Complete mathematics for JEE Main 2019 9387572560, 9789387572560

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MAIN JEEJEEMAIN CO CM OP MLPELTEET E

MATHEMATICS MATHEMATICS

MAIN JEEJEEMAIN CO CM OP MLPELTEET E

MATHEMATICS MATHEMATICS Ravi Prakash Retired Associate Professor, Rajdhani College University of Delhi, Delhi

Ajay Kumar Professor of Mathematics University of Delhi, Delhi

Usha Gupta Retired Associate Professor, Rajdhani College University of Delhi, Delhi

McGraw Hill Education (India) Private Limited CHENNAI

McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San  Juan Santiago  Singapore Sydney Tokyo Toronto

Published by McGraw Hill Education (India) Private Limited, 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai-600116 Complete Mathematics—JEE Main Copyright © 2018, McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. Price: `895/1  2 3 4 5 6 7 8 9   7085462   22 21 20 19 18

Printed and bound in India ISBN (13):  978-93-87572-56-0 ISBN (10):  93-87572-56-0 Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Sri Krishna Graphics, Delhi and printed at Cover Designer: Neeraj Dayal

visit us at: www.mheducation.co.in

To Our Readers… How to Crack the JEE

This new edition of our book, Complete Mathematics JEE Main, is primarily meant for the candidates due to appear in CBSE’s JEE (Main). It thoroughly covers all the topics prescribed in the Mathematics syllabus of this examination. The book is divided into 28 chapters. Each chapter is organised as follows: 1. It begins with a review of important definitions, formulae, theorems, tips, tricks and techniques to solve problems. Difficult concepts have been explained by giving appropriate illustrations in sufficient numbers. 2. This theory is followed by examples. These examples (solved in detail) are divided into three categories: (a) Concept Based examples to further strengthen and enrich your conceptual understanding (b) Level 1 examples carefully selected to groom you for the JEE Main examination. This section contains both Straight Objective Type Questions and Assertion-Reason Type Questions. (c) Level 2 examples added to further boost your confidence by initiating you into tougher questions 3. The Solved Examples are followed by Exercises. These have been divided into five parts. (i) Concept Based Questions (ii) Level 1 Questions (iii) Level 2 Questions (iv) Previous Years’ AIEEE/JEE (Main) Questions (v) Previous Years’ B. Architecture Entrance Examination Questions. The philosophy behind (i), (ii) and (iii) is the same as that in Solved Examples. We have added (iv) and (v) to give you glimpse of the questions that have been asked in the previous years (till 2016). Questions from online JEE (Main) papers from 2013 and 2016 have also been added. As the Mathematics syllabus for B. Architecture Entrance Examination Questions is the same as that of JEE (Main) examination, questions from these papers have also been added for further practice. 4. Each exercise has been provided with answers. 5. Hints and Solutions to all the questions in (i), (ii), (iii), (iv) and (v) have been given at the end of the chapters.

How to Use this Book Step 1 Read the theory given in the chapter carefully. Take meaningful notes in a notebook. Step 2 Go to solved examples. First attempt the questions given in this section (Concept-based). Try to solve each example yourself. If you fail to do it within a reasonable time limit, go through the solution carefully. The key to improvement is NOT merely understanding the explanation given in the solution, but to see exactly what clues in the question lead us to the correct answer. Step 3 Repeat Step 2 with Concept-based questions in the exercise. Step 4 Repeat Step 2 and Step 3 with Solved Examples and questions in Exercises at Level-1 and Level-2. Important: Leave Previous Years’ Questions for the future.

vi  To Our Readers…

Facing the Examination The analysis of previous fourteen years’ papers shows that it is not possible to leave any topic. You get questions from each and every topic of mathematics. Thus, it is essential that you study each and every topic thoroughly. Beginning the Preparation The best time to begin preparation for JEE (Main) is when you have studied almost fifty percent of your Class XI course, that is, almost one and a half year before the day you wish to appear in the JEE (Main). Effective Studying You need to set priorities so that you can get the best return for time invested. Write down the time you wish to devote to Physics, Chemistry and Mathematics. As you study, you will develop a list of your problem areas, which will direct you further. This may require you to rearrange your priorities. Do not overestimate the time you have, extra preparation always helps. The key is to set up a workable plan and stick to it.

Some Tips for Preparation ∑ Identify your strengths and weaknesses. It is important to identify your problem areas early on. Your study plan would need to evolve as you identify these areas. Equally important is to know your strength areas. You should ensure that these areas are further strengthened and that you are not a victim of overconfidence. Your study plan should strike a balance between the time kept for strength areas and for problem areas. Though you need to concentrate more on your problem areas, mastering your strength areas in between would ensure that you do not get dejected. ∑ Make notes and continually review each section. Take meaningful notes in a notebook which will be of a great help at a later date. Good notes are short and crisp, but cover the key concepts. Very long notes are redundant as you never have time to go through them and too short notes are useless as they do not trigger any recall. Rapidly review the section you have just studied. Recall all the major ideas of this portion. If you have any difficulties, check your notes. ∑ Study for short durations. Your brain needs time to assimilate your thoughts and develop a deep understanding of each concept that you are learning. Try to study in short durations with rest breaks in between. During these rests, first blank out your mind so as to relax and then go over the key concepts that you have learnt in the just completed leg of your study. ∑ Do each exercise keeping time in mind and look for clues. This is very important. Before proceeding to another section, rework all the questions about which you were not sure or which you could not solve correctly. And make sure that you complete them in the time allocated for each section. ∑ Always have a positive attitude. Be motivated. Don’t be discouraged if you are unable to solve a problem. When you cannot understand a problem you tend to think in different directions. This helps you to develop your mental faculties. ∑ Always keep your goal in mind. ∑ Get your body clock in rhythm with the examination timings a few days before the examination. Some students are in the habit of studying at night and sleeping during the day time. This habit should not be continued till the day of the exam. You should develop a habit of working during the same clock hours as the actual exam at least 15-20 days before the exam. This will help you to work effectively during the exam.

To Our Readers…  vii

Test Taking Strategies

∑ Read the instructions in the question paper carefully and scrupulously follow these instructions. ∑ Read the entire question before attempting to answer it, and recognise the key concepts in the same. Each question typically contains certain key concepts; certain pieces of information that, if you recognise accurately, your ability to come to the right answer increases significantly. The correct answer will typically incorporate all the key concepts contained in the question. ∑ Read each question carefully. Answer the questions asked, not the one you may have expected. ∑ Budget your time. Work swiftly. Don’t devote too much time to any difficult problem. There may be some simple ones waiting for you. Concentrate on those questions which you are absolutely or reasonably certain you know. Leave those questions for later where you have some doubts. Go through the exam answering questions that you can answer and skip questions you can’t answer. Then go through the paper again, deliberating and answering the questions you had left unattended earlier. Best wishes for your endeavours! Ravi Prakash Ajay Kumar Usha Gupta

Please log on to www.mhhe.com/jeemaincompletemathematics for Seven Sectional Test Papers, Five Practice Test Papers and Miscellaneous Questions from the entire book.

26. 27. 28.

4 8 4 8 4 4 8

8 4 4 4 4

4 8 4 4 4 8 4 12

4 4 8 4 4 8

4

4

4 4 120

4 4 120

4 4 4 4 4 12 4 4 8 4 4 4 120

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4 8 4 4 4 4 4 4 4

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4 4 8 4

4 12 4 4 4 4 4 8

12

4 8 4 4 4

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2016 (Architecture)

4 8 4

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Online 10/4/16

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Online 9/4/16

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4 4 4 8 4 4

JEE (Main) (Offline) 2016

17. 18. 19. 20. 21. 22. 23. 24. 25.

8

Architecture 2017

8. 9. 10. 11. 12. 13. 14. 15. 16.

Sets, Relations and Functions Complex Numbers Quadratic Equations Determinants Matrices Permutations and Combinations Mathematical Induction and Binomial Theorem Progressions Limits and Continuity Differentiability and Differentiation Applications of Derivatives Indefinite Integration Definite Integrals Areas by Integration Differential Equations Cartesian System of Rectangular Coordinates and Straight Lines Circles and Systems of Circles Parabola Ellipse Hyperbola Three Dimensional Geometry Vectors Statistics Probability Trigonometrical Ratios, Identities and Equations Inverse Trigonometric Functions Heights and Distances Mathematical Reasoning Total

Online 9/4/17

1. 2. 3. 4. 5. 6. 7.

Chapter Name

Online 8/4/17

Chap. No.

JEE (Main) (Offline) 2017

Trend Analysis

8 4 4 8 4 4 8 4 4 4 8 4 4 4 4 8 4 4 12 4 4

4 4

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About JEE Main 1.  Introduction and Scheme of Examination The Joint Entrance Examination from the year 2013 for admission to the undergraduate programmes in Engineering is being held in two parts, JEE-Main and JEE-Advanced. Only the top 2,00,000 candidates (including all categories) based on performance in JEE Main will qualify to appear in the JEE Advanced examination. Admissions to IITs will be based only on category-wise All India Rank (AIR) in JEE Advanced, subject to condition that such candidates are in the top 20 percentile categories. Admission to NITs will be based on 40% weightage for performance in Class XII board marks (normalized) and the remainder 60% weightage would be given to performance in JEE Main and a combined All India Rank (AIR) would be decided accordingly. In case any State opts to admit students in the engineering Colleges affiliated to state Universities where States require separate merit list to be provided based on relative weightages adopted by the states, then the merit list shall be prepared with such relative weightages as may be indicated by States.

2.  Eligibility Criteria and List of Qualifying Examinations for JEE(Main) Exam The minimum academic qualification for appearing in JEE(Main) is that the candidate must have passed in final examination of 10+2 (Class XII) or its equivalent referred to as the qualifying examination (see below). However, admission criteria in the concerned institution/university will be followed as prescribed by concerned university/institution and as per the guidelines & criteria prescribed by AICTE.

Qualifying Examinations List of Qualifying Examinations (i) The +2 level examination in the 10+2 pattern of examination of any recognized Central/State Board of Secondary Examination, such as Central Board of Secondary Education, New Delhi, and Council for Indian School Certificate Examination, New Delhi (ii) Intermediate or two-year Pre-University Examination conducted by a recognized Board/University. (iii) Final Examination of the two-year course of the Joint Services Wing of the National Defence Academy. (iv) Any Public School/Board/University Examination in India or in foreign countries recognized by the Association of Indian Universities as equivalent to 10+2 system. (v) H.S.C. Vocational Examination. (vi) A pass grade in the Senior Secondary School Examination conducted by the National Open School with a minimum of five subjects. (vii) 3 or 4-year diploma recognized by AICTE or a State Board of Technical Education.

xii  About JEE Main

3.  Pattern of Examination Subject combination for each paper and type of questions in each paper are given below: Subjects

Type of Questions

Duration

Paper 1

Physics, Chemistry & Mathematics

Objective type questions with equal weightage to Physics, Chemistry & Mathematics

3 Hours

Paper 2

Mathematics – Part I

Objective type questions

3 Hours

Aptitude Test – Part II &

Objective type questions

Drawing Test – Part III

Questions to test Drawing Aptitude

Requirement of papers for different courses is given in the table below: Course

Papers

B.E/B.TECH

Paper – 1

B.ARCH/B. PLANNING

Paper – 2

Scoring and Negative Marking There will be objective type questions with four options having single correct answer. For each incorrect response, one fourth (1/4) of the total marks allotted to the question would be deducted. No deduction from the total score will, however, be made if no response is indicated for an item in the answer sheet.

Syllabus

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 determinants, 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 A.M. and G.M. Sum upto n terms of special series: Ân, Ân2, Ân3. 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,

xiv Syllabus

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



Ú x 2 ± a2 , Ú



Ú



Ú



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.

dx

dx 2

x ±a

dx ax 2 + bx + c

,Ú

a2 ± x 2 dx and

2

,Ú

dx 2

a -x

2

( px + q ) dx 2

ax + bx + c

Ú

,Ú

,Ú

dx 2

a -x

2

,Ú

dx 2

ax + bx + c

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

x 2 - a2 dx

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: Co-ordinate Geometry

Cartesian system of rectangular co-ordinates 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.



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.



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.

Syllabus  xv

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, or, implies, implied by, if and only if. Understanding of tautology, contradiction, converse and contrapositive.

Contents To Our Readers… Trend Analysis About JEE Main Syllabus Format of Questions in this Book 1. Sets, Relations and Functions

v ix xi xiii xxix 1.1–1.40

Finite Set and Infinite Set  1.1 Algebra of Sets  1.1 Some Properties of Operations on Sets  1.2 Some Definitions  1.3 Algebraic Operations on Functions  1.4 Domains and Ranges of Some Functions  1.4 Types of Functions  1.4 Graphs of Some Functions  1.5 Inverse of a Function  1.9 Direct and Inverse Images  1.9 Solved Examples:  Concept-based: Straight Objective Type Questions  1.10 LEVEL 1: Straight Objective Type Questions  1.12 Assertion-Reason Type Questions  1.19 LEVEL 2:  Straight Objective Type Questions  1.20 Exercises:  Concept-based: Straight Objective Type Questions  1.24 LEVEL 1: Straight Objective Type Questions  1.24 Assertion-Reason Type Questions  1.27 LEVEL 2:  Straight Objective Type Questions  1.27 Previous Years’ AIEEE/JEE Main Questions  1.28 Previous Years’ B-Architecture Entrance Examination Questions  1.30 Answers 1.31 Hints and Solutions  1.32 2. Complex Numbers Definitions  2.1 Algebraic Operations with Complex Numbers  2.1 Conjugate of Complex Number  2.1 Modulus of a Complex Number  2.2 Geometrical Representation of Complex Numbers  2.2 Argument of a Complex Number  2.2 Polar Form of a Complex Number  2.4 Vector Representation of Complex Numbers  2.4 Geometrical Representation of Algebraic Operations on Complex Numbers  2.4

2.1–2.59

xviii Contents

Some Important Geometrical Results and Equations  2.6 Recognizing Some Loci by Inspection  2.11 Geometric Interpretation of Multiplying a Complex Number by eia.  2.12 De Moivre’s Theorem and its Applications  2.13 Solved Examples:  Concept-based: Straight Objective Type Questions  2.15 LEVEL 1: Straight Objective Type Questions  2.17 Assertion-Reason Type Questions  2.27 LEVEL 2: Straight Objective Type Questions  2.30 Exercises:  Concept-based: Straight Objective Type Questions  2.34 LEVEL 1: Straight Objective Type Questions  2.35 Assertion-Reason Type Questions  2.39 LEVEL 2: Straight Objective Type Questions  2.39 Previous Years’ AIEEE/JEE Main Questions  2.41 Previous Years’ B-Architecture Entrance Examination Questions  2.43 Answers 2.44 Hints and Solutions  2.45 3. Quadratic Equations

3.1–3.58

Quadratic Equations  3.1 Nature of Roots  3.1 Relation between Roots and Coefficients  3.1 Quadratic Expression and its Graph  3.1 Position of Roots of a quadratic Equation  3.2 Conditions for a Number k to Lie Between the Roots of a Quadratic Equation  3.3 Conditions for Exactly one Root of a Quadratic Equation to Lie in the Interval (k1, k2) where k1 < k2  3.3 Conditions for Both the Roots of a Quadratic Equation to Lie in the Interval (k1, k2) where k1 < k2.  3.4 Conditions for a Quadratic Equation to Have a Repeated Root  3.5 Condition for two Quadratic Equations to Have a Common Root  3.5 Condition for Two Quadratic Equations to Have the Same Roots  3.5 Equations of Higher Degree  3.5 Cubic and Biquadratic Equations  3.5 Transformation of Equations  3.6 Descartes Rule of Signs for the Roots of a Polynomial  3.6 Some Hints for Solving Polynomial Equations  3.6 Use of Continuity and Differentiability to Find Roots of a Polynomial Equation  3.7 Solved Examples:  Concept-based: Straight Objective Type Questions  3.7 LEVEL 1: Straight Objective Type Questions  3.10 Assertion-Reason Type Questions  3.24 LEVEL 2: Straight Objective Type Questions  3.27 Exercises:  Concept-based: Straight Objective Type Questions  3.31 LEVEL 1: Straight Objective Type Questions  3.32 Assertion-Reason Type Questions  3.37 LEVEL 2: Straight Objective Type Questions  3.38 Previous Years’ AIEEE/JEE Main Questions  3.39 Previous Years’ B-Architecture Entrance Examination Questions  3.42 Answers 3.42 Hints and Solutions  3.43 4. Determinants Evaluation of Determinants  4.1

4.1–4.60

Contents  xix

Minors and Cofactors  4.1 Properties of Determinants  4.2 Some Tips for Quick Evaluation of Determinants  4.3 Linear Equations 4.4 Cramer’s Rule 4.4 Solved Examples:  Concept-based: Straight Objective Type Questions  4.4 LEVEL 1: Straight Objective Type Questions  4.8 Assertion-Reason Type Questions  4.24 LEVEL 2: Straight Objective Type Questions  4.26 Exercises:  Concept-based: Straight Objective Type Questions  4.31 LEVEL 1: Straight Objective Type Questions  4.33 Assertion-Reason Type Questions  4.38 LEVEL 2: Straight Objective Type Questions  4.39 Previous Years’ AIEEE/JEE Main Questions  4.40 Previous Years’ B-Architecture Entrance Examination Questions  4.42 Answers 4.43 Hints and Solutions  4.43 5. Matrices

5.1–5.46

The Algebra of Matrices  5.1 Transpose of a Matrix  5.2 Adjoint and Inverse of a Matrix  5.2 Some Definitions and Results  5.3 Rank of a Matrix  5.3 System of Linear Equations  5.5 Solution of a System of Equation AX = B when A is a square matrix  5.6 Solution of a System of Homogeneous Linear Equations Ax = 0  5.6 Solved Examples:  Concept-based: Straight Objective Type Questions  5.7 LEVEL 1: Straight Objective Type Questions  5.9 Assertion-Reason Type Questions  5.20 LEVEL 2: Straight Objective Type Questions  5.22 Exercises:  Concept-based: Straight Objective Type Questions  5.24 LEVEL 1: Straight Objective Type Questions  5.25 Assertion-Reason Type Questions  5.28 LEVEL 2: Straight Objective Type Questions  5.28 Previous Years’ AIEEE/JEE Main Questions  5.30 Previous Years’ B-Architecture Entrance Examination Questions  5.33 Answers 5.33 Hints and Solutions  5.34 6. Permutations and Combinations Fundamental Principles of Counting  6.1 Permutations and Combinations  6.1 Some Important Results  6.2 Circular Permutation  6.2 Some Important Identities  6.2 Division of Identical Objects  6.2 Multiplication of Two Infinite Series  6.3 Distribution into Groups  6.3 Selection  6.4

6.1–6.40

xx Contents

Exponent of Prime p in n!  6.4 Solved Examples:  Concept-based: Straight Objective Type Questions  6.4 LEVEL 1: Straight Objective Type Questions  6.6 Assertion-Reason Type Questions  6.17 LEVEL 2: Straight Objective Type Questions  6.19 Exercises:  Concept-based: Straight Objective Type Questions  6.21 LEVEL 1: Straight Objective Type Questions  6.22 Assertion-Reason Type Questions  6.25 LEVEL 2: Straight Objective Type Questions  6.26 Previous Years’ AIEEE/JEE Main Questions  6.26 Previous Years’ B-Architecture Entrance Examination Questions  6.29 Answers 6.29 Hints and Solutions  6.30 7. Mathematical Induction and Binomial Theorem

7.1–7.51

Principle of Mathematical Induction  7.1 Binomial Theorem (for a positive integral index)  7.1 Properties of the Binomial Expansion  7.1 Middle Term(s)  7.1 Some Properties of the Binomial Coefficients  7.2 Reversing Technique  7.2 Multinomial Theorem  7.3 Some useful tips and tricks   7.3 Some Particular Expansions  7.3 Solved Examples:  Concept-based: Straight Objective Type Questions  7.3 LEVEL 1: Straight Objective Type Questions  7.6 Assertion-Reason Type Questions  7.20 LEVEL 2: Straight Objective Type Questions  7.22 Exercises:  Concept-based: Straight Objective Type Questions  7.28 LEVEL 1: Straight Objective Type Questions  7.28 Assertion-Reason Type Questions  7.31 LEVEL 2: Straight Objective Type Questions  7.32 Previous Years’ AIEEE/JEE Main Questions  7.33 Previous Years’ B-Architecture Entrance Examination Questions  7.35 Answers 7.36 Hints and Solutions  7.36 8. Progressions Sequences  8.1 Arithmetic Progression  8.1 Geometric Progression  8.2 Arithmetico-Geometric Sequence  8.2 Harmonic Progression  8.3 Summation of Some Series of Natural Numbers  8.3 Sum of the Products of Two Terms of a Sequence  8.3 Method of Difference for Summation of Series  8.4 Solved Examples:  Concept-based: Straight Objective Type Questions  8.4 LEVEL 1: Straight Objective Type Questions  8.6 Assertion-Reason Type Questions  8.22 LEVEL 2: Straight Objective Type Questions  8.24

8.1–8.55

Contents  xxi

Exercises:  Concept-based: Straight Objective Type Questions  8.27 LEVEL 1: Straight Objective Type Questions  8.28 Assertion-Reason Type Questions  8.32 LEVEL 2: Straight Objective Type Questions  8.32 Previous Years’ AIEEE/JEE Main Questions  8.33 Previous Years’ B-Architecture Entrance Examination Questions  8.37 Answers 8.38 Hints and Solutions  8.38 9. Limits and Continuity

9.1–9.43

Limits  9.1 One-sided Limits  9.1 Frequently Used Limits  9.2 Some Theorems on Limits  9.2 Computation of Limits  9.2 L’Hôpital’s Rules for Calculating Limits  9.3 Use of Series Expansion in Finding Limits  9.4 Sequential Limits  9.4 Some Useful Results on Limits  9.5 Continuity  9.5 Functions Continuous on a Closed Interval  9.6 Solved Examples:  Concept-based: Straight Objective Type Questions  9.6 LEVEL 1: Straight Objective Type Questions  9.8 Assertion-Reason Type Questions  9.16 LEVEL 2: Straight Objective Type Questions  9.18 Exercises:  Concept-based: Straight Objective Type Questions  9.24 LEVEL 1: Straight Objective Type Questions  9.25 Assertion-Reason Type Questions  9.27 LEVEL 2: Straight Objective Type Questions  9.28 Previous Years’ AIEEE/JEE Main Questions  9.30 Previous Years’ B-Architecture Entrance Examination Questions  9.32 Answers 9.33 Hints and Solutions  9.33 10. Differentiability and Differentiation Differentiability  10.1 Differentiability on An Interval  10.2 Higher Order Differentiation  10.2 Some Formulae of Differentiation  10.2 Differentiation of Some Elementary Functions  10.2 Differentiation of Implicit Functions  10.3 Differentiation of Functions Represented Parametrically  10.3 Logarithmic Differentiation  10.3 Leibnitz Theorem and nth Derivatives  10.4 Solved Examples:  Concept-based: Straight Objective Type Questions  10.4 LEVEL 1: Straight Objective Type Questions  10.6 Assertion-Reason Type Questions  10.15 LEVEL 2: Straight Objective Type Questions  10.16 Exercises:  Concept-based: Straight Objective Type Questions  10.22 LEVEL 1: Straight Objective Type Questions  10.23

10.1–10.46

xxii Contents

Assertion-Reason Type Questions  10.26 LEVEL 2: Straight Objective Type Questions  10.26 Previous Years’ AIEEE/JEE Main Questions  10.28 Previous Years’ B-Architecture Entrance Examination Questions  10.31 Answers 10.32 Hints and Solutions  10.32 11. Applications of Derivatives

11.1–11.51

The Derivative as a Rate of Change  11.1 Tangent and Normal  11.2 Angle Between Two Curves  11.3 The Rolle’s and Lagrange’s Theorems  11.3 Monotonicity  11.3 Maxima and Minima  11.4 Solved Examples: Concept-based: Straight Objective Type Questions  11.5 LEVEL 1: Straight Objective Type Questions  11.7 Assertion-Reason Type Questions  11.16 LEVEL 2: Straight Objective Type Questions  11.18 Exercises:  Concept-based: Straight Objective Type Questions  11.23 LEVEL 1: Straight Objective Type Questions  11.23 Assertion-Reason Type Questions  11.26 LEVEL 2: Straight Objective Type Questions  11.26 Previous Years’ AIEEE/JEE Main Questions  11.28 Previous Years’ B-Architecture Entrance Examination Questions  11.32 Answers 11.33 Hints and Solutions  11.34 12. Indefinite Integration

12.1–12.51

Table of Basic Formula  12.1 A List of Evaluation Techniques  12.2 Some Tips for Simplifying Computations  12.7 Some Tips for Partial Fractions  12.7 Solved Examples:  Concept-based: Straight Objective Type Questions  12.9 LEVEL 1: Straight Objective Type Questions  12.11 Assertion-Reason Type Questions  12.21 LEVEL 2: Straight Objective Type Questions  12.23 Exercises:  Concept-based: Straight Objective Type Questions  12.29 LEVEL 1: Straight Objective Type Questions  12.30 Assertion-Reason Type Questions  12.33 LEVEL 2: Straight Objective Type Questions  12.33 Previous Years’ AIEEE/JEE Main Questions  12.36 Previous Years’ B-Architecture Entrance Examination Questions  12.38 Answers 12.39 Hints and Solutions  12.40 13. Definite Integrals The Newton-Leibnitz Formula  13.1 Definite Integral as the Limit of a Sum  13.1 Properties of Definite Integrals  13.1

13.1–13.54

Contents  xxiii

Integrals with Infinite Limits  13.2 Reduction Formulae for

p /2

Ú0

sin n x d x and Ú

p /2

0

sin p x cosq xd x   13.3

Solved Examples:  Concept-based: Straight Objective Type Questions  13.3 LEVEL 1: Straight Objective Type Questions  13.5 Assertion-Reason Type Questions  13.17 LEVEL 2: Straight Objective Type Questions  13.18 Exercises:  Concept-based: Straight Objective Type Questions  13.25 LEVEL 1: Straight Objective Type Questions  13.26 Assertion-Reason Type Questions  13.29 LEVEL 2: Straight Objective Type Questions  13.29 Previous Years’ AIEEE/JEE Main Questions  13.32 Previous Years’ B-Architecture Entrance Examination Questions  13.35 Answers 13.37 Hints and Solutions  13.37 14. Areas by Integration

14.1–14.25

Solved Examples:  Concept-based: Straight Objective Type Questions  14.4 LEVEL 1: Straight Objective Type Questions  14.5 Assertion-Reason Type Questions  14.9 LEVEL 2: Straight Objective Type Questions  14.9 Exercises:  Concept-based: Straight Objective Type Questions  14.12 LEVEL 1: Straight Objective Type Questions  14.12 Assertion-Reason Type Questions  14.13 Previous Years’ AIEEE/JEE Main Questions  14.14 Previous Years’ B-Architecture Entrance Examination Questions  14.15 Answers 14.15 Hints and Solutions  14.16 15. Differential Equations Formation of a Differential Equation  15.1 Methods of Solving Differential Equations Solution or Integral of a Differential Equation  15.2 Equations of First Order and First Degree  15.2 Orthogonal Trajectory  15.3 Differential Equations of First Order but not of First Degree  15.3 Second Order but of Degree One 15.4 Solved Examples:  Concept-based: Straight Objective Type Questions  15.4 LEVEL 1: Straight Objective Type Questions  15.7 Assertion-Reason Type Questions  15.14 LEVEL 2: Straight Objective Type Questions  15.15 Exercises:  Concept-based: Straight Objective Type Questions  15.19 LEVEL 1: Straight Objective Type Questions  15.20 Assertion-Reason Type Questions  15.22 LEVEL 2: Straight Objective Type Questions  15.22 Previous Years’ AIEEE/JEE Main Questions  15.24 Previous Years’ B-Architecture Entrance Examination Questions  15.26 Answers 15.27 Hints and Solutions  15.27

15.1–15.40

xxiv Contents

16. Cartesian System of Rectangular Coordinates and Straight Lines

16.1–16.56

Results Regarding Points in a Plane  16.1 Standard Forms of the Equation of a Line  16.2 Some Results for Two or More Lines  16.3 Some Useful Points  16.4 Locus of a Point  16.4 Change of Axes  16.5 Equation of Family of Lines  16.5 Solved Examples:  Concept-based: Straight Objective Type Questions  16.6 LEVEL 1: Straight Objective Type Questions  16.9 Assertion-Reason Type Questions  16.21 LEVEL 2: Straight Objective Type Questions  16.23 Exercises:  Concept-based: Straight Objective Type Questions  16.29 LEVEL 1: Straight Objective Type Questions  16.30 Assertion-Reason Type Questions  16.34 LEVEL 2: Straight Objective Type Questions  16.34 Previous Years’ AIEEE/JEE Main Questions  16.36 Previous Years’ B-Architecture Entrance Examination Questions  16.40 Answers 16.41 Hints and Solutions  16.42 17. Circles and Systems of Circles

17.1–17.56

Definition of a Circle  17.1 Equations of a Circle  17.1 Some Results Regarding Circles  17.2 Special Forms of Equation of a Circle  17.3 Systems of Circles  17.3 Common Tangents to Two Circles  17.4 Solved Examples:  Concept-based: Straight Objective Type Questions  17.5 LEVEL 1: Straight Objective Type Questions  17.8 Assertion-Reason Type Questions  17.19 LEVEL 2: Straight Objective Type Questions  17.20 Exercises:  Concept-based: Straight Objective Type Questions  17.29 LEVEL 1: Straight Objective Type Questions  17.31 Assertion-Reason Type Questions  17.34 LEVEL 2: Straight Objective Type Questions  17.34 Previous Years’ AIEEE/JEE Main Questions  17.37 Previous Years’ B-Architecture Entrance Examination Questions  17.39 Answers 17.40 Hints and Solutions  17.41 18. Parabola Conic Section  18.1 Parabola  18.1 Solved Examples:  Concept-based: Straight Objective Type Questions  18.5 LEVEL 1: Straight Objective Type Questions  18.8 Assertion-Reason Type Questions  18.13 LEVEL 2: Straight Objective Type Questions  18.15 Exercises:  Concept-based: Straight Objective Type Questions  18.18

18.1–18.37

Contents  xxv

LEVEL 1: Straight Objective Type Questions  18.19 Assertion-Reason Type Questions  18.21 LEVEL 2: Straight Objective Type Questions  18.22 Previous Years’ AIEEE/JEE Main Questions  18.23 Previous Years’ B-Architecture Entrance Examination Questions  18.24 Answers 18.25 Hints and Solutions  18.26 19. Ellipse

19.1–19.37

Definition 1  19.1 Definition 2  19.1 Solved Examples:  Concept-based: Straight Objective Type Questions  19.5 LEVEL 1: Straight Objective Type Questions  19.8 Assertion-Reason Type Questions  19.14 LEVEL 2: Straight Objective Type Questions  19.17 Exercises:  Concept-based: Straight Objective Type Questions  19.20 LEVEL 1: Straight Objective Type Questions  19.21 Assertion-Reason Type Questions  19.23 LEVEL 2: Straight Objective Type Questions  19.24 Previous Years’ AIEEE/JEE Main Questions  19.25 Previous Years’ B-Architecture Entrance Examination Questions  19.26 Answers 19.27 Hints and Solutions  19.27 20. Hyperbola

20.1–20.35

Solved Examples:  Concept-based: Straight Objective Type Questions  20.4 LEVEL 1: Straight Objective Type Questions  20.7 Assertion-Reason Type Questions  20.13 LEVEL 2: Straight Objective Type Questions  20.16 Exercises:  Concept-based: Straight Objective Type Questions  20.18 LEVEL 1: Straight Objective Type Questions  20.20 Assertion-Reason Type Questions  20.21 LEVEL 2: Straight Objective Type Questions  20.22 Previous Years’ AIEEE/JEE Main Questions  20.23 Previous Years’ B-Architecture Entrance Examination Questions  20.25 Answers 20.25 Hints and Solutions  20.25 21. Three Dimensional Geometry Coordinates  21.1 The Plane  21.2 The Straight Line  21.4 Vectorial Equations  21.6 Solved Examples:  Concept-based: Straight Objective Type Questions  21.7 LEVEL 1: Straight Objective Type Questions  21.10 Assertion-Reason Type Questions  21.19 LEVEL 2: Straight Objective Type Questions  21.20 Exercises:  Concept-based: Straight Objective Type Questions  21.27

21.1–21.56

xxvi Contents

LEVEL 1: Straight Objective Type Questions  21.28 Assertion-Reason Type Questions  21.31 LEVEL 2: Straight Objective Type Questions  21.32 Previous Years’ AIEEE/JEE Main Questions  21.33 Previous Years’ B-Architecture Entrance Examination Questions  21.38 Answers 21.39 Hints and Solutions  21.40 22. Vectors

22.1–22.39

Linear Combinations  22.1 Scalar or Dot Product  22.2 Vector or Cross Product  22.2 Scalar Triple Product  22.2 Vector Triple Product  22.3 Geometrical and Physical Applications  22.3 Solved Examples:  Concept-based: Straight Objective Type Questions  22.3 LEVEL 1: Straight Objective Type Questions  22.5 Assertion-Reason Type Questions  22.12 LEVEL 2: Straight Objective Type Questions  22.13 Exercises:  Concept-based: Straight Objective Type Questions  22.17 LEVEL 1: Straight Objective Type Questions  22.18 Assertion-Reason Type Questions  22.20 LEVEL 2: Straight Objective Type Questions  22.20 Previous Years’ AIEEE/JEE Main Questions  22.22 Previous Years’ B-Architecture Entrance Examination Questions  22.25 Answers 22.26 Hints and Solutions  22.27 23. Statistics Measures of Central Tendency  23.1 Arithmetic Mean  23.1 Properties of Arithmetic Mean  23.1 Geometric Mean  23.2 Harmonic Mean  23.2 Median  23.2 Mode  23.2 Dispersion  23.3 Coefficient of Dispersion (C.D.)  23.3 Solved Examples:  Concept-based: Straight Objective Type Questions  23.4 LEVEL 1: Straight Objective Type Questions  23.5 Assertion-Reason Type Questions  23.8 LEVEL 2: Straight Objective Type Questions  23.9 Exercises:  Concept-based: Straight Objective Type Questions  23.10 LEVEL 1: Straight Objective Type Questions  23.11 Assertion-Reason Type Questions  23.12 LEVEL 2: Straight Objective Type Questions  23.12 Previous Years’ AIEEE/JEE Main Questions  23.13 Previous Years’ B-Architecture Entrance Examination Questions  23.15 Answers 23.15 Hints and Solutions  23.16

23.1–23.20

Contents  xxvii

24. Probability

24.1–24.50

Definitions  24.1 Classical Definition of Probability  24.1 Notations  24.1 A Few Theorems on Probability  24.1 Conditional Probability   24.2 Multiplication Theorem   24.2 Total Probability Theorem  24.3 Bayes’ Rule  24.3 Theorem (Bayes’ Rule)  24.3 Independent Events  24.3 Random Variables  24.3 Binomial Distributions  24.4 Solved Examples:  Concept-based: Straight Objective Type Questions  24.5 LEVEL 1: Straight Objective Type Questions  24.7 Assertion-Reason Type Questions  24.22 LEVEL 2: Straight Objective Type Questions  24.24 Exercises:  Concept-based: Straight Objective Type Questions  24.27 LEVEL 1: Straight Objective Type Questions  24.28 Assertion-Reason Type Questions  24.33 LEVEL 2:Straight Objective Type Questions  24.33 Previous Years’ AIEEE/JEE Main Questions  24.34 Previous Years’ B-Architecture Entrance Examination Questions  24.37 Answers 24.38 Hints and Solutions  24.39 25. Trigonometrical Ratios, Identities and Equations Introduction  25.1 Domain and Range of Trigonometrical Functions  25.1 Some Basic Formulae  25.1 Allied or Related Angles  25.2 Compound Angles  25.2 Some Important Results  25.3 Identities  25.3 Conditional Identities  25.3 General Solutions of Trigonometrical Equations  25.4 Solved Examples:  Concept-based: Straight Objective Type Questions  25.4 LEVEL 1: Straight Objective Type Questions  25.8 Assertion-Reason Type Questions  25.16 LEVEL 2: Straight Objective Type Questions  25.17 Exercises:  Concept-based: Straight Objective Type Questions  25.23 LEVEL 1: Straight Objective Type Questions  25.25 Assertion-Reason Type Questions  25.27 LEVEL 2: Straight Objective Type Questions  25.28 Previous Years’ AIEEE/JEE Main Questions  25.30 Previous Years’ B-Architecture Entrance Examination Questions  25.32 Answers 25.32 Hints and Solutions  25.33

25.1–25.47

xxviii Contents

26. Inverse Trigonometric Functions

26.1–26.25

Inverse Trigonometric Functions and Formulae  26.1 Solved Examples:  Concept-based: Straight Objective Type Questions  26.5 LEVEL 1: Straight Objective Type Questions  26.6 Assertion-Reason Type Questions  26.11 LEVEL 2: Straight Objective Type Questions  26.12 Exercises:  Concept-based: Straight Objective Type Questions  26.14 LEVEL 1: Straight Objective Type Questions  26.14 Assertion-Reason Type Questions  26.16 LEVEL 2: Straight Objective Type Questions  26.16 Previous Years’ AIEEE/JEE Main Questions  26.17 Previous Years’ B-Architecture Entrance Examination Questions  26.18 Answers 26.18 Hints and Solutions  26.19 27. Heights and Distances

27.1–27.37

Angles of Elevation and Depression  27.1 Solved Examples:  Concept-based: Straight Objective Type Questions  27.2 LEVEL 1: Straight Objective Type Questions  27.3 Assertion-Reason Type Questions  27.13 LEVEL 2: Straight Objective Type Questions  27.14 Exercises:  Concept-based: Straight Objective Type Questions  27.20 LEVEL 1: Straight Objective Type Questions  27.21 Assertion-Reason Type Questions  27.23 LEVEL 2: Straight Objective Type Questions  27.23 Previous Years’ AIEEE/JEE Main Questions  27.25 Previous Years’ B-Architecture Entrance Examination Questions  27.26 Answers 27.26 Hints and Solutions  27.26 28. Mathematical Reasoning

28.1–28.16

Some Definitions  28.1 Solved Examples:  Concept-based: Straight Objective Type Questions  28.2 LEVEL 1: Straight Objective Type Questions  28.3 Assertion-Reason Type Questions  28.6 LEVEL 2: Straight Objective Type Questions  28.7 Exercises:  Concept-based: Straight Objective Type Questions  28.8 LEVEL 1: Straight Objective Type Questions  28.8 Assertion-Reason Type Questions  28.9 LEVEL 2: Straight Objective Type Questions  28.9 Previous Years’ AIEEE/JEE Main Questions  28.10 Previous Years’ B-Architecture Entrance Examination Questions  28.11 Answers 28.12 Hints and Solutions  28.12 Architecture Entrance – 2017 JEE (Main) 2017 Questions with Solutions Mathematics (8th April - online)

1–8 9–16

JEE (Main) 2017 Questions with Solution (9th April – online)

17–24



25–32

JEE Main 2017 (Offline)

Format of Questions in this Book  xxix

Format of Questions in this Book Straight Objective Type Questions Each Question has four choices—(a), (b), (c) and (d) out of which ONLY ONE is correct

Assertion-Reason Type Questions Each Question has four choices—(a), (b), (c) and (d) out of which ONLY ONE is correct Write (a), (b), (c) or (d) according to the following rules. (a) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1 (b) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1 (c) STATEMENT-1 is True, STATEMENT-2 is False (d) STATEMENT-1 is False, STATEMENT-2 is True

CHAPTER ONE

By a set we mean any collection of objects. For example, we may speak of the set of all living Indians, the set of all letters of the English alphabet or the set of real numbers less than 5. The objects constituting a set are elements or members of the set. If X is a set and an element x is a member of, or belongs to the set X, then this is expressed, as x Œ X. There are two methods of representing a set. (a) Roster method (or tabular method) In this method, a set is represented by listing all its elements in curly brackets and separating them by commas. For example, set having elements 1, 2, 3, 5 only is written as {1, 2, 3, 5}. (b) Property method In this method a set is represented by stating all the properties which are satisfied by the elements of the set and not by other elements outside the set. If X contains all values of ‘x’ for which the condition P(x) is true, then we write X = {x : P(x)}. Illustration

1

The set of all real roots of the equation x4 Р2x2 Р3 = 0 is denoted by {x : x ΠR, x4 Р2x2 Р3 = 0} or equivalently by Illustration

{

}

3, - 3 .

2

The set of integers strictly between 2 and 8 is represented by {x : x ΠI, 2 < x < 8} or equivalently by {3, 4, 5, 6, 7}. Illustration

3

The set of Presidents of India can be represented by {x : x is/was President of India} or equivalently A = {Rajendra Prasad, Sarvepalli Radhakrishnan, Zakir Hussain, V.V. Giri, …….}

The number of elements of A or the cardinality of A is denoted by n(A). In Illustration 3, n(A) = 13.

FINITE SET AND INFINITE SET A set is called a finite set if it contains only finite number of elements. A set which does not contain finite number of elements is called infinite set. Null set A set containing no element is called a null set or empty set or void set and is denoted by ‘f’. Equivalently P(x) is a property satisfied by no object at all. Singleton set A set containing exactly one element is called a singleton set. If a set X has ‘r’ elements then we write n(X) (number of elements in X) = r. Subset A set ‘A’ is said to be subset of the set X if every element of A is an element of X and we write A Õ X. A is said to be a proper subset of X if A Õ X. and A π X, this will be written as A Ã X . The denial of A Ã B is written as π

A Ã/ B. For two sets A and B, A = B if and only if A Õ B and B Õ A. The set of all subsets of X is called Power set of X denoted by P(X) i.e. P(X) = {A : A Õ X}.

Some Basic Properties: (i) (i) (iii) (iv) (v) (vi)

AÕA A Õ B, B Õ C then A Õ C The only subset of f is f itself. The subsets of {x} are f and {x}. The subsets of {x, y} are f, {x}, {y}, {x, y} If n(X) = r then number of all subsets of X will be 2r i.e. n(P(X)) = 2r.

ALGEBRA OF SETS Union of two Sets Union of two sets A and B is denoted by A » B

Complete Mathematics—JEE Main

1.2

A » B = {x : x Œ A or x Œ B} Clearly A Õ A » B, BÕA»B

A

and,

«

B

A

Ê n ˆ n 4. A ~ Á « Ai ˜ = » ( A ~ Ai ). Ë i =1 ¯ i =1 B

Ê n ˆ n 5. A ~ Á » Ai ˜ = « ( A ~ Ai ). Ë i =1 ¯ i =1

A » B = B » A, A » A = A, Fig. 1.1 A » B = B if and only if A Õ B, (A » B) » C = A » (B » C) (Associativity).

and

Intersection of Two Sets Intersection of two sets A and B is denoted by A « B and, A « B = {x : x Œ A and x Œ B} Trivially, A « B Õ A, A « B Õ B and A « B = B « A, (A « B) « C = A « (B « C) (Associativity).

A

B A«B

Fig. 1.2

Complement of a Set Complement of a set B in a set A is written as A ~ B = {x : x Œ A, x œ B} e.g., I ~ N = º – 3, – 2, – 1, 0}. The set of irrational numbers = R ~ Q Where I is the set of integers N is the set of natural numbers R is the set of real number and Q is the set of rational numbers In any discussion involving sets and their operations, we presume that all these sets are subsets of a parent set called Universal set, U. The complement of A in U is denoted by Ac or A¢ = {x : x œ A}. A

A « Ac = f A » Ac = X fc = X Xc = f (Ac )c = A A ~ B = A « Bc A Õ B if and only if Bc Õ Ac. A ~ A = f. (A ~ B) ~ C = (A ~ C) ~ B A ~ (B » C) = (A ~ B) « (A ~ C) If A and B are finite sets, then n(A » B) = n(A) + n(B) – n(A « B) 17. If A, B and C are finite sets then n(A » B » C) = n(A) + n(B) + n(C) – n(A « B) – n(B « C) – n(A « C) + n(A « B « C)

6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

Symmetric Difference of Two Sets A D B= (A ~ B) » (B ~ A)

A

B

Fig. 1.4

A D A = f, A D B = B D A

Cartesian Product

B

A¢ A

Fig. 1.3

SOME PROPERTIES OF OPERATIONS ON SETS 1. A « (B1 » B2 » ... » Bn) = (A « B1) » (A « B2) » … » (A « Bn) (Intersection is distributive over union) 2. A » (B1 « B2 « … « Bn) = (A » B1) « (A » B2) « … « (A » Bn) (Union is distributive over intersection) 3. (A « B)c = Ac » Bc and (A » B)c = Ac « Bc. (De Morgan’s Laws)

The cartesian product of sets A and B is denoted by A ¥ B and, A ¥ B = {(a, b): a Œ A, b Œ B} In general A ¥ B π B ¥ A, moreover, A ¥ (B » C) = (A ¥ B) » (A ¥ C) and A ¥ (B « C) = (A ¥ B) « (A ¥ C) Illustration

4

If A = {0, 1, 2, 3, 5}, B = {0, 1, 2, 3), C = {0, 1, 4, 5} then B Õ A, C Õ/ A, A « C = {0, 1, 5}, B « C = {0, 1}, A ~ B = {5}, A ~ C = {2, 3}, B ~ C = {2, 3}, C ~ B = {4, 5}, A D B = {5}, A D C = {2, 3, 4} and B D C = {2, 3, 4, 5} Illustration

5

Simplify (A » B » C ) « (A « Bc « C c )c « C c

Sets, Relations and Functions 1.3

(A » B » C ) « (A « Bc « C c )c « C c c

Equivalence Relation

= [(A » B » C ) « (A » B » C )] « C

c

A relation R on a set X (i.e. R Ã X ¥ X) is said to be

= [(A « Ac) » (B » C] « C c

(i) reflexive if x R x for all x ŒX. (ii) symmetric if x R y fi y R x, where x, y Œ X (iii) transitive if x R y, y R z fi x R z, where x, y, z Œ X

= [f » (B » C)] « C c = (B » C) « C c = (B « C c ) » (C « C c ). = (B « C c) » f = B « C c = B ~ C.

Further, a relation R in a set X is said to be an equivalence relation if it is reflexive, symmetric and transitive. R is called anti-symmetric if x R y and y R x fi x = y.

Relations A relation from a set of X to a set Y is a subset R of X ¥ Y. If (a, b) Œ R, we say a is related to b and often write it as a R b. If X = Y, we say that R is relation on X. Let R be a relation from X to Y. For each subset A of X, we write

Y

6

Illustration R

For x, y Œ I, write x R y if x – y divisible by 6; this is an equivalence relation and is usually written as x ∫ y (mod. 6). X

Fig. 1.5

7

Illustration

The relation R on R defined as R = {(a, b): a £ b} is not symmetric as (1, 3) Œ R but (3, 1) œ R but is transitive and reflexive.

R(A) = {y Œ Y: x R y for some x Œ A} and call it the (direct) image of A under R, and for each subset B of Y, we write R–1 (B) = {x Œ X : x R y for some y Œ B} and call it the inverse image of B under R. Y B

8

Illustration

If A = {1, 2, 3}; let R = {1, 1), (2, 2) (3, 3), (1, 2) (2, 1)} then R is reflexive, symmetric and transitive. The number of equivalence relations that can be defined on a set containing k elements is given k -1

Bk =

 ( k n-1 )Bn

n=0

E.g.

B0 = 1, B1 = 1, B2 = 2, B3 =

( 20 ) B0 + ( 12 ) B1 + ( 22 ) B2

=1+2+2=5

R(A)

Functions R-1(B)

A

X

Fig. 1.6

One can think of R–1 as a relation from Y to X, where yR –1 x ¤ xRy that is,

R –1 = {(y, x) : (x, y) Œ R}

R–1 is called the reverse of the relation R. Domain of a relation R from a set X to a set Y is the set of all first components of the elements of R, i.e., dom R = {a Œ X: (a, b) Œ R for some b Œ R} Let R be a relation from X to Y and S from Y to Z, then SoR = {(x, z)| x Œ X, z Œ Z and $ y Œ Y such that and (x, y) Œ R and (y, z) Œ S. If n(A), n(B) < •, then the number of relation from A to B will be 2n(A) n(B)

A relation F from A to B is said to be a function if for each a Œ A there exists a unique b Œ B such that (a, b) Œ F. b is called image of ‘a’ under F and is denoted by F(a). Thus every function is a relation but the converse may not be true. If n(A) = n and n(B) = m then total number of functions from A to B is nm.

SOME DEFINITIONS Let A and B be two non-empty sets. A function f from A to B can also be defined as a rule that assigns to each element in the set A, one and only one element of the set B. In general, the sets A and B need not be sets of real numbers. However, we consider only those functions for which A and B are both subsets of the real numbers. We shall denote by R, the set of all real numbers. The set A in the above definition is called the domain of the function f. We usually denote it by dom f. If x is an element in the domain of a function f, then the element that

1.4

Complete Mathematics—JEE Main

f associates with x is denoted by the symbol f (x), and is called the image of x under f, or the value of f at x. The set of all possible values of f (x) as x varies over the domain is called the range of f. If f : A Æ B, then the range of f is a subset of B and the set B is called co-domain of f. Remark If x is an element in the domain of a function f a function requires that f assigns one and only one value to x. This means that a function cannot be multiple-valued. For function example, the expression ± x of x, since it assigns two values to each positive x.

ALGEBRAIC OPERATIONS ON FUNCTIONS 1. If f and g are two functions, then sum of the functions f + g, is defined for all x Œ dom f « dom g by ( f + g) (x) = f (x) + g(x). 2. If k is any real number and f is a function, then k f is defined for all x Œ dom f by (k f ) (x) = k f (x). 3. If f and g are two functions, then the pointwise product f g is defined for all x Œ dom f « dom g by ( f g) (x) = f (x) g(x). 4. If f and g are functions, then f/g is defined for all x Œ dom (f) « dom(g) « {x: g(x) π 0} by (f/g)(x) = f(x)/g(x). 5. Composition of functions Let f : A Æ B and g : B Æ C be functions, then gof : A Æ C defined by (gof ) (x) = g ( f (x)). We have the following formulae for domains of functions. 1. dom ( f + g) = dom f « dom g 2. dom ( f g) = dom f « dom g 3. dom (f/g) = dom f « dom g « {x : g(x) π 0} 4. dom

f = dom f « {x : f (x) ≥ 0}

Note Note that if gof fog example if f : A Æ B and g : B Æ C then gof but fog is not. Let f : R Æ R and g : R Æ R f (x) = cos x and g(x) = x3. Then gof (x) = g( f (x)) = g(cos x) = cos3 x and fog (x) = f (x3) = cos x3. Thus even if fog and gof fog = gof.

DOMAINS AND RANGES OF SOME FUNCTIONS 1. Constant functions A function that assigns the same value to every member of its domain is called a constant

function. The domain of the constant function f (x) = c is R and its range is {c}. 2. Polynomial functions A function of the form c xn, where c is a constant and n is a non-negative integer, is called a monomial in x. Examples are 2 x3, 5x4, – 6x and x8. The function 4 x1/2 and x–3 and not monomials because the powers of x are not non-negative integers. A function that is expressible as the sum of finitely many monomials in x is called a polynomial in x. Thus f (x) =an xn + an – 1 xn – 1 + º + a1 x + a0 is a polynomial. The domain of a polynomial is R and its range is a subset of R. 3. The domain of f (x) = loge x (= ln x) is (0, •) = {x Œ R, x > 0} = R+, and its range is (– •, •) (i.e., the whole of R). 4. The domain of f (x) = ex is R and its range is R+.

TYPES OF FUNCTIONS 1. Rational function. This function is defined as the ratio of two polynomials y=

a0 x n + a1 x n - 1 +  + an b0 x m + b1 x m - 1 +  + bm

For example, y = (x2 + 3)/(x3 + 4) is rational function. 2. Irrational function. If in the function y = f (x), the operations of addition, subtraction, multiplication, division and raising to a power with rational non-integral exponents are performed on the right-hand side, the function y = f (x) is said to be irrational. Examples are y = (3x2 + x )/ 2 + 4 x and y = x . 3. Even function. A function y = f (x) is said to be an even function if f (–x) = f (x) " x Œ dom ( f ). Examples are y = | x |, y = cos x and y = x 2n. The graph of an even function will be symmetric about y-axis 4. Odd function. A function y = f (x) is said to be an odd function if f (– x) = – f (x) " x Œ dom (f). Examples are y = sin x and y = x2n + 1. Clearly, y = f (x) + f (– x) is always an even function, and y = f (x) – f (– x) is always odd. Any function y = f (x) can be expressed uniquely as the sum of an even and an odd function as follows: 1 1 f (x) = ( f (x) + f(–x)) + ( f (x) – f (– x)) 2 2 5. Periodic function. A function y = f (x) is said to be periodic if there exists a number T > 0 such that f (x + T) = f (x) for all x in the domain of f. The least such T is called the period of f. For example, the period of sin x and cos x is 2p, and that of tan x is p. If f (x) is a periodic function with period T, then the function f (a x + b), a > 0, is periodic with period T/a. For

Sets, Relations and Functions 1.5

example, sin 2 x has a period p, cos 3x has a period 2p/3 and tan (2 x + 4) has a period p/2. The function Ï1 if x ŒQ f (x) = Ì Ó-1 if x Œ R ~ Q is a periodic function without any period. The sum of two periodic functions may not be periodic e.g., f (x) = {x}, the fractional part of x and g (x) = sin x. 6. Onto function (or Surjective Function). If a function f : A Æ B is such that each element in B is the f-image of at least one element in A, then we say that f is a function of A ‘onto’ B. Equivalently a function f is an onto function if co-domain of f = Range of f.

that | f (x)| £ k for all x Œ I. Equivalently there is m and 1 M such that m £ f (x) £ M for all x Œ I. E.g. f (x) = is x 2 not bounded on (0, 1) and f (x) = x is bounded on [0, 1]. f (x) = sin x or cos x are bounded functions on R whereas f (x) = tan x is unbounded on (– p /2, p /2).

GRAPHS OF SOME FUNCTIONS 1. Constant function f (x) = c represents a constant function (Fig. 1.8). y y=c

(0, c ) Not onto

Onto

x

O

Fig. 1.7

For example, f : R Æ [–1, 1] defined by f (x) = sin x is an onto function but f : R Æ R defined by f (x) = sin x is not onto since Range of f = [– 1, 1] and co-domain of f = R. In order to show that a function f : A Æ B is onto we start with any y Œ B and try to find x Œ A such that f (x) = y. 7. One-to-one function (or injective function). A function f is said to be one-to-one if it does not take the same value at two distinct points in its domain. For example, f (x) = x3 is one-to-one, whereas f (x) = x2 is not, as f (1) = 1 and f (–1) = 1. Note that a periodic function f : R Æ R cannot be one-to-one as f (x + T) = f (x) for some T > 0. In order to show that a function f : A Æ B is one-to-one, we may take any x, y Œ A such that f (x) = f ( y) and try to show that x = y. If n(A) = m and n(B) = n(m £ n), then the number of injections (or one-one functions) from A to B is n

n! . Pm = (n - m)!

8. Bijective function (One-to-One and Onto). If a function f is both one-to-one and onto, then f is said to be a bijective function. For example an identity function iA : A Æ A defined by iA (x) = x is trivially a bijective function. The function f : [–p/2, p/2] Æ [–1, 1] defined by f (x) = sin x is a bijective function. The function f : (– p/2, p/2) Æ R defined by f (x) = tan x is also a bijective function. If A and B are finite sets and f: A Æ B is a bijection then n(A) = n(B). If n(A) = n, then the number of bijections from A to B is the total number of arrangements of n items taken all at a time which is n! 9. Bounded and unbounded functions A function f defined on an interval I is said bounded on I if there is k > 0 such

Fig. 1.8

2. Proportional values. If variables y and x are direct proportional, then the functional dependence between them is represented by the equation: y = kx, where k is a constant a factor of proportionality. A graph of a direct proportionality is a straight line, going through an origin of coordinates and forming with an x-axis an angle a, a tangent of which is equal to k: tan a = k. Therefore, a factor of proportionality is called also a slope. These are shown in three graphs with k = 1/3, k = 1 and k = –3 on Fig.1.9 k=–3

Y k=1 k = 1/3 a O

X

Fig. 1.9

3. Linear function. If variables y and x are related by the 1-st degree equation: Ax + By = C, (at least one of numbers A or B is non-zero), then a graph of the functional dependence is a straight line. If C = 0, then it goes through an origin of coordinates, otherwise – not. Graphs of linear functions for different combinations of A, B, C are represented on Fig.1.10

1.6

Complete Mathematics—JEE Main

parabola – a curve going through an origin of coordinates. Every parabola has an axis of symmetry OY, which is called an axis of parabola. The point O of intersection of a parabola with its axis is a vertex of parabola. Y

y = 1/2 x2

3 2 1

Fig. 1.10

4. Inverse proportionality. If variables y and x are inverse proportional, then the functional dependence between them is represented by the equation: y = k/x, where k is a constant. A graph of an inverse proportionality is a curve, having two branches (Fig. 1.11). This curve is called a hyperbola. These curves are received at crossing a circular cone by a plane. As shown on Fig. 1.11, a product of coordinates of a hyperbola points is a constant value, equal in this case to 1. In general case this value is k, as it follows from a hyperbola equation: xy = k Y

–3 –2 –1 O –1

1

2

3

X

Fig. 1.12

A graph of the function y = ax2 + bx + c is also a quadratic parabola of the same shape, that y = ax2, but its vertex is not an origin of coordinates, this is a point whose coordinates: Ê b b2 ˆ c , ÁË 2a 4a ˜¯ The form and location of a quadratic parabola in a coordinate system depends completely on two parameters: the coefficient a of x2 and discriminant D : D = b2 – 4ac. These properties follow from analysis of the quadratic equation roots. All possible different cases for a quadratic parabola are shown on Fig. 1.13

2 1 –2

–1

X

O –1 –2

Fig. 1.11

The main characteristics and properties of hyperbola: ∑ the function domain: x π 0, and codomain: y π 0; ∑ the function is monotone (decreasing) at x < 0 and at x > 0, but it is not monotone on the whole, because of a point of discontinuity x = 0 ∑ the function is unbounded, discontinuous at a point x = 0, odd, non-periodic; ∑ there are no zeros of the function. 5. Quadratic function. This is the function: y = ax2 + bx + c, where a, b, c – constants, a π 0. In the simplest case we have b = c = 0 and y = ax2. A graph of this function is a quadratic

Fig. 1.13

The main characteristics and properties of a quadratic parabola: ∑ the function domain: – • < x < + • (i.e. x is any real number) ∑ the function is not monotone on the whole, but to the right or to the left of the vertex it behaves as a monotone function; ∑ the function is unbounded, continuous in everywhere, even at b = c = 0, and non-periodic; ∑ the function has no zeros at D < 0.

Sets, Relations and Functions 1.7

6. Power function. This is the function: y = axn where a, n are constants. At n = 1 we obtain the function, called a direct proportionality: y = ax; at n = 2 – a quadratic parabola; at n = –1 – an inverse proportionality or hyperbola. So, these functions are particular cases of a power function. We know, that a zero power of every non-zero number is 1, thus at n = 0 the power function becomes a constant: y = a, i.e. its graph is a straight line, parallel to an x-axis, except an origin of coordinates. All these cases (at a = 1) are shown on Fig. 1.14 (n ≥ 0) and Fig. 1.15 (n < 0). Negative values of x are not considered here, because then some of functions:

that this is the two-valued function (the sign ± before the square root symbol says about this). Such functions are not studied in an elementary mathematics, therefore we consider usually as a function one of its branches: either an upper or a lower branch. Y y = x2

O X

y = x1/2, y = x1/4 y = x3

lose a meaning. Y 2

n=4 n=2

n=1

Fig. 1.16

n

y = x ,n > 0

Y

n = 1/2 n = 1/4 n=0

1

y=± x

O

O

1

2

X

Fig. 1.17

Fig. 1.14 Y 2 – 1/2 – 1 – 2

y = xn,n 1 an exponential function increases, and at 0 < a < 1 – decreases. Y

a = 1/2

a=2

5

3 1

–3 –2 –1

1

Fig. 1.18

2

3

X

Complete Mathematics—JEE Main

y = ex y=

y = e|x|

e–x

Fig. 1.19

Fig. 1.20

y y = |x|

y=

-

x

x

0

Fig. 1.22

Greatest Integer Function and Fractional part function f(x) = [x] will denote the integral part of x or greatest integer less than or equal to x. E.g. [5] = 5, [5.2], = 5, [–5.1] = – 6, [p] = 3, [e] = 2. The domain of this function is R and the range is I. For positive x, this function might represent, for example, the legal age of a person is a function of his chronological age x. The fractional part of x is the fractional part {x} of x so {x} = x – [x] e.g. {5.04} = 0.04, {–4} = 0, {–1.7} = 0.3, {p} = p – 3. The domain of this function is R whereas the range is [0, 1). y

8. Logarithmic function. The function y = logax, where a is a positive constant number, not equal to 1 is called a logarithmic function. This is an inverse function relatively to an exponential function; its graph (Fig.1.19) can be obtained by rotating a graph of an exponential function around of a bisector of the 1-st coordinate angle.

3 2 2

1

1 0

x¢

Y

x

The main characteristics and properties of an exponential function: ∑ the function domain: –– • < x < + • (i.e. x is any real number) and its codomain: y > 0; ∑ this is a monotone function: it increases at a > 1 and decreases at 0 < a < 1; ∑ the function is unbounded, continuous in everywhere, non-periodic; ∑ the function has no zeros.

y=

1.8

1

a>1

1

2

x

3

graph of [x]

2 3

O

X

y¢

1

Fig. 1.23 y

0< a 0 and its codomain: – • < y < + • (i.e. y is any real number); ∑ this is a monotone function: it increases for a > 1 and decreases for 0 < a < 1; ∑ the function is unbounded, continuous in everywhere, non-periodic; ∑ the function has one zero: x = 1. 9. Absolute-value function is given by Ï x , x≥0 y = | x | = max {x, – x} = Ì Ó- x , x < 0 It is depicted in Fig. 1.22.

1

x¢

2

3

x

graph of {x}

y¢

Fig. 1.24

Some Properties 1. 2. 3. 4.

x – 1 < [x] £ x [{x}] = 0 ={[x]} If n £ x < n + 1, n Œ I, then [x] = n and conversely. If x Œ I, then [x] = x otherwise [x] < x.

, if x Œ I Ï-[ x ] 5. [–x] = Ì Ó-[ x ] - 1 , if x œ I Ï0, if x Œ I in other word, [ x ] + [ - x ] = Ì Ó-1, if x œ I

Sets, Relations and Functions 1.9

6. [x + y] = [x] + [y], if x or y ΠI.

Figure 1.26 shows the graph. For positive value of x, this function might represent for example, the cost of parking x hours in a parking lot which charges Rs 10 for each hour or part of an hour in this case f (x) = 10 [x]¢.

if {x} + {y} < 1 Ï[ x ] + [ y], 7. [x + y] = Ì Ó[ x ] + [ y] + 1, if {x} + {y} ≥ 1 Moreover, [x + y] ≥ [x] + [y]. Ï x, 8. {x} = Ì Ó0,

INVERSE OF A FUNCTION

if 0 < x < 1 x ŒI

The function f (x) = x1/3 and g(x) = x3 have the following property: f (g(x)) = f (x3) = (x3)1/3 = x g( f (x)) = g(x1/3) = (x1/3)3 = x Similarly, the functions f (x) = loge x and g(x) = ex cancel the effect of each other. If two functions f and g satisfy f (g(x)) = x for every x in the domain of g, and g( f (x)) = x for every x in the domain of f, we say that f is the inverse of g and g is the inverse of f, and we write f = g–1, or g = f –1. (The symbol f –1 does not mean 1/f.) To find the inverse of f, write down the equation y = f (x) and then solve x as a function of y. The resulting equation is x = f –1 ( y). If f is one-to-one, then f has an inverse defined on its range and, conversely, if f has an inverse, then f is one-toone. f –1 is defined on the range of f. If f is one-to-one from A to B, and g is one-to-one from B to C, then f o g is one-to-one from A to C, and ( f o g)–1 = g–1 o f –1. If f : X Æ Y is bijective then f –1 : Y Æ X is bijective and

Ï0, x Œ I 9. {x} + {–x} = Ì Ó1, if x œ I n - 1˘ 1˘ È È 10. [ x ] + Í x + ˙ + ... + Í x + = [n x ] . n˚ n ˙˚ Î Î 11. Signum function is given by if x > 0 Ï1 Ô sgn (x) = Ì0 if x = 0 Ô-1 if x < 0 Ó

and it is depicted in Fig. 1.25 y y = sgn(x )

0

x 1

( f -1 )

12. The least integer function: The function whose value at any number x is the smallest integer greater than or equal to x is called the least integer function, e.g. if f (x) = [x]¢ = smallest integer, greater than or equal to x. then [1.5]¢ = 2 [1.99]¢ = 1 [1.4]¢ = 2. y

1

9

{ }

ax + b d . The domain of f is R ~ . cx + d c ax + b b - dy so cxy + dy = ax + b fi x = . Let y = cx + d cy - a b - dx Hence f –1(x) = . cx - a Let f(x) =

DIRECT AND INVERSE IMAGES

y = [(x]

2

= f. Moreover, f –1 ({ y }) = {f –1( y )} for all y Œ Y.

Illustration

Fig. 1.25

3

-1

1

2

3

4

x

Let f be a function with domain X and co-domain Y. If A Õ X, then the direct image of A under f is the subset of Y (denoted by f (A)) is defined to be { f (x) : x Œ A}. For example, let f : R Æ R be a function defined by f (x) = x2. If A = {–3, –1, 0, 1, 3}, then f (A) = {0, 1, 9}. Note that if f (X) = Y, then the function f : X Æ Y is onto. The following relations can be verified easily. (i) A à B fi f (A) à f (B) (ii) f (A » B) = f (A) » f (B) (iii) f (A « B) à f (A) « f (B). The inclusion may be strict.

Fig. 1.26

Let f be a function with domain X and range Y and let B be a subset of Y. The inverse image of B under f is a subset of X (denoted by f –1 (B)) is defined to be {x : f (x) Œ B}.

1.10

Complete Mathematics—JEE Main

Illustration

Note that if f is one-to-one, then f –1({ x }) is a singleton set. The following relations can be verified easily for A, B Ã Y.

10

If f : R Æ R is a function defined by f (x) = | x | and A = (– •, 0), then f –1(A) = {x: f (x) Œ A} = {x: | x | Œ (– •, 0)} = f (the empty set). If B = (–3, 3), then f –1 (B) = (–3, 3).

(i) A Ã B = f –1 (A) Ã f –1 (B) (ii) f –1 (A » B) = f –1 (A) » f –1 (B) (iii) f –1 (A « B) = f –1 (A) « f –1 (B).

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Example 1: Let I be set of integers, N = the set of nonnegative integers, Np = the set of non-positive integers. Then the sets A and B satisfying A « B = f are (a) A = I ~ Np, B = N ~ Np (b) A = I ~ N, B = I ~ Np (c) A = N D Np, B = I ~ Np (d) A = N D Np, B = (I ~ N) » {0} Ans. (b) Solution: I ~ N = {… –3, –2, –1}, I ~ Np = {1, 2, 3…}, N ~ Np = {1, 2, …}, N D Np = (N ~ Np) » (Np ~ N) = {0, 1, 2, …} (I ~ N) » {0} = {…, –3, –2, –1, 0} The disjoint sets are I ~ N and I ~ Np. Example 2: Which of the following equality is not true. (a) A « (B ~ C) = A « B ~ (A « C) (b) A ~ (A « B) = A ~ B (c) A ~ (B ~ C) = (A ~ B) » (A « C) (d) A ~ (B D C ) = (A ~ B) D (A ~ C) Ans. (d) Solution: For equality (a), A « (B ~ C) = A « (B « C¢) = f » (A « B « C¢) = (A « B « A¢) » (A « B « C¢) = A « B « (A¢ » C¢) = A « B ~ (A « C) For equality (b), A ~ (A « B) = A « (A¢ » B¢) = (A « A¢) » (A « B¢) = f » (A « B¢) = A « B¢ = A ~ B For equality (c) A ~ (B ~ C) = A ~ (B « C¢) = A « (B¢ » C) = (A « B¢) » (A « C) = (A ~ B) » (A « C) For (d) Let A = {1, 2, 3, 4, 5}, B = {3, 4, 5}, C = {1, 2, 3} So, B D C = {4, 5} » {1, 2} = {1, 2, 4, 5} Thus A ~ (B D C) = {3}

A ~ B = {1, 2}; A ~ C = {4, 5} Therefore (A ~ B) D (A ~ C) = ({1, 2} ~ {4, 5}) » ({4, 5} ~ {1, 2}) = {1, 2} » {4, 5} = {1, 2, 4, 5). Hence A ~ (B D C) π (A ~ B) D (A ~ C) Example 3: A boating club consists of 82 members, each member is either a sailboat owner or a powerboat owner. If 53 members owned sailboats and 38 members owned powerboats, the number of members owned both sailboat and powerboat is (a) 6 (b) 7 (c) 9 (d) 4 Ans. (c) Solution: Let S = the set of all members owning sailboats and P = the set of all members owning powerboats n(S » P) = n(S) + n(P) – n(S « P) 82 = 53 + 38 – n(S « P) fi n(S « P) = 91 – 82 = 9. Example 4: If, B Ã A¢, then which of the following sets π is equal to A¢. (a) (A « B) » B (b) (A « B) » A¢ (c) (A » B) « A¢ (d) (A » B) « B Ans. (b) Solution: For (a), (A « B) » B = (A » B) « (B » B) = (A » B) « B = B. For (b), (A « B) » A¢ = (A » A¢) « (B » A¢) = X « A¢ = A¢ For (c), (A » B) « A¢ = (A « A¢) » (B « A¢) = f » (B « A¢) = B « A¢ = B. For (d) (A » B) « B = B. È x ˘ È x ˘ 5x Example 5: If Í ˙ + Í ˙ = , then x is any term of Î2˚ Î3˚ 6 the following (a) 3, 6, 9, 12,… (c) 6, 12, 18, 24,… Ans. (c)

(b) 9, 18, 27, 36, … 6 12 18 (d) , , ,... 5 5 5

Sets, Relations and Functions 1.11

Èx˘ Èx˘ Èx˘ Èx˘ Solution: Since Í ˙ , Í ˙ Œ I , so Í ˙ + Í ˙ Œ I Î2˚ Î3˚ Î2˚ Î3˚ 5x 6 Œ I fi x = n, n Œ I . Thus 6 5 È x ˘ È x ˘ 5x Substituting this value in Í ˙ + Í ˙ = , we have Î2˚ Î3˚ 6 È3 ˘ È2 ˘ ÍÎ 5 n ˙˚ + ÍÎ 5 n ˙˚ = n

{ } { } { }{ } { } { }

3 3 2 2 n- n + n- n =n 5 5 5 5 3 2 n + n =0 5 5

fi fi

3 2 n =0= n 5 5

fi

(Since 0 £ {x} < 1)

2n ◊ 3n 5 5m1 ¥ 5m2 x ◊ 5m1 = = 5m1 m2 fi = 5m1 m2 5 3 fi x = 3m2 x ◊ 5m2 Similarly = 5m1 m2 fi x = 2 m1 2 Hence x is multiple of 2 and 3 so of 6 and x Œ I

Thus 3n = 5m1, 2n = 5m2, Therefore xn =

Example 6: The relation R defined by ‘>’ on the set N is (a) reflexive (b) symmetric (c) transitive (d) equivalence relation Ans. (c) / so (3, 2) Œ R but (2, 3) œ R. Thus R is not symmetric. If (a, b) Œ R and (b, c) Œ R then a > b, b > c fi a > c so (a, c) Œ R. R is not an equivalence relation. Example 7: The relation a R b defined by a is factor of b on N is not (a) reflexive (b) transitive (c) anti symmetric (d) symmetric Ans. (d) Solution: For a Œ N, a is a factor of a so R is reflexive. If a is factor b and b is factor of c then a is factor of c so R is transitive. If a is factor of b and b is factor of a then a = b so R is anti symmetric, 2 is factor of 4 but 4 is not a factor of 2. Example 8: The domain of the function f(x) = log2 sin x is (a) R (b) R ~ {n p : n Œ I} (c) R ~ {np : n Œ N}

(d) » (2np , (2n + 1)p ) nŒI

Ans. (d) Solution: f(x) = log2 sin x is defined for all x for which sin x > 0. But sin x > 0 if x Œ (0, p) » (2p, 3p) » … = » (2np , (2n + 1)p ) . nŒI

Example 9: The domain of y = cos-1 È 3 5˘ (a) Í - , ˙ Î 2 2˚ (c) [0, 2] Ans. (a)

1 - 2x is 4

(b) [–1, 1] È 1 3˘ (d) Í - , ˙ Î 2 2˚

Solution: The given function is defined if 1 - 2x -1 £ £ 1 i.e. if - 4 £ 1 - 2 x £ 4 , 4 5 -3 fi – 5 £ – 2x £ 3 fi £x≥ 2 2 Example 10: Which of the following functions is bounded (a) y = 1 – log10x (b) y = e–2x –1 (c) y = sin (2x + 1) (d) y = tan (4x + 1) Ans. (c) Solution: The range log10x is (–•, • ) so y = 1 – log10x is unbounded. y = e–2x is unbounded from below as x Æ – •, È -p p ˘ yÆ •. The range of sin–1x is Í , ˙ so y = sin–1(2x + 1) is Î 2 2˚ a bounded function. The range of tan x is R so y = tan (4x + 1) is unbounded Example 11: A function out of the following whose period is not p is (a) sin2x (b) cos2x (c) tan (2x + 3) (d) y = |sin x| Ans. (c) 1 Solution: y = sin2x = [1 - cos 2 x ] . Since period of 2 cos x is 2p so period of cos 2x is p. 1 y = 1 + cos2x = 1 + (1 + cos 2 x ) . The period of this is 2 again p. The period of tan x is p so period of tan (2p + 3) is p/2. If f(x) = |sin x| then f(x + p) = |sin (x + p)| = |–sin x| = Ásin x Á = f(x). Thus the period of f is p. Example 12: Which of the following functions is an odd function (a) y = x4 – 2 x2 (b) y = x –x2 3 5 (c) y = cos x (d) y = x - x + x 6 40 Ans. (d) Solution: If f(x) = x4 – 2x2 then f(–x) = (–x)4 –2(–x)2 = x4 – 2x2 = f(x). Hence f is an even function. If u(x) = x – x2 then u(–x) = x – x2 so u is neither even nor an odd function. p(x) = cos x, p(–x) = cos (–x) = cos x = p(x) so p is even x3 x 5 x3 x 5 + then s(–x) = - x = function. If s(x) = x 6 40 6 40 3 5 Ê x x ˆ -Á x + ˜ = – s(x), so s is an odd function. 6 40 ¯ Ë

1.12

Complete Mathematics—JEE Main

LEVEL 1 Straight Objective Type Questions Example 13: Let X = {x : x = n3 + 2n + 1, n Œ N} and Y = {x : x = 3n2 + 7, n Œ N} then (a) X « Y is a subset of {x : x = 3n + 5, n Œ N} (b) X « Y Õ {x : x = n2 + n + 1, n Œ N} (c) 34 Œ X « Y (d) none of these Ans. (c)

n

r, .

Solution: If n3 + 2n + 1 = 3n2 + 7 fi n3 – 3n2 + 2n – 6 = 0 fi (n – 3) (n2 + 2) = 0 fi n = 3 as n Œ N So, x = 3 ¥ 32 + 7 = 34 Œ X « Y. In (a) and (b) x π 34, for any n Œ N. Example 14: A, B, C are the sets of letters needed to spell the words STUDENT, PROGRESS and CONGRUENT, respectively. Then n (A » (B « C )) is equal to (a) 8 (b) 9 (c) 10 (d) 11 Ans. (b) Solution:

A= B= C= So A » (B « C) = = and n [(A » (B « C)]

{D, E, N, S, T, U ] {E, G, O, P, R, S } {C, E, G, N, O, R, T, U } {D, E, N, S, T, U } » {E, G, O, R} {D, E, G, O, N, R, S, T, U } = 9.

Example 15: Let A = {x : x is a prime factor of 240} B = {x : x is the sum of any two prime factors of 240}. Then (a) 5 œ A « B (b) 7 Œ A « B (c) 8 Œ A « B (d) 8 Œ A » B Ans. (d) Solution: 240 = 2 ¥ 3 ¥ 5 ¥ 8 So A = {2, 3, 5}, B = {5, 7, 8}. Clearly 8 Œ A » B. Example 16: A, B, C are three sets such that n(A) = 25, n(B) = 20, n(c) = 27, n(A « B) = 5, n(B « C) = 7 and A « C = f then n(A » B » C) is equal to (a) 60 (b) 65 (c) 67 (d) 72. Ans. (a)

Solution: A « C = f fi A « B « C = f. \ n(A » B » C) = n(A) + n(B) + n(C)– n(A « B) – n(B « C) – n(A « C) + n(A « B « C) = 25 + 20 + 27 – 5 – 7 – 0 + 0 = 60. Example 17: Let X = {(x,y,z) | x,y,z Œ N. x + y + z = 10, x < y < z} and Y = {(x,y,z) | x,y,z Œ N, y = |x – z|} then X « Y is equal to (a) {(2,3,5)} (b) {1,4,5} (c) {5,1,4} (d) {(2,3,5), (1,4,5)} Ans. (d) Solution: X = {(1,2,7), (1,3,6), (1,4,5), (2,3,5)}. Elements of X which belong to Y are (1,4,5) and (2,3,5) both so they belong to X « Y. Example 18: If A,B,C are three non-empty sets such that A « B = f, B « C = f, then (a) A = C (b) A à C (c) C à A (d) none of these Ans. (d) Solution: Let A = {1,2,3,4,5}, B = {6,7,8,9} and C = {11,12,13} which satisfy the given conditions but none of (a), (b) or (c). Example 19: Two finite sets have m and n elements respectively. The total number of subsets of first set is 56 more than the total number of subsets of the second set. The values of m and n respectively are (a) 7, 6 (b) 6, 3 (c) 5, 1 (d) 8, 7 Ans. (b) Solution: According to the given condition, we have 2m = 2n + 56 fi 2m –3– 2n –3 = 7 fi 2n –3 (2m – n – 1) = 7. Since 7 is a prime number so we must have n – 3 = 0 (clearly m π n). Thus n = 3. Therefore, 2m = 23 + 56 = 64 = 26 fi m = 6. Example 20: Among employee of a company taking vacations last years, 90% took vacations in the summer, 65% in the winter, 10% in the spring, 7% in the autumn, 55% in winter and summer, 8% in the spring and summer, 6% in the autumn and summer, 4% in the winter and spring, 4% in winter and autumn, 3% in the spring and autumn, 3% in the summer, winter and spring, 3% in the summer, winter and autumn, 2% in the summer, autumn and spring, and 2% in the winter, spring and autumn. Percentage of employee that took vacations during every season:

Sets, Relations and Functions 1.13

(a) 4 (c) 2 Ans. (c)

(b) 3 (d) 8

Solution: Suppose that number of employee taking vacations is 100. Su – set of employee taking leave in Summer W – set of employee taking leave in Winter Sp – set of employee taking leave in Spring A – set of employee taking leave in Autumn n(Su) = 90, n(W) = 65, n(Sp) = 10, n(A) = 7 n(W « Su) = 55, n(Sp « Su) = 8, n(A « Su) = 6 n(W « Sp) = 4, n(W « Au) = 4, n(Sp « A) = 3 n(Su « A) = 3, n(Su « W « A) = 3 n (Su « W « Sp) = 3, n(Su « A « Sp) = 2 n(W « Sp « A) = 2 n(Su « Sp « W « A) = n(Su) + n(Sp) + n(W) + n(A) – n(Su « Sp) – n(Sp « W) – n(W « A) – n(Su « A) – n(Su « W) – n(Sp « A) + n(Su « Sp « W) + n(Su « W « A) + n(W « A « Su) + n(Su « Sp « A) –n(Sp » Su » A » W) = 90 + 65 + 10 + 7 – 55 – 8 – 6 – 4 – 4 – 3 + 3 + 3 + 2 + 2 – 100 = 2 Example 21: If A = {1, 2, 3, 4}, B = {3, 4, 5} then the number of elements in (A » B) ¥ (A « B) ¥ (A D B) is (a) 5 (b) 30 (c) 10 (d) 4 Ans. (b) Solution: A » B = {1, 2, 3, 4, 5}. n(A » B) = 5 A « B = {3, 4}, n(A « B) = 2 A D B = (A ~ B) » (B ~ A) = {1, 2} » {5} = {1, 2, 5} n(A D B) = 3. Hence n((A » B) ¥ (A « B) ¥ (A D B)} = 5 ¥ 2 ¥ 3 = 30. Example 22: Let I be the set of integers. For a, b Œ I, a R b if and only if |a – b| < 1, then (a) R is not reflexive (b) R is not symmetric (c) R = {(a,a); a Œ I} (d) R is not an equivalence relation. Ans. (c) Solution: For any integers a, b, |a – b| < 1 if and only if |a – b| = 0 so a = b. Hence R = {(a,a); a Œ I}.Thus R is reflexive, symmetric and transitive. Example 23: Let W denote the words in the English Dictionary. Define the relation R by R = {(x, y) Œ W ¥ W : the words x and y have at least one letter common}, then R is

(a) (b) (c) (d) Ans. (c)

reflexive, not symmetric and transitive not reflexive, symmetric and transitive reflexive, symmetric and not transitive reflexive, symmetric and transitive

Solution: (x, x) Œ R V x Œ W as all letters in both are common. If (x, y) Œ R then x and y have a letter in common fi (y, x) Œ R. Next, let x = fix, y = six and z = son then(x, y) Œ R, (y, z) Œ R but (x, z) œ R So R is reflexive, symmetric but not transitive Example 24: If the relation R : A Æ B, where A = {1, 2, 3, 4} and B = {1, 3, 5} is defined by R = {(x, y); x < y, x Œ A, y Œ B} then RoR –1 is (a) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} (b) {(3, 1), (5, 1), (5, 2), (5, 3), (5, 4)} (c) {(3, 3), (3, 5), (5, 3), (5, 5) (d) none of these Ans. (c) Solution: R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} and R –1 = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}. Thus RoR –1 = {(3, 3), (3, 5), (5, 3), (5, 5)}. Example 25: Let A = {x Œ R : [x + 3] + [x + 4] £ 3} and x -3 • ÏÔ ¸Ô 3 ˆ xÊ B = Ìx ŒR : 3 Á  r ˜ < 3-3 x ˝ then Ë r =1 10 ¯ ÓÔ ˛Ô (a) A = B

(b) A Ã B

(c) B Ã A

(d) A « B = f

π

π

Ans. (a) Solution: Let x Œ A, [x + 3] + [x + 4] £ 3 fi [x] + 3 + [x] + 4 £ 3 fi 2[x] £ – 4 fi [x] £ –2 fi x Œ (– •,–1) A = (– •, –1) x

If x ΠB then 3 3

x–3 Ê

•

1 ˆ ÁË Â 10r ˜¯ r =1

fi

Ê 1 10 ˆ 32 x -3 Á Ë 1 - 1 10 ˜¯

fi

32 x -3 3-2

( )

x -3

x -3

< 3-3 x x -3

< 3-3 x

< 3-3 x

33 < 3–3x fi 3 < –3x so x Œ (– •, –1) Hence B = (– •, –1). Thus A = B. fi

Example 26: The range of the function f(x) = 7 – x Px – 3 is (a) {1, 2, 3, 4} (b) {1, 2, 3, 4, 5, 6} (c) {1, 2, 3} (d) {1, 2, 3, 4, 5} Ans. (c)

1.14

Complete Mathematics—JEE Main

Solution: 7 – x ≥ 1, x – 3 ≥ 0 and 7–x≥x–3 fi x £ 6, x ≥ 3, x £ 5. Thus 3£x£5 \ Range = {4P0, 3P1, 2P2} = {1, 3, 2} Example 27: The domain of the function f(x) =

sin -1 ( x - 3) 9 - x2

(a) [1, 2] (c) [1, 3) Ans. (b)

is

Solution: Since x – x + 2 = 2 which is an irrational number so x R x for all x Œ R. Hence R is reflexive. R is not symmetric as ( 2 , 1) Œ R but (1, 2 ) œ R. Again R is not transitive since ( 2 , 1) ŒR and (1, 2 2 ) ŒR but

( 2 , 2 2 ) œR. Example 32: If n Áq and A = {z ŒC : zn = 1}, B = {z : zq = 1} then (a) A = B (b) A « B = {1} (c) B Ã A (d) A Ã B π

(b) [2, 3) (d) [1, 2]

Solution: – 1 £ x – 3 £ 1 and 9 – x2 > 0 fi 2 £ x £ 4 and – 3 < x < 3. So domain of f is [2, 3). Example 28: The solution of 8x ∫ 6 (mod 14) is (a) [8], [6] (b) [6], [14] (c) [6], [13] (d) [8], [14], [16] where [a] = {a + 14 k : k Œ I} Ans. (c) Solution: We need to solve 14 Á(8x – 6) i.e., 8x – 6 = 14k, for k Œ I. The values 6 and 13 satisfy this equation, while 8, 14 and 16 do not. Example 29: Let R = {(x, y): x, y Œ A, x + y = 5}, where A = {1, 2, 3, 4, 5} then (a) R is not reflexive, symmetric and not transitive (b) R is an equivalence relation (c) R is reflexive, symmetric but not transitive (d) R is not reflexive, not symmetric but transitive Ans. (a) Solution: R = {(1, 4), (4, 1), (2, 3), (3, 2)}, so R is not reflexive as (1, 1) œ R. R is symmetric by definition and R is not transitive as (1, 4) ŒR, (4, 1) Œ R but (1, 1) œR. Example 30: Let R be a relation on a set A such that R = R –1 then R is (a) reflexive (b) symmetric (c) transitive (d) an equivalence relation Ans. (b) Solution: If (a, b) Œ R then (b, a) Œ R–1= R so R is symmetric. The relation in Example 29 satisfy R = R–1 but is neither reflexive nor transitive. Example 31: For x, y Œ R, define a relation R by x R y if and only if x – y + 2 is an irrational number. Then R is (a) reflexive (b) symmetric (c) transitive (d) none of these Ans. (a)

Ans. (d) Solution: q = p n for some p Œ N z q – 1 = (z n)p –1 = (z n – 1) (z(p – 1)n + ... + zn + 1) Every root of z n –1 is a root of zq – 1 and every root of (p–1)n ... z + + zn + 1 = 0 is also a root of zq – 1. Hence A à B π and A « B = A. Example 33: If A = {z : (1 + 2i) z + (1 – 2i) z + 2 = 0} and B = {z : (3 + 2i) z + (3 – 2i) z + 3 = 0}, then (a) A « B is a singleton set (b) A à B (c) B à A (d) A « B = f. Ans. (a) Solution: Equations in sets A and B represent straight a a line with a1 = 1 +2i and a 2 = 3 + 2i. Since 1 π 2 so the a1 a 2 lines are intersecting, hence A « B is a singleton set. Example 34: Let x, y ŒI and suppose that a relation R on I is defined by x R y if and only if x £ y then (a) R is reflexive but not symmetric (b) R is an equivalence relation (c) R is neither reflexive nor symmetric (d) R is symmetric but not transitive Ans. (a) Solution: Since x £ x for all x Œ I so R is reflexive but is not symmetric as (1, 2) Œ R and (2, 1) œ R. Also R is transitive as x £ y, y £ z fi x £ z. Example 35: If f : R Æ R is defined by f(x) = x2 + 1, then value of f –1 (17) and f –1 (–3) are, respectively, (a) f, {4, – 4} (b) f, {3, –3} (c) {3, –3}, f (d) {4, – 4}, f Ans. (d) f

–1

Solution: For any A Õ R, we have (A) = {x Œ R: f (x) Œ A}. Thus, f –1 (17) = {x : f (x) Œ {17}} = {x : f (x) = 17} = {x : x2 + 1 = 17} = {4, – 4}, and similarly, f –1 (– 3)= {x Œ R : x2 + 1 = – 3} = f.

Sets, Relations and Functions 1.15

Example 36: The functions f and g are given by f(x) = {x}, 1 the fractional part of x and g(x) = sin [x]p, where [x] 2 denotes the integral part of x. Then range of gof is (a) [–1, 1] (b) {0} (c) {–1, 1} (d) [0, 1] Ans. (b) Solution: (gof) (x) = g ( f(x)) = 1/2 sin [{x}]p = 0, for all x Œ R. Hence the range of gof is {0}. Example 37: The period of the function f (x) = cos2 3x + tan 4x is (a) p/3 (b) p/4 (c) p/6 (d) p Ans. (d) Solution: f (x) = (1/2) (1 + cos 6x) + tan 4x. The period of cos 6x is 2p/6 = p/3 and the period of tan 4x is p/4. Hence the period of f is l.c.m. of p/3 and p/4 = p. Example 38: The domain of the function Ê f (x) = sin–1 Á log3 Ë (a) [–1, 9] (c) [–9, 1] Ans. (b)

xˆ ˜ is 3¯ (b) [1, 9] (d) [3, 9]

Solution: The function f is defined only if - 1 £ log3 (x/3) £ 1. This inequality is possible only if 1/3 £ x/3 £ 3 i.e., 1 £ x £ 9. Example 39: The domain of the function f(x) =

- log0.3 ( x - 1) - x2 + 2 x + 8

(a) (1, 4) (c) (2, 4) Ans. (c)

is (b) (–2, 4) (d) none of these

Solution: Since for, 0 < a < 1, loga x < 0 for x > 1 so log0.3 (x - 1) < 0 for x > 2. Also - x2 + 2x + 8 > 0 if and only if x Œ (- 2, 4). Hence the domain of the given function is (2, 4). Example 40: The function f: [–1/2, 1/2] Æ [–p/2, p/2] defined by f(x) = sin–1 (3x – 4x3) is (a) both one-one and onto (b) neither one-one nor onto (c) onto but not one-one (d) one-one but not onto Ans. (a) Solution: Since sin–1 (3x – 4x3) = 3sin–1 x Œ [–p/2, p/2] i.e., sin–1 x Œ [–p/6, p/6] or x Œ [–1/2, 1/2] so f is onto. 3 > 0 for –1/2 < x < 1/2. Therefore, f Also f ¢( x ) = 1 - x2 increases on [–1/2, 1/2] and hence f is one-one.

1

Example 41: Given f(x) = 1

g(x) =

x-|x|

(a) (b) (c) (d) Ans. (a)

|x|- x

and

then

dom f π f and dom g = f dom f = f and dom g π f f and g have the same domain dom f = f and dom g = f

Solution: dom f = {x : | x | > x} and dom g = {x : x > | x |} = f. Thus dom f = R– (the set of all negative real numbers) and dom g = f. Example 42: Which of the following functions is not onto (a) f: R Æ R, f(x) = 3x + 4 (b) f: R Æ R+, f (x) = x2 + 2 (c) f: R+ Æ R+, f(x) = (d) none of these Ans. (b)

x

Solution: The function f (x) = 3x + 4 is onto as for y Œ Ê y-4ˆ = y. The function f : R+ Æ R+, f (x) = x is R, f Á Ë 3 ˜¯ onto as for y Œ R+, f (y2) = y. f : R Æ R+, f (x) = x2 + 2 is not onto as 1 Œ R+ has no pre-image. Example 43: Which of the following functions is not one-one (a) f: R Æ R, f(x) = 2x + 5 (b) f: [0, p] Æ [–1, 1], f(x) = cos x (c) f: [– p /2, p/2] Æ [1, 7] f (x) = 3 sin x + 4 (d) f : R Æ [–1, 1], f (x) = sin x Ans. (d) Solution: The function in (a) is one-one as 2x1 + 5 = 2x2 + 5 is possible only if x1 = x2. The function in (b) is one-one x - x2 as cos x1 = cos x2 if and only if sin 1 = 0 i.e., x1 = x2. 2 Similarly the function in (c) is also one-one. The function in (d) is not one-one as f (p) = f (2p) = 0. Example 44: Which of the following functions is nonperiodic (a) f(x) = x – [x] Ï1 if x is a rational number (b) f(x) = Ì Ó0 if x is an irrational number 8 8 + 1 + cos x 1 - cos x (d) none of these Ans. (d) (c) f(x) =

1.16

Complete Mathematics—JEE Main

Solution: The function in (a) is periodic with period 1 and the function in (b) is also periodic since f (x + r) = f (x) 4 for every rational r. The function in (c) is equal to sin x and thus has period p. ax , x π – 1. Then, for what Example 45: Let f (x) = x +1 value of a is f (f (x)) = x ? (b) - 2 (d) –1

(a) 2 (c) 1 Ans. (d) Solution: f (f (x)) =

a f ( x) a2x = f ( x ) + 1 (a + 1) x + 1

Thus f (f (x)) = x ¤ a2x = (a + 1)x2 + x ¤ (a2 – 1)x = (a + 1) x2 ¤ (a + 1) ((a – 1) – (a + 1)x) = 0 Since (a – 1) – (a + 1) x π 0 for all x so a = – 1. Example 46: Let R = {(x, y): x, y Œ R, x2 + y2 £ 25} 4 R¢ = ( x, y) : x, y Œ R, y ≥ x 2 then 9 (a) dom R « R¢ = [– 4, 4] (b) range R « R¢ = [0, 4] (c) range R « R¢ = [0, 5] (d) R « R¢ defines a function. Ans. (c)

{

}

Solution: The equation x2 + y2 = 25 represents a circle 4 2 with centre (0, 0) and radius 5 and the equation y = x 9 represents a parabola with vertex (0, 0) and focus (0, 1/9). Hence R « R¢ is the set of points indicated in the Fig.1.27 (-3, 4)

(0, 5)

(3, 4)

(3, 0)

Fig. 1.27

= {(x, y ): – 3 £ x £ 3,0 £ y £ 3]}. Thus dom R « R ¢ = [-3, 3] and range R « R¢ = [0, 5] … [0, 4] Since (0, 0) Œ R « R¢ and (0, 5) Œ R « R¢ \ 0 is related to 0 as well as 5. Hence R « R¢ doesn’t define a function. Example 47: In a factory 70% of the workers like oranges and 64% like apples. If x% like both oranges and apples, then

(a) x £ 34 (c) 34 £ x £ 64 Ans. (c)

(b) x ≥ 64 (d) none of these.

Solution: Let the total number of workers be 100. A, the set of workers who like oranges and B, the set of workers who like apples. So n(A) = 70, n(B) = 64, n ( A « B ) = x. n ( A » B ) £ 100. Also fi n(A) + n(B) – n ( A « B ) £ 100 fi 70 + 64 – x £ 100 fi x ≥ 34 n ( A « B ) £ n(B) fi x £ 64 Since Hence 34 £ x £ 64. Example 48: The Cartesian product of A ¥ A has 16 elements. S = {(a, b) Œ A ¥ A Áa < b}. (– 1, 2) and (0, 1) are two elements belonging to S. The remaining elements of S are given by. (a) {(–1, 0), (– 1, 1), (0, 2), (1, 2)} (b) {(–1, 0), (1, 1), (2, –1), (1, 2)} (c) {(0, –1), (1, – 1), (0, 2), (1, 2)} (d) none of these Ans. (a) Solution: ( -1, 2 ) Œ A ¥ A fi –1 Œ A, 2 Œ A and (0, 1) Œ A ¥ A fi 0 Œ A, 1 Œ A So, A = {– 1, 0, 1, 2} as A has four elements and S = {(– 1, 0 ), (–1, 1), (–1, 2), (0, 1), (0, 2), (1, 2)}. Hence the required element of S are given by (a) Example 49: If R and R¢ are two symmetric relations (not disjoint) on a set A, then the relation R « R¢ is (a) reflexive (b) symmetric (c) transitive (d) none of these. Ans. (b) Let fi fi fi

Solution: (a, b) Œ R « R¢ for some a, b Œ A. (a, b) Œ R and (a, b) Œ R¢ (b, a) Œ R and (b, a) Œ R¢ as R and R¢ are symmetric (b, a) Œ R « R¢ showing that R « R¢ is symmetric.

Example 50: a, b, g denote respectively the sets containing the letters in the names Apoorv, Mannan and Manvi of three children playing together. Which of the following is not correct. (a) n (a « g ) Án (a » b » g ) (b) n (b « g ) Án (a » b » g ) (c) n (a » b » g ) = 8 (d) n (a » b » g ) = n (a » g ) Ans. (b) a = {a, o, p, r, v}, b = {a, m, n} g = {a, i, m, n, v} a « b = {a}, a « g = {a, v}, b « g = {a, m, n} a » b »g = {a, i, m, n, o, p, r, v} = a » g.

Solution:

Sets, Relations and Functions 1.17

Example 51: If f : R Æ R, defined by f (x) = x3 + 7, then the value of f –1(71) and f –1 (–1) respectively are (a) {4}, f (b) f, {– 2} (c) {4}, {–2} (d) {2}, {– 4} Ans (c) Solution: f (x) = x3 + 7 = 71 fi x3 = 64 fi x = 4 fi

f –1 (71) = {4}

and f(x) = x3 + 7 = –1 fi x3 = – 8 fi x = –2 fi f –1(–1) = {–2} Example 52: If A = {1, 3, 5, 7, 9, 11, 13, 15, 17}, B = {2, 4, 6, 8, 10, 12, 14, 16, 18} and N, the set of natural numbers is the universal set, then A¢ » [(A » B) « B¢] is (a) A (b) A¢ (c) B (d) N. Ans. (d) Solution: (A » B) « B¢ = A as A « B = f fi A¢ » [(A » B) « B¢] = A¢ » A = N. Example 53: If X = {1, 2, 3, 4,}, then a one-one onto mapping f: X Æ X such that f(1) = 1, f(2) π 2 and f(4) π 4 is given by (a) {(1, 1), (2, 3), (3, 4), (4, 2)} (b) {(1, 1), (2, 4), (3, 1), (4, 2)} (c) {(1, 2), (2, 4), (3, 2), (4, 3)} (d) none of these Ans. (a) Solution: f in (a) is clearly one-one and onto also satisfies f(1) = 1, f(2) = 3 π 2, f(4) = 2 π 4. 2

Example 54: Let f(x) = (x + 1) – 1, x ≥ –1 then the set {x : f(x) = f –1(x)} is equal to (a) {0, – 1} (b) {0, 1} (c) {– 1, 1} (d) {0} Ans. (a) Solution: Let y = (x + 1)2 – 1, x ≥ – 1 fi (x + 1)2 = y + 1 fi

x+1=

fi Thus So fi fi fi

y + 1 as x ≥ – 1

x=–1+ f –1(x) = – 1 +

y +1 , y ≥ – 1 x +1

–1

f (x) = f (x) (x + 1) – 1 = – 1 + x + 1 2

x + 1 = 0 or (x + 1)3/2 = 1 x = – 1 or x = 0

Example 55: Let P = {q : sin q – cos q = 2 cos q} and Q = {q : sin q + cos q = 2 sin q } be two sets, then (a) P Ã Q and Q ~ P π f (b) Q À P (c) P À Q (d) P = Q. Ans. (d)

Solution: sin q – cos q =

2 cos q

¤

sinq = ( 2 + 1 ) cos q

¤

cosq = ( 2 - 1 ) sin q

¤ fi

sinq + cosq = P = Q.

2 sin q

Example 56: If A, B, C are three sets such that A « B = A « C and A » B = A » C, then (a) B = C (b) A « B = f (c) A = B (d) A = C Ans. (a) Solution: Let x Œ B fi x Œ A » B fi x Œ A » C fi x Œ A or x Œ C If x Œ A, then x Œ A « B = A « C fi x Œ C So x Œ B fi x Œ C. Similarly x Œ C fi x Œ B, Hence B = C. Example 57: The domain of the function sin -1 ( x - 3) f(x) = is 9 - x2 (a) [1, 2] (c) [1, 3] Ans. (b)

(b) [2, 3] (d) [1, 2]

Solution: x2 < 9 fi –3 < x < 3 and –1 £ x – 3 £ 1 fi 2 £ x < 3 Example 58: Let X = {1, 2, 3, 4, 5}. The number of different ordered pairs (Y, Z) that can be formed such that Y Õ X, Z Õ X, and Y « Z is empty is (b) 53 (a) 25 2 (c) 5 (d) 35 Ans. (d) Solution: For each x Œ X, we have three choices x Œ Y, x œ Z; x œ Y, x Œ Z; x œ Y, x œ Z So the required number of ordered pairs is 35. Example 59: Let X = {1, 2, 3, 4}. The number of equivalence relations that can be defined on X is (a) 10 (b) 15 (c) 16 (d) 8 Ans. (b) Solution: The number of equivalence relations k -1

Bk =

 ( kn-1 )Bn ;

B0 = 1, B1 = 1, B2 = 2, B3 = 5

n=0

B4 =

( 30 ) B0 + ( 13 ) B1 + ( 32 ) B2 + ( 33 ) B3

= 1 + 3 + 3 ¥ 2 + 1 ¥ 5 = 15 Example 60: The function f(x) = sin

px px - cos is (n + 1)! n!

1.18

Complete Mathematics—JEE Main

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

non periodic periodic with period 2(n!) periodic with period 2 (n + 1)! periodic with period n!

f(x – 1) + f(x + 1) =

f(x – 2) + f(x) =

Solution: Since the period of sin x is 2p so the period px px 2p n! of sin is = 2(n!). The period of cos is ( + 1)! n n! p 2(n + 1)! The period of f(x) = l.c.m. (2(n!), 2(n + 1)!) = 2(n + 1)! Ê1- xˆ = x 3 , x π –1, 1 and Example 61: If f (x) f Á Ë 1 + x ˜¯ f(x) π 0, then {f(–2)} (the fractional part of f(–2)) is equal to 2

(b) 1 3

(c) 1 2

(d) 0

... (i)

2 f ( x - 1)

... (ii)

2 f ( x + 1)

... (iii)

Replace x by x –1 in (i)

Ans. (c)

(a) 2 3

2 f ( x)

Replace x by x +1 in (ii) f(x) + f(x + 2) = Adding (ii) and (iii), we have 2 ( f(x – 1) + f(x + 1))

f(x – 2) + f(x + 2) + 2 f(x) = =

2 f(x) = 2f(x)

2

fi f(x –2) = – f (x + 2) Replacing x by x + 2, we have f(x) = –f(x + 4) = – ( f (x + 8)) = f(x + 8) 10

 f (3 + 8r )

= 11 ¥ f(3) = 11 ¥ 5 = 55.

r =0

Ans. (a) Solution: Replacing x by

Example 63:

1- x in the equation 1+ x

1- xˆ 3 ( f ( x ))2 f ÊÁË ˜ =x 1+ x¯

sin 2 q ...(i)

Let f(q) =

We have 2

Ê Ê1- xˆˆ Ê1- xˆ ÁË f ÁË ˜¯ ˜¯ f ( x ) = ÁË ˜ 1+ x¯ 1+ x

3

... (ii)

Solving (i) and (ii), we have 2

1- xˆ3 Ê x3 ˆ = ÊÁ f x ( ) ÁË ( f ( x ))2 ˜¯ Ë 1 + x ˜¯ ( f ( x ))3

fi

x6

Ê1+ xˆ = Á Ë 1 - x ˜¯

then f is (a) (b) (c) (d) Ans. (c)

2

sin q

1 + cos q

4 sin 4q

1 + sin 2 q

cos2 q

4 sin 4q

3

Solution: f(q) =

-1 -1

0 1

1 0

1 + sin 2 q

cos2 q

4 sin 2q

(R1 Æ R1 - R3

4 Ê -1ˆ f(–2) = 4 Á ˜ = Ë 3¯ 3 {f(–2)} = -

R2 Æ R2 - R3 )

4 È 4˘ 4 2 - Í- ˙ = - + 2 = . 3 Î 3˚ 3 3

Example 62: Let f(x) be a function such that f(x – 1) + 10

f(x + 1) =

2 f ( x ) for all x ΠR. If f(3) = 5 then

is equal to (a) 50 (c) 0

Â

r =0

(b) 55 (d) 10

Ans. (b) Solution: The given equation is

1 + 4 sin 4q

2

a non periodic function periodic with period p periodic with period p 2 odd function

Ê1+ xˆ f(x) = x 2 Á Ë 1 - x ˜¯

fi

cos2 q

f (3 + 8 r )

=

0 -1 2

1 + sin q + 4 sin 4q

0 1

1 0

cos2 q

4 sin 4q

(C1 Æ C1 + C3)

= – (cos2q + 1 + sin2q + 4 sin 4 q) = – 2 (1 + 2 sin 4 q) 2p p = . which is periodic function with period 4 2

Sets, Relations and Functions 1.19

Assertion-Reason Type Questions

Example 64: Let R be the real line. Consider the following subsets of the plane R ¥ R: S = {(x, y) : y = x + 1 and 0 < x < 2} T = {(x, y) : x – y is an integer} Statement-1 : T is an equivalence relation on R but S is not an equivalence relation on R. Statement-2 : S is neither reflexive nor symmetric but T is reflexive, symmetric and transitive. Ans. (a) Solution: Since x π x + 1, (x, x) œ S, so S is not reflexive. Next x, y Œ S, fi y = x + 1 fi x = y – 1 fi (y, x) œ S, so is not symmetric. Since x – x = 0 is an integer (x, x) Œ T " x Œ T fi T is reflexive. Again (x, y) Œ T fi x – y is an integer fi y – x is also an integer fi (y, x) Œ T So T is symmetric. Also (x, y) Œ T, (y, z) Œ T. fi x – y and y – z are integers fi x – z = (x – y) – (y – z) is also an integer fi (x, z) Œ T So T is Transitive. Which shows that statement-2 is true and hence statement-1 is also true. Example 65: Consider the following relations. R = {(x, y) | x, y are real numbers and x = wy for some rational number w} ÏÊ m p ˆ ¸ S = ÌÁ , ˜ ˝ m, n, p, q, are integer such that n◊q π 0 ÓË n q ¯ ˛ and qm = pn} Statement-1: S is an equivalence relation but R is not an equivalence relation. Statement-2: R and S both are symmetric. Ans. (c) Solution: Since (0, 1) Œ R but (1, 0) œ R, R is not symmetric and hence is not an equivalence relation so statement-2 is false. m p Next, For the relation S, qm = pn fi = n q m p Ê m pˆ Thus Á , ˜ Œ S fi which shows that S is = Ë n q¯ n q reflexive and symmetric Ê m pˆ Ê p rˆ Again, Á , ˜ ŒS and Á , ˜ Œ S Ë n q¯ Ë q s¯

m p r m r = = fi ÊÁ , Œ S ˆ˜ Ën s ¯ n q s Thus S is transitive and hence S is an equivalence relation. So, statement 1 is true. fi

Example 66: Let R be a relation on the set N of natural numbers defined by n Rm ¤ n is a factor of m (i.e. n | m). Statement-1: R is not an equivalence relation Statement-2: R is not symmetric Ans. (a) Solution: Statement-2 is true as 2 | 6 fi 2R6 but 6 does not divide 2 so R is not symmetric fi R is not an equivalence relation and the statement-1 is also true. Example 67: Let A = {1, 2, 3} and B = {3, 8} Statement-1: (A » B) ¥ (A « B) = {(1, 3), (2, 3), (3, 3), (8, 3)} Statement-2: (A ¥ B) « (B ¥ A) = {(3, 3)} Ans. (b) Solution: A » B = {1, 2, 3, 8}, A « B = {3} (A » B) ¥ (A « B) = {(1, 3), (2, 3), (3, 3), (8, 3)} Statement-1 is True. (x, y) Œ (A ¥ B) « (B ¥ A) fi (x, y) Œ A ¥ B and (x, y) Œ B ¥ A fi x Œ A « B, y Œ A « B fi {(3, 3)} = (A ¥ B) « (B ¥ A) fi Statement-2 is also true but is not a correct explanation for statement-1. fi fi

Example 68: Statement-1: The number of bijective functions from the set A containing 100 elements to itself is 2100. Statement-2: The total number of bijections from a set containing n elements to itself is n! Ans. (d) Solution: Statement-2 is true and so, statement-1 is False. Example 69: Statement-1: f : R Æ R is a function x-3 defined by f(x) = 5x + 3. If g = f –1, then g(x) = . 5 Statement-2: If f : A Æ B is a bijection and g : B Æ A is the inverse of f, then fog is the identity function on A. Ans. (c) y-3 Solution: Let y = 5x + 3 fi x = . 5 x-3 fi is the inverse of f, so statement-1 is True. g(x) = 5

1.20

Complete Mathematics—JEE Main

Statement-2 is false because g : B Æ A and f : A Æ B fi fog : B Æ B and g = f –1 fi fog is an identity function on B. Example 70: Let X and Y be two sets. Statement-1: X « (Y » X)¢ = f Statement-2: If X » Y has m elements and X « Y has n elements then symmetric difference X D Y has m – n elements Ans. (b) Solution: X « (Y » X)¢ = X « (Y ¢ « X ¢) = X « X ¢ « Y ¢ = f. fi Statement-1 is True. X D Y = (X ~ Y) » (Y ~ X) = (X » Y) ~ (X « Y) fi number of elements in X D Y = m – n. fi Statement-2 is True but does explain statement-1. Example 71: Let f be a function defined by f(x) = (x – 1)2 + 1, (x ≥ 1) Statement-1: The set {x : f(x) = f –1(x)} = {1, 2} Statement-2: f is a bijection and f –1(x) = 1 + x - 1 , x ≥ 1 Ans. (a) Solution: Let y = f(x) = (x – 1)2 + 1 fi y – 1 = (x – 1)2 fi x = 1 + y - 1 , y ≥ 1 Thus f –1(x) = 1 +

x - 1 , x ≥ 1. So statement-2 is true.

–1

Now f(x) = f (x) fi

(x – 1)2 =

x -1

fi x - 1 [( x - 1)3 / 2 - 1] = 0 x = 1, 2. fi So statement-1 is true and statement-2 is a correct explanation for statement-1. Example 72: Let R be the set of real numbers Statement-1: A = {(x, y) Œ R ¥ R : y + x is an integer} is an equivalence relation on R. Statment-2: B = {(x, y) Œ R ¥ R : y = a x for some rational number a} is not equivalence relation on R.

Ans. (d) Solution: A is neither reflexive nor transitive as x + x may not be integer " x ŒR and if x + y and y + z are integers, x + z may not be an integer for x, y, z ŒR. So statement-1 is false. Statememt-2 is true as B is not symmetric, because ( 3 , 0) Œ B as 0 =

3 ¥ 0 for a = 0, but (0,

3 ) œ B.

Example 73: 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-1: R is an equivalence relation. Statement-2: For any two invertible 3 ¥ 3 matrices M and N, (MN)–1 = N –1M –1. Ans. (b) Solution: Statement-2 in true (See Text.) In statement-1, A = I–1 A I for all real square matrices A of order 3. fi (A, A) Œ R fi R is reflexive, Next, let (A, B) Œ R fi $ a invertible matrix P of order 3. such that A = P –1 B P fi B = P A P–1 = (P –1)–1A (P –1) fi R in symmetric If Now (A, B) Œ R and (B, C) Œ R Then $ invertible matrices P and Q of order 3 such that A = P –1 B P and B = Q –1 C Q. fi A = P –1 Q –1 C Q P = (QP)–1 C QP (From statement-2) fi (A, C) Œ R and thus R in transitive. Hence R is an equivalence relation and the statement-1 in also true but statement-2 is not a correct explanation for it.

LEVEL 2 Straight Objective Type Questions Example 74: From 50 students taking examinations in Mathematics, Physics and Chemistry, 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at most 29 Mathematics and Chemistry and at most 20 Physics and Chemistry. The largest possible number that could have passed all three exams is (a) 10 (b) 12 (c) 9 (d) none of these. Ans. (d)

Solution: The given conditions can be expressed as n (M » P » C) = 50, n (M) = 37, n(P) = 24, n(C) = 43, n(M « P) £ 19, n(M « C) £ 29 and 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) fi

50 = 37 + 24 + 43 - n (M « P) - n(P « C)

Sets, Relations and Functions 1.21

- n(M « C) + n(M « P « C) fi

n (M « P « C) £ n (M « P) + n (M «C) + n (P « C) - 54. Therefore, the number of students that could have passed all three exams is at most 19 + 29 + 20 - 54 = 14. Example 75: Suppose A1, A2, º, A30 are thirty sets each having 5 elements and B1, B2 º, Bn are n sets each with 3 30

elements, let

∪

n

Ai =

i =1

∪

Bj = S and each element of S

j =1

belongs to exactly 10 of the Ai’s and exactly 9 of the Bj’s. Then n is equal to (a) 15 (b) 3 (c) 45 (d) none of these Ans. (c) 30 1 Solution: S = ∪ Ai, so n(S) = (5 ¥ 30) = 15 (since 10 i =1 element in the union S belongs to exactly 10 of the Ai’s). n

Again S = ∪ Bi so i =1

n (S) = 1/9(3 ¥ n) = n/3 = 15 fi n = 45 Example 76: Let R = {(3, 3), (6, 6), (9, 9) (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12}. The relation is (a) an equivalence relation. (b) reflexive and symmetric only. (c) reflexive and transitive only (d) reflexive only. Ans. (c) Solution: R is reflexive as (3, 3), (6, 6), (9, 9), (12, 12) Œ R. R is not symmetric as (6, 12) Œ R but (12, 6) œ R. R is transitive as the only pair which needs verification is (3, 6) and (6, 12) Œ R fi (3, 12) Œ R. Example 77: Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}, the relation R is (a) not symmetric (b) transitive (c) a function (d) reflexive Ans. (a) Solution: R is not symmetric as (2, 3) Œ R but (3, 2) œ R. It is not transitive as (1, 3), (3, 1) Œ R. but (1, 1) œ R so is not reflexive also. Again as (2, 4) and (2, 3) Œ R, it is not a function. Example 78: Let I be the set of integers, N the set of non-negative integers; Np the set of non-positive integers ; E is the set of even integers and P is set of prime numbers. Then

(a) N « Np = f (c) N D Np = I – {0} Ans. (c)

(b) I – N = Np (d) E « P = f.

Solution: N « Np = {0}, I – N = {..., – 2, – 1} N D Np = {N – Np) » (Np – N) = {1, 2, 3 - -} » { ..., – 3, – 2, – 1} = I ~ {0} and E « P = {2}. Example 79: If n(A) = n then n{(x, y, z); x, y, z Œ A, x π y, y π z, z π x}= (b) n(n – 1)2 (a) n3 2 (c) n (n – 2) (d) n3 – 3n2 + 2n Ans. (d) Solution: There are n choices for the first coordinate, n – 1 choices for second coordinate and n – 2 choices for the third coordinate, hence n ({(x,y, z); x, y, z Œ A, x π y π z}) = n(n – 1) (n – 2) = n3 – 3n2 + 2n. Example 80: If A = {x : x Œ I, – 2 £ x £ 2}, B = {x Œ I, 0 £ x £ 3}, C = {x: x Œ N, 1 £ x £ 2} and D = {x, y) Œ N × N; x + y = 8}. Then (a) n(A » (B » C) = 5 (b) n(D) = 6 (c) n(B » C) = 5 (d) none of these Ans. (d) Solution: A = {– 2, – 1, 0, 1, 2}, B = {0, 1, 2, 3}, C = {1, 2} so B » C = {0, 1, 2, 3}, A » (B » C) = {–2, –1, 0, 1, 2, 3} so n(A » (B » C) = 6, n (B » C) = 4 and D = {(1, 7), (2, 6), (3, 5), (4, 4), (5,3), (6, 2), (7, 1)} so n(D) = 7. Example 81: If A = {4n – 3n – 1| n Œ N} and B = {9n – 9: n Œ N}, then A»B is equal to (a) B (b) A (c) N (d) none of these Ans. (a) Solution: It can be shown by induction that 9 Á4n – 3n – 1 for every n Œ N. Thus A Õ B. Clearly A π B as 27 Œ B but 27 œ A. Thus A » B = B. Example 82: A and B are two sets having 3 and 4 elements respectively and having 2 elements in common. The number of relations which can be defined from A to B is (a) 25 (b) 210 – 1 12 (c) 2 (d) none of these. Ans. (c) Solution: The number of elements in A ¥ B is 12. Hence the number of subsets of A ¥ B is 212, which includes the empty set

1.22

Complete Mathematics—JEE Main

Example 83: If R and S are two symmetric relations then (a) R o S is a symmetric relation (b) S o R is a symmetric relation (c) R o S –1 is a symmetric relation (d) R o S is a symmetric relation if and only if RoS = SoR. Ans. (d) Solution: Since R and S are symmetric relations so R –1 = R and S –1 = S. But (R o S)–1 = S –1 o R –1 = S o R. Thus RoS is symmetric if and only if RoS = SoR. Example 84: Let A be the set of all determinants of order 3 with entries 0 or 1 only, B the subset of A consisting of all determinants with value 1, and C the subset consisting of all determinants with value –1. Then if n(B) and n(C) denote the number of elements in B and C, respectively, we have (a) C = f (b) n(B) = n(C) (c) A = B » C (d) n(B) = 2n(C) Ans. (b) Solution: C cannot be the empty set because, for instance, 0 1 0 1 0 1 – 1 = 1 0 0 Œ C. We also have 1 1 0 = 2, 0 0 1 0 11 so A π B » C. In general, the determinant a11 a12 a13 a21 a22 a23 = a31 a32 a33

a11a22 a33 + a12 a23 a31 + a13 a21a32 - a11a22 a32 - a12 a21a33 - a13 a31a22 ,

with the a’s being 0 or 1, equals 1 only if a11a22a33 = 1 and the remaining terms are zero; if a12 a23 a31 = 1 and the remaining terms are zero; or if a13 a21 a32 = 1 and the remaining terms are zero. Since there are three similar relations for determinants that equal - 1, we must have n (B) = n (C). Example 85: The domain of the function 1 ˆ ˆ Ê Ê f(x) = log2 Á - log1 / 2 Á 1 + 4 ˜ - 1˜ is Ë Ë x¯ ¯ (a) 0 < x < 1 (c) x ≥ 1 Ans. (a)

(b) 0 < x £ 1 (d) x > 1

Solution: For f to be defined we must have log1/2 1 1 ˆ Ê –1 –1 ÁË 1 + 4 ˜¯ < -1 ¤ 1 + 4 > (2 ) = 2 which is possible x x 1 if and only if 4 > 1 i.e., 0 < x < 1. x Hence the domain of the given function is {x : 0 < x < 1}. Example 86: The domain of definition of f(x) =

1 Ê x - 1ˆ log0. 4 Á ¥ 2 is Ë x + 5 ˜¯ x - 36

(a) (b) (c) (d) Ans. (c)

{x : x < 0, x π – 6} {x : x > 0, x π 1, x π 6} {x : x > 1, x π 6} {x : x ≥ 1, x π 6}

Ê x -1 ˆ log0 ◊ 4 Á to be defined, we must Ë x + 5 ˜¯ 1 x -1 have 0< < 1, which is true if x > 1. Morever, 2 x - 36 x+5 is defined for x π ± 6. Hence the domain of f is {x : x > 1, x π 6}. Solution: For

Example 87: The set of all x for which f(x) = log x - 2 2 x+3 1 are both not defined is and g(x) = x2 - 9 (a) (– 3, 2) (b) [– 3, 2) (c) (– 3, 2] (d) [–3, 2] Ans. (d) Solution: The function f is not defined for - 3 £ x £ 2 and g is not defined for those x for which x2 - 9 £ 0 i.e., x Œ [– 3, 3]. Thus f and g are not defined on [-3, 2]. Example 88: If f (x) is a polynomial satisfying f (x) f (1/x) = f (x) + f (1/x) and f (3) = 28, then f (4) is given by (a) 63 (b) 65 (c) 67 (d) 68 Ans. (b) Solution: Any polynomial satisfying the functional equation f (x) ◊ f (1/x) = f (x) + f (1/x) is of the form xn + 1 or - xn + 1. If 28 = f (3) = - 3n + 1 then 3n = - 27 which is not possible for any n. Hence 28 = f (3) = 3n + 1 fi 3n = 27 fi n = 3. Thus f (x) = x3 + 1, so f (4) = 43 + 1 = 65. Example 89: Part of the domain of the function cos x - 1 / 2

lying in the interval [–1, 6] is 6 + 35 x - 6 x 2 (a) [–1/6, p/3] » [5p/3, 6] (b) (–1/6, p/3] » [5p/3, 6) (c) (–1/6, 6) (d) none of these Ans. (a) f (x) =

Solution: The function f is meaningful only if cos x - 1/2 ≥ 0, 6 + 35 x - 6x2 > 0 or cos x - 1/2 £ 0, 6 + 35x - 6x2< 0 i.e., cos x ≥ 1/2, (6 - x) (1 + 6x) > 0 or cos x £ 1/2, (6 - x) (1 + 6x) < 0. These inequalities are satisfied if x Œ (-1/6, p/3] » [5p/3, 6). Example 90: Let f: R Æ R be a function defined by e| x| - e- x f (x) = x . Then e + e- x (a) f is both one-one and onto (b) f is one-one but not onto

Sets, Relations and Functions 1.23

(a) f(A « B) = f(A) « f(B) (b) f(A « B) = [0, 1]

(c) f is onto but not one-one (d) f is neither one-one nor onto Ans. (d) Solution: f is not one-one as f (0) = 0 and f (-1) = 0. f is also not onto as for y = 1 there is no x Œ R such that f (x) = 1. If there is such a x Œ R then e|x| – e–x = ex + e–x, clearly x π 0. For x > 0, this equation gives - e–x = e–x which is not possible. For x < 0, the above equation gives ex = - e–x which is also not possible. Example 91: Let f (x) = x2 and g(x) = 2x then the solution set of fog (x) = g o f (x) is (a) R (b) {0} (c) {0, 2} (d) none of these Ans. (c) x

x 2

2x

Solution: fog (x) = f (g (x)) = f (2 ) = (2 ) = 2 and gof x2

2

x2

2x

(x) = g (f (x)) = g (x ) = 2 . Thus the solution of 2 = 2 is given by x2 = 2x which is x = 0, 2. Example 92: A function f : R Æ R satisfies the equation f (x) f (y) – f (xy) = x + y for all x, y Œ R and f (1) > 0, then (a) f (x) = x + 1/2 (b) f (x) = (1/2)x + 1 (c) f (x) = (1/2)x – 1 (d) f(x) = x + 1 Ans. (d) Solution: Taking x = y = 1, we get f (1) f (1) – f (1) = 1 + 1 fi f (1)2 – f (1) – 2 = 0 fi (f (1) – 2) (f (1) + 1) = 0 fi f (1) = 2 (∵ f (1) > 0) Taking y = 1, we get f (x) f (1) – f (x) = x + 1 fi 2f (x) – f (x) = x + 1 fi f (x) = x + 1. Example 93: If f : [1, •) Æ [2, •) is given by f (x) = x + 1/x then f –1(x) equals (a)

(c) Ans. (a)

x + x2 - 4 2

(b)

x - x2 - 4 2

2 (d) 1 + x - 4

x 1 + x2

Solution: y = x + 1/x fi x2 – xy + 1 = 0 x=

y ± y2 - 4 2

x=

y + y2 - 4 . 2

f – 1(x) =

x + x2 - 4 . 2

fi Since x Œ [1, •) Hence

so

Example 94: Consider the function f = {(x, sin x) | – • < È p˘ È p˘ x < •}. Let A = Í0, ˙ and B = Í0, ˙ then Î 6˚ Î 2˚

(c) f(A) « f(B) = [0, 1] Ans. (a)

È 1˘ (d) f(A) » f(B) = Í0, ˙ Î 2˚

Solution: f(x) = sin x is an increasing function on 1 p [0, p / 2] so f(A) = ÈÍ0, sin ˘˙ = ÈÍ0, ˙˘ and f(B) = [0, 1]. 6˚ Î 2˚ Î 1 È È p˘ ˘ Thus f(A) « f(B) = Í0, ˙ . Also A « B = Í0, ˙ , so f(A « B) Î 2˚ Î 6˚ 1 p È 1˘ È ˘ ˘ È = Í0, sin ˙ = Í0, ˙ . Thus f(A) « f(B) = Í0, ˙ = f(A « B), 6 ˚ Î 2˚ Î 2˚ Î Also f(A) » f(B) = [0, 1]. Example 95: Let f(x) be a polynomial of even degree Ê Ê 1 ˆˆ satisfying f(2x) Á 1 - f Á ˜ ˜ + f 16 x 2 y = f(–2) – f(4xy) Ë Ë 2x ¯ ¯ for all x, y Œ R ~ {0} and f(4) = –255, f(0) = 1. Then the f (2 ) + 1 is value of 2 (a) 4 (b) 5 (c) 7 (d) 6 Ans. (c) 1 Solution: Replacing y by in the given functional 8x2 equation, we obtain Ê Ê 1 ˆˆ Ê 1ˆ f (2x) Á 1 - f Á ˜ ˜ + f (2) = f (–2) – f Á ˜ . Ë Ë 2x ¯ ¯ Ë 2x ¯

(

)

Since f is an even function so f (2) = f (–2), Ê 1ˆ Ê 1ˆ so f (2 x ) - f (2 x ) f Á ˜ = - f Á ˜ Ë 2x ¯ Ë 2x ¯ Ê 1ˆ Ê 1ˆ fi f (2 x ) + f Á ˜ = f (2 x ) f Á ˜ Ë 2x ¯ Ë 2x ¯ x Replacing x by , we have 2 Ê 1ˆ Ê 1ˆ f ( x) + f Á ˜ = f ( x) f Á ˜ Ë x¯ Ë x¯ n Since f is a polynomial, so f (x) = ± x + 1 (see Example 88) But –255 = f(4) = ± 4n + 1. Only negative sign is possible, thus 4n = 256 fi n = 4. i.e. f (x) = –x4 + 1. f (2) = –24 + 1 = –15 f (2 ) + 1 14 = = 7. 2 2 1 Example 96: The domain of the function f(x) = sin x x-5 + 3 sin x + log10 2 is x - 10 x + 24 (a) {(2 k p, (2k + 1)p) : k Œ I} (b) {(2 k p, (2k + 1)p ) : k Œ N} (c) (6, •) » (4, 5) (d) {(2 k p, (2 k + 1)p) : k Œ I} » (6, •) Ans. (b)

1.24

Complete Mathematics—JEE Main

Solution: The domain of 3 sin x is R. 1 = {x : sin x > 0} The domain of sin x = {(2kp, (2k + 1)p) : k ΠI} The domain of log10

x-5 2

x - 10 x + 24

= log10

x-5 ( x - 6)( x - 4)

= {x : x π 5, x > 5, x > 6, x > 4 or x π 5, x < 5, 4 < x < 6} = (4, 5) « (6, •) Thus the domain of f (x) = {(2 k p, (2k + 1)p) : k Œ I} « {(4, 5) » (6, •)} = {(2 k p, (2k + 1)p) : k Œ N} (since (4, 5) does not intersect {(2 k p, 2k + 1)p : k Œ I})

EXERCISE Concept-based Straight Objective Type Questions 1. Let U be a Universal set and n(U ) = 12. If A, B Õ U are such that n(B) = 6 and n(A « B) = 2 then n(A » B¢) is equal to (a) 6 (b) 10 (c) 7 (d) 8 2. Let R be a relation on R defined as a R b if |a| £ b. Then, relation R is (a) reflexive (b) symmetric (c) transitive (d) not antisymetric 3. Let f(x) = ax + b, x Œ R, and g(x) = x + d, x Œ R, then fog = gof if and only if (a) f(a) = g(c) (b) f(d) = g(b) (c) f(b) = g(d) (d) f(c) = g(a) sin x is 4. The domain of the function f(x) = |x|- x (a) R (c) R+

(b) R ~ {0} (d) R–

(R+ is the set of positive real numbers and R – is the set of negative real numbers) 5. The domain of y = cos–1 (1 – 2x) is (a) [– 1, 1] (b) [0, 1] 1 1 (c) [–1, 0] (d) È- , ˘ ÍÎ 2 2 ˙˚ 6. If f(x) = sin x – cos x is written as f1(x) + f2(x) where f1(x) is even and f2(x) is odd then (b) f1(x) = – cos x (a) f1(x) = cos x (c) f2(x) = – sin x + cos x (d) f2(x) = sin (2p – x) 7. The range of y = 1 – sin x is (a) [–1, 1] (b) [0, 1] (c) [–1, 2] (d) [0, 2] 1- x is 8. The function f(x) = x2 log 1+ x (a) a periodic function (c) an odd function

(b) a bounded function (d) an even function

LEVEL 1 Straight Objective Type Questions 9. If A = {2, 3, 4, 5, 7}, B = {1, 2, 4, 7, 9} then ((A ~ B) » (B ~ A)) « A is equal to (a) {3, 5} (b) {2, 4} (c) {3, 7} (d) {2, 7} 10. If A = {2x : x Œ N}, B = {3x : x Œ N} and C = {5x : x Œ N) then A « (B « C) is equal to (a) {15, 30, 45, ... } (b) {10, 20, 30 ... } (c) {30, 60, 90, ... } (d) {7, 14, 21 ... }. 11. If X and Y are two sets such that X « Y = X » Y, then

(a) X à Y, X π Y (b) Y à X, Y π X (d) none of these (c) X = Y 12. If X, Y and A are three sets such that A « X = A « Y and A » X = A » Y then (a) X à Y (b) Y à X (c) X = Y (d) none of these 13. If A = {x : x Œ R and satisfy x2 – 15x + 56 = 0} B= {x : x Œ N and x + 5 £ 14} and C = {x : x Œ N and x/112}. Then A » (B « C) is equal to (a) {1, 2, 3,....8} (b) {8, 7} (c) {1, 2, 8, 7} (d) {1, 2, 4, 7, 8}

Sets, Relations and Functions 1.25

14. If A is the set of letters needed to spell “MATHEMATICS” and B is the set of letters needed to spell STATISTICS, then (a) A Ã B (b) A ~ B = f (c) A D B = A ~ B (d) none of these 15. If A and B are two sets such that n(A » B) = 36, n(A « B) = 16 and n(A ~ B) = 15, then n(B) is equal to (a) 21 (b) 31 (c) 20 (d) 52 16. The maximum number of sets obtainable from A and B by applying union and difference operations is (a) 5 (b) 6 (c) 7 (d) 8 17. If A and B both contain same number of elements and are finite sets then (a) n(A » B) = n(A « B) (b) n(A ~ B) = n(B ~ A) (c) n(A D B) = n(B) (d) n(A ~ B) = n(A) 18. If A D B = A » B then (a) A = B (c) A D B = f

(b) A « B = f (d) A D B = A ~ B

19. In a class 60% passed their Physics examination and 58% passed in Mathematics. Atleast what percentage of students passed both their Physics and Mathematics examination? (a) 18% (b) 17% (c) 16% (d) 2% 20. If the relation R: A Æ B, where A = {1, 2, 3} and B = {1, 3, 5} is defined by R = {(x, y): x < y, x Œ A, y Œ B}, then (a) R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5)} (b) R = {(1, 1), (1, 5), (2, 3), (3, 5)} (c) R–1 = {(3, 1), (5, 1), (3, 2), (5, 3)} (d) R–1 = {(1, 1), (5, 1), (3, 2), (5, 3)} 21. The relation R defined on the set A = {1, 2, 3, 4, 5} by R = {(x, y): |x2 – y2| < 16} is given by (a) R1 = {(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)} (b) R2 = {(2, 2), (3, 2), (4, 2), (2, 4)} (c) R3 = {(3, 3), (4, 3), (5, 4), (3, 4) (d) none of these 22. Let L be the set of all lines in a plane and R be a relation on L defined by l1R l2 if and only if l1 ^ l2 then R is (a) reflexive (b) symmetric (c) transitive (d) an equivalence relation 23. For non-empty subsets A and B, (a) Any subset of A × B defines a function from A to B. (b) Any subset of A × B defines an equivalence relation

(c) Any subset of A × A defines a function on A (d) none of these. 24. Let f (x) = (x + 1)2 – 1 (x ≥ –1). Then the set S = {x : f (x) = f –1(x)} contains Ï -3 + i 3 -3 - i 3 ¸ , (a) Ì0, - 1, ˝ 2 2 Ó ˛ (b) {0, 1, –1} (c) {0, –1} (d) none of these 25. If a set A has n elements then the number of all relations on A is 2 2 (b) 2 n – 1 (a) 2 n n (c) 2 (d) none of these 26. The function f: R Æ R given by f (x) = 3 – 2 sin x is (b) onto (a) one-one (c) bijective (d) none of these 27. Which of the following are functions? (a) {(x, y): y2 = 4ax, x, y Œ R} (b) {(x, y): y = |x|, x, y Œ R} (c) {(x, y): x2 + y2 = 1, x, y Œ R} (d) {(x, y): x2 – y2 = 1, x, y Œ R} 28. If f: R Æ R defined by f (x) = x4 + 2 then the value of f –1 (83) and f –1(– 2) respectively are (a) f, {3, –3} (b) {3, –3}, f (c) {4, – 4}, f (d) {4, – 4}, {2, –2}. 29. The minimum number of elements that must be added to the relation R = {(1, 2), (2, 3)} on the sub set {1,2,3} of natural numbers so that it is an equivalence relation is (a) 4 (b) 7 (c) 6 (d) 5 30. Let X be a non-empty set and P(X) be the set of all subsets of X. For A, B Œ P(X), ARB if and only if A « B = f then the relation (a) R is reflexive (b) R is symmetric (c) R is transitive (d) R is an equivalence relation 31. Let f be a function satisfying 2f (x) – 3f (1/x) = x2 for any x π 0. Then the value of f (2) is (a) – 2 (b) – 7/4 (c) – 7/8 (d) 4 1 x ( x + 1) 32. If f (x) = 2x x( x - 1) ( x + 1) x 3 x( x - 1) x( x - 1)( x - 2) x( x - 1)( x + 1) then

1.26

Complete Mathematics—JEE Main

f (50) + f (51) + …… + f (99) is equal to (a) 0 (b) 1275 (c) 3725 (d) none of these 33. The domain of definition of the functions y(x) given by the equation ax + ay = a (a > 1) is (a) 0 < x £ 1 (b) 0 £ x £ 1 (c) – • < x < 1 (d) – • < x £ 0 x - [ x] , where [x] denotes the greatest 1 + x - [ x] integer less than or equal to x, then the range of f is (a) [0, 1/2] (b) [0, 1) (c) [0, 1/2) (d) [0, 1]

34. Let f (x) =

35. If 2f (x2) + 3 f (1/x2) = x2 – 1, then f (x2) is (b) (1 – x2)/5x (a) (1 – x4)/5x2 2 4 (c) 5x /(1 – x ) (d) none of these 36. If f (x + 3y, x – 3y) = 12xy, then f (x, y) is (a) 2xy (b) 2(x2 – y2) 2 2 (d) none of these (c) x – y 37. If the function f : [1, •) Æ [1, •) is defined by f (x) = 2x(x–1) then f –1(x) is (a) (1/2)x (x –1) (c)

1 1 - 1 + 4 log2 x 2

(

(b)

)

1 1 + 1 + 4 log2 x 2

(

)

(d) not defined

38. If n (A) = 3 and n (B) = 5 then number of one-one functions that can be defined from A to B is (a) 30 (b) 40 (c) 120 (d) 60 39. Let f: {x, y, z} Æ {1, 2, 3} be a one-one function. If it is given that exactly one of the following statements is true, Statement-1: f(x) = 1, Statement-2: f(y) π 1, Statement-3: f(z) π 2. then f –1 (1) is (a) x (b) y (c) z (d) none of these 40. The value of n Œ z for which the function f (x) = (a) 2 (c) 4

sin nx has 4p as its period is sin ( x / n) (b) 3 (d) 5

41. Let R be a relation on N defined by R = {(m, n): m, n ΠN and m = n2}.

Which of the following is true. (a) (n, n) Œ R . " n Œ N (b) (m, n) Œ R fi (n, m) Œ R (c) (m, n) Œ R, (n, p) Œ R fi (m, p) Œ R (d) none of these 42. Of the number of three athletic teams in a school, 21 are in the basketball team, 26 in hockey team and 29 in the football team, 14 play hockey and basketball, 15 play hockey and football, 12 play football and basketball and 8 play all the games. The total number of members is (a) 42 (b) 43 (c) 45 (d) none of these 43. E, I, R, O denote respectively the sets of all equilateral, isosceles, right angled and obtuse angled triangles in a plane, then which of the following is not true. (a) R « E = f, R « I π f (b) E « O = f, O « I π f (c) E « I = f, E « O π f (d) E « I π f, Eà I 44. If A = {3n : n Œ N, n £ 6}, B = {9n : n Œ N, n £ 4} then which of the following is false (a) A D B = {6561} (b) A ~ B = {3, 27, 243} (c) A « B = {9, 81, 729} (d) A » B = {3, 9, 27, 81, 243, 729, 6561} 45. If f : R Æ R given by f(x) = ax + sin x + a, then f is one-one and onto for all (a) a Œ R (b) a Œ R ~ [–1, 1] (c) a Œ R ~ {0} (d) a Œ R ~ {–1} n

46. If f: (0, p) Æ R is given by f(x) =

 [1 + sin kx ] , [x] k =1

denotes the greatest integer function, then the range of f(x) is (a) {n–1, n + 1} (b) {n} (c) {n, n + 1} (d) {n – 1, n} 47. If the number of elements in (A ~ B) ~ C, (B ~ C) ~ A, (C ~ A) ~ B and A « B « C is 10, 15, 20, and 5 respectively then the number of elements in (A D B) D C is (a) 35 (b) 50 (c) 40 (d) 45 x-3 , x π - 1. Then f 2010(2014) (where 48. Let f(x) = x +1 f n(x) = fof . . . of (x) (n times)) is (a) 2010 (c) 4028

(b) 4020 (d) 2014

Sets, Relations and Functions 1.27

Assertion-Reason Type Questions 49. Let R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} be a relation on set A = {1, 2, 3} Statement-1: R is not an equivalence relation on A. Statement-2: R is function from A to A. 50. Statement-1: If A is a set with 5 elements and B is a set containing 9 elements, then the number of injective mappings from A to B to 9! – 4!. Statement-2: The total number of injective mappings from a set with m elements to a set with n elements, n! m £ n, is . (n - m)! sin x . 51. Let f(x) = sin x + cos x, g(x) = 1 - cos x Statement-1: f is neither an odd function nor an even function.

Statement-2: g is an odd function. 52. Statement-1: A function f : R Æ R satisfies the equation f (x) – f (y) = x – y " x, y Œ R and f (3) = 2, then f (xy) = xy – 1 Statement-2: f (x) = f (1/x) " x Œ R, x π 0, and f(2) = 7/3 if f (x) =

x2 + x + 1 x2 - x + 1

.

53. Statement-1: Let A = {2, 3, 7, 9} and B = {4, 9, 49, 81} f : A Æ B is a function defined as f (x) = x2. Then f is a bijection from A to B. Statement-2: A function f from a set A to a set B is a bijection if f (A) = B and f (x1) π f(x2) if x1 π x2 for all x1, x2 Œ A and n(A) = n(B).

LEVEL 2 Straight Objective Type Questions 54. The range of the function f(x) = cos [x] for – p/2 < x < p/2 contains (a) {–1, 1, 0} (b) {cos 1, 1, cos 2} (c) {cos 1, – cos1, 1} (d) [– 1, 1]. 55. The domain of the function x-5 - 3 x + 5 is f (x) = log10 2 x - 10 x + 24 (a) (– 5, •) (b) (5, •) (c) (2, 5) » (5, •) (d) (4, 5) » (6, •) 56. If g (x) = 1 +

3

x then a function f such that 3

f (g(x)) = 3 – 3 ( x ) + x is (a) f (x) = x3 – 3x2 + x + 5 (b) f (x) = x3 + 3x2 – x – 5 (c) f (x) = x3 – 3x2 + 5 (d) f (x) = x3 – 3x2 + 3x + 3 x x 57. The function f (x) = x + + 1 is e -1 2 (a) even (b) periodic (c) odd (d) neither even nor odd x 58. Let f and g be two functions defined by f (x) = , x +1 x g (x) = . Then (f o g)–1 (x) is equal to 1- x

(a) x (b) 1 (c) 2x (d) none of these 59. Let f : (–1, 1) Æ (0, p) be defined by 2x Then f (x) = cot -1 1 - x2 (a) f is one-one but not onto (b) f is onto but not one-one (c) f is both one-one and onto (d) f is neither one-one nor onto 60. Let f : X Æ [1, 27] be a function by f (x) = 5 sin x + 12 cos x + 14. The set X so that f is one-one and onto is (a) [– p/2, p/2] (b) [0, p] (c) [0, p/2] (d) none of these 1- x . Then f o f (cos x) is equal to 61. Let f (x) = 1+ x (a) cos 2x (b) cos x (c) tan 2x (d) tan x 62. Let f (x) = sin x , then (a) f (x) is periodic with period

2p

(b) f (x) is periodic with period p (c) f (x) is periodic with period 4p2 (d) none of these

1.28

Complete Mathematics—JEE Main

63. Let X = Y = R ~ {1}. The function f : X Æ Y defined x+2 by f (x) = is x -1 (a) one-one but not onto (b) onto but not one-one (c) neither one-one nor onto (d) one-one and onto a x + a- x and f (x + y) + f(x – y) = K f (x) 2 f (y) then K is equal to (a) 2 (b) 4 (c) – 2 (d) none of these

64. If f (x) =

65. If f (x) = 2 sin–1 (a) [3, 4] (c) [– p/2, p/2]

x - 3 , then the domain of f is (b) [– 1, 1] (d) [6, 8]

66. Let f (x) = x |x| and g(x) =

| x | then the number of

elements in the set {x Œ R : f (x) = g (x)} is (a) 1 (b) 2 (c) 3 (d) infinitely many 67. Which of the following functions are odd functions Ê a x + a- x ˆ (a) f (x) = x Á x Ë a - a - x ˜¯ (b) f (x) =

(c) f (x) =

ax + x ax - x ax - 1 ax + 1

(

2 (d) f (x) = x log2 x + x + 1

)

68. Let g (x) = 1 + x – [x] and f (x) = sgn x. Then for all x, f o g (x) is equal to

(a) x (c) f (x) Note

(b) 1 (d) g (x)

f(x) = sgn x = 1 if x > 0 = 0 if x = 0 = – 1 if x < 0

69. The domain of definition of the function f (x) = log2 (log1/2 (x2 + 4x + 3)) + sin–1 (2[x]2 – 3), [x] denotes the greatest integer £ x is (a) 0 < x £ 1 (b) 0 £ x £ 1 (c) – • < x £ 0 (d) none of these 70. If g (f (x)) = |sin x| and f(g(x)) = (sin x )2, then f and g may be given by (a) f (x) = sin2 x, g(x) = x (b) f (x) = sin x, g (x) = |x| (c) f (x) = x2, g(x) = sin x (d) f and g cannot be determined 71. If f (x) = 3x – 5, then f –1(x) 1 (a) is given by 3x - 5 x+5 (b) is given by 3 (c) does not exist because f is not one-one (d) does not exist because f is not onto 72. Let f (x) = Áx – a Á, a π 0 then (a) f (x2) = (f (x))2 (b) f ( Áx Á) = Áf (x) Á (c) f (x + y) = f (x) + f (y) (d) none of these 73. If n (A) = n (B) = 4 then number of bijections from A to B is (a) 6 (b) 24 (c) 12 (d) 18

Previous Years' AIEEE/JEE Main Questions 1. Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is (a) not symmetric (b) transitive (c) a function (d) reflexive [2004] 2. The range of the function is f(x) = 7 – xPx – 3 is (a) {1, 2, 3, 4} (b) {1, 2, 3, 4, 5, 6} (c) {1, 2, 3} (d) {1, 2, 3, 4, 5} [2004]

3. The domain of the function f(x) = (a) [1, 2] (c) [2, 3]

sin -1 ( x - 3)

(b) [2, 3] (d) [1, 2]

4. If f : R Æ S f(x) = sin x – onto, then the interval of S is (a) [0, 1] (b) [–1, 1] (c) [0, 3] (d) [–1, 3]

9 - x2

is

[2004] 3 cosx + 1 is

[2004]

Sets, Relations and Functions 1.29

5. The graph of the function y = f(x) is symmetrical about x = 2 then (a) f (x) = f (–x) (b) f (2 + x) = f(2 – x) (c) f (x + 2) = f (x – 2) (d) f (x)= –f(–x) [2004] 6. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12} The relation is (a) an equivalence relation (b) reflexive and symmetric only (c) reflexive and transitive only (d) reflexive only [2005] 7. A real valued function f (x) satisfies the functional equation f (x – y) = f (x) f(y) – f(a – x) f (a + y) where a is a given constant and f(0) = 1, f (2a – x) is equal to (a) f (a) + f (a – x) (b) f (–x) (c) –f (x) (d) f (x) [2005] 8. Let W denote the words in the English Dictionary. Define the relation R by R = {(x, y) Œ W ¥ W : the words x and y have at least one letter in common}, then R is (a) reflexive, not symmetric and transitive. (b) not reflexive, symmetric and transitive. (c) reflexive, symmetric and not transition. (d) reflexive, symmetric and transitive [2006] Ê p pˆ 9. The largest interval lying in Ë- , ¯ for which the 2 2 2 Êx ˆ function f(x) = 4–x + cos– 1 Ë - 1¯ + log cos x is 2 defined is

Ê pˆ (a) Ë0, ¯ 2

Ê p pˆ (b) Ë- , ¯ 2 2

È p pˆ (c) Í- , ˜ Î 4 2¯

È pˆ (d) Í0, ˜ Î 2¯

[2007]

10. Let R be the real line. Consider the following subsets of the plane R ¥ R: S = {(x, y): y = x + 1 and 0 < x < 2} T = {(x, y): x – y is an integer} which of the following is true : (a) T is an equivalence relation on R but S is not (b) Neither S nor T is an equivalence relation. (c) Both S and T are equivalence relation on R. (d) S is an equivalence relation but T is not. [2008] 11. Let f : N Æ Y be a function defined as f (x) = 4x + 3 where Y = {y Œ N : y = 4x + 3 for some x Œ N}.

Show that f is invertible and its inverse is 3y + 4 y+3 (b) g(y) = 4 + (a) g(y) = 4 3 y+3 y-3 (c) g(y) = (d) g(y) = [2008] 4 4 12. If A, B and C are three sets such that A « B = A « C and A » B = A » C, then (a) B = C (b) A « B = f (c) A = B (d) A = C [2009] 13. Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational number w} ÏÊ m p ˆ ¸ S = ÌÁ , ˜ ˝ {m, n, p, q are integers such that ÓË n q ¯ ˛ n◊q π 0 and qm = pn} Then (a) S is an equivalence relation but R is not an equivalence relation (b) R and S both are equivalence relations. (c) R is an equivalence relation but S is not an equivalence relation. (d) Neither R nor S is an equivalence relation. [2010] 14. Let R be the set of real numbers. Statement-1: A = { (x,y) Œ R ¥ R : y – x is an integer} is an equivalence relation on R. Statement-2: B = { (x, y) Œ R ¥ R : x = a y for some rational numbers a} in an equivalence relation on R. [2011] 15. Let f be a function defined by f (x) = (x – 1)2 + 1, (x ≥ 1) Statement-1: The set {x : f (x) = f –1 (x)} = {1, 2} Statement-2: f is a bijection and f –1(x) = 1+

x - 1 , x ≥ 1.

[2011]

16. Consider the following relation R on the set of real square matrices of order 3. R = {(A,B): A = P –1 B P for some invertible matrix P} Statement-1: R is an equivalence relation. Statement-2: For any two invertible 3 ¥ 3 matrices M and N, (M N)–1 = N –1M –1 [2011] 17. Let X = {1, 2, 3, 4, 5}. The number of different ordered pairs (Y, Z) that can be formed such that Y Õ X, Z Õ X and Y « Z is empty is

1.30

Complete Mathematics—JEE Main

(a) 25 (c) 52

(b) 53 (d) 35.

[2012]

18. Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A ¥ B having 3 or more elements is (a) 211 (b) 256 (c) 220 (d) 219 [2013] 19. Let P be the relation defined on the set of all real numbers such that P = {(a, b) : sec2a – tan2b = 1}. Then P is (a) reflexive and symmetric but not transitive. (b) reflexive and transitive but not symmetric. (c) symmetric and transitive but not reflexive. (d) an equivalence relation. [2014] 1 3 n È ˘ 20. Let f(n) = Í + n, where [n] denotes the greatest Î 3 100 ˙˚ 56 integer less than or equal to n. Then to (a) 689 (b) 1399 (c) 1287 (d) 56

 f ( n)

is equal

n =1

[2014]

21. The function f(x) = |sin 4x| + |cos 2x| is a periodic function with period (a) p/2 (b) 2p (c) p (c) p/4 [2014] | x | -1 22. Let f : R Æ R be defined by f(x) = then f is | x | +1 (a) onto but not one-one (b) both one-one and onto (c) one-one but not onto (d) neither one-one nor onto [2014] 23. A relation on the set A ={x: |x|< 3, x Œ Z}, where Z is the set of integers is defined by R = {(x, y) : y = |x|, x π –1}. Then the number of elements in the power set of R is (a) 32 (b) 16 (c) 8 (d) 64 [2014] 24. Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A ¥ B, each having atleast three elements is (a) 219 (b) 256 (c) 275 (d) 510 [2015]

25. In a certain town, 25% of the families own a phone and 15% own a car, 65% families own neither a phone nor a car and 2000 families own both a car and a phone. Consider the following three statements: (a) 5% families own both a car and a phone (b) 35% families own either a car or a phone (c) 40,000 families live in town. Then, (a) only (a) and (b) are correct (b) only (a) and (c) are correct (c) only (b) and (c) are correct (d) all (a), (b) and (c) are correct [2015, online] 26. Let A = {x1, x2, … x7} and B = {y1, y2, y3} be two sets containing seven and three distinct elements respectively. Then the total number of functions f : A Æ B that are onto, if there exist exactly three x in A such f(x) = y2 is equal to (b) 16.7C3 (a) 14.7C2 7 (c) 12. C2 (d) 14.7C3 [2015, online] Ê 1ˆ 27. If f(x) + 2f Á ˜ = 3x, x π 0 and S = {x Œ R : f(x) = Ë x¯ f(–x)}; then S (a) is an empty set (b) contains exactly one element (c) contains exactly two elements (d) contains more than two elements [2016] 28. If the function f : [1, •[ Æ [1, •[ is defined by f(x) = 3x(x – 1), then f–1(x) is 1 1 (a) 1 - 1 + 4 log3 x 1 + 1 + 4 log3 x (b) 2 2

(

(c) not defined

)

(

1 x ( x -1) 2(d) ÊÁ ˆ˜ Ë 3¯

)

[2016]

1 and fn+1 1- x (x) = f0(fn(x)), n = 0, 1, 2, … . Then the f100(3) + f1 Ê 2 ˆ + f Ê 3 ˆ is equal to 2Á Ë 2 ˜¯ Ë 3¯ 8 4 (a) (b) 3 3 5 1 (c) (d) [2016, online] 3 3

29. For x Œ R, x π 0, x π 1, let f0(x) =

Previous Years' B-Architecture Entrance Examination Questions 1. The domain of the function f(x) =

2 x - 3 + sin x +

x - 1 is

(a) (– •, 1] È3 ˆ (c) Í , •˜ Î2 ¯

(b) [0, 1] (d) [1, •]

[2006]

Sets, Relations and Functions 1.31

2. Let f : (1, •) Æ (1, •) be defined by f(x) = Then (a) f is 1 – 1 and onto (b) f is 1 – 1 but not onto (c) f is not 1 – 1 but onto (d) f is neither 1 – 1nor onto

x+2 . x -1

[2008] 2

2

3. Let A = {(x, y) : x > 0, y > 0, x + y = 1} and let B = {(x, y) : x > 0, y > 0, x6 + y6 = 1} Then A « B (a) A (b) B (c) f (d) {(0, 1), (1, 0)} [2008] 4. A school awarded 38 medals in football, 15 in basketball and 20 in cricket. Suppose these medals went to a total of 58 students and only three students got medals in all three sports. If only 5 students got medals in football and basketball, then the number of medals received in exactly two of three sports is (a) 7 (b) 9 (c) 11 (d) 13 [2008] 5. Let Q be the set of all rational numbers and R be the relation defined as R = {(x, y) : 1 + xy > 0, x, y Œ Q} Then relation R is (a) symmetric and transitive (b) reflexive and transitive (c) an equivalence relation (d) reflexive and symmetric 6. The domain of the function f(x) = (a) (– •, 30) (c) (3, 30) » (30, •)

[2009] 1 3 - log3 ( x - 3)

is

(b) (– •, 30) » (30, •) (d) (4, •) [2009]

7. Let f : R Æ R be a function defined by f(x) = x2009 + 2009x + 2009 Then f(x) is (a) one-one but not onto (b) not one-one but onto (c) neither one-one nor onto (d) one-one and onto [2010] È p p˘ 8. Let f be a function defined on Í- , ˙ by Î 2 2˚ 4 3 f(x) = 3 cos x – 6 cos x – 6 cos2 x – 3 Then the range of f(x) is (a) [–12, –3] (b) [–6, –3] (c) [–6, 3) (d) (–12, 3] 2 –x2

9. Statement-1: The function f(x) = x e

10. Consider the following relations R1 = {(x, y) : x and y are integers and x = ay or y = ax for some integer a} R2 = {(x, y): x and y are integres and ax + by = 1 for some integers a, b} Then (a) R1, R2 are not equivalence relations (b) R1, R2 are equivalence relation (c) R1 is an equivalence relation but R2 is not (d) R2 is an equivalence relation but R1 is not [2012] 1 11. Let f and g be functions defined by f(x) = xŒ x +1 R, x π – 1 and g(x) = x2 + 1, x Œ R. Then go f is (a) one-one but not onto (b) onto but not one-one (c) both one-one and onto (d) neither one-one nor onto [2013] 12. Let N be the set of natural number and for a Œ N, aN denotes the set {ax : x Œ N}. If bN « cN, = dN, where b, c, d are natural numbers greater than 1 and the greatest common divisor of b and c is 1 then d equals (a) max{b, c} (b) min{b, c} (c) bc (d) b + c [2014] 13. Let f(x) = (x + 1)2 – 1, x ≥ – 1, then the set {x : f(x) = f – 1 (x)}: (a) is an empty set (b) contains exactly one element (c) contains exactly two elements (d) contains more than two elements [2014] 14. Let f : R Æ R e|x| - e - x , then f is e x + e- x (a) one-one and onto (b) one-one but not onto (c) onto but not one-one (d) neither onto nor one-one [2015]

f(x) =

15. If f is a function of real variable x satisfying f(x + 4) – f(x + 2) + f(x) = 0, then f is periodic with period (a) 8 (b) 10 (c) 12 (d) 6 [2016]

Answers Concept-based

[2010]

1. (d)

2. (c)

3. (b)

4. (d)

sin|x| is even

5. (b)

6. (b)

7. (d)

8. (c)

10. (c)

11. (c)

12. (c)

Statement-2: Product of two odd function is an even function [2011]

Level 1 9. (a)

1.32

Complete Mathematics—JEE Main

13. (d)

14. (c)

15. (a)

16. (d)

17. (b)

18. (b)

19. (a)

20. (a)

21. (d)

22. (b)

23. (d)

24. (c)

25. (a)

26. (d)

27. (b)

28. (b)

29. (b)

30. (b)

31. (b)

32. (a)

33. (c)

34. (c)

35. (d)

36. (c)

37. (b)

38. (d)

39. (b)

40. (a)

41. (d)

42. (b)

43. (c)

44. (a)

45. (b)

46. (c)

47. (b)

48. (d)

49. (c)

50. (d)

51. (b)

52. (b)

3.

4.

5.

53. (a)

Level 2

6.

54. (b)

55. (d)

56. (c)

57. (a)

58. (a)

59. (c)

60. (d)

61. (b)

62. (d)

63. (d)

64. (a)

65. (a)

66. (b)

67. (c)

68. (b)

69. (d)

70. (a)

71. (b)

72. (d)

73. (b)

Previous Years' AIEEE/JEE Main Questions 1. (a)

2. (c)

3. (b)

4. (d)

5. (b)

6. (c)

7. (b)

8. (c)

9. (d)

10. (a)

11. (d)

12. (a)

13. (a)

14. (c)

15. (a)

16. (b)

17. (d)

18. (d)

19. (c)

20. (b)

21. (a)

22. (d)

23. (b)

24. (a)

25. (d)

26. (d)

27. (c)

28. (b)

29. (c)

Previous Years' B-Architecture Entrance Examination Questions 1. 5. 9. 13.

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

2. 6. 10. 14.

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

3. 7. 11. 15.

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

so R is not symmetric. If |a| £ b and |b| £ c then |a| £ b £ |b| £ c so |a| £ c, hence a is related to c. Thus R is transitive. If |a| £ b, and |b| £ a then a £ |a| £ b £ |b| £ a. So a = b, R is anti symmetric. f(g(x)) = g(f(x)) for all x Œ R ¤ f(cx + d) = g(ax + b) ¤ a(cx + d) + b = c(ax + b) + d ¤ ad + b = cb +d ¤ f(d) = g(b) For f(x) to be defined, we have |x| > x. Since |x| ≥ x " x Œ R and |x| = x for x Œ [0, •) so |x| > x if x Œ R = (– •, 0) The function is defined if – 1 £ 1 – 2x £ 1 ¤ – 2 £ – 2x £ 0 ¤ 0 £ x £ 1. f ( x ) + f (- x ) 1 f1(x) = = [sinx – cosx – sinx – cosx] 2 2 = – cosx f ( x ) - f (- x ) 1 = [sinx – cosx + sinx + cosx] f2(x) = 2 2 = sinx

7. – 1 £ sinx £ 1 ¤ – 1 £ – sin x £ 1 ¤ 0 £ 1 – sinx £2 -1 1+ x Ê1 - xˆ 8. f(– x) = (– x)2 log = x2 log Á = Ë 1 + x ˜¯ 1- x 1- x = – f(x). – x2 log 1+ x

Level 1 9. ((A ~ B) » (B ~ A)) « A = ((A ~ B) « A) » (B ~ A) « A = (A ~ B) » f = (A ~ B) = {3, 5} 10. (A « B) « C = {2 ¥ 3 ¥ 5 x : x Œ N} = {30x : x Œ N} 11. X à Y fi X « Y = X, X » Y = Y Y à X fi X « Y = Y, X » Y = X. So X « Y = X » Y fi X = Y 12. If X à Y, let y Œ Y and y œ X then either y œ A. -/

4. (b) 8. (a) 12. (c)

Hints and Solutions Concept-based 1. n(A » B¢) = n(A) + n(B¢) – n(A « B¢) = n(A) + n(U) – n(B) – (n(A) – n(A « B)) = n(U) – n(B) + n(A « B) = 12 – 6 + 2 = 8. 2. If a = – 1 than a is not related to a. So R is not reflexive. If a = – 1, b = 2 then |a| £ b but |b| > a,

or y Œ A So if y Œ A, then y Œ A « Y but y œ A « X and thus A « X π A « Y. If y œ A, then y Œ A » Y but y œ A » X and thus A » X π A » Y So X Õ Y, similarly Y Õ X. fi X = Y. 13. A = {7, 8}, B = {1, 2, 3, .... 9} C = {1, 2, 4, 7, 8, 14, 16, ....} B « C = {1, 2, 4, 7, 8} A » (B « C) = {1, 2, 4, 7, 8} 14. A = {M, A, T, H, E, I, C, S} B = {S, T, A, I, C} B ~ A = f, So A D B = (A ~ B) » (B ~ A) = A ~ B. Verify (a) and (b) are not correct.

Sets, Relations and Functions 1.33

15. n(A ~ B) = n(A) – n(A « B) fi 15 = n(A) – 16 fi n(A) = 31 n(A » B) = n(A) + n(B) – n(A « B) fi 36 = 31 + n(B) – 16 fi n(B) = 21. 16. A » B, A ~ B, B ~ A (A » B) ~ (A ~ B) = B « (A » B) ~ (B ~ A) = A A D B = (A ~ B) » (B ~ A) (A » B) ~ (A D B) = A « B and (A ~ B) ~ A = f Thus, the required no is 8. 17. n(A ~ B) = n(A) – n(A « B) = n(B) – n(A « B) = n(B ~ A) 18. A » B = A D B » (A « B) So A » B = A D B fi A « B = f. 19. n(P) = 60, n(M) = 58, n(P » M) = 100, N(P « M) = n(P) + M(M) – n(P » M) 20. R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5)} 21. (1, 2) Œ R but (1, 2) œ R1 or R2 or R3 so R π R1 or R2 or R3. 22. L1 is not ^ L1, so R is not reflexive L1 ^ L2 fi L2 ^ L1 fi R is symmetric. L1 ^ L2 and L2 ^ L3 fi L1 || L3 fi R is not transitive. 23. See the definition of function 24. f (x) = f –1(x) fi f (f (x)) = x. fi [(x + 1)2 – 1 + 1]2 – 1 = x. fi (x + 1)2 = x + 1 fi x = 0 or x = –1 So S = {0, – 1}. 25. A ¥ A has n ¥ n = n2 elements and any subset of A ¥ A is a relation on A. The number of such subsets 2 is 2 n . 26. – 1 £ sin x £ 1 fi 1 £ f(x) £ 5 fi f(x) is not onto, f (x) = f (x + 2p) fi f is not one-one and f is not bijective. 27. In (a) a Æ 2a and a Æ – 2a, so it is not a function In (b) for each x Œ R there is unique | x | = y Œ R, so it is a function. 28. f(x) = x4 + 2 = y fi x = (y – 2)1/4. fi f –1(x) = (x – 2)1/4 fi f –1 (83) = (81)1/4 = ± 3 and f –1(– 2) = (– 4)1/4 = f. 29. R is reflexive if (x, x). Œ R for all x Œ {1, 2, 3} fi (1, 1), (2, 2), (3, 3) Œ R R is symmetric if (1, 2), (2, 3) ŒR fi (2, 1), (3, 2) ŒR. R is transitive if (1, 2), (2, 3) ŒR fi (1, 3) ŒR also (3, 1) ŒR as R is symmetric. So the total numbers of elements is 9. 30. A « A = A π f if A π f, So R is not reflexive. A«B=f fi B«A=f ARB fi BRA fi R is symmetric Note R is not transitive and not an equivalence relation.

31. 2f(2) – 3f(1/2) = 22 = 4 2f(1/2) – 3f(2) = (1/2)2 = 1/4 fi 5(f(2) – f(1/2)) = 4 – 1/4 = 15/4 fi f(1/2) = f (2) – 3/4. So 2f (2) – 3[f(2) – 3/4] = 4 fi

f(2) =

7 9 -4=- . 4 4

32. Applying C3 Æ C3 – (C1 + C2) we get f(x) = 0 for all x. 33. ax–1 + ay–1 = 1 fi ay–1 = 1 – ax–1 fi (y – 1)log a = log(1–ax–1) R.H.S is defined if 1 – ax–1 > 0 fi ax–1 < 1 = a° fi x – 1 < 0 as a > 1 fi x < 1 and the required domain is – • < x < 1. 34. If x is an integer, f(x) = 0 p where p and q are q p 1 positive integers and p < q, so that f(x) = < p+q 2

if x is not an integer x – [x] =

1 1 and the required range is [0, ] 2 2 35. 2f(x2) + 3f(1/x2) = x2 – 1 fi 2f(1/x2) + 3f(x2) = 1/x2 – 1 Subtracting, f(x2) – f(1/x2) = 1/x2 – x2 fi f(x2) = 1/x2. 36. f(x + 3y, x – 3y) = 12xy = (x + 3y)2 – (x – 3y)2 fi f(x, y) = x2 – y2. 37. Let f(x) = y = 2x(x – 1) fi log y = x(x – 1) log2 fi x2 – x – log2y = 0 so 0 £ f(x)
1, f¢(x) > 0 i.e. f is increasing If a < –1, f¢(x) < 0 i.e. f is decreasing Thus f is monotonic if a Œ R ~ [–1, 1]. n

46. f(x) =

 1 + [sin kx] =

n + [sin x] + [sin 2x] + ...

k =1

p + [sin nx]. If kx π for any k = 1, 2 ... n then 0 2 < kx < p and k x π p/2 so 0 < sin kx < 1. Hence p [sin kx] = 0, k = 1, 2 ... n. i.e. f(x) = n. If kx = 2 p for some k then x = , hence sin x, sin 2 x, ..., 2k sin(k–1)x will lie between 0 and 1 so [sin j x] = 0 1 £ j £ k–1; sin k x = 1 so f(x) can be n + 1 or n. 47. (A D B) D C is disjoint union of (A ~ B) ~ C, (B ~ C) ~ A, (C ~ A) ~ B and A « B « C. Threfore, number of elements is (A D B) D C is 10 + 15 + 20 +5 = 50. Ê x - 3ˆ 48. f 2(x) = f o f(x) = f( f(x)) = f Á Ë x + 1 ˜¯ x-3 -3 x+3 = x +1 =x-3 x -1 +1 x +1 -x - 3 -3 3 x + Ê ˆ x -1 = x. = f 3(x) = f Á Ë x - 1 ˜¯ - x - 3 +1 x -1

f 3k(x) = x

Hence

f 2010(2014) = f 3.670 (2014) = 2014. 49. (1, 2) Œ R, But (2, 1) œ R fi R is not symmetric and hence not an equivalence relation. fi Statement-1 is True. Staement-2 is False as 1 Æ 1, 2, 3. 50. Statement-2 is True. Because if A has m elements, then first elements can be mapped to n elements. For the 2nd element the choice is n – 1 and so on. So the total number of injective mappings is n(n – 1) (n – 2) ..... (n – m + 1) = n! / (n – m)! Which shows that statement-1 is False. 51. f (– x) = – sin x + cos x π f(x) or – f(x), fi f is neither odd nor even. So statement-1 is True - sin x g(– x) = = – (g(x)) fi g is an odd function 1 - cos x fi Statement-2 is also True but does not lead to statement-1. 52. Taking y = 3, f (x) – f (3) = x – 3 fi f (x) = x – 3 + f (3) = x – 3 + 2 = x – 1 fi f (xy) = xy – 1 fi Statement –1 is True f (x) =

x2 + x + 1 x2 - x + 1

fi f (1/x) =

1 + x + x2 1 - x + x2

= f (x)

and f (2) = 7/3 fi Statement-2 is also True, but does not lead to statement-1. 53 . Statement-2 is true by definition of a bijective mapping, using which statement-1 is also true.

Level 2 54. – p /2 < x < p /2 fi [x] = –2, –1, 0, 1 fi f (x) = cos[x] = cos (–2), cos (–1), cos (0), cos 1 = cos 2, cos 1, 1 cos 1 55.

x-5 > 0 and x + 5 > 0 x - 10 x + 24 fi x > 5, x > 6 fi x > 6 fi x Œ (6, •) 2

or x < 5, x > 4 fi 4 < x < 5 fi x Œ (4, 5) 1/3 x =yfix =y–1 so f ( g(x)) = 3 – 3x1/3 + x

56. Let g(x) = 1 +

3

fi f ( y) = 3 – 3( y – 1) + ( y – 1)3 = y3 – 3y2 + 5 57. f (–x) =

=

-x x - +1 e -1 2 x

- xe x x - +1 1 - ex 2

Sets, Relations and Functions 1.35

=

x(e x - 1 + 1) x - +1 2 ex - 1

fix=

fi f is onto

x x = + + 1 = f ( x) x 2 e -1

Also f (x1) = f (x2) fi x1 = x2 fi f is one-one. 64. f (x + y) + f (x – y)

Ê x ˆ 58. (fog) (x) = f (g(x)) = f Á Ë 1 - x ˜¯ x x = 1- x = =x x +1 x +1- x 1- x

=

a x + y + a -( x + y ) a x - y + a -( x - y ) + 2 2

=

a x (a y + a - y ) + a - x (a - y + a y ) 2

=

fi (fog)–1 (x) = x 59. f (x) = cot–1

y+2 , x ΠX = R ~ {1} y -1

2x is clearly one-one as 1 - x2

f (x1) = f (x2)

( a x + a - x )( a y + a - y )

2 = 2. f (x) f (y) fi k = 2. 65. x – 3 ≥ 0 fi x ≥ 3 and – 1 £ fi x £ 4. fi x Œ [3, 4].

fi x1 = x2

66. x[x] =

Next, let y = f (x) fi cot y =

2x 1 - x2

fi

| x | fi x2 = x if x is an integer

x = 0, 1 x

fi y = p/2 – q = p/2 – 2tan–1x. Taking x = tan(q / 2) so y Œ (0, p) fi 0 < y < p

f (x) =

ax + x 1 - xa x fi f (– x) = π –f (x) ax - x 1 + xa x

fi f is not odd

fi – p / 4 < tan–1 x < p / 4 f (x) =

fi – 1 < x < 1 fi x Œ (–1, 1)

ax - 1 1 - ax fi f (– x) = = –f (x) fi f is odd. ax + 1 1 + ax

and thus f is onto. 60. As – 52 + 122 £ 5 sinx + 12 cosx £ all x

-x

67. f (x) = x a + a fi f (–x) = f (x) fi f is not odd. a x - a- x

2 tan(q / 2) = = tanq 1 - tan 2 (q / 2)

52 + 122 for

fi 14 – 13 £ 5 sin x + 2 cos x £ 14 + 13

f (x) = x log2

(x +

(

x2 + 1

)

fi f (–x)

)

= – x log - x + x 2 + 1 π f (–x). 68. g (x) = 1 + x – [x] > 0 for all x

fi 1 £ f (x) £ 27 f is not one-one as it is periodic.

Ê 1 - cos x ˆ = (tan 2 ( / )) f x 2 61. fof (cos x) = f Á Ë 1 + cos x ˜¯ 2

=

x -3 £ 1

1 - tan ( x / 2) = cos x. 1 + tan 2 ( x / 2)

62. Since f(x + T) = sin T.

x + T π sin x for any value

f (x) is not periodic. x+2 , y ΠY = R Р{1} 63. Let y = f (x) = x -1

fi (fog) (x) = f ( g(x)) = 1. 69. The function f (x) = log2 (log1/2(x2 + 4x + 3) + sin–1[x] is defined if 0 < x2 + 4x +3 < 1 and –1 £ x £ 2

(I)

2

x + 4x + 3 > 0 fi x > – 1 or x < –3 2

x + 4x + 3 < 1 fi –2 – 2 < x < – 2 +

(II) 2

(III)

So the domain of the function is –1 < x < – 2 + 2 . 70. g ( f (x)) = |sin x| =

sin 2 x

which is satisfied if f (x) = sin2 x and g (x) =

x

Complete Mathematics—JEE Main

1.36

Also it satisfies f ( g (x)) = f

( x)

= (sin x )2

y+5 71. f (x) = 3x – 5 = y fi x = =f 3 x+5 fi f – 1(x) = 3 2 2 72. f (x ) = | x – a | π ( f (x))2

–1

f (a)2 = 12 – 1 = 0 fi f (a) = 0

fi

Putting x = a, y = a, we get f (0) = [f (a)]2 – f (0) f (2a) 1 = 0 – f(2a) fi f (2a) = –1 Next, we put x = 0, y = a to obtain f (–a) = f (0) f (a) – f (a) f(2a) = 0 Now, putting x = 2a, y = x, we get

( y)

f (|x|) = | |x| – a| π | f (x)|

f(2a – x) = f(2a) f(x) – f(–a) f(a + x)

f (x + y) = |x + y – a| π |x – a| + |y – a|.

= (–1) f(x) – (0) f(a + x)= – f (x)

73. Required number is 4! = 24.

8. Let w ŒW, then (w, w) Œ R.

Previous Years’ AIEEE/JEE Main Questions 1. R is not symmetric as (2, 3) Œ R but (3, 2) œ R 2. We must have 7 – x ≥ 1, x – 3 ≥ 0 and 7 – x ≥ x – 3 fi x £ 6, x ≥ 3 and x £ 5 Thus, 3£x£5 \ Range of f = {4P0, 3P1, 2P2) = {1, 3, 2}

Thus, R is not transitive. 9. 4–x

4. We have a 2 + b 2 £ a sin x + b cos x £

2

x ΠR.

cos–1 Ê x - 1ˆ Ë2 ¯ x -1 £ - 1 £ 1 or for 0 £ x £ 4 2 and log cos x

3. We have –1 £ x – 3 £ 1 and 9 – x2 > 0 fi 2 £ x £ 4 and –3 < x < 3 \ domain of f is [2, 3,)

–

\ R Also, if w1, w2, Œ W and (w1, w2) Œ R, then (w2, w1) Œ R. \ R is symmetric. Next, let w1 = ink w2 = link and w3, = let, then (w1, w2) Œ R, (w2, w3) Œ R but (w1, w3) œ R.

cos x > 0 or for 2np –

a 2 + b2

where n ΠN.

Thus, –

1 + 3 £ sin x –

3 cos x £

\

– 2 + 1 £ f (x) £ 2 + 1

fi

Thus, domain of f is 0 £ x
2

[2002]

3. The locus of the centre of a circle which touches the circles |z – z1| = a and |z – z2| = b, a π b, externally is (a) an ellipse (c) a circle

(b) a hyperbola (d) a pair of straight lines [2002]

x

1 + iˆ 4. If ÊÁ = 1 , then Ë 1 – i ˜¯ (a) (b) (c) (d)

x = 2n, where n is any positive integer x = 4n + 1, where n is any positive integer x = 2n + 1, where n is any positive integer x = 4n, where n is any positive integer [2003]

5. If z and w are two non-zero complex numbers such p , then z w that |zw| = 1, and Arg(z) – Arg(w) = 2 is equal to (a) – 1 (c) – i

(b) i (d) 1

[2003]

6. Let z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. If the origin, z1 and z2 form an equilateral triangle, then (a) a2 = 2b (c) a2 = 4b

(b) a2 = 3b (d) a2 = b

[2003]

7. Let z, w be two complex numbers such that z + i w = 0 and arg(zw) = p, then arg(z) equals (a)

3p 4

(b)

[2004]

8. If z = x – iy and z 1/3 = p + iq, then

2. If |z – 4| < |z – 2|, then (a) Re(z) > 0 (c) Re(z) > 3

5p 4

p 2

2

is equal to (a) 2 (c) 1 2

(b) – 1 (d) – 2

[2004]

2

9. If |z – 1| = |z| + 1, then z lies on (a) a circle (c) the real axis

(b) the imaginary axis (d) an ellipse [2004]

10. If the cube roots of unity 1, w, w2, then the roots of the equation (x – 1) 3 + 8 = 0, are (a) (b) (c) (d)

– 1, 1 – 2w, 1 – 2w2 – 1, 1 + 2w, 1 + 2w2 – 1, – 1 + 2w, – 1 – 2w2 – 1, – 1, – 1

[2005]

z

and |w| = 1, then z lies on 1 z– i 3 (a) straight line (b) a parabola (c) an ellipse (d) a circle [2005]

11. If w =

12. If z1 and z2 are two non-zero complex numbers such that |z1 + z2| = |z1| + |z2|, then arg z1 – arg z2 is equal to p (a) 0 (b) – 2 (c)

p 2

(d) p 10

13. The value of

 ÊËsin

k =1

(a) – i (c) 1

[2005]

2 kp 2 kp ˆ is + i cos 11 11 ¯ (b) i (d) – 1

[2006]

2.42

Complete Mathematics—JEE Main

14. If z 2 + z + 1 = 0, where z is a complex number, then value of 2 1 ˆ2 1ˆ 2 Ê 2 1 ˆ Ê Ê + Á z + 2 ˜ + Á z3 + 3 ˜ + z+ Ë Ë Ë z¯ z ¯ z ¯

22. If z is a complex number of unit modulus and argu1+ zˆ ment q, then arg Ê equals Ë1+ z ¯ p (a) -q (b) q 2 (c) p – q (d) –q [2013]

+

2 Ê 6 1ˆ ÁË z + 6 ˜¯ is z

(a) 12 (c) 54

(b) 18 (d) 6

[2006]

15. If |z + 4| £ 3, then the maximum value of |z + 1| is (a) 4 (c) 6

(b) 10 (d) 0

–1 i –1

(b)

1 i +1

(c)

–1 i +1

(d)

1 i –1

(c)

(b) 2 + 3 +1

(d)

2

5 +1

[2009]

(b) • (d) 1

[2010]

(b) (0,1) (d) (1, 0)

[2011]

20. Let a, b be real and z be a complex number. If z2 + az + b = 0 has two distinct roots on the line Re (z) = 1, then it is necessary that (a) b Œ (1, •) (c) b (–1, 0)

[2013, online]

Ê 1 + z2 ˆ 24. Let a = Im Á , where z is any non-zero comË 2iz ˜¯ plex number. The set A = {a : |z| = 1 and z π ±1} is equal to: (a) (–1, 1) (c) [0, 1)

(b) [–1, 1] (d) (–1, 0]

[2013, online]

25 . Let z ΠC satisfy |z| = 1 and z = 1 Рz .

19. If w (π1) is a cube root of unity and (1 + w)7 = A + Bw, there (A, B) (a) (–1, 1) (c) (1, 1)

(b) 5 (d) 1

†

18. The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals (a) 2 (c) 0

(a) 2 (c) 3

[2008]

4 = 2, then the maximum value of |z| is 17. If z – z equal to (a) 1

23. If z1 π 0 and z2 be two complex numbers such that 2 z1 + 3z2 z2/z1 is purely imaginary number, then 2 z1 - 3z2 is equal to:

[2007]

1 16. The conjugate of a complex number is . Then i – 1 that number is (a)

(d) either on the real axis or on a circle passing through the origin. [2012]

(b) b Π(0, 1) (d) |b| = 1

[2011]

z2 is real, then the point represented z -1 by the complex number z lies:

21. If z π 1 and

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

Statement-1: z is a real number Statement-2: Principal argument of z is ± p/3. [2013, online] 26. If a complex number z satisfies z + = 0, then |z| is equal to: (a) 2 (c) 5

(b) 3 (d) 1

2 |z + 1| + i

[2013, online]

27. If z is a complex number such that |z| ≥ 2, then 1 minimum value of z + : 2 (a) (b) (c) (d)

is strictly greater than 5/2 is strictly greater that 3/2 but less than 5/2 is equal to 5/2 lies in the interval (1, 2) [2014]

28. Let w (Im w π 0) be a complex number. Then the set of all complex numbers z satisfying the equation w – wz = k (1 – z), for some real number k, is: (a) {z : |z| = 1} (c) {z : z π 1}

(b) {z : z = z} (d) {z : |z| = 1, z π 1} [2014, online]

29†. If z1, z2 and z3, z4 are 2 pairs of complex conjugate numbers and z1, z3 œ R, then Êz ˆ Êz ˆ arg Á 1 ˜ + arg Á 2 ˜ equals: Ë z4 ¯ Ë z3 ¯ † Slightly modified version.

Complex Numbers 2.43

(a) 0

(b)

p 2

(a)

3p (d) p [2014, online] 2 z-i 30. Let z π –i be any complex number such that z+i 1 is a purely imaginary number. Then z + is: z (a) 0 (b) any non-zero real number other than 1 (c) any non-zero real number (d) a purely imaginary number [2014, online] (c)

31. For all complex numbers z of the form 1 + ia, a ΠR if z2 = x + iy, then (a) y2 Р4x + 2 = 0 (c) y2 Р4x + 4 = 0

(b) y2 + 4x – 4 = 0 (d) y2 + 4x + 2 = 0 [2014, online] 32. A complex number z is said to be unimodular if |z| = 1. Suppose z1 and z2 are complex numbers such z - 2 z2 is unimodular and z2 is not unimoduthat 1 2 - z1 z2 lar. Then the point z1 lies on a: (a) (b) (c) (d)

straight line parallel to the x-axis. straight line parallel to the y-axis. circle of radius 2. circle of radius 2 .

17

(b)

3 2 (d) 2 34. If z is a non-real complex Im z 5 mum value of is: (Im z )5 (a) – 1 (b) (c) – 4 (d) (c)

2 5 2 [2015, online] 2 number, then the mini-

–2 –5 [2015, online] 2 + 3i sin q is purely 35. A value of q for which z = 1 - 2i sin q imaginary, is p p (a) (b) 3 6

Ê 3ˆ Ê 1 ˆ (c) sin 1 Á ˜ (d) sin 1 Á ˜ [2016] Ë 3¯ Ë 4 ¯ 36. The point represented by 2 + i in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there 2 2 units in the south-westwards direction. Then its new position in the Argand plane is at the point represented by: (a) 1 + i (c) –2– 2i

(b) 2 + 2i (d) –1 – i

[2016, online]

37. Let z = 1 + ai be a complex number, a > 0, such that z3 is a real number. Then the sum 1 + z + z2 + … + z11 is equal to: [2015]

33. The largest value of r for which the region represented by the set {w Œ C: |w – 4 – i| £ r} is contained in the region represented by the set {z ŒC: |z – 1| £ |z + i|}, is equal to:

(a) 1365 3 i

(b) – 1365 3 i

(c) –1250 3 i

(d) 1250 3 i [2016, online]

Previous Years’ B-Architecture Entrance Examination Questions 1.

5 + i sin q is a real number when 5 - 3i sin q (a) q = p/4 (b) q = –p (c) q = –p/2 (d) q = p/2

(a) 1 (c) 3 [2006]

2. Two points P and Q in the Argand diagram represent complex numbers z and 3z + 2 + i. If P moves arround the circle with centre at the origin and radius 2, then Q moves on the circle, whose centre and radius are (a) –2 – i, 6 (c) 2 + i, 6

(b) 2 – i, 3 (d) 2 + i, 3

[2007]

3. Let z be a complex number such that |z| = 2, then 2 is maximum possible value of z + z

(b) 2 (d) 4

Ê 1 3ˆ 4. If i = -1 , then 4 + 3 ÁË - + i ˜¯ 2 2 is equal to

[2008] 127

Ê 1 3ˆ + 5 ÁË - + i ˜¯ 2 2

(a) 4 3 i

(b) 2 3 i

(c) 1 - 3 i

(d) 1 + 3 i

124

[2009]

5. The real part of a complex number z having minimum principal argument and satisfying |z – 5i| £ 1 is 2 6 (a) (b) 0 5 2 1 6 (c) (d) [2010] 5 5

2.44

Complete Mathematics—JEE Main

6. Area of a triangle with vertices given by z, iz, z + iz, where z is a complex number, is 1 2 z (a) 0 (b) 2 (c) | z |2 (d) 2| z |2 [2011] 7. Two circles in the complex plane are C1 : |z – i| = 2 C2 : |z – 1 – 2i| = 4 then (a) C1 and C2 touch each other (b) C1 and C2 intersect at two distinct points (c) C1 lies within C2 (d) C2 lies within C1 [2012] 8. If z = i (i + 2 ) , then value of z4 + 4z3 + 6z2 + 4z is (a) –5 (c) 6

(b) 3 (d) – 9

[2013]

9. Suppose z is a complex number such that z π –1, z(1 - z ) , then Re(w) |z| = 1 and arg(z) = q. Let w = z (1 + z ) is equal to Êq ˆ (a) 1 + cos Ë ¯ 2

Êq ˆ (b) 1 - sin Ë ¯ 2

Êq ˆ Êq ˆ (c) -2 sin 2 Ë ¯ (d) 2 cos2 Ë ¯ 2 2 10. If |z1| = |z2| = |z3| = 1 and z1 + z2 + z3 =

[2014]

Level 1 21. 25. 29. 33. 37. 41. 45. 49. 53. 57. 61. 65. 69. 73. 77. 81. 85.

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

22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62. 66. 70. 74. 78. 82.

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

23. 27. 31. 35. 39. 43. 47. 51. 55. 59. 63. 67. 71. 75. 79. 83.

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

24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 64. 68. 72. 76. 80. 84.

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

Level 2 86. (d)

87. (b)

88. (d)

89. (d)

90. (b)

91. (c)

92. (d)

93. (c)

94. (a)

95. (b)

96. (c)

97. (a)

98. (b)

99. (a)

100. (a)

101. (a)

102. (c)

103. (b)

104. (a)

105. (b)

Previous Years’ AIEEE/JEE Main Questions

2 + i, then the number

z1 z2 + z2 z3 + z3 z1 is:

1. (b)

2. (c)

3. (b)

4. (d)

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

5. (c)

6. (b)

7. (a)

8. (d)

9. (b)

10. (a)

11. (a)

12. (a)

[2015]

13. (a)

14. (a)

15. (c)

16. (c)

11. Let S = {z Œ C: z(iz1 – 1) = z1 + 1, [z1] < 1}. Then, for all z ŒS, which one of the following is always true?

17. (d)

18. (d)

19. (c)

20. (a)

21. (d)

22. (b)

23. (d)

24. (a)

25. (d)

26. (c)

27. (d)

28. (d)

29. (a)

30. (no answer)

32. (c)

33. (d)

36. (a)

37. (a)

a positive real number a negative real number always zero a purely imaginary number

(a) Re z + Im z < 0 (c) Re z – Im z > 1

(b) Re z < 0 (d) Re z – Im z < 0 [2016]

Answers Concept-based 1. 5. 9. 13. 17.

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

2. 6. 10. 14. 18.

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

3. 7. 11. 15. 19.

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

4. 8. 12. 16. 20.

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

31. (b) 34. (c)

35. (d)

Previous Years’ B-Architecture Entrance Examination Questions 1. (b)

2. (c)

3. (c)

4. (a)

5. (a)

6. (b)

7. (c)

8. (b)

9. (c)

10. (c)

11. (*)

Complex Numbers 2.45

Hints and Solutions

\

z=

Concept-based 1. |(1 + i)z| = | i | z || fi fi

2 |z| = |z| |z| = 0

fi

fi

=

( 2 – 1) |z| = 0

z=0

2. z = a + (b + ic)2015 + (b – ic)2015 = a + (b – ic) \

+ (b + ic)

2015

fi

fi | x + iy|2 |c + id| = |a + ib| fi (x2 + y2)

2

we get z = –1 fi z = ± i 2 (1 + i ) =k 4 z+ (1 + i )

c 2 + d 2 = a 2 + b2

1 z1 1 8. D = 1 z2 - 4i 1 z3

4. We can write the equation as z-

(1)

z1 z2 = D z3

fi D is purely real fi Im(D) = 0 9. (a + ib) (x + iy) = (a + ib) (a – ib) i As a π 0

2 12 - i 2 = =1–i 1+ i 1+ i 4 and = 2(1 – i) 1+ i Therefore, (1) can be written as But

x + iy = (a – ib) i = b + ia fi (x, y) = (b, a) 10. w 2017 = (w 3)672 w = w and w 2225 = (w 3)741 w 2 = w 2 \ tan [(w 2017 + w 2225)p – p/3] (2)

This will not represent a circle if k = 1. When k = 1, (2) represents perpendicular bisector of the segment joining –2(1 – i) and 1 – i. 1 1 £ . Now 5. As |z| ≥ 5, z 5 1 1 24 1 1 z- ≥ z = z - ≥ 5- = z z z 5 5 The least value is attained when z = 5. 2p ˆ 2p Ê 6. i Ë 1 - cos ¯ + sin 7 7 p p p = 2 i sin 2 + 2 sin cos 7 7 7 p˘ Êpˆ È p = 2 sin Ë ¯ Ícos + i sin ˙ 7 Î 7 7˚

3p ˆ Ê 3p 2 Ëcos + i sin ¯ 4 4

17p 28

fi |(x + iy)2| |c + id| = |a + ib|

If |z| = 0, we get z = 0. If |z| = 1,

=

()

7. (x + iy)2 (c + id) = a + ib

|z| = 0 or |z| = 1

i ˆ Ê 1 + Also, i –1 = 2 ÁË ˜ 2 2¯

È Ê 17p ˆ Ê 17p ˆ ˘ + i sin Ë cos Ë Í ¯ p Î 28 28 ¯ ˙˚ 2 sin 7 2

=z

|z|2 = |z|

z - (1 - i ) =k z + 2(1 - i )

()

Thus, arg (z) =

z is real and thus, Im(z) = 0

3. |z2| = | – |z|| fi

2015

È Ê 3p p ˆ Ê 3p p ˆ ˘ cos Ë - ¯ + i sin Ë Í p Î 4 7 ¯ ˙˚ 4 7 2 sin 7 2

= tan [(w + w 2)p – p/3] = tan (–p – p/3) = – tan(p + p/3) = –tan (p/3) = – 3 11. Let z = –i t

where t > 0, then

i z = i(i t) = – t \ arg(i z) + arg(z) = p – p/2 = p/2 12. The inequality a + ib < c + id is true if and only if b = d = 0 and a > c. 13. Let z = x + iy, then z2 = x2 – y2 + 2ixy \

Re(z2) = 0

fi

|Re(z)| = |Im(z)|

fi

x2 – y2 = 0

Êz ˆ Êz ˆ 14. arg Á 1 ˜ + arg Á 2 ˜ Ë z4 ¯ Ë z3 ¯

Ê z 2ˆ Êzz ˆ = arg Á 1 2 ˜ + 2kp = arg Á 1 2 ˜ + 2kp = 2kp Ë z4 z3 ¯ Ë z3 ¯ where k = 0 or 1 [ z1 = z3 = i, gives answer 2p and z1 = z3 = 1, gives answer 0]

2.46

15.

Complete Mathematics—JEE Main

z-4 x-4 y = + 2i 2i 2 For

i (– 1 + 2i) 2

Ê z - 4ˆ p 0 £ sin -1 Ë £ , we must 2i ¯ 2 y £1 2 0 £ y £ 2.

0£

x – 4 = 0, fi

or z = (2 – i) –

x = 4,

-3i fi z is purely imaginary. z +a 17. Let z = bi, b Œ R, b π 0, 16. z =

Fig. 2.56

Then z2 = –b2 Œ R,

3 1 fi z = 1 - i or z = 3 - i 2 2

Therefore Im(z2) = 0 18. We have zk = wk where Êpˆ Êpˆ w = cos Ë ¯ + i sin Ë ¯ 10 10 Thus, z1 z2 z3 z4 = w . w2. w3. w4 = w10 Ê 10 p ˆ Ê 10 p ˆ = cos Ë + i sin Ë ¯ 10 10 ¯ [by the De Movire’s Theorem] = cos p + i sin p = –1.

\ equation whose roots are a19, b 7 is x 2 + x + 1 = 0. 25. Using w10 = w, w 23 = w2, we get

{

sin (w 10 + w 23 ) p -

z1 + z2 1 / z1 + 1 / z2 z +z = 1 2 =z = 1 + z1 z2 1 + 1 / z1 z2 1 + z1 z2 Thus, z is purely real.

20. Let z = x + iy, then z2 = x2 – y2 – 2ixy Re(z2) = k2 fi

}

p 1 p p = sin Ê -p – ˆ = sin = Ë ¯ 4 4 4 2

26. (1 + w)2017 = (– w 2)2017 = – w 4034 = – w2 \ A + Bw = 1 + w

19. z =

\

a = w, b = w 2 so that a 19 = w and b 7 = w 2.

24. Let

27. Use

A = 1, B = 1.

R2 Æ R2 – R1 – R3

28. (1 + w – w 2)7 = (– w 2 – w 2)7 = – 27w14 = – 128w2 29. Use

x2 – y2 = k2

fi

C2 Æ C2 + 3i C3.

30. Let a = arg z < 0, then arg(– z) = p + a,

which represents a hyperbola.

Level 1 21. |1 – w| = |w – w 2| = |w2 – 1| Alternatively plot the points on an argand diagram. 3

( )

22. x3 = p = p 3 1

1

1

Let

1

1

fi x = p 3 , p 3 w, p 3 w 2. 1

1

xa + yb + zg x + yw + w 2 z 1 = = = w2 2 x b + yg + za xw + yw + z w 1 3

1 3

1 3

2

If a = p w, b = p , g = p w , then xa + yb + zg 1 = = w. x b + yg + za w 2 23.

Fig. 2.57

a = p 3 , b = p 3 w, g = p 3 w 2.

z - (2 - i ) z - (2 - i ) 1 AM ± p i 2 = (± i ) = fi e - 1 + 2i 1 + i - (2 - i ) MD 2 fi z = (2 – i) +

i (– 1 + 2i) 2

31. fi

|z1| = |z2| = |z3| = 1 z1 z1 = z2 z2 = z3 z3 = 1

\ |z1 + z2 + z3| =

1 1 1 1 1 1 + + + + = =1 z1 z2 z3 z1 z2 z3

32. nth roots of unity are given by Ê 2mp ˆ Ê 2mp ˆ 2mp i/n cos ÁË for m = 0, 1, 2, ˜¯ + i sin ÁË ˜=e n n ¯

, n – 1.

2m p i / n 2m p i / n Let z1 = e 1 and z2 = e 2

where 0 £ m1, m2 < n, m1 π m2. As the join of z1 and z2 subtend a right angle at the origin z1/z2 is purely imaginary we get

Complex Numbers 2.47 e2 m1p i / n e2 m2p i / n

= il for some real l fi

e2(m1 - m2 )p i / n = il

È 2 ( m1 - m2 ) p ˘ È 2 ( m1 - m2 ) p ˘ fi cos Í ˙ + i sin Í ˙ = il n n Î Î ˚ ˚ 2 ( m1 - m2 ) p p È 2 ( m1 - m2 ) p ˘ fi cos Í =0 fi = ˙ n 2 n Î ˚ fi n = 4(m1 – m2) Thus, n must be of the form 4k. 1- i 3 z1 - z3 z -z = 1 3 = 2 z2 - z3 z2 - z3

33.

= fi Also, fi fi fi

1 (1 + 3) = 1 4

|z1 – z3| = |z2 – z3| z1 - z3 1- i 3 –1= –1 2 z2 - z3 z1 - z2 -1 - i 3 = z2 - z3 2 | z1 - z2 | = | z2 - z3 |

1 (1 + 3) = 1 4

|z1 – z2| = |z2 – z3|

z 3 = z fi |z3| = | z | fi |z3| = | z |

38. fi

|z| = 0 or |z| = 1 fi z = 0 or z = 1/z.

\

z3 = z fi z 3 =

fi z = 1,–1, i, –i. Thus, solutions of z 3 = z are 0, 1, –1, i, –i 39. Let z = x + iy so that x = 5 + t,

0

0 2

D = 1 w –1

w –1

1 w2 – 1

w –1

2

= (w – 1) – (w 2 – 1) 2 = (w + w2 – 2) (w – 1 – w2 + 1) = (– 3) (w – w 2) = 3w (w – 1) 35. |z – i| = |z + i| represents the real axis. As z = i satisfies |z – i| < |z + i|, we get |z – i| < |z + i|, represents Im (z) > 0 36. iz 3 + z2 – z + i = 0 fi iz 3 – i2z2 – (z – i) = 0 fi iz 2 (z – i) – (z – i) = 0 fi (iz 2 – 1) (z – i) = 0 fi z 2 = 1/i or z = i In any case |z| = 1. 2(1 - cos q ) 37. 2x = (x + iy) + (x – iy) = (1 - cos q )2 + 4 sin 2 q fi

1 x= 5 + 3 cosq

Thus, maximum value of x is 1/2. It is attained at q = p.

25 – t 2

¤ x [bc + ca + ab + 2(a + b +c)x + 3x2] = 2[abc + (bc + ca + ab)x + (a + b + c)x 2 + x3] ¤ x 3 – (bc + ca + ab)x – 2abc = 0. If a is the third root of this equation, then a + w + w2 = 0 fi a = 1. 41. z = x + iy, |z – i Re(z)| = |z – Im(z)| fi | x + iy – ix|2 = |x + iy – y|2 fi x2 + (y – x)2 = (x – y)2 + y2 fi x = ± y. 42. We have

Hence, z1, z2 and z3 are the vertices of an equilateral

1

y=

fi (x – 5) 2 + y2 = 25, This clearly passes through 2 + 4i 40. Note that w, w2 are roots of 1 1 1 2 + + = a+x b+x c+x x

Thus, |z1 – z3| = |z2– z3| = |z2 – z1| triangle. 34. Using 1 + w 2 = – w, w4 = w and applying C2 Æ C2 – C1, C3 Æ C3 – C1 we get

1 fi z4 = 1 z

(r – w) (r – w2) = r2 – (w + w2 ) r + 1 = r 2 + r + 1 = (r + 1)2 – r 12

12

\

 ( r – w ) (r – w 2 )

=

r =1

 ÈÎ(r + 1)2 – r ˘˚

r =1

1 1 = (13) (13 + 1) (26 + 1) – 1 – (12 ) (13) = 740 6 2

43. Let ix (x ΠR) be root of z2 + az + b = 0, then (1) Рx 2 + aix + b = 0 fi Рx 2 Рa ix + b = 0 Subtracting we get (a + a ) ix + b Рb = 0 fi

x= –

i (b – b ) b–b = a+a i (a + a )

Putting this in (1), we get

(b – b )2 2

(a + a )

–

a (b – b ) +b=0 a+a

fi (b – b )2 – a(b – b ) (a + a ) + b(a + a )2 = 0 fi (b – b )2 – (a + a ) {ab – a b – ab – a b} = 0 fi (b – b )2 + (a + a ) (a b + a b) = 0 44. The circle |z – (– 1 + i)| = 2 completely lies inside the circle |z| = 3. 45. 8iz 3 + 12z2 – 18z + 27i = 0 fi 8iz 3 – 12i2z2 – 18z + 27i = 0 fi 4iz2 (2z – 3i) – 9 (2z – 3i) = 0

Complete Mathematics—JEE Main

2.48

(4iz2 – 9) (2z – 3i) = 0

fi

9 3i z2 = or z = 4i 2

fi

In any case |z| = 3/2. 46. As – 1 lies on the circle |z – 4| = 5, the real number |z + 1| is maximum when z is the other end point of the diameter. y

–1

4

9

52. As |w| = 1 fi |w |2 = 1 fi ww = 1 fi Thus, Now,

1 = w. w

z = w + 1/w = w + w = 2 Re(w) |Re(z)| = |2 Re(w)| = |2 Re(w)| £ 2|w | = 2

53. | z | = 1 fi z = cos q + i sin q for some q Π[0, 2p) Now, | z | = 1 fi |z|2 = 1 fi zz = 1. z z 1 + = z2 + 2 z z z 2 = (cos q + i sin q) + (cos q Рi sin q)2 = 2(cos2q + i2 sin2q) = 2 (cos 2q)

Thus,

x

Fig. 2.58

47.

cosq + 2 1 = + i sinq 3 x + iy 3 cosq + 2 sinq x – iy fi 2 = + i 3 3 x + y2

x

fi

2

x +y

2

=

cosq + 2 3

Fig. 2.59

9 = Also, x + y = 2 2 (2 + cos q ) + sin q 5 + 4 cos q 2

9

2

\ 4x – (x2 + y2) 4 (cosq + 2 ) ˆ 2 = Ê - 1¯ ( x + y 2 ) Ë 3

=

4 cos q + 5 9 ◊ =3 3 5 + 4 cos q

48. Affix of G is

1 ( z1 + z2 + z3 ) . 3

As origin is the mid-point of AG, 1 1 0 = ÈÍ ( z1 + z2 + z3 ) + z1 ˘˙ ˚ 2 Î3 fi 4z1 + z2 + z3 = 0. az + bz2 1 49. z3 = – ( az1 + bz2 ) = 1 c a+b 50.

z –1 z –1 = it where t Œ R. fi = – it z +1 z +1 z –1 z –1 fi + =0 z +1 z +1

fi (z – 1) ( z + 1) + ( z – 1) (z + 1) = 0 fi 2(z z – 1) = 0 fi |z| = 1. 51.

z z

–

z z

£ |arg (z) – arg( z )|

fi |z – z | £ |z| |arg (z) – arg( z )|

z z 1 + = 1 fi |cos 2q| = z z 2 fi cos 2q = ± 1/2 fi 2q = p/3, 2p/3, 4p/3, 5p/3, 7p/3, 8p/3, 10p/3, 11p/3 fi q = p/6, p/3, 2p/3. 5p/6, 7p/6, 4p/3, 5p/3, 11p/6 Hence, there are 8 values of z \

54. |z1 + z2|2 + |z1 – z2|2 = 2|z1|2 + 2|z2|2 fi 3 + |z1 – z2|2 = 2(2) + 2(2) fi |z1 – z2|2 = 5 fi |z1 – z2| = 5 . 55. z1 z2 = z2 z3 = z3 z1 fi

|z1| |z2| = |z2| |z3| = |z3| |z1| fi |z1| = |z2| = |z3| = r(say) [ z1, z2, z3 π 0] Thus, z1 z1 = z2 z2 = z3 z3

Now,

z1 z2 = z2 z3 = z3 z1

fi

z z1 z = 2 = 3 z2 z3 z1

fi

z12 = z2z3,

z22 = z3 z1 , z32 = z1 z2

Hence, z12 + z22 + z32 = z2z3 + z3z1 + z1z2 fi z1, z2, z3 are vertices of an equilateral triangle.

56. |z1 + z2 + z3| ≥ 0 fi |z1|2 + |z2|2 + |z3|2 + 2Re ( z1 z2 + z2 z3 + z3 z1 ) ≥ 0 fi

Re ( z1 z2 + z2 z3 + z3 z1 ) ≥ –

3 2

Now, |z2 – z3|2 + |z3 – z1|2 + |z1 – z2|2 = 6 – 2Re ( z1 z2 + z2 z3 + z3 z1 )

Complex Numbers 2.49

\ A = (–1, 1) 61. |z| = 1 fi zz = 1. Therefore z = 1 – 2z fi z = 1 – 2/z fi z2 – z + 2 = 0

£ 6 – (–3) = 9 57. z1 + z2 + z3 = 0 fi z1 + z2 + z3 = 0 1 1 1 + + =0 z1 z2 z3

fi

[ 2

Now, 0 = (z1 + z2 + z3)

= z12

z22

+

+

z32

z1 z1 = 1 etc.]

+ 2 (z2z3 + z3z1 + z1z2)

fi

= z12 + z22 + z32

63. z = i (1 + 3 ) = - 1 + 3 i = 2 w

= 0+

58. Note that

z22

= z2

2

where w π 1 is a cube root of unity. \

=1

1 = z1 etc. Thus, z1

z = (z1 + z2 + z3) ( z1 + z2 + z3 ) = |z1 + z2 + z3|2 £ (|z1| + |z2| + |z3|)2 = 9

Then z4 = a = b4(–1) p pˆ Ê fi z = b ˱ cos ± i sin ¯ 4 4 \

z2 - 1 z - 1 1- z = z -1+ = z +1 z +1 1+ z 1 zˆ Ê Re(w) = Re( z ) - 1 + Re Ë1+ z¯

\

Re

z4 + 2z3 + 4z2 + 5 = (2w)4 + 2 (2w)3 + 4 (2w)2 + 5 = 16w4 + 16w3 + 16w2 + 5 = 16 (w + 1 + w2) + 5 = 16 (0) + 5 = 5

64. Let a = – b4 where b > 0.

The maximum value is obtained when z1 = z2 = z3 = 1. z-zz z - 1 ( z - 1)z = = 59. w = [ |z| = 1] z +1 z + z z 1 z +1

But

but then |z| π 1.

2iz1 + 5z2 2i + 5 ( z2 / z1 ) 2i + 5a = =1 = 2iz1 - 5z2 2i - 5 ( z2 / z1 ) 2i - 5a

Now,

z12 + 2 z22 + z32 = z12 + z22 + z32 + z22

Thus,

1± 7 i , 2

62. Suppose z2/z1 = a, where a ΠR.

Ê1 1 1ˆ 0 = z12 + z22 + z32 + 2z1z2z3 Á + + Ë z1 z2 z3 ˜¯

fi

z=

z12 + z22 + z32 + z42 È Êp ˆ Êp ˆ Êp ˆ Êp ˆ˘ = 2b2 Ícos Ë ¯ - i sin Ë ¯ + cos Ë ¯ + i sin Ë ¯˙ Î 2 2 2 2 ˚ = 0 = a + |a| [ a < 0]

Ê1- zˆ 1 Ê1- z 1- z ˆ = + Ë1+ z¯ 2 Ë1+ z 1+ z ¯ 1 Ê 1 - z z - 1ˆ =0 + = 2 Ë 1 + z z + 1¯

Re(w) = cosq –1 = –2 sin2(q/2)

z+z is a positive real number, 1 + zz Ê z+z ˆ arg = 0. Ë 1 + zz ¯ 66. Using (cos a + i sin a) (cos b + i sin b) 65. As

= cos (a + b) + i sin (a + b), 2

60. Let w =

1+ z 1 . Im(w) = (w - w ) 2iz 2 2

For |z| = 1,

w=

= Thus, Im(w) = = -

we get cos (25q) = Р1, sin (25 q) = 0 fi 25q = (2k+ 1)p, k ΠI.

2

1+ z 1 + (1 / z ) = -2iz -2i (1 / z ) z2 + 1 = –w -2iz

1( w + w ) = -iw 2i

fi

k = 0, 1, 2, …, 12

67. log(1/7)|z – 2| > log(1/7)|z| fi

1Ê 1ˆ 1 = - (z + z ) z+ Ë ¯ 2 z 2

1 = - (2 cosq ) = – cos q 2 [ |z| = 1 fi z = cos q + i sin q] As z π 1, q π 0 and as z π –1, q π p,

fi

(2 k + 1)p £p 25 1 £ (2k +1) £ 25

Now, 0
1. 68.

1 È 2z + 1 2 z + 1 ˘ =5 2i ÍÎ iz + 1 -i z + 1 ˙˚

2.50

Complete Mathematics—JEE Main

= (–1)n+2 wn+2 + 1

fi

(2z + 1) (i z –1) + (2 z + 1) (iz +1) = 10 i (iz + 1) (i z – 1) fi 4iz z + (–2 + i) z + (2 + i) z = 10 i (–z z – iz + i z – 1) fi 14iz z + (8 + i) z + (–8 + i) z – 10 i = 0 5 1 1 fi z z + (1 - 8i )z + (1 + 8i ) z - = 0 14 14 7 This represents a circle. 69. Take z1 and z2 as two real numbers. 70.

(1 + i ) (1 - i )

n

n-2

Ê1+ iˆ =Ë 1- i¯

n-2

2

Ê -i + i ˆ (1 + i )2 = ÁË ˜ 1- i ¯

( 2i )

Thus, range of f is {0, 3}. 1 72. z + = 2 cos q fi z2 – 2z cos q + 1 = 0 z fi z = cos q ± i sin q Now, z2n – 2 zn cos (nq) = zn [zn – 2 cos (nq)] = zn [cos(nq) ± i sin(nq) – 2 cos(nq)] = –zn z n = –1 73. Using wk + wk+1 + wk+2 = 0 " k Œ I, 60

Âw

we get

=0

Next,

’w

k

fi

t = w, w2 where w π 1 is a cube root of unity. z =w z -1 |z| = |z – 1| |w | = |z –1|

Thus,

fi z lies on the perpendicular bisector of the segment joining z = 0 and z = 1, that is, z lies on Re(z) = 1/2. Let z = 1/2 + ai where a Œ R. Note that a π 0 since z = 1/2 does not satisfy z2 + z|z – 1| + |z –1|2 = 0 Putting z = 1/2 + ai in z2 + z|z –1| + |z –1|2 = 0 and equating imaginary parts of both sides, we get a + a|z –1| = 0 fi |z – 1| = –1. A contradiction. Thus z2 + z |z – 1| – |z + 1|2 = 0 has no solution in C, that is, statement-1 is true. Statement-2 is false as aw, aw2 are solutions of z2 + az + a2 = 0. 77. z = 2i + z fi

fi

z – z = 2i

2i Im(z) = 2i

fi

Im(z) = 1

As |z| = 1, we get z = ± i As arg(i) = p/2 and arg (–i) = –p/2,

m

\

= w = where

k =1

Statement-2 is false.

78. Note that

1 m = 1 + 2 + … + 30 = (30)(31) = 465 2 30

\

t = z/|z –1|

Thus, statement-1 is true.

k =1

30

76. Clearly z π 0, 1. We can write the given equation as t2 + t + 1 = 0 where

fi n-2

= i n –2 (2i) = 2i n –1 = –2i n +1 71. If n = 3k, k Œ I, then f(n) = 1 + 1 + 1 = 3 If n = 3k + 1, k Œ I, then f(n) = 1 + w + w2 = 0 If n = 3k + 2 k Œ I, then f(n) = 1 + w2 + w = 0

k

= 0 if n = 6 k + 1

’w k

8p ˆ Ê8p ˆ Ê a = cos Ë ¯ + i sin Ëp - ¯ 11 11 Ê8p ˆ Ê8p ˆ = cos Ë ¯ + i sin Ë ¯ 11 11

= w465 = 1

k =1

74. x2 + x + 1 = (x – w) (x – w2) Therefore, w, w2 are zeros of P(x)

\ a11 = cos(8p) + i sin(8p) = 1 [De Moivre’s Theorem]

0 = g(1) + w h(1) and 0 = g(1) + w2h(1)

Now,

10

Âak =

k= 0

fi g(1) = 0, h (1) = 0.

\

2

Therefore, –w, –w2 must be zeros of xn +1– xn+1. Now, (–w)

n+1

Statement-2 is true

2

75. x – x + 1 = (x + w) (x + w )

n

– (–w) + 1 = (–1)

n+1

n

w (w + 1) + 1

= (–1)n+1 wn (–w2) + 1

1 - a 11 =0 1-a

We have a =

aa 1 = a10 etc. = a a

Thus, Re (a + a2 + a3 + a4 + a5)

Complex Numbers 2.51

1 (a + a 2 + a 3 + a 4 + a 5 +a + a 2 + a 3 + a 4 + a 5 ) 2

= =

1 10 k 1 1 a = (-1) = Â 2 k =1 2 2

[Use statement-2]

\ Statement-1 is also correct and Statement-2 is a correct explanation for it. Statement-2 is true as it is formula for sun of a G.P. 79. zk = cos (k q) + i sin(k q) 15

f(q) =

 Im (z2 k -1 ) k =1

Ê 15 ˆ = Im Á  z 2 k -1 ˜ Ë k =1 ¯ Ê z(1 - z 30 ˆ = Im Á Ë 1 - z 2 ˜¯ But z30 = cos (30 q) + i sin (30 q) When q = a = p/60, z30 = 0 + i = i Ê z (1 - z 30 ) ˆ \ f(a) = Im Á Ë 1 - z 2 ˜¯ = Im

Ê 1- i ˆ Ë 1/ z - z¯

Ê 1- i ˆ = Im Á Ë - 2i sin a ˜¯ 1 1 = Im (1 + i ) = 2 sin a 2 sin a 80. Statement-2 is true. See theory of chapter 5 on matrices. È 1 a b˘ Let A = Í w 1 c ˙ , Í ˙ ÍÎw 2 w 1 ˙˚ w c w 1 1 c -a 2 +b 2 w 1 w w w 1 2 = 1 – cw – a(w – cw ) = (1 – aw) (1– cw) Note that |A| = 0 if a = w2 or c = w2 Thus, |A| π 0 if a = c = w and b = w or w2 |A| = 1

\

S contains exactly two distinct elements.

81. |z1 – z2| –

1 1 z1 z2

z2 - z1 z1 z2 = |z1 – z2| – |z2 – z1| = 0 = z1 - z2 -

\

Statement-1 is true.

Statement-2 is also true but a correct explanation for truth of statement-1. 82. Statement-2 is a true statement. Suppose f (z) = (z2 + 1) q (z) + r (z). If r (z) π 0, then r (z) = az + b where a, b Œ C. We have i = f (i) = (i2 + i) q(i) + ai + b fi ai + b = i Also, 1 + i = f (–i) = ((–i)2 + 1) q (–i) – ai + b fi – ai + b = 1 + i 1 1 \ b = + i, a = i 2 2 1 Thus, az + b = (iz + 1) + i 2 83. Statement-2 is true, as z3 = a3 fi |z|3 = |a|3 fi |z| = |a| Next, (z + ab)3 = a3 fi z + ab = a, aw, aw2 Let z1 = a – ab, z2 = aw – ab and z3 = aw2 – ab We gave |z2 – z1| = |a| |w –1| = |z3 – z2| =

3 a , |z1 – z3| =

3a,

3a

Thus, statement-1 is also true, but statement-2 is not a correct explanation of statement-1. 84. Statement-2 is true. See Theory. Let a be a root of z2 – z + p = 0 and suppose |a| < 1. We have p = |p| = |a – a2| £ |a| + |a|2 < 2 A contradiction. Thus, statement-1 is also true and statement-2 is a correct explanation for it. 85. We know that |sin q | £ q

" q ≥ 0.

If q < 0, then |sin (–q)| £ – q fi |sin q | £ |q | Thus, |sin q | £ |q | " q Œ R and statement-2 is true. Now, |z1 – z2|2 = (r1 cos q – r2 cosj )2 + (r1 sin q – r2 sinj )2 = r12 + r22 - 2r1r2 cos(q - j ) Êq - j ˆ 2 = (r1 - r2 ) + 4r1r2 sin 2 Ë 2 ¯ 2

Êq - f ˆ £ (r1 – r2)2 + 4(1) (1) Ë 2 ¯ Thus, Statement-1 is also true and Statement-2 is a correct explanation for it.

2.52

Complete Mathematics—JEE Main y

Level 2

2

86. As ABCD is a parallelogram, mid point of AC = mid point of BD -2

1 1 fi ( z1 + z3 ) = ( z2 + z4 ) 2 2 fi z4 = z1 + z3 – z2

x

2 -2

87. |a + bw + cw2|2 = (a + bw + c w ) (a + b w + cw) = a2 + b2 + c2 – bc – ca – ab 1 2 2 2 = ÈÎ(b - c ) + (c - a ) + (a - b ) ˘˚ 2 As a, b, c are integers and at least two of them are unequal, we get,

0

Fig. 2.60

This represents a square. See Fig. 2.60. 91. Let z = x + iy, then x = 1 – t, y =

t2 + t + 2

(b – c)2 + (c – a)2 + (a – b) ≥ 2.

fi t = 1 – x and y2 = t2 + t + 2 = (t + 1/2)2 + 7/4

Thus, |a + bw + cw2 |2 ≥ 1 fi |a + bw + cw2| ≥ 1

fi y2 = (x – 3/2)2 + 7/4.

Least value 1 is attained when a = 2, b = 1, c = 1.

This represents a hyperbola.

88. We have AP = AB = AQ = 2 Thus, for the shaded region |z + 1| > 2

p Ê 2 ˆ = Also, –BAQ = tan -1 Á ˜ 4 Ë 2 - 1 + 1¯ and –BAP = – p / 4 Hence, for the shaded region |z + 1| > 2 and |arg (z + 1)| < p / 4 w - wz 89. As is purely real, 1- z w - wz w - wz = 1- z 1- z fi (1 – z ) (w – wz ) = (1 – z) ( w – wz ) fi (w – w ) (1 – z z ) = 0 As w π w , we get z z = 1 Thus, set of values of z is {z: |z| = 1, z π 1}. 90. zw = |z|2 fi zw = z z fi w = z Thus, |z – z | + |w + w | = 4 fi |z – z | + | z + z| = 4 fi |2iy| + |2x| = 4 fi |x| + |y| = 2

92. w + w = =

=

az1 - bz2 az1 - bz2 + az1 + bz2 az1 + bz2

(az1 - bz2 ) (az1 + bz2 ) + (az1 - bz2 ) (az1 + bz2 ) az1 + bz2

2

a2 z1 z1 - b2 z2 z2

= 0 [ a |z1| = b |z2|] 2 az1 + bz2 fi w lies on the imaginary axis. 93. |z| + |1 – z| + |z – 2| ≥ max {|z|, |(1 – z) + (z – 2)|, |z + (1 – z)| + |z – 2|, |1 – z| + |z – (z – 2)|} = 2 The value 2 is attained when z = 1 94. See Theory. 95. |z2 – z3|2 + |z1|2 = |z2 – z3|2 + | –z2 – z3|2 fi |z2 – z3|2 + 1 = 2|(|z2|2 + |z3|2) = 4 fi |z2 – z3| = 3 Similarly, |z3 – z1| = |z1 – z2| =

3

Thus, z1, z2, z3 are vertices of an equilateral triangle. 96. 1 + w + w2 + º + wn – 1 = = =

1 - wn 1 - cos p + i sin p = 1- w 1 - cos (p / n) - i sin (p / n)

2 2

2 sin (p 2n) - 2 i sin (p 2n) cos (p 2n)

2 - 2 i sin(p 2 n) [cos (p 2 n) + i sin (p 2 n)]

Complex Numbers 2.53

101. |z – a2| + |z – 2a| = 3 will represent an ellipse if

cos (p 2n) - i sin (p 2n) Êpˆ = 1 + i cot Á ˜ Ë 2n ¯ - i sin (p 2n)

=

|a2 – 2a| < 3

97. |z1 + z2|2 = |z1 – z2|2

¤ –3 < a2 – 2a < 3

fi |z1|2 + |z2|2 + z1 z2 + z1 z2

¤ –2 < (a – 1)2 < 4

= |z1|2 + |z2|2 – z1 z2 – z1 z2

¤ (a – 1)2 < 4 ¤ –1 < a < 3 ¤ aŒ(0, 3)

z1 z2 + =0 z1 z2 98. z1 lies on the circle |z| = 2 and z2 lies on the line x+y= 4 2 fi 2(z1 z2 + z1 z2) = 0 fi

Distance of x + y = 4 2 from (0, 0) is 4 Thus, minimum distance between z1 and z2 is 2.

102. As AB = BC = CA, we get 2|z| = |1| = |1 – 2z| 1 1 1 and |z – | = fi |z| = 2 2 2 fi z is the point of intersection of circles |z| = 1/2 and |z – 1/2| = 1/2 1 1 ± 3i fiz= 4 103. |z + 1| + |z – 3| £ 10 represents the ellipse with focii at (–1, 0), and (3, 0) and length of major axis 10. Its centre is (1, 0), and its equation is

(

z2 z1

( x - 1)2 25

+

)

y2 =1 21

Any point on the ellipse is P (1 + 5 cosq,

21 sin q ).

Its distance from A (7, 0) is given by Fig. 2.61

AP2 = (5cosq + 8)2 + 21 sin2q

Alternative Solution

= (2cos q + 20)2 – 202 + 85

Distance of z1 = 2(cosq + isinq) from x + y = 4 2, is

fi 182 – 202 + 85 £ AP2 £ 222 – 202 + 85

2 cos q + 2 sin q - 4 2

fi 3 £ AP £ 13

2 =

104. |z – 1|2 + 2| z – w|2 = 3|z – w2|2

2 ( 2 2 – (cosq + sinq))

fi |z|2 + 1 – z – z + 2[|z|2 + 1 – z w – z w]

Maximum possible value of cosq + sinq is 2 .

= 3[|z|2 + 1 – z w 2 – z w2] fi (3w – 2w2 – 1) z + (3w2 – 2w – 1) z = 0

99. Let z1 = r (cosq + isinq). Equation of tanget to x2 + y2 = r2 at (rcosq, rsinq) is x cosq + y sinq = r Ê z + z ˆ Ê z1 + z1 ˆ Ê z - z ˆ Ê z1 - z1 ˆ fiÁ = r2 + Ë 2 ˜¯ ÁË 2 ˜¯ ÁË 2i ˜¯ ÁË 2i ˜¯ fi z z1 + z z1 = 2r 2

which represents a straight line. 105. As R.H.S is real, L.H.S must be real. Ê 1 + i ˆ Ê i - i2 ˆ Also, Á = Ë 1 - i ˜¯ ÁË 1 - i ˜¯

n

= in is real when n is even.

1 + x2 1 Ê 1ˆ = Á x + ˜ > 1 for x π 1. Ë x¯ 2x 2 Thus, we get only possible value of x is 1. As x > 0 and

z z fi + =2. z1 z1 100. Using |z1| + |z2| ≥ |z1 – z2|, we get |z – 2 + 3i| + |z – 1 + i| ≥ |–1 + 2i| =

fi 9 £ AP2 £ 169

5

\ RHS = 1, thus, least value of n is 4.

Complete Mathematics—JEE Main

2.54

02 + z21 + z22 = 0(z1) + 0(z2) + z1z2

Previous Years’ AIEEE/JEE Main Questions 1. Let |z| = |w| = r and Arg (w) = q, so that Arg (z) = p – q. We have z = r[ cos (p – q) + i sin (p – q)] = r[– cos q + i sin q] = – r ( cos q – i sin q) = – w

7. z + iw = 0 fi z + iw = 0 Now, arg (zw) = p

fi z – iw = 0

Ê z2 ˆ arg Á ˜ = p Ë i ¯ 2 fi arg (z ) – arg (i) = p

fi

Y

2.

(z1+ z2)2 = 3z1z2 fi a2 = 3b.

fi

fi 2 arg (z) – p/2 = p fi O

2

3

X

4

8. z

1/3

arg (z) = 3p/4 = p + iq

fi x – iy = (p + iq)3

If z satisfies |z – 4| = |z – 2|, then z lies on the perpendicular bisector of the segment joining z = 2 and z = 4. i.e., |z – 4| = |z – 2| fi Re(z) = 3. As z = 0 does not satisfy |z – 4| < |z – 2|, we get |z – 4| < |z – 2| represents the region Re(z) > 3. 3. Suppose |z – w| = r touches |z – z1| = a and |z – z2| = b externally.

fi

= p3 + 3p2(iq) + 3p (iq)2 + (iq)3

fi

x = p3 – 3pq2 and – y = 3p2q – q3

-y x = 3 p2 - q2 = p 2 - 3q 2 and q p x y fi + = -2( p 2 + q 2 ) p q fi

\

Ê x yˆ ÁË p + q ˜¯

( p 2 + q 2 ) = -2

9. |z2 – 1| = |z|2 + 1 can be written as |z2 +(–1)| = |z2| + |–1|

Then |w – z1| = a + r, |w – z2| = b + r

¤

fi fi

¤

z2 is a non-negative real number. -1 z2 is a non-positive real number.

¤

z lies on the imaginary axis.

|w – z1| – |w – z2| = a – b w lies on a hyperbola with foci at z1 and z2

Alternative solution Let z = x + iy then |z2 – 1| = |z|2 + 1, we get |(x2 – y2 – 1) + 2ixy| = x2 + y2 + 1

4. As 1 = – i2, x

x x 2 Ê 1 + i ˆ = Ê -i + i ˆ = Ê i(1 - i ) ˆ = i x 1 = ÁË ˜ Á ˜¯ ÁË 1 - i ˜¯ Ë 1- i ¯ 1- i fi x = 4n for some n Œ N.

5. | zw | = | z || w | = |z||w| = |zw| = 1 Arg ( zw) = arg (w) + arg ( z ) = arg (w) – arg (z) = – p/2 \ zw = | zw|[cos(– p/2) + i sin (–p/2)] = – i

fi

( x 2 - y 2 - 1)2 + 4 x 2 y 2 = x 2 + y 2 + 1

fi

(x2 – y2)2 + 1 – 2 (x2 – y2) + 4x2y2 = (x2 + y2 + 1)2

fi

(x2 + y2)2 + 1– 2(x2 – y2) = (x2 + y2)2 + 1 + 2(x2 + y2)

fi

\ z lies on the imaginary axis. 10. (x –1)3 = – 8 3

fi

Ê - x - 1ˆ = 1 Ë 2 ¯

fi

-

6. z1 + z2 = – a, z1Z2 = b As 0 , z1, z2 for an equilateral triangle,

– 4x2 = 0 fi x = 0

x -1 = 1, w , w 2 2

Complex Numbers 2.55

fi

x = – 1, 1 – 2w, 1– 2w2

As – 1 is an end point of a diameter of the circle, maximum possible value of |z + 1| is 6 which is attained when z = – 7

1 11. |w| = 1 fi |z| = z - i 3 1 fi z is equidistant from z = 0 and z = i. Thus, z 3 lies on the perpendicular bisector of the segment joining z = 0 and z = i/3. Therefore, z lies on a straight line.

16. z =

z1, z2 lie on a ray through the origin O and same side of the origin

¤

arg (z1) = arg (z2)

2 kp 2 kp + i cos 11 11 2 2k p ˆ k p = i Ê cos - i sin Ë 11 11 ¯ k = iw 2p 2p where w = cos - i sin 11 11 10 2k p 2k p ˆ Thus, S = Â Ê sin + i cos Ë 11 11 ¯ k =1

13. sin

11

Êw -w ˆ = iÂwk = iÁ Ë 1 - w ˜¯ k =1 10

w11 = cos 2p – i sin 2p = 1

But \

S=–i

14. z2 + z + 1 = 0

fi z = w or w2

where w is complex cube root of unity. Let

z = w, so that

1 = w2 z

Thus, 2 2 2 2 Ê z + 1 ˆ + Ê z 2 + 1 ˆ + Ê z3 + 1 ˆ +  + Ê z6 + 1 ˆ ˜ ÁË ÁË ˜ ÁË ˜ ÁË ˜ z¯ z2 ¯ z3 ¯ z6 ¯ = (–1)2 + (–1)2 + (2)2+ (–1)2 + (–1)2 + (2)2 = 12. 15. |z + 4| £ 3 represents the interior and boundary of the circle with centre at (–4, 0) and radius = 3. See Figure.

fi

4 4 £ |z| – £ |z| |z| |z|2 – 2|z| – 4 £ 0

fi

(|z| – 1)2 £ 5

17. |z| –

12. |z1 + z2| = |z1| + |z2| ¤

1 1 1 -1 = = fi z = i -1 i - 1 -i - 1 i + 1

fi

z-

4 =2 z

|z| £ 5 + 1

18. z is equidistant from A(1 + 0i), B(–1 + 0i) and C(0 + i). Thus, z is circumcentre of DABC, that is, there is exaclty one such z. (1 + w)7 = (– w2)7 = (– w2)6(– w2) = – w2

19.

=1+w \ A = 1, B = 1 20. As a, b Œ R, the roots of z2 + a z + b = 0 are of the form 1 + ia, 1 – ia, where a Œ R, a π 0. Now, b = (1 + ia)(1 – ia) = 1 + a2 > 1 fi b Œ (1, •) z2 is real, we get z -1 z2 z2 = z -1 z -1 ¤ z 2 ( z - 1) = z 2 ( z - 1) ¤ zz ( z - z ) - ( z - z )( z + z ) = 0 ¤ ( z - z )( zz - z - z ) = 0 ¤ z - z = 0 or zz - z - z = 0

21. As

fi

z lies on the real axis or z lies on a circle through the origin.

22. |z| = 1 fi zz = 1 1+ z ˆ Ê 1+ z ˆ arg ÊÁ = arg Á Ë 1 + z ˜¯ Ë 1 + 1/ z ˜¯ = arg (z) = q 23. Let z2/z1 = ik where k is a real number. 2 z1 + 3z2 2 / 3 + z2 / z1 2 / 3 + ik = = =1 2 z1 - 3z2 2 / 3 - z2 / z1 2 / 3 - ik 1 + z2 1 . Im (w ) = (w - w ) 2iz 2 1 + z 2 1 + (1 / z ) 2 For |z| = 1, w = = -2iz -2i (1 / z )

24. Let w =

=

z2 +1 = -w -2iz

2.56

Complete Mathematics—JEE Main

1 (w + w ) = -iw 2i 1 = - (z + z ) 2

Thus, Im(w) =

1Ê 1ˆ = - Áz + ˜ Ë 2 z¯ 1 = - (2 cosq ) = - cosq 2 [

|z| = 1 fi z = cos q + isin q]

As z π 1, q π 0 and as z π –1, q π p \

a Œ (–1, 1)

25. z = 1 - z

fi z + z =1

fi 2 Re(z) = 1 fi

Re (z) =

1 p = cos 2 3

As |z| = 1, Re(z) = cos(p/3), Thus, arg (z) = ± p/3 \ statement-1 is false and statement-2 is true . 26. Im(z) + 1 = 0 Im(z) = –1 Let z = a – i Now, z + 2 | z + 1 | +i = 0 2

fi a - i + 2 (a + 1) + 1 + i = 0 fi a2 = 2 (a2 + 2a + 2) \ a=–2 Thus, z = – 2 – i fi 27. z +

fi

(a + 2)2 = 0

|z| =

5

1 1 3 ≥|z|- ≥ 2 2 2

Minimum value 3/2 of |z + 1/2| is attained when z = Р2, and 3/2 Π(1, 2).

Also, z π 1 for otherwise w = w fi

Im(w) = 0

Êz ˆ 29. arg Ê z1 ˆ + arg Á 2 ˜ ÁË z ˜¯ Ë z3 ¯ 4 2 = arg Ê z1z2 ˆ = arg Ê | z1 | ˆ = 0 ÁË | z |2 ˜¯ ÁË z z ˜¯ 3 4 3 z i 30. Let = ik , where k Œ R. z +i fi z – i = ikz – k

fi

z(1– ik) = – k + i

-k + i 1 - ik 2 k +1 Note that | z |2 = =1 1+ k2 fi zz = 1 fi z = 1 z 1 Thus, z + = z + z , which is a real number. z Also, z + z = 0 fi

Im (z) = ± sin (p/3)

fi a2 + 4a + 4 = 0

\ |z| = 1

z=

fi

2Re(z) = 0

fi

fi

z = ai for some a ΠR.

Re(z) = 0

But in this case z -i is a real number z +i Therefore, z + z π 0 . 31. (1+ ai)2 = x + iy fi

1 – a2 + 2ai = x + iy

fi

1 – a2 = x, 2a = y

fi fi

1 - ( y / 2) 2 = x 4 – y2 = 4x

fi

y2 + 4x – 4 = 0

32. z1 - 2 z2 = 1 2 - z1z2 ¤ | z1 - 2 z2 |2 = | 2 - z1z2 |2 ¤ | z1 |2 + 4 | z2 |2 - 2 z1z2 - 2 z1z2 w - wz = k(1 – z)

28. fi

w – k = z(w - k )

fi

|w – k| = | z || w - k |

= |z||w – k| As Im (w) π 0, |w – k| π 0

= 4 + | z1 |2 | z2 |2 - 2 z1z2 - 2 z1z2 ¤ |z1|2 |z2|2 – |z1|2 – 4|z2|2 + 4 = 0 ¤ (|z1|2 – 4) (|z1|2 – 1) = 0 As |z2| π 1, |z1|2 – 4 = 0 fi |z1| = 2

Complex Numbers 2.57

fi z1 lies on a circle of radius 2.

=0 fi 2(2 – 6 sin2q) = 0 fi sin q = ±

33. |z – 1| = |z + i| fi

z is equidistant from A(1) and B(–i)

fi

z lies on the line y = – x

1

3 Thus, a value of q for which z is purely imaginary -1 Ê 1 ˆ is sin Á ˜ Ë 3¯

Coordinates of A are (2, 1) that of B are (3, 1) and C are (3, 3). Coordinates of D are (3 + 2 2 cos (5p/4), 3 + 2 2 sin (5p/4)) = (3 – 2, 3 – 2) = (1, 1) Thus, D is represented by 1 + i y

As z = 1 satisfies the inequality

3 + 3i C

|z – 1| £ |z + i|, 2÷2

we get |z – 1| £ |z + i| represents the region lying above the line x + y = 0 Largest value of r is length of perpendicular from (4, 1) to the line x + y = 0, that is, | 4 + 1| 5 5 = = 2. 1+1 2 2 34. Let z = r (cos q + isin q), so that largest r =

D

= 1 + 3ai – 3a2 – ia3 As z3 is real 3a – a3 = 0 fi a( 3 – a) (a +

z = r [cos (5q) + isin (5q)] 5

fia=

5

fi

(Im(z)) = r cos q

and

Im (z5) = r5 cos(5q)

p pˆ Ê 3 i = 2 Á cos + i sin ˜ Ë 3 3¯

Thus, z = 1 +

We have S = 1 + z + z2 + … + z11

Im( z 5 ) cos(5q ) = (Im( z ))5 cos5 q But cos (5q) = 16 cos5 q – 20 cos3 q + 5 cos q

1 - z12 1- z

=

But z12 = 212 (cos (4p) + i sin (4p))

cos(5q ) = 16 – 20 sec2 q + 5 sec4 q cos5 q = 5(sec4 q – 4sec2 q + 4) – 4

= 212(1 + i0) = 212 and 1 – z = –

3i 12

= 5(sec2 q – 2)2 – 4 ≥ – 4

Thus, S =

Thus, least value of cos(5q ) is – 4 which is attained when q = p/4. cos5 q 2 + 3i sin q is purely imaginary, Re(z) = 0 35. As z = 1 - 2i sin q that is, z + z = 0 2 + 3i sin q 2 - 3i sin q fi + =0 1 - 2i sin q 1 + 2i sin q fi (2 + 3isinq)(1 + 2isinq) + (2 – 3isinq)(1 – 2isinq)

3) = 0

3

Thus,

fi

x

37. z3 = (1 + ai)3

5

5

B 3+i

O

fi As z is non-real complex number, Im (z) π 0 5

2+i

=

1- 2

- 3i 4095 3

i = 1365 3 i

Previous Years’ B-Architecture Entrance Examination Questions 1.

5 + i sin q is a real number 5 - 3i sin q ¤

5 + i sin q 5 - i sin q = 5 - 3i sin q 5 + 3i sin q

2.58

Complete Mathematics—JEE Main

¤

(5 + i sin q)(5 + 3 isin q) = (5 – 3 isin q)(5 – i sin q)

¤

25 + 20 isinq – 3sin2q = 25 – 20 isinq – 3sin2q

¤ 40 isin q = 0 ¤ sin q = 0 This is possible when q = – p. TIP Amongst the choices given, – p is the only value which makes imaginary part of the numerator and denominator 0. 2. |z| = 2 Let w = 3z + 2 + i, then z = (w – (2 + i))/3 fi

|z| = 2

6. Area of triangle is z 1 D = | iz 4 z + iz

z 1 -iz 1 | z - iz 1

Using R3 Æ R3 – R1 – R2, we get z z 1 1 D = | iz -iz 1| 4 0 0 -1 1 1 | -izz - izz |= | z |2 4 2 7. As |i – (1 + 2i)| = 2 < | 4 – 2 | = 2, =

C1 lies inside C2

|w – (2 + i)| = 6

fi w lies on a circle with centre at (2 + i) and radius 6. 3. Let z = 2(cos q + i sin q). Now z+

2 = 2(cos q + isin q) + (cos q – isin q) z = 3 cos q + isin q

fi

z+

22 = 9 cos2q + sin2q z

= 1 + 4(1 + cos 2q) £ 9 2 is 3 which \ maximum possible value of z + z is attained when z = 1 1 3 then w3 = 1 4. Let w = – + i 2 2

(

)

8. z = i i + 2 = –1 + fi

2

(z + 1) = –2

2i fi z2 + 2z + 3 = 0

We now divide z4 + 4 z3 + 6 z

2

+ 4 z by z2 + 2z + 3.

z = 4 + 3w127 + 5w124 = 4 + 3(w3)42w + 5(w3)41w = 4 + 3w + 5w = 4 + 8w = 4 - 4 + 4 3i = 4 3i 5. Principal argument is least at point P. sin(p / 2 - q ) =

AP OA

fi

cos q = 1/5

fi

sin q = 24 / 5

Now, z = 24(cosq + i sin q ) fi

Re(z) =

24 2 = 6 5 5

\

z4 + 4z3 + 6z2 + 4z = (z2 + 2z + 3) (z2 + 2z – 1) + 3 =0+3=3

9. |z| = 1, arg(z) = q fi z = cos q + isin q Now, w =

z (1 - z ) z - zz = z (1 + z ) z + zz

Complex Numbers 2.59

=

w= Now,

z -1 z +1

2

10.

2 + i = z1 + z2 + z3

2

fi 3 = |z1|2 + |z2|2 + |z3|2 + 2Re ( z1 z2 + z2 z3 + z3 z1 )

z -1 z +1

fi

2Re(w) = w + w =

z -1 z -1 + z +1 z +1

z2 -1 + z 2 -1 = ( z + 1)( z + 1)

3 = 1 + 1 + 1 + 2 Re ( z1 z2 + z2 z3 + z3 z1 )

fi Re ( z1 z2 + z2 z3 + z3 z1 ) = 0 11. z (iz1 – 1) = z1 + 1 fi z1(iz – 1) = 1 + z

2 cos 2q - 2 = zz + z + z + 1

fi z1 =

2(cos 2q - 1) = 2(cosq + 1)

fi

1+ z iz - 1

1+ z = |z1| < 1 iz - 1

fi |1 + z|2 < |1 – iz|2

2

=

-2 sin (q ) qˆ = - 4 sin 2 ÊÁ ˜ 2 Ë 2¯ 2 cos (q / 2)

fi 1 + z + z + zz < 1 – iz + iz + zz fi 2 Re(z) < – i(2i Im(z))

fi

Êqˆ Re(w) = -2 sin 2 Á ˜ Ë 2¯

fi Re(z) – Im(z) < 0

Quadratic Equations 3.1

CHAPTER THREE

5. If equation (1) is satisfied by more than two distinct complex numbers, then equation (1) becomes an identity, that is a = b = c = 0.

QUADRATIC EQUATIONS An equation of the form ax 2 + bx + c = 0

(1)

where a π 0, a, b, c Œ C, the set of complex numbers, is called a quadratic equation. A root of the quadratic equation (1) is a complex number a such that aa2 + ba + c = 0 The quantity D = b2 – 4ac is known as the discriminant of the equation (1). The roots of equation (1) are given by the quadratic formula x=

-b ± D 2a

If a and b are the roots of the quadratic equation (1), then b c a+b=– and a b = a a Note that a quadratic equation whose roots are a and b is given by (x – a) (x – b) = 0

QUADRATIC EXPRESSION AND ITS GRAPH

NATURE OF ROOTS 1. If a, b, c Œ R and a π 0. Then the followings hold good: (a) The equation (1) has real and distinct roots if and only if D > 0. (b) The equation (1) has real and equal roots if and only if D = 0. (c) The equation (1) has complex roots with nonzero imaginary parts if and only if D < 0. (d) p + iq (p, q Œ R, q π 0) is a root of equation (1) if and only if p – iq is a root of equation (1). 2. If a, b, c Œ Q and D is a perfect square of a rational number, then equation (1) has rational roots. 3. If a, b, c Œ Q and p +

RELATION BETWEEN ROOTS AND COEFFICIENTS

q (p, q ΠQ) is an irrational

root of equation (1) then p Рq is also a root of equation (1). 4. If a = 1, b, c ΠI, the set of integers, and the roots of equation (1) are rational numbers, then these roots must be integers.

Let We have

f (x) = ax2 + bx + c, where a, b, c Œ R and a π 0. b È f (x) = a Í x 2 + x + a Î

c˘ a ˙˚

È b b2 c b2 ˘ = a Í x2 + x + 2 + - 2 ˙ a a 4a ˚ 4a Î ÈÊ b ˆ 2 4 ac - b2 ˘ = a ÍÁ x + ˜ + ˙ 2a ¯ 4a2 ˚ ÎË

(2)

When is a Quadratic Expression always positive Or always negative? It follows from (2) that f (x) > 0 (< 0) "x Œ R if and only if a > 0 (< 0) and D = b2 – 4ac < 0. See Fig. 3.1 and (Fig. 3.2). Also, it follows from (2) that f (x) ≥ 0 (£ 0) "x Œ R if and only if a > 0 (< 0) and D = b2 – 4ac = 0. In this case f (x) > 0 (< 0) for each x Œ R, x π – b/2a. Also, in this case the graph of y = f (x) will touch the x-axis at x = – b/2a.

3.2

Complete Mathematics—JEE Main

If D = b2 – 4ac > 0 and a < 0 then Ï< 0 for x < b or x > a Ô f (x) Ì> 0 for b < x < a Ô= 0 for x = a , b Ó See Fig. 3.4

Fig. 3.1

Fig. 3.4

Note that if a > 0, then f (x) attains the least value at x = – b/2a. This least value is given by 2 Ê b ˆ 4 ac - b f Á- ˜ = Ë 2a ¯ 4a

Fig. 3.2

If a < 0, then f (x) attains the a greatest value at x = – b/2a. This greatest value is given by

Sign of a Quadratic Expression If D = b2 – 4ac > 0, then (2) can be written as 2 È 2 Ê b2 - 4ac ˆ ˘ b Ê ˆ f (x) = a ÍÁ x + ˜ - Á ˜ ˙ ÍË 2a ¯ 2a Ë ¯ ˙˚ Î ÈÊ b + b2 - 4ac ˆ Ê b - b2 - 4ac ˆ ˘ = a ÍÁ x + ˜ Áx + ˜˙ 2a 2a ÍÎË ¯Ë ¯ ˙˚ = a (x – a) (x – b)

where a = If

- b - b2 - 4ac 2a

and b =

D = b2 – 4ac > 0 and a > 0 then Ï> 0 for x < a or x > b Ô f (x) Ì< 0 for a < x < b Ô= 0 for x = a , b Ó

- b + b2 - 4ac 2a

2 Ê b ˆ 4 ac - b f Á- ˜ = Ë 2a ¯ 4a

POSITION OF ROOTS OF A QUADRATIC EQUATION Let f (x) = ax2 + bx + c, where a, b, c Œ R be a quadratic expression and let k be a real number. Conditions for Both the Roots to be more than a Real Number k. If a > 0 and b2 – 4ac > 0 then the parabola y = ax2 + bx + c opens upwards and intersect the x-axis in a and b where - b ± b2 - 4ac 2a In this case both the roots a and b will be more than k if k lies to left of both a and b. See Fig. 3.5 a, b =

See Fig. 3.3 Y

Fig. 3.3

Fig. 3.5

Quadratic Equations 3.3

From the Fig. 3.5, we note that both the roots are more than k if and only if -b (i) D > 0 (ii) k < (iii) f (k) > 0 2a In case a < 0, both the roots will be more than k (see Fig. 3.6) if and only if -b (i) D > 0 (ii) k < (iii) f (k) < 0 2a

(i) D > 0

(ii) af (k) < 0

Fig. 3.7

Fig. 3.6

Combining the above two sets, we get both the roots of ax2 + bx + c = 0 are more than a real number k if only if -b (iii) af (k) > 0 (i) D > 0 (ii) k < 2a Conditions for Both the Roots to be less than a Real Number k Both the roots of ax2 + bx + c = 0 are less than a real number k if and only if -b (iii) af (k) > 0 (i) D > 0 (ii) k > 2a Note

Fig. 3.8

CONDITIONS FOR EXACTLY ONE ROOT OF A QUADRATIC EQUATION TO LIE IN THE INTERVAL (k1, k2) WHERE k1 < k2 If a > 0, then exactly one root of f (x) = ax2 + bx + c = 0 lies in the interval (k1, k2) if and only if f(k1) > 0 and f (k2) < 0. Also, exactly one root lies in the interval (k1, k2) if and only if f (k1) < 0 and f (k2) > 0. See Fig. 3.9. Thus, if a > 0, exactly one root of f (x) = ax2 + bx + c = 0 lies in the interval (k1, k2) if and only if f (k1) f (k2) < 0.

1. Both the roots of ax2 + bx + c = 0 are positive if and only if b c D ≥ 0, - > 0 and > 0 a a 2. Both the roots of ax2 + bx + c = 0 are negative if and only if c b D ≥ 0, - < 0 and > 0 a a Fig. 3.9

CONDITIONS FOR A NUMBER k TO LIE BETWEEN THE ROOTS OF A QUADRATIC EQUATION

Similarly, if a < 0, exactly one of the roots of f (x) = ax2 + bx + c = 0 lies in the interval (k1, k2) if f (k1) f (k2) < 0.

The real number k lies between the roots of the quadratic equation f (x) = ax2 + bx + c = 0 if and only if a and f(k) are of opposite signs, that is, if and only if (i) a > 0

(ii) D > 0

(i) a < 0

(ii) D > 0

(iii) f (k) < 0 [See Fig. 3.7] or (iii) f (k) > 0 [See Fig. 3.8]

Combining, we may say k lies between the roots of f (x) = ax2 + bx + c = 0 if and only if

Fig. 3.10

Hence, if f (k1) f (k2) < 0, then exactly one root of f (x) = 0 lies in the interval (k1, k2).

3.4

Complete Mathematics—JEE Main

CONDITIONS FOR BOTH THE ROOTS OF A QUADRATIC EQUATION TO LIE IN THE INTERVAL (k1, k2) WHERE k1 < k2.

Illustration

1

1 is a root of (b – c) x2 + (c – a) x + a – b = 0 as (b – c) + (c – a) + (a – b) = 0.

If a > 0, both the roots of f(x) = ax2 + bx + c = 0 lie in the interval (k1, k2) if and only if b < k2 (i) D > 0 (ii) k1 < 2a (iii) f(k1) > 0 and f(k2) > 0 See Fig. 3.11

Finding Range of a Rational Function Let f(x) = ax2 + bx + c, g(x) = px2 + qx + r where a, b, c, p, q, r Œ R and one of a, p π 0. To find range of f ( x) , x Œ R, put R(x) = g( x )

fi

ax 2 + bx + c

, px 2 + qx + r (a – py)x2 + (b – qy)x + (c – r y) = 0. Since x is real, (b – qy)2 – 4(a – py) (c – r y) ≥ 0 We use this inequality to obtain range of R(x). y=

Fig. 3.11

Illustration

In case a < 0, the conditions read as b < k2 (i) D > 0 (ii) k1 < 2a (iii) f(k1) < 0 and f(k2) < 0

Find range of x2 - 2 x + 5 y= 2 x + 2x + 7

See Fig. 3.12

fi fi

fi fi fi fi

Fig. 3.12

¤ D ≥ 0, -

(x2 + 2x + 7)y = x2 – 2x + 5 ( y – 1)x2 + 2x(y + 1) + 7y – 5 = 0 As x is real, 4(y + 1)2 – 4(y –1) (7y – 5) ≥ 0 (y2 + 2y + 1) – (7y2 – 12y + 5) ≥ 0 3y2 – 7y + 2 £ 0 (3y – 1) (y – 2) £ 0 1/3 £ y £ 2

Some Useful Tips

Some Useful Tips Let quadratic equation be f(x) = ax2 + bx + c = 0 where a, b, c Œ R, a π 0. Let roots of (1) be a, b, and D = b2 – 4ac. 1. Both the roots are positive

2

(1)

c b = a + b > 0, = ab > 0 a 2a

ax 2 + bx + c

Suppose range of

px 2 + qx + r

real values, then range of Illustration

a(h ( x ))2 + b (h( x )) + c p(h ( x ))2 + q(h ( x )) + r

Range of

tan 2 x - 2 tan x + 5 2

tan x + 2 tan x + 7

3. Both the roots lie in the interval (p, q). b ¤ D ≥ 0, p < 0, af(q) > 0 2a

4. If one of a, b is real, then other must be real. 5. If sum of the coefficients of a quadratic equation is 0, then 1 is a root of the quadratic equation.

is also S.

3

2. Both the roots are negative c b = a + b < 0, = ab > 0 ¤ D ≥ 0, a 2a

is the set S and h(x) takes all

However, range of

is

1 £ y £ 2. 3

sin 2 x - 2 sin x + 5 sin 2 x + 2 sin x + 7

is not

[1/3, 2] as – 1 £ sin x < 1 will curtail its range. For instance try y = 2.

Quadratic Equations 3.5

CONDITIONS FOR A QUADRATIC EQUATION TO HAVE A REPEATED ROOT The quadratic equation f (x) = ax2 + bx + c = 0, a π 0 has a as a repeated if and only if f (a) = 0 and f ¢(a) = 0. In this case f(x) = a(x – a)2. In fact a = – b/2a. See Fig. 3.13 and Fig. 3.14.

Fig. 3.15

TIP Fig. 3.13

How to Obtain the Common Root? Make coefficients of x2 in both the equations same and subtract one equation from the other to obtain a linear equation in x. Solve it for x to obtain the common root.

CONDITION FOR TWO QUADRATIC EQUATIONS TO HAVE THE SAME ROOTS Fig. 3.14

CONDITION FOR TWO QUADRATIC EQUATIONS TO HAVE A COMMON ROOT Suppose that the quadratic equations ax2 + bx + c = 0 and a¢ x2 + b¢ x + c¢ = 0 (where a, a ¢ π 0 and ab¢ – a¢b π 0) have a common root. Let this common root be a. Then aa2 + ba + c = 0 and a¢a2 + b¢a + c¢ = 0 Solving the above equations, we get a2 a 1 = = bc ¢ - b ¢c a ¢ c - ac ¢ ab ¢ - a ¢b fi

bc ¢ - b ¢c a ¢ c - ac ¢ and a = a2 = ab ¢ - a ¢b a b ¢ - a ¢b

Eliminating a we get (a ¢ c - ac ¢ )2 (ab ¢ - a ¢b) fi

2

=

bc ¢ - b ¢c a b ¢ - a ¢b

(a ¢c - ac ¢ )2 = (bc ¢ - b ¢c) (ab ¢ - a ¢b)

This is the required condition for two quadratic equation to have a common root. See Fig. 3.15

Two quadratic equations ax2 + bx + c = 0. and a¢x2 + b¢x + c¢ = 0 have the same roots if and only if a b c = = a¢ b¢ c¢

EQUATIONS OF HIGHER DEGREE The equation f (x) = a0 xn + a1xn – 1 + a2xn– 2 + … + an–1 x + an = 0 where a0, a1, …, an – 1, an Œ C, the set of complex numbers, and a0 π 0, is said to be an equation of degree n. An equation of degree n has exactly n roots. Let a1, a2, …, an Œ C be the n roots of (1). Then f (x) = a0(x – a1) (x – a2) … (x – an) Also

a1 + a2 + … + an = -

and

a1a2 … an = (–1)n

a1 a0

an . a0

CUBIC AND BIQUADRATIC EQUATIONS If a, b, g are the roots of ax3 + bx2 + cx + d = 0, then b c d a + b + g = – , bg + g a + ab = and a b g = – . a a a

3.6

Complete Mathematics—JEE Main

Also, if a, b, g, d are the roots of the equation ax4 + bx3 + cx2 + dx + e = 0, then –b c , (a + b ) (g + d ) + ab + g d = a+b+g+d= a a –d e ab (g + d) + gd ( a + b) = , abgd = . a a

TRANSFORMATION OF EQUATIONS We now list some of the rules to form an equation whose roots are given in terms of the roots of another equation. Let given equation be (1) a0 xn + a1 xn–1 + … + an–1 x + an = 0 Rule 1: To form an equation whose roots are k(π 0) times x roots of the equations in (1), replace x by in (1). k Rule 2: To form an equation whose roots are the negatives of the roots in equation (1), replace x by – x in (1). Alternatively, change the sign of the coefficients of xn–1, xn–3, xn–5, … etc. in (1). Rule 3: To form an equation whose roots are k more than the roots of equation in (1), replace x by x – k in (1). Rule 4: To form an equation whose roots are reciprocals of 1 in (1) and the roots in equation (1), replace x by x n then multiply both the sides by x . Rule 5: To form an equation whose roots are square of the roots of the equation in (1) proceed as follows: x in (1)

Step 1

Replace x by

Step 2 Step 3

Collect all the terms involving x on one side. Square both the sides and simplify.

For instance, to form an equation whose roots are squares of the roots of x3 + 2x2– x + 2 = 0, replace x by x to obtain x x + 2x – x + 2 = 0 fi

x (x – 1) = –2(x + 1)

Squaring we get x(x – 1)2 = 4(x + 1)2 or x3 – 6x2 – 7x – 4 = 0 Rule 6: To form an equation whose roots are cubes of the roots of the equation in (1) proceed as follows: Step 1 Replace x by x1/3. Step 2 Collect all the terms involving x1/3 and x2/3 on one side. Step 3 Cube both the sides and simplify.

DESCARTES RULE OF SIGNS FOR THE ROOTS OF A POLYNOMIAL Rule 1: The maximum number of positive real roots of a polynomial equation

f(x) = a0 xn + a1 xn–1 + a2 xn–2 + … + an–1 x + an = 0 is the number of changes of the signs of coefficients from positive to negative and negative to positive. For instance, in the equation x3 + 3x2 + 7x – 11 = 0 the signs of coefficients are +++– As there is just one change of sign, the number of positive roots of x3 + 3x2 + 7x – 11 = 0 is at most 1. Rule 2: The maximum number of negative roots of the polynomial equation f(x) = 0 is the number of changes from positive to negative and negative to positive in the signs of coefficients of the equation f(– x) = 0.

SOME HINTS FOR SOLVING POLYNOMIAL EQUATIONS 1. To solve an equation of the form (x – a)4 + (x – b)4 = A a+b put y =x– 2 In general to solve an equation of the form (x – a)2n + (x – b)2n = A a+b where n is a positive integer, we put y = x – . 2 2. To solve an equation of the form (1) a0 f(x)2n + a1(f(x))n + a2 = 0 we put (f(x))n = y and solve a0 y2 + a1y + a2 = 0 to obtain its roots y1 and y2. Finally, to obtain solutions of (1) we solve, (f(x))n = y1 and (f(x))n = y2 3. An equation of the form (ax2 + bx + c1) (ax2 + bx + c2) … (ax2 + bx + cn) = A where c1, c2, …, cn , A Œ R, can be solved by putting ax2 + bx = y. 4. An equation of the form (x – a) (x – b) (x – c) (x – d) = Ax2 where ab = cd, can be reduced to a product of two ab . quadratic polynomials by putting y = x + 2 5. An equation of the form (x – a) (x – b) (x – c) (x – d) = A where a < b < c < d, b – a = d – c can be solved by change of variable y=

( x - a ) + ( x - b) + ( x - c) + ( x - d )

4 1 = x - (a + b + c + d) 4 6. A polynomial f (x, y) is said to be symmetric if f (x, y) = f (y, x) ⁄ x, y. A symmetric polynomials can be represented as a function of x + y and xy.

Quadratic Equations 3.7

Equations Reducible to Quadratic 4

2

1. To solve an equation of the type ax + bx + c = 0, put x2 = y. 2. To solve an equation of the type a (p(x))2 + bp(x) + c = 0 (where p(x) is an expression in x) put p(x) = y. 3. To solve an equation of the type b +c=0 ap(x) + p ( x) where p(x) is an expression in x, put p(x) = y. This reduces the equation to ay2 + cy + b = 0. 4. To solve an equation of the form 1ˆ 1ˆ Ê Ê a Á x2 + 2 ˜ + b Á x + ˜ + c = 0 Ë Ë ¯ x¯ x 1 put x+ =y x and to solve 1ˆ 1ˆ Ê Ê a Á x2 + 2 ˜ + b Á x - ˜ + c = 0 Ë ¯ Ë x¯ x 1 put x– =y x 5. To solve reciprocal equation of the type ax4 + bx3 + cx2 + bx + a = 0, a π 0 we divide the equation by x2, to obtain 1ˆ 1ˆ Ê Ê a Á x2 + 2 ˜ + b Á x + ˜ + c = 0 Ë Ë ¯ x¯ x 1 = y. and then put x+ x 6. To solve equation of the type (x + a) (x + b) (x + c) (x + d) + k = 0 where a + b = c + d, put x2 + (a + b)x = y 7. To solve equation of the type ax + b = cx + d or ax 2 + bx + c = dx + e square both the sides and solve for x. 8. To solve equation of the type

ax + b ± cx + d = e proceed as follows. Step 1 Transfer one of the radical to the other side and square both the sides. Step 2 Keep the expression with radical sign on one side and transfer the remaining expression on the other side. Step 3 Now solve as in 7 above.

USE OF CONTINUITY AND DIFFERENTIABILITY TO FIND ROOTS OF A POLYNOMIAL EQUATION Let f (x) = a0 xn + a1 xn – 1 + a2xn – 2 + … + an – 1 x + an, then f is continuous on R. Since f is continuous on R, we may use the intermediate value theorem to find whether or not f has a real root in an interval (a, b). If there exist a and b such that a < b and f (a) f (b) < 0, then there exists at least one c Œ (a, b) such that f (c) = 0. Also, if lim f ( x ) and f (a) are of opposite signs, then at xÆ - •

least one root of f (x) = 0 lies in the interval (– •, a). Also, if f(a) and lim f ( x ) are of opposite signs, then at least one xÆ •

root of f(x) = 0 lies in the interval (a, •). Result 1 If f (x) = 0 has a root a of multiplicity r (where r > 1), then we can write f (x) = (x – a)r g(x) where g(a) π 0. Also, f ¢(x) = 0 has a as a root with multiplicity r – 1. Result 2 If f (x) = 0 has n real roots, then f ¢(x) = 0 has (n – 1) real roots. It follows immediately using Result 1 and Rolle’s Theorem. Result 3 If f(x) = 0 has n distinct real roots, we can write f(x) = a0 (x – a1) (x – a2) … (x – an) where a1, a2, … an are n distinct roots of f(x) = 0. We also have n f ¢ (x) 1 = Â f (x) k = 1 x - ak

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Example 1: Solution set of 3 - x = – x2 – x – 1, x Œ R is (a) (– 1, •) (b) (0, 1) (c) (–1, 3/4) (d) f

Ans. (d) Solution: Note that LHS = Ê and RHS = – Ëx -

2

3 - x ≥ 0 ⁄ x £ 3.

1ˆ 5 - < 0 ⁄ x. ¯ 2 4

3.8

Complete Mathematics—JEE Main

Thus, the given equation has no solution. Example 2: Number of solutions of x–

5 5 =2– x-2 x-2

(1)

is (a) 0 (c) 2 Ans. (a)

(b) 1 (d) infinite fi

Example 3: The number of real solutions of x2 + 5|x| + 4 = 0 is (a) 0 (b) 1 (c) 2 (d) 4 Ans. (a) Solution: If x Œ R, then x2 + 5|x| + 4 ≥ 4. Thus, the given equation has no solution. Example 4: If

1 is a root of ax2 + bx + c = 0, (a, b, 3 - 4i

c Œ R, a π 0), then (a) b + 6c = 0 (c) a + 25c = 0 Ans. (a)

(b) b = 6c (d) b2 = 6c

Solution: TIP: As a, b, c ΠR, the other root must Thus,

1 . 3 + 4i

1 1 b + =- , 3 - 4i 3 + 4i a

1 1 c ◊ = 3 - 4i 3 + 4 i a

b 6 1 c = = - and a 25 25 a b Ê cˆ Now, 6 Ë ¯ = - fi b + 6c = 0. a a Also, a – 25c = 0 fi

Example 5: If 1 – p is a root of the quadratic equation x + px + 1 – p = 0, then its roots are (a) 0, 1 (b) –1, 1 (c) 0, – 1 (d) –2, 1 Ans. (c) 2

2

1 p (1 – p) < 0 3 p(1 – p) < 0 fi 0 < p < 1.

Solution: We must have

Solution: The equation (1) is defined when x π 2. But when x π 2. We can cancel 5/(x – 2) from both the sides of (1) to obtain x = 2 Thus, (1) has no solution.

and

Example 6: Suppose p Œ R. If the roots of 3x2 + 2x + p(1 – p) = 0 are of opposite signs, then p must lie in the interval (a) (– •, 0) (b) (0, 1) (c) (1, •) (d) (2, • ) Ans. (b)

Solution: (1 – p) + p(1 – p) + (1 – p) = 0 fi (1 – p) [1 – p + p + 1] = 0 fi 2(1 – p) = 0 fi p =1 Thus, equation becomes x2 + x = 0 fi x = 0 or – 1

Example 7: If a, b are the roots of (x – a) (x – b) + c = 0, c π 0, then roots of (ab – c) x2 + (a + b ) x + 1 = 0 are (a) 1/a, 1/b (b) – 1/a, – 1/b (c) 1/a, –1/b (d) –1/a, 1/b Ans. (b) Solution: a, b are roots of x2 – (a + b)x + ab + c = 0. \ a + b = a + b, ab = ab + c The equation (ab – c) x2 + (a + b) x + 1 = 0 becomes abx2 + (a + b)x + 1 = 0 fi (ax + 1) (bx + 1) = 0 fi x = – 1/a, – 1/b Example 8: If a, b are roots x2 + px + q = 0, then value of a3 + b3 is (a) 3pq + p3 (b) 3pq – p3 (c) 3pq (d) p3 – 3pq Ans. (b) Solution: a + b = – p, ab = q. Now, a3 + b3 = (a + b)3 – 3ab(a + b ) = – p3 + 3pq = 3pq – p3 Example 9: If a Œ R and both the roots of x2 – 6ax + 9a2 + 2a – 2 = 0 exceed 3, then a lies in the interval (a) (1, •) (b) (2, •) (c) (11/9, •) (d) f Ans. (d) Solution: Given equation can be written as (x – 3a)2 = 2(1 – a) We must have 1 – a ≥ 0 or a £ 1. Also, x = 3a ± 2 1 - a. Both the roots will exceed 3, if smaller of the two roots viz 3a - 2 1 - a exceeds 3, that is, 3a –

2 1- a > 3

(1) fi 3(a – 1) > 2 1 - a. For a = 1, this is not possible. For a < 1, we can write (1) as –3 which is no possible

(

2

1- a) > 2 1- a

Quadratic Equations 3.9

Thus, there is not real value of a. 7 - 4 3 , then x +

Example 10: If x = (a) 2

(b) 3 7

(c) 4 Ans. (c)

(d) 4 7

1 is equal to: x

2 Solution: 7 – 4 3 = 22 + ( 3 ) - 4 3

= (2 - 3 ) \

x=

and

2

Solution: For the equation to be defined, x ≥ 3, and x2 + 7x + 10 ≥ 40 ⁄ x ≥ 3. Therefore, the given equation has exactly one root, viz. 3. Example 14: Suppose a Œ R. If 3x2 + 2(a2 + 1)x + (a2 – 3a + 2) = 0 possesses roots of opposite signs, then a lies in the interval: (a) (– •, –1) (b) (–1, 1) (c) (1, 2) (d) (2, 3) Ans. (c) Solution: As the roots are of opposite signs, the product of roots must be negative, that is,

7-4 3 =2- 3

1 1 =2+ 3 = x 2- 3

1 =4 x Example 11: Suppose a, b are roots of ax2 + bx + c = 0 then roots of a(x – 2)2 – b(x – 2) (x – 3) + c(x – 3)2 = 0 are 2+a 2+ b 2+a 3+a , , (b) (a) 1+ b 1+a 3+b 2+b

Thus, x +

2 + 3a 2 + 3b , (c) 2+a 2+ b Ans. (d)

2 + 3a 2 + 3b , (d) 1+a 1+ b

fi

Example 15: Suppose a2 = 5a – 8 and b2 = 5b – 8, then a b equation whose roots are and is b a 2 (a) 6x – 5x + 6 = 0 (b) 8x2 – 9x + 8 = 0 (c) 9x2 – 8x + 9 = 0 (d) 8x2 + 9x + 8 = 0 Ans. (b) Solution: a, b are roots of x2 – 5x + 8 = 0. \ a + b = 5, ab = 8 Now,

Solution: Write the equation as 2 Ê x - 2ˆ Ê x - 2ˆ + bÁ a Á+c=0 ˜ Ë x - 3 ˜¯ Ë x - 3¯

x-2 = a, b x-3 2 + 3a 2 + 3b , fi x= 1+a 1+ b Example 12: If x, y, z ΠR then the least value of the expression E = 3x2 + 5y2 + 4z2 Р6x + 20y Р8z Р3 is: (a) Р15 (b) Р30 (c) Р45 (d) Р60 Ans. (b)

Now, –

Solution: We can write E = 3(x2 – 2x + 1) + 5 (y2 + 4y + 4) + 4(z2 – 2z + 1) – 30 = 3(x – 1)2 + 5(y + 2)2 + 4(z – 1)2 – 30 ≥ –30 The least value is attained when x = 1, y = – 2, z = 1. Example 13: The number real of roots of x - 3 (x2 + 7x + 10) = 0, where x Œ R is (a) 0 (b) 1 (c) 2 (d) 3 Ans. (b)

a 2 - 3a + 2 < 0 fi (a Р1) (a Р2) < 0 3 a Π(1, 2)

a b a 2 + b2 ( a + b )2 - 2 ab + = = b a ab ab

9 25 -2= 8 8 \ required equation is x2 – (9/8)x + 1 = 0 or 8x2 – 9x + 8 = 0 =

Example 16: Sum of the roots of the equation x2 + |2x – 3| – 4 = 0 is (a) 2 (b) –2 (d) - 2 (c) 2 Ans. (c) Solution: Case 1: When x < 3/2 In this case the equation becomes x2 – (2x – 3) – 4 = 0 fi x2 – 2x + 1 = 2 fi fi

(x – 1)2 = 2 fi x – 1 = ± x =1 ±

2

2

As x < 3/2, we take, x = 1 – Case 2: When x ≥ 3/2 In this case the equation becomes x2 + (2x – 3) – 4 = 0 fi

(x + 1)2 = 8 fi x + 1 ± 2 2

As x ≥ 3/2, x = –1 + 2 2

2

3.10

Complete Mathematics—JEE Main

Sum of the roots is (1 –

2 ) + (– 1 + 2 2 ) =

Thus, f(x) = 0 has exactly one root in (–1, 1). Hence, ax2 + bx + c = 0 has real and distinct roots.

2

Example 17: If the quadratic equation a(b – c) x2 + b(c – a) x + c (a – b) = 0, where a, b, c are distinct real numbers and abc π 0, has equal roots, then a, b, c are in (a) A.P. (b) G.P. (c) H.P. (d) A.G.P. Ans. (c) Solution: TIP If sum of the coefficients of a polynomial equation is zero, then 1 is a root of the equaiton.

As 1 is a root of the equation and equation has equal roots, the other root must be 1. Thus we have c (a - b) a (b - c ) a(b – c) = c (a – b)

1 = product of roots = fi fi

b=

fi

Example 19: If both the roots of the quadratic equation x2 + 2(a + 2) x + 9a – 1 = 0 are negative, then a lies in the set (a) [1, •) (b) (1/9, 1] » [4, •) (c) (–2, 4) » (6, •) (d) f Ans. (b) Solution: TIP: Both the roots are negative, if and only if D ≥ 0, sum of roots < 0 and product of roots > 0 Thus, 4(a + 2)2 – 4(9a – 1) ≥ 0, – 2 (a + 2) < 0, 9a – 1 > 0 fi a2 – 5a + 4 ≥ 0, a > –2, a > 1/9 fi (a – 1) (a – 4) ≥ 0, a > 1/9 fi a £ 1 or a ≥ 4 and a > 1/9 fi a Œ (1/9, 1] » [4, •) Example 20: Suppose p Œ R and a, b are roots of 1 = 0, then the minimum value of a 4 + b 4 is x2 – px + 2 p2 (a)

2ac a+ c a, b, c are in H.P.

(c) Ans. (c)

Example 18: Suppose a, b, c Œ R, a π 0, and (a + c)2 < b . Then the quadratic equation ax2 + bx + c = 0 has (a) real and distinct roots (b) real and equal roots (c) purely imaginary roots (d) none of these Ans. (a)

2

(b) 2 - 2

2 -2

(d) – 2

Solution: a + b = p, ab = 1/2p2 fi a2 + b2 = (a + b)2 – 2ab = p2 – 1/p2

2

1 ˆ2 1 Ê a4 + b 4 = (a2 + b2)2 – 2a2b2 = Á p2 - 2 ˜ - 4 Ë 2p p ¯

\

= p4 +

Solution:

= p4 +

TIP Suppose f(x) = ax2 + bx + c, a, b, c Œ R, a π 0. If f(a) f(b) < 0, then exactly one root of the equation f(x) = 0 lies between a and b and the equation f(x) = 0 has real roots.

1 4

p 1

-2-

2 p4

1 2 p4

–2

1 ˆ Ê ≥ 2 Á p4 ◊ 4 ˜ Ë 2p ¯

1/ 2

-2= 2 -2 [

Let f(x) = ax2 + bx + c. Then f(1) = a + b + c, f(–1) = a – b + c. We have f(1) f(–1) = (a + c)2 – b2 < 0

4

4

Thus, minimum possible value of a + b is is attained when p = 1

AM ≥ GM] 2 - 2 , which

LEVEL 1 Straight Objective Type Questions Example 21: The number of real solutions of x+ is:

x +

(a) 0 (c) 2 Ans. (a)

x-2 =3

Solution: For (1)

(b) 1 (d) 4

fi

x,

x - 2 to be defined x ≥ 0, x – 2 ≥ 0

x ≥ 2. But for x ≥ 2

x + x + x-2 ≥ 2 + Thus, (1) has no real solution.

2 >3

Quadratic Equations 3.11

Example 22: In Fig. 3.16 graph of y = ax2 + bx + c is given. Which one of the following is not true? (a) a < 0 (b) c > 0 (c) b2 – 4ac > 0 (d) b < 0 Ans. (d)

Solution: We have tan 25° + tan 20° = – 2p and tan 25° tan 20° = q Now, 1 = tan 45° = tan (25° + 20°)

fi a

0

X

b

(a) tan a = c - a b

Solution: The given parabola opens downwards, a < 0. Also, y(0) = c > 0. As roots of ax2 + bx + c = 0 are distinct, b2 – 4ac > 0. However nothing can be said about b. It can be positive, zero or negative. For instance, consider equations – x2 – 5x + 6 = 0, – x2 + 3x + 4 = 0 and – x2 + 9 = 0. Example 23: Let a and b be the roots of px2 + qx + r = 0, 1 1 + = 4, then value of p π 0. If p, q, r are in A.P. and a b |a – b | is 1 2 (b) (a) 61 17 9 9 1 34 9

(d)

2 13 9

Ans. (d) Solution: 4 = fi Also, Thus,

a +b q - q/ p = =ab r/p r

q + 4r = 0 2q = p + r [ p, q, r are in A.P.] p = – 9r, q = – 4r |a – b|2 = (a + b)2 – 4ab - q 2 4r = ÊÁ ˆ˜ Ë p¯ p =

fi

|a – b| =

1 – q = – 2p fi 2p – q = – 1

Example 25: Suppose 0 < a < b, and 2a + b = p/2. If tan a, tan b are roots of ax2 + bx + c = 0, then

Fig. 3.16

(c)

tan(25∞) + tan(20∞) -2 p = 1 - tan(25∞) tan(20∞) 1 - q

=

Y

(c) tan a =

(b) tan b = c + a b

b c-a

(d) tan b = –

b c+a

Ans. (a) Solution: tan a + tan b = – b/a and tan a tan b = c/a. tan a + tan b Now, tan (a + b ) = 1 - tan a tan b fi

tan (p/2 – a) =

fi

cot a =

- b/a 1- c/a

-b a-c

c-a . b Example 26: The value of a for which one root of the equation x2 – (a + 1) x + a2 + a – 8 = 0 exceeds 2 and other is less than 2, are given by (a) 3 < a < 10 (b) a ≥ 10 (c) – 2 < a < 3 (d) a £ – 2 Ans. (c) Thus,

tan a =

Solution: As the roots are real and distinct (a + 1)2 – 4(a2 + a – 8) > 0 3a2 + 2a – 33 < 0 11 < a < 3 (1) fi (3a + 11) (a – 3) < 0 fi 3 Y

16 Ê 1 ˆ 52 - 4Ë - ¯ = 81 9 81

2 13 9

Example 24: If tan 25° and tan 20° are roots of the quadratic equation x2 + 2px + q = 0, then 2p – q is equal to (a) – 2 (b) – 1 (c) 0 (d) 1 Ans. (b)

0

2

Also, for 22 – 2(a + 1) + a2 + a – 8 < 0 fi a2 – a – 6 < 0 fi (a – 3) (a + 2) < 0 fi –20 ( x + 7) ( x - 2 ) fi – 7 < x < 2 or x > 5. Therefore the least integral value a of x is – 6. This value of a satisfies the relation a2 + 5a – 6 = 0. Thus,

Example 28: Let p and q be two non-zero real numbers and a, b are two numbers such that a3 + b3 = – p, ab = q, then the quadratic equation whose roots are a 2/b and b 2/a is (b) qx2 + px + q2 = 0 (a) px2 – qx + p2 = 0 (c) px2 + qx + p2 = 0 (d) qx2 – px + q2 = 0 Ans. (b) Solution:

and

a2 b2 a3 + b3 -p + = = q b a ab

Ê a2 ˆ Ê b2 ˆ ÁË b ˜¯ ÁË a ˜¯ = ab = q

Thus, required equation is p x2 – ÊÁ - ˆ˜ x + q = 0 Ë q¯ qx2 + px + q2 = 0.

or

Example 29: The set of values of a for which the quadratic equation (a + 2) x2 – 2ax – a = 0 has two roots on the number line symmetrically placed about the point 1 is (a) {– 1, 0} (b) {0, 2} (c) f (d) {0, 1} Ans. (c) Solution: Let roots of the equation be 1 – k and 1 + k, where k > 0 Then

2 = (1 – k) + (1 + k) =

2a a fi1= a +2 a +2

Example 30: Suppose k Œ R and the quadratic equation x2 – (k – 3)x + k = 0 has at least one positive roots, then k lies in the set: (a) (– •, 0) » [9, 16] (b) (– •, 0) » [16, 8) (c) (– •, 0) » [9, •) (d) (– •, 0) » (1, •) Ans. (c) Solution: Both the roots a, b will be non-positive if D ≥ 0, a + b £ 0, ab ≥ 0 fi (k – 3)2 – 4k ≥ 0, (k – 3) £ 0, k ≥ 0 fi (k – 1) (k – 9) ≥ 0, k £ 3, k ≥ 0 fi 0 £ k £ 1. Thus, quadratic equation will have at least one positive root if k < 0 or k > 1 and (k £ 1 or k ≥ 9) fi k Œ (– •, 0) » [9, •) Example 31: Two non-integer roots of the equation (1) (x2 + 3x)2 – (x2 + 3x) – 6 = 0 are (a)

1 1 (- 3 + 11), (- 3 - 11) 2 2

(b)

1 1 (- 3 + 7 ), (- 3 - 7 ) 2 2

1 1 (- 3 + 21), (- 3 - 21) 2 2 (d) none of these Ans. (c) (c)

Solution: TIP: It is an equation which is reducible to quadratic. Put x2 + 3x = y. The equation (1) becomes y2 – y – 6 = 0 fi (y + 2) (y – 3) = 0 fi y = – 2, y = 3 When y = 2, x2 + 3x = – 2 fi x2 + 3x + 2 = 0 fi (x + 1) (x + 2) = 0 fi x = – 1, – 2 When y = 3, x2 + 3x = 3 fi x2 + 3x – 3 = 0 1 1 fi x= (- 3 ± 9 + 12 ) = (- 3 ± 21) 2 2 Example 32: Two non-integer roots of 4

are

2

Ê 3x - 1 ˆ Ê 3x - 1 ˆ ÁË 2 x + 3 ˜¯ – 5 ÁË 2 x + 3 ˜¯ + 4 = 0

(a) – 5/7, – 2/5 (c) 5/7, 7/5 Ans. (a)

(b) – 2/5, 7/5 (d) – 2/5, 3/5

Solution: TIP: It is an equation which is reducible to 2

This is not possible for any value of a.

(1)

Ê 3x - 1 ˆ quadratic. Put Á = y. The equation (1) becomes Ë 2 x + 3 ˜¯

Quadratic Equations 3.13

y2 – 5y + 4 = 0 fi (y – 1) (y – 4) = 0 fi y = 1, 4

Solution: Put x +

2

When

3x - 1 Ê 3x - 1 ˆ =1fi =±1 y = 1, Á Ë 2 x + 3 ˜¯ 2x + 3

fi

x = 4, – 2/5

When

3x - 1 Ê 3x - 1ˆ y = 4, Á =4fi =±2 ˜ Ë 2 x + 3¯ 2x + 3

1 1 = y, so that x2 + = y2 – 2. Then x x

equation (1) becomes 2(y2 – 2) – 9y + 14 = 0 fi 2y2 – 9y + 10 = 0 fi fi y = 5/2, 2

2

(2y – 5) (y – 2) = 0

fi x = – 7, – 5/7 Thus, required roots are – 2/5, – 5/7.

When x +

1 5 1 = , we get x = 2, x 2 2

Example 33: Sum of the roots of the equation is 4x – 3(2x + 3) + 128 = 0 (a) 5 (b) 6 (c) 7 (d) 8 Ans. (c)

When x +

1 = 2, we get x = 1. x

(1)

Example 36: The non-integral roots of x4 – 3x3 – 2x2 + 3x + 1 = 0 are

Example 34: The only real value of x satisfying the equation is

where x ΠR (a) 4/35 (c) 16/3 Ans. (d)

Solution: For (1) to be defined either x < – 4 or x > 0. x = y. Equation (1) becomes Put x+4

fi As

Thus, fi

(6y + 1) (y – 2) = 0 fi y=

6y2 – 11y – 2 = 0 y = – 1/6, y = 2.

x 1 ≥ 0, we reject y = – . 6 x+4

x = y2 = 4 x+4 x = 4x + 16 fi x = – 16/3 < – 4 y =2

Ans. (a) Solution: Dividing (1) by x2, we get 1 3 + 2 =0 x2 – 3x – 2 + x x

(1)

(b) – 4/35 (d) none of these.

6y – 2/y = 11 fi

1 1 (3 + 13 ), (3 - 13 ) 2 2 1 1 (3 - 13 ), (- 3 - 13 ) (b) 2 2 1 1 (3 + 17 ), ( 3 - 17 ) (c) 2 2 (d) none of these. (a)

Solution: Put 2x = y. Equation (1) becomes y2 – 3(8y) + 128 = 0 fi y2 – 24y + 128 = 0 fi (y – 8) (y – 16) = 0 fi y = 16, 8 fi 2x = 16, 8 fi x = 4, 3 \ Sum of the roots is 7.

x x+4 6 -2 = 11 x+4 x

(1)

fi

Example 35: The number of real values of x satisfying the equation 1ˆ 1ˆ Ê Ê 2 Á x 2 + 2 ˜ – 9 Á x + ˜ + 14 = 0 (1) Ë Ë x¯ x ¯ is (a) 1 (b) 2 (c) 3 (d) 4 Ans. (c)

1ˆ Ê Ê 2 1ˆ ÁË x + 2 ˜¯ – 3 ÁË x - ˜¯ – 2 = 0 x x Put x – 1/x = y, so that 1 1 x2 + 2 – 2 = y2 or x2 + 2 = y2 + 2 x x Equation (2) now becomes y2 + 2 – 3y – 2 = 0 fi y2 – 3y = 0 fi y(y – 3) = 0 fi y = 0 or y = 3

or

When y = 0, we get x –

1 =0 x

When y = 3, we get x –

1 =3 x

fi

x2 – 3x – 1 = 0

fi

fi

(2)

x=±1

x=

1 (3 ± 13 ) 2

Example 37: Suppose a, b, c Œ R and b π c. If a, b are roots of x2 + ax + b = 0 and g, d are roots of x2 + ax + c = 0, then value of (a - g ) (a - d ) is independent of: (b - g ) (b - d ) (a) a, b (b) b, c (c) c, a (d) a, b, c

3.14

Complete Mathematics—JEE Main

Solution: Note that we must have 3x2 + x + 5 ≥ 0 and x – 3 ≥ 0 or x ≥ 3. Squaring both sides of (1), we get 3x2 + x + 5 = x2 – 6x + 9

Ans. (d) Solution: x2 + ax + c = (x – g ) (x – d ) Thus,

=

-b+c =1 -b+c

Example 38: Number of real solutions of (x – 1) (x + 1) (2x + 1) (2x – 3) = 15

(1)

is (a) 0 (c) 3 Ans. (b)

(b) 2 (d) 4

fi

When y = 6, 2x – x = 6

fi (y – 6) (y + 2) = 0

fi

2

fi

2x – 4x + 3x – 6 = 0

fi

2x – x – 6 = 0 (2x + 3) (x – 2) = 0

x = – 3/2, 2

When y = – 2, 2x2 – x = – 2 or 2x2 – x + 2 = 0 As D = 1 – 4 (2) (2) = – 15 < 0, we get 2x2 – x + 2 = 0 does not have real roots. Example 39: If p, q ŒR and 2 + 3i a root of x2 + px + q = 0, then (b) p = – 4, q = 7 (a) p = – 2, q = 3 (c) p = 3, q = 2 (d) p = – 4, q = 2 Ans. (b) Solution: Other root of the equation is 2 – – p = (2 + 3i ) + (2 – 3i ) and fi

3i . Thus (2)

q = (2 + 3i )(2 – 3i ) p = – 4, q = 7

Example 40: The number of solutions of 3 x 2 + x + 5 = x Р3, where x ΠR is: (a) 0 (c) 2 Ans. (a)

(b) 1 (d) 4

is: (a) 0 (c) 2 Ans. (c)

(b) 1 (d) 3

x+9 =5–

4-x

x + 9 = 25 – 10 4 - x + 4 – x fi 10 4 - x = 20 – 2x fi 5 4 - x = 10 – x Squaring both the sides we get 25 (4 – x) = 100 – 20x + x2 fi x2 + 5x = 0 fi x(x + 5) = 0 fi x = 0 or x = – 5 Both x = 0 and x = – 5 satisfy (1).

2

fi

(1)

Squaring both the sides we get

y = 6, – 2. 2

Example 41: The number of solutions of 4-x + x+9 =5

We can write (1) as

fi (2x2 – x – 1) (2x2 – x – 3) = 15 (2) 2 Put 2x – x = y, so that (2) becomes (y – 1) (y – 3) = 15 fi y2 – 4y + 3 = 15 y2 – 4y – 12 = 0

(2x – 1) (x + 4) = 0

Solution: Note that 4 – x ≥ 0 and x + 9 ≥ 0 fi – 9 £ x £ 4.

Solution: We write (1) as (x – 1) (2x + 1) (x + 1) (2x – 3) = 15

fi

fi

fi x = 1/2, – 4 None of these satisfy the inequality x ≥ 3. Thus, (1) has no solution.

a, b are roots of x2 + ax + b = 0]

[

2x2 + 7x – 4 = 0

fi

a 2 + aa + c (a - g ) (a - d ) = 2 b + ab + c (b - g ) (b - d )

(1)

Example 42: Suppose a ΠR and a, b are roots of x2 Р4ax + 5a2 Р6a = 0. The maximum possible distance between a and b is (a) 6 (b) 5 (c) 3 (d) 1 Ans. (a) Solution: (a Рb)2 = (a + b)2 Р4 ab = 16a2 Р4(5a2 Р6a) = 4(6a Рa2) = 4[9 Р(3 Рa)2] As a ΠR, (a Рb)2 is maximum when a = 3. Thus, maximum possible distance between a and b is 6. Example 43: The value of a for which one root of the quadratic equation. (1) (a2 Р5a + 3) x2 + (3a Р1)x + 2 = 0 is twice the other, is (a) Р2/3 (b) 1/3 (c) Р1/3 (d) 2/3 Ans. (d) Solution: Let a and 2a be the roots of (1), then (a2 Р5a + 3) a2 + (3a Р1)a + 2 = 0 2

2

(2)

and (a – 5a + 3) (4a ) + (3a – 1) (2a) + 2 = 0 (3) Multiplying (2) by 4 and subtracting it form (3) we get

Quadratic Equations 3.15

(3a – 1) (2a) + 6 = 0

E = (c – a) (c – b) (– d – a) (– d – b) = (c2 + pc + 1) [(– d)2 – pd + 1] [∵ a + b = – p] 2 2 = (c + pc + 1) (d – pd + 1)

Clearly a π 1/3. Therefore, a = – 3/(3a – 1) Putting this value in (2) we get (a2 – 5a + 3) (9) – (3a – 1)2 (3) + 2(3a – 1)2 = 0 fi

9a2 – 45a + 27 – (9a2 – 6a + 1) = 0

fi

– 39a + 26 = 0

But c2 + qc + 1 = 0 and d 2 + qd + 1 = 0 \

fi a = 2/3. 2 For a = 2/3, the equation becomes x + 9x + 18 = 0, whose roots are – 3, – 6. x2 + x + 2

Example 44: Range of function f(x) = 2 , x + x +1 x Œ R is (a) (1, •) (b) (1, 3/2) (c) (1, 7/3] (d) (1, 7/5] Ans. (c) Solution: Let y =

x2 + x + 2 x2 + x + 1

=1+

=1+ 1 = x2 + x + 1 = y -1

fi

1 x2 + x + 1 1

( x + 1/ 2 ) 2 + 3 / 4

Ê ÁË x +

g(x) = – x2 – 2cx + b2 = b2 + c2 – (x + c)2

fi

2

c > 2b fi |c| >

3 x 3/2 2 4 8 Ê 2 ˆ Ê 4ˆ Ê 4ˆ = = Á 1/ 2 ˜ = Á ˜ ÁË ˜¯ = ¥ Ë 3¯ 3 3 3 3 3 Ë3 ¯ x = 3/2

Example 48: If x2 + 2ax + 10 – 3a > 0 for each x Œ R, then (a) a < –5 (b) – 5 < a < 2 (c) a > 5 (d) 2 < a < 5 Ans. (b) fi fi fi fi

Solution: x2 + 2ax + 10 – 3a > 0 ⁄ x Œ R (x + a)2 – (a2 + 10 – 3a) > 0 ⁄ x Œ R a2 + 3a – 10 < 0 (a + 5) (a – 2) < 0 –5 b + c 2

3x 4x = 3x 3 + 2 3 Ê 3 ˆ 4x = Ê 4 ˆ 3x ÁË ˜¯ ÁË ˜¯ 2 3

4x +

fi

max g(x) = b2 + c2 2

fi

Solution: 4x – 3x – 1/2 = 3x + 1/2 – 22x – 1 fi 4x + 22x – 1 = 3x + 1/2 + 3x – 1/2

fi

Solution: f(x) = (x + b)2 + 2c2 – b2 fi min f(x) = 2c2 – b2

[∵ cd = 1]

Example 47: If 4x – 3x – 1/2 = 3x + 1/2 – 22x – 1, then value of x is equal to (a) 5/2 (b) 2 (c) 3/2 (d) 1 Ans. (c)

2

Example 45: If f(x) = x2 + 2bx + 2c2 and g(x) = – x2 – 2cx + b2 are such that min f(x) > max g(x), then relation between b and c, is: (a) no relation (b) 0 < c < b/2 |b| (d) |c| > 2 |b|. (c) |c| < 2 Ans. (d)

fi

= cd (q2 – p2) = q2 – p2

>1

Thus, y > 1 and y – 1 £ 4/3 fi y £ 7/3 \ 1 < y £ 7/3

Also

= cd(q – p) (q + p)

fi

1ˆ 3 3 ˜¯ + ≥ 2 4 4

E = (– qc + pc) (– qd – pd)

2

2 |b|

Example 46: If a, b are the roots of x2 + px + 1 = 0, and c, d are the roots of x2 + qx + 1 = 0, then the value of E = (a – c) (b – c) (a + d) (b + d) is: (a) p2 – q2 (b) q2 – p2 2 2 (c) q + p (d) none of these Ans. (b) Solution: We have x2 + px + 1 = (x – a) (x – b) Thus,

x +1 - x -1 = 1

(x ΠR)

is (a) 1 (c) 4 Ans. (a)

(b) 2 (d) infinite

Solution: For the equation to make sense we must have x + 1 ≥ 0 and x – 1 ≥ 0 fi x ≥ – 1, x ≥ 1 i.e. x ≥ 1. We rewrite equation as x +1 =1+ x -1 and square both the sides to obtain x + 1 = 1 + x – 1 + 2 x -1

3.16

fi

Complete Mathematics—JEE Main

1 1 5 = x -1 fi = x – 1 or x = 2 4 4 Also, x = 5/4 satisfies the given equation.

