statement I The derivative of

53 The equation of the curve passing through the point

 1 π  and having slope of tangent at any point ( x , y ) as  ,   2 8 ( y / x ) − cos 2 ( y / x ) is equal to e (a) y = x tan−1  log   2x  2x  (c) y = x 2 tan−1    e 

e (b) x = y tan−1  2  x  (d) None of these

54 If x dy = y (dx + y dy ), y (1) = 1 and y ( x ) > 0, then y ( − 3) is equal to (a) 3

55

∫

π /4 0

(b) 2

(c) 1

(d) 0

sin x + cos x dx is equal to 9 + 16 sin 2x

1 log 3 20 1 (c) log 5 20

(a)

e

56 If f ( x ) =  (a) 0

Statement II The solution of a linear equation is obtained by multiplying with its integrating factor. x 2 −1 x2 +1 61 Statement I ∫ 2 dx = sec −1 +C 4 x 2 ( x + 1) x + 1

(b) log 3

3 sin x , | x | ≤ 2 , then ∫ f ( x ) dx is equal − 2 2, otherwise

(c) 2

2

(b)

π is 4

2 +1

(c)

2 −1

(d)

t t −a 2

=

1 t sec −1 +C a a

| x − 1| + | x − 2 | + | x − 3 | is 2. Statement II The function | x − 1| + | x − 2 | + | x − 3 | is differentiable on R ~ {1, 2}.

63 Statement I If f ( x ) = max {x 2 − 2x + 2,| x − 1| }, then the greatest value of f ( x ) on the interval [0, 3] is 5. Statement II Greatest value, f ( 3) = max {5 , 2) = 5

(d) 3

57 The area bounded by curves y = cos x and y = sin x and ordinates x = 0 and x =

dt

62 Statement I The absolute minimum value of

(d) None of these

(b) 1

 3 cos t 2dt , ( x > 0) at x = 1 is   cos 1.  2

d φ (x ) f (t ) dt = f ( φ ( x )) − f ( ψ ( x )) dx ∫ ψ ( x ) 60 Statement I The solution of the equation dy 1 x + 6y = 3xy 4 / 3 is y ( x ) = . dx ( x + cx 2 ) 3

cos x



x 1/ x

Statement II

Statement II ∫

to

(a)

f (x ) = ∫

2 ( 2 − 1)

Direction (Q. Nos. 58-66) Each of these questions contains two statements : Statement I (Assertion) and Statement II (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes ( a), (b), (c) and (d) given below. (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

64 Statement I The point of contact of the vertical tangents to x = 2 − 3 sin θ, y = 3 + 2 cos θ are ( − 1, 3) and ( 5 , 3). Statement II For vertical tangent, dx /d θ = 0

65 Let f : R → R be differentiable and strictly increasing function throughout its domain. Statement I If | f ( x ) | is also strictly increasing function, then f ( x ) = 0 has no real roots. Statement II At ∞ or − ∞ f ( x ) may approach to 0, but cannot be equal to zero.

66 Statement I The area by region | x + y | + | x − y | ≤ 2 is 4 sq units. Statement II Area enclosed by region | x + y | + | x − y | ≤ 2 is symmetric about X-axis.

ANSWERS 1 11 21 31 41 51 61

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

2 12 22 32 42 52 62

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

3 13 23 33 43 53 63

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

4 14 24 34 44 54 64

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

5 15 25 35 45 55 65

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

6 16 26 36 46 56 66

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

7 17 27 37 47 57

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

8 18 28 38 48 58

(a) (b) (a) (b) (b) (a)

9 19 29 39 49 59

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

10 20 30 40 50 60

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

203

DAY

Hints and Explanations 1 Given, f : (2, 3) → (0, 1)

=

and f ( x) = x − [ x] ∴ f ( x) = y = x − 2 ⇒ x = y + 2 = f −1 ( y ) −1 ⇒ f ( x) = x + 2

=

dx

π π   2  sin x ⋅ sin − cos x ⋅ cos + 1  4 4  1 dx = ∫  x π 2 1 − cos 2  +  2 8 1 π 2  x = ∫ cosec  2 + 8  dx 2 2

 x π − cot  +  2 1 8 = +C 1/2 2 2 =−

1 2

n (n − 1 )

n( n − 1 )

−1 dy dx

3π −1  dy  = ⇒   =1  dx  (3, 4 ) 4  dy     dx  (3, 4 )

∴

f ′ (3) = 1

= Area of rectangle OABC − Area of curve OBCO p 0, 2

O

B A

1 x= Ö2

−1

f (θ) is an odd function of θ . I =∫

∴

π /2

−π /2

 2 − sin θ  log   dθ = 0  2 + sin θ 

x = X + h, y = Y + k dY 2 X − Y + 2 h − k + 1 = dX X − 2 Y + h − 2k − 1

Again, put 2 h − k + 1 = 0 and h − 2k − 1 = 0 On solving, h = − 1, k = − 1 dY 2 X − Y ∴ = dX X − 2Y On putting Y = vX , we get dv 2 X − vX 2−v ⇒ v+ X = = dX X − 2 vX 1 − 2 v ⇒X ∴

∴

dv 2 − 2 v + 2 v 2 (v − v + 1) = = dX 1 − 2v 1 − 2v 2

X

x3 3 ⇒ f ′ ( x) = 4 x 3 − x 2 For increasing, 4 x 3 − x 2 > 0 ⇒ x 2 (4 x − 1) > 0 Therefore, the function is increasing for x > 1 / 4. 1 Similarly, decreasing for x < . 4

8 f ( x) = x 4 −

9 Given, x2 x3 xn + +K+ 2! 3! n! dy x2 xn −1 ⇒ =0+1+ x+ + ... + dx 2! (n − 1) ! y =1 + x +

⇒

⇒

2

dX (1 − 2 v) = dv X 2 (v2 − v + 1) v2 − v + 1 = t (2 v − 1) dv = dt dt dX =− X 2t

On integrating, log X = log t −1 /2 + log C1 /2 ∴ X = t −1 / 2 C 1 / 2 ⇒ X = (v2 − v + 1)−1 /2 C1 /2 2 ⇒ X (v2 − v + 1) = C

Y

p 4 C

⇒ x 2 + y2 − xy + x + y = C

 2 − sin θ  = − log   = − f (θ)  2 + sin θ 

Put ⇒

5 Required area

0,

⇒ y2 + x 2 − xy + x + y = C − 1

+ [cos y] π0 / 4

 2 − sin θ    2 + sin θ 

Put

  1 1 +  n ( n − 1 ) e  = lim  = −1 = e 2 n( n − 1 ) n→ ∞ e   1 1 −  n ( n − ) 1  

⇒ tan

⇒ ( y + 1) 2 − ( y + 1) ( x + 1) + ( x + 1) 2 =C

sin y dy

6 f (− θ) = log 

⇒

4 Slope of the normal =

π /4 0

(2 x − y + 1) dx + (2 y − x + 1) dy = 0 dy 2 x − y + 1 , ⇒ = dx x − 2y − 1

n (n − 1 )

 n (n − 1) + 1  = lim   n → ∞  n (n − 1) − 1 

4 2

∫

7. Given,

 x π cot  +  + C 2 8

 n2 − n + 1  3 lim  2  n→ ∞  n − n − 1

4 2 π

−

 π  1 = + − 1 sq units 4 2  2

2 Let I =∫

π

 ( y + 1) 2 ( y + 1)  ⇒ ( x + 1) 2  − + 1 2 ( x + 1)  ( x + 1)  =C

dy x n x2 + =1+ x+ +K dx n ! 2! xn −1 xn + + (n − 1) ! n ! dy xn = y− dx n!

1 −3 / 4 1 −4 / 5 x − x x1 / 4 − x1 / 5 4 5 10 lim = lim x →1 x →1 x3 − 1 3 x2 1 1 − 1 4 5 = = 3 60

11 Given, f ( x) = | x − 1| + | x − 3| x 3  4 − 2 x , x < 1  =  2, 1 < x < 3 2 x − 4 , x ≥ 3 

In the neighbourhood of x = 2, f ( x) = 2 Hence, f ′ ( x) = 0

12 The given function is periodic with period 2π . So, the difference between the greatest and least values of the function is the difference between these values on the interval [0, 2π ]. Now, f ′ ( x) = − (sin x + sin 2 x − sin 3 x) = − 4 sin x sin(3 x/ 2) sin( x/2)

204 Hence, x = 0, 2 π /3, π and 2π are the critical points. 1 1 7 Also, f (0) = 1 + − = 2 3 6 13 1  2π  f   =− , f (π ) = –  3 12 6 7 and f (2 π ) = 6 ∴Required difference 7  13  27 9 = − −  = = 6  12  12 4

13 From mean value theorem, f ′ (c ) =

f (b ) − f (a) b −a

Given, a = 0 ⇒ f (a) = 0 1 3 and b = ⇒ f (b ) = 2 8 f ′ ( x) = ( x − 1)( x − 2) + x( x − 2) + x( x − 1) f ′ (c ) = (c − 1)(c − 2) + c (c − 2) + c (c − 1) = c 2 − 3 c + 2 + c 2 − 2c + c 2 − c ⇒ f ′ (c ) = 3 c 2 − 6 c + 2 According to mean value theorem, f (b ) − f (a) f ′ (c ) = b −a (3 /8) − 0 3 ⇒ 3 c2 − 6 c + 2 = = (1 / 2) − 0 4 5 ⇒ 3 c2 − 6 c + = 0 4 6 ± 36 − 15 6 ± 21 ⇒ c= = 2 ×3 6 =1 ±

21 =1 − 6   Q  1 +  

21 6 21   1    ∉  0,   6   2 

∴

da db dc + + cos A cos B cos C = 2 R (dA + dB + dC) = 0 1.5

15 ∴ ∫ = =

∫ ∫

[x2 ] dx

0 1 0 1 0

[x2 ] dx + 0 dx + ∫

a = 2 R sin A On differentiating, we get da = 2 R cos A dA da = 2 R dA cos A db Similarly, = 2 R dB cos B dc and = 2 RdC cos C

2 1

∫

2 1

[x 2 ] dx +

1 dx +

∫

1. 5 2

∫

1. 5 2

[ x 2 ] dx

2 dx

∫

k 1−k

16 The differential equation is dy − y tan x = − y2 sec x dx This is Bernoulli’s equation, which can be reducible to linear equation. On dividing the equation by y2 , we get 1 dy 1 …(i) − tan x = − sec x y2 dx y 1 1 dy dY = Y ⇒− 2 = y y dx dx

Put

Eq. (i) reduces to dY − − Y tan x = − sec x dx dY ⇒ + Y tan x = sec x, which is a dx linear equation. tan x dx Hence, IF = e ∫ = sec x 2

17 A = ∫ [2 x − (2 x − x2 )] dx 0

2

 2x x3  3 4 = − x2 + = − 3  0 log 2 3  log 2

18 y = x − 1, if x > 1 and y = − ( x − 1) ; if x 1

x→ π / 2

X

discontinuous at all integer numbers.

x ≥1

∴ f (0 − ) = f (0 + ) = f (0) = − 1 Also, f is differentiable at x = 0, therefore Lf ′ (0) = Rf ′ (0) f (0 − h) − f (0) lim ⇒ h→ 0 −h f (0 + h) − f (0) = lim h→ 0 h −2 h e − 1 + 1 ⇒ lim   h→ 0   −h

Y

X¢

x ≤0

29 Since, f is continuous at x = 0.

30 y = lim

equation of parabola whose vertex is  −1 −1  V  , .  2 4

31 Using the fact that [ x] is

0 < x 1   x = 1  not exist,

2

1 1  ⇒  x +  = y + , which is an  2 4

x ≥ 1,

35

f ( x) =

∫

x − 10

(t 4 − 4) e −4 tdt

⇒ f ′ ( x) = ( x 4 − 4) e −4 x Now, f ′ ( x) = 0 ⇒ x = ± 2

Now, f ′ ′ ( x) = − 4 ( x 4 − 4) e −4 x + 4 x 3e −4 x At x = 2 and x = − 2 , the given function has extreme value.

36 f ′ ( x) = 3 kx2 − 18 x + 9 = 3 [ kx 2 − 6 x + 3] > 0, ∀x ∈ R ∴ ∆ = b 2 − 4 ac < 0, k > 0 i.e. 36 − 12 k < 0 or k > 3 . 3



 x    x2 + 1 

37 Let I = ∫ tan −1  −1 

 x2 + 1   + tan −1   dx  x 

206

∫ {tan 3

=

−1

−1

= ∫ e x [ f ( x) + f ′ ( x)] dx

 x   2   x + 1 + cot

∫

=

3 −1

−1

π dx = 2 π 2

π   Q tan −1 x + cot −1 x = , ∀ x ∈ R  2  n

38 lim

n→ ∞

∑

r =1

 r 99   100  = n 

∫

1 0

  x −1 2 ∴ f ( x) = x + 1 , f ′ ( x) = ( x + 1) 2      − x 1 = ex   +C  x + 1

 x   2  dx  x + 1 

99

x dx

100 1

x  1 =  = 100 100  0 − cos 2 x sin 2 x 39 I = + + k 2 2 1  π π =  sin 2 x cos − cos 2 x sin  + k  4 4 2 1 π   = sin  2 x −  + k  4 2 π ∴ c = ;a=k 4 (cos x − sin x)(cos x + sin x) 40 ∫ dx (cos x + sin x) 2 cos x − sin x =∫ dx cos x + sin x

44 I1 = ∫ sin −1 x dx. Let sin −1 x = θ ⇒ x = sin θ ⇒ dx = cos θ dθ ∴ I1 = ∫ θ cos θ dθ

= θ sin θ − ∫ sin θ dθ = θ sin θ + cos θ = x sin −1 x + 1 − x 2

+

∴

dx

∫ sin( x − a) sin( x − b )

Put

1 sin{( x − b ) − ( x − a)} = dx sin(a − b ) ∫ sin( x − a) sin( x − b ) 1 = [ cot( x − a) dx sin(a − b ) ∫

and

43

=

1 sin( x − a) log +C sin(a − b ) sin( x − b )

∫

e ( x − 1 + 2) dx ( x + 1) 2 x

2

x −1  2 = ∫ ex  + dx 2  + + x 1 ( x 1 )  

I1 + I2 = x (cos −1 x + sin −1 x) π x = 2

I =

∫

π /4 0

sec2 x sec2 x dx 1 − tan2 x + tan 4 x

Put tan x = t ⇒ sec2 x dx = dt 1 1 + t2 I =∫ 4 dt ∴ 0 t − t2 + 1 1  1 1 + 2  1+ 2 1 1  t  t =∫ dt = ∫ dt 2 0 2 0 1 1  t −1 + 2 t − + 1   t  t

1 1 sin 4 x dx = − cos 4 x + C 2∫ 8 1 ∴ k =− 8

− ∫ cot ( x − b ) dx

∫ − cos φ dφ

= φ cos φ − sin φ = x cos −1 x − 1 − x 2

∴

=

42

∫ φ sin φ dφ = φ cos φ

by cos 4 x,

= ∫ cos 2 x ⋅ sin 2 x dx

⇒ dx = 6 t 5 dt t3 6 t 5dt Then, I = ∫ 3 =6 ∫ dt 2 ( t − 1) (t − t ) ( t 3 − 1) + 1 = 6∫ dt ( t − 1) =6

] 46 Given,

z=t −



∫ t

2

+t +1+

1   dt t − 1

t3 t2  =6  + + t + ln|t − 1 | + C 3  2

45 Divide numerator and denominator

1 ∫ t dt = log t + C = log (sin x + cos x) + C 2 cos2 2 x 41 ∫ ⋅ sin x cos x dx cos2 x − sin2 x ∴

Put 1 + x = t 6

1 − x2 dx

Let cos φ = x ⇒ − sin φ dφ = dx ∴ I2 = −

x 3

2

 1 d  1 − cos      x  dx  x  1 cos x  1 1 = + cos  2  2 x  x 2 x 1 ⇒ f ′ (1) = cos 1 + cos 1 2 3 = cos 1 2

60 Given equation can be rewritten as x y 4 /3

dy + 6 y −1 / 3 = 3 x dx y −1 /3 = v dy dv y −4 / 3 = −3 dx dx dv 2 − v = −1 dx x

Put ⇒ ∴

Here, IF = e ∫

−

2 dx x

= x −2 1 ∴Solution is x −2 v = + C x ⇒ v = x + Cx 2 1 ⇒ y( x ) = ( x + Cx 2 ) 3

61

∫ (x =

x2 − 1 2

∫

+ 1)

x4 + 1

dx

1 x2

1−

dx

1 1 2 x +  x + 2  x x dt 1  =∫ put t = x + 2  x  t t − 2 =

1 2

sec −1

x +1 2

x 2

+C

y y=3x–6

6 5 4 3 2 1 X¢

–6 –5 –4 –3 –2 –1 O –1 –2 –3 –4 –5 –6

y=x

1 2 3 4 5 6

π , x =2 −3 2 = − 1, y = 3 + 0 = 3 i.e. (− 1, 3) and 3π At θ = , x =2 + 3 =5 2 and y = 3 + 0 = 3 i.e. (5 , 3).

At θ =

65 Suppose f (x) = 0 has a real root say

x

y=4–x y=6–3x

Y¢

x = a, then f (x) < 0 for all x < a. Thus,| f (x)| becomes strictly decreasing on (− ∞, a). So, Statement I is true.

66 As the area enclosed by | x | + | y | ≤ a is the area of square i.e. 2a2 . ∴ Area enclosed by | x + y| +| x − y| ≤ 2 is area of square shown as

Clearly, the function has absolute minimum at x = 2. So, the absolute minimum is equal to 2. Also, the curve is taking sharp turn at x = 1, 2 and 3. ∴ f is not differentiable at x = 1, 2 and 3.

63 Given,

f (x) = max {(x − 1) 2 + 1,| x − 1|} = (x − 1) 2 + 1 [say] ∴ f ′ (x) = 2 (x − 1) = 0 ⇒ x = 1 ∈[0, 3] Now, f (0) = 2, f (1) = 1, f (3) = 5 ∴Greatest value of f (x) = max {f (0), f (1), f (3)} = 5 dx 64 For vertical tangent, =0 dθ ∴ − 3 cos θ = 0 ⇒ cos θ = 0 π 3π ⇒ θ= , 2 2

Y y=x 1 –1 X 1 1 y=–x

∴ Required area 1  = 4  × 2 × 1 = 4 sq units 2 

Also, the area enclosed by | x + y| +| x − y| ≤ 2 is symmetric about X-axis, Y-axis, y = x and y = − x . Hence, both the statements are true but Statement II is not the correct explanation of Statement I.

DAY TWENTY

Trigonometric Functions and Equations Learning & Revision for the Day u

u

u

u

Angle on Circular System Trigonometric Functions Trigonometric Identities Trigonometric Ratios/Functions of Acute Angles

u

u

u

Trigonometric Ratios of Compound and Multiple Angles Transformation Formulae Conditional Identities

u

u

u

Maximum and Minimum Values Trigonometric Equations Summation of Some Trigonometric Series

Angle on Circular System If the angle subtended by an arc of length l at the centre of a circle of radius r is θ, l then θ = r

l

If the length of arc is equal to the radius of the circle, then the angle subtended at the centre of the circle will be one radian. One radian is denoted by 1c and 1c = 57°16′ 22′′ approximately.

q B

O

(Figure shows the angle whose measure are one radian.) 2 π C = 360°, 1C =

180°  π  , 1° =    180 π

C

Also, let P( x, y) be any on the circle with ∠AOP = θ radian, i.e. length of arc AP = θ

Your Personal Preparation Indicator u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

Y B

Trigonometric Functions Let X ′ OX and YOY′ be the coordinate axes. Taking O as the centre and a unit radius, draw a circle, cutting the coordinate axes at A, B, A′ and B′ as shown in the figure

PRED MIRROR

A

P(x, y)

1 X¢

A¢

q y x M

O

B¢ Y¢

q A

X

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

210 Then, the six trigonometric functions, can be defined as OM =x OP PM (ii) sinθ = =y OP OP 1 (iii) sec θ = = , x ≠0 OM x OP 1 (iv) cosec θ = = , y ≠0 PM y (i) cos θ =

Trigonometric Ratios/Functions of Acute Angles The ratios of the sides of a triangles with respect to its acute angles are called trigonometric ratios or T-ratios. In a right angled ∆ ABC, if ∠CAB = θ, then BC Perpendicular = AC Hypotenuse

C

po te n

us

e

AB Base 2. cos θ = = AC Hypotenuse BC Perpendicular = AB Base 1 AC Hypotenuse 4. cosec θ = = = sin θ BC Perpendicular

PM y = , x ≠0 OM x OM x (vi) cot θ = = , y ≠0 PM y

Hy

3. tan θ =

(v) tan θ =

1 AC Hypotenuse = = cos θ AB Base 1 AB Base 6. cot θ = = = tan θ BC Perpendicular

An equation involving trigonometric functions which is true for all those angles for which the functions which is true for all those angles for which the functions are defined is called trigonometric identity. Some identities are given below

90°

q A

Base

5. sec θ =

Trigonometric Identities

Perpendicular

1. sin θ =

B

Sign for Trigonometric Ratios in four Quadrants sin θ cos θ

Quadrant

(iii) cosec2 θ − cot2 θ = 1

cot θ

sec θ

cosec θ

+

+

+

+

+

+

II. (90 ° , 180 ° )

+

−

−

−

−

+

III. (180 ° , 270 ° )

−

−

+

+

−

−

IV. (270 ° , 360 ° )

−

+

−

−

+

−

I. (0, 90° )

(i) sin2 θ + cos2 θ = 1 (ii) sec2 θ − tan2 θ = 1

tan θ

Angle

Trigonometric ratios of some useful angles between 0°and 90° 0°/0

sin θ

0

cos θ

1

tan θ

0

cot θ

∞

sec θ

cosec θ

π 15°/ 12

π 18°/ 10

3−1

5−1 4

2 2 3+1 2 2

10 + 2 5 4

3−1

5−1

3+1

10 + 2 5

3+1

10 + 2 5

3−1

5−1

1

2 2 3+1

10 + 2 5

∞

2 2 3−1

4

5+ 1

22.5°/

π π 30°/ 8 6

2− 2

36°/

π 5

1 2

10 − 2 5

3 2

5+ 1 4

2−1

1 3

10 − 2 5

2+1

3

4−2 2

4+ 2 2

2 2+

2

2

4

5+1

5+ 1

45°/

π 4

54°/

3π 10

60°/

π 3

67.5°/

3π 8 2

1 2

5+ 1 4

3 2

2+

1 2

10 − 2 5

1 2

2− 2

10 − 2 5

3

2+1

10 − 2 5

1 3

2−1

4+ 2 2

1

4 5+ 1

10 − 2 5

1

2 3

5−1

2

10 − 2 5

2

2

10 − 2 5

2

5−1

2 3

5+ 1 4

4

2

2

4−2 2

72°/

2π 5

10 + 2 5 4 5−1 4

75°/

5π π 90°/ 12 2

3+1 2 2 3−1 2 2

10 + 2 5

3+1

5−1

3−1

1

0

∞

5−1

3−1

10 + 2 5

3+1

5+ 1

2 2 3−1

∞

2 2 3+1

1

4 10 + 2 5

0

211

DAY Trigonometric ratios of allied angles cosec θ

cos θ

sec θ

tan θ

cot θ

−θ

θ

− sin θ

sin θ

− cosec θ

cos θ

secθ

− tan θ

− cot θ

90° − θ

cos θ

sec θ

sin θ

cosec θ

cot θ

tan θ

90° + θ

cos θ

sec θ

− sin θ

− cosec θ

− cot θ

− tan θ

180° − θ

sin θ

cosec θ

− cos θ

− sec θ

− tan θ

− cot θ

180° + θ

− sin θ

− cosec θ

− cos θ

− sec θ

tan θ

cot θ

270° − θ

− cos θ

− sec θ

− sin θ

− cosec θ

cot θ

tan θ

270° + θ

− cos θ

− sec θ

sin θ

cosec θ

− cot θ

− tan θ

360° − θ

− sin θ

− cosec θ

cos θ

sec θ

− tan θ

− cot θ

Trigonometric Ratios of Compound and Multiple Angles Compound Angles

Transformation Formulae (i) 2 sin A cos B = sin( A + B) + sin ( A − B) (ii) 2 cos A sin B = sin ( A + B) − sin ( A − B) (iii) 2 cos A cos B = cos ( A + B) + cos ( A − B)

(i) sin ( A + B) = sin A cos B + cos A sin B

(iv) 2 sin A sin B = cos ( A − B) − cos ( A + B)

(ii) sin ( A − B) = sin A cos B − cos A sin B

C+D C−D cos 2 2 C+D C−D (vi) sin C − sin D = 2 cos sin 2 2 C+D C−D (vii) cos C + cos D = 2 cos cos 2 2 C+D C−D (viii)cos C − cos D = − 2 sin sin 2 2

(iii) cos ( A + B) = cos A cos B − sin A sin B (iv) cos ( A − B) = cos A cos B + sin A sin B tan A + tan B (v) tan ( A + B) = 1 − tan A tan B (vi) tan ( A − B) =

tan A − tan B 1 + tan A tan B

(vii) cot ( A + B) =

cot A cot B − 1 cot A + cot B

cot A cot B + 1 (viii) cot ( A − B) = cot B − cot A 1 + tan A (ix) (a) = tan 1 − tan A

π   + A 4 

1 − tan A (b) = tan 1 + tan A

π   − A 4 

Multiple Angles (i) sin 2 A = 2 sin A cos A =

2 tan A 1 + tan2 A

(ii) cos 2 A = cos2 A − sin2 A =

1 − tan2 A 1 + tan2 A

= 2 cos2 A − 1 = 1 − 2 sin2 A 2 tan A (iii) tan 2 A = 1 − tan2 A (iv) sin 3 A = 3 sin A − 4 sin3 A (v) cos 3 A = 4 cos3 A − 3 cos A 3 tan A − tan3 A (vi) tan 3 A = 1 − 3 tan2 A

(v) sin C + sin D = 2 sin

(ix) sin ( A + B)sin ( A − B) = sin2 A − sin2 B (x) cos ( A + B)cos ( A − B) = cos2 A − sin2 B (xi) cos A cos 2 A cos 4 A cos 8 A … cos 2 n − 1 A =

1 sin (2 n A) 2 n sin A

Conditional Identities If A + B + C = 180 °, then (i) sin 2 A + sin 2 B + sin 2 C = 4sin A sin B sin C (ii) cos 2 A + cos 2 B + cos 2 C = − 1 − 4cos A cos B cos C A B C (iii) sin A + sin B + sin C = 4 cos cos cos 2 2 2 A B C (iv) cos A + cos B + cos C = 1 + 4 sin sin sin 2 2 2 (v) tan A + tan B + tan C = tan A tan B tan C (vi) tan

A B B C C A tan + tan tan + tan tan = 1 2 2 2 2 2 2

Maximum and Minimum Values 1. −1 ≤ sin x ≤ 1,|sin x| ≤ 1

2. −1 ≤ cos x ≤ 1,|cos x| ≤ 1

3. |sec x| ≥ 1,|cosec x| ≥ 1 4. tan x,cot x take all real values

212 NOTE

• Maximum value of a cos θ ± b sin θ = a2 + b2 • Minimum value of a cos θ ± b sin θ = − a2 + b2 • Maximum value of a cos θ ± b sin θ + c = c +

a2 + b2

• Minimum value of a cos θ ± b sin θ + c = c − a2 + b2

Trigonometric Equations An equation involving one or more trigonometrical ratios of unknown angle is called a trigonometrical equation. e.g. sin θ + cos2 θ = 0

Principal Solution The value of the unknown angle (say θ) which satisfies the trigonometric equation is known as principal solution, if 0 ≤ θ < 2 π.

General Solution Since, trigonometrical functions are periodic function, solution of trigonometric equation can be generalised with the help of the periodicity of the trigonometrical functions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution.

Some general trigonometric equations and their solutions Equations π  π sin x = sin α  − ≤ α ≤   2 2 cos x = cos α (0 ≤ α ≤ π ) π  π tan x = tan α  − < α <   2 2

Solutions

Equations

x = nπ + (− 1)n α n ∈I

sin2 x = sin2 α

x = 2 nπ ± α n ∈I

cos2 x = cos2 α

x = nπ + α, n ∈ I

tan2 x = tan2 α

Solutions         

x = nπ ± α, n ∈ I

sin x = 0

x = n π, n ∈ I

sin x = 1

x = (4n + 1) π / 2, n ∈ I

cos x = 0

x = (2 n + 1) π / 2, n ∈ I

cos x = 1

x = 2 nπ, n ∈ I

tan x = 0

x = nπ, n ∈ I

cos x = − 1

x = (2 n + 1) π, n ∈ I

sin x = sin α and cos x = cos α

x = 2 nπ + α, n ∈ I

Summation of Some Trigonometric Series (i) sin α + sin (α + β) + sin (α + 2 β) + K to n terms  n β sin    2   β  = ⋅ sin α + (n − 1)     2  β    sin    2

(ii) cos α + cos (α + β) + cos (α + 2 β) + … to n terms  n β sin    2    β  = ⋅ cos α + (n − 1)     2  β    sin    2

213

DAY DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 If

2 sin θ 1 − cos θ + sin θ is equal to = x , then 1 + cos θ + sin θ 1 + sin θ

(a) x

2 If

(b)

1 x

(c) 1 − x

(d) 1 + x

3π < α < π , then cosec2 α + 2 cot α is equal to 4

(a) 1 + cotα (c) −1 − cotα

(b) 1 − cotα (d) −1 + cotα (b) 26

(c) 28

(d) 36

4 If tan A + sin A = m and tan A − sin A = n, then

(m 2 − n 2 )2 mn

is equal to (a) 4

(b) 3

(c) 16

(d) 9

2

a sin3 α + 3a cos 2 α sin α = n, then (m + n )2 / 3 + (m − n )2 / 3 is equal to (a) 2 a

2

1/ 3

(b) 2 a

(c) 2 a

(b) cos

7 If sin θ = −

π 8

(c)

1 8

(d) 2 a

(d)

3

1+ 2 2 2

4 θ and θ lies in the third quadrant, then cos is 5 2

equal to 1 (a) 5

1 (b) − 5

8 The expression

2 5

(c)

NCERT Exemplar 2 (d) − 5

tan A cot A can be written as + 1 − cot A 1 − tan A JEE Mains 2013

(a) sin A cos A + 1 (c) tan A + cot A

(b) secA cosec A + 1 (d) secA + cosec A

9 If tan θ, 2 tan θ + 2 and 3 tan θ + 3 are in GP, then the value of (a)

12 5

7 − 5 cot θ

9 − 4 sec2 θ − 1 (b) −

33 28

is 33 100

(d)

12 13

1 , then 2 tan (α + 2 β) tan ( 2 α + β) is equal to

10 If sin (α + β) = 1 and sin (α − β) = (a) 1 (c) zero

(b) –1 (d) None of these

sec A + sec B + sec C is (b) 4

(a) b = a + c (c) c = a + b

(c) 5

(c) 45°

(d) 22

1 ° 2

(b) b = c (d) a = b + c

4 5 π and sin (α − β ) = , where 0 ≤ α , β ≤ . 5 13 4 Then, tan 2α is equal to AIEEE 2010

14 If cos (α + β) = 25 16

(b)

56 33

(c)

19 12

(d)

20 7

rational function is z2 − x2 − y2 xy z2 + x2 + y2 (c) xy

(d) 6

z2 − x2 − y2 2 xy z2 + x2 + y2 (d) 2 xy

(a)

(b)

16 Let α and β be such that π < α − β < 3π. If 21 27 and cos α + cos β = − , then the 65 65  α − β value of cos   is AIEEE 2004  2 

sin α + sin β = −

(a) −

3 130

3 130

(b)

(c)

6 65

(d) −

6 65

17 If A + B + C = π and cos A = cos B cos C, then tan B tan C is equal to (a)

1 2

(b) 2

(c) 1

(d) −

1 2

18 If tan α = (1 + 2− x )−1, tan β = (1 + 2x +1 )−1, then α + β equals π 6

(b)

π 4

(c)

19 If 0 < x < π and cos x + sin x = (a)

11 In an acute angled triangle, the least value of (a) 3

(b) 30°

π P  Q  2  2 2 of ax 2 + bx + c = 0, where a ≠ 0, then

(a) (c)

1 , then A is equal to 2

15 If sin α = x , sin β = y , sin(α + β) = z , then cos(α + β) as a 2/ 3

π  3π   5π   7π   6 1 + cos  1 + cos  1 + cos  is  1 + cos        8 8 8 8 equal to (a) 1

(a) 15°

(a)

5 If a cos α + 3a cos α sin α = m and 3

cos A + cos B =

3 and 2

13 In a ∆ PQR , ∠R = . If tan   and tan   are the roots

3 The least value of cosec2 x + 25 sec2 x is (a) 0

12 If 0 < A < B < π, sin A + sin B =

− (4 + 7 ) 1+ 7 (b) 3 4

(c)

π 3

(d)

π 2

1 , then tan x is equal to 2 1− 7 4

(d)

4− 7 3

20 cosec10° − 3 sec10° is equal to (a) 1

(b) 1/2

(c) 2

(d) 4

21 The value of sin12° sin 48° sin 54° is equal to (a)

1 16

(b)

1 32

(c)

1 8

(d)

1 4

214 22 The value of1 + cos 56° + cos 58° − cos 66° is equal to (a) 2 cos 28 ° cos 29° cos 33 ° (b) 4 cos 28 ° cos 29° sin 33 ° (c) 4 cos 28 ° cos 29° cos 33 ° (d) 2 cos 28 ° cos 29° sin 33 °

π π n + cos = . Then, 2n 2n 2

(a) n = 6 (c) n = 5

(c) cos A

 

25 The maximum value of sin  x +  π interval  0,  is attained at  2 π (a) x = 12

π (b) x = 6

(b)

(d) sinA

π (d) x = 2

m+1 m

(d) None of these

2π  4π     = z cos θ +  , then the   3 3 1 1 1 value of + + is equal to x y z (b) 2

(c) 0

(d) 3 cosθ

π  π  28 The maximum value of cos  − x  − cos 2  + x  is 3  3  2

(a) −

3 2

(b)

1 2

(c)

3 2

(d)

3 2

29 If sum of all the solution of the equation

2 3

(b)

13 9

(c)

8 9

(d)

4xy 30 sin θ = is true, if and only if ( x + y )2 2

(a) x − y ≠ 0 (c) x + y ≠ 0

20 9 AIEEE 2002

(b) x = − y (d) x ≠ 0, y ≠ 0

(b) 1

(c) 3

32 If A = sin x + cos x , then for all real x 13 ≤ A≤1 16 3 13 (c) ≤ A ≤ 4 16 (a)

4

(b) 1 ≤ A ≤ 2 (d)

3 ≤ A≤1 4

(b) 2 π

(c)

5π 2

(d) None of these

36 The number of solution of cos x = |1 + sin x |, 0 ≤ x ≤ 3π is (a) 2

(b) 3

(c) 4

(d) 5

37 If sin 2x + cos x = 0, then which among the following is/are true? I. cos x = 0

II. sin x = − π ,n ∈ Z 2

1 2

7π ,n ∈ Z 6

IV. x = nπ + (– 1)n

(a) I is true (c) I,II and III are true

(b) I and II are true (d) All are true

38 The number of solutions of the equation sin 2 x − 2 cos x + 4 sin x = 4 in the interval [ 0 , 5π ] is (a) 3

(b) 5

(c) 4

JEE Mains 2013 (d) 6

39 The possible values of θ ∈( 0, π ) such that sin (θ ) + sin ( 4θ ) + sin (7θ ) = 0 are 2π , 9 2π (c) , 9 (a)

(a) 6

equation cos( p sin x ) = sin( p cos x ) in x has a solution in [ 0, 2π ] is 2

(a) π

π , 4 π , 4

4π π , , 9 2 π 2π , , 2 3

3π , 4 3π , 4

8π 9 35 π 36

π 5π π 2π 3π 8π , , , , , 4 12 2 3 4 9 2π π π 2π 3π 8π (d) , , , , , 9 4 2 3 4 9

(b)

the equation 2 sin2 x + 5 sin x − 3 = 0 is

31 The smallest positive integral value of p for which the

(a) 2

(d) [−1, 3]

(c) [0,1]

40 The number of values of x in the interval [ 0, 3π ] satisfying

 π  π  1 8 cos x ⋅  cos  + x  ⋅ cos  − x  −  6  6  2  = 1 in [ 0, π ] is kπ, then k is equal to (a)

3 and less than or equal to 3 2

 5π 5π  35 If x ∈ − , , then the greatest positive solution of  2 2  1 + sin4 x = cos 2 3x is

III. x = ( 2n + 1)

27 If x cos θ = y cos θ +

(a) 1

(b) [−11 ,]

(a) [0, 3]

A+B B−A 26 If cos A = m cos B and cot = λ tan , then λ is 2 2 m m−1 m+1 (c) m−1

3 2

then the interval of S is

π π   + cos  x +  in the  6 6

π (c) x = 3

(a)

(b) less than or equal to

34 If f : R → S , defined by f ( x ) = sin x − 3 cos x + 1, is onto,

(b) n = 1, 2 , 3, . . . , 8 (d) None of these (b) tanA

3 2

(d) None of the above

24 The value of tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A is (a) cot A

(a) greater than −

(c) greater than or equal to −

23 Let n be a positive integer such that sin

33 cos 2 θ + 2 cos θ is always

(d) 5

(b) 1

(c) 2

(d) 4

41 If α is a root of 25 cos 2 θ + 5 cos θ − 12 = 0, then sin 2 α is equal to (a)

24 25

(b) −

42 Statement I cos

24 25

(c)

13 18

π < α < π, 2

(d) −

13 18

π 2π 4π ⋅ cos ⋅ cos = − 1/ 8 7 7 7

Statement II cos θ cos 2 θ cos 4 θ . . . 1 π cos 2 n − 1θ = − n for θ = n 2 2 −1 (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

215

DAY DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 (sink x + cos k x ), where x ∈ R , k ≥ 1, then k JEE Mains 2014 f4 ( x ) − f6 ( x ) is equal to

1 If fk ( x ) = (a)

1 6

1 3

(b)

(c)

1 4

(d)

1 12

2 If tan α , tan β and tan γ are the roots of the equation x 3 − px 2 − r = 0, then the value of (1 + tan2 α ) (1 + tan2 β) (1 + tan 2 γ ) is equal to

(b) 1 + (p − r ) 2 (d) None of these

(a) (p − r ) 2 (c) 1 − (p − r ) 2

∞ ∞ π 3 For 0 < φ < , if x = ∑ cos 2 n φ, y = ∑ sin2 n φ and 2 n = 0 n = 0

z=

∞

∑

cos

2n

φ sin

2n

φ , then

n= 0

(a) xyz = xz + y (c) xyz = x + y + z

(b) xyz = xy − z (d) xyz = yz + x

π 4 If sin x + cos y + 2 = 4 sin x cos y and 0 ≤ x , y ≤ , 2 then sin x + cos y is equal to 4

4

(a) –2

(b) 0

(c) 2

(d) 3/2

5 If 0 ≤ x < 2π, 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

(b) 1 − 2a 2 − 2b 2 (d) 2 − a 2 − b 2

7 If α = sin θ + cos θ, then which of the following is true? (a) α > 1

14

(b) α ≤ 1

(c) α = 0

8 Let n be an odd integer. If sin nθ =

n

∑

(d) α < 0

br sin θ for all real r

r=0

| sin θ ⋅ cos θ | + (a) 0

2 + tan2 θ + cot 2 θ = 3,θ ∈[ 0, 4π] is/are

(b) 1

(c) 2

(d) 3

 π π 11 The number of values of θ in the interval  − ,   2 2 sec 2 θ 4 satisfying the equation ( 3 ) = tan θ + 2 tan2 θ is (a) 1 these

(b) 2

(c) 3

(d) None of

12 The equation e sin x − e − sin x − 4 = 0 has (a) (b) (c) (d)

AIEEE 2012

infinite number of real roots no real roots exactly one real root exactly four real roots

13 Find the general solution of the equation π 5π − 4 12 3π (c) 2n π ± π − 12

(b) 2n π ±

π 5π + 4 12

(d) None of these

14 At how many points the curve y = 81sin

2

+ 81cos will intersect X-axis in the region − π ≤ x ≤ π ? (a) 4 these

(b) 6

(c) 8

x

2

x

− 30

(d) None of

15 In a ∆PQR, if 3 sin P + 4 cos Q = 6 and

4 sin Q + 3 cos P = 1, then the angle R is equal to

θ, then (a) b0 = 1, b1 = 3 (c) b0 = − 1, b1 = n

(b) cos a , cos b , cos c (d) cos 2 a , cos 2 b,cos 2 c

10 The number of solutions of the equation

(a) 2n π ±

cos 2 (α − β ) − 4ab cos(α − β ) is equal to

8

(a) sina , sinb, sinc (c) sin 2 a , sin 2 b, sin 2 c

( 3 − 1 ) cos θ + ( 3 + 1 ) sin θ = 2.

6 If sin (θ + α ) = a and sin (θ + β) = b, then (a) 1 − a 2 − b 2 (c) 2 + a 2 − b 2

9 If x sin a + y sin 2 a + z sin 3 a = sin 4 a x sin b + y sin 2 b + z sin 3 b = sin 4 b x sin c + y sin 2 c + z sin 3 c = sin 4 c Then, the roots of the equation z  y + 2 z − x t 3 −   t 2 −  t +  = 0, a , b , c ≠ nπ, are 2  4   8 

(b) b0 = 0, b1 = n (d) b0 = 0, b1 = n 2 − 3n − 3

5π (a) 6

π (b) 6

π (c) 4

AIEEE 2012 3π (d) 4

216

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

SESSION 1

1. 11. 21. 31. 41.

SESSION 2

1. (d) 11. (b)

2. 12. 22. 32. 42.

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

3. 13. 23. 33.

2. (b) 12. (b)

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

4. 14. 24. 34.

3. (c) 13. (b)

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

4. (c) 14. (c)

5. 15. 25. 35.

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

5. (c) 15. (b)

6. 16. 26. 36.

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

7. 17. 27. 37.

6. (b)

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

8. 18. 28. 38.

7. (b)

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

8. (b)

9. 19. 29. 39.

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

10. 20. 30. 40.

9. (b)

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

10. (a)

Hints and Explanations SESSION 1 1 Given, ∴ x=

2 sin θ = x 1 + cos θ + sin θ 2 sin θ (1 + sin θ − cos θ) (1 + sin θ)2 − cos2 θ

2 sin θ (1 + sin θ − cos θ) = 2 1 + sin2 θ + 2 sin θ − (1 − sin2 θ) =

1 + sin θ − cos θ 1 + sin θ

2 We have,

cosec2 α + 2 cot α

= 1 + cot2 α + 2 cot α = |1 + cot α | 3π But 0

8 Given, sin n θ =

n

∑

r=0

b r sin r θ = b 0

+ b1 sin θ + b2 sin θ + K + b n sin n θ …(i) 2

b r sin r θ

r =1

sin n θ = sin θ

n

∑

b r sin r − 1 θ

r =1

On taking limit as θ → 0, we get sin n θ lim = b1 θ → 0 sin θ  sin nθ   θ  ⇒ lim n     = b1 θ→ 0  nθ   sin θ  ⇒ n = b1 Hence, b 0 = 0; b1 = n

9 Equation first can be written as x sin a + y × 2 sin a cos a + z × sin a (3 − 4 sin2 a) = 2 × 2 sin a cos a cos 2 a ⇒ x + 2 y cos a + z (3 + 4 cos2 a − 4) = 4 cos a (2 cos2 a − 1 ) as sin a ≠ 0 ⇒ 8 cos3 a − 4z cos2 a − (2 y + 4) cos a + (z − x) = 0  z 3 ⇒ cos a −   cos2 a  2  y + 2  z − x −  cos a +   =0  4   8 

+ 2 ab (1 − a2 − b 2 + a2b 2 )

Now, cos 2 (α − β) − 4 ab cos (α − β) = 2 cos2 (α − β) − 1 − 4 ab cos (α − β) = 4 a2b 2 + 2 − 2 a2 − 2b 2

n

∑

which shows that cos a is a root of the equation  z  y + 2  z − x t3 −   t2 −  t +   =0  2  4   8  Similarly, from second and third equation we can verify that cos b and cos c are the roots of the given equation.

10 Given,|sin θ.cos θ| + 2 + tan2 θ + cot2 θ = 3 ⇒ |sin θ ⋅ cos θ | + (tan θ + cot θ)2 = 3 ⇒ | sin θ ⋅ cos θ | + | tan θ + cot θ| = 3 ⇒ | sin θ ⋅ cos θ | +

sin 2 θ + cos2 θ = 3 sin θ ⋅ cos θ

⇒ |sin θ ⋅ cos θ| +

1 = 3 | sin θ ⋅ cos θ |

We know that, |sin θ ⋅ cos θ | +

1 ≥2 | sin θ ⋅ cos θ |

Hence, there is no solution of this equation.

221

DAY 2

11 ( 3)sec

θ

= (tan2 θ + 1 )2 − 1

= (sec2 θ )2 − 1 Put sec θ = x

(x ≥ 1 )

2

Then, ( 3)x = x2 − 1 Let

y = ( 3)x = (x 2 − 1 )

(x > 1 )

Now, graphs of y = ( 3)x and y = x 2 − 1 intersect at one point y y=( 3)x

y=x2–1

1 x¢

O

1

x

_1 y¢

So, esin x = 2 + 5 is not possible for any x ∈ R and esin x = 2 − 5 is also not possible for any x ∈ R Hence, we can say that the given equation has no solution.

13 Given, ( 3 − 1) cos θ + ( 3 + 1) sin θ = 2 ...(i) ( 3 − 1) = r cos α

and

( 3 + 1) = r sin α

Therefore, there are two values of θ  π π in  − ,  .  2 2

12 Given equation is esin x − e − sin x = 4 ⇒ e sin x −

Now, let y = e sin x Then, we get 1 y− =4 ⇒ y ∴

y=

⇒ e sin x

y2 − 4y − 1 = 0

4 ± 16 + 4 2 =2 ± 5

1 =4 e sin x

⇒ y =2 ± 5

Since, sine is a bounded function i.e. − 1 ≤ sin x ≤ 1. Therefore, we get e −1 ≤ e sin x ≤ e 1  e sin x ∈ , e ⇒  e  Also, it is obvious that 2 + 5 > e and 2 − 5
a, c + a > b (iv) a − b < c, b − c < a, c − a < b (v) a > 0, b > 0, c > 0

Your Personal Preparation Indicator

Relations between the Sides and Angles of Triangle For a triangle ∆ABC with sides a, b , c and opposite angles are respectively A, B and C, then a b c (i) Sine Rule In any ∆ ABC, = = sin A sin B sin C (ii) Cosine Rule (a) cos A =

b +c −a 2 bc 2

2

2

a2 + b 2 − c2 (c) cos C = 2 ab

(b) cos B =

c + a −b 2ca 2

2

2

u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

223

DAY (iii) Projection Rule (a) a = b cos C + c cos B

(b) b = c cos A + a cos C

The length of medians AD, BE and CF of a ∆ABC are (shown as in given figure)

(c) c = a cos B + b cos A

A

(iv) Napier’s Analogy B −C b −c A C−A c−a B (a) tan = cot (b) tan = cot 2 b +c 2 2 c+a 2 (c) tan

A − B a −b C = cot 2 a+b 2

B

(v) Half Angle of Triangle A (s − b ) (s − c) (a) sin = 2 bc

B (s − c) (s − a) (b) sin = 2 ca

(c) sin

C (s − a) (s − b ) = 2 ab

(d) cos

(e) cos

B = 2

s(s − b ) ca

(f) cos

(g) tan

A (s − b ) (s − c) = 2 s(s − a)

(i) tan

C (s − a) (s − b ) = 2 s(s − c)

A = 2 C = 2

(h) tan

s(s − a) bc s(s − c) ab

B (s − a) (s − c) = 2 s(s − b )

E

F

D

1 2 b 2 + 2 c2 − a2 , 2 1 CF = 2 a2 + 2 b 2 − c2 2 AD =

and

C

BE =

1 2 c2 + 2 a2 − b 2 2

Circles Connected with Triangle 1. Circumcircle The circle passing through the vertices of the ∆ABC is called the circumcircle. (shown as in given figure) A

1 1 1 bc sin A = ca sin B = a b sin C 2 2 2 = s (s − a) (s − b ) (s − c)

(vi) Area of a Triangle ∆ =

Some Important Theorems 1. m-n Theorem (Trigonometric Theorem) If in a ∆ABC, D divides AB in the ratio m : n, then (shown as in given figure)

O

R

B

C

Its radius R is called the circumradius, a b c abc and R= = = = 2 sin A 2 sin B 2 sin C 4 ∆ NOTE

• The mid-point of the hypotenuse of a right angled triangle is equidistant from the three vertices of the triangle.

C

• The mid-point of the hypotenuse of a right angled triangle is the circumcentre of the triangle.

a b

A

A m

• Distance of circumcentre from the side AC is R cos B. • Radius of circumcircle of a n-sided regular polygon with

q D

B n

each side a is R = B

a π cosec . 2 n

2. Incircle

(i) (m + n) cot θ = n cot A − m cot B

The circle touching the three sides of the triangle internally is called the inscribed circle or the incircle of the triangle. Its

(ii) (m + n) cot θ = m cot α − n cot β

A

2. Appolonius Theorem If in ∆ABC, AD is median, then A

r B C

D

B

AB + AC = 2 ( AD + BD 2 ) 2

2

2

C

radius r is called inradius of the circle. ∆ (i) r = s A B C (ii) r = (s − a) tan = (s − b ) tan = (s − c) tan 2 2 2

ONE

224 A B C sin sin 2 2 2 B C C A A B a sin sin b sin sin c sin sin 2 2 2 2 2 2 (iv) r = = = A B C cos cos cos 2 2 2

p/n

Similarly, r2 and r3 denote the radii of the escribed circles opposite to angles B and C, respectively. r1 , r2 and r3 are called the exradius of ∆ABC. B C a cos cos ∆ A A B C 2 2 (i) r1 = = s tan = 4R sin cos cos = A s−a 2 2 2 2 cos 2 C A b cos cos ∆ B B C A 2 2 (ii) r2 = = s tan = 4R sin cos cos = B s −b 2 2 2 2 cos 2 A B c cos cos ∆ C C A B 2 2 (iii) r3 = = s tan = 4R sin cos cos = C s −c 2 2 2 2 cos 2

r

NOTE

(iii) r = 4R sin

NOTE Radius of incircle of a n-sided regular polygon with each side

a is r =

a π cot . (shown as in given figure) 2 n

O

p/n

• r1 + r2 + r3 = 4 R + r

• r1 r2 + r2 r3 + r3 r1 =

a

• ( r1 − r ) ( r2 − r ) ( r3 − r ) = 4 Rr 2 •

3. Orthocentre and Pedal Triangle ●

The point of intersection of perpendiculars drawn from the vertices on the opposite sides of a triangle is called orthocentre. (shown as in given figure)

1 1 1 1 + + = r1 r2 r3 r

Angle of Elevation and Depression Let O be the observer’s eye and OX be the horizontal line through O. (shown as in following figures) If a object P is at a higher level than eye, then ∠POX is called the angle of elevation.

A

P

E

F

Horizontal line

O

O

●

●

●

D

e Lin

C

The ∆DEF formed by joining the feet of the altitudes is called the pedal triangle.

O

of

X

q

t

B

r1 r2 r3 r

h sig

Lin

eo

fs

igh

t

q

X

Horizontal line

P

Orthocentre of the triangle is the incentre of the pedal triangle.

If a object P is at a lower level than eye, then ∠POX is called the angle of depression.

Distance of the orthocentre of the triangle from the angular points are 2R cos A, 2R cos B, 2R cos C and its distances from the sides are 2 R cos B cos C, 2 R cos C cos A, 2 R cos A cos B.

Important Results on Heights and Distances Results shown by the following figures. (i) a = h (cot α − cot β)

4. Escribed Circle

D

The circle touching BC and the two sides AB and AC produced of ∆ABC, is called the escribed circle opposite to A. Its radius is denoted by r1 . (shown as in given figure)

B r1

A

C

h

A

b

a a

B

C r

225

DAY  α + β (ii) If AB = CD, then x = y tan    2 

(iv) H = A

h cot β cot α D

y B

D

E

b

a

C

E

B

x

H

h a

b

A

H sin (β − α ) h cot α (iii) h = and H = cos α sin β cot α − cot β

C

(v) a = h (cot α + cot β), h = a sin α sin β cosec (α + β) and d = h cot β = asin α cos β cosec (α + β)

A

A

E h

H

a

h

B

a

B

b

b C

C

D

D

d

a

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 6 In a ∆ABC, (c + a + b ) (a + b − c ) = ab. The measure of ∠C

1 If a , b and c are sides of a triangle, then (a) a + b > c (b) a − b > c (if c is smallest) (c) a + b < c (d) None of the above

is

2 If in a ∆ABC, A = 30 °, B = 45 ° and a = 1, then the values of b and c are respectively (a)

2,

(c)

3,

3 +1 2 3 −1 2

(b)

2,

(d)

2,

3 −1 2 3 + 2 2

3 If A = 75 °, B = 45 °, then b + c 2 is equal to (a) 2a

(b) 2a + 1

(c) 3a

4 If in a ∆ ABC, 2 b 2 = a 2 + c 2 , then (a)

c2 −a2 2 ca

c 2 − a 2  (c)    ca 

(b) 2

(d) 2 a − 1

sin 3B is equal to sin B

c2 −a2 ca

c 2 − a 2  (d)    2 ca 

2

(b) 90°

π . Then, the greatest 2

(c) 120°

π 6

(d) None of these

7 In ∆ ABC, if tan

A C 1 tan = , then a , b and c are in 2 2 2

(a) AP (c) HP

(b) GP (d) None of these

8 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

s  s  s  − 1  − 1  − 1 is equal to a  b  c 

9 In any ∆ ABC, 4  r R 3r (c) R

(b)

2r R

(d) None of these

10 In a ∆ ABC,R = circumradius and r = inradius.

angle of the triangle is (a) 60°

(b)

(a)

5 The sides of a triangle are sin α , cos α and 1 + sin α cos α for some 0 < α
0 ∴

a+

b> c

= −1 +

2 Given, A = 30 °, B = 45°, a = 1

⇒

=

3+1 2

3 Given, A = 75° and B = 45° ⇒

C = 60 ° a b c , we By sine rule, = = sin A sin B sin C a b c get = = sin 75° sin 45° sin 60 ° sin 45° sin 60 ° Now, b + c 2 = a+ 2 a sin 75° sin 75° 1 3 2 a+ 2 2 = a 3+1 3+1 =

4

2 2 2 2 2 2 3a a+ = 2a 3+1 3+1

sin 3B 3 sin B − 4 sin3 B = sin B sin B = 3 − 4 sin2 B = 3 − 4 (1 − cos 2 B )

(a + c 2 ) 2 4 (ac ) 2 2

2

5 Let a = sinα, b = cos α

b = 2

and c = 2 sin 105° = 2cos 15°  3 + 1 ∴ c =2  =  2 2 

2

(a2 + c 2 )2 − 4a2c 2  c 2 − a2  =  2 4 (ac )  2 ac 

and c = 1 + sin α cos α Here, we can see that the greatest side isc. a2 + b 2 − c 2 ∴ cos C = 2ab sin2 α + cos2 α − 1 − sin α cos α ⇒ cos C = 2ab sin α cos α ⇒ cos C = − 2sin α cos α 1 ⇒ cos C = − = cos 120° 2 ⇒ ∠C = 120°

6 We have, 2s (2s − 2c ) = ab ⇒

s (s − c ) 1 = ab 4

C 1 = 2 4 C 1 we get, cos = 2 2 since, C must be acute  2   ⇒

cos 2

C = 60° ⇒ C 2 7 We have, tan A tan 2 (s − b )(s − c ) ⇒ s(s − a) ⇒

2π 3 C 1 = 2 2 (s − a)(s − b ) 1 = s(s − c ) 2 =

s−b 1 = ⇒ 2 s − 2b − s = 0 s 2 ⇒ a + c − 3b = 0 8 We have, R = abc and r = ∆ 4∆ s R abc s abc = ⋅ = ∴ r 4∆ ∆ 4(s − a)(s − b )(s − c ) ⇒

[Q2b 2 = a2 + c 2 (given)]

[Qa + b > c , as sides of triangle] ∴ C = 105° Using sine rule, a b c , we get = = sin A sin B sin C 1 b c = =  1  1  sin 105°      2  2

4 (a2 + c 2 − b 2 )2 4 (ac )2

Since, a : b : c = 4 : 5 : 6 a b c (say) ⇒ = = =k 4 5 6 R (4k )(5k )(6k ) Thus, = 15k 15k r  4 − 4k   − 5k   2   2   15k − 6k     2  3 120k ⋅ 2 16 = 3 = k ⋅ 7 ⋅ 5⋅ 3 7

9 We have, 4  s − 1  s − 1  s − 1 a  b  c  (s − a) (s − b ) (s − c ) abc s( s − a) (s − b ) (s − c ) = 4⋅ s ⋅ abc 4∆ ∆ ∆2 r = 4⋅ = ⋅ = s ⋅ abc abc s R a cos A + b cos B + c cos C 10 We have, 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 2B + sin 2C ) 2s R = ⋅ 4 sin A sin B sin C 2s 2R abc abc = ⋅ = s 8 R3 4sR2 4 ∆R r  abc ∆ = = QR = ,r =   ∆ 2 R 4∆ s  4⋅ ⋅ R r = 4⋅

229

DAY 11 Key Idea Orthocentre, centroid and circumcentre are collinear and centroid divide orthocentre and circumcentre in 2:1 ratio. We have orthocentre A(−3, 5) and centroid B(3,3). Let C be the circumcentre.

C

B(3, 3)

A(–3, 5)

…(ii) ⇒ d = 36 tanθ From Eqs. (i) and (ii), we get d 2 = 36 × 64 ⇒ d = 48 m

15 Let CD be a tower of height h and AB is building of height a.

Clearly, AB = (3 + 3)2 + (3 − 5)2

D

= 36 + 4 = 2 10 We know that, AB : BC = 2:1 ⇒ BC = 10 AC = AB + BC = 2 10 + 10 = 3 10 Since, AC is a diameter of circle. AC 3 10 5 ∴ r = =3 ⇒ r = 2 2 2    12 Since,  1 − r1   1 − r1  = 2 r2   r3   1 − s − b  1 − s − c  = 2 ∴     s − a  s − a (b − a)(c − a) =2 ⇒ (s − a)2 2 bc − ab − ac + a =2 ⇒ 2  a + b + c − a     2

2bc − 2ab − 2ac + 2a2 = b 2 + c 2 + a2 + 2bc − 2ab − 2ac 2 2 ⇒ a = b + c2 So, triangle is right angled. 13 Given, cot α = 12 and sin β = 3 5 5 C

30°

h L

Now,

2(bc − ab − ac + a2 ) ⇒ =1 (b + c − a)2

(100 − 64) d

In ∆CDE, tan(90°− θ) =

B a

45°

C

A

(h − a) ...(i) = 3 (h − a) tan30° h In ∆ACD, tan 45° = ⇒ h = CA ⇒ CA h = LB [Q LB = CA] ...(ii)

∴

LB =

From Eqs. (i) and (ii), we get h ( 3 − 1) = 3 a ∴

h=

3a 3−1

=

3 ( 3 + 1) a 2

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

From Eqs. (i) and (ii) x+ y = 3y 3 ⇒ x + y = 3y ⇒

h−a LB

In ∆BL D, tan30° =

Now, from ∆ACD and ∆BCD, we have h tan30° = x+ y h and tan 60° = y x+ y ⇒ h= 3 and h = 3y

x − 2y = 0 ⇒ y =

x 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 A B 18 In ∆OA 1 B 1 , tan 45° = 1 1 OB1 20 = 1 ⇒ OB1 = 20 ⇒ OB1 A1

A2

3 + 3 =  a  2 

⇒

20 m

16 Let OP be the pole of height x m. P 45°

30°

45°

O

20

B1

x

B2

In ∆OA 2 B2 , A

15 m b B D d In ∆DAC and ∆DBC , AD = 15 cot α, BD = 15 cot β ⇒ d = 15 (cot α + cot β ) 12 4  = 15  +  = 56 m  5 3 64 14 In ∆DAB, tanθ = d

75°

30° B

1 km

a

A

Using sine rule in APB, ⇒ PB = 1000 ×

1/2

tan30° =

O

sin 30° PB = sin 45° AB

= 500 2 m

1/ 2

3+1 2 2

⇒

B1 B2 = 20 3 − 20

⇒

B1 B2 = 20( 3 − 1)m Distance 20( 3 − 1) = Time 1 = 20( 3 − 1) m/s

= 250 ( 3 + 1) m

17 According to given information, we have the following figure

C

⇒ B1 B2 + OB1 = 20 3

Now, Speed =

In ∆ PBO, x = 500 2 ⋅ sin 75° = 500 2 ×

20 ⇒ OB2 = 20 3 OB2

19 Let OP be a tower with height ( 3 + 1 ) m and AB = 2 m. P

D

90°–q q d

D

100 m

(Ö3 + 1) m

E

q

⇒

a

30° B

O

In ∆AOP,

64 m A

A

Pillar h

d

d = 64cotθ

B

A

…(i)

30° x

B

3+1 OA OA = ( 3 + 1 ) 3

tan 30 ° =

60° y

C

⇒

ONE

230 3+1 OB OB = ( 3 + 1 ) cot α

and in ∆BOP, ⇒

tan α =

Now, OA − OB = (3 +

3 ) − ( 3 + 1 ) cot α

2=3+

⇒ ⇒ ∴

3 − ( 3 + 1 ) cot α

cot α = 1 α = 45°

20 According to the given information the figure should be as follows. Let the height of tower = h E

⇒ BD = 100 cot (45° − A ) ⇒ BC + CD = 100 cot (45° − A ) …(ii) ∴ From Eqs. (i) and (ii), we get CD = 100 [cot (45° − A ) − cot(45° + A )]  1 + tan A 1 − tan A  = 100  −   1 − tan A 1 + tan A   4 tan A  = 100  2  = 200 tan 2 A  1 − tan A  ⇒ 200 = 200 tan 2 A ⇒ tan2 A = 1 ⇒ tan 2 A = tan 45° ⇒ 2 A = 45° 1° ∴ A = 22 2 22 Let PQ and RS be the poles of height 20 m and 80 m subtending angles α and β at R and P , respectively. Let h be the height of the point T , the intersection of QR and PS.

h

S

30° A

45° B

60° C

D

ED In ∆EDA, tan30° = AD 1 ED h ⇒ = = ⇒ AD = h 3 AD AD 3 h In ∆EDB, tan 45° = ⇒ BD = h BD h h In ∆EDC, tan 60° = ⇒ CD = CD 3 AB AD − BD AB h 3−h Now, = ⇒ = h BC BD − CD BC h− 3 h ( 3 − 1) AB AB 3 ⇒ = ⇒ = BC BC 1 h ( 3 − 1) 3 AB : BC = 3 :1

∴

Q

80 m T a R

h

b

V

20 m

D a b

…(i) d = 60cotβ DC tanα = EC ⇒ DC = d tanα ⇒ 60 − h = d tanα (Q BC = EA = h ) ⇒ 60 − h = 60cot β tan α [from Eq. (i)] cos β sin α   ⋅ ⇒ h = 60  1 −   sin β cos α  60sin(β − α ) ⇒ h= cos α sin β 60sin (β − α ) 60sin (β − α ) (given) ⇒ = x cos α sin β In ∆DEC,

x = cos α sin β

30°

45°+A B

200 m 100 BC ⇒ BC = 100 cot (45° + A ) and in ∆ADB, 100 tan (45° − A ) = BD In ∆ACB, tan(45° + A ) =

…(i)

8

75°

C

0 ft

15°

B

In ∆ABC, on applying sine rule BC AB = sin75° sin30° 80sin 30° 40 × 2 2 ⇒ AB = = sin 75° 3+1 = 40( 6 − 2 ) ft

b a

26 Clearly, a1 = tan  α + β  P

2h

and

Now, in ∆ABP, AB h 1 tan(α + β ) = = = AP 2h 2

C

Tower 75°

C

A

100 m

B

⇒

h/2 h

h/2

D

d

A

B

45°–A

45°–A

b

A

h = 80 − 4 h h = 16 m

A

60

tower.

h 2 Again, let ∠CPA = α

45°+A

C d

h

Again, h cot α + h cot β = 20 cotα ⇒ (h − 20)cot α = − hcotβ cot α h ⇒ = =4 cot β 20 − h ⇒ ∴

a

E

25 Let BC be the declivity and BA be the

Then, PR = h cot α + h cot β = 20cot α = 80cot β cot α ⇒ cotα = 4cotβ ⇒ =4 cot β

AC = BC =

100 m. Distance between the objects, CD = 200 m

d

⇒

P

23 Let AB = h, then AP = 2h and

21 Let AB be the tower with height

24 In ∆ABD, tanβ = 60

h AC 1 Also, in ∆ACP, tanα = = 2 = AP 2h 4 Now, tan β = tan[(α + β ) − α] 1 1 1 − tan(α + β ) − tan α 2 4 = = = 4 9 1 + tan(α + β )tan α 1 + 1 × 1 2 4 8 2 = 9

…(i)

 2  b1 a2 β + γ = tan    2  b2

…(ii) G b1 F b2 E

A

a2

g B

a

b a1

C

D

231

DAY Since, a1 a2 = b1 b2 a b α + β ∴ 1 = 2 ⇒ tan   =  2  b1 a2

1 β + γ  tan    2 

⇒

α + β β + γ  = 1  tan    2  2  α+β β+ γ π + = 2 2 2 α + β + γ = π −β < π

∴ ⇒

27 Let AB = x D θ

p

C

√p 2 +q

p

π–(θ+α) A

sin θ cos α + cos θ sin α ( p2 + q 2 )sin θ = q sin θ + p cos θ

2

SESSION 2 1

P

a π a π and = sin = tan 2R n 2r n r π = cos R n r 1 n = 3 gives, = R 2 r 1 n = 4 gives, = R 2 r 3 n = 6 gives, = R 2

∴

R

C

OP ⇒ OP = OA tanα OA OP = R tanα (QOA = R, given)

In ∆OAP, tanα = ⇒

29 Let h be the height of a tower C

Alternate Solution Applying sine rule in ∆ABD,  a = b = c     sin A sin B sin C 

O

90º 30º

60º

P

C

a

p

2

a

p +q AB = sin θ sin{ π − (θ + α)}

⇒ ⇒

s−c 1 a+ b −c 1 = ⇒ = s 3 a+ b + c 3

⇒ 3 a + 3b − 3c = a + b + c ⇒ a + b = 2c Hence, a, c and b are in AP. A 0 A 1 = 2 × 1 cos 60 ° = 1 = A1 A2 [using cosine rule]] A4 A3

30° a

a

B 2

⇒

3 Clearly,

D 30°

A

⇒

⇒ Since, ∠AOB = 60° Also, OB = OA = radii ∴ ∠OBA = ∠OAB = 60° So, ∆OAB is an equilateral ∴ OA = OB = AB = a h In ∆OAC, tan30° = a 1 h a ⇒ = ⇒ h= a 3 3

30 Let PD be a pole.

q

⇒

A

a

B

A+ B A−B C cos = 4 sin2 2 2 2 A−B A+ B cos = 2 cos 2 2 Qcos A + B ≠ 0   2 A B A B cos cos + sin ⋅ sin 2 2 2 2 A B A B  = 2  cos cos − sin ⋅ sin   2 2 2 2 A B A B 3 sin ⋅ sin = cos cos 2 2 2 2 A B 1 tan ⋅ tan = 2 2 3 (s − b ) (s − c ) (s − c ) (s − a) 1 ⋅ = s (s − a) s (s − b ) 3

2 2 cos

h

( p + q )sin θ p cos θ + q sin θ

π–(θ+α)

a

B

2

Öp 2 +q

O

a R

p nr

R 2p n

∴

− pq cos θ + p2 sin θ p cos θ + q sin θ

2

2

R

p x −q

 q cos θ − p sin θ  = q − p   q sin θ + p cos θ  q 2 sin θ + pq cos θ

q

  p + q  p

2

make equal angles at the vertices of the triangle, therefore foot of tower is at the circumcentre.

B

 q Qin ∆BDC ,cot α = p     q cot θ − p  = q − p .  q + p cot θ 

2

, sin α =

2

a

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 

D

p +q

x

In ∆DAM, tan( π − θ − α ) =

AB =

q 2

A

q

x=

 Qcos α = 

α

x–q M

⇒

p2 + q 2 sin θ

AB =

DP =

⇒

28 Let OP be the tower. Since, the tower

q

α

a 3 DP In ∆PDB, tan θ = BD a/ 3 1 tan θ = = ⇒ 2a 6

p2 + q 2 AB = sin θ sin(θ + α )

tan  

⇒

⇒

⇒

C

A q

a

a

O

A5

B

DP In ∆DAP, tan 30° = AD

A2

60°

0°

12 A0

A1

ONE

232 and

Let height of tower TM be h. TM In ∆PMT , tan 45° = PM h ⇒ 1= PM ⇒ PM = h h ; QM = 3h In ∆TQM, tan30° = QM

( A 0 A1 )2 + ( A1 A2 )2 – ( A 0 A2 )2 cos 120 ° = 2 ⋅ A 0 A 1 ⋅ A1 A2 1 + 1 − ( A 0 A2 )2 2⋅1⋅1 ⇒ A 0 A2 = 3 =

∴ A 0 A1 × A 0 A2 = 1 × 3 = 3 ships at the end of 3 h. Then, OA = (24 × 3) = 72 km and OB = (32 × 3) = 96 km

km

x km

O

72

75° 96 km

E

B

S

Hence, the distance between the ships at the end of 3 h is 86.53 km ≈ 86.4 km (approx.).

5 In ∆APC, α r α sin = ⇒ AC = r cosec 2 2 AC

r C a

H

P

A

b B

In ∆ ABC , sin β =

BC AC

⇒ H = AC sin β ⇒ H = r sin β cosec

6

α 2

9 Let PAQ be the hill,AB be the tree and [say]

45º

200 m

T 30º 90º Q

30º M

R

PD be the horizontal. Let P be the point of observation. B

35 m A C 60°

A

D

45° 2 km 30°

B

O

Let C be the point where elevation is 60°. Then, ∠BAC = 30° and AC = 2 km BC ∴ = sin30° AC 1 ⇒ BC = 2 × = 1 2 AB i.e. BC = 1 km and = cos 30° AC 2 3 ⇒ AB = = 3 2 i.e. AB = 3 km PD = OP − OD = OP − BC = h − 1 and CD = BO = AO − AB = h − 3 h−1 PD In ∆DCP, = tan 60° ⇒ = 3 CD h− 3 ⇒

3 h − 3 = h − 1 ⇒ ( 3 − 1) h = 2 h=

2 × 3−1

3+1 3+1

= 3+1

= 1.732 + 1 = 2732 . km

200 m

GH 5 1 = = TG 5 3 3 θ = sin −1 (1 / 3 )

H

∴

P

H B

Then, sinθ = ⇒

Now,

A

T

Let θ be the required angle of elevation of G at T .

P

Let AB = x km We have, ∠NOA = 45° and ∠SOB = 75° ∴ ∠AOB = 180°−(45°+75° ) = 60° Using cosine formula on ∆AOB, we get (72)2 + (96)2 − x2 cos 60° = 2 × 72 × 96 ⇒ x2 = 14400 − 6912 = 7488 ⇒ x = 7488 = 86.53 km

G C

q

4h2 = (200)2 ⇒ h = 100 m

the mountain. Then, ∠OAP = 45° ∴ ∠OPA = 45° Hence, ∆OAP is an isosceles. Let AO = OP = h km height of mountain.

Q D

7 Let P be the summit and A be the foot of

A 45°

W

⇒

R

P

In ∆PMQ, PM 2 + QM 2 = PQ 2 h2 + ( 3h )2 = (200)2

4 Let A and B be the positions of the two

N

S

8 Let H be the mid-point of BC. Since, ∠ TBH = 90° , therefore, TH 2 = BT 2 + BH 2 = 52 + 52 = 50 Also since, ∠ THG = 90° , TG 2 = TH 2 + GH 2 = 50 + 25 = 75

Q

45 ° 15°

P

C D Produce BA to meet PD at C. Let AB = H m. Then, ∠DPA = 15° , PA = 35m ∠CPB = 60° and ∠PCA = 90°.

∴

∠APB = (60°−15° ) = 45°

In ∆PAC, ∠PAC = 180°− (15° + 90° ) = 75° ∴ ∠PAB = (180°−75° ) = 105° and ∠PBA = 180°−(45° + 105° ) = 30° Applying sine rule on ∆PAB, we get PA AB 35 H = ⇒ = sin ∠PBA sin ∠APB sin 30° sin 45° ∴

35 × 2 = H × 2 ⇒ H = 35 2 m

Hence, the height of the tree is 35 2 m.

10 Let OP be the flag staff of height h standing at the centre O of the field. P 45° 45° D C h F 90° 15° 90° E O A

B

In ∆ OEP, OE = h cot15° = h (2 +

3)

and in ∆ OFP, OF = h cot 45° = h ∴ EF = h 1 + (2 + 3 ) 2 = 2h 2 + Since, BD = 1200 m ⇒

2EF = 4 h (2 +

∴ h=

300 2+

3 ) = 1200

= (300 2 − 3 ) m 3

3

DAY TWENTY TWO

Inverse Trigonometric Function Learning & Revision for the Day u

Inverse Trigonometric Function

u

Properties of Inverse Trigonometric Function

Inverse Trigonometric Function Trigonometric functions are not one-one and onto on their natural domains and ranges, so their inverse do not exists in the whole domain. If we restrict their domain and range, then their inverse may exists. y = f ( x) = sin x. Then, its inverse is x = sin −1 y. NOTE

 1  y

• sin −1 y ≠ (sin y ) −1

• sin −1 y ≠ sin  

The value of an inverse trigonometric functions which lies in its principal value branch is called the principal value of that inverse trigonometric function.

Domain and range of inverse trigonometric functions Function

Domain

x

[− 1, 1]

cos −1 x

[−1, 1]

tan −1 x

R

sin

cot

−1

−1

x

sec −1 x −1

cosec x

Range (Principal Value Branch)

[0, π ]  π π − ,   2 2

R

(0, π )

R − (−1, 1)

π  [0, π ] −   2 

R − (−1, 1)

PRED MIRROR

 π π − ,  2 2 

 π π − 2 , 2  − {0}

Your Personal Preparation Indicator u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

234

Properties of Inverse Trigonometric Functions π ; (−1 ≤ x ≤ 1) 2 π (ii) tan −1 x + cot −1 x = ; x ∈R 2 π (iii) sec −1 x + cosec −1 x = ; (x ≤ − 1 or x ≥ 1) 2 −1 −1 2. (i) sin (− x) = − sin x; (− 1 ≤ x ≤ 1) (ii) cos −1 (− x) = π − cos −1 x; (− 1 ≤ x ≤ 1) (iii) tan −1 (− x) = − tan −1 ( x); (− ∞ < x < ∞) (iv) cot −1 (− x) = π − cot −1 x; (− ∞ < x < ∞) (v) sec −1 (− x) = π − sec −1 x; x ≤ − 1 or x ≥ 1) (vi) cosec −1 (− x) = − cosec −1 x; (x ≤ − 1 or x ≥ 1) 3. (i) sin −1 (sin x) is a periodic function with period 2π.   π π x∈ − ,  x,  2 2    π − x, x ∈  π , 3 π   2 2   −1 sin (sin x) =  3π 5π  x − 2 π, x ∈  ,   2 2    5π 7π  3 π − x, x ∈  ,  2 2   (ii) cos –1 (cos x) is a periodic function with period 2π. x ∈ [0, π ]  x, 2 π − x, x ∈ [π, 2 π ] cos −1 (cos x) =   x − 2 π, x ∈ [2 π, 3 π] 4π − x, x ∈ [3 π, 4π ] −1 −1 1. (i) sin x + cos x =

(iii) tan −1 (tan x) is a periodic function with period π.   π π x∈ − ,  x,  2 2    π 3π   x − π, x∈ ,  2 2   −1 tan (tan x) =  3π 5π    x − 2 π, x ∈ ,   2 2   5π 7π   x − 3 π, x ∈  ,  2 2   (iv) cot −1 (cot x) is a periodic function with period π. cot −1 (cot x) = x; 0 < x < π (v) sec −1 (sec x) is a periodic function with period 2π . π π sec −1 (sec x) = x; 0 ≤ x < or 0  1   cot x, (iii) tan −1   =   x  − π + cot −1 x, if x < 0

5. (i) sin −1 x = cos −1 1 − x2  1 − x2   x   = tan −1  = cot −1    1 − x2  x    1  −1  1  = sec −1   = cosex   , if x ∈(0, 1) 2  x  1− x  (ii) cos −1 x = sin −1 1 − x2  1 − x2   x  −1  1   = cot −1  = tan −1   = sec    x  1 − x2  x    1  = cosec −1   , if x ∈(0, 1)  1 − x2   x  (iii) tan −1 x = sin −1    1 + x2   1  −1  1  = cos −1   = cot    x  1 + x2   1 + x2   = sec −1 ( 1 + x2 ), if x ∈ (0, ∞) = cosec −1  x   6. (i) sin −1 x + sin −1 y sin −1 ( x 1 − y2 + y 1 − x 2 ); | x|,| y| ≤ 1 and  x2 + y2 ≤ 1 or ( xy < 0 and x 2 + y2 > 1)   −1 2 2 π − sin ( x 1 − y + y 1 − x ); 0 < x, y ≤ 1 = and x 2 + y2 > 1   −1 2 2 − π − sin ( x 1 − y + y 1 − x ); 2 2  − 1 ≤ x, y < 0 and x + y > 1  (ii) sin −1 x − sin −1 y  sin −1 ( x 1 − y2 − y 1 − x 2 ); x , y ≤ 1  2 2 2 2  and x + y ≤ 1 or ( xy > 0 and x + y > 1)  −1 2 2  π − sin ( x 1 − y − y 1 − x ); = 2 2  0 < x ≤ 1, − 1 ≤ y < 0 and x + y > 1  −1 2 2  − π − sin ( x 1 − y − y 1 − x );  − 1 ≤ x < 0, 0 < y ≤ 1 and x 2 + y2 > 1  (iii) cos −1 x + cos −1 y cos −1 { xy − (1 − x2 ) (1 − y2 )}; | x |,| y | ≤ 1  and x + y ≥ 0  = 2 2 −1 2 π − cos { xy − (1 − x ) (1 − y )};  | x |,| y | ≤ 1 and x + y ≤ 0  −1 −1 (iv) cos x − cos y cos −1 { xy + (1 − x2 ) (1 − y2 )}; | x |,| y | ≤ 1  and x ≤ y = −1 2 2 − cos { xy + (1 − x ) (1 − y )};   − 1 ≤ y ≤ 0, 0 < x ≤ 1 and x ≥ y

235

DAY (v) tan −1 x + tan −1 y  −1  x + y  xy < 1  tan  1 − xy  ;   x + y  x > 0, y > 0, xy > 1 =  π + tan −1  ;  1 − xy    −1  x + y   ; x < 0, y < 0, xy > 1  − π + tan   1 − xy   (vi) tan −1 x − tan −1 y  −1  x − y  tan  1 + xy  ; xy > − 1   −1  x − y  xy < − 1, x > 0, y < 0 = π + tan  ;  1 + xy    −1  x − y   ; xy < − 1, x < 0, y > 0 − π + tan   1 + xy   7. (i) 2 sin −1 x 1 1  − ≤x≤ sin −1 {2 x (1 − x 2 )};  2 2  1 2 −1 =  π − sin (2 x 1 − x ); ≤ x ≤1 2  − π − sin −1 (2 x 1 − x 2 ); − 1 ≤ x < − 1  2 −1 2  cos (2 x − 1); 0 ≤ x ≤1 (ii) 2 cos −1 x =  −1 2 2 π cos ( 2 x 1 ); 1 ≤ x 1 1 + x 2     2x  − π − sin −1   ; x < −1 1 + x 2    −1  1 − x 2  cos  1 + x2  ; 0 ≤ x < ∞  x= 2 − cos −1  1 − x  ; − ∞ < x ≤ 0     1 + x2  

• If sin −1 x + sin −1 y = θ, then cos −1 x + cos −1 y = π − θ • If cos −1 x + cos−1 y = θ, then sin −1 x + sin −1 y = π − θ

−1 1  sin −1 (3 x − 4 x3 ), if ≤x≤  2 2  1 −1 −1 3 (i) 3 sin x = π − sin ( 3 x − 4 x ), if < x ≤ 1  8. 2  − π − sin −1 (3 x − 4 x3 ), if − 1 ≤ x < −1 2  1  cos −1 (4 x3 − 3 x), if ≤ x ≤ 1  2  −1 1 (ii) 3 cos −1 x =  2 π − cos −1 (4 x3 − 3 x), if ≤x≤ 2 2  1 2 π + cos −1 (4 x3 − 3 x), if − 1 ≤ x ≤ 2  3  −1 1 −1  3 x − x  if

 1 − 3 x2  3  3  −1 −1  3 x − x  − π + tan  1 − 3 x2  , if x < 3 

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE `

1 The principal value of sin (a)

3π 5

20

2 If

∑

(b)

7π 5

sin−1 xi = 10π , then

i =1

(a) 20

−1

33π    cos  is  5  j NCERT Exemplar (c)

π 10

(d) −

π 10

20

∑

xi is equal to

i =1

(b) 10

(c) 0

(d) None of these

3 The domain of the function defined by f ( x ) = sin−1 (a) [1, 2] (c) [0, 1]

x − 1 is

NCERT Exemplar (b) [– 1, 1] (d) None of these j

4 The value of cos ( 2 cos (a) 1

−1

(b) 3

x + sin

−1

x ) at x =

1 is 5

(d) −

(c) 0

2 6 5

5 The value of cos[tan−1 {sin(cot −1 x )}] is (a)

1 x +2 2

(b)

x2 + 2 x2 + 1

(c)

x2 + 1 x2 + 2

(d)

1 x2 + 1

 1   has  3

6 The equation tan−1 x − cot −1 x = tan−1  (a) (b) (c) (d)

no solution unique solution infinite number of solutions two solutions

j

NCERT Exemplar

236

1  2x  . Then,  , where | x | < 1 − x 2  3

7 Let tan−1 y = tan−1 x + tan−1  a value of y is

j

3x − x 3 (a) 1− 3x 2 3x − x 3 (c) 1+ 3x 2

JEE Mains 2015

3x + x 3 (b) 1− 3x 2 3x + x 3 (d) 1+ 3x 2

(a) 0

(b)

π 4

(c)

π 2

π2 2 (c) π 2

5 π2 4 (d) None of these

(a)

(b)

19 The trigonometric equation sin−1 x = 2 sin−1 a , has a solution for

8 If θ = tan− 1 a , φ = tan− 1 b and ab = − 1, then (θ − φ) is equal to

18 The maximum value of ( sec −1x )2 + (cosec −1x )2 is

(d) None of these

 

20 If x ∈  −

9 The range of (a) [− π, π)

(b) (− π, π)

(c) [ − π, π]

(a)

10 The number of solutions of the equation cos (cos −1 x ) = cosec (cosec −1x ) is (a) 2

(b) 3

(c) 4

 

11 The value of cot  cosec−1 5 (a) 17

(d) 1

5 2 + tan−1  is 3 3

6 (b) 17

1 2

12 If tan (cos −1 x ) = sin  cot −1  , then x is equal to (a) ±

5 3

(b) ±

5 3

(c) ±

5 3

(d) None of these

π and tan−1 x − tan−1 y = 0, then 2 x 2 + xy + y 2 is equal to

13 If cos −1 x + cos −1 y = 1 (b) 2

(a) 0

14 If sin

−1

3 (c) 2

1 (d) 8

π x −1  5    + cosec   = , then the value of x is  5  4 2 j AIEEE 2007

(a) 1

(b) 3

(c) 4

(d) 5

3π 15 If sin−1 x + sin−1 y + sin−1 z = , then the value of 2 101 101 202 202 ∑ ( x303 + y 303) ( x 404 + y 404) is (x + y ) (x +y ) (a) 0

(b) 1

(c) 2

(d) 3

16 The root of the equation  x − 1 −1  2 x − 1 −1  23  tan−1   + tan   = tan   is  36  x + 1  2x + 1 3 (a) − 8

1 (b) − 2

3 (c) 4

17 The number of solutions of tan−1 x ( x + 1) + sin−1 x 2 + x + 1 = (b) 1

(c) 2

1 2

π π ,  , then the value of 2 2

x 2

(b) 2 x

(c) 3x

(d) x

π   π  + θ + tan  − + θ = a tan 3 θ, then a 3   3 

21 If tan θ + tan  (a) 1/3 (c) 3

(b) 1 (d) None of these

22 If cos −1 x = tan−1 x, then sin (cos −1 x ) is equal to (a) − x

(b) x 2

23 The real solution of tan−1 x ( x + 1) + sin−1 x (a) 2, 3

x  5

(a) 1

2

+ x +1 =

 5  4

(d) 3, 1

π , then the value of x is j 2 AIEEE 2007

(b) 3 −1

1 x2

π is 2

(c) −1, 0

(b) 1, 0

24 If sin−1   + cosec −1   =

(d) −

(c) x 3

−1

(c) 4

(d) 5

−1

25 If cosec x + cos y + sec z

≥ α 2 − 2π α + 3π , where α is a real number,

then (a) x = 1, y = −1 (c) x = 2, y = 1

(b) x = −1, z = −1 (d) x = 1, y = −2

26 The solution of sin−1 x ≤ cos −1 x is 1  (a)  −1,   2 1  (c) 1,  2 

1  (b)  −1,  2  1  (d)  1,   2

27 If m and M are the least and the greatest value of 4 (d) 3

π is 2 j

(a) 0

(d) | a | ≥

is equal to AIEEE 2008 4 (d) 17 j

3 (c) 17

 

AIEEE 2003

(b) all real values of a

3 sin 2 x   tan x  −1  tan− 1   + tan   is  4   5 + 3 cos 2 x 

f ( x ) = | 3 tan−1 x − cos −1( 0) | − cos −1( − 1) is

π π (d)  − ,   2 2 

j

1 1 (a) < | a | < 2 2 1 (c) | a | ≤ 2

NCERT Exemplar (d) infinite

(cos −1 x )2 + (sin−1 x )2 , then (a) 10

(b) 5

M is equal to m (c) 4

(d) 2

28 The number of real solutions of the equation 1 + cos 2x = 2 cos −1 (cos x ) in

π  , π is  2  j

(a) 0

(b) 1

(c) 2

NCERT Exemplar (d) infinite

237

DAY

31 The value of x for which sin[cot −1(1 + x )] = cos (tan−1 x ),

29 The sum of the infinite series  2 − 1  3 − 2  1  −1 sin−1    + sin−1   + sin   2 6  12     n − (n − 1)   is + K + sin−1  n (n + 1)   π (a) 8

π (b) 4

π (c) 2

(d) π

30 A root of the equation

10 17

(b) −1

(c) −

7 17

j

1 (a) − 2

(b) 1

(c) 0

JEE Mains 2013 1 2

(d)

32 If 0 < x < 1, then 1 + x 2 [{x cos (cot −1 x ) + sin(cot −1 x )} 2 − 1]1/ 2 is equal to (a)

x 1+ x2

(c) x 1 + x 2

(b) x

1+ x2

(d)

33 If x, y and z are in AP and tan−1 x, tan−1 y and tan−1 z are

  1 π  17x 2 + 17x tan 2 tan−1   −  − 10 = 0 is  5 4   (a)

is

also in AP, then (d) 1

j

(a) x = y = z (c) 6x = 3 y = 2 z

AIEEE 2012

(b) 2 x = y = 6z (d) 6x = 4 y = 3 z

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE  

1 If sin−1  x −

   x2 x3 x4 x6 + −... + cos −1  x 2 − + −...    2 4 2 4

π = , where 0 < | x | < 2, then x is equal to 2 (a)

1 2

(c) −

(b) 1

1 2

(d) −1

2 If the mapping f ( x ) = ax + b, a > 0 maps [ − 1, 1] onto −1

−1

−1

[ 0, 2], then cot [cot 7 + cot 8 + cot 18 ] is equal to (a) f (− 1)

(b) f (0)

(c) f (1)

(d) f (2)

   1 1 −1  3 If S = tan−1  2  + tan  2  +K  n + n + 1  n + 3 n + 3   1 + tan−1  , then tan S is equal to 1 + ( n + 19 ) ( n + 20 ) j   JEE Mains 2013 20 401 + 20n 20 (c) 2 n + 20n + 1

n n 2 + 20n + 1 n (d) 401 + 20n

(a)

4 If f ( x ) = e

π cos −1 sin  x +   3

7π (a) f  −  =e  4 

π 11

3π

7π (c) f  −  = e 12  4 

(b)

, then 13 π

8π (b) f   = e 18  9 

11 π

7π (d) f  −  = e 13  4 

3π and f (1) = 2 , 2 f ( p + q ) = f ( p ) ⋅ f (q ), ∀ p, q ∈ R , then (x + y + z ) is equal to x f (1) + y f ( 2 ) + z f ( 3 ) − f (1) x + y f (2 ) + z f (3 )

5 If sin−1 x + sin−1 y + sin−1 z =

(a) 0 (c) 2

(b) 1 (d) 3

6 If [cot −1 x ] + [cos −1 x ] = 0, where x is a non-negative real number and [.] denotes the greatest integer function, then complete set of values of x is (a) (cos 1, 1] (c) (cos 1, cot 1)

(b) (cot 1, 1) (d) None of these

7 cot −1 ( cos α ) − tan−1 ( cos α ) = x , then sin x is equal to AIEEE 2002 α (d) cot    2 j

α α (a) tan   (b) cot 2   (c) tan α  2  2 2

8 If (tan−1 x ) 2 + (cot −1 x ) 2 = (a) −2

(b) −3

5π 2 , then the value of x is 8 (c) −1

(d) 2

−1

9 The solution set of tan (sin x ) > 1 is 2

1   1  (a)  − 1, − , 1  ∪  2  2  (c) (− 1, 1 ) ~ {0}

1 1  (b)  − ,  ~ {0 }  2 2 (d) None of these

10 If θ and φ are the roots of the equation 8x 2 + 22x + 5 = 0, then (a) (b) (c) (d)

both sin−1 θ and sin−1 φ are equal both sec−1 θ and sec−1 φ are real both tan−1 θ and tan−1 φ are real None of the above

11 2 tan−1( − 2) is equal to −3 (a) cos−1    5  π 3 (c) − + tan−1  −   4 2

3 5 3 (d) − π + cot −1  −   4

(b) π + cos−1

12 Let x ∈( 0, 1 ). The set of all x such that sin−1 x > cos −1 x , is the interval

1 1  (a)  ,  2 2 (c) (0, 1)

j

1  (b)  , 1  2   3 (d)  0,  2  

JEE Mains 2013

238

ANSWERS SESSION 1

1 (d)

2 (a)

3 (a)

4 (d)

5 (c)

6 (b)

7 (a)

8 (c)

9 (a)

10 (a)

11 (b)

12 (b)

13 (c)

14 (b)

15 (d)

16 (d)

17 (b)

18 (b)

19 (c)

20 (d)

21 (c)

22 (b)

23 (c)

24 (b)

25 (a)

26 (b)

27 (a)

28 (a)

29 (c)

30 (d)

31 (a)

32 (c)

33 (a)

1 (b) 11 (c)

2 (d) 12 (b)

3 (c)

4 (b)

5 (c)

6 (b)

7 (a)

8 (c)

9 (a)

10 (c)

SESSION 2

Hints and Explanations SESSION 1 1 cos  33 π  = cos  6 π + 3 π  = cos 3 π  5   5 5 π 3π  π   = sin  −  = sin  −  2  10  5 33 π  ∴ sin −1  cos   5  π  π  = sin −1  sin  −  = −  10    10 2 Since, − π ≤ sin −1 x ≤ π 2

2 π , 1 ≤ i ≤ 20 2 x i = 1, 1 ≤ i ≤ 20

sin −1 x i =

∴ ⇒

20

Thus,

∑

x i = 20

i =1

3 Given, f ( x ) = sin −1 x − 1 For domain of f ( x ) ⇒ ∴

− 1≤

x−1≤1

0 ≤ (x − 1)≤ 1 ⇒1 ≤ x ≤ 2 x ∈ [1, 2]

4 cos (2 cos −1 x + sin −1 x ) = cos [2(cos −1 x + sin −1 x ) − sin −1 x] = cos ( π − sin −1 x ) = − cos (sin −1 x ) 1   Q x = 1  = − cos sin −1      5  5 2  1  = − cos cos −1 1 −     5     2 6 −1 2 6  = − cos  cos =− 5  5 

5 We have, cos[tan −1 {sin(cot −1 x )}] Let cot −1 x = α ⇒ cotα = x ⇒ cosec α = 1 + x2

⇒

sinα =

⇒

1

7 Given,

1 + x2

α = sin

−1

1

1 + x2 Hence, cos[tan {sin(cot −1 x )}]      1 = cos  tan −1 sin  sin −1   2 1 + x      −1

  1 = cos  tan −1 2  1 + x    x2 + 1  −1 = cos cos  x2 + 2      1 −1   =β  Q let tan   1 + x2    1   = tan β   2 + x 1   2  1 x + 2 = + = sec β 1   1 + x2 x2 + 1     x2 + 1 cos β =  x2 + 2   =

x2 + 1 x2 + 2

6 Given, tan −1 x − cot −1 x = tan −1  1   

3 π …(i) ⇒ tan x − cot x = 6 π But tan −1 x + cot −1 x = …(ii) 2 On adding Eqs. (i) and (ii), we get π π 2π 2 tan −1 x = + = 6 2 3 π −1 ⇒ tan x = 3 π ⇒ x = tan ⇒ x = 3 3 It has unique solution. −1

−1

2x  tan −1 y = tan −1 x + tan −1  ,  1 − x2  1 where| x | < 3

  2x  x + 1 − x2  ⇒ tan y = tan    1 − x  2 x   2   1 − x    −1 −1 −1  x + y  Q tan x + tan y = tan  1 − xy  ,    −1

−1

 xy < 1   3 −1  x − x + 2 x  = tan    1 − x2 − 2 x2   3 x − x3  tan −1 y = tan −1    1 − 3 x2  3 x − x3 1 − 3 x2 Alternate Method 1 1 1 | x |< ⇒− < x< 3 3 3 π π Let x = tanθ ⇒ − < θ < 6 6 ∴ tan −1 y = θ + tan −1 (tan 2θ) = θ + 2θ = 3θ ⇒ y = tan3θ 3 tan θ − tan3 θ ⇒ y = 1 − 3 tan θ 3 x − x3 y = ⇒ 1 − 3 x2 ⇒

y =

8 Given that, θ = tan −1 a and φ = tan −1 b and ab = −1 ∴tan θ tan φ = ab = − 1

DAY ⇒ ⇒ ⇒

tan θ = − cot φ π tan θ = tan  + φ 2  π θ−φ= 2 −1

−1

−1

9 f ( x ) =| 3 tan x − cos (0 ) | − cos (−1 )

⇒ ⇒

−1

π π < tan −1 x < 2 2 −3 π 3π < 3 tan −1 x < 2 2 π −2 π < 3 tan −1 x − < π 2 π 0 ≤ 3 tan −1 x − < 2π 2 π − π ≤ 3 tan −1 x − − π< π 2 −1

cosec (cosec x ) = x, ∀ x ∈ (−∞, − 1] ∪ [1, ∞ ) ⇒ cos (cos −1 x ) = cosec (cosec −1 x ) for x = ± 1 only. Hence, there are two roots.

11 Since, cosec  5 = tan −1  3   3  4 2  −1 3 + tan −1  ∴ cot  tan  4 3   3 + 2     = cot  tan −1  4 3   1   1 −    2  −1

 = cot tan −1 

⇒ sin

−1

⇒ sin

−1

⇒ sin −1 ⇒ sin

sin φ =

1 + cot φ 2

⇒

19 Given that, sin −1 x = 2sin −1 a Q ⇒

x −1  3   = sin    5 5

⇒

x=3

sin −1 x = sin −1 y = sin −1 z =

π 2

2

⇒ ⇒ ⇒

x= y = z=1 ( x101 + y 101 ) ( x202 + y 202 ) ∴ Σ 303 ( x + y 303 ) ( x 404 + y 404 ) (1 + 1) (1 + 1) =Σ = Σ1 = 3 (1 + 1) (1 + 1)

π π ≤ sin −1 x ≤ 2 2 π π − ≤ 2sin −1 a ≤ 2 2 π π − ≤ sin −1 a ≤ 4 4 π π  sin  −  ≤ a ≤ sin  4 4 1 1 − ≤ a≤ 2 2 1 |a| ≤ 2 −

tan x  3 sin 2 x  −1  20 tan −1    + tan  

= tan −1

 

 x − 1 2x − 1 +  x + 1 2x + 1  x − 1 2 x − 1     1 −       x + 1   2 x + 1  

=

23 = tan −1    36 

2 5

tan x  6 tan x  −1  = tan −1    + tan  2  4   8 + 2 tan x 

2 x2 − 1 23 = 3x 36

⇒

3 tan x  tan x  −1  = tan −1    + tan  2  4   4 + tan x 

⇒ 24 x − 12 − 23 x = 0 4 3 x= ,− ⇒ 3 8 But x cannot be negative. 4 ∴ x= 3 2

1 − x2

1 − 1 ⇒ tan θ = x2

x 1  2

1 − x2    x  2   = sin  sin −1   5

17 For existence, x ( x + 1) ≥ 0 and ⇒ ⇒

x2 + x + 1 ≤ 1 x2 + x ≤ 0 x( x + 1) ≤ 0

4   5 + 3 cos 2 x   tan x  + tan −1    4    6 tan x   2 1 + tan x   2  3 ( 1 − tan x )  + 5   1 + tan2 x  

16 tan −1 

Now, tan (cos −1 x ) = sin  cot −1   ⇒ tan  tan −1  

∴ I max

⇒

1 Let cos x = θ ⇒ sec θ = x ⇒ tan θ = sec2 θ − 1 tan θ =

5 4 2 x π −1  4   + sin   =  5 5 2 x π   = −1  4  − sin      5  5 2  x  = cos −1  4       5  5   

  

π2  + 2 (sec −1 x )2  4  π π2 π2  − (sec −1 x ) + − 2 16 16  2  π2 π  = + 2   sec −1 x −   8 4   2 2  9π  π 5π2 = +2 =   8 4  16  =

15 Given, sin x + sin −1 y + sin −1 z = 3 π

−1

⇒

−1

2

1

= (sec −1 x + cosec −1 x )2 − 2 sec −1 x cosec −1 x π2 π = − 2 sec −1 x  − sec −1 x  2  4

−1

12 Let cot −1 1 = φ ⇒ 1 = cot φ ⇒

x − tan

∴

  17      12        1      2     17   = 6    6   17

2

18 Let I = (sec −1 x )2 + (cosec −1 x )2

−1

14 Q sin −1  x  + cosec −1  5 = π    

10 cos (cos −1 x ) = x, ∀ x ∈[− 1, 1] and

  = cot  tan −1   

From Eqs. (i) and (ii), we get x( x + 1) = 0 ⇒ x = 0, − 1 But x = −1 is not satisfied the given equation.

y = 0⇒ x = y π −1 Also, cos x + cos −1 y = 2 π π ⇒ 2 cos −1 x = ⇒ cos −1 x = 2 4 1 1 2 ⇒ ⇒ x = x= 2 2 3 2 2 Hence, x + xy + y = 3 x2 = 2

We know that, −

⇒

2 = ⇒ (1 − x2 )5 = 2x x 5 On squaring both sides, we get (1 − x2 )5 = 4 x2 5 ⇒ 9 x2 = 5 ⇒ x = ± 3

13 Qtan

π = 3 tan −1 x −   − π  2

⇒

1 − x2

⇒

= tan −1

…(i)

…(ii)

 tan x 3 tan x  +   4 4 + tan2 x   2   3 tan x  1−  4(4 + tan2 x )     tan x 3 tan x   < 1 ⋅  as  4 tan2 x    4

 16 tan x + tan3 x  = tan −1   16 + tan2 x   = tan −1 (tan x ) = x

240 21 tan θ + tan  π + θ + tan  − π + θ = a tan 3 θ 3 + tan θ

x 4 sin −1   = cos −1    5  5 x 3 ⇒ sin −1   = sin −1    5  5

1 − 3 tan θ

∴

3

⇒ tan θ + +





tan θ − 3 1+

3 tan θ



3

RHS = α2 − 2 π α + 3 π

⇒ ⇒

x = tan

sin θ cos θ = cos θ

cos θ = tan θ ⇒

⇒ cos 2 θ = sin θ ⇒ sin2 θ + sin θ − 1 = 0 −1± 1+ 4 ⇒ sin θ = 2 5−1 ⇒ sin θ = 2 5−1 2 2 ∴ x = cos θ = sin θ = 2 and sin (cos −1 x ) = sin θ 5−1 = = x2 2

π x + x+1= 2 2

1 1 + (x + x ) x2 + x + 1 =

1

⇒ cos −1

1 + ( x2 + x )

=

π 2

− sin −1

⇒

π − 1 ≤ x ≤ sin    4 1   x ∈ − 1,  2 

x2 + x + 1

x2 + x + 1 1

x2 + x + 1

=

⇒

x2 + x + 1

x2 + x + 1

x + x + 1 = 1 ⇒ x = − 1, 0 2

 5  4 x 4 π     ⇒ sin −1   + sin −1   =  5  5 2

x π 4 sin −1   = − sin −1    5  5 2

2

= tan



π2 + 2 (sin −1 x )2 − π sin −1 x 4 2 π π2 = + 2 sin −1 x −    8 4  π2 5π2 Here, m = ,M = 8 4 M ∴ = 10 m =

28 Given, 1 + cos 2 x = 2 cos −1 (cos x ) ∴

2cos 2 x = 2 x

⇒

2 |cos x | = 2 x

π For x ∈  , π  , |cos x| = − cos x  2  − 2 cos x = 2 x ⇒

− cos x = x

−1

∴ S ∞ = tan

 r − r −1    1 + r r − 1

−1

r − tan −1 (r − 1)]

−1

n − tan −1

0

n −0 ∞ =

π 2

1 2×  −1  1   5 = 5 30 Now, tan 2 tan    = 1   5 12   1− 25 Given equation can be rewritten as π 1 17 x2 − 17 x tan  − 2 tan −1    − 10 = 0  5  4 5 1− 2 12 − 10 = 0 ⇒ 17 x − 17 x ⋅ 5 1+ 12 ⇒ 17 x2 − 7 x − 10 = 0 ⇒ ( x − 1) (17 x + 10) = 0 Hence, x = 1 is a root of the given equation. 



1  31 sin  sin −1  (1 + x )2 + 1     1 = cos  cos −1  2  1+ x  1 1 = 2 (1 + x ) + 1 1 + x2

2

27  π − sin −1 x  + (sin −1 x )2 2

24 Since, sin −1  x  + cosec −1  5 = π

⇒

π 2

1

−1

= cos −1 ⇒

Qsin −1 x + cos −1 x = π , ∀x ∈ [−1,1] 2   π π π −1 −1 ⇒ sin x ≤ ⇒ − ≤ sin x ≤ 4 2 2  − π , π  −1 Q range of sin x is     2 2  

2

+ sin −1

⇒ cos

π ≥ 2 sin −1 x 2

⇒

n

∑ [tan

= tan −1

…(i)

π π QLHS, cosec x ∈  − ,  − {0}  2 2  cos −1 y ∈ [0, π] π and sec −1 z ∈ [0, π] −   2 5π LHS ≤ …(ii) ∴ 2 From Eqs. (i) and (ii), we get only possibility is sign of equality x = 1, y = − 1, z = − 1

⇒

−1

r =1

26 Given, cos −1 x ≥ sin −1 x

23 Given, tan −1 x ( x + 1 ) + sin −1

⇒ cos −1

=

−1

x=θ

n

∑ tan r =1

2

π 5π 5π ≥  + 2 2 2

 = α − 

x = cos θ = tan θ

∴ Sn =

π π π α + + 3π − 2 2 2

= α2 − 2

⇒ 3 tan 3 θ = a tan 3 θ ⇒ a=3

22 Let cos

x=3

cosec −1 x + cos −1 y + sec −1 z ≥ α2 − 2π α + 3π

3(3 tan θ − tan3 θ ) ⇒ = a tan 3 θ 1 − 3 tan2 θ

−1

 r − (r − 1)   r (r + 1)    r − (r − 1)  = tan −1    1 + r ( r − 1 )

29 Q T r = sin −1 

25 Given,

= a tan 3 θ

8 tan θ = a tan 3 θ ⇒ tan θ + 1 − 3 tan2 θ

−1

∴ cos x = − x Hence, no solution exist.

⇒

⇒ ⇒ ⇒

(1 + x )2 + 1 = 1 + x2 2x + 1 = 0 1 x=− 2

32 We have, 0 < x < 1 Now, 1 + x2 [{ x cos(cot −1 x ) + sin(cot −1 x )}2 − 1]1 /2 = 1+ x

2

 x   x +   1 + x2 

1 /2

2   − 1   1 + x2  

1

1 /2

2   1 + x2  = 1 + x  − 1   1 + x2     2

= 1 + x2 [1 + x2 − 1)1 /2 = x 1 + x2

DAY 33 Since, x, y and z are in AP. ∴ 2y = x + z Also, tan −1 x, tan −1 y and tan −1 z are in AP. ∴2 tan −1 y = tan −1 x + tan −1 z ⇒ ⇒

 2y   x+ z tan −1  = tan −1   2 1 − y   1 − xz  x+ z x+ z = 1 − y 2 1 − xz

[Q2y = x + z] ⇒ y 2 = xz Since x, y and z are in AP as well as in GP. ∴ x= y =z

 = cot tan −1 

 65      195 

= 1 + 2 = f (2)

[Q f ( x ) = x + 1]

(n + 1 ) − (n + 0)   1  + (n + 0) (n + 1 )

 (n + 2) − (n + 1 )  + tan    1 + (n + 1 ) (n + 2)  (n + 20) − (n + 19)  + K + tan −1    1 + (n + 19) (n + 20) −1

2

3

2

4 x

2x = x 2+ x 1+ 2 4 x x6 2 and x − + −K 2 4 2 x 2 x2 = = 2 x 2 + x2 1+ 2 π Q sin −1 α + cos −1 α = 2 2x 2 x2 ∴ = ,x ≠ 0 2 + x 2 + x2 =

2 + x2 = 2 x + x2 x=1

= tan −1 (n + 1 ) − tan −1 n

⇒

f (− 1) = 0 and f (1) = 2

⇒

− a+ b = 0

∴

f ( x) = x + 1

Now, cot[cot −1 7 + cot −1 8 + cot −1 18] 1 1   tan −1   + tan −1     7  8  = cot   1  + tan −1      18      1+ 1     −1 −1  1   7 8 = cot  tan   + tan     18  1 − 1⋅ 1     7 8   15 1   = cot tan −1   + tan −1     55  18   3 1   = cot tan −1   + tan −1     11   18  

and 0 < cot −1 x < π Given, [cot −1 x] + [cos −1 x] = 0 ⇒ [cot −1 x] = 0 and [cos −1 x] = 0 ⇒ 0 < cot −1 x < 1 and 0 ≤ cos −1 x < 1 ∴ x ∈ (cot 1, ∞ ) and

x ∈ (cos 1, 1 ) ⇒ x ∈(cot 1, 1 )

7 Given that, cot −1 ( cos α ) − tan −1 ( cos α ) = x …(i) We know that, cot −1 ( cos α ) + tan −1 ( cos α ) =

 n + 20 − n  = tan −1    1 + n(n + 20)

Qcot −1 x + tan −1 

  20 = tan −1  2   n + 20n + 1     20 ⇒ tan S = tan  tan −1  2   n + 20n + 1    20 n2 + 20n + 1

4 Given, f ( x ) = e ⇒

 π cos −1 sin  x +   3

cos 8π  f   =e  9 

=e

a+ b =2 a=b =1

0 ≤ cos −1 x ≤ π

= tan −1 (n + 20) − tan −1 n

∴ f ′( x ) = a > 0 So, f ( x ) is an increasing function.

3 =3−1=2 1+ 1+ 1

+ tan −1 (n + 2) − tan −1 (n + 1 ) + ... + tan −1 (n + 20) − tan −1 (n + 19)

⇒ tan S =

2 Q f ( x ) = ax + b

Then,

6Q

3 S = tan −1 

1 Now, x − x + x − …

and

=1+ 1+ 1−

1   = cot tan −1   = cot(cot −1 3) = 3   3  

SESSION 2

⇒ ∴

Then, f (2) = f (1 )f (1 ) = 2 ⋅ 2 = 4 and put p = 1, q = 2 Then, f (3) = f (1 )f (2) = 2 ⋅ 22 = 8 x+ y + z ∴ x f(1 ) + y f(2 ) + z f(3 ) − f(1 ) x + y f(2 ) + z f(3 )

  3 + 1     −1 = cot  tan  11 18   3 1 1 − ⋅    11 18   

−1

 11 π  cos −1 sin   9 

cos 8π  f   =e  9 

⇒

 8π π  sin  +   9 3

−1

 13 π  cos   18 

cos 7π  Also, f  −  =e  4

=e

−1

=e

13 π 18

 7π π  sin  − +   4 3

 17 π  cos −1 sin  −  12 

=e

cos −1 cos

π

π

12

= e 12

5 Q− π ≤ sin −1 x ≤ π , − π ≤ sin −1 y ≤ π 2

2 2 π π ≤ sin −1 z ≤ 2 2 Given that,

2

and −

sin −1 x + sin −1 y + sin −1 z =

3π 2

which is possible only when sin −1 x = sin −1 y = sin −1 z = ⇒ x= y = z=1 Put p = q = 1

π 2

π 3 2 …(ii) π x=  2 

On adding Eqs. (i) and (ii), we get π 2cot −1 ( cos α ) = + x 2 π x ⇒ cos α = cot  +  4 2 x cot − 1 2 cos α = ⇒ x 1 + cot 2 x x cos − sin 2 2 ⇒ cos α = x x cos + sin 2 2 On squaring both sides, we get x x cos 2 + sin2 2 2 ⇒ cos α = x x cos 2 + sin2 2 2 x x − 2sin cos 2 2 x x + 2sin cos 2 2 1 − sin x ⇒ cos α = 1 + sin x α 1 − tan2 2 = 1 − sin x ⇒ 2 α 1 + sin x 1 + tan 2 On applying componendo and dividendo rule, we get α sin x = tan2    2

242 8 Given, (tan −1 x )2 + (cot −1 ⇒ (tan −1 x + cot −1

5π2 x )2 = 8 x )2 − 2 tan −1 x ⋅ cot −1 x =

or − ⇒ 5π2 8

⇒

π π < sin −1 x < − 2 4 1   1   x∈ , 1 or x ∈  − 1, −   2   2 1    1  x ∈  − 1, − , 1  ∪   2  2

⇒

2

π π ⇒   − 2 tan −1 x  − tan −1 x   2 2  5π2  π −1 −1 = Q tan x + cot x =  8 2   π2 π ⇒ − 2 ⋅ tan −1 x 4 2 5π2 + 2 (tan −1 x )2 = 8 3 π2 2 −1 −1 ⇒ 2 (tan x ) − π tan x − =0 8 π 3π ⇒ tan −1 x = − , 4 4 π ⇒ tan −1 x = − 4 ⇒ x= −1     neglecting tan −1 x     3π as principal value of  =   4 π π   tan −1 x ∈  − ,     2 2

9 tan2 (sin −1 x) > 1 ⇒

π π < sin −1 x < 4 2

10 8x2 + 22x + 5 = 0 1 5 x= − ,− ⇒ 4 2 1 5 Q−1 < − < 1 and − < − 1 4 2 1 5 ∴ sin −1  −  exists but sin −1  −   4  2 does not exist. 5 1 sec −1  −  exists but sec −1  −  does  2  4

−3 − 2θ = cos −1    5 3 = π − cos −1 5 3 5 −1 4 π + tan 3 3 π + cot −1 4 π 3 π+ − tan −1 2 4 π −1 3 − tan 2 4 π 3 −1  + tan  −   4 2

⇒

2θ = − π + cos −1

⇒

2θ = − =− =− =− =−

12 Y

not exist. So, and

tan −1 tan −1

 − 1    4  − 5 both exist.    2

11 Let tan −1 (− 2) = θ ⇒ ⇒

π y = cos –1 x

X′

(–1, 0)

π 2

y = sin–1 x

π 4

O

tan θ = − 2 π θ ∈  − , 0  2 

⇒ 2θ ∈ (− π, 0 ) Now, cos ( − 2θ) = cos 2θ 1 − tan2 θ = 1 + tan2 θ −3 = 5

X 1,0 (1, 0) 2

Y′

 1  ∴ sin −1 x > cos −1 x, ∀ x ∈  , 1  2 

243

DAY

DAY TWENTY THREE

Unit Test 3 (Trigonometry)

1 The range of f ( x ) = sin−1 x + tan−1 x + sec −1x , is π (a)  , 4 π (c)  , 4

3π  4  3π  4 

π 3π (b)  ,  4 4 

(a) 3 (c) 5

(d) None of these

2 From a point a m above a lake the angle of elevation of a cloud is α and the angle of depression of its reflection is β. The height of the cloud is a sin (α + β) m sin (α − β) a sin (β − α) (c) m sin (α + β)

(b)

(a)

2n

a sin(α + β) m sin( β − α)

(d) None of these

i =1

(a) n n (n + 1) (c) 2

(d) None of these

4 The value of x for which cos (cos 4) > 3x − 4x, is 2

 2 + 6π − 8   (a)  0, 3    2 − 6π − 8  (b)  , 0 3  

(a) α = 0

x + cos value of y is

9 If cos (a)

1 2

(b)

−1

(c) positive

1 − x + cos

3 2

−1

(d) zero

1 − y = π, then the

(c) 1

(d)

1 4

9 40

π (c) α = 4

(b)

9 41

(c)

3 10

(d)

21 31

11 The angular depressions of the top and the foot of a chimney as seen from the top of a second chimney, which is 150 m high and standing on the same level as 4 the first are θ and φ, respectively. If tan θ = and 3 5 tan φ = , then the distance between their tops is 2 (b) 110 m (d) None of these

60° and at a place Q due South of it is 30°. If the distance PQ be 200 m, then the height of the hill is π (d) β = 3

6 The angles of elevation of the top of a tower at the top and the foot of a pole of height 10 m are 30° and 60°, respectively. The height of the tower is (a) 10 m (c) 20 m

(b) negative

−1

12 The elevation of the hill from a place P due West of it is

6π − 8   3 

5 If α ≤ sin−1 x + cos −1x + tan −1x ≤ β, then π (b) β = 2

(a) − 3

(a) 120 m (c) 100 m

(c) (− 2, 2) 6π − 8 2 + , 3

when α , β and γ are real numbers satisfying α + β + γ = π , is

(a)

(b) 2n

−1

2 − (d)  

8 The minimum value of the expression sin α + sin β + sin γ,

5 cos θ + 4 sin θ = 3, then cos (α + β ) is equal to

3 If ∑ cos xi = 0, then ∑ xi is equal to i =1

(b) 4 (d) None of these

10 If α and β are the roots of the equation

2n

−1

7 The maximum value of 12 sin θ − 9 sin2 θ is

(b) 15 m (d) None of these

(a) 109.54 m (c) 110.6 m

(b) 108.70 m (d) None of these

7π 2π    1  − sin  cos   is equal to 5 5   2

13 cos −1 

13 π 20 33 π (c) 20 (a)

(b)

21 π 20

(d) None of these

244

14 sec 2 θ =

4 ab , where a , b ∈ R is true if and only if (a + b ) 2

(a) a + b ≠ 0 (c) a = b

(b) a = b, a ≠ 0 (d) a ≠ 0, b ≠ 0

(a) 5

15 If x = cos −1(cos 4) and y = sin−1(sin 3), then which of the following conditions holds? (a) x − y = 1 (c) x + 2 y = 2

(b) x + y + 1 = 0 (d) tan(x + y) = − tan 7



16 If log2 x ≥ 0 , then log1/ π sin−1 

 2x + 2 tan−1 x  is equal 1+ x2  (b) 0 (d) None of these

5 sin 2A + 3 sin A + 4 cos A is equal to 24 5 48 (d) 5 (b) −

(a) 0 24 5 2

(b)

3 3 2 (c) 9 3

x x is equal to = cosec x − sin x , then tan2 2 2

(a) 2 − 5 (c) (9 − 4 5 ) (2 +

5)

(b) 5 − 2 (d) (9 + 4 5 ) (2 − 5 )

θ (a) 2 cos  n  2  θ (c) 2 cos  n + 1  2 

θ (b) 2 cos  n − 1  2  (d) None of these

sin2 A : sin2 B : sin2 C is equal to 9 5 8 : : 10 10 10 9 8 7 (c) : : 10 10 10 (a)

1 3 3

(d) None of these

(b)

9 7 8 : : 10 10 10

(d) None of these

29 If 0 ≤ x ≤ 3π, 0 ≤ y ≤ 3π and cos x ⋅ sin y = 1, then the

19 If sin x + a sin x + 1 = 0 has no real number solution, 2

possible number of values of the ordered pair ( x , y ) is (a) 6

then (a) | a | ≥ 2 (c) | a | < 2

(b) | a | ≥ 1 (d) None of these

n+1 (a) tan α n −1 n (c) tan α 1+ n

(a)

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

21 sin6 x + cos 6 x lies between

(c) 0 and 1

1 and 2 4

π 22 If n = , then tan α ⋅ tan 2 α ⋅ tan 3 α K tan ( 2n − 1) α is 4α equal to

23 The ratio of the greatest value of 2 − cos x + sin2 x to its least value is (b)

9 4

(c)

13 4

(d)

2

3

2

sin θ, then the value of (m + n ) 2 / 3 + (m − n ) 2 / 3 is (b) a 2 / 3

(c) a 3 / 2

nπ 2

(c) 3nπ

(d) None of these

x π sin2 x = x 2 + x −2 , 0 < x ≤ has 2 2

no real solution a unique real solution finitely many real solutions infinitely many real solutions

subtend the same angle α at a point O on the road. If the height of the largest pole is h and the distance of the foot of the smallest pole from O is a, the distance between two consecutive poles is h cos α − a sin α (n − 1) sin α h cos α − a sin α (c) (n + 1) sin α

(d) 4 a 2 / 3

h cos α + a sin α (n − 1) sin α a cos α − h sin α (d) (n − 1) sin α (b)

33 The solution of the equation

17 4

24 If m = a cos θ + 3 a cos θ sin θ, n = a sin θ + 3 a cos θ 3

(b)

(a)

(b) −1 (d) None of these

1 4

(d) 15

32 The n poles standing at equal distance on a straight road

(d) None of these

(a) 1 (c) ∞

nπ 3

31 The equation 2 cos 2

(d) None of these

(b)

(c) 8

1 + sin4 2x = cos 2 6x is

1+ n (b) tan α 1− n

1 and 1 4

(b) 12

30 The general solution of the equation

20 If sin θ = n sin ( θ + 2 α ), then tan ( θ + α ) is equal to

(a) 2a 2 / 3

(d) None of these

28 In ∆ABC, tan A + tan B + tan C = 6, tan B tan C = 2 , then

18 The minimum value of 27 cos x + 81 sin x is

(a)

a 2 − 5a + 6 a2

a 2 + 5a + 6 a2

(where, number of 2’ s is n) is equal to

17 If A lies in the third quadrant and 3 tan A − 4 = 0, then

(a)

(b)

27 If 0° < θ < 180° , then 2 + 2 + 2 + ... + 2 (1 + cos θ )

(a) log1 / π (4 tan−1 x) (c) −1

(a)

(c)

26 If tan

to

(c)

π 1 cosec 2 α = 0, 0 < α < , then a 2 cos 2 α + 5 sin α cos α + 6 sin2 α is equal to

25 If a sin2 α −

 8 tan−1( x + 1) + tan−1 ( x − 1) = tan−1   is  31 1 4 (c) − 8 (a)

1 or − 8 4 (d) None of these

(b)

245

DAY π 2

I. The general value of θ satisfying the equations sin2 θ = sin2 α, cos 2 θ = cos 2 α and tan2 θ = tan2 α is given by θ = n π ± α II. The general value of θ satisfying equations sin θ = sin α and cos θ = cos α simultaneously is given by θ = 2nπ ± α , n ∈ Z .

34 The solution of cos −1 x 3 + cos −1 x = , is (a) − (c)

1 2

1 2

(b) −

1 1 or 2 2

(d) None of these

35 The number of solutions of the equation

tan x + sec x = 2 cos x lying in the interval [ 0, 2 π ], is (a) 0

(b) 1

(c) 2

(d) 3

36 If cot 2 x + cosec x − a = 0 has atleast one solution, then complete set of value of a is (a) [− 1, ∞ )

(b) (− 1, 1)

(c) (− ∞, 1)

(d) (− ∞, 1]

37 From the top of the cliff 400 m high the top of a tower was observed at an angle of depression 45° and from the foot of the tower the top of the cliff was observed at an angle of elevation 60°. The height of the tower is (a) 169.06 m (c) 180.0 m

(b) 179. 2 m (d) None of these

38 If cos 5 θ = P cos θ − 20 cos 3 θ + Q cos 5 θ + R, then P + Q + R is equal to (a) 21

(b) 18

(c) 15

(d) 31

(b) obtuse angled (d) isosceles

40 If the area of a ∆ ABC be λ, then a 2 sin 2 B + b 2 sin 2A is equal to (a) 2 λ (c) 4λ

(c) 2

(d) −1

42 If sin x (1 + sin x ) + cos x (1 + cos x ) = A and

sin x (1 − sin x ) + cos x (1 − cos x ) = B, then the value of A 2 − 2A − sin 2x = B 2 + 2B − sin 2x is (a) 1

(b) 2

(c) 3

(d) 0

43 If the sides a, b and c of a ∆ABC are roots of the equation x 3 − 15x 2 + 47x − 82 = 0, then the value of cos A cos B cos C is + + a b c (a)

225 164

(b)

131 164

(c)

131 82

(d)

169 82

Statement I The number of solutions of the pair of equations 2 sin2 θ − cos 2θ = 0 and 2 cos 2 θ − 3 sin θ = 0 in the interval [ 0, 2π ] is two. Statement II If 2 cos 2 θ − 3 sin θ = 0, then θ cannot lie in III or IV quadrant.  θ  2

49 Statement I If 2 sin   = 1 + sin θ + 1 − sin θ, θ π 3π lies between 2nπ + and 2nπ + . 2 4 4 π 3π θ Statement II If ≤ θ ≤ , then sin > 0 . 4 4 2

50 Suppose a person is standing on a ground and see the

angle c is (a) 30° or 150° (c) 60° or 120°

tower, then they make an angle of elevation.

(b) 45° or 135° (d) None of these

45 Let f (θ, α ) = 2 sin2 θ + 4 cos(θ + α )

π sin θ sin α + cos 2 (θ + α ). Then, the value of f  , 3 (b) 1

(c) 2

π  is 4

(d) 3

46 Which among the following is/are correct statement/ statements?

 1 1  Statement II sin−1( 2x 1 − x 2 ) = 2 sin−1 x , if x ∈ − , .  2 2 

then

44 If in a ∆ABC, a 4 + b 4 + c 4 − 2b 2c 2 − 2c 2a 2 = 0. Then, the

(a) 0

1 ≤ x ≤ 1, then 2 x 3 − 3x 2  π cos −1 x − sin−1  +  is equal to − . 2 2 3  

said to be solution of an equation.

is (b) 1

(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

48 Suppose any angle is satisfied both equations, then it is

(b) λ (d) None of these

41 If cos 2x + 2 cos x = 1, then the value of sin2 x ( 2 − cos 2 x ) (a) 0

Directions (Q. Nos. 47-50) Each of these questions contains two statements : Statement I and Statement II. Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a ), (b), (c) and (d ) given below.

47 Statement I If

cos A cos B cos C , then the triangle is 39 In ∆ABC, if = = a b c (a) right angled (c) equilateral

(a) I is correct (b) II is correct (c) Both I and II are correct (d) Both I and II are incorrect

Statement I A tower subtends angles α , 2α and 3 α respectively at points A , B and C all lying on a horizontal line through the foot of tower, then AB AB : BC = sin 3 α : sin α. a b c Statement II In ∆ABC, if , then ∆ABC = = sin A sin B sin C is an equilateral triangle.

246

ANSWERS 1 (c)

2 (b)

3 (b)

4 (d)

5 (a)

6 (b)

7 (b)

8 (c)

9 (c)

10 (b)

11 (c)

12 (a)

13 (d)

14 (b)

15 (d)

16 (c)

17 (a)

18 (c)

19 (c)

20 (b)

21 (a)

22 (a)

23 (c)

24 (a)

25 (d)

26 (c)

27 (a)

28 (a)

29 (a)

30 (b)

31 (a)

32 (a)

33 (a)

34 (c)

35 (c)

36 (a)

37 (a)

38 (a)

39 (c)

40 (c)

41 (b)

42 (d)

43 (b)

44 (b)

45 (a)

46 (a)

47 (b)

48 (a)

49 (b)

50 (c)

Hints and Explanations ⇒ 3 x2 − 4 x − (2 π − 4) < 0

1 f ( x ) = sin −1 x + tan −1 x + sec −1 x, Hence, domain of f ( x ) is ± 1. So, the π 3π  range is { f (1), f (− 1)} i.e.  , . 4 4 

2 In ∆ PMC, tanα =

h−a PM

⇒

am

h

M

β

Q

3

< x
c 2 − 4 (d) None of these

7 If the line 3x + 4y = 7 touches the ellipse 3x 2 + 4y 2 = 1 then the point of contact is 1 1  (a)  ,   7 7 1 − 1 (c)  ,   7 7

1 − 1 (b)  ,   3 3 (d) None of these

8 The equation of common tangent of the curves x 2 + 4y 2 = 8 and y 2 = 4x are (a) x − 2 y + 4 = 0, x + 2 y + 4 = 0 (b) 2 x −y + 4 = 0, 2 x + y + 4 = 0 (c) 2 x − y + 2 = 0, 2 x + y + 2 = 0 (d) None of the above

(a) x + y − 6y − 7 = 0 (c) x 2 + y 2 − 6y − 5 = 0 2

2

2

2

11 If tangent at any point P on the ellipse 7x 2 + 16y 2 = 12 cuts the tangent at the end points of the major axis at the points A and B, then the circle with AB as diameter passes through a fixed point whose coordinates are (a) (± (c) (0, ±

a 2 − b 2 , 0)

(b) (±

a2 − b 2 )

a 2 + b 2 , 0)

(d) (0, ±

a2 + b 2 )

12 The length of the common tangent to the ellipse x2 y2 + = 1 and the circle x 2 + y 2 = 16 intercepted by 25 4 the coordinate axes is (a) 5

(b) 2 7

(c)

7 3

(d)

14 3

13 The distance of the centre of ellipse x 2 + 2y 2 − 2 = 0 to those tangents of the ellipse which are equally inclined from both the axes, is (a)

3 2

(b)

3/2

(c)

2 /3

(d)

3 2

14 PQ is a chord of the ellipse through the centre. If the square of its length is the HM of the squares of major and minor axes, find its inclination with X -axis. π 4 2π (c) 3 (a)

(b)

π 2

(d) None of these

15 If straight line ax + by = c is a normal to the ellipse 4x 2 + 9y 2 = 36, then 4a 2 + 9b 2 is equal to 169a 2 b 2 c2 13a 2 b 2 (c) c2 (a)

(b)

25a 2b 2 c2

(d) None of these

305

DAY 16 If the normal at the point P(θ) to the ellipse 5x 2 + 14y 2 = 70 intersects it again at the point Q(2θ), then cos θ is equal to (a)

2 3

(b) −

2 3

(c)

1 3

(d) −

1 3

17 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 as its semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is (a) 4 x 2 + y 2 = 4 (c) 4 x 2 + y 2 = 8

(b) x 2 + 4 y 2 = 8 (d) x 2 + 4 y 2 = 16

18 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the latusrectum to the x2 y2 JEE Mains 2015 ellipse + = 1 is 9 5 (a)

27 4

(b) 18

(c)

27 2

(d) 27

19 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 (a) (x − y ) = 6x + 2 y (c) (x 2 + y 2 )2 = 6x 2 + 2 y 2 2

2 2

2

2

JEE Mains 2014 (b) (x − y ) = 6x 2 − 2 y 2 (d) (x 2 + y 2 )2 = 6x 2 − 2 y 2 2

2 2

x2 + y 2 = 1 at 27 ( 3 3 cos θ , sin θ ) (where, θ ∈( 0, π /2)). Then, the value of θ such that the sum of intercepts on axes made by this tangent is minimum, is

20 Tangent is drawn to the ellipse

(a)

π 3

(b)

π 6

(c)

π 8

(d)

π 4

21 If the tangent at a point (a cos θ, b sin θ ) on the ellipse x2 y2 + 2 = 1 meets the auxiliary circle in two points, the 2 a b chord joining them subtends a right angle at the centre, then the eccentricity of the ellipse is given by (a) (1 + cos2 θ)−1 / 2 (c) (1 + sin2 θ)−1 / 2

(b) (1 + sin2 θ) (d) (1 + cos2 θ)

22 Chord of contact of tangents drawn from the point P (h, k ) to the ellipse x 2 + 4y 2 = 4 subtends a right angle at the centre. Locus of the point P is (a) x 2 + 16y 2 = 20 (c) 16x 2 + y 2 = 20

(b) x 2 + 8 y 2 = 10 (d) None of these

23 Tangent at a point P on the ellipse x 2 + 4y 2 = 4 meets the X -axis at B and AP is ordinate of P. If Q is a point on AP produced such that AQ = AB, then locus of Q is (a) x 2 + xy − 4 = 0 (c) x 2 + xy − 1 = 0

(b) x 2 − xy + 4 = 0 (d) x 2 − xy + 1 = 0

24 A parabola is drawn whose focus is one of the foci of the x2 y2 + = 1 (where, a > b ) and whose directrix a 2 b2 passes through the other focus and perpendicular to the major axis of the ellipse. Then, the eccentricity of the

ellipse

ellipse for which the latusrectum of the ellipse and the parabola are same, is 2 −1

(a)

(b) 2 2 + 1

2+1

(c) 2

(d) 2 2 − 1

2

x y + = 1 tangent PQ is a 2 b2 drawn. If the point Q be at a distance 1/p from the point P, where p is distance of the tangent from the origin, then the locus of the point Q is

25 At a point P on the ellipse

x2 y2 1 + = 1+ 2 2 a2 b 2 a b x2 y2 1 (c) 2 + 2 = 2 2 a b a b

x2 y2 1 − = 1− 2 2 a2 b 2 a b x2 y2 1 (d) 2 − 2 = 2 2 a b a b

(a)

(b)

26 If CP and CD are semi-conjugate diameters of an ellipse x2 y2 + = 1, then CP 2 + CD 2 is equal to 14 8 (a) 20

(b) 22

(c) 24

(d) 26

27 If θ and φ are eccentric angles of the ends of a pair of conjugate diameters of the ellipse

x2 y2 + 2 = 1, then θ − φ 2 a b

is equal to π 2

(a) ±

(b) ± π

(c) 0

(d) None of these

28 The coordinates of all the points P on the ellipse x2 y2 + = 1, for which the area of the ∆PON is maximum, a 2 b2 where O denotes the origin and N, the foot of the perpendicular from O to the tangent at P, is  a2 (a)  ± 2  a + b2  a2 (b)  ±  a2 − b 2  2a 2 (c)  ±  a2 + b 2  2a 2 (d)  ,  a2 − b 2

  a +b   b2 ,±  2 2 a −b  2b 2  ,±  a2 + b 2  2b 2   a2 − b 2  b2

,±

2

2

29 If α, β, γ are the eccentric angles of three points on the ellipse 4x 2 + 9y 2 = 36, the normals at which are concurrent, then sin(α + β) + sin(β + γ ) + sin( γ + α) is equal to (a)

2 3

(b)

4 9

(c) 0

(d) None

30 A triangle is drawn such that it is right angled at the x2 y2 + = 1(where, a > b ) and its a 2 b2 other two vertices lie on the ellipse with eccentric angles  α + β 1 − e 2 cos 2    2  is equal to α and β. Then,  α − β cos 2    2  centre of the ellipse

(a)

a2 a + b2 2

(b)

a2 + b 2 a2

(c)

a2 a − b2 2

(d)

a2 − b 2 a2

306 Directions (Q. Nos. 31-35) Each of these questions contains two statements : Statement I (Assertion) and Statement II (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a ), (b), ( c) and (d) given below. (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

31 Let S1 ≡ ( x − 1)2 + ( y − 2)2 = 0 and

S 2 ≡ ( x + 2)2 + ( y − 1)2 = 0 be the equations of two circles.

Statement I Locus of centre of a variable circle touching two circles S1 and S 2 is an ellipse. Statement II If a circle S1 = 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. 3 3  ,1 is a point on the ellipse  2 

32 Statement I If P 

4x 2 + 9y 2 = 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.

33 Statement I 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.

Statement II If y = mx and y = m1x are two conjugate b2 diameters of an ellipse, then mm1 = − 2 . a 34 Statement I The condition on a and b for which two x2 y2 + = 1 passing 2 2a 2b 2 through (a, − b ) are bisected by the line x + y = b is a 2 + 6ab − 7b 2 ≥ 0 . distinct chords of the ellipse

Statement II Equation of chord of the ellipse x2 y2 + = 1 whose mid-point is ( x 1, y 1), is T = S1. a2 b 2 35 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 3 x and the ellipse 2 x 2 + y 2 = 4, then m satisfies m 4 + 2m 2 = 24. Statement II If the line y = mx +

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 The eccentricity of an ellipse whose centre is at origin is 1 . If one of its directrices is x = − 4, then the equation of 2  3 the normal to it at 1,  is  2 JEE Mains 2017 (a) 4 x + 2 y = 7 (c) 2 y − x = 2

(b) x + 2 y = 4 (d) 4 x − 2 y = 1

2 The locus of the mid-points of a focal chord of the ellipse x2 y2 + = 1 is a 2 b2 (a)

x 2 y 2 ex + = a2 b 2 a

(c) x 2 + y 2 = a 2 + b 2

(b)

x 2 y 2 ex − = a2 b 2 a

(d) None of these

3 Foot of normal to the ellipse 4x 2 + 9y 2 = 36 having slope 2 may be 9 − 8 (a)  ,  5 5  − 9 8 (c)  ,   5 5

9 8 (b)  ,  5 5  (d) None of these

4 On the ellipse 4x 2 + 9y 2 = 1, the points at which the tangents are parallel to the line 8x = 9y are − 2 1 2 1 (a)  ,  ,  ,   5 5  5 5 − 3 − 1  3 1  (c)  , ,  ,   5 5   5 5

− 2 1   2 − 1 (b)  , ,  ,   5 5  5 5 

(d) None of these

5 The normal at a point P on the ellipse x 2 + 4y 2 = 16 meets the X -axis at Q. If M is the mid-point of the line segment PQ, then the locus of M intersects the latusrectums of the given ellipse at the points  3 5 2 (a)  ± ,±  2 7  1 (c)  ± 2 3 , ±   7

 3 5 19  (b)  ± ,±  2 4    4 3 (d)  ± 2 3 , ±  7  

6 The line passing through the extremity A of the major axis and extremity B of the minor axis of ellipse x 2 + 9y 2 = 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at AM and the origin O is (a)

31 10

(b)

29 10

(c)

21 10

(d)

27 10

307

DAY 1 1 1 + 4  a 2 b 2  1 1 (c) 2 + 2 a b

7 The ellipse x 2 + 4y 2 = 4 is inscribed in a rectangle

(a)

alingent with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point ( 4, 0). Then the equation of the ellipse is (a) x 2 + 12 y 2 = 16 (c) 4 x 2 + 64 y 2 = 48

(b) 4 x 2 + 48 y 2 = 48 (d) x 2 + 16y 2 = 16

(a)

(b) x 2 + 2 3 y = 3 − 3 (d) None

x y + = 1 is inscribed in a rectangle R 9 4 whose sides are parallel to the co-ordinate axes. Another ellipse E 2 passing through the point ( 0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E 2 is 3 (b) 2

1 (c) 2

3 (d) 4

π in points whose eccentric angles differ by , then r 2 is 3 equal to 3 (4 p 2 + q 2 ) 4 2 (c) (4 p 2 + q 2 ) 3

(b)

4 (4 p 2 + q 2 ) 3

(d) None of these

11 The sum of the square of the reciprocals of two perpendicular diameters of an ellipse

(b)

b a

(c)

a2 b2

(d)

b2 a2

α  β chord of the ellipse 16x 2 + 25y 2 = 400, then tan   tan    2  2 is equal to (a) − 4

(b)

−1 4

(c)

−3 8

(d) None

14 The tangent and normal drawn to the ellipse x 2 + 4y 2 = 4 at the point P(θ) meets the X -axis at the points A and B. If AB = 2, then cos 2 θ is equal to

10 If the line px + qy = r intersects the ellipse x 2 + 4y 2 = 4

(a)

a b

13 If α, β are the eccentric angles of the extremities of a focal

2

9 The ellipse E 1 :

2 (a) 2

(d) None of these

constant ratio of the area of triangle formed by joining the corresponding points on the auxiliary circle of the vertices of the first triangle. This ratio is

points of the latusrectum of the ellipse x 2 + 4y 2 = 4. The equations of parabolas with latusrectum PQ are

2

1 1 1 + 2  a 2 b 2 

12 The area of a triangle inscribed in an ellipse bears a

8 Let P ( x1, y1 ) and Q ( x 2 , y 2 ), y1 < 0, y 2 < 0, be the end (a) x 2 − 2 3 y = 3 + 3 (c) both (a) and (b)

(b)

x2 y2 + = 1, is a 2 b2

(a)

4 9

(b)

8 9

(c)

2 9

(d) None

x2 y2 + = 1 (a > b ) with foci at S and S′ and a 2 b2 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 feet 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

15 Given an ellipse

(a) e : 2 (c) e : 1

(b) e : 3 (d) e : 4

ANSWERS SESSION 1

SESSION 2

1 (d)

2 (b)

3 (b)

4 (d)

5 (c)

6 (c)

7 (a)

8 (a)

9 (b)

10 (a)

11 (a)

12 (d)

13 (b)

14 (a)

15 (b)

16 (b)

17 (d)

18 (d)

19 (c)

20 (b)

21 (c)

22 (a)

23 (a)

24 (a)

25 (a)

26 (b)

27 (a)

28 (a)

29 (c)

30 (b)

31 (d)

32 (a)

33 (b)

34 (a)

35 (b)

1 (d) 11 (a)

2 (a) 12 (b)

3 (b) 13 (b)

4 (b) 14 (a)

5 (c) 15 (c)

6 (d)

7 (a)

8 (c)

9 (c)

10 (a)

308

Hints and Explanations SESSION 1 2

1

It passes through p(−31 , ) and e =

2

y x + = 1 will represent an 8− a a−2 ellipse if 8 − a > 0, a − 2 > 0, 8 − a ≠ a − 2 ⇒ 2 < a < 8, a≠ 5 ⇒ a ∈ (2, 8) − {5}

2 2b = 2ae ⇒

a2e 2 = b 2 = a2 (1 − e 2 ) 1 ⇒ e2 = 2 2b 2 = 10 ⇒ b 2 = 5a a ⇒ a2 (1 − 1 / 2) = 5a ⇒ a= 10 ∴ b 2 = 50, a2 = 100 Hence, equation of ellipse is x2 y2 + =1 100 50 2 2 ⇒ x + 2 y = 100

2 . 5

9 1 …(i) + =1 a2 b 2 b2 and e2 = 1 − 2 a 2 b2 b2 3 ⇒ = 1− 2 ⇒ 2 = 5 a a 5 9 5 From Eq. (i), 2 + 2 = 1 a 3a 27 + 5 = 1 ⇒ 3a2 32 32 and then b 2 = ⇒ a2 = 3 5 ∴Equation of ellipse is 3 x2 5y 2 + = 1 ⇒ 3 x2 + 5y 2 = 32 32 32 ∴

5 Let equation of ellipse be 2

2

y x + =1 a2 b 2

B

3 Let A = (a, 0), B(0, b ). B(0,b) Y

F2 O

2:1

O

P(h,k)

A(a,0)

X

∴

x2 y2 + =1  64   16       9  9

4 Let the equation of ellipse is

(-3,1) Y P X

Y¢

4 x2 + 9(mx + c )2 − 36 = 0 ⇒ (9m2 + 4)x2 + 18cmx + 9c 2 − 36 = 0 Roots are real, so 18 × 18c 2 m2 − 4 ( 9m2 + 4) (9c 2 − 36)≥ 0 ⇒ 9m2 − c 2 + 4 ≥ 0 ⇒ 9m2 ≥ c 2 − 4

7 Equation of ellipse is 3 x2 + 4 y 2 = 1

y2 x2 + 2 = 1, (a > b ) 2 a b

O

a2e 2 = b 2 = a2 (1 − e 2 ) 1 1 e2 = ⇒e = 2 2

y = mx +

1 m

It touches the ellipse

x2 y2 + =1 8 2

1 = 8m2 + 2 [Qc 2 = a2 m2 + b 2 ] m2 ⇒ 8m 4 + 2m2 − 1 = 0 1 1 ⇒ m2 = i.e. m = ± 4 2 Hence, required tangent are x − 2 y + 4 = 0, x + 2 y + 4 = 0 ⇒

9 Any point on the line can be taken as

P (h, 6 − h ) Equation of chord of contact of P w.r.t. x2 + 2 y 2 = 4 is hx + 2(6 − h )y = 4 [T = 0] ⇒ h( x − 2 y ) + 4(3 y − 1) = 0 which passes through the fixed point 2 1 1 2 y = , x = i.e., through  ,  .  3 3 3 3

10 Given equation of ellipse is x2 y2 + = 1. 16 9 Here, a = 4, b = 3, e = 1 − ∴ Foci is (± ae , 0)  7  = ± 4 × , 0 = (± 4  

9 7 = 16 4

7, 0)

Y r

we get

16 64 3 3 = (1 − e 2 ) ⇒ e 2 = i.e. e = 9 9 4 2

X¢

⇒

8 Any tangent to parabola y 2 = 4 x is

6 Solving 4 x2 + 9 y 2 = 36 and y = mx + c , [Q AB = 4]

⇒ Locus of P is

Then, F1 = (ae , 0), F2 = (− ae , 0), B = (0,b ). π ∠F1 BF2 = 2 − b   − b  = − 1 ⇒      ae   − ae  ⇒

P divides AB in the ratio 1:2. 2a b h = ,k= 3 3 9h2 2 2 ∴ a +b = + 9k 2 = 16 4

∴

F1

1 1  Clearly, P  ,  lies on the ellipse,  7 7 therefore it is a point of contact.

Tangent at P ( x1 , y 1 ) is ...(i) 3 xx1 + 4 yy 1 = 1 On comparing Eq. (i) with the given 4y 3x 1 tangent 3 x + 4 y = 7, 1 = 1 = ⇒ 3 4 7  1 1  P ( x1 , y 1 ) = P  ,   7 7

X¢

X

Y¢

∴

Radius of the circle, r = (ae )2 + b 2 = 7 + 9 = 16 = 4

Now, the equation of circle is ( x − 0)2 + ( y − 3)2 = 16 ⇒ x2 + y 2 − 6 y − 7 = 0

11 Equation of tangent at any point y x cos θ + sin θ = 1. a b The equation of tangents at the end

P (acos θ,b sin θ) is

DAY

309 1 cos 2 θ sin2 θ = + r2 a2 b2

points of the major axis are x = a, x = − a. ∴The intersection point of these tangents are θ θ A =  a,b tan  , B =  − a,b cot  .   2 2

∴

Equation of circle with AB as diameter θ ( x − a)( x + a) +  y − b tan   2

From Eqs. (i) and (ii), we get

 y − b cot θ  = 0    2 θ θ ⇒ x2 − a2 + y 2 − by  tan + cot   2 2 + b2 = 0 ⇒ x2 + y 2 − a2 + b 2 − 2by cosec θ = 0 which is the equation of family of circles passing through the point of intersection of the circle x2 + y 2 − a2 + b 2 = 0 and y = 0 So, the fixed point is (± a2 − b 2 , 0). 2 2 12 The tangent to x + y = 1 is

25 4 y x cos θ + sin θ = 1. If it is also tangent 5 2 to the circle, then 1 100 16 = = cos 2 θ sin2 θ 4 + 21sin2 θ + 25 4 3 25 2 , cos 2 θ = ⇒ sin θ = 28 28 If the tangents meets the axes at A and B, then 5 2  A =  , 0 and B =  0,   cos θ   sinθ  25 4 + cos 2 θ sin2 θ 4 196 14 = 28 + ⋅ 28 = ⇒ AB = 3 3 3

∴ AB 2 =

13 Given, equation of ellipse is x2 y2 + = 1. General equation of tangent 2 1 to the ellipse of slope m is y = mx ± 2m2 + 1 Since, this is equally inclined to axes, so m = ± 1. Then, tangents are y = ± x± 2+ 1 = ± x± 3 Distance of any tangent from origin =

|0 + 0 ± 12 + 12

3|

= 3/2

14 The straight line x = r cos θ , y = r sin θ meets the ellipse given.

y2 x2 + 2 = 1, where r is 2 a b

But PQ 2 = 4 r 2 = HM of 4 a2 , 4b 2 1 1 1 ∴ = + 2 2b r 2 2a2

…(i)

…(ii)

cos 2 θ sin2 θ 1 1 + = 2 + 2 a2 b2 2a 2b π It is possible, when θ = 4 2 2 15 Given ellipse is x + y = 0, (a = 3, b = 2)

9 4 Equation of normal is 3 x sec θ − 2 y cosec θ = 9 − 4 = 5 Comparing it with given normal ax + by = c , 3sec θ − 2 cosec θ 5 = = a b c 3c 2c ⇒ cos θ = , sinθ = − 5a 5b 4c 2 9c 2 + =1 ⇒ 25a2 25b 2 25a2b 2 ⇒ 9b 2 + 4a2 = c2

2 2 16 Given ellipse is x + y = 1

∴

14 5 P(θ) is ( 14 cos θ, 5 sin θ)

Equation of normal at P(θ) is 14 x 5y − = 14 − 5 = 9 cos θ sin θ Q(2θ) lies on this normal, therefore 14 cos 2θ 5sin 2θ − =9 cos θ sin θ ⇒ 14 cos 2θ sin θ − 10sin θ cos 2 θ = 9 cos θ sin θ ⇒ 18 cos 2 θ − 9 cos θ − 14 = 0 ⇒ (3 cos θ + 2) (6 cos θ − 7) = 0 −2 7 or ⇒ cos θ = 6 3 −2 7 as cos θ ≠ ⇒ cos θ = 3 6

17 Given, (i) An ellipse whose semi-minor axis coincides with one of the diameters of the circle ( x − 1)2 + y 2 = 1. (ii) The semi-major axis of the ellipse coincides with one of the diameters of circle x2 + ( y − 2)2 = 4. (iii) The centre of the ellipse is at origin. (iv) The axes of the ellipse are coordinate axes. Now, diameter of circle ( x − 1)2 + y 2 = 1 is 2 units and that of circle x2 + ( 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, the equation of the ellipse is y2 x2 x2 y2 + 2 =1 ⇒ + =1 2 a b 16 4 ⇒ x2 + 4 y 2 = 16

18 Area of quadrilateral formed by tangents at the ends of latusrectum =

2a2 . e

2 3

Here, a2 = 9, b 2 = 5, e = 3 ∴ Area = 2 × 9 × = 27 2

19 x2 + 3 y 2 = 6 x2 y 2 ...(i) + = 1 (a2 = 6, b 2 = 2) 6 2 Equation of any tangent to Eqs. (i) is ...(ii) y = mx ± 6m2 + 2

⇒

Equation of perpendicular line drawn from centre (0, 0) to Eqs. (ii) is 1 ...(iii) y=− x m Eliminating m from Eqs. (ii) and (iii), required locus of foot of perpendicular is   x  x2 y =  −  x ±  6 2 + 2  y  y  ⇒

( y 2 + x2 )2 = 6 x2 + 2 y 2

20 The equation of tangent from the point (3 3 cos θ,sin θ) to the curve is x cos θ y sin θ + = 1. 3 3 1 Thus, sum of intercepts = 3 3secθ + cosecθ = f (θ)

[say]

3 3 sin θ − cos θ sin2 θ cos 2 θ Put f ′ (θ) = 0 1 ⇒ sin3 θ = 3 /2 cos 3 θ 3 1 π ⇒ tanθ = ⇒θ= 6 3 Now, f ′ ′ (θ) > 0, for θ = π / 6 [i.e. minimum] π Hence, value of θ is . 6 ⇒

f ′ (θ) =

3

3

21 Equation of tangent at (acos θ,b sin θ) to y x cos θ + sin θ = 1 …(i) a b The joint equation of the lines joining the points of intersection of Eq. (i) and the auxiliary circle x2 + y 2 = a2 to the origin, which is the centre of the circle, 2 y x is x2 + y 2 = a2  cos θ + sin θ b  a 

the ellipse is

Since, these lines are at right angles.  cos 2 θ   sin2 θ  ∴ 1 − a2  2  + 1 − a2  2  = 0  a   b  [Qcoefficient of x2 + coefficient of y 2 = 0]

310

⇒

 a2  sin2 θ  1 − 2  + 1 = 0 b  

⇒ sin2 θ (b 2 − a2 ) + b 2 = 0 ⇒ (1 + sin2 θ)(a2e 2 ) = a2 ⇒ e = (1 + sin2 θ)−1 /2 x 2 + 4 y 2 = 4 is hx + 4ky − 4 = 0 ...(i) Making x2 + 4 y 2 − 4 = 0 homogeneous with Eq. (i), we get 2 hx + 4ky  x2 + 4 y 2 − 4   =0   4 Chord of contact subtends right angles at the centre, so coefficient of x2 + coefficient of y 2 = 0 1 1 + 4 − h2 − 4k 2 = 0 ∴ 4 ⇒ locus of P (h, k ) is x2 + 16 y 2 = 20

23 Ellipse is x2 + 4 y 2 = 4. P = (2 cos θ,sin θ) Equation of tangent at P is 2cos θx + 4 y sin θ = 4 It meets X -axis at B. ∴ B = (2sec θ, 0), A = (2 cos θ, 0) 2sin2 θ AQ = AB = 2 |sec θ − cos θ| = cos θ

y − b sin θ x − acos θ = asin θ −b cos θ

2 y2 24 Equation of the ellipse is x2 + 2 = 1. a b Equation of the parabola with focus S (ae , 0) and directrix x + ae = 0 is y 2 = 4 aex. Now, length of latusrectum of the ellipse 2b 2 and that of the parabola is 4 ae. is a For the two latusrectum to be equal,

Y

(ae, 0)

Y¢

2b 2 2a2 (1 − e 2 ) = 4 ae ⇒ = 4 ae a a 2 2 ⇒ 1 − e = 2e ⇒ e + 2e − 1 = 0

X

sin θ sin φ = − cos θ cos φ cos(θ − φ) = 0 π θ − φ= ± 2

⇒ (a cos q, b sin q) P

X¢

28 Equation of tangent at P is

X

O

x cos θ y sin θ + = 1. a b Y

Y¢

The distance of the tangent from the origin is ab p= b 2 cos 2 θ + a2 sin2 θ 1 = p

2

2

P (a cos q, b sin q) X¢

x = acos θ −

and

y = b sin θ +

2

2

X

O

2

L

b cos θ + a sin θ ab

⇒

⇒

N

b cos θ + a sin θ ab 2

1 = = p

∴ Locus of Q is xy = 4 − x2 . ⇒ x2 + xy − 4 = 0

S

1/p

Q

Now, the coordinates of the point Q are given as follows, x − acos θ − asin θ b 2 cos 2 θ + a2 sin2 θ y − b sin θ b cos θ = b 2 cos 2 θ + a2 sin2 θ

 1 − h2  ⇒ hk = 4    4 

⇒ ⇒

Y

⇒

Let Q = (h, k ). Then, 2sin2 θ 2(1 − cos 2 θ) = h = 2cos θ, k = cos θ cos θ

S' (–ae, 0) O

b2 a2 b sin θ − 0 b sin φ − 0 b2 =− 2 ⇒ × a a cos θ − 0 a cos φ − 0 Then, m1 m2 = −

25 Equation of the tangent at P is

22 Chord of contact of P (h, k ) to the ellipse

X¢

Q (acos φ, b sin φ) be ends of these two diameters.

−2 ± 8 = −1 ± 2 2 Hence, e = 2 − 1 as 0 < e < 1 for ellipse. Therefore, e =

2

2

asin θ ab

b cos θ ab 2 2  x +  y  = 1 + 1      a b a2b 2

is the required locus.

26 CP and CD are semi-conjugate diameters x2 y2 + = 1 and let 14 8 eccentric angle of P is φ, then eccentric π angle of D is + φ, therefore the 2 coordinates of P and D are (acos φ,b sin φ) and π π   14 cos  + φ  , 8 sin  + φ   2  2   of an ellipse

i.e.CP 2 + CD 2 = (a2 cos 2 φ + b 2 sin2 φ) + ( a2 sin2 φ + b 2 cos 2 φ) = a2 + b 2 = 14 + 8 = 22

27 Let y = m1 x and y = m2 x be a pair of conjugate diameters of an ellipse y2 x2 + 2 = 1 and let P (acos θ,b sin θ) and 2 a b

Normal Y¢ ON =

=

1 cos θ sin2 θ + a2 b2 ab 2

b 2 cos 2 θ + a2 sin2 θ Equation of the normal at P is axsecθ – by cosec θ = a2 − b 2 . a2 − b 2 OL = 2 2 a sec θ + b 2 cos ec2 θ =

(a2 − b 2 )sin θ cos θ

a2 sin2 θ + b 2 cos 2 θ where, L is the foot of perpendicular from O on the normal. 1 Area of ∆PON = × ON × OL 2 [QNP = OL ] (a2 − b 2 ) ab tan θ (a2 − b 2 ) ab = = 2 a2 tan2 θ + b 2 a tan θ + b 2 cot θ which is minimum when a2 tan θ + b 2 cot θ is maximum. Thus, for area to be minimum, tanθ = ∴ cos θ = ±

a a2 + b 2

, sinθ = ±

b a2 + b 2

So, the required coordinates is   a2 b2 ,± ± ⋅ 2 2 2 2  a +b a +b 

29 If α,β, γ, δ are eccentric angles of four co-normal points. Then, α + β + γ + δ = (2n + 1)π

b a

DAY

311

and Σsin(α + β ) = 0 Now, sin (α + β ) = sin [2nπ + π − ( y + δ )] = sin(γ + δ) Similarly, sin(β + γ ) = sin(α + δ ) and sin(γ + α) = sin(β + δ) ∴2[sin(α + β ) + sin(β + γ ) + sin(γ + α)] = 0

30. Equation of the chord with eccentric

angles of the extremities as α and β is x α + β y α + β cos    + sin   2  b  2  a α −β = cos    2 

α+β α −β Let = ∆ 1 and = ∆2 , 2 2 y x so that cos ∆ 1 + sin ∆ 1 = cos ∆ 2 a b As the triangle is right angled, homogenising the equation of the curve, we get 2

y 2  x cos ∆ 1 y sin ∆ 1  x2 + 2 − +  =0 2 a b  acos ∆ 2 b cos ∆ 2  1 cos 2 ∆ 1  ⇒ 2 − 2  a cos 2 ∆ 2  a 1 sin2 ∆ 1  + 2 − 2  =0 b cos 2 ∆ 2  b [as coefficient of x2 + coefficient of y 2 = 0] 2 2 2 ⇒ b (cos ∆ 2 − cos ∆ 1 ) + a2 (cos 2 ∆ 2 − sin2 ∆ 1 ) = 0 2 2 ⇒ cos ∆ 2 (a + b 2 ) − b 2 cos 2 ∆ 1 − a2 + a2 cos 2 ∆ 1 = 0 ⇒ cos 2 ∆ 2 (a2 + b 2 ) = a2 (1 − e 2 cos 2 ∆ 1 ) α + β 1 − e 2 cos 2    2  a2 + b 2 ⇒ = α −β a2 cos 2    2 

31 Let C1 and C2 be the centres and R1 and R2 be the radii of the two circles. Let S 1 = 0 lie completely inside in the circle S 2 = 0. Let ‘C’ and ‘r’ be the centre and radius of the variable centre. Then, CC2 = R 2 − r and C1 C = R1 + r [constant] ∴ C1 C + C 2C = R1 + R2 So, the locus of C is an ellipse. Therefore, Statement II is true. Hence, Statement I is false (two circles are intersecting). 2 2 32 The ellipse is x + y = 1. 9 4 So, auxiliary circle is x2 + y 2 = 9 and (− 5, 0) and ( 5, 0) are foci.

Hence, Statement I is true, then Statement II is also true.

33 Let PCP ′ and DCD ′ be the conjugate diameters of an ellipse and let the eccentric angle of P is φ, then coordinate of P is (acos φ, b sin φ). So, coordinate of D is (− asin φ, b cos φ). Let S and S ′ be two foci of the ellipse. Then, SP ⋅ S ′ P = (a − ae cos φ) ⋅ (a + ae cos φ)

P (a cos φ, b sin φ)

D

S¢

C

S

D¢  ⇒ b 2 = a2 (1 − e 2 ) = a2 − a2e 2 cos 2 φ  2 2 2 2  ⇒ a − b = a e  = a2 − (a2 − b 2 )cos 2 φ = a2 sin2 φ + b 2 cos 2 φ = CD 2 P¢

34 Let (t ,b − t ) be a point on the line x + y = b, then equation of chord whose mid-point is (t ,b − t ), is (b − t )y tx + −1 2a2 2b 2 (b − t )2 t2 = 2 + − 1 …(i) 2a 2b 2 Since, point (a, − b ) lies on Eq. (i), then b(b − t ) (b − t )2 ta t2 − = 2 + 2 2 2a 2b 2a 2b 2 2 2 2 ⇒ t (a + b ) − ab (3 a + b )t + 2a2b 2 = 0 Since, t is real. ∴ B 2 − 4 AC ≥ 0 2 2 ⇒ a b (3 a + b ) 2 − 4(a2 + b 2 ) 2a2b 2 ≥ 0 ⇒ 9 a2 + 6 ab + b 2 − 8 a2 − 8b 2 ≥ 0 ∴ a2 + 6 ab − 7b 2 ≥ 0

35. Statement I Given, a parabola y 2 = 16 3 x and an ellipse 2 x2 + y 2 = 4. To find the equation of common tangent to the given parabola and the ellipse. This can be very easily done by comparing the standard equation of tangents. Standard equation of tangent with slope ‘m’ to the parabola y 2 = 16 3 x is 4 3 …(i) m Standard equation of tangent with slope x2 y2 ‘m’ to the ellipse + = 1 is 2 4 …(ii) y = mx ± 2m2 + 4 y = mx +

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) and 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, comparing Eqs. (i) and (ii), we get

4 3 = ± 2m2 + 4 m On squaring both sides, we get m2 (2m2 + 4) = 48 4 ⇒ m + 2m2 − 24 = 0 ⇒ (m2 + 6)(m2 − 4) = 0 ⇒ m2 = 4 [Qm2 ≠ −6] ⇒ m=±2 On substituting m = ± 2 in Eq. (i), we get the required equation of the common tangent as y = 2 x + 2 3 and y = −2 x − 2 3 Hence, Statement I is correct. Statement II We have already seen 4 3 that, if the line y = mx + is a m common tangent to the parabola x2 y2 + = 1, y 2 = 16 3 x and the ellipse 2 4 then it satisfies the equation m 4 + 2m2 − 24 = 0. Hence, Statement II is also correct but is not able to explain the Statement I. It is an intermediate step in the final answer.

SESSION 2 1 Let equation of ellipse be x2 y 2 + =1 a2 b 2

Equation of directrix is x = ±

a e

a 1 = 4 ⇒ a = 4e ⇒ a= 4 × = 2 Qe =  e 2 b2 2 2 e = 1 − 2 ⇒b = 3 a x2 y 2 ∴ Equation of ellipse = + =1 4 3 Equation of normal a2 x b 2 y 4x 3y − = a2 − b 2 , − =1 x1 y1 1  3    2 ∴

1 2 

3   Q( x1 , y 1 ) =  1,    2   ⇒

4x −2y = 1

2 Let P (h, k ) be the mid-point of the focal chord. Then its equation is hx ky h2 k2 ...(i) + 2 = 2 + 2 2 a b a b Eq. (i) is focal chord so it passes through (ae , 0) h ea h2 k 2 ∴ = 2 + 2 a2 a b y 2 ex x2 ∴ Locus of P is 2 + 2 = a b a 3 Let foot of normal be P ( x1 , y 1 ). Then a2 y slope of normal = 2 1 = 2 ⇒ 9 y 1 = 8 x1 b x1 P lies on the ellipse 4 x2 + 9 y 2 = 36

312  64 x12  81 2 ⇒ 4 x12 + 9   = 36 ⇒ x1 = 25  81  9 8 On taking x1 = , we get y 1 = 5 5 9 8 ∴One possible foot of normal is  ,   5 5 4 Slope of tangent m = 8 9 Let P ( x1 , y 1 ) be the point of contact. Then, equation of tangent is 4 xx1 + 9 yy 1 = 1 4 x1 8 ∴ − = ⇒ x1 = − 2 y 1 9 y1 9 4 x + 9 y = 1 ⇒ 16 y + 9 y = 1 1 y1 = ± 5 ∴Required points of contact are  − 2 , 1  and  2 , − 1       5 5 5 5  2 1

2 1

2 1

2 2 7 x2 + 4 y 2 = 4 ⇒ x + y = 1

Y

4 1 ⇒ a = 2, b = 1 ⇒ P = (2, 1) y2 x2 Required Ellipse is 2 + 2 = 1 a b x2 y2 + = 1 ( 2 , 1 ) lies on it ⇒ 42 b2

(–3,2)

P(2,1)

(–3,–2)

1

V′ A′

2

4 1 + =1 16 b 2 1 1 3 4 = 1 − = ⇒ b2 = ⇒ b2 4 4 3 x2 y2 ∴ + =1 16  4     3 ⇒

5 Normal is 4 x sec φ − 2 y cosec φ = 12 Q ≡ (3cos φ, 0), M = (α,β) 3 cos φ + 4 cos φ 7 α= = cos φ 2 2

⇒

P (4 cos φ, 2 sin φ)

x2 + 12 y 2 = 16

2 2 8 x + y = 1 and b2 = a2 (1 − e 2 )

4

1

M

R Q(x2, y2)

2 ⇒ cos φ = α and β = sin φ 7 cos 2 φ + sin2 φ = 1 4 2 4 2 α + β2 = 1 ⇒ x + y2 = 1 ⇒ 49 49 ⇒ Latusrectum x = ± 2 3

6 Equation of auxiliary circle is x2 + y 2 = 9 x y Equation of AM is + = 1 3 1

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

X

O

(3,0)

(3,–2)

9 4 9 1 + 2 = 1, 2 + = 1 ⇒ a2 = 12 a2 b a 4 12 2 2 2 = 1 − e2 a = b (1 − e ) ⇒ 16 12 14 1 1 ⇒ e2 = 1 − = = ⇒e = 16 16 4 2 2 2 10 Ellipse is x + y = 1

4 1 Let A(θ) = (2 cos θ, sin θ) π  Then, B =  2 cos  θ +  , sin   3

θ + π      3 

Equation of chord AB is x π y π cos  θ +  + sin  θ +    2 6 1 6 π 3 = 6 2 or given chord is px + qy = r Comparing the coefficient, we get π π cos  θ +  sin  θ +    3 6 6 = = 2p q 2r

e =

less than 0) Co-ordinates of mid-point of PQ are 1 R ≡  0, −   2 PQ = 2 3 = length of latusrectum. ⇒ Two parabolas are possible whose  3 1 vertices are  0, − −  and 2 2 

⇒ ⇒

3 p2 3q 2 + 2 =1 r2 4r 3(4 p2 + q 2 ) = 4r 2 3 r 2 = (4 p2 + q 2 ) 4

11 Let POP ′, QOQ ′ be two perpendicular diameters. Let ∠AOP = θ, then ∠AOQ =

Q

B(0,1)

A(3,0)

x2 + 2 3 y = 3 − 3

9 Let required ellipse is On solving Eqs. (i) and (ii), we get − 12 9  M  ,   5 5 Now, area of ∆AOM 1 27 sq unit. = ⋅ OA ⋅ MN = 2 10

θ

y2 x2 E2 : 2 + 2 = 1 a b It passes through (0, 4) 16 0 + 2 = 1 ⇒ b 2 = 16 b It also passes through (± 3, ± 2)

A

O

Hence, the equations of the parabolas are x2 − 2 3 y = 3 + 3 and

P′

π +θ 2

P

 3 1 −   0, 2 2 

M

O

P(x1, y1)

3 1 ⇒ P  3, −   2 2 1  and Q =  − 3, −  (given y 1 and y 2  2 ⇒

48 1 1 + y 2 = 1 ⇒ y = ±  ± 2 3, ±  49 7 7

N

(3, 2)

= cos

Q (3 cos φ, 0)

12 9 – , 5 5

2

V A 2 (4, 0)

2 1

⇒

4

Q′

Then, P = (OP cos θ, OP sin θ) OP 2 cos 2 θ OP 2 sin2 θ + =1 ∴ a2 b2 2 2 cos θ sin θ 1 ...(i) ⇒ = + OP 2 a2 b2 Also Q = (OQ cos(90° + θ), OQ sin(90° + θ))

DAY

313

sin2 θ cos 2 θ 1 = 2 + OQ 2 a b2 1 1 1 1 + = 2 + 2 ⇒ OP 2 OQ 2 a b 1 1 1 1 1 ∴ + =  2 + 2  (POP ′ )2 (QOQ ′ )2 4 a b  [QPOP ′ = 2OP, QOQ ′ = 2OQ] ∴

12 Let P(θ1 ), Q(θ2 ), R(θ3 ) be the vertices of ∆PQR, inscribed in the ellipse y2 x2 + 2 =1 2 a b Then, P ′, Q ′, R ′, corresponding points on auxiliary circle x2 + y 2 = a2 are P ′(a cos θ1 , a sin θ1 ), Q ′(acos θ2 , a sin θ2 ) and R ′(acos θ3 , asinθ3 ) ∆ 1 = Area of ∆PQR a cos θ1 b sin θ1 1 1 = a cos θ2 b sin θ2 1 2 a cos θ3 b sin θ3 1 cos θ1 1 = ab cos θ2 2 cos θ3

sin θ1 sin θ2 sin θ3

∆ 2 = area of P ′Q ′ R ′ cos θ1 sin θ1 1 = a2 cos θ2 sin θ2 2 cos θ3 sin θ3 ∴

1 1 1

Equation of the line S ′ R ′ is 1 y = − ( x + ae ) m ⇒ x + my + ae = 0

⇒ ⇒ ⇒ ⇒

∆1 b = = Constant ∆2 a

 2 − 3 cos 2 θ = 4 cos 2 θ     2 9 cos 4 θ − 40 cos 2 θ + 16 = 0 (9 cos 2 θ − 4) (cos 2 θ − 4) = 0 4 cos 2 θ = as cos 2 θ ≠ 4 9

y = mx +

13 Equation of chord P(α), Q(β) is

...(i)

Here, a = 5, b = 4. PQ is a focal chord. a2 − b 2 9 3 e2 = = ⇒e = a2 25 5 One of the foci = (ae, 0) = (3, 0) Eq. (i) passes through (3, 0)

am +b 2

2

2

− aem +

Y¢

a2 m2 + b 2

1 + m2 1 (S ′ R ′+SR ) × SD 2

Then,

∆1 =

⇒

 a2 m2 + b 2   ∆ 1 = 2ae  2   1+ m 

Area of A′ B ′ BA is ∆2 =

1 ( A′ B ′+ AB ) × AE 2

Equation of A′ B ′ is y =−

1 ( x + a) m

x + my + a = 0

Therefore,

and

AB =

…(i)

∴ S ′ R ′RS is a trapezium and its area 1 ∆ 1 = (SR + S ′ R ′ ) × SD 2 Y B¢ R¢ P R B E D X¢ X O A¢ S S¢ A (a, 0) (ae, 0)

a2 m2 + b 2 1 + m2

and S ′ R ′ =

⇒

15 Equation of tangent at P is

1 + m2

aem +

SR =

…(ii)

ae + ae

SD =

Therefore,

2

1 1 1

α + β x α + β + y cos  sin     2  b  2  a α − β = cos    2 

3 α + β =  α −β  cos   cos    2   2  5 α β α β ⇒ 5 cos cos + sin sin   2 2 2 2  α β α β  = 3 cos cos − sin sin  2 2 2 2   α β α β ⇒ 8sin sin = − 2 cos cos 2 2 2 2 α β 1 ⇒ tan tan = − . 2 2 4 2 2 14 Any point P(θ) on the ellipse x + y = 1 4 1 is P(2 cos θ,sin θ) Equation of tangent at P(θ) is 2x cos θ + 4 y sin θ = 4 ∴ A = (2 sec θ, 0) Equation of normal of P(θ) is 2x y − = 4−1=3 cos θ sin θ 3 ∴ B =  cos θ, 0 2  3 AB = 2 ⇒ 2 secθ − cos θ = 2 2 ∴

and A ′ B ′ =

…(iii)

a+ a

AE =

1 + m2

ma +

a2 m2 + b 2 1 + m2

− ma +

a2 m2 + b 2 1 + m2

1 ( A′ B ′+ AB ) × AE 2  a2 m2 + b 2   ∆ 2 = 2a  2   1+ m 

Then, ∆ 2 = ⇒ Hence,

∆1 : ∆2 = e : 1

DAY TWENTY NINE

Hyperbola Learning & Revision for the Day u

u

Concept of Hyperbola Equations of Hyperbola in Standard Form

u

Tangent to a Hyperbola

u

u

Normal to a Hyperbola

u

u

Pole and Polar

u

Director Circle Asymptotes Rectangular or Equilateral Hyperbola

Concept of Hyperbola Hyperbola is the locus of a point in a plane which moves in such a way that the ratio of its distance from a fixed point (focus) in the same plane to its distance from a fixed line (directrix) is always constant which is always greater than unity.

M

N T

P S

SP Mathematically, = e, where e > 1. PN M¢

Terms Related to Hyperbola Some important terms related to hyperbola are given below. 1. 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. 2. Transverse and conjugate axes Transverse axis is the one which lie along the line passing through the foci and perpendicular to the directrices and conjugate axis and conjugate axis is the one which is perpendicular to the transverse axis and passes through the mid-point of the foci i.e. centre. 3. Centre The mid-point C of AA′ bisects every chord of the hyperbola passing through it and is called the centre of the hyperbola. 4. Focal chord A chord of a hyperbola which is passing through the focus is called a focal chord of the hyperbola. 5. Directrix A line which is perpendicular to the axis and it lies between centre and a vertex. The equation of directrix is x = ± . e 6. Double ordinates If Q be a point on the hyperbola draw QN perpendicular to the axis of the hyperbola and produced to meet the curve again at Q′. Then, QQ′ is called a double ordinate of Q.

PRED MIRROR Your Personal Preparation Indicator u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

315

DAY 7. Latusrectum The double ordinate passing through focus is called latusrectum. Y Directrix

M'

S' (–ae, 0) A' (– a, 0) Latusrectum

Centre

X¢

Directrix

Latusrectum

M

P

Q

A S(ae, 0) (a, 0)

C

(viii) Length of conjugate axis, 2b (ix) Equation of transverse axis, y = 0 (x) Equation of conjugate axis, x = 0

X N Double ordinate

(xi) Focal distances of a point on the hyperbola is ex ± a. (xii) Difference of the focal distances of a point on the hyperbola is 2a.

Q¢

x = –a/e

x=a/e Y¢

NOTE

(vii) Length of transverse axis, 2a

• The vertex divides the join of focus and the point of intersection of directrix with axis internally and externally in the ratio e : 1. x2 y 2 • Domain and range of a hyperbola 2 − 2 = 1 are x ≤ − a or a b x ≥ a and y ∈ R, respectively. • The line through the foci of the hyperbola is called its transverse axis. • The line through the centre and perpendicular to the transverse axis of the hyperbola is called its conjugate axis.

x 2 y2 2. Conjugate Hyperbola − 2 + 2 = 1 a b 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. The conjugate hyperbola of the hyperbola x 2 y2 x 2 y2 − 2 = 1 is − 2 + 2 = 1. 2 a b a b Y

l X¢

x 2 y2 1. Hyperbola of the Form 2 − 2 = 1 a b When the hyperbola is in the given form, then it is also called the equation of auxiliary circle. Y

S1

L A1

A

S

O

X

X

O

If the centre of the hyperbola is at the origin and foci are on the X-axis or Y-axis, then that types of equation are called standard equation of an ellipse.

X¢

L

A

Equations of Hyperbola in Standard Form

L¢

S

L1

l¢ A1 L¢1

S1

L¢

Y¢

(i) Centre, O(0, 0) (ii) Foci, S (0, be), S1 (0, − be) (iii) Vertices, A (0, b ), A1 (0, − b ) b b (iv) Directrices l : y = , l ′ : y = − e e (v) Length of latusrectum LL1 = L′ L1 ′ =

2 a2 b

2

L¢1

l¢

l

L1

Y¢

(vii) Length of transverse axis, 2b

(i) Centre, O(0, 0)

(viii) Length of conjugate axis, 2a

(ii) Foci : S (ae, 0), S1 (− ae, 0)

(ix) Equation of transverse axis, x = 0

(iii) Vertices : A(a, 0), A1 (− a, 0) a a (iv) Directrices l : x = , l ′ : x = − e e (v) Length of latusrectum, LL 1 = L′ L′1 = b (vi) Eccentricity, e = 1 +    a

 a (vi) Eccentricity, e = 1 +   as a2 = b 2 (e2 − 1) b

(x) Equation of conjugate axis, y = 0 (xi) Focal distances of a point on the hyperbola is ey ± b . 2 b2 a

2

or b 2 = a2 (e2 − 1)

(xii) Difference of the focal distances of a point on the hyperbola is 2b. NOTE If the centre of the hyperbola is ( h, k ) and axes are parallel to

the coordinate axes, then its equation is

( x − h) 2 ( y − k ) 2 − = 1. a2 b2

316

Results on Hyperbola

x2 y2 − =1 a2 b2

or T = 0.

1. The foci of a hyperbola and its conjugate are concyclic and form the vertices of a square. 2. The points of intersection of the directrix with the transverse axis are known as foot of the directrix. 3. Latusrectum (l ) = 2e (distance between the focus and the foot of the corresponding directrix). 4. The parametric equation of a hyperbola is x = a sec θ and y = b tan θ, where θ ∈(0, 2 π ). 5. The position of a point (h, k ) with respect to the hyperbola S lie inside, on or outside the hyperbola, if x2 y2 S1 > 0, S1 = 0 or S1 < 0 where, S1 ≡ 12 − 12 − 1 a b

A line which intersects the hyperbola at only one point is called the tangent to the hyperbola.

(ii) In slope ‘ m’ form, y = mx ± (iii) In parametric form,

a m −b 2

2

x y sec θ − tan θ = 1 at (a sec θ, b tan θ). a b

(iv) The line y = mx + c touches the hyperbola, iff c2 = a 2 m 2 − b 2 and the point of contact is

1. Two tangents can be drawn from a point to a hyperbola. 2. The point of intersection of tangents at t 1 and t 2 to the curve  2 c t1 t2 2c  xy = c2 is  , .  t1 + t2 t1 + t2  3. The tangent at the point P(a sec θ1 , b tan θ1 ) Q (a sec θ2 , b tan θ2 ) intersect at the point

and

  θ + θ2    θ1 − θ2   b sin  1   a cos   2   2 . R ,  θ1 + θ2   θ + θ2    cos  1     cos   2    2  4. The equation of pair of tangents drawn from an external point P( x1 , y1 ) to the hyperbola is SS1 = T 2

and

2 1 2

2 1 2

x y x y − 2 − 1, S1 = − 2 a b a b xx1 yy1 T = 2 − 2 −1 a b

where S =

A line which is perpendicular to the tangent of the hyperbola is called the normal to the hyperbola. a2 x b 2 y + = a2 + b 2 . x1 y1 m(a2 + b 2 )

(ii) In slope ‘m’ form y = mx m

a2 − m2b 2

a2 a −b m 2

2

2

,m

and the point of

  a −b m  b2 m

2

2

2

(iii) In parametric form, ax cos θ + by cot θ = a2 + b 2 at (a sec θ, b tan θ)

Results on Normals

2. Four normals can be drawn from any point to a hyperbola.

Results on Tangent

2

Normal to a Hyperbola

1. If the straight line lx + my + n = 0 is a normal to the x2 y2 a2 b2 (a 2 + b 2 )2 hyperbola 2 − 2 = 1, then 2 − 2 = a b l m n2

 a2 m b2  , ±  , where c = a2 m2 − b 2 . ±  c c

2

7. Equation of chord joining the points (a sec α , b tan α ) and (a sec β, b tan β) is x  α − β y  α + β  α + β cos   − sin   = cos  .  2  b  2   2  a

 intersection is  ± 

xx1 yy − 21 = 1 a2 b 2

6. The equation of chord of the hyperbola, whose xx yy x2 y2 mid-point is ( x1 , y1 ) is T = S1 i.e. 21 − 21 = 12 − 12 . a b a b

(i) In point ( x1 , y1 ) form

Tangent to a Hyperbola (i) In point ( x1 , y1 ) form,

xx1 yy − 21 = 1 2 a b

5. The equation of chord of contact is

−1

3. The line y = mx + c will be a normal to the hyperbola x 2 y2 m2 (a2 + b 2 )2 . − 2 = 1, if c2 = 2 2 a b a − b 2 m2 4. If the normals at four points P ( x1 , y1 ), Q ( x2 , y2 ), R ( x3 , y3 ) x2 y2 and S ( x 4 , y4 ) on the hyperbola − 2 = 1 are 2 a b concurrent, then 1 1 1 1 ( x1 + x2 + x3 + x 4 )  + + +  = 4.  x1 x2 x3 x4 

Pole and Polar 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 point 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.

317

DAY (i) Polars of two points P( x1 y1 ) and Q( x2 , y2 ) of hyperbola x 2 y2 xx yy xx yy − = 1 are 21 − 21 = 1 and 22 − 22 = 1 a2 b 2 a b a b (ii) If the polar of P( x1 , y1 ) passes through Q( x2 , y2 ), then x1 x2 y1 y2 − 2 = 1 and the polar of Q( x2 , y2 ) passes through a2 b x x y y P( x1 , y1 ), then 1 2 2 − 1 2 2 = 1 a b

Conjugate Points and Conjugate Lines (i) Two points are said to be conjugate points with respect to a hyperbola, if each lies on the polar of the other. (ii) Two lines are said to be conjugate lines with respect to a x 2 y2 hyperbola 2 − 2 = 1, if each passes through the pole of a b the other. (iii) Two lines l1 x + m1 y + n1 = 0 and l2 x + m2 y + n2 = 0 are x 2 y2 conjugate lines with respect to the hyperbola 2 − 2 = 1, a b if a2 l1 l2 − b 2 m1 m2 = n1 n2 .