Example 50: If a, b, c are real and a π b, then the roots of the equation (1) 2(a – b)x2 – 11(a + b + c)x – 3(a – b) = 0 are (a) real and equal (b) real and unequal (c) purely imaginary (d) none of these Ans. (b) Solution: The discriminant D of the quadratic equation (1) is given by D = 121(a + b + c)2 + 24(a – b)2 As a, b, c are real, 121(a + b + c)2 ≥ 0 Also, as a π b, (a – b)2 > 0 Thus, D > 0. Therefore, equation (1) has real and unequal roots. Example 51: Let a > 0, b > 0 and c > 0. Then both the roots of the equation 2ax2 + 3bx + 5c = 0 (1) (a) are real and negative (b) have negative real parts (c) have positive real parts (d) none of these Ans. (b) Solution: The discriminant D of the equation (1) is given by D = (3b)2 – 4(2a) (5c) = 9b2 – 40ac If D ≥ 0, then both the roots of (1) are real. Also, since ac > 0, D < 9b2. D < 3b [∵ b > 0] fi Therefore, in this case both the roots

- 3b - D and 4a

- 3b + D are negative. 4a If D < 0, both the roots of (1) are imaginary and are given by - 3b ± i 40ac - 9b2 2 Both these roots have negative real parts. x=

Example 52: If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 (1) are always (a) positive (b) negative (c) real (d) none of these Ans. (c) Solution: We can write (1) as

3x2 – 2(a + b + c)x + bc + ca + ab = 0

(2)

The discriminant D of (2) is given by D = 4(a + b + c)2 – 4 ¥ 3(bc + ca + ab) = 4[a2 + b2 + c2 + 2bc + 2ca + 2ab – 3bc – 3ca – 3ab] = 4[a2 + b2 + c2 – bc – ca – ab] = 2[(b2 + c2 – 2bc) + (c2 + a2 – 2ca) + (a2 + b2 – 2ab)] = 2[(b – c)2 + (c – a)2 + (a – b)2] As a, b, c are real, we get D ≥ 0. Thus, roots of (1) are real. Example 53: If the roots of the equation (4a – a2 – 5) x2 – (2a – 1)x + 3a = 0, a Œ R, are real and lie on opposite sides of unity, then the a lies in the interval: (a) (1, 5) (b) (1, 4) (c) (3, •) (d) (– •, 4) Ans. (b) Solution: TIP: If roots of a quadratic equation f(x) = 0 are on the opposite sides of a real number p, then f(x + p) = 0 has roots of opposite sings. Put y = x – 1, so that x = y + 1 The equation now becomes: (4a – a2 – 5) (y + 1)2 – (2a – 1) (y + 1) + 3a = 0 fi (4a – a2 – 5)y2 + (8a – 2a2 – 10 – 2a + 1)y + (4a – a2 – 5 – 2a + 1 + 3a) = 0 fi (4a – a2 – 5)y2 + (6a – 2a2 – 9)y + (5a – a2 – 4) = 0 This equation must have roots of opposite signs. This is possible if and only if 5a - a 2 - 4 4a - a2 - 5

< 0 or

a 2 - 5a + 4 a 2 - 4a + 5

0 ⁄ a, we get (a – 1) (a – 4) < 0 fi 1 < a < 4. Example 54: If a, b are roots of ax2 + bx + c = 0, then roots of a3x2 + abcx + c3 = 0 are (a) ab, a + b (c) ab, a2b2 Ans. (b)

(b) a2b, ab2 (d) a3, b3

–b c , ab = . a a 3 2 3 as a x + abcx + c = 0

Solution: a + b = Write

fi

c 3 x +Ê ˆ = 0 Ë a¯ a 2 x – (a + b) abx + a3b3 = 0

fi fi fi

x2 – (a2b + ab2)x + (a2b) (ab2) = 0 (x – a2b) (x – ab2) = 0 x = a2b, ab2

x2 +

bc 2

Quadratic Equations 3.17

Example 55: If P(x) = ax2 + bx + c and Q(x) = – ax2 + dx + c, where a, b, c Œ R, then P(x) Q(x) = 0 has (a) no real root (b) exactly two real roots (c) at least two real roots (d) none of these Ans. (c) Solution: Let D1 = b2 – 4ac and D2 = d2 + 4ac We have D1 + D2 = b2 + d2 ≥ 0 fi at least one of D1, D2 ≥ 0 fi one of P(x) = 0 or Q(x) = 0 has real roots. Thus, P(x) Q(x) = 0 has at least two real roots. Example 56: If the product of the roots of the equation x2 – 5kx + 2e4lnk – 1 = 0 is 31, then sum of the root is (a) – 10 (b) 5 (c) – 8 (d) 10 Ans. (d) Solution: We have product of the roots = 2e4lnk – 1 = 31 (given) fi e4lnk = 16 fi k4 = 16 fi k4 – 16 = 0 fi (k – 2) (k + 2) (k2 + 4) = 0 fi k = 2, – 2 As k > 0, we get k = 2. Sum of the roots = 5k = 10 Example 57: Let a, b be the roots of the equation x2 – a px + r = 0 and , 2b be the roots of the equation x2 – qx + r 2 = 0. Then the value of r is : 2 2 (a) (p – q) (2q – p) (b) (q – p) (2p – q) 9 9 2 2 (q – 2p) (2q – p) (d) (2p – q) (2q – p) (c) 9 9 Ans. (d) Solution: a + b = p, ab = r a + 2b = q. and 2 But a + b = p and a + 4b = 2q 1 2 fi b = (2q – p) and a = (2p – q) 3 3 Thus, ab = r 2 (2p – q) (2q – p) = r fi 9 Example 58: Sum of all the values of x satisfying the equation log17 log11 x + 11 + x = 0 (1) is: (a) 25 (b) 36 (c) 171 (d) 0

(

)

Ans. (a) Solution: Equation (1) is defined if x ≥ 0. We can rewrite (1) as log11 x + 11 + x = 170 = 1 1 fi x + 11 + x = 11 = 11 fi x + 11 = 11 – x Squaring both the sides we get x + 11 = 121 – 22 x + x

(

)

fi 22 x = 110 fi x = 5 or x = 25 This clearly satisfies (1). Thus, sum of all the values satisfying (1) is 25. 1

1 and are the roots of the a b equation ax2 + bx + 1 = 0 (a π 0, a, b, Œ R), then the equation, x(x + b3) + (a3 – 3abx) = 0 has roots: (a) a3/2 and b3/2 (b) ab–3/2 and a1/2 b Example 59: If

(c) Ans. (a)

(d) a–3/2 and b –3/2

ab and a b

Solution: The equation x(x + b3) + (a3 – 3abx) = 0 can be written as x2 + (b3 – 3ab)x + a3 = 0 We have 1 1 b 1 1 + = =– , a a b a a b a + b = - b, a b = a Now – (b3 – 3ab) = (–b)3 – 3a(–b) =

(

3

a + b) -3( a + b) a b

= ( a )3 + ( b )3 = a 3 2 + b 3 2 and a3 = a3/2 b3/2 \ The roots of the equation x2 + (b2 Р3ab)x + a3 = 0 are a3/2, b3/2 Example 60: If p, q are roots of x2 + px + q = 0, then (a) p = 1 (b) p = 1 or 0 (c) p = Р2 (d) p = Р2 or 0 Ans. (b) Solution: We have p + q = Рp, pq = q As pq = q, we get q(p Р1) = 0 fi q = 0 or p = 1 If q = 0, we get p = 0 If p = 1, we get p + q = Рp fi q = Р2. Thus, p = 1 or p = 0. Example 61: The equation x + 1 - x - 1 = 4 x - 1 , (x ΠR)

3.18

Complete Mathematics—JEE Main

(a) (b) (c) (d) Ans. (a)

no solution one solution two solutions more than two solutions

Solution: The given equation is valid if x + 1 ≥ 0, x – 1 ≥ 0 and 4x – 1 ≥ 0 i.e. if x ≥ 1. Squaring both the sides we get x + 1 + x – 1 – 2 ( x + 1) ( x - 1) = 4x – 1 fi 1 – 2x = 2 ( x + 1) ( x - 1) Squaring again, we get 1 – 4x + 4x2 = 4(x2 – 1) fi 4x = 5 or x = 5/4. Putting this value of x in the given equation, we get 5 5 Ê 5ˆ +1 - 1 = 4Á ˜ - 1 Ë 4¯ 4 4 3 1 - = 2 or 1 = 2 2 2

fi

which is not true. Thus, the given equation has no solution. Example 62: The sum of all the real roots of the equation (1) |x – 2|2 + |x – 2| – 2 = 0 is (a) 7 (c) 1 Ans. (b)

fi As \ fi fi \

(b) 4 (d) none of these

Solution: Putting y = |x – 2|, we can rewrite (1) as y2 + y – 2 = 0 or y2 + 2y – y – 2 = 0 y(y + 2) – (y + 2) = 0 fi (y – 1) (y + 2) = 0 y = |x – 2| ≥ 0, y + 2 ≥ 2 y–1=0fiy=1 |x – 2| = 1 fi x – 2 = ± 1 x = 2 ± 1 fi x = 3 or 1 Sum of the roots = 4

Example 63: Let p and q be the roots of x2 – 2x + A = 0 and let r and s be the roots of x2 – 18x + B = 0. If p < q < r < s are in A.P. then ordered pair (A, B) is equal to (a) (– 3, 77) (b) (77, – 3) (c) (– 3, – 77) (d) none of these Ans. (a) Solution: We have p + q = 2, pq = A (1) and r + s = 18, rs = B (2) As p, q, r, s are in AP, we take p = a – 3d, q = a – d, r = a + d, s = a + 3d. As p < q < r < s, we have d > 0 Now, 2 = p + q = 2a – 4d and 18 = r + s = 2a + 4d

Solving above equations, we get a = 5 and d = 2 \ p = – 1, q = 3, r = 7 and s = 11 Thus,

A = pq = – 3 and B = rs = 77.

Example 64: In a triangle PQR, –R = p/2. If tan (P/2) and tan (Q/2) are the roots of the equation ax2 + bx + c = 0 where a π 0, then (a) a + b = c (b) b + c = a (c) a + c = b (d) b = c Ans. (a) Solution: We have -b Ê Pˆ Ê Qˆ tan Á ˜ + tan Á ˜ = Ë 2¯ Ë 2¯ a P Q c Ê ˆ Ê ˆ and tan Á ˜ tan Á ˜ = Ë 2¯ Ë 2¯ a Now,

P+Q= =

fi fi

p Êpˆ Ê P Qˆ fi 1 = tan Á ˜ = tan Á + ˜ Ë ¯ Ë 2 2¯ 2 4 tan ( P / 2 ) + tan (Q / 2 ) 1 - tan ( P / 2 ) tan (Q / 2 )

- b/a -b = 1 - c/a a - c a – c =– b fi c = a + b 1=

Example 65: If a and b (a < b), are the roots of the equation x2 + bx + c = 0 where c < 0 < b, then (a) 0 < a < b (b) a < 0 < b < |a | (c) a < b < 0 (d) a < 0 < |a | < b Ans. (b) Solution: We have a + b = – b, ab = c As c < 0, b > 0, we get a < 0 < b Also, b = – b – a < – a = |a | Thus,

a < 0 < b < |a |

Example 66: For the equation 3x2 + px + 3 = 0, p > 0, if one of the roots is square of the other, than p is equal to (a) 1/3 (b) 1 (c) 3 (d) 2/3 Ans. (c) Solution: Let a, a2 be the roots of 3x2 + px + 3 = 0. Then a + a2 = – p/3 and a ◊ a2 = 1 As a3 = 1, we get a = 1, w or w2. If a = 1, then p = – 3(a + a2) = – 6 < 0. Not possible as p > 0. Thus, a π 1. -p We take a = w and a2 = w2, so that = w + w2 = – 1 3 fi p =3 Example 67: Let a, b be roots of ax2 + bx + c = 0, where ac π 0. Roots of cx2 – bx + a = 0 are

Quadratic Equations 3.19

(a) 1/a, 1/b (c) 1 / a , 1 / b Ans. (b)

(b) –1/a, – 1/b (d) -1 / a , - 1 / b

When

b ba a+b =– =c ab ca

a(1 + r) = 1 fi a = 1/3

In this case

a 1 Ê 1ˆ Ê 1ˆ = = c ab Ë a ¯ ÁË b ˜¯

Thus, roots of cx2 – bx + a = 0 are - 1 a , - 1 b . Example 68: If b > a, then the equation (x – a) (x – b) – 1 = 0 has (a) both roots in [a, b] (b) both roots in (– •, a) (c) both roots in (b, •) (d) one root in (– •, a) and other in (b, •). Ans. (d) Solution: Graph of y = (x – a) (x – b) – 1 is given in Fig. 3.17. It is a parabola which open upwards. Also, y < 0 for x = a and x = b. \ y = (x – a) (x – b) – 1 meets the x-axis at two points once in (– •, a) and once in (b, •). Thus, one root lies in (– •, a) and one in (b, •).

and

q = gd = (ar2) (ar3) = a 2 r 5 = – 32

Hence,

p = – 2, q = – 32.

Example 70: If a, b, and c are not all equal and a and b be the roots of the equation ax2 + bx + c = 0, then value of (1 + a + a2) (1 + b + b2) is (a) 0 (b) positive (c) negative (d) non-negative Ans. (d) Solution: We have a + b = – b/a, ab = c/a Now, (1 + a + a2) (1 + b + b2) = 1 + a + b + a 2 + b 2 + ab + a 2b + ab 2 + a 2b 2 = 1 + (a + b) + (a + b)2 – 2ab + ab [1 + a + b + ab ] = 1=

b

x axis

p = ab = a(ar) = a2r

1 2 = (2 ) = 9 9 which is not an integer. Thus, r = – 2. In this case, a (1 + r) = 1 fi a = – 1. \ p = a 2r = (– 1)2 (– 2) = – 2

y

a

a r 2 (1 + r ) = 4 fi r2 = 4 fi r = ± 2 a (1 + r ) r = 2,

we get

Ê 1ˆ Ê 1ˆ = Ë- ¯ + Á- ˜ Ë b¯ a And

ar2 (1 + r) = 4

fi Thus,

Solution: a + b = – b/a, ab = c/a. Now,

g + d = 4 fi ar2 + ar3 = 4

=

b b 2 2c c c + + + a a2 a a a

1 2

a 1

2a2

2 Ê bˆ c + ÁË ˜ a ¯ a2

{a2 + b2 + c2 – bc – ca – ab} [(b – c)2 + (c – a)2 + (a – b)2] ≥ 0

Example 71: If a, b, c are in A.P. and if the equations (1) (b – c)x2 + (c – a)x + (a – b) = 0

-1

Fig. 3.17 2

Example 69: Let a, b be the roots of x – x + p = 0 and g, d be the roots of x2 – 4x + q = 0. If a, b, g, d are in G.P. then the integral values of p and q respectively, are (a) – 2, – 32 (b) – 2, 3 (c) – 6, 3 (d) – 6, – 32 Ans. (a) Solution: We have a + b = 1, ab = p, g + d = 4, gd = q Let r be the common ratio of the GP a, b, g, d. Then a + b = 1 fi a + ar = 1 fi a(1 + r) = 1

and 2(c + a)x2 + (b + c)x = 0 have a common root, then (a) a2, b2, c2 are in A.P. (b) a2, c2, b2 are in A.P. (c) c2, a2, b2 are in A.P. (d) none of these Ans. (b)

(2)

Solution: Clearly x = 1 is a root of (1). If a is the other root of (1), then a = 1◊a = =1 [

a-b [product of roots] b-c

a, b, c are in A. P.]

3.20

Complete Mathematics—JEE Main

Thus, the roots of (1) are 1, 1. Now, (1) and (2) will have a common root if 1 is also a root of (2). fi 2(c + a) + b + c = 0 fi 2(2b) + b + c = 0 fi c = – 5b Also, a + c = 2b fi

[∵ a, b, c are in AP]

a = 2b – c = 2b + 5b = 7b

\ a2 = 49b2, c2 = 25b2 This, show that a2, c2, b2 are in A.P.

a b a 2 + b 2 (a + b )2 + = = –2 b a ab ab 25 19 = –2= 3 3 Êaˆ Ê bˆ and ÁË b ˜¯ Ë ¯ = 1. a a b Thus, the quadratic equation whose roots are , is b a Next,

x2 –

Example 72: Value of 6 + 6 + 6 +  up to •

x=

(1)

is (a) 3 (c) 1 Ans. (a)

(b) 2 (d) none of these

Solution: We can write (1) as x = 6 + x fi x2 = 6 + x fi x2 – x – 6 = 0 fi (x – 3) (x + 2) = 0 fi x = 3, – 2 As x > 0, we take x = 3 Example 73: Two complex numbers a and b are such that a + b = 2 and a4 + b4 = 272, then the quadratic equation whose roots are a and b can be (b) x2 – 2x + 12 = 0 (a) x2 – 2x – 16 = 0 2 (d) none of these (c) x – 2x – 8 = 0 Ans. (c) Solution: a + b = 2 fi a2 + b2 + 2ab = 4 fi a2 + b2 = 4 – 2ab fi (a 2 + b 2)2 = 16 – 16ab + 4a 2b 2 fi a 4 + b 4 + 2a2b 2 = 16 – 16ab + 4a 2b 2 fi 272 + 2(ab)2 = 16 – 16ab + 4(ab)2 fi 2(ab)2 – 16(ab) – 256 = 0 fi (ab)2 – 8(ab) – 128 = 0 fi (ab – 16) (ab + 8) = 0 fi ab = 16 or ab = – 8 Thus, required equation is either x2 – 2x + 16 = 0 2

or x – 2x – 8 = 0 \ Required answer is (c). Example 74: If a π b and a 2 = 5a – 3, b 2 = 5b – 3, then a b and is the equation whose roots are b a (a) 3x 2 – 25x + 3 = 0 (b) x 2 – 5x + 3 = 0 (d) 3x 2 – 19x + 3 = 0 (c) x 2 + 5x – 3 = 0 Ans. (d) Solution: a, b are roots of x2 = 5x – 3 or x2 – 5x + 3 = 0. Thus, a + b = 5, ab = 3.

19 x + 1 = 0 or 3x 2 – 19x + 3 = 0. 3

Example 75: If a π b and difference between the roots of the equations x2 + ax + b = 0 and x2 + bx + a = 0 is the same, then (a) a + b + 4 = 0 (b) a + b – 4 = 0 (c) a – b + 4 = 0 (d) a – b – 4 = 0 Ans. (a) Solution: Let a, b be the roots of x2 + ax + b = 0 and g, d be the roots of x2 + bx + a = 0. We are given |a – b | = |g – d | fi

|a – b |2 = |g – d |2

fi

(a + b)2 – 4ab = (g + d)2 – 4gd

fi

a2 – 4b = b2 – 4a

fi

a2 – b2 + 4(a – b) = 0

fi

(a – b)(a + b + 4) = 0

As

a π b, a + b + 4 = 0.

Example 76: Product of real roots of the equation t2x2 + |x| + 9 = 0. t Œ R is always (a) positive (b) negative (c) zero (d) does not exist. Ans. (d) Solution: Note that t2x2 + |x| + 9 ≥ 0 + 9 > 9 ⁄ x Œ R, Thus, t2x2 + |x| + 9 = 0 does not have real roots. Example 77: Let f(x) = of f(x) is (a) [0, •) (c) [3, •) Ans. (b) Solution: Let y = each x Œ R. fi

x2 - 2 x + 4 x2 + 2 x + 4

, x ΠR. The range

(b) [1/3, 3] (d) [0, 3] x2 - 2 x + 4 x2 + 2 x + 4

=

( x - 1)2 + 3 > 0 for ( x + 1)2 + 3

(y – 1)x2 + 2(y + 1)x + 4(y – 1) = 0

As x is real, 4(y + 1)2 – 16(y – 1)2 ≥ 0

Quadratic Equations 3.21

fi (y + 1 + 2y – 2) (y + 1 – 2y + 2) ≥ 0 fi 3(y – 1/3) (y – 3) £ 0 fi 1/3 £ y £ 3. Example 78: The values of a Œ R for which one root of the equation f(x) = x2 – (a + 1)x + a2 + a – 8 = 0 exceeds 2 and the other is less than 2 are given by (a) 3 < a < 10 (b) a > 0 (c) – 2 < a < 3 (d) a £ – 2 Ans. (c) TIP: If f(x) = ax2 + bx + c, a, b, c Œ R with a > 0 and f(a ) < 0 for some a Œ R then D = b2 – 4ac > 0, and hence f(x) = 0 has real and distinct roots. Solution: As one of the roots is less than 2 and other is more than 2, f(2) < 0. fi 22 – (a + 1) (2) + a2 + a – 8 < 0 a2 – a – 6 < 0 fi (a + 2) (a – 3) < 0 – 2 < a < 3.

fi fi

Example 79: If one root of the equation x2 + px + 12 = 0 is 4, while equation x2 + px + q = 0 has equal roots, then the value of q is (a) 3 (b) 12 49 (c) (d) 4 4 Ans. (c) Solution: 42 + 4p + 12 = 0 fi p = – 7 Since x2 – 7x + q = 0 has equal roots, we get 49 – 4q = 0 49 fiq= 4 Example 80: Suppose a is an integer and x1 and x2 are positive real roots of x2 + (2a – 1) x + a2 = 0, then value of x1 - x2 is (a) 1 (b) 2 (c) a (d) 1 – 4a Ans.(a) Solution: Two real roots are positive implies 2a – 1 £ 0, that is, a £ 1/2. As a is an integer, we get a £ 0. Now, x1 - x2

2

=

(

x1 - x2

2

)

= x1 + x2 – 2 x1 x2 = (1 – 2a) – 2 a 2 = 1 – 2a – 2 |a| = 1 – 2(a + |a|) = 1 fi

x1 - x2

=1

Example 81: If a, b are two real number satisfying the relations 2a2 – 3a – 1 = 0 and b2 + 3b – 2 = 0 and ab π 1, then ab + a + 1 is value of b

(a) – 1 (c) 1 Ans. (c)

(b) 0 (d) 2

3 2 Solution: b2 + 3b – 2 = 0 fi 1 + - 2 = 0 b b 2 3 fi = 0 1 b2 b Thus, a and 1/b are roots of 2x2 – 3x – 1 = 0 3 a -1 1 \ a+ = and = 2 b 2 b ab + a + 1 1 a 3 1 Now, = a + + = - =1 b b b 2 2 Example 82: If a, b are roots of x2 – 2x – 1 = 0, then value of 5a4 + 12b3 is (a) 153 (b) 169 (c) 183 (d) none of these Ans. (b) Solution: We have a 2 = 2a + 1 fi a4 = (2a + 1)2 = 4a 2 + 4a + 1 = 4(2a + 1) + 4a + 1 = 12a + 5, and b2 = 2b + 1 fi Thus,

b3

= 2b2 + b = 2(2b + 1) + b = 5b + 2

5a4 + 12b3= (60a + 25) + (60b + 24) = 60(a + b ) + 49 = 60 (2) + 44 = 169

Example 83: Suppose a, b, c are the lengths of three sides of a DABC, a > b > c, 2b = a + c and b is a positive integer. If a2 + b2 + c2 = 84, then value of b is (a) 7 (b) 6 (c) 5 (d) 4 Ans. (c) Solution: We have 1 1 ac = [(a + c)2 – (a2 + c2)] = [4b2 – (84 – b2)] 2 2 = 5b2/2 – 42 Thus, a and c are the roots of the equation x2 – 2bx + (5b2/2 – 42) = 0 As a and c are distinct real numbers the discriminant of the above must be positive, that is, 4b2 – 4(5b2/2 – 42) > 0 fi 6b2 < 168 or b2 < 28. Also, ac > 0 fi 5b2 > 84. \ 84/5 < b2 < 28. As b is a positive integer, we get b = 5. Example 84: Suppose a, b Œ R, a π 0 and 2a + b π 0. A root of the equation (a + b) (ax + b) (a – bx) = (a2x – b2) (a + bx) is a + 2b 2a + b (a) (b) 2a + b a + 2b

3.22

Complete Mathematics—JEE Main

(c) –

a - 2b a + 2b

(d) -

Solution: As roots of x2 – ax + b = 0 are real and distinct, a2 – 4b > 0 fi 4b < a2.

a + 2b 2a + b

Ans. (d) Solution: We can write the quation as 2

[a b + (a + b) ab]x2 + [(a3 – b3) – (a + b) (a2 – b2)]x – ab2 – ab(a + b) = 0

Thus, a2 – 1 £ 4b < a2.

fi (2a + b)x2 – (a – b)x – (a + 2b) = 0 Since the sum of coefficients is 0 one of the roots is 1 and the a + 2b other root is – . 2a + b Example 85: Let p and q be real numbers such that p π 0, p3 π q and p3 π – q. If a and b are nonzero complex numbers satisfying a + b = – p and a3 + b3 = q, then a a and b as its roots is quadratic equation having b a 3 2 3 (a) (p + q)x – (p + 2q)x + (p3 + q) = 0 (b) (p3 + q)x2 – (p3 – 2q)x + (p3 + q) = 0 (c) (p3 – q)x2 – (5p3 – 2q)x + (p3 – q) = 0 (d) (p3 – q)x2 – (5p3 + 2q)x + (p3 – q) = 0 Ans. (b) Solution: q = a3 + b3 = (a + b)3 – 3ab (a + b) = – p3 + 3abp p3 + q fi ab = 3p We have a b a2 + b2 + = b a ab 2

= = = and

If a, b are roots of x2 – ax + b = 0, then |a – b | £ 1 fi (a – b)2 £ 1 fi (a + b)2 – 4ab £ 1 fi a 2 – 4b £ 1 fi a2 – 1 £ 4b.

Example 87: If a is the minimum root of the equation x2 – 3|x| – 2 = 0, then value of – 1/a is (a)

p

3 p3 - 2 p3 - 2q p3 + q

|x| = 1 3 ± 17 2 |x| ≥ 0, we get |x| = 1 3 + 17 2 1 x=± 3 + 17 2

(

fi

or

)

(

As

(

fi \

a=–

)

)

1 2 17 - 3 1 = 3 + 17 fi - = a 4 17 + 3 2

(

)

Example 88: Let f(x) = The range of f is

x2 - 2 x + 3 x2 - 2 x - 8

, x ΠR Р{Р2, 4}

Ê -2 ˘ (a) Á , 1˙ Ë 9 ˚

-2 (b) R – Ê , 1ˆ Ë 9 ¯

-2 ˘ Ê (c) Á - •, ˙ Ë 9˚

Ê -2 ˘ (d) R – Á , 1˙ Ë 9 ˚

=

p3 - 2q

Solution: Let y =

p3 + q

( p3 - 2 q )

x+1=0 p3 + q (p3 + q)x2 – (p3 – 2q)x + (p3 + q) = 0

Example 86: Suppose a, b Œ R. If the roots of x2 – ax + b = 0 are real and distinct and differ by at most 1, then 4b lies in the interval (b) [a2 – 1, a2) (a) (a2 – 1, a2) 2 (d) (a2, •) (c) (a – 1, •) Ans. (b)

x2 - 2 x + 3 x2 - 2 x - 8

, x ΠR Р{Р2, 4}

Note that y π 1. Now, (y – 1)x2 – 2x(y – 1) – (8y + 3) = 0

Thus, required quadratic equation is x –

(d) – 3

Ans. (d)

a b ◊ =1 b a

2

)

17 + 3 4

=0

2

-2

(

(b)

Solution: x2 – 3|x| – 2 = 0 can be written as |x|2 – 3|x| – 2

2

( p + q )/ 3 p

)

17 - 3 4

(c) 2 Ans. (a)

(a + b ) - 2ab (a + b ) = -2 ab ab 3

(

As x is real fi fi fi

4(y – 1)2 + 4(y – 1) (8y + 3) ≥ 0 (y – 1) [y – 1 + 8y + 3] ≥ 0. (y – 1) (y + 2/9) ≥ 0 y £ – 2/9 or y > 1. [ y π 1]

Thus,

y ΠR Р(Р2/9, 1].

Example 89: Suppose a, b, c are three non-zero real numbers. The equation

Quadratic Equations 3.23

x2 + (a + b + c)x + (a2 + b2 + c2) = 0

(1)

Example 93: Let for a, a1 π 0, a π a1 f(x) = ax2 + bx + c, g(x) = a1x2 + b1x + c1

has (a) (b) (c) (d) Ans. (d)

two negative real roots, two positive real roots, two real roots with opposite signs, no real roots

and

p(–2) = 2, then the value of p(2) is (a) 3 (b) 9 (c) 6 (d) 18 Ans. (d)

= Р{(b2 + c2 Р2bc) + (c2 + a2 Р2ca) + (a2 + b2 Р2ab) + (a2 + b2 + c2)} = Р[(b Рc)2 + (c Рa)2 + (a Рb)2 + (a2 + b2 + c2)] < 0 Thus, (1) cannot have real roots. Example 90: Suppose a, b ΠR. If the equation x2 Р(2a + b)x + (2a2 + b2 Рb + 1/2) = 0 has two real roots, then 1 1 (a) a = , b = Р1 (b) a = Р, b = 1 2 2 (c) a = 2, b = 1 (d) a = Р2, b = Р1 Ans. (a)

(1)

Solution: As (1) has real roots, (2a + b)2 – 4 (2a2 + b2 – b + 1/2) ≥ 0 fi

4a2 + 3b2 – 4ab – 4b + 2 £ 0

fi

(2a – b)2 + 2(b – 1)2 £ 0 fi b = 1, a =

1 2

Example 91: The equation esin x – e– sin x = 4 has: (a) no real roots (b) exactly one real root (c) exactly four real roots (d) infinite number of real roots. Ans. (a) Solution: Put esin x = y. Note that 1 e £ y £ e. Also, the given equation can be written as fi

or

y2 – 4y – 1 = 0

y = 2 ± 5.

As 1 e £ y £ e, none of the two values of y is possible. 2

4

If p(x) = 0 only for x = –1

and

Solution: Discriminant D of (1) is given by D = (a + b + c)2 – 4(a2 + b2 + c2)

y– 1 y =4

p(x) = f(x) – g(x).

Solution: As a π a1, p(x) = f(x) – g(x) is a quadratic polynomial. As p(x) = 0 only for x = –1, we get p(x) = k(x +1)2 where k = a –a1 As Thus,

Example 94: If a, b, c are non-zero rational numbers such that a + b + c = 0, then the roots of the equation (b + c – a) x2 + (c + a – b) x + (a + b – c) = 0 are (a) both rational (b) both irrational (c) both purely imaginary (d) one rational and one irrational Ans. (a) Solution: As a + b + c = 0, we can write the equation 2ax2 + 2bx + 2c = 0 or ax2 + bx + c = 0. Note that x = 1 satisfies this equation and its other root is c/a. Example 95: If a and b are the roots of the equation ax2 + bx + c = 0, then roots of ax2 – bx(x – 1) + c(x – 1)2 = 0 (1) are 1 1 , (b) 1 , b (a) a b a 1 a b , (c) (d) a , 1+a 1+ b b Ans. (c) Solution: We can write (1) as

2

Example 92: If x – 3x + 2 is a factor of x – ax + b then the equation whose roots are a and b is (b) x2 – 9x + 20 = 0 (a) x2 – 9x – 20 = 0 2 (c) x2 + 9x + 20 = 0. (c) x + 9x – 20 = 0 Ans. (b) Solution: As x2 – 3x + 2 = (x – 1) (x – 2) is a factor of x4 – ax2 + b, x = 1 and x = 2 are zeros of x4 – ax2 + b, therefore, 1 – a + b = 0, 16 – 4a + b = 0 fi a = 5, b = 4. Thus, equation whose roots a and b is x2 – 9x + 20 = 0.

p(–2) = 2, we get 2 = k(–2 + 1)2 = k. p(2) = 2(2 + 1)2 = 18

x ˆ2 Ê Ê -x ˆ aÁ+ bÁ +c=0 ˜ Ë x - 1¯ Ë x - 1˜¯ fi

-x = a, b x -1

fi

x=

a b , 1+a 1+ b

Example 96: If 3p2 = 5p + 2 and 3q2 = 5q + 2 then the equation whose roots 3p – 2q and 3q – 2p is (a) x2 – 5x + 100 = 0 (c) 3x2 + 5x + 100 = 0 Ans. (b)

(b) 3x2 – 5x – 100 = 0 (d) 5x2 – x + 7 = 0

3.24

Complete Mathematics—JEE Main

Solution: Note that p, q are roots of 3x2 = 5x + 2 or 3x2 – 5x – 2 = 0 p + q = 5/3, pq = –2/3. a = 3p – 2q, b = 3q – 2p,

Let then and

a + b = p + q = 5/3 a b = (3p – 2q) (3q – 2p)

˘ È 1 1 1 1 ...., ˙ y = 6 + log3/2 Í 444Í3 2 ˙ 3 2 3 2 3 2 Î ˚ is (a) 2 (b) 3 (c) 4 (d) 8 Ans. (c)

= – 6(p2 + q2) + 13pq = – 6(p + q)2 + 25pq = –100/3. Thus, required quadratic equation is

Solution: Let x = 1 3 2

4-

1 3 2

4-

2

x – (5/3)x – 100/3 = 0 fi

2

3x – 5x – 100 = 0

Example 97: The number of distinct real roots of the equation (x + 3)4 + (x + 5)4 = 16 (1) is (a) 1 (b) 2 (c) 3 (d) 4 Ans. (b) Solution: Put x + 4 = t, so that (1) becomes (t – 1)4 + (t + 1)4 = 16 fi 2(t4 + 6t2 + 1) = 16 fi t4 + 6t2 – 7 = 0 fi t2 = 1, –7 fi t = ±1, 7i Thus, the equation (1) has two real roots. Example 98: If a, b, c, Œ R and the equation x2 + (a + b)x + c = 0 has no real roots, then c(a + b + c) more than (a) 2 (b) – 2 (c) 0 (d) none of these. Ans. (c) 2

Solution: We have f(x) = x + (a + b) x + c > 0 " x ΠR Thus, f(0) > 0, f(1) > 0 fi c(a + b + c) > 0 Example 99: The value of

then fi fi fi As Thus,

1 3 2

4-

1 ...., 3 2

1

4-x 3 2 18x2 + x – 4 = 0 (9x – 4) (2x + 1) = 0 x = 4/9, –1/2 x > 0, we get x = 4/9. y = 6 + log3/2 (4/9) = 6 + log3/2 (3/2)–2 = 4 x=

Example 100: The sum of the roots of the equation x + 1 – 2 log2 (2x + 3) + 2 log4 (10 – 2–x) = 0 (1) is (a) log211 (c) log213 Ans. (a)

fi fi fi fi

(b) log212 (d) log214

Solution: Put 2x = t, then (1) can be written as 2 1ˆ Ê 1 + log2t – 2 log2 (t + 3 ) + log2 ÁË 10 - ˜¯ = 0 t 2 1 + log2t – log2 (t + 3)2 + log2(10t – 1) – log2t = 0 log2(10t – 1) / (t + 3)2 = –1 1 10t – 1 = (t + 3)2 2 t2 – 14t + 11 = 0 (2) If t1 = 2 x 1 , t2 = 2 x2 are roots of (2),

then, t1t2 = 2 x1 . 2 x2 = 11 fi x1 + x2 = log211

Assertion-Reason Type Questions

Example 101: Suppose a, b, c, p Œ R a π 0, c π 0 and b2 – 4ac > 0. Statement-1: If the Roots of f(x) = ax2 + bx + c = 0 are symmetrically placed on the real line about the point p, then p = b/2a and a(ap2 + bp + c) < 0 Statement-2: If the roots of ax2 + bx + c = 0 are equal in magnitude but opposite in signs, then b = 0 and ac < 0.

Ans. (a) Solution: If a, – a are roots of ax2 + bx + c = 0, then 0 = a + (– a) = –b/a fi b = 0 Also, a(– a) = c/a. As c π 0, – a π 0 and therefore c/a = – a2 < 0 fi ac < 0. Thus, Statement-2 is true.

Quadratic Equations 3.25

If p – a and p + a are roots of f(x) = ax2 + bx + c = 0, then a, – a are roots of f(x + p) = a(x + p)2 + b(x + p) + c = 0, that is, a, – a are roots of ax2 + (2ap + b)x + ap2 + bp + c = 0. By Statement-2: 2ap + b = 0 fi p = –b/2a and a(ap2 + bp + c) < 0. Example 102: Suppose a, b, c Œ R, a π 0, b2 – 4ac > 0. Statement-1: If a, b are roots of f(x) = ax2 + bx + c = 0, such that a < – 1 and b > 1, then a + |b| + c < 0. Statement-2: If a, b are roots of f(x) = ax2 + bx + c = 0 such that a < – 1, b > 1, then ac < 0. Ans. (d) Solution: If a < –1, b > 1, then a < 0, b > 0 c < 0 ¤ ac < 0. fi ab < 0 ¤ a Thus, Statement-2 is true. b c As ax2 + bx + c = 0 and x2 + x + = 0 have the same a a roots, roots of x2 +

b c x+ = 0 are a, b. a a

b c x + = (x – a) (x – b). a a Note that g(x) < 0 for a < x < and a < – 1 < 1 < b \ g(– 1) < 0, g(1) < 0 \

g(x) = x2 +

fi

b c b c 1 – + < 0, 1 + + < 0 a a a a

fi

1+

b c + < 0. a a

We cannot conclude a + |b| + c < 0 unless a > 0. Example 103: Suppose a, b, c ΠR Statement-1: If a > 0, b2 Р4ac < 0, and f(x) = a + bx + cx 2 then domain of f is R Statement-2: ax2 + bx + c > 0 " x ΠR if a > 0 and b2 Р4ac > 0. Ans. (a) Solution: Statement-2 is true. See Theory. As ax2 + bx + c > 0 " x ΠR, we get c = a(0)2 + b(0) + c > 0. Thus, cx2 + bx + a > 0 " x ΠR. Therefore, domain of f is R. Hence, Statement-1 is true and Statement-2 is a correct explanation for it. Example 104: Suppose a, b, c, d ΠR. Statement-1: If a < b < c < d, then the quadratic equation (x Рa) (x Рc) + 2(x Рb) (x Рd) = 0 has real and distinct roots. Statement-2: Let f(x) = ax2 + bx + c: If f(a) f(b ) < 0 for some a, b ΠR, then f(x) = 0 has real and distinct roots.

Ans. (a) Solution: As f(x) = ax2 + bx + c is a continuous function and f(a) f(b) < 0, there is at least one g ΠR lying between a and b such that f(g ) = 0. As a, b, c ΠR and one root of f(x) = 0 is real, other must be real. Also, roots are distinct. Therefore, Statement-2 is true. Now, let f(x) = (x Рa) (x Рc) + 2 (x Рb) (x Рd), then f(a) = 2(a Рb) (a Рd) > 0, f(b) = (b Рa) (b Рc) > 0 Thus, f(x) = 0 has a real root lying in the interval (a, b). As coefficients are real and one of the roots is real, the quadratic equation must have real and distinct roots. Example 105: Suppose a, b, c a ΠR Statement-1: The quadratic equation (x Рsin a) (x Рcos a) + 3 = 0 has imaginary roots. Statement-2: If a > 0 and b2 Р4ac < 0, then the graph of y = ax2 + bx + c lies above the x-axis. Ans. (a) Solution: For truth of Statement-2, see Theory. We can write (x Рsin a) (x Рcos a) Р3 = 0 as f(x) = 0 where

f(x) = x2 – (sin a + cos a) + sin a cos a +

3

coefficient of x2 = 1 > 0, and b2 – 4ac = (sin a + cos a)2 – 4(sin a cos a + = (sin a – cos a)2 – 4 3

3)

2 £ ( 2) – 4 3 < 0

as maximum possible value of sin a – cos a is 2 . Thus graph of y = f(x) lies above the x-axis and hence f(x) = 0 has imaginary roots. Example 106: Suppose a, b, c Œ R and a π 0. Statement-1 If ac > 0, ab < 0, b2 – 4ac > 0 the equation ax4 + bx2 + c = 0 has four distinct roots. Statement-2 If ac > 0, ab < 0, then the equation ax2 + bx + c = 0 has distinct positive roots. Ans. (c) Solution: Statement-2 is false as x2 – x + 1 = 0 satisfies the condition ac > 0 and ab < 0 but x2 – x + 1 = 0 does not have positive roots. Statement-1 is true. Put x2 = y, so that ay2 + by + c = 0. As b2 – 4ac > 0. This equation has two distinct roots. b ab c ac = - 2 > 0 and = 2 > 0 both the roots of ay2 + by a a a a + c = 0 are positive. If the roots are a, b > 0, then x2 = a, b

As -

fi x=±

a, ± b .

3.26

Complete Mathematics—JEE Main

Thus, ax4 + bx2 + c = 0 has four distinct roots. Example 107: Suppose a, b, c Œ R, a π 0, and let f(x) = ax2 + bx + c. Statement-1: If f(x) = 0 has imaginary roots and a + c > 0, then f(x) > 0 " x Œ R. Statement-2: If f(x) > 0 " x Œ R, then g(x) = f(x) + f ¢(x) + f ¢¢(x) > 0 " x Œ R. Ans. (b) ¤

Solution: f(x) > 0 " x ΠR, a > 0, b2 Р4ac < 0.

We have g(x) = ax2 + bx + c + (2ax + b) + 2a = ax2 + (b + 2a)x + c + b + 2a Note that (b + 2a)2 – 4a (c + b + 2a) = (b2 – 4ac) – 4a2 < 0 and a > 0, thus g(x) > 0 " x Œ R \ Statement-2 is true. As f(x) = 0 has imaginary roots, f(x) > 0 " Œ R or f(x) < 0 "xŒR We have f(1) + f(–1) = 2(a + c) > 0, therefore f(x) > 0 " x Œ R. Thus, statement-1 is true but statement-2 is not a correct explanation for it.

Example 109: Suppose a, b, c Œ I, a π 0 and f(x) = ax2 + bx + c. Statement-1: If f(x) = 0 has no rational roots, then 1 Ê pˆ f Á ˜ ≥ 2 " p, q Œ I, q π 0 and Ë q¯ q Statement-2: If a +

f(x), then its other zero must be a –

( x - b) ( x - c) ( x - c) ( x - a) +b (a - b) (a - c ) (b - c ) (b - a )

+ c

( x - a ) ( x - b) = x " x ΠR. (c - a ) (c - b )

Statement-2: If Ax2 + Bx + C = 0 for x = a, b, c, then A = B = C = 0. Ans. (a) Solution: If three distinct values of x satisfy a quadratic equation, then it is identically equal to zero. Thus, Statement-2 is true. Let f(x) = a

Let

Solution: Statement-2 is true. (See Theory) x = p/q, p, q Œ I and q π 0.

Then

Ê pˆ 1 f Á ˜ = 2 (ap2 + bpq + cq2) Ë q¯ q f(x)= 0 has no rational roots, f(p/q) π 0 fi ap2 + bpq + cq2 π 0

As

As ap2 + bpq + cq2 is an integer, we get |ap2 + bpq + cq2| ≥ 1. 1 Ê pˆ Ê bˆ Thus, q2 f Á ˜ ≥ 1 fi f Á ˜ ≥ 2 . Ë q¯ q Ë q¯ Example 110: Suppose a, b, c Œ R and a < b < c. Let f : R Æ R be defined by Ï ( x - a )( x - c) if x π b Ô x-b f ( x) = Ì ÔÓ 0 if x = b Statement-1: Range of f is R. Statemetnt-2: For each y Œ R, y = f(x) has exactly two distinct solutions. Ans. (c) Solution: Statement-2 is false as f(a) = f(b) = f(c) = 0 fi f(x) = 0 has three distinct solutions. Note that Statement-1, put t = x – b, so that f(x) =

( x - a ) ( x - b) – x. (c - a ) (c - b )

Then f(x) = 0 is a quadratic and f(a) = f(b) = f(c) = 0. Thus, f(x) ∫ 0. Therefore, Statement-1 is true and statement-2 is a correct reason for it.

(t + b - a ) (t + b - c ) t

= t – (a + c – 2b) +

( x - b) ( x - c) ( x - c) ( x - a) +b (a - b) (a - c ) (b - c ) (b - a )

+ c

b.

Ans. (b)

Example 108: Suppose a, b, c ΠR are three distinct real numbers. Statement-1: a

b , a, b Œ Q and b π 0 is a zero of

= x – (a + c – b) +

( a - b ) (c - b ) t

( a - b ) (c - b ) x-b

note that f is continuous on (– •, b). Also, lim f(x) = – • xÆ - •

and lim f(x) = • xÆ b-

Thus, range of f is (–•, •) = R. \ Statement-1 is true.

Quadratic Equations 3.27

LEVEL 2 Straight Objective Type Questions Example 111: Suppose a, b, c, d Œ R and 2ac = b + d. Consider the quadratic equations x2 + 2ax + b = 0 and x2 + 2cx + d = 0. Then (a) none of these equations have real roots (b) both the equations have real roots (c) exactly one of the equation has real roots (d) at least one of the equation has real roots Ans. (d) Solution: Let D1 be the discriminant of x2 + 2ax + b = 0 and D2 be the discriminant of x2 + 2cx + d = 0 Then D1 + D2 = 4 (a2 – b) + 4(c2 – d) = 4(a2 + c2) – 8ac = 4(a – c)2 ≥ 0 fi at least one of D1, D2 is non-negative. Thus, at least one of the equations has real roots. Example 112: If the roots of the equation 1 1 1 (1) + = x+a x+b c are equal in magnitude but opposite in sign, then their product is 1 2 1 2 (a + b2) (b) – (a + b2) (a) 2 2 1 1 (c) ab (d) – ab 2 2 Ans. (b) Solution: The equation (1) can be written as c(x + b + x + a) = (x + a) (x + b) (2) or x2 + (a + b – 2c)x + ab – ac – bc = 0 Let a and – a be the roots of (2) then 1 (a + b) 0 = a + (– a) = a + b – 2c fi c = 2 Also, 2a ( – a) = 2ab – 2 (a + b)c = 2ab – (a + b)2 = – (a2+ b2) 1 fi a (– a) = - (a 2 + b2 ) 2 Example 113: If c, d are roots of x2 – 10ax – 11b = 0 and a, b are root of x2 – 10cx – 11d = 0, then value of a + b + c + d is (a) 1210 (b) – 1 (c) 2530 (d) – 11 Ans. (a)

Solution: c + d = 10a (1) a + b = 10c (2) Subtracting (1) from (2) we get (a – c) + (b – d) = 10 (c – a) fi b – d = 11(c – a) (3) As c is a root of x2 – 10ax – 11b = 0, we get c2 – 10ac – 11b = 0 (3) 2 (4) Similarly, a – 10ac – 11d = 0 Subtracting (4) from (3), we get c2 – a2 = 11(b – d) fi (c – a) (c + a) = (11)11(c – a) fi c + a = 121 \ a + b + c + d = 10(a + c) [from (1) and (2)] = 10(121) = 1210 Example 114: If a, b are the roots of the equation ax2 + bx + c = 0, then the value of a3 + b3 is (a) (c) Ans. (c)

3abc + b3 a3 3abc - b3 a

3

(b) (d)

a 3 + b3 3abc

(

- 3abc + b3 a

)

3

Solution: We have -b c a+b= , ab = a a Now, a3 + b3 = (a + b)3 – 3ab(a + b) 3 3c Ê - b ˆ 3abc - b3 Ê - bˆ = Á ˜ - Á ˜= Ë a ¯ aË a ¯ a3

Example 115: Suppose a Œ I and the equation (x – a) (x – 5) = 3 has integral roots, then the set of values which a can take is: (a) f (b) {– 11, – 13} (c) {3, 7} (d) {– 3, – 7} Ans. (c) Solution: Let m Œ I be a roots of (x – a) (x – 5) = 3, then (m – a) (m – 5) = 3 As m – a and m – 5 are integers, m – 5 = ± 1 or m – 5 = ± 3 fi m = 2, 4, 6, 8. Thus, a = 3, 7.

3.28

Complete Mathematics—JEE Main

Example 116: If the ratio of the roots of the equation x + bx + c = 0 is the same as that of the ratio of the roots of x2 + qx + r = 0, then (b) cq2 = rb2 (a) br2 = qc2 2 2 2 2 (c) q c = b r (d) bq = rc Ans. (b) 2

Solution: Let a, b be the roots of the equation x2 + bx + c = 0 and g, d be the roots of the equation x2 + qx + r = 0. We are given a g a - b g -d = = fi b d a + b g +d 2

2

fi

Êa - bˆ Êg -dˆ ÁË a + b ˜¯ = ÁË g + d ˜¯

fi

(a + b )2 - 4ab (g + d )2 - 4gd = (a + b )2 (g + d )2

fi fi

ab 2

(a + b ) c 2

(- b)

=

=

gd

But f ¢(x) = 3ax2 + 2bx + c 2 Hence, 3ax + 2bx + c = 0 has a root in [0, 1]. Example 119: If a < b < c < d, then the equation 3(x – a) (x – c) + 5(x – b) (x – d) = 0 has (a) real and distinct roots (b) real and equal roots (c) purely imaginary roots (d) none of these Ans. (a) Solution: Let f(x) = 3(x – a) (x – c) + 5(x – b) (x – d) Since f is a polynomial, f is continuous on R. Also, since a < b < c < d, f(a) = 5(a – b) (a – d) > 0 f(b) = 3(b – a) (b – c) < 0 f(c) = 5(c – b) (c – d) < 0 f(d) = 3(d – a) (d – c) > 0

(g + d )2 r 2

(- q )

fi cq2 = rb2

Example 117: If a and b are the non-zero distinct roots of x2 + ax + b = 0, then the least value of x2 + ax + b is (a) 2/3 (b) 9/4 (c) – 9/4 (d) 1 Ans. (c) Solution: We have a + b = – a, ab = b As b π 0, we get a = 1 \ 1+b=–1fib=–2 Thus, x2 + ax + b = x2 + x – 2 Ê = Áx + Ë \

1ˆ 2 9 9 ˜¯ - ≥ 2 4 4

least value of x2 + ax + b is – 9/4 which is attained at x = – 1/2.

Example 118: If a + b + c = 0, then the quadratic equation 3ax2 + 2bx + c = 0 has (a) at least one root in [0, 1] (b) one root in [2, 3] and other is [ – 2, – 1] (c) imaginary roots (d) none of these Ans. (a) Solution: Let f(x) = ax3 + bx2 + cx, x Œ [0, 1]. Since f is a polynomial function, f is differentiable on the whole real line and in particular on [0, 1]. Also, f(0) = 0 and f(1) = a + b + c = 0. Thus, all the conditions in the hypothesis of the Rolle’s theorem are satisfied. By the Rolle’s theorem there exists at least one a Œ (0, 1) such that f ¢(a) = 0

Fig. 3.18

As f is continuous on R, y = f(x) crosses x-axis at least once between a and b and once between c and d. See Fig. 3.18. Thus, f has two distinct real roots, one lying between a and b and one lying between c and d.

( x - a) ( x - c) Example 120: For real x, the function x-b will assume all real values provided (a) a < b < c (b) b < c < a (c) c < a < b (d) none of these Ans. (a) x 2 - ( a + c ) x + ac x-b x-b 2 fi y(x – b) = x – (a + c)x + ac 2 (1) fi x – (a + c + y)x + ac + by = 0 Since x is real, the discriminant of (1) (a + c + y)2 – 4(ac + by) ≥ 0 fi y2 + 2(a + c)y + (a + c)2 – 4ac – 4by ≥ 0 2 (2) fi y + 2(a + c – 2b)y + (a – c)2 ≥ 0 Solution: Let y =

( x - a) ( x - c)

=

Since y takes all real values, (2) is possible if and only if 4(a + c – 2b)2 – 4(a – c)2 < 0 [∵ coeff. of y2 = 1> 0] ¤ (a + c – 2b + a – c) (a + c – 2b – a + c) < 0

Quadratic Equations 3.29

¤ ¤ ¤

(2a – 2b) (2c – 2b) < 0 (a – b) (c – b) < 0 b lies between a and c. Thus, one of the possibilities a < b < c.

Example 121: Let a, b, c Œ R and a π 0. If a is a root of a2 x2 + bx + c = 0, b is a root of a2 x2 – bx – c = 0 and 0 < a < b, then the equation a2 x2 + 2bx + 2c = 0 has a root g that always satisfies 1 1 (a + b) (b) g = a + b (a) g = 2 2 (c) g = a + b (d) a < g < b Ans. (d) Solution: By the hypothesis, (1) a2 a 2 + ba + c = 0 2 2 (2) and a b – bb – c = 0 where 0 < a < b, and a π 0. Let f(x) = a2 x2 + 2bx + 2c Since f is a polynomial function, f is continuous on R and hence on [a, b]. Also, f(a) = a2 a 2 + 2ba + 2c = 2(a2a2 + ba + c) – a2 a 2 = 2(0) – a2 a2 = – a2 a 2 < 0 2

[using (1)]

2

f ( b ) = a b + 2bb + 2c = 3a2 b 2 – 2(a2 b 2 – bb – c) = 3 a2 b 2 – 2(0) = 3a2 b 2 > 0 [using (2)] Since f is continuous on [a, b] and f(a) < 0, f(b) > 0, f must vanish at least once in (a, b).

and

Example 122: Suppose p, q, r, s Œ R and a, b be the roots of x2 + px + q = 0 and a 4, b 4 be the roots of x2 – rx + s = 0, then the equation x2 – 4qx + 2q2 – r = 0 has always (a) two imaginary roots (b) two positive roots (c) two negative roots (d) one positive and one negative root Ans. (d) Solution: We have a + b = – p, ab = q, a 4 + b 4 = r, a 4 b 4 = s Discriminant D of the equation (1) x2 – 4qx + 2q2 – r = 0 2 2 2 2 is given by D = 16q – 4(2q – r) = 8q + 4r = 4[2q + r] = 4[2a 2 b 2 + a 4 + b 4] = 4(a 2 + b 2)2 ≥ 0 Thus, the equation (1) has real roots. The roots of (1) are given by 4q ± D x= = 2ab ± (a 2 + b 2) 2 = (a + b)2, – (a – b)2 Hence, (1) has one positive and one negative root.

Example 123: The equation x (3 / 4 )(log2 x )

2

+ log2 x - 5 / 4

=

2

(1)

has (a) (b) (c) (d) Ans. (c)

exactly two real roots no real root one irrational root none of these

Solution: Taking log of the sides in (1), we get 5˘ 2 È3 ÍÎ 4 ( log2 x ) + log2 x - 4 ˙˚ log2 x = log2 ( 2 ) 5ˆ 1 Ê3 2 fi ÁË y + y - ˜¯ y = log2 2 where y = log2 x 4 4 2 fi 3y3 + 4y2 – 5y – 2 = 0 fi 3y3 – 3y2 + 7y2 – 7y + 2y – 2 = 0 fi

(y – 1) (3y2 + 7y + 2) = 0

fi

(y – 1) [3y2 + 6y + y + 2] = 0

fi

(y – 1) (y + 2) (3y + 1) = 0

fi

y = 1, – 2, – 1/3

fi

log2 x = 1, – 2, – 1/3 fi x = 2,

1 –1/3 ,2 4

Thus, (1) has one irrational root viz. 2–1/3. Example 124: Let f(x) be a quadratic expression which is positive for all x. If g (x) = f (x) + f ¢(x) + f ¢¢(x) then for all real x, (a) g(x) < 0 (b) g(x) > 0 (c) g(x) = 0 (d) g(x) ≥ 0 Ans. (b) Solution: Let f(x) = ax2 + bx + c As f(x) > 0 " x Œ R, we must have a > 0 and b2 – 4ac < 0. Also, f ¢(x) = 2ax + b and f ¢¢(x) = 2a Thus, g(x) = ax2 + (2a + b)x + 2a + b + c. Since a > 0 and discriminant (2a + b)2 – 4a(2a + b + c) = 4a2 + 4ab + b2 – 8a2 – 4ab – 4ac [ b2 – 4ac < 0] = – 4a2 + (b2 – 4ac) < 0 we get g(x) > 0 " x Œ R Example 125: If a, b are the roots of the quadratic equation ax2 + bx + c = 0, then the quadratic equation whose roots are a 3, b 3 is (a) a3 y2 + (b3 – 3abc)y + c3 = 0 (b) a3 y2 + (3abc – b3)y – c3 = 0 (c) a2 y2 + 2aby + c2 = 0 (d) none of these Ans. (a) Solution: Put y = a3 fi a = y1/3. As a is a root of ax2 + bx + c = 0, we get ay2/3 + by1/3 + c = 0 fi y1/3 (ay1/3 + b) = – c

(1)

3.30

Complete Mathematics—JEE Main

Cubing both the side, we get y(ay

1/3

3

+ b) = – c

3

3

fi

y[a y + b3 + 3aby1/3 (ay1/3 + b)] = – c3

fi

y[a3y + b3 + 3ab (– c)] = – c3 [using (1)]

fi

a3y2 + (b3 – 3abc)y + c3 = 0

Example 126: If a, b, c ΠR and the equations ax2 + bx + c = 0 and x3 + 3x2 + 3x + 2 = 0 have two common roots, then (a) a = b = Рc (b) a = Рb = c (c) a = b = c (d) none of these Ans. (c) Solution: x3 + 3x2 + 3x + 2 = 0 fi (x + 1)3 = Р1 fi x + 1 = Р1, Рw, Рw2 fi x = Р2, Р1, Рw, Р1 Рw2 As a, b, c ΠR, the roots of ax2 + bx + c = 0 are both real or both imaginary. \ roots of ax2 + bx + c = 0 must be Р1 Рw, Р1 Рw2. b Thus, Р1 Рw Р1 Рw2 = Рa c and (Р1 Рw) (Р1 Рw2) = a b c fi 1= and 1 = a a fi a=b=c

\ f(x) is partly positive and partly negative on [1, 2]. fi there exist a, b Π[1, 2] such that f(a) > 0 and f(b) < 0. As f is continuous on [1, 2] there exists g lying between a and b (and hence between 1 and 2) such that f(g) = 0 fi (1 + cos8 g) (ag2 + bg + c) = 0 fi

ag2 + bg + c = 0 [ 1 + cos8 g ≥ 1] Thus, ax2 + bx + c = 0 has at least one root in [1, 2].

Example 128: The number of real solutions of the equation (1) 271/x + 121/x = 2(81/x ) is (a) 0 (b) 1 (c) infinite (d) none of these Ans. (a) Solution: We can write (1) as 1/ x 1/ x Ê 27 ˆ Ê 12 ˆ =2 + ÁË ˜¯ ÁË ˜¯ 8 8 or As and

Ê 3ˆ ÁË ˜¯ 2

3/ x

Thus,

Ê 3ˆ ÁË ˜¯ 2

3/ x

and

Example 127: Let a, b, c be non-zero real number such that 1

Ú0 (1 + cos =

2

Ú0

8

2

x) (ax + bx + c) dx

(1 + cos8 x) (ax2 + bx + c) dx

Then the quadratic equation ax2 + bx + c = 0 has (a) no root in (0, 2) (b) at least one root in (1, 2) (c) a double root (0, 2) (d) none of these Ans. (b) Solution: Let f(x) = (1 + cos8 x) (ax2 + bx + c) We are given 1

Ú0 fi

1

2

f(x) dx =

Ú f(x) dx 0 1

2

Ú0 f(x) dx = Ú0 f (x) dx + Ú1 2

f(x) dx

Ú1

If

f(x) > 0 (f(x) < 0) ⁄ x Œ [1, 2],

then

Ú1

But

Ú1

2

f(x) dx = 0

1/ x

Ê 3ˆ +Á ˜ Ë 2¯

1/ x

< 2 if x < 0

> 2 if x > 0.

Therefore, (1) is possible if 1/x = 0. But this is not true for any real value of x. Example 129: If 0 < a < b < c < d, then the quadratic equation (1) ax2 + {1 – a(b + c)}x + abc – d = 0 has (a) real and distinct roots out of which one lies between c and d. (b) real and distinct roots out of which one lies between a and b (c) real and distinct roots out of which one lies between b and c (d) non-real roots Ans. (a)

f(x) dx = 0

f(x) dx > 0

Ê 3ˆ +Á ˜ Ë 2¯

(1)

Solution: We can rewrite (1) as ax2 – a(b + c)x + abc + x – d = 0

fi

2

3/ x 1/ x Ê 3ˆ Ê 3ˆ =2 + ÁË ˜¯ ÁË ˜¯ 2 2 3/2 > 1, (3/2)t > 1 if t > 0 (3/2)t < 1 if t < 0.

(Ú

2

1

f (x) d x < 0

)

or

a(x – b) (x – c) + x – d = 0

Let f(x) = a(x – b) (x – c) + x – d. As a > 0, y = f(x) represents a parabola which open upwards. See Fig. 3.19.

Quadratic Equations 3.31

b

Solution: As f(x) = ax2 + 2bx – 3c = 0 has no real roots, f(x) > 0 " x Œ R or f(x) < 0 " x Œ R Since 4a + 4b – 3c > 0, f(2) > 0. \ f(x) = ax2 + 2bx – 3c > 0 fi f(0) = – 3c > 0 fi c < 0 Also, a > 0 and b2 + 3ca < 0 fi c < – b2/3a fi c Œ (– •, – b2/3a)

c d

Fig. 3.19

Also,

f(b) = b – d < 0

b x + c, x Œ R – {0}, assume all real values, then a and b satisfy the relation (a) ab £ 0 (b) ab ≥ 0 (c) ab ≥ 1 (d) ab £ 1 Ans. (a) b Solution: Let y = ax + + c, then x ax2 + (c – y)x + b = 0. As x is real, (c – y)2 – 4ab ≥ 0 This is true for each y Œ R if ab £ 0 Example 131: Suppose a, b, c Œ R and a π b. If ax +

f(c) = c – d < 0, and f(d) = a(d – b) (d – c) > 0 Thus, f(x) = 0 has a root between – • and b and a root between c and d. Example 130: If ax2 + 2bx – 3c = 0 has no real roots and 4 c< (a + b), then range of c is 3 (a) (0, b) (b) (–1, b) (c) (– •, – b2/3a) (d) (– •, – b/12a) Ans. (c)

EXERCISE Concept-based Straight Objective Type Questions 1. The number of real solutions of x2 – 2x + 2 + |x – 1| = 0 is: (a) 0 (b) 1 (c) 2 (d) infinite

6. The number of real solutions of x2 – 4|x| – 2 = 0 is

2. If roots of 7x2 – 11x + k = 0, k π 0 are reciprocal of each other, then k is equal to

7. Suppose a and b are roots of the equation x2 + px 3 + p = 0. If |a – b| = 10 , then p belongs to the 4 set (a) {–2, 5} (b) {–3, 2} (c) {2, –5} (d) {3, –5}

(a) –1 (c) 7

(b) 7/11 (d) 11/7

3. If a + b π 0 and the roots of x2 – px + q = 0 differ by –1, then p2 – 4q equals: (a) –1 (c) 1

(b) 0 (d) 2 2

2

4. If the equations x – ax + b = 0 and x + bx – a = 0 have a common root then (a) a = b (c) a – b = – 1

(b) a + b = – 1 (d) a – b = 1 2

5. If both the roots of x + x + a = 0 exceed a, then a belongs to: (a) (– •, –1) (c) (0, 1)

(b) (– •, – 2) (d) (1, •)

(a) 1 (c) 3

(b) 2 (d) 4

8. The number of solutions of is (a) 0 (c) 2

5+ x + x = 2

(b) 1 (d) infinite

9. Suppose a Œ R. The set of values of a for which the quadratic equation x2 – 2(a + 1)x + a2 – 4a + 3 = 0 has two negative roots is (a) (– •, – 1) (b) (1, 3) (c) (– •, 1) » (3, •) (d) f 10. Suppose a > 0, b > 0 and a + b = p/4. If tan a, tan b are roots of x2 – ax + b = 0, then

3.32

Complete Mathematics—JEE Main

(a) a + b = 1 (c) a = b

y

(b) a + b = 1, 0 < b < 1 (d) 0 < a + b < 2

11. If b > a, and c > 0 then the equation (x – a) (x – b) – c = 0 has: (a) (b) (c) (d)

2

12. If the quadratic equation x + 2 (k + 1)x + 9k – 5 = 0 has exactly one positive root, then k lies in the set (a) [5/9, •) (c) (– •, 5/9]

13. If k ΠR lies between the roots of ax + 2bx + c = 0, then ak2 + 2bk + c < 0 a2k2 + 2abk + ac < 0 a2k2 + 2abk + ac > 0 ak2 + 2bk + c > 0

14. If both the roots of the quadratic equation x2 – 4ax + 2a2 – 3a + 5 are less than 2, then a lies in the set (a) (b) (c) (d)

(9/2, •) (– •, 9/2) (–1, •) (2, •)

15. Fig. 3.20 Shows graph of y = ax2 + bx + c. Then which one of the following is not true. (a) (b) (c) (d)

0

a>0 c 0 b>0

x2

x

Fig. 3.20

16. Greatest value of the expression (a) 2

(b) (– •, 1) » (6, •) (d) [1, 6] 2

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

x1

both roots in (– •, a) both roots in (a, b] one root in (– •, a) and other root in (b, •) one root in (– •, a) and other root in [a, b]

8 2

9x - 6x + 5

is

(b) 5

1 (d) 9.2 3 17. If a, b Π{1, 2, 3, 4} the number of quadratic equation of the form ax2 + bx + c = 0 which have non-real complex roots is: (a) 27 (b) 35 (c) 52 (d) 56 18. The number of real roots of the equation (x Р1)2 + (x Р2)2 + (x Р3)2 = 0 is (a) 0 (b) 2 (c) 3 (d) infinite (c) 8

19. Suppose 0 < a < b < c. If the roots a, b of ax2 + bx + c = 0 are imaginary, then c a (b) |b | = (a) |a| = a c (c) a + b = 0 (d) a – b = – b/2a 20. If a > 0 and both the roots of ax2 + bx + c = 0 are more than 1, then (a) a + b + c > 0 (b) a + b + 4c = 0 (c) a + b + c < 0 (d) a + 4b + c = 0

LEVEL 1 Straight Objective Type Questions 21. In Fig. 3.21 graph of y = ax2 + 2bx + c is given. Which one of the following is not true? y

22. If ax2 + 2bx + c = 0 and ax2 + 2cx + b = 0, b π a+b+c is equal to c have a common root, then a (a) – 2 (c) 3/4

y = ax2 + bx + c x

0

Fig. 3.21

(a) a > 0 (c) c > 0

(b) b > 0 (d) b2 < ac

(b) – 1 (d) –1/4

23. Suppose a, b, c Œ R, a π 0 and 4a – 6b + 9c < 0. If ax2 + bx + c = 0 does not have real roots, then b+c is less than a (a) 0 (b) 1 (c) –1 (d) –2

Quadratic Equations 3.33

24. Suppose a, b are roots of 4x2 Р16x + c = 0, where c ΠN, the set of natural numbers. The number of values of c for which 1 < a < 2 < b < 3 is (a) 2 (c) 4

(b) 3 (d) 9

(a) (–7, –8) (c) (7, 8)

(b) (–8, –7) (d) (8, 7) a 1 1 26. Suppose a, b Œ R. If the equation = + x x-b x+b is not satisfied by any real value of x, then (b) 0 < a < 2 (d) 0 < b < a < 2

27. Suppose a Œ R, a π –1/2. Let a, b be roots of (2a + 1)x2 – ax + a – 2 = 0 If a < 1 < b, then (a)

1 (6 - 2 23 ) < a < 1 (b) 1 (6 - 2 23 ) < a < 1 7 7 2

1 1 < a < (6 + 2 23 ) (d) none of these 2 7 28. Suppose a, b are roots of 8x2 – 10x + 3 = 0, then (c) •

 (a

n

+b

n

)

is

n=0

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

(b) 2 (d) 0

34. If 2 + 5 i is a root of x2 – px + q = 0 where p and q are real, then the ordered pair (p, q) is equal to (a) (4, 9) (c) (3, 3)

(b) (9, 4) (d) (2, 3)

35. If the quadratic equation 2x2 + ax + b = 0 and 2x2 + bx + a = 0, (a π b) have a common root, the value of a + b is (a) – 3 (c) – 1

(b) – 2 (d) 0

36. If a, b, c, d and p are distinct real numbers such that (a2 + b2 + c2)p2 + 2(ab + bc + cd)p + (b2 + c2 + d2) £ 0, then a, b, c, d (a) are in A.P. (b) are in G.P. (c) are in H.P. (d) none of these 37. The number of real roots of the equation sin (ex) = 5x + 5–x is

30. Suppose 0 < b < c and f(x) =

(a) 2 (c) 0

(b) 1 (d) infinitely many

(b) – 2 (d) none of these

39. The roots of the equation |x2 – x – 6| = x + 2 are (a) – 2, 1, 4 (c) 0, 1, 4

(b) 0, 2, 4 (d) – 2, 2, 4

40. The number of real roots of the equation 2

x - bc , x ΠR, 2 x - (b + c )

then f(x) cannot lie in (b) (– •, b) (d) (0, b) » (b, c)

31. If a, b are the roots of x2 + px + q = 0 and g, d are the roots of x2 + rx + s = 0, then value of (a – g) (a – d) (b – g) (b – d) is (a) (b) (c) (d)

(b) 2 (d) 0

38. The value of a for which the equations x3 + ax + 1 = 0 and x4 + ax2 + 1 = 0 have a common root is

real and distinct real and equal purely imaginary non-real complex numbers

(a) (b, c) (c) (c, •)

(a) 4 (c) 1

(a) 0 (c) 2

7/4 3/7 6 7

29. Suppose a, b, c Œ R a π 0. If a + |b| + 2c = 0, then roots of ax2 + bx + c = 0 are (a) (b) (c) (d)

(a) 4 (c) 1

33. The number of real solutions of x2 – 3|x| + 2 = 0 is

25. Suppose a, b Œ R, ab π 0. If all the three quadratic equations x2 + ax + 12 = 0, x2 + bx + 15 = 0 and x2 + (a + b)x + 36 = 0 have a common negative root, then (a, b) is equal to

(a) 0 < a < b < 2 (c) 0 < b < 2

32. The number of real solutions of x2 + 5|x| + 4 = 0 is

(r – p)2 – (q – s)2 (r – p)2 + (q – s)2 (r – p)2 + (q – s)2 – 2rp(r – p) (q – s) none of these

x2 – 3|x| + 2 = 0 (a) 4 (c) 2

(b) 1 (d) infinite

41. The least value of n ΠN for which (n Р4)x2 + 8x + n + 2 > 0 " x ΠR, is (a) 11 (c) 8

(b) 10 (d) 7

42. Let f(x) be a quadratic expression such that f(x) < 0 " x Œ R. If g(x) = f(x) + f ¢(x) + f ¢¢(x) then for x Œ R. (a) g(x) < 0 (c) g(x) > 0

(b) g(x) £ 0 (d) g(x) ≥ 0

3.34

Complete Mathematics—JEE Main

43. If x + 1 is a factor of x4 + (p – 3)x3 – (3p – 5)x2 + (2p + 9)x + 12, then value of p is (a) – 2 (c) 1

(b) 2 (d) –1

44. If both the roots of x2 – (p – 4)x + 2e2log p – 4 = 0 are negative, then p belongs to 2 , 4) (c) (4, •)

(a)

(–

2 , 4) (d) none of these (b)

(

45. Let f and g be two real valued functions and S = {x|f(x) = 0} and T = {x|g(x) = 0}, then S « T represent the set of roots of (a) f(x) g(x) = 0

(b) f(x)2 + g(x)2 = 0

(c) f(x) + g(x) = 0

(d)

f (x) =0 g (x)

46. If domain of f(x) = x 2 + bx + 4 is R, then maximum possible integral value of b is (a) 2 (c) 4

(b) 3 (d) 5

47. If p Œ (–1, 1), then roots of the quadratic equation (p – 1)x2 + px + (a) (b) (c) (d)

1 - p2 = 0 are

48. If a, b, c are positive real numbers, then the number of positive real roots of the equation ax2 + bx + c = 0 is (b) 1 (d) infinite 2

2

49. If the roots of the equation x + p = 8x + 6p are real, then p belongs to the interval (a) [2, 8] (c) [– 2, 8]

(b) [– 8, 2] (d) [– 8, – 2]

50. If sum of the roots of the equation (a + 1)x2 + (2a + 3)x + (3a + 4) = 0 is – 3, then the product of the roots is (a) 1 (c) 2

(b) 4 (d) – 2

51. If 3 Р4i is a root of x2 Рpx + q = 0 where p, q 2p - q is ΠR, then value p+q (a) Р12/31 (c) Р15/31

x2 + (1 + i)x + 1 + i = 0 x2 + (1 + i)x + i = 0 x2 + 2(1 + i)x – 2 = 0 none of these

53. If a and b be the roots of the equation x2 + px Р1 = 0, where p ΠR. Then the minimum possible 2 p2 value of a2 + b 2 is (a) 2 (c) 2 +

54. The equation = 1 has (a) (b) (c) (d)

(b) 2 2 (d) none of these

2

x + 3- 4 x -1 + x +8 - 6 x -1

no solution exactly one solution exactly two solutions more than two solution

55. The equation |x – x2 – 1| = |2x – 3 – x2| has (a) (b) (c) (d)

no solution exactly one solution exactly two solutions more than two solutions

56. If sin a, cos a are the roots of the equation ax2 + bx + c = 0, (a π 0), then

purely imaginary non-real complex numbers real and equal real and distinct.

(a) 0 (c) 2

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

(b) – 13/31 (d) none of these

52. If x = 1 + i is a root of x3 – ix + 1 – i = 0, then the quadratic equation whose roots are the remaining two roots of x3 – ix + 1 – i = 0 is

a2 Рb2 + 2ac = 0 a2 + b2 Р2ac = 0 (a Рc)2 = b2 + c2 none of these x2 - x + 1 57. If x ΠR, and k = 2 , then x + x +1 (a) (b) (c) (d)

(a) 1/3 £ k £ 3 (c) k £ 0

(b) k ≥ 5 (d) none of these

58. If the equation x2 + bx + ca = 0 and x2 + cx + ab = 0 have a common root and b π c, then their other roots will satisfy the equation (a) x2 + (b + c)x + bc = 0 (b) x2 – ax + bc = 0 (d) none of these (c) x2 + ax + bc = 0 59. If the inequality

mx 2 + 3 x + 4 x2 + 2 x + 2

5 (c) m < 71/24 (d) m > 71/24 60. If a, b, c are distinct real numbers, then the quadratic expression

( x - b) ( x - c) ( x - c) ( x - a ) ( x - a ) ( x - b) + + ( a - b ) ( a - c ) (b - c ) (b - a ) (c - a ) (c - b ) is identically equal to

Quadratic Equations 3.35

(a) 1 (c) x2

69. The number of real solution of 2 sin = 7 is

(b) x (d) none of these

61. If ax2 + bx + c, a, b, c Œ R, a π 0 has no real zero and a – b + c < 0, then value of ac is (a) positive (c) negative

(b) zero (d) non-negative

62. Suppose p Œ R. Let a, b be roots of x2 – 2x – (p2 – 1)= 0 and g, d (g < d ) be roots of x2 – 2(p + 1) x + p(p – 1) = 0. If a, b Œ (g, d ), then p lies in the interval (a) (–1/4, 1)

(b) (–1, 1)

(c) (0, •)

(d)

(

2, • )

63. If a, b are the roots of the equation ax2 + 2bx + c = 0 and a + h, b + h are the roots of the equation Ax2 + 2Bx + C = 0, then b2 - ac

a (a) 2 = B - AC A (c) h =

bA - aB 2 Aa

2

2

64. The quadratic equation x + 7x = 14 (q + 1), where q is an integer has (a) (b) (c) (d)

real and distinct roots integral roots imaginary roots none of these

65. Let a, b, c Œ R and a > 0. If the quadratic equation ax2 + bx + c = 0 has two real roots a and b such c b that a < – 1 and b > 1, then value of + is a a (a) less than 2 (b) less than 1 (c) less than 0 (d) less than – 1 66. Let a, b, c Œ R and a π 0 be such that (a + c)2 < b2, then the quadratic equation ax2 + bx + c = 0 has (a) (b) (c) (d)

imaginary roots real roots two real roots lying between (– 1, 1) none of these

(x – a) (x – 10) + 1 = 0 has integeral roots are (a) – 1, 3 (b) 2, 3 (c) 12, 8 (d) – 8, – 12 68. The number of real solution of 4 (10) (6x) is (a) zero (c) two

(b) one (d) infinite

(a) 0 (c) 2

(b) 1 (d) infinitely many

71. If roots of the quadratic equation ax2 + bx + c = 0 a a +1 are real and are of the form , then value , a -1 a 2 of (a + b + c) is (a) 4ac – b2 (c) c2 + a2 – 2b2

+ 9

x + 0.5

(b) b2 – 4ac (d) none of these

(b) 1 (d) any of (a), (b) and (c)

73. If a, b are the roots of the equation x2 + ax + b = 0, 1 then maximum value of – x2 + ax + b + (a – 4 b)2 is (a)

1 2 (a – 4b) 4

(b)

(c)

a2 2

(d) none of these

(a) b = – 1, c = 2 (c) b = – 5, c < 2

(b) b > – 2, c < 1 (d) none of these

75. Let a, b, c ΠR be such that a + b + c < 0, a Рb + c < 0 and c > 0. If a and b are roots of the equation ax2 + bx + c = 0, then value of [a] + [b] is (b) 1 (d) 0

76. If roots of the equation x2 – 2mx + m2 – 1 = 0 lie in the interval (– 2, 4), then (a) – 1 < m < 3 (c) 1 < m < 3 77. The value of

=

1 2 (b – 4a) 4

74. If both the roots of the equation x2 + bx + c = 0 lie in the interval (0, 1), then

(a) 2 (c) – 1

67. The integral values of a for which the quadratic equation

x + 1.5

(b) 1 (d) infinitely many

70. The number of values of k for which the equation x2 – 2x + k = 0 has two distinct roots lying in the interval (0, 1) is

(a) 0 (c) 2

Ac + aC Aa + Bb

2

2

+ 5(2 cos x )

72. Let x be an integer and x2 + x + 1 is divisible by 3. When x is divided by 3, it leaves remainder

2

(b) b – ac = B – AC (d) h =

(a) zero (c) finitely many

2x

(a) 10 (c) 8

(b) 1 < m < 5 (d) – 1 < m < 5 8 + 2 8 + 2 8 + 2 8 +  is (b) 6 (d) none of these

78. The number of solutions of the equation Ê px ˆ = x2 – 2 3 x + 4 is sin Á Ë 2 3 ˜¯

3.36

Complete Mathematics—JEE Main

(a) 1 (c) 0

(b) 2 (d) infinite

79. The number of solutions of |x + 2| = 2(3 – x) is (a) 1 (c) 3

(b) 2 (d) 0

80. If the equation ax2 + bx + c = 0, a > 0 has two distinct roots a and b such that a < – 2 and b > 3, then (a) c > 0 (c) c < 0

(b) c = 0 (d) c = a – b

81. Two non-integer roots of (x2 – 5x)2 – 7(x2 – 5x) + 6 = 0 are

(a) 0 (c) 3

(b) 2 (d) 4

87. The number of irrational roots of the equation 1ˆ 1 2 Ê 4 ÊÁ x - ˆ˜ + 8 Á x + ˜ = 29 Ë Ë ¯ x¯ x is (a) 0 (b) 2 (c) 4 (d) infinite 88. Irrational roots of the equation 2x4 + 9x3 + 8x2 + 9x + 2 = 0 are (a) - 2 - 3 , 2 + 3

(b) 2 - 3 , 2 + 3

(c) - 2 + 3 , - 2 - 3 (d) none of these (a) (b) (c) (d)

1 1 (5 + 29 ), (5 - 29 ) 2 2 1 1 (- 5 + 29 ), (- 5 + 29 ) 2 2 1 1 (- 5 + 14 ), (- 5 - 41) 2 2 none of these

89. Sum of the roots of the equation

is (a) 5 (c) – 5/2

2

Ê x - 1ˆ Ê x - 1ˆ ÁË x + 1˜¯ – 13 ÁË x + 1˜¯ + 36 = 0, x π – 1

83. The number of negative roots of 9x + 2 – 6(3x + 1) + 1 = 0 (b) 1 (d) 4

(a) 1 (c) 3

is (a) 2 (c) 7

2

+ 4 = 0, x π 1/3

(x2 + 3x + 2)2 – 8(x2 + 3x) – 4 = 0 (b) 2 (d) 4

86. The number of roots of the equation

is

3x - 6 = 2 (b) 5 (d) 10

92. The number of real roots of

85. The number irrational roots of

x + x-3

(b) 2 (d) 4

x–

2 2 x + 1 = 2x – 1 is (a) 1 (c) 3

x-3 5 = , x π 0, x π 3 2 x

(b) 2 (d) 4

93. Product of roots of the equation

(b) 2 (d) 4

is (a) 0 (c) 3

(b) 1 (d) – 1

is (a) 0 (c) 3

84. The number of rational roots of 4

1ˆ ˜ +1=0 x¯

91. Product of roots of the equation

(b) 2 (d) 4

Ê 2 x - 5ˆ Ê 2 x - 5ˆ 81 Á – 45 Á Ë 3 x + 1 ˜¯ Ë 3 x + 1 ˜¯ is

Ê ÁË x -

(x – 1) (x – 2) (3x – 2) (3x + 1) = 21

4

is (a) 0 (c) 2

1ˆ 2 ˜ –4 x¯

90. The number of irrational roots of the equation

82. The number of real roots of

is (a) 0 (c) 3

Ê 4Áx Ë

13 - x 2 = x + 5 is (a) – 6 (c) 6

(b) 7 (d) – 7

94. The number of roots of the equation x 2 - 4 - ( x - 2) = x 2 - 5 x + 6 is (a) 0 (b) 1 (c) 2 (d) 3 95. The product of the roots of the equation

Quadratic Equations 3.37

98. If a is a root of x4 + x2 – 1 = 0, the value of (a6 + 2a4)2012 is

x 2 - 4 x + 3 + x 2 - 7 x + 12 = 3 x - 3 is (a) 15 (b) – 15 (c) 20 (d) – 20 96. Suppose a and b satisfy the equations 18a2 + 77a + 2 = 0 and 2b2 + 77b + 18 = 0 then ab + a + 1 is value of b (a) – 25 (b) – 25/6 (c) 6/25 (d) – 1/25 97. Suppose a, b are roots of x2 – 7x + 8 = 0, with a 16 + 3b 2 - 19b is > b, then value of a (a) – 10 (b) 10 (c) – 23 (d) 17

(a) 0 (b) – 1 (c) 1 (d) none of these 99. Sum and product of all the roots of the equation (x2 – x – 1) (x2 – x – 2) (x2 – x – 3) … (x2 – x – 2012) = 0 is (a) 2012, 2012! (b) – 2012, 2012! (c) – 2012, – 2012! (d) 2012, – 2012! 100. Suppose three distinct non-zero real numbers satisfy a2(a + k) = b2(b + k) = c2(c + k), where k is some 1 1 1 real number, then value of + + is a b c (a) 0 (b) k (c) – k (d) 2k

Assertion-Reason Type Questions 101. Suppose a, b, c, a, b Œ R a π 0 and 0 < a < b. Let f(x) = a2x2 + bx + c, g(x) = a2x2 – bx – c

Statement-1: If ax2 + bx + c = x does not have real roots, then

Statement-1: If a is a root of f(x) = 0 and b is a root of g(x) = 0, then there exists g Π(a, b ) such that g is a root of a2x2 + 2bx + 2c = 0.

a(ax2 + bx + c)2 + b(ax2 + bx + c) + c = x has two real and two imaginary roots.

Statement-2: If a function h : [a, b] Æ R is continuous on [a, b] and h(a) h(b) < 0, then there exists g Œ (a, b) such that h(g ) = 0. 102. Suppose p Œ R and f(x) = x2 + 2px + p Statement-1: If f(x) < 0 " x Œ [1, 2] then p Œ (– •, –4/5). Statement-2: a lies between the roots of f(x) = 0, if and only if f(a) < 0.

106. Suppose a, b, c Œ R, a π 0 and a – b + c < 0. Statement-1: If ax2 + bx + c = 0 has imaginary roots, and yˆ Ê a Ë x 2 + ¯ + (b + 1) x + c = ax 2 + bx + c + x + y a then (x, y) lies in the half plane {(x, y)| x + y £ 0} Statement-2: |a + b| = |a| + |b| ¤ ab ≥ 0

103. Suppose a Œ R. Statement-1: The equation a (2 x - 2 ) + 1 = 1 – 2x has a real solution for all a Œ (0, 1]. Statement-2: ax2 + bx + c = 0, a π 0 has two positive roots if ab > 0 and ac > 0.

(

104 Suppose a, b, c ŒR and f(x) = 2(a – x) x + x 2 + b2 x Œ R. 2

)

2

Statement-1: Maximum value of f(x) is a + b . Statement-2: If a < 0, the maximum value of f(x) = ax2 + bx + c is (b2 – 4ac)/4a. 105. Suppose a, b, c Œ R, a π 0.

Statement-2: If a + ib, where a, b Œ R, b π 0 is a root of ax2 + bx + c = x, then its other root must be a – ib

107. a, b, c ΠR and c < min {a, b}. ( x + a ) ( x + b) , x > Рc. Let f(x) = x+c Statement-1: Minimum value of f is

(

2

a - c + b - c)

Statement-2: If a < 0, then ax2 + bx + c has no minimum value. 108. Suppose a, b, c, p, q, r Œ R, a, p π 0. Let f(x) = ax2 + bx + c and g(x) = px2 + qx + r Statement-1: If f(x) = g(x) for three distinct real values of x then a = p, b = q and c = r.

3.38

Complete Mathematics—JEE Main

Statement-2: If a π p, then f(x) – g(x) = 0 has two real roots.

110. Suppose a, b, c Œ R a π 0 and c > 0. Let f(x) = ax2 + bx + c. Statement-1: If f(x) = 0, does not have real roots, then ¸ Ï 2 a > max Ì b , b - c ˝ ˛ Ó 4c 2 Statement-2: If b – 4ac < 0 then f(x) > 0 " x Œ R.

109. Suppose p Œ R. Statement-1: If sin4 x + psin2 + 1 = 0 has a solution, then p Œ (– •, –2] Statement-2: y +

1 ≥ 2 " y > 0. y

LEVEL 2 Straight Objective Type Questions 111. If a Π(0, p/2), then the expression

1 1 1 + + = 0, has x – sin a x – sin b x – sin g

2

x2 + x +

tan a

x2 + x is always greater than or equal to (a) 2 tan a (b) 2 (c) 1 (d) sec2 a 112. If a, b Œ R, and the equation x2 + (a – b)x – a – b + 1 = 0 has real roots for all b Œ R, then a lies in the interval (a) (1, •) (b) (0, •) (c) (– •, 1) (d) (– 1, 1)

118.

119.

113. If a and b (a < b) are the roots of the equation x2 + bx + c = 0, where c < 0 < b, then (a) 0 < a < b (c) a < b < 0

(b) a < 0 < b < |a | (d) none of these.

120.

114. Suppose a, b, c ΠR and the equation x2 + (a + b)x + c = 0 has no real roots, then which one of the following is not true. (a) (b) (c) (d)

c + c (a + b + c) > 0 c – c (a + b – c) > 0 (a + b)2 – 4c < 0 c (a + b + c) < 0

121.

115. If [x] denotes the greatest integer £ x, then number of solutions of the equation x2 – 2 – 2[x] = 0 is (a) 4 (c) 3

(b) 2 (d) none of these

116. If [x] denotes the greatest integer £ x, and a, b are two odd integers, then number of solutions of [x]2 + a[x] + b = 0 is (a) 1 (c) 2

(b) 0 (d) infinite

117. Let a, b, g be distinct real numbers lying in (0, p/2), then the equation

122.

(a) two distinct real roots (b) two equal roots (c) two imaginary roots (d) one real and one imaginary root. If roots of the equation x2 + ax + b = 0 are a, b, then the roots of x2 + (2a + a)x + a2 + aa + b = 0 are (a) 1, b – a (b) 0, a – b (c) 0, b – a (d) 0, 1. Let a, b, c be the sides of a triangle with a π c and l Œ R. If the roots of x2 + 2(a + b + c)x + 3l (ab + bc + ca) = 0 are real, then l lies in (a) (– •, 4/3) (b) (5/3, •) (c) (1/3, 5/3) (c) (4/3, 5,3) If tan q and cot q are roots of x2 + 2ax + b = 0, then least value of |a| is 1 (b) 1 (a) 2 (c) 2 (d) cannot be found. Let a, b, g be the roots of x3 + x2 – 5x – 1 = 0, then value of [a] + [b] + [g], where [x] denotes the greatest integer £ x, is (a) 1 (b) 2 (c) – 2 (d) – 3. [Hint: Show that f (–3) < 0, f (– 2) > 0, f (– 1) > 0, f (0) < 0, f (1) < 0, f (2) > 0] If tan A and tan B are roots of the quadratic equation x2 – px + q = 0, then the value of sin2 (A + B) is (a)

(c)

p2 p2 + q 2 p2

(1 – q )2 + p2

(b)

p2

(q + p )2

(d) 1 –

p2

(1 – q )2

Quadratic Equations 3.39

123. The equation x + 3 + 4 x – 1 + x + 8 + 6 x – 1 = 1 has (a) no solution (b) only one solution (c) only two solutions (d) infinite number of solutions 124. The number of irrational solutions of the equation x 2 + x 2 + 11 + x 2 – x 2 + 11 = 4 is (a) 0 (b) 2 (c) 4 (d) infinite 125. Let a, b, c, p, q be five different non-zero real numbers and x, y, z be three numbers satisfying the system of equations x y z + + =1 a a– p a–q

and

x y z + + =1 b b– p b–q x y z + + =1 c c– p c–q then x equals abc pq (a) (b) pq abc abc (c) (d) none of these. p+q

126. Let f (x) = ax2 + bx + c, where a, b, c Œ R. Suppose |f (x)| £ 1 " x Œ [0, 1], then |a| cannot exceed (a) 5 (b) 6 (c) 7 (d) 8 127. If a(p + q)2 + 2bpq + c = 0 and a(p + r)2 + 2bpr + c = 0, then |q – r| equals (a)

2 a

(2a + b)bp2 - ac

(b)

2 a

p2 - 4ac

(c) p2 + c (d) none of these. a 128. Let a, b, c Œ R be such that b2 ≥ 4ac. All the four roots of the equation ax4 + bx2 + c = 0 will be real if (a) a > 0, b < 0, c > 0 or a < 0, b > 0, c < 0 (b) a > 0, b > 0, c > 0 or a < 0, b < 0, c < 0 (c) a > 0, b < 0, c > 0 or a > 0, b > 0, c < 0 (d) none of these. 129. If the equations x2 + mx + 1 = 0 and (b – c) x2 + (c – a) x + (a – b) = 0 have a common root, then (a) m = – 2 (b) m = – 1 (c) m = 0 (d) m = 1 130. If x, y, z Œ R, x + y + z = 4 and x2 + y2 + z2 = 6, then the maximum possible value of z is (a) 1 (b) 2 (c)

3 2

(d)

4 3

Previous Years' AIEEE/JEE Main Questions 1. If a π b and a2 = 5a – 3, b2 = 5b – 3, then the a b equation whose roots are and is b a (a) 3x2 – 25x + 3 = 0 (b) x2 – 5x + 3 = 0 (d) 3x2 – 19x + 3 = 0 (c) x2 + 5x – 3 = 0 [2002] 2. If a π b and differences between the roots of the equations x2 + ax + b = 0 and x2 + bx + a = 0 is the same, then (a) a + b + 4 = 0 (b) a + b – 4 = 0 (c) a – b + 4 = 0 (d) a – b – 4 = 0 [2002] 3. If a, b, c Œ R and 2a + 3b + 6c = 0, then the equation ax2 + bx + c = 0 has (a) at least one root in [0, 1]

(b) at least one root in [2, 3] (c) at least one root in [– 1, 0] (d) at least one root in (– •, 1) [2002] 4. Product of real roots of the equation t2x2 + |x| + 9 = 0, t Œ R is always (a) positive (b) negative (c) zero (d) does not exit [2002] 5. The value of a for which one root of the quadratic equation (a2 – 5a + 3)x2 + (3a – 1)x + 2 = 0 is twice as large as the other is 2 1 (b) (a) – 3 3 1 2 (c) – (d) [2003] 3 3

Complete Mathematics—JEE Main

3.40

6. If sum of the roots of the quadratic equation ax2

7.

8.

9.

10.

11.

12.

+ bx + c = 0 is equal to the sum of the squares of a b c their reciprocals, then , and are in c a b (a) G.P. (b) H.P. (c) A.G.P. (d) A.P. [2003] The number of real solution of the equation x2 – 3|x| + 2 = 0 is (a) 4 (b) 1 (c) 3 (d) 2 [2004] Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are roots of the quadratic equation (a) x2 + 18x – 16 = 0 (b) x2 – 18x + 16 = 0 (c) x2 + 18x + 16 = 0 (d) x2 – 18x – 16 = 0 [2004] If 1 – p is a root of quadratic equation x2 + px + 1 – p = 0, then its roots are (a) 0, – 1 (b) – 1, 1 (c) 0, 1 (d) – 1, 2 [2004] If one root of the equation x2 + px + 12 = 0 is 4, while equation x2 + px + q = 0 has equal roots, then the value of q is (a) 3 (b) 12 49 (d) 4 [2004] (c) 4 If 2a + 3b + 6c = 0, then at least one root of the equation ax2 + bx + c = 0 lies in the interval (a) (2, 3) (b) (1, 2) (c) (0, 1) (d) (1, 3) [2004] p P . If tan Ê ˆ and In a triangle PQR, –R = Ë2 ¯ 2 Q tan Ê ˆ are the roots of ax2 + bx + c = 0, a π 0, Ë2 ¯ then

(a) b = c (b) b = a + c (c) a = b + c (d) c = a + b [2005] 13. The value of a for which sum of the squares of the roots of the equation of the roots of the equation x2 – (a – 2)x – a – 1 = 0 assume least value is (a) 3 (b) 2 (c) 1 (d) 0 [2005] 2 14. If the roots of the equation x – bx + c = 0 be two consecutive integers then b2 – 4c equals (a) 2 (b) 1 (c) – 2 (d) 3 [2005] 15. If both the roots of the quadratic equation x2 – 2kx + k2 + k – 5 = 0 are less than 5, then k lies in the interval (a) (– •, 4) (b) [4, 5] (c) (5, 6) (d) (6, •) [2005]

16. If the equation anxn + an – 1xn – 1 + º + a1x = 0, a1 π 0, n ≥ 2 has a positive root x = a, then the equation n an xn – 1 + (n – 1) an – 1 xn – 2 + º + a1 = 0 has a positive root, which is (a) greater than or equal to a (b) equal to a (c) greater than a (d) smaller than a [2005] 2 17. If the roots of the quadratic equation x + px + q = 0 are tan 30° and tan 15° respectively, then the value of 2 + q – p is (a) 1 (b) 2 (c) 3 (d) 0 [2006] 18. All the values of m for which both the roots of the equation x2 – 2mx + m2 – 1 = 0 are greater than – 2 but less than 4, lie in the interval (a) 1 < m < 4 (b) – 2 < m < 0 (c) m > 3 (d) – 1 < m < 3 [2006] 3 x 2 + 9 x + 17 19. If x is real, the maximum value of 3x2 + 9 x + 7 is 17 1 (b) (a) 7 4 (c) 41 (d) 1 [2006] 20. If the difference between the roots of x2 + ax + 1 = 0 is less than 5 , then set of possible values of a lie in the interval (a) (– 3, 3) (b) (– 3, •) (c) (3, •) (d) (– •, – 3) [2007] 21. The quadratic equations x2 – 6x + a = 0 and x2 – cx + 6 = 0 have one common root. The other roots of first and second equation are integers in the ratio 4 : 3. Then common root is (a) 1 (b) 4 (c) 3 (d) 2 [2008] 22. If the roots of the equation bx2 + cx + a = 0 be imaginary, then for all real values of x, the expression 3b2x2 + 6bcx + 2c2 is (a) greater then – 4ab (b) less then – 4ab (c) greater than 4ab (d) less than 4ab [2009] 23. If a and b are the roots of the equation x2 – x + 1 = 0, then a2009 + b2009 equals (a) 1 (b) 2 (c) – 2 (d) –1 [2010] 24. Let for a, a1 π 0, a π a1 f(x) = ax2 + bx + c, g(x) = a1x2 + b1x + c1 and p(x) = f(x) – g(x). If p(x) = 0 only for x = –1 and p(–2) = 2, then value of p(2) is: (a) 3 (b) 9 (c) 6 (d) 18 [2011]

Quadratic Equations 3.41

25. The equation esin x – e– sin x = 4 has (a) no real roots (b) exactly one real root (c) exactly four real roots (d) infinite number of real roots. [2012] 2 2 26. If the equations x + 2x + 3 = 0 and ax + bx + c = 0, a, b, c Œ R have a common root, then a: b: c is (a) 3 : 2 : 1 (b) 1 : 3 : 2 (c) 3 : 1 : 2 (d) 1 : 2 : 3 [2013] 27. If a and b are the roots of the equation x2 + px + 3 p = 0, such that |a – b | = 10 , then p belongs 4 to the set (a) {2, – 5} (b) {–3, 2} (c) {–2, 5} (d) {– 3, 5} [2013, online] 28. If p and q are non-zero real number such that a3 + b 3 = – p and ab = q, then a quadratic equation a2 b2 whose roots are and is b a (a) px2 – qx + p2 = 0 (b) qx2 + px + q2 = 0 (c) px2 + qx + p2 = 0 [2013, online] (d) qx2 – px + q2 = 0 29. The values of a for which one root of the equation x2 – (a + 1)x + a2 + a – 8 = 0 exceeds 2 and the other is less than 2 are given by (a) 3 < a < 10 (b) a > 10 (c) –2< a < 3 (d) a £ –2 [2013, online] 30. The least integral value a of x such that x-5 > 0, satisfies 2 x + 5 x - 14 (a) a 2 + 3a – 4 = 0 (c) a 2 – 7a + 6 = 0

(b) a 2 – 5a + 4 = 0 (d) a 2 + 5a – 6 = 0 [2013, online] 2

31. Let a and b be the roots of the equation px + qx 1 1 + r = 0, p π 0. If p, q, r are in A.P. and + = a b 4, then value of |a – b | is 1 2 (a) (b) 61 17 9 9 1 2 34 13 (d) [2014] 9 9 32. If equations ax2 + bx + c = 0, (a, b,c Œ R, a π 0) and 2x2 + 3x + 4 = 0 have a common root, then a : b : c equals: (c)

(a) 1 : 2 : 3 (c) 4 : 3 : 2

(b) 2 : 3 : 4 (d) 3 : 2 : 1 [2014, online] 2

33. The sum of the roots of the equation x + |2x – 3| + 4 = 0, is:

(a) 2 (c)

(b) –2 (d) - 2

2

[2014, online]

2

34. If a and b are the roots of x – 4 2 kx + 2e4lnk – 1 = 0 for some k, and a2 + b 2 = 66, then a 3 + b 3 is equal to: (a) 248 2

(b) 280 2

(c) -32 2

(d) -280 2 [2014, online]

35. The equation 3 x 2 + x + 5 = x – 3, where x is real, has: (a) has no solution (b) exactly one solution (c) exactly two solutions (d) exactly four solutions [2014, online] 36. Let a and b be the roots of equation x2 – 6x – 2 = 0. If an = a n – b n, for n ≥ 1, then the value of a10 - 2a8 is equal to: 2a9 (a) 6 (b) – 6 (c) 3 (d) –3 [2015] 37. If 2 + 3i is one of the roots of the equation 2x3 – 9x2 + kx – 13 = 0, k Œ R, then the real root of the equation: (a) does not exist. (b) exists and is equal to 1 2 (c) exists and is equal to – 1 2 (d) exists and is equal to 1 [2015, online] 38. If the two roots of the equation, (a – 1)(x4 + x2 +1) + (a + 1)(x2 + x + 1)2 = 0 are real and distinct, then the set of all values of a is: (a) ÊÁ - 1 , 0ˆ˜ Ë 2 ¯

(b) (–•, –2) » (2, •)

1 1 1 (c) ÊÁ - , 0ˆ˜ » ÊÁ 0, ˆ˜ (d) ÊÁ 0, ˆ˜ Ë 2¯ Ë 2 ¯ Ë 2¯

[2015, online]

39. The sum of all real values of x satisfying the equa2

tion ( x 2 - 5 x + 5) x + 4 x - 60 = 1 is (a) 3 (b) –4 (c) 6 (d) 5 [2016] 40. If the equations x2 + bx – 1 = 0 and x2 + x + b = 0 have a common root different from –1, then |b| is equal to: (a) 2 (b) 3 (d) 2 [2016, online] (c) 3 41. If x is a solution of the equation, 2 x + 1 - 2 x - 1 1ˆ Ê = 1, Á x ≥ ˜ , then 4 x 2 - 1 is equal to: Ë 2¯

3.42

Complete Mathematics—JEE Main

(a)

3 4

(b)

(c) 2 2

1 2

(d) 2

[2016, online]

42. Let x, y, z be positive real numbers such that x + y + z = 12 and x3y4z5 = (0.1)(600)3. Then x3 + y3 + z3 is equal to (a) 342 (b) 216 (c) 258 (d) 270 [2016, online]

Previous Years' B-Architecture Entrance Examination Questions 8. If the quadratic equation 3x2 + 2(a2 + 1)x + a2 – 3a + 2 = 0 posseses roots of opposite signs, then a lies in the interval:

1. If the roots of the quadratic equation x2 + 2px + q = 0 are tan 30° and tan 15° respectively, then q is (a) 1 + p (c) 1 – 2p

(b) 1 – p (d) 1 + 2p

[2006]

2. The set of values of a for which the quadratic equation (a + 2) x2 – 2ax – a = 0 has two roots on the number line symmetrically placed about the point 1 is (a) {–1, 0} (c) f

(b) {0, 2} (d) {0, 1}

[2007] 2

3. The number of solutions of the equation x – 4|x| – 2 = 0 is: (a) 1 (c) 3

(b) 2 (d) 4

[2008]

4. The quadratic equation whose roots are a/b and b/a, a π b π 0, where a2 = 5a – 3, and b2 = 5b – 3, is: (a) (b) (c) (d)

3x2 – 19x + 3 = 0 3x2 + 19x – 3 = 0 3x2 + 19x + 3 = 0 3x2 – 19x – 3 = 0

[2009] 2

5. If the roots of the quadratic equation ax + bx + c = 0 are a, b, then the roots of the quadratic equation ax2 – bx(x – 1) + c(x – 1)2 = 0, are a +1 b +1 a b , (b) , (a) a b a -1 b -1 (c)

a b , a +1 b +1

(d) 1 – a, 1 – b

[2010]

6. If x2 – 3x + 2 is a factor of x4 – ax2 + b = 0 then the equation whose roots are a and b is (a) x2 + 9x + 20 = 0 (c) x2 – 9x + 20 = 0

(b) x2 – 9x – 20 = 0 (d) x2 + 9x – 20 = 0 [2011]

7. Let a, b, c Œ R, a > 0 and the function f : R Æ R be defined by f(x) = ax2 + bx + c. Statement-1: b2 < 4ac fi f(x) > 0 for every value of x. Statement-2: f is strictly decreasing in the interval (– •, – b/2a) and strictly increasing in the interval (–b/2a, •). [2012]

(a) (– •, –1) (c) (1, 2)

(b) (–1, 1) (d) (2, 3) [2013] 1 1 1 9. If the roots of the equation + = are x+ p x+q r equal in magnitude and opposite in sign, then product of roots is: 1 2 (b) (p + q2) (a) p2 + q2 2 1 2 1 2 (c) – (p + q2) (d) – (p – q2) [2014] 2 2 10. The values of k for which each root of the equation, x2 – 6kx + 2 – 2k + 9k2 = 0 is greater than 3, always satisfy the inequality: (a) 7 – 9y > 0 (b) 11 – 9y < 0 (c) 29 – 11y > 0 (d) 29 – 11y < 0 [2015] 11. The number of integral values of m for which the equation, (1 + m2)x2 – 2(1 + 3m)x + (1 + 8m) = 0, has no real root, is: (a) 2 (b) 3 [2016]

Answers Concept-based 1. 5. 9. 13. 17.

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

2. 6. 10. 14. 18.

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

3. 7. 11. 15. 19.

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

4. 8. 12. 16. 20.

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

22. 26. 30. 34. 38.

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

23. 27. 31. 35. 39.

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

24. 28. 32. 36. 40.

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

Level 1 21. 25. 29. 33. 37.

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

Quadratic Equations 3.43

41. (d) 45. (b) 49. (c) 53. (a) 57. (a) 61. (a) 65. (d) 69. (d) 73. (c) 77. (d) 81. (a) 85. (d) 89. (b) 93. (c) 97. (a) 101. (a) 105. (d) 109. (a)

42. 46. 50. 54. 58. 62. 66. 70. 74. 78. 82. 86. 90. 94. 98. 102. 106. 110.

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

43. 47. 51. 55. 59. 63. 67. 71. 75. 79. 83. 87. 91. 95. 99. 103. 107.

112. 116. 120. 124. 128.

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

113. 117. 121. 125. 129.

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

44. 48. 52. 56. 60. 64. 68. 72. 76. 80. 84. 88. 92. 96. 100. 104. 108.

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

Hints and Solutions Concept-based 1. (x – 1)2 + 1 + |x – 1| ≥ 1 " x Œ R k Ê 1ˆ 2. 1 = a Ë ¯ = a 7 3. a – b = –1 = fi (a + b )2 – 4a b = 1 or p2 – 4q = 1 4. If a is a common root, then a 2 – aa + b = 0, a2 + ba – a = 0 Subtracting, we get – (a + b) a + b + a = 0 fi a = 1. Thus, 1 – a + b = 0 fi a – b = 1

Roots of the equation are – These will exceed a, if -

Level 2 111. (a) 115. (d) 119. (a) 123. (a) 127. (a)

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

114. 118. 122. 126. 130.

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

Previous Years' AIEEE/JEE Main Questions 1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 41.

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

2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42.

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

2

1ˆ 1 = –a ¯ 2 4 1 1 We must have –a≥0fia£ . 4 4

Ê 5. x2 + x + a = 0 fi Ëx +

3. 7. 11. 15. 19. 23. 27. 31. 35. 39.

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

4. 8. 12. 16. 20. 24. 28. 32. 36. 40.

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

Previous Years' B-Architecture Entrance Examination Questions

fi -

1 1 ± -a 2 4

1 1 -a >a 2 4

1 1 -a >a+ 4 2

This is possible if a + 1/2 < 0 and

1 1ˆ Ê - a < Ëa + ¯ 4 2

1 1 – a < a2 + a + 4 4 2 fi a + 2a > 0 fi a < – 2 as a < 0. fi

6. (|x| – 2)2 = 6 fi |x| = 2 ± 6 . As |x| ≥ 0, |x| = 2 +

6 fi x = ± (2 +

6)

2

7. 10 = ( 10 ) = |a – b |2 = (a + b)2 – 4ab = p2 – 3p fi p2 – 3p – 10 = 0 fi (p – 5) (p + 2 ) = 0 fi p = – 2, 5 8. For the equation to be defined, x ≥ 0, 5 + x ≥ 0. For x ≥ 0, x + 5 ≥ 5 fi x + x + 5 ≥ 5 > 2 9. D = 4(a + 1)2 – 4(a2 – 4a + 3) ≥ 0, 2(a + 1) < 0 and a2 – 4a + 3 > 0 First two imply a ≥ 1/3, a < – 1. Not possible. 10. tan a + tan b = – a, tan a tan b = b. tan a + tan b a = 1 - tan a tan b 1 - b Also, 0 < a < p /4, 0 < b < p /4 1 = tan (a + b) =

1. (d)

2. (c)

3. (b)

4. (a)

5. (c)

6. (c)

7. (a)

8. (c)

9. (c)

10. (b)

11. (c)

3.44

11.

12.

13.

14.

Complete Mathematics—JEE Main

fi 0 < tan a < 1, 0 < tan b < 1 fi 0 < tan a tan b < 1 fi 0 < b < 1. Let f(x) = (x – a) (x – b) – c Use coefficient of x2 > 0, f(a) = – c = f(b) < 0. D ≥ 0 and product of roots £ 0. fi (k + 1)2 – (9k – 5) ≥ 0 and 9k–5 £ 0 fi k2 – 7k + 6 ≥ 0 and £ 5/9 fi (k – 1) (k – 6) ≥ 0 and k £ 5/9 Thus, k £ 5/9 Use : a and a (k – a) (k – b) have the opposite signs, where a, b are roots of ax2 + bx + c = 0. Write the equation as (x – 2a)2 = 4a2 – (2a2 – 3a + 5) = (2a + 5) (a – 1) fi x = 2a ± (2a + 5) ( a - 1) We must have (2a + 5) (a – 1) ≥ 0 and 2a + (2a + 5) ( a - 1) < 2 fi a £ – 5/2 or a ≥ 1 and

15.

16. 17.

18. 19.

(2a + 5) (a - 1) < – 2(a – 1)

Clearly, a ≥ 1 is not possible For a £ – 5/2, we must have - (2a + 5) (1 - a ) < 2 (1 – a) fi – (2a + 5) < 4(1 – a) fi a < 9/2 As parabola open upwards, a > 0. Also, for x = 0 ax2 + bx + c = c < 0. Since the equation has distinct roots, b2 – 4ac > 0. However, we cannot be sure of sign of b. If can be positive, negative or zero. For instance, consider x2 + 2x – 3 = 0, x2 – 2x – 3 = 0 and x2 – 3 = 0. 9x2 – 6x + 5 = (3x – 1)2 + 4 attains least value when x = 1/3. ax2 + bx + c = 0 has real roots if b2 – 4ac ≥ 0, i.e. if b2 ≥ 4ac. Minimum possible value of b is 2. For b = 2, (a, c) = (1, 1) For b = 3, ac £ 9/4 fi (a, c) = (1, 1), (1, 2) or (2, 1) For b = 4, ac £ 4 fi (a, c) = (1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1), (2, 2) Thus, there 43 – 12 = 52 such quadratic equations. LHS > 0 " x Œ R. As the roots are imaginary b2 – 4ac < 0 and 1 - b ± i 4ac - b2 a, b = 2a

(

|a | =

1 b2 + 4 ac - b2 = 2a

)

c = |b| a

20. Suppose a, b > 1, a < b be roots of ax2 + bx + c = 0, then ax2 + bx + c > 0 if x < a or x > b < 0 if a < x < b Thus, a + b + c > 0.

Level 1 21. a > 0, c > 0, 4b2 – 4ac < 0. However nothing can be said about b. For instance, consider parabolas y = x2 – x + 1, y = x2 + x + 1 or y = x2 + 1 22. If a is a common root, then aa2 + 2ba + c = 0, aa2 + 2ca + b = 0 Subtracting, we get 2(b – c) a + c – b = 0 1 1 fi a = . Thus a + b + c = 0 2 4 a+b+c 3 fi = a 4 2 23. Let f(x) = ax + bx + c. As f(x) = 0, does not have real roots, f(x) > 0 " x Œ R or f(x) < 0 " x Œ R. But Ê 2ˆ 1 f Ë - ¯ = (4a – 6b + 9c) < 0, therefore 3 9 f(x) < 0 x " x Œ R. fi f(1) < 0 fi a + b + c < 0. b+c fi < -1 a 24. We have 162 – 16c > 0 fi c < 16 If f(x) = 4x2 – 16x + c, then 1 < a < 2 < b < 3 implies f(1) > 0, f(2) < 0, f(3) > 0. fi c > 12, c < 16, c > 12 Thus, c = 13, 14, 15. 25. If a is a common root of the three equations, then a2 + aa + 12 = 0, a2 + ba + 15 = 0 and a2 + (a + b)a + 36 = 0 fi ba = – 24, aa = – 21 Thus, a2 – 21 + 12 = 0 fi a = ± 3 As a > 0, we get a = 3. Therefore, a = – 7, b = – 8. a 2x = 2 26. fi a (x2 – b2) = 2x2 2 x x -b fi (a – 2)x2 – ab2 = 0 This equation will have no solution if a – 2 and a have opposite signs, that is , a(a – 2) < 0 fi 0 < a < 2. a2 4 (a - 2) 27. As the roots are real and distinct, 2 2a + 1 (2a + 1) >0 fi a2 – 4(a – 2) (2a + 1) > 0 fi

6 - 2 23 6 + 2 23 0, and p – 4 < 0, 2p2 – 4 > 0, and (p – 4)2 – 4(2p2 – 4) ≥ 0

30. Let y =

q–s ˘Èq – s ˘ = (r – p)2 È ÍÎ r – p – a ˙˚ ÍÎ r – p – b ˙˚ 2

È q – sˆ ˘ Ê q – sˆ = (r – p)2 ÍÊÁ + q˙ + pÁ ˜ ˜ Ë r – p¯ ÎË r – p ¯ ˚ = (q – s)2 + p(q – s) (r – p) + q(r – p)2 32. Use x2 ≥ 0, |x| ≥ 0 33. x2 – 3|x| + 2 = 0 fi |x| = 1, 2 fi x = ±1 , ±2 34. As p, q Œ R, other root of the equation is 2 – i 5. Thus, p = 4, q = (2 + 5i ) (2 – 5i ) = 9 35. Subtract one equation from the other to obtain x = 1. 36. Write the expression as (ap + b)2 + (bp + c)2 + (cp + d)2 £ 0. fi ap + b = 0, bp + c = 0, cp + d = 0 fi

b c d = = =– p. a b c

45. 46. 47. 48. 49.

fi

p < 4, p >

fi

2 < p
0 ax2 + bx + c > 0 " x > 0. x2 – 8x + p2 – 6p = 0 has real roots fi 64 – 4(p2 – 6p) ≥ 0 fi p2 – 6p – 16 £ 0 fi (p – 8) (p + 2) £ 0 fi p Œ [– 2, 8]

50. –

2a + 3 =–3 a +1

fi

\ Product of roots =

a = 0. 3a + 4 =4 a +1

51. Use that other root of the equation is 3 + 4i to obtain p = 6, q = 25 52. Let other roots be b, g then (1 + i) + b + g = 0 and (1 + i)bg = – (1 – i) fi b + g = – (1 + i), bg = i 1 53. a + b = – p, ab = - 2 . 2p 1 2 2 2 fi a + b = (a + b) – 2ab = p2 + 2 p 1ˆ2 Ê 2 2 fi a + b = Áp– ˜ +2≥2. Ë p¯ 54. Put \

x - 1 = t so that x = t2 + 1 x + 3 – 4 x –1 + x +8 – 6 x –1 =1

3.46

Complete Mathematics—JEE Main

fi

t 2 - 4t + 4 + t 2 - 6t + 9 = 1

fi (1 – p)2 – 2(1 + p) (1 – p) + p(p – 1) < 0 and (1 + p)2 – 2(1 + p)2 + p(p – 1) < 0

fi |t – 2| + |t – 3| = 1 fi 2 £ t £ 3 fi2£

fi –1/4 < p < 1 and p > –

x -1 £ 3

fi 4 £ x – 1 £ 9 fi 5 £ x £ 10. 55. |x – x2 – 1| = |2x – 3 – x2| fi |x2 – x + 1| = |x2 – 2x + 3| 2 3 = |(x – 1)2 + 2| fi Êx - 1 ˆ + Ë 4 2¯ fi x2 – x + 1 = x2 – 2x + 3 fi x = 2 c b 56. sin a + cos a = - , sin a cos a = a a Now 1 = sin2a + cos2a = (sin a + cos a)2 – 2sin a cos a fi

1=

fi

2

57. k =

b2 a

2

-

2c a

fi

a2 = b2 – 2ac

a – b2 + 2ac = 0. x2 - x + 1

1 3

Ê 1 ˆ Thus, p Œ Ë- , 1¯ 4 b B 63. a + b = - , a + b + 2h = a A 2 64. D = 49 + 56(q + 1) > 0. 65. Note that a, b are roots of b c =0 x2 + x + a a As a < – 1, b > 1, we get b c 1– + 0. x2 – 2x – p2 + 1 = 0 fi (x – 1)2 = p2 fix=1±p Let a = 1 – p, b = 1 + p Let f(x) = x2 – 2(p + 1)x + p(p – 1) As f(x) = 0 has distinet real roots, (p + 1)2 – p(p – 1) > 0 1 fi 3p + 1 > 0 fi p > – 3 Also, f(1 – p) < 0, and f(1 + p) < 0 fi

58.

59.

60. 61.

62.

Fig. 3.22

66. Let f (x) = ax2 + bx + c. Now, (a + c)2 < b2 ¤ (a – b + c) (a + b + c) < 0 fi f (– 1) f (1) < 0. Thus, ax2 + bx + c = 0 has a real root between – 1 and 1. The other root also must be real. 67. If x is an integer and (x – a) (x – 10) = – 1 fi x – 10 = 1 and x – a = – 1 or x – 10 = – 1 and x–a=1 fi x = 11 and a = x + 1 = 12 or x = 9 and a = x–1=8 x 2 x 3 68. 41.5 Ê ˆ + 90.5 Ê ˆ = 10 Ë 3¯ Ë2 ¯

Put Ê 3 ˆ Ë2 ¯

x

= t, so that

3t2 – 10t + 8 = 0 4 fi (3t – 4) (t – 2) = 0 fi t = , 2 3 x 4 3ˆ Ê fi = ,2 Ë2 ¯ 3 3t + 8/t = 10

fi

Quadratic Equations 3.47

fix=

log ( 4 / 3) log 2 , log (3 / 2 ) log (3 / 2 ) 2

69. Put 2sin x = t, so that equation becomes 5¥2 t+ = 7 fi t2 – 7t + 10 = 0 fi t = 2, 5 t But 1 £ t £ 2. Thus, sin2x = 1 70. Let the two distinct roots lying between 0 and 1 be a, b such that a < b. As f (x) = x2 – 2x + k is a differentiable and f (a) = f (b) = 0, therefore by the Roll’s theorem there exists g Œ (a, b) à (0, 1) such that f¢ (g) = 0 fi 2g – 2 = 0 fi g = 1. Not possible. 71.

a a +1 a a +1 c b + . = = - , and a -1 a a -1 a a a b c 2 Now, (a + b + c)2 = a 2 Ê 1 + + ˆ Ë a a¯ a + 1 ¸ a + 1˘ È Ï a + = a 2 Í1 - Ì ˝+ 1 a a ˛ a - 1 ˙˚ Î Ó

77. y =

2

fi fi

È a a + 1ˆ 2 a a + 1˘ = a ÍÊÁ + -4 . ˙ ˜ a ¯ a -1 a ˚ ÎË a - 1 2

È 2 ˘ = a 2 Í b - 4c ˙ = b2 - 4 ac 2 a˚ Îa 72. Let x = 3m + r, 0 £ r £ 2. x2 + x + 1 = (9m2 + 6mr + 3m) + (r2 + r + 1) Note that x2 + x + 1 will be divisible by 3 if and only if r = 1.

= =

1 (a – b )2 – x2 – (a + b )x + ab 4

(a - b )2 + 4ab 4

{

- x-

{ }

1 a (a + b )2 - x + 2 2

8 + 2 8 + 2 8 + 2 8 + y2 = 8 + 2y fi y2 – 2y – 8 = 0 y = 4, – 2. As y > 0, y = 4

Ê px ˆ 2 = 78. sin Á +1≥1 Ë 2 3 ˜¯ ( x – 3 )

a a + 1ˆ 2 = a 2 ÊÁ Ëa -1 a ˜¯

73. f (x) =

are not sufficient to guarantee that the roots are real. For instance, x2 + 0x + 0.5 = 0 does not have real roots. 75. Let f (x) = ax2 + bx + c. Note that f (0) = c > 0, f (– 1) = a – b + c < 0 and f (1) = a + b + c < 0 Thus, y = ax2 + bx + c represents a parabola that Fig. 3.24 open downwards. See Fig.3.21 Thus [a] = – 1 and [b] = 0 fi [a] + [b] = – 1 76. x2 – 2mx + m2 – 1 = 0 fi (x – m)2 = 1 fix–m=±1 fi x=m±1 Therefore, – 2 < m – 1 < m + 1 < 4 fi – 1 < m < 3.

a +b 2

}

2

+

(a + b )2 4

2

2

£

a 2

a2 Thus, maximum value f (x) is which is 2 a attained when x = – 2 74. Let a, b be roots of f(x) = x2 + bx + c, then 0 < a, b < 1 fi 0 < a + b < 2 and ab < 1 fi – b < 2 and c < 1 fi b > – 2 and c < 1 However, these conditions

Ê px ˆ But sin Á £ 1. Thus, sin ÊÁ p x ˆ˜ = 1 Ë 2 3 ˜¯ Ë 2 3¯ Therefore, x = 3 79. 2(3 – x) = |x + 2| ≥ 0 fi x £ 3 If x £ – 2, the equation becomes 2(3 – x) = – x – 2 fi x = 8. Not possible If – 2 < x £ 3, the equation becomes 2(3 – x) = x + 2 fi 3x = 4 fi x = 4/3. 80. Use product of roots < 0 81. Put x2 – 5x = y; so that y2 – 7y + 6 = 0 fi y = 1, 6 \ x2 – 5x = 1 or x2 – 5x = 6 fi x2 – 5x – 1 = 0 or x2 – 5x – 6 = 0 But x2 – 5x – 6 = 0 has integral roots and roots of x2 – 5x – 1 = 0 are 1 x = (5 ± 29 ) 2 2 Ê x - 1ˆ 82. Put Á = y to obtain Ë x + 1˜¯ y2 – 13y + 36 = 0 fi y = 4, 9 x -1 = ± 2, ± 3 \ x +1 Thus, the given equation has four real roots.

83. 9x + 2 – 6(3x + 1) + 1 = 0 Put 3x + 1 = t so that t2 – 6t + 1 = 0 Fig. 3.23

fit=

6 ± 32 =3±2 2 2

3.48

Complete Mathematics—JEE Main

fi x + 1 = log3 (3 + 2 2 ) > 1 or x + 1 = log3 (3 – 2 2 ) < 0 2 Ê 2 x – 5ˆ = t, so that equation becomes 84. Put Á Ë 3 x + 1 ˜¯ 81t2 – 45t + 4 = 0 fi 81t2 – 36t – 9t + 4 = 0 fi (9t – 1) (9t – 4) = 0 2x – 5 1 2 1 4 =± ,± fit= , fi 3x + 1 3 3 9 9 Thus, the given equation has four rational roots. 85. Put x2 + 3x + 2 = t to obtain t2 – 8(t – 2) – 4 = 0 fi t2 – 8t + 12 = 0 fi (t – 2) (t – 6) = 0 fi t = 2, 6 But both x2 + 3x + 2 = 2 and x2 + 3x + 2 = 6 have integral roots.

x 1 5 = t to obtain t + = x-3 t 2 1 fi t = 2, 2 x x 1 1 \ = 2, fi = 4, x 3 2 4 x-3

89.

90.

86. Put

87. Put x + 1/x = t to obtain 4{t2 – 4} + 8t = 29 or 4t2 + 8t – 45 = 45 45 fi t2 + 2t = fi (t + 1)2 = +1= 4 4 7 9 5 fi t = -1 ± = - , 2 2 2 Thus,

x+

91.

92. 0 2 Ê7 ˆ Ë2 ¯

1 9 5 = - , x 2 2

9 5 x + 1 = 0, x 2 - x + 1 = 0 2 2 The first equation has irrational roots and the second equation has rational roots. 88. Divide the given equation by x2 to obtain fi

x2 +

fit=1± 3. But t ≥ 0, therefore t = 1 + 3 1 fi 2x + 1 = 1 + 2 + 2 3 fi x = (3 + 2 3 ) 2 93. Note that – 13 £ x £ 13 and 13 – x2 = x2 + 10x + 25 fi x2 + 5x + 6 = 0 fi x = – 2, – 3. 94.

fix=2

x-3

fi 3x2 – 6x – 25 = 0 28 28 fi x = 1± fi (x – 1)2 = 3 3

1 1 ,= - –4 x 2

Note that 1 –

2

x +1 1 x +1 = - , =–4 x x 2

fi 2x2 + x + 2 = 0,

x+2 - x-2 =

x-3

fi x + 3 = 2 x 2 - 4 fi x2 + 6x + 9 = 4(x2 – 4)

1 fit= - ,–4 2

fi

or

x-2

fi x + 2 + x – 2 – 2 x2 - 4 = x – 3

1 Put x + = t to get 2(t2 – 2) + 9t + 8 = 0 x

2

x 2 - 4 - ( x - 2) = x 2 - 5x + 6 (1) is defined for x = 2 or x £ – 2 or x ≥ 3. Write (1) as x - 2 [ x + 2 - x - 2] =

1 1 2 ÊÁ x 2 + 2 ˆ˜ + 9 Ê x + ˆ + 8 = 0 Ë ¯ Ë x¯ x

\ x+

The first equation has non-real complex roots and the second equation has irrational roots. viz. –2± 3. 1 Put x - = t to obtain 4t2 – 4t + 1 = 0 x 1 1 fi t= , 2 2 1 x2 - 1 1 1 Now, x - = fi = 2 2 x x 2 fi 2x – x – 2 = 0 1 1 \ Sum of roots of the given equation = + = 1. 2 2 (3x – 2)(x – 1) (3x + 1)(x – 2) = 21 fi (3x2 – 5x + 2) (3x2 – 5x – 2) = 21 fi (3x2 – 5x)2 – 4 = 21 fi 3x2 – 5x = ± 5 fi 3x2 – 5x – 5 = 0 or 3x2 – 5x + 5 = 0 The first equation has irrational roots and the second has imaginary roots. Note that x ≥ 2 and fi x – 2 = 3 x - 6 fi x2 – 4x + 4 = 3x – 6 fi x2 – 7x + 10 = 0 fi x = 2, 5 Both of these satisfy the given equation. 1 Note that x ≥ - . Put 2 x + 1 = t, so that 2 2t = t2 – 1 – 1 or t 2 – 2t + 1 = 3

x2 + 4x + 1 = 0

95.

28 < – 2 and 1 + 3

28 >3 3

x 2 - 4 x + 3 + x 2 - 7 x + 12 = 3 x - 3 is defined for x = 3 or x ≥ 4.

(1)

Quadratic Equations 3.49

Write (1) as x - 3 [ x - 1 + x - 4] = 3 x - 3 fi x = 3 or

x -1 + x - 4 = 3

fi x = 3 or

x -1 = 3 - x - 4

fi x = 3 or x – 1 = 9 - 6 x - 4 + x - 4 fix = 3 or 6 x - 4 = 6 fi x = 3 or x = 5. Thus, product of roots = 15. 96. Note that 2b2 + 77b + 18 = 0 fi 2 + 77/b + 18(1/b)2 = 0 Thus, a and 1/b are roots of the equation 18x2 + 77x + 2 = 0 a 2 1 \ a + 1/b = – 77/18 and = = b 18 9 1 a 77 1 ab + a + 1 25 Now, = a+ + =+ =b b b 18 9 6 97. b2 = 7b – 8. 16 16 Thus, + 3b 2 - 19b = + 3(7b - 8) - 19b a a 16 + 2b - 24 = a Also, a =

\ Thus

7 + 17 7 - 17 ,b= 2 2

(

)

2 7 - 17 1 2 7 - 17 = = = a 49 - 17 16 7 + 17 16 + 3b 2 - 19b a

(

) (

= a2a2 – 2a2a2 < 0 and a2b2 + 2bb + 2c = a2b2 + 2a2b2 > 0 As a2x2 + 2bx +2c is continuous on [a, b], by statement-2 there exists g such that a2g 2 + 2bg + 2c = 0 102. As y = f(x) open upwards, a lies between the roots of f(x) = 0, if and only if f(a) < 0. Therefore, statement-2 is true. If f(x) < 0 for 1 £ x £ 2, then 1 and 2 lie between roots of f(x) = 0. Thus, we must have 4p2 – 4p > 0 fi p(p – 1) > 0 fi p < 0 or p > 1. Also, f(1) = 1 + 2p + p < 0 fi p < – 1/3 and f(2) = 4 + 4p + p < 0 fi p < – 4/5. \ p Œ (– •, –4/5) 103. Statement-2 is false as x2 + 5x + 4 = 0 meets both the conditions but does not have positive roots. As LHS ≥ 0, 1 – 2x ≥ 0 fi 2x £ 1 fi x £ 0. Squaring, we get a(2x – 2) + 1 = 1 – 2(2x ) + 22x fi y2 – (a + 2)y + 2a = 0 fi (y – a) (y – 2) = 0 fi y = 2, a. y = 2x = 2 fi x = 1. Not possible. Thus, y = a. This implies 0 < 2x = a £ 1 Therefore 0 < a £ 1, and x = log2a 104. Put t = x + fi

)

= 7 - 17 + 7 - 17 - 24 = – 10 98. As a is a root of x4 + x2 – 1 = 0, we get a4 + a2 – 1 = 0 fi a6 + a4 – a2 = 0 fi a 6 + a 4 – (1 – a 4) = 0 fi a 6 + 2a 2 = 1 \ (a 6 + 2a 2)2012 = 1 99. Use the fact that sum and product of the roots of x2 – x – k = 0 are 1 and – k respectively. 100. Suppose a2(a + k) = b2(b + k) = c2(c + k) = t, then a, b, c are roots of x3 + kx2 – t = 0 thus, bc + ca + ab = 0 and abc = t 1 1 1 bc + ca + ab fi + + = =0 a b c abc 101 For truth of Statement-2, see theory f(a) = a2a2 + ba + c = 0 and g(b ) = a2b 2 – bb – c = 0 We have a2a2 + 2ba + 2c

1 = t

\t–

x 2 + b2 1

x 2 + b2 + x

=

x 2 + b2 - x b2

b2 = 2x t

Thus, b2 )t = b2 + 2at – t2 t = a2 + b2 – (t – a)2 Therefore, maximum value of f(x) is a2 + b2 which is f(x) = (2a – t +

b2 ˆ 1Ê ÁË a - ˜¯ a 2 Statement-2 is false as maximum value of f(x) = ax2 + bx + c, a < 0 is (4ac – b2)/4a. 105. For truth of statement-2 see theory Let P(x) = ax2 + bx + c. As P(x) = x does not have real roots, P(x) > x " x Œ R or P(x) < x " x Œ R. attained when t = a or x =

3.50

Complete Mathematics—JEE Main

Suppose P(x) < x " x Œ R. fi P(P(x)) < P(x) < x ⁄ x Œ R fi a(ax2 + bx + c)2 + b(ax2 + bx + c) + c – x = 0 cannot have real roots. 106. Statement-2 is true as |a + b| = |a| + |b| ¤ a, b are both non-negative or both non-positive. As ax2 + bx + c = 0 has imaginary roots, ax2 + bx + c > 0 " x Œ R or ax2 + bx + c < 0 " x Œ R. Since a(–1)2 + b(–1) + c = a – b + c < 0, we get ax2 + bx + c < 0 " x Œ R. Now, |ax2 + bx + c| + |x + y| = a Ê x 2 + y ˆ + (b + 1) x + c Ë a¯ 2 = |(ax + bx + c) + (x + y)| ¤ (ax2 + bx + c) (x + y) ≥ 0 ¤ x + y £ 0 as ax2 + bx + c < 0 " x Œ R. 107. Statement-2 is true. See Theory Put x + c = t, so that [ t + ( a - c )] [ t + ( b - c )] f(x) = t ( a - c ) (b - c ) =t+ + [(a – c) + (b – c)] t 2

È 2 ( a - c ) (b - c ) ˘ = Í t˙ + [ a - c + b- c] Î ˚ t Thus, minimum value of f(x) is and it is attained when t = x=–c+

(

2

a - c + b - c)

1 ≥ 2. y

112.

113.

114.

115.

1

È tan 2 a ˘ 2 x +x+ ≥ 2 Í x2 + x ˙ ÍÎ x2 + x x 2 + x ˙˚ = 2 tan a (a – b)2 + 4(a + b – 1) ≥ 0 " b ŒR. fi b2 – 2b(a – 2) + a2 + 4a – 4 ≥ 0 " b ŒR. fi 4(a – 2)2 – 4(a2 + 4a – 4) < 0 fi – 8a + 8 < 0 fi a > l \ a Œ (1, •) a + b = –b < 0 and ab = c < 0. As ab < 0, and a < b, a < 0 < b < – a = |a| As x2 + (a + b) x + c = 0 has no real roots, and the coefficient of x2 = 1 > 0, f (x) = x2 + (a + b) x + c > 0 x ŒR. fi (a + b)2 – 4c < 0. Also f (0) = c > 0, f (1) = a + b + c + 1 > 0 fi c (a + b + c) + c > 0 Next, f (0) f (–1) > 0 fi c (1 – a – b + c) > 0 fi c – c(a + b – c) > 0 As x – 1 < [x] £ x, we get –x £ – [x] < – x + 1 = x2 – 2x – 2 £ x2 – 2[x] – 2 < x2 – 2x = (x – 1)2 –3 £ x2 –2[x]–2 < (x – 1)2 – 1 2

(a - c ) (b - c ) or

108. If a π p, f(x) – g(x) = 0 is a quadratic equation whose coefficient of x2 π 0. \ f(x) – g(x) = 0 has two real roots Thus, statement-2 is true Let h(x) = f(x) – g(x) = (a – b) x2 + (b – q) x + (c – r) If h(x) = f(x) – g(x) = 0 for three distinct real values of x, then a – p = 0, b – q = 0, c – r = 0 fi a = p, b = q, c = r. \ Statement-1 is also true. However, statement-2 is not a correct explanation for Statement-1. 109. As A.M. ≥ G.M, we get

fiy+

Level 2 111. As A.M ≥ G.M, we get

(a - c ) (b - c ) > – c.

1Ê 1ˆ Ê 1ˆ y + ˜ ≥ yÁ ˜ = 1 Á Ë y¯ y¯ 2Ë

Thus, statement-2 is true. Since, sin2 x = 0, does not satisty sin4 x + p sin2 x + 1 = 0, we get 0 < sin2 x £ 1, for all x satisfying sin4 x + p sin2 x + 1 = 0 For such a value of sin x, 1 – p = sin2x + ≥2 sin 2 x fi p £ – 2 or p Œ (– •, –2] Let f(t) = t 2 + pt + 1 As f(0) = 1 > 0 and f(1) = 1 – p + 1 £ 0, we get there is at least one value of t Œ (0, 1] such that f(t) = 0. 110. As b2 – 4ac < 0, f(x) > 0 " x Œ R or f(x) < 0 " x Œ R. As f(0) = c > 0, f(x) > 0 " x Œ R. Thus, statement-2 is true. As f(x) = 0 does not have real roots, b2 – 4ac < 0. By the statement-2, f(x) > 0 " x Œ R. fi f(–1) = a –b + c > 0 fi a > b – c. Also b2 – 4ac < 0 fi a > b2/4c ¸ Ï b2 Hence, a > max Ì , b - c ˝ . Ó 4c ˛

tan 2 a

Quadratic Equations 3.51

Note that (x – 1)2 – 3 £ 0 for – 3 + 1 £ x £ 3 + 1 and (x – 1)2 – 1 > 0 for x < 0 or x > 2, The roots of x2 – 2[x] – 2 = 0

( If x Π( -

) 3 + 1, 0) , [x] = –1,

lie in - 3 + 1, 0 or (2, 1 +

3)

and x2 Р2[x] Р2 = 0 fi x2 = 0 or x = 0. This is not possible. Suppose x Π(2, 1 + 3 ), then [x] = 2, x2 Р2[x] Р2 = 0 fi x2 = 6 or x =

6.

Note that x = 6 satisfies x2 – 2[x] – 2 = 0 116. If [x] is odd, then [x]2 + a[x] + b is odd and if [x] is even, then [x]2 + a[x] + b is also odd Thus, [x]2 + a[x] + b can never take value 0. 117. Let a = sina, b = sinb, c = sing . Note that a, b, c are distinct and 0 < a, b, c < l. We can write the given equation as f (x) = (x – b) (x – c) + (x – c) (x – a) + (x – a)(x – b) = 0. Assume that a < b < c. Note that f (a) = (a – b) (a – c) > 0 f (b) = (b – c) (b – a) < 0 and f (c) = (c – a) (c – b) > 0. Thus, f (x) = 0 has a root in (a, b) and a root in (b, c). 118. Let f (x) = x2 + (2a + a) x + a 2 + aa + b. f (0) = a2 + aa + b = 0 Let other root of f (x) = 0 be r, then 0 + r = –(2a + a) = –2a + a + b = b – a. Thus, two roots are 0 and b – a. 119. As roots are real (a + b + c)2 – 3l (ab + bc + ca) ≥ 0 (a + b + c)2 = 3l £ ab + bc + ca Since, a, b, c are sides of triangle. b + c – a > 0, c + a – b > 0, a + b – c > 0 fi 2(ab + bc + ca) > a2 + b2 + c2 fi (a + b + c)2 < 4 (ab + bc + ca) Thus, 3l < 4 fi lŒ (–•, 4/3). 120. tan q + cot q = 2a and tanq cotq = b Now, 2a = tanq + cotq ≥ 2 if tanq > 0 and 2a = tanq + cot q £ – 2 if tanq < 0. fi 2|a| ≥ 2 fi |a| ≥ 1 Thus, least value of |a| is 1.

121. Let f (x) = x3 + x2 – 5x – 1, then f (–3) < 0, f (–2) > 0, f (–1)> 0, f (0) < 0, f (1) < 0, f (2) > 0. Thus, a root of f (x) = 0 lies in each of the three intervals: (–3, –2), (–1, 0),(1, 2). \ [a] + [ b ] + [g ] = –3 + (–1) + 1 = –3 122. tan A tan B = p, tan A tan B = q. tan A + tan B p = , we get As, tan (A + B) = 1 - tan A tan B 1- q sin2 (A + B) =

=

tan 2 ( A + B) 1 + tan 2 ( A + B)

p2 p 2 /(1 - q)2 2 2 = 1 + p /(1 - q) (1 - q )2 + p 2

123. Put x - 1 = t fi x – 1 = t2, so that equation becomes; t 2 + 1 + 3 + 4t + t 2 + 1 + 8 + 6t = 1 fi |t + 2| + |t + 3| = 1 Not possible as t ≥ 0. Thus, the equation has no solution. 124. Put x 2 + 11 = t fi x2 + 11 = t2, so that equation becomes t 2 + t - 11 + t 2 - t - 11 = 4. Since (t2 + t – 11) – (t2 – t – 11) ∫ 2t, we get t t 2 + t - 11 - t 2 - t - 11 = 2 t \ 2 t 2 + t - 11 = 4 + 2 1 fi 4(t2 + t – 11) = 16 + t 2 + 4t fi t2 = 16 4 fi x2 +11 = 16 fi x = ± 5 125. Note that a, b, c are roots of x y z + + =1 t t- p t-q ¤ t3 – (x + y + z + p + q) t2 +[(p + q) x + qy + pz] t – pqx = 0. \ product of roots = pqx abc pq 126. Putting x = 0, 1, 1/2, we get –1 £ c £ 1 , – 1 £ a + b + c £ 1, fi abc = pqx fi x =

-1 £

1 1 a + b + c £1, 4 2

3.52

Complete Mathematics—JEE Main

fi – 4 £ – a – 2b – 4c £ 4 As – 4 £ 4a + 4b + 4c £ 4, we get – 8 £ 3a + 2b £ 8. Also, – 8 £ a + 2b £ 8 fi –16 £ 2a £ 16 fi |a| £ 8 127. p + q and p + r are roots of at2 + 2bp (t – p) + c = 0 or at2 +2bpt + c – 2bp2 = 0. We have |q – r|2 = |(p + q) – ( p + r)|2 = [( p + q) + ( p + r)]2 – 4( p + q)( p + r)

(

2 4 c - 2bp 2 Ê -2bp ˆ = Á Ë a ˜¯ a

)

4 = 2 ÈÎb 2 p 2 - ac + 2abp 2 ˘˚ a 4 = 2 [(2a + b) bp2 – ac] a fi |q – r| 2 (2a + b)bp 2 - ac = |a| 128. Put x2 = y. The given equation will have four real roots if f ( y) = ay2 + by + c = 0 has two non-negative roots. This is possible if b – ≥ 0 , af (0) ≥ 0, b2 – 4ac ≥ 0 a fi ab £ 0, ac ≥ 0. This condition is met if a > 0, b < 0, c > 0 or a < 0, b > 0, c < 0 129. As x = 1 satisfies (b – c) x2 + (c – a) x + (a – b ) = 0, x = 1 must satisfy x2 + mx + 1 = 0 fi 1 + m + 1 = 0 fi m = –2. 130. Eliminating x from x + y + z = 4, and x2 + y2 + z2 = 6, we get y2 + z2 + (4 – y – z)2 = 6 fi 2y2 – 2(4 – z)y + (4 – z)2 + z2 – 6 = 0 As y is real, we get (4 – z)2 –2[(4 – z)2 + z2 – 6] ≥ 0 fi 3z2 – 8z + 4 £ 0 fi (3z – 2) (z – 2) £ 0 fi 2/3 £ z £ 2. Thus, maximum value of z is 2.

Previous Years’ AIEEE/JEE Main Questions 1. a, b are root of x2 = 5x – 3 or x2 – 5x + 3 = 0 fi a + b = 5, ab = 3. 2 2 2 Now, a + b = a + b = (a + b ) - 2ab b a ab ab

=

25 - 6 19 = 3 3

Êaˆ b and Á ˜ Ê ˆ = 1 Ë b¯ Ëa¯ Thus, equation whose roots are a/b and b/a is x2 – (19/3)x + 1 = 0 3x2 – 19x + 3 = 0

or

2. |a – b| = |a1– b1| fi (a – b)2 = (a1 - b1)2 fi (a + b)2 - 4ab = (a1 + b1)2 - 4a1b1 fi a2 - 4b = b2 – 4a fi (a – b)(a + b + 4) = 0 fi a + b + 4 = 0 [ a π b] 1 3 1 2 ax + bx + cx. 3 2 As f f is continuous on [0, 1] and f is differentiable on (0, 1). Also

3. Let f(x) =

f(0) = 0 and f(1) =

1 (2a + 3b + 6c) = 0 6 a Π(0, 1)

such that f ¢(a) = 0. Thus, ax2 + bx + c = 0 has at least one root in [0, 1]. 4. For each x Œ R, t Œ R t2x2 + |x| + 9 ≥ 9 > 0 Thus, t2x2 + |x| + 9 = 0 does not have real roots. 5. Let a and 2a be the roots of the given equation, then (a2 – 5a + 3)a2 + (3a – 1)a + 2 = 0

(1)

(a2 – 5a + 3)(4a2) + (3a – 1)(2a) + 2 = 0

(2)

get (3a – 1) (2a) + 6 = 0

Quadratic Equations 3.53

a π 1/3. Therefore,

a = – 3/(3a – 1)

Putting this value in (1) we get (a2 – 5a + 3)(9) – (3a –1)2(3) + 2(3a – 1)2 = 0 2

2

fi

9a – 45a + 27 – (9a – 6a + 1) = 0

fi

–39a + 26 = 0

fi

a = 2/3.

For a = 2/3, the equation becomes x2 + 9x + 18 = 0, whose roots are –3, –6. 6. Let a, b be the roots of ax2 + bx + c = 0

1 1 + 2 2 a b

a + b=

a = –1 and – (1– p) = 1– p

fi

a = 1 and 1 – p = 0

Thus, equation is x2 + x = 0 fi

x = 0, –1

10. Let a be the other root of x2 + px + 12 = 0 \

(a)(4) = 12 and a + 4 = p

fi

a = 3 and 3 + 4 = p

As x2 + px + q = 0 has equal roots, we must have p2 – 4q = 0

a 2 + b 2 (a + b ) 2 - 2ab = (ab ) 2 (ab ) 2

=

( - b / a ) 2 - 2( c / a ) -b = a (c / a ) 2 2

2

b c b 2c - ◊ 2= 2 a a a a

fi fi

fi

Thus, p = 7,

We are given

fi

a + (1 – p) = – p and a (1– p) = 1 – p

1 2 49 p = 4 4 11. See solution to Question No. 3 fi q=

12. We have Pˆ Ê Q ˆ -b and tan ÊÁ P ˆ˜ tan ÊÁ Q ˆ˜ = c tan ÊÁ ˜ + tan Á ˜ = Ë 2¯ a Ë 2¯ Ë 2¯ a Ë 2¯

2a2c = ab2 + bc2

Also

abc to obtain

fi

a c b = + b a c c a b , , are in A.P. a b c a b c , , are in H.P. c a b x2 – 3|x| + 2 = 0

P + Q = p – R = p/2 P Q p + = 2 2 4 P Q p tan Ê + ˆ = tan Ê ˆ Ë 2 2¯ Ë 4¯

2

fi fi 7.

fi

tan( P / 2) + tan(Q / 2) =1 1 - tan( P / 2) tan(Q / 2) -bla -b =1fi =1 1 - cla a-c

fi fi

2

fi

|x| – 3|x| + 2 = 0

fi

(|x| – 1) (|x| – 2) = 0

fi

|x| = 1, 2

fi

x = ± 1, ± 2

8. Let two numbers be a and b. We are given

fi

–b = a – c

or

c=a+b

13. Let a, b be roots of x2 – (a – 2)x – a – 1 = 0 Then

a + b = a – 2, ab = – (a + 1)

Now,

a2 + b2 = (a + b)2 - 2ab = (a – 2)2 + 2(a + 1)

a+b = 9, ab = 4 2 fi a + b = 18 and ab = 16

= a2 – 2a + 1 + 3

\

= (a – 1)2 + 3 x2 – (a + b)x + ab = 0

or

= a2 – 4a + 4 + 2a + 2

Thus, a2 + b2 is least when a = 1

2

x – 18x + 16 = 0

9. Let other root be a. Then

14. We know that if a and b are roots of x2 – bx + c = 0, then (a – b)2 = discriminant = b2 – 4c.

3.54

Complete Mathematics—JEE Main

In the present case b = a + 1.

fi

1 = b2 – 4c

\

3(y – 1)x2 + 9(y – 1)x + 7y – 17 = 0

As x is real 81(y – 1)2 – 12(y – 1) (7y – 17) ≥ 0

15. We can write the given equation as (x – k)2 = 5 – k. Thus, for roots to be real we must have 5 – k ≥ 0 or k £ 5. Also, for k £ 5, x = k ± 5-k For both the roots to be less than 5, we must have that larger of the two roots is less than 5. Therefore.

fi

3(y – 1) [27(y – 1) – 4(7y – 17)] ≥ 0

fi

3(y – 1) (41 – y) ≥ 0

fi

(y – 1)(y – 41 ) £ 0

fi

1 £ y £ 41

20. a + b = – a, ab = 1 |a – b|
Р4ab " x ΠR. 23. x2 Рx + 1 = 0

c2 –

Quadratic Equations 3.55

fi (p – 5) (p + 2) = 0

1 3 i x = – w, – w2 where w = - + 2 2 Let a = -w and b = – w2

fi

We have a

2009

+b

= (– w)2009 + (– w2)2009 = (– 1) [w2007 w2 + (w2)2007 w4] = (– 1) (w2 + w) = (– 1) (– 1) = 1 24. As a π a1, p(x) = f(x) – g(x x = –1, we get p(x) nomial. As p(x must be of the form p(x) = k(x + 1)2 where k = a – a 1. p(–2) = 2, we get

As

2 = k(– 2 + 1)2 fi Thus,

p(x) = 2(x + 1)2

\

p(2) = 2(2 + 1)2 = 18

Thus, p lies in the set {–2, 5} 28.

2009

k = 2.

p = – 2, 5

a 2 b 2 a3 + b 3 - p + = = q b a ab Ê a2 ˆ Ê b2 ˆ and Á ˜ Á ˜ = ab = q Ë b ¯Ë a ¯ Thus, required equation is Ê pˆ x2 - Á - ˜ x + q = 0 Ë q¯ qx2 + px + q2 = 0

or

29. As the roots are real and distinct (a + 1)2 – 4(a2 + a – 8) > 0 3a2 + 2a – 33 < 0 fi

(3a + 11)(a – 3) < 0 fi -

11 ln(4) > 1 This is not possible. Thus, equation has no real roots. 26. x2 + 2x + 3 = 0 fi fi

Also, for 22 – 2(a + 1) + a2 + a – 8 < 0 fi

a2 – a – 6 < 0 fi (a – 3) (a + 2) < 0

fi

–2 5.

Therefore the least integral value a of x is – 6. a2 + 5a – 6 This value of a = 0.

3.56

Complete Mathematics—JEE Main

31. 4 = fi

fi (k2 – 4)2 = 0 fi k = 2

a + b -q / p -q = = ab r/ p r q + 4r = 0

Also, Thus,

We have

2q = p + r

[

a3 + b3 = (a + b)3 – 3ab (a + b)

p, q, r are in A.P.]

= (a + b) [(a + b)2 - 3ab]

p = – 9r, q = – 4r

(

)

35. Note that we must have 3x2 + x + 5 ≥ 0 and x – 3 ≥ 0 or x ≥ 3. Squaring both sides of (1), we get

16 1 52 - 4Ê - ˆ = Ë ¯ 81 9 81 2 fi |a - b | = 13 9 32. As roots of 2x2 + 3x numbers, the given quadratic equations have both roots in common. [ See Solution Question No. 26.] =

a b c fia:b:c=2:3:4 = = 2 3 4 33. Case 1: When x < 3/2 Thus,

3x2 + x + 5 = x2 – 6x + 9 fi

2x2 + 7x – 4 = 0

fi

(2x – 1)(x + 4) = 0

fi

x = 1/2, – 4 x ≥ 3. Thus,

(1) has no solution. 36. a + b = 6, ab = - 2. a10 – 2a8 = a10 + b10 + ab (a8 + b8)

In this case the equation becomes

= (a + b) (a9 + b9) = 6a9

x2 – (2x – 3) – 4 = 0

fi

fi

(x – 1)2 = 2

fi

x=1±

fi

x–1=± 2

2

As x < 3/2, we take x = 1 –

}

= 280 2

2

4r Ê -q ˆ =Á ˜ Ë p¯ p

x2 – 2x + 1 = 2

{

= 4 2 ( 2) ÈÎ66 - 2(24 ) - 1 ˘˚

|a - b|2 = (a + b)2 - 4ab

fi

k > 0]

[

a10 - 2a8 =3 2a9

2x3 Р9x2 Рkx Р13 = 0 jugate pair, so let the roots of (1) be 2 + 3i, 2 Р3i and a, where a ΠR. Now,

2

Case 2: When x ≥ 3/2

(2 + 3i) + (2 – 3i) + a = 9/2 fi

a = 1/2

In this case the equation becomes \ real root of the equation is 1/2.

x2 + (2x – 3) – 4 = 0 fi

i) (2 – 3i)a = 13/2 fi

(x + 1)2 = 8 fi x + 1 ± 2 2

38. Write

As x ≥ 3/2, x = –1 + 2 2

(

) (

)

sum of roots = 1 - 2 + -1 + 2 2 = 2 34. a + b = 4 2 k, ab = 2e4lnk – 1 = 2k4 –1 Now, 66 = a2 + b2 = (a + b)2 – 2ab = 32k2 – 2(2k4 – 1) fi 33 = 16k2 – 2k4 + 1 fi 2k4 – 16k2 + 32 = 0 fi k4 – 8k2 + 16 = 0

a =

1/2]

x4 + x2 + 1 = x4 + 2x2 + 1 – x2 = (x2 + 1)2 – x2 = (x2 + x + 1) (x2 – x + 1) The given equation, now can be written as (x2 + x + 1) [(a – 1)(x2 – x + 1) + (a + 1) (x2 + x + 1)] = 0 fi (x2 + x + 1) [2ax2 + 2x + 2a] = 0

Quadratic Equations 3.57 3 4 5

42. x + y + z = 12, x y z = (0.1)(600)3

As x is real, we get

We have

ax2 + x + a = 0.

x3y4z5 = (0.1)(600)(600)2

This equation will have real and distinct roots if 2

= (60)(600)2

2

a π 0 and 1 – 4a > 0 fi a π 0, a < 1/4

= 3 32 85 5

fi a Œ ( – 1/2, 0) » (0, 1/2) 2

\

2

39. If x – 5x + 5 π 1, then x + 4x – 60 = 0

x = 3, y = 4, z = 5 Thus, x3 + y3 + z3 = 33 + 43 + 53 = 27 + 64 + 125 = 216

fi (x + 10)(x – 6) = 0 fi x = –10, 6 When x = –10 and 6, x2 – 5x + 5 π 1 If x2 – 5x + 5 = 1, ( x 2 - 5 x + 5) x

2

+ 4 x - 60

=1

Previous Years' B-Architecture Entrance Examination Questions

2

In this case, x – 5x + 4 = 0 fi (x – 1) (x – 4) = 0 fi x = 1, 4 If x2 – 5x + 5 = –1, then x2 + 4x – 60 must be even integer.

1.

tan 30º + tan 15º = – 2p tan 30º tan 15º = q

2

But x – 5x + 6 = 0 fi x = 2, 3 For x = 2, x2 + 4x – 60 = – 48 is even.

1 = tan 45º =

tan 30∞ + tan 15∞ 1 - tan 30∞ tan 15∞

-2 p 1- q

1 – q = – 2p

For x = 3, x2 + 4x – 60 = – 39 is odd.

fi

Thus, sum of desired values is –10 + 6 + 1 + 4 + 2 = 3.

1=

fi

q = 1 + 2p.

40. Let a be common roots of

fi

2. Let roots of the equation be 1– k and 1 + k, where k > 0.

x2 + bx – 1 = 0 and x2 + x + b = 0. Then a2 + ba – 1 = 0 and a2 + a + b = 0 fi (a2 + ba –1) – (a2 + a + b) = 0 fi (b – 1)a – (b + 1) = 0 b +1 fia= if b π 1 b -1 As a π –1, b π 0

Then

2 = ( 1 – k) + (1 + k) =

fi

1=

2a a+2

a a+2 a.

3. |x|2 – 4|x| – 2 = 0

2

Ê b + 1ˆ Ê b + 1ˆ Also, Á –1=0 + bÁ ˜ Ë b - 1¯ Ë b - 1˜¯

fi (|x| – 2)2 = 6

fi (b + 1)2 + b(b + 1)(b – 1) – (b – 1)2 = 0

fi |x| = 2 ± 6

fi 4b + b(b + 1)(b – 1) = 0

As | x | > 0, | x | = 2 +

2

fi x = ± (2 +

3

2x + 1 - 2x - 1 = 1 Also, (2x + 1) – (2x – 1) = 2 2x + 1 + 2x - 1 = 2

(1) (2)

3 , 2

4 x2 - 1 =

6)

(3)

AIEEE/JEE Questions. 5. ax2 – bx(x – 1) + c(x – 1)2 = 0 fi

From (1) and (3), we get

\

6

\ there are two values of x.

41. We are given

2x + 1 =

|x| – 2 = ± 6

2

As b π 0, 4 + b – 1 = 0 fi b = 3 fi |b| =

fi

2x - 1 = 3 4

1 2

fi fi

x ˆ2 x ˆ Ê Ê + bÁ aÁ+c=0 ˜ Ë x - 1˜¯ Ë x - 1¯ x = a,b x -1 a b x= , a +1 b +1

Complete Mathematics—JEE Main

3.58

x4– ax2 + b = (x2 – 3x + 2) p (x)

6.

9. The given equation can be written as

= (x – 1)(x – 2) p(x)

r(2x + p + q) = (x + p) (x + q)

where p(x \1–a+b=0 fi

and

16 – 4a + b = 0

Let roots of this equation be a , -a.

a = 5, b = 4

Then

Thus, equation whose roots are a and b is

0 = a + (-a) = - (p + q – 2r) fi 2r = p + q

2

x – 9x + 20 = 0

and

bˆ Ê 7. f ¢(x) = 2ax + b = a Ë x + ¯ 2a As a > 0,

(x – 3k)2 = 2(k – 1) fi

f ¢(x) > 0 for x > – b/2a

Thus, f - b/2a) (– b/2a, •).

a(–a) = pq – r(p + q)

1 1 2 2 2 = pq - ( p + q) = - ( p + q ) 2 2 10. Write the equation as

f ¢(x) < 0 for x < – b/2a and

x2 + (p + q – 2r) + pq – r(p + q) = 0

or

•,

x = 3k ± 2 k - 1

Both the roots will be greater than 3 if smaller root is greater than 3, that is, if 3k - 2 k - 1 > 3

-b 4ac - b 2 Also, Min f(x) = f Ê ˆ = >0 Ë 2a ¯ 4a 2 if b < 4ac

fi

3(k - 1) > 2 k - 1

fi

k > 1 and 9(k – 1) > 2 fi

As Min f(x) > 0, f(x) > 0 " x ΠR

fi

11 – 9k < 0.

\ Both the statements are true and statement- 2 is

k > 11/9

11. (1 + 3m)2 – (1 + m2)(1 + 8m) < 0 fi 1 + 6m + 9m2 – (1 + m2 + 8m + 8m3) < 0

8. As the roots are of opposite signs, the product of roots must be negative, that is, 2

a - 3a + 2 < 0 fi (a - 1)(a - 2) < 0 3 fi a Π(1, 2)

fi –2m + 8m2 – 8m3 < 0 fi 2m(4m2 – 4m + 1) > 0 fi m(2m – 1)2 > 0 which is true for each m Œ N.

CHAPTER FOUR

Determinants

EVALUATION OF DETERMINANTS

MINORS AND COFACTORS

A determinant of order two is written as

Minor

a11 a21

a12 a22

(ai j ΠC " i, j)

and is equal to a11 a22 – a12 a21. A determinant of order three is written as a11 a21 a31

a12 a22 a32

a13 a23 a33

a22 a32

a23 a - a12 21 a33 a31

1

Illustration (ai j ΠC " i, j)

and is equal to a11

Let A = (aij)n × n be a square matrix of order n. Then the minor Mij of the element aij of the matrix A is the determinant of the square sub-matrix of order (n – 1) obtained by deleting ith row and jth column of matrix A.

a23 a + a13 21 a33 a31

a22 a32

= a11 (a22 a33 – a23 a32) – a12 (a21 a33 – a23 a31) + a13(a21 a32 – a22 a31) = a11 a22 a33 + a12 a23 a31 + a13 a32 a21 – a13 a31 a22 – a32 a23 a11 – a12 a21 a33 A determinant of order 3 can also be evaluated by using the following diagram, due to Sarrus:

Minor of element a23 in the determinant a11 a12 a13 a21 a22 a23 a31 a32 a33 is a a M23 = 11 12 a31 a32

Cofactor Let A = (aij)n ¥ n be a square matrix of order n. Then the cofactor of the element aij of the matrix A is denoted by Cij and is equal to (- 1)i + j Mij where Mij is the minor of the element aij of the matrix A. Note that a11 a12 a13 a21 a22 a23 = a11 M11 – a12 M12 + a13 M13 = a11 C11 + a12C12 + a13C13 a a a 31

The product of the three terms on each of the three single arrows are prefixed by a positive sign and the product of the three terms on each of the three double arrows are prefixed by a negative sign.

32

33

a11 a12 a13 D = a21 a22 a23 a31 a32 a33 D = ai1 Ci1 + ai2 Ci2 + ai3 Ci3 = a1j C1j + a2j C2j + a3j C3j

If then

i = 1, 2, 3 j = 1, 2, 3

Remark Note This method does not work for higher order determinants.

and

ai1 Cj1 + ai2 Cj2 + ai3 Cj3 = 0 a1i C1j + a2i C2j + a3i C3j = 0

iπj iπj

4.2

Complete Mathematics—JEE Main

The above results remain true for determinants of every order.

To evaluate 1+ a b c D= a 1+ b c a b 1+ c D = D1 + aD2 where

PROPERTIES OF DETERMINANTS write

1.

Reflection Property

The determinant remains unaltered if its rows are changed into columns and the columns into rows. In other words, if A is a square matrix, then |A| = |A¢| where A¢ is transpose of A.

2.

1 b c D1 = 0 1 + b c 0 b 1+ c = (1 + b) (1 + c) – bc = 1 + b + c and

All-zero Property

1 b c c D2 = 1 1 + b 1 b 1+ c

If all the elements of a row (column) are zero, then the determinant is zero.

3.

Proportionality [Repetition] Property

If the elements of a row (column) are proportional [identical] to the element of the some other row (column), then the determinant is zero.

4.

Switching Property

The interchange of any two rows (columns) of the determinant changes its sign.

5.

Scalar Multiple Property

If all the elements of a row (column) of a determinant are multiplied by a non-zero constant, then the determinant gets multiplied by the same constant.

6.

Property of Invariance a1 a2 a3

b1 b2 b3

c1 a1 + a b1 + b c1 c2 = a2 + a b2 + b c2 c3 a3 + a b3 + b c3

b1 b2 b3

c1 c2 c3

d1 a1 d2 = a2 d3 a3

c1 c2 c3

d1 b1 d2 + b2 d3 b3

8.

Factor Property

If a determinant D becomes zero when we put x = a, then (x – a) is a factor of D.

9.

Triangle Property

If all the elements of a determinant above or below the main diagonal consists of zeros, then the determinant is equal to the product of diagonal elements. That is, a1 0 0

Sum Property a1 + b1 a2 + b2 a3 + b3

Using C2 Æ C2 – bC1 and C3 Æ C3 – cC1, we get 1 0 0 D2 = 1 1 0 = 1 1 0 1 \ D =1 + a + b + c

c1 c2 c3

That is, a determinant remains unaltered under an operation of the form Ci Æ Ci + a Cj + b Ck, where j, k π i, or an operation of the form Ri Æ Ri + a Rj + b Rk, where j, k π i.

7.

2

Illustration

Remark

c1 c2 c3

d1 d2 d3

Remark It is one of the most under used property. But evaluation of some of the determinants beame very easy when we use it.

10.

a2 b2 0

a3 a1 b3 = a2 c3 a3

0 b2 b3

0 0 = a1 b2 c3 c3

Product of Two Determinants a1 a2 a3

b1 b2 b3

c1 c2 c3

a1 a2 a3

a1a1 + b1b1 + c1g 1 = a2a1 + b2 b1 + c2g 1 a3a1 + b3 b1 + c3g 1

b1 b2 b3

g1 g2 g3

a1a 2 + b1b2 + c1g 2 a2a 2 + b2 b2 + c2g 2 a3a 2 + b3 b2 + c3g 2 a1a 3 + b1b3 + c1g 3 a2a 3 + b2 b3 + c2g 3 a3a 3 + b3 b3 + c3g 3

Determinants 4.3

Here we have multiplied rows by rows. We can also multiply rows by columns, or columns by rows, or columns by columns.

11.

where a21, a22, a23, a31, a32 and a33 are constants, then a11 ¢ ( x ) a12 ¢ ( x ) a13 ¢ ( x) D ¢(x) = a21 and a22 a23 a31 a32 a33

Conjugate of a Determinant

If ai, bi, ci ΠC (i = 1, 2, 3), and a1 Z = a2 a3

12.

b1 b2 b3

a1 c1 c2 then Z = a2 a3 c3

b1 b2 b3

Ú a11 ( x)d x Ú a12 ( x)d x Ú a13 ( x)d x

c1 c2 c3

Differentiation of a Determinant

Ú D( x)dx =

a1 ( x ) a2 ( x ) a3 ( x ) a4 ( x )

D(x) =

then

a ¢ ( x ) a2 ( x ) a ( x ) a2¢ ( x ) D¢(x) = 1 + 1 a3¢ ( x ) a4 ( x ) a3 ( x ) a4¢ ( x )

If we write D(x) = [C1, C2], where Ci denotes ith column, then D¢(x) = [C 1¢ , C2] + [C1, C 2¢ ] where C 1¢ denotes the column which contains the derivative of all the functions in the ith column Ci. Similarly, if È R¢ ˘ È R ˘ ÈR ˘ D(x) = Í 1 ˙ then D¢(x) = Í 1 ˙ + Í 1 ˙ R Î R2 ˚ Î R2¢ ˚ Î 2˚ a11 ( x ) a12 ( x ) a13 ( x ) D(x) = a21 ( x ) a22 ( x ) a23 ( x ) a31 ( x ) a32 ( x ) a33 ( x ) D¢(x)

Next, if then =

a11 a11 ( x ) a12 ¢ ( x ) a12 ( x ) a13 ( x ) ¢ ( x ) a13 ( x ) a21 ¢ ( x ) a22 ( x ) a23 ( x ) + a21 ( x ) a22 ¢ ( x ) a23 ( x ) a31 a31 ( x ) a32 ¢ ( x ) a32 ( x ) a33 ( x ) ¢ ( x ) a33 ( x )

a11 ( x ) a12 ( x ) a13 ¢ ( x) + a21 ( x ) a22 ( x ) a23 ¢ ( x) a31 ( x ) a32 ( x ) a33 ¢ ( x) = [C 1¢ , C2, C3] + [C1, C 2¢ , C3] + [C1, C2, C ¢3] Similarly, if

Corollary

If

È R1 ˘ È R1¢ ˘ È R1 ˘ È R1 ˘ Í ˙ D(x) = R2 then D¢(x) = Í R2 ˙ + Í R2¢ ˙ + Í R2 ˙ Í ˙ Í ˙ Í ˙ Í ˙ ÍÎ R3 ˙˚ ÍÎ R3 ˙˚ ÍÎ R3 ˙˚ ÍÎ R3¢ ˙˚ (Differentiation and Integration of Determinant) a11 ( x ) a12 ( x ) a13 ( x ) a22 a23 D(x) = a21 a31 a32 a33

a22 a32

a23 a33

In general, for any positive integer m m m m a11 ( x ) a12 ( x ) a13 ( x) m D (x) = a21 a22 a23 a31 a32 a33

If each ai(x) is differentiable function and

a21 a31

13.

Determinant of Cofactor Matrix

a11 If D = a21 a31

C11 C12 a13 a23 then D1 = C21 C22 C31 C32 a33

a12 a22 a32

C13 C23 = D2 C33

where Ci j denotes the co-factor of the element ai j in D.

SOME TIPS FOR QUICK EVALUATION OF DETERMINANTS 1. If D is a skew symmetric determinant of odd order, then D = 0 Illustration

3

0 a b D = -a 0 c -b -c 0 Using the reflection property, write 0 - a -b D = a 0 -c b c 0 Taking –1 common from R1, R2 and R3 we get 0 a b 3 D = (–1) - a 0 c = –D -b -c 0 fi 2D = 0 fi D = 0 2. If a1, a2, a3 are in A.P.; b1, b2, b3 are in A.P. and c1, c2, c3 are also in A.P. Then a1 a2 a3 D = b1 b2 b3 = 0 c1 c2 c3 Use C1 Æ C1 + C3 – 2C2 Let

Complete Mathematics—JEE Main

4.4

3. If a1, a2, a3 are in G.P.; b1, b2, b3 are also in G.P., with the same common ratio, then a1 a2 a3 D = b1 b2 b3 = 0 c1 c2 c3 [c1, c2, c3 can be any three complex numbers.]

Some Frequently used Determinants 1.

1 a

1 b

a2

b2

1 1 1 1 c = a b c c 2 bc ca ab

a1 a D= 2 a3

2.

1 b

3

3

a

b

1 c = (a – b) (b – c) (c – a) (a + b + c) c

c1 c2 c3

If D π 0, then the only solution of the above system of equations is x = 0, y = 0 and z = 0. Corollary If at least one of x, y, z is non-zero and x, y and z are connected by the three given equations, then the elimination of x, y and z leads to the relation a1 a2 a3

= (a – b) (b – c) (c – a) 1 a

b1 b2 b3

b1 b2 b3

c1 c2 = 0 c3

CRAMER’S RULE

3

1

1

1

3. a 2

b2

c 2 = (a – b) (b – c) (c – a) (bc + ca + ab)

a3

b3

c3

If

a b c 4. b c a = 3abc – a3 – b3 – c3 c a b = (a + b + c) (bc + ca + ab – a2 – b2 – c2) 1 = - (a + b + c)[(b - c)2 + (c - a)2 + (a - b)2 ] 2

LINEAR EQUATIONS The system of linear homogeneous equations a1 x + b1 y + c1 z = 0 a2 x + b2 y + c2 z = 0 a3 x + b3 y + c3 z = 0 has a non-trivial solution (i.e., at least one of the x, y, z is different from zero) if and only if D = 0, where

a1 D = a2 a3

b1 b2 b3

c1 c2 π 0 c3

then the solution of the system of linear equations a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 a3 x + b3 y + c3 z = d3 is given by x =

D D1 D ,y= 2 ,z= 3 D D D

where

d1 D1 = d2 d3

b1 b2 b3

c1 a1 c2 , D 2 = a2 c3 a3

a1 D 3 = a2 a3

b1 b2 b3

d1 d2 d3

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Example 1: Let 1 x D(x, y) = 1 x + y 1 x

y y x+y

Then D(–3, 2) equals (a) 13 (c) 12 Ans. (b)

(b) – 6 (d) – 5

d1 d2 d3

c1 c2 and c3

Determinants 4.5

Solution: Using R2 Æ R2 – R1, R3 Æ R3 – R1 we get 1 x y D(x, y) = 0 y 0 = xy 0 0 x \

D(–3, 2) = – 6 Example 2: Suppose a, b, c are distinct real numbers and a a2 D= b b

2

c c2 Then a + b + c equals (a) –1 (c) 0 Ans. (b)

b+c c+a =0 a+b (b) 2 (d) –5

Solution: Using C3 Æ C3 + C1 and taking (a + b + c) common from C3 we get (1) D = (a + b + c) D1 where D1 = b b 2 1 c2

1

Using R1 Æ R1 – R2 and R2 Æ R2 – R3, we get a - b a 2 - b2 D1 = b - c c

Solution: Interchanging the rows and columns, we get 0 a-b a-c 0 b-c D = b-a c-a c-b 0 Taking –1 common from each of R1, R2, R3 we get 0 b-a c-a 3 0 c-b D = (–1) a - b a-c b-c 0 fi D = –D fi 2D = 0 or D = 0 Alternative Solution D is a skew symmetric determinant of odd order, therefore D = 0

a a2 1 c

Example 4: Let 0 b-a c-a D = a-b 0 c - b , then a-c b-c 0 D equals: (a) 0 (b) abc (c) a2 + b2 + c2 (d) bc + ca + ab Ans. (a)

2

b -c

0

2

0

c2

1

1 a+b 0 = (a – b) (b – c) 1 b + c 0 c = (a – b) (b – c)

c2

1

1 a+b 1 b+c

[Expand along C3] = (a – b) (b – c) (c – a) (2) From (1) and (2) D = (a + b + c) (a – b) (b – c) (c – a) As D = 0 and a, b, c are distinct, we get a+b+c=0 Example 3: Suppose A = (aij)3 × 3, where aij Œ R. If det (adj A) = 25, then |det (A)| equals: (a) 5 (b) 12.5 (d) 52/3 (c) 5 5 Ans. (a) Solution: Using det (adj A) = (det (A))2, we get (det (A))2 = 25 fi |det (A)| = 5

Example 5: Let A = (aij)3 × 3, where aij Œ C the set of complex numbers. If det (A) = 2 – 3i, then det (A–1) equals: 1 1 (a) (2 – 3i) (b) (2 + 3i) 13 13 (c) 2 – 3i (d) 2 + 3i Ans. (b) 1 1 = Solution: det (A–1) = det ( A) 2 - 3i 2 + 3i 1 = = (2 + 3i ) 2 2 13 2 +3 Example 6: In a triangle ABC, if 1 a b D= 1 c a =0 1 b c 2

then sin A + sin B + sin2C is: 3 3 2 9 (c) 4 Ans. (a) (a)

2

(b)

5 4

(d) 2

Solution: Evaluating along C1, we get D = (c2 – ab) + (b2 – ac) + (a2 – bc) = 0 fi (a – b)2 + (b – c)2 + (c – a)2 = 0 fi a = b = c fi A = B = C = p /3 \

sin2A + sin2B + sin2C =

3 3 . 2

Complete Mathematics—JEE Main

4.6

Example 7: If

a - b a 2 - b2

2

x +x

x +1

2

2 x + 3x - 1

x-2 3 x - 3 = Ax + B,

3x

2

x + 2x + 3

D1 = b - c c

2x - 1 2x - 1

then A is equal to: (a) 12 (c) 24 Ans. (c)

Ax + B =

x-2 0

x2 + 2 x + 3 2 x - 1 2 x - 1 x +1 x - 2 2x - 1 2x - 1

3 x-2 0 2x - 1 = 4(3) (2x – 1) = 24x – 12 A = 24 =4

Thus,

Let

p + a3

D = b b2

p + b3

c2

p + c3

c

Now,

1

D1 = b b 2 1 c

c2

a a and

2

b3

c2

c3

c

write a a2 1

D2 = abc 1 b b2 = abc(-1)2 b b2 1 1 c

c2

a

3

D2 = b b 2

1

1 a a2

0 -51 -71 – c = - P(0) = - 51 0 -73 71 73 0 =0 [ Skew symmetric determinant of odd order]

Example 10: Let

Solution: Write D = pD1 + D2 where a a

TIP Suppose P(x) = x3 + ax2 + bx + c, then product of zeros of P(x) is – c = –P(0).

If D = 0, then which one of the following is not true (a) a = –1, b = 1 (b) b = 1, c = p (c) a = 0, c = p (d) abc + p = 0 Ans. (c)

2

1

Solution:

Example 8: Suppose a, b, c are three integers such a < b < c and p is a prime number. a a2

0

Example 9: Suppose x -51 -71 P(x)= 51 x -73 71 73 x Product of zeros of P(x) is (a) 0 (b) 195 (c) –195 (d) –264333 Ans. (a)

Expanding along R2, we get Ax + B = (-1)(- 4)

c2

0

1 a+b 1 b+c = (a – b) (b – c) (c – a) As a < b < c, D1 π 0. Therefore, D = 0 fi p + abc = 0 fi p = –abc As p is prime and a, b, c are integers such that a < b < c, we must have – a = 1, b = 1, c = p. fi a = –1, b = 1, c = p.

Solution: Applying R2 Æ R2 – R1 – R3, we get x +1 0

b -c

2

= (a - b)(b - c)

(b) 18 (d) 30

x2 + x -4

2

c

= abc D1 Thus, D = (p + abc) D1. Using R2 Æ R2 – R3, R1 Æ R1 – R2, we get

c2

1

x -3 + 4i 3 - 4i -7i 5 + 6i P(x) = x - x 7 - 2i -7 - 2i The number of values of x for which P(x) = 0 is (a) 0 (b) 1 (c) 2 (d) 3 Ans. (b) Solution: Using R2 Æ R2 – R1, R3 Æ R3 + R1, we get x -3 + 4i 3 - 4i P(x) = 0 3 - 11i 2 + 10i 0 4 + 2i -4 - 6i As P(x) is a linear polynomial, P(x) = 0 for exactly one value of x.

Determinants 4.7

Example 11: Let

Thus, sin q 1 - sin q

1 D (q) = - sin q -1 Solution of D(q) = 3 is p 3p (a) , 2 2 (c)

{ } { }

1 sin q , 0 £ q £ 2p 1

{ {

(b)

p 3p , 4 4

(d)

p 3p 5p 7p , , , 4 4 4 4 p p 3p , , ,p 4 2 4

}

= (1 + a2 + a4) x2 – x3 As x π 0, D(x) = 0

1 sin q 1

= 2 ( 1 + sin2q ) D(q) = 3 fi 2 sin2q = 1 p 3p 5p 7p 1 sin q = ± fi q= , , , 4 4 4 4 2

\ fi

a

a2

a

a2 - x

a3

D(x) =

a2 a3 a4 - x Number of values of x for which D(x) = 0 is (a) 0 (b) 1 (c) 2 (d) 3 Ans. (b) Solution: Using the sum property, write D(x) = D1 –xD2 where a

a2

a2 - x

a3

a2

a3

a4 - x

1

a

a2

1 D1 = a

and

2

D2 = 0 a - x 0

a

a

3

2

3

4

a -x 4

= (a – x) (a – x) – a6 = – (a2 + a4)x + x2 In D1, use C2 Æ C2 – aC1, C3 Æ C3 – a2C1 to obtain 1 D1 = a

0 -x

0 0 = x2

a2

0

-x

a2

b2

c2

( a + l )2

( b + l )2

( a - l )2

( b - l )2

a2 ( c + l )2 = k l a 1 ( c - l )2

l π 0, then k is equal to: (a) 4 labc (c) 4 l2 Ans. (c)

b2 b 1

c2 c 1

(b) – 4 labc (d) – 4 l2

Solution: Using R3 Æ R3 – R2 and R2 Æ R2 – R1, we get a2 D = 2 al + l 2 - 4 al

Example 12: Suppose a Œ R and x π 0. Let 1- x

1

x=

Example 13: If

Ans. (b) 2 sin q 1 D(q) = 0 0 - sin q

fi

. 1 + a + a2 Thus, D(x) = 0 for exactly one value of x.

}

Solution: Using C1 Æ C1 + C3, we get

D(x) = x2 + (a2 + a4) x2 – x3

b2

c2

2 bl + l 2 - 4 bl

2cl + l 2 - 4cl

Take – 4l common from R3, and applying R2 Æ R2 – 2lR3, we get a 2 b2 c 2 D = - 4l 3 1 1 1 a b c a2 = l (4l ) a 1 2

\

b2 b 1

c2 c 1

k = 4l2 Example 14: Let 1 f (q) = - sin q -1

cos q 1 sin q

-1 - cos q 1

Suppose A and B are respectively maximum and minimum value of f (q). Then (A, B) is equal to: (a) (2, 1) (b) (2, 0) (d) ÊÁ 2, 1 ˆ˜ (c) ( 2 , 1) Ë 2¯ Ans. (b) Solution: Using C1 Æ C1 + C3, we get 0 cos q 1 f (q) = - (sin q + cos q ) 0 sin q Evaluating along C1, we get

-1 - cos q 1

Complete Mathematics—JEE Main

4.8

f (q) = (sin q + cos q )

cos q sin q

has a non-trivial solution, then ab + bc + ca equals: (a) a + b + c (b) a b c (c) 1 (d) –1 Ans. (b)

-1 1

= (sin q + cos q)2 2 = È 2 ÊÁ 1 sin q + 1 cos q ˆ˜ ˘ ÍÎ Ë 2 ¯ ˙˚ 2

Êp ˆ = 2 sin 2 Ë + q ¯ 4 As

and Thus,

Êp ˆ 0 £ sin 2 Ë + q ¯ £ 1, we get 4 0 £ f (q) £ 2 p A = 2 for q = 4 p B = 0 for q = 4 (A, B) = (2, 0)

Example 15: If a, b, c are non-zero real numbers and if the system of equations (a – 1)x = y + z, (a – 1)y = z + x, (a – 1)z = x + y,

Solution: As the given system of equations has a nontrivial solution, a - 1 -1 -1 D = -1 b - 1 -1 = 0 -1 -1 c - 1 Write a -1 -1 1 -1 -1 D = 0 b - 1 -1 - 1 b - 1 -1 0 -1 c - 1 1 -1 c - 1 1 0 0 = a [(b - 1)(c - 1) - 1] - 1 b 0 1 0 c [use C2 Æ C2 + C1 C 3 Æ C 2 + C 1] = a(bc – b – c) – bc As D = 0, we get ab + bc + ca = abc

LEVEL 1 Straight Objective Type Questions Example 16: Let x - 10 7 6 P(x) = 2 x - 10 5 x - 10 3 4 sum of zeros of P(x) is (a) 30 (b) 28 (c) 27 (d) 25 Ans. (a) Solution: Using R1 ´ R3, write P(x) = –

x - 10 3 4 2 x - 10 5 7 6 x - 10

=–

coefficient of x 2

coefficient of x 3 Example 17: Let

P(x) =

x 2 - 13

4

3

x 2 - 13

6

5

= 30.

2 7 2

x - 13

If x = – 2 is a zero of P(x), then sum of the remaining five zeros is (a) –2 (b) 0 (c) 2 (d) 3 Ans. (c) Solution:

TIP TIP Observe P(x) is of the form P(x) = – (x –10)3 + a (x – 10) + b where a, b are some real numbers.

\ sum of zeros of P(x)

Observe P(x) = (x2 –13)3 + a (x2 – 13) + b. where a, b are some real numbers.

As coefficient of x5 in P(x) is 0, sum of six zeros of P(x) is 0. fi sum of the remaining five zeros + (–2) = 0

Determinants 4.9

fi

sum of the remaining five zeros = 2.

=

a, b are two real numbers and

Example 18: Suppose f(n) = a n + b n. Let 3 D = 1 + f (1) 1 + f (2 )

1 + f (1) 1 + f (2) 1 + f (2) 1 + f (3) 1 + f (3) 1 + f (4)

If D = k (a – 1)2 (b – 1)2 (a – b)2, then k is equal to (a) 1 (b) 4 ab (c) 9 (d) a2 b2 Ans. (a) Solution: 1+a + b

1+ a2 + b2

1+ a2 + b2

1 + a3 + b3

1+1+1 D = 1+a + b 1+ a2 + b2 1 = 1

1 a

1 a2 where

1 D1 = 1

1 a

2

1 a

b2 1 a2

1 a

1 b = D12

0 a -1

\ D = (a + b + c) (a2 + b2 + c2) = 0 As a, b, c are distinct real numbers, a2 + b2 + c2 π 0, therefore a + b + c = 0. fi the line a (x – 5) + b(y – 2) + c = 0 passes through (6, 3). Example 20: Suppose a, b, c and x are real numbers. Let 1 + a 1 + ax 1 + ax 2 D = 1 + b 1 + bx 1 + bx 2 1 + c 1 + cx 1 + cx 2

2

Applying C3 Æ C3 – C2 and C2 Æ C2 – C1, we get 1 D1 = 1

a c -b 2 2 2 D1 = b 0 a = a (a + b + c ) c -a 0

b2

1 b b

a a+c a-b D1 = b b a+b c c-a c Using C2 Æ C2 – C1, C3 Æ C3 – C1, we get

1 + a3 + b3 1 + a 4 + b4

1 1 b 1

0 b -a

Then D is independent of (a) a, b, c (c) a, b, c, x Ans. (c)

1 1 + ax 1 + ax 2 D1 = 1 1 + bx 1 + bx 2

1 1 = (a - 1) ( b - a ) a +1 b +a = (a –1) (b –1) (b – a) D = (a –1)2 (b –1)2 (a – b)2 k =1

Example 19: Suppose a, b and c are distinct real numbers. Let a a+c a-b D = b-c b a+b = 0 c+b c-a c Then the straight line a(x – 5) + b(y – 2) + c = 0 passes through the fixed point (a) (5, 2) (b) (6, 2) (c) (6, 3) (d) (5, 3) Ans. (c) Solution: Applying C1 Æ aC1 + bC2 + cC3, we get Let

a (a + b + c) a + c a - b 1 b a+b D = b (a + b + c) a c (a + b + c) c - a c

(b) x (d) none of these

Solution: Write D = D1 + D2, where

1 a2 -1 b2 - a2

Thus, \

1 a (a + b + c) D1 , where a

1 1 + cx 1 + cx 2 a ax ax 2 and

D2 = b bx bx 2 = 0

cx 2 [ C1 and C2 are propotional] In D1, use C2 Æ C2 – C1, C3 Æ C3 – C1 to obtain c

cx

1 ax ax 2 D1 = 1 bx bx 2 = 0 cx 2 [ C2 and C3 are proportional] D = 0 and hence independent of a, b, c, x. 1 cx

Thus,

Example 21: Suppose a, b, c > 1 and a- x

ax

-3 x

3x

f (x)= b

c -5 x

b

c5 x

x 3x3 , x ΠR 5x5

4.10

Complete Mathematics—JEE Main

then f is (a) (b) (c) (d) Ans. (d)

1 b c D1 = 0 c - b a - c 0 a-b b-c = – (b – c)2 – (a – b) (a – c) = – (a2 + b2 + c2 – bc – ca – ab) 1 fi D1 = – [(b – c)2 + (c – a)2 + (a – b)2] < 0 2 As a + b + c > 0, we get D = (a + b + c) D1 < 0

a constant function a polynomial of degree 5 an odd function an even function

Solution: ax

a- x

3x

-3 x

-3 x 3

c -5 x

-5 x 5

f (–x) = b

b

c5 x

-x

Example 24: Suppose A, B, C are angles of a triangle, and let e2iA e-iC e-iB

a- x

ax

x

= (-1)(-1) b -3 x

b3 x

3x3

-5 x

5x

5

c

c

5x

= f(x) Thus, f is an even function. Example 22: Suppose n, m are natural numbers and 1 f(x) = (1 + mx )n (1 + nx )m

(1 + x )m

(1 + mx )mn

1

(1 + nx )mn

(1 + x )n

1

constant term of the polynomial f (x) is: (a) 1 (b) m + n (c) m – n (d) 0 Ans. (d) Solution: Constant term of polynomial f (x) is f (0), and 1 1 1 f (0) = 1 1 1 = 0

Solution: Using C1 Æ C1 + C2 + C3, we get D = (a + b + c) D1 where 1 b c D1 = 1 c a 1 a b Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get

e2iB

e-iA

e-iB

e-iA

e2iC

Then value of D is (a) – 1 (b) – 4 (c) 0 (d) 4 Ans. (b) Solution: Taking eiA, common from R1, eiB from R2 and iC e from R3, we get D = ei(A + B + C) D1 where eiA e-i( A + C ) e-i( A + B) D1 = e - i ( B + C ) e-i( B + C )

eiB

e-i( A + B)

e-i( A + C )

eiC

But A + B + C = p, so that ei(A + B + C) = eip = cos p + i sin p = –1. Also, A + C = p – B fi e–i(A + C) = e–pi eiB = –eiB. eiA

1 1 1 Example 23: Suppose a, b, c are sides of a scalene triangle. Let a b c D= b c a c a b Then (a) D £ 0 (b) D < 0 (c) D > 0 (d) D ≥ 0 Ans. (b)

D = e -iC

Thus,

-eiB

-eiC

D1 = -e- iA

eiB

-eiC

-eiA

-eiB

eiC

1 =e -1 -1 Using C1 Æ C1 + C2, we get 0 -1 D1 = (-1) 0 1 -2 -1 Therefore, D = (–1) D1 = – 4 i ( A + B +C )

-1 -1 1 -1 -1 1 -1 -1 = (–1) (–2) (2) = 4 1

Example 25: Suppose x1, x2, x3 are real numbers such that x1 x2 x3 π 0. Let x1 + a1b1 a1b2 a1b3 x2 + a2 b2 a2 b3 D = a2 b1 a3 b1 a3 b2 x3 + a3 b3

Determinants 4.11

Then

D - 1 equals: x1 x2 x3

1 -4 Example 27: Let D = 1 - 2

(b) – 1 (c)

1 2x

D = 0 is (a) {– 2, 3} (c) {4, – 6} Ans. (d)

ab ab a b (a) 1 1 + 2 2 + 3 3 x1 x2 x3 a1a2 a3 + b1b2 b3 x1 x2 x3

1 D= 0

Solution: Using the sum property, write D = x1 D1 + b1 D2 where 1 a1b2 a1b3 a2 b3 D1 = 0 x2 + a2 b2 0 a3 b2 x3 + a3 b3

(b) {– 3, 4} (d) {– 2, – 1}

-4 2

20 - 15

= x2 x3 + x2 a3 b3 + x3 a2 b2 a1b2 x2 + a2 b2 a3 b2

a1b3 a2 b3 x3 + a3 b3

Now,

0 x2 0

= 10(x + 2) (x + 1) D = 0 fi x = – 2, – 1.

Example 28: Let D = 1 b b2 - ca , then D is equal to 1 c c 2 - ab

(a) 0 1 2 (a + b2 + c2) (c) 2 Ans. (a)

D = x1[x2 x3 + x2 a3 b3 + x3 a2 b2] + a1 b1 x2 x3 fi

D ab ab a b -1 = 1 1 + 2 2 + 3 3 x1 x2 x3 x1 x2 x3 4

3

(d) none of these

1 a a 2 - bc D = (b – a) (c – a) 0 1 a + b + c = 0 0 1 a+b+c [

R2 and R3 are identical]

Example 29: Suppose a, b, c > 0 and a, b, c are the pth, qth, rth terms of a G.P. Let 1 p log a D= 1 q

2

Example 26: Let pl + ql + rl + sl + t l 2 + 3l = l -1 l -3

(b) a + b + c

Solution: Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get

0 0 = a1 x2 x3 x3

Thus,

1 - 15 1 5 ( x - 2)

1 a a 2 - bc

Using C2 Æ C2 – b2 C1, C3 Æ C3 – b3 C1, we get a1 D2 = a2 a3

= 2(x + 2)

0 2 ( x + 2) 5 ( x 2 - 4)

= (x2 + a2 b2) (x3 + a3 b3) – a3 b2 a2 b3 a1 D2 = a2 a3

5x2

Solution: Applying R2 Æ R2 – R1, R3 Æ R3 – R1 we get

(d) 0 Ans. (a)

and

20 5 . Solution set of

l -1 l - 3 - 2l l - 4 l + 4 3l

where p, q, r, s and t are constants. Then value of t is (a) 0 (b) – 1 (c) 2 (d) 3 Ans. (a) Solution: Putting l = 0, we obtain 0 -1 3 t= 1 0 -4 = 0 -3 4 0 as it is a skew symmetric determinant of odd order.

log b 1 r log c then numerical value of D is (a) – 1 (b) 2 (c) 0 (d) none of these Ans. (c) Solution: Let a = ARp – 1, b = AR q – 1 and c = ARr – 1 fi

log a = a + (p – 1)b, log b = a + (q – 1)b and log c = a + (r – 1)b where a = log a, b = log R. Now, 1 p a + ( p - 1) b D = 1 q a + (q - 1) b

1 r

a + (r - 1) b

Using C3 Æ C3 – (a – b)C1 – bC3, we get D = 0.

4.12

Complete Mathematics—JEE Main

1 – 1 + 3i and 2

(

Example 30: Let w = 1

)

1

1 2

D= 1 -1-w w

1

w 2 then D equals

2

w

(a) 3w (c) 3w2 Ans. (b)

Solution: If n is a multiple of 3, we get each element of D becomes 1. \ D = 0. If n = 3k + 1, then

4

(b) 3w (w – 1) (d) 3w (1 – w)

1

D= 0

w

D= w

w2

If

w 2 = 3(w2 – w) = 3w (w – 1)

Then numerical value of D is (a) 0 (b) – 1 (c) 1 (d) none of these Ans. (a) 1 1 1 , , are the pth, qth, rth terms of an A.P. a b c

1 1 = A + (p – 1)D, = A + (q – 1)D, Let a b

1 r A + (r - 1)D

w 2n

D = wn

w 2n

1

1

wn

w 2n Then D equals (a) 0 (c) w Ans. (a)

(b) 1 (d) w2

1 = 0 [use C1 Æ C1 + C2 + C3]

w

1

w2 y+z x a p z + x and D1 = y b q x+y z c r,

then (a) D = 2D1 (c) D = 4D1 Ans. (a)

(b) D = – 2D1 (d) D = – 4D1

Solution: Using R1 Æ R1 + R2 + R3, we get 2(a + b + c) 2( p + q + r ) 2( x + y + z ) D= c+a r+p z+x a+b p+q x+y Taking 2 common from R1 and applying R2 Æ R2 – R1, R3 Æ R3 – R1, we obtain

a+b+c -b D =2 -c

p+q+r -q -r

x+y+z -y -z

Applying R1 Æ R1 + R2 + R3, we get

a p x a D = 2(– 1) (– 1) b q y = – 2 b c r z c

x p y q z r

x a p = 2 y b q = 2D1 z c r

Example 32: Let w π 1 be complex cube root of unity and n be a natural number and wn

w

b+c q+r D = c+a r+ p a+b p+q

Using R3 Æ R3 – (A – D)R1 – DR2, we get D = 0.

1

w

Example 33: Let

Now

1 q A + (q - 1)D

w

1

2

D= w

1 and = A + (r – 1)D. c

1 D = abc p A + ( p - 1)D

1 = 0 [use C1 Æ C1 + C2 + C3]

w2

1

w

Example 31: Let a, b, and c be respectively the pth, qth and rth terms of a harmonic progression and 1 1 1 D= p q r bc ca ab

2

w2

n = 3k + 2, then

1

0 w2

Solution:

w

w

Solution: Using w3 = 1 and 1 + w + w2 = 0, and applying C1 Æ C1 + C2 + C3, we get 3

1

x+y x x Example 34: If x = – 2, and D = 5 x + 4 y 4 x 2 x 10 x + 8 y 8 x 3 x then numerical value of D is

(a) 8 (c) 4 Ans. (b)

(b) – 8 (d) – 4

Determinants 4.13

Solution: Taking x common from C2 and C3 we obtain 1 1 1 x+y 1 1 3 5 4 2 D = x 5x + 4 y 4 2 = x 10 8 3 10 x + 8 y 8 3 [using C1 Æ C1 – yC2] Next using C1 Æ C1 – C2, C2 Æ C2 – C3, we obtain

Solution: Using change of base formula log b logab = , a, b > 0, a, b π 1, log a

2

we can write

log x log y log z 1 log x log y log z = 0. D= log x log y log z log x log y log z

0 0 1 1 2 D = x 1 2 2 = x3 = x3 2 5 2 5 3

Example 37: Let w π 1 be a cube root of unity and

3

As

Example 35: If a = w π 1, is a cube root of unity, b = – 785, c = 2008 i, and a a+b a+b+c D = 2 a 3a + 2b 4 a + 3b + 2c 3a 6a + 3b 10a + 6b + 3c then D equals (a) – i (b) i (c) 1 (d) 1 – wi Ans. (c) 1 a+b a+b+c D = a 2 3a + 2b 4a + 3b + 2c 3 6a + 3b 10a + 6b + 3c

and apply C2 Æ C2 – bC1, C3 Æ C3 – cC1 to obtain 1 a a+b D = a 2 3a 4a + 3b 3 6a 10 a + 6c

w - w2 - 1

2

then D equals (a) – w (c) 0 Ans. (c)

2w

2 2w 2

2

w -1-w

(b) 3w (1 – w) (d) 1 – w2

Solution: Using R1 Æ R1 + R2 + R3 and 1 + w + w2 = 0, we obtain D = 0. Example 38: Suppose x = – y = cos

1 1 + 7i and 3

(

)

p p + isin . 4 4 1 x D= 1 x + y 1 x

Let

x y , x+y

then D equals

Take a common from C2 and apply C3 Æ C3 – bC2 to obtain 1 1 1 1 1 a 3 D = a 2 3 4a = a 2 3 4 3 6 10 3 6 10a 2

Apply C3 Æ C3 – C2, C2 Æ C2 – C1 to obtain 1 0 0 3 D = a 2 1 1 = a3 = w3 = 1. 3 3 4 Example 36: Let x, y, z be positive and x, y, z π 1. Let

1 log x y log x z 1 log y z , D = log y x then numerical value of D is (a) – 1 (c) 1 Ans. (b)

2w 2w

Solution: Write

log z y

2

D=

x = – 2, D = – 8.

log z x

1 - w - w2

1

(b) 0 (d) none of these

(a) - 7 (c) i Ans. (c)

(b) 7 (d) – 1.

Solution: Using R2 Æ R2 – R1 and R3 Æ R3 – R1, we get 1 x D= 0 y 0 0

x p p y - x = y2 = cos + i sin = 0 + i(1) = i 2 2 y [Using De Moivre’s Theorem]

Example 39: Let x = cos

p p + i sin and 3 3

1

x

x2

D = x2

1

x

2

1 x 1 then numerical value of D is (a) 0 (b) – 1 (c) 8 (d) – 4 Ans. (c)

4.14

Complete Mathematics—JEE Main

Example 42: Let f : N Æ N be defined by

Solution: Use C3 Æ C3 – xC2 and C2 Æ C2 – xC1. 1 D = x2

0

0

1 – x3

0

f (x) = ( x + 1)2 + x - ÈÎ ( x + 1)2 + ( x + 1) ˘˚

([ ] denotes the greatest integer function). Suppose a, b, c are three distinct natural numbers. Let

2

x – x 1 – x3

1

= (1 – x3)2 = [1 – cosp – isinp]2 = 8 [De Moivre’s Theorem] Example 40: Let f(x) = [2 - x [2 x ]], x Œ R. ([ ] denotes the greatest integer function.) Let x1 = 0, x2 = log23 and x3 = 2 . Suppose 0 < x4 < 1. 2

f ( x1 ) D = f ( x4 ) f ( x2 )

f ( x2 ) f ( x4 ) f ( x3 )

2

f ( x3 ) f ( x2 ) , then D is equal to: f ( x1 )

(a) –1 (c) 1 Ans. (b)

(b) 0 (d) 2 log23 2

2

[2 ] 2x

range f (x) is {0, 1} Also

x2

x2

2

£ 1. Therefore,

x2

[2 ] = 2 ¤ 2 is a positive integer.

Thus, f (x1) = 1, f (x2) = 1, f (x3) = 1 and f (x4) = 0. Therefore, 1 1 1 D= 0 0 1 = 0 1 1 1

w

w

w2

1

x +w

D=

1 + x = 0, then value of x is w2

(b) 1 (d) none of these

Solution: Applying C1 Æ C1 + C2 + C3, we get x + w2 + w +1 2

D = w +w +1+ x 1+ x + w + w2

x

w

= x

w2

x

x +w

w w

2

x +w

1 1+ x w2

1 1 + x using 1 + w + w2 = 0 w2

D is clearly equal to 0 for x = 0.

D = f (b ) b

b

f (c ) c

2

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

Solution: If n ΠN, then n2 < n2 + n < (n + 1)2 n
0, y > 0, z > 0 and x log 2 3 15 + log (a x ) D(a, b, c) = y log 3 5 25 + log (b y )

Statement-1: If ak > 0 " k ≥ 1 and a1, a2, a3, . . . are in G.P. then Dn = 0 " n ≥ 1. Statement-2: If a1, a2, a3 . . . are in A.P. then Dn = 0 " n ≥ 1. Ans. (b) Solution: Let ak = ar

k–1

" k ≥ 1, then

D n = a na n + 3 a n + 6

1 r

1 r

1 r

2

2

2

=0

r r r Next, if ak = b + (k – 1)d, then using C2 Æ C3 – C1 and C2 Æ C2 – C1, we get an Dn = an + 1 an + 2

Statement-1: D(8, 27, 125) = 0 Ê 1 1 1ˆ Statement-2: D Á , , ˜ = 0 Ë 2 3 5¯ Ans. (b) Solution: Using log (bc) = C3 – 5C2 we get x log 2 3 D(a, b, c) = y log 3 5 z log 5 7

x log a y log b z log c

Example 84: Let a π 0, p π 0 and a b c

3d 3d C2 and C3 are identical]

0 cos x - sin x 0 cos x Example 82: Let f(x) = sin x cos x sin x 0 Statement-1: If sin 2x = 1, then f(x) = 2/3 Statement-2: f(x) = 0 if sin x = cos x Ans. (d)

c log b and applying C3 Æ

D(8, 27, 125) = D(23, 33, 53) = 0 as in this case C1 and C3 are proportional. Similarly, D(1/2, 1/3, 1/5) = D(2–1, 3–1, 5–1) = 0.

3d 3d 3 d 3d = 0 [

2

.

Solution: Multiplication of two determinants leads us 1 -y y f(x) = - y 1 y y y 1 where y = sin x cos x Using C1 Æ C1 – C2, C2 Æ C2 + C3, we get f(x) = (1 + y)2

z log 5 7 35 + log (c z )

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

= (1 + y)2 (1 – 2y) When sin 2x = 1, y = 1/2 and f(x) = 0 When sin x = cos x, 1 – 2y = 1 – 2 sin2x = cos 2x = 0 \ f(x) = 0.

D= 0 p q p q 0 Statement-1: If the equations ax2 + bx + c = 0 and px + q = 0 have a common root, then D = 0. Statement-2: If D = 0, then the equations ax2 + bx + c = 0 and px + q = 0 have a common root. Ans. (b) Solution: If l is a common root of ax2 + bx + c = 0 and px + q = 0, then al2 + bl + c = 0, pl + q = 0 and pl2 + ql = 0 Eliminating l, we obtained D = 0. For statement-2, expanding D along C1 we obtain – aq2 + p(bq – cp) = 0 2

or

Ê qˆ Ê qˆ aÁ - ˜ + bÁ - ˜ + c = 0 Ë p¯ Ë p¯

Thus, ax2 + bx + c = 0 and px + q = 0 have a common root. Example 85: Statement-1: sin p cos( x + p / 4) tan( x - p / 4) log( x / y) =0 D = sin( x - p / 4) - cos(p / 2) cot(p / 4 + x ) log( y / x ) tan p

Determinants 4.25

Statement-2: A skew symmetric determinant of odd order equals 0. Ans. (a) Solution: For statement-2, see theory. Now, using pˆ Èp Ê p ˆ˘ Ê cos Á x + ˜ = cos Í - Á - x˜ ˙ Ë ¯˚ Ë 4¯ 2 4 Î pˆ Êp ˆ Ê = sin Á - x˜ = - sin Á x - ˜ ; Ë4 ¯ Ë 4¯

0 -1 3 Solution: e = f (0) = 1 0 - 4 -3 4 0 =0 [skew symmetric determinant of odd order] \ Statement-2 is true. 1 To obtain a, replace x by , so that x 2 3 Ê 1ˆ Ë x¯ + x 1 +1 x 1 -3 x

Êp ˆ Èp Ê p ˆ˘ cot Á + x˜ = cot Í - Á - x˜ ˙ Ë4 ¯ Ë ¯˚ 4 Î2 p p = tan ÊÁ - xˆ˜ = - tan ÊÁ x - ˆ˜ , Ë4 ¯ Ë 4¯ and log (x/y) = – log (y/x) we find D is a skew symmetric determinant of odd order. Example 86: Let f : Q Æ [–1, 1] by f (x) = sin x Let x1, x2, x3 be three distinct rational numbers. Let a = f(x1), b = f(x2), c = f(x3). Statement-1: 1 a a2 D= 1 b b

2

π0

1 c c Statement-2: f is a one-to-one function. Ans. (a) Solution: Let x1, x2 Œ Q be such that f(x1) = f(x2) fi sin x1 = sin x2 fi x1 = np + (–1)n x2, n Œ I fi x1 + (–1)n+1 x2 = np, n Œ I. But LHS is rational and RHS is irrational except when n = 0. \ x1 = x2 Thus, f is one-to-one. Also, D = (a – b) (b – c) (c – a) π 0 as x1, x2, x3 are distinct and hence a, b, c are distinct. Example 87: Let x2 + 3x f (x) = x + 1 x-3

x -1 x + 3 -2 x x - 4 x + 4 3x

= ax4 + bx3 + cx2 + dx + e Statement-1: a = – 1 Statement-2: e = f (0) Ans. (a)

1 +3 x 1 -4 x 3 x

4 3 2 Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ aË ¯ + bË ¯ + cË ¯ + d Ë ¯ + e x x x x 2 Take x common from C1, x from C2 and C3 1 + 3x 1 - x 1 + 3x 1 1 - 4x = 4 x + x2 -2 x 2 x - 3x 1 + 4 x 3x

1

[a + bx + cx 2 + dx 3 + ex 4 ] x4 Cancel 1/x4 and put x = 0 to obtain 1 1 1 0 -2 1 = a 0 1 0 fi a = –1 =

2

1 -1 x 2 x 1 +4 x

Example 88: Statement-1: The system of line equations x + (sin a) y + (cos a) z = 0 x + (cos a) y + (sin a) z = 0 x – (sin a) y – (cos a) z = 0 has a non-trivial solution for only one value of a lying in the interval (0, p /2). Statement-2: The equation in a cos a sin a cos a D = sin a cos a sin a = 0 cos a - sin a - cos a has only one solution lying in the interval (0, p /2). Ans. (b) Solution: Using C1 Æ C1 – C3 in D, we get 0 0 D= 2 cos a

sin a cos a - sin a

cos a sin a - cos a

= 2 cos a (sin2 a – cos2 a)

4.26

Complete Mathematics—JEE Main

= –2 cos a cos 2 a D = 0 for a = p/4 Œ (0, p /2). This is the only value of a lying in (0, p /2) for which D = 0. The system of linear equations will have a non-trivial solution if and only if 1 sin a cos a sin a = 0 D1 = 1 cos a 1 - sin a - cos a Using R2 Æ R2 + R1, we get

2 0 D1 = 1 cos a 1 - sin a

0 sin a = 0 - cos a

fi 2 (– cos2 a + sin2 a) = 0 fi –2 cos 2a = 0 This is true for only one value of a Œ (0, p /2) viz, a = p /4. Thus, statement-1 is also true. However, statement-2 is not a correct reason for statement-1.

LEVEL 2 Straight Objective Type Questions Example 89: Consider the system of equations in x, y, z: (sin 3q) x – y + z = 0 (cos 2q) x + 4y + 3z = 0 2x + 7y + 7z = 0 The values of q for which the system has a non-trivial solution are given by 1 Ê ˆ (a) p Á n + (- 1)n ˜ Ë ¯ 3

1 (b) p ÊÁ n + (- 1)n ˆ˜ Ë ¯ 4

1 Ê ˆ (b) p Á n + (- 1)n ˜ Ë ¯ 6

(d)

np 2

Example 90: D(x) =

Let

then

p /2

Ú- p / 2 x D( x)dx

2 sin 4 x

3 + 2 cos4 x

sin 2 2 x

2 sin 4 x

2 cos4 x

3 + sin 2 2 x

equals (b) p (p – 1) (d) 0

1

2 cos4 x

D(x) = f(x) 1 3 + 2 cos4 x equations to have a non1 3 =0 7

sin 3q D = cos 2 q + 4 sin 3q 2 + 7 sin 3q

fi

sin 2 2 x

Solution: Using C1 Æ C1 + C2 + C3, we get

Using R2 Æ R2 + 4R1 and R3 Æ R3 + 7R1 we get

fi fi fi fi fi

2 cos4 x

(a) p2 (c) 1 Ans. (d)

where n is an integer Ans. (c) Solution: For the system of trivial solution we must have sin 3q - 1 D = cos 2q 4 2 7

3 + 2 sin 4 x

-1 1 0 7 =0 0 14

2(cos 2q + 4 sin 3q) – (2 + 7 sin 3q) = 0 sin 3q + 2 cos 2q – 2 = 0 3 sin q – 4 sin3 q – 4 sin2 q = 0 sin q(2 sin q + 3) (2 sin q – 1) = 0 sin q = 0 or sin q = 1/2 [ |sin q | £ 1] Ê (- 1)n ˆ q = m p or q = n Á p + where m 6 ˜¯ Ë and n are integers

1

2 cos4 x

sin 2 2 x sin 2 2 x 3 + sin 2 2 x

f(x) = 3 + 2 sin4x + 2 cos4x + sin2 2x

where

= 3 + 2 (sin4x + cos4x + 2 sin2x cos2 x) = 3 + 2 (cos2x + sin2x)2 = 5 Applying C2 Æ C2 – (2 cos4 x) C1 and C3 Æ C3 – (sin2 2x) C1, we get 1 0 0 D(x) = 5 1 3 0 = 45 1 0 3 Thus,

p /2

p /2

Ú- p / 2 x D( x)dx = 45Ú- p / 2 xdx = 0

Example 91: If f (x), g(x)and h(x) are three polynomials of degree 3 then f ¢( x ) g¢ (x) h¢ (x) f(x) =

f ¢¢ ( x ) g ¢¢ ( x ) h ¢¢ ( x ) f ¢¢¢ ( x ) g ¢¢¢ ( x ) h ¢¢¢ ( x )

Determinants 4.27

is a polynomial of degree (a) 3 (c) 5 Ans. (d)

(b) 4 (d) none of these

Solution: The function D(x) is continuous on [0, p/2] and differentiable on (0, p/2). Also D(0) = 0 and D(p/2) = 0. Thus, by the Rolle’s theorem there exists at least one c e (0, p/2) such that D¢(c) = 0. cos2 x

Solution: We have f ¢¢( x ) f¢(x) =

g ¢¢ ( x ) h ¢¢ ( x ) f ¢¢ ( x ) g ¢¢ ( x ) h ¢¢ ( x ) + f ¢¢¢ ( x ) g ¢¢¢ ( x ) h ¢¢¢ ( x )

f ¢( x ) g ¢ ( x ) h ¢ ( x ) f ¢( x ) g¢ (x) h¢ (x) f ¢¢¢ ( x ) g ¢¢¢ ( x ) h ¢¢¢ ( x ) + f ¢¢ ( x ) g ¢¢ ( x ) h ¢¢ ( x ) f ¢¢¢ ( x ) g ¢¢¢ ( x ) h ¢¢¢ ( x ) f iv ( x ) giv ( x ) hiv ( x )

f ¢( x ) g ¢ ( x ) h ¢ ( x ) = 0 + 0 + f ¢¢ ( x ) g ¢¢ ( x ) h ¢¢ ( x ) = 0 0 0 0 since f, g, h are polynomials of degree 3, f iv(x) = giv(x) = hiv(x) = 0 fi f(x) must be a constant.

equals (a) 9b2(a + b) (c) 9(a + b)3 Ans. (a)

(b) 9a2(a + b) (d) 9ab(a + b)

then

p /2

Ú0

Applying R2 Æ R2 – R1 and R3 Æ R3 – R1, 1 a + b a + 2b -b D = 3(a + b) 0 -b 0 b -2b -b -b = 3(a + b) = 9b2(a + b) b -2b Example 93: Let sin x cos x sin 2 x + cos 2 x 1 1 D(x) = 0 1 0 -1 then D¢(x) vanishes at least once in (a) (0, p/2) (b) (p/2, p) (c) (0, p/4) (d) (– p/2, 0) Ans. (a)

cos x 0

[D(x) + D¢(x)] dx equals

(a) p/3 (c) 2p Ans. (b)

(b) p/2 (d) 3p/2

Solution: Applying C1 Æ C1 – sin x C3 and C2 Æ C2 + cos x C3, we get 1 0 - sin x D(x) = 0 1 cos x 0 sin x - cos x Applying R3 Æ R3 – sin x R1 + cosx R2, we get - sin x cos x

1 0 D(x) = 0 1

2

=1 2

0 0 cos x + sin x fi

D¢(x) = 0

Thus,

p /2

Ú0

[D(x) + D¢(x)] dx =

p /2

Ú0

dx =

p . 2

Example 95: The determinant

Solution: Applying C1 Æ C1 + C2 + C3, we get 1 a + b a + 2b D = 3(a + b) 1 a a+b 1 a + 2b a

sin 2 x - cos x

Example 94: Let D(x) = cos x sin x sin x

Example 92: The determinant a a + b a + 2b a a+b D = a + 2b a + b a + 2b a

cos x sin x - sin x

D=

13 + 3

2 5

15 + 26

5

10 equals

3 + 65

15

5

(a) 15 2 - 25 3

5

(b) 25 3 - 15 2 (d) -15 2 + 7 3

(c) 3 5 Ans. (a) Solution: Taking

5 common from C2 and C3, we get 13 + 3

D=

2

( 5)

Applying C1 Æ C1 – D = ( 5)

= 5

(

1

15 + 26

5

2

3 + 65

3

5

13 C3 – - 3

2

3 C2, we get

2

1

(

)(

0

5

2 = 5 - 3 5- 6

0

3

5

)

18 - 25 3 = 15 2 - 25 3

)

4.28

Complete Mathematics—JEE Main

Example 96: The values of l for which the system of equations x + y – 3= 0

Example 98: If a + b + c π 0, the system of equations (b + c) (y + z) – ax = b – c (c + a) (z + x) – by = c – a (a + b) (x + y) – cz = a – b

(1 + l)x + (2 + l)y – 8 = 0 x – (1 + l)y + (2 + l) = 0 has a non-trivial solution, are (a) – 5/3, 1 (b) 2/3, – 3 (c) – 1/3, – 3 (d) 0 Ans. (a) Solution: The given system of equations has a non trivial solution if 1 1 -3 2+l -8 = 0 D = 1+ l 1 - (1 + l ) 2 + l Applying C2 Æ C2 – C1 and C3 Æ C3 + 3C1, we get 1 D = 1+ l 1

fi fi fi fi

0 1 -2 - l

0 -5 + 3l = 0 5+l

(5 + l) + (2 + l) (3l – 5) = 0 5 + l + 6l – 10 + 3l2 – 5l = 0 2 3l + 2l – 5 = 0 fi (3l + 5) (l – 1) = 0 l = – 5/3 or l = 1.

Example 97: The values of m for which the system of equations 3x + my = m and 2x – 5y = 20 has a solution satisfying the conditions x > 0, y > 0 are given by the set (a) {m : m < – 13/2} (b) {m : m > 17/2} (c) {m : m < – 13/2 or m > 17/2} (d) none of these Ans. (d) Solution: Here D = Dx =

3 m = – 15 – 2m 2 -5 m m = – 25 m 20 -5

3 m and Dy = = 60 – 2m 2 20 If D = 0, then m = – 15/2. But for this value of m, Dx π 0 and Dy π 0. Thus, in this case, the system of equations is not consistent. 25m 2m - 60 and y = , by the Cramer’s If D π 0, then x = 2m + 15 2m + 15 rule. Now, x > 0, y > 0 ¤ 25m > 0, 2m – 60 > 0, 2m + 15 > 0 or 25m < 0, 2m – 60 < 0, 2m + 15 < 0 fi m > 30 or m < – 15/2.

has (a) (b) (c) (d) Ans. (a)

a unique solution no solution infinite number of solutions finitely many solutions

Solution: We can write the above system of equations (a + b + c) (y + z) – a(x + y + z) = b – c (a + b + c) (z + x) – b(x + y + z) = c – a (a + b + c) (x + y) – c(x + y + z) = a – b Adding the above equations, we obtain 2(a + b + c) (x + y + z) – (a + b + c) (x + y + z) = 0 fi fi fi

(a + b + c) (x + y + z) = 0 x+y+z =0 y+z =–x

[

\

(b + c) (– x) – ax = b – c fi x =

c-b a+b+c

Similarly, y =

a + b + c π 0]

a-c b-a ,z= . a+b+c a+b+c

Example 99: Let a, b, c be positive real numbers. The following system of equations in x, y and z.

x2 a2 has

+

y2 b2

-

(a) (b) (c) (d) Ans. (d)

z2 c2

= 1,

x2 a2

-

y2 b2

+

z2 c2

= 1,

- x2 a2

+

y2 b2

+

z2 c2

=1

no solution unique solution infinitely many solutions finitely many solutions

Solution: Adding all the equations, we obtain

x2 a2

+

y2 b2

+

z2 c2

=3

Subtracting first equation from it we get

2z2 2

= 2 fi z2 = c2

c fi z = ± c. Similarly, x = ± a, y = ± b. Thus, the given system of equations has eight solutions. Example 100: If the system of equations x – ky – z = 0, kx – y – z = 0, x + y – z = 0 has a non-zero solution, then the possible values of k are (a) – 1, 2 (b) 1, 2 (c) 0, 1 (d) – 1, 1 Ans. (d)

Determinants 4.29

Solution: As the given system has a non-zero solution, 1 - k -1 1+ k 0 = k -1 -1 = 1 + k 1 1 -1 0

- k - 1 -1 -2 -1 0 -1

[using C1 Æ C1 – C2, C2 Æ C2 + C3] fi

0 = (– 1) [(1 + k) (– 2) – (1 + k) (– k – 1)]

fi

0 = (1 + k) (– 2 + k + 1) fi k = – 1, 1

Example 101: If the system of equations lx1 + x2 + x3 = 1, x1 + lx2 + x3 = 1, x1 + x2 + lx3 = 1 is inconsistent, then l equals (a) 5 (b) – 2/3 (c) – 3 (d) – 2 Ans. (d) l 1 1 l +2 1 1 Solution: Let D = 1 l 1 = l + 2 l 1 1 1 l l +2 1 l [C1 Æ C1 + C2 + C3] 1 1 = (l + 2) 1 l 1 1

1 0 0 1 0 1 = (l + 2) 1 l - 1 1 0 l -1 l

[using C2 Æ C2 – C1 and C3 Æ C3 – C1] = (l + 2) (l – 1)2 If D = 0, then l = – 2 or l = 1. But when l = 1, the system of equation becomes x1 + x2 + x3 = 1 which has infinite number of solutions. When l = – 2, by adding three equations, we obtain 0 = 3 and thus, the system of equations is inconsistent. Example 102: If p π a, q π b, r π c and the system of equations px + ay + az = 0 bx + qy + bz = 0 cx + cy + rz = 0 has a non-trivial solution, then the value of p q r + + is p-a q-b r-c (a) – 1 (c) 1 Ans. (d)

(b) 0 (d) 2

Solution: As the given system of equations has a nontrivial solution p a a D= b c

q b =0 c r

Applying C3 Æ C3 – C2 and C2 Æ C2 – C1, we get

p a- p a- p 0 =0 D= b q-b 0 c r-c Expanding along C3, we get p a- p b q-b (a – p) + (r – c) =0 b q-b c 0 fi (a – p) (– c) (q – b) + (r – c) {p(q – b) – b(a – p)} = 0 fi (p – a) (q – b) c + p (r – c) (q – b) + b (r – c) (p – a) = 0 Dividing by (p – a) (q – b) (r – c) we get c p b + + =0 r-c p-a q-b fi

p q r q-b r-c + + + = =2 p-a q-b r-c q-b r-c

Example 103: Let l and a be real. Relation between l and a for which the system of equations lx + (sin a)y + (cos a)z = 0 x + (cos a)y + (sin a)z = 0 – x + (sin a)y – (cos a)z = 0 has a non-trivial solution is (a) l = sin 2a + cos 2a (b) l = |sin 2a| (c) l = |sin 2a – cos 2a| (d) l = cos 2a Ans. (a) Solution: The system of equations will have a nontrivial solution if and only if l sin a cos a 0 = 1 cos a sin a -1 sin a - cos a 0 (l + 1)sin a (1 - l )cos a = 0 cos a + sin a sin a - cos a -1 sin a - cos a [using R1 Æ R1 + lR3, R2 Æ R2 + R3] = (l + 1) sin a (sin a – cos a) + (l – 1) cos a(cos a + sin a) = l(sin2a + cos2a) + [sin2a – cos2 a – 2 sin a cos a] Thus, l = sin 2a + cos 2a Example 104: If fr(x), gr(x), hr(x), r = 1, 2, 3 are polynomials in x such that fr(a) = gr(a) = hr(a), r = 1, 2, 3 f1 ( x ) f2 ( x ) f3 ( x ) F (x) = g1 ( x ) g2 ( x ) g3 ( x ) h1 ( x ) h2 ( x ) h3 ( x ) then F ¢(a) is (a) – 1 (b) a (c) 0 (d) none of these Ans. (c) and

4.30

Complete Mathematics—JEE Main

Solution: Differentiating w.r.t. x we get F ¢(x)

f1¢( x ) f2¢ ( x ) f3¢ ( x ) f1 ( x ) f2 ( x ) f3 ( x ) = g1 ( x ) g2 ( x ) g3 ( x ) + g1¢ ( x ) g2¢ ( x ) g3¢ ( x ) h1 ( x ) h2 ( x ) h3 ( x ) h1 ( x ) h2 ( x ) h3 ( x ) f1 ( x ) f2 ( x ) f3 ( x ) + g1 ( x ) g2 ( x ) g3 ( x ) h1¢ ( x ) h2¢ ( x ) h3¢ ( x ) Since fr(a) = gr(a) = hr(a) for r = 1, 2, 3, we get each of the three determinants on the right side becomes zero when x is replaced by a. (In each case two rows become identical.) Thus, F ¢(a) = 0.

1 1 (y – w), z = (y + w), y ≥ 0 7 3 (d) x = 1, y = 0, z = 0, w = 0. Ans. (d) (c) x =

Solution: We have 1 – (x + 2y) = z ≥ 0 and

2x – 3y – 2 = – w £ 0.

fi

x + 2y £ 1 and 2x – 3y £ 2, x ≥ 0, y ≥ 0

The only point where the two shaded region intersect is (1, 0). Thus, x = 1, y = 0 for these values z = 0, w = 0. 1

Example 105: The number of real values of a for which the system of equations x + ay – z = 0, 2x – y + az = 0, ax + y + 2z = 0 has a non-trivial solution, is (a) 3 (b) 1 (c) 0 (d) infinite Ans. (a) Solution: Since the given system of equations has a non-trivial solution, 1 a -1 D = 2 -1 a = 0 a 1 2 Using C1 Æ C1 + C3, C2 Æ C2 + aC3, we get 0

2

D = 2 + a -1 + a 2 + a 1 + 2a

a 2

fi

(2 + a) (1 + 2a + 1 – a2) = 0

Example 106: The solution set of x + 2y + z = 1 subject (a) x =

1 (y – w), 7

y ≥ w ≥ 0, z ≥ 0

(b) x =

1 (y + w), 8

z=

2x

y –3

=2

1

x

–2 3

Fig. 4.1

Example 107: The number of values of k for which the system of equations (k + 1)x + 8y = 4k

Solution: Let D =

fi a = – 2, 1 ± 3 . Thus, there are three real values of a for which the system of equations has a non-trivial solution.

2x – 3y + w = 2 x ≥ 0, y ≥ 0, z ≥ 0, w ≥ 0

=1

O

2 + a -1 + a 2 =0 2 + a 1 + 2a

(-1)

2y

has infinitely many solutions is (a) 0 (b) 1 (c) 2 (d) infinite Ans. (b)

=0

fi

x+

kx + (k + 3)y = 3k – 1

-1

0

1 2

y

is

1 (y – w), y ≥ w ≥ 0 3

k +1 8 k k +3

= k2 – 4k + 3 = (k – 1) (k – 3) If D π 0, the system of equations has a unique solution. For the system of equations to have an infinite number of solutions D = 0 fi k = 3, 1 For k = 3, the system equations becomes 4x + 8y = 12 and 3x + 6y = 8 8 3 Thus, in this case the system of equations has no solution. For k = 1, the system of equations becomes 2x + 8y = 4 and x + 4y = 2 fi

x + 2y = 3, x + 2y =

fi x + 4y = 2, x + 4y = 2 This system has infinite number of solutions.

Determinants 4.31

Example 108: Suppose a, b, c ΠR and let 0 a-x b-x f(x) = - a - x 0 c-x -b-x -c-x 0

(k – 1)x – 2y = 0. If k = – 1, we can choose y in infinite number of ways, corresponding to which we can choose x and z in infinite number of ways. If k = 1, then y = 0 and x and hence z can be chosen in infinite number of ways.

2

Then coefficient of x in f(x) is (a) – (a + b + c) (b) a + b + c (c) 0 (d) ab + bc + ca Ans. (c)

Example 110: Let a2, a3 ΠR be such that | a2 Рa3 | = 6. Let 1 a3 a2 f(x) = 1 a3 2a2 - x , x ΠR 1 2a3 - x a2

Solution: Expanding along C1 a-x b-x a-x b-x - (b + x ) -c-x 0 0 c-x = (a + x) (b – x) (c + x) – (b + x) (a – x) (c – x)

f(x) = (a + x)

= – [(x + a) (x – b) (x + c) + (x – a) (x + b) (x – c)] 3

The maximum value of f (x) is (a) 6 (b) 9 (c) 12 (d) 36 Ans. (b)

2

= – [2x + x (a – b + c – a + b – c) + ...] = – 2x3 + 0x2 + ... \ coefficient of x2 in f(x) is 0.

Solution: Using R2 Æ R2 – R1, R3 Æ R3 – R1, we get

Example 109: If the system of equations x – ky – z = 0, kx – y – z = 0, x + y – z = 0 has a non-zero solution, then possible values of k are (a) – 1, 2 (b) 1, 2 (c) 0, 1 (d) – 1, 1 Ans. (d) Solution: Subtracting the last equation from the first two equations, we get – (k + 1)y = 0

1 a3 a2 f(x) = 0 0 a2 - x 0 a3 - x 0

= – (a2 – x) (a3 – x) = – [x2 – (a2 + a3)x + a2a3] =

1 a + a3 ˆ 2 (a2 – a3)2 – Ê x - 2 £9 Ë 4 2 ¯

f(x) attains maximum value 9 when x=

1 (a2 + a3). 2

EXERCISES Concept-based Straight Objective Type Questions 1. Suppose A = (aij)n × n, where aij Œ R. If det (adj(A)A–1) = 3, then det (adj(A)) equals: (a)

3

(c) 3 3

(a) – p3

p+2 p+5 p+8

(d) 0

(b) 3

1

x

x2

(d) 9

D = x2

1

x

x

2

1

2. If a, b, c are in A.P. and p is a real number, and p+c D = p+b p+a then D equals:

(c) p3 – 2 abc 3. Suppose x π 1 and

p+a p+b p+c (b) p3

x

then D = 0 (a) for exactly two distinct complex numbers (b) for exactly four distinct complex numbers (c) for exactly two distinct real numbers (d) none of these

4.32

Complete Mathematics—JEE Main

10! 11! 12! D 4. Let D = 11! 12! 13! , then – 260 equals: (10!)3 12! 13! 14! (a) 1 (c) 3

(b) 2 (d) 4

x +1 2 3 x+2 3 , P(x) = 1 1 2 x+3 the product of zeros of P(x) is (b) 6 (d) 12

6. Let A(x1, y1), B(x2, y2) and C(x3, y3) be vertices of an equilateral triangle whose side is 4 units. Let x1 D = x2 x3

(b) 128 (d) 256

7. Suppose a, b, c are in A.P. If p, q, r are also in A.P., then value of x2 + a

x+p c

D = x2 + b

x+q

b ,

x2 + c is dependent on (a) x (c) p, q, r

x+r

a

(a + 2016)2

D = (a - 2016)2

(b - 2016)2

(c - 2016)2

b2

c2

If D = k (2016)3 (a – b) (b – c)(c – a), then k is equal to: (a) – 1 (b) – 4 (c) 4 (d) 1 x -6 -1 11. Let P(x) = 2 -3 x x + 3 , -3 2 x x + 2 sum of the zeros of P(x) is (a) – 6 (b) – 7 (c) 13/5 (d) –12/5

a+b

a+b

a 2 + b2

x n+2

x n +3

D = y n +1

yn+2

y n +3

z n +1 z n + 2 z n +3 If D = (x – y) (y – z)(z – x) x2 y2 z2, then n is equal to: (a) –1 (b) 0 (c) 1 (d) 2 13. Suppose a, b, c are distinct real numbers. Let x3 - a

x4 - b

0

x5 + c

P(x) = x 3 + a

(b) a, b, c (d) none of these

2

x n +1

0

8. Suppose a, b are two non-zero numbers. Let D=

(b + 2016)2

12. Let n be an integer and x, y, z > 1. Suppose

y1 1 y2 1 , y3 1

then D2 is equal to (a) 64 (c) 192

(a + 2016)2 a2

5. Let

(a) 0 (c) – 6

10. Suppose a, b, c ΠR. Let

a 2 + b2 a 3 + b3 ,

a 2 + b 2 a 3 + b3 a 4 + b 4 then D is equal to: (a) 0 (b) ab (d) a3b5 + a5b3 (c) a6 + b6 9. Suppose a, b, c > 1. Let log a log b log c D = log (2007a) log (2007 b) log (2007 c) log (2017 a) log (20117 b) log (2017 c) then D is equal to: (a) 0 (b) log (4024 abc) Ê 2017 ˆ (d) none of these (c) log Ë 2007 ¯

0 x4 + b x5 - c A value of x satisfying P(x) = 0 is (a) –(a + b + c) (b) a + b + c (c) a + b – c (d) 0 14. Suppose a, b, c Œ R and abc π 0. Let 1+ a 1 1 D = 1 + b 1 + 2b 1 1 + c 1 + c 1 + 3c 1 1 1 + + is equal to: a b c (a) 0 (b) –1 (c) –2 (d) –3 15. Suppose n and m are natural numbers such that If D = 0, then

D=

xm

x m +2

x2m

1

xn

2n

x m + 5 x n +6 x 2 m + 5 Then a possible relationship between n and m is (a) n = m + 2 (b) n = m + 1 (c) n = m (d) n = m – 1

Determinants 4.33

LEVEL 1 Straight Objective Type Questions Œ R and a + b + c π 0. Let c+a a+b a+b b+c b+c c+a

16. Suppose a, b, c b+c D= c+a a+b If D = 0, then (a) a = b = c (c) a = b + c

3

3

3

(b) a + b – c = 0 (d) a = b = c = 0

p +2

+ 2a q + 2

n

2 +a

D= 2

2

n

2

p

2

2

q

D= b

e

yz

zx

zy

z2 + 1

a D= x x and f(x) = (x

=5

(b) –1 (d) –3 x x b x x c – a) (x – b) (x – c)

Determinant D is equal to: (a) f(x) – x3 (c) xf ¢(x) – f(x)

(b) f ¢(x) (d) f ¢(x) – x f ≤(x)

24. Straight line n+2

n +1

2

+ 3b r + a

+b

n +1

2b 2

20. Suppose a, b, c, d, e and f are in G.P. with common ratio > 1. Let p, q, r be three real numbers. Let d2

y +1

23. Let

p +2 +a q + 2 + 2b r - c Then D equals: (a) –1 (b) 0 (d) p2 q2 r2 – 4 (a + b + c) (c) p2 q2 r2 – 3abc

a2

yx

(a) 0 (c) –2

19. Suppose a, b, c are in A.P. Let 2

xz

-1 cos C cos B -1 cos A D = cos C cos B cos A -1 then D equals:

3 (a + b + c) (bc + ca + ab) a+b+c 3 (a + b + c) (a2 + b2 + c2) 0 n +1

xy 2

22. Suppose A, B, and C are angles of a triangle. Let

3a -a + b -a + c 3b -b + c D = -b + a -c + a -c + b 3c Then D equals

2

x2 + 1

Then (x, y, z) lies on a (a) plane (b) straight line (c) sphere (d) none of these

17. Distance of line x +1 x x x+2 x y= x x x x+3 from the origin is 6 7 (b) (a) 11 13 6 7 (c) (d) 122 122 18. Suppose a, b, c ΠR. Let

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

21. Suppose point (x, y, z) in space satisfies the equation

c2 f 2 r Then D depends on (a) a, b, c (b) d, e, f (c) p, q, r (d) none of these

2-x-y 2x 2y

4 4 x-y-2 2x =0 2y y-2-x

passes through the fixed point (a) (–2, –2) (c) (0, –2)

(b) (–2, 0) (d) ( –1, –1)

25. Suppose a ΠR. Let x+a x x f (x) = x x+a x x x x+a Then f (2x) Рf (x) is equal to (a) 3xa2 (b) 3x2a 2 (d) a2x (c) xa 26. If a, b, g are the roots of x3 + ax2 + b = 0, then the determinant

4.34

Complete Mathematics—JEE Main

a b g D = b g a equals g a b (a) – a3 (c) a2 – 3b

(b) a3 – 3b (d) a3

27. If a, b, g are the roots of x3 + bx + c = 0, then the determinant a b g D = b g a equals g a b (a) – b3 (c) b2 – 3c 28. If a, b, c are distinct bc ca D = ca ab ab bc (a) (b) (c) (d)

1 cos x D = -1 1 - cos x

(b) b3 – 3c (d) 0 and different from zero and ab bc = 0, then ca

a–1 + b–1 + c–1 = 0 a–1 + b–1 – c–1 = 0 a–1 – b–1 + c–1 = 0 a–1 – b–1 – c–1 = 0 l a l 2 + a2 1

29. The determinant D = l b l 2 + b2 1 equals lc

l 2 + c2

sin 2 a

cos 2a

cos2 a

D = sin 2 b

cos 2b

cos2 b equals

sin 2 g

cos 2g

cos2 g

2

3

1 b

1 c2

b

1 b

1 c

a2

b2

c2

the value of l is (a) – 1 (c) 9

(b) 0 (d) 18

equals

1 - 2 sin ( x + p / 4 )

(a) 0 (c) 1

(b) (d) x a 34. The factors of D = a x a b

–1 none of these b b are x

(a) x – a, x – b and x + a + b (b) x – a, x – b and x – a – b (c) x + a, x + b and x – a – b (d) none of these 35. If q e R, maximum value of 1 1 1 1 is D = 1 1 + sin q 1 1 1 + cos q (a) 1/2

3 /2

(b)

(d) 3 2 /4 x+a b c a x+b c = 0, then x equals 36. If D = a b x+c (a) a + b + c (b) – (a + b + c) (c) 0, a + b + c (d) 0, – (a + b + c) 37. The determinant (c)

2

sec 2 q tan q 12

1 =l a

c3

-1

2

(a) 0 (b) – 1 (c) sin2 a + sin2 b + sin2 g (d) none of these 31. If bc + ca + ab = 18, and a3

0

0 sin x + cos x

1

(a) l(a – b) (b – c) (c – a) (b) l(a2 + b2 + c2) (c) l(a + b + c) (d) l2(a – b) (b – c) (c – a) 30. If a, b, g are real numbers, then the determinant

1 a2

32. If x π 0, the determinant a0 a1 a2 D = -x x 0 0 -x x vanishes if (b) a0 + a1 = 2a2 (a) a0 + a1 + a2 = 0 (c) a0 + a2 = 2a1 (d) none of these 33. If x Œ R, the determinant

tan 2 q 2

sec q 10

1 -1 equals 2

2 sin2q 12 sec2 q – 10 tan2 q 12 sec2 q – 10 tan2 q + 5 0 - a 2b 0 38. If D = 0 - a 2b = 0, then 2b 0 - a (a) (b) (c) (d)

(a) 1/b is a cube root of unity (b) a is one of the cube roots of unity

Determinants 4.35

(c) b is one of the cube roots of 8 (d) a/b is a cube root of 8 39. The determinant 1 1+ i i D = 1+ i i 1 equals i

1

6i -3i w 45. If D = 4 3i - w = x + iy, then 20 3 iw

(a) x = 3, y = 1 (c) x = 0, y = 3 p p £x£ 46. If – 4 4 of D = 0 sin x D = cos x cos x

1+ i

(a) 7 + 4i (b) – 7 + 4i (c) – 7 – 4i (d) 2(i – 1) 40. If a, b, c are non-zero real numbers, then 1 1 1 ab + a b 1 1 + equals D = 1 bc b c 1 1 + 1 ca c a (a) 0 (c) a–1 + b–1 + c–1

(b) bc + ca + ab (d) abc – 1

x + iy = -i

42. If D =

then l equals (a) 0 (c) x

1 43. Let D = - sin q -1

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

-w

2

1 sin q , 0 £ q £ 2p.The 1

(b) D Œ (2, •) (d) D Œ [2, 4] bc - ac

ab - a 2

a - b b2 - ab

bc - ac

c - a ab - a 2

(b – c) (c – a) (a – b) abc (b – c) (c – a) (a – b) (a + b + c) (b – c) (c – a) (a – b) 0

then

1

(a) x = – 1, y = 0 (c) x = 1, y = 1

(b) x = 1, y = – 1 (d) none of these

48. If eix = cosx + isinx and

(a) x = – 1, y =

b2 - ab b - c 44. The determinant D =

w

e -p i / 3

(b) 1 (d) 1 – x2

(a) D = 0 (c) D Π(2, 4)

w2

1

ep i / 4

x + iy = e-p i / 4

p + qx a c p px + q = l b d q w u v w

sin q 1 - sin q

cos x cos x sin x cos x is cos x sin x

1

0 loga b + logb c + logc a logabc (a + b + c) none of these a + bx c + dx ax + b cx + d u v

, the number of distinct real roots

(a) 0 (b) 2 (c) 1 (d) 3 47. If w π 1 is a complex cube root of unity, and 1 i -w

log a ( abc ) log a b log a c 41. If a, b, c > 1, then D = logb ( abc ) logb c 1 logc ( abc ) logc b 1 equals

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

(b) x = 1, y = 3 (d) x = 0, y = 0

ep i / 3 e2p i / 3 , then

1 e-2p i / 3

e-2p i / 3

2

(b) x = 1, y = - 2 (c) x = - 2 , y = 2 (d) none of these 49. If a, b, c e R, the number of real roots of the equation x c -b -c x a = 0, is b -a x (a) 0 (b) 1 (c) 2 (d) 3 1

x

x2

50. If D = x

x2

1

2

1

x

x

D1 =

= – 7 and

x3 - 1

0

0

x - x4

x - x4

x3 - 1

(a) D = 7 (c) D = – 49

x - x4 x 3 - 1 , then 0

(b) D = 343 (d) D = 49

4.36

Complete Mathematics—JEE Main

sin a 51. If D = sin b sin g

cos a cos a cos a

sin a + cos b sin b + cos b sin g + cos b

57. For a fixed positive integer n, if then D equals

(a) sin a sin b sin g (b) cos a sec b tan g (c) sin a sin b sin g + cos a cos b cos g (d) 0 52. Suppose a, b, c ΠR and a, b, c > 0,

D=

2

sin 2 q

4 sin 4q

2

cos q

1 + sin q

4 sin 4q

cos2 q

sin 2 q

1 + 4 sin 4q

=0

are given by (a) p/24, 5p/24 (b) 7p/24, 11p/24 (c) 5p/24, 7p/24 (d) 11p/24, p/24 54. The number of distinct roots of the equation x2 - 1

x2 + 2 x + 1

P(x) = 2 x 2 + x - 1 2 x 2 + 5 x - 3

2 x2 + 3x + 1 4 x2 + 4 x - 3 = 0

6 x 2 - x - 2 6 x 2 - 7 x + 2 12 x 2 - 5 x - 2 is (a) 6 (c) 3

(b) 5 (d) 4 2rp 2rp + i sin , then value of the 55. If ar = cos 9 9 determinant 1 a8 a7 D = a3 a2 a1 is a6

a5

a4

(a) – 1 (b) 1 (c) 0 (d) – 2 56. If a π p, b π q, c π r and the system of equations px + by + cz = 0 ax + qy + cz = 0 ax + by + rz = 0 has a non-zero solution, then value of p+a q+b r+c + + is p-a q-b r-c (a) 2 (c) 1

(b) – 2 (d) 1

D is equal to n! ( n + 1)! ( n + 2 )! (a) – 4 (b) – 2 (c) 2 (d) 4 x+2 x+7 a 58. If a, b, c are in A.P., and D = x + 5 x + 11 b x + 8 x + 15 c then D equals (a) 0 (b) 1 (c) – (a + b + c) (d) a + b + c. then

log a log b log c Let D = log (7 a) log (49 b) log (343 c) log (3 a ) log (9 b) log (27 c) then D is equals to (a) 0 (b) –1 (c) 1 (d) 30 53. The values of q lying between q = 0 and q = p/2 and satisfying the equation 1 + cos2 q

n! (n + 1)! (n + 2)! D = ( n + 1)! ( n + 2 )! ( n + 3)! (n + 2)! (n + 3)! (n + 4)!

1+ y 1- y 1- y 59. If D = 1 - y 1 + y 1 - y = 0, then value of y are 1- y 1- y 1+ y (a) 0, 3 (c) – 1, 3 60. The determinant

(b) 2, – 1 (d) 0, 2

al + a ¢l ¢ am + a ¢m ¢ an + a ¢n ¢ D = bl + b ¢l ¢ bm + b ¢m ¢ bn + b ¢n ¢ cl + c ¢l ¢ cm + c ¢m ¢ cn + c ¢n ¢ is equal to (a) (abc + a¢b¢c¢) (lmn + l¢m¢n¢) (b) abc lmn + a¢b¢c¢ l¢m¢n¢ (c) (a2 + b2 + c2) (l2 + m2 + n2) + (a¢2 + b¢2 + c¢2) (l¢2 + m¢2 + n¢2) (d) 0 61. If a = i, b = w, c = w2 where w is complex cube root of unity, then a a+b a+b+c D = 3a 4 a + 3b 5a + 4b + 3c is equal to 6 a 9a + 6b 11a + 9b + 6c

(a) – w (c) i 1 62. If D =

m

C1

1 m +3

C1

(b) – w2 (d) – i 1 m +6

C1

= 2a 3b 5g, then

C2 m + 3 C2 m + 6 C2 a + b + g is equal (a) 3 (b) 5 (c) 7 (d) none of these 63. Suppose a, b, c, x, y ΠR. Let 1 2 + ax 3 + ay D = 1 2 + bx 3 + by 1 2 + cx 3 + cy m

Determinants 4.37

Then D is independent of (a) a, b, c (b) x, y (c) a, b, c, y (d) a, b, c, x, y 64. If A, B and C are the angles of a triangle, then the determinant 0 cos C cos B -1 cos A is equal to D = cos C cos B cos A -1 (a) sin2A (c) sin2C 65. Let f(x) = given by

cos x

(b) sin2B (d) 0 x 1

2 sin x tan x

x2 x

f (x) is 2 x , then lim 2 xÆ0 x 1

(a) 0 (b) – 1 (c) 2 (d) 3 66. If w is a complex cube root of unity, then value of a1 + b1w

D = a2 + b2w

a1w 2 + b1

c1 + b1w

2

c2 + b2w

a2w + b2 2

a3 + b3w a3w + b3 c3 + b3w is (a) 0 (b) – 1 (c) 2 (d) none of these 67. If pqr π 0 and the system of equations (p + a)x + by + cz = 0 ax + (q + b)y + cz = 0 ax + by + (r + c)z = 0

a b c has a non-trivial solution, then value of + + p q r is (a) – 1 (b) 0 (c) 1 (d) 2 68. The system of equations ax + by + (aa + b)z = 0 bx + cy + (ba + c)z = 0 (aa + b)x + (ba + c)y = 0 has a non-zero solutions if a, b, c are in (a) A.P. (b) G.P. (c) H.P. (d) A.G.P. 69. If the system of equations ax + ay – z = 0 bx – y + bz = 0 – x + cy + cz = 0 (where a, b, c π – 1) has a non-trivial solution, then 1 1 1 is value of + + 1+ a 1+ b 1+ c (a) 2 (b) – 1 (c) – 2 (d) 0

70. The values of l for which the system of equations (l + 5)x + (l – 4)y + z = 0 (l – 2)x + (l + 3)y + z = 0 lx + ly + z = 0 has a non-trivial solution is (are) (a) – 1, 2 (b) 0, – 1 (c) 0 (d) none of these 71. Given the system of equations (b + c) (y + z) – ax = b – c (c + a) (z + x) – by = c – a (a + b) (x + y) – cz = a – b (where a + b + c π 0); then x : y : z is given by (a) b – c : c – a : a – b (b) b + c : c + a : a + b a b c (c) a : b : c (d) : : b c a 72. If a, b, c e R and a + b + c π 0 and the system of equations ax + by + cz = 0 bx + cy + az = 0 cx + ay + bz = 0 has a non-zero solution, then a : b : c is given by (a) 1 : a : b where a, b are roots of ax2 + bx + c = 0 (b) 1 : r : r2 where r is some positive real number (c) 1 : k : 2k where k is some positive real number (d) none of these 73. If p is a constant and x3 f(x) = 1 p

x3 3x 2 -6 4 , then f ¢≤(x) is p2

p3

(a) proportional to x3 (b) proportional to x2 (c) proportional to x (d) a constant a1 74. If D = a2 a3

b1 b2 b3

a1 + pb1 D = a2 + pb2 a3 + pb3

c1 c2 c3

and

b1 + qc1 b2 + qc2 b3 + qc3

c1 + ra1 c2 + ra2 c3 + ra3

then

(a) D = D (b) D = D (1 – pqr) (c) D = D(1 + pqr) (d) D = D(1 + p + q + r) 75. Number of real values of l for which the system of equations (l + 3)x + (l + 2)y + z = 0 3x + (l + 3)y + z = 0 2x + 3y + z = 0 has a non-trivial solutions is (a) 0 (b) 1 (c) 2 (d) infinite

4.38

Complete Mathematics—JEE Main

Assertion-Reason Type Questions

76. Statement-1: If

D(x) =

ax + by + cz = 0

sin x - cos x

pˆ Ê sin Á x - ˜ Ë 3¯

pˆ Ê cos Á x - ˜ Ë 3¯

Êp ˆ sin Á - x˜ Ë3 ¯

pˆ Ê tan Á x - ˜ Ë 4¯

pˆ Ê sec Á x - ˜ Ë 4¯

Ê 2p ˆ cos Á + x˜ Ë 3 ¯

Ê 2p ˆ sec Á + x˜ Ë 3 ¯

pˆ Ê cot Á x + ˜ Ë 4¯

Êpˆ then D Á ˜ = 0 Ë 4¯ Statement-2: If A is a skew symmetric matrix of odd order, then |A| = 0. 77. Statement-1: Let p < 0 and a1, a2, . . . a9 be the nine roots of x9 = p, then a1 D = a4 a7

a2 a3 a5 a6 = 0 a8 a 9

Statement-2: If two rows of a determinant are identical, then determinant equals zero. 78. Statement-1:

6 w =

2i

12

3 + 8i

18

2 + 12i

3 + 6i 3 2 + 6i 27 + 2i

is a purely imaginary number. Statement-2: |z| = | z | for each complex number z. 79. Statement-1: If ai, bi ΠN, for i = 1, 2, 3 and (1 + x )a1 b1

(1 + x )a1 b2

(1 + x )a1 b3

a b D(x) = (1 + x ) 2 1

(1 + x )a2 b2

(1 + x )a2 b3 ,

(1 + x )a3 b1

(1 + x )a3 b2

(1 + x )a3 b3

then coefficient of x in expansion of D(x) is 0. Statement-2: If P(x) = (1 + x)n, n Œ N then coefficient of x in the expansion of P(x) is P¢(0). 80. Statement-1: If a + b + c = 0 and a2 + b2 + c2 π bc + ca + ab, then the system of homogenous equations

bx + cy + az = 0 cx + ay + bz = 0 has infinite number of solutions. Statement-2: If |A| = 0, the system of equations AX = B has infinite number of solutions. 81. Suppose a, b, c Œ R. Consider the system of linear equations: ax + by + cz = 0 bx + cy + az = 0 cx + ay + bz = 0 Statement-1: If a + b + c π 0 and a2 + b2 + c2 = bc + ca + ab, then the system of equations has infinite number of solutions. Statement-2: If a + b + c = 0 and a2 + b2 + c2 π bc + ca + ab, then the system of equations has unique solutions. 82. Suppose a1, a2, a3, b1, b2, b3 Œ R. Let a1 b1 1 D1 = a2 b2 1 a3 b3 1 2 a1b1 D2 = a1b2 + a2 b1 a1b3 + a3 b1

a1b2 + a2 b1 2 a2 b2 a2 b3 + a3 b2

a1b3 + a3 b1 a2 b3 + a3 b2 2a3 b3

Statement-1: D2 π 0 for some values of a1, a2, a3, b1, b2, b3 Œ R. Statement-2: If (a1, b1), (a2, b2), (a3, b3) are noncollinear than D1 π 0. 83. Suppose a1, b1, c1, d1 a2, b2, c2, d2 are eight integers. Statement-1: ( a12 + b12 + c12 + d12 ) ( a22 + b22 + c22 + d22 ) can be written in the form a2 + b2 + c2 + d2 where a, b, c, d are some integers. Statement-2: If a, b, c, d Œ Z, then a + ib c + id a2 + b2 + c2 + d2 = -c + id a - ib

Determinants 4.39

LEVEL 2 Straight Objective Type Questions 84. If l12 + m12 + n12 = 1 etc., and l1l2 + m1m2 + n1n2 = 0 etc., and l1 m1 n1 D = l2 m2 n2 l3 m3 n3 then (a) | D | = 3 (b) | D | = 2 (c) | D | = 1

(d) D = 0

85. If a, b, c are non-zero real number and b2 c 2

bc

b+c

D = c2 a2

ca

c+a

2 2

a b then D equals (a) abc (c) bc + ca + ab

ab a + b

86. Let D(x) =

2

3x + x + 9

2x - 5

3

6x + 1

9

7 x 2 - 6 x + 9 14 x - 6 21 = ax3 + bx2 + cx + d, then a equals (a) – 1 (b) 0 (c) 2 (d) none of these 87. The number of distinct values of t for which the system (a + t)x + by + cz = 0 ax + (b + t)y + cz = 0 ax + by + (c + t)z = 0 has a non-trivial solution is (a) 1 (b) 2 (c) 3 (d) none of these 88. If a2 + b2 + c2 = 1, then a 2 + (b2 + c 2 ) cos q

ab (1 - cos q ) 2

ba (1 - cos q ) ca(1 - cos q )

2

2

b + (c + a ) cos q cb(1 - cos q ) ac (1 - cos q ) bc (1 - cos q ) 2

2

2

c + (a + b )cos q (a) cos2q (c) 1

(b) 0 (d) sin2q

A(a, b, g, q) =

cos (b + q ) sin (b + q ) 1 cos (g + q ) sin (g + q ) 1

p p 2p Numerical value of A Ê - , 0, , ˆ is Ë 2 2 13 ¯ (a) 0 (b) – 1 (c) 2 (d) none of these 90. If a, b, c are positive integers such that a > b > c and 1 1 1 a b c =–2

a 2 b2 c 2 then 3a + 7b Р10c equals (a) 10 (b) 11 (c) 12 (d) 13 91. If A, B, C, P, Q, R ΠR, and

(b) a2 b2 c2 (d) 0

x2 - 5x + 3

89. Suppose a, b, g, q ΠR and cos (a + q ) sin (a + q ) 1

equals

cos ( A + P ) cos ( A + Q ) cos ( A + R) D = cos ( B + P ) cos ( B + Q) cos ( B + R) cos (C + P ) cos (C + Q) cos (C + R) (a) D depends on P, Q, R (b) D depends on A, B, C (c) D depends on A, B, C, P, Q, R (d) none of these 2 cos x 1 0 then 92. Let f (x) = 1 2 cos x 1 0 1 2 cos x (a) f Ê p ˆ = 1

(b) f ¢ Ê p ˆ = - 3

p (c) f Ê ˆ = – 1 Ë2 ¯

(d) none of these

Ë 3¯

Ë 3¯

93. Consider the set A consisting of all determinants of order 3 with entries 0 and 1 only. Let B be the subset of A consisting of all the determinants with value 1. Let C be the subset of A consisting of all the determinants with value –1. Then (a) C = f (b) B has as many elements as C (c) A = B « C (d) A = B » C 94. let w = e2pi/3 and consider the system of linear equations

4.40

Complete Mathematics—JEE Main

x+y+z=a x + w y +w2 z = b x + w2 y + w z = c If x, y, z is a solution of the above system of equa| a |2 + | b |2 + | c |2 is tions, then value of | x |2 + | y |2 + | z |2 (a) 9 (b) 6 (c) 3 (d) 1 a a 2 a3 - 1 b b2

b3 - 1

c c2 (b) 1 (d) –2

c3 - 1

95. If a, b, c are distinct and then abc equals (a) 0 (c) –1

=0

96. If the adjoint Ê1 Á2 Á Ë1

of 4 1 1

a 3 ¥ 3 matrix P is 4ˆ 7˜ then ˜ 3¯

the possible value(s) of determinant P is (are) (a) ± 2 (b) ± 3 (d) ± 1 (d) 0 1 -3 4 = 0, then x equals 97. If -5 x + 2 2 4 1 x-6 (a) 17, 21 (b) 0,19 (c) 0, 35 (d) 21, 35

Previous Years' AIEEE/JEE Main Questions 1. If the equation ax2 + 2bx + c = 0 has equal roots then the determinant D=

a b ax + b b c bx + c ax + b bx + c 0

is (a) positive (c) 0

(b) negative (d) dependent on a.

[2002] 2. If l, m, n > 0 and l, m, n are the pth, qth, rth terms of a G.P., then determinant log l p 1 [2002] D = log m q 1 equals log n

w

1

log an +1

log an + 2

log an + 1

log an + 2

log an + 3

log an + 2

log an + 3

log an + 4

1 + a2 x

f(x) = (1 + a 2 ) x

1

equals

wn (a) 1 (b) 2 (d) 0 [2003] (c) w2 4. If the system of linear equations x + 2ay + az = 0, x + 3by + bz = 0, x + 4cy + cz = 0 has a non-zero solution, then a, b, c (a) are in G.P (b) are in H.P. w

2n

2n

log an

is (a) 2 (b) 1 (c) 0 (d) – 2 6. If a2 + b2 + c2 = – 2 and

r 1

(a) 0 (b) – 1 (c) p + q + r (d) none of these 3. If 1, w, w2 are the cube roots of unity then 1 w n w 2n D = wn

(c) satisfy a + 2b + 3c = 0 (d) are in A.P. [2003] 5. If a1, a2, a3, ... , an, ... are in G.P., then the value of the determinant

[2004, 2005]

(1 + b2 ) x (1 + c 2 ) x 1 + b2 x

(1 + c 2 ) x

(1 + a 2 ) x (1 + b2 ) x

then f(x) is a polynomial of degree (a) 3 (b) 2 (1) 1 (d) 0 7. The system of equations ax + y + z = a – 1

1 + c2 x

[2005]

x + ay + z = a – 1 has (a) (b) (c) (d)

x + y + az = a – 1 no solution, if a is not – 2 1 –2 either – 2 or 1

[2005]

Determinants 4.41

1 1 1 8. If D = 1 1 + x 1 for x π 0, y π 0, then D is 1 1 1+ y (a) (b) (c) (d) 9. Let

divisible by neither x nor y divisible by both x and y divisible by x but not y divisible by y but not x a, b, c be such that b(a + c) π 0. If a a +1 a -1 D = -b b +1 b -1 c c -1 c +1

+

a +1 a -1

b +1 b -1

n+ 2

n+1

(-1)

a (-1)

[2007]

c -1 c +1

=0

n

b (-1) c

[2009] of a scalene triangle, then the c a is: b

[2014] 13. If a, b, c are non-zero real numbers and if the system of equations (a – 1)x = y + z, (b – 1)y = z + x, (c – 1)z = x + y, (a) a + b + c (b) abc (c) 1 (d) –1 [2014 online] 14. Let for i = 1, 2, 3, pi(x) be a polynomial of degree 2 in x, pi¢(x) and pi≤(x) be the first and second order derivatives of pi(x) respectively. Let, p1¢ ( x ) p2¢ ( x ) p3¢ ( x )

È p1 ( x ) Í A(x) = p2 ( x ) Í ÍÎ p3 ( x )

p1¢¢( x ) ˘ p2¢¢( x )˙ ˙ p3¢¢( x ) ˙˚

and B(x) = [A(x)]T A(x). Then determinant of B(x). (a) (b) (c) (d)

is a polynomial of degree 6 in x is a polynomial of degree 3 in x is a polynomial of degree 2 in x does not depend on x [2014 online] a2

has a non-trivial solution for only one value of a lying in the interval (0, p /2). Statement-2: The equation in a sin a cos a - sin a

(d) ab

15. If

[2013 online] 11. Statement-1: The system of linear equations x + (sin a)y + (cos a)z = 0 x + (cos a)y + (sin a)z = 0 x – (sin a)y – (cos a)z = 0

cos a D = sin a cos a

(b) 1

has a non-trivial solution, then ab + bc + ca equals:

then the value of n is (a) any odd integer (b) any integer (c) zero (d) any even integer 10. If a, b, c are sides a b value of D = b c c a (a) non-negative (b) negative (c) positive (d) non-positive

1 ab (c) –1 (a)

cos a sin a = 0 - cos a

has only one solution lyining in the interval (0, p /2). [2013 online] n n 12. If a, b π 0, and f(n) = a + b and 3 1 + f (1) 1 + f (2) 1 + f (1) 1 + f (2) 1 + f (3) 1 + f (2) 1 + f (3) 1 + f (4) = K(1 – a)2 (1– b)2 (a – b)2, then K is equal to:

b2

c2

a 2 b2 c 2 (a + l )2 (b + l )2 (c + l )2 = kl a b c , l π 0, 1 1 1 ( a - l )2 ( b - l )2 ( c - l )2 then k is equal to: (a) 4labc (c) 4l2

(b) – 4labc (d) – 4l2 [2014 online]

16. If Dr =

r n 2

2r - 1

3r - 2

n -1

a

1 n(n - 1) (n - 1)2 2

1 (n - 1)(3n + 4) 2

n -1

then the value of

 Dr : r =1

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

depends only on a depends only on n depends both on a and n is independent of both a and n. [2014 online]

Complete Mathematics—JEE Main

4.42

17. The set of all values of l for which the system of linear equations: 2x1 – 2x2 + x3 = lx1 2x1 – 3x2 + 2x3 = lx2 – x1 + 2x2 = lx3 has a non-trivial solution, (a) is an empty set (b) is a singleton (c) contains two elements (d) contains more than two elements x2 + x 18. If

x +1

2

D = 2 x + 3x - 1

3x

2

x + 2x + 3 then a is equal to: (a) 12 (c) – 12 *

19 . The least value of x determinant D = 1 1

[2015]

x-2 3 x - 3 = ax - 12,

2x - 1 2x - 1

(b) 24 (d) – 24 the 1 y 1

(a) -2 2 (b) -16 2 (c) – 8 (d) – 1 [2015 online] 20. The system of linear equations x + ly – z = 0 lx – y – z = 0 x + y – lz = 0 has a non-trivial solution for l (b) exactly one value of l (c) exactly two values of l (d) exactly three values of l [2016] 21. The number of distinct real roots of the equation. cos x sin x sin x D = sin x cos x sin x = 0 sin x sin x cos x È p p˘ in the interval Í - , ˙ is: Î 4 4˚ (a) 1 (b) 4 (c) 2 (d) 3

[2015 online]

product xyz for which the 1 1 is non-negative, is:

[2016 online]

z

Previous Years’ B-Architecture Entrance Examination Questions 1. If the system of equations x + y + z= 0 ax + by + z = 0 bx + y + z = 0 has a non-trivial solution, then (a) b2 = 2b + 1 (c) b – a = 0

(b) b2 = 2b – 1 (d) b2 = 2b

n

then the value of

k =1

(a) (b) (c) (d) [2006]

only on a and b not on g on all a, b and g on none of a, b and g only on a, not on b and g

[2007]

4. Let A = (aij) be a 3 × 3 matrix whose determinant is 5. The determinant of the matrix B = (2i – j aij) is (a) 5 (b) 10 (c) 20 (d) 40 [2008] r -1 r -1 r -1 2 2(3 ) 4(5 ) , for b g 5. Let Dr = a

2. The system of linear equations (l + 3)x +(l + 2)y + z = 0 3x + (l + 3)y + z = 0 2x + 3y + z = 0 has a non-trivial solution (a) if l = 1 (b) if l = –1 (c) for no real value of l (d) if l = 0

 D k depends

2n - 1

3n - 1

5n - 1

n

r = 1, 2,…, n. Then  D r is

[2007]

r =1

3. If

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

2(3k -1 ) 3(4 k -1 ) 4(5k -1 ) Dk = a b g 3n - 1

4n - 1

5n - 1 *

independent of a, b, g and n independent of n only depends on a, b, g and n independent of a, b, g only

Question is incorrect as xyz can take any real value.

[2010]

Determinants 4.43

6. If x1, x2, x3,…, x13 are in A.P. then the value of e x1

e x4

e x7

e x4

e x7

e x10 is:

Concept-based

e x7 e x10 e x13 (a) 27 (b) 0 (c) 1 (d) 9 7. The value of the determinant

[2011]

2 5

15 + 26

5

10 is equal to:

3 + 65

15

5

5

(b) 5 3 ( 6 - 5 )

(c) 5 ( 6 - 5) (d) 3 ( 6 - 5 ) [2012] 8. If the system of linear equations, x + 2ay + az = 0, x + 3by + bz = 0 and x + 4cy + cz = 0 has a non-zero solution, then a, b, c satisfy: (a) 2b = a + c (b) b2 = ac (c) 2ac = ab + bc (d) 2ab = ac + bc [2013] 9. In a DABC, if 1 a b 1 c a = 0,

1 b c then sin2A + sin2B + sin2C is 3 9 3 (a) (b) 2 4 5 (c) (d) 2 [2014] 4 10. The system of linear equations x – y + z= 1 x + y – z= 3 x – 4y + 4z = a has: (a) a unique solution when a = 2 (b) a unique solution when a π – 2 (c) an infinite number of solutions, when a = 2 (d) an infinite number of solutions, when a = –2 [2015] Ê pˆ 11. For all values of q Œ Á 0, ˜ , the determinant of the Ë 2¯ È -2 Í matrix D = Í- sin q Í -3 Î

tan q + sec2 q cos q -4

lies in the interval: (a) [3, 5]

(b) (4, 6)

Ê 5 19 ˆ (c) Á , ˜ Ë2 4 ¯

È 7 21 ˘ (d) Í , ˙ Î2 4 ˚

1. 5. 9. 13.

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

2. 6. 10. 14.

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

3. 7. 11. 15.

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

4. (d) 8. (a) 12. (c)

17. 21. 25. 29. 33. 37. 41. 45. 49. 53. 57. 61. 65. 69. 73. 77. 81.

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

18. 22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62. 66. 70. 74. 78. 82.

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

19. 23. 27. 31. 35. 39. 43. 47. 51. 55. 59. 63. 67. 71. 75. 79. 83.

85. 89. 93. 97.

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

86. (b) 90. (d) 94. (c)

Level 1

13 + 3

(a) 5 3 ( 6 - 5)

Answers

3 ˘ ˙ sin q ˙ always 3 ˙˚

16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 64. 68. 72. 76. 80.

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

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

Level 2 84. 88. 92. 96.

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

87. (b) 91. (d) 95. (b)

Previous Years’ AIEEE/JEE Main Questions 1. 5. 9. 13. 17. 21.

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

2. 6. 10. 14. 18.

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

3. 7. 11. 15. 19.

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

4. 8. 12. 16. 20.

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

Previous Years’ B-Architecture Entrance Examination Questions 1. (b) 5. (a) 9. (b)

2. (c) 6. (b) 10. (d)

3. (c) 7. (c) 11. (a)

4. (a) 8. (c)

Hints and Solutions Concept-based 1. det (adj (A)) A–1) = det (adj(A)) det(A–1)

[2016] = (det(A))2

1 = det(A) det( A)

4.44

Complete Mathematics—JEE Main

2. Use R1 Æ R1 + R3 – 2R2 2

3. Using R2 Æ R2 – x R1 and R3 Æ R3 – xR1, we get 1

x

a2

D = 0 1 - x3 0

x(1 - x 3 ) 1 - x3

0

= (1 – x3)2 = (1 – x)2 (1 + x + x2)2 As x π 1, D = 0 fi 1 + x + x2 = 0 fi x = w, w2 4. Taking 10!, 11! and 12! common from C1, C2, C3 respectively, we get D = (10! 11! 12!) D1 where 1 1 1 D1 = 11 12 13 (11)(12) (12)(13) (13)(14) Using C3 Æ C3 – C2 and C2 Æ C2 – C1, we get 1 0 0 D1 = 11 1 1 =2 (11)(12) 2(12) 2(13) Thus,

D (10!)3

= 2(11) (12) = 264

D2 = 192

fi

8. Write 1+1

a+b

a 2 + b2

a+b

a 2 + b2

a 3 + b3

a 2 + b2

a 3 + b3

a 4 + b4

1 = a

1 b

0 0

1 a

1 b

0 0 =0

a2

b2

0

a2

b2

0

9. Use R2 Æ R2 – R1, R3 Æ R3 – R1 and log x – log y = log (x/y) to obtain log a log b log c D = log (2007) log (2007) log (2007) = 0 log (2017) log (2017) log (2017) [

R2 and R3 are proportional]

10. Using R2 Æ R2 – R1, we get

Applying R1 Æ R1 – 2 (2016) R2 – R3, we get 1 1 1 3 D = - 4 (2016) a b c a2

b2

c2

= – 4 (2016)3 (a – b) (b – c) (c – a) 11. Using R2 Æ R2 + R1, R3 Æ R3 – R1, we get x -6 -1 P(x) = x + 2 -3( x + 2) x + 2 -3 - x 2( x + 3) x + 3 x = (x + 2) (x + 3) 1 -1 Using R1 Æ R1 + R3, R2 Æ

-6 -3 2 R2 –

-1 1 1 R3, we get

x -1 -4 0 -5 0 P(x) = (x + 2) (x + 3) x -1 2 1

12. Write D as 1 x

x2

D = xn+1 yn+1 zn+1 1 y

y2

1 z z2 = (xyz)n+1 (x – y) (y – z)(z – x)

7. Use R1 Æ R1 + R3 – 2R2.

D=

c2

sum of zeros of P(x) is –2 – 3 + 13/5 = –12/5.

1 3 2 ( 4) = 4 3 6. D = area of triangle = 2 4 |D| = 8 3

b2

= (x + 2) (x + 3) (– 5x + 13)

5. Product of zeros = –P(0) = 0

fi

(a + 2016)2 (b + 2016)2 (c + 2016)2 a b c D = - 4(2016)

\ n+1 = 2 fi n = 1. 13. When x = 0, P(0) is a skew-symmetric determinant of odd order. 14. Write D = D1 + aD2 where 1 1 1 D1 = 1 + b 1 + 2 b 1 1 + c 1 + c 1 + 3c 1 0 0 1 and D2 = 1 + b 1 + 2b 1 + c 1 + c 1 + 3c = (1 + 2b) (1 + 3c) – (1 + c) = 2 (b + c + 3c) In D1, apply C2 Æ C2 – C1 and C3 Æ C3 – C1 to obtain 1 0 0 D1 = 1 + b b -b = 2bc 1 + c 0 2c

Determinants 4.45

Thus, D = 2(bc + ac + ab + 3abc) 1 1 1 \ D =0 fi + + = –3 a b c m

15. Taking x common from R1 and x get 1 x2 xm D = x 2m +5 1

xn n-m

19. Using R1 Æ R1 – R3 – R2 we get R1 consists of all zeros. m+5

from R3, we

2n

20. If R is the common ratio of the G.P., then b = aR, c = aR2, d = aR3, e = aR4, f = aR5, and 1 2

2

6

D = (a )(a R ) R

Level 1 16. Using C1 Æ C1 + C2 + C3, we get 1 c+a a+b D = 2(a + b + c) 1 a + b b + c 1 b+c c+a Using R2 Æ R2 – R1, R3 Æ R3 – R1, we get 1 c+a a+b D = 2(a + b + c) 0 b - c c - a 0 b-a c-b 2

= – 2(a + b + c) [(b – c) + (b – a) (c – a)] = – (a + b + c) [(b – c)2 + (c – a)2 + (a – b)2] Now, use D = 0, a + b + c π 0. 17. Write y = D1 + D2 where 1 x x x = 5x + 6 D1 = 0 x + 2 0 x x+3 1 x x 1 0 0 x = x 1 2 0 = 6x and D2 = x 1 x + 2 1 x x+3 1 0 3 Thus, y = 11x + 6 6 Its distance from the origin = 122 18. Using C1 Æ C1 + C2 + C3, we get 1 -a + b -a + c 3b -b + c D = (a + b + c) 1 1 -c + b 3c Now, use R2 Æ R2 – R1, R3 Æ R3 – R1, we get 1 -a + b -a + c D = ( a + b + c ) 0 2b + a - b + a 0 - c + a a + 2c D = (a+b+c) [(a + 2b) (a + 2c) – (a – c) (a – b)] = 3(a + b + c) (bc + ca + ab)

R

R4 [

m

1 x x If n – m = 2, then R1 and R3 are identical and hence D = 0.

1

2

p

2

q =0

R4 r C1 and C2 are identical]

21. Multiply C1 by x, C2 by y and C3 by z to obtain

5=

( x 2 + 1) x

xy 2

xz 2

yx 2

( y 2 + 1) y

yz 2

1 xyz

zx 2 zy 2 z( z 2 + 1) Taking x common from R1, y from R2 and z from R3, we get x2 + 1 5=

xyz xyz

x2

y2

z2

y2 + 1 z 2

x2

y2

z2

Using C1 Æ C1 + C2 + C3, we get 5 = (x2 + y2 + z2 + 1) D1 1

y2

where D1 = 1 y 2 + 1 1

y2

z2 z2 z2 + 1

Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get 1 y2 D1 = 0 1 0 0

z2 0 =1 1

Thus, x2 + y2 + z2 = 4, which represents a sphere. 22. Apply R1 Æ aR1 + bR2 + cR3 and use a = b cos C + c cos B etc. 23. Write D = D1 + xD2 where a-x x x D1 = 0 b x = (a - x )(bc - x 2 ) 0 x c 1 x x and D2 = 1 b x 1 x c 1 x x 0 = 0 b-x 0 0 c-x = (b – x) (c – x)

[use R2 Æ R2 – R1, R 3 Æ R 3 – R 1]

4.46

Complete Mathematics—JEE Main

Thus, D = (a – x) (bc – x2) + x(b – x) (c – x) = 2x3 – (a + b + c)x2 + abc Also, x f ¢(x) – f(x) = x [(x – b) (x – c) + (x – a) (x – c) + (x – a) (x – b)] – f (x) = 2x3 – (a + b + c)x2 + abc 24. Applying R1 Æ R1 + R2 + R3, we get 1 0 = ( x + y + 2) 2 x 2y

1 1 x-y-2 2x 2y y-2- x

1 0 0 0 0 = ( x + y + 2) 2 x - ( x + y + 2 ) 2y 0 - ( x + y + 2) 0 = (x + y + 2)3

D = l b b2 1 c2 1 Using R3 Æ R3 – R2, R2 Æ R2 – R1, we get c

25. Write f(x) = D1 + xD2 where x x+a x 2

x x = a[(x + a)2 – x2] x+a

a2

a c-b

x x x+a

2

0

2

2

0

c -b

1 b+a 1 c+b = l(b – a) (c – b) (c – a) = l(b – a) (c – b)

= l(a – b) (b – c) (c – a) 30. Use C2 Æ C2 – C3 + C1 to show that C2 consists of all zeros. 31. We know 1 1 1 a b c = (a – b) (b – c) (c – a)

1 x x = 0 a 0 = a2 0 0 a f (x) = 2xa2 + a3 + xa2 = 3xa2 + a3

a2

\ f(2x) – f(x) = 3xa2 26. a + b + g = – a, b g + g a + a b = 0, a b g = – b. Applying C1 Æ C1 + C2 + C3, we get 1 b g D = (a + b + g ) 1 g a 1 a b

(1)

Using R3 Æ R3 – R2 and R2 Æ R2 – R1, we get 1 b D= – a 0 g -b 0 a -g

1

2

D= l b-a b -a

3

= 2 xa + a 1 x and D2 = 1 x + a 1 x

Thus,

a=

a a2 1

fi x+y+2=0 It passes through (–1, –1). a D1 = 0 0

1 1 1 ,b= ,g= a b c From (2) in solution question 26, we have D = (abc)2 (a + b + g) [(b – g)2 + (g – a)2 + (a – b)2] fi a + b + g = 0 fi a–1 + b–1 + c–1 = 0 29. Using C2 Æ C2 – l2C3, we can write where

Using C2 Æ C2 – C1, C3 Æ C3 – C1, we get

fi

= – a[b g + g a + a b – (a 2 + b 2 + g 2)] (2) = – a[0 – a2] = a3 27. In this case a + b + g = 0. Thus, from (1) in solution to question 26, we get D = 0. 28. We can write a b g 2 D = (abc) b g a g a b

g a -g b -a

= – a[(g – b ) (b – a) – (a – g )2]

b2

c2

1 a2

a3

Let D1 = 1 b2

b3

1 c2

c3

Using R3 Æ R3 – R2 and R2 Æ R2 – R1, we get 1

a2

a3

D1 = 0 b 2 - a 2

b3 - a 3

0 c 2 - b2

c 3 - b3

= (b – a) (c – b)

b + a b2 + a 2 + ba c+b

c 2 + b2 + cb

Determinants 4.47 3

Thus, D = 0 fi Ê a ˆ = 8

Applying C2 Æ C2 – bC1, we get D1 = (b – a) (c – b) = (b – a) (c – b)

Ë b¯

b + a a2 c+b

c

2

a2

b+a

c - a c2 - a2 [using R2 Æ R2 – R1]

b + a a2 c+a 1 = (a – b) (b – c) (c – a) (bc + ca + ab) Thus, l = 18 32. Show that D = (a0 + a1 + a2)x2 33. Using R2 Æ R2 + R1, we get = (b – a) (c – b) (c – a)

1 cos x 0 D= 0 1 sin + cos x 0 - 1 1 - (sin x + cos x )

Using R3 Æ R3 + R2, we get 1 cos x 0 D= 0 1 sin + cos x = 1 0 0 1 34. Using C1 Æ C1 + C2 + C3, we get 1 a b D = (x + a + b) 1 x b 1 b x Using R3 Æ R3 – R2, R2 Æ R2 – R1, we get 1 a D = (x + a + b) 0 x - a 0 b-a

b 0 x-b

= (x + a + b) (x – a) ( x – b) 35. Use R2 Æ R2 – R1, R3 Æ R3 – R1to obtain D = sinq cosq =

1 sin(2q) 2

36. Use C1 Æ C1 + C2 + C3 to obtain 1 b c D = (x + a + b + c) 1 x + b c 1 b x+c

Now use R2 Æ R2 – R1, R3 Æ R3 – R1 to obtain D = x2(x + a + b + c) Thus, D = 0 fi x = 0 or x = – (a + b + c) 37. Use C1 Æ C1 – C2. 38. Show D = – a3 + 8b3

fi a/b is a cube root of 8. 39. Using C1 Æ C1 + C2 + C3, we can write D = 2(1 + i)D1 where 1 1+ i i 1 1+ i i 1 i 1 D1 = = 0 -1 1- i = i 1 1 1+ i 0 -i 1

[using R2 Æ R2 – R1 and R3 Æ R3 – R1] Thus, D = 2(i – 1) 1 1 1 + c a b 1 1 1 40. Write D = abc 1 + a b c 1 1 1 1 + b c a and use C2 Æ C2 + C1, to show that D = 0 1

41. Applying C1 Æ C1 – C2 – C3 and reduce D to determinant in Example 36. 42. Applying R1 Æ R1 – xR2 we get a (1 - x 2 ) c (1 - x 2 ) D = ax + b cx + d u v Take 1 – x2 common – xR1 to obtain a c 2 D = (1 – x ) b d u v

p (1 - x 2 ) px + q w

from R1 and apply R2 Æ R2 p q w

43. Applying R1 Æ R1 + R3 to obtain 0 D = - sin q -1

0 1 - sin q

2 sin q = 2(sin2q + 1) 1

As 0 £ sin2q £ 1, we get D Œ [2, 4]. 44. Write D = D1 – D2 where b2 - ab b D1 = ab - a 2

bc - ac

a b2 - ab and c

ab - a 2

b2 – ab c

bc – ac

bc - ac

2

b b2 – ab

bc – ac

a ab – a 2

D2 = ab – a

In D1 use C1 Æ C1 – (b – a)C2 to show that D1 = 0.

4.48

Complete Mathematics—JEE Main

In D2 use C3 Æ C3 – (b – a)C2 to show that D2 = 0. Thus, D = 0. 45. We can write 6i 1 1 D = – 3iw 4 - 1 -1 = 0 20 i i \ x = 0, y = 0 46. Using C1 Æ C1 + C2 + C3, we get

49. Show that the determinant equals x3 + (a2 + b2 + c2) x = 0. fi x = 0 as x Œ R \ Number of real roots of the equation is one. 50. Note that C11 C12 C13 D1 = C21 C22 C23 C31 C32 C33 where Cij cofactor of (i, j)th element of È1 Í Íx Í 2 Îx

1 cos x cos x D = (sin x + 2cos x) 1 sin x cos x 1 cos x sin x

Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get 1 cos x cos x D = (sin x + 2cos x) 0 sin - cos x 0 0 0 sin x - cos x = (sin x + 2cos x) (sin x – cos x)2 D = 0 fi tan x = – 2, tan x = 1. p p £x£ , – 1 £ tan x £ 1. As – 4 4 Thus, tan x = 1 fi x = p/4. 47. Applying C1 Æ C1 + iC2, we get 0

i

-w

0

1

w2

w - iw 2

- w2

1

x + iy =

e- p i / 4

e- p i / 3

1

e - 2p i / 3

e2p i / 3

e 2p i / 3

x – iy = ep i / 4 ep i / 3

ep i / 4

ep i / 3

1

e2p i / 3

e - 2p i / 3

e 2p i / 3

1

= e -p i / 4 e -p i / 3

\ D1 = D = 49 51. Applying C3 Æ C3 – C1, we get sin a D = sin b sin g

D=

e

1

e

- pi 3

- 2p i 3

e

cos b cos b cos b

=0

log a

log b

log 7

2

log c

log(7 ) log(73 )

1 + cos2 q cos2 q

D = (2 + 4sin4q)

sin 2 q

cos2 q

sin 2 q

1 + cos2 q -1 D = 2(1 + 2sin4q) 0

sin 2 q 1 -1

D = 0 fi sin4q = – 1/2

0 e

1

Applying R3 Æ R3 – R2, R2 Æ R2 – R1, we get

0 - 2p i 3

1

1 + sin 2 q 1

= 2(1 + 2sin4q)

pi 4

= e

cos a cos a cos a

log(3) log(32 ) log(33 ) = (log 7) (log 3) D1 log a log b log c where D1 = =0 1 2 3 1 2 3 \ D =0 53. C3 Æ C3 + C1 + C2, gives

2iy = (x + iy) – (x – iy) 1

1

2

Now,

- pi 4

x2

52. Using R2 Æ R2 – R1 and R3 Æ R3 – R1, and Ê xˆ log x – log y = log Á ˜ , we get Ë y¯

= (w – iw2) (iw2 + w) = w2 – i2w4 = w2 + w = – 1 fi x = – 1, y = 0. 48. Taking conjugate, we get 1

x2 ˘ ˙ 1˙ ˙ x˚

x

-e

= (e – 2pi/3 – e2pi/3) [1 – 1] = 0 fi y = 0.

Now, 0 £ q £ 2p i 3

\

4q =

p fi 0 £ 4q £ 2p 2

7p 11p , 6 6

fi

q=

7p 11p , 24 24

1 0 0

Determinants 4.49

54. 2

( x -1)( x +1) ( x +1) (2 x +1)( x +1) P(x) = (2 x -1)( x +1) (2 x -1)( x + 3) (2 x + 3)(2 x -1) (2 x +1)(3 x - 2) (2 x -1)(3 x - 2) (4 x +1)(3 x - 2) x - 1 x + 1 2x + 1 = ( x + 1)(2 x - 1)(3 x - 2) x + 1 x + 3 2 x + 3 2x + 1 2x - 1 4x + 1 Using C2 Æ C2 – C1, C3 Æ C3 – 2C1, we get x -1 2 3 P(x) = ( x + 1)(2 x - 1)(3 x - 2) x + 1 2 1 2 x + 1 -2 -1 Using R2 Æ R2 – R1, R3 Æ R3 + R1, we get x -1 2 3 P(x) = ( x + 1)(2 x - 1)(3 x - 2) 2 0 -2 3x 0 2 P(x) = – 4(x + 1) (2x – 1) (3x – 2) (3x + 2) \ P(x) = 0 has four distinct roots. 2p 2p + i sin . 9 9

55. Put ar = wr where w = cos Note that w9 = 1 fi 1 D= w

3

w6

1 = w8. Now, w

w8

w7

2

w

w5

w4

w

=0

as C2 and C3 are proportional. 56. As the system has a non-zero solution

p b c D= a q c = 0 a b r Applying R3 Æ R3 – R2, R2 Æ R2 – R1, we get

p b c 0 =0 D= a- p q-b a- p 0 r-c Expanding along R3, we get – (a – p) (q – b)c + (r – c) [p(q – b) – b(a – p)] = 0. Dividing (p – a) (q – b) (r – c), we get

c p b + + = 0. r-c p-a q-b fi

c-r r p b-q q + + + + =0 r-c r–c p-a q-b q-b

p q r + + = 2. p-a q-b r-c Now use p+a 2 p - ( p - a) 2p = = - 1 etc. p-a p-a p-a 57. We can write D as 1 n + 1 (n + 1) (n + 2) D = n! (n + 1)! (n + 2)! 1 n + 2 (n + 3) (n + 2) 1 n + 3 (n + 4) (n + 3) fi

Applying R3 Æ R3 – R3, R2 Æ R2 – R1, we get 1 n + 1 (n + 1) (n + 2) D = n! (n + 1)! (n + 2)! 0 1 2 ( n + 2) 0 1 2 (n + 3) fi

D =2 n!(n + 1) ! (n + 2)!

58. Use R1 Æ R1 + R3 – 2R2 59. Using C1 Æ C1 + C2 + C3, we get

1 1- y 1- y D = (3 – y) 1 1 + y 1 - y 1 1- y 1+ y Applying R3 Æ R3 – R2, R2 Æ R2 – R1, we get 1 1- y 1- y D = (3 – y) 0 2 y 0 = (3 - y ) ( 4 y 2 ) 0 -2 y 2y \ D = 0 fi y = 0 or y = 3. 60. Write D = lD1 + l¢D2, where a am + a¢ m¢ an + a ¢ n¢ D1 = b bm + b¢ m¢ bn + b¢ n¢ and c cm + c¢ m ¢ cn + c¢ n¢ a¢ am + a ¢ m¢ an + a¢ n¢ D2 = b¢ bm + b¢ m¢ bn + b¢ n¢ c¢ cm + c¢ m¢ cn + c¢ n¢ In D1 apply C2 Æ C2 – mC1, C3 Æ C3 – nC1 to obtain a a ¢ m ¢ a ¢ n¢ D1 = b b ¢ m ¢ b ¢ n ¢ = 0 c c ¢ m ¢ c ¢ n¢ Similarly D2 = 0. Thus, D = 0 61. Imitate Example 35. 62. We have

4.50

D=

Complete Mathematics—JEE Main

1 m

1 m+3

1 m+6

1 1 1 (m + 3) (m + 2) (m + 6) (m + 5) m (m - 1) 2 2 2 Applying C3 Æ C3 – C2, C2 Æ C2 – C1, we get 1 m

D=

0 3

0 3

1 m (m - 1) 3 (m + 1) 3(m + 4) 2 2

Using C1 Æ C1 + C2 + C3, we get 1

0 cos C cos B 1 64. Write D = - 1 cos A a a cos C a cos B cos A - 1 Applying C1 Æ C1 + bC2 + cC3, we get

a b cˆ Ê D = pqr Á 1 + + + ˜ Ë p q r¯ \

lim

x Æ0

x2

1 1 1 1 0 0 = 2 1 2 = 2 1 2 =–1 0 1 1 0 1 1

a b aa +b D= b c ba + c = 0 aa + b ba + c 0 Applying R3 Æ R3 – aR1 – R2 to obtain

66. Applying C2 Æ C2 – w2C1 we find the second column of D becomes 0. 67. The system will have a non-trivial solution if p+a b c

q+b c =0 b r+c

a a

0 0 – (a a 2 + 2ba + c)

-1

D = pqr

a p a p

a p

b q 1+ b q

b q

c r c r 1+

c r

c

c

Using C3 Æ C3 – C2, C2 Æ C2 – C1, we get a 0 - (1 + a) D = b - (1 + b) (b + 1) -1 c +1 0 = – a( b + 1) (c + 1) – (1 + a)[b(c + 1) – (1 + b)] 1 a b =0 \ D=0fi + 1+ a 1+ b 1+ c 1 1 1 = 2. + + 1+ a 1+ b 1+ c 70. The system will have a non-trivial solution if D = 0 where l+5 l-4 1 D= l -2 l +3 1 l l 1 Applying C2 Æ C2 – lC3, C1 Æ C1 – lC3, we get

fi

Write 1+

aa +b ba + c

a b D= b c

[using R1 Æ R1 – R3]

D=

a b c + + = – 1. p q r

= – (aa2 + 2ba + c) (ac – b2) Note that D = 0 if a, b, c are in G.P. 69. The system will have a non-trivial solution if a a -1 D= b -1 b = 0.

cos x 1 1 sin x f ( x) = 2 1 2 2 x x tan x 1 1

f ( x)

D = 0, pqr π 0 fi

(1)

68. The system will have a non-zero solution if

a cos C cos B 1 D= 0 - 1 cos A = sin2A a 0 cos A - 1 65. For x π 0,

c r c r

a b cˆ b Ê D = pqr Á 1 + + + ˜ 1 1 + Ë p q r¯ q b c 1 1+ q r Apply R2 Æ R2 – R1, R3 Æ R3 – R1, we get

3

= 3 (m + 4 – m – 1) = 3 \ a+b+g=3 63. Use C2 Æ C2 – 2C1, C3 Æ C3 – 3C1 to obtain C2 and C3 are proportional.

b q

Determinants 4.51

5 -4 1 D= -2 3 1 = 7 π 0 0 0 1 Thus, the system cannot have a non-trivial solution. 71. Adding above the system of equations, we get (a + b + c) (x + y + z) = 0 fi x + y + z = 0 \ (b + c) (– x) – ax = b – c b-c etc. fi x=– a+b+c 72. The above system of equations will have a nontrivial solution if a b c D= b c a = 0 c But D = –

1 (a + b + c) 2 [(b – c)2 + (c – a)2 + (a – b)2]

6 6 1 -6

0 4

p2

p3

p

77. Statement-2 is true, see theory. The roots of x9 = p are p1/9 wr 2p 2p where w = cos + i sin . 9 9 value of determinant D depends on a1, a2, . . . a9. If we put ak = p1/9 w k, then 1 w8

a1 D1 = a2 a3

b1 + q c1 c1 + ra1 b2 + qc2 c2 + ra2 b3 + qc3 c3 + ra3

b1 D2 = b2 b3

b1 + q c1 c1 + ra1 b2 + qc2 c2 + ra2 b3 + qc3 c3 + ra3

w7

D = (p1/9)3 1

w

w2 = 0

1

w

w2

However, if a1 = p1/9, a2 = p1/9 w8, a3 = p1/9 w7, a4 = p1/9 w, a5 = p1/9 w5 a6 = p1/9 w4, a7 = p1/9 w2, a8 = p1/9 w3, a9 = p1/9 w6, then 1 w8 w7 D = (p1/9)3 w w2

\ f ¢¢¢(x) is a constant. 74. Write D = D1 + pD2 where

w5 w4 w3

w6

= p1/3 (2w2 – w7 – w6) π 0. 78. Statement-2 is true. and

Now show that D1 = D and D2 = pqr D. 75. The system will have a non-trivial solution if l +3 l +2 1 D= 3 l + 3 1 = 0. 2 3 1 Using R1 Æ R1 – R2 and R2 Æ R2 – R3, we get l D= 1 2

[using statement-2]

a b

\ D = 0, a + b + c π 0 fi a = b = c. Thus, a : b : c = 1 : 1 : 1 73. f ¢¢¢(x) =

If n is odd, then |A| = – |A| fi 2|A| = 0 fi |A| = 0. We have 0 sin(p / 12) cos(p / 12) Êpˆ D Á ˜ = - sin(p / 12) 0 sec(p / 12) = 0 Ë 4¯ - cos(p / 12) - sec(p / 12) 0

-1 0 2 l 0 =l +1 3 1

Note that there is no real value of l for which D = 0. 76. Let A be a skew symmetric matrix of order n, then |A¢| = (– 1)n |A| fi |A| = (– 1)n |A|

Using R2 Æ R2 – get 6 w=

2 R1 and R3 Æ R3 – 2i

3 R1, we

3 + 6i

0

3 ( 6 - 2 3 )i

0

2

(2 - 3 2 )i

6 ÎÈ2 3 – 3 6 – 2 3 + 2 6 ˘˚ i = – 6i 79. We first show statement-2 is true. Ê nˆ Ê nˆ Ê nˆ P(x) = 1 + Á ˜ x + Á ˜ x 2 + . . . + Á ˜ x n Ë1 ¯ Ë 2¯ Ë n¯ =

fi

Ê nˆ Ê nˆ Ê nˆ P¢(x) = Á ˜ + 2 Á ˜ x + . . . + n Á ˜ x n - 1 Ë1 ¯ Ë 2¯ Ë n¯

Ê nˆ P¢(0) = Á ˜ = coefficient of x in the expanË1 ¯ sion of P(x) \ Statement-2 is true. Note that D(x) consists of 6 terms of the form (1 + x)n. Thus, coefficients of x in D(x) = D¢(0) a1b1 a1b2 a1b3 1 1 1 But D¢(0) = 1 1 1 + a2 b1 a2 b2 a2 b3 1 1 1 1 1 1

fi

4.52

Complete Mathematics—JEE Main

1 1 1 + 1 1 1 =0 a3 b1 a3 b2 a3 b3 80. Statement-2 is not always true. For instance, the system of equations x + 2y + 3z = 1 2x + 3y + 4z = 2 3x + 4y + 5z = 4 has no solution but |A| = 0 For statement-1, let a b c D= b c a c a b Using C1 Æ C1 + C2 + C3, we obtain 1 b c D = (a + b + c) 1 c a = 0 1 a b Also, note that x = y = z satisfies each of the three equations. Thus, the system of equations has infinite number of solutions. 81. Let a b c D= b c a

c

a b

1 = - (a + b + c)[(b - c)2 + (c - a)2 + (a - b)2 ] 2 If a + b + c π 0 and a2 + b2 + c2 = bc + ca + ab, then (b – c)2 + (c – a)2 + (a – b)2 = 0 fi a=b=cπ0 [ a + b + c π 0] Thus, the system of equation reduces to x + y + z = 0 which is satisfied by infinite number of solutions. \ Statement-1 is true. If a + b + c = 0 and a2 + b2 + c2 π bc + ca + ab, then at least one following is true: b2 π ca, c2 π ab. a2 π bc, Suppose b2– ac π 0 Write first two equations as: ax + by = (a + b)z bx + cy = (b + c)z Eliminating y, we get (ac – b2)x = (ac – b2)z fi x= z Thus, from (1) equation, we get by = bz fi y = z \ x= y=z Therefore, statement-2 is false. 82. If (a1, b1), (a2, b2) (a3, b3) are non-collinear, then D1 π 0. Therefore statement-2 is true.

We can write D2 as a1 b1 0 b1 a1 0 D2 = a2 b2 0 b2 a2 0 = 0 a3 b3 0 a3 b3 0 for all values of a1, a2, a3, b1, b2, b3 ΠR. a + ib c + id 83. As -c + id a - ib = (a + ib) (a Рib) + (c Рid) (c + id) = a2 + b2 + c2 + d2 we get statement-2 is true. Now,

(a12 + b12 + c12 + d12 ) (a22 + b22 + c22 + d22 ) =

a1 + ib1 -c1 + id1

c1 + id1 a2 + ib2 a1 - ib1 - c2 + id2

c2 + id2 a2 - ib2

a + ib c + id = a 2 + b2 + c2 + d 2 -c + id a - ib where a = a1a2 – b1b2 + c1c2 – d1d2, b = a 2b 1 + a 1b 2 + c 1d 2 + c 2d 1, c = – a 2c 1 + b 2d 1 – b 2c 1 + a 2d 1, d = – b 2c 1 + a 2d 1 – b 1c 2 – b 1d 2 =

Level 2 84. We have l1 m1 2 D = l2 m2 l3 m3 a1 = b12 b13

b21 a2 b23

n1 n2 n3

l1 l2 l3

m1 m2 m3

n1 n2 n3

b31 b32 a3

where ak = lk2 + mk2 + nk2 = 1 for k = 1, 2, 3 and bij = li lj + mimj + ninj = 0 " i π j 1 0 0 Thus, D = 0 1 0 = 1 fi |D| = 1. 0 0 1 85. Using R1 Æ aR1, R2 Æ bR2, R3 Æ cR3, we get 2

ab2 c 2

abc ab + ac 1 2 2 a bc abc bc + ab D= abc 2 2 a b c abc ac + bc bc 1 ab + ac a2 b2 c 2 = ca 1 bc + ab abc ab 1 ac + bc

Determinants 4.53

using C3 Æ C3 + C1, we get C2 and C3 are propotional. Thus, D = 0 86. Using C1 Æ C1 – (x2/3)C3, C2 Æ C2 – (2x/3) C3, we get - 5x + 3 - 5 3 D(x) = x+9 1 9 - 6 x + 9 - 6 21 Using C1 Æ C1 – xC2, we get 3 -5 3 D(x) = 9 1 9 = a constant. 9 -6 21 Thus, a = 0. Alternate Solution Replacing x by 1/x, we get 1 Ê 1ˆ Ê 1ˆ -5 Á ˜ + 3 2 Á ˜ - 5 2 Ë x¯ Ë x¯ x Ê 1ˆ 3Á 2 ˜ Ëx ¯

Ê 1ˆ = aÁ 3˜ Ëx ¯

1 - 5x + 3x fi 3 + x + 9x

2

7 - 6x + 9x Putting 1 a= 3 7

x= 2 6 14

2

2 - 5x 6+x

b2 (1 - cos q )

a2 (1 - cos q )

b2 + (c 2 + a2 )cos q

a2 (1 - cos q )

b2 (1 - cos q )

D=

c 2 (1 - cos q ) c 2 + (a2 + b2 )cos q Using C1 Æ C1 + C2 + C3 and a2 + b2 + c2 = 1, we get b 2 (1 - cos q )

1

2 2 D = 1 b + (1 - b ) cos q

3

b 2 (1 - cos q )

1

1 Ê 1ˆ +9 6Á ˜ +1 9 Ë x¯ x Ê 1ˆ Ê 1ˆ -6 Á ˜ + 9 14 Á ˜ - 6 21 Ë x¯ Ë x¯ Ê 1ˆ Ê 1ˆ + bÁ 2 ˜ + cÁ ˜ + d Ëx ¯ Ë x¯ 2

a2 + (b2 + c 2 )cos q

c 2 (1 - cos q )

c 2 (1 - cos q ) c 2 (1 - cos q ) c 2 + (1 - c 2 ) cos q

Using R2 Æ R2 – R1, R3 Æ R3 – R1, we get

+

Ê 1ˆ 7Á 2 ˜ Ëx ¯

88. First multiply C1 by a, C2 by b, C3 by c, followed by multiplying R1 by 1/a, R2 by 1/b and R3 by 1/c, we get

1 b2 (1 - cos q ) c 2 (1 - cos q ) cos q D= 0 0 = cos2 q cos q 0 0 89. We have - sin(a + q ) sin(a + q ) 1

3 2

9 = a + bx + cx + dx

3

14 - 6 x 21

0, we get 3 9 =0 21 [ C1 and C2 are proportional]

87. The given system of equations will have a nontrivial solution if a+t b c D= a b+t c =0 a b c+t Using C1 Æ C1 + C2 + C3, we get 1 b c c =0 D = (a + b + c + t) 1 b + t 1 b c+t Using C2 Æ C2 – bC1, C3 Æ C3 – cC1, we get D = (a + b + c + t)t2 = 0 fi t = 0, – (a + b + c) Thus, there are just two distinct values of t.

dA = - sin(b + q ) sin(b + q ) 1 dq - sin(g + q ) sin(g + q ) 1

cos(a + q ) cos(a + q ) 1 + cos(b + q ) cos(b + q ) 1 cos(g + q ) cos(g + q ) 1 =0+0=0 fi A is independent q. Thus, A(a, b, g, q ) = A (a, b, g , 0 ) p 2p ˆ p ˆ Ê p Ê p fi A Á - , 0, , ˜ = A Á - , 0, , 0˜ Ë 2 ¯ Ë 2 13 2 2 ¯ 0 -1 1 = 1 0 1 =2 0 1 1 90. We have D = (a – b) (b – c) (c – a) = – 2. As a > b > c, a – b, b – c are positive integers and c – a is a negative integers. Only possibilities are a – b = 2, b – c = 1, c – a = –1

(1)

4.54

Complete Mathematics—JEE Main

or a – b = 1, b – c = 2, c – a = –1

(2)

or a – b = 1, b – c = 1, c – a = –2

(3)

(1) and (2) lead us to 0 = 2. \ a – b = 1, b – c = 1, c – a = –2. Now, 3a + 7b – 10c = 3(a – c) + 7(b – c) = 13 91. Write cos A sin A 0 D = cos B sin B 0 cos C sin C 0

cos P - sin P 0 cos Q - sin Q 0 cos R - sin R 0

=0

93. For each D ΠB, there exist D1 ΠC where D1 is obtained by interchanging 1st and 2nd row of D, similarly, for each D ΠC there exists D1 ΠB. \ B and C has same number of elements. 94. |a|2 = |x|2 + |y|2 + |z|2 + x y + x y + x z + x z + yz + yz \ |a|2 + |b|2 + |c|2 = 3(|x|2 + | y|2 + |z|2) + ( x y + x y + x z + x z + y z + y z) (1 + w + w2) fi

| a |2 + | b |2 + | c |2

95. Write the determinant as = abc D1 – D2

Alternate Solution

1 a a2

Write D = cos P D1 – sin P D2 cos A cos( A + Q) cos( A + R) where D1 = cos B cos( B + Q) cos( B + R) cos C cos(C + Q) cos(C + R) sin A cos( A + Q) cos( A + R) and D2 = sin B cos( B + Q) cos( B + R) sin C cos(C + Q) cos(C + R) In D1, use C2 Æ C2 – cosQ C1 C3 Æ C3 – cosR C1

=0

[

C2 and C3 are propotional]

Similarly, D2 = 0 D=0

92. Expanding along C1, we get f(x) = 2cosx

where D1 = 1 b b2 1 c c

= (a – b) (b – c) (c – a)

2

a a2 1 and D2 = b b2 1 = D1 c c2

1

\ (a – b) (b – c) (c – a) (abc – 1) = 0 Since, a, b, c are distinct, we get abc – 1 = 0 or abc = 1

cos A sin A sin Q sin A sin R D1 = cos B sin B sin Q sin B sin R cos C sin C sin Q sin C sin R

Thus,

=3

| x |2 + | y |2 + | z |2

2 cos x 1 1 0 – 1 2 cos x 1 2 cos x

1 4 4 96. | P | = |adj P| = 2 1 7 = 4 1 1 3 2

fi |P| = ± 2 97. Using R2 Æ R2 + 5R1, R3 Æ R3 – 4R1, we get 1 -3 4 0 x - 13 22 0 13 x - 22

=0

= 8cos3 x – 4cos x = 4cos x cos2x

fi (x – 13) (x – 22) – (13)(22) = 0

Êpˆ Êpˆ fi f Á ˜ = –1, f Á ˜ = 0. Ë 2¯ Ë 3¯

fi x(x – 35) = 0 fi x = 0, 35

f¢(x) = – 4[sinx cos2x + 2cosx sin2x] = – 4[sin3 x + cosx sin2x] Êpˆ È Êpˆ Ê 2p ˆ ˘ fi f ¢ Á ˜ = - 4 Ísin p + cos Á ˜ sin Á ˜ ˙ ¯ Ë 3 ¯˚ Ë 3¯ Ë 3 Î =–

3

Previous Years’ AIEEE/JEE Main Questions 1. As ax2 + 2bx + c = 0 has equal roots, b2 – ac = 0 Using R3 Æ R3 – xR1 – R2, we get a b ax + b D = b c bx + c 0 0 a

Determinants 4.55

where a = 0 – x(ax + b) – (bx + c) = – (ax2 + 2bx + c) \

fi

D = a (ac – b2) = 0

2. As l, m, n are pth, qth, rth terms of a G.P., log l, log m, log n are pth, qth, rth terms of an A.P. Let a and its common difference be d. Now, a + ( p - 1)d D = a + (q - 1)d a + (r - 1)d

p 1 q 1 r 1

Using C1 Æ C1 – dC2 – (a – d)C3, we get 0 p 1 D= 0 q 1=0 0 r 1

3. If n is a multiple of 3, we get each element of D becomes 1. \

D=0

If

n = 3k + 1, then

a, c, b are in H.P.

5. Let R be the common ratio of the G.P. Applying C3 Æ C3 – C2 and C2 Æ C2 – C1, we get that the given determinant log an = log an +1 log an + 2

log R log R log R log R = 0 log R log R

6. Applying C1 Æ C1 + C2 + C3, we get 1 + 2 x + (a 2 + b 2 + c 2 ) x (1 + b 2 ) x (1 + c 2 ) x f(x) = 1 + 2 x + (a 2 + b 2 + c 2 ) x

1 + b2 x

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

(1 + c 2 )xx 1 + c2 x

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

1 + b2 x

(1 + c 2 ) x

1 (1 + b 2 ) x

1 + c2 x

Applying R2 Æ R2 – R1 and R3 Æ R3 – R1, we get

1

w

D= w

w2

w2

1

w2 1 =0

[use C1 Æ C1 + C2 + C3]

1 (1 + b 2 ) x (1 + c 2 ) x f ( x) = 0 1- x 0 0 0 1- x = (1 – x)2

w

which is a polynomial of degree 2.

If n = 3k + 2, then 1

w2

w

D = w2

w

1 =0

w

1

7. If A [use C1 Æ C1 + C2 + C3]

w2

4. As the system has a non-zero solution 1 2a a 1 3b b = 0 1 4c c 1/ a 2 1 1/ b 3 1 = 0 1/ c 4 1 Using R1 Æ R1 – 2R2 + R3, we get fi

1 2 1 - + =0 a b c

1/ a - 2 / b + 1/ c 0 0 1/ b 3 1 =0 1/ c 4 1

a 1 1 1 1 1 | A | = 1 a 1 = (a + 2) 1 a 1 1 1 a 1 1 a [using C1 Æ C1 + C2 + C3] 1 1 1 = (a + 2) 0 a - 1 0 0 0 a -1 = (a + 2)(a – 1)2 If |A| π 0, the system has a unique solution. If |A| = 0, then a = – 2 or a = 1. For a = 1, the system of equations becomes x + y + z = 0 which For a = – 2, the system of equation becomes – 2x + y + z = – 3 x – 2y + z = – 3

4.56

Complete Mathematics—JEE Main

Applying R2 Æ R2 – R1, R3 Æ R3 – R1, we get

x + y – 2z = – 3 which on adding becomes 0 = – 9 Thus, the system has no solution if a = – 2. 1 1 1 8. D = 1 1 + x 1 1 1 1+ y

1 b c D1 = 0 c - b a - c 0 a-b b-c = – (b – c)2 – (a – b)(a – c) = – (a2 + b2 + c2 – bc – ca – ab) 1 D1 = - [(b - c)2 + (c - a)2 + (a - b)2 ] < 0 2 As a + b + c > 0, we get

Applying R2 Æ R2 – R1 and R3 Æ R3 – R1, we get

fi

1 1 1 D = 0 x 0 = xy 0 0 y

D = (a + b + c) D1 < 0 11. Using C1 Æ C1 – C3 in D, we get

which is divisible by both x and y. 9. Let D 2 =

a +1 a -1

b +1 b -1

0 D= 0 2 cos a

c -1 c +1

( -1)n + 2 a ( -1)n +1 b ( -1)n c

sin a cos a - sin a

cos a sin a - cos a

= 2 cos a (sin2 a – cos2 a) a + 1 a - 1 ( -1)n + 2 a

= –2 cos a cos 2a

D 2 = b + 1 b - 1 ( -1)n +1 b c -1 c +1

D = 0 for a = p/4 Π(0, p/2).

( -1)n c

This is the only value of a lying in (0, p/2) for which D = 0.

(-1) n+2 a a + 1 a - 1 = (-1) 2 (-1) n+1 b (-1) n c

The system of linear equation will have a non-trivial solution if and only if

b +1 b -1 c -1 c +1

Thus, n+ 2 a a + 1 a - 1 ( -1) a a + 1 a - 1 D = -b b + 1 b - 1 + ( -1)n+1 b b + 1 b - 1 c c -1 c +1 ( -1)n c c - 1 c + 1

a + ( -1)n + 2 a

a +1 a -1

n +1

b b +1 b -1

c + ( -1)n c

c -1 c +1

= -b + ( -1)

1 sin a D1 = 1 cos a 1 - sin a

cos a sin a = 0 - cos a

Using R2 Æ R2 + R1, we get 2 0 D1 = 1 cos a 1 - sin a fi fi

0 sin a = 0 - cos a

2(– cos2 a + sin2 a) = 0 – 2 cos 2a = 0

This is true for only one value of a Π(0, p/2) n is any odd integer. \

D = 0 if n is any odd integer.

10. Using C1 Æ C1 + C2 + C3, we get D = (a + b + c) D1 where 1 b c D1 = 1 c a 1 a b

viz, a = p/4. Thus, statement-1 is also true. However statement-2 is not a correct reason for statement-1. 1+1+1 12. D = 1 + a + b 1+ a2 + b2

1+ a + b

1+ a2 + b2

1+ a2 + b2

1 + a3 + b3

1 + a3 + b3 1 + a 4 + b4

Determinants 4.57

1 =1

1 a

1 a

2

1 1 b 1

1 a

2

b 1 a

1 where D1 = 1

1 a

we get

1 b = D12

2

b

b1 x + c1 | A( x ) | = 2 b2 x + c2 b3 + c3

2

1 b

2

Applying C1 Æ C1 – xC2, we get

0 a -1

c1 | A( x ) | = 2 c2 c3

0 b -a

\

1 a2 -1 b2 - a2 = (a - 1)(b - a )

1 1 a +1 b +a

\

b1 b2 b3

a1 a2 a3

|B(x)| is independent of x.

15. Using R3 Æ R3 – R2 and R2 Æ R2 – R1, we get a2 D = 2 al + l 2 -4al

= (a – 1) (b – 1) (b – a) Thus,

a1 a2 a3

2

1 a b Applying C3 Æ C3 – C2 and C2 Æ C2 – C1, we get 1 D1 = 1

b1 b2 b3

D = (a – 1)2 (b – 1)2(a – b)2

b2

c2

2 bl + l 2 -4bl

2cl + l 2 -4cl

Take – 4l common from R3, and applying R2 Æ R2 – 2lR3, we get

K=1

13. As the given system of equations has a non-trivial solutions.

a2 D = -4l 3 1 a

a - 1 -1 -1 D = -1 b - 1 -1 = 0 -1 -1 c - 1

a2 = l (4l 2 ) a 1

Write a -1 -1 1 -1 -1 D = 0 b - 1 -1 - 1 b - 1 -1 0 -1 c - 1 1 -1 c - 1 1 0 0 = a[(b - 1)(c - 1) - 1] - 1 b 0 1 0 c

b2 1 b

\

c2 1 c b2 b 1

c2 c 1

k = 4l2 n -1

16. As  r = r =1

n -1 1 n(n - 1), Â (2r - 1) = (n - 1)2 2 r =1

n -1

and  (3r - 2) = r =1

[use C2 Æ C2 + C1, C3 æ C3 + C1]

=

= a (bc – b – c) – bc

3 n(n - 1) - 2(n - 1) 2 1 (n - 1)(3n - 4) 2

Thus,

As D = 0, we get ab + bc + ca = abc

n(n - 1)/ 2 (n - 1)2 n -1 Â Dr = n / 2

n -1

14. We have |B(x)| = |(A(x))TA(x)| T

r =1

n(n - 1)/ 2 (n - 1)2

(n - 1)(3n - 4)/ 2 a (n - 1)(3n + 4)/ 2

= |A(x) | |A(x)| = |A(x)| |A(x)| = |A(x)|2 Let pi(x) = ai x2 + bi x + ci (i = 1, 2, 3) Applying C1 Æ C1 -

1 2 x C3 , C2 Æ C2 - xC3 . 2

Using R3 Æ R3 – R1, we get n(n - 1)/ 2 (n - 1)2 n -1 Â Dr = n / 2 r =1 0 0

n -1

(n - 1)(3n - 4)/ 2 a 4(n - 1)

4.58

Complete Mathematics—JEE Main

Expanding along R3, we get n -1

 D r = 4(n - 1) r =1

n(n - 1)/ 2 (n - 1)2 n/2 n -1

2 2 -1 4 3 0 = a - 12 6 1 1 fi 12 = a – 12

n -1 n -1 Ê nˆ = 4(n - 1) Á ˜ (n - 1) =0 Ë 2¯ 1 1 17. The system will have a non-trivial solution if 2-l D= 2 -1

-2 -3 - l 2

1 2 =0 -l

Using R1 Æ R1 + R3 , R2 Æ R2 + 2R3, 1- l 0 1- l D= 0 1 - l 2 - 2l = 0 -1 2 -l 1 0 1 (1 - l ) 0 1 2 = 0 -1 2 - l 2

fi

Using C3 Æ C3 – C1, we get 1 0 0 2 (1 - l ) 0 1 2 =0 -1 2 - l + 1

fi a = 24 19. Let x be any real number and y = 1, z = 1, then x 1 1 D= 1 1 1 =0≥0 1 1 1 and xyz = x can take any real value, thus, minimum value of xyz does not exist. 20. The given system of equations has a non-trivial solution if 1 l -1 l -1 -1 = 0 1 1 -l fi l – l – l – 1 + 1 + l3 = 0 fi l (l – 1) (l + 1) = 0 fi l = 0, –1, 1 When l = 0, x = –y = z = k

0

When l = –1, x = 0, y = –z = k When l = 1, y = 0, x = –z = k

0 0

2

fi (1 – l) [ – l + 1 – 4] = 0 fi l = 1, – 3 Thus, the set contains two elements 18. Using R2 Æ R2 – R1 – R3, we get x2 + x -4

D=

x +1 0

x-2 0

x2 + 2 x + 3 2 x - 1 2 x - 1

Thus, there are three values of l, for which the system of equations has a non-trivial solutions. 21. Using C1 Æ C1 + C2 + C3, we get 1 sin x sin x D = (2 sin x + cos x) 1 cos x sin x 1 sin x cos x Using C2 Æ C2 – (sin x) C1

Expanding along R2, we get

and C3 Æ C3 – (sin x) C1,

x +1 x - 2 2x - 1 2x - 1 Using C2 Æ C2 – C1, we get

we get

D = -( -4)

D=4

x + 1 -3 2x - 1 0

= 4(3) (2x – 1) = 24x – 12

1 0 0 D = (2 sin x + cos x) 1 cos x - sin x 0 1 0 cos x - sin x = (2 sin x + cos x) (cos x – sin x)2 1 \ D = 0 fi tan x = – , 1 2 As –p/4 £ x £ p/4, –1 £ tan x £ 1

\ a = 24

and tan x is one-to-one in the interval [– p/4, p/4].

TIP : Put x = 1 to obtain

Thus, there are two values of x.

Determinants 4.59

Thus, the system has non-trivial solution for no value of l.

Previous Years’ B-Architecture Entrance Examination Questions

a a

2

3. Â D k =

(b – 1)x = 0

k =1

n

In this case y + z = 0, by + z = 0

where a = Â 2(3k -1 )

fiy=0

(b – 1)y = 0

k =1

Thus, z = 0

2(3n - 1) = 3n - 1, 3 -1

=

That is, if b π 1, then the system has only trivial solution. \

b=1

n

b = Â 3(4 k -1 ) = k =1

3(4 n - 1) = 4n - 1 4 -1

n

and c = Â 4(5k -1 ) =

(a – 1) x = 0.

k =1

If a π 1, x = 0 and y + z = 0

a Thus, Â D k = a k =1 a n

tion. If a = 1, all the three equations become identical to x + y + z of solutions. b = 1 fi (b – 1)2 = 0

2 -1 a12

2 -2 a13

4. det(B) = 2a21

a22

2 -1 a23

22 a31

2a32

a33

a11

fi b2 = 2b – 1

22 a11

l +3 l +2 1 D= 3 l +3 1 =0 2 3 1

n

5. Â D r = r =1

a12 a22 a32

2

a33 a13 a23 a33

a a

b b

c g

2 n - 1 3n - 1 5n - 1 n

where

4x + 3y + z = 0

(i)

3x + 4y + z = 0

(ii)

2x + 3y + z = 0

(iii)

a = Â 2r -1 = 2 n - 1 r =1 n

b = Â 2(3r -1 ) = r =1

From (i) and (iii) x = 0

y = 0, z = 0

2a32

= det (A) = 5

For l = 1, the system becomes

fi

a23

a11 = (2 )(2 )(2) a21 a31

fil=1

4y + z = 0

= (2 -2 )(2 -1 ) 22 a21 2a22

-3

fi D = l(l + 1) – (l – 1) = 0

and

a13

2 a31

l +1 l -1 0 D= 1 l 0 =0 2 3 1

From (i) and (ii) 3y + z = 0

2a12

2

Using R1 Æ R1 – R3, R2 Æ R2 – R3, we get

fi (l – 1) = 0

4(5n - 1) = 5n - 1 5 -1

b c b g =0 b c

2. As the system has a non-trivial solution,

2

c g

3n - 1 4 n - 1 5n - 1

If b π 1, x = 0. fi

b b

2(3n - 1) 3 -1

= 3n – 1 n

and

c = Â 4(5r -1 ) = r =1

= 5n – 1

4(5n - 1) 5 -1

4.60

Complete Mathematics—JEE Main

a  Dr = a r =1 a n

Thus,

b c b g =0 b c

x1

x4

6. Taking e common from R1, e common from R2 and ex7 from R3, we get

fi

1 2 1 abc ÊÁ - + ˆ˜ ( -1) = 0 Ë a b c¯

fi

2ac = ab + bc

9. Using R1 Æ R1 – R2, R3 Æ R3 – R2, we get 0 a-c b-a 1 c a =0 0 b-c c-a

D = e x 1 + x 4 + x 7D 1 where 1 e3 d

e6 d

fi

(a – c)2 + (b – a) (b – c) = 0

e6 d = 0

fi

a2 – 2ac + c2 + b2 – ab – bc + ac = 0

fi

a2 + b2 + c2 – bc – ca – ab = 0

where d is common difference of A.P.

fi

(a – b)2 + (b – c)2 + (c – a)2 = 0

\D=0

fi

a=b=c

fi

sin A = sin B = sin C =

D1 = 1 e3d 1 e

3d

e

6d

5 common from C2 and C3, we get

7. Taking

13 + 3 where D1 =

2

x = 2.

1

15 + 26

5

2

\

y – z = 1 and – 4y + 4z = a – 2

3 + 65

3

5

fi

4y – 4z + (– 4y + 4z) = 4 + (a – 2)

fi

0=a+2

Applying, C1 Æ C1 - 3C2 - 13C3 , we get

D1 =

3 2

Thus, sin2 A + sin2 B + sin2 C = 9/4

D = 5D1

- 3

fiA=B=C

2

1

0

5

2

0

3

5

= - 3 (5 - 6 ) Thus, D = 5 3 ( 6 - 5) 8. As the given system has a non-zero solution 1 2a a 1 3b b = 0 1 4c c 1/ a 2 1 fi abc 1/ b 3 1 = 0 1/ c 4 1

Thus, the system has no solution for a π –2. If a number of solutions, since y – z number of solutions. 11. Using C1 Æ C1 + C3, we get 1 tan q + sec2 q D= 0 cos q 0 -4

3 sin q 3

= 3 cos q + 4 sin q dD = – 3 cos q + 4 sin q dq dD sin q cos q 1 =0fi = = dq 4 3 5 fi sin q = 4/5, cos q = 3/5 Max D = max {D(0), D(sin–1 (4/5)), D(1)}

Using R1 Æ R1 – 2R2 + R3, we get 1 2 1 - + 0 0 a b c abc 1/ b 3 1 =0 1/ c 4 1

Ï 24 ¸ 24 = max Ì3, , 4˝ = Ó 5 ˛ 5 min D = 3. \ D Œ[3, 5]

CHAPTER FIVE

Matrices

THE ALGEBRA OF MATRICES A matrix is a rectangular array of numbers. The numbers may be real or complex. It may be represented as È a11 a12  a1n ˘ Ía a22  a2 n ˙ ˙ A = Í 21 Í  ˙ Í ˙ Î am1 am 2  amn ˚ or as A = (aij)m ¥ n. A matrix with m rows and n columns is called as m ¥ n matrix and the size (or dimension) of this matrix is said be m ¥ n. Two matrices are said to be equal provided they are of the same size and corresponding elements are equal. For example È-1 2 5 ˘ Èa b c ˘ Íd e f ˙ = Í 7 3 11˙ Î ˚ Î ˚ if and only if a = – 1, b = 2, c = 5, d = 7, e = 3 and f = 11. Definitions A matrix A = (aij ) m ¥ n is said to be a (i) square matrix if m = n (ii) row matrix if m = 1 (iii) column matrix if n = 1 (iv) null or zero matrix if aij = 0 " i and j (v) diagonal matrix if m = n and aij = 0 "iπj (vi) Scalar matrix if m = n and aij = 0 " i π j and aii = l "i (vii) Unit or identity matrix if m = n and aij = 0 " i π j and aii = 1 " i (viii) Upper triangular matrix if m = n and aij = 0 "i>j (ix) Lower triangular matrix if m = n and aij = 0 "i 1, then a n × n matrix cannot have inverse in the above form.

Èa b ˘ Example 2: Let A = Í ˙ , a, b, c, d Œ R Îc d ˚ If A5 = A3 + I, then A is (a) a symmetric matrix (b) a skew symmetric matrix (c) an invertible matrix (d) none of these Ans. (c) Solution: A5 = A3 + I fi A(A4 – A2) = (A4 – A2) A = I fi A is invertible and A–1 = A4 – A2.

Example 3: Let A and B be two 3 × 3 invertible matrices. If A + B = AB, then (a) A–1 + B –1 = O (b) A–1 + B –1 = B –1 A–1 (c) I – A–1 is invertible (d) B –1 + I is invertible Ans. (c) Solution: A + B = AB fi A = AB – B = (A – I)B. Premultiplying by A–1 we get I = A–1A = (I – A–1)B –1 fi I – A is invertible. Example 4: Let A be a 3 × 3 matrix and ÏÊ xˆ ¸ ÔÁ ˜ Ô S = Ì y x, y, z Œ R ˝ ÔÁË z ˜¯ Ô Ó ˛ Define f : S Æ S by Ê xˆ Ê xˆ Á ˜ f y = A Á y˜ Á ˜ Á ˜ Ë z¯ Ë z¯ Ê x ˆ Ê 0ˆ Suppose f Á y˜ = Á 0˜ fi x = y = z = 0. Á ˜ Á ˜ Ë z ¯ Ë 0¯ Then (a) f is one-to-one (b) f cannot be onto (c) A is not invertible (d) A = O Ans. (a) Ê x1 ˆ Ê x2 ˆ Á ˜ Solution: Let y1 and Á y2 ˜ be such Á ˜ Á ˜ Ë z1 ¯ Ë z2 ¯ that

Ê x1 ˆ Ê x2 ˆ Á ˜ f y1 = f Á y2 ˜ Á ˜ Á ˜ Ë z1 ¯ Ë z2 ¯

fi

Ê x1 ˆ Ê x2 ˆ A Á y1 ˜ = A Á y2 ˜ Á ˜ Á ˜ Ë z1 ¯ Ë z2 ¯

TIP If P(x) is a polynomial such that P(0) π 0 and P(A) = O, then A is invertible.

fi

Ê x1 - x2 ˆ Ê 0ˆ A Á y1 - y2 ˜ = Á 0˜ Á ˜ Á ˜ Ë z1 - z2 ¯ Ë 0¯

Complete Mathematics—JEE Main

5.8

Ê x1 - x2 ˆ Ê 0ˆ f Á y1 - y2 ˜ = Á 0˜ Á ˜ Á ˜ Ë z1 - z2 ¯ Ë 0¯ x1 – x2 = 0, y1 – y2 = 0, z1 – z2 = 0 x 1 = x 2, y 1 = y 2, z 1 = z 2

fi fi fi

Ê x1 ˆ Áy ˜ = or Á 1˜ Ë z1 ¯ Thus, f is one-to-one.

Ê x2 ˆ Áy ˜ Á 2˜ Ë z2 ¯

Ê a bˆ , where a, b, c, d Œ R. Example 5: Let A = Á Ë c d ˜¯

Example 8: For 1 £ i, j £ 3, let p /2

and let (a) (b) (c) (d) Ans. (b)

A = (aij)3×3. Then A is a singular matrix AX = B has a unique solution for every 3×3 matrix B A is a skew-symmetric matrix. A2 = I

Solution: For 1 £ i £ 3, p /2

cos(ix )cos(ix ) dx

0

[integrand is an even function] p /2

Ê a bˆ Example 6: Let A = Á , where a, b, c, d Œ R. Ë c d ˜¯

=

Ú

[1 + cos(2ix )] dx

0 p /2

sin(2ix ) ˆ ˘ Ê = Ëx+ 2i ¯ ˚˙0 For 1 £ i, j £ 3, i π j,

Then

=

p p +0= 2 2

p /2 2

2

2

2

det (A) £ a + b c + d det (A) £ (a + b) (c + d) det (A) £ ac + bd det (A) £ (| a | – | b |) (| c | – | d |)

aij = 2

Solution: det(A) = ad – bc £ |ad – bc| £ | a || d | + | b || c | Now, (a2 + b2) (c2 + d2) – (| a || d | + | b || c |)2

=

=

| a || d | + | b || c | £

a +b

2

c +d

2

Example 7: Let A be a 3×3 matrix such that det(A) = –2. Then det (–2 A–1) is equal to (a) 4 (b) – 4 (c) 8 (d) – 2 Ans. (a) Solution: As A–1 is a 3 × 3 matrix, det (–2 A–1) = (–2)3 det (A–1) = (–2)3 (det(A))–1 1 = (-8) Ê- ˆ = 4 Ë 2¯

Ú

[cos((i + j ) x ) + cos((i - j ) x )] dx

0

\

Thus,

sin((i + j )p / 2) sin((i - j )p / 2) + i+ j i- j

1 2 a12 = - + 1 = = a21 3 3 a13 = 0 = a31, and a23 =

= (| ac | – | bd |)2 ≥ 0 2

cos(ix )cos( jx ) dx

0

– (a2d2 + b2c2 + 2 | a || d | | b || c |) 2

Ú

p /2

= a 2c 2 + b 2c 2 + a 2d 2 + b 2d 2

fi

Ú

aii = 2

(b) det(A) ≥ k2 (d) det(A) £ k

Solution: det(A) = ad – bc £ |ad – bc| fi det (A) £ | a || d | + | b || c | £ 2k2

(a) (b) (c) (d) Ans. (a)

cos(ix )cos( jx ) dx

-p / 2

If a , b , c , d £ k, where k > 0, then (a) det(A) ≥ 2k2 (c) det(A) £ 2k2 Ans. (c)

Ú

aij =

6 1 + 1 = = a32 5 5

Ê 1 2/3 0 ˆ A = Á 2 / 3 1 6 / 5˜ Á ˜ Ë 0 6/5 1 ¯

2 Using C2 Æ C2 - C1 , we get 3 1 0 0 det (A) = 2 / 3 5 / 9 6 / 5 0 6/5 1 5 36 199 =π0 9 25 225 As det (A) π 0, A–1 exists. Therefore, AX = B has a unique solution viz. X = A–1B. =

Matrices 5.9

Example 9: The number of values of l for which there exist a non-zero 3×3 matrix A such that A¢ = lA is: (a) 0 (b) 1 (c) 2 (d) infinite Ans. (c) Solution: We have A = (A¢)¢ = (lA)¢ = lA¢ = l2A. As A π 0, l2 = 1 fi l = ±1. For l = 1 all non-zero symmetric matrices can work as A. For l = –1, all non-zero skew symmetric matrices can work as A. È1 a˘ Example 10: Let A = Í ˙ , where a > 0. Sum of the Î0 1˚ 1 1 1 series S = trace (A) + trace Ê Aˆ + trace ÊÁ A2 ˆ˜ + trace ÊÁ A3 ˆ˜ Ë2 ¯ Ë 23 ¯ Ë 22 ¯ + … is

(a) 3 (c) 6 Ans. (b)

(b) 4 (d) 8

È0 a ˘ Solution: A = I + B, where B = Í ˙ Î0 0 ˚ As B2 = O, we get Br = O " r ≥ 2. Ar = (I + B)r = I + rB

Thus,

È1 ra ˘ =Í ˙ Î0 1 ˚

" r ≥ 1.

1 1 ˆ Ê 1 trace Á r Ar ˜ = r (2) = r -1 ¯ Ë2 2 2

fi \

S = 2 +1+

1 2

2

+

" r ≥ 1. 1

2

3

+ =

2 1-

1 2

=4

LEVEL 1 Straight Objective Type Questions 2 È1 4 ˘ È x y ˘ Example 11: If Í = Í ˙ , y < 0 then x– y + z ˙ Î2 0 ˚ Î z 0 ˚ is equal to (a) 5 (b) 2 (c) 1 (d) – 3 Ans. (a)

2

Solution: By the equality of two matrices, x = 1, y = 4, z=2 fi x = 1, y = – 2, z = 2 as y < 0. \ x – y + z =1 + 2 + 2 = 5 È 2˘ Í Example 12: If A = [1 – 2 3], B = -3˙ , then AB is equal Í ˙ ÎÍ -1˚˙ to È 2˘ (a) Í-3˙ Í ˙ ÍÎ -1˙˚ (c) [2 6 – 3] Ans. (d)

È 2˘ (b) Í 6 ˙ Í ˙ ÍÎ-3˙˚ (d) none of these

È 2˘ Solution: AB = [1 -2 3] Í-3˙ = [2 + 6 - 3] = [5] Í ˙ ÍÎ -1˙˚ È -i 0 ˘ Example 13: If A = Í ˙ , then A¢ A is equal to Î0 i˚ (a) I (b) – iA (c) – I (d) iA

Ans. (c) Solution: We have È-i 0 ˘ È-i 0 ˘ È-1 0 ˘ A¢ A = Í ˙=Í ˙ =–I ˙Í Î 0 i ˚ Î 0 i ˚ Î 0 -1˚ È cos a sin a ˘ Example 14: If Aa = Í ˙ , then Aa Ab is Î- sin a cos a ˚ equal to (b) Aab (a) Aa + b (c) Aa - b (d) none of these Ans. (a) Solution: We have È cos a sin a ˘ È cos b sin b ˘ Aa Ab = Í ˙ ˙Í Î- sin a cos a ˚ Î- sin b cos b ˚ È cos a cos b - sin a sin b = Í Î- sin a cos b - cos a sin b cos a sin b + sin a cos b ˘ - sin a sin b + cos a cos b ˙˚ È cos (a + b ) sin (a + b ) ˘ = Í ˙ = Aa + b Î- sin (a + b ) cos (a + b )˚ Example 15: Let A and B be two 2 × 2 matrices. Consider the statements (i) AB = O fi A = O or B = O (ii) AB = I2 fi A = B –1 (iii) (A + B)2 = A2 + 2AB + B2

5.10

Complete Mathematics—JEE Main

Then (a) (i) is false, (ii) and (iii) are true (b) (i) and (iii) are false, (ii) is true (c) (i) and (ii) are false, (iii) is true (d) (ii) and (iii) are false, (i) is true Ans. (b) Solution: (i) is false. È0 1˘ È1 1 ˘ and B = Í If A= Í ˙ ˙ , then Î0 -1˚ Î0 0 ˚ È0 0 ˘ AB = Í ˙=O Î0 0 ˚ Thus, AB = O fi / A = O or B = O (iii) is false since matrix multiplication is not commutative. (ii) is true as product AB is an identity matrix, if and only if B is inverse of the matrix A. È-2 5˘ È1 5˘ Example 16: If A – 2B = Í and 2A – 3B = Í ˙, ˙ Î 0 7˚ Î3 7˚ then matrix B is equal to È 0 6˘ È- 4 -5˘ (b) Í (a) Í ˙ ˙ Î-3 7˚ Î - 6 -7˚ È2 -1˘ (c) Í ˙ Î3 2˚ Ans. (a)

È6 -1˘ (d) Í ˙ Î0 1˚

Solution: We have È - 4 - 5˘ B = (2 A - 3 B ) - 2 ( A - 2 B ) = Í ˙ Î - 6 - 7˚ Example 17: If A and B two are 3 × 3 matrices, then which one of the following is not true: (a) (b) (c) (d) Ans. (b)

(A + B) ¢= A ¢ + B ¢ (AB)¢ = A¢ B¢ det (AB) = det (A) det (B) A (adj A) = |A| I3

Solution: If A and B are two 3 × 3 matrices, then (AB)¢ = B¢A¢ [Reversal Law] and not (AB)¢ = A¢ B¢. Ê cos q - sin q ˆ Example 18: If A = ÁË , then sin q cos q ˜¯ (a) A is an orthogonal matrix (b) A is a symmetric matrix (c) A is a skew-symmetric matrix (d) none of these Ans. (a) Solution: We have Ê cos q AA¢ = Á Ë sin q

- sin q ˆ Ê cos q cos q ˜¯ ÁË - sin q

sin q ˆ cos q ˜¯

Ê cos2 q + sin 2 q = Á Ë cos q sin q - sin q cos q

cos q sin q - sin q cos q ˆ ˜ ¯ sin 2 q + cos2 q

1 0ˆ = ÊÁ = I2 Ë 0 1˜¯ Similarly, A¢ A = I2 Thus, A is an orthogonal matrix. Example 19: If Èa 2 ab ac ˘ c -b˘ È0 Í ˙ 2 Í A = Íab b bc ˙ and B = -c 0 a ˙ then the product Í ˙ Í 2˙ Í ˙˚ 0 b a Î ac bc c ÍÎ ˙˚ AB is equal to (a) O (b) A (c) B (d) I Ans. (a) Solution: We have È0 - abc + abc a 2 c + 0 - a 2 c - a 2 b + a 2 b + 0 ˘ Í ˙ AB = Í 0 - b2 c + b2 c abc + 0 - abc - ab2 + ab2 + 0 ˙ = O Í ˙ 2 2 2 2 ÍÎ 0 - bc + bc ac + 0 - ac - abc + abc + 0 ˙˚ Example 20: If A is an invertible matrix and B is an orthogonal matrix, of the order same as that of A, then C = A –1 BA is (a) an orthogonal matrix (b) symmetric matrix (c) skew-symmetric matrix (d) none of these Ans. (d) Ê cos (p / 2) sin (p / 2) ˆ Ê 0 1ˆ Solution: Let B = Á = Ë - sin (p / 2) cos (p / 2)˜¯ ÁË -1 0˜¯ and

Ê 1 3ˆ A= Á , Ë 0 1˜¯

Ê 1 -3ˆ A-1 = Á Ë 0 1 ˜¯

Note that B is an orthogonal matrix. Ê 1 -3ˆ Ê 0 1ˆ Ê 1 3ˆ C = A–1 BA = Á Ë 0 1 ˜¯ ÁË -1 0˜¯ ÁË 0 1˜¯ Ê 3 10 ˆ =Á Ë -1 -3˜¯ Note that C is neither symmetric, nor skew-symmetric and nor-orthogonal. È cos2 a cos a sin a ˘ Example 21: Let E(a) = Í ˙ . If ÍÎcos a sin a sin 2 a ˙˚ a and b differs by an odd multiple of p/2, then E(a) E(b) is a (a) null matrix (b) unit matrix (c) diagonal matrix (d) orthogonal matrix

Matrices 5.11

Ans. (a) Solution: We have E(a) E(b) È cos2 a cos b sin b ˘ cos a sin a ˘ È cos2 b = Í Í ˙ ˙ ÍÎcos a sin a sin 2 b ˙˚ sin 2 a ˙˚ ÍÎcos b sin b cos a cos b cos (a - b ) cos a sin b cos (a - b )˘ = È Í sin a cos b cos (a - b ) sin a sin b cos (a - b ) ˙ Î ˚ As a and b differ by an odd multiple of p/2, a –b = (2n + 1)p/2 for some integer n. Thus, cos [(2n + 1) p/2] = 0 \

E(a) E(b) = O 2 ˘ È1 0 ˘ È2 1 ˘ È-3 AÍ Example 22: If Í ˙ ˙=Í ˙ , then Î 7 4 ˚ Î 5 - 3˚ Î 0 1 ˚ matrix A equals 5˘ È 7 È2 1˘ (b) Í (a) Í ˙ ˙ Î-11 - 8˚ Î 5 3˚ È 7 1˘ È 5 3˘ (c) Í (d) Í ˙ ˙ Î34 5˚ Î13 8˚ Ans. (a) Solution: If XAY = I, then A = X –1 Y –1 = (YX)–1 5˘ È-3 2 ˘ È2 1 ˘ È 8 =Í In this case YX = Í ˙ ˙ Í ˙ Î 5 -3˚ Î7 4 ˚ Î-11 -7˚ 5 ˘ -1 È 7 5˘ È 8 =Í \ A= Í ˙ ˙ Î-11 -8˚ Î-11 -7˚ Example 23: The matrix A satisfying È1 5˘ È3 -1˘ AÍ ˙ is ˙ =Í Î0 1˚ Î6 0 ˚ È3 2 ˘ (a) Í ˙ Î6 -3˚

È3 -16 ˘ (b) Í ˙ Î6 -30 ˚

È3 -16 ˘ È3 -3˘ (c) Í (d) Í ˙ ˙ Î6 30 ˚ Î6 2 ˚ Ans. (b) Solution: We know that if AC = B, then A = BC –1. \

-1 È3 -1˘ È1 -5˘ È3 -1˘ È1 5˘ = Í A= Í ˙Í ˙ ˙ Í ˙ Î6 0 ˚ Î0 1 ˚ Î6 0 ˚ Î0 1˚

È3 -16 ˘ =Í ˙ Î6 -30 ˚ È1 1 ˘ È3 2 ˘ is the matrix Í Example 24: If product A Í ˙ ˙, Î2 0 ˚ Î1 1 ˚ –1 then A is given by È0 -1˘ (a) Í ˙ Î2 - 4˚ 1˘ È0 (c) Í ˙ Î2 - 4˚

È 0 -1˘ (b) Í ˙ Î-2 - 4˚ (d) none of these

Ans. (c) Solution: If AB = C, then B –1 A –1 = C –1 fi A –1 = BC –1 È1 1 ˘ È3 2 ˘ Here AÍ ˙ ˙ = Í Î2 0 ˚ Î1 1 ˚ fi

È1 A –1 = Í Î2 È1 = Í Î2

1 ˘ È3 2 ˘ -1 0 ˙˚ ÍÎ1 1 ˙˚ 1 ˘ È 1 -2 ˘ È0 1 ˘ = 0 ˙˚ ÍÎ-1 3 ˙˚ ÍÎ2 - 4 ˙˚

Example 25: If A and B are two skew-symmetric matrices of order n, then (a) AB is a skew-symmetric matrix (b) AB is a symmetric matrix (c) AB is a symmetric matrix if A and B commute (d) none of these Ans. (c) Solution: We are given A¢ = – A and B¢ = –B Now, (AB)¢ = B¢ A¢ = (– B) (– A) = BA = AB if A and B commute. Example 26: Which of the following statements is false: (a) If |A| = 0, then |adj A| = 0 (b) Adjoint of a diagonal matrix of order 3 × 3 is a diagonal matrix (c) Product of two upper triangular matrices is a upper triangular matrix (d) adj (AB) = adj (A) adj (B) Ans. (d) Solution: We have adj (AB) = adj (B) adj (A) and not adj (AB) = adj (A) adj (B) Example 27: If A and B are symmetric matrices, then AB – BA is a (a) symmetric matrix (b) skew-symmetric matrix (c) diagonal matrix (d) null matrix Ans. (b) Solution: We are given A¢ = A, B¢ = B Now (AB – BA)¢ = (AB)¢ – (BA)¢ = B¢ A¢ – A¢ B¢ = BA – AB = – (AB – BA) i.e. (AB – BA)¢ = – (AB – BA) Hence, AB – BA is a skew-symmetric matrix. Example 28: Let A and B be two 3 × 3 matrices, such that A + B = 2B¢ and 3A + 2B = I, then (a) A – B = O (b) A + B = I (c) A – B = I (d) A + 2B = O Ans. (a)

5.12

Complete Mathematics—JEE Main

Solution: A + B = 2B¢ 3A + 2B = I fi B = 6B¢ – I fi B¢ = (6B¢ – I)¢ = 6B – I Therefore, B = 6(6B – I) – I fi (1 – 36)B = – 7I 1 fi B= I 5 2 3 \ 3A = I - I = I 5 5 1 fi A= I 5 \ A =B Example 29: Let A and B be two non-zero 3 × 3 matrices such that AB = O. Then (a) Both A and B are non-singular (b) Exactly one of A, B is singular (c) Both A and B are singular (d) Both A + B and AB are singular Ans. (c) Solution: Suppose A is non-singular, then A–1 (AB) = A–1(O) fi B = O A contradiction. \ A is singular Similarly, B is singular. Example 30: Suppose A is a 3×3 skew-symmetric matrix. Let B = (I + A)–1 (I – A). Then (a) B is orthogonal (b) B is skew symmetric (d) B is a diagonal matrix (c) B2 = O Ans. (a) Solution: We have BB¢ = (I + A)–1 (I – A) [(I + A)–1 (I – A)]¢ = (I + A)–1 (I – A) (I – A)¢ ((I + A)¢]–1 = (I + A)–1 (I – A) (I + A) (I – A)–1 [∵ A¢ = –A] = (I + A)–1 (I + A) (I – A) (I – A)–1 [I + A and I – A commute] = (I) (I) = I Thus, B is orthogonal. Example 31: Let an = 3n + 5n, n Œ N, and let Ê an A = Á an +1 Á Ë an + 2

an +1 an + 2 an + 3

an + 2 ˆ an + 3 ˜ ˜ an + 4 ¯

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

0 is a root of the equation det (A – xI) = 0 det(A) = an an + 2 an + 4 det(A) < 0 det(A) = an + an + 2 + an + 4

Ans. (a) Solution: Using C2 Æ C2 – 3C1, C3 Æ C3 – 32C1, we get an det(A) = an +1

2(5n )

16(5n )

2(5n +1 ) 16(5n +1 )

an + 2 2(5n + 2 ) 16(5n + 2 ) =0 [∵ C2 and C3 are proportional] Thus, x = 0 is a root of det (A – xI) = 0 Example 32: First row of a matrix A is [1 3 2]. If È -2 4 a ˘ adj A = Í -1 2 1 ˙ Í ˙ ÎÍ3a -5 -2˚˙ then a possible value of det(A) is (a) 1 (b) 2 (c) –1 (d) –2 Ans. (a) Solution: We know that det(A) = a11 A11 + a12 A12 + a13 A13 = (1) (–2) + (3) (–1) + (2) (3a) = 6a – 5 and (det(A))2 = det (adj A) = 11a – 10 Thus, (6a – 5)2 = 11a – 10 fi 36a2 – 71a + 35 = 0 fi a = 1, 35/36 Therefore, a possible value of det(A) is 1. Example 33: Suppose ABC is a triangle with sides a, b, c and semiperimeter s. Then matrix s-c È s ˘ Í ˙ Ès - a ˘ 2 2 A = Ís(s - b) (s - a) (s - c)˙ Ís - b ˙ Î ˚ 2 ¥1 Í s( s - c ) (s - a)2 ˙˚3 ¥ 2 Î bc È ˘ Í - ca (s - a )(s - b)˙ Í ˙ ÎÍ ab (s - a) ˚˙3 ¥1 is equal to: Èa ˘ (a) Íb ˙ Í ˙ ÎÍc ˚˙ Ès ˘ (c) Ís˙ Í ˙ ÍÎs˙˚ Ans. (b)

È0˘ (b) Í0˙ Í ˙ ÎÍ0˚˙ Ès - a ˘ (d) Ís - b ˙ Í ˙ ÍÎs - c ˙˚

Solution: Ès(s - a) + (s - c)(s - b) - bc ˘ A = Í(s - a )(s - b) (s(s - b) + (s - a )(s - c) - ca)˙ Í ˙ ÍÎ(s - a )(s(s - c) + (s - a )(s - b) - ab) ˙˚

Matrices 5.13

But s(s – a) + (s – c) (s – b) – bc = s(s – a) + s(s – b – c) = s(2s – a – b – c) = 0 etc. È0˘ Thus, A = Í0˙ Í ˙ ÍÎ0˙˚ Example 34: The number of matrices Èa b ˘ A= Í ˙ (where a, b, c, d Œ R) Îc d ˚ such that A–1 = – A is : (a) 0 (b) 1 (c) 2 (d) infinite Ans. (d) Solution: As A–1 exists, det(A) = ad – bc π 0. È- a - b ˘ Also, det (–A) = det Í ˙ = ad – bc Î -c - d ˚ = det (A) 1 , A–1 = –A implies since det(A–1) = det( A) 1 = ad – bc ad - bc fi (ad – bc)2 = 1 fi ad – bc = ±1. If ad – bc = 1, then A–1 = –A gives È d - b ˘ È- a - b ˘ A–1 = Í ˙=Í ˙ Î- c a ˚ Î - c - d ˚ fi a + d = 0. \ a(–a) – bc = 1 fi bc = – (1 + a2) fi bc π 0 and b =–

1 + a2 c

È (1 + a 2 ) ˘ a Í ˙ \ A= c ˙ , where c π 0. Í ÍÎc - a ˙˚ Thus, there are infinite number of such matrices. TIP It is unnecessary to consider the case ad – bc = –1

Example 35: Let A be a 3×3 matrix with entries from the set of real numbers, If the system of equations A2X = 0 has a non-trivial solution, then (a) AX = 0 has a non-trivial solution. (b) AX = 0 does not have a non-trivial solution. (c) A is a non-singular matrix. (d) none of these. Ans. (a)

Solution: As A2X = 0 has a non-trivial solution, det (A2) = 0 fi (det(A))2 = 0 fi det(A) = 0 \ AX = 0 has a non-trivial solution. Example 36: Suppose a, b Œ R and a π b. Èa b ˘ Let A= Í ˙. Îb - a ˚ Let M be a matrix such that MA = A2m for some m Œ N, then M is equal to (b) (a2 + b2)m–1 A (a) (a2 + b2)m I 2 2 m–1 (d) (a2 + b2)m A (c) –(a + b ) A Ans. (b) Solution: We have Èa 2 + b 2 0 ˘ 2 2 A2 = Í ˙ = (a + b )I 2 ÍÎ 0 a + b2 ˙˚ fi A2m = ((a2 + b2) I )m = (a2 + b2)m I Also,

A–1 = =

Now, fi

È- a - b ˘ Í ˙ a + b Î- b a ˚ -1

2

2

Èa b ˘ Í ˙ a + b Îb - a ˚ 1

2

2

MA = A2m = (a2 + b2)m I M = (a2 + b2)m A–1

= (a2 + b2)m–1 A Èa b ˘ Example 37: Let A = Í ˙ , be a 2 × 2 matrix where Îc d ˚ a, b, c, d Œ {0, 1}. The number of such matrices which have inverse is (a) 5 (b) 6 (c) 7 (d) 8 Ans. (b) Solution: det(A) = ad – bc Note that det(A) can take value –1, 0 or 1. We have det(A) = 1 ¤ ad = 1, bc = 0 ¤ a = 1, d = 1 or (b = 0, c = 0, b = 0; c = 1; b = 1, c = 0) and det(A) = –1 ¤ ad = 0 or bc = 1 This is also possible in 3 cases. \ A–1 exists in 6 cases. Example 38: If D = diag (d1, d2, …., dn) where di π 0, for i = 1, 2, . . . ., n, then D –1 is equal to (a) D (b) 2D (c) diag (d1–1, d2–1, …., d –1 n) (d) Adj D Ans. (c) Solution: See Theory.

5.14

Complete Mathematics—JEE Main

Example 39: The inverse of a symmetric matrix (if it exists) is (a) a symmetric matrix (b) a skew-symmetric matrix (c) a diagonal matrix (d) none of these Ans. (a)

È1 0 2 ˘ Example 43: If A = Í5 1 x ˙ is a singular matrix, Í ˙ ÍÎ1 1 1 ˙˚ then x is equal to (a) 3 (b) 5 (c) 9 (d) 11 Ans. (c)

Solution: Let A be an invertible symmetric matrix. We have AA–1 = A–1 A = In fi (AA–1)¢ = (A–1 A)¢ = (In)¢ fi (A–1)¢ A¢ = A¢(A–1)¢ = In fi (A–1)¢ A = A(A–1)¢ = In (A–1)¢ = A–1 [inverse of a matrix is unique] Example 40: The inverse of a skew-symmetric matrix (if it exists) is (a) a symmetric matrix (b) a skew-symmetric matrix (c) a diagonal matrix (d) none of these Ans. (b) Solution: We have A¢ = – A Now, AA–1 = A–1 A = In fi (AA–1)¢ = (A–1 A)¢ = (In)¢ fi (A–1)¢ A¢ = A¢(A–1)¢ = In fi (A–1)¢ (–A) = (– A) (A–1)¢ = In Thus,

(A–1)¢ = – (A–1)

Solution: As A is a singular matrix, 1 0 0 5 1 x - 10 = 0 |A| = 0 fi 1 1 -1 fi fi

Solution: As A is an orthogonal matrix, A¢ A = AA¢ = In fi |A¢ A| = |AA¢| = |In| fi |A¢| |A| = 1 fi |A| |A| = 1 fi |A|2 = 1 fi |A| = + 1

fi

– 1 – x + 10 = 0

x = 9.

2 ˘ È2 x -1 Í ˙ x 2 x 2 ˙ is singular is A= Í 1 Í 1 1x 2 ˙ Î ˚ (a) ± 1 (b) ± 2 (c) ± 3 (d) none of these Ans. (a) Solution: We have 2 1 x 1 2 x2 Ê 2ˆ x 2x + +2 |A| = Á ˜ Ë x¯ 1 x 2 1 1x 1 2

Example 41: The inverse of a skew-symmetric matrix of odd order is (a) a symmetric matrix (b) a skew-symmetric matrix (c) diagonal matrix (d) does not exist Ans. (d)

Example 42: If A is an orthogonal matrix, then |A| is (a) 1 (b) – 1 (c) ± 1 (d) 0 Ans. (c)

=0

Example 44: The value of x for which the matrix

[inverse of a matrix is unique]

Solution: Let A be a skew-symmetric, matrix of order n. By definition A¢ = – A fi | A¢| = |– A| fi | A| = (– 1)n |A| fi | A| = – |A| [∵ n is odd] fi 2|A| = 0 fi |A| = 0 \ A–1 does not exist.

[using C3 Æ C3 – 2C1]

1 x - 10 1 -1

=

Now,

2 Ê1 ˆ (0 ) + 2 - 2 x 2 + 2 Á - x ˜ Ë ¯ x x

2 2 2 = 2 x (1 - x ) + 2(1 - x ) = 2( x + 1) (1 - x ) x x |A| = 0 fi x = + 1.

Example 45: If square matrix A is such that 3A3 + 2A2 + 5A + I = O, then A–1 is equal to (a) 3A2 + 2A + 5I (b) – (3A2 + 2A + 5I) (c) 3A2 – 2A – 5I (d) none of these Ans. (b) Solution: We have A(3A2 + 2A + 5I) \

=–I

A–1 = – (3A2 + 2A + 5I). [\ Inverse of a matrix is unique]

È1 0 ˘ È1 0 ˘ Example 46: If A = Í and I = Í ˙ ˙ then which Î1 1 ˚ Î0 1 ˚ one of the following holds for all n ≥ 1, by the principle of mathematical induction.

Matrices 5.15

(a) (b) (c) (d) Ans. (c)

An = nA + (n – 1) I An = 2n – 1 A + (n – 1) I An = nA – (n – 1) I An = 2n – 1 A – (n – 1) I.

È-1 0 1 ˘ È 2 6 4 ˘ È-3 -5 -5˘ \ A = Í 1 1 3˙ Í 1 0 1 ˙ = Í 0 9 2 ˙ ˙ Í ˙Í ˙ Í ÍÎ 2 0 2 ˙˚ ÍÎ-1 1 -1˙˚ ÎÍ 2 14 6 ˙˚ Example 49: If w is a complex cube root of unity, then È 1 w2 w ˘ Í ˙ 1 ˙ is a the matrix A = Íw 2 w Í ˙ 1 w 2 ˙˚ ÍÎ w (a) singular matrix (b) non-singular matrix (c) skew-symmetric matrix (d) none of these Ans. (a) –1

È1 0 ˘ Solution: For n = 2, A2 = Í ˙ Î2 1 ˚ È3 0 ˘ 2 For n = 2, RHS of (a) = 2A + I = 3 Í ˙ πA Î2 3 ˚ For n = 2, RHS of (b) = 2A + I π A2 So possible answer is (c) or (d). È1 0 ˘ In fact An = Í ˙ which equals nA – (n – 1)I. În 1 ˚

Solution: We have

Alternatively. Write A = I + B È0 0 ˘ where B = Í ˙ Î1 0 ˚ As B2 = 0, we get Br = 0 " r ≥ 2. By the binomial theorem An = I + nB = I + n(A – I) = nA – (n – 1)I. Example 47: Let A, B, C be three square matrices of the same order, such that whenever AB = AC then B = C, if A is (a) singular (b) non-singular (c) symmetric (d) skew-symmetric Ans. (b) Solution: If A is non-singular, A–1 exists. Thus, AB = AC fi A–1 (AB) = A–1 (AC) fi (A–1 A) B = (A–1 A) C fi IB = IC fi B = C. È 2 6 4˘ Example 48: Let B = Í 1 0 1 ˙ . If the product BA Í ˙ ÍÎ -1 1 -1 ˙˚

È -3 5 5 ˘ (b) Í 0 0 9 ˙ Í ˙ ÍÎ 2 14 16 ˙˚

È -3 -5 -5 ˘ (c) Í 0 0 2 ˙ Í ˙ ÍÎ 2 14 6 ˙˚ Ans. (c)

È -3 -3 -5 ˘ (d) Í 0 9 2 ˙ Í ˙ ÍÎ 2 14 6 ˙˚

Solution: We have (BA)–1 = C fi A–1 = CB

fi

A–1 B–1 = C

w2

w

|A| = w 2

w

1 = w2 + w +1

w

2

1

w

1

w

2

1+ w2 + w w2 w +1+w

w 1 =0 w2

using C1 Æ C1 + C2 + C3 and 1 + w + w 2 = 0 \ A is a singular matrix. È0 1 2 ˘ Example 50: If A = Í1 2 3˙ and Í ˙ ÎÍ3 x 1 ˚˙ È1/ 2 -1/ 2 1/ 2 ˘ Í y ˙ , then A = -4 3 Í ˙ ÍÎ5 / 2 -3 / 2 1/ 2 ˙˚ (a) x = 1, y = – 1 (b) x = – 1, y = 1 (c) x = 2, y = – 1/2 (d) x = 1/2, y = 1/2 Ans. (a) –1

Solution: We have È1 0 0 ˘ È0 1 2 ˘ È1/ 2 -1/ 2 1 / 2 ˘ Í0 1 0 ˙ = AA-1 = Í1 2 3˙ Í -4 3 y ˙ Í ˙ Í ˙Í ˙ ÍÎ0 0 1 ˙˚ ÍÎ3 x 1 ˙˚ ÍÎ5 / 2 -3 / 2 1/ 2 ˙˚ 0 y +1 ˘ È 1 Í 0 1 2( y + 1)˙ = Í ˙ ÎÍ4(1 - x ) 3( x - 1) 2 + xy ˙˚

È -1 0 1 ˘ has inverse C = Í 1 1 3 ˙ , then A–1 equals Í ˙ ÍÎ 2 0 2 ˙˚ È -3 -5 5 ˘ (a) Í 0 9 14 ˙ Í ˙ ÍÎ 2 2 6 ˙˚

1

fi

1 – x = 0, x – 1= 0, y +1 = 0, y + 1= 0, 2 + xy = 1 \ x = 1, y = – 1. Èa b ˘ Example 51: Let A = Í ˙ , where a, b, c, d Œ R. If Îc d ˚ A – a I is invertible for all a Œ R, then (a) bc > 0 (b) bc = 0 Ê 1 ˆ (d) a = 0 (c) bc > min Ë0, ad ¯ 2

5.16

Complete Mathematics—JEE Main

Ans. (c) Solution: As A – a I is invertible for all a Œ R. det (A – a I) π 0 " a Œ R. fi

(a – a) (d – a) – bc π 0

" a ΠR.

fi a – (a + d)a + ad – bc π 0

" a ΠR.

2

Solution: A2 – A + I = 0 fi I = A – A2 = A(I – A). Thus, A–1 = I – A. Ê 1 -1 1 ˆ Example 55: Let A = Á 2 1 -3˜ and Á ˜ Ë1 1 1 ¯ Ê 4 2 2ˆ (10)B = Á - 5 0 a ˜ Á ˜ Ë 1 2 3¯

Therefore (a + d)2 – 4(ad – bc) < 0 fi (a – d)2 + 4 bc < 0 Therefore, bc < 0. Also, a2 + d2 – 2ad + 4bc < 0 fi 0 £ a2 + d2 < 2ad – 4bc 1 ad. 2 1 Thus, bc < min Ê0, ad ˆ Ë 2 ¯ È1 2 ˘ 2 Example 52: If A = Í ˙ , then A – 5A – I equals 3 4 Î ˚ (a) O (b) I (c) 2I (d) none of these Ans. (b) fi

If B is the inverse of matrix A, then a equals (a) 2 (b) –1 (c) –2 (d) 5 Ans. (d) Solution: 10I = 10(AB) = A(10B)

bc
0 and abc = 2. Let Èa b c ˘ Í A = b c a˙ Í ˙ ÎÍ c a b ˙˚ If A2 = I, then value of a3 + b3 + c3 is (a) 7 (b) 2 (c) 0 (d) –1 69. If A is a 3 ¥ 3 skew-symmetric matrix with real entries and trace of A2 equals zero, then

(AB)k = AkBk for k = 2015, 2016, 2017, then (a) AB = O (b) BA = O (c) AB = BA (d) AB + BA = O 71. Let A be a square matrix of order 3 such that |Adj A | = 100, then | A | equals (a) ±10 (b) –100 (c) 100 (d) 25 72. Let M be a 3 ¥ 3 matrix satisfying È0 ˘ È – 1˘ È 1 ˘ È 1˘ È1˘ È 0 ˘ Í ˙ Í ˙ Í ˙ Í ˙ M 1 = 2 , M – 1 = 1 , and M Í1˙ = Í 0 ˙ Í ˙ Í ˙ Í ˙ Í ˙ Í˙ Í ˙ ÍÎ0 ˙˚ ÍÎ 3 ˙˚ ÍÎ 0 ˙˚ ÍÎ – 1˙˚ ÍÎ1˙˚ ÍÎ12 ˙˚ then sum of the diagonal entries of M is (a) 0 (c) 6

(b) –3 (d) 9

73. If A, B and A + B are non-singular matrices, then (A–1 + B –1) [A – A (A + B)–1 A] equals (a) O (c) A

(b) I (d) B

74. If A + B is a non-singular matrix, then A – B – A (A + B)–1A + B (A + B)–1 B equals (a) O (c) A

(b) I (d) B

75. If A and B are two matrices such that A + B = AB, then (a) A = I (b) B = I (c) AB = BA (d) AB = I a 76. Let A = ÊÁ Ëc (a) ad (c) 1 Ê0 77. Let A = Á y Á Ë0

bˆ such that A3 = O, then a + d equals ˜ d¯ (b) bc (d) 0 x 0ˆ 0 - x ˜ , then A3 equals ˜ y 0¯

(a) O (c) (x2 + y2) I

(b) x2I (d) none of these

5.30

Complete Mathematics—JEE Main

Previous Years' AIEEE/JEE Main Questions

Èa b ˘ Èa b ˘ 1. If A = Í and A2 = Í ˙ , then ˙ Îb a ˚ Îb a ˚ (a) (b) (c) (d)

a = a2 + b2, b = 2ab a = a 2 + b 2, b = a 2 – b 2 a = 2ab, b = a2 + b2 a = a2 + b2, b = 2ab

Ê0 2. Let A = Á 0 Á Ë -1 about matrix

Ê 1 2ˆ Ê a 0ˆ and B = Á a, b Œ N. Then 7. Let A = Á ˜ Ë 3 4¯ Ë 0 b˜¯

[2003]

0 -1ˆ -1 0 ˜ . The only correct statement ˜ 0 0¯ A is

A–1 does not exist A = (–1)I, where I is a unit matrix A is a zero matrix [2004] A2 = I 1 1 1 4 2 2ˆ Ê Ê ˆ 3. Let A = Á 2 1 -3˜ and (10)B = Á -5 0 a ˜ . Á Á ˜ ˜ Ë1 1 1 ¯ Ë 1 -2 3 ¯ (a) (b) (c) (d)

If B is the inverse of matrix A, then a is (a) 2 (c) –2

(b) –1 (d) 5

4. If A – A + I = O, then the inverse of A is (b) I – A (d) A

An = nA + (n – 1)I An = (2n – 1) A + (n – 1)I An = nA – (n – 1)I An = (2n – 1) A – (n – 1)I

[2005]

[2005]

6. If A and B are square matrices of size n ¥ n such that A2 – B2 = (A – B) (A + B), then which of the following will be always true? (a) (b) (c) (d)

either A or B is an identity matrix A=B AB = BA either A or B is a zero matrix

(a) 52 (c) 1/5

a˘ 5a ˙ . If | A2 | = 25 then | a | ˙ 5 ˙˚ (b) 1 (d) 5

[2007]

9. Let A be a 2 ¥ 2 matrix with real entries. Let I be the 2 ¥ 2 identity matrix. Denote by tr(A), the sum of diagonal entries of A. Assume that A 2 = I. Statement-1: If A π I and A π – I, then det (A) = – 1. Statement-2: If A π I and A π – I, then tr(A) π O. [2008] 10. Let A be a 2 ¥ 2 matrix.

È1 0 ˘ È1 0 ˘ and I = Í 5. If A = Í ˙ ˙ , then which one of Î1 1 ˚ Î0 1 ˚ the following holds for all n ≥ 1, by the principle of mathematical induction (a) (b) (c) (d)

È5 5a 8. Let A = Í0 a Í ÍÎ0 0 equals

[2004]

2

(a) A – I (c) A + I

(a) there exist infinitely many B’s such that AB = BA (b) there cannot exist B such that AB = BA (c) there exist more than one but finite number of B’s such that AB = BA (d) there exists exactly one B such that AB = BA [2006]

Statement-1: Statement-2:

adj (adj A) = A |adj A| = |A|

11. Let A be a 2 ¥ 2 matrix with non-zero entries and let A2 = I, where I is 2 ¥ 2 identity matrix. Define Tr(A) = Sum of diagonal elements of A, |A| = determinant of matrix A. Statement-1: Tr (A) = 0 Statement-2: |A| = 1

[2010]

12. The number of 3 ¥ 3 non-singular matrices with four entries as 1 and all other entries 0, is (a) 6 (c) less than 4

(b) at least 7 (d) 5

13. Consider the system of linear equations x1 + 2x2 + x3 = 3 2x1 + 3x2 + x3 = 3 3x1 + 5x2 + 2x3 = 1

[2006]

[2009]

The system has

[2010]

Matrices 5.31

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

a unique solution no solution infinite number of solutions exactly 3 solutions

[2010]

14. The number of values of k for which the linear equations 4x + ky + 2z = 0 kx + 4y + z = 0 2x + 2y + z = 0 possess a non zero solution is (a) 0 (c) 2

(b) 3 (d) 1

[2011]

Èw 0 ˘ 15. If w π 1 is cube root of unity and H = Í ˙ Î0 w ˚ then H70 equals to (a) O (c) H2

(b) –H (d) H

[2011]

16. If trivial solution is the only solution of the system of linear equations: x – ky + z = 0 kx + 3y – kz = 0 3x + y – z = 0 then set of all values of k is (a) R – {2, –3} (c) R –{–3}

(b) R –{2} (d) {2, –3}

[2011]

17. Let A and B be two symmetric matrices of order 3. Statement-1: A(BA) and (AB)A are symmetric matrices Statement-2: AB is symmetric if matrix multiplication of A with B is commutative [2011] 18. Let P and Q be 3 ¥ 3 matrices such that P π Q. If P3 = Q3 and P2Q = Q2P, then det (P2 + Q2) is equal to (a) 1 (b) 0 (c) –1 (d) –2 [2012] Ê 1 0 0ˆ 19. Let A = Á 2 1 0˜ . If u1 and u2 are column maÁ ˜ Ë 3 2 1¯ Ê 1ˆ Ê 0ˆ trices such that Au1 = Á 0˜ and Au2 = Á 1˜ , then Á ˜ Á ˜ Ë 0¯ Ë 0¯ u + u is equal to 1

2

Ê -1ˆ (a) Á 1˜

Á ˜ Ë -1¯

Ê 1ˆ (c) Á -1˜ Á ˜ Ë -1¯

Ê -1ˆ (b) Á -1˜

Á ˜ Ë 0¯ Ê -1ˆ

(d) Á 1˜

Á ˜ Ë 0¯

[2012]

È1 a 20. If P = Í1 3 Í ÍÎ2 4 A and |A| = 4, (a) 11 (c) 0

3˘ 3 ˙ is the adjoint of a 3 × 3 matrix ˙ 4 ˙˚ then a is equal to (b) 5 (d) 4

[2013]

21. The number of values of k for which the system of equations (k + 1)x + 8y = 4 k kx + (k + 3)y = 3 k – 1 has no solution, is (a) 1 (b) 2 (c) 3 (d) infinite [2013] 22. If the system of linear equations x1 + 2x2 + 3x3 = 6 x1 + 3x2 + 5x3 = 9 2x1 + 5x2 + ax3 = b is consistent and has infinite number of solutions, then (a) a = 8, b can be any real number (b) b = 15, a can be any real number (c) a Œ R – {8} and b Œ R – {15} (d) a = 8, b = 15 [2013, online] 23. If p, q, r are 3 real numbers satisfying the matrix equation, È3 4 1 ˘ [ p q r ] Í3 2 3˙ = [3 0 1] Í ˙ ÎÍ2 0 2 ˚˙ then 2p + q – r equal (a) –3 (b) –1 (c) 4 (d) 2 [2013, online] 24. Consider the system of equations: x + ay = 0, y + az = 0, and z + ax = 0. Then the set of all values of a for which the system has a unique solution is (a) R – {1} (b) R – {–1} (c) {1, –1} (d) {1, 0, –1} [2013, online] a12 ˆ ÏÊ a ¸ 25. Let S = ÌÁ 11 ˜¯ aij Œ{0, 1, 2}, a11 = a22 ˝ Ë a a Ó 21 ˛ 22 Then the number of non-singular matrices in the set S is (a) 27 (b) 24 (c) 10 (d) 20 [2013, online] 26. The matrix A2 + 4A – 5I, where I is the identity È1 2 ˘ matrix and A = Í ˙ , equals: Î4 -3˚

5.32

Complete Mathematics—JEE Main

È0 -1˘ (b) 4 Í ˙ Î2 2 ˚

È2 1 ˘ (a) 4 Í ˙ Î2 0 ˚ È2 1 ˘ (c) 32 Í ˙ Î2 0 ˚

È1 1 ˘ (d) 32 Í ˙ Î1 0 ˚ [2013, online] 27. Let A, other than I or –I, be a 2 × 2 real matrix such that A2 = I, I being the unit matrix. Let Tr(A) be the sum of diagonal elements of A. Statement-1: Tr(A) = 0 Statement-2: det(A) = –1 [2013, online] 28. If A is an 3 × 3 non-singular matrix such that AA¢ = A¢A and B = A–1 A¢, then BB¢ equals: (a) I (b) B –1 (c) (B –1)¢ (d) I + B [2014] 29. If B is a 3 × 3 matrix such that B2 = O, then det [(I + B)50 – 50B] is equal to: (a) 1 (c) 3 30. Let A be È1 A Í0 Í ÍÎ0

(b) 2 (d) 50 a3 2 2 1

× 3 matrix 3˘ È 0 0 3˙ = Í 1 0 ˙ Í 1˙˚ ÍÎ0 1

[2014, online]

such that 1˘ 0˙ ˙ 0 ˙˚

then A–1 is : È3 1 2 ˘ (a) Í3 0 2 ˙ ˙ Í ÎÍ1 0 1 ˙˚ È 0 1 3˘ (c) Í0 2 3˙ Í ˙ ÍÎ1 1 1˙˚

È3 2 1 ˘ (b) Í3 2 0 ˙ ˙ Í ÎÍ1 1 0 ˙˚

È 1 2 3˘ (d) Í0 1 1˙ Í ˙ ÍÎ0 2 3˙˚ [2014, online] È y˘ È1 2 x ˘ Í ˙ 31. If A = Í ˙ and B = Í x ˙ be such that AB Î3 -1 2 ˚ ÎÍ1 ˚˙ È6 ˘ = Í ˙ , then: Î8 ˚ (a) y = 2x (b) y = –2x (c) y = x (d) y = –x [2014, online] 32. Let A and B be any two 3×3 matrices. If A is symmetric and B is skew-symmetric, then the matrix AB – BA is: (a) skew-symmetric (b) symmetric (c) neither symmetric nor skew-symmetric (d) I or –I, where I is an identity matrix. [2014, online]

2˘ È1 2 Í 33. If A = 2 1 -2 ˙ is a matrix satisfying the equa˙ Í b ˙˚ ÍÎ a 2 tion AAT = 9I, where I is 3 ¥ 3 identity matrix, then the ordered pair (a, b) is equal to: (a) (2, –1) (b) (–2, 1) (c) (2, 1) (d) (–2, –1) [2015] 34. If A is 3 ¥ 3 matrix such that |5adj A| = 5, then |A| is equal to: 1 (a) ± (b) ± 5 5 1 [2015, online] (c) ± 1 (d) ± 25 0 -1˘ 35. If A = È , then which one of the following Í 1 0˙ Î ˚ statement is not correct? (a) A4 – I = A2 + I (b) A3 – I = A(A – I) 2 2 (c) A + I = A(A – I) (d) A3 + I = A(A3 – I) [2015, online] È5a -b ˘ 36. If A = Í ˙ and Aadj A = AA¢, then 5a + b Î3 2˚ is equal to (a) –1 (b) 5 (c) 4 (d) 13 [2016] È 3 1 ˘ Í ˙ È1 1˘ 2 2 ˙ ,A= Í 37. If P = Í ˙ and Q = PAP ¢, then Í 1 3˙ Î0 1˚ Í˙ Î 2 2 ˚ P ¢Q2015 P is: 0 ˘ È2015 È0 2015˘ (a) Í (b) Í ˙ 0 ˚ 2015˙˚ Î 1 Î0 È1 2015˘ (c) Í 1 ˙˚ Î0

1 ˘ È2015 (d) Í 2015˙˚ Î 0 [2016, online]

È-4 -1˘ 38. If A = Í ˙ , then the determinant of the matrix Î3 1˚ (A2016 – 2A2015 – A2014) is: (a) –175 (b) 2014 (c) 2016 (d) –25 [2016, online] 39. Let A be a 3 ¥ 3 matrix such that A2 – 5A + 7I = 0. 1 (5I – A) Statement-1: A–1 = 7 Statement-2: The polynomial A3 –2A2 – 3A + I can be reduced to 5(A + 4I). Then (a) Both the statements are true. (b) Both the statements are false. (c) Statement-1 is true, but Statement-2 is false. (d) Statement-1 is false, but Statement-2 is true. [2016, online]

Matrices 5.33

Previous Years' B-Architecture Entrance Examination Questions 1. If A and B are square matrices of the same order then which one of the following is always true? (a) (A + B)–1 = A–1 + B –1 (b) adj (AB) = (adj B) (adj A) (c) A and B are non-zero and |AB| = 0 ¤ |A| = 0 and |B| = 0 [2006] (d) (AB)–1 = A–1 B –1 È1 1 0 ˘ 2. Let A = Í0 1 0 ˙ , and let I denote the 3 × 3 ˙ Í ÎÍ0 0 1 ˚˙ identity matrix. Then 2A2 – A3 = (a) A + I (b) A – I (c) I – A (d) A [2008] 3. Let A and B be 2 × 2 matrices with real entries. Let I be the 2 × 2 identity matrix. Denote by tr(A) the sum of diagonal entries of A. Statement-1: AB – BA π I Statement-2: tr(A + B) = tr(A) + tr(B) and tr(AB) = tr(BA) [2009] Èa b ˘ 4. Let A = Í ˙ be a 2 × 2 real matrix. If A – a I Îc d ˚ is invertible for every real number a, then (a) bc > 0 (b) bc = 0 (d) a = 0 [2010] (c) bc < min Ê0, 1 ad ˆ Ë 2 ¯ 5. Let A and B be two 2 × 2 matrices. Statement-1: Statement-2:

A (adj A) = |A| I2 adj(AB) = (adj A)(adj B)

Èa b ˘ 6. Let A = Í ˙ , be a 2 × 2 matrix, where a, b, Îc d ˚ c, d take values 0 or 1 only. The number of such matrices which have inverses is: (a) 8 (c) 7 (c) 6 (d) 5 [2012] Èa b ˘ 7. Let S be the set of all real matrices A = Í ˙ Îc d ˚ such that a + d = 2 and A¢ = A2 – 2A. Then S: is an empty set has exactly one element has exactly two elements has exactly four elements

(a) S2n + k –1

(b) S2n + k -1

(c) S2n - k

(d) S2n – k

1 9. Let A = ÈÍ Î0 10 Adj(A ) = B,

[2014]

È b1 b2 ˘ 1˘ 10 and B = Í ˙ . If 10A + ˙ 1˚ Îb3 b4 ˚ then b1 + b2 + b3 + b4 is equal to:

(a) 91 (c) 111

(b) 92 (d) 112

[2015]

È 1 -2 4˘ 10. If for a matrix A, |A| = 6 and adj A = Í 4 1 1 ˙ , Í ˙ ÍÎ-1 k 0 ˙˚ then k is equal to (a) 0 (b) 1 (c) 2 (d) –1 [2016]

Answers Concept-based 1. (c) 5. (b) 9. (c)

2. (a) 6. (c) 10. (b)

3. (d) 7. (a)

4. (c) 8. (a)

Level 1 [2011]

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

È1 k ˘ 8. Let Sk = Í ˙ k Œ N, where N is the set of Î0 1 ˚ natural numbers, then (S2)n (Sk)–1, for n Œ N is:

[2013]

11. 15. 19. 23. 27. 31. 35. 39. 43. 47. 51. 55. 59.

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

12. 16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 60.

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

13. 17. 21. 25. 29. 33. 37. 41. 45. 49. 53. 57.

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

14. 18. 22. 26. 30. 34. 38. 42. 46. 50. 54. 58.

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

5.34

Complete Mathematics—JEE Main

Level 2

5. Let X1, X2 X3 be solutions of AX = B, when

61. (b)

62. (b)

63. (a)

64. (a)

65. (b)

66. (c)

67. (d)

68. (a)

69. (a)

70. (c)

71. (a)

72. (d)

73. (b)

74. (a)

75. (c)

76. (d)

77. (a)

Previous Years' AIEEE/JEE Main Questions 1. (a)

2. (d)

3. (d)

4. (b)

5. (c)

6. (c)

7. (a)

8. (c)

9. (c)

10. (b)

11. (c)

12. (b)

13. (b)

14. (c)

15. (d)

16. (a)

17. (b)

18. (b)

19. (c)

20. (a)

21. (a)

22. (d)

23. (a)

24. (b)

25. (d)

26. (a)

27. (a)

28. (a)

29. (a)

30. (a)

31. (a)

32. (b)

33. (d)

34. (a)

35. (c)

36. (b)

37. (c)

38. (d)

39. (a)

Ê1 ˆ B = Á 0˜ , Á ˜ Ë 0¯

2. (d) 6. (c) 10. (c)

3. (a) 7. (a)

4. (c) 8. (d)

Hints and Solutions

fi A¢ is invertible. Thus A¢Y = O fi Y = O. 6. det(A) = cos2q + sin2q = 1 π 0 \ A–1 exist. Now, AB = A(C – D) –1 fi A (AB) = A–1 (A(C – D)) B=C–D fi C=B+D fi C¢ = (B + D)¢ = B¢ + D¢ 7. (A – B) A–1 (A + B) = (I – BA–1) (A + B) = A + B – BA–1 A – BA–1 B = A – BA–1 B = A + B – BA–1 A – BA–1 B = (A + B) – BA–1 (A + B) = (A + B) (I – BA–1) = (A + B)A–1 (A – B)

1. If f (x) = 1 + x, then det f (A) = 1 + det(A) may not hold. Take A = –I. 2. |det(A)| = |ad – bc| £ |a||d| + |b||c| £ (|a| + |b|) (|c| + |d|) Also, |det(A)| £ (|a| + |c|) (|b| + |d|) For truth of (c), see Example 6

[

È 5 -1˘ ˘ (a) does not hold in general. Take A = Í ˙˙ Î10 2 ˚ ˚ 3. (a), (b), (c) cannot hold in general unless AB = BA. 4. det(ABA) = 1 fi

det(A) det(B) det(A) = 1

fi

det(A) π 0, Both A

–1

det(B) π 0

and B

–1

and

\

AA¢ = A2

fi

A¢ = A

A2 = I fi

A–1 (AA¢) = A–1A2

9. As A, B are orthogonal matrices, AA¢ = BB¢ = I, det(A) π 0, det(B) π 0. Let C = A + B.

B2 = (A–2)2 = A– 4

exist fi

fi AC ¢B = B + AB ¢B = B + A Now, det(A + B) = det(AC ¢B) = det(A) det(C ¢) det(B) = det(A) det(C) (– det(A)) fi det(C) = – (det(A))2 det(C) = – det(C) fi det(C) = 0 2 Ê a b ˆ Ê a b ˆ Ê a + bc b(a + d )ˆ = 10. A2 = Á Ë c d ˜¯ ÁË c d ˜¯ ÁË c(a + d ) bc + d 2 ˜¯

\

(a + d)A – A2 = A¢

fi

0 ˆ Ê a bˆ Ê ad - bc = fi ad = a, d ÁË 0 ad - bc˜¯ ÁË c d ˜¯

As ad π 0, we get a = d = 1.

Also ABA = I fi BA = A–1 fi B = A–1A–1 = A–2 fi

8. AA¢ = I

C ¢ = A¢ + B ¢ fi AC ¢ = AA¢ + AB ¢ = I + AB ¢

Concept-based

fi

Ê 0ˆ Á 0˜ respectively. Let C be the 3 × 3 Á ˜ Ë1 ¯

matrix [X1 X2 X3], then AC = I fi A is invertible.

Previous Years' B-Architecture Entrance Examination Questions 1. (b) 5. (c) 9. (d)

Ê 0ˆ Á1 ˜ , Á ˜ Ë 0¯

A 4B 2 = I

Level 1 Ê 1 2ˆ =I+B 11. S2 = Á Ë 0 1˜¯

where

Matrices 5.35

\

Ê 1 0ˆ Ê 0 2ˆ and B = Á I= Á ˜ Ë 0 1¯ Ë 0 0˜¯ Note that B2 = O, therefore, Br = O

14. In A + B two rows are identical. " r ≥ 2.

(S2)n = (I + B)n = I + nB Ê 1 2 nˆ =Á Ë 0 1 ˜¯

Thus,

Ê 1 -k ˆ Sk-1 = Á Ë 0 1 ˜¯

Also,

Ê 1 2 nˆ Ê 1 - k ˆ \ (S2) (Sk) = Á Ë 0 1 ˜¯ ÁË 0 1 ˜¯ n

Èa b ˘ Èa b ˘ 12. A2 = Í ˙Í ˙ Îc d ˚ Îc d ˚ È a 2 + bc b(a + d )˘ = Í ˙ ÍÎc(a + d ) bc + d 2 ˙˚ Èa + bc 3b ˘ = Í ˙ ÍÎ 3c bc + d 2 ˙˚ Èa 2 - 3a + bc ˘ 0 ˙ A – 3A = Í 2 ÍÎ 0 d - 3d + bc ˙˚ 2

A¢ = A2 – 3A, we get

b = 0, c = 0,

fi a (a – 4) = 0 fi

a = a2 – 3a, d = d2 – 3d

and d(d – 4) = 0

a = 0 or a = 4 and d = 0 or d = 4

But then a + d cannot be 3. \

I= A

{

}

1 (3I - A) 2

1 (3I – A) 2 18. | A2 | = | 2A | = 23 | A | fi | A |2 = 8 | A | 19. (A – A¢)¢ = A¢ – (A¢)¢ = A¢ – A = – (A – A¢) \ A – A¢ is skew-symmetric matrix. Ê cos 2 cos 3 cos 4ˆ 20. A = Á cos 3 cos 4 cos 5˜ = A¢ Á ˜ Ë cos 4 cos 5 cos 6¯ fi A –1 =

22. (A + B)2 = (A + B) (A + B) = A2 + BA + AB + B2

2 ˘ 0 Èa c ˘ Èa - 3a + bc = ˙ Íb d ˙ Í Î ˚ ÍÎ 0 d 2 - 3d + bc ˙˚

fi

fi

21. A3 – 3A + I = 0 fi I = A(3I – A2) fi A is invertible and hence non-singular.

2

Since

15. B = (cos q)I + (sin q)J. 16. A is a diagonal matrix, aij = 0 " i π j As A is a skew-symmetric matrix, aii = 0 " i \ A=O 17. A2 – 3A + 2I fi 2I = A(3I – A)

–1

Ê 1 2n - k ˆ = S2 n - k =Á Ë0 1 ˜¯

\

B –1 does not exist.

S =f

13. Let Ê a pˆ A = Á b q ˜ , then Á ˜ Ëc r¯ Ê a 2 + p2 ab + pq ac + pr ˆ Á ˜ B = AA¢ = Á ab + pq b2 + q 2 bc + qr ˜ ÁË ac + pr bc + qr c 2 + r 2 ˜¯ a p 0 a b c det (B) = b q 0 p q r = 0 c r 0 0 0 0

\ (A + B)2 = A2 + 2AB + B2 if and only if AB = BA 2 ˘ È i 0 ˘ È0 2 ˘ Èi -Í =Í 23. 2X = Í ˙ ˙ ˙ Î3 4 + i ˚ Î3 -i ˚ Î0 4 + 2i ˚ fi

1 ˘ È0 X= Í ˙ Î0 2 + i ˚

È0 -i ˘ È1 0 ˘ È0 i ˘ 24. AB = Í ˙Í ˙=Í ˙ Î i 0 ˚ Î0 -1˚ Î i 0 ˚ È1 0 ˘ È0 -i ˘ È 0 -i ˘ and BA = Í ˙=Í ˙Í ˙ Î0 -1˚ Î i 0 ˚ Î-i 0 ˚ Thus, AB + BA = O È0 0 0 ˘ È 1 2 3 ˘ È 1 2 3˘ Í ˙ Í ˙ 1 2 3 = Í0 0 0 ˙ 25. A = 1 2 3 Í ˙Í ˙ Í ˙ ÍÎ-1 -2 -3˙˚ ÍÎ-1 -2 3˙˚ ÍÎ0 0 0 ˙˚ \ A is a nilpotent matrix of index 2 2

Èa 26. Let A = Í Îc AA* = I fi

b˘ Èa c ˘ , A* = Í ˙ d ˙˚ Îb d ˚ | A | | A*| = 1

fi (ad Рbc) (a d - b c) = 1 fi | ad Рbc |2 = 1 fi ad Рbc = eiq for some q ΠR.

5.36

Complete Mathematics—JEE Main

Ê -1 / 2 - 3 / 2ˆ Ê cos(4p / 3) sin (4p / 3) ˆ 27. A = Á ˜ ˜ =Á Ë 3 / 2 -1 / 2 ¯ Ë - sin (4p / 3) cos(4p / 3)¯ Note that A3 = I fi A2 – A–1 = O 28. C2 + C = I fi C(C + I) = I fi C –1 = C + I Thus, C –2 = (C + I)2 = C2 + 2C + I = I – C + 2C + I = 2I + C 29. CA3C–1 = (CAC –1) (CAC –1) (CAC –1) = B3 30. A2 = (I – X(X¢X) –1 X¢)2 = I – 2X (X¢X)–1 X¢ + X(X¢X)–1 X¢ X(X¢X)–1 X¢ = I – 2X(X¢X)–1 X¢ + X(X¢X)–1 X¢ = I – X(X¢X)–1 X¢ = A 31. AA¢ = I p - q ˆ Ê p q ˆ Ê 1 0ˆ fi ÊÁ = Ë q p ˜¯ ÁË -q p˜¯ ÁË 0 1˜¯ Ê p2 + q 2 fi Á Ë 0

ˆ Ê 1 0ˆ ˜ ˜ =Á p 2 + q 2 ¯ Ë 0 1¯ 0

\ p2 + q2 = 1 32. AA¢ = I Êl 0 l ˆ Êl fi Á l 0 -l ˜ Á 0 Á ˜Á Ë 0 1 0 ¯ Ël Ê 2l 2 Á fi Á 0 Á 0 Ë \ 2l2 = 1 33. A¢ = A

0 2l 0

2

fi

l 0ˆ Ê 1 0 0ˆ 0 1˜ = Á 0 1 0˜ ˜ Á ˜ - l 0¯ Ë 0 0 1¯

0 ˆ Ê 1 0 0ˆ ˜ 0 ˜ = Á 0 1 0˜ Á ˜ 1˜¯ Ë 0 0 1¯

fi a + 1 = a2 – 1 and 4a = a2 + 4 fi a + 1 = 4a – 4 – 1 fi 6 – 3a fi a = 2. 34. Since At–1 does not exist, | At | = 0 fi (–30 – t(7 – t)) – 3(–12 – 4t) + 2(14 – 2t – 20) =0 fi t2 + t – 6 = 0 fi t = –3, 2 Èa - ib -c - id ˘ 35. A–1 = Í ˙ Îc - id a + ib ˚ 1 [(eix + e–ix)2 – (eix – e–ix)2] 4

1 (4eix e–ix) = 1 π 0 = 4 Thus, A–1 exists for all x Œ R.

b2 ˘ È ab ˙Í - ab ˙˚ ÍÎ- a 2

Èa 2 b2 - a 2 b2 = Í ÍÎ - a3 b + a3 b

b2 ˘ ˙ - ab ˙˚

ab3 - ab3 ˘ ˙ =O - a 2 b2 + a 2 b2 ˙˚

Èa 2 + bc ab + bd ˘ Èa b ˘ 2 38. Let A = Í ˙ , then A = ÍÍ ac + cd bc + d 2 ˙˙ Îc d ˚ Î ˚ Now, A2 = O fi a2 + bc = 0, (a + d) b = 0 (a + d) c = 0, bc + d2 = 0 Suppose a + d π 0, then b = 0, c = 0. \ a2 = 0, d2 = 0 fi a = 0, d = 0 Contradiction. Thus, a + d = 0 we get tr(A) = 0 39. Suppose A–1 exists, then A–1 A2 – (a + d)A–1 A = O fi A – (a + d)I = O È- d b ˘ fi Í ˙ =O Î c -a˚ fi a = b = c = d = 0 fi |A| = 0 A contradiction. Thus, A–1 does not exist. 40. We have A–1 =

1 È1 -2 ˘ 3 ÍÎ0 3 ˙˚

È1 -2 ˘ È1 -2 ˘ È1 -2 ˘ Í0 3 ˙ Í0 3 ˙ Í0 3 ˙ Î ˚Î ˚Î ˚ 1 8 1 2 1 È 1 È1 -26 ˘ ˘È ˘ = = Í ˙ Í ˙ 27 Î0 9 ˚ Î0 3 ˚ 27 ÍÎ0 27 ˙˚

fi A–3 =

l = ± 1/ 2

a2 - 1 a + 1 -3ˆ Ê a -3 ˆ Ê a Á ˜ Á 2 2 4 a˜ = Á a + 1 2 a 2 + 4˜ fi Áa -1 ˜ ÁË -3 -1 ˜¯ 4a a 2 + 4 -1˜¯ ÁË -3

36. | A | =

È ab 37. A2 = Í ÍÎ- a 2

1 27

41. A* = –A = fi ( A)¢ = –A fi aii = –aii " i fi 2Re(aii) = 0 fi aii is purely imaginary. È1 2 1 ˘ È1 2 1 ˘ È 4 42. A = Í0 1 -1˙ Í0 1 -1˙ = Í-3 Í ˙Í ˙ Í ÍÎ3 -1 1 ˙˚ ÍÎ3 -1 1 ˙˚ ÍÎ 6 È 4 3 0 ˘ È1 2 1 ˘ È 4 3 A = Í-3 2 -2 ˙ Í0 1 -1˙ = Í-9 Í ˙Í ˙ Í ÍÎ 6 4 5 ˙˚ ÍÎ3 -1 1 ˙˚ ÍÎ 21 Now, A3 – 3A2 – A + 9I È 4 11 1 ˘ È4 3 0˘ Í ˙ = -9 -2 -7 - 3 Í-3 2 -2 ˙ Í ˙ Í ˙ ÍÎ 21 11 7 ˙˚ ÍÎ 6 4 5 ˙˚ 2

3 0˘ 2 -2 ˙ ˙ 4 5 ˙˚ 11 1 ˘ -2 -7˙ ˙ 11 7 ˙˚

È1 2 1 ˘ È 9 0 0 ˘ – Í0 1 -1˙ + Í0 9 0 ˙ = O Í ˙ Í ˙ ÍÎ3 -1 1 ˙˚ ÎÍ0 0 9˙˚

Matrices 5.37

È1 2 x ˘ 43. (3A)¢ = Í2 1 2 ˙ Í ˙ ÍÎ2 -2 y ˙˚ Now, 9I = (3A) (3A)¢

t t +1 t -1 t t+2 D = t +1 t -1 t + 2 t Applying R3 Æ R3 – R2, R2 Æ R2 – R1, we get t t +1 t -1 2t + 1 t + 1 t - 1 1 1 3 = 0 -1 3 D= -2 2 -2 0 2 -2

È 1 2 2 ˘ È1 2 x ˘ = Í2 1 -2 ˙ Í2 1 2 ˙ Í ˙Í ˙ ÍÎ x 2 y ˙˚ ÎÍ2 -2 y ˙˚ È 9 0 x + 4 + 2y ˘ Í ˙ 0 9 2 x + 2 - 2 y˙ = Í Í x + 4 + 2 y 2 x + 2 - 2 y x 2 + 4 + y2 ˙ Î ˚ fi x + 2y + 4 = 0 2x – 2y + 2 = 0 x2 + y2 + 4 = 9 fi x = –2, y = –1 44. Solving x + 2y = 3, 3x + 4y = 7, we get x = 1, y = 1 \ a + 1 = 3 fi a = 2. 45. If p π –2,

3 -1 then z = p +2 2

1 and y = corresponding to this value we can p+2 get x. If p = – 1/2, we get z = 2. Corresponding to which we get infinite number of value of x and y. For p = –2, the system of equations is inconsistent. 46. The system of equations will have a unique solution if 1 1 1 1 0 0 2 1 -1 π 0 fi 2 -1 -3 π 0 3 2 k 3 -1 k - 3 [using C1 Æ C1 – C1, C3 Æ C3 – C1] fi (k – 3) (–1) –3 π 0 fi k π 0. 47. | A | π 0 fi 4(x + 1) – (2x – 3) (x + 2) π 0. fi 2x2 – 3x – 10 π 0 fi x π 3 ± 89 4 48. See Theory. 49. Subtracting first equation from the second and third, we get y + 3z = a – 1 3y + 9z = a2 – 1 \ a2 – 1 = 3(a – 1) fi a = 1 or a = 2 50. The system of equations will have a non-trivial solution if D = 0 where

[using C1 Æ C1 + C2] = (2t + 1) (2 – 6) = – 4(2t – 1) Note that D = 0 for t = –1/2 i.e., for just for one value of t. 51. Choice (d) is not true as | A + A | = | 2A | = 8| A | 52. A = iB Ê 1 -1ˆ Ê 1 -1ˆ = –2B fi A2 = i2B2 = (–1) Á Ë -1 1 ˜¯ ÁË -1 1 ˜¯ fi A4 = (–2B)2 = 4B2 = 4(2B) = 8B fi A8 = (8B)2 = 64B2 = 64(2B) = 128B 53. We have 3+i i ˆ Ê 2 7 - i˜ A¢ = Á 3 - i p Á ˜ Ë -i 7 + i e ¯ 3 - i -i ˆ Ê 2 Á 7 - i˜ = A fi A*= ( A¢ ) = 3 + i p Á ˜ Ë i 7+i e ¯ Thus A is a Hermitian matrix. 54. Statement-2 is true since |A| π 0, implies A–1 exists. \ AX = B fi A– 1 (AX) = A– 1 B fi (A–1 A)X = A–1 B fi IX = A–1 B fi X = A–1 B That statement-1 is false can be seen by the following example. È0 0 1 ˘ È0 1 0 ˘ Í ˙ Let A = 0 0 0 and B = Í0 0 0 ˙ then Í ˙ Í ˙ ÍÎ0 0 0 ˙˚ ÍÎ0 0 0 ˙˚ AB = O but neither A = O nor B = O 55. Since B commutes with I, we can use binomial theorem to obtain Ê nˆ Ê nˆ Ê nˆ (I + B)n = I + Á ˜ B + Á ˜ B2 + º + Á ˜ Bn Ë1 ¯ Ë 2¯ Ë n¯ Since B2 = O, we get Br = O " r ≥ 2. Thus, (I + B)n = I + nB Now,

5.38

Complete Mathematics—JEE Main

È0 p ˘ A = I + B where B = Í ˙ Î0 0 ˚ Since B2 = O, we get È1 100p ˘ A100 =I + 100B = Í 1 ˙˚ Î0 n n –1 56. Since a0 A + a1 A + º + an – 1 A + an I = O, and an π 0, we get AB = I where an – 1 a a I B = – 0 An – 1 – 1 An – 2 –  – an an an fi B = A–1. 57. If |A| π 0, A is invertible and we can write AX = B as X = A–1 B. \ AX = B has a unique solution and hence is consistent. Subtracting (2) from (3) and (1) and (2), we get the system of equations as 3x + 4y + 5z = a (4) x+y+z= b–a (5) x+y+z= c–b (6) As a, b, c are in A.P. b – a = c – b \ the last two equations are identical. From (4) and (5) we obtain x = 4b – 5a + k y = 4a – 3b – 2k z= k where k is an arbitrary complex number. Thus, the system of equations in statement-1 is consistent. 58. Suppose A is symmetric, then A¢ = A Since X¢AY is a 1 ¥ 1, matrix, X¢AY = (X¢AY)¢ = Y¢A¢(X¢)¢ = Y¢AX Next, suppose X¢AY = Y¢AX for each pair of X and Y. È1 ˘ È0 ˘ = Í ˙ , E2 = Í ˙ . Let E1 Î0 ˚ Î1 ˚

\ Statement-2 is true Also, BX = X fi (I – B)X = O fi X = (I – B)–1 O = O Thus, both the statements are true but statement-2 is not a correct reason for it.

Level 2 Ê a -bˆ Ê 1 - tan q 61. Á = Ë b a ˜¯ ÁË tan q 1

fi

1 = abc

a +1 b c a b +1 c a b c +1

Using C1 Æ C1 + C2 + C3, we get b c 1 1 (a + b + c + 1) 1 b + 1 c | A| = abc b c +1 1 Using R2 Æ R2 – R1, R3 Æ R3 – R1, we get 1 b c 1 | A| = (a + b + c + 1) 0 1 0 abc 0 0 1 1 (a + b + c + 1) π 0 abc 63. A2 = 5A – 7I =

fi A4 = (A2)2 = (5A – 7I ) (5A – 7I ) = 25 A2 – 70A + 49I = 25 (5A – 7I ) – 70A + 49I

a12 = a21

Thus, A is symmetric. 59. For truth of statement-2, see theory As det (Adj A) = (det A)n – 1 = O if det A = O Thus, statement-1 is true and statement-2 is correct reason for it. 60. We have (I + B + B2 + … + Bk) (I – B) = (I – B) + (B – B2) + (B2 – B3) + … + (Bk – Bk + 1) k +1 =I – B fi

- sin q cos q ˆ ˜ cos2 q ¯

Ê cos 2q - sin 2q ˆ =Á Ë sin 2q cos 2q ˜¯ fi a = cos2q, b = sin2q 1 + 1/ a 1 1 62. 1 1 +1 / b 1 1 1 1 + 1/ c

Taking X = E1 and Y = E2, then E1¢AE2 = E2¢AE1

2 ˆ Ê cos q ˜¯ Á Ë sin q cos q

I + B + B2 + … + Bk = (I – Bk + 1) (I – B)–1

= 55A – 126I 8

fi A = (A4)2 = (55A – 126I ) (55A – 126I ) = 3025 A2 – 13860A + 15876 I = 3025 (5A – 7I) – 13860A + 158766I = 1265A – 5299 I Thus, a = 1265. 1 cos ( b - a ) cos (g - a ) 1 cos (g - b ) 64. |A| = cos (a - b ) cos (a - g ) cos ( b - g ) 1

Matrices 5.39

cos a = cos b cos g

sin a sin b cos g

0 cos a 0 cos b 0 cos g

sin a sin b sin g

We have

0 0 =0 0

a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – bc – ca – ab) =1

fi A is singular.

fi a3 + b3 + Èa 69. Let A = Íb Í ÍÎ c

65. | A(q)| = sin2q – i2 cos2q = 1. Ê sin q \ A(q) – 1 = Á Ë - i cos q

- i cos q ˆ sin q ˜¯

= A(p – q ) n

66. A B = A

n–1

c3 = 7 b c˘ d e˙ ˙ e f ˙˚

Trace of A2 = (a2 + b2 + c2) + (b2 + d2 + e2)

(AB) = An – 1 (BA)

+ (c2 + e2 + f 2)

= An – 2 (AB) A

fi a2 + 2b2 + 2c2 + d2 + 2e2 + f 2 = 0

= An – 2 (BA) A = An – 2 BA2

As a, b, c, d, e, f are real numbers, we get

Continuing in this way, we get

a = b = c = d = e = f = 0.

AnB = BAn

\A=0

Similarly, ABn = BnA, (AB)n = AnBn

70. Let k = 2015. We are given

and AnBn = BnAn

(AB)k = Ak Bk and (AB)k + 1 = Ak

However, AnB = BnA does not hold.

+1

Bk

+1

We have

67. AX = l X

Ak + 1 B

k+1

= (AB)k + 1 = (AB)k (AB) = (AkBk) (AB)

fi (A – l I) X = 0

fi Ak + 1 Bk + 1 = Ak (Bk A) B

As X π 0, |A – l I | = 0 1- l 2 1 0 1 l 1 =0 fi 3 1 1- l

As A, B are non-singular matrices, we get

fi (1 – l)

ABk = Bk A. Similarly, using (AB)k + 1 = Ak + 1 Bk + 1 and (AB)k + 2 = Ak + 2 Bk + 2

1- l -1 2 +3 1 1- l 1- l

we get AB k + 1 = B k + 1 A

1 =0 -1

Now, AB k + 1 = B k + 1 A = B(B k A)

= (1 – l) [(1 – l)2 + 1] + 3[– 2 – (1 – l)] = 0 = (1 – l)3 – 2(1 – l) – 6 = 0

fi

fi l3 – 3l2 + 5l + 7 = 0

(1)

If l1, l2, l3 are roots of (1), then l1 + l2 + l3 = 3.

Èa b c ˘ Èa b c ˘ Èa 68. A = ÍÍb c a ˙˙ ÍÍb c a ˙˙ = Í b Í ÍÎ c a b ˙˚ ÍÎ c a b ˙˚ ÍÎ b where a = a2 + b2 + c2, 2

b = bc + ca + ab. 2

As A = I, we get a2 + b2 + c2 = a = 1 bc + ca + ab = 0 Now, (a + b + c)2 = a + 2b = 1 fia+b+c=1

= B(AB k)

b b˘ a b˙ ˙ b a ˙˚

(AB) B k = (BA) B k

As B is invertible, we get AB = BA. 71. |adj A| = |A|2 = 100 = |A|2 fi |A| = ± 10 È1 ˘ 72. M Í0 ˙ = M Í ˙ ÍÎ0 ˙˚

Ê È 1 ˘ È0 ˘ˆ Á Í-1˙ + Í1 ˙˜ Á Í ˙ Í ˙˜ Ë ÍÎ 0 ˙˚ ÍÎ0 ˙˚¯

È-1˘ È 1 ˘ È0 ˘ = Í 2 ˙ + Í 1 ˙ = Í3 ˙ Í ˙ Í ˙ Í ˙ ÍÎ 3 ˙˚ ÍÎ-1˙˚ ÍÎ2 ˙˚ È0 ˘ Ê È1˘ È1 ˘ È0 ˘ˆ Í ˙ and M 0 = M Á Í1˙ - Í0 ˙ - Í1 ˙˜ Í ˙ Á Í ˙ Í ˙ Í ˙˜ Ë ÍÎ1˙˚ ÍÎ0 ˙˚ ÍÎ0 ˙˚¯ ÍÎ1 ˙˚ È 0 ˘ È0 ˘ È-1˘ È 1 ˘ = Í 0 ˙ - Í3˙ - Í 2 ˙ = Í-5˙ Í ˙ Í ˙ Í ˙ Í ˙ ÍÎ12 ˙˚ ÍÎ2 ˙˚ ÍÎ 3 ˙˚ ÍÎ 7 ˙˚

5.40

Complete Mathematics—JEE Main

Ê 0 0 -1ˆ Ê 0 0 -1ˆ 2. We have A = Á 0 -1 0 ˜ Á 0 -1 0 ˜ = I Á ˜Á ˜ Ë -1 0 0 ¯ Ë -1 0 0 ¯

È1 ˘ È 0 ˘ M ÍÍ0˙˙ = ÍÍ3˙˙ fi m11 = 0 ÍÎ0˙˚ ÍÎ0˙˚

2

È0˘ È-1˘ Similarly, M ÍÍ1˙˙ = ÍÍ 2 ˙˙ fi m22 = 2 ÍÎ0˙˚ ÍÎ 3 ˙˚

3. As B is inverse of A, AB = I \

Ê 1 -1 1 ˆ Ê 4 2 2 ˆ Á 2 1 -3˜ Á -5 0 a ˜ = 10 I Á ˜Á ˜ Ë 1 1 1 ¯ Ë 1 -2 3 ¯

fi

Ê10 0 5 - a ˆ Ê10 0 0ˆ Á 0 10 a - 5˜ = Á 0 10 0˜ Á ˜ Á ˜ Ë 0 0 5 + a ¯ Ë 0 0 10¯

and m33 = 7 Thus, m11 + m22 + m33 = 9 73. (A–1 + B –1) [A – A (A + B)–1 A] = B –1 (B + A)A–1 A[I – (A + B)–1 A] –1

–1

= B (A + B) (A + B) [A + B – A]

fi

a=5

2

4. A – A + I = O fi I = A – A2 = A(I – A) fi A–1 = I – A

= B –1 I B = I 74. A – B – A (A + B)–1 A + B (A + B)–1 B = A[I – (A + B)–1 A] – [I – B (A + B)–1] B –1

= A(A + B) [A + B – A] – [A + B – B] (A + B)–1 B

5. For n = 1, all (a), (b), (c) and (d) hold good. We have È1 0 ˘ È1 A2 = Í ˙Í Î1 1 ˚ Î1 If (a) holds for some n

0˘ È1 0˘ = 1 ˙˚ ÍÎ 2 1 ˙˚ ≥ 1, then

= A (A + B)–1 B – A (A + B)–1 B

An + 1 = AnA = [nA + (n – 1)I] A

=O

= nA2 + (n – 1)A È1 0 ˘ È1 0˘ = nÍ - (n - 1) Í ˙ ˙ Î2 1˚ Î1 1 ˚ 2n - 1 0 ˘ = ÈÍ ˙ Î 3n - 1 2n - 1˚

75. A + B = AB fi I – (A + B) + AB = I fi (I – A) (I – B) = I fi (I – A)–1 = I – B fi I – (B + A) + BA = I

È1 0˘ È1 0˘ π (n + 1) Í + nÍ ˙ ˙ = (n + 1) A + nI Î1 1 ˚ Î0 1˚

fi A + B = BA

Similarly (b) does not hold for n + 1.

Thus, AB = BA

For (c),

fi (I – B) (I – A) = I

An +1 = AnA = [nA – (n – 1)I ]A

76. Use

= nA2 – (n – 1)A

O = A3 = (a + d)2 A Ê xy 2 77. A = Á 0 Á ÁË y 2

0 - x2 ˆ 0 0 ˜ ˜ 0 - xy ˜¯

È1 0 ˘ È1 0˘ = nÍ - (n - 1) Í ˙ ˙ Î2 1˚ Î1 1˚ 1 0˘ = È Í n + 1 1˙ = (n + 1) A - nI Î ˚ Thus, (c) holds.

A3 = O

Previous Years' AIEEE/JEE Main Questions 2

Èa b ˘ Èa b ˘ Èa + b 1. A2 = Í ˙Í ˙=Í Î b a ˚ Î b a ˚ Î 2ab \ a = a2 + b2, b = 2ab

2

2ab ˘ ˙ a 2 + b2 ˚

Alternatively È0 0˘ 2 Put B = Í ˙ and note that A = I + B and B = 1 0 Î ˚ O. As I and B commute.

Matrices 5.41

An = (I + B)n = I + nB

[∵ B2 = O]

= n(B + I) – (n – I) I = nA – (n – I)I

Now, A2 = I fi |A2| = |I|

A2 – B2 = (A – B)(A + B)

6.

= A2 – BA + AB – B2 fi

BA = AB

Ê a 0ˆ commutes with 7. Each matrix of the form Á Ë 0 a ˜¯ A. 8.

|A| = 25a fi |A2| = |A|2 = 25 fi (25a)2 = 25

fi

|A|2 = 1 fi |A| = ± 1.

Suppose |A| = 1. In this case, A2 = I fi A = A–1 Ê a b ˆ Ê d -b ˆ fiÁ = Ë c a ˜¯ ÁË -c a ˜¯ fi

1 fi a = 5 Ê a bˆ 9. Let A = Á Ë c d ˜¯

a = d, b = 0, c = 0 Tr(A) = 0 fi a + d = 0

fi

2a = 0 fi a = 0

Ê 0 0ˆ In this case A = Á Ë 0 0˜¯ Thus, in this case, if we assume statement-2 is true then we get statement-2 is false.

Now, A2 = I fi det(A2) = 1 fi (det A)2 = 1 fi det A = ± 1. Also, A2 = I fi A = A–1

In case |A| = –1, then

Ê a bˆ Ê d -b ˆ fi Á = det A Á ˜ Ëc d¯ Ë -c a ˜¯ If det A = 1, then

A = – A–1

a = d, b = – b, c = – c fi a = d, b = c = 0. Ê a 0ˆ In this case A = Á Ë 0 a ˜¯ 2 |A| = 1 fi a = 1 fi a = ± 1.

Ê a b ˆ Ê d -b ˆ fiÁ = fia+d =0 Ë c d ˜¯ ÁË -c a ˜¯ \ |A| = –1 fi Tr(A) = 0.

Therefore Statement-1 is true and Statement-2 is false.

\ A = I or A = –I. A contradiction. Thus, det (A) = –1. Ê a bˆ Ê d -b ˆ Ê - d = -1Á = \Á ˜ Ëc d¯ Ë -c a ˜¯ ÁË c

a b˘ 11. Let A = ÈÍ ˙ Îc d ˚

bˆ - a ˜¯

\ a = –d fi Tr(A) = a + d = 0 \ Statement-1 is true and Statement-2 is false. Ê a bˆ 10. Let A = Á Ë c d ˜¯ Ê d -b ˆ adj A = Á Ë -c a ˜¯ |adj A| = ad – bc = |A| \ Statement-2 is true. Ê a bˆ =A adj (adj A) A = Á Ë c d ˜¯ Statement-1 is also true but Statement-2 is not a correct reason for it.

12. The matrix È1 Íc Í ÍÎ e

a 1 f

b˘ d˙ ˙ 1 ˙˚

where exactly one of a, b, c, d, e, f is 1 and rest of them are zeros, is invertible. There are six such matrices. È1 0 Also, the matrix Í 0 1 Í ÍÎ 1 0 Thus, there are at least invertible.

1˘ 0 ˙ is invertible. ˙ 0 ˙˚ 7 such matrices which are

13. Adding the first two equations and subtracting the third from the sum, we obtain (x1 + 2x2 + x3) + (2x1 + 3x2 + x3) – (3x1 + 5x2 + 2x3) = 3 + 3 – 1 fi0=5

5.42

Complete Mathematics—JEE Main

Thus, the system of equation has no solution. 14. The system of equations will have a non-zero solution if and only if 4 k 2

k 2 4 1 =0 2 1

fi4

k 1 k 4 4 1 -k +2 =0 2 1 2 1 2 2

19. We have Ê 1 ˆ Ê 0ˆ Ê 1 ˆ A(u1 + u2 ) = Au1 + Au2 = Á 0˜ + Á 1˜ = Á 1˜ Á ˜ Á ˜ Á ˜ Ë 0¯ Ë 0¯ Ë 0¯ We solve the above equation for u1 + u2. We consider the augmented matrix

fi 4(2) – k(k – 2) + 2(2k – 8) = 0 fi k2 – 6k + 8 = 0 fi k = 2, 4

Ê 1 0 0 1ˆ ( A / B) = Á 2 1 0 1˜ Á ˜ Ë 3 2 1 0¯ Applying R3 Æ R3 – 2R2 + R1 and R2 Æ R2 – 2R1, we get Ê 1 0 0 1ˆ Ê 1ˆ ( A / B) ∼ Á 0 1 0 -1˜ fi u1 + u2 = Á -1˜ Á ˜ Á ˜ Ë 0 0 1 -1¯ Ë -1¯

15. H = wI fi H70 = w70I But w70 = (w3)23w = w \ H70 = wI = H 16. The system of equations will have only trivial solution if and only if 1 -k k 3 3 1 ¤ (1)

fi (1) (0) – a (4 – 6) + (3)(4 – 6) = 16 fi 2a – 6 = 16 fi a = 11

1 -k π 0 -1

21. As the system of equations has no solution

3 -k k -k k 3 +k + (1) π0 1 -1 3 -1 3 1

fi 2(k2 + k – 6)

20. |P| = |A|2 = 16

0fik

–3, 2

8 4k k +1 = π k k + 3 3k - 1 k +1 8 = fi k 2 + 4 k + 3 = 8k k k +3 fi (k – 1) (k – 3) = 0 fi k = 1, 3 But

17. We have A¢ = A and B¢ = B Now, (A(BA))¢ = (BA)¢ A¢ = (A¢B¢)A¢ = (AB)A = A(BA) fi A(BA) is symmetric. Similarly, (AB) A is a symmetric matrix Also, (AB)¢ = B¢A¢ = BA Note that AB is symmetric if and only if AB = BA. Thus, both Statement-1 and Statement-2 are true but Statement-2 is not a correct explanation for the statement-1. 18. P3 = Q3 and P2Q = Q2P gives 2

8 4k = k + 3 3k - 1

8 8 4 4k 3 = = and = k +3 6 3 3k - 1 2 Thus, there is exactly one such value of k. and for k = 3,

[∵ matrix multiplication is associative]

3

For k = 1,

3

22. Adding first two equations and subtracting from third, we get (a – 8)x3 = b – 15 8, we get value of x3, and substituting in If a first two we get values of x1 and x2. When a 8, we get the system has a unique solution. Thus, a = 8 and hence b = 15.

2

P –PQ=Q –QP 2

fi P (P – Q) = – Q2(P – Q) fi (P2 + Q2) (P – Q) = 0 If det (P2 + Q2) 0, then P2 + Q2 is invertible and hence P = Q. Therefore, det (P2 + Q2) = 0.

When a = 8 and b = 15, we solve first two equations in terms of x3 to obtain infinite number of solutions. 23. 3p + 3q + 2r = 3

(i)

4p + 2q = 0

(ii)

Matrices 5.43

p + 3q + 2r = 1

(iii)

From (i) and (iii) 2p = 2 or p = 1 When p = 1, q = –2 [from (ii)]

a bˆ 27. Let A = ÊÁ Ë c d ˜¯ A2 = I fi det (A)2 = 1 fi det (A) = ± 1.

Now, 2p + q – r

Now, A2 = I fi A–1 = A

1 = 2 p + q - [1 - p - 3q ] 2 1 = 2(1) + ( -2) - [1 - 1 - 6] = -3 2 24. As the system of equation has a unique solution.

Ê d - bˆ Ê a b ˆ fi det( A) Á = Ë -c a ˜¯ ÁË c d ˜¯

(1)

If det (A) = 1, then a = d, –b = b, –c = c

1 a 0 D= 0 1 a π0 a 0 1

fi a = d, b = 0, c = 0.

Using C1 Æ C1 + C2 + C3, we get

Thus, in this case A = I or A = –I.

\ det (A) = 1 fi a2 = 1 fi a = ± 1.

But A π I, –I.

1 a 0 D = (1 + a ) 1 1 a 1 0 1

Therefore det(A) = –1. fi Statement-2 is true.

Using R2 Æ R2 – R1, R3 Æ R3 – R1, we get

Now, from (1)

1 a 0 D = (1 + a ) 0 1 - a a 0 -a 1

Ê d -b ˆ Ê a b ˆ ( -1) Á = Ë -c a ˜¯ ÁË c d ˜¯ fi a + d = 0 fi tr(A) = 0

= (1 + a) (1 – a + a2) = 1 + a3 D

0fia

Thus, Statement-1 is true and Statement-2 is correct reason for it.

–1 [∵ a Œ R]

a bˆ 25. Let A = ÊÁ ; a, b, c Œ {0, 1, 2}. Ë c a ˜¯

BB¢ = (A–1A¢ )(A–1A¢ )¢ = (A–1A¢ )[(A¢ )¢(A–1)¢] = A–1(A¢A)(A–1)¢

|A| = a2 – bc When a = 0, bc

28.

0 fi b, c Π{1, 2}

There are four such values of b, c, viz. (b, c) = (1, 1), (1, 2), (2, 1) and (2, 2) When a = 1, both b and c cannot be 1. In this case there are 8 such pairs of (b, c). When a = 2, both b and c cannot be 2. In this case also, there are 8 such pairs of (b, c). 2˘ È 1 2˘ È 9 - 4˘ È1 26. A2 = Í ˙ Í ˙=Í ˙ Î 4 - 3˚ Î 4 - 3˚ Î - 8 17 ˚ 2

A + 4A – 5I 8˘ È - 5 0 ˘ È 9 - 4˘ È 4 =Í +Í ˙ ˙+Í ˙ Î - 8 17 ˚ Î16 - 12 ˚ Î 0 -5˚ È8 4 ˘ È2 1˘ =Í = 4Í ˙ ˙ Î8 0 ˚ Î 2 0˚

= A–1(AA¢ ) (A¢ )–1 = (A–1A)[A¢(A¢)–1] = (I)(I) = I 29. As B2 = 0, Br = 0 " r ≥ 2. Now, (I + B)50 = I + 50 C50 B50

50

C 1B +

= I + 50B fi (I + B)50 – 50B = I fi det(I + B)50 – 50B = 1 30. Applying C2 ´ C3, we get È1 3 2˘ È0 1 0˘ A Í0 3 2˙ = Í1 0 0˙ Í ˙ Í ˙ ÍÎ 0 1 1 ˙˚ ÍÎ 0 0 1 ˙˚

50

C 2B 2 +  +

[∵ Br = 0 " r ≥ 2]

5.44

Complete Mathematics—JEE Main

Applying C1 ´ C2, we get È3 1 2 ˘ È1 A Í3 0 2 ˙ = Í 0 Í ˙ Í ÍÎ1 0 1 ˙˚ ÍÎ 0 È3 1 -1 fi A = Í3 0 Í ÎÍ1 0 È6 ˘ 31. AB = Í ˙ Î8 ˚ fi

0 0˘ 1 0˙ ˙ 0 1 ˙˚ 2˘ 2˙ ˙ 1 ˙˚

Now, A4 – I = O = A2 + I and A3 – I = –A – I = –A + A2 = A(A – I) and A3 + I = –A + I = A4 – A = A(A3 – I) But A2 + I = O and A(A2 – I) = – 2A. \ A2 + I

A(A2 – I).

36. We know A(Adj A) = |A|I2 \ AA¢ = |A|I2

y + 2x + x = 6 3y – x + 2 = 8

fi

fi A3 = –A and A4 = I

3x + y = 6 –x + 3y = 6

fi

3x + y = –x + 3y

fi

y = 2x

32. Let C = AB – BA, then C¢ = (AB – BA)¢ = (AB)¢ – (BA)¢ = B¢A¢ – A¢B¢ = (–B)A – A(–B) = –BA + AB = C fi AB – BA is symmetric. 33. AAT = 9I È 1 2 2 ˘ È1 2 a ˘ È1 0 0˘ Í ˙ Í ˙ fi 2 1 -2 2 1 2 = 9 Í 0 1 0 ˙ Í ˙Í ˙ Í ˙ ÍÎ a 2 b ˙˚ ÎÍ 2 -2 b ˙˚ ÍÎ 0 0 1 ˙˚ È 9 0 a + 4 + 2b ˘ È 9 0 0 ˘ Í ˙ 0 9 2a + 2 - 2b ˙ = Í 0 9 0 ˙ fiÍ Í ˙ Í a + 4 + 2b 2a + 2 - 2b a 2 + 4 + b 2 ˙ ÍÎ 0 0 9 ˙˚ Î ˚

È5a -b ˘ È 5a 3˘ È1 0 ˘ fi Í = (10a + 3b) Í Í ˙ ˙ ˙ Î 3 2 ˚ Î -b 2 ˚ Î0 1 ˚ 2 2 È1 0 ˘ È ˘ fi Í25a + b 15a - 2b ˙ = (10a + 3b) Í ˙ Î0 1 ˚ 13 ˚ Î 15a - 2b

fi 25a2 + b2 = 10a + 3b, 15a – 2b = 0, 10a + 3b = 13 Now, b = (15/2)a 100 +

45 a = 13 2

fi a = 2/5, b = 3 \ 5a + b = 2 + 3 = 5 Ê cos a 37. P = Á Ë - sin a Ê cos a P¢ = Á Ë sin a

sin a ˆ where a = p/6 cos a ˜¯ - sin a ˆ cos a ˜¯

Ê 1 0ˆ We have PP¢ = Á Ë 0 1˜¯ –1 fi P¢ = P Thus, Q = PAP–1 fi P–1QP = A fi A2 = (P–1QP)2 = P–1Q2P fi A3 = P–1Q3P Continuing like this we get,

fi a + 2b = –4, a – b = –1, a2 + b2 = 5

A2015 = P–1Q2015P = P¢Q2015P

fi (a, b) = (–2, –1)

Ê 0 1ˆ We have A = I + B where B = Á Ë 0 0˜¯

34. |5 adj A| = 5 fi 53 |adj A| = 5 fi |adj A| = 1/25 2

fi |A| = 1/25 fi |A| = ± 1/5 È 0 -1˘ È 0 -1˘ 2 35. A = Í ˙Í ˙ = -I Î 1 0˚ Î 1 0˚

Ê 0 1ˆ Ê 0 1ˆ Ê 0 0ˆ Also, B2 = Á = Á ˜ Á ˜ Ë 0 0¯ Ë 0 0¯ Ë 0 0˜¯ Ê 0 0ˆ fi Br = Á "r≥2 Ë 0 0˜¯ Ê 1 2015ˆ Thus, A2015 = I + 2015B = Á 1 ˜¯ Ë0

Matrices 5.45 2016

– 2A

=A

2014

2

(A – 2A – I)

=A

2014

(A – ( 2 + 1)I)(A + ( 2 – 1)I)

38. Let B = A

2015

–A

2014

Note that B2 = O. A2 = (I + B)2 = I2 + 2IB + B2 = I + 2B

-4 -1 We have |A| = = –1 3 1 |A – ( 2 + 1)I| =

3

A = AA2 = (I + B)(I + 2B) = I2 + BI + 2(IB) + 2B2

-4 - ( 2 + 1)

-1

3

1 - ( 2 + 1)

= I + 3B Thus,

[∵

B2 = O]

2A2 – A3 = 2(I + 2B) – (I + 3B) = I + B = A.

= (–4 – ( 2 + 1)(1 – ( 2 + 1) + 3

3. Statement-2 is true.

= ( 2 + 1)2 + 3( 2 + 1) – 1 = 5 + 5 2 |A + ( 2 – 1)I| =

-4 + ( 2 - 1)

-1

3

1 + ( 2 - 1)

= (–4 + ( 2 – 1))(1 + ( 2 – 1)) + 3 2

= ( 2 – 1) – 3( 2 – 1) – 1

tr(AB – BA) = tr(AB) – tr(BA) = 0 But tr(I) = 2. \

AB – BA

I.

Thus, both statements are true and Statement-2 is a correct explanation for Statement-1. 4. As A РaI is invertible for all a ΠR.

=5–5 2 Thus det (B) = (–1)2014 (5 + 5 2 )(5 – 5 2 ) = –25 39. A2 – 5A = 7I = O fi 7I = A(5I – A) = (5I – A)A

det (A – aI)

" a ΠR.

0 0

" a ΠR.

fi a2 – (a + d)a + ad – bc

0 " a ΠR.

fi (a – a)(d – a) – bc

Therefore

I = AB = BA 1 where B = (5I – A) 7 \ A–1 = B =

[See Theory]

(a + d)2 – 4(ad – bc) < 0 (a – d)2 + 4bc < 0

fi

1 (5I –A) 7

Therefore, bc < 0.

fi Statement-1 is true.

Also, a2 + d2 – 2ad + 4bc < 0

We have

fi

0 £ a2 + d2 < 2ad – 4bc

= A(5A – 7I) – 2(5A – 7I) – 3A + I

fi

bc
n 1 2 n (c)

(d)

5

Statement-2: For every natural number n ≥ 2, n ( n + 1) < n + 1.

[2008]

16. The sum to infinity of the series S = 1 + 10 14 + + ... , is 33 34 (a) 4 (c) 2

2 6 + 3 32

+

(b) 6 (d) 3

[2009]

17. A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1 = a2 = ... a10 = 150 and a10, a11, ... are in A.P. with common difference – 2, then the time taken by him to count all the notes is (a) 125 minutes (b) 135 minutes (c) 24 minutes (d) 34 minutes [2010] 18. A man saves ` 200 in each of the first three months of his service. In each of the subsequent months his saving is ` 40 more than the saving of the immediately previous month. His total saving from the start of the service will be ` 11040 after. (a) 21 months (b) 18 months (c) 19 months (d) 20 months [2011] 19. Let an be the nth term of an A.P. 100

If

 a 2r r =1

100

= a and

 a 2r -1

= b, then

r =1

common difference of the A.P. is 1 (a – b) (a) a - b (b) 100

a then 6 equals a21 (a)

[2006]

7 2

13. If a1, a2, ... an are in H.P., then the expression a1a2 + a2a3 + ... + an – 1 an is equal to (b) n(a1 – an) (a) (n – 1) a1 an (d) n a1 an [2006] (c) (n – 1) (a1 – an)

(a)

(b) 1

(c) 0

(c)

(b)

41 11

(c) b - a

(d)

1 (a – b) 200

[2011]

Progressions

20. If 100 times the 100th term of an A.P. with non zero common difference equals the 50 times its 50th term, then the 150 term of this A.P. is (a) 150 times its 50th term (b) 150 (c) zero (d) –150 [2012] 21. Statement-1 : The sum of the series 1 + (1+2 +4) + (4 + 6 + 9) + (9 + 12 + 16) + ... + (361 + 380 + 400) is 800. n

Statement-2 :

 (k

k =1

3

3

)

- ( k - 1) = n , for any natural

number n

3

[2012]

22. If x, y, z are in A.P. and tan–1 x, tan–1 y and tan–1 z are also in A.P., then (a) 2x = 3y = 6z (b) 6x = 3y = 2z (c) 6x = 4y = 3z (d) x = y = z [2013] 23. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, . . . is 7 7 (b) (179 + 10–20) (a) (99 – 10–20) 9 81 7 (c) (99 + 10–20) 9

7 (d) (179 – 10–20) [2013] 81

24. Given sum of the first n terms of an A.P. is 2n + 3n2. Another A.P. is formed with the same first term and is double the common difference, the sum of n terms of the new A.P. is (b) 6n2 – n (a) n + 4n2 (c) n2 + 4n (d) 3n + 2n2 [2013, online] 25. The sum 3 5 7 + 2 + 2 + . . . upto 11 terms 2 2 1 1 +2 1 + 22 + 32 is 7 11 (a) (b) 2 4 11 60 (c) (d) [2013, online] 2 11 26. Given a sequence of 4 numbers, first three of which are in G.P., and the last three are in A.P. with common difference six. If first and last term of the sequence are equal, then the last term is (a) 16 (b) 8 (c) 4 (d) 2 [2013, online] 27. The value of 12 + 32 + 52 + . . . + 252 is (a) 2925 (b) 1469 (c) 1728 (d) 1456

[2013, online]

8.35

28. Let a1, a2, a3, . . . be an A.P. such that a1 + a2 + . . . + a p a1 + a2 + . . . + aq Then (a)

=

p2 q2

;pπq

a6 is equal to a21

41 11

11 41 29. The sum of the series (c)

(b)

121 1681

(d)

121 1861

[2013, online]

22 + 2(42) + 3(62) + . . . upto 10 terms is (a) 11300 (b) 11200 (c) 12100 (d) 12300 [2013, online] 1 1 Ê ˆ Ê ˆ 30. If S = tan–1 Á 2 + tan–1 Á 2 Ë n + 3n + 3 ˜¯ Ë n + n + 1˜¯

1 ˆ Ê + . . . + tan–1 Á , then Ë 1 + (n + 19) (n + 20 ) ˜¯ tan (S) is equal to 20 (a) 401 + 20n (c)

20 2

n + 20 n + 1

(b) (d)

n 2

n + 20 n + 1 n 401 + 20 n [2013, online]

31. If a1, a2, a3, . . ., an . . . are in A.P. such that a4 – a7 + a10 = m, then sum of the first 13 terms of the A.P. is (a) 10 m (b) 12 m (c) 13 m (d) 15 m [2013, online] 32. Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is (a) 3 + 2

(b) 2 - 3

(c) 2 + 3

(d)

2+ 3

[2014]

33. If (10)9 + 2(11)1 (10)8 + 3(11)2 (10)7 + . . . + 10 (11)9 = k (10)9, then k is equal to 441 (a) (b) 100 100 221 [2014] 10 34. Given an A.P. whose terms are all positive integers. The sum of its nine terms is greater than 200 and less than 220. If the second term in it is 12, then its 4th term is (c) 110

(d)

8.36

Complete Mathematics—JEE Main

(a) 8 (c) 20

(b) 16 (d) 24

[2014, online]

35. If the sum 3 5 7 + 2 + 2 + . . . + upto 2 2 1 1 +2 1 + 22 + 32 k 20 terms is equal to , then k is equal to 21 (a) 120 (b) 180 (c) 240 (d) 60 [2014, online] 36. In a geometric progression, if the ratio of the sum of first 5 terms to the sum of their reciprocals is 49, and the sum of the first and the third term is 35. Then the first term of this geometric progression is (a) 7 (b) 21 (c) 28 (d) 42 [2014, online] 37. The sum of the first 20 terms common between the series 3 + 7 + 11 + 15 + . . . and 1 + 6 + 11 + 16 + . . ., is (a) 4000 (b) 4020 (c) 4200 (d) 4220 [2014, online] 38. Let G be the geometric mean of two positive numbers 1 1 and . a and b, and M be the arithmetic mean of a b 1 If : G is 4 : 5, then a : b can be M (a) 1 : 4 (b) 1 : 2 (c) 2 : 3 (d) 3 : 4 [2014, online] 2 2 39. The least positive integer n such that 1 – - 2 – 3 3 2 1 . . . – n -1 < , is 100 3 (a) 4 (b) 5 (c) 6 (d) 7 [2014, online] 40. The number of terms in an A.P. is even: the sum of the odd terms in it is 24 and that the even terms is 1 30. If the last terms exceeds the first term by 10 , 2 then the number of terms in the A.P. is (a) 4 (b) 8 (c) 12 (d) 16 [2014, online] È 1 3n ˘ 41. Let f(n) = Í + n , where [n] denotes the greatest Î 3 100 ˙˚ 56

integer less than or equal to n. then

 f (n) is equal to

n =1

(a) 56 (c) 1287

(b) 689 (d) 1399

[2014, online]

42. If m is the A.M. of two distinct real number l and n (l, n > 1) and G1, G2 and G3 are three geometric between l and n, then G14 + 2G24 + G43 equals, (b) 4lm2n (a) 4l2 mn 2 (d) 4l2m2n2 [2015] (c) 4lmn 43. The sum of first 9 terms of the series 13 13 + 23 13 + 23 + 33 + + + 1 1+ 3 1+ 3 + 5 (a) 71 (c) 142

is

(b) 96 (d) 192

[2015]

30

4. The value of  (r + 2)(r - 3) is equal to r =16

(a) 7785 (c) 7775

(b) 7780 (d) 7770 rd

[2015, online]

th

45. The sum of the 3 and the 4 terms of a G.P. is 60 and the product of its first three terms is 1000. If the first term of this G.P. is positive, then its 7th term is: (a) 7290 (b) 320 (c) 640 (d) 2430 [2015, online] 5

k 1 = , then k is equal to 3 n =1 n( n + 1)( n + 2)( n + 3)

46. If Â

55 336 1 (c) 6

17 105 19 (d) [2015, online] 112 47. If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is 8 4 (b) (a) 5 3 (a)

(b)

(c) 1

(d)

2

2

7 4

2

[2016]

2

16 Ê 3ˆ Ê 2ˆ Ê 1ˆ 2 Ê 4ˆ ÁË1 ˜¯ + ÁË 2 ˜¯ + ÁË 3 ˜¯ + 4 + ÁË 4 ˜¯ + ,is m, 5 5 5 5 5 then m is equal to (a) 102 (c) 100

(b) 101 (d) 99

[2016, online]

49. Let a1, a2, a3, …, an, … be in A.P. If a3 + a7 + a11 + a15 to (a) 306 (b) 204 (c) 153 (d) 612 [2016, online]

Progressions

8.37

Previous Years' B-Architecture Entrance Examination Questions 1 1 , a, b, are 16 6 in G.P. and the last three are in H.P., then the values of a and b respectively are

1. If the first three terms of a sequence

(a)

1 4 , 12 9

(b)

4 3 , 7 4

1 (d) - , 1 4

1 1 (c) , 9 12

[2006]

2. If a, x, b are in H.P. and a, y, z, b are in G.P., then yz is the value of 3 x ( y + z3 ) (a) ab

(b)

1 2ab

1 ab (d) 2ab [2006] 2 3. If x = 1 + a + a2 + . . . (|a| < 1) and y = 1 + b + b2 + . . . (|b| < 1), then sum of the series (c)

1 + ab + a2b2 + . . . is xy (a) x - y -1 (c)

xy x + y +1

(b)

xy x - y +1

(d)

xy x + y -1

[2007]

4. If three distinct positive numbers a, b, c are in A.P. such that abc = 4, then value of b is always (a) greater than (2)2/3 (b) less than (2)2/3 2/3 (c) equal to (2) (d) equal to (2)3/2 [2007] 5. The value of the sum 20

1

1

1

1

 i ÈÍÎ i + i + 1 + i + 2 + . . . + 20 ˘˙˚

9. A tree, in each year grows 5 cm less than it grew in the previous year. If it grew half a metre in the first year, then the height of the tree (in metres) when it ceases to grow, is (a) 3.00 (b) 2.75 (c) 2.50 (d) 2.00 [2013] 48 47 46 2 + + + . . . + + 10. If (2)(3) (3)(4) (4)(5) (48)(49) 1 1 1ˆ 1 51 Ê = + K Ë 1 + + + . . . + ¯ , then 2 3 50 (49)(50) 2 K equals (a) –1

13. If  f(k ) = [2008]

6. The sum of the numbers between 200 and 400 that are divisible by 7 is (a) 8729 (b) 7511 (c) 6328 (d) 5712 [2009] 7. Let a, b and c be distinct real numbers. If a, b, c are in geometric progression and a + b + c = xb, then x lies in the set

(b) –

1 2

(c) 1 (d) 2 [2014] 11. If log10 2, log10 (2x – 1) and log10 (2x + 3) are three consecutive terms of an A.P. for (a) no real x (b) exactly one real x (c) exactly two real x (d) more than two real x [2014] 12. Let a, b, c, d and and e be real numbers such that 1 1 1 a > b > 0 and c > 0. If , , are in A.P., b, c, d a b c ab 2 are in G.P. and c, d, e are in A. P., then is ( 2a - b) 2 equal to

n

(b) 105 (d) 115

(b) (–1, 0) » (1, 2) (d) (0, 1) [2010]

8. If the sum of first n terms of two A.P.’s are in the ratio 3n + 8 : 7n + 15, then the ratio of 12th term is (a) 8 : 7 (b) 7 : 16 (c) 74 : 169 (d) 13 : 47 [2012]

(a) c (c) e

i =1

is (a) 100 (c) 110

(a) (1, 3) (c) (– •, –1) » (3, •)

k =1

(b) de (d) d

[2015]

10 1 2n , then  is equal to n +1 k =1 f( k )

11 (b) 220 20 55 (d) 110 [2015] (c) 18 14. Let a, b, c, d and e be distinct positive number. If a, 1 1 1 b, c and , , both are in A.P. and b, c, d are in c d e G.P. then (a)

8.38

Complete Mathematics—JEE Main

(a) a, b, c are in G.P (c) a, c, e are in A.P.

(b) a, b, c are in A.P. (d) a, c, e are in G.P.

2. 6. 10. 14.

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

3. 7. 11. 15.

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

4. (a) 8. (c) 12. (c)

Level 1 16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 64.

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

17. 21. 25. 29. 33. 37. 41. 45. 49. 53. 57. 61. 65.

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

43. (b)

44. (b)

45. (b)

46. (a)

47. (b)

48. (b)

Previous Years' B-Architecture Entrance Examination Questions

Concept-based (a) (d) (b) (a)

42. (b)

49. (a)

Answers 1. 5. 9. 13.

41. (d)

18. 22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62. 66.

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

19. 23. 27. 31. 35. 39. 43. 47. 51. 55. 59. 63. 67.

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

1. (d)

2. (b)

3. (d)

4. (a)

5. (d)

6. (a)

7. (c)

8. (b)

9. (b)

10. (a)

11. (b)

12. (c)

13. (b)

14. (d)

Hints and Solutions Concept-based 1. a1, a2, a3, . . . is a G.P. fi log5 a1, log5 a2, log5 a3, . . . are in A.P. 2. If r > 1 is the common ratio of the G.P., then 1 > 2. r and d = an – an – 1.

b + r

b + br = xb fi x – 1 = r + 3. Use : an = Sn – Sn – 1

4. Middle terms are nth and (n + 1)th terms.

68. (c)

69. (b)

70. (b)

71. (a)

72. (c)

73. (d)

74. (c)

75. (d)

a – b = tn = a + (n – 1)d

76. (d)

77. (d)

78. (a)

79. (b)

and a + b = tn + 1 = a + nd

80. (c)

81. (d)

82. (a)

83. (d)

84. (b)

85. (c)

2n [2a + (2n – 1)d] = 2na 2 5. Use : 3 + 7(n – 1) = 63 + 2(n – 1) S2n =

Level 2 86. (c)

87. (a)

88. (a)

89. (c)

90. (b)

91. (c)

92. (b)

93. (a)

94. (a)

95. (a)

96. (b)

Previous Years' AIEEE/JEE Main Questions

6. For example, if a = 2, b = b + 1, c forms a G.P.

2 – 1, c = 1, then a,

7. alog x = blog y = clog z = k (say) fi a = k1/log x, b = k1/log y, c = k1/log z Now, a, b, c are in G.P.

1. (b)

2. (b)

3. (b)

4. (a)

5. (b)

6. (c)

7. (d)

8. (c)

9. (b)

10. (c)

11. (a)

12. (a)

13. (a)

14. (d)

15. (b)

16. (d)

fi log x, log y, log z are in H.P.

17. (d)

18. (a)

19. (b)

20. (c)

However, nothing can be said about x, y, z.

21. (a)

22. (d)

23. (b)

24. (b)

25. (c)

26. (b)

27. (a)

28. (c)

29. (c)

30. (c)

31. (c)

32. (c)

33. (b)

34. (c)

35. (a)

36. (c)

37. (b)

38. (a)

39. (d)

40. (b)

fi

1 1 1 are in A.P. , , log x log y log z

8. S3 =

a (1 - r 3 ) a (1 - r 6 ) , S6 = 1- r 1- r

S3 1 - r 3 125 125 = = fi S6 1 - r 6 152 152

Progressions

fi 1 + r3 =

152 27 Ê 3 ˆ 3 fi r3 = = 125 125 Ë 5 ¯

1 Ê 1 3ˆ 3a + b + ¯ = Ë 4 a b 4ab

A3 =

a, a – 18, a – 9, a

1 15 (3a + b) (a + b) (a + 3b) = 64 2 1 5 and (a + 3b) (a + b) (3a + b) = 3 18 64(ab)

As a, a – 18, a – 9 are in G.P.

Thus, (ab)3 =

(a – 18)2 = a(a – 9)

fi ab = 3.

fi r = 3/5

Now,

9. As the common difference is 9, we let the four numbers be

2

2

fi a – 36a + 324 = a – 9a

a+b b+c = –b 1 - ab 1 - bc

13. b –

fi 27a = 324 fi a = 12 Thus, sum of the four numbers is 21

b - ab2 - a - b b + c - b + b2 c = 1 - ab 1 - bc

fi

10. Let two positive numbers be a, b. Then a + b = 2(7) = 14 and

ab = 5 or ab = 25

fi–

Now, a2 + b2 = (a + b)2 – 2ab = 196 – 50 = 146

8 (1 - r )3 1 - r3

= 24 fi

1 - 2r + r 2 1 + r + r2

fi 2abc = a + c 2b = \

=3

fi 2r2 + 5r + 2 = 0 fi r = –2, –1/2

1 1 , b, are in A.P. a c

S=

Now, t5 = ar4 = 3/16 12. Let three arithmetic means between a and b be a1, a2, a3. Then a, a1, a2, a3, b are in A.P. Let common difference of this A.P. be d. Then 1 d = (b – a) 4 1 a1 = a + b = (3a + b) 4 1 a2 = a + 2d = (a + b) 2 1 a3 = a + 3d = (a + 3b) 4 If A1, A2, A3 are arithmetic means between 1/a and 1/b, then 1 Ê 3 1ˆ a + 3b + ¯ = A1 = Ë 4 a b 4ab

1 1 + a c

14. Let

As |r| < 1, r = –1/2, a = 3.

1 Ê 1 1ˆ a+b A2 = + ¯ = Ë 2ab 2 a b

c (1 + b2 ) a (b2 + 1) = 1 - bc 1 - ab

fi – a + abc = c – abc

1 2 (a + b2) = 73. 2 a a3 = 2, = 24 11. 1- r 1 - r3 fi

\

15 2 = 27 5 18

5 55 555 + + +... 17 (17)2 (17)3

1 5 55 + + ... S = 2 17 (17) (17)3 Subtracting, we get 1 ˆ = 5 + 50 + 500 + 5000 + . . . Ê 1 Ë 17 ¯ S 17 (17)2 (17)3 (17)4 16 5 17 5 85 S = = fiS= 17 1 - 10 17 7 112 2016

15.

 (-1)r

r =1

2016

But

 (-1)r

r =1

2016

2016

r =1

r =1

(a + rd) = a  ( -1)r + d  ( -1)r (r ) 2016

= 0 and

 (-1)r (r )

= 1008

r =1

Level 1 16. 2(2a + b) = (5a – b) + (a + 2b) fi b = 2a. Also, (a – 1)2 (b + 1)2 = (ab + 1)2 fi (a – 1) (b + 1) = ± (ab + 1) fi (a – 1) (2a + 1) = ± (2a2 + 1)

8.39

Complete Mathematics—JEE Main

8.40

2a2 – a – 1 = 2a2 + 1 or 2a2 – a – 1 = – 2a2 – 1 fi a = – 2, a = 1/4

fi t = 4, 48. \ g = 4, l = 48 Let a = 4a1, b = 4b1, where hcf (a1, b1) = 1.

1 1 1 1 + + =0 17. + a c-b c a-b fi

Also, 4a1 b1 = 48 fi a1 b1 = 12. fi a1 = 1, b1 = 12 or a1 = 3, b1 = 4

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

Thus, smaller of a, b, is 4 or 12. 99 2 2 2 21. < + + . . . + n -1 100 3 32 3

As a + c – b π 0, a(c – b) + c(a – b) = 0 fib=

2ac fi a, b, c are in H.P. a+c

18. 0 = Sp = Sp + q = =

p 2 [2a + (p – 1)d ] fi d = 1- p 2 p+q [2a + (p + q – 1)d ] 2

1 (p + q) [2a + (p – 1)d + qd ] 2

fi ma + nb =

ab

22. a, b, c, d are in A.P.

(1)

n

Thus, n = k a =

4 gl

=

fi (g + l)2 =

H.M. = 2 ab a+ b

2a b a+ b

169 48 169 (4) (192) = 132 (4)2 48

fi g + l = 52 Note that g, l are roots of

23. a + b = (2)

20. We have gl = ab = 192. 2gl 1 = 169 : 48 Also, (g + l ): g+l 2

( g + l )2

1 1 1 1 , , , are in H.P. a b c d 1 1 1 1 , , , are in H.P. fi d c b a Now multiply each term by abcd

ab (m + n)

= = k(say) b a From (1) and (2) 1 k ( a + b ) = ab fi k = 2

fi

1 1 > n - 1 fi 3n – 1 > 100 100 3

24.

or t – 52t + 192 = 0

4+ 2 5+ 2

, ab =

8+2 2 5+ 2

16 + 4 2 2a b = =4 a +b 4+ 2

2 1 1 + = cos x cos ( x - y) cos ( x + y) fi 2(cos2x – sin2y) = 2cosx cosx cosy. fi cos2x(1 – cosy) = sin2y = 1 – cos2y y fi cos2x = 1 + cosy = 2cos2 Ê ˆ . Ë2 ¯ y fi cos x sec Ê ˆ = ± 2 . Ë2 ¯

25. a(2n) = 1 + >1+

1 1 1 1 + ... + 2n + + 2 3 4 2 –1

1 Ê 1 1ˆ 1 1 1 1 + + Ê + + + ˆ + ... + Ë 5 6 7 8¯ 2 Ë 3 4¯

1 1 ˆ 1 Ê + ... + 2 n ˜ – + Á 2n – 1 Ë2 ¯ +1 2 22 n

t2 – (g + l)t + gl = 0 2

100

Thus, least value of n is 7.

fi m a ( a - b) = n b ( a - b) fi

fi

1-1 3

fi

ma + nb 1 = (m + n) = 2 m+n

m

(2 3) ÈÎ1 - (1 3)n - 1 ˘˚ 99 >

fin–1>5fin>6

( p + q)q = 1- p 19.

fi

>

1 1 2 4 22 n - 1 n 1 1 + + + + ... + 2 n + ÊÁ - 2 n ˆ˜ > if n > 1 Ë2 2 ¯ 2 2 2 4 8 2

Progressions

\ a(200) > 100. 4 44 444 + + + ... 26. S = 19 192 193

=1+ (1)

1 4 44 S = 2 + 3 + ... 19 19 19 Subtracting (2) from (1), we get Ê 18 ˆ S = 4 + 40 + 400 + ... Ë 19 ¯ 19 192 193 = fi

S=

fi (2)

x

y

fi 164 = 3m + 5m – [3(m – 1)2 + 5(m – 1) = 6m + 2 fi m = 27. b 34. G1 = ar, G2 = ar 2 where r = Ê ˆ Ë a¯

19 4 38 ¥ = 18 9 81

35. Use A1 + A2 = a + b and G1 G2 = ab. 36. Let n be the number of terms. Then n 2s . s = (a + l ) fi n = 2 a+l Also,

u

log a log b log c log d = = = 1/ z 1/ u 1/ x 1/ y Now, use log a, log b, log c, log u are in A.P.

fi

d= 37. x = 2 S

29. (a2p2 – 2abp + b2) + (b2p2 – 2bcp + c2) + (c2p2 – 2cdp + d 2) £ 0

where S =

14 1 1 1 1 1 S= + + ... = + + 1–1 2 2 4 8 16 32

ap – b = 0, bp – c = 0, cp – d = 0

30. a, b, c are in A.P. fi2

ax + 1

,2

bx + 1

,2

=

2ax, 2bx, 2cx are in G.P.

cx + 1

fi \

are in G.P.

357 357 + + ... 31. 2 ◊ 357 = 2 + 1000 (1000)2 =2+

38.

357 1000 357 2355 =2+ = 999 999 1 – 1 1000

4 7 10 + + + ... 5 52 53 1 1 4 7 S = + 2 + 3 + ... 5 5 5 5 Subtracting (2) from (1), we get 4 3 3 3 35 S = 1 + + 2 + 3 + ... = 1 + 5 5 5 1 – 15 5

32. S = 1 +

1 1 ¥2= 4 2 S = 1. x=2

a, b, c, d are in G.P. fi

1 2 3 4 + + + + ... 4 8 16 32

fi

As a, b, c, d and p are real this is possible if and only if

fi

l – a (l – a ) (l + a ) l 2 – a2 = = 2 s – (l + a) n – 1 2 s – (l + a)

1 1 2 3 + + ... S= + 2 8 16 32 Subtracting, we get

fi (ap – b)2 + (bp – c)2 + (cp – d)2 £ 0

b c d = = a b c

.

= a + ar 3 = a + b = 2A

fi x log a = y log b = z log c = u log d

fi

13

G12 G2 2 G 3 + G23 a 3r 3 + a 3r 6 + = 1 = G2 G1 G1 G2 a2 r 3

log a, log b, log c º are in A.P.

28. a = b = c = d

– 1

2

4 19 4 = 9 1 – 10 19

z

35 . 16

33. Tm = Sm – Sm

27. a1/x = b1/y = c1/z = ... = k(say) log a log b log c = = = º log k, fi x y z Now, use

S=

3 7 = 4 4

a a2 = 20 and = 100. 1– r 1 – r2 fi

(1) (2)

400 (1 - r )2 1 - r2

= 100

1– r 1 = fi 4 – 4r = 1 + r 1+ r 4 fi r = 3/5

fi

Thus,

a = 20 fi a = 8 25

39. ar 7 = 128 = 27 and r = 2 fi a = 1.

8.41

8.42

Complete Mathematics—JEE Main

Product of first five terms = a5 r10 = 1024. 40. l = ARp

– 1

, m = ARq

– 1

a + ( p - 1)d a + (q - 1)d a + (r - 1)d

D=

, n = ARr

– 1

p q r

fi (a2 + c2 + 4ac) (a2 + c2 – 2ac) = 0 1 1 1

where a = log A, d = log R. Use C1 Æ C1 – (a – d)C3 – dC3 to show D = 0. p a+ [2a + ( p - 1)d ] p2 2 41. = 2 fi q q a+ [2a + (q - 1)d ] 2 Now, put p = 11 and q = 41. 42. a =

2, r =

p –1 d p 2 = q –1 q d 2

fi [(a + c)2 + 2ac] (a – c)2 = 0 fi (a + c)2 + 2ac = 0

2 (35 - 1) 3 -1

2 (242) ( 3 + 1)

= 121 ( 6 + 2 ) ( 3 - 1) ( 3 + 1) 1 43. n(n + 1) (2n + 1) = 1015 6 fi (2n) (2n + 1) (2n + 2) = (28) (29) (30)

1 44. tr = 1 + 2 + ... + r = r (r + 1) (2r + 1) 6 1 = (2r3 + 3r2 + r) 6 2

2

n

1 1 \ Â tr = ◊ n2(n + 1)2 3 4 r =1 1 1 1 n(n + 1) + ◊ n(n + 1) (2n + 1) + 2 6 12 =

1 n(n + 1)[n2 + n + 2n + 1 + 1] 12

=

1 n(n + 1) (n2 + 3n + 2) 12

=

1 n(n + 1)2 (n + 2) 12

22

1 (22) (23)2 (24) = 23276 \ Â tr = 12 r =1 45. 2b = a + c, b2 =

fi

(a + c) 4

2

2a2 c2 a2 + c2 2

=

a π c]

fi 4b + 2ac = 0 a , b, c are in G.P. 2 46. am = n, an = m fi –

fi (m – n)d = an – an = – (m – n) fi d = – 1. Now, ar – am = (r – m) (– 1) 47. As b2 = ac, ax2 + 2bx + c = 0 can be written as ax2 + 2 ac x + c = 0 fi ( a x + c )2 = 0 fix= –

2a c

2

a2 + c2

c ,– a

c . a

c is a root of dx2 + 2ex + f = 0 a

\–

fi 2n = 28 or n = 14. 2

[∵

2

fi ar = am – r + m = m + n – r.

3

a (r10 - 1) S10 = = r -1 =

fi (a2 + c2)2 + 2ac(a2 + c2) – 8a2c2 = 0

c c fi d Ê ˆ – 2e + f = 0. Ë a¯ a fi

1 d f – 2e + = 0. a ac c

fi2

e d f + = b a c

fi

d e f , , are in A.P. a b c

48. 2, 2 x + 1, 2 x + 3 are in G.P. fi (2 x + 1)2 = 2(2x + 3) fi 22x = 5 fix=

1 log25 2

1 x, |x + 1|, |x – 1| are in A.P. 2 1 2|x + 1| = x + |x – 1| 2

49. As

(1)

If x < – 1, then (1) give us 1 – 2(x + 1) = x – (x – 1) 2 fix=–2 In this case common difference is 2 and sum to 10 terms is 10 [2(– 1) + (9) (2)] = 85 2

Progressions

If – 1 £ x < 1, 1 2(x + 1) = x – (x – 1) fi x = – 2/5 2 In this case common difference is 4/5, and sum to 10 terms is 10 È Ê 1 ˆ Ê 4ˆ ˘ 2 - + (9) Ë ¯ ˙ = 34 5 ˚ 2 ÍÎ Ë 5 ¯ If x ≥ 1, then (1) becomes 1 x+x–1 2x + 2 = 2 1 x=–3 fi 2 Not possible as x ≥ 1. 50.

2ab : ab = 4 : 5 fi a+b fi fi fi

a+b ab

3

3

2

1 + 2 + ... + r r (r + 1) = 1 + 3 + ... (2r - 1) 4r 2 16

 tr

r =1

But (x – a) (x – b) is least when x = \ T1 T2 T7 is least when d =

2

=

(r + 1) 4

2

1 Ê 17 2 ˆ = Á Â r – 1˜ 4 Ër =1 ¯

1 Ê1 ˆ (17) (18) (35) - 1¯ = 446 Ë 4 6 52. Given equation implies

t 81 = 28 where t = 32sin2q + 3 t fi t2 – 84t + 243 = 0 fi t = 3, 81 But 1/9 £ t £ 9. Thus, t = 3.

\ a1 = 1, d = 13

fi

a5 = a1 + 4d = 53.

56. Write the given series as Ê 1 ˆ + Ê 1 + 1 ˆ + Ê 1 + 1 ˆ ... + Ê 1 + 1 ˆ ˜ ÁË ˜ ˜ Á Ë1 + 2 ¯ ÁË 22 ¯ Ë 23 ¯ 2n ¯

fi S2m

2ac fi a, b, c are in H.P. a+c

53. Required sum = (2 + 4 + ... + 100) + (5 + 10 + ... + 100) – (10 + 20 + ... + 100) 50 20 10 (2 + 100) + (5 + 100) – (10 + 100) = 2 2 2 = (50) (51) + (10) (105) – 550 = 2550 + 1050 – 550 = 3050 54. T1 T2 T7 = (T7 – 6d) (T7 – 5d)T7 = 9(9 – 6d) (9 – 5d)

= m(2m + 1)2 – 2(2m)2 = m(2m – 1)2

1 ˆ Ê r +1– r ˆ Ê 58. ar = tan–1 Á = tan–1 Á Ë 1 + (r + 1)r ˜¯ Ë 1 + r (r + 1) ˜¯ = tan–1(r + 1) – tan–1(r) n

Thus, Sn =

 ar

= tan–1(n + 1) – tan–1(1)

r =1

Ê n +1–1 ˆ Ê n ˆ = tan–1 Á = tan–1 Á Ë 1 + (n + 1) (1) ˜¯ Ë n + 2 ˜¯

fi (a + c)2 – (a – c)2 = 2b(a + c) 4ac = 2b(a + c)

– 1

2m (2m + 1)2 = m(2m + 1)2 2

Now, put n = 2m – 1.

(a + c) (a + c – 2b) = (a – c)2

fib=

1 Ê 3 9ˆ 33 + ¯ = Ë 2 2 5 20

55. We have

57. We have S2m =

=

fi

1 (a + b). 2

1Ê 1 1- nˆ Ë 2 2 ¯ =n+1– 1 =n+ 1–1 2 2n

a b 5 a 1 fi + = = 2, b a 2 b 2 a : b = 4 : 1 or 1 : 4

Thus,

9 3 = 729 Êd – ˆ Ê d – ˆ Ë ¯ Ë 5¯ 2

5 2

=

3

51. t r =

2 ab 4 = a+b 5

8.43

59. tr=

1 1 1 1 – + = r (r + 1) (r + 2) 2r r + 1 2(r + 2)

[split into partial fractions] 1 ˆ 1Ê 1 1 ˆ 1 Ê1 – = Á – – Á ˜ ˜ 2 Ë r r + 1¯ 2 Ë r + 1 r + 2 ¯ n

fi

Sn =

 tr

r =1

= Thus, S30 =

=

1 ˆ 1 Ê1 1 ˆ 1Ê ÁË 1 – n + 1˜¯ – ÁË 2 – n + 2 ˜¯ 2 2

1 n n – 2 n + 1 4 ( n + 2) 1 30 30 30 È 64 – 31 ˘ 495 – = = Í ˙ 2 31 4(32) 1984 4 Î (31) (32) ˚

60. 9(n!)2 = n!(n+ 1)!

fi

9=n+1fin=8

This shows that 8!, 5(8!), 9(8!) are in A.P.

8.44

Complete Mathematics—JEE Main

2 . 3 = ar n

61. Here a = 81, r = Also, Sn – Sn

– 1

65. Coefficient of x99 – 1

= – (1 + 3 + 5 + ... 199) = – 1002 = – 10000

.

y = (Sn – Sn + 3 (Sn = ar

n – 1

= arn

– 4

– 1) – 2

– 3(Sn

– Sn

– 3ar

– 1

– 3)

n – 2

– Sn

– (Sn

+ 3ar

n

– 2)

– 3

n – 3

– Sn

– ar

 tr

– 4)

(r3 – 3r 2 + 3r – 1) 67. x =

=

Â

r +

r =1

1 n(n + 1) (n + 2) 6

y – 1 yx xy – x = ◊ y–x y–x y

68.

 [(2r – 1) (2r ) – (2r )2 ]

fi

x + yz = x(y + z).

a a3 108 = 3, = where | r | < 1. 3 1– r 13 1– r

r =1

1 = (2m – 1) (2m) (4m – 1) – 2 6

m –1

Â

\

r

27(1 - r )3 1- r

3

=

1 m(2m – 1) (4m – 1) – (m – 1)m 3

=

1 Ê n + 1ˆ n + 1ˆ Ê n - 1ˆ n (2n + 1) - Ê Ë ¯ Ë 3 2 2 ¯Ë 2 ¯

fi

1 – 2r + r 2 1 + r + r2

Â

[(r + 1)log a – (r – 1)log b]

r =1

=

n

n 2 L = lim (1 – 3–2) (1 + 3–2) (1 + 3–4) ... (1 + 3 –2 ) nÆ• 3 n

= lim (1 - 3-2 ) (1 + 3–4) º (1 + 3 –2 ) nÆ•

) =1

L = 3/2

64. Note that 1 1 1 + + ... + a1 a2 an 1 1 1 1 ˆ < ÊÁ 1 + + 2 + ...ˆ˜ ÊÁ 1 + + 2 + ...˜ ¯ Ë ¯ Ë 3 3 5 5 15 Ê 1 ˆÊ 1 ˆ = Á = is a G.P. with first term equal to 5 and common ratio 1/2. È = s Í1 – Ê 1 – Î Ë

Progressions •

 a2 n – 1

\

=

n =1

73.

74.

a1

=

1 – r2

2ab : ab = 12 : 13 a+b

5 = 20/3. 1–1 4

fi

fi fi

a 2 3 = or fi a : b = 4 : 9 or 9 : 4 b 3 2

5 ab, a + b + ab = 26 8

13 ab = 26 fi ab = 16 and a + b = 10 8 Thus, numbers are 2 and 8.

\

=

+ a 2n – 2 = (an – an – 1) (an + an – 1)

– 1

1 1– r

a

,B=

fi (a2 + a25) + (a6 + a21) + (a9 + a18) = 165 fi 3(a1 + a26) = 165 \

a1 + a2 + ... + a26

80. We are given Sn = cn2. Now, tn = Sn – Sn – 1 = cn2 – c(n – 1)2 = (2n – 1)c We have n

t12 + t22 + ... + tn2 = n

a log (1 - A ) = b log (1 - B -1 )

77. Note that ak – ak –1 = ak – 1 – ak – 2 " k ≥ 3 implies a1, a2, a3 ..., a12 are in A.P. Also, d = – 4. We have 12S = (a1 – a2) (a1 + a2) + (a3 – a4) (a3 + a4) + ... + (a11 – a12) (a11 + a12) = 4(a1 + a2 + ... + a12) 12 =4¥ [2a1 + (12 – 1) (– 4)] 2 S = 2[2(45) – 44] = 92

78. We have 1 – (– x )2013 1 + x 2013 P(x) = = 1 – (– x ) 1+ x

 (4k 2 – 4k + 1)

= c2

k =1

1 – rb

-1

 c2 (2k - 1)2

k =1

1

b log r = log(1 – 1/B)

fi a1 + a26 = 55 26 = (a1 + a26) 2

= 13 ¥ 55 = 715

4 È4 ˘ = c2 Í n(n + 1)(2n + 1) - n(n + 1) + n ˙ 2 Î6 ˚

1 1 fi 1–r = , 1 – rb = A B fi a log r = log(1 – 1/A) and

fi

1 2013 (3 + 1) 4

a2 + a6 + a9 + a18 + a21 + a25 = 165

a

\

[∵y = 1 = x – 2]

Thus,

– (an – 1 – an – 2) (an – 1 + an – 2) = d(an + an – 1 – an – 1 – an – 2) = d(2d) = 2d2 76. A =

= Q(1) = P(3)

a1 + a26 = a2 + a25 = a3 + a24 = ... = a9 + a18.

2ab 16 = , a + b + ab = 26 a+b 5

75. a n2 – 2a2n

 ai

i=0

79. We have

b 13 = a 6

a+b=

2012

Also

2 ab 12 = a + b 13

a + b

fi

8.45

=

c2 n È2 2 n2 + 3n + 1 – 6 n – 6 + 3˘˚ 3 Î

(

)

1 n 4n2 – 1 c 2 3 81. Sum to n terms of an A.P. is of the form an + bn2 and not of the form c + bn + an2 where c π 0.

(

=

82. |2x – 1| < 1 fi

fi

)

– 1 < 2x – 1 < 1

0 0]

8.48

Complete Mathematics—JEE Main

fi 3x = 3/4 fi x = 1 – log34 3

3

3

3

3

4. 1 – 2 + 3 – 4 +

–8 +9

= (13 – 23) + (33 – 43) +

3

+ (73 – 83) + 93

= (1 – 2) (12 + 22 + 2) + (3 – 4) (32 + 42 + (3)(4)) + (5 – 6) (52 + 62 + (5) (6)) + (7 – 8) (72 + 82 + (7)(8)) + 93 = – (12 + 22 + 32 +

8 2)

– (2 + 12 + 30 + 56) + 93 1 (8)(9)(17) – 100 6 = 729 – 204 – 100 = 425

= 93 –

5. Let S1 = {2x | 1 £ x £ 50} S2 = {5x | 1 £ x £ 20} S1 « S2 = {10x | 1 £ x £ 10}

12 + 2 ◊ 22 + 32 + 2 ◊ 42 + 52 + 2 ◊ 62 + … + (2m – 1)2 + 2(2m)2 1 (2m)(2m + 1)2 2 fi 12 + 2 ◊ 22 + 32 + 2 ◊ 42 + =

x ŒS2

= m(4m2 + 4m + 1) – 8m2

x ŒS1nS2

Ê 50 ˆ (51) + 5 Ê 20 ˆ (21) - 10 Ê 10 ˆ (11) = 2 Ë 2¯ Ë 2¯ Ë 2¯ = (50) (51) + 5(10) (21) – 5(10) (11) = (50) [51 + 21 – 11] = 3050

a ˆ fi 20 Ê = 100 Ë1+ r ¯

fi

=

1 1 1 ,y= ,z= 1- b 1- c 1- a a, b, c are in A.P.

10. We have x =

1 1 1 are in H.P. , , 1- a 1- b 1- c 1 È 1 - 1 ˘ 1 = 11. (b - a) x ÍÎ1 - b 1 - ax ˙˚ (1 - ax)(1 - bx) fi

a =5 1+ r

a /(1 - r ) 20 1 + r =4 = fi a /(1 + r ) 5 1- r

=

fi 1 + r = 4 – 4r fi 5r = 3 fi r = 3/5. 7. Let f(x) = Ax2 + Bx + C

¥

2

\ f(x) = Ax + C fi f¢(x) = 2Ax Now, a, b, c are in A.P. fi 2Aa, 2Ab, 2Ac are in A.P. fi f ¢(a), f ¢(b), f ¢(c) are in A.P.

1 È(1 - bx)-1 - (1 - ax) -1 ˘˚ (b - a) x Î 1 1 xn in is (b - a) x (1 - ax)(1 - bx) n+1 n x in (1 – bx)

xn + 1 in (1 – ax)n]

As f(1) = f(–1), B = 0

1 1 and Tn = m n

1 2 n (n + 1) 2

fi 1 – a, 1 – b, 1 – c are in A.P.

2

a a 6. = 100 = 20 , 1- r2 1- r

8. Tm =

+ (2m – 1)2

Putting n = 2m – 1, we get desired sum.

= Â x+ Â x- Â x

\

1 1 1 fid= . mn n m 1 Thus, a - d = - md n 1 1 = - =0 n n 9. Let n = 2m, then ( m - n) d =

= m(2m – 1)2

Required sum x ŒS1

1 1 and a + (n - 1)d = n m Subtracting we get

fi a + (m - 1)d =

=

1 {b n+1 - a n+1} b-a

12. Let d be the common difference of the A.P., then p [2a + ( p - 1)d ] p 2 2 1 = 2 q q [2a1 + (q - 1)d ] 2 1 a1 + ( p - 1)d p 2 fi = 1 q a1 + (q - 1)d 2

Progressions

Putting p = 11 and q = 41, we get a6 11 = a21 41 an are in H.P. 13. a1, a2, fi fi fi fi

1 1 1 are in A.P. , , a1 a2 an 1 1 1 1 1 1 - = = = d (say) a2 a1 a3 a2 an an-1 1 1 a1a2 = (a1 - a2 ), a2 a3 = (a2 - a3 ) , d d 1 an–1an = (a1 - an ) d

Thus,

a 1a 2 + a 2a 3 +

+ an –1an

1 [a1 - a2 + a2 - a3 + + a n – 1 – a n] d 1 = (a1 - an ) d 1 1 a -a But = + (n - 1)d fi 1 n = (n - 1)d an a1 an a1 =

\

a 1a 2 + a 2a 3 + n

14. ar = ar fi

n +1

+ ar

1=r+r

2

n+2

an

–1

an = (n – 1) a1an

, a > 0, r > 0

fi

2

r +r–1=0

1 r = ( 5 + 1) as r > 0 2 15. As n < n + 1 " n ≥ 2 fi

fi

n(n + 1) < (n + 1)2 " n ≥ 2

fi

n(n +1) < n + 1 " n ≥ 2

Thus, Statement-2 is true.

8.49

Subtracting, we get 2 1 4 4 4 S =1+ + 2 + 3 + 4 + 3 3 3 3 3 1 4 / 32 4 2 = 1+ + = + =2 3 1 - 1/ 3 3 3 fi

S=3

17. Suppose he takes n minutes to count 4500 notes. We have a1 + a2 + + a10 = 10(150), a11 = 148 and a11, a12, ..., an is an A.P. with common difference d = –2. We are given. a1 + a2 + ... + a10 + a11 + ... + an = 4500 fi fi fi

a11 + a12 +... + an = 3000 n - 10 [a11 + an ] = 3000 2 1 (n - 10)[148 + 148 + (n - 11)( -2)] = 3000 2

fi

(n – 10) (148 – n + 11) = 3000

fi

(n – 10) (159 – n ) = 3000

fi

n2 – 169n + 4590 = 0

fi

n2 – 135n – 34n + 4590 = 0

fi

(n – 135) (n – 34) = 0

fi

n = 135, 34

All the notes get counted in 34 minutes. 18. Suppose his total saving is ` 11040 after n months. His savings are 200, 200, 200, 240, 280, ... upto n terms Note that

Also, for n ≥ 2 1 1 1 + + + n 1 2

200, 240, 280, ... upto (n – 2) terms is an A.P. whose sum is 10640.

1 1 1 > + ++ n  nn

Thus, 1 (n - 2)[2(200) + (n - 3)(40)] = 10640 2 fi (n – 2) (n + 7) = 532

1 1 1 + + + > n 1 2 n \ Statement-1 is true.

fi n2 + 5n – 546 = 0 fi n = – 26, 21

n times

fi

Thus, both the statements are correct but Statement-2 is not a correct explanation of Statement-1. 2 6 10 14 16. Let S = 1 + + 2 + 3 + 4 + 3 3 3 3 1 1 2 6 10 S= + 2+ 3+ 4+ 3 3 3 3 3

As n > 0, n = 21. 19. Let d the common difference of the A.P. then 100

100

r =1

r=1

a – b = Â ( a2 r - a2 r -1 ) = Â d = 100d fi d = 1 (a - b ) . 100

8.50

Complete Mathematics—JEE Main

20. We have

7 È179 + 10-20 ˘˚ 81 Î 24. Sn = 2n + 3n2 =

100a100 = 500a50 fi

100 [a + 99d] = 50[a + 49d]

fi

(100 – 50)a = 50(49 – 2 ¥ 99)d

fi

a = – 149d

or

a150 = 0

fi

a = S1 = 2 + 3 = 5 a + (a + d) = S2 = 2(2) + 3(22) = 16

a + 149d = 0

\ d = 16 – 10 = 6 Sum of required A.P.

21. Statement-2 is true since n

3 3 3 3 Â (k 2 - (k - 1)3 ) = (1 – 0 ) + (2 – 1 ) +

k =1

fi

+

(n3 – (n – 1)3) = n3 n

 (k - (k - 1)) (k 2 + k (k - 1) + (k - 1)2 ) = n3

k =1 n

fi

 (k 2 + k (k - 1) + (k - 1)2 ) = n3

n [2a + (n - 1)(2d )] 2 n = [ 2(5) + (n - 1)(2)(6) ] 2 = 6n2 – n =

25. ar =

k =1

Putting n = 20, we get

2r + 1 1 + 22 + ... + r 2

=

20

 ((k - 1)2 + k (k - 1) + k 2 ) = 203

k =1

fi

1 + (1 + 2 + 4) + (4 + 6 + 9) +

+

(361 + 380 + 400) = 8000 22. 2y = x + z

=

2

(

1

(2r + 1) 6 ) r ( r + 1)( 2r + 1)

6 1 ˆ Ê1 = 6Á Ë r r + 1˜¯ r (r + 1)

11 1 11 \ Â ar = 6 Ê1 - ˆ = Ë ¯ 12 2 r=1

and 2 tan– 1 y = tan– 1x + tan– 1 z fi

Ê 2y ˆ –1 Ê x + z ˆ tan Á ˜ ÁË 2 ˜ = tan Ë1- y ¯ 1 - xz ¯

fi

2y x+ z = 2 1 - xz 1- y

fi

y2 = xz

fi

1 2 ( x + z )2 = xz fi (x – z) = 0 4 x=z

fi

–1

[

2y = x + z]

\ 2y = 2x fi y = x fix=y=z 23. S = 0.7 + 0.77 + 0.777 +

+ upto 20 terms

=

7 [0.9 + 0.99 + 0.999 + 9

=

7 [20 – (0.1 + 0.01 + 0.001 + 9

( 7È = Í20 9Í Î

1 10

) (1 - ( 110)20 ) ˘ 1 - 110

˙ ˙ ˚

upto 20 terms] upto 20 terms)]

a/r, a, ar and last three numbers be a, ar, 2ar – a We are given ar – a = 6 and 2ar – a = a/r fi 2r 2 – r = 1 fi (2r + 1) (r – 1) = 0 fi r = – 1/2, 1 r π 1 is not possible, for then a (1) – a = 6 or 0 = 16 Let r = –1/2. Thus, a(–1/2) – a = 6 fi a = – 4 a/r = 8 \ 27. 12 + 32 + 52 + 13

2

+ 252 13

= Â (2r - 1) = Â (4r 2 - 4r + 1) r =1

r =1

4 Ê 1ˆ = 4 Á ˜ (13)(14)(27) - (13)(14) + (13) Ë 6¯ 2 = 2925

Progressions

28. See Solution to Question 12. 10

10

= 10 Ê 11ˆ Ë 10 ¯

k =1

k =1

k = 100

29.  k (2k )2 = 4  k 3 4 (10) 2 (10 + 1)2 = 12100 4 1 ˆ 30. tk = tan– 1 ÊÁ Ë 1 + (n + k - 1)(n + k ) ˜¯

10

11 10 - 10 - 10 Ê ˆ Ë 10 ¯

9 [2a + (9 - 1)d ] = 9(a + 4d) 2 200 < 9(a + 4d) < 220

34. S9 =

As a + 4d is a positive integer

Ê (n + k ) - (n + k - 1) ˆ = tan– 1 Á Ë 1 + (n + k )(n + k - 1) ˜¯

a + 4d = 23, 24

= tan– 1 (n + k) – tan–1 (n + k – 1)

Also, a + d = 12 \ 3d = 11, 12

20

\ S = Â tk = tan– 1(n + 20) – tan–1 (n) k =1

As d is an integer, d = 4 Thus, t4 = a + 3d = (a + 4d) – d = 20

20 Ê ˆ = tan– 1 Á Ë 1 + n(n + 20) ˜¯ 20 fi tan (S) = 2 n + 20n + 1 31. Let d be the common difference of the A.P., then m = a4 – a7 + a10 = a4 + (a10 – a7)

[a + 4d = 23 is not possible] 35. From Question 25, 1 ˆ Ê1 ar = 6 Ë r r + 1¯ 20 120 k 1ˆ Ê = fi  ar = 6 1 = Ë 21¯ 21 21 r=1

= a1 + 3d + 3d = a1 + 6d

\ k = 120

Now, S13 = 13 [2a1 + (13 - 1)d ] = 13m 2

a a , , a, ar, ar2 r2 r Reciprocals are

32. Let three positive numbers in G.P. be

r2 r 1 1 1 , , , , a a a ar ar 2

a , a , ar, a > 0, r > 1 r Now, 2(2a) = a + ar r fi r 2 – 4r + 1 = 0

We are given

fi a2 = 49 fi a = ± 7

fi r = 2 + 3 as r > 1. 2 Ê 11ˆ Ê 11ˆ 33. k = 1 + 2 Á ˜ + 3 Á ˜ + Ë 10 ¯ Ë 10 ¯

11 11 11 2 k= + 2 ÊÁ ˆ˜ + Ë 10 ¯ 10 10 Subtracting, we get

a / r 2 + a / r + a + ar + ar 2 = 49 r / a + r / a + 1/ a + 1/ ar + 1/ ar 2 2

9

11 11 + 9 ÊÁ ˆ˜ + 10 ÊÁ ˆ˜ Ë 10 ¯ Ë 10 ¯

11 Ê 11ˆ 2 1 - k = 1+ +Á ˜ + 10 10 Ë 10 ¯

a + a = 35 r2 As r2 > 0, a = 7 Also,

11 + 10 ÊÁ ˆ˜ Ë 10 ¯ 10

11 9 11 10 + ÊÁ ˆ˜ - 10 ÊÁ ˆ˜ Ë 10 ¯ Ë 10 ¯

and \

1 1 2 +1 = 5 fi r = 2 4 r 2 a/r = 28

37. Two A.P. are 3, 7, 11, 15, 19, 23, 27, 31,

=

10

(1110)

-1

11 - 1 10

11 - 10 Ê ˆ Ë 10 ¯

10

and

1, 6, 11, 16, 21, 26, 31,

Common terms are

8.51

8.52

Complete Mathematics—JEE Main

11, 31, 51,

6(2n - 1) 21 2n - 1 7 fi = = n 4 2n 2 \ number of terms in the A.P. is 8. fi

Thus, sum to 20 common terms is 20 [2(11) + (20 - 1)(20)] = 4020 2 38. G =

41.

ab ,

1 Ê 1 1ˆ a+b + = Ë ¯ 2 a b 2ab 1 5 4 M fi MG = = 4 G 5 a b 5 a+b 5 + = fi = fi b a 2 2 ab 4

3n 2 fi 9n < 200 < 100 3 fi n < 200/9 fi n £ 22 Next,

M=

2 3n 5 £ < 3 100 3 fi

23 £ n £ 55.

n = 23

= 0 + Â n + 2(56) n = 23

=

39. Let Sn denote the sum to n terms of the given geometric series, that is, 2 2 2 2 Sn = 1 - - 2 - 3 - n -1 3 3 3 3 1 n -1 ˘ È 3 ) Î1 - ( 3 ) ˚ 1 - 13

Ê 1ˆ Now, Á ˜ Ë 3¯

55

55

\a:b=1:4

1 = 1-1+ Ê ˆ Ë 3¯

n =1

n =1

1 a = 2, b 2 1 a fi = 4, 4 b

= 1-

22

56

\ Â f (n) = Â f (n) + Â f (n) + f (56)

fi

(2

fi 200 £ 9n < 500

n -1

n -1


100 fi n – 1 ≥ 5

fin≥6

Thus, least value of n is 6. 40. Let the A.P. be a, a + d, ..., a + (2n – 1)d It is given a + (a +2d) +

+ (a + (2n – 2)d) = 24(1)

and (a + d) + (a + 3d) + ...+ (a + (2n – 1)d) = 30 (2) From (1) and (2), we get d +d+ + d = 6 fi nd = 6   n times

Also, 21 1 21 a + (2n – 1)d – a = 10 = = 2 2 2 21 (2n – 1)d = 2

43. tr =

13 + 23 + + r 3 1 + 3 + + (2r - 1)

r 2 (r + 1)2 / 4 r2 1 2 = (r + 1) 4 9 1 9 fi  tr =  (r + 1)2 4 r =1 r=1 =

=

1 È 10 2 ˘ Â r - 1˙ 4 ÍÎr =1 ˚

=

1 È1 ˘ (10)(10 + 1)(20 + 1) - 1˙ Í 4 Î6 ˚

= 96

Progressions 30

44. Â (r + 2)(r - 3)

fi

r =16 15

3d 4d = [d = common difference of the A.P.] a2 a5

fir=

= Â (r + 15 + 2)(r + 15 - 3) r =1

4 a5 = 3 a2

48. Note that 3 2 1 4 1 , 2 , 3 , 4, 4 ,... 5 5 5 5

15

= Â (r + 17)(r + 12) r =1 15

= Â (r 2 + 29r + 204)

a = 1

r =1

45. ar2 + ar3 = 60

4 5 Sum to 10 terms 10

= Â [a + (k - 1)d ]2

(1)

and (a)(ar)(ar2) = 1000

k =1

= 10a2 + 2ad(1 + 2 + … + 9) + d2(12 + 22 + … + 92) Ê 1ˆ = 10a2 + (ad)(9)(10) + d2 Á ˜ (9)(10)(19) Ë 6¯

(2)

(2) fi (ar)3 = 103 fi ar = 10 \ (1) gives us

Ê 8ˆ = 10 Á ˜ Ë 5¯

2

10r + 10r = 60 fi r + r – 6 = 0 fi (r + 3)(r – 2) = 0 fi r = – 3, 2.

49. For A.P. ar + as = 2a1 + (r + s – 2)d

5 1 1 1 ˘ È1 + = ÂÍ 2(n + 1) 2(n + 2) 6(n + 3) ˙˚ n =1 Î 6n

\ a3 + a15 + 2a1 + 16d = 2a9 Similarly a7 + a11 = 2a9.

5

1 1 1 ˆ 1Ê 1 1 ˆ =  ÈÍ Ê Ë ¯ Ë n + 1 3 n + 1 n + 2¯ n =1 Î 6 n 1 1 1 ˆ˘ + Ê 6 Ë n + 2 n + 3 ¯ ˙˚ 1 1 1 1 1 1 1 k 1 fi = Ê1 - ˆ - Ê - ˆ + Ê - ˆ Ë ¯ Ë ¯ 3 6 6 3 2 7 6 Ë 3 8¯

=

5 È 1 1 1 ˘ 5 Ê 11 ˆ 55 - + = = 2 ÍÎ 6 7 24 ˙˚ 2 Ë 168 ¯ 336

47. Let an denotes the nth term of the A.P., then a5 a9 = = r (say) a2 a5 fi

a5 - a2 a9 - a5 = a2 a5

2

Ê 4ˆ Ê 8ˆ Ê 4ˆ + Á ˜ Á ˜ (9)(10) + Á ˜ (15)(19) Ë 5¯ Ë 5¯ Ë 5 ¯

16 16 [8 + 36 + 57] = (101) 5 5 \ m = 101

Thus, t7 = ar6 = (ar)r5 = 10(25) = 320 5 1 46. k = Â 3 n=1 n(n + 1)(n + 2)(n + 3)

2

=

As a > 0, ar = 10, therefore, r = 2.

1 5 5 1 5 fi k= Ê ˆ- + Ê ˆ 2 Ë 6 ¯ 14 2 Ë 24 ¯

3 and common 5

difference d =

1 29 = (15)(16)(31) + (15)(16) + (204)(15) 6 2 = 7780

2

8.53

Now, 72 = 4a9 fi a9 = 18 17 [2a1 + (17 – 1)d] 2 = 17a9 = (17)(18) = 306 =

Previous Years' B-Architecture Entrance Examination Questions 1.

1 , a, b are in G.P. 16 2 fi a = b/16 Also, as a, b, 1/6 are in H.P. 2(a)(1/ 6) 2a = b= a + 1/ 6 6a + 1 From (1) and (2) 16a2 =

2a 6a + 1

(1)

(2)

8.54

Complete Mathematics—JEE Main

fi 8a (6a + 1) = 1

[

a π 0]

20

1 (1 + 2 + k =1 k

= Â

2

fi 48a + 8a – 1 = 0 fi (12a – 1) (4a + 1) = 0

1 20 + k ) = 2 Â ( k + 1) k =1

1 1 ◊ (20)(2 + 21) = 115 2 2 6. Required numbers are =

fi a = 1/12, a = –1/4 When a = 1/12, b = 1/9 When a = –1/4, b = 1

203, 210,

2ab a+b Let common ratio of G.P. a, y, z, b

Now, 399 = 203 + 7 (n – 1)

2. x =

be r, then r3 = b/a and

\ yz = ab and y3 + z3 = a3r3 +

Thus, required sum is 29 [203 + 399] = 8729 2 7. a + b + c = xb

b3 r3

fi b + b + br = xb , where r is common r ratio of the G.P and r π 1

3 Ê bˆ 3 Ê aˆ = a Ë ¯ +b Ë ¯ a b = ab (a + b)

2ab (ab)(a + b) = 2a2b2 a+b ab 1 yz \ = = 2 3 3 2ab 2(ab) x( y + z )

3. x = 1 + a + a2 +

1 = (x – 1) r 1 1 But r + < -2 or r + > 2 r r

fi r+

fi x(y3 + z3) =

=

1 1- a

1 1- b fi a = 1 – 1/x, b = 1 – 1/y y=

Now, 1 + ab + a2b2 + ... 1 1 = = 1 - (1 - 1/ x)(1 - 1/ y ) 1- ab xy = x + y -1 4. 3b = a + b + c ≥ 3(abc)1/3 fi b ≥ 41/3= 22/3 20 1 1 1 1 5. Â i Ê + + + ... + ˆ Ë i +1 i + 2 20 ¯ i=1 i

+

fi 57 = 29 + (n – 1) fi n = 29

=

y = ar, z = b/r

Ê1 1 = ÁË + + 1 2

, 399

1ˆ ˜ 20 ¯

1ˆ 1ˆ 1 Ê1 Ê1 1 +2 Á + +  + ˜ + 3 Á +  + ˜ +20 Ê ˆ Ë ¯ ¯ Ë2 3 Ë 20 ¯ 20 3 20 1 1 = (1)(1) + (1 + 2) + (1 + 2 + 3) + 2 3 1 + (1 + 2 + + 20) 20

\ x – 1 < –2 or x – 1 > 2 fi x < – 1 or x > 3 fi x Œ ( – •, – 1) » (3, •) 8. Let two A.P.’s be a, a + d, a + 2d, ... and A, A + D, A + 2D, ... We are given n [2a + (n - 1)d ] 3n + 8 2 = n [2 A + (n - 1) D ] 7n + 15 2 n -1 a+ d 3n + 8 2 fi = n -1 7 n + 15 A+ D 2 Put (n – 1)/2 = 11 or n = 23. \

77 7 3(23) + 8 a + 11d = = = A + 11D 7(23) + 15 176 16

9. Let an denote the growth of height in the nth year. We have a1 = 50 cm, ar = 50 – 5(r – 1) = 55 – 5r The tree ceases to grow in the rth year where ar = 0 fi r = 11

Progressions

So height of tree is a1 + a2 +

+ a9

10 [50 + 5]cm = 275 cm = 2.75 m 2 10. rth term of the series is =

tr = =

49 - r (r + 1)(r + 2)

50 51 r +1 r + 2

1 1 ˆ 1 = 50 Ê Ë r + 1 r + 2¯ r + 2

1 1 1 , , are in A.P. a b c 1 2 1 2r 1 2r - 1 fi = - = - = a b c c c c 2 1 1 Also, - = b a c Also,

1 2a - b = c ab ab 2 c2 ab 2 \ = = 2 (ab / c) a ( 2a - b) 2 fi

= c(2r – 1) = e n

48

48 1 ˆ Ê 1 - 1 ˆ -Â Ê fi  tr = 50 Â Ë ¯ Ë r + r + 1 2 r + 2¯ r =1 r =1 r=1 48

48 1 1 1 = 50 Ê - ˆ - Â Ë 2 50 ¯ r =1 r + 2

=

\K=–1 11. 2, 2x – 1, 2x + 3 are in G.P. fi (2x – 1)2 = 2(2x + 3) fi 22x – 2(2x) + 1 = 2(2x) + 6 fi 22x – 4(2x) – 5 = 0 fi (2x – 5) (2x + 1) = 0 [

k =1

2x + 1 > 0]

Thus, there is exactly one value of x. 12. Let r be the common ratio of the G.P. b, c, d, then c , c = c, d = cr r As c, d, e, are in A.P.

b=

e = 2d – c = 2cr – c = c(2r – 1)

2n n +1

fi f(n) = Sn – Sn–1 = =

51 50 Ê 1 ˆ -Â 2 r =1Ë r ¯

fi 2x = 5

13. Let Sn = Â f(k ) =

fi

2n 2(n - 1) n +1 n

2 2[n 2 - (n 2 - 1)] = n(n + 1) (n + 1)n 1 n(n + 1) = f(n) 2 1 10 1 = Â (k 2 + k ) 2 k =1 k =1 f( n) 10

fi  =

1 1 1 1 ◊ (10)(11)(21) + ◊ (10)(11) 2 6 2 2

= 220 14. 2b = a + c,

2 1 1 = + , c2 = bd d c e

2ce c2 =d= b c+e e c c fi = = c+e 2b a+c

Now,

fi ae + ce = c2 + ce fi c2 = ae fi a, c, e are in G.P.

8.55

Limits and Continuity 9.1

CHAPTER NINE

LIMITS Let f be a function defined in some neighbourhood of c, but not necessarily at the point c. We are interested in the behaviour of f (x) when the distance between x and c is small, i.e., | x – c | is small. Specifically, we would like to know whether there is some real number l such that f (x) approaches l as x approaches c. We write

for each e > 0, there exists a d > 0 such that if c – d < x < c, then | f (x) – l| < e. Similarly we define, the right-hand limit written as lim f (x) = l

xÆc+

Y

lim f (x) = l

l1

xÆc

to indicate that the difference between f (x) and l can be made arbitrarily small (smaller than any pre-assigned positive number) simply by requiring that x be sufficiently close to c but different from c. Take a positive number e. limxÆ c f (x) = l, if | f (x) – l| < e for all x sufficiently close to c but different from it. That is, there exists a positive number d such that if 0 < | x – c | < d, then | f (x) – l| < e. y l+e l l-e a-d a a+d

x

Fig. 9.1

ONE-SIDED LIMITS

l2

X

c lim f (x) = l1 x ® c–

lim f (x) = l2 x ® c+

Fig. 9.2

If limx Æ c + f (x) = lim x Æ c – f (x), then the limit of f (x) as x Æ c exists and is written as limx Æ c f (x). The limit of a function f (x) may not exists in any of the following situations: 1. limx Æ c + f (x) does not exist, i.e., f (x) becomes infinitely small or large as x Æ c+. For example, as x Æ 0 +, the function f (x) = 1/x becomes infinitely large, as in this case x crosses over values which are positive but sufficiently small. 2. limx Æ c – f (x) does not exist as for example, in f(x) = 1/x as x Æ 0–, where x crosses over values which are negative, so that f (x) becomes infinitely small. 3. Both limx Æ c + f (x) and limx Æ c – f (x) exist but are Ï 1, x > c unequal. for example, in f ( x ) = Ì In this Ó-1, x < c case, lim x Æ c+ f (x) = 1 and lim xÆc– f (x) = – 1.

The left-hand limit is written as lim f (x) = l

xÆc-

If x approaches c from its left, then it crosses over values of the form c – h, h > 0, the difference between f (x) and l can be made arbitrarily small. More precisely, lim f ( x ) = l if x Æ c-

x f(x)

c + .001 1

c + .01 1

c + .1 1

c – .001

c – .01

c – .1

–1

–1

–1

9.2

Complete Mathematics—JEE Main

FREQUENTLY USED LIMITS 1. lim

xÆ0

sin x tan x = 1 = lim cos x = lim xÆ0 xÆ0 x x

sin -1 x tan -1 x = lim . xÆ0 xÆ0 x x 1ˆ x Ê 2. lim (1 + x)1/x = e = lim Á 1 + ˜ . xÆ0 xÆ• Ë x¯

1. If f, g are Polynomial, Trigonometric, Logarithmic or Exponential function such that g(a) π 0. f ( x ) f (a ) = provided f (a) and g(a) are Then lim x Æ a g (x) g (a ) finite.

= lim

3. lim

xÆ0

log a (1 + x ) = loga e x

(a > 0, a π 1).

xÆ0

log (1 + x ) = 1. x

ax - 1 = log a (a > 0). xÆ0 x ex - 1 = log e = 1. In particular, lim xÆ0 x (1 + x )m - 1 5. lim = m, m Œ R. xÆ0 x log x 6. lim = 0 (m > 0). x Æ • xm

4. lim

xn - an = nan – 1. 7. lim xÆa x - a

SOME THEOREMS ON LIMITS The calculation of limits is based on the following theorems. If lim f (x) and lim g (x) exist, then xÆc

1. lim (f (x) ± g (x)) = lim f (x) ± lim g (x). xÆc

xÆc

xÆc

xÆc

x3 + 1 Since the numerator and denominator are polynomials and if f (x) = x2 + x + 2 and g(x) = x3 + 1 then f(2) = 4 + 2 + 2 = 8 and g(2) = 8 + 1 = 9, thus lim

xÆ2

f ( x ) f (2 ) 8 = = . g ( x ) g (2 ) 9

Illustration

3 + sin x x Æ 0 cos x

Since the numerator and denominator are trigonometric function and if f(x) = 3 + sin x and g(x) = cos x then f(0) = 3 and g(0) = 1, thus 3 + sin x 3 + 0 = =3 x Æ 0 cos x 1 lim

2. If g(a) = 0 and f(a) = 0 and f, g are polynomials then (x – a) is a factor of both f and g. In such a case we factorise the numerator and denominator and cancel out the common factor and then the situation is as above. Illustration

xÆc

xÆc

lim f ( x )

4. lim

xÆc

f ( x) xÆc = (provided lim g(x) π 0). g( x) xÆc lim g ( x ) xÆc

5. If f(x) > 0 and lim f ( x ) = L then L ≥ 0 xÆ a

For many elementary functions (e.g., polynomial, trigonometric, logarithmic and exponential), we have lim f (x) = f Ê lim xˆ = f (c), Ë xÆc ¯ in general, for continuous functions. xÆc

COMPUTATION OF LIMITS f (x) Suppose we have find lim x Æ a g (x)

2

Evaluate lim

3. lim (f (x) ◊ g (x)) = lim f (x) ◊ lim g(x). xÆc

x2 + x + 2

xÆ2

2. lim [kf (x)] = k lim f(x) where k is a scalar. xÆc

1

Evaluate lim

In particular taking a = e, lim

xÆc

Illustration

lim

x Æ1

3

x3 - 1 x2 - 1

If f(x) = x3 –1 and g(x) = x2 –1 then f(1) = g(1) = 0. Therefore x –1 is a factor of x3 –1 and x2 –1, hence

( x - 1) ( x 2 + x + 1) x Æ1 x2 - 1 x Æ1 ( x - 1) ( x + 1) lim

x3 - 1

= lim

x2 + x + 1 x Æ1 x +1 1+1+1 3 = = . 1+1 2 = lim

3. If f and g are of the form p ( x ) – q ( x ) and r ( x ) - s( x ) where p(x), q(x), r(x), s(x) are polynomials such that f (a) = g(a) = 0 then we use rationalisation of the denominator and numerator, e.g.

Limits and Continuity 9.3

L’HÔPITAL’S RULES FOR CALCULATING LIMITS

4

Illustration

4a + 3 x - x + 6a

lim

1. ( 0 0 form) If f and g are differentiable functions on (0, d ) (see the definition of differentiability in the next chapter) such that (i) g¢(x) π 0 for any x Œ (0, d )

2 a + 5 x - 3a + 4 x

xÆa

( 2 x - 2 a ) ( 2 a + 5 x + 3a + 4 x )

= lim

( x - a)

xÆa

(

4a + 3 x + x + 6a

)

(ii)

(Rationalizing the numerator and denominator) 2a + 5 x + 3a + 4 x

= 2 lim

4a + 3 x + x + 6a

xÆa

4. lim

a

=

= 2.

-1 , where f(x) Æ 0 as x Æ 0 g (x) f (x) exist g (x)

and lim

xÆ0

then lim

xÆ0

a f (x) - 1 a f (x) - 1 f ( x ) = lim ◊ xÆ0 g (x) f (x) g (x) xÆ0

f (x) g (x)

5

Illustration

asin x - 1 x cos x

= lim

xÆ0

a

- 1 sin x 1 = log a.1.1 = log a ◊ x cos x sin x

xÆ0

lim (1 + f(x))1/g(x) = lim (1 + f(x))1/f(x).( f (x) /g (x)) xÆ0

f ( x) Ê f ¢( x ) ˆ = L, then lim = L. (ii) lim Á ˜ x Æ • Ë g ¢( x ) ¯ x Æ • g( x ) 3. (•/• form). If f and g are differentiable functions on (0, d ) such that (i) g¢(x) π 0 for any x Œ (0, d ) (ii) f (x) Æ •, g (x) Æ • as x Æ 0+ f ¢( x ) f ( x) (iii) lim = L, then lim =L g( x ) x Æ 0 + g ¢( x ) xÆ0+

7

What is wrong with the following application of L’Hôpital’s rule? lim

x3 + 3x - 4 2 x2 + x - 3

3x2 + 3 6x 3 = lim = . x Æ1 4 x + 1 x Æ1 4 2

= lim

3x2 + 3 0 is not of form as x Æ 1. 0 4x + 1 Therefore, it is not correct to apply L’Hôpital rule. In fact

)

lim 3 x 2 + 3 3x2 + 3 x Æ1 6 lim = . = x Æ1 4 x + 1 lim ( 4 x + 1) 5

6

x Æ1

To find lim (1 + (x2 + 3x))1/sinx, write the required limit as xÆ0

(1 + ( x 2 + 3 x )[1 /( x

2

+ 3 x )] ( x ( x + 3) / sin x )

= lim (1 + (x2 + 3x))[(1/(x

2

4. The following illustrations will show the methods of converting 0°, 1•, 0•, ••, • – • etc., to the forms 0 • and . 0 •

+ 3x)] (x(x +3) / sin x)

xÆ0

Illustration 2

= lim (1 + (x + 3x)) = e1.3 = e3

xƕ

(

= el

xÆ0

f ( x) = L. g( x )

The expression

xÆ0

lim

f ¢( x ) = L, then g ¢( x )

2. (L¢ Hôpital’s rule when x Æ •). If f and g are differentiable functions on [a, •) such that (i) lim f (x) = lim g (x) = 0

x Æ1

f (x) lim = l exist x Æ 0 g (x) then

xÆ0

lim

xÆ0+

Illustration

sin x

5. lim (1 + f (x))1/g(x), where f(x) Æ 0 as x Æ 0 and

Illustration

lim

xÆ0+

xÆ0+

xƕ

= log a lim

xÆ0

2 7a

(iii)

f (x)

xÆ0

lim

2 (2 7 a )

lim f (x) = 0 = lim g (x)

xÆ0+

8

(1/(x2 + 3x)) (x/sin x) (x + 3)

lim (sin x )

xÆ0 +

x

9.4

Complete Mathematics—JEE Main

It is of the form 00. If L is the limit sought, then log sin x Ê • ˆ log L = lim x log sin x = lim Á ˜ x Æ 0+ x Æ 0+ 1/ x Ë • ¯

Illustration lim

0 - cos x x cos x = lim = =0 1 x Æ 0 (sin x ) / x 1 sin x ÊÁ 2 ˆ˜ Ëx ¯

= lim

x Æ 0+

(1 + x )1 / x - e

x Let y = (1 + x)1/x xÆ0

fi log y =

Thus L = 1 Illustration

10

˘ x2 x3 1 1È + - ˙ log (1 + x) = Í x xÎ x 2 3 ˚ È ˘ x x2 = Í1 - + - ˙ Î 2 3 ˚

9

lim (1 – 3x)4/x (1• form)

xÆ0

fiy=

If L is the required limit then

È ˘ x x2 - ˙ Í1 - + 2 3 ˚ eÎ

= e◊e

2 ˘ ˆ 1 Ê x x2 ˙ +  +  ˜¯ 2 ÁË 2 3 ˙˚

Thus L = e–12

y-e x È 1 x = e Í- + + terms containing Î 2 3

USE OF SERIES EXPANSION IN FINDING LIMITS

Thus

The following series expansions are useful in evaluating limits p ( p - 1) 2 p ( p - 1) ( p - 2 ) 1. (1 + x)p = 1 + px + x + 3! 2! x3 + º (for |x| < 1 and p Œ Q, the set of rational numbers) x2 x3 + + 2! 3! 2

3. ax = 1 + x (loge a) + 4. log (1 + x) = x –

x (loge a)2 + º 2!

x2 x3 x 4 + +  (|x| < 1) 2 3 4

Ê ˆ x2 x3 5. log (1 – x) = – Á x + + + ˜ (|x| < 1) 2 3 Ë ¯ 6. sin x = x -

x3 x 5 + - 3! 5!

7. cos x = 1 -

x2 x4 + - 2! 4!

8. tan x = x +

x3 2 x 5 + + 3 15

9. sin–1 x = x + 10. tan–1 x = x -

x x2 + - 2 3

È Ê x x2 ˆ = e Í1 + Á - + - ˜ + ¯ Î Ë 2 3

4 log (1 - 3 x ) Ê 0 ˆ 4 ( -3) log L = lim = – 12 ÁË ˜¯ = lim xÆ0 x Æ 0 1 - 3x x 0

2. ex = 1 + x +

-

1 x3 1 3 x 5 + ◊ +  (–1< x 0 such that |an – a| < Œ holds true for all n > N. Illustration

1 1 are , 2 nƕ nƕ n n n 2n - 1 convergent sequences. If xn = then lim nƕ 2n + 1 xn = 1. lim

1 n

11 = 0,

lim

1

2

= 0 so

lim f(x) = l if and only if for every sequence xn (of values

xÆa

of x) such that lim xn = a we have lim f(xn) = l. n Æ•

nƕ

The above criterion is very useful in showing lim f(x) (for a x Æa

Limits and Continuity 9.5

given f) does not exist. It is enough to give a sequence xn with

For x Π(0, 1), [x] = 0 so

lim xn = a such that lim f(xn) does not exist. Alternatively,

nƕ

nƕ

give two sequences xn and x¢n such that lim xn = a = lim x¢n nÆ•

Hence lim

nƕ

xÆ 0+

[ x] =0 x

[ x] = 0. x

but lim f(xn) π lim f(x¢n). nÆ•

nƕ

CONTINUITY

12

Illustration

1 does not exist let f (x) = x -1 1 2 1 sin and choose xn = 1 + , xn¢ = 1 + , (4n + 1)p ( x - 1) np so lim xn = 1 = lim x¢n. Now f (xn) = sin np = 0

To show that lim sin xÆ1

nƕ

nƕ

4n + 1 p = 1. Hence nÆ• 2 lim f(x¢n) = 1. Thus lim f(x) does not exist.

so lim f(xn) = 0. f(xn) = sin nƕ

nƕ

SOME USEFUL RESULTS ON LIMITS

In ordinary language, to say that a certain process is continuous implies that it goes on without interruption and without abrupt changes. In mathematics, the word continuous has much the same meaning. We shall first introduce the idea of continuity at a point c, and then that of continuity on an interval. The function f is said to continuous at x = c if limxÆ c f (x) = f (c) or, equivalently, limh Æ 0 f (c + h) = f (c). The function f is said to be continuous on an open interval (a, b) if it is continuous at each point of (a, b). The function f is said to be continuous on a closed interval [a, b] if (i) f is continuous at each point of (a, b), (ii) lim f (x) = f (a), (i.e., right hand side continuity xÆa+

1. If lim f(x) = l with l π 0 and lim g(x) = 0 then xÆc

lim

x Æc

at a)

xÆc

f (x) does not exist. g (x)

(iii) lim f (x) = f (b). (i.e., left hand side continuity xÆb-

2. (Sandwich theorem) If f, g and h are functions having a common domain D and defined near a. Suppose that h(x) £ f(x) £ g(x) for x ŒD and lim h(x) = lim g(x) = l then lim f(x) = l. xÆa

xÆa

xÆa

x

x

3. lim e = 1, lim e = •, lim e xÆ0

xƕ

xƕ

lim ex = 0, lim

xÆa

xÆ-•

–x

= 0,

e x - ea = e a. x-a

13

Illustration

To show lim x sin xÆ 0

1 =0. x

1 £ x and lim x = 0 xÆ 0 x 1 so by Sandwich theorem lim x sin = 0 . xÆ 0 x

For values of x near 0, 0 £ x sin

Illustration Find lim

14

xÆ 0+

[ x] x

at b) Graphically, f (x) is continuous if its graph can be drawn on paper without raising the pencil. A function f is said to be discontinuous at c if it is not continuous at that point. A function f is said to be discontinuous on an interval if it is discontinuous at least at one point of the interval. A function f is discontinuous at c under any of the following circumstances: (i) f is not defined at c (e.g., f (x) = 1/x at x = 0). (ii) limx Æ c f (x) does not exist. (iii) f is defined at c and limx Æ c f (x) exists but limx Æ c f (x) π f (c). For example.

But

Ï x3 - 1 Ô x π1 f(x) = Ì x - 1 ÔÓ 4 x =1

lim f ( x ) = lim( x 2 + x + 1) = 3 π f (1) x Æ1

x Æ1

One can prove that if f and g are continuous at x = c and a is a scalar, then f + g, a f, f g, max ( f, g) and min ( f, g) are continuous at c. Moreover, if g (c) π 0, then the quotient f / g is also continuous at c. If g is continuous at c and f is continuous at g (c), then the composition f o g is continuous at c. All elementary functions (e.g., polynomial, trigonometric, logarithmic, exponential and inverse trigonometric functions) are continuous at each point of their domains.

Complete Mathematics—JEE Main

9.6

Illustration

3. Intermediate Value Theorem If m = min f (x) and M = max f (x) for a £ x £ b, then for any A satisfying the inequalities m £ A £ M, there exists a point x0 Œ [a, b] for which f (x0) = A, i.e., a continuous function assumes every value between any two of its values. 4. If f is monotonically increasing on an open interval I and x0 Œ I then

15

Examine the continuity at origin Ï| x | , xπ0 Ô f(x) = Ì x ÔÓ 1 , x = 0 lim

x Æ0 +

x |x| |x| = lim = 1 but lim x Æ 0 + x Æ 0 x x x = – lim

x Æ0 -

x = – 1. So f is not continuous at x = 0. x

lim

lim

f(x) and

lim

f(x) £ f(x0)£ lim

x Æ x0 + x Æ x0 -

x Æ x0 -

f(x) are finite and

x Æ x0 +

f(x).

5. If f is monotonically increasing and continuous on an interval I then f –1 exist and is continuous and monotonically increasing. 6. Let f be a function which is continuous at c and f(x) π 0. Then there exists an interval (c – d, c + d ) throughout which f(x) is of the same sign. 7. Let f be a continuous function on [a, b]. If f(a) and f(b) are of opposite sign, then there exists a point c Œ (a, b) such that f(c) = 0

FUNCTIONS CONTINUOUS ON A CLOSED INTERVAL A function f (x), continuous on the interval [a, b], possesses the following properties: 1. f (x) is bounded on [a, b], i.e., there exist k, K Œ R such that k £ f (x) £ K for all x Œ [a, b]. 2. f (x) has minimum and maximum values on [a, b].

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Ïx - | x | , xπ0 Ô Example 1: Let f(x) = Ì x then ÔÓ 1 , x=0

È sin x ˘ Example 2: Let f(x) = Í , x π 0, where [.] denotes Î x ˙˚ the greatest integer function then lim f ( x ) x Æ0

(a) lim f(x) = 1 xÆ 0 +

(a) does not exist

(b) is equal to 1

(c) is equal to 0

(d) lim f ( x ) = 1

(b) lim f ( x ) = 0 x Æ0 -

(c) lim f ( x ) π lim f ( x ) x Æ0 +

Solution: For x π 0

x Æ0 -

(d) lim f ( x ) does not exist x Æ0 +

Ans. (c) Solution: If x > 0, |x| = x, so x-|x| x-x = lim =0 x Æ0 + x x for x < 0, |x| = – x, so lim

x Æ0 +

x-|x| x+x lim = lim =2 x Æ0 x Æ0 x x Thus lim f ( x ) π lim f ( x ) x Æ0 +

x Æ0 +

Ans. (c)

x Æ0 -

sin x sin x < 1 . For – 1 < x < 0, >0, x x

sin x tends to 1 as x Æ 0 but through values which are less x È sin x ˘ = 0 as x tends to 0. than 1. Hence Í Î x ˙˚ Thus lim f ( x ) = 0. so

x Æ0

Example 3: lim

x Æ•

is equal to (a) 100 (c) 1 Ans. (a) Solution: f(x) =

( x + 1)10 + ( x + 2 )10 +  + ( x + 100 )10 x10 + 1010 (b) 0 (d) 10

( x + 1)10 + ( x + 2 )10 +  + ( x + 100 )10 x10 + 1010

Limits and Continuity 9.7

( ) ( ) ( ( ) 1 x

1+

=

10

+ 1+

2 x

10

++ 1+

100 x

)

(a) 1

(b) –1 1 (d) 2

(c) 2 Ans. (b)

1 3 1 - x 1 - x3 1 È 3 ˘ 1= 2 Í (1 - x ) Î 1 + x + x ˙˚

Solution: For x π 1, f(x) =

x2 + x - 2 1 ( x + 2 ) ( x - 1) ◊ = 2 (1 - x ) (1 + x + x ) (1 - x ) (1 + x + x 2 )

=-

3 2

(b) 0

(c) 1

(d) -

3 2

x Æ0 +

lim f ( x ) = lim ( mx + x + n ) = n 2

x Æ0 -

Hence

x Æ0 -

m = n.

Also lim f ( x ) = lim ( nx + m ) = n + m x Æ1-

x Æ1-

and lim f ( x ) = lim (2nx 3 + x 2 - 2 x + m ) x Æ1+

x Æ1+

= 2n + 1 – 2 + m = 2n + m – 1 Hence n + m = 2n + m – 1 fi n = 1 so m = 1 p Ï if x£Ô -2 sin x 2 Ô Ô p p Example 7: If f(x) = Ì A sin x + B if - < x < 2 2 Ô Ô p if x≥ ÔÓ cos x 2 is a continuous function then (a) A = 1, B = – 1 (b) A = – 1, B = 1 (c) A = 0, B = 2 (d) A = – 2, B = 2 Ans. (b)

x+2

1 + x + x2 x+2 so lim f ( x ) = – lim = -1. x Æ1 x Æ1 1 + x + x 2 sin 3 x Example 5: lim is equal to x Æp sin 2 x (a)

x Æ0 +

10

10 1+ x 1 1 + 1  + 1 - (100 times) As Æ 0, for x Æ • so lim f ( x ) = x Æ• x 1 = 100. 3 ˆ Ê 1 Example 4: lim Á is equal x Æ1 Ë 1 - x 1 - x 3 ˜¯

=

Solution: lim f ( x ) = lim ( nx + m ) = m

10

Ê pˆ Ê pˆ Solution: f Ë- ¯ = – 2 sin Ë- ¯ = 2 2 2 lim f ( x ) = lim ( A sin x + B ) = – A + B

x Æ- p 2 +

x Æ- p 2

so

2 = – A + B … (i) p Êp ˆ Also f Ë ¯ = cos =0 2 2 lim f ( x ) = lim ( A sin x + B ) = A + B p xÆ 2

x Æp / 2 -

Hence A + B = 0 (ii) Solving (i) and (ii), we get A = – 1, B = 1.

Ans. (d) sin 3 (p - u ) sin 3 x Solution: lim = lim (u = x – p) uÆ 0 sin 2 (p - u ) x Æp sin 2 x 3 sin 3u 2u sin 3u = - lim ¥ = lim uÆ 0 - sin 2u 2 3u sin 2u 3 = - . 2 Ï , mx 2 + x + n x1 Ó2nx + x - 2 x + m , and lim f ( x ) and lim f(x) exist then x Æ0

xÆ1

(a) m = 2, n = 1 (c) m = 1, n = 2 Ans. (d)

(b) m = 0, n = 1 (d) m = 1, n = 1

x2 - 1 Example 8: The function f(x) =

is not defined for x3 - 1 x = 1. The value of f(1) so that the function extended by this value is continuous is 2 1 (b) (a) 3 3 (c) 1 (d) 0 Ans. (a) Solution: For f to be continuous at x = 1, f(1) = lim f ( x ) = lim x Æ1

x Æ1

x2 - 1 x3 - 1

( x - 1) ( x + 1) x Æ1 ( x - 1) ( x 2 + x + 1)

= lim = lim

x Æ1

x +1 2

x + x +1

=

2 . 3

9.8

Complete Mathematics—JEE Main 1

1

Example 9: Let y = –

2x - 1 1 2x

(a) lim y = - 1

lim f ( x ) = - lim

, then

x Æ0 -

x Æ0 -

+1 x Æ0

(c) lim y = - 1

= – (– 1) = 1.

+1

Example 10: The value of lim

x Æ0 +

x Æ0 +

x Æ0 -

1

1 Solution: As x Æ 0 +, Æ • so 2 x Æ 0 Hence x 1

lim f ( x ) = – lim

x Æ0 +

2x - 1 1 2x

= - lim

1- 2

x Æ0 +

1+ 2

+1

-

1 x

-

1 x

is 6x (b) 0 (d) not a finite number

(a) 1 (c) – 1 Ans. (b) Solution: lim

x Æ•

=–1

x6 Ê• ˆ 6 x Ë• ¯ 6 x5

Ê• ˆ (L¢ Hopital Rule) x Æ• 6 ( log 6 ) Ë • ¯

= lim

x

1

x Æ 0–,

x6

x Æ•

(d) lim y = lim y

Ans. (c)

As

1 2x

(b) lim y = 1

x Æ0

x Æ0 +

2x - 1

1 Æ - • so 2 x Æ 0 x

= lim

6. 5. 4. 3. 2. 1

x Æ•

6 x ( log 6 )6

= 0.

LEVEL 1 Straight Objective Type Questions Ê x + 1ˆ lim Á ˜ xÆ• Ë x + 2¯

Example 11:

(using L’Hôpital Rule)

2x + 1

is

(b) e–2 (d) 1

(a) e (c) e–1 Ans. (b) Solution: Ê x + 1ˆ lim Á ˜ xÆ• Ë x + 2¯

2x + 1

- ( x + 2) È ˘ 1 ˆ Ê = Í lim Á 1 ˙ ˜ x + 2¯ ÍÎ x Æ • Ë ˙˚

- ( x + 2) È ˘ 1 ˆ Ê = Í lim Á 1 ˙ ˜ x + 2¯ ÍÎ x Æ • Ë ˙˚

Example 12: (a) 1/6 (c) 2/3 Ans. (b) Solution: lim

xÆ0

lim

xÆ0

-

2x + 1 x+2

2+ 1 / x 1+ 2/ x

(cos x )1 / 2 - (cos x )1 / 3 sin 2 x (b) – 1/12 (d) 1/3

-

= e–2

is

Example 14:

(cos x )1 / 2 - (cos x )1 / 3 Ê 0 sin 2 x

1 1 -1 / 2 -2/3 - (cos x ) + (cos x ) 3 = lim 2 xÆ0 2 cos x 1 È1 1˘ 1 = - Í - ˙=- . 2 Î2 3˚ 12 1 + log x - x Example 13: lim equals x Æ1 1 - 2 x + x2 (a) 1 (b) 0 (c) – 1 (d) – 1/2 Ans. (d) 1 + log x - x 1/ x - 1 Solution: lim = lim 2 x Æ1 1 - 2 x + x x Æ1 - 2 + 2 x (using L’Hôpital Rule) 1 (1 - x ) 1 1 1 = - lim = = lim x Æ1 2 2 x (1 - x ) 2 x Æ1 x

ˆ ÁË form˜¯ 0

1 1 -1 / 2 -2/3 sin x + (cos x ) - (cos x ) sin x 2 3 = lim xÆ0 2 sin x cos x

lim

xÆ0

(a) 1/3 (c) 1/2 Ans. (a) Solution: lim

xÆ0

tan x - x

equals to x 2 tan x (b) 2/3 (d) 1

tan x - x 2

x tan x

= lim

xÆ0

sin x - x cos x Ê 0 ˆ ÁË form˜¯ 2 0 x sin x

Limits and Continuity 9.9

= lim

cos x - cos x + x sin x 2

2 x sin x + x cos x (using L’Hôpital Rule)

x Æ0

= lim

xÆ0

sin x 1 (sin x ) / x = lim = 2 sin x + x cos x x Æ 0 ( 2 (sin x ) / x ) + cos x 3

Example 18: -1 ÈÊ x4 - 1 ˘ 4 1 - 3x + x2 ˆ ˙ is lim ÍÁ 2 + 3 x Æ 1 ÍË x - x - 1 x 3 - x -1 ˙˚ 1 - x 3 ˜¯ Î (a) 3 (b) 2 (c) 4 (d) 28/3 Ans. (a)

È 1 8 n3 ˘ lim Í  + + + ˙ is n Æ • Î1 - n 4 1 - n4 1 - n4 ˚ (b) 1/8 (d) none of these

Example 15: (a) 1/4 (c) 1/2 Ans. (d)

Solution:

È 1 8 n3 ˘ Solution: lim Í + ++ ˙ 4 4 n Æ • Î1 - n 1- n 1 - n4 ˚ = lim

nƕ

1 1 - n4

[13 + 23 +

ÈÊ 4 1 - 3x + x2 ˆ lim ÍÁ 2 x Æ 1 ÍË x - x - 1 1 - x 3 ˜¯ Î

-1

ÈÊ 4 x 1 - 3x + x2 ˆ = lim ÍÁ 3 x Æ 1 ÍË x - 1 1 - x 3 ˜¯ Î

-1

ÈÊ 4 x + 1 - 3 x + x 2 ˆ = lim ÍÁ ˜¯ x Æ 1 ÍË x3 - 1 Î

+ n3 ]

2 sin ( x - p / 3) is 1 - 2 cos x

(a) = 2 (c) = 1 Ans. (b)

(b) 2/ 3 (d) 1/3

(a) 1/ 2 (c) 2/3 Ans. (b)

So, lim

x Æ3 +

2 cos ( x - p / 3) x Æp / 3 2 sin x

= cos 0/sin (p/3) = 2/ 3

1 - cos 2 ( x - 3) = x-3

1 - cos 2 ( x - 3) = x-3

lim

x Æ3 -

(a) 11/4 (c) 1/2 Ans. (b) Solution:

2 - cot x - cot 3 x

is

exists and lim

x Æ5

(b) 3/4 (d) none of these

lim

3 = lim = x Æ p / 4 cot 2 x + cot x + 2 4

x-5

2 lim

x Æ3 -

- sin ( x - 3) =- 2 x-3 x Æ5

=0

(b) 1 (d) 3

Solution: As lim

(1 - cot x ) (1 + cot x + cot x ) x Æ p / 4 (1 - cot x ) (cot 2 x + cot x + 2 ) 1 + cot x + cot 2 x

sin ( x - 3) = 2 x-3

x Æ5

2 - cot x - cot 3 x

lim

( f ( x ))2 - 9

(a) 0 (c) 2 Ans. (d)

2

=

x Æ3 +

Then lim f ( x ) equals

1 - cot 3 x

x Æp / 4

2 lim

Example 20: Let f : R Æ [0, •) be such that lim f ( x )

3

lim

2 |sin (x – 3)|

whereas

[using L’Hôpital Rule]

1 - cot x

(b) does not exist (d) = – 2

1 - cos 2 ( x - 3) =

fi

= lim

x Æp / 4

˘ + 3x ˙ ˙˚

Solution: 1 – cos 2 ( x - 3) = 2 sin2(x – 3)

2 sin ( x - p / 3) Ê 0 ˆ Solution: lim form˜ ¯ x Æp / 3 1 - 2 cos x ÁË 0

Example 17:

)

1 - cos 2 ( x - 3) x-3

Example 19: lim

x Æ3

lim

(

3x x 4 - 1 ˘ ˙ x 4 - 1 ˙˚

x Æ1

2

1 ( n ( n + 1)) 1 1 (1 + 1/ n) = lim = lim =4 4 nÆ• 4 n Æ • 4 4 1- n 1/ n - 1

x Æp / 3

+

x4 - 1 ˘ ˙ x 3 - x -1 ˙˚

= lim [x – 1 + 3x] = 3

2

Example 16:

-1

+3

( f ( x ))2 - 9

= 0, so | x - 5| (f(x))2 – 9 = K |x – 5|a P(x), where a > 1/2 and P(x) is a function of x such that lim P( x ) exist x Æ5

x Æ5

fi lim ( f ( x )) = lim ÈÎ K x - 5 a P( x ) + 9˘˚ = 9 2

x Æ5

x Æ5

The only possible answer is lim f ( x ) = 3 x Æ5

9.10

Complete Mathematics—JEE Main

Ï sin (1 + [ x ]) for [ x ] π 0 Ô Example 21: If f(x) = Ì where [ x] ÔÓ 0 for [ x ] = 0 [x] denotes the greatest integer £ x, then lim f(x) equals

1 ax = lim ÈÎ(1 + a x )( ) ˘˚ xÆ0

fi

Example 25:

(b) 0 (d) none of these

ax - 1

lim

xÆ0-

lim

xÆ0

sin (1 + [ x ]) sin 0 = = 0. -1 [ x]

ax - 1

Solution: = lim

(

(

2

))

xπ0

a+x - a

xÆ0

xƕ

(a) 0 (c) 1 Ans. (c)

= lim 2 ax log a

(b) – 1/2 (d) none of these

xÆ0

(

È sin (1 x ) 1 sin 1 x 2 Solution: lim f(x) = lim Í + ◊ x Æ • Í 1/ x xÆ• x 1/ x 2 Î Ê sin u sin u2 ˆ Ê = Á lim + lim u Áu = uÆ0 Ë uÆ0 u u2 ˜¯ Ë

) ˘˙ ˙˚

1ˆ ˜ x¯

(a) 50/3 (c) 13 Ans. (d)

Example 23: The value of

= lim

hÆ0

1 Ê 1 ˆ + +  + up to n terms˜ is lim Á ¯ n Æ • Ë 1◊ 3 3◊5

Solution: L = lim

1 ÈÊ 1ˆ Ê 1 1ˆ 1 ˆ˘ Ê 1 ÍÁË 1 - 3 ˜¯ + ÁË 3 - 5 ˜¯ + … + ÁË 2n - 1 - 2n + 1˜¯ ˙ Î ˚

nƕ 2

1 nƕ 2

= lim

Example 24:

f (1 - h ) - f (1) - 1 ◊ 2 -h h +3 2

Example 27: lim

e x - cos x

xÆ0

(b) 3/2 (d) 2

e x - 1 + 1 - cos x x2

xÆ0

Solution: e4 = lim (1 + ax)b/x

is equal to

Solution: lim

(b) a = 8, b = 4 (d) none of these

x2

(a) 1/2 (c) 2/3 Ans. (b)

= 1+

xÆ0

xÆ0

h3 + 3 h

= f ¢(1) (–1/3) = 53/3.

If lim (1 + ax)b/x = e4, where a and b are

natural numbers then (a) a = 4, b = 2 (c) a = 16, b = 8 Ans. (d)

f (1 - h ) - f (1)

2

1 ˘ 1 È Í1 - 2n + 1 ˙ = 2 Î ˚

)

a + x = 2 a log a.

(b) 22/3 (d) none of these

hÆ0

(b) 1/2 (d) none of these

(

Example 26: Let f(x) = 3x10 – 7x8 + 5x6 – 21x3 + 3x2 – 7. f (1 - h ) - f (1) The value of lim is hÆ0 h3 + 3h

Solution: = lim

= 1 + 0.1 = 1.

(a) 1/4 (c) 1 Ans. (b)

a x log a Ê 0 ˆ ÁË form˜¯ 1 1 0 ◊ 2 a+x

= lim

x=0

is

(b) a log a (d) none of these

xÆ0

ÏÔ x sin (1/ x ) + sin 1/ x Example 22: Let f(x) = Ì 0 ÓÔ lim f(x) equals then

a+x - a

(a) 2 a log a (c) log a Ans. (a)

Solution: For – 1 < x < 0, [x] = – 1, so

= eab

ab = 4. But none of the values in (a), (b), (c) satisfy ab = 4.

xÆ0-

(a) 1 (c) – 1 Ans. (b)

ab

2

Ê e x - 1 2 sin 2 x / 2 ˆ = lim Á + 2 ˜ xÆ0 Ë x2 4 ( x / 2) ¯

1 3 = 2 2

Example 28: Let f(x) = < x >*, where < x >* is the distance from x to the integer nearest to x then lim f(x) is xÆ2

(a) 2 (c) 0 Ans. (c)

(b) 1 (d) none of these

Limits and Continuity 9.11

Solution: and Hence

lim f(x)= lim * = lim h = 0

xÆ2+

hÆ0+

hÆ0+

lim f (x) = lim * = lim h = 0.

xÆ2-

hÆ0+

hÆ0+

lim f (x) = 0.

xÆ2

Example 29: lim

(1 - cos 2 x ) (3 + cos x ) x tan 4 x (b) 1 (d) – 1/4

xÆ0

(a) 1/2 (c) 2 Ans. (c) Solution: For x π 0,

2 =

(1 - cos 2 x ) (3 + cos x ) x tan 4 x sin 2 x ¥ (3 + cos x ) x2 tan 4 x 4 4x

x Æ0

(a) – p (c) p/2 Ans. (b)

sin (p cos2 x ) x2

is equal to

(b) p (d) 1

sin (p cos2 x ) Ê 0 ˆ Solution: lim Ë 0¯ x Æ0 x2 - cos (p cos2 x ) 2p cos x sin x = lim x Æ0 2x sin x = – p lim cos (p cos x ) lim lim cos x x Æ0 x Æ0 x x Æ0 =p 2

Example 31: Let f(x) = sgn (sgn (sgn x)). Then lim f(x) is

xÆ0

(a) 1 (c) 0 Ans. (d)

(b) 2 (d) none of these

Solution: By definition we have for x π 0, sgn (sgn x) = x/ x x Ê xˆ sgn Á ˜ = = sgn x. Thus, sgn [sgn) (sgn x)] = = Ë x¯ x/x x Ï 1 x>0 Ô sgn x = Ì 0 x = 0 Ô- 1 x < 0 Ó Therefore,

Ê x 3 È x ˘3 ˆ a3 lim Á -Í ˙ ˜ = = a2 . xÆa - Ë a a Îa˚ ¯

So

Ï1 + tan x ¸ Example 33: lim Ì ˝ x Æ 0 Ó 1 + sin x ˛

cos ec x

xÆ0 -

is equal to

(b) e–1 (d) none of these

(a) e (c) 1 Ans. (c)

cos ec x

Ï1 + tan x ¸ Solution: lim Ì ˝ x Æ 0 Ó 1 + sin x ˛

cosec x

Ï 1 ¸ = lim {1 + tan x}cot x sec x Ì . ˝ xÆ0 Ó1 + sin x ˛ 1 = lim {1 + tan x}cot x ◊ sec x ◊ xÆ0 (1 + sin x )cosec x 1 = e1 ◊ = 1. e Example 34: lim (6n + 5n)1/n is equal to nÆ•

(a) 6 (c) 5/6 Ans. (a)

(b) 5 (d) e

Solution: lim (6n + 5n)1/n = 6 lim (1 + (5/6)n)1/n nƕ

nƕ

= 6 lim

nƕ

È 6 / 5)n .(1 / n )(5 / 6 )n ˘ ˚

(1 + (5/ 6)n )Î(

= 6e0 = 6. x Ê x2 + 5x + 3ˆ Example 35: If f(x) = Á 2 then lim f (x) is xÆ• Ë x + x + 2 ˜¯ (a) e–4 (c) e2 Ans. (d)

(b) e3 (d) e4

Ê x2 + 5 x + 3ˆ Solution: lim f (x) = lim Á 2 xÆ• xÆ•Ë x + x + 2 ˜ ¯ Ê 4x +1 ˆ = lim Á 1 + 2 xÆ•Ë x + x + 2 ˜¯

lim f (x) = 1 but lim f(x) = – 1.

xÆ0 +

Èx˘ Solution: For, 0 < x < a, Í ˙ = 0, Îa˚

is equal to

2 ¥1¥ 4 =2 Therefore, the required limit = 4 ¥1 Example 30: lim

Ê x 3 È x ˘3 ˆ lim Á - Í ˙ ˜ (a > 0), where [x] dexÆa - Ë a Îa˚ ¯ notes the greatest integer less than or equal to x is (b) a2 – 1 (a) a2 – 3 2 (d) none of these (c) a Ans. (c) Example 32:

x

x

9.12

Complete Mathematics—JEE Main 2

where

a

x +x+2˘ È 4 x +1 ˆ 4x +1 ˙ ÍÊ = lim ÍÁ 1 + 2 (1) ˙ xÆ• Ë x + x + 2 ˜¯ Í ˙ Î ˚ x ( 4 x + 1) 4 + 1/ x = a= 2 Æ 4 as x Æ • x + x + 2 1 + 1/ x + 2 / x 2

(1) fi lim f (x) = e4.

\

xƕ

Ê ˆ lim Á x + x + x - x ˜ is given by Ë ¯

Example 36:

xƕ

(a) 0 (c) log 2 Ans. (b)

(b) 1/2 (d) none of these

Ê ˆ Solution: lim Á x + x + x - x ˜ ¯ xÆ•Ë x+ x

= lim

xƕ

x+ x+ x + x 1 + x - 1/ 2

= lim

xƕ

1 + x-1 + x- 3/ 2 + 1

Example 37: If lim

xÆ-a

(a) – 7 (c) 7 Ans. (b)

1 . 2

=

x9 + a9 = 9, then the value of a is x+a (b) – 1 (d) none of these

Example 39: Let f be a continuous function on [1, 3]. If f takes only rational values for all x and f(2) = 10 then f(1.5) is equal to f (1) + f (3) (a) 8 (b) 3 (c) 20 (d) none of these Ans. (d) Solution: Since f is continuous, so it must take all real values between f(1) and f(3). But since f takes only rational values so f must be a constant function. Hence f(1.5) = f(2) = 10. Ï| x | cos (1/ x ) + 9 x 2 x π 0 Example 40: Let f(x) = Ì k x=0 Ó then f is continuous if the value k is (a) 9 (b) 6 (c) 0 (d) none of these Ans. (c) Solution: Since 0 £ ||x| cos (1/x)| £ |x|, so lim |x| cos 1/x = 0. xÆ0 Hence k = lim f (x) = 0. xÆ0

Example 41: If f(x) = tan (p/4 – x)/cot 2x for x π p/4. The value of f(p/4) so that f is continuous at x = p/4 is (a) 1/3 (b) 1/2 (c) 1/4 (d) 1/6 Ans. (b) Solution: For f to be continuous at x = p/4, we must have f (p/4) = lim f(x) = lim [tan (p/4 – x)/cot 2x] x Æp / 4

9

x 9 - (- a ) x9 + a9 = 9(– a)8 = lim xÆ-a x + a x Æ - a x - (- a ) 9(– a)8 = 9 fi a8 = 1 fi (a – 1) (a + 1) (a2 + 1) (a4 + 1) = 0 a = 1 or – 1.

Solution: So fi

lim

Example 38: Let f(x) = [x] + [– x], where [x] denotes the greatest integer not greater than x. Then for any integer m (a) f is continuous at x = m (b) lim f (x) exists and is equal to f (m) x Æ m+

(c) lim f (x) exists but is not equal to f (m) xÆm

(d) f is differentiable at x = m Ans. (c) Solution: For, m – 1 < x < m, [x] = m – 1 and [– x] = – m and for m < x < m + 1, [x] = m and [– x] = – m – 1. Thus lim f(x) = m – 1 – m = – 1 and lim f(x) = m – m – 1 = – 1

xÆm

xÆm +

=

x Æp / 4

È 2 tan x ˘ 1 È1 - tan x 2 tan x ˘ = lim Í lim Í ◊ ˙= . ˙ 2 x Æ p / 4 Î1 + tan x 1 - tan x ˚ x Æ p / 4 (1 + tan x )2 Î ˚ 2

Example 42: Let a function f be defined by f(x) = x-|x| for x π 0 and f(0) = 2. Then f is x (a) continuous nowhere (b) continuous everywhere (c) continuous for all x except x = 1 (d) continuous for all x except x = 0 Ans. (d) Ï2 if x < 0 Ô Solution: f (x) = Ì0 if x > 0 Ô2 x = 0 Ó Thus lim f (x) = 2 π 0 = lim f (x). Hence, f is continuous x Æ0 -

x Æ0 +

and so lim f(x) = – 1. But f(m) = [m] + [– m] = m – m = 0.

everywhere except at x = 0.

Hence lim f(x) exists but is not equal to f(m).

Example 43: If f(x) is a continuous function satisfying f(x) f(1/x) = f(x) + f(1/x) and f(1) > 0 then lim f(x) is equal to

xÆm

xÆm

x Æ1

Limits and Continuity 9.13

(a) 2 (c) 3 Ans. (a)

(b) 1 (d) none of these

Solution: Since f is continuous function so lim f(x) = f(1). x Æ1

2

Put x = 1 in the given equation we have (f(1)) = 2f(1) so f(1) = 0 or 2. Since f(1) > 0 so f(1) = 2. Example 44: The function f(x) = (x – 1)1/(2 – x) is not defined at x = 2. The value of f(2) so that f is continuous at x = 2 is (a) 1 (b) e (c) 1/e (d) 1/e2 Ans. (c) Solution: For f to be continuous at x = 2, f(2) = lim (x – 1)1/(2 – x) = lim (1 + (x – 2)–1/(x – 2) = e–1.

xÆ2

xÆ2

x + x 2 + + x n - n , x π 1, the x -1 value of f(1) of that f is continuous is n +1 (a) n (b) 2 n ( n + 1) n ( n - 1) (c) (d) 2 2 Ans. (c) Example 45: Let f(x) =

Ï0 Example 47: Let f (x) = [x] and g (x) = Ì 2 Óx Then (a) g is continuous at x = 1 (b) f is continuous at x = 2 (c) g o f is a continuous function (d) f o g is a continuous function Ans. (c)

1 + 2 x + + n x x Æ1 1

Ê ax + bx ˆ lim Á xÆ0 Ë 2 ˜¯

Example 46: Let y =

Solution:

u +u-2 is discontinuous only at x = (a) 1, 2 (b) 1, – 2 (c) 1, 1/2, 2 Ans. (c)

1 is discontinuous at x = 1, x -1 1 1 y = f(u) = 2 = is discontinuous at u (u + 2) (u - 1) u +u-2 1 = – 2, u = 1. If u = – 2 then – 2 = x -1 1 fi x = 1/2. If u = 1 then 1 = fi x = 2. Hence the comx -1 posite function is discontinuous only at x = 1, 1/2, 2. Solution: u = f (x) =

2/ x

Ê ax + bx ˆ ÁË 2 ˜¯

Ê ( b / a ) - 1ˆ ( b / a ) = a2 Á1 + ˜ 2 Ë ¯ Ê ax + bx ˆ lim Á x Æ0 Ë 2 ˜¯

1 then y x -1

(d) none of these

(b) 2/49 (d) 4/49

2

=1+2+º+n

where u =

= 7 IS

x

So 1

2

2/ x

(a) 1/7 (c) 4/7 Ans. (b)

n -1

n ( n + 1) . = 2

, x ŒR ~ I

Solution: f is not continuous at every x ŒI and g is also not continuous at every x Œ I ~ {0}. g o f (x) = g (f (x)) = g([x]) = 0, since [x] e I if x ŒI Ï 0 f o g (x) = f (g (x)) = Ì 2 Ó [ x ] , x ŒR ~ I Thus g o f is a constant function and hence continuous but f o g is not continuous at every x e I~{0}. Example 48: If a, b are chosen from {1, 2, 3, 4, 5, 6, 7} randomly with replacement. The probability that

x + x 2 + + x n - n Ê 0 ˆ Solution: f(1) = lim ÁË form˜¯ x Æ1 x -1 0 = lim

, x ŒI

x

-1

¥

(b / a ) x - 1 x

2/ x

1 / u ◊ lim

= a 2 ◊ lim (1 + u )

(b / a ) x - 1

xÆ0

uÆ0

x

= a 2 elog b / a = ab Thus ab = 7. The favourable outcomes are (1, 7) and (7, 1). Therefore, required probability is 2/49. Example 49: 1 16 1 (c) 4 Ans. (a) (a)

lim

x Æp /2

cot x - cos x

(p - 2 x )3 1 8 p (d) 2

(b)

Solution: Put x – p/2 = q, so that

equals:

9.14

Complete Mathematics—JEE Main

lim

cot x - cos x

x Æp /2

3

(p - 2 x ) =

=

= lim

cot (p / 2 + q ) - cos (p / 2 + q ) 3

(- 2q )

q Æ0

tan q - sin q 1 lim 8 q Æ0 q3

(a)

(b)

sec 2 q - cos q 1 lim 8 q Æ0 3q 2

n2 6

(c)

n3 6

(d) 1/6

xÆ0

equal to (a) 0

(b)

lim f ( x ) = 0

Solution:

fi fi

2n

Example 53: Let f(x) =

xÆ0

sin nx tan x ˘ È ¥ ( n ) Í( a - n ) n =0 x Æ 0 nx x ˙˚ Î n((a – n) n – 1) = 0 1 (a – n) n – 1 = 0 fi a = n + . n n

Example 51: The function f(x) = [x] – [x ], n ≥ 2 (where [y] is the greatest integer less than or equal to y), is discontinuous at all points of (a) I (b) I ~ {0, 1} (c) I ~ {0} (d) I ~ {1}. Ans. (d) Solution: f(x)= 0 for x Œ I If 0 < x < 1, then 0 < xn < 1, so [x] = 0 and [xn] = 0 fi f(x) = 0 for 0 < x < 1 1/n If 1 < x < 2 then 1 < xn < 2 fi [x] = 1 and [xn] = 1 Thus f(x) = [x]n – [xn] = 0 if 1 < x < 21/n So f(x) = 0 if 0 £ x < 21/n Therefore, f is continuous at x = 1 To the left of any integral value m π 1 but close to m, f(x) π 0 but to the right of m and close to m, f(x) = 0. Hence f is discontinuous for all m Œ I ~ {1}. Example 52: The value of

sin n ( x / a ) ( log (1 + ( x / a )))

n

for

Solution: lim f ( x ) = xÆ0

lim

n

(enx - 1)

x π 0. If f(0) = 16n and f is a continuous function, then the value of a is (a) 16 (b) 2 (c) 8 (d) 4 Ans. (d)

n n +1

(d) n + (1/ n )

(c) n Ans. (d)

n3 3

n3 3 n4 4 x + x + n3 24 lim 6 = . xÆ0 6 x3

((a - n) nx - tan x ) sin nx

,xπ0 x2 and lim f ( x ) = 0, n is a non-zero real number, then a is

(n > 0) is

Ans. (c) Solution: Using series expansion of enx, the required limit is equal to

1 Ê 2 + 1ˆ 1 Á ˜= . 24 Ë 2 ¯ 16

Example 50: If f(x) =

fi

x3

xÆ0

1 2 sec 2 q tan q + sin q Ê 0 ˆ = lim form¯ Ë q Æ 0 24 2q 0 =

lim

Ê n2 2 ˆ enx - Á 1 + nx + x ˜ 2 Ë ¯

Ê e x - 1ˆ lim Á xÆ0 Ë x ˜¯

2n

¥

( x / a )n ( x / a )n ¥ ◊ a2n n n sin ( x / a ) ( log (1 + ( x / a )))

= a2n f(0) = lim f ( x ) so a2n = 16n = 42n

Since

xÆ0

Thus

a = 4.

Ïx + a ; x < 0 Example 54: Let f(x) = Ì and Ó x -1 ; x ≥ 0 ; if x < 0 Ï x +1 g(x) = Ì 2 ÓÔ( x - 1) + b ; x ≥ 0 If gof is continuous (a > 0) then (a) a = 2, b = 0 (b) a = 2, b = 1 (c) a = 1, b = 0 (d) a = 1, b = 1 Ans. (c) Solution: Ï x + a +1 Ô 2 Ô( x + a - 1) + b (go f) (x) = Ì x2 + b Ô Ô 2 Ó ( x - 2) + b

;

x < -a

; -a £ x < 0 ;

0 £ x 0 and q Œ [–p, p ], then the value of q is p p (a) ± (b) ± 4 3 p p (c) ± (d) ± 6 2 Ans. (d) Solution: lim (1 + x log (1 + b2))1/x xÆ0

1 ¥ log (1+ b2 ) 2 x log (1+ b2 )

= lim (1 + x log (1 + b ) xÆ0

= elog (1 + b

2

)

(a) 0 (c) 2 Ans. (c)

(b) 1 (d) 3

Solution: Since, lim

= 1 So | x -3| for Œ = 1 there is d > 0 such that 2 0 < ( f ( x ) - 9) < 2 whenever 0 < | x – 3 | < d xÆ3

| x -5| fi 0 < f (x) – 2 < 2

xÆ3

fi

xÆ3

Ê È 11 x ˘ È 21sin x ˘ˆ Example 58: The value of lim Ë Í , + xÆ0 Î sin x ˙˚ ÍÎ x ˙˚¯ where [x] is the greatest integer less than or equal to x is (a) 32 (b) 31 (c) 11 (d) 21 Ans. (b) x Solution: Since x > sin x, for x > 0 and lim =1 x Æ 0 + sin x 11x 11x > 11. Thus so Æ 11 as x Æ 0 + but sin x sin x È 11x ˘ ÍÎ sin x ˙˚ = 11 for value x Æ 0 +. Similarly 21 sin x Æ 21 as x Æ 0 + but 20 £ 21 sin x < 21 x x 21 sin x È ˘ ÍÎ x ˙˚ = 20 . Hence Ê È 11x ˘ È 21 sin x ˘ˆ lim Ë Í = 31 + x Æ 0 + Î sin x ˙ ˚ ÍÎ x ˙˚¯ È 11x ˘ = 11 Similarly x < sin x for x < 0, so Í Î sin x ˙˚ È sin x ˘ = 20 as x Æ 0 –. Thus as x Æ 0 – and Í21 Î x ˙˚ lim

Ê È 11x ˘

Í sin x ˚˙ x Æ0 - Ë Î

Thus,

fi

| x -3| whenever 0 1 1

(

5 + 1) x 2 +

(

5 - 2 + f ( x )) x + 2 5 - 5 f ( x ) = 0

satisfies for x ΠR then f ( 5 ) is equal to (a) 2 - 5

(b) 5 + 5

(c) 3 - 5 Ans. (d)

x n - sin x n

1 - sin 1 = sin 1 – 1 For n = 0, lim = lim x Æ 0 x - sin n x x Æ0 x - 1

(d)

Solution: For x π

x n - sin x n

x - sin x For n = 1, lim = lim =1 x Æ 0 x - sin n x x Æ 0 x - sin x

f(x) = –

n

Ê 1 ˆ Example 60: lim  tan -1 Á 2 ˜ is Ë 2k ¯ nÆ• k =1 (a) p/8 (c) p/4 Ans. (c)

=–

(b) p/2 (d) p/3 -1

x 3 - ( 5 + 1) x 2 + ( 5 - 2 ) x + 2 5 x- 5

( x + 1) ( x - 2 ) ( x - 5 ) x- 5

= – (x + 1) (x – 2) f ( 5 ) = lim f ( x ) = – (1 + 5 ) ( 5 - 2 )

2 Ê 1 ˆ -1 ÁË 2 ˜¯ = tan 1 + (2k - 1) (2 k + 1) 2k

= tan -1

= – (3 - 5 ) = 5 - 3 Example 62: If x = u is a point of discontinuity of f(x) = lim cos2 n x , then the value of cos u is

(2k + 1) - (2k - 1) 1 + (2k - 1) (2 k + 1)

–1

nƕ

–1

= tan (2k + 1) – tan (2k – 1)

1

tan–1(2n + 1) – tan–1(2n – 1) + tan–1  tan -1 ÊÁË 2k 2 ˆ˜¯ = (2n – 1) – tan–1 (2n – 3) + ... + tan–1

k =1

5,

xÆ 5

Solution: tan

n

5-3

3 – tan–1 1 = tan–1(2n + 1) – tan–1 1 p = tan–1 (2n + 1) – 4

(a) 0 (c) (– 1)n Ans. (c)

(b) 1/2 (d) 1

n Solution: f(x) = lim (cos2 x ) nƕ

ÏÔ0 , 0 £ cos2 x < 1 =Ì cos2 x = 1 ÓÔ1 , Ï0 , 0 π np =Ì Ó1 , x = np

n

p p p p Ê 1 ˆ tan -1 Á 2 ˜ = tan–1 • – = - =  ¯ Ë nÆ• 4 2 4 4 2k k =1 lim

Example 61: If f is a continuous function and x3 –

Hence f(x) is discontinuous when u = np, n ΠI So cos u = cos n p = (Р1)n.

Assertion-Reason Type Questions

Example 63: Let f(x) =

Ê1 + 1 1 + 0 Ê 1 ˆ ˆ - 1 Ê1 + 1 + 0 Ê 1 ˆ ˆ Á ÁË 2 ˜¯ ˜¯ ÁË ÁË 2 ˜¯ ˜¯ 2 x2 x x1 / 3 Ë x 3x2 = Ê1 + 1 1 + 0 Ê 1 ˆ ˆ - 1 Ê1 + 1 + 0 Ê 1 ˆ ˆ Á ÁË 4 ˜¯ ˜¯ ÁË ÁË 4 ˜¯ ˜¯ 4 x4 x x1 / 5 Ë x 5x 4

x2 + 1 – 3 x2 + 1 4

x4 + 1 – 5 x4 + 1

Statement-1: lim f (x) = 1 xƕ

Statement-2: lim

x Æ•

1 x

n

Since lim

= 0 for n > 0.

x Æ•

Ans. (a) Solution: f (x) =

2 1/ 2

2 1/ 3

( ) - x (1 + 1 / x ) 1/ 5 1/ 4 x (1 + 1 / x 4 ) - x 4 / 5 (1 + 1 / x 4 ) x 1 + 1/ x

2/3

1 x

n

= 0 for all n > 0, so lim f (x) =

Example 64: Let f(x) =

xƕ

1 - 0.1 = 1. 1 - 0.1

1 3 ,xπ1 1 - x 1 - x3

Statement-1: The value of f (1) so that f is a continuous function is 1

Limits and Continuity 9.17

Statement-2: g(x) =

x+2 1 + x + x2

is a continuous function

Ans. (d)

Example 67: Let f(x) = and

˘ 1 È 3 Solution: For x π 1, f(x) = Í1 ˙ 2 1- x Î 1+ x + x ˚ = So,

Ê g(x) = Á 1 + Ë

x+2 ( x - 1) ( x + 2 ) =– 2 (1 - x ) (1 + x + x ) 1 + x + x 2

f (1) = lim f (x) = – lim

x+2

x Æ1 1 +

xÆ1

x + x2

xÆ0

xƕ

Statement-2: Both the limits are equal to 1. Ans. (a)

= – 1.

Solution: lim f(x) xÆ0

1 + sin x - 1 + sin x 1 = lim ¥ x Æ0 tan x 1 + sin x + 1 - sin x 1 = lim 2 cos x ¥ x Æ0 1 + sin x + 1 - sin x

xÆa

=2¥1¥

Ans. (b) sin x cos 2 x + Solution: lim f(x) = lim

cos x sin 2 x cos 2 x

2x sin x sin 2 x cos x 1 = lim lim cos 2 x + lim lim x Æ0 2 x x Æ0 2 x Æ0 x x Æ0 cos2 x 3 1 = +1= . 2 2 Statement-2 is also true since f (x) = f (– x). Example 66: x-a p xˆ a Ê lim Á sin tan ˜ = , a π 0 xÆa Ë 2 2a ¯ p Statement-2: Trigonometric functions are continuous on their domain of definition. Statement-1:

1 = 1. 1+1 Ê1

1ˆ

1 ˆ x ¥ ËÁ x + x 2 ¯˜ Ê lim g(x) = lim Á 1 + ˜ = e0 = 1 xÆ• x Æ• Ë x¯

x Æ0

xÆ0

x +1 x

Statement-1: lim f (x) = lim g(x)

Example 65: Let f (x) = 1 - cos x cos 2 x , x π 0 x2 Statement-1: The value of f (0) so that f is a continuous function is 3/2. Statement-2: For an even function defined near a point a, lim f(x) = lim f(x) xÆa +

1ˆ ˜ x¯

1 + sin x - 1 - sin x ,xπ0 tan x

Example 68: Define F(x) as the product of two real Ïsin 1 / x ; if x π 0 functions f1(x) = x, x Œ R, and f2(x) = Ì ; if x = 0 Ó 0 as follows: Ï f ( x ) f2 ( x ) ; x π 0 F(x) = Ì 1 0 ; x=0 Ó Statement-1: Statement-2: Ans. (c)

F(x) is continuous on R f1(x) and f2(x) are continuous on R.

Solution: Since x sin 1 x £ | x | so lim F( x ) = 0 = xÆ0

Ans. (d)

F(0). Thus F is a continuous function. Since lim sin 1 x

p xˆ x-a Ê Solution: lim Á sin tan ˜ xÆa Ë 2 2 a¯ = lim sin xÆa

x Æ0

doesn’t exist so f2 is not continuous at x = 0.

x-a pÊ cot Á 1 2 2Ë

Example 69:

xˆ ˜ a¯

pÊ xˆ x-a x-a ÁË 1 - ˜¯ 2 a 2 ¥ 2 ¥ p a-x x-a x pÊ sin Á 1 - ˆ˜ 2 2 a 2Ë a¯

sin = lim

xÆa

pÊ ¥ cos Á 1 2Ë a a = 1¥ – ¥1¥1= p p Statement 2 is clearly true.

xˆ ˜ a¯

È tan -1 x ˘ Statement-1: lim Í ˙ = 0, where [x] represents xÆ0 Î x ˚ greatest integer £ x. Statement-2:

tan -1 x < 1 for all x π 0. x

Ans. (a) Solution: Let f(x) = tan–1 x – x f ¢ (x) =

1

-1=

- x2

< 0. Hence f is decreasing func1 + x2 1 + x2 tion so for x > 0, tan–1 x < x and for x < 0, f(x) > f(0) = 0 tan -1 x fi tan–1 x > x fi < 1 (x < 0) x

9.18

Complete Mathematics—JEE Main

Statement-1: f(x) = sgn(x2 – 5x + 6) is continuous for all x Statement-2: ax2 + bx + c = 0 has no real roots if b2 – 4ac < 0 Ans. (d) Solution: f(x) = sgn (x – 2) (x – 3). Hence f is not continuous at x = 2 and x = 3 Thus statement 1 is not true but statement 2 is true.

È tan -1 x ˘ Thus lim Í ˙ = 0 and x Æ0 + Î x ˚ È tan -1 x ˘ lim Í ˙ =0 x Æ0 - Î x ˚ Example 70:

LEVEL 2 Straight Objective Type Questions 1 - cos (a x 2 + bx + c)

Example 71: The value of lim

( x - a )2

x Æa

where a and b are the roots of ax2 + bx + c = 0 is (a) (a – b)2 (c)

(b)

1 2 a (a - b )2 2

(a - b )2 2

(d) none of these

Ans. (c)

(

)

1 - cos a x 2 + b x + c Ê 0 ˆ ÁË form˜¯ 2 x Æa 0 (x - a )

Solution: lim

(2 a x + b) sin (a x + bx + c ) = lim x Æa 2(x - a )

ÊÊ ˆ ˆ ÈÊ ˆ x2 x4 x2 x4 + - ˜ - 1˜ ÍÁ 1 + - ˜ = Á Á1 2! 4! 2! 4! ËË ¯ ¯ ÎË ¯ 2 Ê x x3 ˆ ˘ - Á1 + x + +  ˙ 2! 3! ˜¯ ˚ Ë Ê 1 x2 ˆ  = x2 Á - + Ë 2 4! ˜¯

Ê x3 ˆ 2  x x ÁË 3! ˜¯

Ê 1 x2 ˆ  = x3 Á - + Ë 2 4! ˜¯

Ê x2 ˆ  1 x ÁË 3! ˜¯

fi

(cos x - 1) (cos x - e x ) x3

2

[using L’ Hôpital Rule] = lim

x Æa

= lim

x Æa

(2 a x + b) sin [a ( x - a ) ( x - b )] 2(x - a )

=

a2 (2a - a - b ) (a - b ) 2

=

a 2 (a - b ) . 2

2

Example 72: The integer n for which x lim (cos x - 1)(cos x - e ) is a finite non-zero number is xÆ0 xn (a) 1 (b) 2 (c) 3 (d) 4 Ans. (c)

Solution: (cos x – 1) (cos x – ex)

(cos x - 1) (cos x - e x ) 1

= . So n = 3. 2 x3 In case n = 1, 2, 4, the limit is zero or does not exist. lim

xÆ0

Example 73: Let f(x) = lim

nƕ

(2 a x + b) sin [a ( x - a ) ( x - b )] a ( x - b) 2a ( x - a ) ( x - b )

a2 Ê bˆ = ÁË 2a + ˜¯ (a - b ) 2 a

Ê1 ˆ + terms containing x ˜ = Á2 ÁË and higher powers of x ˜¯

x2n - 1 x2n + 1

, then

(a) f (x) = 1 for |x| = 1 Ï 1 for x > 1 (b) f (x) = Ì Ó- 1 for x < 1 Ï 1 for x > 1 (c) f(x) = Ì Ó- 1 for x £ 1 (d) f is not defined for any value of x Ans. (b) Solution: For |x| < 1, x2n Æ 0 as n Æ • and for |x| > 1, 1 x2n

Æ 0 as n Æ •. For |x| = 1 we have x2 = 1. Thus x =1 Ï 0, Ô 1 - x - 2n , x >1 x 2 n - 1 ÔÔ = Ì 1 + x - 2n 2n x + 1 Ô 2n Ô x - 1, x < 1 ÔÓ x 2 n + 1

Limits and Continuity 9.19

x =1 x >1 x 0

9.20

Complete Mathematics—JEE Main

(a) a a (c) –

(b)

a

a

= lim

(d) – a a

Ans. (c) Solution: Note that f is continuous for all x except possibly at x = 0. In order that f becomes continuous at x = 0, we must have

sin x 1 È ˘ lim Í(b - 2a) - a cos x ˙ 6 xÆ0 Î x ˚

=

xÆ0

Rationalizing the numerator and denominator, we have

xÆ0

2 x È a 2 - ax + x 2 + a 2 + ax + x 2 ˘ Î ˚

= - a.

2 - (256 - 7 x )1 / 8

(x π 0); then for (5 x + 32)1 / 5 - 2 f to be continuous on [– 1, 1], f(0) is equal to (a) – 1 (b) 1 (c) 2 6 (d) none of these Ans. (d) Solution: f is continuous everywhere except possibly at x = 0. For f to be continuous everywhere,

Solution:

1/ 8

f (0) = lim f (x) = lim x Æ0

x Æ0

2 - (256 - 7 x ) 1/ 5

(5 x + 32)

lim f(x) = lim

-2

xÆ0 -

7 x ˆ 1/ 8 1 7x 1 - ÊÁ 1 1 - ÈÍ1 - ◊ +0 ˜ Ë 256 ¯ Î 8 256 = lim = lim 1/ 5 1 5x x Æ0 È x Æ0 2 Ê1 + 5 x ˆ - 1 ˜¯ ÁË ÍÎ1 + 5 ◊ 32 + 0 x 32

x ˘˙ ˚ ˘ ˙˚ - 1

( )

( )

2

È1 ◊ 7 - 0 x ˘ ( )˙ 1 7 32 Í 7 ˚ = ◊ ¥ = lim Î 8 256 = . 1 8 256 1 64 x Æ0 + 0 (x) 32 x (1 + a cos x ) - b sin x

,xπ0 x3 and f(0) = 1. The value of a and b so that f is a continuous function are (a) 5/2, 3/2 (b) 5/2, –3/2 (c) –5/2, –3/2 (d) none of these Ans. (c) Example 80: Let f(x) =

Solution: Note that f is continuous everywhere except possibly at x = 0. For f to be continuous at x = 0, x (1 + a cos x ) - b sin x Ê 0 ˆ 1 = f (0) = lim ÁË form˜¯ xÆ0 0 x3 = lim

xÆ0

(1 + a cos x ) - x a sin x - b cos x 3 x2

1 [b - 3a] fi b – 3a = 6 6 Solving a - b = - 1 and b - 3a = 6, we get b = -3/2 and a = -5/2. Ï(3 / x 2 ) sin 2 x 2 if x < 0 Ô 2 1 Ô x + 2x + c if x ≥ 0, x π Example 81: f(x) = Ì 2 3 Ô 1- 3 x Ô x = 1/ 3 0 Ó then in order that f be continuous at x = 0, the value of c is (a) 2 (b) 4 (c) 6 (d) 8 Ans. (c) =

- 2ax ÈÎ a + x + a - x ˘˚

Example 79: If f(x) =

3 x2

This limit will exist only if 1 + (a - b) = 0. In this case limit is equal to - ( a - b ) sin x - a x cos x - a sin x = lim xÆ0 6x (using L’Hôpital Rule)

f (0) = lim f (x)

f (0) = lim

1 + ( a - b ) cos x - x a sin x

xÆ0

xÆ0 -

lim f (x) = lim

xÆ0 +

3 x2

sin 2 x 2 = 6 lim

xÆ0

x2 + 2 x + c 1 - 3x

xÆ0 +

sin 2 x 2

2

=

2 x2

=6

c =c 1

Hence for f to be continuous c = 6. 2

Example 82: The function f(x) = (sin 2x) tan 2x is not defined at x = p/4. The value of f(p/4) so that f is continuous at x = p/4 is e

(a)

(b) 1

(c) 2 Ans. (d)

(d) none of these

Solution: For f to be continuous at x = p/4, f (p/4) =

lim

x Æp / 4

= lim

x Æp / 4

f(x) =

lim

x Æp / 4

{

(1 – cos2 2x)

(1 - cos2 2 x )-(1 / cos

Example 83: Let f(x) =

(

2

2 x)

(1 / 2 ) tan 2 2 x

- (sin 2 2 x ) / 2

} )

(

=

1 e

log 1 + x + x 2 + log 1 - x + x 2

.

),

sec x - cos x x π 0. Then the value of f(0) so that f is continuous at x = 0 is (a) 1 (b) 0 (c) 2 (d) none of these Ans. (a)

Limits and Continuity 9.21

Solution: For f to be continuous at x = 0,

(

log 1 + x + x

f (0) = lim f (x) = lim xÆ0

((

= lim

1 - cos x

( (

log 1 + x 2 + x 4 sin 2 x

xÆ0

= lim

- x2

2

xÆ0

= lim

2

)

( (

log 1 + x 2 + x 4 x2 + x4

xÆ0

Ans. (a)

) + log (1 - x + x )

=

)) ◊ cos x sin 2 x

(

¥ 1+ x

(e x - 1)4

sin ( x 2 / k 2 ) log {1 + ( x 2 / 2)} may be continuous function is (a) 1 (b) 4 (c) 2 (d) 3 Ans. (c)

2

) cos x

= 1.

Solution: f(0) = lim

= lim

)

(

fi

= lim

f (0) = lim f (x) = xÆ0

È log (1 + a x ) log (1 - b x ) ˘ lim Í ˙˚ xÆ0 Î x x a log (1 + a x ) ( - b ) log (1 - b x ) = a + b. - bx ax

Example 86: Let f(x) = (a) (b) (c) (d)

( (

sin p 1 - sin 2 x x

(

sin p sin 2 x x

(

log 1 + x 2 4

2

)

x - 26 x + 25

f is continuous on [6, 10] f is continuous on [– 2, 2] f is continuous on [– 6, 6] f is continuous on [1, 7]

))

2

)

2

(

)

= 1 ◊ p ◊ 1 = p.

k = ± 2.

Example 85: The function f(x) = [log (1 + ax) – log (1 – bx)]/x is not defined at x = 0. The value, which should be assigned to f at x = 0 so that it is continuous at x = 0 is (a) a – b (b) a + b (c) log a + log b (d) none of these. Ans. (b)

xÆ0

)

2

sin p sin 2 x p sin 2 x ◊ xÆ0 p sin 2 x x2

)

= 2k2

= lim

x

xÆ0

4

Solution

(

xÆ0

= lim

)

sin p cos2 x

xÆ0

Ê e x - 1ˆ x2 / k 2 x2 / 2 lim Á ¥ ¥ 2 k2 ¥ ˜ 2 2 2 xÆ0Ë x ¯ sin x / k log 1 + x / 2 2k2 = 8

)( ) log (1 + x 2 )

- 25 x - 1

( x - 5) ( x + 5) ( x - 1) ( x + 1)

x π 0; f (0) = 8

,

xÆ0

(

)

2

Since log (1 + x2) is continuous on (- •, •), so f is continuous on R ~ {- 5, -1, 1, 5} which clearly contains the interval [6, 10]. sin p cos2 x Example 87: Let f(x) = , x π 0. The value x2 of f(0) so that f is a continuous function is (a) – p (b) p (c) p/2 (d) 1 Ans. (b)

Solution: We have lim f (x) =

So

(x

2

(

x2

Example 84: The value of k (k > 0) for which the function f (x) =

Solution: f (x) =

) ◊cos x

)) ¥

(

log 1 + x 2

2

sec x - cos x

xÆ0

log 1 + x 2

2

. Then

Ï ea x - e x - x xπ0 Ô Example 88: Let f(x) = Ì x2 Ô 32 x=0 Ó The value of a so that f is a continuous function is (a) 1 (c) 4 Ans. (d) Solution:

(b) 0 (d) 2 3 ea x - e x - x Ê 0 ˆ = lim f ( x ) = lim ÁË form˜¯ 2 x Æ 0 x Æ 0 2 0 x

a ea x - e x - 1 xÆ0 2x For the last limit to exist, we must have lim (a eax – ex – 1) = 0 = lim

xÆ0

\ a – 1 – 1= 0 fi a = 2. In this case, the last limit is equal to a 2 ea x - e x a2 -1 3 = = . xÆ0 2 2 2 lim

9.22

Complete Mathematics—JEE Main

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

Example 89: Let f be a continuous function on R such 2

that f(1/2n) = (sin en) e- n + (a) 1 (c) 2 Ans. (c)

2n

2

. Then the value of f(0) is n2 + 1 (b) 1/2 (d) none of these

f is a continuous function f is continuous except x = 1/2 f is continuous except for finitely many points The set of points where f is not continuous is infinite.

Ans. (d) n

Solution: As f is continuous so f(0) = lim f ( x ) = xÆ0

lim f ( xn ) for any sequence xn such that lim xn = 0.

nƕ

nƕ

Ê 1ˆ Solution: lim f ( x ) = Á ˜ whereas Ë 2¯ n x Æ (1 2 ) lim f ( x ) = (1/2)n – 1 as f (x) = (1/2)n – 1 for (1/2)n £ x < x Æ(1 2 )n + n –1

Thus 2

2 Ê 2n ˆ f (0) = lim f 1 2 n = lim Á sin en e- n + 2 ˜ nÆ• nÆ• Ë n + 1¯

( )

(

)

2 Ê 2 ˆ = lim Á sin en e- n + = 0 + 2 = 2. nÆ• Ë 1 + 1 n2 ˜¯

(

)

Example 90: Let f(x) = lim

nƕ 1 +

continuous at (a) p (c) p/4 Ans. (d)

x

(2 sin x )2 n

then f is dis-

(b) p /3 (d) p /6

Solution: Ï Ôx Ô Ôx f (x) = Ì Ô2 Ô ÔÓ0

if if if

p p + np < x < + np 6 6 1 p sin x = i.e. x = ± + n p 2 6 1 p 5p sin x > i.e + np < x < + np 2 6 6 sin x < 1 / 2 i.e -

Thus points of discontinuities of f are of the form x = p ± + np . In particular, f is discontinuous at x = p/6. 6 Alternatively since the choices are in [0, p] so one may look at the graph of sin x and decide about the point of discontinuity. Example 91: Let f be a non-zero continuous function satisfying f(x + y) = f(x) f(y) for all x, y ŒR. If f(2) = 9 then f(3) is (a) 1 (b) 27 (c) 9 (d) none of these Ans. (b) Solution: Any non-zero continuous function satisfying the given functional equation is of the form ax for some a > 0. Since f(2) = 9 so a2 = 9 i.e., a = 3. Thus f(x) = 3x and so f(3) = 33 = 27. Example 92: Let f be a function on [0, 1] defined by f(x) = (1/2)n, (1/2)n + 1 £ x < (1/2)n, n = 0, 1, 2, º Then

(1/2) . So f is discontinuous at 1/2, (1/2)2…. Thus f has infinitely many discontinuities. Ï Ô sin ( a + 1) x + sin x Ô x Ô Example 93: If f(x) = Ì c Ô 2 Ô x +bx - x Ô bx 3 2 Ó

,x0

is continuous at x = 0, then (a) (b) (c) (d) Ans. (c)

a = – 3/2, b = 0, c = 1/2 a = – 3/2, b = 1, c = 1/2 a = – 3/2, b e R, c = – 1/2 none of these

Solution:

sin ( a + 1) x + sin x xÆ0 x

lim f(x) = lim

x Æ 0-

sin ( a + 1) x sin x 2 ˘ È = lim Í(a + 1) + x ˙ =a+2 xÆ0 Î x (a + 1) x ˚ Also

lim f(x) = lim

xÆ 0 +

x + bx 2 - x bx 3 2

xÆ0+

= lim

xÆ0

= lim

xÆ0

1 + bx - 1 Ê 0 = Á formˆ˜ Ë0 ¯ bx

1 1 2 1 + bx b 2 b

◊

Thus we must have a + 2 = c = 1/2. So a = – 3/2, c = 1/2 and b is any real number. Ï 72 x - 9 x - 8 x + 1 , xπ0 Ô Example 94: If f(x) = Ì 2 - 1 + cos x Ô K log 2 log 3, x=0 Ó is a continuous function then K is equal to (b) 24 (a) 2 (c) 18 3

(d) 24 2

Limits and Continuity 9.23

Ans. (d) Solution: K log2 log 3 = f(0) = lim f(x)

x Æ0

xÆ0

= lim

72 x - 9 x - 8 x + 1 2 - 1 + cos x

xÆ0

=

9 x - 1) (8 x - 1) ( lim

= lim

xÆ0

=

16

x2

x2

xÆ0

2 (1 - cos x 2 ) 2

9x - 1 8x - 1 ◊ x x

16 ( x 4 )

2 ◊ 2 sin 2 x 4

log 9 log 8 =

2 2

8 2

Example 97: For x π 1, f is defined by f(x) = 1 1 log x x - 1 The value of f(1), so that f is a continuous function is 1 (a) 1 (b) 2 (c) 0 (d) 2 Ans. (b)

6 log 3 log 2

Thus K = 24 2 . Example 95: Let f be a function defined on R by f (x) = x - [ x ] then (a) f is not continuous at every x ΠI (b) f is not continuous at every x ΠR ~ I (c) f is a continuous function (d) none of these Ans. (b)

[x] +

x-n ,n£x –1)

lim f ( x ) = lim |5x – 2| [x – 2] = – lim |5x – 2| =– 3

x Æ1+

xÆ1+

xÆ1+

Hence f is discontinuous at x = 1. x=0 Ï1 , Ô For 0 £ x £ 1, [cos p x] = Ì 0 , 0 < x £ 1 / 2 Ô-1 , 1 / 2 < x £ 1 Ó so [cos p x] is discontinuous at x = 0, 1/2 For 1 < x £ 2, |5x – 2| [x – 2] = (5x – 2) [x – 2] Ï - (5 x - 2 ) , 1 < x < 2 = Ì 0 , x=2 Ó

9.24

Complete Mathematics—JEE Main

This function is discontinuous at 2. Thus f is discontinuous at 0, 1/2, 1, 2.

Example 100: lim

x Æ1

2

Ï x + Ax + 5 x ŒQ Example 99: If f(x) = Ì x ŒR ~ Q Ó 1+ x is continuous at exactly two points, then the possible values of A are in (a) (1, •) (b) (–3, •) (c) (5, •) » (– •, – 3) (d) (– •, •) Ans. (c) 2

Solution: For f to be continuous, x + Ax + 5 = 1 + x should have real roots fi x2 + (A – 1) x + 4 = 0 i.e. (A – 1)2 – 16 £ 0 fi A – 1 ≥ 4 or A – 1 £ – 4 fi A ≥ 5 or A £ – 3 fi A Œ [5, •) » (– •, –3]

n x n +1 - n x n + 1

(e x - e2 ) sin p x

n(n + 1) (a) (c)

is equal to

n (b)

2

e -e n

(d)

(e - e2 ) p

(e

2

- e) p

n2 (n + 1)

(e2 - e ) p

n x n +1 - n x n + 1 Ê 0 ˆ x Æ1 (e x - e2 ) sin p x Ë 0 ¯

Solution: lim

= lim

x Æ1

n(n + 1) x n - n2 x n + 1 e x sin p x + (e x - e2 ) p cos p x n

=

(e2 - e ) p

.

EXERCISE Concept-based Straight Objective Type Questions 1

1. lim

n2 (a) 0 1 (c) 2 x Æ•

(1 + 2 +

+ n) is equal to (b) 1 1 (d) 4

1 1 1 ˆ 2. lim ÊÁ is equal to + + x Æ• Ë 1.2 2.3 (n - 1)n ˜¯ 1 (a) 0 (b) 2 1 (c) (d) 1 3 3. lim

x Æ1

xm - 1 xn - 1

is equal to (m and n are positive integers)

(a) m n

(b) 1

(c) 0

(d)

4. lim

a Æ0

sin (a n )

(sin a )m

(a) 1 if m > n (c) 1 if > m

m -1 n -1

(m, n ΠI +) is equal to (b) 0 if n > m (d) 1 if m = n

k ˆ mx Ê 5. lim Ë 1 + ¯ is equal to x Æ• x

(a) mk (c) emk

6. Let

If f (a) (c)

(b) em (d) em / k

Ïa x 2 - x - 2 , x < 2 Ô ,x=2 f(x) = ÔÌb Ô x - [x] ,x>2 ÔÓ x - 2 is continuous at x = 2 then a = 1, b = 1 (b) a = 1, b = 2 a = 2, b = 1 (d) a = 2, b = 2

7. Let f(x) =

2 x 5 - 8 x 2 + 11 x4 + 4 x 3 + 8x2 + 8x + 4

. Then

A = {x : f is continuous at x} is equal to (a) R (b) R ~ {0} (c) R ~ {–1} (d) R ~ {– 1, – 2} x £1 Ï ax - b , Ô 8. Let f(x) = Ì 3 x , 1< x < 2 Ôbx 2 - a , x≥2 Ó If f is continuous function then (a, b) is equal to (a) (1, 2) (b) (1, 3) (c) (3, 6) (d) (6, 3) p Ï -1 ÔÔcos (cot x ) , x < 2 9. If f(x) = Ì Ô a (p [ x ] - 1) , x ≥ p ÔÓ 2

Limits and Continuity 9.25

p The value of a for f to be continuous at x = is 2 p (b) p – 1 (a) 2(p - 1) (c)

p p +1

(d)

p p -1

Ï sin 4 x , xπ0 Ô 10. Let f(x) = Ì log(1 + 3 x ) ÔÓ A + 1 , x=0 The value of A for f to be continuous at x = 0 is 2 1 (b) (a) 3 3 3 (c) 1 (d) 4

LEVEL 1 Straight Objective Type Questions (a) 1 (c) 0

x +1 x ˆ Ê - tan -1 11. The value of lim x Á tan -1 ˜ is xÆ• Ë x+2 x + 2¯ (a) 1 (c) 1/2

(b) 0 (d) 1/4

(

12. The value of lim 1 + tan 2 x xÆ0

(a) e (c) e1/4 13. If f (x) = lim f (x) is

1 2x

)

(b) e1/2 (d) e–1/2 sin ( a + x ) - sin ( a - x ) , x π 0 then tan ( a + x ) - tan ( a - x )

19.

1 cos a 2 (d) none of these

20.

xÆ0

(a) cos a

(b)

2

(c) cos a sin ( x - p 6 ) is equal to 14. lim x Æp 6 3 2 - cos x (a) 2 (b) 1 (c) – 1 (d) 3 /2 cos x 15. lim is equal to x Æp 2 3 (1 - sin x )2 (a) 1 (b) 2 (c) 21/3 (d) none of these 16. If x2 + 2x3 £ x + f (x) £ x5 – 2x3 for value of x near 0, then lim

xÆ0

18.

f (x) is x

(a) – 1 (c) 1

(b) 0 (d) none of these

sin -1 ([ x ] + x ) , [x] π 0 [ x] = 0, [x] = 0 where [x] denotes the greatest integer less than or equal to x, then lim f (x) is

17. If f (x) =

xÆ0

21.

22.

(b) – 1 (d) none of these 2x - 1 p, where [ ] deThe function f(x) = [x] cos 2 notes the greatest integer function is discontinuous at (a) all x (b) all integer points (c) no x (d) x which is not integer y-a py Let f (y) = sin tan , y π a. The value of 2 2a f(a) so that f is a continuous function is (a) p/a (b) – a/p (c) p/2a (d) none of these Let f (x) = (1 + sin x)cosec x, the value of f(0) so that f is a continuous function is (a) e (b) e1/2 2 (c) e (d) 1 Ï x +1 , x £1 Let f (x) = Ì 2 Ó3 - a x , x > 1 The value of a so that f is continuous is (a) 1/2 (b) 1 (c) 2 (d) 3 Let f (x) = [x2 + 1], ([x] is the greatest integer less than or equal to x). Then f is continuous (a) on [1, 3] (b) for all x in [1, 3] except four points (c) for all x in [1, 3] except seven points (d) for all x in [1, 3] except eight points

1 - cos2 x is not continuous at 2 (a) x = p/4 (b) x = 3p/4 (c) x = 5p/4 (d) none of these 24. The number of points at which the function f (x) = 1/(x – [x]) is not continuous is (a) 1 (b) 2 (c) 3 (d) none of these 23. The function y =

9.26

Complete Mathematics—JEE Main

25. Let f(x) = x, x ŒQ; f(x) = 1 – x, x Œ R ~ Q then f is continuous only at x = (a) 1/2 (b) 1 (c) 0 (d) 2 26. lim

x tan 2 x - 2 x tan x 2

(1 - cos 2 x )

xÆ0

(a) 2 (c) 1/2

xƕ

36.

is (b) – 2 (d) – 1/2

27. Let f (x) =

35. The value of lim x3/2

37.

1 - cos 2 ( x - 2 ) , x π 2. The lim f ( x ) x-2 xÆ2

(a) exists and is equal to

2

(b) does not exist because lim f ( x ) doesn't exist xÆ2+ (c) equal to 1

38.

(d) doesn't exist because lim f ( x ) π lim f ( x ) xÆ2+

xÆ2-

28. The value of f (0) so that the function f(x) = 3

1+ x - 1+ x becomes continuous is equal to x (a) 1/6 (b) 1/4 (c) 1/3 (d) 3 29. The set of all points of continuity of fo fo f, where f (x) = sgn x is (a) R ~ {0} (b) R ~ {1, 0, 1} (c) R ~ {– 1, 1} (d) none of these 30. The set of all points of discontinuity of f (x) = x –1 x 3 + 6 x 2 + 11x + 6 (a) f (b) {– 1} (c) {– 1, – 2, – 3} (d) {1, 2, 3} 31. The number of continuous functions on R which satisfy (f(x))2 = x2 for x Œ R is (a) 2 (b) 4 (c) 1 (d) infinitely many x2 + 1 - 1

32. The value of lim

x 2 + 16 - 4 (b) 4 (d) 2

xÆ0

(a) 3 (c) 1

x 2 - a2

xÆa

(a) (c)

1 4a

(b) 1

(d)

2a a - b

34. The value of lim

xƕ

(a) – 1 (c) 0

(

(a > b) is

1 a a-b 1 4a a - b

)

x 2 + 1 - x 2 - 1 is (b) 1 (d) none of these

)

x 3 + 1 - x 3 - 1 is

(a) 1 (b) – 1 (c) 0 (d) none of these The value of f (1) so that f(x) = (ex – e)/(x – 1) is continuous at x = 1 is (b) e2 (a) e–1 1/2 (c) e (d) e The function f (x) = a[x + 1] + b[x – 1], (a π 0, b π 0) where [x] is the greatest integer function is continuous at x = 1 if (a) a = 2b (b) a = b (c) a + b = 0 (d) a + 2b = 0 Let f (x) = [2x3 – 6], where [x] is the greatest integer less than or equal to x. Then the number of points in (1,2) where f is discontinuous is (a) 5 (b) 7 2n (c) 13 (d) 12 È Êp x ˆ˘ Let f be defined on R by f (x) = lim Ísin Ë ¯˙ nÆ•Î 2 ˚ then (a) f is continuous on [0, 2] (b) f is continuous on R ~ {2n, n Œ I} (c) f is continuous on R ~ {(2n + 1), n Œ I} (d) lim f (x) = 1 xÆ0+

40. If f(x) = (1/2)x – 1, then on the interval [0, p], 1 are both continuous (a) tan f(x) and f (x) (b) tan f(x) and

1 are both discontinuous f (x)

(c) tan f(x) and f–1(x) are both continuous 1 is not (d) tan (f (x)) is continuous but f (x) 41. If lim [cot(p / 4 + x )]1 x = Ae2 then the value of A is xÆ0 2

(b) e–4 (d) e3 [1 2 + x ] - [1 2] 42. Let f (x) = , – 1 £ x £ 2 and f (0) x = 0 ([x] is greatest integer less than or equal to x) then (a) f is not continuous at x = 0 (b) f is a continuous function on [–1/2, 1/2] (c) f is continuous on [– 1, 0] (d) f is continuous on [0, 2] 43. Let f and g functions such that f + g is a continuous function then (a) f and g are continuous function (b) f or g is a continuous function (c) f may be discontinuous but g is continuous (d) both f and g may be discontinuous (a) e (c) e

is

x-b - a-b

33. The value of lim

39.

(

Limits and Continuity 9.27 13

44.

(1 + x2 ) lim

xÆ0

14

- (1 - 2 x )

is

x + x2

(a) 2 (b) 1/2 (c) 3/2 (d) cannot be determined 45. The set of all points for which f(x) = |x| |x – 1| + 1/[x + 1] ([x] is the greatest integer function) is continuous is (a) R (b) R ~ I (c) R ~ (I » [– 1, 0)) (d) none of these

((

[ x]

{x}

)

)

-x

Ï e -e e + A , x0 46. Let f(x) = Ì tan {x} Ô Ô 2 , x=0 Ó The value of A so that f is continuous at x = 0 is ([x] is greatest integer function and {x} is the fractional part of x) is (b) 3 – e–1 (a) e–1 –1 (c) 2 – e (d) 2

bˆ Ê sin 2 x 47. If lim Á 3 + a + 2 ˜ = 0 then the value of 3a xÆ0 Ë x x ¯ + b is (a) 2 (b) – 2 (c) – 1 (d) 0 48. For continuous functions f and g on R, let f(a) = 4, f¢(a) = 6, g(a) = 2, g¢(a) = 1. Then the value of lim

xÆa

f ( x ) g( a ) - g( x ) f ( a ) ( x - a)

(

f ( x ) g( a ) + g( x ) f ( a )

(a) 0

(b) 3

(c) 1/4

(d) 2/3

)

Ï 1 + ax - 1 - ax , -1£ x < 0 ÔÔ x 49. Let f(x) = Ì 2x + 1 Ô ,0 £ x £1 ÔÓ x-2 The value of a so f is continuous on [– 1, 1] is (a) 1/2 (b) –1/2 (c) 1 (d) – 1

Assertion-Reason Type Questions

n

50. Let f(x) =

m

x –1 x –1

, x π 1, and g(x) =

sin mx ,xπ0 sin nx

and n π m. Statement-1: lim f(x) = lim g(x) x Æ1

xÆ0

Statement-2: Both the limits are equal to 51. Let f (x) =

n m

cos x - sin x p ,xπ . 4 cos 2 x

Êpˆ Statement-1: The value of f Á ˜ so that f is a Ë 4¯ 1 continuous is 2 1 1 Statement-2: f(x) = Ê 2 sin x + p ˆ ˜ ÁË 4¯ 52. Let f (x) =

( x + 1)10 + ( x + 2 )10  + ( x + 50)10 x10 + 1010

Statement-1: lim f(x) = 10 xƕ

Statement-2: f(x) = 10

50

 Â

r =1 a =1

10

Cr x

r – 10 10 – r

a

Ê Ê 10 ˆ 10 ˆ Á 1 + ÁË x ˜¯ ˜ Ë ¯

–1

53. Let f(x) = x if x is rational and f(x) = 1 – x if x is irrational Statement-1: f is discontinuous on [0, 1] Statement-2: f satisfies the intermediate value property. Ê x + yˆ 54. Let f be a continuous function satisfying f Á Ë 2 ˜¯ f ( x ) + f ( y) = , f(0) = 0, f(1) = 1 2 Statement-1: There is only one function satisfying the above conditions. Statement-2: A continuous function satisfying f(x + y) = f(x) + f(y) is of the form f(x) = ax. 55. Statement-1: lim cos–1 {x} does not exist, {x}bexÆ0

ing the fractional part of x Statement-2: {x} is discontinuous 56. Statement-1: The functions |x2 + ex – log x|, x2 + e x – log x are continuous functions Statement-2: Continuity of | f | implies the continuity of f 57. Statement-1: If |f(x)| £ |x| for all x Œ R then |f | is continuous at x = 0 Statement-2: If f is continuous then |f | is also continuous 58. Statement-1: f(x) = cos x + 2[x] is discontinuous at x = 0, where [x] denotes the greatest integer £ x.