Director Circle Y The locus of the point of intersection of the tangents P (h, k) to the hyperbola 90° 2 2 x y − 2 = 1, which are X¢ X 2 C a b perpendicular to each other, is called a director circle. Y¢ The equation of director circle is x2 + y2 = a2 − b 2 . The circle x2 + y2 = a2 is known the auxiliary circle of both hyperbola.

Asymptotes An asymptote to a curve is a straight line, which touches on it two points at infinity but which itself does not lie entirely at infinity.

Results on Asymptotes 1. A hyperbola and its conjugate hyperbola have the same asymptotes. 2. The angle between the asymptotes of hyperbola x 2 y2 b − = 1 is 2 tan −1   .  a a2 b 2 3. Asymptotes always passes through the centre of the hyperbola. 4. The bisectors of the angle between the asymptotes are the coordinate axes. 5. The asymptotes of a hyperbola are the diagonals of the rectangle formed by the lines drawn through the extrimities of each axis parallel to the other axis.

Rectangular or Equilateral Hyperbola A hyperbola whose asymptotes include a right angle is said to be rectangular hyperbola or if the lengths of transverse and conjugate axes of any hyperbola is equal, then it is said to be a rectangular hyperbola.

Rectangular Hyperbola of the Form x 2 − y 2 = a2 1. Asymptotes are perpendicular lines i.e. x ± y = 0 2. Eccentricity, e = 2,

Y L¢

3. Centre, O (0, 0) 4. Foci, S and S1 (± 2 a, 0) a 5. Directrices, x = ± 2

X¢

L

A1

A

S1

6. Latusrectum = 2a

L¢1

7. Point form, x = x1 , y = y1

S

O l¢

l

X

L1

Y¢

Equation of tangent, xx1 − yy1 = a2 x y Equation of normal, 1 + 1 = 2 x y 8. Parametric form, x = a sec θ, y = a tan θ Equation of tangent, x sec θ − y tan θ = a x y Equation of normal, + = 2a sec θ tan θ

Rectangular Hyperbola of the Form xy = c 2 1. Asymptotes are perpendicular lines i.e. x = 0 and y = 0 2. Eccentricity, e = 2 Y

3. Centre, (0, 0) 4. Foci S ( 2c, 2c), S1 (− 2c, − 2c) 5. Vertices, V1 (c, c), V2 (− c, − c)

V1 X¢

X

6. Directrices, x + y = ± 2 c 7. Latusrectum = 2 2 c

S

V2 S1

8. Point form, x = x1 , y = y1 Equation of tangent, xy1 + yx1 = 2c2 x y ⇒ + =2 x1 y1 Equation of Normal, xx1 − yy1 = x12 − y12 c 9. Parametric form : x = ct , y = t Equation of tangent, x + yt 2 = 2 ct 1  Equation of normal, t 2 x − y = c t 3 −   t

Y¢

318

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 If the vertices of hyperbola are ( 0 , ± 6) and its eccentricity is

5 , then 3

j

10 Length of the latusrectum of the hyperbola xy − 3x − 4y + 8 = 0 is

NCERT Exemplar

y2 x2 − = 1. 36 64 II. The foci of hyperbola are ( 0, ± 10). (a) Both I and II are true (c) Only II is true

(b) Only I is true (d) Both I and II are false

3 latusrectum is 8 and eccentricity is , is 5

(b) 4 x 2 − 5 y 2 = 100 (d) − 5 x 2 + 4 y 2 = 100

x2 y2 − = 1 which a 2 b2 passes through the points ( 3, 0) and ( 3 2, 2), is

3 The eccentricity of the hyperbola

j

(a)

13 3

13 3

(b)

4 Eccentricity of hyperbola (a)

1+ k

(b) 1− k

(c)

13 9

NCERT Exemplar 13 (d) 3

1+

1 k

(d) 1 −

1 k

5 The equation of the hyperbola whose foci are ( − 2 , 0) and ( 2 , 0) and eccentricity is 2, is given by (a) − 3 x 2 + y 2 = 3 (c) 3 x 2 − y 2 = 3

j

AIEEE 2011

(b) x 2 − 3 y 2 = 3 (d) − x 2 + 3 y 2 = 3

(b)

16 2 3

(c)

3 32

(d)

64 3

7 The difference between the length 2a of the transverse axis of a hyperbola of eccentricity e and the length of its latusrectum is (a) 2a (3 − e 2 ) (c) 2a (e 2 − 1)

(b) 2a | 2 − e 2 | (d) a (2e 2 − 1)

x2 y2 8 If eccentricity of hyperbola 2 − 2 = 1 is e and e′ is the a b eccentricity of its conjugate hyperbola, then (a) e = e ′ 1 1 (c) 2 + e (e ′)2

(a)

4 3

4 3

(b)

2 3

(c)

12 A general point on the hyperbola

(d) None of these

x2 y2 − = 1 is a 2 b2

(a) (a sin θ,b cos θ) (where, θ is parameter) (b) (a tan θ,b sec θ) (where, θ is parameter)  et + e −t et − e −t  (c)  a ,b  (where, t is parameter) 2 2   (d) None of the above

x2 y2 − = 1. Which of the cos 2 α sin2 α following remains constant when α varies?

13 For the hyperbola

(b) Abscissae of vertices (d) Eccentricities

14 The product of the perpendicular from two foci on any tangent to the hyperbola (a) a 2

x2 y2 − = 1, is a 2 b2 (c) −a

(b) b 2

2

(d) −b

2

15 If a line 21x + 5y = 116 is tangent to the hyperbola (a) (− 6, 3)

3x 2 − 4y 2 = 32 is 8 2 3

(d) None of these

7x 2 − 5y 2 = 232, then point of contact is

6 The length of transverse axis of the hyperbola (a)

(c) 8

(a) Directrix (c) Abscissae of foci

x2 y2 + 2 = 1 (k < 0) is k k (c)

(b) 4 2

and conjugate axis is equal to half of the distance j NCERT Exemplar between the foci is

2 The equation of the hyperbola, the lenght of whose (a) 5 x 2 − 4 y 2 = 100 (c) −4 x 2 + 5 y 2 = 100

(a) 4

11 The eccentricity of the hyperbola whose latusrectum is 8

I. The equation of hyperbola is

(b) ee′ = 1 (d) None of these

9 A hyperbola, having the transverse axis of length 2 sin θ, is confocal with the ellipse 3x 2 + 4y 2 = 12. Then, its equation is

(a) x 2 cosec2 θ − y 2 sec2 θ = 1 (b) x 2 sec2 θ − y 2 cosec2 θ = 1 (c) x 2 sin 2 θ − y 2 cos2 θ = 1 (d) x 2 cos 2 θ − y 2 sin2 θ = 1

(b) (6, − 2)

(c) (8, 2)

(d) None of these

x2 y2 16 If e1 is the eccentricity of the ellipse + = 1 and e 2 is 16 25 the eccentricity of the hyperbola passing through the foci of the ellipse and e1 e 2 = 1, then equation of the hyperbola is x2 y2 − =1 9 16 2 2 x y (c) − =1 9 25 (a)

(b)

x2 y2 − = −1 16 9

(d) None of these

17 If the line 2x + 6y = 2 touches the hyperbola x 2 − 2y 2 = 4, then the point of contact is (a) (−2, 6)

1 1  (b) (−5, 2 6) (c)  ,   2 6

(d) (4, −

6)

18 If a hyperbola passes through the point P( 2, 3 ) and

has foci at ( ± 2, 0), then the tangent to this hyperbola at j JEE Mains 2017 P also passes through the point (a) (3 2 , 2 3 ) (b) (2 2 , 3 3 ) (c) ( 3 , 2 )

(d) (−

2, − 3)

19 The common tangent to 9x − 4y = 36 and x + y = 3 is 2

2

2

2

(a) y − 2 3 x −

39 = 0

(b) y + 2 3 x +

(c) y − 2 3 x +

39 = 0

(d) None of the above

39 = 0

319

DAY 20 The locus of a point P(α , β) moving under the condition

that the line y = α x + β is a tangent to the hyperbola x2 y2 − = 1, is j AIEEE 2005 a 2 b2 (a) a hyperbola (b) a parabola (c) a circle (d) an ellipse

21 The straight line x + y = 2p will touch the hyperbola 4x 2 − 9y 2 = 36, if (a) p 2 = 2

(b) p 2 = 5

(c) 5 p 2 = 2

(d) 2 p 2 = 5

22 The locus of the point of intersection of perpendicular tangents to the hyperbola (a) x 2 + y 2 = 2 (c) x 2 − y 2 = 3

x2 y2 − = 1 is 3 1

(b) x 2 + y 2 = 3 (d) x 2 + y 2 = 4

x2 y2 − = 1 and N is foot 9 4 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

23 If P is a point on the hyperbola

(a) 9

(b) 4

(c) e 2

(d) None of these

points of intersection of perpendicular 24 The locus of the 2 2 x y tangents to − = 1 is 16 9 (a) x 2 − y 2 = 7 (c) x 2 + y 2 = 25

(b) x 2 − y 2 = 25 (d) x 2 + y 2 = 7

25 Tangents are drawn from points on the hyperbola

x2 y2 − = 1 to the circle x 2 + y 2 = 9. The locus of the 9 4 mid-point of the chord of contact is x2 y2 x2 y2 (a) x + y = (b) (x 2 + y 2 ) 2 = − − 9 4 9 4  x2 y2   x2 y2  2 2 2 2 2 2 (c) (x + y ) = 81  −  (d) (x + y ) = 9  −  4  9 4  9 2

2

x2 y2 − = 1 meets X -axis at 4 2 P and Y -axis at Q. Lines PR and QR are drawn such that OPRQ is a rectangle (where, O is the origin). Then, R j lies on JEE Mains 2013

26 A tangent to the hyperbola

4 2 + 2 =1 x2 y 2 4 (c) 2 + 2 = 1 x y (a)

2 4 − 2 =1 x2 y 4 2 (d) 2 − 2 = 1 x y (b)

29 Let P (a sec θ, b tan θ ) and Q (a sec φ , b tan φ ), where

π x2 y2 , be two points on 2 − 2 = 13. If (h, k ) is the 2 a b point of intersection of normals at P and Q, then k is

θ+φ=

a2 + b 2 a a2 + b 2 (c) b

30 If the chords of contact of tangents from two points ( x1, y1 ) and ( x 2 , y 2 ) to the hyperbola 4x 2 − 9y 2 − 36 = 0 x x are at right angles, then 1 2 is equal to y 1y 2 (a)

9 4

(b) −

 3 − 1 sq unit 2   3 + 1 sq units 2 

28 Given the base BC of ∆ABC and if ∠ B − ∠C = K, a constant, then locus of the vertex A is a hyperbola. (a) No

(b) Yes

(c) Both

81 16

(d) None

(d) −

81 16

(a) 9x 2 − 8 y 2 + 18 x − 9 = 0 (b) 9x 2 − 8 y 2 − 18 x + 9 = 0 (c) 9x 2 − 8 y 2 − 18 x − 9 = 0 (d) 9x 2 − 8 y 2 + 18 x + 9 = 0

32 If chords of the hyperbola x 2 − y 2 = a 2 touch the

parabola y 2 = 4ax . Then, the locus of the middle points of these chords is (a) y 2 = (x − a)x 3 (c) x 2 (x − a) = x 3

(b) y 2 (x − a) = x 3 (d) None of these

33 The locus of middle points of chords of hyperbola 3x 2 − 2y 2 + 4x − 6y = 0 parallel to y = 2x is (a) 3 x − 4 y = 4 (c) 4 x − 3 y = 3

(b) 3 y − 4 x + 4 = 0 (d) 3 x − 4 y = 2

34 The equation of the chord joining two points

( x1, y1 ) and ( x 2 , y 2 ) on the rectangular hyperbola xy = c 2 j AIEEE 2002 is x + x1 + x 2 y1 x (c) + y1 + y 2 x1

(a)

y =1 + y2 y =1 + x2

x y + =1 x1 − x 2 y1 − y 2 x y (d) + =1 y1 − y 2 x1 − x 2 (b)

35 Match the vertices (v ) and foci (f ) of hyperbola given in Column I with their corresponding equation given in Column II and choose the correct option from the codes given below. Column I

x − 2y − 2 2x − 4 2y − 6 = 0 with vertex at the point A. Let B be one of the end points of its latusrectum. If C is the focus of the hyperbola nearest to the point A , then the area of the ∆ABC is  (b)    (d)  

(c)

x 2 − y 2 = 9, then the equation of the corresponding pair of tangents is

2

 2 (a)  1 −  sq unit 3   2 (c)  1 +  sq units 3 

9 4

31 If x = 9 is the chord of contact of the hyperbola

27 Consider a branch of the hyperbola 2

(a 2 + b 2 ) a (a 2 + b 2 ) (d) − b (b) −

(a)

Column II 2

A.

v (± 2, 0), f (± 3, 0)

1.

y x2 − =1 25 39

B.

v (0, ± 5), f (0, ± 8)

2.

y2 x2 − =1 9 16

C. v (0, ± 3), f (0 ± 5)

3.

x2 y2 − =1 4 5

Codes A (a) 3 (c) 3

B 1 2

C 2 1

A (b) 1 (d) 2

B 3 1

C 2 3

320

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 If chords of the hyperbola x 2 − y 2 = a 2 touch the

parabola y 2 = 4ax . Then, the locus of their middle point is (a) y (x − a) = 2 x (c) y 2 (x − a) = x 4 2

(b) y (x − a) = x (d) y 2 (x + a) = x 3

2

2

3

2 The equation of a tangent to the hyperbola 3x 2 − y 2 = 3 parallel to the line y = 2x + 4 is (a) y = 3 x + 4 (c) y = 2 x − 2

(b) y = 2 x + 1 (d) y = 3 x + 5

π x2 y2 be two points on the hyperbola 2 − 2 = 1. If 2 a b (h, k ) is the point of intersection of normals at P and Q, then k is equal to

θ+φ =

a 2 + b 2  a 2 + b 2  a2 +b2 (b) −  (c) (d) −    a b    b 

4 The equation of the asymptotes of the hyperbola 3x 2 + 4y 2 + 8xy − 8x − 4y − 6 = 0 is

(a) 3 x 2 + 4 y 2 + 8 xy − 8 x − 4 y − 3 = 0 (b) 3 x 2 + 4 y 2 + 8 xy − 8 x − 4 y + 3 = 0 (c) 3 x 2 + 4 y 2 + 8 xy − 8 x − 4 y + 6 = 0 (d) 4 x 2 + 3 y 2 + 2 xy − x + y + 3 = 0

(b)

π 3

(c)

π 4

(d)

π 6

x2 y2 − =1 a 2 b2 such that OPQ is an equilateral triangle, O being the centre of the hyperbola. Then, the eccentricity e of the hyperbola satisfies

6 If PQ is a double ordinate of the hyperbola

(a) 1 < e
3 (b) e =

3 (c) e = 2

intersects its transverse and conjugate axes at L and M respectively. If locus of the mid-point of LM is hyperbola, then eccentricity of the hyperbola is (b)

e (e 2 − 1)

(c) e

(d) None of these

8 Tangents are drawn to the hyperbola 4x 2 − y 2 = 36 at the points P and Q. If these tangents intersect at the point T (0, 3), then the area (in sq units) of ∆PTQ is j

(a) 45 5 (c) 60 3

(b) 54 3 (d) 36 5

y 2 = 8x and xy = −1 is (a) 3 y = 9x + 2 (c) 2 y = y + 8

(b) y = 2 x + 1 (d) y = x + 2

11 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) (y 2 − x 2 )2 = 4 x 2 (x 2 − a 2 ) (b) (x 2 − y 2 )2 = x 2 (x 2 − a 2 ) (c) (x 2 − y 2 ) = 4 x 2 (x 2 − a 2 ) (d) None of these

 5+1  (a)  , 2  2   5 −1  (c)  , 1   2

(b) (1, 2]  5 −1   5 + 1 (d)  , 1 ∪  1, 2     2

13 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) r 2 (c) 3 r 2

(b) 2 r 2 (d) 4 r 2

14 A point P moves such that sum of the slopes of the

7 The normal at P to a hyperbola of eccentricity e ,

 e + 1 (a)    e − 1

(b) x 2 + y 2 = a 2 (d) None of these

a tangent to the ellipse x 2 + α 2 y 2 = α 2 and the portion of the tangent intercepted by the hyperbola α 2 x 2 − y 2 = 1 subtends a right angle at the center of the curves is

( y − mx )(my + x ) = a 2 and (m 2 − 1)( y 2 − x 2 ) + 4mxy = b 2 is π 2

(a) x 2 + y 2 = b 2 (c) x 2 + y 2 = (a 2 − b 2 )

12 The exhaustive set of values of α 2 such that there exists

5 The angle between the rectangular hyperbolas

(a)

x2 y2 − = 1 (a > b ) so that the fourth vertex a 2 b2 S of parallelogram PQSR lies on circumcircle of ∆PQR, then locus of P is the hyperbola

10 The equation of the common tangent to the curves

3 Let P (a sec θ, b tan θ ) and Q (a sec φ, b tan φ ), where

a2 +b2 (a) a

9 If tangents PQ and PR are drawn from variable point P to

JEE Mains 2018

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

15 A triangle is inscribed in the rectangular hyperbola

xy = c 2 , such that two of its sides are parallel to the lines y = m1x and y = m2 x . Then, the third side of the triangle touches the hyperbola c 2 (m1 + m2 )2  (a) xy =    4m1 m2 

c 2 (m1 − m2 )2  (b) xy =    4m1 m2 

c 2 (m1 − m2 )2  (c) xy =   m1 m2  

c 2 (m1 + m2 )2  (d) xy =   m1 m2  

321

DAY

ANSWERS SESSION 1

SESSION 2

1. (a) 11. (c) 21. (d)

2. (b) 12. (c) 22. (a)

3. (d) 13. (c) 23. (a)

4. (d) 14. (b) 24. (d)

5. (c) 15. (b) 25. (c)

31. (b)

32. (b)

33. (a)

34. (a)

35. (a)

1. (b) 11. (a)

2. (b) 12. (a)

3. (d) 13. (d)

4. (a) 14. (a)

5. (a) 15. (a)

6. (a) 16. (b) 26. (d)

7. (b) 17. (d) 27. (b)

8. (c) 18. (b) 28. (b)

9. (a) 19. (a) 29. (d)

10. (b) 20. (a)

6. (d)

7. (b)

8. (a)

9. (c)

10. (d)

30. (d)

Hints and Explanations SESSION 1

5

Y

E = 2a −

1 Since, the vertices are on the Y-axis (with origin as the mid-point), the y2 x2 equation is of the form 2 − 2 = 1 a b As vertices are (0, ± 6) 25 ∴ a = 6, b 2 = a2 (e 2 − 1) = 36  − 1  9  = 64 So, the required equation of the y2 x2 hyperbola is − = 1 and the foci 36 64 are (0, ± ae ) = (0, ± 10).

2 Let the equation of the hyperbola be x2 y 2 …(i) − =1 a2 b 2 we have, length of the latusrectum = 8 2b 2 ⇒ = 8 ⇒ b 2 = 4a a ⇒ a2 (e 2 − 1) = 4a ⇒ a(e 2 − 1) = 4 9 ⇒ a  − 1 = 4 ⇒ a = 5 5  Putting a = 5 in b 2 = 4a, we getb 2 = 20 Hence, the equation of the required x2 y 2 hyperbola is − = 1. 25 20 2 y2 3 Given that the hyperbola x2 − 2 = 1 a b is passing through the points (3, 0) and (3 2, 2), so we get a2 = 9 and b 2 = 4. Again, we know that b 2 = a2 (e 2 − 1). This gives 4 = 9 (e 2 − 1) 13 13 ⇒ e2 = ⇒e = 9 3 4 Given equation can be rewritten as y2 x2 − = 1(− k > 0) 2 k (− k ) (− k ) k e2 = 1 + 2 = 1 − 2 k k 1 ⇒ e = 1− k

2b 2 2 = 2a2 − a2e 2 | a a

∴ Difference = 2a | 2 − e 2 |

8 Given equation of hyperbola is X′

O

(–2, 0)

X

(2, 0)

Y′ Let equation of hyperbola be y2 x2 − 2 =1 2 a b where, 2ae = 4 and e = 2 ⇒ a=1 Q a2e 2 = a2 + b 2 ⇒ 4 = 1 + b 2 ∴ b2 = 3 Thus, equation of hyperbola is x2 y2 − = 1 or 3 x2 − y 2 = 3 1 3

6 The given equation may be written as 2

2

2

2

x y x y − = 1⇒ − =1 2 32 8 (2 2 )2  4 2   3  3  On comparing the given equation with the standard equation x2 y 2 − = 1, we get a2 b 2  4 2 a2 =    3 

2

and b 2 = (2 2 )2

∴ Length of transverse axis of a 4 2 8 2 hyperbola = 2 a = 2 × = 3 3

7 Let equation of hyperbola be y2 x2 − 2 =1 2 a b Length of transverse axis is 2a and 2b 2 Length of latusrectum is . a Now, difference

x2 y 2 − = 1 and equation of conjugate a2 b 2 x2 y 2 hyperbola is 2 − 2 = 1 b a Since, e and e ′ are the eccentricities of the respective hyperbola, then b2 a2 e 2 = 1 + 2 ,(e ′ )2 = 1 + 2 a b 1 1 a2 b2 + =1 ∴ 2 + 2 = 2 e e′ a + b 2 a2 + b 2

9 Here, a = sinθ Since, foci of the ellipse are (± 1, 0). ∴ ±1 =

a2 + b 2 ⇒ b 2 = cos 2 θ

Then, equation is

x2 y2 − =1 2 sin θ cos 2 θ

⇒ x2 cos ec2θ − y 2 sec2 θ = 1 10 Given equation can be rewritten as ( x − 4) ( y − 3) = 4 which is a rectangular hyperbola of the type xy = c 2 . ∴ c=2 Then, a= b = c 2 = 2 2 ∴ Length of latusrectum 2b 2 = = 2a = 4 2 a 11 Let equation of hyperbola be x2 y 2 − = 1. a2 b 2 2 2b b2 Given, = 8⇒ =4 a a 1 and 2b = (2ae ) ⇒ 2b = ae 2  b2  ⇒ 4b 2 = a2e 2 ⇒ 4  2  = e 2 a  ⇒ 4 (e 2 − 1) = e 2 [Qb 2 = a2 (e 2 − 1)] 4 2 ⇒ 3e 2 = 4 ⇒ e 2 = ⇒ e = 3 3

322

12 Now, taking option (c).

13 The given equation of hyperbola is x2 y2 − =1 cos 2 α sin2 α 2 2 Here, a = cos α and b 2 = sin2 α So, the coordinates of foci are (± ae , 0). ∴

e = 1+

y = −x + 2 p

17 As we know, equation of tangent at

2x e t + e −t Let x = a ⇒ = e t + e − t …(i) 2 a 2y and …(ii) = e t − e −t b On squaring and subtracting Eq. (ii) from Eq. (i), we get 4 x2 4 y 2 x2 y 2 − 2 = 4 ⇒ 2 − 2 =1 a2 b a b

b2 sin2 α ⇒ e = 1+ 2 a cos 2 α

⇒ e = 1 + tan2 α = sec α Hence, abscissae of foci remain constant when α varies.

14 Let equation of tangent to hyperbola y2 x2 − 2 = 1 is y = mx + a2 m2 − b 2 2 a b i.e. mx − y + a2 m2 − b 2 = 0 ∴ Required product mae + a2 m2 − b 2 =  m2 + 1  

( x1 , y 1 ) is xx1 − 2 yy 1 = 4, which is same as 2 x + 6 y = 2 2y x1 4 =− 1 = ⇒ x1 = 4 ∴ 2 2 6 and y 1 = − 6

18 Let the equation of hyperbola be 2

2

y x − 2 = 1. a2 b ∴ ae = 2 ⇒ a2e 2 = 4 ⇒ a2 + b 2 = 4 ⇒ b 2 = 4 − a2 y2 x2 ∴ − =1 a2 4 − a2 Since, ( 2, 3 ) lie on hyperbola. 2 3 ∴ − =1 a2 4 − a2 ⇒ 8 − 2a2 − 3a2 = a2 (4 − a2 ) ⇒ 8 − 5a2 = 4a2 − a4 ⇒ a4 − 9a2 + 8 = 0 ⇒ (a2 − 8)(a2 − 1) = 0 ⇒ a2 = 8, a2 = 1 ∴ a=1 Now, equation of hyperbola is x2 y2 − = 1. 1 3 ∴ Equation of tangent at ( 2, 3 ) is given by

− mae +

a2 m2 − b 2

m +1 2

a2 m2 − b 2 − m2 a2e 2 =  m2 + 1   m2 a2 (1 − e 2 ) − b 2 − m2b 2 − b 2 = =   2 m2 + 1    m +1  [Qb 2 = a2 (e 2 − 1)] 2 =b 15. Here, a2 = 232 , b2 = 232 7 5 21 116 21 and y = − with slope − . x+ 5 5 5 2 116   2 2 2 Now, a m − b =    5  [since, line is tangent] If ( x1 , y 1 ) is the point of contact, then tangent is S 1 = 0 ∴ 7 x1 x − 5y 1 y = 232 On comparing it with 21 x + 5y = 116, we get x1 = 6, y 1 = − 2 So, the point of contact is (6, − 2). 16 Here, e 1 = 1 − 16 = 3 25 5 5 Q e1 e2 = 1 ⇒ e2 = 3 Since, foci of ellipse are (0, ± 3). Hence, equation of hyperbola is x2 y2 − = − 1. 16 9

y 3y = 1 ⇒ 2x − =1 3 3 which passes through point (2 2, 3 3 ). 2x −

19 Suppose the common tangent is

y = mx + c to 9 x2 − 4 y 2 = 36 and x2 + y 2 = 3 …(i) ∴ c 2 = a2 m2 − b 2 = 4 m2 − 9 and c 2 = 3 + 3 m2 …(ii) From Eqs. (i) and (ii), we get 4 m2 − 9 = 3 m2 + 3 ⇒ m2 = 12 ⇒ m=2 3 ∴

c = 3 + 3 × 12 = 39

Hence, the common tangent is y = 2 3 x + 39

20 Since, the line y = αx + β is tangent to 2

2

y x − 2 = 1. a2 b ∴ β2 = a2α2 − b 2. . So, locus of (α, β ) is y 2 = a2 x2 − b 2 ⇒ a2 x2 − y 2 − b 2 = 0 Since, this equation represents a hyperbola, so locus of a point P(α, β ) is a hyperbola. the hyperbola

21 Given equation of hyperbola is x2 y 2 − = 1. 9 4 Here, a2 = 9, b 2 = 4 and equation of line is

...(i)

If the line y = mx + c touches the hyperbola x2 y 2 − = 1, then c 2 = a2 m2 − b 2 …(ii) a2 b 2 From Eq. (i), we get m = −1,c = 2 p On putting these values in Eq. (ii), we get ( 2 p )2 = 9(1) − 4 ⇒ 2 p2 = 5

22 We know that the locus of the point of intersection of perpendicular tangents to the hyperbola x2 y 2 − = 1 is a circle x2 + y 2 = a2 − b 2 a2 b 2 Thus, locus of the point of intersection of perpendicular tangents to the hyperbola x2 y 2 − = 1 is a circle 3 1 x2 + y 2 = 3 − 1 ⇒ x2 + y 2 = 2

23 The point on the hyperbola is P ( x1 , y 1 ), then N is ( x, 0). xx1 yy 1 − =1 9 4  9  This meets X-axis at T  , 0  x1  9 ∴ OT ⋅ ON = ⋅ x1 = 9 x1 ∴ Tangent at ( x1 , y 1 ) is

24 The equation of tangent in slope form to x2 y2 − = 1 is 16 9 16 m2 − 9

the hyperbola y = mx +

Since, it passes through (h, k ). ∴

k = mh +

16 m2 − 92

⇒ (k − mh )2 = (16 m2 − 92 ) ⇒ k 2 + m2 h2 − 2m kh − 16 m2 + 9 2 = 0 It is quadratic in m and let the slope of two tangents be m1 and m2 , then k2 + 9 m1 m2 = 2 h − 16 ⇒

−1=

k2 + 9 ⇒ h2 + k 2 = 7 h2 − 16

The required locus is x2 + y 2 = 7

25 Any point on the hyperbola is

(3sec θ , 2 tan θ). The chord of contact to the circle is 3 x sec θ + 2 y tan θ = 9 …(i) If ( x1 , y 1 ) is the mid-point of the chord, then its equation is …(ii) xx1 + yy 1 = x12 + y 12 From Eqs. (i) and (ii), we get 3sec θ 2 tan θ 9 = = 2 x1 y1 x1 + y 12

1 2 x2 y 2 ( x1 + y 12 ) 2 = 1 − 1 81 9 4 Hence, the locus is  x2 y2  ( x2 + y 2 ) 2 = 81  −  4  9 Eliminating θ,

323

DAY tan B − tan C 1 + tan B ⋅ tan C y y − λ+ x λ− x = y2 1+ 2 λ − x2

2 2 26 Given hyperbola is x − y = 1.

∴ tan K =

4 2 Here, a2 = 4 , b 2 = 2 ⇒ a = 2, b = 2 The equation of tangent is y x sec θ − tan θ = 1 a b x y ⇒ sec θ − tan θ = 1 2 2 So, the coordinates of P and Q are P (2cos θ, 0) and Q (0, − 2 cot θ),

⇒ λ2 − x2 + y 2 = − 2 xy cot K ⇒ x2 − 2 cot K ⋅ xy − y 2 = λ2 which is a hyperbola. ax by θ is + = a2 + b 2 and normal sec θ tan θ π ax by at φ = − θ is + = a2 + b2 2 cos ecθ cot θ Eliminating x, we get 1 1  2 2 by  −  = (a + b )  sin θ cos θ   1 − 1     cos θ sin θ  (a2 + b 2 ) ⇒ by = −(a2 + b 2 ) or k = − b

30 The equation of hyperbola is

( x − 2 )2 ( y + 2 )2 − =1 4 2 For A( x, y ), Y

2

9 x1 4 x2 x x 9 81 ⋅ × ⋅ = − 1 ⇒ 1 2 = −   = −  4 4 y1 9 y2 y1 y2 16

B X¢

X

C

A Y¢

2 3 = 4 2 x− 2=2⇒ x=2+ e = 1+

∴

For C ( x, y ), x − 2 = ae = ∴

x=

6+

2 6

2

Now, AC =

6 + 2−2− 2 = 6 −2 b2 2 and BC = = =1 a 2 1 ∴ Area of ∆ABC = × AC × BC 2  3  1 = × ( 6 − 2) × 1 =  − 1 sq unit 2  2 

28 Let B (−λ , 0), C (λ , 0) and A ( x, y ). Given, K = ∠B − ∠ C

Y

x2 y 2 ...(i) − =1 9 4 The equation of the chords of contact of tangents from ( x1 , y 1 ) and ( x2 , y 2 ) to the given hyperbola are x x1 y y1 ...(ii) − =1 9 4 x x2 y y 2 ...(iii) and − =1 9 4 Lines (ii) and (iii) are at right angles. 4 x2 − 9 y 2 = 36 ⇒

27 The given equation can be rewritten as

31 Let (h, k ) be the point whose chord of contact w.r.t. hyperbola x2 − y 2 = 9 is x = 9. We know that chord of (h, k ) w.r.t. hyperbola x2 − y 2 = 9 is T = 0 ⇒ hx − ky − 9 = 0 But it is the equation of line x = 9. This is possible only whenh = 1, k = 0. Again, equation of pair of tangents is T 2 = SS 1 ⇒ ( x − 9)2 = ( x2 − y 2 − 9)(1 − 9) ⇒ x2 − 18 x + 81 = ( x2 − y 2 − 9)(−8) ⇒ 9 x2 − 8 y 2 − 18 x + 9 = 0

32 Equation of chord of hyperbola x2 − y 2 = a2 with mid-point as (h, k ) is given by h (h2 − k 2 ) xh − yk = h2 − k 2 ⇒ y = x − k k This will touch the parabola y 2 = 4ax, if  h2 − k 2  a ⇒ ak 2 = − h3 + k 2 h −  =  k  h/k

A (x, y)

∴ Locus of the mid-point is x 3 = y 2 ( x − a)

B -l

O

l

C

X

33 Let (h, k ) is mid-point of chord. Then, its equation is

⇒ 3h − 4k = 4 Thus, locus of mid-point is 3 x − 4 y = 4

34 The mid-point of the chord is

29 Equation of normal at

respectively. Let coordinates of R are (h, k). ∴ h = 2cos θ, k = − 2cotθ k − 2 − 2h ⇒ = ⇒ sin θ = h sinθ 2k On squaring both sides, we get 2h2 2h2 sin2 θ = ⇒ 1 − cos 2 θ = 4 k2 4 k2 2h2 h2 h2 2h2 ⇒ 1− + =1 = ⇒ 2 2 4k 4 4 4k h2  2 2 4 ⇒ + 1 = 1 ⇒ 2 + 1 = 2   4  k2 k h 4 2 − =1 ⇒ h2 k 2 4 2 Hence, R lies on 2 − 2 = 1. x y

3hx − 2 ky + 2( x + h ) − 3( y + k ) = 3h2 − 2 k 2 + 4h − 6k ⇒ x(3h + 2) + y (−2 k − 3) = 3h2 − 2 k 2 + 2 h − 3k Since, this line is parallel to y = 2 x. 3h + 2 = 2 ⇒ 3h + 2 = 4k + 6 ∴ 2k + 3

 x1 + x2 , y 1 + y 2  .     2 2 The equation of the chord T = S 1 . y + y2   x1 + x2  ∴ x  1  + y       2 2 x1 + x2   y 1 + y 2   =2        2 2 ⇒ ⇒

x ( y 1 + y 2 ) + y ( x1 + x2 ) = ( x1 + x2 )( y 1 + y 2 ) y x + =1 x1 + x2 y1 + y2

35 A. Since, vertices (± 2, 0) and foci (± 3, 0) lie on X -axis, as coefficient of Y-axis is zero. Hence, equation of hyperbola will be of the form y2 x2 …(i) − 2 =1 2 a b Here, it is given that a = 2 and c = 3 Q c 2 = a2 + b 2 ⇒ 9 = 4 + b2 ⇒ b2 = 5 Put the values of a2 = 4 and b 2 = 5 in Eq. (i), we get x2 y2 …(ii) − =1 4 5 B. Since, vertices (0, ± 5) and foci (0, ± 8) lie on Y-axis as coefficient of X-axis is zero. Hence,equation of hyperbola will be of the form y 2 x2 …(i) − =1 a2 b 2 Here, it is given that (0, ± a) = (0, ± 5) and foci (0, ± c ) = (0, ± 8) ⇒ a = 5and c = 8 Q c 2 = a2 + b 2 ⇒ b 2 = 39 Put a2 = 25 and b 2 = 39 in Eq. (i), we get y2 x2 − =1 25 39 C. Since, vertices (0, ± 3) and foci (0, ± 5) lie on Y- axis as coordinate of x is zero. Hence, equation of hyperbola will be of the form y2 x2 …(i) − 2 =1 2 a b

324

Here, it is given that vertices (0, ± 3) = (0, ± a) and foci (0, ± 5) = (0, ± c ) ⇒ a = 3 and c = 5 Q c 2 = a2 + b 2 ⇒ 25 = 9 + b 2 ⇒ b 2 = 16 2 Put a = 9 and b 2 = 16 in Eq. (i), we get y2 x2 − =1 9 16

SESSION 2

− 4 (− 4)2 − λ ( 4)2 = 0 ⇒

12λ + 64 − 12 − 64 − 16λ = 0

⇒

− 4λ − 12 = 0 ⇒ λ = −3

∴ Required equation is 3 x2 + 4 y 2 + 8 xy − 8 x − 4 y − 3 = 0

5 Given equations are …(i) ( y − mx )(my + x ) = a2 and (m2 − 1)( y 2 − x2 ) + 4 mxy = b 2 …(ii) On differentiating Eq.(i), we get dy ( y − mx )  m + 1  dx 

1 If (h, k ) is the mid-point of the chord,

then its equation by T = S 1 is hx − ky = h2 − k 2 k 2 − h2 h ⇒ y = x+ k k If it touches the parabola y 2 = 4ax, then we get k 2 − h2 a⋅ k = ⇒ ak 2 = hk 2 − h3 k h ⇒ ay 2 = xy 2 − x3 So, required locus is y 2 ( x − a) = x3 .

dy + (my + x )  − m  = 0  dx  dy (my + x + my − m2 x ) dx

⇒

+ y − mx − m2 y − mx = 0 dy − y + m2 y + 2mx ⇒ = = m1 [say] dx 2my + x − m2 x On differentiating Eq.(ii), we get dy (m2 − 1)  2 y − 2 x    dx

2 Equation of the hyperbola is x2 y2 − =1 1 3 Equation of tangent in terms of slope y = mx ± (m2 − 3 )

dy + 4 m  x + y  = 0  dx 

3 x2 − y 2 = 3 ⇒

m=2 y = 2x ± 1

Given, ∴

⇒

dy [2 y (m2 − 1) + 4mx] dx = − 4 my + 2 x (m2 − 1) dy −2my + m2 x − x = dx m2 y − y + 2mx

⇒

3 Equation of the tangents at y x sec θ − tan θ = 1 a b ∴ Equation of the normal at P is ax + b cosecθ y = (a2 + b 2 )sec θ ...(i) Similarly, the equation of normal at Q (asec φ,b tan φ) is

= m2 m1 m2 = − 1

P (asec θ,b tan θ) is

ax + b cosecφ y = (a + b )sec φ 2

2

...(ii)

On subtracting Eq. (ii) from Eq. (i), we get a2 + b 2 sec θ − sec φ y = ⋅ b cosecθ − cosec φ a2 + b 2 So, k = y = b π secθ − sec  − θ 2  ⋅ π  cosecθ − cosec − θ  2 =

Q

So, angle between the hyperbola =

2

2

MP = MQ = l . Given that ∆OPQ is an equilateral, then OP = OQ = PQ ⇒ (OP )2 = (OQ )2 = (PQ )2 2 a a2 ⇒ 2 (b 2 + l 2 ) + l 2 = 2 (b 2 + l 2 ) + l 2 = 4l 2 b b a2 2 2 ( b + l ) = 3l 2 ⇒ b2 a 2 2 Y b Öb + l , l P

2

X¢

O

4 The equation of the asymptotes of the hyperbola 3 x2 + 4 y 2 + 8 xy − 8 x − 4 y − 6 = 0 is 3 x2 + 4 y 2 + 8 xy − 8 x − 4 y + λ = 0 It should represent a pair of straight lines. ∴ abc + 2 fgh − af 2 − bg 2 − ch2 = 0 3 ⋅ 4 ⋅ λ + 2 ⋅ (−2) (−4) 4 − 3 (−2)2

π ⋅ 2

6 Q PQ is the double ordinate. Let

a + b sec θ − cosecθ a + b  ⋅ =−  b  b cosecθ − sec θ   2

[say]

Y¢

⇒ ⇒

M

X

Q a 2 2 b Öb + l , –l

 a2  a2 = l 2  3 − 2  b   a2b 2 2 l = >0 (3b 2 − a2 )

∴ ⇒ ⇒

3b 2 − a2 > 0 3b 2 > a2 3a2 (e 2 − 1) > a2

2 3 7 Equation of normal at P (asec φ,b tan φ) is a xcos φ + by cot φ = a2 + b 2 . Then, coordinates of L and M are  a2 + b 2    a2 + b 2 ⋅ sec φ,0 and  0, tan φ  b  a    ⇒

e 2 > 4/3 ∴ e >

respectively. Let mid-point of ML is Q (h, k ), (a2 + b 2 ) then h= sec φ 2a 2 ah ...(i) ∴ sec φ = 2 (a + b2 ) 2 2 (a + b ) and k = tan φ 2b 2bk ...(ii) ∴ tan φ = 2 (a + b2 ) From Eqs. (i) and (ii), we get 4a2 h2 4b 2 k 2 sec2 φ − tan2 φ = 2 − (a + b 2 )2 (a2 + b 2 )2 Hence, required locus is y2 x2 − =1 2 2 2 2 2 a +b   a + b2       2a   2b  Let eccentricity of this curve is e 1 . 2

2

 a2 + b 2   a2 + b 2  2 ⇒   =  (e 1 − 1)  2b   2a  ⇒ a2 = b 2 (e 12 − 1) ⇒a2 = a2 (e 2 − 1)(e 12 − 1) [Qb 2 = a2 (e 2 − 1 )] 2 2 2 2 ⇒ e e1 − e − e1 + 1 = 1 e ⇒ e 12 (e 2 − 1) = e 2 ⇒ e 1 = (e 2 − 1) 8 Tangents are drawn to the hyperbola 4 x2 − y 2 = 36 at the point P and Q. Tangent intersects at point T (0, 3) Y T (0, 3)

O (–3Ö5, –12)Q

S(0, –12)

X P(3Ö5, –12)

Clearly, PQ is chord of contact. ∴Equation of PQ is − 3 y = 36 ⇒ y = − 12 Solving the curve 4 x2 − y 2 = 36 and y = − 12, we get x = ± 3 5 1 Area of ∆PQT = × PQ × ST 2 1 = (6 5 × 15) = 45 5 2

325

DAY 9 Fourth vertex of parallelogram lies on circumcircle ⇒ Parallelogram is cyclic. ⇒ Parallelogram is a rectangle. ⇒ Tangents are perpendicular ⇒ Locus of P is the director circle i.e., x2 + y 2 = a2 − b 2

10 Any point on parabola y 2 = 8 x is (2t 2 , 4 t ). The equation of tangent at that point is ...(i) yt = x + 2t 2 Given that, xy = −1 ...(ii) On solving Eqs. (i) and (ii), we get y ( yt − 2t 2 ) = −1 ⇒ ty 2 − 2t 2 y + 1 = 0 Q It is common tangent. It means they are intersect only at one point and the value of discriminant is equal to zero. i.e. 4 t 4 − 4 t = 0 ⇒ t = 0, 1 ∴ The common tangent is y = x + 2, (when t = 0, it is x = 0 which can touch xy = −1 at infinity only)

11 The equation of a hyperbola of the y2 x2 − 2 = 1 where, λ is a 2 a λ parameter. The asymptotes of this y x hyperbola = ± ⋅ Suppose ( x ′ , y ′ ) is a λ a point P on the hyperbola which is equidistant from the transverse axis and asymptote. y ′2 x ′2 Then, ...(i) − 2 =1 2 a λ x ′/ a − y ′/λ and ...(ii) y′ = 1 1 + a2 λ2 2 2 y′ x′ i.e. = 2 − 1 [from Eq. (i)] ...(iii) λ2 a y ′2 2 x ′ y ′ 1 1 x ′2 and y ′2  2 + 2  = 2 + 2 − a λ  a λ aλ series is

[from Eq. (ii)] ...(iv) On simplification the second relation gives 4 x ′2 y ′2 a2 ( y ′2 − x ′2 )2 = λ2 = 4 x ′2 ( x ′2 − a2 ) [using Eq. (iii)] So, the locus of P is ( y 2 − x2 )2 = 4 x2 ( x2 − a2 ).

12 The equation of tangent at point

P (α cos θ,sin θ) x y cos θ + sin θ = 1 ...(i) α 1 Let it cut the hyperbola at points P and Q. Homogenising hyperbola α2 x2 − y 2 = 1 with the help of Eq. (i), we get 2 x α2 x2 − y 2 =  cos θ + y sin θ α 

This is a pair of straight lines OP OQ. Given ∠POQ = π/2. Coefficient of x2 + Coefficient of y 2 = 0

cos 2 θ − 1 − sin2 θ = 0 α2 cos 2 θ or α2 − − 1 − 1 + cos 2 θ = 0 α2 α2 (2 − α2 ) or cos 2 θ = α2 − 1 2 Now, 0 ≤ cos θ ≤ 1 α2 (2 − α2 ) or 0≤ ≤1 α2 − 1  5+ 1  After solving, we get α2 ∈  , 2  2 

13 Let equation of the rectangular

hyperbola be xy = c 2 …(i) and equation of circle be …(ii) x2 + y 2 = r 2 . From Eq. (i) and Eq.(ii) eliminating y, we get ...(iii) x 4 − r 2 x2 + c 4 = 0 Let x1 , x2 , x3 and x 4 are the roots of Eq. (iii). 4

∑x

∴ Sum of roots =

i

=0

i =1

Sum of products of the roots taken two at a time = Σ x1 x2 = − r 2 From Eq. (i) and Eq. (ii) eliminating x, we get ...(iv) y 4 − r 2 y2 + c 4 = 0 Let y 1 , y 2 , y 3 and y 4 are the roots of Eq. (iv). ∴

4

∑y

i

= 0 and

Σ

y1 y2 = − r 2

i =1

Now, CP 2 + CQ 2 + CR2 + CS 2 = x1 + y 1 + x2 + y 2 2

2

2

2

+ x + y 2 3

2 3

+ x4 + y 4 2

2

= ( x1 + x2 + x3 + x 4 ) 2

2

2

+ ( y1 + y 2

+ y3 + y 4 ) 2

2

4

[Q ( ∑ x i ) = 0, Σ x1 x2 = − r 2 ,

∑y

i

From Eqs. (ii) and (iii), we get h2 t 12 + t 22 + t 32 + t 24 = 4 h2 Given that, =k 4 Hence, locus of (h, k ) is x2 = 4 ay , which is a parabola.       15 Let P  ct 1 , c  , Q  ct 2 , c  , R  ct 3 , c  be 

= 0, Σ y 1 y 2 = − r 2 ]

i =1

= 4r2

14 Any point on the hyperbola xy = 4 is  2t , 2  . Now, normal at  2t , 2  is       t t 2 y − = t 2 ( x − 2t ). [its slope is t 2 ] t 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 feet of normals passing through (h, k ). Let roots be t 1 , t 2 , t 3 and t 4 , then

t1 



t2 



t3 

the vertices of a ∆PQR inscribed in the rectangular hyperbola xy = c 2 such that the sides PQ and QR are parallel to y = m1 x and y = m2 x, respectively. ∴ m1 = Slope of PQ and m2 = Slope of QR 1 m1 = − ⇒ t1 t2 1 t2 t3

and

m2 = −

∴

m  m1 t = 3 ⇒ t3 =  1  t1 m2 t 1  m2 

2

2 2

 4  = ( ∑ x i )2 − 2 Σ x1 x2  i = 1    4  + ( ∑ y i )2 − 2 Σ y 1 y 2   i =1  = (0 + 2r 2 ) + (0 + 2r 2 ) i =1 4

h …(ii) 2 t1 t2 + t2 t3 + t1 t3 + t1 t 4 + t 2 t 4 + t 3 t 4 = 0 …(iii) t1 t2 t3 + t1 t2 t 4 + t1 t3 t 4 k …(iv) + t2 t3 t 4 = − 2 and t 1 t 2 t 3 t 4 = − 1 …(v) On dividing Eq. (iv) by Eq. (v), we get 1 1 1 1 k + + + = t1 t2 t3 t4 2 2 ⇒ y 1 + y 2 + y 3 + y 4 = k Q y =  t   t1 + t2 + t3 + t 4 =

or α2 −

The equation of PR is x + yt 1 t 3 = c (t 1 + t 3 )  m  m  ⇒ x + y  1  t 12 = c  t 1 + 1 t 1  m2   m2   m   m  ⇒ x + y  1  t 12 = c  1 + 1  t 1 m2   m2   ⇒

 m1  x+ y  t1   m2 

2

 c (m1 + m2 ) m1  = 2 ⋅ t1  m 2 m m 2 1 2   ⇒ x + yt 2 = 2λt , where, c (m1 + m2 ) m1 t = ⋅ t 1 and λ = m2 2 m1 m2 Clearly, it touches the hyperbola, 2

or

 c (m1 + m2 ) xy =    2 m1 m2   c 2 (m1 + m2 )2  xy =    4 m1 m2 

326

DAY THIRTY

Unit Test 4

(Coordinate Geometry) 1 The parabola y 2 = 4x and the circle ( x − 6)2 + y 2 = r 2 will have no common tangent, then (a) r >

(b) r
18

20

(d) r ∈( 20, 28 )

nx + ly + m = 0 are concurrent, if

substends 60° angle at the other focus, then its eccentricity e is 2

(b)

3

(c)

5

(d)

6

4 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  (a)  −∞, −  

2 − 1  ∪ 2  

2 − 1  ,∞  2 

(d) R

5 If ω is one of the angles between the normals to the

x2 y2 + 2 = 1 at the points whose eccentric angles 2 a b π 2 cot ω are θ and + θ, then is equal to 2 sin 2θ ellipse

1− e

2

(b)

e2 1+ e

2

(c)

e2 1 − e2

(d)

e2 1 + e2

6 If in a ∆ABC (whose circumcentre is origin), a ≤ sin A ,

then for any point ( x , y ) inside the circumcircle of ∆ABC (a) xy < (c)

1 8

1 1 < xy < 8 2

(b) | xy | >

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

(− ∞,5 2 ) (4 2 − 14 , 5 2 ) (4 2 − 14 , 4 2 + None of the above

14 )

Tangents drawn to parabola at A and B meet Y -axis at A 1 and B 1 , respectively. If the area of trapezium AA1B1B is equal to 24 a 2 , then angle subtended by A1B1 at the focus of the parabola is equal to

(c) (−1, 1)

e2

r r   ,− 3 + − 5 +  is an interior point of the major  2 2 segment of the circle x 2 + y 2 = 16, cut off by the line x + y = 2, is

9 AB is a double ordinate of the parabola y 2 = 4 ax .

(b) (−∞,1) ∪ (1, ∞)

(a)

(b) 1 (d) 3

8 The range of values of r for which the point

(b) l + m − n = 0 (d) None of these

3 If the latusrectum of a hyperbola through one focus

(a)

the quadrilateral formed by the lines x = 0, y = 0, 3x + y − 4 = 0 and 4x + y − 21 = 0, then the possible number of positions of the point A is (a) 0 (c) 2

2 The lines lx + my + n = 0, mx + ny + l = 0 and (a) l + m + n = 0 (c) l − m + n = 0

7 If A (n, n 2 ) (where, n ∈ N) is any point in the interior of

1 8

(d) None of these

(a) 2 tan−1 (3) (c) 2 tan−1 (2)

(b) tan−1 (3) (d) tan−1 (2)

10 If two tangents can be drawn to the different branches of hyperbola

x2 y2 − = 1 from the point (α , α 2 ), then 1 4

(a) α ∈ (− 2, 0) (c) α ∈ (− ∞, − 2)

(b) α ∈ (− 3, 0) (d) α ∈ (− ∞, − 3)

11 A hyperbola has the asymptotes

x + 2y = 3 and x − y = 0 and passes through ( 2, 1). Its centre is (a) (1, 2) (c) (1, 1)

(b) (2, 2) (d) (2, 1)

327

DAY 12 The equation of the ellipse having vertices at ( ± 5 , 0) and foci at ( ± 4, 0) is

( 3, 3 3 ), then the third vertex is

x2 y2 (a) + =1 25 16 2 (c) 9x + 25 y 2 = 225

(b) 4 x 2 + 5 y 2 = 20 (d) None of these

13 The foci of an ellipse are ( 0 , ± 1) and minor axis is of unit length. The equation of the ellipse is (a) 2 x 2 + y 2 = 2 (c) 4 x 2 + 20y 2 = 5

(b) x 2 + 2 y 2 = 2 (d) 20x 2 + 4 y 2 = 5

14 The radius of the circle passing through the foci of the ellipse

x2 y2 + = 1 and having its centre at (0, 3) is 16 9

(a) 3 (c)

(b) 4 7 (d) 2

12

ax + by − 1 = 0 form a triangle with origin as orthocentre, then (a, b ) is given by (a) (− 3, 3) (c) (− 8, 8)

(b) (6, 4) (d) (0, 7)

16 If P (1, 0), Q ( − 1, 0) and R ( 2, 0) are three given points, then the locus of S satisfying the relation SQ 2 + SR 2 = 2 SP 2 is

1 (b) a ∈ (− ∞, − 3) ∪  , 1 3  1 (d) a ∈  , ∞  3 

x = 2y is

32 x 9 32 y (d) x = 9 (b) y =

(b) x 2 + 4 y 2 = 9 (d) 4 x 2 + y 2 = 81

(b) 6x − y + 32 = 0 (d) x + 6y − 32 = 0

21 The length of the latusrectum of the parabola (c)

12 13

between the pair of tangents drawn from (a, a ) to the π  circle x 2 + y 2 − 2x − 2y − 6 = 0 lies in the range  , π  3  is (a) (0, ∞) (c) (− 2, − 1) ∪ (2, 3)

(b) − 2

(b) (− 3, − 1) ∪ (3, 5) (d) − (3, 0 ) ∪ (1, 2)

(c) 2

(d) 1

anti-clockwise direction about origin, then coordinates of P in new position are 1 7  (a)  ,   2 2 1 7  (c)  − ,   2 2

7 1  (b)  − ,−   2 2 1 7  (d)  ,−   2 2

28 The number of integral values of λ for which the equation x 2 + y 2 − 2λx + 2λy + 14 = 0 represents a circle whose radius cannot exceed 6, is (b) 10

(c) 11

(d) 12

(d)

which passes through the focus of the parabola y 2 = 16x are (a) ± 2 (c) –1/2, 2

(b) 1/2, –2 (d) ± 1

both the parabolas y 2 = | x | is

(b) (− 1, 1) (d) None of these

(a) (0, 1) (c) (− 1, 0)

31 The parameters t and t′ of two points on the parabola y 2 = 4ax , are connected by the relation t = k 2t ′. The tangents at these points intersect on the curve (a) y 2 = ax

169 [( x − 1)2 + ( y − 3)2 ] = ( 5x − 12 y + 17)2 is 28 13

25 The exhaustive range of values of a such that the angle

30 The range of values of n for which (n, − 1) is exterior to

by it with the points (1, 5) and (3, – 7) is 21 sq units. Then, locus of the point is

(b)

(b) (x − q) 2 = 4 py (d) (x − q) 2 = 4 px

29 The slopes of tangents to the circle ( x − 6) + y 2 = 2

20 A point moves such that the area of the triangle formed

14 13

the X-axis. The locus of the outer end of the diameter through A is

2

of tangent to the hyperbola x 2 − y 2 = 9 is

(a)

(b) (1, 2) (d) None of these

24 A variable circle through the fixed point A ( p, q ) touches

(a) 9

19 The locus of poles with respect to the parabola y 2 = 12x

(a) 6x + y − 32 = 0 (c) 6x − y − 32 = 0

(a) (3, 6) (c) (4, 8)

27 If a point P ( 4, 3) is rotated through an angle 45° in

18 The diameter of 16x 2 – 9y 2 = 144 which is conjugate to

(a) 4 x 2 + y 2 = 36 (c) x 2 + 4 y 2 = 36

A ( − 1, 4), B ( 6, − 2) and C ( − 2, 4). If 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

(a) 0

3x − y + 1 = 0 and x + 2y − 5 = 0 containing the origin, if

16x 9 16y (c) x = 9

23 Let ABC is a triangle with vertices

are concyclic, if k is equal to

17 The point (a 2 , a + 1) lies in the angle between the lines

(a) y =

(b) (− 3, 3) (d) None of these

26 The four distinct points ( 0, 0), ( 2, 0), ( 0, − 2) and (k , − 2)

a straight line parallel to X-axis a circle through the origin a circle with centre at the origin a straight line parallel toY-axis

1 (a) a ∈ (− 3, 0) ∪  , 1 3  1 (c) a ∈  − 3,   3

(a) (3, − 3) (c) (− 3, 3 3 )

(a) (x − p) 2 = 4qy (c) (x − p) 2 = 4qx

15 If the straight lines 2x + 3y − 1 = 0 , x + 2y − 1 = 0 and

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

22 If two vertices of an equilateral triangle are ( 0, 0) and

16 13

1 (c) y 2 = ax  k +   k

(b) y 2 = k 2 x 2

(d) None of these

328

32 Triangle ABC is right angled at A. The circle with centre A and radius AB cuts BC and AC internally at D and E respettvely if BD = 20 and DC = 16 then the length of AC equals. (a) 6 21

(b) 6 26

(c) 30

(d) 32

33 If the line y − 3x + 3 = 0 cuts the parabola y 2 = x + 2 at A and B, and if P ≡ ( 3,0), then PA ⋅ PB is equal to 2+ 3 (a) 2    1  (c)

(b)

4 (2 + 3 ) 3

(d)

4 (2 − 3 ) 3 2 (2 − 3 ) 3

x2 y2 + = 1 and P is 25 16 any point on it, then difference of maximum and minimum of SP ⋅ S ′ P is equal to (b) 9

(c) 15

(d) 25

35 The locus of a point which moves, such that the chord of contact of the tangent from the point to two fixed given circles are perpendicular to each other is (a) circle (c) ellipse

(b) parabola (d) None of these

x + y 2 = 1 at 27   π  ( 3 3 cos θ, sin θ) where,θ ∈  0,  . Then, the value of θ  2   such that sum of intercept on axes made by this tangent is minimum, is

36 Tangent is drawn to the ellipse

π 3

(b)

π 6

(c)

π 8

(d)

π 4

37 The condition for the line px + qy + r = 0 to be tangent to the rectangular hyperbola x = ct , y = (a) p < 0, q > 0 (c) p > 0, q < 0

c is t

(b) p > 0, q > 0 (d) None of these

38 If the line x + 3y + 2 = 0 and its perpendicular line are conjugate w.r.t. 3x 2 − 5y 2 = 15 , then equation of conjugate line is (a) 3 x − y = 15 (c) 3 x − y + 10 = 0

(b) 3 x − y + 12 = 0 (d) 3 x − y = 4

39 The product of the lengths of perpendicular drawn from any point on the hyperbola x 2 − 2y 2 − 2 = 0 to its asymptotes, is (a) 1/2

(b) 2/3

(c) 3/2

(d) 2

x2 y2 40 Tangents at any point on the hyperbola 2 − 2 = 1 cut a b the axes at A and B respectively, if the rectangle OAPB, where O is the origin is completed, then locus of the point P is given by a 2 b2 (a) 2 − 2 = 1 x y (c)

a2 b2 − 2 =1 2 y x

(a) x + y + 1 = 0 (b) 5 x + y + 17 = 0 (c) 4 x + 9y + 30 = 0 (d) x − 5 y − 7 = 0

a2 b2 (b) 2 + 2 = 1 k y (d) None of these

a b c −2 = + where a, b, c > 0 the family of lines c b bc a x + by + c = 0 passes through the point.

(a) (1, 1)

(b) (1, − 2)

(c) (−1, 2)

(d) (− 1, 1)

43 A series of ellipses E 1, E 2 , E 3 ,....., E n are drawn such that E n touches E n −1 at the extremeties of the major axis of E n −1 and the foci of E n coincide with the extremeties of minor axis of E n −1. If eccentricity of the ellipse B independent of x then the value of eccentricity is. 5 −1 2 5+1 4

(a) (c)

2

(a)

3x − 2y + 5 = 0 are equation of on altitude and on angle bisector respectively drawn from B, the equation of BC is.

42 If

34 If S and S′ are the foci of the ellipse

(a) 16

41 In a triangle ABC, if A( 2,−1) and 7x − 10y + 1 = 0 and

(b) (d)

5+1 3 5 −1 4

44 If one of the diagonal of a square is along the line x = 2y and one of its vertices is (3, 0), then its sides through this vertex are given by the equations (a) (b) (c) (d)

y y y y

− 3 x + 9 = 0, 3 y + x − 3 = 0 + 3 x + 9 = 0, 3 y + x − 3 = 0 − 3 x + 9 = 0, 3 y − x + 3 = 0 − 3 x + 3 = 0, 3 y + x + 9 = 0

45 Given A( 0,0) and B( x , y ) with x ∈( 01 , ) and y > 0. Let the

slope of line AB equals m1. Point C lies on the line x = 1 such that the slope of BC equals m2 where 0 < m2 < m1. If the area of the triangle ABC can be expressed as (m1 − m2 )f ( x ), then the largest possible value of f ( x ) is. (a) 1 / 8

(b) 1 / 2

(c) 1 / 4

(d) 1

46 A circle is inscribed into a rhombus ABCD with one angle 60°. The distance from the centre of the circle to the nearest vertex is equal to 1. If P is any point of the circle, then | PA|2 + | PB|2 +| PC| 2 +| PD| 2 is equal to. (a) 8

(b) 9

(c) 10

(d) 11

47 The equation of common tangent touching the circle

( x − 3) 2 + y 2 = 9 and the parabola y 2 = 4x above the X-axis is (a) (c)

3y = 3x + 1 3y = x + 3

(b) (d)

3 y = − (x + 3) 3 y = − (3 x + 1)

x 2 11y 2 + =1 16 256 − 2x − 15 = 0, then φ is equal to

48 If the tangent at the point φ on the ellipse touches the circle x π 2 π (c) ± 3 (a) ±

2

+y

2

π 4 π (d) ± 6 (b) ±

329

DAY Directions (Q.Nos. 49-55) Each of these questions contains two statements : Statement I (Assertion) and Statement II (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below. (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 A cyclic quadrilateral is a square, if its diagonals are the diameters of the circle.

52 If a circle S = 0 intersects a hyperbola xy = c 2 at four points. Statement I If c = 2 and three of the intersection points  2 are (2, 2), (4, 1) and  6,  , then coordinates of the fourth  3 1  point are  , 16 . 4  Statement II If a circle intersects a hyperbola at t1, t 2 , t 3 , t 4 , then t1 ⋅ t 2 ⋅ t 3 ⋅ t 4 = 1.

53 The auxiliary circle of an ellipse is described on the major

49 If p, x1, x 2 , x 3 and q , y1, y 2 , y 3 form two arithmetic progression with common differences a and b. Statement I The centroid of triangle formed by points ( x1, y1 ), ( x 2 , y 2 ) and ( x 3 , y 3 ), lies on a straight line. Statement II The point (h, k ) given by x + x2 + K + xn y + y2 + K + yn and k = 1 always lies h= 1 n n on the line b ( x − p ) = a ( y − q ) for all values of n.

50 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, 1 then slope of AB is . 2 Statement II If normal at (at12 , 2at1 ) cuts again the 2 parabola at (at 22 , 2at 2 ), then t 2 = − t1 − . t1

51 Suppose ABCD is a cyclic quadrilateral inscribed in a circle. Statement I If radius is one unit and AB ⋅ BC ⋅ CD ⋅ DA ≥ 4, then ABCD is a square.

axis of an ellipse. Statement I The circle x 2 + y 2 = 4 is auxiliary circle of an x2 y2 ellipse + 2 = 1 (where, b < 2). 4 b Statement II A given circle is auxiliary circle of exactly one ellipse. x2 y2 + 2 = 1, which 2 a b is not an extremity of major axis, meets a directrix atT .

54 The tangent at a point P on the ellipse

Statement I The circle on PT as diameter passes through the focus of the ellipse corresponding to the directrix on which T lies. Statement II PT subtends a right angle at the focus of the ellipse corresponding to the directrix on whichT lies.

55 Statement I If the perpendicular bisector of the line

segment joining P (1, 4) and Q (k , 3) has y-intercept − 4, then k 2 − 16 = 0. Statement II Locus of a point equidistant from two given points is the perpendicular bisector of the line joining the given points.

ANSWERS 1 (b)

2 (a)

3 (b)

4 (a)

11 (c)

12 (c)

13 (d)

14 (b)

21 (b)

22 (c)

23 (b)

24 (a)

5 (a)

6 (a)

7 (b)

8 (b)

9 (d)

10 (c)

15 (c)

16 (d)

17 (a)

18 (b)

19 (a)

20 (a)

25 (b)

26 (c)

27 (a)

28 (c)

29 (d)

30 (b)

31 (c)

32 (b)

33 (c)

34 (b)

35 (a)

36 (b)

37 (b)

38 (b)

39 (b)

40 (a)

41 (b)

42 (d)

43 (a)

44 (a)

45 (a)

46 (d)

47 (c)

48 (c)

49 (a)

50 (d)

51 (c)

52 (d)

53 (c)

54 (a)

55 (a)

330

Hints and Explanations e2 − 1 2e 3 e 2 − 2e − 3 = 0

1 Any normal of parabola is y = − t x + 2t + t .

⇒

Y

X¢

⇒

y 2 = 4x

A

X

⇒

t = 0, t 2 = 4

Thus,

A ≡ (4, 4)

0 + 0 − 4< 0

Y

 ⇒  m2 − 

X S (ae, 0) L¢ (ae, –b2/a)

2 − 1  2 

 ⇒ m −  

 ⇒m ∈  −∞,−  

 m +  

2 − 1   ∪ 2  

5 The equations of the normals to the 2

2

y x + 2 = 1 at the points whose a2 b π eccentric angles are θ and + θ are 2 ax secθ − by cosec θ = a2 − b 2 and − ax cosec θ − by sec θ = a2 − b 2 , respectively. Since, ω is the angle between these two normals. Therefore, a a tan θ + cot θ b b tan ω = a2 1− 2 b ellipse

=

Y¢

ab (tan θ + cot θ) b 2 − a2

2 ab ⇒ tan ω = sin 2θ (b 2 − a2 ) = ∴

2 cot ω = sin 2θ

2a2 1 − e 2 2ab = 2 2 2 2 (a − b ) sin 2θ a e sin 2θ e2 1 − e2

S

2 − 1  >0 2  2 − 1  ,∞  2 

A (n, n2) 3x + y–

X¢

O

0

L (ae, b2/a)

x=0

2 − 1 >0 2 

=

Y

R

21

focus S (ae , 0) of the hyperbola y2 x2 − 2 = 1. 2 a b It substends angle 60° at the otherfocus S ′ (− ae , 0).

2 − 1  2 1 + 2   m + >0 2   2 

y–

3 Let LSL ′ be a latusrectum through the

We have, ∠LS ′ L ′ = 60° ∴ ∠LS ′ S = 30° LS In ∆LS ′ S , tan30° = S ′S 1 b2 / a ⇒ = 3 2ae 1 b2 = 2 ⇒ 3 2a e

7 Origin is on the left of PS, as

r

r < 20

2 Since, lines are concurrent.

X¢

1

i.e.

sin A 1 ⇒ 2R ≤ 1 ⇒ R ≤ 2 So, for any point ( x, y ) inside the circumcircle, 1 x2 + y 2 < 4  x2 + y 2  Using AM ≥ GM,  ≥ | xy | 2   1 ⇒ | xy | < 8

⇒ 4m 4 + 4m2 − 1 > 0

Thus, for no common tangent,

∴

1 . m

This will be a chord of the circle x2 + y 2 = 4, if length of the perpendicular from the centre (0, 0) is less than the radius.

Y¢

AC =

e = 3

the parabola y 2 = 4 x isy = mx +

If it passes through (6,0), then − 6t + 2t + t3 = 0

⇒

(e − 3 ) ( 3 e + 1) = 0

∴

4 The equation of tangent of slope m to

C (6, 0)

O

6 Given, a ≤ sin A ⇒ a ≤ 1

=

3

4=

0

Q

P

y=0 X

Y¢

∴At point A (n, n2 ), 3 n + n2 − 4 > 0 ⇒ n2 + 3 n − 4 > 0 ⇒ (n + 4) (n − 1) > 0 ⇒ n − 1> 0 …(i) Now, as A and O lies on the same sides of QR. and

4 x + y − 21 = 0 + 0 − 21 < 0

∴At point A (n, n2 ), 4 n + n2 − 21 < 0 ⇒ n2 + 4 n − 21 < 0 ⇒ (n + 7) (n − 3) < 0 [Qn ∈ N ]…(ii) ⇒ 0< n< 3 From Eqs. (i) and (ii), 1< n< 3 ⇒ n = 2 Hence, A (2, 4) is only one point.

8 The given point is an interior point, if 2

2

r  r    − 5+  + − 3 +  − 16 < 0   2 2 ⇒ r 2 − 8 2 r + 18 < 0 ⇒ 4 2 − 14 < r < 4 2 +

14

Since, the point is on the major segment, the centre and the point are on the same side of the line x + y = 2.

331

DAY ⇒

r −3+ 2

− 5+

r − 2< 0 2

⇒ r 0

and

a 2 + 2 (a + 1) − 5 < 0

i.e. a (3 a − 1) > 0 and a 2 + 2a − 3 < 0 i.e. a(3a − 1) > 0 and (a − 1)(a + 3) < 0 1 a ∈ (− ∞, 0) ∪  , ∞  ⇒ 3  and

a ∈ (− 3, 1)

∴

1 a ∈ (− 3, 0) ∪  , 1 3 

18 Diameters y = m1 x and y = m2 x are conjugate diameters of the hyperbola x2 y 2 b2 – = 1, if m1 m2 = 2 . a2 b 2 a 1 2 2 Here, a = 9, b = 16 and m1 = 2 b2 Q m1 m2 = 2 a 1 16 (m2 ) = ⇒ 2 9 32 ⇒ m2 = 9 32 x Thus, the required diameter is y = . 9

19 Let the pole be (h, k ), so that polar is ky = 6( x + h ) 6x 6h y = + k k Since, it is tangent to the hyperbola,

⇒

∴

∴ x+ y =0 Since, AO is perpendicular to BC. a ∴ (− 1)  −  = − 1 ⇒ a = − b  b

11 Given equations of asymptotes are

∴

x2 − y 2 = 9 y–

B

So, the asymptotes are y = ± 2 x. ∴ 2α < α2 ⇒ α < 0 or α > 2 and − 2 α < α2 α < − 2 or α > 0 ∴ α ∈ (−∞ , − 2) or (2, ∞ )

on the same side of both the lines.

A

2 x+

1 4 Since, (α, α2 ) lies on the parabola y = x2 , then (α, α2 ) must lie between the x2 y2 asymptotes of hyperbola − =1 1 4 in 1st and 2nd quadrant.

0

10 Given that, x − y = 1

2 x + 3 y − 1 + λ ( x + 2 y − 1) = 0. Since, it passes through (0, 0), then λ = − 1.

y-1 =

2

2x +3

2

∴ Required radius = ( 7 − 0) 2 + (3 − 0) 2 = 4

17 Since, origin and the point (a 2 , a + 1) lie

⇒ ⇒

2 – a  − 1 − ⋅  = −1 3+ a  2 2 − a = − 6 − 2a a = − 8 and b = 8

16 Let the coordinates of a point S be ( x, y ). Since, SQ 2 + SR2 = 2SP 2 ⇒ ( x + 1) 2 + y 2 + ( x − 2) 2 + y 2 = 2 [( x − 1) 2 + y 2 ] ⇒ 2x + 3 = 0 Hence, it is a straight line parallel to Y-axis.

c 2 = 9 m2 − 9

36 h2 324 = 2 −9 k2 k Qc = 6 h , m = 6   k k  2 2 ⇒ 4 h + k = 36 Hence, the locus is 4 x2 + y 2 = 36. ⇒

20 Let ( x, y ) be the required point. Then,

x y 1 1 1 5 1 = 21 2 3 −7 1

⇒ (5 + 7)x − (1 − 3)y + (−7 − 15) = 42 ⇒ 12 x + 2 y − 22 = 42 ⇒ 6 x + y − 32 = 0

21 Given parabola is 5x − 12 y + 17  ( x − 1) 2 + ( y − 3) 2 =     13 Focus = (1, 3), directrix is 5x − 12 y + 17 = 0 ∴ Length of latusrectum 5 − 36 + 17 28 =2 = 13 13

2

332

22 Since, ∠AOM is 30°. Y 3

B

M

2a2 − 4 a − 6 > 0 a < − 1 or a > 3 θ 2 2 Now, in ∆PAC, tan = 2 2 2a − 4 a − 6 So, ⇒

A (3, 3√3)

3 3√3

Y

2√2

X

O (0, 0)

P

θ/2 C

Hence, the required point B is (− 3, 3 3 ). O

23 Let ( x1 , y 1 ), ( x2 , y 2 ) and ( x3 , y 3 ) are 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)

(x1, y1) D

E (x2, y2)

B (6, –2)

C (–2, 4)

3 × 6 − 1 × 1 17 = 4 4 −2×3+ 4×1 2 1 y1 = =− =− 4 4 2 5 Similarly, x2 = 0, y 2 = 2 5 and x3 = − , y 3 = 4 4 Let ( x, y ) be the coordinates of centroid of ∆DEF . 1 17 5 + 0 −  = 1 ∴ x =  3 4 4 1 1 5 and y =  − + + 4 = 2  3 2 2 ∴ x1 =

So, the coordinates of centroid are (1, 2).

24 The circle touching the X -axis is

x 2 + y 2 + 2gx + 2 fy + g 2 = 0. Since, it passes through ( p, q ) . ∴ p2 + q 2 + 2gp + 2 fq + g 2 = 0 ...(i) If ( x, y ) is the other end of the diameter, then p + x = − 2g , q + y = − 2 f Now, Eq. (i) gives p2 + q 2 − p ( p + x ) − q (q + y ) ( p + x )2 + =0 4 2 ⇒ ( x + p ) = 4 px + 4qy ⇒ ( x − p ) 2 = 4qy

25 Given that, x + y − 2x − 2y − 6 = 0 Centre = C (1, 1), radius = 2 2 2

2

Since, point (a, a) must lie outside the circle.

X

π As given that, < θ < π 3 π θ π < < ⇒ 6 2 2 2 2 1 ∴ > 3 2a2 − 4 a − 6 ⇒

F (x3, y3)

A (a , a )

∴ ⇒ ⇒ ∴

a − 2a − 3 < 2 3 2

a2 − 2a − 3 < 12 2 a − 2a − 15 < 0 − 3< a< 5 a ∈ (− 3, − 1) ∪ (3, 5)

26 Let the general equation of the circle be

x 2 + y 2 + 2g x + 2 f y + c = 0. QThe equation of circle passing through (0, 0), (2, 0) and (0, − 2). ∴ c=0 …(i) 4 + 4g + c = 0 …(ii) and 4 − 4f + c = 0 …(iii) On solving Eqs. (i), (ii) and (iii), we get c = 0, g = − 1, f = 1 ∴ The equation of circle becomes x2 + y 2 − 2 x + 2 y = 0 Since, it is passes through (k , − 2), k 2 + 4 − 2k − 4 = 0 ⇒ k 2 − 2k = 0 ⇒ k = 0, 2 We have already take a point (0, − 2), so we take only k = 2. 27 Slope of line OP = 3 , let new position is 4 Q ( x, y ). y Slope of OQ = , x also x2 + y 2 = OQ 2 = 25 = (OP 2 ) y 3 − x 4 ∴ tan 45° = 3y 1+ 4x 4y − 3x ⇒ ± 1= 4x + 3y ⇒ or

4x + 3y = 4y − 3x − 4x − 3y = 4y − 3x 1 ...(i) ⇒ x= y 7 or ...(ii) − x = 7y 1 Correct relation is x = y as new 7 point must lies in Ist quadrant.

x 2 + 49 x 2 = 25 1 7 x=+ ,y = 2 2

∴ ⇒

28 Since, ( radius ) 2 ≤ 36 ⇒ λ2 + λ2 − 14 ≤ 36 ⇒ λ2 ≤ 25 ⇒ − 5≤ λ ≤ 5 ⇒ λ = 0, ± 1, ± 2, ± 3, ± 4, ± 5 Hence, number of integer values of λ is 11.

29 Tangent to the circle with slope m is y = m ( x − 6) ±

2 (1 + m2 ).

Since, it passes through (4, 0). ∴ 4 m2 = 2 + 2m2 ⇒ m=±1

30 Since, 1 − |n| > 0 ⇒ |n| < 1 or n ∈ (− 1, 1)

31 Tangents at t and t ′ meet on the point ( x, y ) given by ...(i) x = att ′ = ak 2 t ′ 2 and y = a (t + t ′ ) = a (k 2t ′ + t ′ ) = at ′ (k 2 + 1) ...(ii) From Eqs. (i) and (ii), ak 2 ⋅ y 2 k2 y 2 = x= 2 2 2 a (k + 1) a (k 2 + 1) 2 ⇒ y2 =

ax (k 2 + 1) 2 1 = ax  k +   k2 k

2

32. In ∆ ABC AC 2 + AB 2 = BC 2 AC 2 + r 2 = 362 and CF × CE = BC × CD ⇒ ( AC + r )( AC − r ) = 36 × 16 ⇒ AC 2 − r 2 = 36 × 16 From Eqs. (i) and (ii), we get 2 AC 2 = 36(36 + 16) ⇒ AC 2 = 18 × 52 ⇒ AC = 6 26

...(i)

...(ii)

C 16

E

D 20

r r

A

B

F

33 Let PA = r1 , PB = − r2 Put ( 3 + r cos θ, r sin θ) in y 2 = x + 2 ⇒ r 2 sin2 θ = ( 3 + r cos θ) + 2 ⇒ r 2 sin2 θ − r cos θ − ( 3 + 2) = 0 PA ⋅ PB = − r1 . r2 =

3+2 sin2 θ

= ( 3 + 2)(1 + cot2 θ)

333

DAY 1 = ( 3 + 2) 1 +   3 4(2 + 3 ) PA ⋅ PB = 3 Y

[Qtanθ = 3]

A X¢

X

P( 3,0)

O

(-2,0)

A′∈ (−43 , ) Coordinate of B is intersection point of 7 x − 10 y + 1 = 0 and 3 x − 2 y + 5 = 0 i.e. (−3,−2) ∴Equation of BC is 3+ 2 y −3 = ( x + 4) −4 + 3

35 Let the equations of two given circles are …(i) x2 + y 2 + 2g 1 x + 2 f1 y + c 1 = 0 and x2 + y 2 + 2g 2 x + 2 f2 y + c 2 = 0 …(ii) Now, the equations of the chords of contacts from P (h, k ) to Eqs. (i) and (ii) are x ( h + g 1 ) + y ( k + f1 ) + g 1 h + f1 k + c 1 = 0 and x (h + g 2 ) + y (k + f2 ) + g 2 h + f2 k + c 2 = 0 According to the given condition, (h + g 1 ) (h + g 2 ) × = −1 ( k + f1 ) ( k + f2 ) ⇒ h2 + (g 1 + g 2 ) h + g 1 g 2 + k 2 + k ( f1 + f2 ) + f1 f2 = 0 Hence, the locus of point is x2 + y 2 + (g 1 + g 2 )x + ( f1 + f2 )y + g 1 g 2 + f1 f2 = 0 which is the equation of a circle.

36 Equation of tangent at (3 3 cos θ, sin θ) 2

x + y 2 = 1 is 27

+ y sin θ = 1.

3 3 This intersect on the coordinate axes at (3 3 sec θ, 0) and (0,cosec θ) [say]

On differentiating w.r.t. θ, we get f ′ (θ) = 3 3 sec θ tan θ − cosec θ cot θ 3 3 sin3 θ − cos 3 θ sin2 θ cos 2 θ

⇒

⇒

5x + y + 17 = 0 B(–3,–2)

0

3x–2y+5=0

y+1=

= a2 (1 − e 2 cos 2 θ) = a2 − a2e 2 cos 2 θ = 25 − 9cos 2 θ Maximum = 25 − 9(0) = 25 [θ = 90° ] Minimum = 25 − 9(1) = 16 [θ = 0° ] Maximum–Minimum = 25 − 16 = 9

∴ Sum of intercepts on axes is 3 3 secθ + cosec θ = f (θ)

dy −c dx = 2 and =c dt t dt dy −1 ∴ = 2 dx t But equation of tangent is px + qy + r = 0. p p 1 1 − =− 2 ⇒ = >0 ∴ q t q t2 p ⇒ >0 q

A¢

7x–10

34 (SP )(S ′ P ) = a (1 − e cos θ) a (1 + e cos θ)

=

3 x − 2 y + 5 = 0. A′ is given by x − 2 y + 1 −2(6 + 2 + 5) = = = −2 3 −2 13

Then,

Y¢

x cos θ

41 Image of A(2,−1) with respect to line

37 Given, x = ct , y = c /t B

to the ellipse

For maxima and minima, put f ′ (θ) = 0 3 3 sin3 θ − cos 3 θ = 0 1 ⇒ tanθ = 3 π ⇒ θ= 6 π At θ = , f ′ ′ (θ) > 0. So, f (θ) is 3 π minimum at θ = . 6

p > 0, q > 0 or p < 0, q < 0

38 Since, the lines x + 3 y + 2 = 0 and 3 x − y + k = 0 are conjugate w.r.t. x2 y2 − = 1. 5 3 ∴ 5 (1) (3) − 3 (3) ( − 1) = 2k ⇒ k = 12 Hence, equation of conjugate line is 3 x − y + 12 = 0.

42 We have, a − 2 = b + c

39 Given, equation can be rewritten as x2 y2 − =1 2 1 Here, a 2 = 2, b 2 = 1 The product of length of perpendicular drawn from any point on the hyperbola to the asymptotes is a 2 b2 2 (1) 2 = = . a2 + b 2 2 + 1 3

⇒ ⇒ ⇒

c bc a − 2 bc = b + c a = b + c + 2 bc ( a )2 = ( b + c )2

⇒

( b + c )2 − ( a )2 = 0

⇒ or ⇒

b 2 h2 = sin2 θ a 2k2 b 2 h2 h2 + 2 =1 ⇒ 2 2 a k a [Q sin2 θ + cos 2 θ = 1] a 2 b2 ⇒ − 2 =1 h2 k a2 b2 Hence, the locus of P is 2 − 2 = 1. x y

b

b+ c − a=0 b+ c + a=0 ax + by + c = 0

(rejected)

passes through fixed point (−11 , )

43 Here,

b ′ = a, ae ′ =b (b ′ )2 = (a′ )2 − (a′ e )2

⇒

40 The equation of tangent is y x sec θ − tan θ = 1. a b So, the coordinates of A and B are (a cos θ, 0) and ( 0, − b cot θ), respectively. Let coordinates of P are (h, k ) . ∴h = a cos θ, k = − b cot θ k b ⇒ =− h a sinθ

C

A(2,–1)

a2 =

b2 − b2 e2 Y

(0, a¢) a¢e X¢

{

(0,b) X

a

(a, 0) or (b¢, 0)

⇒

Y¢  b2  Q1 − e 2 = 2  a  

⇒

b2 e 2 = 2 (1 − e 2 ) a

⇒

e 2 = (1 − e 2 )(1 − e 2 )

⇒

1 − e2 = ±e

334 ⇒

e 2 − e − 1 = 0 or e 2 + e − 1 = 0

⇒

e =

or

−1 ± 5 2 1± 5 2

5−1 [0 < e < 1] 2 44 Equation of diagonal AC is y − 0 = − 2( x − 3) ⇒ 2x + y = 6 On solving 2 x + y = 6 and x = 2 y , we 6 12 get y = and x = 5 5 ⇒

e =

Y x=2y

C B D X¢ O

45° X A (3, 0)

⇒

(m1 − m2 )x = c − m2

⇒

c = (m1 − m2 )x + m2

Now area of 0 0 1 1 1 ∆ ABC = x m1 x 1 = cx − m1 x 2 2 1 c 1 1 = |((m1 − m2 )x + m2 )x − m1 x| 2 1 = (m1 − m2 )( x − x2 ) 2 [Qx > x2 in (0, 1)] 1 Hence f ( x ) = ( x − x2 ) 2 1 f ( x )max = , 8 1 when x = 2

46 r = 3 sin 30° = 3 2

P ( x, y ) is any point on circle (PA )2 + (PB )2 + ( PC )2 + ( PD )2

Y¢ 12 6 So, the centre of square is  ,  .  5 5

Y

Let slope of side AB or AD is m, then  m − (− 2) = 1  1 + m (−2)

A(0,1)

⇒

30° r

(m + 2) = ± (1 − 2m ) 1 m = − and m = 3 3 Hence, slopes of AB and AD are 3 and 1 − , respectively. 3 ∴ Equations of sides AB and AD are y − 0 = 3( x − 3 ) 1 and y − 0 = − ( x − 3) 3 or y − 3 x + 9 = 0 and 3 y + x − 3 = 0, respectively. ⇒

45 Let the coordinate of C be (1,c) c−y m2 = 1− x c − m1 x m2 = 1− x

⇒

Qm = y    1  x

C(0, –1) Y¢ = x2 + ( y − 1)2 + ( x + 3 )2 + y 2 + x2 + ( y + 1)2 + ( x − 3 )2 + y 2 [Q x2 + y 2 = r 2 ]

3 = 4  + 2 = 11 4 

47 Any tangent to y 2 = 4 x is of the form C(1, c)

y)m 2 B(x, m1 m 1 > m2

A(0,0) Y′

X D (Ö3, 0)

= 4( x2 + y 2 + 2)

Y

X′

30°

X¢ B (–Ö3, 0)

= 4(r 2 + 2)

X

y = mx +

which is the required equation of tangent.

48 Given equation is x2 y2 + = 1. 16 (16 / 11 ) 2 Thus, the parametric coordinates are 16   sin φ . The equation of  4cos φ,   11 tangent at this point is x cos φ 11 y sin φ + = 1. 4 16 This touches the circle x 2 + y 2 − 2 x − 15 = 0 cos φ −1 4 ∴ =4 cos 2 φ 11sin2 φ + 16 256 ⇒ cos 2 φ + 16 − 8cos φ  cos 2 φ 11sin2 φ  = 256  +  256   16 ⇒ 15 cos 2 φ + 11 (1 − cos 2 φ) + 8cos φ − 16 = 0 ⇒ 4cos 2 φ + 8cos φ − 5 = 0 1 Q cos φ ≠ 5 ⇒ cos φ =  2 2  π π or − ⇒ φ= 3 3

49 Since, p, x1 , x2 , ... and q, y 1 , y 2 , K are in AP with common differences a and b, respectively. ⇒ x i = p + ai and y i = q + ib x1 + x2 + ... + x n n y1 + y2 + K + y n and k = n

∴

1 , (Q a = 1), m

This touches the circle ( x − 3)2 + y 2 = 9, whose centre is (3, 0) and radius is 3. 1 m (3) + −0 m So, =3 2 m +1 3m + 1 = ± 3m m + 1 2

9 m 4 + 1 + 6 m2 = 9 m2 (m2 + 1) [on squaring both sides] ⇒ 3 m2 = 1 1 ⇒ m= ± 3 If the tangent touches the parabola and circle above X-axis, then slope m should be positive. 1 and the equation is ∴ m= 3 x y = + 3 3 ⇒ 3y = x + 3 ⇒

2

⇒

h=

nh =

n

∑x

i

and nk =

i =1

⇒

nh =

i =1

n

∑( p +

ia)

i =1

and nh =

n

∑ (q +

i =1

n

∑y

ib )

i

335

DAY n (n + 1) a 2 n (n + 1) and nk = nq + b 2 h− p n+1 k −q n+1 and ⇒ = = a 2 b 2 h− p k −q ∴ = a b Hence, locus of (h, k ) is b ( x − p) = a (k − q ). Hence, Statement II is true and since for Statement I, n = 3 So, Statement I is true and Statement II is a correct explanation of Statement I. ⇒ nh = np +

50 If t 1 and t 2 are parameters of a and b, then t1 t2 = − 4  2 Also, t 1  − t 1 −  = − 4 t1  

…(i) …(ii)

On solving Eqs. (i) and (ii), we get t1 = 2 2 Also, m AB = = − t1 = − 2 t2 + t1

51 Clearly, ac + bd = AC ⋅ BD ≤ 4 [using Ptolemy’s theorem]

D

c

d

C

b O

A ⇒

a

ac + bd = 4 and AC ⋅ BD = 4

but ac + bd ≥ 4 ⇒ ⇒

B

[Q AM≥ GM]

AC = BD = 2 and ac = bd = 2 [Qabcd ≥ 4] a=b =c =d = 2

52 Statement II is true. For the point (2, 2), t 1 = 1 For the point (4, 1), t 2 = 2 For the point (6, 2/3), t 3 = 3 1 1 For the point  , 16 , t 4 = 4  8 3 Now, t1 ⋅ t2 ⋅ t3 ⋅ t 4 = ≠ 1 4 Hence, Statement I is false.

53 The auxiliary circle of an ellipse x2 y2 + 2 = 1, b < 2 is x2 + y 2 = 4 4 b

54 The equation of tangent to the ellipse is y x cos θ + sinθ = 1 and it meets the a b a directrix x = at e  a b (e − cos θ) T  , . e sin θ  e Since, focus is S (ae, 0). ∴

Slope of SP =

and slope of ST =

b sin θ a (cos θ − e ) b (e – cos θ) a sin θ (1 − e 2 )

Now, as product of slopes = −

b2 a (1 − e 2 ) 2

= −1, therefore PT substends a right angle at the focus. Hence, circle with PT as diameter passes through the focus.

55 Statement II is true, using in Statement

I, ( x − 1) 2 + ( y − 4) 2 = ( x − k ) 2 + ( y − 3) 2 ⇒ 2 (k − 1)x − 2y = k 2 − 8 k2 − 8 y-intercept = − =–4 [given] 2 ⇒ k 2 = 16 ⇒ k 2 − 16 = 0

DAY THIRTY ONE

Vector Algebra Learning & Revision for the Day u

u

u

Vectors and Scalars Types of Vectors Addition, Subtraction and Scalar Multiplication of Vectors

u

u

u

u

u

Angular Bisectors

u

Position Vector (PV) Components of a Vector in 2D and 3D

u

u

Scalar (Dot) Product Vector (Cross) Product

Scalar Triple Product Vector Triple Product Linear Combination, Linear Independence and Dependence

Vectors and Scalars Physical quantities are divided into two categories Scalar Quantities and Vector Quantities. Those quantities which have only magnitude and not related to any fixed direction in space are called scalar quantities or simply scalars. Examples of scalars are mass, volume, density, work, temperature etc., Vectors are those quantities which have both magnitude as well as direction. Displacement, velocity, acceleration, momentum, weight, force etc., are examples of vector –→

→

quantities. A directed line-segment is a vector, denoted as AB (or AB) or simply a (or a). A (Initial point) ●

a

B (Terminal point)

Magnitude (or length) of a vector a is denoted by|a| and it is always a non-negative scalar.