9.28

Complete Mathematics—JEE Main

Statement-2: If u(x) is continuous and v(x) is discontinuous at x = a then u(x) + v(x) is discontinuous at x = a 59. Statement-1: f(x) = {sin x} – {cos x} is discontinuous at x = p/2, {x} is the fractional part of x.

Statement-2: Sum of two discontinuous function is a discontinuous function Ï e1 x - 1 ¸ 60. Statement-1: lim [ x ]2 Ì 1 x ˝ does not exist x Æ0 Ó e + 1˛ 1x Ï e1 x - 1 ¸ - 1¸ 2 Ïe Statement-2: lim Ì 1 x ˝ = lim [ x ] Ì 1 x ˝ x Æ0 + Ó e + 1 ˛ x Æ0 + Ó e + 1˛

LEVEL 2 Straight Objective Type Questions 61. If a and b are the roots of ax2 + bx + c = 0, c π 0 then

(

1 - cos cx 2 + bx + a

lim

is

2

c Ê 1 1ˆ (b) 2 a ÁË a b ˜¯

(a) c(a – b) 1 Ê 1 1ˆ b ÁË a b ˜¯

(d)

Ê 1 1ˆ c ÁË a - b ˜¯ 2a

62. Let f be a continuous function satisfying f(x) f(y) = f(x) + f(y) + f(xy) – 2 for all x, y Œ R and f(2) = 5 then lim f(x) is xÆ4

(a) 5 (c) – 5

(b) 17 (d) 21

63. The value of lim

e1 x - e-1 x

tan (1/x) is + e-1 x 1 (b) 0 –1 (d) none of these f be a continuous function satisfying f(x + y) = + f(y) for all x, y Œ R and f(1) = 5 then lim xÆ4 is equal to x Æ • e1 x

(a) (c) 64. Let f(x) f(x)

(a) 4 (b) 80 (c) 0 (d) none of these 65. Let f(x) = ex sgn (x + [x]), where sgn is the signum function and [x] is the greatest integer function. Then (a) lim f ( x ) = 0

(b) lim f ( x ) = – 1

xÆ0+

xÆ0+

(c) lim f ( x ) = 1

(d) lim f ( x ) = 1

xÆ0+

xÆ0-

66. The value of lim

x Æ -4

(b) less than 2 (d) equal to 1 sin x

2(1 - a x )

x Æ1 a

(c)

)

(a) greater than 3 (c) equal to 0

tan p x + lim (1 + 1/x2)x is x + 4 xƕ

Ê sin x ˆ x -sin x 67. The value of lim Á is ˜ xÆ0Ë x ¯ (a) e (c) e2 68. lim

xÆ0

(b) 1 (d) 1/e

(sin x - tan x )2 + (1 - cos 2 x )4 + x 5 7 tan 7 x + sin6 x + 2 sin 5 x

is equal to

(a) 0 (b) 1 (c) 1/2 (d) 2 69. Let f(x) = x – 1 and g(x) = 1/x. Then the set of points where gof og is continuous is (a) R ~ {0} (b) R ~ {1} (c) (– •, •) (d) (– •, •) ~ {0, 1} ÈÎ x 2 ˘˚ + È(2 x )2 ˘ +  + È( nx )2 ˘ Î ˚ Î ˚ then 70. Let f (x) = lim nÆ• n3 the set of all points of continuity of f ([x] denotes the greatest integer function) (a) (– •, •) ~ {0} (b) (– •, •) ~ I (c) (– •, •) (d) (– •, •) ~ {0, 1} Ï1 x is rational , then 71. Let f(x) = Ì Ó0 x is irrational (a) (b) (c) (d)

f is discontinuous for every real x f is continuous on R f is continuous at the points where x is rational f is continuous at the points where x is irrational. 72. Let f: R Æ R be defined by x irrational Ï x, f (x) = Ì Ó1 - x, x is rational then f is

Limits and Continuity 9.29

(a) discontinuous at x = 1/2 (b) continuous at x = 1/2 (c) continuous everywhere (d) discontinuous everywhere 73. Let f: R Æ R be any function. Define g: R Æ R by g(x) = |f(x)| for all x. Then g is (a) one-one if f is one-one (b) discontinuous if f is discontinuous (c) continuous if f is continuous (d) onto if f is onto 74. Let f be a function defined on R by log (3 + x ) - x 2 n sin x

then 1 + x2n f is continuous on R f is continuous on R ~ {–1, 1} f is continuous on R ~ {0} none of these

f(x) = lim

nƕ

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

x

Ê x2 + 5x + 3ˆ , then lim f(x) is 75. If f(x) = Á 2 xÆ• Ë x + x + 2 ˜¯ (a) e4 (c) e2

(b) e3 (d) 24

76. Let f(x) be defined for all x > 0 and be continuous. Ê xˆ Let f(x) satisfy f Á ˜ = f(x) – f(y) for all x, y and Ë y¯ f(e) = 1, then (a) f is bounded (b) f (1/x) Æ 0 as x Æ 0 (c) x f (x) Æ 1 as x Æ 0 (d) f (x) = log x 77. Given the function f(x) = 1/(1 – x), the numbers of discontinuities of f3n = f o f º of (3n times) is (a) 1 (b) 2 (c) 3 (d) infinite 1 + sin x - cos x , x π 0. The value of f(0) 78. If f(x) = 1 - sin x - cos x so that f is a continuous function is (a) 1 (b) – 2 (c) – 1 (d) 2 Ï a x +1 , x £ p 2 is continuous then 79. If f (x) = Ì Ósin x + b , x > p 2

(a) a = 1, b = 0 (c) b = a

p 2

(b) a = b

p +1 2

(d) a = b = p/2

Ï x 3 + x 2 - 16 x + 20 , if x π 2 Ô 80. Let f (x) = Ì ( x - 2 )2 Ô , if x = 2 K Ó then the value of K so that f is a continuous function is (a) 5 (b) 7 (c) 6 (d) 4 Ï 2 x + 2 - 16 , if x π 2 Ô 81. If f (x) = Ì 4 x - 16 Ô A , if x = 2 Ó then the value of A so that f is continuous is (a) 1/2 (b) 1 (c) 2 (d) e 82. If the function f(x) =

cos2 x - sin 2 x - 1

x2 + 1 - 1 ous at x = 0 then f(0) is equal to (a) – 2 (b) – 1 (c) 0 (d) – 4 83. If a and b are positive integers then (a) (c)

x Èb˘ a = ˙ Í x Æ 0+ a Î x ˚ b lim

(b)

lim

a Èx˘ b = (d) Í ˙ x Îb˚ a

x Æ 0+

is continu-

x Èb˘ = ab x Æ 0+ a Í Î x ˙˚ lim

lim

x Æ 0+

x Èb˘ b = Í ˙ a Îx˚ a

tan ÈÎe2 ˘˚ x 2 - tan ÈÎ-e2 ˘˚ x 2

x π 0. Then the sin 2 x value of f(0) so that f is a continuous function is (a) 15 (b) 10 (c) 7 (d) 8

84. Let f(x) =

85. The value of f (0) so that the function f (x) = cos a x - cos bx , x π 0 is continuous is given by x2 (a) a – b (c)

b2 - a 2 2

(b) a2 – b2 (d)

a 2 + b2 2

Complete Mathematics—JEE Main

9.30

Previous Years' AIEEE/JEE Main Questions

Ï x if x is rational 1. f is defined on [– 5, 5] as f (x) = Ì Ó- x if x is irrational (a) f (x) is continuous at every x, except x = 0 (b) f (x) is discontinuous at every x, except x = 0 (c) f (x) is continuous everywhere (d) f (x) is discontinuous everywhere [2002] 2.

1 - cos 2 x

lim

xÆ0

2x

(b) – 1 (d) does not exist

Ê x2 + 5x + 3ˆ 3. lim Á 2 x Æ •Ë x + x + 3 ˜ ¯ (a) e 4 (c) e3

[2002]

x

(b) e2 (d) e

[2003]

log (3 + x ) - log (3 - x ) = k , the value of k is 4. If lim xÆ0 x 1 2 (a) – (b) 3 3 2 (c) – (d) 0 [2003] 3

( (

(a) – (c)

is

(a) 1 (c) 0

8. Let a and b be the distinct roots of ax2 + bx + c = 0, 1 - cos ( ax 2 + bx + c ) then lim is equal to x Æa ( x - a )2

a2 (a – b )2 2

(a) 0 (c) •

(b) – 1 (d) 1

[2007]

10. Let f: R Æ R be a positive increasing function with f (2 x ) f (3 x ) lim = 1 . Then lim x Æ• f ( x) x Æ• f ( x ) (a) 3/2 (b) 3 (c) 1 (d) 2/3 [2010] 11. The values of p and q for which function

(b)

[2003]

is continuous for all x in R, are: 1 3 (a) p = , q = 2 2

1 3 (b) p = , q = 2 2

5 1 (c) p = , q = 2 2

3 1 (d) p = - , q = 2 2 [2011]

12. Let f : R Æ [0, •) be such that lim f ( x ) exists x Æ5

p continuous in È0, ˘ then f Êp ˆ is ÍÎ 2 ˙˚ Ë4 ¯ 1 (b) 2 (d) – 1

[2005]

Ï sin ( p + 1) x + sin x ,x0 ÔÓ x3 / 2

a b 2x 6. If lim ÊÁ 1 + + 2 ˆ˜ = e2, then the values of a and x Æ •Ë x x ¯ b are (a) a Œ R, b = 2 (b) a = 1, b Œ R (c) a Œ R, b Œ R (d) a = 1 and b = 2 [2004] 1 – tan x p p , x π , x Œ È0, ˘ . If f (x) is 7. Let f (x) = Í Î 4 ˙˚ 4x – p 4

1 (a) – 2 (c) 1

1 (a – b )2 2

(d) 0

defining f (0) as (a) 2 (c) 0

) )

1 32 1 (d) 8

(b)

9. The function f : R ~ {0} Æ R given by f (x) = 1 2 – x e2 x – 1 can be made continuous at x = 0 by

x (1 - sin x ) 2 is 5. lim p x 3 xÆ 2 x ( p ) tan 1 + 2 2 1 - tan

a2 (a – b )2 2

2

and (a) 0 (c) 2 [2004]

lim

x Æ5

( f ( x )) - 9 | x -5|

= 0. Then lim f ( x ) equals. x Æ5

(b) 1 (d) 3

[2011]

Limits and Continuity 9.31

13. Define F(x) as the product of two real functions f1(x) Ïsin 1 x , x π 0 = x, x Œ R, and f2 (x) = Ì as follows: , x=0 Ó 0 Ï f ( x ). f2 ( x ) , if x π 0 F(x) = Ì 1 0 , if x = 0 Ó Statement-1: F(x) is continuous on R. Statement-2: f1 (x) and f2 (x) are continuous on R. [2011] 1 - cos 2( x - 2) 14. lim x Æ2 x-2 1 (a) equals (b) does not exist 2 (d) equals - 2

(c) equals 2

[2011]

15. If f: R Æ R is function defined by f(x) = [x] cos 2x - 1 p whose [x] denotes the greatest integer func2 tion, then f is: (a) discontinuous only at x = 0. (b) discontinuous only at non-zero integral value of x. (c) continuous only at x = 0. (d) continuous for every real x. [2012] 16. lim

(1 - cos 2 x ) (3 + cos x ) x tan 4 x

x Æ0

(a) 1/2 (c) 2 17. If the function

sin (p cos2 x )

x Æ0

(a) p/2 (c) – p

x

2

tan( x - 2) { x 2 + (k - 2 ) x - 2 x} x2 - 4 x + 4

x Æ2

equal to (a) 0 (c) 2 21. If lim

x Æ•

e

(b) 1 (d) 3 x2

- cos x

sin 2 x

x Æ0

(b)

(a) 4

(b) 3 1 (c) 2 (d) 2 23. Let k be a non-zero real number. If Ï (e x - 1)2 Ô f(x) = ÔÌ sin ÊÁ x ˆ˜ log ÊÁ 1 + x ˆ˜ Ë Ë k¯ 4¯ Ô ÔÓ 12

(

(a) 2

, xπp

is equal to

(b) 1 (d) p [2014] 2 Ê9 ˆ 19. If f(x) is continuous and f Ë ¯ = , then 9 2 Ê 1 - cos 3 x ˆ lim f Á ˜¯ is equal to x Æ0 Ë x2 9 2 (a) (b) 2 9 8 (c) 0 (d) [2014, online] 9

[2015]

,xπ0 ,x=0

is a continuous function, then the value of k is (a) 1 (b) 2 (c) 3 (d) 4 [2015]

(c)

, x=p then k equals (b) 1/4 (d) 0 [2014, online]

[2015]

(1 - cos 2 x )(3 + cos x ) is equal to x tan 4 x

xÆ0+

[2013]

[2014, online]

3 2 (d) 2

(a) 3 5 (c) 4 22. If lim

= 5, then k is

is equal to

24. Let p = lim 1 + tan 2 x (b) 1 (d) – 1/4

Ï 2 + cos x - 1 Ô f(x) = Ì (p - x )2 Ô k Ó is continuous at x = p (a) 2 (c) 1/2 18. lim

is equal to

20. If lim

)

1 2x

, then log p is equal to

(b) 1

1 2

(d)

Ê a 4ˆ 25. If lim Á1 + - 2 ˜ x Æ• Ë x x ¯

1 . 4

2x

= e3. Then a is equal to

(a) 2

(b)

3 2

1 2

(d)

2 3

(c)

[2016]

[2016, online]

26. Let a, b Œ R(a π 0). If the function f defined as Ï 2 x2 , 0 £ x 0 for all x. So f is 1 – 1.

11.18

Complete Mathematics—JEE Main

LEVEL 2 Straight Objective Type Questions Example 81: The point M (x, y) of the graph of the function y = e–|x| so that area bounded by the tangent at M and the coordinate axes is greatest is (a) (1, e–1) (b) (2, e–2) 2 (d) (0, 1) (c) (– 2, e ) Ans. (a) Solution: For x ≥ 0, y = e–x. The equation of tangent is Y – y = – e–x (X – x). This will intersect coordinate axes at (x + yex, 0) and (0, y + xe–x ). Hence the area of the required 1 (y + xe–x) (x + yex) triangle A is 2 1 (1 + x)2 e–x [∵ y = e–x ] = 2 dA 1 = [– (1 + x)2 e–x + 2(1 + x)e–x] Now dx 2 1 (1 + x)e–x (1– x) = 2 dA Note that = 0 fi x = 1, -1 dx dA dA Also, > 0, if 0 £ x < 1 and < 0 if x < 1. Hence A is dx dx maximum when x = 1 so y = e–1. Since y is even function other possibility of M is (–1, e–1). Example 82: The abscissa of the point on the curve 9y2 = x3, the normal at which cuts off equal intercepts on the coordinate axes is (a) 2 (b) 4 (c) – 4 (d) – 2 Ans. (b) dy x 2 Solution: Differentiating 9y2 = x3 we have . = dx 6 y dy Any point on the curve is of the form (t2, t3/3) and so at dx this point is t/2. Thus an equation of normal is y – t3/3 = (– 2/t) (X – t2) This will intersect coordinate axes at (0, 2t + t3/3) and Ê t4 2 ˆ ÁË 6 + t , 0˜¯ . Hence we must have t3 t3 t4 t2 2t + = t2 + fi 2+ =t+ 3 6 3 6 Clearly t = 2 satisfies, the last equation. Hence the abscissa of the required point is 4. (For t = 0, the normal meets both the axes only at origin.)

Example 83: A curve passes through the point (2, 0) and the slope of the tangent at any point (x, y) is x2 – 2x for all value of x. The point of maximum ordinate on the curve is (a) (0, 4/3) (b) (0, 2/3) (c) (1, 2/3) (d) (2, 4/3) Ans. (a) ˆ dy d Ê x3 = x 2 - 2 x = Á - x 2 + c˜ Solution: dx dx Ë 3 ¯ x3 - x2 + c 3 Since the curve passes through (2, 0), we get 0 = 8/3 – 4 + c, i.e., c = 4/3 Hence the equation of the curve is y=

4 x3 - x2 + 3 3 Now, from dy/dx = 0, we get x = 0 or 2. Also y=

d2 y dx 2

= 2x - 2 fi

d2 y dx 2

= -2 and x =0

d2 y dx 2

=2 x =2

Hence at x = 0, y has a maximum. Thus the required point is (0, 4/3). Example 84: If the tangent to the curve 2y3 = ax2 + x3 at the point (a, a) cuts off intercepts a and b on the coordinate axes, where a 2 + b 2 = 61 then the value of |a| is (a) 16 (b) 28 (c) 30 (d) 31 Ans. (c) Solution: The slope of the tangent is 2ax + 3 x 2 dy = 6 y2 dx and the value of this slope at (a, a) is 5/6. Therefore, the equation 5 x y + = 1, y – a = ( x - a) fi -a / 5 a / 6 6 represents the tangent. Thus the x-intercept a is – a /5, and the y-intercept b is a/6. From a 2 + b 2 = 61, we now get a2 a2 = 61 + 25 36

fi

a2 = 25 × 36

fi |a| = 30.

Example 85: The coordinates of the point on the parabola y2 = 8x, which is at minimum distance from the circle x2 + (y + 6)2 = 1 are

Applications of Derivatives 11.19

(a) (2, – 4) (c) (2, 4) Ans. (a)

(b) (18, – 12) (d) none of these

Solution: A point on the parabola is at a minimum distance from the circle if and only if it is at a minimum distance from the centre of the circle. Any point on the parabola y2 = 8x is of the form P(2t2, 4t). The centre of the circle x2 + (y + 6)2 = 1 is O (0, - 6) OP2 = 4t4 + (- 6 - 4t)2 = 4 (t4 + 4t2 + 12t + 9) Let A = t4 + 4t2 + 12t + 9 dA = 4t3 + 8t + 12 = 4 (t3 + 2t + 3) dt = 4(t + 1) (t2 - t + 3) dA = 0 if t = - 1. Moreover, So dt

(b) (1/2) (a + b) (d) none of these

Solution: Let ABCD be the given rectangle of sides a and b and EFGH be any rectangle, whose sides pass through A, B, C, D. A = Area EFGH = (b sin q + a cos q) (a sin q + b cos q) = ab + (a2 + b2) sin q cos q. G

C

D

H

d2 A dt 2

(a) (1/2) (ab)2 (c) (1/2) (a + b)2 Ans. (c)

= 4 (3 (- 1)2 + 2) > 0.

a

a

b A

F

B

q

t = -1

Hence required point is P(2, - 4). E

Example 86: The equation ex – 8 + 2x – 17 = 0 has (a) two real roots (b) one real root (c) eight real roots (d) four real roots Ans. (b) Solution: Clearly x = 8 satisfies the given equation. Assume that f (x) = ex – 8 + 2x - 17 = 0 has a real root a other than x = 8. We may suppose that a > 8 (the case for a < 8 is exactly similar). Applying Rolle’s theorem on [8, a], we get b Œ (8, a), such that f ¢(b) = 0. But f ¢(b) = e b – 8 + 2, so that e b – 8 = -2 which is not possible. Hence there is no real root other than 8. Example 87: The maximum and minimum value of f(x) = ab sin x + b 1 - a 2 cos x + c lie in the interval ( assuming | a | < 1, b > 0 ) (a) [b – c, b + c] (b) (b – c, b + c) (c) [c – b, b + c] (d) none of these Ans. (c)

Fig. 11.4

dA/dq = (a2 + b2) cos 2q so dA/dq = 0 fi

d2 A dq 2

= - 2 (a2 + b2) sin 2q, so

fi

q = p/4

d2 A dq 2

0. Hence f has maximum at p/2 - f and minimum at 3p/2 - f. Moreover, max f = ab sin (p/2 - f)

The critical points of f (x) are given by f ¢(x) = 12x2 - 12 = 12 (x2 - 1) = 0. That is, x = ± 1, so that f (1) = - 8, f (- 1) = 8 and f (3) = 72.

+ b 1 - a 2 cos (p/2 - f) + c = a2b + b(1 - a2) + c = b + c and min f = c - b.

Hence the image of [- 1, 3] under the mapping f (x) is [- 8, 72].

Alternatively f (x) = b sin (x + f) + c, f = cos–1 a so max f (x) = b.1 + c and min f(x) = – b + c = c – b Example 88: The maximum area of the rectangle whose sides pass through the angular points of a given the rectangle is of sides a and b is

\

max f (x) = f (3) = 72 and min

x Œ[ - 1,3]

x Œ[ - 1, 3]

f (x) = f (1) = - 8

Example 90: The difference between the greatest and least values of the function f (x) = cos x + (1/2) cos 2x – (1/3) cos 3x is (a) 3/8 (b) 2/3 (c) 8/7 (d) 9/4 Ans. (d)

11.20

Complete Mathematics—JEE Main

Solution: The given function is periodic, with period 2p. So the difference between the greatest and least values of the function is the difference between these values on the interval [0, 2p]. We have f ¢(x) = - (sin x + sin 2x - sin 3x) = - 4 sin x sin (3x/2) sin (x/2). Hence x = 0, 2p/3, p and 2p are the critical points. Also , f (0) = 1 + 1/2 - 1/3 = 7/6, f (2p/3) = - 13/12, f (p) = - 1/6 and f (2p) = 7/6. Hence the greatest value is 7/6 and the least value is - 13/12. Thus the difference is 7/6 – (–13/12) = 27/12 = 9/4. Example 91: The longest distance of the point (a, 0) from the curve 2x2 + y2 – 2x = 0 is (a)

1 - 2a + 2a2

(b)

1 - 2a + a2

(c) Ans. (a)

1 + 2a + 2a2

(d)

1 + a + a2

Solution Let (x, y) be any point on the curve 2x2 + y2 2x = 0. Its distance S from the point (a, 0) is given by T = S2 = (x - a)2 + y2 = (x - a)2 + (2x - 2x2) (1) = - x2 + 2x (1 - a) + a2 For S to be maximum, we have dT/dx = 0 fi - 2x + 2 (1 – a) = 0 fi x = 1 - a. Since d2T/dx2 < 0, so S2 is maximum when x = 1 - a and this maximum value is given by - (1 - a)2 + 2(1 - a)2 + a2 = 1 - 2a + 2a2. Hence Smax = (1 - 2a + 2a2)1/2. Example 92: The sides of the rectangle of the greatest area, that can be inscribed in the ellipse x2 + 2y2 = 8, are given by (a) 4 2 , 4 (c) 2, Ans. (b)

(b) 4, 2 2

2

(d) 2 2 , 2

Solution: Any point on the ellipse ( 2 2 cos q, 2 sin q). [see Fig. 11.5]

x 2 y2 = 1 is + 8 4

2

Also

d A dq 2

dA = 16 2 cos 2q = 0 dq

fi

q = p/4

= - 32 2 sin 2q < 0 for q = p/4.

Hence, the inscribed rectangle is of largest area if the sides are 4 2 cos p/4 and 4 sin (p/4) i.e. 4 and 2 2 . Example 93: An equation of the circle that is tangent to y = x3 at (1, 1) and has the same second derivative there is (a) x2 + y2 + 24x – 28y + 2 = 0 (b) 2(x2 + y2) + 12x – 8y – 8 = 0 (c) 3(x2 + y2) – 24x + 10y + 8 = 0 (d) none of these Ans. (d) dy d2 y dy = 3 and 2 =6 Solution: = 3x2 so dx (1,1) dx (1, 1) dx Let the required circle be x2 + y2 + 2gx + 2fy + c = 0. Since this should pass through (1, 1) so we have 2 + 2g + 2f + c = 0 (1) Also

2x + 2y

dy dy + 2g + 2f =0 dx dx

Putting x = 1, y = 1 and

dy = 3, we have dx

1 + 3 + g + 3f = 0 Again differentiating (1), we have 2 d2 y d2 y Ê dy ˆ 1+ Á ˜ +y 2 + f 2 =0 Ë dx ¯ dx dx

d2 y dy = 3 and 2 = 6, we have dx dx 16 + 6f = 0 fi f = – 8/3 so g = 4 and c = – 14/3. Thus required circle is Putting y = 1,

3 (x2 + y2) + 24x – 16y – 14 = 0.

y (2 2 cos q, 2 sin q ) x

0

= 8 2 sin 2q

Fig. 11.5

A = Area of the inscribed rectangle = 4 ( 2 2 cos q) (2 sin q)

Example 94: Let f (x) = 6x4/3 – 3x1/3, x Œ [– 1, 1]. Then (a) The maximum value of f (x) on [– 1, 1] is 3 (b) The maximum value of f (x) on [– 1, 1] is 9 (c) The minimum value of f (x) on [– 1, 1] is 0 (d) none of these Ans. (b) 8x - 1 Solution: f ¢ (x) = 2 / 3 . Thus f ¢ (x) = 0 when x = 1/8 x and f ¢ (x) does not exist when x = 0. Now f (– 1) = 9, f (0) = 0, f (1/8) = – 9/8 and f (1) = 3. The maximum value of f (x) is 9 and the minimum value is – 9/8 on [–1, 1].

Applications of Derivatives 11.21

Example 95: Let g (x) = (log (1 + x)) – 1 – x – 1, x > 0 then (a) 1 < g (x) < 2 (b) – 1 < g (x) < 0 (c) 0 < g (x) < 1 (d) none of these Ans. (c)

i.e. Y = (-1/2) x + 3/2. The intersection of this line with the given curve is given by (- x/2 + 3/2)2 = x (2 - x)2

Solution: For x > 0, consider the function f (x) = log (1 + x) on the interval [0, x]. Since the function is differentiable on (0, x), by Lagrange’s theorem, there is a x Œ (0, x) such that

fi

log (1 + x ) - log (1 + 0 ) 1 = f ¢(x) = x-0 1+x

fi

log (1 + x ) = 1 < 1 1+x x log (1 + x) < x fi g(x) > 0

Also

x Π(0, x)

fi

fi fi

fi

x 1+x 1+ x

For curves to intersect at right angles, we must have at the

Example 96: The maximum value of 2 for all x + x +1 real values of x is (a) 1/2 (b) 1 (c) 2 (d) 3 Ans. (d) x2 - x + 1 2x Solution: Let S = 2 =1- 2 x + x +1 x + x +1

( x2 + x + 1) - x (2 x + 1) 2 ( x2 + x + 1) - x2 + 1 2

( x 2 + x + 1)

S¢(x) = 0 ¤ x = ± 1. Since S¢(x) > 0 for x < - 1 and S¢(x) < 0 for - 1< x 0 for x > 1. So S is maximum when x = - 1. Hence Smax = 1 + 2/1 = 3. 2

Example 97: If the tangent at (1, 1) on y = x (2 – x) meets the curve again at P, then P is (a) (4, 4) (b) (– 1, 2) (c) (9/4, 3/8) (d) none of these Ans. (c) dy Solution: 2y = (2 - x)2 - 2x (2 - x) dx dy = 3x2 - 8x + 4. So = - 1/2. d x (1, 1)

Thus x = 1, 9/4, - 1. But x = -1 cannot lie on the given curve so required point is (9/4, 3/8).

Solution: The intersection of the two curves is given by 9x2 + 6bx – x = x (i) 3 d y = . Differentiating y2 = 6x, we have dx y dy 9x =- . Differentiating 9x2 + 6y2 = 16, we have dx by

x2 - x + 1

=-2

x2 - 6x + 9 = 16x + 4x3 - 16x2. So, 4x3 - 17x2 + 22x - 9 = 0 (x - 1) (4x - 9) (x + 1) = 0.

fi

Example 98: If the curves y2 = 6x, 9x2 + by2 = 16, cut each other at right angles then the value of b is (a) 2 (b) 4 (c) 9/2 (d) none of these Ans. (c)

log (1 + x ) > 1 1+ x x x log (1 + x) > fi g(x) < 1. 1+ x

S ¢(x) = - 2

An equation of tangent at (1, 1) is Y - 1 = (-1/2) (X - 1)

2

points of intersection

3 Ê -9 x ˆ = –1 y ÁË by ˜¯

fi

27x = by2. Thus

we must have 9x2 + 9y2 = 16 fi 9x2 + 27x – 16 = 0 (i) and (ii) must be identical so 27 = 6b fi b = 9/2.

(ii)

Example 99: The distance of that point on y = x4 + 3x2 + 2x which is nearest to the line y = 2x – 1 is (a) 4/ 5

(b) 3/ 5

(c) 2/ 5 Ans. (d)

(d) 1/ 5

Solution: Distance of any point (x, y) from y = 2x - 1 is: y - 2x +1 . If (x, y) is on y = x4 + 3x2 + 2x then this distance 5 is S =

x4 + 3 x2 + 1 5 dS 4 x3 + 6 x dS = fi = 0 fi x = 0. dx dx 5

Also, S ¢(x) < 0 for x < 0 and S¢(x) > 0 for x > 0. Thus S is minimum when x = 0, and min. S is 1/ 5 . Example 100: A given right circular cone has a volume p, and the largest right circular cylinder that can be inscribed in the cone has a volume q. Then p : q is (a) 9 : 4 (b) 8 : 3 (c) 7 : 2 (d) none of these Ans. (a)

11.22

Complete Mathematics—JEE Main

Solution: Let H be the height of the cone and a be its semi vertical angle. Suppose that x is the radius of the inscribed cylinder and h be its height h = QL = OL - OQ = H - x cot a V = volume of the cylinder 1 p (H tan a)2 H 3

dV = p (2Hx - 3x2 cot a) dx dV = 0 ¤ x = 0, so dx 2 d 2V x= H tan a, 3 d x2

a x Q

L

= px2 (H - x cot a) Also p =

O

M

P

Fig. 11.6

(i)

Example 103: If f (x) =

a sin x + b cos x decreases for c sin x + d cos x

all x if (a) ad – bc < 0 (c) ab – cd > 0

(b) ad – bc > 0 (d) ab – cd < 0

Ans. (a) = - 2pH < 0. so 2 x = H tan a 3

2 4 2 H tan a and q = Vmax = p H V is maximum when x = 3 9 1 4 tan2 a H = p. [using (i)] 3 9 Hence p : q = 9 : 4. Example 101: The set of all values of a for which f (x) = (a2 – 3a + 2) (cos2 (x/4) – sin2 (x/4)) + (a – 1)x + sin 1 doesn’t possess critical points is (a) [1, •) (b) (– 2, 4) (c) (1, 3) » (3, 5) (d) (0, 1) » (1, 4) Ans. (d) Solution: The given function can be written as f (x) = (a - 1) (a - 2) cos (x/2) + (a - 1) x + sin 1. It is clearly differentiable, so its critical points are given by f ¢(x) = (–1/2) (a - 1) (a - 2) sin (x/2) + (a - 1) = 0 If a = 1, f ¢(x) = 0 for all x, while for values of a other than 1, f ¢(x) will be zero if (a - 2) sin (x/2) = 2 In order not to have critical point, a must therefore satisfy 2 >1 which is the same as saying a Œ (0, a π 1 and a-2 1) » (1, 4). Example 102: Let x, p Œ R, x + 1 > 0, p π 0, 1. Then (a) (1 + x) p > 1 + px for p > 0 (b) (1 + x) p > 1 + px for p Œ (– •, 0) » (1, •) (c) (1 + x) p > 1 + px for 0 < p < 1 (d) (1 + x)p < 1 + px for p < 1 Ans. (b)

Solution: Let f(x) = (1 + x)p– (1+ px) so f(0) = 0. Now f ¢(x) = p (1 + x)p – 1 – p and f ≤ (x) = p (p – 1) (1 + x)p – 2. Let p Œ (– • , 0) » (1, •) then f ≤(x) > 0 so f ¢ is increasing. This implies f ¢ < 0 on (– 1, 0) and f ¢ > 0 on (0, •) as f ¢ (0) = 0. Therefore f decreases on (– 1, 0) and increases on (0, •) but f (0) = 0 so f (x) > 0 if x > – 1 (x π 0). Thus (1 + x )p > 1 + px for p Œ (– •, 0) » (1, •). Similarly it follows that (1 + x) p < 1 + px, if 0 < p < 1.

Solution: f ¢ (x) =

= =

(a cos x - b sin x ) (c sin x + d cos x ) (a sin x + b cos x ) (c cos x - d sin x ) (c sin x + d cos x )2

ad (cos2 x + sin 2 x ) - bc (cos2 x + sin 2 x ) (c sin x + d cos x )2 ad - bc (c sin x + d cos x )2

f decreases for all x if and only if f ¢(x) < 0 for all x i.e. ad – bc < 0. Example 104: On the interval [0, 1] the function 25 x (1 – x)75 takes its maximum value at the point (a) 0 (b) 1/3 (c) 1/2 (d) 1/4 Ans. (d) Solution: Let f (x) = x25 (1 – x)75. The critical points of f are given by f ¢(x) = 0. But f ¢ (x) = x24 (1 – x)74 [25 – 25x – 75x] = 25x24 (1 – x)74 (1 – 4x) Thus the critical points are 0, 1, 1/4 Since f (0) = 0, f (1) = 0 and f (1/4) = (1/4)25 (3/4)75 so f takes its maximum value at x = 1/4. Example 105: For a >0, the value of a for which the equation ax2 = log x possess a single root is (a) 1/2 (b) 1/2e (c) 1/e (d) 2e–1 Ans. (b) Solution: For a > 0, the curves y1 = ax2 and y2 = log x can have only one point in common if they touch each other. At the point of tangency y¢1(x) = y¢2 (x) fi 2ax = 1/x fi x = 1/ 2a (clearly x cannot be negative). Putting this value in ax2 = log x, we have (1/2) = log (2a)–1/2 fi log 2a = – 1 fi a = 1/2e.

Applications of Derivatives 11.23

EXERCISE Concept-based Straight Objective Type Questions 1. The length of the tangent to the curve x = a sin3 t, y = a cos3 t (a > 0) at an arbitrary is (a) a cos2t (c)

a sin 2 t cos t

(b) a sin2 t (d)

a cos2 t sin t

2. Equation of normal to x = 2et, y = e–t at t = 0 is (a) x + y – 4 = 0 (b) x + 2y – 4 = 0 (c) 2x – y – 3 = 0 (d) x – 2y – 3 = 0 2 p 3. A point moves according s = sin t + s0 . The ac9 2 celeration at the end of first second is (a) -

p 18

(b) -

p2 18

p p2 (d) 18 18 4. Let f(x) = x log x – x + 1 then the set {x : f (x) > 0} is equal to (a) (1, •) (b) (1/e, •) (c) [e, •) (d) (0, 1) » (1, •) (c)

5. On the curve y = x3, the point at which the tangent line in parallel to the chord through the point (–1, –1) and (2, 8) is 1 1 (a) (1, 1) (b) ÊÁ , ˆ˜ Ë 2 8¯

Ê1 1 ˆ (c) Á , ˜ Ë 3 27 ¯

1 1 (d) ÊÁ - , - ˆ˜ Ë 2 8¯ 6. Let f (x) = 2x2 – log x, then (a) f increases on (0, •) Ê1 ˆ (b) f decrease on Á ,•˜ Ë2 ¯ Ê1 ˆ (c) f increases on Á ,•˜ Ë2 ¯ (d) f decreases on (0, 1) 3 4 x - x 3 - 9 x 2 + 7 , then the number of 4 critical points in [– 1, 4] is (a) 4 (b) 3 (c) 2 (d) 1

7. Let f(x) =

8. On the curve x3 = 12y, the values of x for which the abscissa changes at a faster rate than the ordinate is (a) (–2, 2) ~ {0} (b) (–3, 3) ~ {0} (c) (1, 4) (d) (2, 4) x 2 y2 =1 9. The value of k > 0 for which the curves 2 + 4 k and y2 = 16x cut each other orthogonally is 2 3 3

(a) 1

(b)

(c) 3 3

(d) 5 5

10. The least value of g(t) = 8t – t4 on [–2, 1] is (a) –16 (b) –20 (c) –32 (d) 7

LEVEL 1 Straight Objective Type Questions 11. Let (a) (b) (c) (d)

x f (x) = tan–1 x and g (x) = , x > 0 then 1 + x2 f (x) < g (x), on (0, •) f (x) £ g (x) on [1, •) g (x) < f (x) on (0, •) none of these

12. Let f (x) = (x – 2) (x – 3) (x – 4) (x – 5) (x – 6) then (a) f ¢(x) = 0 has five real roots (b) four roots of f ¢(x) = 0 lie in (2, 3) » (3, 4) » (4, 5) » (5, 6)

(c) the equation f ¢ (x) has only three roots (d) four roots of f ¢(x) = 0 lie in (1, 2) » (2, 3) » (3, 4) » (4, 5) 13. Let (a) (b) (c) (d)

f (x) = (x – 3)5 (x + 1)4 then x = – 1 is point of minima x = –1 is point of maxima x = 7/9 is a point of maxima x = – 1 is neither a point of maxima and minima.

11.24

Complete Mathematics—JEE Main

14. The normal to the curve represented parametrically by x = a (cos q + q sin q) and y = a (sin q – q cos q) at any point q, is such that it (a) makes a constant angle with the x-axis (b) is at a constant distance from the origin (c) does not touch a fixed circle (d) passes through the origin. 15. The value of a for which the equation x3 – 3x + a = 0 has two distinct roots in [0, 1], is given by (a) – 1 (b) 1 (c) 3 (d) none of these 16. If the sum of the squares of the intercepts on the axes cut off by the tangent to the curve x1/3 + y1/3 = a1/3 (a > 0) at (a/8, a/8) is 2, then a has the value (a) 1 (b) 2 (c) 4 (d) 8 17. The value of m for which the area of the triangle included between the axes and any tangent to the curve xm y = bm is constant, is (a) 1/2 (b) 1 (c) 3/2 (d) 2 18. If the tangent at any point on the curve x4 + y4 = a4 cuts off intercepts p and q on the coordinate axes, the value of p–4/3 + q–4/3 is (b) a–1/2 (a) a–4/3 1/2 (c) a (d) none of these 19. The interval of increase of the function y = x – 2 sin x if 0 £ x £ 2p, is (a) (0, p) (b) (0, p) (c) (p/2, p) (d) (p/3, 5p/3) 20. The greatest value of y = (x + 1)1/3 – (x – 1)1/3 on [0, 1] is (a) 1 (b) 2 (c) 3 (d) 21/3 21. Let xπ (a) (b) (c) (d)

f be a function defined by f (x) = 2x2 – log |x|, 0 then f increase on [– 1/2, 0] » [1/2, •) f decreases on (– •, 0) f increases on (– •, – 1/2) f decreases on [1/2, • ]

22. The shortest distance of (0, 0) from the curve y = e x + e- x is 2 (a) 1/2 (b) 1/3 (c) 2 (d) none of these 23. The normal to the circle x2 + y2 – 2x – 2y = 0 passing through (2, 2) is (a) x = y (c) x + 2y – 6 = 0

(b) 2x + y – 6 = 0 (d) x + y – 4 = 0

24. If f (0) = 0 and f ¢¢(x) exists and > 0, for all x > 0 then f (x)/x (a) decreases on (0, •) (b) increases on (0, •) (c) decreases on (1, •) (d) neither increases nor decreases on (0, •) 25. The value of k so that the equation x3 – 12x + k = 0 has distinct roots in [0, 2] is (a) 4 (b) 2 (c) –2 (d) none of these 26. Let (a) (b) (c) (d)

f (x) = 6x4/3 – 3x1/3 defined on [– 1, 1] then maximum value of f is 7 maximum value of f is 5 maximum value of f is 9 minimum value of f is – 3/2

27. An equation of tangent line at an inflection point of f (x) = x4 – 6x3 + 12x2 – 8x + 3 is (a) y = 3x + 4 (b) y = 4 (c) y = 3x + 2 (d) none of these 28. The number of real roots of the equation 2x3 – 3x2 + 6x + 6 = 0 is (a) 1 (b) 2 (c) 3 (d) none of these 29. Let f (x) = (x – 2) (x4 – 4x3 + 6x2 – 4x + 1) then value of local minimum of f is (a) – 2/3 (b) – (4/5)4 4 5 (c) – 4 /5 (d) – (4/5)5 30. Let f (x) = x2 – 2|x| + 2, x Œ [– 1/2, 3/2] then (a) min f (x) = 1/2 (b) min f (x) = 1 x Œ [– 1/2, 3/2] x Œ [– 1/2, 3/2] (c) max f (x) = 3/2 (d) none of these x Œ [– 1/2, 3/2] x -1 is 31. A critical point of the function f (x) = x2 (a) – 1 (b) 3 (c) 2 (d) 1/2 32. The function f (x) = xx decreases on the interval (a) (0, e) (b) (0, 1) (c) (0, 1/e) (d) none of these 33. The interval of increase of the function y = x – ex + tan (p/7) is (a) (– •, 1) (b) (0, •) (c) (– •, 0) (d) (1, •) 34. Let f (x) = x2 + px + q. The value of (p, q) so that f (1) = 3 is an extreme value of f on [0, 2] is (a) (– 2, 2) (b) (1, 4) (c) (– 2, 4) (d) (– 2, 3) 35. The number of inflection points of a function given by a third degree polynomial is exactly

Applications of Derivatives 11.25

(a) 2 (c) 3

(c) f (6) ≥ 10 (d) f (6) £ 5 47. An extremum value of the function

(b) 1 (d) 0

36. Let f (x) = 2 tan–1 x + sin–1

2x 2

then

1+ x (b) min f (x) = p/4 (d) none of these

(a) max f (x) = p/2 (c) max f (x) = p

37. If the normal to the curve x3 = y2 at the point (m2, – m3) is y = mx –2m3, then the value of m2 is (a) 1 (b) 1/2 (c) 1/3 (d) 2/3 38. Let x+ (a) (b) (c)

f (x) = 2 sin x + cos 2x (0 £ x £ 2p) and g (x) = cos x then g is a decreasing function f increases on (0, p/2) f increases on (0, p/6) » (p/2, 5p/6)

(d) f decreases on (0, p/2) 39. The minimum value of the function f (x) = tan x + cot x in the interval (0, p/2) is (a) 1 (b) 0 (c) 2 (d) 1/2 40. The number of points of extremum of the function f (x) = 3x4 – 4x3 + 6x2 + b for any value of b is (a) 4 (b) 3 (c) 1 (d) 2 41. The shortest distance of the line y – x – 1 = 0 from x = y2 is (a) 3/8 (b) 3 2 /4 (c) 3/4

(d) 3 2 /8

42. The value of a for which the extremum of the function f (x) = x3 – 3ax2 + 3(a2 – 1)x + 1 lie in the interval (– 2, 4) lie in (a) (–1, 0) (b) (–2, 4) (c) (– 1, 5) (d) (– 1, 3) 43. If A > 0, B > 0 and A + B = p/3 then the maximum value of tan A tan B is (a) 1/3 (b) 1/2 (d) 3 /2 (c) 1/ 2 44. The maximum value of |x log x| for 0 < x £ 1 is (a) 0 (b) 1/e (d) none of these (c) 2e–1 45. The greatest value of the function logx 1/9 – log3 x2 (x > 1) is (a) 2 (b) 0 (c) – 4 (d) – 2 46. Let f be differentiable for all x. If f (1) = –2 and f ¢(x) ≥ 2 for x Œ [1, 6] then (a) f (6) < 8

(b) f (6) ≥ 8

f (x) = (sin–1 x)3 + (cos–1 x)3 (–1 < x < 1) is (a) 7p3/8

(b) p3/8

(c) p3/32

(d) p3/16

48. Let (a) (b) (c) (d)

f (x) = x log x + 3x. Then f increases on (e–4, •) f increases on (0, •) f decreases on (0, •) f decreases on (0, e–2)

49. Let f (x) = x2 e–x then (a) max f (x) = e–1 (b) max f (x) = 4e–2 –1 (c) min f (x) = e (d) min f (x) > 0 50. The minimum value of f (x) = |3 – x| + |2 + x| + |5 – x| is (a) 0 (b) 7 (c) 8 (d) 10 51. Let f (x) = 2 + 2x – 3x2/3 on [– 1, 10/3]. Then f has (a) Absolute maximum at an end point (b) Absolute minimum at an interior point (c) Absolute minimum is f (10/3) (d) Absolute minimum is f(– 1) 52. If f and g are defined on [0, •) by f (x) = lim

nƕ

xn - 1 n

x +1

and g(x) =

x

Ú0

f (t ) dt .

Then (a) g has local maximum at x = 1 (b) g has local minimum at x = 1 (c) g is an increasing function on (0, •) (d) g is a decreasing function on (0, •). 53. Let (a) (b) (c) (d)

f(x) = sin x + cos x then x = 17p/4 is a point of minima x = 13p/4 is a point of maxima x = 21p/4 is a point of minima x = 29p/4 is a point of maxima

54. If f(x) = x ex(1 – x), then f(x) is (a) increasing on [–1/2, 1] (b) decreasing on R (c) increasing on R (d) decreasing on [–1/2, 1] 55. The tangent to the curve y = ex drawn at the point (c, ec) intersects the line joining the points (c – 1, ec – 1) and (c + 1, ec + 1) (a) on the left of x = c (b) on the right of x = c (c) at no point (d) at all points

11.26

Complete Mathematics—JEE Main

Assertion-Reason Type Questions 56. Statement-1: ep > p e Statement-2: The function x1/x (x > 0) has local maximum at x = e. 57. Statement-1: |cot x – cot y | £ |x – y | for all x, y Œ (– p/2, p/2) Statement-2: If f is differentiable on an open interval and | f ¢(x) | £ M then | f (x) – f (y) | £ M | x – y |. 58. Statement-1: The function f (x) = 2 sin x + cos 2x (0 £ x £ 2p) has minimum at x = p /3 and maximum at 5p /3. Statement-2: The function f (x) above decreases on (0, p /3), increases on (p /3, 5p /3) and decreases on (5p /2, 2p)

59. Let f (x) = tan - 1

1- x 1+ x

Statement-1: The difference between the greatest and smallest value of f (x) on [0, 1] is p/4 Statement-2: If a function g decreases on [a, b] then the greatest value of g = g(a) and least value of g is g(b). x 60. Let f (x) = log x Statement-1: The minimum value of f (x) is e Statement-2: log x > 1 for x > e and < 1 for x < e.

LEVEL 2 Straight Objective Type Questions 61. The critical points of the function f (x) = (x – 2)2/3 (2x + 1) are (a) – 1 and 2 (b) 1 (c) 1 and – 1/2 (d) 1 and 1/2 62. The function f (x) = (1/4)x3 – sin px + 3 on [– 2, 2] takes the value (a) 1 (b) 16/3 (c) 6 (d) 8 1 log x on 63. The greatest value of f(x) = tan–1 x – 2 [1 / 3, 3 ] is (a) p/2 + (1/2) log 3 (b) p/6 + (1/4) log 3 (c) p/6 + (1/2) log 3 (d) p/4 – (1/4) log 3 64. Equations of those tangents to 4x2–9y2=36 which are perpendicular to the straight line 2y + 5x = 10; are

(

(a) 5(y – 3) = x - 117 / 4

)

(b) 5(y Р2) = 2(x Р18 ) (c) 5(y + 2) = 2(x Р18 ) (d) none of these 65. If a,b,c, ΠR, then f(x) =

x + a2

ab

ac

ab

x + b2

bc

ac

bc

(a) (– (2/3) (a2 + b2 + c2), 0) (b) (0, (2/3) (a2 + b2 + c2))

x+c

decrease on 2

(c) ((1/3) (a2 + b2 + c2), 0) (d) none of these 66. A channel 27 m wide falls at a right angle into another channel 64 m wide. The greatest length of the log that can be floated along this system of channels is (a) 120 (b) 125 (c) 100 (d) 110 67. For a Œ [p, 2p], the function 1 a-2 sin a tan3 x + (sin a – 1) tan x + f (x) = 3 8-a has (a) x = np (n Œ I ) as critical points (b) no critical points (c) x = 2n p (n Œ I ) as critical points (d) x = (2n + 1) p (n Œ I ) as critical points. 68. The value of a for which the function f (x) = (4a – 3) (x + log 5) + 2(a – 7) cot (x/2) sin2 (x/2) does not possess critical point is (a) (– •, – 4/3] (b) (– •, – 1) (c) [1, •) (d) (0, • ) 69. The interval to which b may belong so that the function Ê 21 - 4b - b2 ˆ 3 f (x) = Á 1 ˜ x + 5x + 6 b +1 Ë ¯ is increasing at every point of its domain is (a) [– 7, 0] (b) [– 6, 0] (c) [1, 4] (d) [2, 3]

Applications of Derivatives 11.27

70. A tangent is drawn at a variable point on an ellipse x2/a2 + y2/b2 = 1, then minimum area of the triangle formed by the tangent and the coordinate axes is (a) a b (b) (a2 + b2)/2 2 (d) 2 ab (c) (a + b) /4 71. The set of all x for which log (1 + x) £ x is (a) (1, •) (b) (0, •) (c) (– 1, •) (d) none of these 2

3

72. The minimum value of 2( x -3) + 27 is (b) 2 (a) 227 (c) 1 (d) none of these Ï x for 0 < | x | £ 2 73. Let f (x) = Ì Ó1 for x = 0 Then at x = 0, f has (a) a local maximum (b) no local maximum (c) a local minimum (d) no extremum 74. If f(x) = xe x(1 – x), then f(x) is (a) increasing on [–1/2, 1] (b) decreasing on R (c) increasing on R (d) decreasing on [–1/2, 1] Ï xa log x if x > 0 75. Let f(x) = Ì if x = 0 Ó0 Rolle's theorem can be applied to f on [0, 1] then value of a can be (a) –1 (b) –1/2 (c) 0 (d) 1/2 76. A cone is made from a circular sheet of radius 3 by cutting out a sector and gluing the cut edges of the remaining piece together. The maximum volume attainable for the cone is (a) p/3 (b) p/6 (c) 2p/3 (d) 3 3p 77. The dimensions of the rectangle of maximum area that can be inscribed in the ellipse (x/4)2 + (y/3)2 = 1 are (a)

8, 2

(c) 2 8 , 3 2

(b) 4, 3 (d) none of these

80. The point in the interval [0, p] for which the curve y = (1/2)x and y = sin x are farthest apart is (a) p/2 (b) p/4 (c) p/6 (d) p 81. The points at which the tangents to the curve ax2 + 2hxy + by2 = 1 is parallel to y-axis is (a) (0, 0) (b) where hx + by = 0 meets it (c) where ax + hy meets it (d) none of these ax 2 (0 < a < 82. If the point on y = x tan a – 2u2 cos2 a p/2) where the tangent is parallel to y = x has an ordinate u2/4a then the value of a is (a) p/2 (b) p/6 (c) p/3 (d) none of these Ï x - 1 + a, x £ 1 83. Let f (x) = Ì x >1 Ó 2 x + 3, If f (x) has local minimum at x = 1 and a ≥ 5 then the value of a is (a) 5 (b) 6 (c) 11/2 (d) 15/2 84. Let g (x) =

x

Ú0

f (t ) dt and f (x) satisfies the equation

f (x+y) = f (x) + f (y) +2xy – 1 for all x, y Œ R and f ¢(0) = 2 then (a) g increases on (0, •) and decreases on (– •, – 0) (b) g increases on (0, •) (c) g decreases on (0,• ) and increases (– •, 0) (d) g decreases on (– •, • ) 85. The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle x2 + y2 = 4 at (1, 3 ) is (a) 2 3

(b)

(c) 4 3

(d) 3

3

x -1 86. The interval into which the function y = 2 x - 2x + 3 transforms the entire real line is (a) [1/3, 2] (b) [– 1/3, 2] (c) [–1/3, 1] (d) none of these

78. The condition for y = ax + bx3 + cx2 + dx + e to have points of inflection is (b) 3b2 – 8ac = 0 (a) b2 – 4ac > 0 2 (d) 3b2 – 8ac < 0 (c) 3b – 8ac > 0

87. The angle at which x2 + y2 = 16 can be seen from the point (8, 0) is (a) p/6 (b) p/4 (c) p/2 (d) p/3

79. The largest value of m such that |x2 – 3x + 2| ≥ m for all x in the interval [3/2, 7/4] is (a) 3/4 (b) 3/8 (c) 3/16 (d) 7/4

88. The critical points of the function f (x) = (x + 2)2/3 (2x – 1) are (a) – 1 and 2 (b) 1 (c) 1 and – 1/2 (d) –1 and –2

4

11.28

Complete Mathematics—JEE Main

log (p + x ) is log (e + x ) increasing on [0, •) decreasing on [0, •) increasing on [0, p/e) and decreasing on [p/e, •) decreasing on [0, p/e) and increasing on [p/e, •)

89. The function f (x) = (a) (b) (c) (d)

90. A rectangle with perimeter 32 cm has greatest area if its length is (a) 12 (b) 10 (c) 8 (d) 14 91. The greatest value of the function f (x) = cot–1 x + (1/2) log x on [1, 3 ] is (a) (p/6) + 0.25 log 3

(b) (p/3) – 0.25 log 3

(c) p/4

(d) tan–1 e – 1/2

92. A particle is moving along the parabola y2 = 4(x + 2). As it passes through the point (7, 6) its y-coordinate is increasing at the rate of 3 units per second. The rate at which x-coordinate change at this instant is (in units/sec) (a) 4 (b) 6 (c) 8 (d) 9 93. The perimeter of a rectangle is fixed at 24 cm. If the length l of the rectangle is increasing at the rate of 1 cm per second, the value of l for which the area of rectangle start to decrease is (a) 2 cm (b) 6 cm (c) 4 cm (d) 8 cm 94. The rate at which fluid level inside vertical cylindrical tank of radius r drop if we pump fluid out at the rate of 3cm3 l/min is

1

(a) (c)

(b)

2

pr 2

pr

(d)

2

3 p r2 4 pr

95. The length l of a rectangle is decreasing at the rate of 2 cm/sec while the widths w is increasing at the rate of 2 cm/sec. When l = 12 and w = 5, the rate of change of area is (in cm2/sec) (a) 14 (b) 12 (c) 8 (d) 4 96. Let f be twice differentiable function such that f(x) = x2, x = 1, 2, 3. Then (a) f ¢¢(x) = 2 " x Œ (1, 3) (b) f ¢¢(x) = 2 for some x Œ (1, 3) (c) f ¢¢(x) = 3 " x Œ (2, 3) (d) f ¢¢(x) = f ¢(x) for some x Œ (2, 3) 97. A tangent drawn to the curve y = f(x) at P(x, y) cuts the x-axis and y-axis at A and B respectively such that BP : AP = 3 : 1, given that f(1) = 1 then dy – 3y = 0. (a) equation of the curve is x dx (b) normal at (1, 1) is x + 3y = 4. (c) curve passes through (2, 1/8) dy (d) equation of the curve is x + 4y = 0. dx 98. If 0 < b2 < c then f(x) = x3 + bx2 + cx + d (a) has no local minima (b) has no local maxima (c) is strictly increasing on R (d) is strictly decreasing on R

Previous Years' AIEEE/JEE Main Questions 1. If 2a + 3b + 6c = 0(a, b, c Œ R) then the quadratic equation ax2 + bx + c = 0 has (a) at least one root in [0, 1] (b) at least one root in [2, 3] (c) at least one root. (d) none of these [2002, 2004] 2. The maximum distance from origin of a point on the Ê at ˆ curve x = a sin t – b sin Á ˜ , y = a cos t – b cos Ë b¯ Ê at ˆ ÁË ˜¯ , both a, b > 0 is b

(a) a – b (c)

2

a +b

(b) a + b 2

(d)

a 2 - b2

[2002]

3. If the function f(x) = 2x3 – 9ax2 + 12a2x + 1 where a > 0, attains its maximum and minimum at p and q respectively such that p2 = q, then a equals (a) 1 (b) 2 (c) 1/2 (d) 3 [2003] 4. If u =

a 2 cos2 q + b2 sin 2 q + a 2 sin 2 q + b2 cos2 q

Applications of Derivatives 11.29

then the difference between the maximum and minimum values of u2 is given by (a) (a + b)

2

(c) 2(a2 + b2)

[2004]

5. A function y = f(a) has a second order derivative f ¢¢(x) = 6(x – 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x – 5, then the function is (b) (x – 1)3 (a) (x + 1)3 2 (c) (x – 1) (d) (x + 1)2 [2004] 6. The normal to the curve x = a(1 + cos q), y = a sin q at q always passes through the fixed point (a) (0, 0) (b) (0, a) (c) (a, 0) (d) (a, a) [2004] 7. A function is matched below against an interval where it is supported to be increasing which is of the following pairs in incorrectly matched? Interval Function (a) (–•, 1/3) 3x2 – 2x + 1 (b) (–•, –4) x3 + 6x2 + 6 (c) (–•, •) x3 – 3x2 + 3x + 3 (d) (2, •) 2x3 – 3x2 – 12x + 6 [2005] 8. The normal to the curve x = a (cos q + q sin q), y = a(sin q – q cos q) at any point Q is such that Ê ap ˆ , - a˜ (a) it passes through Á Ë 2 ¯ (b) it is at constant distance from origin (c) it passes through origin p + q with the x-axis [2005] (d) it makes angle 2 9. Let f be differentiable for all x. If f(1) = -2 and f ¢(x) ≥ 2 for x Œ [1, 6] then (a) f(6) < 5 (b) f(6) = 5 (c) f(6) ≥ 8 (d) f(6) < 8 [2005] 10. A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is (a) 1/54p cm/min (b) 5/6p cm/min (c) 1/36p cm/min (d) 1/8p cm/min [2005] 11. The function f(x) = (a) x = 1 (c) x = – 2

(b) (1/2)log 3 (d) log 3.

[2007]

–1

(b) 2 a 2 + b2 (d) (a – b)2

(a) 2 log3 e (c) log3 c

x 2 + has a local minimum at 2 x (b) x = 2 (d) x = 0 [2006]

12. A value of c for which the conclusion of mean value theorem holds for the function f(x) = log x on the interval [1, 3] is

13. The function f (x) = tan (sin x + cos x) is an increasing function is (a) (p/4, p/2) (b) (–p/2, p/4) (c) (0, p/2) (d) (–p/2, p/2) [2007] 14. Suppose the cube x 3 – px + q has three distinct real roots where p > 0 and q > 0. Then which one of the following holds? (a) The cubic has minima at

(b) The cubic has minima at –

p p and maxima at – 3 3 p and maxima at 3

(c) The cubic has minima at both

(d) The cubic has maxima at both

p 3

p p and – 3 3 p p and – 3 3 [2008]

15. Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of P¢(x) = 0. If P(– 1) < P(1), then in the interval [– 1, 1] (a) P(– 1) is the minimum but P(1) is not the maximum of P (b) neither P(– 1) is the minimum nor P(1) is the maximum of P (c) P(– 1) is the minimum and P(1) is the maximum of P (d) P(– 1) is not minimum but P(1) is the maximum of P. [2009] 16. Let f : R Æ R be defined by Ïk - 2 x, if f(x) = Ì Ó 2 x + 3, if

x £ -1 x > -1

If f has a local minimum at x = – 1, then a possible value of k is (a) –1/2 (b) – 1 (c) 1 (d) 0 [2010] 17. The curve that passes through the point (2, 3), and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by: (a) 2y – 3x = 0 (b) y = 6/x (c) x2 + y2 = 13

x 2 y 2 (d) ÊÁ ˆ˜ + ÊÁ ˆ˜ = 2 Ë 2¯ Ë 3¯ [2011]

11.30

Complete Mathematics—JEE Main

25. Statement-1: The equation x log x = 2 – x is satisfied by at least one value of x lying between 1 and 2 Statement-2: The function f(x) = x log x is an increasing function in [1, 2] and g(x) = 2 – x is a decreasing function in [1, 2] and the graphs represented by these functions intersect at a point in [1, 2]. [2013, online]

18. Let f be a function defined by Ï tan x , xπ0 Ô f(x) = Ì x ÔÓ 1 , x=0 Statment-1: x = 0 is point of maxima of f [2011] Statement-2: f ¢(0) = 0. 19. Let a, b Œ R be such that the function f given by f(x) = log |x| + bx2 + ax, x π 0 has extreme values at x = –1 and x = 2 Statement-1: f has local maximum at x = –1 and at x=2 Statement-2: a = 1/2 and b = –1/4. [2012]

26. Statement-1: The function x2(ex + e–x) is increasing for all x > 0 Statement 2: The function x2 ex and x2 e–x are increasing for all x > 0 and sum of two increasing functions in any interval (a, b) is an increasing function in (a, b) [2013, online] 27. If f and g are differentiable function in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f (1) = 6 then for some c Œ (0, 1) (a) 2 f ¢(c) = g¢(c) (b) 2 f ¢(c) = 3 g¢(c) (c) f ¢(c) = g¢(c) (d) f ¢(c) = 2 g¢(c) [2014]

20. A spherical balloon is filled with 4500 p cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72 p cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is (a) 7/9 (b) 2/9 (c) 9/2 (d) 9/7 [2012] bˆ Ê 21. The cost of running a bus from A to B is Rs. Á av + ˜ , Ë v¯ where v km/h is the average speed of bus. When the bus travels at 30 km/h, the cost comes out to be Rs. 75 while at 40 km/h, it is Rs. 65. Then the most economical speed in km/h of the bus is (a) 45 (b) 50 (c) 60 (d) 40 [2013, online]

28. If x = –1 and x = 2 are extreme points of f(x) = a log | x | + bx2 + x then: 1 1 (b) a = –6, b = (a) a = –6, b = 2 2 1 1 (c) a = 2, b = (d) a = 2, b = [2014] 2 2 29. If the volume of a spherical ball is increasing at the rate of 4p cc/sec, then the rate of increase of its radius (in cm/sec), when the volume is 288p cc is

22. If the surface area of a sphere of radius r is increasing uniformly at the rate of 8 cm2/s then the rate of change of its volume is (a) constant (b) proportional to r 2 (c) proportional to r (d) proportional to r [2013, online]

1 1 (d) [2014, online] 24 36 30. If non-zero real number b and c are such that min f(x) > max g(x) where f(x) = x2 + 2bx + 2c2 and c g(x) = –x2 – 2cx + b2 (x Œ R) then lies in the b interval:

23. The real number k for which the equation 2x3 + 3x + k = 0 has two distinct real roots in [0, 1] (a) lies between 2 and 3 (b) lies between –1 and 0 (c) does not exist (d) lies between 1 and 2 [2013] 24. The maximum area of a right angled triangle with hypotenuse h is (d) (c)

h2 2 2 h2 2

(b)

h2 2

(d)

h2 4

[2013, online]

(a)

1 9

(b)

1 6

(c)

È 1 ˘ , 2˙ (a) Í Î 2 ˚

1 (b) È0, ˘ ÍÎ 2 ˙˚

È1 1 ˘ (c) Í , (d) ÎÈ 2,• ˘˚ [2014, online] Î 2 2 ˙˚ 31. Let f and g be two differentiable functions on R such that f ¢(x) > 0 and g¢(x) < 0 for all x Œ R. Then for all x (a) g( f(x)) > g( f(x – 1)) (b) f(g(x)) > f(g(x + 1) (c) f( g(x)) > f(g(x – 1)) (d) g( f(x)) < g ( f(x + 1)) [2014 online]

Applications of Derivatives 11.31

32. If Rolle’s theorem holds for the function f(x) = 2x3 + 1 ax2 + bx in the interval [–1, 1] for the point c = , 2 then the value of 2a + b is (a) 1 (c) 2

(a) 4 3

(b) 3 3

(c) 3 2

(d) 4 2

39. Consider

(b) – 1 (d) – 2

Ê 1 + sin x ˆ Ê pˆ f(x) = tan -1 Á , x ŒÁ 0, ˜ . ˜ Ë 2¯ Ë 1 - sin x ¯

[2014, 2015 online] 33. Let f(x) be a polynomial of degree four having extreme f ( x) ˘ È value at x = 1 and x = 2. If lim Í1 + 2 ˙ = 3, then xÆ0 Î x ˚ f(2) is equal to (a) –8 (b) – 4 (c) 0 (d) 4 [2015] 34. The equation of a normal to the curve sin y = Êp ˆ x sin Á + y˜ at x = 0 is Ë3 ¯ (a) 2x +

3y = 0

(b) 2y –

3x = 0

(c) 2y +

3x = 0

(d) 2x –

3y = 0

A normal to y = f(x) at x =

35. Let k and K be the minimum and maximum values (1 + x)0.6 in [0, 1] respectively, of the function f(x) = 1 + x 0.6 then the order pair (k, K) is equal to (b) (2–0.4, 20.6) (a) (1, 20.6) –0.6 (c) (2 , 1) (d) (2–0.4, 1) [2015, online] 36. From the top of a 64 metres high tower, a stone is thrown upward vertically with the velocity of 48m/s. The greatest height (in metres) attained by stone, assuming the value of the gravitational acceleration g = 32 m/s2, is: (a) 100 (b) 88 (c) 128 (d) 112 [2015, online] 37. The distance, from the origin, of the normal to the curve, x = 2 cos t + 2t sin t, y = 2 sin t – 2t cos t at p , is t= 4 (a) 4

(b) 2 2

(c) 2

(d)

p also passes through 6

the point: (a) (0, 0)

Ê 2p ˆ (b) Á 0, ˜ Ë 3¯

Êp ˆ (c) Á , 0˜ Ë6 ¯

Êp ˆ (d) Á , 0˜ Ë4 ¯

[2016]

40. If m and M are are the minimum and the maximum values of 4 + 1 sin22x Р2cos4x, x ΠR, then M Рm 2 is equal to

[2015, online]

2

[2015]

(a)

9 4

(b)

15 4

7 1 (d) [2016, online] 4 4 41. If the tangent at a point P, with parameter t, on the curve x = 4t2 + 3, y = 8t3 – 1, t Œ R meets the curve again at a point Q, then the coordinates of Q are: (a) (16t2 + 3, –64t3 – 1) (b) (4t2 + 3, –8t3 – 1) (c) (t2 + 3, t3 – 1) (d) (t2 + 3, –t3 – 1) [2016, online] (c)

42. Let f(x) = sin4x + cos4x. Then f is an increasing function in the interval ˘ 5p 3p È (a) ˙ , Í ˚ 8 4 Î

˘ p 5p È (b) ˙ , Í ˚2 8 Î

˘p p È (c) ˙ , Í ˚4 2Î

˘ pÈ (d) ˙ 0, Í [2016, online] ˚ 4Î

3 . 4 If P is a point on C, such that the tangent at P has 2 slope , then a point through which the normal at 3 P passes, is (a) (1, 7) (b) (3, –4) (c) (4, –3) (d) (2, 3) [2016, online]

43. Let C be a curve given by y(x) = 1 + 4 x - 3 , x > [2015, online]

38. Let the tangents drawn to the circle, x2 + y2 = 16 from the point P(0, h) meet the x-axis at points A and B. If the area of DAPB is minimum, then h is equal to:

11.32

Complete Mathematics—JEE Main

Previous Years' B-Architecture Entrance Examination Questions 1. The slope of the normal to curve y = x3 – 4x2 at (2, –1) is 1 1 (b) (a) 4 2 (c) 4 (d) –4 [2006] 2. For the curve x = t2 – 1, y = t2 – t, the tangent line is perpendicular to the x-axis when (a) t = 0 (b) t = 1 (c) t =

1

(d) t =

3

1 2

[2006]

3. If f(x) = 4 sin x satisfies the Rolle's theorem on [0, p], then the value of c Œ (0, p) for which f ¢(c) = 0 is p p (b) c = 6 4 p p (c) c = (d) c = [2006] 2 3 4. The value of c for which the conclusion of Lagrange's mean value theorem holds for the function f (x) = (a) c =

25 - x 2 on the interval [1, 5] is (a)

3

(b)

(c)

15

(d) 2

[2007]

Ï x - 1 + a if x £ 1 5. Let f(x) = Ì if x > 1 Ó2 x + 3 If f (x) has a local minimum at x = 1, then (a) a > 5 (b) 0 < a £ 5 (c) a £ 5 (d) a = 5

[2007] 2y

6. If m is the slope of a tangent to the curve e = 1 + 4x2 then (a) Ám Á £ 1 (b) Ám Á > 1 (c) Ám Á ≥ 1 (d) Ám Á < 1 [2007] 7. Let f : (– •, •) Æ (– •, •) be a continuous and differentiable function and let f ¢(.) denote the derivative of f(.). If f (0) = –2 and f ¢(x) £ 3 for each x Œ [0, 2], then the largest possible value of f(2) is (a) 1 (b) 2 (c) 3 (d) 4 [2008]

f (x) =

x 4 + 3x2 + 1 2

x +1

21 28 (d) [2008] 5 5 9. Let y = f(x) be a curve which passes through (3, 1) and is such that normal at any point on it passes through (1, 1). Then y = f (x) describes (a) a circle of area p (b) an ellipse of area 2p (c) an ellipse of area 3p (d) a circle of area 4p [2008] (c)

p Ï Ô x sin 10. Let f(x) = Ì x ÔÓ0

, 0 < x £1 , x=0

then f ¢(x) = 0 for (a) exactly two values of x (b) no value of x (c) infinitely many values of x (d) exactly one value of x

[2009]

2

5

8. Let f: [–1, 2] Æ (– •, •) be given by

Then the maximum possible value of f (.) on [–1, 2] is 29 (a) 1 (b) 5

11. Let f(x) = [1 – x ], x Œ R, where [ ] is the greatest integer function. Then (a) f is increasing (b) x = 0 is the point of maxima of f (c) f is continuous at x = 0 (d) f is decreasing [2009] 12. A particle is constrained to move along the curve y = x starting at the origin at time t = 0. The point on the curve where the abscissa and the ordinate are changing at the same rate is: Ê1 1 ˆ (a) Á , ˜ Ë2 2¯

Ê1 1 ˆ (b) Á , Ë 8 2 2 ˜¯

Ê 1 1ˆ (c) Á , ˜ (d) (1, 1) [2009] Ë 4 2¯ 13. If the tangent and normal to the hyperbola x2 – y2 = 4 at a point cut off intercepts a1 and a2 respectively on x-axis and b1 and b2 respectively on y-axis then the value of a1 a2 + b1 b2 is (a) –1 (b) 0 (c) 4 (d) 1 [2010] 14. Let f be a differentiable function defined on R such that f(0) = –3. If f ¢(x) £ 5 for all x then

Applications of Derivatives 11.33

(a) f(2) > 7 (c) f(2) > 8

(b) f(2) £ 7 (d) f(2) = 8

[2010]

p p 15. Let f be a function defined on È - , ˘ by f(x) = 3 ÍÎ 2 2 ˙˚ cos4x – 6 cos3x – 6 cos2 x – 3. Then the range of f (x) is (a) [–12, –3] (b) [–6, –3] (c) [– 6, 3] (d) (–12, 3] [2010] 16. The function f (x) = xe – x has (a) neither maximum nor minimum at x = 1 (b) a minimum at x = 1 (c) a maximum at x = 1 (d) a maximum at x = –1

[2011]

18. The rate of change of the volume of a sphere with respect to its surface area when the radius is 2 units is: (a) 4 (b) 3 (c) 2 (d) 1 [2013] 19. If m is the slope of a tangent to curve ey = 1 + x2 at the point (x, y) on the curve then all possible values of m lie in the interval: (a) [0, 1] (b) (1, •) (c) (– •, –1) (d) [–1, 1] [2013] 20. f(x) = | x log x |, x > 0 is monotonically decreasing in: È1 ˘ (b) Í , 1˙ Îe ˚

(c) (1, e)

(d) (e, •)

[2014]

22. The maximum value of f(x) = 2 sin x + sin 2x, in the È 3 ˘ interval Í0, p ˙ is Î 2 ˚ 2 +1

[2016]

24. Let p(x) be a real polynomial of degree 4 having p ( x) extreme values at x = 1 and x = 2. If lim 2 = 1, x Æ0 x then p(4) is equal to: (a) 16 (b) 32 (c) 64 (d) 8 [2016] Reasoning Type

Statement-1: f is a 1 – 1 function. Statement-2: f is a strictly decreasing function on R. [2009] 26. Statement-1: The function f defined on R as f (x) = min{x, x2 ) is not differentiable at x = 1. Statement-2: The smaller angle between the tangents 1 to the curve y = x and y = x2 at x = 1 is tan–1 . 3 [2010] 27. Statement-1: If f (x) = e(x – 1) (x – 3) then Rolle's theorem is applicable to f (x) in the interval [1, 3]. Statement-2: Mean value theorem is applicable to [2011] f (x) = e(x – 1) (x – 3) in the interval [1, 4]. 28. Let f (x) = x2 – 8x +12, x Œ [2, 6]. Statement 1: f ¢(c) = 0 for some c Œ (2, 6). Statement 2: f is continuous on [2, 6] and differentiable on (2, 6) with f (2) = f (6). [2012] 29. Let a, b, c Œ R, a > 0 and the function f : R Æ R be defined by f(x) = ax2 + bx + c

21. Let f (x) = | x – x1| + | x – x2|, where x1 and x2 are distinct real numbers. Then the number of points at which f(x) is minimum (a) 1 (b) 2 (c) 3 (d) more than 3 [2014]

(a)

p 2 (d) 0

(b)

25. Let b, c be two non-zero real numbers such that b2 £ 3c. Let f (x) = x3 + bx2 + cx + d, x Œ R.

17. Each side of a square is increasing at the uniform rate of 1m/sec. If after sometime the area of the square is increasing at the rate of 8 m2/sec, then the area of square at that time in sq. meters is (a) 4 (b) 9 (c) 16 (d) 25 [2012]

1 (a) ÊÁ 0, ˆ˜ Ë e¯

p 4 (c) p (a)

Statement-1: b2 < 4ac fi f(x) > 0 for every value of x. Statement-2: f is strictly decreasing in the interval bˆ Ê ÁË -•, - ˜¯ and strictly increasing in the interval 2a Ê b ˆ ÁË - , •˜¯ 2a

[2012]

Answers

(b) 2 3

3 3 (c) (d) 3 [2015] 2 23. The abscissa of a point, tangent at which to the curve y = ex sin x, x Π[0, p], has maximum slope, is:

Concept-based 1. (a)

2. (c)

3. (b)

4. (d)

5. (a)

6. (c)

7. (c)

8. (a)

9. (b)

10. (c)

11.34

Complete Mathematics—JEE Main

Level 1 11. (c)

12. (b)

13. (b)

14. (b)

15. (d)

16. (c)

17. (b)

18. (a)

19. (d)

20. (c)

21. (a)

22. (d)

23. (a)

24. (b)

25. (d)

27. (d)

28. (a)

31. (c)

Previous Years' B-Architecture Entrance Examination Questions

26. (c)

1. (a) 5. (c) 9. (d)

2. (b) 6. (a) 10. (c)

3. (c) 7. (d) 11. (d)

4. (c) 8. (b) 12. (c)

29. (c)

30. (b)

13. (b)

14. (b)

15. (a)

16. (c)

32. (c)

33. (c)

34. (c)

17. (c)

18. (d)

19. (d)

20. (b)

35. (b)

36. (c)

37. (d)

38. (c)

21. (b)

22. (c)

23. (b)

24. (a)

39. (c)

40. (c)

41. (d)

42. (d)

25. (c)

26. (b)

27. (b)

28. (a)

43. (a)

44. (b)

45. (c)

46. (b)

29. (b)

47. (c)

48. (a)

49. (b)

50. (b)

51. (d)

52. (b)

53. (c)

54. (a)

55. (a)

56. (d)

57. (d)

58. (a)

59. (a)

60. (a)

Hints and Solutions Concept-based

Level 2 61. 65. 69. 73. 77. 81. 85. 89. 93.

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

97. (c)

1. 62. 66. 70. 74. 78. 82. 86. 90. 94.

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

63. 67. 71. 75. 79. 83. 87. 91. 95.

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

64. 68. 72. 76. 80. 84. 88. 92. 96.

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

Length of the tangent = y = 2.

1. (a) 5. (b) 9. (c)

2. (b) 6. (c) 10. (d)

3. (b) 7. (a) 11. (b)

4. (a) 8. (b) 12. (a)

13. (b)

14. (a)

15. (d)

16. (b)

17. (b)

18. (b)

19. (a)

20. (b)

21. (c)

22. (d)

23. (c)

24. (c)

25. (a)

26. (c)

27. (d)

28. (c)

29. (c)

30. (d)

31. (b)

32. (b)

33. (c)

34. (a)

35. (d)

36. (a)

37. (c)

38. (d)

39. (b)

40. (a)

41. (d)

42. (c)

43. (a)

1+

sin 2 t cos2 t

a cos3 t = a cos2 t cos t

1 dy e- t = - t = - e-2 t . If t = 0 then X = 2, y = 1 2 dx 2e Equation of normal is Y – 1 = 2 e2 t

98. (c)

Previous Years' AIEEE/JEE Main Questions

dy - a cos2 t + sin t cos t = =2 dx sin t a sin t cos t

fi

t =0

Y – 1 = 2(X – 2) 3.

( X - 2) fi 2X – Y – 3 = 0.

d 2 s -p 2 p ds p p = sin t cos t and a = = 2 18 2 dt dt 9 2

p2 18 4. f ¢(x) = 1 + log x – 1 = log x > 0 for x > 1. So f is increasing for x > 1 fi f(x) > f(1) = 0 for x > 1. f ¢(x) < 0 for 0 < x < 1 fi f is decreasing for 0 < x < 1 fi f (x) > f(1) = 0 for 0 < x < 1. a(t = 1) = –

8 +1 9 dy = 3x2, we need x such that 3 x 2 = = =3 2 +1 3 dx fi x2 = 1 so x = 1 or –1. Thus two points are (–1, –1) and (1, 1). 1 6. f ¢(x) = 4 x x 1 The function f increases if 4 x - > 0 i.e. x > 1 2 x 5.

Applications of Derivatives 11.35

1 1 < 0 i.e. 0 < x < x 2 7. f is differentiable function, so the critical points are given by f ¢(x) = 0 f ¢(x) = 3x3 – 3x2 – 18x = 3x (x + 2) (x – 3) The function f decreases if 4 x -

The critical points of f are 0, –2, 3 but only 0, 3 Œ [–1 4]. dx 8. 3 x 2 dx = 12 fi dx = 4 . We need >1 dy dy dy x 2 4

fi

x2

9. For

2 > 1 fi x < 4, x π 0

x2 k

2

+

y2 dy -4 x = 1, = 4 dx k 2 y

fi x Œ (–2, 2) ~ {0}.

and for y3 = 16x,

Ê -4 x ˆ Ê 16 ˆ 16 dy = . For orthogonality Á 2 ˜ Á 2 ˜ = -1 2 Ë k y ¯ Ë 3y ¯ 3y dx fi

64x = 3 k2 y3 = 48 k2 x k2 =

Level 1 –1

11. Let h(x) = tan +

2x

2

( x 2 + 1)2

x–

x

(a/8, a/8) is y + x = a/4. The x and y intercepts of this line both equal a/4, so 2=

a2 a2 fi a = 4. + 16 16

17. The slope of tangent to xm y = bm is

dy my =. So dx x

my the equation of tangent is Y – y = – (X – x) x X Y + which simplifies to = 1. Thus the x + mx y + my base of the triangle is the X-intercept, x + x/m and the height is the y-intercept, y + my. So the area of the triangle is xˆ Ê (1/2) Á x + ˜ (y + my) Ë m¯ xy (1 + m)2 b m (1 + m)2 = 2m 2m n m -1 which is constant if m = 1. 18. The slope of the tangent is –x3/y3 and its equation X Y is 4 3 + 4 3 = 1. a /y a /x =

4 2 3 fi k= 3 3 3 10. g¢(t) = 0 fi 8 – 4t = 0 fi t3 = 2 so t = 21/3 but 21/3 œ [–2, 1]. Now g(–2) = –32, g(1) = 7 So –32 in the least value. fi

Ê dy ˆ 16. Á ˜ (a/8, a/8) = –1. The equation of tangent at Ë dx ¯

1

1

, h¢(x) = – 2 x +1 1 + x2 1 + x2 > 0 for all x. For x > 0 h(x) > h(0) = 0

fi f (x) > g(x) on (0, •) 12. f (2) = f (3) = f (4) = f (5) = f (6) = 0, so, by Rolle's theorem applied on [2, 3], [3, 4], [4, 5], [5, 6] there are x1 Œ (2, 3), x2 Œ (3,4), x3 Œ (4, 5), x4 Œ (5, 6) such that f ¢(xi) = 0, i = 1, 2, 3, 4. Since f ¢ is polynomial of degree 4 so cannot have five roots. 13. f ¢(x) = (x – 3)4 (x + 1)3 (9x – 7) For x < – 1, f ¢(x) > 0 and for –1 < x < 0, f ¢(x) < 0. x = –1 is a point of maxima. dy 14. = tan q, so equation the equation of normal is dx y – a(sin q – q cos q) = –cot q(x – a(cos q + q sin q) which simplifies to x cos q + y sin q = a. The distance of the normal from origin is | a |. 15. Apply Rolle's theorem to show that the given equation cannot have two real distinct roots.

Thus the required intercepts are p = a4/x3 and q = a4/y3 so that p–4/3 + q–4/3 = a–16/3 (x4 + y4) = a–16/3 a4 = a–4/3 19. 1 – 2 cos x > 0 if cos x < 1/2 fi x Œ ÊÁ p , 5p ˆ˜ . Ë3 3 ¯ 20. For x π –1, 1,

dy ( x - 1)2 / 3 - ( x + 1)2 / 3 = 0 only = dx 3( x 2 - 1)2 / 3

at x = 0. y(0) = 2, y(–1) = y(1) = 21/3, so the greatest value of y is 2. 21. f ¢(x) = 4x –

1 4 x 2 - 1 (2 x - 1) (2 x + 1) x = = x x x2 2x – 1 2x + 1

x

f ¢(x)

x < – 1/2

–

–

–

–

–1/2 < x < 0

–

+

–

+

0 < x < 1/2

–

+

+

–

x > 1/2

+

+

+

+

Thus f increases on [–1/2, 0] » [1/2, •]. 22. Let (x, y) be any point on y =

e x + e- x 2

11.36

Complete Mathematics—JEE Main

U = S2 = square of distance of (x, y) from (0, 0) = x2 +

e

2x

+e 4

-2 x

+

1 2

31. The function is not differentiable at x = 0 and x = 1. Now Ïx -1 ÔÔ x 2 x ≥ 1 f(x) = Ì Ô1 - x , x < 1 ÓÔ x 2 x π 0

2x -2 x dU = 2x + e - e = 0 for x = 0 dx 2

23.

24.

25.

26.

27.

28.

29.

d 2U

> 0 at x = 0. Hence S 2 is min for x = 0 dx 2 1 1 and min S 2 = + = 1. 2 2 dy dy dy –2–2 =0fi = –1 2x + 2y dx dx dx ( 2 ,2 ) Required equation of normal is y – 2 = (x – 2) fi y = x. f ( x) xf ¢( x ) - f ( x ) ◊ g¢(x) = . Let g(x) = x x2 Let h(x) = x f '(x) – f(x), h¢(x) = xf ¢¢(x) > 0 So h(x) > h(0) = 0 for x > 0¢. Hence g¢(x) > 0, x > 0 g increases on (0, •). If a and b are real and distinct roots of x3 – 12x + k = 0. Then f(x) = x3 – 12x + k satisfies f(a) = f(b) = 0 then there g Œ [a, b] Õ [0, 2] such that f ¢(g) = 0 i.e., 3v2 – 12 = 0 fi g = ± 2. Not possible. 8x - 1 f ¢(x) = 8x1/3 – x–2/3 = 2 / 3 = 0 if x = 1/8 x Ê 1ˆ f(1) = 3, f(–1) = 6 + 3 = 9, f Á ˜ = –9/8 thus Ë 8¯ max f = 9. 3 2 f ¢(x) = 4x – 18x + 24x – 8, f ¢¢(x) = 12x2 – 36x + 24 = 12(x2 – 3x + 2) = 12(x – 1) (x – 2), f ¢¢(x) = 0 fi x = 1, 2 If x = 1 then y = 2. If x = 2, y = 3. So equation of tangent at (1, 2) is y – 2 = f ¢(1) (x – 1) = 2(x – 1). Equation of the tangent at (2, 3) is y – 3 = f ¢(2) (x – 2) = 0. The odd degree has at least one real root if the equation has at two real roots then by Rolle's theorem there is g Œ [a, b] such that 6g2 – 6g + 6 = 0 fi g2 – g + 1 = 0 which is not possible for g Œ R. f (x) = (x – 2) (x – 1)4 fi f ¢(x) = (x – 1)4 + 4(x – 1)3 (x – 2) = (x – 1)3(5x – 9) so for 1 < x < 9/5, f ¢(x) < 0 and for x > 9/5, f ¢(x) > 0. Thus x = 9/5 1 Ê 4ˆ 4 f has local minimum f(9/5) = – Á ˜ . 5 Ë 5¯ and

2 ÔÏ x + 2 x + 2, -1 / 2 £ x £ 0 30. f(x) = Ì ÔÓ x 2 - 2 x + 2, 0 < x £ 3 / 2 and f ¢(1) = 0 f ¢(0) doesn't exist. Since f (0) = 2, f (–1/2) = 5/4, f (3/2) = 5/4, f(1) = 1 so Max. f (x) = 2 and min f (x) = 1.

fi

Ï-x + 2 ÔÔ x 3 if x > 1 f ¢(x) = Ì Ô x - 2 , if x < 1 xπ0 ÓÔ x 3

so we get a third critical point x = 2. 32. The derivative of the function, xx (log x + 1), is negative when log x < –1 i.e., when x < e–1. 33. y¢(x) = 1 – ex is positive when ex < 1 i.e., when x 1 Ô -1 ÔÓ4 tan x + p if x < -1 So max f(x) = 4 37.

p –p=p 2

dy 3m =The equation of normal at (m2, dx ( m2 - m3 ) 2 – m3) is Y + m3 = (2/3m) (X – m2). Simplifying we have Y = (2/3m) X – (2m/3 + m3) which is identical with y = 2mx – 2m3 if m2 = 2/3.

38. f¢(x) = 2 cos x – 2 sin 2x = 2 cos x(1 – 2 sin x) > 0 if (i) cos x > 0, sin x < 1/2 (ii) cos x < 0, sin x > 1/2 i.e., if x Œ (0, p/6) » (3p 2 , 2p ) or x Œ (p/2, 5p/6). So f increases on (0, p/6) » (p/2, 5p/6). 39. f¢(x) = sec2 x – cosec2 x =

sin 2 x - cos2 x

= sin 2 x cos2 x f¢(p/4) = 0, f¢(3p/4) = 0. For 0

- cos 2 x

sin 2 x cos2 x < x < p/4, f¢(x) < 0 and for p/4 < x < p/2, f¢(x) > 0. Min f(x) = f(p/4) = 2.

40. f¢(x) = 12x(x2 – x + 1), f ¢(x) = 0 only for x = 0. 41. Distance of (t2, t) on y2 = x from y – x – 1 = 0 is | t - t2 - 1 |

=

t2 - t + 1

+ 1. This is minimum if t = 2 2 1/2. So the shortest distance = 3 2 /8.

Applications of Derivatives 11.37

42. f¢(x) = 3(x – (a + 1)) (x – (a – 1)) = 0 fi x = a ± 1. a – 1 Œ (–2, 4) fi a Œ (–1, 5), a + 1 Œ (–2, 4) fi a Œ (–3, 3) x( 3 - x ) , where x = 43. y = tan A tan (p/3 – A) = 1 + 3x tan A. dy -( 3 x - 1) ( x + 3 ) = . As 0 < A < p/3, x = dx (1 + 3 x )2 tan A > 0 dy 1 dy 1 ◊ =0fix= > 0 for 0 < x < and dx 3 dx 3 1 . So ymax = y = x = 1 3 = 1/3. is < 0 for x > 3

(

)

44. f(x) = –x log x, for 0 < x < 1, and f¢(x) = –(1 + log x). So f¢(x) = 0 fi x = e–1. It is easy to see max f(x) = f(e–1) = e–1. log 9 2 log x 45. f(x) = logx (1/9) – log3 x2 = log x log 3 f¢(x) =

2((log 3)2 - (log x )2 ) x(log x )2 log 3

so f¢(x) = 0 fi log x

= ± log 3 fi x = 3 or 1/3. One can see by first derivative test that max f(x) = f(3) = – 4. 46. Applying Lagrange's mean value theorem there is f (6) - +(1) = f¢(c) ≥ 2 c Œ (1, 6) such that 5 fi f(6) ≥ f(1) + 10 = 8. 47. f¢(x) = 3[(sin–1 x)2 – (cos–1 x)2] 1 - x 2 f¢(x) = 0 fi sin–1 x = cos–1 x or sin–1 x = – cos–1 x but sin–1 x + cos–1 x = p/2 so sin–1 x = cos–1 x fi sin–1 x = p/4 fi x = 1/ 2 . Thus f(1/ 2 ) = p3/32. 48. f¢(x) = log x + 4 so f¢(x) > 0 if log x > –4 i.e., x > e–4. 49. f¢(x) = (2 – x)xe–x. f¢(x) = 0 fi x = 0, 2 For 0 < x < 2, f¢(x) > 0 and for x > 2, f¢(x) < 0 So max f(x) = f(2) = 4e–2. 50. f(x) = Ï6 - 3 x Ô10 - x Ô Ì Ô4 + x ÔÓ3 x - 6

, x £ -2 , 2< x£3 ,3< x£5 ,x > 5

min f ( x ) = f (3) = 7.

51. f is not differentiable at x = 0. For x π 0, f ¢(x) = 2 – 3x – 1/3. So f ¢(x) = 0 fi x=

27 10 > , hence f ¢(x) is never zero, for x π 0. 8 3

Now

f (– 1) = – 3, f(0) = 2

f(10/3) =

2/3 26 Ê 10 ˆ - 3 Á ˜ > 0 but is less then 2. Ë 3¯ 3

Hence absolute minimum maximum is f(0). Ï - 1, xn - 1 Ô = Ì 1, 52. f(x) = lim n nÆ• x + 1 Ô 0, Ó

is f(– 1) and absolute if if if

0 £ x 1 x =1

g¢(x) = f(x). Hence g has local minimum at x = 1. 53. f ¢(x) = cos x – sin x = 0 fi x = np + p/4 f ¢¢(x) = – sin x – cos x = – (sin x + cos x) pˆ pˆ Ê Ê p f ¢¢ Á np + ˜ = (–1)n + 1 Á sin + cos ˜ Ë Ë 4¯ 4 4¯ = (– 1) n + 1 2 If n is even then f(x) has maxima and if n is odd then f(x) has minima. 54. Let f(x) = xe x (1 - x ) , we can write f ¢(x) = – (x – 1) (2x + 1) ex(1 – x) since ex(1 – x) > 0 " x Œ R so f ¢(x) < 0 for x < –1/2; f ¢(x) > 0 for –1/2 < x < 1 and is < 0 for x > 1. Thus f(x) increases on [–1/2, 1] 55. Equation tangent at P(c, ec) is y – ec = ec (x – c) (1) The equation of line joining (c + 1, ec+1) and (c–1, ec–1) is y – ec

+1

=

1 c+1 (e – ec – 1) (x – c – 1) 2

(2)

Subtracting (2) from (1) ec (e – 1) =

{

}

1 1 È ˘ ec Í( x - c ) 1 - e - e- 1 + e - e-1 ˙ 2 2 Î ˚ È 1 ˘ 1 fi e – 1 = ( x - c ) Í1 - e - e-1 ˙ + e - e- 1 Î 2 ˚ 2

(

)

(

fic–x=

e + e- 1 - 2 e - e- 1 - 2

>0

)

fi

(

(

)

)

x e and f decreases on (e, •) so f(p) < f(e) fi p e < ep. 57. Applying Lagrange’s mean value on [x, y], we have c Œ (x, y) such that

11.38

Complete Mathematics—JEE Main

f ( y) - f ( x ) y-x | f(y) – f (x) | = | f ¢(c) | | y – x | £ M | y – x | (if | f ¢(x) | £ M) and | f ¢(y) – f (x) | ≥ M | x – y | (if |f ¢(x) | ≥ M) Putting f(x) = cot x, f ¢(x) = – cosec2 x, | f ¢(x) | ≥ 1 Hence | cot x – cot y | ≥ | x – y | 58. f ¢(x) = 2 cos x – 2 sin 2x = 2 cos x (1 – 2 sin x) Now determine the sign of f ¢(x) on [0, 2p]. 1 < 0 for 59. f(x) = p/4 – tan–1 x fi f ¢(x) = – 1 + x2 all x. So the greatest value is f(0) = p/4 and least value is f(1) = 0. f ¢(c) =

60. f ¢(x) =

1 – log x

(log x )2

.

Level 2 61. f ¢(x) = (2/3) (x – 2)–1/3 (2x + 1) + 2(x – 2)2/3

1

0

65. f ¢(x) = ab x + b ac

bc

x + a2 + 0

0 2

bc x + c2

ab 1

ac 0

bc

x + c2

ac x + a2 +

ab 0

ab x+b 0

(x + b2) (x + c2) – b2c2 + (x + a2) (x + c2) a2 c2 + (x + a2) (x + b2) – a2 b2 3x2 + 2 (a2 + b2 + c2) x x (3x + 2 (a2 + b2 + c2)) 2 f decreases if x < 0 and x > - (a2 + b2 + c2) 3 2 2 2 i.e. on ((– 2/3) (a + b + c ), 0).

66. Let 'a' be the part of log in the channel of width 27 m and 'b' be the part of log in the channel of width 64 m. So l = a + b

f ¢(x) = 0 fi x = 1 also f ¢(x) does not exist at x = 2, hence the critical points are x = 1 and

dl = – 27 cosec q cot q + 64 sec q tan q dq 3 dl = 0 fi tan q = 4 dq

= 27 cosec q + 64 sec q

x = 2. 62. f (– 2) = 1 2

- (1 - x ) 1 = 0 i.e. f decreases on È1 / 3 , 3 ˘ Î ˚ Therefore, the greatest value is f 1 / 3 1 1 1 p 1 = + log 3. = tan–1 - log 2 3 3 6 4

(

2 3 One can verify that d l < 0 for q = tan–1 2 4 dq 5 5 max l = 27 ¥ + 64 ¥ = 125. 3 4

)

4x 2 x y dy dy x2 y2 64. =1fi =0fi = 9y 9 2 dx dx 9 4 The tangent will be perpendicular to 2y + 5x = 10 4 x Ê -5 ˆ ¥ Á ˜ = – 1 fi 10x = 9y if 9y Ë 2 ¯

Substituting the value of x in

bc 1

= – = =

Ê 10 ˆ = Á ˜ (x – 1) (x – 2)–2/3 Ë 3¯

(

ac 2

x2 y2 = 1, we have 9 4

1ˆ Ê 9 - ˜ y2 = 1, which is not possible. Thus there ÁË 100 4 ¯ is no point on the curve at which the tangent is perpendicular to 2y + 5x = 10.

q

q

Fig. 11.7

67. f ¢(x) = sin a tan2 x sec2 x + (sin a – 1) sec2 x = sec2 x (sin a tan2 x + (sin a – 1)) 1 - sin a Since sec2 x π 0 so tan2 x = sin a Now a Œ [p, 2p] so sin a < 0. Hence L.H.S. is ≥ 0 but R.H.S. < 0. Thus there is no critical point. 68. The derivative of f (x) is (4a – 3) + (a – 7) cos x. It is 4a - 3 4a - 3 fi £1 zero when cos x = 7-a 7-a

Applications of Derivatives 11.39

fi–1£

4a - 3 £ 1 fi a – 7 £ 4a – 3 £ 7 – a 7-a

i.e. when a ≥ – 4/3 or a £ 2. Hence the interval when f has no critical points are (– • – 4/3] and [2, •). Ê 21 - 4b - b2 ˆ 2 69. f ¢(x) = 3 Á 1 ˜ x + 5 > 0 if b +1 Ë ¯ x2 >

1–

-5 Ê 21 - 4b - b2 ˆ 3 Á1 ˜ b +1 Ë ¯

if

21 - 4b - b2 >0 b +1

72. 2

(x2 – 3)2

+ 27 ≥ 28 and the value 28 is

attained at x = 2

(x2 – 3)2

3 . Hence the minimum value of

+ 27 is 28.

Ï- x -2 < x < 0 Ô x=0 73. f (x) = Ì 1 Ôx 0 < x f (a) for a Œ (–1/2, 1/2) ~ {0}. f has local maximum at x = 0. 74. f ¢ (x) = ex(1 – x) + x (1 – 2x) ex(1 – x) = (1 + x – 2x2) ex(1 – x) = – (x – 1) (2x + 1)ex(1 – x)

If b + 1 > 0 then

Since ex(1 – x) > 0 for all x, so f ¢(x) > 0 if and only if (x – 1) (1 + 2x) < 0 i.e –1/2 = min (1, – 1/2) < x < max (1, –1/2) = 1.

(b + 1)2 > 21 – 4b – b2 fi (b – 2) (b + 5) > 0

Thus f increases on [– 1/2, 1].

which is trivially true if b + 1 < 0 i.e. b < – 1

i.e. b > 2 or b < – 5 so b Œ [2, 3]. 70. Any point on the ellipse is given by (a cos q, b sin q ) 2 dy -b = –b x = cot q Also 2 dx a a y The equation of tangent at (a cos q. b sin q) is Y – b -b cotq (X – a cos q) sin q = a This tangent cuts the coordinate axes at the points b ˆ Ê Ê a ˆ , 0˜ ÁË 0, ˜¯ and ÁË cos q ¯ sin q The area of the triangle formed by the tangent and the coordinate axes is 1 ab A= = ab cosec 2q 2 sin q cos q But cosec 2q is smallest when q = p /4 Amin = ab. 71. Let f (x) = x – log (1 + x). The domain of f is 1+ x -1 1 x = = . Since 1 + x 1+ x 1+ x 1 + x > 0 for x Œ dom f so f ¢ (x) > 0 if x > 0 and f ¢(x) < 0 for (– 1, 0). Thus f increases (0, •). Morever, f (0) = 0. Hence f (x) ≥ f (0) = 0 fi x ≥ log (1 + x) for x Œ (0, •).

(– 1, •). f ¢(x) = 1 –

75. Clearly f (0) = f (1) = 0. For f to be continuous on [0, 1], we must have lim xa log x = 0. This is xÆ 0 +

possible only if a > 0. Thus a = 1/2 is only such value. 76. Let h be the height of cone, then the radius r of the cone satisfies r2 = 3 – h2 1 dV 1 V = p (3 – h2)h so = p (– 3h + 3) 3 3 dh dV d 2V = 0 fi h = 1. Also =–p 0 i.e. 3b2 – 8ac > 0. 79. f (x) = |x2 – 3x + 2|

For x Π(Р1, 0), f decreases so

= – (x2 – 3x + 2), 3/2 £ x £ 7/4

f (x) ≥ f (0). Hence x ≥ log (1 + x) on (– 1, •).

f ¢ (x) = – (2x – 3). So the critical points are

Complete Mathematics—JEE Main

11.40

3/2, 7/4 . f (3/2) =

f (x) ≥

1 3 , f (7/4) = . Thus 4 16

3 3 . So m = 16 16

80. We are looking for x Œ [0, p] for which |(1/2)x – sin x| is maximum. Ï(1 / 2 ) x - sin x if (1 / 2 ) x ≥ sin x S= Ì Ósin x - (1 / 2 ) x if sin x < (1 / 2 ) x dS = 1/2 – cos x = 0 fi cos x = 1/2 fi x = p /3 dx Thus critical points are x = 0, p/3, p. S(0) = 0,

fi sin2a = 3/4 Ï1 - x + a, x £ 1 83. f (x) = Ì Ó 2 x + 3, x > 1 If f (x) has local minimum at x = 1, then f (1) £ f (1 + h), h > 0 i.e. a £ 2 (1 + h) +3 Hence a £ 5. So a = 5. f ( x + h) - f ( x ) 84. f ¢(x) = lim hÆ0 h f ( x ) + f (h) + 2 xh - 1 - f ( x ) = lim hÆ0 h f (h) - f (0) = lim + 2x hÆ0 h (Putting x = y = 0 we obtain f (0) = 1)

3 p p . Thus S(p/3) = - , S (p) = 2 6 2 Smax = p/2 at x = p. dy Ê dy ˆ + y˜ + 2 by =0 81. 2ax + 2h Á x Ë dx ¯ dx

= f ¢(0) + 2x = 2 (x + 1) f (x) = (x + 1)2 + C. Putting x = 0, we obtain C = 0. Thus f (x) = (x + 1)2. But g¢(x) = f (x) = (x + 1)2 > 0 so g increases on (– •, •) in particular on (0, •).

dy ax + hy fi =– dx hx + by The tangent will be || to Y-axis if hx + by = 0

82.

dy ax = tan a – 2 . The tangent will be parallel dx u cos2 a to y = x if tan a –

fix=

ax u2 cos2 a

85.

x dy dy 1 =- , =y dx (1, 3 ) dx 3 Equation of tangent at (1, 3 ) is Y – 3 = 1 3 (X – 1)

(

This will intersect X – axis at (4, 0). Equation of the normal at (1, 3 ) is Y–

=1

a

2

u (sin a – cos a) cos a a The corresponding ordinate is given by u2 (sina – cosa) cosa tana y= a =

-

u4

2 u2 cos2 a a2

2

u = a

3 =

3 (X – 1)

This will intersect X-axis at (0, 0) so the required area of the triangle 1 = 4 3 = 2 3. 2

(tan a - 1) u2 cos2 a

a

)

2

86. y ¢ =

x 2 - 2 x + 3 - 2( x - 1)2 ( x 2 - 2 x + 3)2

- ( x - 1)2 + 2

=

( x 2 - 2 x + 3)2

y ¢ = y 0 for x = 1 ± 2 y ¢ > 0 for 1 < x < 1 + 2

(sina – cosa) cos a

2 and y ¢ < 0 for x > 1 +

2

(

)

(

ymax = y 1 + 2 . Similarly ymin = y 1 - 2

1 È 2 ˘ ÍÎsin a - sin a cos a - 2 (1 - 2 sin a cos a )˙˚

(

y 1+ 2

)

=

2 2 and y 1 - 2 = – . 4 4

(

)

2

=

u (sin2a – 1/2) a

87. Equation of tangent at any point (x1, y1) is xx1 + yy1 = 16

According to given condition u2 (sin2a – 1/2) = a

u2 . 4a

)

Applications of Derivatives 11.41 (2, 2 ÷3)

= q (8,0)

92. 2y

Fig. 11.8

2¥6¥3=4

This passes through (8, 0) if 8x1 = 16 i.e. x1 = 2 so y1 = 2 3 . The slope of tangent is – 3 . 88. For x π – 2, 2 f ¢(x) = (x + 2)–1/3 (2x – 1) + (x + 2)2/32 2 3

then

(10 / 3) ( x + 1) ( x + 2 )1 / 3

at x = – 2. Thus the critical points are – 1, – 2.

=

db dl db dA =l +b = – 1. =b–l dt dt dt dt

dA will start decreasing if b – l £ 0 but b + l = 12 dt so l = 6. dV dh = p r2 94. V = p r2h fi dt dt

f ¢(x) = 0 fi x = – 1 clearly f is not differentiable

89. f ¢(x) =

dx dx fi = 9 unit/s. dt dt

93. 2(l + b) = 24 fi l + b = 12 dl db dl = 0. If is 1 cm/s + dt dt dt

Therefore q = p /6 and 2q = p /3.

log(e + x ) ¥

dx dy = 4 . According to the given condition dt dt

dy = 3. So dt (7,6 )

(2, - 2 ÷3)

=

p 1 + log 3 . 6 4

1 1 - log(p + x ) ¥ p+x e+x

fi

(log(e + x ))2 95.

log(e + x ) ¥ (e + x ) - (p + x ) log (p + x ) 2

(p + x ) (e + x ) ( log(e + x ))

Since log function is an increasing function and e < p, log (e + x) < log (p + x) fi (e + x) log (e + x) < (e + x) log (p + x) < (p + x) log (p + x) for all x > 0. Thus f ¢(x) < 0 for all x > 0.

dh 3 =– 2 . dt pr

db dl dA = l +b dt dt dt dA = 12 ¥ 2 + 5 ¥ (– 2) = 14. dt l = 12,b = 5

96. Let g (x) = f (x) – x2. As f is twice differentiable so g is also twice differentiable. Also g (1) = f (1) – 12 = 1 – 1 = 0

fi f decreases on [0, •).

g (2) = f (2) – 22 = 22 – 22 = 0

90. 2(l + b) = 32 fi l + b = 16. Area of rectangle A = lb = l(16 – l) dA dA = 16 – 2l so =0fil=8 dl dl

g (3) = f (3) –32 = 32 – 32 = 0

dA = – 2 < 0. Thus A is maximum when l = 8. dl

Again applying Rolle’s theorem $ g Œ (a, b )

91. f ¢(x) = –

1 1 + x2

2

+

By Rolle’s theorem $ a Œ(1, 2) and b Œ (2, 3) such that g ¢(a ) = 0, g ¢( b ) = 0.

1 (1 - x ) = > 0 for x Π1, 3 2x 1 + x2 2 x

(

)

(

such that g ¢¢ (g ) = 0

)

f increases on ÎÈ1, 3 ˘˚ . The greatest value of f is equal 1 to f 3 = cot -1 3 + log 3 2

( )

As a Œ (1, 2), b Œ (2, 3), (a, b ) à (1, 3) g ¢¢(g ) = f ¢¢(g ) – 2 = 0 fi f ¢¢(g ) = 2 So there is some x Œ (1, 3) such that f ¢¢(x) = 2. 97. Equation of tangent at P(x, y) Y – y = f ¢ (x) (X – x)

Complete Mathematics—JEE Main

11.42

1 Ê ˆ so A = Á x - y , 0 and B = (0, y – x f ¢(x)). Ë f ¢ ( x ) ˜¯

Êa ˆ t = cot Á + 1˜ Ëb ¯ 2

3:1 = BP:AP

Equation tangent at any point is

=

x 2 + x 2 ( f ¢( x ))2 :

y2 ( f ¢ ( x ))2

at ˆ Ê Y - Á a cos t - b cos ˜ Ë b¯

+ y2

9 x 2 (1 + ( f ¢( x ))2 ) x 2 ( f ¢( x ))2 2 = 2 (f ¢(x)) = 1 y (1 + ( f ¢( x ))2 ) y2 fi9 fi

y2 x2

= ( f ¢(x)2) fi 3

(i)

Distance of origin from (i) is

y dy =± x dx

Ê a cos t - b cos at ˆ - cot Ê Ê a + 1ˆ t ˆ ÁË ÁË ˜ ˜ ˜ ÁË b¯ b ¯ 2¯

x3 = Const or x3y = C y

Ê a sin t - b sin a t ˆ ÁË ˜ b ¯

S=

Since f (1) = 1, so Const = 1 3

ÊÊ a ˆ t ˆ Ê a ˆˆ Ê = cot Á Á + 1˜ ˜ Á X - Á a sin t - b sin t ˜ ˜ Ë Ë Ë b ¯ 2¯ Ë b ¯¯

a t 1 + cot 2 ÊÁ + 1ˆ˜ Ëb ¯ 2

3

Thus the curve is either x = y or x y = 1. The last curve passes through (2, 1/8) 98. f is a differentiable function and f ¢(x) = 3x2 + 2bx + c ÈÊ b ˆ 2 c b2 ˘ = 3 ÍÁ x + ˜ + - ˙ 3¯ 3 9˚ ÎË È 3c - b2 ˘ 2 = 3 Í( x + b / 3) + ˙ 9 ˚ Î Since b2 < c < 3c so 3c – b2 > 0 and hence f ¢(x) > 0 i.e. f is strictly increasing on R.

Previous Years' AIEEE/JEE Questions 1. Let f (x) = 1 ax3 + 1 bx 2 + cx . 3 2 Note that f is continuous and differentiable on R. We have a b f (0) = 0 and f (1) = + + c = 0. 3 2 \ By the Rolle’s theorem $ a Œ (0, 1) such that f ¢(a) = 0. i.e. aa2 + ba + c = 0. 2.

at ˘ È a Ê at ˆ dx = a cos t - b ◊ cos Á ˜ = a Ícos t - cos ˙ Ë b¯ b˚ b Î dt a Ê at ˆ dy È at ˘ = - a sin t + b ◊ sin Á ˜ = a Ísin - sin t ˙ Ë ¯ b b b dt Î ˚

ÊÊ a ˆ t ˆ ÊÊ a ˆ t ˆ at 2 cos Á Á + 1˜ ˜ sin Á Á - 1˜ ˜ sin - sin t ¯ Ë Ë Ë b ¯ 2¯ Ë ¯ b 2 dy b = fi = at dx ÊÊ aˆ t ˆ ÊÊ a ˆ t ˆ cos t - cos -2 sin Á Á + 1˜ ˜ sin Á Á1 - ˜ ˜ b Ë Ë b ¯ 2¯ Ë Ë b ¯ 2¯

ÊÊ a ˆ t ˆ Ê at ˆ = sin ÁË ÁË b + 1˜¯ 2 ˜¯ ÁË a cos t - b cos b ˜¯ ÊÊ a ˆ t ˆ Ê a ˆ - cos Á Á + 1˜ ˜ Á a sin t - b sin t ˜ Ë Ë b ¯ 2¯ Ë b ¯ ÊÊ a ˆ t ˆ Ê a Êa ˆ tˆ = a sin Á Á + 1˜ - t ˜ - b sin Á - t + Á + 1˜ ˜ ËË b ¯ 2 ¯ Ë b Ë b ¯ 2¯ Êt a tˆ Ê t a tˆ = a sin Á - + + b sin Á Ë 2 b 2 ˜¯ Ë 2 b 2 ˜¯ Êa t tˆ = (a + b)sin Á Ë b 2 2 ˜¯ So the maximum value of S is a + b. 3. f ¢(x) = 6x2 – 18ax + 12a2 = 6(x2 – 3ax + 2a2) f ≤(x) = 6(2x – 3a). According to the given condition f ¢(p) = 0 f ¢(q) and f ≤(p) < 0, f ≤(q) > 0. p + q = 3a, pq = 2a2. Since p2 = q, so p3 = 2a2 and 3a -1 ± 1 + 12a 3a p + p2 = 3a and p < and q > .p= 2 2 2 3

Ê -1 ± 1 + 12a ˆ 2 thus Á ˜ = 2a . Clearly a = 2 satisfy this 2 Ë ¯ equation as for a = 1, 1/2, 3 the L.H.S. is irrational and R.H.S. is a rational number. 4. We have u = a 2 + b2 2

+

a 2 - b2 a 2 + b2 a 2 - b2 cos 2q + cos 2q 2 2 2

Applications of Derivatives 11.43

f(x) increases on (–•, •)

Squaring, we get 2

If f(x) = 2x3 – 3x2 – 12x + 6, then

2

Ê a 2 + b2 ˆ Ê a 2 - b2 ˆ 2 u2 = a2 + b2 + 2 Á ˜ Á 2 ˜ cos 2q Ë 2 ¯ Ë ¯

f ¢(x) = 6x2 – 6x – 12 = 6(x2 – x – 2) = 6(x – 2) (x + 1) > 0 for x > 2

Ê a 2 + b2 ˆ Thus, max (u ) = a + b + 2 Á ˜ Ë 2 ¯ = 2(a2 + b2) 2

2

2

2

Ê a2 + b2 ˆ Ê a2 - b2 ˆ and min (u ) = a + b + 2 Á ˜ Á 2 ˜ Ë 2 ¯ Ë ¯ 2

2

fi f(x) increases on [2, •) 8. x = a(cos q + q sin q), y = a(sin q – q cos q) 2

2

= a2 + b2 + 2ab = (a + b)2 \ max (u2) – min (u2) = (a – b)2. 5. f≤(x) = 6(x – 1) fi f ¢(x) = 3(x – 1) + C1 dy ˘ =3 dx ˙˚ ( 2,1) 2

dy = a(cosq – cos q + q sin q) = aq sin q dq

dy sin q = dx cos q Equation of normal at q cos q {x – a(cos q + q sin q)} sin q

fi x cos q + y sin q = a{cos2 q + q cos q sin q + sin2q – q cos q sin q}

fi f(x) = (x – 1)3 + C2

fi x cos q + y sin q = a

As y = f(x) passes through (2, 1)

This line is at a distance a from the origin.

1 = (2 – 1)3 + C2 fi C2 = 0

9. By the Lagrange’s mean value theorem,

\ f(x) = (x – 1)3 6.

and

y – a(sin q – q cos q) = –

\ 3 = 3(2 – 1)2 + C1 fi C1 = 0 Thus, f ¢(x) = 3(x – 1)

dx = a(– sin q + sin q + q cos q) = aq cos q dq

\ 2

But f ¢(x)](2, 1) =

fi

dx dy = –a sin q, = a cos q dq dq \

dy -a cos q sin q = fi slope of normal at q is . dx a sin q cos q

Thus, equation of normal at q is

f (6) - f (1) = f ¢(a) for some a Œ (1, 6) 6 -1 fi f(6) + 2 = 5f ¢(a) ≥ 10 fi f(6) ≥ 8 10. Let thickness of ice at time t be r cm. Then volume of ice:

sin q [x – a(1 + cos q)] cos q This clearly passes through (a, 0)

y – a sin q =

10 cm

7. If f(x) = 3x2 – 2x + 1, then 1ˆ Ê 1 f ¢(x) = 6x – 2 = 6 Á x - ˜ < 0 for x < Ë 3¯ 3

Fig. 11.9

Thus, f(x) decreases on the interval (–•, 1/3]. V=

If f(x) = x3 + 6x2 + 6, f ¢(x) = 3x2 + 12x = 3x (x + 4) > 0 if x < –4 fi f(x) increases on (–•, –4] If f(x) = x3 – 3x2 + 3x + 3, then 2

2

f ¢(x) = 3x – 6x + 3 = 3(x – 1) ≥ 0 " x

4 4 p(10 + r)3 – p(10)3 3 3

dV dr = 4p(10 + r)2 dt dt fi –50 = 4p(10 + r)2

dr dt

r

Complete Mathematics—JEE Main

11.44

15. P(x) = x4 + ax3 + bx2 + cx + d

-50 dr fi = 4p (10 + r )2 dt fi

fi P ¢(x) = 4x3 + 3ax2 + 2bx + c As P ¢(0) = 0, we get c = 0

-50 dr ˘ = ˙ 4p (15) 2 dt ˚ r =5 = -

\ P ¢(x) = x(4x2 + 3ax + 2b) As x = 0 is the only real root of P ¢(x) = 0, roots of 4x2 + 3ax + 2b = 0 must be imaginary, therefore

1 18p

\ thickness is decreasing at the rate

1 cm/min. 18p

1 2 ( x - 2)( x + 2) 11. f ¢(x) = - 2 = 2 x 2 x2

4x2 + 3ax + 2b > 0

xŒR

Thus, P ¢(x) < 0 for x < 0 > 0 for x > 0.

For local maximum/minimum, set f ¢(x) = 0

\ x = 0 is a point of local minimum at x = 0.

fi x = ± 2.

Graph of y = P(x) is given in Fig. 11.10.

As f ¢(x) < 0 for 0 < x < 2

As P(–1) < P(1), we get P(1) is maximum but P(–1) is not minimum of P on [–1, 1].

and f ¢(x) > 0 for 2 < x < •,

y

f(x) has a local minimum at x = 2. 12.

f (3) - f (1) = f ¢(c) 3 -1

P(-1) P(1)

1 fi loge 3 = c 2

-1

O

fi c = 2 log3e 13. f ¢(x) =

Fig. 11.10

cos x - sin x 1 + (sin x + cos x) 2

16. f will be continuous at x = –1 if

f ¢(x) > 0 if and only if cos x > sin x

fi lim f ( x ) = lim f ( x ) = f(–1)

Ê p pˆ This is true if x Œ Á - , ˜ Ë 2 4¯

fi k + 2 = 2(–1) + 3 = k + 2

14. As f(x) = x – px + q has three real and distinct, roots, f ¢(x) = 3x2 – p has two distinct zeros by Rolle’s theorem. p Now f ¢(x) = 0 fi x = ± 3

> 0 if x >

For this value of k, f is continuous at x = –1, f ¢(–1) does not exist and f ¢(x) < 0 for x < – 1 and f ¢(x) > 0 for x > –1, therefore, f has a local minimum at x = –1. 17. Equation of tangent at (x, y) to the curve y = f(x) is given by

p We have f ¢(x) > 0 if x < 3 p 0 if x < –1 < 0 if –1 < x < 0 > 0 if 0 < x < 2 < 0 if x > 2

fi log |xy| = c fi xy = ± ec = A.

Thus f(x) has a local maximum at x = –1 and x = 2

As it passes through (2, 3), we get 6 = A

\ statement-1 and statement-2 are both true and statement-2 is a correct explanation for the statement-1.

\ xy = 6 or y = 6/x 18. Let g(x) = tan x – x, – p/2 < x < p/2 fi g¢(x) = sec2x – 1 = tan2x > 0 for –p/2 < x < p/2.

20. Let r be the radius of balloon and V be its volume. dV = –72p dt

Thus, g(x) is a strictly increasing function on (–p/2, p/2).

It is given that

Therefore, tan x < x for –p/2 < x < p/2

fi

d Ê 4 3ˆ Á p r ˜¯ = –72p dt Ë 3

tan x < 1 for –p/2 < x < p/2, x π 0. x \ x = 0 is a point of maxima.

fi

4 3 p r = –72p t + k 3

fi f (x) =

When t = 0, k =

Also, f ( x) - f (0) x

f ¢(0) = lim

xÆ0

\

tan x -1 = lim x x Æ0 x = lim

x Æ0

tan x - x x

2

4 3 p r = 4500p – 72p t 3

When t = 49,

È0 ˘ Í 0 form ˙ Î ˚

sec2 x - 1 xÆ0 2x

Also,

d Ê 4 3ˆ 2 dr = –72p Á p r ˜¯ = –72p fi 4pr dt Ë 3 dt

dr -72p -18 dr = = 2 fi 2 dt r 4p r dt

(tan x)2 1 =0 lim 2 xÆ0 x

21. C = av +

75 = a.30 +

1 + 2bx + a, x π 0. x As x = –1 and x = 2 are extreme values of f so

Solving a =

19. f ¢(x) =

C=

f ¢(–1) = 0 = f ¢(2) fi –1 – 2b = a = 0 and Solving we get, a =

1 + 4b + a = 0 2

1 1 and b = 4 2

1 1 1 1 (x + 1) (x – 2) - x+ = x 2 2x 2

r =9

2 = - . 9

b . According to the given conditions, v

Thus, both the statements are true. However, statement-2 is not a correct explanation of statement-1.

So f ¢(x) =

4 3 p r = 4500p – (72p) (49) = 972p 3

fi r3 = 729 fi r = 9

= lim

=

4 3 p r = V = 4500p 3

b b and 65 = a.40 + 30 40 1 , b = 1800. Thus 2

1 1800 v+ v 2

1 1800 dC dc = - 2 so = 0 fi v = 60 2 v dv dv d 2C dv

2

=

3600 v3

22. S = 4pr2, 8 =

> 0 . Thus C is minimum when v = 60. dS dr dr 1 = 8p r fi = dt dt dt pr

Complete Mathematics—JEE Main

11.46

V=

a + 2bx + 1 x As x = –1 and x = 2 are extreme points so f ¢(–1) = f ¢(2) = 0 i.e.

4 3 pr 3

28. f ¢(x) =

dV dV dr = 4p r 2 = 4r so ar . dt dt dt

– a – 2b + 1 = 0

23. Let f(x) = 2x3 + 3x + k Suppose a, b Π[0, 1], a < b be such that f(a) = 0 = f(b).

a + 4b + 1 = 0 2 1 Solving a = 2, b = - . 2

As f is differentiable on [a, b], By the Rolle’s theorem there exists r Œ (a, b) such that f ¢(r) = 0 fi 6r2 + 3 = 0.

29. V =

But 6r2 + 3 = 0 is not true for any real r. Thus, there is no real value of k, for which 2x3 + 3x + k = 0 has two real and distinct roots in [0, 1]. 1 24. A = pb , p is the perpendicular and b is the base 2 1 p2 + b2 = h2 so A = p h 2 - p 2 2 dA 1 2 1 1 h2 - p 2 - p 2 (-2 p) = h - p2 + p = dp 2 4 2 h2 - p 2 h 2 - p2 dA h =0fip= . It is easy to see that A is maxidp 2 mum for p = =

h 2

, so Amax =

1 h h2 h2 2 2 2

4 3 4 p r . When V = 288p C.C then 288p = p r 3 3 3

fi r = 6. Also fi

dV dr dr = 4p r 2 fi 4p = 4p 36 dt dt dt

dr 1 = . dt 36

30. f(x) = x2 + 2bx + 2c2 = (x + b)2 + 2c2 – b2 so min f(x) = 2c2 – b2. Also g(x) = –x2 – 2cx + b2 = b2 + c2 – (x + c)2 fi max g(x) = b2 + c2 Thus 2c2 – b2 > b2 + c2 fi

c c Π> 2 fi b b

(

)

2, • .

31. Note that f is a strictly increasing function and g is a strictly decreasing function. Now, x < x + 1 fi g(x) > g(x + 1) fi f(g(x)) > f(g(x + 1))

h2 . 4

32. f(–1) = f(1)

25. If f(x) = x log x then f ¢(x) = 1 + log x > 0 for x Œ [1, 2] so f increases on [1, 2]. For g(x) = 2 – x, g¢(x) = –1 so g decreases on [1, 2]. Therefore g(x) = f(x) for some x Œ [1, 2]. Hence the statement-1 follows. 26. Let f(x) = x2ex, f ¢(x) = 2xex + x2ex > 0 for x > 0. 2 –x

g(x) = x e , g¢(x) = 2xe

–x

2 –x

–xe

2

= (2x – x ) e

–x

g¢(x) > 0 for x < 2 and g¢(x) < 0 for x > 2. Hence g decreases for x > 2 and increases on (0, 2). Statement-2 is not true. If h(x) = x2(ex + e–x) then h¢(x) = 2x(ex + e–x) + x2 (ex – e–x) > 0 for x > 0. Hence h increases for x > 0. 27. Let F(x) = f(x) – 2g(x) F(0) = f(0) – 2g(0) = 2 – 0 = 2. F(1) = f(1) – 2g(1) = 6 – 4 = 2 F is continuous on [0, 1] and differentiable on (0, 1) with F(0) = F(1), so by Rolle’s theorem there is c Œ (0, 1) s. t. f(c) = 0 fi f ¢(c) = 2g¢(c).

fi –2 + a – b = 2 + a + b fi b = –2 Also, f ¢(c) = 0 fi 6c2 + 2ac + b = 0 3 1 +a–2=0fia= 2 2 Thus, 2a + b = 1 – 2 = –1

fi

f ( x) ˘ =3 33. lim ÈÍ1 + xÆ0 Î x 2 ˙˚ fi lim f ( x) = 2 x Æ0 x 2 \ f(x) must be of the form f(x) = ax4 + bx3 + 2x2 fi f ¢(x) = 4ax3 + 3bx2 + 4x As f(x) has extreme values at x = 1 and x = 2, 4a + 3b + 4 = 0

Applications of Derivatives 11.47

32a + 12b + 8 = 0 Solving these we get a =

1 , b = –2 2

1 Thus, f(2) = (24) – 2(23) + 2(22) = 0. 2 34. When x = 0, sin y = 0 fi y = 0 (since all lines given pass through origin) cos y

Êp ˆ dy Êp ˆ dy = sin Á + y˜ + x cos Á + y ˜ Ë3 ¯ dx Ë3 ¯ dx

When x = 0, y = 0, therefore dy ˘ Êpˆ sin Á ˜ + 0 = 3 dx ˙˚ (0,0) = Ë 3¯ 2 Thus, slope of normal at (0, 0) is - 2/ 3 and its equation is

(

)

y = -2/ 3 x or 2 x + 3 y = 0 35. Let m = 0.6, f ¢(x) =

=

=

(1 + x m )(m)(1 + x) m -1 - mx m -1 (1 + x) m

dy = 2 cos t – 2 cos t + 2t sin t = 2t sin t dt dy dy ˘ = tan t fi =1 dx dx ˙˚t =p /4

\

fi slope of normal at t = p/4 is –1. Thus, equation of normal at (x1, y1) is y – y1 = – (x – x1) or x + y – (x1 + y1) = 0 Its distance from the origin is d=

| x1 + y1 | 2

Ê 1 ˆÊ pˆ Ê 1 ˆÊ pˆ But x1 + y1 = 2 Á 1+ ˜ + 2Á 1- ˜ Á ˜ Ë 2 ˜¯ ÁË Ë 2¯ Ë 4¯ 4¯ = 2 2 \d=2 38. Let coordinates of A be (k, 0) then equation of tangent PA is y

(1 + x m )2

m(1 + x )m -1 (1 + x m - x m -1 - x m )

P

(o, h)

m 2

(1 + x )

m(1 + x) m -1 (1 - x m -1 )

B

m 2

(1 + x )

O

A

x

For 0 < x < 1, f ¢(x) < 0 as xm–1 = x–0.4 = 1/x0.4 > 1 for 0 < x < 1. Thus, f ¢(x) < 0 for 0 < x < 1 fi f(x) is strictly decreasing on [0, 1] \ k = max f(x) = f(0) = 1 and k = min f(x) = f(1) = 2–0.4 Hence, (k, K) = (2–0.4, 1) 36. If s is the height at any instant, then –2g(s – 64) = v2 – u2 where v is the initial speed and v is the speed at that instant. At the highest point, –2(32) (s – 64) = 0 – 482 fi s = 64 +

482 = 100. 64

37. Let x1 = x(p/4) and y1 = y(p/4). dx = –2 sin t + 2 sin t + 2t cos t = 2t cos t dt

Fig. 11.11

x y + = 1, where |h|, |k| > 4. k h 1

=4

(1)

1 1 + k 2 h2 fi

|h||k | h2 + k 2

=4

Let D = Area of DPAB = 2(area of DOAP) Ê 1ˆ = 2 Á ˜ (|h| |k|) = 4 h 2 + k 2 Ë 2¯ But from (1), k2 =

16h 2 h 2 - 16

Complete Mathematics—JEE Main

11.48

= 4 – cos 2x – cos2 2x

Ê 16h2 ˆ \ D = 16(h + k ) = 16 Á h2 + 2 ˜ h - 16 ¯ Ë 2

=

2

2

2 9 Ê1 ˆ - Á + cos 2 x˜ ¯ 4 Ë2 Note that

=

16h 4 h 2 - 16

m = 0 when cos 2x = 1

But D is minimum if and only if D2 is, we therefore minimize, D2.

and M =

We have È 4h3 (h 2 - 16) - h 4 (2h) ˘ d ( D 2 ) = 16 Í ˙ (h 2 - 16) 2 dh Î ˚ 3

= As

32h (h - 4 2 )(h + 4 2 )

So M – m =

24t 2 dy 24t 2 (x – (4t3 + 3)) as = 8t dx 8t It will meet the curve again at

y – (8t3 – 1) =

d d 2 ( D 2 ) = 0 for h = 4 2 and ( D ) < 0 if dh dh

Q(4t12 + 3, 8t13 – 1) if

4 < h < 4 2 and is > 0 if h > 4 2

8t13 – 8t3 = 3t[4t12 – 4t2]

2

Thus D is minimum when h = 4 2 . 39. For x Π(0, p/2) 1+ sin x = 1 Рcos (p/2 + x) = 2 sin2(p/4 + x/2) and 1 Рsin x = 1 + cos(p/2 + x)

fi 2(t1 – t)(t12 + t2 + tt1) = 3(t1 – t)(tt1 + tt2) fi (t1 – t)(2t12 – tt1 – t2) = 0 1 fi (t1 – t)(t1 – t)(2t1 + t) = 0 fi t1 = t, - t. 2

= 2 cos2(p/4 + x/2)

Thus, coordinates of Q are (t2 + 3, –t3 – 1) 42. f(x) = sin4 x + cos4 x, since f is differentiable and f ¢(x) = 4sin3x cos x – 4cos3x sin x

fi f ¢(x) = 1/2

= – 2 sin x cos x (cos2 x – sin2 x)

An equation of normal to y = f(x) at x = p/6 is

= – sin(4x) > 0

pˆ Êp p ˆ Ê y - Á + ˜ = -2 Á x - ˜ Ë 4 12 ¯ Ë 6¯

Êp pˆ "x Œ Á , ˜ Ë 4 2¯

p p = –2x + 3 3 fi 2x + y = 2p/3

˘p p È fi f(x) increases on ˙ , Í ˚4 2Î

fiy–

It passes through (0, 2p/3) 1 40. Let E = 4 + sin22x – 2cos4x 2 =4+

1 1 sin2 2x – (2cos2x)2 2 2

1 1 = 4+ (1 – cos2 2x) – (1 + cos x)2 2 2 =4+

1 (1 + cos 2x) [1 – cos 2x – 1 – cos 2x] 2

=4–

1 (1 + cos 2x) (2cos 2x) 2

9 . 4

41. An equation of tangent at P(4t2 + 3, 8t3 – 1) is

(h 2 - 16)2

p x È Ê p xˆ˘ \ f(x) = tan -1 Ítan Á + ˜ ˙ = + Î Ë 4 2¯ ˚ 4 2

9 when cos 2x = –1/2. 4

43.

dy 4 = = dx 2 4x - 3 fi

2 4x - 3

=

2 3

4x - 3 = 3 fi x = 3

For x = 3, y = 1 +

(4)(3) - 3 = 4

The equation of normal at (3, 4) is 3 y – 4 = - (x – 3) 2 It passes through (1, 7).

Applications of Derivatives 11.49

Previous Years' B-Architecture Entrance Examination Questions 1.

This passes through (1, 1), so 1 – y = -

1 . 4

fi1–y=–

dx dy dy 2t - 1 = 2t, = 2t – 1 fi = so tangent 2t dt dt dx

fi

line is perpendicular to x-axis if t = 0. 3. f(0) = f(p) = 1. f ¢(x) = cos x 4sin x so f ¢(c) = 0 fi cos p c=0fic= . 2 4. f(5) = 0, f(1) = -x

f ¢(x) =

25 - x 2

24 . f (5) - f (1) = 1 24 = 5 -1 4 so f ¢(c) =

x Æ1+

6. 2y = log(1 + 4x2) fi 2 fi

8x dy = 1 + 4 x2 dx

4x dy = 1 + 4 x2 dx

10. f ¢(x) = sin

f(2) = 2f ¢(c) + f(0) £ 6 – 2 = 4. 8. f ¢(x) =

( x 2 + 1)(4 x 3 + 6 x ) - ( x 4 + 3 x 2 + 1)2 x

=

2

2 x ( x ( x + 2) + 2) ( x 2 + 1)2

f ¢(x) = 0 for x = 0, f ¢(x) < 0, for x < 0, f ¢(x) > 0 for x>0 f(–1) =

5 29 , f(2) = , f(0) = 1 so the maximum value 2 5

of f is

29 . 5

p p = x x

p < • . By drawing graph of x tan x and x, we can see that there are infinitely many solutions. Since 0 < x £ 1 so p £

11. Since 1 – x2 is a decreasing function and [x] is an increasing function so f(x) = [1 – x2] is a decreasing function. 1 dx dy dx dy = . If = then 2 x = 1 dt dt dt 2 x dt or x =

1 1 so y = . 4 2

x dy = . 13. 2x – 2y dy = 0 fi y dx dx Equation of tangent at (x, y)

( x 2 + 1)2 2

p p p - cos , 0 < x £ 1 x x x

f ¢(0) does not exist. f ¢(x) = 0 fi tan

12.

Since 4x £ 1 + 4x2 so |m| £ 1 f (2) - f (0) 7. We have = f ¢(c) for some c Œ (0, 2) 2-0

(1 - x )2 -(1 - y)2 = +c 2 2

fi (x – 1)2 + (y – 1)2 = 4 which is circle of radius 2 so the area is 4p.

x £1 , f(1) = a x >1

So we must have lim f ( x) ≥ a i.e. 5 ≥ a or a £ 5

dx (1 – x) dy

0 = 2 + C fi C = –2

fi 2c2 = 75 – 3c2 fi c2 = 15. For c Œ (1, 5), we have c = 15 . Ï1 - x + a if 5. f(x) = Ì Ó 2 x + 3 if

1 (1 – x) f ¢( x )

Since it passes through (3, 1) so

3 2

c2 3 3 fi 2 = 25 - c 2 2

1 f ¢( x )

(X – x).

dy dy = 3x2 – 8x. = 12 – 16 = –4. So the slope dx dx (2,-1) of the normal is

2.

9. Equation of normal at any point is Y – y = -

Y–y=

x (X – x) y

x2 4 4 y2 y = , b = = +x 1 y y x x Equation of normal at (x, y)

So a1 = -

y (X – x) x So a2 = 2x, b2 = 2y

Y–y=

a 1a 2 + b 1b 2 =

4 4 ¥ 2 x - ¥ 2 y = 0. x y

Complete Mathematics—JEE Main

11.50

14.

f (2) - f (0) = f ¢(c) for some c Œ (0, 2) 2-0 f(2) = f(0) + 2f ¢(c) £ –3 + 2.5 = 7.

p p 15. f ÊÁ ˆ˜ = f ÊÁ - ˆ˜ = –3. Also f ¢(x) = Ë 2¯ Ë 2¯ 6 cos x sin x (2 cos x + 1) (cos x – 2) p p For x Œ È- , ˘ , 2 cos x + 1 π 0, cos x – 2 π 0 so ÍÎ 2 2 ˙˚

= 2 [2cos2x + cos x – 1] = 2 (cos x + 1) (2 cos x – 1) Now, f ¢(x) = 0 fi x = p/3 (

Ê 3ˆ 3 3 Êpˆ We have f(0) = 0, f Á ˜ = 2 Á ˜ + = 3, Ë 3¯ 2 Ë 2 ¯ 2 and f(p) = 0. Thus, maximum value is

f ¢(x) = 0 fi sin x = 0 fi x = 0. f(0) = 3 – 6 – 6 – 3 = –12. Hence range of f(x) is [–12, –3]. 16. f ¢(x) = e

–x

– xe

–x

–x

= e (1 – x), f ¢(x) = 0 fi x = 1

–x

f ≤(x) = –e (1 – x) + e–x(–1), f ≤(1) = –e–1 < 0 f has maximum at x = 1. 17. Area A(x) = x2 fi fi 8 = 2x

dx = 2x dt

18. V =

2

= 1. r =2

dy 2x 19. e = 1 + x fi y = log(1 + x ), so m = = dx 1 + x2 |m| £ 1 fi m Œ [–1, 1] y

2

d2m 2

= 2ex (cos x – sin x)

d2m ˘ = 2ep/2(–1) < 0 2 ˙ dx ˚ x =p / 2 Thus m is maximum when x = p/2 p( x )

= 1, p(x) must be of the from x2 p(x) = x2 + ax3 + bx4 a, b Œ R x Æ0

4 3 dV dS = 4pr2 and S = 4pr2 fi = 8pr pr fi 3 dr dr

dV dV 4p r 1 = = r, dS dS 8p r 2

p dm = 2ex cos x = 0 fi x = 2 dx

24. As lim

fi x = 4 fi A =16

3 3. 2

dy = ex(sin x + cos x) dx

23. m =

dx

dA dx = 2x dt dt

0 < x < p)

2

Ï- x log x, 0 < x £ 1 20. f(x) = |x log x| = x |log x| = Ì x >1 Ó x log x, Ï-(1 + log x), 0 < x < 1 f ¢(x) = Ì x >1 Ó 1 + log x, f ¢(x) < 0 if 1 + log x > 0 for 0 < x < 1 fi log x > –1 fi x > e–1 for 0 < x < 1 1 So x Œ È ,1˘ . ÍÎ e ˙˚ 21. By drawing the graph, we get two points of minimum. 22. f(x) = 2 sin x + sin 2x = 2 sin x + 2 sin x cos x = 2 sin x (1 + cos x) f ¢(x) = 2 cos x (1 + cos x) – 2 sin x sin x = 2 [cos x + cos2x – (1 – cos2 x)]

p ¢(x) = 2x + 3ax2 + 4bx3 As p ¢(1) = 0, p ¢(2) = 0, we get 2 + 3a + 4b = 0 and 4 + 12a + 32b = 0 fi 3a + 4b = –2, 3a + 8b = –1 fi a = –1, b =

1 4

\ p(4) = 42(1 + 4a + 16b) = 16(1 – 4 + 4) = 16 25. f ¢(x) = 3x2 + 2bx + c. Discriminant of f ¢(x) = 4(b2 – 3ac) £ 0 So f ¢(x) > 0 and hence f increases. Since f is continuous and increasing so f is 1 – 1.

Ï x2 26. min (x, x2) = Ì Óx 1+ h -1 lim =1 hÆ0 + h

f (1 + h ) - f (1) 0 < x £1 , lim = h Æ 0 + h x >1

(1 + h )2 - 1 f (1 + h ) - f (1) = lim = 2. So f is hÆ0 hÆ0 h h not differentiable at x = 1. lim

Applications of Derivatives 11.51

For y = x,

dy dy = 1 so tan q1 = 1. For y = x2, = dx dx

dy ˘ 2 -1 1 =2. tan q = = ˙ dx ˚ x =1 1+ 2 ¥1 3 27. f in statement-1 is continuous function on [1, 3] and differentiable on (1, 3) f(3) = 1 = f(1) so Rolle’s theorem is applicable. Again the mean value is applicable so statement-2 is true but is not a correct explanation for statement-1. 2x so

28. f is continuous on [2, 6] and differentiable on (2, 6) being a polynomial also f(2) = 0 = f(6) so by Rolle’s theorem there is c Œ (2, 6) such that f ¢(c) = 0.

ÊÊ b ˆ 2 c b2 ˆ b c 29. f(x) = a ÊÁ x 2 + x + ˆ˜ = a Á Á x + ˜ + - 2 ˜ Ë 2a ¯ a 4a ¯ ËË a a¯

ÊÊ b ˆ 2 4ac - b2 ˆ = aÁÁ x + ˜ + ˜ >0 2a ¯ 4a 2 ¯ ËË f ¢(x) = 2ax + b. So f ¢(x) > 0 if x > f ¢(x) < 0 if x < -

b . 2a

b and 2a

12.1

CHAPTER TWELVE

A function F(x) is an antiderivative of a function f(x) if F¢(x) = f(x) for all x in the domain of f. The set of all antif with respect to derivatives of f x, denoted by

S.No. 13.

Ú f(x) dx*

f(x)

1

tan–1 x

1 + x2

or

– cot–1 x

14.

tan x

x

x is the

15.

cot x

x

have found one antiderivative, the other antiderivative of f differ from F by a constant. We indicate this by

16.

sec x

x x + tan x

Ú f ( x) dx

x

The function f

Ú f ( x) dx

= F(x) + C

Ê x pˆ tan Á + ˜ Ë 2 4¯ 17.

cosec x

x – cot x

The constant C constant.

tan

TABLE OF BASIC FORMULA S.No. 1. 2.

3.

f(x)

Ú f(x) dx*

0

C

x (n π – 1)

19.

1 xn+1 n +1

a2 - x2

21. ex

5.

a (a > 0, a π 1)

ax log a

cos x sin x sec2 x cosec2 x sec x tan x cosec x cot x

sin x – cos x tan x – cot x sec x – cosec x

1 1- x

2

sin–1 x

(a π 0)

Ê xˆ

sin–1 Á ˜ Ë a¯

1 2

x +a

x + x 2 + a2

2

1

20.

x

ex

12.

1

2

x -a

1 x

4.

6. 7. 8. 9. 10. 11.

18.

x 2

x + x 2 - a2

2

a 2 - x 2 (a π 0)

x 2 a - x2 + 2 1 2 -1 Ê x ˆ a sin Á ˜ Ë a¯ 2

or

22.

23.

x 2 x + a2 2 1 2 + a 2

x 2 + a2

x 2

x 2 - a2

– cos–1 x

x 2 - a2 –

24.

1 2

x - a2

(a π 0)

1 2a

x + x 2 + a2

1 2 a x + x 2 - a2 2 x-a x+a

12.2 S.No. 25.

26.

Ú f(x) dx*

f(x)

1 a2 - x2 1 a2 + x2

1 2a

(a π 0)

TIP a+x a-x Ú

1 Ê xˆ tan–1 Á ˜ Ë a¯ a

(a π 0)

dx =

x 2

a2 dx Ú 2 integrand

Note: Use a2 or – a2

v(x) Ú u(x)dx by parts u(x) v(x)

È dv

˘

Ú ÍÎÚ dx Ú u( x) dx ˙˚ dx

A LIST OF EVALUATION TECHNIQUES S.No.

Form of Integrate

Value/Evaluation Technique

1.

Ú f ¢(x + a) dx

f(x + a) + C

2.

Ú f ¢(ax + b)dx, a π 0

1 f(ax + b) + C a

3.

Ú

4.

Ú (f(x))n f ¢(x) dx, n π – 1

1 (f(x))n+1 + C n +1

5.

Ú f ¢(g(x)) g¢(x) dx

f(g(x)) + C

6.

Ú cos mx cos nx dx

Use product to sum identities

f ¢( x ) dx f ( x)

f (x

C

cos A cos B =

7.

1 [cos (A – B) + cos (A + B)] 2

Ú sin mx sin nx dx

sin A sin B =

1 [cos (A – B) – cos (A + B)] 2

Ú sin mx cos nx dx

sin A cos B =

1 [sin (A – B) + sin (A + B)] 2

Ú sin x cosn x dx

If m is odd put cos x = t

where m, n ΠN

If n is odd put sin x = t If both m and n are odd, put sin x = t if m ≥ n and cos x = t otherwise. If both m and n sin2 x =

1 (1 – cos 2x) 2

cos2 x =

1 (1 + cos 2x) 2

12.3 S.No.

Form of Integrate

Value/Evaluation Technique

If m + n put tan x = t 8.

a sin x + b cos x + c

Ú p sin x + q cos x + r dx ae x + be- x

Ú pe x + qe- x

Write Num = a (DEN) + b

Find a, b sin x and constant term (or ex, e– x

dx

(p, q π 0 and at least one of a, b π 0) a Údx + b

9.

a tan x + b

Ú p tan x + q ae2 x + b

Ú pe2 x + q

a sin x + b cos x

or

dx

Ú

dx

or

Q

dx

or

ÚQ

or

Ú

or

Q or

ae x + be- x

Ú be x + qe- x

x,

dx

Ú DEN dx

dx

and use (8) above Use a = r cos a, b = r sin a

dx

Ú a sin x ± b cos x

1 dx . Now, use formula for Ú r sin ( x ± a ) Ú cosec x dx

a, b π 0)

11.

Ú

d ( DEN ) dx dx + g DEN

Ú p sin x + q cos x

dx

(p, q π 0, at least one of a, b π 0) 10.

d (DEN) + g dx

L

Ú

Q L

ÚQ

ÚL

dx

b Write Q = a ÊÁ x 2 + x + Ë a

dx

and put t = x +

cˆ ˜ a¯

b 2a

Q

where L = px + q, p π 0 and Q = ax2 + bx + c, a π 0 12.

13.

L1 dx L2

Ú

Write as

L1

where L1 = ax + b, a π 0

Ú

L2 = px + q, p π 0

to reduce to form

dx

ÚL

Q

L 1L 2

Put L =

dx

Ú

L Q

dx

1 to reduce to form t

Ú

dx Q

12.4 S.No.

Form of Integrate

Value/Evaluation Technique

where L = px + q, p π 0 Q = ax2 + bx + c, a π 0

and 14.

Ú

Q

Q dx L

L Q

where L = px + q, p π 0 2

Divide Q by L to obtain

Q = ax + bx + c,

Q = L1 (L) + a

aπ0

where L1 is a linear expression in x and a is a constant.

Ú 15.

Ú

Q1

Write Q1 ∫ a Q2 + b

dx

Q2

L1 Q

dx + a Ú

dx

.

L Q

d [Q2] + g dx

Find a, b, g

where Q1 = ax2 + bx + c, a π 0

a

Ú

Q2 dx + b

Ú

d dx [Q ] dx + g 2 Q2

Ú

dx Q2

Q2 = px2 + qx + r, p π 0 16.

dx

ÚL

1

Put L2 = t2

L2

where L1 = ax + b and L2 = px + q, with a, p π 0, 17.

ÚQ

1

a π p.

dx Q2

where Q1 = ax2 + b, Q2 = px2 + q with 18. 19.

a, p π 0, p π 0

First put x = 1/t and simplify then put p + qt2 = u2

ÚR(x1/ , x1/q, x1/ ...) dx

Let l = lcm (p, q, r, ...)

where R is a rational function.

and put x = t

x

Ú e (f(x) + f¢(x)) dx

Write as Ú ex f (x) dx + Úex f ¢(x) dx ex as the second function to obtain ex f(x) - Úex f ¢(x) dx + Úex f ¢(x) dx

20.

Ú e cos (bx + c) dx or Ú e sin (bx + c) dx

e as the second function to obtain 1 b e cos (bx + c) + Ú e cos (bx + c) dx a a e as the second function and Ú e cos (bx + c) dx to the left. ax

12.5 S.No.

Form of Integrate

21.

Ú Q( x) dx

Value/Evaluation Technique

P( x )

P(x) ≥

where P(x) and Q(x) are polynomials in x

Q(x

A(x) when

P(x) is divided by Q(x). Now,

P( x ) B( x ) = A(x) + Q( x ) Q( x ) B(x

22.

P( x )

Q(x)

Write

Ú Q( x) dx P(x

Q(x) and Q(x) is a

An A1 A2 P( x ) = +...+ + x - an x - a1 x - a 2 Q( x )

product of distinct linear factors, that is, Q(x) = A(x – a1) (x – a2) . . . (x – ar)

Evaluate Ai¢ s and use

dx

Úx-a

where ai¢ s are distinct and A π 0 23.

24.

P( x)

Ú ( x - a)n dx

Put x – a = t,

where n ≥ 1 and P(x) is a polynomial in x.

express P(x) in terms of ts

P( x) dx

Ú ( x - a ) m ( x - b) n Am A1 A2 + + ... 2 x - a ( x - a) ( x - a)m

where m, n ≥ 1, a π b,

+ P(x) < m + n.

25.

Evaluate Ai¢ s and B ¢j s

P( x) dx

Ú ( x - a) ( x2 + bx + c) where b2 – 4c

26.

Bn B1 B2 + + ... + x - b ( x - b) 2 ( x - b) n

x2 + 1 Ú x4 + kx2 + 1 dx

P(x) < 3

A Bx + C + 2 x - a x + bx + c Divide the numerator by x2 to obtain

(1 + 1 / x 2 ) dx Ú x2 + 1/ x2 + k Now, put x – 1/x = t

x–a

12.6 S.No.

Form of Integrate

27.

Ú x4 + kx2 + 1 dx

x2 - 1

Value/Evaluation Technique

Divide the numerator by x2 to obtain

(1 - 1 / x 2 ) dx

Ú x2 + 1/ x2 + k 28.

xP( x 2 ) Ú Q( x2) dx

Now, put x + 1/x = t

Put x2 = t

where P and Q are polynomials in x 29.

30.

31.

P( x 2 ) Ú Q( x2 ) dx

Put x2 = t

where P and Q are polynomials in x

for partial fractions not for integration

1

Ú ( x2 + a 2)n dx = An

An–1 1 dx ( x 2 + a 2 )n - 1

where n > 1

An–1 = Ú 1.

ÚR (sin x, cos x) dx

Universal substitution

where R is a rational function

tan (x/2) = t.

Special Cases (a) If R (– sin x, cos x)

Put cos x = t

= – R (sin x, cos x) (b) If R (sin x, – cos x)

Put sin x = t

= – R (sin x, cos x) (c) If R (– sin x, – cos x)

Put tan x = t

= R (sin x, cos x) 32.

Úx (a + bxn) dx where m, n, p are rational numbers. (a) p

Expand (a + bxn)

(b) p

Put x = tk

m=

a c ,n= , b d

where k = l cm (b, d)

a, b, c, d ΠI, b > 0, d > 0 (c)

m +1 n

Put a + bxn = t where p = r/s, r, s ΠI, s > 0.

(d)

m +1 +p n

Put a + bxn = xnt where p = r/s, r, s ΠI, s > 0.

12.7 S.No.

Form of Integrate

33.

Ú

Pn ( x) ax 2 + bx + c

Value/Evaluation Technique

Write

dx

where Pn(x

n in x.

Ú Ú

Pn ( x) 2

ax + bx + c dx

dx = Qn–1 (x) ax 2 + bx + c + k

ax 2 + bx + c

where Qn–1 (x

n – 1) in x and k is a

constant. Differentiate both the sides w.r.t. x by

ax 2 + bx + c to 1 Qn–1 (x) (2ax + b) + k 2 Qn–1 (x) and k.

Pn(x) = Qn¢ - 1 (x) (ax2 + bx + c) +

SOME TIPS FOR SIMPLIFYING COMPUTATIONS 1

1 , where Q = ax2 + bx + c, Q

(i)

ax 2 + bx + c = t

b the substitution x + = t may be used instead of 2a Q.

(ii)

ax 2 + bx + c = tx

(iii)

ax 2 + bx + c = t(x – a)

1. In case of

Q

, Q , or

If Q = (x – a) (x – b) with a π b, then x –

1 (a + b) 2

= t can be used. 2. In case of

L Q

, L Q and

L Q

where L = px + q and Q = ax2 + bx + c, p π 0, q π 0, the following steps in unnecessary Express d [Q] + b L=a dx

x a if a > 0 if c > 0

where a is a real root of ax2 + bx + c = 0 4. To evaluate dx I= Ú a + b sin x we rewrite dx I= Ú a - b cos (p / 2 + x) and put p/2 + x = q. This will lead to simpler computation then direct evaluation.

SOME TIPS FOR PARTIAL FRACTIONS b =t 2a 3. To evaluate

1. Partial fractions

x+

(

f(x) =

)

Ú R x, ax 2 + bx + c dx

P ( x) ( x - a1 ) ... ( x - an ) P(x) < n and a1, a2 . . . an are distinct) is

where R is a rational function of x and An A1 A2 + + ... + x - an x - a1 x - a2

ax 2 + bx + c , where

12.8

Ai =

P(ai ) (ai - a1) . . . (ai - ai - 1) (ai - ai + 1) . . . (ai - an )

= value of the expression (x – ai) f(x) at x = ai. In other words, Ai x = ai everywhere in f(x) except in the factor x – ai. For instance, 2x - 1 2(1) - 1 1 ∫ ( x - 1) ( x + 1) ( x - 2) (1 + 1) (1 - 2) x - 1 +

2(-1) - 1 1 (-1 - 1) (-1 - 2) x + 1

+

2( 2) - 1 1 (2 - 1) (2 + 1) x - 2

∫ -

1 1 1 1 1 + 2 x -1 2 x +1 x - 2

Justification x = 1, 2(– 1) – 1 or – 3 at x = – 1 and 2(2) – 1 or 3 at x = 2. Indeed, it is the polynomial* 1 ( x + 1) ( x - 2) (- 3) ( x - 1) ( x - 2) + (-1 - 1) (-1 - 2) (1 + 1) (1 - 2) +

3( x - 1) ( x - 2) (2 - 1) (2 + 1)

2. If P(x) = b0 + b1x + ... + bnxn, the partial fraction of P( x) ( x - a1) . . . ( x - an ) where, a1, a2, ... an An A1 A2 + + ... + x - an x - a1 x - a2 where Ai¢ a, b, c are distinct x3 a3 1 ∫1+ ( x - a ) ( x - b) ( x - c ) ( a - b) ( a - c ) x - a bn +

+

b3 1 (b - a) (b - c) x - b

+

c3 1 (c - a ) (c - b ) x - c

3. To resolve 1 r

( x - a ) ( x - b) where a π b, put x – a = y, so that

1 r

( x - a ) ( x - b)

=

1 y ( y + a - b) r

1 = ( a - b) y r

y ˘ È Í1 + a - b ˙ Î ˚

-1

-1

y ˆ Ê –1 Now expand Á1 + ˜¯ upto y and divide by y Ë a b to obtain Cr + 1 C0 C + r 1+ 1 + . . . + . r y y y 1 . ( x - a)r - k

The value Ck

1 1 is x-b (b - a)r For instance, if x – 1 = y, then 1 1 1 = 3 = 3 (1 – y)–1 3 ( x - 1) ( x - 2) y ( y - 1) y =

1 (1 + y + y2 + ...) y3

Thus, 1 1 1 = ( x - 1) ( x - 2) x - 2 x -1 3

-

1 1 2 ( x - 1) ( x - 1)3

Illustration Resolve 1 ( x + 1) ( x + 2) 2 3

into partial fraction. First, put x + 1 = y, 1 1 = 3 2 ( x + 1) ( x + 2) y ( y + 1)2 3

=

1 (1 – 2y + 3y2 – ...) y3

Next, put x + 2 = z 1 1 1 = - 2 ( 1 – z) – 3 = 2 3 3 2 z ( z - 1) z ( x + 1) ( x + 2) =–

1 z2

(1 + 3z + . . .)

Thus, 1 3 2 1 + = 3 2 x + 1 ( x + 1) ( x + 1) ( x + 2) ( x + 1)3 3

–

3 1 x + 2 ( x + 2)2

12.9

4. To obtain partial fractions of 1 ( x - a ) ( x - b) ( x 2 + c 2 ) where a π b.

Ax + B 1 1 1 ∫ + 2 2 2 x 3( x + 1) x + 2 x( x + 1) ( x + 2) To obtain A and B, we rewrite above expression as 1 1 and can x-a x-b

be obtained as explained in (1). For instance, we can write

1 1 (x + 1) (x2 + 2) – x (x2 + 2) 2 3 + (Ax + B) x(x + 1) Now, A and B x. 1∫

-

SOLVED EXAMPLES Concept-based Straight Objective Type Questions

Example 1: If f(x) =

log x x3

by (a) C (c) C -

1 2x

2

1 2 x2

, then its antiderivative F(x)

Ans. (c) 2

Solution:

log ( x e ) (b) C +

(d) C +

log ( x e)

1 2 x2 1 2 x3

log x

= Ú (sec x + tan x )2 dx = Ú sec2 x + Ú (sec2 x – 1) dx + 2 Ú sec x tan x dx = 2 tan x – x + 2 sec x + C Ê 1 + sin x ˆ = 2Ë - x+C cos x ¯

log ( x e)

Ans. (a)

Ú

Solution: F(x) =

log x x3

dx

x

1

Ú x3 dx

–

( ) ( ) ( ) ( ) ( )

Ê 1 - cos x + p Á 2 = 2Á p ÁË sin x + 2

1Ê 1 ˆ Ú x ÁË Ú x3 dx˜¯ dx 1 1 dx + Constant 2 Ú x3 2x 1 1 = – 2 log x - 2 + C 2x 4x 1 È 1 ˘ = - 2 Ílog x + log e˙ + C Î ˚ 2 2x =-

=-

1

2

1 2 x2

log x +

1 + sin x

Ú 1 - sin x dx

Êx p ˆ (a) 2 tan Ë + ¯ + C 2 4 Êx p ˆ (b) 2 tan Ë + ¯ + x + C 2 4 Êx p ˆ (c) 2 tan Ë + ¯ – x + C 2 4 Êx p ˆ (d) 2 tan2 Ë + ¯ – x + C 2 4

ˆ ˜ ˜ - x +C ˜¯

x p + 2 4 =2 - x +C x p x p 2 sin + cos + 2 4 2 4 2 sin 2

Êx p ˆ = 2 tan Ë + ¯ - x + C 2 4

log x e + C

Example 2: The value of

1 + sin x (1 + sin x ) Ú 1 - sin x dx = Ú 1 - sin2 x dx

Example 3: The value of is

Ú

dx 4 x - 3 - x2

(a) sin–1 (x – 1) + C

( x - 2) + 4 x - 3 - x 2 ( x - 1) + 4 x - 3 - x 2 + C (d) sin–1 (x – 2) + C Ans. (d) Solution:

4 x - 3 - x 2 = - ( x 2 - 4 x + 3)

12.10

- (( x - 2 )2 - 1) =

= So,

Ú

dx 4x - 3 - x

2

=

Ú

1 - ( x - 2 )2

dx 2

(c)

1 - ( x - 2)

Solution: Put x + 1 = t3, we have 2 dx Ú 1 + 3 x + 1 = Ú 31t + dtt = 3 Ú ÊÁË t - 1 + 1 +1 t ˆ˜¯ dt

Ú x 2 + 1 dx

Èt2 ˘ = 3 Í - t + log 1 + t ˙ + C Î2 ˚

x3 (a) + x + tan–1 x + C 3 (b)

x3 – x + tan–1 + C 3

1+ 3 x +1 + C Example 7: 2 x Ê 1- x ˆ e Ú ÁË 1 + x 2 ˜¯ dx is

x3 (d) – x – tan–1 x + C 3 Ans. (b)

(a) ex(1 + x2)2

Solution: =x – 1 +

x4

Ú x 2 + 1 dx

=

(c)

x2 + 1

Ú

2 x - sin -1 x 1- x

2

dx = C – 2 1 - x 2 –

1 - x2 g( x )

C - 2 1 - x2 -

=

-

sin

-1

= x

1 - x2

=

–1

2 (sin -1 x )3 / 2 . Thus f(x) = (sin–1 x)3 3

Example 6: If

dx

Ú1+ 3 x +1

=

3 ( x + 1)2 / 3 - 3 ( x + 1)1 / 3 2

x

(a) (b)

1+ 3 x +1

(d)

2x

x

Ú 1 + x 2 dx - Ú e ( 2 )2 dx 1+ x ex 1+ x

2

ex 1 + x2

Example 8:

(c)

+ f(x) + C then f(x 1+ 3 x +1

.

where g(x) = (1 – x2) h( x ) h¢(x), h(x) = sin

and other part is of the form

1 + x2

2

ex

2x - g ¢( x )

(1 + x 2 )2

Ê1 - xˆ so Ú e Á dx Ë 1 + x ˜¯

(b) 2sin x (d) 3(sin–1x)3

Solution:

ex

(d)

x

–1

(a) sin x (c) (sin–1x)3 Ans. (c)

e x (1 - x )

(1 + x 2 )2

2 2x ˘ x Ê 1- x ˆ È 1 Solution: e Á = ex Í 2 Ë 1 + x 2 ˜¯ x + 1 (1 + x 2 )2 ˙˚ Î

f ( x ) then f(x –1

- xe x

(b)

Ans. (d)

x3 – x + tan–1 x + C 3

Example 5: If 2 3

1

2

x2 + 1 so

3 (x + 1)2/3 – 3(x + 1)1/3 + 2

=

x2 – x + 2tan–1 x + C (c) 2

x4

1 log 1 + 3 x + 1 3

(d)

Ans. (b)

= sin–1 (x – 2) + C

x4

Example 4:

2 log 1 + 3 x + 1 3

Ú

+ Ú ex

2x 2 2

(1 + x )

dx - Ú e x

2x

(1 + x 2 )2

dx + C

+C 3 x +1 - 7 x -1 x

dx = K1 3–x + K27–x + C

21 1 1 K1 = , K2 = 7 log 3 3 log 7 1 1 K1 = , K2 = – 7 log 3 3 log 7 3 1 K1 = , K2 = 7 log 3 log 7 3 -7 k1 = , k2 = log 7 log 3

12.11

Ans. (c)

1 1 [cos 4x + cos 6x] + [cos 2x + cos 8x] 4 4 1 1 1 = cos 2x + cos 4x + cos 6x 4 4 4 1 cos 8x + 4 1 Ú cos x cos 2x cos 5x dx = sin 2x + 8 1 1 1 sin 4x + sin 6x + sin x + C 16 24 32 =

Solution:

Ú

3 x +1 - 7 x -1 21

=– 3 K1 =

x

dx = 3Ú 7- x dx -

1 -x 3 dx 7Ú

7- x 1 3- x + +C log 7 7 log 3

3 1 , K2 = 7 log 3 log 7

Example 9: If Ú cos x cos 2x cos 5x dx = A1 sin 2x + A2 sin 4x + A3 sin 6x + A4 sin 8x + C then 1 1 (a) A1 = , A2 = 2 4 1 1 (c) A2 = , A3 = 16 8

1 (b) A1 = , A2 = 4 1 (d) A1 = , A4 = 8

Example 10: If is

1 8 1 32

dx

Ú 1 + ex

(a) 1 (c) 2 Ans. (b)

(b) –1 (d) –2

Solution: dx

Ans. (d)

Ú 1 + ex

Solution: (cos x cos 2x) cos 5x 1 = (cos x + cos 3x) cos 5x 2

ex) + C then K

=x

ex = t, we have 1 dt dt =Ú =Ú dt - Ú t t (t - 1) t -1 t t+C ex e x) + C x e)+C =x

LEVEL 1 Straight Objective Type Questions Example 11: Let f(x) = 1 + 3x F(x) be its antiderivative which assume the value 7 for x = 2. The value of x for which the curve F(x) cuts the abscissa axis is (a) 0 (b) 1 (c) 2 (d) 3 Ans. (b) Solution: F(x) = 7 so 7 = 2 + 9 + C

Ú f (x) dx + C = x + 3

x

+ C but F(2) =

fi C = – 4. Hence F (x) = x + 3x – 4. Now F (x) = 0 fi x = 1. 1 Example 12: If f(x) = then its anti2 cos x 1 + tan x F(0) = 4) is 2 1 + tan x + 4 (b) (1+ tan x)3/2 3

Solution: F(x) =

=

(c) 2 Ans. (c)

(

)

1 + tan x + 1

(d) none of these

Ú

dx

+C

1 + tan x 1 1+ t

dt + C (t = tan x)

= 2( 1 + t ) + C = 2 1 + tan x + C Since 4 = F(0) = 2 + C fi C = 2. Hence F (x) = 2

(

)

1 + tan x + 1 .

Example 13: Let f (x) =

derivative F(x (a)

Ú cos2 x

1 2

4 - 3 cos x + 5 sin 2 x

and if

its antiderivative F (x) = (1/3) tan–1 (g(x)) + C then g (x) is (a) 3 tan x (c) 2 tan x Ans. (a)

( )

(b) 2 tan x (d) none of these

12.12

Solution: F (x) =

dx

Ú 4 - 3 cos2 x + 5 sin 2 x

(a) tan–1 (x/2)

+C

sec2 x d x

Ú 4 (1 + tan 2 x ) - 3 + 5 tan 2 x

=

dt = Ú 2 +C 9t +1

(t = tan x)

1

(c)

+C

(

2 x +1

( x - 1)2

Solution:

2

( x2 + 1)

=

Therefore g (x) = 3 tan x.

tan x and F(x) is its antisin x cos x

=

derivative, if F(p/4) = 6 then F(x

( 2 (

) tan x + 2 ) tan x + 1

(a) 2 (c)

(

(b) 2

tan x + 3

)

so

Ú

( x - 1)2

(x

2

)

+1

2

dx =

(d) none of these

=

Ú

1

tan x

Ú sin x cos x dx

=

Ú

tan x sec2 x dx tan x (t = tan x)

d t + C = 2 t + C = 2 tan x + C

t 6 = F (p/4) = 2 + C so C = 4. Hence

Since

F (x) = 2

(

tan x + 2

) dx

Ú x 4 + 5x2 + 4

Example 15: The value

is

(a) (1/3) tan–1 x – (1/2) tan–1 (x/2) + C –1

–1

(b) 3 tan (x/3) + 2 tan (x/2) + C (c) (1/3) tan–1 x + (1/6) tan–1 (x/2) + C (d) (1/3) tan–1 x – (1/6) tan–1 (x/2) + C Ans. (d)

2

( x 2 + 1) x2 + 1

g (x) =

Therefore,

= Therefore, I =

dx

Ú x 4 + 5x2 + 4 1È 1

=Ú

1

dx

( x2 + 1) ( x 2 + 4)

˘

Ú 3 ÍÎ x 2 + 1 - x 2 + 4 ˙˚ d x x 1 -1 1 tan x - tan -1 + C . 3 6 2 2

Example 16: If

Ú

( x - 1)

(x

2

2

)

+1

d x = tan–1 x + g (x) + K then

2x

-

2

2

( x2 + 1) ( x 2 + 1) 1 2

x +1

-

2x

(x

+1

dx

1 2

x +1

2

)

2

Ú x2 + 1 - Ú

2x

(x

2

dx

2

)

+1

1

+K

2

x +1

.

Example 17: If the antiderivative of f(x) = sin x is 1 3 tan–1 1 3 g ( x ) + C then 2 2 sin x + 4 cos x g (x (a) sec x (b) tan x (c) sin x (d) cos x Ans. (a)

(

Solution: f (x) =

((

)

)

)

sin x 2

2

=

tan x sec x

tan 2 x + 4 tan x sec x . = sec 2 x + 3 x = t, d x sec x tan x = dt so sin x + 4 cos x

dt

Solution: I =

g(x

x2 - 2 x + 1

=

= tan–1 x +

Ans. (c) Solution: F (x) =

x +1

Ans. (b)

1 = tan–1 3t + C 3

Example 14: Let f(x) =

2

(d) none of these

)

2

1

(b)

Ú f ( x ) dx = Ú t 2 + 3 Example 18: If

=

1 3

tan–1

x

Ú x 2 - 4 x + 8 dx = K

sec x 3

+ C.

x2 – 4x + 8) +

x - 2ˆ + C then the value of K is tan–1 ÊÁ Ë 2 ˜¯ (a) 1/2 (c) 2 Ans. (a)

(b) 1 (d) none of these

Solution: x2 – 4x + 8 = (x – 2)2 + 4 so put x – 2 = u

12.13

x

Ú x 2 - 4 x + 8 dx

u+2

Ê 1 - cos 2 2 x ˆ sin 2x dx cos 2 x ˜¯

=

Ú u2 + 4 du

=

1 2u du du + 2 Ú 2 Ú 2 2 u +4 u +4

=

1 2

u2 + 4) + tan–1

=

1 2

x2 – 4x + 8) + tan–1

= -ÚÁ Ë

Úx

Example 19: by (a)

(b)

1 log 3

1 - x3 + 1

1 log 3

1 - x3 - 1

2 log (c) 3

(d)

1 - x3 - 1

2

u +C 2 x-2 +C. 2

1 - x3

tan Úe

Example 21:

1 - x3

etan

(a) +C

-1

1 + x2

(c) x e tan Ans. (c)

+C

-1

x

+C -1

Solution: tan Úe

+C

-1

x

= du

1 + x2

Ê 1 + x + x2 ˆ u 2 ÁË 1 + x 2 ˜¯ dx = Ú e (1 + tan u + tan u) du

Úe

(

)

2 Ê1 t -1ˆ log +C Á 3 Ë2 t + 1 ˜¯

1 1 - x3 - 1 + C. = log 3 1 - x3 + 1

1 sin 2x + C (c) 2 Ans. (d)

dx

= tan u eu + C = x etan

1 2t d t 2 dt 1 - 3x2 d x = Ú 2 = - Ú Ú 2 3 3 3 1- t t 3 t -1 3 x 1- x

1 (a) cos 4x + C 2

x = u, we have

= Ú eu (sec2 u + tan u) du

Solution: Put 1 - x3 = t2. Then - 3x2dx = 2t dt and the

Example 20:

Ê 1 + x + x2 ˆ ÁË 1 + x 2 ˜¯ dx is

tan x (b) x e +C 1 + x2 (d) none of these

+C

Ans. (b)

=

x

-1

x

1 log 1 - x 3 + C 3

-

-1

dx

1- x +1 1

1 1- t 2 t2 ˆ 1Ê = +C d t log t 2Ú t 2 ÁË 2 ˜¯ 1 1 x - cos2 2x + C. = 2 4 =

cos 4 x - 1

Ú cot x - tan x 1 (b) cos 4x + C 4

Example 22:

Ê - 2 sin 2 2 x ˆ Ú ÁË cos2 x - sin2 x ˜¯ cos x sin x d x

x

+ C.

x

( f ( x ) + f ¢( x ) dx = e f ( x ) + C )

dx

Ú

= fog (x) + C then

2 - 3 x - x2

(a) f (x) = sin-1 x, g(x) =

2x - 3

(b) f (x) = tan–1 x, g(x) =

2x + 3

(c) f (x) = sin-1 x, g(x) =

2x + 3

17 17 17

(d) none of these Ans. (c) Solution:

Ú

dx 2 - 3x - x 2 =

Ú

(d) none of these

Solution: Put cos 2x = t. Then -2 sin 2x d x = dt and the

x

-1

=

Ú

= sin-1 Therefore, g (x) =

dx

=Ú

(

- x 2 + 3x - 2

)

dx

(

2

- ( x + 3 / 2 ) - 17 / 4

)

dx 2

17 / 4 - ( x + 3 / 2 )

x + 3/ 2 17 / 2 2x + 3 17

+ C = sin-1

2x + 3 17

+C

and f (x) = sin-1 x.

12.14

dx

Úx

Example 23: If

= K tan-1 f (x) + C then

5 x2 - 3

(a) f (x) =

5 2 1 x - 1, K = 3 5

(b) f (x) =

5 2 1 x - 1, K = 3 3

=

Hence K = – 1/8.

x2 - 3 = t2 so 5x dx = t dt 1 5t d t dx x dx Ú x 5x2 - 3 = Ú x2 5x2 - 3 = 5 Ú t 2 + 3 t

(

1

=

3

tan - 1

1 3

t

tan - 1

3

Ú

x + cos x) + 1

2 (d) none of these Ans. (b)

Ú

Solution: sin x - cos x 2

)

(sin x + cos x ) - 3 2

sin 2 x - 1 2 sin 2x

d x = -Ú

Ú

e x [1 + 2 sin ( x / 2 ) cos ( x / 2 )]

dx 2 cos2 ( x / 2 ) x xˆ Ê1 = Ú e x Á sec 2 + tan ˜ d x Ë2 2 2¯ x x 1 = Ú e x sec 2 d x + Ú e x tan d x 2 2 2 x x 1 x 1 = Ú e x sec 2 dx + e x tan - Ú e x sec 2 + C 2 2 2 2 2

d x is

C

sin 2 x - 1 2

= e x tan C

Example 27: If

sin 2 x - 1 2

x + cos x) +

e x (1 + sin x ) Ú 1 + cos x dx =

+C

sin x - cos x

x + cos x) +

(c) –

Solution:

5x2 - 3 + C. 3

Example 24: The value of

C

dt 2

t -3 2

Ú

(a) f (x) =

2 sin ( x 2)

(b) f (x) =

2 cos (x /2), K = 2

(c) f (x) =

2 tan (x/2), K = 2

(d) f (x) =

2 sin (x/2), K =

sin 2 x - 1 2

Solution:

=

Ú

1 + cos x 2 cos x 2 = so cos x 1 - 2 sin 2 x 2

1 + sec x dx =

= 2 sin–1

(b) K = – 1/8 (d) none of these

2 cos 2 2 x Ú cos2 x - sin 2 x ◊ sin x cos x dx

1 + sec x =

x/2 = t, we have

C

cos 4 x + 1 Example 25: If Ú = K cos 4x + C, then cot x - tan x

cos 4 x + 1 Solution: Ú dx cot x - tan x

2

Ans. (a)

t + t2 - 3 2 + C x + cos x +

x +C 2

1 + sec x dx = K sin–1 (f (x)) + C then

(t = sin x + cos x)

(a) K = –1/2 (c) K = –1/5 Ans. (b)

1 cos 4x + C 8

e x (1 + sin x ) Ú 1 + cos x dx x C (b) ex tan x/2 + C x+C (c) ex cot x + C Ans. (b)

Solution:

dt

Ú sin 4x d x

Example 26:

1 1 5 x 2 - 3, K = 2 5 (d) none of these Ans. (b)

Ú t2 + 3 =

1 2

= -

(c) f (x) =

=

Ú cos 2x sin 2x d x =

Ú

2 cos x 2 1 - 2 sin 2 x 2

dx

2t + C (t = sin x/2)

–1

= 2 sin ( 2 sin x/2) + C so

f (x) = 2 sin x/2 and K = 2 Example 28: The antiderivative of f (x x)–2 e, e) is –1 x x) ) (a) x (b) x x x)–1) + e

x) +

12.15

x)–1) + 2e

(c) x x (d) none of these Ans. (c)

=

=x

Ú

x

Ú x log x dx + Ú (log x)

x) –

–1

x) – [x

Ú

x) +

–2

-2

dx + C

x) dx] +

Solution: f (x) =

(d) none of these Ans. (c)

=

Ú

sec 2 x d x

dt

Ú a2 + b2 tan2 x = Ú a2 + b2 t 2 dt

(

1 b Êb ˆ = 2 ◊ tan - 1 Á t ˜ + C Ëa ¯ b a

b2 t 2 + a 2 / b2

)

b 1 tan - 1 ÊÁ tan Ëa ab

xˆ˜ + C. ¯

Example 30: If

Úx

x + 1/x) dx = f (x

x + 1) +

Solution:

=

Ê x + 1ˆ x log Á dx = Úx Ë x ˜¯

x2 2 x2 2 1 – Ú 2

(x + 1) –

= -

1 (- 2 x ) 1 - x 2 dx Ú 2

= -

1 (1 – x2)3/2 + C 3

x + 1) dx – Ú x

x dx

1 x2 x2 1 x2 d log dx x x + 2 Ú x +1 2 2Ú x

x2 x 2 1 ˆ 1 Ê ÁË x - 1 + x + 1˜¯ d x + 2 Ú x d x

(b) - 2 2 cos (a/2) + C (c) 2 cos (a/2) + C (d) - 2 cos (a/2) + C Ans. (b) Solution:

sin a

Ú

da = - Ú

dx

(x = 1 + cos a)

1 + cos a x = – 2 (1 + cos a)1/2 + C = - 2 2 cos a /2 + C.

g(x) x2+ Lx + C, then (a) f (x) = (1/2) (x2– 1) (b) g (x x (c) L = 1 (d) none of these Ans. (a)

=

x 1 - x2 d x

(a) 2 2 cos (a/2) + C

t = tan x

=

Ú

So 7/3 = – 1/3 + C fi C = 8/3. Therefore, f (x) = – (1/3) [(1 – x2)3/2 – 8]. sin a Example 32: The value of Ú da is 1 + cos a

(b) tan-1 ((b/a) tan x) + C 1 Ê Ê bˆ ˆ (c) tan -1 Á Á ˜ tan x ˜ + C Ë Ë a¯ ¯ ab

1

1 . 2

(d) f (x) = – (2/3) [(1 - x2)3/2- 8] Ans. (c)

(a) sin-1 ((a/b) tan x) + C

=

x and L =

(c) f (x) = – (1/3) [(1 - x2)3/2 - 8]

Ú a2 cos2 x + b2 sin2 x (a, b

dx

x + 1) + C

(0, 7/3) and whose derivative is x 1 - x 2

dx

Ú a2 cos2 x + b2 sin2 x

x 1 – 2 2

(b) f (x) = (3/2) [sin-1 x + 6]

–2

x) ) x) – x x)–1 + C =x x = e, we have e = 0 – e + C so C = 2e x x)–1) + 2e. Thus F (x) = x

Solution:

x+

(a) f (x) = (1/3) ((1 - x2)3/2 + 7)

x) dx + C

Example 29:

x2 2

x2 1 1 – , g (x) = – 2 2 2 Example 31: The function f

–1

Ú

x + 1) –

fi f (x) =

Solution: An antiderivative of f (x) = F(x) = x x)–2) dx + C Ú =x

x2 2

Example 33: The value of

Ú

dx is 5 + 4 cos x

(a)

1 -1 Ê 1 tan Á tan xˆ˜ + C ¯ Ë3 3

(b)

x 1 -1 Ê 1 tan Á tan ÊÁ ˆ˜ ˆ˜ + C Ë Ë 3 3 2¯ ¯

2 -1 Ê tan Á Ë 3 2 -1 Ê (d) tan Á Ë 3 Ans. (d) (c)

x + 1) –

Solution:

Ú

1 ˆ tan x ˜ + C ¯ 3 1 Ê xˆ ˆ tan Á ˜ ˜ + C Ë 2¯ ¯ 3

2 dt dx =Ú 2 5 + 4 cos x 5 1 + t + 4 1 - t2

(

)

(

)

(t = tan (x/2))

12.16

dt

= 2Ú

9+t

Example 34:

2

Ú

=

2 tan–1 3

1/ 2

Ê1 Ê xˆˆ ÁË tan ÁË ˜¯ ˜¯ + C 3 2

1 È x-a˘ b - a ÍÎ b - a ˙˚ (d) none of these Ans. (d)

dx x ( x 4 + 1)

dx

Ú x ( x 4 + 1)

=

Ú

b - a - ( x - a) +C x-a

=

2 a-b

b-x + C. x-a

sin 3 x log ( 1 + 3 x )

(tan - 1 x ) (e 2

x/2 + p/8) + C

1

Ê x +pˆ ˜ ÁË 2 8¯

Ê x pˆ ÁË - ˜¯ + C 4 8

= = = =

1

Ú

Ú 2

1 1

2 1 2

1

=

2

Ú

sec (x – p/4) dx

(b) 3/5 (d) none of these

xÆ0

sin 3 x log (1 + 3 x ) 2 (tan- 1 x ) (e

3

x

- 1)

3/ 2

( x - a)

(b - x )1/ 2

is

(a) t = cos x (c) t = 2 cos x Ans. (a) Solution:

Ú

+C

3 sin x + sin3 x

(b) t = tan x/2 (d) t = sin x dx 3

3 sin x + sin x =

1/ 2

1

1 t -1 1 2+t | + C then log | |+ log | 6 t + 1 12 2-t

+C

3 Èb-x ˘ 4 (b - a) ÍÎ x - a ˙˚

xÆ0

Example 38: If the primitive of f(x) = is

1

1/ 2

(b)

(x π 0) then lim f ¢(x) is

3 3 Ê x ˆ x = lim 3 sin x ¥ log (1 + 3 x ) ¥ 3 Á ˜ 3 1 x xÆ0 3x Ë tan x¯ e -1 x 2 = 3(1) (1) (1) (1) = 3

Êp + x - pˆ C ˜ ÁË 4 2 8¯ Ê p + xˆ C. ˜ ÁË 8 2¯

1 Èb-x ˘ b - a ÍÎ x - a ˙˚

)

-1

2

Example 36: The primitive of

(a)

x

xÆ0

dx sin (p / 4) sin x + cos (p / 4) cos x dx

3

Solution: lim f ¢ ( x ) = lim

dx sin x + cos x

Ú 2 cos ( x - p / 4 )

+ C.

b-a

(a) 0 (c) 5/3 Ans. (d)

C

Ans. (c) Solution:

(b - a ) t - 1

Example 37: If f (x) is the primitive of

2

2

(b - a ) t - 1

-2 b-a

p/4 + x/8) + C

(d)

dt

=

dx sin x + cos x

2

=-Ú

t

)

1 Ê1 1 ˆ x4 1 dt = log 4 + C. ˜ Á Ú 4 Ë t t + 1¯ 4 x +1

(c)

1 (1/ t )3 / 2 ÊÁË b - a - ˆ˜¯

1/ 2

(t = x4)

(

=

(b)

1 dt t2

= -2

1 dt Ú 4 t (t + 1)

Ú

-

1 4 x3 d x 4 Ú x4 x4 + 1

=

Example 35:

x – a = (1/t), we have dx = (–1/t2) dt

Solution:

Ê x4 ˆ 1 x4 + 1 1 +C log Á 4 log 4 + C (a) (b) 4 Ë x + 1 ˜¯ 4 x 1 (c) x4 + 1) + C (d) none of these 4 Ans. (b) Solution:

+C

(c)

=Ú

sin x d x

(3 + sin2 x ) sin2 x

sin x d x

Ú (4 - cos2 x ) (1 - cos2 x )

12.17

= -Ú

dt

(

4-t

2

)(

1- t

2

Example 41: If

)

Ê 1 1 ˆ =-1 - 2 Ú Á 2 3 Ë t - 4 t - 1˜¯

Example 39: The value of

Ú

cos x dx is sin x + cos x

x 2 1 (b) (x 2

x + cos x

x + cos x (d) none of these Ans. (b) Solution:

Ú

C

x + cos x

Ú

C

x+C

=

cos x dx sin x + cos x

=

=

1 cos x + sin x + cos x - sin x dx 2Ú sin x + cos x

=

1 2Ú

=

cos x - sin x ˆ Ê ÁË 1 + sin x + cos x ˜¯ dx

|g (x C then (a) dom f = R (c) dom f = R ~ {0} Ans. (c)

C.

(e

x

1 + 1/ t 2

Ú

2

t + 1/ t

1 2 1

2

tan - 1 tan–1

2

f (x) =

dt =

Ú

dx =

Ú

1 + t2 d t (t = tan x) 1 + t4

1 + 1/t 2

dt

2

Ê t - 1ˆ + 2 ˜ ÁË t¯

1 Ê 1ˆ Á t - ˜¯ + C t 2Ë tan x - cot x +C 2 tan x - cot x 2 sin 2 x

Ú

cos6 x

p so f ÊÁ ˆ˜ = 0. Ë 4¯

dx is a x x x x

is

2

)

-1

)

tan 4 x + 1

Example 42: The

1

Example 40: If the primitive of

(

sec 2 x 1 + tan 2 x

Thus

x + cos x

= (1/2) (x

f (x

Solution:

Ú

1

(e

x

(b) g (x) = 1 – ex (d) f (x) = 1 – e–x x

2

)

-1

dx = Ú

e dx e

2

(e -1)

x

x

=Ú

Ans. (c) Solution: dt 2

t (t - 1)

(t = e ) =

=

Ú Ú

Ê 1 1 1ˆ + ˜ dt Á 2 t -1 t¯ Ë (t - 1)

1 = 1- t

t

= (1 – ex )– 1

C

t3 t5 tan 5 x tan3 x +C + +C= + 3 5 5 3 x.

7

Example 43: If Ú tan x dx = f (x) + C then (a) f (x (b) f (x (c) f(x) =

|t| + C e– x

x=t

written as Út2 (1 + t2) dt = x

1 È 1 1˘ - ˙ dt Í t - 1 Ît - 1 t ˚

x x

1 1 1 6 tan x - tan4x + tan2x 4 2 6

x +C x

(d) f (x Ans. (c) x = t, we have dx =

Solution:

Hence f (x) = (1 – ex )–1 and g (x) = 1 – e–x The domain of f = R ~ {0}

tan-1 f (x) +

2

Solution: cos4 x

2

(a)

1

=

C then (a) f (x) = tan x - cot x (b) f (p/4) = 0 (c) f (x) is continuous on R 1 (d) f (x) = (tan x - cot x) 2 Ans. (b)

(t = cos x)

1 t -1 1 2+t |+C. log | |+ log | 6 t + 1 12 2-t

=

dx

Ú sin 4 x + cos4 x

Ú

tan 7 x dx =

t7

Ú 1 + t2

dt

dt 1 + t2

, so

12.18

Ê

Ú ÁË t

=

5

- t3 + t -

ˆ dt 2˜ 1+ t ¯ t

Example 47: I =

t6 t 4 t2 1 t 2) + C - + 6 4 2 2 1 1 1 tan6 x – tan4 x + tan2 x = 6 4 2 1 2 – x+C 2 1 1 1 = tan6 x – tan4 x + tan2 x 6 4 2 x

=

(c) f (x) = sec2 x

2 x4 - 2 x2 + 1

(b)

x3

Ú

C.

2 x4 - 2 x2 + 1

(d)

=-

x

2 x2

Solution: I =

x

2-

Put

2 x

2

-

1 x

4

= t2 I=

C

C. Thus f (x) = tan x.

=

Example 45: If polynomials P and Q

Ú ((3x - 1) cos x + (1 - 2x) sin x) dx

=

= P cos x + Q sin x (a) P = 3x - 2 (c) P = 3(x - 1) Ans. (d)

(b) Q = 2 + x (d) Q = 3(x - 1)

+C

((1/ x3 ) - (1/ x5 )) dx Ú 2 - 2 / x 2 - 1/ x 4 ( ) ( )

\

x

+C

Ans. (d)

(d) none of these

= Ú (tan x + cot x) dx

+C

2 x4 - 2 x2 + 1 +C x

(c)

sin 2 x + cos2 x dx =Ú dx sin x cos x sin x cos x

dx

2 x4 - 2 x2 + 1

x2

Ans. (b) Solution:

Ú x3

2 x4 - 2 x2 + 1

(a)

dx Example 44: If Ú f (x C then sin x cos x (a) f (x) = sin x + cos x (b) f (x) = tan x

x2 - 1

Example 48: Let I =

4ˆ Ê 4 ÁË 3 + 5 ˜¯ dx = 2t dt x x

fi

1 t dt 1 = t+C 2Ú t 2 1 2

2-

2 x

2

-

1 x4

2 x4 - 2 x2 - 1 2 x2

+C

+C

ex

Ú e4 x + e2 x + 1 dx , J =

e- x

Ú e- 4 x + e- 2 x + 1 dx . Then for an arbitrary constant C, the

Solution: (3x - 1) sin x - 3 Ú sin x dx - (1 - 2x) cos x - 2 Ú cos x dx = (3x - 1) sin x + 3 cos x - (1 - 2x) cos x - 2 sin x = (3x - 3) sin x + (2 + 2x) cos x Hence Q = 3(x -1) and P = 2(1 + x)

value of I – J (a)

Ê e 4 x - e 2 x + 1ˆ 1 log Á 4 x +C 2 Ë e + e2 x + 1˜¯

(b)

Ê e 2 x + e x + 1ˆ 1 log Á 2 x +C 2 Ë e - e x + 1˜¯

(c)

Ê e 2 x - e x + 1ˆ 1 log Á 2 x +C 2 Ë e + e x + 1˜¯

(d)

Ê e4 x + e2 x + 1ˆ 1 +C log Á 4 x 2 Ë e - e2 x + 1˜¯

2

Example 46: Let f(x) = Ú e x ( x - 2 ) ( x - 3) ( x - 4 ) dx then f increases on (a) (– •, 2) (b) (2, •) (c) (2, 3) (d) (3, •) Ans. (c) 2

Solution: f ¢(x) = e x ( x - 2 ) ( x - 3) ( x - 4 )

Ans. (b)

x2

As e > 0 for all x Œ R, f ¢(x) > 0 if (x – 2) (x – 3) (x – 4) > 0 i.e., if 2 < x < 3 or x > 4 Thus f (x) increases on the interval (2, 3)

Solution: I – J =

ex

Ú e 4 x + e2 x + 1

dx -

e3 x

Ú e4 x + e2 x + 1 dx

12.19

(

e x 1 - e2 x

Put

)

Solution:

Ú e4 x + e2 x + 1 dx

=

Ú

x)dx 1

x) –

Ú x log x . x dx

= x

x) –

Ú log x

= x

x) – x

ex = t, ex dx = dt

1 -1 t2 dt I–J= Ú 4 = Ú 2 1 dt t + t2 + 1 t + 2 +1 t 1 1- 2 du t = – Ú = Ú , 2 1 - u2 Ê t + 1ˆ - 1 ˜ ÁË t¯

(1 - t 2 )

=

1 1+ u log +C 2 1- u

1 e x + e- x + 1 = log +C 2 e x + e- x - 1 2x

=

fi

È

1

˚

x) g(x) =

(a) e x + 2 log (b) -

sin x 2Ú dx is sin ( x - p / 4 )

(c)

(a) x + log cos ( x - p / 4 ) + C (b) x - log sin ( x - p / 4 ) + C

Solution: Put x – p/4 = t sin (p / 4 + t ) I= 2Ú dt = Ú (cot t + 1) dt sin t x–p

C, C = C¢ + p/4.

1

x 8 log 1 - x + + c 3 3

Solution: Put x = Now

Ú f ( x ) dx

(b) f(x

1 x); g(x) = log x x; g(x) =

1 log x

1 ; g(x x) log x 1 1 (d) f(x) = ; g (x) = x log x log x Ans. (a) (c) f(x) =

x2 - 2x + c 2

3t - 4 8 24 dt =1, dx = 3t + 4 3t + 4 (3t + 4 )2 24 Ê 3t - 4 ˆ dt ÁË 3t + 4 ˜¯ (3t + 4 )2

=

Úf

=

Ú (3t + 4)2 (24) dt

˘

Ú Ílog (log x ) + (log x )2 ˙ dx

Î ˚ = x[f(x) – g(x)] + C, then : (a) f(x

is:

8 2 log 1 - x + x + c 3 3

= t + log sin t + C ¢

È

Ú f ( x ) dx

Ans. (b)

Ans. (c)

Example 50: If

1 log x

3x - 4 +c 3x + 4

(d) e[(3 x - 4) (3 x + 4)] -

(c) x + log sin ( x - p / 4 ) + C

= x

˘

Ê 3x - 4 ˆ Example 51: If f Á = x + 2, then Ë 3x + 4 ˜¯

e + e +1 1 + C. log 2 x 2 e - ex + 1

(d) x - log cos ( x - p / 4 ) + C

1 (-1) 1 + x dx log x Ú (log x )2 x

Î

x

Example 49: The value of I =

dx

Ú Ílog (log x ) + (log x )2 ˙ dx = x[f(x) – g(x)] + C

where f(x) and

1 t

u = t+

1

= x

t+2

= 8Ú

3t + 4 + 2

(3t + 4)2

dt

È

˘ 1 2 + dt 2˙ Î 3t + 4 (3t + 4 ) ˚

= 8ÚÍ =

8 2 8 log 3t + 4 + C1 3 3 3t + 4

=

8 8 2 log - (1 - x ) + C1 3 1- x 3

= -

2 8 log 1 - x + x + C 3 3

12.20

Ï -1 1 - x ¸ Ú cos ÌÓ2 tan 1 + x ˝˛ dx

Example 52: I = 1 2 x +C 2 1 x+C (c) 2 Ans. (a)

1 2 x -1 + C 8 1 2 (d) x +1 + C 4

(a)

(b)

)

(

)

fi

(

)

5 sin x

(sin x - 2 cos x ) + 2 (cos x + 2 sin x )

dx

sin x - 2 cos x

=x a=2

\

= - Ú cos q sin q dq =

Ú

I =

dx = – sinq dq

-1 Ú cos 2 tan tan q / 2 (- sin q ) dq

5 tan x

Ú tan x - 2 dx = Ú sin - 2 cos x dx

Let 5 sin x = A (sin x – 2 cos x) + B (cos x + 2 sin x) x and cos x 5 = A + 2 B and 0 = –2A + B fi A = 1, B = 2 \

Solution: Put x = cosq I =

(

Solution: I =

x –2cos x

K

Example 55:

x2 cos2 q +C= +C. 2 2

I=

Example 53: Let f be thrice differential function and if u = – f ¢¢(q) sin q + f ¢ (q) cos q and v = f ¢¢(q) cos q +

Ú

sin10 x - cos8 x sin 2 x + sin8 x cos2 x - cos10 x 1 - 2 sin 2 x cos2 x 1 sin 2x + C 2

(a)

1/ 2

ÈÊ du ˆ 2 Ê dv ˆ 2 ˘ f ¢ (q) sin q then I = Ú ÍÁ ˜ + Á ˜ ˙ Ë dq ¯ ˚ ÎË dq ¯

dq

(c) -

(b) -

1 sin x + C 2

dx is

1 sin 2x + C 2

(d) – sin2 x + C

Ans. (b) (a) (b) (c) (d)

f (q) – f ¢¢ (q) + C f (q) + f ¢¢(q) + C f ¢ (q) + f ¢¢(q) + C f ¢ (q) – f ¢¢(q) + C

Solution: sin10x – cos8x sin2x + sin8 x cos2x – cos10x = sin2x (sin8 x – cos8x) + cos2x (sin8x – cos8x) = (sin2x + cos2x) (sin8x – cos8x)

Ans. (b)

= (sin2x – cos2x) (sin2x + cos2x) (sin4x + cos4x)

Solution:

du =– f ¢¢(q) cos q – f ¢¢¢ (q) sin q +f ¢¢ (q) dq cos q – f ¢ (q) sin q

= (sin2x – cos2x) (1 – 2sin2x cos2x)

= – ( f ¢¢¢(q) + f ¢ (q)) sin q

dn = f ¢¢¢ (q) cos q – f ¢¢ (q) sin q + f ¢¢ (q) dq sin q + f ¢ (q) cos q = (f ¢¢¢(q) + f ¢ (q)) cos q 2 2 Ê du ˆ Ê dv ˆ 2 ÁË ˜¯ + ÁË ˜¯ = (f ¢¢¢ (q) + f ¢ (q)) dq dq

I=

Ú ( f ¢¢¢ (q ) + f ¢ (q )) dq

I=

= – cos 2x (1 – 2sin2x cos2x) sin 2 x I = – Ú cos 2x dx = – +C 2 Example 56: sin 2 x cos2 x

Ú(

(a)

(b)

= f ¢¢(q) + f (q)) +C

5 tan x

Ú tan x - 2 dx = x +

Example 54: x– 2 cos x

a

(a) –2 (c) 2

Ans. (c)

2

sin 5 x + cos3 x sin 2 x + sin3 x cos2 x + cos5 x ) 1 3 (1 - tan 2 x ) 1 3 (1 + tan3 x ) 1

(c)

3 (2 + tan3 x )

(d)

-

K, then a (b) 1 (d) –1

Ans. (d)

+C +C +C

1 3 (1 + tan3 x )

+C

is

12.21

Solution: 2

So

2

sin x cos x

Ú(

=

Ú(

(sin5 x + cos3 x sin2 x + sin3 x cos2 x + cos5 x )2 =

=

=

tan 2 x sec 2 x

I=

sin 2 x cos2 x

(sin3 x + cos3 x )2 (sin2 x + cos2 x )2

2

1 + tan3 x ) t2 2

1 + t3 )

dx

(t = tan x )

dt

1 du (u = 1 + t 3 ) 3 Ú u2 1 1 +C ==+ C. 3u 3 (1 + tan3 x ) =

sin 2 x cos2 x

(sin3 x + cos3 x )2 tan 2 x sec 2 x

(1 + tan3 x )2

Assertion-Reason Type Questions

Example 57: Let f (x) =

xe x

(1 + x )2

I2 = – Ú

,xπ–1

Statement-1: The antiderivative F of f (x is

x) + constant.

F(0) = 1

dt 2dt = 2 Ú 3 + t2 Ê 1 – t2 ˆ 1 + t2 Á 2 + Ë 1 + t 2 ˜¯ (t = tan (x/2)) 2 Ê 1 ˆ tan -1 Á tan ( x / 2 )˜ + C = Ë ¯ 3 3

1 x e 1+ x

I1 =

Statement-2: f (x) is of the form (g(x) + g¢(x))e x, for some g(x) Ans. (a) Solution: F(x) =

xe x

Ú (1 + x )2 dx

Ê 1 1 ˆ x = ÚÁ – ˜ e dx Ë 1 + x (1 + x )2 ¯ =

F(0) = 1

1 x e . 1+ x

2 – sin x 2 + cos x

Statement-1: The antiderivative of f is a periodic function of period p Statement-2: The period of the function x) + A tan –1 (B tan x/2) + C, A, B, C g(x constants is 2p. Ans. (d) Solution: F(x) = 2I1 + I2, I1 =

(

)

x) +

dx

Ú 2 + cos x

and

4

Ê 1 ˆ tan ( x / 2 )˜ + tan –1 Á Ë ¯ 3 3

C which is periodic with period 2p. Example 59: Let f (x) =

C = 0 so F(x) =

Example 58: Let f (x) =

Ú

Hence F(x

1 x e +C 1+ x

fi

sin x dx 2 + cos x

1 1 – sin 4 x

and F be an anti-

derivative of f. 1 1 tan x + tan –1 ( 2 tan x) + C 2 2 2 Statement-2: F is a one-one function of tan x Ans. (b) Statement-1: F(x) =

Solution: f (x) =

So,

F(x) =

1 2

(

2

cos x 1 + sin x

)

1 + t2

Ú 1 + 2t 2 dt (t = tan x )

Ê1 1 1 ˆ = ÚÁ + dt 2 Ë 2 4 t + 1 / 2 ˜¯

=

(

sec 2 x 1 + tan 2 x 2

1 + 2 tan x

)

12.22

= 1 t + 1 2 tan - 1 2 4

(

1 1 tan x + tan - 1 2 2 2 which is a 1 – 1 function of tan x. =

(

1

Example 60: Let f (x) =

ex Example 62: Statement 1: Ú ex sinx dx = (sin x – cos x) 2 +C

)

2 tan x + C

x6 + x 4

)

2 tan x + C

x

e f(x) Ans. (b)

fi

Statement-2: F(x) = tan –1 x +

So

F(x) = – F(1) =

1 x 1

3x

3

(x

+

)

2

+1

=

1 x

4

1

–

x

2

+

1 2

x +1

1 + tan – 1 x + C x

1 1 x 3x3

Example 61: Let F(x) =

Úe

fi

Ê 1 ˆ p FÁ ˜ = . Ë 3¯ 6

– x2 5

x dx

– x2 5

x dx =

fi

1 2

x + 2(a - 1) x + a + 5

is

Solution: The discriminant of x2 + 2(a – 1) x + a + 5 = 0 is 4(a – 1)2 – 4(a + 5) = 4(a2 – 3a dx –1 Ú x 2 + 2(a - 1) x + a + 5 = l tan g(x) + C Since the denominator is not zero for any value of x so f is a continuous function. Example 64: Statement 1: I =

Ú

f ( x ) g ¢ ( x ) - f ¢ ( x ) g( x ) f ( x ) g( x )

1 –t 2 e t dt 2Ú

1 – x2 4 e (x + 2x2 + 2) 2

=C– F(0) = 0

Úe

Example 63: Statement 1: For – 1 < a < 4, dx Ú x 2 + 2(a - 1) x + a + 5 = l g(x C, where l and C are constants

a continuous function Ans. (d)

5 Statement-1: If F (0) = 0, then F(1) = 1 – e – 1 2 Statement-2: F increases on (0, •) Ans. (b) Solution: Put x 2 = t,

ex (sin x – cos x) + C 2

Statement 2: For – 1 < a < 4, f(x)=

2 p + + C so C = 0 3 4

-1 F(x) = tan x +

I=

Statement 2: is also true but is not correct explanation of statement 1.

1 1 – 3 3x x

Ans. (c) 4

2I = ex (sin x – cos x) + C

fi

Ê 1 ˆ p Statement-1: F Á ˜ = Ë 3¯ 6

Solution: f (x) =

Solution: Let I = Ú ex sin x dx = – ex cos x + Ú ex cos x dx = – ex cos x + ex sin x – Ú ex sin x dx + C

and F be a antideriva-

p 2 + 4 3

tive of f such that F(1) =

ex(f(x) + f ¢(x)

Statement 2:

=

g(x

g ( x ) ¸2 1Ï Ìlog ˝ +C f ( x) ˛ 2Ó

Statement 2: Ú(f (x))n f¢(x)dx =

C=1

2 1 Hence F(x) = 1 – e – x (x 4 + 2 x2 + 2) 2

f(x)} dx

(f ( x ))n +1 n +1

+C

Ans. (a) Solution: I =

2

Ê g( x ) ˆ d Ê

Ú log ÁË f ( x) ˜¯ dx ÁË log

F ¢(x) = e – x x5 > 0 x Œ (0, •). = Put f (x) = t

g( x ) ˆ 2 1Ê +C log f ( x ) ˜¯ 2 ÁË

g( x ) ˆ dx f ( x ) ˜¯

12.23

LEVEL 2 Straight Objective Type Questions

Example 65: If

x - 1 (3 x + 2 )

dx

Ú x3 ( x - 1)1 2 =

4 x2

Solution: We want a substitution that will allow us to +

K tan–1 x - 1 + C then the value of K is (a) 1/2 (b) 1 (c) 1/4 (d) 3/4 Ans. (d) Solution: Put x –1 = t 2 so that d x = 2t dt, and dx 2 t dt dt Ú x3 ( x - 1)1 2 = Ú 2 3 = 2Ú 2 3 t +1 t t +1

(

)

Now we use formula of

Ú Ú

dt 3

(t 2 + 1) dt

=

)

4 t2 + 1

+

3

x+ x

2

3 dt and Ú 4 t2 + 1 2

)

1 dt = + Ú 2 2 2 t +1 2 t +1

(

=

)

t

(

+

)

2 t2 + 1

1 tan–1 t 2

(

2

(

2

)

2 t +1 =

1

)

4 t +1 =

2

+

Ú

x + x 4 + 4 x3 + 4 x 2 + 4 x + 4 5

( x 2 + 2)

(b)

3t

(

)

4 t +1

+

3 tan–1t + C 4

(2t + 3t3 + 3t) +

+

2

Example 66: The value of

3 tan–1 t + C 4

4x - 3

+

2

( x2 + 1)

1 3 x x + tan -1 +C 2 8 x +2 2 2

1 2x - 3 3 x 3 x + + tan -1 +C 2 2 12 x 2 + 2 16 x + 2 16 2 2

(

)

2x - 3

(x

2

+2

2

)

+

3 x 1 x + tan -1 +C 8 x2 + 2 2 2 2

(d) none of these Ans. (d) Solution: x5 + x4 + 4x3 + 4x2 + 4x + 4 = x3(x2 + 2) + x2(x2 + 2) + 2x(x2 + 2)

x - 1 (3 x + 2 ) 4x

2

Ú

(a) 3 x + 3( 3 x ) - 6( 6 x )

3 tan–1 4 dx x+3x

x - 1 + C.

is

( 6 x + 1) + C

( 6 x + 1) + C x - 3( 3 x ) + 6( 6 x ) ( 6 x + 1) + C

+ 2(x2 + 2) = (x2 + 2) (x3 + x2 + 2x + 2) = (x2 + 2) (x(x2 + 2) + (x2 + 2)) = (x2 + 2)2 (x + 2). Therefore, f (x) =

x 5 + x 4 + 4 x3 + 4 x 2 + 4 x + 4 5

( x 2 + 2)

(b) 2 x + 6 6 x

(c) 2 (d) none of these Ans. (c)

( 6 x + 1) + C

Example 67: The value of

(c) 2

u + 1) + C

5

dx

t

u3

= 2 x - 3( 3 x ) + 6 ( 6 x )

(a)

Ú x3 ( x - 1)1 2 =

6u5 du

Ú u3 + u2 = 6 Ú u + 1 du

= 2u3 – 3u2 + 6u

to write

(

=

1 ˆ Ê = 6 Ú Á u2 - u + 1 ˜ du Ë u + 1¯

t

(t 2 + 1)

Hence

Ú ( y 2 + k 2 )n 2

dx

Ú

)

dy

t

(

(

fractional exponents. Thus we want a substitution of the form x = uk where k is a multiple of 2 and 3. Let us use the least common multiple 6. Let x = u6, d x = 6u5 du

=

x+2 3

( x 2 + 2)

=

1 2x 2 x2 + 2

(

3

+

2 3

) ( x 2 + 2)

12.24

Now

Ú

dx

1 x 8 x2 + 2

=

3

( x 2 + 2)

(

3 dx Ú 8 x2 + 2

+

2

)

(

1 sin x - 1 1 (a) 4 log sin x + 1 - 2 log

2

)

ˆ Ê dy ÁË using Ú ( y 2 + k 2 )n ˜¯

Ú

dx

( x 2 + 2)

Ú f ( x ) dx = -

\

1

(

4 x2 + 2

x 1 4 x2 + 2

+

2

)

(

=

(

4 x2 + 2

)

2

Example 68: If

Ú

x tan - 1 x 1 + x2

(

1

sin x

dx =

dx

1 + x2

=K

1 dt 1 t = ◊ = t = 1 + x2 2 Ú t 2 1/ 2

=

2

dt

Ú

x tan

1+ x

x

2

=

1 4

=

sin x - 1 1 1 log log sin x + 1 4 2 8

È1 t -1 t - 1/ 2 ˘ 1 log Í log ˙ +C t +1 t + 1/ 2 ˚ 2 Î2

Ú

= (tan-1 x)

= =

1 + x 2 tan-1 x 1+ x

2

tan

-1

Thus f (x) = tan

1 1+ x

Ú

x-

-1

2

◊ 1 + x2 d x

=

23 ( x + 1) 1

Ê ÁË

+

f (x) =

1 + x2

(x +

written as – 2

)

x +1 + C

x, K = -1.

Example 69: The value of

sin x

Ú sin 4 x

dx is

+ C.

= -

1

Ú 3

Ú

2

ˆ 23 + ˜ 26 23 ( x + 1) ¯ 5- x

4-x + ( x + 1)

x + 1 = 1/ y dy 3 y2 - y + 2 dy 2

( y - 1/ 6)2 + ( 23 / 6 )

2

23 Ê 4-x ˆ ÁË ( x + 1) ˜¯ + 16

y 2 + 1, y =

(d) none of these Ans. (d) Solution:

dx

2 sin x + 1

( x + 1) 2 x 2 + 3 x + 4

(c) K = 2, f (x) = y +

1 + x2 - Ú

2 sin x - 1

dx

5- x

(b) K =

dx

(t = sin x)

1 È 1 1 ˘ - 2 Í 2 ˙ dt Ú 4 Î t - 1 t - 1/ 2 ˚

3 fi

)

=

x

x / 1 + x 2 as the second. -1

+C

f (x) + C then 1 (a) K = , f (x) 3

[where 1 + x = t] -1

)(

Ú sin 4 x d x = 4 Ú (1 - t 2 ) (1 - 2t 2 )

Example 70: If

Ú

2 sin x + 1

cos x 1 = , hence 2 4 cos x cos 2 x 4 1 - sin x 1 - 2 sin 2 x

f (x) = tan x, K = -1 f (x) = tan–1 x, K = 1 f (x) = 2 tan-1 x, K = -1 f (x) = 2 tan-1 x, K = 1 x

2 sin x - 1

1

+C

Solution:

-1

Solution:

2 cos x + 1

(c) 8 log sin x + 1 - 4 2 log

1 + x 2 f (x) + K log ( x + x 2 + 1) + C then (a) (b) (c) (d) Ans. (a)

2 cos x - 1

Ans. (c)

3 x 3 x + + tan -1 +C 2 16 x + 2 16 2 2

)

sin x - 1

1

+C

(d) none of these 2

3 È1 x x ˘ 1 tan -1 + Í 2 + +C 4 Î4 x + 2 4 2 2 ˙˚ x -1

2 sin x + 1

1 cos x - 1 1 log log 8 cos x + 1 2 2

(b)

1 x 1 1 x = tan–1 +C + 2 4 x +2 4 2 2

=

2

2 sin x - 1

5- x x +1

12.25

= -

1

2

3

log ( y - 1/ 6 ) + ( y - 1/ 6 ) +

5- x log + = 6 ( x + 1) 3 1

Example 71: If

(

23 / 6

2

2

)

Ê 23 ˆ Ê 5- x ˆ ÁË 6 ( x + 1) ˜¯ + Á 6 ˜ Ë ¯ dx

Ú (1 + x 2 )

1 - x2

+C

(b)

+ C.

= F (x) and F(1) = 0,

Ê 2x ˆ p + tan -1 Á Ë 1 + x 2 ˜¯ 2 2

1

Ê p 1 2x ˆ tan -1 Á ˜ 2 2 Ë 1+ x ¯ 2 2

Ê p 2x ˆ + tan -1 Á ˜ 2 Ë 1- x ¯ 2 2 2 (d) none of these Ans. (b) 1 1 Solution: x = so that dx = - 2 dt t t We have dx - t dt Ú (1 + x 2 ) 1 - x 2 = Ú (1 + t 2 ) t 2 - 1 (c)

1

Ê 1 1 - x2 ˆ =tan Á ˜ +C x ¯ 2 Ë 2 -1

(c) (1/2) x2

x2 + 1)) + tan–1 x + C

Solution: Let f (x) = ax3 + bx2 + cx + d. Since f (0) = 1, so d = 1. Moreover, f (1) = 2 implies that a + b + c = 1 since 0 is a critical point, 0 = f ¢ (0) = 3a .0 + 2b.0 + C i.e., C = 0. Also f ≤(x) = 6ax + 2b and since f (x) does not have extremum at 0 so 0 = f ≤ (0) = b. Hence a = 1 and f ( x) 1 2x 1 ˆ x3 + 1 Ê x d = Ú x2 + 1 Ú x 2 + 1 dx = Ú ÁË x - 2 x 2 + 1 + x 2 + 1˜¯ dx =

1 2 [x 2

(

Ú

1 + x2

Ê 2x ˆ = + tan -1 Á Ë 1 - x 2 ˜¯ 2 2 2 1 ˆ Ê -1 cot q = tan -1 , for q > 0 and ˜ Á q ˜ Á Á cot -1 q = tan -1 1 + p , for q < 0 ˜ ¯ Ë q Example 72: Let f (x such that f(0) = 1, f (1) = 2 and 0 is a critical point of f (x) such that f (x) does not have a local extremum at 0. Then

) dx

= g o f (x) +

Const. then

(x + (x + (x +

) (b) f (x x + 1) and g (x) = x (c) f (x x + 1) and g (x) = x /2 (d) f (x) = x /2 and g (x ( x + x + 1) (a) f (x

x2 + 1 2

2

2

2

2

2

Ans. (c)

(x +

1 + x2

) = t, we have

Ê x ˆ ¥ Á1 + ˜ dx = dt i.e. Ë x + 1+ x 1 + x2 ¯

dx

1

2

[tan–1 q + cot–1 q = p/2] 1

x2 + 1)] + tan–1 x + C.

log x + 1 + x 2 Example 73: If

Solution:

Ê 1 1 - x2 ˆ ˘ 1 Èp Í - cot -1 Á ˜˙ x ¯ ˙˚ 2 ÍÎ 2 Ë 2

p

x2 + 1) – tan–1 x + C

(d) (1/2) (x2 Ans. (d)

Since F(1) = 0, so C = 0. Hence F (x) = -

x2 + 1) – tan–1 x + C

(b) x

= - dx (t 2 - 1 = u2 ) Ú 2 + u2 u 1 = tan -1 + C 2 2 1

x2 + 1) + tan–1 x + C

(a) x

2

then for x > 0, F (x (a)

f ( x)

Ú x 2 + 1 dx

(

log x + 1 + x 2 So

Ú

1+ x =

Thus f (x) =

2

) dx =

1

Ú t dt = 2 t

(

1 log ( x + x 2 + 1 ) 2

(log ( x +

1 + x2 2

=dt

+C

2

)

+C

)

x 2 + 1) and g (x) = x2/2.

Example 74: The value of

cos3 x + cos5 x

Ú sin2 x + sin 4 x

(a) sin x - 6 tan-1 (sin x) + C (b) sin x - 2 (sin x)-1 + C

dx is

12.26

(c) sin x - 2 (sin x)-1 + 5 tan-1 (sin x) + C (d) none of these Ans. (d)

Thus g (x

cos3 x + cos5 x

Ú sin2 x + sin 4 x d x Ú

cos x ÈÎ(1 - sin 2 x ) + (1 - sin 2 x )2 ˘˚ sin 2 x + sin 4 x

Ú

=

Ú ÁË 1 + t 2 - 1 + t 2 ˜¯ d t

2

t +t

Ê

(c)

2 (sin x + tan q ) sec q + C

dx

Solution:

6 ˆ

Ê f (x) ˆ Ú ÁË x 2 ˜¯

Ê 1 + 2 f (x) ˆ 2 Ê gÁ h ˜ 3 ÁË 2 Ë 1 - 2 f (x) ¯

1

x=

3 y2 - 2 1 - 2y

and d x = -

2

1/ 2

Ú ( f ( x ))

So

=–

Ú

= 2Ú

y◊

1 2

(1 - 2 y )

2y 2

(1 - 2 y2 )

1 - 2y

dt

(t = tan x)

t cos q + sin q t cos q + sin q +C cos q

2

=

2

2 ( tan x sec q + tan q sec q ) + C

=

(a) (b) (c) (d) Ans. (c)

dy.

(

◊

1 - 2y

2

3 y2 - 2

dy

Ú

e4 x - 1 e

2x

Ê e2 x + 1 ˆ log Á 2 x dx Ë e - 1 ˜¯

t t2 u2 u2 log t - + log u + C then 2 4 2 4 t = e–x– ex u = ex + e-x t = ex - e-x, u = ex + e-x t = ex + e-x, u = ex - e-x none of these

-

)

(

) (

)

( = (e x + e- x ) (e x - e- x ) log (e x + e - x ) - (e x + e- x ) (e x - e- x ) log (e x - e - x ) .

dy

È 1 2 ˘ - 2 Í 2 ˙ dy Î 2 y - 1 3y - 2 ˚ 1 + 2y

Ú 2

Solution: e2 x - e- 2 x log e x + e- x - e2 x - e- 2 x log e x - e - x

(2 y2 - 1) (3 y2 - 2)

log

tan x cos q + sin q

=

y2

=– 2Ú =

2 2

sec 2 x d x

2

dx x

2y

sin x cos q + sin q cos x

1

Example 77: If

x+2 , we have 2x + 3

y2 = f (x) =

Ú 2

dx

3 f (x) + 2 ˆ ˜ 3 f (x) - 2 ¯

Ú

sec3 / 2 x d x

1

=

1/ 2

2 sin ( x + q ) cos x

2

=

x g(x) = tan-1 x, h(x g(x x h(x) = tan-1x g(x) = h(x) = tan-1 x x h(x x g(x

Solution:

sec x d x

1

=

2 – 6 tan–1 t + C t = sin x – 2 (sin x)– 1 – 6 tan–1 (sin x) + C.

+ C where (a) (b) (c) (d) Ans. (d)

is

(d) none of these Ans. (b)

=t-

is

sin (2 x + q ) + sin q

2 ( tan x + tan q ) sec q + C

Ú

x+2 Example 75: If f (x) = . Then 2x+3

sec x d x

(b)

4

2

Ú

(tan x + tan q ) sec q + C

2

=

x

(a)

(1 - t ) (2 - t ) dt 2

h (x

Example 76: The value of

Solution: R (sin x, – cos x)= – R (sin x, cos x) Therefore, we substitute sin x = t, so that cos x dx = dt.

=

x

2 log 3

3y + 2 3y - 2

+C

Ú t t dt – Ú u u d u ( t = ex + e–x, u = ex – e–x) 2 1 u2 = t – Ú t dt + 2 2 2

u–

1 u du 2Ú

)

12.27 2

=

t2 u2 + 4 2

t–

2

u2 + C, 4

u–

where t = ex + e–x and u = ex– e–x. Example 78: If the primitive of sin

-3/2

x sin

-1/2

= -Ú

(x + q)

f ( x ) + C then

is - 2 cosec q

sin ( x + q ) sin x

(d) f (x) =

tan ( x + q ) tan x

Example 80: If

Solution:

=

=

Ú Ú

====

dx

-1

3

sin x sin ( x + q ) sin 3 x (sin x cos q + cos x sin q ) cos q + cot x sin q 1

sin q

d cot q +

(cot q + t)

f (x

1/2

C then

+C

f (x) = 1 + sin x g(x) = (3p/8) – (x/4) f (x) = 2(1 - sin x) none of these

Solution:

Ú

sin x 1 + sin x

d

=

Ú ( 2 +1) tan -1

=

Ú

du , u

+ C.

sin x

1 + sin x C then

is

x2 + 1 . x

x2 + 1 + C. x

xn - x- n , 0 < x

(c) g(p/4) = -15/8

sin11 x cos x

)

-1 n ( n + 1) 1 ¥ n ( n + 1) n2 x

Ú x f (x) dx = Ú x

(a) (– •, – 2) (c) (1, 2) Ans. (c)

(b) g(x) is continuous for all x ŒR

3

1 n ( n + 1)

decreases in the interval

(a) g(p/4) = 3/2

Ú

(

Example 85: Let f (x) =

dx = g(x) + C (C

f ( x)

1 2 x + C (d) none of these 2

(c) x 2 log x -

2

nƕ

So

( x1/ n - x1/ (n + 1) ) , x > 0

(b) 0

2n

x tan (1/ n ) log (1/ n ) Example 83: If f (x) = lim e , and

sin11 x cos x

x = 0.

Example 84: If f (x) = lim n2

2 x n x 2 n + 2 - ( n + 1) x 2 n + 1

nƕ

f (x)

)

then Úx f (x) dx

hn(x) =g¢n(x) =

3

3 - 8/3 3 - 2/3 t t +C 2 8

)

Clearly g

1 (d) 1 - x2

Now f (x) = lim hn(x) =

Ú

+ t - 5/3) dt

(

x2 n + 2 - 1 Solution: Let gn(x) = 1 + x + x + º + x = x2 - 1

Ú f (x) dx = Ú

- 11/3

2 15 3 1 + 4 tan x Thus g(x) = , g(p/4) = - . 3 8 8 tan 2 x tan 2 x

1

(b)

2

Thus

Ú (t

(

1 (c) 2 x -1 Ans. (d)

so

=

dt (t = tan x)

11/ 3

2 3 1 + 4 tan x = +C 8 tan 2 x 3 tan 2 x

]

(0 < x < 1) then Ú f (x) dx (a) -

Ú

= -

2

1+

=

sin11 x cos x

Example 86: If

Ú

1+ 3 x dx = 2f (x)3/2 +C then f (x) x2 / 3

is (a) 1 + x2/3 (c) 1 – x1/3 Ans. (b)

(b) 1 + x1/3 (d) 1 – x2/3

12.29

Solution: I = Ú x–2/3 (1 + x1/3)1/2 dx. m + 1 ( -2 / 3 + 1) = Here m = – 2/3, n = 1/3, p = 1/2 ; = 1 (see n 1/ 3 theorem). Put 1 + x

1/3

2

= t , (1/3)x

–2/3

dx = 2t dt.

I = 6 Ú t 2 dt = 2t 3 + C = 2(1 + x1 / 3 )3 / 2 + C .

Hence

Example 87: Let f and g be two polynomials then

(a)

f ( x) g ¢ ( x)

(b) f ¢ (x) g (x) – f (x) g¢ (x) (c) f (x) g¢ (x) – f ¢ (x) g (x) (d) f ¢ (x) g¢ (x) + f ¢ (x) g(x) Ans. (c) Solution:

Ú ( f ( x) g ¢¢( x) - f ¢¢( x) g ( x)) dx

Ú ( f ( x) g ¢¢( x) - f ¢¢( x) g ( x)) dx = f ( x ) g ¢ ( x ) - Ú f ¢ ( x ) g ¢ ( x ) dx – [g(x) f(x) – Ú f¢(x)g¢ (x) dx] = f(x) g¢(x) – g(x) f ¢(x)

EXERCISE Concept-based Straight Objective Type Questions 1. If Ú(1 – tan 3x)2 dx = f(x

f(x)] + C then

f(x

1/ 2 Ê x - 1ˆ (c) Á Ë x + 1˜¯

2 Ê x - 1ˆ (d) Á Ë x + 1˜¯

x3 - 1

Ú 4 x3 - x 1 4 1 (b) 4 1 (c) 4 1 (d) 4 (a)

4. Úx5

3

x

x

x

x

x

x

x

x

(1 + x3 )2

1 8/3 1 u – 8 5 3 8/3 3 (b) u – 8 5 3 (c) – u8/3 + 8 (a)

3 16 5 16 7 16 7 16

x x x x

9 16 7 16 9 16 9 16

dx u5/3 + C, u = 1 + x3 u5/3 + C, u = 1 + x3 3 5/3 u + C, u = 1 + x3 5

1 8/3 1 5/3 u + u + C, u = 1 + x3 8 5 dx

Ú sin5 x cos5 x

1 (tan4 x – cot4 x) + 2(tan2 x – cot2x) + C 4 1 (b) (tan4 x – cot4 x) + 2(tan2x – cot2x 4 +C 1 (c) (tan4 x – cot4x) + 2(cot2x – tan2x 4 +C 1 (d) (cot4 x – tan4x) + 2(cot2x – tan2x 4 +C 1 dx 6. If Ú + 2 + f ( x) 5 - 4 sin x + 3 cos x then f(x x (a) tan x (b) tan 2 x (c) – tan (d) tan x – tan2x 2 dx 7. If Ú = x + f(x) + C, then f(x 1 + ex 1 ex 1 + ex (a)

C then f(x x -1 x +1

(b)

(d) – 5.

(b) cos2 3x (d) cos3 3x

(a) cos 3x (c) sin2 3x dx 1 2. If Ú 4 = 2 x x -x by x +1 (a) x -1

3.

1 [tan 3x 3

x

+C

x

+C

x

+C

x

+C

e x)

(d)

x

x

x

C

x + ex 1 + ex

5 8. If Úex cos 4x dx = A e5x (sin 4x + cos 4x) + C 4 then A

12.30

4 41 5 (c) 41

3 41 9 (d) 41 f(x) = 3x cos x

(a)

(b)

(d) 10. If Ú

3 x ( log 3 sin x + cos x )

(1 + (log 3)2 )

+C

1 dx = C - log f(x sin 2 x - 2 sin x 4

1 x 2

8 sin 2

f(x

(a) 3x (sin x + cos x C x + cos x) + C (b) 3x 3 x (sin x + cos log 3) (c) (1 + (log 3)2 )

(a) tan2x (c) tan

(b) 2 tan

x 2

then

x 2

x 2

(d) sin

LEVEL 1 Straight Objective Type Questions 11. If fn (x x then Ú [x f1(x) f2(x) ... fn (x)]–1 dx f n+1 ( x) +C n +1 (d) none of these

(a) fn +1 (x) + C

(b)

(c) nfn (x) + C 12.

Ú3

x 33

x

33 3x dx x

(a)

x3 +C (log 3)3

3x

(b) 33 (log 3)3 + C

3x

33 (c) +C (log 3)3 13. The value of

(d) none of these

Ú log (1 -

x ) dx is

1 3 xx +C 2 2 1 (b) ( x - 1) log (1 - x ) - x - x + C 2 (a) x log (1 - x ) -

(

)

(c) x 2 log 1 - x -

1 x +C 2

(d) none of these

(d)

x+C

1 x(1 + xe x ) log +C 2 x +1

xe x +C (c) log 1 + xe x (d) log

15. If Ú (sin 2x – cos 2x) dx =

1 2

sin (2x + k) + C

then (a) k = p/4 (b) k = 5p/4 (c) k = 3p/4 (d) none of these x + sin x x dx = f ( x) tan + C 16. If Ú 1 + cos x 2 2 (a) f (x) = x (b) f (x) = x (c) f (x) = 1 + cos x (d) none of these sin 2 x dx = tan -1 ( f ( x)) + C then 17. If Ú 4 4 cos x + sin x (b) f (x) = cos x (a) f (x) = x2 (c) f (x) = sin x (d) f (x) = tan2 x p /4, 0)

18. The function f and whose derivative is p 4

x)2 + (c)

1 2

x)2

log (tan x) sin x cos x

1 2 1 (d) 2

(b)

x)2 x)2

19. The function f

( x + 1) dx is 14. The value of Ú x(1 + xe x ) x ex

n times)

( x + 1) e x +C 1 + xe x

20)

and whose derivative is cos ( 4 - x ) (a)

4 - x sin 4 - x + (4 - 6 x) cos 4 - x

(b)

4 - x sin 4 - x + 4 - x cos 4 - x - 20

(c) – 2 ( 4 - x sin 4 - x + cos 4 - x ) – 18 (d) – (2x + 12)

4 - x sin 4 - x

+(4 - 6 x) cos 4 - x

12.31

x = 1 two antiderivative of f (x) = ex differ by 2. The difference of these antiderivatives at x = 100 is

(a) g(x) =

(a) e100 (c) 2

(c) f (x) = x – 1

(b) 100 (d) none of these

(a) –

(b) k = 1/4 (d) k depends on x

1

(b) K =

2

(c) M = –

2

Ú log (

M sin

–1

(b) L = – 2 (d) K = 4/3

1 - x + 1 + x ) dx = x f (x) + L x +

1ˆ

Ê

dx = f (x

26. If

dx

Ú (x2 + 1) ( x2 + 4)

32. If

x + 1) + g(x)x2 +

then (a) K = 2/3 (c) K = – 2/3 4

27. If

cos x

C then

x + C then f (x sin x

(b) x

x

(d) none of these

dx

Ú 2 sin x - cos x + 5 = A tan

-1

1 5

f ( x) + C , then

(a) A = 1/ 5

33. The antiderivative of x + C, 2

1 is sin x + tan 2 x 2

(a) (-1/ 2) ÎÈ tan x + (1/ 2 ) tan -1 (1 / 2 ) tan x ˘˚ + C

(

)

(

)

(b) (-1/ 2) ÈÎcot x + 1/ 2 tan -1 (1/ 2 ) tan x ˘˚ + C

(b) L = 2/3 (d) L = –1/6.

(

)

(c) - ÈÎcot x + (1 / 2 ) tan -1 (1 / 2 ) tan x ˘˚ x

Ú sin 2 x dx = K cot x + M sin 2x + L 2

then (a) L = 1 (b) K = – 2 (c) M = –1/4 (d) none of these. x xe d x = f ( x) 1 + e x g(x 28. If Ú x 1+ e then

x+

(b) A = 5 (c) f (x) = 5 tan (x/2) + 1 (d) f (x) = 2 tan (x/2) + 1

(b) g(x x (d) none of these. = K tan–1 x + L tan–1

x

x)]

x

(c) 2 cos

Lx + C, then (a) f (x) = (1/2) x2 (c) L = 1

x)2

31. If Ú f (x) dx = – 2 cos to (a) sin

x + C, then

Ú x log ÁË1 + x ˜¯

C

g (x) = [0, •) (b) dom f = (0, •) (c) g¢ (x) = – cosec2 x (d) f ¢ (x) = 1/x for all x Œ (0, •)

(a) f (x ( 1 - x + 1 + x) (b) L = – 1/3 (c) M = – 2/3 (d) M = – 1/2. 25. If

2

x

30. If Ú cosec 2x dx = f g(x

(d) K = – 2 2 23. If Ú tan4 x dx = K tan3 x + L tan x + f (x), then

24. If

log ( x + 1) - log x dx is x ( x + 1) 1 – x 2 x 2

C (c) C (d) none of these.

1

(a) K = 2/3 (c) f (x) = x + C

1 + ex + 1

(d) f (x) = 3 (x – 2)

x + 1)}2

1

1

1 + ex - 1

1 2 x

sin 2 x dx = x - K tan -1 ( M tan x) + C then 22. If Ú 1 + sin 2 x

(a) M =

1 + ex - 1

(b) g(x) =

Ú

dx 21. If Ú = tan -1 (k tan 2 x) + C then 6 6 cos x + sin x

(a) k = 1/2 (c) k = 1/6

1 + ex + 1

+ C,

(d) none of these 34. If the antiderivative of then A (a) 2 (c) 3

C, 35. If the antiderivative of C then f (x

1 x + x3 / 2

is A 1 + x + C

(b) 1 (d) 4 e3 x + e x is f (ex – e–x) + e - e2 x + 1 4x

12.32

(b) cos–1 x (d) tan x

(a) sin x (c) tan–1 x

(

36. The value of Ú esecx sec3 x (sin2 x + cos x + sin x + sin x cos x) dx is (a) (b) (c) (d)

e sec x (sec x + tan x) + C esec x (sec2 x + sec x tan x) + C esec x (sec2 x + tan x) + C none of these

x + 5 cos x

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

(b) log e (log e ex - 1) + C (c) log e (log e x - 1) + C

C then

A = 23/41, B = 1/41 A = 23/41, B = 2/41 A = 11/23, B = 2/23 A = 12/23, B = 2/23

(d) log e (1 + log e x ) + C 43.

1 dx = f (3 tan x) + C , then f(x) 38. If Ú 2 2 4 - 3 cos x + 5 sin x 3 –1

2

(a) 3 tan x (c) x2 + 1

(b) x (d) tan–1 x

x2 and g (x) = sin x then 39. If f (x) = 1 + x2

Ú fo g ( x) cos x dx (a) (b) (c) (d)

x3 Ú x3 - x2 dx , Q (x) =

(a) 5/2

41.

Ú

10 4 + log 3 3 x4 - 1

(

4

(a)

(b)

1/ 2

)

dx

x + x +1 +C x

x4 + x2 + 1

dx

1 - 9 x2

1 (b) C + ÊÁ Ë9

3 1 - 9 x 2 + cos - 1 3x ˆ˜ ¯

(c) C -

(

(

)

(

)

(

1 - 9 x 2 + cos - 1 3x

1 1 - 9 x 2 - cos - 1 3x 9

(

+C

(

log x + 1 + x 2 44. If I = 1 Ú x3 - x2 dx and

Ú

1 + x2

)) 3

(b)

(d) none of these

(c) (d)

3

)

+C

) dx

( 1+ x ) + C (1/ 2) (log ( x + 1 + x )) + C x log ( x + 1 + x ) + C log ( x + 1 + x ) + C 2

(a) (1/ 2) log x +

45. If I =

2

x2

2

3 1 - 9 x 2 + cos - 1 3x ˆ˜ ¯

5 4 + log 4 3

(b)

x2 x4 + x2 + 1

)

1 (a) C - ÊÁ Ë9

(d)

(P + Q) (2) = 5/2 then P (3) + Q

(c)

Ú

(

x + cos - 1 3x

is

sin x – tan–1 (sin x) + C cos x + tan–1 (cos x) + C sin x + 2 tan–1 (sin x) + C none of these

40. If P (x) =

1

Ú x logex e dx (a) log e (1 - log e x ) + C

3 cos x + 2 sin x 37. If Ú dx = Ax + 4 sin x + 5 cos x B

+C

x4 + x2 + 1 +C x

(d) 42.

3/ 2

)

4 2 (c) x x + x + 1

2

2

2

2

Ú sec

(a) x sec - 1 (b) sec - 1

-1

x dx, then I

x - log (1 + x ) + C x - tan - 1

x +C

(c) x sec - 1

x-

(d) x sec - 1

x + log (1 + x ) + C

x -1 + C

12.33

Assertion-Reason Type Questions

46. Let I =

Ú

e

x

x

Ú ex

dx and J =

dx 1 – e– 2 x

I

Statement-2: is xex = t.

then

1 and F(x) be an antiderivasin 2 x - 2 sin x tive of f (x)

49. Let f (x) = Statement-1: I + J = 2e

– sin –1 e – x + C

x

Statement-2: Suitable substitution is 47. Let f(x) =

1

x = t in I +J

Statement-1: F(x) is a periodic function with period 2p 1 1 Statement-2: F(x) = C – log cos x + x 4 8 sin 2 2

and F be the antiderivative of

x – x2 F(0) = 0

f

Statement-1: F(1) = p

50. If

Statement-2: F(x) = 4 tan 48. Statement-1:I = Ú

–1

( x + 1)

(

x 1 + xe

x

x dx = log

)

x 2 x Ú xe ( x + 1) dx = f ( x)e 2

2

+C

xn for some n 2 Statement-2: f is an even function

Statement-1: f(x) is of the form

xe x

+C

1 + xe x

LEVEL 2 Straight Objective Type Questions 51. If f (x) = Ú (x3 – 2x2 + 3)e3x dx then the number of critical points of f are (a) 1 (c) 3 52. If f (x) =

(b) 2 (d) 4 3x + 2

Ú x 4 - x3 + x2 - 1

(a) one critical point (c) three critical points 53. The function f (x) = (a) (b) (c) (d)

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

dx, then f (x) has

(b) two critical points (d) no critical point ( x - 2) dx

Ú x 2 - 7 x + 12

Ú

of

Ú 9e x - 4e- x

dx = Ax + B

(

2

) dx

e2x – 4) + C then

A = 3/2, B = 35/36 A = – 5/4, B = 23/24 A = – 3/2, B = 35/36 A = 5/4, B = – 35/36

Ú

x f ( x) 1 + f ( x)

dx is

(a) 2 x 1 + e x - 4 1 + e x - 2 log |

e f ( x ) x cos3 x - f ( x ) 1 - ( f ( x ))

4e x + 6e - x

f (x + y) 56. Let f = f (x) f (y) (x, y ΠR with f (1) = e then the value

decreases on R increases on R ~ (2, 3) increases on [2, 3) » (4, •) decreases on (2, •)

54. If f (x) = sin x then

55. If

is

1 + ex - 1 1 + ex + 1

|+ C

(b) 2 1 + e x - 4 1 + f ( x ) -2 log | 1 + f ( x ) - 1 | +C

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

e f (x) (x – sin x) + C e f (x) (x – sec x) + C e f (x) (x + sec x) + C e f (x) (x + sec2 x) + C

(c) 2 1 + f ( x ) - 4 1 + f ( x ) - 2 log | (d) none of these.

1 + f ( x) - 1 1 + f ( x) + 1

| +C

12.34

57. If

Ú

x7 2

(1 + x 4 )

dx =

1 [log (1 + x 4 ) + f ( x)] + C then 4

(c) f (x) =

1

(

1 + x4

)

(b) C -

2

)

dx

tan -1 x 2

2(1 + x )

+

f ( x) 1 -1 tan x + 2 4 1 + ( f ( x ))

+

tan -1 x x + 4 4(1 + x 2 )

4

–1

(c) C -

4

(d) f (x) = tan (1 + x ) 58. The function f and whose derivative is

(d) C -

cos4 x + sin 4 x cos2 x - sin 2 x

(b) 4

cos x - sin x 1 + sin x cos x + x cos x + sin x 2 (d) none of these (c) log

Ú v (x) dx,

(c)

n v (x) be any function and v1(x) =

v2(x) = Ú v1(x) dx ... vn + 1 (x) =

Ú vn(x) dx.

Ú P(x) v(x) dx

(a) P (x) v1(x) +

2(1 + x 2 ) 1 2

(1 + x )

tan -1 x +

x 1 -1 tan x + 2 2(1 + x 2 )

dx

Ú 4 ( x - 1)3 ( x + 2)5

is

(a) 3( x - 1)1 / 4 - (5 / 3)( x + 2)3 / 4 + C

cos x + sin x + sin 2 x (b) log cos x - sin x

59. Let P (x

tan x

62. The value of

1 1 + tan x 1 log + sin x cos x (a) 4 1 - tan x 2

then

(

1 + x2

(a) cannot be determined explicitly

2

1 1+ x

f ( x ) tan -1 x

Ú

and f (1) = 2 then

(a) f (x) = 1 + x4 (b) f (x) =

61. Let f f (x + y) = f (x) + f (y), for each x, y ΠR

P ¢( x )v2 ( x ) + 2! P ¢¢( x )v3 ( x ) v ( x) º + n +1 (n + 1)! 3!

(b) P(x) v1(x) – P ¢ (x) v2(x) + P≤(x) v3(x) … + (– 1)n n! vn + 1 (x) (c) P(x) v1(x) + P ¢ (x) v2 (x) + P≤ (x) v3(x)… + n vn +1(x) P ¢( x )v2 ( x ) P ¢¢( x )v3 ( x ) + º (d) P(x) v1(x) 2! 3! +(-1)n

4 3

( 4 x - 1 ) - 5 ( 4 ( x + 2 )5 ) + C 4

x -1 +C x+2

(d) none of these cos 4 x - 1 Ú cot x - tan x 1 cos 4x + C 2 1 (c) – sin 2x + C 2

(a) –

64. If f (x) = lim

(b) –

(d) none of these.

xn - x-n

x Æ•

xn + x-n

(

, x > 1 then

x f ( x ) log x + 1 + x 2

Ú

1+ x

2

(

1 cos 4x + C 4

) dx is

)

(a) log x + 1 + x 2 - x + C

( ( ( x - 1) log ( x +

) )

Ê 1ˆ (b) Á ˜ x 2 log x + 1 + x 2 - 1 + C Ë 2¯ vn +1 ( x ) (n + 1)!

(c)

)

1 + x2 + C

(d) none of these 60. If

x +1

Ú ( x 2 + x + 1)

the value of k is (a) 1/2 (c) 2/3

dx 2

=k

x + x +1

x -1 2

x + x +1

(b) 1/3 (d) none of these

+ C then

65. The value of

Ú sin

3

x dx is

(a) {(2 – x2/3) cos x1/3 + 2x1/3 sin x1/3} + C (b) 3 {(2 – x2/3) cos x1/3 + 2x1/3 sin x1/3} + C

12.35

(c) 3 {(2 – x2/3) sin x1/3 – 2 x1/3 cos x1/3} + C (d) none of these f ¢ (x) = f (x) and

66. Let f (x f ( x) f (0) = 2. Then Ú dx 3 + 4 f ( x)

1

f ( x)

tan 3 (a) 3x

–1

2x - 1 3

3 2

(d) (1 / 2 ) f o (g + h) (x) + C

x2 – x + 1) + (a)

+ C then f (x (b)

(b) 3x – 4 1 (d) 1 + x2

(c) 3x – 1

(c)

1

68. If the antiderivative of x

2

1+ x

f ( x) +C x

is –

2

(a) 1 + 1/x2

(b) tan–1 x

(c) 1 + x2

(d) 2 1 + x 2

(d)

sin 2 x

can be

expressed as (a) a rational function of tan x (b) an irrational function tan x (c) a rational function of cos x (d) a rational function of sin x

120 x 5

( f ( x ))3 (( f ( x ))2 - 4) 6 x5

( f ( x ))2 (( f ( x ))3 - 2 ) 20 x 4

( f ( x ))3 ( x 2 - 6) 6 x5

(a) 2 [g o f (x) – g o f o h (x)] + C (b) 2 [ f o g (x) – f o g o h (x)] + C (c) f o g (x) + f o g o h (x) + C (d) 2 [f o g (x) – h o f o g (x)] + C x+3 71. If f (x) = 4 x 2 + 4 x - 3 then Ú dx f ( x)

1 + 2 cos x 1 - 2 cos x

f ( x) x6

dx

+C

+C +C

+C

x + 1 + f (x x + 1 + f (x x + 3) + f (x x + 3) + f (x and g (x) = tan

(b)

g( x ) f ( x)

f (x) + 1) +

log f ( x )

Ú ( f ( x))2

dx

x2 +C 2

f ( x) g (x) + 1) + x + C g( x ) g( x ) f (x) + 1] – x + C (d) f ( x) 75. If f (x) = cos x then Ú (2(f (x))2 – 1) (4(f (x))3 – 3 f (x)) dx 1 1 (a) sin 5x + sin x + C 10 2 1 (b) sin 5x + sin x + C 5 1 1 (c) sin 5x – sin x + C 15 2 (d) none of these (c)

70. If f (x) = x , g (x) = ex – 1 and h (x) = tan–1 x then the antiderivative of f o g (x) is

(1/4) f (x) + (2/3) l (1/4) f (x (1/3) f (x (2/3) f (x

Ú

74. If f (x) = cos x and g (x) = sin x then is f (x) + 1) + x + C (a) f (x

dx

Ú cos3 x

72. If f (x) =

x 2 + 4 then

( f ( x ))3 (( f ( x ))2 - 10)

then f (x

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

g o h (x) + C

2 h o f (x) + h o g (x) + C

73. If f (x) =

dx =

f (x

(c) h o (f + g) (x) + C

(d) none of these

Ú x2 - x + 1

(a) (1 / 2 ) (b)

ex ) + C ex ) + C e2x ) + C 67. If

dx

Ú sin x(2 cos2 x - 1)

x

C C C x and h(x) 2

76. If In = Ú cotn x dx, and I0 + I1 + 2 (I2 + … + I8) Ê u2 u9 ˆ + º + ˜ , where + I9 + I10 = A Á u + 2 9¯ Ë u = cot x then (a) A = 1 (c) A = 1/2

(b) A = – 1 (d) A = – 1/2

12.36

dx

Ú sin 4 x

77. f (x) =

is a x x x x

(a) po

(d) polyno 78. If f (x) = (a) x f (x) +

x2

Ú f ( x) dx

x 2 - a 2 then a2 2

x + f (x

(b)

x2 a2 f (x) – 2 2

(c)

x2 f (x) + a2 2

x – f (x

(d)

x a2 f (x) + 2 2

x + f (x

79. Let f(x) =

(1 + x n )1 / n

f o f ... o f . Then n times

Úx

(b)

1 (1+n xn)1 – 1/n + k n –1

(c)

1 (1 + n xn)1 + 1/n + k n (n - 1)

(d)

1 (1 + n xn)1 + 1/ + k. n –1

80. If Úf(x) dx = f (x) i.e. f is a function such that f ¢(x) = f(x), then Úx9 f (x5) dx (a)

1 5 [x f (x5) – Úx4 f(x5)dx] + C 5

C

(b)

1 5 x f (x5) – 5 Ú x5 f(x5) dx + C 5

C

(c)

1 5 x f (x5) – Ú x4 f(x5) dx + C 5

(d)

1 5 [x f (x5) – Ú x5 f(x5)dx] + C 5

C

for n ≥ 2 and g(x) =

n–2

1 (1 + n xn)1 – 1/n + k n (n - 1)

is

C

x + f (x

x

(a)

g(x) dx

Previous Years' AIEEE/JEE Main Questions

1. If

sin x

Ú sin ( x - a )

dx = Ax + B

x – a) sin +

3.

C then the value of (A, B) is (–sin a, cos a) (cos a, sin a) (sin a, cos a) (–cos a, sin a) dx 2. Ú cos x - sin x (a) (b) (c) (d)

(a) (b) (c)

(d)

1 2 1 2 1 2 1 2

(a)

4.

x ex 1+ x

x 2

+C

log x 2

(log x ) + 1

(b) +C

(d)

(log x )2 + 1 x 2

x +1

+C

+C [2005]

dx

Ú cos x +

3 sin x

1 x p log tan Ê + ˆ + C Ë 2 2 12 ¯ 1 x p log tan Ê - ˆ + C (b) Ë 2 12 ¯ 2 (a)

x +C 2

x p tan Ê - ˆ + C Ë 2 8¯ Ê x 3p ˆ log tan Á + +C Ë 2 8 ˜¯

(c)

[2004]

x 3p tan Ê - ˆ + C Ë2 8 ¯ cot

2 Ï (log x - 1) ¸ ˝ Ì Ú Ó1 + (log x)2 ˛ dx

[2004]

x p (c) log tan Ê + ˆ + C Ë 2 12 ¯ x p (d) log tan Ê - ˆ + C Ë 2 12 ¯

[2007]

12.37

2Ú

5. The value of

sin x is p sin ÊÁ x - ˆ˜ Ë 4¯

Ê Ú Ë1 + x -

pˆ Ê (a) x + log cos Á x - ˜ + C Ë 4¯ p (b) x – log sin ÊÁ x - ˆ˜ + C Ë 4¯

(a) (x – 1) e (b) x e

pˆ Ê (c) x + log sin Á x - ˜ + C Ë 4¯ p (d) x - log cos ÊÁ x - ˆ˜ + C Ë 4¯

(a) –2 (c) 2 7. If

Ú

x -x +1 x2 + 1

e

cot–1x

12. I = to

cot–1x

+C

+C

[2013, online]

1 sin x + C 2 (d) – sin2 x + C

(b) x – (1 + x2) cot–1 x + C (c) – x + (1 + x2) cot–1 x + C (d) x – (1 + x2) tan–1 x + C [2013, online] x6

Ú x + x7

Ú(

dx

x p(x) + C x p(x) + C (c) x – p(x) + C (d) x + p(x) + C

(a)

[2013, online] (c)

3

10. If Ú f(x) dx = y (x), then Ú x f(x ) dx 1 3 1 (b) 3 1 (c) 3 1 (d) 3

1 1 + cot 3 x

(b) –

5

(a)

[2014, online]

(a) – x + (1 + x2) tan–1 x + C

1 - 2 + x2 + C = p(x) then,

dx

Ê 1 - x2 ˆ Ú x cos–1 Á dx (x Ë 1 + x 2 ˜¯

1 - 2 - x2 + C

Ú x + x7

[2014]

1 sin2x + C 2

2 - x2

dx

9. If

+C

1 sin 2x + C 2

+C

1 + 2 - x2 + C

(d) x

1 x

Ú 1 - 2 sin2 x cos2 x

(b) –

1 + 2 + x2 + C

(c) – x

x+

1 x

x dx

Ú 2 - x2 +

+C

(c) –

1+ x

(d)

x+

1 x

sin8 x - cos8 x

[2012]

(b) x

1- x

(d) – x e

(a)

dx = A(x) e

(a) –x (c)

[2008]

(b) 1 (d) –1

2

1 x

(c) (x + 1) e

5 tan x Ú tan x - 2 dx = x – 2cos x k

x+a

x+

x+

1

1 ˆ x+ x e dx x¯

x3 y (x)3 – 3 Ú x3 y (x3) dx + C (d) x3 y (x3) – Ú x2 y (x3) dx + C

2

sin3 x + cos3 x )

+C

3 (1 + tan3 x )

1 + cos3 x

dx

+C

+C

cos3 x 3 (1 + sin3 x )

+C

[2014, online]

15. If m is non zero number and

[x3 y (x3) – Ú x3 y (x3) dx] + C [x3 Y(x3) – Ú x2 Y (x3) dx] + C

sin 2 x cos2 x

1 sin3 x

[2014, online]

x 5m -1 + 2 x 4 m -1

[2013]

Ú(

3

x 2 m + x m + 1)

dx = f(x) + C, then f(x) is

12.38

(a)

(b)

x 5m 2

2m ( x 2 m + xm + 1) x 4m

2

2 m ( x 2 m + xm + 1)

2 m ( x 5m + x 4 m )

(c)

x

(d)

5m

+x

(a)

4m 2

2m ( x 2 m + x m + 1)

dx

Ú x 2 ( x 4 + 1)3/ 4

equals

1

Ê x 4 + 1ˆ 4 (a) Á +C Ë x 4 ˜¯

(

)

(c) - x 4 + 1

1 4

(b)

(

)

x4 + 1

+C

(c) 1 4

+C

(d)

Ê x 4 + 1ˆ 4 (d) - Á + C [2015] Ë x 4 ˜¯ dx

Ú ( x + 1)3/ 4 ( x - 2)5/ 4

x + 1 ˆ 1/ 4 (a) 4 ÊÁ +C Ë x - 2 ˜¯

x - 2ˆ (b) 4 ÊÁ Ë x + 2 ˜¯

is equal to

(

log t + 1 + t 2

)dt = 1 (g(t))

2 1 + t2 is a constant, then g(2) is equal to (2 + 5 ) (c)

1 5

+C

( x 5 + x 3 + 1)2 x10 2( x 5 + x 3 + 1)5 x5 2( x 5 + x 3 + 1)2 - x10 2( x 5 + x 3 + 1)2

log(2 + 5 )

+C

+C

+C

[2016]

1- x 1+ x 1- x

(c) -2

1+ x

+ C, where C

(1 + x ) x - x 2

is equal to (where

+C

(b) -

+C

(d) 2

1- x 1+ x 1+ x 1- x

+C

+C [2016, online]

[2015, online] 2

dx

C is a constant of integration)

+C

21. If I = Ú

dx

= (tan x)A + C (tan x)B +

cos3 x 2 sin 2 x

k, where k is a constant of integration, then A + B + C equals: 16 27 (b) (a) 5 10

(2 + 5 ) (d)

20. The integral I = Ú

(a) -2

1/ 4

4 x + 1 ˆ 1/ 4 4 Ê x - 2 ˆ 1/ 4 (c) - ÊÁ (d) + C ˜ Á ˜ +C 3 Ë x - 2¯ 3 Ë x + 1¯

Ú

(b)

- x5

is equal to

1

17. The integral I =

18. If I =

Ú ( x 5 + x3 + 1)3 dx

( x 2 m + xm + 1)2

[2014, online] 16. The integral

2 x12 + 5 x 9

19. The integral I =

1 log(2 + 5 ) 2

(c)

7 10

(d)

21 5

[2016, online]

[2015, online]

Previous Years' B-Architecture Entrance Examination Questions 2 x Ê 1- x ˆ e Ú ÁË 1 + x 2 ˜¯ dx is

(a) ex (1 + x2)2

(c)

e x (1 - x )

(1 + x 2 )2

(b)

(d)

’

1 (x - l)

where for real number a1

(1 + x 2 )2 1 + x2

10

f(x) =

l =1

- xe x

ex

2. Let f : (11, •) Æ (0, •

an

n

’ al [2007]

l =1

denotes the product a1 ¥ a2

¥ an

12.39 10

Statement 1: Ú f(x) dx =

Â

l =1

( -1)l log | x - l |

(l - 1)! (10 - l )! 10

Statement 2: For x Œ [11, •) f(x) =

Â

l =1 10

’

where Al =

j =1

3.

Ú

2

x +1

-5

6. If

j l = 1, 2, 10 l- j

Al x-l

7. If

[2008]

(b) 2 ( x + 1) - x + C 5 x +1 2x

(d)

5

x +1

+

x

Ú( x

(a)

(b)

a 2 - b2 x 2

a 2 - b2 x 2

5.

Ú( (a)

(b)

(c)

(d)

32

6 x 1 (d) 2 (b) -

[2016]

Answers 1. (b) 5. (b) 9. (c)

+C

+C

+C

1 a

9

Ú x(1 + x 9 ) dx = A log |x| + B log |1 + x | + C,

2. (c) 6. (c) 10. (c)

3. (d) 7. (b)

4. (a) 8. (a)

Level 1

ax

2

[2014]

Concept-based

a 2 - b2 x 2 )

a 2 a 2 - b2 x 2

(d)

[2009]

dx

x

(c)

+ C, then f(x):

cos5 x

1 - x9

1 6 x (c) 2

1 2–x – 2 5–x + C log 5 5 log 2

f ( x)

(b) cosec x (d) – cot x

(a) -

+C

2x

dx =

then the ratio A : B is equal to (a) –2 : 9 (b) 2 : 9 (c) 9 : –2 (d) 9 : 2 [2015] dx = f(x) (1 + x– 6)1/3 + C, where 8. If I = Ú 3 6 2/3 x (1 + x ) C is a constant of integration, then f(x) is equal to:

x -1

2 ( x + 1)

Ú cos5 x sin2 x

(a) – cosec x (c) cot x

dx 10 x 1 2 (a) 2x+1 – 5x–1 + C 5 log 2 log 5

(c)

1 - 5 sin 2 x

+C

2

a - b2 x 2

[2012]

7 x13 + 5 x15 3

x 7 + x 2 + 1) x14

+C

( x 7 + x 2 + 1)2 x14 2

2 ( x 7 + x 2 + 1) x7

x7 2

2 ( x 7 + x 2 + 1)

+C

+C

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

12. 16. 20. 24. 28. 32. 36. 40. 44. 48.

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

13. 17. 21. 25. 29. 33. 37. 41. 45. 49.

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

14. 18. 22. 26. 30. 34. 38. 42. 46. 50.

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

52. 56. 60. 64. 68. 72. 76. 80.

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

53. 57. 61. 65. 69. 73. 77.

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

54. 58. 62. 66. 70. 74. 78.

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

Level 2

+C

( x 7 + x 2 + 1)2

11. 15. 19. 23. 27. 31. 35. 39. 43. 47.

[2013]

51. 55. 59. 63. 67. 71. 75. 79.

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

12.40

Previous Years' AIEEE/JEE Main Questions 1. 5. 8. 12. 16. 20.

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

2. 5. 9. 13. 17. 21.

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

3. 6. 10. 14. 18.

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

4. 7. 11. 15. 19.

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

t8 t 5 +C 8 5 83 53 1 1 = (1 + x 3 ) - (1 + x 3 ) + C. 8 5

=

5. Substitute tan x = t, sec2 x dx = dt fi dx = dx

Ú sin5 x cos5 x

=

Ú

Previous Years' B-Architecture Entrance Examination Questions 1. (d) 5. (b)

2. (c) 6. (d)

3. (d) 7. (c)

4. (b) 8. (d)

Hints and Solutions 1. Ú(1 – tan 3x)2 = Ú (1 + tan2 3x – 2tan 3x) dx = Ú(sec2 3x – 2 tan 3x) dx tan 3 x 2 = x C 3 3 1 2 = [tan 3x 3x] + C 3 1

1

1 1 È 1 1 ˘ 2 Í 2 x Î x - 1 x + 1 ˙˚ x ( x - 1) ( x + 1) x -x 1È 1 1 1 1 ˆ ˘˘ È 1 Ê1 - - 2 -Í 2 -Á = Í Ë x x + 1˜¯ ˙˚ ˙˚ 2 Îx -1 x x Îx 1È 1 2 1 ˘ - 2 = Í x + 1 ˙˚ 2 Îx -1 x 4

2

=

dx

=

2

1˘

1È

Ú x 4 - x 2 = 2 ÎÍlog | x - 1 | - log | x + 1 | + x ˚˙ x -11 2 1 + log + C. x x +1 1 x -1 x3 - 1 1 4 3. = + 4 x(2 x - 1)(2 x + 1) 4 x3 - x 1 1 7 1 9 1 = + 4 x 8 2x - 1 8 2x + 1

Ú ÁË t 5 + t 3 + 6 t + 4t + t

x -1

Ú 4 x3 - x

dx =

1 4 –

9 16

x

1

3ˆ

˜¯ dt

t4 t2 t 4 - 2t -2 + 6 log | t | + 4 + + C 4 2 4

1 (tan4x – cot4x) + 2 (tan2x – cot2x 4 +C x 2dt 6. Put tan = t fi dx = 2 1 + t2 dx Ú 5 - 4 sin x + 3 cos x

x

=

=

Ú

2dt 2 Ê (1 + t 2 ) Á 5 - 8t 2 + 3 (1 - t2 ) ˆ˜ Ë 1+ t 1+ t ¯ dt

= 2Ú

+C

=

2 t 2 - 8t + 8 dt

Ú t 2 - 4t + 4

=–

dt

Ú (t - 2)2

=

1 +C= t-2

1

Thus I = x

dx

t+C e x) + C x e)+C

=x

4. Put 1 + x3 = t3 fi x2 dx = t2 dt

=x

5 3 3 2 3 4 Ú x (1 + x ) dx = Ú (t - 1) t dt

1 1 + ex

+C e4x

have

1ˆ

Ú 1 + e x = Ú t (t - 1) = Ú ÁË t - 1 - t ˜¯ dt t ex

C.

Ê 1

dt

+C x 2 t – 1)

2 - tan

7. Let 1 + ex = t fi ex = t – 1 fi x 1 so dx = dt t -1

7 16

x

5

dt

4

=

3

ˆ Ê 1 ˆ 2˜ Á Ë 1 + t ¯ Ë 1 + t 2 ˜¯

=

Ê1

5

t

(1 + t 2 ) ÊÁ

Ú

t5

1 + t2

dt

=

= -

Concept-based

2.

(1 + t 2 )4

dt

12.41

I = Úe5x cos4x dx =

1 5x 5 5x e sin 4x – Ú e sin 4x dx 4 4

3x

12. Put 33

3

=tfi

1 5x 5 È 5 x cos 4 x 5 5 x ˘ e sin 4x – -e + Ú e cos 4 x dx ˙ Í ˚ 4 4 Î 4 4 1 5x 5 5x 25 = e sin 4x + e cos4x – I 4 16 16 =

=x

Ú 3x sin x dx

= 3x sin x

x

= 3x sin x

x

cos x

x)

x

I= 10.

3 (sin x + (log 3) cos x )

(1 + (log 3)2 ) dx

.

1

dx

x e x) + C

x+x

Ú sin 2 x - 2 sin x = 2 Ú sin x (cos x - 1) 1 = 2

Ú

t 2 dt 1- t

1 ˆ Ê dt (1 - x ) + Ú Á - t - 1 + Ë 1 - t ˜¯ x (1 - x ) - x + C. 2 1 ( x + 1) e x +1. x 1 + x ex

= (x

2

cos x 2 x ) I = 3 (sin x

fi

(1 - x ) +

=x Ú3x cos x dx]

1 x dx Ú 2 1- x

(1 - x ) +

x

4 5x 5 e (sin 4x + cos x) + C 41 4

9. I = Ú(3x cos x dx = 3x sin x

ex

x

sin x dx

Ú (1 - cos2 x ) (cos x - 1)

x e x) + C

xe x + C. 1 + xe x

1 - sin x dx Ú 2 (1 - cos x )2 (1 + cos x ) 1 dt = Ú 2 (1 - t )2 (1 + t ) =

2 Ú cos (2 x + x / 4) dx . (t = cos x)

1 Ê1 t 1 1 1 ˆ + + dt = ÚÁ 4 Ë 2 1 + t 2 1 - t (1 - t )2 ˜¯ 1 È1 1 1 ˘ = Í log | 1 + t | - log | 1 - t | + +C 2 1 - t ˙˚ 4 Î2 =

1 4

1 1 1 + cos x È ÍÎ1 - cos x + 2 log 1 - cos x

2

dt 1 = dx x f1 ( x) ... f n - 1 ( x) -1

Ú [ x f1 ( x) ... f n ( x)] + 1(x)

(

sin 2 x +

+ C.

dx =

Ú

dt t

)

1

1

x

x + sin x

=

1 2

sin x

x + C. 2

= x tan 17.

5p ˆ + C. sin Ê2 x + Ë 4 ¯ 2

x x x˘ È + tan dx ÍÎ2 x tan 2 - 2 Ú tan 2 ˙˚ Ú 2

sin 2 x cos4 x + sin 4 x

=

2 sin x cos x cos4 x + sin 4 x

=

2 tan x sec 2 x 1 + tan 4 x

Now put tan2 x = t. log (tan x )

19. f (x) = =–2

1

x)2 + C

Ú sin x cos x = 2

1 p 0 = f Ê ˆ= Ë4 ¯ 2 t+C

p +C = 4

Ú 1 + cos x dx = 2 Ú x sec2 2 dx + Ú 1 + cos x dx

18. f (x) =

11. Put fn (x) = t,

= fn

16.

1

1

Level 1

So

=–

˘ ˙˚ + C

1 2 sin 2 x 2 - log +C 2 8 2 x cos x 2 2 sin 2 1 1 1 = - log tan x 2 + C 2 8 sin x 2 4 1 = 8

x

(1 - x )

4 5x 5 25 ˆ Ê e (sin 4x + cos 4x) fi Ë1 + ¯ I = 16 4 16 fiI=

dt dx dt t Ú (log 3)3 = (log 3)3 + C.

3x

33 . 33 . 3x =

2

+CfiC=0

Ú cos 4 - x dx + C Ú t cos t dt, t = 4 - x

12.42

=–2

(

)

4 – x sin 4 – x + cos 4 – x + C

= -

1 2Ú

– 20 = f(4) = – 2 + C fi C = – 18.

dx 1 - x2

20. Let the two antiderivates be ex + C1 and ex + C2 At x = 1, C1 – C2 = 2 At x = 100 it is e100 + C1 – e100 – C2 = C1 – C2 = 2. 21.

1 cos6 x + sin6 x =

=

=

1

sec 4 x sec 4 x - 3 tan 2 x

=

(1 + tan 2 x )2 - 3 tan 2 x

x=t

=

du = Ú , u = t – 1/t 2 u +1 = tan–1 (tan x – cot x) + C

22.

= cot = tan

–1

fi f(x) =

)

1 - tan 2 x + C 2

=

4

x2 x + 1) – 2

x + 1) –

Ê 1- x - 1+ xˆ Á ˜ dx 1 - x2 Ë 1 - x + 1 + x ¯

1 = Ú 2

x2 1 1 - , g(x) = – 2 2 2

Ê 2 - 2 1 - x2 ˆ Á ˜ dx ¯ - 2x 1 - x2 Ë

x, L =

x + 1) + C 1 . 2

1 - cos 2 x – 2. 2 1 3 sin 2 x - + C . 4 2

2Ú log (t 2 - 1) dt 1 ˆ ˘ È Ê = 2 Ít log (t 2 - 1) - 2Ú Á1 + 2 dt Ë t - 1˜¯ ˙˚ Î t - 1ˆ ˘ 1 È Ê +C = 2 Ít log (t 2 - 1) - 2 Á t + log Ë t + 1˜¯ ˙˚ 2 Î

= 2x 1 + e x - 4 1 + e x - 2 log

x

x

x+ x 1 – 2 2

ex = t2

1 - x + 1 + x ) ) dx

1 = Ú 2

x2 2

x–

( 1 - x + 1 + x) - I

d

x

1 ˆ 1 1 Ê ÁË x - 1 + x + 1˜¯ dx + Ú x dx Ú 2 2

x2 2

cosec2 x +

x dx = Ú tan 2 x (sec2 x – 1)dx

Ú x dx (log (

1 x2 dx 2 Ú x +1

1 1 x tan–1 x – tan–1 + C. 2 3 6

tan 2 x = – tan x + x + C. 3

where I =

x dx

1 x2 x2 log x + Ú dx 2 2 x

)

tan 3 x - Ú (sec 2 x - 1) dx 3

x

Úx

1Ê 1 1 ˆ - 2 Á 2 3 Ë x + 1 x + 4 ˜¯

sin 2 x 1 Ú 1 + sin 2 x = x – 2 tan–1 ( 2 tan x) + C.

Ú tan

x + 1) -

1 tan 2 x + Const. 2

sin 2 x sec2 x 1 = 1 = 1 1 + sin 2 x 1 + sin 2 x 1 + 2 tan 2 x So

23.

( (

x2 2

-

1 + t2 1 + 1/ t 2 dt = Ú 1 + t4 - t2 Ú t 2 + 1 / t 2 - 1 dt

–1

x + 1)dx –

x2 = 2

(1 + tan 2 x ) sec 2 x

1 1 1 x+C dx = - sin–1 x + Ú 2 2 2

Úx

-

1 - 3 sin 2 x cos2 x

+

So

f(x) = 2(x – 2), g(x) =

1 + ex - 1 1 + ex + 1

1 + ex - 1 1 + ex + 1

.

+C

12.43

1 x log t 1 =– Ú t)2 + C t 2 1 1 =– x + 1)]2 – 2 2 x x + C.

29. Put t = 1 +

d

Ú dx

(esec x (sec x + tan x))dx = esec (sec x + tan x) + C.

x + 2 sin x = A(4 sin x + 5 cos x) + B(4 cos x – 5 sin x

3 x)2

sin x

f(x) =

x

.

dt

5

tan–1

23 2 x+ 41 41

x 2

=

sec 2 x dx

Ú 4 (1 + tan2 x) - 3 + 5 tan2 x

1 dt 3 Ú (t + 1 / 3)2 + ( 5 / 3)2

3 tan x / 2 + 1 5

=

=

+C. sec2 x tan 2 x(2 + tan 2 x)

39. f og(x) =

sin 2 x 1 + sin 2 x

=

1 1 1 ˘ +C = ÈÍ- tan - 1 2Î 2 2 ˙˚ 1 ˆ 1 È ˆ˘ Ê 1 ˆ = Ê– ˆ Ícot x + ÊÁ tan – 1 Á ÊÁ tan x˜ ˙ + C ˜ Ë 2 ¯Î ¯˚ Ë Ë 2 ˜¯ Ë 2¯

Ú

2

Ú

4

+1 = - 2 +1 =

dt t +1

=

Úe

dt

x3 + 1

Ú x3 – x 2

–1

(sin x) + C

dx

2 1 1 Ê - - 2 ˆ˜ dx = Ú Á1 + Ë x -1 x x ¯ = 4 1 + x1/ 2 + C

=x

x

x

1 +C x

Since 1 + 1/ dt - 1 / )2 + 1

(P + Q) (2) =

du

Hence

Ú u2 + 1 (u = t – 1/t)

sec x (sec x tan x + tan2 x + sec x + tan x)dx

sec x

dt

Ú 9t 2 + 1

, put sin x = t

dt

2

Ú(

=

Ú 1 dt – Ú 1 + t 2 = sin x – tan

sec x tan x (sec x + tan x) + esec x (sec2 x + sec x tan x) dx

5 so C 2 10 8 + log . 3 3

(P + Q) (3) =

3 sec x

Ú 1 + t2

40. P(x) + Q(x) =

= tan–1 (e x – e– x ) + C

Úe

t2

reduces to

1 ˆ 1 Ê1 dt Á 2 Ú 2 + 2 ˜¯ 2 Ë

34. Put x1/2 = t t dt = 2 2Ú t 2 + t3 3 t = ex

x

1 dt 1 = tan–1 (3t) + C. Ú 2 2 9 t + (1 / 3) 3

x=t dt

C.

reduces to

3

Ú t 2 (2 + t 2 )

x + 5 cos x 2

t = tan

1

23 2 4 cos x - 5 sin x dx dx + Ú 41 41 Ú 4 sin x + 5 cos x =

Ú 3t 2 + 2t + 2

23 2 and B = 41 41

we obtain A =

1 x C 2 x and g(x) = tan x, dom f = (0, •).

So f(x

=

=

41.

=

x4 - 1 1/ 2

(

)

x2 1 + x2 + x4

( x + 1/ x ) (1 - 1/ x 2 )

(( x + 1/ x)

2

1/ 2

)

-1

=

( x2 - 1) ( x2 + 1) 1/ 2 x3 ( x 2 + 1 + 1/ x 2 )

12.44

Ú

x4 - 1

(

x2 1 + x2 + x4

dx = Ú

1/ 2

)

Level 2

u du , 2 (u - 1)1 2

51. We have

u = x + 1/x

f ¢ (x) = (x3 – 2x2 + 3) e3x = (x + 1) (x2 – 3x + 3) e3x

u2 - 1 + C

=

As e3x and x2 Р3x + 3 > 0 " x ΠR

x2 +

= 42.

=

1

Ú x logex e dx 1

x4 + x2 + 1 +C. x

1 +1 + C = x2 =

f ¢ (x) = 0 fi x = – 1. Thus, number of critical points of f is 1.

1 log c e

52. We have

Ú x loge ex dx

f ¢(x) =

1

Ú x 1 + loge x dx = loge (1 + loge x) + C .

Note that x3 + x + 1 = 0 has exactly one real root, say a.

I1 + I2 where

4 I1 =

1 dx = 2 18 1 - 9x

Ú

Ú

dt

\ f ¢ (x) does not exist for x = 1 and x = a .

(t = 1 - 9 x ) 2

t

-1 1 - 9 x2 + C 9

=

I2 =

x

Ú

(cos-1 3x) 1 - 9x

2

dx =

2

-1 2 t dt , t = cos - 1 3x Ú 3

Also, f ¢ (x) = 0 for x = – 2/3. Thus, f has 3 critical points. x-2 x-2 = 53. f ¢ (x) = 2 ( x - 3) ( x - 4) x - 7 x + 12 Note that f ¢ (x) < 0 for x < 2 > 0 for 2 < x < 3

3

=

cos - 1 3x ) ( -

(

44. Put log x + 1 + x 45. Put

Ú q 2 sec

2

46. Put

2

< 0 for 3 < x < 4

.

9

dx

) = t, dt =

1 + x2

> 0 for x > 4 Thus, f (x) increases on (2, 3) » (4, •). 54. Let I =

x = sec q

Úe

d

sin x

( x cos3 x - sin x) dx 1 - sin 2 x

2 Ú q dq (sec q ) dq

=

Úe

= q sec2 q - Ú sec 2 q dq

=

Ú xe

= q sec 2 q - tan q + C

= x esin x – Ú esin x dx

q tan q dq =

sin x

( x cos x - tan x sec x) dx.

sin x

cos x dx - Ú esin x tan x sec x dx – esin x sec x + Ú esin x cos x sec x dx

x = t in I and e–x = t in J.

= esin x (x – sec x) + C

47. Put x = sin2 u. 48. Put x ex = t. 49. F(x) = C -

55. Let I =

1 x log tan + 4 2

1 8 sin 2

x 2

(Put tan x/2 = t) 50.

3x + 2 3x + 2 = 3 2 x - x + x -1 ( x - 1) ( x3 + x + 1) 4

x 2 Ú xe ( x + 1) dx

4e x + 6e - x Ú 9e x - 4e- x dx

4e 2 x + 6 dx = Ú 2x 9e - 4 Put e2x = t, so that

= (1/2) Ú et (t + 1) dt (t = x2) 2

= (1/2) e .t + c = e x .

x2 +c. 2

I=

4t + 6 1

Ú 9t - 4 ◊ 2t dt

12.45

=

58. We have

È -3 1 35 1 ˘ + dt t 4 9t - 4 ˙˚

Ú ÍÎ 4

f (x) =

= - 3 ln | t | + 35 ln | 9t - 4 | + C 4 36 = - 3 x + 35 ln 9e2 x - 4 + C 2 36

(

)

\ A = - 3 , B = 35 . 2 36 Alternatively differentiate both the sides w.r.t. x, we

= (cos2x + sin2x)2 + (cos2x – sin2x)2, we can write 1 1 dx f (x) = Ú 2 cos 2 x - sin 2 x + 1 2 =

4 e x + 6 e- x 18 Be2 x = A + 9 e x - 4 e- x 9 e2 x - 4 fi 4ex + 6e–x = A (9ex – 4e – x) + 18 Bex ex and e,– x A + 18 B and 6 = – 4A -3 35 ,B= fiA= 2 36 56. First show that f (x) = ex Let I =

x ex

Ú

1 + ex

Ú

=

1 1 + tan x 1 log + sin x cos x + C 4 1 - tan x 2 C = 0.

Now, use P(n) (x) = n! w · r · t · x, and sim3k x +1 2 x +1 = fik= . 2 3/ 2 3/ 2 2 ( x + x + 1) 3 ( x + x + 1) 2

1 + ex - 1

61. First show that f (x) = 2x. Now, f ( x) tan -1 x 2x dx = Ú I= Ú tan–1x dx 2 2 (1 + x ) (1 + x 2 ) 2

x

1+ e +1

I = 2 x 1 + e x - 4 1 + e x - 2 log 57. Put 1 + x4 = t, so that x7 dx = 1 (t - 1) dt Ú 4 2 4 Ú t2 1+ x

)

t

pˆ 1 1 Ê sin2 x + C log tan Á x + ˜ + Ë 4 4¯ 4

+ ( –1)n ÚP(n) (x) vn (x) dx

Thus,

1 4

=

= P(x) v1 (x) – P¢(x) v2 (x) + P¢¢ (x) v3 (x) – ....

= 2 È + 1 log - 1 ˘ ÍÎ 2 + 1 ˙˚

=

)

x - sin 2 x dx

= P(x) v1 (x) – P¢(x) v2 (x) + Ú P¢¢(x)v2 (x) dx

Put 1 + ex = t2, so that È 1 ˘ t 2t dt = 2 Ú Í1 + 2 I1 = Ú 2 ˙ dt t -1 Î t - 1˚

(

2

ÚP (x) v (x)dx = P(x) v1 (x) – Ú P¢(x) v1 (x)dx

dx

1 + e x dx

x = 2 1 + e + log

Ú (cos

1 1 sec 2 x dx + Ú cos 2 x dx 2Ú 2

Since f

= 2 x 1 + e x – 2I1 where I1 =

cos 4 x + sin 4 x Ú cos2 x - sin 2 x dx 4 x + sin4x)

11 +C 4

1È 1 ˘ 4 Ílog (1 + x ) + ˙+C 4Î 1 + x4 ˚ 1 \ f (x) = 1 + x4

1 + ex - 1 1 + ex + 1

+C

Put x = tan q, so that 2 tan q q sec2q dq I= Ú 4 sec q =

Ú q sin (2q ) dq

=–

1 1 q cos 2q + sin 2q + C 2 4

=C–

1 tan q 1 1 - tan 2 q q + 2 1 + tan 2 q 2 1 + tan 2 q

=C–

1 x 1 1 - x2 tan -1 x + 2 2 1 + x2 2 1+ x

=

12.46

=C–

62. I =

=

Ú4

1 x 1 1 tan–1 x + tan–1 x + . 2 2 1 + x2 2 1+ x dx ( x -1)3 ( x + 2)5

Ê x - 1ˆ Ú ÁË x + 2 ˜¯

-3 / 4

=

1 Ê 2 ˆ + 1 2/3 Á Ë 3 ˜¯ 3 1 + (2 x - 1) 3 2x - 1 1 = 2 2 + 2 x - x +1 2( x - x + 1)

- 2 sin 2 2 x cos x sin x Ú cos2 x - sin 2 x dx .

sin 3 2 x dx = – = –Ú cos 2 x

fi f (x) = 3x – 1.

sin 3 2 x Ú 1 - sin 2 2 x cos 2x dx

Put sin2x = t, to obtain I= -

1 t3 1 t (1 - t 2 ) - t dt dt = 2 Ú 1 - t2 2 Ú 1 - t2

1 2 1 + 4 4 1 = sin 2 2 x + 1 2 4

t2

=

ex ) + C.

f ( x) 3 2x - 1 = 2 2 x2 - x + 1 x - x +1

I = 1 t -3 / 4 dt = 4 t1 / 4 + C . 3 3Ú 63. We can write I=

1 4

w.r.t. x

1 dx ( x + 1) 2

x -1 = t, to obtain x +1

Now, put

66. First show that f (x) = 2ex. Now, 2e x dx I= Ú 3 + 8e x

68. Let I =

1 + x2 Put x = 1/t, to obtain tdt I = –Ú = – 1 + t2 + C 2 1+ t 1 + x2 x Thus, f (x) = 1 + x2. = C-

C x

C.

2n xn - x-n lim = lim 1 - (1 / x ) =1 64. f ¢(x) = n n n Æ• x + x nÆ• 1 + (1 / x )2 n (1/x < 1)

(

x log x + 1 + x 2

Ú

1 + x2 2

( log ( x +

=

1 + x log

=

1 + x2

) dx

x + 1 + x2

Ú 2 cos 4 x

1 + x2

) ) – x + C.

Ú dx

65. Put x1/3 = q so that dx = 3q 2dq \ I = 3Ú q 2 sin q dq = – 3q2 cosq + 6Ú q cosq dq = – 3q2cosq + 6q sinq – 6Ú sinq dq

tan x

Put tan x = t to obtain 1

Ú 2

= 1 2

dx –

dx

1

I=

I=

I=

dx

Ú x2

=

1 + t2

È Í2 Î

t +

1

dt = 2 5

2

5/ 2 ˘

˙ ˚

Ú [t

-1 / 2

+ t 3 / 2 ] dt

+C

1 È ˘ 2 Í tan x + (tan x)5 / 2 ˙ + C. 5 Î ˚

Thus, I is an irrational function of tan x. 70. I =

Ú

e x - 1 dx

Put ex – 1 = t2, so that exdx = 2tdt \I=

t2 + 1 - 1 t 2t = 2 dt Ú t 2 + 1 dt Ú t2 + 1

= 2[t – tan – 1 t] + C

2

= 3(2 – q ) cosq + 6q sinq + C.

x -1 e x - 1˘ + C = 2 ÈÎ e - 1 - tan ˚

12.47

x

x) ] + C. 2

71. Put 2x + 1 = t, so that f (x) = I=

=

1 4

t+5

Ú

1 4

t2 - 4 2

- 4 , then

5 - 4 + log 4

2

+

-4 + C

1 5 x + 1 + f (x f ( x) + 4 4 sin x dx 72. I = Ú 2 (1 - cos x)(2 cos 2 x - 1) Put cos x = t, so that dt I= Ú 2 (1 - t )(1 - 2t 2 )

= = 73. I = Put

È 2 Ú ÍÎ1 - 2

2

-

C.

1+ 2

2

1- 2

x2 + 4

Ú

x

6

=

1 [5x2 (f (x))3 – 3(f (x))5]+ C. 240 x5

= (tan x = (tan x = (tan x

x– x+

Ú (1 + cot

3/ 2 ˘

)

˙ +C ˚

Ú

x2 x2 - a2

+C

Ê - sin x ˆ Ú tan x ÁË cos x ˜¯ dx

Ú (sec

2

x ) cosec2 x dx

dx

x 2 - a 2 dx + a2

dx

=

Ú

=

1 1 2 2 x x 2 - a 2 - a 2 log x + x - a 2 2

+ a2

Ú

x 2 - a2

x + x2 - a2 + c

=

1 1 x f ( x) + a 2 log x + f (x 2 2

- 1) dx

x + 1] – x + C

C

79. We have f ( x)

(fof)(x) = f ( f (x)) =

1/ n

n˘ È Î1 + ( f ( x)) ˚

x log cos x dx

2

+ I8) + I9 + I10

5/ 2

1 Ê 4ˆ ÁË1 + 2 ˜¯ 48 x

Ú sec

1 cot n +1 x n +1

1 Ê ˆ = - Á cot x + cot 3 x˜ + C Ë ¯ 3

=

2

x cos ec 2 x dx = -

Thus, A = – 1

1 + 4t dt

log cos x 74. I = Ú dx = cos 2 x

n

1 1 1 ˆ Ê = - Á u + u 2 + u3 +  + u9 ˜ Ë 2 3 9 ¯ where u = cot x.

78. I =

1 Ê 4ˆ ÁË1 + 2 ˜¯ 80 x

Ú cot

+ (I7 + I9) + (I8 + I10)

77. f (x) =

1 È Ê 2ˆ Ê 1ˆ 5 / 2 Ê 2ˆ Ê 1ˆ = - Í( 4 + 1) Á ˜ Á ˜ - Á ˜ Á ˜ (1 + 4 Ë 5¯ Ë 4¯ Ë 3¯ Ë 4¯ 8 Î –

1 1 sin 5 x + sin x + C . 10 2

1 + 4 / x 2 dx x2 x3

dx = Ú

3/ 2

=

+

x).

Ú (4t + 1 - 1)

1 [cos 5 x + cos x] dx 2Ú

= (I0 + I2) + (I1 + I3) + (I2 + I3)

(- 1) I = Ú t 1 + 4t dt 2

1 8

=

I0 + I1 + 2 (I2 + I3 +

1 = t, so that x2

= -

Ú cos 2 x cos 3x dx

Now,

1+ +C 1-

2 (hof) (x

=

76. In + In + 2 =

1 ˘ ˙ dt 1- 2 ˚

2

f (x) + 1] – x + C.

75. I = Ú (2 cos 2 x - 1) (4 cos3 x - 3 cos x) dx

dt

=

=

g ( x) f ( x)

=

=

x n 1/ n

(1 + 2 x )

Similarly, (fofof) (x) =

x n 1/ n

(1 + 3x )

12.48

x g(x) = (fof  of )( x ) =



(1 + nx n )1 / n

1 = Ú [ f (t ) + f ¢(t )] e t dt where f(t) = 1+

n times

Thus, I =

Úx

n-2

= f(t) et + C 1 +C = x◊ 1 + (log x )2

x dx (1 + nx n )1 / n

Put 1 + nxn = tn to obtain 1 t n-1 1 dt = Ú t n- 2 dt nÚ t n 1 1 = t n -1 + C = 1 + nx n n(n - 1) n (n - 1)

4. I = Ú

I =

(

1 - 1/ n

)

+C

80. Let x5 = t, so Úx9 f(x5) dx 1 = Ú t f (t) dt 5 1 = [t Ú f(t) dt – Ú f(t) dt] 5 1 5 [x f (x5) – 5 Ú f (x5) x4 dx] + C = 5 1 5 = x f (x5) – Ú x4 f (x5) dx + C 5

Previous Years’ AIEEE/JEE Main Questions 1. I = Ú

sin x sin{( x - a ) + a } dx = Ú dx sin( x - a ) sin( x - a )

= Ú [cos a + sin a cot( x - a )]dx = x cos a + sin a log |sin (x – a)| C \ (A, B) = (cos a, sin a) dx dx = Ú 2. Ú p cos x - sin x cos x + cos ÊÁ + xˆ˜ ¯ Ë2 = = =

=

1 dx Ú 2 cos(p / 4) cos( x + p / 4)

Ê x p pˆ log tan Á + + ˜ + C Ë 2 8 4¯ 2

1

Ê x 3p ˆ log tan Á + ˜ + C Ë2 8 ¯ 2

1

3. Put log x = t, so that x = et and dx = etdt. Thus t 2 + 1 - 2t 2

(1 + t 2 )

(1 + t 2 ) 2

et dt = Ú 2

È 1 2t ˘ t = ÚÍ ˙ e dt 2 (1 + t 2 )2 ˚ Î1 + t

cos x + 3 sin x

1 dx Ú 2 1 3 cos x + sin x 2 2 dx

=

1 Ú 2

=

pˆ 1 Ê Ú sec Á x - ˜ dx Ë 2 3¯

=

1 Ê x p pˆ log tan Á - + ˜ + C Ë 2 6 4¯ 2

=

1 Êx pˆ log tan Á + ˜ + C Ë 2 12 ¯ 2

pˆ Ê cos Á x - ˜ Ë 3¯

5. Same as Q1. Take a =

= cos =

p , so Ú 4

sin x dx p sin ÊÁ x - ˆ˜ Ë 4¯

p p pˆ Ê x + sin log sin Á x - ˜ + Const Ë 4 4 4¯

pˆ 1 È Ê x + log sin Á x - ˜ Í Ë 4¯ 2Î

˘ ˙˚ + Const

5 tan x 5 sin x dx = Ú dx tan x - 2 sin x - 2 cos x Let 5 sin x = A(sin x – 2cos x) + B(cos x + 2sin x). Equating coefficients of sin x and cos x, we get 5 = A + 2B and 0 = – 2A + B fi A = 1, B = 2 \ I = Ú sin x - 2 cos x + 2(cos x + 2 sin x ) dx sin x - 2 cos x

pˆ Ê Ú sec Á x + ˜ dx Ë 4¯ 2

(t - 1)2

=

dx

6. Let I = Ú

1

I= Ú

2

et dt

È 2(cos x + 2 sin x ) ˘ dx = Ú Í1 + sin x - 2 cos x ˙˚ Î = x + 2 In |sin x – 2cos x| + k \a=2 x 2 - x + 1 cot -1 x 2 t 7. Ú e dx = - Ú (cot t - cot t + 1)e dt 2 x +1 (x = cot t) = Ú (cot t - (cot 2 t + 1)) e t dt =

Ú (cot t - (cosec

2

t ) et dt

12.49

= cot t.et + Ú cosec2 t . e t dt - Ú cosec2t .e t dt + C = cot t. et + C = x.ecot

x

-1 = 2 Ú x tan xdx

+C - xdx

2 - x2 = t fi

8. Put

Ú

-1

xdx 2

Ê 1 - x2 ˆ 13. I = Ú x cos -1 Á ˜ dx Ë 1 + x2 ¯

2- x + 2- x

2

2- x

= Ú

= dt.

2

= x 2 tan -1 x - Ú

x2 1 + x2

xdx 2

(

2 - x 1+ 2 - x

2

)

dt 1+ t = – log |1 + t| + C

= x 2 tan -1 x - Ú

dx

x2 + 1 - 1

dx 1 + x2 = –x + (x2 + 1) tan–1 x + C

= -Ú

= – log 1 + 2 - x

2

14. Dividing numerator and denominator by cos6x, we +C

1 + x6 - 1

6

1 dx x dx = Ú dx - Ú 9. Ú dx = Ú 6 7 x x(1 + x ) x + x7 x+x 10. Let I = Ú x5 f ( x3 )dx

\I=

1 dt 3

1 Ú tf (t ) dt 3

=

1 1 t Ú tf (t ) dt - Ú (1) [ Ú f (t )dt ] dt 3 3

=

1 1 ty (t ) - Ú y (t ) dt 3 3

=

1 3 x y ( x 3 ) - Ú x 2y ( x 3 ) dx 3

1 Ê x+ 1 1 1 ˆ x+ x ˆ x + x Ê1 1 ˆ x+ x Ê e e Á ˜ dx ÁË ˜ 11. Ú Á1 + x - ˜ e dx = Ú x2 ¯ Ë ¯ Ë x¯

= Ú

Ê x+ 1 ˆ xe x

d Á dx Ë

= xe

x+

1 x

˜ dx ¯

+C

8

12. sin x – cos8x = (sin2x – cos2x) (sin2x + cos2x) (sin4x + cos4x) = (– cos 2x) (1) [(sin2x + cos2x)2 – 2sin2x cos2x] = (– cos 2x) (1 – 2 sin2x cos2x) 1 \ I = - Ú cos 2xdx = - sin 2 x + C 2

(tan 3 x + 1)2

dx

Put 1 + tan3x = t I=

1 d 1 = - +C Ú 2 3 3

= -

1 1 +C 3 1 + tan 3 x

= log |x| – p(x) + C.

Put x3 = t to obtain x2 dx =

tan 2 x sec2 x

have I = Ú

15. I = Ú

( x 5m-1 + 2 x 4 m-1 ) x 6 m (1 + x - m + x -2 m )3

dx

( x - m-1 + 2 x -2 m-1 )

dx (1 + x - m + x -2 m )3 Put 1 + x–m + x–2m = t, fi [(–m) x–m–1 + (–2m) x–2m–1] dx = dt 1 1 dt +C I= - Ú 3 = m t 2mt 2 = Ú

=

x 4m 1 +C 2m ( x 2 m + x m + 1)2

16. I = Ú

dx 2

4

x ( x + 1)

3/ 4

= Ú

dx x x (1 + x -4 )3/ 4 2 3

x -5

dx (1 + x -4 )3/ 4 Put 1 + x–4 = t, so that (-1/ 4) I = Ú 3 / 4 dt = t1/4 + C t = Ú

Ê x 4 + 1ˆ 1 1/ 4 +C=– Á 4 ˜ = – ÊÁ 1 + ˆ˜ Ë x ¯ Ë x4 ¯ dx 17. I = Ú x - 2 ˆ 5/ 4 ( x + 1)2 ÊÁ Ë x + 1 ˜¯ 3 x-2 dt Put =tfi = ( x + 1) 2 x+2 dx

1/ 4

+C

12.50

\I=

1 dt 1 = Ú t -5/ 4 dt Ú 3 t 5/ 4 3

4 3

=–

4 Ê x + 1 ˆ 1/ 4 +C Á ˜ 3 Ë x - 2¯

+C

(

18. Put log t + 1 + t 2

1/ 2 2 ˆ +C = -2 ÊÁ - 1˜ Ë1+ t ¯

) = v, so that

Ê t ˆ = du. 1+ Á ˜ dt t + 1 + t2 Ë 1 + t2 ¯ 1

1

fi

dt = du

1 + t2

(

\ g(t) = u = log t + 1 + t 2

(

1- tˆ = -2 ÊÁ Ë 1 + t ˜¯

)

=

)

fi g(2) = log 2 + 5 .