Types of Vectors (i) A vector whose initial and terminal points coincide is called the zero or null vector and it is denoted as 0. (ii) A vector whose magnitude is 1, is called a unit vector. The unit vector in the direction of a a is given by and is denoted by a$ . Unit vectors parallel to X-axis, Y-axis and Z-axis are a denoted by i, j and k, respectively. (iii) Vectors are said to be like when they have the same sense of direction and unlike when they have opposite directions. (iv) Two vectors a and b are said to be equal, written as a = b, if they have same length and same direction. (v) Vectors which are parallel to the same line are called collinear vectors or parallel vector, otherwise they are called non-collinear vector. If a and b are two collinear vectors, then a = λb for same λ ∈ R. (vi) Vectors having the same initial point are called coinitial vectors.

PRED MIRROR Your Personal Preparation Indicator u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

337

DAY (vii) Vectors having the same terminal point are called coterminous vectors. (viii) A system of vectors is said to be coplanar, if they are parallel to the same plane or lie in the same plane otherwise they are called non-coplanar vectors. (ix) The vector which has the same magnitude as that of a given vector a but opposite direction, is called the negative of a and is denoted by − a. (x) A vector having the same direction as that of a given vector a but magnitude equal to the reciprocal of the given vector is known as the reciprocal of a and is denoted by a −1 .

a 1 , a 2 , a 3 , ... in consecutive order, then their sum is represented by the nth side, but in opposite direction (as shown in the adjoining figure). →

→

→

→

→

→

→

→

b a+

O

→

(v) To subtract b from a, we reverse the direction of b and add it to vector a. (vi) Let PQRS be a parallelogram such that PQ = a = SR and PS = b = QR. Then, diagonal PR = a + b (addition of vectors) and diagonal SQ = a − b (subtraction of vectors). a

S

R

a– b b

(xiii) When a particle is displaced from point A to other point B, then the displacement AB is a vector, called displacement vector of the particle.

There are three laws for vector addition, which are given below. (i) Triangle law If the vectors a and b lie along the two sides of a triangle in consecutive order (as shown in the adjoining figure), then their sum (resultant) a + b is represented by the third side, but in opposite direction.

→

by a− b , i.e., a− b = a+ (− b ).

(xii) Vector whose initial points are not specified are called free vectors.

Addition, Subtraction and Scalar Multiplication of Vectors

→

is defined as the vector sum of a and − b and it is denoted

(xi) A vector which is drawn parallel to a given vector through a specified point in space is called a localised vector.

(xiv) Two vectors are called orthogonal, if angle between the two is a right angle.

→

(iv) If a and b are two vectors, then the subtraction of b from a

P

b

b a+ a

Q

(ii) If a is a vector and λ is a scalar (i.e. a real number), then λa is a vector whose magnitude is λ times that of a and whose direction is the same as that of a if λ > 0 and opposite of a if λ < 0. Thus, λa = λ a

Important Results (i) a + b = b + a (iii) a − b ≠ b − a (v) a ± b ≥ a − b (vii) (λ 1 + λ 2 )a = λ 1 a + λ 2 a

(ii) a + (b + c) = (a + b) + c (iv) a ± b ≤ a + b (vi) λ 1 (λ 2a) = (λ 1 λ 2 )a = λ 2 (λ 1a) (viii) λ (a + b) = λa + λb

NOTE When the sides of a triangle are taken in order, it leads to zero

b

resultant, e.g., In ABC, AB + BC + CA = 0.

Angular Bisectors

a

(ii) Parallelogram law If the vectors lie along two adjacent sides of a parallelogram (as shown in the adjoining figure), then diagonal of the parallelogram through the common vertex represents their sum.

Let a and b are unit vectors, the internal bisector of angle between a and b is along a + b and external bisector of angle is along a − b. b a+b

b a

+

O

b

a

a

(iii) Polygon law If (n − 1) sides of a polygon represents vector

a1 O

+

a2

+

a3

a3 a2

a1

a–b

If a and b are not unit vectors, then above angle bisectors are a b a b along and , respectively. + − |a | | b| |a | | b| These bisectors are perpendicular to each other.

ONE

338

Position Vector (PV)

where Ax , A y and Az are the magnitudes of A x , A y and A z , respectively.

Every point P ( x, y, z) in space is associated with a vector whose initial point is O(origin) and terminal point is P. This vector is called position vector and it is given by

Here Ax , A y and Az are called scalar components of A and Ax , A y and Az are called Vector components of A.

OP (or r) = x i + y j + z k. Using distance formula, the magnitude of OP (or r) is OP = x2 + y2 + z2 ●

If A (a1 , a2 , a3 ) and B (b1 , b2 , b3 ) have position vectors then a and b respectively, then AB = b − a and AB = | b − a | = (a1 − b1 )2 + (a2 − b2 )2 + (a3 − b3 )2

●

●

If a and b are the PV of A and B respectively and r be the PV of the point P which divides the join of A and B in the mb ± na ratio m : n, then r = . m±n Here, ‘+’ sign takes for internal division and ‘−’ sign takes for external division. If a, b and c be the PV of three vertices of ∆ABC and r be the a+b+c PV of the centroid of ∆ABC, then r = 3

Components of a Vector in 2D and 3D The process of splitting a vector is called resolution of a vector. The parts of the vector obtained after splitting the vectors are known as the components of the vector. Let A x is the resolved part of A along X-axis i.e. the projection of A on X-axis. Similarly, A y is the resolved part of A along Y-axis, i.e. the projection of A on Y-axis. Then, by the parallelogram law, Y P

M A

Ay j X¢

Ax O

X

N

i Y¢

A = A x + A y = Ax $i + A y $j

OP = ON + NP or

Where Ax and A y are the magnitudes of Ax and A y . Similarly, in three dimension we can represents vector A. as A = A x + A y + A z = Ax i + A y j + Az k

If two vectors a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) are equal, then their resolved parts will also equal i.e. a1 = b1 , a2 = b2 and a3 = b3 . If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k, then (i) a + b = (a1 + b1 )i + (a2 + b2 )j + (a3 + b3 )k (ii) a − b = (a1 − b1 )i + (a2 − b2 )j + (a3 − b3 )k (iii) λa = λa1 i + λa2 j + λa3 k

Scalar (Dot) Product The scalar product of two vectors a and b is given by a ⋅ b = |a || b|cos θ, 0 ≤ θ ≤ π . Properties of Scalar Product is listed below: 1. a ⋅ b = b ⋅ a [commutative law] 2. a ⋅ b = 0, if a ⊥ b 3. a ⋅ a =|a |2 4. i ⋅ i = j ⋅ j = k ⋅ k = 1 and i ⋅ j = i ⋅ k = j ⋅ k = j ⋅ i = k ⋅ i = k ⋅ j = 0 5. If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k, then a ⋅ b = a1 b1 + a2 b2 + a3 b3 6. a ⋅ (α ⋅ b) = α (a ⋅ b) 7. a ⋅ (b ± c) = a ⋅ b ± a ⋅ c [distributive law] 8. For any two vectors a and b, we have (i) | a + b |2 = | a |2 + | b |2 + 2 (a ⋅ b) (ii) | a − b |2 = | a |2 + | b |2 − 2 (a ⋅ b) (iii) (a + b) ⋅ (a − b) = | a |2 − | b |2 (iv) a ⋅ b < 0 iff a and b are inclined at an obtuse angle. (v) a ⋅ b > 0 iff a and b are inclined at an acute angle. 9. If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k are inclined at an angle θ, then a1b1 + a2b2 + a3b3 cos θ = 2 a1 + a22 + a32 b12 + b22 + b32 10. If r is a vector making angles α , β and γ with OX , OY and OZ respectively, then cos α = r ⋅ i, cos β = r ⋅ j, cos γ = r ⋅ k r = | r |cos α i + | r |cos β j + | r |cos γ k If r is a unit vector, then r = (cos α ) i + (cos β) j + (cos γ) k 11. Projection of a on b(scalar component of a along b) is A

Y a

j Ay k

O Az

Z

Ax

i

X

O

M

OM =

a⋅b b

b

339

DAY 12. Components of a along and perpendicular to b are  a ⋅ b $ a⋅b $ OM = ⋅ b and MA = a −   ⋅ b, respectively. b  b  13. Work done If a particle acted on by a force F has displacement d, then work done = F ⋅ d

Vector (Cross) Product The vector product of two vectors a and b is given by a × b = |a || b|sin θ ⋅ n$ , where n$ is a unit vector perpendicular to a and b such that a, b and n$ form a right handed system and θ (0 ≤ θ ≤ π ) is the angle between a and b. Properties of vector product are listed below: 1. a × b = − (b × a ) 2. (a × b)2 = |a|2 | b|2 − (a ⋅ b)2 3. ma × b = m (a × b) = a × mb 4. a × (b ± c) = a × b ± a × c and (b ± c) × a = b × a ± c × a 5. a × b = 0 ⇔ a || b, where, a and b are non-zero vectors. 6. If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k i j k then, a × b = a1 a2 a3 b1 b2 b3 7. The vector perpendicular to both a and b is given by a × b. 8. The unit vectors perpendicular to the plane of a and b a×b are ± and a vector of magnitude λ perpendicular |a × b| λ (a × b) to the plane of (a and b or b and a ) is . |a × b| 9. If i, j and k are three unit vectors along three mutually perpendicular lines, then i × i = j × j = k × k = 0, i × j = k, j × i = − k, j × k = i, k × j = − i, k × i = j, i × k = − j 10. (i) The area of parallelogram with adjacent sides a and b is a × b . (ii) The area of quadrilateral with diagonals d 1 and d 2 1 is d 1 × d 2 . 2 (iii) The area of triangle with adjacent sides a and b, is 1 a ×b . 2 (iv) If a, b, c are position vectors of a ∆ABC, then the 1 area = (a × b) + (b × c) + (c × a ) . 2 11. Three points with position vectors a, b, c are collinear if a × b + b × c + c × a = 0

Scalar Triple Product If a, b, c are three vectors, then their scalar triple product is defined as the dot product of a and b × c. It is denoted by [a b c]. Thus, [a b c] = a ⋅ (b × c).

Properties of Scalar Triple Product are listed below: 1. [a b c] = [b c a] = [c a b] 2. [a b c] = − [b a c] = − [c b a ] = − [a c b] 3. If λ is a scalar, then [λa b c] = λ [a b c] 4. [a b c1 + c2 ] = [a b c1 ] + [a b c2 ] 5. (a × b) ⋅ c = a ⋅ (b × c) 6. The scalar triple product of three vectors is zero, if any two of them are equal or parallel or collinear. 7. If a, b and c are coplanar, then [a b c] = 0 8. If [a b c] = 0, then any two of the vectors are parallel or a, b and c are coplanar or c = α a + βb. 9. Four points with position vectors a, b, c and d will be coplanar, if [b − a, c – a, d – a] = 0. 10. Volume of parallelopiped, whose coterminous edges are a, b and c is [a b c] . 11. If a = a1 i + a2 j + a3 k, b = b1 i + b2 j + b3 k a1 a2 a3 and c = c1 i + c2 j + c3 k, then [a b c] = b1 b2 b3 c1 c2 c3 a⋅ a 12. [a × b b × c c × a] = [a b c]2 = b ⋅ a c⋅ a

a⋅ b a⋅ c b⋅ b b⋅ c c⋅ b c⋅ c

a⋅ u a⋅ v a⋅ w 13. [a b c] ⋅ [u v w] = b ⋅ u b ⋅ v b ⋅ w c⋅ u c⋅ v c⋅ w 14. If a = a1 l+a2 m+a3 n; b = b1 l+b2 m+b3 n and a1 a2 a3 c = c1 l+c2 m+c3 n, then [a b c] = b1 b2 b3 [l, m, n] c1 c2 c3

Tetrahedron and Its Volume A tetrahedron is a three dimensional figure formed by four triangles, as shown in figure A (a)

(a) O (b)

(c) C (c)

(b) B

Volume of tetrahedron OABC =

1 [a b c] 6

If a, b, c and d are position vectors of vertices A, B, C and D of a tetrahedron ABCD, then its volume 1 = [a − d b − d c − d ]. 6

ONE

340

Vector Triple Product

●

If a, b and c are three vector quantities, then the vectors a × (b × c) and (a × b) × c represents the vector triple product and is given by a × (b × c) = (a ⋅ c) b − (a ⋅ b) c (a × b) × c = (a ⋅ c) b − (b ⋅ c) a Properties of Vector Triple Product are listed below: 1. (a × b) × c = a × (b × c) if some or all a, b and c are zero vectors or a and c are collinear. 2. a × b × a = (a × b) × a = a × (b × a) 3. Vector a × (b × c) is perpendicular to a and lies in the plane of b and c. a ⋅c a ⋅d 4. (a × b) ⋅ (c × d ) = . b ⋅c b ⋅d

If two non-zero vectors a and b are linearly dependent, then it means (i) there are non-zero scalar α and β such that α a + βb = 0 . (ii) a and b are parallel. (iii) a and b are collinear. (iv) a × b = 0 Otherwise, a and b are linearly independent.

●

If three non-zero vectors a, b and c are linearly dependent, then it means (i) there are non-zero scalars α , β and γ such that αa + βb + γc =0 (ii) a, b and c are parallel to same plane. (iii) a, b and c are coplanar. (iv) a = α 1 b + α 2c etc. (v) [a b c] = 0

Linear Combination, Linear Independence and Dependence A vector r is said to be a linear combination of vectors a, b, c, K etc., if there exist scalars x, y, z, ... etc., such that r = x a + yb + z c + K

Otherwise a, b and c are linearly independent. ●

A set of non-zero vectors a 1 , a 2 , a 3 , ……, a n is said to be linearly independent, if x1a 1 + x2a 2 + ...+ x na n = 0 ⇒ x1 = x2 = ... = x n = 0, where x1 , x2 ,... x n are scalars.

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 Let us define the length of a vector a i + b j + c k as

| a | + | b | + | c |. This definition coincides with the usual definition of length of a vector a i + b j + c k iff (a) a = b = c = 0 (b) any two of a, b and c are zero (c) any one of a, b and c are zero

c = –7b. Then, the angle between a and c is π (c) 4

(b) 0

AIEEE 2008 π (d) 2 j

3 If a vector r of magnitude 3 6 is directed along the bisector of the angle between the vectors a = 7 i − 4 j − 4 k and b = − 2i − j + 2k, then r is equal to (a) i − 7 j + 2 k (c) i + 7 j + 2 k

(b) i + 7 j − 2 k (d) i − 7 j − 2 k

then

AIEEE 2005

(b) PA +PB + 2PC =0 (d) PA +PB = 2PC

5 If the vectors AB = 3 i + 4 k and AC = 5 i − 2 j + 4 k are the

sides of a ∆ ABC, then the length of the median through j JEE Mains 2013 A is (b)

72

(c)

33

(a) 4

(b) 9

(c) 8

(d) 6

8 a and c are unit collinear vectors and | b | = 6, then b − 3 c = λa , if λ is (a) − 9, 3 (c) 3, – 3

(b) 9, 3 (d) None of these

9 If a and b are unit vectors, then what is the angle j

j

(a) PA +PB +PC =0 (c) PA +PB =PC

| a − b | 2 + | b − c | 2 + | c − a | 2 does not exceed to

between a and b for 3 a − b to be a unit vector?

4 If C is the mid-point of AB and P is any point out side AB

18

(b) 8 (d) 6

7 If a,b and c are unit vectors, then

2 The non-zero vectors a,b and c are related by a=8b and

(a)

If a parallelogram is constructed with adjacent sides 2a − 3 b and a + b, then its longer diagonal is of length (a) 10 (c) 2 26

(d) a + b + c = 0

(a) π

π 4

6 Let | a | = 2 2, | b | = 3 and the angle between a and b is .

(d)

45

(a) 30° (c) 60°

NCERT Exemplar

(b) 45° (d) 90°

10 If a and b are unit vectors inclined at an angle α , α ∈ [ 0, π ] to each other and | a + b | < 1. Then, α belong to π 2π (a)  ,  3 3  π (c)  0,   3

2π  (b)  , π  3  π 3π  (d)  ,  4 4 

341

DAY 11 If a, b and c are unit vectors satisfying a − 3 b + c = 0, then the angle between the vectors a and c is (a)

π 4

(b)

π 3

(c)

π 6

(d)

π 2

12 The value of a , for which the points, A, B, C with position

vectors 2 i − j + k, i − 3 j − 5 k and a i − 3 j + k respectively are the vertices of a right angled triangle π with C = are j AIEEE 2006 2 (a) –2 and –1 (c) 2 and –1

(b) –2 and 1 (d) 2 and 1

13 If the vectors a = i − j + 2 k, b = 2 i + 4 j + k and

c = λ i + j + µk are mutually orthogonal, then ( λ , µ ) is j AIEEE 2010 equal to (a) (– 3, 2) (c) (– 2, 3)

(b) (2, – 3) (d) (3, – 2)

21 Let u, v

If the projection of v along u is equal to that of w along u and v and w are perpendicular to each other, then | u − v + w | is equal to (a) 4

π (b) 2

π (c) 3

π (d) 4

15 If p, q and r are perpendicular to q + r, r + p and p + q respectively and if | p + q | = 6, | q + r | = 4 3 and | r + p | = 4, then | p + q + r | is (a)5 2 (c)5

(b) 10 (d) 15

16 Let a and b be the position vectors of points A and B with respect to origin and | a | = a, | b | = b. The points C and D divide AB internally and externally in the ratio 2 : 3. If OC and OD are perpendicular, then (a) 9a 2 = 4 b 2 (c) 9a = 4 b

(b) 4 a 2 = 9b 2 (d) 4 a = 9b

i + 2j + k and perpendicular to i + j + k is (b) i − k

(c) i − j

(d) i − 2 j + k

p th, q th and r th terms of GP, then the vectors log a ⋅ i + log b ⋅ j + log c ⋅ k and (q − r )i + (r − p )j + ( p − q )k are (b) parallel (d) None of these

(a)

(b)

5

(c)

6

(d)

8

20 Let a = 2 i − j + k, b = i + 2j − k and c = i + j − 2 k be three vectors. A vector of the type b + λ c for some scalar λ, 2 , is whose projection on a is of magnitude 3 (a) 2 i + j + 5 k (c) 2 i − j + 5 k

l 2 + m2 l + m2 + n2 2

l 2 + m2 l +m +n 2

2

2

(b) (d)

l 2 + m2 + n2 l+m 2

23 Resolved part of vector a along the vector b is a 1 and that perpendicular to b is a 2 , then a 1 × a 2 is equal to (a)

(a × b) b | b |2

(b)

(a × b) a | a |2

(c)

(a ⋅ b) (b × a ) | b| 2

(d)

(a ⋅ b) (b × a ) |b×a |

24 A particle is acted upon by constant forces 4i + j − 3k

and 3i + j − k which displace it from a point i + 2j + 3k to the point 5i + 4j + k The work done in standard units by j the forces is given by AIEEE 2004 (a) 40 units (c) 25 units

(b) 30 units (d) 15 units

25 If u and v are unit vectors and θ is the acute angle (a) (b) (c) (d)

exactly two values of θ more than two values of θ no value of θ exactly one value of θ

j

AIEEE 2007

(b) 2 i + 3 j − 3 k (d) 2 i + 3 j + 3 k

a, c and b form a right handed system, then c is (a) z i − x k (c) y j

(b) 0 (d) − z i + x k

a × c = b and a ⋅ c = 3 is

i + 2 j + 3 k from the line passing through the point A, whose position vector is 4 i + 2 j + 2 k and parallel to the vector 2 i + 3j + 6 k is 10

(d) 2

j

AIEEE 2002

27 If a = i+ j+ k and b = j − k, then a vector c such that

19 The distance of the point B with position vector

(a)

14

26 If the vectors c , a = x i + y j + z k and b = j are such that

18 If the positive numbers a, b and c are the

(a) equal (c) perpendicular

(c)

between them, then 2 u × 3v is a unit vector for

17 A vector of magnitude 2 coplanar with i + j + 2 k and (a) − j + k

7

then the projection of the vector a b (a × b ) along the angle bisector of the l +m +n |a | |b| | a ×b | vector a and b is

14 Let a and b be two unit vectors. If the vectors c = a + 2 b π (a) 6

(b)

22 If a, b and c are three mutually perpendicular vectors,

(c)

and d = 5 a − 4 b are perpendicular to each other, then j AIEEE 2012 the angle between a and b is

w be three vectors such that | u | = 1,| v | = 2, | w | = 3

5 (a) i + 3 5 (c) i + 3

2 j+ 3 2 j+ 3

2 k 3 1 k 3

j NCERT Exemplar 2 5 5 (b) i + j + k 3 3 3

(d) None of these

28 Vectors a and b are not perpendicular and c and d are

two vectors satisfying b × c = b × d and a ⋅ d = 0. Then, j the vector d is equal to AIEEE 2011 a ⋅ c (a) c +  b  a ⋅ b a ⋅ c (c) c −  b  a ⋅ b

b⋅c (b) b +  c  a ⋅ b b⋅c (d) b −  c  a ⋅ b

ONE

342

29 If u, v and w are non-coplanar vectors and p, q are real numbers, then the equality [ 3 u p v p w ] − [ p v w q u] − [ 2 w q v q u] = 0 j AIEEE 2009 holds for (a) (b) (c) (d)

exactly two values of (p, q) more than two but not all values of (p, q) all values of (p, q) exactly one value of (p, q)

the maximum value of [u v w ] is (b) (d)

10 + 60

6

31 If a = i − j, b = j − k, c = k − i and d is a unit vector such that a ⋅ d = 0 = [ b c d ], then d is/are i+ j−k 3 i+ j+k (c)± 3

(b) ±

(a)±

i + j − 2k 6

(c) a × b

(d) b

33 The points with position vectors α i + j + k, i − j − k, i + 2j − k, i + j + βk are coplanar if (a) (1 − α)(1 + β) = 0 (c) (1 + α)(1 + β) = 0

(b) (1 − α)(1 − β) = 0 (d) (1 + α)(1 − β) = 0

34 Let a = i + j + k,b = i − j + 2k and c = xi + ( x − 2) j − k If the vector c lies in the plane of a and b, then x equal to j

(a) 0 (c) – 4

AIEEE 2007

(b) 1 (d) –2

(a) −2 (c) 0

AIEEE 2011

(b) 2 (d) −1

(c) γ

(d) All of these

37 If a , b, c are non-coplanar vectors and λ is a real number, then [ λ ( a + b ) λ2 b λc ] = [ a (a) (b) (c) (d)

b + c b ] for

exactly two values of λ exactly three values of λ no value of λ exactly one value of λ

j

AIEEE 2005

pairwise non-collinear. If a + 3 b is collinear with c and b + 2 c is collinear with a, then a + 3 b + 6 c is equal to (b) a

(c) c

(d) 0

39 If [ a × b b × c c × a ] = λ [ a b c ]2 , then λ is equal to j

(a) 0 (c) 2

no value of λ all except one value of λ all except two values of λ all values of λ

(a) −3 (c) 3

(b) 5 (d) −5

$ b$ and c$ be three unit vectors such that 43 Let a, 3 $ $ = $ then the a$ × (b$ × c) (b+c$ ). If b$ is not parallel to c, 2 j JEE Mains 2016 angle between a$ and b$ is (a)

3π 4

(b)

π 2

(c)

2π 3

(d)

5π 6

44 If a = j − k and c = i − j − k. Then, the vector b satisfying a × b + c = 0 and a ⋅ b = 3, is

j

AIEEE 2010

(b) 2 i − j + 2 k (d) i + j − 2 k

and b = $j+k$ . If u is perpendicular to a and u ⋅ b=24, then 2

is equal to

(a) 336

j

(b) 315

(c) 256

JEE Mains 2018 (d) 84

46 Let a, b and c be three unit vectors such that a is perpendicular to the plane of b and c. If the angle π between b and c is , then | a × b − a × c | 2 is equal to 3 1 3 (c) 1

1 2 (d) 2

(a)

(b)

47 The vectors a and b are non-collinear. The value of x, for

38 Let a, b and c be three non-zero vectors which are (a) a + c

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

u

be three coplanar vectors with a ≠ b and v = i + j + k. Then, v is perpendicular to (b) β

number, then the vectors a + 2b + 3c , λb + 4c and ( 2λ − 1) c are non-coplanar for

45 Let u be a vector coplanar with the vectors a = 2$i + 3$j – k$ j

36 Let α = a i + b j + c k, β = bi + c j + a k and γ = c i + a j + bk (a) α

41 If a, b and c are non-coplanar vectors and λ is a real

(a) − i + j − 2 k (c) i − j − 2 k

35 If the vectors pi + j + k, i + qj + k and i + j + r k ( p ≠ q ≠ r ≠ 1) are coplanar, then the value of pqr − ( p + q + r ) is

(b) 3V (d) 2V

1 1 ( 3 i + k ) and b = ( 2i + 3j − 6 k ), then the value 7 10 j AIEEE 2011 of ( 2a − b ) ⋅ [(a × b ) × (a + 2b )] is

(d) ± k

(b) a

α = (a ⋅ a ) a + (a ⋅ b ) b + (a ⋅ c )c β = (a ⋅ b ) a + (b ⋅ b ) b + (b ⋅ c ) c γ = (a ⋅ c ) a + (b ⋅ c ) b + (c ⋅ c ) c

42 If a =

32 The vector ( i × a ⋅ b ) i + ( j × a ⋅ b ) j + (k × a ⋅ b ) k is equal to (a) b × a

coterminous edges, as a, b and c, then the volume of the parallelopiped having three coterminous edges as

(a) V 3 (c) V 2

30 Let v = 2i + j − k and w = i+ 3 k. If u is a unit vector, then (a) − 1 (c) 59

40 If V is the volume of the parallelopiped having three

(b) 1 (d) 3

JEE Mains 2014

which the vectors, c = ( 2x + 3) a + b and d = ( 2x + 3) a − b are collinear is (a)

1 2

(c) −

(b) 3 2

1 3

(d) None of these

48 It is given that a, b, c are mutually perpendicular vectors of equal magnitudes. Statement I Vector (a + b + c ) is equally inclined to a, b and c.

343

DAY Statement II If α , β and γ are the angles at which (a + b + c ) is inclined to a, b and c, then α = β = γ. (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

1 the volume of the 3 parallelopiped formed by vectors i + a j , a i + j + k and j + a k is maximum. Statement II The volume of the parallelopiped having three coterminous edges a , b andc is | [ a b c ] |.

49 Statement I For a = −

(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

50 Statement I A relation between the vectors r, a and b is a ×b . a ⋅a Statement II r ⋅ a = 0.

r × a = b ⇒r =

(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

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

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE $ b = $i + $j and c be a vector such that 1 Let a= 2$i + $j − 2k, c – a = 3, (a × b ) × c = 3 and the angle between a and j JEE Mains 2017 a × b is 30°. Then, a ⋅ c is equal to (a)

25 8

(b) 2

(c) 5

(d)

1 8

2 Given, two vectors are i − j and i + 2j, the unit vector coplanar with the two vectors and perpendicular to first is 1 (a) (i + j) 2 1 (c) ± (i + j) 2

1 (b) (2 i + j) 5 (d) None of these

If c = 2a × b − 3 b, then the angle between b and c is π 6

(b)

5π 6

(c)

π 3

(d)

2π 3

4 A unit vector d is equally inclined at an angle α with the vectors a = cos θ ⋅ i + sin θ ⋅ j, b = − sin θ ⋅ i + cos θ ⋅ j and c = k. Then, α is equal to 1  (a) cos−1    2 1 (c) cos−1 3

Then, p and q are parallel for atleast one x in

π 3

(b)

π 2

(c)

2π 3

3 i + 2 j + 6 k and is coplanar with the vectors 2 i + j + k and i − j + k, is 2i − 6 j + k 41 3j − k (c) 10

(d)

2π 5

2i − 3j 13 4i + 3j − 3k (d) 34 (b)

9 The values of x for which the angle between the vectors 2x 2 i + 4x j + k and 7i − 2j + x k is obtuse and the angle between the Z-axis and 7i − 2j + x k is acute and less π than is given by 6 1 2

1 < x < 15 2

(b) x >

1 or x < 0 2

(d) No such value for x

10 If a and b are unit vectors, then the greatest value of | a + b | + | a − b | is (a) 2 (c) 2 2

(b) (1, 0) (d) None of these

6 If | a | = | b | = | c | = 1 and a ⋅ b = b ⋅ c = c ⋅ a = cos θ, then the (a)

(b) 2i − j (d) None of these

8 The unit vector which is orthogonal to the vector

(c)

5 Let p = 3 ax 2 i − 2 ( x − 1) j , q = b ( x − 1) i + x j and ab < 0 .

maximum value of θ is

(a) i + 2 j (c) 2i + j

(a) 0 < x
159 Hence, there is no common value for Eqs. (i) and (ii).

c

b

10 Let θ be an angle between unit vectors a

5 Hence, p × q = {3 ax3 + 2b ( x − 1) } k = f ( x ) k, 2

where, f (0) f (1) = 6 ab < 0 ∴ By intermediate value theorem there exists, x in (0, 1) such that f ( x ) = 0. a ⋅a a ⋅ b a ⋅c 6 [a b c ] = b ⋅ a b ⋅ b b ⋅ c c ⋅a c ⋅ b c ⋅c 1 cos θ cos θ = cos θ 1 cos θ cos θ cos θ 1 = 1 − 3cos 2 θ + 2cos 3 θ = (1 − cos θ)2 (1 + 2cos θ) 2π ⇒ 1 + 2cos θ ≥ 0 ⇒ θ ≤ 3

7 Let c = xi + yj Then, b ⊥ c ⇒ b ⋅ c = 0 ⇒ 4 x + 3 y = 0 y x [say] ⇒ = =λ 3 −4 ⇒ ∴

⇒ 4 p + 3 q = 5 and 3 p − 4 q = 10 ⇒ p = 2, q = − 1

12 By expanding a × (b × c ), we get

⇒ 2 x > 3 53 + x2

1 cos α = d ⋅ k = 3 ⇒

Let the required vector be α = p i + q j. Then, the projections of α on b and c are α⋅b α ⋅c , respectively. and |b| |c | α ⋅c b [given] ∴ α⋅ = 1 and =2 |b | |c |

x = 3λ , y = − 4λ c = λ (3 i − 4 j )

and b. Then, a ⋅ b = cos θ

B

Now,|a + b| =|a| + |b| + 2a ⋅ b θ = 2 + 2 cos θ = 4 cos 2 2 2

⇒|a + b | = 2 cos

2

a

C

2

θ 2

Similarly,| a − b | = 2 sin

θ 2

⇒| a + b | + | a − b | θ θ = 2  cos + sin  ≤ 2 2  2 2

11 We have, u ⋅ n = 0 and v ⋅ n = 0 ⇒ n ⊥ u and n ⊥ v u × v n=± ⇒ |u × v | Now, u × v = ( i + j ) × ( i − j ) = − 2 k ∴ n=± k Hence, | w ⋅ n | = | ( i + 2j + 3 k ) ⋅ (± k ) | = 3

Now, as a , b and c are unit vectors and a + b + c = 0, therefore a , b and c represents an equilateral triangle. Hence, a × b = b × c = c × a ≠ 0

16 Taking O as the origin, let the position vectors of A, B and C be a , b and c, respectively. Then, the position vectors G 1 , G 2 b+c c +a and G 3 are , 3 3 a+b and , respectively. 3 1 V1 = [a b c ] ∴ 6 and

V2 = [OG 1 OG 2 OG 3 ]

Now,

V2 = [OG 1 OG 2 OG 3 ] b +c c +a a + b  = 3 3   3

ONE

350

1 [ b + c c + a a + b] 27 2 2 = [a b c]= × 6V1 27 27 9V2 = 4V1 =

⇒

17 Let| BC | = l In ∆ ABC, l = ∴

tanθ =

AB 2 + AC 2 AB AC

B D

q A

⇒ sin θ = ∴

C

AB AC and cos θ = l l

1 1 Resultant vector = i + j AB AC  1  1 = i + j l sin θ l cos θ  

In ∆ ADC, AD = AC sin θ = l sin θ cos θ AB ⋅ AC = l ∴ Magnitude of resultant vector 1  1 1  +   l 2  sin2 θ cos 2 θ  l 1 = = ( AB ) ( AC ) AD =

18 Given, (a × b) × c = 1|b||c|a

To find The vector r in terms of p and q.

3 1 − c × (a × b) = |b||c|a ⇒ 3 1 ⇒ − (c ⋅ b)⋅ a +(c ⋅ a )b = |b||c|a 3  1|b||c| + (c ⋅ b) a = (c ⋅ a ) b  3 

Let E be the foot of perpendicular from B to side AD. AE = Projection of vector q AE = Vector along AE of length AE = | AE| AE (q ⋅ p ) p = | p |2

Since, a and b are not collinear. 1 c ⋅ b + |b||c | = 0 and c ⋅ a = 0 3 1 ⇒ |c|| b |cos θ + |b||c| = 0 3 1 ⇒ |b||c| cos θ +  = 0  3 1 ⇒ cos θ + = 0 3 [Q|b| ≠ 0, |c| ≠ 0] 1 ⇒ cos θ = − 3 8 2 2 ⇒ sinθ = = 3 3

Now, applying triangles law in ∆ABE, we get AB + BE = AE (q ⋅ p ) p ⇒ q+r= | p |2 (q ⋅ p ) p ⇒ r= −q | p |2

19 Given, D pE A

C r q

B

(i) A parallelogram ABCD such that AB = q and AD = p. (ii) The altitude from vertex B to side AD coincides with a vector r.

q ⋅ p = −q+ p  p⋅ p

20 In an isosceles ∆ABC in which

AB = AC, the median and bisector from A must be same line. Statement II is true. u+v Now, AD = 2 1 α 2 and | AD | = ⋅ 2 cos 2 2 2 α So, | AD | = cos 2 Unit vector along AD, i.e. x is given by AD u+v . x= = | AD | 2 cos α 2

DAY THIRTY TWO

Three Dimensional Geometry Learning & Revision for the Day u

u

u

Coordinates of a Point in a Space Section Formula Direction Cosines and Ratios

u

u

u

Equation of Line in Space Skew-Lines

u

u

Coplanar Lines

Plane Angle between a Line and a Plane

Coordinates of a Point in a Space From the adjoining figure, we have The three mutually perpendicular lines in a space which divides the space into eight parts are called coordinates axes. Z C(0, 0, z) A¢(0, y, z) The coordinates of a point are the distances from the origin to the feet of the perpendiculars from the point on the respective coordinate axes. (x, 0, z)B¢ P(x, y, z) X¢ The coordinates of any point on the X , Y and Z-axes will be as ( x, 0, 0) (0, y, 0) and (0, 0, z) respectively and O (0,0,0) Y Y¢ the coordinates of any point P in space will be as B(0, y, 0) Z¢ ( x, y, z). A(x, 0, 0) ●

●

PRED MIRROR

●

Distance between Two Points

X

C¢(x, y, 0)

Your Personal Preparation Indicator u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

The distance between two points A ( x1 , y1 , z1 ) and B ( x2 , y2 , z2 ) is | AB| = ( x2 − x1 )2 + ( y2 − y1 )2 + (z2 − z1 )2

Section Formula If M ( x, y, z) divides the line joining of points P ( x1 , y1 , z1 ) and Q ( x2 , y 2 , z2 ) in the ratio m : n, then

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

352 For Internal Division x=

mx2 + nx1 my2 + ny1 mz2 + nz1 and z = ,y = m+n m+n m+n

For External Division x=

mx2 − nx1 my2 – ny1 mz2 − nz1 and z = ,y= m−n m−n m−n

The coordinates of the mid-point of the line joining

 x1 + x2 y1 + y2 z1 + z2  , ,   2 2 2 

P ( x1 , y1 , z1 ) and Q ( x2 , y2 , z2 ) are 

Some Important Results 1. If A ( x1 , y1 , z1 ), B ( x2 , y2 , z2 ) and C ( x3 , y3 , z3 ) are the vertices of a ∆ ABC, then  x + x2 + x3 y1 + y2 + y3  (i) Centroid of triangle =  1 ,    3 3 j k   i 1 (ii) Area of ∆ ABC =  x2 − x1 y2 − y1 z2 − z1   2  x3 − x1 y3 − y1 z3 − z1  (iii) If area of ∆ ABC = 0, then these points are collinear. 2. Four non-coplanar points A ( x1 , y1 , z1 ), B ( x2 , y2 , z2 ), C ( x3 , y3 , z3 ) and D ( x 4 , y4 , z4 ) form a tetrahedron with vertices A, B, C and D, edges AB, AC, AD, BC, BD and CD, faces ABC, ABD, ACD and BCD, then  x1 + x2 + x3 + x 4 y1 + y2 + y3 + y4 ,  4 4 (i) Centroid  z1 + z2 + z3 + z4   4

 ,   

x2 − x1 y2 − y1 z2 − z1  1 (ii) Volume = x3 − x1 y3 − y1 z3 − z1   6 x 4 − x1 y4 − y1 z4 − z1 

Direction Cosines and Ratios If a vector makes angles α , β and γ with the positive directions of X -axis, Y-axis and Z-axis respectively, then cos α , cos β and cos γ are called its direction cosines and they are denoted by l , m , n , i.e. l = cos α , m = cos β and n = cos γ. If numbers a, b and c are proportional to l , m and n respectively, then a, b and c are called direction ratios. Thus, a, b and c are the direction ratios of a vector, provided l m n = = a b c

Important Results ●

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A line in space can be extended in two opposite directions and so it has two sets of direction cosines. In order to get unique set of direction cosines, we must take the given line as a directed line. Let L is a directed line which makes α,β and γ with positive direction of X, Y and Z-axis, respectively. If we reverse the

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direction of L, then direction angles are replaced by their supplements, i.e. π − α , π − β, π − γ. If the line does not pass through origin, then draw a line through origin and parallel to given line and then find its direction cosines as two parallel lines have same set of direction cosines.

Some Important Deductions (i) Direction ratios of the line joining two points P ( x1 , y1 , z1 ) and Q ( x2 , y2 , z2 ) are x2 − x1 , y2 − y1 , z2 − z1 and its direction cosines are x2 − x1 y2 − y1 z2 − z1 . , , | PQ | | PQ| | PQ | (ii) If P ( x, y, z) is a point in space and OP = r then (a) x = l | r |, y = m | r |, z = n| r | (b) l | r |, m | r | and n| r | are projections of r on OX , OY and OZ, respectively. (c) r =| r |(li + mj + nk) and r$ = l i + mj + nk (d) If r = ai + bj + ck, then a, b, c are DR’s of vector and DC’s a b c are given by l = ,m = ,n = | r| | r| | r| (iii) The sum of squares of direction cosines is always unity, i.e. l 2 + m2 + n2 = 1 (iv) Direction cosines are unique but direction ratio are not unique and it can be infinite. (v) If a, b, c are DR’s of a line and l, m, n are DC’s of a line, then a b l =± , m=± 2 2 2 2 a +b +c a + b 2 + c2 c and n = ± a2 + b 2 + c2 (vi) The DC’s of a line which is equally inclined to the 1 1   1 coordinate axes are  ± ,± ,± .  3 3 3 (vii) If l , m and n are the DC’s of a line, then the maximum 1 value of lmn = . 3 3

Equations of a Line in Space Equation of line passing through point A(a) and parallel to vector (b) is r = a + λ b. If coordinates of A be ( x1 , y1 , z1 ) and the direction ratios of line x − x1 y − y1 z − z1 . = = a b c

be a, b and c, then equation of line is NOTE

x−0 y −0 z−0 or y = 0 , z = 0 = = 1 0 0 x−0 y −0 z−0 • Equation of Y-axis is or x = 0 , z = 0 = = 0 1 0 x−0 y −0 z−0 • Equation of Z-axis is or x = 0 , y = 0 = = 0 0 1

• Equation of X-axis is

353

DAY Equation of a line passing through two given points x − x1 y − y1 z − z1 A ( x1 , y1 , z1 ) and B ( x2 , y2 , z2 ) is = = x2 − x1 y2 − y1 z2 − z1 Its vector form is r = a + λ (b − a). The parametric equations of a line through (a1 , a2 , a3 ) with DC’s l , m and n are x = a1 + lr , y = a2 + mr and z = a3 + nr .

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The shortest distance between the lines x − x1 y − y1 z − z1 x − x2 y − y2 z − z2 and is = = = = a1 b1 c1 a2 b2 c2

SD =

Angle between Two Intersecting Lines 1. If DR’s of two lines are a1 , b1 , c1 and a2 , b2 , c2 , then a1 a2 + b1b2 + c1c2 cos θ = 2 (a1 + b12 + c12 ) (a22 + b22 + c22 ) (i) Condition for parallel lines,

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a1 b1 c1 = = a2 b2 c2

2. Angle between two lines with DC’s l1 , m1 , n1 and l2 , m2 , n 2 is cos −1 (l1 l2 + m1 m2 + n1 n2 )

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(ii) Condition for perpendicular lines, l1 l2 + m1 m2 + n1 n2 = 0

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1 • The angle between any two diagonals of a cube is cos −1   .  3

• The angle between a diagonal of a cube and a face is  2 cos −1   .  3

• The angle between the diagonal of a cube and edge of 1 cube is cos −1   .  3

• If a straight line makes angles α , β , γ and δ with the diagonals of a cube, then 4 cos 2 α + cos 2 β + cos 2 γ + cos2 δ = 3

Skew-Lines Two straight lines in a space which are neither parallel nor intersecting are called skew-lines. Thus, skew-lines are those lines which do not lie in the same plane.

Shortest Distance between Two Skew-Lines ●

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− b2c1 )2

x2 − x1 a1 a2

y2 − y1 b1 b2

z2 − z1 c1 = 0 c2

Distance or shortest distance between two parallel lines

l m n (i) Condition for parallel lines, 1 = 1 = 1 l2 m2 n2

NOTE

∑ (b c

z2 − z1 c1 c2

Two lines r = a + λb and r = c + µd are intersecting if shortest distance between them is zero. (c − a) ⋅ (b × d ) i.e. = 0 ⇒ (c − a) ⋅ (b × d ) = 0 | b × d| or

sin −1 ( Σ(m1 n2 − m2 n1 )2 ).

y2 − y1 b1 b2 1 2

(ii) Condition for perpendicular lines, a1 a2 + b1b2 + c1c2 = 0

or

x2 − x1 a1 a2

If l1 and l2 are two skew-lines, then there is one and only one line perpendicular to each of the line l1 and l2 , which is known as the line of shortest distance. The shortest distance between two lines l 1 and l 2 is the distance PQ between the points P and Q, where the line of shortest distance intersects the two given lines. The shortest distance between two skew-lines r = a + λb (c − a) ⋅ (b × d ) and r = c + µ d is given by SD = |b × d|

Shortest distance between parallel lines will be the perpendicular distance. If the parallel lines are given by r = a + λb and r = c + µb |(c − a) × b| then distance between them is d = | b|

Coplanar Lines Lines which lie in the same plane are called coplanar lines. Any two coplanar lines are either parallel or intersecting.

Condition for Coplanarity of Two non-parallel Lines Two lines r = a + λ b and r = c + µ d are coplanar or intersecting, if (c − a) ⋅ (b × d ) = 0 ⇒ [a b d ] = [c b d ] x − x1 y − y1 z − z1 The lines = = l1 m1 n1 and

x − x2 y − y2 z − z2 are coplanar, = = l2 m2 n2

x2 − x1 y2 − y1 z2 − z1 if  m1 n1   l1  = 0.  l2 m2 n2 

Plane A plane is a surface such that line joining any two points of the plane totally lies in it.

Equation of a Plane in Different Forms 1. The general equation of a plane is a x + by + cz + d = 0 and a2 + b 2 + c2 ≠ 0, where, a, b and c are the DR’s of the normal to the plane.

354 (i) Plane through the origin is a x + by + cz = 0.

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(ii) Planes parallel to the coordinate planes (perpendicular to coordinate axes) x = k parallel to YOZ plane, y = k parallel to ZOX plane and z = k parallel to XOY plane. (iii) Planes parallel to coordinate axes by + cz + d = 0 parallel to X -axis a x + cz + d = 0 parallel to Y-axis a x + by + d = 0 parallel to Z-axis

(i) If this ratio is positive, then A and B are on opposite sides of the plane.

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2. If a, b and c are the intercepts of plane with the coordinate x y z axes, then equation of plane is + + = 1. It meets the a b c coordinate axes at A (a, 0, 0), B (0, b , 0) and C (0, 0, c). 3. (i) If l , m and n are DC’s of normal to the plane, p is the distance of the origin from the plane, then equation of plane is l x + my + nz = p.

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(ii) Coordinates of foot of perpendicular, drawn from the origin to the plane, is (lp, mp, np). (iii) If ON is the normal from the origin to the plane and n$ is the unit vector along ON. Then ON = pn$ and equation of plane is r$ ⋅ n$ = p, where r = xi + yj + zk. 4. Plane through a point ( x1 , y1 , z1 ) is a ( x − x1 ) + b ( y − y1 ) + c (z − z1 ) = 0. where a, b, c are DR’s of normal to the plane. 5. Plane through three non-collinear points ( x1 , y1 , z1 ), ( x2 , y2 , z2 ) and ( x3 , y3 , z3 ) is  x − x1 y − y1 z − z1  x2 − x1 y2 − y1 z2 − z1 = 0   x3 − x1 y3 − y1 z3 − z1

(ii) Its cartesian equation is

x − a1 b1 c1

y − a2 b2 c2

z − a3 b3 = 0 c3

7. Plane parallel to the given plane ax + by + cz + d = 0 is ax + by + cz + k = 0, where k is a constant determined by the given condition. 8. (i) Any plane passing through the line of intersection of the planes a x + by + cz + d = 0 and a1 x + b1 y + c1 z + d1 = 0 is (a x + by + cz + d) + λ (a1 x + b1 y + c1 z + d1 ) = 0 (ii) If r ⋅ n1 = d1 and r ⋅ n2 = d2 are two planes, then their line of intersection is perpendicular to both n1 and n2 , i.e. line is parallel to the vectors n1 × n2 .

Some Important Results on plane ●

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6. (i) Equation of plane, passing through a point A with position vector a and is parallel to given vectors b and c, is (r − a) ⋅ (b × c) = 0 or [r − a b c] = 0

If a x + by + cz + d1 = 0 and a1 x + b1 y + c1 z + d2 = 0 are the equations of any two planes, then a x + bx + cz + d1 = 0 = a1 x + b1 y + c1 z + d2 gives the equation of straight line.

Plane a x + by + cz + d = 0 intersecting a line segment joining A ( x1 , y1 , z1 ) and B ( x2 , y2 , z2 ) divides it in the ratio a x1 + by1 + cz1 + d − a x2 + by2 + cz2 + d

(ii) If this ratio is negative, then A and B are on the same side of the plane. If θ be the angle between the planes a1 x + b1 y + c1 z + d1 = 0 and a2 x + b2 y + c2 z + d2 = 0, then   a1 a2 + b1b2 + c1c2  θ = cos −1   a2 + b 2 + c2 a 2 + b 2 + c2   1 1 1 2 2 2  Two planes are parallel if their normals are parallel and the planes are perpendicular if their normals are perpendicular. If r ⋅ n1 = d1 (or a1 x + b1 y + c1 z = d1 ) and r ⋅ n2 = d2 (or a2 x + b2 y + c2 z = d2 ) are two planes, then they are a b c (i) parallel if n1 = λn2 or 1 = 1 = 1 . a2 b2 c2 (ii) perpendicular if n1 ⋅ n2 = 0 or a1 a2 + b1b2 + c1c2 = 0. Distance of a point ( x1 , y1 , z1 ) from the plane | a x1 + by1 + cz1 + d| a x + by + cz + d = 0 is . a2 + b 2 + c2 |d| Distance of the origin is . a2 + b 2 + c2 The distance between two parallel planes ax + by + cz + d1 = 0 and a x + by + cz + d2 = 0 is |d1 − d2| . a2 + b 2 + c2

Angle between a Line and a Plane x − x1 y − y1 z − z1 and the = = a b c plane a1 x + b1 y + c1 z + d = 0 is θ, then (90° − θ) is the angle between normal and the line, therefore If angle between the line

cos (90 ° − θ) = ●

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If

aa1 + bb1 + cc1 a + b 2 + c2 2

a12 + b12 + c12

a b c = = , then line is perpendicular to plane. a1 b 1 c 1

If a ⋅ a 1 + b ⋅ b 1 + c ⋅ c 1 = 0, then line is parallel to plane. If a ⋅ a 1 + b ⋅ b 1 + c ⋅ c 1 = 0, and a 1 x 1 + b 1 y1 + c 1 z1 + d = 0, then line lies in the plane.

Important Points Related to Line and Plane ●

Projection of a line segment joining the points ( x1 , y1 , z1 ) and ( x2 , y2 , z2 ) on a line with direction cosine l , m, n is |( x2 − x1 )l + ( y2 − y1 )m + (z2 − z1 )n|

355

DAY ●

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Foot of the perpendicular from a point ( x1 , y1 , z1 ) on the plane a x + by + cz + d = 0 is ( x, y, z), where x − x1 y − y1 z − z1 (a x1 + by1 + cz1 + d) . = = =− a b c a2 + b 2 + c2 Image of the point ( x1 , y1 , z1 ) in the plane a x + by + cz + d = 0 is ( x, y, z), where x − x1 y − y1 z − z1 2 (a x1 + by1 + cz1 + d) . = = =− a b c a2 + b 2 + c2 Four points ( x i , yi , zi ), where i = 1, 2, 3 and 4 are coplanar, if x2 − x1 y2 − y1 z2 − z1 x3 − x1 y3 − y1 z3 − z1 = 0.   x 4 − x1 y4 − y1 z4 − z1

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Planes bisecting the angle between two intersecting planes a1 x + b 1 y + c1 z + d1 = 0 and a2 x + b2 y + c2 z + d2 = 0 are given a1 x + b 1 y + c1 z + d1 (a x + b2 y + c2 z + d2 ) by =± 2 2 2 2 a 1 + b1 + c1 a22 + b22 + c22 (i) If a1 a2 + b 1b2 + c1c2 < 0, then origin is in acute angle and the acute angle bisector is obtained by taking positive sign in the above equation. The obtuse angle bisector is obtained by taking negative sign in the above equation. (ii) If a1 a2 + b1b2 + c1c2 > 0,then origin lies in obtuse angle and the obtuse angle bisector is obtained by taking positive sign in above equation. Acute angle bisector is obtained by taking negative sign.

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 If the orthocentre and centroid of a triangle are ( − 3, 5, 1) and ( 3, 3, − 1) respectively, then its circumcentre is (a) (6, 2, − 2) (b) (1, 2, 0)

(d) (6, − 2, 2)

(c) (6, 2, 2)

2 A line makes the same angle θ with each of the

X and Z -axes. If the angle β, which it makes with Y-axis, is such that sin2 β = 3 sin2 θ, then cos 2 θ is equal to AIEEE 2004 2 (d) 5 j

2 (a) 3

1 (b) 5

3 (c) 5

3 A line makes an angle θ with X and Y-axes both. A possible value of θ is in π (a)  0,   4 

π (b)  0,   2 

π π (c)  ,   4 2 

π π (d)  ,   3 6 

4 The projections of a vector on the three coordinate axes are 6, − 3 and 2, respectively. The direction cosines of the j vector are AIEEE 2009 (a) 6, − 3, 2

6 3 2 6 3 2 6 3 2 (b) , − , (c) , − , (d) − , − , 7 7 7 5 5 5 7 7 7

5 If the projections of a line segment on the X , Y and Z -axes in 3-dimensional space are 2, 3 and 6 respectively, then the length of the line segment is j

(a)12

(b) 7

(c) 9

JEE Mains 2013

(d) 6

6 A vector r is inclined at equal angles to OX , OY and OZ . If the magnitude of r is 6 units, then r is equal to (a) 3 (i + j + k) (c) − 2 3 (i + j + k)

(b) − 3 (i + j + k) (d) None of these

7 A line L1 passes through the point 3 i and is parallel to the vector − i + j + k and another line L2 passes through the point i + j and is parallel to the vector i + k, then point of intersection of the lines is (a) i + 2 j + k (b) 2 i + j + k (c) i − 2 j − k

(d) i − 2 j + k

8 The line passing through the points ( 5, 1, a ) and ( 3, b, 1)  17 13 crosses the YZ -plane at the point  0, , −  . Then,  2 2 j

(a) a = 8, b = 2 (c) a = 4, b = 6

AIEEE 2008

(b) a = 2 , b = 8 (d) a = 6, b = 4

9 The angle between the lines 2x = 3y = − z and 6x = − y = − 4z is (a) 30°

j

(b) 45°

(c) 90°

AIEEE 2005

(d) 0°

10 The angle between a diagonal of a cube and an edge of the cube intersecting the diagonal is (a) cos−1 (c) tan−1

1 3 2

2 3 (d) None of these

(b) cos−1

11 The angle between the lines whose direction cosines satisfy the equations l + m + n = 0 and l 2 = m 2 + n 2 is j

π (a) 3

12 If the line,

π (b) 4

π (c) 6

JEE Mains 2014 π (d) 2

x −3 y +2 z +4 lies in the plane, = = 2 −1 3

lx + my − z = 9, then l 2 + m 2 is equal to (a) 26 (c) 5

j

JEE Mains 2016

(b) 18 (d) 2

13 The direction cosines of two lines at right angles are 1 1  (1, 2 , 3) and  −2, ,  , then the direction cosine  2 3 perpendicular to both the given lines are 25 19 , , 2198 2198 1 −7 (c) , − 2, 3 2 (a)

729 2198

(b)

24 , 2198

38 , 2198

(d) None of these

730 2198

356

14 The foot of perpendicular from ( 0, 2, 3) to the line x + 3 y −1 z + 4 , is = = 5 2 3

(a) 3

(a) (−2, 3, 4) (b) (2 , − 1, 3) (c) (2, 3, − 1)

(d) (3 , 2 , − 1)

15 The projection of the line segment joining (2, 5, 6) and (3, 2, 7) on the line with direction ratios 2 , 1, − 2, is (a)

1 2

1 3

(b)

(c) 2

(d) 1

16 The shortest distance between the lines

x +1 y +1 z +1 x − 3 y − 5 z −7 and , is = = = = 7 −6 1 1 −2 1 j NCERT (a)

29 units (b) 29 units

29 units (c) 2

(d) 2 29 units

rectangular parallelopiped whose sides are a , b , c and the edges not meeting it, are

(b) (c)

bc b 2 − c2 bc b 2 + c2 2 bc b −c 2

2

, , ,

ca c2 − a2

,

ca

,

c2 + a2 2 ca c −a 2

2

,

ab

2 9

2 ab a2 − b 2

9 2

(d) 0

t , y = 1 + t , z = 2 − t , with parameters s and t 2 respectively are coplanar, then λ is equal to j AIEEE 2004 x =

(c) −

(b) –1

1 2

(d) 0

20 The distance of the point (1, 0, 2) from the point of intersection of the line

x − 2 y +1 z − 2 and the = = 3 4 12

plane x − y + z = 16 is (a) 2 14

(b) 8

j

(c) 3 21

JEE Mains 2015 (d) 13

21 A vector n is inclined to X -axis at 45°, to Y -axis at 60° and at an acute angle to Z -axis. If n is a normal to a plane passing through the point ( 2, − 1, 1 ), then the equation of j the plane is JEE Mains 2013 (a) 4 2 x + 7 y + z = 2 (c) 3 2 x − 4 y − 3 z = 7

(b) 2 x + y + z = 2 (d) 2 x − y − z = 2

22 Let Q be the foot of perpendicular from the origin to the plane 4x − 3y + z + 13 = 0 and R be a point ( − 1, 1, − 6) on j the plane. Then, length QR is JEE Mains 2013 (a)

14

(b)

19 2

(c) 3

( 3 , − 4 , − 5) and ( 2 , − 3 , 1) crosses the plane passing through three points ( 2 , 2 , 1), ( 3 , 0 , 1) and ( 4 , − 1, 0), is j NCERT Exemplar (b) (−1, 2 , − 7) (d) None of these

(a) (1, 2 , 7) (c) (1, − 2 , 7)

25 The volume of the tetrahedron formed by coordinate planes and 2x + 3y + z = 6, is (a) 5

(b) 4

(c) 6

(d) 0

parallel to the plane 3x − 4y + 5z = 4 is (a) 3 x − 4 y + 5 z − 27 = 0 (c) 3 x − 4 y + 5 z + 26 = 0

(b) 3 x − 4 y + 5 z + 21 = 0 (d) 3 x − 4 y + 5 z + 17 = 0

x−2 y−3 z−4 = = 1 −2 5 x−2 y−3 z−4 (d) = = 1 2 5

(a)

19 If the straight lines x = 1 + s, y = − 3 − λ s, z = 1 + λ s and

(a) – 2

24 The coordinates of the point where the line through

x−2 y−3 z−4 = = −1 2 5 x−2 y−3 z−4 (c) = = −1 −2 5

a2 + b 2

(c)

(d) 20

reflection in the plane x − 2y + 5z = 6, then the equation of the line PQ is

ab

x −1 y + 1 z −1 x − 3 y −k z and 18 If the line = = = = 2 3 4 1 2 1 j AIEEE 2012 intersect, then k is equal to (b)

(c) 12

27 If Q is the image of the point P( 2, 3, 4) under the

a2 − b 2

(d) None of the above

(a) − 1

(b) 6

26 The equation of the plane passing through ( 2 , 1, 5) and

17 The shortest distance between the diagonals of a

(a)

(1, 1, 1) makes intercepts on the coordinate axes and the sum of whose length is

7 2

(d)

3 2

23 The plane passing through the point ( −2, − 2, 2) and containing the line joining the points (1, − 1, 2) and

(b)

28 If the points (1, 2 , 3) and ( 2 , − 1, 0) lie on the opposite sides of the plane 2x + 3y − 2z = k , then (a) k < 1 (c) k < 1or k > 2

(b) k > 2 (d) 1 < k < 2

29 The equation of the plane containing the lines 2x − 5y + z = 3, x + y + 4z = 5 and parallel to the plane j JEE Mains 2015 x + 3y + 6z = 1 is (a) 2 x + 6y + 12 z = 13 (c) x + 3 y + 6z = 7

(b) x + 3 y + 6z = − 7 (d) 2 x + 6y + 12 z = − 13

30 The equation of a plane through the line of intersection of the planes x + 2y = 3, y − 2z + 1 = 0 and j perpendicular to the first plane is JEE Mains 2013 (a) 2 x − y − 10z = 9 (c) 2 x − y + 10z = 11

(b) 2 x − y + 7 z = 11 (d) 2 x − y − 9z = 10

31 An equation of a plane parallel to the plane x − 2y + 2z − 5 = 0 and at a unit distance from the origin j is AIEEE 2012 (a) x − 2 y + 2 z ± 3 = 0 (c) x − 2 y + 2 z − 1 = 0

(b) x − 2 y + 2 z + 1 = 0 (d) x − 2 y + 2 z + 5 = 0

32 Two systems of rectangular axes have the same origin. If a plane cuts them at distances a, b, c and a′ , b′ , c′ from j the origin, then AIEEE 2003 1 a2 1 (b) 2 a 1 (c) 2 a 1 (d) 2 a (a)

1 b2 1 + 2 b 1 − 2 b 1 + 2 b +

1 c2 1 − 2 c 1 − 2 c 1 + 2 c +

1 a ′2 1 + 2 a′ 1 + 2 a′ 1 − 2 a′ +

1 1 + 2 =0 b ′2 c′ 1 1 + 2 − 2 =0 b′ c′ 1 1 − 2 − 2 =0 b′ c′ 1 1 − 2 − 2 =0 b′ c′ +

357

DAY 33 The distance of the point (1, − 5, 9) from the plane

x − y + z = 5 measured along a straight line x = y = z is j

(a) 3 10

(b) 10 3

10 (c) 3

JEE Mains 2016 20 (d) 3

34 Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is 3 (a) 2

5 (b) 2

j

7 (c) 2

JEE Mains 2013 9 (d) 2

35 Find the planes bisecting the acute angle between the planes x − y + 2z + 1 = 0 and 2x + y + z + 2 = 0. (a) x + z − 1 = 0 (c) x − z − 1 = 0

(b) x + z + 1 = 0 (d) None of these

36 The angle between the lines plane x + y + 4 = 0 is (a) 0°

(b) 30°

x −1 y − 2 z + 3 and the = = 2 1 −2 (c) 45°

(d) 90°

y −1 z − 3 and the = 2 λ  5 plane x + 2y + 3z = 4 is cos −1   , then λ is equal to  14  j AIEEE 2011

37 If the angle between the line x =

(a)

3 2

(b)

2 5

(c)

5 3

(d)

2 3

38 The distance between the line r = 2i − 2j + 3k + λ(i − j + 4k ) and the plane r ⋅ (i + 5j + k ) = 5 is 10 (a) 3 3

10 (b) 9

10 (c) 3

3 (d) 10

39 Consider the following statements. Statement I If the coordinates of the points A,B,C,D are (1 , 2 , 3),( 4 , 5 , 7),( − 4 , 3 ,− 6) and ( 2 , 9 , 2) respectively, π then the angle between the lines AB and CD is . 6 x −1 y − 2 z − 3 Statement II The straight lines and = = 1 2 3 x −1 y − 2 z − 3 are parallel. = = 2 2 −2 Choose the correct option. (a) Statement I is true (b) Statement II is true (c) Both statements are true (d) Both statements are false

40 Consider a line is perpendicular to the plane, then DR’s of plane is proportional to the line. x −1 y z +1 Statement I The lines and = = 1 −1 1 x − 2 y +1 z = = are coplanar and equation of the plane 1 2 3 containing them is 5x + 2y − 3z − 8 = 0 x − 2 y +1 z Statement II The line = = is perpendicular 1 2 3 to the plane 3x + 6y + 9z − 8 = 0 and parallel to the plane x + y − z = 0

(a) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I (b) Statement I is true; Statement II is true; Statement II is not a correct explanation of Statement I (c) Statement I is true; Statement II is false (d)Statement I is false; Statement II is true

41 Statement I The point A (1, 0, 7) is the mirror image of the

x y −1 z − 2 . = = 1 2 3 x y −1 z − 2 Statement II The line = bisects the line = 1 2 3 j AIEEE 2011 segment joining A (1, 0, 7) and B (1, 6, 3). point B (1, 6, 3) in the line

(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

42 Consider the following statements Statement I If the line drawn from the point ( −2 , − 1, − 3) meets a plane at right angle at the point (1, − 3 , 3), then the equation of plane is 3x − 2y + 6z − 27 = 0. Statement II The equation of the plane through the points ( 2 ,1, 0),( 3 , − 2 , − 2) and ( 3 ,1, 7) is 7x + 3y − z = 17. Choose the correct option. (a) Statement I is true (b) Statement II is true (c) Both statements are true (d) Both statements are false

43 Statement I The point A (3, 1, 6) is the mirror image of

the point B (1, 3, 4) in the plane x − y + z = 5 . Statement II The plane x − y + z = 5 bisects the line segment joining A (3, 1, 6) and B (1, 3, 4). j AIEEE 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

44 Statement I A point on the straight line 2x + 3y − 4z = 5 and 3x − 2y + 4 z = 7 can be determined by taking x = k and then solving the two equations for y and z, where k is any real number. Statement II If c′ ≠ kc, then the straight line ax + by + cz + d = 0, kax + kby + c′ z + d ′ = 0, does not intersect the plane z = α, where α is any real number. (a) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I (b) Statement I is true; Statement II is true; Statement II is not a correct explanation of Statement I (c) Statement I is true; Statement II is false (d) Statement I is false; Statement II is true

358

45 Consider the lines x+ 1 y+ 2 z+ 1 x− 2 y+ 2 z− 3 . L1 : = = , L2 : = = 3 1 2 1 2 3

Statement I The distance of the point (1, 1, 1) from the plane passing through the point ( −1, − 2, − 1) and whose normal is perpendicular to both the lines L1 and L2 is 13 . 5 3

Statement II The unit vector perpendicular to both the − i − 7j + 5 k lines L1 and L2 is . 5 3 (a) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I (b) Statement I is true; Statement II is true; Statement II is not a correct explanation of Statement I (c) Statement I is true; Statement II is false (d) Statement I is false; Statement II is true

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 The direction ratios of normal to the plane through

π (1, 0 , 0), ( 0 ,1, 0) which makes an angle with the plane 4 x + y = 3 are (a) 1, 2 ,1

(b) 1, 1, 2

(c) 1, 1, 2

(d) 2 ,1, 1

2 The angle between the lines whose direction cosines are given by 2l − m + 2n = 0 ,lm + mn + nl = 0 , is (a)

π 6

(b)

π 4

(c)

π 3

(d)

π 2

3 If L1 is the line of intersection of the planes

2x − 2y + 3z − 2 = 0, x − y + z + 1 = 0 and L2 is the line of intersection of the planes x + 2y − z − 3 = 0, 3x − y + 2z − 1 = 0, then the distance of the origin from the plane, containing the lines L1 and L2 is j JEE Mains 2018 (a)

1 4 2

(b)

1 3 2

(c)

1

1 (d) 2

2 2

4 If the plane x + y + z = 1 is rotated through an angle 90° about its line of intersection with the plane x − 2y + 3z = 0 , the new position of the plane is (a) x − 5 y + 4 z = 1 (c) x − 8 y + 7 z = 2

(b) x − 5 y + 4 z = − 1 (d) x − 8 y + 7 z = − 2

5 A variable plane at a distance of 1 unit from the origin cut the coordinate axes at A , B and C. If the centroid D ( x , y , z ) of ∆ABC satisfies the relation 1 1 1 + 2 + 2 = k , then k is equal to x2 y z (a) 3

(b) 1

1 (c) 3

(d) 9

6 The lines x = py + q , z = ry + s and x = p′ y + q ′ , z = r ′ y + s′ are perpendicular, if (a) pr + p ′r ′+1 = 0 (c) pr + p ′r ′ = 0

NCERT Exemplar (b) pp ′+ rr ′+1 = 0 (d) pp ′+ rr ′ = 0 j

7 A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2 z . The coordinates of each of the points of intersection are given by

(a) (3a , 3a , 3a), (a , a , a) (c) (3a , 2a , 3a), (a , a , 2a)

(b) (3a , 2a , 3a), (a , a , a) (d) (2a , 3a , 3a), (2a , a , a)

8 A parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7), parallel to the coordinate planes. The length of a diagonal of the parallelopiped is (a) 7 units

(b)

(c)

(d) None of these

155 units

38 units

9 The distance of the point (1, 3, −7) from the

plane passing through the point (1, − 1, − 1) having normal perpendicular to both the lines x −1 y + 2 z − 4 x − 2 y +1 z + 7 and , is = = = = 1 −2 3 2 −1 −1 j JEE Mains 2017 20 10 5 10 (a) units (b) units (c) units (d) units 74 83 83 74

10 Find the distance of the plane x + 2y − z = 2 from the point ( 2 , − 1, 3) as measured in the direction with DR’s ( 2 , 2 , 1). (a) 2

(b) − 3

(c) − 2

(d) 3

11 ∆ABC is such that the mid-points of the sides BC, CA and AB are (l, 0, 0), ( 0, m, 0), ( 0, 0, n ), respectively. Then, AB 2 + BC 2 + CA 2 is equal to l 2 + m2 + n 2 (a) 2

(b) 4

(c) 8

(d) 16

12 If α , β , γ and δ are the angles between a straight line with the diagonals of a cube, then sin2 α + sin2 β + sin2 γ + sin2 δ is equal to (a)

5 3

(b)

8 3

(c)

7 4

(d) None of these

13 The equation of the line passing through the points ( 3, 0, 1) and parallel to the planes x + 2y = 0 and j NCERT Exemplar 3y − z = 0, is x − 3 y − 0 z −1 = = −2 1 3 x − 3 y − 0 z −1 (c) = = 3 1 −2 (a)

(b)

x − 3 y − 0 z −1 = = 1 −2 3

(d) None of these

359

DAY 14 The length of the projection of the line segment joining

18 Let L be the line of intersection of the planes

the points (5, −1, 4) and ( 4, − 1, 3) on the plane, j JEE Mains 2018 x + y + z = 7 is (a)

2 3

(b)

2 3

(c)

1 3

(d)

2x + 3y + z = 1 and x + 3y + 2z = 2. If L makes an angle α with the positive X -axis, then cos α is equal to

2 3

(a)

15 Let ABC be a triangle with vertices at points A( 2, 3, 5),

(b) (7, 5)

(c) (7, 10)

(a) 3 5

parallel to the plane 2x − 2y + z = 0. The distance of the point ( −1, 2 , 0) from the plane, is (b) 3

(c) 4

(c)

1 2

(d)

1 3

2x + 3y − 4z + 22 = 0 measured parallel to the line x y z = = is Q, then PQ is equal to j JEE Mains 2017 1 4 5

(d) (5, 7)

16 A plane passes through the point (1, − 2 , 3) and is

(a) 2

(b) 1

19 If the image of the point P(1, − 2, 3) in the plane

B( − 1, 3, 2) and C( λ , 5, µ ) in three dimensional space. If the median through A is equally inclined with the axes, j JEE Mains 2013 then ( λ , µ ) is equal to (a) (10, 7)

1 2

(b) 2 42

(c) 42

(d) 6 5

x −1 y − 3 z − 4 20 The image of the line in the plane = = 3 1 −5 j JEE Mains 2014 2x − y + z + 3 = 0 is the line x + 3 y −5 = 3 1 x + 3 y −5 (b) = −3 −1 x−3 y+5 (c) = 3 1 x−3 y+5 (d) = −3 −1

(d) 5

(a)

17 The equation of the plane through the line intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5, which is perpendicular to plane x − y + z = 0, is (a) x + 2 y + 3 z − 4 = 0 (b) 5 x + 6y + 7 z − 8 = 0 (c) 120 x + 144 y + 168 z − 5 = 0 (d) x − z + 2 = 0

z−2 −5 z+ 2 = 5 z−2 . = −5 z−2 = 5 =

ANSWERS SESSION 1

SESSION 2

1. 11. 21. 31. 41.

(a) (a) (b) (a) (b)

1. (b) 11. (c)

2. 12. 22. 32. 42.

(c) (d) (c) (d) (c)

2. (d) 12. (b)

3. 13. 23. 33. 43.

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

3. (b) 13. (a)

4. 14. 24. 34. 44.

(c) (c) (c) (c) (b)

4. (d) 14. (d)

5. 15. 25. 35. 45.

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

5. (d) 15. (c)

6. 16. 26. 36.

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

6. (d) 16. (d)

7. 17. 27. 37.

(b) (b) (b) (d)

7. (b) 17. (d)

8. 18. 28. 38.

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

8. (a) 18. (d)

9. 19. 29. 39.

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

9. (b) 19. (b)

10. 20. 30. 40.

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

10. (d) 20. (a)

360

Hints and Explanations SESSION 1

1 1   1 = 6 i+ j+ k  3 3 3 

1 Since, S divides OG in the ratio 3 : − 1. 9 + 3 −5 + 9 −3 − 1  Then, S =  , ,   2 2 2 

If α is obtuse, then 1 and |r | = 6 3 r = |r | (li + mj + nk ) 1 1   1 = 6 − i− j– k  3 3 3  l =m=n=−

2 A line makes angle θ with X-axis and

Z-axis and β with Y-axis. ∴ l = cos θ, m = cos β, n = cos θ Q l 2 + m2 + n2 = 1 ∴ cos 2 θ + cos 2 β + cos 2 θ = 1 ⇒ 2cos 2 θ = 1 − cos 2 β …(i) ⇒ 2cos 2 θ = sin2 β But it is given that, …(ii) sin2 β = 3 sin2 θ From Eqs. (i) and (ii), we get

3 sin2 θ = 2cos 2 θ ⇒ 3(1 − cos 2 θ) = 2cos 2 θ ⇒ 3 = 5cos 2 θ 3 ∴ cos 2 θ = 5

∴

11 We know that, angle between two lines is cos θ =

= − 2 3 (i + j + k )

7 Equation of L1 is

⇒ ⇒ ⇒

r = 3 i + λ (− i + j + k ) = (3 − λ ) i + λj + λ k Equation of L2 is r = ( i + j) + µ ( i + k )

⇒

= (1 + µ ) i + j + µ k For point of intersection, we get λ = µ = 1 ⇒ r = 2i + j + k

8 Equation of line passing through (5, 1, a) and (3, b, 1) is x−3 y −b z−1 = = 5− 3 1−b a−1

3 We know that, cos 2 θ + cos 2 θ + cos 2 γ = 1 ⇒ cos 2 γ = − cos 2θ ⇒ cos 2θ ≤ 0 π π θ∈ ,  ∴  4 2 

4 Projection of a vector on coordinate axes are x2 − x1 , y 2 − y 1 , z2 − z1 ⇒ x2 − x1 = 6, y 2 − y 1 = − 3, z2 − z1 = 2 Now, ( x2 − x1 )2 + ( y 2 − y 1 )2 + (z2 − z1 )2 = 36 + 9 + 4 = 7 6 3 So, the DC’s of the vector are , − 7 7 2 and ⋅ 7

5 Given that, the projections of a line segment on the X , Y and Z-axes in 3 D-space are, lr = 2, mr = 3 and nr = 6 Q(lr )2 + (mr )2 + ( nr )2 = (2)2 + (3)2 + ( 6)2 ⇒ (l 2 + m2 + n2 ) r 2 = 4 + 9 + 36 ⇒ r 2 = 49 ⇒ r = 7

6 Let r be inclined at an angle α to each axis, then l = m = n = cos α Since, l 2 + m2 + n2 = 1 ⇒ 3 cos 2 α = 1

|r | = 6 ∴ r = | r |(l i + mj + n k )

and z, then diagonal has DR’s 1, 1, 1 and edge along X-axis has DR’s 1, 0, 0. The angle between them is 1 cos −1 = tan −1 2 3

= 2 3 (i + j + k )

= (6, 2, − 2)

If α is acute, then l = m = n =

10 If three edges of the cube are along x, y

1 and 3

…(i)

 x − x1 y − y1 z − z1  Q x − x = y − y = z − z  2 1 2 1 2 1   17 13   Point  0, , −  satisfies Eq. (i), we  2 2 get 17 13 −b − −1 3 2 2 − = = 2 1−b a−1 ⇒

a−1=

 − 15    2 = 5 ⇒ a= 6  − 3    2

⇒ 3b − 3 = 17 − 2b ⇒ 5b = 20 ⇒ b = 4

9 The given equations of lines can be rewritten as x y z x y z and = = = = 3 2 −6 2 −12 − 3

  a + b + c 

a1 ⋅ a2 + b2 ⋅ b2 + c 1 ⋅ c 2 a +b +c 2 1

2 1

= cos −1 (0) = 90°

2 2

2 2

a22 + b22 + c 22

l+ m+ n=0 l = − ( m + n) (m + n )2 = l2 m2 + n2 + 2mn = m2 + n2 [Ql2 = m2 + n2 , given] 2mn = 0

When, m = 0 ⇒ l = − n Hence, (l, m, n ) is (1, 0, − 1). When n = 0, then l = − m Hence, (l, m, n ) is (1, 0, –1). 1+ 0+ 0 1 π = ⇒ θ= ∴ cos θ = 2 3 2× 2

12 Since, the line

x−3 y + 2 z+ 4 = = 2 −1 3

lies in the plane lx + my − z = 9, therefore we have 2l − m − 3 = 0 [Q normal will be perpendicular to the line] ...(i) ⇒ 2l − m = 3 and 3l − 2m + 4 = 9 [Q point (3, − 2, − 4) lies on the plane] ...(ii) ⇒ 3l − 2m = 5 On solving Eqs. (i) and (ii), we get l = 1 and m = − 1 ∴ l 2 + m2 = 2 perpendicular to two given lines is (l , m, n ), then l + 2m + 3 n = 0 and m n −2l + + =0 2 3 From the above equation, l m =− 1 1 1 − ×3 − 3 × (−2) − 1 × 3 2 3 n = 1 1 × − 2 (−2) 2 l2 m2 n2 1 ⇒ = = = 25 361 81 25 361 81 + + 36 9 4 36 9 4 25 19 729 ,m= ,n = ∴ l = 2198 2198 2198 2×

∴ Angle between the lines is θ = cos −1   3 × 2 + 2(−12) − 6(−3)    32 + 22 + (− 6)2 (2)2 + (− 12)2 + ( − 3)2   

2 1

a + b12 + c 12

13 Let the direction cosine of the line

17 Also, − 3(1 − b ) = 2  − b   2 

 Qcos θ = 

a1 a2 + b1 b2 + c 1 c 2 2 1

2 2

361

DAY 14 Let L be foot of perpendicular from P (0, 2, 3) on the line x − (−3) y − 1 z − (−4 ) = = = t …(i) 5 2 3 Any point on Eq. (i) is L (−3 + 5 t , 1 + 2 t , − 4 + 3 t ). Then, DR’s of PL are (−3 + 5 t − 0, 1 + 2 t − 2, − 4 + 3 t − 3) or ( 5 t − 3, 2 t − 1, 3 t − 7). Since, PL is perpendicular to Eq. (i), therefore 5( 5t − 3) + 2(2t − 1) + 3(3 t − 7) = 0 ⇒ t = 1 So, the coordinate of L is (2, 3,−1).

15 The vector joining the points is

i − 3 j + k. Its projection along the vector 2i + j − 2k |( i − 3 j + k ) ⋅ (2i + j − 2 k )| = 22 + 12 + 22 |2 − 3 − 2| = =1 3

16 The given lines are

x+1 y +1 z+1 = = 7 −6 1 x−3 y − 5 z−7 and = = 1 −2 1 For line Ist DR’s = (7, − 6, 1) and it passes through (− 1, − 1, − 1), then equation of given lines (in vector form) is r1 = − i − j − k + λ ( 7i − 6 j + k ) Similarly,r2 = 3 i + 5 j + 7 k + µ ( i − 2 j + k )

17 Let one vertex of the parallelopiped be at the origin O and three coterminous edges OA, OB and OC be along OX , OY and OZ, respectively. The coordinates of the vertices of the parallelopiped are marked in figure. The edges which do not meet the diagonal OF are AH , AD and BD and their parallels are BE, CE and CH, respectively. The vector equation of the diagonalOF is …(i) r = 0 + λ (ai + bj + ck ) The vector equation of the edge BD is …(ii) r = b j + µ ai We have, (ai + b j + ck ) × ai = ba ( j × i ) + ca ( k × i ) = − ba k + ca j ∴ |(ai × bj × ck ) × ai | = b 2 a2 + c 2 a2 and {(ai + bj + ck ) × ai} ⋅ (bj − 0 ) = (− ba k + ca j) ⋅ bj = abc

Z

C (0, 0, c)

(a,0, c) H

E (0, b, c)

O (0,0,0) D (a, b, 0) A (a, 0, 0)

and a 2 = 3 i + 5j + 7 k, b2 = i − 2 j + k Now, a 2 − a 1 = (3 i + 5 j + 7 k ) − ( − i − j − k) = 4 i+ 6 j+ 8 k i j k and b1 × b2 = 7 −6 1 1 −2 1 = i (−6 + 2) − j (7 − 1) + k (−14 + 6) = −4 i − 6 j − 8 k |b1 × b2|= (− 4)2 + (− 6)2 + (− 8)2 = 16 + 36 + 64 = 116 = 2 29 So, the shortest distance between the given lines d = = =

( b1 × b2 ) ⋅ ( a 2 − a 1 ) |b1 × b2| |(−4 i − 6 j − 8 k )⋅ (4 i + 6 j + 8 k )| 2 29 |(− 4) × 4 + (− 6) × 6 + (− 8) × 8 |

2 29 | − 16 − 36 − 64 | 116 = = 2 29 2 29 58 = = 2 29 units 29

X Thus, the shortest distance between Eqs. (i) and (ii) is given by |{(a i + b j + c k ) × a i} ⋅ (bj − 0 )| SD = |(ai + bj + ck ) × ai | abc bc = = b 2 a2 + c 2 a2 b2 + c 2 Similarly, it can be shown that the shortest distance between OF and AD is ca and that between OF and AH 2 a + c2 ab is . 2 a + b2 x−1 y + 1 z−1 = = = p 2 3 4 x−3 y − k z−0 and L2 : = = =q 1 2 1 ⇒ Any point P on line L1 is of type P (2 p + 1, 3 p − 1, 4 p + 1) and any point Q on line L2 is of type Q (q + 3, 2q + k , q ) Since, L1 and L2 are intersecting each other, hence both point P and Q should coincide at the point of intersection,

18 Let L1 :

19 The given straight line can be rewritten as

and

x−1 y + 3 z−1 = = =s 1 −λ λ x− 0 y −1 z−2 = = =t 1 2 −2

These two lines are coplanar, if x1 − x2 y 1 − y 2 z1 − z2 l1 m1 n1 =0 l2 m2 n2 ⇒

1 − 0 −3 − 1 1 − 2 1 −λ λ =0 1 2 −2

⇒

1 − 4 −1 1 −λ λ = 0 1 2 −2

F (a, b, c) B (0, b, 0) Y

which are of the form r1 = a 1 + λ b1 and r2 = a 2 + µ b2 where, a 1 = − i − j − k, b1 = 7 i − 6 j + k

i.e. corresponding coordinates of P and Q should be same. 2 p + 1 = q + 3, 4 p + 1 = q and 3 p − 1 = 2q + k On solving 2 p + 1 = q + 3 and 4 p + 1 = q , we get the values of p and q as −3 and q = − 5 p= 2 On substituting the values of p and q in the third equation 3 p − 1 = 2q + k , we get −3 9 ∴ 3   − 1 = 2 (− 5) + k ⇒ k =  2  2

⇒ 1(2 λ − 2 λ ) + 4(−2 − λ ) − 1(2 + λ ) = 0 ⇒ − 8 − 4λ − 2 − λ = 0 ⇒ −10 = 5λ ⇒ λ = − 2

20 Given equation of line is x−2 y + 1 z−2 = = = λ [say] …(i) 3 4 12 and equation of plane is x − y + z = 16 …(ii) Any point on the line (i) is (3λ + 2, 4λ − 1, 12λ + 2) Let this point be point of intersection of the line and plane. ∴ (3λ + 2) − (4λ − 1) + (12λ + 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

21 Q cos2 α + cos2 β + cos2 γ = 1 ⇒

cos 2 45° + cos 2 60° + cos 2 γ = 1 1 1 3 1 ⇒ cos 2 γ = 1 − − = 1 − = 2 4 4 4 1 ⇒ cos γ = 2

362 ∴ Direction Ratio’s of normal to the 1 plane is < cos 45° ; cos 60° , > 2 1 1 1 =< , , > 2 2 2 ∴Equation of plane passing through

25 Since, the vertices of the tetrahedron are ( 0, 0, 0),(3, 0, 0), ( 0, 2, 0) and ( 0, 0, 6). ∴ Volume of tetrahedron 3 0 0 1 = 0 2 0 =6 6 0 0 6

( 2, − 1, 1 ) is 1 1 1 ( x − 2 ) + ( y + 1 ) + (z − 1 ) = 0 2 2 2 ⇒ 2( x − 2 ) + 2( y + 1 ) + 2(z − 1 ) = 0 ⇒

2( x − 2 ) + ( y + 1 ) + (z − 1 ) = 0

⇒

2x − 2 + y + 1 + z − 1 = 0

⇒

2x + y + z = 2

22 Let foot of perpendicular Q ( x, y , z ) from

= 1+

1 121 + = 4 4

points in the equation 2 x + 3 y − 2z − k = 0, we get (2 + 6 − 6 − k )(4 − 3 − k ) < 0 ⇒ (k − 1)(k − 2) < 0 ∴ 1< k < 2

29 Let equation of plane containing the 2

126 7 =3 2 2

23 Required equation of plane is x−1 −3 0

y −1 z−1 −3 1 =0 −2 1

⇒

x − 3 y − 6z + 8 = 0 8 8 Since, the intercepts are 8, , . 3 6 So, their sum is 12.

24 Equation of plane through three points

(2, 2, 1), (3, 0, 1) and (4, − 1, 0) is [(r − i + 2j + k )] ⋅ [(i − 2j) × (i − j − k )] = 0 i.e. r ⋅ (2i + j + k ) = 7 or 2x + y + z − 7 = 0 …(i) Equation of line through (3, − 4, − 5) and (2, − 3, 1) is x−3 y + 4 z+ 5 …(ii) = = −1 1 6 Any point on line (ii) is (− λ + 3, λ − 4, 6λ − 5.) This point lies on plane (i). Therefore, 2 (− λ + 3) + (λ − 4) + (6λ − 5) − 7 = 0 ⇒ λ =2 Hence, the required point is (1, − 2, 7).

lines 2 x − 5y + z = 3 and x + y + 4 z = 5be (2 x − 5y + 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 ⇒ 2λ = − 11 11 λ=− ⇒ 2 On taking last two equalities, we get 6λ − 30 = 3 + 12λ 11 ⇒ − 6λ = 33 ⇒ λ = − 2 So, the equation of required plane is  2 − 11  x +  −11 − 5 y       2  2

⇒ ⇒

44 11 +  − + 1 z − 3 + 5 × =0  2  2 7 21 42 49 − x− y − z+ =0 2 2 2 2 x + 3 y + 6z − 7 = 0

30 Intersection of two planes is ( x + 2 y − 3) + λ ( y − 2z + 1 ) ⇒ x + (2 + λ )y − 2λz + λ − 3 = 0 ∴

1(1 ) + 2(2 + λ ) + 0(−2λ ) = 0 ⇒ λ = −

31 Given, a plane P : x − 2 y + 2z − 5 = 0 Equation of family of planes parallel to the given plane P is Q : x − 2 y + 2z + d = 0 Also, perpendicular distance of Q from origin is 1 unit. 0 − 2 (0) + 2 (0) + d

⇒

12 + 22 + 22

=1

d =1 ⇒ d = ± 3 3 Hence, the required equation of the plane parallel to P and at unit distance from origin is x − 2 y + 2z ± 3 = 0.

⇒

32 Consider OX , OY , OZ and Ox, Oy , Oz are

28 On substituting the coordinates of the

2

1 +  −6 +   2

Since, this plane passes through (2, 1, 5). On substituting coordinates (2, 1, 5,) we get 3 × 2 − 4 × 1 + 5 × 5 + k = 0 ⇒ k = − 27 So, the equation of plane is 3 x − 4 y + 5z − 27 = 0. is perpendicular to the plane x − 2 y + 5z = 6. ∴ Required equation of line is x −2 y −3 z− 4 = = 1 −2 5

x y z −13 1 ⇒ = = = =− 4 −3 1 26 2 3 1 x = − 2, y = , z = − 2 2 3 1 ∴ Q  − 2, , −   2 2

∴ PQ =

plane 3 x − 4 y + 5z = 4 is 3 x − 4 y + 5z + k = 0

27 Since Q is the image of P, therefore PQ

O(0, 0, 0) x−0 y −0 z−0 = = 4 −3 1 {4(0) − 3(0) + 1(0) + 13} =− 42 + 32 + 12

3 (−1 + 2)2 +  1 −   2

26 The equation of the plane parallel to the

5 ( x + 2 y − 3) − ( y − 2z + 1) = 0 2 ⇒ 2 x + 4 y − 6 − 5y + 10z − 5 = 0 ⇒ 2 x − y + 10z − 11 = 0 ⇒ 2 x − y + 10z = 11

∴

5 2

two systems of rectangular axes. Let their corresponding equations of plane be y x z + + =1 …(i) a b c y x z and + + =1 …(ii) a′ b ′ c ′ Length of perpendicular from origin to Eqs. (i) and (ii) must be same. 1 1 = ∴ 1 1 1 1 1 1 + + + + a2 b 2 c 2 a ′2 b ′2 c ′2 ⇒

1 1 1 1 1 1 + 2 + 2 = 2 + 2 + 2 a2 b c a′ b′ c′

⇒

1 1 1 1 1 1 + + − − − =0 a2 b 2 c 2 a ′2 b ′2 c ′2

33 Equation of PQ is x−1 y + 5 z− 9 = = =λ 1 1 1 So, x = λ + 1, y = λ − 5 and z = λ + 9 lies on the plane x − y + z = 5. ⇒ λ + 1 − λ + 5+ λ + 9 = 5 ∴ λ = − 10 So, the coordinate of Q is (− 9, − 15, − 1) and coordinate of P is (1, − 5, 9). ∴ | PQ | = (10)2 + (10)2 + (10)2 = 10 3

34 Given planes are, 2 x + y + 2z − 8 = 0 5 2 x + y + 2z + = 0 2 Distance between two planes and

=

|c1 − c2 | a2 + b 2 + c 2

21 7 = 2 = 3 2

=

5 2 22 + 12 + 22 −8 −

363

DAY 35 Now, a1 a2 + b1 b2 + c 1 c 2 = 2 − 1 + 2 > 0. The acute angle bisecting plane is x − y + 2z + 1 = − (2 x + y + z + 2) i.e. x+ z+ 1= 0

36 DR’s of line are 2, 1, −2 and DR’s of normal to the plane are 1, 1, 0. 2 1 2 ∴ Therefore, their DC’s are , ,− and 3 3 3 1 1 , ,0 respectively. 2 2 Now, let θ be the angle b/w line and the plane, then 2 1 1 1 2 cos(90° − θ) = ⋅ + ⋅ +  −  ⋅ 0 3 2 3 2  3 1 sinθ = ⇒ θ = 45° ⇒ 2

37 Angle between straight line r = a + λ b and plane r ⋅ n = d, b⋅n sin θ = |b || n| ∴ sin θ = ⇒ sin θ =

(i + 2 j + λ k )⋅ (i + 2 j + 3 k ) 1 + 4 + λ2

1+ 4+ 9

5 + 3λ λ + 5 ⋅ 14 2

5 14 3 sin θ = 14 3 = 14

⇒

cos θ =

(3i+3 j+4k ) ⋅ (6i+ 6 j+8k )

32 + 32 + 42 62 + 62 + 82 18 + 18 + 32 68 = = =1 2 × 34 34 136 −1 ∴ θ = cos 1 = 0 Hence, Statement I is false. II. Given, a1 = 1, b1 = 2, c 1 = 3 a2 = 2,b2 = 2,c 2 = −2 a1 1 b 2 = , 1 = a2 2 b2 2 a1 b ≠ 1 Q a2 b2 Hence, given lines are not parallel. Therefore, Statement II is false.