=

=

2 x12 + 5 x 9

Ú x15 (1 + x -2 + x -5 )3 dx

Put 1 + x–2 + x–5 = t fi –(2x–3 + 5x–6)dx = dt dt 1 Thus, I = - Ú 3 = 2 + C t 2t 1 = +C -2 2(1 + x + x -5 )2 =

x10 2( x 5 + x 3 + 1)2

20. I =

Ú (1 +

Put \I=

+C

sec 4 x

Ú

4 sin x cos x cos2 x

x = t,

2 x

= dt

1 5 1 ,B= ,C= 2 2 5

Ú (1 + t )

1- t

2

Put 1 + t = 1/u fi dt = (–1/u2) du Thus, I =

1. Ú e x ÊÁ Ë

2 2 1- x ˆ x Ê 1 + x - 2x ˆ e = dx Ú Á ˜ dx ˜ Ë (1 + x 2 ) 2 ¯ 1 + x2 ¯

1 2x ˆ xÊ = Úe Á ˜ dx 2 Ë1+ x (1 + x 2 ) 2 ¯

=

1 1+ x

2

ex 1 + x2

2. f(x) =

ex + Ú ex

1 - (1/u - 1)2

1

’ x-l

where A =

=

2x 2 2

(1 + x )

dx - Ú e x

2x (1 + x 2 )2

+C

=

l =1

2(-1/u2 )du

Ú (1/u)

16 5

Previous Years’ B-Architecture Entrance Examination Questions

10

2dt

1 (1 + tan 2 x )sec 2 x 2Ú tan x

1 tan x + (tan x )5 / 2 + k 5

Thus, A + B + C =

x) 1- x 1

dx =

1 È(tan x )-1/ 2 + (tan x )3 / 2 ˘˚ sec 2 xdx 2ÚÎ

\A=

=

dx / x

+C

1- x

2 x -3 + 5 x -9

Ú (1 + x -2 + x -5 )3 dx

1/ 2

+C 1+ x 21. By dividing numerator and denominator by cos4 x, we can write I as = -2

I=

1 2 u +C 2

Thus, I = Ú udu =

19. I =

2u - 1

2 = - (2u – 1)1/2(2) + C 2

=–

-1/ 4

du

= -2 Ú

A A1 A + 2 +  + 10 x -1 x - 2 x - 10

1 (i - 1) (i - (i + 1)) (i - 10)

( -1) ( - 1)!(10 - )!

dx + C

12.51

Ú

10 f ( x )dx = Â Ú Ai dx = i =1 x - i

10

 Ai log | x - i| i=1

So statement 1 is true but not statement 2. x ÊÊ 2 ˆ x 2 x +1 - 5 x -1 Ê 5 ˆ 1ˆ dx = 3. Ú ◊ 2 ÁË ˜¯ ˜ dx Ú ÁË ÁË 10 ˜¯ 10 x 10 5 ¯ x

1 1 1 = 2 Ú ÊÁ ˆ˜ dx - Ú ÊÁ ˆ˜ dx Ë 5¯ 5 Ë 2¯ = 2

dx

=

=

sin q

1

a2 b 1 - sin 2 q

+C

x 2

a - b2 x 2

7 x13 + 5 x15

Ú ( x 7 + x 2 + 1)3 dx

=

=

Ú

( x 7 + x 2 + 1)3

ˆ d Ê x14 Ú dx ÁË 2( x 7 + x 2 + 1)2 ˜¯ dx 2( x 7 + x 2 + 1)2

5

cos x

+C

cos x

dx - 5Ú

1 cos5 x

dx

+ Ú cot x(-5 cos -6 x)(- sin x)dx - 5Ú

cot x 5

5

cos x

+ 5Ú

dx 5

cos x

- 5Ú

dx cos5 x

1 cos5 x

dx

+C

1 - x9

dx = A log |x| + B log |1 + x9| + C x(1 + x 9 ) Differentiating both the sides, we get 1 - x9

+C

dx

=

A 9 Bx8 + x 1 + x9

fi 1 – x9 = A (1 + x9) + 9B x9 fi A = 1, –1 = A + 9B 2 fi A = 1, B = 9 Thus, A : B = 9 : –2 dx dx 8. I = Ú = Ú 3 6 2 /3 3 6 2 / 3 -6 x (1 + x ) x ◊ ( x ) ( x + 1)2 / 3 =

7 x13 ( x 7 + x 2 + 1) - x14 (7 x 6 + 2 x )

x14

cot x

x(1 + x9 )

bx 1 a = +C = 2 2 2 a b a b 1 - 2 x2 a

=

cos x sin x

=-

7. Ú

a cos q dq 1 = 2 tanq + C b Ú a3 cos3 q a b

1

5.

2

cosec2 x

cot x +C cos5 x So f(x) = – cotx

a a sin q fi dx = cos q dq b b

Ú (a2 - b2 x 2 )3/ 2

5

dx = Ú

= -

-x

5 1 2 + +C - log 5 5 log 2

4. Put x =

1 - 5 sin 2 x

= -

x

-x

6. Ú

x -7

Ú ( x -6 + 1)2 /3 dx

1 . 3(x–6 + 1)1/3 + C 6 \ f(x) = –1/2 = -

[Put x–6 + 1 = t]

THE NEWTON-LEIBNITZ FORMULA If F(x) is one of the antiderivatives of a continuous function f (x) on [a, b] i.e., F¢(x) = f (x) (a < x < b), then we have the following formula due to Newton and Leibnitz: (FUNDAMENTAL THEOREM OF CALCULUS) b

Úa

b

f ( x ) dx = F(x)

PROPERTIES OF DEFINITE INTEGRALS

Úa

2.

Úa f ( x) dx = - Úb

3.

Úa f (u) du = Úa f (t ) dt

= F(b) – F(a) a

DEFINITE INTEGRAL AS THE LIMIT OF A SUM terval [a, b]. Then

dx =

particular,

lim

b

a

b

f ( x ) dx

b

b

Úa f ( x)

b

f ( x ) d x > 0.

Úa

h  f (a + rh). In

nÆ•, h Æ 0 nh = b - a r = 1

b

b

a

a

5. If f (x) £ g(x) on [a, b], then Ú f ( x ) dx £ Ú f ( x ) dx.

r=n

1 Ê rˆ lim  f Á ˜ = Ë n¯ nÆ• r =1 n

1

Ú0 f (x) dx

6.

b

Úa

(f1(x) + f2(x)) dx = Ú

b

a

f1(x) dx + Ú

b

a

f2(x) dx

and Illustration

1

b

Úa

n˘ 2 È1 lim Í 2 + 2 +  + 2 ˙ nÆ• Î n n n ˚ 1

x2 1r = Ú x dx = = lim  nÆ• 2 r =1 n n 0 n

Illustration

dx

≥ 0. If f (x) > 0 for all points of [a, b]. Then

n

Úa f ( x)

f ( x ) dx = 0

4. If f (x) ≥ 0 on the interval [a, b], then

Let f (x) be a continuous function defined on the closed inb

a

1.

1 0

1 = 2

b

1

x3 1 Ê r ˆ2 = lim Â Ë ¯ = Ú x 2 dx = nÆ• n 3 r =1 n 0

1 0

b

7.

Úa f ( x)dx

8.

Úa f ( x) dx = Úa

9.

Ú0

£

b

a

2 È 12 22 (n - 1)2 ˘ lim Í 3 + 3 +  + ˙ nÆ• Î n n n3 ˚

a f (x) dx = a

c

f (x) dx =

b

10.

Úa f ( x) dx

11.

Ú0

a

Úa

b

Úa

f(x) dx

f ( x ) dx f (x)dx +

a

Ú0

b

Úc

f (x)dx.

f (a – x)dx

b

=

f (x) dx =

Úa f (a + b - x) dx a/2

Ú0

f (x) +

a/2

Ú0

f (a – x)dx. In

particular, 1 = 3

a

Ú0

if f (a - x ) = - f ( x ) ÏÔ0 f ( x ) dx = Ì a / 2 2 f ( x ) d x if f (a - x ) = f ( x ) ÓÔ Ú0

13.2

Complete Mathematics—JEE Main

12. If f (–x) = f (x) (i.e., f is an even function), then a

a

Ú- a f (x) dx = 2 Ú0

b

Úa f ( x)g( x) d x

f (x) dx

13. If f (–x) = – f (x) (i.e., f is an odd function), then a

(Ú

b

a

f 2 ( x) d x

x0 within this interval then

14. If f is continuous on [a, b], then the integral function g defined by g(x) =

x

Úa

f (t ) dt for x Π[a,

b] is derivable on [a, b], and g¢(x) = f (x) for all x Œ [a, b]. 15. If m and M are the smallest and greatest values of a function f (x) on an interval [a, b], then

1/ 2

b 2 a

b

Úa

f(x) dx > 0.

2Ê

Ú0 ÁË

2 2 1 2 2 x dx - 7 Ú x dx + 5Ú dx Ú 0 0 4 0

1 x3 = 4 3

16. If f (x) is continuous on [a, b], then there exb

Úa f (x) dx

=

2

-7 0

x2 2

2

2

+ 5x 0

0

b 1 f ( x )dx is f (c) (b – a). The number f (c) = Ú b-a a

=

1 È8 ˘ 7 - 0 ˙ - [ 4 - 0 ] + 5 [2 - 0 ] Í ˚ 4 4 Î3

called the mean value of the function f (x) on the

=

2 10 - 14 + 10 = 3 3

interval [a, b]. The above result is called the first mean value theorem for integrals. b + nT

Úa + nT nT

Ú0

b

Úa

f(x) dx =

f(x) dx, where n is an integer. In particular,

Find the average value of f (x) = 4 – x2 on [0, 3].

f ( x ) dx = n Ú f ( x )dx.

b 1 f ( x ) dx Ú b-a a

0

18. If the function j (x) and y (x) are defined on [a, b] and differentiable at every point x Œ (a, b), and f (t) is continuous for j (a) £ t £ j (b), then Ê y ( x)

ˆ d dy dj – f (j (x)) . Á Ú f (t ) d t ˜ = f (y (x)) d x Ë j ( x) dx dx ¯ 19. Change of variables If the function f (x) is continuous on [a, b] and the function x = j (t) is continuously differentiable on the interval [t1, t2] and a = j (t1), b = j (t2), then

Úa

f (x)dx =

t2

Út

f (j (t)) j¢(t)dt

b

b

Úa f (x, a)dx, then I¢(a) = Úa f ¢ (x, a)dx, where

I¢ (a) is the derivative of I(a) w.r.t. a, and f ¢(x, a) is the derivative of f (x, a) w.r.t. a, keeping x constant. 21. If f 2(x) and g2(x) are integrable on [a, b], then

=

1 3 (4 - x 2 ) dx Ú 0 3-0

=

x3 1 4x 3 3

The value of For

3 0

1È 27 ˘ = Í12 - ˙ = 1 3Î 3˚

5

Illustration 1

Ú0 sin x

2

dx cannot be 2

x Œ [0, 1], 0 £ sin x2 £ 1, so

1

20. Let a function f (x, a) be continuous for a £ x £ b and c £ a £ d. For any a Œ [c, d], if I (a) =

4

Illustration

average (f ) =

T

b

.

ˆ x2 - 7 x + 5˜¯ dx 4 The given integral is equal to

Evaluate

Úa f ( x ) dx £ M(b – a).

17. If f(x) is periodic with period T then

1/ 2

3

Illustration

b

ists a point c Π(a, b) such that

) (Ú g ( x) d x )

22. Let f be continuous and non-negative at all points of [a, b] and f(x0) > 0 at least at one point

Ú- a f (x)dx = 0.

m(b – a) £

£

0£

1

Ú0 sin x

2

dx £

1

Ú0 1. dx

= 1.

INTEGRALS WITH INFINITE LIMITS If a function f (x) is continuous for a £ x < •, then by definition,

Definite Integrals 13.3 •

Úa

f (x)dx = lim

bƕ

b

Úa f

(x)dx

Ïn - 1 Ô n ¥ Ô = Ì Ôn - 1 ¥ ÔÓ n

(1)

If there exists a finite limit on the right-hand side of (1), then the improper integral is said to be convergent; otherwise it is divergent. Geometrically, the improper integral (1) for f (x) > 0, is the area of the figure bounded by the graph of the function y = f (x), the straight line x = a and the x-axis. Similarly, we can define b

Ip, q =

p /2

Ú0

b

•

•

a

Ú- • f (x)dx = Ú- • f (x)dx + Úa

f (x)dx

where

p /2

sin n x d x and Ú

p /2

0

p /2

Ú0

cosn x d x = Ú

p /2

0

1 and q +1

Ïq - 1 q - 3 2 Ô q q - 2  3 if q is odd Ô I0, q = Ì Ô q - 1 ◊ q - 3  1 ◊ p if q is even ÔÓ q q - 2 2 2

sin p x cosq xd x sin n x dx

sin px cosq x dx

I1, q =

REDUCTION FORMULAE FOR

Ú0

n-3 2 ¥  if n is odd n-2 3

2 p-3 Ï p -1 Ô p + q ¥ p + q - 2 ¥  ¥ 3 + q ◊ I1, q if p is odd Ô = Ì Ô q - 1 ¥ p - 3 ¥  1 I if p is even 0, q ÔÓ p + q p + q - 2 2+q

f (x)dx Ú- • f (x) dx = a lim Æ - • Úa and

n-3 1 p ¥  ¥ ◊ if n is even n-2 2 2

Note that if p is odd then it does not matter whether q is even or odd.

SOLVED EXAMPLES Concept-based Straight Objective Type Questions 1

Example 1: The value of

x dx

Ú ( x 2 + 1)2

is

Solution: Substitute

0

1 (a) 2 1 (c) 3 Ans. (d)

When x =

(b) 1 (d)

1 4

Solution: Substituting x2 + 1 = t. When x = 0, t = 1 and when x = 1, t = 2 1

x dx

Ú ( x 2 + 1)2 0

(c) –1 Ans. (a)

x

2

dx = - Ú

p /2

p

Example 3: If equal to

p /2

sin t dt = cos t ex

1

Ú0

= -(-1) = 1 p

dx = log

e x + e- x

e+ A 1+ 2

then A is

2

=

1 dt 112 1 È1 ˘ 1 = = - Í - 1˙ = Ú 2 21t 2t1 2 Î2 ˚ 4

Example 2: The value of (a) 1

p 1 2 , t = p and when x = , t = . Thus p p 2

sin 1 / x

2/p

Ú1 / p

1 1 = t fi - 2 dx = dt x x

2/p

Ú1 / p

sin 1 / x

1 2 1 (d) 4 (b)

x2

dx is equal

(a) e2 + 1 (c) e2 Ans. (b)

(b) e2 + 1 (d) e2 – 1

1

Solution:

Ú

1

ex x

-x

0 e +e the last integral is equal to e

Ú1

dt t2 + 1

dx = Ú

= log t + t 2 + 1

0 e 1

ex e

2x

dx. Putting ex = t +1

13.4

Complete Mathematics—JEE Main

= log So

e + e2 + 1 1+ 2

2

A=

e +1 dx

x

Úlog 2

ex - 1

=

p is 6

Ú

max (sin x, cos x) dx =

0

dx = p = 6

2t dt t2 + 1

p /4

ex - 1 x

= 2Ú

p /2

=Ú

dt

e -1

2

t +1

1

2 tan -1 e x - 1 =

2t dt (t 2 + 1)t

1

e x -1

= 2 tan -1 t 1

= 2 tan -1 e x - 1 fi

e x -1

p 2

e x - 1 = tan

fi

ex – 1 = 3

Example 7: If equal to

(a) 0 p (c) 4 Ans. (a)

p 8 p 8

Ú

0

1 2

= 2.

= log K then K is

6 - 5 sin x + sin 2 x 4 3 5 (d) 3

1

1

dt

Ú 6 - 5t + t 2 0

=

dt

ÚÊ 0

5ˆ 2 Ê 1ˆ 2 ÁË t - ˜¯ - ÁË ˜¯ 2 2

5 1 1 2 2 = log 1 5 1 t- + 2 2 2 2

1

t-

(b) 1 p 10 (d) ÊÁ ˆ˜ Ë 8¯

= log

t -3 t-2

1

= log 0

0

4 3

1

Ú

Example 8:

x dx is equal to

-1

f ( x )dx = 0 . Thus

(a) 1

(b) 0 1 (d) 2

-a

Ú- p / 8

+

(b)

a

p /8

1

cos x dx

2 3 1 (c) 3 Ans. (b)

x10 sin9 x dx is equal to

Solution: Let f (x) = x10 sin9 x, f (–x) = (–x)10 sin9(–x) = –x10 sin9 x = – f (x) Hence f is an odd function. Now using Prop. 13 Page 13.2,

Ú

Ú

(a)

p = 3 3 i.e. ex = 4 fi x = log 4.

Example 5: The value of

we have

p /4

Solution: Putting sin x = t, cos x dx = dt. When x = 0, sin x = 0 and when x = p/2, sin x = 1. So the given integral reduces to

2p 3

fi

- cos x 0

=

sin x dx p /2

= sin x

, so dx

Ú

p /4

2

x

Úlog 2

Ú

p /2

cos x dx +

0

(a) log 2 (b) 3 log 2 (c) 2 log 2 (d) 2 Ans. (c) Solution: Substituting Ex – 1 = t2, ex dx = 2t dt fi

0£ x £

p /4

p /2

Example 4: Value of x satisfying

p 4 p p £x£ 4 4

Ï ÔÔcos x Solution: max (sin x, cos x) = Ì Ô sin x x Œ [0, p/2] ÔÓ

2 = log (e + e + 1) - log (1 + 2 )

(c) 2

x10 sin 9 x dx = 0

Example 6:

p /2

Ú0

(a) 1 (c) 2 - 2 Ans. (d)

Ans. (a)

max (sin x, cos x) dx is equal to

1

(b) 2 (d)

Solution: The integrand f(x) = |x| is an even function so

2

1

1

x2 Ú x dx = 2 Ú x dx = 2 Ú x dx = 2 2 -1 0 0

1

=1 0

Definite Integrals 13.5

1 1 È + + Example 9: Thevalue of lim Í nÆ• Î 4n2 - 1 4n2 - 22 1 ˘ + is equal to 2 2 ˙ 4n - n ˚ (a) p

4n - 1

2

4n - 2

2

r n

dx

Ú

4 - x2

0

10p

+ +

2

x1 p = . 20 6

1 - sin 2 x dx is equal to

(a) 0 (c) 10p Ans. (d)

1

+

2

()

Example 10: Ú0

Solution: =

4-

=

= sin -1

p (d) 4

1

1

1

p 6

(b)

p (c) 3 Ans. (b)

1 n  nƕ n r =1 lim

(b) 10 (d) 20

Solution:

1

10p

2

4n - n

10p

1 - sin 2 x dx = Ú0 | cos x | dx

Ú0

2

p

1 1 1 1 = È + + + 2 nÍ 1 2 n Í 44- 2 4ÍÎ n2 n n

= 10 Ú | cos x | dx (Prop. 16) 0 p /2

()

2

= 20 Ú | cos x | dx (Prop. 10) 0 p /2

= 20 Ú0 cos x dx = 20 sin x p0 / 2 = 20.

So, required limit is equal to

LEVEL 1 Straight Objective Type Questions

Example 11: The value I = (a) p (c) p/4 Ans. (c)

= Considering 2I =

p /2

(a) p 2/12 (c) p 2/6 Ans. (a)

sin (p / 2 - x ) sin (p / 2 - x ) + cos (p / 2 - x )

p /2

I= p /2

Ú0

f (x) dx +

a

0

0

p /2

Ú0

cos x dx cos x + sin x

2

4 cos x + 9 sin 2 x

is

(b) p 2/4 (d) p 2/3

2

dx

p

a

x dx

p

Ú0

(p - x ) d x 4 cos (p - x ) + 9 sin 2 (p - x )

p

Ú0

= p Ú0

f (p/2 – x) dx

using Ú f ( x )dx = Ú f (a - x ) dx

sin x dx + cos x + sin x

p 2

Solution: Using Property 9, we have

cos x cos x + sin x

Ú0

1 ◊ dx =

Ú0

So I = p/4

1 sin x = , so 1 + cot x sin x + cos x

( p /2

sin x + cos x dx = cos x + sin x

Example 12: The value of I =

f (p/2 – x) =

Ú0

p /2

Ú0

=

(b) p/2 (d) p/3

Solution: Let f(x) =

=

1 dx is 1 + cot x

p /2

Ú0

)

2

4 cos x + 9 sin x

2I = p Ú

-Ú

x dx

p

0

2

4 cos x + 9 sin 2 x

dx

p 2

4 cos x + 9 sin 2 x

0

= 2p

2

dx

p /2

Ú0

2

4 cos x + 9 sin 2 x (using Property 9)

Complete Mathematics—JEE Main

13.6

= 2p = 2p 2p = 9

sec 2 x d x

p /2

Ú0

Solution: Using Property 18, we have f ¢(x) =

4 + 9 tan 2 x dt

•

Ú0

and thus the given equation reduces to x2 – 2 - x 2 = 0 fi (x2 + 2) (x2 – 1) = 0. Thus the real roots are given by x

(t = tan x)

4 + 9t 2

= ± 1.

• 2p 3 p2 -1 3 = ◊ tan t = 2 0 6 t2 + 4/9 9 2

dt

•

Ú0

Hence I = p 2/12. Example 13: If

x

Úe

t f (t) dt = sin x – x cos x –

all x ΠR ~ {0} then the value of f (p/6) is (a) 0 (b) 1 (c) Р1/2 (d) 3/2 Ans. (c)

x2 for 2

1/ 2

1 + xˆ Ê Example 16: The integral Ú Á [ x ] + 2 log dx Ë 1 - x ˜¯ equals -1 / 2 (a) – 1/2 (b) 0 (c) 1 (d) 2 log (1/2) Ans. (a) 1/ 2 -1 / 2

0

=

2

2 + 1]

x[x

Ú0

1

=

1

Ú0 2

Ú3

x

2 + 1]

=

Ú

(– 1) dx = –

-1 / 2

1 . 2

Ê -1 x 7 - 3x 5 + 7 x3 - x ˆ + cos x ÁË ˜¯ dx is cos2 x

(a) p/2 (c) 2p Ans. (d)

(b) 0 (d) p

7 5 3 Solution: Let f(x) = x - 3 x + 7 x - x then f(– x) = cos2 x

dx

1

[x2 + 1]

x[x

2 + 1]

Ú

-1

[0, 2] = [0, 1] » ÎÈ1, 2 ˘˚ » ÈÎ 2 , 3 ˘˚ » ÎÈ 3, 2 ˘˚ . x[x

0

[x] dx =

0

I=

Solution: For x Π[0, 2], x + 1 Π[1, 5], we must break

Ú

Ú [ x ] dx

-1 / 2

Example 17: The value

dx, where [x] is

2

2

Ú

1/ 2

1+ x is an odd function) 1- x

1/ 2

[x] dx +

-1 / 2

the greatest integer less than or equal to x is (a) 2 (b) 8/3 (c) 4 (d) none of these Ans. (d)

Ú0

1 + xˆ Ê ÁË [ x ] + 2 log 1 - x ˜¯ dx = (since log

Hence f (p/6) = sin p/6 – 1 = – 1/2. Example 14: The value of

Ú

Solution:

Solution: Differentiating both the sides and using Property 17, we have xf (x) = cos x – cos x + x sin x – x, so f(x) = sin x – 1.

Hence

2 - x2

Ú1

x

[x2 + 1]

Ú

dx +

3 2

x

[x2 + 1]

dx +

–f(x) so

Ú

f(x) dx = 0

(Property 12)

-1 1

dx

1

Ú0

dx +

2

Let

x dx + Ú

1

2

x2 d x + Ú

3 2

x3 d x + Ú

2 3

g(x) = cos–1 x then I1 =

Ú

cos–1 x dx

-1

x4 d x

1

=

1 1 1 1 = + ÈÎ23 / 2 - 1˘˚ + [9 - 4 ] + ÈÎ32 - 35 / 2 ˘˚ 2 3 4 5

Ú

cos–1 (– x) dx

(Property 8)

-1 1

=

469 1 3 / 2 1 5 / 2 = + 2 - 3 . 60 3 5

Ú

(p – cos–1 x) dx = 2p – I1

-1

Thus I = I1 = p. Example 15: Let f (x) =

x

Ú1

2 - t 2 dt. Then the real

p /2

Example 18: The value of I =

roots of the equation x2 – f ¢(x) = 0 are

is

(a) ± 1 (c) ± 1/2 Ans. (a)

(a) 0 (c) 4/3 Ans. (c)

(b) ± 1/ 2 (d) 0 and 1

Ú

-p / 2

(b) 2/3 (d) 1/3

cos x - cos3 x dx

Definite Integrals 13.7 np + t

p /2

Ú

Solution: I =

Example 22: The value of the integral Ú (|cos x| 0 + |sin x|) dx is (a) n (b) 2n + sin t + cos t (c) cos t (d) sin t – cos t + 4n + 1 Ans. (d)

cos x |sin x| dx

-p / 2 p /2

Ú

= 2

cos x | sin x | dx

-p / 2

Solution: Since the period of |sin x| + |cos x| is p/2 so

(the integand is an even function)

np + t

p /2

= 2

Ú

Ú0 t Ú0

cos x sin x dx

0

= - 4 (cos x )3 / 2

3

p /2 0

n (n + 1) (a) 2 (c) n (n – 1) Ans. (b) Solution:

Ú0

[x] dx = Â Ú

Ú0 =

Â

i =1

[x] dx = Â Ú

i -1

i =1

n

n

i

i =1

i

i -1

x

3

I1 =

(b) 2 (d) 3/2

nƕ

5

(c) – 1 + Ans. (b)

n2 + r 2

(d) 1 +

Solution: Required limit = lim

nƕ

=

2

Ú0

x 1 + x2

dx =

x

Ú0 cos

4

equals 5 2

p

=

Ú0

cos4 t dt

4

x

(p + u) du = Ú cos4 u du = g( x ) . 0

dx

p /2

Ú0

1 + tan3 x

is

(b) 1 (d) p/2 dx

p /2

Ú0

=

3

1 + tan x

dx

p /2

Ú0

3

1 + tan (p / 2 - x )

1 + cot 3 x tan3 x

p /2

3

1 + tan x

dx =

ˆ 1 p ÁË 1 - 1 + tan3 x ˜¯ d x = 2 - I .

p /2 Ê

Ú0

Thus I = p/4.

r /n 1 2n  n r = 1 1 + r 2 / n2

1 + x2

Ú0

x +p

dx

p /2

=

cos4 t dt

t dt + Ú

(a) 0 (c) p /4 Ans. (c) Solution: I =

(b) – 1 + 2

p

Ú0 cos

Example 24: The value of

3

r

x +p

= g (p) + I1 t = p + u, so that

In I1, put

Thus F¢(p/6) = 2 – 1/2 = 3/2. 1 2n  n r =1

(sin x + cos x) dx

cos4 t dt then g(x + p) equals

0

=

Solution: F(x) = Ú (2 – sin t)dt so F ¢(x) = 2 – sin x.

t

Ú0

(b) g(x) – g(p) (d) g(x)/g(p)

Solution: g (x + p) = Ú

(i - 1) dx

x

Example 21: lim

x

Ú0

(a) g(x) + g(p) (c) g(x) g(p) Ans. (a)

d Ê ˆ ÁË 2 + cos t ˜¯ dt then dt

Ú

(sin x + cos x) dx +

Example 23: If g(x) =

n ( n - 1) (i - 1) = . 2

Example 20: If F(x) = Êpˆ F ¢ Á ˜ is equal to Ë 6¯ (a) 1/2 (c) 3/4 Ans. (d)

(a) 1 +

p/2

Ú0

= (2n) (2) + sin t - cos t + 1 = (4n + 1) + sin t - cos t.

[x] dx (where

n (n - 1) (b) 2 (d) none of these n

n

= 2n n

(|sin x| + | cos x|)dx +

(|sin x| + |cos x|) dx

4 = . 3

Example 19: If n ΠN , the value of [x] is the greatest integer function) is

p/2

Ú0

(|sin x| + |cos x|) dx = 2n

Example 25: The value of

=

2x + 1

dx is

(a)

6 Ê 4 ˆ 66 (b) 13 log 3 ˜¯ log 3 log 3 ÁË

(c)

6 Ê 5 ˆ (d) none of these 13 log 3 ˜¯ log 3 ÁË

2 0

4

Ú0 3

5 - 1.

Complete Mathematics—JEE Main

13.8

Ans. (d)

Example 29: If f(x) = (1 + tan x) (1 + tan (p/4 – x)) and 2

Solution: Putting 2x + 1 = t , we have dx = t dt, so fi

4

Ú0 3 =

dx =

2x +1

3

Ú1

3

t ◊ 3t 3 t dt = log 3 t

0

1 log 3

3

Ú0

g(x) is a function with domain R, then

3t dt

24 (3) ◊ 33 - 3 78 1 3 1 [3 3 ] = (log 3)2 log 3 log 3 (log 3)2

1

x3 g o f (x) dx is

Ú0

(a)

1 g (p /4) 2

(b)

1 g(2) 4

(c)

1 g(1) 4

(d) none of these

Ans. (b) Example 26: The value of

1

99

Ú0

x(1 – x) dx is

(a)

1 10100

(b)

(c)

1 10010

(d) none of these

11 10100

2 ˆ Ê = (1 + tan x) Á = 2. Ë 1 + tan x ˜¯

1

Ú0

1

Ú0

So

Ans. (a) Solution:

Ê 1 - tan x ˆ Solution: f (x) = (1 + tan x) Á 1 + Ë 1 + tan x ˜¯

x (1 - x)99 dx =

1

Ú0

1

x3 g o f (x) dx =

(1 - x) (1 - (1 - x))99 dx

x 4 g (2 ) 4

=

(Property 9) =Ú

101 1

1

(x99 - x100) dx =

0

x100 x 100 101 1000

n

Â Ú n -1

Example 27: The value of

= 0

1 . 10100

ex–[x] dx is ([x] is

n =1

(a)

e1000 - 1 1000

e1000 - 1 e -1

(b)

n

1000

 Ún - 1 ex - [x] dx = Ú0 n =1

= 1000

1

Ú0

ex – [x] dx = 1000

nƕ

Sn =

1 + 2n

1

1

Ú0

4n - 1

p /2

Ú-p / 2

4n - 4

is

2

3n + 2 n - 1

(b) 2 (d) p /6

1È 1 + Í nÆ• n Í 4 - 0 Î

1

= lim

4 - 1/ n

1

+

2

4 - 4 / n2

+… 1

1 n -1 1 =  2 nƕ n r = 0 4 - (r / n )

= lim

Example 31: The value of (a) 1 (c) 2 Ans. (b)

F(- x) = (f (- x) + f (x)) (g (- x) - g(x)) = - F(x). (using Property 11)

dx

1

Ú0

[(f (x) + f (–x))

4-x 3p

Ú-p

2

˘ ˙ n - 1ˆ ˙ 4 - ÊÁ Ë n ˜¯ ˙˚

= sin -1

2

x 2

1

= 0

Solution: I =

3p

Ú- p

3p

=

Ú- p

=

Ú- p

3p

p . 6

log (sec q – tan q) dq is

(b) 0 (d) none of these

Solution: Let F(x) = ( f (x) + f (- x)) (g(x) - g(- x)) so

Ú- p / 2 f (x) dx = 0.

1

+…+

2

+

ex dx = 1000 (e – 1).

functions then the value of the integral (g(x) – g(–x))] dx is (a) 1 (b) 0 (c) – 1 (d) p Ans. (b)

p /2

1

+

2

e x – [x] dx

Example 28: If f : R Æ R, g: R Æ R are continuous

Hence

0

nƕ

Solution: Since the period of the function x - [x] is 1 so 1000

1 g(2). 4

=

Solution: lim Sn

e -1 (d) 1000

(c) 1000 (e – 1) Ans. (c)

1

Example 30: The lim Sn if

(a) p /2 (c) 1 Ans. (d)

the greatest integer function)

x3 g(2) dx

Ú0

log (sec q - tan q) dq

log (sec (2p - q) - tan (2p - q)) dq log (sec q + tan q) dq.

Definite Integrals 13.9

Thus

3p

Ú- p

2I =

3p

Ú- p

=

Solution: Since f ¢(x) = f (x) and f (0) = 2 so f (x) = 2ex, thus f (x) = x + 2ex. Hence

[log (sec q - tan q) + log (sec q + tan q)] dq 2

1

Ú0

2

log (sec q - tan q)dq 3p

Ú- p

=

dx

•

value of k is (a) 1/60 (c) 1/40 Ans. (a) dx

=

=

1 5

(a) 1 (c) – 1 Ans. (a)

1 1 ˘ - 2 Í 2 ˙dx Î x + 4 x + 9˚

\

lim

xÆ0

lim

xÆ0

Ú0 cos t

2

3/ 2

1

| x sin p x | dx = k/p2, then the value

= 2Ú

|x sin p xdx| +

= lim

x

xÆ0

Example 36: If f (x) = to

(b) 2p + 1 (d) 4

1

0

3/ 2

x sin p x dx – Ú

3/ 2

cos t dt, then f ¢(1) is equal (b) 2 cos 1 (d) none of these d d 1 (x2) – cos 1/ x 2 dx d x x2 (Using Property 18)

cos (1/x). Hence f ¢(1) = 4 cos 1. x3 Example 37: The least value of the function

x sin p x dx

1

2

2

= 2 x cos x +

1

x

Ú0

F (x) = 3/ 2

È - x cos p x sin p x ˘ + ÍÎ p p 2 ˙˚1

1 1 1 ˘ 3p + 1 = 2 È ˘ - È= , so k = 3p + 1. ÍÎ p ˙˚ ÍÎ p 2 - p ˙˚ p2 1

x2

Ú1/ x

(a) cos 1 (c) 4 cos 1 Ans. (c)

È - x cos p x sin p x ˘ =2 Í – + p Î p 2 ˙˚0

Ú0

cos x 2 ◊1 - cos 0 ◊ 0 1

|x sin p x|dx

Ú1

is

cos t2 dt and g(x) = x. Then

Solution: f ¢(x) = cos x 2

Ú- 1

x

2 = lim cos x = cos 0 = 1. xÆ0 1

Solution: I=

x

Ú0

dt

1 Èp p ˘ p 1 = Í - ˙ = , so k = . 60 5 Î4 6˚ 60

Ú- 1

dt

f (x) f ¢ (x) = lim x Æ 0 g (x) g¢ ( x)

x

fi

2

(b) 0 (d 2

Solution: Let f (x) = f (0) = g(0) = 0.

• 1 È1 1 -1 x -1 x ˘ tan tan 3 ˙˚0 5 ÍÎ 2 2 3

Example 33: If of k is (a) 3p + 1 (c) 1 Ans. (a)

Ú0 cos t

xÆ0

•È

Ú0

x

Example 35: The value of lim

Solution: •

(xex + 2e2x)dx

= kp then the

(b) 1/80 (d) 1/20

Ú0 ( x 2 + 4) ( x 2 + 9)

1

Ú0

= 2 [xex - ex|10 + e2x|10] = 2 [([e - e] + 1) + e2 - 1] = 2e2.

log1 dq = 0.

Ú0 ( x 2 + 4) ( x 2 + 9)

Example 32: If

f (x) f (x) dx = 2

(3 sin u + 4 cos u) du

on the interval (5p/4, 4p/3] is (a) 3 / 2 - 3 / 2 (c)

7-4 3 2

(b)

5-4 3 2

(d)

9-4 3 2

Example 34: If f (0) = 2, f ¢(x) = f (x), f (x) = x + f (x) then

Ans. (d)

f (x) f (x) dx is

Solution: We have F¢(x) = 3 sin x + 4 cos x. Since sin x and cos x assume negative values in the third quadrant, we have F¢(x) < 0 for all x Œ (5p/4, 4p/3) so F(x) assumes the least value at the point x = 4p/3. Thus the least value is

(a) e2 (c) 2e Ans. (b)

(b) 2e2 (d) 2e – 3/2

F(4p/3) =

4p /3

Ú0

(3 sin u + 4 cos u)du

13.10

Complete Mathematics—JEE Main

= (– 3 cos u + 4 sin u)| 04p/3 = –3 cos

4p 4p + 4 sin – (– 3) 3 3

9 4 3 9-4 3 = . = 2 2 2 1/ a

Ú

Example 38: The integral

0

equal to p 8 1 p (c) (log 2) a 8 Ans. (c)

log (1 + ax ) 1 + a2 x2

Im,n = -

Hence

5 I5,5 = - I 5,4 6 5Ê 4 ˆ = - Ë - I 5,3 ¯ 6 6

dx (a > 0) is

= (-1)3

1 p (log 2) a 4 1 (d) 2 (log 2) a

(a) a (log 2)

(b)

1 log(1 + t ) 1 dt = Ú 2 a 0 1+ t a =

=

=

=

1 a 1 a 1 a 1 a 2 a

fi

1 a

fi

p /4

Ú 0 p /4

Ú 0 p /4

Ú 0 p /4

Ú 0 p /4

Ú

1 + tan 2 u

Ú 0 p /4

Ú 0

5! 6

6

=-

5! 66

1

Ú (cos

-1

x )n dx then I6 – 360 I2 is

0 3

p p (a) 6 Ê ˆ - 24 Ê ˆ Ë2 ¯ Ë2 ¯ 5

5

3

p p (b) 6 Ê ˆ - 120 Ê ˆ Ë2 ¯ Ë2 ¯ 5

p (c) 6 Ê ˆ Ë2 ¯ Ans. (b)

3

p p (d) 6 Ê ˆ - 4 Ê ˆ Ë2 ¯ Ë2 ¯

Solution: Integrating by parts, we obtain [log 2 - log(1 + tan u)] du 1 p log(1 + tan u) du = (log 2) a 4 1 p log(1 + tan u) du = (log 2) a 8

0

dx, then I5,5 is given by 5! (a) - 5 6 5! (c) - 6 6 Ans. (c)

(b) (d)

1

5

5 5!

0

Ú (cos x) dx = x cos x 0 1

= nÚ 0

x 1- x

2

0

66

n x m (log x )n -1 dx m + 1 Ú0

1

+ Ún 0

(cos-1 x )n -1 x 1 - x2

dx

(cos-1 x )n -1 dx

= n [- 1 - x (cos x ) -1

2

1 n -1 0

1

- Ú (n - 1)(cos-1 x )n - 2 dx ]

n -1

- n (n - 1) I n - 2 5

Since lim x m (log x )n = lim ( x m / n log x )n = 0 x Æ0 +

-1

n

p I6 = 6. Ê ˆ - 6.5 I 4 Ë2 ¯

1

-

-1

Êp ˆ = nË ¯ 2

5!

Solution: Integrating by parts, we obtain x m +1 (log x )n m +1

1

1

In =

Example 39: For m > 0 , n > 0, let Im,n = Ú x m (log x) n

x Æ0 +

I5,5 = (-1)5

5

1

Im,n =

1

1 1 x 6 log x - Ú x 6 dx 6 6 x 0 0

62

Example 40: If In = given by

Ê 1 - tan u ˆ log Ë 1 + du 1 + tan u ¯

I 5,1

1

sec u du (t - tan u)

Ê Êp ˆˆ log Ë 1 + tan Ë - u ¯¯ du 4

64 1

2

log(1 + tan u) du

5.4.3.2

I 5,2 = (-1)4

5 Ú x log x dx =

= -

0 p /4

6

3

0

So log(1 + tan u)

5.4.3

1

I5,1 =

Solution: Put ax = t, the given integral reduces to 1

n Im, n –1 m +1

So,

5 È p 3 ˘ p = 6. Ê ˆ - 30 Í4 Ê ˆ - 12 I 2 ˙ Ë2 ¯ Ë ¯ Î 2 ˚ 5

3

p p I6 – 360 I2 = 6. Ê ˆ - 120 Ê ˆ Ë2 ¯ Ë2 ¯

0

Definite Integrals 13.11

Example 43: The difference between the greatest and

p /2

x n cos x dx then I8 + 56I6 is

Ú

Example 41: Let In =

0

equal to 6

8

Êp ˆ (b) Ë ¯ 2

6

Êp ˆ (d) Ë ¯ - 1 2

Êp ˆ (a) Ë ¯ 2

8

p (c) Ê ˆ - 1 Ë2 ¯

the least value of the function F (x) = [2, 3] is (a) 3 (c) 7/2 Ans. (c)

x n cos x dx

Ú

(b) 2 (d) 3/2

This is positive for all x Π[2, 3], so F is an increasing function in this interval. Therefore its greatest value is

p /2

In =

p /2

p /2

= x n sin x

3

2

0

0

F(3) = Ú (t + 1) dt and its least value is F(2) = Ú (t + 1) dt,

0

-n 0

x n -1 sin x dx

Ú 0

p /2 p /2 ˘ È Êp ˆ = Ë ¯ - n Í- x n -1 cos x + (n - 1) Ú x n - 2 cos x dx ˙ 2 0 ˙˚ ÎÍ 0 n

n

p = Ê ˆ – n(n –1) In – 2 Ë2 ¯ 8

p I8 + 8.7 I6 = Ê ˆ Ë2 ¯

so that the required difference between these values is 7 3 2 3 Ú0 (t + 1)dt - Ú0 (t + 1)dt = Ú2 (t + 1)dt = 2 . Example 44: The value of the integral p / 4 sin x + cos x Ú0 3 + sin 2 x dx is (a) log 2 (b) log 3 (c) (1/4) log 3 (d) (1/8) log 3 Ans. (c) Solution: The integral can be written -Ú

Example 42: If a function f : [0, 27] Æ R is differentiable then for some 0 < a < b < 3,

Ú f ( x)dx

3

2

3

(a) 3[a f (a ) + b f (b )]

Ans. (c) 3

Ú f (t ) dt

so g(0) = 0

0

2

g¢(x) = 3x f (x3) 27

Ú f (t ) dt 0

= g(3) =

1 g ¢( b ) 2 for some a Œ (1, 3), b Œ (0, 1) Õ (0, 3). but g¢(a) = 3a2 f (a3) and g¢(b) = 3b2 f (b )3 27

Ú 0

1 f (t ) dt = 3[a 2 f (a 3 ) + b 2 f (b 3 )] 2

dt

0

-1 t 2

-4

= -

1 4

0

t-2˘ È Ílog t + 2 ˙ Î ˚- 1

1 1 (log 1 - log 3) = log 3. 4 4 Example 45: The inflection points on the graph of =-

x

Ú0

(t – 1) (t – 2)2 dt are

(a) x = – 1 (c) x = 4/3 Ans. (c) Solution:

g(3) - g(1) 1 g(1) - g(0) + 3 -1 2 1- 0

= g ¢(a ) +

Therefore,

-Ú

function y = x

dx.

Now put t = sin x - cos x. Then dt = (cos x + sin x)dx, and the integral becomes

(b) 3[a2 f (a) + b 2 f (b )] 1 (c) 3[a2 f (a3) + b 2 f (b 3)] 2 1 (d) 3[a2 f (a) + b 2 f (b )] 2

Solution: Let g(x) =

(sin x - cos x )2 - 4

is equal to

0

2

sin x + cos x

p /4

0

27

fi

(t + 1) dt on

Solution: Differentiating the given function, we get dx d0 F¢(x) = t + 1]t = x - [t + 1]t = 0 = x + 1. dx dx

Ans. (b) Solution:

x

Ú0

(b) x = 3/2 (d) x = 1

dy d2 y = (x – 1) (x – 2)2 so = (x– 2) (3x – dx d x2

4). The points of inflection are given by x = 4/3 are points of inflection.

d2 y d x2

Example 46: The value of the integral is

= 0 so x = 2,

p 2

Ú0

dx 1 2 1 + sin x 6

Complete Mathematics—JEE Main

13.12

(a)

p 6 2 7

(c)

p 6

(b) p 3 p (d) 2

(Property 10)

2 3 \

Ans. (a) Solution: I =

p 2

Ú0

dx = 1 2 1 + sin x 6

p 2

Ú0

•

-1

2I =

Úp 4

=

Úp 4

p2 4

Ú0

= -Ú

-1

3

Ú-2

t dt + - Ú

0

0

t2 + 2 -1

1 5 = +2 = 2 2

0

3

Ú-2 =

f(x) dx = 2

Ú-2

(

= 0 + 2 = 2 since Example 49:

Ú-2

a

Ú- a 3p 4

Úp 4

3

Ú2

3

Ú2

Úp 4

x = t so that dx = 2t dt p 2

0

Úp 4

t sin t dt

Example 51: The value of the integral Ú 0 dx is (b) 200 2 (a) 100 2 (c) 0 (d) 100 p Ans. (b) Solution: We have

100 p

Ú0

1 - cos 2 x = 100 p

Ú0

sin x dx = 100 2

(a) b – a

)

dx is equal to 1 + cos x

3p 4

x dx is

p

Ú0

(c) |b| – |a| Ans. (c)

1 - cos 2 x

2 |sin x|. Since

1 - cos 2 x dx = sin x dx = 200 2 .

Example 52: Whenever a < b, the value of

2 dx

g ( x ) d x = 0 if g ( - x ) = - g ( x )

dx = 1 + cos x

sin

= 2[0 + sin t|p/20] = 2

f(x) dx

b

Úa

(b) a – b

dx 1 + cos (p - x )

x dx is x

(d) |b| + |a|

x = 1, therefore, x f(x) dx = b – a. If a < b £ 0 then f(x) = – 1 and so

Solution: If 0 £ a < b, then f(x) = b

Úa

b

b

Úa f(x) dx = a – b. Finally if a < 0 < b then Úa 0 b Úa f(x) dx + Ú0 f(x) dx

(b) – 2 (d) – 1/2 3p 4

Ú0

p /2 = 2 È-t cos t ]p0 / 2 + Ú cos t dt ˘ 0 ˚˙ ÎÍ

2

f(x) dx +

ecosx sin x dx +

(a) 2 (c) 1/2 Ans. (a) Solution: I =

2

p2 4

(b) 1 (d) none of these

the period of |sin x| is p, so

(b) 1 (d) 3

Solution:

dx

100 p

2

Ïecos x sin x for x £ 2 Example 48: If f(x) = Ì then 2 otherwise, Ó f(x) dx =

(a) 0 (c) 2 Ans. (c)

1 - cos2 x

sin x dx = 2 Ú

1 log x e2 log x log x Solution: Ú -1 dx = - Ú -1 dx + Ú dx e e 1 x x x (since log x < 0 for x Œ [e–1, 1] and log x > 0 for x Œ (1, e2))

t2 t dt = 2

2

3p 4

Solution: Putting

(b) 5/2 (d) 5

2

1 1 ˆ ÁË 1 + cos x + 1 - cos x ˜¯ dx

= – 2 cot x p3p44 = 4. Hence I = 2

e2

0

3p 4 Ê

(a) 0 (c) 2 Ans. (c)

log x dx is x

e2

Úe

Example 47: The value of (a) 3/2 (c) 3 Ans. (b)

0

Úp 4

dx 1 - cos x

Example 50: The value of 6 p. ◊ 7 2

=

=

sec 2 x d x 1 sec 2 x + tan 2 x 6

6 • dt = Ú (t = tan x) 7 0 6 7 + t2 7 = 6 ¥ 7 tan–1 t 6 7 6

3p 4

= – (0 – a) + (b – 0) = b – (– a) The above three cases can be represented by b x Úa x dx = |b| – |a|.

f(x) dx =

Definite Integrals 13.13

Example 53: Let f be an odd function then 1

Ú-1

(|x| + f(x) cos x) dx is equal to (a) 0 (b) 1 (c) 2 (d) none of these Ans. (b) Solution: The function g(x) = |x| is an even function and h(x) = f(x) cos x is an odd function so 1

Ú0

=2

|x| dx = 2

1

Ú0

4 + 4T

Ú4

T

Ú0

Ú-1 (|x| + f(x) cos x) dx

1 = 3

12 + 12T

ÚK T

fi If

T

f ( y ) dy = Ú f (u ) du ,

and

Ú12 T

Hence

f ( y ) dy =

Ú0

f (u ) du

12 1È T f ( y ) dy + 11I + Ú f (u ) du ˘ Ú 0 ˚˙ 3 ÎÍ 12

I1 =

x

(c) x = ± Ans. (c)

(b) x = ±

2

2

x -

Ú- 3 / 2 ([ x ] + x 3/ 2

6 - x 2 , the equation x2 – f ¢ (x) 4

2

fi

x +x –6=0

fi (x2 – 2) (x2 + 3) = 0

fi

x2 – 2 = 0 (x2 + 3 π 0)

x= ± 2.

, if cos x π 0 then

3

(

)

cos x π 0 so f 1/ 3 = 3.

))

(

x 2 + 1 dx is equal

+ log a x +

(a) 0 (c) 1 Ans. (b)

(b) – 3/2 (d) 3/2

(

Solution: Let f (x) = x 3 + log a x +

(

- x 3 + log a

x2

1 x+

x2 + 1

(

x 2 + 1 = – f(x)

)

= - x 3 + log a x + \

) + 1)

x2 + 1

f(–x) = - x 3 + log a - x +

3

=0

fi

6-x

2

sin 2 x

=

(d) x = ± 1

Solution: Since f ¢(x) = = 0 becomes

1

6 - u2 du . Then the real

Ú0

6

(d) 3

sin x = 1/ 3

roots of the equation x2 – f ¢ (x) = 0 are (a) x = ±

(b) 1/ 3

3

f(sin x) =

1 T 1 = È Ú f ( y ) dy + 11I ˘ = ¥ 12 I = 4 I . Í ˙ 0 ˚ 3Î 3 Example 55: Let f(x) =

(

f (t ) dt = 1 – sin x, then f 1/ 3

Example 58: The integral I =

0

12

2

Solution: Differentiating both the sides, we obtain – sin2 x f(sin x) cos x = – cos x

(Put KT + u = y or use Property 17) 12 + 12 T

1

Úsin x t

Example 57: If

f ( y ) dy

Ú12

( K + 1)T

2

Since 1 - e16 + 8 x < 0 for all x so f increases for x < 0.

(c) Ans. (d)

11 È T ˘ 12 + 12 T ( K + 1)T f ( y ) dy + Ú f ( y ) dy ˙ Í Ú12 f ( y ) dy + Â ÚK T 12 T ÍÎ ˙˚ K =1

But

2

is equal to (a) 1/3

(a) I (b) 2I (c) 3I (d) 4I Ans. (d) Solution: Put 3x = y in I1 1 I1 = 3

(b) (0, •) (d) (– 2, •)

2 - x2 + 4) È1 - e16 + 8 x ˘ = 2x e ( Î ˚

f ( x ) dx Then the value of I1 =

f (3 x ) dx is

then the function

2

x dx = 1.

with period T > 0. If I =

x2 + 4 - t 2 e dt , x2

4 - x2 + 4) - 2 x e- x Solution: f ¢(x) = 2 x e (

1

Example 54: Let f be a periodic continuous function

Ú

Example 56: If f(x) = f(x) increases in (a) (– •, 0) (c) (– 1, 2) Ans. (a)

3/ 2

Ú- 3 / 2 f ( x ) dx = 0

Thus I = =

3/ 2

Ú- 3 / 2 [ x ] dx -1

0

1

Ú- 3 / 2 (- 2) dx + Ú- 1 - 1 dx + Ú0 0 dx

+

3/ 2

Ú1

1 dx

= (– 2) [– 1 + 3/2] + (– 1) [0 + 1] + 0 + 1 [3/2 – 1] = – 1 – 1 + 1/2 = – 3/2.

)

13.14

Complete Mathematics—JEE Main

Example 59:

p /4

(

)

log 1 + tan 2 q + 2 tan q dq =

Ú0

(a) p log 2 (c) (p log 2)/4 Ans. (c)

=

fi

p /4

Ú0

p /4

(b) (p log 2)/2 (d) log 2 p /4

Ú0

Solution: Let I =

=

Ê 1 - tan q ˆ log Á 1 + dq Ë 1 + tan q ˜¯

=

Ú0

=

Ú0 [log 2 - log (1 + tan q )] dq

and b

Úa

a=

(a) I >

2 and J > 2 3

(b) I


2 and J < 2 3

\

I= J=

2

(a) p(b – a) /8 (c) p(b – a)2 Ans (a) Solution: Let t =

p log 2 4

( x - a ) (b - x ) dx (b > a ) is equal to

2

(b) p(b + a) /8 (d) p(b + a)2

1 sin x

Ú0

1 cos x

Ú0

0

x

a - t dt

=

Ê t ˆ˘ a 2 - t 2 + a 2 sin - 1 Á ˜ ˙ Ë a ¯ ˚0

dx = 2

where [◊] denotes the

p /2

Ú ([cot x ] + [cot (p - x)]) dx 0

Ú ([cot x ] + [- cot x ]) dx 0

Put cot x = t, so that •

I=

2 = p a = p (b - a ) . 2 8

dt

Ú [t ] + [ - t ] 1 + t 2 0

k

n

1 1ˆ Ê1 Example 61: lim Á + +…+ ˜ = nÆ• Ë n n+2 3n ¯

= lim

nƕ

dt Â Ú ([t ] + [- t ]) 1 + t 2 k =1 k -1

But [t] + [– t] = – 1 for k – 1 < t < k, therefore

(b) log 3 (d) 0

k

dt Ú ([t ] + [- t ]) 1 + t 2

k -1

1 1ˆ Ê1 Solution: lim Á + +…+ ˜ nÆ• Ë n n+2 3n ¯

k

=

1 1 ˆ Ê1 = lim Á + +…+ nÆ• Ë n n +1 n + 2 n ˜¯ 2n

1 1 1 = lim   nƕ nƕ n k =0 n + k k = 0 1 + k /n

= lim

x

2 3

p /2

2

2

2n

1

Ú [cot x ] dx ,

Solution: I = Ú [cot x ] dx =

a

(a) log 2 (c) log 5 Ans. (b)

1

0

0

= 2Ú =t

dx < Ú

p

2

x dx =

greatest integer function, is equal to (a) – 1 (b) – p/2 (c) p/2 (d) 1 Ans. (b)

(t + a ) (a - t )dt a

1

0

b-a , so that 2

Úa

dx .

p

x-a+x-b 1 = x – (a + b) 2 2

( x - a ) (b - x )dx =

dx < Ú

0

x

Example 63: I =

a

x

Solution: Since sin x < x for x > 0, we get sin x < x for 0 < x < 1 x

p /4

Úa

1 cos x

Ú0

Ans. (b)

2 I = p/4 log 2.

Example 60:

dx and J =

x Then which one of the following is true

log (1 + tan (p / 4 - q )) dq

b

1 sin x

Ú0

Example 62: Let I =

log (1 + tan q ) dq

Required integral = 2I =

2

dx Ú0 1 + x = log (1 + x ) 0 = log 3. 2

Ú

(- 1)

k -1

dt 1 + t2

n

\

(

= ÈÎtan - 1 k - tan - 1 ( k - 1)˘˚

I = - lim  tan - 1 k - tan - 1 ( k - 1) n Æ • k =1

)

Definite Integrals 13.15

= - lim ÈÎtan - 1 n - tan - 1 0 ˘˚ = nÆ•

p . 2

Example 64: Let p(x) be a function defined on R such that p¢(x) = p¢(1 – x), for all x Œ [0, 1], p(0) = 1 and p(1) = 41. Then

1

Ú0 p ( x ) dx

equals

(a) 41 (c) 41

(b) 42 (d) 21

Ans. (d) Solution: 1

1

Ú0 p ( x ) dx = Ú0 1 . p ( x ) dx = [ xp ( x )]0 - I1 = p (1) - I1 where I1 = =

1

1

Ú0 xp¢ ( x ) dx = Ú0 (1 - x ) p¢ (1 - x ) dx 1

Ú0 (1 - x ) p¢ ( x ) dx = Ú0 p ¢ ( x ) dx – I1

= p (1) =

Example 67: Let [.] denote the greatest integer function then the value of

1 8 log(1 +

Ú0

x)

1 + x2

I=

dx is

(b) p log2

p log 2 8

(d)

p log 2 2

Solution: Let I = 8Ú

0

= 8Ú

p /4

0

log

x)

1 + x2

sec 2 q

(b) 3/2 (d) 5/4

sec2q dq (x = tan q )

=

2 2◊25 1È 0 + Ú dt + 2 Ú dt ˘ ˙˚ Í 1 2 2Î

=

1 3 [1 + 2 ¥ .25] = 2 4

Solution: g(x + p) =

p

=

I = 8Ú log(1 + tan q ) dq

1 dt 2

2 2◊25 1 2◊25 1 1 t ] dt = Ú [t ]dt + Ú [t ] dt + Ú [t ]dt [ Ú 1 1 2 0 2 0

x

Ú0 cos 4t dt, , then g(x + p) equals

(a) g(x) + g(p) (c) g(x) g(p) Ans. (a), (b)

dx

(1 + tan q )

x ÈÎ x 2 ˘˚ dx is

Example 68: If g(x) =

Ans. (b) 1 log(1 +

1◊5

Ú0

Solution: Put x2 = t, so that x d x =

1 1 ( p (1) + p (0)) = (41 + 1) = 21 2 2

(a) log 2

So

x sin x, so f ¢(x) = 0 1 fi x = p, 2p but f ¢¢(x) = x cos x sin x, so f ¢¢(p) = 2 x - p < 0 and f ¢¢(2p) = 2p > 0. Hence f has local maximum at x = p and local minimum at x = 2p.

1 ( p (1) - p (0)) 2

Example 65: The value of

(c)

I = p log 2 x

(a) 0 (c) 3/4 Ans. (c)

1

1

fi

Example 66: For x Œ (0, 5p/2), define f(x) = Ú t sin t dt 0 Then f has (a) local maximum at p and local minimum at 2p. (b) local maximum at p and 2p (c) local minimum at p and 2p (d) local minimum at p and local maximum at 2p Ans. (a)

1

2I1 = [ p ( x )]0 = p (1) - p (0 )

Ú0 p ( x ) dx

p 4

Solution: f ¢(x) =

1

fi Thus,

2I = (8 log 2 )

fi

0

(b) g(x) – g(p) (d) g(x) g(p) x+p

Ú0

cos 4t dt

p

x +p

Ú0 cos 4t dt + Úp

cos 4t dt

= g(p) + I = 8Ú

p /4

0

(Ú

a

0

= 8Ú

p /4

p /4

0

a

f ( x )dx = Ú f ( a - x ) dx

0

= 8Ú

Êp ˆ log(1 + tan Á - q ˜ dq Ë4 ¯ 0

)

Ê 1 - tan q ˆ dq log Á 1 + Ë 1 + tan q ˜¯ log

where

p 2 dq = (8 log2 ) - I 4 1 + tan q

So

Úp

I=

Ú0 cos(4p + 4q )dq

cos 4t dt , Put t = p + q, so that

x

=

x

Ú0 cos 4q dq

g(x + p) = g(p) + g(x) but g(p) =

\

x +p

I=

p

1

p

Ú0 cos 4t dt = 4 (sin 4t 0 = 0.

g(x + p) = g(x) – g(p) also.

= g(x)

13.16

Complete Mathematics—JEE Main

Example 69: Let f : [–1, 2] Æ [0, •] be a continuous function such that f(x) = f(1 – x) for all x Œ [–1, 2]. Let R1 = 2

Ú-1 x f ( x) dx, , and R2 be the area of the region bounded by

Example 71: Let f be a continuous function satisfying t2

Ú ( f ( x) + x

(a) p (c)

Solution:

= fi 2R1 =

2

Ú-1 x f ( x)dx = Ú-1 (2 + (-1) - x) f (2 + (-1) - x)dx 2

2

Ú-1 (1 - x) f (1 - x)dx = Ú-1 (1 - x) f ( x)dx 2

Ú-1 f ( x)dx

= R 2.

I=

(b)

p Êp ˆ 2 Ë4 ¯

p Êp ˆ 2 Ë4 ¯

2

(d) p -

4

p4 16

Solution: Differentiating both sides, we have (f (t2) + t4) 2t = 4t2 f (t2) + t4 = 2t fi f(t2) = 2t – t4 Ê Êp ˆ2ˆ p Êp ˆ4 Êp2 ˆ f ÁË ˜¯ = f ÁË Ë ¯ ˜¯ = 2 - Ë ¯ 2 2 4 2

so

Ú

log 3

x sin x 2

log 2

sin x 2 + sin log 6 - x 2

1 3 log 4 2

(

(b)

1 3 log 2 2

1 3 log (d) 6 2

3 (c) log 2 Ans. (a)

Solution: Put x2 = t, so that I= =

=

1 log 3 sin t dt Ú log 2 sin t + sin(log 6 - t ) 2

)

p4 . 16 Example 72: For a continuous function f, the value = p-

dx is

•

Ú f (x

n

+ x-n )

0

p 2 (c) –p Ans. (a)

=

1 [log 3 - log 2] 2

I=

1 1 = t fi dx = - 2 dt, so x t

•

Ú f (x

n

+ x - n ) log x

0

dx x

0

1ˆ dt Ê = - Ú f (t - n + t n ) Ë log ¯ t 2 t t • •

= - Ú f (t - n + t n ) log t

log 3 sin t

1 3 log . 4 2

(d) 2p

I=

1 log 3 sin(log 6 - t ) dt Ú log 2 sin t + sin (log 6 - t ) 2 1 + sin (log 6 - t ) dt 2 Úlog 2 sin t + sin (log 6 - t )

(b) 0

Solution: Putting

1 log 3 sin(log 2 + log 3 - t ) dt 2 Úlog 2 sin(log 2 + log 3 - t ) + sin t

=

log x 1 + dx is x 1 + x2

(a)

2I = I + I

fi

p4 8

Ans. (d)

2

Example 70: The value of

(a)

4 3 t for all t, then f (p 2/4) is equal to 3

) dx = p +

-p

y = f(x), x = –1 and x = 2 and the x-axis. Then (a) R1 = 2R2 (b) R1 = 3R2 (c) 2R1 = R2 (d) 3R1 = R2 Ans. (c)

R1 =

2

0

fi 2I = 0 fi I = 0 The given integral is equal to •

1

Ú 1 + x2 0

•

dx = tan -1 x

= 0

p 2

dt = -I t

Definite Integrals 13.17

Assertion-Reason Type Questions

Example 73: If

• – x2 e dx 0

Ú

• e– x

Ú0

Statement-1:

=

Ans. (a)

p then 2

dx = p

x

2

Statement-2: lim e- x = 0 xƕ

Ans. (b) Solution: Put x = t

2

• e– t • 2 Ú0 x dx = 2Ú0 t t dt = 2Ú0 e– t dt • sin x p Example 74: If Ú dx = then 0 x 2 • sin ax cos bx Statement-1: Ú dx = p /2 (a > b > 0) 0 x sin ax cos bx Statement-2: lim =a xÆ0 x Ans. (b) 1 Solution: sin ax cos bx = [sin (a + b) x + sin (a – b)x] 2

cos bx dx x

Example 75: Statement-1:

p /3

10

Ú– p / 3 x

sin9 x dx = 0

Statement-2: f (x) = x 2n is an even function and g(x) = sin2m + 1x is an odd function, m and n are integers. Ans. (a) Solution: x10 sin9 x is an odd function so x10 sin9 x dx = 0.

Example 76: Suppose that f is an odd function and F(x) =

x

Úa

–x

a

Ú– a f (t ) dt = 0

–a

–a

f (u ) du

f (u ) du + Ú

–x

a

(u = – t) f (u ) du

cos x

18

Ú10

1 + x4

dx

cos x 1 + x4

< 0.1

Ans. (b) Solution: Since cos x £ 1, the condition x > 10 yields cos x 1+ x

4

< 10– 2 < 10– 1 | I | < 8.10– 2 < 10–1. 2p

Example 78: Statement-1:

Ú cos

2 m +1

x dx = 0 (m > 0)

0

Ú

a

f ( x ) dx = 2 Ú f ( x ) dx if f (2a – x) = f (x)

0

0

Ans. (a) 2p

Solution:

p

Ú cos

2 m +1

x dx = 2 Ú cos2 m +1 x dx

0

0

= 0 (since cos (p – x) = – cos x) Example 79: p /4

Statement-1:

Ú

log (1 + tan q ) dq =

0

p log 2 8

p /2

Statement-2:

Ú

log sin q dq = - p log 2

0

Ans. (c) Solution: I=

Statement-1: F is an even function

a

–x

f (u ) du = F ( – x )

p /4

f (t ) dt .

Statement-2:

Úa

Statement-2:

Statement-2:

• sin ( a - b ) x 1 È • sin ( a + b ) x ˘ dx + Ú dx ˙ Ú Í 0 2Î 0 x x ˚ • sin u 1 È • sin t ˘ = ÍÚ dt + Ú du ˙ 0 2Î 0 t u ˚ = p /2

Ú– a

=

f (u ) du = Ú

2a

=

a

Ú– a

f (t ) dt = – Ú

Statement-1: | I | < 0.1

• sin ax

Ú0

–x

=

Example 77: Let I =

2

• e– x

x

Úa

Solution: F(x) =

Ú 0

p /4

log (1 + tan q ) dq =

Ú 0

Ê Êp ˆˆ log Ë 1 + tan Ë - q ¯¯ dq 4

13.18

Complete Mathematics—JEE Main p /4

=

Ú 0 p /4

=

Ú 0 p /4

=

Ú

Ê log Ë

log 2 dq -

Ú

fi

Ú

I=

0

Ú 0

I2 =

Ú

Ú

log sin 2q dq =

1 log sin t dt 2 Ú0

p

1 = .2 2

p /2

=

p p log 2 = I 2 - log 2 2 2

0

Êp ˆ log sin Ë - q ¯ dq 2

Hence, I1 = -

log cos q dq

Ê sin 2q ˆ log Ëlog dq 2 ¯

log sin 2q dq -

p /2

but

p log 2 8

p /2

log sin q dq =

Ú 0

log (1 + tan q ) dq

Ú 0

p /2

=

0

p log 2 4

Ú

p /2

log (sin q cos q ) dq =

0

p /4

p /2

I1 =

2I1 =

2 ˆ dq 1 + tan q ¯

0

2I =

p /2

Ê 1 - tan q ˆ log Ë 1 + dq 1 + tan q ¯

p /2

Ú

log sin t dt = I1

0

p log 2 . 2

0

LEVEL 2 Straight Objective Type Questions e

x

Example 80: Let f: (0, •) Æ R and F(x) =

Ú

1 2u du = - I1 + Ú 2 1 / e 1 + u2

f(t) dt.

0

If F(x2 ) = x2(1 + x) then f (4) equals (a) 5/4 (b) 7 (c) 4 (d) 2 Ans. (c)

So

I1 + I2 =

x2

2

Solution: F(x ) = x

È Ê e 2 + 1ˆ ˘ 2 log e + log 1 Í ÁË e2 ˜¯ ˙ Î ˚

=

1 2

=

1 ¥ 2 =1. 2

(

0

2

Ú

Ú

f(t) dt, therefore, x2(1 + x) =

1 e log(u2 + 1)]1/e 2

sin 2 x

Example 82: If f(x) =

f(t) dt. Differentiating both sides w.r.t. x using Property

tan x

1+ t

1/ e

then the value of I1 + I2 is (a) 1/2 (c) e/2 Ans. (b)

2

dt and I2 =

Ú 1/ e

dt t (1 + t 2 )

(b) 1 (d) (1/2) (e + 1/e)

I2 = –

Ú e

u du 1+ u

2

tan x

=-

Ú 1/ e

t dt then the value of f(x) + g(x) is (b) p/4 (d) sin2 x + sin x + x

Solution: f ¢(x) + g¢(x) = sin–1 (sin x) 2 sin x cos x – cos–1 (cos x) 2 sin x cos x (Property 17) = x sin 2x – x sin 2x = 0 for all x Œ R. Hence f (x) + g(x) = constant = C (say) 1/ 2

Ú

Putting x = p/4, C =

sin -1

1/ 2

t dt +

0

u du 1+ u

cos

-1

(a) p (c) p/2 Ans. (b)

Solution: Putting t = 1/u in I2 we have tan x

sin–1 t dt and g(x) =

0

cot x

t

Ú

Ú

Ú 0

cos2 x

0

17, we have 2x + 3x2 = f (x 2 ) ◊ 2x fi f (x2) = 1 + (3/2)x Putting x = 2, we have f (4) = 1 + 3 = 4. Example 81: If I1 =

)

2

e

+

Ú 1/ e

u du 1+ u

2

1/ 2

=

Ú 0

Ú cos 0

p p dt = 4 2

Hence f (x) + g(x) = p /4.

-1

t dt

Definite Integrals 13.19 x

dx

Ú

Example 83: Solution of the equation

log 2

ex - 1

=

p 6

are (a) x = log 6 (c) x = 3 Ans. (b)

Solution: Putting ex – 1 = t2, we have ex dx = 2t dt x

log 2

Ú

=

x

e -1

1

e -1

2t dt

(

)

2

t t +1

=2

Ú 1

k

Ú1 - k

dt

I1 =

2

t +1

Thus the given equation reduces to p p p fi e x - 1 = tan = 3 tan–1 e x - 1 - = 4 12 3 so ex = 4 fi x = 2 log 2

Ú0

Example 84: The value of lim

p /2

mƕ

(a) 0 (c) 2 Ans. (d)

Ú0

p /2

Ú0

sin 2 m x d x

p /2

Ú0

I2n + 1 =

sin2n x dx

I2 m +1 I2 m -1

Also

I2 m +1

Hence lim

I2 m -1

m Æ • I2 m +1

=

≥

I2 m I2 m +1

≥1

= lim 2 m + 1 = 1. mƕ 2 m

From (i) and using sandwitch theorem we have lim

m Æ • I2 m +1

= 1.

Ú1 - k

=

Ú1 - k

k

x

0

2m +1 . 2m

I2 m

=

= bÚ

sin2n + 1 x dx

I2 m -1

b

Úa

f (a + b - x) dx, we

(1 - x) f ((1 - x)x) dx k

Ú1 - k

f((1 - x)x) dx -

x f ((1 - x)x) dx

bt cos 4t - a sin 4t

a sin 4 x –1 x t holds for all x such that 0 < x < p/4 then a and b are given by (a) a = 1/4, b = 1 (b) a = 2, b = 2 (c) a = –1, b = 4 (d) a = 2, b = 4 Ans. (a) x b t cos 4 t - a sin 4 t Solution: Ú dt 0 t2

= bÚ

x

0

For all x Œ (0, p/2), sin2m - 1 x > sin2m x > sin 2m + 1 x Integrating from 0 to p/2, we get I2m - 1 ≥ I2m ≥ I2m + 1 hence

k

x

2m I2m - 1 . 2m +1

I2m + 1 =

f (x) dx =

(k + 1 - k - x) f ((k + 1 - k - x)

Ú0

sin 2 m + 1 x d x

2n 2n - 2 2 ¥ … ¥ and = ¥ 2n +1 2n -1 3 Also,

k

Ú1 - k

b

Úa

Example 86: If the equality

2n -1 2n - 3 1 p ¥…¥ ¥ , ¥ 2 2 2n 2n - 2

=

f (x (1 – x)) dx

= I 2 - I 1. 2I1 = I2 and I1/I2 = 1/2.

So

(b) 1/2 (d) 1

Solution: We know that I2n =

k

Ú1 - k

(1 - (k + 1 - k - x)) dx.

pˆ Ê = 2 Á tan - 1 e x - 1 - ˜ Ë 4¯

p /2

x f (x (1 – x)) dx, I2 =

Solution: Since have

x

e -1

dx

Ú

so

I1 =

where 2k – 1 > 0 then I 1/I2 is (a) 2 (b) k (c) 1/2 (d) 1 Ans. (c)

(b) x = 2 log 2 (d) x = 1/2

x

Example 85: Let f be a positive function and

dt =

x sin 4 t cos 4 t dt - a Ú0 2 dt t t

È sin 4 t cos 4 t dt - a Ít t Î

x 0

+4Ú

x

0

cos 4 t ˘ dt ˙ t ˚

x

cos 4 t a sin 4 x dt + - 4a x t

x

cos 4 t a sin 4 x dt + - 4a x t

= (b - 4a)

Ú0

(b - 4a)

Ú0

Thus (i)

2

a sin 4 x -1 x x cos 4 t i.e. (b - 4a) Ú dt = 4a - 1. 0 t Since R.H.S. is independent of x, so we must have b - 4a = 0 and 4a - 1 = 0 i.e., a = 1/4, b = 1. =

Example 87: The value of (a) –1/2 (c) 1/2

p /2

Ú0

1 + 2 cos x (2 + cos x )2

(b) 2 (d) none of these

dx is

Complete Mathematics—JEE Main

13.20

Ans. (c)

Ans. (d)

= 2Ú

p /2

0

2

(2 + cos x)

dt

Ú0 3 + t 2

dt

1

Ú0

2

(3 + t 2 )

(cos x + 2) - 3 dx (2 + cos x)2

Ú0 3 + t 2

+ 12

Ú0

dt

2 2

(3 + t )

È 1 t + 12 Í 2 Í 2.3 t + 3 Î

(

1

+

)0

1 2.3

˘ dt ˙ Ú0 3 + t 2 ˙ ˚

3

Ú -1

{|x – 2| + [x]} dx, where

3

1

Ú0

0

Ú- 1 {|x - 2| + [x]}dx = Ú- 1 {|x - 2| + [x]}dx +

Solution:

{|x - 2| +[x]}dx +

2

Ú1

{|x - 2| + [x]}dx +

0

1

Ú- 1 (2 - x - 1) dx + Ú0 2

Ú1 x2 2

Ú0

f (x) dx = -

cos 3 x 3

p /2

= 0

1 . 3

Example 90: If a is a positive integer, then the number of values of a satisfying

1

[x] denotes the greatest integer less than or equal to x is (a) 5 (b) 7 (c) 4 (d) 3 Ans. (b)

= x-

p /2

p /2

Ú0

{

}

Ê cos 3 x 3 ˆ a2 Á + cos x˜ + a sin x - 20 cos x dx Ë 4 ¯ 4

a2 are 3 (a) one (c) three Ans. (d) £

Example 88: The value of

=

= sin x (3 - 4 sin2 x) = sin3 x So

1 . 2

=

Solution: By applying the operation C1 Æ C1 - C2 - C3, f (x) can be written as sin x sin 2 x sin 3 x 0 3 4 sin x f (x) = 0 sin x 1

(t = tan x/2)

dt

1

Ú0 3 + t 2

=-2

0

1 + t2

1

-6

dt

1

=-2

p /2 2

dx = Ú

p /2 dx dx - 3 Ú0 (2 + cos x )2 (2 + cos x )

1

=4

1 + 2 cos x

p /2

Ú0

Solution:

0

+ 2x -1

Ê = - Á-1Ë

1ˆ Ê ˜ + Á2 2¯ Ë

x2 2

3

Ú2

{|x - 2| + [x]}dx

(2 - x + 0)dx +

(2 - x + 1)dx + 1

+ 3x 0

x2 2

2

+ 1

1ˆ Ê ˜¯ + (6 - 2 ) - ÁË 3 2

3

Ú2

x2 2

(x - 2 + 2) dx 3

2

1ˆ 9 ˜ + - 2 = 7. 2¯ 2

(b) two (d) four

Solution: The L.H.S. of the above inequality is equal p /2 Ê sin 3 x 3 ˆ to a2 Á = + sin x˜ – a cos x - 20 sin x 0 Ë 12 ¯ 4 2 a2 Ê 1 3ˆ + ˜ - a (0 - 1) - 20 = + a - 20. a2 Á Ë 12 4 ¯ 3 Thus the given inequality is (2a2/3) + a - 20 £ - a2/3 i.e., a2 + a - 20 £ 0 ¤ - 5 £ a £ 4 Since a is a positive integer so a = 1, 2, 3, 4. Example 91: The area bounded by the curve y = f(x) = x4 – 2 x3 + x2 + 3, x-axis and ordinates corresponding to minimum of the function f(x) is (a) 1 (b) 91/30 (c) 30/9 (d) 4 Ans. (b) Solution: f ¢(x) = 4x3 - 6x2 + 2x = 2x(2x2 - 3x + 1) = 2x (2x - 1) (x - 1). Since f is a differentiable function, so extremum points of f (x), we must have f ¢(x) = 0 so x = 0, 1/2, 1. Now f ≤(x) = 12x2

Example 89: If f (x) = sin x + sin 2 x + sin 3 x sin 2 x sin 3 x 3 + 4 sin x 3 4 sin x 1 + sin x sin x 1 then the value of (a) 3 (c) 2/3

p /2

Ú0

- 12x + 2, f ≤(0) = 2, f ≤(1) = 2 and f ≤(1/2) = 3 - 6 + 2 = - 1. Thus the function has minimum at x = 0 and x = 1. Therefore, the required area =

1

Ú0

(x4 - 2x3 + x2 + 3) dx (See Chapter 14

for area as definite integral) f (x) dx is (b) 0 (d) 1/3

1

Ê x 5 x 4 x3 ˆ 1 1 1 91 + + 3x ˜ = - + + 3 = . = Á 2 3 Ë 5 ¯0 5 2 3 30

Definite Integrals 13.21

Example 92: The value of (a) (c)

(2m + 1) ! p

sinn x cos2m + 1 x dx is

(b)

(n !)2 cos2m – 1 x dx

Ú0

p

Ú0

(a) 1 (c) p /2 Ans. (d)

(2m + 1) ! n!

(d) none of these

p

Ú0

=

p

Ú0

sinn x cos2m + 1x dx

sin n (p - x) cos 2m + 1 (p - x) dx p

sin nx cos

Ú0

= –

2m + 1

x dx = – I. p

Ú0

So 2I = 0 which implies that I = 0 Also, by a similar reasoning.

cos

So 2m – 1

x dx = 0

p /2

Ú0

p /2

=

Ú0

=

Ú0

2I =

Ú0

p /2

p /2

x

1/ 3

Ú -1 /

1- x

3

4

cos- 1

(a) 0 (c)

2x 1+ x (b)

p 3 +1 + log 3 2 3 -1

1

p 3

=

x4

1/ 3

Ú- 1/ 1/ 3

Ú- 1/

p = 2

31-

x

4

3 -1

(d) none of these

cos- 1

2x 1 + x2

sin (p - 2 q ) sin (p/2 - q)dq sin 2 q cos q d q sin 2 q (sin q + cos q) dq

Ú-1

1 1 Èt ˘ 1 - t 2 dt = 2 Í 1 - t 2 + sin -1 t ˙ 2 Î2 ˚0

p p = 4 2

= 2

Example 95: Given Im =

p . 4

fi

I=

e

(log x)m dx. If

Ú1

= e then values of K and L are 1 1 (a) 1 – m, (b) ,m 1- m m

dx

(c)

x4

Êp 2x ˆ dx - sin - 1 Á 31- x Ë 2 1 + x 2 ˜¯

Im Im - 2 + K L

m (m - 2) 1 m , (d) ,m–2 1- m m -1 m -1

Ans. (a)

4

x

1/ 3

Ú- 1/

Solution: Im =

4

dx x4 (sin–1 (–x) = – sin–1 x, so the last integral is zero)

Ú0

= p

Ú0

m -1 È = e - m Í x ( log x ) Î

3

p 2

È1 3 +1 p ˘ + ˙. Í log 3 -1 6˚ Î2

1

-

e 1

e

- ( m - 1) Ú ( log x ) 1

m-2

d x˘ ˚˙

= e - me + m (m - 1) Im - 2 = (1 - m)e + m (m - 1) Im - 2 Im 1 So + m Im - 2 = e. Thus K = 1 - m and L = . 1- m m

1Ê 1 1 ˆˆ ÁË - 1 + 2 ÁË 1 - x 2 + 1 + x 2 ˜¯ ˜¯ dx

+

e

1

1/ 3 Ê

p

(log x)m dx = x (log x)m

e

1 ˆ ÁË - 1 + 1 - x 4 ˜¯ d x

È p p È1 1+ x ˘ + Í log + tan - 1 x ˙ = Í1- x 3 2 Î2 ÍÎ ˚

e

Ú1

m–1 m Ú (log x) dx

31-

1/ 3 Ê

= p

= -

2I =

3 +1

+ log

Ans. (d) Solution:

1

\

dx is

2

sin 2 q sin q dq

In 2I, put sin q – cos q = t, so that t2 = 1– sin 2q fi 1 – t2 = sin 2q. Also dt = (cos q + sin q) dq.

Example 93: The value of the integral 4

sin 2q sin q dq is

(b) 0 (d) p/4

Solution: I =

Ans. (c) Solution: I =

p /2

Ú0

Example 94: The value of

1/ 3 ˘

Example 96: The numbers P, Q and R for which the function f (x) = Pe2x + Qex + Rx satisfies f (0) = – 1,

0

f ¢(log 2) = 31 and

˙ ˙˚

(a) (b) (c) (d) Ans. (d)

log 4

Ú0

[f (x) – Rx] dx = 39/2 are given by

P = 2, Q = – 3, R = 4 P = – 5, Q = 2, R = 3 P = 5, Q = – 2, R = 3 P = 5, Q = – 6, R = 3

13.22

Complete Mathematics—JEE Main

Solution: We have f ¢(x) = 2Pe2x + Qex + R, so that 31 = f ¢(log 2) = 8P + 2Q + R. Also, - 1 = f (0) = P + Q. Further, log 4 log 4 39 = Ú [f (x) - Rx] dx = Ú [Pe2x + Qex] dx 0 0 2 P 2x e + Qex = 2

log 4 0

15 P + 3Q 2 Solving the above equations, we get P = 5, Q = - 6 and R = 3.

Ú0

(b) tan (sin a) (d) (a/2) (sin a)

(a) a / (2 sin a ) (c) a sin a Ans. (a)

=

dx

1

( x + cos a ) + 1 - cos a

=

1 x + cos a tan - 1 sin a sin a

1

=

1 sin a

aˆ Ê -1 Ê ˆ -1 ÁË tan ÁË cot ˜¯ - tan cot a ˜¯ 2

=

1 Êp a p a ˆ sin a. Á - - + a ˜¯ = sin a Ë 2 2 2 2

Example 98: The value of the integral is (a) 7 (c) 5 log 13 Ans. (b)

=

1

Ú0

x log x

•

Ú0

2 2

(1 + x )

dx

x log x 2 2

(1 + x )

1

I3 =

Ú0

x log x 2 2

(1 + x )

x log x

1

Ú0

dx -

2

(1 + x2 )

1

e– x cos2 x dx, 2

e– x dx,

1

I2 =

Ú0

I4 =

Ú0

1

2

2

e– x /2 dx

1

Ú0

2

e- x cos2 x dx >

e- x cos2 x dx. Also cos2 x £ 1, therefore

1

e- x cos2 x dx £

2

1

Ú0

2

e- x dx
0 is (a) log c (b) 2 log (c + 1) (c) 3 log c (d) log (c + 1) Ans. (d) Solution: Note that the integral is not an elementary function. 1 xc - 1 dx so Let I(c) = Ú 0 log x

0

Ê - 1 Ê 1 + cos a ˆ - 1 cos a ˆ ÁË tan ÁË sin a ˜¯ - tan sin a ˜¯

(1 + x2 )

dx = Ú

dy

2

Hence I4 is the greatest integral.

1 sin a

Ú0

(1 + x ) Ú0

Ú0

=

x log x

2 2

I1 =

Ú0

dx

Ú0 ( x + cos a )2 + sin2 a

•

(1 + y2 )

- x2 > - x, so that e- x > e–x. Hence

2

=

Solution:

y log y

2

2

1

1

0

(b) I = I2 (a) I = I1 (c) I = I3 (d) I = I4 Ans. (d) Solution: For 0 < x < 1, we have (1/2) x2 < x2 < x i.e.,

Ú0 x 2 + 2 x cos a + 1

Ú0

Ú1

1◊ log y - 1 Ê 1 ˆ - 2 ˜ dy 2 Á Ê 1ˆ Ë y ¯ y Á1 + 2 ˜ Ë y ¯

then

dx

1

0

Example 99: If I is the greatest of the definite integrals

dx/(x + 2x cos a +1) is equal to –1

x log x

•

Ú0

fi

2

Solution:

2

(1 + x2 )

dx =

= -Ú

Example 97: For 0 < a < p. The value of the definite integral

Ú1

x log x

P P = ¥ 16 + 4Q -Q 2 2

=

1

•

\

•

Ú1

x log x 2

(1 + x2 )

dx 2

Put x = 1/y in the second integral, so that dx = (- 1/y )dy. If x Æ •, then y Æ 0, and if x = 1, then y = 1.

(a) p/2 (c) – p/2 Ans. (c)

(b) p/4 (d) none of these

Definite Integrals 13.23

Solution:

fi

d dx

d Ê -1 1 ˆ Ú- 1 d x ÁË tan x ˜¯ 1

I=

= -

-1 Ê 1ˆ - 2˜ = Á 1 Ë x ¯ 1 + x2 1+ 2 x 1 -d x 1 dx = Ú = - tan-1 x -1 -1 1 + x2

Ê -1 1 ˆ ÁË tan ˜ = x¯

1

is (a) p (c) 3p Ans. (a)

0

sin ( n + 1/ 2 ) x dx (n ΠN) sin ( x / 2)

= sin

Ê1 ˆ ÁË + cos x + cos 2 x + … + cos n x˜¯ 2

x x x x + 2 sin cos x + 2 sin cos 2x + … + 2 sin 2 2 2 2 cos nx

= sin

x 3x x 5x 3x + sin - sin + sin - sin +… 2 2 2 2 2 1ˆ 1ˆ 1ˆ Ê Ê Ê + sin Á n + ˜ x - sin Á n - ˜ x = sin Á n + ˜ x Ë ¯ Ë ¯ Ë 2 2 2¯

\

fi

p

p

+…+ 0

sin n x n

p

ˆ ˜ =p 0¯

Example 103: The equation of the tangent to the curve x3

Úx

2

at x = 1 is

2y + 1 = x

˘ d 3 È 1 ◊ x -Í ˙ ˙˚t = x3 d x ÍÎ 1 + t 2

( )

˘ d ◊ x2 ˙ ˙˚t = x 2 d x

( )

3x

2

1+ x

6

-

2x 1 + x4

Therefore the slope of the required tangent is

This tangent passes through the point x = 1 and y(1) = 1 dt 1 Ú1 1 + t 2 = 0, so that its equation is y - 0 = 2 (x - 1) fi

2 y + 1 = x.

Example 104: The mean value of the function f(x) = 1 on the interval [1, 3/2] is x2 + x (a) log (6/5) (b) 2 log (6/5) (c) 4 (d) log 3/5 Ans. (b) b 1 Solution: Mean value = f ( x ) dx (Property 14) Ú a b-a =

1 3 2 -1

1

32

Ú1

2

x +x

dx = 2 Ú

3 2 È1

1

1 ˘ dx ÍÎ x - x + 1 ˙˚

32

= 2 (log x – log (x + 1)

1

= 2[log (3/2) – log (5/2) – (log1 – log2)] = 2 log (6/5).

p p Ê p1 ˆ d x + Ú cos x d x + … + Ú cos nx d x˜ = 2 ÁÚ 0 0 Ë 0 2 ¯

y=

(d)

=

1 sin ÊÁ n + ˆ˜ x Ë 2¯ dx sin ( x / 2 )

Êp = 2 Á + sin x Ë2

3 x+1=y

(using Property 17 )

1 sin ÊÁ n + ˆ˜ x Ë 1 2¯ + cos x + cos 2x + … + cos nx = x 2 sin ( / 2) 2

Ú0

(c)

Ê d yˆ 3 2 1 = = ÁË d x ˜¯ 2 2 2 x =1

Solution: We have, x 2

3 x–1=y

dy È = Í 1 dx ÍÎ 1 + t 2

(b) 2p (d) none of these

2 sin

(b)

Solution: Differentiating the given equation, we get

p Ê pˆ p Note that I = tan - 1 (1/x)|1- 1 = - Á - ˜ = is incorrect, 4 Ë 4¯ 2 -1 since the function tan (1/x) is not an antiderivative of (d/dx) [tan - 1 (1/x)] on the interval [- 1, 1]. p

2 y–1=x

Ans. (d)

p p Ê pˆ + Á- ˜ = – Ë ¯ 2 4 4

Example 102: The value of Ú

(a)

Example 105: The value of 16 p + 2 3 3 4 (c) p + 2 3 3 Ans. (d)

1 + t2

tan–1

x - 1 dx is

4 p – 2 3 3 16 p -2 3 (d) 3

(a)

dt

16

Ú1

(b)

Solution: Integrating by parts, the given integral is equal to x tan–1

16

x -1 1

-Ú

16

1

x x4 x

1

dx x -1

Complete Mathematics—JEE Main

13.24

=

1 16 p – 4 3

dx

16

Ú1

x -1

(

4t 1 + t 2

) dt (

16 1 = p – 3 4

Ú0

16 p– = 3

16 3+ 3 = -2 3. 3

(

3

t

and

x = 1 + t 2)

)

x

Ú0

Example 106: Let g(x) =

f(t) dt, where f is such that

(a) – 3/2 £ g(2) < 1/2 (b) 0 £ g(2) < 2 (c) 3/2 < g (2) £ 5/2 (d) 2 < g(2) < 4 Ans. (b) 2

Ú0

1

Ú0

f(t) dt =

f(t) dt +

Since 1/2 £ f (t) £ 1 for t Œ [0, 1] so 1/2 £ Ú

1

0

2

Ú1

dt £ 1. Hence 0 + 1/2 £

Ú0

f (t) dt +

2

Ú1

23

f(t) dt

Ú1

0, and the area of the region bounded by P (x), the x-axis, and the vertical lines x = 0 and x = K is K2(K + 3)/3. The polynomial P (x) is (a) x2 + 2x (b) x 2 + x + 1 2 (d) x 3 + 1 (c) x + 2x + 1 x

Ú sin t xÆ0

sin x

(a) 1 (c) 2 x

e- t

2

/2

f ≤(x) dx is

(a) 1 -

1 3

p 12

(b) -

p 12

(d)

3 –1

1- t dt is 1+ t (a) an even function (b) an odd function (c) a periodic function (d) none of these 1 [(n + 1) (n + 2) … (n + n)]1/n, then 19. If f (n) = n lim f (n) equals nÆ•

18. The function F(x) =

(a) e (c) 2/e

x

Ú0

2

log

(b) 1/e (d) 4/e b

is

(b) 0 (d) none of these

Ú1

b

Úa

dt

0

15. The value of lim

16. Let f(x) =

2

f has maximum at x = 0 f has minimum at x = – 1 f has maximum at x = – 1 f has no critical point

17. A line tangent to the graph of the function y = f(x) at the point x = a forms an p/3 with y-axis and at x = b an angle p/4 with x-axis then

(c)

f ¢ (x) dx is equal to

(a) 2 (c) 3

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

(1 – t 2) dt then

20. The value of the integral b > a, is (a) sin–1 a/b (c) sin–1 b/2a

Ú

a

dx (x - a ) (b - x )

(b) p/2 (d) p.

for

Definite Integrals 13.27

21. The value of the integral

x sin x

p /3

Ú- p / 3

2

cos x

(a) 1 (c) 0

dx is

3p ˆ Êp (a) Á - log tan ˜ Ë3 2 ¯

30. The value of

)

pˆ Êp (c) 3 Á - log tan ˜ Ë2 12 ¯ (d) none of these. tan x cot x dt t 22. The value of Ú is d + t Ú 2 1/ e 1 / e t (1 + t 2 ) 1+ t (a) 1/2 (c) p /4

(b) 1 (d) none of these

Úx

y=

1 + t2

(a)

2y+1=x

(c)

3x+1+

24. The value of

at x = 1 is (b)

Ú- 1 x | x | dx is

(a) 8 (c) 6

(b) 2 (d) 11 2

(Ú e d x ) lim

xƕ

Ú0

)

|sin 2p x|dx is equal to (b) – 1/p (d) 2/p

x x 0 x 2 x2 e 0

Ú

(a) 1 (c) 3 27. The absolute value of (a) less than 10–7 (c) less than 10–9

Ú10

(a) 11/15 (c) 2/5

is

3

2

)

x dx is

x2 + 1 (b) p 2/2 (d) p t Æ•

(b) 3 – e–1 (d) 0 2

Ú-2

x cosp x dx = k p then the (b) 8 (d) none of these

a

dx

Ú0 1 - cos a cos x

(0 < a < p /2) is

(a)

p sin a

(b)

p 2 cos a

(c)

p cos a

(d)

p 2 sin a

37. The value of

sin x

Ú0

-1

e- x dx, then lim A (t) is equal to

35. If the value of value of k is (a) 4 (c) 12

dx

dx x + 1 + 5x + 1

Ê1 ˆ n n n lim Á + + ++ ˜ is 2 2 2 nÆ• Ë n (n + 1) (n + 2) (2n - 1) ¯ (a) 1 (c) 1/2 is

(b) 1/3 (d) 3/2

38. The value of Ê 1 22 n2 ˆ lim Á 3  is + + + n Æ • Ë 1 + n3 23 + n3 n3 + n3 ˜¯

(b) 14/15 (d) none of these

29. Let f (x) = {x}, the fractional part of x then

(a)

1 3

(b)

1 log 2 3

(c)

1 log 3 2

(d)

1 log 3 3

1

Ú-1 f (x) dx is equal to

t

2

dx is 1 + x8 (b) more than 10–7 (d) none of these

28. The value of the integral

xƕ

Ú-1

36. The value of

(b) 2 (d) 0 19

Ú (tan lim 0

(a) 2 – e–1 (c) 4

(t + 1) dt on [1, 3] is

26. The value of

32.

1

34. If A (t) =

1

25. The difference between the greatest and least values of the function x

3 /2

(c)

(a) p/4 (c) p 2/4

3x+1=y

(b) 1 (d) none of these

Ú0

(

(a) 0

33. The value of

3 = y (d) none of these

(a) 2 (c) 0

F(x) =

•

Ú0

(b) 1/3 (d) - 2 /3 x log x dx is 2 1 + x2 (b) 1 (d) 5 /2

x

dt

4

31. The value of

cos t sin (2t – p/4) dt is

(a) 0 (c) 1/p

23. The equation of the tangent to the curve x6

p 2

Ú-p 2

(a) – 1/3 (c) 2 /3

Ê 2p ˆ (b) 2 Á - log 2 + 3 ˜ Ë 3 ¯

(

(b) 2 (d) 1/2

13.28

Complete Mathematics—JEE Main

39. The value of

p

Ú0

sin 2n x dx is sin x

(a) 0 (c) 1/2

px + B, f ¢(1/2) = 2 and 2 1 2A Ú0 f (x) dx = p then the constants A and B are respectively (b) 2/p, 3/p (d) 4/p, 0

41. If f ¢(x) = g (x) for a £ x £ b then is equal to

(c)

2

(d)

2 b

(

x

Úa Úa

Úa

(a) 4 (c) 5

Úx

f (t ) d t dx and I2 =

equal to (a) h(a) – h(b) (c) h(a) + h(b)

2 b

Úa

(b) 6 (d) 9

(b – x) f (x)

50. The value of

44. The greatest value of the function F (x) = x

Ú4

1 [x] 2 log 2

(d)

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

(b) 1/3 (d) 2/3

45. If for a continuous function f, a

Ú- a

f (x) dx = K

a

Ú0

(f (x) + f(– x)) dx then the value

of K is (a) 1 (c) 2 46. If Jm = (a) 2 (c) 2e

(b) 1/2 (d) a e

Ú1

logm x dx then J8 + 8J7 is equal to (b) e8 (d) e

Ú

[x] log 2 1 [x] 4 log 2

p / 2 sin 2

nx dx then a2 – a1, a3 – a2, sin x

Ú0

a4 – a3 are in

(t2 – 8t + 16) dt on [0, 5] is

(a) 5 (c) 4

is

[ x ] 2t dt 0 2[t ]

(b)

51. If an =

b

(b) h(b) – h(a) (d) b h(b) – a h(a)

(a) [x] log2

(c)

log1/3 t dt, x Π[1/10, 4] is at x = (b) 4 (d) none of these

f (b )

Ú f (a) g ( x ) dx + Úa f ( x ) dx

and h(x) = x f(x), then

2

I1 = I2 I1 = (b – a) + I2 I1 = (b – a) I2 none of these

(a) 1/10 (c) 1

f (t) (x – t)2 dt then

49. If y = f(x) be an invertible function with inverse g

43. The least value of the function F (x) = 2

f1 (t) dt and f3(x)

(b) 1/2 (d) none of these

dx then (a) (b) (c) (d)

x

Ú0

x

Ú0

48. If the variables x and y are related by the equation y dx d2 y x= Ú and is proportional to y then 0 dx 2 1+9x 2 the constant of proportionality is

( g ( b )) - ( g ( a ))

)

f2(t) dt if f3(x) = A

(a) 1 (c) 2

f (x) g(x) dx

2

( f ( b ) ) - ( f ( a ))

42. Let I1 =

b

(b) f (b) – f(a)

2

f (t) dt, f2(x) =

the value of A is

40. If f (x) = A sin

(a) g (b) – g(a)

x

Ú0

=

(b) 1 (d) p/3

(a) p/2, p/2 (c) 0, – 4/p

x

Ú0

47. Let f 1(x) =

52.

A.P. G.P. H.P. Arith – Geo progression

21

4 Ú0 [ x ] dx

is equal

(a) 44100 (c) 48400

(b) 2100 (d) 42400

53. If x2 f(x) + f(1/x) = 0 for all x ΠR ~ {0} then sec q

Úcosq

f ( x ) dx =

(a) sin2q (c) secq – cosq

(b) 1 (d) 0

Definite Integrals 13.29

Assertion-Reason Type Questions

Statement-1: I = p

dx

•

54. Let I =

Ú– • 1 + x 2

Statement-2: The integrand can be expressed as X[0, p/2] X[p/2, p]; XA being the characteristic function of A.

Statement-1: I = p Statement-2: The integrand is even and lim F (b ) bƕ

= p, F(b) = 55. Let I =

57. Let I =

dx

b

Ú0 1 + x 2

Ú0

p

Ú0 x

sin x dx

Statement-2: I = 2 J 58. Let I =

1 1 1 £ . Statement-2: 8 £ 2 5 + 3 cos x 5 cos x

p

Ú0

cos x dx and J =

5 + 3 cos2 x

p p £I£ Statement-1: 16 10

56. Let I =

2

Statement-1: I = – 2p

dx

p /2

p

Ú0 x

1 - sin 2 x

2

Ú1

10 x 2

(x

3

2

)

+1

dx and J =

9

dx

Ú2 x 2

35 27 Statement-2: 3I = 10 J Statement-1: I =

dx

LEVEL 2 Straight Objective Type Questions 59. The value of (a) 3/4 (c) 2/3 –

p 3

Úp 4

2 /2

3

(b) 1/2 (d) none of these

63. The value of

60. If f is integrable on [a, b] then Ú [f (x) – fave] d x a is equal to 1 (f (b) – f (a)) (a) f (b) – f (a) (b) 2 (c) f (a) – f(b) (d) 0 61. If m π n, m, n Œ N then the value of cos nx dx is

2p

Ú0

cos mx

(b) 2p (d) dependent on m and n

1 a0 + a1 cos x + b1 sin x + a2 cos 2x 62. If Q (x) = 2 + º + an cos nx + bn sin nx, then the value of 2p

Ú0

Q(x) sin kx dx (k = 1, 2, º n)

Ú - 3

b

(a) 0 (c) p

(b) pa0 (d) pbk

(a) pak (c) pbn

dx is sin x + tan x

(a) p/2 (c) 1 64. The value of (a) sin 1 (c) 2 sin 1

Ê d Ê -1 1 ˆ 3 ˆ ˜ + x ˜¯ dx is ÁË d x ÁË tan x¯ (b) p/4 (d) none of these

1

Ú1/2

1 1ˆ Ê ÁË 2 x sin - cos ˜¯ dx is x x (b) cos 1 (d) none of these

65. Let f be a continuous function [a, b] such that f (x) > 0 for all x Π[a, b]. If F(x) = (a) (b) (c) (d)

x

Úa

f (t) dt then

F is differentiable but not increasing on [a, b] F is differentiable and increasing on [a, b] F is continuous and decreasing on [a, b] F is neither differentiable nor increasing on [a, b]

13.30

Complete Mathematics—JEE Main

66. Let f be a continuous function on R satisfying f (x + y) = f (x) + f (y) for all x, y ΠR with f (1) = 2 and g be a function satisfying f (x) + g (x) = ex then the value of the integral (a)

1 –4 e

(c) 2/3

1

Ú0

f (x) g (x) dx is

(b)

1 (e – 2) 4

(d)

1 (e – 3) 2

Ú0

value of

Ú

(b) – 2 log 3

(c) log 3

(d) 1/4 1

68. The equal to integral Ú log 0

(a)

(

)

dx

x 2

x x -1

(b) 2 (d) 3

(a) log

1Ê pˆ 1Ê pˆ (c) ÁË log 4 - 1 + ˜¯ (d) ÁË log 3 - 1 + ˜¯ . 3 8 4 2 1 - x2 n

(a) less than 1 (c) more than 1

for n ≥ 1, the value of I is

p 2

(b) more than 1/2 (d) less than 1/2

f ¢(1) = 8, f (2) + f ¢¢(2) = 33 and are (a) A = 1, B = – 4, C = 2 (b) A = 7, B = – 6, C = 3 (c) A = 8, B = – 6, C = 3 (d) none of these.

1

Ú0

Ú0

x

2

Úx

log t dt

at x = 2 is

y – 6 log 2 = 7 log 2(x – 2) y – log 2e1/3 = (log 2)x y – 8 log 2e– 1/3 = 5 log 2 x y + 8 log 2 + 2 = (7 log 2)x

72. The value of

p /2

Ê 2 - sin q ˆ

Ú- p / 2 log ÁË 2 + sin q ˜¯ dq is (b) 1 (d) none of these

78.

p

e +1

(

e +1

Ú0

2 2

x log (sin x) dx is (given that p log 2) 2

(a) -

p2 log 2 2

(b) -

(c) -

p2 log 2 8

(d) none of these

p

Ú-p

)

(d) 2 + log e2 + 1

log sin x dx = -

f (x) dx = 7/3,

71. The equation of the tangent to the curve

(a) 0 (c) 2

2 2

77. The value of

form f (x) = A x2 + Bx + C satisfies the conditions

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

(b) 1 + log

e +1

(c) 2 + log

70. The numbers A, B and C such that a function of the

y=

2 2

dx

1/ 2

Ú0

p is given by 12

=

2

2 76. The mean value of the function f(x) = x on e +1 the interval [0, 2] is

1Ê 1 pˆ 1Ê pˆ ÁË log 2 - + ˜¯ (b) ÁË log 2 - 1 + ˜¯ 2 4 2 2 2

69. If I =

(b) I1 = I2 (d) I2 = (p/2) – tan–1 x

(a) 1 (c) 3

1 - x + 1 + x dx is

sin x + cos x dx is 3 + sin 2 x

75. The solution of the equation

f (x) dx is

(a) 2 log 3

p /4

Ú0

(a) log 2 (b) log 3 (c) (1/4) log 3 (d) (1/8) log 3. 1 dt 1/ x dt 74. If I1 = Úx for x > 0, 2 and I2 = Ú 1 1+ t 1 + t2 then (a) I1 > I2 (c) I2 > I1

67. If f (x) = cosec (x + p/3) cosec (x + p/6), then the p /2

73. The value of the integral

p2 log 2 4

(cos px – sin qx)2 dx, where p and q are in-

tegers is equal to (a) – p (c) p

(b) 0 (d) 2p

79. Let f, g and h be continuous functions on [0, a] such that f(x) = f(a – x), g(x) = – g(a – x) and a

Ú0

3h(x) – 4h(a – x) = 5. Then is equal to (a) 5/4 (c) 1 80. If

x

Úp 2

f(x) g(x) h(x) dx

(b) 3/4 (d) none of these 3 - 2 sin 2 z dz +

dy at (p/2, p) is dx

y

Ú0

cos t dt = 0 then

Definite Integrals 13.31

(a) 1 (b) 2 (d) = 0 (c) 3 81. If P(x) is a polynomial of least degree that has a maximum equal to 6 at x = 1, and a minimum equal 1

Ú0

to 2 at x = 3, then

P (x) dx equals

(a) 17/4 (c) 19/4

3

(b) 13/4 (d) 5/4

1

Ú0

83. If Sn =

e1 n

n2 lim Sn is

(b) 2 –1 (d) none of these +

2 n2

3

e2 n +

n2

e3 n

1 +  + e then n

nƕ

(b) 0 (d) none of these 1

x k +2 2k  k ! dx is k =0 n

Ú0 n Æ • lim

(a) e2 – 1 e2 - 1 (c) 2

Úa

cos2 x

Ú-p 1 + a x dx , a > 0, is (b) ap (d) 2p

91. Let f : (0, •) Æ R and let F(x) =

Ú0

-1 2n - 1 dx, if a = is 2 log x (b) 2 log n (d) (1/2) log n

5 4 (c) 4

(b) 7

(c) ±

, 1 £ x < 4 then

9

Ú0

g (x) dx is

(b) 9 3 (d) none of these

88. The value of lim

x Æ0

Ú

-x 2x

Ú0

2 - t 2 dt . Then the real roots of the

(b) ±

1 2

(d) 0 and 1 1

Ú0

t m (1 + t)n dt, then the expression

(a)

n 2n l ( m + 1, n – 1) – m +1 m +1

(b)

n l(m + 1, n – 1) m +1

(c)

n 2n + l(m + 1, n – 1) m +1 m +1

(d)

m l(m + 1, n – 1) n +1

, 4£ x 0 the definite integral

0

t2

2 5 Ê 4ˆ Ú0 x f (x) dx = 5 t , then f Ë 25 ¯ is equal to 2 2 (a) (b) – 5 5 2 2 (c) (d) – 5 5

Ú–2

(x 3 + 3x2 + 3x + 3 + (x + 1) cos(x + 1)) dx

equals (a) – 4 (c) 4

(b) 0 (d) 6

Previous Years' AIEEE/JEE Main Questions

1.

Ú0

2

0

(a) 2 – 2

(b) 2 + 2

(c)

(d)

2 –1 p

2. I n =

Ú0

4

10p

[2002]

nƕ

1 2 (c) •

Úp

2–2

tan n x dx , then lim n(In + In – 2 ) equals

(a)

3.

t

6. If f (y) = ey, g(y) = y, y > 0 and F (t) = Ú f(t – y)

[x 2] dx is

(b) 1 (d) 0

[2002]

|sin x| dx is

(a) 20 (c) 10

(b) 8 (d) 18

[2002]

4. If y = f (x) makes positive intercept of 2 and 1 unit on x-axis and y-axis and encloses an area of 3 2 square unit with axes then Ú x f ¢(x) dx is 4 0 3 (a) (b) 1 2 3 5 (c) (d) – [2002] 4 4 5. Let f (x) be a function satisfying f ¢(x) = f (x) with f (0) = 1 and g(x) be a function that satisfies f (x) + g(x) = x2. Then the value of the integral 1

Ú0

g(y) dy, then (a) F(t) = et – (1 + t) (b) F(t) = tet (c) F(t) = te –t (d) F(t) = 1 – et(1 + t) 7. If f (a + b – x) = f (x), then to a+b b (a) f ( x ) dx 2 Úa (b)

b–a b f ( x ) dx 2 Úa

(c)

a+b b f ( a + b – x ) dx 2 Úa

(d)

a+b b f (b – x ) dx 2 Úa

e2 3 – 2 2

e2 5 (c) e + + 2 2

e2 5 (d) e – – 2 2

[2003]

Ú0

n

(b)

1 1 – n +1 n + 2

(c)

1 1 + n +1 n + 2

(d)

1 n +1

(a) 2 (c) 0

Ú lim 0

xÆ0

is equal

x (1 – x ) dx is

1 n+2

9. The value of

e2 3 – 2 2

1

(a)

x2

(b) e –

b

Úa x f ( x ) dx

[2003]

8. The value of the integral I =

f (x) g(x) dx is

(a) e +

[2003]

[2003]

sec 2 t dt

x sin x (b) 1 (d) 3

is

[2003]

Definite Integrals 13.33 3

Ú–2 1 – x 2 dx

10. The value of

is

(a)

7 3

(b)

14 3

(c)

28 3

(d)

1 3

p

Ú0

11. The value of I =

2

[2004]

-p

1 + sin 2 x (b) 1 (d) 3 p

p

Ú0 x f (sin x ) dx

= AÚ

p 4 (c) 0

2

0

[2004]

f (sin x ) dx , then A is

(b) p

(a)

20.

(d) 2p e

13. If f (x) =

f (a )

x

1 + ex

Ú

, I1 =

x g ( x (1 – x )) dx , and

f (– a)

I2 is I1

(a) – 1 (b) – 3 (c) 2 (d) 1 [2004] 1 1 2 4 1 14. lim ÍÈ 2 sec 2 2 + 2 sec 2 2 + 2 sec 2 1˘˙ equals n Æ • În ˚ n n n n (a) tan 1 1 (c) sec 1 2

(b) tan 1 1 (d) cosec 1 2 1 . Then lim xÆ0 48

(a) 12 (c) 24

I4 =

1 x2 2 0

Ú

2 x3

Ú1 2

dx , I2 =

Ú 6

3

4t dt equals x-2

17. The value of p a

(c) ap

1 x3 2 dx 0

Ú

[2005] , I3 =

2 x2 2 1

Ú

dx and

dx then

(a) I3 = I4 (c) I2 > I1

(a)

f ( x)

(b) 18 (d) 36

16. If I1 =

(b) I3 > I4 (d) I1 > I2 p

2

ÈÎ( x + p )3 + cos2 ( x + 3p )˘˚ dx is equal to

Êpˆ (a) Á ˜ - 1 Ë 4¯

(b)

p4 32

p Ê 4ˆ (c) Á p ˜ + Ê ˆ Ë ¯ Ë 32 2¯

(d)

p 2

p

Ú0 x f (sin x ) dx p

2

(a) p Ú

0

[2005]

cos2 x

f (cos x ) dx

(c) p f (sin x ) dx Ú

p

(b) p Ú f (cos x ) dx 0

(d)

0

p p2 f (sin x ) dx 2 Ú0 [2006]

21. The value of

a

Ú1 [ x ] f ¢ ( x ) dx , a > 1, where [x] de-

notes the greatest integer not exceeding x is (a) (b) (c) (d)

a f ([a]) – { f (1) + f (2) + º + f (a)] a f (a) – { f (1) + f (2) + º + f ([a])} [a] f (a) – { f (1) + f (2) + º + f ([a])} [a] f ([a]) – { f (1) + f (2) + º + f (a)}

1 22. Let F(x) = f (x) + f Ê ˆ where f (x) = Ë x¯ Then F(e) equals 1 (a) (b) 0 2 (c) 1 (d) 2

Ú1

[2006] log t dt. 1+ t

[2007]

(b) – 2

(c)

3 2

(d) 2 2 1 sin x

Ú0

dx and J =

1 cos x

Ú0

x x which one of the following is true? (a) I > 2/3, J > 2 (c) I < 2/3, J > 2

[2005]

x

23. The solution for x of the equation x p dt Ú 2 t t 2 – 1 = 2 is

24. Let I =

(b) 2p p 2

[2006]

is equal to

(a) 2

Ú-p 1 + a x , a > 0 is

(d)

2

dx is

[2006]

[2005]

15. Let f : R Æ R be a differentiable function having f (2) = 6, f ¢(2) =

-3p

p

f (– a)

g ( x (1 – x )) dx , then the value of

Ú

Ú

[2004]

f (a )

I2 =

19.

dx is

9– x + x

(b)

3 2

(c)

x

6

Ú3

1 2 (d) 2

(a) 1

(sin x + cos x )2

(a) 2 (c) 0 12. If

18. The value of the integral,

[2007] dx . Then

(b) I < 2/3, J < 2 (d) I > 2/3, J < 2

[2008]

13.34

25.

Complete Mathematics—JEE Main p

Ú0 [cot x ] dx , where [ ] denotes the greatest integer function, is equal to (a) – 1 (c) p/2

Ú0 p ( x ) dx

= 41. Then (a) 41 (c) 41

(b) – p/2 (d) 1

(c)

p log 2 8

x)

1 + x2

(d)

x

Ú0

(c)

3 2 5 (d) 4 (b)

3 4

30. If g(x) =

x

Ú0

1 + t2

(b) g(x) – g(p)

(c) g(x) g(p)

(d)

y

(c)

1+ y

1 + y2 [2013, online]

g( x ) g(p )

tan 2 x dx is equal to

Ú

35. The integral

(b) log 2 (d) log 2 [2013, online]

(a) log 2 2 (c) 2 log 2 p

36. The integral

Ú

0

line y = 2x, are equal to (a) ± 2 (b) ± 3 (c) ± 4 (d) ± 1 32. Statement-1: The value of the integral

[2013]

2p -4 3

(b)

(c) 4 3 - 4

(d) 4 3 - 4 1/ 2

Ú 0

curve, y = Ú t dt , x Œ R, which are parallel to the

x x - 4 sin dx equals 2 2

(a) p – 4

37. The integral

[2012]

1 + 4 sin 2

0

[2011]

x

p Ú 1 + tan x = 6 p /6

is equal to

(d) y2

2

log (1 + 2 x ) 1 + 4 x2

p 3

[2014]

dx equals

(a)

p log 2 4

(b)

p log 2 8

(c)

p log 2 16

(d)

p log 2 [2014, online] 32

31. The intercepts on x-axis by the tangents to the

dx

dx 2 (b)

cos 4t dt , then g(x + p) equals

(a) g(x) + g(p)

p /3

d2 y

7p / 4

x[ x 2 ] dx is

(a) 0

, then

[2013, online]

7p / 3

t sin t dt then

local maximum at p and local minimum at 2p. local maximum at p and 2p local minimum at p and 2p local minimum at p and local maximum at 2p [2011] 29. Let [.] denote the greatest integer function then the 1◊5

dt

(a) y

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

Ú0

y

0

[2011]

dx is

(b)

34. If x = Ú

p log 2 2

x

p 2 p (d) 4

(a) p

[2010]

dx is

28. For x Π(0, 5p/2), define F(x) = f has

value of

-p / 2 1 + 2

(c) 4p

(b) p log 2

(a) log 2

Ú

sin 2 x

equals

1 8 log(1 +

Ú0

33. The value of

[2013]

a

p /2

[2009]

(b) 42 (d) 21

27. The value of

Ú

b

f ( x ) dx = Ú f (a + b - x ) dx

a

26. Let p(x) be a function defined on R such that p¢(x) = p¢(1 – x), for all x Œ [0, 1], p(0) = 1 and p(1) 1

b

Statement-2:

e

38. If for n ≥ 1, Pn = Ú (log x )n dx , then P10 – 90P8 is equal to 1 (a) –9 (b) 10e (c) – 9e (d) 10 [2014, online] 39. If [,] denotes the greatest integer function, then the p

integral

Ú [cos x] dx

is equal to

0

p 2 (c) –1 (a)

(b) 0 (d) -

p 2

[2014, online]

Definite Integrals 13.35 t

Ú ( f ( x) + x) dx

(a)

3 2

Ê pˆ = p2 – t2, for all t ≥ – p, then f Ë - ¯ is equal to: 3

(c)

3 2

40. If for a continuous function f (x),

-p

(b)

p 3

(c)

[2014, online] (a) x

et dt x > 0 then t 1

41. Let function F be defined as F(x) = Ú x

e

t

Ú t + a dt where a > 0 is

the value of the integral

1

[2014, online] dx is

(b) 4 (d) 6

[2015]

2 50 Ú0 f ( x) dx = 5. Then the value of Ú10 f ( x) dx is:

(b) 100 (d) 200

[2015, online]

44. Let f: (–1, 1) Æ R be a continuous function. If

Ú

-1

(1 – x + x2)dx is equal to (b) log 2 (d) log 4

[2016, online]

1/ n

43. Let f : R Æ R be a function such that f(2 – x) = f(2 + x) and f(4 – x) = f (4 + x) for all x Œ R and

sin x 0

1

p + log 2 2 p (c) – log 4 2

2

equal to

(a) 80 (c) 125

1 (log x)2 2 1 (d) log x2 [2015 online] 4

(b)

(a)

Ú2 log x 2 + log(36 - 12 x + x 2 )

(a) 2 (c) 1

1 (log x)2 4

Ú0 tan

log x

4

[2015, online]

46. If 2Ú tan -1 x dx = Ú cot -1 (1 – x + x2)dx, then 0 0 1

ea [F(x) – F(1 + a)] e– a [F(x + a) – F(a)] ea [F(x + a) – F(1 + a)] e– a [F(x + a) – F(1 + a)]

42. The integral

(d) 1 2

(c) log x

1

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

3

Ê 1ˆ x log t 45. For x > 0, let f(x) = Ú dt . Then f(x) + f ÁË ˜¯ 1 1+ t x is equal to

p 2 p (d) 6

(a) p

(b)

(n + 1)(n + 2)  (3n) ˆ is equal to: 47. lim ÊÁ ˜¯ Ë nÆ• n2n 18 27 (a) 4 (b) 2 e e 9 (c) 2 (d) 3 log 3 – 2 [2016] e 48. The value of the integral [ x2 ] 10 dx , where [x] denotes the Ú4 2 [ x - 28 x + 196] + [ x 2 ] greatest integer less than or equal to x, is 1 3 (c) 7

Ê 3ˆ 3 f (t ) dt = x , then f Á ˜ is equal to 2 Ë 2 ¯

(a)

(b) 6 (d) 3

[2016, online]

Previous Years' B-Architecture Entrance Examination Questions -5

1.

Úe

( x + 5 )2

-4

2/3

dx + 3

Ú

( ) dx e 9 x-

2 3

2

(a) f (1.5) – f ( 2 ) – f (1) is

(b) f (1.5) + f ( 2 ) + f (1)

1/ 3

(a) 0 (b) –2 (c) 1 (d) 2 [2006] 2. If f is continuously differentiable function then

(c) 2 f (1.5) + f ( 2 ) + f (1) (d) 2 f (1.5) – f ( 2 ) – f (1) a

1.5

Ú [x 0

2

] f ¢( x ) dx is

3.

Ú log( x +

-a

x 2 + 1) dx

[2006]

13.36

Complete Mathematics—JEE Main

(a) 2 log (a2 + 1)

10. Using the fact that 0 £ f (x) £ g(x), c < x < d

(b) 2 log ( a 2 + 1 - a )

d

(d) 2 log ( a + a 2 + 1 ) [2007] 1 1 1 when 4. If f (x) = < x £ n , n = 0, 1, 2… n n +1 2 2 2 (c) 0

1

lt

then (a)

nƕ

Ú

f ( x ) dx equals

1 / 2n

1 3

2 3 1 (d) 2

(b)

(c) 0

[2007]

1

5. The value of

Úx

2

(1 - x )9 dx is

0

1 (a) 610 1 (c) 640

1 630 1 (d) 660

c

c

3

Ú

3 + x 3 lies in the interval

1

Ê1 ˆ (a) Ë , 3¯ 2

(b)

(2,

Ê3 ˆ (c) Ë , 5¯ 2

(d)

(4, 2

Ú max (e , e ) 1- x

p 4

(d) 2 (e + 1)

12. If f (x) = x |x|, then for any real number a and b

(b)

1 3 b - a3 3

(c)

1 3 ( a + b3 ) 3

(d)

1 3 ( a - b3 ) 3

min (sin x, cos x) dx equals to

log 7

(b)

(a) 2 2

13. The integral

2

Ú log 5

(c) 2 - 2

(d) 2 + 2

[2011] 1

8. Let f : R Æ R be defined by f (x) =

Ú 0

Then the curve y = f (x) is (a) an ellipse (b) a straight line (c) a parabola (d) a hyperbola

2

x +t dt. 2-t

1 5 log 4 7

(b)

1 5 log 2 7

(c)

1 7 log 4 5

(d)

1 7 log 2 5

14. If f (x) =

ex 1 + ex

Ú

I2 =

, I1 =

Ú

(d) 0

1 2

(c) 1

3 [2011]

x g{x(1 - x )} dx and

f (- a)

g {x(1 - x )} dx where g is not an identity

f (- a)

(a) (b) -

[2013]

f (a)

is: 1

dx is

f (a)

function. Then the value of

1

3

cos(log 35 - x 2 ) + cos x 2

(a)

[2011]

3

1

x cos x 2

[2013]

equal to

2

9. The equation of a curve is given by y = f (x), where f ¢(x) is a continuous function. The tangent at points p p (1, f (1)), (2, f (2)) and (3, f (3)) make angles , 6 3 p and respectively with positive x-axis. Then 4

Ú f ¢( x) f ¢¢( x) dx + Ú f ¢¢( x) dx

equals

1 3 ( b - a 3) 3

0

(c)

Ú f ( x)dx

(a) [2010]

p /2

(a) 1

[2012]

a

(c) 2(e + e )

2

p 6

(d)

with a < b, the value of

dx equals

(b) 2(e - e )

3

[2012]

b x

(a) 2 (e – 1)

Ú

30 )

˘ n2 n2 n2 1È + + 2 1+ 2 2 + 2 Í ˙ 2 2 nÆ• n Î n +1 n +2 n + (n - 1) ˚ is equal to p p (a) (b) 2 3

0

7.

30 )

11. lim

(c) [2008]

d

Ú f ( x) dx £ Ú g( x) dx , we can conclude that

(b)

1

6. The value of

fi

I2 is I1

(b) 2 (d) –1

[2014]

Definite Integrals 13.37

e x (2 - x 2 )

1/ 2

15. The integral Ú0

(1 - x)3/ 2 (1 + x)1/ 2

dx is equal to

(a)

3e

(b)

3e - 1

(c)

e 3

(d)

e -1 3

91. (c)

92. (a)

95. (a)

96. (c)

93. (a)

94. (c)

Previous Years' AIEEE/JEE Main Questions [2015]

2

16. The integral I = Ú0 [ x 2 ]dx ([t] denotes the greatest integer less than or equal to t) is equal to:

1. (c)

2. (b)

3. (d)

4. (d)

5. (b)

6. (a)

7. (a)

8. (b)

9. (b)

10. (c)

11. (a)

12. (b)

13. (c)

14. (b)

15. (b)

16. (d)

17. (d)

18. (c)

19. (d)

20. (a)

(a) 5 - 2 3

(b) 5 - 2 - 3

21. (c)

22. (a)

23. (b)

24. (b)

(c) 6 - 2 - 3

(d) 3 - 2

25. (b)

26. (d)

27. (b)

28. (a)

29. (c)

30. (a)(b)

31. (d)

32. (c)

33. (b)

34. (a)

35. (a)

36. (d)

37. (c)

38. (c)

39. (d)

40. (a)

41. (d)

42. (a)

43. (b)

44. (b)

45. (b)

46. (b)

47. (b)

48. (d)

Answers Concept-based 1. (b) 5. (a) 9. (d)

2. (d) 6. (d) 10. (b)

3. (b) 7. (c)

4. (c) 8. (c)

11. (d)

12. (c)

13. (c)

14. (a)

15. (b)

16. (b)

17. (a)

18. (a)

19. (d)

20. (d)

21. (b)

22. (b)

23. (d)

24. (c)

25. (c)

26. (d)

27. (a)

28. (d)

29. (a)

30. (d)

31. (a)

32. (d)

33. (c)

34. (a)

35. (b)

36. (a)

37. (c)

39. (b)

39. (a)

40. (d)

41. (c)

42. (a)

43. (c)

44. (b)

45. (a)

46. (d)

47. (b)

48. (d)

49. (b)

50. (b)

Previous Years' B-Architecture Entrance Examination Questions

Level 1

51. (c)

52. (a)

53. (d)

54. (c)

55. (a)

56. (d)

57. (c)

58. (a)

1. (a)

2. (d)

3. (c)

4. (b)

5. (d)

6. (b)

7. (c)

8. (c)

9. (b)

10. (d)

11. (c)

12. (b)

13. (c)

14. (b)

15. (b)

16. (b)

Hints and Solutions Concept-based p /2

1.

Ú

p /4

0

cot q cosec 2q dq = - Ú u du 1

=

nƕ

59. (d)

60. (d)

61. (a)

62. (d)

63. (d)

64. (d)

65. (b)

66. (c)

67. (b)

68. (b)

69. (a)

70. (b)

71. (d)

72. (a)

73. (c)

74. (a)

75. (b)

76. (c)

77. (a)

78. (d)

79. (d)

80. (a)

81. (c)

82. (d)

83. (a)

84. (d)

85. (c)

86. (c)

87. (c)

88. (d)

89. (c)

90. (a)

Ú u du = 0 n

2. Reqd. limit = lim

Level 2

(u = cot q )

1

u2 2

1

0

p /2

3. I = Ú-p / 2

x sin x x

e +1

= 0

1 2

rp 1 sin  n n =1 n

= Ú sin p x dx = =

1

cos p x 1 p 0

2 -1 [ -1 - 1] = p p

dx = p /2

=Ú -p / 2

p /2

Ú-p / 2

(- x )sin(- x ) e- x + 1

( x sin x)e x ex + 1

dx

dx (Prop. 11)

Complete Mathematics—JEE Main

13.38

p /2

2I = Ú -p / 2 p /2

x sin x ex + 1

p /2

+ Ú-p / 2

e x ( x sin x) ex + 1

3

dx

dt

= 2Ú

t2 + 1

1

3 -1

= 2 tan t 1

x

=Ú -p / 2

(e + 1)( x sin x)

p /2

p /2

Èp p ˘ p = 2Í - ˙ = . Î3 4˚ 6

dx

ex + 1

=Ú x sin x dx = 2Ú0 -p / 2

x sin x dx

2.5

8.

(Prop. 12)

Ú [x

2

] f ¢ ( x ) dx

0 p /2 = 2 È - x cos x p0 / 2 + Ú cos x dx ˘ 0 Î ˚ p /2 sin x 0

=2 fi

1

=

2

2 Ú [ x ] f ¢( x) dx + 0

=2

1

11

sin x is an odd function,

p /2

so

x14 sin11x dx = 0. Thus the given integral is

Ú

p /2

equal to

Ú

p /2

sin 2 x dx = 2

-p / 2

Ú 0

1 p p sin 2 x dx = 2. . = 2 2 2

1

0

2dt 2

Ê (1 - t ) ˆ (1 + t ) Á 2 + 3˜ Ë 1 + t2 ¯ 2

1

0

= p /4

Ú 0

x sin x cos3 x

= 2Ú 0

2 5

dt

Ú

2 - 2t + 3 + 3t 2 =

(f(

1

t

tan -1

2

=

5

5

0

tan -1

1 5

x tan x sec 2 x dx

x + 1(2 + x )

=

3 2.5

6

6

5

2 ) - f (1)) + 2 ( f ( 3 ) - f ( 2 ))

+ (f ( 5 ) + f ( 6 )) 9. Given integral is equal to 1 3

11

Ú (1 - x) (1 - (1 - x)) 1 2

Ú

2t dt

(1 - x )3 dx

1

tan x dx =

0

11

Úx

(1 - 3 x + 3 x 2 - x 3 )dx

0 2

(sec x - 1) dx

1

= Ú ( x11 - 3 x12 + 3 x13 - x14 ) dx

0

Ú t (t 2 + 1) , t

11

Úx 0

2

p /4

Ú

dx =

0

p /4

1 p 1È p˘ p 1 ◊ - 1= 2 4 2 ÎÍ 4 ˚˙ 4 2

1

2

2

1

7. The given integral in equal to dx

3

= 6 f(2.5) – (f (1) + f ( 2 ) + f ( 3 ) + f (2)

0

1 p 1 = ◊ - [tan x - x ]p / 4 2 4 2

3

] f ¢ ( x ) dx

+ 5 [ f ( 6 ) - f ( 5 )] + 6 [ f (2.5) - f ( 6 )]

0

=

2

f ¢( x ) dx + 2 Ú f ¢( x ) dx + 3 Ú f ¢( x ) dx

2

2

p /4

dx =

Ú

Ú [x 6

5

t2 + 5

1 p 1 = ◊ 2 4 2

0

2.5

] f ¢( x ) dx

+ 3 ( f (2 ) - f ( 3 ) ) + 4 ( f ( 5 ) - f (2 ) )

p /4

Ú

6

2

+ 4 Ú f ¢( x ) dx + 5 Ú f ¢( x ) dx + 6 Ú f ¢( x ) dx

dt

tan 2 x = x 2 0

2

2

1

1

= 2Ú

6.

3

2

= 0+

5. Put tan x/2 = t. The given integral reduces to

Ú

Ú [x

5

-p / 2

] f ¢( x ) dx

5

2 Ú [ x ] f ¢( x) dx +

+

2

2

2 Ú [ x ] f ¢( x) dx +

+

4. The function f (x) = x

Ú [x

2

I = 1. 14

3

2 Ú [ x ] f ¢( x ) +

2

= x +1

=

1 3 3 1 - + 11 12 13 14

Ï2 - x , 0 £ x £ 2 10. 2 - x = Ì Óx - 2 , 2 < x £ 3 3

2

2

Ú 2 - x dx = Ú (2 - x) dx + Ú ( x - 2) dx 0

0

2

Definite Integrals 13.39 2

= 2x - x

2 0

3

ˆ Ê x2 + ÁË - 2 x˜¯ 2 2

19. A = lim f (n) nÆ0

1

n + 1ˆ Ê n + 2 ˆ Ê n + n ˆ ˘ n = lim ÍÈÊ  Ë nÆ•Î n ¯ Ë n ¯ Ë n ¯ ˙˚

8 Ê4 ˆ - 4 - Ë - 4¯ 2 2

= 4-4+

1

n n 1 2 = lim ÍÈÊ 1 + ˆ Ê 1 + ˆ  Ê 1 + ˆ ˘˙ n Æ • ÎË n¯ Ë n¯ Ë n¯ ˚

=2

Level 1

1 1 n r  log ÊË1 + n ˆ¯ = Ú0 log (1 + x ) dx nÆ• n r =1

Ï1 – x , x £ 1 11. |1 – x| = Ì so Óx – 1 , x > 1 2

Ú–2

1

Ú–2

1 – x dx =

log A = lim 2

1 – x dx + Ú 1 – x dx 1

1

2

Ú–2 (1 – x ) dx + Ú1 ( x – 1) dx

=

= 5. x+p

Ú0

12. f (x + p) = =

p

Ú0 sin

8

sin8 t dt

t dt t Ú

x+p

p

(

8

sin t dt

x

13. f (0) = 4, f (2) = 7 and 7 – 4 = 3. K

Ú0

x

15.

Ú sin t lim 0

xÆ0

2

= lim

a +b b 2 sin -1 = sin –1(1) – sin –1(–1) = p b -a 2 a 21. The integrand is an even function, so the given integral is equal to p

2Ú

sin x 2 1 lim . x . cos x = 0 2 x Æ 0 x2

log

1- u du = F(x) 1+ u

p

3

0

sin x 2

1 p = and f ¢(b) = 17. f ¢(a ) = tan Ê p – p ˆ = tan Ë 2 3¯ 6 3 p = 1, so tan 4 1 b 1– ( ) f dx = f ¢(b) – f ¢(a) = . ¢¢ x Úa 3 -x x 1- t 1+ u 18. F (– x) = Ú log dt = - Ú log du (u = - t ) 0 0 1+ t 1- u

Ú0

cos x

p

dx = 2 [ x sec x ]0 3 - Ú sec x dx ] p p = 2. .2 - 2 [ log (sec x + tan x )]0 3 3

4p – 2 log (2 + 3 ) 3

= 22.

t 1 1 = – . So the required integral is t (1 + t 2 ) t 1 + t2 equal to

2

x

2

0

16. f ¢(x) = e–x /2(1 – x2) > 0 for –1 < x < 0 and is negative for x < –1. f (x) has minimum at x = –1.

=

x sin x

3

x Æ 0 cos x 2 .2 x

=

)

2

x-

K 2 2K + ( K + 3) = K 2 + 2K. 3 3

dt

sin x 2

= f (2) – f (0) =

K 2 ( K + 3) . Differentiating 3

P(t )dt =

w.r.t K, we get P(K) =

2

Ú0 f ¢ ( x ) dx

) ( 2

B-a a +b - x2 2 So the value of the required integral is

= f ( x ) + Ú sin8 (p - u ) du = f (p ) + f ( x ) . 0

14. We have

= log 4 e 4 fi A= e 20. The integrand can be written as 1

Ú

t dx

tan x 1

1+ t

e

cot x

+ Ú1

2

e

cot x t dt dt - Ú1 t e 1 + t2

tan x 1 cot x x + log t cot - 2 log (1 + t 2 ) 1 log (1 + t 2 ) 1 e e 1 2 e = log |sec x| + log |cot x| + 1 – log |cosec x| = 1.

=

23.

dy = dx

1 1+ t =

2

t = x3

3x2 1+ x

dy dx

x =1

=

1 2

.

6

1 d 3 (x ) – dx 1 + t2 –

2x 1 + x4

t = x2

d 2 (x ) dx

Complete Mathematics—JEE Main

13.40

Also y(1) = 0. So the equation of tangent is y – 0 1 = (x – 1). 2 24. The integrand is an odd function, so the given integral is zero. 25. F ¢(x) = x + 1 > 0 for x Œ [1, 3]. Therefore, the greatest value is F(3) and the least value is F(1). The required difference is 3

1

Ú0 (t + 1) dt – Ú0 (t + 1) dt = 26.

2 x x 0 x 2 2 e x dx 0

(Ú e dx) lim

xƕ

= lim

xƕ

Ú

(e

e2 x

xƕ

Ú

p cos t sin Ê2t - ˆ dt Ë 2 4¯

2

-p

=

x

2

- 1)

Ú 2

= 2 lim

1 - e- x

2 x Æ • e2 x - 2 x

= 0.

19 sin x -8 Ú10 1 + x8 dx £ Ú10 10 dx

1

2 2

= –

1 Ê 1 1 2 2 ˆ . =– . + 1˜ = – ÁË – ¯ 3 3 2 2 3

p

Ú0

2

(cos 3t + cos t) dt

x log x

•

u log u

0

Ú0 (1 + x 2 ) dx = Ú• ( 2 )2 du = – I 1+ u

Ú0

|sin 2p x| dx = 1/2

Ú0

=–

x + 1 – 5x + 1 . – 4x

1/2

Ú0

| sin 2p x| dx +

sin 2p x dx –

Ú

1 1

Ú

1 1

|sin 2p x | dx 2

sin 2p x dx

2

2 cos 2p x 2 cos 2p x 1 + = . 2p 0 2p 1 2 p

x

33. lim

2

Ú0 (tan -1 x )

dx

x2 + 1

xƕ

1 ˆ 1 2Ê 1+ 2 dt Á Ú t – 1˜¯ 2 1Ë

2

34.

A(t) =

0

Ú–1 e

x

2

( -1 ) x + 1 = lim tan x xÆ• x (L’ Hôpital Rule)

= lim (tan -1 x )2 1 + xƕ

2 2 Êp ˆ = p . = 4 x2 Ë 2 ¯

1

t

dx + Ú e – x dx = 1 – e–1 – (e–t – 1) 0

lim A (t ) = 2 – e . –1

tƕ

35. Since the integrand is an even function so required 2

integral is equal to 2 Ú x cosp x dx

2

0

=

1 4 2u du (u2 = 5 x + 1) 4 Ú1 u2 – 1

=

1 3 1 u -1ˆ 1Ê log ÁË 3 + 2 log 5 - 2 ulim Æ1 u + 1 ˜¯ 2

1

1 3 Thus I1 + I2 = 1 + log . 4 5 1

cos t cos 2t dt

2

1

1 3 5x + 1 dx 4 Ú0 x

29. f (x) = x – [x] so

2

–p

=–

=

–1 3 x + 1 1 2 t2 dx = – Ú 2 dt Ú 4 0 x 2 1 t –1

I2 =

p

Ú 2

So I = 0.

19

1 t –1ˆ 1 = – ÊÁ 1 – lim log t + 1 ˜¯ 2 t Æ1 2Ë (x + 1 = t2) and

1

I=

So the given integral is of the form I1 + I2

= –

cos t (sin 2t - cos 2t ) dt .

2

31. Putting u =1/x

32.

28. The integrand can be written as

2

-p

= 9(10 –8) < 10 –7

I1 =

p

dx

27. For x ≥ 10, we have |sin x| £ 1 and 1 + x 8 ≥ 108 1 fi £ 10 –8 1 + x8 Therefore, 19 sin x 1Ú dx £ 10 1 + x 8

1

=–

Ú1 (t + 1) dt = 6.

x 2 x2

2

p

30.

3

Ú0 e

2 (e x - 1) e x

= lim

1 1 = 0 – Ê – 1ˆ + = 1 Ë2 ¯ 2

0

1

Ú–1 f ( x ) dx = Ú–1 ( x + 1) dx + Ú0 x dx

3

2 2 2 = 2 È Ú x cos p x dx - Ú 1 x cos p x dx + Ú3 x cos p x dx ˘ ÍÎ 0 ˙˚ 2 2 1

È sin p x cos p x 2 Ê sin p x cos p x ˆ = 2 Íx + -Áx + ˜ p p p2 0 Ë p2 ¯ ÍÎ

3

1

2

2

2 Ê x sin p x cos p x ˆ ˘ +Á + ˙ ˜ Ë p p 2 ¯ 32 ˚

Definite Integrals 13.41

1 1 3 1 ˆ 3 ˘ 8 È 1 = 2Í – 2 – Ê– + 2 – Ê– ˆ˙ = – Ë Ë ¯ Î 2p p 2p 2p 2p ¯ ˚ p p x = t, the given integral reduces to 2

36. Putting tan

= 2Ú

Ú 2a 0

a 2

t 2 + tan 2

2

tan -1

a tan 2

a a tan 2 2 1 p p . = = a a 4 2 sin a sin cos 2 2 cos2

n -1

I1 =

nƕ

( ) r n

2

1 1 = Ú =– . 0 (1 + x )2 1+ x 0

nƕ

45.

3

1+

f (x) dx =

3

so A =

fi

1

2

a

f ( x ) dx

a

a

0

0

e

e

m m Ú1 log x dx = x log x 1e - Ú1 m x

log m - 1 x dx x

d d f3 (x) = f 2(x), f2(x) = f(x) and dx dx

x

f 3(x) = t f2 (t ) 0x – t f1 (t ) dt Ú0

p

x

0

0

2

(x – t)

1 x ( x – t )2 f (t ) dt Ú 0 2 2 1 x 1 = Ú ( x – t )2 f (t) dt. Thus A = . 2 0 2

Ap 1 2 = f¢Ê ˆ = Ë2 ¯ 2 2

Ú0 f ( x ) dx = – A p cos 2 x + Bx 10

x

= x f2 ( x ) – t f1 (t ) dt = ( x – t ) f1 (t ) dt Ú Ú =–

4 . Also p

2A = p

0

Ú– a f ( x ) dx + Ú0

d f1(x) = f (x). Integrating by parts dx

x2

p p cos x 2 2

x =

0

39. For the integrand f (p – x) = – f (x). 40. f ¢(x) = A

1/3

a

47. Note that

1 dx = log 2 . = Ú0 3 3 1+ x 1

b

Úa (b – x ) f ( x ) dx = I2 .

= e – m Jm – 1 So J8 + 8 J7 = e.

+ r3

2

r n

a

Ú–a

46. Jm =

() () r n

n

1 = lim  nƕ n n =1

Â

r =1n

b

b

= – Ú f ( – x ) dx + Ú f ( x ) dx = Ú ( f ( - x ) + f ( x )) dx . r2

n

b

Úa F ( x ) dx = x F( x ) a – Úa x f ( x ) dx

44. F ¢(x) = x 2 – 8x + 16 = (x – 4)2 > 0 for x Œ [0, 5] ~ {4} So the greatest value of F is F(5) 5 1 = Ú (t – 4 )2 dt = . 4 3

1 1 = – +1= 2 2 38. Required limit = lim

then F ¢(x) = f (x).

log x 1 . So F ¢(x) < 0 for log 3 10 < x < 1 and F¢(x) > 0 for x > 1. Hence f has least value at x = 1.

1 1+

x

Úa f (t ) dt

43. f ¢(x) = – log

dx

1

=

n

1 n -1 Â nr=0

g¢(x) g(x) dx

b

0

r=0

= lim

b

Úa

B = 0.

= bF (b ) – Ú x f ( x ) dx a

 (n + r )2 nƕ

37. Required limit = lim

fi

1 ((g(b))2 – (g(a))2) 2

42. Let F(x) =

tana 2

t

1

.

2A p

f (x) g(x) dx = =

dt

tana 2

1

=

b

Úa

(1 – cos a ) + t 2 (1 + cos a )

1 cos

41.

dt

tana 2

0

=

= B+

48.

dx = dy

1 1 + 9y

2

f1 (t ) 0x +

fi

dy = dx

1 + 9 y2

Complete Mathematics—JEE Main

13.42

1 18 y dy = 2 1 + 9 y 2 dx = 9y. Thus the constant of dx 2 proportionality is equal to 9. 49. Put g(x) = t fi x = f(t) fi dx = f ¢(t) dt d2 y

I=

b

55. Since 0 £ cos2 x £ 1, so 1 1 1 £ £ 8 5 + 3 cos2 x 5

b

Úa t f ¢ (t ) dt + Úa f ( x ) dx b a

b

b

a

a

cos x

56.

= [t f (t )] - Ú 1. f (t ) dt + Ú f ( x ) dx

1 - sin 2 x

b

= h (t )]a = h(b) – h(a) 50. I =

[ x ] 2t dt = 0 2[t ]

[x]

Ú

Ú0

– [t]

1

1

I = [ x ] Ú 2t - [t ] dt = [ x ] Ú 2t dt =

p

0

Ú0 x

[ x ] 2t 1 [ ]0

J =Ú

p / 2 sin

Ú0

p / 2 sin - 2

Ú0

nx - sin 2 ( n - 1) x dx sin x

(2n - 1) x sin x

=

1 2n - 1

dx =

=

cos (2n - 1) x 2n - 1 0

r +1

dx

Ú0 1 + x 2

 Úr [ x ]

Ú1

10 x 2 2

( x3 + 1)

fi

10 9 du 10 È = Í– 3 Ú2 u2 3 Î

dx =

1 ˘ 9 35 = u ˙˚ 2 27 .

cos x dx sin x(cos x + 1) cos x sin x

20 ¥ 21ˆ ˜ = 44100 2 ¯ 1

Ê 1ˆ f Á ˜ dx x Ë x¯ 2

sec q

cos q

b

bƕ 0

dx 1+ x

2

1 ˆ Ê ÁË Put = t ˜¯ x

(

1/ 2 È 1

p 2

˘ 1 ˆ 1 Ê 1 + Í Á ˙ dt ˜ 2 Î 4 Ë 1 + t 1 - t ¯ 2 (1 + t ) ˚

Ú1 / 2

=

È 1 log 1 + t + 1 ˘ ÍÎ 4 1 - t 2(1 + t ) ˙˚1 / 2

=

ÈÊ 2 + 1ˆ 1 log ÍÁ ˜ 4 ÎË 2 - 1¯

1/ 2

60.

)

(1 + t )2 (1 - t )

=

f (t ) dt = - I

= lim tan - 1 b - 0 = bƕ

t dt

1/ 2

Ú1 / 2

2

f (t ) dt = - Ú

Ú

x sin x dx

J = p.

Úp / 4 (1 - cos2 x) (1 + cos x) dx

I=

sec q

cos q

p /3

Úp / 4

59. I =

dx

f ( x ) dx = - Úcos q

= lim

p

0

Put cos x = t, so that

4

Hence 2I = 0 fi I = 0 54. The integrand is even, so • • dx dx Ú– • 1 + x 2 = 2 Ú0 1 + x 2 •

p sin x dx – Ú

p

p /3

 r 4 = ÊÁË

Úsec q

0

2 J = p ( – cos x ) 0 = 2p

fi

=

r=0

Úcosq

0

2

r=0

sec q

p

x sin x dx = Ú

p

- Ú 2 x sin x dx = – 2 J

Level 2

sin (2n - 1) x dx = -

Ú0

20

53. I =

0

p /2

=

=

p

dx = 0

dx

sin x

20

Ú0 [ x ]

p

p

p /2

58. Set x + 1 = u

which is a Harmonic Progression. 52.

dx - Ú

3

51. n > 1, an – an – 1 =

4

p /2

0

cos x = x 2 sin x

0

fi

log 2

21

2

log 2

[x] . =

p /2

=Ú

2

1 - sin x 57. Integrating by parts,

is periodic with period

0

x Œ[0, p / 2 ] cos Ï1 =Ì cos x Ó- 1 , x Œ(p / 2, p )

=

cos x

p

Ú0

2t - [t ] dt

The function g(t) = 2t 1, Therefore

=

p /2 p dx p £Ú £ 2 0 16 5 + 3 cos x 10

fi

Ê 2ˆ ˘ + 1 - 1 . ÁË ˜¯ ˙ 3 ˚ 2+ 2 3

b

Úa [ f ( x) - fave ] dx b

=

Úa

=

Úa

b

b

f ( x) dx - ( fave)Ú dx a

b È 1 ˘ f ( x) dx - Í f ( x) dx ˙ ( b - a ) Ú a Îb - a ˚

Definite Integrals 13.43

= 2Ú

= 0. 61.

2p

Ú0

0

= 2 log

2p

1È 1 1 ˘ sin(m + n) x + sin(m - n) x ˙ m-n 2 ÍÎ m + n ˚0

=

= 0.

Ú0

1

Ú0 log (

68. I =

= x log

sin k x sin r x

1 2p = Ú [cos(k + r ) x + cos(k - r ) x ] dx = 0 2 0 1 2p sin kx sin kx dx = Ú [1 - cos(2kx )] dx 2 0 1 = (2p ) = p 2

2p

Ú0

and

2p

Ú0

Also,

sin kx cos(rx ) dx = 0 " r, k ŒN

2p

Ú0

Thus,

Q ( x ) sin(kx ) dx = pak.

63. Imitate Example 101. 1 È Ê 1ˆ Ê 1ˆ ˘ 64. Ú Í2 x sin Á ˜ - cos Á ˜ ˙ 1/ 2 Î Ë x¯ Ë x¯ ˚ 1 Ê 1 ˆ ˘ - 1 x 2 Ê cos Ê 1 ˆ ˆ Ê - 1 ˆ dx 2 x sin = Á ÁË ˜¯ ˜¯ ÁË 2 ˜¯ ÁË ˜¯ ˙ x ˚1 / 2 Ú1 / 2 Ë x x

-Ú

1

1/ 2

Ê 1ˆ cos Á ˜ dx Ë x¯

1 sin (2). 4 65. F¢(x) = f(x) > 0 = sin (1) –

fi

sin ( x + p / 6 ) ˘ sin ( x + p / 3) ˙˚0

È Ê 3 / 2ˆ Ê 1/ 2 ˆ ˘ = 2 Ílog Á - log Á ˜ Ë 3 / 3 ˜¯ ˙˚ Ë 1/ 2 ¯ Î = 2 log 3.

62. For r π k, r, k Œ N, 2p

[cot ( x + p / 6) - cot ( x + p / 3)] dx p /2

cos mx cos nx dx

1 2p [cos(m + n) x + cos(m - n) x ]dx 2 Ú0

=

p /2

–

Ê

1

Ú0 x ÁË

(

)

1 - x + 1 + x dx

)

1

1 - x + 1 + x ˘˚ 0

ˆ Ê -1 1 ˆ dx + ˜ Á 1 - x + 1 + x ¯ Ë 2 1 - x 2 1 + x ˜¯ 1

= log ( 2 ) +

Ê 1+ x - 1- xˆ 1 1 x Á ˜ dx Ú 2 0 1- x + 1+ x Ë ¯ 1 - x2

=

1 log 2 2

+

1 2

=

1 1 log 2 + 2 2

x 1 + x + 1 - x - 2 1 - x2 dx Ú0 (1 + x) - (1 - x) 1 - x2 1

1Ê

ˆ - 1˜ dx ¯ 1 - x2 1

Ú0 ÁË

p ˆ 1Ê Á log 2 + - 1˜¯ . 2 2Ë 69. For n ≥ 1, 0 £ x £ 1/2, x2n £ x2, 1 1 £ fi 0 £ 1 – x2 £ 1 – x2n fi 2n 1- x 1 - x2 =

fi

F is differentiable and increasing on [a, b].

1

Ú0

2

dx 1 - x2n

1

£Ú

0

1

dx

2

2 = sin -1 x ˘˚ 0

1 - x2

66. First show that f(x) = 2x, x ΠR 1

Now, =

Ú0 f ( x) g ( x) dx

1

Ú0 2 x(e

x

- 2 x) dx

4 3 1 x ˘ 3 ˚0 4 2 = 2e – 2 (e – 1) – = . 3 3 1

1

0

0

= 2 xe x ˘˚ - 2Ú e x dx -

67. I =

=

p /2

Ú0

cos ec ( x + p / 3) cos ec ( x + p / 6 ) dx

p / 2 sin [( x + p / 3) - ( x + p / 6 ) 1 = dx Ú sin (p / 6 ) 0 sin ( x + p / 3) sin ( x + p / 6 )

p < 1. 6

70. 8 = f ¢(1) = 2A + B 33 = f (2) + f ≤(2) = 6A + 2B + C 7 1 1 = A+ B+C 3 3 2 fi 14 = 2A + 3B + 6C Also,

Solving we obtain A = 7, B = – 6, C = 3 71. y =

x2

Úx

x2

log t dt = (t log t - t )]x

13.44

Complete Mathematics—JEE Main

= (2x2 – x) log x – x2 + x

p

Ú0 [2 + cos 2 px - cos 2qx] dx

=

fi y(2) = 6 log 2 – 2

[∵ third term in the integrand is odd]

Also, y¢ (2) = 7 log 2 Thus, equation of tangent at x = 2 is y – (6 log 2 – 2) = (7 log 2) (x – 2)

79. I =

a

Ú0 a

=

Ú0

=

Ú0

or y + 8 log 2 + 2 = (7 log 2) (x)

a

Ê 2 - sin q ˆ 72. Ú log Á dq = 0, as the integrand is an -p / 2 Ë 2 + sin q ˜¯ odd function. p /2

3 f ( x) {- g ( x)} ÈÍ h ( x) Î4

7 5 I= 4 4 fiI=

dt I = Ú-1 3 + 1 - t2 0

1 2+t ˘ 1 log = log 3 . 2(2) 2 - t ˙˚ -1 4

74. In I2, put t = 1/u to obtain x

Ú1

75.

Ú

(- 1 u ) du 2

1+1 u

2

=

1

dt

Úx 1 + t 2

= I1

2

p = 2 12 t t -1

fi sec–1 x – sec–1

5 I1 7

=

Ú0

a

a

Ú0

f ( x) g ( x) dx

f (a - x) g (a - x) dx f ( x) [- g ( x) ] dx = – I1

fi 2I1 = 0 fi I1 = 0

dy cos y = 0 dx Putting x = p /2 and y = p, we get 3 - 2 sin 2 x +

2 = p /12

fi x = sec (p /3) = 2. 1 2 2 dx = 76. Average = Ú0 x 2 e +1

)

Ú0

f ( x) g ( x) dx

80. Differentiating w·r·t· x, we get

fi sec–1 x = p /3

(

a

=

5˘ dx 4 ˙˚

Thus, I = 0.

dt

x

a

Ú0

where I1 =

0

I2 =

f ( x) g ( x)h( x) dx

f (a - x) g (a - x)h(a - x) dx

73. Put sin x – cos x = t, so that

=

= 2p.

2

e

-x

Ú0 1 + e- x dx

2

dy ˘ = 1. dx ˙˚(p 2, p ) 81. We have

= - log 1 + e - x ˘ ˚0 = log 2 – log (1 + e–2)

P¢(x) = a(x – 1) (x – 3)

= 2 + log [2/(1 + e2)].

1 fi P(x) = a ÈÍ ( x - 2)3 - x ˘˙ + b Î3 ˚ 4 As 6 = P(1) = – a + b 3

77. I = =

p

Ú0

= a[(x – 2)2 – 1]

x log (sin x) dx

p

Ú0 (p - x) log [sin(p - x)] dx

2 = P(3) = –

p

fi 2I = p Ú log sin x dx 0

p 2

p 2

= p ÈÍ Ú0 log sin x dx + Ú0 log sin (p - x) dx ˘˙ Î ˚ fi I = pÚ

p 2

0

78. I =

p

Ú-p ÈÎcos

2

p2 log (sin) dx = log 2. 2 px + sin 2 qx - 2 cos px sin qx ˘˚ dx

8 a+b 3

fi a = 3, b = 10 Thus P(x) = (x – 2)3 – 3x + 10 19 . 4 82. We have P ¢¢(x) = a (x + 1) x (x – 1) fi

1

Ú0 P( x) dx

=

Definite Integrals 13.45

= a (x3 – x) 1 1 fi P ¢(x) = a ÊÁ x 4 - x 2 ˆ˜ + b Ë4 2 ¯ 1 5 1 3ˆ fi P(x) = a Ê x - x + bx + c Ë 20 6 ¯ As 1 1ˆ + – 1 = aÊ–b+c Ë 20 6 ¯ 1 1 1 = aÊ - ˆ + b + c Ë 20 6 ¯

Thus, I = log (2a + 1) = log (2n).

1 (sin 4a – sin 2a ) = 0 2 fi sin a [1 + cos(3a )] = 0 As – p < a < 0, sin a π 0, therefore cos (3a ) = – 1 fi a = –p /3. 87.

9

= 3 =b

(

1

88. lim

1 1 Ú0 P( x) dx = - 30 a + 2 b 1

)

Ú- x f (2t ) dt 2x

Ú0

Â

k =1

k k e n 1

Ú0 x e

lim Sn =

= lim

f (2 x) + f ( - 2 x) 2 f ( 2 x + 4)

=

f (0) + f (0) 2 f ( 4)

=

f (0) f (4)

dx = 1.

x k + 2 2k  k ! = x2 n ƕ k =0 n

84. As lim

•

Â

k =0

(2 x)k = x2 e2x k!

89.

f (t + 4) dt

(1) f (2 x) - ( f ( - 2 x))( - 1) 2 f ( 2 x + 4)

n

x

3

2

(

Ú

=

1 2 (e –1). 4

85. Let I =

e

90. Let I =

)

I=

1 x 2a

Ú0

-1 dx log x 1

fi I = log (2a + 1) + C When a = 0, I = 0 fi C = 0

2

p

Ú- p

(1)

cos 2 q ( - 1)dq = 1 + a -q

-p

Úp

p

Ú- p

a x cos 2 x dx 1 + ax

Adding (1) and (2), we get

1 2 x 2a log x dI 2 ˘ fi dx = x 2a + 1 ˙ =Ú 0 da log x 2a + 1 ˚0

2 = 2a + 1

3

sin x dx + Ú 2 dx

cos 2 x dx 1 + ax Put x = – q, to obtain

1 1 dx = 2 x 2 - 2 x + 1 e2 x ˘˙ 4 ˚0

=

cos x

= 0 + 2 = 2. [integrand of first integral is odd]

1

2x

[∵ f is continuous]

Ú-2 f ( x) dx = Ú-2 e

Ê n x k + 2 2k ˆ lim Ú0 n Æ • ÁË kÂ=0 k ! ˜¯ dx

1 2 x 0

Ê 0 formˆ ˜¯ ÁË 0

= lim

xÆ0

n

n Æ•

Thus,

1 4 2 (4 – 1) + (93 / 2 - 43 / 2 ) 4 3

xÆ0

3 2 . 1- 3 + 7 2

83. Sn = 1 n

x dx

929 . 12

xÆ0

(

9

x

)

3 -1 7

We have

=

4 3

Ú0 g ( x) dx = Ú0 dx + Ú1 x dx + Ú4 =1+

7 and 1 = – a – a + b 60

fi a = 60

cos 2 x dx = 0

fi sin a

fic=0

Also,

2a

Úa

86. sin a +

2I=

p

Ú-p cos

2

fi I = p/2. 91. F(x) =

p

Ú0

p

x dx = 2Ú cos 2 x dx = p

f (t ) dt

fi F ¢(x) = f (x) \ f (4) = F ¢ (2)

0

(2)

13.46

Complete Mathematics—JEE Main

Also, 2xF ¢ (x2) = 2x (1 + x) + x2 fi F ¢ (x2) = 1 + 3x /2

p /4

= Ú 0

2

\ f (4) = F ¢(2 ) = 1 + 3 = 4.

tan n - 2 x sec 2 x dx – In – 2

fi In + In – 2

Now, x2 – f ¢(x) = 0 fi x2 =

tan n - 2 x (sec2 x - 1) dx

p /4

2 - x2

92. f ¢ (x) =

p /4

2. In = Ú 0

tan n -1 x = n -1 0

fi n(In + In – 2) =

2 - x2

n = n -1

fix=±1 1

93. I (m, n) = -

Ú0

94. I =

1

Ú0

)

9p

10p

9p

p

= 9Ú | sin u | du (|sin u| has period p) 0

1- x 1 - x2

p

= 9 Ú sin u du = – 9 cos u p0 0 = 18

dx

p -1. = sin -1 x + 1 - x 2 ˘ = ˚0 2 95. Differentiating both the sides of t2 2 5 Ú0 x f ( x) = 5 t we get

(

1 n

3. Ú | sin x | dx = Ú | sin(p + u ) | du = Ú | sin u | du p 0 0

n 2n I (m + 1, n – 1) m +1 m +1

1- x dx = 1+ x

1-

nƕ

n I (m + 1, n – 1) m +1

1

1

fi lim n(In + In – 2) = 1.

1 m +1 ˘ (1 + t ) n ˙ t m +1 ˚0

fi I (m, n) =

1 n -1

=

1

2

2

4. Ú x f ¢( x)dx = x f ( x) 02 - Ú f ( x)dx 0 0 2

= 2 f(2) – Ú f ( x)dx 0 =0–

3 3 =4 4

(2t) (t2) f (t2) = 2t4

dy dy =yfi = dx y dx fi log y = x + const.

fi f (t2) = t

fi y = Cex. Since y(0) = 1 so C = 1

2 Ê 4ˆ fi f ÁË ˜¯ = . 25 5

Hence f(x) = ex,

0

Ú-2 [ x

96. I = =

3

1

+ 3x 2 + 3x + 3 + ( x + 1) cos( x + 1)] dx

0

Ú-2 ÈÎ( x + 1)

3

+ 2 + ( x + 1) cos( x + 1) ˘˚ dx

Put x + 1 = t, therefore, 1

Ú-1[t

I=

3

+ 2 + t cos t ] dt 3

= 4 [∵ t + t cos t is an odd function].

Previous Years’ AIEEE/JEE Main Questions 2

1

2 –1

1

1

= Ú x 2 e x dx - Ú e2 x dx 0 0 2x 1

1 È 2 x1 ˘ e x = Í x e - 2Ú0 x e dx ˙ 0 Î ˚ 2

0

1 1 È ˘ 1 = Í e - 2 ÈÍ xe x - Ú0 e x dx ˘˙ ˙ - (e2 – 1) 0 Î ˚˚ 2 Î

= [e – 2e + 2e – 2] – =e–2–

2

= Ú 0 dx + Ú 1 dx 0 1 =

1

x 2 x Ú0 f ( x) g ( x) dx = Ú0 e ( x - e ) dx

2

1. Ú [ x 2 ]dx = Ú [ x 2 ]dx + Ú [ x 2 ]dx 0 0 1 1

5. f ¢(x) = f(x) fi

t

1 2 (e – 1) 2

1 3 1 2 1 e + = e - e2 2 2 2 2 t

6. F(t) = Ú f (t - y ) g ( y ) dy = Ú et - y y dy 0 0

Definite Integrals 13.47

12. Let I = Ú xf (sin x)dx = Ú (p - x) f {sin(p - x)}dx 0 0 p

= p Ú f (sin x)dx - I 0

t t = e È - y e - y + Ú e - y dy ˘ 0 ÍÎ ˙˚ 0 = et [– t e–t – [e–t – 1]

t

p /2

2I = p Ú [ f (sin x) + f {sin(p - x)}]dx 0 p /2

= – t – 1 + et = et – (t + 1). b

I=p Ú 0

f (sin x) dx

b

7. Ú x f ( x) dx = Ú (a + b - x) f (a + b - x) dx a a b (a a

= Ú

b

b x a

fi2 Ú fi Ú

I1 =

b a

f ( x) dx = (a + b) Ú f ( x) dx

1

1- b

= Ú (1 - x) g((1 – x)x)dx b

dx - Ú

14. The given limit can be written as 1 k Êkˆ sec2 Á ˜ Â Ë n¯ nÆ• n n lim

2

Ú sec t dt

=

x2

= lim

x Æ0

x sin x

2

1

= Ú x sec2 x 2 dx 0

1

x sin x tan t 0

I2 = 2. I1

fi 2I1 = I2 fi

1 n +1 x dx 0

1 1 . n +1 n + 2

xÆ0

xg ( x (1 - x )) dx

= I2 – I1

1

9. lim

1- b

Úb

– b + x))dx

1

2

=1

1 + e- a

1- b

= Ú (1 - x) x n dx 0

x 0

1 + ea

e- a

= Ú (1 - b + b - x) g((1 – b + b – x) (1 – 1 + b b

8. I = Ú x(1 - x)n dx = Ú (1 - x)(1 - (1 - x)) n dx 0 0

= Ú

+

Let f(– a) = b, so f(a) = 1 – b

b

a+b b f ( x) dx = Ú f ( x) dx 2 a

b x a

1 n x 0

ea

13. f(a) + f(– a) =

+ b - x) f ( x) dx

= (a + b) Ú f ( x) dx - Ú x f ( x) dx a a

=

p

p

t

= et Ú y e - y dy 0

= lim

xÆ0

tan x 2 x

◊

2

x =1 sin x

1 1 tan x 2 = tan 1. 2 2 0

3 f x 4t 15. lim Ú ( ) dt xÆ 2 6 x-2

3

10. Ú | 1 - x 2 | dx -2

f ( x)

-1 |1 -2

= Ú

1 |1 -1

2

x | dx + Ú

-1

3 |1 1

2

x | dx + Ú

1

2

x | dx

3

= - Ú-2 (1 - x 2 ) dx + Ú-1 (1 - x 2 ) dx - Ú1 (1 - x 2 ) dx = -x

-1 -2

x3 + 3

-1

+

1 x -1

-2

1 4˘ t = lim xÆ 2 x - 2 ˙ ˚6

x3 3

1

3

-1

[ f ( x)]4 - 64 xÆ 2 x-2

= lim

3 x1

x3 + 3 1

7 2 26 28 +2- -2+ = . 3 3 3 3 11. We have 1 + sin 2x = (sin x + cos x)2. Thus, =–1+

p /2

I = Ú (sin x + cos x) dx 0 p /2 0

= ( - cos x + sin x )]

= – cos p + sin p – (– cos 0 + sin 0) = 2 2 2

Ê f ( x) - 6 ˆ 2 = lim Á ˜ (f(x) + 6) ([f(x)] + 36) xÆ 2 Ë x - 2 ¯ = f ¢(2) (f(2) + 6) ([f(2)]2 + 36) =

1 (6 + 6) [36 + 36) = 18. 48

16. For 0 < x < 1, x2 > x3 2

3

1

2

1

3

fi 2 x > 2 x fi Ú 2 x dx > Ú 2 x dx or I1 > I2 0 0 For 1 < x < 2, x2 < x3

13.48

Complete Mathematics—JEE Main 2

2

p /2

3

2

fi Ú 2 x dx < Ú 2 x dx or I3 < I4 1 1

fi 2I = p Ú [ f (sin x) + f (sin(p - x))] dx 0 p /2

17. Putting x = – y p

I = Ú-p

cos 2 x 1+ a

x

= 2p Ú 0

p /2

fiI=p Ú 0

dx

p /2

=p Ú 0

we get -p

I= Ú p

[cos( y )]2 1+ a

f (sin x)dx

p

(– 1)d y = Ú -p

-y

a x cos 2 x 1+ a

x

dx

(2)

f {sin(p /2 - x)}dx

f (cos x)dx

21. Let m < a < m + 1, m ΠN, so that [a] = m a

Ú1 [ x] f ¢( x)dx

Adding (1) and (2), we get

2

x

p

2I = Ú -p

2

(1 + a ) cos x 1+ a

x

p

m

3

= Ú f ¢( x)dx + Ú 2 f ¢( x)dx +  + Ú (m - 1) f ¢( x)dx 1 2 m -1

p

dx = Ú-p cos 2 x dx

a

+ Ú m f ¢( x)dx m

p

= 2 Ú cos 2 x dx = Ú (1 + cos 2 x) dx 0 0

= (f(2) – f(1)) + 2 (f(3) – f(2)) + …

1 ˘ = ÊÁ x + sin 2 xˆ˜ ˙ = p Ë ¯ 2 ˚0

+ (m – 1) (f(m) – f(m – 1)) + m(f(a) – f(m))

fi I = p/2.

= [a] f(a) – (f(1) + … + f([a])).

p

x

6

18. Let I = Ú 3

9-x + x

= mf(a) – [f(1) + f(2) + … + f(m)]

3+6- x

6

= Ú3

9 - (3 + 6 - x) + 3 + 6 - x 9- x

6

Ú3

dx x + 9- x Adding (1) and (2), we get =

(1)

dx

Ê 1ˆ 22. F(e) = f(e) + f Á ˜ Ë e¯ e log t

= Ú 1

(2)

e log (1/u) Ê 1ˆ log t dt = Ú ÁË - 2 ˜¯ du 1 1+ t 1 + 1/u u

1/ e

Ú1

e

= Ú1

6

2I = Ú dx = 6 – 3 = 3 3 fiI=

1/ e log t

dt + Ú1

dt 1+ t 1+ t In the second integral, put t = 1/u to obtain

dx

3 . 2

log u du u (1 + u )

log t ˘ e È log t + \ F(e) = Ú1 Í ˙ dt Î 1 + t t (1 + t ) ˚

19. Put x + p = q, so that e log t

= Ú1

p /2

I= Ú [q 3 + cos 2 (q + 2p )] dq -p / 2 p /2

p /2

= Ú cos 2 q dq = 2 Ú0 cos 2 q dq -p / 2

= Ú

Êp ˆ p = 2Á ˜ = Ë 4¯ 2

=

e log t 1 t

= Ú

p

- x) f {sin(p - x)} dx

= p Ú f (sin x) dx - I 0

23.

Ú

e

dt =

1 ˘ (log t )2 ˙ 2 ˚1

1 . 2

p

20. Let I = Ú xf (sin x) dx 0 p (p 0

Ê 1ˆ Á1 + ˜ dt 1+ t Ë t¯

x 2

dt

=

2

t t -1

fi sec -1 t ˘ ˚

x 2

p 2 =

p 2

Definite Integrals 13.49

fi sec–1 x – sec–1

2=

p 2

1

1 Ú0 p( x)dx = p(1) – 2 (p(1) – p(0))

1 1 (p(1) + p(0)) = (41 + 1) = 21 2 2 1 log (1 + x ) 27. Let I = 8Ú dx 0 1 + x2

fi sec–1 x = p + p = 3p 2 4 4

=

Ê 3p ˆ fi x = sec Á ˜ = - 2 Ë 4¯

Put x = tan q, so that 24. For 0 < x < 1, sin x < x, so 1 sin x

Ú0

I=

x

1 cos x

I = Ú0

x

dx < Ú

x

1

0

x 1

1

dx < Ú0

x

2a

3/ 2 1

1

dx = Ú

x dx =

0

dx = 2 x

2x 3

= 0

2 3

1

=2

0

a

0

0

write p

sec q

È p /4 Êp ˆ˘ = 8Ú0 log Í1 + tan Á - q ˜ ˙dq Ë ¯˚ 4 Î

p /4 È 1 - tan q ˘ = 8 Ú0 log Í1 + ˙ dq Î 1 + tan q ˚

2 ˆ p /4 Ê = 8 Ú0 log Á dq Ë 1 + tan q ˜¯

p /2

I = Ú [cot x]dx = Ú ([cot x] + [cot (p - x)])dx 0 p /2

sec2 q dq

È∵ a f ( x)dx = a f (a - x)dx ˘ Ú0 Î Ú0 ˚

Ú f ( x)dx = Ú [ f ( x) + f (2a - x)]dx , we can

25. Using

p / 4 log(1 + tan q ) 2

I = 8Ú 0

0

p /4 p /4 = 8 È Ú (log 2)dx - Ú log(1 + tan q )dq ˘ ˙˚ ÍÎ 0 0

= Ú ([cot x] + [ - cot x])dx 0

Put cot x = t, so that •

I = Ú ([t ] + [ -t ]) 0

n

dt 1+ t

k

= lim Â Ú ([t ] + [ -t ]) nÆ•

2

k =1 k -1

dt 1 + t2

But [t] + [– t] = – 1 for k – 1 < t < k, therefore k

Ú ([t ] + [ -t ]) k -1

k

dt 1+ t

= Ú ( -1)

2

k -1

dt 1+ t

2

=

Èp ˘ = 8 Í log 2 ˙ - I Î4 ˚ fi 2I = 2p log 2 fi I = p log 2. x sin x

28. F¢(x) =

For local maximum or local minimum, set f ¢(x) = 0 fi x = p, 2p as x Œ (0, 5p/2)

– [tan–1 k – tan–1 (k – 1)] As

x > 0 for 0 < x < 5p/2

n

\ I = - lim  (tan -1 k - tan -1 (k - 1)) nƕ k =1

p = - lim [tan– 1 n – tan–1 0] = - . nÆ• 2 1

1

26. Ú p( x)dx = Ú 1. p( x)dx = [xp(x)]10 – I1 = p(1) – I1 0 0 where

f ¢(x) > 0 for 0 < x < p < 0 for p < x < 2p > 0 for 2p < x < 5p/2 f has a local maximum at x = p and a local minimum at x = 2p. 1.5

1

1

I1 = Ú xp ¢( x)dx = Ú (1 - x) p ¢(1 - x)dx 0 0 1

1

= Ú (1 - x) p ¢( x)dx = Ú p ¢( x)dx - I1 0 0 fi 2I1 = [p(x(x)]10 = p(1) – p(0) Thus,

2 1 2 1.5 29. Ú0 x[ x ]dx = Ú0 x[ x 2 ]dx + Ú1 x[ x 2 ]dx + Ú 2 x[ x 2 ]dx 2

= 0 + Ú1 x dx + 2Ú x2 =0+ 2

2

1

x2 +2 2

1.5 x dx 2 1.5

= 2

1 1 3 + = 2 4 4

13.50

Complete Mathematics—JEE Main x+p

30. g(x + p) = Ú 0

tan x ˆ 1 p /3 Ê = Úp / 6 Á + ˜ dx Ë 1 + tan x 1 + tan x ¯

cos(4t )dt = I1 + I2

p

Where I1 = Ú cos(4t )dt = g(p) and 0

p /3

= Úp / 6 dx =

x +p cos(4t )dt p

I2 = Ú

In I2, put t = p + q, so that x cos(4p 0

I2 = Ú

x cos(4q )dq 0

+ 4q )dq = Ú

However, g(p) = Ú

1 + 2x x

31.

p /2

p

p /2

p /2

= Ú0 (1 - cos 2 x) dx = =

dy = |x|, so |x| = 2 fi x = ±2. If x = 2, then y dx

fiI=

= Ú

| dt = Ú

2

t2 = 2

34. = 2. Hence the equation

-2 |t 0

If x = – 2, then y = Ú

-2 tdt 0

| dt = - Ú

t2 =2

-2

0

=–2

p 4

dx 1 dy = fi = 1 + y2 2 dy dx 1+ y

0

tangent is Y – 2 = 2(X – 2). The intercept on x-axis is given by 0 – 2 = 2(X – 2) fi X = 1.

p sin 2 x 2 2 0

p 2

Therefore, g(x + p) = g(x) + g(p) and g(x + p) = g(x) – g(p) both are correct options.

2 tdt 0

dx

2I = I + I = Ú sin 2 x dx = 2Ú0 sin 2 x dx -p / 2

1 ˘ = (sin 4t ) ˙ 4 ˚0

1 (sin 4p – sin 0) = 0. 4

2 |t 0

1 + 2- x

2

p /2

=

sin 2 x

p /2

dx = Ú-p / 2

= Ú p / 2 2 sin x dx -p / 2 1 + 2x

= g(x). p cos 4tdt 0

sin 2 x

p /2

33. I = Ú -p / 2

p 6

fi

d2y dx

2

1

=

2 1+ y

2

◊ 2y

7p / 3

dy = y. dx

7p / 3

35. Ú tan 2 x dx = Ú7p / 4 | tan x | dx 7p / 4 2p

7p / 3

= Ú7p / 4 | tan x | dx + Ú2p | tan x | dx 2p

7p / 3

Hence the equation of tangent is

= - Ú7p / 4 tan x dx + Ú2p tan x dx

Y + 2 = 2(X + 2)

= - log | sec x | 7p / 4 + log | sec x | 2p

The x-axis intercept is given by 0 + 2 = 2(X + 2)

= - ÈÎ log 1 - log 2 ˘˚ + (log 2 - log 1)

2p

fi X = – 1. p /3 p /6

32. I = Ú

=

= log 2 2 dx

1 + tan x

=Ú

dx ˆ Êp p 1 + tan Á + - x˜ ¯ Ë3 6

dx 1 + cot x

2I = I + I

p

36. Ú 1 + 4 sin 2 0

x x x p - 4 sin dx = Ú0 | 2 sin - 1 | dx 2 2 2

p /2

= 2Ú | 2 sin t - 1 | dt 0 p /6 p /2 = 2 È Ú | 2 sin t - 1 | dt + Ú | 2 sin t - 1 | dt ˘ 0 p /6 Î ˚

p 1 + tan ÊÁ - xˆ˜ ¯ Ë2

p /3

= Ú p /6

p /3 p /6

dx

p /3

Úp /6

7p / 3

p /3

= Úp / 6

tan x 1 + tan x

p /6 p /2 = 2 È - Ú (2 sin t - 1) dt + Ú (2 sin t - 1)dt ˘ p /6 Î 0 ˚

dx

p p˘ È p /6 p /2 = 2 Í 2 cos t 0 + + -2 cos t p / 6 - ˙ 6 3˚ Î p p˘ È = 2 Í 3-2+ + 3- ˙ 6 3˚ Î

Definite Integrals 13.51

p˘ È = 2 Í2 3 - 2 - ˙ 6˚ Î = 4 3-41/ 2 log(1 +

37. Ú0

p 3

2 x)

1 + 4x

2

log( x 2 ) + log((6 - x) 2 )

dx =

1 p /4 Ú log(1 + tan t )dt 2 0

1 p /4 1 p /4 Ê Ê p ˆˆ log(1 + tan t )dt = Ú log Á 1 + tan Á - t ˜ ˜ dt Ú 0 0 Ë Ë 4 ¯¯ 2 2

1 p /4 Ú (log 2 - log(1 + tan t ))dt 2 0 1p = log 2 – I 24 p log 2 I= 16 e

log((6 - x) 2 ) + log( x 2 )

(2)

Adding (1) and (2), we get 4 log( x

2

) + log((6 - x) 2 )

log( x 2 ) + log((6 - x) 2 )

dx

43. f(4 + x) = f(4 Рx) = f(2 + (2 Рx)) = f(2 Р(2 Рx)) = f(x) " x ΠR e e n -1 dx 1 - n 1 (log x )

Ú

Thus, f is a periodic function with period 4. Now,

= e – n Pn – 1

2

So Pn = e – n[e – (n – 1) Pn – 2] Putting n = 10, P10 = e – 10e + 10.9 P8 fi P10 – 90 P8 = – 9e p

dx

2I = 2 fi I = 1

=

38. Pn = Ú (logx) n dx = x(log x)n 1

log((6 - x) 2 )

2I = Ú2

1 p /4 Ê 1 - tan t ˆ = Ú0 log Á1 + dt Ë 1 + tan t ˜¯ 2

(1)

log((2 + 4 - x) 2 ) + log((6 - (2 + 4 - x)) 2 )

4

= Ú2

dx

log((2 + 4 - x)2 )

4

= Ú2

(2x = tan t) I=

log( x 2 )

4

42. I = Ú2

p /2

p

39. Ú [cos x]dx = Ú [cos x]dx + Ú [cos x]dx 0 0 p /2 p =0+ Ú =2 40. Differentiating both sides, we have p ( -1)dx p /2

f(t) + t = – 2t fi f(t) = – 3t, Ê pˆ so f Á - ˜ = p. Ë 3¯ et dt t+a Put t + a = u, then

2

2

5 = Ú f ( x)dx = Ú f (2 - x)dx = Ú f (2 + x)dx 0 0 0 4

4

= Ú f (u )du = Ú f ( x)dx 2 2 4

Ú0

\

2

4

0

2

f ( x )dx = Ú f ( x )dx + Ú f ( x )dx = 10

As f is periodic with period 4, 50

Ú10

f ( x )dx = Ú

10 + 4 (10 )

10

4

f ( x )dx = 10 Ú f ( x )dx 0

= 10(10) = 100. 44. Differentiating both the sides, we get (cos x) f(sin x) =

3 2

x

41. Let I = Ú 1

x+a

I= Ú 1+ a

eu - a du u

u È x + a eu ˘ 1+ a e du -Ú1 du ˙ = e– a Í Ú1 u u Î ˚

= e– a [F(x + a) – F(1 + a)]

Ê 3ˆ p To obtain f Á ˜ , put x = , so that 3 Ë 2 ¯ 1 2

Ê 3ˆ 3 fÁ ˜= 2 Ë 2 ¯

Ê 3ˆ fi f Á ˜ = 3. Ë 2 ¯ 1/ x log t Ê 1ˆ dt 45. f Á ˜ = Ú1 Ë x¯ 1+ t

13.52

Complete Mathematics—JEE Main

Put t = 1/u, so that dt = (– 1/u2) du, then 1ˆ x log(1/u ) Ê Ê 1ˆ f Á ˜ = Ú1 ÁË - 2 ˜¯ du Ë x¯ 1 + 1/u u x

= Ú1

Ú0

10

Ú4

2I =

dx

(2)

-5 ( x + 5)2

Ú-4 e

1. I1 =

=

1 p 1 2x - 2 x tan -1 x ˘˚ + Ú0 dx 0 2 1 + x2

2

0

Ú

2 / 3 ( 3 x - 2 )2 e dx 1/ 3

=Ú

(3x – 2 = t)

1 0 t2 1 0 2 e dt = - Ú et dt Ú 3 -1 3 1

=

p 1 - 2Ú0 tan -1 x dx 2

1

dx = - Ú et dt (t = x + 5)

2 / 3 9 ( x - 2 / 3 )2 e dx 1/ 3

I2 =

=

dx = 6 fi I = 3

Previous Years' B-Architecture Entrance Examination Questions

1Ê p -1 2 ˆ = Ú0 ÁË - cot (1 - x + x )˜¯ dx 2

1 1 t2 Ú e dt 3 0 I1 + 3I2 = 0 =

2. Ú1.5 [ x 2 ] f ¢( x)dx 0

1 p = - 2 tan -1 (1) + In (1 + x 2 ) ˘˚ 0 2

1

1.5

2

= Ú [ x 2 ] f ¢( x)dx + Ú [ x 2 ] f ¢( x)dx + Ú [ x 2 ] f ¢( x)dx 0 1 2

p Êp ˆ - 2 Á ˜ + log 2 = Ë 4¯ 2

=0+ Ú 1

= log 2 Ê (n + 1)(n + 2) … (n + 2n) ˆ 47. Let y = lim Á ˜¯ x Æ• Ë n2n

2

f ¢( x)dx + 2Ú

1.5 2

f ¢( x)dx

= f ( 2 ) – f(1) + 2 [f(1.5) – f ( 2 ) ]

1/ n

= 2f(1.5) – f ( 2 ) – f(1)

kˆ 1 2n Ê log Á 1 + ˜ Â Ë h¯ nÆ• n k =1

3. I =

log y = lim = Ú

[ x 2 ] + [(14 + x)2 ]

dx

Adding (1) and (2), we get

tan -1 (1 - x + x 2 )dx

2 log(1 + 0

[(14 - x)2 ]

10

1ˆ log t xÊ Ê 1ˆ dt \ f(x) + f Á ˜ = Ú1 Á1 + ˜ Ë x¯ Ë t¯ 1+ t x log t dt = Ú1 t x 1 1 = (log t ) 2 ˘˚ = (log x)2 . 1 2 2 1

[(14 - (10 + 4 - x))2 ] + [(10 + 4 + x)2 ]

I = Ú4

log u x log t du = Ú1 dt u (1 + u ) t (1 + t )

46. Let I =

[(10 + 4 - x)2 ]

10

Ú4

a

(Ú

a -a

x)dx

x 2 + 1)dx =

Ú- a log( x + a

f ( x)dx = Ú- a f ( - x)dx

a

Ú- a log(- x +

x 2 + 1)dx

)

a Ê - x 2 + x 2 + 1ˆ 2 log Ú- a ÁË x + x 2 + 1 ˜¯ dx = - Ú- a log( x + x + 1)dx a

= x log (1 + x )]02

2 x - Ú0 dx 1+ x

=–I

= 2 log 3 – Ú 2 x + 1 - 1 dx 0 x +1

fiI=0

= 2 log 3 – 2 + log( x + 1) 02

4.

1

Ú1/ 2

= 2 log 3 – 2 + log 3 2

= log 27 – log e = log 10

48. Let I = Ú4

27 e2

fiy=

27 e2

= . =

2

[ x ]dx [(14 - x) 2 ] + [ x 2 ]

(1)

n

f ( x )dx

1/ 2 n - 1

Ú1/ 2

n

f ( x )dx + Ú

1/ 2 n - 2

1/ 2

n -1

1 È 1 1˘ 1 + 2n -1 ÍÎ 2n -1 2n ˙˚ 2n - 2

f ( x )dx +  + Ú

1

1/ 2

f ( x )dx

1 ˘ È 1 È 1˘ Í 2n - 2 - 2n -1 ˙ +  + 1 Í1 - 2 ˙ Î ˚ Î ˚

Definite Integrals 13.53

=

1 22 n -1

+

1 22 n -3

lim Ú

1

n nƕ 1/ 2

5.

1 2

Ú0 x

1 1 È1 - (1/ 22 )n ˘ 2 È Ê 1 ˆ n ˘ = Í ˙ = Í1 - Á ˜ ˙ 2 2 Î 1 - 1/ 22 ˚ 3 Î Ë 4 ¯ ˚

f ( x )dx =

2 2 pˆ pˆ ˘ 1 ÈÊ p p Ê tan tan Í = ÁË ˜¯ ÁË ˜¯ ˙ + tan - tan 3 ˙˚ 4 6 2 ÍÎ 4 1 1 1 = [1 – 3] + 1 – =2 3 3

2 3 0

1

1

2

- 2t )t 9 dt

t10 t12 2t11 + = 10 12 11

1 [(f ¢(3))2 – (f ¢(2))2] + f ¢(3) – f ¢(1) 2

=

(1 - x )9 dx = - Ú (1 - t )2 t 9 dt (1 – x = t)

Ú0 (1 - t

=

+ +

1

fi4
1), t = x and axis of abscissa f (t) = t et (t0 > 0) between t = a, then the area bounded by g(t) = t + t0 t = x and axis of abscissa is (a) et0 [F (x + t0) – 2 F (a + t0)] (b) et0 [F (x + t0) – F (a + t0)] (c) e -t0 [F (x + t0) – F (a + t0)] (d) 2e -t0 [F (x + t0) – F (a + t0)] Ans. (c)

4 3/ 2 1 32 x + p ◊ 16 = + 4p 3 4 3 0

Example 14: The ratio in which the area bounded by the curves y2 = 12x and x2 = 12y is divided by the line x = 3 is (a) 15 : 49 (b) 13 : 48 (c) 12 : 37 (d) 16 : 35 Ans. (a) Solution: The point of intersection of y2 = 12x and x2 = 12y is given by (12, 12) 2 3 3x A1 = Area OAB = Ú0 12 x dx - Ú0 dx 12 3

3

x3 2 45 = = 12 x3/ 2 3 36 0 4 0

14.8

Complete Mathematics—JEE Main 12

A2 = Area ABC = Ú3

12

12 x dx - Ú3

x2 dx 12

147 4 A1 : A2 = 45 : 147 = 15 : 49 =

( x + 1,

)

1 - x for all x < 1 is

7 3 11 (c) 6 Ans. (b)

7 6 1 (d) 6 (b)

(a)

x2 = 12y

(12, 12) C

Ï x + 1, x £ -1 Ô x + 1 , 1 x f (x) = min = Ì x + 1, -1 < x < 0 Ô Ó 1 - x, 0 £ x £ 1

y2 = 12x

A O

Example 16: Area bounded by y = f (x) = min

(

B (3, 0)

)

y y = 1- x

y=x+1

Fig. 14.30

Example 15: The area of the region bounded by the curves y + |x| = 3 and y = |x – 1| is (a) 6 (b) 2 (c) 4 (d) 3 Ans. (c)

1 -

+

x

y=

3

(2, 1)

(-3, 0) (-1, 0)

1

1

2 x2 + x - (1 - x)3/ 2 2 3 0 -1 1ˆ 2 7 Ê = - Á -1 + ˜ + = Ë 2¯ 3 6

x-

x

0

0

y=

(10, 3) y= 3

x

-

Fig. 14.32

=

Ï x - 1, x ≥ 0 y = |x – 1| = Ì Ó1 - x, x < 0

1

x

O

Required area = Ú-1 ( x + 1)dx + Ú0 1 - x dx

Ï3 + x, x < 0 Solution: y = 3 – |x| = Ì Ó3 - x, x ≥ 0

y=

(-1, 0)

Example 17: The area of the region bounded by x = 1, x = 2, y = 4x + 1, y = ex is (a) 7 + e – e2 (b) 5 + e – e2 (c) 7 – e + e2 (d) 7 – 2e + e2 Ans. (a) Solution: For 1 < x < 2, 4x + 1 is always greater than ex. So the required area

(1, 0)

2

2

= Ú1 ((4 x + 1) - e x )dx = 2 x 2 + x -e x 1 = 7 + e – e2

Fig. 14.31

Required area 0

1

= Ú-1 ((3 + x) - (1 - x))dx + Ú0 ((3 - x) - (1 - x))dx 2

+ Ú1 ((3 - x) - ( x - 1))dx 0

1

2

= Ú-1 (2 + 2 x)dx + Ú0 2dx + Ú1 (4 - 2 x)dx = 2 x + x2

0 -1

1

+ 2 x 0 + 4 x - x2

2 1

= – (– 2 + 1) + 2 + (4 – 3) = 1 + 3 = 4. Fig. 14.33

Areas by Integration 14.9

Assertion-Reason Type Questions

Example 18: Let y = f (x) be symmetrical about y-axis. Statement-1 : Area of the region bounded by x = – a, y = f (x), x = a and x-axis is twice the area of the region bounded by x = 0, y = f (x), x = a and x-axis. Statement-2 : f is an even function. Ans. (a) Solution: y = f (x) is symmetrical about y-axis so y = f (x) = f (– x). In this case a -a

a 0

Ú f ( x)dx = 2Ú f ( x)dx Example 19: Let An be the area bounded by y = x and y = xn. 1 Statement-1 : A2 A3 … An = n - 2 2 n(n + 1) Statement-2 : An =

n 2(n + 1)

Ans. (c) Solution: The straight line y = x and the curve y = xn meet at (0, 0) and (1, 1), so

=

1 1 n -1 = 2 n + 1 2(n + 1)

1 È1 2 3 n - 1˘ ◊ ◊ ◊ Í n + 1 ˙˚ 2 Î3 4 5 1 1 = n-2 n(n + 1) 2

A2 A3 … An =

n -1

Example 20: Statement-1 : The area between y = 2 – x2 1 17 x from x = 0 to x = 1 is . and y = 2 6 1 Statement-2 : The area between y = 2 – x2 and y = x from 2 1 ˆ 1Ê x = 0 to x = 1 is given by Ú0 Á 2 - x 2 - x˜ dx . Ë 2 ¯ Ans. (d) Solution: y = 2 – x2 is parabola with vertex (0, 2) opening downwards. 1 ˆ 1Ê Required area = Ú0 Á 2 - x 2 - x˜ dx Ë 2 ¯

1

An = Ú0 ( x - x n )dx

x3 x 2 = 2x – 3 4

1

=20

1 1 17 - = 3 4 12

LEVEL 2 Straight Objective Type Questions Example 21: The area between the curve y = – x2 + 4x and y = x2 – 6x + 5 over the interval 0 < x < 1 is (a)

52 + 5 15 3

52 (c) - + 3 15 3 Ans. (b)

(b) -

52 + 5 15 3

Since 0 < x < 1 so required point of intersection is a=

5 - 15 . 2 y = x2 - (x + 5) (2, 4)

47 (d) + 5 15 3

Solution: We find the intersection of the two curve by equating – x2 + 4x = x2 – 6x + 5 2 fi 2x – 10x + 5 = 0 10 ± 100 - 40 5 ± 15 fi x= = 2 2

y = - x2 + 4x

(0, 0)

(1, 0)

Fig. 14.34

14.10

Complete Mathematics—JEE Main

The required area a = Ú0 [ x 2 - 6 x + 5 - ( - x 2 + 4 x)]dx

Solution: Required area 2

1

1

+ Úa [ - x 2 + 4 x - ( x 2 - 6 x + 5)]dx a

x3 = 3

1

= Ú0 [2 x 2 - 10 x + 5]dx + Úa ( -2 x 2 + 10 x - 5)dx a

1

2 x3 Ê 2 ˆ = - 5 x 2 + 5 x + Á - x3 + 5 x 2 - 5 x ˜ Ë 3 ¯a 3 0 = -

52 + 5 15 (on simplification) 3 2

Example 22: The area of the figure bounded by y = 9x and y = 3x is (a) 1 (b) 1/4 (c) 1/2 (d) 2 Ans. (c) Solution: Required area (1, 3)

2

= Ú0 ( - x 2 + 2 x + 4)dx - Ú0 x dx - Ú1 x 2 dx 1

+x

2 2 0

2

1

+

2 4x0

0

2 x3 19 - x3/ 2 = . 3 3 1 3 0

Example 24: The value of ‘b’ (b > 0) for which the area x 1 bounded by the curves y = + 2 , x-axis, x = b and x = 2b 6 x has the least value is (a) 1 (b) 2 (d) 21/ 3

(c) 2 Ans. (d)

x 1 Solution: The area bounded by the curves y = + 2 , 6 x x-axis, x = b and x = 2b is given by 1ˆ 2b Ê x f (b) = Úb Á + 2 ˜ dx Ë6 x ¯ 2b

Ê x2 1 ˆ b2 1 = Á - ˜ = + 4 b Ë 12 x ¯ b

Q

P

2b 1 1/3 - 2 =0fib=2 4 b 1 2 f ≤(b) = + 3 > 0 2 b

f ¢(b) = R(1, 0)

O

so f (b) is minimum for b = 21/3. Fig. 14.35

= area OPQRO – area DOQR 1 1 2 3 1 1 = Ú0 9 x dx - ¥ 1 ¥ 3 = 3 x3/ 2 - = 0 2 3 2 2 Example 23: The area of the plane figure bounded by lines y = x , x Œ [0, 1], y = x2, x Œ [1, 2] and y = – x2 + 2x + 4, x Œ [0, 2] is (a) 10/7 (b) 10/3 (c) 3/5 (d) 4/3 Ans. (b)

Example 25: The area bounded by the curve y = x(x2 – 1) and x-axis is 1 (a) 0 (b) 2 1 (c) 1 (d) 4 Ans. (b) Solution: The given curve is a cubic curve and cuts x-axis at – 1, 0 and 1. y

y = - x2 + 2x + 4

-1

1

O

x

y = x2

Fig. 14.37 0

1

2

y= x

Fig. 14.36

Required area 0

1

= Ú-1 x( x 2 - 1)dx + Ú0 x( x 2 - 1)dx

Areas by Integration 14.11 0

È 1 4 x2 ˘ = Í x - ˙ + 2 ˚ -1 Î4

1

49

4

2 Ú1 ( x - x)dx = 3

È 1 4 x2 ˘ Í x - ˙ 2 ˚0 Î4

So the total area between y = x2, y = 1 49 50 = . [0, 4] is + 3 3 3

1 1 1 = + = 4 4 2 Example 26: The area between the curves x = 1 – y2 and x = y2 – 1 is 4 1 (b) (a) 3 3 8 5 (c) (d) 3 3 Ans. (c)

y y = x2

y= x

Solution: The two curves are parabolas with x = 1 – y2 opening to the left and x = y2 – 1 opening to the right.

1

1 x = y2 - 1

3

4

x

Example 28: The area of the region enclosed by y = x3 – 2x2 + 2 and y = 3x + 2 is

(1, 0)

71 6 29 (c) 3 Ans. (a)

(b) 14

(a)

-1

Fig. 14.38

Required area 1

= Ú-1 (Right - left )dy 1

2

Fig. 14.39

x = 1 - y2

(- 1, 0)

x on the interval

1

= Ú-1 (1 - y 2 - ( y 2 - 1))dy = 2 Ú-1 (1 - y 2 )dy

71 3

(d)

Solution: The intersection of two curves is given by x3 – 2x2 + 2 = 3x + 2 fi x3 – 2x2 – 3x = 0 fi x(x + 1) (x – 3) = 0

1

È y3 ˘ 1 ˘ 8 È2 = 2 Í y - ˙ = 2 Í - ÊÁ -1 + ˆ˜ ˙ = . Ë 3 3 3¯ ˚ 3 Î Î ˚ -1 Example 27: The area bounded by y = x2, y = 0 < x < 4 is 20 10 (a) (b) 3 3 50 (c) 10 (d) 3 Ans. (d) Solution: On [0, 1], x2 < curves is

x , so the area between two 1

2 Ú0 ( x - x )dx = 3 1

On [1, 4], x2 >

x,

x , so the area is

x+

3 y= y=

3

x

2

-2

2

x

+2

-1 1

2

3

Fig. 14.40 0

Required area = Ú-1[( x3 - 2 x 2 + 2) - (3x + 2)]dx 3

+ Ú0 [( x + 2) - ( x3 - 2 x 2 + 2)]dx =

7 45 71 + = . 12 4 6

14.12

Complete Mathematics—JEE Main

EXERCISE Concept-based Straight Objective Type Questions 1. The area below y = ex and above y = x between x = 0 and x = 2 is (b) e2 + e (a) e2 – 1 (c) e2 – 3 (d) e + 2 2. The area between y = x and y = x2, 0 < x < 2 is 2 (a) 1 (b) 3 1 4 (c) (d) 3 3 3. The area of the region bounded by y = x2 – x and y = 2x + 4 is 175 31 (a) (b) 6 6

(c) 12

(d)

125 6

4. The area of the region enclosed by the parabola y = 2 – x2, and line y = – x and x-axis is (a)

8 2 +3 7

(b)

8 2 +6 7

7 2 -1 7 2 +1 (d) 5 3 5. The area of the region bounded by y = x2 and y = – x2 + 2 is 5 4 (a) (b) 3 3 8 (c) (d) 1 3 (c)

LEVEL 1 Straight Objective Type Questions 6. The total area between y = 4x – x2 and y = x from x = 0 and x = 4 is (a) 17/3 (b) 37/6 (c) 16/3 (d) 4 7. The area between the curves y = x2 and y = x1/3, – 1 < x < 1 is 1 (a) (b) 2 2 3 3 (c) (d) 4 2 8. The area of the region bounded by the parabola (y – 2)2 = (x – 1), the tangent to the parabola at the point (2, 3) and the x-axis is (a) 9 (b) 12 (c) 3 (d) 6 9. The area bounded by the curves y = cos x and 3p is y = sin x between the ordinates x = 0 and x = 2 (a) 4 2 - 1 (b) 4 2 + 1 (c) 4 2 - 2

(a)

(b)

2 (8 + 3p) 3

1 5 3 2p (4 + 3p) (d) + 3 2 3 11. The area of the region bounded by y = 2 cos x, y = 3 tan x and the y-axis is 3 3 (a) 1 + log 3 – log 2 (b) log 2 2 2 3 (c) 1 + 3 log (d) 1 + log 3 – log 8. 3 2 p 12. The area bounded by y = sin–1 x, y-axis and |y| = 2 is (c)

(a) 2 (c) 2p

(b) p (d) 1

13. The area of the region bounded by x = 1, x = 2, y = log x and y = 3x is 9 e 6 + log (a) (b) – 2 log 2 + 1 log 3 4 log 3

(d) 4 2 + 2

10. The area included between x2 + y2 = 4x and y2 = x above x-axis is

4 (3p – 8) 3

(c)

3 – 2 log 2 log 3

(d)

9 – 2 log 2 + 1 log 3

Areas by Integration 14.13 2

14. The area inside the parabola y = 5x but outside y = 2x2 + 9 is (a) 4 3

(b) 6 3

(c) 12 3

(d) 8 3

15. The area bounded by the curve y = 2 log x, x-axis, y-axis and y = log 5 is equal to (a) 3( 5 - 1)

(b) 4

(c) 2( 5 - 1)

(d)

5 -1

2

16. Let A = {(x, y): y < 4x, y – 2x > – 2, y > 0}. The area of the region A is 2 (a) (1 + 5 )3/ 2 - 2 3

4 (3 + 5 )3/ 2 3 2 1 (3 + 5)3/ 2 - (7 + 3 5) (c) 2 33 4 (d) (3 + 5 )3/ 2 - (5 + 4 5 ) 3 (b)

17. The area of the region above x-axis bounded by y = p p £x£ is cot x, 12 4 (a) 1

(b) log( 2 + 1)

(c) log( 5 + 1)

(d) log( 3 + 1)

Assertion–Type Questions 1 ) 4 Statement-1 : The area of the curve y = f (x), 3 – 1 < x < 1 is equal . 8 Statement-2 : The area of the curve y = f (x) is twice 1 the area of a trapezium of parallel sides 1 and . 4 19. Let (a, b) and (c, d) be two points on the curve y = f (x).

18. Let f (x) = min (|x|, 1 – |x|,

Statement-1 : Area bounded by the y = f ¢(x), x-axis and x = a, x = c is d – b. x0¢

Statement-2 : Úx0 f ¢( x)dx = f (x¢0) – f (x0) 20. Statement-1 : The area bounded by y=

x , 2y – x + 3 = 0, x-axis lying in the first

quadrant is 9. Statement-2 : The area in statement 1 is given by 9Ê Ê x - 3ˆ ˆ Ú0 ÁË x - ÁË 2 ˜¯ ˜¯ dx

LEVEL 2 21. The area between x = y + 3 and x = y2 from y = – 1 to y = 1 is 4 16 (a) (b) 3 3 5 (c) (d) 4 3

22. The area of the region between y = 3x3 – x2 – 10x and y = – x2 + 2x is (a) 10 (b) 12 (c) 18 (d) 24 23. The area of the region bounded by y = x – 1 and x = 3 – y2 is 5 7 (a) (b) 2 2 9 (c) (d) 4 2 24. The area between the curves y = x2 + 3 and y = – x2 – 1 on [–2, 2] is 80 29 (b) (a) 3 3 47 (c) 21 (d) 3 25. The area the region enclosed by f (x) = x3 – 10x and g(x) = 6x, x ≥ 0, y ≥ 0 is (a) 24 (b) 39 (c) 42 (d) 84 26. The area bounded by y = x2 and y = 1 – x2 is (b) 16/3 (a) 8 / 3 (c) 32/3 (d) 16/3 27. The area of the figure bounded by the lines x = 0, x = p/2, f (x) = sin x and g(x) = cos x is 3 -1

(a) 2( 2 - 1)

(b)

(c) 2( 3 - 1)

(d) 2( 2 + 1)

28. The area of the figure bounded by the lines y2 = 2x + 1 and x – y – 1 = 0 is (a) 16/3 (b) 4/3 (c) 8/3 (d) 11/3

14.14

Complete Mathematics—JEE Main

Previous Years' AIEEE/JEE Main Questions 1. The area of the region bounded by the curves y = |x – 2|, x = 1, x = 3 and the x-axis is (a) 3 (b) 2 (c) 1 (d) 3 [2004] 2. The area enclosed between the curve y = loge (x + e) and the coordinate axes is (a) 3 (b) 4 (c) 1 (d) 2 [2005] 3. The area enclosed between the curves y2 = x and y = |x| is 2 (a) (b) 1 3 1 1 (c) (d) [2007] 6 3 4. The area of the plane region bounded by the curves x + 2y2 = 0, x + 3y2 = 1 is equal to (a) 5/3 (b) 1/3 (c) 2/3 (d) 4/3 [2008] 5. The area of the region bounded by the parabola (y – 2)2 = (x – 1), the tangent to the parabola at the point (2, 3) and the x-axis is (a) 9 (b) 12 (c) 3 (d) 6 [2009] 6. The area bounded by the curves y = cos x and y = 3p is sin x between the ordinates x = 0 and x = 2 (a) 4 2 - 1

(b) 4 2 + 1

(c) 4 2 - 2 (d) 4 2 + 2 [2010] 2 7. The area bounded by the curves y = 4x and x2 = 4y is 32 16 (a) (b) 3 3 8 (c) (d) 0 [2011] 3 y 8. The area bounded between the parabola x2 = and 4 x2 = 9y, and the straight line y = 2 is (a)

10 2 3

(c) 10 2

(b)

20 3 3

(d) 20 2

[2012]

9. The area bounded by the curves y = x , 2y – x + 3 = 0, x-axis and lying in first quadrant is

(a) 36 (b) 18 27 (c) (d) 9 [2013] 4 10. The area under the curve y = |cos x – sin x|, p , and above x-axis is 0 1 – x} is p 4 p 4 + (a) (b) 2 3 2 3 p 2 p 2 + (c) (d) [2014] 2 3 2 3 13. Let A = {(x, y): y2 < 4x, y – 2x > – 4}. The area (in square units) of the region A is (a) 8 (b) 9 (c) 10 (d) 11 [2014, online] 14. The area of the region above x-axis bounded by the p curve y = tan x, 0 < x < and the tangent to the 4 p curve at x = is: 4 1Ê 1ˆ (a) ÁË log 2 - ˜¯ 2 2 (b)

1Ê 1ˆ ÁË log 2 + ˜¯ 2 2

1 (1 - log 2) 2 1 (1 + log 2) (d) 2 (c)

[2014, online]

15. The area (in square units) of the region bounded by the curves y + 2x2 = 0 y + 3x2 = 1 is equal to 3 3 (a) (b) 5 4 1 4 (c) (d) [2015, online] 3 3 16. The area (in square units) of the region described by {(x, y): y2 < 2x and y > 4x – 1} is

Areas by Integration 14.15

7 5 (b) 32 64 15 9 (c) (d) [2015] 64 32 17. The area (in square units) of the region {(x, y): y2 > 2x and x2 + y2 < 4x, x > 0, y > 0} is 4 8 (a) p – (b) p – 3 3 (a)

4 2 p 2 2 (d) [2016] 3 3 2 18. The area (in square units) of the region described by A = {(x, y): y > x2 – 5x + 4, x + y > 1, y < 0} is 19 17 (a) (b) 6 6 7 13 (c) (d) [2016, online] 2 6 (c) p –

Previous Years' B-Architecture Entrance Examination Questions 1. The area enclosed by the parabola y = 3(1 – x2) and the x-axis is (a) 4 (b) 3 (c) 9 (d) 2 [2006] 2. The line y = x + 1 divided the area the curves y = È p p˘ cos x, Í - , ˙ and the x-axis into two regions which Î 2 2˚ are in the ratio (a) 2 : 1 (b) 1 : 3 (c) 2 : 3 (d) 1 : 1 [2006] 3. The area enclosed by the curves x2 = y, y = x + 2 and x-axis is 5 5 (a) (b) 4 2 15 5 (c) (d) [2007] 4 6 4. The area of the region described by the set {(x, y): x2 < y < |x|} is 1 1 (a) 2 (b) 1 6 3 5 1 (c) (d) 2 [2009] 6 3 5. The area of the region {(x, y): 0 < y < x2 + 1, 0 < y < x + 1, 0 < x < 2} is 23 21 (a) (b) 3 6 23 15 (c) (d) [2010] 6 2 6. The area of the region bounded by curves y = 1 – x2, x + y + 1 = 0 and x – y – 1 = 0 is 10 7 (a) (b) 3 3 8 (c) (d) 3 [2011] 3

7. The area bounded by the curves y2 = 12x and x2 = 12y is divided by the line x = 3 in two parts. The area (in square units) of the larger part is: 147 45 (a) (b) 4 4 137 245 (c) (d) [2014] 4 4 8. The area (in sq. units) of region bounded by the curve y = x and the lines y = 0, y = x – 2 is 10 8 (a) (b) 3 3 4 16 (c) (d) [2015] 3 3 9. If the line x = a bisects the area under the curve y = 1 , 1 < x < 9 then a is equal to x2 9 5 (a) (b) 5 9 9 4 (c) (d) [2016] 4 9

Answers Concept-based 1. (c)

2. (a)

3. (d)

4. (b)

6. (b)

7. (d)

8. (a)

9. (c)

10. (d)

11. (d)

12. (a)

13. (b)

14. (c)

15. (c)

16. (c)

5. (c)

Level 1

17. (d)

Complete Mathematics—JEE Main

14.16

Assertion Type 18. (a)

19. (a)

20. (c)

y = 2x + 4

y = x2 - x

Level 2 21. (b)

22. (d)

23. (c)

25. (b)

26. (a)

27. (a)

-1

24. (a) 28. (a)

Fig. 14.41

Previous Years’ AIEEE/JEE Main Questions 2

1. (c)

2. (c)

3. (c)

4. (d)

5. (a)

6. (c)

7. (b)

8. (b)

9. (d)

10. (b)

11. (d)

12. (b)

13. (b)

14. (a)

15. (d)

16. (d)

17. (b)

18. (a)

2 4. Reqd. area = Ú (2 - x )dx -1

= 2x –

x3 3

= 2 2-

Previous years’ B-Architecture Entrance Examination Questions

=

1. (a)

2. (b)

3. (d)

4. (c)

5. (c)

6. (b)

7. (a)

8. (a)

2

-1

1 ¥1¥1 2

1 2

2 2 Ê 1ˆ 1 - Á -2 + ˜ Ë 3 3¯ 2

4 2 5 1 8 2 +7 + - = 3 3 2 6

9. (a)

Hints and Solutions

(-1, 1)

2, 0

Concept-based 1. Since ex > x on [0, 2], so required area is 2 x (e 0

Ú

x2 - x)dx = e 2

2

x

5. Intersection of y = x2 and y = – x2 + 2 is x = – 1, 1.

0

2. For 0 < x < 1, x > x2 and for 1 < x < 2, x2 > x. So the required area is 1 (x 0

Ú

2

2 2 (x 1

- x )dx + Ú

- x)dx 1

2

Ê x 2 x3 ˆ Ê x3 x 2 ˆ = Á - ˜ +Á - ˜ 3¯ 2¯ Ë 2 Ë 3 0 1

2

= Ú-1 ((2 x + 4) - ( x 2 - x))dx = Ú-1 ( - x 2 + 3x + 4)dx . =

125 6

1 4 ¥ 3 ¥ 3 + Ú3 (4 x - x 2 )dx 2

x3 ˆ 9 Ê = + Á 2 x2 - ˜ 2 Ë 3¯

4

=

9 5 37 + = 2 3 6

3 1 | x1/ 3 -1

7. Required area = Ú 0 -1

= Ú

1 x3

Ê

0 x2 -1 Á

= Ú

Ë

1 0

2

- x dx + Ú

-

1ˆ x3

1

2

Reqd. area = Ú 8 = 3

6. Required area =

1ˆ 1 Ê 3. The intersection of y = x2 – x = Á x - ˜ - is x2 – x Ë 2¯ 4 = 2x + 4 fi x = 4, – 1, so the required area is 4

Ê 2 ˆ + 2) - x )dx = Á - x3 + 2 x˜ Ë 3 ¯

1 (( - x 2 -1

Level 1

1 Ê8 ˆ 1 + Á - 2˜ + = 1. ¯ 6 6 Ë3

4

Fig. 14.42

= e2 – 2 – 1 = e2 – 3.

1 x3

- x 2 | dx - x 2 dx

Ê 1 1 x3 0Á

˜ dx + Ú ¯ Ë

ˆ - x 2 ˜ dx ¯

-1

Areas by Integration 14.17 4 0

x3 3 3 = - x 3 4

4

-1

3 x3 + x3 4 3

p /6

1

Reqd. area = Ú0 (2 cos x - 3 tan x)dx = [2 sin x – 3 log | sec x |]p0 / 6

0

Ê -1 3 ˆ Ê 3 1 ˆ 3 = 0 – Á - ˜ +Á - ˜ = Ë 3 4 ¯ Ë 4 3¯ 2

= 1 – 3 log

2 3

2

3

Required area = Ú0 [( y - 2) 2 + 1 - (2 y - 4)]dy = 9 9. Required area = Ú =

3p / 2

2 Ú0

3 log 3 – log 8. 2

y = 3 tan x

8. Equation of tangent at (2, 3) to the parabola y2 – 4y – x + 5 = 0 is x – 2y + 4 = 0 3p / 2 | sin x 0

+ 3 log 1 = 1 +

1

y = 2cos x

- cos x | dx p 6

5p / 4

| sin( x - p / 4) | dx = 2 Ú-p / 4 | sin q | dq

p 2

0 p 5p / 4 = 2 ÈÚ ( - sin q )dq + Ú0 sin q dq + Úp ( - sin q )dq ˘ Î -p / 4 ˚

Fig. 14.44

= 4 2-2. 10. The intersection x2 + y2 = 4x and y2 = x is x2 = 3x fi x = 0, 3 3

= x

0

4 3

+Ú

p /2

= 2 Ú0 sin y dy = 2.

4

Reqd. area = Ú0 x dx + Ú3 4 x - x 2 dx 3/ 2 3

p /2

12. Reqd. area = 2 Ú0 | sin y | dy y

2

y = p/2

4 - ( x - 2) dx 4

È ( x - 2) 1 Ê x - 2ˆ ˘ 4 - ( x - 2) 2 + 4 sin -1 Á =3 + Í Ë 2 ˜¯ ˙˚ 2 Î 2 3 1 1 Ê ˆ = 33/2 + 2 sin–1 1 – 3 - 2 sin -1 Á ˜ Ë 2¯ 2

y = sin-1x

3/2

= =

x

y = - p/2

1ˆ p p Ê 3 Á3 - ˜ + 2 - 2 Ë 2¯ 2 6 Fig. 14.45

5 3 2p . + 2 3

2

13. Reqd. area = Ú1 (3x - log x)dx y2 = x

2

x2 + y2 = 4x 1

2

3

4

3x 2 = - ( x log x - x) 1 log 3 1 =

1 [9 – 3] – [2(log 2 – 1) + 1] log 3

=

6 – 2 log 2 + 1 log 3 y

Fig. 14.43

11. The intersection of two curves is given by 2 cos x = 3 tan x fi 2(1 – sin2 x) = 3 sin x fi 2 sin2 x + 3 sin x – 2 = 0 -3 ± 9 + 16 -3 ± 5 1 fi sin x = – 2, = 4 4 2 1 p So sin x = fix= 2 6

y = log x

y = 3x

x=1 x=2

Fig. 14.46

x

14.18

Complete Mathematics—JEE Main

fi (2x – 2)2 = 4x fi x2 – 2x + 1 = x fi x2 – 3x + 1 = 0

14. The intersection of two curves is given by 5x2 = 2x2 + 9 fi x2 = 3 fi x =

3, - 3

y = 2x2 + 9

- 3,15

=x=

3,15

3+ 5 2 0

y = 5x2

Ê 73 ˆ ÁË 0, ˜¯ 8

3± 5 2

Reqd. area = Ú 3+ 5 2

4 = x3/ 2 3 0 Fig. 14.47

=

Ê 73 ˆ y = 2x2 + 9 is a parabola with focus at Á 0, ˜ . Both Ë 8¯ the curves are symmetrical about y-axis.

6 2 2 3 3

)(

1 7+3 5 2 1 - 7+3 5 2

(3 + 5 )3/ 2 -

(

)

(3 + 5 )3/ 2

(

)

)

p /4

˜¯

= log

= 12 3 = log

log 5 y/ 2 e dy 0

15. Reqd. area = Ú = 2e y/ 2

(

= log sin x p /12

3ˆ 0

1 2 + 5 1+ 5 2

p /4

3

3

1 1+ 5 ¥ 2( 2 + 5 ) 2 2

17. Reqd. area = Úp /12 cot x dx

Reqd. area = 2Ú0 ((2 x 2 + 9) - 5 x 2 )dx = 2Ú0 (9 - 3x 2 )dx = 2[9 3 - x3

=

4

-

4 x dx -

1 2

6- 2 4

- log 2 2

( 6 - 2)

= log

( 6 + 2) 2

= log( 3 + 1)

log 5 0

= 2( 5 - 1) . y y = 2 log x

y = log 5

p p p 12 4 2

p

x

Fig. 14.50

Assertion-based Questions 1

Fig. 14.48

16. The intersection of y2 = 4x and y – 2x = – 2 is given by Ê3+ 5 ˆ Á 2 ,2 2+ 5 ˜ Ë ¯

(

y2 = 4x

1

2

3+ 5 2

y – 2x = – 2

)

18. Reqd. area = Ú-1 f ( x)dx = 2 area of trapezium OABC Ê 1 Ê 1ˆ 1ˆ 3 = 2 Á Á1 + ˜ ˜ = Ë 2 Ë 2¯ 4¯ 8

y = -x A

B

y = 1/4 (3/4, 2/4)

(-1, 0)

Fig. 14.49

y=x

0

y=1+x

Fig. 14.51

C (1, 0) y=1-x

Areas by Integration 14.19 c

19. Reqd. area = Úa f ¢( x)dx = f (c) – f (a) = d – b. 20. The intersection two curves is given by 2 x = x – 3

(

fi

x -3

)(

)

x +1 = 0 fi x = 3 fi x = 9

23. Intersection is given by y = – 2 and 1 1

Reqd. are = Ú-2 [(3 - y 2 ) - ( y + 1)]dy 9 = . 2 2

24. Reqd. area = Ú-2 ( x 2 + 3 - ( - x 2 - 1))dx =

y= x

x y=

80 3

–3

2

3

9

x = -2

(2, 0)

Fig. 14.52 Fig. 14.55

Reqd. area = 9

= Ú1

x dx -

1 ¥6¥3 2

25. Reqd. area =

=9

4 1 3 ¥ 4 ¥ 24 - Ú 10 x - 10 x dx 2

(

)

= 128

Level 2 16 . 3

1

21. Reqd. area = Ú-1 ( y + 3 - y 2 )dy =

y = x3 - 10x -4

x = y2

y = 6x

4

x=y+3

Fig. 14.56

26. The two curves y = x2 and y = 1 – x2 intersect at Ê 1 1ˆ Ê 1 1ˆ , ˜ and Á , ÁË ¯ Ë 2 2 ˜¯ 2 2

Fig. 14.53

22. Intersection is given by 3x3 – x2 – 10x = – x2 + 2x fi 3x3 – 12x = 0 fi x = – 2, 0, 2

y y = x2

1

0

Reqd. area = Ú-2 [3x3 - x 2 - 10 x] - ( - x 2 + 2 x)]dx

y = 1- x2 -1/ 2

O

2

+ Ú0 (( - x 2 + 2 x) - (3x3 - x 2 - 10))dx

-1/ 2

= 24

x

Fig. 14.57 y = 3x2 - x2 - 10 x

Thus, required area 1/ 2

(2, 0)

(-2, 0)

y = -x2 + 2x

2 2 = Ú-1/ 2 [(1 - x ) - x ]dx 1/ 2

= 2Ú0

(1 - 2 x 2 )dx =

8 3

27. Required area Fig. 14.54

p /4

p /2

= Ú0 [cos x - sin x]dx + Úp / 4 [sin x - cos x]dx

14.20

Complete Mathematics—JEE Main p /4

= (sin x + cos x )]0

p /2

+ ( - cos x - sin x )]p / 4

0

2. Required area: A = Ú1- e log( x + e)dx Put x + e = t, so that e

A = Ú1 log t dt = t(log t – 1)]1c = e(log e – 1) – 1(log 1 – 1) = 1.

y = cos x y = sin x

y y = log (x + e)

0

p/4

p/2

Fig. 14.58 -e

= 2( 2 - 1)

x

1-e 0

28. Two curves intersect at (0, – 1) and (4, 3) y

y2 = 2x + 1

Fig. 14.61 x=y +1 x

1

-1/2 -1

3. Required area 1

= Ú0 ( x - x)dx 1

Ê 2 3/ 2 1 2 ˆ ˘ = Á x - x ˜˙ Ë3 2 ¯ ˚0

Fig. 14.59

Required area

=

Ê y 2 - 1ˆ ˘ 3 È = Ú-1 Í y + 1 - Á ˜ ˙dy Ë 2 ¯ ˙˚ ÍÎ 1 3 = Ú-1[4 - ( y - 1)2 ]dy 2

2 1 1 - = . 3 2 6 y y = |x|

y=

y2 = x

|x|

(1.1)

3

=

1È 1 ˘ 4 y - ( y - 1)3 ˙ = 16/3. Í 2Î 3 ˚ -1

O

x

Previous Years’ AIEEE/JEE Questions 1. Required area 3

= Ú1 | x - 2 | dx 1 |t -1

= Ú

| dt

1 |t 0

=2 Ú

Fig. 14.62

[put x – 2 = t] 1 t dt 0

| dt = 2Ú

1 2˘

= t ˚ = 1. 0 y = |x - 2|

0

1

2

3

Fig. 14.60

4. The two curves intersect at x = – 2, y = ±1 È 1 1 ˘ 1 0 Reqd. area = 2 Í Ú-2 (1 - x) dx - Ú-2 - x dx ˙ 3 2 Î ˚ 1 1 2 È 2 ˘ = 2 Í(1 - x)3/ 2 u du ˙ Ú 0 -2 2 Î 3 3 ˚

4˘ 4 È = 2 Í2 - ˙ = . 3˚ 3 Î

Areas by Integration 14.21

= 2y2 = - x

p

0

5p / 4

2 Ú-p / 4 ( - sin q )dq + 2 Ú0 sin q dq + 2 Úp

(– sin q) dq 3y2 = 1 - x

(1, 0)

-2

=

2[cos q ]0-p / 4 + 2[ - cos q ]p0 + 2[ - cos q ]p5p / 4

=

1 ˘ È È 1 ˘ 2 Í1 + 1˙ ˙ + 2[1 + 1] + 2 Í 2˚ 2 ˚ Î Î

= 4 2-2 y y = sin x

1

Fig. 14.63

5. (y – 2)2 = x – 1

3p/2

p

fi y2 – 4y – x + 5 = 0

0

Equation of tangent at (2, 3) is 3y – 2(y + 3) –

1 (x + 2) + 5 = 0 2

-1

y = cos x

fi x – 2y + 4 = 0

Fig. 14.65 y

x2 ˆ 4Ê 7. Reqd. area = Ú Á 2 x - ˜ dx 0 4¯ Ë

(2, 3) (0, 2)

x¢

-4

O

(1, 2)

(1, 0)

4

(5, 0)

=

4 3/ 2 4 x3 x 0 3 12 0

=

4.8 64 64 16 = = 3 12 12 3

x

Fig. 14.64

y

Area of shaded region

y2 = 4y (4, 4)

3

2 = Ú [( y - 2) + 1 - (2 y - 4)]dy

y2 = 4x

0

x

3

2 = Ú [( y - 2) - 2 y - 5]dy 0 3

˘ Ê1 3 2 = Á ( y - 2) - y + 5 y ˙ Ë3 ˚0 =

Fig. 14.66

1 {1 – (– 2)3} – 9 + 15 = 9 3

6. Required area 3p / 2

A = Ú0

3p / 2

| sin x - cos x | dx = Ú0

Put x – p/4 = q so that 5p / 4

A = Ú-p / 4 2 | sin q | dq

| 2 sin( x - p / 4) | dx

8. The required region is shaded below. Area 1 È 2 ˘ y ˙ dy = 2 Í Ú0 3 y 2 Î ˚ 2

˘ 2 20 = 5 ÊÁ ˆ˜ y 3/ 2 ˙ = 2 Ë 3¯ ˚0 3

14.22

Complete Mathematics—JEE Main y y=

2

4x2

y=

y2 = 1 - x

x2/9

x2 + y2 = 1

x

0

Fig. 14.67

Fig. 14.69

9. Same as Q20 10. Reqd. area p /4

p /2

= Ú0 (cos x - sin x)dx + Úp / 4 (sin x - cos x)dx p /4

= sin x + cos x 0 =

p /2

+ ( - cos x - sin x ) p / 4

13. We first solve y2 = 4x, y – 2x = – 4 fi y2 = 2(y + 4) fi y2 – 2y – 8 = 0 fi (y – 4) (y + 2) = 0 Required area y2 ˆ 4 Ê y+4 - ˜ dy = 9(unit)2 = Ú-2 Á 4¯ Ë 2

2 - 1 + ( -1 + 2 )

y

= 2( 2 - 1) 4

3

11. Reqd. area = OABC – Ú1 log x dx = 3 log 3 – ( x log x -

y - 2x = - 4

3 x)1

O

= 3 log 3 – (3 log 3 – 3 + 1)

-2

y2 = 4x

=2 y = log 3 A

0

Fig. 14.70

B

D

Fig. 14.68

12. Reqd. area 1 ˘ Èp = 2 Í - Ú0 1 - x dx ˙ 4 Î ˚

C (3, 0)

14. Area of shaded region 1 Ê 1ˆ p /4 = Ú0 tan x dx - (1) Á ˜ 2 Ë 2¯ = log(sec x )]p / 4 - 1 0 4 1Ê 1ˆ = Á log 2 - ˜ 2Ë 2¯ y

y = tan x y = 1 + 2 (x - p/4)

1˘ Èp 2 = 2 Í + (1 - x )3/ 2 ˙ 0˚ Î4 3

Èp 2 ˘ p 4 =2 Í - ˙= - . Î 4 3˚ 2 3

x

2

1 O

p 1 p 4 2 4

x

Fig. 14.71

15. The curve y = – 2x2 and y = 1 – 3x2 are shown in the figures. Also, the required area is shaded.

Areas by Integration 14.23

For point of intersection of two curves, we set – 2x2 = 1 – 3x2 fi x = ±1

2

-2 2

x3/ 2 ˘ ˙ 3/ 2 ˚0

= – 2 sin–1(– 1) – 1 A

y

0

B

8 8 =p– 3 3

P C

B

y 2 = 2x

A 4

2

O

Fig. 14.72

As the two curves are symmetrical about the y-axis required area = area (ABOCA) = 2 area (AOCA)

Q

1

= 2 Ú0 [(1 - 3x 2 ) - ( -2 x 2 )]dx

Fig. 14.74

1

1 ˆ˘ Ê 1ˆ 4 Ê = 2 Á x - x 3 ˜ ˙ = 2 Á1 - ˜ = . Ë 3¯ 3 Ë 3 ¯ ˚0 16. For point of intersection of two curves, we set y y2 = 2x

B

1

y = 4x - 1 O -1/2

18. For point of intersection of y = x2 – 5x + 4 and x + y = 1 is given by 1 – x = x2 – 5x + 4 fi x2 – 4x + 3 = 0 fi x = 1, x = 3. Required area is shaded in the figure 3

A

= Fig. 14.73

1 (y + 1) 2 fi 2y2 – y – 1 = 0 fi y = 1, – 1/2

1 1 1 1 Required area = Ú-1/ 2 ( y + 1)dy - Ú-1/ 2 y 2 dy 4 2 1 1 1 1 = ( y + 1)2 ˘˚ - y3 ˘˚ 1 2 / -1/ 2 8 6 1 È 2 1 ˘ 1 È 1˘ 9 2 - 2 ˙ - Í1 + ˙ = 8 ÍÎ 2 ˚ 6 Î 8 ˚ 32 17. Required area = area (OAPBO)

3 ˘4

9 ˙ + ˙˚3 4

=

3 3 17 1 ÏÔÊ 3 ˆ Ê 1 ˆ ¸Ô - ÌÁ ˜ - Á ˜ ˝ Ë 2¯ Ô 4 3 ÓÔË 2 ¯ ˛

=

19 . 6 y

4 3

=

2 1

2

= Ú0 4 x - x 2 dx - Ú0 2 x dx 2

2 3 1 5ˆ 9˘ 4 ÈÊ (1 - x)2 ˘˚ - Ú3 Í Á x - ˜ - ˙ dx 1 2 2¯ 4 ˙˚ ÍÎ Ë

1 1Ê 5ˆ = ( 4 - 0) - Á x - ˜ Ë 2 3 2¯

y2 = 2x =

2

4

A = A1 + A2 = Ú1 -(1 - x)dx + Ú3 ( - x 2 + 5 x - 4)dx

x

O

2

1 2

3 4 A1 A2

= Ú0 4 - (2 - x) 2 dx - 2 Ú0 x1/ 2 dx 2

È (2 - x) 4 x - x 2 4 Ê 2 - xˆ˘ = -Í ˙ + sin -1 Á Ë 2 ˜¯ ˙ 2 2 ÍÎ ˚0

Fig. 14.75

x

14.24

Complete Mathematics—JEE Main

Previous Years’ B-Architecture Entrance Examination Questions

C

(2, 4) x2 = y

1

1. Reqd. area = 2 Ú0 3(1 - x 2 )dx

B

1

È x3 ˘ = 6 Íx - ˙ 3 ˚0 Î =6◊

E A(-1, 0) y=x+2

2 = 4. 3

D(2, 0)

Fig. 14.78 1

4. Reqd. area = 2 Ú0 ( x - x 2 )dx y = 3 (1 - x2)

(-1, 0)

(1, 0)

1

È x 2 x3 ˘ =2 Í - ˙ 3 ˚0 Î2 3 - 2˘ 1 = 2 ÈÍ ˙= Î 6 ˚ 3 (1, 1)

x

y=

-

x

y=

Fig. 14.76

2. Let A1 = ABDA

x2 = y

0

= Ú -p cos x dx – area of DBDO 2

= sin x

0

-

p 2

Fig. 14.79

1 1 ¥1¥1 = 2 2

5. Reqd. area 1

2

A2 = DBCD

= Ú0 ( x 2 + 1)dx + Ú1 ( x + 1)dx

= 1 + Ú p / 2 cos x dx = 3 2 0 2

1 1 + 1 + [4 – 1] + [2 – 1] 3 2 4 3 23 = + +1 = 3 2 6 =

A1 : A2 = 1 : 3 y=x+1 B y = cos x

A - p D(-1, 0) 0 2

Fig. 14.77

0

= area of DBEA + Ú-1 x 2 dx 0

= -1

Fig. 14.80

6. Reqd. area 1 2 = 2 È Ú0 (1 - x )dx + area of DOAB ˘ Î ˚

3. Required area

1 x3 = + 2 3

C p 2

1 1 5 + = 2 3 6

1 ˘ È 1 x3 Í =2 x+ ¥ 1 ¥ 1˙ 3 0 2 Í ˙ Î ˚ È 2 1 ˘ 14 7 = =2 Í + ˙= Î3 2˚ 6 3

Areas by Integration 14.25 y y = 1 - x2

x0

y

= -1

A(1, 0) B(0, -1) x+ y+ 1= 0

Fig. 14.81

2

4

x

-2

Fig. 14.83

Required area 4

4

4

2 3/ 2 4 1 ˘ x ˘˚ - ( x - 2) 2 ˙ 0 3 2 ˚2 2 3/2 1 = [4 – 0] – [(22) – 0] 3 2 16 10 (unit)2 = -2= 3 3 9 1 a 1 9. Ú1 2 dx = Úa 2 dx x x =

2ˆ

x 12 Ê = Ú3 Á 12 x - ˜ dx 12 Ë ¯ 12 2 1 12 x3/ 2 - [123 - 33 ] 3 3 36

= 84 –

0

= Ú0 x dx - Ú0 ( x - 2)dx

7. Reqd. area

=

y= x

0

567 441 147 = = . 12 12 4 x2 = 12y y2 = 12x

x = 12 x=3

a

9

-1 ˘ -1 ˘ = ˙ fi x ˙˚1 x ˚a 1 1 1 fi1– =+ a a 9 10 2 9 fi = fia= 9 a 5 y

Fig. 14.82

8. For point of intersection of curves, we solve x =y=x–2 fix– fi

A1

x –2=0

x =2fix=4

A2 1

9

Fig. 14.84

x

Differential Equations 15.1

CHAPTER FIFTEEN

An equation involving the derivatives of a dependent variable with respect to one or more independent variables is called a differential equation. The order of a differential equation is the order of the highest differential coefficient involved. When an equation is a polynomial in all the differential coefficients, the power to which the highest differential coefficient is raised is known as the degree of the equation. Illustration

1

Any differential equation of order 1 is of the form f(x, y, y ¢) = 0 e. g. y ¢ = y, xy¢ = 2y + x are 1 order differential equations. The order of the equation y ¢¢ + 5y = 0 is 2. The order of x2 y ¢¢¢ y¢ + exy¢¢ = (x2 + 3) y3 is order 3. The degree of the equations dy d 2 y Ê dy ˆ 2 + + y2 = 0, +y=0 dx dx 2 Ë dx ¯ 2

Ê d2 yˆ dy + y = x2 are and Á 2 ˜ + Ë dx ¯ dx 1, 1 and 2 respectively.

FORMATION OF A DIFFERENTIAL EQUATION To obtain the differential equation whose solution is the equation f (x, y, c1, c2, L, cn) = 0, where x and y are variables and c1, c2, L, cn are arbitrary constants we differentiate the above equation n times successively, so that n + 1 equations are obtained. From these n + 1 equations, eliminate the constants c1, c2, L cn. Illustration

2

Find the differential equation of family of circles of radius 5cm having centre on x-axis. The equation of such a circle is (1) (x – a)2 + y2 = 25

Differentiating this equation with respect to x, we get dy =0 2(x – a) + 2 y dx dy fi x – a = -y dx Substituting this value in (1), we get 2

dy ˆ y Á ˜ + y 2 = 25 Ë dx ¯ 2Ê

which is the required differential equation. Illustration

3

Find the differential equation corresponding to the equation y = aex + be2x + ce–3x (i) where a, b, c are arbitrary constants. (ii) y¢ = aex + 2be2x – 3ce–3x (iii) y≤ = aex + 4be2x + 9ce–3x (iv) y¢¢¢ = aex + 8be2x – 27ce–3x (v) y¢ – y = be2x – 4ce–3x (vi) y≤ – y¢ = 2be2x + 12ce–3x (vii) y¢¢¢ – y≤ = 4be2x – 36ce–3x Putting (v) in (vi) y≤ – y¢ = 2(y¢ – y) + 8ce–3x + 12ce–3x fi y≤ – 3y¢ + 2y = 20ce–3x Putting this value in (v), be2x = (y¢ – y) + 4ce–3x 1 = (y¢ – y) + (y≤ – 3y¢ + 2y) 5 1 2 3 = y ¢¢ + y ¢ - y 5 5 5 Substituting all these values in (vii), we have 36 4 y¢¢¢ – y≤ = ( y ¢¢ + 2 y ¢ - 3 y ) - ( y ¢¢ - 3 y ¢ + 2 y ) 5 20 4 9ˆ Ê Ê 8 27 ˆ Ê 12 18 ˆ y ¢¢¢ + Á -1 - + ˜ y ¢¢ + Á - - ˜ y ¢ + Á + ˜ y Ë Ë 5 5¯ Ë 5 5¯ 5 5¯ =0 y¢¢¢ – 7y¢ + 6y = 0

15.2

Complete Mathematics—JEE Main

METHODS OF SOLVING DIFFERENTIAL EQUATIONS SOLUTION OR INTEGRAL OF A DIFFERENTIAL EQUATION It is a relation between the variables, not involving the differential coefficients such that this relation and the derivatives obtained from it satisfy the given differential equation. The solution of a differential equation is also called its primitive. A solution which involves a number of essentially distinct arbitrary constants equal to the order of the equation is known as the general solution. A solution of a differential equation obtained from a general solution by giving particular values to one or more arbitary constants is called a particular solution. A solution which cannot be obtained from any general solution by any choice of the arbitrary constants is called a singular solution.

EQUATIONS OF FIRST ORDER AND FIRST DEGREE A differential equation is said to be linear if the dependent variable and its differential coefficients occur in it in the first degree only and are not multiplied together.

Equations in which the Variables are Separable These are the equations which can be so expressed that the coefficients of dx is only a function of x and that of dy is only a function of y. The general form of such an equation is f (x) dx = g( y)dy. Integrating both the sides, we get the solution. Illustration

4

Solve 9yy ¢ + 4x = 0. Separating variables, we have 9y dy = – 4x dx Integrating both sides, we obtain 9 fi

y2 = – 2x2 + Const 2 x 2 y2 = Const + 9 4

Linear Equations A linear equation of first order and first degree is either of the form dy + Py = Q (i) dx where P and Q are functions of x, or dx + P 1x = Q 1 (ii) dy

where P1 and Q1 are functions of y. In order to solve equation (i), we multiply both sides by the integrating factor is eÚ Pd x. After the multiplication, left hand side becomes the differential coefficient of yeÚPdx and now integrating both the sides, we have yeÚPdx =

Ú

QeÚPdx dx + C

as the solution. Similarly we solve equation (ii) Equations reducible to linear form. If the given equation is of the form dy/dx + Py = Qyn where P, Q are functions of x, we can reduce it to a linear equation by dividing both the sides by yn and then substituting y–n+1 = z. The given equation will be linear in z. Illustration

5

Solve y¢ – y = e2x This is a linear equation with P = – 1, Q = e2x. The I.F. = e–Ú1dx = e–x. Multiplying with I.F., we have d (ye–x) = e–x y¢ – ye–x = ex dx Integrating both sides ye–x = Úex dx + C = ex + C fi y = e2x + cex Illustration

6

y¢ – Ay = – By2, A, B are constants. This is a differential equation reducible to linear equation with P = – A, Q = – B, n = 2. Dividing by y2, we have 1 1 y¢ – A = – B 2 y y Put –

1 dy dZ 1 =Zfi 2 = dx y y dx

dZ + AZ = – B dx This is a linear equation with P = A, Q = – B I.F. = eAx, so d (ZeAx)= – BeAx dx B Ax fi ZeAx = – e +C A 1 B fi– =– + Ce–Ax y A

Homogeneous Equations It is a differential equation of the form

dy f ( x, y ) , where = dx f ( x, y )

f (x, y) and f (x, y) are homogeneous functions of x and y of

Differential Equations 15.3

the same degree. A function f (x, y) is said to be homogeneous of degree n if it can be written as xn f1( y/x). Such an equation can be solved by putting y = Vx. After substituting y = Vx, the given equation will have variables separable in V and x. The equations of the form dy/dx = (a x + b y + c)/ (A x + B y + C) can be reduced to a homogeneous equation by changing x = X + h and y = Y + k, where h and k are the constants to be chosen so that it makes the given equation homogeneous provided aB – bA π 0. If aB – bA = 0 then the given equation can be written as a b ÊÁ x + yˆ˜ + c ¯ Ëb A B ÊÁ x + yˆ˜ + C ¯ ËB

dy = dx

Now put (a/b) x + y = z so that the given equation reduces to an equation whose variables are seperable in z and x.

7

Illustration

x+y x-y y¢ = f(x, y), f(x, y) is a homogeneous function of dy dv =V + x degree 1. Putting y = Vx, we have dx dx The given equation reduces to dV x(1 + V ) 1 + V = = V+ x dx x(1 - V ) 1 - V Solve y¢ =

fi x fi

dV 1+V 1+ V - V + V2 1+ V2 = -V = = dx 1-V 1-V 1-V

1-V 1+V

2

dV =

dx x

1 2V ˆ dx Ê 1 fi Á dV = ˜ 2 2 2 1+V ¯ x Ë1 + V fi tan–1V – fi tan–1 fi tan

–1

1 log(1 + V2) = log x + C 2

y = log x (1 + V2)1/2 + C x y = log (x2 + y2)1/2 + C x

Procedure for finding the orthogonal trajectory (i) Let f (x, y, c) = 0 be the equation, where c is an arbitrary parameter. (ii) Differentiate the given equation w.r.t. x and then eliminate c. dy dx by – in the equation obtained (iii) Replace dx dy in (ii) (iv) Solve the differential equation in (iii).

DIFFERENTIAL EQUATIONS OF FIRST ORDER BUT NOT OF FIRST DEGREE The equations which are of first order but not of the first degree, the following types of equations are discussed. (i) Equations solvable for p = d y d x (ii) Equations solvable for y (iii) Equations solvable for x (iv) Clairut’s Equations (i) If there is a quadratic (or third degree) equation in p, solve the equation for p to get two (or three) first order and first degree equations in x, y and p. These equations can be solved as above. (ii) Suppose that the given differential equation on solving for y, gives y = f (x, p) (1) d pˆ Ê dy = Q Á x, p, ˜ , Ë dx¯ dx so that we obtain a new differential equation with variables x and p. Suppose that it is possible to solve this equation. Let the solution be f (x, p, c) = 0 (2) where, c is the arbitrary constant. We may either eliminate p between (1) and (2) or we may solve (1) and (2) for x, y. (iii) In this case differentiate w.r.t. y and proceed as in case (ii). (iv) The equation of the form y = p x + f ( p) is known as Clairut’s equation. Differentiating w.r.t. x, we obtain p =

Differentiating w.r.t. x, we get p = p + x [x + f ¢( p)]

ORTHOGONAL TRAJECTORY Any curve which cuts every member of a given family of curve at right angle is called an orthogonal trajectory of the family. For example, each straight line y = mx passing through the origin, is an orthogonal trajectory of the family of the circles x2 + y2 = a2.

dp =0 dx

fi

dp dp + f ¢(p) dx dx

dp = 0, x + f ¢( p) = 0 dx

dp = 0, we have p = constant = c (say). Eliminating p dx we have y = c x + f (c) as a solution. If x + f ¢( p) = 0, then by eliminating p, we will obtain another solution. This solution is called singular solution. Remark If u = rf (q), then du = rf¢ (q)dq + f (q) dr If

15.4

Complete Mathematics—JEE Main

SECOND ORDER BUT OF DEGREE ONE

Solve y ¢¢ + y = 0 Consider the equation m2 + 1 = 0 fi m = ± i are two complex roots with a = 0, b = 1. Hence required solution is y = C1 cos 0.x + C2 sin x = C1 + C2 sin x.

If there is a linear second order equations with constant coed2 y

+ a1 dy + a2y = 0 then we solve the quaddx dx ratic equation m2 + a1 m + a2 = 0. There will be three cases Case I Discriminant of the quadratic equation above is positive say roots are a1 and a2 then y = C1 ea1x + C2 ea2x is a solution. Case II If a1 = a2 = a(say) then a solution is y = (C1 x + C2)eax. Case III a1 = a + ib (complex roots) then a solution is y = C1 cos ax + C2 sin bx. fficients say

Illustration

2

8

Illustration

Equation of the form

d2 y d x2

= f(x) then inte-

dy = F(x) + C1, where F(x) dx f(x) dx and C1 is a parameter.

grating both the sides, we get

Ú

Integrating again, we have y = G(x) + C1 x + C2

Ú

where G(x) = Illustration

F(x) dx and C1, C2 are parameter.

11

9

Solve y ¢¢ + 2y ¢ + y = 0 Consider the equation m2 + 2m + 1 = 0 (m + 1)2 = 0, so m = –1 are coincident roots. Hence required solution y = (C1 + C2x)e–x.

d2 y = f (x) d x2

If the given differential equation is

=

Solve y ¢¢ + y ¢ – 2y = 0 Consider the equation m2 + m – 2 = 0 fi (m + 2) (m – 1) = 0 fi m = –2, 1 Hence the required solution y = C1e–2x + C2ex

10

Illustration

Integrating, we get dy = dx

Ú

d2 y = sin2 x d x2

sin2 x dx + C1 =

1 2

Ú

(1 – cos2 x) dx + C1

sin 2 x ˆ 1Ê ˜ + C1 ÁË x 2 2 ¯ Integrating again, we have =

y=

x2 1 + cos 2x + C1 x + C2. 4 8

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Example 1: The solution of the equation y¢ = given by (a) y = Cx + 1 (c) y = Cx2 + 1 Ans. (b)

y +1 is x

(b) y = Cx – 1 (d) y = Cx – 2

Solution: The equation is with separable variables, so can be written as

dy dx = x y +1 Integrating, we have log(y + 1) = log x + Conts fi y + 1 = Cx fi y = Cx – 1. Example 2: The solution of the equation xy ¢ - y y = tan is given by x x

Differential Equations 15.5

y = Cx x y (c) tan = Cx2 x Ans. (a)

y = Cx x y (d) tan = Cx2 + x x

(a) sin

Example 4: The solution of y ¢ + y = –

(b) cos

1 = – x + ex y 1 (b) y2 = – x2 + (1 + e–x) 2 (a)

Solution: The given equation can be written as y ¢ = y y tan + x x This is a homogeneous equation of degree 1 so put y = Vx. dV = tan V + V V+x dx dV dx = tan V fi cot V dV = x dx log sin V = log x + Const fi sin y/x = Cx. x

fi

Example 3: The solution of y¢

2 xy 1 - x2

=1+x

1+ x + sin–1 x 1- x

(b) y =

1 1- x È 2 Î2 + x 1 - x ˘˚ 2 1+ x

(c) y =

1 1+ x È 2 -1 Î2 + x 1 - x + sin x ˘˚ 2 1- x

(d) y =

(c) y2 = – x + (d)

1 (1 + e–2x) 2

1 1 3 = – x + e–x y 2 2

Ans. (c) Solution: The given equation is reducible to linear equation yy ¢ + y2 = – x y2 = Z fi 2y

Put

d (Ze2x) = – 2xe2x dx È xe2 x e2 x ˘ Ze2x = – 2 Ú xe2x dx + C = – 2 Í -Ú dx ˙ + C Î 2 ˚ 2

fi

Ê x 2 x3 x 4 ˆ x + + - ˜¯ 1 Á 2 3 4 1 - x2 Ë 1

= – xe2x +

Ans. (d) Solution: The equation is a linear equation with integrating factor e

-Ú

dx 1 - x2

= e

-Ú

2x dx 1 - x2

2

= elog(1 – x ) = (1 – x2)

y2 = – x +

fi For

d (y(1 – x2)) = (1 – x2) (1 + x) dx y(1 – x2) = Ú(1 – x2 + x – x3)dx + C x3 x 2 x 4 + +C 3 2 4

y= since

Ê x3 x2 x 4 ˆ C x + - ˜¯ + Á 2 Ë 3 2 4 1- x 1 - x2 1

y(0) = 1 so C = 1. Hence y=

Ê x2 x3 x 4 ˆ x + + - ˜¯ 1 Á 2 3 4 (1 - x 2 ) Ë 1

Thus

e2 x +C 2

1 + C e–2x 2

x = 0, y(0) = 1 so 1 =

Multiplying with I.F., we have

=x –

dy d Z = . Thus dx dx

1 dZ + Z =– x 2 dx dZ fi + 2Z = – 2x dx which is a linear equation with I.F. = e2x, so

y|x = 0 = 1 is given by (a) y =

x is given by y

y2 = – x +

1 1 +CfiC= . 2 2

1 (1 + e–2x) 2

Example 5: The solution of y¢ = (a) (b) (c) (d) Ans. (b)

2x - y + 1 is given by x - 2y + 1

x2 – xy + y2 + x + y = C x2 – xy + y2 + x – y = C x2 – xy + y2 – x + y = C – x2 + 2xy – y2 – x + 2y = C

Solution: The given equation is reducible to homogeneous equation. Put y = Y + k, x = X + h dy dy dY dX dY = ◊ = dx dY dX dx dX

Complete Mathematics—JEE Main

15.6

Select h, k such that 2h – k + 1 = 0, h – 2k + 1 = 0 so 1 1 h = – and k = . Thus 3 3 dY 2X - Y = which is a homogeneous equation dX X - 2Y Putting Y = vX, we have dv 2-v = v+X dX 1 - 2v fi fi

X

dv 2 - v - v + 2v2 2 (1 - v + v 2 ) = =– 1 - 2v dx 2v - 1

2v - 1 v2 - v + 1

dV = – 2

2

fi

Êy y ˆ 2 ÁË 2 - + 1˜¯ X = Const x x

fi

Y2 – YX + X2 = Const

fi

1ˆ 2 Ê 1ˆ Ê 1ˆ Ê 1ˆ 2 Ê y y x + + x + = Const Ë 3¯ Ë 3¯ Ë 3¯ Ë 3¯

fi

y2 – +

fi

1 2 1 Ê 1 y + – Ë xy - x + y 3 9 3 3

2 1ˆ + x2 + x ¯ 3 9

1 = Const 9 y2 + x2 – xy – y + x = C

Example 6: A particular solution of dy log = 4x + 5y, y(0) = 0 is given by dx (a) 5e4x + 4e–5y = 9 (b) 5e4x + 4e5y = 9 4x –4x (c) 4e + 5e = 20 (d) 4e4x + 5e–4y = 9 Ans. (a) dy Solution: = e4x + 5y = e4x. e5y dx fi e–5ydy = e4x dx Integrating, we have

e e + 4 5

C=

Putting

x = 0 and y(0) = 0, C =

Thus

9 = 5e4x + 4e–5y

Êx ˆ fi Ë + 1¯ a

13

Ê yˆ = cos t, Ë ¯ a

13

= sin t

Example 8: The solution of initial value problem xy¢ + y = 0, y(2) = – 2 is given by (a) xy + 4 = 0 (b) (y + 2) = 0 (c) x – 2= y + 2 (d) (x – 2) (y + 2)2 = 4 Ans. (a) dy dy dx Solution: x =–yfi =0 + dx y x fi log y + log x = Const fi xy = Const Since y(2) = – 2 so Const = 2(–2) = –4 Thus xy + 4 = 0 is the required solution Example 9: The solution of y¢ + xy = xy–1, y(0) = 2 is given by 2 2 (a) y2 = 2 + 2e–x (b) y2 = 3 + ex 2 –x (c) y2 = 3 + e–x (d) y2 = 2 + 2e Ans. (c) Solution: y¢ + xy = xy–1 fi yy¢ + xy2 = x This is reducible to linear equation, so put y2 = z fi y =

1 dz 2 dx

dy dx

1 dz + xz = x 2 dx dz + 2xz = 2x dx

2 2 xdx = ex . Multiplying with I.F., we have I.F. = e Ú

-5 y

fi

Solution: x = – 3aÚ cos2t sin t dt + C = a cos3 t + C For t = 0, C = – a y = aÚsin2t cos t dt = a sin3 t + G, at t = 0, y = 0 so C1 = 0 Thus x = a(cos3 t – 1), y = a sin3 t

fi

e-5 y e4 x = +C -5 4 4x

(x + a)2/3 + (y – a)2/3 = 1 (x – a)2/3 + (y + a)2/3 = 2a2/3 x2/3 + y2/3 = a2/3 (x + a)2/3 + y2/3 = a2/3

23 23 Êx ˆ Ê yˆ + Ë ¯ = 1. fi (x + a)2/3 + y2/3 = a2/3 fi Ë + 1¯ a a

dX X

log (v2 – v + 1) + logX2 = Const

fi

(a) (b) (c) (d) Ans. (d)

1 1 9 + = 4 5 20

dx Example 7: The motion of a particle is given by = dt dy = 3a sin2t cos t and passes through at t – 3a cos2 t sin t, dt = 0, then the path is given by

2 2 d (zex ) = 2x ex dx 2 2 2 fi z ex = Ú2x ex dx + C = ex + C 2 2 fi y2 ex = ex + C Since y(0) = 1 so C = 3 2 fi y2 = 3 + e–x

Example 10: The solution of y¢ + cos(x + y) = cos(x – y) Êp ˆ p is given by and y Ë ¯ = 4 4

Differential Equations 15.7

(a) sin y = 2 sin x cos2 y (b) cos y = 2 sin x sin2 y (c) cos y = 2cos x sin2 y (d) sin x = 2 cos y sin2 y Ans. (b) dy = cos(x – y) – cos(x + y) = 2sin x sin y dx

Solution:

fi

– cosec y cot y = – 2 cos x + C Êp ˆ p Since y Ë ¯ = so C = 0. Thus 4 4 cosec y cot y = 2 cos x cos y = cos x sin2y

fi

dy = 2 sin x dx sin y

fi

LEVEL 1 Straight Objective Type Questions Example 11: The degree of the differential equation 2

Ê d2 y ˆ d2 y 3 + 5 = x log is ÁË d x 2 ˜¯ d x2 d x2 d3 y

(a) 1 (c) 3 Ans. (d)

(b) 2 (d) none of these

Solution: Since the equation is not a polynomial in all the differential coefficients so the degree is not defined. Example 12: The degree of the differential equation y23/2 – y11/2 – 4 = 0 is (a) 6 (b) 3 (c) 2 (d) 4 Ans. (a) Solution: y23/2 = y11/2 + 4. Squaring both sides, we have y23 = y1 + 16 + 8y11/2 fi (y23 - y1 - 16)2 = 64y1 fi y26 - 32y23 - 2y23 y1 + y12 - 32y1 + 256 = 0. Hence the degree of the given equation is 6. Example 13: Which of the following equations is a linear equation of order 3 ? (a) (b)

d3 y d x3 d3 y d x3

(c) x

+ +

d3 y dx

d2 y

3

Example 14: The order and degree of the differential

d2 y d y +y=x d x2 d x d2 y d x2 +

2

+y =x

d2 y dx

2

2

equation x2 =

Ê Ê dy ˆ 2 ˆ ÁË 1 + ÁË dx ˜¯ ˜¯ d 2 y / dx 2

(a) 2, 1 (c) 2, 2 Ans. (c)

3/2

are (respectively) (b) 2, 3 (d) 2, 6

Solution: The given equation can be written as 2 Ê Ê dyˆ 2ˆ Ê d2 y ˆ x Á 2 ˜ = Á1 + Á ˜ ˜ Ë dx ¯ Ë Ë dx¯ ¯

3

4

which is clearly of order 2 and degree 2. Example 15: An equation of the curve in which subnormal varies as the square of the ordinate is (k is constant of proportionality) (a) y = Aekx (b) y = e kx 2 (c) y /2 + kx = A (d) y2 + kx2 = A Ans. (a) dy Solution: According to the given condition y = ky2 dx dy fi = kdx (variables separable equation) y fi log |y| = kx + C fi |y| = Bekx fi y = Aekx where A = + B and k is the constant of proportionality. Example 16: A solution of the differential equations

x

=e

dy = log x dx

2

dy Ê dy ˆ ÁË d x ˜¯ - x d x + y = 0 is

Ans. (c)

(a) y = 2 (c) y = 2x – 4 Ans. (c)

Solution: The equations in (a), (b) and (c) are of order 3 and in (d) is of order 2. The equations in (c) and (d) are linear, see theory for definition.

dy so that the given equation can be dx written as p2 – xp + y = 0. Therefore y = xp – p2 which is a

(d)

dx

2

+

Solution: Let p =

(b) y = 2x (d) y = 2x2 – 4

15.8

Complete Mathematics—JEE Main

Clairut’s equation see theory. Differentiating with respect to x, we have p=

dy dp =p+x – 2p dp dx dx dx

(x – 2p) dp = 0 fi dp = 0 so p = const = C (say) dx dx So y = Cx – C 2 is a solution, in particular C = 2. Example 17: The solution of (a) (b) (c) (d)

1 dy = is given by dx 2 x - y 2

fi fi

dx = k dy y x k A|x| = |y|

Solution: According to the given condition

= e–2y.

= \

fi

2

(b) u = ez (d) u = (log z)2

dz 1 1 1 + = 2 x z ( log z ) d x log z x 1

(i)

2

Ú

ye–2y dy + Const

y 2 –2y ye – 2 y 1 e + – 2 2 2

Ú

e–2y dy + Const

y 2 –2y y –2y e- 2 y e + e + + C1 2 4 2

2 x = C1 e2y + y + y + 1 2 2 4

Example 18: The curves whose subtangents are proportional to the abscissas of the point of tangency (the proportionality factor is equal to k) is (a) yk = Cx (c) yk/2 = Cx3 Ans. (a)

=C

(a) u = log z (c) u = (log z)–1 Ans. (c)

Integrating, we have

=

2

Ê d xˆ y + y Á ˜ = C2 Ë d y¯ 2

Solution: Dividing the given equation by z(log z)2, we have

d (xe–2y) = – y2 e–2y dy - y2 e- 2 y – -2

2

Example 20: Which of the following transformation dz z z reduce the differential equation + log z = (log z)2 2 dx x x into the form du + uP( x ) = Q( x ) dx

dx Solution: = 2x – y2 (This is a linear equation in x). dy

xe–2y =

yk = + Ax = Cx

The degree of this equation is 2.

Ans. (d)

So

fi

Ê d xˆ y 1+ Á ˜ Ë d y¯

1 1 1 y = Ce + x 2 + x + 4 2 4 1 1 1 x = Ce–y + y 2 + y + 4 4 2 1 1 x = Ce y + y 2 + y + 4 2 1 1 1 x = Ce2y + y 2 + y + 2 2 4

2 dy

log A + log |x| = k log |y|

Example 19: The degree of the differential equation of all curves having normal of constant length C is (a) 1 (b) 3 (c) 4 (d) none of these Ans. (d)

–2x

The integrating factor is e Ú

fi

(b) yk = Cx2 (d) none of these

dx Solution: Subtangent = y so according to the given d y dx = kx condition y dy

Putting

1 du 1 dz = u, we have = - (log z)- 2 log z dx z dx

So (i) can be written as -

du u 1 + = dx x x2

fi

du u -1 - = dx x x2

which is in the required form with P(x) = -1/x and Q(x) = - 1/x2. dy Example 21: A particular solution of log = 3x + dx 4y, y (0) = 0 is (a) e3x + 3e– 4y = 4 (b) 4e3x – e– 4y = 3 3x 4y (d) 4e3x + 3e–4y = 7 (c) 3e + 4e = 7 Ans. (d) Solution:

dy = e3x + 4y = e3x e4y dx

fi

e– 4y dy = e3x dx

e - 4 y e3 x = Const. But y(0) = 0, so, - 1 - 1 = C. -4 3 4 3 -4y 3x - 4y Hence (e /(- 4)) - (e /3) = - 7/12 fi 3e + 4e3x = 7. Thus

Differential Equations 15.9

dy ax + b Example 22: The solution of represents = dx cy + d a parabola if (a) a = 0, c = 0 (b) a = 1, b = 2 (c) a = 0, c π 0 (d) a = 1, c = 1 Ans. (c)

will pass through (3, 9) if 9 = 9/2 - 1/3 + C fi C = 29/6. Hence the required equation is 6xy = 3x3 + 29x -6.

Solution: The given equation has separable variables so c y2 (cy + d)dy = (ax + b) dx. Integrating we have + dy + 2 2 ax K= + bx, K being the constant of integration. The last 2 equation represents a parabola if c = 0, a π 0 or a = 0, c π 0.

(a) y = x cot (c – x) (c) y = x tan (c – x) Ans. (c)

Example 23: The differential equation corresponding to the family of curves y = ex (a cos x + b sin x), a and b being arbitrary constants is (a) 2y2 + y1 – 2y = 0 (b) y2 – 2y1 + 2y = 0 (c) 2y2 – y1 + 2y = 0 (d) none of these Ans. (b) Solution: y1 = ex (- a sin x + b cos x) + ex (a cos x + b sin x) fi y1 = ex (- a sin x + b cos x) + y fi y2 = y1 + ex (- a cos x - b sin x) + ex (- a sin x + b cos x) fi y2 - 2y1 + 2y = 0. dy Example 24: The solution of y5 x + y – x = 0 is dx (a) x4/4 + (1/5) (x/y)5 = C (b) x5/5 + (1/4) (x/y)4 = C (c) (x/y)5 + x4/4 = C (d) (x y )4 + x5/5 = C Ans. (b) Solution: The given differential equation can be written as y5 xdx + ydx - xdy = 0. Multiplying by x3/y5, we have x4 dx +

x3 Ê y d x - x d y ˆ ˜¯ = 0 y2 y3 ÁË 4

x 3 Ê ydx - xdy ˆ 3 ˜¯ = u du y3 ÁË y2

where u =

dx is

x . y

Example 25: The equation of the curve passing through (3, 9) which satisfies dy/dx = x + 1/x2 is (a) 6xy = 3x2 – 6x + 29 (b) 6xy = 3x2 – 29x + 6 (c) 6xy = 3x3 + 29x – 6 (d) none of these Ans. (c) Solution: The given differential equation has variable separable so integrating, we have y = x2/2 - 1/x + C. This

ˆ Ê y =Á 2 - 1˜ 2 ¯ Ë x +y x +y x dy

2

2

(b) cos–1 (y/x) = –x + c (d) y2/x2 = x tan (c – x)

Solution: The given equation can be written as x dy- ydx 2

x +y fi

fi

2

= - dx fi

x dy- ydx x

2

¥

1 1 + y2 / x 2

= - dx

d Ê yˆ Á ˜ = – dx. Integrating we have 1 + y /x dx Ë x¯ tan-1 (y/x) = - x + c 1

2

2

y = x tan (c - x).

Example 27: The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy = 0 is (a) log |2x + y –1| = C + x + y (b) log (4x + 2y – 1) = C + 2x + y (c) log (2x + y + 1) + x + 2y = C (d) log |2x + y – 1| + x + 2y = C Ans. (d) dy dX Solution: Put 2x + y = X fi 2 + = . Theredx dx fore, the given equation is reduced to (see theory the case when aB – bA = 0) dX dX 3 ( X - 1) X +1 -2= fi = dx d x 2X -1 2X -1 fi fi

Integrating, we get x5/5 + (1/ 4) ( x / y ) = C, since

Example 26: The solution of

fi fi

2X -1 dX = dx 3 ( X - 1)

fi

1 3

1 ˘ È Í2 + X - 1 ˙ dX = dx Î ˚

1 [2X + log |X - 1| ] = x + Const 3 2(2x + y) + log |2x + y - 1| = 3x + Const x + 2y + log | 2x + y - 1| = C.

Example 28: Solution of the differential equation dy = 2 sin x cos x – y2 cos x satisfying y(p/2) = 1 2y sin x dx is given by (b) y = sin2 x (a) y2 = sin x 2 (d) y2 sin x = 4 cos2 x (c) y = cos x + 1 Ans. (a) Solution: The given equation can be written as dy 2y sin x + y2 cos x = sin 2x dx

15.10

Complete Mathematics—JEE Main

fi

d (y2 sin x) = sin 2x dx

fi so fi Hence fi

y2 sin x = (–1/2) cos 2x + C. (y(p/2))2 sin (p / 2) = (–1/2) cos (2p/2) + C C = 1/2. y2 sin x = (1/2) (1 – cos 2x) = sin2 x y2 = sin x.

Example 29: Solution of the differential equation x 2 + y 2 dx = 0 is

xdy – y dx – (a) y –

x 2 + y 2 = Cx2 (b) y +

x 2 + y 2 = Cx2

(c) x +

x 2 + y 2 = Cy2 (d) x –

x 2 + y 2 = Cy2

Ans. (b) dy Solution: Writing the given equation as = dx dv y + x 2 + y2 and putting y = vx, we have v + x dx x d v d x = v + 1 + v2 fi = x 1 + v2 Integrating, we have log (v + fi

1 + y 2 / x 2 = Cx

y/x +

1 + v2 ) = log x + Const. fi

x 2 + y 2 = Cx2.

y+

Solution: Putting y = vx in the given equation, we have dv dv = v(log v + 1) fi x = v log v v+x dx dx fi

dv dx = v log v x

log |log v| = log |x| + Const

fi

log v = + Ax = cx

y = xecx. dy Example 32: The solution of x3 + 4x2 tan y = x d x e sec y satisfying y (1) = 0 is

fi

(a) (b) (c) (d) Ans. (b)

fi

tan y = (x – 2) ex log x sin y = ex (x – 1) x–4 tan y = (x – 1) ex x – 3 sin y = ex (x – 1) x–3

Solution: Rewriting the given equation in the form dy d 4 + 4x3 sin y = xex fi (x sin y) = xex x cos y dx dx 4

fi

x4 sin y =

Ú

xex dx

= (x - 1)ex + C Since y (1) = 0, so C = 0. Thus sin y = x- 4 (x - 1)ex. Example 33: If for the differential equation y¢ = y x Ê xˆ then f (x/y) + f Á ˜ the general solution is y = Ë y¯ log Cx x is given by

Example 30: An equation of the curve satisfying x 2 - y 2 dx and y(1) = 0 is

xdy – y dx =

(a) y = x2 log |sin x| (b) y = x sin (log |x|) (d) y = 2x2 (x – 1) (c) y2 = x (x – 1)2 Ans. (b) Solution: The equation can be written as x2

x dy - ydx x

fi

d (y / x) 2

1 - (y / x)

2

=

dx x

2

= x 1 - (y / x) fi

dx

sin-1 y/x = log |x| + Const

(a) – x2/y2 (c) x2/y2 Ans. (d)

x dv dy +v= , we have dx dx dv dx = f (1/ v ) x

Solution: Putting v = y/x so that

x dv + v = v + f (1/v) fi dx dv fi log |Cx| = Ú (C being constant of integration.) f (1/ v ) x But y = is the general solution so log (Cx )

Since y(1) = 0 so Const = 0. Hence y = x sin (log |x| ). Note that one can also solve the given equation as a homogenous equation also, i.e. by putting y = vx Example 31: A solution of the equation dy x = y (log y – log x + 1) is dx (a) y = xecx (b) y2/x = cx 2 (d) log y = cx (c) y = cx log x Ans. (a)

(b) y2/x2 (d) – y2/x2

x 1 dv = =Ú y v f (1/ v ) fi

fi f (1/v) = – v2

f (x/y) = – y2/x2.

Example 34: The solution y(x) of the differential equad2 y tion = sin 3x + ex + x2 when y1(0) = 1 and y(0) = 0 is d x2 sin 3 x x4 1 + ex + + x -1 (a) 9 12 3

Differential Equations 15.11

(b) -

sin 3 x x4 1 + ex + + x 9 12 3

t 2 ( f ( x ) - f (t )) + lim (t + x ) f (t ) tÆ x tÆ x t-x 2 = – x f ¢(x) + 2x f(x) Thus f(x) is a solution of the differential equation 1 dy dy 2 – 2xy = –1 fi – y = – 2 (1) x2 x dx dx x = lim

cos 3 x x4 1 + ex + + x +1 3 12 3 (d) none of these Ans. (a) (c) -

Solution: Integrating the given differential equation, we have dy x3 cos 3 x =+ ex + + C1 dx 3 3 1 1 but y1(0) = 1 so 1 = - + 1 + C1 fi C1 = 3 3 Again integrating, we get 4

but

sin 3 x 1 x y= + ex + + x + C2 9 12 3 y(0) = 0 so 0 = 1 + C2 fi C2 = – 1. Thus y= -

sin 3 x x4 1 + ex + + x - 1. 9 12 3

Example 35: The differential equation

dy = dx

1 – y2 y

determines a family of circles with (a) variable radii and a fixed centre (0, 1) (b) variable radii and a fixed centre (0, –1) (c) fixed radius 1 and a variable centres along the x-axis (d) fixed radius 1 and variable centres along the y-axis Ans. (c) y

Solution: dx = fi

c + x= –

1 – y2

fi (x + c)2 + y2 = 1 which represents a family of circles of fixed radius 1 and variable centre on the x-axis. Example 36: Let f(x) be a differentiable on the interval t 2 f ( x ) - x 2 f (t ) (0, •) such that f(1) = 1, and lim = 1 for tÆ x t-x each x > 0. Then f(x) is

2

2

Since f(1) = 1

so

=

1 x2

.

2

C=

fi 2 3

\

y

=

1

+C 3x3 2 1 y = x2 + . 3 3x

x

2

Example 37: Suppose y = y(x) satisfies the differential equation ydx + y2dy = xdy. If y(x) > 0 " x Œ R and y(1) = 1 then y(–3) equals (a) 1 (b) 2 (c) 3 (d) 5 Ans. (c) Solution:

ydx – xdy y2

Ê xˆ d Á ˜ = – dy Ë y¯

fi since

= – dy fi x/y = – y + c

y(1) = 1 so c = 2 Thus x/y = – y + 2 3 x = – 3, - = – y + 2 y

when

Example 38: The equation of the curve whose tangent at any point (x, y) makes an angle tan–1 (2x + 3y) with x-axis and which passes through (1, 2) is: (a) 6x + 9y + 2 = 26e3(x – 1) (b) 6x – 9y + 2 = 26e3(x – 1) (c) 6x + 9y – 2 = 26e3(x – 1) (d) 6x – 9y – 2 = 26e3(x – 1) Ans. (a)

fi

2

t f ( x ) - t f (t ) + t f (t ) - x f (t ) tÆ x t-x

Solution: 1 = lim

( 1 x ) we get

1 d Ê yˆ ÁË 2 ˜¯ = – 4 dx x x

Solution:

2 (b) – 1 + 4 x 3x 3 1 (d) x

2

Multiplying (1) with

2 – Ú dx x

fi – 3 = – y2 + 2y fi (y – 1)2 = 4 fi y = 1 ± 2 = 3 or – 1 As y(x) > 0 for all x Œ R so y(–3) = 3.

dy

1 2y dy = – 1 – y 2 2 Ú 1 – y2

2 (a) 1 + 2 x 3x 3 1 2 (c) – + x x2 Ans. (a)

This is a linear equation with I.F. = e

dy = tan[tan–1 (2x + 3y)] = 2x + 3y dx

dy – 3y = 2x dx I.F. = e–3x e–3x

Multilplying (1) by e–3x, we get

dy – 3e–3xy = 2xe–3x fi dx

d [ye–3x] = 2xe–3x dx

15.12

fi

Complete Mathematics—JEE Main

ye–3x =

Ú 2 xe

–3 x

2 –3 x 2 xe + Ú (1)e –3 x dx 3 3

dx = –

2 2 = – xe –3 x – e –3 x + C 3 9 As this curve passes through (1, 2), we get 2 26 –3 e 2 = – (3 + 1) + Ce3 fi C = 9 9 Thus, required curve is 2 26 3( x – 1) e y = – (3 x + 1) + 9 9

Solution: Equation of any parabola whose axis is x-axis is y2 = 4a(x + b) 2y

dy = 4a dx

fi

y

d2 y

2

Ê dy ˆ +Á ˜ =0 2 Ë dx ¯ dx

which is a differential equation of order 2 and degree 1. Example 40: The solution of the differential equation dy x+y = satisfying the condition y(1) = 1 is dx x (a) y = logx + x (b) y = xlogx + x2 (x – 1) (c) y = xe (d) y = xlogx + x Ans. (d) Solution: xdy = xdx + ydx xdy – ydx

dx x x fi d(y/x) = d(log x) fi Putting x = 1, we get C = 1, so y = x(log x + 1) 2

=

y = x(log x + C)

Example 41: The differential equation which represents the family of curves y = c1ec2 x , where c1 and c2 are arbitrary constants, is (a) yy≤ = y¢ (b) yy≤ = y¢2 2 (c) y≤ = y (d) y≤ = y¢y Ans. (b) Solution: y = c1ec2 x fi fi

fi

Solution: The given equation can be written as dy y(sin x - y) = dx cos x Put

–

fi

1 dy 1 – tan x = – secx y 2 dx y

1 = z, we have y

dz (tan x)z = – secx dx tan x I.F. = e Ú dx = elogsecx = secx

fi 6x + 9y + 2 = 26e3(x – 1)

Example 39: The degree and order respectively of the differential equation of all parabolas whose axis is x-axis, are: (a) 2, 1 (b) 1, 2 (c) 2, 2 (d) 1, 1 Ans. (b)

fi

(a) y tanx = secx + c (b) tanx = (secx + c)y (c) secx = (tanx + c)y (d) y secx = tanx + c Ans. (c)

d ( z sec x ) = – sec2x dx fi zsecx = – tanx + c1 fi secx = (c + tanx)y, c = – c1. dy Example 43: If = y + 3 > 0 and y(0) and y(0) = 2, dx y(log 2) is equal to (1) reduces to

(a) – 2

(b) 7

(c) 5 Ans. (b)

(d) 13

dy = dx fi log(y + 3) = x + c when x = 0, y+3 y = 2, therefore log 5= C Thus log (y + 3) = x + log 5 When x = log 2, we have log (y + 3) = log 2 + log 5 = log 10 fi y + 3 = 10 fi y = 7. Solution:

Example 44: Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years. The value V(t) depreciates at a rate given by differential dV (t ) = – k (T – t), where k > 0 is a constant and equation dt T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is: (a) e–kT (c) I -

(b) T 2 kT 2 2

I k

(d) I – K(T)

Ans. (c)

y¢ = c1c2 ec2 x

y≤ = c1 c22 ec2 x

dv = – K(T – t), K > 0 dt dV = – K(T – t) dt

Solution:

(

yy≤ = c12 c22 ec2 x ec2 x = c1c2 ec2 x

2

)

= y¢2

Example 42: Solution of the differential equation cosx dy = y(sinx – y)dx, 0 < x < p/2 is

(1)

fi

2 Integrating, we have V(t) = k (T - t ) + C 2

Differential Equations 15.13

We have V(0) = I, therefore I=

KT 2 +C 2

Scrap value V(T) = C = I -

fi

C= I-

k 2 T 2

pˆ Ê (a) y tan Ëx + ¯ = x + C 4

K 2 T . 2

Example 45: The population p(t) at time t of a certain dp(t ) mouse species follows the differential equation = 0.5 dt p(t) – 450. If p(0) = 850, then the time at which the population becomes zero is 1 (a) log 9 (b) log 18 2 (c) log 18 (d) 2 log 18 Ans. (d) Solution: The given differential equation in a linear equation with I.F. = e- Ú(0.5) dt = e –t/2. Multiplying with I.F. we have d -t / 2 –t/2 fi e–t/2 p(t) = 900 e–t/2 + C e p = – 450 e dt When t = 0, p = 850, 850 = 900 + C fi C = – 50 Thus p(t) = 900 – 50 e t/2 When p(t) = 0, we get 900 – 50e t/2 = 0 fi e t/2 = 18 fi t 2 = log 18 i.e. t = 2 log 18

(

Example 47: The general solution of the differential Ê dy 1 + tan x ˆ – y = 0 is given by equation cos 2x Á Ë dx 1 - tan x ˜¯

pˆ Ê (b) y cot Ëx + ¯ = x + C 4 pˆ Ê (c) y tan Ëx + ¯ = tan x + C 4 pˆ Ê (d) y cot Ëx + ¯ = cot x + C 4 Ans. (b) Solution: The given equation can be written as pˆ dy Ê – sec 2x y = tan Ëx + ¯ 4 dx –Úsec2x dx This is a linear equation whose I.F. = e

)

Example 46: If the differential equation representing the family of all circles touching y-axis at the origin is x2 – y2 dx = f(x)y then f(x) is equal to dy (a) 2x (c) x2 Ans. (a)

(b) x (d) 3x

Solution: Equation of circles touching y-axis at origin is (i) x2 + (y – a)2 = a2 Differentiating, we get dy 2x + 2(y – a) =0 dx x dx fi y – a =– fia=y+x dy dy dx Substituting this value in (i), we have

=e

p ˆˆ Ê Ê = Ë tan Ëx + ¯¯ 4

fi

x2 – y2 = 2xy

dx . So f(x) = 2x dy

-1 2

Multiplying with I.F., we have d Ê 1 y Á dx p tan x + ÁË 4

( )

ˆ =1 ˜ ˜¯

pˆ Ê fi y cot Ëx + ¯ = x + C 4 Example 48: The curve y = f(x) (f(x) ≥ 0, f(0) = 0) bounding a curvilinear trapezoid with the base [0, x], whose area is proportional to the (n + 1)th power of f(x) f(1) = 1 is given by (a) x = yn (b) y = xn n (c) y + 1 = (x + 1) (d) x = yn + 1 Ans. (a) Solution: According to the given condition x

Ú0

f ( x ) dx = k(f(x))n + 1, K being

the constant of proportionality. Differentiating we have f(x) = k(n + 1) (f(x))n f ¢(x) dy dx dx = k(n + 1)yn – 1 dy y = k(n + 1) yn

2 dx ˆ 2 Ê Ê dx ˆ x2 + x2 Á ˜ = Á y + x ˜ Ë Ë dy ¯ dy ¯ 2 Ê Ê dx ˆ 2 ˆ dx Ê dx ˆ fi x 2 Á 1 + Á ˜ ˜ = y2 + x2 Á ˜ + 2xy Ë Ë dy ¯ ¯ Ë dy ¯ dy

( )

1 p - log tan x + 2 4

fi

x+C=

k (n + 1) y n n

since y(0) = 0 so C = 0. Also y(1) = 1, so k = x = y n.

n . Thus n +1

15.14

Complete Mathematics—JEE Main

Assertion-Reason Type Questions

Example 49: Consider the differential equation dy y = . 2 y log y + y - x dx Statement-1: xy = y2 log y + C is a solution of the given differential equation Statement-2: The differential equation is a linear equation in y and x. Ans. (a) dx 2 y log y + y - x x Solution: = = 2 log y + 1 dy y y dx x fi + = 2 log y + 1 which is linear in x and y. dy y 1

I.F. = e So fi

Ú y dy

= y.

d ( xy ) = 2 y log y + y dy y2 +C xy = 2 Ú y log y dy + 2 È y2 y2 ˘ y2 +C = 2 Í log y - ˙ + 4˚ 2 Î2 = y2 log y + C.

Solution: The equation of a circle contains three independent constants if it passes through three non-collinear points. Example 51: Statement-1: Curve satisfying the differential equation y¢ = y/2x passing through (2, 1) is a parabola with focus (1/4, 0) Statement-2: The differential equation y¢ = y/2x is of variable separable. Ans. (d)

fi

fi

log y 2 = log x + const 2

fi

2dy dx = y x y2 = Cx, this passes through

(2, 1) if C = 1/2. Thus y = 1/2 x which represents a parabola with focus (1/8, 0).

Ans. (b) x-y x+y y x - sin = - 2 sin cos . 2 2 2 2 fi cosec y/2 dy = – 2 cos (x/2) dx fi 2 log | tan (y/4) | = – 4 sin (x/2) + Const fi log | tan (y/4) | = – 2 sin (x /2) + Const fi | tan (y/4) | = Const e – 2 sin (x/2) Since y(0) = p so Const = 1. Thus | tan (y/4)| = e – 2 sin x/2 so y = 4 tan –1 (± e – 2 sin x/2) which is periodic with period 4p.

Solution: y¢ = sin

Example 53: Let (xy2 + x)dx + (y – x2y)dy = 0 satisfy y(0) = 0. Statement-1: The curve represented by the solution of the given differential equation is a circle. Statement-2: It is circle with radius 1 and centre (0, 0). Ans. (c) Solution: x(1 + y2)dx + y(1 – x2)dx = 0

Example 50: Statement-1: The differential equation of all circles in a plane must be of order 3. Statement-2: There is only one circle passing through three non-collinear points. Ans. (a)

dy y Solution: = dx 2 x

x+y x–y = sin 2 2 Statement-1: A solution satisfying y(0) = p is a periodic function with period 4p. Statement-1: y can be explicitly represented in terms of x. Example 52: Let y¢ + sin

x

fi

1– x

2

dx +

y 1 + y2

dy = 0

– log (1 – x 2) + log (1 + y2) = Const 1 + y2 = C(1 – x 2) Since y(0) = 0 so C = 1 x 2 + y 2 = 0 which is a point circle.

fi fi fi

Example 54: Let a solution y = y(x) of the differential equation x x 2 – 1 dy – y y 2 – 1 dx = 0 satisfy y(2) = 2 3 Statement 1: y(x) = sec(sec–1 x – p/6) Statement 2: y(x) is given by

1 2 3 1 = – 1– 2 y x x

Ans. (c) Solution: The given equation can be written as dy y y2 – 1

–

dx x x2 – 1

Integrating sec–1y – sec–1x = Const

=0

Differential Equations 15.15

Putting x = 2, Const = sec–1 2 / 3 – sec–12 = so

p p p – =– 6 3 6

tion. Put

y = sec(sec–1x – p/6).

Example 55: Statement 1: The differential equation of all circles passing through one fixed point is of order 2. Statement 2: General equation of a circle involves three arbitrary constants. Ans. (a) Solution: General equation of a circle is x2 + y2 + 2gx + 2fy + C = 0 where g, f, c are arbitrary constant. Since the circle passes through a fixed point so the arbitrary constants are two. Hence required differential equation is of order 2. Example 56: Statement 1: The solution of xy¢ + y = y2 log x is given by y(1 + log x + Cx) = 1. Statement 2: The given differential equation is a linear equation whose I.F. is x. Ans. (c) Solution: xy¢ + y = y2 log x fi

1 y

2

y¢ +

1 1 log x = x y x

This is reducible to linear equation but not a linear equa-

–

1 1 dy dZ =Zfi– 2 , so = y y dx dx

dZ 1 log x dZ 1 log x fi + Z= - Z=– dx x x dx x x 1 - Ú dx e x

The I.F. is

=

1 . Multiplying with I.F. x

d Ê 1ˆ log x Z ◊ ¯ =– Ë dx x x2 Integrating, we have Z

1 =– x

1 1 ◊ =– y x

fi

Ú

log x x2

dx + C

1 ˘ È 1 +C ÍÎ- x log x + Ú x 2 dx ˙˚

= – È- 1 log x - 1 ˘ + C ÍÎ x x ˙˚ 1 = [1 + log x] + Cx y

fi fi

1 = y(1 + log x + Cx)

LEVEL 2 Straight Objective Type Questions

Example 57: The degree of the differential equation satisfying

(

)

1 + x 2 + 1 + y 2 = A x 1 + y 2 - y 1 + x 2 is (a) 2 (c) 4 Ans. (d)

(b) 3 (d) none of these

Solution: Put x = tan q and y = tan f. Then = sec q,

1+ y

2

1 + x2

sec q + sec f = A (tan q sec f - tan f sec q) cos f + cos q Ê sin q - sin f ˆ =A Á Ë cos q cos f ˜¯ cos q cos f

fi

cos f + cos q = A (sin q - sin f)

fi

2 cos

fi

q +f q -f q -f q +f cos = 2A sin cos 2 2 2 2

q - f = 2 cot - 1 A

cot

fi

tan-1 x - tan-1 y = 2 cot-1 A.

Differentiating this, we get

= secf, and the equation becomes

fi

q -f =A 2

fi

Ê 1 ˆ dy = 0, -Á Ë 1 + y 2 ˜¯ d x 1+ x 1

2

which is a differential equation of degree 1. Example 58: The solution of differential equation f ( y / x) dy y = +2 is f ¢( y / x ) dx x (b) y2 f ( y/x) = k (a) x2 f ( y/x) = k 2 (d) f ( y/x) = k y2 (c) f ( y/x) = k x Ans. (c) dy y du Solution: Putting = u we have =u+x . The dx x dx given differential equation can be written as u + x

du =u dx

15.16

Complete Mathematics—JEE Main

f (u ) f ¢ (u )

+2

fi

x

du f (u ) = 2 dx f ¢ (u )

fi

f ¢ (u ) dx du = 2 f (u ) x

Integrating, we get log f (u) = log x2 + log k, so f (u) = kx2 i.e. f (y/x) = kx2, k being an arbitrary constant. Example 59: The orthogonal trajectories of the family of curves an – 1 y = xn are given by (a) xn + n2 y = const (b) ny2 + x2 = const (c) n2 x + yn = const (d) n2 x – yn = const Ans. (b) Solution: Differentiating, we have (see theory) dy = nxn-1 fi dx

dx a = nx a dy Putting this value in the given equation, we have dx nxn-1 y = xn dy dx dy dx Replacing by , we have ny = – x dy dx dy n-1

n-1

dy Example 61: The solution of the equation = dx cos (x – y) is x - yˆ x - yˆ = c (b) x + cot ÊÁ =c (a) y + cot ÊÁ Ë 2 ˜¯ Ë 2 ˜¯ x - yˆ = c (d) none of these (c) x + tan ÊÁ Ë 2 ˜¯ Ans. (b) Solution: Putting u = x - y, we have given equation can be written as 1 fi

n -1

du

Ú 1 - cos u 1 2

fi

Alternate Solution We have 2

dx dx d 2 x = -2 x 2 dt dt dt

2

fi

Ê dx ˆ 2 ÁË ˜¯ = – x + C dt

Since x¢ (0) = 0 and x (0) = 1 so C = 1. Hence fi fi fi When Hence

dx

dx = ± 1 - x2 dt

=±t 1 - x2 sin–1 x = t + C1 or cos–1 x = t + C2 x = sin (t + C1) or x = cos (t + C2) t = 0, x = 1 so C1 = p/2 and C2 = 0 x = cos t.

= dx

Ú

Example 62: A solution of the differential equation dy 1 = is 2 dx xy[ x sin y 2 + 1] ( C is an arbitrary constant) 2

(a) x2 (cos y2 – sin y2 – 2C e–y ) = 2

2

Solution: Consider the equation m2 + 1 = 0, m = ± i. Hence x = C1 cos t + C2 sin t (see theory) Now x (0) = 1 fi C1 = 1 and x¢ = – C1 sin t + C2 cos t so C2 = 0. Hence x = cos t is the required solution which is a periodic function.

du = cos u dx

cosec2 u du = Ú dx + Const 2 u x-y x + cot = Const, i.e. x + cot = c. 2 2

fi

fi ny dy + x dx = 0 fi ny2 + x2 = const. which is the required family of orthogonal trajectories. Example 60: The solution of the differential equation d x + x = 0; x(0) = 1, x¢(0) = 0 dt 2 (a) approaches infinity as t Æ • (b) is a periodic function (c) is always greater than or equal to unity (d) does not exist Ans. (b)

du dy =1. The dx dx

2

(b) y2 (cos x2 – (sin y2 – 2C e–y ) = 2 2

(c) x2 (cos y2 – sin y2 – e- y ) = 4 (d) none of these Ans. (a) Solution: The given differential equation can be written dx as = xy [x2 sin y2 + 1] dy 1 dx 1 y = y sin y2. This equation is reducible to x3 d y x 2 linear equation, so putting - 1/x2 = u, the last equation can be written as du + 2uy = 2y sin y2 dy 2 The integrating factor of this equation is e y . So required solution is 2 2 ue y = Ú 2y sin y 2.e y dy + C fi

=

Ú

(sin t).et dt + C

(t = y2)

2

= (1/2) e y (sin y2 - cos y2) + C fi fi

2u = (sin y2 - cos y2) + Ce-y

2 2

2 = x2 [cos y2 - sin y2 - 2 Ce-y ].

Differential Equations 15.17

Example 63: The solution of (y (1 + x–1) + sin y) dx + (x + log x + x cos y) dy = 0 is (a) (1 + y–1 sin y) + x–1 log x = C (b) ( y + sin y) + xy log x = C (c) xy + y log x + x sin y = C (d) none of these Ans. (c)

Example 66: The solution of y2 – 7 y1 + 12y = 0 is (a) y = C1 e3x + C2e4x (b) y = C1 x e3x + C2 e4x (c) y = C1e3x + C2 x e4x (d) none of these. Ans. (a) Solution: The given equation can be written as Ê d ˆ Êdy ˆ ÁË d x - 3˜¯ ÁË d x - 4 y˜¯ = 0

(i)

Solution: The given equation can be written as y(1 + x–1)dx + (x + log x)dy + sin y dx + x cos ydy = 0 fi d(y(x + log x)) + d(x sin y) = 0 fi y(x + log x) + x sin y = C

fi

Example 64: If f (x) is a differentiable function then the solution of dy + ( y f¢(x) – f (x) f¢(x)) dx = 0 is

- 4y = C1e3x which is a linear equation whose I.F. is

(a) y = (f (x) – 1) + C e–f (x)

If

(b) yf (x) = (f (x)) + C (c) yef (x) = f (x) ef (x) + C

d (yef(x)) = f(x) f¢(x) ef(x) dx

= tet - et + C Hence y = (f (x) - 1) + C e-f(x).

ye– 4x = - C1 e–x + C2

(c) y = 2 c (x + 1) Ans. (d)

(b) y = 2 c x2 + c2 (d) y = 2 c x + c2

dy Solution: Writing p = and differentiating w.r.t. x, dx we have p = 2p + 2x

dp dp + 2x p4 + 4p3 x2 dx dx

0 = p(1 + 2x p3) + 2x dp =0 dx

dp (1 + 2p3 x) dx

fi

p + 2x

fi

2 log p + log x = const

fi

2

dp dx = p x fi

p2x = c or p =

c x

Substituting this value in the given equation, we get

Solution: The given equation can be written as

y = 2 c x + c 2. (i)

dy du - y = u then (i) reduces to -u=0 fi u= dx dx dy c1 ex. Therefore, we have - y = c1ex which is a linear dx d equation whose I.F. is e–x. So (ye–x) = c1 fi ye–x = c1x dx If

fi

(a) y = 2 c1/2 x1/4 + c

fi

Example 65: The solution of y2 – 2y1 + y = 0 is (a) y = x2 ex + c1 xex + c2 (b) y = (c1 + c2 x) ex (c) y = c1 x2 ex + ex + c2 (d) none of these Ans. (b)

+ c2

d (ye– 4x) = C1 e–x fi dx

dy dx

Ê dyˆ Ê dyˆ Example 67: A solution of y = 2x Á ˜ + x 2 Á ˜ is Ë dx¯ Ë dx¯

tet dt + C, (where t = f (x))

Ê d ˆ Êdy ˆ ÁË d x - 1˜¯ ÁË d x - y˜¯ = 0

u = C1e3x. Therefore, we have

4

Solution: The given equation can be written in the linear form as follows: dy + yf¢(x) = f (x) f¢(x) dx f ¢ ( x) d x The integrating factor of this equation is e Ú = ef(x).

Ú

fi

fi y = C1e3x + C2e4x (For a short cut see theory).

(d) ( y – f (x)) = (f (x)) e–f (x). Ans. (a)

Integrating, we have yef(x) =

du = 3dx u

e– 4x. So

2

Hence

dy du - 4y = u then (i) reduces to - 3u = 0 dx dx

y = (c1x + c2)ex. (For a short cut see theory).

Example 68: The equation of the curve not passing through origin and having the portion of the tangent included between the coordinate axes is bisected at the point of contact is (a) a parabola (b) an ellipse or a straight line (c) a circle or an ellipse (d) a hyperbola Ans. (d)

Complete Mathematics—JEE Main

15.18

Solution: The equation of tangent at any point P(x, y) is Y - y =

POM =

dy (X - x) dx

d yˆ Ê . According to the given condition Y-axis at B Á 0, y - x Ë d x ˜¯ mid point of AB = (x, y) dx dy fi x-y = 2 x and y - x = 2y dy dx dx dy fi x+ y = 0 and y + x =0 dy dx

fi fi

x

2 Ú y dx = xy 0

(b) y = –K x2/2 + C (a) y = K x3/3 + Cx (c) y = – K x3/2 + Cx (d) y = K x3/3 + Cx2/2 (K is constant of proportionality) Ans. (c) Solution: The portion of y-axis cut off between the dy origin and the tangent is y - x . According to the given dx dy = Kx3 (K is constant of proportionality) dx

dy 1 - y = - Kx2. This is a linear equation whose dx x I.F. is 1/x.

Hence

d (y/x) = - Kx dx

fi

y = - K x3/2 + Cx.

Example 70: Through any point (x, y) of a curve which passes through the origin, lines are drawn parallel to the coordiante axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of (a) circles (b) parabolas (c) hyperbolas (d) straight lines Ans. (d) Solution: Let P(x, y) be the point on the curve passing through the origin O(0, 0), and let PN and PM be the lines parallel to the x- and y-axes, respectively (Fig. 15.1). If the equation of the curve is y = y(x), the area

x

Ú0 y dx

x

3 Ú y dx = xy.

fi

0

dy dx = 2 y x

fi log |y| = 2 log |x| + C fi y = Cx2, with C being a constant. This solution represents a parabola. We will get a similar result if we had started instead with 2(PON) = POM. y

P (x, y)

N

O

Example 69: An equation of the curve for which the portion of y-axis cut off between the origin and the tangent varies as the cube of the abscissa of the point of contact is

fi

Ú y dx Assuming 0

Differentiating both sides of this gives dy dy 3y = x + y fi 2y = x fi dx dx

dx dy + =0 x y log(xy) = log c xy = c, which is a hyperbola.

condition y - x

Ú0

x

y dx and the area PON = xy –

that 2 (POM) = PON, we therefore have

dx ˆ Ê This intersects cut the X-axis at A Á x - y , 0 and the Ë d y ˜¯

fi

x

M

x

Fig. 15.1

Example 71: The solution of (y + x + 5)dy = (y – x + 1) dx is y+3 =C (a) log ((y + 3)2 + (x + 2)2) + tan–1 x+2 y-3 =C (b) log ((y + 3)2 + (x – 2)2) + tan–1 x-2 y+3 =C (c) log ((y + 3)2 + (x + 2)2) + 2 tan–1 x+2 y+3 =C (d) log ((y + 3)2 + (x + 2)2) – 2 tan–1 x+2 Ans. (c) Solution: The intersection of y - x + 1 = 0 and y + x + 5 = 0 is (- 2, -3). Put x = X - 2, y = Y - 3. The given equation dY Y-X reduces to = . This is a homogeneous equation, dX Y+X so putting Y = vX, we get

Ê v 1 ˆ v2 + 1 dv - 2 = fi Á- 2 dv = d X Ë v + 1 v + 1˜¯ v+1 dX X 1 fi - log (v2 + 1) - tan-1 v = log | X| + Const 2 Y fi log (Y 2 + X2) + 2 tan-1 = Const X y+3 fi log ((y + 3)2 + (x + 2)2) + 2 tan-1 = C. x +2 X

Example 72: General solution of the differential equady dx tion y = x + represent dx dy

Differential Equations 15.19

(a) (b) (c) (d) Ans. (b)

a straight line or a hyperbola a straight line or a parabola a parabola or a hyperbola circles

fi

1- x

dy , the given equation can be dx written as y = px + 1/p. Differentiating w.r.t. x, we have Solution: Putting p =

p= fi

1

y

dy dp 1 dp = p+x dx d x p2 d x dp dp 1 ( x - (1 / p2 )) = 0 fi p2 = or =0 dx dx x

dp = 0 then p = constant = c putting this value in given dx equation, we get y = cx + 1/c which represents a straight

x dx 1 - x2

( Its integrating factor is e = e 1 = if – 1 < x < 1 and if x2 > 1 then I.F. = 2 1- x d dx

Ê 1 ˆ ax = Áy ˜ Ë 1 - x2 ¯ 1 - x2

(

3/2

)

- (1 / 2 ) log 1 - x 2

(c) 1 +

+ C fi y = a + C 1 - x2

1 e1 / y y e

(d) 1 -

1 e1 / y + y e

Ans. (c) Solution: The given differential equation can be written as

1 dx 1 + 2 x = 3 which is a linear equation with I.F. = dy y y

e Ú(

) = e-1 / y . Multiplying with the integrating factor

1 / y 2 dy

d 1 x e-1 / y = 3 e-1 / y dx y

(

x e–1/y =

fi

1

Ú y3 e

)

-1 / y

dy + C

= Ú -u eu du + C (u = - 1 y ) = –u eu +

Úe

u

du + C

= e–1/y (1 + 1 y ) + C

) 1 x2 - 1

- 2x 1 =- a 2 1 - x2 3 / 2

(

1- x

2

Example 74: Consider the differential equation y2 dx + 1ˆ Ê ÁË x - y ˜¯ dy = 0. If y (1) = 1, then x is given by: 1 e1 / y 2 e1 / y (a) 4 - (b) 3 - + y e y e

Solution: The given equation is linear in y and can be written as dy x ax + y= dx 1 - x 2 1 - x2 Ú

a

=

fi (y – a)2 = C2(1 – x2) fi (y – a)2 + C2 x2 = C2 Thus if –1 < x < 1 the given equation represents an ellipse. If x2 > 1 then the solution is of the form – (y – a)2 + C2 x2 = C 2 which represents a hyperbola.

If

line. If p2 = 1/x then y2 = (px + 1/p)2 = p2 x2 + 1/p2 + 2x = 1 2 x + x + 2x = 4x, which represents a parabola. x Example 73: The curves satisfying the differential equation (1 – x2) y¢ + xy = ax are (a) ellipses and hyperbolas (b) ellipses and parabola (c) ellipses and straight lines (d) circles and ellipses Ans. (a)

2

1 + C e1/ y y

Thus,

x = 1+

When

x = 1, y = 1 1 = 1 + 1 + C e = fi C = -1 e .

)

x = 1+

\

1 e1 y . y e

EXERCISE Concept-based Straight Objective Type Questions 1. The general solution of y¢ + xy = 4x is given by 2 2 (a) y = e(-1 2) x + Cx (b) y = e(1 2) x + Cx

2 2 (c) y = Ce(-1 2) x – 4 (d) y = e(1 2) x + Cx + 4

2. The solution of x ¢ + x tan t = sec t, x(0) = 1 is (a) x = cot t + sin t (b) x = cos t + 2 sin t (c) x = cos t + sin2 t (d) x = cos2 t + sin t 3. A particular solution of the initial value differential equation

15.20

Complete Mathematics—JEE Main

yˆ y dx + x cos dy = 0, ¯ x x

dy = 7x + 8y, y(0) = 0 is dx (a) 15 y = 8e–8y + 7e–7y (b) 15x = 7e–8y + 8e7x (c) 8e–8y + 7e7x = 15 (d) 7e–8y + 8e 7x = 15 4. If x(t) is a solution of (t + 1)dx = (2x + (t + 1)4) dt, x(0) = 2 then lim x(t ) is

Ê 7. The solution of Ë x - y cos y(1) = 0 is given by y (a) sin + log|x| = 0 (b) x

(a) 8 (b) 10 (c) 14 (d) 12 dy + x = 5. The degree of the differential equation dx -4 dy ˆ Ê Ë y + x dx ¯ is (a) 2 (b) 3 (c) 4 (d) 5 1 - 2y - 4x is given by 6. The solution y¢ = 1 + y + 2x

8. The solution yn–1 (ay ¢ + y) = x is

log

(c) tan

t Æ1

1 (a) (y + 2x) + (y + 2x)2 = 2x + C 2 1 (b) (y + (y + 2x)2) = x + C 2 1 (c) x + (2x + y)2 = y + C 2 (d) 2x +

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

y cos

y + log|x| = 0 x

y y + log|x| = 0 (d) y sin + log|x| = 0 x x

(a) (n – 1)y (b) nyn–1 = (c) nyn =

n–1

= Ce

nx Ce a

nx Ce a

(d) nyn–1 = Ce

-

-

nx a

+ nx + a

+ nx

+ nx – a nx a

+ nx + a

9. A curve passing through origin, all the normals to which pass through (1, 2) is (a) a circle with centre origin (b) a straight line (c) a parabola (d) a circle with centre (1, 2) 10. The solution of y¢ = (x + y)2, y(0) = 0 is given by (a) tan x = x + y (b) sin 2y = x + y (c) tan y = x + y (d) sin y = x + y

LEVEL 1 Straight Objective Type Questions 11. y = ae–1/x + b is a solution of (a) a = 1, b = 0 (c) a = 1, b = 1

dy y = 2 when dx x

(b) a = 3, b = 0 (d) a = 2, b = 2

12. The solution of y – xy¢ = 2(x + yy¢), y(1) = 1 is (a) tan–1 (y/x) + log (x2 + y2) = log 2 (b) tan–1 (y/x) + log (x2 + y2) = p/4 + log 2 (c) tan–1 (y/x) + log (x2 + y2) = log 3 (d) none of these 13. The solution of (a) (b)

2

2

2

2

x +y x +y

x 3 dx + yx 2 dy x 2 + y2

= Cx + y/x = C

= ydx – xdy is

(c)

x 2 + y 2 + y/x2 = C

(d) (x2 + y2)2 + xy2 = C 14. Let f (x), g(x) be twice differentiable function on [0, 2] satisfying f ¢¢(x) = g¢¢(x), f ¢(1) = 2g¢(1) = 4 and f (2) = 3, g(2) = 9, then the value of f (4) – g(4) is (a) 0 (c) 8

(b) 2 (d) – 2

15. Which of the following differential equation is not degree 1 (a) x3 y2 + (x +x2) y12 + ex y3 = sin x (b) y21/2 + sin xy1 + xy = x (c)

y1 + y = x + 1

(d) y = 2y1 +

y1 + y

Differential Equations 15.21

16. The general solution of y¢ =

( (

2

2

) )

y x f y /x + x y f ¢ y2 / x 2 2

2

2

(a) y = K f (y /x ) (c) x = K f (y2/x2)

dy = 25. Solution of the differential equation (x – y)2 dx 2 a is

is 2

2

2

(b) x = K f (y /x ) (d) y = Kx f (y2/x2)

dy = 3x + 4y, y(0) = 0 is 17. The solution of log dx (b) 4e3x – e– 4y = 3 (a) e3x + – 4y = 4 3x 4y (d) 4e3x + 3e– 3x =7 (c) 3e + 4e = 7 18. The solution of the equation (x + y) dy – (x – y) dx = 0 is 2

2

2

2

(a) y + 2xy + x = K (b) y + 2xy – x = K (c) x3 + 2xy – y2 = K (d) y2 – 2xy + x2 = K 19. The solution of (1 + y + x2y) dx + (x + x3 ) dy = 0 is (a) y + tan–1 x = C (c) y2 + tan–1 x = C

(b) xy + tan–1 x = C (d) x2 + tan–1 y/x = C

20. The equation of curves which intersect the hyperbola xy = 4 at an angle p/2 is (a) y = 2

x2 2 2

+C

(c) y – x = C

(b) y2 =

x3 +C 3

(d) xy = x2 + C

21. The differential equation of a curve such that the initial ordinate of any tangent at the point of contact is equal to the corresponding subnormal is (a) (b) (c) (d)

a linear equation not a homogeneous equation an equation with separable variables none of these

22. The curve satisfying y dx – x dy + log x dx = 0 (x > 0) passing through (1, –1) is (a) y + log x + 1 = 0 (b) – y2 + log x + 1 = 0 3 2 (c) y + (log x) + 1 = 0 (d) none of these

(a) y =

a x-y-a log +c 2 x-y+a

(b) x =

a x-y+a log +c 2 x-y-a

x-y+a +c x-y-a (d) none of these (c) y2 = a log

26. Solution of the differential equation dy = sin (x + y) + cos (x + y) is dx Ê x + yˆ (a) log 1 + tan Á =y+c Ë 2 ˜¯ Ê x + yˆ (b) log 2 + sec Á =x+c Ë 2 ˜¯ (c) log |1 + tan (x + y) | = y + c (d) none of these 27. Equation of the curve through the origin satisfying dy = (sec x + y tan x) dx is (a) y sin x = x (c) y tan x = x

(b) y cos x = x (d) none of these

28. The solution of (x dx + y dy) (x2 + y2) + (xdy – ydx) = 0 is (a) x2 + y2 +

y =C x

Ê yˆ (b) x2 + y2 + 2 tan–1 Á ˜ = C Ë x¯ 2 Ê xˆ x +y (c) + tan–1 Á ˜ = C Ë y¯ 2 (d)

Ê xˆ x 2 + y2 + tan–1 Á ˜ = C Ë y¯ 2

23. A differential equation associated with the primitive y = a + be5x + ce–7x is (a) y3 + 2y2 – y1 = 0 (b) 4y3 + 5y2 – 20y1 = 0 (c) y3 + 2y2 – 35y1 = 0 (d) none of these

29. General solutions of y¢ = ey/x + y/x is

24. The differential equation of the family of circles passing through the fixed points (a, 0) and (– a, 0) is

(a) log |tan ( y / 2) | = C + sin x/2 (b) log |tan ( y / 2) | = C – 2 sin x (c) log |tan ( y / 4) | = C – 2 sin x/2 (d) none of these 31. The order and the degree (respectively) of the differential equation representing the family of curves y2 = 2c(x + ÷c), c > 0 is a parameter are

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

y1 ( y2 – x2) + 2 xy + a2 = 0 y1 y2 + xy + a2 x2 = 0 y1( y2 – x2 + a2) + 2xy = 0 none of these

(a) Cx = e–y/x (c) log |Cx| = ey/x

(b) log |Cx| = –e–y/x (d) log |Cx| = e–y/x

30. The general solution of y¢ + sin

x+y x-y = sin is 2 2

15.22

Complete Mathematics—JEE Main

(a) 3, 1 (b) 1, 3 (c) 2, 4 (d) 4, 1 32. Let f be a real-valued on R such that f(1) = 1. If the y-intercept of the tangent at any point {(x, y) on the curve y = f(x) is equal to cube of the abscissa of P, then the value of f (– 3) is equal to (a) 3 (b) 8 (c) 12 (d) 9 33. The curves given by 2xyy¢ = y2 – x2 represent (a) family of circles with centre on y-axis (b) family of parabolas passing through origin (c) family of circles with centre on x-axis (d) family of hyperbola

34. Solution to 9yy¢ + 4x = 0 represent (a) family of circles (b) family of ellipses (c) family of straight lines (d) family of hyperbolas 35. The solution of initial value problem y¢ = – 2xy, y(0) = 1 represents (a) a circle with centre on x-axis (b) an ellipse (c) bell-shaped curve (d) a circle with centre on y-axis

Assertion-Reason Type Questions 36. Let y ¢ = cos (x – y) such that y(0) = – p then Statement-1: y can be expressed explicitly in terms of x Statement-2: x + sec (x – y) = – 1 37. Let y = f (x) be a curve having the following property: The segment of the tangent between the point of tangency and the x-axis is bisected at the point of intersection with the y-axis. Statement-1: Such curves represent hyperbola Statement-2: Curves have latus rectum parallel to y-axis. 38. Let y¢ = 3x – 2y + 5

Statement-1: The solution of the above equation is 4y – 6x – 7 = Ce– 2x Statement-2: The given equation is linear in y and x with I.F. e2x 39. Let a solution y = y(x) of the differential equation y sin x + y¢ cos x = 1 satisfy y(0) = 1 Statement-1: y(x) = sin (x + p/4) Statement-2: The integrating factor of the given differentiable equation is sec x. 40. Let xy¢ + y – e x = 0, y(a) = b Statement-1: The solution is given by yx = ex + ab – e a Statement-2: The given equation is a linear function with I.F. x.

LEVEL 2 Straight Objective Type Questions (c) (x2 + y2)2 = C(y2 + 2x2) (d) (x2 + y2) = x3 + xy2

41. A curve such that the area of the trapezoid formed by coordinate axes, ordinate at an arbitrary point and the tangent at this point equals half the square of its abscissa is

44. The general solution of xy5 = y4 is given by

(b) y = (1/2)x + Cx2 (a) y = Cx2 2 (c) 3y = x + Cx (d) y = x + Cx2 42. A curve passing through origin, all the normals to which pass through (x0, y0) is (b) x2 + y2 = 2(xx0 + yy0) (a) yy0 = x2 + y2 2 2 2 2 (c) x + y = x 0 + y 0 (d) none of these

45. The general solution of yy¢¢ = (y¢)2 is

2

3

43. The trajectories orthogonal to (2a – x)y = x is (a) x2 + y2 = C(y2 + 2x2) (b) (x2 + y2)2 = Cxy

(a) y = C1 x5 + C2 x3 + C3 x2 + C4 x + C5 (b) y = C1 x5 + C2 x4 + C3 x2 + C4 x + C5 (c) y = C1 + C2 x + C3 x2 + C4 x3 + C5 x4 (d) none of these (C1, C2, C3, C4, C5 being arbitrary constants) (a) y = C1 x + C2 (c) y = C2 + eC1x

(b) y = C2 eC1x (d) y = eC2x + eC1x

Differential Equations 15.23

46. The curve such that the ratio of the subnormal at any point to the sum of its abscissa and ordinate is equal to the ratio of the ordinate of this point to its abscissa is (a) y = log |Cx| (b) y = x2 + Cx 2 2 (c) y = x log C(x + y ) (d) y = x log |Cx| 47. The curve y = f (x) (f(x) ≥ 0, f(0), = 0, f(1) = 1) bounding a curvilinear trapezoid with the base [0, x], whose area is proportional to the (n + 1)th power of f(x) is n+1

=x (b) y (d) xn + 1 = y

(a) y¢ = x (c) xn = y

48. The curve satisfying the equation y1 =

y 2 - 2 xy - x 2 y 2 + 2 xy - x 2

passing through (1, –1) is a (a) straight line (c) ellipse

(b) circle (d) none of these

49. The curve whose subtangent is n times the abscissa of the point of contact and passes through (2, 3) represent (a) a straight line for n = 2 (b) a parabola with vertex (2, 3) for n = 1 (c) a circle with centre (2, 3) for n = 2 (d) a parabola with vertex origin and axes coincide with x-axis for n = 2 2 dy = y is 50. Solution of y ÊÁ dy ˆ˜ + 2x dx Ë dx ¯ (a) y = 2 Cx + C2 (b) x2 = 2 Cy + C2 (c) y2 = 2 Cx + C2 (d) xy = Cx2 + C2 51. If the area of the figure bounded by a curve, the x-axis, and two ordinates, one of which is constant, the other variable is equal to the ratio of the cube of the variable ordinate to the variable abscissa, then the curve is (a) (2y2 – x2)3 = Cx2 (b) (y2 – x2)3 = Cx2 (c) (2y – x2)2 = Cx3 (d) (2y + x2)2 = Cx3 52. The curve such that the initial ordinate of any tangent is less than the abscissa of the point of tangency by two unit is (a) y = Cx + log |x| – 2 x2 log |x| – 2 2 (c) y = Cx2 + log |x| – 2 (d) y = Cx – x log |x| – 2

(b) y = Cx –

53. The solution of y¢(y – x – 4) = x + y – 2 is given by (a) x2 + 2xy + y2 – 4x + 16y = C (b) x2 + 2xy – y2 – 4x + 8y = C

(c) x2 – 2xy + y2 – 2x + 4y = C (d) x2 – 2xy + y2 – 4x + 8y = C 54. A solution of y = xy¢ – 3y¢3 is (a) y = 2x – 3 (c) y = x – 3

(b) y = 3x – 9 (d) y = 2x – 6

55. The trajectories orthogonal to x2 + y2 = 2ax is (b) y = C(x2 + y2) (a) y = x2 2 2 (d) y2 = Cx (c) y = C(x + 2y ) 56. The solution of y≤ – 4y¢ + 3y = 0, y(0) = 6 and y¢(0) = 10 is (a) y = 4ex + 2e3x (b) y = ex + 6e3x (c) y = (2x + 6)ex (d) y = 2ex + 4e3x 57. The order of the differential equation whose general solution is given by y = (C1 + C2) cos (x + C3) – C4 e x +C5 , where C1, C2, C3, C4, C5 are arbitrary constant is (a) 5 (b) 4 (c) 3 (d) 2 dy = sec (x + y) is 58. The solution of the equation dx given by (a) y – tan (x + y) = C 1 (b) y – tan (x + y) = C 2 1 (x + y) = C (c) y – tan 2 1 tan (x + y) = C (d) y + 2 59. The differential equation corresponding to y = C1em1x + C2em2x + C3em3x, where Ci are arbitrary constants and m1, m2, m3 are roots of the equation m3 – 7m + 6 = 0 is (a) y3 – 7y1 + 6y = 0 (b) y3 – 7y2 + 6y1 + y = 0 (c) y3 – 6y2 + 7y1 + y = 0 (d) y3 + 7y2 – 6y = 0 60. The solution of the differential equation d2 y dx

2

= x + e3x when y1(0) =

x 3 e3 x 1 1 + + x6 9 3 9 1 (b) y = (3x3 + 2e3x – 2) 18 (a) y =

x 3 e3 x 2 x 1 + + + 6 9 9 3 (d) none of these (c) y =

1 and y(0) = 0 is 3

15.24

Complete Mathematics—JEE Main

d2 y 61. The solution of the differential equation = ex 2 d x Êpˆ Êpˆ sin x when y1 Á ˜ = 0, y Á ˜ = 0 is Ë 2¯ Ë 4¯

(a) y = ex sin (x – p/4) + x + (b) y = (1/2) ex sin (x + p/2)

p 2

(c) y = (1/ 2 ) ex sin (x – p/2) (d) y = (1/2) ex sin (x – p/2)

Previous Years' AIEEE/JEE Main Questions

1. The solution of the differential equations -1 dy (1 + y2) + ( x - etan y ) = 0 is dx (a) 2x etan (b) x etan

-1

-1

(c) x e2 tan

y

-1

y

+k

(a) x2 = y2 + xy

= tan–1 y + k

y

-1

= e2 tan

6. The differential equation of all circles passing through the origin and having their centres on the x-axis is

y

= etan

(d) (x – 2) = k e- tan

-1

-1

y

(c) y2 = x2 + 2xy

+k

y

(a) (x2 – y2)y ¢ = 2xy

(b) 2(x2 + y2)y¢ = xy

(c) 2(x2 – y2)y¢ = xy

(d) (x2 + y2)y¢ = 2xy [2004]

3. The solution of the differential equation y dx + (x + x2y) dy = 0 is 1 + log y = C xy

(c) –

1 =C xy

(b) –

1 + log y = C xy

(d) log y = Cx

[2004]

dy = y(log y – log x + 1), then the solution dx of the equation is

4. If x

y (a) log Ê ˆ = Cx Ë x¯

dy dx

(b) x2 = y2 + 3xy

dy dx

(d) y2 = x2 – 2xy

dy dx

[2003]

2. The differential equation for the family of curves x 2 + y 2 – 2ay = 0, where a is an arbitrary constant is

(a)

dy dx

Ê xˆ (b) log Á ˜ = Cy Ë y¯

y Ê xˆ (c) y log Á ˜ = Cx (d) x log Ê ˆ = Cy [2005] Ë x¯ Ë y¯ 5. The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants is of (a) second order and first degree (b) second order and second degree (c) first order and second degree (d) first order and first degree [2006]

[2007] 7. The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is (a) (x – 2)y¢2 = 25 – (y – 2)2 (b) (y – 2)y¢2 = 25 – (y – 2)2 (c) (y – 2)2y¢2 = 25 – (y – 2)2 (d) (x – y)2 y¢ 2 = 25 – (y – 2)2 [2008] 8. The differential equation which represents the family of curves y = c1ec2 x , where c1 and c2 are arbitrary constants is (a) yy≤ = y¢ (c) y≤ = y2

(b) yy≤ = y¢2 (d) y≤ = y¢y

[2009]

9. Solution of the differential equation cosx dy = y(sinx – y)dx, 0 < x < p/2 is (a) y tan x = sec x + C (b) tan x = (sec x + C)y (c) sec x = (tanx + C)y (d) y sec x = tan x + C [2010] dy 10. If = y + 3 > 0 and y(0) = 2, y(log 2) is equal dx to: (a) –2 (c) 5

(b) 7 (d) 13

[2011]

11. Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years. The value V(t) depreciates at a rate given by difdV (t ) = –k(T – t), where k > 0 ferential equation dt is a constant and T is the total life in years of the

Differential Equations 15.25

equipment. Then the scrap value V(T) of the equipment is: I (b) T 2 (a) e–kT K (c) I -

kT 2 2

2

(d) I -

K (T - t ) 2

[2011] 1ˆ Ê 12. Consider the differential equation y2 dx + Á x - ˜ Ë y¯ dy = 0. If y(1) = 1, then x is given by: (a) 4 -

2 e1 / y y e

(b) 3 -

1 e1 / y + y y

(c) 1 +

1 e1 / y y e

(d) 1 -

1 e1 / y + y y

[2011]

13. The population p(t) dt time t of a certain mouse species satisfies the differential equation d p(t) = 0.5 p(t) – 450. If p(0) = 850, then the dt time at which the population becomes zero is : 1 log 18 (a) log 9 (b) 2 (c) log 18 (d) 2 log 18 [2012] 14. At present, a firm is manufacturing 2000 items. It is estimated that the change of production P w.r.t. dP = additional number of workers is given by dx 100 – 12 x . If the firm employs 25 more workers, then the new level of production of items is (a) 3000 (b) 3500 (c) 4500 (d) 2500 [2013] 15. Consider the differential equation dy y3 = dx 2( xy 2 - x 2 ) Statement 1: The substitution z = y2 transforms the above equation into first order homogeneous differential equation Statement 2:2 The solution of this differential equa[2013, online] tion is y2 e–y /x = C. 16. Let the population of rabbits surviving at a time t be d p( t ) 1 = p(t ) governed by the differential equation 2 dt – 200. If p(0) = 100, then p(t) equals (b) 300 – 200 e–t/2 (a) 400 – 300 et/2 t/2 (c) 600 – 500 e (d) 400 – 300 e–t/2 [2014] 17. If the differential equation representing the family of all circles touching x-axis at the origin is

(x2 – y2)

dy = g(x)y, then g(x) equals dx

1 x 2 (c) 2x

(b) 2x2 1 2 (d) x [2014, online] 2 18. If the general solution of the differential equation y Ê xˆ + f Á ˜ , for some function f is given by y¢ = Ë y¯ x y log |cx| = x, where C is an arbitrary constant, then f (2) is equal to 1 (a) 4 (b) 4 1 (c) – 4 (d) – [2014, online] 4 19. The general solution of the differential equation Ê dy ˆ sin 2x Ë - tan x ¯ – y = 0 is dx (a)

(a) y tan x = x + C (b) y cot x = tan x + C (c) y tan x = cot x + C (d) y cot x = x + C

[2014, online]

dy + y tan x = sin 2x and y(0) = 1, then y(p) dx is equal to (a) 1 (b) – 1 (c) – 5 (d) 5 [2014, online]

20. If

21. Let y(x) be the solution of the differential equation dy + y = 2x log x (x ≥ 1). (x log x) dx Then y(e) is equal to (a) e (c) 2

(b) 0 (d) 2e

[2015]

22. The solution of the differential equation y dx – (x + 2y2)dy = 0 is x = f(y). If f(–1) = 1, then f(1) is equal to (a) 4 (c) 2

(b) 3 (d) 1

[2015, online]

23. If y(x) is the solution of the differential equation dy = x2 + 4x – 9, x π –2 and y(0) = 0, (x + 2) dx then y(–4) is equal to (a) 0 (c) –1

(b) 1 (d) 2

[2015, online]

15.26

Complete Mathematics—JEE Main

24. For x Œ R, x π 0 if y(x) is a differentiable function such that x

x sec x + tan x

(c) y = 1 -

x sec x + tan x

x

x Ú y(t ) dt = ( x + 1)Ú t y(t ) dt , then y(x) 1

1

equals: (where C is a constant) C -1/ x e (b) (a) C x3e1/x x2 (c)

(b) y = 1 +

C -1/ x e x

(d)

C x

3

e-1/ x [2016, online]

25. The solution of the differential equation tan x dy y p , where 0 £ x £ , and y(0) = + sec x = 2 y 2 dx 2 0 is given by

(d) y2 = 1 -

[2016, online] 26. If a curve y = f(x) passes through the point (1, –1) and differential equation, y(1 + xy)dx = x dy, then Ê 1ˆ f Á - ˜ is equal to Ë 2¯ (a) (c)

x (a) y = 1 + sec x + tan x 2

x sec x + tan x

2 5

(b) -

2 5

(d)

4 5

4 5

[2016]

Previous Years' B-Architecture Entrance Examination Questions 1. A particular solution of the initial value differential equation dy = 3x + 4y, y(0) = 0 log dx (a) 16y = 3 (4x – 3 + 3e4x) (b) 3e–4y – 4e3x = 1 (c) 4e3x + 3e–4y = 7 [2006] (d) 16y = –3(4x + 3 – 3e4x) 2. The equation of motion of a particle are given by dx dy 1 = t(t + 1), = dt dt t + 1 where the particle is at (x(t)), y(t) at time t. If the particle is at the origin at t = 0 then (a) 6x = (ey + 1) (ey – 1)2 (b) 6x = (2ey – 1) (ey + 1)2 (c) 6x = (ey – 1) (ey + 1)2 (d) 6x = (2ey + 1) (ey – 1)2 [2007] 3. The degree of the differential equation which has a solution y2 = 4a (x + a2) where a is arbitrary constant (a) 2 (b) 3 (c) 4 (d) 1 [2009] 4. Let u(t) and v (t) be two solutions of the differential 2 dy = et y(t) + sin t with u(2) < v(2) equation dt Statement 1: u(t) < v(t) for all t

Statement 2: u – v is proportional to a positive function of t [2010] dy 5. If y(x) is a solution of the differential equation dx + 3y = 2, then lim y( x ) is equal to x Æ•

(a) 1 3 (c) 2

(b) 0 2 (d) 3

6. The differential equation

[2011]

dy 1 - y2 determines = dx y

a family of circles with (a) variable radius and fixed centre (b) variable radius and variable centre (c) fixed radius and variable centre on x-axis (d) fixed radius and variable centre on y-axis [2012] dy 7. The general solution of the differential equation dx x+y x-y = sin is: + sin 2 2 y + 2 sin x = C (a) log tan 2 (b) log tan

y x + 2 sin =C 4 2

Differential Equations 15.27

y + 2 sin x = C 2 y x + 2 sin =C [2013] (d) log cot 4 2 8. Consider the differential equation, ydx – (x + y2) dy = 0. If for y = 1, x takes value 1, then the value of x when y = 4 is (a) 9 (b) 16 (c) 36 (d) 64 [2014] 9. The general solution of the differential equation (c) log cot

y dy + (a) (b) (c) (d) 10. The

1. 5. 9. 13. 17. 21. 25.

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

[2015]

12. 16. 20. 24. 28. 32. 36. 40.

5. (d)

6. (c)

7. (a)

8. (b)

9. (c)

10. (a)

Concept-based 1. The given equation is a linear equation with I.F. =

(

4. (c) 8. (c)

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

=

1 - x2 e 2

1 - x2 e 2

fiy e

1 - x2 2

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

) = 4x e

1 - x2 2

= 4Úxe

1 - x2 2

dx + C = –4 e

1 - x2 2

+C

1 - x2

13. 17. 21. 25. 29. 33. 37.

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

14. 18. 22. 26. 30. 34. 38.

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

so

e-8 y e7 x 1 1 + C. Since y(0) = 0 so – = + C = 8 7 -8 7

fiC=– 42. 46. 50. 54. 58.

. so

so y = – 4 + Ce 2 2. I.F. = eÚtan t dt = elog sec t = sec t. Multiplying with d I. F. we have (x sec t) = sec2 t dt fi x sec t = tan t + C Since x(0) = 1 so C = 1. Thus x sec t π tan–1 t + 1 fi x = sin t + cos t dy = e7x + 8y = e7x e8y fi e–8y dy = e7x dx 3. dx

Level 2 41. 45. 49. 53. 57. 61.

–Ú xdx

d y dx

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

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

4. (a)

e

Level 1 (a) (c) (b) (c) (b) (b) (c) (d)

4. 8. 12. 16. 20. 24.

3. (b)

Concept-based

11. 15. 19. 23. 27. 31. 35. 39.

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

2. (d)

Answers 3. (d) 7. (a)

3. 7. 11. 15. 19. 23.

Hints and Solutions

(a) x3 = 3y3 (1 – e–xy) (b) x3 = 3y3(–1 + exy) (c) x3 = 3y3(1 – exy) (d) x3 = 3y3(–1 + e–xy) [2016]

2. (a) 6. (b) 10. (a)

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

1. (c)

x 2 e xy ydx + xdy = , satisfying y(0) = 1, is y4 ydx - xdy

1. (c) 5. (d) 9. (d)

2. 6. 10. 14. 18. 22. 26.

Previous Years' B-Architecture Entrance Examination Questions

1 + y 2 dx = 0 represent a family of

circles ellipses other than circles hyperbolas parabolas solution of differential equation

Previous Years' AIEEE/JEE Main Questions

43. 47. 51. 55. 59.

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

44. 48. 52. 56. 60.

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

fi –7e–8y

1 1 15 e-8 y e7 x 15 = - =. Thus 8 7 56 -8 7 56 = 8e7x – 15 fi 7e – 8y + 8e7x = 15. dt

4.

-2 Ú dx 2 1 x = (t + 1)3. The I. F. is e t + 1 = dt t + 1 (t + 1)2 d Ê 1 ˆ The equation redues to = (t + 1) x Á dt Ë (t + 1)2 ˜¯

15.28

Complete Mathematics—JEE Main

fi

x

=

2

(t + 1)

(t + 1)2 2

n

+ C. Since x(0) = h 2,

( )= n

x d ue a dx

2

So,

3 1 (t + 1) 3 . Thus = + 2 2 2 2 (t + 1)

C=

fix=

(t + 1)4 2

fi

1+ u du = dx 3

Ê u2 ˆ fi ÁË u + ˜¯ = 3x + C 2 i. e. (y + 2x +

n x u ea

3 + (t + 1)2. lim x(t ) = 8 + 6 = 14. t Æ1 2

dy ˆ 4 Ê dy Ê ˆ + x¯ = 1 5. Ë y + x ¯ Ë dx dx fi the degree is 5. dy du 6. Put u = y + 2x fi – 2, so the given equa= dx dx tion reduces to du 1 - 2u + 2 + 2u 1 - 2u du = –2= fi dx 1+ u 1+ u dx

( y + 2 x )2 2

dy dV =V+x dx dx dV ˆ Ê cos V Ë V + x = (V cos V – 1) dx ¯ dv =–1 dx dx fi cos V dV + =0 x fi sin V + log |x| = C y i. e. sin + log |x| = C. Since y(1) = 0 so C = 0 x y Thus sin + log |x| = 0 x 8. a yn – 1 y¢ + yn = x, Put yn = u dy du = . The given equation reduces to fi n yn – 1 dx dx fi x cos V

a du du n n + +u=xfi u= x n dx dx a a

n

n È ax˘ Î xe ˚ a

n

So u = x –

- x a + Ce a n

n - x

fi n yn = nx – a + Ce a 9. Let the curve be y = f(x) so f(0) = 0. Equation of normal at any point (x, y) is 1 (X – x) Y–y=– f ¢( x ) This passes through (1, 2) if 1 (1 – x) 2–y=– f ¢( x )

) = 3x + C

Put y = Vx fi

x

È nx n ˘ x˙ 1 n Í x ea = Í ea ˙ + C 2 aÎ na (n a ) ˚

2–y=–

1 y+ (y + 2x)2 = x + C 2 7. Dividing by x, we get y yˆ y Ê Ë 1 - x cos x ¯ dx + cos x dy = 0

n

x

I. F. = e a . Multiplying with e a , we have

dx (1 – x) dy

fi (2 – y)dy = – dx (1 – x) 2y –

y2 x2 =–x+ +C 2 2

Since y(0) = 0, so C = 0. Hence x 2 y2 – (x + y) = 0 + 2 2 x2 + y2 – 2x – 4y = 0 which is a circle with centre (1, 2). du . The equation reduces 10. Put x + y = u fi 1 + y¢ = dx du du – 1 = u2 fi =x to dx 1 + u2 fi tan –1 (x + y) = x + C But y(0) = 0 so C = 0 fi x + y = tan x.

Level 1 1 dy 1 = 2 ae-1 x = 2 (y – b), so b = 0 and a = 1 x dx x dy y – 2x 12. = . Putting y = vx, we get 2y + x dx 11.

v+x

dv v–2 2v + 1 dx = fi dv = – 2 2 2v + 1 dx x 1+ v

fi log (1 + v2) + tan–1 v = – log x2 + C

Differential Equations 15.29

y =C x p Putting y(1) = 1, we obtain C = + log 2. 4 13. The given equation can be written as x dx + y dy x2 = – (x dy – y dx) x 2 + y2 fi log (x2 + y2) + tan–1

fi

1 d ( x 2 + y2 ) ( x dy - y dx ) =– = – d Ê yˆ Ë x¯ 2 x 2 + y2 x2

y +C x 14. f ¢¢ (x) – g¢¢(x) = 0 fi f ¢(x) – g¢(x) = C. Putting x = 1, we obtain c = 2. Thus f(x) – g(x) = 2x + C1. Putting x = 2, we get C1 = – 10. Hence f(4) – g(4) = 8 – 10 = – 2. 15. The degree of the equation in (a), (b), (c) is clearly 1. The equation in (a) can be written as (y – 2y1)2 = 9(1 – y1)2 which is of degree 2. Integrating we have,

x 2 + y2 = –

y dy du = u, we have = u + x . The x dx dx given differential equation can be written as u + x

16. Putting

du 1 f (u2 ) =u+ u f (u2 ) dx fi

u f ¢(u2 ) 2

f (u )

du =

dx x

Ê y2 ˆ fi log f (u2) = log x2 + const fi x2 = k f Á ˜ . Ë x2 ¯ dy 17. = e3x + 4y = e3x . e4y fi e–4y dy = e3x dx dx e – 4 y e3 x – Thus = Const. But y(0) = 0, so –4 3 – 1 – 1 = C. Hence 3e–4y + 4e3x = 7. 4 3 18. The given equation can be written as x dy + y dx + y dy – x dx = 0 fi d(xy + y2/2 – x2/2) = 0 fi 2xy + y2 – x2 = const. 19. Write the given equation as dx + (1 + x2) (y dx + x dy) = 0 dx + d(xy) = 0 fi tan–1 x + xy = c. fi 1 + x2 dx dy dy , = 0 fi dy = – y . Replace by – dy dx dx dx x dx y = fi y2 – x2 = Const. we have dy x

20. y + x

21. Equation of tangent at (x, y) to curve y = f(x) is Y – y = f ¢(x) (X – x). Putting X = 0, the initial ordinate of the tangent is y – x f ¢(x). The subnormal at this dy , so we have point is given by y dx y dy dy dy =y–x fi = y x + y dx dx dx dx x fi = 1 which is a linear equation. – dy y 22. Multiplying the given equation by 1/x2, we get y dx x dy log x – 2 + 2 dx = 0 x2 x x 1 1 – y d Ê ˆ – Ê ˆ dy – log x d Ê 1 ˆ = 0 Ë x¯ Ë x¯ Ë x¯ y log x ˆ dx =0 fi dÊ ˆ + dÊ Ë x¯ Ë x ¯ x2 y log x 1 + + + C = 0. x x x The curve passes through (1, – 1), so C = 0 Thus the required curve is y + log x + 1 = 0. (i) 23. y1 = 5 b e5x – 7c e–7x (ii) y2 = 25 b e5x + 49 c e–7x (iii) and y3 = 125 b e5x – 343 c e–7x Multiplying (i) by 7 and the result to (ii) gives y2 + 7y1 = 60 b e5x (iv) Multiplying (i) by 5 and substracting the result from (ii), we get (v) y2 – 5y1 = 84 C e–7x 5x –7x use (iv) and (v) to replace to e and C e in (iii) fi

y + 7 y1 ˆ y – 5 y1 ˆ y3 = 125 Ê 2 – 343 Ê 2 Ë 60 ¯ Ë 84 ¯ fi y3 + 2y2 – 35y1 = 0. 24. Let the equation of the circle be x2 + y2 + 2gx + 2fy + C = 0, it passes through (a, 0) and (– a, 0) of a2 + 2ga + C = 0 and a2 – 2ga + C = 0. Substracting we obtain g = 0 and so C = – a2. Hence the equation of circle reduces to x2 + y2 + 2fy – a2 = 0. Differentiating, we have x + yy1 + fy1 = 0 x + yy1 . Thus, we have y1 2 x( x + yy1 ) =0 x2 + y2 – a2 – y1 2 xy fi x2 – y2 – a2 – =0 y1 fi y1(y2 – x2 + a2) + 2xy = 0. du du 25. Put x – y = u. Then , and the given =1– dx dx equation becomes u2 Ê 1 – du ˆ = a2 Ë dx ¯ fif=–

15.30

Complete Mathematics—JEE Main

fi

Ê Ê u2 ˆ a2 ˆ du dx = Á 2 = du 1 + ÁË Ë u – a 2 ˜¯ u2 – a 2 ˜¯

Integrating, we obtain u–a a2 x=u+ log +C u+a 2a =x–y+ fi

y=

x-y-a a +C log x-y+a 2

x-y-a a + C. log x-y+a 2

26. Putting x + y = u, the given equation becomes

dy = – 2 cos x/2 dx. sin y / 2 Integrating, we have y 2 log tan = const – 4 sin x/2 4 y = C – 2 sin x/2. fi log tan 4 31. Differentiating the given equation y2 = 2c (x + c ) (1) dy dy 2y = 2c fi c = y dx dx Putting this value in (1) fi

du = dx 1 + sin u + cos u fi

du

Ú 1 + sin u + cos u

y2 = 2y

=x+C

Substituting tan (u/2) = t, to calculate the integral on the L.H.S, we obtain its value as log | 1 + t | so that the required solution is x+y = x + C. 2 27. The given equation can be written as dy – y tan x = sec x dx which is linear equation with integrating factor log 1 + tan

e Ú = e – log sec x = cos x. Multiplying with the integrating factor and integrating we obtain y cos x = x + C. Since it passes through (0, 0) so we have C = 0. 28. The given equation can be written as x dy - y dx 1 d Ê ( x 2 + y 2 )ˆ + =0 Ë2 ¯ x 2 + y2 – tan x dx

d( y / x) 1 =0 d Ê ( x 2 + y 2 )ˆ + Ë2 ¯ 1 + y2 / x 2 Integrating, we obtain 1 2 (x + y2) + tan–1 y/x = C 2 du dy =x +u, 29. Put y/x = u, so that dx dx From the given equation, we have dx du + u = eu + u fi e–u du = x x dx –u fi – e = log x + Const fi log | Cx | = – e–y/x x–y x+y x y - sin 30. y¢ = sin = – 2 cos sin 2 2 2 2

fi y – 2x

dy Ê dy ˆ x+ y ˜ Á dx Ë dx ¯

3/2 dy Ê dy ˆ = 2Á y ˜ Ë dx ¯ dx

3 dy ˆ 2 Ê 3 Ê dy ˆ ÁË y – 2 x ˜¯ = 4 y ÁË ˜¯ dx dx Thus the order of the differential equation is 1 and the degree is 3. 32. Equation of tangent a P(x, y) to the curve y = f(x) is Y – y = f ¢(x) (X – x) Intercept on Y-axis is y – xf ¢(x) We are given, y – xf ¢(x) = x3

fi

dy 1 – y = – x2 dx x 1 This is a linear equation whose I.F. is . x Multiplying with I.F., the above equation reduces to d Ê1 ˆ 1 3 Á y˜ = – x fi f(x) = y = – x + Cx dx Ë x ¯ 2 1 3 As f(1) = 1 so C = 3/2. Thus f(x) = – x 3 + x, 2 2 so f(3) = 9. 33. Putting y = Vx, the given equation reduces fi

2 dx = –V +1 dx 2V

fi

2VdV 1+V

2

= –

dx x

fi (1 + V2)x = Const fi x2 + y2 = Cx fi (x – C/2)2 + y2 = C2/4 which represents circle with centre (C/2, 0) i.e., on x-axis. 9 2 y = – 2x2 + const 34. 9ydy = – 4xdx fi 2 x2 y2 fi + = Const, so represents a family of 9 4 ellipses.

Differential Equations 15.31

35.

dy = – 2xdx y

fi

y = Ce – x

For

x = 0, y = 1

u x–y = x + C i.e. – cot =x+C 2 2 Putting y (0) = – p, we have C = 0 y = x – 2 cot – 1 (– x).

fi – cot

37. Let (x, y) be the point of tangency. Equation of tangent is Y – y = f ¢(x) (X – x). This inter1 Ê ˆ y, 0˜ . The mid point is sects x-axis at Á x – Ë ¯ f ¢ (x) yˆ 1 Ê ÁË x – 2 f ¢ ( x ) y, 2 ˜¯ is on Y-axis so x– fi

2

1 y =0 2 f ¢ (x)

dy dx = y x

fi

y 2 = Cx

which represents parabola with latus ractum parallel to Y-axis. dy + 2y = 3x + 5 fi ye2x = Ú (3x + 5) e2x + C 38. dx e2 x 3 = (3x + 5) – e2 x + C 2 4 Ê 3x + 5 3 ˆ fi y= Á – ˜ + Ce – 2x Ë 2 4¯ fi 4y – 6x – 7 = Ce – 2x tan x dx 39. It is linear equation with I.F. e Ú = sec x, Required solution is y = sin x + cos x. 1

Ú x dx

d (xy) = ex dx fi xy = ex + C Put y(a) = b ab – e a = C, so xy = e x + ab – e a.

40. I.F. = e

dy ˆ Ê = ÁË y + y - x ˜¯ x dx

so C = 1

– x2

which is bell shaped. dy du = . The given equa36. Put x – y = u fi 1 – dx dx du tion reduces to 1 – = cos u dx du 1 = dx fi Ú cosec 2 (u 2 ) du = x + C fi 2 1 – cos u Thus

1 2 x = Area of the trapezoid 2

2

fi

y= e

log y = – x2 + C

= x so

Level 2 41. The tangent at any point (x, y) meets the Y-axis at Ê 0, y - x dy ˆ ˜ . According to the given condition ÁË dx ¯

fi

1 dy y -2 = - (which is a linear equation) 2 dx x

I. F. = e

-2 Ú

1 dx x

= x–2

Multiplying with I.F. and integrating, we have yx–2 = =

1 -2 x dx + C 2Ú

1 -1 x +C 2

1 x + Cx2. 2 42. Equation of normal at any point (x, y) 1 Y–y=– (X – x) f ¢( x) This passes through (x0, y0) if 1 y0 – y = (x0 – x) f ¢( x) (y0 – y) dy + (x0 – x) dx = 0 fiy=

( y0 - y ) 2 ( x0 - x) 2 = C + 2 2 2 (y – y0) + (x – x0)2 = Const Since y(0) = 0 so Const = y02 + x02 Thus y2 – 2yy0 + y02 + x2 +x02 – 2xx0 = x02 + y02 x2 + y2 = 2(xx0 + yy0). dy – y2 = 3x2 (Differentiating given dx equation) dy = 3x2 + y2 2 (2a – x)y dx Eliminating a between the last equation and given equation i.e. (2a – x) y2 = x2, we have dx y (3x2 + y2) = 2x3 dy

43. (2a – x) 2y

Replacing

dy dx by – we have dy dx

2x3 + y(3x2 + y2)

dy =0 dx

dy - 2 x3 = , which is a homogeneous equadx (3x 2 + y 2 ) y tion.

15.32

Complete Mathematics—JEE Main

Putting y = Vx V+x fix

y = f (x) = K (n + 1) yn

dV 2 =– dx (3 + V 2 )V

fi dx = K (n + 1) yn–1

È 2 + 3V 2 + V 4 ˘ dV = -Í ˙ 2 dx Î (3 + V ) V ˚

È 2V V ˘ (3 + V 2 ) V dx = = -Í ˙ 2 2 2 (1 + V )(2 + V ) 2 +V2 ˚ x Î1 + V fi log x2 = – log

(1 + V 2 ) 2 + Const 2 +V2

x 2 (1 + y 2 x 2 ) 2 = Const fi 2 + y 2 x2 fi (x2 + y2)2 = C(y2 + 2x2).

(n + 1) n y n Since f (0) = 0, C = 0 n K (n + 1) f (1) = 1, 1 = fiK= n +1 n n Thus x = y .

fix+C=K

48. Putting y = Vx dV V 2 - 2V - 1 = 2 dx V + 2V - 1

V+x fix

44. xy5 = y4 fi x d y4 = y4 dx dy4 fi = dx fi y4 = (Const) x y4 x

dy dx dy

dV V 2 - 2V - 1 - V 3 - 2V 2 + V = dx V 2 + 2V - 1

V 2 + 2V - 1 dx = - 3 dV x V +V2 +V +1

So y3 = const x2 + const

Ê 1 2V ˆ dV + 2 = - ÁË V + 1 V + 1˜¯

fi y 2 = C 1¢ x 3 + C 2¢ x + C 3¢

Integrating both sides, we get

fi y1 = C1¢¢ x4 + C2¢¢ x2 + C3¢¢ x + C4¢¢

log x – log (V + 1) + log (V2 + 1) + log C = O

5

3

2

fi y = C 1 x + C 2 x + C 3 x + C 4 x + C 5. y ¢¢ y¢ 45. yy ¢¢ = y ¢2 fi = Integrating, we get y¢ y log y ¢ = log y + const fi y¢ = Cy

If this curve passes through (– 1, 1), we get 2C = 0 i.e.

dy = C dx fi log y = Cx + C¢ y

C = 0. Hence the required curve is y + x = 0 which is a straight line.

fi

y = C2 eC1x . 46. According to the given condition dy y dx = y x+y x dy x+ y = dx x Putting y = Vx, V +

dV =1+V dx

dx x fi V = log x + const

Ê x(V 2 + 1) ˆ C = 1 fi C (y2 + x2) = y + x fi Á Ë V + 1 ˜¯

49. According to the given condition y = nx f ¢( x) dy dx 1 dx = nx fi = y dy n x n Integrating y = Cx. This passes through y

(2, 3) if 3n/2 = C So y n =

fi dV =

fi y = x log Cx. 47.

x

Ú0

f (t ) dt = K y n+1

Differentiating, we have

y fi ÊÁ ˆ˜ Ë 3¯

3n x 2

n

x . For n = 2. This represents a 2 parabola whose axes coincide with x-axis. =

pˆ Ê 1 - ˜ y . Differentiating both sides 50. x = Á Ë 2p 2¯

Differential Equations 15.33

dy w·r·t y ÊÁ p = ˆ˜ Ë dx ¯ Ê 1 1 ˆ dp 1 1 p - ˜ = + yÁ2 2 ¯ dy Ë 2p p 2p 2 2 Ê 1 + p ˆ dp 1 p + = - yÁ Ë 2 p 2 ˜¯ dy 2p 2 fi

Ê 1 + p 2 ˆ dp 1 + p2 = - yÁ Ë 2 p 2 ˜¯ dy 2p

dp dy = fi py = C p y i.e. y dy = c dx fi y2 = 2 Cx + C2. -

51. According to the given condition y3 Ú0 x Differentiating, we get x

y=

f (t ) dt =

3 y 2 dy y 3 x dx x 2

x2 y + y3 3 y 2 dy = x2 x dx x2 y + y3 dy fi = 3 y2 x dx Putting y = Vx V +V3 = V + x dV 3V 2 dx fi

1+V2 - V = x dV 3V dx

fi

2 dy 1 - y = -1+ x dx x

This is a linear equation with I.F.

1 x

d Ê 1ˆ 1 2 ÁË y . ˜¯ = - + 2 x x dx x 1 y ◊ = – log|x| – 2x – 1 + C x y = Cx – x log |x| – 2. x+ y-2 53. y¢ = . Put x = X + h, y = Y + K y-x-4 dY = X + Y + (h + K - 2) Y - X + (- h + K - 4) dX Select h, K such that h + K – 2 = 0, – h + K – 4 =0 fi K = 3, h = – 1 dY X +Y = dX Y-X Putting Y = v X, we have v+ X

1+V dV = V -1 dX

1+V 1 + 2V - V 2 dV V = = fi X V -1 V -1 dX - 1 - 2V + 2 dX dV = 2 1 + 2V - V 2 X Integrating, we have X2 (1 + 2V – V2) = Const X2 – Y2 + 2XY = Const (x + 1)2 – (y – 3)2 + 2 (x + 1) (y – 3) = Const

2

fi 1 - 2V = x dV 3V dx 3V dx dV = fi 2 1 - 2V x Integrating, we have -3 log (1 – 2V2) = log x + Const fi 4 fi x4 (1 – 2V2)3 = Const fi (2y2 – x2)3 = Cx2. 52. The initial ordinate of tangent is equal to dy , so y–x dx dy y–x =x–2 dx

x2 + 2xy – y2 – 4x + 8y = Const. dy 54. y = xp – 3p3, p = . Differentiating w · r · t · x, dx dp dp - 9 p2 we have p = p + x dx dx fi (x – 9p2)

dp =0 dx

dp = 0 fi p = Const so y = Cx – 3C3 dx y = x – 3 is solution with C = 1. dy dy = 2a fi a = x + y 55. 2x + 2y dx dx Eliminating a, we have dy x2 + y2 = 2 ÊÁ x + y ˆ˜ x Ë dx ¯

15.34

Complete Mathematics—JEE Main

Replacing

dy dx by dx dy

59. Clearly y3 – 7y1 + 6y = 0 is the required equation 60.

dx ˆ Ê x2 + y2 = 2 Á x 2 - xy ˜ Ë dy ¯

V +x

(

V 1+V2

)

V = const x (1 + V 2 )

fi y = C (x2 + y2). 56. m2 – 4m + 3 = 0 fi (m – 1) (m – 3) = 0

1 + C1 fi C1 = 0 3 x 3 e3 x 1 1 Hence y = (3x3 + 2e3x – 2). + - = 6 9 9 18

61.

dy = dx

0=y

6 = y(0) = C1 + C2, 10 = y¢(0) = C1 + 3C2 C2 = 2, C1 = 4. Thus y = 4ex + 2e3x. 57. Since C1 + C2 is one constant C4 ex + C5 = C1 eC5 ex Thus there are three arbitrary constants. So the order of the differential equation is 3. du dy = . The given equation 58. Put x + y = u fi 1 + dx dx reduces to du - 1 = sec u dx du fi = dx 1 + sec u 1 ˆ Ê du = dx fi Á1 Ë 1 + cos u ˜¯ fi u – tan u/2 = x + C x+ y =x+C 2 x+ y fi y – tan = C. 2

fi x + y – tan

e x sin ( x - p / 4)

x Ú e sin x dx =

2

() () p 2

0 = y1

2

1 e x sin ( x - p 2)

fiy=

so m = 1, 3 Hence y = C1 ex + C2 e3x

1 1 and = y1 (0) 9 3

=

d V = dx x

È1 2V ˘ dx dV = Í 2˙ x ÎV 1 + V ˚ fi

Since y (0) = 0 so C2 = -

dV = 2 V 1-V2 dx

1-V2

x 3 e3 x + C1 x + C2 + 6 9

fiy=

2 xy dy = 2 . Putting y = Vx, we have x - y2 dx

fi

Ê dy ˆ = x + e3x. Integrating, we have ÁË ˜¯ dx

dy x 2 e3 x = + C1 + 2 3 dx

y 2 - x2 dx = - 2 xy dy

fi

d dx

+ C 1x + C 2

2

= 0 + C1

p 4

+ C1

p + C2 2

1 p 4È p e Í - sin + cos p 4 ˘˙ + C1 2 4 Î ˚

=

fi C1 = 0 So C2 = 0 1 p Hence y = ex sin x . 2 2

( )

Previous Years’ AIEEE/JEE Main Questions

(

1. (1 + y2) + x - etan

(

2 fi 1+ y

+x ) dx dy

-1

) dxdy

y

= e tan

-1

=0 y

-1 1 dx 1 + x = e tan y 2 2 dy 1 + y 1+ y

fi

1

I.F. = e

Ú 1+ y 2 dy

(

-1 d xe tan y dy

fi xe =

)

tan -1 y

=

= Ú

= e tan e2 tan

-1

-1

y

y

1 + y2 e tan

-1

y

1 + y2

dy + const

1 2 tan -1 y e + const 2

fi 2 xe tan

-1

y

= etan

-1

y

+k

Differential Equations 15.35

2. x2 + y2 – 2ay = 0 fi 2a =

2

x2 + y 2 y

Ê d2 yˆ Ê dy ˆ fi –Ax = Bx Á ˜ + Bxy Á 2 ˜ Ë dx ¯ Ë dx ¯

Differentiating, we get (2 x + 2 yy ¢ ) y - ( x 2 + y 2 ) y ¢ y2 2

Ê d2 yˆ Ê dy ˆ dy + Bx Bxy fi By = ÁË ˜¯ Á 2 ˜ [from (1)] dx Ë dx ¯ dx

=0

fi xy

2

fi (x – y )y¢ = 2xy

d2y dx

2

+ ( x - y)

dy =0 dx

3. We can write the given differential equation as (ydx + xdy) + fi

d ( xy ) ( xy )

2

+

1 (xy)2 dy = 0 y

6. Equation of any circle through origin and having centre on the x-axis is (x – a)2 + y2 = a2

1 dy = 0 y

(1)

Differentiating w.r.t. x, we get

1 fi - + log y = C. xy

2(x – a) + 2yy¢ = 0

4. The given differential equation can be written as

fi a = x + yy¢ y

yÈ y ˘ dy = Ílog + 1˙ x x dx Î ˚

(1) (a, 0)

Put y/x = n, so that fi

O

x

dy dn =n+x dx dx

Putting this in (1), we get Fig. 15.2

dn n+x = n(log n + 1) dx fi

Putting this in (1), we get (–yy¢)2 + y2 = (x + yy¢)2

dn dx = n log n x

fi y2 = x2 + 2xy

dn fi Ú = Ú dx n log n x fi log log n = log x + log c

dy dx

7. Let the centre be at (a, 2). Equation of circle of radius 5 is (x – a)2 + (y – 2)2 = 25

Ê yˆ fi log n = cx or log Á ˜ = cx Ë x¯

Differentiating, we have

5. Differentiating w.r.t. x, we get dy =0 2Ax + 2By dx

2(x – a) + 2(y – 2)

dy fi –Ax = By dx Again differentiating w.r.t. x, we get

fi x – a = – (y – 2)

2 Ê d2 yˆ Ê dy ˆ – A = B Á ˜ + By Á 2 ˜ Ë dx ¯ Ë dx ¯

(1)

dy =0 dx dy dx

Putting this value in (1), we have Ê dy ˆ (y – 2)2 Á ˜ Ë dx ¯

2

+ (y – 2)2 = 25

(y – 2)2 y¢2 = 25 – (y – 2)2.

(1)

15.36

Complete Mathematics—JEE Main

8. y = c1ec2 x

I=

fi y¢ = c1c2 ec2 x

k 2 k T + C fi C = I – T2 2 2

Also, scrap value = V(T) = C

fi y≤ = c1c22 ec2 x

=I–

\ yy≤ = c12 c22 ec2 x ec2 x

(

= c1c2 ec2 x

)

2

Ê 1ˆ 12. y2dx + Á x - ˜ dy = 0 Ë y¯

= (y¢)2

9. We can write the given differential equation as

fi

1 dy 1 dy y (sin x - y ) = fi 2 = – sec x y dx y dx cos x Put -

I.F. = e Ú

tan xdx

fi

(1)

fi xe–1/y = Ú

= elog sec x = sec x

Thus, xe–1/y = e–1/y(1 + 1/y) + c

dy =y+3 dx

fix=1+

Hence, 1 = 1 + 1 + ce fi c = –1/e

Thus, ln (y + 3) = x + ln 5 \x=1+

When x = ln 2, we get

fi dV = –k(T – t)dt fi ÚdV = –kÚ(T – t)dt fi V(t) =

k (T – t)2 + C 2

We have V(0) = I, therefore

1 + ce1/y y

When x = 1, y = 1.

dy = Údx fi ln (y + 3) = x + c y+3

dV = – k(T – t), k > 0 dt

e-1/ y dy = I (say)

= et(1 – t) = e–1/y(1 + 1/y).

When x = 0, y = 2, therefore, ln 5 = c

11.

y

3

I = Ú(–t)et dt = –tet + Úet dt

- sec x = c1 – tan x fi sec x = (c + tan x) y y

In (y + 3) = ln 2 + ln 5 fi y + 3 = 10 or y = 7

1

To evaluate I, put –1/y = t, so that

d [(sec x)z] = – sec2 x fi (sec x)z = c1 – tan x dx

fi Ú

= e–1/y

1 -1/ y d ÈÎ xe-1/ y ˘˚ = 3 e y dx

Where c = – c1. 10.

(1/ y 2 ) dy

Multiplying (1) by e–1/y, we get

Multiplying (1) by sec x, we get fi

dx 1 1 + 2x = 3 dy y y

I.F. = e Ú

1 = z, so that the above equation becomes y

dz + (tan x) z = – sec x dx

k 2 T 2

13.

1 e1/ y y e

d p(t) – 0.5p(t) = – 450. I.F. = e–0.5t dt fi

d (p(t) e–0.5t) = – 450 e–0.5t dt

fi p(t)e–0.5t = 450 ¥ 2. e–0.5t + C Putting t = 0, 850 = 900 + C fi C = –50 fi p(t) = 900 e–0.5t – 50

(1)

Differential Equations 15.37

p(t) = 0 fi e–0.5t =

1 fi – 0.5t = – log 18 18

1 1 ˆ - t d Ê - 2t 2 200 e ( ) = e p t Á ˜ dt Ë ¯

fi t = 2 log18.

fi

dP = 100 – 12 x 14. dx

( ) Ú (100 - 12 x ) dx

fi C = – 300 Thus p(t) = 400 – 300

We have P(0) = 2000,

1 t e2

17. An equation of circle touching the x-axis at the origin is

Therefore, c = 2000. Thus, P(25) = 100 (25) – 8(25)3/2 + 2000 = 3500

x2 + (y – a)2 = a2

(1)

where a is parameter.

dz dy = 2y dx dx

Differentiating (1) w.r.t. x,

3

4

y 2y dy dy = fi 2y = 2 2 2( xy - x ) 2( xy 2 - x 2 ) dx dx

we get 2x + 2(y – a)y¢ = 0

z2

dz fi = dx

+C

100 = p(0) = 400 + C

P(x) = 100 x – 8x3/2 + c

15. z = y2 fi

=

1 - t 400 e 2

Putting t = 0

fi dP = 100 - 12 x dx fiP=

1 - t e 2 p(t )

( xz - x ) 2

fi a = y+

x y¢ y

dz du tion. Put z = ux fi =u+ x dx dx u 2 x2

du = dx

fi x

u2 - u2 + u u du = = u -1 u -1 dx

fi

(x u - x ) 2

2

=

O

u -1 dx du = u x

fi u – log x = log x + C fi u = log ux + C

Putting this in (1), we get 2

Ê xˆ = Áy+ ˜ Ë y¢ ¯

fi ux = ceu fi ze–z/x = c

fi x2 = y2 + (2xy)

2

/x

=c

1 d p (t ) - p(t ) = – 200. 2 dt I.F. is e

1 - t 2

, so

x

Fig. 15.3

Ê xˆ x +Á ˜ Ë y¢ ¯

fi y 2 e- y 16.

a

u2 u -1

u+x

2

fi (x2 – y2)

2

1 y¢

dy = (2x)y dx

\ g(x) = 2x Ê 1ˆ 18. y¢ ln |cx| + y ÁË ˜¯ = 1 x

15.38

Complete Mathematics—JEE Main

Ê xˆ y fi y¢ Á ˜ + =1 Ë y¯ x y Ê yˆ fi y¢ = - Á ˜ x Ë x¯ fi f (u) = -

2

fi

fi (log x)y = 2(x log x – x) + C y Ê yˆ = +fÁ ˜ Ë x¯ x

When x = 1, we get

1

0 = 2(0 – 1) + C fi C = 2

u2

When x = e, we get

1 4 19. Given differential equation is

(log e) y(e) = 2(e log e – e) + 2

\ f (2) = -

=e

fi y(e) = 2 22. Write the differential equation as

dy – cosec (2x)y = dx I.F. = e

d [(log x)y] = 2 log x dx

tan x

(1) dx Ê 1 ˆ + x = 2y dy ÁË y ˜¯

- Ú cosec( 2 x ) dx

–(1/2)log(tan x)

=

Multiplying (1)

I.F. = e Ú(–1/y)dy = e–log y =

1 tan x 1 tan x

1 y

Multiplying (1) by 1/y, we get , we get

1 dx Ê 1 ˆ + x =2 y dy ÁË y 2 ˜¯

d Ê y ˆ =1 dx ÁË tan x ˜¯ fi y cot x = x + c 20. I.F. = eÚ(tan x)dx = elog sec x = sec x Multiplying the given differential equation by sec x, we get

dx Ê x ˆ =2 dy ÁË y ˜¯

fi

x = 2y + c y

fi x = 2y2 + cy When y = –1, x = 1,

d ((sec x)y) = 2 sin x dx

Therefore, 1 = 2 – c fi c = 1

fi (sec x) y = –2 cos x + C

Thus, f(y) = 2y2 + y fi f(1) = 3.

Putting x = 0, y = 1, we get

23. We can write differential equation as

1 = –2 + C fi C = 3

dy 13 x2 + 4 x - 9 = =x+2– dx x+2 x+2

When x = p, we get (–1)y = –2 cos p + 3 fi y = – 5

1 (x + 2)2 – 13 log |x + 2| + C 2 As y(0) = 0, we get fiy=

21. We can write the given differential equation as 1 dy + y =2 dx x log x

fi

(1)

0 = 2 – 13 log |2| + C or C = 13 log 2 – 2

I.F. = eÚdx/x log x = elog (log x) = log x

Also, y(–4) = 2 – 13 log 2 + 13 log 2 – 2

Multiplying (1) by log x, we get

= 0.

(log x)

dy 1 + y = 2 log x dx x

x

24. x Ú1 y(t ) dt = (x + 1)

x

Ú1 ty(t ) dt

Differentiating both the sides, w.r.t. x, we get

Differential Equations 15.39 x x Ú1 y (t )dt + xy ( x) = Ú1 ty (t ) + ( x + 1) xy ( x) x

dt 1 + t =1 dx x dt fi x +t = x dx dt (tx) = x fi dx 1 fi tx = x 2 + C 2 1 1 fi - x = x2 + C y 2

x

fi Ú y (t )dt = Ú ty (t )dt + x 2 y ( x) 1 1 Differentiating both the sides w.r.t. x, we get y(x) = xy(x) + 2xy(x) + x2y¢(x) fi (1 – 3x) y(x) = x2y¢(x) fi

y ¢ ( x) 1 3 = 2y ( x) x x

Integrating we get 1 – 3 ln |x| + A x

fi ln|y(x)| =

fi ln |x3y(x)| = A –1/x

3

As it passes through (1, –1), 1 =

x y(x) = ± e e

\ -

1 +A x = Ce

c x3

1 1 1 1 x = x 2 + . When x = - , y 2 2 2

1 Ê 1ˆ - Á- ˜ = 1 + 1 = 5 y Ë 2¯ 8 2 8

–1/x

where C is a constant fi y(x) =

fiy=

e–1/x

4 . 5

Previous Years’ B-Architecture Entrance Examination Questions

tan x dy 1 + y sec x = 25. 2y dx 2

I.F. = eÚsec x dx = eln(sec x + tan x)

dy = 3x + 4y dx dy = e3x + 4y = e3x e4y fi dx

= sec x + tan x

fi e–4y dy = e3x dx

d 2 ( y ) + (sec x)y2 = tan x fi dx

1. log (1)

Multiplying (1) by sec x + tan x, we get

fi -

d [y2 (sec x +tan x)] = tan x (sec x + tan x) dx

e -4 y e3 x + const = 4 3

y(0) = 0 fi -

2

fi y (sec x + tan x) = sec x + tan x – x + C

fi 4e3x + 3e–4y = 7

1=1+CfiC=0 2.

x Thus, y = 1 sec x + tan x 2

26. We can write the differential equation as y -2

Put –y

–1

=tfi y

-2

We can write (1) as

(1) dt dy = dx dx

1 7 1 = + const fi const = 4 12 3

so –3e–4y = 4e3x – 7

When x = 0, y = 1, so that

dy 1 -1 - y =1 dx x

1 1 +C fi C = 2 2

dx t3 t2 = t(t + 1) fi x = + + C1 3 2 dt But at t = 0, x = 0 so C1 = 0 Thus x =

t3 t2 + 3 2

dy 1 = fi y = log (t + 1) + C2 dx t +1 But at t = 0, y = 0 so C2 = 0 Thus y = log (t + 1) fi t = ey – 1.

(1)

15.40

Complete Mathematics—JEE Main

Putting in (1), we have 6x = t2 (2t + 3) = (ey – 1)2 (Zey + 1). 3. Differentiating y2 = 4a(x + a2)

(1)

yy¢ dy = 4a fi a = 2 dx

We have, 2 y Putting in (1) y2 = 4

dy x- y x+ y = sin - sin dx 2 2 x y = -2 cos sin 2 2 dy x fi = -2 cos dx sin y /2 2 x fi 2log tan y/4 = - 4 sin + c 2 y x fi log tan + 2 sin = C 4 2 8. y dx – (x + y2)dy = 0

7.

1 2 2ˆ yy ¢ Ê ÁË x + y y ¢ ˜¯ 2 4

= 2xy y¢ + y3 y¢3 fi

So the degree is 3. 2

2

4. u¢(t) = et u(t) + sin t, v¢(t) = et v(t) + sin t

u¢(t) – v¢(t) = et (u(t ) - v(t )) fi log (u – v) = u(t) – v(t) = C e Ú

2

t Ú e dt + c

fi

x =y+c y

Since y(1) = 1 so c = 0 fi x = y2

2

et dt

so u – v is proportional to a positive function of t. Since u(2) < v(2) so C < 0.

x = 16, when y = 4. 1 + y 2 dx = 0

9. y dy + fi dx +

Hence u(t) < v(t) for all t. 5.

dy + 3y = 2 fi I.F. = e3x. So dx d ye3 x dx

(

fi ye

3x

)

= 2e3x

2 2 = e3 x + C fi y = + Ce -3 x 3 3

lim y( x ) =

x Æ•

6.

dy = dx

= dy

y2

Ê xˆ fi d Á ˜ = dy Ë y¯

Subtracting, we have 2

ydx - xdy

2 . 3

1 - y2 1 -2 y fi dy = x y 2 1 - y2

fi (x + c)2 + y2 = 1, which represent a circle with radius 1 and centre on x x-axis.

1 + y2

dy = 0

fi x + 1 + y2 = c fi (c – x)2 = 1 + y2 fi (x – c)2 – y2 = 1, which represents a hyperbola. 10. Write the given differential equation as 2

Ê xˆ Ê xˆ e–xy d(xy) = Á ˜ d Á ˜ Ë y¯ Ë y¯ Integrating, we get 3

– e–xy =

fi – 1 - y2 = x + c fi 1 – y2 = (x + c)2

y

1Ê xˆ +C 3 ÁË y ˜¯

When y(0) = 1, we get – e0 =

1 (0) + C fi C = –1 3

\ 1 – e–xy =

1Ê xˆ 3 ÁË y ˜¯

3

fi 3y3(1 – e–xy) = x3

CHAPTER SIXTEEN

Cartesian System of Rectangular Coordinates and Straight Lines RESULTS REGARDING POINTS IN A PLANE Distance formula The distance between two points P(x1, y1) and Q(x2, y2) is given by PQ =

( x1 - x2 )2 + ( y1 - y2 ) 2 . The distance from the origin

O (0, 0) to the point P(x1, y1) is OP = Illustration

x12 + y12 .

2

The line x – y + k = 0 passes through the point which divides the segment joining the points (2, 3) and (4, 5) in the ratio 2 : 3. Find the value of k. Solution: Using the section formula, we get the coordinates of the point of division as Ê 2 ¥ 4 + 3 ¥ 2 2 ¥ 5 + 3 ¥ 3 ˆ Ê 14 19 ˆ , Ë ¯ =Ë 5 , 5¯ 5 5 Since it lies on the line x – y + k = 0 14 19 +k=0fik=1 5 5

1

Show that the triangle with vertices (3, 0), (–1, –1) and (2, 4) is isosceles and right angled.h Solution: Let the vertices of the triangle be A (3, 0), B (–1, –1) and C (2, 4) then using distance formula, we have AB =

(3 + 1)2 + (0 + 1)2 = 17

BC =

(-1 - 2 )2 + (-1 - 4 )2 = 9 + 25 = 34

and CA =

Illustration

(2 - 3)2 + (4 - 0 )2 = 17 .

Since AB = AC = 17 , the triangle ABC is isosceles. Also (AB)2 + (AC)2 = (BC)2 shows the triangle is right angled.

Section formula If R(x, y) divides the join of P(x1, y1) and Q(x2, y2) in the ratio m : n (m, n > 0, m π n) then my2 ± ny1 mx2 ± nx1 x= and y = m±n m±n The positive sign is taken for internal division and the negative sign for external division. The mid-point of P(x1, y1) and Q(x2, y2) is Ê x1 + x2 y1 + y2 ˆ , ÁË ˜ which corresponds to internal division 2 2 ¯ when m = n. Note that for external division m π n.

Centroid of a triangle If G(x, y) is the centroid of the triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3), then x=

Illustration

x1 + x2 + x3 3

and

y=

y1 + y2 + y3 3

3

Show that the centroid of the triangle with vertices (5 cos q, 4 sin q), (4 cos q, 5 sin q) and (0, 0) lies on the circle x2 + y2 = 9. Solution: Coordinates of the centroid of the given triangle are, Ê 5 cosq + 4 cosq + 0 4 sin q + 5 sin q + 0 ˆ , Ë ¯ 3 3 = (3cos q, 3sin q ) which lies on the circle x2 + y2 = 9 as (3 cos q )2 + (3 sin q)2 = 9.

Note When A, B, C are taken as vertices of a triangle, it is assumed that they are not collinear.

16.2

Complete Mathematics—JEE Main

Incentre of a triangle

Illustration

If I(x, y) is the incentre of the triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3), then x=

a x1 + bx2 + c x3 a+b+c

and

a y1 + b y2 + c y3 y= a+b+c a, b and c being the lengths of the sides BC, CA and AB, respectively of the triangle ABC.

Illustration

5

If the area of the triangle with vertices at the points whose coordinates are (2, 5), (0, 3) and (4, k) is 4 units, then find the value of k. Solution: We have 2 5 1 1 | 0 3 1 |=4 2 4 k 1 fi 2(3 – k) + 4(5 – 3) = ± 8 fi 7 – k = ± 4 fi k = 3 or 11

4

Find the coordinates of the in center of the triangle with vertices A (–1, 12), B (–1, 0) and C (4, 0). Solution: We have A(–1, 12)

b

c

B(–1, 0)

a

Three points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear if and only if x1 y1 1 x2 y2 1 = 0 x3 y3 1

Slope of a line C(4, 0)

Fig. 16.1 a = BC =

(-1 - 4 )2 + 0 = 5

b = CA =

(4 + 1)2 + (0 - 12 )2 =13 2

Condition of collinearity

2

c = AB = ( -1 + 1) + (12 - 0 ) = 12 So using the formula for the coordinates of the in-center we get the required coordinates as Ê 5(-1) + 13(- 1) + 12(4) 5(12) + 13(0) + 12(0) ˆ , ÁË ˜¯ 5 + 13 + 12 5 + 13 + 12 = (1, 2)

Let A(x1, y1) and B(x2, y2) (x1 π x2) be any two points. Then the slope of the line joining A and B is defined as y - y1 = tan q m= 2 x2 - x1 where q is the angle which the line makes with the positive direction of the x-axis, 0° £ q £ 180°, except at q = 90°. Which is possible only if x1 = x2 and the line is parallel to the y-axis. Condition for the points Zk = xk + iyk (k = 1, 2, 3) to form an equilateral triangle is Z12 + Z 22 + Z 32 = Z1Z2 + Z2Z3 + Z3Z1 (See Complex numbers)

STANDARD FORMS OF THE EQUATION OF A LINE 1. An equation of a line parallel to the x-axis is y = k and that of the x-axis itself is y = 0.

Area of triangle ABC with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is x1 1 | x2 2 x3

y1 1 y2 1 | y3 1

1 | x1 ( y2 – y3) + x2( y3 – y1) + x3( y1 – y2)| 2 and is generally denoted by D. Note that if one of the vertex 1 (x3, y3) is at O(0, 0), then D = | x1 y2 – x2 y1 |. 2 =

Illustration

6

Find the equation of the line parallel to x-axis passing through the intersection of the lines 3ax + 2by +7b = 0 and 3bx – 2ay – 7a = 0,where (a, b) π (0, 0) Solution: Equation of the line parallel to x-axis is y = k. So we need to find the y-coordinates of the point of intersection of the given lines. Multiplying the first equation by b and the second by a and subtracting we get. 2(b2 + a2)y + 7(b2 + a2) = 0 7 which is the required equation. fiy=– 2

Cartesian System of Rectangular Coordinates and Straight Lines 16.3

2. An equation of a line parallel to the y-axis is x = h and that of the y-axis itself is x = 0. 3. An equation of a line passing through the origin and (a) making an angle q with the positive direction of the x-axis is y = x tan q; (b) having a slope m is y = m x; and (c) passing through the point (x1, y1) is x1 y = y1 x. 4. Slope-Intercept form An equation of a line with slope m and making an intercept c on the y-axis is y = m x + c. 5. Point-Slope form An equation of a line with slope m and passing through (x1, y1) is y – y1 = m(x – x1). 6. Two-Point form An equation of a line passing through the points (x1, y1) and (x2, y2) is y - y1 x - x1 = y2 - y1 x2 - x1

x y + =1 a b

c 2

a +b

sin a = ±

2

, cos a = ± b

2

a + b2

a 2

a + b2

and

, the positive sign being taken if

(iv) If p1 denotes the length of the perpendicular from (x1, y1) on this line, then p1 =

7

Find the equation of the straight line passing through the point (10, –7) and making intercepts on the coordinate axes whose sum is 12. x y Solution: Let the equation of the line be + = 1 where a b a + b = 12. Since it passes through (10, – 7),

(iii) p =

c is negative and vice versa.

7. Intercept form An equation of a line making intercepts a and b on the x-axis and y-axis respectively, is

Illustration

10. General form In general, an equation of a straight line is of the form ax + by + c = 0, where a, b and c are real numbers and a and b cannot both be zero simultaneously. From this general form of the equation of the line, we can calculate the following: (i) The slope is –a/b (b π 0). (ii) The intercept on the x-axis is –c/a (a π 0), and the intercept on the y-axis is –c/b (b π 0).

10 7 - =1 a b

fi 10b – 7a = ab fi 10 (12 – a) – 7a = a (12 – a) fi a2 – 29a + 120 = 0 fi a = 5, 24. When a = 5, b = 7; the required equation is

x y + =1 5 7

When a = 24, b = – 12, the required equation is

x y =1 24 12

a x1 + b y1 + c a 2 + b2

(v) The points (x1, y1) and (x2, y2) lie on the same side of the line if the expressions ax1 + by1 + c and ax2 + by2 + c have the same sign, and on the opposite side if they have the opposite signs.

SOME RESULTS FOR TWO OR MORE LINES 1. Two lines given by the equations ax + by + c = 0 and a¢ x + b¢y + c¢ = 0 are (i) parallel (i.e., the slopes are equal), if ab¢ = a¢b (ii) perpendicular (i.e., the product of their slopes is –1), if aa¢ + bb¢ = 0 (iii) identical if ab¢c¢ = a¢ b¢c = a¢c¢ b. (iv) not parallel, then (a) angle q between them at their point of intersection is given by tan q = ±

m - m¢ a ¢b - ab ¢ =± 1 + mm ¢ aa ¢ + bb ¢

8. Parametric form An equation of a line passing through a fixed point A(x1, y1) and making an angle q, 0 £ q £ p, q π p/2 with the positive direction of the x-axis is x - x1 y - y1 = =r cos q sin q

m, m¢ being the slopes of the two lines. (b) the coordinates of their point of intersection are

where r is the distance of any point P(x, y) on the line from the point A(x1, y1). Note that x = x1 + r cos q and y = y1 + r sin q. 9. Normal form An equation of a line such that the length of the perpendicular from the origin on it is p and the angle which this perpendicular makes with the positive direction of the x-axis is a, is x cos a + y sin a = p.

(c) An equation of any line through their point of intersection is (a x + b y + c) + l(a¢x + b¢y + c¢) = 0 where l is a real number.

Ê bc ¢ - b ¢c ca ¢ - c ¢a ˆ ÁË ab ¢ - a ¢b , ab ¢ - a ¢b ˜¯

16.4

Complete Mathematics—JEE Main

2. An equation of a line parallel to the line ax + by + c = 0 is ax + by + c¢ = 0, and the distance between these lines is | c - c¢ | a 2 + b2 Illustration

The line L is given by 2x + by = 7, passes through (8, 3). The line K is parallel to L and has the equation cx + 2y = c. Find the distance between L and K. Solution: Equation of L is 2x – 3y = 7 as it passes through (8, 3) fi slope of L is (2/3). c 2 4 fi slope of K = – = fi c = – . 2 3 3

So the required distance is

7-2 2

2 + (-3)

2

=

13

4. Equations of the bisectors of the angles between two intersecting lines a x + by + c = 0 and a¢x + b¢y + c¢ = 0 are 2

=±

a ¢x + b ¢ y + c ¢ a ¢2 + b¢2

Any point on the bisectors is equidistant from the given lines. If f is the angle between one of the bisectors and one of the lines a x + by + c = 0, such that | tan f | < 1, i.e., – p/4 < f < p/4, then that bisector bisects the acute angle between the two lines, i.e., it is the acute angle bisector of the two lines. The other equation then represents the obtuse angle bisector between the two lines. 5. Equations of the lines through (x1, y1) and making an angle f with the line a x + by + c = 0, b π 0 are tan q - tan f y – y1 = m1 (x – x1) where m1 = 1 + tan q tan f and

y – y1 = m2 (x – x1) where m2 =

If P(4, 7), Q(7, 2) R(a, b) and S(3, 8) are the vertices of a parallelogram then find the value of a and b. Solution: Since the diagonals of a parallelogram bisect each other. Mid points of PR and QS are same. 4 + a 7 + bˆ , so ÊË = ÊË 7 + 3 , 8 + 2 ˆ¯ 2 2 ¯ 2 2 fi a = 6, b = 3

Rhombus

5

3. Three lines a1 x + b1 y + c1 = 0, a2 x + b2 y + c2 = 0 and a3 x + b3 y + c3= 0 are concurrent (intersect at a point) if and only if a1 b1 c1 a2 b2 c2 = 0 a3 b3 c3

ax + b y + c

9

Show that the diagonals AC and BD bisect each other and a pair of adjacent sides, say, AB and BC are equal.

and thus the equation of K is 2x – 3y = 2.

a +b

Show that the diagonals AC and BD bisect each other. Illustration

8

2

Parallelogram

tan q + tan f 1 - tan q tan f

Square Show that the diagonals AC and BD are equal and bisect each other, a pair of adjacent sides say AB and BC are equal.

Rectangle Show that the diagonals AC and BD are equal and bisect each other.

LOCUS OF A POINT To obtain the equation of a set of points satisfying some given condition(s) called locus, proceed as follows. 1. Let P (h, k) be any point on the locus. 2. Write the given condition involving h and k and simplify. If possible draw a figure. 3. Eliminate the unknowns, if any. 4. Replace h by x and k by y and obtain an equation in terms of (x, y) and the known quantities. This is the required locus. Illustration

10

Find the locus of the mid point of the portion between the axes of the line x cos a + y sin a = sin a tan a. Solution: The given line meets x-axis at the point A (tan2 a, 0) and y-axis at the point B (0, tan a) Let P(h, k) be the mid point of AB, then Y

B(0, tana)

where tan q = – a/b is the slope of the given line. Note that m1 = tan (q – f), m2 = tan (q + f) and when b = 0, q = p/2. A(tan2a,0)

O

SOME USEFUL POINTS

Fig. 16.2

To show that A, B, C, D are the vertices of a

2

h=

tan a tana ,k= . 2 2

Cartesian System of Rectangular Coordinates and Straight Lines 16.5 Eliminating a, we get 2k2 = h

So the required locus is 2y2 – x = 0.

CHANGE OF AXES Rotation of Axes If the axes are rotated through an angle q in the anticlockwise direction keeping the origin fixed, then the coordinates (X, Y ) of a point P (x, y) with respect to the new system of coordinate are given by X = x cos q + y sin q and Y = y cos q – x sin q.

Translation of Axes The shifting of origin of axes without rotation of axes is called Translation of axes. If the origin (0, 0) is shifted to the point (h, k) without rotation of the axes then the coordinates (X, Y ) of a point P (x, y) with respect to the new system of coordinates are given by X = x – h, Y = y – k. Illustration

11

If origin is shifted to the point (2, 3) and the axes are rotated through an angle p /4 in the anticlockwise direction, then find the coordinates of the point (7, 11) in the new system of coordinates. Solution: When origin is shifted to the point (2, 3), the coordinates of P(7, 11) are (7 – 2, 11 – 3) = (5, 8). Now when the axes are rotated through an angle p /4 in the anticlockwise direction, the coordinates of P are 13 3 ˆ (5 cos p/4 + 8 sin p/4, 8 cos p/4 – 5 sin p/4.) = ÊÁ , Ë 2 2 ˜¯ which are the required coordinates.

EQUATION OF FAMILY OF LINES A first degree equation ax + by + c = 0 represents a straight line involving three constants a, b and c which can be reduced to two by dividing both the sides of the equation by a non zero constant. For instance if a π 0, we can write the equation as b c x+ y+ =0 a a or x + By + C = 0 So in order to determine an equation of a line we need two conditions on the line to determine these constants. For instance, if we know two points on the line or a point on the line and its slope etc., we know the line. But if we know just one condition, we have infinite number of lines satisfying the given condition. In this case the equation of the line

contains an arbitrary constant and for different values of the constant we have different lines satisfying the given condition and the constant is called a parameter.

Some Equations of Family of Lines 1. Family of lines with given slope (Family of parallel lines) y = mx + k, where k is a parameter represents a family of lines in which each line has slope m. 2. Family of lines through a point y – y0 = k(x – x0), where k is a parameter represents a family of lines in which each line passes through the point (x0, y0). 3. Family of lines parallel to a given line ax + by + k = 0, where k is a parameter represents a family of lines which are parallel to the line ax + by + c = 0. 4. Family of lines through intersection of two given lines a1 x + b1 y + c1 + k(a2 x + b2 y + c2) = 0; where k is a parameter represents a family of lines, in which each line passes through the point of intersection of two intersecting lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. Illustration

12

Two sides of a triangle given by 2x + 3y – 5 = 0 and 5x – 4y + 10 = 0 intersect at A. Centroid of the triangle is at the origin. find the equation of the median through A. Solution: Equation of any line through A is (2x + 3y – 5) + l (5x – 4y + 10) = 0. This will represent the median through A if it passes through the centroid (0, 0) fi – 5 + 10 l = 0 fi l = 1/2 and the required equation is 2(2x + 3y – 5) + (5x – 4y + 10) = 0

fi 9x + 2y = 0 5. Family of lines perpendicular to a given line bx – ay + k = 0, where k is a parameter represents a family of lines, in which each line is perpendicular to the line ax + by + c = 0. 6. Family of lines making a given intercept on axes x y = 1, where k is a parameter, represents a (a) + a k family of lines, in which each line makes an intercept a on the axis of x. x y = 1, where k is a parameter, represents a (b) + k b family of lines, in which each line makes an intercept b on the axis of y. 7. Family of lines at a constant distance from the origin x cos a + y sin a = p, where a is a parameter, represents a family of lines, in which each line is at a distance p from the origin.

16.6

Complete Mathematics—JEE Main

Circumcentre of a triangle The circumcentre of a triangle is the centre of the circle passing through the vertices of the triangle, so it is equi distant from the vertices of the triangle. For example–The circumcentre (x, y) of the triangle with vertices (0, 0), (3, 4) and (4, 3) is given by x2 + y2 = (x – 3)2 + (y – 4)2 = (x – 4)2 + (y – 3)2 Solving, we get (x, y) = (25/14, 25/14) as the required circumcentre of the triangle.

Orthocentre of a triangle The orthocentre of a triangle is the point of intersection of any two altitudes of the triangle. For example–The orthocentre (p, q) of the triangle with vertices (0, 0), (3, 4) and (4, 3) is the point of intersection of the

altitudes y – x = 0 through (0, 0) and 4x + 3y = 24 through Ê 24 24 ˆ (3, 4) which is given by (p, q) = Ë , ¯ 7 7 Note We know from geometry that circumcentre, centroid and orthocentre of a triangle lie on a line.

Ê 25 25 ˆ Ê 7 7ˆ We observe that the centroid Ë , ¯ circumcentre Ë , ¯ 3 3 14 14 Ê 24 24 ˆ and orthocentre Ë , ¯ of the triangle in the above exam7 7 ple lie on the line y = x

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Example 1: The base of an equilateral triangle with side 2a lies along the y-axis such that the midpoint of the base is at the origin if the centroid of the triangle lies on the + ve x-axis, coordinates of the vertex are. (a) (a, 0) (b) (– a, 0) (d) (– a 3 , 0) (c) (a 3 , 0) Ans. (c) Solution: Coordinates of the base BC of the triangle are B (0, a) and C (0, – a) as origin is the mid-point. Since the triangle is equilateral vertex A lies on the perpendicular bisector of the base which is x-axis. Let the coordinates of A be (x, 0).

0

Example 2: If three points (h, 0), (a, b) and (o, k) lie on a line then a b + =1 (a) ah + bk = 1 (b) h k a b (c) ak + bh = 1 (d) + =1 k h Ans. (b) Solution: If given points lie on a line, then h 0 1 a b 1 =0 0 k 1 Expanding along first row h(b – k) + 1. (ak – b ¥ 0) = 0 fi bh + ak = hk a b + =1 fi h k Alternatively: Equation of the line x y joining (h, 0) and (0, k) is + =1 as h k it makes intercepts h on x-axis and k on

Fig. 16.3

Since the triangle is equilateral AB = BC fi x2 + a2 = (2a)2 fi x2 = 3a2 fi x = ± 3a But since the centroid of the triangle lies on positive x-axis, coordinates of the vertex A are (a 3 , 0)

y-axis If the point (a, b) also lies on it. then a b + =1 h k

0, k

0

h, 0

Fig. 16.4

Cartesian System of Rectangular Coordinates and Straight Lines 16.7

Example 3: The line joining the points (2, x) and (3, 1) is perpendicular to the line joining the points (x, 4) and (7, 5). The value of x is (a) 2 (b) 3 (c) 4 (d) 7 Ans. (c)

(x – 7)2 + 62 = (x – 3)2 + 42 fi x = 15/2 and 72 + (y – 6)2 = 32 + (y – 4)2 fi y = 15 So coordinates of P are (15/2, 0) and of Q are (0, 15)

Solution: Slope of the line joining (2, x) and (3, 1) is 1- x 5-4 and that of the line joining (x, 4) and (7, 5) is . 7- x 3-2 Since the lines are perpendicular the product of the slopes is – 1 1- x 1 ¥ =–1 fi 1 7- x

Example 6: The slope of a line is 3 times the slope of the other line and the tangent of the angle between them is 4/13, sum of their slopes is equal to (a) 2 (b) 4 (c) 6 (d) 8 Ans. (d)

fi

x = 4.

Example 4: If the point (a, b) divides a line between the axes in the ratio 2 : 3, the equation of the line is (a) ax + by = 5 (b) bx + ay = 5 2 x 3y (c) + =5 (d) 2 ax + 3by = 5 a b Ans. (c) Solution: Let the line meet x-axis at (h, 0) and y-axis at x y (o, k), then the equation of the line is + = 1 h k (Intercept form)

2

fi

PQ =

15 5 Ê15 ˆ ( )2 Ë 2 ¯ + 15 = 2

Solution: Let the slopes of the given lines be m and 3 m. if q is the angle between them then tan q =

3m - m 4 = 13 1 + (3m)(m)

± 2m ¥ 13 = 4 (1 + 3m2) 6m2 ± 13m + 2 = 0 1 1 fi m= or 2 and m = – or – 2. 6 6 For m = 2, the sum of the slopes is 4m = 8 fi fi

Example 7: A line perpendicular to the line segment joining the points (7, 3) and (3, 7) divides it in the ratio 1 : 3, the equation of the line is (a) x + y – 10 = 0 (b) x – y + 4 = 0 (c) x – y + 2 = 0 (d) x – y – 2 = 0 Ans. (d) 7-3 Solution: Slope of the given line is = – 1 and the 3-7 coordinates of the point which divides the given line segment in the ratio 1 : 3 (7, 3)

Fig. 16.5

Since P(a, b) divides BA in the ratio 2 : 3 we get 2h + 0 ¥ 3 2 ¥ 0 + 3k = a, =b 5 5 5a 5b 2x 3y ,k= and the required equation is + =5 2 3 a b Example 5: P is a point on x-axis, Q is a point on y-axis. Both are equidistant from the points (7, 6) and (3, 4). Distance between P and Q is (b) 15 5 (a) 2 5 fih=

(c) 15 Ans. (c)

5 /2

(d) 5

15 / 2

Solution: Let the coordinates of P be (x, 0) and of Q be (0, y), then

is

1 : 3

(3, 7)

Ê1 ¥ 3 + 3 ¥ 7 1 ¥ 7 + 3 ¥ 3 ˆ = (6, 4) , Ë ¯ 4 4

So equation of the line passing through the point (6, 4) and perpendicular to the given line is y – 4 = x – 6 fi x – y – 2 = 0. Example 8: If the distance between the parallel lines 3 x + 4y + 7 = 0 and ax + y + b = 0 is 1, the integral value of b is (a) 1 (b) 2 (c) 3 (d) 4 Ans. (c) Solution: Since the given lines are parallel. Slopes of the 3 lines are equal so a = and the two lines are 3x + 4y + 7 = 0 4 and 3x + 4y + 4b = 0.

16.8

Complete Mathematics—JEE Main

Distance between them is

7 - 4b 32 + 42

=1

fi 7 – 4b = ± 5 fi b = 1/2 or 3 Example 9: If P and Q are the lengths of the perpendiculars from the origin to the lines x cos q + y sin q = k cos 2 q and x sec q – y cosec q = k respectively then (b) p2 + 4q2 = k2 (a) p2 – 4q2 = k2 (c) 4p2 + q2 = k2 (d) p2 + q2 = k2 Ans. (b) Solution: p =

q=

k cos 2 q cos2 q + sin 2 q k sec 2 q + cos es2 q

= |k cos 2q |

= |k sin q cos q |

= | k (1/2) sin2q | So

p2 + 4q2 = k2 (cos2 2q + sin2 2q) = k2.

Example: 10: If the lines 3x – y + 1 = 0 and x – 2y + 3 = 0 are equally inclined to the line y = mx, then the value of m is given by (b) 7m2 –7m – 2 = 0 (a) 2m2 – 7m – 7 = 0 2 (d) 2m2 – 7m – 2 = 0 (c) 7m – 2m – 7 = 0 Ans. (c) Solution: If q is the angle between the lines then m-3 m - (1 2) = tan q = 1 + 3m 1 + m (1 2 ) fi fi fi

(m – 3) (2 + m) = ± (2m – 1) (3m + 1) m2 – m – 6 = ± (6m2 – m – 1) 7m2 – 2m – 7 = 0 or m2 + 1 = 0

Example 11: Equation of the line which makes an intercept of length 2 on positive x-axis and an intercept of length 3 on the negative y-axis is (a) 2x – 3y + 6 = 0 (b) 3x – 2y – 6 = 0 (c) 2x – 3y – 6 = 0 (d) 3x – 2y + 6 = 0 Ans. (b) x y Solution: Equation of the required line is + =1 2 -3 fi 3x – 2y = 6

Fig. 16.6

Example 12: The perpendicular from the origin to a line L meets it at the point (3, – 9), equation of the line L is (a) x – 3y = 30 (b) 3x – y = 18 (c) x + 3y + 24 = 0 (d) 3x + y = 18 Ans. (a) -9 Solution: Slope of the perpendicular line is =–3 3 fi slope of L is (1/3) and as it passes through (3, – 9) equation of L is y + 9 = (1/3) (x – 3) fi

x – 3y = 30

Example 13: The distance of the point (2, 3) from the line 4x – 3y + 26 = 0 is same as its distance from the line 3x – 4y + p = 0. The value of p can be (a) 5 (b) 25 (c) 31 (d) – 31 Ans. (c) Solution: We have 4 ¥ 2 - 3 ¥ 3 + 26 42 + 32 fi fi fi

=

3¥2- 4¥3 + p 42 + 32

± 25 = 6 – 12 + p p = ± 25 + 6 p = 31 or – 19.

Example 14: If the segment of the line between the lines x – y + 2 = 0 and x + y + 4 = 0 is bisected at the origin, equation of the line is (a) y + 3x = 0 (b) x + 3y = 0 (c) y – 3x = 0 (d) x – 3y = 0 Ans. (c) Solution: Let the required line meet the given lines at (a1, b1) and (a2, b2) respectively. then a1 – b1, + 2 = 0, a2 + b2 + 4 = 0 a1 + a 2 b + b2 = 0, 1 =0 2 2 fi a2 = – 1, b2 = – 3 and a1 = 1, b1 = 3 Equation of the line passing through (1, 3) and (0, 0) is y = 3x fi y – 3x = 0 Also

Example 15: A ray of light passing through the point (3, 7) reflects on the x-axis at a point A and the reflected ray passes through the point (2, 5), the coordinates of A are Ê 29 ˆ Ê1 ˆ (b) Ë , 0 ¯ (a) Ë , 0 ¯ 12 2 1 Ê ˆ Ê 29 ˆ (c) Ë- , 0 ¯ (d) Ë- , 0 ¯ 2 12 Ans. (a)

Cartesian System of Rectangular Coordinates and Straight Lines 16.9 Reflected line ray

Fig. 16.7

Solution: Let P be the given point (3, 7) and the coordinates of A be (x, 0) so, if the ray PA makes an angle q with the positive direction of x-axis, the reflected ray AQ , where

Q is (2, 5), will make the same angle q with the negative direction of x-axis. So the slope of PA = – (the slope of QA) 7 5 =– fi 3- x 2-x fi 7 (2 – x) + 5(3 – x) = 0 29 fi x= 12 Ê 29 ˆ fi coordinates of A are Ë , 0 ¯ . 12

LEVEL 1 Straight Objective Type Questions Example 16: If a vertex of a triangle is (1, 1) and the mid points of two sides through this vertex are (– 1, 2) and (3, 2), then centroid of the triangle is (a) (1, 7/3) (b) (1/3, 7/3) (c) (– 1, 7/3) (d) (– 1/3, 7/3) Ans. (a) Solution: Let (x1, y1) and (x2, y2) be the other two vertix +1 y +1 = – 1, 1 =2 ces of the triangle, then 1 2 2 fi (x1, y1) = (– 3, 3); similarly (x2, y2) = (5, 3) Ê 1 - 3 + 5 1 + 3 + 3ˆ centroid of the triangle is Ë = (1, 7/3) , 3 3 ¯ Example 17: The points (a, b + c), (b, c + a) and (c, a + b) are (a) vertices of an equilateral triangle (b) concyclic (c) vertices of a right angled triangle (d) none of these Ans. (d) Solution: As the given points lie on the line x + y = a + b + c, they are collinear. Example 18: If the lines x + 2ay + a = 0, x + 3by + b = 0 and x + 4cy + c = 0 are concurrent, then a, b, c are in (a) A.P. (b) G.P. (c) H.P. (d) none of these Ans. (c) Solution: Since the given lines are concurrent 1 2a a 1 3b b = 0 fi – bc + 2ac – ab = 0 1 4c c

fi

b=

2ac a+c

fi

a, b, c are in H.P.

Example 19: If p is the length of the perpendicular from x y the origin on the line + = 1 and a2, p2, b2 are in A.P., a b then a4 – 2p2a2 + 2p4 = (a) – 1 (b) 0 (c) 1 (d) none of these Ans (b) ab a 2b2 p2 = 2 fi Solution: We have p = a + b2 a 2 + b2 Also, since a2, p2, b2 are in A.P., fi

a2 + b2 = 2p2 thus a2b2 = p2(a2 + b2) = 2p4 2p4 = a2 (2p2 – a2) fi a4 – 2p2a2 + 2p4 = 0

Example 20: If a, x1, x2 are in G.P. with common ratio r, and b, y1, y2 are in G.P. with common ratio s where s – r = 2, then the area of the triangle with vertices (a, b), (x1, y1) and (x2, y2) is (a) | ab (r2 – 1) | (c) ab (s2 – 1) Ans. (a)

(b) ab (r2 – s2) (d) abrs

Solution: Area of the triangle a b 1 1 1 = | ar bs 1 |= | ab (r - 1)( s - 1)( s - r ) | 2 2 ar 2 bs 2 1 = |ab (r – 1) (r + 1)| = |ab (r2 – 1)| Example 21: The line joining A (b cos a, b sin a) and B (a cos b, a sin b ) is produced to the point M (x, y) so that a +b a +b + y sin = AM : MB = b : a, then x cos 2 2

16.10

Complete Mathematics—JEE Main

(a) – 1 (c) 1 Ans. (b)

(a) 4 : 1 (c) 8 : 3

(b) 0 (d) a2 + b2

(b) 2 : 1 (d) 4 : 3

Ans. (c)

Solution: As M divides AB externally in the ratio b : a x=

b ( a cos b ) - a (b cos a ) and b-a

Solution: Taking the coordinates of vertices O, P, Q, R as (0, 0), (a, 0), (a, a), (0, a) respectively we get the coordinates of M as (a, a/2) and of N as (a /2, a)

b ( a sin b ) - a (b sin a ) y= b-a

R(0, a)

a +b

N (a/2, a)

a -b

2 sin sin x cos b - cos a 2 2 = = y a+b b -a sin b - sin a

fi

2 cos

2

sin

M(a, a/2)

2

a+b a+b + y sin = 0. 2 2 Example 22: If the circumcentre of a triangle lies at the origin and the centroid is the middle point of the line joining the points (a2 + 1, a2 + 1) and (2a, – 2a); then the orthocentre lies on the line (b) y = 2ax (a) y = (a2 + 1)x (c) x + y = 0 (d) (a – 1)2 x – (a + 1)2y = 0 Ans. (d) fi

Solution: We know from geometry that the circumcentre, centroid and orthocentre of a triangle lie on a line. So the orthocentre of the triangle lies on the line joining the Ê (a + 1)2 (a - 1)2 ˆ , circumcentre (0, 0) and the centroid Á 2 ˜¯ Ë 2 (a + 1)2 (a - 1)2 y = x i.e. 2 2 or(a – 1)2x – (a + 1)2 y = 0. Example 23: If p1, p2 denote the lengths of the perpendiculars from the origin on the lines x sec a + y cosec a = 2a and x cos a + y sin a = a cos 2a respectively, 2 Ê p1 p2 ˆ + ˜ is equal to then Á Ë p2 p1 ¯

Solution: p12 + p22 =

and

(b) 4 cos2 4a (d) 4 sec2 4a 4a 2 a 2 cos 2 2a + sec 2a + cosec 2a cos 2 a + sin 2 a

= a2(sin2 2a + cos2 2a) = a2 p21 p22 = a4 sin2 2a cos2 2a = (1/4)a4 sin2 4a

(

2 p12 + p22 Ê p1 p2 ˆ + \Á = Ë p2 p1 ˜¯ p12 p22

)

2

=

4 sin 2 4a

2

= 4 cosec 4a .

Example 24: OPQR is a square and M, N are the middle points of the sides PQ and QR respectively then the ratio of the areas of the square and the triangle OMN is

P(a, 0)

O(0, 0)

x cos

(a) 4 sin2 4a (c) 4 cosec2 4a Ans. (c)

Q (a, a)

Fig. 16.8

\

\

1 Area of the DOMN = | 2

0 0 1 3a 2 a a / 2 1 |= 8 a/2 a 1

Area of the square = a2 the required ratio is 8 : 3.

Example 25: The locus of the point of intersection of the lines x sin q + (1 – cos q ) y = a sin q and x sin q – (1 + cos q ) y + a sin q = 0 is (a) x2 – y2 = a2 (b) x2 + y2 = a2 2 (c) y = ax (d) none of these Ans. (b) Solution: From the given equations we have 1 - cos q 1 + cos q a + x a-x = = and y sin q sin q y Multiplying we get

1 - cos 2 q a 2 - x 2 = fi sin 2 q y2

x2 + y2 = a2

Example 26: Two points (a, 3) and (5, b) are the opposite vertices of a rectangle. If the other two vertices lie on the line y = 2x + c which passes through the point (a, b) then the value of c is (a) – 7 (b) – 4 (c) 0 (d) 7 Ans. (a) Solution: Mid point of the line joining the given points lie on the given line 3+b a + 5ˆ +c = 2 ÊÁ Ë 2 ˜¯ 2 fi 2a + 2c – b + 7= 0 Also since the given line passes through (a, b) b = 2a + c Solving (i) and (ii) we get c = – 7

(i) (ii)

Cartesian System of Rectangular Coordinates and Straight Lines 16.11

Example 27: If every point on the line (a1 – a2)x + (b1 – b2)y = c is equidistant from the points (a1, b1) and (a2, b2) then 2c = (b) a21 + b21 + a22 + b22 (a) a21 – b21 + a22 – b22 (c) a21 + b21 – a22 – b22 Ans. (c)

(d) none of these

Solution: Let (h, k) be any point on the given line then (h – a1)2 + (k – b1)2 = (h – a2)2 + (k – b2)2 fi 2(a1 – a2)h + 2 (b1 – b2)k = a21 + b21 – a22 – b22 fi (a1 – a2)h + (b1 – b2)k = (1/2) (a12 + b21 – a22 – b22) (i) Since (h, k) lies on the given line (a1 – a2)h + (b1 – b2)k = c (ii) 2 2 2 Comparing (i) and (ii) we get c = (1/2) (a 1 + b 1 – a 2 – b22). Example 28: Equations of the straight lines passing through the point (4, 3) and making intercepts on the coordinate axes whose sum is – 1 are (a) x/2 + y/3 = 1 and x/2 + y/1 = 1 (b) x/2 – y/3 = – 1 and x/(– 2) + y/1 = – 1 (c) x/2 + y/3 = –1 and x/(– 2) + y/1 = – 1 (d) x/2 – y/3 = 1 and x /(– 2) + y/1 = 1 Ans. (d) Solution: Let the equation of the line be Since it passes through (4, 3),

x y + = 1. a -1 - a

4 3 + =1 a -1 - a

fi a2 = 4 fi a = ± 2 and the required equations are x y x y + = 1. + = 1 and -2 1 2 -3

Example 30: The line parallel to x-axis passing through the intersection of the lines ax + 2by + 3b = 0 and bx – 2ay – 3a = 0 where (a, b) π (0, 0) is (a) above x-axis at a distance 3/2 from it. (b) above x-axis at a distance 2/3 from it. (c) below x-axis at a distance 3/2 from it. (d) below x-axis at a distance 2/3 from it. Ans. (c) Solution: Eliminating x, we get (2b2 + 2a2)y + 3b2 + 3a2 = 0 fi y = – 3/2 which is the required line and hence below x-axis at a distance 3/2 from it. Example 31: A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is (a) 3x + 4y = 25 (b) x + y = 7 (c) 3x – 4y + 7 = 0 (d) 4x + 3y = 24 Ans. (d) x y a Solution: Let the equation be + = 1 where = 3, a b 2 b =4 2 fi a = 6, b = 8 and the required equation is 8x + 6y = 48 or 4x + 3y = 24 Example 32: Let A(h, k), B(1, 1) and C(2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which k can take is given by (a) {1, 3} (b) {0, 2} (c) {– 1, 3} (d) {– 3, – 2} k

Example 29: If non zero numbers a, b, c are in H.P, x y 1 then the straightline + + = 0 always passes through a b c a fixed point. That point is (a) (1, – 2) (b) (1, – 1/2) (c) (– 1, 2) (d) (– 1, – 2) Ans. (a) Solution: a, b, c are in H.P. fi

1 1 b a

=

1 1 c b

1 2 1 - + =0 a b c which shows that the given line passes through the point (1, – 2).

fi

Fig 15.9

Ans. (c) Solution: Equation of BC is y = 1. AB is perpendicular to BC through B(1, 1) so in equation is x = 1 fi h = 1. 1 AB ¥ BC = 1 Area of the DABC = 2 fi fi

AB = 2 fi |k – 1| = 2 k = – 1 or 3

16.12

Complete Mathematics—JEE Main

Example 33: Let A(2, – 3) and B(– 2, 1) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line (a) 3x + 2y = 5 (b) 2x – 3y = 7 (c) 2x + 3y = 9 (d) 3x – 2y = 3 Ans. (c) Solution: Let C(h, k) be the vertex, then the centroid h + 2 - 2 k - 3 + 1ˆ , is ÊÁ ˜ i.e. (h/3, (k – 2)/3) which lies on Ë 3 3 ¯ 2x + 3y = 1 h 3(k - 2) fi 2 + =1 3 3 fi 2h + 3k = 9 and the locus of (h, k) is 2x + 3y = 9.

Example 37: A square of side a lies above the x-axis and has one vertex at the origin. The side passing through the origin makes an angle a (0 < a < p /4) with the positive direction of x-axis. The equation of its diagonal not passing through the origin is (a) y (cos a + sin a) + x (sin a – cos a) = a (b) y (cos a + sin a) + x (sin a + cos a) = a (c) y (cos a + sin a) + x (cos a – sin a) = a (d) y (cos a – sin a) – x (sin a – cos a) = a Ans. (c) B C

Example 34: If the sum of the slopes of the lines given by x2 – 2cxy – 7y2 = 0 is four times their product, then the value of c is (a) 2 (b) – 1 (c) 1 (d) – 2 Ans. (a) Solution: If m1, m2 are the slopes, then m1 + m2 = – 2c/7, m1 m2 = – 1/7 m1 + m2 = 4 m1 m2 fi c=2 Example 35: Locus of mid point of the portion between the axes of x cos a + y sin a = p, where p is constant is (a) x2 + y2 = 4/p2 (b) x2 + y2 = 4p2 (c) 1/x2 + 1/y2 = 2/p2 (d) 1/x2 + 1/y2 = 4/p2 Ans. (d) Solution: If (h, k) is the mid-point, then h = p/2 cosa, k = p/2 sina so ( p/2h)2 + ( p/2k)2 = cos2 a + sin2 a = 1 fi 1/h2 + 1/k 2 = 4/p2 Locus of (h, k) is 1/x 2 + 1/y 2 = 4/p2 Example 36: The lines x + y = a and ax – y = 1 intersect each other in the first quadrant. Then the set of all possible values of a is in the interval. (a) (0, •) (b) (1, •) (c) ( – 1, •) (d) ( – 1, 1) Ans.(b) Solution: Adding x + y = lal and ax – y = 1 a +1 We obtain (1 + a) x = a + 1 fi x = ,aπ–1 a +1 a a -1 Ê a + 1ˆ –1= Also y = a Á Ë a +1 ˜¯ a +1 As x > 0, y > 0, a + 1 > 0, we get a + 1 > 0 and a a – 1 > 0 fi a > – 1 and a a > 1 As a ≥ 0, and a a > 1, we get a > 0 and thus a2 > 1 or a > 1

0

a

A

Fig. 16.10

Solution: Coordinates of A are (a cos a, a sin a) and of C are (a cos (a + p/2), a sin (a + p/2)) i.e. (– a sin a, a cos a) So the equation of the diagonal AC is a(sin a - cos a ) (x – a cos a) y – a sin a = a(cos a + sin a ) or y (cos a + sin a) + x (cos a – sin a) = a (sin2 a + sin a cos a – sin a cos a + cos2 a) =a Example 38: Locus of the centroid of the triangle whose vertices are (a cos t, a sin t), (b sin t, – b cos t) and (1, 0); where t is a parameter is (a) (3x – 1)2 + (3y)2 = a2 + b2 (b) (3x + 1)2 + (3y)2 = a2 + b2 (c) (3x + 1)2 + (3y)2 = a2 – b2 (d) (3x – 1)2 + (3y)2 = a2 – b2 Ans. (a) Solution: If (h, k) is the centroid, then h=

a cos t + b sin t + 1 , 3

k =

a sin t - b cos t + 0 3

fi (3h – 1)2 + (3k)2 = (a cos t + b sin t)2 + (a sin t – b cos t)2 = a2 + b2 Locus of (h, k) is (3x – 1)2 + (3y)2 = a2 + b2

Cartesian System of Rectangular Coordinates and Straight Lines 16.13

Example 39: If x1, x2, x3 and y1, y2, y3 are both in G.P. with the same common ratio, then the points (x1, y1), (x2, y2) and (x3, y3) (a) lie on an ellipse (b) lie on a circle (c) are vertices of a triangle (d) lie on a straight line Ans. (d) Solution: Let x2 = x1 r, x3 = x1 r 2 y2 = y1 r, y3 = y1 r 2 The given points lie on the line x1 y – y1 x = 0 for all values of r. Example 40: Q, R and S are the points on the line joining the points P (a, x) and T (b, y) such that PQ = QR = RS = Ê 5a + 3b 5 x + 3 y ˆ ST, then Á , ˜ is the mid point of the segment Ë 8 8 ¯ (a) PQ (c) RS Ans. (b)

(b) QR (d) ST

Ê 5a + 3b 5 x + 3 y ˆ Solution: The point L Á , ˜ divides PT Ë 8 8 ¯ in the ratio 3 : 5 and hence is the middle point of QR.

1 1 1ˆ Ê + + Á a + b + g a b g ˜ Ê 3 p ab + bg + ga ˆ , ˜= Á , Á ˜¯ 3 3 3abg ˜ Ë 3 Á ¯ Ë = (p, q) [∵ a + b + g = 3p, ab + bg + ga = 3q, abg = 1] Example 43: The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the inter1 1 1 1 cepts of the line L on the new axes, then 2 - 2 + 2 - 2 a p b q is equal to (a) – 1 (b) 0 (c) 1 (d) none of these Ans. (b) Solution: Equation of the line L in the two coordinate x y X Y + =1 systems is + = 1, p q a b where (X, Y) are the new coordinates of a point (x, y) when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed. 1 1 1 1 1 1 = fi + 2 = 2+ 2 2 p q a b 1 1 1 1 + + a 2 b2 p2 q2 or

Example 41: If a, b, c form a G.P. with common ratio r, the sum of the ordinates of the points of intersection of the line ax + by + c = 0 and the curve x + 2y2 = 0 is (a) – r2/2 (b) – r/2 (c) r/2 (d) r2/2 Ans. (c) Solution: The equation of the given line is ax + by + c = 0 (i) fi ax + ary + ar2 = 0 fi x + ry + r2 = 0 2 (i) intersects the curves x + 2y = 0 at the points whose ordinates are given by – 2y2 + ry + r2 = 0 or 2y2 – ry – r2 = 0 Therefore required sum of the ordinates = r/2. Example 42: If a, b, g are the real roots of the equation x3 – 3px2 + 3qx – 1 = 0, then the centroid of the triangle with 1 Ê 1ˆ Ê 1ˆ vertices ÊÁ a , ˆ˜ , Á b , ˜ and Á g , ˜ is at the point Ë a¯ Ë b¯ Ë g¯ (a) (p, q) (c) (p + q, p – q) Ans. (a)

(b) (p/3, q/3) (d) (3p, 3q)

Solution: The centroid of the given triangle is the point

1 1 1 1 = 0. + a 2 p 2 b2 q 2

Example 44: Lines ax + by + c = 0 where 3a + 2b + 4c = 0 a, b, c ΠR are concurrent at the point. (a) (3, 2) (b) (2, 4) (c) (3, 4) (d) (3/4, 1/2) Ans. (d) Solution: 3a + 2b + 4c = 0 3 1 a+ b+c=0 4 2 fi ax + by + c passes through (3/4, 1/2) for all values of a, b, c. fi

Example 45: If P (1, 0), Q (– 1, 0) and R (2, 0) are three given points. The point S satisfies the relation SQ2 + SR2 = 2SP2. The locus of S meets PQ at the point (a) (0, 0) (b) (2/3, 0) (c) (– 3/2, 0) (d) (0, – 2/3) Ans. (c) Solution: Let S be the point (x, y) then (x + 1)2 + y2 + (x – 2)2 + y2 = 2[(x – 1)2 + y2 ] fi 2x + 3 = 0, the locus of S and equation of PQ is y = 0. So the required points is (– 3/2, 0).

16.14

Complete Mathematics—JEE Main

Example 46: If algebraic sum of distances of a variable line from points (2, 0), (0, 2) and (– 2, – 2) is zero, then the line passes through the fixed point (a) (– 1, – 1) (b) (0, 0) (c) (1, 1) (d) (2, 2) Ans. (b)

Since (3, 4) lies on the locus, we get 9 + 16 = a2 + b2 i.e. a2 + b2 = 25 Also, (a, b) lies on 3x – 4y = 0 so 3a – 4b = 0 fi b = (3/4) a From (i), a2 + (9/16)a2 = 25 fi a2 = 16 | a + b| = 7 So that | a + b|2 = (7/4)2 a2 = 49

Solution: Let the equation of the variable line be ax + by + c = 0 then according to the given condition

Example 49: Equations to the sides of a triangle are x – 3y = 0, 4x + 3y = 5 and 3x + y = 0. The line 3x – 4y = 0 passes through the (a) incentre (b) centroid (c) circumcentre (d) orthocentre of the triangle Ans. (d)

2a + c

2

+

2b + c 2

2

+

-2a - 2b + c

=0 a +b a +b a 2 + b2 fi c =0 which shows that the line passes through (0, 0) for all values of a and b. 2

Example 47: An equation of a straight line passing through the inter-section of the straight lines 3x – 4y + 1 = 0 and 5x + y – 1 = 0 and making non-zero, equal intercepts on the axes is (a) 22x + 22y = 13 (b) 23x + 23y = 11 (c) 11x + 11y = 23 (d) 8x – 3y = 0 Ans. (b) Solution: Equation of any line through the point of intersection of the given lines is (3x – 4y + 1) + k (5x + y – 1) = 0 (1) or

(3 + 5k)x + (k – 4)y + 1 – k = 0 x y + =1 or (k - 1) /(3 + 5k ) (k - 1) /(k - 4) Since x-intercept = y-intercept k -1 k -1 fi = fi (k – 1) (3 + 5k – k + 4) = 0 3 + 5k k - 4 fi k = 1 or k = – 7/4 For k = 1, (1) becomes 8x – 3y = 0 which makes zero intercepts on the axes. \ k = – 7/4 fi The required equation is 4(3x – 4y + 1) – 7(5x + y – 1) = 0 fi 23x + 23y = 11. Example 48: If the point (3, 4) lies on the locus of the point of intersection of the lines x cos a + y sin a = a and x sin a – y cos a = b (a is a variable), the point (a, b) lies on the line 3x – 4y = 0 then |a + b| is equal to (a) 1 (b) 7 (c) 12 (d) 5 Ans. (b) Solution: Squaring and adding the given equations of the lines we get x2 + y2 = a2 + b2 as the locus of the point of intersection of these lines.

(i) (ii)

Solution: Sides x – 3y = 0 and 3x + y = 0 of the triangle being perpendicular to each other, the triangle is right angled at the origin, the point of intersection of these sides. So that origin is the orthocentre of the triangle and the line 3x – 4y = 0 passes through this orthocentre. Example 50: Locus of the mid-points of the intercepts between the coordinate axes by the lines passing through (a, 0) does not intersect (a) x-axis (b) y-axis (c) y = x (d) y = a Ans. (b) Solution: Equation of any line through (a, 0) be x y + = 1 , where b is a parameter. This line meets y-axis a b at (0, b) and if (h, k) denotes the mid-point of the intercept of the line between the coordinate axes, then h = a/2, k = b/2 and then the locus of (h, k) is x = a/2. This clearly does not intersect y-axis. Example 51: If h denotes the arithmetic mean and k denotes the geometric mean of the intercepts made on the coordinate axes by the lines passing through the point (1, 1), then the point (h, k) lies on (a) a circle (b) a parabola (c) a straight line (d) a hyperbola Ans. (b) x y Solution: Let the equation of the line be + = 1 If it a b 1 1 passes through (1, 1), then + = 1 a b fi a + b = ab (i) Since the line makes intercepts of lengths a and b on the coordinate axes, we have a+b h= and k = ab 2

Cartesian System of Rectangular Coordinates and Straight Lines 16.15

so that from (i) we get 2h = k2 and thus the point (h, k) lies on 2x = y2 which is a parabola. Example 52: If a, b, c are in A.P., a, x, b are in G.P. and b, y, c are in G.P., the point (x, y) lies on (a) a straight line (b) a circle (c) an ellipse (d) a hyperbola Ans. (b) Solution: We have 2b = a + c, x2 = ab, y2 = bc so that x + y2 = b(a + c) = 2b2 which is a circle. Example 53: If one of the diagonals of a square is along the line x = 2y and one of its vertices is (3, 0), then its side through this vertex nearer to the origin is given by the equation. (a) y – 3x + 9 = 0 (b) 3y + x – 3 = 0 (c) x = 3y – 3 = 0 (d) 3x + y – 9 = 0 Ans. (b)

(a) (3, 1) (c) (3, 4) Ans. (a)

Solution: Let the coordinates of the other vertices lying on x = 3 be (3, y1) and (3, y2). Since the diagonals of a rectangle are equal and bisect each other, we have y1 + y2 5+2 = fi y1 + y2 = 7 2 2

2

Solution: The point (3, 0) does not lie on the diagonal x = 2y. Let the equation of a side through the vertex (3, 0) be y – 0 = m (x – 3) Since the angle between a side and a diagonal of a square is p /4, we have m - 1/ 2 2m - 1 p = ± tan = (i) 1 + m(1/ 2) 2 + m 4 fi m = 3, – 1/3 Thus the equation of a side through (3, 0) is 1 y = 3(x – 3) or y = ÊÁ - ˆ˜ (x – 3) and the one nearer to the Ë 3¯ origin is 3y + x –3 = 0 Example 54: A line which is parallel to x-axis and crosses the curve y = (a) x = 1/ 4 (c) y = 1/2 Ans. (c)

x at angle of 45º is (b) y = 1/4 (d) y = 1

tween the tangent to the curve y = x at (k2, k) and the line y = k is 45º. The line y = k being parallel to x-axis. We have È dy ˘ = + 1 on the curve y = x . ÍÎ dx ˙˚ 2 (k , k ) fi

È 1 ˘ =+1 ÍÎ 2 x ˙˚ 2 (k , k ) k > 0, as y =

fi

|y2 – y1|=

and

k=

1 2

x lies in the first quadrant

Example 55: A rectangle has two opposite vertices at the points (1, 2) and (5, 5). If the other vertices lie on the line x = 3, the coordinates of the vertex nearer the axis of x are

(5 - 1) 2 + (5 - 2) 2 = 16 + 9 = 5

If y2 > y1 then y2 = 6, y1 = 1 So that the coordinates of the vertex nearer the axis of x are (3, 1). Example 56: A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively. The point O divides the segment PQ in the ratio (a) 1 : 2 (c) 2 : 1 Ans. (b)

(b) 3 : 4 (d) 4 : 3

Solution: It is clear that the lines lie on opposite side of the origin O. Let the equation of any line through O be x y . If OP = r1 and OQ = r2 then the coordinates = cos q sin q of P are (r1 cos q, r1 sin q ) and that of Q are (– r2 cos q, – r2 sin q ) Since P lies on 4x + 2y = 9, 2r1(2 cos q + sin q) = 9 and Q lies on 2x + y + 6 = 0, – r2 (2 cos q + sin q) = – 6 so that

Solution: Equation of a line parallel to x-axis is y = k. Which meets the curve y = x at the point (k2, k). Since the line y = k crosses the curve at an angle of 45º, the angle be-

(b) (3, 2) (d) (3, 6)

r1 9 3 = = r2 12 4

and the required ratio is thus 3 : 4. Alternately Let the equation of the line through O be y = mx, then coordinates of P and Q are 9m ˆ Ê -6 -6m ˆ Ê 9 , , and Á so that respectively Á Ë 2 + m 2 + m ˜¯ Ë 4 + 2m 4 + 2m ˜¯ OP 9 |2 + m| 3 = = ¥ 6 4 OQ | 4 + 2m | Example 57: Let P = (– 1, 0), Q = (0, 0) and R = (3, 3 3 ) be three points. Then the equation of the bisector of the angle PQR is (a) (c) Ans. (c)

3 +y=0 2 3x + y = 0

(b) x +

3y = 0 3 (d) x + y =0 2

16.16

Complete Mathematics—JEE Main

Solution Let the equation of QS, the bisector of angle PQR be y = mx. Slope of QR = fi

3 = tan 60º

PQR = 120º fi PQS = 60º

fi m = – tan 60º = – 3 and thus the required equation of the bisector is

3x + y = 0.

Example 60: A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O in the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is (a) – 2 (b) – 1/2 (c) – 1/4 (d) – 4 Ans. (a) Solution: Equation of the line be y = mx + c. Since it passes through (1, 2), 2 = m + c. P (0, c), Q (– c/m, 0). If A is the area of the triangle, then A=

2 c 1 c2 m2 - 4m + 4 c = = (2 - m ) = m 2 2m 2m 2m m 2 = -2+ 2 m

2 dA 1 - 2 = fi m2 = 4 fi m = ± 2 =0fi m dm 2 Fig. 16.11

Example 58: The number of integer value of m, for which the x-coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer is (a) 2 (b) 0 (c) 4 (d) 1 Ans. (a) Solution: x-coordinate of the point of intersection is given by 3x + 4 (mx + 1) = 9 5 fi x= 3 + 4m As x is an integer, 3 + 4m must be a divisor of 5 so 3 + 4m = ± 1 or ± 5 fi m = – 1 or – 2 (considering the integer value only) Example 59: If the line 2x + y = k passes through the point which divides the line segment joining the points (1,1) and (2,4) in the ratio 3:2, then k equals (a) 6 (b) 11/5 (c) 29/5 (d) 5 Ans. (a) Solution: Let A (1,1) and B(2,4) If P (x,y) divides the line segment AB in the ratio 3:2, then 3(2) + 2(1) 3(2) + 2(1) x= ,y= 5 5 P (8/5, 14/5) As the line 2x + y = k passes through P 2 × (8/5) + (14/5) = k fi k = 6

Example 61: Let PQR be a right angled isosceles triangle right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equations representing the pair of lines PQ and PR is (a) x + 3y = 5, 3x – y = 5 (b) x + y = 3, x–y=1 (c) 2x + 3y = 7, 3x – 2y = 4 (d) 3x + y = 7, x – 3y = –1 Ans. (a) Solution: Let the slopes of PQ and PR be m and – 1/m respectively. Since PQR is an isosceles triangle PQR = PRQ 1 +2 m fi 2 1+ m [∵ slope of QR = – 2] fi m + 2 = ± (1 – 2m) fi m = 3 or – 1/3 So the equations of PQ and PR are (y – 1) = 3(x – 2) and y – 1 = (– 1/3) (x – 2) m+2 = 1 - 2m

-

R 2x

+y

=3

P(2, 1)

Q

Fig. 16.12

fi

3x – y = 5, x + 3y = 5

Cartesian System of Rectangular Coordinates and Straight Lines 16.17

Example 62: The orthocentre of the triangle formed by the lines xy = 0 and 2x + 3y – 5 = 0 is (a) (2, 3) (b) (3, 2) (c) (0, 0) (d) (5, – 5) Ans. (c) Solution: The given triangle is right angled at (0, 0) which is therefore the orthocentre of the triangle. Example 63: The equations of the sides of a parallelogram are x = 2, x = 3 and y = 1, y = 5. Equations to the pair of diagonals are (a) 4x + y = 7, 4x – y = 13 (b) x + 4y = 13, x – 4y = 7 (c) x + 4y = 7, x – 4y = 13 (d) 4x + y = 13, 4x – y = 7 Ans. (d) Solution: The vertices of the parallelogram are A(2, 1), B(2, 5), C(3, 5) and D(3, 1) Equation of the diagonal AC is 5 -1 y –1= (x – 2) fi 4x – y = 7 3-2 Equation of the diagonal BD is 1- 5 y–5= (x – 2) fi 4x + y = 13 3-2 Example 64: The lines joining the origin to the points of intersection of 3x2 + lxy – 4x + 1 = 0 and 2x + y – 1 = 0 are at right angles for (a) l = – 4 (b) l = 4 (c) l = 7 (d) all values of l Ans. (d) Solution: Equation of the lines joining the origin to the points of intersection of the given lines is 3x2 + lxy – 4x(2x + y) + 1◊(2x + y)2 = 0 (Making the equation of the pair of lines homogeneous with the help of the equation of the line) fi x2 – lxy – y2 = 0 which are perpendicular for all values of l as the product of the slopes is – 1. Example 65: The lines p(p2 + 1)x – y + q = 0 and (p + 1)2x + (p2 + 1)y + 2q = 0 are perpendicular to a common line for (a) exactly two values of p (b) more than two values of p (c) no value of p (d) exactly one value of p Ans. (d) 2

Solution: Two lines will be perpendicular to a common line if these two are parallel.

\

p(p2 + 1) = -

( p 2 + 1) 2 p2 + 1

fi (p + 1) (p2 + 1) = 0 fi p = –1 x y + = 1 passes 5 b through the point (13, 32). The line K is parallel to L and x y = 1. Then the distance between has the equation + c 3 L and K is Example 66: The line L is given by

(a) 17/ 15

(b) 23/ 17

(c) 23/ 15

(d)

15

Ans. (b)

x y Solution: As L : + = 1 passes through (13, 32), 5 b we get 13 32 = 1 fi b = – 20 + b 5 x y As K : + = 1 is parallel to L c 3 b 3 3 =– fic=– . – 5 c 4 Now equation of L and K are respectively 4x – y – 20 = 0 and 4x – y + 3 = 0 and the distance between them is | -20 - 3 | 23 = . 16 + 1 17 Example 67: Consider three points P(cos a, sin b), Q(sin a, cos b) and R(0, 0). Where 0 < a, b < p/4, Then (a) P lies on the line segment RQ (b) Q lies on the line segment PR (c) R lies on the line segment QP (d) P, Q, R are non-collinear. Ans. (d) cos a sin b 1 Solution: Consider D = sin a cos b 1 0 0 1 = cos a cos b – sin a sin b = cos (a + b) π 0 as 0 < a + b < p/2 Hence the given points are non-collinear. Example 68: Consider the lines given by L1 : x + 3y – 5 = 0, L2 : 3x – ky – 1 = 0 L3 : 5x + 2y – 12 = 0. If one of L1, L2, L3 is parallel to at least one of the other two, then k satisfies (a) k2 + 4k – 45 = 0 (b) 5k2 + 51k + 45 = 0 (c) 5k2 – 19k – 6 = 0 (d) 5k2 + 51k + 54 = 0 Ans. (d) Solution: L1 and L3 are not parallel.

16.18

Complete Mathematics—JEE Main

-1 3 = fik=–9 3 k -5 3 -6 = fik = L2 is parallel to L3 if . 2 k 5 So the required values of k are given by (k + 9) (k + 6/5) = 0 or 5k2 + 51k + 54 = 0

= 10 2 + 30 + 20 + 20 2 + 20 + 30

L2 is parallel to L1 if

= 100 + 30 2

Example 70: If a line through P(a, 2) making an angle p/4 with the positive direction of x-axis, meets the curve 4x2 + 9y2 = 36 at A and D and meets the axes at B and C, so that PA, PB, PC, PD are in G.P, then a is equal to

Example 69: Consider the lines L1 : x + y = 10, L2 : x + y = 60 L4 : y = 40. L3 : x = 40, L1 meets x-axis and y-axis at A and B respectively. L4 meets y-axis at C and L2 at D L3 meets L2 at E and x-axis at F. Perimeter of the hexagon ABCDEF is (a) 100 + 30 2 (c) 100 Ans. (a)

(b) 50 + 40 2 (d) none of these

Solution:

(a)

2 13

(b)

13 2

(c)

13 5

(d)

5 13

Ans. (b)

Solution: Equation of any line through P(a, 2) making x-a an angle q = p/4 with the +ve direction of x-axis is 1/ 2 y-2 = . Any point on this line at a distance r from P is 1/ 2 (a + r/ 2 , 2 + r/ 2 ). If it lies on the curve, then

(

4 a +r/ 2

)

2

(

+ 9 2+r/ 2

2

)

= 36

This gives two values of r representing PA and PD 4 a 2 ¥ 2 8a 2 . = 4+9 13 Since it meets the axes at B and C So that

PA ◊ PD =

So

PB = – a 2 , PC = – 2 2 PB ◊ PC = 4a

As PA, PB, PC, PD are in G.P PA ◊ PD = PB ◊ BC 8a 2 13 = 4a fi a = . 13 2

fi

Fig. 16.13

From the figure it is clear that OF = 40, OA = 10 fi AF = 30 Similarly BC = 30. Coordinates of D are (20, 40) so CD = DG = 20 Similarly GE = EF = 20 AB =

102 + 102 = 10 2

ED =

(20) 2 + (20) 2 = 20 2

Hence the required perimeter is AB + BC + CD + DE + EF + FA

Example 71: The quadrilateral formed by the lines y = ax + c, y = ax + d and y = bx + c and y = bx + d has area 18. The quadrilateral formed by the lines y = ax + c, y = ax – d, y = bx + c and y = bx – d has area 72. If a, b, c, d are positive integers then the minimum value of the sum a + b + c + d is (a) 13 (b) 14 (c) 15 (d) 16. Ans. (d) Solution: Coordinates of A are (0, c) and of C are (0, d) So Length of AC = |c – d| Length of the perpendicular from B on AC (x = 0) is equal to c-d the x-coordinate of B. Thus area of this quadrilateral a-b is equal to

(c - d )2

|a-b|

= 18

Cartesian System of Rectangular Coordinates and Straight Lines 16.19

Solution: x + 2y – 3 = 0, 2x + 3y – 4 = 0 intersect at ( – 1, 2). 4x + 5y – 6 = 0 passes through this point but 3x + 4y – 7 = 0 does not pass through it, so only three of them are concurrent

Fig. 16.14

Similarly area of the other quadrilateral is

(c + d )2

= 72 |a-b| As a, b, c, d, are positive integers minimum value of |a – b| = 2 c+d =±2 c-d fi c = 3d or d = 3c. We can take a = 3 or b = 1 or a =1, b = 3 and c = 3, d = 9 or d = 3, c = 9. In all the cases a + b + c + d = 3 + 1 + 3 + 9 = 16. Example 72: A straight line L through the point (3, – 2) is inclined at an angle 60° to the line 3 x + y = 1. If L also intersects the x-axis, then the equation of L is (a) y +

3x+2–3 3 =0

(b) y –

3x+2+3 3 =0

(c)

3y–x+3+2 3 =0

(d) Ans. (b)

3y+x–3+2 3 =0

Solution: Let the slope of L be m m+ 3

then

1 - 3m

fi

m+

3 =

or

m+

3 =–

= tan 60∞ = 3

3 (1 –

3 m)

3 (1 –

3 m)

fi m = 0 or m = 3 but m π 0 as L intersects x- axis Hence equation of L is y + 2 = 3 (x – 3) y–

3 x+2+3

Ê 1 7 ˆ , ˜ (a) ÁË 2 2¯ Ê 1 7 ˆ , ˜ (c) Á Ë 2 2¯ Ans. (c)

(b)

(-

(d)

(

2, 7 2 ) 2, 7 2 )

Solution: Reflection of the point (4, 1) about the line y = x is (1,4) which is transfered to the point (3,4) at a distance 2 units along the positive direction of x- axis. Finally a rotation through an angle p/4 takes it to (3 cos (p/ 4) – 4 sin (p/4)), (3sin (p/4) + 4 cos (p/4)) Ê -1 , 7 ˆ = Ë 2 2¯ Example 75: If the sum of the distances of a point from two perpendicular lines is 1 then its locus is (a) square (b) circle (c) straight line (d) two intersecting lines. Ans. (a) Solution: Let us take the two perpendicular line as the axes of coordinates. If (h, k) is any point of the locus, then h + k =1 Locus is | x | + | y | = 1 This consists of four line segment x + y = ± 1, x – y = ± 1 which form a square y

3 =0

y=

1

1

-

x+

y=

y=

Example 73: Given four lines with equations x + 2y – 3 = 0, 3x + 4y – 7 = 0, 2x + 3y – 4 = 0 4x + 5y – 6 = 0 (a) they are all concurrent (b) they are sides of a qudrilateral (c) only three lines are concurrent (d) none of these Ans. (c)

x+

x+

or

Example 74: The point (4, 1) undergoes the following three transformations successively. (i) reflection about the line y = x (ii) transformation through a distance 2 units along the positive direction of x- axis (iii) rotation through an angle p/4 about the origin in the counter clockwise direction. Then the final position of the point is given by the coordinates

-1

x-

y=

Fig. 16.15

-1

x

16.20

Complete Mathematics—JEE Main

Example 76: The diagonals of a parallelgram PQRS are along the lines x + 3y = 4 and 6x – 2y = 7. Then PQRS must be a (a) rectangle (b) square (c) cyclic quadrilateral (d) rhombus Ans.(d) Solution: Since the diagonals of the parallelogram are at right angles, it is a rhombus. Example 77: Let A0 A1 A2 A3 A4 A5 be a regular hexagon inscribed in a circle of unit radius then the product of the lengths of the line segments A0A1, A0A2 and A0A4 is (a) 3/4

(b) 3 3 (d) 3 3 / 2

(c) 3 Ans. (c)

Solution: The vertices of the hexagon be as shown in the figure 1 3 (A0 A1)2 = + = 1 4 4 9 3 + =3 4 4 (A0 A4)2 = (A0 A2)2 = 3 (A0 A2)2 =

So (A0 A1) (A0 A2) (A0 A4) =

1¥ 3 ¥ 3 = 3 y

A2 - 1 2

3 2

A1

A3 (-1, 0)

1 2

3 2

A0 (1, 0)

A4 -1 - 3 2 2

x

A5 - 1 + 3 2 2

Fig. 16.16

Example 78: The incentre of the triangle with vertices (1,

3 ) (0,0) and (2,0) is (a) (1, 3 / 2)

(c) (2 3 , 3 / 2 ) Ans. (d)

Solution: Let q be the angle which the ray 2x – 3y + 5 = 0 makes with the line x + y = 2 then tan q =

2/3 +1 1 - 2 / 3 (1)

= |5| = 5 Reffected ray makes angle – q with x + y = 2, so if the slope of the reflected ray is m then

m +1 = tan(– q ) = – 5 1- m

fi m = 3/2. Also, the reflected ray passes through (1/5, 9/5), the point of intersection of the ray and the line Hence equation of the reflected ray is (y – 9/5) = 3/2 (x – 1/5) fi 3x – 2y + 3 = 0 Example 80: Let 0 < a < p/2 be a fixed angle. If P = (cos q, sin q ) and Q = (cos (a – q ), sin (a – q )) Then Q is obtained from P by (a) clockwise rotation around origin through an angle a (b) anti clockwise rotation around origin through an angle a (c) reflection in the line through the origin with slope tan a (d) reflection in the line through the origin with slope tan a /2 Ans. (d) Solution: Clockwise rotation of P through an angle a takes it to the point (cos (q – a), sin (q – a)) and anticlockwise takes it to (cos (a + q ), sin (a + q )) Now slope of PQ

(b) (2 3 , 1 / 3 )

=

sin q - sin (a - q ) cos q - cos(a - q )

=

2 cos(a / 2)sin (q - a / 2) - 2 sin(a / 2)sin (q - a / 2)

(d) (1, 1 / 3 )

Solution: Length of each side of the triangle is 2 so it is an equilaltral triangle and hence it incentre is the centroid (1, 1 / 3 ) of the triangle.

Example 79: A ray of light travels along the line 2x – 3y + 5 = 0 and strikes a plane mirror lying along the line x + y = 2. The equation of the straight line containing the reflected ray is (a) 2x – 3y + 3 = 0 (b) 3x – 2y + 3 = 0 (c) 21x – 7y + 1 = 0 (d) 21x + 7y – 1= 0 Ans. (b)

= – cot (a / 2) fi PQ is perpendicular to the line with slope tan (a / 2) Hence Q is the reflection of P in the line through the origin with slope tan (a / 2) .

Cartesian System of Rectangular Coordinates and Straight Lines 16.21

Assertion-Reason Type Questions

Example 81: Consider the lines L1 : p2x + py – 1 = 0 L2 : q2x + qy + 6 = 0 L1 passes through the point (3, 2) and L2 passes through the point (2, 7) Statement-1: If the product of the slopes of L1 and L2 is 2, then they intersect at the point (– 4, – 5) Statement-2: L1 and L2 are neither parallel nor perpendi cular. Ans. (b) Solution: we have 3p2 + 2p – 1 = 0 and 2q2 + 7q + 6 = 0 1 -3 fi p = – 1 or and q = – 2 or 3 2 Slope of L1 = – p, slope of L2 = – q, product of the slopes = pq π – 1 for any values of p and q. So L1 and L2 are not perpendicular. Also as p π q for any value, they are not parallel. Thus the statement-2 is true. Next, if the product of the slope is 2 then p = – 1, q = – 2 and the equations of L1 and L2 are respectively x – y – 1 = 0 and 4x – 2y + 6 = 0 which intersect at (– 4, – 5) so the statement-1 is also true but does not follow from statement-1. Example 82: Statement-1: Consider the point A(0, 1) and B(2, 0) and P be a point on the line 4x + 3y + 9 = 0, then the coordinates of P such that |PA – PB| is maximum is (– 24/5, 17/5) Statement-2: If A and B are two fixed points and P is any point in a plane, then |PA – PB| £ AB. Ans. (a) Solution: From geometry |PA – PB| < AB and |PA – PB| = AB if P lies on extended line segment AB so statement-2 is true. Using in statement-1, |PA – PB| is maximum if P lies on AB whose equation is x + 2y = 2 The given line is 4x + 3y + 9 = 0, on solving we get the required coordinates of P as (– 24/5, 17/5) and hence statement-1 is also true. Example 83: Statement-1: The points A(3, 4), B(2, 7), C(4, 4) and D(3, 5) are such that one of them lies inside the triangle formed by the other points. Statement-2: Centroid of a triangle lies inside the triangle. Ans. (a) 3+ 2+ 4 4+7+4 Solution: = 3, =5 3 3 fi D(3, 5) is the centroid of the triangle ABC.

Using statement-2 which is true, statement-1 is also true. Example 84: Statement-1: If the circumcentre of a triangle lies at the origin and centroid is the mid point of the line joining the points (2, 3) and (4, 7), then its orthocentre lies on the line 5x – 3y = 0 Statement-2: Circumcentre, centroid and orthocentre of a triangle lie on the same line Ans. (a) Solution: From geometry, statement-2 is True. Using it in statement-1, orthocentre lies on the line joining (0, 0) 4 + 2 7 + 3ˆ , and ÊÁ ˜ i.e. (3, 5) which is 5x – 3y = 0 and so Ë 2 2 ¯ the statement-1 is also true. Example 85: Statement-1: The points A(– 2, 2), B(2, – 2) and C(1, 1) are the vertices of an obtuse angled isosceles triangle. Statement-2: Every obtuse angled triangle is isosceles Ans. (c) Solution: Equation of AB is y + x = 0 in statement-1 and y = x is the perpendicular bisector of AB. C(1, 1) lies on it so the triangle ABC is isosceles, AC = BC slope of AC = – 1/3, slope of AB = – 1 tan A =

-1 3 + 1 1 = 1+1 3 2

fi

A < p /4.

(A cannot be obtuse as A = B) fi A + B < p/2 fi C > p/2 and the triangle is obtuse angled. So statement-1 is True. Statement-2 is false as a triangle with angles 120°, 40°, 20° is obtuse angled but not isosceles. Example 86: Statement-1: The lines (a + b)x + 2(a – b)y = 2a are concurrent at the point (1, 1/2) Statement-2: The lines x + 2y – 2 = 0 and x – 2y = 0 intersect at the point(1, 1/2) Ans. (a) Solution: Statement-2 is true. Lines in statement-1 can be written as a(x + 2y – 2) + b(x – 2y) = 0 which passes through the point of intersection(1, 1/2) of the lines x + 2y – 2 = 0, x – 2y = 0, follows from statement-1 Example 87: Statement-1: Lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect the line L3 : y + 2 = 0 at P and Q respectively. The bisector of the acute angle between L1 and L2 intersect L3 at R. Statement-1: The ratio PR : RQ equals 2 2 : 5 .

16.22

Complete Mathematics—JEE Main

Statement-2: The bisector of an angle of a triangle divides the triangle into two similar triangles. Ans. (c) Solution: Vertices of the triangle in statement-1 are O(0, 0), P(– 2, – 2), Q(1, – 2). Since the bisector of an angle of a triangle divides the opposite side. In the ratio of the side containing the angle. PR : RQ = OP : OQ = 2 2 : 5 . So statement-1 is true Statement-2 is false, because for example if A = 50°, B = 60°, C = 70° then bisector of angle B does not divide the triangle into two equiangular and hence similar triangles. Example 88: Statement-1: If the perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept equal to – 4, then k2 – 16 = 0 Statement-2: Centroid of an isoceles triangle ABC lies on the perpendicular bisector of the base BC where B = C. Ans. (b) Solution: Any point L(x, y) on the perpendicular bisector in statement-1 is equidistant from P and Q. Locus of L is (x – 1)2 + (y – 4)2 = (x – k)2 + (y – 3)2 fi 2(k + 1)x – 2y = k 2 – 8 k2 - 8 = – 4 fi k2 – 16 = 0 2 So statement-1 is true but does not follow from statement-2 which is also true, as the perpendicular bisector of BC is also the median through A. y-intercept = -

Example 89: Statement-1: Circumcentre of the triangle formed by the lines x + y = 0, x – y = 0 and x – 7 = 0 is (7, 0) Statement-2: Circumcentre of a triangle lies inside the triangle. Ans. (c) Solution: In statement-1, the triangle is right angled with hypotenuse x – 7 = 0 and the vertices of the hypotenuse are (7, 7) and (7, – 7), circumcentre is the mid-point (7, 0) of the hypotenuse. So statement-1 is True. Statement-2 is false as the circumcentre of an obtuse angled triangle lies outside the triangle and that of the right angled is on the hypotenuse. Example 90: Statement-1: Equation of the pair of lines through the origin perpendicular to the pair of lines 7x2 – 55xy – 8y2 = 0 is 8x2 – 55xy – 7y2 = 0 Statement-2: Equation of the pair of lines through the origin perpendicular to the pair to lines ax2 + 2hxy + by2 = 0 is bx2 – 2hxy + ay2 = 0 Ans. (a) Solution: Statement-2 is True because if m1 and m2 are

the slopes of ax2 + 2hxy + by2 = 0, then m1 + m2 = m1 m2 =

a . b

2h , b

1 1 2h 1 b + == . Equation of the lines and m1 m2 a m1m2 a

fi

with slope -

1 1 ˆ 1 1 Ê 1 and is y 2 + Á xy + + ˜ Ë ¯ m1 m2 m1 m2 m1m2

x2 = 0 2h b xy + x 2 = 0 fi bx 2 – 2hxy + ay 2 = 0 a a Using it, statement-1 is also true.

fi y2 –

Example 91: Statement-1: The area enclosed with in the curve x + y = 1 is 2 square units. Statement-2: The curve x + y = 1 is a square of each side

2 units

Ans. (a) Solution: The statement-2 is true as it represents the sides x + y = ± 1 and x – y = ± 1, of a square of each side 2 units which shows that the statement-1 is also true Example 92: x1 y1 1

a1 b1 1

Statement-1 If x2 y2 1 = a2 b2 1 x3 y3 1

a3 b3 1

then the two triangle with vertices (x1,y1), (x2,y2), (x3,y3) and (a1,b1), (a2,b2), (a3,b3) must be congruent Statement-2 Two congruent triangles have the same area Ans. (d) Solution: Statement-2 is true, but two triangles having the same area may not be congruent for example, the triangles with vertices (0, 0), (4, 0), (4, 2) and (0, 0), (2, 0) and (1, 4) have same area but are not congruent and thus statement-1 is false. Example 93: Let q1 be the angle between two lines 2x + 3y + c1 = 0 and – x + 5y + c2 = 0 and q2 be the angle between two lines 2x + 3y + c1 = 0 and – x + 5y + c3 = 0 where c1, c2, c3 are any real number. Statement 1: If c1 and c2 are proportional then q1 = q2 Statement 2: q1 = q2 for all c2 and c3. Ans. (a) Solution: Lines – x + 5y + c2 = 0 and – x + 5y + c3 = 0 are parallel for all values of c2 and c3 so both will make the same angle with the line 2x + 3y + c1 = 0 and q1 = q2 for all c2 and c3 showing that statement-2 is true and hence statement-1 is also true.

Cartesian System of Rectangular Coordinates and Straight Lines 16.23

Example 94: L1 : x – y + 1 + l1 (2x – y – 2) = 0 and L2 : 5x + 3y – 2 + l2 (3x – y – 4) = 0 where l1, l2 are arbitrary numbers. Statement 1: 5x – 2y – 7 = 0 is the equation of a line belonging to both the family of lines L1 and L2. Statement 2: There is only one line belonging to both the family of lines L1 and L2 Ans. (a) Solution: L1 represent the lines passing through the point (3, 4), the point of intersection of x – y + 1 = 0 and 2x – y – 2 = 0 and L2 represents the lines passing through the point (1, – 1), the point of intersection of the lines 5x + 3y – 2 = 0 and 3x – y – 4 = 0. So the line belonging to both the families will be the only one line joining (3, 4) and (1, – 1) which is 4 +1 (x – 1) y+1= 3 -1

or 5x – 2y – 7 = 0. So statement-2 is true and is a correct explanation for statement-1 to be true. Example 95: Statement-1: The line 2x + y + 6 = 0 is perpendicular to the line x – 2y + 5 = 0 and second line passes through (1, 3) Statement-2: Product of the slopes of any two parallel lines is equal to – 1 Ans. (c) Solution: Statement-2 is false as the product of the slopes of two perpendicular line is – 1. Product of the slopes 1 = – 1 so they are perof the lines in statement-1 is – 2 ¥ 2 pendicular and the second line passes through (1, 3) so the statement-1 is true.

LEVEL 2 Straight Objective Type Questions Example 96: If a, b, c are unequal and different from 1 3

2

3

2

Ê a a - 3ˆ Ê b b - 3ˆ , , , such that the points Á and Ë a - 1 a - 1 ˜¯ ÁË b - 1 b - 1 ˜¯ Ê c3 c 2 - 3 ˆ ÁË c - 1 , c - 1 ˜¯ , are collinear, then (a) (b) (c) (d) Ans. (d)

bc + ca + ab + abc = 0 a + b + c = abc bc + ca + ab = abc bc + ca + ab – abc = 3 (a + b + c)

Solution: Suppose the given points lie on the line lx + my + n = 0 then a, b, c are the roots of the equation. lt3 + m(t2 – 3) + n(t – 1) = 0 3 or lt + mt2 + nt – (3m + n) = 0 fi a + b + c = – m/l bc + ca + ab = n/l abc = (3m + n)/l Eliminating l, m, n, we get abc = – 3(a + b + c) + bc + ca + ab fi bc + ca + ab – abc = 3(a + b + c)

Example 97: If two vertices of a triangle are (–2, 3) and (5, –1), orthocentre lies at the origin and centroid on the line x + y = 7, then the third vertex lies at (a) (7, 4) (b) (8, 14) (c) (12, 21) (d) none of these Ans. (d) Solution: Let O(0, 0) be the orthocentre; A(h, k) the third vertex; and B(– 2, 3) and C(5, – 1) the other two vertices. Then the slope of the line through A and O is k/h, while the line through B and C has the slope – 4/7. By the property of the orthocentre, these two lines must be perpendicular, so we have Ê k ˆ Ê - 4ˆ = – 1 fi k = 7 (i) ˜ ÁË ˜¯ ÁË 7¯ h h 4 5 - 2 + h -1 + 3 + k + =7 Also 3 3 fi h + k = 16 (ii) Which is not satisfied by the points given in (a), (b) or (c). Example 98: The points A (2, 3); B (3, 5), C (7, 7) and D (4, 7) are such that (a) ABCD is a parallelogram (b) A,B,C and D are collinear (c) D lies inside the triangle ABC (d) D lies on the boundary of the triangle ABC

16.24

Complete Mathematics—JEE Main

Ans. (d) Solution: A, B, C are not collinear as slope of BA π slope of BC slope of AD = slope of BD fi D lies on the side AB of the triangle ABC Example 99: If p, x1, x2 … xi, … and q, y1, y2 …, yi, … are in A.P., with common difference a and b respectively, then the centre of mean position of the points Ai (xi, yi), i = 1, 2 … n lies on the line (a) ax – by = aq – bp (b) bx – ay = ap – bq (c) bx – ay = bp – aq (d) ax – by = bq – ap ÈNote. Centre of Mean Position Ê Sxi , Syi ˆ ˘ ˜ ÁË n n ¯ ˚˙ ÎÍ

fi fi

x > 8/15 y < – 8/5.

Example 101: The area enclosed by 2|x| + 3|y| £ 6 is (a) 3 sq. units (b) 4 sq. units (c) 12 sq. units (d) 24 sq. units Ans. (c) Solution: The given inequality is equivalent to the following system of inequalities. 2x + 3y £ 6, when x ≥ 0, y ≥ 0 2x – 3y £ 6, when x ≥ 0, y £ 0 – 2x + 3y £ 6, when x £ 0, y ≥ 0 – 2x – 3y £ 6, when x £ 0, y £ 0 which represents a rhombus with sides 2x + 3y = 6 and 2x + 3y = – 6

Ans. (c)

B

Solution: Let the coordinates of the centre of mean 2x

position of the points Ai, i = 1, 2, … n be (x, y), then x1 + x2 +  + xn y + y2 +  + yn x= ,y= 1 n n fi

n p + a (1 + 2 +  + n) , x= n y=

n q + b (1 + 2 +  + n) n

fi

x=p+

fi

x=p+

fi

n ( n + 1) n ( n + 1) a, y = q + b 2n 2n

n +1 n +1 a,y=q+ b 2 2 ( x - p) ( y - q) 2 =2 fi bx – ay = bp – aq. a b

Example 100: If P is a point (x, y) on the line. y = – 3x such that P and the point (3, 4) are on the opposite sides of the line 3x – 4y = 8, then (a) x > 8/15, y < – 8/5 (c) x = 8/15, y = – 8/5 Ans. (a)

(b) x > 8/5, y < – 8/15 (d) none of these

Solution: Let k = 3x – 4y – 8 then the value of k at (3, 4) = 3 × 3 – 4 × 4 – 8 = – 15 < 0 \ For the point P (x, y) we should have k > 0 fi 3x – 4y – 8 > 0 fi 3x – 4(– 3x) – 8 > 0 [∵ P (x, y) lies on y = – 3x]

C

2x

-

= 3y

2x

-6

+3

2

y=

6

3 O

+3

y=

A

-6

2x

y -3

=6

D

Fig. 16.17

Length of the diagonals is 6 and 4 units along x-axis and y-axis. \ The required area 1 = × 6 × 4 = 12 sq. units. 2 Example 102: Let O be the origin, A (1, 0) and B (0, 1) and P (x, y) are points such that xy > 0 and x + y < 1, then (a) P lies either inside the triangle OAB or in the third quadrant (b) P cannot lie inside the triangle OAB (c) P lies inside the triangle OAB (d) P lies in the first quadrant only Ans. (a) Solution: Since xy > 0, P either lies in the first quadrant or in the third quadrant. The inequality x + y < 1 represents all points below the line x + y = 1. So that xy > 0 and x + y < 1 imply that either P lies inside the triangle OAB or in the third quadrant. Example 103: A line has intercepts a and b on the coordinate axes. When the axes are rotated through an angle a, in

Cartesian System of Rectangular Coordinates and Straight Lines 16.25

the anti clockwise direction keeping the origin fixed, the line makes equal intercepts on the coordinate axes, then tan a = a+b a-b (b) (a) a-b a+b 2 2 (c) a – b (d) none of these Ans. (b) Solution: Let the equation of the line in the original x y coordinate system be + = 1 . If (X, Y ) denote the coora b dinates of any point P (x, y) in the new coordinate system obtained by rotation of the axes through an angle a, then x = X cos a – Y sin a, y = X sin a + Y cos a So that the equation of the line with reference to new system of coordinates is X cos a - Y sin a X sin a + Y cos a + =1 a b or

cos a sin a ˆ Ê cos a - sin a ˆ = 1 + X ÊÁ ˜ ˜¯ + Y ÁË Ë a b b a ¯

Since it makes equal intercepts on the coordinates axes. b cos a + a sin a = a cos a – b sin a fi (a – b) cos a = (a + b) sin a fi

tan a

=

a-b a+b

Example 104: The equations to the sides of a triangle are x – 3y = 0, 4x + 3y = 5 and 3x + y = 0. The line 3x – 4y = 0 passes through the (a) incentre (b) centroid (c) circumcentre (d) orthocentre of the triangle Ans. (d) Solution: Two sides x – 3y = 0 and 3x + y = 0 of the triangle being perpendicular to each other, the triangle is right angled at the origin, the point of intersection of these sides. So that origin is the orthocentre of the triangle and the line 3x – 4y = 0 passes through this orthocentre. Example 105: A line through P(1, 2) is such that it makes unequal intercepts on the axes, and the intercept between the axes is trisected at P, an equation of the line is (b) 4x + y = 6 (a) x + y = 3 (c) 4x – y = 6 (d) 3x + 2y = 1 Ans. (b) x y Solution: Let the equation of the line be + = 1 , a b which meets x-axis at A (a, 0) and y-axis at B (0, b). Then the coordinates of the points which divide AB in the ratio 2a b a 2b 1 : 2 and 2 : 1 are respectively ÊÁ , ˆ˜ and ÊÁ , ˆ˜ Ë3 3 ¯ Ë 3 3¯

2a b Now if ÊÁ , ˆ˜ represents P (1, 2) then a = 3/2, b = 6 and Ë 3 3¯ the required equation of the line is

2x y + = 1 fi 4x + y = 6 3 6

which is given in (b) a 2b If ÊÁ , ˆ˜ represents P (1, 2) then a = b = 3 and the equaË3 3 ¯ tion of the line is x + y = 3 which makes equal intercepts on the axes and so we do not consider this. Example 106: The point (4, 1) undergoes the following successive transformations: (i) reflection about the line y = x (ii) translation through a distance 2 units along the positive x-axis. then, the final coordinates of the point are (a) (4, 3) (b) (3, 4) (c) (1, 4) (d) (4, 4) Ans. (b) Solution: Let Q (x, y) be the reflection of P(4, 1) about the line y = x, then mid-point of PQ lies on this line and y +1 x + 4 = PQ is perpendicular to it. So we have and 2 2 y -1 = – 1. x-4 fi x – y = – 3 and x + y = 5 fi x = 1, y = 4 Therefore reflection of (4, 1) about y = x is (1, 4). Next, this point is shifted 2 units along the positive x-axis, the new coordinates are (1 + 2, 4 + 0) = (3, 4) Example 107: Two adjacent sides of a parallelogram are 4x + 5y = 0 and 7x + 2y = 0. If an equation to one of the diagonals is 11x + 7y – 9 = 0, then an equation of the other diagonal is (b) 7x – 11y = 0 (a) x + y = 0 (c) x – y = 0 (d) none of these Ans. (c) Solution: Since the given lines intersect at the origin O, one of the vertex is O (0, 0). Let A and B be the points of intersection of the sides 4x + 5y = 0 and 7x + 2y = 0 respectively with the diagonal. 11x + 7y – 9 = 0, then the coordinates of A 2 7 5 -4 and B are respectively ÊÁ , ˆ˜ and ÊÁ - , ˆ˜ the coordinates Ë 3 3¯ Ë3 3 ¯ 1 1 of the mid point of AB are ÊÁ , ˆ˜ . Since the other diagonal Ë 2 2¯ 1 1 passes through the vertex (0, 0) and the mid-point ÊÁ , ˆ˜ Ë 2 2¯ of AB, its equation is y = x.

16.26

Complete Mathematics—JEE Main

Example 108: If a line joining two points A (2, 0) and B (3, 1) is rotated about A in anticlockwise direction through an angle 15º, then equation of the line in the new position is 3x +y= 2 3

(a)

(c) x +

3x –y= 2 3

(b)

3y = 2 3

3y = 2 3

(d) x –

Ans. (b) Solution: Slope of AB = fi

1- 0 =1 3- 2

–BAX = 45º (Ref. Fig. 16.18)

Fig. 16.19

Y

\ Equation of the line through (a, b) perpendicular to AC is a y–b= (x – a) b

C B (3, 1)

fi

15°

fi ± 2bx – by = b2 from (i)

45°

X

A (2, 0)

O

± 2x – y = b

If AC is the new position of the line AB then –CAX = 45° + 15º = 60º. and thus its equation is

3 (x – 2)

fi

3x - y = 2 3

Example 109: If the medians AD and BE of the triangle with vertices A(0, b), B (0, 0) and C (a, 0) are perpendicular, then an equation of the line through (a, b) perpendicular to AC is (a) y =

2x + b

(b) y = – 2x + b

(c) y =

2x – b

(d) x =

Slope of BE =

2b b-0 =0 - a/2 a

lying in the positive quadrant such that OAn = nOAn – 1, O (2520 2 , 2520 2 ), then n = (a) 5 (c) 7 Ans. (c)

(b) 6 (d) 8

Solution: We have OAn = n. OAn–1 = n (n–1) OAn–2 … = n (n – 1) … 2(OA1) = n! OA1 = n! 2 (2520 2 ) = n! 2520 × 2 = n! 7 × 6 × 5 × 4 ×3 × 2 = n! n=7 Example 111: If x1, x2, x3 are the abcissa of the points

b/ 2 b = a /2 a

Since AD and BE are perpendicular -2b b . = -1 a a fi a2 = 2b2 Slope of AC is – b/a

2x – b

Example 110: A1, A2 … An are points on the line y = x

fi fi fi fi

2y – a

Ans. (c) Solution: Slope of AD =

y = ± 2x – b

being the origin. If OA1 = 1 and the coordinates of An are

y = tan 60º (x – 2) y=

fi

So an equation of required line is y =

Fig. 16.18

fi

ax – by = a2 – b2

(i)

A1, A2, A3 respectively where the lines y = m1 x, y = m2 x, y = m3 x meet the line 2x – y + 3 = 0 such that m1, m2, m3 are in A.P., then x1, x2, x3 are in (a) A.P. (b) G.P. (c) H.P. (d) none of these Ans. (c) 3 3 3 Solution: We have x1 = , x2 = , x3 = m1 - 2 m2 - 2 m3 - 2 fi

m - 2 + m3 - 2 m + m3 - 4 1 1 = 1 = 1 + 3 3 x1 x3

Cartesian System of Rectangular Coordinates and Straight Lines 16.27

Solution: The intercepts between the axes made by the

2 2m2 - 4 = x2 3 (as m1, m2, m3 are in A.P.) x1, x2, x3 are in H.P. =

fi

given lines are a 2 , ar So the required sum = =

Example 112: If the lines x = k; k = 1, 2 º, n meet the line y = 3x + 4 at the points Ak(xk, yk), k = 1, 2, º, n then the ordinate of the centre of Mean position of the points Ak, k = 1, 2, º, n is n +1 3n + 11 (a) (b) 2 2 3 ( n + 1) (c) (d) none of these 2 Ans. (b) Solution: We have yk = 3k + 4, the ordinate of Ak, the point of intersection of x = k and y = 3x + 4. So the ordinate of the centre of Mean position of the points Ak, k = 1, 2, º n is 1 n

n

 yk k =1

1 = n

n

Â

k =1

3 (3k + 4) = n

n

Âk

+4

2

Example 115: The distance between the parallel lines given by (x + 7y)2 + 4 2 (x + 7y) – 42 = 0 is (a) 4/5

(b) 4 2

(c) 2 Ans. (c)

(d) 10 2

Solution: The lines given by the equation are (x + 7y – 3 2 ) (x + 7y + 7 2 ) = 0 fi

x + 7y – 3 2 = 0

the family of lines representing the sides of an equilateral triangle with one vertex at the origin, then the product of the slopes of all the lines of this family is (a) a3

(b) a(a2 – 3)

(c) a (1 – 3a2)

(d)

(

)

a a2 - 3 1 - 3a

2

Ans. (d) Solution: Let a = tan q, then the slopes of the lines making an equilateral triangle with one vertex at the origin are: tan q, tan (q + 60∞), tan (q + 120∞), tan (q + 180∞) = tan q So the product of the slopes is tan q tan (q + 60∞) tan (q + 120∞)

distance between these lines

= a◊

).

Example 114: The sum of the intercepts cut off by the axes on the lines x + y = a, x + y = ar, x + y = ar2, º where a π 0 and r = 1/2 is (a) 2a (c) 2 2a Ans. (c)

=

(

7 2 - -3 2

)

12 + 7 2 10 2 5 2

= 2.

Example 116: If the lines joining the origin to the intersection of the line y = mx + 2 and the curve x2 + y2 = 1 are at right angles, then (b) m2 = 3 (a) m2 = 1 (d) 2m2 = 1 (c) m2 = 7 Ans. (c) Solution: Joint equation of the lines joining the origin and the point of intersection of the line y = mx + 2 and the curve x2 + y2 = 1 is y - mx ˆ 2 x2 + y2 = ÊÁ Ë 2 ˜¯ fi x2 (4 – m2) + 2mxy + 3y2 = 0 Since these lines are at right angles 4 – m2 + 3 = 0 fi m2 = 7. Example 117: If one of the lines given by the equation

a a2 - 3

◊ = 1 - 3a 2 1- a 3 1+ a 3

x + 7y + 7 2 = 0

=

= tan q tan (q + 60∞) tan (q – 60∞)

(

and

k =1

3n ( n + 1) 3n + 11 +4= . n◊2 2 Example 113: If y = a x is one of the lines belonging to

a- 3

2 [a + a r + ar2 + º]

2a a = = 2 2a . 1- r 1 - (1 2)

=

a+ 3

2 , ar2 2 , º

(b) a 2 (d) a/ 2

2

2x + axy + 3y2 = 0 coincide with one of those given by 2x2 + bxy – 3y2 = 0 and the other lines represented by them be perpendicular, then (a) a = – 5, b = 1 (b) a = 5, b = – 1 (c) a = 5, b = 1 (d) none of these Ans. (c) Solution: Let and

2 2 a x + x y + y2 = (y – mx) (y – m¢x) 3 3

2 2 b 1 x + x y + y2 = ÊÁ y + xˆ˜ (y – m¢x) Ë m ¯ -3 -3

16.28

Complete Mathematics—JEE Main

then

fi

2 a m + m¢ = - , mm¢ = 3 3

(i)

2 1 -b m¢ – m¢ = , = 3 m 3 m

(ii)

m2 = 1fi m = ± 1 2 fi a = – 5, b = – 1 If m = 1, m¢ = 3 If m = – 1, m¢ = -

2 fi a = 5, b = 1. 3

Example 118: If the equation ax3 + 3bx2y + 3cxy2 + dy3 = 0 (a, b, c, d, π 0) represents three coincident lines then (a) a = d (b) b = c a b c = = (d) ac = bd (c) b c d Ans. (c)

6 ± 36 - 16 = 3± 5 2 which gives two real values of d. fi

d=

Example 120: The line x + y = 1 meets the lines represented by the equation y3 – xy2 – 14x2y + 24x3 = 0 at the points A, B, C. If O is the point of intersection of the lines represented by the given equation then OA2 + OB 2 + OC2 is equal to (a) 22/9 (b) 85/72 (c) 181/72 (d) 221/72 Ans. (d) Solution: The given cubic can be written as (y – 2x) (y – 3x) (y + 4x) = 0 \ The three lines given by this equation are y = 2x, y = 3x and y = – 4x, they intersect at O (0,0) and meet the line x + y = 1 at the points A (1/3, 2/3), B (1/4, 3/4) and C (–1/3, 4/3)

Solution: Let the given equation represent three coincident lines y – mx = 0, then 3c 3b 2 a 3 y3 + y 2 x + yx + x = (y – mx)3 d d d

fi

= y3 – 3y2 (mx) + 3y(mx)2 – m3x3 b a c m3 = - , m2 = , m = d d d a b c = - = b c d

fi

m=–

fi

a b c = = . b c d Example 119: Two of the lines represented by x3 – 6x2y

+ 3xy2 + dy3 = 0 are perpendicular for (a) all real values of d (b) two real values of d (c) three real values of d (d) no real value of d Ans. (b) Solution: Let m1, m2, m3 be the slopes of the three lines represented by the given equation such that m1 m2 = – 1 1 1 We have m1 m2 m3 = - so that m 3 = d d Since y = m3 x i.e. x = dy satisfies the given equation, we get d3 – 6d2 + 3d + d = 0 fi d(d2 – 6d + 4) = 0 If d = 0, the given equation represents the lines x = 0 and x2 – 6xy + 3y2 = 0 which are not perpendicular. \ d π 0 and d2 – 6d + 4 = 0

\

OA2 + OB2 + OC 2 =

5 10 17 221 + + = 9 16 9 72

Example 121: A line passing through the point P (2, 3) meets the lines represented by x2 – 2xy – y2 = 0 at the points A and B such that PA ◊ PB = 17, the equation of the line is (a) x = 2 (b) y = 3 (c) 3x – 2y = 0 (d) none of these Ans. (b) Solution: Let the equation of the line through P(2, 3) making an angle q with the positive direction of x-axis be x-2 y-3 = . cosq sinq Then the coordinates of any point on this line at a distance r from P are (2 + r cos q, 3 + r sin q). If PA = r1 and PB = r2, then r1, r2 are the roots of the equation. (2 + r cos q )2 – 2(2 + r cos q) (3 + r sin q ) – (3 + r sin q )2 = 0 fi r 2 (cos 2q – sin 2q ) – 2r (cos q + 5 sin q) – 17 = 0 17 fi 17 = PA . PB = r1 r2 = cos 2q - sin 2q fi cos 2q – sin 2q = 1 which is satisfied by q = 0 and thus the equation of the line is y = 3. Example 122: If (l, 2) is an interior point of the triangle formed by the lines x + y = 4, 3x – 7y = 8 and 4x – y = 31, then (a) 2 < l < 22/3 (b) 2 < l < 33/4 (c) 22/3 < l < 33/4 (d) l > 9 Ans. (c) Solution: The point (l, 2) lies on the line y = 2 which meets the line x + y = 4 at a point outside the triangle and the line 3x – 7y = 8 at D = (22/3, 2) and the line 4x – 3y = 31 at E = (33/4, 2).

Cartesian System of Rectangular Coordinates and Straight Lines 16.29

So x-coordinate of the point on the line y = 2, lying inside the triangle lies between 22/3 and 33/4. y

A

B

=8 31 7y = y 3 3x y = 2 E 4x -

D

(a) (1, – 1) (c) (1, 1)

Solution: Let the equation of the line be ax + by + c = 0 2a + c 2b + c a+b+c then + + =0 2 2 2 2 a +b a +b a 2 + b2

C x+

y=

(b) ( – 1, 1) (d) ( – 1, – 1)

Ans. (c)

x

O

Example 124: Let the algebraic sum of the perpendicular distances from the points (2, 0) (0, 2) and (1, 1) to a variable straight line be zero, then the line passes through a fixed point whose coordinates are

fi 3a + 3b + 3c = 0 fi a+b+c=0 So the equation of the line can be ax + by – (a + b) = 0 fi a(x – 1) + b (y – 1) = 0 which passes through (1,1)

4

Fig. 16.20

Example 123: The lines given by L1: x + 3y – 5 = 0, L2: 3x – k y – 1 = 0 L3: 5x + 2y – 12 = 0 form a triangle if k is (a) 5 (b) 5/6 (c) – 6/5 (d) – 9 Ans. (b) Solution: L1 and L3 intersect at (2,1) The lines will form a triangle if no two of them are parallel and all of them are not concurrent. Lines are concurrent at (2,1) if k = 5 L2 is parallel to L1 if k = – 9 L3 is parallel to L1 if k = – 6/5 So the lines form a triangle for any value of k other than 5, – 9, – 6/5

Example 125: The locus of the point P(h, k), when the area of the triangle formed by the lines y = x, x + y = 2 and the line through P (h,k) and parallel to the x-axis is 4h2 is (a) x + 2y – 1 = 0 (c) 2x – y – 1 = 0 Ans. (b)

(b) 2x + y – 1 = 0 (d) x – 2y + 1 = 0

Solution: Let the line through P (h, k) be y = k Vertices of the triangle are (1,1), (k,k), (2 – k, k) 1 1 1 1 2 k k 1 fi 4h2 = (k – 1)2 4h = 2 2 -k 1 = 2h = ± (k – 1) Hence locus of P is 2x ± (y – 1) = 0 i.e 2x + y – 1 = 0 or 2x – y + 1 = 0

EXERCISE Concept-based Straight Objective Type Questions 1. A is a point on the positive x-axis at a distance 3 units from the origin and B is a point on the positive y-axis at a distance 4 units from the origin. If P divides AB in the ratio 1 : 2, the coordinates of P are (a) (1, 8/3) (b) (2, 4/3) (c) 8/3, 1) (d) (4/3, 2) 2. The slopes of the line which passes through the origin, and the mid-point of the line segment joining the points. P(3, – 4) and Q (– 5, – 2) is (a) 3 (b) – 3 (c) – 1 (d) 1

3. Distance between P (x1, y1) and Q (x2, y2) when PQ is parallel to y-axis is (a) x1 – x2 (b) |x1 – x2| (d) |y1 – y2| (c) y1 – y2 4. The lines parallel to the axes and passing through the point (4, – 5) are (a) x = – 5, y = 4 (b) x = 5, y = – 4 (c) x = 4, y = – 5 (d) x = – 4, y = 5 5. The equation of the line whose perpendicular distance from the origin is 3 units and the angle which the normal makes with the positive direction of x-axis is 30° is

16.30

Complete Mathematics—JEE Main

(a) x + (c)

3y=3

3x+y=1

(b)

3x+y=6

(d) x +

3y=6

6. Points (8, 2), (– 2, – 2) and (3, 0) are the vertices of (a) an equilateral triangle (b) an isosceles triangle (c) right angled triangle (d) none of these 7. If the angle between the lines 3 y – x + 4 = 0 and x + y – 6 = 0 is q, then tan q is equal to. (a)

3 +1

(c) 2 +

3

(b)

3 –1

(d) 3 +

2

8. Equation of the line passing through the point (a – 1, a + 1) and making zero intercept on both axes is (a) ax + ay – 1 = 0 (b) (a + 1)x + (a – 1)y = 0 (c) (a – 1)x – (a + 1)y = 0 (d) (a + 1)x – (a – 1)y = 0 9. The angle which the normal to the line x – 3 y + 8 = 0 passing through the origin, makes with the positive x-axis is (a) 30° (b) 60° (c) 120° (d) 150° 10. If the line through the points (h, 7) and (2, 3) intersects the line 3x – 4y – 5 = 0 at right angles, then the value of h is (a) – 1 (b) 1 (c) 5 (d) – 5

11. Equation of a line passing through the intersection of the lines 7x – y + 2 = 0 and x – 3y + 6 = 0 parallel to x-axis is (a) y = 2 (b) y = – 2 (c) y = 3 (d) y = – 3 12. The value of p for which the lines 2x + y – 3 = 0, 3 x – y – 2 = 0 and x – py + 5 = 0 may intersect at a point is (a) 2 (b) 3 (c) 5 (d) 6 13. Equation of the line equidistant from the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0 is (a) 6x + 4y + 5 = 0 (b) 18x + 12y + 11 = 0 (c) 18x + 12y – 11 = 0 (d) 12x + 8y + 7 = 0 14. Area of the triangle formed by the lines y – x = 0, x + y = 0 and y = k in square units is (a) 2k (b) k2 2 (c) 2k (d) k2/2. 15. The distance of the line 2x + 3y – 5 = 0 from the point (3, 5) along the line 5x – 3y = 0 in units is (a)

2 34 21

(b)

16 21

(c)

4 34 21

(d)

3 43 21

LEVEL 1 Straight Objective Type Questions 16. The points (k – 1, k + 2), (k, k + 1), (k + 1, k) are collinear for (a) any value of k (b) k = – 1/2 only (c) no value of k (d) integral values of k only 17. The area of the triangle formed by the points (k, 2 – 2k), (– k + 1, 2 k) and (– 4 – k, 6 – 2 k) is 70 units. For (a) four real values of k (b) no integral value of k (c) two integral values of k (d) only one integral value of k 18. The quadrilateral ABCD formed by the points A (0, 0); B (3, 4), C (7, 7) and D (4, 3) is a (a) rectangle (b) square (c) rhombus (d) parallelogram 19. The triangle with vertices A (2, 7), B (4, y) and C (– 2, 6) is right angled at A if

(a) y = – 1 (c) y = 1

(b) y = 0 (d) none of these

20. The join of the points (– 3, – 4) and (1, – 2) is divided by y-axis in the ratio. (a) 1 : 3 (b) 2 : 3 (c) 3 : 1 (d) 3 : 2 21. The straight lines x + y – 4 = 0, 3x + y – 4 = 0 and x + 3y – 4 = 0 form a triangle which is (a) isosceles (b) right angled (c) equilateral (d) none of these 22. The points P (a, b + c), Q (b, c + a) and R (c, a + b) are such that PQ = QR if (a) a, b, c are in A.P. (b) a, b, c are in G.P. (c) a, b, c are in H.P. (d) none of these 23. If a, b, c are in A.P. then the points (a, x), (b, y) and (c, z) are collinear if (b) x = z2 (a) x2 = y 2 (c) y = z (d) x, y, z are in A.P.

Cartesian System of Rectangular Coordinates and Straight Lines 16.31

24. The centroid of a triangle lies at the origin and the coordinates of its two vertices are (– 8, 7) and (9, 4). The area of the triangle is (a) 95/6 (b) 285/2 (c) 190/3 (d) 285 25. The mid points of the sides AB and AC of a triangle ABC are (2, - 1) and (– 4, 7) respectively, then the length of BC is (a) 10 (b) 20 (c) 25 (d) 30 26. If the vertices of a triangle ABC are A (– 4, – 1), B (1, 2) and C (4, – 3), then the coordinates of the circumcentre of the triangle are, (a) (1/3, – 2/3) (b) (0, – 4) (c) (0, – 2) (d) (– 3/2, 1/2) 27. The extremities of a diagonal of a parallelogram are the points (3, – 4) and (– 6, 5). If third vertex is (– 2, 1) then the coordinates of the fourth vertex are (a) (1, 0) (b) (0, 0) (c) (1, 1) (d) none of these 28. If the vertices A and B of a triangle ABC are given by (2, 5) and (4, – 11) and C moves along the line L1 : 9x + 7y + 4 = 0, the locus of the centroid of the triangle ABC is a straight line parallel to (a) AB (b) BC (c) CA (d) L1 29. The number of lines that can be drawn through the point (4, – 5) at a distance 12 from the point (– 2, 3) is (a) 0 (b) 1 (c) 2 (d) infinite 30. If O is the origin and the coordinates of A and B are (x1, y1) and (x2, y2) respectively then OA × OB cos –AOB is equal to (b) x1 x2 + y1 y2 (a) x1 y1 + x2 y2 (c) x1 y2 + x2 y1 (d) x1 x2 – y1 y2 31. If the lines x + ay + a = 0, bx + y + b = 0 and cx + cy + 1 = 0 (a, b, c being distinct and π 1) are a b c + + concurrent, then the value of is a -1 b -1 c -1 (a) – 1 (c) 1

(b) 0 (d) none of these

x y + = 1 meets the axis of y and axis of x 3 4 at A and B respectively. A square ABCD is constructed on the line segment AB away from the origin, the coordinates of the vertex of the square farthest from the origin are (a) (7, 3) (b) (4, 7) (c) (6, 4) (d) (3, 8)

32. The line

33. A line 2x + 3y – 1 = 0 intersects the three sides BC, CA and AB of a triangle ABC in P, Q, R respectively BP CQ AR ¥ ¥ = then PC QA RB (a) – 1 (c) 2

(b) 1 (d) 3

34. If the vertices P, Q, R of a triangle are rational points, which of the following points of the triangle PQR is (are) always rational point (s)? (a) centroid (b) incentre (c) circumcentre (d) orthocentre (A rational point is a point both of whose coordinates are rational number) 35. Let A0 A1 A2 A3 A4 A5 be a regular hexagon with vertex A0 and A3 at (1, 0) and (– 1, 0) respectively. The equations representing A1 A4 and A2 A5 are (a) y = ± 3x (c) y = + x

(b) x = ± 3y (d) none of these

36. Let PS be the median of the triangle with vertices P (2, 2), Q (6, – 1) and R (7, 3). The equation of the line passing through (1, – 1) and parallel to PS is (a) 2x – 9y – 7 = 0 (b) 2x – 9y – 11 = 0 (c) 2x + 9y – 11 = 0 (d) 2x + 9y + 7 = 0 37. Two equal sides of an isosceles triangle are given by 7x – y + 3 = 0 and x + y –3 = 0, the slope m of the third side is given by (b) m2 – 3 = 0 (a) m2 – 1 = 0 2 (c) 3m – 1 = 0 (d) 3m2 + 8m – 3 = 0 38. Two sides of a rhombus ABCD are parallel to the line y = x + 2 and y = 7x + 3. If the diagonals of the rhombus intersect at the point (1, 2) and the vertex A is on Y-axis at a distance a from the origin, then a is equal to (a) 2 (b) 5 (c) 2/5 (d) 5/2 39. A line which makes an acute angle q with the positive direction of x-axis is drawn through the point P (3, 4) to meet the lines x = 6 and y = 8 at R and S respectively then RS = 8 – 2 3 , if (a) q = p/3 (b) q = p/4 (c) p/6 (d) p/12 40. In a right angled triangle ABC right angled at C: CA = a, CB = b. If the angular points A and B slide along x-axis and y-axis respectively then C lies on (a) bx + ay = 0 (b) ax + by = 0 x y x y (d) + = 1 (c) ± = 1 a b b a

16.32

Complete Mathematics—JEE Main

41. ABCD is a quadrilateral. P (3, 7) and Q (7, 3) are the middle points of the diagonals AC and BD respectively. The coordinates of the mean point (or the centre of mean position) of the vertices of the quadrilateral are (a) (0, 0) (b) (3, 3) (c) (5, 5) (d) (7, 7) 42. A line meets x-axis at A and y-axis at B such that incentre of the triangle OAB is (1, 1). Equation of AB is (a) 2x + y = 2 (b) 3x + 4y = 12 (c) 2x + y = 6 (d) 2x + y = 4 43. ABCD is a rectangle in the clockwise direction. The coordinates of A are (1, 3) and of C are (5, 1), vertices B and D lie on the line y = 2x + c, then the coordinates of D are (a) (2, 0) (b) (4, 4) (c) (0, 2) (d) (2, 4) 44. The area of a triangle, two of whose vertices are (2, 1) and (3, – 2) is 5. The coordinates of the third vertex can not be (a) (6, – 1) (b) (4, 5) (c) (– 1, 20) (d) (2, 9) 45. The diagonals of a parallelogram PQRS are along the lines x + 3y = 4 and 6x – 2y = 7, the PQRS must be a (a) rectangle (b) square (c) cyclic quadrilateral (d) rhombus 46. If a, b, c are in A.P., then ax + by + c = 0 represents (a) a single line (b) a family of concurrent lines (c) a family of parallel lines (d) none of these 47. If area of the triangle formed by the line L perpendicular to 5x – y = 1 and the coordinate axes is 5, then the distance of L from the origin is (b) 5 13 (a) 5 2 (c) 5 13

(d) none of these

48. The area enclosed by the lines |x| + |y| = 2 is (a) 1 sq unit (b) 2 sq units (c) 4 sq units (d) none of these 49. If the circumcentre of a triangle lies at the point (a, a) and the centroid is the mid-point of the line joining the points (2a + 3, a + 4) and(a – 4, 2a – 3); then the orthocentre of the triangle lies on the line (a) y = x (b) (a – 1) x + (a + 1) y = 0 (c) (a – 1) x – (a + 1) y = 0 (d) (a + 1) x – (a – 1) y = 2a

50. If A (at2, 2at), B(a/t2, – 2a/t) and S (a, 0) are three 1 1 points, then + is independent of SA SB (a) a (b) t (c) both a and t (d) none of these 51. qx + py + (p + q – r) = 0 is the reflection of the line px + qy + r = 0 in the line (a) x – y = p + q (b) x – y = p – q (c) x + y + 1 = 0 (d) x + y – 1 = 0 52. Coordinates of the vertices B and C of the base of a triangle ABC are (– a, 0) and (a, 0) respectively. If C – B = p/3, the vertex A lies on the curve (a) x2 – y2 + 2 3 xy – a2 = 0 (b) x2 + y2 + 2 3 xy – a2 = 0 (c)

3 (x2 – y2) + 2xy –

3 a2 = 0

(d)

3 (x2 + y2) – 2xy +

3 a2 = 0

53. If the line 5x = y meets the lines x = 1, x = 2, … x = n at points A1, A2, …, An respectively, then (0 A1)2 + (0 A2)2 + … + (0 An)2 is equal to (b) 2n3 + 3n2 + n (a) 3n2 + 3n 3 2 (c) 3n + 3n + 2 (d) (3/2) (n4 + 2n3 + n3) 54. One diagonal of a square is the portion of the line 3x + 2y = 12 intercepted between the axes. The coordinates of the extremity of the other diagonals not lying in the first quadrant are (a) (1, – 1) (b) (– 1, – 1) (c) (– 1, 1) (d) none of these 1 (i = 1, 2, 3) represents three straight mi lines whose slopes are the roots of the equation. 2m3 – 3m2 – 3m + 2 = 0, A and B are the algebraic sum of the intercepts made by the lines on x-axis and y-axis respectively, then aA + bB = 0 if (a, b) is (a) (4, 7) (b) (2, 7) (c) (7, 2) (d) (– 1,– 7)

55. If y = mix +

56. If u = a1x + b1y + c1, v = a2x + b2y + c2 = 0 and a1 b1 c1 = π then u + kv = 0 represents a2 b2 c2 (a) a family of concurrent lines (b) a family of parallel lines (c) u = 0 or v = 0 (d) none of these 57. Reflection of the line x + y + 1 = 0 in the line lx + my + n = 0 is (a) (x + y + 1) (l + m) – 2(l2 + m2) (lx + my + n) = 0 (b) (x + y + 1) (l2 + m2) – 2(l + m) (lx + my + n) = 0 (c) (l + m + 1) (x + y) – 2(lx + my) (l2 + m2) = 0 (d) none of these

Cartesian System of Rectangular Coordinates and Straight Lines 16.33

58. ABCD is a square in which A lies on the positive y-axis and B lies on the positive x-axis. If D is the point (12, 17), the coordinates of C are (a) (17, 12) (b) (17, 5) (c) (14, 16) (d) (15, 3) 59. ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD = 2AC. If the coordinates of D and M are (1, 1) and (2, – 1) respectively, the coordinates of A are (a) ( 3, 1/ 2)

(b) (1, – 3/2)

(c) (3/ 2, 1)

(d) (1/ 2, 3)

(a) 3/2, 2/3 (c) 1/3, 3

(b) 1/2, 2/1 (d) – 1/3, – 3

68. The equations of the pairs of opposite sides of a rectangle are x2 – 7x + 6 = 0 and y2 – 14y + 40 = 0, the equation of the diagonal nearer the origin is (a) 5x – 6y + 14 = 0 (b) 6x – 5y + 14 = 0 (c) 6x + 5y – 56 = 0 (d) 6x + 5y – 14 = 0 69. The equation of a line bisecting the join of (2010, 1600) and (– 1340, 1080) and having intercept on the axes in the ratio 1 : 2 is (a) 2x + y = 1680 (b) x + 2y = 1680 (c) 2x + y = 2010 (d) none of these

60. A straight line passes through the point (1, 1) and the portion of the line intercepted between x and y axes is divided at the point in the ratio 3 : 4. An equation of the line is (a) 3x + 4y = 7 (b) 4x + 3y = 7 (c) 4x – 3y = 1 (d) 3x – 4y + 1 = 0

70. Let the coordinates of P be (x, y) and of Q be (a, b) where a is the geometric and b is the arithmetic mean of the coordinates of P. If the mid point of PQ is (42, 31) the coordinates of P are (a) (61, 21) (b) (49, 25) (c) (31, 31) (d) none of these

61. The equation ax2 + 2hxy + by2 = 0 represents a pair of perpendicular lines if (a) a = 3, b = 4 (b) a = 4, b = – 3 (c) h = – 1 (d) a = 11, b = – 11

71. Image of the point (–1, 3) with respect to the line y = 2x is (a) (7/5, 14/5) (b) (1, 2) (c) (3, 1) (d) (5, 1)

62. The equation ax2 + 2hxy + ay2 = 0 represent a pair of coincident lines through the origin if (a) h = 2a (b) 2h = a (d) h2 = a2 (c) h2 = a

72. The point (a2, a) lies between the straight lines x + y = 6 and x + y = 2 for (a) all values of a (b) no value of a (c) |2a – 3| < 1 (d) |2a + 5| > 1

63. The equation ax2 + by2 + cx + cy = 0 represents a pair of straight lines if (a) a = 0 (b) b = 0 (c) a + b = 0 (d) none of these 64. The equation x3 – 6x2y + 11xy2 – 6y3 = 0 represent three straight lines passing through the origin, the slops of which form (a) an A.P. (b) a G.P. (c) an H.P. (d) none of these 65. If the slopes of the lines given by 8x3 + ax2y + bxy2 + y3 = 0 are in G.P., then (a) a = b (b) 2a = b (c) a = 2b (d) a + b = 0 66. If the slope of one of the lines given by 6x2 + axy + y2 = 0 exceeds the slope of the other by one, then a is equal to (a) ± 2 (b) 5 (c) – 5 (d) ± 5 67. If the slope of one of the lines represented by ax2 + (3a + 1) xy + 3y2 = 0 be reciprocal of the slope of the other, then the slopes of the lines are

73. Perimeter of the quadrilateral bounded by the coordinate axis and the lines x + y = 50 and 3x + y = 90 is (a) 80 + 20 2 (b) 80 + 10 10 (c) 80 + 20 2 + 10 10 (d) 110. 74. If the sum of the slopes of the lines given by 3x2 – 2cxy – 5y2 = 0 is twice their product, then the value of c is (a) 2 (b) 3 (c) 6 (d) none of these 75. If one of the lines given by 2cx2 + 2xy – (c2 – 1)y2 = 0 is 2x + 3y = 0, then the integral value of c is (a) 2 (b) 3 (c) 4 (d) 8

16.34

Complete Mathematics—JEE Main

Assertion-Reason Type Questions 76. Let A(2, – 3) and B(– 2, 1) be the vertices of a triangle ABC. Statement-1: If the centroid of the triangle moves on the line x + y = 5, the vertex moves on the line x + y = 17. Statement-2: If the centroid of the triangle moves the line x – y + 1 = 0 (x π 0), the triangle is either isosceles or equilateral. 77. Statement-1: x2y – 3xy – 2x2 + 6x – 4y + 8 = 0 represents three straight lines two of which are parallel and the third is perpendicular to the other two Statement-2: xy – 2x + y – 2 = 0 represents a pair of straight lines one of which is common to the pair of straight lines xy + 2x – y – 2 = 0 78. Statement-1: The locus represented by the equation (2x + 3y –1)2 + (x + y + 1)2 = 0 is a pair of straight lines 2x + 3y – 1 = 0 and x + y + 1 = 0. Statement-2: The equation (x + 4y – 2)2 + (4x + y – 8)2 + (x ± y – k)2 = 0 represents a point if k = 2. 79. Statement-1: Points P = (– sin (b – a), – cos b), Q = (cos (b – a), sin b) and R = (cos (b – a + q), sin (b – q)) where 0 < a, b, q < p /4 are non collinear Statement-2: If the slopes of the lines PQ and QR are not equal the points P, Q and R are non collinear. 80. Statement-1: If x + ky = 1 and x = a are the equations of the hypotenuse and a side of a right angled isosceles triangle then k = ± a. Statement-2: Each side of a right angled isosceles triangle makes an angle p/4 with the hypotenuse. 81. Statement-1: Locus of the centroid of the triangle whose vertices are (a cos t, a sin t), (b sin t, – b cos t) and (1, 0) where t is a parameter is (3x – 1)2 + (3y)2 = a2 + b2 Statement-2: The centroid of a triangle is equidistance from the vertices of the triangle.

82. Statement-1: If non zero numbers a, b, c are in Harx y 1 monic progression, then the straight line + + = 0 a b c always passes through the fixed point (1, 1) Statement-2: If a, b, c are in Harmonic progression. 1 1 1 then , , are in arithmetic progression. a b c 83. Statement-1: The straight line. (sin q + 3 cos q)x + ( 3 sin q – cos q)y + (5 sin q – 7 cos q) = 0 for all values of q except q = np /2, n is an integer; passes through the point of intersection of the lines x + 3 y + 5 = 0 and 3 x – y – 7 = 0. Statement-2: L1 + l L2 = 0 represents a line through the point of intersection of the lines L1 = 0, L2 = 0 for all non-zero finite values of l. 84. Statement-1: If the area of the triangle formed by the lines y = x, x + y = 2 and the lines through P(h, k) parallel to x-axis is 4h2, then P lies on the lines 4x2 + 2y – y2 – 1 = 0. Statement-2: Area of the triangle formed by the lines y = x, x + y = 2 and the axis of x is equal to half the area of the triangle formed by the line x + y = 2 and the coordinate axes. 85. Statement-1: Reflexion of the point (6, 1) in the line x + y = 0 is (– 1, – 6). Statement-2: Reflexion of a point P(a, b) in the line a + a ¢ b + b¢ˆ , ax + by + c = 0 is Q(a ¢, b¢) if R ÊÁ ˜ Ë 2 2 ¯ lies on the line. 86. Statement-1: 4x2 + 12xy + 9y2 = 0 represents a pair of perpendicular lines through the origin. Statement-2: ax2 + 2hxy + by2 = 0 represents a pair of coincident lines if h2 = ab. 87. Statement-1: x2y2 – x2 – y2 + 1 = 0 represents the sides of a square of area 4 square units. Statement-2: 3x2 + lxy – 3y2 = 0 represents a pair of perpendicular lines for all values of l

LEVEL 2 Straight Objective Type Questions 88. If (0, 1), (1, 1) and (1, 0) are the mid points of the sides of a triangle, the coordinates of its incentre are (a) (2 + 2 , 2 + 2 ) (b) ((2 +

2 ), – (2 +

2 ))

(c) ((2 –

2 ), (2 –

(d) ((2 –

2 ), – (2 –

2 )) 2 ))

Cartesian System of Rectangular Coordinates and Straight Lines 16.35

89. The vertices of the triangle ABC are A(1, 2), B (0, 0) and C(2, 3), then the greatest angle of the triangle is (a) 75° (b) 105° (c) 120° (d) None of these 90. The points (0, 8/3), (1, 3) and (82, 30) are the vertices of (a) obtuse angled triangle (b) acute angled triangle (c) right angled triangle (d) none of these 91. Area of the rhombus enclosed by the lines ax ± by ± c = 0 is (b) 2b2/ca (a) 2a2/bc 2 (c) 2c /ab (d) none of these 92. If x cos a + y sin a = – sin a tan a be the equation of a line, then the length of the perpendiculars on the line from the points (a2, 2a), (ab, a + b) and (b2, 2b) are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 93. The coordinates of the points A and B are respectively (– 3, 2) and (2, 3). P and Q are points on the line joining A and B such that AP = PQ = QB. A square PQRS is constructed on PQ as one side, the coordinates of R are (a) (– 4/3, 7/3) (b) (0, 13/3) (c) (1/3, 8/3) (d) (2/3, 1) 94. If A (x1, y1), B (x2, y2), C (x3, y3) are the vertices of a triangle, then the equation x x1 x2

y 1 x y1 1 + x1 y2 1 x3

y 1 y1 1 = 0 y3 1

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

the median through A the altitude through A the perpendicular bisector of BC the line joining the centroid with a vertex

95. Given four lines whose equations are x + 2y – 3 = 0, 2x + 3y – 4 = 0, 3x + 4y – 7 = 0 and 4x + 5y – 6 = 0, then the lines are (a) concurrent (b) the sides of a quadrilateral with one vertex at (3, 0) (c) the sides of a cyclic quadrilateral (d) none of these 96. A ray of light coming from the point (1, 2) is reflected at a point A on the axis of x and then passes through the point (5, 3). The coordinates of the point A are

(a) (5/13, 0) (c) (13/5, 0)

(b) (–7, 0) (d) (15, 0)

97. If D1, D2, D3 are the areas of the triangles with vertices (0, 0), (a tan a, b cot a), (a sin a, b cos a); (a, b), (a sec2 a, b cosec2 a), (a + a sin2 a, b + b cos2 a) and (0, 0), (a tan a, – b cot a), (a sin a, b cos a), then D1, D2, D3 are in G.P. for (a) all values of a (b) only one value of a (c) finite number of values of a (d) no value of a 98. The orthocentre of the triangle formed by the lines y = 0, (1 + t)x – ty + t (1 + t) = 0 and(1 + u) x – uy + u (1 + u) = 0 (t π u) for all values of t and u lies on the line (b) x + y = 0 (a) x – y = 0 (c) x – y + 1 = 0 (d) x + y + 1 = 0 99. If A (3, 0) and B(6, 0) are two fixed points and U (x1, y1) is a variable point in the plane. AU and BU meet y-axis at C and D respectively and AD meets OU at V, then for all positions of U in the plane, CV passes through the point (a) (0, 0) (b) (0, 2) (c) (2, 0) (d) none of these 100. The incentre of the triangle with vertices (1, (0, 0) and (2, 0) is (a) (1, (c) (2/3,

3/2 ) 3/2 )

3 ),

(b) (2/3, 1/ 3 ) (d) (1, 1/ 3 )

101. If x + 2y + 3 = 0, x + 2y - 7 = 0 and 2x - y - 4 = 0 form three sides of a square, the equation of the fourth side nearer the point (1, - 1) is (a) 2x - y - 6 = 0 (b) 2x - y + 6 = 0 (c) 2x - y - 14 = 0 (d) 2x - y + 14 = 0 102. The distance between the orthocentre and the circumcentre of the triangle with vertices (0, 0), (0, a) and (b, 0) is (a)

a 2 - b2 2

(c) a - b

(b) a + b (d)

a 2 + b2 2

103. The centroid of a triangle lies at the origin and the coordinates of its two vertices are (- 8, 0) and (9, 11), the area of the triangle in sq. units is (a) 11/8 (b) 8/11 (c) 88 (d) none of these 104. The line 3x + 2y = 24 meets the y-axis at A and the x-axis at B; C is a point on the perpendicular bisector of AB such that the area of the triangle ABC is 91 sq. units. The coordinates of C can be

16.36

Complete Mathematics—JEE Main

(a) (29/2, - 1) (c) (- 13/2, 1)

(b) (29/2, 13) (d) (- 13/2, 13)

105. The straight lines 4x - 3y - 5 = 0, x - 2y = 0, 7x + y - 40 = 0 and x + 3y + 10 = 0 form (a) a rectangle (b) a parallelogram (c) a cyclic quadrilateral (d) none of these 106. If the straight lines x + 2x - 9 = 0, 3x + 5y - 5 = 0 and ax + by + 1 = 0 are concurrent, then the straight line 35x - 22y - 1 = 0 passes through (a) (a, b) (b) (b, a) (c) (a, - b) (d) (- a, b)

107. If the slope of one of the lines represented by ax2 + 2hxy + by2 = 0 be the square of the other, then a + b 8h 2 is equal to + h ab (a) 0 (b) 1 (c) 6 (d) 8 108. The locus represented by the equation (x – y + c)2 + (x + y – c)2 = 0 is (a) a line parallel to x-axis (b) a point (c) a pair of straight lines (d) a line parallel of y-axis

Previous Years' AIEEE/JEE Main Questions 1. If the pair of lines ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 intersect on y-axis then (b) bg2 π ch2 (a) 2fgh = bg2 + ch2 (c) abc = 2fgh (d) none of these [2002] 2. Locus of mid point of the portion between the axes of x cos a + y sin a = p where p is constant is (a) x2 + y2 = 4/p2 (b) x2 + y2 = 4p2 (c) 1/x2 + 1/y2 = 2/p2 (d) 1/x2 + 1/y2 = 4/p2 [2002] 3. A triangle with vertices (4, 0), (- 1, - 1), (3, 5) is (a) isosceles and right angled (b) isosceles but not right angled (c) right angled but not isosceles (d) neither right angled nor isosceles. [2002] 4. The sides of a triangle are 3x + 4y = 0, 4x - 3y = 0 and 5x + 5y = 1 then the triangle is (a) right angled (b) obtuse angled (c) equilateral (d) none of these [2002] 5. A square of side a lies above the x-axis and has one vertex at the origin. The side passing through the origin makes an angle a (0 < a < p/4) with the positive direction of x-axis. The equation of the diagonal not passing through the origin is (a) y (cos a + sin a) + x (sin a - cos a) = a (b) y (cos a + sin a) + x (sin a + cos a) = a (c) y (cos a + sin a) + x (cos a - sin a) = a (d) y (cos a – sin a) - x (sin a - cos a) = a [2003] 6. If the pair of straight lines x2 - 2pxy - y2 = 0 and x2 - 2qxy - y2 = 0 be such that each pair bisects the angle between the other pair, then

(a) p = – q (c) pq = – 1

(b) pq = 1 (d) p = q

[2003]

7. Locus of centroid of the triangle whose vertices are (a cost, a sint),(b sint, - b cost) and (1, 0), where t is a parameter is (a) (3x - 1)2 + (3y)2 = a2 + b2 (b) (3x + 1)2 + (3y)2 = a2 + b2 (c) (3x + 1)2 + (3y)2 = a2 - b2 (d) (3x - 1)2 + (3y)2 = a2 - b2 [2003] 8. If x1, x2, x3 and y1, y2, y3 are both in G.P. with the same common ratio, then the points (x1, y1), (x2, y2) and (x3, y3) (a) lie on an ellipse (b) lie on a circle (c) are vertices of a triangle (d) lie on a straight line [2003] 9. If the equation of the locus of point equidistant from the points (a1, b1) and (a2, b2) is (a1 - a2) x + (b1 b2) y + c = 0, then c = (a) a 12 - a22 + b 21 - b22 (b) (1/2) (a21 + a22 + b21 + b22) (c)

a12 + b12 - a22 - b22

(d) (1/2) (a22 + b22 - a21 - b21)

[2003]

10. Let A(2, - 3) and B(- 2, 1) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line (a) 3x + 2y = 5 (b) 2x - 3y = 7 (c) 2x + 3y = 9 (d) 3x - 2y = 3 [2004]

Cartesian System of Rectangular Coordinates and Straight Lines 16.37

11. The equation of the straight line passing through the point (4, 3) and making intercepts on the coordinate axes whose sum is - 1 is x y x y + = 1 and + = 1 (a) 2 3 2 1 (b)

x y x y - = - 1 and - =-1 2 3 -2 1

x y x y + = - 1 and - =-1 2 3 -2 1 x y x y - = 1 and + =1 (d) 2 3 -2 1 (c)

[2004]

12. If the sum of the slopes of the lines given by x2 - 2cxy - 7y2 = 0 is four times their product, then c has the value (a) 2 (b) – 1 (c) 1 (d) - 2 [2004] 13. If one of the lines given by 6x2 - xy + 4cy2 = 0 is 3x + 4y = 0, then c equals (a) 3 (b) – 1 (c) 1 (d) - 3 [2004] 14. The line parallel to x-axis passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0 where (a, b) π (0, 0) is (a) above x-axis at a distance 3/2 from it (b) above x-axis at a distance 2/3 from it (c) below x-axis at a distance 3/2 from it (d) below x-axis at a distance 2/3 from it. [2005] 15. If a vertex of a triangle is (1, 1) and the mid points of two sides through this vertex are (- 1, 2) and (3, 2), then the centroid of the triangle is (a) (1, 7/3) (b) (1/3, 7/3) (c) (- 1, 7/3) (d) (- 1/3, 7/3) [2005] 16. If non zero numbers a, x y 1 + straight line + a b c a fixed point. That point (a) (1, - 2) (c) (- 1, 2) 2

b, c are in H.P., then the = 0 always passes through is (b) (1, - 1/2) (d) (- 1, - 2)

[2005] 2

17. If the pair of lines ax + 2(a + b) xy + by = 0 lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sector is thrice the area of the another sector, then (a) (b) (c) (d)

3a2 + 10ab + 3b2 = 0 3a2 + 2ab + 3b2 = 0 3a2 - 10ab + 3b2 = 0 3a2 -2ab + 3b2 = 0

18. A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is (a) 3x + 4y = 25 (b) x + y = 7 (c) 3x - 4y + 7 = 0 (d) 4x + 3y = 24 [2006] 19. Let A(h, k), B(1, 1) and C(2, 1) be the vertices of a right angled with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which ‘k’ can take is given by (a) {1, 3} (b) {0, 2} (c) {- 1, 3} (d) {- 3, - 2} [2007] 20. Let P = (- 1, 0), Q = (0, 0) and R = ( 3 , 3), be three points. The equation of the bisector of angle PQR is (b) x + ( 3/2) y = 0 (a) 3 x + y = 0 (c) ( 3/2) x + y = 0

2

3 y=0

[2007]

21. If one of the lines of my + (1 - m ) xy - mx2 = 0 is a bisector of the angle between the line xy = 0 then m is (a) - 1/2 (b) - 2 (c) 1 (d) 2 [2007]

2

22. The perpendicular bisector of the lines segment joining P(1, 4) and Q(k, 3) has y-intercept-4. Then a possible value of k is (a) – 4 (b) 1 (c) 2 (d) – 2 [2008] 23. The lines p(p2 + 1) x – y + q = 0 and (p2 + 1)2x + (p2 + 1) y + 2q = 0 are perpendicular to a common line for (a) exactly two values of p (b) more than two values of p (c) no value of p (d) exactly one value of p [2009] x y + = 1 passes through the 5 b point (13, 32). The line K is parallel to L and has the x y equation + = 1. Then the distance between L and c 3 K is

24. The line L is given by

(a) (c)

[2005]

(d) x =

17 15 23

(b)

23 17

(d) 15 [2010] 15 25. The line L1 : y – x = 0 and L2 : 2x + y = 0 intersect the line L3: y + 2 = 0 at P and Q respectively The bisector of the acute angle between L1 and L2 intersect L3 at R.

16.38

Complete Mathematics—JEE Main

Statement-1 : The ratio PR:RQ equals 2 2 :

5

Statement-2 : In any triangle, bisector of an angle divides the triangle into two similar triangles [2011] 26. The lines x + y = a and ax – y = 1 intersect each other in the first quadrant then the set of all possible values of a is in the interval. (a) (0, •) (b) (1, •) (c) ( – 1, •) (d) (–1, 1) [2011] 27. If A (2, – 3), B (–2, 1) are two vertices of a triangle and the third vertex moves on the line 2x + 3y = 9, then the locus of the centroid of the triangle is (a) x – y = 1 (b) 2x +3y = 1 (c) 2x + 3y = 3 (d) 2x – 3y = 1 [2011] 28. If the line 2x + y = k passes through the point which divides the segment joining the points (1, 1) and (2, 4) in the ratio 3:2, then k equals (a) 6 (b) 11/5 (c) 29/5 (d) 5 [2012] 29. A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is (a) – 2 (b) –1/2 (c) –1/4 (d) – 4 [2012] 30. A ray of light along x + 3 y = 3 gets reflected upon reaching the x-axis, an equation of the reflected ray is (b) y = 3x – 3 (a) 3y = x – 3 (c)

3y = x – 1

(d) y = x + 3

[2013]

34. The x-coordinates of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1), (1, 1) and (1, 0) is (a) 2 – 2 (b) 1 + 2 (c) 1 –

2

(d) 2 +

2

[2013]

35. If x-intercept of some line L is double as that of the line 3x + 4y = 12 and the y-intercept of L is half as that of the same line, then the slope of L is: (a) – 3 (b) – 3/8 (c) – 3/2 (d) – 3/16 [2013, online] 36. If the extremities of the base of an isosceles triangle are the points (2a, 0) and (0, a) and the equation of one of the sides is x = 2a, then the area of the triangle, in square units, is: 5 2 5 2 a (b) a (a) 4 2 25 2 (d) 5a2 [2013, online] (c) a 4 37. Let q1, be the angle between two lines 2x + 3y + c1 = 0 and – x + 5y + c2 = 0 and q2 be the angle between two lines 2x + 3y + c1 = 0 and – x + 5y + c3 = 0, where c1, c2, c3 are any real numbers: Statement 1: If c2 and c3 are proportional, then q1 = q2 Statement 2: q1 = q2 for all c2 and c3. (a) Statement-1 is true, statement-2 is true and statement-2 is a correct explanation for statement-1. (b) Statement-1 is true, statement-2 is true but statement-2 is Not a correct explanation for statement-1 (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is false. [2013, online]

31. Equation of the line passing through the points of intersection of the parabola x2 = 8y and the ellipse x2 + y2 = 1 is: 3 (a) y – 3 = 0 (b) y + 3 = 0 (c) 3y + 1 = 0 (d) 3y – 1 = 0 [2013, online]

38. If the image of point P (2, 3) in a line L is Q (4, 5) then the image of point R(0, 0) in the same line is: (a) (2, 2) (b) (4, 5) (c) (3, 4) (d) (7, 7) [2013, online]

32. A light ray emerging from the point source placed at P(1, 3) is reflected at a point Q in the axis of x. If the reflected ray passes through the point R(6, 7), then the abscissa of Q is: (a) 1 (b) 3 (c) 7/2 (d) 5/2 [2013, online]

39. Let A(– 3, 2) and B(– 2, 1) be the vertices of a triangle ABC. If the centroid of this triangle lies on the line 3x + 4y + 2 = 0, then vertex C lies on the line. (a) 4x + 3y + 5 = 0 (b) 3x + 4y + 3 = 0 (c) 4x + 3y + 3 = 0 (d) 3x + 4y + 5 = 0 [2013, online]

33. If three lines x – 3y = p, ax + 2y = q and ax + y = r form a right-angled triangle, then: (b) a2 – 6a – 12 = 0 (a) a2 – 9a + 18 = 0 2 (d) a2 – 9a + 12 = 0 (c) a – 6a – 18 = 0

40. Let a, b, c and d be non zero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5 bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then: (a) 2bc – 3ad = 0 (b) 2bc + 3ad = 0 (c) 3bc – 2ad = 0 (d) 3bc + 2ad = 0 [2014]

[2013, online]

Cartesian System of Rectangular Coordinates and Straight Lines 16.39

41. Let PS be the median of the triangle with vertices P(2, 2), Q(6, – 1) and R(7, 3). The equation of the line passing (1, – 1) and parallel to PS is: (a) 4x – 7y – 11 = 0 (b) 2x + 9y + 7 = 0 (c) 4x + 7y + 3 = 0 (d) 2x – 9y – 11 = 0 [2014] 1 1 42. Let a and b be any two numbers satisfying 2 + 2 a b 1 = . Then, the foot of perpendicular from the origin 4 x y on the variable line + = 1, lies on: a b (a) (b) (c) (d)

a hyperbola with each semi-axis = 2 a hyperbola with each semi-axis = 2 a circle of radius = 2 [2014, online] a circle of radius = 2

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

y – 2ax = 0 y – (a2 + 1)x = 0 y+x=0 (a – 1)2 x – (a + 1)2 y = 0

[2014, online]

48. If a line L is perpendicular to the line 5x – y = 1, and the area of the triangle formed by the line L and the coordinate axes, is 5, then the distance of L from the line x + 5y = 0 is (a) 7

5

(b) 5

13

(c) 7

13

(d) 5

7

[2014, online]

49. The number of points having both coordinates as integer, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0) is (a) 901 (b) 861 [2015] (c) 820 (d) 780

43. Given three points P, Q, R with P(5, 3) and R lies on the x-axis. If equation of RQ is x – 2y = 2 and PQ is parallel to x-axis, then centroid of D PQR lies on the line (a) 2x + y – 9 = 0 (b) x – 2y + 1 = 0 (c) 5x – 2y = 0 (d) 2x – 5y = 0 [2014, online]

50. Locus of the image of the point (2, 3) in the line (2x Р3y + 4) + k (x Р2y + 3) = 0 , k ΠR, is a: (a) straight line parallel to x-axis (b) straight line parallel to y-axis (c) circle of radius 2 (d) circle of radius 3 [2015]

44. The base of an equilateral triangle is along the line given by 3x + 4y = 9. If a vertex of the triangle is (1, 2), then length of a side of the triangle is:

51. A straight line L through the point (3, –2) is inclined at an angle of 60∞ to the line 3 x + y = 1. If L also intersects x-axis, then equation of L is:

(a)

2 3 15

(b)

4 3 15

4 3 2 3 (d) [2014, online] 5 5 45. If a line intercepted between the coordinates axes is trisected at a point A (4, 3), which is nearer to the x-axis, then its equation is (a) 4x – 3y = 7 (b) 3x + 2y = 18 (c) 3x + 8y = 36 (d) x + 3y = 13 [2014, online] (c)

46. If three distinct lines x + 2ay + a = 0, x + 3by + b = 0 and x + 4ay + a = 0 are concurrent, then the point (a, b) lies on a: (a) circle (b) hyperbola (c) straight-line (d) parabola [2014, online] 47. The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points (a2 + 1, a + 1) and (2a, – 2a), a π 0. Then for any a, the orthocentre of this triangle lies on the line:

(a) y + 3x + 2 - 3 3 = 0 (b) y - 3x + 2 + 3 3 = 0 (c)

3y - x + 3 + 2 3 = 0

(d)

[2015, online] 3y + x - 3 + 2 3 = 0 52. The points Ê 0, 8 ˆ , (1, 3) and (82, 30) Ë 3¯ (a) (b) (c) (d)

form an obtuse angled triangle form an acute angled triangle form a right angled triangle lie on a straight line.

[2015, online]

53. Let L be the line passing through the point P(1, 2) such that its intercepted segment between the coordinate axes is bisected at P. If L1 is the line perpendicular to L and passing through the point (–2, 1), then the point of intersection of L and L1 is: Ê 11 29 ˆ Ê 4 , 12 ˆ (a) (b) Ë , ¯ Ë5 5 ¯ 20 10 (c) Ê 3 , 17 ˆ Ë 10 5 ¯

(d) Ê 3 , 23 ˆ [2015,online] Ë 5 10 ¯

16.40

Complete Mathematics—JEE Main

54. Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals intersect at (– 1, – 2), then which one of the following is a vertex of this rhombus? (a) (– 3, – 9) (b) (– 3, – 8) Ê 10 7 ˆ Ê 1 8ˆ (d) Á - , - ˜ [2016] (c) Á - ˜ Ë 3 Ë 3 3¯ 3¯ 55. If a variable line drawn through the intersection on x y x y the lines + = 1 and + = 1 , meets the coordi3 4 4 3 nate axes at A and B, (A π B), then the locus of the midpoint of AB is: (a) 7xy = 6(x + y) (b) 4(x + y)2 – 28(x + y) + 49 = 0 (c) 6xy = 7(x + y) [2016, online] (d) 14(x + y)2 – 97(x + y) + 168 = 0 56. The point (2, 1) is translated parallel to the line L : x – y – 4 = 0 by 2 3 units. If the new point Q lies

in the third quadrant, then the equation of the line passing through Q and perpendicular to L is: (a) x + y = 2 –

6

(c) x + y = 3 – 3 6

(b) 2x + 2y = 1 –

6

(d) x + y = 3 – 6 [2016, online]

57. A straight line through origin O meets the lines 3y = 10 – 4x and 8x + 6y + 5 = 0 at points A and B respectively. Then O divides the segment AB in the ratio: (a) 2 : 3 (b) 1 : 2 (c) 4 : 1 (d) 3 : 4 [2016, online] 58. A ray of light is incident along a line which meets another line 7x – y + 1 = 0, at the point (0, 1). The ray is then reflected from this point along the line, y + 2x = 1. Then the equation of the line of incidence of the ray of light is: (a) 41x – 25y + 25 = 0 (b) 41x + 25y – 25 = 0 (c) 41x – 38y + 38 = 0 (d) 41x + 38y – 38 = 0 [2016, online]

Previous Years' B-Architecture Entrance Examination Questions 1. The y-axis and the lines (a5 – 2a3)x + (a + 2)y + 3a = 0 and (a5 – 3a2)x + 4y + a – 2 = 0 are concurrent for (a) Two values of a (b) Three values of a (c) Five values of a (d) no value of a [2007] 2. If the point of intersection of the lines 2px + 3qy + r = 0 and px – 2qy – 2r = 0 lies strictly in the fourth quadrant and is equidistant from the two axes, then (a) 5p + 4q = 0 (b) 4p – 5q = 0 (c) 4p + 5q = 0 (d) 5p – 4q = 0 [2009] 3. Consider a triangle ABC with vertices at (0, – 3), (– 2 3 , 3) and (2 3 , 3) respectively. The incentre of the triangle with vertices at the mid points of the sides of triangle ABC is (a) (0, 0) (b) (0, 1) (c) ( - 3, - 3 )

(d)

(

3, 3 )

[2009]

4. The equation of a straight line belonging to both the families of lines x – y + 1 + l1 (2x – y – 2) = 0 and 5x + 3y – 2 + l2 (3x – y – 4) = 0, where l1, l2 are arbitrary numbers is: (a) 5x – 2y – 7 = 0 (b) 2x + 5y – 7 = 0 (c) 5x + 2y – 7 = 0 (d) 2x – 5y – 7 = 0 [2010] 5. Statement-1: The equation |x| + |y| = 2 represents a parallelogram.

Statement-2: Lines x + y = 2 and x + y = – 2 are parallel. Also lines x – y = 2 and – x + y = 2 are parallel. (a) Statement-1 is false, Statement-2 is true; (b) Statement-1 is true, Statement-2 is true; statement-2 is not correct explanation for statement-1. (c) Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for statement-1. (d) Statement-1 is true, Statement-2 is false. [2011] 6. If the line joining the points A(2, 0) and B(3, 1) is rotated about A in anti-clockwise direction through an angle of 15°, then the equation of the line in new position is (a)

3x–y=2 3

(c) x +

3y=2

(b)

3x+y=2

(d) x –

3y=2

3 [2012]

7. If m1 and m2 are the roots of the equation x2 + ( 3 + 2) x + ( 3 – 1) = 0, then the area of the triangle formed by the lines y = m1 x, y = – m2 x and y = 1 is (a)

1 Ê 3 + 2ˆ 2 ÁË 3 - 1 ˜¯

(b)

1 Ê 3 + 2ˆ 2 ÁË 3 + 1 ˜¯

(c)

1 Ê - 3 + 2ˆ 2 ÁË 3 - 1 ˜¯

(d)

1 Ê - 3 + 2ˆ 2 ÁË 3 + 1 ˜¯

[2012]

Cartesian System of Rectangular Coordinates and Straight Lines 16.41

8. Statement-1: The line 2x + y + 6 = 0 is perpendicular to the line x – 2y + 5 = 0 and second line passes through (1, 3). Statement-2: Product of the slopes of any two parallel lines is equal to – 1. (a) Statement-1 is true, statement-2 is true; statement2 is a correct explanation for statement-1 (b) Statement-1 is true, statement-2 is true; statement-2 is not correct explanation for statement-1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statsement-2 is true [2013] 9. If a variable line, passing through the point of intersection of the lines x + 2y – 1 = 0 and 2x – y – 1 = 0, meet the coordinate axes in A and B, then the locus of the mid-point of AB is (a) x + 3y = 0 (b) x + 3y = 10 (c) x +3y = 10xy (d) x + 3y + 10xy = 0 [2013] 10. If the point (p, 5) lies on the line parallel to the yaxis and passing through the intersection of the lines 2(a2 + 1) x + by + 4 (a3 + a) = 0 and (a2 + 1)x – 3by + 2 (a3 + a) = 0, then p is equal to (a) –2a (b) –3a (c) 2a (d) 3a [2014] 11. If the points (x, –3x) and (3, 4) lie on the opposite side of the line 3x – 4y = 8, then -8 8 -8 8 (b) x > , y > (a) x > , y < 5 15 15 5 -8 8 -8 8 (c) x < , y > (d) x = , y = [2015] 15 5 15 5 12. Two vertices of a triangle are (3, – 2) and (–2, 3) and its orthocentre is (–6, 1). Then the third vertex of this triangle can NOT lie on the line: (a) 4x + y = 2 (b) 5x + y = 2 (c) 3x + y = 3 (d) 6x + y = 0 [2016] 13. A line passing through the point P(1, 2) meets the line x + y = 7 at the distance of 3 units from P. Then the slope of this line satisfies the equation (b) 16x2 – 39x + 16 = 0 (a) 7x2 – 18x + 7 = 0 2 (d) 8x2 – 9x + 1 = 0 [2016] (c) 7x – 6x – 7 = 0

Answers Concept-based 1. 5. 9. 13.

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

2. 6. 10. 14.

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

3. 7. 11. 15.

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

4. (c) 8. (d) 12. (d)

Level 1 16. (a)

17. (d)

18. (c)

19. (a)

20. (c)

21. (a)

22. (a)

23. (d)

24. (b)

25. (b)

26. (c)

27. (d)

28. (d)

29. (a)

30. (b)

31. (c)

32. (b)

33. (a)

34. (a)

35. (a)

36. (d)

37. (d)

38. (d)

39. (c)

40. (b)

41. (c)

42. (b)

43. (a)

44. (d)

45. (d)

46. (b)

47. (b)

48. (d)

49. (d)

50. (b)

51. (c)

52. (c)

53. (b)

54. (c)

55. (b)

56. (b)

57. (b)

58. (b)

59. (b)

60. (b)

61. (d)

62. (d)

63. (c)

64. (c)

65. (c)

66. (d)

67. (d)

68. (b)

69. (c)

70. (b)

71. (c)

72. (c)

73. (c)

74. (b)

75. (c)

76. (b)

77. (c)

78. (d)

79. (a)

80. (d)

81. (c)

82. (d)

83. (a)

84. (b)

85. (c)

86. (d)

87. (b)

88. (c)

89. (d)

90. (d)

91. (c)

92. (b)

93. (d)

94. (a)

95. (d)

96. (c)

97. (d)

98. (b)

99. (c)

100. (d)

101. (b)

102. (d)

103. (d)

104. (b)

105. (c)

106. (a)

107. (c)

Level 2

108. (b)

Previous Years' AIEEE/JEE Main Questions 1. (a)

2. (d)

3. (a)

4. (a)

5. (c)

6. (c)

7. (a)

8. (d)

9. (d)

10. (c)

11. (d)

12. (a)

13. (d)

14. (c)

15. (a)

16. (a)

17. (b)

18. (d)

19. (c)

20. (a)

21. (c)

22. (a)

23. (d)

24. (b)

25. (c)

26. (b)

27. (b)

28. (a)

29. (a)

30. (a)

31. (d)

32. (d)

33. (a)

34. (a)

35. (d)

36. (b)

37. (a)

38. (d)

39. (b)

40. (c)

41. (b)

42. (c)

43. (d)

44. (b)

45. (b)

46. (c)

47. (d)

48. (b)

16.42

Complete Mathematics—JEE Main

49. (d)

50. (c)

53. (a)

54. (c)

57. (c)

58. (c)

51. (a) 55. (a)

52. (d) 56. (d)

a +1 a +1- 0 x as the slope of the line is a -1 a -1- 0 or (a + 1) x – (a – 1) y = 0

4. (a) 8. (c) 12. (b)

9. We write the equation of the given line in the normal 1 3 8 form as – x + y= 2 2 2 -1 or x cos a + y sin a = p where cos a = , 2 3 and p = 4, a being the required angle. sin a = 2 fi a = 120°

Previous Years' B-Architecture Entrance Examination Questions 1. 5. 9. 13.

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

2. (d) 6. (a) 10. (a)

3. (b) 7. (a) 11. (a)

Hints and Solutions Concept-based 1. Coordinate of A are (3, 0) and of B are (0, 4) so coorÊ 1 ¥ 0 + 2 ¥ 3 1 ¥ 4 + 2 ¥ 0 ˆ Ê 4ˆ dinate of P are Ë , ¯ = Ë 2, 3 ¯ . 3 3 2. Coordinate of the mid-point of the line segment Ê3 - 5 - 4 - 2 ˆ = (– 1, –3) joining P and Q are Ë , 2 ¯ 2 0 - ( - 3) =3 so slope of the required line is 0 - ( - 1) 3. When PQ is parallel to y-axis x1 = x2 and the required distance is |y1 – y2| 4. Line parallel to x-axis is y = – 5 and parallel to y-axis is x = 4. 5. Equation of the line is x cos a + y sin a = p where Ê 3ˆ p = 3, a = 30°. So the required equation is ÁË ˜¯ x 2 Ê1 ˆ + Ë ¯ y = 3 or 3 x + y = 6 2 2 and 5 2 the slope of the line joining (8, 2), (3, 0) is also , 5 so the points are collinear.

6. Slope of the line joining (8, 2), (– 2, – 2) is

7. Slopes of the given lines are

1

and – 1, so

3 È 1 + 1˘ Í 3 ˙ È 3 + 1˘ tan q = Í ˙ = Í ˙ =2+ 3. Í1 - 1 ˙ Î 3 -1˚ ÍÎ 3 ˙˚ 8. Since the line makes zero intercept on both the axes, it must pass through the origin. So equation of the

line is y =

10. Product of the slopes of the given lines is – 1. 7- 3 3 fi ¥ =–1 h -2 4 h=–1 11. Equation of any line through the intersection of given line is 7x – y + 2 + l (x – 3y + 6) = 0 It will be parallel to x-axis if 7 + l = 0 fi l = – 7 and the required equation is y = 2. 12. First two lines intersect at (1, 1) and the third line passes through (1, 1) if p = 6 È1 - p 5 ˘ Alternatively Í2 1 - 3 ˙ = 0. Í ˙ ÍÎ3 - 1 - 2 ˙˚ 13. The given lines are 3x + 2y + 6 = 0 and 3x + 2y –

7 = 0. 3

Let the required line be 3x + 2y + k = 0. Since it is equidistant from the given lines 7 k+ 11 3 = 6- k fi k= . 6 9+ 4 9+ 4 14. Vertices of the triangle are (0, 0), (k, k), (– k, k) Ê È 0 0 1˘ˆ 1 Area = Á Í k k 1 ˙˜ = k 2 ˙˜ Á2Í Ë ÎÍ- k k 1 ˚˙¯ Ê 5 25 ˆ 15. Two lines intersect at the point Á , . DisË 7 21˜¯ tance of this point from the given point (3, 5) is 2 25 ˆ 2 4 34 Ê 5ˆ Ê ÁË 3 - ˜¯ + ÁË 5 - ˜¯ = 7 21 21

Level 1 16. Slope of the line joining any two points is independent of k

Cartesian System of Rectangular Coordinates and Straight Lines 16.43

17.

18. 19. 20. 21.

k 2 - 2k 1 - k +1 2k 1 = 140 - 4 - k 6 - 2k 1 which is true for only one integral value of k. Opposite sides are parallel, all sides are equal. AB is not perpendicular to AD. Product of the slopes of AB and AC is - 1 l -3 = 0 fi l = 3, ratio is 3 : 1 l +1 Vertices are (0, 4), (4, 0), (1, 1) and the sides are 10 , 10 and

32 .

22. PQ = 2 |a - b|, QR = 2 |b - c| so PQ = QR if a - b = b - c i.e. a, b, c are in A.P. y-x z-y = fi y–x=z–y 23. b-a c-b (∵ b – a = c – b) fi x, y, z are in A.P. 24. Area is thrice the area of the triangle with third vertex at the centroid. 25. Length of BC is twice the length of the line joining the mid-point of AB and AC. 26. Triangle is right angled at B so mid-point of AC is the circumcentre. 27. Diagonals of a parallelogram bisect each other. h+2+4 , 28. Let C (h, k), then Centroid is (x = 3 k + 5 -1 y = ) and 9h + 7k + 4 = 0 fi 9(3x – 6) 3 + 7(3y – 4) + 4 = 0 is the locus of the centroid parallel to L1. 29. The distance between the given points is less than 12. 30. 2OA ¥ OB cos AOB = (OA)2 + (OB)2 – (AB)2 31.

x + y + 1 = 0, a 1 x+y+ =0 c c -1 a x= . , a -1 c x+y+

1 =0 c

fi

x+

y + 1 = 0, b

y=

c -1 b . b -1 c

a b 1 + + =0 a -1 b -1 c -1

a b c + + =1 a -1 b -1 c -1 32. Coordinates of D are (5 cos a, 4 + 5 sin a) cos a = 4/5, sin a = 3/5 check the coordinates of C. fi

Fig. 16.21

33. Let P divide BC in line Ê l x + x2 ˆ +3 2 Á 3 Ë l + 1 ˜¯ fi

l= -

Similarly

the ratio l : 1. As P lies on the Ê l y3 + y2 ˆ ÁË l + 1 ˜¯ - 1 = 0

2 x2 + 3 y2 - 1 BP = 2 x3 + 3 y3 - 1 PC

2 x + 3 y3 - 1 CQ = - 3 2 x1 + 3 y1 - 1 QA

2 x + 3 y1 - 1 AR = - 1 2 x2 + 3 y2 - 1 RB Sum of rational numbers is a rational number and a rational number divided by 3 is again a rational number. A1 A4 and A2 A5 pass through the mid point of A0 A3 i.e. (0, 0) and make an angle of 60° with A 0 A3 which is x-axis. Slope of the line joining P(2, 2) and S(13/2, 1) is - 2/9. m-7 m - (- 1) =– 1 + 7m 1- m and

34.

35.

36. 37.

38. The diagonals of the rhombus are parallel to the angle bisectors of the given lines. Slope of the diagonal through A(0, a) is 2 - a. x-3 y-4 = =r 39. cosq sinq PR = 3/cos q, PS = 4/sin q RS = PS – PR 40. Let A(p, 0), B(0, q) and C(h, k), then (h - p)2 + k2 = a2, h2 + (k - q)2 = b2 and p2 + q2 = a2 + b2 Eliminate p, q 41. Centre of mean position is the mid point of PQ. 42. Length of the perpendicular from (1, 1) to the line is equal to 1. 43. Mid point (3, 2) of AC lies on BD, c = - 4. Let the coordinates of D be (x, 2x - 4). AD ^ CD fi x = 4, 2. The required coordinates are (4, 4) or (2, 0). As D lies below A, its coordinates are (2, 0)

16.44

44. 45. 46.

47.

Complete Mathematics—JEE Main

x y 1 2 1 1 = 10 fi 3x + y - 7 = 10 3 -2 1 Diagonals of a rhombus are at right angles. (a + c) y+c=0 ax + 2 a(2x + y)+ c(y + 2) = 0 which represents a family of lines passing through the intersection of 2x + y = 0 and y + 2 = 0 L: y = - (1/5)x + c, meets axes at (0, c) and (5c, 0), (1/2) ¥ 5c ¥ c = 5 fi c2 = 2 Distance of L from the origin =

c

=

1 + 1/25

5 2

fi - 21a + 6b = 0 which is satisfied by (b) 56. u = 0, v = 0 are parallel lines but not identical. u + kv = 0 represents a family of lines parallel to these lines. 57. Required line is (x + y + 1) + l(l x + my + n) = 0 such that lx + my + n = 0 is equally inclined to x + y + 1 = 0 and this line. 58. Triangles NDA, MBC and OAB are congruent. fi MB = OA = ND = 12 and CM = OB = NA = ON - OA = 5 OM = OB + BM = 5 + 12 = 17 coordinates of C = (OM, CM) = (17, 5)

26 12

48. The lines enclose a square, the length of each side being 2 2 . 49. Circumcentre, centroid and the orthocentre lie on the same line. 50. SA = a(1 + t2), SB = a(1 + t2)/t2 51. Equation of the bisectors of the angle between the given lines are px + qy + r p2 + q2

=±

qx + py + ( p + q - r ) p2 + q2

Taking the negative sign we get x + y + 1 = 0 so one is the reflection of the other line, in the line x + y + 1 = 0. 52. Let the coordinates of A be (x, y), then y y , tan B = tan C = a-x a+x tan (C - B) = tan (p/3) =

3

53. An (n, n 5 ), (0 An)2 = 6n2, required 6n (n + 1) (2n + 1) . 6 54. Line meets the coordinate axes at (4, 0) and (0, 6). If (x, y) are the coordinates of an extremity of the other diagonal, then (x - 4)2 + y2 = x2 + (y - 6)2 fi 2x - 3y + 5 = 0 which is satisfied by the coordinates in (c). 55. (m + 1) (2m - 1) (m - 2) = 0 fi m1 = - 1, m2 = 1/2, m3 = 2 Sum = 6Sn2 =

A = –Â B=

21 1 = 2 4 mi 1

Âm

i

–

=

3 2

21 3 a+ b =0 4 2

5

12 5

12

Fig. 16.22

59. Let the coordinates of A be (x, y) AM is perpendicular to MD and AM = (1/2) MD. y +1 1+1 ¥ = - 1 fi x - 2y - 4 = 0, then x - 2 1- 2 MD = 5 . which is satisfied by (b) only. x y + = 1. Meets a b the coordinate axes at A(a, 0) and B(0, b) (1, 1) divides AB in the ratio 3 : 4. 61. Lines are perpendicular if a + b = 0 62. Lines are coincident if (h)2 = a . a 63. If a + b = 0, the lines are (x + y)(ax – ay + c) = 0

60. Let the equation of the line be

64. We can write (x - y) (x - 2y) (x - 3y) = 0 65. Let the slopes be m/r, m, mr, then product = m3 = - 8 fi m = - 2 and m satisfies 8 + am + bm2 + m3 = 0 66. Let the slopes be m and m + 1, then 2m + 1 = - a, m(m + 1) = 6. 67. Product of the slopes is 1 = a/3 fi a = 3 68. Vertices of the rectangle are A(1, 4) B(6, 4), C(6, 10) and D(1, 10) Equation of AC is 6x - 5y + 14 = 0 Equation of BD is 6x + 5y - 56 = 0 AC is nearer the origin.

Cartesian System of Rectangular Coordinates and Straight Lines 16.45

69. Mid point is P(335, 1340). Let the required x y + = 1, it passes through P if equation be k 2k k = 1005. x+ 70. Q = (a, b) = ÊÁ xy , Ë 2

yˆ ˜¯ ,

x+ y Ê + yˆ Á x + xy ˜ 2 , mid-point of PQ = Á ˜ 2 2 ÁË ˜¯ fix+

Statement -2 is false as the two sets of lines are x + 1 = 0, y – 2 = 0 and x – 1 = 0, y + 2 = 0 78. Statement -2 is true, the given equation represents points which satisfy x + 4y – 2 = 0, 4x + y – 8 = 0 and x ± y – k = 0. Point of intersection of first two is (2, 0) which also satisfies the third equation if k = 2. Statement -1 is false, as the locus represents a point, the point of intersection of the lines given. 79. Statement-2 is true and using it statement-1 is also sin b + cos b , true as the slope of PQ = cos ( b - a ) + sin ( b - a )

xy = 84, x + 3y = 124. both are satisfied

by (b) (49, 25) 71. Equation of the line through (–1, 3) perpendicular to y = 2x is x + 2y = 5 which meets y = 2x at a -1 (1, 2). Required point is (a, b) such that = 1, 2 b +3 = 2. 2 72. (a2, a) lies between the given lines ... if a2 + a – 6 > 0 and a2 + a – 2 < 0 ... or a2 + a – 6 < 0 and a2 + a – 2 > 0 (1) does not hold for any value of a From (2) – 3 < a < 2 and [a < – 2 or a > 1] fi – 3 < a < – 2 or 1 < a < 2. fi |2a + 5| < 1 or |2a – 3| < 1 73. Coordinates of the vertices are (0, 0), (30, (20, 30), (0, 50). Lengths of the sides are

Slope of QR =

80.

(1) (2)

81.

0), 30,

82.

m1 + m2 2c =2fi =2fic=3 m1m2 3

75. Let the slope of the other line be m then and fi -

2 2c ¥m= 3 - (c 2 - 1)

2 -2 +m= 3 - (c 2 - 1)

2 3c 2 + 2 = 2 3 c -1 c -1

fi 2c2 – 9c + 4 = 0 76. In statement-1, let the vertex C be (a, b), then a + 2 - 2 b - 3 + 1ˆ centroid is ÊÁ , ˜¯ which lies on Ë 3 3 x + y = 5 fi a + b = 17 In statement-2, x – y + 1 = 0 is the equation of the perpendicular bisector of AB which is also the median through C. 77. Statement -1 is true, as the lines are x + 1 = 0, y – 2 = 0, x – 4 = 0

cos ( b - (q 2)) sin ( b - a + (q 2)) Statement-2 is true, using it in statement-1 hypotenuse makes an angle p/4 with x = a which is parallel to y-axis, so it makes an angle p/4 with x-axis also and then the slope of the hypotenuse is ±1 fi k = ± 1, statement-1 is false Statement-2 is false, In statement-1 a cos t + b sin t - 1 a sin t - b cos t ,y= x= 3 3 fi (3x – 1)2 + (3y)2 = a2 + b2, statement-1 is true. 2 1 1 = + Statement-2 is true, using it fi the b a c x y 1 line + + = 0 passes through the point a b c (1, – 2), so the statement-1 is false. Statement-2 is true, equation in statement-1 is =

10 10 , 20 2 , 50. 74.

sin b - sin ( b - q ) cos ( b - a ) - cos ( b - a + q )

83.

( x + 3. y + 5) sin q + ( 3 x - y - 7) cos q = 0 or ( x + 3 y + 5) + tan q ( 3x - y - 7 ) = 0 (cos q π 0) Using statement-2, statement-1 is true. 84. Statement-1, equation of the third line is y = k, vertices are (1, 1),(k, k) and (2 – k, k) 1 1 1 1 2 k k 1 fi (k – 1)2 fi 4h = 2 2–k k 1 So statement-1 is true. Area of the first triangle in 0 0 1 1 statement-2 is 1 1 1 2 2 0 1

= 1 and the area of the

1 second is ÊÁ ˆ˜ (2) (2) = 2 and statement-2 is also Ë 2¯ true but does not lead to statement-1

16.46

Complete Mathematics—JEE Main

85. Statement-2 is false, PQ should be perpendicular to the line and mid point of PQ lies on the line. Using this correct statement, statement-1 is true. 86. Statement-1 is false, the lines are coincident as the statement-2 is true and statement-1 satisfies the condition in statement-2. 87. In statement-1, (x2 – 1) (y2 – 1) = 0 fi x = ± 1, y = ± 1 which enclose a square of each side equal 2, so statement-1 is true. Statement-2 is also true as the sum of the coefficients of x 2 and y2 is zero but does not imply statement-1

1 2c 2c 2 c2 = ¥ ¥ ab 2 a b 92. l1 = a2 cosa + 2a sina + sina tana l2 = ab cosa + (a + b) sina + sina tana l3 = b2 cosa + 2b sina + sina tana fi l1l3 = l22 fi l1, l2, l3 are in G.P. =

93. Length of AB =

26

P

Q

A (-3, 2)

B (2, 3)

Level 2 88. The given triangle is right-angled with vertices A(0, 0), B(2, 0) and C (0, 2). Coordinates of the incentre are (0, 2) C

R (x, y)

Fig. 16.24

Length of PQ =

(0, 1)

2 ¥ 2 - 3 2 ¥ 3 + 2ˆ , Coordinates of Q are ÊÁ ˜¯ Ë 3 3 1 8 = ÊÁ , ˆ˜ Ë 3 3¯

(1, 1)

A (0, 0)

26 3

(1, 0)

R (x, y) lies on the line through Q perpendicular to AB fi y – (8/3) = –5 (x – (1/3))

B (2, 0)

Fig. 16.23

Ê a x1 + b x2 + c x3 a y1 + b y2 + c y3 ˆ , ÁË ˜¯ a+b+c a+b+c

fi 5x + y = 13/3 Equation of AB is x – 5y + 13 = 0 Distance of R from AB is equal to PQ.

Ê 2 2 ¥ 0 + 2 ¥ 2 + 2 ¥ 0 2 2 ¥ 0 + 2 ¥ 0 + 2 ¥ 2ˆ = Á , ˜ 2 2 +2+2 2 2 +2+2 Ë ¯

fi

(

= 2- 2,2- 2

) 5

b2 + c2 - a 2 2 + 5 - 13 -3 = = 2 bc 2¥ 2 ¥ 5 10 90. The given points are collinear. cos A =

91. Area of the rhombus 1 = ¥ product of the diagonals 2 (0, c/b) ax 0 + c by +

(2)

94. The given equation can be written as x( y1 – y2 + y1 – y3) – y(x1 – x2 + x1 – x3) + x 1y 2 – x 2y 1 + x 1y 3 – y 1x 3 = 0 fi x (2y1 – y2 – y3) –y (2x1 – x2 – x3) + y1 (2x1 – x2 – x3) – x1(2y1 – y2 – y3) = 0 2 y1 - y2 - y3 ( x - x1 ) 2 x1 - x2 - x3

which represents the median through A -

=

0

(c/a, 0)

95. Three lines x + 2y – 3 = 0, 2x + 3y – 4 = 0 and 4x + 5y – 6 = 0 are concurrent at (–1, 2). But the fourth line 3x + 4y – 7 = 0 does not pass through this point.

c

=

0

c

c

=

by

+

-

by

0 (0, -c/b)

ax

ax (-c/a, 0) ax +

by

26 3

1 + 25 fi x – 5y + 13 = ± 26/3

fi y – y1 =

=

-

=

Solving (1) & (2) we get R (x, y) = (0, 13/3) or (2/3, 1)

89. Sides of the triangle are a = BC = 4 + 9 = 13 , b = 2 , c = The greatest angle A is given by

x - 5 y + 13

(1)

Fig. 16.24

96. Slope of the incident ray PA is 2/(1 – x) and that of the reflected ray QA is 3/(5 – x)

Cartesian System of Rectangular Coordinates and Straight Lines 16.47

13 13 2 -3 fi A ÊÁ , 0ˆ˜ = fix= Ë5 ¯ 5 5- x 1- x

so

Q (5, 3)

P (1, 2)

and y =

-t ( x + u) 1+ t

fi [(1 + u) – (1 + t)] y + (u – t) x = 0 fi (u – t) (x + y) = 0 fi x + y = 0 as u π t

A (x, 0)

99. Equation of AU is

Fig. 16.26

0 1 97. D1 = | a tan a 2 a sin a

y – y1 =

0 1 b cot a 1 | b cos a 1

0 - y1 ( x - x1 ) 3 - x1

so coordinates of C are (0, 3y1/(3 – x1)) C

1 ab (sin a - cos a ) 2 a b 1 1 2 2 D2 = a sec a b cos es a 1 2 a + a sin 2 a b + b cos 2 a 1 =

=

=

1 ab |(sin2 a – cos2 a)| 2 0 1 a tan a 2 a sin a

But for these values of a, the triangles are not defined. Hence D1, D2, D3 are in G.P. for no value of a. 98. The orthocentre (x, y) lies on the lines.

(1

+

x

-

B (- t, 0)

-

y=0

Fig. 16.27

y=

-u (x + t) 1+ u

Fig. 16.28

Equation of CV is y=

fi a = ± p /4 or a = ± p /2

t)

B (6, 0)

6 y1 ˆ Ê 6 x1 V= Á , Ë 6 + x1 6 + x1 ˜¯

fi |sin2 a – cos2 a | = 0 or 1

1 t(

A (3, 0)

Solving, we get the coordinates of

Now D1 D3 = (1/4) a2 b2 |sin2a – cos2a | = D 22

+

(2, 0)

Equation of OU is y1x = x1y

1 ab | (sin a + cos a ) | 2

ty

O

Similarly coordinates of D are (0, 6y1/(6 – x1)) x y (6 - x1 ) =1 Equation of AD is + 3 6 y1

0 1 - b cot a 1 b cos a 1

( 0 1+ = u) ) A x +t

U (x1, y1) V

1 ab [cosee2 a – 1 – cos2 a + 1 + sin2 a – 2 sec2 a + sec2a + 1 – cosec2a – 1]

and D3 =

=

D

uy

+

u(

1+

u)

=

0

C (- u, 0)

3 y1 3 y1 3 y1 Ê x= Á1 3 - x1 2(3 - x1 ) 3 - x1 Ë

xˆ ˜ 2¯

which passes through (2, 0) for all values of (x1, y1) 100. The triangle is equilateral, so the incentre is the centroid (1, 1 / 3 ) of the triangle 101. Equation of the fourth side is 2x – y – k = 0 4-k 3+7 such that = ± fi k = – 6 or 14. 5 5 So the required equation is 2x – y + 6 = 0, as it is nearer to (1, –1) 102. The triangle is right angled at (0, 0) so the orthocentre is (0, 0) and circumcentre is (a/2, b/2), the mid point of the hypotenuse.

16.48

Complete Mathematics—JEE Main

103. Third vertex of the triangle is (–1, –11) fi a + b – 6h +

8h3 =0 ab

104. A(0, 12), B(8, 0), C(x, y) lies on the line x y 1 1 2x – 3y + 10 = 0 and 0 12 1 = ± 91 2 8 0 1 Solving we get (x, y) = (–13/2, – 1) or (29/2, 13)

a + b 8h 2 = 6. + h ab 108. From the given equation we have

1 +7 = –3 105. tan A = 2 7 12

Previous Years’ AIEEE/JEE Main Questions

abc + 2fgha – af 2 – bg2 – ch2 = 0 as they intersect on y-axis, by2 + 2fy + c = 0

4x - 3y - 5 = 0 D C

+

3y

7x +

fi +

10

=

0

x - 2y = 0

B

af 2 = abc and from (1) we have 2fgh = bg2 + ch2.

2. Let (h, k) be mid-point of the portion, then p p ,k= h= 2cosa 2sina fi

fi tan (A + C) = 0

fi

fi A + C = 180°

1 = cos2a + sin2a = 1 1 4 + 2 = 2 2 h k p

fi ABCD is a cyclic quadrilateral Note: ABCD cannot be a rectangle or a parallelogram, so verify for cyclic quadrilateral. 106. If given lines are concurrent at (–35, 22), then 35a – 22b – 1 = 0 fi The line 35x – 22y – 1 = 0 passes through (a, b). y 2 2h Ê y ˆ a 107. We can write ÊÁ ˆ˜ + Á ˜+ =0 Ë x¯ b Ë x¯ b If the slopes of the lines are m and m2 then a 2h m + m2 = and mm2 = b b 1/ 3

a + ÊÁ ˆ˜ Ë b¯

2/3

a a 2 a fi + ÊÁ ˆ˜ + 3 ÊÁ ˆ˜ Ë b¯ b Ë b¯

2h = b 3 Ê - 2 h ˆ = - 8h ˜ ÁË b¯ b3

Locus of (h, k) is

8h a [b + a - 6 h ] + 3 = 0 2 b b

p2 p2 + 4h 2 4k 2

1 1 4 + 2 = 2 2 x y p

3. Let A (4, 0), B (–1, – 1), C (3, 5) (AB)2 = 26, (BC)2 = 52, (AC)2 = 26 so the triangle is right angled. 4. 3x + 4y = 0 and 4x – 3y = 0 are at right angle so the triangle is right angled. 5. Coordinates of A are (a cos a, a sin a) and of C are (a cos(a + p/2)a sin(a + p/2)) i.e. (–a sina, a cosa) So the equation of the diagonal AC is y – a sina = fi

a(sin a - cos a ) ( x - a cos a ) a(cos a - sin a )

y(cosa + sina + x (sina – cosa) = a B C

A a

3

fi

(2)

(2) gives a unique value of y if 4f = 4bc x

Fig. 16.29

a fi ÊÁ ˆ˜ Ë b¯

(1)

2

y-

40 =

0

x – y + c = 0 and x + y – c = 0 which represents the point (0, c.)

1. Since the given equation represents a point lines we have

4 1 + tan C = 3 3 = 3 4 1 1- ¥ 3 3

A

fi

O

Fig. 16.30

x

Cartesian System of Rectangular Coordinates and Straight Lines 16.49 2

2

6. Equation of the bisectors of x – 2pxy – y = 0 is xy x2 - y 2 = -p 2 or x 2 +

2 xy - y 2 = 0 p

x y x y - = 1 and + =1 2 3 -2 1 12. Slopes m1, m2 of two lines are given by 1 – 2cm – 7m2 = 0 We are given

Comparing with x2 – 2qxy – y2 = 0 we get pq = –1 7. Let (h, k) be the centroid of the triangle, then h=

a cos t + b sin t + 1 a sin t - b cos t ,k= 3 3

fi (3h – 1)2 + (3k)2 = a2 +b2 Locus of (h, k) is (3x – 1)2 + (3y)2 = a2 + b2 8. Let x2 = x1r, x3 = x1r2 y2 = y1r, y3 = y1r2 then x1 y1 1 x1 x2 y2 1 = x1r x3 y3 1 x1r 2

y1 y1r

1 1=0

y1r 2 1

13. Slopes of the two lines are connected by 6 – m + 4cm2 = 2 3 As y = - x is one of the lines 4 3 3 2 6 + + 4c Ê - ˆ = 0 Ë 4¯ 4 9 -27 c= fi c = -3 4 4 14. A straight line passing through the point of intersection of given lines is

(a1 + a2 ) (b + b ) + (b1 - b2 ) 1 2 + c = 0 2 2 1 2 c= a2 + b22 - a12 - b12 2

(

)

10. Let the third vertex be C(h, k). Centroid of DABC is Ê h k - 2ˆ 1 2 a2 + b22 - a12 - b12 = G Ë 3 , 3 ¯

( 2

ax + 2by + 3b + k(bx – 2ay – 3a) = 0 or

(a1 - a2 )

2h k - 2ˆ = 1 fi 2h + 3k = 9 + 3Ê Ë 3 3 ¯ 11. Let the intercepts on the axes be a and – a – 1. Equation of such a straight line x y =1 a a +1 fi

As it pass through (4, 3) we get fi 4a + 4 – 3a = a(a + 1)

fi a2 = 4 fi a = ± 2 Thus, required lines are

(a + kb)x + 2(b – ka)y + 3(b – ka) = 0

This line will be parallel to the x-axis if a + kb = 0 or k = –a/b Thus, the line parallel to the x-axis is 2 2 3 (a + a 2 ) y + (b 2 + a 2 ) = 0 b b fi y = –3/2

or 2y + 3 = 0

This line is below the x-axis at a distance of (3/2) from it. 15. Let coordinates of B be (x, y). Then x +1 y +1 = -1, =2 2 2

)

which lies on 2x + 3y = 1.

4 3 =1 a a +1

-2c Ê -1ˆ = 4Á ˜ fi c = 2 Ë 7¯ 7

fi

So the given points lie on a straight line. 9. Mid-point Ê a1 + a2 , b1 + b2 ˆ of the given points lies Ë 2 2 ¯ on the locus, so

fi

fi

m1 + m2 = 4m1m2

fi

x = –3, y = 3

\ Coordinates of B( – 3, 3) Similarly coordinates of C are (5, 3) Thus, centroid of DABC is Ê 1 - 3 + 5 , 1 + 3 + 3 ˆ = Ê1, 7 ˆ Ë 3 3 ¯ Ë 3¯ 16. a, b, c are in H.P. fi fi

1 1 1 , , are A.P. a b c 1 2 1 - + =0 a b c

Fig. 16.31

Complete Mathematics—JEE Main

16.50

Thus, x + y + 1 = 0 passes through (1, –2). a b c 17. As area of one of the sectors thrice that of the other, p – q = 3q or q = p/4.

20. RQX = 60∞ Let QS be the bisector of PQR , then RQS = 60∞ fi

SQX = 120∞

fi Slope of QS = - 3 \ Equation of QS is y = – 3 x or

3x + y = 0

Fig. 16.32

Thus, angle between two lines is p/4 or 3p/4 2 (a + b) 2 - ab a+b

\

±1 =

fi

(a + b)2 = 4(a2 + b2 + ab)

fi

3a2 + 2ab + 3b2 = 0

18. Let equation of straight line be x y + =1 a b It meets the axes in P(a, 0) and Q(0, b). As the midpoint of PQ is A(3, 4), we get Ê a , b ˆ = (3, 4) ˜ ÁË 2 2¯ fi a = 6, b = 8 Thus, required line is x y + =1 6 8 or

4x + 3y = 24 2

2

21. Bisectors of lines xy = 0 i.e. x = 0, y = 0 are y = ± x. For m = 1, the given equation reduces to y2 – x2 = 0 k +1 7ˆ 22. Mid-point of PQ is Ê , Ë 2 2¯ Equation of the perpendicular bisector of PQ in 7 k -1 Ê k - 1ˆ y- = xË 2 3- 4 2 ¯ y-intercept = fi

k2 = 16

(k - 1)(k + 1) 7 + =4 2 -2 fik=±4

23. Slope of two lines are 2

19. AB + BC = AC fi (h – 1)2 + (k – 1)2 + 1 = (h – 2)2 + (k – 1)2 fi h=1 Area of DABC = 1 1 (1) |k - 1| = 1 fi 2 fi |k – 1| = 2 fi k – 1= ± 2 fi

Fig. 16.34

k = 3, –1

Fig. 16.33

m1 =

1 1 1 =- 2 , m2 = 2 2 +1 p + 1 p p ( ) p( p + 1)

Two lines will be perpendicular to a common line if these two lines are parallel. 1 1 \ =- 2 2 p +1 p p +1

(

)

fi p=–1 For p = –1, two lines become – 2x – y + q = 0 and 4x + 2y + 2q = 0 which are parallel. 24. As L: x + y = 1 passes through (13, 32), we get 5 b 32 13 8 13 32 = 1- = + =1 fi b 5 5 5 b fi b = – 20 x y \ equation of line L is =1 5 20

Cartesian System of Rectangular Coordinates and Straight Lines 16.51

As K: x + y = 1 is parallel to L. c 3 1 -1 5 = 20 fi c = -3 fi c = -3 1 1 5 20 4 c 3 -4 x y Thus, equation of K is + =1. 3 3 \ Equation of L and K are 4x – y – 20 = 0 and 4x – y + 3 = 0 Distance between L and K is | -20 - 3 | 23 = 16 + 1 17 25. We have OP = 2 2 and OQ =

28. Coordinates of the point dividing the join of the points (1, 1), (2, 4) in the ratio 3 : 2 are Ê 3 ¥ 2 + 2 ¥ 1 , 3 ¥ 4 + 2 ¥ 1ˆ = Ê 8 , 14 ˆ ˜ ÁË 3+ 2 ¯ Ë5 5 ¯ 3+ 2 2x + y = k passes through this points. if 2 ¥ (8/5) + (14/5) = k

fi

k=6

29. Let equation of line through (1, 2) be y – 2 = m(x – 1). Ê m - 2 , 0ˆ It meets the coordinate axes in P Á ˜¯ Ë m Q(0, 2–m). Let A = Area of DOPQ =

5.

and

1 m-2 |m-2| 2 m

1 (m - 2) 2 2 |m| 1 4 4m = |m|+ |m| |m| 2 =

R divides PQ is the ratio 2 2 : 5 , that is, PR: RQ = 2 2 : 5 However, statement-2 is false.

2 m ˆ˘ 1 ÈÊ 2 ˆ Ê + 4 Á1 = ÍÁ | m | ˙ ˜ Ë | m | ˜¯ ˙ 2 ÍÎË | m |¯ ˚

Note A is least if |m| = 2 or m = ± 2. For m = 2, the line passes through the origin, and P and Q coincide. \ Fig. 16.35

26. If a > 0, the given lines intersect at the point (1, a a – 1 > 0 fi a > 1. If a < 0, then the equation ax – y = 1 gives y = ax – 1 which is negative for all positive values of x. Hence a < 0 is not possible and the required values of a are in the interval (1, •).

required value of m is –2.

30. Ray of light x + 3 y = 3 meets the x-axis at ∞ with ( 3 the x-axis. Its equation is 1 y-0= ( x - 3) fi 3 y = x - 3 3

27. Let G(h, k) be centroid of D ABC, then coordinates of C are (3h – 2 + 2, 3k + 3 – 1) = (3h, 3k + 2) As C lies on 2x + 3y = 9, we get 2(3h) + 3 (3k + 2) = 9 fi

2h + 3k = 1

Thus, locus of centroid G is 2x + 3y = 1

Fig. 16.37

31. Solving the given equation we get 8y + y2 = 1 3 1 fi y= 3 Fig. 16.36

fi

3y2 + 8y – 3 = 0

or

y = – 3 but y π –3.

1 and the line 3y – 1 = 0 passes through the 3 point of intersection of the the two curves. so y =

16.52

Complete Mathematics—JEE Main

parallel. Hence q1 = q2 for all c1 and c2. So statement-2 is true and thus it follows that statement-1 in also true.

32. Slope of PQ = – (Slope of RQ) fi

3 -7 = 1- x 6 - x

fi x=

38. Slope of PQ = 1. Mid point of PQ is (3, 4). So equation of L is y – 4 = – (x – 3)

5 2 Fig. 16.38

33. Slopes of the given lines are

fi

x+y–7=0

1 a , - , -a 3 2 Two of the lines are perpendicular. if

1 Ê aˆ = -1 3 Ë 2¯

fi

a=6

fi

a

or

or

1 (- a ) = -1 3 Fig. 16.40

a=3

Let the image of q (0, 0) be R (h, k) the slope of OR = slope of PQ

a2 – 9a + 18 = 0 34. Vertices of the triangle are A(0, 0), B(2, 0), C(0, 2). AB = AC = 2 and BC = 2 2 So x-coordinates of the incentre is

and the required point is (7, 7)

2¥0+2¥2+2 2 ¥0

39. See Question No. 10 (2004)

2+2+2 2

40. Let the point of intersection be (h, – h) then

4 = = 2- 2 4+2 2 35. 3x + 4y = 12 fi

4ah – 2ah + c = 0 and 5bh – 2bh + d = 0 fi

x y + =1 4 3

Equation of L is

4 x + =1 8 3/ 2

BC = AC fi (2a)2 + (y – a)2 = y2 5a 2 Area of the triangle 1 1 = AC ¥ 2a = y ¥ 2a = ay 2 2

h=

-c - d = 2a 3b

fi

3bc – 2ad = 0

41. Coordinates of S are ÊÁ 13 , 1ˆ˜ . Ë2 ¯ 2 -1 2 Slope of PS is =13 9 22 Required equation of the live is 2 y + 1 = - ( x - 1) 9

3 1 3 Slope of L = - ¥ = 2 8 16 36. Let A(2a, 0), B(0, a) and C(2a, y)

fi

k =1 fi k = h h h k Also mid-point ÊÁ , ˆ˜ of OR lies on L Ë 2 2¯ \ h + k -7=0 fi h = k = 7 2 2

fi

fi

2x + 9y + 7 = 0

42. Let (h, k) be the foot of perpendicular from (0, 0) to x 4 the line + = 1. If this length of perpendicular is p, then a b

y=

Fig. 16.39

5a 2 sq. units 2 37. 2x + 3y + c1 – x + 5y + c2 = 0 and – x + 5y + c1 = 0 as they are

p=

=

0 0 + -1 a b (1 + a )2 + (1/ b) 2

=2

Now, h2 + k 2 = p = 2 Fig. 16.41

Cartesian System of Rectangular Coordinates and Straight Lines 16.53

fi h2 + k2 = 4

fi

Thus (h, k

Thus, (a, b) lies on a straight line.

43. Coordinates of R are (2, 0). As PQ is parallel to the x-axis, and meets x – 2y = 2 in Q, coordinates of Q are (8, 3). Centroid of DPQR is (5, 2) and it lies on 2x – 5y = 0

– 2a(b – a) = 0 fi a = b

of a triangle lie on a straight line, say L. centroid of triangle is Ê a 2 + 1 + 2a a 2 + 1 - 2a ˆ = Ê 1 1 , (a + 1) 2 , (a - 1) 2 ˆ ÁË ˜ Ë ¯ 2 2 2 2 ¯ Equation of line L is x-0 y-0 = 2 (a + 1) / 2 - 0 (a - 1) 2 / 2 - 0 fi

48. An equation of line perpendicular to 5x – y = 1 is x + 5y = c

Fig. 16.42

44. If a is the side of the equilateral triangle, then 3 a = AM 2 | 3(1) + 4(2) - 9 | = 32 + 42 fi

a=

45. We have

2 2 4 ◊ = 3 5 3 15

(a – 1)2 x – (a + 1)2 y = 0

Now, 5 = Area of DOAB =

1 c |c| 2 5

fi

c2 = 50

fi

|c| = 5 2

Fig. 16.45

Distance between x + 5y = c and x + 5y = 0 is Fig. 16.43

|c| 1+ 5

Ê 2h , k ˆ = (4, 3) ˜ ÁË 3 3¯ fi h = 6, k = 9

2

=

5 2 5 = 26 13

49. For 1 £ k £ 40, the number of integral points on the line x = k and lying in the interior of triangles is k – 1.

Equation of line is x y + =1 6 9

40

or 3x + 2x = 18

Thus, required number of points = Â (k - 1) k =1

1 (39)(40) = 780 2

Fig. 16.44

46. As the lines are distinct a π 0. As the lines are concurrent. 1 2a a 1 3b b = 0 1 4a a R3 ÆR3 – R1, 1 1 0

we get 2a a 3b b = 0 2a 0

Fig. 16.46

50. (2x – 3y + 4) + k(x – 2y + 3) = 0, k ŒR, represents a family of lines passing through the intersection of the lines 2x – 3y + 4 = 0 and x – 2y + 3 = 0, that is, through (1, 2). If (x, y) is image of (2, 3) in (1), then both (x, y) and (2, 3) are equidistant from (1, 2), thus, (x – 1)2 + (y – 2)2 = (2 –1)2 + (3 – 2)2 = 2. Thus, (x, y) lies on a circle of radius

2

Complete Mathematics—JEE Main

16.54

51. Let slope of required line L be m, then ± tan 60º = fi ± 3

and x – y – 3 = 0 Ê 1 8ˆ Thus a vertex of rhombus is Á , - ˜ Ë 3 3¯

( ) 1 + m (- 3 ) m- - 3

55. An equation of line through intersection of 1 1 1 1 x + y - 1 = 0 and x + y - 1 = 0 is 3 4 4 3

= m+ 3

fi m = 0 or m = - 3 As the desired line intersects the x-axis, m = - 3 .

Ê1 or Á + Ë3

Thus, equation of required line is y + 2 = - 3 ( x - 3) or

52. Slope of line joining A(0, 8/3) and B(1, 3) is 3 - 8/ 3 1 = 1- 0 3

Let (a, b) be the mid-point of AB, therefore

and slope of line joining B(1, 3) and C(82, 30) is m2 =

30 - 3 1 = 82 - 1 3

53. Suppose L meets the axes at A(a, 0) and B(0, b). As P(1, 2) is the mid-point of AB, b a = 1 and = 2 2 2

7 ( k + 1) 1 1 7 = + = 2a 2b 12 ( k + 1) 12

fi 6(a + b) = 7a b Thus, locus of mid-point AB is 6(x + y) = 7xy 56. An equation of line through (2, 1) and parallel to x – y = 4 is (x – 2) – (y – 1) = 0 or x – y – 1 = 0.

Its distance from (2, 1) is d=

Fig. 16.47

x y (1) + = 1 or 2x + y = 4 2 4 and the equation of L1 is (x + 2) – 2(y – 1) = 0 or

12 ( k + 1) 12 ( k + 1) , 2b = 4 + 3k 3 + 4k

A point on this line is (t, t – 1).

a = 2, b = 4.

Thus, equation of line L is

2a = fi

As m1 = m2, AB || AC. But B is common, therefore A, B, C are collinear.

fi

kˆ Ê 1 kˆ ˜¯ x + ÁË + ˜¯ y - (1 + k ) = 0 4 4 3

Ê 12 ( k + 1) ˆ , 0˜ and the It meets the x-axis is in A Á Ë 4 + 3k ¯ 12 k + 1 ( ) Ê ˆ y-axis in B Á 0, Ë 3 + 4 k ˜¯

y + 3x + 2 - 3 3 = 0

m1 =

1 1 Ê1 ˆ Ê1 ˆ ÁË x + y - 1˜¯ + k ÁË x + y - 1˜¯ = 0 3 4 4 3

x – 2y + 4 = 0

(2)

solving (1) and (2) we get the required point as Ê 4 , 12 ˆ Ë5 5 ¯ 54. A vertex of the rhombus is (1, 2), the point of intersection of x – y + 1 = 0 and 7x – y – 5 = 0

Now

(t - 2)2 + (t - 1 - 1)2 = 2 t - 2 2 t-2 =2 3

fi t – 2 = ± 6fit =2± 6 As the point lies in the third quadrant, we take t=2– 6 \ point Q is (2 –

6)

An equation of required line through Q and perpendicular to L is

Another vertex of the rhombus is

(x – (2 –

(2(– 1) – 1, 2(– 2) – 2) = (– 3, – 6)

or x + y = 3 – 2 6

Other two sides to the rhombus are

6,1–

6 )) + (y – (1 –

57. Two given lines are

(x + 3) – (y + 6) = 0 and 7(x + 3) – (y + 6) = 0

4x + 3y – 10 = 0

or x – y – 3 = 0 and 7x – y + 15 = 0.

and 8x + 6y + 5 = 0

A vertex can be obtained by solving 7x – y – 5 = 0

These lines are parallel.

6 )) = 0

Cartesian System of Rectangular Coordinates and Straight Lines 16.55

2. Let the point of intersection be (h, – h); h > 0 then 2ph – 3qh + r = 0 and ph + 2qh – 2r = 0, A C

4x

O

D

+ 10

+ 8x

fi

p + 2q + 2(2p – 3q) = 0

fi

5p – 4q = 0

=

+ 6y

0

5 = 0

Fig. 16.48

Thus OA : OB = OC : OD 10 5 : =4:1 4 8 58. An equation of normal AN is (x – 0) + 7(y – 1) = 0 =

3. Mid-points of the sides of the triangle ABC are P(0, 3), Q - 3, 0 , R 3, 0

(

)

(

)

PQR is an equilateral triangle with each side of length 2 3 . So its in-centre is same as its centroid and hence its coordinates are (0, 1). 4. (1 + 2l1)x – (1 + l1)y + (1 – 2l1) = 0 (5 + 3l2)x + (3 – l2)y – (2 + 4l2) = 0 They represent the same line.

or x + 7y – 7 = 0

if

N Q

2x

a

y+

a

=

1

P

A 7x - y + 1 = 0 (0, 1)

Fig. 16.49

Let an equation of PA be y = mx + 1, then –PAN = –NAQ

mˆ Ê ˆ Ê1 fi Á + m˜ (9) = 13 Á 1 - ˜ Ë Ë7 ¯ 7¯ 13 ˆ 9 Ê fi Á 9 + ˜ m = 13 Ë ¯ 7 7 fi 76m = 82 fi m = 41/38 Thus, required equation is 41 y= x +1 38 or 41x – 38y + 38 = 0

Previous Years' B-Architecture Entrance Examination Questions 1. Let the lines intersect at the point k. then (a + 2)k + 3a = 0

and 4k + a – 2 = 0

a-2 3a = 4 a+2

a2 – 12a + 4 = 0

fi

-2 - 4l2 5 + 3l2 3 - l2 = = 1 - 2l1 1 + 2l1 -1 - l1

Solving, we get l1 = – 3, l2 = – 25 and the required equation of the straight line is 5x – 2y – 7 = 0 5. Statement-2 is true and the statement-1 represents the lines given in statement so statement-1 is also true. 6. See Example 108. Page 15.26. 7. m1 + m2 = -

(

)

3 + 2 , m 1m 2 =

3 -1

Ê 1 ˆ Ê -1 ˆ vertices the triangle are (0, 0), Á ,1˜ , Á ,1 Ë m1 ¯ Ë m2 ˜¯ so the area of the triangle is

- 1 7 - m -2 - ( -1 7) = 1- m 7 1 + (2 7)

fi

r = 2 p - 3q

3y

B

-2r p + 2q

fi

which gives us two values of a.

0 1 1 | 2 m1

0 1 1 1 |=

1È 1 1 ˘ + Í 2 Î m1 m2 ˙˚

-1 1 1 m2 =

1 m1 + m2 1 - ( 3 + 2) 1 È 3 + 2˘ = = Í ˙ 2 m1m2 2 3 -1 2 Î 3 -1˚

8. Statement-1 is true but the statement-2 is false as the slopes of parallel lines are same. 9. Equation of a line through the point of intersection of the given lines is (1 + 2l)x + (2 – l)y – ( 1 + l) = 0 1+ l ˆ coordinates of A are Ê , 0 and of B are Ë 1 + 2l ¯

16.56

Complete Mathematics—JEE Main

fi Slope of AC = 3

Ê 0, 1 + l ˆ Ë 2- l¯

\ Equation of AC is

If (h, k) is the mid-point of AB, then

y – 3 = 3(x + 2) or 3x – y + 9 = 0

h =

1+ l , 1+ l 2(1 + 2l ) 2(2 - l )

Let m2 = slope of CH =

1- 3 1 = -6+2 2

fi Slope of AB = – 2

1 + 2h 4k - 1 fi l= = 4h - 1 2k + 1

\ Equation of AB is y + 2 = – 2(x – 3)

fi

(1 – 2h) (2k + 1) = (4h – 1)(4k – 1)

fi

2h + 6k – 20 hk = 0

or 2x + y – 4 = 0

(2)

As A lies on (1) and (2), coordinates of A are (– 1, 6) and it does not lie on 5x + y = 2.

Locus of (h, k) is x + 3y = 10xy. 10. The point of intersection of the given lines is (– 2a, 0) and the equation of the line parallel to y-axis passing through this point is x = – 2a , (p, 5) lies on it so p = – 2a. 11. [3x – 4 (– 3x) – 8] [3(3) – 4(4) – 8] < 0

13. An equation of line through (1, 2) is y – 2 = m(x – 1) A point on this line is of the form Q(x, mx – m + 2). This point will lie on the line x + y = 7 if x + mx – m + 2 = 7 fi x =

fi (15x – 8) (– 17) < 0 fi 15x – 8 > 0

As PQ = 3, PQ2 = 9

8 . 15 Also, y = – 3x < – 8/5.

fi (x – 1)2 + (mx – m)2 = 9

fi

x>

m+5 m +1

fi (x – 1)2 1 + m2 = 9 fi 16m2 + 16 = 9m2 + 18m + 9

12. Let m1 = slope of

fi 7m2 – 18m + 7 = 0

1 - ( -2 ) -1 = BH = -6 - 3 3

fi m satisfies the equation 7x2 – 18x + 7 = 0 A

H (-6, 1) B(3, -2)

C(-2, 3)

Fig. 16.50

(1)

CHAPTER SEVENTEEN

Circles and Systems of Circles

DEFINITION OF A CIRCLE A circle is the locus of a point which moves in a plane, so that its distance from a fixed point in the plane is always constant. The fixed point is called the centre of the circle and the constant distance is called its radius.

EQUATIONS OF A CIRCLE 1. An equation of a circle with centre (h, k) and radius r is (x – h)2 + ( y – k)2 = r2 Illustration

1

A circle of area 9p square units has two of its diameters along the lines x + y = 5 and x – y = 1. Find the equation of the circle. Solution: The centre of the circle is the point of intersection of the diameters x + y = 5 and x – y = 1, which is (3, 2). If r is the radius of the circle, then p r2 = 9p fi r = 3 and the equation of the circle is (x – 3)2 + (y – 2)2 = 32 fi x2 + y2 – 6x – 4y + 4 = 0 2. An equation of a circle with centre (0, 0) and radius r is x2 + y2 = r2 3. An equation of the circle on the line segment joining (x1, y1) and (x2, y2) as diameter is (x – x1) (x – x2) + ( y – y1) ( y – y2) = 0 Illustration

2

The line 3x + 4y – 12 = 0 meets the coordinate axes at A and B. Find the equation of the circle drawn on AB as diameter. Solution: 3x + 4y – 12 = 0 meets x-axis at A(4, 0) and y-axis at B(0, 3). Equation of the circle on the line joining A and B as diameter is (x – 0) (x – 4) + (y – 0) (y – 3) = 0 fi x2 + y2 – 4x – 3y = 0 4. General equation of a circle is

x 2 + y2 + 2g x + 2f y + c = 0 where g, f and c are constants. (i) Centre of this circle is (– g, –f ). g 2 + f 2 - c , ( g2 + f 2 ≥ c).

(ii) Its radius is

(iii) Length of the intercept made by this circle on the x-axis is 2 g 2 - c if g2 – c ≥ 0, and that on the y-axis is 2

f 2 - c if f 2 – c ≥ 0.

Illustration

3

Show that x + y – 1 = 0 is the equation of a diameter of the circle x2 + y2 – 6x + 4y + c = 0 for all values of c. Ê 1 1ˆ Solution: Centre of the circle is Ë- , - ¯ 2 2 i.e (3, – 2) which lies on the line x + y – 1 = 0 and any line passing through the centre of the circle is a diameter of the circle. Hence x + y – 1 = 0 is a diameter of the circle. 5. General equation of second degree a x2 + 2hxy + by2 + 2gx + 2f y + c = 0 in x and y, represents a circle if and only if (i) coefficient of x2 equals coefficient of y2, i.e., a = b π 0. (ii) coefficient of x y is zero, i.e., h = 0. (iii) g2 + f 2 – ac ≥ 0 Illustration

4

Find the centre and radius of the circle 3x2 + (a + 1)y2 + 6x – 9y + a + 4 = 0 Solution: Since the given equation represents a circle, co-efficient of x2 = co-efficient of y2 fi 3 = a + 1 fi a = 2 and the equation of the circle becomes 3x2 + 3y2 + 6x – 9y + 6 = 0 or x2 + y2 + 2x – 3y + 2 = 0 whose centre is (– 1, 3/2) and radius =

2

1 + (3 / 2 ) - 2 =

5 . 2

17.2

Complete Mathematics—JEE Main

SOME RESULTS REGARDING CIRCLES 1. Position of a point with respect to a circle. Point P(x1, y1) lies outside, on or inside a circle S ∫ x2 + y2 + 2g x + 2 f y + c = 0, according as S1 ∫ x 21 + y21 + 2g x1 + 2 f y1 + c >, = or < 0. Illustration

Solution: Let the radius of the circle be r, so its equation is x2 + y2 = r2. As the line y = 3 x + 4 touches this circle, (4)2 = r2 (1 + ( 3 )2) so the required radius is 2. 8. An equation of the chord of the circle S = x2 + y2 + 2gx + 2fy + c = 0, whose mid-point is (x1, y1), is T = S1, where

5

Find the values of a for which the point (a, a), a > 0, lies outside the circle x2 + y2 – 2x + 6y – 6 = 0 Solution: The point (a, a) lies outside the given circle if a2 + a2 – 2a + 6a – 6 > 0 fi 2 (a2 + 2a – 3) > 0 fi (a + 3) (a – 1) > 0 fi a > 1 as a > 0 2. Parametric coordinates of any point on the circle (x – h)2 + ( y – k)2 = r2 are given by (h + r cos q, k + r sin q), with 0 £ q < 2p. In particular, parametric coordinates of any point on the circle x2 + y2 = r2 are (r cos q , r sin q) with 0 £ q < 2p. 3. An equation of the tangent to the circle x2 + y2 + 2g x + 2f y + c = 0 at the point (x1, y1) on the circle is x x1 + y y1 + g (x + x1) + f ( y + y1) + c = 0 4. An equation of the normal to the circle x2 + y2 + 2gx + 2 f y + c = 0 at point (x1, y1) on the circle is y - y1 x - x1 = y1 + f x1 + g

T ∫ xx1 + yy1 + g (x + x1) + f ( y + y1) + c and S1 ∫ x12 + y12 + 2gx1 + 2fy1 + c In particular, an equation of the chord of the circle x2 + y2 = r 2, whose mid-point is (x1, y1), is xx1 + yy1 = x12 + y12. 9. An equation of the chord of contact of the tangents drawn from a point (x1, y1) outside the circle S = 0, is T = 0. (S and T are as defined in (8) above.) 10. Length of the tangent drawn from a point (x1, y1) outside the circle S = 0, to the circle, is in (8) above.) Illustration

x x1 + y y1 = r 2

and

4

2

are tangents to the circle

2

x + y = r , for all finite values of m. If m is infinite, the tangents are x ± r = 0. Illustration

C(2, -3)

Fig. 17.1

Solution: Centre of the circle is (2, – 3) and radius

6. The line y = m x + c is a tangent to the circle x2 + y2 = r2 if and only if c2 = r2 (1 + m2). 2

P(3, 4)

x y = x1 y1

Find the equation of the tangent to the circle x2 + y2 = 25 at the point in the first quadrant where the diameter 4x – 3y = 0 meets the circle. Solution: The diameter meets the given circle at the point (3, 4) in the first quadrant. Equation of the tangent to the circle at this point is x(3) + y(4) = 25 fi 3x + 4y = 25

2

8 T

6

7. The lines y = m x ± r 1 + m

S1 . (S and S1 are as defined

Find the length of a tangent drawn from the point (3, 4) to the circle x2 + y2 – 4x + 6y – 3 = 0.

5. Equations of the tangent and normal to the circle x2 + y2 = r2 at the point (x1, y1) on the circle are, respectively,

Illustration

fi r2 = 4 fi r = 2

7

The line y = 3 x + 4 touches a circle with centre at the origin. Find the radius of the circle.

= 22 + (-3)2 + 3 = 4. So if PT is the tangent from P(3, 4) to the circle, then (PT)2 = (PC)2 – (CT)2 = (3 – 2)2 + (4 + 3)2 – 42 = 34. The required length of the tangent is

34 .

Note If S ∫ x2 + y2 – 4x + 6y – 3 then length of the tangent from P(3, 4) to this circle is S = (3)2 + (4)2 - 4(3) + 6(4) - 3 = 34 . 11. Two circles with centres C1(x1, y1) and C2(x2, y2), and radii r1, r2 respectively, (i) touch each other externally if |C1 C2| = r1 + r2. The point of contact is Ê r1 x2 + r2 x1 r1 y2 + r2 y1 ˆ , ÁË r + r r1 + r2 ˜¯ 1 2

Circles and Systems of Circles 17.3

9

Illustration

Show that the circles x2 + y2 – 2x = 0 and x2 + y2 – 8x + 12 = 0 touch each other. Find the point of contact. Solution: Centre of the first circle is (1, 0) and its radius is 1. Centre of the second circle is (4, 0) and its radius in 2. Since the distance between the centres = 3, which is equal to the sum of the radii. The two circles touch each other externally and the point of contact is the point (2, 0) which divides the join of (1, 0) and (4, 0) in the ratio 1 : 2. Y

(1, 0)

(4, 0)

X

O

Fig. 17.2

(ii) touch each other internally if | C1 C2 | = | r1 – r2 |, r1 π r2. The point of contact is Ê r1 x2 - r2 x1 r1 y2 - r2 y1 ˆ , ÁË r - r r1 - r2 ˜¯ 1 2 12. An equation of the family of circles passing through the points (x1, y1) and (x2, y2) is x y 1 (x – x1) (x – x2) + (y – y1) (y – y2) + l x1 x2

y1 1 = 0 y2 1

13. An equation of the family of circles which touch the line y – y1 = m(x – x1) at (x1, y1) for any finite value of m, is (x – x1)2 + (y – y1)2 + l[(y – y1) – m(x – x1)] = 0 If m is infinite, equation becomes (x – x1)2 + (y – y1)2 + l(x – x1) = 0. 14. Let QR be a chord of a circle passing through the point P(x1, y1), and let the tangents at the extremities Q and R of this chord intersect at the point L(h, k) (Fig. 17.3). Then locus of L is called as polar of P with respect to the circle, and P is called the pole of its polar. R

(ii) If the polar of P with respect to a circle passes through Q, then the polar of Q with respect to the same circle passes through P. Two such points P and Q are called conjugate points of the circle. 15. If lengths of the tangents drawn from a point P to the two circles S1 ∫ x2 + y2 + 2g1x + 2f1y + c1 = 0 and S2 ∫ x2 + y2 + 2g2x + 2f2y + c2 = 0 are equal, then the locus of P is called the radical axis of the two circles S1 = 0 and S2 = 0, and its equation is S1 – S2 = 0, i.e., 2(g1 – g2)x + 2(f1 – f2)y + c1 – c2 = 0 (i) Radical axes of two circles is perpendicular to the line joining their centres. (ii) Radical axes of three circles, taken in pairs, pass through a fixed point called the radical centre of the three circles, if the centres of these circles are non-collinear.

SPECIAL FORMS OF EQUATION OF A CIRCLE 1. An equation of a circle with centre (r, r), radius |r| is (x – r)2 + (y – r)2 = r2. This touches the coordinate axes at the points (r, 0) and (0, r). Illustration

10

A circle of radius 3 units touches both the axes. Find its equation. Solution: As the circle touches both the axes, the distance of the centre from both the axes in 3 units. So the centre of the circle can be (± 3, ± 3) and hence there are four circles with radius 3, touching both the axes and their equations are x2 + y2 ± 6x ± 6y + 9 = 0. 2. An equation of a circle with centre (x1, r), radius |r| is (x – x1)2 + (y – r)2 = r 2. This touches the x-axis at (x1, 0). 3. An equation of a circle with centre (r, y1), radius |r| is (x – r)2 + ( y – y1)2 = r2. This touches the y-axis at (0, y1). 4. An equation of a circle with centre (a/2, b/2) and radius

(a 2 + b2 ) / 4

is

x2 + y2 – ax – by = 0 This circle passes through the origin (0, 0), and has intercepts a and b on the x and y axes, respectively.

L (h, k)

Q P (x1, y1)

Fig. 17.3

(i) Equation of the polar of P (x1, y1) with respect to the circle S ∫ x2 + y2 + 2gx + 2fy + c = 0, is T = 0, where T is as defined in (8) above.

SYSTEMS OF CIRCLES Let S ∫ x2 + y2 + 2gx + 2fy + c, S ¢ ∫ x2 + y2 + 2g¢x + 2f ¢y + c¢ and L ∫ ax + by + k ¢. 1. If two circles S = 0 and S ¢ = 0 intersect at real and distinct points, then S + lS¢ = 0 (l π –1) represents a family of circles passing through these points (l being a parameter),

17.4

Complete Mathematics—JEE Main

and S – S¢ = 0 (for l = –1) represents the common chord of the circles. 2. If two circles S = 0 and S ¢ = 0 touch each other, then S – S ¢ = 0 represents equation of the common tangent to the two circles at their point of contact. 3. If two circles S = 0 and S ¢ = 0 intersect each other orthogonally (the tangents at the point of intersection of the two circles are at right angles), then 2gg¢ + 2ff ¢ = c + c¢. 4. If the circle S = 0 intersects the line L = 0 at two real and distinct points, then S + lL = 0 represents a family of circles passing through these points.

11

Illustration

Find the equation of the circle described on the chord 3x + y + 5 = 0 of the circle x2 + y2 = 16 as diameter. Solution: Equation of any circle passing through the points of intersection of the chord and the circle is x2 + y2 – 16 + l(3x + y + 5) = 0 The chord 3x + y + 5 = 0 is a diameter of this circle if the -3 l - l , of the circle lies on the chord. centre 2 2

( ) ( )

l -3l - +5=0 2 2 fi l = 1 and the required equation of the circle is x2 + y2 + 3x + y – 11 = 0. fi

3

5. If L = 0 is a tangent to the circle S = 0 at P, then S + lL = 0 represents a family of circles touching S = 0 at P, and having L = 0 as the common tangent at P. 6. Coaxial Circles A system of circle is said to be coaxial if every pair of circles of the system have the same radical axis. The simplest form of the equation of a coaxial system of circles is x2 + y2 + 2gx + c = 0, where g is a variable and c is constant, the common radical axis of the system being y-axis and the line of centres being x-axis. The Limiting points of the coaxial system of circles are the members of the system which are of zero radius. Thus the limiting points of the coaxial system of circles 2

2

(

)

x + y + 2gx + c = 0 are ± c, 0 if c ≥ 0. The equation S + lS ¢ = 0 (l π –1) represents a family of coaxial circles, two of whose members are given to be S = 0 and S ¢ = 0. Conjugate systems (or orthogonal systems) of circles Two system of circles such that every circle of one system cuts every circle of the other system orthogonally are said to be conjugate system of circles. For instance, and x2 + y2 + 2fy – c = 0, x2 + y2 + 2gx + c = 0

where g and f are variables and c is constant, represent two systems of coaxial circles which are conjugate.

COMMON TANGENTS TO TWO CIRCLES If (x – g1)2 + (y – f1)2 = a12 and (x – g2)2 + (y – f2)2 = a22 are two circles with centres C1 (g1, f1) and C2 (g2, f2) and radii a1 and a2 respectively, then we have the following results regarding their common tangents. 1. When C1 C2 > a1 + a2 i.e., distance between the centres is greater than the sum of their radii, the two circles do not intersect with each other and four common tangents can be drawn to two circles. Two of them are direct common tangents and the other two are transverse common tangents. The points T1, T2 of intersection of direct common tangents and transverse common tangents respectively, always lie on the line joining the centres of the two circles and divide it externally and internally respectively in the ratio of their radii. Direct Transverse common tangent

comm on tan gent

C1 T2

T1

C2

gent on tan m m o c Direct

Fig. 17.4

C1T1 a1 CT a = (externally) 1 2 = 1 internally T1C2 a2 T2C2 a2 2. When C1 C2 = a1 + a2 i.e., the distance between the centres is equal to the sum of the radii, the two circles touch each other externally, two direct common tangents are real and distinct and the transverse common tangents coincide.