5 + 3λ λ2 + 5 ⋅ 14

containing them is x−1 y z+ 1 1 −1 1 =0 1 2 3 ⇒

−(5x + 2 y − 3z − 8) = 0 1 2 3 = = 3 6 9 1 1 1 ⇒ = = 3 3 3 and 1(1) + 2(1) + 3(−1) = 0 ∴ Statement II is true. Statement II Here,

which lies on

2

38 Clearly, given line is parallel to the plane. Given point on the line is A (2, − 2,3) and a point on the plane is B (0, 0, 5) ∴ AB = (2 − 0)i + (−2 − 0) j + (3 − 5)k = 2i − 2 j − 2k Now, required distance = Projection of AB or i + 5j + k (2i − 2j − 2k ) ⋅ (i + 5j + k ) = 1 + 25 + 1 2 − 10 − 2 = 27 10 = 3 3

39 I. AB = OB − OA = (4i+5j+7k ) − ( i+2 j+3k ) = 3i+3 j+4k CD = OD − OC = (2i+9 j+2k ) − (−4i+3 j − 6k )

x y −1 z−2 . = = 1 2 3 A

M

x 1

=

y–1 2

=

z–2 3

B 1 3 − 1 5− 2 = = ∴ 1 2 3 ⇒ 1=1=1 Hence, Statement II is true. Also, direction ratio of AB is (1 − 1, 6 − 0, 3 − 7) = (0, 6, − 4) ...(i) and direction ratio of straight line is (1, 2, 3) ... (ii) These two lines are perpendicular, if 0 (1) + 6 ( 2) − 4 (3) = 12 − 12 = 0 Hence, Statement I is true.

42 I. N = OB − OA

B (1, –3, 3)

Equation of a plane is given by A( x − x1 ) + B ( y − y 1 ) + C ( z − z1 ) = 0 ⇒ 3( x − 1) + (−2)( y + 3) + 6(z − 3) = 0 ⇒ 3 x − 3 − 2 y − 6 + 6z − 18 = 0 ⇒ 3 x − 2 y + 6z − 27 = 0 II. Equation of any plane through (2,1, 0) is a( x − 2) + b( y − 1) + c (z − 0) = 0 …(i) Since, it passes through the points (3,-2,-2) and (3,1,7). Then, we get a − 3b − 2c = 0 …(ii) and a + 0b + 7c = 0 …(iii) On solving Eqs.(ii) and (iii) by cross-multiplication, we get a = 7λ,b = 3λ,c = − λ On substituting the value of a,b,c in Eq. (i), we get 7λ( x − 2) + 3λ( y − 1) − λz = 0 ⇒ 7 x + 3 y − z = 17 This is the required equation of the plane.

43 The image of the point (3, 1, 6) with

41 Since, mid-point on AB is M (1, 3, 5.)

⇒ 9 (λ + 5) = 9 λ + 30 λ + 25 ⇒ 9 λ2 + 45 = 9 λ2 + 30 λ + 25 ⇒ 30 λ = 20 2 λ= ∴ 3 2

A (–2, –1, –3)

40 Statement I The equation of the plane

Given, cos θ = ∴

= 6i+ 6 j+8k AB ⋅ CD Qcos θ = | AB||CD|

= ( i − 3 j + 3k ) − ( − 2i − j − 3k ) = 3i − 2 j + 6k

respect to the plane x − y + z = 5is x −3 y −1 z− 6 = = −1 1 1 − 2(3 − 1 + 6 − 5) = 1+ 1+ 1 x −3 y −1 z− 6 = = −2 ⇒ = 1 −1 1 ⇒ x = 3−2 = 1 y = 1+ 2 = 3 z = 6−2 = 4 which shows that Statement I is true. We observe that the line segment joining the points A (3, 1, 6) and B (1, 3, 4) has direction ratios 2, – 2, 2 which are proportional to 1, – 1, 1 the direction ratios of the normal to the plane. Hence, Statement II is true. Thus, the Statements I and II are true and Statement II is correct explanation of Statement I.

44 Statement I 3 y − 4z = 5 − 2k − 2 y + 4z = 7 − 3 k ∴ x = k , y = 12 − 5 k and 31 − 13 k is a point on the line z= 4 for all real values of k. Statement I is true. Statement II Direction ratios of the straight line are

364 < bc ′ − kbc , kac − ac ′ , 0 > and direction ratios of normal the plane are < 0, 0, 1 >. Now, 0 × (bc ′ − kbc ) + 0 × (kac − ac ′ ) + 1× 0= 0 Hence, the straight line is parallel to the plane.

45 Statement II Lines L1 and L2 are parallel to the vectors a = 3 i + j + 2 k and b = i + 2j + 3 k, respectively. The unit vector perpendicular to both L1 and L2 is a ×b − i − 7j + 5k = |a × b| 1 + 49 + 25 =

− i − 7j + 5k

5 3 Hence, Statement II is true. Statement I Plane is − ( x + 1) − 7( y + 2) + 5(z + 1) = 0, whose 13 distance from (1, 1, 1) is . 5 3 Hence, Statement I is true. Thus, statement I is true, statement II is true; Statement II is a correct explanation of Statement I.

SESSION 2 1 Let the equation of plane in intercept x y z + + = 1. a b c Since, it passes through the point (1, 0, 0) and (0,1,0). x y z ∴ Equation of plane is + + = 1 1 1 c 1 DR’s of normal are 1,1, and of given c plane are 1,1,0. 1 1⋅1 + 1⋅1 + ⋅ 0 π c Now, cos = 4  1   2 + 2 2  c  1 2 ⇒ =  2  1  2 + 2 2 c   1 1 2 ⇒ + 2 = 4 ⇒ c = c2 2 1 ∴ c = 2 So, the DR’s of normal are 1,1, 2. form is

2 On eliminating m from given equations, we get 2 (l + n ) 2 + nl = 0 [Q put m = 2 l + 2 n] ⇒ (2l + n ) (l + 2n ) = 0 ⇒ n = − 2l ⇒ m = − 2l or l = − 2n ⇒ m = − 2n The DR’s are 1, − 2, − 2 and −2, − 2, 1. Now, 1 (−2) − 2 (−2) − 2 (1) = 0

Hence, lines are perpendicular. So, angle between them is π / 2.

3 L1 is the line of intersection of the plane 2 x − 2 y + 3z − 2 = 0 and x − y + z + 1 = 0 and L2 is the line of intersection of the plane x + 2 y − z − 3 = 0 and 3 x − y + 2z − 1 = 0 Since L1 is parallel to $i $j k$ 2 −2 3 = $i + $j   1 −1 1 L2 is parallel to k$  $i $j 1 2 −1 = 3$i − 5$j − 7k$   3 −1 2  5 8 Also, L2 passes through  , , 0 . 7 7  [put z = 0 in last two planes] So, equation of plane is x − 5 y − 8 z    7 7 1 0 = 0  1  −5 −7  3    ⇒ 7 x − 7 y + 8z + 3 = 0 Now, perpendicular distance from origin is   3 =  2 2 2 7 + 7 + 8  

3 1 = 162 3 2

4 The new position of plane is x − 2 y + 3z + λ( x + y + z − 1) = 0 ⇒ (1 + λ )x + (λ − 2)y + (λ + 3)z − λ = 0 Since, it is perpendicular to x + y + z − 1 = 0. ∴1+ λ + λ −2+ λ + 3= 0 2 ⇒λ = − 3 Hence, required plane is x − 8 y + 7z = − 2. y 5 Let plane x + + z = 1 cuts the axes at a b c A (a, 0, 0), B (0, b, 0) and C (0, 0, c ). a b c Centroid of plane ABC is D  , ,  .  3 3 3 Distance of the plane from the origin 1 [given] d = =1 1 1 1 + + a2 b2 c2 1 1 1 1= 2 + 2 + 2 ⇒ a b c ∴ D ( x, y , z ) ⇒ a = 3 x, b = 3 y , c = 3z 1 1 1 ⇒ + 2 + 2 =9 x2 y z Hence, the value of k is 9.

6 Given lines are x = py + q, z = ry + s x −q z−s = y = p r

⇒

x = p ′ y + q ′, z = r ′ y + s′ x − q′ z − s′ ⇒ = y = p′ r′ y − y1 x − x1 z − z1 Two lines = = a1 b1 c1 y − y2 x − x2 z − z2 and are = = a2 b2 c2 perpendicular, if a1 a2 + b1 b2 + c 1 c 2 = 0 ∴ Given lines are perpendicular, if pp ′+ r r ′ = 0 and

7 Let the equation of line AB be

x−0 y + a z−0 = = =k 1 1 1

A

C

E

F

[say] B

D

Any point on the line is F (k , k − a, k ). Also, the equation of other line CD is x+ a y −0 z−0 = = =λ 2 1 1

[say]

Any point on the line is E (2 λ − a, λ, λ ). Direction ratios of EF are [(k − 2 λ + a), (k − a − λ ), (k − λ )]. Since, it is given that direction ratios of EF are proportional to 2, 1, 2. k − 2λ + a k − λ − a k − λ ∴ = = 2 1 2 On solving first and second fractions, we get k − 2λ + a = 2k − 2λ − 2a ⇒ k = 3a …(i) On solving second and third fractions, we get 2k − 2λ − 2a = k − λ ⇒ k − λ = 2a ⇒ λ = 3a − 2a [from Eq. (i)] ∴ λ=a Hence, coordinates of E are (3 a, 2 a, 3 a) and coordinates of F are (a, a, a).

8 A parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7), parallel to the coordinate planes. Let a, b and c be the lengths of edges, then a = 5 − 2 = 3, b = 9 − 3 = 6 and c = 7 − 5 = 2 So, the length of diagonal of a parallelopiped

365

DAY =

12 Qcos2 α + cos2 β + cos2 γ + cos2 δ = 4

a2 + b 2 + c 2

=

9 + 36 + 4

=

49 = 7 units

3 and cos 2 α = 1 − sin2 α, similarly for all other angles. ∴ cos 2 α + cos 2 β + cos 2 γ + cos 2 δ = 4 − (sin2 α + sin2 β + sin2 γ + sin2 δ ) ⇒ sin2 α + sin2 β + sin2 γ 4 8 + sin2 δ =  4 −  =  3 3

9 Given, equations of lines are x−1 y + 2 z− 4 = = 1 −2 3 x−2 y + 1 z+ 7 and = = 2 −1 −1 $ $ $ Let n = i − 2 j + 3k $ n2 = 2$i − $j − k

∴ Any vector n perpendicular to both n1 , n2 is given by n = n 1 × n2 $i $j $ k ⇒

n= 1 −2 3 2 −1 −1 $ = 5$i + 7 $j + 3k

∴ Equation of a plane passing through (1, − 1, − 1) and perpendicular to n is given by 5 ( x − 1) + 7( y + 1) + 3(z + 1) = 0 ⇒ 5x + 7 y + 3z + 5 = 0 ∴ Required distance = =

5 + 21 − 21 + 5 52 + 72 + 32 10 units 83

10 Consider the line through (2, − 1, 3) with 2 2 1 DC’s , , is 3 3 3 x−2 y + 1 z−3 [say] = = =r 2/3 2/3 1 /3 2r 2r r ∴ x=2+ ,y = −1 + ,z = 3 + 3 3 3 Since, it lies on the plane x + 2 y − z = 2. 2r 4r r ∴ 2+ −2+ −3− =2 3 3 3 ⇒ r =3

the required line. Then, its equation is x−3 y − 0 z−1 ...(i) = = a b c Since, Eq. (i) is parallel to the planes x + 2 y + 0z = 0 and 0 x + 3 y − z = 0. Therefore, normal to the plane is perpendicular to the line. ∴ a(1) + b (2) + c (0) = 0 and a (0) + b(3) + c (− 1) = 0 On solving these two equations by cross-multiplication, we get a b = (2) (− 1) − ( 0) (3) ( 0) ( 0) − (1) (− 1) c = (1) (3) − ( 0) (2) a b c [say] = = =λ ⇒ −2 1 3 ⇒ a = − 2λ, b = λ and c = 3λ On substituting the values of a, b and c in Eq. (i), we get the equation of the required line as x−3 y − 0 z−1 = = −2 1 3

14 Key idea length of projection of the line segment joining a 1 and a 2 on the plane (a − a 1 ) × n r ⋅ n = d is 2 |n| Length of projection the line segment joining the points (5, −1, 4) and (4, −1, 3) on the plane x + y + z = 7 is B (4, –1, 3) (a2)

11 Given, mid-points of sides are D (l , 0, 0), E (0, m, 0) and F (0, 0, n )

^

(5, –1, 4)

B

E

D

A(a1)

C

(a − a1 ) × n AC =  2  |n |  

C

BC 2 [by mid-point theorem] 4 2 ⇒ BC = 4 (m2 + n2 ) Similarly, AB 2 = 4(l 2 + m2 ) and CA2 = 4(l 2 + n2 ) AB 2 + BC 2 + CA2 =8 ∴ l 2 + m2 + n2

^

^

n=i+j+k

A F

=

13 Let a, b and c be the direction ratios of

1

and

Alternative Method Clearly, DR’s of AB are 4 − 5, − 1 + 1, 3 − 4, i.e. −1, 0, − 1 and DR’s of normal to plane are 1, 1, 1. Now, let θ be the angle between the line and plane, then θ is given by −1+ 0−1 sin θ = (−1)2 + (−1)2 12 + 12 + 12

Also, EF 2 =

= AC = ⇒

|(− ^i − ^ k ) × (^i + ^j + ^ k )| |i + j + k| |^i − ^ k| 3

AC =

2 = 3

2 3

2 = 2 3

2 3 B (4, –1, 3)

A (5, –1, 4)

q C

q

Plane : x + y + z = 7

2 ⇒ cos θ = 1 − sin2 θ 3 2 1 = 1− = 3 3 Clearly, length of projection 1 = AB cos θ = 2 [Q AB = 2] 3 2 = 3

⇒ sinθ =

15 Centroid of ∆ABC, 2 − 1 + λ 3 + 3 + 5 5+ 2 + µ G =  , ,    3 3 3 1 + λ 11 7 + µ   = , ,   3 3 3  Since, median is always passes through centroid and they are equally inclined. 1+ λ 7+ µ 11 ∴ −2= −3= −5 3 3 3 λ−5 2 µ −8 ⇒ = = 3 3 3 ⇒ λ = 7, µ = 10

16 Let parallel plane be

2 x − 2 y + z + λ = 0. It passes through (1, − 2, 3). ∴ λ = −9 The distance of (−1, 2, 0) from the plane 2x − 2y + z − 9 = 0 −2 − 4 − 9 = 5 is 3  

17 Equation of any plane through the intersection of planes can be written as x + y + z − 1 + λ(2 x + 3 y + 4z − 5) = 0 ⇒ (1 + 2λ )x + (1 + 3λ )y + (1 + 4λ )z − 1 − 5λ = 0 …(i)

366

The direction ratios, a1 ,b1 ,c 1 , of the plane are (1 + 2λ),(3λ + 1) and (4λ + 1). The plane in Eq. (i) is perpendicular to x − y + z = direction ratios, a2 ,b2 ,c 2 are 1,−1 and 1. Since, the planes are perpendicular. ∴ a1 a2 + b1 b2 + c 1 c 2 = 0 ⇒ 1(1 + 2λ) − 1(1 + 3λ) + 1(1 + 4λ) = 0 ⇒ 1 + 2λ − 1 − 3λ + 1 + 4λ = 0 1 ⇒ 3λ = −1 ⇒ λ = − 3 On substituting this value of λ in Eq. (i), we obtain the required plane as 1 − 2 x + 1 − 3 y       3 3 4 5 +  1 −  z − 1 + = 0  3 3 1 1 2 ⇒ x− z+ = 0 3 3 3 ⇒ x−z+2= 0 This is the required equation of the plane.

18 The two normal vectors are

m = 2i + 3 j + k and n = i + 3 j + 2k i j k The line L is along, m × n = 2 3 1 1 3 2 = 3(i − j + k ) and DC’s of X-axis are (1, 0, 0).

∴ cos α =

3 (i − j + k ) ⋅ ( i ) 32 (1 + 1 + 1) 1

=

1 3

20 Here, plane, line and its image are

x−1 y + 2 z−3 (say) = = =λ 1 4 5 Any point on above line can be written

parallel to each other. So, find any point on the normal to the plane from which the image line will be passed and then find equation of image line. Here, plane and line are parallel to each other. Equation of normal to the plane through the point (1, 3, 4) is x−1 y −3 z− 4 [say] = = =k 2 −1 1 Any point on this normal is (2k + 1, − k + 3, 4 + k ). Then,  2k + 1 + 1 3 − k + 3 4 + k + 4  , ,   lies   2 2 2

as (λ + 1, 4λ − 2, 5λ + 3).

on plane.

19 Any line parallel to x = y = z and 1 4 5 passing through P(1, − 2, 3) is P (1, –2, 3)

x y z = = 1 4 5 R

2x + 3y – 4z + 22 = 0

Q

6 − k ⇒ 2(k + 1) −    2 

∴ Coordinates of R are (λ + 1, 4λ − 2, 5λ + 3). Since, point R lies on the above plane. ∴

2(λ + 1) + 3(4λ − 2) − 4(5λ + 3) + 22 = 0

⇒ λ =1 So, point R is (2, 2, 8). Now, PR = (2 − 1)2 + (2 + 2)2 + (8 − 3)2 = ∴

42

PQ = 2PR = 2 42

8 + k +  +3= 0  2  ⇒ k = −2 Hence, point through which this image pass is (2k + 1, 3 − k, 4 + k ) i.e. [2(−2) + 1, 3 + 2, 4 − 2] = (−3, 5 , 2) Hence, equation of image line is x+ 3 y − 5 z−2 = = 3 1 −5

367

DAY

DAY THIRTY THREE

Unit Test 5 (Vectors and 3D Geometry) 1 The equation of plane perpendicular to 2x + 6y + 6z = 1 and passing through the points ( 2, 2,1) and ( 9, 3, 6), is (a) (b) (c) (d)

3x 3x 3x 3x

+ + + +

4y 4y 4y 4y

+ 5z − 9 = 0 − 5z + 9 = 0 − 5z − 9 = 0 + 5z + 9 = 0

2 Let a, b and c be the unit vectors such that a and b are mutually perpendicular and c is equally inclined to a and b at an angle θ. If c = x a + y b + z (a × b ), then (a) (b) (c) (d)

z 2 = 1 − 2x 2 z2 = 1− x2 + y2 z 2 = 1+ 2y 2 None of the above

19 57 17 (a)  − , ,   8 16 16  19 57 17  (c)  , , −   8 16 16 

19 57 17  (b)  , − ,   8 16 16  19 57 17 (d)  , ,   8 16 16 

7 Vectors a and b are inclined at an angle θ = 120° . If | a | = 1, | b | = 2, then [(a + 3 b ) × ( 3 a − b )]2 is equal to (b) 325

(c) 275

(d) 225

8 A point moves so that the sum of the squares of its

a + b + c = α d and b + c + d = β a , then a + b + c + d is equal to (b) αa (d) (α + β) c

(a) 0 (c) β

4 The plane ax + by = 0 is rotated through an angle α

about its line of intersection with the plane z = 0. Then, the equation to the plane in new position is (a) ax − by ± z a 2 + b 2 cotα = 0 (b) ax + by ± z a 2 + b 2 cotα = 0 (c) ax − by ± z a + b tan α = 0 2

(d) ax + by ± z a 2 + b 2 tan α = 0

5 If the axes are rectangular, the distance from the point

x −3 y −4 z −5 ( 3, 4, 5) to the point, where the line = = 1 2 2 meets the plane x + y + z = 17 is (a) 1 (c) 3

forming a triangle. If AD, the bisector of ∠BAC meets BC in D, then coordinates of D are

(a) 300

3 If a, b and c are three non-coplanar vectors such that

2

6 Let A ( 3, 2, 0), B ( 5, 3, 2), C ( − 9, 6, − 3) are three points

(b) 2 (d) None of these

distances from the six faces of a cube given by x = ± 1, y = ± 1, z = ± 1 is 10 units. The locus of the point is (a) x 2 + y 2 + z 2 = 1 (c) x + y + z = 1

(b) x 2 + y 2 + z 2 = 2 (d) x + y + z = 2

9 If (a × b )2 + (a ⋅ b )2 = 144 and | a | = 4 , then | b | is equal to (a) 3

(b) 8

(c) 12

(d) 16

10 The values of x for which the angle between

a = 2x 2 i + 4x j + k, b = 7 i − 2 j + x k is obtuse, is (a) x > 1/ 2 or x < 0 (c) 1/ 2 < x < 15

(b) 0 < x < 1/ 2 (d) None of these

11 Let a = 3 i + 2 k and b = 2j + k. If c is a unit vector, then the maximum value of the vector triple product [a b c ], is (a) (c)

61 3 ⋅ 36

(b) 59 (d) None of these

12 The ratio of lengths of diagonals of the parallelogram

constructed on the vectors a = 3 p − q, b = p + 3 q is (given that | p | = | q | = 2 and the angle between p and q π is ) 3 (a)

7: 3

(b)

3: 2

(c)

5: 7

(d)

5: 3

368

13 Let p, q and r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies equation p × {( x − q ) × p} + q × {( x − r ) × q} + r × {( x − p ) × r} = 0 Then, x is given by 1 (p + q − 2r) 2 1 (c) (2 p + q − r) 3 (a)

(b)

1 (p + q + r) 2

each of the lines given by the equation x = y + 2 = z ; x + 2 = 2y = 2z . The coordinates of the point of intersection are (b) (6, 6, 6), (2, 6, 2) (d) None of these

15 The vector B satisfying the vector equation A + B = a ,

A × B = b and A ⋅ a = 1, where a and b are given vectors is (b × a ) + a (a 2 − 1) a2 a (a 2 − 1) + b (b 2 − 1) (c) a2 (a)

(b)

5 (5 i + 5 j + 2 k) 3 5 (d) ± (− 5 i + 5 j + 2 k) 3 (b) ±

17 Let a, b and c be three non-zero and non-coplanar vectors and p, q and r be three vectors given by p = a + b − 2 c , q = 3 a − 2b + c and r = a − 4 b + 2c. If the volume of the parallelopiped determined by a, b and c is V1 and the volume of the parallelopiped determined by p, q and r is V2 , then V2 : V1 is equal to (a) 7 : 1 (c) 11 : 1

18 A non-zero vector a is parallel to the line of intersection of the plane determined by the vectors i , i + j and the plane determined by the vectors i − j , i + k. Then, the angle between a and the vector i − 2 j + 2 k is π (b) 4 π (d) 2

and 2, respectively. If a × (a × c ) + b = 0, then the acute angle between a and c is π 6 π (c) 3

(b)

π 4

triple product [ 2 a − b 2b − c 2c − a ] is equal to (b) 1

(c) −

2p 3 3 lmn 6p 3 (d) lmn (b)

23 If a variable plane forms a tetrahedron of constant volume 64 k 3 with the coordinate planes, then the locus of the centroid of the tetrahedron is (b) xyz = 2k 3 (d) xyz = 6 k 3

r = (i + 2 j − k ) + λ ( 2 i + j + k ) and r = ( 2 i + 6 j + k ) + µ (i + 2 j + 3 k ) is

(a) r = (i + 2 j − k) + λ (− i + 5 j − 3 k) (b) r = (i + 3 j + 2 k) + λ (i − 5 j + 3 k) (c) r = (i + 3 j + 2 k) + λ (i + 5 j + 3 k) (d) None of the above

25 The orthogonal projection A′ of the point A with position vector (1, 2, 3) on the plane 3x − y + 4z = 0, is (a) (− 1, 3, − 1)

1 5 (b)  − , , 1  2 2 

1 5 (c)  , − , − 1 2  2

(d) (6, − 7, − 5)

26 The equation of the plane containing the points A(1, 0, 1) (a) x + y – z = 0 (c) x − y + z = 0

(b) x + y + z = 0 (d) x − y − z = 0

27 The planes 3x − y + z + 1 = 0 and 5x + y + 3z = 0 intersect in the line PQ. The equation of the plane through the point ( 2, 1, 4) and perpendicular to PQ is (b) x + y − 2 z = − 5 (d) x + y + 2 z = − 5

28 The line of intersection of the planes r ⋅ ( 3 i − j + k ) = 1 and r ⋅ (i + 4 j − 2 k ) = 2 is parallel to the vector (a) − 2 i + 7 j + 13 k (c) 2 i + 7 j − 13 k

(b) − 2 i − 7 j + 13 k (d) None of these

29 A line with direction cosines proportional to 2, 1, 2 meets

(d) None of these

20 If a, b and c are unit coplanar vectors, then the scalar (a) 0

p3 6 lmn 3 p3 (c) lmn

(a)

(a) x + y − 2 z = 5 (c) x + y + 2 z = 5

19 Let a, b and c be three vectors having magnitudes 1, 1

(a)

by the equation r ⋅ (l i + m j ) = 0, r ⋅ (m j + n k ) = 0 , r ⋅ (nk + p i) = 0 and r ⋅ (l i + m j + nk ) = p, then the volume of the tetrahedron is

and B ( 3, 1, 2) and parallel to the line joining the origin to the point C(1, − 1, 2) is

(b) 3 : 1 (d) None of these

π (a) 6 π (c) 3

(c) 2

(b) 1 1 (d) 2

24 The line through i + 3 j + 2 k and perpendicular to the line

(d) None of these

between the vectors a = 7 i − 4 j − 4 k, b = − 2 i − j + 2 k and | c | = 5 6 is 5 (i − 7 j + 2 k) 3 5 (c) ± (i + 7 j + 2 k) 3

(a) 0

(a) xyz = k 3 (c) xyz = 12k 3

(a × b) + a a

16 The vector c, directed along the bisectors of the angle

(a) ±

a = α i + β j + γ k, k × (k × a ) = 0. Then, γ is equal to

22 If the four plane faces of a tetrahedron are represented

(d) None of these

14 A line with direction cosines proportional to 2, 1, 2 meets

(a) (6, 4, 6), (2, 4, 2) (c) (6, 4, 6), (2, 2, 0)

21 Point (α , β, γ ) lies on the plane x + y + z = 2. Let

3

(d)

3

each of the lines x = y + a = z and x + a = 2y = 2z . The coordinates of each of the points of intersection are given by (a) (3 a, 2a, 3 a), (a, a, 2a) (c) (3 a, 2 a, 3 a), (a, a, a)

(b) (3 a, 3 a, 3 a), (a, a, a) (d) None of these

369

DAY 30 The sides of a parallelogram are 3 i + 4 j − 6 k and 2 i − 3 j + 5 k. The unit vector parallel to one of the diagonals is 5i + j −k 27 i − 2 j + 4k (c) 21 (a)

(b)

4i + 3 j − 2k 29

(d) None of these

31 Let p = 8 i + 6 j and q be two vectors perpendicular to each other in the xy-plane. Then, the vector in the same plane having projections 2 and 4 along p and q respectively is (a) ± 3 (i − 2 j ) (c) ± 2 (2 i − j )

(b) ± (i + 2 j ) (d) None of these

32 The equation of the plane containing the line

2x − y + z − 3 = 0 and 3x + y + z = 5 at a distance of 1 from the point ( 2, 1, − 1) is 6 (a) (b) (c) (d)

x+ y+ z−3=0 2x − y − z − 3 = 0 2x − y + z + 3 = 0 62 x + 29y + 19z − 105 = 0

(a) 8 x + 14 y + 13 z + 37 = 0 (b) 8 x − 14 y + 13 y + 37 = 0 (c) 8 x + 14 y − 13 z + 37 = 0 (d) None of these

r = − (i + j + k ) + λ ( 2 i + 3 j + 4 k ) r = − i + µ ( 3 i + 4 j + 5 k ) is 1 (c) 3

and C( − 1, 1, 2). The angle between the faces OAB and ABC is 19 (a) cos−1    35 

17 (b) cos−1    31

(c) 30°

(d) 90°

36 If p = 2 i − 3 j + 3 k and q = 4 i − 2 j + k be two vectors and r is a vector perpendicular to p and q and satisfying the condition. r ( 2 i − 4 j + 2 k ) = −12, then r is equal to

(c)

20 j + 16k 3

1 (3 i − 10 j + 8 k) 3

(b)

2 (3 i + 10 j + 8 k) 3

(d) None of these

37 The direction ratios of a normal to the plane through (2, 0, 0) and (0, 2, 0) that makes an angle 2x + 3y = 5 is (a) 1 : 1 : 2 (c) 2 : 1 : 3

and L2 =

x y −7 z + 7 is = = 1 −3 2

(a) (− 3, 2, 1) (c) (1, − 3, 2)

x +1 y − 3 y + 2 = = −3 2 1

(b) (2, 1, − 3) (d) None of these

40 If a and b are unit vectos, then the greatest value of | a + b| + | a − b| is (a) 2

(b) 4

(c) 2 2

(d) 2

41 If a, b and c are non-coplanar vectors and r is a real number, then the vectors a + 2b + 3c , λb + 4c and ( 2λ − 1) c are non-coplanar for

π with the plane 3

(b) 1 : 1 : 3 (d) 1 : 1 : 5 .7

perpendicular to the plane of b and c. If the angle π between b and c is , then | a × b − a × c | is equal to 3 (a) 1/3

(b) 1/2

(c) 1

(d) 2

43 The distance of the point A( − 2, 3, 1) from the line PQ

through P( − 3, 5, 2) which make equal angles with the axes is 2 3

(b)

14 3

(c)

16 3

(d)

5 3

44 The plane passing through the point (5, 1, 2) 1 (d) 6

35 A tetrahedron has vertices at O( 0, 0, 0), A(1, 2, 1), B( 2, 1, 3)

(a) 2 i −

(b) 2.01 units (d) None of these

39 The intersecting point of lines, L1 =

(a)

34 The shortest distance between the lines

(a) 1

(a) 1.14 units (c) 3.16 units

42 Let a, b and c be three unit vectors such that a is

x −1 y + 2 z and parallel to the lines and = = 3 2 −4 x y −1 z − 2 is = = 2 −3 2

1 (b) 2

x − 2 y −1 z − 3 x −1 y − 2 z − 4 and is equal to = = = = 3 2 2 2 3 4

(a) no value of λ (b) all except one value of λ (c) all except two values of λ (d) all values of λ

33 The equation of the plane through the point ( 2, − 1, − 3)

and

38 The shortest distance between the lines

perpendicular to the line 2( x − 2) = y − 4 = z − 5 will meet the line in the point (a) (1, 2, 3) (c) (1, 3, 2)

(b) (2, 3, 1) (d) (3, 2, 1)

Direction (Q. Nos. 45-48) Each of these questions contains two statements : Statement I (Assertion) and Statement II (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a ), (b), (c) and (d) given below. (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

45 Consider vectors a and c are non-collinear, then Statement I The lines r = 6 a − c + λ ( 2c − a ) and r = a − c + µ (a + 3 c ) are coplanar. Statement II There exist λ and µ such that the two values of r become same.

370

46 Consider u and v are unit vectors inclined at an angle α

Statement I The equation of the plane through the intersection of the planes P1 and P2 and the point ( 4, 4, 4) is 29x + 23y + 17y = 276. Statement II Equation of the plane through the line of intersection of the planes P1 = 0 and P2 = 0 is P1 + λP2 = 0, λ ≠ 0.

and a is a unit vector bisecting the angle between them, u+ v Statement I Then, a = . 2 cos (α / 2) Statement II If ABC is an isosceles triangle with AB = AC = 1, then vector representing bisector of ∠ A is AB + AC . 2

49 The two adjacent sides of a parallelogram are 2i − 4j + 5k and i − 2j − 3 k. Statement I The unit vector parallel to its diagonal is 3 6 2 i − j + k. 5 5 5 Statement II Area of parallelogram is11 5 sq units.

47 Suppose π : x + y − 2z = 3, P : ( 2, 1, 6), Q : ( 6, 5 , − 2) Statement I The line joining PQ is perpendicular to the normal to the plane π. Statement II Q is the image of P in the plane π.

(a) Only Statement I is true (b) Only Statement II is true (c) Both statements are ture (d) Both statements are false

48 Consider the equation of planes P1 = x + y + z − 6 = 0 and P2 = 2x + 3y + 4z + 5 = 0.

ANSWERS 1. 11. 21. 31. 41.

(c) (a) (c) (c) (c)

2. 12. 22. 32. 42.

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

3. 13. 23. 33. 43.

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

4. 14. 24. 34. 44.

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

5. 15. 25. 35. 45.

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

6. 16. 26. 36. 46.

(d) (a) (d) (d) (a)

7. 17. 27. 37. 47.

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

8. 18. 28. 38. 48.

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

9. 19. 29. 39. 49.

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

10. 20. 30. 40.

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

Hints and Explanations 1 The plane passing through (2, 2, 1), is a( x − 2) + b( y − 2) + c (z − 1) = 0 Since, it passes through (9, 3, 6). …(i) ∴ 7a + b + 5c = 0 Since, it is perpendicular to 2 x + 6 y + 6z − 1 = 0. …(ii) ∴ 2 a + 6b + 6c = 0 a b c = = ⇒ −24 −32 40 [from Eqs. (i) and (ii)] a b c = = ⇒ 3 4 −5 The required plane is 3 ( x − 2) + 4( y − 2) − 5(z − 1) = 0 ⇒ 3 x + 4 y − 5z − 9 = 0

2 Now, a ⋅ c = x ( a ⋅ a ) + y (a ⋅ b ) + za ⋅ (a × b ) ⇒ x = cos θ Similarly, y = cos θ Now, |c |2 = x2 |a |2 + y 2 |b|2 + z2|a × b|2 ⇒ 1 − 2 cos 2 θ = z2 ⇒ 1 − 2 x2 = z2 where, z = |a b c |

3 Given, a + b + c = α d and b + c + d = βa

∴ a + b + c + d = (α + 1) d and a + b + c + d = ( β + 1) a ⇒ (α + 1) d = ( β + 1) a If α ≠ − 1, then (α + 1) d = ( β + 1) a β+1 ⇒ d = a α +1 ∴ ⇒

 β + 1 a + b + c = α a  α + 1  α ( β + 1) 1 − a + b+c =0 α +1  

Hence, a , b and c are coplanar, which is a contradiction to the given condition. ∴ α = −1 ⇒ a + b+c +d =0

4 Equation of any plane passing through the line of intersection of given plane, is ax + by + kz = 0 …(i) ∴ DC’s of Eq. (i) are a b , , a2 + b 2 + k 2 a2 + b 2 + k 2 k a2 + b 2 + k 2

The DC’s of a normal to the given plane is a b , , 0. 2 2 2 a +b a + b2 ∴ cos α = =

a⋅ a + b ⋅ b + k ⋅ 0 a2 + b 2 + k 2

a2 + b 2

a2 + b 2 a + b2 + k 2 2

⇒ k 2 cos 2 α = a2 (1 − cos 2 α ) + b 2 (1 − cos 2 α ) 2 2 (a + b ) sin2 α ⇒ k2 = cos 2 α ∴ k = ± a2 + b 2 tan α From Eq. (i), ax + by ± z

a2 + b 2 tan α = 0

5 Any point on the line is

(r + 3, 2r + 4, 2r + 5) which lies on the plane x + y + z = 17. Q (r + 3) + (2r + 4) + (2r + 5) = 17 ∴ r =1 Thus, the point of intersection is (4, 6, 7). So, the required distance

371

DAY = (4 − 3)2 + (6 − 4)2 + (7 − 5)2

Similarly,|d 2 | = 4 3

= 1+ 4+ 4=3

∴

(5 − 3) + (3 − 2) 2

6 Here, AB =

=3

(−9 − 3) + (6 − 2) 2

and AC =

13 |p| = |q | = |r| = c [say]

2

+ (2 − 0)2

and p ⋅ q = 0 = p ⋅ r = q ⋅ r Given that, p × {(x − q ) × p} + q × {(x − r ) × q} + r × {(x − p) × r} = 0 ⇒ (p ⋅ p ) (x − q ) − {p ⋅ (x − q )}p + K = 0 ⇒ c 2 (x − q + x − r + x − p ) − (p ⋅ x ) p − (q ⋅ x )q − (r ⋅ x ) r = 0 ⇒ c 2 {3 x − ( p + q + r )} − {( p ⋅ x ) p + (q ⋅ x )q + (r ⋅ x ) r} = 0 1 which is satisfied by x = ( p + q + r ) 2

2

+ (−3 − 0)2

= 13

Since, AD is the bisector of ∠ BAC. BD AB 3 = = ∴ DC AC 13 Since, D divides BC in the ratio 3 : 13. ∴ The coordinates of D are  3 (−9) + 13 (5) 3 (6) + 13 (3) , ,  3 + 13 3 + 13  3 (−3) + 13 (2)  3 + 13  19 57 17 =  , ,   8 16 16 

7 [(a + 3 b ) × (3a − b ) ]2 = [10 (b × a )] 2 = 100 [|a |2 |b|2 − (a ⋅ b )2 ] = 100 [1 × 4 − (1 × 2 × cos 120° )2 ] = 100 (4 − 1) = 300

8 Let P ( x, y , z ) be any point on the locus, then the distances from the six faces are | x + 1 |, | x − 1 |,| y + 1 |, | y − 1 |, |z + 1 |, |z − 1 | According to the given condition, |x + 1|2 + | x − 1|2 + | y + 1|2 + | y − 1|2 + |z + 1 |2 + |z − 1 |2 = 10 ⇒ 2 ( x2 + y 2 + z2 ) = 10 − 6 = 4 ∴ x2 + y 2 + z2 = 2 2

2

2

2

+ |a | |b| cos θ = |a |2 |b|2 144 = (4)2 |b|2 ⇒ |b| = 3 2

⇒

10 Since, ⇒ ⇒ ⇒ ∴

2

2

a ⋅b< 0

14 x2 − 8 x + x < 0 14 x2 − 7 x < 0 7 x (2 x − 1) < 0 1 0< x < 2

Now, [a b c ] = (a × b ) ⋅ c = (− 4 i − 3 j + 6 k ) ⋅ c ∴ The maximum value of [a b c ] = |− 4 i − 3 j + 6 k ||c | =

⇒ A ⋅ a + B⋅ a = a ⋅ a ⇒ 1 + B ⋅ a = a2 ⇒ B ⋅ a = a2 − 1 Also, A ×B=b ⇒ a × (A × B) = a × b ⇒ (a ⋅ B) A − (a ⋅ A ) B = a × b

61

and d 2 = a − b = 2 p − 4q ⇒ d 21 = 16 p2 + 4q 2 + 16 p ⋅ q π = 16 (4) + 4 ( 4) + 16  2 × 2 × cos   3

1 1 −2 Then, V2 = 3 − 2 1 [a b c ] 1 −4 2 = 15 [a b c ] ∴ V2 : V1 = 15 : 1

18 The normal to the first plane is along

i × ( i + j ) = k and the normal to the second plane is along (i − j) × (i + k) = − i − j + k Since, a is perpendicular to the two normals. So, a is along k × (− i − j + k ) = i − j Hence, the angle between a and the vector i − 2 j + 2 k is a ⋅ ( i − 2j + 2 k ) cos −1 |a ||i − 2 j + 2 k | ( i − j ) ( i − 2j + 2k) = cos −1 ⋅ 3 2 1 π −1 = cos = 2 4

19 Since, b = (a × c ) × a ⇒ ∴

|b| = |a × c ||a | ⇒ 1 = 2 sin θ π θ= 6

20 If a , b and c lie in a plane, then …(ii)

⇒ (a − 1) A − B = a × b 2

From Eqs. (i) and (iii), we get (a × b ) + a A = a2 (a × b ) + a  and B = a −   a2   (b × a ) + a (a2 − 1) ∴ B= a2  a b  c =±λ +  |b|   |a |

12 Now, d 1 = a + b = 4 p + 2q

= 112 |d 1 | = 4 7

general coordinates of points on the two given lines. ∴ DR’s of PQ are (r − 2k + 2, r − k − 2, r − k ) r − 2k + 2 r − k − 2 r − k ∴ = = 2 1 2 ⇒ r = 6, k = 2 So, the points of intersection are (6, 4, 6) and (2, 2, 2). 15 Given, A + B = a …(i)

16 The required vector c is given by

11 Here, a × b = − 4 i − 3 j + 6 k

⇒

14 Let P (r , r − 2, r ) and Q (2k − 2, k, k ) are the

[from Eq. (ii)] ...(iii)

9 (a × b ) + (a ⋅ b ) = |a | |b| sin θ 2

17 Given, V1 = [a b c ] and V2 = [ p q r]

d1 : d2 = 7 : 3

1 1 = ± λ  (7i − 4j − 4k ) + (−2i − j + 2k ) 3 9  λ ⇒ c = ± (i − 7 j + 2 k ) 9 λ 5 6= 1 + 49 + 4 ⇒ 9 λ = 54 [Q|c | = 5 6] 9 ⇒ λ = 15 15 ∴ c =± (i − 7 j + 2 k ) 9 5 = ± ( i − 7j + 2k) 3

2a − b, 2 b − c and 2c − a , also lie in the same plane. So, their scalar triple product is ‘0’.

21 Since,

k × (k × a ) = 0

⇒ ⇒ Q ∴

(α i + β j ) = 0 α =β = 0 α + β + γ =2 γ =2

22 The first three planes meet at the point whose position vector is (0, 0, 0). The first two and the fourth planes meet at the point whose position p −p p vector is  , ,  . Similarly, the  l m n other two vertices of the tetrahedron have position vectors  − p , p , p  and  p , p , − p  .      l l m n  m n ∴ Volume of the tetrahedron p/l − p/m p/n 1 = − p/l p/m p/n 6 p/l p/m − p/n =

1 −1 1 p3 −1 1 1 6 lmn 1 1 −1

=

0 0 2 p3 0 2 0 6 lmn 1 1 −1

=

|− 4 | p3 2 p3 = 6 lmn 3 lmn

 R1 → R1 + R2 , R → R + R   2 2 3 

372

23 Let the variable plane intersects the coordinate axes at A (a, 0, 0), B (0, b, 0) and C (0, 0, c ). Then, the equation of the plane will be y x z + + =1 …(i) a b c Let P (α, β, γ ) be the centroid of tetrahedron OABC, then a b c α = , β = and γ = 4 4 4 ∴ a = 4 α, b = 4β, c = 4 γ Now, volume of tetrahedron = (Area of ∆AOB ) OC 1 1 abc 3 64 k =  ab  c = ⇒  3 2 6 (4 α ) (4 β ) × (4 γ ) 3 ⇒ 64 k = 3×2 αβγ 3 ∴ k = 6 Hence, locus of P (α, β, γ ) is xyz = 6 k 3 .

24 The required line passes through the

point i + 3 j + 2 k and is perpendicular to the lines r = (i + 2 j − k ) + λ (2 i + j + k ) and r = (2 i + 6 j + k ) + µ ( i + 2 j + 3 k ) So, it is parallel to the vector. ∴ b = (2 i + j + k ) × ( i + 2 j + 3 k ) = ( i − 5j + 3 k ) The required equation is r = ( i + 3 j + 2 k ) + λ ( i − 5j + 3 k )

25 Let n = 3 i − j + 4 k Line through A and parallel to n is r = i + 2 j + 3 k + λ (3 i − j + 4 k ) = (3λ + 1) i + (2 − λ ) j + (3 + 4λ ) k …(i) Eq. (i) must satisfy the plane 3 x − y + 4z = 0.

A (1, 2, 3) n

A′ ∴ 3(3λ + 1) − (2 − λ ) + 4(3 + 4λ ) = 0 ⇒ 26λ + 13 = 0 1 ⇒ λ=− 2 1 5 Hence, A′ is  − , ,1 which is the foot  2 2  of the perpendicular from A on the given plane.

26 DR’s of OC are (1, − 1, 2). Let the equation of plane passing through (1, 0, 1) is a( x − 1) + b( y − 0) + c (z − 1) = 0 …(i) Since, its normal is perpendicular to OC

∴ 1 ⋅ a + (−1) b + 2c = 0 ⇒ a − b + 2c = 0 …(ii) As Eq. (i) passes through (3, 1, 2). …(iii) ∴ 2a + b + c = 0 a b c = = ⇒ −1 1 1 [from Eqs. (ii) and (iii)] Hence, required equation of plane be x − y − z = 0.

27 Let DC’s of PQ be l , m and n. ∴ 3 l − m + n = 0 and 5l + m + 3 n = 0 l m n ∴ = = −3 − 1 5 − 9 3 + 5 l m n ⇒ = = 1 1 −2 Thus, the equation of plane perpendicular to PQ will have x + y − 2z = λ. It passes through (2, 1, 4), therefore λ = − 5. Hence, the required equation of plane be x + y − 2z = − 5

28 The line of intersection of the planes

r ⋅ (3i − j + k ) = 1 and r ⋅ ( i + 4j − 2k ) = 2 is perpendicular to each of the normal vectors. Here, n 1 = 3 i − j + k and n2 = i + 4 j − 2 k ∴ It is parallel to the vector n 1 × n 2 = (3 i − j + k ) × ( i + 4 j − 2 k ) = − 2 i + 7 j + 13 k + 29 Since, lines are x = y a = z 1 1 1 x+ a y z and = = 2 1 1 Let P ≅ (r , r − a, r ) and Q ≅ (2λ − a, λ, λ ) be the points of I and II lines. So, DR’s of PQ are r − 2λ + a, r − λ − a, r − λ . According to the given question, r − 2λ + a r − λ − a r − λ = = 2 1 2 From I and II terms, r − a = 2a ⇒ r = 3a From II and III terms, λ = a ∴ P ≡ (3 a, 2 a, 3 a) and Q ≡ (a, a, a)

30 Let a = 3 i + 4 j − 6 k and b = 2i − 3j + 5 k Q Diagonals of a parallelogram in terms of its sides are p = a + b and q = b − a ⇒ p = 5i + j − k and q = − i − 7 j + 11 k The unit vectors along the diagonals are 5i + j − k − i − 7j + 11 k and 25 + 1 + 1 (−1)2 + 49 + 121 ⇒

5i + j − k 27

and

− i − 7j + 11 k 171

31 Since, p and q are perpendicular. ∴ p ⋅q = 0 Let q = x i + y j, then (8i + 6j )( xi + yj ) = 0 ⇒ 8 x + 6 y = 0 8x 4x ∴ y =− =− 6 3 3x i − 4x j −4 x   q = xi +  j =  3  3 x = (3 i − 4 j ) 3 Again, the projection of vector r = ± ( x1 i + x2 j ) on vector p is 2 and on q is 4. 8 x1 + 6 x2 6 x1 − 8 x2 and 4 = ∴2= 10 10 ⇒ and ⇒ and ⇒ ∴

8 x1 + 6 x2 = 20 6 x1 − 8 x2 = 40 4 x1 + 3 x2 = 10 3 x1 − 4 x2 = 20 x1 = 4 and x2 = − 2 r = ± (4 i − 2 j ) = ± 2 (2 i − j )

32 The plane is (2 + 3λ )x + (λ − 1)y + (λ + 1)z − 5λ − 3 = 0 1 Its distance from (2, 1, − 1) is . 6 (4 + 6λ + λ − 1 − λ − 1 − 5λ − 3)2 1 = ∴ (2 + 3λ )2 + (λ − 1)2 + (λ + 1)2 6 −24 or 0 ⇒ (5λ + 24)λ = 0 ⇒ λ = 5 The planes are 2 x − y + z − 3 = 0 and 62 x + 29 y + 19z − 105 = 0

33 Equation of a plane passing through the

point ( 2, − 1, − 3) and parallel to the given line is x−2 y + 1 z+ 3 3 2 −4 = 0 2 −3 2

⇒ ( x − 2)(4 − 12) − ( y + 1)(6 + 8) + (z + 3)(−9 − 4) = 0 ⇒ 8 x + 14 y + 13z + 37 = 0 i j k 34 The common normal is 2 3 4 3 4 5 r = − i + 2j − k ∴ Shortest distance = ( j + k ) ⋅

r 1 = |r | 6

35 Normal to OAB is OA × OB i j k = 1 2 1 = 5i − j − 3 k 2 1 3 Normal to ABC is i j k AB × AC = 1 −1 2 = i − 5j − 3 k −2 −1 1

373

DAY If θ is angle between the planes, then 5 + 5 + 9 19 cos θ = = 35 ⋅ 35 35

36 Let r = a1 i + a2 j + a3 k Since, r ⊥ p and r ⊥ q ∴ r ⋅ p = 0 and r ⋅ q = 0 ⇒ 2a1 − 3 a2 + 3 a3 = 0 and 4 a1 − 2a2 + a3 = 0 ⇒ 2a1 − 3 a2 + 3 a3 = 0 and 4 a1 − 2a2 + a3 = 0 a1 − a2 a3 = = =k ∴ −3 − (−6) 12 − 2 −4 − (−12) [say] a1 a2 a = = 3 =k 3 10 8 ∴ a1 = 3 k , a2 = 10k , a3 = 8 k …(i) Again, (a1 i + a2 j + a3 k ) (2 i − 4 j + 2 k ) = −12 ...(ii) ⇒ 2a1 − 4 a 2 + 2a3 = − 12 From Eqs. (i) and (ii), we get 6 k − 4 (10 k ) + 2 ( 8 k ) = − 12 6 k − 40 k + 16 k = − 12 12 2 −18 k = − 12 ⇒ k = = 18 3 20 16 ∴ a1 = 2, a2 = , a3 = 3 3 20 16 ∴ r = 2i + j+ k 3 3

38 Shortest distance

=

=

⇒

37 Plane passing through a point ( x1 , y 1 , z1 ) is A ( x − x1 ) + B ( y − y 1 ) + C ( z − z1 ) = 0 ∴ Plane through (2, 0, 0) is a( x − 2) + b ( y − 0) + c (z − 0) = 0 …(i) contains (0, 2, 0), if −2a + 2b = 0 ⇒ − a + b = 0 …(ii) Since, plane a( x − 2) + b ( y − 0) + c (z − 0) = 0 π makes an angle with the plane 3 2 x + 3 y = 5. 2a + 3b π ∴ cos = 3 a2 + b 2 + c 2 4 + 9 2a + 3 b

⇒

1 = 2

⇒

(2a + 3 b )2 1 = 4 [ (a2 + b 2 + c 2 )(13)] 2

(a2 + b 2 + c 2 )(13)

⇒ 13 (a2 + b 2 + c 2 ) = 100 a2 Q a=b ∴ 13 ( 2a2 + c 2 ) = 100 a2 ⇒ 26 a2 + 13 c 2 = 100 a2 74 a ⇒ 13c 2 = 74 a2 ⇒ c = 13 ∴

a:b :c = a: a:

74 a 13

= a : a : 5.7 a = 1 : 1 : 5.7

= = =

x2 − x1 a1 a2

y2 − y1 b1 b2

i a1 a2 −1 1 1 3 2 2 2 3 4

j b1 b2

z2 − z1 c1 c2

k c1 c2

Now,|a × b − a × c |2 = |a × b|2 + |a × c|2 − 2 (a × b ) ⋅ (a × c ) 1 0 =1+ 1−2 =1 0 1/2

43 Here, α = β = γ cos 2 α + cos 2 β + cos 2 γ = 1 1 ∴ cos α = 3 1 1 1  DC’s of PQ are  , ,   3 3 3

Q

A(–2, 3, 1)

i j k 3 2 2 2 3 4 −1 (8 − 6) + 1 (4 − 12) + 1 (9 − 4) i (8 − 6) + j (4 − 12) + k (9 − 4) −2 − 8 + 5 93 5 = 0.52 unit 93

P(–3, 5, 2)

PM = Projection of AP on PQ 1 1 = (− 2 + 3) + (3 − 5) ⋅ 3 3

39 Any point on the lines L1 and L2 are (−3 r1 − 1, 2r1 + 3, r1 − 2) and (r2 , − 3 r2 + 7, 2r2 − 7) Since, they intersect each other, therefore −3 r1 − 1 = r2 ,2r1 + 3 = − 3r2 + 7 and r1 − 2 = 2r2 − 7 On solving, we get r2 = 2 and r1 = − 1 Hence, the required point is (2, 1, − 3).

40 Let θ be an angle between unit vectors a

and b. Then, a ⋅ b = cos θ Now, |a + b|2 = |a |2 + |b|2 + 2a ⋅ b θ = 2 + 2cos θ = 4cos 2 2 θ θ ⇒ |a + b | = 2cos |a − b | = 2sin 2 2 ⇒ |a + b | + |a − b | θ θ = 2  cos + sin  ≤ 2 2  2 2

41 Let α = a + 2b + 3c , β = λb + 4c and γ = (2λ − 1)c 1 2 3 Then, [α β γ ] = 0 λ 4 [a b c ] 0 0 (2λ − 1) [α β γ ] = λ (2λ − 1) [abc ] ⇒ [α β γ ] = 0, 1 If λ = 0, [Q [a b c ] ≠ 0] 2 Hence, α, β and γ are non-coplanar for 1 all value of λ except two values 0 and . 2 1 42 Since, a ⋅ b = a ⋅ c = 0, b ⋅ c = 2 ∴ |a × b | = |a × c | = 1

Q

M

+ (1 − 2) ⋅

1 2 = 3 3

and AP = (− 2 + 3)2 + (3 − 5)2 + (1 − 2)2 =

6

AM = ( AP )2 − (PM )2 =

6−

4 = 3

14 3

44 Equation of the plane through (5, 1, 2) is a( x − 5) + b ( y − 1) + c (z − 2) = 0 …(i) Given plane (i) is perpendicular to the line x−2 y − 4 z− 5 …(ii) = = 1/2 1 1 ∴ Equation of normal of Eq. (i) and straight line (ii) are parallel a b c i.e. = = = k (say) 1/2 1 1 k ∴ a = , b = k, c = k 2 From Eq. (i), k ( x − 5) + k ( y − 1) + k (z − 2) = 0 2 ⇒ x + 2 y + 2z = 11 Any point on Eq. (ii) is  2 + λ , 4 + λ, 5 + λ      2 which lies on Eq. (iii), then λ = − 2. ∴ Required point is (1, 2, 3).

45 If the lines have a common point, then there exists λ and µ such that 6− λ =1+ µ and −1 + 2λ = − 1 + 3 µ ⇒ λ = 3, µ = 2 ∴ r = 3a + 5c

374

46 In an isosceles ∆ABC in which AB = AC the median and bisector from A must be same line, so Statement II is true. u + v Now, AD = 2 1 2 ⇒ |AD| = [|u|2 + |v|2 + 2u ⋅ v] 4 1 α = (2 + 2cos α ) = cos 2 4 2 ∴ Unit vector along 1 AD = (u + v) α 2cos 2

47 Direction ratios of PQ are 6 − 2, 5 − 1,−2 − 6 i.e. 4, 4, − 8 which are proportional to the direction ratios of the normal to the plane π, so PQ is perpendicular to π. Hence, Statement I is false and Statement II is true.

Since, it passes through (4, 4, 4), then (4 + 4 + 4 − 6) + λ(8 + 12 + 16 + 5) = 0 6 ⇒ 6 + 41λ = 0 ⇒ λ = − 41 From Eq. (i), we get 41( x + y + z − 6) −6(2x + 3y + 4z + 5) = 0 ∴ 29x + 23y + 17z = 276

49 Adjacent sides of a parallelogram are

given as a = 2 i − 4 j + 5 k and b = i − 2j − 3 k. Then, the diagonal of a parallelogram is given by v = a + b. [since, from the figure, it is clear that resultant of adjacent sides of a parallelogram is given by the diagonal] ∴ v = 2 i − 4 j + 5k + i − 2j − 3 k = (2 + 1) i + (− 4 − 2)j + (5 − 3) k =3i − 6j+ 2k D

C

48 The equation of plane through the line of intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0 is ( x + y + z − 6) + λ(2x + 3y + 4z + 5) = 0 …(i)

v a

x2 + y 2 + z2

= (3)2 + (− 6)2 + (2)2 =

9 + 36 + 4

= 49 = 7 Thus, the unit vector parallel to the diagonal is 3i − 6j + 2k v = |v| 7 3 6 2 = i − j+ k 7 7 7 Also, area of parallelogram i j k ABCD =|a × b | = 2 −4 5 1 −2 −3 = | i (12 + 10) − j (− 6 − 5) + k (− 4 + 4)| = |22 i + 11 j + 0 k | = (22)2 + (11)2 + 02 = (11)2 + (22 + 12 ) = 11 5 sq units

b

A

∴ | v| =

B

DAY THIRTY FOUR

Statistics Learning & Revision for the Day u

Measures of Central Tendency

u

Measure of Dispersion

Measures of Central Tendency A value which describes the characteristics of entire data, is called an average or a central value. Generally, an average lies in the central part of the data and therefore such values are called the measure of central tendency. There are five measures of central tendency, which are given below :

1. Mean (Arithmetic Mean) The sum of all the observations divided by the number of observations, is called mean and it is denoted by x. The most stable measure of central tendency is mean. If x1, x2,..., x n be the n observations, then mean is given by x + x2 + ... + x n ∑ x i x= 1 = n n ●

n

or x = A +

●

Σ di i =1

, where di = x i − A and A is assumed mean. n If corresponding frequencies of n observations are f1 , f2 ,K , fn, then n

x=

f1 x1 + f2 x2 + K + fn x n = f1 + f2 + K + fn

Σ fi x i

i =1 n

Σ fi

PRED MIRROR

n

or x = A +

Σ fi d i

i =1

N

Your Personal Preparation Indicator

,

i =1 n

where di = x i − A and N = ●

●

Σ fi i =1

If corresponding weights are w 1 , w 2 ,K , w n, then weighted mean, x w =

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

∑ w i xi . ∑ wi

If x1 , x2 ,..., x k be the means of k sets of observations of size n1, n2 ,K , nk respectively, then their combined mean, n x + n2 x2 + K + nk x k x1k = 1 1 n1 + n2 + K + nk Here, x1 = Mean of first set of observations n1 = Number of observations in first set

u

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

376 x2 = Mean of second set of observations n2 = Number of observations in second set and so on. NOTE

to N/2 or just greater than N/2 is called median class and in that case, N −c Median = l + 2 × h, f

• The sum of the deviations of the individual values from AM is always zero, i.e. Σ( xi − x ) = 0.

where, l = Lower limit of median class f = Frequency of median class h = Size of median class c = Cumulative frequency of class preceding the median class

• The sum of squares of deviations of the individual values is least when taken from AM i.e. Σ( xi − x ) 2 is least.

2. Geometric Mean (GM) If x1 , x2, ... x n are n observations, then nth root of the product of all observations is called geometric mean. If x1 , x2 , K , x n are n non-zero positive observations, then 1 n  GM = ( x1 ⋅ x2K x n)1 / n = antilog  log x i   n i =1    ●

Σ

●

If corresponding frequencies of each observation are f1 , f2 ,K , fn, then 1 1 n  f f GM = [ x1 1 ⋅ x2 2 K x nfn ] N = antilog  fi log x i   N i = 1 

Σ

5. Mode Mode is the observation which has maximum frequency. ●

●

f1 − f0 × h, 2 f1 − f0 − f2 where, l = Lower limit of modal class f1 = Frequency of modal class f0 = Frequency of the class preceding the modal class f2 = Frequency of the class succeeding the modal class h = Class size

Σ fi i =1

3. Harmonic Mean (HM) The harmonic mean of any set of non-zero observations, is the reciprocal of the arithmetic mean of the reciprocals of the observations. The harmonic mean of n items x1 , x2 ,..., x n is defined as

●

●

HM =

●

n 1 1 1 + +…+ x1 x2 xn

=

n

Σ

1 n i =1

1 n 1  =  1  n i = 1 x i  xi

Σ

−1

If corresponding frequencies of each observation are 1 f1 , f2 ,K , fn, then HM =   N

●

1

n

Σ

i =1

fi   xi  

−1

●

Relation between Mean, Median and Mode Mode = 3 Median − 2 Mean (Emperical formula) It is not necessary that a distribution has unique mode.

Measure of Dispersion A measure of dispersion is designed to state the extent to which the individual observations vary from their average. The commonly used measures of dispersion are :

Range

Relation between AM, GM and HM (GM ) = ( AM ) ⋅ (HM ) 2

4. Median If the observations are arranged in ascending or descending order, then the value of the middle observation is defined as the median. Let x1 , x2,.., x n be n observations, arranged in ascending or descending order, then n+1 If n is odd, then Median = th observation 2  n n    2  th +  2 + 1 th  observation  If n is even, then Median =  2 ●

●

In a continuous distribution the interval which has maximum frequency called modal class and in that case, mode = l +

n

where, N =

If x1 , x2 , … , x n are the n observations and corresponding frequencies are f1 , f2 ,… , fn , then the observation of maximum frequency is a modal value.

If in a continuous distribution the total frequency is N, then the class whose cumulative frequency is either equal

The difference between the maximum and the minimum observations is called range. i.e. Range = L − S. where, L = Maximum observation and S = Minimum observation L −S Coefficient of range = L+S

Mean Deviation (MD) The mean, of the absolute deviations of the values of the variable from a measure of their average is called Mean Deviation (MD). n ●

If x1, x2,..., x n are n observations, then MD = where, z = mean or mode or median

Σ | xi − z|

i =1

n

377

DAY ●

If corresponding frequencies of each observation are f1 , f2 ,K , fn, then

●

n

MD =

●

●

Σ fi| xi − z| i =1

n

, where N =

N Mean deviation Coefficient of MD = Corresponding average

Σ fi i =1

σ2 =

Mean deviation is least when deviations are taken from median.

NOTE

• Standard deviation is always less than range.

• Standard deviation of n natural numbers is σ = 

The square root of the arithmetic mean of the squares of deviations of the observations from their arithmetic mean is called standard deviation and it is denoted by σ.

4 • Mean deviation = σ. 5

n

●

2

n 1 n  1 x2i −  x σ= =  n i = 1 i n i =1 n   If corresponding frequencies of each observation are f1 , f2 ,K , fn, then

Σ

Σ

n

Σ fi ( x i − x )

2

i =1

σ=

N

n

where N =

Σ

i =1

=

1 N

 1   fi x i2 –   N i =1    n

Σ

n



Σ fi xi i =1 

2

,

fi

1 2  ( n − 1)  12 

1/2

.

Effect of average and dispersion on change of origin and scale

If x1 , x2 ,K , x n are n observations, then

Σ ( xi − x)2 i =1

n1 σ21 + n2 σ22 + n1d12 + n2d22 , n1 + n2

where d1 = ( x1 − x12 ), d2 = ( x2 − x12 ) n x + n2 x2 and x12 = 1 1 n1 + n2

Standard Deviation (SD)

●

Two different series having n1 and n2 observations and whose corresponding means and variances are x1, x2 and σ21 , σ22 . Then, their combined variance,

●

Change of origin

Change of scale

Mean

Dependent

Dependent

Median

Dependent

Dependent

Mode

Dependent

Dependent

Standard deviation

Independent

Dependent

Variance

Independent

Dependent

If x, me , mo and σ represent the mean, median, mode and standard deviation, respectively, of x1 , x2 , ..., x n. Then, (i) Mean of ax1 + b , ax2 + b , K , axn + b , is ax + b (ii) Median of ax1 + b , ax2 + b , K , ax n + b , is ame + b

Variance The square of SD is called variance and it is denoted by σ2 . σ Coefficient of variation = × 100% x

(iii) Mode of ax1 + b , ax2 + b , K , axn + b , is amo + b (iv) Standard deviation of ax1 + b , ax2 + b , K , ax n + b , is| a|⋅ σ .

●

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 The mean of a data set consisting of 20 observations is 40. If one observation 53 was wrongly recorded as 33, j then the correct mean will be JEE Mains 2013 (a) 41

(b) 49

(c) 40.5

(d) 42.5

99 2 If the variance of 1, 2, 3, 4, 5, ..., 10 is , then the 12 standard deviation of 3, 6, 9, 12, ..., 30 is 297 (a) 4

3 (b) 33 2

3 (c) 99 2

99 (d) 12

3 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 (a) x + n (c) x +

n+1 2

n (b) x + 2 (d) None of these

4 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 j

(a) 16.8

(b) 16.0

(c) 15.8

JEE Mains 2015

(d) 14.0

5 If the sum of deviation of a set of values x1, x 2,K xn measure from 59 is 20 and from 54 is 70, then sample size (n ) and the sample mean is (a) 10, 61

(b) 10, 55.67 (c) 6, 55.67

(d) 6, 44

6 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 in the class is (a) 40%

(b) 20%

(c) 80%

(d) 60%

378 17 If the mean deviation of number 1, 1 + d , 1 + 2d, ...,

7 The weighted mean of first n natural numbers whose

1 + 100d from their mean is 255, then d is equal to

weights are equal to the squares of corresponding numbers is (a)

n+1 2

(b)

3n (n + 1) (n + 1) (2n + 1) n (n + 1) (c) (d) 2 (2n + 1) 6 2

8 A distribution consists of three components with frequencies 20, 25 and 30 having means 25, 10 and 15 respectively. The mean of the combined distribution is (a) 14

(b) 16

(c) 17.5

(d) 20

9 A car completes the first half of its journey with a velocity v1 and the rest half with a velocity v 2. Then the average velocity of the car for the whole journey is (a)

v1 + v 2 2

(b) v1v 2

(c)

2v1v 2 v1 + v 2

(d) None of these

10 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 is (a) increased by 2 (b) decreased by 2 (c) two times the original median (d) remains the same as that of original set

(c) 32

7 5 1 1 , α − , α − 3, α − 2, α + , α − , α + 5 2 2 2 2 ( where, α > 0), then the median is α + 4, α −

5 4

13 Median of

(b) α − 2n

C0,

2n

C1,

1 2 2n

C2,

(d) α +

(c) α − 2 2n

C3, ...,

2n

2n

(b)

Cn

2n

Cn

2

(c)

2n

Cn

5 4

{x , y , 2x + y , x − y }, where 0 < y < x < 2y, are 10 and 28 respectively, then the mean of the numbers is j

JEE Mains 2013

(d) 14 j

Frequency

NCERT Exemplar

(b) 7

0-6

6-12

12-18

18-24

24-30

4

5

3

6

2

(c) 7.1

(d) 7.05

16 If the mean deviations about the median of the numbers a, 2a, … , 5 a is 50, then | a | is equal to (a) 3

j

(b) 4

(c) 5

JEE Mains 2016

(b) 3a − 32a + 84 = 0 (d) 3a 2 − 23a + 44 = 0 2

20 Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the j NCERT Exemplar correct standard deviation. (b) 10.24

(c) 10.29

(d) 10.27

60 and their arithmetic means are 30 and 25, respectively. Difference of their standard deviation is j NCERT Exemplar (a) 0

(b) 1

(c) 1.5

(d) 2.5

22 If MD is 12, the value of SD will be (a) 15

(b) 12

(c) 24

(d) None of these

a σx p

(b)

p σx a

(c)

p σx a

(d)

10 + 20 10

a σx p

(b) 10 + 10 (d) None of these

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 2 g. The correct mean and standard deviation (in j gram) of fishes are respectively AIEEE 2011 (b) 32, 2

(c) 32, 4

(d) 28, 2

26 The variance of first 50 even natural numbers is j

Class interval

(a) 7.08

(a) 3a − 26a + 55 = 0 (c) 3a 2 − 34a + 91 = 0 2

(a) 28, 4

15 Find the mean deviation from the median of the following data.

3.5, then which of the following is true?

25 A scientist is weighing each of 30 fishes. Their mean

14 If the median and the range of four numbers

(c) 5

(d) x 201

19 If the standard deviation of the numbers 2, 3, a and 11 is

(a) (c)

+1

(d) None of these

(b) 10

(c) x101

−5 , − 4 , − 3, − 2, − 1, 0, 1, 2, 3, 4, 5 is 10. The standard deviation of the observations 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 will be

Cn (where, n is

2

(a) 18

(b) x1

24 If the standard deviation of the observations

2 −1

(a) x110

(a)

even) is (a)

(d) 20.2

ax + b 23 If SD of variate x is σ x , then the SD of , ∀ a, b, p ∈ R p is

(d) 34

12 If a variable takes the discrete values

(a) α −

(c) 10.1

21 Coefficient of variation of two distributions are 50 and

observations with values 8 and 32 are further included, then the median of the new group of 21 observations will be (b) 30

(b) 20.0

It is given that x1 < x 2 < ... < x 200 < x 201. Then, the mean deviation of this set of observations about a point k is minimum when k is equal to

(a) 10.20

11 The median of 19 observations of a group is 30. If two

(a) 28

(a) 10.0

18 Consider any set of 201 observations x1, x 2,K , x 200, x 201.

(d) 2

833 (a) 4

(b) 833

(c) 437

JEE Mains 2014

(d)

437 4

27 Mean of 5 observations is 7. If four of these observations are 6, 7, 8, 10 and one is missing, then the variance of all j the five observations is JEE Mains 2013 (a) 4

(b) 6

(c) 8

(d) 2

379

DAY 28 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? (a) a = 3, b = 4 (c) a = 5, b = 2

(b) a = 0, b = 7 (d) a = 1, b = 6

29 In an experiment with 15 observations on x, the following results were available Σx = 2830, Σx = 170. One observation that was 20 was found to be wrong and was replaced by the correct value 30. Then, the corrected variance is 2

(a) 78.0

(b) 188.66

(c) 177.33

(d) 8.33

30 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 (a)

5 2

(b)

11 2

(c) 6

(d)

13 2

31 If a variable x takes values xi such that a ≤ xi ≤ b, for i = 1, 2 ,..., n, then

(a) a ≤ Var (x) ≤ b a2 (c) ≤ Var (x) 4 2

2

(b) a ≤ Var (x) ≤ b (d) (b − a) 2 ≥ Var (x)

32 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 after the grace j marks were given? JEE Mains 2013 (a) Mean

(b) Median

(c) Mode

(d) Variance

33 Let x1, x 2,K , xn be n observations. Let w i = l ⋅ xi + k for

i = 1, 2,..., n, where l and k are constants. If the mean of xi ’s is 48 and their standard deviation is 12, the mean of w i ’s is 55 and standard deviation of w i ’s is 15. The values j NCERT Exemplar of l and k should be (a) l = 1. 25, k = −5 (c) l = 2 .5, k = −5

(b) l = −1. 25, k = 5 (d) l = 2 .5, k = 5

34 If n is a natural number, then Statement I The mean of the squares of first n natural (n + 1)( 2n + 1) . 6 n (n + 1) Statement II Σn = 2

number is

(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

35 Let x1, x 2,… , xn be n observations and let x be their arithmetic mean and σ 2 be the variance.

Statement I Variance of 2x1, 2x 2,… ,2xn is4σ 2 . Statement II Arithmetic mean 2x1 , 2x 2,… , 2xn is 4x . j AIEEE 2012 (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

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 The mean age of a combined group of men and women is 25 yr. If the mean age of the group of men is 26 and that of the group of women is 21, then the percentage of men and women in the group is (a) 40, 60

(b) 80, 20

(c) 20, 80

formed using each of the digits 3, 5, 7 and 9 exactly once in each number, is (b) 5555

(c) 6666

(d) 7777

3 Suppose a population A has 100 observations 101, 102, ......, 200 and another population B has 100 observations 151, 152, ......, 250. If VA and VB represent the variances V of the two populations respectively, then A is equal to VB 9 (a) 4

4 (b) 9

2 (c) 3

frequency 1 and n is even, then the mean deviation from mean equals to (a) n

(d) 60, 40

2 The average of the four-digit numbers that can be

(a) 4444

4 If the value observed are 1, 2, 3, ..., n each with

(d) 1

(b)

n 2

(c)

n 4

(d) None of these

5 In a class of 19 students, seven boys failed in a test. Those who passed scored 12, 15, 17, 15, 16, 15, 19, 19, 17, 18, 18 and 19 marks. The median score of the 19 students in the class is (a) 15

(b) 16

(c) 17

(d) 18

6 If G is the geometric mean of the product of r sets of observations with geometric means G1, G 2, G 3, K , Gr respectively, then G is equal to (a) log G1 + log G2 + K + log Gr (b)G1 ⋅ G2 K Gr (c) log G1 ⋅ log G2 ⋅ log G3 K log Gr (d) None of the above

380 7 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 marks of the girls? (a) 73

(b) 65

(c) 68

were raised to a maximum of 100 and variance of new marks was calculated. The new variance is (a) 81

measure 100 mile 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 the aeroplane around the square is (b) 195 m/h

(c) 192 m/h

(a) 12

(c) 6

(b)

n (n + 1) 2

(c)

(a) 2

(d) 3.52

n (n − 1) 2

(d)

(d) 16

(b)

2

(c) 4

9

9

i =1

i =1

(d) 2 2

14 If ∑ ( xi − 5) = 9 and ∑ ( xi − 5)2 = 45, then the standard deviation of the 9 items xi , x 2, .., x 9 is (a) 9

frequencies proportional to the binomial coefficients n C0,n C1,...,n Cn , then mean of the distribution is n 2

(c) 14

and the remaining half are equal to − a. If the standard deviation of all the observations is 2, then the value of | a | j JEE Mains 2013 is

10 If a variable takes the values 0 , 1, 2 , ... ,n with

(a)

(b) 9

13 In a set of 2n observations, half of them are equal to a

(d) 200 m/h

SD = 3. If the whole group has 250 items with mean 15.6 and SD = 13.44, the SD of the second group is (b) 5

(d) 125

is

9 The first of two samples has 100 items with mean 15 and

(a) 4

(c) 144

Σxi = 80. Then, a possible value of n among the following

(d) 74

8 An aeroplane flies around a squares the sides of which

(a) 190 m/h

(b) 122

12 If x1, x 2,..., xn are n observations such that Σxi2 = 400 and

(b) 4

j

(c) 2

JEE Mains 2018

(d) 3

15 If x1 and x 2 are the means of two distributions such that x1 < x 2 and x is the mean of the combined distribution, then

2 n

(a) x < x1 x + x2 (c) x = 1 2

11 The marks of some students were listed out of 75. The SD of marks was found to be 9. Subsequently the marks

(b) x > x 2 (d) x1 < x < x 2

ANSWERS (a) (b) (a) (d)

SESSION 1

1. 11. 21. 31.

SESSION 2

1. (b) 11. (c)

2. 12. 22. 32.

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

2. (c) 12. (d)

3. 13. 23. 33.

(c) (a) (a) (a)

3. (d) 13. (a)

4. 14. 24. 34.

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

4. (c) 14. (c)

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

6. (c) 16. (b) 26. (b)

7. (b) 17. (c) 27. (a)

8. (b) 18. (c) 28. (a)

9. (c) 19. (b) 29. (a)

10. (d) 20. (b) 30. (b)

5. (a) 15. (d)

6. (b)

7. (b)

8. (c)

9. (a)

10. (a)

5. 15. 25. 35.

Hints and Explanations SESSION 1 1 Given; Σx incorrect = 40 20

Σx incorrect = 20 × 40 = 800 Σxcorrect = 800 − 33 + 53 = 820 Σxcorrect 820 ⇒ = 20 20 ∴Correct mean = 41 ⇒ ∴

2 Given, σ = 99 = 33 ⇒ σ = 33 12 4 2 Clearly, SD of required series 3 33 = 3σ = 2 2

3 Let the observations be x1 , x2 , ..., x n .

x1 + x2 + K + x n n Now, when the first term is increased by 1, second term by 2 and so on, then the observations will be ( x1 + 1), ( x2 + 2), ..., ( x n + n ). Then, x =

∴ New mean ( x1 + 1) + ( x2 + 2) + K + ( x n + n ) n ( x + x2 + K + x n ) (1 + 2 + K + n ) = 1 + n n n (n + 1) n+1 = x+ = x+ 2n 2 =

4 Given, ⇒

16

x1 + x2 + x3 + ... + x16 = 16 16

∑ x i = 16 × 16 i =1

Sum of new observations =

18

∑ y i = (16× 16 − 16) + (3 + 4 + 5) = 252 i =1

Number of observations = 18 18

∴ New mean =

5 We have,

∑ yi i =1

18

=

252 = 14 18

n

∑ ( x i − 59) = 20

i =1

…(i)

381

DAY n

∑ ( x i − 54) = 70

and

…(ii)

11 Since, there are 19 observations. So, the middle term is 10th. After including 8 and 32, i.e. 8 will come before 30 and 32 will come after 30. Here, new median will remain 30.

i =1

From Eq. (i), we get n

⇒

∑

x i − 59n = 20

∑

x i = 20 + 59n

i =1 n

…(iii)

ascending order. 7 5 1 α − , α − 3, α − , α − 2, α − , 2 2 2 1 α + , α + 4, α + 5 2 1 ∴ Median = [ Value of 4th item 2 + Value of 5th item] 1 5 α −2+ α − 2α − 5 2 2 = = =α − 2 2 4

and similarly, from Eq. (ii), we get n

54n

…(iv)

i =1

From Eqs. (iii) and (iv), we get 70 + 54n = 20 + 59n ⇒ 5n = 50 ⇒ n = 10 Now, from Eq. (iii), we get n

∑

x i = 610

13 Total number of terms = n + 1

i =1

∴ Sample mean = 61

n + 1 + 1 ∴ Median =   th term   2 n =  + 1 th term = 2 nC n /2 2 

6 Let the number of boys and girls be x

14 First we arrange four numbers according to the condition 0 < y < x < 2 y i.e. x − y , y , x, 2 x + y 2nd term + 3rd term Median = = 10 2 ⇒ y + x = 20 …(i) Range = (2 x + y ) − ( x − y ) = 28 ⇒ x + 2 y = 28 …(ii) On solving Eqs. (i) and (ii), we get x = 12, y = 8 So, four numbers are 4, 8, 12, 32. 4 + 8 + 12 + 32 56 = = 14 ∴ Mean = 4 4

7 Required mean =

1 ⋅ 12 + 2 ⋅ 22 + K + n ⋅ n2 12 + 22 + K + n2

n2 (n + 1)2 13 + 23 + K + n3 4 = 2 = n (n + 1) (2n + 1) 1 + 22 + K + n2 6 n (n + 1) 3n (n + 1) 6 = × = 4 2n + 1 2 (2n + 1)

8 Here, x1 = 25, x2 = 10, x3 = 15 and n1 = 20, n2 = 25, n3 = 30 Now, combined mean n x + n2 x2 + n3 x3 x13 = 1 1 n1 + n2 + n3 500 + 250 + 450 1200 = = = 16 75 75

9 Clearly, average velocity = HM of (v 1 , v 2 ) =

2v 1 v 2 v1 + v2

10 After arranging the terms in ascending order median is the  

n + 1  th term, 2 

i.e. 5th term. Here, we increase largest four observations of the set which will come after 5th term. Hence, median remains the same as that of original set.

50

15 Class Mid f cf |x − M d| f|x − M d| interval value x i 0-6

3

4 4

11

44

6-12

9

5 9

5

25

12-18

15

3 12

1

3

18-24

21

6 18

7

42

24-30

27

2 20

13

26

Total

20

140

N 20 = = 10, which lies in the 2 2 interval 12-18. l = 12, cf = 9, f = 3 10 − 9 ∴ Md = 12 + ×6 3 N   − cf   2 Q M = l + × h d   2     = 12 + 2 = 14 Σf | x − Md | 140 ∴Mean deviation = i i = =7 20 N Now,

=

∑ x i − median

i =1

n 1 ⇒ 50 = 50 {2|a|( ⋅ 0.5 + 1.5 + 2.5 + … + 24.5)} 25 ⇒ 2500 = 2|a|⋅ (2 × 0.5 + 24 × 1 ) 2 25 = 2 a ⋅ ( 25) 2 ∴ |a|= 4

17 Clearly,

(which is odd)

and y. ∴ 52 x + 42 y = 50 ( x + y ) ⇒ 52 x + 42 y = 50 x + 50 y ⇒ 2x = 8y ⇒ x = 4y ∴ Total number of students in the class = x + y = 4 y + y = 5y ∴ Required percentage of boys  4y  = × 100 % = 80%  5y 

25a + 26 a = (25.5) a 2 Mean deviation about median

12 Firstly arrange the given data in

i =1

∑ x i = 70 +

16 Median of a, 2 a, 3 a, 4 a, …, 50 a is

n (a + l ) Sum of quantities = 2 n n 1 = [1 + 1 + 100d ] = 1 + 50d 2 1 Now, MD = Σ| x i − x| ⇒ 255 n 1 = [50d + 49d + 48d 101 + K + d + 0 + d + K + 50d ] 2d  50 × 51  =    101  2 255 × 101 ∴d = = 101 . 50 × 51 ( x) =

18 Given that, x1 < x2 < x3 < K < x201 Hence, median of the given observation 201 + 1  =   th item = x101   2 Now, deviation will be minimum of taken from the median. Hence, mean deviation will be minimum, if k = x101 .

19 We know that, if x1 , x2 , ..., x n are n observations, then their standard deviation is given by

1 Σx Σx2i −  i   n  n

2

(22 + 32 + a2 + 112 ) 4 2 2 + 3 + a + 11   −    4 2 49 4 + 9 + a2 + 121  16 + a  = − ⇒   4  4 4 2 2 256 + a + 32a 49 134 + a ⇒ = − 4 4 16 49 4a2 + 536 − 256 − a2 − 32a ⇒ = 4 16 ⇒ 49 × 4 = 3a2 − 32a + 280 ⇒ 3a2 − 32a + 84 = 0 We have, (3.5)2 =

20 ∴ New mean, x=

100 × 40 + 3 + 27 − 30 − 70 100

382 4000 − 70 3930 = = 393 . 100 100 2 2 2 Q Σx = N ( σ + x ) =

+ (10 − 7 )2 + (4 − 7 )2 5 12 + 0 + 12 + 32 + 32 20 = = =4 5 5 =

∴ SD = 100 (100 + 1600) = 170000 New Σx2 =170000 − (30)2 − (70)2 + (3)2 +(27)2

= 170000 − 900 − 4900 + 9 + 729 = 164938

∴ New SD =

New Σx2 − (New x )2 N

164938 − (393 . )2 100 = 164938 . − 1544.49 = 104.89 =

= 10.24

21 Given, coefficient of variation, C1 = 50 and coefficient of variation, C2 = 60 We have, x1 = 30 and x2 = 25 σ σ Q C = × 100 ⇒ 50 = 1 × 100 x 30 σ ⇒ σ 1 = 15 and 60 = 2 × 100 25 ⇒ σ 2 = 15 ∴ Required difference, σ 1 − σ 2 = 15 − 15 = 0 22 We know that MD = 4 SD 5 5 5 ∴ SD = MD = × 12 = 15 4 4 ax + b ax + b 23 Let u = , then u = p p ∴ SD = =

Σ (u − u )2 = Σf a2 p2

σ 2x =

a2 Σ ( x − x )2 Σf p2

a σx p

24 The new observations are obtained by adding 20 to each previous observation. Hence, the standard deviation of new observations will be same i.e. 10.

25 Correct mean = Old mean +2 = 30 + 2 = 32 As standard deviation is independent of change of origin. ∴ It remains same. ⇒ Standard deviation = 2 26 Here, x = Σx i n 2 + 4 + 6 + 8 + K + 100 = 50 50 × 51 = = 51 50 [Q Σ2n = n(n + 1), here n = 50] 1 Variance, σ 2 = Σx2i − ( x )− 2 n 1 2 2 σ2 = (2 + 4 + K + 1002 ) − (51)2 = 833 50 6 + 7 + 8 + 10 + x 27 Mean, 7 = 5 ⇒

x=4

⇒

Variance (6 − 7 )2 + (7 − 7 )2 + (8 − 7 )2

28 According to the given condition, (6 − a) 2 + (6 − b ) 2 + (6 − 8) 2    + (6 − 5) 2 + (6 − 10) 2  6.80 =  5 2 2 ⇒ 34 = (6 − a) + (6 − b ) + 4 + 1 + 16 ⇒

(6 − a) 2 + (6 − b ) 2 = 13 = 9 + 4

⇒

(6 − a) 2 + (6 − b ) 2 = 32 + 22 a = 3, b = 4

∴

2 x1 + 2 x2 + 2 x3 +…+ 2 x n n x1 + x2 + x3 +…+ x n   =2  = 2x   n

and AM =

Since, one observation 20 was replaced by 30, then Σ x2 = 2830 − 400 + 900 = 3330 and Σ x = 170 − 20 + 30 = 180

SESSION 2 1 Let x be the number of men and y be

2

2

2

3330  180  −   15  15 3330 − 2160 = = 78.0 15

=

30 Here, σ 21 = 4, σ 22 = 5, x1 = 2, x2 = 4, n1 = n2 = 5 Clearly, combined mean n x + n2 x2 x12 = 1 1 n1 + n2 1 = (2 + 4) = 3 2 Combined variance,

1 11 = [4 + 5 + 1 + 1] = 2 2

31 Since, SD < Range ⇒ σ ≤ (b − a) σ 2 ≤ (b − a)2 ⇒ (b − a)2 ≥ Var ( x )

32 Since, variance is independent of change of origin. So, variance will not change whereas mean, median and mode will increase by 10.

33 Given, w i = lx i + k x = 48, σ ( x i ) = 12 55 and σ (w ) = 15 IM ( x i ) + M (k ) l × 48 + k lσ ( x i ) + σ (k )

the number of women. Then, we have x ⋅ 26 + y ⋅ 21 25 = x+ y  n1 x1 + n2 x2  QCombined mean = n1 + n2   ⇒ 25x + 25y = 26 x + 21 y ⇒ x = 4y Now, percentage of men x = × 100 x+ y 4y = × 100 = 80% 5y and hence percentage of women = 20%

n1 σ 21 + n2 σ 22 + n1 d 12 + n2 d 22 n1 + n2

M( xi ) = M (w ) = M (w i ) = 55 = and σ (w i ) =

and Statement II is also a true Statement. 22 ⋅ σ 2 = 4σ 2 .

and Σ x = 170

σ2 =

Σn2 n n(n + 1)(2n + 1) (n + 1)(2n + 1) = = 6n 6

34 Required mean =

35 Clearly, variance of 2 x1 , 2 x2 , ..., 2 x n is

29 Given, n = 15, Σ x2 = 2830

Σ x2  Σ x  − ∴ Variance, σ =   n  n

15 = l(12) + 0 15 ⇒ l = = 1.25 12 From Eq. (i), 55 = 1.25 × 48 + k ⇒ k = 55 − 60 ∴ k = −5

…(i)

2 Number of required four-digit numbers = 4 × 3 × 2 × 1 = 24 and the sum of all the required four-digit numbers  104 − 1  = (3 + 5 + 7 + 9) × (4 − 1)! ×    10 − 1  9999 9 Now, required average 24 × 6 × 9999 = = 6666 9 × 24

= 24 × 6 ×

3 Since, variance is independent of change of origin. So, variance of observations 101, 102, 200 is same as variance of 151, 152, ..., 250. ∴ V A = VB VA ⇒ =1 VB

383

DAY Total marks of 30 girls

4 Clearly, mean

1+ 2+ 3+ K+ n ( x) = n n(n + 1) n + 1 = = 2n 2 Now, mean deviation from mean 1 = Σ| x i − x| n n+1 n+1 1  = 1− + 2− n  2 2

n+1  2  3− n n−1 1 1 − n = + +K+ n  2 2 2  1 n − 1 n − 3 = + n  2 2 n − 3 n − 1 +K+ + 2 2  1 1 = [(n − 1) + (n − 3) + K + 1] n 2 1 n n [n − 1 + 1] = 2n  2  4 +K+ n−

5 Clearly, median score = score of  19 + 1  = 10th student.    2  Since, seven boys are failed out of 19, therefore according to given information there score will be less than 12. Now, on arranging these scores in ascending order, we get score of 10th student is 15.

6 Taking X as the product of variates X 1 , X 2 , ..., X r , we get log X = log X 1 + log X 2 + log X 3 + K + log X r ⇒ Σ log X = Σ log X 1 + Σ log X 2 + Σ log X 3 + K + Σ log X r 1 1 Σ log X = Σ log X 1 ⇒ n n 1 1 + Σ log X 2 + Σ log X 3 n n 1 + K + Σ log X r n ⇒ log G = log G1 + log G2 + K + log G r ⇒ G = G1 ⋅ G2 ⋅ K G r

7 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

= 7200 − 5250 = 1950 ∴ Average marks of 30 girls

12 Given that, Σ x2i = 400 and Σ x i = 80, since σ 2 ≥ 0 2

1950 = 65 30

⇒

Σ x2i Σ xi  −   ≥0  n  n

8 Using the harmonic mean formula,

⇒

400 6400 − ≥0 n n2

=

1 1 n fi 1 = ∑ ⇒H= 1 n f H N i = 1 xi ∑ i N i = 1 xi ∴ Average speed =

400 1 1 1 1   100  + + +   100 200 300 400 

= 192 m / h

∴

n ≥ 16 (a + a + ... + n times ) 13 Mean = + (− a − a − ... − n times) 1444442444443 2n =0 ∴

9 We know,

(a)2 2n = |a | 2n

2=

14 Since, standard deviation is remain unchanged, if observations are added or subtracted by a fixed number 9

∑ (x1

− 5) = 9

∑ (x1

− 5)2 = 45

We have,

i =1 9

and

i =1

9

SD =

10 Here, N =

a2 + a2 + ... + 2n times 2n

=

n (σ 2 + d 12 ) + n2 (σ 22 + d 22 ) , σ2 = 1 1 n1 + n2 where d 1 = m1 − a, d 2 = m2 − a, a being the mean of the whole group. 100 × 15 + 150 × m2 ∴ 15.6 = 250 ⇒ m2 = 16 Thus, (100 × 9 + 150 × σ 2 ) + 100   × (0.6)2 + 150 × (0.4)2  13.44 =  250 ⇒ σ =4

( x i − x )2 2n

SD =

n

∑ fi = k (

i =1

n

constant of proportionality. and Σ f i x i = k (1 ⋅ nC1 + 2 ⋅ nC2 + K + n ⋅ nC n ) (n − 1)(n − 2) = kn 1 + (n − 1) + + K + 1   2! = kn 2n − 1 1 2n

i =1

9

C 0 + nC1 + ... + nC n )

= k (1 + 1)n = k 2n where, k is a

∴ Mean, x =

2 ∑ (x1 − 5)

( n ⋅ 2n − 1 ) =

n 2

11 Given, σ = 9 Let a student obtains x marks out of 75. 4x Then, his marks out of 100 are . Each 3 4 observation is multiply by . 3 4 ∴ New SD, σ = × 9 = 12 3 Hence, variance is σ 2 = 144.