C1

T2 C2

T1

Fig. 17.5

3. When C1 C2 < a1 + a2 i.e., the distance between the centres is less than the sum of the radii, the circles intersect at two real and distinct points, the two direct common tangents are real and distinct while the transverse common tangents are imaginary.

Circles and Systems of Circles 17.5

C1

T1

C2

Fig. 17.6

two direct common tangents are real and coincident while the transverse common tangents are imaginary. (Fig 16.7) 5. When C1 C2 < |a1 – a2| a1 π a2 i.e., the distance between the centres is less than the difference of the radii, one circle with smaller radius lies inside the other and the four common tangents are all imaginary. (Fig 16.8) Illustration

12

Find the number of common tangents to the circles x2 + y2 = 16 and x2 + y2 – 2y = 0. Solution: Centre of the circle x2 + y2 = 16 is (0, 0) and its radius is 4. Centre of the circle x2 + y2 – 2y = 0 is (0, 1) and its radius is 1. Distance between the centre = 1 which is less than the difference between the radii.

Fig. 17.7

(0, 4) C1 C2 O

Fig. 17.8

4. When C1 C2 = |a1 – a2| (a1 π a2) i.e., the distance between the centres is equal to the difference of their radii, the circles touch each other internally,

Fig. 17.9

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Example 1: If each of the lines 5x + 8y = 13 and 4x – y = 3 contains a diameter of the circle x2 + y2 – 2(a2 – 7a + 11) x – 2 (a2 – 6a + 6)y + b3 + 1 = 0, then (a) a = 5 and b œ (– 1,1) (b) a = 1 and b œ (– 1, 1) (c) a = 2 and b Œ ( –•, 1) (d) a = 5 and b Œ (–•, 1) Ans. (d) Solution: The point of intersection (1, 1) of the given lines is the centre of the circle. fi a2 – 7a + 11 = 1, a2 – 6a + 6 = 1 fi a = 5 and the equation of the circle is x2 + y2 – 2x – 2y + b3 + 1 = 0 so radius of the circle = 1 + 1 - (b3 + 1) = 1 - b3 For radius to be real b3 < 1 fi b Œ(– •, 1)

Example 2: If a circle touches the axis of x at (5, 0) and passes through the point (4, – 1), it also passes through the point (a) (–1, 6) (b) (6, –1) (c) (1, 6) (d) (6, 1) Ans. (b) Solution: Centre of the circle is on the line x = 5 at a distance equal to the radius of the circle from the axis of x. Let the coordinates of the centre be (5, k), so the equation of the circle is (x – 5)2 + (y – k)2 = k2 As it passes through the point (4, – 1) (4 – 5)2 + (– 1 – k)2 = k2

fi k=–1

17.6

Complete Mathematics—JEE Main

and the equation of the circle is (x – 5)2 + (y + 1)2 = 1 which passes through the point (6, – 1) Note (6, – 1) is the other end of the diameters through (4, –1) of the circle. (5, 0)

(4, -1)

(6, -1)

(5, -1)

Fig. 17.10

Example 3: For all values of a, equation of a diameter of the circle x2 + y2 – 2ax + 2ay + a2 = 0 is (a) x + y = a (b) x – y = a (c) x + y = 2a (d) x – y = 2a. Ans. (d) Solution: Centre of the circle is (a, – a) and the line x – y = 2a passes through it. So it is a diameter of the given circle. Example 4: S: x2 + y2 – 6x + 4y – 3 = 0 is a circle and L : 4x + 3y + 19 = 0 is a straight line. (a) L is a chord of S. (b) L is a diameter of S (c) L is a tangent to S (d) none of these Ans. (d) Solution: Centre of S is (3, – 2) and its radius is 32 + (-2)2 + 3 = 4 Length of the perpendicular of (3, – 2) from the line L is 4(3) + 3(-2) + 19 of S

(4)2 + (3)2 L

=5 S 2

2

Example 5: Circles x + y + 2x – 8y – 8 = 0 and x + y2 + 2x – 6y – 6 = 0 (a) touch each other internally at (–1, – 1) (b) touch each other externally at (–1, –1) (c) intersect each other at (–1, –1) (d) none of these. Ans. (a) Solution: Centre of the first circle is (–1, 4) and its radius is 1 + 16 + 8 = 5. Centre of the second circle is 2

(– 1, 3) and its radius is 1 + 9 + 6 = 4. Distance between the centres = 1 = difference between the radii, so the two circles touch each other internally at a common point (– 1, –1)

Example 6: Vertices of an isosceles triangle of area a2 are (– a, 0) and (a, 0). Equation of the circumcircle of the triangle is (a) x2 + y2 + 2ax – 2ay + a2 = 0 (b) x2 + y2 – 2ax + 2ay + a2 = 0 (c) x2 + y2 = a2 (d) none of these Ans. (c) Solution: Since the triangle is isosceles, third vertex lies on the line x = 0, perpendicular to the base and passing through the mid-point (0, 0) of the base. As the area is a 2, distance of the vertex from the base is a as the length of the base is 2a. So vertex of the triangle is (0, ± a) and let the equation of the circle passing through the vertices of the triangle x2 + y2 + 2gx + 2fy + c = 0, then a2 + 2ga + c = 0, a2 – 2ga + c = 0 and a2 ± 2fa + c = 0. fi c = – a2, g = f = 0 and the equation of the required circle is x2 + y2 = a2. Example 7: If the circle x2 + y2 – 6x – 8y + (25 – a2) = 0 touches the axis of y, then a equals. (a) 0 (b) ± 4 (c) ± 2 (d) ± 3 Ans. (d) Solution: If the circle touches the axis of y, distance of its centre (3, 4) from the axis of y is equal to the radius (3)2 + (4)2 - (25 - a 2 ) of the circle. fi

3 = a 2 fi a = ±3

Example 8: A circle is described on the line joining the points (2, – 3) and (– 4, 7) as a diameter. The circle (a) passes through the origin (b) touches the axis of x. (c) touches the axis of y (d) origin lies inside the circle Ans. (d) Solution: Equation of the circle is (x – 2) (x + 4) + (y + 3) (y – 7) = 0 fi x2 + y2 + 2x – 4y – 29 = 0. Centre is (1, – 2), radius = 34 Distance of the centre from the origin is So the origin lies inside the circle.

5 < 34 .

Example 9: A circle with centre at the centroid of the triangle with vertices (2, 3), (6, 7) and (7, 5) passes through the origin, radius of the circle in units is (a) 5 (c) 4

(b) 5 2 (d) none of these

Circles and Systems of Circles 17.7

Ans. (b)

Ê 2 + 6 + 7 3 + 7 + 5ˆ , Solution: Centroid of the triangle is Ë 3 3 ¯ 2 = (5, 5). Let the equation of the circle be (x – 5) + (y – 5)2 = r2, r being the radius. fi x2 + y2 – 10x – 10y + 50 – r2 = 0 which passes through the origin if r2 = 50 fi r = 5 2. Example 10: S : x2 + y2 – 8x + 10y = 0 and L : x – y – 9 = 0 are the equations of a circle and a line. (a) L is a normal to the circle S. (b) S is the only circle having radius 41 and a diameter along L. (c) L is a tangent to the circle S. (d) L does not intersect the circle S. Ans. (a) Solution: L passes through the centre (4, –5) of the circle S and hence is a diameter and every diameter is a normal to the circle. Example 11: A circle has its centre on the y-axis and passes through the origin, touches another circle with centre (2, 2) and radius 2, then the radius of the circle is (a) 1 (b) 1/2 (c) 1/3 (d) 1/4. Ans. (b) Solution: Let the centre of the circle be (0, k). As it passes through the origin, its radius is k. Since it touches the second circle ( 0 - 2 )2 + ( k - 2 )2 = k + 2 1 fi (k + 2)2 – (k – 2)2 = 4 fi k = 2 Example 12: If the point (3, 4) lies inside and the point (– 3, – 4) lies outside the circle x2 + y2 –7x + 5y – p = 0, then the set of all possible values of p is (a) (24, 25) (b) (25, 26) (c) (24, 26) (d) (0, 24) Ans. (c) Solution: (3, 4) lies inside the circle fi (3)2 + (4)2 – 7(3) + 5(4) – p < 0 fi 24 < p (–3, – 4) lies outside the circle fi (–3)2 + (–4)2 –7 (–3) + 5 (–4) – p > 0. fi p < 26. So p Œ (24, 26) Example 13: The mid point of chord by the circle x2 + y2 = 16 on the line x + y + 1 = 0 is

( ) ( )

1 1 , (a) 2 2 1 -3 , (c) 2 2 Ans. (b)

Ê 1 1ˆ (b) Ë- , - ¯ 2 2 (d)

( ) 3 7 ,4 4

Solution: The line meets the circle at points whose x-coordinates are the roots of the equation x2 + (x + 1)2 = 16 fi 2x2 + 2x – 15 = 0 fi x1 + x2 = – 1 So if (h, k) is the required point. x + x2 1 h= 1 = - and h + k + 1 = 0 2 2 Ê 1 1ˆ and the required point is Ë- , - ¯ 2 2

fi k= -

1 2

Example 14: If a chord of a circle x2 + y2 = 25 with one extremity at (4, 3) subtends a right angle at the centre of this circle, then the coordinates of the other extremity of this chord can be. (a) (– 3, –4) (b) (4, –3) (c) (3, 4) (d) (–3, 4) Ans. (d) Solution: Let P be the point (4, 3) and Q (h, k) be the other extremity of the chord through P. Centre of the circle is the origin 0. Then OP is perpendicular to OQ. 4 k 3 k ¥ = -1 fi = h 4 h 3 which is satisfied by (h, k) = (–3, 4) fi

Example 15: Equation of a common chord of the circles x2 + y2 + 6x – 10y + 9 = 0 and x2 + y2 – 10x + 6y + 25 = 0 is (a) x + y + 4 = 0 (b) x – y + 4 = 0 (c) x + y + 1 = 0 (d) x – y – 1 = 0 Ans. (d) Solution: Equation of the common chord is x2 + y2 + 6x – 10y + 9 – (x2 + y2 – 10x + 6y + 25) = 0 fi x–y–1=0 Example 16: x + ay = a2 + 1 is a tangent to the circle x + y2 = 10 for (a) any values of a (b) only one value of a (c) two values of a (d) no value of a. Ans. (c) 2

Solution: If x + ay = a2 + 1 touches the circle, length of the perpendicular of the centre (0, 0) of the circle from the line is 10 , the radius of the circle. fi fi

0 + 0.9 - (a 2 + 1) 1 + a2

= 10

1 + a 2 = 10

fi a2 = 9 fi a = ± 3

Example 17: If the circles x2 + y2 + 5x – 6y – 1 = 0 and x + y2 + ax – y + 1 = 0 intersect orthogonally (the tangents at the point of intersection of the circles are at right angles), the value of a is 2

17.8

Complete Mathematics—JEE Main

(a) 6/5 (c) – 6/5 Ans. (c)

(b) 5/6 (d) – 5/6

Ê 5 ˆ Solution: Let C1 Ë- , 3¯ be the centre and r1 = 2 2 5 + (3)2 + 1 be the radius of the first circle and 2 Ê a 1ˆ C2 Ë- , ¯ be the centre and 2 2

( )

2

r1

2

Ê a ˆ Ê1 ˆ Ë- 2 ¯ + Ë 2 ¯ - 1 be the radius of the second circle. If P is a point of intersection of the two circles then (C1 P)2 + (C2 P)2 = (C1 C2)2 r2 =

C1

P

r2

C2

Fig. 17.11

2 2 2 2 2 5 1 -5 a 1 a fi Ê- ˆ + (3)2 + 1 + Ê- ˆ + Ê ˆ - 1 = Ê + ˆ + Ê3 - ˆ Ë 2¯ Ë 2 ¯ Ë2 ¯ Ë 2 2¯ Ë 2¯

fi

a= -

6 5

Note We could also use 2gg¢ + 2ff ¢ = c + c¢

Example 18: Length of a tangent drawn from the origin to the circle x2 + y2 – 6x + 4y + 8 = 0 in units is (a) 4 (b) 6 (c) 8 Ans. (d)

(d) 2 2

Example 19: A circle passing through the intersection of the circles x2 + y2 + 5x + 4 = 0 and x2 + y2 + 5y – 4 = 0 also passes through the origin. The centre of the circle is

( )

5 5 , 2 2 Ê 5 5ˆ (c) Ë- , - ¯ 4 4 Ans. (c) (a)

( )

5 5 , 4 4 Ê 5 5ˆ (d) Ë- , - ¯ 2 2

(b)

Solution: Let the equation of the circle be x2 + y2 + 5x + 4 + l (x2 + y2 + 5y – 4) = 0 It passes through the origin if 4 – 4l = 0 fi l = 1 and the equation of the circle is 2x2 + 2y2 + 5x + 5y = 0 Ê 5 5ˆ Centre of the circle is Ë- , - ¯ 4 4 Example 20: The locus of the point from which mutually perpendicular tangents can be drawn to the circle x2 + y2 = 36 is (b) x2 + y2 = 48 (a) x2 + y2 = 42 2 2 (c) x + y = 60 (d) x2 + y2 = 72. Ans. (d) Solution: y = mx ± 6 1 + m 2 is the equation of any tangent to the circle. If it passes through (h, k), then k = mh ± 6 1 + m 2 (k – mh)2 = 36 (1 + m2) (36 – h2) m2 + 2 m h k + (36 – k2) = 0 which gives two values of m say m1 and m2, slopes of two tangents passing thought (h, k). These tangents are perpendicular fi fi

P 4 O (0, 0)

Solution: If P is the point of contact of the tangent from O and C is the centre of the circle, then (OP)2 = (OC)2 – (CP)2 = 32 + (–2)2 – (32 + (– 2)2 – 8) = 8

C(3, -2)

m1 m2 = –1

If Fig. 17.12

fi

h2 + k2 = 72

36 - k 2

= -1 36 - h 2 fi Locus of (h, k) is x2 + y2 = 72. fi

LEVEL 1 Straight Objective Type Questions Example 21: Four distinct points (2, 3), (0, 2), (4, 5) and (0, t) are concyclic if the value of t is (a) – 2 (b) 2 (c) 17 (d) – 17 Ans. (c)

Solution: An equation of a circle through (2, 3) and (4, 5) is x y 1 (x – 2) (x – 4) + (y – 3) (y – 5) + l 2 3 1 = 0 4 5 1

Circles and Systems of Circles 17.9

It will pass through (0, 2) if

fi fi

0 2 1 (– 2) (– 4) + (– 1) (– 3) + l 2 3 1 = 0 4 5 1 11 + 2l = 0 l = – 11/2 and it will pass through (0, t) if

0 t 1 11 (– 2) (– 4) + (t – 3) (t – 5) – 2 3 1 =0 2 4 5 1 fi t2 – 19t + 34 = 0 fi t = 2, 17 Example 22: The locus of the middle points of the chords of the circle x2 + y2 = 4a2 which subtend a right angle at the centre of the circle is (a) x + y = 2a (b) x2 + y2 = a2 (d) x2 + y2 = x + y (c) x2 + y2 = 2a2 Ans. (c)

distances from A and B of the tangent to the circle at the origin O be m and n, then the diameter of the circle is (a) m (m + n) (b) m + n (c) n (m + n) (d) m2 + n2 Ans. (b) Solution: Let the coordinates of A be (a, 0) and of B be (0, b), then AOB being a right angled triangle AB is a diameter of the circle, so equation of the circle is (x – a) (x – 0) + (y – b) (y – 0) = 0 fi x2 + y2 – ax – by = 0

Solution: Let M(h, k) be the middle point of the chord AB of the given circle x2 + y2 = 4a2 , with centre at O (0, 0) and radius equal to 2a. Then OM is perpendicular to AB Since AOB is a right angled triangle 4 (AM)2 = (AB)2 = (OA)2 + (OB)2 = (2a)2 + (2a)2 = 8a2 fi AM = 2 a Also (OA)2 = (OM)2 + (AM)2 fi (2a)2 = h2 + k2 + 2a2 fi h2+ k2 = 2a2 fi Locus of (h, k) is x2 + y2 = 2a2. Example 23: Two tangents are drawn from the origin to a circle with centre at (2, – 1). If the equation of one of the tangents is 3x + y = 0, the equation of the other tangent is (a) 3x – y = 0 (b) x + 3y = 0 (c) x – 3y = 0 (d) x + 2y = 0 Ans. (c) Solution: Let the equation of the other tangent from the origin be y = mx, then length of the perpendiculars from the centre (2, –1) on the two tangents is same. fi

2 m +1 1 + m2

=

6 -1 9 +1

=

5 10

fi 10 (2m + 1)2 = 25(1 + m2) fi 3m2 + 8m - 3 = 0 fi (3m – 1) (m + 3) = 0 fi m = – 3 or 1/3. m = – 3 represents the given tangent hence the slope of the required tangent is 1/3 and its equation is y = (1/3) x fi

x – 3y = 0.

Example 24: A line meets the coordinate axes in A and B. A circle is circumscribed about the triangle OAB. If the

Fig. 17.13

Equation of the tangent at the origin is ax + by = 0 (i) Let AL and BM be the perpendicular from A and B on (i) then fi

AL =

a2 a 2 + b2

m+n=

= m and BM =

b2 a 2 + b2

=n

a 2 + b2 = diameter of the circle.

Example 25: A variable chord of the circle x2 + y2 – 2ax = 0 is drawn through the origin. Locus of the centre of the circle drawn on this chord as diameter is (a) x2 + y2 + ax = 0 (c) x2 + y2 – ax = 0 Ans. (c)

(b) x2 + y2 + ay = 0 (d) x2 + y2 – ay = 0

Solution: Let (h, k) be the centre of the required circle then (h, k) is the middle point of the chord say OA of the given circle x2 + y2 – 2ax = 0, O being the origin. fi A (2h, 2k) lies on the circle fi h2 + k2 – ah = 0 fi Locus of (h, k) is x2 + y2 – ax = 0. Example 26: If a circle passes through the point (3, 4) and cuts the circle x2 + y2 = a2 orthogonally, the equation of the locus of its centre is (a) 3x + 4y – a2 = 0 (b) 6x + 8y = a2 + 25 (c) 6x + 8y + a2 + 25 = 0 (d) 3x + 4y = a2 + 25

17.10

Complete Mathematics—JEE Main

Ans. (b) Solution: Let the equation of the required circle be x2 + y2 + 2gx + 2fy + c = 0 (i) Since it passes through (3, 4) 9 + 16 + 6g + 8f + c = 0 fi 6g + 8f + c = – 25 (ii) 2 2 2 As (i) cuts the circle x + y = a orthogonally 2g ¥ 0 + 2f ¥ 0 = c – a2 fi c = a 2. So from (ii) we get 6g + 8f + a2 + 25 = 0 Hence locus of the centre (– g, –f) is 6x + 8y - (a2 + 25) = 0. Example 27: An equation of the circle passing through the origin, having its centre on the line x + y = 4 and cutting the circle x2 + y2 – 4x + 2y + 4 = 0 orthogonally is (a) x2 + y2 – 2x – 6y = 0 (b) x2 + y2 – 6x – 2y = 0 (c) x2 + y2 – 4x – 4y = 0 (d) x2 + y2 – 8x = 0 Ans. (c) Solution: Let the centre of the circle lying on the line x + y = 4 be (g, 4 – g). Since the required circle passes through the origin, equation of the circle is x2 + y2 – 2gx – 2(4 – g)y = 0 Since it cuts the circle x2 + y2 - 4x + 2y + 4 = 0 orthogonally, - 2g(– 2) – 2(4 – g) (1) = 4 fi 6g = 12 fi g = 2 Hence an equation of the required circle is x2 + y2 – 4x – 4y = 0.

Example 29: The circle passing through three distinct points (1, t), (t, 1) and (t, t) passes through the point (a) (1, 1) (b) (– 1, – 1) (c) (– 1, 1) (d) (1, – 1) for all values of t Ans. (a) Solution: Equation of a circle passing through (1, t) and (t, 1) is x y 1 (x – 1) (x – t) + (y – t) (y –1) + l 1 t 1 = 0 t 1 1 t t 1 It passes through (t, t) if 0 + 0 + l 1 t 1 = 0 t 1 1 fi l=0 fi The circle through the given points is (x – 1) (x – t) + (y – t) (y – 1) = 0, which clearly passes through (1,1) Example 30: If OA and OB are tangents from the origin O, to the circle x2 + y2 + 2gx + 2fy + c = 0, c >0 and C is the centre of the circle, then area of the quadrilateral OACB is (a)

(

(c) c g 2 + f 2 - c

)

(

(b)

c g2 + f 2 - c

(d)

g2 + f 2 - c c

)

Ans. (b) Solution: Since OA = OB =

c (length of the tangent

from the origin to the circle) and CA = CB = Area of the quadrilateal OACB

g2 + f 2 - c

A

Example 28: If O is the origin and OP, OQ are the tangents from the origin to the circle x2 + y2 – 6x + 4y + 8 = 0, the circumcentre of the triangle OPQ is (a) (3, – 2) (b) (3/2, – 1) (c) (3/4, – 1/2) (d) (– 3/2, 1) Ans. (b) Solution: We note that PQ is the chord of contact of the tangents from the origin to the circle (i) x2 + y2 – 6x + 4y + 8 = 0 Equation of PQ is 3x – 2y – 8 = 0 (ii) Equation of a circle passing through the intersection of (i) and (ii) is (iii) x2 + y2 – 6x + 4y + 8 + l (3x – 2y – 8) = 0 If this represents the circumcircle of the triangle OAB, it passess through O (0, 0), so l = 1 and the equation (iii) becomes x2 + y2 – 3x + 2y = 0. So that the required coordinates of the centre are (3 / 2, - 1)

1 c g2 + f 2 - c 2

O (0, 0)

C

B

Fig. 17.14

= 2 Area of the triangle OAC = 2 ¥ (1/2) OA ¥ CA =

c

g2 + f 2 - c

Example 31: If the line y = x + 3 meets the circle x2 + y2 = a2 at A and B, then equation of the circle on AB as diameter is (a) x2 + y2 + 3x – 3y – a2 + 9 = 0 (b) x2 + y2 – 3x + 3y – a2 – 9 = 0 (c) x2 + y2 + 3x + 3y + a2 – 9 = 0 (d) x2 + y2 – 3x + 3y – a2 + 9 = 0 Ans. (a)

Circles and Systems of Circles 17.11

Solution: Equation of any circle passing through the intersection of the given circle and the given line is (i) x2 + y2 – a2 + l (y – x – 3) = 0 If AB is a diameter of the circle, the centre

lˆ Êl ÁË 2 , - 2 ˜¯

of (i) lies on the given line y – x – 3 = 0 l l - - 3 = 0 fi l = – 3. 2 2 So that equation of the required circle is x2 + y2 + 3x – 3y – a2 + 9 = 0. fi

-

Example 32: The abscissae of two points A and B are the roots of the equation x2 + 2ax – b2 = 0, and their ordinates are the roots of the equation x2 + 2px – q2 = 0. The radius of the circle with AB as diameter is (a)

a 2 + b2 + p2 + q 2

(b)

a 2 + p2

(a) a1a2 + b1b2 = 0 (c) a1b1 + a2b2 = 0 Ans. (b)

Solution: The given lines meet x-axis at (– c1/a1, 0) and (– c2/a2, 0), y-axis at (0, – c1/b1) and (0, – c2/b2). If an equation of the circle passing through these points is x2 + y2 + 2gx + 2 f y + c = 0, then – c1/a1, – c2/a2 are the roots of x2 + 2gx + c = 0 fi c = c1c2/a1a2. Similarly c = c1c2/b1b2 fi a1a2 – b1b2 = 0. Example 35: The length of the tangent drawn from any point on the circle x2 + y2 + 2gx + 2fy + a = 0 to the circle x2 + y2 + 2gx + 2fy + b = 0 is (a)

b -a

(b)

a -b

(c)

ab

(d)

a /b

Ans. (a) Solution: Let (h, k) be any point on the first circle, then h2 + k2 + 2gh + 2 f k + a = 0 (1) length of the tangent from (h, k) to the second circle is

(c) b2 + q 2 (d) none of these

h 2 + k 2 + 2 gh + 2 f k + b =

Ans. (a) Solution: Let the coordinates of A and B be (x1, y1) and (x2, y2), respectively. From the given conditions, we then have x1 + x2 = – 2a, x1x2 = – b2, y1 + y2 = – 2p and y1y2 = – q2. Therefore the required radius is (1/2) AB = (1/ 2) = (1/ 2)

(

)

a 2 + b2 + p2 + q 2 .

Example 33: If two circles x2 + y2 + 2gx + 2fy = 0 and x + y2 + 2g1x + 2f1y = 0 touch each other, then 2

(a) f1g = fg1 (c) f 2 + g2 = f 21 + g 21 Ans. (a)

(b) ff1 = gg1 (d) none of these

Solution: Both the circles clearly pass through the origin (0, 0). They will therefore touch each other if they have a common tangent at the origin. Now the tangent to the first circle at (0, 0) is gx + fy = 0, and that of the second circle is g1x + f1y = 0. If these two equations are to represent the same line, we must have g/g1 = f/f1, i.e., f1g = fg1. Example 34: If two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 cut the coordinate axes at concyclic points, then

( (1 + 4 / (1 + 8 /

) 5) 5)

(a) 2 + 8 / 5 , 1 + 4 / 5 (b)

( x1 + x2 )2 + ( y1 + y2 )2 - 4 ( x1 x2 + y1 y2 )

b -a .

Example 36: The coordinates of the point on the circle x2 + y2 – 2x – 4y – 11 = 0 farthest from the origin are

( x1 - x2 )2 + ( y1 - y2 )2

2 2 2 2 = (1/2) 4a + 4 p + 4 b + q

=

(b) a1a2 – b1b2 = 0 (d) a1b1 – a2b2 = 0

5, 2 + 8 /

5, 2 + 4 / (c) (d) none of these Ans. (b)

Solution: The required point lies on the normal to the circle through the origin, i.e on the line 2x = y which gives x2 + 4x2 – 2x – 8x – 11 = 0 fi 5x2 – 10x – 11 = 0 fi

x = 1 ± 4 / 5 and y = 2 (1 ± 4 / 5 )

and the coordinates of the required point farthest from the origin are (1 + 4 / 5 , 2 + 8 / 5 ) . Example 37: C1 is a circle with centre at the origin and radius equal to r and C2 is a circle with centre at (3r, 0) and radius equal to 2r. The number of common tangents that can be drawn to the two circles are (a) 1 (b) 2 (c) 3 (d) 4 Ans. (c) Solution: The given circles C1 and C2 touch each other externally as the distance between the centres of the two

17.12

Complete Mathematics—JEE Main

circles is equal to the sum of their radii, hence the number of common tangents to these circles is 3.

Example 40: The lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 square units. An equa-

Example 38: An equilateral triangle is inscribed in the circle x2 + y2= a2 with the vertex at (a, 0). The equation of the side opposite to this vertex is (a) 2x – a = 0 (b) x + a = 0 (c) 2x + a = 0 (d) 3x – 2a = 0 Ans. (c)

tion of this circle is (p = 22 / 7) (a) x2 + y2 + 2x – 2y = 62 (b) x2 + y2 + 2x – 2y = 47 (c) x2 + y2 – 2x + 2y = 47 (d) x2 + y2 – 2x + 2y = 62 Ans. (c)

Solution: A (a, 0) be the vertex of the equilateral triangles ABC inscribed in the circle x2 + y2 = a2 Let M be the middle point of the side BC, then MOA is perpendicular to BC and O being the centroid of the triangle OA = 2 (OM)

Solution: The centre of the circle is the point of intersection of the given diameters 2x – 3y = 5 and 3x – 4y = 7, which is (1, – 1) and if r is the radius of the circle, then 7 pr2 = 154 fi r2 = 154 × 22 fi r = 7. Hence an equation of the required circle is

Y

(x –1)2 + (y + 1)2 = 72

B

(h, k) M

A (a, 0)

O

X

Fig. 17.15

(Circumcentre and Centroid of an equilateral triangle are same.) So if (h, k) be the coordinates of M, 2h + a 2k + 0 then = 0 and =0 3 3 fi h = - (a/2) and k = 0 and hence the equation of BC is x = – a/2 or 2x + a = 0. Example 39: Length of the common chord of the circles (x – 1)2 + (y + 1)2 = c2 and (x + 1)2 + (y – 1)2 = c2 is (a)

1 2 c -2 2

(c) 2 c 2 - 2 Ans. (c)

(b)

c2 - 2

(d) c + 2

Solution: Equation of the common chord AB of the given circles is (x –1)2 + (y + 1)2 – (x + 1)2 – (y – 1)2 = 0 fi y=x Let C1 (1, – 1) be the centre of the first circle and M be the mid-point of AB, then 1+1 = 2 C1A = c, C1 M = 2 and AB = 2AM = 2 (C1 A)2 - (C1 M )2 = 2 c 2 - 2 .

x2 + y2 – 2x + 2y = 47.

Example 41: The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is (a) x2 + y2 = 9a2 (c) x2 + y2 = 4a2 Ans. (c)

C

fi

(b) x2 + y2 = 16 a2 (d) x2 + y2 = a2

Solution: As the centre of the circumcircle of an equilateral triangle is its centroid and the distance of the centroid from the vertex of the triangle is 2/3 of its median through that vertex. So the distance of the centre from a vertex of the triangle is (2/3) ¥ 3a = 2a and hence the equation of the circumcircle is x2 + y2 = 4a2. Example 42: The circles x2 + y2 – 10x + 16 = 0 and x2 + y2 = r 2 intersect each other in two distinct points if (a) r < 2 (b) r > 8 (c) 2 < r < 8 (d) 2 £ r £ 8 Ans. (c) Solution: Centres of the given circles are (5, 0) and (0, 0) and their radii are 3 and r respectively. The two circles will intersect in two distinct points if the distance between their centres is greater than the difference and less than the sum of their radii fi |3 – r| < 5 < 3 + r fi 2 < r < 8. Example 43: A line is drawn through the point P(3, 11) to cut the circle x2 + y2 = 9 at A and B. Then PA ◊ PB is equal to (a) 9 (b) 121 (c) 205 (d) 139 Ans. (b) Solution: From geometry we know PA . PB = (PT)2 where PT is the length of the tangent from P to the circle. Hence PA . PB = (3)2 + (11)2 – 9 = 112 = 121.

Circles and Systems of Circles 17.13

Solution: Let r be the radius of the circle. Since it

Alternate Solution Equation of any line through (3, 11) is x -3 y - 11 = = r (say) cos q sin q Then the coordinates of a point on this line at a distance r from (3, 11) are (3 + r cos q, 11 + r sin q) and if this lies on the given circle x2 + y2 = 9 then (3 + r cos q)2 + (11 + r sin q)2 = 9 fi 9 + 121 + 2r (3 cos q + 11 sin q) + r2 = 9 fi r2 + 2r (3 cos q + 11 sin q) + 121 = 0 which is quadratic in r, gives two values of r say r1 and r2 and hence the distances of the points A and B from P. Thus, PA◊PB = r1 r2 = 121. Example 44: If the line x cos a + y sin a = p represents the common chord APQB of the circles x2 + y2 = a2 and x2 + y2 = b2 (a > b) as shown in the Fig. 17.16, then AP is equal to

O p P

A

Q

L

B

touches the coordinate axes and the line x – y = a 2 , the coordinates of the centre of the circle can be (r, r), (–r, –r) or (r, – r) (As r >0 and the line x – y = a 2 meets the coordinates axes at (a 2 , 0) and (0, – a 2 )). If the centre is (r, r) or (– r, – r), then -a 2 1+1

2

2

2

2

(a)

a +p + b +p

(b)

a 2 - p2 + b2 - p2

(c)

a 2 - p2 - b2 - p2

(d) Ans. (c)

a 2 + p2 - b2 + p2

Solution: The given circles are concentric with centre at (0, 0) and the length of the perpendicular from (0, 0) on the given line is p. Let OL = p then

AL =

(O A)2 - (OL )2 = a 2 - p2

and

PL =

(OP )2 - (OL )2 = b2 - p2

fi

AP =

a 2 - p2 -

b2 - p2

Example 45: A circle touches both the coordinate axes and the line x – y = a 2 (a > 0), the coordinates of the centre of the circle can be (a) (a, a) (b) (a, – a) (c) (– a, a) (d) none of these Ans. (a)

r=a

So (a, a) can be the coordinates of the centre of the circle, If the centre is (r, – r) we have r +r-a 2 2

fi

=r

r=

(

)

2 ±1 a.

Example 46: If the tangent at the point P on the circle x2 + y2 + 6x + 6y = 2 meets the straight line 5x – 2y + 6 = 0 at a point Q on the y-axis, then the length of PQ is (b) 2 5

(a) 4

(c) 5 (d) 3 5 Ans. (c) Solution: The line 5x – 2y + 6 = 0 meets y-axis at (0, 3). So the coordinates of Q are (0, 3) and PQ is the length of the tangent from Q (0, 3) to the circle x2 + y2 + 6x + 6y – 2 = 0. Hence PQ =

Fig. 17.16

fi

=r

0 + 32 + 6 ¥ 0 + 6 ¥ 3 - 2 = 25 = 5 .

Example 47: A triangle PQR is inscribed in the circle x2 + y2 = 25. If Q and R have coordinates (3, 4) and (– 4, 3) respectively, then –QPR is equal to (a) p/2 (b) p/3 (c) p/4 (d) p/6 Ans. (c) 1 –QOR; O being 2 the centre (0, 0) of the given circle x2 + y2 = 25. Let m1 = slope of OQ = 4/3 and m2 = slope of OR = – 3/4 As m1 m2 = –1, –QOR = p /2 fi –QPR = p/4 Solution: We know that –QPR =

Q(3, 4) R(- 4, 3) O

P

Fig. 17.17

17.14

Complete Mathematics—JEE Main

Example 48: Two vertices of an equilateral triangle are (– 1, 0) and (1, 0) and its third vertex lies above the x-axis, an equation of the circumcircle is 2

Both the circles clearly touch x = 0 and y = 4 3 ¥ 1 + 4 ¥ 3 - 10

Now

32 + 42

2

(a) 3x + 3y – 2 3y = 3 (b) 2x2 + 2y2 – 3 2y = 2 (c) x2 + y2 – 2y = 1 (d) none of these Ans. (a) Solution: Let A (– 1, 0), B (1, 0) and C (0, b) be the vertices of the triangle, as C lies on the locus of points equidistance from A(–1, 0) and B(1, 0) i.e., y-axis. Then

AB = AC

fi

1 + b2 = 2

fi b2 = 3 fi b= 3 [∵ b > 0] Since the triangle is equilateral, the centre of the circumcircle is at the centroid of the triangle which is (0,

3 /3). Thus the equation of the circumcircle is (x – 0)2 + (y – 1/ 3 )2 = (1 – 0)2 + (0 – 1/ 3 )2 2

2

fi

x + y – (2/ 3 ) y + 1/3 = 1 + 1/3

fi

3x2 + 3y2 – 2 3 y = 3.

Example 49: An isosceles triangle is inscribed in the circle x2 + y2 – 6x – 8y = 0 with vertex at the origin and one of the equal sides along the axis of x. Equation of the other side through the origin is (a) 7x – 24y = 0 (b) 24x – 7y = 0 (c) 7x + 24y = 0 (d) 24x + 7y = 0 Ans. (d) Solution: Centre of the circle is (3, 4) and it passes through the origin. If y = mx is the equation of the required line, then length of the perpendicular from the centre on this line is equal to the length of the perpendicular from the centre on the axis of x. 3m - 4 = +4 fi 1 + m2 9m2 – 24m + 16 = 16(1 + m2) m = – 24/7 (∵ m = 0 corresponds to x-axis) and hence the required equation is 24x + 7y = 0.

fi fi

3(-3) + 4(1) - 10

=3 32 + 42 Shows that both the circles touch 3x + 4y = 10 also. and

Example 51: A circle whose radius is 5 and which touches externally the circle x2 + y2 – 2x – 4y – 20 = 0 at the point (5, 5) intersects in real distinct points the line (b) y = 0 (a) x = 0 (c) y = x (d) none of these Ans. (c) Solution: Centre of the given circle is A (1, 2) and its 2

radius is 1 + (2) + 20 = 5 . Point of contact P is (5, 5). Let B (h, k) be the centre of the required circle of radius 5, then P is the mid-point of AB, so that h +1 k+2 = 5 and =5 2 2

fi

h = 9, k = 8

and an equation of the required circle is x2 + y2 – 18x – 16y + 120 = 0 If x = 0, y2 – 16y + 120 = 0 does not give real values of y If y = 0, x2 – 18x + 120 = 0 does not give real values of x If y = x, 2x2 – 34x + 120 = 0 or x2 – 17x + 60 = 0 fi x = 5, 12 This shows that the circle intersects the line y = x at two real distinct points. Example 52: A point moves such that the sum of the squares of its distances from the sides of a square of side unity is equal to 9. The locus of such a point is a circle (a) inscribed in the square (b) circumscribing the square (c) inside the square (d) containing the square Ans. (d) Solution: Let P (x, y) be any point on the locus and OABC be the square of side unity.

Example 50: Equations of the common tangents of the circles x2 + y2 – 2x – 6y + 9 = 0 and x2 + y2 + 6x – 2y + 1 = 0 are (a) x = 0 (b) y = 4 (c) 3x + 4y = 10 (d) all of these Ans. (d) Solution: Equations of the given circles can be written as C1: (x – 1)2 + (y –3)2 = 1 having centre at (1, 3) and radius 1 and C2: (x + 3)2 + (y –1)2 = 9 having centre at (–3, 1) and radius 3.

=1

Fig. 17.18

Circles and Systems of Circles 17.15

Then, according to the given condition 2

fi

2

2

2

Solution: The centre of the given circle is (1, –2)

x + y + (x – 1) + (y – 1) = 9

and its radius is

2(x2 + y2) – 2x – 2y – 7 = 0

tices of the square inscribed in the given circle are

fi x2 + y2 – x – y – 7/2 = 0 which is a circle with centre (1/2, 1/2) coinciding with the centre D of the square and the radius of the circle is

Solution: Let the equation of the required circle be (x + 4)2 + (y – 3)2 = r2 (i) (ii) If (i) touches the circle x2 + y2 = 1 internally the distance between the centres (- 4, 3) and (0, 0) of these circles is equal to the difference of their radii fi

42 + 32 = r – 1 fi r = 5 + 1 fi r = 6

So that an equation of the required circle is x2 + y2 + 8x – 6y – 11 = 0. Example 54: The tangents drawn from the origin to the circle x2 + y2 – 2px – 2qy + q2 = 0 are perpendicular if (a) p2 + q2 = 1 (c) p2 – q2 = 0 Ans. (c)

(b) p2 – q2 = 1 (d) none of these

Solution: Equation of the given circle can be written as (x - p)2 + (y – q)2 = p2 (i) This has (p, q) as the centre and p as the radius showing that it touches y-axis. So one of the tangents from the origin to the circle is y-axis. fi Other tangent from the origin to the circle must be xaxis. which is possible if q = ± p. fi

p2 – q2 = 0.

Example 55: A square is inscribed in the circle x2+ y2–2x + 4y –93 = 0 with its sides parallel to the axes of coordinates. The coordinates of the vertex farthest from the origin are (a) (9, – 8) (b) (8, – 9) (c) (8, 5) (d) (– 6, 9) Ans. (b)

98 cos q, – 2 +

(1 +

98 sin q) where q = ± p/4, ± 3p/4.

2

S = square of the distance of a vertex from the origin. = 5 + 98 + 2 98 (cos q – 2 sin q) which is maximum when

1 1 7 + + = 2 , showing that the circle contains the square. 4 4 2 Example 53: Equation of a circle with centre (– 4, 3) touching internally and containing the circle x2 + y2 = 1 is (a) x2 + y2 + 8x – 6y + 9 = 0 (b) x2 + y2 – 8x + 6y + 9 = 0 (c) x2 + y2 + 8x – 6y – 11 = 0 (d) x2 + y2 – 8x + 6y – 11 = 0 Ans. (c)

98 , so the coordinates of the ver-

q = – p/4, so the required point is (1 + 7, – 2 – 7) = (8, – 9) Example 56: An equation of a circle touching the axes of coordinates and the line x cos a + y sin a = 2 is x2 + y2 – 2gx + 2gy + g2 = 0 where g = (a) 2 (cos a + sin a + 1)–1 (b) 2 (cos a – sin a + 1)–1 (c) 2 (cos a + sin a – 1)–1 (d) – 2 (cos a – sin a – 1)–1 Ans. (b) Solution: Centre of the circle is (g, –g) and radius is |g|. If it touches the line x cos a + y sin a = 2, the g cos a – g sin a – 2 = + g fi g (cos a – sin a + 1) = 2 2 cos a - sin a ± 1 which is satisfied by (b). fi

g=

Example 57: If common chord of the circle C with centre at (2, 1) and radius r and the circle x2 + y2 – 2x – 6y + 6 = 0 is a diameter of the second circle, then the value of r is (a) 3 (b) 2 (c) 3/2 (d) 1 Ans. (a) Solution: Equation of C is (x – 2)2 + (y – 1)2 = r2 or x2 + y2 – 4x – 2y = r2 – 5 Equation of the common chord is x2 + y2 – 4x – 2y – (r2 – 5) – [x2 + y2 – 2x – 6y + 6] = 0 fi 2x – 4y + r2 + 1 = 0 Since it is a diameters of the second circle, the centre (1, 3) lies on it. fi 2 – 12 + r2 + 1 = 0 fi r = 3. Example 58: The centres of a set of circles, each of radius 3, lie on the circle x2 + y2 = 25. The locus of any point in the set is (b) x2 + y2 £ 25 (a) 4 £ x2 + y2 £ 64 (c) x2 + y2 ≥ 25 (d) 3 £ x2 + y2 £ 9. Ans. (a)

17.16

Complete Mathematics—JEE Main

Ans. (d) O 2 (0,0) A

3

C3

B

Fig. 17.19

Solution: Distance of any point (x, y) on a circle with centre C lying on the circle x2 + y2 = 25 and radius 3 from the origin lies between OA (= 2) and OB (= 8) Hence 4 £ x2 + y2 £ 64. Example 59: If a circle passes through (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is (a) 2ax – 2by + (a2 + b2 + 4) = 0 (b) 2ax + 2by – (a2 + b2 + 4) = 0 (c) 2ax + 2by + (a2 + b2 + 4) = 0 (d) 2ax – 2by – (a2 + b2 + 4) = 0 Ans. (b) Solution: Let the equation of the circle be x2 + y2 – 2gx – 2fy + c = 0. As it passes through (a, b) a2 + b2 – 2ga – 2fb + c = 0 Since it intersects the circle x2 + y2 = 4 orthogonally 2g ¥ 0 + 2f ¥ 0 = c – 4 fi c = 4 and the locus of (g, f ), the centre is 2ax + 2by – (a2 + b2 + 4) = 0 Example 60: Consider a family of circles which are passing through the point (– 1, 1) and are tangent to x-axis. If (h, k) are the coordinates of the centre of the circles, then the set of values of k is given by the interval (a) 0 < k < 1/2 (b) k ≥ 1/2 (c) – 1/2 £ k £ 1/2 (d) k £ 1/2. Ans. (b) Solution: Equation of the circle with centre (h, k) and touching x-axis is (x – h)2 + (y – k)2 = k2 which passes through (– 1, 1) if h2 + 2h – 2k + 2 = 0 or if (h + 1)2 = 2k – 1 For real values of h, 2k – 1 ≥ 0. fi k ≥ 1/2 Example 61: If the circles x2 + y2 + 2ax + cy + a = 0 and x + y2 – 3ax + dy – 1 = 0 intersect in two distinct points P and Q then the line 5x + by – a = 0 passes through P and Q for (a) Infinitely many values of a (b) Exactly two values of a (c) Exactly one value of a (d) No value of a 2

Solution: Equation of the line passing through the points P and Q is x2 + y2 + 2ax + cy + a – (x2 + y2 – 3ax + dy – 1) = 0 fi 5ax + (c – d)y + a + 1 = 0 which coincides with 5x + by – a = 0 a +1 5a c–d if = = . –a b 5 or if a2 + a + 1 = 0 which does not give any real value of a. Example 62: A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre of the circle is (a) a hyperbola (b) a parabola (c) an ellipse (d) a circle Ans. (b) Solution: Let (h, k) be the centre of the circle. Since it touches x-axis, its radius is k. As it also touches the circle with centre (0, 3) and radius 2 (externally as x-axis lies outside the given circle) distance between the centres is equals to the sum of their radii. So h2 + (k – 3)2 = (k + 2)2 fi h2 = 5(2k – 1) Locus of the centre (h, k) is x2 = 5(2y – 1) which is a parabola. Example 63: Three distinct points A, B, C, are given in the two dimensional coordinate plane such that the ratio of the distance of anyone of them from the point (1, 0) to the distance from the point (–1, 0) is equal to 1/3. Then the circumcentre of the triangle ABC is at the point (a) (5/2, 0) (b) (5/3, 0) (c) (0, 0) (d) (5/4,0) Ans. (d) Solution: Let Q = (1, 0) and R = (– 1, 0) Let P(x, y) be any point such that 1 PQ = fi 9(PQ)2 = (PR)2 3 PR fi fi

9[(x – 1)2 + y2]= (x + 1)2 + y2 8x + 8y2 – 20x + 8 = 0 2

A, B, C lie on this locus of P which is a circle circumscribing the triangle ABC. So the circumcentre of the triangle ABC is the centre (5/4, 0) of this circle. Example 64: Let A(1,2) B(3, 4) be two points and C(x, y) be a point such that area of the triangle ABC is 3 sq. units and (x – 1) (x – 3) + (y – 2) (y – 4) = 0. Then maximum number of positions of C, in the xy plane is

Circles and Systems of Circles 17.17

(a) 2 (c) 8 Ans. (d)

(b) 4 (d) none of these

Solution: C lies on the circle on AB as diameter of length 2 2 . So radius of the circle is 2 . Area of the triangle ABC = 3 = (1/2) AB ¥ altitude from C on AB 6 3 fi altitude = = > 2. 2 2 2

Solution: Given equations represent the lines x = 2, y = 9, x = 6, y = 5 which forms a square of each side equal to 4, vertex A(2, 5), C(6, 9). Centre of the circle is the mid point of AC and the radius is (1/2) AC. Hence the equation of the circle is x=6

D y=5

C y=9

A

x=2

B

Fig. 17.21

Maximum altitude from C on AB is equal to the radius of the circle. So no such C exists.

(

(x – 4)2 + (y – 7)2 = 2 2

Example 65: A circle touches y-axis at (0, 3) and makes

fi

x2 + y2 – 8x – 14y + 16 + 49 – 8 = 0

an intercept of 2 units on the positive x-axis. Intercept made

fi

x2 + y2 – 8x – 14y + 57 = 0.

by the circle on the line (a) 3 (c) 2 10 Ans. (c)

Example 67: If the centre of the circle passing through the origin and the points of intersection of the pair of straight lines xy – 7x + 3y – 21 = 0 with the coordinate axes lies on the x + y = k, then k is equal to. (a) 0 (b) 1 (c) 2 (d) 4 Ans. (c)

10 x – 3y = 1 in units is (b) 6 (d) 10.

Solution: AB = 2, AD = 1 CD = 3 = OE AC = =

2

Solution: Equation the circle passing through the points (– 3, 0) and (0, 7) where the lines meet the axes, and the origin is x2 + y2 + 3x – 7y = 0. Whose centre (– 3/2, 7/2) lies on the line x + y = k

2

( AD ) + (CD ) 1 + 32 =

So the coordinates of C are

(

10 = CE 10 , 3

)

fi

Fig. 17.20

The line 10 x – 3y = 1 passes through the centre C

2

)

(

10 , 3

)

of the circle and hence is a diameter and the required intercept is twice the radius of the circle. Example 66: Equation of the circle which circumscribes the square formed by the lines xy – 9x – 2y + 18 = 0 and xy – 5x – 6y + 30 = 0 is (a) x2 + y2 – 4x – 10y + 21 = 0 (b) x2 + y2 – 8x – 14y + 57 = 0 (c) x2 + y2 – 12x – 10y + 53 = 0 (d) none of these. Ans. (b)

k=2

Example 68: If P and Q are the points of intersection of the circles x2 + y2 + 3x + 7y + 2p – 5 = 0 and x2 + y2 + 2x + 2y – p2 = 0, then there is a circle passing through P, Q and (1, 1) for (a) all except two values of p (b) exactly one value of p (c) all values of p (d) all except one value of p. Ans. (d) Solution: Equation of a circle passing through P and Q is x2 + y2 + 3x + 7y + 2 p – 5 + l (x2 + y2 + 2x + 2y – p2) =0 which passes through (1, 1) if (7 + 2p) – l (p2 – 6) = 0 fi fi fi

l =

7 + 2p p2 - 6

p2 – 6 + 7 + 2p π 0 p π–1

fi

π–1 (p + 1)2π 0

17.18

Complete Mathematics—JEE Main

Example 69: Equation of circle passing through (1, 5) and (4, 1) and touching y-axis is x2 + y2 – 5x – 6y + 9 + l(4x + 3y – 19) = 0 where l is equal to (a) 0, – 40/9 (b) 0 (c) 40/9 (d) – 40/9 Ans. (a) Solution: Equation of any circle through the given points is (x – 1) (x – 4) + (y – 5) (y – 1) + l (4x + 3y – 19) = 0 where 4x + 3y – 19 = 0 is the equation of the line joining (1, 5) and (4, 1) fi x2 + y2 – 5x – 6y + 9 + l (4x + 3y – 19) = 0 Which touches y-axis if the roots of the equation y2 – 6y + 9 + l (3y – 19) = 0 are equal. (Taking the intersection of the circle with x = 0) fi (3l – 6)2 = 4(9 – 19l) fi 9l2 + 40l = 0 fi l = 0, – 40/9.

Example 70: The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if (a) 15 < m < 65 (b) 35 < m < 85 (c) – 85 < m < – 35 (d) – 35 < m < 15 Ans. (d) Solution: The line intersects the circle at two distinct points if length of the perpendicular from (2, 4), the centre of the circle, to the line is less than the radius. (3) (2) - 4 (4) - m fi < 4 + 16 + 5 (3)2 + (4)2 fi |m+10| < 25 fi – 25 < m + 10 < 25 fi – 35 < m < 15. Example 71: The equation of the circle passing through the points (1, 0) and (0, 1) and having the smallest radius is (a) x2 + y2 – 2x –2y + 1 = 0 (b) x2 + y2 – x – y = 0 (c) x2 + y2 + 2x + 2y –7 = 0 (d) x2 +y2+ x + y– 2 = 0 Ans. (b) Solution: Let the equation of the circle be x2 + y2 +2gx + 2f y + c = 0 As it passes through (1, 0) and (0, 1), we get 1 + 2g + c = 0, 1 + 2f + c = 0 If r is the radius of the circle, then r2 =g2 + f 2 – c =

1 1 (1 + c)2 + (1 + c)2 – c 4 4

1 (1 + c2) 2 Now r2 will be least if c = 0 and the required equation is 2 x + y2 – x – y = 0 Example 72: The circles x2 + y2 = ax and x2 + y2 = c2 (c > 0) touch each other (a) | a | = 2c (b) 2 | a | = c (c) | a | = c (d) a = 2c Ans. (c) Solution: Given circles can touch each other internally see the figures. This is possible if c =a or c = –a fi c = | a |. =

Fig. 17.22

Example 73: The circle passing through the point (–1, 0) and touching y-axis at (0, 2) also passes through the point (a) (–3/2, 0) (b) (–5/2, 2) (c) (–3/2, 5/2) (d) (–4, 0) Ans. (d) Solution: Let he equation of the circle be (x – k)2 +(y – 2)2 = k2 which passes through the point (–1, 0) if (–1 – k)2 + 4 = k2 fi k = –5/2 and the equation of the circle is x2 + y2 + 5x – 4y + 4 = 0 which passes through (–4, 0) Example 74: The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line 4x –5y = 20 to the circle x2 + y2 = 9 is (a) 20(x2 + y2) –36x + 45y = 0 (b) 20(x2 + y2) +36x – 45y = 0 (c) 36(x2 + y2) –20x + 45y = 0 (d) 36(x2 + y2) +20x – 45y = 0 Ans. (a) Solution: Equation of the chord contact in terms of the mid-point (a, b ) is xa + yb = a 2 + b 2 Let P (5t, 4t – 4) be a point on 4x – 5y = 20 Equation of the chord of contact of P is x(5t) + y (4t – 4)= 9 comparing the equations we get 4 (5t ) - 5 ( 4t - 4 ) 5t 4t - 4 9 = = 2 = 2 a b 4a - 5b a +b

Circles and Systems of Circles 17.19

20 (a 2 + b 2) –36a + 45b = 0 Locus of (a, b ) is 20(x2 + y2) –36x + 45y = 0

fi

Example 75: The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3) is

(a) 6/5 (c) 10/3 Ans. (c)

(b) 5/3 (d) 3/5

Solution: Let the equation of the circle be (x–1)2 + (y – k)2 = k2. Since it passes through (2, 3) 1 + (3 – k)2 = k2 fi k = 5/3 fi diameter = 2k =10/3.

Assertion-Reason Type Questions

Example 76: Statement-1: Limiting points of a family of coaxial circles are (1, 2) and (2, 1). No circle of this family passes through the origin. Statement-2 : Equation of a circle passing through (1, 2) and (2, 1), the centre of which does not lie on the join of these points is x2 + y2 – 3x – 3y + 4 = 0 Ans. (c) Solution: In statement 1, two members of the family are (x – 1)2 + (y – 2)2 = 0 and (x – 2)2 + (y – 1)2 = 0. Equation of any member of this family is x2 + y2 – 2x – 4y + 5 + l (x2 + y2 – 4x – 2y + 5) = 0 which passes through the origin if l = – 1, in which case the equation does not represent a circle. So the statement is true. In statement-2, Equation represents the circle on the join of the given points as diameter. So the statement is false. Example 77: Statement-1: Equation of a tangent to the circle x2 + y2 = 50 at a point which has positive integral coordinates (a, b) (a π b) is x + 7y = 50 or 7x + y = 50. Statement-2 : There are 12 points on the circle x2 + y2 = 50 with integral coordinates. Ans. (a) Solution: Statement-2 is true as the required points are, (± 1, ± 7), (± 7, ± 1) (± 5, ± 5) out of which the points satisfying the conditions in statement are (1, 7) and (7, 1) only and the tangents at these points are respectively x + 7y = 50 and 7x + y = 50 so statement–1 is also true.

Statement-2 is false as the line joining the given points may not be a diameter for each circle passing through these points. Example 79: Statement-1: The circle x2 + y2 – 8x – 4y + 16 = 0 touches x-axis at the point (4, 0) Statement-2 : The circle (x – a)2 + (y – r)2 = r2 touches xaxis at the point (a, 0) Ans. (a) Solution: Statement-2 is true as the centre of the circle is (a, r) and its radius is r, centre is at a distance r from xaxis. In statement-2, the circle is (x – 4)2 + (y – 2)2 = 4 = 22, using statement-2, statement-1 is also true. Example 80: Statement-1 : Point (3, – 1) lies outside the circle 2x2 + 2y2 – 3x + 5y – 7 = 0 Statement-2 : A point (a, b) lies outside the circle x2 + y2 + 2gx + 2fy + c = 0 if a2 + b2 + 2ga + 2fb + c > 0 Ans. (d) Solution: Statement-2 is true because the distance between the point (a, b) and the centre (– g, – f) of the circle is greater than its radius. fi (a + g)2 + (b + f )2 > g2 + f 2 – c fi a2 + b2 + 2ga + 2fb + c > 0. Using it in statement-1, 2 × 9 + 2 × 1 – 3 × 3 + 5(– 1) – 7 = – 1 < 0 so statement-1 is false.

Example 78: Statement-1: The centre of the circle passing through the points (3, 8) and (5, 4) and having smallest radius is (4, 6) Statement-2 : The centre of a circle passing through two given points lies at the mid-point of the line joining the points. Ans. (c)

Example 81: Tangents are drawn from the point (17, 7) to the circle x2 + y2 = 169 Statement-1: Tangents are mutually perpendicular. Statement-2: The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is x2 + y2 = 338 Ans. (a)

Solution: In statement-1, the radius is smallest when the line joining the given points is a diameter of the circle Ê 3 + 5 4 + 8ˆ and hence its centre is the mid-point Á , ˜ i.e (4, 6) Ë 2 2 ¯ of the line joining the given points. So statement-1 is True.

Solution: Statement-2 is true, because equation of any tangent to the circle is y = mx ± 13 1 + m 2 . If it passes through (h, k), then k = mh ± 13 1 + m 2 fi (169 – h2 )m2 + 2mhk + (169 – k2) = 0

17.20

Complete Mathematics—JEE Main

which gives two values of m say m1, m2, the slopes of the tangents drawn from (h, k) to the circle m 1m 2 = – 1

fi

169 – k 2 169 – h

2

=–1

fi

h2 + k2 = 338

fi Locus of (h, k) is x2 + y2 = 338 Since the point (17, 7) in statement-1 satisfies the above equation, using statement-2, statement-1 is also true. Example 82: Statement-1: Limiting points of a family of co-axial system of circles are (1, 1) and (3, 3). The member of this family passing through the origin is 2x2 + 2y2 – 3x – 3y = 0 Statement-2: Equation of the tangent to the circle 2x2 + 2y2 – 3x – 3y = 0 at the origin is x + y = 0 Ans. (b) Solution: In statement-1, two members of the co-axial system are circles with centres at (1, 1), (3, 3) and radius zero. So the equation of the system of circles is (x – 1)2 + (y – 1)2 + l[(x – 3)2 + (y – 3)2] = 0 If it passes through the origin. 1 + 1 + l[9 + 9] = 0 fi l = – 1/9 and the required circle is 2x2 + 2y2 – 3x – 3y = 0 So statement-1 is true. Statement-2 is also true but does not lead the statement-1. Example 83: Statement 1: x2 + y2 – 6x – 10y – 2 = 0 is the only circle of radius 6 units having a diameter along the line 5x – 2y – 5 = 0. Statement-2: 5x – 2y – 5 = 0 is a normal to the circle x2 + y2 – 6x – 10y – 2 = 0. Ans. (d) Solution: Given circle is (x – 3)2 + (y – 5)2 = 36 fi centre of the circle is (3, 5) and radius is 6. Since the line 5x – 2y – 5 = 0 passes through (3, 5) it is a diameter of the circle. So this is a circle of radius 6 units having 5x – 2y – 5 = 0 as a diameter. But any other point on this can also be taken as the centre, giving us another circle of radius 6 with the same diameter. So statement-1 is false. Since every diameter is normal to the circle, statement-2 is true.

Example 84: Statement-1: The circle x2 + y2 – 8x + 10y + 32 = 0 does not touch or intersect any axis. Statement-2: A circle with centre (a, b) and radius r does not touch or intersect any axis if r < min [|a|, |b|] Ans. (a) Solution: We find by fig. r < |a| and r < |b| fi r < min (|a|, |b|) So statement-2 is true, using it statement-1 is also true, as the centre of the circle is (4, – 5) and r = 3.

r C

(a, b)

P

b a

Fig. 17.23

Example 85: C1: x2 + y2 + 2x – 2y + 2 – a2 = 0 C2: x2 + y2 – 10x – 14y + 74 – a2 = 0 L: x + y – 6 = 0 Statement 1: L is a common tangent to both the circles C1 and C2 if a = 3 2 . Statement 2: L is a common chord of both the circles C1 and C2 for all values of a. Ans. (c) Solution: C1 : (x + 1)2 + (y – 1)2 = a2 C2 : (x – 5)2 + (y – 7)2 = a2 Centre of C1 is (– 1, 1) and of C2 is (5, 7) and both have same radius equal to a. Distance between the centre = 6 2 = 2a (when a = 3 2 ) = sum of the radii, so the circle touch each other externally and the equation of a common tangent is (x2 + y2 + 2x – 2y + 2 – a2) – (x2 + y2 – 10x – 14y + 74 – a2) = 0 fi 12x + 12y – 72 = 0 or x + y – 6 = 0 a < 3 2 , circles C other, L

C

LEVEL 2 Straight Objective Type Questions Example 86: Let ABCD be a quadrilateral with area 18, side AB parallel to the side CD and AB = 2CD. Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateral ABCD touching all the sides, then its radius is (a) 3 (b) 2 (c) 3/2 (d) 1

Ans. (b) Solution: Let CD = a, then AB = 2a and r be the radius of the circle, then AD = 2r. Let A be the origin and AB and AD as x-axis and y-axis respectively. The coordinates of A, B, C, D are respectively

Circles and Systems of Circles 17.21 D (0, 2, r)

(0, 0), (2a, 0), (a, 2r), (0, 2r) Area (ABCD) = (1/2) (a + 2a) (2r) = 18 fi ar = 6. Equation of BC is 2rx + ay – 4ar = 0 and the coordinates of the centre of the circle are (r, r) Since the circle touches BC,

r r,r)

(

B (2a,0)

A (0, 0)

Fig. 17.24

2r 2 + ar – 4ar 4r 2 + a 2 fi fi

C (a,2r)

Solution: Since the hypotenuse of a right angled triangle inscribed in a circle is a diameter of the circle, if the coordinates of the end C of the hypotenuse BC are (a, b), the coordinates of B are (– a, – b). y b Equation of BC is = . If A is the vertex of the x a isosceles triangle then OA is perpendicular to BC and y –a = which meets the circle the equation of AO is x b x2 + y2 = r2 at points for which Ê a2 ˆ 2 2 2 2 ÁË b2 + 1˜¯ x = r = a + b [∵ (a, b) lies on x2 + y2 = r2]

=r

4r4 – 72r2 + 324 = 4r4 + 36 r = 2.

Example 87: An equation of the chord of the circle x2 + y2 = a2 passing through the point (2, 3) farthest from the centre is (a) 2x + 3y = 13 (b) 3x – y = 3 (c) x – 2y + 4 = 0 (d) x – y + 1 = 0 Ans. (a) Solution: Let P (2, 3) be the given point, M be the middle point of a chord of the circle x2 + y2 = a2 through P. Then the distance of the centre O of the circle from the chord is OM. and (OM)2 = (OP)2 - (PM)2 which is maximum when PM is minimum, i.e., M coincides with P i.e., P is the middle point of the chord. Hence the equation of the chord is 2◊x + 3◊y– a2 2 = (2) + (3)2 – a2 fi 2x + 3y =13. Fig. 17.25 Example 88: An isosceles right angled triangle is inscribed in the circle x2 + y2 = r2. If the coordinates of an end of the hypotenuse are (a b), the coordinates of the vertex are (a) (– a, – b) (b) (b, – a) (c) (b, a) (d) (– b, – a) Ans. (b)

fi x2 = b2 fi x = ± b fi y = ∓ a. coordinates of A are (– b, a) or (b, – a). Example 89: Two rods of lengths a and b slide along the x-axis and y-axis respectively in such a manner that their ends are concyclic. The locus of the centre of the circle passing through the end points is (a) 4 (x2 + y2) = a2 + b2 (b) x2 + y2 = a2 + b2 (c) 4 (x2 – y2) = a2 – b2 (d) x2 – y2 = a2 – b2 Ans. (c) Solution: Let C (h, k) be the centre of the circle passing through the end points of the rod AB and PQ of lengths a and b respectively, CL and CM be perpendiculars from C on AB and PQ respectively. (Fig. 17.27)

Q

P O

B (- a, - b)

Fig. 17.26

L

B

then

AL = (1/2) AB = a/2 PM = (1/2) PQ = b/2 and CA = CP (radii of the same circle) fi

O

A

Fig. 17.27

C (a, b)

A

C (h, k)

M

k2 +

a2 b2 = h2 + 4 4

fi 4 (h2 – k 2) = a2 – b2 so that locus of (h, k) is 4(x2 – y2) = a2 – b2 Example 90: If the point (1, 4) lies inside the circle x2 + y2 – 6x – 10y + p = 0 and the circle does not touch or intersect the coordinate axes, then

Complete Mathematics—JEE Main

17.22

(a) 0 < p < 34 (c) 9 < p < 25 Ans. (b)

(b) 25 < p < 29 (d) 9 < p < 29

its centre (a, 0) on this line is equal to the radius b of the circle, which gives ma - b 1 + m 2

Solution: Since the circle does not touch or intersect the coordinates axes, the absolute values of x and y coordinates of the centre are greater than the radius of the circle. Coordinates of the centre of the circle are (3, 5) and the radius is 9 + 25 - p

=±b 1 + m2 Taking negative value on R.H.S. we get m = 0, so we neglect it. Taking the positive value on R.H.S. we get ma = 2b 1 + m 2

9 + 25 - p fi p > 25

(1)

fi m2(a2 – 4b2) = 4b2

5 > 9 + 25 - p fi p > 9 and the point (1, 4) lies inside the circle fi 1 + 16 – 6 – 10 ¥ 4 + p < 0 fi p < 29 From (1), (2), (3) we get 25 < p < 29.

(2)

fi

so that

3>

(3)

Example 91: The lengths of the intercepts made by any circle on the coordinates axes are equal if the centre lies on the line (s) represented by (a) x + y +1 = 0 (b) x + y = 1 (d) x – y = 1 (c) x2– y2 = 0 Ans. (c)

m =

2b 2

a - 4b 2

.

Example 93: Let PQ and RS be tangents at the extremities of a diameter PR of a circle of radius r. Such that PS and RQ intersect at a point X on the circumference of the circle, then 2r equals. PQ + RS (b) (a) PQ.RS 2 (c)

2PQ.RS PQ + RS

(d)

PQ 2 + RS 2 2

Ans. (a) Solution: From Fig. 17.28, we have

S

Solution: Let the equation of any circle be (i) x2 + y2 + 2gx + 2 fy + c = 0 For intercept made by the circle on x-axis, put y= 0 in (i) fi x2 + 2gx + c = 0 If x1, x2 are roots of (ii), then length of the intercept on x-axis is |x1 – x2 | = ( x1 + x2 )2 - 4 x1 x2 = 2 g 2 - c

Q

X

p -q 2

q P

Similarly length of the intercept of the y-axis is 2 f 2 - c

O

R

Since the lengths of these intercepts are equal g2 - c

f2 -c

=

fi g2 = f 2 = (– g)2 = (– f)2 Therefore, centre lies on x2 – y2 = 0 Example 92: If a > 2b > 0 then the positive value of m for which y = mx – b 1 + m 2 is a common tangent to x2 + y2 = b2 and (x – a)2 + y2 = b2 is 2b (a)

2

a - 4b

2b (c) a - 2b Ans. (a)

2

(b)

a 2 - 4b 2 2b

b (d) a - 2b 2

Solution: y = mx – b 1 + m is a tangent to the circle x + y2 = b2 for all values of m. If it also touches the circle (x – a)2 + y2 = b2, then the length of the perpendicular from 2

Fig. 17.28

and fi fi fi fi

PQ = tan (p/2 –q ) = cot q . PR RS = tan q PR PQ RS . =1 PR PR (PR)2 = PQ . PS (2r)2 = PQ . PS 2r =

PQ.PS

Example 94: If two distinct chords, drawn from the point (p, q) on the circle x2 + y2 = px + qy (where pq π 0) are bisected by the x-axis, then

Circles and Systems of Circles 17.23

(a) p2 = q2 (c) p2 < 8q2 Ans. (d)

(b) p2 = 8q2 (d) p2 > 8q2

Solution: Let PQ be a chord of the given circle passing through P (p, q) and the coordinates of Q be (x, y). Since PQ is bisected by the x-axis, the mid-point of PQ lies on the x-axis which gives y = – q. Now Q lies on the circle x2 + y2 – px – qy = 0 so x2 + q2 – px + q2 = 0 fi x2 – px + 2q2 = 0 (i)

So (b) is the correct answer Example 96: Let A0 A1 A2 A3 A4 A5 be a regular hexagon inscribed in a unit circle with centre at the origin. Then the product of the lengths of the line segments A0 A1, A0 A2 and A0 A4 is (b) 3 3 (d) 3 3 /2

(a) 3/4 (c) 3 Ans. (c) Solution:

Y

X

O

Fig. 17.29

which gives two values of x and hence the coordinates of two points Q and R (say), so that the chords PQ and PR are bisected by x-axis. If the chords PQ and PR are distinct, the roots of (i) are real distinct. fi the discriminant p2 – 8q2 > 0 fi p2 > 8q2. Example 95: Let L1 be a straight line passing through the origin and L2 be the straight line x + y = 1. If the intercepts made by the circle x2 + y2 – x + 3y = 0 on L1 and L2 are equal, then L1 can be represented by (a) x + y = 0 (b) x – y = 0 (c) 7x + y = 0 (d) x – 7y = 0 Ans. (b) Solution: Let the equation of L1 be y = mx. Since the intercepts made by the circle on L1 and L2 are equal, their distances from the centre of the circle are also equal. Centre of the given circle is (1/2, – 3/2), so that we have 1 3 1 3 - -1 m¥ + 2 2 2 2 = 1+1 m2 + 1 fi

2

=

m+3 2

2 2 m +1 fi 8(m2 + 1) = (m + 3)2 fi 7m2 – 6m – 1 = 0 fi (m – 1) (7m + 1) = 0 fi m=1 or m = – 1/7 So the equations representing L1 are y =x or y = (– 1/7) x fi x – y = 0 or x + 7y = 0

Fig. 17.30

Let O be the centre of the circle of unit radius and the coordinates of A0 be (1, 0). Since each side of the regular hexagon makes an angle of 60º at the centre O. Ê 1 3ˆ Coordinates of A1 are (cos 60º, sin 60º) = Á , ˜ Ë2 2 ¯ Ê 1 3ˆ A2 are (cos 120º, sin 120º) = Á - , ˜ Ë 2 2 ¯ A3 are (– 1, 0) Ê 1 Ê1 3ˆ 3ˆ A4 are Á - , and A5 are Á , - ˜ ˜ 2 ¯ 2 ¯ Ë 2 Ë2 2

Now

A0 A1 =

2 Ê 3ˆ 1 3 Ê 1ˆ + =1 ÁË 1 - ˜¯ + Á ˜ = 2 4 4 Ë 2 ¯

A0 A2 =

2 Ê 3ˆ Ê 1ˆ 1 + ÁË ˜¯ + Á ˜ = 2 Ë 2 ¯

2

= So that

9 3 + 4 4

3 = A0 A4 (A0 A1) (A0 A2) (A0 A4) = 3

Example 97: For each natural number k, let Ck denote the circle with radius k centimeters and centre at the origin O. On the circle Ck a particle moves k centimetres in the counter-clockwise direction. After completing its motion on Ck, the particle moves to Ck +1 in the radial direction. The motion of the particle continues in this manner. The particle starts at (1, 0). If the particle crosses the positive direction of x-axis for the first time on the circle Cn, then n = (a) 4 (b) 5 (c) 6 (d) 7

17.24

Complete Mathematics—JEE Main

Ans. (d) Solution: The motion of the particle on the first four circles is shown with bold line in the figure. Note that on every circle the particle travels just one radian. The particle crosses the positive direction of x-axis first time on Cn, where n is the least positive integer such that n ≥ 2p fi n = 7.

As the centre of C3, lies above the x-axis, we take l = - 3 and thus an equation of C3 is x2 + y2 – x - 3y = 0 Since C1 and C3 intersect and are of unit radius, their common tangents are parallel to the line joining their centres (0, 0) and (1/2, 3 /2). So, let the equation of a common tangent be 3x–y+k=0 It will touch C1, if k =1fi k=+2 3 +1 From the figure (16.32), we observe that the required tangent makes positive intercept on the y-axis and negative

(1, 0)

on the x-axis and hence its equation is

3 x – y + 2 = 0. 2

Fig. 17.31

Example 98: C1 and C2 are circles of unit radius with centres at (0, 0) and (1, 0) respectively. C3 is a circle of unit radius, passes through the centres of the circles C1 and C2 and have its centre above x-axis. Equation of the common tangent to C1 and C3 which does not pass through C2 is (a) x – (c) Ans. (b)

3y+2=0

(b)

3x–y–2=0

3x–y+2=0

(d) x + 3 y + 2 = 0

Solution: Equation of any circle (1, 0) is x (x – 0) (x – 1) + (y – 0) (y – 0) +l 0 1

through (0, 0) and y 1 0 1 =0 0 1

fi x2 + y2 – x + ly = 0 If it represents C3, its radius = 1 fi

1 = (1/4) + (l2/4)

fi

l= ± 3

Y

Example 99: A chord of the circle x + y2 – 4x – 6y = 0 passing through the origin subtends an angle tan–1 (7/4) at the point where the circle meets positive y-axis. Equation of the chord is (a) 2x + 3y = 0 (b) x + 2y = 0 (c) x – 2y = 0 (d) 2x – 3y = 0 Ans. (c) Solution: The given circle passes through the origin O and meets the positive Y-axis at B (0, 6). Let OP be the chord of the circle passing through the origin subtending an angle q at B, where tan q = 7/4 fi –OBP = q Equation of the tangent OT at O to the given circle is 2x + 3y = 0 fi

slope of the tangent = – 2/3

So that, if –XOT = a, tan a = 2/3 From geometry, –POT = –OBP = q fi –POX = q – a tan q - tan a and tan (q – a) = 1 + tan q tan a 7 2 13 1 = 4 3 = = 7 2 26 2 1+ ¥ 4 3 Y

C3

B (0, 6) q P

C1 (0, 0)

C2 (1, 0)

X X

O a q

T

Fig. 17.32

Fig. 17.33

Circles and Systems of Circles 17.25

Hence the equation of OP is y = x tan (q – a) fi x – 2y = 0

or y = tan (a /2) (x + a) Similarly equation of BQ is

Example 100: On the line joining the points A (0, 4) and B (3, 0), a square ABCD is constructed on the side of the line away from the origin. Equation of the circle having centre at C and touching the axis of x is (a) x2 + y2 – 14x – 6y + 49 = 0 (b) x2 + y2 – 14x – 6y + 9 = 0 (c) x2 + y2 – 6x – 14y + 49 = 0 (d) x2 + y2 – 6x – 14y + 9 = 0 Ans. (a) Solution: Let –ABO = q then –CBL= 90º –q, CL being perpendicular to x-axis. The coordinates of C are (OL, LC) A (0, 4) C (7, 3)

y=

(i)

a sin b ( x - a) a (cos b - 1)

or y = – cot (b /2) (x – a) (ii) We now eliminate a, b from (i) and (ii) y a-x , tan (b/2) = From (i) and (ii) tan (a /2) = a+x y Now a – b = 2g y a-x tan(a / 2) - tan ( b / 2 ) y a+x = fi tan g = y a-x 1 + tan (a / 2 ) tan(b / 2) 1+ . a+x y y2 - (a 2 - x 2 ) x 2 + y2 - a2 = fi tan g = (a + x ) y + (a - x ) y 2 ay fi x2 + y2 – 2ay tan g = a2 which is the required locus.

5

4

R P

5 90° - q

q 3

O

B (3, 0)

L

a (- a, 0) A

Q (a cos b, a sin b ) b

O

B (a, 0)

Fig. 17.34

OL = OB + BL = 3 + 5 sin q = 3 + 5 × (4/5) = 7 CL = 5 cos q = 5 × (3/5) = 3 So the coordinate of C are (7, 3) and the equation of the circle having C as centre and touching x-axis is (x – 7)2 + (y –3)2 = (CL)2 = 9 fi

x2 + y2 – 14x – 6y + 49 = 0.

Example 101: A circle with centre at the origin and radius equal to a meets the axis of x at A and B. P (a) and Q (b) are two points on this circle so that a – b = 2g, where g is a constant. The locus of the point of intersection of AP and BQ is (a) x2 – y2 – 2ay tan g = a2 (b) x2 + y2 – 2ay tan g = a2 (c) x2 + y2 + 2ay tan g = a2 (d) x2 – y2 + 2ay tan g = a2 Ans. (b) Solution: Coordinates of A are (– a, 0) and of P are (a cos a, a sin a) \ Equation of AP is y=

a sin a (x + a) a (cos a + 1)

Fig. 17.35

Example 102: Equation of a circle having radius equal to twice the radius of the circle x2 + y2 + (2p + 3)x + (3 – 2p)y + p – 3 = 0 and touching it at the origin is (a) x2 + y2 + 9x – 3y = 0 (b) x2 + y2 – 9x + 3y = 0 (c) x2 + y2 + 18x + 6y = 0 (d) x2 + y2 + 18x – 6y = 0 Ans. (d) Solution: Since the given circle passes through the origin p – 3 = 0 fi p = 3 and the equation of the given circle is x2 + y2 + 9x – 3y = 0 Equation of the tangent at the origin to this circle is 9x – 3y = 0 (i) Let the equation of the required circle which also passes through the origin be x2 + y2 + 2gx + 2fy = 0. Equation of the tangent at the origin to this circle is gx + fy = 0 If (i) and (ii) represent the same line, then

(ii)

17.26

Complete Mathematics—JEE Main

c a

b C1

O (0, 0)

(a + b, 0)

C2 (a + 2b + c, 0)

Fig. 17.36

g f = k (say) = 9 -3 We are given that

(iii)

2 2 Ê 9ˆ Ê -3 ˆ g 2 + f 2 = 2 Á ˜ + Á ˜ = 81 + 9 Ë 2¯ Ë 2¯

From (iii) we get | k | 92 + 32 = 90

fi

k = ±1

For k = 1, g = 9, f = – 3 and the equation of the required circle is x2 + y2 + 18x – 6y = 0. Example 103: A circle C1 of radius b touches the circle x2 + y2 = a2 externally and has its centre on the positive x-axis; another circle C2 of radius c touches the circle C1 externally and has its centre on the positive x-axis. Given a < b < c, then the three circles have a common tangent if a, b, c are in (a) A.P. (b) G.P. (c) H.P. (d) none of these Ans. (b) Solution: Refer Fig. 17.36. The centre of C1 is (a + b, 0) and the centre of C2 is (a + 2b + c, 0) Let y = mx + k be a tangent common to the three circles. Since it touches x2 + y2 = a2, C1 and C2 k 1 + m2

= ±a ,

m( a + b) + k 1 + m2

=±b

m ( a + 2b + c ) + k

=±c 1 + m2 As the centre of the three circles lie on the same side of the line y = mx + k, taking the same sign, say positive, in the three relations we get, k m +1 =a=b– = c – m Ê 1 , 9ˆ 2 Ë 5 5¯ 1- m 1+ m and

fi fi

a+b a + 2b + c = (eliminating m) c-a b-a (a + b) (c – a) = (b – a) (a + 2b + c)

fi ac – a2 + bc – ba = ba – a2 + 2b2 – 2ab + bc – ac

fi fi

2ac = 2b2 a, b, c are in G.P.

fi

ac = b2

Example 104: If two circles, each of radius 5 units, touch each other at (1, 2) and the equation of their common tangent is 4x + 3y = 10, then equation of the circle, a portion of which lies in all the quadrants is (a) x2 + y2 – 10x – 10y + 25 = 0 (b) x2 + y2 + 6x + 2y – 15 = 0 (c) x2 + y2 + 2x + 6y – 15 = 0 (d) x2 + y2 + 10x + 10y + 25 = 0 Ans. (b) Solution: The centres of the two circles will lie on the line through P (1, 2) perpendicular to the common tangent 4x + 3y = 10. If C1 and C2 are the centres of these circles then PC1 = 5 = r1, PC2 = – 5 = r2. Also C1, C2 lie on the x -1 y - 2 line = = r where tan q = 3/4. When r = r1 the cos q sin q coordinates of C1 are (5 cos q + 1, 5 sin q + 2) or (5, 5) as cos q = 4/5, sin q = 3/5. When r = r2, the coordinates of C2 are (– 3, – 1) The circle with centre C1 (5, 5) and radius 5 touches both the coordinate axes and hence lies completely in the first quadrant. Therefore the required circle is with centre (–3, –1) and radius 5, so its equation is (x + 3)2 + (y + 1)2 = 52 2 or x + y2 + 6x + 2y – 15 = 0 Since the origin lies inside the circle, a portion of the circle lies in all the quadrants. Example 105: The locus of the point of intersection of the tangents at the extremities of a chord of the circle x2 + y2 = a2 which touches the circle x2 + y2 – 2ax = 0 is (a) y2 = a(a – 2x) (b) x2 = a(a – 2y) (c) x2 + y2 = (x + a)2 (d) x2 + y2 = (y + a)2 Ans. (a)

Circles and Systems of Circles 17.27

Solution: Let P (h, k) be the point of intersection of the tangents at the extrimities of the chord AB of the circle x2 + y2 = a2. Then AB is the chord of contact of the tangents from P to the circle and so its equation is hx + ky = a2 (i) If (i) touches the circle x2 + y2 – 2ax = 0 (h ) ( a ) + ( k ) (0 ) - a 2

then

h2 + k 2

=±a

fi

(h – a)2 = h2 + k2

fi

locus of (h, k) is (x – a)2 = x2 + y2

y2 = a(a – 2x). Example 106: If q is the angle subtended by the circle S ∫ x2 + y2 + 2gx + 2fy + c = 0 at a point P(x1, y1) outside the circle and S1 ∫ x21 + y21 + 2gx1 + 2fy1 + c, then cos q is equal to

fi

(a)

S1 + c - g 2 - f 2

(c)

S1 + c + g 2 - f 2

S1 - c + g 2 + f 2 S1 - c + g 2 - f 2

(b)

S1 - c + g 2 + f 2

(d)

S1 - c + g 2 + f 2

S1 + c - g 2 + f 2 S1 + c + g 2 - f 2

Ans. (a) Solution: Let PA and PB be the tangents from P(x1, y1) to the given circle with centre C (– g, – f) such that –APB = q then –APC = –CPB = q/2 A (x1, y1) P

q /2 q /2

C (- g, -f )

B

Ans. (b) Solution: Let OA, OB be the tangents from the origin to the given circle with centre C(- 3, 5) and radius 9 + 25 - c = 34 - c. Then area of the quadrilateral OACB = 2 ¥ area of the triangle OAC = 2 ¥ (1/2) × OA ¥ AC Now OA = length of the tangent from the origin to the and

q CA tan = = 2 PA Now

cos q =

=

1 - tan 2 q / 2 2

1 + tan q / 2

g2 + f 2 - c S1 =

( ) 2 2 S1 + ( g + f - c ) S1 - g 2 + f 2 - c

S1 + c - g 2 - f 2 S1 - c + g 2 + f 2

Example 107: If the area of the quadrilateral formed by the tangent from the origin to the circle x2 + y2 + 6x – 10y + c = 0 and the pair of radii at the points of contact of these tangents to the circle is 8 square units, then c is a root of the equation (b) c2 – 34c + 64 = 0 (a) c2 – 32c + 64 = 0 2 (d) c2 + 34c – 64 = 0 (c) c + 2c – 64 = 0

34 - c

so that c 34 - c = 8 (given) fi c (34 – c) = 64 fi c2 – 34c + 64 = 0 Example 108: An equation of the circle in which the chord joining the points (1, 2) and (2, – 1) subtends an angle of p/4 at any point on the circumference is (a) x2 + y2 – 15 = 0 (b) x2 + y2 – 6x – 2y + 5 = 0 (c) x2 + y2 + 6x + 2y – 15 = 0 (d) x2 + y2 – 2x – 4y + 4 = 0 Ans. (b) Solution: Let P(x, y) be any point on the circle passing y-2 through A (1, 2) and B (2, –1) the slope of AP = and x -1 y +1 slope of BP = x-2 Since they include an angle of p/4 so y +1 y -2 p x - 2 x -1 = ± tan =±1 y - 2 y +1 4 1+ ¥ x -1 x - 2 fi

(y + 1) (x – 1) – (x – 2) (y – 2) = ± [(x – 1) (x – 2) + (y – 2) (y + 1)]

fi

3x + y – 5 = ± [x2 + y2– 3x – y]

fi

x2 + y2 – 5 = 0

Fig. 17.37

From right angled triangle PAC

given circle = c AC = radius of the circle =

or

x2 + y2 – 6x – 2y + 5 = 0.

Example 109: An equation of a common tangent to the circles x2 + y2 + 14x – 4y + 28 = 0 and x2 + y2 – 14x + 4y – 28 = 0 is (a) x – 7 = 0 (b) y + 7 = 0 (c) 28x + 45y + 371 = 0 (d) None of these Ans. (c) Solution: By subtracting the equation of one of the circles from that of the other, we get the equation of their common chord as 7x – 2y + 14 = 0. This chord intersects the circles at two real, distinct points, so the given circles have two common tangents. Let PA1 A2 be a common tangent to these circle, whose centres are O1 (– 7, 2) and O2 (7, – 2), and radii 5 and 9, respectively. Then triangles PO1 A1 and PO2 A2 are similar, so

17.28

Complete Mathematics—JEE Main

P O2 P O1 = O2 A2 O1 A1 fi

fi

n

P O1 P O2 = . 5 9

+

i =1

P divides O1O2 externally in the ratio 5 : 9, so its coordinates are (– 49/2, 7). Thus any line through P is given by

fi

9

5 P

(– 7, 2)

n Ê1 n ˆ which is a circle with centre Á Â xi , Â yi ˜ the point of Ë n i =1 i =1 ¯ mean position of the given points.

(7, –2)

O2

O1

Fig. 17.38

49 ˆ Ê y - 7 = m Áx + ˜ . Ë 2¯

(1)

This line will touch the circle of radius 5 centred at O1 if 49 m ÊÁ - 7 + ˆ˜ - 2 + 7 Ë 2¯ 1 + m2 fi

35 m +5=±5 2

fi

Ê 7m ˆ + 1˜ = 1 + m2 ÁË ¯ 2

Ê1 n ˆ Ê1 n ˆ x + y2 – 2 Á Â xi ˜ x - 2 Á Â yi ˜ y Ë n i =1 ¯ Ë n i =1 ¯ 2

n Ê n ˆ + 1 Á Â xi2 + Â yi2 - k 2 ˜ = 0 ¯ n Ë i =1 i =1

A2 A1

 yi2 – k2 = 0

Example 111: An isosceles triangle ABC is inscribed in a circle x2 + y2 = a2 with the vertex A at (a, 0) and the base angles B and C each equal to 75º, then length of the base BC is (a) a/2 (b) a (c) 2a/ 3 Ans. (b) fi fi fi fi fi

=±5

1 + m2

2

(d)

3 a/2

Solution: –B = –C = 75º –BAC = 30º –BOC = 60º BOC is an equilateral triangle BC = OB = the radius of the circle BC = a. y

45 m 2 + 7m = 0, 4 fi m = 0 or m = – 28/45. Substituting m = – 28/45 in (1) we get the equation given in (c).

fi

Example 110: The locus of the point, the sum of the squares of whose distances from n fixed points Ai (xi, yi ), i = 1, 2, …, n is equal to k2 is a circle (a) passing through the origin (b) with centre at the origin (c) with centre at the point of mean position of the given points (d) none of these Ans. (c) Solution: Let (x, y) be any point on the locus, then

B x¢

A (a, 0) M

C

O (0, 0)

x

y¢

Fig. 17.39

Example 112: If the locus of a point which moves so that the line joining the points of contact of the tangents drawn from it to the circle x2 + y2 = b2 touches the circle x2 + y2 = a2, is the circle x2 + y2 = c2, then a, b, c are in (a) A.P. (b) G.P. (c) H.P. (d) none of these Ans. (b)

n

Â

[(x – xi)2 + (y – yi)2] = k2

i =1 n

fi

n(x2 + y2) – 2 x  xi – 2 y i =1

n

 yi +

i =1

n

 xi2

i =1

Solution: Let P (h, k) be any point on the locus. Equation of the chord of contact of P with respect to the circle x2 + y2 = b2 is hx + ky = b2. If it touches the circle x2 + y2 = a2, then

Circles and Systems of Circles 17.29

-b2

=a

h2 + k 2

Ans. (b)

a2 (h2 + k2) = b4

fi

So that the locus of P (h, k) is x2 + y2 = (b2/a)2 \

Ê b2 ˆ c = Á ˜ Ë a¯

2

2

fi

ac = b2

fi a, b, c are in G.P. Example 113: If the line 3x – 4y – k = 0, (k > 0) touches the circle x2 + y2 – 4x – 8y – 5 = 0 at (a, b), then k + a + b is equal to (a) 20 (b) 22 (c) – 30 (d) – 28 Ans. (a) Solution: Since the given line touches the given circle, the length of the perpendicular from the centre (2, 4) of the circle to the line 3x – 4y – k = 0 is equal to the radius 4 + 16 + 5 = 5 of the circle. fi

3¥2-4¥4-k 9 + 16

fi fi

g 1g 2 =

then a = 3l + 2, b = 4 – 4l

and

2a + 4b + 5 = kl

fi

2(3l + 2) + 4(4 – 4l) + 5 = 15l

fi

l=1

(1)

(∵ k = 15)

a = 5, b = 0 and k + a + b = 20.

Example 114: The circles x2 + y2 + 2g1 x – a2 = 0 and x + y2 + 2g2 x – a2 = 0 cut each other orthogonally. If p1, p2 are perpendiculars from (0, a) and (0, – a) on a common tangent of these circles then p1 p2 = (b) a2 (a) a2/2 2 (c) 2a (d) a2 + 2 2

a 2 (l 2 + m 2 ) - 1

m2 a2 (l2 + m2 ) = 1 – a2 m2 p 1p 2 =

= ±5

b-4 a-2 2a + 4b + 5 = = = l (say) -4 3 k

fi

fi m2 g21 – 2lg1 + a2(l2 + m2) – 1 = 0 Similarly m2 g22 – 2lg2 + a2 (l2 + m2 ) – 1 = 0 So that g1, g2 are the roots of the equation m2g2 – 2lg + a2(l2 + m2 ) – 1 = 0

Now

fi k = 15 [∵ k > 0] Now equation of the tangent at (a, b) to the given circle is xa + yb – 2(x + a) – 4(y + b) – 5 = 0 fi (a – 2)x + (b – 4)y – (2a + 4b + 5) = 0. If it represents the given line 3x – 4y – k = 0 then

Solution: Since the given circles cut each other orthogonally g1 g2 + a2 = 0 (1) If lx + my = 1 is a common tangent of these circles, then -lg1 - 1 = ± g12 + a 2 2 2 l +m fi (lg1 + 1)2 = (l2 + m2) (g21 + a2)

=

ma - 1 2

l +m

2

1 - m2 a2 l 2 + m2

= - a 2 by (1) (2)

.

- ma - 1 l 2 + m2 = a 2 by (2)

Example 115: A circle C touches the x-axis and the circle x2 + (y – 1)2 = 1 externally, then locus of the centre of the circle C is given by (a) {(x, y) : x2 = 4y} » {(0, y) : y £ 0}. (b) {(x, y) : y = x2} » {(0, y) : y £ 0} (c) {(x, y) : x2 + (y – 1)2 = 0} » {(0, y) : y £ 0} (d) {(x, y) : x2 + 4y = 0} » {(0, y) : y £ 0} Ans. (a) Solution: Let the centre of C be (h, ± r) and radius r because it touches x-axis. Since it touches the given circle with radius 1 and centre (0, 1), externally h2 + (1 ± r)2 = (r + 1)2. fi h2 = (r + 1)2 – (r ± 1)2 fi h2 = 0 or h2 = 4r. \ Locus of (h, r) is {(x, y) : x2 = 4y} » {(0, y) : y £ 0}.

EXERCISE Concept-based Straight Objective Type Questions 1. Equation of the circle passing through the origin and having its centre on the line y = 3x at a distance 10 from the origin is

(a) x2 + y2 – 2x + 6y = 0 (b) x2 + y2 + 2x – 6y = 0 (c) x2 + y2 – 2x – 6y = 0 (d) none of these

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Complete Mathematics—JEE Main

2. The radius of the circle 3x2 + by2 + 4bx – 6by + b2 = 0 is (a) 1 (b) 3 (d) 11 (c) 10 3. Equation of the circle described on the line segment of 3x + 4y = 12 intercepted between the axes as a diameter is (a) x2 + y2 – 4x – 3y = 0 (b) x2 + y2 + 4x – 3y = 0 (c) x2 + y2 – 4x + 3y = 0 (d) x2 + y2 + 4x + 3y = 0 4. The point (1, 2) lies inside and (3, 4) outside the circle x2 + y2 – 7x + 15y – c = 0, if (a) c = 25 (b) c = 35 (c) c = 65 (d) c takes any real value 5. S : L: (a) (c)

x2 + y2 + 6x – 14y – 6 = 0 is a circle and 7x + 3y + 58 = 0 is a straight line L is a diameter of S (b) L is a chord of S L is a tangent to S (d) none of these

6. The angle between the two tangents from the origin to the circle (x – 7)2 + ( y + 1)2 = 25 is (b) p 4 (a) p 6 (c) p 3

(d) p 2

7. A line passes through the point P(5, 6) outside the circle x2 + y2 = 12 and meets the circle at A and B. The value of PA . PB is equal to (a) 25 (b) 36 (c) 49 (d) 61 8. The tangent to the circle x2 + y2 = 5 at (1, – 2) touches the circle x2 + y2 – 8x + 6y + 20 = 0 at the point (a) (3, 1) (b) (3, – 1) (c) (– 3, 1) (d) (1, – 3) 9. Two circles of equal radius of 5 units have their centres at the origin and the point (2, – 3). Equation of the common chord of these circles is (a) 4x – 6y – 13 = 0 (b) 2x – 3y + 13 = 0 (c) 4x – 6y + 12 = 0 (d) 2x – 3y – 12 = 0 10. Two circles touch each other externally at the point (0, k) and y-axis is the common tangent to these circles. Centres of these circle lie on the line (a) x = k (b) y = k (c) x + y = k (d) x – y = k 11. A circle has radius 3 units and its centre lies on the line y = x – 1. If the circle passes through the point (7, 3), then an equation of the circle is (a) x2 + y2 – 8x – 6y + 16 = 0 (b) x2 + y2 + 6x + 8y + 16 = 0 (c) x2 + y2 – 14x – 12y – 76 = 0 (d) x2 + y2 + 12x +14y – 76 = 0

12. The line 3x – y – 17 = 0 meets the circle x2 + y2 – 8x + 10y + 5 = 0 at the point A and B. P is any point on the circle other than A or B, then the triangle APB is (a) equilateral (b) isosceles (c) right angled (d) obtuse angled 13. A circle passes through the origin and its centre is on the line y = x. If it cuts the circle x2 + y2 – 4x – 6y + 10 = 0 orthogonally, then the equation of the circle is (a) x2 + y2 + 2x + 2y = 0 (b) x2 + y2 – 2x – 2y = 0 (c) 2x2 + 2y2 – x – y = 0 (d) 2x2 + 2y2 + x + y = 0 14. Equation of the circle on the common chord of the circles x2 + y2 – ax = 0 and x2 + y2 – ay = 0 as a diameter is (a) 2x2 + 2y2 – ax – ay = 0 (b) x2 + y2 – ax – ay = 0 (c) 2x2 + 2y2 + ax + ay = 0 (d) x2 + y2 – x – y + a = 0 15. A circle touches the lines x – y – 1 = 0 and x – y + 1 = 0, the centre of the circle lies on the line (a) x + y – 1 = 0 (b) x + y + 1 = 0 (c) x – y = 0 (d) none of these 16. The number of common to the circles x2 + y2 x2 + y2 + 2x + 2y + 1 = (a) 0 (c) 2

tangents that can be drawn – 4x – 6y – 3 = 0 and 0 is (b) 1 (d) 3

17. If the circle (x – 2)2 + (y – 3)2 = a2 lies entirely with in the circle x2 + y2 = b2, then (a) a = b (b) a – b > 13 (c) b – a >

13

(d) b – a =

13

18. There are four circles each of radius 1 unit touching both the axis. The equation of the smaller circle touching all these circles is

(

(a) x2 + y2 = 2

(b) x2 + y2 =

(c) x2 + y2 = 4

(d) x2 + y2 = 1

2

2 - 1)

19. The locus of the point which moves so that the ratio of the lengths of the tangents to the circle x2 + y2 – 4y + 3 = 0 and x2 + y2 + 6y + 5 = 0 is (a) 5x2 + 5y2 + 70x – 33 = 0 (b) 5x2 + 5y2 + 60y + 7 = 0 (c) 5x2 + 5y2 + 70y + 33 = 0 (d) 5x2 + 5y2 + 60x + 7 = 0 20. A circle has two of its diameters along the lines 2x + 3y – 18 = 0 and 3x – y – 5 = 0 and touches the line x + 2y + 4 = 0. Equation of the circle is x + y + 6x – 8y x + y – 6x + 8y x + y – 6x – 8y x + y – 6x – 8y

Circles and Systems of Circles 17.31

LEVEL 1 Straight Objective Type Questions Ê a bˆ 21. Equation of the circle with centre Á , ˜ and radius Ë 2 2¯ a 2 + b2 is 4 (a) x2 + y2 – ax – by = (a + b)2 (b) x 2 + y2 + ax + by = (a – b)2 (a 2 + b2 ) 4 (d) x2 + y2 – ax – by = 0 (c) x2 + y2 – ax – by =

22. The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a circle if (a) a = h = 2, b = 0 (b) b = h = 2, a = 0 (c) a = b = 2, h = 0 (d) none of these 23. The radius of the circle passing through the point (6, 2), two of whose diameter, are x + y = 6 and x + 2y = 4 is (b) 2 5 (d) 4

(a) 10 (c) 6

24. An equation of the circle through (1, 1) and the points of intersection of x2 + y2 + 13x – 3y = 0 and 2x2 + 2y2 + 4x – 7y – 25 = 0 is (a) 4x2 + 4y2 – 30x – 10y – 32 = 0 (b) 4x2 + 4y2 + 30x – 13y – 25 = 0 2

2

(c) 4x + 4y – 43x + 10y + 25 = 0 (d) none of these 25. An equation of the normal at the point (2, 3) to the circle x2 + y2 – 2x – 2y – 3 = 0 is (a) 2x + y – 7 = 0 (b) x + 2y – 3 = 0 (c) 2x – y – 1 = 0 (d) x – 2y + 1 = 0 26. The line x + y tan q = cos q touches the circle x2 + y2 = 4 for (a) q = p/6 (b) q = p/3 (c) q = p/2 (d) no value of q 27. If (x, 3) and (3, 5) are the extremities of a diameter of a circle with centre at (2x, y), then the values of x and y are (a) x = 1, y = 4 (b) x = 4, y = 1 (c) x = 8, y = 2 (d) none of these 28. The circle (x – r)2 + (y – r)2 = r2 touches (a) x-axis at the origin (b) y-axis at the origin

(c) both the coordinate axes (d) none of these 29. The number of circles touching the coordinate axes and the line x + y = 1 is (a) exactly one (b) two (c) three (d) four 30. The centre of circle passing through the points (0 0), (1, 0) and touching the circle x2 + y2 = 9 is the point whose coordinates are 2)

(a) (1/ 2 , 2)

(b) (1/2,

(c) ( 2 , 1/2)

(d) (2, 1/ 2 )

31. Equation of a circle which touches y-axis at (0, 2) and cuts off an intercept of 3 units from the x-axis is x2 + y2 – 2ax – 4y + 4 = 0 where a2 = (a) 5/2 (b) 25/4 (c) 4/25 (d) 29 32. The circles x2 + y2 + 2x – 2y + 1 = 0 and x2 + y2 – 2x – 2y + 1 = 0 (a) touch each other externally (b) touch each other internally (c) intersect on the y-axis (d) intersect on x-axis 33. The tangent at any point to the circle x2 + y2 = r2 meets the coordinate axes at A and B. If lines drawn parallel to the coordinate axes through A and B intersect at P, the locus of P is (b) x – 2 + y – 2 = r 2 (a) x2 + y2 = r – 2 1 1 1 1 1 (c) 2 + 2 = 2 (d) 2 - 2 = r2 x y r x y 34. Equation of a circle with centre C(h, k) and radius 5 such that h2 – 3h + 2 = 0 and k2 + 5k – 6 = 0 is (a) x2 + y2 – 2x – 4y = 0 (b) x2 + y2 – 2x – 2y – 6 = 0 (c) x2 + y2 – 4x – 2y – 20 = 0 (d) x2 + y2 – 4x – 6y – 12 = 0. 35. If S ∫ x2 + y2 – 2x – 4y – 4 = 0, L ∫ 2x +2y +15 = 0 and P(3, 4) represent a circle, a line and a point respectively then (a) L is a tangent to S at P (b) L is polar of P with respect to S (c) L is the chord of contact of P with respect to S (d) P is inside and L is outside S

17.32

Complete Mathematics—JEE Main

36. If PQR is the triangle formed by the common tangents to the circles x2 + y2 + 6x = 0 and x2 + y2 – 2x = 0, then the centroid of the triangle is at the point (a) (1, 0) (b) (0, 0) (c) (– 1, 0) (d) none of these 37. A rectangle ABCD is inscribed in the circle x2 + y2 + 3x + 12y + 2 = 0. If the coordinates of A and B are respectively (3, – 2) and (– 2, 0), then the coordinates of the mid-point of CD are (a) (–3/2, – 6) (b) (– 7/2, 11) (c) (5/2, – 1) (d) (– 5/2, – 1) 38. A line makes equal intercepts of length a on the coordinate axes. A circle is circumscribed about the triangle which the line makes with the coordinate axes. The sum of the distance of the vertices of the triangle from the tangent to this circle at the origin is (b) a (a) a / 2 (c) a 2

(d) a +

2

39. a, b and g are the parametric angles of three points P, Q and R, respectively, on the circle x2 + y2 = 1, and A is the point (– 1, 0). If the lengths of the chords AP, AQ and AR are in G.P., then cos (a/2), cos (b/2) and cos (g / 2) are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 40. If P is a point with integral coordinates on the circle x2 + y2 = 9, Q is a point on the line 7x + y + 3 = 0, and the line x – y + 1 = 0 is the perpendicular bisector of PQ, the coordinates of P are (a) (3, 0) (b) (0, 3) (c) (– 72, 21) (d) (– 1, 1) 41. If the circle C1: x2 + y2 = 16 intersects another circle C2 with centre at (a, b) and radius 5 in such a manner that the common chord is of maximum length, then a2 + b2 = (a) 9 (b) 12 (c) 15 (d) 25 42. If OA and OB are equal chords of the circle x2 + y2 – 2x + 4y = 0 perpendicular to each other passing through the origin, then the slopes of OA and OB are the roots of the equation (a) 2m2 – 5m – 2 = 0 (b) 3m2 – 8m – 3 = 0 2 (d) m2 + 1 = 0 (c) 6m + 5m – 6 = 0 43. Equation of the circle passing through (1, 0) and (0, 1) and having the smallest possible radius is (a) x2 + y2 – x – y = 0 (b) x2 + y2 – 2x – 2y + 1 = 0 (c) x2 + y2 = 1 (d) none of these

44. From the point A (0, 3) on the circle x2 + 4x + (y – 3)2 = 0, a chord AB is drawn and extended to a point M such that AM = 2AB. Equation of the locus of M is (a) x2 + 2x + (y – 3)2 = 0 (b) (x – 3)2 + 4y + y2 = 0 (c) x2 + 8x + (y – 3)2 = 0 (d) x2 + 4x + (y + 3)2 = 0 45. If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, radius of the circle is (a) 3/4 (b) 1 (c) 7/8 (d) 11/10 46. If the distances from the origin of the centres of three circles x2 + y2 – 2lix = c2 (i = 1, 2, 3) are in G.P., then the lengths of the tangents drawn to them from any point on the circle x2 + y2 = c2 are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 47. The distance between the chords of contact of the tangents to the circle x2 + y2 + 2gx + 2fy + c = 0 from the origin and the point (g, f ) is (a) 2 g 2 + f 2 - c (b) (g2 + f 2 – c)/ 2 g 2 + f 2 (c) (1/2) (g2 + f 2 + c) (d) g2 + f 2 – c 48. A variable circle passes through the fixed point A (p, q) and touches the axis of x, the locus of the other end of the diameter through A is (b) (x – q)2 = 4py (a) (x – p)2 = 4 qy 2 (d) (y – q)2 = 4px (c) (y – p) = 4 qx 49. The locus of a point, which moves such that the lengths of the tangents from it two concentric circles of radii a and b are inversely proportional to their radii is a circle with radius (a) a + b (c)

a 2 + b2

(b)

a 2 - b2

(d) none of these

50. The circle that can be drawn to touch the coordinate axes and the line 4x + 3y = 12 cannot lie in (a) first quadrant (b) second quadrant (c) third quadrant (d) fourth quadrant 51. Area of an equilateral triangle inscribed in a circle of radius a is (a)

3 a2/4

(c) 3 3 a2/4

(b) p a2/3 (d)

3 pa2/4

52. A point P moves such that PA/PB = p where A and B are two fixed points with AB = a, the locus of P is a circle with radius

Circles and Systems of Circles 17.33

(a) a/(1 – p2) (c) ap/(1 – p2)

(b) p/(1 – a2) (d) ap/(1 – a2)

53. If the lengths of the tangents from two points A and B to a circle are l and l¢ respectively and points are conjugate with respect to the circle (i.e. each passes through the polar of the other with respect to the circle), then (AB)2 = (a)

l 2 - l¢2

(c) l 2 + l ¢ 2 54. If two circles which (a, 0) and (– a, 0) touch orthogonally then (a) c2 = a2 (1 + m2) (c) c2 = a2(1 + m)2

(b) (l + l¢ )2 (d) (l – l¢ )2 pass through the points the line y = mx + c and cut (b) c2 = a2(1 + 2m2) (d) c2 = a2 + m2

55. No portion of the circle x2 + y2 – 16x + 18y + 1 = 0 lies in the (a) first quadrant (b) second quadrant (c) third quadrant (d) fourth quadrant 56. The greatest distance of the point P(10, 7) from the points on the circle x2 + y2 – 4x – 2y – 20 = 0 is (a) 5 (b) 5 3 (c) 10

(d) 15

57. The length of the longest ray drawn from the point (4, 3) to the circle x2 + y2 + 16x + 18y + 1 = 0 is equal to (a) the radius of the circle (b) the diameter of the circle (c) circumference of the circle (d) the distance of the centre of the circle from the origin 58. If y = + a is a pair of tangents to the circle x2 + y2 = a2 meeting the tangent at any point C on the circle at P and Q, then CP.CQ = (a) 1 (b) a2 2 (c) 1/a (d) none of these 59. The circle x2 + y2 = 9 is contained in the circle x2 + y2 – 6x – 8y + 25 = c2 if (a) c = 2 (b) c = 3 (c) c = 5 (d) c = 10 60. The line (x – 1) cos q + (y – 1) sin q = 1, for all values of q touches the circle (b) x2 + y2 – 2x = 0 (a) x2 + y2 = 1 2 2 (d) x2 + y2 – 2x – 2y + 1 = 0 (c) x + y – 2y = 0 61. Equation of the circle which cuts each of the circles x2 + y2 + 2gx + c = 0, x2 + y2 + 2g1x + c = 0 and x2 + y2 + 2hx + 2ky + a = 0 orthogonally is (a) h (x2 + y2) + (a – c)x – ch = 0 (b) k (x2 + y2) + (a – c)y – ck = 0

(c) x2 + y2 + (a – c)x + c = 0 (d) x2 + y2 + (a – c)y – c = 0 62. The point at which the circles x2 + y2 + 2x + 6y + 4 = 0 and x2 + y2 + 6x + 2y + 7 = 0 subtend equal angles lies on the circle (a) x2 + y2 + 10x – 2y + 10 = 0 (b) x2 + y2 – 2x + 10y + 10 = 0 (c) x2 + y2 + 10x – 2y – 10 = 0 (d) x2 + y2 – 10x + 10y – 10 = 0 63. Equation of a circle which passes through the point (2, 0) and whose centre is the limit of the point of intersection of the lines 3x + 5y = 1 and (2 + a )x + 5a2y = 1 as a tends to 1 is (a) (b) (c) (d)

25(x2 + y2) + 20x – 2y – 140 = 0 25(x2 + y2) – 20x + 2y – 60 = 0 9(x2 + y2) – 20x + 2y + 4 = 0 9(x2 + y2) – 2x – 20y + 4 = 0

64. If the limiting points of the system of circles x2 + y2 + 2gx + l (x2 + y2 + 2fy + k) = 0, where l is a parameter, subtend a right angle at the origin, then k/f 2 = (a) – 1 (b) 1 (c) 2 (d) none of these 65. If the chord of the circle x2 + y2 – 4y = 0 along the line x + y = 1 subtends an angle q at a point on the circumference of the larger segment then cos q = (a) 1/2

(b) 1/ 2

(c)

(d) 1/2 2

3 /2

66. Locus of the centre of the circle touching the line 3x + 4y + 1 = 0 and having radius equal to 5 units is (a) a pair of perpendicular lines (b) a pair of parallel lines (c) a pair of straight lines (d) none of these 67. Chord of the circle x2 + y2 = 81 bisected at the point (–2, 3) meets the diameter x + 5y = 0 at a point (a) on the circle (b) inside the circle (c) outside the circle (d) none of these 68. An equilateral triangle is inscribed in the circle x2 + y2 = 1 with one vertex at the point (1, 0). Length of each side of the triangle is (a) 1 (c)

3 /2

(b)

2

(d)

3

69. The curve parametrically described by the equations x = 2 + 3cosq, y = 3 + 3sinq represents a circle which touches (a) x-axis at (2, 0) (b) y-axis at (0, 3) (c) x + y = 2 at (1, 1) (d) x + y = 3 at (1, 2)

17.34

Complete Mathematics—JEE Main

70. Consider two circles x2 + y2 = a2 – l and x2 + y2 – 2axcosq – 2aysinq + l = 0. Each circle passes through the centre of the other for (a) all values of l (c) only one value of l

(b) no value of l (d) more than one value of l

71. Tangents at two points A and B on the circle x2 + y2 – 6x – 4y – 11 = 0 intersect at the point (1, 8). Equation of the chord AB is (a) x + 3y – 15 = 0 (c) 3x – y + 15 = 0 (c) x – 3y + 15 = 0 (d) 3x + 2y + 11 = 0

72. Equation of the circle on AB as diameter, where A and B are the points of intersection of the circle x2 + y2 – 8x = 0 and the curve 4x2 – 9y2 = 36 is (a) x2 + y2 – 12x + 24 = 0 (b) x2 + y2 + 12x + 24 = 0 (c) x2 + y2 + 24x – 12 = 0 (d) x2 + y2 – 24x – 12 = 0 73. A polygon of nine sides, each of length 2, is inscribed in a circle with centre at the origin. Equation of the circle is x2 + y2 = r2, where 1/r is equal to (a) cos 20° (b) sin 20° (c) cos 40° (d) sin 40°

Assertion-Reason Type Questions 74. Statement-1: The radius of the circle 2x2 + 2y2 – 4x cosq + 4y sinq – 1 – 4 cosq – cos2q = 0 is 1 + cosq Statement-2: The radius of the circle (x – cosq)(x – sinq) + (y – cosq)(y – sinq) = 0 (q π p/4) is |cosq – sinq|

77. Statement-1: The centre of a circle passing through the points (0, 0), (1, 0) and touching the circles C : x2 + y2 = 9 lies inside the circle C.

75. Statement-1: The points of intersection of the two curves whose equations are S1: x2 + 2y2 – 6x – 12y + 23 = 0 and S2: 4x2 + 2y2 – 20x – 12y + 35 = 0 lie on a circle

78. C1 : (x – 3)2 + (y – 4)2 = a2

with centre at (8/3, 3) and radius equal to 47 / 3 2 . Statement-2: Equation of a curve passing through the intersection of S1 = 0 and S2 = 0 in Statement-1 is S1 + lS2 = 0 which represents a circle if there exists a real value of l for which the coefficients of x2 and y2 are equal. 76. Statement-1: The locus of the point of intersection of the tangents to the circle x = a cos q, y = a sin q at points whose parametric angles differ by p/2 is x2 + y2 = 2a2 Statement-2: Tangents at the extremities of a diameter of a circle are parallel.

79. Statement-1: The common chord of the circles x2 + y2 – 10x + 16 = 0 and x2 + y2 = r2 is of maximum length if r2 = 34.

Statement-2: If a circle C1 passes through the centre of the circle C2 and also touches the circle, the radius of the circle C2 is twice the radius of the circle C1. Statement-1: C1 touches the axis of x if a = 4 Statement-2: C1 touches the line y = x if a = 3.

Statement-2: The common chord of two circles is of maximum length if it passes through the centre of the circle with smaller radius. 80. Statement-1: The equation x2 – y2 – 4x – 4y = 0 represents a circle with centre (2, 2) passing through the origin. Statement-2: The equation x2 + y2 + 4x + 6y + 13 = 0 represents a point.

LEVEL 2 Straight Objective Type Questions 81. The circles x2+ y2 + ax = 0 and x2 + y2 = c2 (c > 0) touch each other if (b) a2 – c2 = 0 (a) a2 + c2 = 1 2 2 (d) none of these (c) a – c = 1 82. If (a cos qi, a sin qi) i = 1, 2, 3 represent the vertices of an equilateral triangle inscribed in a circle, then

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

cos q1 + cos q2 + cos q3 = 0 sin q1 + sin q2 + sin q3 π 0 tan q1 + tan q2 + tan q3 = 0 cot q1 + cot q2 + cot q3 = 0

83. A circle C is drawn on the line joining the centres of the circles C1: x2 + y2 – 4 = 0 and C2: x2 + y2 – 8x + 7 = 0 as a diameter. The length of the intercept made

Circles and Systems of Circles 17.35

on the common chord of the circles C1 and C2 by the circle C is (a)

231 /8

(c)

231 /2

(b)

231 /4

(d) 11/8 2

84. If the equation ax + 2(a + ab – 16)xy + by2 + 2ax + 2by – 4 2 = 0 represents a circle, the radius of the circle is (a) 2 (b) 2 2 (c)

2

2

(d) 4 2

85. If a circle passes through the points of intersection of the coordinate axes with the line x – ly + 1 = 0 (l π 0) and x – 2y + 3 = 0, then the value of l is a root of the equation (b) 7l2 – 6l – 2 = 0 (a) 6l2 – 7l + 2 = 0 2 (d) 2l2 – 7l + 6 = 0 (c) 2l – 6l + 1 = 0 86. If a square with length of each side equal to a is inscribed in the circle x2 + y2 + 4x + 10y + 21 = 0, then a is equal to (a)

2

(b) 2

(d) 4 (c) 2 2 87. The locus of a point which divides the join of A (– 1, 1) and a variable point P on the circle x2 + y2 = 4 in the ratio 3 : 2 is a circle whose centre is at the point (a) (2/5, 2/5) (b) (–2/5, 2/5) (c) (2/5, – 2/5) (d) (– 5/14, 5/14) 88. If (a, b ) are the roots of the equation 15x2 – 22x + 8 = 0 and (a¢, b¢ ) are the roots of the equation 8x2 – 22x + 15 = 0, then the equation of the circle on A (a, a¢ ) and B (b, b¢ ) as diameter is (a) 120 (x2 + y2) – 22 (8x + 15y) + 289 = 0 (b) 120 (x2 + y2) – 22 (15y + 8x) + 289 = 0 (c) 23 (x2 + y2) + 22 (8x + 15y) – 120 = 0 (d) none of these 89. If two circles cut a third circle orthogonally, the radical axis of the two circles (a) touches the third circle (b) passes through the centre of the third circle (c) lies out side the third circle (d) none of these 90. If a circle of constant radius 3k passes through the origin and meets the axes at A and B, the locus of the centroid of the triangle OAB is the circle (a) x2 + y2 + kx + ky = 0 (b) x2 + y2 = 4k2 (c) x2 + y2 = k2 (d) x2 + y2 + 2kx – 2ky + k2 = 0

91. The locus of a point which moves such that the sum of the squares of its distances from the three vertices of a triangle is a constant, is a circle whose centre is the (a) incentre of the triangle (b) circumcentre of the triangle (c) orthocentre of the triangle (d) centroid of the triangle 92. The tangents to the circle x2 + y2 = 48, which are inclined at angle of 60º with the axis of x form a rhombus, the length of whose sides is (a) 2 3

(b) 3

(c) 3 3

(d) 4

93. An equation of the circle described on the chord x cos a + y sin a – p = 0 of the circle x2 + y2 = a2 as diameter is (a) x2 + y2 + 2px cos a + 2py sin a – (2p2 + a2) = 0 (b) x2 + y2 – 2px cos a – 2py sin a + 2p2 – a2 = 0 (c) x2 + y2 – 2px cos a + 2py sin a – 2p2 + a2 = 0 (d) x2 + y2 + 2px cos a – 2py sin a + 2p2 – a2 = 0 94. The locus of a point which moves in a plane so that the sum of the squares of its distances from the line ax + by + c = 0 and bx – ay + d = 0 is r2, is a circle of radius (a) r

(b) r a 2 + b2

(c) r ab

(d) none of these

95. Two circles are such that one is inscribed in and the other is circumscribed about a square A1 A2 A3 A4. If the length of each side of the square is a and P, Q are two points respectively on these circles, then 4

4

| Â ( PAi )2 - Â (QAi )2 |= i =1

i =1

2

(b) a2/2 (d) 2a2

(a) a /4 (c) a2

96. Two tangents T1 and T2 are drawn from (– 2, 0) to the circle C : x2 + y2 = 1. Equation of a circle touching C, having T1, T2 as a pair of tangents from (–2, 0) and radius greater than the radius of C is (a) x2 + y2 – 6x + 5 = 0 (b) x2 + y2 – 8x + 7 = 0 (c) 9x2 + 9y2 + 24x + 15 = 0 (d) none of these 97. The locus of the point of intersection of tangents to the circle x = a cos q, y = a sin q at points whose parametric angles differ by p/4 is (a) x2 + y2 = 2

(

2

)

2 - 1 a2

(b) x2 + y2 = 2(2 –

2 )a2

17.36

Complete Mathematics—JEE Main

(c) x2 + y2 =

(

2

)

2 +1

a2

(d) none of these 98. The locus of the centre of the circle which cuts the circles x2 + y2 + 4x – 6y + 9 = 0 and x2 + y2 – 4x + 6y + 4 = 0 orthogonally is (a) 12x + 8y + 5 = 0 (b) 8x – 12y + 5 = 0 (c) 5x – 12y + 8 = 0 (d) none of these 99. An equation of the tangent at the point (5, 2) of a circle is given by 3x – 2y – 11 = 0. If the circle passes through the origin, an equation of the circle is (a) x2 + y2 – 23x – 102y = 0 (b) 11x2 + 11y2 – 102x – 23y = 0 (c) 11x4 + 11y2 – 23x – 102y = 0 (d) x2 + y2 – 102x – 23y = 0 100. A circle passes through the origin O and cuts the axes at A (a, 0) and B (0, b). The reflection of origin O in the line AB is the point Ê 2ab2 2a2 b ˆ , (a) Á 2 Ë a + b2 a 2 + b2 ˜¯

Ê 2a 2 b 2ab2 ˆ , (b) Á 2 Ë a + b2 a 2 + b2 ˜¯

(c) ÊÁ 2ab , 2ab ˆ˜ Ë a 2 + b2 a 2 + b2 ¯

(d) (a, b)

101. If a circle cuts x2 + y2 = a2 orthogonally and passes Ê a2 p a2 q ˆ , through the point Á 2 , then it will also Ë p + q 2 p2 + q 2 ˜¯ pass through (a) (p, 0) (c) (p, q)

(b) (0, q) (d) (ap, aq)

102. Two circles intersect at the point P (2, 3) and the line joining the other extremity of the two diameter through P makes an angle p /6 with x-axis, then the equation of the common chord of the two circles is (a) x +

3y – (2 + 3 3 ) = 0

(b) x +

3y – ( 2 3 + 2) = 0

(c)

3x + y – ( 2 3 + 3) = 0

(d)

3x + y – (2 + 3 3 ) = 0

103. A system of circles is drawn through two fixed points (–1, 0) and (1, 0); tangents are drawn to these circles parallel to the line y = x, the locus of the points of contact is (a) x2 – y2 – 2xy = 1 (b) x2 + y2 = 1

(c) x2 + y2 + 2xy = 0 (d) x2 – y2 + 2xy + 1 = 0 104. The radius upto one place of decimal of the smallest circle which touches the line 3x – y = 6 at (1, – 3) and also touches the line y = x is (a) 1.2 (b) 1.6 (c) 1.8 (d) 2.1 105. Of the two concentric circles the smaller one has the equation x4 + y2 = 4. If each of the two intercepts on the line x + y = 2 made between the two circles is 1, the equation of the larger circle is (a) x2 + y2 = 5 (c) x2 + y2 = 7 + 2 2

(b) x2 + y2 = 5 + 2 2 (d) x2 + y2 = 11

106. An equation of the circle situated symmetrically opposite to the circle x2 + y2 – 2x = 0 with respect to the line x + y = 2 is (a) x2 + y2 + 2y – 4 = 0 (b) x2 + y2 – 4x – 2y + 4 = 0 (c) x2 + y2 – x + y – 2 = 0 (d) x2 + y2 – 3x – y + 2 = 0 107. If the chord along the line y – x = 3 of the circle x2 + y2 = k2 subtends an angle of 30º in the major segment of the circle cut off by the chord then k2 = (a) 3 (b) 6 (c) 9 (d) 36 108. Three concentric circles of which the biggest is x2 + y2 = 1 have their radii in A.P. with common difference d (> 0). If the line y = x + 1 cuts all the circles in real distinct points, then (a) d
1 +

(b) d >

1

2+ 2 4

(d) d is any real number

2

109. Radius of the circle centred at (3, – 2), of maximum area contained in the circle x2 + y2 – 4x + y = 0 is (a) (1/2)

(

17 - 13

)

(b) 1/ 2

(c) 2 (d) none of these 110. Equation of a family of circles, such that each member of the family passes through the origin and makes an intercept on the line y = 2x which is twice the intercept made on the line x = 2y is (l being a parameter) (a) x2 + y2 – 2l y = 0 (b) x2 + y2 – 2lx = 0 (c) x2 + y2 – 2lx – 2ly = 0 (d) none of these

Circles and Systems of Circles 17.37

Previous Years' AIEEE/JEE Main Questions 1. Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are (a) x = ± (y + 2a) (b) y = ± (x + 2a) (c) x = ± (y + a) (d) y = ± (x + a) [2002] 2. If the chord y = mx + 1 of the circle x2 + y2 = 1 subtends an angle of measure 45º at the major segment of the circle then the value of m is (a) 2 (b) t – 2 (c) – 1 (d) none of these [2002] 3. The centres of a set of circles, each of radius 3, lie on the circle x2 + y2 = 25. The locus of any point in the set is (b) x2 + y2 £ 25 (a) 4 £ x2 + y 2 £ 64 2 2 (d) 3 £ x2 + y2 £ 9 [2002] (c) x + y £ 25 4. The centre of the circle passing through (0, 0) and (1, 0) and touching the circle x2 + y2 = 9 is (a) (1/2, 1/2)

(b) (1 / 2, - 2 )

(c) (3/2, 1/2) (d) (1/2, 3/2) [2002] 5. The equation of a circle with origin as centre and passing through equilateral triangle whose medium is of length 3a is (b) x2 + y2 = 16a2 (a) x2 + y2 = 9a2 2 2 2 (c) x + y = 4a (d) x2 + y2 = a2 [2002] 2 2 2 2 2 6. If two circle (x – 1) + (y – 3) = r and x + y – 8x + 2y + 8 = 0 intersect in two distinct points then (a) r < 2 (b) r = 2 (c) r > 2 (d) 2 < r < 8 7. The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle having area as 154 sq. units. Then the equation of the circle is (a) (b) (c) (d)

x2 + y2 + 2x – 2y = 47 x2 + y2 – 2x + 2y = 47 x2 + y2 – 2x + 2y = 62 x2 + y2 + 2x – 2y = 62

[2003]

8. If a circle passes through (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is (a) 2ax – 2by + (a2 + b2 + 4) = 0 (b) 2ax + 2by – (a2 + b2 + 4) = 0 (c) 2ax + 2by + (a2 + b2 + 4) = 0 [2004] (d) 2ax – 2by – (a2 + b2 + 4) = 0 9. A variable circle passes through the fixed point A(p, q) and touches x-axis. The locus of the other end of the diameter through A is

(a) (y – p)2 = 4qx (b) (x – q)2 = 4py 2 (d) (y – q)2 = 4px [2004] (c) (x – p) = 4qy 10. If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10p, then the equation of the circle is (a) x2 + y2 + 2x + 2y – 23 = 0 (b) x2 + y2 – 2x – 2y – 23 = 0 (c) x2 + y2 – 2x + 2y – 23 = 0 [2004] (d) x2 + y2 + 2x – 2y – 23 = 0 11. The intercept on the line y = x by the circle x2 + y2 – 2x = 0 is AB. Equation of the circle on AB as diameter is (a) x2 + y2 + x + y = 0 (b) x2 + y2 – x + y = 0 (c) x2 + y2 – x – y = 0 [2004] (d) x2 + y2 + x – y = 0 12. If the circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 – 3ax + dy – 1 = 0 intersect in two distinct points P and Q then the line 5x + by – a = 0 passes through P and Q for (a) infinitely many values of a (b) exactly two values of a (c) exactly one value of a (d) no value of a [2005] 13. If a circle passes through the point (a, b) and cuts the circle x2 + y2 = p2 orthogonally, then the equation of the locus of its centre is (a) x2 + y2 – 2ax – 3by + (a2 – b2 – p2) = 0 (b) 2ax + 2by – (a2 + b2 + p2) = 0 (c) x2 + y2 – 3ax – 4by + (a2 + b2 – p2) = 0 [2005] (d) 2ax + 2by – (a2 – b2 + p2) = 0 14. A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre of the circle is (a) a hyperbola (b) a parabola (c) an ellipse (d) a circle [2005] 15. Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords of the circle C that subtends an angle 2p/3 at its centre is (b) x2 + y2 = 3/2 (a) x2 + y2 = 9/4 2 2 (d) x2 + y2 = 27/4 [2006] (c) x + y = 1 16. If the lines 3x – 4y – 7 = 0 and 2x – 3y – 5 = 0 are two diameters of a circle of area 49p square units, the equation of the circle is

17.38

Complete Mathematics—JEE Main

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

x2 + y2 – 2x + 2y – 47 = 0 x2 + y2 + 2x – 2y – 47 = 0 x2 + y2 + 2x – 2y – 62 = 0 x2 + y2 – 2x + 2y – 62 = 0

(a) 6/5 (c) 10/3 [2006]

17. Consider a family of circles with are passing through the point (– 1, 1) and are tangent to x-axis. If (h, k) are the coordinates of the centre of the circles, then the set of values of k is given by the interval (a) 0 < k < 1/2 (b) k ≥ 1/2 (c) – 1/2 £ k £ 1/2 (d) k £ 1/2 [2007] 18. The point diametrically opposite to the point P(1, 0) on the circle x2 + y2 + 2x + 4y – 3 = 0 is (a) (3, 4) (b) (3, – 4) (c) (– 3, 4) (d) (– 3, – 4) [2008] 19. Three distinct points A, B and C are given in the 2-dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point (– 1, 0) is equal to 1/3. Then the circumcentre of the triangle ABC is at the point (a) (5/2, 0) (b) (5/3, 0) (c) (0, 0) (d) (5/4, 0) [2009] 20. If P and Q are the points of intersection of the circles x2 + y2 + 3x + 7y + 2p – 5 = 0 and x2 + y2 + 2x + 2y – p2 = 0, then there is a circle passing through P, Q and (1, 1) for (a) all except two values of p. (b) exactly one value of p (c) all values of p (d) all except one value of p [2009] 21. The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if (a) 15 < m < 65 (b) 35 < m < 85 (c) – 85 < m < – 35 (d) – 35 < m < 15 [2010] 22. The equation of the circle passing through the points (1, 0) and (0, 1) and having the smallest radius is (a) x2 + y2 – 2x – 2y + 1 = 0 (b) x2 + y2 – x – y = 0 (c) x2 + y2 + 2x + 2y – 7 = 0 [2011] (d) x2 + y2 + x + y – 2 = 0 23. The two circles x2 + y2 = ax, x2 + y2 = c2 (c > 0) touch each other (a) | a | = 2c (b) 2 | a | = c (c) | a | = c (d) a = 2c [2011] 24. The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3) is

(b) 5/3 (d) 3/5

[2012]

25. A circle passing through (1, – 2) and touching the axis of x at (3, 0) also passes through the point (a) (2, – 5) (b) (5, – 2) (c) (– 2, 5) (d) (– 5, 2) [2013] 26. If each of the lines 5x + 8y = 13 and 4x – y = 3 contains a diameter of the circle x2 + y2 – 2(a2 – 7a + 11) x – 2 (a2 – 6a + 6) y + b3 + 1 = 0, then (a) a = 5 and b œ (– 1, 1) (b) a = 1 and b œ (– 1, 1) (c) a = 2 and b Œ (– •, 1) (d) a = 5 and b Œ (– •, 1) [2013, online] 27. If a circle C passing through (4, 0) touches the circle x2 + y2 + 4x – 6y – 12 = 0 externally at a point (1, – 1), then the radius of the circle C is: (a) 5

(b) 2 5

(c) 4

(d)

57

[2013, online]

28. If two vertices of an equilateral triangle are A (– a, 0) and B(a, 0), a > 0, and the third vertex C lies above x-axis, then the equation of the circumcircle of D ABC is: (a) (b) (c) (d)

3x2 + 3y2 – 2 3 ay = 3a2 3x2 + 3y2 – 2ay = 3a2 x2 + y2 – 2ay = a2 x2 + y2 – 3 ay = a2 2

[2013, online]

2

29. If the circle x + y – 6x – 8y + (25 – a2) = 0 touches the axis of x, then a equals (a) 0 (b) ± 4 (c) ± 2 (d) ± 3 [2013, online] 30. Statement-1: The only circle having radius and a diameter along line 2x + y = 5 is x2 + y2 – 6x + 2y = 0.

10

Statement-2: 2x + y = 5 is a normal to the circle x2 + y2 – 6x + 2y = 0 (a) Statement-1 is false, Statement 2 is true. (b) Statement-1 is true, Statement 2 is true, Statement-2 is a correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false (d) Statement-1 is true, Statement-2 is true. Statement-2. is not a correct explanation for Statement-1. [2013, online] 31. If a circle of unit radius is divided into two parts by an arc of another circle subtending an angle 60° on the circumference of the first circle, then the radius of the arc is: (b) 1/2 (a) 3 [2013, online] (c) 1 (d) 2

Circles and Systems of Circles 17.39

32. Let C be the circle with circle with centre at (1, 1) and radius = 1. If T is the circle centre at (0, y), passing through origin and touching the circle C externally, then radius of T is equal to (a)

3

(c) 1/2

2

(b)

3 2

(d) 1/4

[2014] 2

2

33. If a point (1, 4) lies inside the circle x + y – 6x – 10 y + p = 0 and the circle does not touch or intersect the coordinate axes, then the set of all possible values of p is the interval: (a) (0, 25) (b) (25, 39) (c) (9, 25) (d) (25, 29) [2014, online] 34. The set of all real values of l for which exactly two common tangents can be drawn to the circle x2 + y2 – 4x – 4y + 6 = 0 and x2 + y2 – 10x – 10y + l = 0 is the interval (a) (12, 32) (b) (18,42) (c) (12,24) (d) (18, 48) [2014, online] 35. For the two circles x2 + y2 = 16 and x2 + y2 – 2y = 0 there is/are (a) one pair of common tangents (b) two pairs of common tangents (c) three common tangents (d) no common tangent [2014, online] 36. The equation of the circle described on the chord 3x + y + 5 = 0 of the circle x2 + y2 = 16 as diameter is (a) x2 + y2 + 3x + y – 11 = 0 (b) x2 + y2 + 3x + y + 1 = 0 (c) x2 + y2 + 3x + y – 2 = 0 [2014, online] (d) x2 + y2 + 3x + y – 22 = 0 37. The number of common tangents to the circles x2 + y2 –4x – 6y – 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0 is: (a) 1 (b) 2 (c) 3 (d) 4 38. If incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0 than the equation of the circumcircle of this triangle is:

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

x2 + y2 – 2x – 2y – 2 = 0 x2 + y2 – 2x – 2y – 14 = 0 x2 + y2 – 2x – 2y + 2 = 0 x2 + y2 – 2x – 2y – 7 = 0

[2015, online]

39. If a circle passing through the point (– 1, 0) touches y-axis at (0, 2), then length of the chord of the circle x 5 3 (a) (b) 2 2 (c) 3 (d) 5 [2015, online] 40. If y + 3x = 0 is the equation of a chord of the circle, x2 + y2 – 30x = 0, then the equation of the circle with their chord as diameter is: (a) x2 + y2 + 3x + 9y = 0 (b) x2 + y2 – 3x + 9y = 0 (c) x2 + y2 – 3x – 9y = 0 (d) x2 + y2 + 3x + 9y = 0 [2015, online] 41. The centres of those circles which touch the circle, x2 + y2 – 8x – 8y – 4 = 0, externally and also touch the x-axis, lie on (a) a circle (b) an ellipse which is not a circle (c) a hyperbola (d) a parabola [2016] 42. If one of the diameters of the circle, given by the equation, x2 + y2 – 4x + 6y – 12 = 0, is a chord of a circle S, whose centre is at (–3, 2), then the radius of S is: (b) 5 3 (a) 5 2 (c) 5 (d) 10 [2016] 43. A circle passes through (–2, 4) and touches the y-axis at (0, 2). Which one of the following equations can represent a diameter of this circle? (a) 2x – 3y + 10 = 0 (b) 3x + 4y – 3 = 0 (c) 4x + 5y – 6 = 0 (d) 5x + 2y + 4 = 0 [2016, online] 44. Equation of the tangent to the circle, at the point (1, –1), where centre is the point of intersection of the straight lines x – y = 1 and 2x + y = 3 is: (a) x + 4y + 3 = 0 (b) 3x – y – 4 = 0 (c) x – 3y – 4 = 0 (d) 4x + y – 3 = 0 [2016, online]

Previous Years' B-Architecture Entrance Examination Questions x sin a – y cos a a a x +y (a) a Œ [0, p] (b) a Œ [– p, p] È p p˘ (c) a can have any value (d) a Œ Í- , ˙ Î 2 2˚

p x – 3y [2006]

x + y – 4x + 6y (b) x2 + y2 – 4x – 6y – 3 = 0

x + 3y

17.40

Complete Mathematics—JEE Main

(c) x2 + y2 – 4x – 6y – 13 = 0 [2006] (d) x2 + y2 – 4x + 6y – 3 = 0 3. The circle passing through the distinct points (1, t), (t, 1) and (t, t) for all values if t passes through the point (a) (1, – 1) (b) (– 1, 1) (c) (–1, – 1) (d) (1, 1) [2006] 2 2 4. The value of k for which the circle x + y – 4x + 6 y + 3 = 0 will bisect the circumference of the circle x2 + y2 + 6x – 4y + k = 0 is (a) 53 (b) – 53 (c) 47 (d) – 47 [2007] 2 5. If the point (2, k) lies outside the circles x + y2 = 13 and x2 + y2 + x – 2y – 14 = 0, then (a) K Œ ] – •, – 2 [U] 3, • [ (b) K Œ ] – 3, – 2 [U] 3, 4[ (c) K Œ ] – 3, – 4[ (d) K Œ ] – •, – 3 [U] 4, • [ [2007] 2 6. The shortest distance between the circles x + y2 = 1 and (x – 9)2 + (y – 12)2 = 4 is (a) 9 (b) 11 (c) 12 (d) 14 [2008]

Statement-1: If L1 is a chord of the circle C, then the line L2 is not always a diameter of the circle C. Statement-2: If L1 is a diameter of the circle C, then the line L2 is not a chord of the circle C. then (a) both the statements are true (b) both the statements are false (c) Statement-1 is true and statement-2 is false (d) statement-2 is true and statement-1 is false [2013] 12. A circle has two of its diameters along the lines x + y = 5 and x – y = 1 and has area 9p, then equation of the circle is (a) x2 + y2 – 6x – 4y – 3 = 0 (b) x2 + y2 – 6x – 4y – 4 = 0 (c) x2 + y2 – 6x – 4y + 3 = 0 [2014] (d) x2 + y2 – 6x – 4y + 4 = 0 13. The number of interger values of k for which the equation x2 + y2 + (k – 1)x – ky + 5 = 0 represents a circle whose radius can not excced 3, is: (a) 10 (b) 11 (c) 4 (d) 5 [2015]

7. The midpoint of the chord intercepted by the circle x2 + y2 = 16 on the line through the point (1, – 2) and (0, – 1) is Ê 1 1ˆ Ê1 3 ˆ (b) Ë , - ¯ (a) Ë- , - ¯ 2 2 2 2 Ê3 1 ˆ (d) Ë , ¯ 4 4

Ê 1 1ˆ (c) Ë- , - ¯ 4 4 2

ax + by +y 4y -4 3

-5 3 1 4

-1 3

2

9. The equation of a circle of area 22p square units for which each of the two lines 2x + y = 2 and x – y = – 5 is diameter, is: (a) x2 + y2 – 2x + 8y – 5 = 0 (b) x2 + y2 – 2x – 8y – 5 = 0 (c) x2 + y2 + 2x – 8y – 5 = 0 [2011] (d) x2 + y2 + 2x + 8y – 5 = 0

x x +y –

a+b

[2009]

8. The circle x + y – 6x – 10y + p = 0 does not touch or intersect the axes and the point (1, 4) lies inside the circle for all p in the interval (a) (25, 35) (b) (25, 29) (c) (0, 25) (d) (0, 29) [2010]

aπ

x

[2015]

Answers Concept-based 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Level 1

10. If a chord of a circle x2 + y2 = 4 with one extremity at (1, 3 ) subtends a right angle at the centre of this circle, then the coordinates of the other extremity of this chord can be: (a) (– 1 3 ) (b) (1, – 3 ) (d) ( 3 , – 1) [2012] (c) (– 3 , –1)

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

11. Consider L1 : 3x + y + a – 2 = 0 and L2 : 3 x + y – a + 3 = 0, where a is a positive real number, and C : x2 + y2 – 2x + 4y – 4 = 0.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

Circles and Systems of Circles 17.41

53.

54.

55.

56.

57.

58.

59.

60.

61.

62.

63.

64.

65.

66.

67.

68.

69.

70.

71.

72.

73.

74.

75.

76.

77.

78.

79.

80.

2. Given equation represents a circle if b = 3 and the equation is x2 + y2 + 4x – 6y + 3 = 0 whose radius =

10 .

3. Line meets the axes at (4, 0) and (0, 3). Equation of the circle on the line joining this points as a diameter is (x – 0) (x – 4) + (y – 0) (y – 3) = 0. fi x2 + y2 – 4x – 3y = 0.

Level 2

4. (1, 2) lies inside the circle

81.

82.

83.

84.

85.

86.

87.

88.

89.

90.

91.

92.

93.

94.

95.

96.

fi (3)2 + (4)2 – 7(3) + 15(4) – c > 0 fi c < 64.

97.

98.

99.

100.

5. Centre of the circle S is (– 3, 7) and its radius is 8

101.

102.

103.

104.

105.

106.

107.

108.

109.

fi 1 + (2)2 – 7 + 15(2) – c < 0 fi c < 28 (3, 4) lies outside the circle

distance of (– 3, 7) from the line is =

110.

Previous Years' AIEEE/JEE Main Questions 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

49 + 9

58 < 8

So the line L is a chord of S. 6. Let y = mx be a tangent through the origin fi

m(7) - (- 1) 1+ m

2

= 25 fi m =

3 -4 or 4 3

Product of the slopes = – 1 fi The required angle is p . 2 7. From Geometry PA . PB = (PC)2 = (5)2 + (6)2 – 12 = 49. C O P (5, 6)

Previous Years' B-Architecture Entrance Examination Questions

-21 + 21 + 58

A

B

Fig. 17.40

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

Hints and Solutions Concept-based 1. Centre of the circle is (1, 3) or (– 1, – 3) and radius is 10 .

8. Equation of the tangent is x – 2y = 5, which meets the second circle at points for which (2y + 5)2 + y2 – 8(2y + 5) + 6y + 20 = 0 fi 5y2 + 10y + 5 = 0 fi y = – 1 and x = 3 so the required point of contact is (3, – 1) 9. Two circles are x2 + y2 = 25 and (x – 2)2 + (y + 3)2 = 25. So the equation of the required chord is (x – 2)2 + (y + 3)2 – (x2 + y2) = 0. fi 4x – 6y – 13 = 0 10. Centres will lie on the line through (0, k) perpendicular to y-axis (x = 0) required line y = k.

17.42

Complete Mathematics—JEE Main Y

=

1 (2 2 – 2) = 2

2 –1

So its equation is x2 + y2 = ( 2 – 1)2 19. If (h, k) is any point on the locus,

X

O

h 2 + k 2 - 4k + 3

Fig. 17.41

2

2

=

h + k + 6k + 5

11. Let the centre be (h, h – 1) then the equation of the circle is (x – h)2 + (y – h + 1)2 = 9. As it passes through (7, 3), (7 – h)2 + (3 – h + 1)2 = 9. fi h = 4 or 7 and the equation of the circle is x2 + y2 – 8x – 6y + 16 = 0 or x2 + y2 – 14x – 12x + 76 = 0. 12. The line passes through the centre of the circle, so AB is a diameter of the circle and APB being an angle in a semi circle is a right angle. 2

3 2

fi 5h2 + 5k2 + 70k + 33 = 0 Required locus is 5x2 + 5y2 + 70y + 33 = 0 20. Diameters intersect at the point (3, 4) so the centre of the circle is (3, 4) and the distance of the centre 3 + 2(4) + 4 from the line x + 2y + 4 = 0 is = 3. 32 + 42 So the equation of the circle is (x – 3)2 + (y – 4)2 = 9 or x2 + y2 – 6x – 8y + 16 = 0

2

13. Let the equation of the circle be x + y + 2gx + 2gy = 0. As it cuts the given circle orthogonally. 2g (– 2) + 2g (– 3 ) = 10 fi g = – 1. 14. Equation of the common chord is y = x and the end points of which are (0, 0) and (a/2, a/2). Equation of the circle on this chord as diameter is x(x – a/2) + y(y – a/2) = 0.

Level 1 x–a

a +b y

xy

x

2x2 + 2y2 – ax – ay = 0.

+ (y – b x

l(x + y

y + 4x – 7y

x – 3y

15. Centre will lie on the line parallel to both the given lines and equidistant from both. 16. Centre of the first circle is (2, 3) and radius 4-centre of the second circle is (– 1, – 1) and radius, distance between the centre = 5 = 4 + 1. So they touch externally and hence 3 common tangents. 17. r1 – r2 > C1 C2 where r1 = b, r2 = a. C1 (0, 0) , C2 (2 , 3) fib–a>

2

2

(0 - 2 ) + (0 - 3) = 13

cos q

from

1 + tan 2 q

of the ± 2

fi cos q

3+5 x+3 = 2 x, =y 2 2

18. Equation of the given circle can be (x ± 1)2 + (y ± 1)2 = 1

r r x+y

r (-1,1)

A2

(-1,-1) A 3

A1

(1,1)

A4 (1,-1)

Fig. 17.42

Centre of the required circle is the origin and radius 1 (A1 A3 – (1 + 1)) = 2

±r ±r –1 2 fi r

±rfi

r fi

r

r

1/ 2

r or r r

1/ 2

±4 ± 16 – 8 , which 4 r cir one

Circles and Systems of Circles 17.43

l , l , l3 2li c cos q (i

li x –x

g, f fy + g + f +

a, x x

x +x

gx + fy + c

gx

c x cos q + y sin q r, r r r ˆ ˆ Ê Ê , B Á 0, P(x, y ,P Ë sin q ˜¯ ÁË cos q sin q ˜¯

ˆ Ê r ,0 A Á Ë cos q ˜¯ fi

1 x

2

+

1 y

h

2

1

=

r2

h, k Ê p + h k + qˆ centre of the circle is Á , ˜ Ë 2 2 ¯ k+q x 2 2

k

h, k x +y

hx

ky + h + k L, S

P L

x 2 + y2 – a2

S

2

2

x +y –b P on x y

1+ m

fi x +y

b a

2

a +b

mx + c

– 3m + c

x

m+c

2

1+ m

fi

2

a2 + a2 – x 2

c/m

2a2

P

x fix

P C A

D B

3 2 x 4

a

52. Let A(0, 0), B(a, 0) and P(h, k), then h2 + k 2

x + y – ax – ay x+y a

a a

AP

+ sin a

p

( h - a )2 + k 2

a

x +y +

a c

d

A (x , y a

b

l a

C :x +y

ax

OA, OB m+2 1 – 2m

fi

2 ap2 1 – p2

x–

a 2 p2 1 – p2

=0

B(x , y

x +y –a,l A B

x +y x +y –a y

by + a + b ax by – a – b a +b p

a

xx +yy

x f

f

f1 f2 ¥ a a fi

ff

a y

3m – 8m f1 – c

mx + c

a +f

1 + m2 B

2

P is

x

on x P(– c/m

2

p + hˆ Ê k + q ˆ Êk + q ˆ + q– = so Ê p – Ë ¯ Ë 2 2 ¯ Ë 2 ¯

AM fi

mf

cf f

f

m a –c f

two

Complete Mathematics—JEE Main

17.44

l (f – k

(1 + m 2 ) a 2 - c 2

f f

a

m2

fi

c

m

kl + g

OA

a

fi

l1 f l2 f ¥ g g

y

a B

A a

line y

a

y q

C

a

a

Fig. 17.44

g2 2

f –k

=–

g2 f

fi

2

PA

PC

QB

3x + 4 y + 1 9 + 16 QC

fi

PC QC

a y

B

is A

(1, 0) A

AB

q + (y

(x

q

C

x +y l

l+c

Gg l

G Fk

Gx

Fy +

Gg

c l+a

fi

F

AC

1

( x - 1) 3 which intersect the circle x + y fi x x fix x y

Fig. 17.43

+ (y

x – 3y x + 5y 25 + 1 < 9, the

c

x

Gh

fi

x

c,

is O c > OA

fi

f2

1/ 2 1 = 2 2 2

so cosq

a (1 + sin q ) cos q

QB

k

from the centre on the line is 1 / 2

q

a (1 - sin q ) cos q

PA

B

L

f2

C fi

A

g2

fil l PC

O (0, 0) r

OB

a–c k

Ê 1 3ˆ Ê 1 3ˆ B Á– , , CÁ- , ˜ 2 ˜¯ Ë 2 2 ¯ Ë 2

x

P(h, k 3 P

P to two (x

+ (y

h 2 + k 2 + 6h + 2 k + 7

3 = 2 2 h + k + 2h + 6 k + 4 6

fi

h +k

h

x l AB

k

(3a – a

x a fi (3a 2 1 lim fi x = , y = a Æ1 5 25

6

A

x

a

aπ

– l1 f ˆ g Ê ÁË – 1 + l , 1 + l ˜¯ 1 1

– l2 f ˆ g Ê , B Á– where l , l Ë 1 + l2 1 + l2 ˜¯

x y fi x – 3y

x

y

4x – 9(8x – x fi x x fi (x x x

x

Circles and Systems of Circles 17.45

fiy

r

3 x

(x

y

y

3

r r

3

2p 9

(x AB

cos2 q + sin 2 q +

r

Level 2

Fig. 17.45

81. x2 + y2 + ax = 0 passes through the centre (0, 0) of the circle x2 + y2 = c2. Circles will touch each other if the radius a/2 of the first circle is equal to half the radius of the 2nd circle

1 + 4 cos q + cos 2q 2

fi c = ± a fi c2 – a2 = 0

(1/ 2) (3 + 4 cos q + 2 cos2 q - 1)

82. Since each side subtends an angle of 120° at the centre (0, 0) of the circle.

(1 + cos q )2 fi q, cos q

(cos q, sin q

B

1 (cos q - sin q )2 + (sin q - cos q )2 2

x

y – 6x

l (4x

y

y

x l

S

So cos q 1 + cosq 2 + cosq 3 = cosq 1 + cos (120° + q 1) + cos (240° + q 1)

47 / 3 2 x cos q + y sin q a, x cos (p q y sin (p a fi (x cos q + y sin q + (– x sin q + y cos q fi x +y a fi

C

Fig. 17.47

q 2 = 120° + q 1, q 3 = 240° + q 1

x – 36y

of the circle C

C

y+ l

S

C C

A (a cos q1, a sin q1)

120° q1

fil 6(x + y

a

Fig. 17.46

fi

r

a

(0, 0)

fi AL AOL

AL

+ (y

= cosq 1 + 2 cos (180° + q 1) cos (–120°) q

=cosq 1 + 2 ¥ cosq 1 (– 1/2) = 0 Similarly sinq 1 + sinq 2 + sinq 3 = 0

a

83. Equation of the circle C : x (x – 4) + y2 = 0 Equation of the common chord of C1 and C2 is 8x – 11 = 0 which meets the circle C at points for

C

which x = 11/8 and y2 =

4 ¥ 11 Ê 11ˆ 2 231 -Á ˜ = Ë 8¯ 8 64

Hence the required intercept 231 231 = 64 4 84. We have a = b and a2 + ab – 16 = 0 fia=b= ± = 2¥

from x 4–3

afia

2 x

1

2

Equation of the circle is x2 + y2 + 2x + 2y ± 2 = 0 r

Radius of the circle is

(-1) 2 + (-1) 2 ± 2 = 0 or 2.

17.46

Complete Mathematics—JEE Main

85. Circle passing through the points (–3, 0), (–1, 0) and (0, 3/2) is x2 + y2 + 4x – (7/2)y + 3 = 0

91. Let the vertices of the triangle be Ai (xi, yi), i = 1, 2, 3. If (x, y) is any point on the 3

which passes through the point (0, 1/l) if 1 7 + 3 = 0 fi 6l2 – 7l + 2 = 0 2 2l l

locus then

 ÈÎ( x - xi )2 + ( y - yi )2 ˘˚

= c2, c being

i =1

constant.

(

3

)

3

fi 3 x 2 + y 2 - 2 x  xi - 2 y  yi +

86. 2a2 = (2r)2

i =1

where r 2 = (–2)2 + (–5)2 – 21

coordinates (x, y) of R which divides the join of A and P in the ratio 3:2 are 6 cosq - 2 6 sinq + 2 y= x= 5 5 2 Locus of R is (5x + 2) + (5y – 2)2 = 36 which is a circle with centre (–2/5, 2/5) 88. Equation of the circle is (x – a) (x – b ) + ( y – a ¢ ) (y – b ¢ ) = 0 fi x2 + y2 – (a + b)x – (a ¢ + b ¢)y + ab + a ¢ b ¢ = 0 22 22 8 15 fi x2 + y2 – + =0 xy+ 15 8 15 8 fi 120 (x2 + y2) –22 (8x + 15y) + 289 = 0 89. Let C1 : x2 + y2 +2g1x + 2f1y + c1 = 0

3

 xi2 +  yi2 - c2 = 0

fi a2 = 2 ¥ 8 = 16 fia=4 87. A (–1, 1), P(2cos q, 2sin q)

i =1

3 i =1

i =1

Which represents a circle with centre at 3 ˆ Ê 3 x yi ˜ Â Â i Á Á i = 1 , i = 1 ˜ , the centroid of the triangle A A A 1 2 3 Á 3 3 ˜ ˜ Á ¯ Ë

92. Equation of a tangent is y = ± 3x ± 4 3 1 + 3 which meets the axis of x at x = ± 8 Length of the tangent from (± 8, 0) of the circle x2 + y2 = 48 is

(± 8) 2 - 48 =

16 = 4

which is the required length of the side of the rhombus. 93. Equation of any circle on this chord is

C2: x2 + y2 + 2g2x + 2f2y + c2 = 0

x2 + y2 – a2 + l (x cosa + y sina – p) = 0

C : x2 + y2 + 2gx + 2 fy + c = 0

Chord is a diameter of the circle if the centre (– l cosa /2, – l sina /2) lies on the chord.

If C1 and C2 cut C orthogonally. then 2gg1 + 2 ff1 = c + c1 2 gg2 + 2 ff2 = c + c2 fi 2(g1 – g2)g + 2 ( f1 – f2) f = c1 – c2  I Radical axis of C1 and C2 is 2(g1 – g2)x + 2( f1 – f2)y + c1 – c2 = 0 which passes through (–g, –f ), the centre of C (using I) 90. Let the equation of the circle be x2 + y2 + 2gx + 2fy = 0 s.t

g + f

= 3k

centroid of the triangle OAB is - 2g - 2 f ˆ (x, y) = ÊÁ , ˜ Ë 3 3 ¯ g2 + f 2 = 9k2 fi 9(x2 + y2) = 4 ¥ 9k2 fi x2 + y2 = 4k2

fi l = – 2p 94. (ax + by + c)2 + (bx – ay + d)2 = r2 (a2 + b2) fi (a2 + b2) (x2 + y2) + 2(ac + bd)x + 2(bc – ad) y + c2 + d2 = r2 (a2 + b2) which is a circle with radius r 95. Vertices of the square be (± a/2, ± a/2) Equation of the inscribed circle is x2 + y2 = a2/4

(1)

A4

A3 (a/2, a/2)

A1

A2

Fig. 17.48

Circles and Systems of Circles 17.47

Equation of the circumscribed circle is 2

2

2

x + y = a /2

(2)

x2 + y2 + 2gx + 2fy + c = 0

If P(x, y) is a point on (1), then

then 2g(2) + 2f (–3) = c + 9

4

= [x ± (a/2)]2 + [ y ± (a/2)]2

 ( PAi )2

and 2g(–2) + 2 f (3) = 4 + c

i =1

2

2

Êa +a ˆ = 4(x2 + y2) + 4 Á = a2 + 2a2 = 3a2 Ë 4 ˜¯

fi 8g – 12 f = 5 So locus of the centre (– g, – f ) is 8x –12y + 5 = 0

4

99. Let the equation of the circle be

= 2a2 + 2a2 = 4a2

 (QAi )2

Similarly

x2 + y2 – 2gx – 2fy = 0

i =1

96. Equation of a tangent from (–2,0) to the circle x2 + y2 = 1 is

98. Let the equation of the circle cutting the given circle orthogonally be

then (g – 5)2 + ( f – 2)2 = g2 + f 2

3y = x + 2 M

(-2, 0)

O

(1, 0)

(0, 0)

r

C (g, f ) r

C (r + 1, 0)

T (5, 2)

Fig. 17.50

fi 10g + 4f – 29 = 0

Fig. 17.49

Let the centre of the required circle be C (r + 1, 0) and radius r. r +1+ 2 - 0 =r Then CM = 1+ 3 fi r = 3 and the equation of the circle is (x – 4)2 + y2 = (3)2 fi x2 + y2 – 8x + 7 = 0

fi x[(

fi

) (cosq + sinq)] = 0

Taking the –ve sign, we get hb + ak = 2ab

x

=

cos q - ( 2 - 1) sin q

y Solving, h =

( 2 - 1) cos q + sin q

x2 + y 2

( 2 - 1) + 1 then from x cosq + y sinq = a, we get + k sinq [(

101. Let the equation of the circle be x2 + y2 +2gx + 2fy + c = 0 As it cuts the circle x2 + y2 = a2 orthogonally

– 1) sinq ]

– 1) cosq + sinq] = a

c = a2 and since it passes through the given point

fik=a 2

2

fi x + y = [(

ab a b ,k= a +b a +b

Note + ve sign gives (h, k) = (0, 0)

2

fi k cosq [cosq – (

(2)

then OP is perpendicular to AB and AB is equidistant from O and P. k a - ab hb + ak - ab = and = ± So 2 2 b h a +b a +b

) (cosq – sinq)]

– 1) cos q + sin q] – [cos q –

=k=

so 2g + 3f – 16 = 0

If P(h, k) is the required point of reflection

– 1) sinq ] y = 0

(

tangent 3x – 2y – 11 = 0

100. Equation of AB is x/a + y/b = 1 or bx + ay = ab

x cosq + y sinq = a and x cos ( p/4 + q) + y sin(p /4 + q ) = a

+ y [sinq – (1/

C ( g, f ) lies on the line through T (5, 2) perpendicular to the

23 51 Solving (1) and (2) we get g = ,f= 22 11

97. Let the equations of the two tangents be

fi x [cosq – (1/

(1)

2

2

– 1) + 1] a = 2(2 –

)a

2

Ê a q ˆ Ê a p ˆ ga p fa q ÁË p + q ˜¯ + ÁË p + q ˜¯ + p + q + p + q + a =0

17.48

Complete Mathematics—JEE Main

fi a2 + 2gp + 2fq + p2 + q2 = 0

105. Let the equation of the larger circle be x2 + y2 = a2 and the line x + y = 2 meet the smaller circle at A and B, the larger circle at C and D.

fi p2 + q2 + 2 pg + 2qf + a2 = 0 Showing that the circle passes through ( p, q).

D

102. Let the other extremities of the diameter through P (2, 3) be A (a1, b1), B (a2, b2) then equations of the two circles are (x – a1) (x – 2) + ( y – b1) ( y – 3) = 0 and (x – a2) (x – 2) + (y – b2) ( y – 3) = 0 So equation of the common chord is

B

B

L O

A

C

Fig. 17.52

(x – 2) (a2 – a1) + ( y – 3) (b2 – b1) = 0 Since AB makes an angle p /6 with x–axis

fi OA = OB = 2 and OC = OD = a

b2 - b1 1 = a2 - a1 3

Let OL be perpendicular to AB 2 then OL = = , LD = LB + BD 1+1

103. Equation of the circle passing through (–1, 0) and (1, 0) is x2 + y2 + 2fy – 1 = 0 ( f being a parameter) Let (h, k) be the point of contact of the tangent parallel to y = x Equation of the tangent at (h, k) is

= (1 / 2) 4 + 4 + 1 fi a2 = (OD)2 = (OL)2 + (LD)2 = 2 + = 5+2 2.

(

)

2 +1

2

106. Centre of the required circle is the point of reflection of the centre (1, 0) in the line x + y = 2 and its radius is equal to the radius 1 of the given circle. So its equation is (x – 2)2 +( y –1)2 = 1

hx + (k + f ) y + f k – 1 = 0 -h so that = 1 fi f = – (h + k) k+ f

107. If the chord AB makes an angle of 30° in the major segment, it makes an angle of 60° at the centre so that the length of the chord is equal to the radius of the circle. Let OL be perpendicular to AB.

since (h, k) lies on the circle h2 + k2 + 2fk – 1 = 0 fi h2 + k2 – 2k (h + k) –1 = 0 fi h2 – k2 – 2hk – 1 = 0 Locus of (h, k) is x2 – y2 – 2xy = 1

O

104. Line y = x meets the y=

60°

x A C r B (1, -3)

3x - y = 6

Fig. 17.51

line 3x – y = 6 at A(3, 3), B be (1, –3), C the centre of the circle, and CAB = q /2 Then the radius r = A B tan q/2 = Also tan q = fir=

40 tan (q /2)

3 -1 1 = fi tan (q /2) = – 2+ 5 1+ 3 2

40 (- 2 + 5 ) = 1.6.

B

y-x=3

Fig. 17.53

q/2 A (3, 3)

L

3 ˆ2 9 then (OL)2 = ÊÁ = Ë 2 ˜¯ 2 2 2 2 (AL) = (OA) – (OL) = k2 – (9/2) k2 = (AB)2 = 4 (AL)2 = 4(k2 – (9/2)) fi k2 = 6 108. Let the radii of the three circles be 1, 1 – d, 1 – 2d Line y = x + 1 will cut all the three circles in real distinct points if the distance of the centre (0, 0)

Circles and Systems of Circles 17.49

from the line is less than the radius of the circle with smallest radius. 1 2

< 1 - 2d fi d
0]

1 is a tangent to the parabola y2 = 4x m

| m(0 - 3 / 2) - 0 | 1 + m2

=

5 2

2

which will touch the parabola x = –32y if the root fi (3m)2 = 5(1 + m2) fi m = ±

5 . 2

Parabola

\ Required slope is

(

5 2

S = CP2 = t 4 + 2 2t + 6 ds = 4t 3 + 4 2 2 2t + 6 dt

(

y

5/2

(3/2, 0)

O

18.35

)

2

)

= 4 ÈÎt 3 + 4t + 6 2 ˘˚

x

y

Fig. 18.26

x

O

19. Let coordinates of Q be (x¢, y¢) and that of P be (x, y), then

P(t2, 2 2/t) A -6 C

1 1 x = x¢, y = y¢ 4 4

Fig. 18.27

As (x¢, y¢) lies on x2 = 8y, (4x)2 = 8(4y) fi x2 = 2y

For least value of S,

20. Let coordinates of P be (–t2, 2t), where t > 0, then coordinates of Q are (–t2, – 2t). If coordinates of R are (x, y), then x=

fi t3 + 4t + 6 2 = 0 t 3 + 2t 2 - 2t 2 - 2t + 6t + 6 2 = 0

(

-t 2 - 2t 2 = –t2 2 +1

Also,

2

fi

or 9y2 = –4x

y – 6 = x2 at (2, 10) is

(x – (–4)) + 4(y – 1) = 0 (2)

The point (a, b) is the point of intersection of (1) and (2). Solving (1) and (2), we get 8 2 ,b= . 17 17

d 2s = 4[3t2 + 4] 2 dt

d 2s ˘ ˙ dt 2 ˚t = -

>0 2

An equation of required circle is (1)

An equation of normal at the point of contact (a, b) of (1) and x2 + y2 + 8x – 2y = k is

a= -

[Q t is real]

fi S is least when t = - 2 . Thus, coordinates of P are (2, –4).

21. An equation of tangent to

or x + 4y = 0

)

fit= - 2

Eliminating t, we get

(y + 10) – 12 = 2(2)x or 4x – y + 2 = 0

)(

fi t + 2 t 2 - 2t + 6 = 0

2 1(2t ) + 2(-2t ) and y = = - t 3 2 +1

Ê 3 ˆ –x = Á - y˜ Ë 2 ¯

ds =0 dt

(x – 2)2 + (y + 4)2 = (0 – 2)2 + (–6 + 4)2 or x2 + y2 – 4x + 8y + 12 = 0 23. Let coordinates of a point P on y = x2 – 4 is (t, t2 – 4). Let d = distance of P from the origin, then d2 = (t – 0)2 + (t2 – 4 – 0)2 = t2 + t4 + 16 – 8t2 = t4 – 7t2 + 16

18.36

Complete Mathematics—JEE Main 2

7ˆ 15 Ê = Át2 - ˜ + Ë 2¯ 4 Note that d is least if t2 = 7/2. Least value of d is 15 / 2 . 24. An equation of normal at P(t) is y = –tx + 2t + t

3

It passes through Q(t1) i.e. (t12, 2t1) if 2t1 = – tt12 + 2t + t3 fi 2(t1 – t) = t(t2 – t12)

2 t

fi t1 = t +

3. Equation of a tangent to the parabola y2 = 4x is 1 y = mx + which passes through (1, 4) m 1 fi m2 – 4m + 1 = 0 m

fi m1m2 = –1. So the required angle is

4

2

2

Which minimum for t = 1. So statement-1 is also true but does not follow from statement-2.

if 4 = m +

fi 2 = –t(t + t1) fi t1 = – t –

1 1 t - t2 -1 | (t - 1) 2 + 3 | 2 4 = 4 2 2

4ˆ t Á 2 ˜ + 4 = 6 [AM ≥ Ët ¯ 2Ê

+4≥

t2

GM]

p . 2

4. Tangent to the parabola x2 = 4y at (4, 4) and (–4, 4) are x(4) = 2(y + 4) and x(–4) = 2(y + 4) which intersect on x = 0, i.e. y-axis. So statement-1 is true.

Previous Years' B-Architecture Entrance Examination Questions 1. Equation of the side AB of the triangle ABC is 1 2 y= x which meets the parabola y = 8x at the 3

(

point B 24, 8 3

)

Extremities of the latus rectum are (±2, 1) and the tangents at these points are 2x = 2(y + 1) and (–2) x = 2(y + 1). Which again intersect on y-axis, the axis of the parabola and statement-2 is also true but does justify statement-1. 5. Coordinates of L are (4, 8).

B

Equation of the chord through L with slope –1, is x + y = 12 (1)

A

Any point on the parabola is (4t2, 8t) which lies on (1) if C

4t2 + 8t – 12 = 0 fi t = 1, –3.

Fig. 18.28

t = 1 coordinates to L, so other end M of the 2

2

( )

fi (AB) = (24) + 8 3

2

fi AB = 16 3

Ê 1 1ˆ 2. Tangent at Á , ˜ to the parabola y2 = x is Ë 4 2¯ 1Ê 1ˆ 1 Ê 1ˆ yÁ ˜ = Á x + ˜ fi y = x + Ë ¯ Ë 2¯ 2 4 4 Which is parallel to y = x + 1, so statement-2 is true.

chord is (36, – 24) and length of the chord is (36 - 4)2 + (-24 - 8)2 = 32 2 6. Let (h, k) be the mid point of the chord of the parabola x2 = 4 py with slope m then its equation is hx – 2p(y – k) = h2 – 4pk. fim=

Ê1 1 ˆ Any point on the parabola y2 = x is Á t 2 , t ˜ Ë4 2 ¯ whose distance from the line y = x + 1 is

h fi h = 2pm. 2p

Locus of (h, k) is x = 2pm.

Parabola

Which is parallel to y-axis at a distance |2pm| from it. 7. Let coordinates of P be (t2, 2t), then coordinates Ê 1 -2 ˆ of Q are Á 2 , ˜ . Centre of the circle with PQ Ët t ¯ Ê1Ê 1ˆ 1ˆ as diameter is Á Á t 2 + 2 ˜ , t - ˜ . ¯ Ë 2Ë t¯ t As it lies on t–

5y + 4 = 0,

2 2 ˘ Ê 1ˆ ÈÊ 1ˆ = Á t + ˜ ÍÁ t - ˜ + 4˙ Ë t ¯ ÍË t ¯ ˚˙ Î

˘ ÈÊ 1ˆ 2 = ÍÁ t - ˜ + 4˙ ˙˚ ÍÎË t ¯

Ê 1ˆ fi PQ = Á t - ˜ Ë t¯ =

1 -4 = t 5 2

1ˆ Ê 1ˆ Ê Also, PQ2 = Á t 2 - 2 ˜ + 4 Á t + ˜ Ë t¯ Ë t ¯

2

2

16 36 +4= 5 5

2

+4

18.37

CHAPTER NINETEEN

Ellipse

Definition 1:

fi

An ellipse is the locus of a point which moves in a plane such that the sum of its distances from two fixed points is always constant and the constant is greater than the distance between the fixed points.

cx ˆ Ê fi (x – c)2 + y2 = Á a - ˜ Ë a¯

Y

O F2

x2 a2

+

y2 a2 - c2

2

X

4cx 2cx = 2a a

=1

Let a2 – c2 = b2, then locus of P(x, y) is

P(x, y) F1

fi

( x + c) 2 + y 2 - ( x - c) 2 + y 2 =

x2 a2

+

y2 b2

=1

where a > b > 0.

Definition 2:

Fig. 19.1

(i) Two fixed points are called the foci of the ellipse. (ii) Mid-point of the line joining the foci is called the centre of the ellipse. (iii) The line segment through the foci of the ellipse is called the major axis of the ellipse. (iv) The line segment through the centre and perpendicular to the major axis is called the minor axis of the ellipse. (v) The end points of the major axis are called the vertices of the ellipse. (vi) A line segment through a focus perpendicular to the major axis is called a Latus rectum of the ellipse. (vii) Major axis and minor axis are called the principal axis of the ellipse.

Standard Equation of the Ellipse Let F1 (– c, 0) and F2 (c, 0) be the foci of the ellipse. Then O (0, 0), the mid-point of F1 F2 is the centre of the ellipse. Let P (x, y) be any point on the ellipse. Then by definition PF1 + PF2 = 2a (constant) fi

( x + c)2 + y 2 + ( x - c) 2 + y 2 = 2a. (2a > 2c)

Also [(x + c)2 + y2] – [(x – c)2 + y2] = 4cx.

An ellipse is the locus of a point which moves in a plane in such a way that the ratio of its distance from a fixed point in the plane to its distance from a fixed line in the plane (not passing through fixed point) is a constant and equal to e, where 0 < e < 1. (i) Fixed point is called a focus of the ellipse. (ii) Fixed line is called a directrix of the ellipse. (iii) The constant ratio e is called the eccentricity of the ellipse. Standard Equation of an Ellipse referred to its principle axes along the coordinate axes is

x2 a2

+

y2 b2

= 1,

b2 = a2 (1 – e2) Y B (0, - b) L M¢ S¢(-ae, 0) A¢ (-a,0)

M

S(ae, 0) X

A (a,0)

C(0, 0)

B¢(0, - b)

P

L¢

x = - a/e

x = a/e Ellipse (e , - b)

Fig. 19.2

19.2

Complete Mathematics—JEE Main

Let P (x, y) be any point on the ellipse. S = (ae, 0) be a focus and x =

a be a directrix. e

Then according to the definition Êa ˆ (x – ae)2 + y2 = e2 Á - x˜ Ëe ¯ fi

x2 a2

+

y2 a 2 (1 - e2 )

x

+

y

(iv) e =

=1

a 2 - b2

is the eccentricity of the ellipse. a2 From symmetry we observe that if -a S ¢ = (– ae, 0) be taken as a focus and x = is taken as e a directrix, same ellipse is described. So (v) S ¢ (– ae, 0) and S (ae, 0) are the two foci of the ellipse. -a a (vi) x = and x = are the two directrices of the e e ellipse. (vii) x = ae are the two latera recta of the ellipse. (viii) Latus rectum x = ae meets the ellipse at point Ê b2 ˆ 2b 2 ae , ± , so length of each latus rectum is . Á a a ˜¯ Ë

Find the length and equations of the latera recta of the ellipse x2 y 2 =1 + 36 25 x2 y 2 Solution: Let the equation of the ellipse be 2 + 2 = 1, a b

Equations of latera recta are x = fix=

6¥

11 = ± 11 6

( x - 1) 2 ( y + 1)2 + =1 16 9

Which can be written as

25 11 = 36 36 ae

X2 Y2 + =1 a 2 b2

where X = x – 1, Y = y + 1, a = 4, b = 3. centre is X = 0, Y = 0 i.e. (1 , –1) ae, Y = 0 fi x – 1 =

foci are X = fix=1

4

1-

b2 = ± 7 a2

7 ; y = –1

Length of the major axis = 2a = 8 Length of the minor axis = 2b = 6. eccentricity =

1-

7 b2 = 4 a2

Equations of the directrices are X = 4¥4 fix=1 7

fix–1=

a . e

16 7

3

Illustration

1 ,a 2 focus is (2, 3) and a directrix is x = 7. Find the length of the major and minor axes of the ellipse. Solution: Find the equation of the ellipse whose eccentricity is

(x – 2)2 + (y – 3)2 =

1

where a2 = 36, b2 = 25 fi e2 = 1 –

2

Find the centre, foci, the length of the axis, eccentricity and the equation of the directrices of the ellipse. 9x2 + 16y2 – 18x + 32y – 119 = 0 Solution: The equation of the ellipse can be written as 9(x – 1)2 + 16 (y + 1)2 = 144 fi

2

Illustration

25 units. 3

Illustration

=1 a b2 In the figure (i) A¢ A is the major axis of the ellipse along x-axis of length 2a; A(a, 0), A¢ (–a, 0) are the vertices of the ellipse; (ii) BB¢ is the minor axis of the ellipse along y-axis of length 2b. (iii) O(0, 0) is the centre of the ellipse. 2

=

2b 2 2 ¥ 25 = a 6

2

Since e < 1, 1 – e2 > 0 so let a2 (1 – e2) = b2 and the required equation is 2

and length of each latus rectum =

1 (7 – x)2 4

fi 3x2 + 4y2 – 2x – 24y + 3 = 0 which can be written as 2

1ˆ Ê 1 100 3 ÁË x - ˜¯ + 4 (y – 3)2 = 3 ¥ +4¥9–3= 3 9 3 2

1ˆ Ê x- ˜ ËÁ ( y - 3) 2 3¯ or =1 + 100 100 9 12

Ellipse 19.3 So length of the major axis = 2 Length of the minor axis = 2

x y cos q + sin q = 1 7 5 Given equation can be written as

100 20 = . 9 3 100 20 10 = = . 12 2 3 3

x 3 y =1 + 14 10 Comparing (1) and (2) we get

Some Properties and Standard Results for the Ellipse

x2

+

y2

=1

a 2 b2 1. The parametric equations of the ellipse or the coordinates of any point on the ellipse are x = a cos q, y = b sin q. The point is denoted by “q”. q is called the eccentric angle of the point. 2. An equation of the tangent at the above point “q ” is xx ¢ yy ¢ x y cos q + sin q = 1 and at (x¢, y¢) is 2 + 2 = 1. a b a b

Illustration

4

Find an equation of the tangent to the ellipse

x2 y 2 =1 + 81 49

at the point P whose eccentric angle is p coordinates of P. p pˆ Ê Solution: Coordinates of P are Á 9 cos , 7 sin ˜ Ë 6 6¯ Ê 9 3 7ˆ Á 2 , 2˜ Ë ¯

y 3x + = 1 fi 7 3x + 9y = 126. 9¥2 7¥2

3. An equation of the normal at P(q) is ax by = a2 – b2 cos q sin q and at (x¢, y¢) is x - x¢ x ¢/ a

2

cos q =

(2)

3 1 , sin q = fi q = 30° 2 2

Ê 7 3 5ˆ So the coordinates of P are Á 2 , 2 ˜ . Ë ¯ Equation of the normal at P is 7x 5y =1 cos q sin q fi 14x – 10 3 y =

3

4. The condition that the line y = mx + c is a tangent to the ellipse is c2 = a2m2 + b2. So equation of any tangent to the ellipse (not parallel to y-axis) can be written as y = mx a 2 m2 + b2 Illustration

=

An equation of the tangent at P to the ellipse is x p y p cos + sin = 1 9 6 7 6 fi

(1)

6

If y = mx + 5 is a tangent to the ellipse 4x2 + 25y2 = 100, then m2. Solution: Equation of the ellipse is x2 y 2 =1 + 25 4 If y = mx + 5 touches the ellipse. Then (5)2 = 25 m2 + 4 fi 25m2 = 21 fi 100 m2 = 84.

5. An equation of the chord of contact of the point (x¢, y¢) joining the points of contact of the tangents drawn from (x¢, y¢) to the ellipse is xx ¢ a

2

+

yy ¢ b2

=1

y - y¢ =

Illustration

y ¢/ b 2

5

x2 y 2 = 1 at + 49 25 the point P. Find the coordinates of P and the equation of the normal at P. Solution: Let the coordinates of P be (7 cos q, 5 sin q) Equation of the tangent at P is 5 3x + 7y = 70 is a tangent to the ellipse

Note This equation is same as the tangent at (x¢, y¢) to the ellipse. When P (x¢, y¢) lies on the ellipse the two points of contact coincide with P and the tangent at P and the chord of contact of P are same).

6. An equation of the chord of the ellipse whose midpoint is (x¢, y¢) is T = S¢, where T

xx ¢ a2

+

yy ¢ b2

– 1 and S¢

x¢2 a2

+

y ¢2 b2

= 1.

19.4

Complete Mathematics—JEE Main

7. An equation of pair of tangents from a point P(x¢, y¢) outside the ellipse to the ellipse is SS¢ = T2, x2 y 2 + – 1. S a 2 b2 Illustration

7

Find the equation of the chord of contact of the point (3, 1) to the ellipse x2 + 9y2 = 9. Also find the mid-point of this chord of contact. Solution: Equation of the ellipse is x2 y 2 =1 + 9 1 Equation of the chord of contact of (3, 1) is x(3) y (1) = 1 fi x + 3y = 3 (1) + 9 1 If (h, k) is the mid-point of this chord, then its equation is hx ky h2 k 2 – 1 –1= + + 9 1 9 1 From (1) and (2) we get

(2)

8

Find the equation of the ellipse whose auxiliary circle is the x2 y 2 director circle of the ellipse = 1 and the length of a + 36 13 latus rectum is 2 units. Solution: Equation of the director circle of the ellipse is x2 + y2 = 36 + 13 = 49. x2 y 2 So if the required equation of the ellipse is 2 + 2 = 1, a b 2 2 2 2 then auxiliary circle is x + y = a = 49 fi a = 49 fi a = 7. Also length of a latus rectum =

2b 2 =2 a

fi b2 = a = 7 and the required equation is x2 y 2 = 1. + 49 7

10. A diameter of an ellipse is the locus of the mid points of a system of parallel chords of the ellipse and its equation is

h 9k h 2 + 9k 2 = = 1 3 3 fih=

Illustration

y= -

3 1 ,k= . 2 2

Ê 3 1ˆ So the mid point of the chord of contact is Á , ˜ . Ë 2 2¯

8. Director circle of the ellipse is the locus of the point of intersection of the tangents to the ellipse which intersect at right angles and its equation is x 2 + y 2 = a 2 + b 2. So a pair of tangent draw from any point on the director circle to the ellipse are at right angles. 9. Auxiliary circle of the ellipse is the circle on the major-axis as a diameter and its equation is x2 + y2 = a2. If P is a point on the ellipse and Q is a point on the auxiliary circle such that Q lies on the ordinate produced of the point P. If C is the centre of the ellipse and CA is the semi-major axis of the ellipse then ACQ = q is called the eccentric angle of the point P and the coordinate of P are (a cos q, b sin q). Also PQ = (a – b) sin q. Q P C q (0, 0)

Fig. 19.3

b2 a2m

where m is the slope of the parallel chords of the ellipse which are bisected by it. This is a line through the centre of the ellipse. Two diameters of an ellipse are said to be conjugate when each bisects that chords parallel to the others. Thus two diameters y = mx and b2 y = m¢x of the ellipse are conjugate if mm¢ = – 2 . a Illustration

9

x2 y 2 =1 + 36 16 bisecting the chords parallel to the line y = x – 1. Also find the equation of the conjugate diameter bisecting the chords parallel of this diameter. Solution: Slope of the given line is m = 1. So equation of the diameter bisecting the chords with slope 1 16 x fi 4x + 9y = 0. Slope of the conjugate diameter is y = 36 Find the equation of the diameters of the ellipse

is m¢ such that – A (a, 0)

x,

16 16 m¢ = – fi m¢ = 1 and the required 36 36

equation of the conjugate diameter is y = x.

11. The tangent and normal at any point of an ellipse bisect the angle between the focal radii to that point. 12. The locus of the feet of the perpendiculars from the foci on any tangent to an ellipse is the auxiliary circle.

Ellipse 19.5

13. If W, W¢ are the feet of the perpendiculars from the foci S and S¢ respectively on the tangent at any point P of an ellipse with centre C, then CW is parallel to S¢ P and CW¢ is parallel to SP.

14. If the normal at any point P of the ellipse x2/a2 + y2/ b2 = 1 meet the major and minor axes in G and g respectively and CF is perpendicular from the centre C on the normal then PF . PG = b2 and PF. Pg = a2

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Example 1: Equation of a directrix of the ellipse x y2 + = 1 is 36 4 (a) 9x – 8 = 0 (b) 8x – 9 = 0 2

(c) Ans. (c)

2x+9=0

(d) x + 9 2 = 0

(d)

x2 y 2 + =1 29 25

2 2

x2 a2

+

y2 b2

2 5

4ˆ Ê b2 = a2 (1 – e2) = 25 ÁË1 - ˜¯ = 21 25

9

=

2

Required equation of the ellipse is

9=0

(d)

25 9

Ans. (c) Solution: Let the equation of the ellipse be 1 where e =

x2 y 2 + =1 25 16

ae = 2 fi e =

6¥3

25 3

(c)

where 2a = 10 fi a = 5

Example 2: The locus of a point whose distance from the point (3, 0) is 3/5 times its distance from the line x = p is an ellipse with centre at the origin. The value of p is (a) 5 (b) 7 (c)

x2 y 2 + =1 25 4

Solution: Let the equation of the ellipse be

Equation of the directrices are

2x

(b)

=1

36 - 4 32 8 2 2 = = fie= e = 3 36 36 9 2

a = e

x2 y 2 + =1 25 21

Ans. (a)

Solution: Let a2 = 36, b2 = 4 then if e is the eccentricity of the ellipse

x=

(a)

x2 a2

+

y2 b2

Example 4: If F1 = (3, 0), F2 = (– 3, 0) and P is any point on the curve 16x2 + 25y2 = 400, then P F1 + P F2 equals (a) 8 (b) 6 (c) 10 (d) 12 Ans. (c) Solution: Equation of the curve is

=

3 is the eccentricity Focus is (3, 0) = (ae, 0) 5

fia=5 Equation of the directrix is x =

x2 y 2 + =1 25 21

a 5¥5 25 = = which 3 e 3

25 is same as x = p fi p = . 3 Example 3: An equation of the ellipse whose length of the major axis is 10 and foci are ( 2, 0) is

Here

a2 = 25, b2 = 16, e2 =

x2 y 2 + =1 25 16

25 - 16 9 = 25 25

3 ˆ Ê Foci are Á ± 5 ¥ , 0˜ = ( 3, 0) Ë 5 ¯ fi So

F1, F2 are the foci of the ellipse. P F1 + P F2 = e ¥ distance between the directrices = e¥

2a = 2a = 10 e

19.6

Complete Mathematics—JEE Main

Example 5: Equation of a common tangent to the circle x2 y 2 x2 + y2 = 16, parabola x2 = y – 4 and the ellipse = + 25 16 0 is (a) x = 4 (b) x = – 4 (c) y = 4 (d) y = 5 Ans. (c) Solution: From the figure we find y = 4 is a common tangent to the three curves.

Solution: If 2a is the length of the major axis and e is the eccentricity of the ellipse then F1 F2 = 2ae fi 26 = 2ae Also OF1 + OF2 = 2a, O being the origin. (As O lies on the ellipse) fi 25 + 29 = 2a

(2)

From (1) and (2) e=

Parabola y=4

(1)

26 5 + 29

x2 y 2 x2 y 2 + 2 = 1 and E2: 2 + = 1 are 81 b 49 b two ellipses having the same eccentricity if b2 is equal to (a) 32 (b) 63 (c) 65 (d) 64 Ans. (b) Example 8: E1:

Circle

Ellipse

Fig. 19.4

Example 6: If y = x + c is a normal to the ellipse x2 y 2 = 1, then c2 is equal to + 9 4 13 (b) (a) 25 25 (c) (d) 9 Ans. (b)

fi 25 13 13 4

Example 9: If the normal at any point P on the ellipse x y2 = 1 meets the major axis at G1 and the minor axis + 64 36 at G2 then the ratio of PG1 and PG2 is equal to (a) 6 : 8 (b) 8 : 6 (c) 9 : 16 (d) 16 : 9 Ans. (c) 2

Solution: Any point on the ellipse is P (3 cos q, 2 sin q) equation of the normal at P is 3.x sec q – 2.y cosec q = 9 – 4 Comparing it with y = x + c, we get

Solution: Equation of the normal at P(8 cos q, 6 sin q) is 8x sec q – 6y cosec q = 64 – 36 = 28. Ê -28 ˆ Ê 28 ˆ So G1 is Á cos q , 0˜ and G2 is Á 0, sin q ˜ Ë ¯ Ë 8 ¯ 6

1 1 -c = = 3secq 2cosecq 5

fi

-3c -2c , sin q = . 5 5 9c2 + 4c2 = 25 as cos2 q + sin2 q = 1

fi

c2 =

fi

b 2 - 49 81 - b 2 = b2 81 b4 = 49 ¥ 81 fi b2 = 63.

Solution: We have

cos q =

2

28 ˆ Ê (PG1)2 = Á 8 - ˜ cos2 q + 36 sin2 q Ë 8¯ =

25 13

36 (36 cos2 q + 64 sin2 q) 64 2

Example 7: If F1 (– 3, 4) and F2 (2, 5) are the foci of an ellipse passing through the origin, then the eccentricity of the ellipse is (a)

(c) Ans. (c)

29 5 + 26 26 5 + 29

(b)

(d)

28 ˆ Ê (PG2)2 = 64 cos2 q + ÁË 6 + ˜¯ sin2 q 6 =

21 5 + 29 29 5 + 21

Ê PG1 ˆ ÁË PG ˜¯ 2

64 (36 cos2 q + 64 sin2 q) 36

2

=

PG1 36 36 36 9 fi = = . ¥ PG2 64 64 64 16

Ellipse 19.7

Ê 4 3 ˆ , Example 10: The tangent at the point P Á to Ë 2 2 ˜¯ the ellipse 9x2 + 16y2 = 144 meets the axis of x at A and the axis of y at B. If C is the centre of the ellipse, then area of the D ABC is (in sq. units) (a) 12 (b) 16 (c) 9 (d) 24 Ans. (a)

Solution: Centre of the ellipse is the origin and focus

x2 y 2 + Solution: Ellipse is = 1. 16 9

is (ae, 0) =

16 2

+

y ◊3 9 2

(a

2

)

- b 2 , 0 = (3, 0). Required equation of the

parabola is

Ê 4 3 ˆ , Equation of the tangent at P Á is Ë 2 2 ˜¯ x◊4

Example 12: The equation of the parabola whose vertex x2 y 2 + = 1 and the focus is at the centre of the ellipse 25 16 coincide with the focus of the ellipse on the positive side of the major axis of the ellipse is (a) y2 = 3x (b) y2 = 4x 2 (c) y = 5x (d) y2 = 12 x Ans. (d)

y2 = 4 ¥ 3x fi y2 = 12x. Example 13: Consider the ellipses E1:

=1

x2 + 5 (a) (c) Ans. (d) E2:

coordinate of A are (4 2 , 0) and of B are (0, 3 2 ) C is (0, 0) 1 ¥ 4 2 ¥ 3 2 = 12 sq. units 2 Example 11: A point on the ellipse 4x2 + 9y2 = 36. Where the normal is parallel to the line 4x – 2y – 5 = 0 is:

Area of the triangle ABC =

Ê 9 8ˆ (a) Á , ˜ Ë 5 5¯

Ê 8 9ˆ (b) Á , - ˜ Ë 5 5¯

Ê 9 8ˆ (c) Á - , ˜ Ë 5 5¯

Ê 8 9ˆ (d) Á , ˜ Ë 5 5¯

y2 = 1. Both the ellipses have 9 the same foci (b) same major axis the same minor axis (d) the same eccentricity

Solution: Major axis and minor axis of E1 are respectively the minor axis and major axis of E2 x2 y3 + 5 9

x2 y 2 + =1 9 5

E1

Ans. (a) Solution: Slope of the normal is 2, so slope of the tan1 gent is – . 2

fi x + 2y = 5 (1) If it touches at (x1, y1), then equation of the tangent at (x1, y1) to the ellipse is

x1 y 1 = 1 = ± 9 8 5 fi

Ê 9 8ˆ (x1, y1) = ÁË , ˜¯ 5 5

E2

Fig. 19.5

1 1 Equation of the tangent is y = - x ± 9 ¥ + 4 2 4

xx1 yy1 =1 + 9 4 Comparing (1) and (2) we get

x2 y 2 = 1 and + 9 5

(2)

Foci of E1 are ( 2, 0) Foci of E2 are (0, 2) Eccentricity of both =

9-5 2 = 9 3

Example 14: If the area of the ellipse

x2 a2

+

y2 b2

= 1 is

4p, then the maximum area of a rectangle inscribed in the ellipse in sq. units is (a) 4 (b) 8 (c) 16 (d) 32 Ans. (b) Solution: Area of the ellipse

19.8

Complete Mathematics—JEE Main

Hence the maximum area of the rectangle is 2ab = 8 (sq. units)

a

b 2 a - x 2 dx 0a

= 4Ú

Example 15: Equation of a circle described on the Latus b a = Ú 4 a 2 - x 2 dx a 0 =

b (Area of the circle of radius a) a

=

b ¥ p a2 = p ab = 4p (given) a

on the + ve x-axis is (a) 25x2 + 25y2 – 150y – 31 = 0 (b) 25x2 + 25y2 – 150x + 31 = 0 (c) 25x2 + 25y2 – 150x – 31 = 0 (d) 25x2 + 25y2 – 150y + 31 = 0 Ans. (c)

fi ab = 4 B(0, b)

x2 y 2 = 1 as a diameter with centre + 25 16

rectum of the ellipse

Solution: Coordinates of the centre which is the focus (a cos q, b sin q)

of the ellipse on the +ve x-axis is ( 25 - 16 , 0) i.e (3, 0) and radius of the circle is equal to the length of the semi-latus rectum i.e. 16 . So the required equation of the circle is 5

A(a, 0)

0

Fig. 19.6

Area of a rectangle inscribed in the ellipse is 4a cos q ¥ b sin q = 2ab sin 2 q which is maximum when sin 2 q = 1

Ê 16 ˆ (x – 3)2 + y2 = Á ˜ Ë 5¯

2

or 25x2 + 25y2 – 150x – 31 = 0

LEVEL 1 Straight Objective Type Questions Example 16: The eccentricity of an ellipse with its centre at the origin is ½. If one of the directrix is x = 4, then the equation of the ellipse is (a) 4x2 + 3y2 = 12 (b) 3x2 + 4y2 = 12 2 2 (d) 4x2 + 3y2 = 1 (c) 3x + 4y = 1 Ans. (b) Solution: Let the equation of the ellipse be x2

y2

a 2 - b2

1 = e2 a b a 4 and directrix x = a/e = 4 fi a = 2, b2 = 3 and the ellipse is x2/4 + y2/3 = 1 2

+

2

= 1 then

2

=

(b) 3/5 (d) 4/5

Solution: Let the equation of the ellipse be where b = 4.

Example 18: The locus of the point of intersection of the tangents to the ellipse x2/a2 + y2/b2 = 1 which are at right angles is (a) a circle (b) a parabola (c) an ellipse (d) a hyperbola Ans. (a) Solution: Equation of the tangent to the given ellipse with slope m is

Example 17: In an ellipse, the distance between the foci is 6 and minor axis is 8, then the eccentricity is (a) 1/ 5 (c) 1/2 Ans. (b)

If e is the eccentricity then 2ae = 6 fi a 2e 2 = 9 fi a2 – b2 = 9 fi a2 = 25 fi a = 5 and e = 3/5

a 2 m2 + b2 and the equation of tangent perpendicular to (1) is y = mx +

(2) a 2 + b2 m2 Squaring and adding (1) and (2) to eliminate m, we get (y – mx)2 + (my + x)2 = a2m2 + b2 + a2 + b2 m2 fi (x2 + y2) (1 + m2) = (a2 + b2) (1 + m2) fi x2 + y2 = a2 + b2 which is a circle. my + x =

x2 a2

+

y2 b2

=1

(1)

Ellipse 19.9

Ans. (b)

Note The circle x2 + y2 = a2 + b2 is called the Director Circle of the ellipse x2/a2 + y2/b2 = 1, whose centre is at the centre of the ellipse and whose radius is equal to the length of the line joining the end points of the major axis minor axis of the ellipse.

Example 19: The normal at an end of a latus rectum of the ellipse x2/a2 + y2/b2 = 1 passes through an end of the minor axis if (a) e4 + e2 = 1 (b) e3 + e2 = 1 2 (d) e3 + e = 1 (c) e + e = 1 Ans. (a)

(

)

Solution: Let an end of a latus rectum be ae, b 1 - e2 , then the equation of the normal at this end is x - ae ae / a 2

=

y - b 1 - e2 b 1 - e2 / b2

It will pass through the end (0, – b) if

(

- b2 1 + 1 - e2 2

–a =

1 - e2

) or b

or (1 – e2) ÈÍ1 + 1 - e2 ˙˘ = Î ˚ or

2

a2

=

1 - e2 1 + 1 - e2

1 - e2

2 4 2 1 - e2 + 1 – e = 1 or e + e = 1.

Example 20: The locus of the middle points of the portions of the tangents of the ellipse x2/a2 + y2/b2 = 1 included between the axis is the curve. (a) x2/a2 + y2/b2 = 4 (b) a2/x2 + b2/y2 = 4 2 2 2 2 (c) a x + b y = 4 (d) b2 x2 + a2 y2 = 4 Ans. (b) Solution: Equation of a tangent to the ellipse can be x y written as cosq + sin q = 1 which meets the axes at A a b (a/cos q, 0) and B (0, b/sin q). If (h, k) is the middle point of AB, then h = a/2 cos q, k = b/2 sin q Eliminating q we get (a/2h)2 + (b/2k)2 = 1 fi locus of P (h, k) is a2/x2 + b2/y2 = 4. Example 21: In a model, it is shown that an arc of a bridge in semi-elliptical with major axis horizontal. If the length of the base is 9 m and the highest part of the bridge is 3 m from the horizontal; the best approximation of the height of the arch at 2 m from the centre of the base is (a) 11/4 m (b) 8/3 m (c) 7/2 m (d) 2 m

Solution: Let the equation of the semi elliptical arc be x2/a2 + y2/b2 = 1 (y > 0). Length of the major axis = 2a = 9 fi a = 9/2 Length of the semi minor axis = b = 3. So the equation of the arc becomes If x = 2 then y2 =

4 x2 y 2 + =1 81 9

65 1 8 fiy= 65 = approximately. 9 3 3

Example 22: The locus of the foot of the perpendicular drawn from the centre to any tangent to the ellipse x2/a2 + y2/b2 = 1 is (a) a circle (b) an ellipse (c) a hyperbola (d) none of these Ans. (d) Solution: Equation of any tangent to the ellipse is (1) a 2 m2 + b2 Equation of the line through the centre (0, 0) perpendicular to (1) is y = (– 1/m) x (2) Eliminating m from (1) and (2) we get the required locus of the foot of the perpendicular y = mx +

as

y= -

x2 x2 + a 2 2 + b2 y y

or (x2 + y2)2 = a2x2 + b2y2. which does not represent a circle, an ellipses or a hyperbola. Example 23: Sum of the focal distance of any point on the ellipse x2/a2 + y2/b2 = 1 is equal to the length of the (a) major axis (b) minor axis (c) latus rectum (d) none of these Ans. (a) Solution: If P be any point on the ellipse, then distance of P from a focus is e times its distance from the correspondfocal distance is equal to e times the distance between the two directrices of the ellipse, x = a/e. Hence the required distance = e ¥ 2a/e = 2a = the length of the major axis of the ellipse. Example 24: The line passing through the extremity A of major axis and extremity B of the minor axes of the ellipse 9x2 + 16y2 = 144 meets the circle x2 + y2 = 16 at the point P. Then the area of the triangle OAP, O being the origin (in square units) is (a) 96/25 (b) 192/25 (c) 48/25 (d) 96/50 Ans. (b)

19.10

Complete Mathematics—JEE Main

Solution: Coordinates of A are (4, 0) and of B are (0, 3). x y So the equation of AB is + = 1. Which meets the circle 4 3 x2 + y2 = 16 at points whose y coordinate is given by

(a) (0, – 1/2) (c) (0, 1/2) Ans. (c)

2

Ê 12 - 4 y ˆ = 16 y + Á Ë 3 ˜¯

(b) (0, 0) (d) (0, 1)

Q( − 3, 1 / 2)

2

P ( 3, 1 / 2

S

fi

y = 0 or y = 96/25 y = 0 corresponds to the point A. So the y-coordinate of P is 96/25 Area of the triangle OAP = (1/2) OA ¥ (y-coordinate of P) = (1/2) ¥ 4 ¥ (96/25) = 192/25.

Example 25: The ellipse x2 + 4y2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then equation of the ellipse is (b) 4x2 + 6y2 = 48 (a) 4x2 + 48y2 = 48 (c) x2 + 16y2 = 16 (d) x2 + 12y2 = 16 Ans. (d) Solution: Let the equation of the required ellipse by x2/a2 + y2/b2 = 1 Y

A1

A2

Fig. 19.8

Solution: Eccentricity of the ellipse is e2 = fi

e=

a2 - b2 a

2

=

4 -1 3 = 4 4

3/2

fi coordinates of the foci of the ellipse are (±2 ¥ 3 / 2, 0) = (± 3, 0) Length of the latus rectum of the ellipse is

2 ¥1 = 1. So 2

the coordinates of P are ( 3,1/ 2) and of Q are (- 3,1/ 2) .

B

P(2, 1) A

O

Focus of the parabola is the mid point of PQ i.e., (0, 1/2) X

(4, 0)

Note There are two such parabola.

Example 27: Tangents are drawn from the point P(3, 4) Fig. 19.7

It passes through (4, 0) and (2, 1) as OB = 1 and OA = 2 are respectively the lengths of the semi-minor axis and semimajor axis of the given ellipse, coordinates of P are (2, 1) So,

16 a

2

= 1 and

4 a

2

+

1 b2

=1

fi a2 = 16 and b2 = 4/3. and the required equation is x2 3 y 2 + =1 16 4 fi

x2 + 12y2 = 16.

Example 26: A parabola has its latus rectum along PQ, where P(x1, y1) and Q(x2, y2), y1 > 0, y2 > 0 are the end points x2 y 2 = 1. Coordinates + of the latus rectums of the ellipse 4 1 of the focus of the parabola are

x2 y 2 = 1 touching the ellipse at point A + 9 4 and B. The equation of the locus of the point whose distances from the point P and the line AB are equal is (a) 9x2 + y2 – 6xy – 54x – 62y + 241 = 0 (b) x2 + 9y2 + 6xy – 54x + 62y – 241 = 0 (c) 9x2 + 9y2 – 6xy – 54x – 62y – 241 = 0 (d) x2 + y2 – 2xy + 27x + 31y – 120 = 0 Ans. (a) to the ellipse

Solution: AB being the chord of contact of the ellipse from P(3, 4) has its equation. 3x 4 y = 1 fi x + 3y = 3 + 9 4 If (h, k) is any point on the locus, then (h - 3) 2 + (k - 4) 2 =

h + 3k - 3 1+ 9

Ellipse 19.11

fi 10(h2 + k2 – 6h – 8k + 25) = (h + 3k – 3)2 Locus of (h, k) is 9x2 + y2 – 6xy – 54x – 62y + 241 = 0

It also passes through the point (3, 2) fi

Example 28: Equation of the ellipse whose axes are the axes of coordinates, which passes through the point (– 3, 1) and has eccentricity 2 / 5 is (a) 5x2 + 3y2 – 32 = 0 (c) 5x2 + 3y2 – 48 = 0 Ans. (b)

(b) 3x2 + 5y2 – 32 = 0 (d) 3x2 + 5y2 – 15 = 0

a 2

2

2

+

2

y2 b

2

a

+

2

2

1 b

2

=1fi

9 a

2

+

5

=1

3a 2

4 =1 16

a fi a2 = 12

Equation of the ellipse is

x2 y 2 =1 + 12 16 1-

So eccentricity of the ellipse is

(c)

fi a2 = 9 + 5/3 = 32/3. fi b2 = 32/5 So the required equation is 2

+

a2 2

=

1-

3 1 = 4 2

Example 31: A rod of length 12 cm moves with its ends always touching the coordinate axes. The locus of a point P on the rod, which is 3 cm from the end in contact with x-axis is an ellipse whose eccentricity is: 2 2 (a) (b) 3 3

=1

We have b = a (1 – e ) = a (1 – 2/5) = 3a2/5 Since it passes through the point (– 3, 1) 9

2

b Note Fig ia Similar to the Fig of Example 25.

Solution: Let the equation of the ellipse be x2

9

2 2 3

2 3 2

(d)

Ans. (c)

2

x y + = 1 or 3x2 + 5y2 = 32 32 / 3 32 / 5

Y B

Example 29: An ellipse is drawn by taking a diameter of the circle (x – 1)2 + y2 = 1 as its semi-minor axis and a diameter of circle x2 + (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) 4x2 + y2 = 8 (b) x2 + 4y2 = 16 2 2 (d) x2 + 4y2 = 8 (c) 4x + y = 4 Ans. (b) Solution: Diameter of (x – 1)2 + y2 = 1 is 2 ¥ 1 = 2 = b and of x2 + (y – 2)2 = 4 is 2 ¥ 2 = 4 = a So the required equation of the ellipse is x2 (4) 2

+

y2 (2) 2

= 1 fi x2 + 4y2 = 16

x2 y 2 = 1 is inscribed + 9 4 in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E2 passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E2 is Example 30: The ellipse E1:

(a)

2 /2

(c) 1/2 Ans. (c)

(b)

3/2

(d) 3/4

Solution: Let the required ellipse be it passes through (0, 4) fi b2 = 16.

x2 a

2

+

y2 b

2

= 1 since

Q O

P(x, y) Rq

A

X

q

Fig. 19.9

Solution: Let AB be the rod making an angle q with x-axis and P(x, y) be any point on the rod such that AP = 3 cm. fi PB = AB – AP = 12 – 3 = 9 cm. PQ x cos q = = PB 9 PR y sin q = = PA 3 x2 y 2 so eliminating q we get the required locus as =1 + 81 9 81 - 9 2 2 = . 81 3 Example 32: Eccentricity of the ellipse whose latus rectum is half of its major axis is: 1 1 (b) (a) 2 2

which is an ellipse with eccentricity

(c) Ans. (b)

2 3

(d)

1 3

19.12

Complete Mathematics—JEE Main

Solution: Let the equation of the ellipse be

x2 a2

+

y2 b2

Ans. (c) = 1.

2b 2 Length of the latus rectum = = a, half of major axis a 2b2 = a2 fi e2 =

fi fi

e=

1 2

a 2 - b2 a

=

2

1 2

Solution: Let the equation of the ellipse be

l - 2l + 1 = e Since 0 < e < 1 fi 0 < |l Р1| < 1 fi lΠ(0, 2) ~ {1} Example 34: If the line lx + my + n = 0 cuts the ellipse x y2 p + at points whose eccentric angles differ by , then 2 2 a b 2 a 2l 2 + b2 m2

(-ae, 0)

F¢

0

F

=1

is equal to (b) 2

(ae, 0)

Fig. 19.10

fi

b b ¥ = – 1 fi b 2 = a 2e 2 - ae ae

fi a2(1 – e2) = a2e2 fi 2e2 = 1 fi e =

2

(a) 1

b2

B(0, b)

2

n2

y2

.

Solution: Given equation represents an ellipse with ec-

the value of

a2

+

F(ae, 0), F¢ (–ae, 0) BF is perpendicular to BF ¢

Example 33: If the equation 5[(x Р2)2 + cy Р3)2] = (l2 Р2l + 1) (2x + y Р1)2 represents an ellipse then (a) lΠ(0, 2) (b) lΠ(Р1, 1) (c) lΠ(0, 2) Р{1} (d) lΠ(Р1, 1) Р{0} Ans. (c) centricity

x2

1

2 Example 36: The locus of the foot of the perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent is (a) (x2 – y2)2 = 6x2 + 2y2 (b) (x2 – y2)2 = 6x2 – 2y2 (c) (x2 + y2)2 = 6x2 + 2y2 (d) (x2 + y2)2 = 6x2 – 2y2 Ans. (c)

Ans. (b)

Solution: Equation of any tangent to the ellipse x y2 = 1 is y = mx + 6m 2 + 2 . If (h, k) is the foot of 6 2

Solution: Let the line meet the ellipse at P(a cos q, Êp ˆ Êp ˆ b sin q) and Q(a cos Á + q ˜ ), b sin Á + q ˜ . Since they Ë2 ¯ Ë2 ¯

the perpendicular from (0, 0), the centre of the ellipse on this k tangent then k = mh ¥m=–1 6m 2 + 2 and h

lie on the line lx + my + n = 0 la cos q + m b sin q + n = 0 and – l a sin q + m b cos q + n = 0 Eliminating q by squaring and adding we get a2l2 + b2m2 = 2n2

Eliminating m, we get (h2 + k2)2 = 6h2 + 2k2

(c)

fi

1 2

a 2l 2 + b2 m2 n

2

(d)

3 . 2

= 2.

Example 35: An ellipse has OB as semi minor axis F and F ¢ its foci and the angle FBF ¢ is a right angle. The eccentricity of the ellipse is 1 1 (a) (b) 3 4 1 1 (d) (c) 2 2

2

So, required locus is (x2 + y2)2 = 6x2 + 2y2 Example 37: The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse

x2 y 2 = 1 is: + 9 5

(a)

27 4

(b) 18

(c)

27 2

(d) 27

Ans. (d)

Ellipse 19.13

fi fi fi

(0, 3) B L

fi

0 A (9/2, 0)

C L¢

Fig. 19.11

2 -1.

Example 39: The tangent to the ellipse 3x2 + 16y2 = 12,

(a) no point (c) two distinct point Ans. (b)

Solution: we have a2 = 9, b2 = 5 4 5 e2 = 1 – = 9 9 e=

e=

Ê 3ˆ at the point Á1, ˜ intersects the curves y2 + x = 0 at: Ë 4¯

D

fi

b2 = 2a2e a (1 – e2)= 2a2e e2 + 2e = 1 fi (e + 1)2 = 2 2

2 3

Ê b2 ˆ Ê 5ˆ Coordinates of L are Á ae, ˜ = Á 2, ˜ Ë 3¯ a Ë ¯ Equation of the tangent at L is

(b) exactly one point (d) more than two points.

Ê 3ˆ yÁ ˜ Ë 4¯ Ê 3ˆ x ◊1 Solution: Tangent at Á1, ˜ to the ellipse is + Ë 4¯ 3 4 4 = 1 fi x + 4y = 4 which intersects the curve y2 + x = 0 at points for which y2 + (4 – 4y) = 0 fi (y – 2)2 = 0 fi y = 2 and the point of intersection is (– 4, 2) i.e. exactly one point. Example 40: The line 2x + y = 3 intersects the ellipse 4x + y2 = 5 at two points. The tangents to the ellipse at these two points intersect at the point:

2x 5 Ê y ˆ + Á ˜ = 1 fi 2x + 3y = 9. 9 3Ë 5¯

2

Ê9 ˆ It meets the axes at A Á , 0˜ and B(0, 3) Ë2 ¯

Ê 5 5ˆ (a) Á , ˜ Ë 6 3¯

Ê 5 5ˆ (b) Á , ˜ Ë 6 6¯

From symmetry, we observe that the quadrilateral is a rhombus whose area is 4 ¥ area of the D AOB.

Ê 5 5ˆ (c) Á , ˜ Ë 3 6¯

Ê 5 5ˆ (d) Á , ˜ Ë 3 3¯

Hence the required area is 4 ¥

9 1 ¥3¥ 2 2

= 27 sq. units. Example 38: If the distance between the foci of an ellipse is half the length of its latus rectum, then eccentricity of the ellipse is: (a)

(c)

1 2 2 -1

2 2 -1 2

fi

(d)

2 -1 2

fi

Solution: Let the equation of the ellipse be x a

2

+

y

2

b2

= 1, then

2ae =

1 Ê 2b 2 ˆ 2 ÁË a ˜¯

Solution: Let the line intersect the ellipse at P and Q and the tangents at P and Q to the ellipse intersect at R(h, k), then PQ is the chord of contact of R w.r.t. the ellipse; so its equation is 4hk + ky = 5 But the equation of the chord of contact is given as 2x + 3y = 3

(b)

Ans. (c) 2

Ans. (a)

4h k 5 = = 2 1 3 h=

5 5 ,k= 6 3

Ê 5 5ˆ and the required point is Á , ˜ Ë 6 3¯

19.14

Complete Mathematics—JEE Main

Assertion-Reason Type Questions

Example 41: E: 2x2 + 3y2 = 6, e: x2 + y2 = 2x + 4y + 4 = 0 Statement 1: The area of the ellipse E is more than the area of the circle C.

S1 c1

Statement 2: The length of the semi-major axis of E is more than the radius of C. Ans. (b) Solution: Equation of the ellipse is

c2 S

x2 y 2 =1 + 3 2

whose area is p 3 2 = p 6 . sq. units. Equation of the circle is (x – 1)2 + (y + 2)2 = 1 whose area is p sq. units So area of E > area of C and thus Statement 1 is true. 3 > 1, the radius of C so statement 2 is also true but does not justify statement 1.

Length of the semi major axis of E is

Example 42: Statement 1: The line 2x + 3y = 1 x2 y 2 intersects the ellipse = 1 at P and Q. The tangents + 4 2 to the ellipse at P and Q. intersect at the point (8, 6). x2

xx1

+

a2 yy1

a2 b2 Ans. (a)

+

y2 b2

Solution: Let C1, C2 be the centres and r1, r2 the radii of the circles S1 and S2 respectively. C be the centre and r the radius of the circles touching S1 and S2. Then CC1 = r1 – r and CC2 = r2 + r fi CC1 + CC2 = r1 + r2 (constant) So distance of C C1 and C is constant C is an ellipse. In statement 1, the circle C2 lies inside the circle C1 and hence by statement 2, the required locus is an ellipse and thus the statement 1 is also true.

(6,1)

(2,1) (4,1)

= 1 from an external point (x1, y1) is Fig. 19.13

= 1.

Solution: Statement 2 is true (see theory). In statement 1, let the coordinates of the required point be (h, k), then 2x + 3y = 1 is the chord of contact of the ellipse hx ky = 1. Comparing and by statement 2, its equation is + 4 2 the two equations, we get h = 8, k = 6 and the statement 1 is also true. Example 43: Statement 1: The locus of the centre of a variable circle touching two circles C1: (x – 2)2 + (y – 1)2 = 16 and C2: (x – 4)2 + (y – 1)2 = 1 is an ellipse. Statement 2: If a circle S2 = 0 lies completely inside the circle S1 = 0, then the locus of the centre of a variable circle touching both the circles is an ellipse. Ans. (a)

c

Fig. 19.12

Statement 2: Equation of the chord of contact to the ellipse

S2

Example 44: Statement 1: If the line touches the ellipse

x2 a2

+

y2 b2

x y + = a b

2

= 1, and q is the eccentric angle

of the point of contact, then sin2q =

1 . 2

Statement 2: If the length of the semi major axis of an el2 times the length of the semi-minor axis, then the eccentricity e of the ellipse satisfies 2e2 – 1 = 0. Ans. (b) lipse is

Solution: Let the coordinates of the point of contact in statement 1 be (a cos q, b sin q). Equation of the tangent at this point to the ellipse is x cos q y sin q =1 + a b

Ellipse 19.15 2

(Note x + y = 150 is the director circle of the ellipse

Comparing with the given line 1 1 fi sin2q = cos q = sin q = 2 2

x2 y2 = 1) + 100 50

and the statement 1 is true.

Example 46: The tangent at a point P(a cos a, b sin a)

In statement 2, let the equation of the ellipse be s.t a =

x2 a2

+

y2 b2

=1

2 b

fi a2 = 2b2 and e2 =

2

a 2 - b2

=

a2

1 thus statement 2 is also 2

true but does not justify statement 1. Example 45: Statement 1: If L and M are the feet of the x2 y2 perpendiculars draw from the foci of the ellipse + 100 50 = 1 on any tangent to it, then L and M lie on the circle x2 + y2 = 150. Statement 2: The locus of feet of the perpendiculars from the foci upon any tangent is the auxiliary circle of the ellipse. Ans. (d) Solution: Let the equation of the ellipse be

x2 a2

+

y2 b2

to the ellipse

x2 a2

+

y2 b2

p , O being the origin. 2 Statement 1: Equation of the directrices of the ellipse are

B such that AOB =

a 1 + sin 2 a . Statement 2: Eccentricity e + e2 – 1 = 0. Ans. (a) x=

Y A

= 1.

P B O

x y cos q + sin q = 1 (1) a b Equation of the perpendiculars from ( ae, 0) to the tangent (1) is x sinq y cosq = ± ae sinq b a b

(2)

Eliminating q from (1) and (2) we get the required locus by squaring and adding 2 È cos 2 q sin 2 q ˘ cos 2 q ˘ a 2 e2 sin 2 q 2 È sin q ˙ x2 Í 2 + 2 ˙ + y Í 2 + = 1 + b ˚ a2 ˚ b2 Î a Î b 2

2

2

2

Ê b cos q + a sin q ˆ (x2 + y2) Á ˜ a 2b2 Ë ¯ È b 2 + (a 2 - b 2 )sin 2 q ˘ 2 = Í ˙a a 2b2 Î ˚ 2

e2 sin2a

Solution: Equation of the tangent at P to the ellipse is x y (1) cos a + sin a = 1 a b

Equation of any tangent is

fi

= 1 meets the auxiliary circle at A and

2

2

X

Fig. 19.14

Equation of the auxiliary circle is x2 + y2 = a2 (2) Equation of the pair of lines OA and OB is x2 + y2 = a2 y Êx ˆ ÁË cos a + sin a ˜¯ a b

2

(Making (2) homogeneous with the help of (1)) Since OA OB Ê a2 ˆ (1 – cos2 a) + Á1 - 2 sin 2 a ˜ = 0 Ë b ¯ fi b2 = (a2 – b2) sin2 a fi 1 – e2 = e2 sin2 a fi e2 sin2 a + e2 – 1 = 0 So statement 2 is true. x= ±

2

È b cos q + a sin q ˘ 2 = Í ˙a a 2b2 Î ˚ fi x2 + y2 = a2 which is the auxiliary circle of the ellipse. Hence the state-

a = e

± a 1 + sin 2 a and the statement 1 is also true. Example 47: The tangent and normal at a point P(x1, y1), x2 y 2 x1 > 0, y1 > 0 on the ellipse 2 + 2 = 1 meets the major a b axis at T and N respectively.

19.16

Complete Mathematics—JEE Main

Solution: Equation of the ellipse in statement 1 is:

Statement 1: PN is the internal bisector of F1 PF2 ; F1, F2 being the foci of the ellipse. Statement 2: PT is constant for all positions of the point P. Ans. (c) Solution: Equation of the tangent at xx1 yy1 + = 1 which meets the major axis at a 2 b2

P(x1, y1) is Ê a2 ˆ Á x , 0˜ Ë 1 ¯

x 2 ( y - 3) 2 = 1 whose centre is (0, 3) and the ends of + 5 9 the major axis are (0, 0) and (6, 0) as the length of semi major axis is 3. Y (0, 6)

2

Ê a2 ˆ (PT)2 = Á x1 - ˜ + y12 x1 ¯ Ë

So

=

( x12 - a 2 ) 2 x12

(0, 3)

+ y12

O (0, 0)

Which shows that PT depends on the position of P and thus the statement 2 is false. Y P

F1 0

N F2

T

X

Fig. 19.15

For statement 1, equation of the normal at P(x1, y1) is 2

2

a x b y = a 2 – b 2 = a 2e 2 y1 x1 So coordinates of N are (e2x1, 0) F1 are (– ae, 0), F2 are (ae, 0) a Also PF1 = e + x1 e = a + ex1 PF2 = a – ex1 NF1 = e2x1 + ae = ePF1 NF2 = ae – e2x1 = e(PF2) PF1 NF1 fi = which shows that PN bisects the F1 PF2 PF2 NF2 Example 48: Statement 1: Equations of the tangents drawn at the ends of the major axis of the ellipse 9x2 + 5y2 – 30y = 0 is y = 0, y = 6. Statement 2. The tangents drawn at the ends of the major x2 y 2 axis of the ellipse 2 + 2 = 1 are always parallel to the a b y-axis.

X

Fig. 19.16

The tangents at these ends are y = 6 and y = 0. Statement 1: is true. Statement 2: is false as the tangents at the ends of the major axis are parallel to y-axis only when a > b. Example 49: Statement 1: If the length of the latus 1 of the major axis, then the rectum of an ellipse is 3 eccentricity of the ellipse is

2 . 3

Statement 2: If a focus of an ellipse is at the origin, direc2 , then the trix is the line x = 4 and the eccentricity is 3 length of the semi major axis is 4 6 . Ans. (b) Solution: In statement 1, if the equation of the ellipse is

fi

x2 a2

+

b2 a

2

fie=

and the statement 1 is true.

Ans. (c)

Minor axis

y2 b2 =

= 1, then

2b 2 1 = ( 2a ) a 3

1 = 1 – e2, e being the eccentricity 3 2 and the statement 1 is true 3

In statement 2, the equation of the ellipse is 2 (4 – x)2 x2 + y2 = 3 fi 3(x2 + y2) = 2(16 – 8x + x2) fi x2 + 16x + 3y2 = 32 fi

( x + 8) 2 y 2 =1 + 96 32

Ellipse 19.17

Length of the semi major axis =

96 = 4 6 . Thus statement 2 is also true but does not justify statement 1.

And to

x2 y 2 = 1 is x = m1 y + + 2 4 y=

1 2 1 x - 4 + 2 ,m = m1 m1 m1

Example 50: Statement 1: An equation of a common tangent to the parabola y2 = 16 3x and the ellipse 2x2 + y2 = 4

or

is y = 2x + 2 3 .

È Ê 4 3ˆ 2 ˘ and Á = Í- 4 + 2 ˙ ˜ m1 ˙˚ Ë m ¯ ÎÍ

2

Statement 2: If the line y = mx + 4 3/m , (m

0) is a com-

mon tangent to the parabola y2 = 16 3 x and the ellipse 2x2 + y2 = 4, then m Ans. (a)

m4 + 2m2 = 24

Solution: Equation of a tangent to y2 = 16 3 x is y = mx + 4 3/m .

48

fi

m2

4m12 + 2

2

= 4 + 2m2

fi m4 + 2m2 – 24 = 0 2 fi m = 4 fi m = 2. Showing that both the statement are true and statement 2 is a correct explanation for statement 1.

LEVEL 2 Straight Objective Type Questions

Example 51: If p is the length of the perpendicular from a focus upon the tangent at any point P of the ellipse x2/a2 + y2/b2 = 1 and r is the distance of P from the focus, then

2a b 2 - 2 = r p

r2 = (ae – a cos q)2 + b2 sin2 q = a2 [(e – cos q)2 + (1 – e2)sin2 q] = a2 [e2 cos2 q – 2e cos q + 1] = a2 (1 – e cos q)2 r = a(1 – e cos q)

Now

fi

(a) – 1 (c) 1 Ans. (c)

b2 2 1 + e cos q 2a Now – 2 = =1 p 1 - e cos q 1 - e cos q r

(b) 0 (d) 2

Solution: The equation of the tangent at P(a cos q, x y cos q + sin q = 1 a b length of the perpendicular from the focus (ae, 0) on the line is b sin q) to the ellipse x2/a2 + y2/b2 = 1 is

Example 52: If y = x and 3y + 2x = 0 are the equations of a pair of conjugate diameters of an ellipse, then the eccentricity of the ellipse is 2/3

(b) 1/ 3

(c) 1/ 2

(d) 2 / 5

(a)

Ans. (b) p=

e cos q - 1 cos 2 q sin 2 q + 2 a2 b

=

ab(e cos q - 1) b 2 cos 2 q + a 2 (1 - cos 2 q )

ab(e cos q - 1)

=

2

2 2

2

a - a e cos q fi

b2 p2

=

1 + e cos q 1 - e cos q

= b

1 - e cos q 1 + e cos q

Solution: Let the equation of the ellipse be x2/a2 + y2/b2 = 1 Slope of the given diameters are m1 = 1, m2 = – 2/3. fi m1m2 = – 2/3 = – b2/a2 [using the condition of conjugacy of two diameters] 3b2 = 2a2 fi 3a2(1 –e2) = 2a2, 1 – e2 = 2/3 fi e2 = 1/3 fi e = 1/ 3 Example 53: If a, b are the eccentric angles of the extremities of a focal chord of the ellipse x2/16 + y2/9 = 1, then tan (a/2) tan (b/2) =

19.18

Complete Mathematics—JEE Main

(a)

(c)

7+4

(b) -

7-4 5-4

(d)

5+4

9 23

Solution: We can write x2 + 4y2 = 4 as Equation of a tangent to the ellipse (i) is

8 7 - 23 9

x cos q + y sin q = 1 2

Ans. (d) Solution: The eccentricity e =

9 7 1= . 16 4

Let P (4 cos a, 3 sin a) and Q (4 cos b, 3 sin b) be a focal chord of the ellipse passing through the focus at ( 7 , 0) . Then

3 sin b 4 cos b - 7

=

3 sin a

sin(a - b ) 7 = sin a - sin b 4

fi

cos[(a - b )/ 2] 7 = cos[(a + b )/ 2] 4

fi

Êaˆ Ê bˆ tan Á ˜ tan Á ˜ = Ë 2¯ Ë 2¯

Equation of the ellipse x2 + 2y2 = 6 can be written as x2 y 2 =1 + 6 3

(iii)

Suppose (ii) meets the ellipse (iii) at P and Q and the tangents at P and Q to the ellipse (iii) intersect at (h, k), then (ii) is the chord of contact of (h, k) with respect to the ellipse hx ky =1 + 6 3 Since (ii) and (iv) represent the same line

(iv)

h/ 6 k /3 = =1 (cos q )/ 2 sin q 7-4 7+4

=

23 - 8 7 -9

Example 54: If an ellipse slides between two perpendicular straight lines, then the locus of its centre is (a) a parabola (b) an ellipse (c) a hyperbola (d) a circle Ans. (d) Solution: Let 2a, 2b be the length of the major and minor axes respectively of the ellipse. If the ellipse slides between two perpendicular lines, the point of intersection P of these lines being the point of intersection of perpendicular tangents lies on the Director circle of the ellipse. This means that the centre C of the ellipse is always at a constant distance a 2 + b 2 from P. Hence the locus of C is a circle.

fi h = 3 cos q, k = 3 sin q. and the locus of (h, k) is x2 + y2 = 9 Example 56: The sum of the squares of the perpendiculars on any tangent to the ellipse x2/a2 + y2/b2 = 1 from two points on the minor axis each at a distance a 2 - b 2 from the centre is (a) 2a2 (c) a2 + b2 Ans. (a)

(b) 2b2 (d) a2 – b2

Solution: The eccentricity e of the given ellipse is given by e2 = 1 – b2/a2 fi a2 – b2 = a2e2. So the point on the minor a 2 - b 2 from the centre (0, 0)

axis, i.e. y-axis at a distance

of the ellipse are (0, ae). The equation of the tangent at any point (a cos q, b sin q) on the ellipse is x y cos q + sin q = 1 a b So the required sum is

B C 90° P

(ii)

(iii) and thus its equation is

4 cos a - 7

fi

x2 y 2 = 1 (i) + 4 1

A

Fig. 19.17

Example 55. The locus of the points of intersection of the tangents at the extremities of the chords of the ellipse x2 + 2y2 = 6 which touch the ellipse x2 + 4y2 = 4 is (a) x2 + y2 = 4 (b) x2 + y2 = 6 2 2 (d) none of these (c) x + y = 9 Ans. (c)

È Í ae sin q - 1 Í b Í 2 2 Í cos q + sin q ÍÎ a2 b2 =

=

2

˘ È ˙ Í - ae sin q - 1 ˙ +Í b ˙ Í 2 2 ˙ Í cos q + sin q ˙˚ ÍÎ b2 a2

(ae sin q - b)2 + (ae sin q + b)2 2

2

2

2

(b cos q + a sin q ) 2a 2 (a 2 e2 sin 2 q + b 2 ) b 2 cos 2 q + a 2 sin 2 q

˘ ˙ ˙ ˙ ˙ ˙˚

2

¥ a2

Ellipse 19.19

=

2a 2 [(a 2 - b 2 )sin 2 q + b 2 )

= 2a2

b 2 cos 2 q + a 2 sin 2 q

Example 57. y = mx + c is a normal to the ellipse x /a2 + y2/b2 = 1 if c2 is equal to 2

(a)

(c)

(a 2 - b 2 )2

(b)

a 2 m2 + b2 (a 2 - b 2 )2 m 2

(d)

a 2 + b2 m2

(a 2 - b 2 )2 a 2 m2 (a 2 - b 2 )2 m 2 a 2 m2 + b2

Ans. (c) Solution: Equation of a normal to the ellipse x2/a2 + y2/b2 = 1 is ax by = a2 – b2, if it represents, mx – y = – c, then cos q sin q a b -(a 2 - b 2 ) = = sinq m cosq c fi

cos q =

ac 2

m( a - b 2 )

, sin q = –

bc 2

a - b2

c2

È a2 ˘ + b2 ˙ = 1 2 2 2 Í 2 (a - b ) Î m ˚

fi

c2 =

fi

Example 59: P is a point on the ellipse E:

a2

the coordinate axes at A and B, the length of AB is

(c) 4 7 units

(d) 14 units

(d) (0, - 5 )

=1

Example 60: If the common tangent in the first quadrant x2 y 2 of the circle x2 + y2 = 16 and the ellipse = 1 meets + 25 4

a 2 + b2 m2

(c) (2, 1)

b2

Solution: P(a cos q, b sin q), P¢ (a cos q, a sin q) Equation of the normal at P is ax sec q – by cosec q = a2 – b2 (1) Equation of the normal at P¢ is x sec q – y cosec q = 0 (2) From (1) and (2) we get (a – b) y cosec q = a2 – b2 fi y = (a + b) sin q. From (2) we get x = (a + b) cos q. So point of intersection of (1) and (2) lie on a circle of radius a + b.

(b) 14 / 3 units

(b) ( 5 , 0)

y2

x2 + y2 = a2. The normal at P to E and at P¢ to C intersect on circle whose radius is (a) a + b (b) a – b (c) 2a (d) 2b. Ans. (a)

(a) 7 / 3 units

(a) (0, 5 )

+

and P¢ be the corresponding point on the auxiliary circle C:

(a 2 - b 2 )2 m 2

Example 58: A tangent at any point to the ellipse 4x2 + 9y2 = 36 is cut by the tangent at the extremities of the major axis at T and T ¢. The circle on TT ¢ as diameter passes through the point.

Ans. (b) Solution: An equation of the tangent to the ellipse x2 y 2 + = 1 is 25 4 y = mn +

25m 2 + 4

(1) m < 0.

Ans. (b) Solution: Any point on the ellipse is P(3 cos q, 2 sin q). x y Equation of the tangent at P is cos q + sin q = 1 which 3 2

B 2

meets the tangents x = 3 and x = – 3 at the extremities of the major axis at

-5

-4

2(1 + cos q ) ˆ Ê 2(1 - cos q ) ˆ Ê T Á 3, ˜¯ and T ¢ ÁË -3, ˜ Ë sin q ¯ sin q

2(1 - cos q ) ˆ Ê 2(1 + cos q ) ˆ Ê (x – 3) (x + 3) + Á y ˜ =0 ˜¯ ÁË y Ë sin q sin q ¯ 4 y – 5 = 0, which passes through ( 5 , 0) sinq

0

4

5 A

-2

Equation of the circle on TT¢ as diameter is

fi x2 + y2 –

x2

Fig. 19.18

The line will touch the circle x 2 + y 2 = 16 if m(0) - 0 + 25m 2 + 4 m2 + 1

=4

Complete Mathematics—JEE Main

19.20

25m2 + 4 = 16 (m2 + 1) -2 4 fi m2 = fim= 3 3 So equation of the tangent is fi

y= fi

2

Ê 4 7ˆ Ê 4 7ˆ so that (AB) = Á +Á ˜ ˜ Ë 2 ¯ Ë 3¯ = (4 7 )2 ¥

-2

Ê 4ˆ x + 25 Á ˜ + 4 Ë 3¯ 3

2x +

2

2

fi

3y = 4 7

AB =

14 3

7 12

units

Ê4 7 ˆ , 0˜ and which meets the coordinate axes at A Á Ë 2 ¯ Ê 4 7ˆ B Á 0, ˜ 3¯ Ë

EXERCISE Concept-based Straight Objective Type Questions 1. Equation of an ellipse with centre at the origin passing through (5, 0) and having eccentricity 2/3 is: (b) 9x2 + 5y2 = 225 (a) 4x2 + 9y2 = 100 2 2 (c) 5x + 9y = 125 (d) 6x2 + 4y2 = 150 x2 y 2 = 1, 2. Equation of a tangent to the ellipse + 36 25 passing through the point where a directrix of the ellipse meets the + ve x-axis is (a) 5 y + 11x = 36

(b) 6 y + 11x = 25

(c) 6 y + 11x = 36

(d) 5 y - 11x = 36

6. Length of a latera recta of the ellipse is (in units) (a) 9 (c) 14

4. If the eccentricity of the ellipse x2 a2

+

x2 y 2 + = 1 and 25 a 2

y2 = 1 is same, then the value of a2 is: 16

(a) 9 (c) 15

(b) 41 (d) 20

5. If y = mx + 4 is a tangent to the ellipse then 625m4 + 25m2 – 156 is equal to (a) – 1 (b) 0 (c) 1 (d) 4

x2 y 2 = 1, + 25 4

(b) 7 (d) 18

(

)

7. The normal at the point 3, 2 3 on the ellipse x2 y 2 = 1 meets the major axis of the ellipse at + 36 16 (a, 0), the value of a is: (a)

5 3

(b)

3 5

(c)

2 3

(d)

3 2

3. Equation of a line joining a foci of the ellipse x2 y 2 x2 y 2 = 1 to a foci of the ellipse = 1 is, + + 25 9 9 25 (a) x + y = 4 (b) x + y = 5 (c) x + y = 3 (d) 5x + 3y = 1

x2 y 2 = 1 + 81 63

8. In an ellipse, the distance between the foci is 6 and length of semi minor axis is 4, then eccentricity of the ellipse is (a)

2 5

(b)

(c)

2 3

(d)

3 5 1 2

9. If F1 and F2 are the foci of the ellipse 4x2 + 9y2 = 36, P is a point on the ellipse such that PF1 : PF2 = 2 : 1; then area of D PF1F2 is: (a) 1 sq. unit (b) 2 sq. unit (c) 3 sq. unit (d) 4 sq. unit

Ellipse 19.21

10. If the eccentric angles of two points P and Q on the x

2

y

2

p are a, b such that a + b = , then a b2 2 the locus of the point of intersection of the normals at P and Q is (a) ax + by = 0 (b) ax – by = 0 (c) x + y = 0 (d) x + y = a + b ellipse

2

+

11. P1(q1) and P2(q2) are two points on the ellipse x

2

+

y

2

= 1 such that tan q1 tan q2 =

-a

2

. The a b b2 chord joining P1 and P2 of the ellipse subtends a right angle at the (a) focus (ae, 0) (b) focus (– ae, 0) (c) centre (0, 0) (d) vertex (a, 0) 2

2

12. Coordinates of the foci of the ellipse 2x2 + 3y2 – 4x – 12y + 13 = 0 are 1 ˆ Ê , 2˜ (a) Á1 ± Ë 6 ¯

1 ˆ Ê (b) Á 2,1 ± ˜ Ë 6¯

(c) (1, 2)

Ê 1 1 ˆ , (d) Á Ë 2 3 ˜¯

lines through P and Q parallel to y-axis and x-axis respectively meet at the point R, then the locus of R is (b) x2 – y2 = a2 – b2 (a) x2 + y2 = a2 + b2 (c)

x2 a2

+

y2 b2

=1

(d) x2

x2 b2 +

+

y2 a2

=1

y2

= 1 (a > b) and Q a 2 b2 is the point corresponding to P on the auxiliary circle x2 + y2 = a2. N is the foot of the perpendicular from PN is equal to: P on the major axis of the ellipse. PQ

14. P is a point on the ellipse

(a)

b a-b

(b)

a a-b

(c)

b a+b

(d)

a a+b

15. If the normal at any point P on the ellipse

13. C1: x2 + y2 = a2 and C2: x2 + y2 = b2 are two contentric circles. A line through the centre of these circles intersects them at P and Q respectively. If the

x2

+

y2

=1 a 2 b2 meets the axis of x at G and the axis of y at g, then PG : pg is: (b) a : b (a) a2 : b2 (c) b : a (d) b2 : a2

LEVEL 1 Straight Objective Type Questions 16. On the ellipse 4x2 + 9y2 = 1, the point at which the tangent is parallel to the line 8x = 9y is (a) (2/5, 1/5) (b) (– 2/5, 1/5) (c) (– 2/5, – 1/5) (d) none of these 2

2

17. The coordinates of the foci of the ellipse 4x + 9y = 1 are (a) (± 5 / 3, 0)

(b) (± 5 / 6, 0)

(c) (0, ± 5 / 3)

(d) (0, ± 5 / 6)

18. The distance of the point ( 6 cos q , 2 sin q ) on the ellipse from the centre of the ellipse is 2 if q = (a) p/6 (b) p/4 (c) p/3 (d) none of these 19. An equation of the normal to the ellipse x2/a2 + y2/b2 = 1 with eccentricity e at the positive end of the latus

rectum is (a) x + ey + e3 a = 0 (c) x – ey + e3 a = 0

(b) x – ey – ae3 = 0 (d) x + ey – e3 a = 0

20. The circle x2 + y2 = c2 contains the ellipse x2/a2 + y2/b2 = 1 if (a) c < a (b) c < b (c) c > max{a, b} (d) c > b 21. In an ellipse, if the lines joining a focus to the extremilites of the minor axis make an equilateral triangle with the minor axis, the eccentricity of the ellipse is (a) 3/4

(b)

(c) 1/2

(d) 2/3

3/2

22. An equilateral triangle is inscribed in the ellipse x2 + 3y2 = 3 such that one vertex of the triangle is (0, 1) and one altitude of the triangle is along the y-axis. The length of its side is

19.22

Complete Mathematics—JEE Main

(a) 4 3 / 5

(b) 3 3 / 5

(c) 6 3 / 5

(d) 2 3

(b) 25x2 + 16y2 = 9 (c) 16x2 + 25y2 = 1 (d) x2 + y2 = 41

x2 y2 = 1 represents an + 10 - 2a 4 - 2a ellipse, then a lies in the interval (a) (– •, 5) (b) (2, 5) (c) (– •, 2) (d) (5, •)

23. If the equation

24. The curve represented by x = 3 (cos t + sin t), y = 4 (cos t – sin t) is an ellipse whose eccentricity is e, such that 16e2 + 7 is equal to: (a) 14 (b) 12 (c) 11 (d) 21 25. The eccentricity of an ellipse, with centre at the origin is 2/3. If one of directrices is x = 6, then equation of the ellipse is x2 y 2 =1 + 9 4 (c) 5x2 + 9y2 = 80

x2 y 2 =1 + 9 5 (d) 3x2 + 2y2 = 6 (b)

(a)

26. The locus of the foot of the perpendicular from a x2 y 2 focus of the ellipse 2 + 2 = 1 on any tangent is a b (a) x2 + y2 = a2 + b2 (c) x2 + y2 = b2 27. If the ellipse

x2 2

+

(b) x2 + y2 = a2 (d) x2 + y2 = 2a2

y2 2

=1 and the circle x2 + y2 = r2

a b where b < r < a intersect in four points and the slope of a common tangent to the ellipse and the circle is b , then r2 is equal to a (a)

(c)

2b 2 a 2 + b2 2a 2 2

a +b

2

(b)

(d)

(a) 144 (c) 122

the ellipse

a 2 + b2 2a 2 b 2 2

a -b

2

x2 y2 = 1 on any tangent is: + 144 100 (b) 100 (d) 200

29. The locus of the foot of the perpendicular drawn from x2 y 2 =1 the centre on any tangent to the ellipse + 25 16 is: (a) 25x2 + 16y2 = (x2 + y2)2

x2 y 2 = 1 at Q, S1 and S2 are the foci + 16 9

of the ellipse, then S1Q : S2Q is equal to: (a)

(c)

8

(b)

7 8+ 7

(d)

8- 7

31. Chord of the ellipse

7+ 8 7- 8 7 4

x2 y 2 = 1 whose middle point + 25 16

Ê 1 2ˆ is Á , ˜ meets the minor axis at A and major axis Ë 2 5¯ at B, length of AB (in units) is: (a)

41 5

(b)

2 41 5

(c)

3 41 5

(d)

7 41 5

32. If B is an end of the minor axis of the ellipse

x2

+

y2

= 1, a 2 b2 F1 and F2 are the foci such that the triang BF1F2 is equilateral, then eccentricity of the ellipse is: 1 1 (a) (b) 2 2

2a 2 b 2

28. The product of the perpendiculars from the foci of the ellipse

Ê 3 3ˆ 30. If the normal at P Á 2, 2 ˜ meets the major axis of Ë ¯

(c)

1 3

(d)

3 2

33. Length of the major axis of the ellipse (5x – 10)2 (3x - 4 y + 7) + (5y + 15)2 = is: 4 (a)

10 units 3

(b) 5 units

(c)

20 units 3

(d)

5 units 3

x2 y 2 = 1 at any point + 25 16 meets the line x = 0 at a point Q. R is the image of Q in the line y = x. The circle on QR as a diameter

34. A tangent to the ellipse

Ellipse 19.23

passes through a fixed point whose coordinate are: (a) (5, 5) (b) (4, 5) (c) (1, 1) (d) (0, 0) 35. P is a point on a directrix of an ellipse S is the corresponding focus and C is the centre of the ellipse. The line PC meets the ellipse at A. The angle which the tangent at A makes with PS is a. If the eccentricity of the ellipse is e, then a is equal to (b) tan–1 (1/e) (a) tan–1 e (c)

p 2

(d)

p 4

(a) x =

36. If the tangent with slope –2 of the ellipse

x

+

y

2

=1 a b2 is a normal to the circle x2 + y2 – 4x + 1 = 0, then the maximum value of ab is (a) 1 (b) 2 (c) 3 (d) 4 2

2 37. If a line 3 px + 2 1 - p y

E for all p Π[Р1, 1], then equation of a directrix of the ellipse is:

(b) y =

3 5 10

5 5 (d) y = 3 3 38. If a and b are the natural numbers such that a + b = ab, then equation of the chord of the ellipse x2 + 4y2 = 4 with (a, b) as the mid point is: (a) x + 4y = 2 (b) x + 4y = 10 (c) 4x + y = 2 (d) 4x + y = 10 (c) x =

39. If 2

3 5 10

x2

+

y2

= 1 represents an ellipse with ecsec2 q tan 2 q centricity e and length of the major axis l then (a) e is independent of q (b) l is independent of q (c) el is independent of q (d) l2 – e2 is independent of q

40. 3x2 + 4y2 – 6x + 8y + k = 0 represents an ellipse with eccentricity 1/2, (a) for all value of k (b) for k > 7 (c) k < 7 (d) k = 7

Assertion-Reason Type Questions

41. The tangent at a point P on the ellipse, which is not an extremity of major axis meets a directrix at T. Statement 1: The circle on PT as diameter passes through the focus of the ellipse corresponding to the directix on which T lies. Statement 2: PT subtends a right angle at the focus of the ellipse corresponding to the directix on which T lies. 42. Statement 1: If the normal at an end of a latus rectum x2

y2

+ = 1 meets the major axis at G, a 2 b2 O is the centre of the ellipse, then OG = ae3, e being the eccentricity of the ellipse. of the ellipse

Statement 2: Equation of the normal at a point (a cos q, b sin q) on the ellipse ax by = a2 + b2 + cos q sin q

x2 a2

+

y2 b2

= 1 is

43. Statement 1: In a triangle PQR, if the base QR is fixed and perimeter of the triangle is constant, then vertex P traces an ellipse. Statement 2: If the sum of the distances of a point A from two fixed points B and C is constant > BC, then A traces an ellipse. 44. Statement 1: The locus of a moving point (x, y) satisfying

( x - 2) 2 + y 2 + ( x + 2) 2 + y 2 = 4 is an

ellipse. Statement 2: The distance between (–2, 0) and (2, 0) in 4 units. 45. Statement 1: Product of the perpendiculars drawn x2 y 2 from the foci on any tangent to the ellipse =1 + 15 7 is 7. Statement 2: Foot of the perpendiculars drawn from the foci on any tangent lies on the circle x2 + y2 = 15.

19.24

Complete Mathematics—JEE Main

46. Statement 1: Origin is the centre of the conic x2 + y2 + xy = 1. Statement 2: A point is the centre of a conic if all chords of the conic through this point are bisected at this point. 2 47. For all real p, the line 2 px + y 1 - p = 1 touches a fixed ellipse whose axes are the coordinate axes. Statement 1: Equation of the director circle of the ellipse is 4x2 + 4y2 = 5.

Statement 2: Length of the major and minor axes of the ellipse are 2 and 1 units respectively. 48. Statement 1: If the extremities of the latus rectum of the x2 y 2 ellipse 2 + 2 = 1 (a > b), having positive ordinates a b lies on the parabola x2 = – 2(y – 2), then a = 2. Statement 2: If the length of the latus rectum of the x2 y 2 ellipse 2 + 2 = 1 is equal to the distance between a b

the foci, then the eccentricity e e2 + e – 1 = 0. 49. Statement 1: Two diameters y = mx and y = m¢x x2

+

y2

= 1 are conjugate if mm¢ = -

a2

. a 2 b2 b2 Statement 2: Two diameters of an ellipse are said to be conjugate when each bisects the chords parallel to the other. 50. E:

x2 2

+

y2 2

= 1 and P : y2 = 4bx, a > b.

a b Statement 1: The tangent at the positive end of the minor axis of the ellipse. E passes through the positive end of the latus rectum of the parabola P. Statement 2: If the latus rectum of the parabola P is same as that of the ellipse E, then eccentricity of E is 1/ 2 .

LEVEL 2 Straight Objective Type Questions 51. If chords of contact of the tangent from two points x2 y 2 (x1, y1) and (x2, y2) to the ellipse 2 + 2 = 1 are at a b x1 x2 y1 y2 right angles, then ¥ 2 is equal to a2 b (a) (c)

a

2

b

2

-a b

(b) 4

4

(d)

-b

2

a2 b

4

a4

x2 y 2 + 52. If the normal at the point P(q) to the ellipse =1 14 5 intersect it again at the point Q(2q), then cos q = (a)

-2 3

(b)

2 3

(c)

-6 7

(d)

6 7

53. Let E be the ellipse x2/9 + y2/4 = 1 and C be the circle x2 + y2 = 9. Let P (1, 2) and Q (2, 1) be two points, then (a) Q lies inside C but outside E

(b) Q lies outside both C and E (c) P lies inside both C and E (d) P lies inside C but out side E 54. If CF is perpendicular from the centre of the ellipse x2 y 2 = 1 to the tangent at P, G is the point where + 25 9 the normal at P meets the major axis, then CF ◊ PG = (a) 9 (b) 18 (c) 25 (d) 34 55. If the tangent at P(q) on the ellipse 16x2 + 11y2 = 256 touches the circle x2 + y2 + 2x – 15 = 0, then q = p p (b) (a) 6 3 (c)

2p 3

(d)

5p 6

56. If lx + my + n = 0 is an equation of the line joining the extremities of a pair of semi-conjugate diameters 9l 2 + 4m 2 x2 y 2 = 1, then is equal of the ellipse + n2 9 4 to

Ellipse 19.25

(a) – 1 (c) 1

(b) 0 (d) 2

59. Let d be the perpendicular distance from the centre

57. Let f

R such

that f(x) > 0 " x ΠR, and

x2 f (a 2 + 5a + 3)

+

y2 f (3a + 15)

= 1 represents an ellipse with major axis along the y-axis, then (a) a (– •, – 6) (b) a (2, •) (d) a > 0 (c) a (– 6, 2) 58. Number of points from which two perpendicular tangents can be drawn to both the ellipses E1: x2 a2 + 2

+

(a) 0

y2 b2

= 1 and E2:

x2 a2

+

y2 b2 + 1

= 1 is

of the ellipse

x2

+

y2

= 1 to the tangent at a point a 2 b2 P on the ellipse. If F1 and F2 are two foci of the ellipse then (PF1 – PF2)2 = Ê a2 ˆ (a) 4b2 Á1 - 2 ˜ Ë d ¯

Ê b2 ˆ (b) 4d2 Á1 - 2 ˜ Ë a ¯

Ê b2 ˆ 1 (c) 4a Á ˜ Ë d2 ¯

(d) 4a2

2

60. If from a point (a, 0) three distinct chords of the ellipse x2 + 2y2 = 1 are drawn, which are bisected by the parabola y2 = 4x, then (a) 4 < a < 8

(b) 1

(c) 0 < a
0, 4 – 2a > 0 fi a < 5, a < 2. fi a Œ (– •, 2) 2

So PG : Pg = b2 : a2 fi

Level 1 16. Let the point be ((1/2) cos q, (1/3) sin q), then the slope of the tangent is -

1/ 3 8 cot q = 1/ 2 9

2

Ê xˆ Ê yˆ 24. Á ˜ + Á ˜ = 2(cos2 t + sin2 t) = 2 Ë 3¯ Ë 4¯ x2 y 2 + =1 18 32

which is an ellipse with e2 = fi 16e2 – 7 = 0 fi 16e2 + 7 = 14

32 - 18 7 = 32 16

19.30

Complete Mathematics—JEE Main

25. Let the equation of the ellipse be e=

x2 a2

+

y2 b2

= 1 then

b2 2 a 80 , = 6 fi a = 4 e2 = 1 - 2 fi b2 = . a 3 e 9

fi 25h2 + 16k2 = (h2 + x2)2 Locus of (h, k) is 25x2 + 16y2 = (x2 + y2)2 30. e2 =

Hence the required equation is

7 7 fie= 16 4

x2 9 y 2 = 1 fi 5x2 + 9y2 = 80 + 16 80 (1) a 2 m2 + b2 Equation of the perpendicular from the focus (ae, 0) on it is 1 (2) y = - (x – ae) or my + x = ae m Squaring and adding (1) and (2), eliminating m, (y – mx)2 + (my + x)2 = a2m2 + b2 + a2e2 fi (1 + m2) (x2 + y2) = (1 + m2) a2 fi x2 + y2 = a2 is the required locus 27. An equation of the tangent to the ellipse with slope b is a

P

⎛ 3 3⎞ ⎜ 2, 2 ⎟ ⎝ ⎠

26. Equation of a tangent is y = mx +

b b2 x ± a 2 ¥ 2 + b2 a a As it touches the circle x2 + y2 = r2

S1

x=

Q S2

16 7 Directrix x=−

16 7

Directrix

Fig. 19.22

Coordinates of S1 are (- 7 , 0) and S2 are ( 7 , 0) PQ is the internal bisector of S1 PS2 16 +2 S1Q PS1 7 So = = S2Q PS2 16 -2 7

y=

Ê b2 ˆ 2 2 1+ 2b = r Á ˜ Ë a2 ¯ fi r2 =

2a 2 b 2

. a 2 + b2 28. The product of perpendiculars from foci upon any tangent to the ellipse is equal to square of the semiminor axis (see theory). So the required product is 100. 29. Let P(h, k) be the foot of the perpendicular, O the k and the slope of the centre then slope of OP = h h tangent is - . Equation of the tangent is k y–k= -

h (x – h) k

h h2 + k 2 fiy= - x+ k k which touches the given ellipse 2

2 Ê h2 + k 2 ˆ Ê hˆ if Á k ˜ = 25 ÁË ˜¯ + 16 Ë ¯ k

=

8+ 7

8- 7 31. Equation of the chord is 1/ 4 4 / 25 (1/ 2) x (2 / 5) y + -1 = + - 1 (T = S1) 25 25 16 16 fi 4x + 5y = 4 Ê 4ˆ which meets minor axis at A ÁË 0, ˜¯ and major axis at 5 B (1, 0). So the length of AB =

1+

16 = 25

41 units. 5

32. F1 (– ae, 0), F2 (ae, 0), B (0, b) (F1F2)2 = (B F1)2 fi 4a2e2 = a2e2 + b2 fi 3a2e2 = a2 (1 – e2) 1 1 fie= fi e2 = 4 2 33. Equation of the ellipse is 2

Ê 1 ˆ È 3x - 4 y + 7 ˘ (x – 2)2 + (y + 3)2 = Á ˜ Í ˙ Ë 2¯ Î 5 ˚

2

1 Centre is (2, – 3), eccentricity e = and a directrix 2 is 3x – 4y + 7 = 0.

Ellipse 19.31

Length of the perpendicular from the centre on the directrix is 3 ¥ 2 - 4(-3) + 7 =5 5 a fi = 5 fi 2a = 5 e Hence the length of the major axis is 5 units. 34. The equation of the tangent to the ellipse at P(5 cos q, 4 sin q) is x cos q y sin q =1 + 5 4 which meets x = 0 at Q (0, 4 cosec q) Image of Q in y = x is R (4 cosec q, 0) Equation of the circle on QR as a diameter is x(x – 4 cosec q) + y(y – 4 cosec q) = 0 fi x2 + y2 – 4(x + y) cosec q = 0 which passes through the point of intersection of x2 + y2 = 0 and x + y = 0 i.e the point (0, 0). 35. Let the equation of the ellipse be

x2 a2

+

y2 b2

=1

A be the point (a cos q, b sin q) P

A C

S

x=

a e

Fig. 19.23

4a 2 + b 2 fi 4a2 + b2 = 16 Using A.M. > G. M.

0=–4

4a 2 + b 2 ≥ 4a 2 b 2 2 8 > 2ab. fi ab < 4. 37. Let p = cos q, 0 £ q £ p. The equation of the line becomes (3 cos q) x + (2 sin q)y = 1 or

x y cos q + sin q = 1 1/ 3 1/ 2

which touches the ellipse x2 y 2 + =1 1 1 9 4 eccentricity of the ellipse is 1 1 4 9 = 5 1 3 4 and the equation of a directrix is 1/ 2 5 /3

=

3 5 . 10

x2 y 2 = 1 + 4 1 x(2) y (2) having (a, b) as the mid point is + -1 = 4 1 22 22 + - 1 (T = S1) 4 1 Equation of the chord of the ellipse

-b . a tanq

S is (ae, 0) b tanq b tan q b tanq e So slope of PS = = = a(1 - e2 ) a b2 - ae a¥ 2 e a a tan q. b

4a 2 + b 2 This is a normal to the circle x2 + y2 – 4x + 1 = 0 if it passes through the centre (2, 0) of the circle.

y = – 2x

38. Since a and b are natural numbers such that a + b = ab, a = b = 2

which meets the directrix Êa b ˆ a at P Á , tanq ˜ x= Ë ¯ e e e

=

p 2

36. Equation of a tangent with slope-2 is

y=

Equation of AC is b tan q x. y= a

slope of the tangent at A =

So PS is perpendicular to the tangent at A. fi a =

fi x + 4y = 10. 39. l = 2 sec q, e =

sec2 q - tan 2 q sec 2 q

= cos q

fi el = 2 which is independent of q.

19.32

Complete Mathematics—JEE Main

40. Given equation can be written as 3(x – 1)2 + 4(y + 1)2 = 7 – k which represent an ellipse if k < 7

fi

( x - 1) 2 ( y + 1) 2 =1 + 7-k 7-k 3 4

7-k 7-k 4 = e2 = 1 for all k < 7 So 3 7-k 4 3 41. Tangent at P(a cos q, b sin q) be

46. Let y = mx be a chord of the conic x2 + y2 + xy = 1 through (0, 0), which meets the conic at points for which x2 + m2x2 + mx2 – 1 = 0. fi (1 + m + m2) x2 – 1 = 0 fi x1 + x2 = 0 fi y1 + y2 = 0 Showing that (0, 0) is the mid point of the chord and hence statement-1 is true using statement-2 which is true. 47. Let the equation of the ellipse be

x2 a2

+

y2 b2

=1

a 2 m2 + b2 . Comparing with given equation we get Equation of any tangent is y = mx

x y cos q + sin q = 1 a b

2p

and a2m2 + b2 =

1

a It meets the directrix x = at the point T e Ê a b(e - cos q ) ˆ ÁË , ˜. e e sin q ¯

m= -

Focus S(ae, 0)

fi p2(4a2 – b2) + b2 – 1 = 0 which is true for all p if b2 = 1, 4a2 = b2 = 1 so the equation of the ellipse is

b sin q , Slope of SP = a (cos q - e) Slope of ST =

b(e - cos q ) b2 a 2 (1 - e2 )

4 p2 1- p

2

+ b2 =

1 - p2

1 1 - p2

x2 y 2 + 1 1 =1 4

2

a e sin q (1 - e )

Product of the slope = -

fi a2 ¥

1 - p2

=–1

fi Statement-2 is true using which statement-1 is also true. 42. Statement-2 is false. Equation of the normal is ax by = a2 – b2 cos q sin q In statement-1 L(ae, b2/a) = (a cos q, b sin q) fi cos q = e. ax by So normal at L, = a 2e 2 2 e 1- e which meets the major axis y = 0 at x = ae3 and the statement-1 is true. using it statement-1 is also true. 44. Statement-2 is true but statement-1 is false as the locus represents a straight line. 45. Statement-1 is true as p1p2 = b2 = 7. Statement-2 is also true as the foot of the perpendicular lie on the auxiliary circle which is x2 + y2 = 15 but does not justify statement-1.

Length of the major axis = 2b = 2 and of minor axis = 2a = 1. Thus statement-2 is true. Using it in statement-1 equation of the director circle is x2 + y2 = a2 + b2 or 1 5 = fi 4x2 + 4y2 = 5 and thus x2 + y2 = 1 + 4 4 statement-1 is also true. Ê b2 ˆ ± ae , since 48. Extremities of the latus rectum are Á a ˜¯ Ë they lie on the parabola x2 = – 2 (y – 2) Ê b2 ˆ We have a2e2 = – 2 Á - 2˜ , b2 = a2 (1 – e2) Ë a ¯ fi a2e2 – 2a e2 + 2a – 4 = 0 fi (ae2 + 2) (a – 2) = 0 fi a = 2 Thus statement-1 is true. 2b 2 = 2ae fi b2 = a2e a fi a2(1 – e2) = a2e fi e2 + e – 1 = 0 and the statement-2 is also true but does not justify statement-1. In statement-2

Ellipse 19.33

49. Statement-2 is true by definition of conjugate diameters. Let y = mx and y = m¢x be two conjugate diameters of x2 y 2 the ellipse 2 + 2 = 1. Let (h, k) be the mid point a b of chord whose step is m then hx a2

+

ky

–1=

b2

a2

+

k2

–1

b2

-b 2

-b 2

= m¢ fi m¢m =

a2m

a2

1 1 fie= . 2 2

Thus statement-2 is true.

51. Chords of contacts are a2

yy1 b2

= 1 and

xx2 a2

(0, 2)

(3, 0)

E

Fig. 19.24

So P lies outside E and inside C. For Q(2, 1), C: 4 + 1 – 9 < 0, E:

a

2

4 1 + –10 9 4

For P(1, 2), C: 1 + 4 – 9 < 0, E:

fi CF = +

C

P

x cos q y sin q + =1 5 3

Level 2

+

(0, 3)

. (using statement-2)

so statement-1 is false. In statement-2 if x = b and x = ae represent the same line then b = ae b = e fi a2(1 – e2) = a2e2 fi a

xx1

fi 18 cos2q – 9 cos q – 14 = 0 fi (3 cos q + 2) (6 cos q – 7) = 0 fi cos q = – 2/3. 53. C: x2 + y2 – 9 = 0 E: x2/9 + y2/4 – 1 = 0

Q

and thus statement-1 is false. 50. Tangent at the positive end of the minor axis of E is y = b which meets the parabola y2 = 4bx at the point Êb ˆ ÁË , b˜¯ which is not an end of the latus rectum of P. 4

fi e2 =

14(2 cos 2 q - 1) 5 ¥ 2 sin q cos q =9 cos q sin q

b2 h - b2 fi Locus of (h, k) is y = x a2 k a2 m

fim=–

fi

h2

fi

F

x -b 2 x ¥ 2 ¥ 2 ¥ 1 =–1 y2 y1 a

P C G

x1 x2 -a 4 fi = 4 y1 y2 b

Fig. 19.25

52. Normal at P ( 14 cos q , 5 sin q ) to the ellipse is 14 x 5y = 14 – 5 = 9. cos q sin q Since it passes through Q (2q) 14 ¥ 14 cos 2q 5 ¥ 5 sin 2q =9 cos q sin q

9 cos 2 q + 25 sin 2 q

Equation of the normal at P is 5x 3y = 25 – 9 = 16 cos q sin q Ê 16 cos q ˆ , 0˜ Coordinates of G are ÁË ¯ 5 fi PG =

3 9 cos 2 q + 25 sin 2 q 5

19.34

Complete Mathematics—JEE Main

Hence CF ◊ PG = 9. 55. Centre of the circle is (– 1, 0) and radius is 4. Equation of the tangent at P(q) to the ellipse is x y cos q + sin q = 1 4 16 / 11 If this touches the circle, then - cos q -1 4 =4 cos 2 q sin 2 q + 16 256 /11

1

d=

Ê cos 2 q 11sin 2 q ˆ fi (cos q + 4) = 256 Á 16 + 256 ˜ Ë ¯ 2

2

fi 4 cos q – 8 cos q – 5 = 0 fi (2 cos q – 5) (2 cos q + 1) = 0 fi cos q = – 1/2 fi q = 2p/3 56. Let the extremities of a pair of semi-conjugate diameters of the ellipse be P(3 cosq, 2 sinq) and Q (3 cos a, 2 sin a), C being the centre. (3 cos a, 2 sin a) Q

fi (a + 6) (a – 2) > 0 fi a œ (– 6, 2). 58. The director circle of E1 is x2 + y2 = a2 + 2 + b2 and of E2 is x2 + y2 = a2 + b2 + 1. Both are concentric circle such that one lies inside the other so that they have no point in common. 59. Equation of the tangent at P(a cos q, b sin q) is x y cos q + sin q = 1 a b

P(3 cos q, 2 sin q)

cos 2 q sin 2 q + 2 a2 b

fi

1 d2

=

cos 2 q a2

+

sin 2 q b2

(1)

F1 (– ae, 0), F2 (ae, 0) aˆ Ê fi P F1 = e Á a cosq + ˜ = a (1 + e cos q) Ë e¯ P F2 = a (1 – e cos q) So (P F1 – P F2)2 = 4 a2e2 cos2 q 1 1 - 2 2 2 2 d b = 4 (a – b ) ¥ 1 1 - 2 2 a b Ê b2 ˆ 1 =4a Á ˜. Ë d2 ¯ 2

C(0,0)

60. Let the middle point of the chord be (t2, 2t) It must lie inside the ellipse So t4 + 8t2 – 1 < 0

Fig. 19.26

Then slope of CP ¥ slope of CQ = fi

4 9

4 2 2 tan q ¥ tan a = 9 3 3

fi 1 + tan q tan a = 0 fi a – q = p/2 fi coordinates of Q are (– 3 sin q, 2 cos q) and the equation of PQ is 2(sin q – cos q) x – 3 (cos q + sin q) y + 6 = 0 which represents lx + my + n = 0, then l m n = = 2(sin q - cos q ) -3(cos q + sin q ) 6 So

9l 2 + 4m 2 n2

= (sin q – cos q)2 + (cos q + sin q)2 = 2.

57. We must have f(a2 + 5a + 3) < f(3a + 15) fi a2 + 5a + 3 > 3a + 15 [∵ f is strictly decreasing] fi a2 + 2a – 12 > 0

fi t2 Œ (0, –4 +

17 ) Equation of the chord of the ellipse with (t2, 2t) as the mid-point is t2x + 4ty = t4 + 8t2 (T = S1) Since it passes through (a, 0) at2 = t4 + 8t2 fi t4 + (8 Рa)t2 = 0 fi t2 = 0 or t2 = a Р8 fi a = t2 + 8 fi a Π(8, 4 + 17 )

Previous Years’ AIEEE/JEE Main Questions 1. Equation one of directrix of ellipse with centre at the origin is x = a/e a \ = 4 fi a = 2 [∵ e = 1/2] e Ê 1ˆ Also, b2 = a2(1 – e2) = 4 Á1 - ˜ = 3 Ë 4¯

Ellipse 19.35

Thus, equation of required ellipse is

fi 3x2 + 8x + 4y2 = 16

x2 y 2 = 1 or 3x2 + 4y2 = 12 + 4 3

4ˆ 4 Ê 16 16 64 = fi Á x + ˜ + y2 = + Ë 3¯ 3 3 9 9

2

2. AB = 2 b sin q AD = 2 a cos q Area of rectangle = 4ab sin q cos q

2

4ˆ Ê ÁË x + ˜¯ y2 3 + fi =1 64 3 64 ¥ 9 4 9

y A(a cos q, b sin q)

D

x

0 C

6. The required ellipse passes through (4, 0) and (2, 1). Let equation of ellipse be

B

y

Fig. 19.27

= 2ab sin 2q Greatest area is obtained when q = p/4 and greatest area = 2ab. -b 3. Slope of BF is m1 = ae

a2 F (ae, 0)

Fig. 19.28

-b b and that of BF¢ is m2 = = - ae ae As BF ^ BF¢, m1 m2 = – 1 fi

b

2

2 2

a e

=1

fi a2(1 – e2) = a2 e2 1 fi 2e2 = 1 or e = 2 4. We have 2ae = 6 and 2b = 8 fi ae = 3, b = 4 Now, b2 = a2 (1 – e2) fi 16 = a2 – 9 fi a = 5 \ e = 3/5 5. Equation of the ellipse is x2 + y2 =

1 (4 – x)2 4

x

1

(2, 1)

O

2

(4, 0)

x

Fig. 19.29

x2

B (0, b)

F¢ (-ae, 0)

64 8 = . 9 3

Length of the semi-major axis =

+

y2 b2

=1

As it pass through (4, 0) and (2, 1) 16 4 1 = 1 and 2 + 2 = 1 2 a a b fi a2 = 16 and b2 = 4/3 Thus, required ellipse is x2 y 2 + =1 fi x2 + 12y2 = 16 16 4 / 3 7. Let the equation of the ellipse be x2 a2

+

y2 b2

=1

Since it passes through (– 3, 1), we have 9 1 3 Ê 2ˆ + 2 = 1, also b2 = a2 Á1 - ˜ = a2 2 Ë 5¯ 5 a b fi9+

32 2 32 5 = a2 fi a2 = ,b = . 3 5 3

and the required equation is x 2 y 2 = 1 fi 3x2 + 5y2 = 32 + 32 32 3 5 8. Minor axis is along the x-axis and its length is 2(1) = 2. Major axis is along the y-axis and its length is 2(2) = 4.

19.36

Complete Mathematics—JEE Main

Equation of the ellipse is x

2

2

+

y

2 2

= 1 fi 4x2 + y2 = 4

1 2 9. We have x2 + 4y2 = 4c2 as the ellipse. Points of intersection with the circle x2 + y2 = 9a2 are given by 9a2 – y2 + 4y2 = 4c2 fi 3y2 = 4c2 – 9a2 fi 4c2 > 9a2 fi 3a < 2c fi 9a2 + 2c2 < 6c2, 9ac < 6c2 fi 9ac – (9a2 + 2c2) < 0 1 3- 2 16 - b 2 10. e12 = = , e22 = 3 3 16 (e1e2)2 =

Ê 1ˆ 16 - b 2 = Á ˜ Ë 2¯ 48

x2 + 3y2 = 6 is y = mx ± 6m 2 + 2 If (h, k) is the foot of the perpendicular from (0, 0), the centre of the ellipse on this tangent then Ê kˆ k = mh ± 6m 2 + 2 and Á ˜ m = –1 Ë h¯ Eliminating m we get (h2 + k2)2 = 6h2 + 2k2 Required locus is (x2 + y2)2 = 6x2 + 2y2 14. Coordinates of F1, F2 and B are (ae, 0), (– ae, 0) and (0, b) respectively. Slope of B F1 = – b/ae and slope of B F2 is b/ae. As F1 B ^ F2 B, Ê -b ˆ Ê b ˆ 2 2 2 ÁË ˜¯ ÁË ˜¯ = – 1 fi b = a e ae ae

2

fi a2 (1 – e2) = a2e2 fi e2 = 1/2 15. An equation of tangent to the ellipse

fi 16 – b2 = 12 fi b2 = 4 fi b = 2 and the required length is 4. 11. Let P(x1, y1) be a common point. Tangents to the curves at Pare xx yy yy1 = 8(x + x1) and 1 + 1 = 1 a 4

x2 y 2 = 1 is + 16 81 x y cos q + sin q = 1 4 9 y

8 Ê -4 x1 ˆ ¥ Product of the slopes = =–1 y1 ÁË a y1 ˜¯ fi

y12

B 2

32 x1 = a

(1)

(4 cosq, 9 sin q) 1

As P lies on the parabola y2 = 16x. fi

y12

= 16x1

(2)

From (1) and (2) we get a = 2 1 12. Slope of the tangent is - , so its equation is 2 1 1 y = - x ± 9 ¥ + 4 fi x + 2y = ±5 2 4 x2 y 2 = 1 at (x1, y1) If it meets the parabola + 9 4 its equation is So

x

O

A

Fig. 19.30

Coordinates of A and B are (4 sec q, 0) and (0, 9 cosec q) respectively. 1 36 36 Area of DOAB = = 2 |sin q cos q | sin 2q So minimum value of area 36. 16. We have a2 = 9, b2 = 5, e2 = 1 -

Coordinates of L are (ae, b2/a) = (2, 5/3)

xx1 yy1 =1 + 9 4

Y B

x1 y1 1 = =± 9 8 5

Ê 9 8ˆ fi (x1, y1) = Á , ˜ taking +ve sign. Ë 5 5¯ 13. Equation of any tangent to the ellipse

5 4 2 = fie= . 9 9 3

L C

O

A L¢ D

Fig. 19.31

Ellipse 19.37

Equation of tangent at L is 2x 5 Ê y ˆ + Á ˜ =1 9 3Ë 5¯

1. 2

or 2x + 3y = 9 It meets the axes at A(9/2, 0) and B(0, 3). Area of quadrilateral = 4[area of DOAB] È1 Ê 9ˆ ˘ = 4 Í Á ˜ (3)˙ = 27 (unit)2 Î 2 Ë 2¯ ˚ 17. We have distance between the foci = 2ae 2b 2 length of the latus rectum = a 1 Ê b2 ˆ So 2 ae = Á 2 ˜ fi b2 = 2 a2e 2Ë a ¯ fi a2 (1 – e2) = 2 a2e fi e2 + 2e = 1fi (e + 1)2 = 2 fi e = 2 -1 18. An equation of tangent at (3 3 cos q , 3 sin q ) to the ellipse

1 2 1 2 x + y = 1 is 27 3

x cos q + 3 y sin q = 3 3 . It meets the axes in A (3 3 sec q , 0) , and B (0, 3 cosec q ) Let A = area of DOAB is =

Previous Years’ B-Architecture Entrance Examination Questions

1 |(3 3 sec q )( 3cosecq )| 2

9 ≥9 | sin 2q |

Thus, least value of A is 9 which is attained when q = p/4, 3p/4, 5p/4, or 7p/4.

a 1 = 4 ¥ 2ae fi e = e 2

Ê 1ˆ a = 2, so b2 = 4 Á1 - ˜ = 3 Ë 4¯ Hence the required equation is x2 y 2 = 1 fi 3x2 + 4y2 = 12 + 4 3 Ê 3ˆ 2. Tangent at Á1, ˜ to the ellipse 3x2 + 16y2 = 12 is Ë 4¯ Ê 3ˆ yÁ ˜ Ë 4¯ x ◊1 =1 + 3 4 4 fi x + 4y = 4 which intersects the curve y2 + x = 0 at points for which y2 + (4 – 4y) = 0 fi (y – 2)2 = 0 fi y = 2 and the points of intersection is (– 4, 2) exactly one point. 3. Equation of the tangent with slope tan 135° to the x2 y 2 = 1 is y = – x + 16(1) + 9 ellipse + 16 9 Ê 16 9 ˆ fi x + y = 5 which meets the given ellipse at Á , ˜ Ë 5 5¯ 4. Suppose the tangent intersect at (h, k), then equation of chord of contact is 4hx + ky = 5. But equation of chord of contact is: 2x + y = 3 \

4h k 5 = = 2 1 3

fi h = 5/6, k = 5/3. Ê 5 5ˆ Thus, required point is Á , ˜ . Ë 6 3¯

Hyperbola

Definition A hyperbola is the locus of a point which moves in a plane in such a way that the ratio of its distance from a fixed point in the plane to its distance from a fixed line in the plane (not passing through the fixed point) is a constant and equal to e, where e > 1. (i) Fixed point is called a focus of the hyperbola. (ii) Fixed line is called a directrix of the hyperbola. (iii) Constant ratio e, is called the eccentricity of the hyperbola. Standard equation of a hyperbola referred to its princix2 y 2 pal axes along the coordinate axies is 2 - 2 = 1 where a

b

b2 = a2 (e2 – 1). y

x = a/e L B (0, b) M

(-ae, 0) (-a, 0)

O (0, 0)

A¢

S¢

P A (a, 0)

B¢ (0, -b) x = - a/e

L¢

Hyperbola (e >1)

In the figure (i) A¢A is the transverse axis of the hyperbola along the axis of x and is of length 2a. A(a, 0), A1(–a, 0) are the vertices of the hyperbola. (ii) b¢b is the conjugate axis of the hyperbola along the axis of y and is of length 2b. (iii) O(0, 0) is the centre of the hyperbola. a 2 + b2

is the eccentricity of the hyperbola. a2 From symmetry we observe that if S ¢(–ae, 0) be taken as

(iv) e =

the focus and x = -

bola is described. So (v) S¢(–ae, 0) and S(ae, 0) are the two foci of the hyperbola. a a (vi) x = - and x = are the two directrices of the hye e perbola. (vii) x = ae and x = –ae are the two latera recta of the hyperbola. (viii) Latus rectum x = ae meets the hyperbola at the points Ê b2 ˆ Á ae,± a ˜ so length of each latus rectum is 2b2/a. Ë ¯

Fig. 20.1

Let P(x, y) be any point on the hyperbola. S (ae, 0) be a focus and x =

a be a directrix then according e

to the definition aˆ Ê (x – ae)2 + y2 = e2 Á x - ˜ Ë e¯ fi

x2 a2

-

Illustration

1

Find the eccentricity, length of a latus rectum, equations of the latera recta of the hyperbola

a 2 (e2 - 1)

=1

Since e > 1, e2 – 1 > 0 so a2(e2 – 1) = b2 and the required 2 2 equation is x 2 - y2 = 1.

b

x2 y 2 = 1. 16 9

2

Solution Let the equation of the hyperbola be where a2 = 16, b2 = 9.

y2

a

a is taken as the directrix, same hypere

fi e2 =

16 + 9 5 fie= . 16 4

x2 y 2 = 1, a 2 b2

20.2

Complete Mathematics—JEE Main 2b 2 2¥9 9 = = . a 2 4

Lengths of a latus rectum =

Equations of the latera recta are x = ± ae fi x = ± 4¥

Illustration

5 fi x = ± 5. 4

2

( x - 2) 2 ( y - 1) 2 =1 25 11

which can be written as

fie=

X2 Y2 =1 a 2 b2

4

Illustration

If y = mx + 1 is a tangent to the hyperbola 4x2 – 25y2 = 100, find the value of 25m4 + 5m2 + 1.

5

Find the equation of the normal to the hyperbola 4x2 – 9y2 = 144 at the point whose eccentric angle q is p /3. Solution: Equation of the hyperbola is

So the coordinates of the foci are (X = ± ae, Y = 0) = (x – 2 = ± 6, y – 1 = 0) = (8, 1) and (– 4, 1).

3

e1 is the eccentricity of the ellipse eccentricity of the hyperbola

x2 y 2 =1 36 16

Coordinates of the point whose eccentric angle is p/3 are (6 sec (p/3), 4 tan (p/3)) = (12, 4 3 ) The required equation of the normal is x2 y 2 + = 1 and e2 is the 25 b 2

x2 y 2 = 1 such that e1e2 = 1. 16 b 2

Find the value of b. 25 - b 2 16 + b 2 and e2 = Solution: e1 = 25 16 2

x2 y 2 =1 25 4

y = mx + 1 is a tangent to the hyperbola fi (1)2 = 25m2 – 4 fi 5m2 = 1 fi 25m4 + 5m2 + 1 = 3.

Illustration

25 + 11 25

6 . 5

Illustration

x - x¢ y - y ¢ =0 + x¢ / a 2 y ¢ / b 2

Solution: Equation of the hyperbola is

where X = x – 2, Y = y – 1, a2 = 25, b2 = 11. So centre of the hyperbola is (X = 0, Y = 0) = (2, 1) eccentricity e =

and at (x ¢, y ¢) is

ax by = a2 + b2 + secq tan q

4. The condition that the line y = mx + c is a tangent to the hyperbola is c2 = a2m2 – b2. So equation of any tangent to the hyperbola (not parallel to y-axis) can be written as y = mx ± a 2m2 - b2

Find the centre, foci and the eccentricity of the hyperbola. 11x2 – 25y2 – 44x + 50y – 256 = 0. Solution: The equation of the hyperbola can be written as 11(x – 2)2 – 25(y – 1)2 = 275 fi

3. An equation of the normal at the point ‘q’is

2

e1e2 = 1 fi (25 – b ) (16 + b ) = 25 ¥ 16. fi 9b2 – b4 = 0 fi b2 = 9 fi b = ± 3. Some Properties and Standard Results for the Hyperbola x 2 y2 =1 a 2 b2 1. The parametric equation of the hyperbola or the coordinates of any point on the hyperbola are x = a secq, y = b tan q. The point is denoted by ‘q’. q is called the eccentric angle of the point. 2. An equation of the tangent at the above point ‘q’ is

x y x x ¢ y y¢ sec q - tan q = 1 and at (x¢, y¢) is 2 - 2 = 1. a b a b

6x 4 y = 36 + 16 + 2 3

fi 3 3 x + 4y = 52 3 .

5. An equation of the chord of contact of the point (x ¢ , y ¢ ) joining the points of contact of the tangents drawn xx ¢ yy¢ from (x ¢, y ¢) to the hyperbola is 2 - 2 = 1 a b 6. An equation of the chord of the hyperbola whose midpoint is (x ¢, y ¢) is T = S ¢, where T∫

xx ¢ a2

-

yy¢ b2

2 2 – 1 and S ¢ ∫ x ¢2 - y ¢2 – 1.

a

b

7. An equation of pair of tangents drawn from a point P(x ¢, y ¢) outside the hyperbola to the hyperbola is SS ¢ = T 2, where. S∫

x2 y2 –1 a2 b2

Illustration

6

Find the equation of the chord of contact of the point (5, 1) to the hyperbola, x2 – 4y2 = 16. Also find the mid-point of this chord.

Hyperbola 20.3 2 2 Solution: Equation of the hyperbola is x - y = 1 16 4

Illustration

Equation of the chord of contact of (5, 1) is 5 x - y(1) = 16 4 fi 5x – 4y = 16. (1) If (a, b) is the mid-point of this chord then its equation is ax by a2 b2 – 1 –1= 16 4 16 4

(2)

If y = mx + 2 5 is a tangent to the hyperbola 4x2 – y2 = 16. Find the equation of a diameter of the hyperbola bisecting the chords parallel of the tangent. Solution: Equation of the hyperbola can be written as x2 y 2 =1 4 16 y = mx + 2 5 touches the hyperbola.

From (1) and (2) we get 80 16 ,b= . 21 21 So that coordinate of the mid-point of the chord are a=

Ê 80 16 ˆ ÁË , ˜¯ . 21 21 8. Director circle of a hyperbola is the locus of the point of intersection of the tangents to the hyperbola which intersect at right angles and its equation is x2 + y2 = a2 – b2. 9. A diameter of a hyperbola is the locus of mid-points of a system of parallel chords of the hyperbola and its equation is b2 y = 2 x where m is the slope of the parallel chords of the a m hyperbola which are bisected by it. Two diameters of a hyperbola are said to be conjugate when each bisects the chords parallel to the other. Thus two diameters y = mx and y = m¢x of the hyperbola are conjugate b2 if mm¢ = 2 . a

Illustration

8

fi (2 5 )2 = 4m2 – 16 fi m2 = 9 fi m = ± 3 So the required equation of a diameter is 16 y= x fi 4x ± 3y = 0. 4(±3) 10. Asymptotes If the length of the perpendicular let fall from a point on a hyperbola. to a straight line tends to zero as the point on the hyperbola moves to infinity along the hyperbola then the straight line is called an asymptote of the hyperbola. y y = -ab x

y = ab x

A¢(-a, y)

A(a, 0)

Fig. 20.2 (i) Equation of the asymptotes is

7

Find the points on the hyperbola

x2 y 2 = 2 from which a 2 b2

two perpendicular tangents can be drawn to the circle x2 + y2 = a2. Solution: Director circle of x2 + y2 = a2 is x2 + y2 = 2a2 or x2 y2 + 2 =1 2 2a 2a

(1)

Equation of the hyperbola can be written as x2 y2 = 1 2a 2 2b 2

(2)

Since this represents the director circle of the given circle. Subtracting (2) from (1) we get 1 ˆ 2 Ê 1 ÁË 2 + 2 ˜¯ y = 0 fi y = 0 2a 2b

(

)

So the required points are ± 2a, 0 .

x

0

i.e.

x2 a2

-

y2 b2

x y =0 ± a b

=0

Ê bˆ

(ii) Angle between two asymptotes is 2 tan–1 Á ˜ . Ë a¯ Asymptotes are at right angle if and only if a = b. (iii) If angle between two asymptotes is 2q, then eccentricity of the hyperbola is sec q. 11. Rectangular hyperbola A hyperbola in which the lengths of the transverse and conjugate axes are equal is called a rectangular hyperbola. (i) Equation of a rectangular hyperbola is x2 – y2 = a2. (ii) Eccentricity of the rectangular hyperbola is 2 . (iii) Asymptotes of a rectangular hyperbola are at right angles. (iv) Rotation of the system of coordinates through an angle p/4 in the clockwise direction (i.e. the axes are taken along the asymptotes of the hyperbola) gives another form of the equation of the rectangular hyperbola i.e. xy = c2. (v) Some facts about rectangular hyperbola xy = c2 (a) Vertices (c, c) and (– c, – c) (b) Foci ( 2c, 2c) and (- 2c, - 2c)

20.4

Complete Mathematics—JEE Main (c) Directrices x + y = ±

(b) If e1 and e2 are the eccentricities of the hyperbola and its

2c

conjugate, then

(d) length of the latus rectum = 2 2 c (e) parametric equation: x = ct, y = c/t.

9

Illustration

If the circle x2 + y2 = a2 intersects the hyperbola xy = 25 in four points, then find the product of the ordinates of these points. Solution: The ordinates of the points of intersection of x2 + y2 = a2 and the hyperbola xy = 25 are given by Ê 25 ˆ ËÁ y ¯˜

2

+ y2 = a2 or y4 – a2y2 + (25)2 = 0

If y1, y2, y3, y4 are the roots of this equation then y1 y2 y3 y4 = (25)2 = 625. 12. Conjugate hyperbola Two hyperbolas such that the transverse and conjugate axes of one hyperbola are, respectively, the conjugate and transverse axis of the other are called conjugate hyperbolas of each other. (a) Equation of the hyperbola conjugate to

x

2

b

2

-

y

x2 a2

-

y 2 = 1 is b2

2

=1

a2

1 e12

+

1 e22

= 1.

(c) The foci of a hyperbola and its conjugate are concentric and form the vertices of a square. 13. Similar hyperbolas Two hyperbolas are said to be similar if they have the same eccentricity. 14. Equal hyperbola two hyperbolas are equal if they are similar and have the same latus rectum. 15. Some more properties of the hyperbola x2/a2 – y2/b2 = 1 (i) If PN be the ordinate of a point P on the hyperbola and the tangent at P meets the tranverse axis in T, then ON. OT = a2, O being the origin. (ii) If PM be drawn perpendiculars to the conjugate axis from a point P on the hyperbola and the tangent at P meets the conjugate axis in T, then OM. OT = –b2; O, being the origin. (iii) If the normal at P on the hyperbola meets the transverse axis in G, then SG = eSP; S being a foci and e the eccentricity of the hyperbola. (iv) The tangent and normal at any point of a hyperbola bisect the angle between the focal radii to that point. (v) The locus of the feet of the perpendiculars from the foci on a tangent to a hyperbola is the auxillary circle. (vi) The product of the perpendiculars from the foci on any tangent to a hyperbola is constant and equal to b2.

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Example 1: If e1 and e2 are respectively the eccentricities of 2

the conics

2

2

2

x y x y = 1 and = 1 then e1e2 is equal to + 25 11 16 7

(a)

10 9

(b)

4 3

(c)

9 10

(d)

8 5

(a) ±3 5

(b) ± 5

(c) ±2 5 Ans. (b)

(d) none of these

Solution: Equation of the tangent is y =

Ans. (c) Solution: e1 = fi e 1e 2 =

Example 2: If x – 5 y + c = 0 is a tangent to the x 2 y2 = 1, then the value of c is hyperbola 25 4

6 25 + 11 = , e2 = 5 25

6 3 9 ¥ = 5 4 10

3 16 - 7 = . 4 16

1 5

2

x+

c 5

2

Ê c ˆ Ê 1 ˆ so Á ˜ = 25 Á ˜ – 4 fi c = ± 5 Ë 5¯ Ë 5¯ Example 3: If the tangent at the point P(a sec q, b tan q) x2 y 2 to the hyperbola 2 - 2 = 1 passes through the point a b

Hyperbola 20.5

where a directrix of the hyperbola meets the positive side of the transverse axis, then q is equal to (b) tan–1(1/e) (a) cos–1(1/e) –1 (c) cot (1/e) (d) sec–1(1/e) Where e is the eccentricity of the hyperbola. Ans. (a) Solution: Equation of the tangent at P(a secq, b tanq) to the hyperbola is x y (1) sec q - tan q = 1 a b which passes through the point of intersections of the direca and y = 0, the transverse axis of the hyperbola. trix x = e Êa ˆ i.e. Á , 0˜ . So from (1) we have Ëe ¯ a 1 secq = 1 fi secq = e fi cosq = a¥e e fi q = cos–1(1/e). Example 4: The circle described on the line joining x2 y 2 the foci of the hyperbola = 1 as a diameter passes 16 9 through an end of the latus rectum of the parabola y2 = 4ax, the length of the latus rectum of the parabola is (a) 2 5 units

(b) 5 units

(d) 5 5 units (c) 4 5 units Ans. (c) Solution: Eccentricity of the hyperbola is given by 16 + 9 5 = . e= 16 4 5 ˆ Ê Coordinates of the foci are Á ± 4 ¥ , 0˜ = (± 5, 0). Ë 4 ¯ So the circle described on the line joining the foci is x2 + y2 = 25. If it passes through an end (a, ±2a) of the latus rectum of the parabola y2 = 4ax then a2 + 4a2 = 25 fi a2 = 5 fi a = 5 . Length of the latus rectum = 4a = 4 5 . Ê 1ˆ Example 5: The curve represented by x = 5 Á t + ˜ , Ë t¯ Ê 1ˆ y = Á t - ˜ , t π 0 is Ë t¯ (a) (b) (c) (d)

a point of straight lines an ellipse a hyperbola a rectangular hyperbola.

Ans. (c) 2

Ê xˆ Solution: Á ˜ – y2 = 4 Ë 5¯ 2 2 fi x - y =1 100 4 which is a hyperbola.

Example 6: If the distance between two directrices of a rectangular hyperbola is 15, then the distance between its foci in units is: (a) 15 2 (b) 30 (c) 60 (d) 45 Ans. We know that the eccentricity of a rectangular hyper2a , bola is 2 = e. Distance between the directricess = e where 2a is the length of the transverse axis. So

2a = 15 fi 2a = 15 2 e

Distance between the foci = 2ae = 15 2 ¥

2 = 30.

Example 7: If two perpendicular tangents can be drawn from the point (a, b) to the hyperbola x2 – y2 = a2, then (a, b) lies on (a) y = ±x (b) x2 + y2 = a2 2 2 2 (c) x + y = 2a (d) y2 = 4ax Ans. (a) Solution: An equation of a tangent to the hyperbola x2 – y2 = a2 is y = mx + a m 2 - 1 Equation of the tangent perpendicular to (1) is y= -

(1)

1 1 x+a -1 m m2

fi my + x = a 1 - m 2

(2)

We must have m2 – 1 ≥ 0 and 1 – m2 ≥ 0 fi m2 = 1 fi m = ±1 and thus (a, b) lies on y = ± x. Example 8: The tangent at an extremity (in the first x2 y 2 quadrant) of latus rectum of the hyperbola = 1 4 5 meets x-axis and y-axis at A and B respectively. Then (OA)2 – (OB)2, where O is the origin, is equal to (a) (c) 4 Ans. (a)

20 9

(b)

16 9

(d) -

4 3

Complete Mathematics—JEE Main

20.6

Solution: Coordinates of the extremity of latus rectum Ê b2 ˆ Ê 5ˆ in the first quadrant is P Á ae, ˜ = Á 3, ˜ . Equation of the a Ë 2¯ Ë ¯ tangent at P is

3x 5 Ê y ˆ - Á ˜ = 1 or 3x – 2y = 4. 4 2 Ë 5¯

Solution: Let the coordinates of a point P on the hyperbola be (x1, y1). Equation of the tangent at P to the hyperbola is 4 4 xx1 – yy1 = 4 fi a1 = , b1 = - . x1 y1 Equation of the normal is

Ê4 ˆ which meets x-axis at A Á , 0˜ and y-axis at B(0, –2). Ë3 ¯ So (OA)2 – (OB)2 =

16 20 -4 = - . 9 9

Example 9: If PQ is a double ordinate of the hyperbola y2 = 1 such that OPQ is an equilateral triangle, O a 2 b2 being the centre of the hyperbola, then eccentricity e of the hyperbola satisfies. 2 3 (a) e = (b) e = 2 3 x2

(c) e >

2

(d) 1 < e
a2 2

fi 3a (e – 1) > a fi 3e > 4 fi e >

2

.

3 Example 10: If the tangent and normal to the hyperbola x2 – y2 = 4 at a point cut off intercepts a1 and a2 respectively on the x-axis, and b1 and b2 respectively on the y-axis, then the value of a1a2 + b1b2 is (a) –1 (b) 0 (c) 4 (d) 1 Ans. (b)

through the focus of the ellipse

x2 b2 x2 a2

eccentricity of the hyperbola is 2

(b)

3

(c) 3/2

(d)

3/ 2

(a)

Ans. (b)

+

y2 a2 y2 b2

= 1 passes = 1, then

Hyperbola 20.7

Solution: Let e1 and e2 be the eccentricities of the ellipse and hyperbola respectively. Since the hyperbola passes through the focus (± ae1, 0) fi b = ± ae1 For the ellipse b2 = a2(1 – e12) fi e1 = 1/ 2 For the hyperbola a2 = b2(e22 – 1) fi e2 =

Ans. (c) Solution: e12 = fi

3.