⇒

SD =

⇒

SD =

45  9  −   9 9 5− 1 =

  9  ∑ ( x1 − 5) i = 1  −   9    

2

2

4=2

15 Let n1 and n2 be the number of observations in two distributions having means x1 and x2 respectively. Then, n x + n2 x2 x= 1 1 n1 + n2 Now consider, n x + n2 x2 x − x1 = 1 1 − x1 n1 + n2 n ( x − x1 ) = 2 2 > 0 [Q x2 > x1 ] n1 + n2 ⇒

x > x1

…(i)

n ( x − x2 ) Similarly, x − x2 = 1 1 0 n

and

∑p

i

= 1, where i = 1, 2, n

i =1

If a random variable X assumes values x1 , x2 , x3 , … , x n with the respective probabilities p1 , p2 ,… , pn , then variance of X is 2 n  n  given by Var ( X ) = ∑ pi x2i −  ∑ pi x i  .  i =1  i =1

Bernoulli Trials and Binomial Distribution Bernoulli Trials Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions: (i) There should be a finite number of trials (ii) The trials should be independent (iii) Each trial has exactly two outcomes; success or failure (iv) The probability of success (or failure) remains the same in each trial.

Binomial Distribution

Mean If a random variable X assumes values x1 , x2 , x3 , … , x n with respective probabilities p1 , p2 , p3 , … , pn , then the mean X of X is defined as n

X = p1 x1 + p2 x2 + ... + pi x i or X = ∑ pi x i i =1

The mean of a random variable X is also known as its mathematical expectation and it is denoted by E (X).

The probability of r successes in n independent Bernoulli trials is denoted by P ( X = r ) and is given by P( X = r ) = nCr prq n − r where, p = Probability of success and q = Probability of failure and p + q = 1 (i) Mean = np (ii) Variance = npq (iii) Mean is always greater than variance.

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 If three distinct numbers are chosen randomly from the

4 The probability that atleast one of the events A and B

first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is

occur is 3/5. If A and B occur simultaneously with probability 1/5 , then P ( A′ ) + P (B ′ ) is equal to

(a)

4 55

(b)

4 35

(c)

4 33

(d)

4 1155

2 If the letters of the word ‘MATHEMATICS’ are arranged arbitrarily, the probability that C comes before E, E before H, H before I and I before S, is (a)

1 75

(b)

1 24

(c)

1 120

(d)

1 720

3 If A and B are two events, then the probability that exactly one of them occurs is given by (a) (b) (c) (d)

P (A) + P (B) − 2 P (A ∩ B) P (A ∩ B ′) − P (A′ ∩ B) P (A ∪ B) + P (A ∩ B) P (A′) + P (B ′) + 2P (A′ ∩ B ′)

(a)

2 5

(b)

4 5

(c)

6 5

(d)

7 5

5 A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then, P ( A ∪ B ) is j

2 (a) 5 (c) 0

AIEEE 2008

3 (b) 5 (d) 1

6 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

387

DAY 1 , then the probability that 16 j JEE Mains 2017 atleast one of the events occurs, is occur simultaneously) =

(a)

7 32

(b)

7 16

(c)

7 64

(d)

3 16

7 In a leap year, the probability of having 53 Sunday or 53 Monday is 2 (a) 7

j

3 (b) 7

4 (c) 7

NCERT Exemplar 5 (d) 7

8 A number is chosen at random among the first 120 natural numbers. The probability of the number chosen being a multiple of 5 or 15 is (a)

1 8

(b)

1 5

(c)

1 24

(d)

1 6

9 Consider two events A and B. If odds against A are as 1 ≤ P (B) ≤ 2 1 (c) ≤ P (B) ≤ 4

3 4 3 5

(b)

5 3 ≤ P (B) ≤ 12 4

P ( A ∪ B ) = 0.5 . Then, P (B / A ) is equal to j

3 (c) 10

NCERT Exemplar 1 (d) 5

1 2

(a)

1 12

(b)

3 4

(c)

j

1 4

NCERT Exemplar 3 (d) 16

1 4

12 It is given that the events A and B are such that P ( A ) = , P( A / B ) =

1 2 and P (B / A ) = . Then, P (B ) is equal to 2 3

AIEEE 2008 2 (d) 3 j

1 (a) 2

1 (b) 6

1 (c) 3

13 Let A and B be two events such that P ( A ) = 0 . 6,

P (B ) = 0 . 2 and P ( A / B ) = 0 . 5. Then, P ( A′ /B′ ) equals j

1 (a) 10

3 (b) 10

3 (c) 8

NCERT Exemplar 6 (d) 7

14 If two events A and B are such that P ( A′ ) = 0.3, P (B ) = 0.4  B  and ( A ∩ B′ ) = 0.5, then P   is equal to  A ∪ B′  (a)

1 4

(b)

1 5

(c)

3 5

(d)

2 5

then the correct statement among the following is (b) P (C / D) < P (C) (d) P (C / D) = P (C)

17 An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the sum of the numbers obtained is noted. If the result is a tail, a card from a well-shuffled pack of eleven cards numbered 2, 3, 4, ..., 12 is picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is 192 401

(b)

193 401

3 10 3 (c) 5

(c)

193 792

(d)

17 75

2 5 (d) None of these

(a)

(b)

 A B

19 For two events A and B, if P ( A ) = P   =

1 and 4

A′ 3 (a) A and B are independent (b) P   = B 4 B′ 1 (c) P   = (d) All of these  A′  2

1 6

20 Let A and B be two events such that P ( A ∪ B ) = , 1 1 and P ( A ) = , where A stands for the 4 4 complement of the event A. Then, the events A and B are

P( A ∩ B ) =

j

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

JEE Mains 2014

independent but not equally likely independent and equally likely mutually exclusive and independent equally likely but not independent

1 , i +1 i = 1, 2, .., n. Then, the probability that none of the events will occur is

21 For independent events A1, ..., An ⋅ P ( Ai ) =

(a)

15 If C and D are two events such that C ⊂ D and P (D ) ≠ 0, (a) P (C / D) ≥ P (C) P (D) (c) P (C / D) = P (C)

1 6 1 2

B 1 P   = , then  A 2

1 3

11 If A and B are two events such that P ( A ) = , P (B ) = ,  A 1 P   = , then P ( A′ ∩ B′ ) is equal to B 4

1 , P (F ) = 2 1 (d) P (E ) = , P (F ) = 3 (b) P (E ) =

drawn one-by-one. The probability that they are of different colours, is

(d) None of these

1 (b) 2

1 1 , P (F ) = 3 4 1 (c) P (E ) = 1, P (F ) = 12 (a) P (E ) =

18 A bag contains 3 red and 3 white balls. Two balls are

10 A and B are events such that P ( A ) = 0.4, P (B ) = 0.3 and

2 (a) 3

P (E ) > P (F ). The probability that both E and F happen is 1 1 and the probability that neither E nor F happens is , 12 2 then

(a)

2 : 1 and those in favour of A ∪ B are as 3 : 1, then (a)

16 Let E and F be two independent events such that

n (n + 1)

(b)

(n − 1) (n + 1)

(c)

1 (n + 1)

 1  (d) n +    (n + 1) 

22 Let 0 < P ( A ) < 1, 0 < P (B ) < 1 and P ( A ∪ B ) = P ( A ) + P (B ) − P ( A ) ⋅ P (B ). Then,

B (a) P   = P (B) − P (A)  A

(b) P (Ac ∪ B c ) = P (Ac ) + P (B c )

A (c) P (A ∪ B)c = P (Ac ) ⋅ P (B c ) (d) P   = P (B / A) B 

FIVE

388 23 Let two fair six-faced dice A and B be thrown

30 A person goes to office either by car, scooter, bus or

simultaneously. If E 1 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 E 2 are independent (b) E 2 and E 3 are independent (c) E1 and E 3 are independent (d) E1 , E 2 and E 3 are not independent

(a)

24 Let A, B, C be three mutually independent events. Consider the two Statements S1 and S 2 S1 : A and B ∪ C are independent S 2 : A and B ∩ C are independent Then, (a) Both S1 and S2 are true (c) Only S2 is true

(b) Only S1 is true (d) Neither S1 nor S2 is true

1 and the 10 3 probability that neither of them occurs is . Then, the 10 probability of occurrence of event B is (b)

4+ 7 3

(c)

4+ 6 2

(d)

4− 6 10

26 A committee of 4 students is selected at random from a group consisting 8 boys and 4 girls. Given that there is atleast one girl on the committee, the probability that there are exactly 2 girls on the committee, is j

(a)

7 99

(b)

13 99

(c)

14 99

NCERT Exemplar

(d) None of these

27 Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only, if the first misses the target. The probability that the target is hit by the second plane, is (a) 0.06

(b) 0.14

(c) 0.32

(d) 0.7

28 Three machines E 1, E 2 and E 3 in a certain factory produce 50%, 25% and 25%, respectively, of the total daily output of electric tubes. It is known that 4% of the tubes produced on each of machines E 1 and E 2 are defective and that 5% of those produced on E 3 are defective. If on tube is picked up at random from a day’s production, the j probability that it is defective, is NCERT Exemplar (a) 0.025

(b) 0.125

(c) 0.325

(d) 0.0425

29 A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is j red, is JEE Mains 2018 (a)

3 10

(b)

2 5

(c)

1 5

(b)

1 9

(c)

2 11

(d)

1 7

31 A pack of playing cards was found to contain only

2 3

(a)

probability that both A and B occurs is

4− 7 3

1 5

51 cards. If the first 13 cards which are examined are all red, then the probability that the missing card is black, is

25 Two independent events namely A and B and the

(a)

1 3 2 1 , , and , 7 7 7 7 respectively. The probability that he reaches office late, if 2 1 4 1 he takes car, scooter, bus or train is , , and , 9 9 9 9 respectively. If he reaches office in time, the probability that he travelled by car is train the probabilities of which being

(d)

3 4

(b)

15 26

(c)

16 39

(d)

37 52

32 A discrete random variable X has the following probability distribution. X

1

2

3

4

5

6

7

P( X )

C

2C

2C

3C

C2

2C 2

7C 2 + C

The value of C and the mean of the distribution are j NCERT Exemplar 1 (b) and 2.66 20

1 (a) and 3.66 10 1 (c) and 1.33 15

(d) None of these

33 For a random variable X , E ( X ) = 3 and E ( X 2 ) = 11. Then, variable of X is (a) 8

(b) 5

(c) 2

(d) 1

34 A random variable X has the probability distribution X

1

2

3

4

5

6

7

8

P(X)

0.15

0.23

0.12

0.10

0.20

0.08

0.07

0.05

For the events E = {X is a prime number} and F = {X < 4}, then the probability P (E ∪ F ) is (a) 0.87

(b) 0.77

(c) 0.35

(d) 0.50

35 Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards, then the mean of the number of aces is (a)

1 13

(b)

3 13

(c)

2 13

(d) None of these

36 Consider 5 independent Bernoulli’s trials each with probability of success p. If the probability of atleast one failure is greater than or equal to 31/32, then p lies in the interval  3 11 (a)  ,   4 12 

1 (b)  0,   2 

11 (c)  , 1  12 

1 3 (d)  ,   2 4 

37 A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to getting 9 heads, then the probability of getting 2 heads is (a)

15 28

(b)

2 15

(c)

15 213

(d) None of these

389

DAY 38 A fair coin is tossed n times. If X is the number of times heads occur and P ( X = 4), P ( X = 5) and P ( X = 6) are in AP, then n is equal to (a) 13

(b) 7

(c) 11

(d) None of these

39 A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn one-by-one with replacement, then the variance of the number of green balls drawn is j

(a)

12 5

(b) 6

(c) 4

(d)

6 25

40 A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is j JEE Mains 2013 (a)

JEE Mains 2017

17 35

(b)

13 35

(c)

11 35

(d)

10 35

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 7 white balls and 3 black balls are placed in a row at random. The probability that no two black balls are adjacent is 1 (a) 2

7 (b) 15

2 (c) 15

(a) 0.50

1 (d) 3

3x + 1 1−x 1 − 2x and P (C ) = . Then, the , P (B ) = 3 4 2 set of possible values of x are in the interval

P( A ) =

1 2 (b)  ,   3 3 

1 13 (c)  ,   3 3 

probability that one of the boxes contains exactly 3 balls, j is JEE Mains 2015 55  2    3  3

11

2 (b) 55    3

10

1 (c) 220   3

12

1 (d) 22    3

11

4 If 6 objects are distributed at random among 6 persons, the probability that atleast one person does not get any object is (a)

313 324

(b)

315 322

(c)

317 324

(d)

319 324

5 10 apples are distributed at random among 6 persons. The probability that atleast one of them will receive none is (a)

6 143

14

(b)

15

C4 C5

(c)

137 143

(d)

135 143

6 A draws two cards at random from a pack of 52 cards. After returning them to the pack and shuffling it, B draws two cards at random. The probability that their draws contain exactly one common card is (a)

25 546

(b)

50 663

(c)

25 663

(c) 0.70

(d) 0.90

asked. Suppose a students tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers true, if it falls tails, he answers false. The probability that he answers atleast 12 questions correctly is 20

(d) [0, 1]

3 If 12 balls are to be placed in 3 identical boxes, then the

(a)

(b) 0.20

8 In an examination, 20 questions of true-false type are

2 Events A , B and C are mutually exclusive events such that

1 1 (a)  ,   3 2 

the player’s expected profit per throw over a long series j NCERT Exemplar of throws?

(d) None of these

7 In a dice game, a player pays a stake of ` 1 for each throw of a die. She receives ` 5, if the die shows a 3, ` 2, if the die shows a 1 or 6 and nothing otherwise. What is

1 (a)   (20 C12 + 20 C13 + ...+ 20 C20 )  2 10 1 (b)   (20 C11 + 20 C12 + ...+ 20 C20 )  2 20 1 (c)   (20 C11 + 20 C12 + ...+ 20 C20 )  2 (d) None of the above

9 In a multiple choice question there are four alternative answers of which one or more than one is correct. A candidate will get marks on the question only if he ticks the correct answer. The candidate decides to tick answers at random. If he is allowed up to three chances of answer the question, then the probability that he will get marks on it is (a)

1 3

(b)

2 3

(c)

1 5

(d)

2 15

10 Three natural numbers are taken at random from the set A = {x |1 ≤ x ≤ 100, x ∈ N}. The probability that the AM of the numbers taken is 25, is 77

(a)

C2 C3

100

25

(b)

C2 C3

100

74

(c)

C72 C97

100

75

(d)

C2 C3

100

11 Given that x ∈[ 0, 1] and y ∈[ 0, 1]. Let A be the event of ( x , y ) satisfying y 2 ≤ x and B be the event of ( x , y ) satisfying x 2 ≤ y . Then,

1 (b) A, B are exhaustive 3 (c) A, B are mutually exclusive (d) A, B are independent (a) P (A ∩ B) =

FIVE

390 12 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 j

(a)

6 55

(b)

12 55

(c)

14 45

JEE Mains 2017 7 (d) 55

13 One Indian and four American men and their wives are to

adjacent to his wife given that each American man is seated adjacent to his wife, is (a)

1 2

(b)

1 3

(c)

2 5

(d)

1 5

14 If the integers m and n are chosen at random from 1 to 100, then the probability that a number of the form 7n + 7m is divisible by 5 equals

be seated randomly around a circular table. Then, the conditional probability that the Indian man is seated

(a)

1 4

(b)

1 2

(c)

1 8

(d)

1 3

ANSWERS SESSION 1

SESSION 2

1 (d)

2 (c)

3 (a)

4 (c)

5 (d)

6 (b)

7 (b)

8 (b)

9 (b)

10 (b)

11 (c)

12 (c)

13 (c)

14 (a)

15 (a)

16 (a)

17 (c)

18 (c)

19 (d)

20 (a)

21 (c)

22 (c)

23 (d)

24 (a)

25 (d)

26 (d)

27 (c)

28 (d)

29 (b)

30 (d)

31 (a)

32 (a)

33 (c)

34 (b)

35 (c)

36 (b)

37 (c)

38 (b)

39 (a)

40 (c)

1 (b) 11 (a)

2 (a) 12 (a)

3 (a) 13 (c)

4 (d) 14 (a)

5 (c)

6 (b)

7 (a)

8 (a)

9 (c)

10 (c)

Hints and Explanations SESSION 1 1 Here, n(S ) =

4 Here, P ( A ∪ B ) = 3 and P ( A ∩ B ) = 1

100

C3

Let E = All three of them are divisible by both 2 and 3. ⇒ Divisible by 6 i.e. {6, 12, 18, ..., 96} Thus, out of 16 we have to select 3. ∴ n(E ) = 16C3 16 C 4 ∴ Required probability = 100 3 = C3 1155

2 Total number of arrangement is 11! 11! = 2!2!2! 8 Number of arrangement in which C, E, H, I and S appear in that order 6! 11! = (11 C 5 ) = 2!2!2! 8.5 ! ∴ Required probability 11! 11! 1 1 = ÷ = = 8.5 ! 8 5! 120

3 P(Exactly one of the events occurs) = P ( A ∩ B ' ) ∪ (B ∩ A ' ) = P ( A ∩ B ' ) + P (B ∩ A ' )

A

B

= P ( A ) + P (B ) − 2 P ( A ∩ B )

5 5 From addition theorem, we get 3 1 = P ( A ) + P (B ) − 5 5 4 ⇒ = 1 − P ( A ′ ) + 1 − P (B ′ ) 5 4 6 ∴ P ( A ′ ) + P (B ′ ) = 2 − = 5 5 5 Clearly, A = {4, 5, 6} and B = {1, 2, 3, 4} ∴ A ∩ B = {4} Now, by addition theorem of probability ∴ P ( A ∪ B ) = P ( A ) + P (B ) − P ( A ∩ B ) 3 4 1 ⇒ P (A ∪ B) = + − = 1 6 6 6

6 We have, P (exactly one of A or B

occurs) = P ( A ∪ B ) − P ( A ∩ B ) = P ( A ) + P (B ) − 2P ( A ∩ B ) According to the question, 1 …(i) P ( A ) + P (B ) − 2P ( A ∩ B ) = 4 1 …(ii) P (B ) + P (C ) − 2P (B ∩ C ) = 4 1 and P (C ) + P ( A ) − 2P (C ∩ A ) = …(iii) 4 On adding Eqs. (i), (ii) and (iii), we get 2 [P ( A ) + P (B ) + P (C ) − P ( A ∩ B ) − P (B ∩ C ) 3 − P (C ∩ A )] = 4

⇒ P ( A ) + P (B ) + P (C ) − P ( A ∩ B ) − P (B ∩ C ) − P (C ∩ A ) =

3 8

∴ P (atleast one event occurs) = P( A ∪ B ∪ C ) = P ( A ) + P (B ) + P (C ) − P ( A ∩ B ) − P (B ∩ C ) − P (C ∩ A ) + P ( A ∩ B ∩ C ) 3 1 7  1  = + = Q P( A ∩ B ∩ C ) = 8 16 16  16 

7 Since, a leap year has 366 days and hence 52 weeks and 2 days. The 2 days can be (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). Therefore, 3 P (53 Sunday or 53 Monday) = 7

8 In first 120 natural numbers, total

number of multiples of 5, n ( A ) = 24 and total number of multiples of 15, n (B ) = 8 and n ( A ∩ B ) = 8. ∴ n ( A ∪ B ) = n ( A ) + n (B ) − n ( A ∩ B ) = 24 + 8 − 8 = 24 24 1 ∴ Required probability = = 120 5 9 Here, P ( A ) = 1 and P ( A ∪ B ) = 3 3 4 ∴ P ( A ∪ B ) = P ( A ) + P (B ) − P ( A ∩ B ) ≤ P ( A ) + P (B )

391

DAY 3 1 5 ≤ + P (B ) ⇒ P (B ) ≥ 4 3 12 Also, B ⊆ A ∪ B 3 ⇒ P (B ) ≤ P ( A ∪ B ) = 4 5 3 ∴ ≤ P (B ) ≤ 12 4 ⇒

10 Given, P ( A ) = 0.4, P (B ) = 03, . P ( A ∪ B ) = 0.5 B P( A ∩ B ) ∴ P   =  A P( A ) P ( A ) + P (B ) − P ( A ∪ B ) = P( A ) 0.4 + 03 . − 0.5 1 = = 0.4 2 1 1 11 Given, P ( A ) = , P (B ) = 2 3 A 1  and P   = B 4 ∴ P ( A ′∩ B ′ ) = 1 − P ( A ∪ B ) = 1 − P ( A ) − P (B ) + P ( A ∩ B ) A = 1 − P ( A ) − P (B ) + P   ⋅ P (B ) B 1 1 1 1 1 1 1 = 1 − − + ⋅ = 1− − + 2 3 4 3 2 3 12 12 − 6 − 4 + 1 3 1 = = = 12 12 4

12 Given that, P ( A ) = 1 , P  A  = 1 4 B 2 B 2  and P  =  A 3 We know that, P(A ∩ B) A P   = B P (B ) B  P (B ∩ A )  and P   =  A P(A) B 2 1  P   ⋅ P ( A )      A  3  4 1 ∴ P (B ) = = = A 3  1 P     B  2

13 Given that, P ( A ) = 0. 6, P (B ) = 0.2 P ( A / B ) = 0. 5 P ( A ∩ B ) = P ( A / B ) ⋅ P (B ) = (0. 5)(0.2) = 0.1 P ( A ′∩ B ′ ) P [( A ∪ B )′ ] P ( A ′/ B ′ ) = = P (B ′ ) P (B ′ ) 1 − P( A ∪ B ) = 1 − P (B ) 1 − P( A) − P( B ) + P( A ∩ B ) 3 = = 1 − 0.2 8

14 P (B / A ∪ B ′ ) = P {B ∩ ( A ∪ B ′ )}

P( A ∪ B ′ ) P( A ∩ B ) = P ( A ) + P (B ′ ) − P ( A ∩ B ′ ) P ( A ) − P ( A ∩ B ′ ) 07 . − 0.5 1 = = = 07 . + 0.6 − 0.5 0.8 4

15 As, P  C  = D

=

P (C ∩ D ) P (D ) QC ⊂ D  ∴ P (C ∩ D ) = P (C )   P (C ) P (D )

…(i)

Also, as P (D ) ≤ 1 P (C ) 1 ∴ ≥ 1 and ≥ P (C ) P (D ) P (D )

and

20 Given, P ( A ∪ B ) = 1 , P ( A ∩ B ) = 1 and

…(ii)

∴

From Eqs. (i) and (ii), we get P (C ) C P   = ≥ P (C )  D  P (D )

16 P (E ∩ F ) = P (E ) P (F ) = 1

12

1 2 1 …(ii) (1 − P (E )) (1 − P (F )) = ⇒ 2 On solving Eqs. (i) and (ii), we get 1 1 P (E ) = and P (F ) = , 3 4 as P (E ) > P (F ) P (E C ∩ F C ) = P (E C ) ⋅ P (F C ) =

17 The probability of getting the sum 7 or 8 6 5 11 from two dice is + = . 36 36 36 The probability of getting the card with 2 number 7 or 8 is . 11 ∴ Required probability 1 11 1 2 11 2 193 = ⋅ + ⋅ = + = 2 36 2 11 72 22 792

18 Let A be event that drawn ball is red and B be event that drawn ball is white. Then, AB and BA are two disjoint cases of the given event. ∴ P ( AB + BA ) = P ( AB ) + P ( BA ) B A = P ( A ) P   + P (B ) ⋅ P    A B 3 3 3 3 3 = ⋅ + ⋅ = 6 5 6 5 5

19 Given that, P (B ∩ A ) 1 B 1 P   = ⇒ =  A 2 P(A) 2 1 1 1 ⇒ P (B ∩ A ) = × = 2 4 8 P(A ∩ B) 1 A 1 P   = ⇒ = B 4 P (B ) 4 1 ⇒ P (B ) = 4P ( A ∩ B ) ⇒ P (B ) = 2 1 1 1 Q P ( A ∩ B ) = = ⋅ = P ( A ) ⋅ P (B ) 8 2 4 ∴ Events A and B are independent. P(A′ ∩ B ) A′ Now, P   =  B P (B ) P ( A ′ ) P (B ) 3 = = P (B ) 4

6 1 4 P(A ∪ B) = 1 − P (A ∪ B) 1 5 =1− = 6 6

4

P(A) =

1 3 = 4 4 P ( A ∪ B ) = P ( A ) + P (B ) − P ( A ∩ B ) 5 3 1 1 ⇒ = + P (B ) − ⇒ P (B ) = 3 6 4 4 ⇒ A and B are not equally likely. 1 Also, P ( A ∩ B ) = P ( A ) ⋅ P (B ) = 4 So, events are independent. and

…(i)

P (B ′ ∩ A ′ ) B′ P   =  A′ P(A′ ) P (B ′ ) P ( A ′ ) 1 = = P(A′ ) 2

P( A) = 1 − P( A) = 1 −

21 P (non-occurrence of ( A1 )) =1−

1 i = (i + 1) (i + 1)

∴ P (non-occurrence of any of events) 1 2 1  n  =   ⋅   K  =  2  3 ( n + 1 ) ( n + 1)  

22 P ( A ∩ B ) = P ( A ) ⋅ P (B ) It means A and B are independent events, so AC and B C will also be independent. Hence, P ( A ∪ B )C = P ( AC ∩ B C ) [De-Morgan’s law] = P ( AC ) P ( B C ) As A is independent of B, A P   = P ( A ) B A   Q P ( A ∩ B ) = P (B ) ⋅ P    B   

23 Clearly, E1 = {(4, 1),(4, 2),(4, 3),(4, 4), ( 4, 5),( 4, 6)} E2 = {(1, 2),(2, 2),(3, 2),(4, 2),( 5, 2),( 6, 2)} and 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 6 1 ⇒ P ( E1 ) = = , P ( E2 ) = = 36 6 36 6 18 1 and P (E3 ) = = 36 2 Now, P (E1 ∩ E2 ) = P [getting 4 on die A and 2 on die B] 1 = = P ( E1 ) ⋅ P ( E2 ) 36 P ( E2 ∩ E3 ) = P [getting 2 on die B and sum of numbers on both dice is odd] 3 = = P ( E2 ) ⋅ P ( E3 ) 36

FIVE

392 P ( E1 ∩ E3 ) = P [getting 4 on die A and sum of numbers on both dice is odd] 3 = = P ( E1 ) ⋅ P ( E3 ) 36 and P (E1 ∩ E2 ∩ 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.

24 Consider,

P { A ∩ (B ∩ C )} = P ( A ∩ B ∩ C ) = P ( A ) ⋅ P (B ) ⋅ P (C ) = P ( A ) ⋅ (B ∩ C ) Now, consider 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 ) ⋅ P (B ) + P ( A ) P (C ) − P ( A ) P (B ) P (C ) = P ( A )[P (B ) + P (C ) − P (B ) P (C )] = P ( A ) ⋅ P (B ∪ C ) B ∪ C is independent to A, so S 1 is true. B ∩ C is also independent to A, so S 2 is true.

25 We have, 1 3 and P ( A ∪ B ) = . 10 10 7 Then, P ( A ∪ B ) = 1 − P ( A ∪ B ) = 10 Q P ( A ∪ B ) = P ( A ) + P (B ) − P ( A ∩ B ) ∴ P ( A ) + P (B ) = P ( A ∪ B ) + P ( A ∩ B ) 7 1 4 …(i) + = ⇒ P ( A ) + P (B ) = 10 10 5 1 Q P ( A ) ⋅ P (B ) = 10 1 P (A) = ⇒ 10P (B ) 1 4 From Eq. (i), P (B ) + = 10 {P (B )} 5 P (A ∩ B)=

⇒

10 {P (B )}2 + 1 = 8 P (B )

Let P (B ) = t , then 10 t 2 + 1 = 8t ⇒ ∴ t =

10 t 2 − 8 t + 1 = 0 8±

So, P (B ) =

64 − 4 × 10 × 1

8± 2 6 = 20

2 × 10 4− 6 is possible. 10

26 Let A denote the event that atleast one girl will be chosen and B the event that exactly 2 girls will be chosen. We require P (B | A ). Since, A denotes the event that atleast one girl will be chosen. A denotes that no girl is chosen i.e. 4 boys are chosen. Then, 8 C 70 14 P ( A ′ ) = 12 4 = = C4 495 99

14 85 = 99 99 Now, P ( A ∩ B ) = P (2 boys and 2 girls) 8 C ⋅ 4C2 6 × 28 56 = 122 = = C4 495 165 P( A ∩ B ) Thus, P (B / A ) = P( A ) 56 99 168 = × = 165 85 425 P( A ) = 1 −

27 Let the events be A = Ist aeroplane hit the target B = IInd aeroplane hit the target and their corresponding probabilities are P ( A ) = 03 . and P (B ) = 0.2 ⇒ P ( A ) = 07 . and P (B ) = 0.8 ∴ Required probability = P ( A ) P ( B ) + P ( A ) P (B ) P ( A ) P (B ) + K = (0.7)(0.2) + (0.7)(0.8)(0.7)(0.2) + (0.7) (0.8)(0.7)(0.8) (0.7)(0. 2) + ... = 0.14 [1 + (0.56) + (0.56)2 + ...] 1  = 0.14 = 0.32 = 014 .    1 − 0.56  0.44

28 Let D be the event that the picked up tube is defective. Let A1 , A2 and A3 be the events that the tube is produced on machines E1 , E2 and E2 , respectively. P (D ) = P ( A1 ) P (D | A1 ) + P ( A2 ) P (D | A2 ) + P ( A3 ) P (D | A3 ) …(i) 50 1 1 1 P ( A1 ) = = , P ( A2 ) = , P ( A3 ) = 100 2 4 4 4 1 Also, P (D | A1 ) = P (D | A2 ) = = 100 25 5 1 P (D | A3 ) = = . 100 20 On putting these values in Eq. (i), we get 1 1 1 1 1 1 P (D ) = × + × + × 2 25 4 25 4 20 1 1 1 17 = + + = = 0.0425 50 100 80 400

29 Key idea Use the theorem of total probability Let E1 = Event that first ball drawn is red E2 = Event that first ball drawn is black A = Event that second ball drawn is red  A 4 6 P ( E1 ) = ,P  = 10  E1  12  A 6 4 ⇒ P ( E2 ) = , P  = 10  E2  12 By law of total probability  A  A P ( A ) = P ( E1 ) × P   + P ( E2 ) × P    E2   E1 

4 6 6 4 24 + 24 × + × = 10 12 10 12 120 48 2 = = 120 5 =

30 Let C , S , B and T be the events of the person going by car, scooter, bus and train, respectively. 1 3 2 1 P (C ) = , P (S ) = , P (B ) = , P (T ) = 7 7 7 7 Let L be the event that the person reaching the office in time. Then, L be the event that the person reaching the office in late. 7 8 P (L / C ) = , P (L / S ) = , 9 9 5 8 P (L / B ) = , P (L / T ) = 9 9 ∴ Required probability = P (C ) ⋅ P (L / C ) P (C / L ) =  P (C ) ⋅ P (L /C )   + P (S ) ⋅ P (L /S )    + P (B ) ⋅ P (L / B )    + P (T ) ⋅ P (L /T ) 1 7 ⋅ 7 1 7 9 = = = 1 7 3 8 2 5 1 8 49 7 ⋅ + ⋅ + ⋅ + ⋅ 7 9 7 9 7 9 7 9

31 Let B stand for the event that black card 1 is missing, then P (B ) = P (B ) = . 2 Let E be the event that all the first 13 cards are red. 26 25 14 ∴ P ( E /B ) = ⋅ ... , 51 50 39 25 24 13 P ( E /B ) = ⋅ K 51 50 39 P ( B ) ⋅ P ( E /B ) P ( B /E ) = P ( B ) ⋅ P ( E /B ) + P ( B ) ⋅ P ( E / B ) 26 ⋅ 25 K 14 = 26 ⋅ 25 K 14 + 25⋅ 24 K 13 26 26 2 = = = 26 + 13 39 3

32 Since, ΣPi = 1, we have C + 2C + 2C + 3C + C 2 + 2C 2 + 7C 2 + C = 1 i.e. 10C 2 + 9C − 1 = 0 i.e. (10C − 1) (C + 1) = 0 1 C = ,C = − 1 ⇒ 10 Therefore, the permissible value 1 of C = 10 Mean =

n

∑xp i

i =1

=1×

i

=

7

∑xp i

i

i =1

1 2 + 2× 10 10 2 2 3 1 + 3× + 4× + 5   10  10 10

393

DAY 2 2  1 1 1 + 6 × 2   + 7  7   +   10    10 10   1 4 6 12 5 = + + + + 10 10 10 10 100 12 49 7 + + + 100 100 10 = 3.66

33 Given that, E ( X ) = 3 and (E ( X 2 )) = 11 Variance of X = E ( X 2 ) − [E ( X )]2 = 11 − (3)2 = 11 − 9 = 2

34 Given, E = { X is a prime number} = {2, 3, 5, 7} P (E ) = P ( X = 2) + P ( X = 3) + P ( X = 5) + P ( X = 7) ⇒ P (E ) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62 and F = { X < 4} = {1, 2, 3} ⇒ P (F ) = P ( X = 1) + P ( X = 2) + P ( X = 3) ⇒ P (F ) = 0.15 + 0.23 + 0.12 = 0.5 and E ∩ F = {X is prime number as well as < 4} = {2, 3} P (E ∩ F ) = P ( X = 2) + P ( X = 3) = 0.23 + 0.12 = 0.35 ∴ Required probability P (E ∪ F ) = P (E ) + P (F ) − P (E ∩ F ) = 0.62 + 0.5 − 0.35= 0.77 ∴

35 Let X denote the number of aces. Probability of selecting a ace, 4 1 p= = 52 13 And probability of not selecting ace, 1 12 q = 1− = 13 13 2 12 P ( X = 0 ) =   ,  13  1 12 24 P ( X = 1 ) = 2 ⋅   ⋅   =  13   13  169 2

0

1 12 1 P ( X = 2) =   ⋅   =  13   13  169 24 Mean = Σ Pi X i = 0 + 1 × +2 169 1 24 2 2 × = + = 169 169 169 13

36 We have, n = 5and r ≥ 1

⇒ ∴ p≤

p5 ≤ 1 −

7

1 1 ⇒ C7 ⋅   ⋅    2  2

n −7

⇒

n

9

1 1 = C 9 ⋅   ⋅    2  2

n− 9

n

n

C7 = nC 9 ⇒ n = 16 [Q nC x = nC y ⇒ x + y = n ] 2

1 1 ∴ P ( X = 2) = 16C2 ⋅   ⋅    2  2

14

C2 16 ⋅ 15 15 = = 13 216 217 2 38 Since, nC 4 1n , nC 5 1n and nC 6 1n are in 2 2 2 AP. ∴ nC 4 , nC 5 and nC 6 are also in AP. ∴ 2 ⋅ nC 5 = nC 4 + nC 6 ∴ On dividing by nC 5 both sides, we get n n n−5 C C 5 2= n 4 + n 6 = + C5 C5 n − 4 6 n2 − 9 n + 50 = 6(n − 4) =

16

⇒ n2 − 21 n + 98 = 0 ⇒ n = 7,14

39 Given box contains 15 green and 10 yellow balls. ∴ Total number of balls = 15 + 10 = 25 15 3 P (green balls) = = = p 25 5 = Probability of success 10 2 P (yellow balls) = = =q 25 5 = Probability of failure and n = 10 = Number of trials. 3 2 12 ∴ Variance = npq = 10 × × = 5 5 5

40 Clearly, probability of guessing correct 1 and probability of 3 2 guessing a wrong answer, q = 3 ∴ The probability of guessing a 4 or more correct answer 4 5 1 2 1 = 5C 4   ⋅ + 5C 5    3 3  3 2 1 11 = 5⋅ 5 + 5 = 5 3 3 3 answer, p =

SESSION 2

∴ P ( X = r ) = nC r p n − r q r , ∴ P ( X ≥ 1) = 1 − P ( X = 0) = 1 − 5C 0 . p 5 .q 0 ≥

2 Since, 0 ≤ P ( A ) ≤ 1, 0 ≤ P (B ) ≤ 1,

According to the given condition, ∴ P ( X = 7) = P ( X = 9)

B

B

B

B

B

B

B

31 1 = 32 32

1 1 and p ≥ 0 ⇒ p ∈  0,   2  2

37 Let the coin was tossed n times and X be the random variable representing the number of head appearing in n trials.

10! 8! n (E ) = 8C3 = (7!) (3!) (3!)(5!) [because there are 8 places for 3 black balls] 8! (3!)(5!) (8!)(7!) P (E ) = = 10! (10!)(5!) (7!)(3!) 7⋅ 6 7 = = 10 ⋅ 9 15 n (S ) =

∴

3 There seems to be ambiguity in this question. It should be mentioned that boxes are different and one particular box has 3 balls. Then, the required probability 11 12 C 3 × 29 55  2  = =   12 3 3  3

4 Number of ways of distributing 6 objects to 6 persons = 66 Number of ways of distributing 1 object to each person = 6! ∴ Required probability 6! 5! 319 =1− 6 =1− 5 = 6 6 324

5 The required probability = 1 − probability of each receiving n(E ) atleast one = 1 − n(S ) Now, the number of integral solutions of x1 + x2 + x3 + x 4 + x 5 + x 6 = 10 such that x1 ≥ 1, x2 ≥ 1, ..., x 6 ≥ 1 gives n(E ) and the number of integral solutions of x1 + x2 + K + x 5 + x 6 = 10 such that x1 ≥ 0, x2 ≥ 0, ..., x 6 ≥ 0 gives n(S ). ∴ The required probability 10 − 1 9 C C 137 = 1 − 10 + 6 − 1 6 − 1 = 1 − 15 5 = C 6 −1 C 5 143

6 The probability of both drawing the B

1 −W −W −W −W −W −W −W − 31 32

0 ≤ P (C ) ≤ 1 and 0 ≤ P ( A ) + P (B ) + P (C ) ≤ 1 3x + 1 1 2 …(i) ∴ 0≤ ≤ 1⇒− ≤ x ≤ 3 3 3 1− x …(ii) 0≤ ≤ 1 ⇒ −3 ≤ x ≤ 1 4 1 − 2x 1 1 …(iii) 0≤ ≤ 1⇒ − ≤ x ≤ 2 2 2 3x + 1 1 − x 1 − 2x and 0 ≤ + + ≤1 3 4 2 ⇒ 0 ≤ 13 − 3 x ≤ 12 1 13 …(iv) ⇒ ≤ x≤ 3 3 From Eqs. (i), (ii), (iii) and (iv), we get 1 1 ≤ x≤ 3 2

common card x, P ( X ) = (Probability of A drawing the card x and any other card y) × ( Probability of B drawing the card x and a card other than y) 51 50 C C = 52 1 × 52 1 ∀ x, C2 C2

where x has 52 values. ∴ Required probability = Σ P ( X ) 51 × 50 × 4 50 = 52 × = 52 × 51 × 52 × 51 663

FIVE

394 7 Let X be the money won in one throw. Money lost in 1 throw = ` 1

1 6

Also, probability of getting 3 =

Probability of getting 1 or 6 1 1 2 ⇒ + = 6 6 6 Probability of getting any other number i.e. 2 or 4 or 5 1 1 1 3 = + + = 6 6 6 6 Then, probability distribution is 5 1 6

X P(X)

2 2 6

0 3 6

answer given by the student. The repeated tosses of a coin are Bernoulli trials. Since, head on a coin represent the true answer and tail represents the false answer, the correctly answered of the question are Bernoulli trials. ∴ p = P (a success) = P(coin show up a 1 1 1 head) = and q = 1 − p = 1 − = 2 2 2 So, X has a binomial distribution with 1 1 n = 20, p = and q = 2 2 r 20 − r 1  1  20 ∴ P (X = r ) = Cr ⋅      2  2

to the area OTQPO B = the event of ( x, y ) belonging to the area OSQRO Y

= C12 p q + +

T S

+

C15 p15q 5 +

20

C17 p17q 3 + 20C18 p18q 2 +

20

= ( C12 + +

C13 +

20

C15 +

20

C16 p16q 4

20

C19 p19q 1 20

20

C20 p

20

C14

C16 +

20

C17 +

20

C18 +

20

C19 1 220

20

(20 C12 +

C13 + K +

20

20

C20 )

9 The total number of ways of ticking one or more alternatives out of 4 is 4 C1 + 4C2 + 4C3 + 4C 4 = 15. Out of these 15 combinations only one

=

x

ar (OTQPO ) ar(OPQRO )

∫

1 0

x dx

1×1

1

2 2 =  x3 /2  =  3  0 3

ar (OSQRO ) P (B ) = = ar (OPQRO ) ar (OTQS ) P (A ∩ B) = ar (OPQRO )

20

+ 20 C20 ) ⋅ 1 =    2

P (A) =

20

+ 20

P 1

O

=

13 7

C13 p q

C14 p14q 6 +

20

x2 = y y2 = x Q

1 R

+ P (17) + P (18) + P (19) + P (20) 20

C3

11 A = the event of ( x, y ) belonging

Hence, P (atleast 12 questions are answered as true) = P ( X ≥ 12) = P (12) + P (13) + P (14) + P (15) + P (16) 8

100

As the AM of three numbers is 25, their sum = 75 ∴ n(E ) = the number of integral solutions of x1 + x2 + x3 = 75, where, x1 ≥ 1, x2 ≥ 1, x3 ≥ 1 75−1 C3 −1 = 74C2 = 74C72 74 74 C C ∴ P (E ) = 100 72 = 100 72 C3 C 97

8 Let X denote the number of correct

12

Thus, the probability that the candidate will get marks on the question, if he is allowed up to three trials is 1 1 1 1 + + = 15 15 15 5

10 Here, n(S ) =

Then, expected money that player can be won 5 4 9 E ( X ) = + + 0 = = ` 1. 5 6 6 6 Thus, player’s expected profit = ` 1.5 − ` 1 = 0.50

20

combination is correct. The probability of ticking the alternative correctly at the 1 first trial is that at the second trial is 15  14   1  = 1 and that at the      15  14  15 third trial is  14   13   1  = 1        15  14   13  15

∫

1 0

x dx −

∫

1 0

∫

1 0

y dy

1×1

x2 dx

1×1 2 1 1 = − = 3 3 3 2 2 Q P ( A ) + P (B ) = + ≠ 1 3 3 So, A and B are not exhaustive. 2 2 4 P ( A ) ⋅ P (B ) = ⋅ = ≠ P ( A ∩ B ) 3 3 9

2 = 3

So, A and B are not independent. P ( A ∪ B ) = 1, P ( A ) + P (B ) 2 2 = + ≠ P(A ∪ B) 3 3 So, A and B are not mutually exclusive.

12 Total number of ways of selecting 2 different numbers from {0, 1, 2, ..., 10} = 11 C2 = 55 Let two numbers selected be x and y. Then, x + y = 4m …(i) and x − y = 4n …(ii) ⇒ 2 x = 4(m + n ) and 2 y = 4(m − n ) ⇒ x = 2(m + n ) and y = 2(m − n ) Thus, x and y both are even numbers. x 0 2 4 6 8 10

y 4, 8 6, 10 0, 8 2, 10 0, 4 2, 6

∴ Required probability =

6 55

13 Let E = event when each American man is seated adjacent to his wife and A = event when Indian man is seated adjacent of his wife. Now, n ( A ∩ E ) = (4!) × (2!)5 QWhile grouping each couple, we get 5 groups which can be arranged in (5 − 1)! ways, and each of the group can be arranged in 2! ways. and n(E ) = (5!) × (2!)4 QWhile grouping each American man with his wife, we get 4 groups. These 4 groups together with indian man and his wife (total 6) can be arranged in (6–1)! ways and each of the group can be arranged in 2! ways. n( A ∩ E ) (4!) × (2!)5 2 A = = ∴ P   = E n(E ) (5!) × (2!)4 5

14 We observe that 71 , 72 , 73 and 74 ends in 7, 9, 3 and 1 respectively. Thus, 7l ends in 7, 9, 3 or 1 according as l is of the form 4k + 1, 4k + 2, 4k − 1 or 4k, respectively. If S is the sample space, then n(S ) = (100)2 . 7m + 7n is divisible by 5 if (i) m is of the form 4k + 1 and n is of the form 4k − 1 or (ii) m is of the form 4k + 2 and n is of the form 4k or (iii) m is of the form 4k − 1 and n is of the form 4k + 1 or (iv) m is of the form 4k and n is of the form 4k + 1. Thus, number of favourable ordered pairs (m, n ) = 4 × 25 × 25. 1 Hence, required probability is . 4

DAY THIRTY SIX

Mathematical Reasoning Learning & Revision for the Day u

u

Statement (Proposition) Truth Value and Truth Table

u

u

Elementary Operations of Logic Converse, Inverse and Contrapositive of an Implication

u

u

u

Tautology Contradiction (Fallacy) Algebra of Statement

Sentence A sentence is a relatively independent grammatical unit. It can stand alone or it can be combined with other sentences to form a text, a story etc. Sentences can be divided into different types as declarative, interrogative, imperative, exclamatory and operative sentences.

Statement (Proposition) A statement is a sentence which is either true or false but not both simultaneously i.e. ambiguous sentence are not considered as statements. The working nature of statement in logic is same as nature of switch in circuit. ON (1)

i.e. Switch

OFF (0)

True (T) and Statement

False (F)

PRED MIRROR Your Personal Preparation Indicator u

No. of Questions in Exercises (x)—

u

No. of Questions Attempted (y)—

u

Types of statements 1. Simple statement A statement, which cannot be broken into two or more sentences, is a simple statement. Generally, small letters p, q , r , . . . denote simple statements. 2. Compound statement A statement formed by two or more simple statements using the words such as “and”, “or”, “not”, “if then”, “if and only if”, is called a compound statement.

No. of Correct Questions (z)— (Without referring Explanations)

u

Accuracy Level (z / y × 100)—

u

Prep Level (z / x × 100)—

In order to expect good rank in JEE, your Accuracy Level should be above 85 & Prep Level should be above 75.

SIX

396 3. Substatements Simple statements which when combined to form a compound statement are called substatements or components. NOTE

(ii) If p is statement, then negation of p is denoted by ‘ ~ p.’ (iii) The truth table for NOT is given by

• A true statement is known as a valid statement. • A false statement is known as an invalid statement. • Imperative, exclamatory, interrogative, optative sentences are not statements. • Mathematical identities are considered to be statements because they can either be true or false but not both.

4. Open statement A sentence which contains one or more variables such that when certain values are given to the variable it becomes a statement, is called an open statement.

Truth Value and Truth Table ●

●

●

Two or more statements are combined to form a compound statement by using symbols. These symbols are called logical connectives. Logical connectives are given below

T

F

F

T

A compound sentence formed by two simple sentences p and q using connective “AND” is called the conjunction of p and q and is represented by p ∧ q .

A1 A 2

The truth table for operation ‘AND’ is given by

A table that shows the relationship between the truth values of compound statement, S ( p, q , r , . . .) and the truth values of its substatements p, q , r , . . . etc., is called the truth table of statement S.

Logical Connectives or Sentencial Connectives

~p

2. Conjunction (AND)

A statement can be either ‘true’ or ‘false’, which are called truth values of a statement and it is represented by the symbols ‘T’ and ‘F’, respectively.

If a compound statement has simply n substatements, then there are 2 n rows representing logical possibilities.

p

NOTE

p

q

p ∧q

T

T

T

T

F

F

F

T

F

F

F

F

• The statement p ∧ q is true, if both p and q are true. • The statement p ∧ q is false, if atleast one of p and q or both are false.

3. Disjunction (OR) A compound sentence formed by two simple sentences p and q using connective “OR” is called the disjunction of p and q and is represented by p ∨ q . The truth table for operation ‘OR’ is given by

Words

Symbols

and

∧

p

q

p ∨q

or

∨

T

T

T

implies that (if ..., then)

⇒

T

F

T

If and only if (implies and is implied by)

⇔

F

T

T

F

F

F

Elementary Operations of Logic Formation of compound sentences from simple sentence using logical connectives are termed as elementary operation of logic.

NOTE

• The statement p ∨ q is true, if atleast one of p and q or both are true.

• The statement p ∨ q is false, if both p and q are false.

There are five such operations, which are discussed below.

1. Negation (Inversion) of Statement (i) A statement which is formed by changing the truth value of a given statement by using word like ‘no’ or ‘not’ is called negation of a given statement. It represents the symbol ‘~’.

4. Implication (Conditional) A compound sentence formed by two simple sentences p and q using connective “if ... then ...” is called the implication of p and q and represented by p ⇒ q which is read as “ p implies q”.

DAY Here, p is called antecedent or hypothesis and q is called consequent or conclusion.

Tautology

The truth table for if … then is given by

A compound statement is called a tautology, if it has truth value T whatever may be the truth value of its compounds.

~p

~p ∨ q

T

F

T

F

F

F

F

F

T

T

T

T

F

F

T

T

T

p

q

T

T

T

p⇒q

Statement ( p ⇒ q ) ∧ p ⇒ q is a tautology. The truth table of above statement is prepared as follows.

It is clear from the truth table that column III is equal to column V. i.e. statement p ⇒ q is equivalent to ~ p ∨ q .

5. Biconditional Statement Two simple sentences connected by the phrase “if and only if,” form a biconditional statement. It is represented by the symbol ‘⇔’. The truth table for if and only if is given by p

q

p⇔q

~p

T

T

T

F

F

T

F

F

F

F

T

F

F

F

T

NOTE

~ q ~p ∨ q

p ∨~q

(~ p ∨ q ) ∧ (p ∨ ~ q)

T

T

T

T

F

T

F

T

F

T

F

F

T

T

T

T

T

• It is clear from the truth table that column III is equal to

column VIII. i.e. statement p ⇔ q is equivalent to (~ p ∨ q) ∧( p∨ ~ q). • The statement p ⇔ q is true, if either both are true or both are false. • The statement p ⇔ q is false, if exactly one of them is false.

Converse, Inverse and Contrapositive of an Implication For any two statements p and q, (i) Converse of the implication ‘if p, then q’ is ‘if q, then p’ i.e. q ⇒ p (ii) Inverse of the implication ‘if p, then q’ is ‘if ~ p, then ~q ’ i.e. ~ p →~ q . (iii) Contrapositive of the implication ‘if p, then q’ is ‘if ~q, then ~ p’ i.e. ~ q →~ p. NOTE

• ~ ( p ⇒ q) ≡ ~ (~ pvq) = { p ∧ (~ q)} ∴ ~ ( p ⇔ q) ≡ { p ∧ (~ q)} • ~ ( p ⇒ q) ≡ ( p ∧ ~ q) ∨ ( q∧ ~ p) • p⇒q=~p∨q • ( p ⇔ q) ⇔ r = p ⇔ ( q ⇔ r ) • p ⇔ q = ( p ⇒ q) ∧ ( q ⇒ p)

p

q

p ⇒q

p ⇒q ∧ p

( p ⇒ q) ∧ p ⇒ q

T

T

T

T

T

T

F

F

F

T

F

T

T

F

T

F

F

T

F

T

Contradiction (Fallacy) A compound statement is called contradiction, if its truth value is F whatever may be the truth value of its components. Statement ~ p ∧ p is a contradiction. The truth table of above statement is prepared as follows p

~p ∧ p

~p

T

F

F

F

T

F

A statement which is neither a tautology nor a contradiction is a contigency.

Algebra of Statement Some of the important laws considered under the category of algebra of statements are given as 1. Idempotent Laws For any statement p, we have

(a) p ∧ p ≡ p

(b) p ∨ p ≡ p

2. Commutative Laws For any two statements p and q , we have

(a) p ∧ q ≡ q ∧ p

(b) p ∨ q ≡ q ∨ p

3. Associative Laws For any three statements p, q and r , we have

(a) ( p ∧ q) ∧ r ≡ p ∧ (q ∧ r) (b) ( p ∨ q) ∨ r ≡ p ∨ (q ∨ r) 4. Distributive Laws

(a) p ∨ (q ∧ r) ≡ ( p ∨ q) ∧ ( p ∨ r) (b) p ∧ (q ∨ r) ≡ ( p ∧ q) ∨ ( p ∧ r)

5. Involution Laws For any statement p, we have ~(~ p) ≡ p 6. De-morgan’s Laws

(a) ~ ( p ∧ q) = ~ p ∨ ~ q (b) ~ ( p ∨ q) = ~ p ∧ ~ q

SIX

398 7. Complement Laws For any statement p ,we have

(a) p ∨ ~ p ≡ T (c) ~ T ≡ F

●

(b) p ∧ ~ p ≡ F (d) ~ F ≡ T

8. Identity Laws For any statement p, we have

(a) p ∧ T ≡ p (c) p ∨ T ≡ T

9. Duality Two compound statements S1 and S2 are said to be duals of each other, if one can be obtained from the other by replacing ∧ by ∨ and ∨ by ∧. The connectives ∧ and ∨ are also called duals of each other. Symbolically, it can be written as , if S ( p, q ) = p ∧ q , then its dual is S * ( p, q ) = p ∨ q . ●

(b) p ∨ F ≡ p (d) p ∧ F ≡ F

●

where, T and F are the true and false statement.

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE 1 Which of the following is not a statement?

j

NCERT Exemplar

(a) p ∧ q is true (c) p ⇒ q is true j

NCERT Exemplar

(a) 72 is not divisible by 2 or 72 is not divisible by 3 (b) 72 is not divisible by 2 and 72 is not divisible by 3 (c) 72 is divisible by 2 and 72 is not divisible by 3 (d) 72 is not divisible by 2 and 72 is divisible by 3

3 The dual of the statement ( p ∨ q ) ∨ r is (a) (p ∧ q) ∨ r (c) (p ∨ q) ∧ r

(b) (p ⇒ q) ∧ (q ⇒ p) (d) (p ∧ q) ⇒ (p ∨ q)

8 If both p and q are false, then

2 The negation of the statement “72 is divisible by 2 and 3” is

7 The logically equivalent proposition of p ⇔ q is (a) (p ∧ q) ∨ (p ∧ q) (c) (p ∧ q) ∨ (q ⇒ p)

(a) Smoking is injurious to health (b) 2 + 2 = 4 (c) 2 is the only even prime number (d) Come here

(b) (p ∧ q) ∧ r (d) None of these

4 Choose disjunction among the following sentences (a) It is raining and the Sun is shining (b) Ram and Shyam are good friends (c) 2 or 3 is a prime number (d) Everyone who lies in India is an Indian

5 Which among the following is not a conjunction? (a) Gautam and Rahul are good friends (b) The Earth is round and the Sun is hot (c) 9 > 4 and 12 > 15 (d) None of the above

6 If p : a natural number n is odd and q : natural number n is not divisible by 2, then the biconditional statement j NCERT p ⇔ q is (a) A natural number n is odd if and only if it is divisible by 2 (b) A natural number n is odd if and only if it is not divisible by 2 (c) If a natural number n is odd, then it is not divisible by 2 (d) None of the above

(b) p ∨ q is true (d) p ⇔ q is false

9 The negation of ‘12 > 4’ is (a) 12 ≤ 4

(b) 13 > 5

(c) 12 > 3

(d) 12 ≥ 4

10 Which among the following is not a negation of ‘The Sun is a star’? (a) The Sun is not a star. (b) It is not the case that Sun is a star. (c) It is false that the Sun is a star. (d) It is true that the Sun is a star.

11 Which of the following is not equivalent to p ↔ q ? (a) (b) (c) (d)

p if and only if q p is necessary and sufficient for q q if and only if p None of the above

12 For integers m and n, both greater than 1, consider the following three statements

j

P : m divides n Q : m divides n (a) Q ∧ R → P (c) Q → R

2

JEE Mains 2013

R : m is prime, then

(b) P ∧ Q → R (d) Q → P

13 If p and q are two statements such that p : the questions paper is easy q : we shall pass, then the symbolic statement ~ p → ~ q means (a) If the question paper is easy, then we shall pass (b) If the question paper is not easy, then we shall not pass (c) The question paper is easy and we shall pass (d) The question paper is easy or we shall pass

DAY 14 For the following three statements p : 2 is an even number q : 2 is a prime number. r : Sum of two prime numbers is always even, then the symbolic statement ( p ∧ q ) → ~ r means (a) 2 is an even and prime number and the sum of two prime numbers is always even (b) 2 is an even and prime number and the sum of two prime numbers is not always even (c) 2 is an even and prime number, then the sum of two prime numbers is not always even (d) 2 is an even and prime number, then the sum of two prime numbers is always even

15 The converse of the statement “If x > y , then x + a > y + a” is

j

NCERT Exemplar

(a) If x < y, then x + a < y + a (b) If x + a > y + a, then x > y (c) If x < y, then x + a > y + a (d) If x > y, then x + a < y + a

16 Consider the following statements p : If a number is divisible by 10, then it is divisible by 5. q : If a number is divisible by 5, then it is divisible by 10. Then, the correct option is (a) q is converse of p (c) p is not converse of q

(b) p is converse of q (d) Both (a) and (b)

17 The negation of the statement. “If I become a teacher, then I will open a school”, is

j

AIEEE 2012

(a) I will become a teacher and I will not open a school (b) Either I will not become a teacher or I will not open a school (c) Neither I will become a teacher nor I will open a school (d) I will not become a teacher or I will open a school

18 Find the contrapositive of “If two triangles are identical, then these are similar”. (a) If two triangles are not similar, then these are not identical (b) If two triangles are not identical, then these are not similar (c) If two triangles are not identical, then these are similar (d) If two triangles are not similar, then these are identical

19 The contrapositive of the statement “If 7 is greater than 5, then 8 is greater than 6” is j NCERT Exemplar (a) If 8 is greater than 6, then 7 is greater than 5 (b) If 8 is not greater than 6, then 7 is greater than 5 (c) If 8 is not greater than 6, then 7 is not greater than 5 (d) If 8 is greater than 6,then 7 is not greater than 5

20 The conditional statement of “You will get a sweet dish after the dinner” is NCERT Exemplar (a) If you take the dinner, then you will get a sweet dish j

(b) If you take the dinner, you will get a sweet dish (c) You get a sweet dish if and only if you take the dinner (d) None of the above

21 Let S be a non-empty subset of R. Consider the following statement P : There is a rational number x ∈ S such that x > 0 . Which of the following statements is the negation of the j AIEEE 2010 statement P? (a) There is a rational number x ∈ S such that x ≤ 0 (b) There is no rational number x ∈ S such that x ≤ 0 (c) Every rational number x ∈ S satisfies x ≤ 0 (d) x ∈ S and x ≤ 0 ⇒ x is not rational

22 The contra positive of p → (~ q →~ r ) is (a) (~ q ∧ r ) →~ p (c) p → (~ r ∨ q)

(b) (q ∧ ~ r ) →~ p (d) p ∧ (q ∧ r )

23 p ∨ ~ ( p ∧ q ) is a (a) contradiction (c) tautology

(b) contingency (d) None of these

24 If p, q and r are simple propositions with truth values T, F, T respectively, then the truth value of (~ p ∨ q ) ∧ ~ q → p is (a) true (c) true, if r is false

(b) false (d) None of these

25 The statement p → (q → p ) is equivalent to j

(a) p → q (c) p → (p → q)

JEE Mains 2013

(b) p → (p ∨ q) (d) p → (p ∧ q)

26 The statement ( p → q ) → [(~ p → q ) → q ] is j

JEE Mains 2017

(a) a tautology (b) equivalent to ~ p → q (c) equivalent to p → ~ q (d) a fallacy

27 Let p and q be two statements. Then, (~ p ∨ q ) ∧ (~ p ∧ ~ q ) is a (a) tautology (b) contradiction (c) neither tautology nor contradiction (d) both tautology and contradiction

28 The proposition ( p ⇒ ~ p ) ∧ (~ p ⇒ p ) is (a) contigency (b) neither tautology nor contradiction (c) contradiction (d) tautology

29 If p and q are two statements, then ( p ⇒ q ) ⇔ (~ q ⇒ ~ p ) is a (a) contradiction (c) neither (a) nor (b)

(b) tautology (d) None of these

30 The proposition S : ( p ⇒ q ) ⇔ (~ p ∨ q ) is (a) a tautology (c) either (a) or (b)

(b) a contradiction (d) neither (a) nor (b)

SIX

400 36 The proposition ~ ( p ⇒ q ) ⇒ (~ p ∨ ~ q ) is

31 The statement ~ ( p ↔~ q ) is (a) equivalent to p ↔ q (c) a tautology

(b) equivalent to ~ p ↔ q (d) a fallacy

32 The false statement in the following is

(a) a tautology

(b) a contradiction

(c) either (a) or (b)

(d) neither (a) nor (b)

37 The negation of the compound proposition is p ∨ (~ p ∨ q )

(a) p ∧ (~ p) is a contradiction (b) (p → q) ↔ (~ q → ~ p) is a contradiction (c) ~ (~ p) ↔ p is a tautology (d) p ∨ (~ p) is a tautology

(a) (p ∧ ~ q) ∧ ~ p

(b) (p ∨ ~ q) ∨ ~ p

(c) (p ∧ ~ q) ∨ ~ p

(d) None of these

38 The negation of ~ s ∨ (~ r ∧ s ) is equivalent to

33 Among the following statements, which is a tautology? (a) p ∧ (p ∨q) (c) [p ∧ (p → q)] → q

(b) p ∨ (p ∧ q) (d) q → [p ∧ (p → q)]

P : Suman is brilliant. Q : Suman is rich. R : Suman is honest. The negative of the statement.‘Suman is brilliant and dishonest if and only if Suman is rich.’ can be expressed j AIEEE 2011 as (b) ~ (Q ↔ P ∧ R) (d) ~ P ∧ (Q ↔~ R)

35 The statement ( p ⇒ q ) ⇔ (~ p ∧ q ) is a (a) tautology (c) neither (a) nor (b)

(a) s ∧ ~ r (c) s ∨ (r ∨ ~ s)

JEE Mains 2015

(b) s ∧ (r ∧ ~ s) (d) s ∧ r

39 ~ S ( p, q ) is equivalent to

34 Consider the following statements

(a) ~ (Q ↔ (P ∧ ~ R)) (c) ~ (P ∧ ~ R) ↔ Q

j

(b) contradiction (d) None of these

(a) S * (~ p, ~ q) (c) S * (~ p, q)

(b) S * (p, ~ q) (d) None of these

40 Let p : 25 is a multiple of 5. q : 25 is a multiple of 8. Statement I The compound statement “p and q” is false. Statement II The compound statement “p or q” is false. Choose the correct option (a) Only Statement I is correct (b) Only Statement II is correct (c) Both statements are correct (d) Both statements are incorrect

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE 1 If p : 4 is an even prime number, q : 6 is divisor of 12 and r : the HCF of 4 and 6 is 2, then which one of the following is true? (a) (p ∧ q) (c) ~ (q ∧ r ) ∨ p

(b) (p ∨ q) ∧ ~ r (d) ~ p ∨ (q ∧ r )

contains neither the biconditional nor the conditional is (~ p ∨ q ∧ r ) ∨ ((~ r ∨ s) ∧ (r ∨ ~ s)) (~ p ∧ q ∧ r ) ∨ ((~ r ∨ s) ∧ (r ∨ ~ s)) (~ p ∨ q ∧ r ) ∧ ((~ r ∨ s) ∨ (r ∨ ~ s)) None of the above

3 If ( p ∧ ~ r ) ⇒ (~ p ∨ q ) is false, then the truth values of p, q and r respectively (a) T, F and F (c) F, T and T

4 The Boolean expression ( p ∧ ~ q ) ∨ q ∨ (~ p ∧ q ) is (a) ~ p ∧ q (c) p ∨ q

j

(b) p ∧ q (d) p ∨ ~ q

(b) (p → q) ∨ (p → q) (d) None of these

and ‘4 divides 7’, respectively. Then, what are the truth values for following biconditional statements? (i) p ↔ q (iii) ~ q ↔ p

(ii) ~ p ↔ q (iv) ~ p ↔ ~ q

(a) T T T T (c) F T F F

(b) F T T T (d) F T T F

7 The only statement among the followings that is a tautology is (a) B → [A ∧ (A → B)] (c) A ∨ (A ∧ B)

(b) F, F and T (d) T, F and T

equivalent to

(a) (p → q) ∧ (p → q) (c) (p → q) ∨ (q → p)

6 Let p and q stand for the statements ‘2 × 4 = 8’

2 An equivalent expression for ( p ⇒ q ∧ r ) ∨ (r ⇔ s ) which (a) (b) (c) (d)

5 Which of the following is a tautology?

JEE Mains 2016

j AIEEE 2011 (b) A ∧ (A ∨ B) (d) [A ∧ (A → B)] → B

8 Which of the following is not correct? (a) (b) (c) (d)

~ (p ∧ q) = (~ p) ∨ (~ q) Truth value of p ∧ q = truth value of q ∧ p ~ (~ p) = p p ⇔ q ≡ (p ⇒ q) ∨ (q ⇒ p)

DAY 9 ~ ( p ⇒ q ) ⇔~ p ∨ ~ q is

14 Consider Statement I ( p ∧ ~ q ) ∧ (~ p ∧ q ) is a fallacy. Statement II ( p → q ) ↔ (~ q → p ) is a tautology.

(a) a tautology (b) a contradiction (c) neither a tautology nor a contradiction (d) cannot come to any conclusion

j

10 Which of the following is wrong statement? (a) p → q is logically equivalent to ~ p ∨ q (b) If the truth values of p, q, r are T, F, T respectively, then the truth value of (p ∨ q) ∧ (q ∨ r ) is T (c) ~ (∨ q ∨ q ∨ r ) ≅ ~ p ∧ ~ q ∧ ~ r (d) The truth value of p ∧ ~ (p ∨ q) is always T

11 If p, q and r are simple propositions, then ( p ∧ q ) ∧ (q ∧ r ) is true, then (a) p, q and r are true (b) p, q are true and r is false (c) p is true and q, r are false (d) p, q and r are false

12 ~ p ∧ q is logically equivalent to (a) p ⇒ q (c) ~ (p ⇒ q)

15 Statement I The statement A → (B → A ) is equivalent to A → ( A ∨ B ). Statement II The statement ~ {( A ∧ B ) → (~ A ∨ B )} is j JEE Mains 2013 tautology. (a) Statement I is true, Statement II is false (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; Statement II is a correct explanation for Statement I (d) Statement I is false, Statement II is true

(b) q ⇒ p (d) ~ (q ⇒ p)

13 Which of the following is true for any two statements p and q? (a) ~ [p ∨ ~ q] ≡ ~ p ∧ q (c) p ∨ ~ q is a tautology

JEE Mains 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

(b) ~ p ∧ q is a fallacy (d) p ∨ ~ p is a contradiction

ANSWERS SESSION 1

SESSION 2

1 (d)

2 (a)

3 (b)

4 (c)

5 (a)

6 (b)

7 (b)

8 (c)

9 (a)

10 (d)

11 (d)

12 (a)

13 (b)

14 (c)

15 (b)

16 (d)

17 (a)

18 (a)

19 (c)

20 (a)

21 (c)

22 (a)

23 (c)

24 (a)

25 (b)

26 (a)

27 (c)

28 (c)

29 (b)

30 (a)

31 (a)

32 (b)

33 (c)

34 (a)

35 (c)

36 (a)

37 (a)

38 (d)

39 (a)

40 (a)

1 (d) 11 (a)

2 (a) 12 (d)

3 (a) 13 (a)

4 (c) 14 (b)

5 (c) 15 (c)

6 (d)

7 (d)

8 (d)

9 (c)

10 (d)

SIX

402

Hints and Explanations SESSION 1

17 Let us assume that p : ‘I become a teacher’ and q : I will open a school. Then, we can easily ascertain that Negation of ( p → q ) is ~ ( p → q ) = p ∧ ~ q , which means that ‘I will become a teacher and I will not open a school.’

1 In option (a), (b) and (c). It is a declarative sentence, which is clearly true, therefore it is a true statement. In option (d), it is an imperative sentence, therefore it is not a statement.

18 Consider the following statements p : Two triangles are identical. q : Two triangles are similar. Clearly, the given statement in symbolic form is p ⇒ q . So, its contrapositive is given by ~ q ⇒ ~ p.

2 The negation of the given statement is “72 is not divisible by 2 or 72 is not divisible by 3”.

3 According to definition let s ( p ,q ) = p ∧ q be a compound statement. Then, s * ( p,q ) = p ∨ q , Therefore, the dual of ( p ∨ q ) ∨ r is ( p ∧ q ) ∧ r .

19 The contrapositive of the given statement is “If 8 is not greater than 6, then 7 is not greater than 5.

20 The conditional statement of given statement is “If you take the dinner, then you will get a sweet dish”.

4 In disjunction, two sentences are connected with ‘OR’. 5 In the sentence, ‘Gautam and Rahul are good friends’, the word

21 P : There is rational number x ∈ S such that x > 0. ~ P : Every rational number x ∈ S satisfies x ≤ 0.

‘and’ is not a connective. So, this sentence is not a conjunction.

6 Given, p : A natural number n is odd and q : natural number n is

22 The contrapositive of p → q is ~ q →~ p ∴Contrapositive of p → (~ q →~ r ) is ~ (~ q → ~ r ) → ~ p

not divisible by 2. The biconditional statement p ⇔ q i.e. “A natural number n is odd if and only if it is not divisible by 2”.

7 ( p ⇒ q ) ∧ (q ⇒ p ) means p ⇔ q. Hence, option (b) is correct. 8 If both p and q are false, then p ⇒ q is true.

 Q~ ( p → q ) ≡ p ∧ ~ q  ∴~ (~ q →~ r ) ≡~ q ∧ r   

≡ (~ q ∧ r ) →~ p

23

9 Negation is a denial of a statement. So, 12 ≤ 4 is correct option.

10 The denial of any statement is called its negation. ‘It is true that the Sun is a star’ is an assertion of the given statement.

p

q

p∧q

~ ( p ∧ q)

p ∨ ~ ( p ∧ q)

T

T

T

F

T

T

F

F

T

T

F

T

F

T

T

F

F

F

T

T

~ p ~q

~p∨q

(~ p ∨ q ) ∧~ q

(~ p ∨ q ) ∧~ q → p

F

F

T

11 Options (a), (b) and (c) are equivalent to p ↔ q. 2 12 P : n ; Q : n ; R : m is prime.

m

24

p

q

r

T

F

T

m

Let m = 5, p = 10, n2 = 100, m = 3, n = 12, n2 = 144 m = 7, n = 14, n = 196 2

F

T

25

So, P ,Q and R are true statements.

p

q

q→ p

p → ( q → p)

p∨q

p → ( p ∨ q)

∴

T

T

T

T

T

T

T

F

T

T

T

T

F

T

F

T

T

T

F

F

T

T

F

T

Q ∧R = T∧T→ T = P

13 Given, p : The question paper is easy and q : We shall pass given. The symbolic representation of given option are (a) : p → q (b) : ~ p → ~ q (c) : p ∧ q (d) : p ∨ q

14 Given, p : 2 is an even number q : 2 is a prime number r : sum of two prime numbers is always even. The symbolic representation of given option (s) are (a) : p ∧ q ∧ r (b) : p ∧ q ∧ ~ r (c) : p ∧ q ⇒ ~ r (d) : p ∧ q ⇒ r

So, statement p → (q → p ) is logically equivalent to p → ( p ∨ q ).

26 The truth table of the given expression is given below : p

q x ≡ p → q ~ p ~p → q

y ≡ (~ p → q ) → q

x→ y

T

T

T

F

T

T

T

15 Converse statement is “If x + a > y + a, then x > y ”.

T

F

F

F-

T

F

T

16 The converse of the statement “if p then q” is “if q then p”.

F

T

T

T

T

T

T

F

F

T

T

F

T

T

Given statements are p : if a number is divisible by 10, it is divisible by 5. q : if a number is divisible by 5, then it is divisible by 10. It is clear that q is converse of p and p is converse of q.

Hence, it is a tautology.

DAY

27

(~ p ∨ q ) ∧ (~ p ∧ ~ q )

(~ p ∨ q ) (~ p ∧ ~ q )

p

q

~p

~q

T

T

F

F

T

F

F

T

F

F

T

F

F

F

F

T

T

F

T

F

F

F

F

T

T

T

T

T

33 p ∧ ( p ∨ q ) is F, when p ≡ F p ∨ ( p ∧ q ) is F, when p ≡ F, q ≡ F and q → [ p ∧ ( p → q )] is F, when p ≡ F, q ≡ T So, for [ p ∧ ( p → q )] → q ≡ [ p ∧ (~ p ∨ q )] → q ≡ [{ p ∧ (~ p )} ∨ ( p ∧ q )] → q

34 Suman is brilliant and dishonest, if and only if Suman is rich, is expressed as, Q ↔ (P ∧ ~ R ) So, negation of it will be ~ (Q ↔ (P ∧ ~ R )).

35 Hence, it is neither tautology nor contradiction.

28 p

~p

p ⇒~p ~p ⇒ p

( p ⇒ ~ p ) ∧ (~ p ⇒ p )

F

T

T

F

F

T

F

F

T

F

p

q

~p

p⇒q

~p∧q

( p ⇒ q ) ⇔ (~ p ∧ q )

T

T

F

T

F

F

T

F

F

F

F

T

F

T

T

T

T

T

F

F

T

T

F

F

36 By truth table

∴Statement is contradiction.

~ p ~ q p ⇒ q ~( p ⇒ q ) ~ p ∨~ q

p T

q

~ p ~q ⇒~ p

~q

T

F

F

T

p⇔q

q

( p ⇒ q) ⇔ (~ q ⇒ ~ p )

T

T

F

F

T

F

F

T

T

F

F

T

F

T

T

T

T

F

T

T

F

T

F

T

T

F

F

T

T

T

F

T

T

T

T

F

T

F

F

F

T

F

T

F

T

T

T

T

F

F

T

T

T

T

T

Hence, given proposition is a tautology.

37 Since, S : ~ ( p ∨ (~ p ∨ q )) ⇒ S : ~ p ∧ ~ (~ p ∨ q ) (De-Morgan’s Law) ⇒ S : ~ p ∧ ( p ∧ ~ q ) (De-Morgan’s law)

Hence, it is tautology.

30 p

q

p⇒q

T

T

T

T

F

F F

~ ( p ⇒ q) ⇒ (~ p ∨~q )

p

29

38 ~ (~ s ∨ (~ r ∧ s)) ≡ s ∧ (~ (~ r ∧ s)) ≡ s ∧ (r ∨ ~ s)

(~ p ∨ q )

( p ⇒ q ) ⇔ (~ p ∨ q )

F

T

T

F

F

F

T

39 Q

T

T

T

T

T

40

F

T

T

T

T

~p

≡ (s ∧ r ) ∨ (s ∧ ~ s) ≡ (s ∧ r ) ∨ F ≡ s∧r

[Q s ∧ ~ s is false]

~ S ( p, q ) = ~ ( p ∧ q ) = (~ p ) ∨ (~ q ) = S * (~ p, ~ q ) I. Compound statement with ‘AND’ is 25 is a multiple of 5 and 8. This is a false statement. Since, p is true but q is false.

Since, all values of given proposition is true, hence it is a tautology.

31 q ↔ q p ↔ ~ q ~ p ↔ q ~( p ↔ ~ q )

p

q

~p

~q

T

F

F

T

F

T

T

F

F

T

T

F

F

T

T

F

T

T

F

F

T

F

F

T

F

F

T

T

T

F

F

T

~ ( p ↔ ~ q ) is equivalent to ( p ↔ q ).

32 p → q is equivalent to ~ q → ~ p ∴ ( p → q ) ↔ (~ q → ~ p ) is a tautology but not a contradiction.

[since, 25 is divisible by 5 but not divisible by 8] II. Compound statement with ‘OR’ is 25 is a multiple of 5 or it is a multiple of 8. This is a true statement. Since, p is true and q is false.

SESSION 2 1 Given that, p : 4 is an even prime number. q : 6 is a divisor of 12 and r : the HCF of 4 and 6 is 2. So, the truth value of the statements p,q and r are F, T and T, respectively. Hence, ~ p ∨ (q ∧ r ) is true.

2 ( p ⇒ q ∧ r ) ∨ ( r ⇔ s) ≡ ( p ⇒ q ∧ r ) ∨ [(~ r ∨ s) ∧ (r ∨ ~ s)] ≡ (~ p ∨ q ∧ r ) ∨ [(~ r ∨ s ) ∧ (r ∨ ~ s )] [Q p ⇒ q ∧ r ≡ ~ p ∨ (q ∧ r )]

SIX

404 3 Truth table p

q

8 ~ p ~r

r

p ∧~ r

~p∨q

( p ∧ ~r ) ⇒ (~p ∨ q )

p

q

~p

~q

( p ∧ q)

~ ( p ∧ q)

T

T

F

F

T

F

T

T

T

F

F

F

T

T

T

F

F

T

F

T

T

T

F

F

T

T

T

T

F

T

T

F

F

T

T

F

T

F

F

F

F

T

F

F

T

T

F

T

T

F

F

F

T

T

F

F

F

T

T

T

F

F

T

T

F

T

F

T

T

F

T

F

F

T

T

F

F

F

F

F

T

T

F

~ p∨ ~q

q∧ p

p⇔q

p⇒q

q⇒ p

( p ⇔ q) ∨ ( q ⇒ p)

T

F

T

T

T

T

T

T

T

T

F

F

F

T

T

T

T

T

F

F

T

F

T

T

F

T

T

F

T

Since, ( p ∧ ~ r ) ⇒ (~ p ∨ q ) is F. Then, p = T, q = F, r = F

It is clear from the table that false statement is p ⇔ q ≡ ( p ⇒ q ) ∨ (q ⇒ p )

4 Consider, ( p ∧ ~ q ) ∨ q ∨ (~ p ∧ q ) ≡ [( p ∧ ~ q ) ∨ q] ∨ (~ p ∧ q ) ≡ [( p ∨ q ) ∧ (~ q ∨ q )] ∨ (~ p ∧ q ) ≡ [( p ∨ q ) ∧ t ] ∨ (~ p ∧ q ) ≡ ( p ∨ q ) ∨ (~ p ∧ q ) ≡ ( p ∨ q ∨ ~ p) ∧ ( p ∨ q ∨ q ) ≡ (q ∨ t ) ∧ ( p ∨ q ) ≡ t ∧ ( p ∨ q ) ≡ p ∨ q

Hence, it is clear from the table that p ⇔ q and ( p ⇒ q ) ∨ (q ⇒ p ) is not logically equilibrium.

9

5 Truth Table p

q

T T F F

p→ q q→ p

T F T F

T F T T

T T F T

( p→ q ) ∧ ( p→ q )

( p→ q ) ∨ ( p→ q )

( p→ q ) ∨ ( q→ p )

T F T T

T F T T

T T T T

So, only ( p → q ) ∨ (q → p ) is a tautology.

6 Since, p is true and q is false ⇒ ~ p is false and ~ q is true. p ↔ q is F ~ p ↔ q is T ~ q ↔ p is T ~ p ↔ ~ q is F

[since, p is true, q is false] [since, ~ p is false, q is false] [since, ~ q is true, p is true] [since, ~ p is false, ~ q is true]

p

q

p⇔q

~ ( p ⇔ q)

T

T

T

F

T

F

F

T

F

T

F

T

F

F

T

F

~p

~q

~ p∨ ~q

~ ( p ⇔ q) ⇔ ~ p ∨~ q

F

F

F

T

F

T

T

T

T

F

T

T

T

T

T

F

Last column shows that result is neither a tautology nor a contradiction.

10 The truth tables of p → q and ~ p ∨ q are given below 7

A

B

A∨B

A∧B

A ∨ ( A ∨ B)

A ∨ ( A ∧ B)

p

q

~p

p→ q

~p ∨q

T

T

T

T

T

T

T

T

F

T

T

T

F

F

F

F

F

T

T

T

T

F

F

T

T

T

T

F

T

F

T

T

F

T

T

F

T

F

F

F

F

F

F

F

A→ B

A ∧ (A → B)

A ∧ ( A→ B ) → B

B→ ( A ∧ ( A → B ))

T

T

T

T

F

F

T

T

T

F

T

F

T

F

T

T

Since, the truth value of all the elements in the column A ∧(A → B)→ B So, A ∧ ( A → B ) → B is tautology.

Clearly, truth tables of p → q and ~ p ∨ q are same. So, p → q is logically equivalent to ~ p ∨ q . Hence, option (a) is correct. If the truth value of p, q, r are T, F, T respectively, then the truth values of p ∨ q and q ∨ r are each equal to T. Therefore, the truth value of ( p ∨ q ) ∧ (q ∨ r ) is T. Hence, option (b) is correct. We know that, ~ ( p ∨ q ∨ r ) ≅ (~ p ∧ ~ q ∧ ~ r ) So, option (c) is correct. If p is true and q is false, then p ∨ q is true. Consequently ~ ( p ∨ q ) is false and hence p ∧ ~ ( p ∨ q ) is false. Hence, option (d) is wrong.

DAY 11

12.

p

q

r

p∧q

p∧r

( p ∧ q) ∧ ( q ∧ r )

T

F

F

F

F

F

T

F

T

F

T

F

T

T

F

T

F

F

T

T

T

T

T

T

F

F

F

F

F

F

F

F

T

F

F

F

F

T

F

F

F

F

F

T

T

F

F

F

p

q

T

T

~p ~p ∧q p ⇒ q F

F

14 Statement II ( p → q ) ↔ (~ q → ~ p) ≡ (p → q) ↔ (p → q) which is always true, so Statement II is true. Statement I ( p ∧ ~ q ) ∧ (~ p ∧ q ) ≡ p ∧ ~q ∧ ~ p ∧ q ≡ p ∧ ~ p ∧ ~q ∧ q ≡ f ∧ f ≡ f Hence, it is a fallacy statement. So, Statement I is true. Alternate Method Statement II ( p → q ) ↔ (~ q → ~ p) ~ q → ~ p is contrapositive of p → q Hence, ( p → q ) ↔ ( p → q ) will be a tautology. Statement I ( p ∧ ~ q ) ∧ (~ p ∧ q ) p

q

~p

~q

p ∧ ~q

~p ∧q

( p ∧ ~ q) ∧ (~p ∧ q)

F

T

T

F

F

F

F

F

q ⇒ p ~ ( p⇒q) ~( q⇒ p)

T

T

F

T

F

F

F

F

T

T

F

T

F

F

T

T

F

F

F

T

T

T

T

F

F

T

F

T

T

F

F

T

F

F

F

T

F

T

T

F

F

F

F

T

T

F

F

F

It is clear from the above table that columns 4 and 8 are equal. Hence, ~ p ∧ q is equivalent to ~ (q ⇒ p).

13 ~ p ~q

~p∧ q

p ∨ ~ q p ∨ ~ p ~ [ p ∨ ~ q]

p

q

T

T

F

F

F

T

T

F

T

F

F

T

F

T

T

F

F

T

T

F

T

F

T

T

F

F

T

T

F

T

T

F

So, ~ p ∧ q ≡ ~ [ p ∨ ~ q] and p ∨ ~ p is a tautology.

Hence, it is a fallacy.

15 A

B

A∨B

B→ A

A∧B

~ A ~ A ∨ B A → ( A ∨ B)

T

T

T

T

T

F

T

T

T

F

T

T

F

F

F

T

F

T

T

F

F

T

T

T

F

F

F

T

F

T

T

T

406

DAY THIRTY SEVEN

Unit Test 6 (Statistics, Probability & Mathematical Reasoning) 1 While shuffling a pack of playing cards, four are accidently dropped. The probability that the cards are dropped one from each suit is 1 (a) 256

2197 (b) 20825

3 (c) (d) None of these 20825

2 If p : Ajay works hard, q : Ajay gets good marks, then proposition ~ p ⇒ ~ q is equivalent to (a) Ajay does not work hard and yet he gets good marks (b) Ajay work hard if and only if he gets good marks (c) If Ajay does not work hard, then he does not get good marks (d) None of the above

3 ~ [( p ∨ q ) ∧ ~ ( p ∧ q )] is equivalent to (a) p ⇔ q (c) ~ (p ⇔ q)

(c) 1/4

(d) 2/5

5 For a frequency distribution consisting of 18 observations, the mean and the standard deviation were found to be 7 and 4, respectively. But on comparison with the original data, it was found that a figure 12 was miscopied as 21 in calculations. The correct mean and standard deviation are (a) 6.7, 2.7 (c) 6.34, 2.34

(b) 6.5, 2.5 (d) None of these

6 A boy is throwing stones at a target. The probability of hitting the target at any trial is 1/2. The probability of hitting the target 5th time at the 10th throw is (a)

5 210

(b)

63 29

10

(c)

C5 210

10

(d)

30 + 25 + 50 km/h 3 3 (c) km/h 1 1 1 + + 30 25 50

(b)

(a)

25 + 35 + 15 km/h 3

(d) None of these

given below

8 without repeating the digits. If a number so formed is chosen at random, probability that it is divisible by 20 is (b) 1/3

120 km distant at an average speed of 30 km/h. Then, he makes the return trip at an average speed of 25 km/h. He covers another 120 km distance on plane at an average speed of 50 km/h. His average speed over the entire distance of 360 km will be

8 The marks obtained by 60 students in a certain test are

(b) ~ p ∧ q (d) None of these

4 Five-digit numbers are formed using the digits 0, 2, 4, 6,

(a) 1/2

7 An automobile driver travels from plane to a hill station,

C4 210

Marks

10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

No. of Students

2

3

4

5

6

12

14

10

4

Find the median of the above data. (a) 68.33 (c) 71.11

(b) 70 (d) None of these

9 Assuming ( p ∨ q ) is true and ( p ∧ q ) is false, state which of the following proposition have true values? (a) ~ p ∧ q

(b) ~ p ∨ ~ q (c) p ⇔ q (d) None of these

10 Let S be the universal set and n ( X ) = k . The probability of selecting two subsets A and B of the set X such that B = A , is (a)

1 2

(b)

1 2k − 1

(c)

1 2k

(d)

1 3k

11 The probability that when 12 balls are distributed among three boxes, the first box will contain three balls, is (a)

29 312

12

(b)

C3 ⋅ 2 9 312

12

(c)

C3 ⋅ 212 312

(d) None

407

DAY 12 A fair coin is tossed 100 times. The probability of getting tails an odd number of times is 1 (a) 2

1 (b) 8

month is a Saturday, is

3 (c) 8

(d) None of these

13 One mapping 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 to one is given by 1 (a) n n (n − 1)! (c) n n −1

1 (b) n! (d) None of these

X = {x : 1 ≤ x ≤ 100}. The probability that the number satisfies the inequation x 2 − 13x ≤ 30 is 9 50

(b)

3 20

(c)

2 11

(d) None of these

15 If 0 < P ( A ) < 1, 0 < P (B ) < 1

(A /B) + P (B / A) = 1

throws a die and reports that it is a six. The probability that it is actually a six is 1 (b) 5

3 (c) 4

(d) None of these

17 A letter is taken at random from the letters of the word ‘STATISTICS’ and another letter is taken at random from the letters of the word ‘ASSISTANT’. The probability that they are the same letters is (a)

1 45

(b)

13 90

(c)

19 90

(d)

5 18

2 2 is inscribed in a circle and 3 a point within the circle is chosen at random. The probability that this point lies outside the ellipse is

18 An ellipse of eccentricity

1 (a) 3

2 (b) 3

1 (c) 9

2 (d) 9

19 Four-digit numbers are formed using each of the digit 1, 2, ..., 8 only once. One number from them is picked up at random. The probability that the selected number contains unity is 1 2 1 (c) 8 (a)

(b)

1 4

(d) None of these

20 The mean of 10 numbers is 12.5, the mean of the first six is 15 and the last five is 10. The sixth number is (a) 12 (c) 18

(b) 15 (d) None of these

21 The geometric mean of numbers 7, 72 , 73 , ..., 7n is (a) 7 7 / n (c) 7 ( n − 1) / 2

(b) 7 n / 7 (d) 7 ( n + 1) / 2

1 7

(d) None of these

23 The mean of the numbers 30

30 C0 30C2 30C4 C30 , , ,......, equals to 1 3 5 31

2 30 31 2 26 (c) 31 × 15

(b)

2 26 31

(d) None of these

24 If the mean of a binomial distribution is 25, then its standard deviation lies in the interval (a) [0, 5) (c) [0, 25)

(b) (0, 5] (d) (0, 25)

symbolic form of “it is not true that she is not intelligent or she is not studious” is

B  = 0    A

16 A man is known to speak the truth 3 out of 4 times. He

3 (a) 8

(b)

25 Let p : She is intelligent and q : She is studious. The

and P ( A ∪ B ) = P ( A ) + P (B ) − P ( A ) P (B ) , then A (a) P   = 0 (b) P B  (c) P (A′ ∩ B ′) = P (A′) P (B ′) (d) P

1 12 1 (c) 84 (a)

(a)

14 A natural number is selected at random from the set

(a)

22 The probability that the 14th day of a randomly chosen

(a) p ∧ ~ q (c) p ∧ q

(b) ~ p ∧ q (d) None of these

26 The negation of p → (~ p ∨ q ) is (a) p ∨ (p ∨ ~ q) (c) p → q

(b) p →~ (p ∨ q) (d) p ∧ ~ q

27 The proposition of ( p ∨ r ) ∧ (q ∨ r ) is equivalent to (a) (p ∧ q) ∨ r (c) p ∧ (q ∨ r )

(b) (p ∨ q) ∧ r (d) p ∨ (q ∧ r )

28 Which of the following statement has the truth value ' F ' ? (a) A quadratic equation has always a real root. (b) The number of ways of seating 2 persons in two chairs out of n persons is P (n, 2). (c) The cube roots of unity are in GP. (d) None of the above

29 The variates x and u are related by hu = x − a, then correct relation between σ x and σu is (a) σ x = hσu (c) σ x = a + hσu

(b) σu = hσ x (d) σu = a + hσ x

30 Given that, x ∈[ 0 , 1] and y ∈[ 0 , 1]. If A is the event of ( x , y ) satisfying y 2 ≤ x and B is the event of ( x , y ) satisfying x 2 ≤ y . Then, 1 3 (b) A, B are exhaustive (c) A, B are mutually exclusive (d) A, B are independent

(a) P (A ∩ B) =

31 There are two independent events A and B. The 1 and the 8 1 probability that neither of them occurs is . Then, the 4 probability of the two events are, respectively probability that both A and B occurs is

408 7 ± 17 2 , 16 7 ± 17 1 1 1 1 (c)  ,  ,  ,   3 2  2 3 (a)

(b)

5 ± 14 3 , 16 5 ± 14

Direction (Q. Nos. 37-40) Each of these questions contains

(d) None of these

32 If p ⇒ (q ∨ r ) is false, then the truth values of p, q , r are

two statements : Statement I (Assertion) and Statement II (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a ), (b), (c) and (d ) given below. (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

respectively. (a) T, F, F

(b) F, F, F

(c) F, T, T

(d) T, T, F

33 In a factory, workers work in three shifts say shift A, shift B and shift C and they get wages in the ratio 4 : 5 : 6 depending on the shift A, B and C, respectively. Number of workers in the shifts are in the ratio 3: 2 :1. If total number of workers is 1500 and wages per worker in shift A is ` 400. The mean wage of a worker is (a) ` 467

(b) `500

(c) ` 600

(d) ` 400

34 The odd in favour of standing first of three students appearing in an examination are 1:2, 2:5 and 1:7, respectively. The probability that either of them will stand first, is (a)

125 168

(b)

35 Median of (a)

2n

75 168

(c)

32 168

(d)

4 168

C0 ,2 n C1,2 n C2 ,......, 2 nCn (when n is odd) is

 1  2n  C( n − 1) + 2 nC( n + 1)  2  2 2

A = ( x , x + 2, x + 4) and B = ( x − 2, x + 2, x + 6) Statement I Group B has more variability than group A. Statement II The value of mean for group B is more than that of group A. 38 Let p : He is poor. q : He is happy. Statement I The symbolic form of the statement “It is not true that if he is poor, then he is happy” is ~ ( p ⇒ q ). Statement II The negation of the above statement is ( p ⇒ ~ q ). 39 Let A and B be two independent events. Statement I If P ( A ) = 0 . 3 and P ( A ∪ B ) = 0.8, then P (B ) 2 is . 7 Statement II P (E ) = 1 − P (E ) , where E is any event. 40 Statement I The statements ( p ∨ q ) ∧ ~ p and ~ p ∧ q are logically equivalent.

(b) 2 nCn / 2 (c) 2 nCn (d) None of the above

36 The contrapositive of ( p ∨ q ) → r is (a) ~ r → (p ∨ q) (c) ~ r → (~ p ∧ ~ q)

37 Suppose two groups of scores A and B are such that

(b) r → (p ∨ q) (d) p → (q ∨ r )

Statement II The end columns of the truth table of both statements are identical.

ANSWERS 1 (b)

2 (c)

3 (a)

4 (c)

5 (b)

6 (b)

7 (c)

8 (a)

9 (b)

10 (b)

11 (b)

12 (a)

13 (c)

14 (b)

15 (c)

16 (a)

17 (c)

18 (b)

19 (a)

20 (b)

21 (d)

22 (c)

23 (b)

24 (a)

25 (c)

26 (d)

27 (a)

28 (a)

29 (a)

30 (a)

31 (a)

32 (a)

33 (a)

34 (a)

35 (a)

36 (c)

37 (c)

38 (c)

39 (b)

40 (a)

DAY

409

Hints and Explanations 4 1 Required probability = 5213 = 2197

C4

20825

120 + 120 + 120 120 120 120 + + 30 25 50 3 km/h = 1 1 1 + + 30 25 50

2 Given, proposition is equivalent to “If Ajay does not work hard, then he does not get good marks”.

3 It is clear from the table that column IIIrd and IVth are identical. p

q

p⇔q

~ [( p ∨ q ) ∧ ~ ( p ∧ q )]

T

T

T

T

T

F

F

F

F

T

F

F

F

F

T

4 The total number of numbers formed is

6 Here, p = 1 and q = 1

2 2 ∴ Required probability 4 5 1 1 1 = 9C 4     ×  2  2 2 10

1 = 9C 4    2 9× 8 ×7 × 6 63 = 9 = (1 × 2 × 3 × 4) × 210 2

of the three boxes, so that total number of ways in which 12 balls can be put into three boxes is 312 . The three balls can be chosen in 12 C3 ways and remaining 9 balls can be put in the remaining 2 boxes in 29 ways. ∴Required probability 12 C 3 ⋅ 29 = 312

8

T

5! − 4! = 96. If a number is divisible by 20, then the last digit is 0 and tens digit must be even. The number of such numbers with zero at the end is 4! = 24. 24 1 = ∴ Required probability = 96 4 Σ xi 5 Given, =7 18 ⇒ Σ x i = 126 Now, correct Σ x i = 126 − 21 + 12 = 117 Σ x i 117 = = 6. 5 ∴ True mean = 18 18 2 Σ xi Since, − (Mean)2 = 42 18 Σ x2i = 42 + (7) 2 ∴ 18 ⇒ Σ x2i = 1170 Now, correct Σ x2i = 1170 − 212 + 122 = 873 ∴True variance Σ x2i = − (Mean)2 18 873 = − (6.5)2 18 = 48.5 − 42.25= 6.25 ∴ True standard deviation = True variance = 6.25 = 2.5

11 Since, each ball can be put into any one

7 Average speed =

Class

fi

cf

10-20

2

2

20-30

3

5

30-40

4

9

40-50

5

14

50-60

6

20

60-70

12

32

70-80

14

46

80-90

10

56

90-100

4

60

12 P ( X = odd number) = P ( X = 1) + P ( X = 3) + K + P ( X = 99) 1 = 100C1    2

= (100 C1 + 1 = 299 .    2

~ p ∨~ q

p⇔ q F

T

F

F

T

F T

T

F

T

T

1 C3    2

C3 + K +

100

100

=

100

+K

100

1 C 99    2

100

100

1 C 99 )    2

100

100

1 2

13 Total number of cases = n n

9

T F

+

+

N Here, N = 60 ⇒ = 30 2 ⇒ median class is 60-70. N −c ×h ∴ Median = l + 2 f 30 − 20 = 60 + × 10 12 100 = 60 + 12 = 68.33 p q p ∨ q p ∧ q ~p ∧q

100

F k

10 Total number of subsets of X is 2 .

∴The number of favourable cases = n(n − 1) …2 ⋅ 1 = n! ∴ Required probability n ! (n − 1)! = n = n n n −1 14 Total number of ways = 100 Given, x2 − 13 x ≤ 30 2  x − 13  ≤ 289 ⇒    2 4 17 13 17 ⇒− ≤ x− ≤ 2 2 2 ⇒ −2 ≤ x ≤ 15 Q x∈N ∴ x ∈ {1, 2, 3, …,15} ∴ Required probability

∴ Total number of possible out comes k = 2 C2 Let n(E ) = The number of selections of two non-intersecting subsets whose union is X . 1 = ( k C 0 + kC1 + kC2 + … ) 2 1 = × 2k 2 ∴ Required probability 1 × 2k 2k −1 2 = k = 2  2k − 1  C2 2k    2  1 = k 2 −1

=

15 3 = 100 20

15 Given,

P ( A ∪ B ) = P ( A ) + P (B ) − P ( A ) P (B ) ∴ P ( A ∩ B ) = P ( A ) P (B ) Q A and B are independent events. ⇒ P ( A ′∩ B ′ ) = P ( A ′ ) P ( B ′ )

16 Let E = The event that six occurs and A = The event that the man reports that it is a six P ( E ) P ( A /E ) E ∴ P   =  A  P (E ) P ( A /E ) + P (E ′ ) P ( A /E ′ ) (1 /6)(3 /4) 3 = = (1 /6)(3 /4) + (5/6)(1 /4) 8

410

17 Letters of the word STATISTICS are A, C, I, I, S, S, S, T, T, T. Letters of the word ASSISTANT are A, A, I, N, S, S, S, T, T. Common letters are A, I, S and T. Probability of choosing A is 1 2 2 . × = 10 9 90 Probability of choosing I is 2 1 2 . × = 10 9 90 Probability of choosing S is 3 3 9 . × = 10 9 90 Probability of choosing T is 3 2 6 . × = 10 9 90 ∴ Probability of required event 2 2 9 6 19 = + + + = 90 90 90 90 90

18 Let equation of ellipse be y2 x2 + 2 = 1 (where, a > b ) 2 a b

Let the sixth number be x, then 35 1× x+ 4× 4 10 = 1+ 4 ⇒ ⇒

26 Clearly, p → (~ p ∨ q ) ≡ (~ p ) ∨ ((~ p ) ∨ q ) ≡ ((~ p ) ∨ (~ p )) ∨ q [by associative law] ≡ (~ p ) ∨ q [by idempotent law]

50 = x + 35 x = 15

Now, ~ ( p → (~ p ∨ q )) ≡ ~ ((~ p ) ∨ q )

21 GM = (7 ⋅ 72 K 7n )1 /n  n( n + 1 )  =  7 2   

1/n

≡ ~ (~ p ) ∧ (~ q ) n +1

=7

≡ p∧ ~ q

2

27 Using Distributive law,

1 can be chosen with probability = 12 Now, as there are 7 possible ways in which the month can start and it will be a Saturday on 14th day if the first day of the month is Sunday. 1 ∴ Its probability = 7 Hence, the required probability 1 1 1 . = × = 12 7 84

23 Consider (1 + x ) = C 0 + 30

A

a

C

a

B

πa 8 = 9 3 and area of circle = π a2 Now, required probability π a2 π a2 − 3 =2 = π a2 3

19 The total number of four-digit numbers formed with the digits 1, 2,K , 8 is 8 C 4 × 4!. Now, the total number of four-digits numbers formed with the digits 1, 2, …, 8 and containing unity as one of the digits is 7 C3 × 4!. ∴ Required probability 7 C × 4! 1 = 8 3 = C 4 × 4! 2

20 Let the mean of the last four digit be A 2 . 6 × 15 + 4 × A 2

⇒

125 = 90 + 4 A 2 35 A2 = 4

⇒

C0 +

6+ 4

C1 x +

C2 x + K +

30

2

 30 C0x +  

2

12.5 =

30

30

2

30

28 Clearly, the roots of a quadratic equation can be imaginary. So, this statement has truth value F. x a 29 Here, u= − h h We know that standard deviation is not dependent on change of origin. σ ∴ σ u = x ⇒ σ x = hσ u h

30 A = The event of ( x, y ) belonging to the area OTQPO and B = The event of ( x, y ) belonging to the area OSQRO

y2 = x

30

C30 x

x2 = y

Y

1

R Q

1 = [(1 + x )30 + (1 − x )30 ] 2 Now, on integrating over 0 to 1, we get

∴ Area of an ellipse = π ab = πa × a 1 − e 2

Then,

30

C2 x + ....+ 30C30 x30 ...(i) and (1 − x )30 = 30C 0 −30C1 x + 30C2 x2 − ... + 30C30 x30 ...(ii) On adding Eqs. (i) and (ii), we get 30

= π a2 1 −

( p ∨ r ) ∧ (q ∨ r ) ≡ ( p ∧ q ) ∨ r

22 Clearly any month, out of 12 months,

C0 + 1

C2 x3 + ....+ 3

1  (1 + x ) 2  31

31

= 30

30

30

−

1

C30 x31  31  0

30

1

(1 − x )  31  0 31

30

C2 C30 + ....+ ⇒ 3 31 1 230 = [231 ] = 2 × 31 31 Now, as there are 16 terms, therefore required mean 30 30 C 0 30C2 C30 + + ....+ 1 3 31 = 16 230 230 226 = 31 = = 16 31 × 16 31

24 Since, ⇒ ⇒

0≤

npq
8

(d) π

418

10 The two consecutive terms in the expansion of ( 3 + 2x )74 whose coefficients are equal, are (a) 11, 12 (c) 30, 31

(b) 7, 8 (d) None of these

11 A parabola is drawn with its focus at (3, 4) and vertex at the focus of the parabola y 2 − 12x − 4y + 4 = 0. The equation of the parabola is (a) y − 8 x − 6y + 25 = 0 (c) x 2 − 6x − 8 y + 25 = 0 2

(b) y − 6x + 8 y − 25 = 0 (d) x 2 + 6x − 8 y − 25 = 0 2

12 If p, p′ denote the lengths of the perpendiculars from the 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 + 1

(d) ap = rp ′

cuts axes at A, B and C, then the locus of centroid of tetrahedron OABC is 1 1 1 16 (a) 2 + 2 + 2 = 2 x y z P 1 1 1 3 (c) 2 + 2 + 2 = 2 x y z P

1 1 1 1 (b) 2 + 2 + 2 = 2P 2 x y z (d) None of these

14 The equation of the locus of the pole with respect to the x2 y2 + 2 = 1, of any tangent line to the auxiliary 2 a b x2 y2 circle, is the curve 4 + 4 = λ2 , where a b

ellipse

1 a2 1 (d) λ 2 = 2 b

(a) λ 2 = a 2 (c) λ = b

(b) λ 2 =

2

15 The solution set for

curve y = f ( x ), which cuts the last two lines above the 2 first line for all a > 1, is equal to [( 2a )3 / 2 − 3a + 3 − 2 2 ].. 3 Then, f ( x ) equals (a) 2 2 x , x > 1 (c) 2 x , x > 1

(b) 2 x , x > 1 (d) None of these

20 Area of a triangle with vertices (a, b ), ( x1, y1 ) and ( x 2 , y 2 ),

1 1 ab (r − 1) (s − 1) (s − r ) (b) ab (r + 1) (s + 1) (s − r ) 2 2 (c) ab (r − 1) (s − 1) (s − r ) (d) None of these

( 2 x + 3 ) ( 4 − 3 x )3 ( x − 4 ) ≤0 ( x − 2 )2 x 5

16 A parallelogram is constructed on the vectors a = 3 α − β , b = α + 3 β, if | α | = | β | = 2 and angle between α and β is π , then length of a diagonal of the parallelogram is 3 (b) 4 3 (d) None of these

17 f ( x ) = x − cot −1 x − log( x + 1 + x 2 ) is increasing in (b) (−∞, 2)

(c) (2, 5)

AC of a ∆ABC are x − y + 5 = 0 and x + 2y = 0 , respectively. If the coordinates of vertex A are (1, − 2), then equation of BC is (a) 14 x + 23 y − 40 = 0 (c) 23 x + 14 y − 40 = 0

(b) 14 x − 23 y + 40 = 0 (d) 23 x − 23 y + 40 = 0

sin θ 1   1  22 Let A = − sin θ sin θ  , where 0 ≤ θ ≤ 2π. 1   1  − sin θ  −1 Then, the range of | A | is (b) {2, 4} (d) None of these

23 If | z − 25 i | ≤ 15, then | max amp ( z ) − min amp ( z )| is equal to 3 (a) cos−1   5 π 3 (c) + cos−1   5 2

3 (b) π − 2 cos−1   5 3 3 (d) sin−1   − cos−1   5 5

24 If ABCD is a cyclic quadrilateral such that

(c) (−∞, 0) ∪ (2, ∞) (d) None of the above

(a) 4 5 (c) 4 17

21 The equation of perpendicular bisectors of sides AB and

(a) 0 (c) [2, 4]

3 4 (a)  − ∞, −  ∪  0,  ∪ (4 , ∞)  2  3 3 4 (b)  − , 0 ∪  , 4   2  3 

(a) (−∞, ∞)

19 The area bounded by the lines y = 2, x = 1, x = a and the

(a)

13 A variable plane at a constant distance P, from origin

2

(~ r ∨ ~ s) → (~ p ∧ ~ q) ( r ∧ s) → (~ p ∧ ~ q) (~ r ∧ ~ s) → (~ p ∨ ~ q) ( r ∨ s) → (~ p ∨ ~ q)

where a, x1 and x 2 are in GP with common ratio r and b, y1 and y 2 are in GP with common ratio s, is given by

(b) rp = ap ′

1 (c) ap = rp ′ + 3

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

(d) (−∞, 10)

18 The contrapositive of the statement, ‘‘if x is a prime number and x divides ab, then x divides a or x divides b’’, can be symbolically represented using logical connectives on appropriately defined statements p, q , r and s as

12 tan A − 5 = 0 and 5 cos B + 3 = 0, then the quadratic equation whose roots are cos C and tan D is (a) (b) (c) (d)

39x 2 + 88 x + 48 = 0 39x 2 − 16x − 48 = 0 39x 2 − 88 x + 48 = 0 None of the above

25 A mirror and a source of light are situated at the origin O and a point on OX , respectively. A ray of light from the source strikes the mirror and is reflected. If the DR’s of the normal to the plane of mirror are 1, −1, 1, then DC’s for the reflected ray are 1 2 2 , , 3 3 3 1 2 2 (c) − , − , − 3 3 3 (a)

1 2 2 , , 3 3 3 1 2 2 (d) − , − , 3 3 3 (b) −

419

DAY 26 If both roots of the equation 4x 2 − 2x + a = 0, a ∈R, lies in the interval (−1, 1), then (a) −2 < a ≤ (c) a >

1 4

1 4

27 The value of the expression 1 π cos  − x  4  2 π   (c) 2 sin  − x  4  (a)

(b) −6 ≤ a ≤

1 4

(d) a < − 2

sin3 x cos 3 x is + 1 + cos x 1 − sin x (b)

π 2 cos  + x  4 

(d) None of these

28 Let λ and θ be real numbers. Then, the set of all values of λ for which the system of linear equations λx + (sin θ)y + (cos θ)z = 0 x + (cos θ)y + (sin θ)z = 0 −x + (sin θ)y − (cos θ)z = 0 has a non-trivial solution, is (a) [0, 2]

(b) [− 2,0]

(c) [− 2, 2]

(d) None of these

29 Suppose , f (x ) = x sin x −

1  sin2 x, x ∈  0,  2

π  , then range 2

of f (x ) is (a) (0, 1) (c)  0, 

π − 1  2 

π (b)  0,   2 (d) Cannot be determined

30 Consider the following statements Statement I The number of ways a lawn tennis mixed double to be made up from seven married couples, if no husband and wife play in the same set are 840. Statement II The number of words beginning with T and ending with E on arranging letters of the word ‘TRIANGLE’ are 720. Choose the correct option. (a) (b) (c) (d)

Only Statement I is correct Only Statement II is correct Both I and II are correct Neither I nor II is correct

420

Hints and Explanations 30

∑ n( A

1 (d) Let the weight of teacher be x kg, then

1 35 × 40 + x 40 + = 2 35 + 1 81 × 36 = 35 × 40 + x ⇒ 2 ⇒ 81 × 18 = 1400 + x ⇒ 1458 = 1400 + x ⇒ x = 58 Hence, the weight of teacher is 58 kg.

2 (a) Let I = =

∫

π/6 0

∫

π 0

[2sin x] dx + +∫

= 0+∫

π /2 π /6

5π / 6 π /2

∫

π /2 π /6

[2sin x] dx

[2sin x] dx +

1 dx +

∫

5π / 6 π /2

∫

π 5π / 6

[2sin x] dx

1 dx + 0

= [ x] ππ //26 + [ x] 5ππ/2/ 6 π π 5π π 2π = − + − = 2 6 6 2 3 1 3 (a) We have, f ( x ) = [sin x] Clearly, sin x ∉ [0,1) [Qif 0 ≤ sin x < 1,[sin x] = 0] π ⇒ x ∉ [2nπ,(2n + 1)π] − (4n + 1) , n ∈ I 2 Thus, f ( x ) is not continuous if x ∈ (2nπ,2nπ + π ), n ∈ I .

4 (a) Given that, ∫ f ( x ) sin x cos x dx 1 = log [ f ( x )] + C 2(b 2 − a2 ) On differentiating both sides, we get d  log[ f ( x )] f ( x )sin x cos x = + C   dx  2(b 2 − a2 )  1 1 ⇒ f ( x )sin x cos x = . f ′( x ) 2(b 2 − a2 ) f ( x ) f ′( x ) ⇒ 2(b 2 − a2 )sin x cos x = [ f ( x )]2

) = 5 × 30 = 150

j

) = 3n

j

) = 9m

n

∑ n(B

and

⇒

j =1

∴

3n = 9m 9m n= = 3 × 15 = 45 3

⇒

⇒

∫

1

∫ x dx + log k

( x − 3)2 = 8( y − 2) ⇒ x2 − 6 x − 8 y + 25 = 0 2 y2 12 (d ) Tangent to the ellipse x2 + 2 = 1

a b at P (a cos θ, b sin θ) is y x cos θ + sin θ = 1 …(i) a b Now, p = perpendicular distance from focus S (ae , 0) to the line (i)

p+q =1 2 1 ⇒ p = and q = 3 3 Now, required probability

Q

2

…(i)

i =1

If there are m distinct elements in S and each element of S belongs to exactly 10 of the A i ′s, then

3

2 1 2 1 = 4 C2     + 4 C3      3  3  3  3

1

2 + 4C 4    3

2 2 π + sin −1 x ≤ π 2 ⇒ 0 ≤ tan −1 x + cot −1 x + sin −1 x ≤ π ⇒ a = 0 and b = π Hence, a + b = π

9 (b) Check through the options. The condition is true for n ≥ 5.

C r +1 3 = Cr 2

74

( y − 2)2 = 12 x Here, vertex and focus are (0, 2) and (3, 2). ∴Vertex of the required parabola is (3, 2) and focus is (3, 4). The axis of symmetry is x = 3 and latusrectum = 4 ⋅ 2 = 8 Hence, required equation is

7 (b) Given, p = 2q

2

74

11 (c) Given, equation can be rewritten as

y φ  = kx  x

⇒

C r 374 − r 2r = 74C r + 1 374 − ( r + 1 )2r + 1

74 − r 3 = r+1 2 ⇒ 148 − 2r = 3 r + 3 ∴ r = 29 Hence, two consecutive terms are 30 and 31.

y log φ   = log x + log k  x

⇒

74

⇒

y φ  dy y  x 6 (b) Given, − = y dx x φ′    x y xdy − y dx  φ′      x  1 x2 ⇒ = dx y x  φ   x y y φ′   d    x  x = y φ   x

10 (c) General term of (3 + 2 x )74 is T r + 1 = 74C r (3)74 − r 2r x r Let two consecutive terms be T r +1 th and T r +2 th terms. According to the question Coefficient of T r +1 = Coefficient of T r +2

j =1

∴ 0≤

elements each, so i

∑ n(B

8 (d) We know, − π ≤ sin −1 x ≤ π

5 (c) Given, A’s are 30 sets with five 30

n

Similarly,

4 × 8 1 × 16 4 + + 81 81 81 72 8 = = 81 9

1 a2 sin2 x + b 2 cos 2 x

∑ n( A

…(ii)

=

4

= 6×

On integrating both sides, we get 1 − b 2 cos 2 x − a2 sin2 x = − f ( x) ∴ f ( x) =

) = 10 m

From Eqs.(i) and (ii), we get m = 15

⇒

[2sin x] dx

i

i =1

=

ae cos θ + 0 − 1 a cos 2 θ sin2 θ + a2 b2 1 − e cos θ

…(ii)

cos 2 θ sin2 θ + a2 b2

Also, p′ = perpendicular distance from centre (0, 0) to the line (i) . 1 …(iii) = cos 2 θ sin2 θ + a2 b2 Again, r = SP = a(1 − e cos θ) ∴

ap =

a − ae cos θ cos 2 θ sin2 θ + a2 b2

= r p′

[using Eqs. (ii) and (iii)]

421

DAY 13 (a) Let the equation of plane be

D

Let the centroid of tetrahedron OABC is (α,β, γ ), then 0+ a+ 0+ 0 0+ 0+ b + 0 , β= , α= 4 4 0+ 0+ 0+ c γ= 4 ⇒

Since, distance of plane from origin is P, 1 P= 1 1 1 + + a2 b 2 c 2 1 1 1 1 + + = ⇒ a2 b 2 c 2 P 2 On putting the values of a, b and c, then 1 1 1 1 + + = 16α2 16β2 16 γ 2 P 2

b

is x + y = a . Let (h, k ) be the pole, then equation of the polar of (h, k ) with respect to the given ellipse is ky hx + 2 =1 a2 b Since, this is tangent to the circle. |0 + 0 − 1| ⇒ =± a 2 2  h +  k  2  2 a  b 

A

B

a

⇒ |AC |2 = {|3 α − β |2 + |α + 3 β|2 + 2(3 α − β ) ( α + 3 β )}

π 3 1 ⇒ |AC|2 = 64 + 16 + 16 × 2 × 2 × 2 ⇒ |AC|2 = 64 + 16 + 16 |α ||β |cos

f ( x ) = x − cot −1 x − log ( x + 1 + x2 ) On differentiating w.r.t. x, we get 1 1 f ′( x ) = 1 + − 1 + x2 ( x + 1 + x2 )  x  1 +  1 + x2   = 1+

1 1 − 1 + x2 1 + x2

2+ x

− 1+ x 1 + x2

2

2

18 (c) Let p = x is a prime number q = x divides ab r = x divides a s = x divides b

⇒ (2 x + 3)(3 x − 4) ( x − 4) x ≥ 0; x ≠ 0,2 [Q( x − 2)2 > 0] 3 4 x = − , , 4, 0 2 3

The given statement becomes in logical form is p ∧ q → r ∨ s Its contrapositive is ~ (r ∨ s) → ~ ( p ∧ q ) ⇒ (~ r ∧ ~ s ) → (~ p ∨ ~ q )

–

–

+ 0

4/3

3 4 ⇒ x ∈(−∞, −  ∪ (0,  ∪ [4, ∞ ) 2  3 

+ 4

⇒

f (a) = 2 2a, a > 1

⇒

f ( x ) = 2 2 x, x > 1

common ratio r. ∴ x1 = ar , x2 = ar 2 Also, b, y 1 and y 2 are in GP with common ratio s. ∴

y 1 = bs, y 2 = bs2

The area of triangle is given by a b 1 a b 1 1 1 ∆ = x1 y 1 1 = ar bs 1 2 2 x2 y 2 1 ar 2 bs2 1 =

1 1 ab r 2 r2

=

1 ab r 2 2 r

1 1 s 1 s2 1 0 0 s−r 1− r s2 − r 2 1 − r 2 [applying C2 → C2 − C1 and C3 → C3 − C1 ]

So, f ( x ) is an increasing function in (− ∞ , ∞ ) .

and

–3/2

2 3 2a ⋅ 2 − 3  3  2

f (a) − 2 = 2 2a − 2

1 + x2 + 1 − 1 + x2 1 + x2

⇒ (2 x + 3) (3 x − 4)3 ( x − 4) ( x − 2)2 x 5 ≥ 0, x ≠ 0, 2

5

f (a) − 2 = ⇒

19 (a) We have, a

2

∫ [ f ( x ) − 2]dx = 3 [(2a) 1

x=a

x=1

20 (a) Since, a, x1 and x2 are in GP with

17 (a) Given that,

=

(2 x + 3)(3 x − 4)3 ( x − 4) ≥0 ( x − 2)2 x 5

Y′

⇒ |AC|2 = 16 α2 + 4 β2 + 16 α .β

=

15 (d) Given inequality can be rewritten as

X

X′

= 9 α2 + β2 − 6 αβ + α2 + 9 β2 + 6 α ⋅ β + 6 α2 − 6 β2 + 16 α ⋅ β

2

y2 h2 1 ⇒ + 4 = 2 4 a b a Hence, locus of (h, k ) is y2 x2 1 + 4 = 2 4 a b a

+

y = f(x) x=2

Similarly,|BD | = |a − b| = 4 3

14 (b) The equation of the auxiliary circle

3

Y

a+b

b

⇒ |AC|= 4 7

Hence, the locus of centroid is 1 1 1 16 + + = x2 y 2 z2 P 2 2

C

⇒ |AC |2 = |a |2 + |b|2 + 2a ⋅ b

a= 4α, b = 4β, c = 4 γ

2

On differentiating both sides w.r.t a, we get

AC = a + b

16 (b) Clearly,

x y z + + = 1. a b c

3 /2

−3a + 3 − 2 2]

ab {(s − r )(1 − r 2 ) − (1 − r )(s2 − r 2 )} 2 ab = (s − r )(1 − r ){1 + r − ( s + r )} 2

=

=

ab (s − r )(1 − r )(1 − s ) 2

=

ab (s − r )(r − 1)( s − 1) 2

21 (a) Let the coordinates of B and C are ( x1 , y 1 ) and ( x2 , y 2 ) respectively. x + 1 y1 − 2  Then, P  1 ,  lies on  2 2  perpendicular bisector x − y + 5 = 0.

422 x1 + 1 y 1 − 2 − =−5 2 2

∴ ⇒

x1 − y 1 = − 13

…(i)

Also, PN is perpendicular to AB. y1 + 2 ∴ ×1= −1 x1 − 1

Clearly, max amp (z ) = amp (z2 ) and min amp (z ) = amp (z1 ) Now, 15 3 amp (z1 ) = θ1 = cos −1   = cos −1    25  5 and amp (z2 ) =

x+ 2y = 0

A (1, –2) x−y +5 =0 P

=

Q

M

x1 + y 1 = −1

⇒

…(ii)

On solving Eqs. (i) and (ii), we get x1 = −7, y 1 = 6 So, the coordinates of B are (−7, 6). Similarly, the coordinates of C are  11 , 2  .    5 5 Hence, equation of BC is 2 −6 y−6= 5 ( x + 7) 11 +7 5 ⇒

=

C (x2, y2)

y−6= −

π 3 + sin −1    5 2

π π 3 3 + − cos −1 − cos −1 2 2 5 5

3 = π − 2 cos −1    5

y 1 + 2 = − x1 + 1

⇒

 15    25

∴ |max amp(z ) − min amp (z )| π 3 3 = + sin −1 − cos −1 2 5 5

N B (x1, y1)

π π + θ2 = + sin −1 2 2

24 (b) Clearly, tan A = 5 = − tan C ,

12 3 = − cos D 5 [Qin cyclic quadrilateral, A + C = π and B + D = π] 5 Now, tan C = − 12 12 (say) cos C = − =α ⇒ 13 3 4 and cos D = ⇒ tan D = = β (say) 5 3 ∴ Required equation is cos B = −

x2 − (α + β )x + αβ = 12 4  x2 −  − + x+  13 3 

14 ( x + 7) 23

⇒ 14 x + 23 y − 40 = 0

0  − 12 × 4  = 0    13 3 

⇒ 39 x2 − 16 x − 48 = 0

22 (c) Clearly,

1 sin θ 1 | A| = − sin θ 1 sin θ −1 − sin θ 1

∴

cos

Let l , m and n be the DC’s of the reflected ray OB. l+1 1 m+ 0 1 Then, = , =− θ θ 3 3 2cos 2cos 2 2 n+ 0 1 and = θ 3 2cos 2 2 ⇒ l = − 1, 3 2 2 m = − ,n = 3 3 1 2 2 ⇒ l = − ,m = − ,n = 3 3 3 Hence, DC’s of the reflected ray are 1 2 2 − ,− , . 3 3 3

26 (a) Let f ( x ) = 4 x2 − 2 x + a Since, both roots of f ( x ) = 0 lie in the interval (−1, 1), we can take D ≥ 0, f (−1) > 0 and f (1) > 0 1. Consider D ≥ 0, ⇒

B

O (0, 0, 0)

25 q2 X

...(iii)

(sin3 x + cos3 x ) + (cos 4 x − sin 4 x ) (1 + cos x )(1 − sin x ) (sin3 x + cos3 x ) + (cos x + sin x )

q/2 q/2

q1

a> −2

27 (a) Let

X

θ (say) 2 DR’s of OA are (a, 0, 0) and so its DC’s are (1, 0, 0). 1 1 1 DC’s of ON are ,− , 3 3 3

Then,

…(ii)

From Eqs. (i), (ii) and (iii), we get 1 − 2< a≤ 4

=

15 q1 z1

…(i)

3. Consider f (1) > 0, 4(1)2 − 2(1) + a > 0

A=

Y

15

1 4

2. Consider f (−1) > 0, 4(−1)2 − 2(−1) + a > 0 ⇒ a> − 6

23 (b) We have,

z2

a≤

⇒

A (a, 0, 0), where a ≠ 0. Let AO be the incident ray and OB be the reflected ray, ON is the normal to the mirror at O. A(a, 0, 0) N(1, –1, 1)

(−2)2 − 4 ⋅ 4 ⋅ a ≥ 0

⇒

25 (d) Let the source of light be situated at

⇒ | A | = 2(1 + sin2 θ) Now, 0 ≤ sin2 θ ≤ 1, for all θ ∈ [0, 2 π]. ⇒ 2 ≤ 2 + 2sin2 θ ≤ 4 , for all θ ∈ [0, 2 π] So, the range of| A| is [2, 4].

θ 1 = 2 3

∠AON = ∠NOB =

(cos2 x + sin2 x ) (cos x − sin x ) (1 + cos x )(1 − sin x )

(sin x + cos x ){(1 − sin x cos x ) + (cos x − sin x )} = 1 + cos x − sin x − sin x cos x ⇒

A = sin x + cos x 1 π = sin  + x  4  2

Again from Eq. (i), we get A=

1 π cos  − x  4  2

…(i)

423

DAY 28 (c) Since the system of given equation has a non-trivial solution, therefore λ sin θ cos θ 1 cos θ sin θ = 0 −1 sin θ − cos θ ⇒ λ[− cos 2 θ − sin2 θ] − sinθ[− cos θ + sinθ] + cos θ [sinθ + cos θ] = 0 ⇒ −λ + sinθcos θ − sin2 θ + sinθcos θ + cos 2 θ = 0 ⇒ −λ + 2sinθcos θ + cos 2 θ − sin2 θ = 0 ⇒ ⇒ Q

−λ + sin2θ + cos 2θ = 0 π ...(i) λ = 2 cos  2θ −   4 π −1 ≤ cos  2θ −  ≤ 1 ∀θ ∈ R  4

π ∴ − 2 ≤ 2 cos  2θ −  ≤ 2 ∀θ ∈ R  4 ⇒ − 2≤ λ≤ 2

[using Eq. (i)]

Hence, λ ∈ [− 2, 2].

29 (c) We have, f ( x) = x sin x − 1 sin2 x

2 f ′( x) = x cos x + sin x − sin x cos x = sin x(1 − cos x) + xcos x π For x ∈  0,  , sin x > 0, (1 − cos x) > 0,  2 cos x > 0 ⇒

π f ′ ( x ) > 0, ∀ x ∈  0,   2

π ⇒ f ( x) is strictly increasing in  0,  .  2 Now,

lim f ( x ) = 0 x→ 0

and

lim f ( x ) = x→

π 2

π −1 2

π − 1 ∴ Range of f ( x) =  0,   2 

30 (c) I. We have seven married couples.

∴Two husbands can be selected in 7 × 6 = 42 ways. Two wives can be selected in 5 × 4 = 20 ways. [as wives of husband’s selected cannot play in same set] ∴ Required number of ways = 42 × 20 = 840 II. The number of words beginning with T and ending with E = 6! [rest 6 alphabets can be arranged in 6! ways] = 720 ways

424

DAY FOURTY

Mock Test 3 Instruction u

u

u

The test consists of 30 questions. Candidates will be awarded marks for correct response of each question. 1/4 (one-fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response.

1 Two circles in complex plane are C1 : | z − i | = 2 C 2 : | z − 1 − 2i | = 4. Then, (a) (b) (c) (d)

2 If ∫

4 Sixteen metre of wire is available to fence off a flower bed in the form of a sector. If the flower bed has the maximum surface then radius is

C1 and C2 touch each other C1 and C2 intersect at two distinct points C1 lies within C2 C2 lies within C1

cos 2x dx = − log | cot x + sin x

cot 2 x − 1 | + A + C,

(b) (d)

21 26

with the coordinate axes and which touches the straight lines 3x − 2y − 20 = 0 and x + 6y − 20 = 0, is x2 y2 + =1 5 8 2 2 x y (c) + =1 40 10 (a)

x2 y2 + = 10 40 10 2 2 x y (d) + =1 10 40 (b)

7 A variable straight line through the point of intersection of x y x y + = 1 and + = 1 meets the coordinate a b b a axes in A and B. the lines

3 If m1 and m2 are the roots of the equation

x 2 + ( 3 + 2)x + 3 − 1 = 0, then the area of the triangle formed by the lines y = m1x , y = − m2 x and y = 1 is (b)

latusrectum of the parabola y 2 = 8x . It is given that, this circle also touches the directrix of the parabola. The radius of this circle is equal to

6 The equation of the ellipse whose axes are coincident

 2 + 1 − tan2 x  1 log  2 2  2 − 1 − tan x   2 + 1 − sec2 x  1 (b) log  2 2  2 − 1 − sec x   2 + 1 − sin2 x  1 (c) log  2 2  2 − 1 − sin x  (d) None of the above (a)

1  3 + 2   2  3 −1  1  − 3 + 2 (c)   2  3 −1 

(b) 8 (d) 4

5 A circle is drawn to pass through the extremities of the

(a) 4 (c) 3

then A is equal to

(a)

(a) 5 (c) 10

1  3 + 2   2  3 + 1

(d) None of these

I. The locus of mid-point of AB is the curve 2xy (a + b ) = ab ( x + y ). II. The locus of mid-point of AB is the curve 2xy ( x + y ) = ab (a + b ). (a)Both I and II are true (c) Only II is true

(b) Only I is true (d) Both I and II are false

425

DAY 2x − sin−1 x , x ≠ 0 is continuous at 2x + tan−1 x each point of its domain, then the value of f ( 0) is

8 If the function f ( x ) = 1 (b) 3

(a) 2

2 (c) 3

1 (d) − 3

9 The number of permutations of the letters a, b, c, d such that b does not follow a, c does not follow b and d does not follow c, is (a) 10 (c) 15

(b) 13 (d) 11

circle x 2 + y 2 = px + qy , (where pq ≠ 0) are bisected by the X-axis, then (a) p = q (c) p 2 < 8 q 2

(b) p = 8 q (d) p 2 > 8 q 2

2

2

2

11 If A and B are different matrices satisfying A 3 = B 3 and A 2B = B 2 A , then (a) det (A 2 + B 2 ) must be zero (b) det (A – B) must be zero (c) det (A 2 + B 2 ) as well as det (A – B) must be zero (d) Atleast one of det (A 2 + B 2 ) or det (A – B) must be zero

12 A fair coin is tossed 100 times. The probability of getting tails 1, 3, …, 49 times is (a)

1 2

(b)

1 4

1 8

(c)

 −2  π  is  −  7  2 

(a)



2

(b)

3 5

2 3

(d)

1 16

 2x + 3 dy at x = 1 is  , then  3 − 2x  dx

13 If f ′ ( x ) = sin (log x ) and y = f 

4500 3 4300 (c) 3

4500 ( 3 − 1) 3 4500 (d) ( 3 + 1) 3

20 The equation of a line of intersection of planes

4x + 4y − 5z = 12 and 8x + 12y − 13z = 32 can be written as x −1 y + 2 z = = 2 −3 4 x y+1 z−2 (c) = = 2 3 4 (a)

21 If f ( x ) is a function satisfying f ( x + y ) = f ( x ) f ( y ) for all n

x , y ∈ N such that f (1) = 3 and

Σ f ( x ) = 120,

x =1

then the

value of n is (a) 4 (c) 6

(b) 5 (d) None of these

a 1 1

22 If the value of the determinant 1 b 1 is positive, then

(b)

3 10

(c)

10 3 3

(d)

10 9

15 The negation of the compound preposition p ∨ (~ p ∨ q ) is (a) (p ∧ ~ q)∧ ~ p (c) (p ∧ ~ q)∨ ~ p

(b) (p ∧ ~ q)∨ ~ p (d) (p ∧ q) ∧ q

x −1  , for x ≠ 1  2 , then f ′ (1) is equal to 16 If f ( x ) =  2x − 7x + 5 1  , for x = 1 −  3 (b) − (d)

(b) abc > − 8 (d) abc > − 2

and B alternately draw a ball from the bag, replacing the ball each time after the draw. A begins the game.

and the plane r ⋅ (i + 5 j + k ) = 5 is

2 9

1 3

17 If x + y + z = 1 and x, y, z are positive numbers such that (1 − x )(1 − y )(1 − z ) ≥ kxyz , then k equals (a) 2 (c) 8

x −1 y − 2 z = = 2 3 4 x y z−2 (d) = = 2 3 4 (b)

23 A bag contains a white and b black balls. Two players A

r = 2 i − 2 j + 3 k + λ ( i − j + 4 k)

1 9 1 (c) − 3

4 5

(b)

(a) abc > 1 (c) abc < − 8

(b) 5 sin log (6) (d) 5 sin log (12)

14 The distance between the line

(a) −

(d)

1 1 c

(a) 6 sin log (5) (c) 12 sin log (5)

10 3

1 5

At the top of the cliff the angle of depression of boat changes from 60° to 45° in 2 min. Then, the speed of the boat (in m/h) is

equal to

(a)

(c)

19 A boat is being rowed away from a cliff of 150 m height.

(a)

10 If two distinct chords, drawn from the point ( p, q ) on the

2



18 The value of tan cos −1 

(b) 4 (d) 16

If the probability of A winning (that is drawing a white ball) is twice the probability of B winning (that is drawing a black ball), then the ratio a : b is equal to (a) 1 : 2 (c) 1 : 1

(b) 2 : 1 (d) None of these

24 Number of solutions of the equation sin x = [ x ], where [.] denotes the largest integer function, is (a) 0 (c) 2

(b) 1 (d) None of these

25 If ( 5 + 2 6 )n = I + f ; n, I ∈ N and 0 ≤ f < 1 , then I equals (a)

1 −f f

26 The value of ∫ (a) 0 (c)

22 −π 7

(b) 1 0

1 −f 1+ f

(c)

1 +f 1+ f

x 4 (1 − x )4 dx is 1+ x2 2 105 71 3 π (d) − 15 2 (b)

(d)

1 −f 1− f

426

27 The area inside the parabola 5x 2 − y = 0 but outside the parabola 2x − y + 9 = 0 is 2

(a) 18 3

(b) 10 3

(c) 11 3

(d) 12 3

28 The median of a set of 11 distinct observations is 20.5. If each of the last 5 observations of the set is increased by 4, then the median of the new set (a) (b) (c) (d)

is increased by 2 is decreased by 2 is two times the original a median remains the same as that of the original set

29 Solution of the differential equation

x dx + y dy = x dx − y dy

is given by (a)

3 y x3/2 + y3/2 y log   + log + tan−1   x x 2 x3/2

3/2

+c = 0

2 y x3/2 + y3/2 y log   + log + tan−1   + c = 0 x x 3 x3/2  y3/2  2 y x y (c) log   + log  +  + tan−1  3 / 2  + c = 0 x  x  3 x  (b)

(d) None of the above

30 Statement I If r ⋅a = 0 ,r ⋅b = 0 ,r ⋅c = 0 for some non-zero vector r, then a,b and c are coplanar vectors.

Statement II If a, b and c are coplanar, then a + b + c = 0. y3 x3

(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

427

DAY

Hints and Explanations 1 (c) Given equations can be rewritten as

x2 + ( 3 + 2) x + ( 3 − 1) = 0.

x + ( y − 1) = 2 and ( x − 1)2 + ( y − 2)2 = 42 Here, centres are C1 (0, 1) and C2 (1, 2) and radii are r1 = 2, r2 = 4 . 2

Now,

2

2

Y

y=–m2x

y = m1x A

B

X¢

C1 C2 = (1 − 0)2 + (2 − 1)2

y=1 X

O

Q

2

|r1 − r2 | = |2 − 4| = 2 C1 C2 < |r1 − r2|

2 (a) Let I = =

∫

=

∫ ∫

cos 2 x dx sin2 x cos x − sin x dx sin2 x 2

2

cot x − 1 dx 2

sec θ tan2 θ ∫ 1 + sec 2 θ dθ

sin2 θ =−∫ dθ cos θ + cos 3 θ 1 − cos 2 θ =−∫ dθ cos θ + cos 3 θ =−

∫

(1 + cos 2 θ) − 2cos 2 θ dθ cos θ (1 + cos 2 θ)

= − ∫ sec θ dθ + 2∫ = − log |sec θ + +

cot x − 1| 2

+ A + C [given]  2 + 1 − tan2 x  1 log  ∴ A=  2 2  2 − 1 − tan x 

3 (a) Since, m1 and m2 are the roots of the equation

 1   1  A , 1 and B  − , 1  m1   m2 

∴ Area of ∆OAB = −

=

1 2

x2 + y 2 − 4 x − 12 = 0

∴ Radius = (2)2 + 12 = 4

to the ellipse is

and O(0, 0). 1 m1 −1 m2 0

⇒

6 (c) The general equation of the tangent

1 1 1 1 0 1

1 1 1 1 m1 + m2 + = 2 m1 m2 2 m1 m2

12 + 3 =   2 3 −1

…(i) y = mx ± a2 m2 + b 2 3 Since, y = x − 10 is a tangent to the 2 ellipse, therefore its comparing with Eq. (i), we get 3 m= and a2 m2 + b 2 = 100 2 …(ii) ⇒ 9 a2 + 4b 2 = 400 Similarly, line 1 10 x+ 6 3 is a tangent to the ellipse, therefore its comparing with Eq. (i), we get 1 100 m=− and a2 m2 + b 2 = 6 9 …(iii) ⇒ a2 + 36b 2 = 400 y =−

4 (d) Let r be the radius of the sector and angle subtended at the centre be θ. 16-2r

r

r q

cot2 x − 1 |

 2 + 1 − tan2 x 1 log  + C 2 2  2 − 1 − tan x

But I = − log |cot x +

⇒ m1 < 0, m2 < 0 So, the point of intersections are

sec2 θ − 1 |

 2 + 1 − cos 2 θ 1 log  + C 2 2  2 − 1 − cos θ

= − log |cot x + +

d (sin θ) 1 + cos 2 θ

3)

m1 m2 = 3 − 1

and

On putting cot x = sec θ ⇒ − cosec2 x dx = sec θ tan θ dθ, we get sec θ tan θ I = ∫ sec2 θ − 1 × dθ − cos ec2 x =−

Y¢ m1 + m2 = − (2 +

∴

5 (a) Extremities of the latusrectum of the parabola are (2, 4) and (2, – 4). Since, any circle drawn with any focal chord at its diameter touches the directrix, thus equation of required circle is ( x − 2)2 + ( y − 4)( y + 4) = 0

= 1 +1 = 2 2

dS = 8 − 2r dr Now, for area to be maximum, dS =0 dr ⇒ r =4 ⇒

Then, S = surface area of sector θ = × πr 2 360 length of arc We know, θ = radius 16 − 2r = r θ 16 − 2r 2 S = πr 2 = ⋅r ∴ 2π 2r 2 ⇒ S = (8 − r ) ⋅ r = 8r − r

On solving Eqs. (ii) and (iii), we get a2 = 40 and b 2 = 10 Hence, required equation of the ellipse is x2 y2 + = 1. 40 10

7 (b) Any line through the point of intersection of given lines is  x + y − 1 + λ  x + y − 1 = 0     b  a  b a 1 λ 1 λ   x  +  + y  +  = (1 + λ ) a b b a b + aλ  a + bλ    ⇒x  + y   = (1 + λ )  ab   ab 

428

⇒

y x =1 +  ab (1 + λ )  ab (1 + λ )      b + aλ   a + bλ  Y B

X¢

O

A

Y¢

X

This meets the X -axis at ab (1 + λ )  A ≡  , 0  b + aλ  and meets the Y-axis at ab (1 + λ ) B ≡  0,   a + bλ  Let the mid-point of AB is M ( x1 , y 1 ). Then, ab (1 + λ ) x1 = 2(b + aλ ) y1 =

and ∴

ab (1 + λ ) 2 ( a + bλ )

1 1 2 (b + aλ ) 2 (a + bλ ) + = + x1 y1 ab (1 + λ ) ab (1 + λ ) 2 = (b + aλ + a + bλ ) ab (1 + λ ) =

⇒ 2 x1 y 1 (a + b ) = ab ( x1 + y 1 ) Hence, the locus of mid-point of AB is 2xy (a + b ) = ab ( x + y )

8 (b) Since, f ( x ) is continuous ∴

∴

through A (p,q) and M (h, 0) be the mid-point of AB. Therefore, the coordinates of B are (–p + 2h, – q). Since, B lies on the circle x2 + y 2 = px + qy , then (– p + 2h ) 2 + (– q ) 2 = p (– p + 2h ) + q(– q ) ⇒ 2 p2 + 2q 2 – 6 ph + 4 h2 = 0 ⇒ 2h2 – 3 ph + ( p2 + q 2 ) = 0 …(i) Y

 2 x − sin −1 x  lim   = f (0) x → 0 2 x + tan −1 x    1   2 −  1 − x2  f (0) = lim x→ 0 1 2+ 1 + x2 1 2− 1 =1 = 1 3 2+ 1 [Qapply L’Hospital’s Rule]

9 (d) We have the following cases. Case I

a

In this case we have only two possibilities, namely, ac b d and ad c b

X¢

A ( p , q) X

O

B C

Y¢

⇒

∴ A − AB =B −B A ⇒ ( A2 + B 2 )( A − B ) = 0 ⇒ det ( A2 + B 2 ) det ( A − B ) = 0 2

3

2

12 (b) Here, p = 1 and q = 1

2 2 ∴P ( X = 1) + P ( X = 3) +…+ P ( X = 49) 1 C1    2

100

100

+

1 C3    2

100

100

+…+

1 C 49    2

100

C 49 )

× 298 =

1 4

 2x + 3 d  2x + 3 dy = f′    dx  3 − 2 x  dx  3 − 2 x 

100

 2x + 3     3 − 2x    (3 − 2 x )(2) − (2 x + 3)(−2)   (3 − 2 x )2  

= ⇒

  2x + 3  12 sin  log   (3 − 2 x )2  3 − 2x    12  dy  = sin log (5)    dx  ( x = 1 ) (3 − 2)2 = 12 sin log (5)

14 (c) Line is parallel to plane as ( i − j + 4k ) ⋅ ( i + 5j + k ) = 0 General point on the line is (λ + 2, − λ − 2, 4λ + 3). For λ = 0 point on this line is (2, –2, 3) and distance from r ⋅ ( i + 5j + k ) = 5or x + 5y + z = 5, is |2 + 5 (−2) + 3 − 5| d = 1 + 25 + 1 d =

|− 10| 3 3

=

10 3 3

15 (a) Negation of p ∨ (~ p ∨ q ) ⇒ ~ [ p ∨ (~ p ∨ q )] ≡~ p ∧ ~ (~ p ∨ q ) ≡~ p ∧ (~ (~ p )∧ ~ q ) ≡~ p ∧ ( p ∧ ~ q ) ≡ ( p ∧ ~ q )∧ ~ p

16 (b) Given,

11 (c) Since, A3 = B 3 and A2 B = B 2 A

=

100

 = sin  log 

⇒

As, there are two distinct chords AB and AC from A ( p, q ) which are bisected on X -axis there must be two distinct values of h satisfying Eq. (i), then D = (b 2 – 4 ac ) > 0, we have (–3 p ) 2 – 4(2)( p2 + q 2 ) > 0 ⇒ p2 > 8q 2

3

100

 2x + 3   3 − 2x 

10 (d) Let AB be a chord of the circle

(h, 0) M

C3 +…+

100

13 (c) Given that, y = f 

In this case we have only three possibilities, namely, d c b a, d ac b and d b ac Hence, the total number of ways = 2 + 3 + 3 + 3 = 11.

( x1 + y 1 ) 2 (a + b ) = x1 y 1 ab

⇒

1 =    2

d

Case IV

2 (b + a) (1 + λ ) ab (1 + λ )

(100 C1 + 100

c

In this case we have only three possibilities, namely, c ad b,c b d a and c b ad

100

Q(100 C1 + C3 +…+ 100C 99 ) = 299  100   C 99 = 100C1  but   100 C 97 = 100C3 ,… , 100C 51 = 100C 49    ∴ 2(100 C1 + 100C3 +…+ 100C 49 ) = 299 

In this case we have only three possibilities, namely, b ad c, b d ac and bd c a Case III

M (x1, y1)

1 =    2

b

Case II

x −1  , x≠1  2 f ( x ) = 2 x − 7 x + 5 1 − , x=1   3  1 , x≠1  f ( x ) = 2 x − 5 1 x=1  − ,  3 f (1 + h ) − f (1) f ′ (1) = lim h→ 0 h 1 1 −  −  2 (1 + h ) − 5  3  = lim h→ 0 h

429

DAY 1 1 + 2h − 3 3 = lim h→ 0 h 3 + 2h − 3 2 = lim =− h → 0 3 h ( 2h − 3) 9

150 ( 3 − 1) 3 AB ∴ Speed of boat = 2 1 150 = × ( 3 − 1) × 60 2 3 4500 = ( 3 − 1) m/h 3

f (1 − h ) − f (1) −h 1 1 −  −  2(1 − h ) − 5  3  = lim h→ 0 −h 2 2 = lim − =− h→ 0 3(2h + 3) 9 2 ∴ f ′ (1) = − 9 Lf ′ (1) = lim h→ 0

17 (c) Clearly, and ∴

x+ y ≥ 2

xy ;

y+z ≥ 2

20 (b) Given equation of planes are

yz

x+ z ≥ xz 2 ( x + y ) ( y + z) ( x + z) ⋅ ⋅ 2 2 2 ≥ xy ⋅ yz ⋅ xz

⇒ (1 − z )(1 − x )(1 − y ) ≥ 8 xyz [Q x + y + z = 1] Hence, k = 8.

18 (a) tan cos −1 

 − 2 − π      7 2

2 π = tan  π − cos −1   −    7 2   2 = tan sin −1     7  

…(i) 4 x + 4 y − 5z = 12 and 8 x + 12 y − 13z = 32 …(ii) Let DR’s of required line be (l , m, n ) . From Eqs. (i) and (ii), we get 4 l + 4 m − 5n = 0 and 8 l + 12 m − 13 n = 0 l m n = = ⇒ 8 12 16 l m n ⇒ = = 2 3 4 Now, we take intersection point with z = 0 is given by …(iii) 4 x + 4 y = 12 and 8 x + 12 y = 32 …(iv) On solving Eqs. (i) and (ii), we get the point (1, 2, 0). x−1 y −2 z = = ⋅ ∴ Required line is 2 3 4

21 (a) f ( x ) = f (1 + 1 + 1 + ... + x times) = f (1) f (1) f (1)........ x times = [ f (1)] x = 3x

 2  = tan  tan −1    3 5   2 = 3 5

∴

Q

150 m A

P

PQ AP 150 AP = 3

In ∆APQ, tan 60° = ⇒ and in ∆BPQ,

PQ tan 45° = AB + AP ⇒

AB +

150 = 150 3

x =1

x =1

x

= 31 + 32 + ...+3 n

…(i)

⇒ ∴ ⇒

( x − 1)2 ( x + 2) > 0 x + 2> 0⇒ x > − 2 x3 > − 8 ⇒ abc > − 8

23 (c) Here, P (W ) = and P (B ) =

a a+ b

b a+ b

∴ Probability of A winning = P ( W ) + P (W ) P ( B ) P ( W ) + K P (W )

=

1 − P (W )P ( B ) a a+ b = b a 1− ⋅ a+ b a+ b =

a(a + b ) = P1 a2 + b 2 + ab

[say]

and probability of B winning a2 + ab = 1 − P1 = 1 − 2 a + b 2 + ab =

b2 = P2 a + b 2 + ab

[say]

2

Given, P1 = 2P2 a2 + ab 2b 2 ⇒ = 2 2 2 a + b + ab a + b 2 + ab ⇒ a2 + ab − 2b 2 = 0 ⇒ (a − b )(a + 2b ) = 0 ⇒ a−b = 0 ⇒ a=b ∴ a:b = 1:1

[Q a + 2b ≠ 0]

24 (c) y = sin x = [ x] Graphs of y = sin x and y = [ x] are as shown.

 a − lr  Q sum = 1 − r   

3n +1 – 3 2

3 n+1 – 3 = 120 2 ⇒ n+1= 5 ∴ n=4 a 22 (b) Let ∆ ′ = 1 1 ∴

45°

B

n

= 60°

60°

n

∑ f ( x) = ∑3

31 – 3n ⋅ 3 = 1–3

19 (b) Let PQ = 150 m

45°

AB =

⇒

⇒ 3 n+1 = 243 = 35

Y

X¢

1

–p 2 –2

1

–1 –1

p 2

2

X

Y¢

Hence, two solutions are x = 0 and π x= . 2

1 1 b 1 1 c

= abc + 2 − a − b − c > 0 or abc + 2 > a + b + c ...(i) 1 a+ b + c Q AM > GM ⇒ > (abc )3 3 1

a + b + c > 3(abc )3 From Eqs. (i) and (ii), abc + 2 > 3(abc )1 /3 Let (abc )1 /3 = x 3 Then, x + 2 > 3 x

...(ii)

25 (d) I + f + f ′ = (5 + 2 6 )n + (5 − 2 6 )n [even integer] = 2k ∴ f + f′=1 Now, (I + f ) f ′ = (5 + 2 6 )n (5 − 2 6 )n = (1)n = 1 ⇒ (I + f ) (1 − f ) = 1 1 I = − f ⇒ 1− f

430

26 (c) Let I = 1

∫

=

∫ (x

=

x 4 (1 − x )4 dx 1 + x2

1 0

Given parabolas intersect at ( 3, 15) and (− 3, 15).

4  x  x2 − 4 x + 5 −  dx  1 + x2 

=

0

∫

4

1

6

0

1

∫ (x 0

1 0

x4 dx (1 + x2 )

(– 3, 15)

(0, a

− 4 x 5 + 5x 4 ) dx − 4

∫

1 0

6

 2 1  x −1+  dx 1 + x2   1

 x3 − 4 − x + tan −1  3 1 4 1 π =  − + 1 − 4  − 1 +  7 6  3 4 22 = − π 7

27 (d) Given parabolas are 5x2 − y = 0 2 and 2x − y + 9 = 0 Now, eliminating y from above equations, we get 5x2 − (2 x2 + 9) = 0 2 ⇒ 3x = 9 ⇒ x = ± 3

)

( 3, 15)

y = 5x 2

O

x  4x = − + x 5 6  7 0 7

y = 2 x2+ 9

− 4 x 5 + 5x 4 ) dx − 4∫

6

Y

X

The two parabolas are 1 1 x2 = y , x2 = ( y − 9) 5 2 3

1

 x 0

∴ Area = 2 ∫ ( y 1 − y 2 ) dx 0

3

= 2 ∫ [(2 x2 + 9) − 5 x2 )] dx 0

3

= 2 ∫ (9 − 3 x ) dx 2

29 (d) We have,

x dx +

y dy

x dx −

y dy

=

y3 x3

d ( x ) + d ( y 3 /2 ) y 3 /2 = 3 /2 d ( x3 /2 ) − d ( y 3 /2 ) x du + dv v ⇒ = du − dv u where u = x3 /2 and v = y 3 /2 ⇒ u du + u dv = v du − v dv ⇒ u du + v dv = v du − u dv u du + v dv v du − u dv ⇒ = u2 + v 2 u2 + v 2 d (u2 + v 2 ) v = −2d tan −1   + c ⇒ u u2 + v 2 On integrating, we get v log (u2 + v 2 ) = −2 tan −1   + c u ⇒

⇒

3 /2

y 1 log( x3 + y 3 ) + tan −1    x 2

3 /2

=

c 2

0

= 2[9 x − x3 ] 03 ∴ Area = 12 3

28 (d) Since, n = 11, then median term 11 + 1  =   th term = 6 th term  2  As, last five observations are increased by 4. Hence, the median of the 6th observations will remain same.

30 (b) We have, r ⋅ a = 0 ⇒ r ⊥ a r ⋅ b = 0 ⇒ r ⊥ b and r ⋅ c = 0 ⇒ r ⊥ c Since, vectors a , b and c are coplanar vectors. Also, a + b + c = 0 ∴ a ×b+ b×c + c ×a = 0 ⇒ a ⋅ (a × b ) + a ⋅ ( b × c ) + a ⋅ (c × a ) = 0 ⇒ 0 + [a b c] + 0 = 0 ⇒ [a b c] = 0 Hence, a , b and c are coplanar vectors.

JEE Main 2019 (April & January Attempt) APRIL ATTEMPT 8 April, Shift-I

3 5

7 If cos(α + β ) = , sin(α − β) =

1 The shortest distance between the

0 < α ,β
1) is

equal to (a) 29 (c) 26

9

(a)

of the chords intercepted on the circle, x2 + y2 = 16, by the lines, x + y = n, n ∈ N , where N is the set of all natural numbers, is

∫

(b) 32 (d) 24

5x 2 dx is equal to x sin 2

sin

has a non-trivial solution, is (b)

63 16 33 (d) 52 (b)

8 The sum of the coefficients of all even

3 The greatest value of c ∈ R for which

(a) −1

π , then tan(2α ) is equal to 4

(a)

sin 2 x equals 2 − 1 + cos x

5 and 13

(where, C is a constant of integration ) (a) (b) (c) (d)

2x + sin x + 2 sin 2x + C x + 2 sin x + 2 sin 2x + C x + 2 sin x + sin 2x + C 2x + sin x + sin 2x + C

10 The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is (a) 45 (c) 48

(b) 49 (d) 40

11 The equation of a plane containing the line of intersection of the planes 2x − y − 4 = 0 and y + 2z − 4 = 0 and passing through the point (1, 1, 0) is (a) x − 3 y − 2z = − 2 (b) 2x − z = 2 (c) x − y − z = 0 (d) x + 3 y + z = 4

12 The magnitude of the projection of $ on the vector the vector 2$i + 3$j + k perpendicular to the plane containing $ and i$ + 2$j + 3k $, the vectors i$ + $j + k is

(b) 105 (d) 210

14 Let A and B be two non-null events such that A ⊂ B. Then, which of the following statements is always correct. (a) P (A /B ) = P (B ) − P (A ) (b) P (A /B ) ≥ P (A ) (c) P (A /B ) ≤ P (A ) (d) P (A /B ) = 1

15 If α and β are the roots of the equation x2 − 2x + 2 = 0, then the least n

α value of n for which   = 1is β (a) 2 (c) 4

(b) 5 (d) 3

16 The area (in sq units) of the region A = {(x, y) ∈ R × R|0 ≤ x ≤ 3, 0 ≤ y ≤ 4, y ≤ x2 + 3x} is 53 6 59 (c) 6 (a)

(b) 8 (d)

26 3

17 If S1 and S2 are respectively the sets of local minimum and local maximum points of the function, f (x) = 9x4 + 12x3 − 36x2 + 25, x ∈ R, then (a) S1 = {−2} ; S2 = {0,1} (b) S1 = {−2, 0} ; S2 = {1} (c) S1 = {−2,1} ; S2 = {0} (d) S1 = {−1} ; S2 = {0, 2}

2

40

3 1 18 If α = cos−1   , β = tan −1   , where

y(0) = 0. If a y(1) =

 5  3 π 0 < α ,β < , then α − β is equal to 2

of ‘a’ is

9  (a) tan −1    5 10 

9  (b) cos−1    5 10 

(c) 1

9 (c) tan −1    14 

9  (d) sin −1    5 10 

19 The sum of the series 2 ⋅20 C0 + 5 ⋅20 C1 + 8 ⋅20 C2 + 11⋅20 C3 + .... + 62 ⋅20 C20 is equal to (a) 226 (c) 2

(b) 225 (d) 224

23

20 The sum of the solutions of the equation | x − 2| + x ( x − 4) + 2 = 0 (x > 0) is equal to (a) 9 (c) 4

(b) 12 (d) 10

4x + y = 8 at the points (1, 2) and (a , b) are perpendicular to each other, then a 2 is equal to 2

128 17 4 (c) 17

(a)

1 4

64 17 2 (d) 17

(b)

differential equation, dy (x2 + 1)2 + 2x(x2 + 1) y = 1 such that dx

2

 3 cos x + sin x    ,  cos x − 3 sin x    π dy is equal to x ∈  0,  then  2 dx

such that 100 < n < 200 and HCF (91, n)>1 is (b) 3303 (d) 3121

24 The length of the perpendicular from

π −x 6 π (c) −x 3

(b) x −

(a)

1−x  ,|x|< 1, then 1 + x

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

(a) 2f (x ) (c) (f (x )) 2

3x + 5 y = 15 which is equidistant from the coordinate axes will lie only in (a) (b) (c) (d)

IV quadrant I quadrant I and II quadrants I, II and IV quadrants

points, then the locus of a point P such that the perimeter of ∆AOP is 4, is (a) 8x 2 − 9 y 2 + 9 y = 18

(b) 2f (x 2 ) (d) −2f (x )

29 Let f : [0, 2] → R be a twice differentiable function such that f ′ ′ (x) > 0, for all x ∈ (0, 2). If φ(x) = f (x) + f (2 − x) , then φ is (a) increasing on (0, 1) and decreasing on (1, 2) (b) decreasing on (0, 2) (c) decreasing on (0, 1) and increasing on (1, 2) (d) increasing on (0, 2)

30 If f (x) =

2 − x cos x and g (x) = log e x, 2 + x cos x

(x > 0) then the value of the integral π/4

∫− π / 4 g (f (x))dx is (a) log e 3 (b) log e e (c) log e 2 (d) log e 1

(b) 9x 2 − 8 y 2 + 8 y = 16

ANSWERS 1. (c) 11. (c) 21. (d)

2. (a) 12. (d) 22. (d)

3. (b) 13. (d) 23. (d)

4. (a) 14. (b) 24. (a)

5. (a) 15. (c) 25. (c)

6. (c) 16. (c) 26. (c)

7. (b) 17. (c) 27. (b)

8. (d) 18. (d) 28. (a)

π 3

28 If f (x) = log e 

 2x   is equal to f  2 1+ x 

greater than 3 but less than 4 less than 2 greater than 2 but less than 3 greater than 4

π 6

(d) 2x −

the point (2, − 1, 4) on the straight x+3 y− 2 z line, = = is 10 −7 1

26 Let O(0, 0) and A(0, 1) be two fixed

22 Let y = y(x) be the solution of the

(d) 8x 2 + 9 y 2 − 9 y = 18

27 If 2 y =  cot −1 

23 The sum of all natural numbers ‘n’

(a) 3203 (c) 3221

(c) 9x 2 + 8 y 2 − 8 y = 16



1 2 1 (d) 16

(b)

25 A point on the straight line,

21 If the tangents on the ellipse 2

(a)

π , then the value 32

9. (c) 19. (b) 29. (c)

10. (c) 20. (d) 30. (d)

For Detailed Solutions Visit : http://tinyurl.com/y4t86zrd Or Scan :

3 8 April, Shift-II 1 If the system of linear equations x − 2 y + kz = 1, 2x + y + z = 2 , 3x − y − kz = 3 has a solution (x, y, z ), z ≠ 0, then (x, y) lies on the straight line whose equation is (a) 3x − 4 y − 4 = 0 (b) 3x − 4 y − 1 = 0 (c) 4x − 3 y − 4 = 0 (d) 4x − 3 y − 1 = 0 20

2 The sum

1

k =1

(a) 2 − (c) 2 −

11

(b) 1 −

219 3

220 21

(d) 2 −

217

1 1 1  and A =  2 b c . If det(A ) ∈ [2, 16],   2 2  4 b c  then c lies in the interval (a) [3, 2 + 23/ 4 ]

(b) (2 + 23/ 4 , 4)

(c) [4, 6]

(d) [2, 3) x

4 Let f (x) = ∫ g (t )dt, where g is a 0

non-zero even function. If f (x + 5) = g (x), then ∫ f (t )dt equals 0

∫ g (t )dt

∫ g (t )dt

(b)

x +5

5

∫ g (t )dt

(d)

∫ g (t )dt

x +5

5

5 If the lengths of the sides of a triangle are in AP and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is (a) 3 : 4 : 5

(b) 4 : 5 : 6

(c) 5 : 9 : 13

(d) 5 : 6 : 7

/ x3 (1 + x6 )23

= xf (x)(1 +

1 x6 )3

+C

where, C is a constant of integration, then the function f (x) is equal to (a) − (c) −

(a) 12

(b) 9

(c) 15

(d) 33

9 If three distinct numbers a, b and c are in GP and the equations ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a common root, then which one of the following statements is correct?

1 6x 3 1 2x 2

(b) − (d)

1

2x 3 3

x2

7 Which one of the following statements is not a tautology? (a) ( p ∧ q) → (~ p ) ∨ q

(a)

6

(b) 2 3

(c)

3

2 3 3

(d)

3 i + (i = −1), then 2 2 (1 + iz + z5 + iz 8 )9 is equal to

15 If z = (a) 1

(b) (−1 + 2i ) 9

(c) −1

(d) 0

16 A student scores the following marks in five tests 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is (a)

10 3

10 3

(b)

(c)

100 3

(d)

100 3

17 The number of integral values of m for

has to toss a fair coin so that the probability of observing atleast one head is atleast 90% is

which the equation (1 + m2 )x2 − 2(1 + 3m)x + (1 + 8m) = 0, has no real root is

(a) 2

(a) 3 (c) 1

(b) 3

(c) 5

(d) 4

11 Suppose that the points (h , k ), (1, 2) and (−3, 4) lie on the line L1 . If a line L2 passing through the points (h , k ) and (4, 3) is perpendicular to L1 , then k /h equals 1 7

(c) 3

(b)

1 3

(d) 0

12 Given that the slope of the tangent to

5

x +5

6 If ∫

of f (f (f (x))) + (f (x))2 at x = 1is

x +5

5

dx

8 If f (1) = 1, f ′ (1) = 3, then the derivative

(a) −

x

(c) 2

of maximum volume inscribed in a sphere of radius 3 is

10 The minimum number of times one

220

3 Let the numbers 2, b, c be in an AP

(a) 5

14 The height of a right circular cylinder

(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

∑ k 2k is equal to

11

(b) ( p ∧ q) → p (c) p → ( p ∨ q) (d) ( p ∨ q) → ( p ∨ (~ q))

a curve y = y(x) at any point (x, y) is 2y . If the curve passes through the x2 centre of the circle x2 + y2 − 2x − 2 y = 0, then its equation is (a) x log e| y| = − 2(x − 1) 2

(b) x log e| y| = x − 1 (c) x log e| y| = 2(x − 1) (d) x log e| y| = − 2(x − 1)

13 Let f : [−1, 3] → R be defined as |x| + [x], −1 ≤ x < 1  f (x) =  x + |x|, 1 ≤ x < 2  x + [x], 2 ≤ x ≤ 3 ,  where, [t ] denotes the greatest integer less than or equal to t. Then, f is discontinuous at (a) (b) (c) (d)

four or more points only two points only three points only one point

(b) infinitely many (d) 2

18 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 latus rectum is (a) 5

(b) 10

(c) 8

(d) 6

19 Let a = 3$i + 2$j + xk$ and b = i$ − $j + k$ , for some real x. Then|a × b| = r is possible if (a) 0 < r ≤ (c) 3

3 2

(b)

3 3 0, then k is equal to (a) 2

10 3

(b) 2 6

(c) 4

5 3

(d)

6

17 Let p, q ∈ R. If 2 − 3 is a root of the quadratic equation, x2 + px + q = 0, then (a) q2 − 4 p − 16 = 0 (b) p 2 − 4q − 12 = 0 (c) p 2 − 4q + 12 = 0 (d) q2 + 4 p + 14 = 0

18 A plane passing through the points (0, − 1, 0) and (0, 0, 1) and making an π angle with the plane y − z + 5 = 0, 4 also passes through the point (a) ( 2 , 1, 4) (c) (− 2 , − 1, − 4)

(b) (− 2 , 1, − 4) (d) ( 2 , − 1, 4)

 1 1  1 2  1 3

 1 n − 1 1 

19 If  . .  ...   0 1  0 1  0 1  0

 1 78 = , 0 1 

12 All the points in the set α + i  S= : α ∈ R (i = α − i  (a) (b) (c) (d)

−1) lie on a

circle whose radius is 2. straight line whose slope is −1. circle whose radius is 1. straight line whose slope is 1.

π π 13 If the function f defined on  ,  by  6 3

 2 cos x − 1 π , x≠  4 is f (x) =  cot x − 1 π  k, x=  4 continuous, then k is equal to 1 (a) 2

(b) 2

(c) 1

(d)

(b) π 13 π (d) 6

 1 0 12 1   1 − 13   (b) 0 1    (c)

 1 0 13 1  

(d)

 1 −12 0 1   

20 Let f (x) = 15 − x − 10 ; x ∈ R. Then, 1 2

S = {θ ∈ [−2 π , 2 π ] : 2 cos2 θ + 3 sin θ = 0}, then the sum of the elements of S is (a) 2π 5π (c) 3

1 n  then the inverse of   is 0 1 (a)

14 Let

value of m is

(b) (2, − 2) (d) (2, − 1)

16 If the standard deviation of the

2

1 − x2 then A is equal to

y = x3 + ax − b at the point (1, − 5) is perpendicular to the line, − x + y + 4 = 0, then which one of the following points lies on the curve ? (a) (−2, 2) (c) (−2, 1)

8 If the function f : R − {1, − 1} → A defined by f (x) =

15 If the tangent to the curve,

the set of all values of x, at which the function, g (x) = f (f (x)) is not differentiable, is (a) (b) (c) (d)

{5, 10, 15, 20} {5, 10, 15} {10} {10, 15}

6

40

21 Let S be the set of all values of x for which the tangent to the curve y = f (x) = x3 − x2 − 2x at (x, y) is parallel to the line segment joining the points (1, f (1)) and (−1, f (−1)), then S is equal to 1 (a)  , − 1 3  1   (c)  − , 1  3 

1 (b)  , 1 3  1 (d)  − , − 1  3 

negation of the expression p ∨ (~ p ∧ q) is

y+1 α α y+β β 1

β 1 is equal to y+α

(a) y ( y 2 − 1) (c) y 3 − 1

(b) y ( y 2 − 3) (d) y 3

A = {(x, y) : x2 ≤ y ≤ x + 2} is 13 6 31 (c) 6

9 2 10 (d) 3

(a)

(b) ~ p ∨ ~ q (d) p ↔ q

23 If the fourth term in the binomial 6

log x 2 expansion of  + x 8  (x > 0) is x  20 × 87 , then the value of x is

(a) 8−2 (c) 8

equation x2 + x + 1 = 0. Then, for y ≠ 0 in R,

26 The area (in sq units) of the region

22 For any two statements p and q, the (a) ~ p ∧ ~ q (c) p ∧ q

25 Let α and β be the roots of the

(b) 83 (d) 82

(b)

cos2 10° − cos10° cos 50° + cos2 50° is 3 3 (1 + cos 20° ) (b) + cos 20° 2 4 (c) 3 / 2 (d) 3 / 4 (a)

(a) m = n = 68 (c) m = n = 78

/ 27 The integral ∫ sec23 x cosec4/3 x dx is

equal to (here C is a constant of integration) (a) 3 tan −1/ 3 x + C

(b) m + n = 68 (d) n = m − 8

29 Let the sum of the first n terms of a non-constant AP a1 , a2 , a3 .....be n (n − 7) A, where A is a 2 constant. If d is the common difference of this AP, then the ordered pair (d , a50 ) is equal to

50n +

(a) (A, 50 + 46A) (c) (50, 50 + 46A)

(b) −3 tan −1/ 3 x + C

24 The value of

m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with atleast 3 females, then

(b) (50, 50 + 45A) (d) (A, 50 + 45A)

r r $ If 30 Let α = 3$i + $j and β = 2$i − $j + 3k. r r r r r β = β1 − β 2, where β1 is parallel to α r r and β 2 is perpendicular to α, then r r β1 × β 2 is equal to 1 $ $) (3i − 9$j + 5k 2 1 $) (b) (−3$i + 9$j + 5k 2 $ (c) −3$i + 9$j + 5k $) (d) 3$i − 9$j − 5k (a)

(c) −3 cot −1/ 3 x + C 3 (d) − tan −4 / 3 x + C 4

28 A committee of 11 members is to be formed from 8 males and 5 females. If

ANSWERS 1. (c) 11. (a) 21. (c)

2. (d) 12. (c) 22. (a)

3. (b) 13. (a) 23. (d)

4. (c) 14. (a) 24. (d)

5. (b) 15. (b) 25. (d)

6. (c) 16. (b) 26. (b)

7. (c) 17. (b) 27. (b)

8. (c) 18. (a) 28. (c)

9. (c) 19. (b) 29. (a)

10. (b) 20. (b) 30. (b)

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7 9 April, Shift-II

8 The area (in sq units) of the region   y2 A =  (x , y ) : ≤ x ≤ y + 4 is 2  

1 Let z ∈ C be such that|z|< 1. If ω=

5 + 3z , then 5(1 − z )

(a) 4 Im(ω ) > 5 (c) 5 Im (ω ) < 1

(b) 5 Re (ω ) > 1 (d) 5 Re(ω ) > 4

2 If a unit vector a makes angles

π 3

π $ with $i, with $j and θ ∈ (0, π ) with k, 4 then a value of θ is 5π 6 5π (c) 12

π 4 2π (d) 3

(a)

(b)

values of p , q, r are respectively (a) T, T, F (c) F, F, F

10 If the two lines x + (a − 1) y = 1and

2 5 2 (c) 5

1 (b) 32 1 (d) 18

2 5 2 (d) 5

the line of intersection of the planes, x + y + z − 6 = 0 and 2x + 3 y + z + 5 = 0 and it is perpendicular to the XY -plane. Then, the distance of the point (0, 0, 256) from P is equal to (b) 205 5 17 (d) 5

in the binomial expansion of (x + 1)n in powers of x are in the ratio 2 : 15 : 70, then the average of these three coefficients is (a) 964

(b) 227

f (2) = 6, then

Water is poured into it at a constant rate of 5 cu m/min. Then, the rate (in m/min) at which the level of water is rising at the instant when the depth of water in the tank is 10 m is (b)

1 5π

(c)

1 15π

(d)

1 10π

6 If the tangent to the parabola y2 = x at a point (α , β ), (β > 0) is also a tangent to the ellipse, x2 + 2 y2 = 1, then α is equal to (a)

2+1

(c) 2 2 + 1

(b)

2 −1

(d) 2 2 − 1

7 A rectangle is inscribed in a circle with a diameter lying along the line 3 y = x + 7. If the two adjacent vertices of the rectangle are (–8, 5) and (6, 5), then the area of the rectangle (in sq units) is (a) 72 (c) 98

(b) 84 (d) 56

lim x→ 2

f ( x)

∫

6

2t dt is (x − 2)

(b) 0

(c) 24f ′ (2)

(d) 2f ′ (2)

13 The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then

y is x

equal to 7 3 8 (c) 3

7 2 9 (d) 4 (b)

14 The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 +... upto 11th term is (a) 915 (c) 916

(b) 946 (d) 945

15 If m is chosen in the quadratic equation(m2 + 1)x2 − 3x + (m2 + 1)2 = 0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is (a) 10 5 (c) 8 3

π2 2 3 π2 (c) − 4 3

π2 2 3 π2 (d) − 2

(b) −

(a)

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 poles, is (a) 5( 3 + 1) 5 (b) (2 + 3 ) 2 (c) 10( 3 − 1) 3)

19 Some identical balls are arranged in

(a) 12f ′ (2)

(a)

dy x − y sin x = 6x,  0 < x <  and  dx 2 π π   y  = 0, then y  is equal to  3  6

(d) 5(2 +

function and

inverted right circular cone, whose 1 semi-vertical angle is tan −1   .  2

2 π

(d) 625

12 If f : R → R is a differentiable

5 A water tank has the shape of an

(a)

(c) 232

(b) 2 34 (d) 6

18 Two poles standing on a horizontal

(b)

11 If some three consecutive coefficients

4 Let P be the plane, which contains

(a) 34 (c) 5 17

17 If cos x

(b) T, F, F (d) F, T, T

(a)

sin 10º sin 30º sin 50º sin 70º is

(a) 63 5 11 (c) 5

9 If p ⇒ (q ∨ r ) is false, then the truth

2x + a 2 y = 1, (a ∈ R − {0, 1}) are perpendicular, then the distance of their point of intersection from the origin is

3 The value of 1 (a) 36 1 (c) 16

(b)

(c) 16

x + 2 y −1 z = = such that 3 0 4 BC = 5 units. Then, the area (in sq units) of this triangle, given that the point A(1, − 1, 2) is the line,

53 3 (d) 18

(a) 30

16 The vertices B and C of a ∆ABC lie on

(b) 8 5 (d) 4 3

rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then, the number of balls used to form the equilateral triangle is (a) 262 (c) 225

(b) 190 (d) 157

20 If ∫ esec x

(sec x tan x f (x) + (sec x tan x + sec2 x))

dx = esec x f (x) + C, then a possible choice of f (x) is (a) x sec x + tan x +

1 2

1 2 1 (c) sec x + x tan x − 2 1 (d) sec x − tan x − 2 (b) sec x + tan x +

8

40

21 Two newspapers A and B are

(a) lim f (x ) exists but lim f (x ) does not

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 (a) 13.5 (b) 13

(c) 12.8

(d) 13.9

22 The area (in sq units) of the smaller of the two circles that touch the parabola, y2 = 4x at the point (1, 2) and the X-axis is (a) 8 π( 3 − 2 2 )

(b) 4 π(3 +

(c) 8 π(2 − 2 )

(d) 4 π(2 − 2 )

2)

23 If the sum and product of the first three terms in an AP are 33 and 1155, respectively, then a value of its 11th term is (a) 25

(b) –36

(c) –25

(d) –35

x 24 If f (x) = [x] −   , x ∈ R where [x]

 4  denotes the greatest integer function, then

x→ 4 +

x→ 4 −

exist (b) f is continuous at x = 4 (c) Both lim f (x ) and lim f (x ) exist x→ 4 −

x→ 4 +

(d) lim f (x ) exists but lim f (x ) does not x→ 4 +

exist

25 The domain of the definition of the function f (x) =

1 4− x

2

+ log10 (x3 − x) is

∫ x cot

−1

(c) (−6, 4)

is continuous at x = 5, then the value of a − b is

1 2 3 (d) 4

−2 π+ 5 2 (c) π −5 (a)

2 π+ 5 2 (d) 5− π (b)

30 The total number of matrices  0 2y 1  A =  2x y −1 , (x, y ∈ R , x ≠ y) for    2x − y 1 

(b)

which AT A = 3I3 is

27 The value of the integral 1

(b) (4, − 2)

a|π − x|+1, x ≤ 5 b  |x − π|+3, x > 5

2x + 3 y − z = 0, x + ky − 2z = 0 and 2x − y + z = 0 has a non-trivial x y z solution (x, y, z ), then + + + k is y z x equal to

1 4

(a) (6, − 2)

29 If the function f (x) = 

26 If the system of equations

(c) −

x2 + y2 = 4 and x2 + y2 + 6x + 8 y − 24 = 0 also passes through the point

(d) (−4, 6)

(a) (−1, 0) ∪ (1, 2) ∪ (3, ∞ ) (b) (−2, − 1) ∪ (−1, 0) ∪ (2, ∞ ) (c) (−1, 0) ∪ (1, 2) ∪ (2, ∞ ) (d) (1, 2) ∪ (2, ∞ )

(a) −4

π 1 − log e 2 2 2 π (d) − log e 2 2 (b)

28 The common tangent to the circles

but are not equal x→ 4 −

π 1 − log e 2 4 2 π (c) − log e 2 4 (a)

(a) 2 (c) 3

(1 − x2 + x4 )dx is

(b) 4 (d) 6

0

ANSWERS 1. (b) 11. (c) 21. (d)

2. (d) 12. (a) 22. (a)

3. (c) 13. (a) 23. (c)

4. (c) 14. (b) 24. (b)

5. (b) 15. (b) 25. (c)

6. (a) 16. (a) 26. (b)

7. (b) 17. (b) 27. (a)

8. (d) 18. (d) 28. (a)

9. (b) 19. (b) 29. (d)

10. (d) 20. (b) 30. (b)

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9 10 April, Shift-I

6 If a directrix of a hyperbola centred at

1 If for some x ∈ R, the frequency distribution of the marks obtained by 20 students in a test is 2

Marks

3

5

7

Frequency (x + 1) 2 2x − 5 x 2 − 3x

x

Then, the mean of the marks is (a) 3.0 (c) 2.5

(b) 2.8 (d) 3.2

point P in the plane 3x − y + 4z = 2 and R is the point (3, − 1, − 2) , then the area (in sq units) of ∆PQR is (a) (c)

(b) 2 13 (d)

(a) 4e4 (b) 4e4 (c) 4e4 (d) 4e4

x2 + y2 + 5Kx + 2 y + K = 0 and 2 (x2 + y2 ) + 2Kx + 3 y −1 = 0, (K ∈ R ), intersect at the points P and Q, then the line 4x + 5 y − K = 0 passes through P and Q, for (b) exactly one value of K (c) exactly two values of K (d) infinitely many values of K

4 Let f (x) = e − x and g (x) = x − x, 2

∀ x ∈ R. Then, the set of all x ∈ R, where the function h (x) = (fog ) (x) is increasing, is 1 (a)  0,  ∪ [1, ∞ )  2  −1 1 (b)  −1,  ∪  , ∞    2   2 (c) [0, ∞ ) −1 (d)  , 0 ∪ [1, ∞ )  2 

dx (x − 2x + 10)2 2

  x − 1 f (x )  + C , = A  tan −1   + 2   3 x − 2 x + 10   where, C is a constant of integration, then (a) A =

1 and f (x ) = 9 (x − 1) 27

(b) A =

1 and f (x ) = 3 (x − 1) 81

(c) A =

1 and f (x ) = 3 (x − 1) 54

(d) A =

1 and f (x ) = 9 (x − 1) 2 54

and|x + y|≤ 2 is bounded by a (a) rhombus of side length 2 units (b) rhombus of area 8 2 sq units (c) square of side length 2 2 units

14 If a > 0 and z =

7 If the line x − 2 y = 12 is tangent to the x2 2

+

y2 2

= 1 at the point

a b  3, −9  , then the length of the    2 latusrectum of the ellipse is (a) 8 3 (c) 5

(a)

(b) 9 (d) 12 2

212

(c)

(b)

(sin θ + 8)12 212

(d)

(sin θ − 4)12

magnitude

15 If the length of the perpendicular from the point (β, 0, β ) (β ≠ 0) to the line,

26 (sin θ + 8)12 212 (sin θ − 8) 6

can be formed using the digits 0, 1, 2,5, 7 and 9 which are divisible by 11 and no digit is repeated, is (b) 72 (d) 36

10 If a1 , a2 , a3 , ..., an are in AP and a1 + a4 + a7 + ... + a16 = 114 , then a1 + a6 + a11 + a16 is equal to (a) 64 (c) 98

x y−1 z + 1 is = = 1 0 −1

11 ABC is a triangular park with AB = AC = 100 m. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot −1 (3 2 ) and cosec−1 (2 2 ) respectively, then the height of the tower (in m) is (b) 20 100 (d) 3 3

(c) 10 5 

12 lim  (n +41/ 3) n → ∞

1/ 3

n

+

(n + 2)1/ 3 n 4/ 3

equal to (b) − 2 (d) 1

(a) 2 (c) − 1

x

16 If ∆1 = − sin θ cosθ

sin θ cosθ −x 1 and 1 x

x sin 2θ cos 2θ , x ≠ 0, ∆ 2 = − sin 2θ −x 1 cos 2θ 1 x π then for all θ ∈  0,   2

+ ...+

(b) ∆ 1 − ∆ 2 = − 2x 3 (c) ∆ 1 + ∆ 2 = − 2x 3 (d) ∆ 1 − ∆ 2 = x(cos 2θ − cos 4θ)

17 Let f (x) = x2 , x ∈ R. For any A ⊆ R, define g (A ) = {x ∈ R : f (x) ∈ A }. If S = [0, 4], then which one of the following statements is not true? (a) f ( g (S )) = S (b) g (f (S )) ≠ S (c) g (f (S )) = g (S ) (d) f(g(S)) ≠ f (S )

(2n )1/ 3   n 4/ 3 

is equal to 4 (a) (2) 4/ 3 3 3 3 (c) (2) 4/ 3 − 4 4

3 , then β is 2

(a) ∆ 1 + ∆ 2 = − 2(x 3 + x − 1)

(b) 76 (d) 38

(a) 25

2 , then z is equal to 5

1 3 − i 5 5 1 3 (b) − − i 5 5 1 3 (c) − + i 5 5 3 1 (d) − − i 5 5

9 The number of 6 digits numbers that

(a) 60 (c) 48

(1 + i )2 , has a−i

(a)

π x2 + x sin θ − 2 sin θ = 0, θ ∈  0,  ,  2 α12 + β12 then −12 is equal to (α + β −12 )(α − β )24

(a) no values of K

5 If ∫

− 24e + 35 = 0

quadratic equation,

3 If the circles

13 The region represented by|x − y|≤ 2

(d) square of area 16 sq units

2

8 If α and β are the roots of the

65 2

x

− 12e2 − 27 = 0 − 24e2 + 27 = 0 + 8e2 − 35 = 0

ellipse

2 If Q(0, − 1, − 3) is the image of the

91 2 91 4

the origin and passing through the point (4, − 2 3 ) is 5x = 4 5 and its eccentricity is e, then

3 4 (b) (2) 4/ 3 − 4 3 4 (d) (2) 3/ 4 3

18 If lim

x→1

4 3 3 (c) 2 (a)

x4 − 1 x3 − k3 , then k is = lim 2 x − 1 x → k x − k2 3 8 8 (d) 3

(b)

10 40 19 All the pairs (x, y) that satisfy the inequality 2

sin 2 x − 2 sin x + 5

⋅

1 4sin

2

y

≤1

also satisfy the equation

point (1, 1). If the circle also passes through the point (1, − 3), then its radius is (b) 2 2 (d) 3

be the vertices of a triangle and M be the mid-point of AC. If G divides BM in the ratio 2 : 1, then cos (∠GOA ) (O being the origin) is equal to

(c)

1 2 15 1 (d) 6 10 (b)

22 Which one of the following Boolean expressions is a tautology ? (a) (b) (c) (d)

( p ∨ q) ∨ ( p ∨ ~ q) ( p ∧ q) ∨ ( p ∧ ~ q) ( p ∨ q) ∧ ( p ∨ ~ q) ( p ∨ q) ∧ (~ p ∨ ~ q)

(b) (− 21, 714) (d) (− 54, 315)

(a) (28, 315) (c) (28, 861)

25 The sum of series +

21 Let A(3, 0, −1), B (2, 10, 6) and C(1, 2, 1)

1 15 1 30

(a) 7 (c) 10

zero, in the expansion of the expression (1 + ax + bx2 ) (1 − 3x)15 in powers of x, then the ordered pair (a , b) is equal to

20 The line x = y touches a circle at the

(a)

not differentiable differentiable if f ′ (c) ≠ 0 not differentiable if f ′ (c) = 0 differentiable if f ′ (c) = 0

24 If the coefficients of x2 and x3 are both

(a) 2|sin x| = 3 sin y (b) sin x = |sin y| (c) sin x = 2 sin y (d) 2 sin x = sin y

(a) 3 2 (c) 2

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

3 × 13 2

1

7 × (13 + 23 + 33 )

5 × (13 + 23 ) 12 + 22

+ .......... + upto

12 + 22 + 32 10th term, is (a) 680 (c) 660

+

(b) 600 (d) 620

26 Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls; is 1 (a) 17

1 (b) 12

1 (c) 10

1 (d) 11

27 If the system of linear equations x+ y+ z=5

and f (c) = 0. If g (x) = |f (x)|, then at x = c, g is

 sin ( p + 1) x + sin x , x< 0  x  28 If f (x) =  q, x=0  x> 0 x + x2 − x  ,  x3 / 2 is continuous at x = 0 , then the ordered pair ( p , q) is equal to 3 1 (a)  − , −   2 2

1 3 (b)  − ,   2 2

5 1 (c)  ,   2 2

3 1 (d)  − ,   2 2

29 If y = y(x) is the solution of the differential equation dy = (tan x − y) sec2 x, x ∈  −  dx  such that y (0) = 0, then y  −  equal to 1 −2 e 1 (c) 2 + e

(b)

(a)

1 −e 2

(d) e − 2 2π

30 The value of ∫ [sin 2x (1 + cos 3x)] dx, 0

x + 3 y + λz = µ,(λ ,µ ∈ R ), has infinitely many solutions, then the value of λ + µ is

(a) − π

(b) 2π

(c) − 2π

(d) π

ANSWERS 1. (b) 11. (b) 21. (a)

2. (a) 12. (c) 22. (a)

3. (a) 13. (c) 23. (d)

4. (a) 14. (b) 24. (a)

5. (c) 15. (c) 25. (c)

6. (d) 16. (c) 26. (d)

7. (b) 17. (c) 27. (c)

8. (a) 18. (d) 28. (d)

π π , , 2 2 π  is 4

where [t ] denotes the greatest integer function, is

x + 2 y + 2z = 6

23 Let f : R → R be differentiable at c ∈ R

(b) 12 (d) 9

9. (a) 19. (b) 29. (d)

10. (b) 20. (b) 30. (a)

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11 10 April, Shift-II 1 The integral ∫

π /3

π /6

sec 2/ 3 x cosec4/3 x dx

is equal to (a) 35/ 6 − 32 / 3 (b) 37/ 6 − 35/ 6 (c) 35/ 3 − 31/ 3 (d) 34/ 3 − 31/ 3

1 + 2 1+ 2 13 + 23 + + 1+ 2 + 1 − (1 + 2 + 2 1+

+

(a) 480 (c) 380

x→1

3

2β | = 19 2β | = 9 6β | = 19 6β | = 11

x − ax + b = 5, then a + b is x−1

(a) − 4 (c) − 7

3

(a) 620 (c) 1240

3 The area (in sq units) of the region bounded by the curves y = 2x and y = | x + 1|, in the first quadrant is 3 (a) 2 1 (c) 2

3 (b) log e 2 + 2 3 1 (d) − 2 log e 2

4 If 5x + 9 = 0 is the directrix of the hyperbola 16x − 9 y = 144, then its corresponding focus is 2

2

5 (a)  − , 0  3 

(b) (− 5, 0)

5 (c)  , 0 3 

(d) (5, 0)

(b) 1 (d) 5

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 (a) 180 (c) 170

(b) 210 (d) 190

system of linear equations x + y + z = 6, 4x + λy − λz = λ − 2 and 3x + 2 y − 4z = − 5 has infinitely many solutions. Then λ is a root of the quadratic equation (a) λ2 − 3λ − 4 = 0 (b) λ2 + 3λ − 4 = 0 (c) λ2 − λ − 6 = 0 (d) λ2 + λ − 6 = 0

in AP and a : b = 1 : 3. If c = 4 cm, then the area (in sq cm) of this triangle is

which touch the circle, x2 + y2 = 1 externally, also touch the Y-axis and lie in the first quadrant, is (a) y = 1 + 2x , x ≥ 0 (b) y = 1 + 4x , x ≥ 0

(a)

2 3

(d)

4 3

$ from the position vector − $i + 2$j + 6k

(d) x = 1 + 4 y , y ≥ 0

6 The tangent and normal to the ellipse 3x2 + 5 y2 = 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 is (a)

16 3

(b)

14 3

(c)

34 15

(d)

68 15

7 If the tangent to the curve y = 3 ), at a point

x , x2 − 3

(b) 4 3 (d) 7

y = α, where 2 y − 1 ≤ x ≤ 1, − 2 ≤ y ≤ 2, x ≤ , then for 2 all x, y, 4x2 − 4xy cosα + y2 is equal to

(a) 2 sin α (c) 4 sin 2 α

equal to (b) − 4 (d) 1

(a) 0 (c) 6

(b) 4 cos α + 2x y (d) 4 sin 2 α − 2x 2 y 2 2

3 from 5 the origin. Then which one of the following points lies on any of these lines?

4x − 3 y + 2 = 0, at a distance

1 2 (a)  − , −   4 3

1 2 (b)  − ,   4 3

1 1 (c)  , −  4 3

1 (d)  , 4

1  3

18 A perpendicular is drawn from a

x−1 y+ 1 z = = to 2 −1 1

(a) (− 1, 0, 4) (c) (2, 0, 1)

(b) (4, 0, − 1) (d) (1, 0, 2)

g (x) = sin − 1 (e− x ), (x ≥ 0). If α is a positive real number such that a = (fog )′ (α ) and b = (fog )(α ), then

13 If cos− 1 x − cos− 1

2

x −6 −1 2 − 3x x − 3 = 0, is − 3 2x x + 2

19 Let f (x) = log e (sin x), (0 < x < π ) and

straight line passing through the point (2, 3, − 4) and parallel to the $ is vector, 6$i + 3$j − 4k (a) 2 13 (c) 6

equation

the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x − y + z = 3. Then, the coordinates of Q are

12 The distance of the point having

(c) x = 1 + 2 y , y ≥ 0

16 The sum of the real roots of the

point on the line

(b) 4 3

(c) 2 3

(b) 15 (d) 9

17 Lines are drawn parallel to the line

10 Let λ be a real number for which the

11 The angles A , B and C of a ∆ABC are

5 The locus of the centres of the circles,

1 2 and units from the 3 3 planes 4x − 2 y + 4z + λ = 0 and 2x − y + 2z + µ = 0, respectively, then the maximum value of λ + µ is equal to the distances

(a) 13 (c) 5

9 Suppose that 20 pillars of the same

(b) 660 (d) 1860

(b) 400 (d) 525

15 If the plane 2x − y + 2z + 3 = 0 has

equal to 3

3 + K + 15) is equal to

x ∈ R, (x ≠ ±

(a) | 6α + (b) | 6α + (c) | 2α + (d) | 2α +

8 If lim

1 + 2 + 3 + ... 1+ 2 + 3 33 + K + 153 3 + K + 15

3

mean of (x1 − 4)2, (x2 − 4)2 , K , (x50 − 4)2 is

2

2 The sum of series 3

(α , β ) ≠ (0, 0) on it is parallel to the line 2x + 6 y − 11 = 0, then

2 2

14 If both the mean and the standard deviation of 50 observations x1 , x2 , K , x50 are equal to 16, then the

(a) aα 2 − bα − a = 0 (b) aα 2 − bα − a = 1 (c) aα 2 + bα − a = − 2α 2 (d) aα 2 + bα + a = 0

20 If the line ax + y = c, touches both the curves x2 + y2 = 1 and y2 = 4 2x, then | c|is equal to (a)

1 2

(b) 2

(c)

2

(d)

1 2

12 40 2

2

21 If ∫ x5 e− x dx = g (x)e− x + C, where C is a constant of integration, then g (− 1) is equal to (a) − 1 1 (c) − 2

(b) 1 (d) −

Then, the common difference of this AP, which maximises the product a1 , a4 , a5 , is 2 (b) 3 6 (d) 5

such that| zw| = 1and π arg(z ) − arg(w) = , then 2

(c) zw = i

(b) zw = (d) zw =

1 36 π

1 (b) 18 π (d)

5 6π

expression ~ s ∨ (~ r ∧ s) is equivalent to (b) ~ s ∧ ~ r (d) r

1− i 2 − 1+ i 2

24 A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3 /min. When the thickness of the ice is 5 cm, then the rate at which

8 6 7 5

differential equation, dy + y tan x = 2x + x2 tan x, dx π π x ∈  − ,  , such that y(0) = 1. Then  2 2

π π (b) y ′   + y ′  −  = −  4  4 π (c) y   + y  −  4  π  (d) y   − y  −  4 

2. (a) 12. (d) 22. (a)

3. (d) 13. (c) 23. (a)

4. (b) 14. (b) 24. (b)

5. (a) 15. (a) 25. (a)

that the coefficient of x in the n 1 expansion of  x2 + 3  is nC23 , is  x 

6. (d) 16. (a) 26. (c)

35 23 58 38

7. (a) 17. (b) 27. (d)

2

π π2 + 2  = 4 2 π  = 2 4

1 ratio r, where a ≠ 0 and 0 < r ≤ . If 2 3a, 7b and 15c are the first three terms of an AP, then the 4th term of this AP is (a) 5a (c) a

2 a 3 7 (d) a 3

(b)

ANSWERS 1. (b) 11. (c) 21. (d)

2

30 Let a, b and c be in GP with common

27 The smallest natural number n, such

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

(b) 3 (d) 2

π π (a) y ′   − y ′  −  = π −  4  4

must be tossed so that the probability of getting atleast one head is more than 99% is (a) (b) (c) (d)

equation 5 + | 2x − 1| = 2x (2x − 2) is

29 Let y = y(x) be the solution of the

25 The negation of the boolean

(a) s ∧ r (c) s ∨ r

28 The number of real roots of the (a) 1 (c) 4

26 Minimum number of times a fair coin

23 If z and w are two complex numbers

(a) zw = − i

1 (a) 9π (c)

5 2

22 Let a1 , a2 , a3 , K be an AP with a6 = 2.

8 (a) 5 3 (c) 2

the thickness (in cm/min) of the ice decreases, is

8. (c) 18. (c) 28. (a)

9. (c) 19. (b) 29. (a)

10. (c) 20. (c) 30. (c)

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13 12 April, Shift-I

(c) log e

1 If A is a symmetric matrix and B is a skew-symmetric matrix such that 2 3  A+ B= , then AB is equal to  5 −1  −4 −2  −1 4     4 −2 (c)  1 −4  

(a)

4  −1   −4 (d) 1 

(b)

1 1 ,  e e2  1 1 , − 2  e e 

120 13

(b)

60 (c) 13

13 (d) 2

4 If the area (in sq units) of the region {(x, y): y ≤ 4x, x + y ≤ 1, x ≥ 0, y ≥ 0} is a 2 + b, then a − b is equal to 2

10 3 8 (c) 3 (a)

(d) −

2 3

integer ≤ x, then the sum of the series − 1 + − 1 − 1  + − 1 − 2  +  3   3 100   3 100  1 99  is … + − −  3 100  (a) − 153 (b) − 133 (c) − 131 (d) − 135

6 The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is (a) 2 − 1 (c) 220

7 The integral ∫

(b) 2

21

(d) 220 + 1

2x3 − 1 x4 + x

dx is equal to

(here C is a constant of integration) (a) (b)

2

|x 3 + 1| 1 log e +C 2 x2 (x 3 + 1) 2 1 log e +C 2 |x 3|

second and third quadrants only first, second and fourth quadrants first, third and fourth quadrants third and fourth quadrants only

9 Let f : R → R be a continuously differentiable function such that 1 f (2) = 6 and f ′ (2) = . If 48 f ( x)

∫6

4t3 dt = (x − 2) g (x), then lim g (x) x→ 2

is equal to (a) 18

(b) 24

(c) 12

(d) 36

18

10 The coefficient of x in the product (1 + x)(1 − x)10 (1 + x + x2 )9 is

(b)

1 5

(c)

3 10

(d)

3 20

12 Consider the differential equation,

 1 y2dx +  x −  dy = 0. If value of y is 1  y when x = 1, then the value of x for which y = 2, is 5 1 + 2 e 1 1 (c) + 2 e (a)

3 1 − 2 e 3 (d) − e 2 (b)

13 Let a = 3$i + 2$j + 2k$ and $ be two vectors. If a b = i$ + 2$j − 2k vector perpendicular to both the vectors a + b and a − b has the magnitude 12, then one such vector is $ ) (b) 4 (2$i − 2$j − k $) (a) 4 (2$i + 2$j + k $ ) (d) 4 (− 2i$ − 2$j + k $) (c) 4 (2i$ + 2$j − k

14 Let a random variable X have a binomial distribution with mean 8 and k variance 4. If P (X ≤ 2) = 16 , then k is 2 equal to (a) 17 (c) 1

(b) 121 (d) 137

16 The equation|z − i| = |z − 1|, i = −1, represents 1 2 (b) the line passing through the origin with slope 1 (c) a circle of radius 1 (d) the line passing through the origin with slope − 1 (a) a circle of radius

17 If the truth value of the statement p → (~ q ∨ r ) is false (F), then the truth values of the statements p , q and r are respectively (a) T, T and F (c) T, F and T

(b) T, F and F (d) F, T and T

π /2

cot x dx = m( π + n ), cot x + cosec x then m ⋅ n is equal to

hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is 1 10

(b) 5 (d) 4

0

11 If there of the six vertices of a regular

(a)

(a) 3 (c) 7

18 If ∫

(b) − 126 (d) 126

(a) 84 (c) − 84

(b) 6

5 For x ∈ R, let [x] denote the greatest

20

8 The equation

2 4

where the two circles with radii 5 cm and 12 cm intersect is 90°, then the length (in cm) of their common chord is 13 5

5π 5π  1 + sin 4 x = cos2 3x, x ∈  − , is  2 2 

+C

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

3 If the angle of intersection at a point

(a)

x

2

equation

y = sin x sin(x + 2) − sin (x + 1) represents a straight line lying in

 dy d 2 y    at x = 0 is equal to , 2  dx dx  (b)  −  (d)  − 

|x 3 + 1|

15 The number of solutions of the

−2 −4

2 If ey + xy = e, the ordered pair

1 1 (a)  , − 2  e e  1 1 (c)  , 2  e e 

(d) log e

x3 + 1 +C x

(a) −

1 2

(b) 1

(c)

1 2

(d) −1

3 19 For x ∈  0,  , let 

2 f (x) = x , g (x) = tan x and 1 − x2 . If φ(x) = ((hof )og )(x), h (x ) = 1 + x2 π then φ   is equal to  3 π 12 7π (c) tan 12

11π 12 5π (d) tan 12

(a) tan

(b) tan

20 The value of sin −1 

12  −1  3   − sin   is  13   5

equal to 63 π 56 (a) π − sin −1   (b) − sin −1    65   65  2 (c)

π 9 33 − cos−1   (d) π − cos−1    65   65  2

21 A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/s, then the rate (in cm/s) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is (a) 25 3 (c)

25 3

(b)

25 3

(d) 25

14 40 22 If α and β are the roots of the equation 375x − 25x − 2 = 0, then 2

n

lim

n→ ∞

∑α

r

+ lim

n→ ∞

r =1

21 (a) 346

29 (b) 358

n

∑β

r

is equal to

r =1

1 (c) 12

7 (d) 116

23 If the normal to the ellipse 3x2 + 4y2 = 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to 5 5 2 221 (c) 2

(a)

(b) (d)

61 2 157 2

terms of an AP. If S4 = 16 and S6 = − 48, then S10 is equal to (a) − 260 (c) − 320

(b) − 410 (d) − 380

the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2000, then the standard deviation of this data is

which the function f (x) = x kx − x is increasing in the interval [0, 3] and M is the maximum value of f in the interval [0, 3] when k = m, then the ordered pair (m, M ) is equal to (b) (4, 3 3) (d) (5, 3 6)

common tangents to the parabola y2 =12x and the hyperbola 8x2 − y2 = 8. If S and S′ denotes the foci of the hyperbola where S lies on the positive X-axis then P divides SS′ in a ratio (a) 13 : 11 (c) 5 : 4

(b) −1 (d) 2

(a) 0 (c) 1

29 If the volume of parallelopiped formed $ $j + λk $ and by the vectors $i + λj$ + k, $ is minimum, then λ is equal to λi$ + k 1 3

(a) −

(b) 2 (d) 2

27 Let P be the point of intersection of the 2

 5 2α 1  2 1  is the inverse of a   α 3 −1

28 If B =  0

3 × 3 matrix A, then the sum of all values of α for which det (A) + 1 = 0, is

26 If the data x1 , x2 , … , x10 is such that

(a) 2 2 (c) 4

24 If m is the minimum value of k for

(a) (4, 3 2) (c) (3, 3 3)

25 Let Sn denote the sum of the first n

(b) 14 : 13 (d) 2 : 1

(c)

3

(b)

1 3

(d) −

3

x− 2 y+ 1 z −1 30 If the line = = 3 2 −1 intersects the plane 2x + 3y − z + 13 = 0 at a point P and the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to (a) 14

(b)

(c) 2 7

(d) 2 14

14

ANSWERS 1. (b) 11. (a) 21. (b)

2. (b) 12. (b) 22. (c)

3. (b) 13. (b) 23. (a)

4. (b) 14. (d) 24. (b)

5. (b) 15. (b) 25. (c)

6. (c) 16. (b) 26. (b)

7. (c) 17. (a) 27. (c)

8. (d) 18. (d) 28. (c)

9. (a) 19. (b) 29. (b)

10. (a) 20. (b) 30. (d)

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15 12 April, Shift-II  sin x − cos x  ,  sin x + cos x 

1 The derivative of tan −1 

x π   with respect to , where  x ∈  0,    2  2 is

(c)

1 2

(d) 2

admission test, a candidate is given fifty problems to solve. If the probability that the candidate can 4 solve any problem is , then the 5 probability that he is unable to solve less than two problem is 201  1    5  5

(c)

54  4    5  5

49

(b)

49

(d)

316  4    25  5 

48

164  1    25  5 

48

3 A value of α such that α+1

∫

α

dx 9 = log e   is  8 (x + α ) (x + α + 1)

(a) − 2

(b)

1 2

(c) −

1 2

(d) 2

4 Let α ∈ (0, π / 2) be fixed. If the integral tan x + tan α

∫ tan x − tan α

dx = A (x) cos 2α + B (x)

sin 2α + C, where C is a constant of integration, then the functions A (x) and B (x) are respectively (a) x + α and log e|sin(x + α )| (b) x − α and log e|sin(x − α )| (c) x − α and log e|cos(x − α )| (d) x + α and log e|sin(x − α )|

(b) [1, 4] (d) [2, 6]

7 A plane which bisects the angle

x + 2 sin x

8 lim

x→ 0

x2 + 2 sin x + 1 −

sin 2 x − x + 1

is (a) 6 (c) 3

(b) 2 (d) 1

9 A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to (a) 28 (c) 25

(b) 27 (d) 24

10 An ellipse, with foci at (0, 2) and (0, − 2) and minor axis of length 4, passes through which of the following points? (a) ( 2 , 2) (c) (2, 2 2 )

(b) (2, 2 ) (d) (1, 2 2 )

11 The boolean expression ~ ( p ⇒ (~ q)) is equivalent to (a) p ∧ q (c) p ∨ q

(b) q ⇒ ~ p (d) (~ p ) ⇒ q

and making a intercept of length 8 on theY -axis passes through the point

vertical tower standing on a horizontal plane is observed to be 45° from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30°, then the distance (in m) of the foot of the tower from the point A is (b) 15 (5 − (d) 15 (1 +

(b) (1, 4, − 1) (d) (2, − 4, 1)

12 A circle touching the X-axis at (3, 0)

5 The angle of elevation of the top of a

(a) 15 (3 + 3 ) (c) 15 (3 − 3 )

(a) R (c) [3, 7]

(a) (1, − 4, 1) (c) (2, 4, 1)

2 For an initial screening of an

(a)

the equation, cos 2x + α sin x = 2α − 7 has a solution. Then, S is equal to

between the two given planes 2x − y + 2z − 4 = 0 and x + 2 y + 2z − 2 = 0, passes through the point

2 (b) 3

(a) 1

6 Let S be the set of all α ∈ R such that

3) 3)

(a) (3, 10) (c) (2, 3)

13 If

20

C1 + (22 )

(b) (3, 5) (d) (1, 5)

C2 + (32 ) 20C3 + .....

20

+ (202 )20 C20 = A (2β ) , then the ordered pair (A , β) is equal to (a) (420, 19) (c) (380, 18)

(b) (420, 18) (d) (380, 19)

14 A value of θ ∈ (0, π / 3), for which 1 + cos2 θ sin 2 θ 4 cos 6θ 2 cos θ 1 + sin 2 θ 4 cos 6θ = 0, is cos2 θ sin 2 θ 1 + 4 cos6θ

π 9 7π (c) 24 (a)

π 18 7π (d) 36 (b)

15 The equation of a common tangent to the curves, y2 = 16x and xy = − 4, is (a) x − y + 4 = 0 (b) x + y + 4 = 0 (c) x − 2 y + 16 = 0 (d) 2x − y + 2 = 0

16 Let z ∈ C with Im (z ) = 10 and it

2z − n = 2i − 1 for some 2z + n natural number n, then satisfies

(a) n = 20 and Re(z ) = − 10 (b) n = 40 and Re(z ) = 10 (c) n = 40 and Re(z ) = − 10 (d) n = 20 and Re(z ) = 10

17 A triangle has a vertex at (1, 2) and the mid-points of the two sides through it are (−1, 1) and (2, 3). Then, the centroid of this triangle is 7 (a)  1,   3 1 (c)  , 1 3 

1 (b)  , 2 3  1 5 (d)  ,   3 3

18 If a1 , a2 , a3 , ... are in AP such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this AP is (a) 200 (b) 280 (c) 120 (d) 150

19 If [x] denotes the greatest integer ≤ x , then the system of liner equations [sin θ]x + [− cosθ]y = 0, [cot θ]x + y = 0 (a) have infinitely many solutions if π 2π  θ ∈  ,  and has a unique 2 3  7π  solution if θ ∈  π , .  6  (b) has a unique solution if π 2π  7π   θ ∈  ,  ∪  π,  2 3   6  π 2π  (c) has a unique solution if θ ∈  ,  2 3  and have infinitely many solutions if 7π  θ ∈  π ,   6  (d) have infinitely many solutions if π 2π   7π  θ ∈  ,  ∪  π,  2 3   6 

16 40 20 A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of 60° with the linex + y = 0. Then, an equation of the line L is (a) x + 3 y = 8 (b) ( 3 + 1) x + ( 3 − 1) y = 8 2 (c) 3x + y = 8 (d) ( 3 − 1)x + ( 3 + 1) y = 8 2

23 A person throws two fair dice. He wins ` 15 for throwing a doublet (same numbers on the two dice), wins ` 12 when the throw results in the sum of 9, and loses ` 6 for any other outcome on the throw. Then, the expected gain/loss (in `) of the person is 1 gain 2 1 (c) loss 2 (a)

21 Let f (x) = 5 −|x − 2|and g (x) = |x + 1|, x ∈ R. If f (x) attains maximum value at α and g (x) attains minimum value (x − 1) (x2 − 5x + 6) of β, then lim is x → − αβ x2 − 6x + 8 equal to (a) 1/2 (b) − 3 / 2 (c) − 1 / 2 (d) 3/2

(b)

1 loss 4

$ , b = 2$i + $j − αk $ a = αi$ + $j + 3k $ $ $ . Then, the set and c = αi − 2 j + 3k

(d) 2 gain

24 The tangents to the curve y = (x − 2)2 − 1 at its points of intersection with the line x − y = 3, intersect at the point 5 (a)  , 1 2 

5 (b)  − , − 1  2 

5 (c)  , − 1 2 

5 (d)  − , 1  2 

S = {α : a , b and c are coplanar} (a) is singleton (b) is empty (c) contains exactly two positive numbers (d) contains exactly two numbers only one of which is positive

differential equation ( y2 − x3 )dx − xydy = 0 (x ≠ 0) is (where, C is a constant of integration) 2x 2 + Cx 3 = 0 2x 3 + Cx 2 = 0 2x 2 + Cx 3 = 0 2x 3 + Cx 2 = 0

28 If the area (in sq units) bounded by the parabola y2 = 4λx and the 1 line y = λx, λ > 0, is , then λ is equal to 9 (a) 2 6 (c) 24

(b) 48 (d) 4 3

29 The length of the perpendicular

terms of a non-constant GP such that the equations αx2 + 2βx + γ = 0 and x2 + x − 1 = 0 have a common root, then, α(β + γ) is equal to (a) 0 (c) αγ

27 The general solution of the

(a) y 2 − (b) y 2 + (c) y 2 + (d) y 2 −

25 If α , β and γ are three consecutive

22 Let α ∈ R and the three vectors

(b) If (A − B ) ⊆ C, then A ⊆ C (c) (C ∪ A ) ∩ (C ∪ B ) = C (d) If (A − C ) ⊆ B, then A ⊆ B

(b) αβ (d) βγ

drawn from the point (2, 1, 4) to the plane containing the lines $) r = ($i + $j) + λ( $i + 2$j − k $ $ $ $ ) is and r = (i + j) + µ (− i + $j − 2k (a) 3

(b)

1 3

(c) 3

2. (c) 12. (a) 22. (b)

3. (a) 13. (b) 23. (c)

4. (b) 14. (a) 24. (c)

5. (a) 15. (a) 25. (d)

φ ≠ A ∩ B ⊆ C. Then, which of the following statements is not true? (a) B ∩ C ≠ φ

6. (d) 16. (c) 26. (d)

7. (d) 17. (b) 27. (b)

8. (b) 18. (a) 28. (c)

9. (c) 19. (a) 29. (c)

1 3

30 The term independent of x in the

26 Let A , B and C be sets such that

 1 x8   expansion of  −  60 81 equal to (a) − 72 (c) − 36

10. (a) 20. (d) 30. (c)

6

3 .  2x2 − 2  is  x 

(b) 36 (d) − 108

ANSWERS 1. (d) 11. (a) 21. (a)

(d)

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17

JANUARY ATTEMPT 9 January, Shift-I 2 3 π 3 1 If cos   + cos−1   =  x >  ,  3x   4x  2  4 then x is equal to −1 

145 10 145 12

(a) (c)

(b) (d)

146 12 145 11

π

0

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

3 For x ≠ nπ + 1, n ∈ N (the set of 2

natural numbers), the integral

∫x

2 sin(x2 − 1) − sin 2(x2 − 1) 2 sin(x − 1) + sin 2(x − 1) 2

2

dx is

equal to (where C is a constant of integration ) (a)

1 log e|sec(x 2 − 1)| + C 2

 x 2 − 1 (b) log e sec   +C  2  (c) log e

1 sec2 (x 2 − 1) + C 2

 x 2 − 1 1 (d) log e sec2   +C 2  2 

4 If y = y(x) is the solution of the dy differential equation, x + 2 y = x2 dx 1 satisfying y(1) = 1, then y   2 is equal to 13 16 49 (c) 16 (a)

1 4 7 (d) 64 (b)

5 The equation of the line passing through (−4, 3, 1), parallel to the plane x + 2 y − z − 5 = 0 and intersecting the line x +1 y − 3 z − 2 is = = −3 2 −1

x+4 y− 3 z−1 = = 3 −1 1 x+4 y− 3 z−1 (b) = = −1 1 1 (a)

numbers in GP and a + b + c = xb, then x cannot be

if if if if

x≤1 1< x < 3 3≤ x < 5 x≥ 5

right circular cone having slant height 3m is

(b) (6, 4 2) (d) (5, 2 6)

4 π 3

(b) 2 3π

(c) 3 3π

(d) 6π

 cosθ − sin θ , then the matrix cosθ 

15 If A =   sin θ

 1  (a)  2 3 −  2

px + qy + r = 0 such that 3 p + 2q + 4 r = 0. Which one of the following statements is true? (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

 3  (c)  2 1 −  2

π , is equal to 12

3 2   1  2  1  2   3  2 

  (b)   

3 2 1 2

1 −  2  3  2 

  (d)   

1 2 3 2

−

3 2   1  2 

16 For x ∈ R − {0, 1}, let f1 (x) =

9 Let α and β be two roots of the equation x + 2x + 2 = 0, then α is equal to

15

+β

15

(b) 512 (d) −512

10 If the fractional part of the number 403

2 k is , then k is equal to 15 15 (a) 14 (c) 4

(a)

A −50 when θ =

8 Consider the set of all lines

(a) 256 (c) −256

(b) (−3, 1, 1) (d) (−3, 0, −1)

14 The maximum volume (in cu.m) of the

its vertex and focus are at distances 2 and 4 respectively from the origin, on the positive X-axis, then which of the following points does not lie on it?

2

(d) −3

(a) (3, 3, −1) (c) (3, 2, 1)

7 Axis of a parabola lies along X-axis. If

(c) (8, 6)

(b) 2

the planes x + y + z = 1and 2x + 3 y − z + 4 = 0 and parallel to Y -axis also passes through the point

continuous if a = − 5 and b = 10 continuous if a = 5 and b = 5 continuous if a = 0 and b = 5 not continuous for any values of a and b

(a) (4, −4)

(a) 4 (c) −2

13 The plane through the intersection of

Then, f is

4 (b) − 3 4 (d) 3

(c) 0

12 If a, b and c be three distinct real

6 Let f: R → R be a function defined as  5, a + bx,  f (x ) =   b + 5x,  30,

2 The value of ∫ |cos x|3 dx is 2 (a) 3

x+4 y− 3 z−1 = = 1 1 3 x −4 y + 3 z + 1 (d) = = 2 1 4 (c)

(b) 6 (d) 8

1 1 , f2 (x) = 1 − x and f3 (x) = x 1−x

be three given functions. If a function, J (x) satisfies (f2 ° J ° f1 )(x) = f3 (x), then J (x) is equal to (a) f2 (x )

(b) f3 (x )

(c) f1 (x )

(d)

1 f3 ( x ) x

17 Consider a class of 5 girls and 7 boys.

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

The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is

(a) 16 (c) 20

(a) 350 (c) 200

11 5 students of a class have an average

(b) 22 (d) 18

(b) 500 (d) 300

18 40 18 If θ denotes the acute angle between

the curves, y = 10 − x2 and y = 2 + x2 at a point of their intersection, then |tan θ|is equal to 7 (a) 17 4 (c) 9

8 (b) 15 8 (d) 17

3 y = 3x + 1 3y = x + 3

(a) (c)



3 + 2i sin θ  1 − 2i sin θ

π 2

23 Let A = θ ∈  − , π  : 



touch each other externally. If they have X-axis as a common tangent, then

3π 4

5π 6 2π (d) 3

(c) π

a , b , c are in AP

(d)

1 1 1 = + b a c

(b)

(c) 9

π π 20 For any θ∈  ,  , the expression  4 2

1 2 2 ( 2 + 1)

T =

(c) has a unique solution for|a| =

(b) (∧, ∧) (d) (∨, ∨)

22 Equation of a common tangent to the circle, x2 + y2 − 6x = 0 and the parabola, y2 = 4x, is

3

3

26 The area (in sq units) bounded by the parabola y = x2 − 1, the tangent at the point (2, 3) to it and theY -axis is 8 3 32 (c) 3 (a)

= 27 and

S − 2T = 75, then a10 is equal to (a) 42

(b) is inconsistent when a = 4 (d) is inconsistent when|a| =

15

∑ a( 2i − 1). If a5

i =1

2x + 3 y + (a − 1)z = a + 1

( p ⊕ q) ∧ (~ p ⋅ q) is equivalent to p ∧ q, where ⊕, ⋅ ∈{∧,∨}, then the ordered pair(⊕, ⋅) is

−

i =1

(a) has infinitely many solutions for a=4

21 If the Boolean expression

y2

= 1 is greater cos2 θ sin 2 θ than 2, then the length of its latus rectum lies in the interval

hyperbola

2

(a) 13 − 4 cos4 θ + 2 sin 2 θ cos2 θ (b) 13 − 4 cos2 θ + 6 cos4 θ (c) 13 − 4 cos2 θ + 6 sin 2 θ cos2 θ (d) 13 − 4 cos6 θ

π 2 x2

and

x+ y+ z=2

+ 4 sin θ equals

(d) exists and equals

30

2x + 3 y + 2z = 5

6

1 2 2

29 Let a1 , a2 , ..... a30 be an AP, S = ∑ ai

25 The system of linear equations

3 (sin θ − cosθ)4 + 6 (sin θ + cosθ)2

(c) exists and equals

3 3 (a) (1, ] (b) (3,∞) (c) ( ,2] (d) (2, 3] 2 2

19 2 17 (d) 2

(a) 8

1 4 2

28 Let 0 < θ < . If the eccentricity of the

vector such that a × c + b = 0 and a ⋅ c = 4, then|c|2 is equal to

(c)

2

y4

y→ 0

(b)

24 Let a = $i − $j, b = $i + $j + k$ and c be a

(a) a , b, c are in AP 1 1 1 (b) = + a b c

1 + 1 + y4 −

(b) does not exist

Then, the sum of the elements in A is (a)

27 lim

(a) exists and equals

 is purely imaginary  

19 Three circles of radii a, b, c(a < b < c)

(a) (∧, ∨) (c) (∨, ∧)

(b) 2 3 y = 12x + 1 (d) 2 3 y = − x − 12

56 3 14 (d) 3

(c) 52

with replacement from a well shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then, P (X = 1) + P (X = 2) equals 25 169 49 (c) 169

52 169 24 (d) 169 (b)

ANSWERS 1. (c) 11. (c) 21. (a)

2. (d) 12. (b) 22. (c)

3. (b) 13. (c) 23. (d)

4. (c) 14. (b) 24. (b)

5. (a) 15. (c) 25. (d)

6. (d) 16. (b) 26. (a)

7. (c) 17. (d) 27. (a)

8. (b) 18. (b) 28. (b)

9. (c) 19. (b) 29. (c)

10. (d) 20. (d) 30. (a)

(d) 47

30 Two cards are drawn successively

(a)

(b)

(b) 57

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19 6 If lines x = ay + b, z = cy + d and

9 January, Shift-II 1 Let the equations of two sides of a triangle be 3x − 2 y + 6 = 0 and 4x + 5 y − 20 = 0. If the orthocentre of this triangle is at (1, 1) then the equation of its third side is (a) 122 y − 26x − 1675 = 0 (b) 26x − 122 y − 1675 = 0 (c) 122 y + 26x + 1675 = 0 (d) 26x + 61y + 1675 = 0

x = a ′ z + b′, y = c′ z + d ′ are perpendicular, then (a) ab′+ bc′+1 = 0 (b) bb′+ cc′+1 = 0 (c) aa ′+ c + c′ = 0 (d) cc′+ a + a ′ = 0

(a) 2

equation, x + x + 1 = 0, If z = 3 + 6iz081 − 3iz093 , then arg z is equal to (a)

π 4

(c)

π 6 π (d) 3

 et e− t cos t  t −t 3 If A =  e − e cos t − e− t sin t  et 2e− t sin t   e− t sin t  −t − e sin t + e− t cos t  then A is  −2e− t cos t  (a) invertible only when t = π (b) invertible for every t ∈ R (c) not invertible for any t ∈ R π (d) invertible only when t = 2

At random a ball is drawn from this pot. If a drawn ball is green then put a red ball in the pot and if a drawn ball is red, then put a green ball in the pot, while drawn ball is not replace in the pot. Now we draw another ball randomnly, the probability of second ball to be red is 27 49 21 (c) 49

26 49 32 (d) 49

(a)

(b)

to

such

R

that

3

| f (x) − f ( y)|≤ 2|x − y|2 , x, y ∈ R. If f (0) = 1, then

for

1 2

(a) 2

(b)

(c) 1

(d) 0

all

1

∫f 0

equal to

x y z the straight line = = and 2 3 4 perpendicular to the plane containing x y z the straight lines = = and 3 4 2 x y z = = is 4 2 3 (a) 5x + 2 y − 4z = 0(b) x + 2 y − 2z = 0 (c) 3x + 2 y − 3z = 0 (d) x − 2 y + z = 0

9 Let a = i$ + $j + 2 k$ , $ and b = b1 $i + b2 $j + 2 k $ be three vectors c = 5 $i + $j + 2 k

(a) 6 (c) 22

2

(x) dx is

(b) 4 (d) 32

10 If x = sin −1 (sin 10) and y = cos−1 (cos 10), then y − x is equal to (a) 0 (c) 7π

(b) 10 (d) π

11 Let a, b and c be the 7th, 11th and 13th terms respectively of a non-constant AP. If these are also the three consecutive terms of a GP, then a is equal to c (a) 2

5 Let f be a differentiable function from R

4 3 2 (d) 3 (b)

such that the projection vector of b on a is a. If a + b is perpendicular to c, then|b |is equal to

4 A pot contain 5 red and 2 green balls.

(c) 4

n

i =1

Σ (xi + 1)2 = 9n and Σ (xi − 1)2 = 5n, (b) 7 (d) 5

(c) 5

8 The equation of the plane containing

(b)

(c) 0

1 3

n

i =1

(a) 2

A = {(x, y); 0 ≤ y ≤ x| x| + 1 and − 1 ≤ x ≤ 1} in sq. units, is

2

observations, x, x2 , ...., xn. If the standard deviation of the data is

7 The area of the region

2 Let z0 be a root of the quadratic

13 In a group of data, there are n

7 (b) 13 1 (d) 2

12 Let A = { x ∈ R : x is not a positive integer}. Define a function f : A → R 2x , then f is as f (x) = x−1 (a) injective but not surjective (b) not injective (c) surjective but not injective (d) neither injective nor surjective

π 2 of x for which sin x − sin 2x + sin 3x = 0, is

14 If 0 ≤ x < , then the number of values

(a) 2 (c) 1

(b) 3 (d) 4

15 If the system of linear equations x − 4 y + 7z = g 3 y − 5z = h − 2x + 5 y − 9z = k is consistent, then (a) 2 g + h + k = 0 (b) g + 2h + k = 0 (c) g + h + k = 0 (d) g + h + 2k = 0

16 Let A (4, − 4) and B(9, 6) be points on the parabola, y2 = 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of ∆ACB is maximum. Then, the area (in sq. units) of ∆ACB, is 1 4 3 (c) 31 4 (a) 31

(b) 32 (d) 30

1 2

17 The number of all possible positive integral values of α for which the roots of the quadratic equation, 6x2 − 11x + α = 0 are rational numbers is (a) 5 (c) 4

(b) 2 (d) 3

18 For each x ∈ R , let [x] be the greatest integer less than or equal to x. Then, lim

x → 0−

x([x] + |x|) sin [x] is equal to |x|

(a) 0 (b) sin 1 (c) − sin 1 (d) 1

19 The coefficient of t 4 in the expansion of 3

 1 − t6    is  1− t  (a) 12 (c) 15

(b) 10 (d) 14

20 40 20 Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is (a) 36 (c) 18

(b) 32 (d) 9

(a) 374 (b) 375 (c) 372 (d) 250

25 If ∫

π /3

0

tan θ 1 dθ =1 − , (k > 0), 2k sec θ 2

then the value of k is

21 The logical statement [~ (~ p ∨ q) ∨ ( p ∧ r )] ∧ (~ q ∧ r ) is equivalent to (a) ~ p ∨ r (c) ( p ∧ r ) ∧ ~ q

using the digits 0, 1, 3, 7, 9 (repitition of digits allowed) is equal to

(b) ( p ∧ ~ q) ∨ r (d) (~ p ∧ ~ q) ∧ r

22 If the circles

1 2 (d) 4

(a) 1

(b)

(c) 2

26 If f (x) = ∫

5x8 + 7x6 (x2 + 1 + 2x7 )2

dx, (x ≥ 0),

x2 + y2 −16x − 20 y + 164 = r 2 and (x − 4)2 + ( y − 7)2 = 36 intersect at two distinct points, then

and f (0) = 0, then the value of f (1) is

(a) 0 < r < 1 (c) 1 < r < 11

(c)

(b) r > 11 (d) r = 11

23 If both the roots of the quadratic equation x2 − mx + 4 = 0 are real and distinct and they lie in the interval [1, 5] then m lies in the interval (a) (4, 5)

(b) (−5, − 4)

(c) (5, 6)

(d) (3, 4)

24 The number of natural numbers less than 7,000 which can be formed by

(a) −

1 2

1 4

(b) − (d)

1 4

1 2

9 (12 + 22 + 32 ) 7 12 (12 + 22 + 32 + 42 ) + 9 15 (12 + 22 + ... + 52 ) + + ... up to 15 11 terms is

1+ 6 +

(a) 7510 (b) 7820 (c) 7830 (d) 7520

29 If x = 3 tan t and y = 3 sec t, then the value of 1 6 1 (c) 3 2 (a)

d 2y dx2

at t =

π , is 4 1 6 2 3 (d) 2 2 (b)

30 A hyperbola has its centre at the

27 Let f : [0, 1] → R be such that f (xy) = f (x). f ( y), for all x, y ∈ [0, 1] and f (0) ≠ 0. If y = y (x) satisfies the dy differential equation, = f (x) with dx 1 3 y(0) = 1, then y   + y   is equal to  4  4 (a) 5 (c) 2

28 The sum of the following series

origin, passes through the point (4, 2) and has transverse axis of length 4 along the X-axis. Then the eccentricity of the hyperbola is (a) 2 (c)

(b) 3 (d) 4

3 2

(b)

2 3

(d) 3

ANSWERS 1 (b) 11 (c) 21 (c)

2 (a) 12 (a) 22 (c)

3 (b) 13 (d) 23 (a)

4 (d) 14 (a) 24 (a)

5 (c) 15 (a) 25 (c)

6 (c) 16 (a) 26 (c)

7 (a) 17 (d) 27 (b)

8 (d) 18 (c) 28 (b)

9 (a) 19 (c) 29 (b)

10 (d) 20 (a) 30 (b)

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21 10 January, Shift-I 1 The equation of a tangent to the hyperbola 4x2 − 5 y2 = 20 parallel to the line x − y = 2 is (a) x − (b) x − (c) x − (d) x −

y− 3= 0 y+ 9= 0 y + 1= 0 y+ 7= 0

2 21 (b) 2 21 (c) 7 3 3

(d)

3 21 2

−π π dy 3 1 + y= , x ∈  ,   3 3 dx cos2 x cos2 x π 4 π and y  = , then y −  equals  4 3  4

3 If

1 + e6 3 1 (c) + e3 3

(b) −

4 3

1 (d) 3

4 Let a = 2i$ + λ1 $j + 3k$ , $ and b = 4$i + (3 − λ 2 )$j + 6k $ be three vectors c = 3$i + 6$j + (λ3 − 1)k

2 3 3 (b) with slope 2 (c) parallel toY -axis (d) parallel to X -axis (a) with slope

(a) Both P (3) and P (5) are true. (b) P (3) is false but P (5) is true. (c) Both P (3) and P (5) are false. (d) P (5) is false but P (3) is true.

10 Let n ≥ 2 be a natural number and

∫

perpendicular to c. Then a possible value of (λ1 , λ 2 , λ3 ) is 1 (c)  − , 4, 0  2 

(b) (1, 5, 1) 1 (d)  , 4, − 2 2 

5 Let f : R → R be a function such that

π . Then, 2

1

(sin n θ − sin θ) n cosθ

dθ is equal to sin n + 1 θ (where C is a constant of integration) (a)

1 1 −    n 2 − 1 sin n + 1 θ 

(b)

1 1 +    n 2 − 1 sin n − 1 θ 

n

1 1 −    n + 1 sin n − 1 θ 

+C

n +1

n

n

+C

n +1

n

(d)

n

(b) − 4 (d) 8

6 If the parabolas y2 = 4b(x − c) and y2 = 8ax have a common normal, then which one of the following is a valid choice for the ordered triad (a , b, c) ?

n

n

+C

a

minimum, then the ordered pair (a , b) is (a) (− 2 , 0)

(b) (0, 2 )

(c) ( 2 , −

(d) (−

2, 2)

12 The sum of all two digit positive

(b) (1, 1, 0)

numbers which when divided by 7 yield 2 or 5 as remainder is

(c) (1, 1, 3)

1 (d)  , 2, 3 2 

(a) 1256 (c) 1356

complex numbers such that 3z 2z 3|z1| = 4|z2|. If z = 1 + 2 , then 2z2 3z1

to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then, the number of students who did not opt for any of the three courses is (a) 42 (c) 38

(b) 102 (d) 1

16 If the third term in the binomial expansion of (1 + xlog 2 x )5 equals 2560, then a possible value of x is (a) 4 2

(b)

1 8

1 4

(d) 2 2

17 If the system of equations

n +1

1 (a)  , 2, 0 2 

7 Let z1 and z2 be any two non-zero

(b) 2 5 (d) 4

15 In a class of 140 students numbered 1

+C

2

2)

(a) 5 (c) 57

(c)

f (x) = x3 + x2f ′ (1) + xf ′ ′ (2) + f ′′′(3), x ∈ R b . 11 Let I = ∫ (x4 − 2x2 ) dx. If I is Then, f (2) equals (a) 30 (c) − 2

13 36 19 (d) 36 (b)

n +1

n

1 1 −  (c) 2   n − 1 sin n − 1 θ 

15 72 19 (c) 72 (a)

(4, 0) touches the circle x2 + y2 + 4x − 6 y = 12 externally at the point (1, − 1), then the radius of C is

‘‘P (n ) : n 2 − n + 41 is prime.’’ Then, which one of the following is true?

0< θ
0), is 1 square unit. Then, k is (a)

3

(b)

1 3

(c)

2 3

(d)

3 2

22 40 20 If the line 3x + 4 y − 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 (a) (4, 3) (c) (4, 4)

(b) (3, 4) (d) (2, 2)

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 (b) 6 : 7 (d) 5 : 8

22 Consider the quadratic equation (c − 5)x2 − 2cx + (c − 4) = 0, c ≠ 5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then, the number of elements in S is (a) 11

(b) 10

(c) 12

(a)

1 4

(b) −

1 4

(c)

1 8

(d)

1 2

25 Let d ∈ R, and

21 The mean of five observations is 5 and

(a) 4 : 9 (c) 10 : 3

B(3, 2, 6) be a point in the space. Then, the value of µ for which the vector AB is parallel to the plane x − 4 y + 3z = 1is

(d) 18

max {|x|, x2 }, |x|≤ 2 23 Let f (x) =  2 < |x|≤ 4  8 − 2|x|, Let S be the set of points in the interval (−4, 4) at which f is not differentiable. Then, S

4+ d  −2 A =  1 (sin θ) + 2   5 (2 sin θ) − d

θ ∈[θ, 2 π ]. If the minimum value of det(A) is 8, then a value of d is (a) −5 (c) 2( 2 + 1)

(b) −7 (d) 2( 2 + 2)

26 If 5, 5r , 5r 2 are the lengths of the sides of a triangle, then r cannot be equal to 5 4 3 (c) 2 (a)

7 4 3 (d) 4 (b)

3π 8 π (c) 2

(b)

(4, − 1, 2) and parallel to the lines x+ 2 y−2 z+1 = = 3 −1 2 x−2 y−3 z−4 also passes = = 1 2 3 through the point and

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

29 The shortest distance between the 3 point  , 0 and the curve 2  y=

x , (x > 0), is

(a)

3 2

(b)

(c)

3 2

(d)

5 4 5 2

30 For each t ∈ R, let [t ] be the greatest

π 27 The sum of all values of θ ∈  0,   2 3 satisfying sin 2 2 θ + cos4 2 θ = is 4 (a)

(a) equals {−2, − 1, 0, 1, 2} (b) equals {−2, 2} (c) is an empty set (d) equals {−2,−1, 1, 2}

(sin θ) − 2  , d  (− sin θ) + 2 + 2d 

28 The plane passing through the point

5π 4

integer less than or equal to t. Then, π (1−|x|+ sin|1 − x|) sin  [1 − x] 2  lim x → 1+ |1 − x|[1 − x] (a) equals 0 (b) does not exist (c) equals − 1 (d) equals 1

(d) π

24 Let A be a point on the line

$ and r = (1 − 3µ )$i + (µ − 1)$j + (2 + 5µ )k

For Detailed Solutions

ANSWERS 1 (c) 11 (d) 21 (a)

2 (a) 12 (c) 22 (a)

3 (a) 13 (c) 23 (a)

4 (c) 14 (a) 24 (a)

5 (c) 15 (c) 25 (a)

6 (c) 16 (b) 26 (b)

7 (*) 17 (a) 27 (c)

8 (a) 18 (a) 28 (c)

9 (a) 19 (b) 29 (d)

10 (c) 20 (d) 30 (a)

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23 10 January, Shift-II 1 The positive value of λ for which the coefficient of x2 in the expression 10 λ x2  x + 2  is 720, is  x  (a) 3

(b) 5 (d) 4

(c) 2 2



2 Let S = (x, y) ∈ R 2 : 

 y2 x2 − = 1, 1+ r 1− r 

where r ≠ ± 1. Then, S represents

7 If the probability of hitting a target by 1 a shooter in any shot, is , then the 3 minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than 5, 6 is (a) 6 (c) 5

along the lines, x + y = 3 and x − y + 3 = 0. If its diagonals intersect at (2, 4), then one of its vertex is

(b) a hyperbola whose eccentricity is 2 , when 0 < r < 1. r+1

(a) (3, 6) (b) (2, 6) (c) (2, 1) (d) (3, 5)

(d) an ellipse whose eccentricity is 1 , when r > 1. r+1

3 With the usual notation, in ∆ABC, if

∠A + ∠B = 120°, a = 3 + 1and b = 3 − 1, then the ratio ∠A : ∠B, is

(a) 7 : 1 (c) 9 : 7

(b) 3 : 1 (d) 5 : 3

4 Let α = (λ− 2) a + b and β = (4λ − 2) a + 3 b be two given vectors where vectors a and b are non-collinear. The value of λ for which vectors α and β are collinear, is (b) −3 (d) −4

(a) 4 (c) 3

5 A helicopter is flying along the curve given by y − x3 / 2 = 7, (x ≥ 0). A soldier 1 positioned at the point  , 7 wants to 2  shoot down the helicopter when it is nearest to him. Then, this nearest distance is 1 7 (a) 3 3 (c)

5 (b) 6

1 7 6 3

(d)

1 2

π/2

dx 6 The value of ∫ , − π / 2 [x] + [sin x] + 4 where [t ] denotes the greatest integer less than or equal to t, is 1 (a) (7 π − 5) 12 3 (c) (4 π − 3) 10

1 (b) (7 π + 5) 12 3 (d) (4 π − 3) 20

π

⋅ cos

π

23

.......cos

1 1024 1 (c) 512 (a)

π 210

(b)

⋅ sin

π 210

is

1 2

(a) one-one but not onto (b) onto but not one-one (c) both one-one and onto (d) neither one-one nor onto

11 If

∑{

Cr ⋅

50 − r

C25

r=0

Then, the minimum value of

det (A ) b

is (a) − 3 (c) 2 3

(b) −2 3 (d) 3 5

5

R (z ) and I (z ) respectively denote the real and imaginary parts of z, then (a) R (z ) > 0 and I (z ) > 0 (b) I (z ) = 0 (c) R (z ) < 0 and I (z ) > 0 (d) R (z ) = − 3

− r}

(4, 3). If its orthocentre is at the origin, then its third vertex lies in which quadrant? (a) Fourth (c) Second

(a) − 4x 3 − 1 (c) − 2x 3 − 1

17 The value of

 19  cot  ∑ cot −1  1 +  n =1   23 (a) 22



p =1



19 (c) 21

(d)

22 23

(1, e) also passes through the point

= K ( C25 ), 50

3 f (x ) , (x > 0) 4 x 1 and f (1) ≠ 4. Then, lim x f    x x→ 0 + 4 7 (c) exists and equals 0 (d) exists and equals 4

21 (b) 19

n

∑ 2 p  is

2

12 Let f be a differentiable function

(b) exists and equals

(b) 4x 3 + 1 (d) − 2x 3 + 1

y = xex passing through the point

(b) 225 − 1 (c) 225 (d) (25) 2

(a) does not exist

1 −4 x 3 e f (x ) + C , 48

where C is a constant of integration, then f (x) is equal to

4 (a)  , 2e 3 

(b) (3, 6e)

(c) (2, 3e)

5 (d)  , 2e 3 

24

such that f ′ (x) = 7 −

(b) Third (d) First

18 The tangent to the curve,

then, K is equal to (a) 2

 2

3

and two functions f and g be defined as f , g : N → N such that  n + 1 ; if n is odd  f (n ) =  2 n ; if n is even   2 and g (n ) = n − (−1)n. Then, fog is

50

b

16 If ∫ x5 e−4 x dx =

1 (d) 256

10 Let N be the set of natural numbers

25

  1

15 Two vertices of a triangle are (0, 2) and

9 The value of 22

1

14 Let z = 

8 Two sides of a parallelogram are

cos

b

 3 i  3 i +  +  −  . If 2 2  2  2

(b) 3 (d) 4

(a) a hyperbola whose eccentricity is 2 , when 0 < r < 1. 1− r

(c) an ellipse whose eccentricity is 2 , when r > 1. r+1

2

13 Let A =  b b2 + 1 b , where b > 0.

19 On which of the following lines lies the point of intersection of the line, x−4 y−5 z−3 and the plane, = = 2 2 1 x + y + z = 2? x−4 y−5 z−5 = = 1 1 −1 x+ 3 4− y z + 1 (b) = = 3 3 −2 x−2 y−3 z+ 3 (c) = = 2 2 3 x−1 y− 3 z+ 4 (d) = = 1 2 −5 (a)

24 40 20 Let f : (−1, 1) → R be a function defined by f (x) = max {− x , − 1 − x2 }. If K be the set of all points at which f is not differentiable, then K has exactly (a) three elements (b) five elements (c) two elements (d) one element

21 The curve amongst the family of curves represented by the differential equation, (x2 − y2 )dx + 2xydy = 0, which passes through (1, 1), is (a) a circle with centre on the Y-axis (b) a circle with centre on the X-axis (c) an ellipse with major axis along the Y-axis (d) a hyperbola with transverse axis along the X-axis. x

22 If ∫ f (t ) dt = x + 2

0

1 f ′   is  2

24 25 6 (c) 25

1 2

∫x t f (t )dt , then 18 25 4 (d) 5

23 The number of values of θ ∈ (0, π) for which the system of linear equations x + 3y + 7z = 0, −x + 4y + 7z = 0, (sin 3 θ)x + (cos2 θ)y + 2z = 0 has a non-trivial solution, is (a) two (c) four

segment joining the points (−3, − 3, 4) and (3, 7, 6) at right angles, passes through which one of the following points ? (a) (4, − 1, 7) (c) (−2, 3, 5)

(b) three (d) one

(b) (2, 1, 3) (d) (4, 1, − 2)

observations x1 , x2 , x3 , x4 , x5 are 10 and 3, respectively, then the variance of 6 observations x1 , x2 , .... x5 and − 50 is equal to (a) 507.5 (c) 582.5

(b) 586.5 (d) 509.5

ai > 0 for i = 1, 2, ......,10 and S be the set of pairs (r, k), r, k ∈ N (the set of natural numbers) for which log e a2r a3k log e a5r a6k log e a8r a9k

log e a3r log e a6r log e a9r

a4k a7k k a10

=0

Then, the number of elements in S, is (a) 4 (b) 2 (c) 10 (d) infinitely many

27 Consider the following three

2. (c) 12. (d) 22. (a)

parabola x2 = 4y having equation x − 2y + 4 2 = 0 is (a) 8 (b) 2 (c) 3 (d) 6

2 11 2 3

inscribed in the circle, x2 + y2 + 10x + 12y + c = 0 is 27 3 sq units, then c is equal to (a) 20 (c) 13

(b) −25 (d) 25

30 The value of λ such that sum of the squares of the roots of the quadratic equation, x2 + (3 − λ)x + 2 = λ has the least value is 4 9 15 (c) 8 (a)

(b) 1 (d) 2

statements: P : 5 is a prime number. Q : 7 is a factor of 192. R : LCM of 5 and 7 is 35.

For Detailed Solutions

ANSWERS 1. (d) 11. (c) 21. (b)

(b) P ∨ (~ Q ∧ R) (c) (~ P ) ∨ (Q ∧ R) (d) (~ P ) ∧ (~ Q ∧ R)

29 If the area of an equilateral triangle

26 Let a1 , a2 , a3 ....., a10 be in GP with

a2k a5k a8k

Then, the truth value of which one of the following statements is true ? (a) (P ∧ Q ) ∨ (~ R)

28 The length of the chord of the

25 If mean and standard deviation of 5

log e a1r log e a4r log e a7r

(b)

(a)

24 The plane which bisects the line

3. (a) 13. (c) 23. (a)

4. (d) 14. (b) 24. (d)

5. (c) 15. (c) 25. (a)

6. (d) 16. (a) 26. (d)

7. (c) 17. (b) 27. (b)

8. (a) 18. (a) 28. (d)

9. (c) 19. (d) 29. (d)

10. (b) 20. (a) 30. (d)

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25 11 January, Shift-I

8 The outcome of each of 30 items was

1 Let [x] denote the greatest integer less than or equal to x. Then, lim

tan( π sin 2 x) + (|x| − sin(x[x]))2 x2

x→ 0

(a) equals π (c) equals 0

(b) equals π + 1 (d) does not exist

−1, −2 ≤ x < 0 and 2 − 1 0≤ x ≤ 2 x , 

2 Let f (x) = 

(a)

(a)

2x + 2 y + 3z = a 3x − y + 5z = b x − 3 y + 2z = c where a , b, c are non-zero real numbers, has more than one solution, then (b) a + b + c = 0 (d) b + c − a = 0

4 The area (in sq units) of the region bounded by the curve x = 4 y and the straight line x = 4 y − 2 is 2

7 8 5 (c) 4

9 8 3 (d) 4

5 If ∫

(b)

1

−1 3x 3

(c)

−1 27x 9

(d)

1 27x 6

f (x) = 3x3 − 18x2 + 27x − 40 on the set S = {x∈R : x2 + 30 ≤ 11x} is (b) − 122 (d) 222

7 Let f : R → R be defined by f (x ) =

x 1+ x

(a) 2 5

1 2

(b)

5 4

(b) 0

7 10

(c)

(c) 4 5

(c) 6

is (b) (−1, 1) − {0}

a9 equals a5

(a) 53

(d)

3 5

(d)

5 2

(d) 8

a3 = 25, a1

(b) 2(52 ) (c) 4(52 ) (d) 54 3

x + iy

13 Let  −2 − i  = (i = −1),  3  27 1

where x and y are real numbers, then y − x equals (a) 91 (c) – 85

(b) 85 (d) – 91

14 If x log e (log e x) − x2 + y2 = 4( y > 0), then (a)

dy at x = e is equal to dx e 4 + e2

(1+ 2e ) 4+e

2

Then, the distance of the vertex of this square which is nearest to the origin is (a) 6

(b) 13

(c)

41

(d)

137

16 If q is false and p ∧ q ←→ r is true, then which one of the following statements is a tautology? (a) p ∨ r (b) ( p ∧ r ) → ( p ∨ r ) (c) ( p ∨ r ) → ( p ∧ r ) (d) p ∧ r

17 If y(x) is the solution of the

which the middle term in the binomial 8  x3 3 expansion of  +  equals 5670 is x  3

(c)

(d) R − [−1, 1]

(d) 2

11 The sum of the real values of x for

, x∈R. Then, the range of f 2

1 1 (a)  − ,   2 2  1 1 (c) R −  − ,   2 2 

2

12 Let a1 , a2 , ...., a10 be a GP. If

6 The maximum value of the function

(a) 122 (c) − 222

(c)

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 circle at the origin is

2 m

(b)

9x 4

(b)

(a) 4

dx = A (x)( 1 − x ) + C, x4 for a suitable chosen integer m and a function A (x), where C is a constant of integration, then (A (x))m equals (a)

2 5

then

1 − x2

5 2

10 The straight line x + 2 y = 1meets the

3 If the system of linear equations

(a)

(b)

from the set { 1, 2, …… , 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is

not differentiable at one point not differentiable at two points differentiable at all points not continuous

(a) b − c − a = 0 (c) b − c + a = 0

2 3

9 Two integers are selected at random

g (x) = |f (x)| + f (|x|). Then, in the interval (−2, 2), g is (a) (b) (c) (d)

observed ; 10 items gave an outcome 1 1 − d each, 10 items gave outcome 2 2 each and the remaining 10 items gave 1 outcome + d each. If the variance of 2 this outcome data is 4, then|d|equals 3

(b) (d)

(2e − 1) 2 4 + e2 (1+ 2e ) 2 4 + e2

15 A square is inscribed in the circle x2 + y2 − 6x + 8 y − 103 = 0 with its sides parallel to the coordinate axes.

differential equation dy  2x + 1 −2x +   y = e , x > 0, dx  x  1 where y (1) = e−2, then 2

1 (a) y (x ) is decreasing in  , 1 2  (b) y (x ) is decreasing in (0, 1) (c) y (log e 2) = log e 4 (d) y (log e 2) =

log e 2 4

18 The value of r for which 20

Cr 20 C0 +

20

Cr −1 20 C1 +

20

Cr − 220 C2 + .... + 20 C0 20Cr

is maximum, is (a) 15 (c) 11

(b) 10 (d) 20

19 Two circles with equal radii are intersecting at the points (0, 1) and (0, −1). The tangent at the point (0,1) to one of the circles passes through the centre of the other circle. Then, the distance between the centres of these circles is (a)

2

(b) 2 2

(c) 1

(d) 2

20 Equation of a common tangent to the parabola y2 = 4x and the hyperbola xy = 2 is (a) x + 2 y + 4 = 0 (b) x − 2 y + 4 = 0 (c) 4x + 2 y + 1 = 0 (d) x + y + 1 = 0

sin 2 x dx −2  x  1 +  π  2

21 The value of the integral ∫

2

(where, [x] denotes the greatest integer less than or equal to x) is (a) 4 − sin 4 (c) sin 4

(b) 4 (d) 0

22 If one real root of the quadratic equation 81x2 + kx + 256 = 0 is cube of the other root, then a value of k is (a) 100 (c) −81

(b) 144 (d) −300

26 40 23 The direction ratios of normal to the

plane through the points (0, − 1, 0) and π (0, 0, 1) and making an angle with 4 the plane y − z + 5 = 0 are (a) 2, − 1, 1 (c) 2, 2 , − 2

(b) 2 , 1, − 1 (d) 2 3 , 1, − 1

two sides is x and the product of the lengths of the same two sides is y. If x2 − c2 = y, where c is the length of the third side of the triangle, then the circumradius of the triangle is (b) (d)

(b) (2, 2, 0) (d) (0, − 2, 2)

0 

2q r   q − r  . If AAT = I3 , then  p −q r   

26 Let A =  p

c 3 y 3

25 The plane containing the line

x −3 y + 2 z − 1 and also = = 2 −1 3 containing its projection on the plane 2x + 3 y − z = 5, contains which one of the following points?

(a) (c)

1 5 1 3

1 2 1 (d) 6 (b)

x2 + 2 y2 = 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 x2 y2 + =1 4 2

(b)

(c)

x2 y2 + =1 2 4

(d)

1 4x 2 1 2x 2

1 12 −1 (c) 12

5 12 1 (d) 4

(a)

(b)

29 The sum of an infinite geometric

27 If tangents are drawn to the ellipse

(a)

1 (sin k x + cosk x) for k k = 1, 2, 3 ... . Then, for all x ∈ R, the value of f4 (x) − f6 (x) is equal to

28 Let fk (x) =

|p|is

24 In a triangle, the sum of lengths of

c (a) 3 3 (c) y 2

(a) (−2, 2, 2) (c) (2, 0, − 2)

+ +

1 2 y2 1 4 y2

series with positive terms is 3 and the 27 sum of the cubes of its terms is . 19 Then, the common ratio of this series is (a)

4 9

(b)

2 3

(c)

2 9

2 (a) 12 (d) 22 (d)

3 (a) 13 (a) 23 (b,c)

4 (b) 14 (b) 24 (b)

5 (c) 15 (c) 25 (c)

6 (a) 16 (b) 26 (b)

7 (a) 17 (a) 27 (d)

8 (c) 18 (d) 28 (a)

9 (a) 19 (d) 29 (b)

1 3

30 Let a = $i + 2$j + 4k$ , b = i$ + λ$j + 4k$ $ be coplanar and c = 2i$ + 4$j + (λ 2 − 1) k

=1

vectors. Then, the non-zero vector a × c is

=1

(a) − 10 $i + 5$j (c) − 14 $i − 5$j

(b) − 10 $i − 5$j (d) − 14 $i + 5$j

ANSWERS 1 (d) 11 (b) 21 (d)

(d)

10 (d) 20 (a) 30 (a)

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27 11 January, Shift-II 1 Let f (x) =

x a 2 + x2

−

d−x b2 + (d − x)2

(a)

,

x ∈ R, where a, b and d are non-zero real constants. Then, (a) (b) (c) (d)

f is an increasing function of x f ′ is not a continuous function of x f is a decreasing function of x f is neither increasing nor decreasing function of x

2 Let K be the set of all real values of x, where the function f (x) = sin| x| − | x| + 2(x − π ) cos| x|is not differentiable. Then, the set K is equal to (b) φ (an empty set) (d) {0, π }

(a) {0} (c) { π }

3 Let z be a complex number such that | z | + z = 3 + i (where i =

− 1).

34 5 (b) 3 3

41 4

(c)

(d)

5 4

4 Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression xm yn is 2m (1 + x ) (1 + y2n ) 1 2 1 (c) 4 (a)

(b) 1 (d)

m+ n 6mn

5 Let a function f : (0, ∞ ) → (0, ∞ ) be defined by f (x) = 1 − (a) (b) (c) (d)

1 . Then, f is x

injective only both injective as well as surjective not injective but it is surjective neither injective nor surjective

6 Let (x + 10)50 + (x − 10)50 = a0 + a1 x + a2x 2 + K + a50 x50 , for all a x ∈ R; then 2 is equal to a0 (a) 12.25 (c) 12.00

(b) 12.50 (d) 12.75

7 Let A and B be two invertible matrices of order 3 × 3. If det(ABAT ) = 8 and det(AB − 1 ) = 8, then det(BA − 1 BT ) is equal to (a) 1

1 (b) 4

1 (c) 16

(d) 16

8 The integral π /4

dx

∫π / 6 sin 2x(tan5 x + cot5 x)

equals

16 Let the length of the latus rectum 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?

1  1  (b) tan − 1    9 3 20 (c)

1 π −1 1    − tan   9 3 10  4

(d)

π 40

(a) (4 2 , 2 3 ) (c) (4 2 , 2 2 )

9 The area (in sq units) in the first quadrant bounded by the parabola, y = x2 + 1, the tangent to it at the point (2, 5) and the coordinate axes is 14 3 8 (c) 3

187 24 37 (d) 24

(b)

(a)

then its (49th term) : (29th term) is (b) 4 : 1 (d) 3 : 1

11 If the point (2, α , β ) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x − 5 y = 15, then 2α − 3β is equal to (a) 17 (c) 5

(b) 7 (d) 12

the X-axis and passes through a point on the Y-axis, distant 2b from the origin. Then, the locus of the centre of this circle, is (b) an ellipse (d) a hyperbola

quadratic equation x2 sin θ − x(sin θ cosθ + 1) + cosθ = 0 (0 < θ < 45º ) and α < β. Then, ∞  (− 1)n  n ∑  α + β n  is equal to n=0 1 1 − 1 − cosθ 1 + sin θ 1 1 (b) + 1 − cosθ 1 + sin θ 1 1 (c) − 1 + cosθ 1 − sin θ 1 1 (d) + 1 + cosθ 1 − sin θ

n

2

a− b− c 2a 2a 18 If 2b b− c− a 2b 2c 2c c−a−b

(a) − (a + b + c) (c) 2(a + b + c)

n

dy = (x − y)2, when y(1) = 1, is dx 2− y = 2( y − 1) 2− x

q + 1  q + 1  q + 1 , Tn = 1 +  +  + K+    2   2   2 

(b) − log e

where q is a real number and q ≠ 1. If 101 C1 + 101C2 ⋅ S1 + K + 101C101 ⋅ S100 = αT100 , then α is equal to

(c) log e

(a) 2100

(b) 202

(c) 200

(d) 299

14 All x satisfying the inequality

(cot − 1 x)2 − 7(cot − 1 x) + 10 > 0, lie in the interval (a) (b) (c) (d)

(− ∞ , cot 5) ∪ (cot 2, ∞ ) (cot 5, cot 4) (cot 2, ∞ ) (− ∞ , cot 5) ∪ (cot 4, cot 2)

15 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 multiple of 4, is (a) (15)! × 6! (c) 5! × 6!

(b) 56 × 15 (d) 65 × (15)!

(b) − 2(a + b + c) (d) abc

19 The solution of the differential equation,

(a) log e

13 Let Sn = 1 + q + q + K + q and 2

17 Let α and β be the roots of the

= (a + b + c) (x + a + b + c)2, x ≠ 0 and a + b + c ≠ 0, then x is equal to

12 A circle cuts a chord of length 4a on

(a) a parabola (c) a straight line

(b) (4 3 , 2 2 ) (d) (4 3 , 2 3 )

(a)

10 If 19th term of a non-zero AP is zero, (a) 1 : 3 (c) 2 : 1

Then,| z |is equal to (a)

1 π −1 1    − tan   3 3 5 4

1+ x − y =x+ y−2 1− x + y

2− x =x− y 2− y

(d) − log e

1− x + y = 2(x − 1) 1+ x − y

20 Contrapositive of the statement “If two numbers are not equal, then their squares are not equal” is (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.

21 lim

x→ 0

x cot(4x) sin 2 x cot 2 (2x)

(a) 0

(b) 1

is equal to (c) 4

(d) 2

28 40 22 A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then   mean of X   is equal to  standard deviation of X  4 3 3 (b) 4 (c) 3 2 (a)

x+1 dx = f (x) 2x − 1 + C, 2x − 1

where C is a constant of integration, then f (x) is equal to (a) (b) (c) (d)

(d) 4 3

2 (x + 2) 3 1 (x + 4) 3 2 (x − 4) 3 1 (x + 1) 3

vertex is at the vertex of the parabola, y2 + 4(x − a 2 ) = 0 and the other two vertices are the points of intersection of the parabola andY -axis, is 250 sq units, then a value of ‘a’ is (a) 5 5 (b) 5 (c) 5(21/ 3 ) (d) (10) 2 / 3

x−3 y+1 z−6 and = = 1 3 −1 x+ 5 y−2 z−3 intersect at the = = 7 −6 4 point R. The reflection of R in the xy-plane has coordinates

24 Two lines

(2, − 4, − 7) (2, − 4, 7) (− 2, 4, 7) (2, 4, 7)

13 12 (b) 2 13 (c) 8 13 (d) 6 (a)

27 Let S = {1, 2, K , 20}. A subset B of S is said to be “nice”, if the sum of the elements of B is 203. Then, the probability that a randomly chosen subset of S is ‘‘nice’’, is

(b) (c)

5 220

28 If in a parallelogram ABDC, the coordinates of A , B and C are respectively (1, 2), (3, 4) and (2, 5), then the equation of the diagonal AD is

29 Let 3 i$ + $j, i$ + 3$j and βi$ + (1 − β )$j

conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is

(a)

(d)

(a) 3x + 5 y − 13 = 0 (b) 3x − 5 y + 7 = 0 (c) 5x − 3 y + 1 = 0 (d) 5x + 3 y − 11 = 0

26 If a hyperbola has length of its

23 If the area of the triangle whose one

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

25 If ∫

6 220 4

respectively be the position vectors of the points A , B and C with respect to the origin O. If the distance of C from the bisector of the acute angle 3 between OA and OB is , then the 2 sum of all possible values of β is (a) (b) (c) (d)

b+ c c+ a a+ b for a = = 11 12 13 ∆ABC with usual notation. If cos A cos B cosC , then the ordered = = α β γ triad (α , β , γ ) has a value

30 Given,

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

20

2 7

220

1 3 4 2

(19, 7, 25) (3, 4, 5) (5, 12, 13) (7, 19, 25)

ANSWERS 1. (a) 11. (b) 21. (b)

2. (b) 12. (a) 22. (d)

3. (b) 13. (a) 23. (b)

4. (c) 14. (c) 24. (a)

5. (d) 15. (a) 25. (b)

6. (a) 16. (b) 26. (a)

7. (c) 17. (b) 27. (d)

8. (c) 18. (b) 28. (c)

9. (d) 19. (d) 29. (a)

10. (d) 20. (b) 30. (d)

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29 12 January, Shift-I 1 Let P (4, − 4) and Q(9, 6) be two points on the parabola, y2 = 4x and let X be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of ∆PXQ is maximum. Then, this maximum area (in sq units) is 125 (a) 2 625 (c) 4

75 (b) 2 125 (d) 4

2 Consider three boxes, each containing 10 balls labelled 1, 2, …, 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni , the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is (a) 82 (c) 240

(b) 120 (d) 164

3 Let y = y(x) be the solution of the differential equation, dy x + y = x log e x, (x > 1). If dx 2 y(2) = log e 4 − 1, then y(e) is equal to e2 (b) − 2 e2 (d) 4

e (a) − 2 e (c) 4

4 The sum of the distinct real values of $ µ, for which the vectors, µ$i + $j + k, $i + µ$j + k $ , $i + $j + µk $ are coplanar, is (a) 2 (c) 1

(b) 0 (d) − 1

5 Let C1 and C2 be the centres of the circles x + y − 2x − 2 y − 2 = 0 and x2 + y2 − 6x − 6 y + 14 = 0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq units) of the quadrilateral PC1QC2 is 2

(a) 8 (c) 6

2

(b) 4 (d) 9

6 If the straight line, 2x − 3 y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, β), then β equals 35 3 35 (c) − 3 (a)

7 A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of 10  1    1 is  23 + 1     2(3)3  (a) 1 : 2(6)1/ 3

(b) 1 : 4(16)1/ 3

(c) 4(36)1/ 3 : 1

(d) 2(36) 3 : 1

1

8 Let S = {1, 2, 3, ... , 100}. The number of non-empty subsets A of S such that the product of elements in A is even, is (a) 250 (250 − 1) (c) 250 + 1

(b) 250 − 1 (d) 2100 − 1

9 The maximum area (in sq. units) of a rectangle having its base on the X-axis and its other two vertices on the parabola, y = 12 − x2 such that the rectangle lies inside the parabola, is (a) 36 (c) 32

(b) 20 2 (d) 18 3

10 The integral ∫ cos (log e x) dx is equal to (where C is a constant of integration) (a)

x [cos(log e x) + sin(log e x)] + C 2

(b) x [cos(log e x ) + sin(log e x )] + C (c) x [cos(log e x ) − sin(log e x )] + C (d)

x [sin(log e x ) − cos(log e x )] + C 2

11 lim

π x→ 4

cot3 x − tan x is π cos  x +   4

(a) 4 2 (c) 8

(b) 4 (d) 8 2

12 An ordered pair (α , β ) for which the system of linear equations (1 + α )x + βy + z = 2 αx + (1 + β ) y + z = 3 ax + βy + 2z = 2 has a unique solution, is (a) (2, 4) (b) (− 4, 2) (c) (1, − 3) (d) (−3, 1)

14 For x > 1, if (2x)2y = 4e2x − 2y , then (1 + log e 2x)2

dy is equal to dx

x log e 2x + log e 2 x x log e 2x − log e 2 (b) x (c) x log e 2x (a)

(d) log e 2x

15 The maximum value of

π 3 cosθ + 5 sin  θ −   6 for any real value of θ is 79 2

(b) 34

(c) 31

(d) 19

(a)

16 The area (in sq units) of the region bounded by the parabola, y = x2 + 2 and lines, y = x + 1, x = 0 and x = 3, is 15 2 21 (c) 2 (a)

17 4 15 (d) 4 (b)

17 The Boolean expression (( p ∧ q) ∨ ( p ∨ ~ q)) ∧ (~ p ∧ ~ q) is equivalent to (a) p ∧ q (c) p ∧ (~ q)

(b) p ∨ (~ q) (d) (~ p ) ∧ (~ q)

18 If a variable line, 3x + 4 y − λ = 0 is such that the two circles x2 + y2 − 2x − 2 y + 1 = 0 and x2 + y2 − 18x − 2 y + 78 = 0 are on its opposite sides, then the set of all values of λ is the interval (a) [13, 23] (c) [12, 21]

(b) (2, 17) (d) (23, 31)

19 The perpendicular distance from the origin to the plane containing the two lines, x+ 2 y−2 z+ 5 = = 3 5 7 x −1 y − 4 z + 4 and , is = = 1 4 7 (a) 11 6 (c) 11

11 6 (d) 6 11

(b)

20 If λ be the ratio of the roots of the

13 If the sum of the deviations of 50

(b) − 5

observations from 30 is 50, then the mean of these observations is

(d) 5

(a) 50 (c) 51

(b) 30 (d) 31

quadratic equation in x, 3m2x2 + m(m − 4)x + 2 = 0, then the least value of m for which 1 λ + = 1, is λ (a) − 2 + 2 (c) 4 − 3 2

(b) 4 − 2 3 (d) 2 − 3

30 40 21 Considering only the principal values of inverse functions, the set π A = x ≥ 0 : tan −1 (2x) + tan −1 (3x) =  4 

succession. The probability that the experiment will end in the fifth throw of the die is equal to

(a) is an empty set (b) is a singleton (c) contains more than two elements (d) contains two elements

(a)

22 If the vertices of a hyperbola be at

(c)

(−2, 0) and (2, 0) and one of its foci be at (−3, 0), then which one of the following points does not lie on this hyperbola? (a) (2 6 , 5) (b) (6, 5 2 ) (c) (4, 15 ) (d) (− 6, 2 10 )

z −α (α ∈ R ) is a purely imaginary z+α number and|z| = 2, then a value of α is

23 If

(a) 2 1 (b) 2 (c) 1 (d) 2

 1 0 0

24 Let P =  3 1 0 and Q = [qij ] be two

(b)

(d)

175 225 65 200

(a) 36 (b) 28 (c) 32 (d) 24

65 150 65

26 Let S be the set of all points in (− π , π ) at which the function, f (x) = min {sin x, cos x} is not differentiable. Then, S is a subset of which of the following? π π , 0,  4 4 π π π π , − , ,  2 4 4 2 3π π 3π π  ,− , ,  4 4 4 4 3π π π 3π  ,− , ,  4 2 2 4 

27 A tetrahedron has vertices P (1, 2, 1),

9 (b) cos−1    35 

7 (a) cos−1    31

2. (b) 12. (a) 22. (b)

(b) ∫

a

0 a

0

(c) 2∫

f (x ) dx

f (x ) dx

a 0

(d) − 3∫

f (x ) dx a 0

f (x ) dx

19 (c) cos−1    35 

For Detailed Solutions

ANSWERS 1. (d) 11. (c) 21. (b)

(b) 301 (d) 303

0

(a) 4∫

(a) 10 (c) 9

rolled until two fours are obtained in

(a) 156 (c) 283

on [0, a] such that f (x) = f (a − x) and g (x) + g (a − x) = 4, then a ∫ f (x) g (x) dx is equal to

   9 3 1 3 × 3 matrices such that Q − P 5 = I3 . q + q31 Then, 21 is equal to q32

25 In a random experiment, a fair die is

1 + 2 + 3 + ... + k . If k 5 2 S12 + S22 + ... + S10 = A, then A is 12 equal to

29 Let Sk =

30 Let f and g be continuous functions

Q (2, 1, 3), R (− 1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is

(b) 135 (d) 15

28 The product of three consecutive terms of a GP is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an AP. Then, the sum of the original three terms of the given GP is

65

(a)  −  (b)  −   (c)  −  (d)  − 

17 (d) cos−1    31

3. (c) 13. (d) 23. (d)

4. (d) 14. (b) 24. (a)

5. (b) 15. (d) 25. (a)

6. (d) 16. (a) 26. (c)

7. (c) 17. (d) 27. (c)

8. (a) 18. (c) 28. (b)

9. (c) 19. (b) 29. (d)

10. (a) 20. (c) 30. (c)

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31 12 January, Shift-II π −

1 lim

x→1−

2 sin − 1 x is equal to 1− x

(a)

π 2

(b)

(c)

π

(d)

2 π 1 2π

such that a plane passing through the points (− λ2 , 1, 1), (1, − λ2 ,1) and (1, 1, − λ2 ) also passes through the point (− 1, − 1, 1). Then, S is equal to 3}

(b) {3, − 3} (d) { 3 }

3 If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is

9 If a curve passes through the point (1, − 2) and has slope of the tangent at x2 − 2 y any point (x, y) on it as , then x the curve also passes through the point



10 lim  n→∞ 

n n 2 + 12

(c) π / 4

n n 2 + 22

+

n n 2 + 32

+... +

1   5n 

(b) tan − 1 (2) (d) π /2

3 5 5 (d) − 3 (b)

5 or 6 on a throw of a fair die and loses ` 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is (a)

400 loss 3

(c) 0

5 The expression ~ (~ p → q) is logically

400 loss 9 400 (d) gain 3

(b)

12 If sin 4 α + 4 cos4 β + 2 = 4 2 sin α cosβ;

equivalent to (b) p ∧ q (d) ~ p ∧ ~ q

6 The set of all values of λ for which the system of linear equations x − 2 y − 2z = λx, x + 2 y + z = λy and − x − y = λz has a non-trivial solution (a) (b) (c) (d)

+

11 In a game, a man wins ` 100 if he gets

x+1 y−2 z−3 and the plane, = = 2 1 −2  2 2 x − 2 y − kz = 3 is cos− 1   , then a  3  value of k is

(a) p ∧ ~ q (c) ~ p ∧ q

(b) (− 1, 2) (d) (3, 0)

(a) ( 3 , 0) (c) (− 2 , 1)

contains exactly two elements. contains more than two elements. is a singleton. is an empty set.

7 Let Z be the set of integers. If ( x + 2) ( x 2 − 5 x + 6 )

A = {x ∈ Z : 2

= 1} and

B = {x ∈ Z : − 3 < 2x − 1 < 9}, then the number of subsets of the set A × B, is (a) 212

(b) 218

(c) 215

(d) 210

α, β ∈[0, π ], then cos(α + β ) − cos(α − β ) is equal to (a) − 1 (c) − 2

(b) 2 (d) 0

13 If the function f given by f (x) = x3 − 3(a − 2) x2 + 3ax + 7, for some a ∈ R is increasing in (0, 1] and decreasing in [1, 5), then a root of f (x) − 14 the equation, = 0 (x ≠ 1) is (x − 1)2 (a) − 7 (c) 7

(b) 6 (d) 5

14 If nC4 , nC5 and nC6 are in AP, then n

that f (1) = 2 and f ′ (x) = f (x) for all x ∈ R. If h (x) = f (f (x)), then h′ (1) is equal to (a) 4e2 (c) 2e

(b) 4e (d) 2e2

observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4, then the absolute value of the difference of the other two observations, is (a) 1 (c) 5

(b) 7 (d) 3

17 The number of integral values of m for which the quadratic expression, (1 + 2m) x2 − 2(1 + 3m)x + 4(1 + m), x ∈ R, is always positive, is (a) 6 (c) 7

(b) 8 (d) 3

18 The total number of irrational terms in the binomial expansion of (71/5 − 31/10 )60 is (a) 49 (c) 54

(b) 48 (d) 55

sin θ 1   1  19 If A =  − sin θ 1 sin θ; then for  − sin θ  − 1 1  3π 5π  all θ ∈  ,  , det(A ) lies in the  4 4 interval 3 (a)  , 3  2 

5 (b)  , 4  2 

3 (c)  0,   2 

5 (d)  1,   2 

20 Let a, b and c be three unit vectors, out of which vectors b and c are non-parallel. If α and β are the angles which vector a makes with vectors b and c respectively 1 and a × ( b × c) = b, then|α − β | 2 is equal to (a) 30º (c) 90º

(b) 45º (d) 60º

21 If the sum of the first 15 terms of the series 3

3

3

3

 3  +  1 1  +  2 1  + 33 +  3 3  + ...          4  2  4  4

is equal to 225 k, then k is equal to

can be (a) 9 (c) 14

15 Let f be a differentiable function such

16 The mean and the variance of five

1 3 5 (d) 6 (b)

(a) tan − 1 (3)

4 If an angle between the line,

5 3 3 (c) − 5

1 6 2 (c) 3

is equal to

(a) (x 2 + y 2 ) 2 = 4R 2x 2 y 2 (b) (x 2 + y 2 ) 3 = 4R 2x 2 y 2 (c) (x 2 + y 2 )(x + y ) = R 2xy (d) (x 2 + y 2 ) 2 = 4Rx 2 y 2

(a)

NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is (a)

2 Let S be the set of all real values of λ

(a) { 3 , − (c) {1, − 1}

8 In a class of 60 students, 40 opted for

(b) 11 (d) 12

(a) 108 (c) 54

(b) 27 (d) 9

32 40 e x 22 The integral ∫   1  e 

2x

 log e x dx is equal to

3 −e− 2 1 (c) −e− 2

1

(a)

25 Let z1 and z2 be two complex numbers

1 1 1 + − 2 e 2e2 3 1 1 (d) − − 2 e 2e2 (b) −

2e2 1 e2

23 The integral ∫

x e  −     x  

3x13 + 2x11

dx is (2x4 + 3x2 + 1)4 equal to (where C is a constant of integration) (a)

x4 6(2x + 3x 2 + 1) 3 4

12

(b) (c) (d)

x

6(2x 4 + 3x 2 + 1) 3 x4 (2x 4 + 3x 2 + 1) 3 x12 (2x 4 + 3x 2 + 1) 3

+C

(a) 1

(b) 2

(c)

2

(d) 0

26 The equation of a tangent to the parabola, x2 = 8y, which makes an angle θ with the positive direction of X-axis, is (a) (b) (c) (d)

y = x tan θ − 2 cotθ x = y cotθ + 2 tan θ y = x tan θ + 2 cotθ x = y cotθ − 2 tan θ

27 If the angle of elevation of a cloud

+C

from a point P which is 25 m above a lake be 30º and the angle of depression of reflection of the cloud in the lake from P be 60º, then the height of the cloud (in meters) from the surface of the lake is

+C +C

24 The tangent to the curve y = x2 − 5x + 5, parallel to the line 2y = 4x + 1, also passes through the point 1 7 (a)  ,   4 2 1 (c)  − , 7  8 

satisfying|z1 | = 9 and |z2 − 3 − 4i | = 4. Then, the minimum value of|z1 − z2 |is

7 1 (b)  ,   2 4 1 (d)  , − 7 8 

(a) 50 (c) 45

(b) 60 (d) 42

28 Let S and S′ be the foci of an ellipse and B be any one of the extremities of its minor axis. If ∆S′ BS is a right angled triangle with right angle at B and area (∆S′ BS) = 8 sq units, then

the length of a latus rectum of the ellipse is (a) 2 2

(b) 4 2

(c) 2

(d) 4

29 There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is (a) (b) (c) (d)

12 11 9 7

30 If a straight line passing through the point P (− 3, 4) is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is (a) x − y + 7 = 0 (b) 4x − 3y + 24 = 0 (c) 3x − 4y + 25 = 0 (d) 4x + 3y = 0

ANSWERS 1. (b) 11. (c) 21. (b)

2. (a) 12. (c) 22. (a)

3. (b) 13. (c) 23. (b)

4. (a) 14. (c) 24. (d)

5. (d) 15. (b) 25. (d)

6. (c) 16. (b) 26. (b)

7. (c) 17. (c) 27. (a)

8. (a) 18. (c) 28. (d)

9. (a) 19. (a) 29. (a)

10. (b) 20. (a) 30. (b)

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