Example 13: Which one of the following is independent of a (0 < a < p/2) for the hyperbola (a) eccentricity (c) abscissa of foci Ans. (c) Solution: Eccentricity =

cos 2 a + sin 2 a cos 2 a

=

x

2 2

-

y

2

2

=1

cos a sin a (b) equation of a directrix (d) abscissa of vertices

sec2 a

cos a = ± cos3a sec a Abscissa of foci = cosa (sec a) = 1 which is independent of a abscissa of vertices = ± cos a. Example 14: If e1 and e2 are the eccentricities of the

Equation of the directrix is x = ±

hyperbolas 1 e12

+

1 e22

x2 y 2 y 2 x2 = 1 and = 1 respectively, then 25 16 25 16

5 4

e12

+

1 e22

= 1.

Example 15: The normal at P(x1, y1) on the hyperbola x2 a2

-

y2 b2

= 1 meets the coordinate axes at A and B. If O, is

the origin and e, the eccentricity of the hyperbola, then (b) OB = e2y1 (a) OA = e2x1 2 (c) OA = e y1 (d) OB = e2x1 Ans. (a) Solution: Equation of the normal at P(x1, y1) to the hyperbola

x2 a2

-

y2 b2

= 1 is

x - x1 y - y1 = 0. + x1 y1 a2 b2 Ê x1 ˆ a 2 + b2 fi OA = b2 Á 2 ˜ + x1 = x1 = e2x1 Ëa ¯ a2 Ê y1 ˆ a 2 + b2 a 2 e2 y y1 OB = a2 Á 2 ˜ + y1 = = 1 Ëb ¯ b2 b2

is equal to

(a)

1

25 + 16 2 25 + 16 , e2 = 25 16

4 5 1 (d) 2 (b)

(c) 1

LEVEL 1 Straight Objective Type Questions Example 16: If the latus rectum of a hyperbola subtend an angle of 60° at the other focus, then eccentricity of the hyperbola is (a) 2

(b)

a2

-

y2 b2

y

3 +1 2

(c) 2 3 (d) 3 Ans. (d) Solution: Let the equation of the hyperbola be x2

F1(–ae, 0), F2(ae, 0) be the foci, e being the eccentricity. Let LF2L¢, the latus rectum subtend an angle of 60° at F1, then slope of LF1 is tan 30°.

F1 (-ae, 0)

Ê b2 ˆ L Á ae, ˜ a¯ Ë

F2 (ae, 0)

30° 0

=1

L¢

Fig. 20.5

x

20.8

Complete Mathematics—JEE Main

So tan 30° =

1

fi

Eliminating q, we get the required locus

b2 / a 2ae =

3

a 2 (e2 - 1) 2

2a e

1

e-

3

e2 - 1 2e

fi b2x2 – a2y2 + 4x2y2 = 0 Example 19: If the slope of a tangent to the hyperbola

2

1 ˆ Ê 1 e fi Áe = 1+ ˜ Ë 3¯ 3 3

e –1=

fi

=

2

2

fi

=

2 3

fie=

x2

3

36m 2 - 16

(1)

2

x x + 36 2 - 16 y y

(x2 + y2)2 = 36x2 – 16y2.

fi

Example 18: The locus of the middle points of the portions of the tangents of the hyperbola. x2 a2

-

y2 b2

= 1 included between the axes is

(a) a2x2 – b2y2 = x2y2 (b) b2x2 – a2y2 = x2y2 (c) b2x2 – a2y2 + 4x2y2 = 0 (d) a2x2 – b2y2 + 4x2y2 = 0 Ans. (c) Solution: Equation of a tangent to the hyperbola is x y sec q - tan q = 1 which meets the axes at A (a cosq, 0) a b and B (0, –b cotq). If (h, k) is the midpoint of AB then h =

(b) (1, 2 2 ]

(c) (1, 3] (d) (1, 4] Ans. (c) Solution: Any tangent to the hyperbola is a 2 m2 - b2

y = mx ± we must have a2m2 – b2 ≥ 0.

Equation of the line through the origin perpendicular to the tangent is 1 y= - x. (2) m Eliminating m from (1) and (2) we get the required locus as 2

y2

(a) (1, 2 ]

(a) (x2 + y2)2 = 36x2 – 16y2 (b) (x2 – y2)2 = 36x2 – 16y2 (c) (x2 + y2)2 = 36x2 + 16y2 (d) (x2 – y2)2 = 36x2 + 16y2 Ans. (a) Solution: Any tangent to the hyperbola is y = mx +

-

= 1 is 2 2 , then the eccentricity e of the hyperbola a 2 b2 lies in the interval.

Example 17: The locus of the foot of the perpendicular drawn from the origin to any tangent to the hyperbola x2 y 2 = 1 is 36 16

y= -

2

2

Ê -b ˆ Ê aˆ ÁË ˜¯ - ÁË ˜¯ = 1 2h 2k

a cosq -b cotq ,k= . 2 2

b2

fi

m2 ≥

fi

m2 ≥ e2 – 1

fi

1 0 and 2 – l < 0 or l + 3 < 0 and 2 – l > 0 fi l > – 3 and l > 2 or l < –3 and l < 2 fi l > 2 and l < –3 fi l œ (–3, 2) Example 21: The normal to the curve at P(x, y) meets the x-axis at G. If the distance of G from the origin is twice the abscissa of P, then the curve is (a) ellipse (b) parabola (c) circle (d) hyperbola or ellipse Ans. (d) Solution: Equation of the normal at (x, y) is Y – y = -

dx dy ˆ Ê (X – x) which meets the x-axis at G Á 0, x + y ˜ , then Ë dy dx ¯

dy = ± 2x dx dy = 2x fi y dy – x dx = 0 fi x2 – y2 = c fix+y dx x+y

or y dy = –3x dx fi 3x2 + y2 = c Thus the curve is either hyperbola or ellipse.

Hyperbola 20.9

Example 22: If the circle x2 + y2 = a2 intersects the hyperbola xy = c2 in four points P(x1, y1), Q(x2, y2) R(x3, y3), S(x4, y4), then which of the following need not hold. (a) x1 + x2 + x3 + x4 = 0 (b) x1 x2 x3 x4 = y1 y2 y3 y4 = c4 (c) y1 + y2 + y3 + y4 = 0 (d) x1 + y2 + x3 + y4 = 0 Ans. (d) Solution: The abcissaes of the points of intersection of the given curves are connected by x2 +

Example 23: If the normal to the rectangular hyperbola xy = c2 at the point (ct, c/t) meets the curve again at (ct ¢, c/t ¢), then (b) t3 t¢ = – 1 (a) t3t¢ = 1 (c) t t¢ = 1 (d) t t¢ = – 1 Ans. (b) Solution: Equation of the tangent at (ct, c/t) to the hyperbola xy = c2 is cx + cty = 2c2 t Slope of the tangent = – 1/t2 and slope of the normal = t2 Equation of the normal at (ct, c/t ) is y – c/t = t2 (x – ct) If it passes through (ct ¢, c/t ¢), then Ê c cˆ 2 ÁË - ˜¯ = t (ct ¢ - ct ) t¢ t (t – t¢) = t3 t¢(t¢ – t) fi t3t¢ = – 1

Example 24: If the normal at P to the rectangular hyperbola x2 – y2 = 4 meets the axes of x and y in G and g respectively and C is the centre of the hyperbola, then 2PC = (a) PG (b) Pg (c) Gg (d) none of these Ans. (c) Solution: Let P(x1, y1) be any point on the hyperbola x2 – y2 = 4 then equation of the normal at P is -y1 (x – x1) y – y1 = x1 fi x1y + y1x = 2x1y1. Then coordinates of G are (2x1, 0) and of g are (0, 2y1) so that

and

(c) e12 + e22 = 1

PG =

(2 x1 - x1 ) 2 + y12 =

x12 + y12 = PC

Pg =

x12 + (2 y1 - y1 ) 2 =

x12 + y12 = PC

Gg =

(2 x1 )2 + (2 y1 )2 = 2 x12 + y12 = 2 PC

(d) none of these

Ans. (b) Solution: Equation of the hyperbola is x2 y2 =1 1/ 2 1/ 2

c4

= a2 fi x 4 – a 2x 2 + c 4 = 0 x2 As x1, x2, x3, x4 are the roots of this equation x1 + x2 + x3 + x4 = 0, x1 x2 x3 x4 = c4 Similarly y1 + y2 + y3 + y4 = 0, y1 y2 y3 y4 = c4

fi

Example 25: If e1, e2 are the eccentricities of the hyperbola 2x2 – 2y2 = 1 and the ellipse x2 + 2y2 = 2 respectively. Then (b) e1e2 = 1 (a) e1 + e2 = 1

So

2 2 e12 = a + b where a2 = b2 = 1 2 a2

fi

e1 =

So

2 2 e22 = a - b = 2 - 1 = 1 2 2 a2

fi

e2 = 1/ 2 hence e1e2 = 1.

2 (Note: It is a rectangular hyperbola) Equation of the ellipse is x2 y 2 =1 + 2 1

Example 26: The line 2x + y = 1 touches a hyperbola and passes through the point of intersection of a directrix and the x-axis. The equation of the hyperbola is (a)

x2 y 2 =1 1 3

(b)

x2 y 2 =2 1 3

(c)

x2 y 2 =1 3 1

(d)

x 2 y2 =2 3 1

Ans. (a) Solution: Let the equation of the hyperbola be x2/a2 – y2/b2 = 1. As the line 2x + y = 1 touches the hyperbola (2)2 a2 – b2 = 1 fi 4a2 – b2 = 1 fi 4a2 + a2 (1 – e2) = 1, e being the eccentricity. fi 5a2 – a2e2 = 1 As the line passes through (a/e, 0). 2 a/e + 0 = 1 fi a = e/2. So that 5

e2 e4 =1 4 4

fi e4 – 5e2 + 4 = 0 fi e2 = 1 or e2 = 4. fi e = 2 as e2 π 1 Hence a = 1 and b2 = 3 and the required equation is x2 y 2 = 1. 1 3 Example 27: The equation of the hyperbola whose foci are (–2, 0) and (2, 0) and eccentricity is 2 is given by

20.10

Complete Mathematics—JEE Main

(a) x2 – 3y2 = 3 (b) 3x2 – y2 = 3 2 2 (d) –3x2 + y2 = 3 (c) –x + 3y = 3 Ans. (b) Solution: Let the equation of the hyperbola be x2

-

(a)

y2

9a 2 = a2 + b2 fi a2 = 2b2 6 Eccentricity of the hyperbola =

3x2 – y2 = 3 Example 28: The eccentricity of the hyperbola

x

2

a

2

-

y

b2

x2 y 2 =1 3 2

(b) a focus of the hyperbola is ( 3 , 0) 5/ 3

(c) the eccentricity of the hyperbola is

(d) the equation of the hyperbola is x2 – 3y2 = 3 Ans. (d) Solution: Eccentricity of the ellipse = 1 - 1 = 3 / 2 4 Foci of the ellipse = (± 3, 0) 1+

b2 a2

=

2 3

.

b 1 = a 3

fi

Since the hyperbola passes through the focus of the ellipse. 3

– 0 = 1 fi a2 = 3 and b2 = 1 a2 and the equation of the hyperbola is x2 y 2 =1 or x2 – 3y2 = 3 3 1 Focus of the hyperbola is (± 2, 0). Example 29: Let P (6, 3) be a point on the hyperbola x

2

y2

1+

b2 a

2

=

3 2

2

= 1 be reciprocal to that of the ellipse x2 + 4y2 = 4. If the hyperbola passes through a focus of the ellipse, then (a) the equation of the hyperbola is

3/ 2

a 2 x b2 y = a2 + b2 + 6 3 As it passes through (9, 0)

x2 y 2 =1 1 3

Eccentricity of the hyperbola =

(b)

(c) 2 (d) 3 Ans. (b) Solution: The equation of the normal at P(6, 3)

=1 a 2 b2 ae = 2 and e = 2 fi a = 1 b2 = a2(e2 – 1) = 1(4 – 1) = 3 so the equation of the hyperbola is

or

5/ 2

= 1. If the normal at the point P intersects the a 2 b2 x-axis at (9, 0) then the eccentricity of the hyperbola is

Example 30: If both the foci of the ellipse and the hyperbola

x2 y 2 + =1 16 b 2

x2 y2 1 = coincide, then the value 144 81 25

of b2 is (a) 9 (b) 7 (c) 41 (d) 12 Ans. (b) Solution: Foci of the ellipse are ± 16 - b 2 , 0

(

)

Ê 144 81 ˆ + , 0 = (± 3, 0) and Foci of the hyperbola are Á ± 25 25 ˜¯ Ë Since they coincide 16 – b2 = 9 fi b2 = 7. Example 31: If the tangent at the point (a sec a, b tan a) x2 y 2 to the hyperbola 2 - 2 = 1 meets the transverse axis at T, a b then the distance of T from a focus of the hyperbola is (a) a(e – cos a) (b) b(e + cos a) (c) a(e + cos a) Ans. (a)

(d)

a 2 e2 + b 2 cot 2 a

Solution: Equation of the tangent is

x y sec a - tan a = 1 a b

Coordinates of T are (a cos a, 0) and of a focus are (ae, 0). Distance of T from the focus (ae – a cos a) = a (e – cos a) (Note e > cos a). Example 32: The distance between the tangent to the hyperbola

x2 y 2 = 1, parallel to the line y = x + 2 is 4 3

Hyperbola 20.11

(a) 2

(b) 2 2

17 17 fim= ± 20 20 Since y = mx + 6 is a tangent to the parabola

fi

(c) 2 (d) 1 Ans. (c) Solution: Equations of the tangents are y = mx ±

y2 = 4ax; 6 =

2

4m - 3 where m = 1

fi

y=x±1

2 = 2 1+1 Example 33: The locus of the middle points of the normal chords of the rectangular hyperbola x2 – y2 = a2 is (a) (y2 – x2)3 = 4a2x2y2 (b) (y2 – x2)2 = 4a2x2y2 (c) (y2 + x2)3 = 4a2x2y2 (d) (y2 + x2)2 = 4a2x2y2 Ans. (a) Solution: Equation of the normal at (a secq, a tanq) to the hyperbola x2 – y2 = a2 is ax ay = a2 + a2 = 2a2 + sec q tan q

(1)

Let (h, k) be the middle point of this normal then its equation is hx – ky – a2 = h2 – k2 – a2 (T = S1) 2 2 (2) fi hx – ky = h – k Comparing (1) and (2) we get h2 - k 2 h secq = – k tanq = 2a h2 - k 2 h2 - k 2 , tanq = 2ah -2ak

fi

secq =

So

1 = sec2q – tan2q =

(h 2 - k 2 )2 Ê 1 1ˆ ÁË 2 - 2 ˜¯ 2 4a h k

fi (h2 – k2)3 + 4a2h2k2 = 0 Locus of (h, k) is (y2 – x2)3 = 4a2x2y2 Example 34: If y = mx + 6 is a tangent to the hyperbola 2

x y2 = 1 and the parabola y2 = 4ax, then the length of 100 49 the latus rectum of the parabola is

(c) 24

17 20 17 20

17 a fi a = 6m = ±6 20 m

Length of the latus rectum of the parabola

Distance between them =

(a) 6

m2 =

(b) 4 (d)

17 20 17 20

Ans. (c) Solution: Since y = mx + 6 is a tangent to the hyperbola, (6)2 = 100m2 – 49.

= 4|a| = 24

17 . 20

x2 y 2 = 1. 81 49 The tangent at P meets the transverse axis at T, N is the foot of the perpendicular from P to the transverse axis. If O is the origin, then ON.OT is equal to. (a) 81 (b) 49 (c) –81 (d) – 49. Ans. (a) Solution: Let the coordinate of P be (9 secq, 7 tanq) then ON = 9 secq. x y Equation of the tangent at P is sec q - tan q = 1 which 9 7 meets the transverse axis at (9 cos q, 0) fi OT = 9 cos q. Example 35: P is a point on the hyperbola

Then ON.OT = 9 secq ¥ 9 cosq = 81 Example 36: The product of the perpendiculars from the foci on any tangent to the hyperbola

x2 y 2 = 1 is 64 9

(a) 8 (b) 9 (c) 16 (d) 18. Ans. (b) Solution: Equation of a tangent at (8 secq, 3 tanq) to the hyperbola is x y sec q - tan q = 1 8 3 If e is the eccentricity of the hyperbola; product of the perpendiculars (e sec q - 1)( -e sec q - 1) = sec2 q tan 2 q + 64 9 =

(1 - e2 sec2 q ) ¥ (64 ¥ 9) 9 sec2 q + 64 tan 2 q

È 64 - 9 2 ˘ Í1 - 64 sec q ˙˚ (64 ¥ 9) = Î 9 sec2 q + 64 tan 2 q =

64 tan 2 q + 9 sec2 q 9 sec2 q + 64 tan 2 q

¥ 9 = 9.

20.12

Complete Mathematics—JEE Main

Example 37: If the normal at P on the hyperbola x y2 = 1 meets the transverse axis at G, S is a foci and a 2 b2 e the eccentricity of the hyperbola then SG : SP is equal to (a) a (b) b (c) e (d) 1/e. Ans. (c) Solution: Equation of the normal at P(a secq, b tanq) is ax by = a2 + b2 + sec q tan q 2

Which meets the transverse axis at Ê (a 2 + b 2 )sec q ˆ , 0˜ whose distance from S (ae, 0) is GÁ a Ë ¯

X2 Y2 = 1 where x = X + 4 2

Since they are at right angles fi

2 x1 2 x2 ¥ = –1 3 y1 3 y2

4x1x2 + 9 y1y2 = 0. Example 39: Consider a branch of the hyperbola

2

x – 2y2 – 2 2 x – 4 2 y – 6 = 0 with vertex at the point A. let B be one of the end points of its latus rectum. If C is the focus of the hyperbola near A, then area of the D ABC is: (a) 1 –

2 3

(b)

3 –1 2

(c) 1 +

2 3

(d)

3 + 1. 2

Ans. (b) Solution: Equation of the hyperbola is ( x - 2 )2 ( y + 2 )2 =1 4 2 Which can be written as

3 2

A

C

O

Fig. 20.6

Area of the DABC =

a˘ È SP = e Ía sec q - ˙ = a[e sec q – 1] e˚ Î

e2 sec q - e = e. e sec q - 1 Example 38: If the chords of contacts of the tangents from the points (x1, y1) and (x2, y2) to the hyperbola 2x2 – 3y2 = 6 are at right angle, then 4x1x2 + 9y1y2 is equal to (a) –1 (b) 0 (c) 6 (d) –12 Ans. (b) Solution: Equation of the chords of contacts are 2xx1 – 3yy1 = 6, 2xx2 – 3yy2 = 6

4+2 = 4 B

=

=

2.

Eccentricity of the hyperbola is

2 2 SG = (a + b )sec q - ae a

[a 2 + a 2 (e2 - 1)]sec q - a 2 e SG = a ¥ a[e sec q - 1] SP

2,y=Y–

1 (AC)(BC) 2

1 b2 (ae – a) ¥ where a2 = 4, b2 = 2 a 2 ˘ 2 1È 3 = Í2 - 2˙ ¥ = 3 – 1 2Î 2 2 ˚ 2

Example 40: Normal at point (5, 3) to the rectangular hyperbola xy – y – 2x – 2 = 0 meets the curve at the point whose coordinates are (a) (0, –2) (b) (–1, 0) Ê 1 -10 ˆ Ê3 ˆ (c) Á , (d) Á , -14˜ . Ë 4 3 ˜¯ Ë4 ¯ Ans. (d) Solution: Equation of the hyperbola can be written as (x – 1) (y – 2) = 4 4 or y =2 + x -1 4 dy =( x - 1) 2 dx Any point on the hyperbola is 2ˆ Ê ÁË1 + 2t , 2 + ˜¯ = (5, 3) fi t = 2 t So slope of the normal at (5, 3) is -

dx dy (5,3) = 4

Equation of the normal at (5, 3) is y – 3 = 4(x – 5) which 2ˆ 2 Ê will pass through Á1 + 2t , 2 + ˜ if 2 + = 4(1 + 2t) – 17. Ë t¯ t 1 fi 8t2 – 15t – 2 = 0 fi t = 2, - . 8 1 For t = - , we get the required point as 8 Ê 2 2ˆ Ê3 ˆ ÁË1 - 8 , 2 - 1/ 8 ˜¯ = ÁË 4 , -14˜¯

Hyperbola 20.13

Assertion-Reason Type Questions

Example 41: Statement-1 An ellipse passes through the foci of the hyperbola 9x2 – 4y2 = 36 and its major and minor axes lie along the transverse and conjugate axes respectively of the hyperbola. If the product of the eccentricities of the 1 two conics is , then the ellipse does not pass through the 2 Ê1 3ˆ i3 , ˜ . point Á 2 ¯ Ë2

Example 42: Statement-1 If a tangent to the hyperbola 2x2 – 4y2 = 8 meets x-axis at P and y-axis at Q. Lines PR and QR are drawn such that OPRQ is a rectangle where O is the origin then locus of R is a hyperbola. Statement-2 The curve described parametrically by 4 4 2 x= + 4t and y = - + represents a hyperbola. 7 t 3 Ans. (d)

Statement-2 Product of the eccentricities of the hyperbola 2 2 1 9x2 – 4y2 = 36 and the ellipse x - y = 1 is . 2 65 60

Solution: Tangent to the hyperbola point (x1, y1) is

xx1 yy1 =1 4 2

Ans. (b) Solution: Eccentricity e1 of the hyperbola 9x2 – 4y2 = 36 2 2 or x - y = 36 is 4 9

(

Foci of the hyperbola ± 13, 0

y

13 . 2

4+9 = 4

1 fi e2 = 2

1 13

(

x2 a2

+

y2 b2

=1

)

= 1 fi a2 = 13 and b2 = a2 (1 - e22 ) = 12

Equation of the ellipse is

x2 y 2 =1 + 13 12

Ê1 3ˆ Which clearly does not pass through Á 13, ˜ 2 ¯ Ë2 So statement-1 is true. In statement-2 if e3 is the eccentricity of the ellipse, then e3 =

65 - 60 = 65

1 13

Q

R

x

Fig. 20.7

Since it passes through ± 13, 0

a2

P

.

Let the equation of the ellipse be

13

O

)

If e2 is the eccentricity of the ellipse then e 1e 2 =

x2 y 2 = 1 at a 4 2

.

1 fi e1e3 = , so statement-2 is also true but does not justify 2 statement-1.

Ê4 ˆ Coordinates of P are Á , 0˜ of Q are Ë x1 ¯

Ê -2 ˆ ÁË 0, y ˜¯ and of R are 1

Ê 4 -2 ˆ ÁË x , y ˜¯ = (h, k) 1 1 x12 y12 = 1 as (x1, y1) lies on the hyperbola 4 2

Also fi

4 h

2

-

2 k2

=1

Locus of (h, k) is

4 x

2

-

2 y2

=1

Which is not a hyperbola and thus statemnet-1 is false. 2ˆ Ê 4ˆ Ê In statement-2 we have Á x - ˜ Á y + ˜ = 16 which repreË 3¯ Ë 7¯ sents a rectangular hyperbola and thus is true. p 2 are two given points. R divides the line segment PQ externally in the ratio 2:3. Example 43: P(2, 3) and Q(2 tanq, 3 secq); 0 < q
2 fi e > 2 and thus statefi b2 > a2 fi e2 = a2 ment-1 is also true.

Example 44: Statement-1: The line bx – ay = 0 does not meet the hyperbola x2/a2 – y2/b2 = 1. Statement-2: The line y = mx + c does not meet the hyperbola x2/a2 – y2/b2 = 1 if c2 = a2m2 – b2. Ans. (c) Solution: Statement-1 is true, as solving y = (b/a)x and x2/a2 – y2/b2 = 1 fi 0 = 1 which is not possible. Statement-2 is false as the line is a tangent to the hyperbola and hence meets the hyperbola at one point only. In statement-1, the line satisfies the condition of being a tangent, so it must touch the hyperbola at a point infinity.

Example 47: Statement-1: The angle between a pair of tangents drawn to the curve 7x2 – 12y2 = 84 from M(1, 2) is p . 2 Statement-2: Equation of the chord of contact of the point M(1, 2) with respect to the hyperbola 7x2 – 12y2 = 84 is 7x – 24y = 84. Ans. (b) x2 y 2 Solution: Equation of the hyperbola is = 1. 12 7

1 1 fi tanq = (6 – x), secq = (9 – y) 4 6 Eliminating q, we get the locus of R as 1 1 ( y - 9) 2 - ( x - 6)2 = 1 36 16 Which is a hyperbola, length of whose transverse axis is 2(6) = 12 So statement-1 is true. 2

2

Ê yˆ Ê xˆ Locus of Q(2 tanq, 3 secq) is Á ˜ - Á ˜ = 1 Ë 3¯ Ë 2¯ fi

Example 45: Statement-1: The rectangular hyperbola xy = 42 does not intersect the parabola y2 = 4x in real point. x2 y 2 Statement-2: If hyperbola 2 - 2 = 1 passes through the b a focus of ellipse perbola is

x2 a2

+

y2 b2

= 1, then eccentricity of the hy-

3.

Ans. (d) Solution: Statement-1 is false as the point (4, 4) lies on both the curves.

Equation of its director circle is x2 + y2 = 12 – 7 = 5 M(1, 2) lies on it. So the tangents drawn from M to the hyperbola are at right angles. Thus statement-1 is true. Equation of the chord of contact of M(1, 2) is x ¥ 1 y (2) = 1. 12 7 fi 7x – 24y = 84. So statement-2 is also true but does not justify statement-1. Example 48: Let H be the rectangular hyperbola xy = c2. Statement-1: If the sum of the slopes of the normal from 1 a point P to the hyperbola H is 2 , then P lies on a pair of c lines parallel to y-axis.

Hyperbola 20.15

Statement-2: Four normals can be drawn to the hyperbola H from any point P. Ans. (c) Ê cˆ Solution: Equation of the normal at any point Ë ct , ¯ t on the hyperbola H is y – c = t2(x – ct) (1) t If it passes through P(c, c), then 1 –

1 = t2(1 – t) t

fi (t – 1) (t3 + 1) = 0. Which gives only two real values of t, 1 and –1. So from the point P only two normals can be drawn and the statement-2 is false. Note: Point P(c, c) lies on the hyperbola. If the point P does not lie on the hyperbola, then there will be four normals from P. For statement-1, let P(a, b) be a point such that sum of 1 the slopes of the normals from P to H is 2 , then from (1) c b–

c = t2(a – ct) t

fi ct4 – a t3 + b t – c = 0 If t1, t2, t3, t4 are the roots of (2). Then

(2)

4

a , Â t 1t 2 = 0 c i=1 Sum of the slopes of the normals

 ti 4

=

= Â ti2 i =1

2

1 a2 Ê 4 ˆ = Á Â ti˜ – 2Ât1t2 = 2 = 2 Ë i=1 ¯ c c

This will represent a real tangent if 4m2 – 16 > 0 fi m2 > 4 fi m Œ (–•, – 2) » (2, •) Showing that statement-2 is true. Using it in statement-1 if any line cuts the hyperbola H at two distinct points, it is not a tangent to H and hence its slope lies in (–2, 2) thus statement-1 is also true. Example 50: Statement-1: If l is the length of the chord of the hyperbola x2 – y2 = 8 whose mid-point is (4, 2) then 3l2 = 80. Statement-2: Length of the chord of contact of the point Ê 8 4ˆ 5 2 2 . ÁË , ˜¯ with respect to the hyperbola x – y = 8 is 4 3 3 3 Ans. (b) Solution: Equation of the chord whose mid-point is (4, 2) of the hyperbola x2 – y2 = 8 is 4x – 2y = 42 – 22 = 12 fi 2x – y = 6 Which meets the hyperbola 2x – y = 6 at points for which. x2 – (2x – 6)2 = 8 fi 3x2 – 24x + 44 = 0 Let (x1, y1) and (x2, y2) be the end points of the chord, then 44 x1 + x2 = 8, x1x2 = . 3 y1 = 2x1 – 6, y2 = 2x2 – 6 l2 = (x2 – x1)2 + (y2 – y1)2 = (x2 – x1)2 + 4(x2 – x1)2 = 5[(x1 + x2)2 – 4x1x2] È 2 Ê 44 ˆ ˘ 80 . = 5 Í8 - 4 Á ˜ ˙ = Ë 3 ¯˚ 3 Î Thus statement-1 is true.

fi a = ±1 fi P lies on the lines x = ±1 and thus statement-1 is true.

Ê 8 4ˆ In statement-2, equation of the chord of contact of Á , ˜ Ë 3 3¯

x2 y 2 = 1 is a hyperbola. 4 16 Statement-1 Every line which cuts the hyperbola H at two distinct points has slope lying in (–2, 2). Statement-2 The slope of the tangents of the hyperbola lies in (–•, –2) » (2, •). Ans. (a) Solution: Equation of the tangent with slope m of the hyperbola H is y = mx ± 4m 2 - 16 .

8 4 to the hyperbola x2 – y2 = 8 is x ¥ - y ¥ = 8 3 3 fi 2x – y = 6. Which is same as the chord in statement-1 and hence the 5 80 required length is = 4 . 3 3

Example 49: H:

Thus statement-2 is also true but does not justify statement-1.

20.16

Complete Mathematics—JEE Main

LEVEL 2 Straight Objective Type Questions Example 51: If (a sec a, b tan a) and (a sec b, b tan b) x2 y 2 be the ends of a chord of 2 - 2 = 1, passing through the a b a b focus (ae, 0) then tan tan is equal to: 2 2 (a)

1+ e 1- e

(b)

e +1 e -1

(c)

1- e 1+ e

(d)

e -1 e +1

Ans. (c) Solution: Equation of the chord joining the given points is x y 1 a sec a b tan a 1 = 0 a sec b b tan b 1 bx(tana – tanb) – ay(seca – secb) + ab(seca tanb – secb tana) = 0 fi bx sin(a – b) – ay (cosb – cosa) + ab(sinb – sina) = 0 fi bx cos

a -b a +b a +b =0 - ay sin - ab cos 2 2 2

x a -b y a +b a +b = cos cos - sin a 2 b 2 2 a -b a +b If it passes through (0e, 0) e cos = cos 2 2

fi

(a) 3x2 + 10xy + 8y2 + 14x + 22y + 11 = 0 (b) 3x2 + 10xy + 8y2 – 14x – 22y + 13 = 0 (c) 3x2 – 10xy + 8y2 + 14x + 22y + 15 = 0 (d) 3x2 + 10xy + 8y2 + 14x + 22y + 15 = 0 Ans. (d) Solution: Since the equation of a hyperbola differs from the combined equation of the asymptotes by a constant. Let the equation of the asymptotes be. 3x2 + 10xy + 8y2 + 14x + 22y + l = 0 This will represent a pair of straight lines if abc + 2fgh – af 2 – bg2 – ch2 = 0 where a = 3, b = 8, c = l, f = 11, g = 7, h = 5 fi 3 ¥ 8 ¥ l + 2 ¥ 11 ¥ 7 ¥ 5 – 3 ¥ 121 – 8 ¥ 49 – l ¥ 25 = 0 fi l = 15. Example 53: The product of the lengths of the perpendiculars drawn from any point on the hyperbola x2 y 2 = 1 to its assumption is a 2 b2 (a)

a 2b2 2

y

a +b 2 =e fi a -b cos 2

y=

cos

M

fi tan a tan b = 1 - e 1+ e 2 2

+

1

bx a

P(a sec a, b tan a) x

y=-

bx a

Fig. 20.8

b Equation of the asymptotes is y = ± x. a Required product = PL.PM =

Example 52: Equations of the asymptotes of the hyperbola 3x2 + 10xy + 8y2 + 14x + 22y + 7 = 0 is

2

0 L

a +b a -b - cos 2 2 = e -1 fi a -b a -b e +1 cos + cos 2 2 cos

e -1 a b fi - tan tan = e +1 2 2

1

(b)

a +b a b2 1 1 (c) (d) a2b2 - 2 2 a b Ans. (a) Solution: Let P(a sec a, b tan a) be a point on the hyperbola. 2

=

a 2b 2 (sec a + tan a )(sec a - tan a ) a 2 + b2 a 2b2 a 2 + b2

Hyperbola 20.17

Example 54: If (5, 12) and (24, 7) are the foci of a hyperbola passing though the origin then the eccentricity of the hyperbola is (a)

386 /12

(b)

386 /13

(c)

386 / 25

(d)

386 / 38

Ans. (a) Solution: Let S(5, 12) and S¢(24, 7) be the two foci and P(0, 0) be a point on the conic then SP = 25 + 144 = 169 = 13;

and

2

2

S ¢P =

(24) + 7

S S¢ =

(24 - 5)2 + (7 - 12) 2 =

=

625 = 25

Therefore equation of the normal at P is y – b tanq = –

fi ax + b cosecq y = (a2 + b2) secq (i) Similarly the equation of the normal at Q(a sec f, b sec f) (ii) is ax + b cosec f y = (a2 + b2) sec f Subtracting (ii) from (i) we get a 2 + b2 sec q - sec f y= ◊ cosecq - cosecf b So that

k =y =

[∵ q + f = p/2]

386 since the conic is a hyperbola, S¢P – SP = 2a, the length of transverse axis and SS¢ = 2ae, e being the eccentricity. e=

SS ¢ = S ¢P - SP

386 . 12

Example 55: An equation of a tangent to the hyperbola. 16x2 – 25y2 – 96x + 100y – 356 = 0 which makes an angle p/4 with the transverse axis is (a) y = x + 2 (b) y = x + 4 (c) x = y + 3 (d) x + y + 2 = 0 Ans. (a) Solution: Equation of the hyperbola can be written as (1) X2/52 – Y2/42 = 1 where X = x – 3 and Y = y – 2. Equation of a tangent which makes an angle p/4. With the transverse axis X = 0 of (1) is

=

y – 2 = x – 3 ± 25 - 16 y = x + 2 or y = x – 4

Example 56: Let P(a secq, b tanq) and Q(a sec f, b tan f) where q + f = p/2, be two points on the hyperbola x2/a2 – y2/b2 = 1. If (h, k) is the point of intersection of normals at P and Q, then k is equal to 2 2 È a 2 + b2 ˘ (b) - Í (a) a + b ˙ a Î a ˚ 2 2 (c) a + b b

È a 2 + b2 ˘ (d) - Í ˙ Î b ˚

Ans. (d) Solution: Equation of the tangent at P(a secq, b tanq) is x y sec q - tan q = 1. a b

È a 2 + b2 ˘ a 2 + b 2 sec q - cosec q = -Í ˙ b cosec q - sec q Î b ˚

Example 57: If P is a point on the rectangular hyperbola x – y2 = a2, C is its centre and S, S¢ are the two foci, then SP.S¢P = (a) 2 (b) (CP)2 2 (c) (CS) (d) (SS¢)2 Ans. (b) Solution: Let the coordinate of P be (x, y). The coordinates of the centre are (0, 0). 2

The eccentricity of the hyperbola is

1+

a2 = b2

2

So the coordinates of the foci are S (a 2 , 0) and S (-a 2 , 0) . (Fig 19.9)

Y = tan p X ± 25 tan 2 p - 16 4 4 fi fi

a 2 + b 2 sec q - sec(p / 2 - q ) cosecq - cosec(p / 2 - q ) b

192 + 52

=

fi

a sinq (x – a sec q) b

S¢ (- a 2, 0)

Y

) P (x, y

C (0, 0)

S (a 2, 0)

x = - a/ 2, 0)

x = a/ 2)

Fig. 20.9

Equation of the corresponding directrices are x = a/ 2 and S = – -a/ 2 . By definition of the hyperbola SP = e(distance of P from x = a/ 2 ) =

2 | x - a/ 2 |

Similarly S¢P =

2 | x + a/ 2 |

So that SP.S¢.P = 2|x2 – a2/2| = 2x2 – a2 = x2 + y2 = (CP)2 (∵ P lies on the hyperbola x2 – y2 = a2)

20.18

Complete Mathematics—JEE Main

Example 58: If P is the length of the perpendicular from a focus upon the tangent at any point P on the hyperbola x2

-

a2 b2 p

2

-

y2 b2

= 1 and r is the distance of P from that focus, then

2a = r

(a) –1 (b) 0 (c) 1 (d) 2 Ans. (c) Solution: Equation of the tangent at P(a secq, b tanq) to the hyperbola is x y sec q - tan q = 1 a b Length of the perpendicular from the focus S(ae, 0) to this tangent is e sec q - 1

P=

2

2

=

sec q tan q + a2 b2

b2

a e2 sec2 q - 1

= b

e sec q - 1 e sec q + 1

e sec q + 1 e sec q - 1 2 r = (ae – a secq)2 + b2 tan2q = a2[e2 – 2e secq + sec2q] + a2(e2 – 1)

p2

Now

=

2

2

tan2q.

2

= a [e sec q – 2e secq + 1 ] = a2(e secq – 1)2 fi

r2 a

2

= (e secq – 1)2 fi

b2 p

2

-

2a 2 e sec q + 1 = = 1. r e sec q - 1 e sec q - 1

Example 59: A hyperbola having transverse axis of length 2 sinq is confocal with the ellipse. 3x2 + 4y2 = 12, then its equation is (a) x2 cosec2q – y2 sec2q = 1 (b) x2 sec2q – y2 cosec2q = 1 (c) x2 sin2q – y2 cos2q = 1 (d) x2 cos2q – y2 sin2q = 1 Ans. (a) x2 y 2 Solution: Foci of the ellipse = 1 is (± 1, 0) + 4 3 Let the equation of the hyperbola be

x2 a2

b2

=1

Example 60: If a hyperbola passing through the origin has 3x + 4y – 1 = 0 and 4x – 3y – 6 = 0 as its assumptions, then the equation of the transverse axis of the hyperbola is (a) x + y – 5 = 0 (b) x – 7y – 5 = 0 (c) x – y – 1 = 0 (d) x + y – 1 = 0 Ans. (b) Solution: Assumptions are perpendicular so the hyperbola is a rectangle hyperbola. Axes are the bisectors of the angles between the asymptotes. Equation of the transverse axis is the bisector of the of the angle which contains the origin and is given by 3x + 4 y - 1 4 x - 3y - 6 = 5 5 x – 7y – 5 = 0

fi

EXERCISE Concept-based Straight Objective Type Questions x2 y 2 =1 18 6 and r is the radius of the circle x2 + y2 – 6x – 18y + 87 = 0, then the value of er is equal to:

y2

x2 cosec2q – y2 sec2q = 1

a 1 = r esecq - 1

1. If e is the eccentricity of the hyperbola

-

Such that 2a = 2 sinq fi a = sinq. and a2 + b2 = a2 e2 = 1 fi b = cosq. So the required equation is x2 y2 =1 sin 2 q cos 2 q or

a 2 (e2 - 1)sec2 q + a 2 tan 2 q ab(e sec q - 1)

fi

b sec2 q + a 2 tan 2 q

ab(e sec q - 1)

=

=

ab(e sec q - 1) 2

So

(a) 2 3 (c) 2

(b) 2 / 3 (d) 3

Hyperbola 20.19

2. If m1 and m2 are two values of m for which the line y = x2 y 2 mx + 2 5 is a tangent to the hyperbola = 1, 4 16 1 then the value of m1 + is equal to m2 8 10 (b) 3 3 (c) 0 (d) 9. 3. Distance between the directrices of the hyperbola x2 y 2 = 1 is 49 16

8. A common tangent to x2 – 2y2 = 18 and x2 + y2 = 9 is (a) y = 2x + 3 5

(b) y =

2x + 3 3

(c) y = 2x + 3 7

(d) y =

2x + 3 5

2

9. If the foci of the ellipse

(a)

(a)

65 7

(b)

(c)

33 4

(d)

49 65 98 65

x2 y 2 25 4 = 1 and the parabola y2 = 20x. Then the value of 25m4 – 4m2 is equal to (a) 29 (b) 21 (c) 25 (d) 4 5. If length of the transverse axis of a hyperbola is 8 and its eccentricity is 5 / 2 then the length of a latus rectum of the hyperbola is (a) 1 (b) 2

4. If A line with slope m touches the hyperbola

(c) 2 5

(d) 8 / 5

2 2 6. A tangent to the hyperbola x - y = 1, meets x-axis 4 2 at 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 lies on 4 2 2 4 (a) 2 + 2 = 1 (b) 2 - 2 = 1 x y x y

(c)

2 x2

+

4 y2

=1

(d)

4 x2

-

2 y2

=1

7. If P(3 secq, 2 tanq) and Q(3 sec f, 2 tan f) where p q + f = , be two distinct points on the hyperbola 2 x2 y 2 = 1, then the ordinate of the point of inter9 4 section of the normals at P and Q is (a) 11/3 (b) –11/3 (c) 13/2 (d) –13/2

bola

2

x y + = 1 and the hyper16 b 2

1 x2 y2 = coincide, then b2 equals 25 144 81

(a) 5 (b) 7 (c) 9 (d) 1 10. Tangent drawn from the point (c, d) to the hyperbola 9x2 – 64y2 = 576 make angle a and b with the x-axis. If tana tanb = 1, then the value of c2 – d2 is equal to (a) 73 (b) 55 (c) 64 (d) 9 11. If the vertex of a hyperbola bisects the distance between its centre and the corresponding focus, then the ratio of the square of its conjugate axis to the square of half the distance between the foci is (a) 4/3

(b) 4 / 3

(c) 2 / 3 (d) 3/4 12. If the chords of contact of the tangents from two points (x1, y1) and (x2, y2) to the hyperbola 4x2 – 5y2 = a2 are at right angles. Then 16x1x2 + 25y1y2 is equal to (a) –1 (b) 0 (c) a2 (d) 1 13. If two perpendicular tangents are drawn from a point (a, b) to the hyperbola x2 – y2 = 16, then the locus of (a, b) is (a) a pair of straight line (b) a circle (c) a parabola (d) an ellipse x2 y2 = 1 represents a hyperbola, then + a+7 5-a (a) a > 5 (b) a < –7 (c) a < 5 (d) a < –7 or a > 5

14. If

15. If e1 is the eccentricity of the hyperbola

x2 y 2 =1 49 36

and e2 is the eccentricity of the hyperbola = 1, then (a) e1e2 = 1 (c)

1 e12

+

1 e22

(b) =1

e1 =1 e2

(d) e12 + e22 = 1

x2 y 2 36 49

20.20

Complete Mathematics—JEE Main

LEVEL 1 Straight Objective Type Questions 16. The line 4 2x – 5y = 40 touches the hyperbola x2 y2 = 1 at the point 100 64 (a) (10, 8 2 ) (c)

(20, 8 3 )

(b) (10 2 , 8) Ê 20 8 ˆ , (d) Á Ë 3 3 ˜¯

(a) 2

17. The line 9 3x + 12y = 234 3 is a normal to the x 2 y2 = 1 at the points. hyperbola 81 36 (a) (18, 6 3 )

(b) (9 2 , 6)

Ê 18 6 ˆ (d) Á , Ë 3 3 ˜¯ 18. Length of the latus rectum of the hyperbola x2 tan a – y2 cota = 1, (0 < a < p/4) is (a) 2 cot a (b) 2 tan a 3/2 (c) 2(tana) (d) 2(cota)3/2 19. If the angle between the asymptotes of the hyperbola Ê x2 y 2 ˆ Á 2 - 2 ˜ is 2q and e is its eccentricity then sinq is b ¯ Ëa

(b)

1 e

1 (c) a (d) 1 e2 be 20. Axymptotes of the hyperbola 2xy = 4x + 3y pass through the point (a) (2, 3/2) (b) (2, 3) (c) (3/2, 2) (d) (0, 0) 21. The curve described parametrically by x = t2 + t + 1, y = t2 – t + 1 represents (a) a pair of straight lines (b) an ellipse (c) a parabola (d) a hyperbola 22. The point (at2, 2bt) lies on the hyperbola for

x2 a2

(b)

2

(c) 2 + 2 (d) none of these 25. If e and e1 are the eccentricities of a hyperbola and its 1 1 conjugate, then 2 + 2 is equal to e e1 (a) –1 (b) 0 (c) 1 (d) none of these 26. Foci of the rectangular hyperbola are (±7, 0), the equation of the hyperbola is (a) x2 – y2 = 49 (b) x2 – y2 = 98 2 2 (c) 2x – 2y = 49 (d) none of these

(c) (9 3, 6)

equal to b (a) a

23. If the coordinates of four concyclic point on the rectangular hyperbola xy = c2 are (cti, c/ti) i = 1, 2, 3, 4 then (a) t1 t2 t3 t4 = –1 (b) t1 t2 t3 t4 = 1 (c) t1 t3 = t2 t4 (d) t1 + t2 + t3 + t4 = c2 24. The eccentricity of a rectangular hyperbola is

-

(a) all real values of t (b) t2 = 2 + 5 (c) t2 = 2 – 5 (d) no real value of t

y2 b2

=1

x2 y 2 = 1, Q is the 16 9 point where the right directrix meets the axis of x and S is the right focus of the hyperbola. If the area of the triangle PQS is 27/10 sq. units, coordinates of P are

27. P is a point on the hyperbola

(a) (4 2 , 3)

(b) (4 3 , 3 2 )

(c) (4 5 , 6)

(d) none of these

28. The normal at a point P to the parabola y2 = 4x is parallel to the tangent at Q ( 2 , 2) to the hyperbola x2 y 2 = 1 and meets the axis of the parabola at R. 1 4 If S is the focus of the parabola, area of the triangle PSR in sq. units is (a) 9 2

(b) 10 2

(c) 18 2 (d) 20 2 29. The difference between the length 2a of the transverse axis of a hyperbola of eccentricity e and the length of its latus rectum is (a) 2a|3 – e2| (b) 2a|2 – e2| 2 (c) 2a|e – 1| (d) a|2e2 – 1| 30. The locus of the point of intersection of the tangents x2 y 2 to the hyperbola = 1 which are at right an16 36 gles is

Hyperbola 20.21

(a) x2 + y2 = 20 (c) x2 + y2 = 52

(b) x2 – y2 = 20 (d) none of these

y2 = 1 are a 2 36 perpendicular to the asymptotes of the hyperbola x2 y 2 = 1 then 49 b 2

31. If the asymptotes of the hyperbola

x2

-

(a) 7a ± 6b = 0 (b) 6a ± 7b = 0 (d) a – b = 1 (c) a2 – b2 = 1 32. P and Q are two points on the rectangular hyperbola xy = c2 such that the abscissa of P and Q are the roots of the equations x2 – 6x – 16 = 0. Equation of the chord joining P and Q is (a) 16x – c2y = 6c2 (b) c2x – 16y = c2 (c) c2x – 16y = 6c2 (d) c2x – 6y = 16c2 33. Normal at (3, 4) to the rectangular hyperbola xy – y – 2x – 2 = 0 meets the curve again at the points (a) (1, 2) (b) (2, 3) (c) (–1, 0) (d) none of these 34. Locus of the mid-points of the chords of the circle x2 y 2 x2 + y2 = 16 which touch the hyperbola = 1 is 16 9 (a) (x2 – y2)2 = 16x2 – 9y2 (b) (x2 + y2)2 = 9x2 – 16y2 (c) (x2 + y2)2 = 16x2 + 9y2 (d) (x2 – y2)2 = 16x2 + 9y2 35. If the eccentricity of the hyperbola

x2 a2

-

y2 b2

= 1 is

5 and the distance between the foci is 12, then b2 – a2 is equal to (3/5) k2 where k is equal to (a) 5 (b) 3 (c) 2 (d) 6

36. If the extremities of the latus rectum of the hyperbola x2 y 2 = 1 with positive coordinates lie on the paa 2 b2 rabola x2 = 3(y + 3), then length of the latus rectum of the hyperbola when its eccentricity is 3 is (a) 3 (b) 6 (c) 12 (d) none of there 37. If the locus of the point of intersection of two straight lines 3 ax + ay – 4 3 = 0 and 3 x – y – 4 3 a = 0 is a hyperbola with eccentricity e; for different values of a, then e2 + 6e – 9 is equal to (a) 0 (b) 7 (c) 3 (d) 4 38. Locus of the mid-points of the chords of the hyperx2 y 2 bola = 1 that are parallel to 3x – 4y = 1 is 16– 3y9 = 0 (a) 4x (b) 3x – 4y = 0 (c) 3x + 4y = 0 (d) 4x + 3y = 0 39. The parametric equation: x = a(secq + tanq), y = b(secq – tanq) represents (a) a parabola (b) an ellipse (c) a hyperbola (d) a rectangular hyperbola 40. If a normal to the hyperbola x2 – 4y2 = 4 having equal positive intercepts on the axes is a tangent to the x2 y 2 ellipse 2 + 2 = 1, then the distance between the a b foci of the hyperbola (a)

x2 a2

10 3

(c) 10 3

(b)

y2 b2

= 1 is 5 3

(d) 5 3

Assertion-Reason Type Questions

41. A line y = x + 4 meets the hyperbola xy = 16 at A and B. Statement 1: A circle on AB as diameter passes through the points (4, 4) and (– 4, – 4). Statement 2: A circle on AB as diameter passes through the intersection of the circle x2 + y2 = 32 and the line y = x. 42. P is a point on the hyperbola are its foci.

x2 a2

-

y2 b2

= 1 S and S¢

Statement-1: Product of the lengths of the perpendiculars from S and S ¢ on the tangent at P is equal to b 2. Statement-2: |PS – PS¢| = 2a. 43. Statement-1: Two tangents drawn from any point on x2 y 2 the hyperbola x2 – y2 = a2 – b2 to the ellipse 2 + 2 a b = 1 make complementary angles with the axis of the ellipse.

20.22

Complete Mathematics—JEE Main

Statement-2: If two lines make complementary angles with the axis of x then the product of their slopes is 1. 44. Statement-1: If the foci of a hyperbola are at the points (4, 1) and (–6, 1), eccentricity is 5/4 then the length of the transverse axis is 8. Statement-2: Distance between the foci of a hyperbola is equal to the product of its eccentricity and the length of the transverse axis. 45. Statement-1: If the angle between two asymptotes x2 y 2 p of a hyperbola 2 - 2 = 1 is , its eccentricity is 3 a b 2 3. Statement 2: Angles between the asymptotes of the Ê bˆ x2 y 2 hyperbola 2 - 2 = 1 are 2 tan–1 Á ˜ or Ë a¯ a b Ê bˆ p – 2 tan–1 Á ˜ . Ë a¯ 46. Statement-1: Equation of a circle on the ends of a x2 y 2 = 1 as a dilatus rectum of the hyperbola 16 9 ameter is 16x2 + 16y2 ± 160x + 319 = 0 Statement-2: Focus of the parabola y2 = 20x coinx2 y 2 cides with a focus of the hyperbola =1 16 9 47. Let P(x, y) be a variable point such that

Statement-1: P traces a hyperbola whose eccentric5 ity is . 3 Statement-2: P traces a hyperbola such that the equation of its conjugate axis is 6x – 8y = 27. 48. Statement 1: A normal to the hyperbola with eccentricity 3, meets the transverse axis and conjugate axis at P and Q respectively. The locus of the mid-point of 3 . PQ is a hyperbola with eccentricity 2 2 Statement 2: Eccentricity of the hyperbola 8x2 – y2 = 3 . 8a2 is 2 2 49. Statement-1: The locus of the point of intersection of the tangents that are at right angles to the hyperbola x2 y 2 = 1 is the circle x2 + y2 = 52. 36 16 Statement-2: Perpendicular tangents to the hyperx2 y 2 bola 2 - 2 = 1 interest on the director circle a b x2 + y2 = a2 – b2 (a2 > b2) of the hyperbola. 50. A hyperbola H:

x2 y 2 = 1 intersects the circle, 9 4

C: x2 + y2 – 8x = 0 at the points A and B. Statement-1: 2x – 5 y + 4 = 0 is a common tangent to both C and H. Statement-2: Circle on AB as a diameter passes through the centre of the hyperbola H.

( x - 3) 2 + ( y - 2)2 - ( x - 6)2 + ( y + 2) 2 = 3.

LEVEL 2 Straight Objective Type Questions 51. Locus of the mid-point of the chord of the hyperbola x2 y 2 - 2 = 1 which is a tangent to the circle x2 + y2 = c2 2 a b is 2

Ê x2 y 2 ˆ Ê x2 y 2 ˆ (a) Á 2 - 2 ˜ = c 2 Á 4 + 4 ˜ b ¯ b ¯ Ëa Ëa 2

Ê x2 y 2 ˆ Ê x2 y 2 ˆ (d) Á 4 - 4 ˜ = c 2 Á 2 + 2 ˜ b ¯ b ¯ Ëa Ëa

52. H1 : xy = c2 and H2 : xy = k2 are two different hyperbolas. From a point on H1, tangents are drawn to H2. Area of the triangle formed by the chord of contact and the asymptote to H2 is

Ê x2 y 2 ˆ Ê x2 y 2 ˆ (b) Á 2 + 2 ˜ = c 2 Á 4 - 4 ˜ b ¯ b ¯ Ëa Ëa

(a)

2

(c)

Ê x2 y 2 ˆ Ê x2 y 2 ˆ (c) Á 2 - 2 ˜ = c 2 Á 2 + 2 ˜ b ¯ b ¯ Ëa Ëa

2

k2 c2 2k 4 c2

(b)

k4 c2

(d) none of these

Hyperbola 20.23

53. e1, e2 are respectively the eccentricities of the hyperbola x2 – y2 cosec2q = 5 and the ellipse x2cosec2q + y2 = 5. If 0 < q < p/2 and e1 = 7 e2, then q is equal to (a)

p 4

(b)

p 6

(c)

p 3

(d) none of these

54. If q is an angle between the two asymptotes of the x2 y 2 Êqˆ hyperbola 2 - 2 = 1, then cos Á ˜ is equal to Ë 2¯ a b b

(a)

a 2 + b2 a

(c)

2

a +b

ab

(b)

2

a + b2 a-b . a+b

(d)

2

55. A and B are two points on the hyperbola

x2

-

y2

= 1, a 2 b2 O is the centre. If OA is perpendicular to OB then 1 1 + is equal to 2 (OA) (OB)2 (a) (c)

1 a

2

1 b

2

+ -

1 b

(b)

2

1

1 a

2

-

1 b2

(d) a2 + b2

a2

56. The coordinates of a point common to a directrix and x2 y 2 = 1 are an asymptote of the hyperbola 25 16

Ê 25 20 ˆ (a) Á , Ë 41 3 ˜¯

Ê 20 -25 ˆ (b) Á , Ë 41 41 ˜¯

Ê 25 20 ˆ (c) Á , ˜ Ë 3 3¯

Ê -25 20 ˆ , (d) Á Ë 41 41 ˜¯

57. If the normals at P, Q, R on the rectangular hyperbola xy = c2 intersect at a point S on the hyperbola, then centroid of the triangle PQR is at (a) an extremity of the latus rectum (b) the centre (c) a focus (d) the point S on the hyperbola. 58. If a diameter of a hyperbola meets the hyperbola in real points then (a) it meets the conjugate hyperbola in imaginary points. (b) the conjugate diameter meets the given hyperbola in real points. (c) the conjugate diameter meets the conjugate hyperbola in imaginary points. (d) none of these. 59. An ellipse has eccentricity 1/2 and a focus at the point P(1/2, 1). One of its directrix is the common tangent near to the point P, to the circle x2 + y2 = 1 and the hyperbola x2 – y2 = 1, the equation of the ellipse is (a) 3x2 + 4y2 – 6x – 8y + 4 = 0 (b) 3x2 + 4y2 – 2x – 8y + 4 = 0 (c) 4x2 + 3y2 – 8x – 6y + 4 = 0 (d) 4x2 + 3y2 – 8x – 2y + 4 = 0 60. The asymptotes of xy = hx + ky are (a) x = h, y = k (b) x = –h, y = –k (c) x = k, y = h (d) x = –k, y = –h

Previous Years' AIEEE/JEE Main Questions

1. The foci of the ellipse

x2 y 2 + = 1 and the hyperbola 16 b 2

x2 y2 1 = coincide then, the value of b2 is: 25 144 81 (a) 5 (b) 7 (c) 9 (d) 1 [2003] 2. The locus of the point P (a, b) moving under the condition that the line y = a x + b is a tangent to the x2 y 2 hyperbola 2 - 2 = 1 is: a b

(a) a parabola (b) a hyperbola (c) an ellipse (d) a circle [2005] 3. The normal to curve at P(x, y) meets the x-axis at G. If the distance of G from the origin is twice the abscissa of P, then the curve is (a) ellipse (b) parabola (c) circle (d) hyperbola [2007] 4. For the hyperbola

x2

y2

= 1, which of the cos 2 a sin 2 a following remains constant when a varies? -

20.24

Complete Mathematics—JEE Main

(a) eccentricity (b) directrix (c) abscissae of vertices (d) abscissae of foci. [2007] 5. The equation of the hyperbola whose foci are (–2, 0) and (2, 0) and eccentricity is 2 is given by (a) x2 – 3y2 = 3 (b) 3x2 – y2 = 3 2 2 (d) –3x2 + y2 = 3 [2011] (c) –x + 3y = 3 2 2 6. A tangent to the hyperbola x - y = 1 meets x-axis 4 2 at 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 lies on 4 2 2 4 (a) 2 + 2 = 1 (b) 2 - 2 = 1 x y x y

(c)

2 x2

+

4

=1

y2

(d)

4 x2

-

2 y2

=1

[2013, online] 7. A common tangent to the conic x = 6y and 2x2 – 4y2 = 9 is 3 (a) x – y = (b) x + y = 1 2 2

(c) x + y =

9 2

(d) x – y = 1 [2013, online]

8. If P(3 secq, 2 tanq) and Q(3 sec f, 2 tan f) where p q + f = , be two distinct points on the hyperbola 2 x2 y 2 = 1. Then the ordinate of the points of in9 4 tersection of the normals at P and Q is 11 11 (a) (b) 3 3 (c)

13 2

(d) -

13 2

[2014, online]

9. The tangent at an extremity (in the first quadrant) of 2 2 latus rectum of the hyperbola x - y = 1 meets 4 5 x-axis and y-axis at A and B respectively. Then (OA)2 – (OB)2, where O is the origin, equals 16 20 (a) (b) 9 9 (c) 4

(d) -

4 3

[2014, online]

10. An ellipse passes through the foci of the hyperbola, 9x2 – 4y2 = 36 and it major and minor axes lie along the transvers and conjugate axes of the hyperbola respectively. If the product of eccentricities of the two 1 conics is , then which of the following points does 2 not lie on the ellipse? (a) ( 13, 0)

Ê 39 ˆ (b) Á , 3˜ Ë 2 ¯

Ê1 3ˆ (c) Á 13, ˜ 2 ¯ Ë2

Ê 13 ˆ (d) Á , 6˜ Ë 2 ¯

[2015, online] 11. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is (a) (c)

4 3

(b)

4 3

2

(d) 3 [2016] 3 12. A hyperbola whose transverse axis is along the major 2 2 axis of the conic, x + y = 4 and has vertices at the 3 4

foci of this conic. If the eccentricity of the hyperbola 3 , then which of the following points does NOT 2 lie on it? is

(a) ( 5 , 2 2 )

(b) (0, 2)

(c) (5, 2 3 )

(d) ( 10 , 2 3 )

[2016, online] 13. Let a and b respectively be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation 9e2 – 18e + 5 = 0. If S(5, 0) is a focus and 5x = 9 is corresponding directrix of this hyperbola, then a2 – b2 is equal to: (a) – 7 (b) – 5 (c) 5 (d) 7 [2016, on line]

Hyperbola 20.25

Previous Years' B-Architecture Entrance Examination Questions

1. If PQ is a double ordinate of the hyperbola

x2

-

Level 1

y2

a 2 b2 = 1 such that OPQ is an equilateral triangle, O being the centre of the hyperbola, then the eccentricity e of hyperbola satisfies 2 3 (b) e = (a) e = 3 2 (c) e >

2

(d) 1 < e
0, 1 – m2 > 0 fi m2 = 1 As (a, b) lies on both, locus of (a, b) is y = ±x, which is a pair of straight lines. 14. We must have

a + 7 > 0 and 5 - a < 0 ¸ a > 5 ˝ fi or a + 7 < 0 and 5 - a > 0 ˛ a < -7

2 2 15. e1 = 49 + 36 and e2 = 49 + 36 . 49 36 1 1 fi 2 + 2 = 1. e1 e2

Hyperbola 20.27

Level 1 16. Equation of the tangent at (10 sec q, 8 tan q) on the x y hyperbola is sec q – tan q = 1 10 8 Comparing with the given equation x 2 y = 1, we get 10 8

9x 6 y = 117 + 2 3 we get sec q = 2, tan q =

3

and the required point is (18, 6 3 ). 18. Length of the latus rectum is 19. Equation of the asymptotes is

2(tan a )

3/2

= 2(tan a) .

cot a

x y ± = 0. a b

b Ê bˆ - Á- ˜ a Ë a¯ So tan 2q = Ê bˆ Ê bˆ 1+ Á ˜ Á- ˜ Ë a¯ Ë a¯

fi q = tan–1 fi sin q =

=

1-

2(b / a) 1 - (b / a)2

1 2

1+

b a2

fi t2 =

4b 2t 2 b

= 1 fi t4 – 4t2 – 1 = 0

2

4 ± 16 + 4 =2± 2

5 But t2 π 2 –

5

so t2 = 2 + 5 . 23. Equation of a circle be x2 + y2 + 2gx + 2fy + k = 0. If the point (ct, c/t) lies on it, then c2t4 + 2gct3 + kt2 + 2fct + c2 = 0 which gives four values of t say t1, t2, t3, t4 and the four concyclic points such that t1 t2 t3 t4 = 1. 24. For a rectangular hyperbola b2 = a2 so b2 = a2(e2 – 1) fi e2 – 1 = 1 fi e2 = 2 fi e = 25. e2 =

a 2 + b2 a

2

2

and e12 =

a 2 + b2 b

2

=

1 e

+

2

1 e12

=1

26. Equation of the hyperbola is x2 – y2 = a2 2.

27. Eccentricity e = 5/4, coordinates of Q(16/5, 0), S(5, 0), QS = 9/5 If P(x, y), then area of the triangle PQS

Ê bˆ = 2 tan–1 Á ˜ Ë a¯

=

2

-

Foci = (± ae, 0) = (± 7, 0) where e =

b b fi tan q = . a a

b ¥ a

a 2t 4 a

so the required point is (10 2 , 8) 17. Equation of the normal at (9 sec q, 6 tan q) on the 9x 6y hyperbola is = 117. + sec q tan q Comparing with the given equation

fi 2q = tan–1

2

21. Eliminating t, we get x = ÊÁ x - y ˆ˜ + x - y + 1 Ë 2 ¯ 2 2 fi (x – y) = 2(x + y – 2) which is a parabola. 22.

2 , tan q = 1

sec q =

Ê3 ˆ Both of them pass through the point Á , 2˜ Ë2 ¯

28. Equation of the tangent at Q ( 2 , 2) is y = 2 2x – 2 so equation of the normal at P is y = 2 2 x - 2 ¥ 2 2 - ( 2 2 )3

b = ae

a 2 (e2 - 1) 2 2

a e

or y = 2 2 x - 20 2 S(1, 0), R(10, 0), P (8, - 4 2 ) Area of the triangle

1 e2

20. Given hyperbola is xy = 2x +

= (1/2) (9/5) |y| = 27/10 fi y = 3, x = 4 2

3 y 2

1 ¥ |10 - 1||- 4 2 | 2

= 18 2 29. Let the equation of the hyperbola be

3ˆ Ê or Á x - ˜ (y – 2) = 3. Ë 2¯

b2 = a2(e2 – 1)

3 Equations of the asymptotes are x – = 0, y – 2 = 0 2

length of the latus rectum = 2b = 2a(e2 – 1)

x2 a2

e2 - 1

-

y2 b2

=1

20.28

Complete Mathematics—JEE Main

30. The locus is the circle x2 + y2 = 16 – 36 which is not possible. So there is no pair of tangents that are at right angles. 31. Asymptotes to the two hyperbola are x y x y ± = 0 and ± = 0 a 6 7 b Ê 6ˆ Ê bˆ If they are at right angles then Á ± ˜ Á ± ˜ = – 1 Ë a¯ Ë 7¯ fi 7a ± 6b = 0 32. Let P(x1, c2/x1) and Q(x2, c2/x2) such that x1 + x2 = 6, x1x2 = – 16. Equation of the chord PQ is c2 c2 c x x1 = 2 (x – x1) y– x1 x2 - x1 2

fi x – x1 = -

x1 x2 Ê c2 ˆ y Á x1 ˜¯ c2 Ë

fi c2x + x1x2 y = c2(x1 + x2) fi c2x – 16y = 6c2 33. Equation of the hyperbola is (x – 1) (y – 2) = 4 4 or y = 2 + x -1 2ˆ Ê Any point on this hyperbola is Á1 + 2t , 2 + ˜ Ë t¯

h k secq - tanq = 2 = 2 2 and h +k h + k2 4 3 Eliminating q, we get (h2 + k2)2 (sec2 q – tan2 q) = (4h)2 – (3k)2 fi (h2 + k2)2 = 16h2 – 9k2. and the required locus is (x2 + y2)2 = 16x2 – 9y2 35. We have e = 5 , 2ae = 12 3 Ê 36 ˆ b2 – a2 = a2e2 – 2a2 = 36 – 2 Á ˜ = (6)2 Ë 5¯ 5 Ê b2 ˆ 36. As Á ae, ˜ a¯ Ë

Ê b2 ˆ , Á - ae, ˜ lies on the parabola a¯ Ë

Ê b2 ˆ x2 = 3 (y + 3), we have a2e2 = 3 Á + 3˜ Ë a ¯ = 3(a (e2 – 1) + 3) fi (a – 3) (ae2 + 3) = 0 fi a = 3. So length of the latus rectum of the hyperbola =

2b 2 2a 2 (e2 - 1) = = 2 ¥ 3 ¥ 2 = 12. a a

37. Point of intersection is (x, y) Ê Ê 1ˆ Ê1 ˆˆ = Á 2Áa + ˜ , 2 3 Á - a˜ ˜ Ë ¯ Ë ¯¯ Ë a a 2

2

Ê y ˆ Ê xˆ fi Á ˜ -Á =4 Ë 2¯ Ë 2 3 ˜¯

4 dy = ( x - 1) 2 dx

fi

2ˆ Ê Slope of the normal at Á1 + 2t , 2 + ˜ is t2. Ë t¯

So e2 = 1 +

x2 y 2 =1 16 48 48 =4fie=2 16

2ˆ Ê Now (3, 4) = Á1 + 2t , 2 + ˜ fi t = 1 Ë t¯

and hence e2 + 6e – 9 = 7 38. Equation of the chord with mid-point (h, k) is

Slope of the normal at (3, 4) is 1. So equation of the normal at (3, 4) is x – y + 1 = 0 2ˆ Ê Point Á1 + 2t , 2 + ˜ lies on this normal. Ë t¯

hx ky h2 k 2 9h = whose slope is 16 9 16 9 16k 9h 3 so = fi 3h – 4k = 0 16k 4 Locus of (h, k) is 3x – 4y = 0. 39. We have bx + ay = 2ab sec q bx – ay = 2ab tan q Eliminating q we get

If 1 + 2t – 2 –

2 + 1 = 0 fi t2 = 1 fi t = ± 1 t

So the required point is for t = – 1, i.e. (–1, 0). 34. Let (h, k) be the mid-point, equation of the chord is hx + ky = h2 + k2 which touches the hyperbola at (4 sec q, 3 tan q). x y So its equation is sec q – tan q = 1. 4 3 Comparing the two equations we get

2

2

Ê bx - ay ˆ Ê bx + ay ˆ ˜ =1 ÁË ˜¯ - ÁË 2ab 2ab ¯ fi 4abxy = 4 a2b2 fi xy = ab which represents a rectangular hyperbola.

Hyperbola 20.29

40. Equation of the normal to the hyperbola at 2x y (2 sec q, tan q) is + = 5. sec q tan q Intercepts on axes are

5 sec q and 5 tan q 2

5 1 p sec q = 5 tan q fi sin q = fiq= 2 2 6

Using it the required points of intersection are (4, 4) and (– 4, – 4). Hence statement-1 is also true. 42. Let P(a sec q, b tan q), equation of the tangent at P is x y sec q - tan q = 1. a b Product of the lengths of the perpendiculars from S(ae, 0) and S¢(– ae, 0) is

and the equation of the normal is 3x + 3 y = 5 5 fiy=–x+ 3

(e sec q - 1)(e sec q + 1) sec2 q tan 2 q + a2 b2

which will touch the ellipse

x2 a

2

+

y2 b

2

=1

=

a 2b 2 (e2 sec2 q - 1) a 2 [(e2 - 1)sec2 q + tan 2 q ]

= b2

2

Ê 5 ˆ if Á ˜ = a2 (– 1)2 + b2 Ë 3¯ Ê 5 ˆ fi a2 + b2 = Á ˜ Ë 3¯

2

Distance between the foci of the hyperbola x2 a2

-

=2¥

y2 b2 5 3

= 1 is 2ae = 2 a 2 + b 2 =

10 3

Ê 4ˆ 41. Any point on the hyperbola xy = 16 is Á 4t , ˜ Ë t¯ Ê 4ˆ Let the coordinate of A and B be Á 4t1 , ˜ and t1 ¯ Ë Ê 4ˆ 1 ÁË 4t2 , t ˜¯ respectively. So slope of AB = - t t = 1 2 12 fi t1t2 = – 1. Equation of the circle on AB as diameter is Ê 4ˆ Ê 4ˆ (x – 4t1) (x – 4t2) + Á y - ˜ Á y - ˜ = 0 t1 ¯ Ë t2 ¯ Ë 4ˆ Ê 4ˆ Ê fi (x – 4t1) Á x + ˜ + Á y - ˜ (y + 4t1) = 0 Ë t1 ¯ Ë t1 ¯ Ê1 ˆ fi x2 + y2 – 32 + 4 Á - t1 ˜ (x – y) = 0 Ë t1 ¯ which shows that the circle passes through the points of intersection of the circle x2 + y2 = 32 and the line x – y = 0 i.e. y = x and thus statement-2 is true.

So statement-1 is true. For statement-2, by definition of hyperbola distance of P from a focus is e times its distance from the corresponding directrix. aˆ Ê So PS = e Á x - ˜ and PS¢ = e Ë e¯

aˆ Ê ÁË x + ˜¯ . e

fi |PS – PS¢| = 2a. Thus statement-2 is also true but does not justify statement-1. 43. If the angle in statement-2 are a and b such that a + b = p/2 then the product of the slopes is tan a tan b = 1, and the statement-2 is true. In statement-1, any tangent to the ellipse is y = mx + a 2 m 2 + b 2 which passes through

(a

2

- b 2 sec q , a 2 - b 2 tan q

)

fi (a2 – b2) (tan q – m sec q)2 = a2m2 + b2 Product of the slopes =

(a 2 - b2 ) tan 2 q - b2

a 2 sin 2 q - b 2

=1 a 2 sin 2 q - b 2 So by statement-2, statement-1 is also true. 44. Statement-2 is true, as the distance between the foci x2 y 2 of 2 - 2 = 1 is 2ae where 2a is the length of the a b transverse axis and e is the eccentricity. Using it in statement-1, length of the transverse axis is

(a 2 - b2 )sec 2 q - a 2

=

(4 + 6)2 10 ¥ 4 = =8 5/ 4 5

and thus statement-1 is also true.

20.30

Complete Mathematics—JEE Main

45. Equation of the asymptotes is

x y ± =0 a b

And if q is an angle between the asymptotes tan q = b a = 2 tan f where f = tan -1 b 1 + tan 2 f a b2 1+ 2 a = tan 2 f b fi q = 2 f = 2 tan–1 and the statement-1 is true. a 2

Using in statement-1. p b b p If = 2 tan -1 fi tan = 3 a a 6 fi 3b2 = a2 fi 3(e2 – 1) = 1 fi e =

fi secq = 3

46. Foci (± 5, 0), length of the latus rectum =

4h 2 2¥9 4

2

4b 2 k 2 81a 4

=1 x2

y2

-

=1 81a2 81a 4 4 4b 2 which is a hyperbola and its eccentricity

Locus of (h, k) is

1+

2

3Ê 9ˆ ÁË x - ˜¯ 4 2

-

81a 2

fi 16(x + y ± 10x + 25) = 81 or 16x2 + 16y2 ± 160x + 319 = 0 and statement-1 is true, statement-2 is also true as the focus of y2 = 20x is (5, 0) but does not justify statement-1. 47. Distance of P from the fixed points F1(3, 2) and F2 (6, –2) is constant 3. So by definition, locus of P is a hyperbola with foci F1 and F2, length of the transverse axes is 3. If e is the eccentricity then F1F2 = 2ae where 2a = 3. 5 1 fie= 9 + 16 = and the statement-1 is true. 3 3 6-3 In statement-2 slope of the conjugate axis is -2 - 2 3 = and it passes through the mid-point of F1F2. 4

y=

2bk 2h , tanq = [∵ a2 + b2 = a2e2 = 9a2] 9a 2 9a

Eliminating q, we get

So the equation of a required circle is (x ± 5)2 + y2 =

So its equation is

a 2 + b2 a 2 + b2 ,k= 2a cosq 2b cotq

2

fi b2 = 3a2 fi e2 – 1 = 3 fi e = 2 Showing that statement-1 is false.

2

Ê a 2 + b2 ˆ Ê a 2 + b2 ˆ , 0˜ , M Á 0, It meets the axes at L Á Ë a cos q ¯ Ë b cot q ˜¯

then h =

p b b p = p - 2 tan -1 fi = tan = 3 a a 3

Ê 9ˆ ÁË ˜¯ 4

ax by + = a2 + b2 sec q tan q

If (h, k) is the mid-point of LM

3 If

fi 6x – 8y = 27 Thus statement-2 is also true but does not lead to statement-1. 48. Equation of the normal at (a sec q, b tan q) to the x2 y 2 hyperbola 2 - 2 = 1 is a b

=

81a 4 4b

2

¥

4 81a

2

= 1+

a2 b

2

= 1+

1 2

e -1

= 1+

1 8

3 2 2

So statement-1 is true. In statement-2, the eccentricity is 8a 2

=3 a2 and the statement-2 is false. 49. Equation of a tangent to the hyperbola is y = mx + a 2 m 2 - b 2 and the equation of the tangent perpen1+

dicular to it is y = -

a2 1 x+ - b2 . 2 m m

Eliminating m we get the required locus as (y – mx)2 + (x + my)2 = a2m2 – b2 + a2 – b2m2 fi x2 + y2 = a2 – b2 and the statement-2 is true using which statement-1 is false. 2 to the hyperbola 50. Equation of a tangent with slope 5 2 2 x y = 1 is 9 4

Hyperbola 20.31

y=

2 5

x+ 9¥

4 - 4 fi 2x – 5

5y+4=0

Next 2x – 5 y + 4 = 0 touches the circle (x – 4)2 + y2 = 16 if the length of the perpendicular from (4, 0) on the line is 4 which is true. Hence statement-1 is true. In statement-2, let A be (3 secq, 2 tanq) A lies on the circle x2 + y2 – 8x = 0 fi 13sec2q – 24 secq – 4 = 0 fi secq = 2 fi tanq = ±

3

So the coordinate of A are (6, 2 3 ) and of B are (6, -2 3 ) and equation of the circle on AB as diameter is (x – 6) (x – 6) + ( y - 2 3 )( y + 2 3 ) = 0. fi x2 + y2 – 12x + 24 = 0 which does not pass through the centre (0, 0) of the hyperbola. Thus statement-2 is false.

Level - 2 51. Let (h, k) be the mid-point of the chord, then its equahx ky h2 k 2 tion is 2 - 2 = 2 - 2 a b a b Since it touches the circle x2 + y2 = c2 2

2 e22 = 5 - 5 sin q = cos2q. 5

e1 =

7 e2 fi 1 + sin2 q = 7 – 7 sin2q

fi sinq =

p 3 fiq= . 3 2

54. Slopes of the asymptotes are ± b Ê bˆ - Á- ˜ a Ë a¯ fi tanq = Ê bˆ Ê bˆ 1+ Á ˜ Á- ˜ Ë a¯ Ë a¯ Êqˆ 2 tan Á ˜ Ë 2¯ 2(b / a) fi = 2 q Ê ˆ Ê bˆ 1 - tan 2 Á ˜ 1- Á ˜ Ë 2¯ Ë a¯

Êqˆ b Êqˆ fi tan Á ˜ = as tan Á ˜ > 0 Ë 2¯ Ë 2¯ a Êqˆ fi cos Á ˜ = Ë 2¯

1

2 2 Ê h ˆ +Ê k ˆ ÁË 2 ˜¯ ÁË 2 ˜¯ a b

=±c

=

b2 a2

1+

2

h k a 2 b2

a a 2 + b2

55. Let OA = r1 and the coordinates of A be (r1 cos a, r1 sin a). If OB = r2 then coordinates of B are

2

Ê h2 k 2 ˆ Ê h2 k 2 ˆ fi Á 2 - 2 ˜ = c2 Á 4 + 4 ˜ b ¯ b ¯ Ëa Ëa

Ê pˆ p ˆˆ Ê Ê ÁË r2 cos ÁË a + 2 ˜¯ , r2 sin ÁË a + 2 ˜¯ ˜¯ As A, B lie on the hyperbola

Required locus is 2

Ê x2 y 2 ˆ Ê x2 y 2 ˆ 2 Á 2 ˜ =c Á 4 + 4˜ b2 ¯ b ¯ Ëa Ëa

Ê cˆ to H2 is (ct)x + Á ˜ y = 2k2. Ët¯

a2

-

y2 b2

=1

Ê sin 2 a cos 2 a ˆ 1 1 + r22 Á 2 2 ˜ =1fi 2 (OA) (OB)2 b ¯ Ë a =

Asymptote of H2 are x = 0 and y = 0 and the chord meets Ê 2k 2 t ˆ Ê 2k 2 ˆ , 0˜ and B Á 0, these asymptotes at A Á c ˜¯ Ë Ë ct ¯

2 5 + 5 sin 2 q = 1 + sin2q. 53. e1 = 5

x2

Ê cos 2 a sin 2 a ˆ r12 Á ˜ =1 b2 ¯ Ë a2

Ê cˆ 52. Equation of the chord of contact from Á ct , ˜ on H1 Ë t¯

Area of the required triangle is

b a

1 2k 2 2k 2 t 2k 4 = 2 . c 2 ct c

1 r12

56. e2 =

+

1 r22

=

1 a

2

-

1 b2

.

25 + 16 fie= 25

Directrix is x = ± Asymptotes are

41 5

25 41

x2 y 2 =0 25 16

aˆ Ê ÁË x = ± ˜¯ e

20.32

Complete Mathematics—JEE Main

Ê 25 20 ˆ , So the required coordinates are Á ˜ Ë 41 41 ¯ 57. Normal at any point (ct, c/t) on the hyperbola xy = c2 is y – (c/t) = t2 (x – ct) or t3 x – ty + c – ct4 = 0 If it passes through another points (ca, c/a) on the hyperbola, then t3 ¥ ca – t ¥

c + c – ct4 = 0 a

Ê c(t1 + t2 + t3 ) c(1/ t1 + 1/ t2 + 1/ t3 ) ˆ , ÁË ˜¯ = (0, 0) 3 3 x2 a2

-

tion of the conjugate hyperbola is

y2 b2 x2

a2

-

a2

b4 x2 b2 a 4 m2

= 1 fi x2 =

and of the hyperbola 2

Ê 12 ˆ Ê 9ˆ ± Á ˜ +Á ˜ Ë 5¯ Ë 5¯

x2 y2 144 81

=

1 25

= 1, so equay2 b2

= –1

a 2 m2 m2 - b2

These points are real if m2 > b2 and for this value of m, the diameter will meet the conjugate hyperbola at -a 2 m2 and these points are point for which x2 = 2 m - b2 imaginary. Note: Conjugate diameter will meet the conjugate hyperbola in real points and the hyperbola in imaginary points. 59. Equation of a common tangent is x = ±1, nearer to P (1/2, 1) is x = 1. So the directrix of the ellipse is x = 1 and the focus is P (1/2, 1), e = 1/2. Thus the required equation is (x – 1/2)2 + (y – 1)2 = 1/4(1 – x)2 fi 3x2 + 4y2 – 2x – 8y + 4 = 0

are

2

If the foci of the two conics coincide then 16 – b2 = 144 + 81 = 9 fi b2 = 7 25 x2 a2

-

y2 b2

is y = mx + a 2 m 2 - b 2 Comparing it with y = a x + b, we get

Equation of a diameter of the hyperbola is y = (b2/a2m) x, which meets the hyperbola at points given by -

x2 y 2 + 2 =1 are (± 16 - b 2 , 0) 16 b

2. Equation of tangent to the hyperbola

which is the centre of hyperbola.

x2

Previous Years’ AIEEE/JEE Main Questions 1. Foci of the ellipse

fi t 3a 2 – t + a – a t 4 = 0 fi (t3a + 1) (a – t) = 0 fi t3a + 1 = 0 as a π t which gives three values of t say t1, t2, t3 and hence three points P, Q, R on the hyperbola the normals at which pass through S. We have t1 + t2 + t3 = 0 = St1t2 and t1t2t3 = –1/a. Centroid of the triangle PQR is

58. Equation of the hyperbola be

60. Equation of the asymptotes differ from the equation of the hyperbola by a constant so the equation of the asymptotes is xy – hx – ky + l = 0 which represents a pair of lines if l = hk and the required asymptotes are x = k and y = h.

m = a, b =

a 2 m2 - b2 fi b 2 = a 2a 2 – b 2 fi a 2a 2 – b 2 = b 2 fi Locus of (a, b) is a2x2 – y2 = b2 which is a hyperbola. 3. Equation of normal at point (x, y) is dx Y–y= (X – x) dy Ê dx ˆ It meets the x-axis at G Á x + y , 0˜ Ë dy ¯ We are given Ê dx ˆ ÁË x + y dy ˜¯ = 2|x| fix+y

dy = ±2x dx

fix+y

dy dy = 2x or x + y = – 2x dx dx

fi y dy = x dx or y dy = – 3x dx fi x2 – y2 = c or 3x2 + y2 = c Thus curve is either a hyperbola or an ellipse.

=1

Hyperbola 20.33

4. b2 = a2(e2 – 1) fi sin2a = cos2 a (e2 – 1) fi e2 – 1 = tan2a fi e = sec a Coordinates of foci are (± ae, 0) = (±1, 0) \ abscissae of foci remains constant.

So equation of a common tangent is x = ± y ± the required equation is x – y =

x2

y2

= 1, a 2 b2 ± ae = ± 2 and e = 2 fi a = 1 and b2 = 1(4 – 1) fi b2 = 3

5. Let the equation of the hyperbola be

-

2 and the required equation is x2 – y = 1 3 or 3x2 – y2 = 3. 6. Equation of the tangent at a point (x1, y1) on the xx yy x2 y 2 hyperbola = 1 is 1 - 1 = 1 4 2 4 2

(0, 0) Ê -2 ˆ ÁË 0, y ˜¯ 1

P Ê 4 , 0ˆ ÁË x ˜¯ 1

O

Q

R

Ê4 ˆ So coordinates of P are Á , 0˜ and of Q are Ë x1 ¯

Ê -2 ˆ ÁË 0, y ˜¯ 1

Let the coordinates of R be (h, k) 4 -2 ,k= h= x1 y1

fi

h2

-

k2

9 = 4 (e12 - 1) fi e1 =

13 / 2

Foci of the hyperbola are (±2e1, 0) = (± 13 , 0) If e2 is the eccentricity of the ellipse, then

=1

Locus of (h, k) is

16 20 -4 = - . 9 9

x2 y 2 =1 4 9 If e1 is the eccentricity of the hyperbola, then

x12 y12 =1 4 2 2

Now, OA2 – OB2 =

10. Equation of hyperbola is

(x1, y1) lies on the hyperbola.

4

3 5 Ê yˆ x - Á ˜ = 1 or 3x – 2y = 4 4 2 Ë 5¯

Ê4 ˆ Coordinates of A are Á , 0˜ and that of B are (0, – 2). Ë3 ¯

Fig. 20.10

So

3 . 2

8. An equation of the normal at P is 3x cosq + 2y cotq = 9+4 fi 3x + 2y cosecq = 13 secq (1) An equation of normal at Q is 3x + 2y cosec f = 13 sec f (2) Subtracting (2) from (1), we get 2y[cosecq – cosec f] = 13(secq – sec f) fi 2y[cosecq – secq] = 13(secq – cosecq) [∵ f = p/2 – q] 13 fiy= 2 9. Coordinates of P, an extremity of latus rectum in the Ê 5ˆ first quadrant are Á 3, ˜ . An equation of tangent at P Ë 2¯ is

(h, k)

3 and 2

4 x

2

-

2 y

2

e1e2 = 1/2 fi e2 = 1/ 13 . =1

7. Equation of a tangent to x2 = 6y is x = my +

3 2m

x2 a2

+

y2 b2

= 1.

As it passes through (± 3, 0) , 13/a2 = 1 fi a2 = 13.

1 3 fiy= xm 2 m2 which touches the hyperbola

Let the equation of ellipse be

x2 y2 =1 9 9 2 4

2 9 1 9 Ê -3 ˆ if Á 2 ˜ = ¥ 2 - fi m2 = 1 Ë 2m ¯ 2 m 4

1ˆ Ê Also, b2 = a2 (1 - e22 ) = 13 Á1 - ˜ = 12. Ë 13 ¯ \ equation of ellipse is

x2 y 2 = 1. + 13 12

Ê1 3ˆ Á 2 13, 2 ˜ does not lie on it. Ë ¯

20.34

11.

Complete Mathematics—JEE Main

Previous Years’ B-Architecture Entrance Examination Questions

2b 2 = 8 fi b2 = 4a a Also, 2b =

1 (2ae) fi 2b = ae 2

1. Equation of OP is y =

2

bola

Ê1 ˆ \ Á ae˜ = 4a fi ae2 = 16 Ë2 ¯

x2 a2

-

y2 b2

3 y2 a2

-

y2 b2

=1

P x

O Q

fi a = 12 \ b2 = 4a fi a2(e2 – 1) = 4a fi a(e2 – 1) = 4

Fig. 20.11 2

fi (3b – a )y = a2b2 For real values of y,

1 3

3 12. Equation of ellipse is x2 y 2 =1 + 12 16 Its major axis is along y-axis, and its foci are (0, 2) and (0, –2). Let equation of hyperbola be

y2 a2

2

2

3b2 > a2

fi 3a2(e2 – 1) > a2 fi 3e2 > 4 fi e >

2

-

x2 b2

= 1, then a =

2 and b2 = a2(e2 – 1) Ê9 ˆ fi b2 = 4 Á - 1˜ = 5 Ë4 ¯ Thus, equation of hyperbola is y 2 x2 = 1 4 5

x which meets the hyper-

p6

b2 = 16 fi a + 4 = 16 fia+ a

fie=

3

= 1 at points for which

Ê b2 ˆ fi a Á1 + 2 ˜ = 16 Ë a ¯

fi 12(e2 – 1) = 4 fi e2 = 1 +

1

2

3 2. Equation of the tangent at (x1, y1) to the hyperbola x2 – y2 = 4 is xx1 – yy1 = 4 -4 4 then a1 = , b1 = . y1 x1 Equation of the normal at (x1, y1) is y1x + x1y = 2x1y1 then a2 = 2x1, b2 = 2y1. Hence a1a2 + b1b2 = 0. 3. Let y = mx + c be a common tangent to x2 – 2y2 = 18 and x2 + y2 = 9. As it touches the circle, c2 = 9(1 + m2) and as it touches the hyperbola, c2 = 18m2 – 9 So 18m2 – 9 = 9 + 9m2 fi m2 = 2, c = 3 3 and the

(1)

Note that ( 5 , 2 2 ) , (0, 2) and ( 10 , 2 3 ) lie on (1) 13. As 9e2 – 18e + 5 = 0 fi 9e2 – 15e – 3e + 5 = 0 fi 3e(3e – 5) – (3e – 5) = 0 fi (3e – 1) (3e – 5) = 0 fi e = 1/3, 5/3 As e > 1, e = 5/3 We have ae = 5 fi a = 3 Also, a2 – b2 = a2 – a2(e2 – 1) = 9(2 – 25/9) = – 7

required equation is y = 2 x + 3 3 . 4. If coordinates of R are (x, y), then x=

(3)(2) - 2(2 tan q ) 3 (3) - 2(3 sec q ) ,y= 3-2 3-2

fi tanq =

1 1 (6 – x), secq = (9 – y) 4 6

As sec2q – tan2q = 1, we get 1 1 (y – 9)2 – (x – 6)2 = 1 36 16

Hyperbola 20.35

It is a hyperbola, length of whose transverse axis is 2(6) = 12. 2

2

5. Foci of x + y = 1 are ( 25 - 9 , 0) and 25 9 ( - 25 - 9 , 0) i.e. (4, 0) and (–4, 0). If a is the length of transverse axis of the hyperbola, then ae = 4 fi a = 2

Also, b2 = a2(e2 – 1) = 4(4 – 1) = 12 \ Equation of hyperbola is

x2 y 2 =1 4 12

An equation of tangent at (4, 6) is or 2x – y = 2.

4x 6 y =1 4 12

CHAPTER TWENTY ONE

Three Dimensional Geometry

Z

COORDINATES

A

Distance Formula

, y, z P (x

Distance between A(x1, y1, z1) and B (x2, y2, z2) is given by r

( x2 - x1 )2 + ( y2 - y1 )2 + ( z2 - z1 )2 .

AB =

)

z

O

Y

Section Formulae The coordinates of the point P which divides the join of A (x1, y1, z1) and B(x2, y2, z2) in the ratio m : n are Ê mx2 + nx1 my2 + ny1 mz2 + nz1 ˆ , , ÁË ˜ m+n m+n m+n ¯ By taking m = n of AB as Ê x1 + x2 y1 + y2 z1 + z2 ˆ , , ÁË ˜ 2 2 2 ¯ Coordinates of any point on the join of two points Ê l x + x1 l y2 + y1 l z2 + z1 ˆ A (x1, y1, z1) and B(x2, y2, z2) are Á 2 , , Ë l +1 l +1 l + 1 ˜¯ where l π – 1. If l takes only positive real values, we get AB (except A and B).

If Ai(xi, yi, zi), i = 1, 2, 3, 4 are the vertices of a tetrahedron,

1 6

y1 y2 y3 y4

x

M

N

y

X

Fig. 21.1

Let P(x, y, z) be any point on A¢OA of OP be r. Let PN P on the plane XOY, and NM N on OX. Then PN = z, NM = y, OM = x x = r cos a, y = r cos b, z = r cos g. and cos2 a + cos2 b + cos2 g = 1.

Direction-Ratios If l, m, n

Volume of a Tetrahedron

x1 x2 x3 x4

A¢

z1 z2 z3 z4

1 1 1 1

Direction-Cosines If a, b, g with the positive directions X¢OX, Y¢OY, Z¢OZ of the coordi nate axes, then cos a, cos b, cos g are called the directioncosines of the line.

OP and a, b, c be l, m, n respectively, then a, b, c are called the direction ratios of OP. If l, m, n cosines of a line and k (π kl, km, kn OP. If a, b, c a, cos b, cos g tions cos a cos b cos g = = a b c = ±

cos2 a + cos2 b + cos2 g 2

2

a +b +c

2

=

±1 2

a + b2 + c 2

21.2

Complete Mathematics—JEE Main

If P be the point (a, b, c), and cos a, cos b, cos g are the direction cosines of the directed line OP, then a

cos a =

2

2

b

, cos b =

2

2

(a + b2 + c 2 )

(a + b + c )

,

tan q = ±

[(a1b2 - a2 b1 )2 + (b1c2 - b2 c1 )2 + (c1a2 - c2 a1 )2 ]1 / 2 a1a2 + b1b2 + c1c2

The lines are parallel to each other if and only if

c

cos g =

2

2

a1 b c = 1 = 1 a2 b2 c2

2

(a + b + c )

-

a 2

2

PO are b

,-

2

2

(a + b + c )

2

2

,-

(a + b + c )

Also, the lines are perpendicular to each other if and only if a1a2 + b1b2 + c1c2 = 0.

c 2

(a + b2 + c 2 )

2

Illustration

1

Illustration

Find the direction cosines of the line whose direction ratios are 12, 4, – 18. Solution: Direction cosines of the given line are 12 2

4

,

2

(12) + (4) + (-18)

2

2

, 2

(12) + (4) + (-18)

Direction ratios of a line L1, are 1, –1, 1 and of L2 are 1, 0, l. Find the value of l so that L1 and L2 are perpendicular to each other. Solution: The line L1 and L2 are perpendicular if the angle q 1(1) + (-1)(0) + (1)(l )

fi cos q =

2

-18

2

(1) + (-1)2 + (1)2 12 + 0 + l 2

fi l = –1

2

(12) + (4)2 + (-18)2 12 4 -18 or , , 22 22 22

Equation of a Plane

6 2 -9 or , , . 11 11 11

by ax + by + cz + d x, y, z represents a plane. f the yz x = 0. zx y = 0. xy z = 0. (d) If l, m, n a plane and p be the length of the perpendicular

Angle Between Two Lines (a) If q cosines are (l1, m1, n1) and (l2, m2, n2) then cos q = l1l2 + m1m2 + n1n2. The expressions for sin q, tan q are given below: m1 m2

sin2 q =

n1 n2

2

n1 n2

+

l1 l2

2

+

l1 l2

m1 m2

2

the plane is lx + my + nz = p.

and tan q = ± [S (l1 m2 – l2 m1)2]1/2/(l1l2 + m1m2+ n1n2) The lines are parallel to each other if and only if l1/l2 = m1/m2 = n1/n2. Also the lines are perpendicular to each other provided l1l2 + m1m2 + n1n2 = 0. (b) Angle q a1, b1, c1 and a2, b2, c2 is given by cos q = ±

a1a2 + b1b2 + c1c2 a12

+ b12 + c12

3

Illustration

4x – 3y + 5z the origin on this plane Solution: 4, –3, 5. So direction cosines are 4 -3 ,m= , l= 2 2 2 2 4 + (-3) + 5 4 + (-3)2 + 52

a22 + b22 + c22

n=

sin q = 2

±

THE PLANE

2

2 1/ 2

or l =

[(a1b2 - a2 b1 ) + (b1c2 - b2 c1 ) + (c1a2 - c2 a1 ) ] a12 + b12 + c12

a22 + b22 + c22

4 5 2

4 5 2

x–

,m=

3 5 2

y+

–3 5 2 5 5 2

,n=

z =

5 5 2

25 5 2

5 2

4 + (-3)2 + 52

Three Dimensional Geometry 21.3 or lx + my + nz = p where p =

25 5 2

=

5 2

a, b, c on the axes of x, y, z x y z + + = 1. a b c

cos q = ±

(a12 + b12 + c12 ) (a22 + b22 + c22 )

Parallelism and Perpendicularity of Two Planes The planes a1x + b1y + c1z + d1 = 0, a2x + b2y + c2z + d2 = 0 are parallel if and only if a1 b1 c1 = = ; a2 b2 c2

Systems of Planes

conditions. If we consider a plane satisfying just two given

a1a2 + b1b2 + c1c2

and perpendicular if and only if a1a2 + b1b2 + c1c2= 0.

Two Sides of a Plane If we consider a plane satisfying one given condition, its

ax + by + cz + k = 0 represents a ax + by + cz + d = 0, k ax + by + cz + k = 0, represents a sys x y z = = . a b c a1x + b1y + c1z + d1) + k (a2x + b 2y + c 2z + d 2 passing through the intersection of the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0, k Illustration

Two points P(x1, y1, z1) and Q(x2, y2, z2 different sides of the plane ax + by + cz + d = 0 according as ax1 + by1 + cz1 + d and ax2 + by2 + cz2 + d or of different signs.

Length of the Perpendicular from a Point to a Plane The perpendicular distance of the point (x1, y1, z1 plane lx + my + nz = p is |p – lx1 – my1 – nz1|, where l, m, n p is The perpendicular distance of the point (x1, y1, z1 the plane ax + by + cz + d = 0 is ax1 + by1 + cz1 + d

(a 2 + b2 + c 2 )

4

Bisectors of the Angles between Two Planes of the planes 3x + 4y + 7z + 5 = 0 and 2x – 3y + 4z – 6 = 0 and x. Solution: the given planes is 3x + 4y + 7z + 5 + l (2x – 3y + 4z – 6) = 0 6l – 5 =1fil=2 3 + 2l

Intercept on x

7x – 2y + 15z – 7 = 0

A(x – x1) + B(y – y1) + C(z – z1) = 0. point (x1, y1, z1), the ratios of A, B, C being the

(A) Let a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 d1 and d2 are both positive. Then a1 x + b1 y + c1z + d1 a x + b2 y + c2 z + d2 = 2 a12 + b12 + c12 a22 + b22 + c22

(

)

(

)

given planes which contains the origin, and a1 x + b1 y + c1z + d1

(a12 + b12 + c12 )

=–

a2 x + b2 y + c2 z + d2

(a22 + b22 + c22 )

given planes which does not contain the origin. Illustration

Angle Between Two Planes Angle q between the planes a1x + b1y + c1z + d1 = 0, a2x + b2y + c2z + d2 = 0, is given by

5

Find the angle between the planes x + y + z + 1 = 0 and x–y+z– the angle between these planes which contain the origin. Solution: If q is the angle between these planes, then

21.4

Complete Mathematics—JEE Main (1) ¥ (1) + (1)(-1) + (1)(1)

fi cos q =

2

1 + 12 + 12 (1)2 + (-1)2 + (1)2

=

1 3

P and Q are two points on a line whose direction ratios are 3, 5, 1. If the coordinates of P are (1, – 4, 2) and (PQ)2 = 35. Find the coordinates of Q. Solution: Direction cosines of the line are 3 5 1 , , 2 2 2 2 2 2 2 3 + 5 +1 3 + 5 +1 3 + 52 + 12

fi q = cos– 1 (1/3) the given planes is x + y + z +1 2

2

2

1 +1 +1 fix+z=0

=

-x + y - z +1 (1)2 + (-1)2 + (1)2

or

gles between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are a1 x + b1 y + c1z + d1 a x + b2 y + c2 z + d2 = ± 2 a12 + b12 + c12 a22 + b22 + c22

(

)

(

6

Illustration

)

3 35

,

5 35

,

1 35

x -1 y+4 z-2 = = = r = 35 = PQ . 3 / 35 5 / 35 1 / 35 and the coordinates of Q are. (3 + 1, 5 – 4, 1 + 2 ) = (4, 1, 3)

If l, m, n

The bisector of the acute angle between the given planes

a straight line, but in this case r x1, y1, z1). the distance of (x, y, z

given planes.

not

x-a y-b z-c = = , l n m

THE STRAIGHT LINE

represent a straight line passing through the point (a, b, c) and having direction ratios l n.

Equations of a Line

Equation of Line through two Given Points nates of every point on the line of intersection of the planes

If A(x1, y1, z1), B(x2, y2, z2 of the line AB is x - x1 y - y1 z - z1 = = x2 - x1 y2 - y1 z2 - z1

er represent that line. Therefore ax + by + cz + d = 0 and a¢x + b¢y + c¢z + d¢ = 0 together represent a straight line.

The coordinates of a variable point on AB can be ex l

The Equations to the Axes of Coordinates x y z

y = 0, z = 0. x = 0, z = 0. x = 0, y = 0.

x=

Symmetrical form of the Equation of a Line If a straight line passes through a given point (x1, y1, z1) and has direction cosines l, m, n, then the coordinates of any

l x2 + x1 l y2 + y1 l z2 + z1 ,y= ,z= , l +1 l +1 l +1

l x, y, z) are the coordinates of the point which divides the join of A and B in the ratio l : 1.

Changing unsymmetrical form to symmetrical form

x - x1 y - y1 z - z1 = (= r) = l m n

ax + by + cz + d = 0, of any point (x, y, z x 1, y 1, z 1 x = x1 + lr, y = y1 + mr, z = z1 + nr. distance form line.

r

a¢x + b¢y + c¢z + d¢ = 0. bd ¢ - b ¢d da ¢ - d ¢a yz ab ¢ - a ¢b ab ¢ - a ¢b = = bc ¢ - b ¢c ca ¢ - c ¢a ab ¢ - a ¢b

x-

Three Dimensional Geometry 21.5

Illustration

7

sin q =

Find the direction ratios of the line x + 2y – z = 3, 2x – y + z

1 2 2 -1

The lines

or 1, –3, –5 For any point on the line, let x = 0, then x = y = 1 and the

x -1 y -1 z - 0 = = 1 -3 -5

x -a y-b z-g = , = l m n

x - a¢ y - b¢ z - g ¢ = = l¢ m¢ n¢ are coplanar if and only if and

(1)

Verification Any point on the line is r + 1, –3r + 1, – 5r r + 1 + 2(–3r + 1) – (–5r) = 3 for all r and 2(r + 1) – (–3r + 1) + (–5r) = 1 for all r. showing that each point on the line

(a2 + b2 + c2 ) (l 2 + m2 + n2 )

Coplanar Lines

Solution: Direction ratios of the line are 2 -1 1 -1 , , -1 1 2 1

al + bm + cn

(1)

a - a¢ b - b¢ g - g ¢ l m n =0 l¢ m¢ n¢ the plane containing the lines is x -a l l¢

Number of Constants in the Equation of a Line x-a y-b z-c = = , m n l of a line can be written as x = (l/n) z + a – (lc/n), y = (m/n) z + b – (mc/n), x = Az + B, y = Cz + D

Illustration

x -a y - b z -g = = , l m n represent a given plane and a straight line respectively. (i) The line is perpendicular to the plane if and only if a b c = = , l m n

z-g n n¢

= 0.

8

Find the value of l, so that the lines x -1 y - 2 z + l and = = 2 3 1 x +1 y -1 z - 3 are coplanar = = 2 3 1

Therefore the equation of a line contains four arbitrary constants.

A Plane and a Straight Line

y-b m m¢

Solution: Given lines are coplanar if 1 + 1 2 - 1 -l - 3 1 2 3 =0 fil=6 2 3 1

ax + by + cz + d = 0,

x -1 y - 2 z + 6 1 2 3 =0 2 3 1 or 7x –5y + z

General Equation of a Plane Containing a Line

(ii) The line is parallel to the plane if and only if al + bm + cn = 0. (iii) The line lies in the plane if and only if al + bm + cn = 0 and aa + bb + cg + d = 0.

Angle between a Line and a Plane The angle q between the line x -a y-b z-g = = , l m n and the plane ax + by + cz + d = 0, is given by

is where

x -a y-b z-g = = , l m n A(x – a) + B(y – b) + C(z – g) = 0, Al + Bm + Cn = 0.

Length of the Perpendicular from a Point to a Line The length p P(x1, y1, z1) to a given line

x -a y-b z-g = = l m n

21.6

Complete Mathematics—JEE Main

(l, m, n are direction cosines of the line), is given by p2 = (x1 – a)2 + (y1 – b)2 + (z1 – g)2 – [l(x1 – a) + m(y1 – b) + n(z1 – g)]2. Another expression for p2 is given by p2 =

x1 - a l

y1 - b m

z1 - g n

+

Illustration

2

x1 - a l

y1 - b m

+

z1 - g n

r = a + tb where t two points with position vectors a and b is r = (1 – t)a + tb where t

2

2

through the point with position vector a and which is parallel to the vectors b and c is r = a + tb + pc where, t and p

9

line x -1 y - 3 z - 4 = = 3 1 -5 Solution: Let P(2, 1, 3) be the given point A(1, 3, 4) is a point on the given line p=PM P on the line. P(2,1,3)

q A(1, 3,4)

through two given points with position vectors a and b and parallel to the vectors c is r = a + t(b – a) + pc = (1 – t) a + tb + pc where t and p with position vectors a, b, c is r ◊ (b ¥ c + c ¥ a + a ¥ b) = [a b c]. 6. Normal form of the vector equation of a plane r ◊ n = p n p is the plane.

M

7. (r – a) ◊ N to the vector N and passing through a point with position vector a.

Fig. 21.2 Let PAM = q then

p2 = (PM)2 = (AP)2 – (AM)2 = (AP)2 – (AP cos q )2 fi p2 = (AP)2 sin2 q Direction ratios of AP are 2 –1, 1 – 3, 3 –4 or 1, –2, –1 and of the given line are 3, 1, –5 so cosq =

Illustration

point (2, 5, –3) and perpendicular to the line with direction ratios 3, 7, 5. Solution: Let a = 2i + 5j – 3k

(1)(3) + (-2)(1) + (-1)(-5) 2

(1) + (-2)2 + (-1)2 (3)2 + (1)2 + (-5)2

=

6 6 35

35

N = 3i + 7j + 5k

and =

6 35

(AP)2 = (2 –1)2 + (1 – 3)2 + (3 – 4)2 = 6 Hence p2 = 6 ¥ 29 fi p =

10

174 35

Equation of a Line through the Intersection of Given Lines If L1 = 0 = L2 and L3 = 0 = L4, where LK = akx + bky + ckz + d k, k for all values of l and m L1 + lL2 = 0 = L3 + mL4 represents a line through their intersection.

VECTORIAL EQUATIONS a point with position vector a and parallel to the vector b is

(r – a) ◊ N = 0 or [r – (2i + 5j – 3k)] (3i + 7j + 5k) = 0

8. Angle between two planes r ◊ N1 = p1 and r ◊ N2 = p2 is cos- 1

N1 ◊ N 2 N1 N 2

Angle between a line and a plane whose vectorial r = a + bt and r ◊ N = q respectively is sin - 1

N ◊b . N b

10. The perpendicular distance of a point with position vector a

r ◊ N = q is q - a ◊ N . N

Three Dimensional Geometry 21.7 r = l (i – j + k) and r = (i – j) + m (–2j + k)

r = a + tb and r = c + pd is (r – a) ◊ b ¥ d =0 fi r ◊ b ¥ d = a ◊ b ¥ d. 12. Condition for the lines r = a + tb and r = c + pd to be coplanar is [cbd] = [abd]. 13. The line LM of shortest distance between the lines r = a + tb and r = c + pd is parallel to the vector b ¥ d. Length of LM is the projection of the line joining the points with position vectors a and c on LM.

(c - a ) ◊ (b ¥ d)

LM =

Thus

[cbd ] - [abd ]

=

b ¥d

b ¥d

Also, the line of shortest distance is the line of in tersection of the two planes through the given lines and the line LM (r – a) ◊ b ¥ (b ¥ d) = 0, (r – c) ◊ d ¥ (b ¥ d) = 0 The line of intersection of these planes is the re

Solution: a = 0, c = i – j,

b=i–j+k

d = – 2j + k

(r – a) ◊ b ¥ (b ¥ d) = 0, (r – c) ◊ d ¥ (b ¥ d) = 0 (b ¥ d) = i – j – 2k b ¥ (b ¥ d) = 3i + 3j d ¥ (b ¥ d) = 5i + j + 2k, c ◊ [d ¥ (b ¥ d)] = 4 r ◊ (3i + 3j) = 0, r ◊ (5i + j + 2k) = 4 r = (i – j So the line of shortest distance is r = (i – j) + l (i – j – 2k) Or

x -1 y +1 z = = 1 -1 -2 (c - a) ◊ (b ¥ d) b¥d

lines. Illustration

11

=

(i - j) ◊ (i - j - 2k ) 2

2

(1) + (-1) + (-2)

2

=

2 6

lines

SOLVED EXAMPLES Concept-based Straight Objective Type Questions

Example 1: The acute angle between the lines such that the direction cosines l, m, n l + m + n = 0 and l2 + m2 – n2 = 0 is

Example 2: y, and z (a) 12

2

2

Solution: l +m =n fi (l + m)2 – 2lm = n2 fi lm = 0 as l + m = – n if l = 0, m = – n, if m = 0, l = – n So direction ratios of the lines are 0, 1, – 1 and 1, 0, – 1 and the angle q

fiq

(b) 7

Ans. (b)

Ans. (c)

cos q =

x,

(0) (1) + (1)(0) + (- 1) (- 1) 2

1 + (-1)

2

2

(1) + (-1)

2

=

1 2

2

Solution: Let r a, b, g respectively with x, y and z r cos a = 2, r cos b = 3, r cos g = 6 fi r 2 (cos2 a + cos2 b + cos2 g ) = 22 + 32 + 62 fi r = 7 as cos2 a + cos2 b + cos2 g = 1 Example 3: of intersection of the planes x + 2y =3, y – 2z + 1 = 0 and

21.8

Complete Mathematics—JEE Main

(a) 2x – y –10z = (c) 2x – y + 10z = 11 Ans. (c)

x – y + 7z = 11 (d) 2x – y – z = 10

Solution: tion of the given planes is x + 2y –3 + l (y – 2z + 1) = 0 or x + (2 + l) y –2 l z + (l – 3) = 0 . (1) (1) + 2(2 + l) + 0 ( –2 l) = 0 fi l = – 5/2 x – y + 10z = 11 x -1 y - 3 z - 4 = = Example 4: 3 1 -5 in the plane 2x – y + z + 3 = 0 is the line x+3 y-5 z-2 x+3 y-5 z-2 = = = = (b) (a) 3 1 -5 -3 -1 5 (c)

x-3 y+5 z-2 x-3 y+5 z-2 = = = = (d) -3 -1 5 3 1 -5

Ans. (a) Solution: Any point P on the given line is (3r +1, r + 3, – 5r + 4) P in the given plane will be the point P ¢ at a dis P the line through P perpendicular to plane. (1, 3, 4)

A

P(3r + 1, r + 3, -5r + 4)

M

P¢

Fig. 21.3

PP ¢ is x - (3r + 1) y - (r + 3) z - (-5r + 4) = = -2 1 -1 Note Direction of PP ¢ the plane.

Length of the perpendicular of P 2(3r + 1) - (r + 3) + ( -5r + 4) + 3 22 + ( -1) 2 + 12 So the coordinates of P ¢ are given by x - (3r + 1) y - (r + 3) = 1 -2 =

fi x = 3r + 1 – 4, y = 2 + r + 3, z = – 2 – 5r + 4 x+3 y-5 z-2 fi = = 3 1 -5 Example 5: the point of intersection of the lines x -1 y - 2 z - 3 x - 3 y -1 z - 2 = = = = and 3 1 2 1 2 3 (a) 7x + 2y + 4z = 54 (c) 4x + 3y + 5z = 50 Ans. (c)

(b) 3x + 4y + 5z = (d) 5x + 4y + 3z = 57

Solution: (3r + 1, r + 2, 2r + 3) which lies on the second line if 3r + 1 - 3 r + 2 - 1 2r + 3 - 2 fir=1 = = 1 2 3 So the point of intersection of the given lines is (4, 3, 5) Which lies on the planes given in (a), (b), (c). 54 49 50 , P2 = , P3 = 69 50 50 P1 < 7, P2, P3 > 7 P2 < P3 x + 3y + 5z = 50 P1 =

Example 6: Let ABC be a triangle with vertices at points A (2, 3, 5), B (–1, 3, 2) and C (l , 5, m A axes then (l , m (a) (10, 7) (b) (7, 5) (c) (7, 10) (d) (5, 7) Ans. (c) Solution: Coordinates of D BC are m + 2ˆ Ê l -1 , 4, ˜ . So direction ratios of AD are ÁË 2 2 ¯ l -1 m+2 l -5 m -8 , 1, . - 2, , 4 – 3, - 5 or 2 2 2 2 Since AD l -5 m -8 fi l = 7, m = 10 and (l, m) = (7, 10) =1= 2 2

= 6

z - (-5r + 4) 2 6 = =2 -1 (-2)2 + (1)2 + (-1) 2

Example 7: The plane containing the line x -1 y - 2 z - 3 = = and parallel to the line 1 2 3 x y z = = passes through the point 1 1 4 (a) (1, –2, 5) (b) (1, 0, 5) (c) (0, 3, – 5) (d) (–1, –3, 0) Ans. (b)

Three Dimensional Geometry 21.9

Solution: is a(x – 1) + b(y – 2) + c(z – 3) where a + 2b + 3c = 0, as it is parallel to the second line a + b + 4c = 0 solving, we get a b c = = +5 -1 -1

Let Q (a, b, g ) given point P (0, 3, –2). Then PQ is perpendicular to the PQ lies on the line. So 2 (a – 0) + (b –3) – (g + 2) = 0 fi 2a + b – g – 5 = 0(1)

5x – y –z = 0 which passes through (1, 0, 5)

fi

Example 8: The distance between the point (–1, –5, –10) and the point of intersection of the line x - 2 y +1 z - 2 = = = with the plane x – y + z = 5 is 3 4 12 (a) 13 (b) 15 (c) 16 (d) 12 Ans. (a) Solution: Any point on the given plane is (3r + 2, 4r – 1, 12r + 2) which lies on the given plane if 3r + 2 – (4r –1) + 12r + 2 = 5 fi r = 0 So the point of intersection of the line and the plane is (2 + 1) 2 + ( -1 + 5) 2 + (2 + 10)2 = 13 Example 9: Let a, b, g with the positive directions of the axes of reference in three q is the acute angle given by cos q =

cos 2 a + cos 2 b + cos 2 g , then q sin 2 a + sin 2 b + sin 2 g

(a)

p 6

(b)

p 3

(c)

p 2

(d)

p 4

Ans. (b) Solution: cos q =

cos 2 a + cos 2 b + cos 2 g 3 - (cos 2 a + cos 2 b + cos 2 g )

1 1 = = 3 -1 2 fi

q=

p 3

Example 10: 1- x in the line = 2 – y = z + 1 is 2 (a) (1, 2, –1) (b) (2, 1, 4) (c) (2, 1, 0) (d) (0, 0, 1) Ans. (c) Solution: x -1 = y - 2 = z +1 1 -1 2

a b +3 g -2 -1 -2 +1 and 2 = 2 = 2 -1 2 1 a - 2 b -1 g = r (say) = = 1 -1 2

(2)

2(2r +2) + r + 1 + r – 5 = 0 fi r = 0 Example 11: A, B, C then the point of intersection of the planes through A, B, C parallel to the coordinate planes lies on 1 1 yz + zx = 6 3 2 (b) yz – 2zx + 3xy = xyz (c) xy – 2yz + 3zx = 3xyz 1 1 (d) xy – yz – zx = 6 3 2 Ans. (b) (a) x y –

Solution: Let the coordinates of A be (a, 0, 0), B be (0, b, 0) and C be (0, 0, c x y z + + = 1. As it passes through (1, –2, 3) a b c 1 2 3 - + =1 a b c or bc – 2ac + 3ab = abc

(1) A, B, C parallel to the coor dinate planes are x = a, y = b, z = c So their point of intersection is (a, b, c) yz – 2zx + 3 xy = xyz Example 12: Let L1 be the line 2x –2y +3z – 2 = 0 = x – y + z + 1 and L2 be the line x + 2y – z – 3 = 0 = 3x – y + 2z L1 and L2 is (a) (c)

1 2 1 3 2

(b) (d)

1 2 2 1 15

Ans. (c) Solution: The plane through L1 is 2x – 2y + 3z – 2 + l (x – y + z + 1) = 0 and the plane through L2 is x + 2y – z – 3 + m (3x – y + 2z –1) = 0

21.10

Complete Mathematics—JEE Main

2+l 2+l 3+ l 2-l so = = = m-2 2m - 1 1 + 3m 3+ m

(c) r ◊ (24i + 45j + 30k) – 17 = 0 (d) r ◊ (45i + 33j + 50k) – 41 = 0 Ans. (a)

3 and l = 5 2 7x – 7y + 8z + 3 = 0.

Solution: tion of the given planes is r ◊ [(1 + 2l)i + (2 + l)j + (3 – l)k + (– 4 + 5l)] = 0

fi m = –

If it perpendicular to the plane r ◊ (5i + 3j – 6k) + 8 = 0 3 2

2

7 + ( -7) + (8)

2

=

1

then (1 + 2l) (5) + (2 + l) (3) + (3 – l) (– 6) = 0 7 fil= 19 place in r ◊ (33i + 45j + 50k) – 40 = 0.

3 2

Example 13: origin to the plane passing through (1, 0, 0), (0, 1, 0) and (0, 0, 1) is (a) (3, 3, 3) 1 1 1 (c) ÊÁ , , ˆ˜ Ë 3 3 3¯ Ans. (c)

Example 15: The points A, B and C with position vec tors a = 3i – 4j – 4k, b = 2i – j + k and c = i – 3j – 5k, respectively are (a) Collinear

1 1 1 (b) ÊÁ , , ˆ˜ Ë 2 2 2¯ (d) (2, 2, 2)

Solution: x + y + z = 1 only the point given in (c) lies on it. Also the line joining the origin 1 1 1 to ÊÁ , , ˆ˜ Ë 3 3 3¯ Example 14: the line of intersection of the planes r ◊ (i + 2j + 2k). = 0, r ◊ (2i + j – k) + 5 = 0 and which is perpendicular to the plane. r ◊ (5i + 3j – 6k) + 8 = 0 is

(c) Vertices of a right angled triangle (d) none of these Ans. (c) Solution: AB = (2 – 3)i + (– 1 + 4)j + (1 + 4)k = – i + 3j + 5k BC = – i – 2j – 6k, CA = 2i – j + k Since AB + BC + CA = 0; A, B, C are the Vertices of a triangle. AB ◊ CA = 0 fi –

(a) r ◊ (33i + 45j + 50k) – 41 = 0 (b) r ◊ (13i + 15j + 25k) – 14 = 0

So the triangle ABC is right angled.

LEVEL 1 Straight Objective Type Questions Example 16: The coordinates of a point which divide the line joining the points P(2, 3, 1) and Q (5, 0, 4) in the ratio 1 : 2 are (a) (7/3, 1, 5/3) (b) (4, 1, 3) (c) (3, 2, 2) (d) (1, –1, 1) Ans. (c) Solution: Ê 2 ¥ 2 + 5 ¥ 1 2 ¥ 3 + 0 ¥ 1 2 ¥ 1 + 4 ¥ 1ˆ , , ÁË 2 + 1 ˜¯ 2 +1 2 +1

or

(3, 2, 2)

Example 17: If a line OP angles a x, y and z axis respectively then the direction cosines of OP are. (a) 1/ 2 , 1/2, 1/2 (c) 1/2, 1/ 2 , 1/2 Ans. (c)

(b) 1/2, 1/2, 1/ 2 (d) none of these

Solution: Direction cosines of OP are cos a 2 2 a + cos2 fi cos2 a + (1/2) + (1/4) = 1

Three Dimensional Geometry 21.11

cos2 a = 1/4 cos a = 1/2 Example 18: The direction cosines of the line joining the points (1, 2, – 3) and (– 2, 3, 1) are. (a) – 3, 1, 4 (b) – 1, 5, – 2 -3 1 4 -1 5 -2 , , , , (d) (c) 26 26 26 30 30 30 Ans. (c) fi fi

Solution: The direction cosines of the given line are proportional to – 2 – 1, 3 – 2, 1 – (– 3) i.e. – 3, 1, 4. -3 , Therefore, the actual direction cosines are 9 + 1 + 16 1 9 + 1 + 16

,

4 9 + 1 + 16

i.e

Example 19: (a) z = 0, x = 0 (c) x = 0, y = 0 Ans. (c)

-3 26

,

1 26

,

4 26

z (b) z = 0, y = 0 (d) x = k, y = – k., (k π 0)

(b) cos–1 (1/ 2 )

(c) cos–1 (1/ 3 ) Ans. (c)

(d) cos–1 ( 3 /2)

Solution: Let the line be inclined at an angle a with each of the three coordinates axes, then the direction co sines of the line are cos a, cos a, cos a and 3 cos2a = 1

z axis is

Solution: Let a with z a. fi

(b) 3 : 2 (d) – 7 : 8

cos

2

2

2

a=1

2

fi

cos a = 1 - 1/4 – 1/4 = 1/2

fi

cos a = ± 1/ 2 so a

Example 25: x, y and z axes are respectively 3, 4 and 12. The length of the

3l - 2 2 =0 fi l= l +1 3 Example 21: a, 0), (0, 0, a) and (0, 0, 0) are. (b) (a/2, a/2, a/2) (d) (2a, 2a, 2a)

(a) 5 (c) 3 17 Ans. (d)

(b) 4 10 (d) 13

Solution: 32 + 42 + 122 =

Solution: (x, y, z) then x2 + y2 + z2 = (x – a)2 + y2 + z2 = x2 + (y – a)2 + z2 = x2 + y2 + (z – a)2 fi x = a/2 = y = z (a/2, a/2, a/2).

9 + 16 + 144 =

169 = 13.

Example 26: If P (x, y, z) is a point in the space at a distance r the line OP are

Example 22: If O is the origin and the line OP of length a with x xz plane, then r the coordinate of P are (a) (r cos a, 0, r sin a) (b) (0, 0, r sin a) (c) (0, 0, r cos a) (d) (r cos a, 0, 0) Ans. (a) Solution: Let the coordinate of P be (x, y, z). Since OP lies in xz with the x p/2 – a with z

(a) cos–1 (1/2)

Ans. (b)

Solution: Let yz (– 2, 4, 7) and (3, – 5, 8) in the ratio l : 1. Then

(a) (a/3, a/3, a/3) (c) (a, a, a) Ans. (b)

Example 23: dinate axes, then the angle of inclination is

Example 24: x and y

Example 20: The ratio in which the yz plane divides the

fi

x = r cos a, y = r cos p/2,

fi cos a = 1/ 3 fi a = cos–1(1/ 3 )

.

Solution: See theory

(a) 2 : 3 (c) 4 : 5 Ans. (a)

is and p/2 with Y z = r cos (p/2 – a are (r cos a, 0, r sin a).

a

(a)

r r r , , x y z

(b) rx, ry, rz

(c)

x y z , , r r r

(d) none of these

Ans. (c) Solution: Let a, b, g be the angle which OP with the coordinates axes, then x = r cos a, y = r cos b and z = r cos g.

21.12

Complete Mathematics—JEE Main

fi

cos a = x/r, cos b = y/r and cos r = z/r. x/r, y/r, z/r.

Example 27: Let N be the foot of the perpendicular of l, m, n be the direc length p tion cosines of ON (a) px + my + nz = l (b) l x + py + nz = m (c) lx + my + pz = n (d) lx + my + nz = p. Ans. (d) Solution: The coordinates of N are (pl, pm, pn) and let P(x, y, z) be any point on the plane. The direction cosines of PN are proportional to x – pl, y – pm, z – pn. Since ON is perpendicular to the plane, it is perpendicular to PN hence l (x – pl) + m(y – pm) + n(z – pn) = 0 fi lx + my + nz = p(l2 + m2 + n2) = p which is the locus of P of the plane. Example 28: which are l, m, n is (a) lx + my + nz = l + 2m + 3n x -1 y - 2 z - 3 + + =0 (b) l m n (c) lx + my + nz = 1 l x m y nz + + =0 (d) 1 2 3 Ans. (a) Solution: l(x - 1) + m(y – 2) + n(z – 3) = 0 fi

lx + my + nz = l + 2m + 3n.

Example 29: A, B, C such that the centroid of the triangle ABC is the point (1, r, r2 (a) x + ry + r2 z = 3r 2 (b) r2x + ry + z = 3r2 2 (c) x + ry + r z = 3 (d) r2x + ry + z = 3 Ans. (b) Solution: x y z + + =1 a b c A (a, 0, 0), B (0, b, 0) and C (0, 0, c). So that the coordinates of the centroid of the triangle ABC are (a/3, b/3, c/3) = (1, r, r2) (given) fi a = 3, b = 3r, c = 3r2

Example 30: the plane x + 3y – 4z + 6 = 0 on the axes is (c) – 22/3 Ans. (a) Solution: x y z =1 + + - 6 - 2 3/ 2 So the intercepts on the coordinate axes are –6, –2, 3/2 and Example 31: the origin and containing the lines whose direction cosines are proportional to 1, –2, 2 and 2, 3, –1 is (a) x – 2y + 2z = 0 (b) 2x + 3y – z = 0 (c) x + 5y – 3z = 0 (d) 4x – 5y – 7z = 0 Ans. (d) Solution: plane be l, m, n the lines lying in the plane, whose direction cosines are proportional to 1, – 2, 2 and 2, 3, –1; we have l – 2m + 2n = 0 and 2l + 3m – n = 0 fi

l m m = = 5 7 -4 lx + my + nz = 0 or 4x - 5y - 7z = 0

Example 32: through the point (1, – 1, 2) and parallel to the plane 3x + 4y – 5z = 0 is (a) 3x + 4y – 5z + 11 = 0 (b) 3x + 4y – 5z = 11 (c) 6x + 8y – 10z = 1 (d) 3x + 4y – 5z = 2 Ans. (a) Solution: 3x + 4y – 5z = 0 is 3x + 4y – 5z = K If it passes through (1, - 1, 2), then 3 – 4 – 5(2) = K fi K = – 11 x + 4y – 5z + 11 = 0. Example 33: 6x – 2 = 3y + 1 = 2z – 2 is (a) r = i – j + 3k + l (i + 2j + 3k) 1 Ê1 ˆ (b) r = i + 2j + 3k + l Á i - j + k ˜ Ë3 ¯ 3 (c) r =

x y z + + 2 =1 3 3r 3r

or

r2x + ry + z = 3r2.

(d) 26/3

1 1 i – j + k + l (i + 2j + 3k) 3 3

(d) r = – 2i + j – 2k + l (6i + 3j + 2k); l, being a Ans. (c)

Three Dimensional Geometry 21.13

Solution:

Example 36: which is parallel to the plane r ◊ (2i – 3j + 5 k) + 7 = 0 is

x - 1/ 3 y + 1/ 3 z -1 = = 3 1 2 So the line passes through the point (1/3, –1/3, 1) and the direction cosines of the line are proportionate to 1, 2, 3 and r=

1 1 i - j + k + l (i + 2 j + 3 k ) 3 3

Example 34: x - 3 y +1 z - 2 = = are 2 7 -3 (a)

x - 2 y +1 z -1 x - 2 y +1 z -1 = = = = (b) -1 2 7 -3 3 2

(c)

x-2 y-7 z+3 x - 3 y +1 z - 2 = = = = (d) -1 -1 2 1 2 1

Ans. (b) Solution: and its direction cosines are proportional to 2, 7, –3 so its x-2 y +1 z -1 = = 2 7 -3 Example 35: If M joining A (4 i + 5j – 10k) and B (– i + 2j + k of the plane through M and perpendicular to AB is (a) r ◊ (– 5 i – 3 j + 11k) + 135/2 = 0 7 9 ˆ 135 Ê3 =0 (b) r ◊ Á i + j - k ˜ + Ë2 2 2 ¯ 2 (c) r ◊ (4i + 5j – 10k) + 4 = 0 (d) r ◊ (– i + 2j + k) + 4 = 0 Ans. (a) Solution: Middle point M of AB is Ê1 ˆ M Á ( 4 i + 5 j - 10 k - i + 2 j + k )˜ Ë2 ¯ 7 9 ˆ Ê3 = M Á i + j - k˜ Ë2 2 2 ¯ Also AB = - i + 2 j + k - ( 4 i + 5 j - 10 k ) = - 5 i - 3 j + 11k So the plane passing through M and perpendicular to the direction AB is 7 9 ˆ˘ È Ê3 ÍÎr - ÁË 2 i + 2 j - 2 k ˜¯ ˙˚ ◊ ( - 5 i - 3 j + 11k ) = 0 r ◊ ( -5 i - 3 j + 11 k ) + 135 / 2 = 0

(a) (b) (c) (d) Ans. (a)

r ◊ (2i – 3j + 5k) + 11 = 0 r ◊ (3i + 4j – k) + 11 = 0 r ◊ (3i+ 4j – k) + 7 = 0 r ◊ (2i – 3j + 5z) –7 = 0

Solution: plane is r ◊ (2 i - 3 j + 5 k ) + l = 0. If r = x i + y j + z k , we get 2x – 3y + 5z + l = 0 This plane passes through the point (3, 4, –1) if 2 ¥ 3 - 3 ¥ 4 + 5(- 1) + l = 0 or if l = 11 r ◊ (2 i - 3 j + 5 k ) + 11 = 0 Example 37: The ratio in which the plane 2x – 1 = 0 divides the line joining (– 2, 4, 7) and (3, – 5, 8) is (a) 2 : 3 (b) 4 : 5 (c) 7 : 8 (d) 1 : 1 Ans. (d) Solution: k : 1, then the co ordinates of the point which divides the join of the points (– 2, 4, 7) and (3, – 5, 8) in this ratio are given by Ê 3 k - 2 - 5 k + 4 8 k + 7ˆ ÁË k + 1 , k + 1 , k + 1 ˜¯ As this point lies on the plane 2x – 1 = 0. fi

3k - 2 1 = k +1 2

fi k= Example 38: A line passes through the points (6, – 7, – 1) and (2, – 3, 1). If the angle a positive direction of x the line are (a) 2/3, – 2/3, – 1/3 (c) 2/3, – 2/3, 1/3 Ans. (a)

(b) 2/3, 2/3, – 1/3 (d) 2/3, 2/3, 1/3

Solution: The direction cosines of the given line are proportional to 2 - 6, - 3 + 7, 1 + 1, i.e. - 2, 2, 1 ∓ 2 ± 2 ±1 the direction cosines are therefore = , , . 3 3 3 Since the angle a tion of x a > 0 fi cos a = 2/3 Thus,

21.14

Complete Mathematics—JEE Main

Example 39:

Example 42: p/4 with the plane

x + y = 3 are

a, b, c and a¢, b¢, c¢

(a) 1, 2 , 1

(b) 1, 1, 2

(c) 1, 1, 2 Ans. (b)

(a)

2 , 1, 1

(d)

(b)

Solution: A(x – 1) + By + Cz = 0. Since it passes through the point (0, 1, 0), – A + B = 0 fi A=B p/4 with x + y = 3 A+B 1 + 1 A2 + B 2 + C 2

= ±

(c) (d)

1

So

2

a 1

a2 1 a2 1 a2

+ – + +

1 2

b 1

b2 1 b2 1 b2

– – + +

1 2

c 1

c2 1 c2 1 c2

+ + – +

1 2

a¢ 1

a ¢2 1 a ¢2 1 a ¢2

1

+ – –

2

b¢ 1

b¢2 1

+

b¢2 1 b¢2

1

– – – +

c ¢2 1 c¢2 1 c¢2 1 c¢2

=0 =0 =0 =0

Ans. (c)

2

Solution:

(A + B) 2 = A 2 + B 2 + C 2 2A2 = C2 A B C = = 1 ± 2 1

fi fi

1

x y z + + =1 a b c and

2.

X Y Z + + = 1 a ¢ b¢ c¢

Example 40: A plane which passes through the point x–4 y–7 z–4 = = is 1 5 4 (a) x – y + z = 1 (b) x + y + z = 5 (c) x + 2y – z = 1 (d) 2x – y + z = 5 Ans. (a)

(3, 2, 0) and the line

Solution: a(x – 3) + b(y – 2) + cz = 0 Since it contains the given line. a(4 – 3) + b(7 – 2) + 4c = 0 fi a + 5b + 4c = 0 Example 41: The lines

x–2 y–3 z–4 = = and 1 1 –k

x –1 y–4 z–5 = = are coplanar if 2 1 k (a) k = 1 or – 1 (b) k = 0 or – 3 (c) k = 3 or – 3 (d) k = 0 or – 1 Ans. (b) Solution: The lines are coplanar if 2 –1 3– 4 4 – 5 1 1 –k = 0 k fi fi

2

1 –1 –1 1 1 –k = 0 k 2 1

1 fi

1 0 0 1 2 1– k = 0 k k + 2 k +1

2(k + 1) – (1 – k) (k + 2) = 0

fi

k = 0 or – 3

–1 So

–1

1 1 1 = + 2 + 2 2 a b c 1

fi

a

2

+

1 b

2

+

1 c

2

=

1 1 1 + 2 + 2 2 a¢ b¢ c¢

1 a¢

2

+

1 b¢

2

+

1 c¢2

.

Example 43: Two lines x = ay + b, z = cy + d and x = a¢y + b¢, z = c¢y + d¢ will be perpendicular, if and only if (a) aa¢ + bb¢ + cc¢ = 0 (b) (a + a¢) (b + b¢) (c + c¢) = 0 (c) aa¢ + cc¢ + 1 = 0 (d) aa¢ + bb¢ + cc¢ + 1 = 0 Ans. (c) Solution: Lines can be written as x–b y z–d x – b¢ y z – d¢ = = and = = a a¢ 1 c 1 c¢ which will be perpendicular if and only if aa¢ + 1 + cc¢ = 0 Example 44: Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is (a) 7/2 (b) 5/2 Ans. (a) Solution: 5/2 + 8 4 +1+ 4

=

7 2

Three Dimensional Geometry 21.15

Example 45: A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1), B(2, 1, 3) and C(– 1, 1, 2). Then the angle be tween the faces OAB and ABC will be (a) cos–1

(a) – 1/2 (c) – 2 Ans. (c)

–1

Solution: OAB be ax + by + cz = 0 where a + 2b + c = 0 and 2a + b + 3c = 0 a b c fi = = 5 –1 – 3 ABC be a¢(x + 1) + b¢(y – 1) + c¢(z – 2) = 0 where 2a¢ + b¢ – c¢ = 0 and 3a¢ + c¢ = 0 b¢ c¢ a¢ = = 1 –5 –3

If q cos q =

5 ¥ 1 + ( –1) ( –5) + ( –3)( –3) 25 + 1 + 9 1 + 25 + 9

Example 46: of the x and z axis. If the angle b is such that sin2b = 3 sin2q, then cos2q (a) 3/5 (b) 1/5 (c) 2/3 (d) 2/5 Ans. (a)

=

19 . 35 q with each y

Solution: We have cos2q + cos2q + cos2b = 1 fi 2cos2q = sin2b = 3sin2q = 3(1 – cos2q) fi 5cos2q = 3 fi cos2q = 3/5. Example 47: A line with directional cosines propor x = y + a = z and x + a = 2y = 2z. The coordinates of each of the points of intersection are given by (a) (3a, 2a, 3a), (a, a, 2a) (b) (3a, 2a, 3a), (a, a, a) (c) (3a, 3a, 3a), (a, a, a) (d) (2a, 3a, 3a), (2a, a, a) Ans. (b) Solution: Let the points be P(r, r – a, r and Q(2r¢ – a, r¢, r¢) on the second line then

fi

x –1 1

Solution:

Ans. (d)

fi

(b) – 1 (d) 0

r – 2r ¢ + a r – a – r¢ r – r¢ = = 2 1 2 r = 3a, r ¢ = a P(3a, 2a, 3a) and Q(a, a, a)

Example 48: If the straight lines. x = 1 + s, y = – 3 – ls, z = 1 + ls and x = t/2, y = 1 + t, s and t are coplanar, then l z=2–t

=

y +3 z –1 x y –1 z – 2 = = and = . Since these are co –l l 2 –2 1

planar 1 1 1

–4 2 –l

–1 –2 =0 l

fi

–4 – 1 –1 = 0 0 6 0 – l + 4 l +1

fi

6(l + 1) + (4 – l) = 0

fi

l=–2

Example 49: The angles between the lines 2x = 3y = – z and 6x = – y = – 4z is

Ans. (d) Solution: Lines can be written as x y z x y z = = and = = 3 2 2 –3 –6 – 12 Since 3 ¥ 2 + 2 ¥ (– 12) + (– 6) (– 3) = 0 The lines are perpendicular. Example 50: If the angle q between the lines x +1 y –1 z–2 = = and the plane 2x – y + 2 2 1 = 0 is such that sin q = 1/3, then the value of l is (a) 3/4 (b) –4/3 (c) 5/3 (d) –3/5. Ans. (c)

lz+4

q with the

Solution: p/2 – q \

2(1) + (- 1) (2) + ( l ) (2) p cos Ê - q ˆ = Ë2 ¯ 1+ 4 + 4 ¥ 4 +1+ l

1 2 l = fi l + 5 = 4l 3 3 l+5 fi l = 5/3. Example 51: The distance between the line r = 2i – 2j + 3k + l (i – j + 4k ) and the plane r. (i + 5 j + k ) = 5 is (a) 3/10 (b) 10/3 3 Ans. (d) fi

21.16

Complete Mathematics—JEE Main

Solution: Direction ratios of the line are 1, – 1, 4 and Since

1 ¥ 1 + (– 1) (5) + 4 ¥ 1 = 0

(a) x – y = 0 = z (c) y – z = 0 = x Ans. (a)

(b) x – z = 0 = y (d) x = y = z

Solution: If q positive direction of z fi cos2q = 0 fi 2(1) + (- 2) (5) + 3 ¥ 1 - 5 1 + 25 + 1

10

=

27

=

x

3 3 1

(d) none of these

y 1

2

=

z 0

c = xi + (x – 2) j – k . If the vector c lies in the plane of a

Solution: A (– 1, 3, 4) and perpendicular to x – 2y = 0 that is on the line y–3 z–4 x +1 = = = t (say) –2 0 1 B(t – 1, – 2t + 3, 4)

and b, then x (a) 0 (c) – 4 Ans. (d)

1 t – 1 + 2t – 6 = 0 2

fi

14 t= 5

x fi

Ê 9 13 ˆ Ë 5 , - 5 , 4¯

l m n = = –1 1 –1

l 2 + m 2 + n2

Example 54: dtrection of each of x

=

–1 ± 1+1+1

Ê 5ˆ z-3 and the plane x + 2y +3z = 4 is cos-1 Á , then l Ë 14 ˜¯

=±

1

(a) 5/3 (c) 3/2 Ans. (b)

(b) 2/3 (d) 2/5

Solution: If q is the angle between the line and the plane then (1)(1) + (2)(2) + (3)(l ) 5 + 3l sin q = = 2 1+ 4 + l 1+ 4 + 9 14 l 2 + 5 since cos q =

3 so

p /4 with the positive y

y -1 2

l

Solution: If l, m, n are the direction ratios of the line L then 2l + 3m + n = 0 and l + 3m + 2n = 0 l m m = fi = 3–6 4 –1 3– 6

l

– 2(– 1 – x) + 2 = 0 x=–2 Example 56: If the angle between the line x =

=

(b) 1/2 (d) 1/ 2

x – 2 –1

1 0 0 =0 1 –2 1 x – 2 –1– x

fi fi

Example 53: Let L be the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = 2. If L angle a with positive x a (a) 1/ 3 (c) 1 Ans. (a)

(b) 1 (d) – 2

Solution: Since the three vectors are coplanar 1 1 1 1 –1 2 =0

t – 2 – 2t + 6 ˆ Mid point M of AB is Ê , 4 ¯ which lies on the Ë 2 , 2 plane x – 2y = 0

and cos a =

2

=

Which can be written as x – y = 0 = z Example 55: Let a = i + j + k , b = i – j + 2k and

(c) (15, 11, 4) Ans. (d)

fi

p /4 + cos2p/4 + cos2 q = 1 cosq = 0 fi q = p /2 and the

10

Example 52: plane x – 2y = 0 is

fi

2

fi fi

3 5 fi sin q = 14 14 5 + 3l 3 = 14 14 l 2 + 5

(l2 + 5) l2 + 30 l + 25 l = 2/3

Three Dimensional Geometry 21.17

Example 57: the plane x –y + z = z is (a) 10 3

(b) 5 3

(c) 3 10

(d) 3 5

x-3 y-k z = = intersect, then k 1 2 1

Ans. (a)

Ans. (a)

Solution: the line x = y = z is x -1 y + 5 z - 9 = = = r (say) 1 1 1 Any point Pon this line is (r +1, r –5, r which lies on the given plane if r +1 – (r – 5) + r fi r = –10 Distances of P =

(-9 - 1)2 + (+ 5 - 15)2 - (-9 - 1)2 = 10 3

Example 58: The length of the perpendicular drawn x y-2 z-3 = = is 2 3 4 (a)

29

(b)

33

(c)

53

(d)

66

Ans. (c) Solution: Any point on the given line is P (2r, 3r + 2, 4r + 3) and the given point be A Direction ratios of AP are 2r – 3, 3r + 2 + 1, 4r + 3 –11 or 2r – 3, 3r + 3, 4r –8 AP is perpendicular to the given line if 2(2r – 3) + 3 (3r + 3) + 4 (4r – 8) = 0 or if r = 1, so coordinates of point (2, 5, 7) and AP =

2

(2 - 3)2 + (5 + 1)2 + (7 - 11)

Example 59: plane x–2y +2z is (a) x – 2y – 2z –1 = 0 (c) x – 2y + 2z –3 = 0 Ans. (c)

=

1+ 4 + 4

Solution: (2r1 +1, 3r1 –1, 4r1 +1) and on the second line is (r2 + 3, 2r2 + k, r2) The lines will intersect when 2r1 +1 = r2 +3, 3r1 –1 = 2 r2 + k, 4r1 +1 = r2 fi 2r1 – r2 = 2, 4r1 –r2 =–1 fi r1 = – 3/2, r2 = –5. and k = 3r1 – 2 r2 . x–2 y –1 z+2 Example 61: If the line = = lies in 3 –5 2 the plane x + 3y – az + b = 0. Then (a, b (a) (5, – 15) (b) (– 5, 5) (c) (6, – 17) (d) (–6, 7) Ans. (d) Solution: As the line lies in the plane. The point (2, 1, – 2) the line, so 2 + 3(1) – a(– 2) + b = 0 and (3) (1) + (3) (– 5) – a (2) = 0 fi a = – 6, b = 7. Example 62: The projection of a vector on the three coordinate axis are 6, – 3, 2 respectively. The direction co sines of the vector are (a) 6/7, – 3/7, 2/7 (b) – 6/7, – 3/7, 2/7 (c) 6, – 3, 2 (d) 6/5, – 3/5, 2/5 Ans. (a) Solution:

2

2

6 2 + ( - 3) + ( 2 ) = ± 7 ±6 ∓3 ±2 , , 7 7 7

53

Example 63: A line AB x (b) x – 2y + 2z + 5 = 0 (d) x – 2y + 2z + 1 = 0

Solution: x –2y +2z + k= 0 k

x -1 y +1 z -1 and = = 2 3 4

Example 60: If the lines

x=y

=±1fik=±3

positive y with the positive z

q

AB q

Ans. (a) Solution: cos2 fi fi

2

2

q=1 cos q = 1 – (1/2) – (–1/2)2 = 1/4 cos q = ± 1/2 fi q 2

21.18

Complete Mathematics—JEE Main

Example 64: A plane L passes through the point

on the plane ABC,

x +1 y + 2 z +1 x-2 y+2 z-3 = = = = and 3 1 2 1 2 3 L is (a) 2

75

(b) 7

(c) 13 / 75 Ans. (c)

75

(d) 23 / 75

Solution: l, m, n; then 3l + m + 2n = 0, l + 2m + 3n = 0 l m n fi = = -1 -7 5 L is – 1(x + 1) – 7(y + 2) + 5(z + 1) = 0 fi x + 7y – 5z 1 + 7 - 5 + 10 2

2

=

1 + 7 + ( -5)

13 75

Example 65: A line with positive direction cosines passes through the point P x+y+ Q PQ z (a) 1 (c) 3 Ans. (c) Solution:

(b)

2

(d) 2

x-2 y +1 z - 2 = = =r 1 1 1 fi x = r + 2, y = r – 1, z = r + 2 The point (r + 2, r – 1, r + 2) lies on the plane 2x + y + z 2(r + 2) + r – 1 + r fir=1 So, the coordinates of Q are (3, 0, 3) and thus PQ =

we get a = 2b = 3g and as H lies

(3 - 2)2 + (0 + 1)2 + (3 - 2)2 = 3 .

Example 66: The vertices of a triangle are A(1, 0, 0), B(0, 2, 0), C(0, 0, 3). If the direction ratios of the line join a, b, – 111, the a + b (a) 5 (b) 10 (c) 15 (d) 25 Ans. (c) Solution: x y z ABC is + + = 1 1 2 3 Let H(a, b, g ) be the orthocenter, then AH ^ BC fi (a – 1) ¥ 0 + 2b – 3g = 0 and BH ^ CA fi a . 1 + (b – 2) ¥ 0 – 3g = 0

fi

a a a 36 + + =1 fi a = . 49 1 4 9 Ê 36 18 12 ˆ H= Á , , ˜ Ë 49 49 49 ¯

linear,

Ê1 2 ˆ 3 3

If G be the centroid, its coordinates are Á , ,1˜ and the direc Ë ¯ tion ratios of the line HG are

36 1 18 2 12 - , - , -1 49 3 49 3 49 so a b = – 44 fi a + b = 15 Example 67: x +1 y - 2 z x-3 y+4 = = and two parallel lines = = 2 1 3 3 2 z -1 is 1 (a) 3x + 2y + z = 14 (b) 8x + y – 26z + 6 = 0 (c) 4x – 6y + z = 53 (d) none of these Ans. (b) Solution: a(x + 1) + b(y – 2) + cz = 0 where 3a + 2b + c = 0 other line will also lie on the plane if the point (3, – 4, 1) lies on the plane a(3 + 1) + b(– 4 – 2) + c = 0 fi 4a – 6b + c = 0 a b c Solving, we get = = 8 1 - 26

8x + y – 26z + 6 = 0. Example 68: The length of projection of the line seg 2x + y + 6z (a)

255 / 41

(b)

237 / 41

(c) 137 / 41 (d) 155 / 41 Ans. (b) Solution: Direction ratios of the line joining A(1, – 1, 0) and B (– 1, 0, 1) are 2, – 1, – 1. If q is the acute angle which the line AB cos q =

2 ¥ 2 + (1) (- 1) + 6(- 1)

=

(4 + 1 + 1) 4 + 1 + 36 AB = Length of the projection =

3 246

.

6. 6 sin q =

6 ¥ 237 246

=

237 . 41

Three Dimensional Geometry 21.19

Example 69: The point P is the intersection of the straight line joining the points Q (2, 3, 5) and R (1, –1, 4) with the plane 5x – 4y – z = 1. If S is the foot of the perpen T (2, 1, 4) to QR, then length PS is (a) 1/ 2

(b)

2

Example 70: the line of intersection of the planes x + 2y + 3z = 2 and x – y + z = 3 and at a distance 2/ 3

(d) 2 2

(c) 2 Ans. (a) Solution:

QR is x-2 y-3 z-5 = = 1 4 1 Let the coordinates of P be (r + 2, 4r + 3, r +5) As P lies on the plane 5x – 4y – z = 1 5(r +2) – 4 (4r + 3) – (r + 5) = 1 Ê 4 1 13 ˆ fi r = –2/3 fi P Á , , ˜ Ë3 3 3¯ Let the coordinates of S be (t + 2, 4t + 3, t + 5) As TS is perpendicular to QR 1 (t + 2 – 2) + 4 (4t + 3 –1) + 1 (t + 5 – 4) = 0 1 9ˆ Ê3 fi t = - fi S Á , 1, ˜ Ë 2 2¯ 2

9 1 = 18 2

=

2x+y=3 2 –1

(a) 5x – 11y + z = 17

(b)

(c) x + y + z =

(d) x +

3

2 y = 1- 2

Ans. Solution: (x + 2y + 3z – 2) + l (x – y + z – 3) = 0 fi (1 + l) x+ (2 – l) y + (3 + l) z – (2 + 3l) = 0 2 3 3 (1 + l ) + (2 - l ) - (3 + l ) - (2 + 3l ) 2

2

2

=

2

(l + 1) + (2 - l ) + (3 + l ) fi

2 2 2 Ê 3 4ˆ Ê 1ˆ Ê 9 13 ˆ Hence PS = Á - ˜ + Á 1 - ˜ + Á - ˜ Ë 2 3¯ Ë 3¯ Ë2 3 ¯

4l 2 2

3l + 4l + 14

=

3

-7 4 fil= 3 2

5x – 11y + z = 17

Assertion-Reason Type Questions

Example 71: Statement-1: The point A(3, 1, 6) is the B(1, 3, 4) in the plane x – y + z = 5. Statement-2: The plane x – y + z = 5 bisects the line seg A(3, 1, 6) and B(1, 3, 4) Ans. (b) Solution: Direction ratios of AB are 1 – 3, 3 – 1, 4 – 6 or 1, – 1, 1 x–y+z fi AB (2, 2, 5) of AB lies on x – y + z = 5. Thus the plane bisects AB AB, A

B

Example 72: Statement-1: If the distance of the point x + 2y – 2z = a, where a > 0, is P P to the plane is (8/3, 4/3/, –7/3) Statement-2: A line through P(1, – 2, 1) and perpendicular to the plane x + 2y – 2z = a intersects the plane at Q. If PQ = 5 then a = 10 Ans. (a)

Solution: to the plane is

P perpendicular

y + 2 z -1 r r x -1 = = . = = 2 -2 1 1+ 4 + 4 3 Coordinates of a point on this line at a distance r

P are

2r -2r ˆ Êr + 1˜ ÁË + 1, - 2, ¯ 3 3 3 If this represents the point Q then it lies on the plane.

r Ê - 2r ˆ Ê 2r ˆ + 1˜ = a + 1 + 2 Á - 2˜ - 2 Á Ë 3 ¯ Ë 3 ¯ 3 Since PQ = 5, r = 5 fi

a=

4¥5 4¥5 5 +1+ -4+ - 2 = 10 3 3 3 Q is now the foot of the per P to the plane and its coordinates are

Example 73: Consider the plane 3x – 6y – 2z = 15 and 2x + y – 2z = 5.

21.20

Complete Mathematics—JEE Main

Statement-1: tersection of the given planes are x = 3 + 14t, y = 1 + 2t, z = 15t; t Statement-2: The vector 14iˆ + 2 ˆj + 15kˆ is parallel to the line of intersection of the given planes. Ans. (d)

which lies on the line

x y -1 z - 2 = = 1 2 3

is

true. Next, direction ratios of AB are 0,6,–4 and the given line are 1, 2, 3. Since 0 × 1 + 6 × 2 + (–4) × 3 = 0, the line AB is perpendicu

Solution: y –1 x–3 z–0 = = = t. This is the line of 14 15 2 intersection of the planes, then the point (3, 1, 0) which lies written as

Example 75: L1:

x –1 y z +1 = = 1 –1 1

x – 2 y +1 z = = 1 2 3 Statement-1: L1 and L2 x + 2y – 3z – 8 = 0 Statement-2: L1 and L2 intersect at a point. Ans. (b) 1– 2 0 +1 –1– 0 1 –1 1 Solution: Since = 0, the 1 2 3 lines L1 and L2 x –1 y z +1 1 –1 1 1 2 3 = – (5x + 2y – 3z 1 –1 1 π L1 and L2 are not parallel as π and are co 1 2 3 L2 :

intersection of the given planes is (– 6) (– 2) – (– 2) (1), (– 2) (2) – (– 2) (3), 3(1) – (2)(– 6) allel to the line of intersection of the planes and thus state Example 74: Statement-1: The point A (1, 0, 7) is the x y -1 z - 2 = = 1 2 3 x y -1 z - 2 = bisects the line = 1 2 3 A (1, 0, 7) and B (1, 6, 3).

Statement-2: The line

Ans. (b) Solution: AB is Ê 1 + 1 0 + 6 7 + 3ˆ , , ÁË ˜ = (1, 3, 5) 2 2 2 ¯

LEVEL 2 Straight Objective Type Questions Example 76: the line

A (1, 0, 0) in x - 1 y + 1 z + 10 = = is 2 -3 8

(a) (3, – 4, – 2) (c) (1, –1, – 10) Ans. (b)

(b) (5, – 8, – 4) (d) (2, – 3, 8)

Solution: Any point P on the given line is (2r + 1, - 3r - 1, 8r - 10) So the direction ratios of AP are 2r, - 3r - 1, 8r - 10. Now AP is perpendicular to the given line if

2 (2r) - 3 (- 3r - 1) + 8(8r - 10) = 0 fi 77r - 77 = 0 fi r = 1 and thus the coordinates of P, the foot of the perpendicular A on the line are (3, - 4, - 2). A in the given line. Then Let B(a, b, c AB P a +1 b c = 3, = - 4, =-2 2 2 2 fi a = 5, b = - 8, c = - 4 - 8, - 4). Example 77: the line of intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0 and passing through (1, 1, 1) is

Three Dimensional Geometry 21.21

(a) 2 x + 3y + 4z (c) x + 2y + 3z = 6

x+y+z=3 (d) 20x + 23y + 26z

Ans. (d) Solution: tersection of the given planes is 2x + 3y + 4z + 5 + l (x + y + z – 6) = 0 It passes through (1, 1, 1) if (2 + 3 + 4 + 5) + l (1 + 1 + 1 - 6) = 0 fil 20x + 23y + 26z

perpendicular to the line joining the points on the plane with position vectors a and c fi (a – c). b × d = 0, which is the Example 80: The shortest distance between the skew lines l1 : r = a1 + l b1 and l 2 : r = a2 + m b2 is (a)

(a2 - a1 ) ◊ b1 ¥ b2

(c)

(a2 - b2 ) ◊ a1 ¥ b1

b1 ¥ b 2

Example 78: A, B, C with position vectors -6i + 3 j + 2k , 3i - 2 j + 4k ,

b1 ¥ b 2

5i + 7 j + 3k is

(b)

(a1 - b1 ) ◊ a2 ¥ b2

(d)

(a1 - b2 ) ◊ b1 ¥ a2

b1 ¥ b 2 b1 ¥ a2

(a) r.(i - j + 7k ) + 23 = 0

Ans. (a)

(b) r.(i + j + 7k ) = 23

Solution: Let PQ be the shortest distance vector be tween l1 and l2. Now l1 passes through A1 (a1) and is parallel to b1 and l2 passes through A2 (a2) and is parallel to b2. Since PQ is perpendicular to both l1 and l2 it is parallel to b1 × b2. Let n be the unit vector along PQ.

(c) r.(i + j - 7k ) + 23 = 0 (d) r. (i - j - 7k ) = 23 Ans. (a) Solution: points A, B, C with position vectors a, b, c is r.(a × b + b × c + c × a) = a . b × c So that if a, b, c represent the given vectors, then (a × b + b × c + c × a) i j k i j k i j k = -6 3 2 + 3 -2 4 + 5 7 3 3 -2 4 5 7 3 -6 3 2 = i (12 + 4 – 6 – 28 + 14 – 10 + 18) + k = -13i + 13 j - 91k and

j

-6 3 2 a.b × c = 3 -2 4 = 299 5 7 3 r. (-13i + 13 j - 91k ) r. (i - j + 7k ) + 23 = 0

or Example 79:

r = a + tb, r = c + t¢d are coplanar if (a) (a – b) . c × d = 0 (b) (a – c) . b ¥ d = 0 (c) (b – c) . a ¥ d = 0 (d) (b – d) . a ¥ c = 0 Ans. (b) Solution: to the plane containing these lines is perpendicular to both

Fig. 21.4

b1 ¥ b 2 . | b1 ¥ b 2 |

Then n =

Let d be the shortest distance between the given lines l1 and l2 | PQ| = d and PQ = dn d nˆ . Next PQ being the line of shortest distance between l1 and l2, is the projection of the line joining the points A1 (a1) and A2 (a2) on n. |PQ| = |A1A2 . n|

fi

d=

(a 2 - a1 ). b1 ¥ b 2 | b1 ¥ b 2 |

Example 81: The length of the shortest distance between the lines r = 3i + 5j + 7k + l (i – 2j + k) and r = – i – j – k + m (7i – 6j + k) is (a) 83 (c) Ans. (d)

3

(b)

6

(d) 2 29

Solution: Taking a1 = 3 i + 5 j + 7 k , b1 = i - 2 j + k

Complete Mathematics—JEE Main

21.22

a2 = - i - j - k , b2 = 7 i - 6 j + k

OABC is

| b1 × b2| = 4i + 6 j + 8k

We get

2

2

0 0 0 20 0 0 0 15 0 0 0 - 12

1 6

2

4 + 6 + 8 = 116

= and

\

| (a2 –a1) . b1 × b2| = | - 4 ¥ 4 - 6 ¥ 6 - 8 ¥ 8 | = 116

hence the length of shortest distance is

= 116 116

= 2 29 .

Example 82: The angle between the lines whose direc l2 + m2 – n2 = 0, l + m + n = 0 is (a) p/6 (b) p/4 (c) p/3 (d) p/2 Ans. (c) Solution: We are given l2 + m2 - n2 = 0 and l + m + n = 0 2n2 = 1

fi

with x (a) 27, – 18, 54 (c) – 27, 18, – 54 Ans. (b)

l+m=-n

and fi

(l + m) = n2 = l2 + m2 fi 2lm = 0

fi

Either l = 0 or m = 0 if l = 0, m + n = 0

fi

m = - n = ±1/ 2

a2 + b2 + c2 = (63)2

(1)

a b c = = = l (say) then 3 -2 6

a = 3l, b = - 2l and c = 6l

2

l2 + 4l2 + 36l2 = (63)2 fi

l=±

So the direction cosines of one of the lines are, 0, ∓ 1/ 2 , ± 1/ 2 and if m = 0, l + n = 0

(b) – 27, 18, – 54 (c) 27, – 18, – 54

Solution: tor be a, b, c, then

also

n = ± 1/ 2

1 ¥ 20 ¥ 15 ¥ ( - 12 ) = 600. 6

Example 84:

also we have l2 + m2 + n2 = 1 So that

1 1 1 1

fi

l = - n

= ±1/ 2 and the direction cosines of the other line are ∓1/ 2 , 0, ± 1/ 2 . 1 1 È Ê 1 ˆ Ê 1 ˆ˘ cos-1 Í0 ¥ ∓ +∓ ¥ 0 + Á± ˜ Á± ˜ Ë Î 2 2 2¯ Ë 2 ¯ ˙˚ 1 p = cos-1 = . 3 2 Example 83: between the plane. 3x + 4y – 5z –60 = 0 and the coordinate planes is (a) 60 (b) 600 (c) 720 (d) none of these Ans. (b) Solution: x y z =1 + + 20 15 - 12 A (20, 0, 0), B (0, 15, 0) and C (0, 0, - 12) and the coordinates of the origin are (0, 0, 0).

fi

l2 = (63)2

63 7

Since a = 3l x l = 18, - 54.

- 27,

Example 85: between the planes 2x – y + 2z + 3 = 0 and 3x – 2y + 6z + 8 = 0 is (a) (b) (c) (d) Ans. (b)

5x – y – 4z – 45 = 0 5x – y – 4z – 3 = 0 23x + 13y + 32z – 45 = 0 23x – 13y + 32z + 5 = 0

Solution: between the given planes are 2x - y + 2z + 3 2

=±

22 + ( - 1) + 22 fi fi

3x - 2y + 6z + 8 2

32 + ( - 2 ) + 62

7(2x - y + 2z + 3) = ± 3 (3x - 2y + 6z + 8) 5x - y - 4z - 3 = 0 taking the + ve sign, and 23x - 13y + 32z + 45 = 0 taking the - ve sign.

Example 86: 3i – 5j + 7k and perpendicular to the plane 3x – 4y + 5z = 8 is (a)

x-3 y+5 z-7 = = -4 3 5

Three Dimensional Geometry 21.23

(b)

x - 6y + 8z + 3 = 0 can be written as 2x - 3y + 4z + 3/2 = 0 2 - 3/ 2 1 So p2 = 2 2 2 = 2 29 2 +3 +4

x-3 y+4 z-5 = = -5 3 7

(c) r = 3i + 5j – 7k + l (3i – 4j – 5k) (d) r = 3i – 4j – 5k + m (3i + 5j + 7k) (l, m

2+6 p3 =

and

Ans. (a) Solution: (3 i - 4 j + 5 k ) ◊ ( x i + y j + z k ) = 8. 3 i - 4 j + 5 k and 3 i - 5 j + 7 k , its r = 3 i - 5 j + 7 k + l (3 i - 4 j + 5 k ) where l Since this line passes through the vector 3 i - 5 j + 7 k i.e. the point (3, -5, 7) and is parallel to 3 i - 4 j + 5 k , its direction ratios are 3, - 4, 5 y+5 x-3 z-7 = = 4 3 5 Example 87: If the perpendicular distance of a point P x+y+z=p

fi

2

2 +3 +4

2

=

8 29

p1 + 8p2 – p3 = 0 Example 89:

and (a) (b) (c) (d) Ans. (a)

r = i + 2j – k + l (i + 2j – k) r = i + 2j – k + m (i + j + 3k) is r ◊ (7i – 4j – k) = 0 7 (x – 1) – 4 ( y – 1) – (z + 3) = 0 r ◊ (i + 2j – k) = 0 r ◊ (i + j + 3k) = 0

Solution: Since both the given lines pass through the point with position vector i + 2 j - k also passes through i + 2 j - k is perpendicular to the vectors i + 2 j - k and i + j + 3 k . If d = a i + b j + c k a + 2b - c = 0 and a + b + 3c = 0.

of P are (b) (0, 2p – p)

(a) (p, 2p, 0) (c) (2p, p, – p) Ans. (c)

b c a = = 4 1 7

fi

(d) (2p, – p, 2p) fi

d = 7i - 4 j - k .

Solution: The perpendicular distance of the origin p -p (0, 0, = x + y + z = p is . 1+1+1 3 If the coordinates of P are (x, y, z x+y+z- p 3 fi

(d) p1 + 2p2 + 3p3 =

Solution: Since the planes are all parallel planes, 2-6 2

2

2 +3 +4

2

=

7i - 4 j - k

4 4 + 9 + 16

is

[r - ( i + 2 j - k )] ◊ ( 7 i - 4 j - k ) = 0 r ◊ (7 i - 4 j - k ) = 1 ¥ 7 + 2 (- 4) + (- 1) (- 1) = 0 i + 2j-k , i.e. the point (1, 2, -1) and the direction ratios of the nor - 4, -

3

Example 88: If p1, p2, p3 denote the distances of the x – 3y + 4z + 6 plane 2x – 3y + 4z = 0, 4x – 6y + 8z + 3 = 0 and 2x – 3y + 4z – 6 = 0 respec tively, then (b) p23 = 16p22 (a) p1 + 8p2 – p3 = 0 (c) 8 p22 = p21 Ans. (a)

i + 2 j - k and the

p

=

|x+y+z–p| =|p|

p 1=

2

=

4 29

29

7 (x - 1) - 4 (y - 2) - (z + 1) = 0 Note: tions. Example 90: gin to the join of A in the ratio (a) 2 : 3 (c) 1 : 1 Ans. (c)

B (11, 0, – 1) divides AB (b) 3 : 2 (d) none of these

21.24

Complete Mathematics—JEE Main

Solution: Let D the origin to the join of A and B and divide AB in the ratio k : 1, then the coordinates of D are Ê 11 k - 9 0 k + 4 - k + 5 ˆ ÁË k + 1 , k + 1 , k + 1 ˜¯ So that the direction cosines of OD are proportional to 11 k - 9, 4, 5 - k and direction cosines of AB are proportional to - 4, - 1 - 5 i.e. 20, - 4, - 6. or 10, - 2, - 3. Since OD is perpendicular to AB. 10 (11 k - 9) - 2 ( 4 ) - 3 (5 - k ) = 0 fi fi

- 8 - 15 + 3k = 0 110k 113k = 113 fi k = 1

Example 91: Cosine of the angle between the lines whose r = 3i + 2j – 4k + l (i + 2j + 2k) and r = 5i – 2k +m (3i + 2j + 6k); l, m (a) -1/ 3 29

(b) 3 / 7 29

Ans. (d) Solution: Direction vectors of the given lines are i + 2 j + 2 k and 3 i + 2 j + 6 k . The angle q between these vectors. ( i + 2 j + 2 k ) (3 i + 2 j + 6 k ) ◊ Hence cos q = 1+ 4 + 4 9 + 4 + 36 =

3 + 4 + 12 19 = 3¥7 21

Example 92: passing through the line of intersection of the planes r ◊(2i – 3j + 4k) =1 and r ◊ (i – j) + 4 = 0 and perpendicular to the plane r ◊ (2i – j + k) + 8 = 0 is (a) (b) (c) (d) Ans. (c)

3x – 4y + 4z = 5 x – 2y + 4z = 3 5x – 2y – 12z + 47 = 0 2x + 3y + 4 = 0

Solution: intersection of the planes r ◊ (2 i - 3 j + 4 k ) = 1 and r ◊ (i - j) + 4 = 0 is 2x - 3y + 4z - 1 + l (x - y + 4) = 0 or (2 + l)x - (3 + l)y + 4z + 4l - 1 = 0 This plane is perpendicular to the plane r ◊ (2 i - j + k ) + 8 = 0 if fi 2 (2 + l) + (3 + l) + 4 = 0. fi 11 + 3l = 0 fi l = –11/3.

fi

3(2x - 3y + 4z - 1) – 11 (x - y + 4) = 0 5x - 2y - 12z + 47 = 0

Example 93: The shortest distances between the lines x - 3 y + 15 x +1 y -1 z - 9 z-9 = = = = and is 2 -7 2 1 -3 5 (a) 4 5 (c) 4 3 Ans. (c)

(b) 4 17 (d) 8 2

Solution: Let l, m, n be the direction cosines of the line of shortest distance, then as it is perpendicular to the given lines 2l - 7m + 5n = 0; 2l + m - 3n = 0 l m n fi = = 16 16 16 fi

l m n = = = 1 1 1

l 2 + m 2 + n2 1+1+1

fi l = m = n = 1/ 3 . Now the shortest distance between the given lines is the projection of the join of the points (3, (1 distance = |(3 + 1) + (- 15 3 = 4 3. Example 94: (2, –1, 1) and the intersection of the lines 2x – y – 4 = 0 = y + 2z, x + 3z – 4 = 0 = 2x + 5z – 8 is (a) x + y + z = 2, x + 2y = 0 (b) x + y + z = 2, x + 2z = 4 (c) x + 2y + z = 1, x + 2z = 4 (d) x + 2y + z = 1, x + 2y = 0 Ans. (b) Solution: the given lines is 2x – y - 4 + l (y + 2z) = 0 = x + 3z - 4 + m (2x + 5z - 8) for all values of l and m. If it passes through the point (2, -1, 1) if 2 ¥ 2 - 1 - 4 + l (- 1 + 2) = 0 = 2 + 3 - 4 + m (4 + 5 - 8) fi l = 1, m = - 1 x+y+z = 2, x + 2z = 4. Example 95: The planes bx – ay = n, cy – bz = l, az – cx = m intersect in a line if (a) al – bm + cn = 1 (b) al + bm + cn = 0 (c) al – bm – cn + 1 = 0 (d) al + bm + cn = 1 Ans. (b)

Three Dimensional Geometry 21.25

Solution:

x, y, z c, a, b respectively and adding we get the re al + bm + cn = 0.

Example 96: The value of k for which the line x-4 y-2 z-k = = lies in the plane 2x – 4y + z = 7 is 1 1 2 (a) 7 (c) no real value Ans. (a)

Solution: a(x – 1) + b(y + 2) + c (z – 1) = 0. Since it is perpendicular to the given planes 2a – 2b + c = 0, a – b + 2c = 0, fi c = 0, a = b, fi x + y + 1 =0

(b) 6 (d) –7

Solution: Since (2) (1) + (–4) (1) + (1) (2) = 0 the line is parallel to the plane.It will lie in the plane if the point (4,2, k) lies on the plane. fi

Ans. (d)

2(4) – 4(2) + k = 7 fi k = 7

1+ 2 +1 1+1

=2 2

Example 99: A line with positive direction cosines x+y+ Q

z

PQ

(a) 1 Example 97: A variable plane at a distance of 1 unit A, B and C. If the centroid G (x, y, z) of the triangle ABC 1 x

2

+

1 y

1

+

2

z

2

(d) 1/3

Solution: Let the coordinates of A, B, C be A(a, 0, 0), B (0, b, 0), C (0, 0, c) then x = a/3, y = b/3, z = c/3 X Y Z + + =1 a b c

1 1 1 + 2+ 2 2 a b c 1 a2

+

1

fi

9x 1

fi

x

2

2

1 b2

+

+

+ 1

9y 1

y

2

2

+

1 c2 + 1 z2

9z

2

=1

2

Ans. (c) Solution: x - 2 y +1 z - 2 = = =k 1 1 1 Let the coordinates of the point Q on this line be Q (k + 2, k –1, k + 2). Since Q lies on the plane 2(k + 2) +(k – 1) + (k fi

k = 1 and PQ =

(3 - 2)2 + (0 + 1)2 + (3 - 2)2 = 3

Then the value of μ for which the vector PQ is parallel to the plane x –4y +3z – 1 is (a) 1/4 (b) –1/4 (c) 1/8 (d) –1/8 Ans. (a) Solution: As Q lies on the given line

=1

=9=k

Example 98: A plane p is perpendicular to the two planes 2x – 2y + z = 0 and x – y + 2z = 4 and passes through the point (1, –2, 1). The distance of p (1, 2, 2) is (a) 0 (b) 1 (c)

(d) 2

r = (i - j + 2k ) + m ( -3i + j + 5k )

=1 1

2

Example 100: Let P (3, 2, 6) be a point in space and Q be a point on the line

0 + 0 + 0 -1

fi

3

(c)

= k , then the value of k is

(c) 1 Ans. (a)

(b)

(d) 2 2

Let

OQ = (i - j + 2k ) + m ( -3i + j + 5k )

Also

OP = 3i + 2 j + 6k

As

PQ is parallel to the plane x – 4y + 3z =1 PQ ◊ (i - 4 j + 3k ) = 0

fi

(OQ - OP ) ◊ (i - 4 j + 3k ) = 0

fi fi

(1 + 4 + 6) + μ (– 3 – 4 + 15) – (3 – 8 + 18) = 0 11 + 8μ – 13 = 0 fi μ = 1/4

21.26

Complete Mathematics—JEE Main

Example 101: x y z straight line = = and perpendicular to the plane con 2 3 4 taining the straight lines. x y z x y z = = and = = 3 4 2 4 2 3 (a) x + 2y – 2z = 0 (c) x – 2y + z = 0 Ans. (c)

(b) 3x + 2y – 2z = 0 (d) 5x + 2y – 4z = 0

and r = 5/3 Ê 8 4 -7 ˆ Hence the coordinates of Q are Ë , , ¯ 3 3 3 Example 103: the plane passing through the point (–1, –2, –1) and whose x +1 y + 2 z +1 = = 3 1 2 x-2 y+2 z+3 and = = is 1 2 3

Solution: Let a, b, c to the plane containing the lines then

x y z x y z = = and = = 3 4 2 4 2 3

3a + 4b + 2c = 0

fi fi

(c) 13 / 75

(d) 23 / 75

Solution: (–1, –2, –1) be a (x + 1) + b (y + 2) + c (z +1) = 0

8a – b – 10g = 0 2a + 3b + 4g = 0 a b g = = 26 -52 26 a b g = = 1 -2 1

fi

3a + b +2c = 0, a + 2b + 3c = 0 a b c = = -1 -7 5

fi

x +7y – 5z +10 = 0 1 + 7 - 5 + 10

x y z = = , the point 2 3 4 (0, 0, 0) lying on it also lies on the plane and hence x – 2y + z = 0 Example 102: If the distance of the point P (1, –2, 1) x + 2y – 2z = a, where, a > 0, is 5, then the P to the plane is Ê 8 4 -7 ˆ (a) Ë , , ¯ 3 3 3

Ê 4 -4 1 ˆ (b) Ë , , ¯ 3 3 3

Ê 1 2 10 ˆ (c) Ë , , ¯ 3 3 3 Ans. (a)

Ê 2 -1 5 ˆ (d) Ë , , ¯ 3 3 2

a +5 ±5 fi a = 10 =r= 9 3

1 + 49 + 25

=

13 75

Example 104: plane passing through the points. (2, 1, 0), (5, 0, 1) and (4, 1, 1) is the point. (a) (–2, – 1, –6) (b) (6, 5, –2) (c) (1, 1, 4) (d) (2, 5, 4) Ans. (b) Solution: points is x - 2 y -1 z - 0 5 - 2 0 -1 1- 0 = 0 4 - 2 1-1 1- 0

Solution: Let Q (x, y, z) be the foot of the perpendicular x -1 y + 2 z -1 = = P PQ is = r (say) 1 2 -2 Let the coordinates of Q be (r +1, 2r – 2, –2r + 1) Since it lies on the plane, r + 1 + 2 (2r – 2) –2(–2r +1) = a r–5=a fi r a) fi Now PQ = 5 fi r 2 + 4r 2 +4r 2 = 25 fi r = ±5 / 3 So

(b) 7 / 75

Ans. (c)

4a + 2b + 3c = 0 a b c fi = = 8 -1 -10 Let a, b, g then

(a) 2 / 75

(∵ a > 0)

fi fi

x - 2 y -1 z - 0 3 -1 1 =0 2 0 1 x + y –2z = 3

P perpendicular to this plane is x - 2 y -1 z - 6 = = = r say 1 1 -2 Let Q (r +2, r +1, –2r P in this plane, r Êr ˆ of PQ lies on the R Ë + 2, + 1, - r + 6 ¯ 2 2 plane.

Three Dimensional Geometry 21.27

r r + 2 + + 1 - 2(-r + 6) = 3 2 2 fi 3r = 12 fi r=4 and the coordinates of Q are (6, 5, –2)

So

Example 105: If the distance between the plane A x – 2y + z= d and the plane containing the lines x -1 y - 2 z - 3 x-2 y-3 z-4 = = = = and is 6 , then 2 3 4 3 4 5 |d| is (a) 3 (b) 4 (c) 5 (d) 6 Ans. (d)

Solution: lines is x -1 y - 2 z - 3 2 3 4 =0 3 4 5 fi x – 2y + z = 0 Ax – 2y + z = d, we get A = 1 As distance between the planes is 6 |d| we get = 6 fi |d| = 6 1+ 4 +1

EXERCISE Concept-based Straight Objective Type Questions 1. If Q to the plane 4x – 3y + z + 13 = 0 and R be the point (–1, 1, –6). Then length QR is (a)

14

(c) 3 7 / 2

(b)

19 / 2

(d) 3 / 2

y-k x +1 y -1 z +1 x+2 2. If the lines = = and = 3 3 2 1 2 z = are coplanar, then the value of k is 4 (a) 11/2

(b) – 11/2

q (0 < q p/2) with both x and y of all values of q is the interval:

p (a) ÊÁ 0, ˘ Ë 4 ˙˚

Èp p ˘ (b) Í , ˙ Î6 3˚

Èp p ˘ (c) Í , ˙ Î4 3˚

Êp p ˘ (d) Á , ˙ Ë 6 2˚

planes x = ay + b and z = cy + d is x - b y -1 z - d (a) = = a 1 c (b)

x - b - a y -1 z - d - c = = a 1 c

x-a y-0 z-c = = b 1 d x - b - a y -1 z - d - c (d) = = 0 d b (c)

5. If the angle between the line 2(x + 1) = y = z + 4 and the plane 2x – y + l is (a) 135/7 (c) 45/7

l z + 4 is p /6, then the value of (b) 45/11 (d) 135/7

6. If the centroid of the triangle with vertices (3c + 2, 2, 0), (2c, –1, –1) and (c + 2, 3c + 1, c + 3) lies in the plane z = c, then the coordinates of the centroid are: Ê 10 5 ˆ Ê 2 1 1ˆ (b) ÁË , , 1˜¯ (a) Á - , - , ˜ Ë 3 3 3¯ 3 3 Ê 4 2 2ˆ (c) Á , , ˜ Ë 3 3 3¯

Ê 2 1 1ˆ (d) Á , , - ˜ Ë 3 3 3¯

a, b, g ). The locus of the foot of the perpendiculars to the plane

(a) a plane inclined at an angle

p with the given plane 3

(b) a straight line (c) a plane perpendicular to the given plane (d) none of these

21.28

Complete Mathematics—JEE Main

8. Let (a, b, c) π does not represent a straight line is (a) ax – by + cz + d = 0, ax + b' y + cz + d = 0 (b π b¢) (b) ax – by + cz + d = 0, ax + by + c¢z + d = 0 (c π c¢) (c) ax + by + cz + d = 0, ax + by + cz – d¢ = 0 (d π d¢) (d) ax + by + cz + d = 0, a¢x + by + cz + d = 0 (a π a¢) z x - 2 y - 5 z +1 = = is 3 2 -5

(a) 1 / 13 (c) 11 / 13

12. Let L: x – 2y + 4z P(2, 1, –1) be a point and O, be the Origin. Q is the foot P on the plane L, then (OP)2 + (PQ)2 (a) 190 (b) 43 21 21 (c) 295 (d) 211 21 21 x – 2y + 3z = 24 on the coordinate axes is (a) 28 (b) 20 (c) 4 (d) 44

(b) 11 / 13 (d) 11 / 13

x-4 y-2 z-l x y+z z and = = = = 1 1 3 1 2 4 intersect each other, then l lies in the interval (a) (– 5, – 3) (b) (13, 15)

10. If the lines

A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6) in units is (b) 6 / 34 (a) 3 / 34 (c) 2 / 17 (d) 3 / 17

14. L1 : 2 (x – 1) + 3 (y + 2) + (z – 7) = 0 and L1 : 2 (x + 1) – 3 (y + 2) + (z + 7) = 0 are two planes. (a) L1 and L2 are parallel (b) L1 and L2 are perpendicular (c) L1 and L2 L1 and L2 is y = 0 15. If the points (1, 1, p r (3i + 4j – 12k) + 13 = 0 then the value of p is (a) –1 (b) 0 (c) 7/3 (d) 3/7

LEVEL 1 Straight Objective Type Questions

the points (–1, – 1, 1) and (–1, 1, –1) are (a) (0, 0, 0) (b) (– 1, 0, 0) (c) (0, –1, 1) (d) (0, 1, – 1). 17. If l, m, n and l ¢, m ¢, n ¢ be the direction cosines of two lines which include an angle q, then (a) cos q = ll¢ + mm¢ + nn¢ (b) sin q = ll¢ + mm¢ + nn¢ (c) cos q = mn¢ + m¢n + nl¢+ n¢l + lm¢ + l¢m. (d) sin q = mn¢ + m¢n + nl¢ + n¢l + lm¢ + l¢m. (a) x = 0 (c) z = 0

XOY plane is (b) y = 0 (d) z = c, c π 0

the point (a, b, c) on z (a) (a, 0, 0) (b) (0, b, 0) (c) (0, 0, c) (d) (a, b, 0)

20. l = m = n = 1 represents the direction ratios of y (a) x (c) z 21. If l1, m1, n1 and l2, m2, n2 are direction cosines of two perpendicular lines, then (a) l1 l2 + m1 m2 + n1 n2 = 0 (b) l1 l2 + m1 m2 + n1 n2 = 1 (c)

l 1 m1 n 1 + + =1 l 2 m2 n 2

(d)

l 1 m1 n 1 + + =0 l 2 m2 n 2

22. P (0, 5, 6), Q (1, 4, 7), R (2, 3, 7) and S (3, 4, 6) the origin O (0, 0, 0) is (a) P (b) Q (c) R (d) S

Three Dimensional Geometry 21.29

23. P (1, 1, 1) and Q (l, l, l) are two points in the space such that PQ = 27 , the value of l can be (a) – 4 (b) – 3 (c) 2 (d) 4

(a) x – y = 0 (c) y – z = 0

(b) x – z = 0 (d) x + y + z = 3 x y-2 z-3 = = are 3 4 2 (b) (– 2, –1, –1) (d) (2, 3, 4)

the point (3, –1, 11) on the line

the vertices of (a) a right angled isosceles triangle

(a) (2, 5, 7) (c) (0, 2, 3)

(c) an isosceles triangle (d) an obtuse angled triangle

the coordinate axes are 12, 4, 3 is

25. If a, b, g coordinate axes, then

(c) 11

(a) sin2 a + cos2 b + cos2 g = 0 2

2

2

(b) cos a + cos b + cos g = 0 (c) cos2 a + cos2 b + cos2 g = 1 (d) sin2 a + sin2 b = 1 – cos2 g 26. If the vector 2i – 3j + 7k is inclined at angles a, b, g with the coordinate axes, then (a) 3cos a = 2 / 62 (b) 2cos b = - 3 / 62 (c) cos g = 7 / 62 (d) 2 cos a = – 3 cos b = 7 cos g A (3, 4, – 7) and B (1, – 1, 6) is (a) r = 3i – 4j + 7k + l (2i – 5j – 13k) (b) r = i – j + 6 k – l (2i – 5j – 13k) x-3 y-4 z+7 = = (c) 2 –5 - 13 (d)

x -1 y +1 z - 6 = = -2 -5 13

28. The lines

x -1 y -1 z +1 = = and -1 3 0

y + 0 z +1 x-4 = = 0 3 2 (a) do not intersect (b) intersect (c) intersect at (4, 0, 4) (d) intersect at (1, 1, – 1) r = 2i + 2j – k + (i + j + k) t and the plane r. (3i – 4j + 5k) = q is (a) 2 6 /15 (c)

201 /15

containing x

(b) 2 3 /15 (d) none of these

(d) none of these

three coordinate axes is (a) 3 (b) 4 (c) 6 (d) 8 point with position vector a and is parallel to the vectors b and c is (a) [r b c]= [a b c] (b) [r c a[ = [b c a] (c) [r a b] = [c a b] (d) none of these 35. If the lines r = (i + j – k) + l(3i – j) and r = (4i – k) + m (2i + 3k) intersect at the point (p, 0, p – 5) then (a) p = 0 (b) p = – 1 (c) p = 4 (d) p = 5 36. The line of intersection of the planes r. (3i – j + k) = 1 and r. (i + 4j – 2k) = 2 is parallel to the vector (a) –2i + 7j + 13k (b) 2i + 7j – 13k (d) 2i + 7j + 13k (c) –2i + 7j – 13k x +1 y - 3 z + 2 = = and the point (0, 7, – 7) is 2 1 -3 (a) x + y + z = 0 (b) x + 2y – 3z = 35 (c) 3x – 2y + 3z + 35 = 0 (d) 3x – 2y – z = 21 38. A line passes through two points A (2, – 3, – 1) and B (8, – 1, 2), the coordinates of a point on this line A are (a) (14, 1, 5) (b) (– 10, – 7, – 7) (c) (10, 7, 7) (d) (– 14, – 1, – 5) x y z + + =1 a b c A, B, C respectively. D and E AB and AC of DE are (a) (a, b/4, c/4) (b) (a/4, b, c/4) (c) (a/4, b/4, c) (d) (a/2, b/4, c/4)

at of

Complete Mathematics—JEE Main

21.30

40. The lines r = a + l (b × c) and r = b + m (c × a) will intersect if (a) a × c = b × c (b) a.c = b.c (c) b × a = c × a (d) none of these 41. The shortest distance between the lines r = (4i - j) + l (i + 2 j - 3k ) and r = (i - j + 2k ) + m (2i + 4 j - 5k ) is 5

(a) 6

(b)

(c) 6/ 5

(d) 6 5

42. The coordinates of a point on the line x = 4y + 5, z = 3y – 6 at a distance 3 26 (5, 0, – 6) are (a) (17, 3, 3) (b) (– 7, 3, – 15) (c) (– 17, – 3, – 3) (d) (7, – 3, 15) 43. If q denotes the acute angle between the line r = (i + 2j – k) + l (i – j + k) and the plane r.(2i - j + k ) = 4 , then sin q + (a) 1/ 2 (c)

2

2 cos q =

(b) 1 (d) 1 + 2

44. The plane containing the lines x +1 y + 3 z + 5 x-2 y-4 z-6 = = = = and passes 3 5 7 1 4 7 through (a) (0, 0, 0) (b) (1, 0, 1) (c) (1, – 1, 1) (d) (– 1, 1, 0) r . (i + 2j + 2k) r. (4i – 3j + 12k) + 3 = 0, passing through the point with position vector i + 7j – k is (a) r .( i + 35 j - 10k ) – 256 = 0 (b) r .( 25i + 17 j + 62k ) - 238 = 0 (c) r .(i + 2 j + 2k ) - 13 = 0 (d) r .(4i - 3 j + 12k ) + 29 = 0 x -1 y - 3 z +1 = = lies in the plane 2 a 3 bx + 2y + 3z – 4 = 0, then (a) a = 11/2, b = 1 (b) a = – 5/2, b = – 7 (c) a = – 11/2, b = 1 (d) a = 1, b = – 11/2

46. If the line

i + j – 2k, 2i – j + k and i + 2j + k is (a) r . (4i + 2j) = 20 (b) r . i + 3j – k) = 14 (c) r. i + 3j – k) = 6 (d) none of these

48. The line of shortest distance between the lines x –1 y+8 z–4 x –1 y–2 = = and = 2 2 –7 5 1 z–6 = –3 (a) (1, 1, 1) (c) (3, 3, 3) r. n = q vector n origin on the plane is (a) q

(b) (– 1, – 1, – 1) (d) (– 3, – 3, – 3)

(c) q|n|

(d)

(b) |n| q n

plane is (a, b, c x y z + + =1 (a) a b c (b) ax + by + cz = 1 (c) ax + by + cz = a2 + b2 + c2 (d) ax + by + cz = 0 51. The plane passing through the points (– 2, – 2, 2) and containing the line joining the points (1, 1, 1) and

(a) 3 (c) 6

(b) 4 (d) 12

position vector is 2i – 3 j + 4k and in the direction of the vector 3i + 4 j – 5k is (a) (b) (c) (d)

4x + 3y = 17, 5y – 4z = 1 4x – 3y = 17, 5y + 4z = 1 4x + 5y = 12, 3y + 4z = 1 5y + 4z = 1, 4x + 3z = 1

53. The lines r = i – j + l (2i + k ) and r= (a) (b) (c) (d)

2i – j + m (i + j – k ) intersect each others do not intersect intersect at r = 3i – j + k are parallel.

54. The points (1, – 2, 3), (2, 3, – 4) (0, – 7, 10) are the vertices of (a) a right angled triangles (b) isosceles triangle (d) none of these

Three Dimensional Geometry 21.31

a, b, c) on the line x = y = z is the point (r, r, r) where (a) r = a + b + c (b) r = 3 (a + b + c) (c) 3r = a + b + c (d) none of these x + py + q = 0 = rz + s on y and z axes is x + py + q + l (rz + s) = 0 where l (a) q/s (b) p/r (c) r/s (d) p/q

at P and Q respectively. Length of PQ is (b)

77

(c)

54

(d)

74

60. The angle between the lines x - 2 y +1 = , z = 2 and 3 -2 x - 1 2y + 3 z + 5 = = is 1 3 2 (a) p/3 (c) p/4

(b) p/6 (d) p/2

x y y + + =3 axes a b c in A, B, C, and the centroid of the triangle ABC is at P(2, 4, 8), then a, b, c are in (a) A.P (b) G.P (c) H.P (d) none of these

58. A line with direction ratios 2, 7, – 5 is drawn to intersect the lines x+3 y-3 z-6 x-5 y-7 z+2 = = and = = 2 4 -3 3 1 -1

78

(b) 2 : 3 (d) 2 : 1

61. If the plane

3x – 6y – 2z – 15 = 2x + y – 2z – 5 = 0 is x-5 y z x -1 y - 5 z -1 = = = = (b) (a) 14 2 15 14 2 15 x - 3 y +1 z = = (c) (d) none of these 14 2 15

(a)

(a) 1 : 2 (c) 3 : 2

62. The variable plane (2l + 1) x + (3 – l) y + z = 4 always passes through the line x y z+4 x y z = = = = (b) (a) 2 1 1 1 2 -3 x y z-4 = = (c) (d) none of these -7 1 2 3x – 4y + 7z = 84 on the axes is (a) 6 (b) 14

Z (3, 5, – 7) and (– 2, 1, 8) in the ratio

Assertion-Reason Type Questions

64. Vertices of a triangle ABC are A (1, 1, 0), B(1, 0, 1) and C(0, 1, 1) Statement-1: triangle ABC is

Staement-2: If R is a point on L such that the per p is

pendicular distance of R

3 then

the coordinates of R are (1, 2, 3) 2/3 .

Statement-2: triangle ABC lies on the plane x + y + z – 2 = 0. 65. Consider the line L :

x y z = = and the plane 1 2 3

p: x + y + z = 0

Statement-1: The unit vector perpendicular to both

Staement-1: If P is a point on L at a distance

14

O and N is the foot of the perpen P to the plane p, then ON =

x +1 y + 2 z +1 = = , 3 1 2 x-2 y+2 z-3 = = . L2 : 1 2 3

66. L1 :

2

L1 and L2 is

– i – 7 j + 5k 5 3

.

Statement-2: the plane passing through the point

21.32

Complete Mathematics—JEE Main

both the lines L1 and L2 is 23 5 3 .

x-2 y-2 z-2 = = 1 1 1 x -1 y z +1 = = L 2: 0 1 2 Statement-1: The shortest distance between L1 and L2 is zero. L 1:

67. Statement-1: The distance between the line r = 2i + 2 j + 3k + l (i – j + 4k ) and the plane r. (i + 5 j + k ) = 5 is 10 3 3 . Statement-2: If a line is parallel to a plane, then the

line to the plane. 68. Statement-1: The direction cosines of the line 6x – 2 = 3y + 1 = 2z x + 3y + z = 14. Statement-2: the plane are p /4, p /4, p /2 and the length of the 2, x + y = 2.

Statement-2: The line L1 and L2 are coplanar. z - 7 . and P be 70. Let L be the line x – 4 = y – 2 = 2 the plane 2x – 4y + z =7. Statement-1: The line L lies in the plane P. Statement-2: The direction ratios of L are l1 = 1 , m1 = 1, n1 are l2 = 2, m2 = – 4, n2 = 1; and l1l2 + m1m2 + n1n2 = 0 holds.

LEVEL 2 Straight Objective Type Questions x-a y-b z-c x - a¢ = = and = c¢ a a¢ b¢ y - b¢ z - c¢ = intersect at the point c b (a) (a – a¢, b – b¢, c – c¢ ) (b) (a + a¢, b + b¢, c + c¢ ) (c) (a, b, c) (d) (a¢, b¢, c¢ )

71. The lines

3x – y + 4z = 2 is (a) (0, 1, – 3) (b) (– 1, 0, – 3) (c) (0, – 1, – 3) (d) (0, – 1, 3) 73. The shortest distance between the lines x + a = 2y = – 12z and x = y + 2a = 6z – 6a is (a) a (b) 2a (c) 4a (d) 6a 74. The P (u, v, w line x = y = z x + y + z = 1 and uv + vw + wu = 0 then u2 + v2 + w2 – 4(u + v + w (a) – 2 (b) 0 (c) 2 (d) 4 75. The angle between the two diagonals of a cube is (a) 30º (b) 45º (c) cos–1 1/ 3 (d) cos–1 (1/3)

(

)

(2, 3, – 1), (0, 8, 5) and (5, 7, 11). The distance of

(a)

74

(b) 2 19

(c)

78

(d)

77. If a, b, g, d the four diagonals of cos2 a + cos2 b + cos2 (a) 1 (c) 2/3

442

a cube, then the value of g + cos2 d is (b) 1/3 (d) 4/3

plane 3x + 4y – 5z – 60 = 0 and the coordinate planes in cubic units is (a) 60 (b) 600 (c) 720 (d) none of these ABC sides are on the x, y, z axes. Lengths of the intercepts are respectively a, b, g. Coordinates of the centroid of the triangle ABC are (a) (– a /3, b/3, g /3) (b) (a /3, – b/3, g /3) (c) (a /3, b/3, – g /3) (d) (a /3, b/3, g /3) x = ± 1, y = ± 1, z = ± 1 of a cube is 10 units. The locus of the point is (a) x 2 + y 2 + z 2 = 1 (b) x2 + y2 + z2 = 2 (c) x + y + z = 1 (d) x + y + z = 2.

Three Dimensional Geometry 21.33

81. The plane 2 x – y + 3z about its line of intersection with the plane 5x – 4y – 2z position is y z – 31 = 0 (a) 6x (b) 27x – 24y – 26z – 13 = 0 (c) 43x – 32y – 2z + 27 = 0 (d) 26x – 43y – 151z – 165 = 0 82. A plane passes through (1, – 2, 1) and is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4. (a) 0 (c) 2

(x, y, z) of DABC z –2 = k, then the value of k is

x –2 + y –2 +

(c) 1 (d) 1/3 84. If t planes 3x – 6y – 2z = 15 and 2x + y – 2z = 5 is (a) x = 3 + 14t, y = 1 + 2t, z = 15t (b) x = 3 + 14t, y = – 1 + 2t, z = 15t (c) x = – 3 + 14t, y = 1 + 2t, z = 15t (d) none of these. 85. Projection of the line 8x – y – 7z = 8, x + y + z = 1 on the plane 5x – 4y – z = 5 is x –1 y – 0 z – 0 x – 0 y –1 z – 0 = = (b) (a) = = 1 2 –3 1 2 –3

(b) 1 (d) 2 2

the coordinates axes at points A, B, C. If the centroid (c)

x – 0 y – 0 z –1 = = (d) none of these 1 2 –3

Previous Years' AIEEE/JEE Main Questions

p /4 with plane x + y = 3 are (a) 1, 2 , 1 (c) 1, 1, 2

(b) 1, 1, 2 (d) 2 , 1, 1

[2002]

2. A plane which passes through the point (3, 2, 0) and x–4 y–7 z–4 = is the line = 1 5 4 (a) x – y + z = 2 (b) x + y + z = 5 (c) x + 2y – z = 1 (d) none of these [2002]

x + 4y + 3z = 327 to the sphere x 2 + y2 + z2 + 4x – 2y – 6z = 155 is 3 (b) 13 (a) 11 4 [2003] 5. The two lines x = ay + b, z = cy + d and x = a¢y + b¢, z = c¢y + d will be perpendicular, if and only if (a) aa¢ + bb¢ + cc¢ = 0 (b) (a + a¢) (b + b¢) (c + c¢) = 0 (c) aa¢ + cc¢ + 1 = 0 (d) aa¢ + bb¢ + c¢ + 1 = 0 [2003] 6. The lines

a, b, c and a¢, b¢, c¢

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

1 a

2

1 a

2

1 a

2

1 a

2

+ – + +

1 b

2

1 b

2

1 b

2

1 b

2

– – + +

1 c

2

1 c

2

1 c

2

1 c

2

+ + – +

1 a¢

2

1 a¢

2

1 a¢

2

1 a¢

2

+ – – +

1 2

b¢

1 b¢

2

1 b¢

2

1 b¢

2

– – – +

1 c ¢2 1 c¢

2

1 c¢2 1 c¢2

=0

x –1 y–4 z–5 = = are coplanar if k 2 1 (a) k = 1 or – 1 (b) k = 0 or – 3 (c) k = 3 or – 3 (d) k = 0 or – 1

[2003]

*7. The radius of the circle in which the sphere x 2 + y2 + z2 + 2x – 2y – 4z plane x + 2y + 2z + 7 = 0 is (a) 2 (b) 3 (c) 4 (d) 1 [2003]

=0 =0 =0

z–4 x–2 y–3 = = and –k 1 1

[2003]

8. A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1), B(2, 1, 3) and C(– 1, 1, 2). Then the angle between the faces OAB and ABC will be

21.34

Complete Mathematics—JEE Main

(a) cos –1

*17. If the plane 2ax – 3ay + 4az + 6 = 0 passes through

–1

[2003]

q with each of the x and b y z such that sin2b = 3sin2q, then cos2q (a) 3/5 (b) 1/5 (c) 2/3 (d) 2/5 [2004] 10. Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is (a) 7/2 (b) 5/2 [2004]

spheres x 2 + y2 + z2 + 6x – 8y – 2z = 13 and x2 + y2 + z2 – 10x + 4y – 2z = 8, then a (a) – 2 (c) – 1

(3a, 2a, 3a), (a, a, 2a) (3a, 2a, 3a), (a, a, a) (3a, 3a, 3a), (a, a, a) (2a, 3a, 3a), (2a, a, a)

(a) 2 (c) 3

*13. The intersection of the spheres x 2 + y 2 + z 2 + 7x – 2y – z = 13 and x 2 + y 2 + z 2 – 3x + 3y + 4z = 8 is the plane (a) x – y – 2z = 1 (b) x – 2y – z = 1 (c) x – y – z = 1 (d) 2x – y – z = 1

[2005]

x – 2y = 0 is

[2006] 21. Let L be the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = z. If L angle a with the positive x a (a) 1/ 3

(b) 1/2

(c) 1

(d) 1/ 2

tion of each of x (a) p/6 (c) p/4

[2004]

14. The angle between the lines 2x = 3y = – z and 6x = – y = – 4z is

(b) 2 (d) 1

x = ay + b, z = cy + d; and x = a¢y+ b¢, z = c¢y + d ¢ are perpendicular to each other if a c + =1 (a) (b) aa¢ + cc¢ = – 1 a¢ c¢ a c (c) aa¢ + cc¢ = 1 (d) + = –1 [2006] a¢ c¢

[2004]

12. If the straight lines x = 1 + s, y = – 3 – ls, z = 1 + ls s and x = t/2, y = 1 + t, z = 2 – t and t are coplanar, then l (a) –1/ 2 (b) – 1 (c) – 2 (d) 0 [2004]

[2005]

*18. The plane x + 2y – z = 4 cuts the sphere x 2 + y2 + z2 – x + z – 2 = 0 in a circle of radius

11. A line with directional cosines proportional to x = y + a = z and x + a = 2y = 2z. The coordinates of each of the points of intersection are given by (a) (b) (c) (d)

(b) 2 (d) 1

[2007]

p/4 with the positive direc y z (b) p/3 (d) p/2 [2007]

* x 2 + y2 + z2 – 6x – 12y – 2z + 20 = 0, then the co

[2005] 15. If the angle q between the lines

x +1 y –1 = 1 2

z–2 and the plane 2 x – y + l z + 4 = 0 is such 2 that sin q = 1/3, the value of l is (a) 3/4 (b) – 4/3 (c) 5/3 (d) –3/5 [2005] =

16. The distance between the line r = 2i – 2j + 3k + l (i – j + 4k) and the plane r . (i + 5j + k) = 5 is (a) 3/10 (b) 10/3 3

[2005]

(c) (4, 3, 5)

(d) (4, 3, – 3)

[2007]

24. 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, the x (a) 0 (b) 1 (c) – 4 (d) – 2 [2007] 25. The line passing through and (3, b, 1) crosses the (0, 17/2, – 13/2) then (a) a = 8, b = 2 (b) (c) a = 4, b = 6 (d)

the points (5, 1, a) yz a = 2, b = 8 a = 6, b = 4

[2008]

Three Dimensional Geometry 21.35

26. If the straight lines x –1 y – 2 z – 3 x – 2 y – 3 z –1 = = = = and in 2 k 2 3 3 k tersect at a point, then the integer k (a) – 2 (b) – 5 (c) 5 (d) 2 [2008] x - 2 y -1 z + 2 = = lie in the plane 3 -5 2 x + 3y – a z + b = 0. Then (a, b (a) (5, – 15) (b) (– 5, 5) (c) (6, – 17) (d) (– 6, 7) [2009]

27. Let the line

28. The projections of a vector on the three coordinate axis are 6, – 3, 2 respectively. The direction cosines of the vector are 6 -3 2 -6 -3 2 , , (b) (a) , , 7 7 7 7 7 7 6 -3 2 (d) , , 5 5 5

(c) 6, – 3, 2

[2009]

AB x y the positive z

q with

AB q

cos-1 ( 5 / 14 ) , then l (a) 5/3 (c) 3/2

(b) 2/3 (d) 2/5

plane x – y + z x = y = z is (a) 10 3

(b) 5 3

(c) 3 10

(d) 3 5

(3, –1, 11) to the line

[2011]

[2011]

x y-2 z-3 = is = 2 3 4

(a)

29

(b)

33

(c)

53

(d)

66

x – 2y +2z origin is (a) x– 2y + 2z – 1 = 0 (b) x– 2y + 2z + 5 = 0 (c) x– 2y + 2z – 3 = 0 (d) x– 2y + 2z + 1 = 0

[2011]

[2012]

x -1 y +1 z -1 = = and 2 3 4 x-3 y-k z = = intersect, then k 1 2 1

36. If the lines [2010] 30. Statement-1: The point A B(1, 3, 4) in the plane x – y + z = 5. Statement-2: The plane x – y + z = 5 bisects the line A(3, 1, 6) and B(1, 3, 4)

not a correct explanation for [2010] 31. Statement-1 : The point A of the point (1, 6, 3) in the line

x y -1 z - 2 = . = 1 2 3

x y -1 z - 2 Statement-2 : The line = = bisects the 1 2 3 A (1, 0, 7) and B (1, 6, 3). [2011] 32. The angle between the line y -1 z - 3 = and the plane x + 2y +3z = 4 is x = l 2

[2012] x-2 y-3 z-4 = = and 1 1 -k x -1 y - 4 z - 5 = = are coplanar, then k can have k 2 1 (a) exactly three values (b) any value (c) exactly one value (d) exactly two values [2013]

37. If the lines

38. Distance between the parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is (a) 5/2 (b) 7/2 [2013] x +1 y -1 z +1 x+2 y-k z = = = = and 2 1 3 2 3 4 are coplanar, then the value of k is (a) 11/2 (b) – 11/2 [2013, online]

40. Let Q to the plane 4x – 3y + z +13 = 0 and R be the point (–1, 1 , –6) on the plane. Then length QR is:

21.36

(a)

Complete Mathematics—JEE Main

(b)

14

19 / 2

(c) 3 7 / 2 (d) 3 / 2 [2013, online] 41. The acute angle between two lines such that the direction cosines l, m, n l + m + n = 0 and l2+ m2 – n2 = 0 is: [2013, online] 42. If the lines L1 and L2 L1 = {x = l y + ( l – 1) z = ( l – 1)y + L2 = {x =

m y + (1 –

l } and

m ), z = (1 – m )y + m }

then L1 is perpendicular to L2 reals l and m, such that: (b) l π m (a) l + m = 1 (c) l + m = 0

(d) l = m

[2013, online] x, y and

z (a) 12

(b) 7 [2013, online]

44. Let ABC be a triangle with vertices at points A(2, 3, 5), B(–1, 3, 2) and (l, 5, m A clined with the axes, then (l, m (a) (10, 7) (b) (7, 5) (c) (7, 10) (d) (5, 7) [2013, online] tion of the planes x + 2y = 3, y – 2z + 1 = 0 and (a) 2x – y – 10z (c) 2x – y + 10z = 11

x – y + 7z = 11 z = 10 (d) 2x – y [2013, online]

x -1 y - 3 z - 4 = = in the 3 1 -5 plane 2x – y + z + 3 = 0 is the line

(a) (c)

x+3 y-5 z-2 x+3 y-5 z+2 = = = = (b) 3 1 -5 -3 -1 5 x-3 y+5 z-2 x-3 y+5 z-2 = = = = (d) 3 1 -5 -3 -1 5 [2014]

47. The angle between the lines whose direction cosines l + m + n = 0 and l2 = m2 + n2 is (a) p /3 (b) p /4 (c) p /6 (d) p /2 [2014]

of intersection of the lines x - 3 y -1 z - 2 = = 1 2 3 the origin is: (a) 7x + 2y + 4z = 54 (c) 4x + 3y + 5z = 50

x -1 y - 2 z - 3 and = = 3 1 2

(b) 3x + 4y + 5z (d) 5x + 4y + 3z = 57 [2014, online]

q (0 < q p/2) with both the x and y axes. Then the set of all values of q is the interval. p p Ê p˘ (b) ÈÍ , ˘˙ (a) Á 0, ˙ Ë 4˚ Î6 3˚

p p Êp p ˘ (c) ÈÍ , ˘˙ (d) Á , ˙ [2014, online] Ë 3 2˚ Î4 2˚ 50. Let A(2, 3, 5), B(–1, 3, 2) and C( l, 5, m) be the ver tices of a D ABC A inclined to the coordinate axes, then: (a) 5l – 8m = 0 (b) 8l – 5m = 0 (c) 10l – 7m = 0 (d) 7l – 10m = 0 [2014, online] x -1 y - 2 z - 3 51. The plane containing the line = = 1 2 3 and parallel to the line x = y = z passes through the 1 1 4 point: (a) (1, –2, 5) (b) (1, 0, 5) (c) (0, 3, –5) (d) (–1, –3, 0) [2014, online] planes x = ay + b and z = cy + d is x - b y -1 z - d (a) = = a 1 c (b)

x - b - a y -1 z - d - c = = c a 1

x-a y-0 z-c = = b 1 d x - b - a y -1 z - d - c (d) = = d b 0 (c)

[2014, online]

53. If the distance between the planes 4x – 2y – 4z + 1 = 0 and 4x – 2y – 4z + d = 0 is 7, then d is (a) 41 or – 42 (b) 42 or – 43 (c) – 41 or 43 (d) – 42 or 44 [2014, online]

Three Dimensional Geometry 21.37

x -1 lines x = y = z and = 0 1 -1 1 (b) (a) x = y = z 1 -1 -2

x -1 y +1 z = = 1 1 1

y +1 z = is: 1 -2

(a) -

x -1 y +1 z = = 1 -2 -1

(c)

1 8

16 19

32 19

[2014, online]

x–y+z

55. If the angle between the line 2(x + 1) = y = z + 4 and p the plane 2x – y + l z + 4 is , then the value of l is 6 135 (b) 45 (a) 7 11 45 (c) (d) 135 [2014, online] 7 11

(a) 3 10

(c)

(d)

x-2 y +1 z-2 = = and 3 4 12 the plane x – y + z = 16 is: (b) 8 (a) 2 14

intersection of the line

(c) 3 21 + z x+ (a) (c)

(d) 13

(c)

58. The shortest distance between the z x + y + 2z – 3 = 0 = 2x + 3y + 4z – 4, is: (a) 1 (b) 2 (c) 3 (d) 4 [2015 online]

(d)

(d)

3

lx + my – z

(a) 3x – 10x + 7 = 0 (c) 3x2 + 10x – 13 = 0

(a) (3, 4] (c) [1, 2)

61. If the shortest distance between the lines

x -1 = a

y +1 z = , (a π –1) and x + y + z + 1 = 0 = 2x – y -1 1 1 , then a value of a is: + z + 3 is 3

[2016] x y z = = 2 2 1

(b) (2, 3] (d) [0, 1)

[2016, online]

passing through the point (1, 2, 2) and perpendicular to the planes x – y + 2z = 3 and 2x – 2y + z = 0, is (a) 2

(b)

(c) 2 2

(d)

2 1 2

[2016, online]

66. ABC is a triangle in a plane with vertices A(2, 3, 5), B(–1, 3 2) and C(l, 5, m A is

lines

(b) 3x + 10x + 7 = 0 (d) 3x2 – 10x + 21 = 0 [2015, online]

[2016]

x+2 y-4 z-5 = = lies in the interval: -1 8 4

(a) 1130 (c) 1077

2

(b) 18 (d) 2

64. The shortest distance between the lines

(a) (0, –3, 1) (c) (0, 7, –10)

2

20 3

l2 + m2

(a) 26 (c) 5

and

[2015, online]

x-3 y+2 z+4 = = lies in the plane, 2 -1 3

(l3 + m3 + 5) is:

60. If the points (1, 1, l x + 4y – 12z + 13 = 0, then l

19 32

(b) 10 3

x -1 y-2 z-3 = = also contains the point: 1 5 4 (b) (0, 7, 10) (d) (0, 3, 1) [2015, online]

19 16

x = y = z is:

10

63. If the line,

[2015]

x – 5y = 3; x + y + 4z = 5, and parallel is the plane, 3y + 6z = 1, is: 2x + 6y + 12z = 13 (b) x + 3y + 6z = –7 x + 3y + 6z = 7 (d) 2x + 6y + 12z = –13. [2015]

(b) -

x -1 y-2 = = 1 2

(b) 1348 (d) 676 z+3 l2

[2016, online]

l for which the y-2 x-3 and = = l2 1

z -1 are coplanar is: 2 (a) 2 (c) 3 Note:

(b) 4 (d) 1

[2016, online]

21.38

Complete Mathematics—JEE Main

Previous Years' B-Architecture Entrance Examination Questions 1. If the centroid of the triangle with vertices (3c + 2, 2, 0), (2c, –1, –1) and (c + 2, 3c + 1, c + 3) coincides with the centre of the sphere x2 + y2 + z2 + 5ax – 4 by – 2cz = 13, then (a) c = 1 (b) c = 2 (c) c = 3 (d) c = 0 [2006]

x -1 y z +1 = = 0 1 2 Statement-1: The shortest distance between L1 and L2 is zero.

L 2:

Statement-2: The lines L1 and L2 are coplanar.

p, q, r). The locus of the foot of the perpendicular to the plane (a) A plane inclined at an angle of p /4 with the given plane (b) A straight line (c) A sphere (d) A plane perpendicular to the given plane. [2006] 2

2

2

2

3. If the section of the sphere x + y + z = a by the plane z = a/2 is a circle of radius 27 units, then its section by the plane z = a/6 is a circle of radius (a)

35 units

(b)

40 units

(c)

45 units

(d)

30 units

[2007]

4. The distance between the point (–1, –5, –10) and the x - 2 y +1 z - 2 = = point of intersection of the line 3 4 12 with plane x – y + z = 5 is (a) 13 (b) 15 (c) 16 (d) 12 [2007] 5. Let L1 be a line with direction ratios (–2, –1, 2) and L2 be the line joining the points (1, 2, 3) and (3, 2, 1). If q is the angle between the lines L1 and L2 then |sin q (a) 1/3 (b) 1/4 (d) 1/2 [2007] (c) 1 / 3 2 6. Let (a, b, c does not represent a straight line is: (a) ax – by + cz + d = 0, ax + b¢y + cz + d = 0 (b b¢) (b) ax – by + cz + d = 0, ax + by + c¢z + d = 0 (c c¢) (c) ax + by + cz + d = 0, ax + by + cz – d¢ = 0 (d d¢) (d) ax + by + cz + d = 0, a¢x + by + cz + d = 0 (a a¢) [2009]

L 1:

x-2 y-2 z-2 = = 1 1 1

not a correct explanation for

[2010] 8. Let a, b and g the positive directions of the axes of reference in q is the acute angle given by cos q =

cos2 a + cos2 b + cos2 g sin 2 a + sin 2 b + sin 2 g

(a) p /6 (c) p /2

, then q

(b) p /3 (d) p /4

[2010]

A, B, C, and D ABC has centroid at the point G (a/2, b/2, c/2), then (a)

x y z 3 + + = a b c 2

(b)

x y z 2 + + = a b c 3

(c)

x y z 1 + + = a b c 2

(d)

x y z 1 + + = a b c 3

[2011]

10. Shortest distance between z x - 2 y - 5 z +1 is = = 3 2 -5 (a) 1 / 13 (c) 11 / 13

(b) 11/13 (d) 11 / 13

[2011]

11. Let L be the line x – 4 = y – 2 = z - 7 and P be the 2 plane 2x – 4y + z = 7 Statement-1: The line L lies in the plane P. Statement-2: The direction ratios of the line L are l1 = 1, m1 = 1, n1 plane P are l2 = 2, m2 = – 4, n2 = 1, and l1l2 + m1m2 + n1n2 = 0 holds

Three Dimensional Geometry 21.39

k, for which the y-6 z -8 x-2 x-4 y -8 lines = = = 2 and 2 = 2 -2k 2k 2 4 z - 10 are coplanar: 2

not a correct explanation for state

[2012] 1- x = 2 – y = z + 1 is 2 (a) (1, 2, –1) (b) (2, 1, 4) (c) (2, 1, 0) (d) (0, 0, 1)

[2015]

14. The angle between the lines 2x = 3y = – z and – 6x = y = 4z is: [2013] x-4 y-2 z-l x y+2 z = = and = 15. If the lines = 1 1 3 1 2 4 intersect each other, then l lies in the interval (a) (–5, –3) (b) (13, 15) [2014] 16. A variable plane is at a constant distance p the origin O OX (i = 1, 2, 3) at points Ai (i = 1, 2, 3), respec tively. If planes are drawn through A1 A2, A3, which are parallel to the coordinate planes, then the locus of their point of intersection is: 1 1 1 1 1 1 1 + + =p (b) 2 + 2 + 2 = 2 (a) x1 x2 x3 x1 x2 x3 p 1 x13

+

1 x23

+

1 x33

=

1 p

3

x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y (a) (3, 0, 1) (b) (–3, 0, 1) (c) (3, 0, –1) (d) (–3, 0, –1) [2016]

[2012]

A, B, C then the point of intersection of the planes through A, B, C parallel to the coordinate planes lies on: (a) xy – (1/2) yz + (1/3) zx = 6 (b) yz – 2zx + 3xy = xyz (c) xy – 2yz + 3zx = 3xyz (d) xy + (1/2) yz – (1/3) zx = 6 [2013]

(c)

(b) is a singleton. (c) contains two points.

(d) x12 + x22 + x32 = p2 [2014]

Answers Concept-based 1. 5. 9. 13.

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

2. 6. 10. 14.

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

3. 7. 11. 15.

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

4. (b) 8. (c) 12. (c)

17. 21. 25. 29. 33. 37. 41. 45. 49. 53. 57. 61. 65. 69.

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

18. 22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62. 66. 70.

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

19. 23. 27. 31. 35. 39. 43. 47. 51. 55. 59. 63. 67.

72. 76. 80. 84.

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

73. 77. 81. 85.

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

74. (a) 78. (b) 82. (d)

Level 1 16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 64. 68.

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

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

Level 2

17. Let P(x, y, z x+3 y -1 z+4 point Q(0, 2, 3) on the line = = . 5 2 3 If R denotes the point (–3, 1, –4) and A denotes the area of the DPQR, then 2A2 [2015]

71. 75. 79. 83.

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

21.40

Complete Mathematics—JEE Main

f with the z

Previous Years' AIEEE/JEE Main Questions 1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 41. 45. 49. 53. 57. 61. 65.

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

2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62. 66.

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

3. 7. 11. 15. 19. 23. 27. 31. 35. 39. 43. 47. 51. 55. 59. 63. 67.

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

4. 8. 12. 16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 64.

(b) (d) (c) (d) none (d) (a) (b) (a) (c) (c) (c) (b) (d) (a) (c)

2. (c)

3. (a)

4. (a)

5. (a)

6. (c)

7. (a)

8. (b)

9. (a)

10. (d)

11. (b)

12. (c)

13. (b) 17. (c)

14. (c) 18. (a)

15. (c) 19. (a)

16. (b)

Concept-based 1. Let the coordinates of Q be (a, b, g ). OQ – to the a b g = = k . As Q lies on the plane. = plane. So 4 -3 1 1 4(4k) – 3(–3k) + 1(k) + 13 = 0 fi k = – 2 3 1ˆ Ê so coordinates of Q are Ë-2, , - ¯ 2 2

-2 + 1 k - 1 0 + 1 11 2 1 3 =0fik= 2. 2 2 3 4

2

cos q + cos q + cos f = 1 fi 2 cos2q = sin2f 1 Èp p ˘ fi q fiqŒ Í , ˙ Î4 2˚ 2 x-b y-0 z-d = = 1 a c fi

y-0 z-d x-b -1= -1= -1 a 1 c

fi

x - b - a y -1 z - d - c = = 1 c a x +1 y z + 4 = = 1 2 2

sin

p = 2(1) + (-1)(2) + ( l )(2) 6 1+ 4 + 4 4 +1+ l

fi

2

l) = 4 (2 l ) fi l =

45 7

7. Let (p, q, r) be the foot of the perpendicular then

Hints and Solutions

and (QR)2 = (– 2 + 1)2 + ÊÁ 3 - 1ˆ˜ Ë2 ¯ 7 63 = fi QR = 3 . 2 2

2

6. Coordinates of the centroid are Ê 6c + 4 3c + 2 c + 2 ˆ , , ÁË ˜ which lies on z = c so 3 3 3 ¯ c+2 =cfic 3 Ê 10 5 ˆ ÁË , ,1˜¯ 3 3

Previous Years' B-Architecture Entrance Examination Questions 1. (a)

2

2

1 + ÊÁ - + 6ˆ˜ Ë 2 ¯

2

p(x – a) + q(y – b) + r(z – g ) = 0 since (p, q, r) lies on it p(p – a) + q(q – b) + r(r – g) = 0 fi p2 + q2 + r2 – a p – bq – g r = 0 Locus of (p, q, r) is x2 + y2 + z2 – a x – b y – g z = 0 which represents neither a line nor a plane. 8. Planes given in (c) are parallel, so they do not repre sent a straight line. r = (2i + 5j – k) + t(3i + 2j – 5k) and of z r = pk (2i + 5 j - k ) ◊ (3i + 2 j - 5k ) ¥ k | (3i + 2 j - 5k ) ¥ k | 11 = (2i + 5 j - k ) ◊ (2i - 3 j) = 13 4+9

Three Dimensional Geometry 21.41

25. See the text. 4-0 2+2 l -0 = 0 fi l = 12 Œ (11, 13) 1 1 3 1 2 4 a(x – 3) + b(y + 1) + c(z – 2) =0 where 2a + 3b + 2c = 0, –4a + 0b + 4c = 0 a b c = fi = 3 -4 3 3x – 4y + 3z (3)(6) - (4)(5) + 3(9) - 19 6 = units 9 + 16 + 9 34

26. cos a = 2 / 62 , cos b = – 3 / 62 , cos g = 7 / 62 x –1 y +1 z–6 = = 3 –1 4 +1 – 7 – 6 r + 1, – r and a point on the second line is (2k r 3k k = 0 and the point of intersection is (4,

169 [2(1) + 1(-2) + (-1)(4) - 9]2 = (PQ) = 21 1 + 4 + 16 (OP)2 + (PQ)2 =

295 . 21 x y z + + =1 24 -12 8

14. L1: 2x + 3y + z – 3 = 0, L2: 2x – 3y + z + 3 = 0 -3 Distance of L1 2 2 + 32 + 12 3

Distance of L2

22 + (-3)2 + 12

3 + ( – 4 )2 + (5)2 2

3–4+5 150

12. (OP)2 = 4 + 1 + 1 = 6 and 2

(3i – 4 j + 5k ). (i + j + k )

q=

=

+ 1, – 1) + 4, 0 , = 1 and 0, – 1).

=

(

1 + 1 + 1)

2 6 . 15 x

x–0 y–0 z–0 = = is 1 0 0 A(x – 0) + B(y – 0) + C(z – 0) = 0 where A¥1+B¥0+C¥0=0 fi By + Cz = 0, since it passes through (1, 1,1) B+C=0 fi 31. Let the foot of the perpendicular be (2r, 3r + 2, 4r + 3) direction ratios of the perpendicular are 2r – 3, 3r + 3, 4r – 8 so 2(2r – 3) + 3(3r + 3) + 4(4r – 8) = 0 fi r = 1 and the foot is (2, 5, 7). (12 )2 + ( 4 )2 + (3)2 = 13. tion is

33. 3 cos2q = 1

fi

cos q = ± 1 3 . So the direction 3 , ± 1/ 3 , ±

1/ 3 ) which gives four lines. r = a + tb + pc .

3x + 4y – 12z + 13 = 0 |3 + 4 – 12p fi 20 – 12p = ± 8 fi p = 1,

7 3

fi

r . b ¥ c = a . b ¥ c + tb . b ¥ c + pc . b ¥ c

fi

[r b c ] = [a b c ]

Level 1 – 1 – 1 – 1 + 1 1 – 1ˆ 16. Coordinates are Ê = (– 1, 0, 0) Ë 2 , 2 , 2 ¯ 17. See the text. 18. In this plane z coordinates is 0. z x = 0, y = 0. 20. Direction ratios of x, y, z axes are respectively 1, 0, 0; 0, 1, 0 and 0, 0, 1. 21. cos q = l1l2 + m1m2 + n1n2 = 0 if q = p /2. 22. OP 2 = 0 + 52 + 62 = 61, OQ 2 = 66, OR2 = 62, OS 2 = 61. 23. 3(l – 1)2 = 27 fi l = – 2 or 4. (8!) (3!) 1 24. Length of the sides are . = 10! 15

35. The lines are

x –1 y –1 z +1 = = and 3 –1 0

x – 4 y – 0 z +1 = = 2 0 3 z = – 1, fi p – 5 = – 1 fi p = 4. 36. Line of intersection is parallel to the vector li + mj + nk where 3l – m + n = 0, l + 4m – 2n = 0 fi

l m n = . = –2 7 13

A(x + 1) + B(y – 3) + C(z + 2) = 0, where – 3A + 2B + C = 0 and A + B(7 – 3) + C(– 7 + 2) = 0.

21.42

Complete Mathematics—JEE Main

x–2 y+3 z +1 = = . Any 8 – 2 –1+ 3 2 +1 A is ((6/7) r + 2, point on this at a distance r (2/7) r – 3, (3/7) r – 1) r = ± 14, and we get (14, 1, 5) or (– 10, – 7, – 7), the one nearer the origin is (– 10, – 7, – 7). A(a, 0, 0), B(0, b, 0), C(0, 0, c), D(a/2, b/2, 0), DE is (a/2, b/4, c/4). E(a/2, 0, c 40. Lines intersect if a + l (b ¥ c) = b + m (c ¥ a) or if a.c + l (b ¥ c) . c = b.c + m (c × a) . c or if a.c = b.c. 41. d =

[(4i – j) – (i – j + 2k )] . (i + 2 j – 3k ) ¥ (2i + 4 j – 5k ) =

(i + 2 j – 3k ) ¥ (2i + 4 j – 5k ) (3i – 2k ) . (2i – j) 6 =

2i – j

42. Line is

x–5 4 / 26

=

1 / 26

=

line. r.

3 / 26

. A point on this

sin q +

3 6

=

2 cos q =

4 18 4 18

+

2 18

6

=

18

= 2

lx + my + mz + d = 0, 3l + 5m + 7n = 0 and l + 4m + 7n = 0. fi l/1 = m/– 2 = n is x – 2y + z + d = 0 since it passes through (– 1, – 3, – 5) and (2, 4, 6), d = 0 and the plane passes through (0, 0, 0). x + 2 y + 2 z – 19 1+ 4 + 4 =±

4 x – 3 y + 12 z + 3 16 + 9 + 144

since it passes through the vec

x + 35y – 10z – 256 = 0

is a unit

n n q/|n|.

a b c = = 1 –3 –6

is 12.

tion is

fi

4x – 3y = 17, 5y + 4z = 1

2 –1 –1+1 53. Since 2 0 1 1 not intersect. 54. Lengths of the sides 2

0 1 = – 1 π 0, the lines do –1 are

(2 – 1) + (3 + 2 ) + ( – 4 – 3)2 = 75 ,

2

75

and

300 . So the triangle is isosceles. 55. Direction ratios of the perpendicular are r – a, r – b, r – c and those of the line are 1, 1, 1. So 1. (r – a) + 1.(r – b) + 1.(r – c) = 0 fi 3r = a + b + c. 56.

1 + 14 – 2 – 19 tor i + 7 j – k or the point (1, 7, – 1), 3 4 – 21 – 12 + 3 which is true for the + ve sign. = ± 13

n

x–2 y+3 z–4 = = 3 4 –5

2 . 18

, cos q =

,

r = 2i –3j + 4k – l (3i + 4j –5k) so the line passes through (2, – 3, 4) and the

(5 ± (3 ¥ 4), ± 3, – 6 ± (– 7, – 3, – 15). 43. Direction ratios of the line are 1, – 1, 1 and that 2 +1+1

q

x y z + + –8 8/3 8/6

line at a distance 3 26

so sin q =

=

vector, 50. Let P(a, b, c a, b, c and as it passes through (a, b, c is a(x – a) + b(y – b) + c(z – c) = 0 a(x + 2) + b(y + 2) + c(z – 2) = 0. As it passes through (1, 1, 1) and (1, –1, 2),

z+6

n n

.

5

y

47. Let the plane be r. ( xi + y j + z k ) = d. x + y – 2z = 2 x – y + z = x + 2y + z = d. x y z = = fi and d = 14. 9 3 –1

– ls – q – ls – q = p lr

fi

p = lr fi l = p/r.

57. If l, m, n are the direction ratios of the line then 3l – 6m – 2n = 0 and 2l + m – 2n = 0 l m n = = fi 14 2 15

or r. (i + 35 j - 10k ) – 256 = 0. 46. b ¥ 1 + 2 ¥ 3 + 3 ¥ (– 1) – 4 = 0 fib=1 2b + 2a fi a = – 11/2.

x -a y - b z -g = = 14 2 15 where (a, b, g ) lies on the line. only (a, b, g )

Three Dimensional Geometry 21.43

58. Let P(3r1 + 5, – r1 + 7, r1 – 2) and Q(– 3r2 – 3, 2r2 + 3, 4r2 + 6), direction ratios of PQ are 2, 7, – 5 3r1 + 3r2 + 8 -r1 - 2r2 + 4 r1 - 4r2 - 8 = = 2 7 -5 fi r1 = r2 = – 1 P(2, 8, – 3), Q(0, 1, 2) fi PQ = 78 . m : n, then x of division is zero -2 m + 3n m 3 fi =0fi = . m+n n 2 60. Direction ratios of the lines are 3, –2, 0 and 3 1, 3/2, 2; 3 ¥ 1 – 2 ¥ +0¥2=0 2 61. Coordinate of A are (3a, 0, 0) B(0, 3b, 0) and C(0, 0, 3c) so the centroid of the triangle ABC is (a, b, c) = (2, 4. 8) 62. Plane passes through the line x + 3y + z – 4 = 2x – y = 0 So

or

x y z-4 = = -7 1 2

is true. In s – (x + 1) – 7(y + 2) + 5(z + 1) = 0 whose distance 13 5 3 lel to the vector i – j + 4k plane is parallel to i + 5j + k. Since 1 – 5 + 4 = 0, the two vectors are perpendicular so the line 2 ¥1+ 2 ¥ 5 + 3 ¥1- 5 1 + 25 + 1

=

10 3 3

and

1, 2, 3 and of the plane are 2, 3, 1 so their direc p /4, cos p/4 ,cos p /2 i.e. 1

2, 1

2

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

2 or x + y = 2, state

2 -1 2 - 0 2 +1 1 1 1 = 0 fi L1 and L2 are coplanar 0 1 2

x y z - + =1 28 21 12

L1 and L2 intersect at (1, 1, 1) so the shortest distance

64. As the triangle ABC is (2/3, 2/3, 2/3), the centroid of the triangle and hence the radius = = 2/3

(1 - 2 / 3)2 + (1 - 2 / 3)2 + (2 / 3)2

since 2(r + 4) – 4(r + 2) + (2r + 7) = 7 for all values of r. L lies in the plane P

circle lies on the plane x + y + z – 2 = 0 through A, B, C. 65. Any point on L is P(r, 2r, 3r) (OP)2 = 14 = r 2 + 4r 2 r2 fi r2 = 1 2 36 Ê r + 2r + 3r ˆ = 12 ˜¯ = 6 1+1+1

(PN)2 = Á Ë

(ON) 2 = OP 2 – PN 2 = 2. fi Let R(r, 2r, 3r) The

r + 2r + 3r 1+1+1

70. Any point on L is (r + 4, r + 2, 2r + 7)

L is parallel to the plane P, so does not explain state

Level 2 (r1a¢ + a, r1 b¢ + b, r1 c¢ + c) and on the second line is

= 3 fi r = ± 1/2

66. L1 and L2 are parallel to the vectors a = 3i + j +2k and b = i + 2j + 3k respectively. The vector perpendicular to both L1 and L2 is a × b and the – i – 7 j + 5k 1 + 49 + 25

(r2 a + a¢, r2 b + b¢, r2 c + c¢) r1 = r2 = 1 A(3, – 2, 1) be B (a, b, g ) then AB is perpendicular to the given plane and the AB lies on the plane. So 3a – b + 4g + 11 = 0

21.44

Complete Mathematics—JEE Main

and

C¢

D¢

a -3 b+2 g -1 = = = r (say) -1 3 4

C

D

Solving we get r (0, –1, –3)

A¢

B¢

r = – ai + l (12 i + 6 j - k ) A

and r = (– 2a j + ak) + t(6i + 6 j + k)

B

Fig. 21.6

(a i - 2 a j + a k ) ◊ (12 i + 6 j - k ) ¥ (6 i + 6 j + k ) (12 i + 6 j - k ) ¥ (6 i + 6 j + k ) (using the shortest distance between the lines r = a + t b and r = c + l d is

(c - a ) ◊ b ¥ d b¥d

be A(0, 0, 0), B(1, 0, 0) C(1, 1, 0), D(0, 1, 0) A¢(0, 0, 1), B¢(1, 0, 1) C¢(1, 1, 1), D¢(0, 1, 1) Direction ratios of the diagonals AC¢ and BD¢ are 1, 1, 1 and –1, 1, 1

)

a -2a a = 12 6 -1 = 12a – 18 (–2a) + 36a = 84a 6 6 1

The angle q between these diagonals is given by cosq =

1 1 (- 1) + (1) (1) + 1 (1) = 3 1+1+1 1+1+1

fi q = cos–1 (1/3) i j k and 12 6 -1 = |12i – 18j+ 36k| 6 6 1

76. Let the vertices of the triangle be Ai (xi, yi, zi), i = 1, 2, 3 x + x3 x + x1 x1 + x2 = 2, 2 = 0, 3 =5 2 2 2 fi x1 + x2 + x3 = 7 then

= 6 4 + 9 + 36 = 42 a. 2

2

2

2

2

fi x1 = 7, x2 = – 3, x3 = 3 2

74. (PM) = (OP) – (OM) = u + v + w – 3a

2

y1 = 2, y2 = 4, y3 = 12

P (u, v, w)

z1 = 5, z2 = –7, z3 = 17

32 + 122 + 172 = 442.

O P (0, 0, 0)

M (a, a, a)

x=y=z

Fig. 21.5

77. Let the direction ratios of the line be l, m, n. The direction ratios of the four diagonals are 1, ± 1, ± 1. So cos2a + cos2b + cos2g + cos2 d

Also PM is perpendicular to the line x = y = z so u – a + v – a + w – a = 0 fi u + v + w = 3a. 4 Now u2 + v2 + w2 –3a 2 = [u + v + w – 1]2 3 fi 3(u2 + v2 + w2) – (u + v + w)2 2

= 4[(u + v + w) –2 (u + v + w) + 1] fi 2(u2 + v2 + w2) – 8(u + v + w) + 4 = 0 fi u2 + v2 + w2 – 4(u + v + w) = –2.

=

=

1 [(l + m + n)2 + (– l + m + n)2 3 (l + m 2 + n 2 ) + (l – m + n)2 + (l + m – n)2] 2

4 (l 2 + m 2 + n 2 ) 4 = . 2 2 2 3 3 (l + m + n )

78. Vertices of the tetrahedron are (0, 0, 0), (20, 0, 0), (0, 15, 0) and (0, 0, 12)

side 1 unit

Three Dimensional Geometry 21.45

0 0 0 1 20 0 0 6 0 15 0 0 0 12

1 1 1 1

84. (See Text) The line of intersection of the given planes is = 600

- 30 + 15 30 + 15 yz 3 + 12 3 + 12 = = = t (say) 3 + 12 -4+6 12 + 2

x-

cubic units.

fi x = 14t + 3, y = 2t –1, z = 15t

x y z + + a b r axes at P(a, 0, 0,), Q(0, b, 0), r(0, 0, g )

is

planes 8x – y – 7z = 8 and x + y + z – 1 = 0 is x + y + z – 1 + l (8x – y – 7z – 8) = 0

P, Q, R ABC. Centroid of the DABC is the centroid of the DPQR (a / 3, b / 3, g / 3)

If this plane is perpendicular to the plane 5x – 4y – z = 5, then 5(1 + 8l) + (1 – l) (– 4) – (1 – 7l) = 0 fi l = 0

P be x

(x, y, z), then the distances of P = ± 1, y = ± 1, z = ± 1 are

x+y+z–1=0 planes x + y + z – 1 = 0 and 5x – 4y – z – 5 = 0

1 + x, 1 – x, 1 + y, 1 – y, 1 + z, 1 – z fi 2(1 + x2 + 1 + y2 + 1 + z2) = 10

x -1 y-0 z-0 = = l m n Then l + m + n = 0

fi x2 + y2 + z2 = 2.

5l – 4m – n = 0 l m n fi = = -1+ 4 5 +1 -4-5

(5x – 4y – 2z + 1) + l(2x – y + 3z + 5) = 0 Since this is perpendicular to the plane 2x – y + 3z + 5 = 0 2(5 + 2l) + (4 + l) –3(2 – 3l) = 0

x -1 y-0 z-0 = = 1 2 -3

fil 27x – 24y – 26z – 13 = 0

Previous Years' AIEEE/JEE Main Questions lx + my + nz = 1, l, m, n it passes through (1, 0, 0) and (0, 1, 0), l = m = 1.

l (x –1) + m ( y + 2) + n(z – 1) = 0 so 2l – 2m + n = 0, l – m + 2n = 0 fi

n l m = = 0 -3 -3

p/4 with the plane x + y

is =3

x+y distance is

1 + 2 +1 1+1

fi

= 2 2

so

1 1 1 + 2 + 2 2 a b c

1 1 1 =1fi 2 + 2 + 2 =1 a b c

Centroid of the DABC = (a/3, b/3, c/3) = (x, y, z) fi so x–2 + y–2 + z–2 = k fi k

2

l +m +n x y z + + =1 a b c

-1

l+m 2

2

= cos 2

1 p = 2 4

fi l2 + m2 + n2 = (l + m)2 fi n2 = 2 fi n = ±

2 2.

2. Planes given in (a) and (c) do not pass through (3, 2, 0) and the plane in (b) does not contain the line with direction ratios 1, 5, 4.

Complete Mathematics—JEE Main

21.46

4. Centre of the sphere is (–2, 1, 3) and radius is 13.

It passes through A, B, C. So l ¢ + 2m ¢ + n ¢ = p ¢

-24 + 4 + 9 - 327 144 + 16 + 9

2l ¢ + m ¢ + 3n ¢ = p ¢

= 26.

– l ¢ + m ¢ + 2n ¢ = p ¢. Solving we get,

plane is 26 – 13 = 13.

l¢ m¢ n¢ = = 1 -5 -3

x-b y z-d x - b¢ y z - d¢ = = and = = a 1 c a¢ 1 c¢

If q

Which are perpendicular if

then cosq =

ll ¢ + mm¢ + nn¢ l 2 + m 2 + n 2 l ¢ 2 + m¢ 2 + n¢ 2

aa ¢ + 1 + cc ¢ = 0 6. Lines are coplanar if

=

2 -1 3 - 4 4 - 5 1 1 -k = 0 k 2 1

5+5+9 2

2

19 35

=

2

5 +1 + 3

cosq, cosb, cosq.

fi k2 + 3k = 0

\ cos2q + cos2b + cos2q = 1

fi k = 0 or –3

fi 2 cos2q + 1 – sin2b = 1

7. Centre of the sphere is (–1, 1, 2) and radius = 5

fi 2 cos2q – 3 sin2q = 0 fi 5 cos2q = 3 fi cos2q = 3/5

-1 + 2 + 4 + 7 1+ 4 + 4

10. Distance between two parallel planes

= 4.

ax + by + cz + d1 = 0 and ax + by + cz + d 2 = 0 |d1 - d 2 |

is

(1, -1, -2) C

a 2 + b2 + c2

Thus, distance between 2x + y + 2z – 8 = 0 and 2x + y + 2z + 5/2 = 0 is |- 8 - 5 / 2|

Fig. 21.7

2

2

2 +1 + 2 2

5 -4

2

2

=

7 2

=3

8. Let l, m, n be the d. r. of the face OAB tion is lx + my + nz = p.

11. Coordinate of point A on L1 are given by (r1, r1 – a, r1) and B on L2 is (2r2 – a, r2, r2). Direction ratio of AB are 2r2 – a – r1, r2 – r1 + a, r2 – r1 L1

Since it passes through O (0, 0, 0), p = 0 and A(1, 2, 1) and B (2, 1, 3) lie on it. So l + 2m + n = 0

L2

DR 2, 1, 2 A

B

2l + m + 3n = 0 fi

l m n = = 5 -1 -3

Next, let l ¢, m ¢, n ¢, be the d. r. of the face ABC and l ¢x + m ¢y, + n ¢z = p ¢

x+a =y=z 2

x=y+a=z

Fig. 21.8

Three Dimensional Geometry 21.47

Thus,

fi 5 + l = 4l fi l = 5/3.

2r2 - r1 - a r -r +a r -r = 2 1 = 2 1 2 1 2 \ r2 = a and r1 = 3a.

the plane

Hence, coordinate of the point of intersection are (3a, 2a, 3a) and (a, a, a).

=

12. As the straight lines =

x -1 y+3 z -1 = = 1 -l l

1 + 25 + 1 |2 - 10 + 3 - 51| 27

=

10 3 3

17. Centres of two spheres are C1 (–3, 4, 1) and C2(5, C1C2 is M(1, 1, 1).

x-0 y -1 z-2 and = = 1/ 2 1 -1 are coplanar

As M lies on 2ax – 3ay + 4az + 6 = 0, we get 2a – 3a + 4a + 6 = 0 or a = –2. 18. Centre of sphere is C (1/2, 0, –1/2) and radius is given by

1 - 0 -3 - 1 1 - 2 1 -l l =0 1/ 2 1 -1

1 1 +0+ +2 = 4 4

1 -5 -1 fi 1 0 l =0 1/ 2 0 -1 fi 5

|(2iˆ - 2 ˆj + 3kˆ) ◊ (iˆ + 5 ˆj + kˆ) - 5|

5 2

C

1 l l = 0 fi –1 – =0 1/ 2 -1 2

M P

fi l = –2

Fig. 21.9

CM = perpendicular distance of the plane x + 2y – z C(1/2, 0, –1/2)

S1 – S2 = 0 fi –10x + 5y + 5z = –5 or 2x – y – z = 1

1 1 + 2(0) + - 4 2 2 = = 1+ 4 +1

3 6

=

3 2

14. The lines can be written as x y z = = 3 2 -6 and x y z = = 2 -12 -3 Since (3) (2) + (2) (–12) + (–6) (–3) = 0,

5 3 - =1 2 2 fi radius of circle = 1.

15. We have

A(–1, 3, 4) and per pendicular to x – 2y = 0, that is on the line

sin q =

a1a2 + b1b2 + c1c2 a12

+ b12

+ c12

a22

+ b22

+ c22

(1)(2) + (2)(-1) + (2)( l ) 1 fi = 1+ 4 + 4 4 +1+ l 3

fi

5+l = 2 l

Now, MP2 = CP2 – CM2 =

x +1 y-3 z-4 = = = t(say) 1 -2 0 fi x = t – 1, y = 3 – 2t, z = 4.

– 2t

B(t – 1, 3 Ê1 ˆ M of AB is Á t - 1, 3 - t , 4˜ and Ë2 ¯ x – 2y = 0.

Complete Mathematics—JEE Main

21.48

fi

x-5 y -1 z-a = = 5-3 1- b a -1

1 14 t – 1 – 6 + 2t = 0 fi t = 2 5 Ê 9 13 ˆ ÁË , - , 4˜¯ 5 5

17 -13 ˆ which passes through ÊÁ 0, , Ë 2 2 ˜¯ x + 2y

21. Let l, m, n be the direction cosines of the line, then 2l + 3m + n = 0

l m n = = 6-3 1- 4 6-3

fi l = –m = n fi l : m : n = \ cosa = 22. l =

1 2

13 -a 2 a -1

fi b = 4 and a = 6.

kr + 1 = 3r ¢ + 2 1 3

:-

1 3

:

1 3

2r + 2 = kr ¢ + 3 3r + 3 = 2r ¢ + 1

1

r and r1 we get

3

,m=

-

26. Any point on the two lines are respectively (kr + 1, 2r + 2, 3r + 3), (3r ¢ + 2, kr ¢ + 3, 2r ¢ + 1)

and l + 3m + 2n = 0 fi

-5 15 / 2 So = = 2 1- b

(k + 5) r = (k + 5) r ¢

1

fi k = –5, is the integral value of k.

2

l2 + m2 + n2 = 1 fi n = 0 fi g =

p 2

angle)

x-2 y -1 z+2 = = lies in the plane x + 3y 3 -5 2 – a z + b = 0, the point (2, 1, –2) lies in the plane

27. As

23. Centre of sphere is (3, 6, 1) Let coordinates of other end point be (a, b, c), then a+2 b+3 c+5 = 3, = 6, =1 2 2 2 fi a = 4, b

c = –3

Thus, 2 + 3(1) – a(–2) + b = 0 and (3) (1) + (3) (–5) + (–a) (2) = 0 fi 2a + b = –5, 2a = –12 fi a = –6, b = 7

   24. As a , b , c are coplanar 1 1 x

1 1 -1 2 =0 x - 2 -1

1 0 0 fi 1 -2 1 =0 x -2 -1 - x fi

-2 1 =0 -2 -1 - x

fi x = –2

28. Here a = 6, b = – 3, c = 2 \ direction cosines are given by ±

a

,±

a 2 + b2 + c2

b a 2 + b2 + c2

,±

c a 2 + b2 + c2

6 3 2 = ± ,∓ ,± 7 7 7 a

b

g=q

Since cos2 a + cos2b + cos2g = 1 fi

2

2

+ cos2q = 1

Three Dimensional Geometry 21.49 2

Ê 1 ˆ Ê 1ˆ + Á- ˜ fi Á ˜ Ë 2¯ Ë 2¯

2

+ cos2q = 1 fi cos2q = 1 -

3 = 4

1 4 1 2

fi cosq =

[∵q is acute]

fiq

x = y = z is x -1 y+5 z-9 = = = r(say) 1 1 1 fi x = r + 1, y = r – 5, z = r The point (r + 1, r – 5, r = 5 if

x–y+z fi r = –10

(r + 1) – (r – 5) + (r

30. Direction ratios of AB are

\

1 – 3, 3 – 1, 4 – 6 or 1, –1, 1 fi AB

x – y + z = 5.

M (2, 2, 5) of AB lies on x – y + z = 5 since 2 – 2 + 5 = 5 Thus, the plane x – y + z AB.

(-9 - 1)2 + (-15 + 5) 2 + (9 + 1) 2 =

300

= 10 3

As x – y + z = 5 is the perpendicular bisector of the AB A B(1, 3, 4) in the plane x – y + z = 5. AB is

34. Any point on x y-2 z-3 = = 2 3 4 is given by P(2r, 3r + 2, 4r + 3)

(1)

A (3, -1, 11)

Ê 1 + 1 0 + 6 7 + 3ˆ , , ÁË ˜ = (1, 3, 5). 2 ¯ 2 2 x y -1 z-2 = = 1 2 3

It lies on

P (2r, 3r + 2, 4r + 3)

Direction ratios of AB are

Fig. 21.10

1 – 1, 6 – 0, 3 – 7. or 0, 6, – 4, As (1) (0) + (2) (6) + (3) (–4) = 0,

Direction ratios of AP are

we get AB is perpendicular to the straight line

2r – 3, 3r + 2 + 1, 4r + 3 – 11 or 2r – 3, 3r + 3, 4r – 8

x y -1 z-2 = = 1 2 3

AP will be perpendicular to (1) if 2(2r – 3) + 3(3r + 3) + 4(4r – 8) = 0 fi

32. If q is angle between the line and the plane, then

fir=1

r

\ Coordinates of P are (2, 5, 7). Thus, AP2 = (2 – 3)2 + (5 + 1)2 + (7 – 11)2

sin q =

(1)(1) + (2)(2) + 3(l ) 1+ 4 + l

As cosq = Thus,

3 14

2

1+ 4 + 9

=

5 , we get sinq = 14 =

5 + 3l 14 5 + l 2

5 + 3l 14 5 + l

= 1 + 36 + 16 = 53 2

fi AP =

53

3

x – 2y + 2z + k = 0.

14

0 - 2(0) + 2(0) + k

fi 3 5 + l 2 = 5 + 3l

25 10 20 10 = fi5+l = + l + l2 fi l 9 3 9 3 fi l = 2/3.

1+ 4 + 4 = 1 fi k = ±3.

2

x – 2y + 2z – 3 = 0.

21.50

Complete Mathematics—JEE Main

36. As the given lines intersect, the lines are coplanar. Thus 3 - 1 k - (-1) 0 - 1 2 k + 1 -1 2 3 4 =0fi 2 3 4 =0 1 2 1 1 2 1

Ê3 ˆ (QR)2 = (–2 + 1)2 + Á - 1˜ Ë2 ¯

fi –10 + 2(k + 1) – 1 = 0 fi k

fi n2 – 2lm = n2 fi lm = 0

So the direction ratios are 0, 1, –1 and 1, 0, –1 and the –1 0 ¥ 1 + 1 ¥ 0 + ( -1)( -1) 2 2

Using C3 Æ C3 – C2 and C2 Æ C2 + C1, we get = cos–1

0 -k - 1 = 0 -1

1 2

x - ( l - 1)

fi –2 + (k + 1) (k + 2) = 0.

l

fi k = 0, –3. 38. One of the points on 2x + y + 2z = 8 is (4, 0, 0). It |4(4) + 5| x + 2y + 4z + 5 = 0 is 16 + 4 + 16 21 7 = = 6 2 Alternately 4x + 2y + 4z = 16 and 4x + 2y + 4z + 5 = 0 is

x - ( m - 1) m

=

y z- l = 1 l -1

=

z- m y = 1 - m +1

If they are perpendicular, then l m + 1 + ( l - 1)(1 - m ) = 0 l + m = 0 fi l = m.

fi

22 + 32 + 62 = 7

21 7 = = 6 2

Ê l - 1 m + 2ˆ , 4, 44. Coordinates of D are Á ˜ Ë 2 2 ¯ D.R. of AD are

-2 + 1 k - 1 0 + 1 11 2 1 3 =0fik= 2 2 3 4

=

40. Let the coordinates Q be (a, b, g ), O the origin. OQ is perpendicular to the plane

l -1 m+2 –5 - 2 , 4 – 3, 2 2

l -5 m -8 , 1, 2 2

AD So d.r. of AD are 1, 1, 1

a b g fi = = =k 4 -3 1 Q lies on the plane

3 1ˆ Ê Q(a, b, g ) = Q Á -2, , - ˜ Ë 2 2¯

63 2

If l = 0, m = –n and if m = 0, l = –n.

1- 2 4 - 3 5 - 4 1 1 -k = 0 k 2 1

fi 4 ¥ 4a –3(–3)k + k + 13 = 0 fi k = –

=

Either l = 0 or m = 0

37. The two lines will be coplanar if

42 + 22 + 42

2

41. l2 + m2 = n2 fi (l + m)2 – 2lm = n2

16 + 5

Ê 1 ˆ + Á - + 6˜ Ë 2 ¯

fi QR = 3 7 / 2

fi 2(3 – 8) – (k + 1) (2 – 4) – (4 – 3) = 0

-1 0 1 2 k 2+k

2

fi

1 2

l -5 m -8 = 1, = 1 fi (l, m) = (7, 10) 2 2

given planes is x + 2y – 3 + l(y – 2z + 1) = 0 fi x + (2 + l) y – 2lz + (l – 3) = 0

Three Dimensional Geometry 21.51

5 1 + 2(2 + l) – 2l(0)= 0 fi l = - . 2

f with the z Direction cosines of the line are cosq, cosq, cosf. We have

2x – y + 10z = 11

cos2q + cos2q + cos2f = 1 fi 2 cos2q = sin2f £ 1

46. Any point on the given line is P(3k + 1, k + 3, –5k + 4)

fi 0 £ cosq £ 1/ 2 fi q Œ[p/4, p/2]

Q in the plane 2x – y + z + 3 = 0 is on the line through this point, perpendicular to the plane, so is given by x - (3k + 1) y - (k + 3) z - (-5k + 4) = = 2 1 -1 -2(6k + 2 - k - 3 - 5k + 4 + 3) = –2 6 fi x = 3k –3, y = k + 5, z = 2 – 5k =

Ê l - 1 m + 2ˆ , 4, 50. Mid point of BC is M ÁË ˜ 2 2 ¯ AM are l -5 m -8 ,1, 2 2 AM l -5 m -8 =1= fi l = 7, m = 10 2 2 \ 10l – 7m = 0

x+3 y-5 z-2 = = 3 1 -5

47. See Question No. 41.

x -1 y-2 z-3 = = is 1 2 3 a(x – 1) + b(y – 2) + c(z – 3) = 0

(1)

48. A point on

where a + 2b + 3c = 0

(2)

Q to double the distance of point P

x -1 y-2 z-3 = = 3 1 2 is (3r + 1, r + 2, 2r + 3).

As

x y z = = is parallel to (1), we get 1 1 4

a + b + 4c = 0

This point will lie on a =b=c -5

x-3 y -1 z-2 = = . 1 2 3 We get 3r + 1 - 3 r + 2 -1 2r + 3 - 2 = = 2 1 3 r +1 2r + 1 fi 3r – 2 = = 3 2 fir=1 Thus, point of intersection of two lies is (4, 3, 5). This point lies on 7x + 2y + 4z – 54 = 0, 3x + 4y + 5z 4x + 3y + 5z – 50 = 0

p1 =

54

49

50

, p2 = and p3 = 5 2 5 2 69 Note that p1 < 7, p2, p3 > 7 and p2 < p3.

Thus, 4x + 3y + 5z

–5(x – 1) + (y – 2) + (z – 3) = 0 or –5x + y + z = 0 It clearly passes through (1, 0, 5) 52. We can write the line as x-b z-d =y= a c or

x-b z-d -1 –1=y–1= a c

or

x-b-a y -1 z-d -c = = a 1 c

(3)

Complete Mathematics—JEE Main

21.52

53. Distance between planes =

|d - 1| 16 + 4 + 16

56. A point of

=7

x-2 y +1 z-2 = = 3 4 12

(3r + 2, 4r – 1, 12r + 2) will lie on x – y + z = 16 if (3r + 2) – (4r – 1) + (12r + 2) = 16

fi |d – 1| = 42 fi d = – 41, 43

fir=1 at A and B.

Thus, point of intersection is (5, 3, 14) and its distance 42 + 32 + 122 = 13.

x −1 x +1 z = = −2 1 0

57. A plane passing through the intersection of 2x – 5y + z = 3 and x + y + 4z = 5 is

B

(2x – 5y + z – 3) + k(x + y + 4z – 5) = 0

A

or (2 + k)x + (–5 + k)y + (1 + 4k)z – 3 – 5k = 0

x y z = = 1 −1 1

It will be parallel to x + 3y + 6z = 1 if

Fig. 21.11

Let coordinates of A be (a, –a, a) and that of B be (1, – 2b – 1, b)

2+k -5 + k 1 + 4k = = 6 1 3 fi k = –11/2

Direction ratios of AB are 1 – a, –2b + a – 1, b – a 7 21 42 49 - x=0 yz+ 2 2 2 2

As AB is perpendicular to both the lines 1(1 – a) + (–1) (–2b + a – 1) + 1(b – a) = 0

or x + 3y + 6z – 7 = 0

and 0(1 – a) + (–2) (–2b + a – 1) + 1(b – a) = 0 fi 3a – 3b = 2 and 3a – 5b = 2

58. Let direction ratios of the line x + y + 2z – 3 = 0 = 2x + 3y + 4z – 4

2 Solving we get a = , b = 0. 3 \ Direction ratio of AB are

be a, b, c. Then a + b + 2c = 0 and 2a + 3b + 4c = 0

1 1 2 , - , - or 1, –1, –2 3 3 3

a b c = = 2 0 -1 To obtain a point on the line put z = 0, so that fi

x -1 y +1 z = = 1 -1 -2

x + y = 3 and 2x + 3y = 4 fi x = 5, y = –2.

x +1 y z+4 = = 1 2 2 As angle between (1) and 2x – y +

p , we get l z + 4 = 0 is 6

Êpˆ | (2)(1) + (-1)(2) + 2 l | sin Á ˜ = Ë 6¯ 1+ 4 + 4 4 +1+ l fi

l) = 4(2 – 2 + 2 l )2

45 = 7l fi l =

45 7

(1)

Thus line (1) is given by r = 5i – 2j + l (2i – k) and z r = mk Here d1 ¥ d2 = (2i – k) ¥ k = –2j \ s.d. = =

4 = 2. 2

|(5i - 2 j) ◊ (d1 ¥ d 2 )| | d1 ¥ d 2 |

(1)

Three Dimensional Geometry 21.53

fi (36l2 + 12l + 1) (3) = 6l2 + 4l + 3 x -1 y-2 z-3 = = 1 5 4

l = 0, –16/51 \a

is a(x – 1) + b(y – 2) + c(z – 3) = 0

A

where

to x = y = z is

a + 5b + 4c = 0.

(1)

It will pass through (3, 2, 0) if

x–1=y+5=z

r (say)

A point on this line is P(r + 1, r – 5, r

2a – 3c = 0.

(2)

This point will lie on x – y + z = 5. If (r + 1) – (r – 5) + (r

a b c = = 15 10 -11

fi r = –10. Distance between A and P is = AP =

15(x – 1) – 11(y – 2) + 10(z – 3) = 0

63. As the line

which passes through (0, 7, 10) 60.

| 3 + 4 - 12l + 13 | 32 + 42 + 122

=

x-3 y+2 z+4 = = 2 -1 3

| 3(-3) + 4(0) - 12 + 13 | 32 + 42 + 122

lies on lx + my + z

2

fi |20 – 12l| = 8 fi (3l – 5) = 4

2l + (–1)m – 3z = 0

2

fi l – 30l + 21 = 0

and 3l + (–2)m

2

fi 3l – 10l + 7 = 0

fi 2l – m = 3, 3l – 2m = 5 2

\l

3 |r| = 3 10

x – 10x + 7 = 0

fi l = 1, m = –1

61. Shortest distance between two lines L1 and L2 to distance between L1 and a plane through L2 and parallel to L1.

fi l2 + m2 = 2 64. Here A1 ∫ (0, 0, 0), A2 ∫ (–2, 4, 5)   d1 = 2iˆ + 2 ˆj + kˆ, d 2 = -iˆ + 8 ˆj + 4kˆ

x + y + z + 1 = 0 and 2x – y + z + 3 = 0 is

We have

x + y + z + 1 + l(2x – y + z + 3) = 0

iˆ ˆj kˆ   d1 ¥ d 2 = 2 2 1 -1 8 4

or (1 + 2l)x + (1 – l)y + (1 + l)z + (1 + 3l) = 0 (1) This will be parallel to x -1 y +1 z = = a -1 1

(2)

if (1 + 2l)a + (1 – l) (–1) + (1 + l) (1) = 0 fia= -

2l 1 + 2l

| (1 + 2l )(1) + (1 - l )(-1) + (1 + 3l ) | 2

2

(1 + 2l ) + (1 - l ) + (1 + l ) 2

2

=

1 3

= 0iˆ + (-9) ˆj + 18kˆ  and A1 A2 = -2iˆ + 4 ˆj + 5kˆ    Now A1 A2 ◊ (d1 ¥ d 2 )   and |d1 ¥ d 2 | = 92 + 182 = 9 5 \ shortest distance =

54 9 5

=

6 5

= 1.2 5

2

fi (6l + 1) (3) = 6l + 4l + 3 a(x – 1) + b(y – 2) + c(z – 2) = 0

(1)

Complete Mathematics—JEE Main

21.54

This plane will be perpendicular to the x – y + 2z = 3 and 2x – 2y + z + 12 = 0 if

Previous Years' B-Architecture Entrance Examination Questions

a – b + 2c = 0 centre of the sphere we get

2a – 2b + c = 0

0 -1+ c + 3 =cfic=1 3 2. Let (a, b, g the origin on the place through (p, q, r)

a = b, c = 0

x–1+y–2=0 or x + y – 3 = 0

=

|1 - 2 - 3| 1+1

(2)

O (0, 0, 0)

= 2 2 P (a, b, g)

Ê l -1 2 + mˆ , 4, D of BC are Á ˜ Ë 2 2 ¯

Fig. 21.13

Direction ratios of AD are A (2, 3, 5)

Then a, b, g a (x – p) + b (y – q) + g (z – r) = 0 Since the plane passes through (a, b, g) a (a – p) + b (b – q) + g (g – r) = 0

B (-1, 3, 2)

D

C (l, 5, m)

Fig. 21.12

fi a2 + b2 + g2 – pa – qb – rg = 0 fi Locus of (a, b, g) is a sphere. 3. OR = a/2, OP = a PR =

l -1 2+m – 2, 4 – 3, –5 2 2

27

a2 4 2 fi a = 36 fi a = 6

So a2 = 27 +

or l – 5, 2, m – 8 As AD l–5=2=m–8 fi l = 7, m = 10 3

O

3

3

3

Thus, l + m + 5 = 7 + 10 + 5 = 1348

a R

67. As the given lines are coplanar 3 -1 2 - 2 1+ 3 D=

1

2

l2

1

l2

2

Fig. 21.14

=0

fi 2[4 – l4) + 4 (l2 – 2) = 0 2

2

fi 2(l – 2) (– l ) = 0 fil=±

P

2, 0

Thus, there three such values.

Let R be the radius of the section of the sphere by the plane z = a/6. Ê aˆ Then R2 = a2 – Á ˜ Ë 6¯

2

fiR=

35 .

x-2 y +1 z-2 = = is (3r + 2, 3 4 12 4r – 1, 12r + 2) which lies on the plane 3r + 2 – 4r + 1 + 12r + 2 = 5

4. Any point on

Three Dimensional Geometry 21.55

fi r = 0 and the point of intersection is (2, –1, 2)

11. Any point on the given line L is (r + 4, r + 2, 2r + 7) Since 2(r + 4) – 4(r + 2) + 2r + 7 = 7 for all r.

2

2

(2 + 1) + (-1 + 5) + (2 + 10)

2

= 13

The line L lies in the plane fi

5. Direction ratios of L2 are 2, 0, –2 (-2)(2) + (-1)(0) + (-2)(2)

So cosq =

4 +1+ 4 4 + 4 +1

=

- 8 3

in a plane parallel to the plane P. So it is not a cor x -1 y-2 z +1 = = 2 1 -1

1 . 3

fi sinq =

6. (c) represents a pair of parallel planes, so does not represent a line.

Let P(a, b, g

2 -1 2 - 0 2 +1 1 1 = 0 fi lines are coplanar 7. 1 0 1 2

fi 2a + b – g – 5 = 0

Also the lines intersect at the point (1, 1, 1).

then 2(a – 0) + 1(b – 3) – 1(g + 2) = 0

1

fi cosq =

2

3 - (cos a + cos 2 b + cos 2 g )

PQ lies of the line) fi

=

1 2

(1)

a b +3 g -2 -1 -2 +1 and 2 = 2 = 2 -1 2 1

So the shortest distance between the lines is zero coplanarity only fi shortest distance is zero. 8. We have cos2a + cos2b + cos2g = 1

Q(0, 3, –2)

a -2 b -1 g = = = r (say) 2 1 -1

(2)

2(2r + 2) + r + 1 + r – 5 = 0 fi r = 0

fi q = p/3

x y z a a 3 + + = 1, then = fia= a a b g 3 2 2 b= is

3 3 b, g = c 2 2

x y z 3 + + = . a b c 2

   r = (2iˆ + 5 ˆj - kˆ) + t (3iˆ + 2 ˆj - 5kˆ) = a + tb x y z = = 0 0 1   fi r = p kˆ = c + p d     (a - c) ◊ (b ¥ d )   Shortest distance = |b¥d |

x y z + + =1 a b c Since it passes through (1, –2, 3) 1 2 3 - + =1 a b c

Let A (a, 0, 0), plane through A parallel to x x = a and through B and C parallel to the axes are y = b and z = c. Point of intersection of these planes is (a, b, c) bc – 2ac + 3ab = abc.

z

()

=

=

(2iˆ + 5 ˆj - kˆ) ◊ (2iˆ - 3 ˆj )

(1)

fi (a, b, c) lies on yz – 2zx + 3xy = xyz. x y z x y z = = and = = 12 3 2 -6 -2 3 Since 3(–2) + 2(12) + (–6)(3) = 0

14. Given lines are

The lines are perpendicular.

4+9 4 - 15 13

=

11 13

.

4-0 2+2 l -0 1 3 fi 1 = 0 fi l = 12 1 2 4

Complete Mathematics—JEE Main

21.56

16. A1(x, 0, 0), A2(0, x2, 0), A3(0, 0, x3) x y z + + =1 x1 x2 x3

fi 2A2 =

1 (121 + 676 + 1) 2 Q (0, 2, 3)

(1)

Plane through A1, A2, A3 parallel to coordinate axes intersect at the point (x1, x2, x3 the origin is p.

fi

P (2, 3, -1)

1

fi

1 1 1 + 2+ 2 2 x1 x2 x3 1 x12

+

1 x22

R (-3, 1, -4)

+

1 x32

=±p Fig. 21.15

1 =

p2

=

18. As the lines are coplanar, 2 - 4 8 - 6 10 - 8

17. Let coordinate of P be (5l – 3, 2l + 1, 3l – 4)

1 2

fi

Direction ratios of QP are

2

2

2k 2

4

5l – 3 – 0, 2l + 1 – 2, 3l – 4 – 3 or 5l – 3, 2l – 1, 3l – 7 As PQ is perpendicular to x+3 y -1 z+4 = = 5 2 3

fi

2

2

2

2

2 -2k 2 = 0

2k 2 2

-2k 2 = 0

4

2

2

2

fi k (k + 3) = 0 But k π 0, therefore, k2(k2 + 3) π 0.

5(5l – 3) + 2(2l – 1) + 3(3l – 7) = 0 fi 38l – 38 = 0 fi l = 1.

k for which the given lines are coplanar.

fi coordinate of P are (2, 3, –1) Area of DPQR is given by 1   A = | PQ ¥ PR | 2 ˆj kˆ iˆ 1 = -2 -1 4 2 -5 -2 -3 =

1 |11iˆ - 26 ˆj - kˆ| 2

planes is x + y + z – 1 + k(2x + 3y – z + 4) = 0 or (1 + 2k)x + (1 + 3k)y + (1 – k)z –1 + 4k = 0 k=0 This will be parallel to y or k = –1/3 1 4 7 x+ z- = 0 3 3 3 or x + 4z – 7 = 0 which passes through (3, 0, 1)

Vectors

4. (a + b) a = aa + ba 0a = 0, 1a = and (–1) a = – a C B

a

c

C

Æ

a , or in bold type, as in a. A scalar is a quantity having magnitude only but no direction, such as mass, length, time, temperature and any real number. A vector whose initial and terminal points are the same is called a zero vector, 0. Two vectors are equal if they have the same magnitude; they lie on the same line or on parallel lines and they have the same direction. Let a be any vector and a a scalar then aa is a vector whose length is equal to | a | |a|, a is a vector on the same line or on a line parallel to the line on which a lies. Moreover, direction of aa is same as that of a if a > 0 and is opposite to that of a if a < 0. If a = AB and b = BC then a + b is the vector AC. C

a+

A

b

a

b B

Fig. 22.1

The following are some elementary properties of addition and scalar multiplication: 1. a + b = b + a (addition is commutative) (Fig. 22.2) 2. (a + b) + c = a + (b + c) (addition is associative) (Fig. 22.3) 3. a (a + b) = aa + a b

b

O

a+

b

a

b

A

(a +

æÆ

symbol AB or AB and the length of the vector AB by |AB|. Graphically, a vector is represented by a directed line segment. We denote it by a letter with an arrow over it, as in

O

Fig. 22.2

a+

c b+c

b) +

A vector is a quantity having both magnitude and direction. For instance displacement, velocity, force and acceleration are vector quantities. Any portion of a straight line, where the two end-points are distinguished as initial and terminal, is called a directed line segment. The directed line segment with initial point A and terminal point B is denoted by the

b

a

B

b A

Fig. 22.3

The position vector r of any point P with respect to the origin of reference O is the vector OP. Two vectors a and b are said to be collinear if they are supported on the same or parallel lines. For such vectors, b = xa for some scalar x. A set of vectors is said to be coplanar if they lie in the same plane, or the planes in which the different vectors lie are all parallel to the same plane. Three vectors a, b and c are coplanar if and only if c = xa + yb for some scalars x and y.

LINEAR COMBINATIONS The vector r = a1 a1 + a2 a2 +  + an an, where a1, a2, , an are scalars, is called a linear combination of a1, a2, , an. The following results are useful in determining coplanar and collinear vectors: 1. If a and b are non-collinear vectors, then xa + yb = x¢a + y¢b ¤ x = x¢, y = y¢ 2. Fundamental theorem in plane. If a and b are non-collinear vectors, then any vector r, coplanar with a and b, can be expressed uniquely as a linear combination of a and b. That is, there exist unique x and y Œ R such that r = xa + yb. 3. If a, b and c are non-coplanar vectors, then xa + yb + zc = x¢a + y¢b + z¢c ¤ x = x¢, y = y¢, z = z¢ 4. Fundamental theorem in space. If a, b and c are non-coplanar vectors in space, then any vector r

22.2

Complete Mathematics—JEE Main

can be uniquely expressed as a linear combination of a, b and c. That is, there exist unique x, y, z Œ R such that r = xa + yb + zc. If i, j and k are three unit vectors along x-axis, y-axis and z-axis respectively, then any vector r can be represented uniquely as r = a1i + a2 j + a3k, where a1, a2 and a3 are the coordinates of r. 5. Section formula. The position vector of a point P which divides the line joining the points A and B with position vectors a and b respectively in the ratio m : n, is na + mb (m π – n) m+n The position vector of mid-point M of AB, is (1/2) (a + b). The point A with position vector a is written A(a). If A(a), B(b) and C(c) are the vertices of a triangle ABC, then the centroid of this triangle is (1/3) (a + b + c). 6. Test of collinearity. Three points A(a), B(b), and C(c) are collinear if and only if there exist scalars x, y and z, not all zero, such that xa + yb + zc = 0, where x + y + z = 0. 7. Test of coplanarity. Four points A(a), B(b), C(c) and D(d) are coplanar if and only if there exist scalars x, y, z and w, not all zero, such that xa + yb + zc + wd = 0, x + y + z + w = 0.

SCALAR OR DOT PRODUCT The scalar product of two vectors a and b is given by | a | | b | cos q, where q (0 £ q £ p) is the angle between the vectors a and b. It is denoted by a ◊ b. Properties of the scalar product 1. a ◊ a = | a |2 = a2. 2. a ◊ (b + c) = a ◊ b + a ◊ c. 3. Two non-zero vectors a and b make an acute angle if a ◊ b > 0, an obtuse angle if a ◊ b < 0 and are inclined at a right angle if a ◊ b = 0. 4. a ◊ b = (projection of a on b) | b |. 5. (a + b) ◊ (a – b) = a2 – b2; (a + b)2 = a2 + b2 + 2a ◊ b, (a – b)2 = a2 + b2 – 2a ◊ b. 6. If a = a1i + a2 j + a3k and b = b1i + b2 j + b3 k, then a ◊ b = a1b1 + a2b2 + a3b3, | a | = a12 + a22 + a32 and

cos q =

a1b1 + a2b2 + a3b3 a12 + a22 + a32 b12 + b22 + b32

7. Vector component of u orthogonal to a vector a is u.a u – proja u where proj a u = 2 a and proja u is a vector component of u along a.

VECTOR OR CROSS PRODUCT The vector product of two vectors a and b, denoted a ¥ b, is the vector c with | c | = | a | | b | sin q, where q is the angle between a and b, with 0 £ q £ p. c is supported by the line perpendicular to a and b, and the direction of c is such that a, b and c form a right-handed system. Properties of the vector product 1. a ¥ b = – b ¥ a. 2. a ¥ a = 0. 3. a ¥ (b + c) = a ¥ b + a ¥ c. 4. (a ¥ b)2 = a2b2 – (a ◊ b)2. 5. Two non-zero vectors a and b are collinear if and only if a ¥ b = 0. 6. If i, j and k are unit vectors along positive x-axis, y-axis and z-axis then i ¥ j = k, j ¥ k = i, k ¥ i = j. 7. If a = a1i + a2j + a3k and b = b1i + b2 j + b3k, then i j a ¥ b = a1 a2 b1 b2

k a3 b3

= (a2b3 – a3b2)i + (a3b1 – a1b3)j + (a1b2 – a2b1)k 8. The area of the parallelogram whose adjacent sides are represented by the vectors OA = a and OB = b, is | a ¥ b |, and the area of the triangle OAB is (1/2) | a ¥ b |. The vector area of the above parallelogram is a ¥ b. 9. A unit vector perpendicular to the plane of a and b is a¥b |a ¥ b| and a vector of magnitude l perpendicular to the plane of a and b is l(a ¥ b) ± | a¥b |

SCALAR TRIPLE PRODUCT The scalar triple product of the three vectors a, b and c is denoted by [a b c], and is defined as a ¥ b ◊ c. Properties of the scalar triple product 1. (a ¥ b) ◊ c = a ◊ (b ¥ c). 2. [a b c] = [b c a] = [c a b] = – [b a c] = – [c b a] = – [a c b]. 3. If a = a1i + a2j + a3k and b = b1i + b2j + b3k, then a1 a2 a3 a ¥ b ◊ c = [a b c] = b1 b2 b3 c1 c2 c3 In particular a ◊ (a ¥ b ) = 0 and b ◊ (a ¥ b ) = 0 .

Vectors

4. The volume of the parallelopiped whose adjacent sides are represented by the vectors a, b and c, is a ¥ b ◊ c. 5. The volume of a tetrahedron ABCD is equal to 1 |AB ¥ AC ◊ AD|. 6 6. Any three vectors a, b and c are coplanar if and only if [a b c] = 0. 7. [a + b c d] = [a c d] + [b c d]. 8. Three vectors a, b and c form a right-handed or left-handed system, according as [a b c] > or < 0. a ◊c b ◊c 9. (a ¥ b) ◊ (c ¥ d) = a ◊d b ◊d a ◊u b ◊u c ◊u 10 [a b c] [u v w] = a ◊ v b ◊ v c ◊ v a◊w b◊w c◊w 11. Four points with position vectors a, b, c and d will be coplanar if [d b c] + [d c a] + [d a b] = [a b c] 12. [a + b b + c c + a] = 2[a b c] 13. [b ¥ c c ¥ a a ¥ b] = 2[a b c]

VECTOR TRIPLE PRODUCT The vector triple product of three vectors a, b and c is the vector a ¥ (b ¥ c), Since a ¥ (b ¥ c) is coplanar with b and c and (a ¥ b) ¥ c is coplanar with a and b, we have a ¥ (b ¥ c) = (a ◊ c)b – (a ◊ b)c (a ◊ c)b – (b ◊ c)a

and

(a ¥ b) ¥ c =

Clearly, a ¥ (b ¥ c) π (a ¥ b) ¥ c in general. In fact, (a ¥ b) ¥ c = a ¥ (b ¥ c) if and only if the vectors a and c are collinear. If a,b,c and d lie in the same plane then (a ¥ b ) ¥ ( c ¥ d ) = 0

22.3

GEOMETRICAL AND PHYSICAL APPLICATIONS Bisector of an angle If a and b are unit vectors along the sides of an angle, then a + b and a – b are respectively the vectors along the internal and external bisectors of the angle. The bisectors of the angles between the lines r = xa and r = yb are given by Ê a bˆ r =l Á ± ˜ (l Œ R) Ë |a| |b| ¯ Equation of a line passing through a point with position vector a and parallel to a vector b is r = a + tb Reciprocal systems of vectors Let a, b and c be a system of three non-coplanar vectors. Then the system a¢, b¢ and c¢, which satisfies a ◊ a¢ = b ◊ b¢ = c ◊ c¢ = 1 and a ◊ b¢ = a ◊ c¢ = b ◊ a¢ = b ◊ c¢ = c ◊ a¢ = c ◊ b¢ = 0 is called the reciprocal system to the vectors a, b and c. Equation of a plane. The equation of a plane passing through the point with position vector a and parallel to the plane containing b and c, is r = a + lb + mc or [r – a b c] = 0. l and m being parameters. The equation of a plane through three points whose position vectors are a, b and c is r = (1 – l – m)a + lb + mc or r ◊ (b ¥ c + c ¥ a + a ¥ b) = [a b c] l and m being parameters. Equation of a plane which is at a distance d from the origin having a unit normal n is r. n = d. Equation of a plane passing through a point with position vector a having a unit normal n is (r – a) ◊ n = 0. Work done. If a force F acts at a point A and displaces it to the point B, then the work done by the force F is F ◊ AB. The moment of a force F applied at B about the point A is the vector AB ¥ F.

SOLVED EXAMPLES Concept Based Straight Objective Type Questions Example 1: If in a triangle OAC, B is the mid point of AC and OA = a, OB = b then 1 (b) OC = 2b – 2a (a) OC = (a + b) 2 (c) OC = 2b – a (d) OC = 3b – 2a Ans. (c) Solution: Let O be the origin of reference. The position vector of A is a and that B is b. If position vector of C is c, then

a+c fi c = 2b – a 2 Example 2: The angle between the vectors i – j + k and –i + j + 2k is (a) 45° (b) 60° (c) 90° (d) 135° Ans. (c) Solution: (i – j + k) · (–i + j + 2k) = –1 – 1 + 2 = 0 So the angle is 90°. b=

22.4

Complete Mathematics—JEE Main

Example 3: A unit vector c perpendicular to a = i – j and coplanar with a and b = i + k is 1 1 (i + j + 2k ) (i - j + k ) (b) (a) 6 3 1 1 (i + j - k ) ( i - j + 2k ) (d) (c) 3 6 Ans. (a) Solution: c = la + mb = l(i – j) + m(i + k) = (l + m)i – lj + mk (i) since c is a unit vector so (l + m)2 + l2 + m2 = 1 2 Also a.c. = 0, but a.c = l |a| + m a.b fi 0 = l. 2 + m .l fi m = –2l (i) fi (1 – 2)2 l2 + l2 + 4l2 = 1 6l2 = 1 1 l2 = 6 1 (i + j + 2k) fic= ± 6 Example 4: If a and b are two non-parallel vectors satisfying |a| = |b|, then the vector (a + b) × (a × b) is parallel to (a) a (b) a – b (c) a + b (d) b Ans. (b) Solution: (a + b) × (a × b) = a × (a × b) + b × (a × b) = (a · b)a – (a · a)b + (b · b)a – (b · a)b = (a · b) (a – b) + |a|2 (a – b) = (a · b + |a|2) (a – b) So (a + b) × (a × b) is parallel to a – b. Example 5: If a = i + j + k and b = i – j + 2k then the projection of a on b is given by 1 1 (b) (i + j + k ) (a) (i - j + 2k ) 2 3 1 1 (c) (i - j - k ) (d) (i - j + 2k ) 3 3 Ans. (d) Solution: Projb a =

a·b b2

b

=

2 ( i - j + 2k ) 6

=

1 ( i - j + 2k ) 3

Example 6: If a, b, c are unit vectors such that a – b + c = 0 then c · a is equal to 1 3 (a) (b) 2 2 1 1 (c) (d) 3 3 Ans. (b) Solution: Taking dot product with a, b, c in the relation a – b + c = 0, we have |a|2 – a · b + a · c = 0 a· b – |b|2 + b · c = 0 a· c – b · c + |c|2 = 0 Adding, we get 2 a· c = –(|a|2 – |b|2 + |c|2) = –1 1 a· c = 2 Example 7: The non-zero vectors a, b and c are related as b = 5a and c = –2b. The angle between a and c is p p (b) (a) 2 4 p (c) p (d) 3 Ans. (c) Solution: cosine of angle between a and c is 1 b ◊ -2b a◊c 5 = = a c 1 b 2b 5 = –1 Hence the angle is p. Example 8: A vector b collinear with a = 2 2 i – j + 4k of length 10 is given by (a) 3(2 2i - j + 4k )

(b) 2(2 2i + j - 4k )

(c) 2(2 2i + j + 4k ) Ans. (d)

(d) 2(2 2i - j + 4k )

Solution: b = la and 10 = |b| = |l||a| but |a|2 = 8 + 1 + 16 = 25 fi |a| = 5 Thus |l| = 2. So

b = ± 2(2 2i - j + 4k )

Example 9: The vector p = (a · c) b – (a · b) c is perpendicular to (a) c (b) b (c) a (d) c + b Ans. (a) Solution: p = a × (b × c) which is perpendicular to a.

Vectors

Example 10: The angle between a + 2b and a – 3b if |a| = 1, |b| = 2 and angle between a and b is 60° is -24 (a) an acute angle (b) cos-1 21 31 1 24 (c) cos-1 (d) cos-1 3 21 31 Ans. (b)

22.5

|a – 3b|2 = |a|2 + 9|b|2 – 6a · b = 1 + 36 – 6 = 31 (a + 2b)·(a – 3b) = |a|2 – 6|b|2 – a · b 1 = 1 – 24 – 1.2. = –24 2 The angle between a + 2b and a – 3b = cos-1

Solution: |a + 2b|2 = |a|2 + 4|b|2 + 4a · b = 1 + 16 + 4 |a||b| cos 60° 1 = 1 + 16 + 4.1.2. = 21 2

(a + 2b) ◊ (a - 3b) a + 2b a - 3b

= cos-1

- 24 21 31

.

LEVEL 1 Straight Objective Type Questions Example 11: Let L1: r = (i + 5 j + 5 k ) + t ( 4i - 4 j + 5k )

1 1 (a) - (6i + 2 j - 11k ) (b) (-6i + 2 j - 11k ) 7 7

and L2: r = ( 2i + 4 j + 5k ) + t (8i - 3j + k ) be two lines then (a) (b) (c) (d) Ans. (c)

1 1 (c) - (6i - 2 j + 11k ) (d) - ( -6i + 2 j + 11k ) 7 7

L1 is parallel to L2 L1 is perpendicular to L2 L1 is not parallel to L2 none of these

Ans. (a)

Solution: The line L1 is parallel to the vector 4i - 4 j + 5 k and the line L2 is parallel to the vector 8i - 3j + k . These vectors are not parallel since neither is a scalar multiple of the other. Also ( 4i - 4 j + 5 k ) . (8i - 3j + k ) = 32 + 12 + 5 =

(a) cos–1 (1/ 3 ) (c) p/6 Ans. (a)

(b) p/4 (d) p/3

Solution: Let a = a1 i, b = a1 j and c = a1k. Then the vector d = a1 (i + j + k ) is a diagonal of the cube. The angle q between d and a is given by cos q =

a12 a.d 1 = = . 2 a d a 3 3a1 1

(

)

Example 13: Let u = 2i - j + 3k and a = 4i - j + 2k . The vector component of u orthogonal to a is

2

5 u – proja u = ( 2i - j + 3k ) - ( 4i - j + 2k ) 7 6 2 11 = - i - j+ k . 7 7 7

49 π 0. So L1 is not perpendicular to L2. Example 12: The angle between a diagonal of a cube and one of its edges is

u.a

5 a = ( 4i - j + 2k ) and vector 7 a component of u orthogonal to a is Solution: Proja u =

Example 14: Volume of the tetrahedron with vertices P(– 1, 2, 0), Q (2, 1, – 3), R (1, 0, 1) and S (3, – 2, 3) is (a) 1/3

(b) 2/3

(c) 1/4 Ans. (b)

(d) 3/4

Solution: PQ = 3i - j - 3k , PR = 2i - 2 j + k and PS = 4i - 4 j + 3k so the required volume =

1 PQ ◊ ( PR ¥ PS ) 6 =

3 -1 -3 1 -4 2 | 2 -2 1 |= = . 6 6 3 4 -4 3

22.6

Complete Mathematics—JEE Main

Example 15: The distance between a point P whose position vector is 5i + j + 3k and the line r = (3i + 7j + k) + t (j + k) is (a) 3 (b) 4 (c) 5 (d) 6 Ans. (d) Solution: Let u = j + k and the point Q whose position vector is 3i + 7 j + k is on the line, so v = QP = (5 – 3)i + (1 – 7)j + (3 – 1)k = 2i - 6 j + 2k d = |v| sin q u¥v 8i + 2 j - 2k 64 + 4 + 4 = = = =6 u j+k 2

Fig. 22.4

Example 16: Let a, b, c be the three vectors such that a ◊ (b + c) = b ◊ (c + a) = c ◊ (a + b) = 0 and |a| = 1, |b| = 4, |c| = 8, then |a + b + c| = (a) 13 (b) 81 (c) 9 (d) 5 Ans. (c) Solution: |a + b + c|2 = (a + b + c) ◊ (a + b + c) Adding a ◊ (b + c) = b ◊ (c + a) = c ◊ (a + b) = 0, we get 2 (a◊ b + b ◊ c + a ◊ c) = 0 fi |a + b + c|2 = |a|2 + | b|2 + | c |2 + 2 (a◊ b + a ◊ c + b ◊ c) = 1 + 16 + 64 = 81. Hence |a + b + c| = 9. Example 17: If a = i + 2j + 3k, b = – i + 2j + k and c = 3i + j, then t such that a + t b is at right angle to c will be equal to (a) 5 (b) 4 (c) 6 (d) 2 Ans. (a) Solution: Since a + tb is at right angle to c so (a + tb) ◊ c = 0. But a ◊c = 5 and b ◊ c = - 1 so 5 - t = 0 fi t = 5. Example 18: If |a| = 2, |b| = 5 and |a ¥ b| = 8 then |a ◊ b| is equal to (a) 4 (b) 6 (c) 5 (d) none of these Ans. (b) |a ¥ b| 8 4 Solution: Since sin q = = = so cos q |a ||b | 10 5 = ± 3/5 Thus a ◊ b = |a| |b| cos q = 10 ◊ (± 3/5) = ± 6.

Example 19: If a ◊ b = b ◊ c = c ◊ a = 0, then [a b c] is equal to (a) 0 (b) 1 (c) – 1 (d) |a| |b| |c| Ans. (d) Solution: Since a ◊ b = c ◊ a = b ◊ c = 0 so a, b and c are mutually perpendicular. Thus a is collinear with b ¥ c. Hence [a b c] = a ◊(b ¥ c) = |a | |b ¥ c| = |a ||b ||c| . Example 20: Let a, b and c be three non-coplanar vectors, and let p, q and r be the vectors defined by the relations b¥c c¥a a¥b , q= and r = p= [a b c] [a b c] [a b c] Then the value of the expression (a + b) ◊ p + ( b + c) ◊ q + (c + a) ◊ r is equal to (a) 0 (b) 1 (c) 2 (d) 3 Ans. (d) a ◊ ( b × c ) [a b c ] Solution: a◊ p = = = 1 = b◊q = c◊ r [a b c ] [a b c ] b◊ (b ¥ c) 0 = = 0 = c◊q = a◊r [a b c ] [a b c ] Therefore, the given expression is equal to 1 + 0 + 1 + 0 + 1 + 0 = 3. b ◊p =

Example 21: The volume of the parallelopiped whose sides are given by OA = 2i – 3j, OB = i + j – k, OC = 3i – k is (a) 4/13 (b) 4 (c) 2/7 (d) none of these Ans. (b) Solution: The volume of the parallelopiped 2 -3 0 = [O A O B O C] = | 1 1 - 1 | = 4. 3 0 -1 Example 22: The points with position vectors 60i + 3j, 40i – 8j, ai – 52j are collinear if (a) a = – 40 (b) a = 40 (c) a = 20 (d) none of these Ans. (a) Solution: Denoting a, b, c by the given vectors respectively. These vectors will be collinear if there is some constant K such that c – a = K (b – a) fi a - 60 = - 20 K and - 55 = - 11K fi a = - 100 + 60 = - 40. Example 23: If |a| = 2, |b| = 3 |c| = 4 and a + b + c = 0 then the value of b ◊ c + c ◊ a + a ◊ b is equal to (a) 19/2 (b) –19/2 (c) 29/2 (d) –29/2 Ans. (d)

Vectors

Solution: 0 = |a + b + c|2 = |a|2 + |b|2 + |c|2 + 2(a ◊ b + a ◊ c + b ◊ c)

Solution: a + b + c = BC + CA + AB = BA + AB So a ¥ (a + b + c) = a ¥ 0 = 0

fi

29 + 2 (a ◊ b + b ◊ c + c ◊ a ) = 0

fi

a¥a+a¥b+a¥c = 0

fi

( a ◊ b + b ◊ c + c ◊ a ) = – 29/2.

Similarly,

b ¥ (a + b + c) = 0

fi

b¥a+b¥c=0

Example 24: If A, B, C, D are four points in a space and |AB ¥ CD + BC ¥ AD + CA ¥ BD| = l (area of the triangle ABC). Then the value of l is (a) 1 (b) 2 (c) 3 (d) 4 Ans. (d) Solution: Let D be the origin of reference and DA = a, DB = b, DC = c AB ¥ CD + BC ¥ AD + CA ¥ BD =

(b - a ) ¥ ( - c ) + (c - b ) ¥ ( - a ) + (a - c ) ¥ ( - b )

=

b¥c-a¥c+c¥a-b¥a+a¥b-c¥b

= 2 a¥b+b¥c+c¥a = 2 (2 area of D ABC) = 4 area of D ABC. Hence l = 4. Example 25: Given a = i + j – k, b = – i + 2j + k and c = – i + 2j – k. A unit vector perpendicular to both a + b and b + c is 2i + j + k (a) (b) j 6 i + j+k (c) k (d) 3 Ans. (c) Solution: (a + b) ¥ (b + c) = 3j ¥ (- 2i + 4 j) = 6 k Hence the required unit vector is k. Example 26: If a, b and c are unit coplanar vectors, then the scalar triple product [2a – b 2b – c 2c – a] = (a) 0 (b) 1 (c) - 3 Ans. (a)

(d)

3

Solution: As a b, c are coplanar vectors, 2a – b, 2b – c and 2c – a are also coplanar vectors. Thus [2a – b 2b – c 2c – a] = 0. Example 27: If the vectors a, b and c form the sides BC, CA and AB respectively, of a triangle ABC, then (a) a ◊ b + b ◊ c + c ◊ a = 0 (b) a ¥ b = b ¥ c = c ¥ a

22.7

fi

a¥b =c¥a

fi a ¥ b = b ¥ c . Thus

a¥b =c¥a=b¥c. Example 28: If a and b are two unit vectors such that a + 2b and 5a – 4b are perpendicular to each other then the angle between a and b is (a) 45º (b) 60º (c) cos–1 (1/ 3 ) Ans. (b)

(d) cos–1 (2/7)

Solution: According to the given conditions |a| = |b| = 1 and (a + 2b) ◊ (5a – 4b) = 0 so 5|a|2 – 8|b|2 + 6a ◊ b = 0 fi a ◊ b = 3/6 = 1/2 fi |a| |b| cos q = 1/2 fi cos q = 1/2 fi q = 60º. Example 29: Let v = 2i + j - k and w = i + 3k . If u is a unit vector, then the maximum value of the scalar triple product [u v w] is (a) – 1 (b) 10 + 16 (c) Ans. (c)

(d)

59

60

Solution: v × w = 3i - 7 j - k Now

[u v w] = u ◊ (3i - 7 j - k ) = |u| 3i - 7 j - k cos q (where q is the

angle between u and v × w) = 59 cos q Thus [u v w] is maximum if cos q = 1 i.e. q = 0 or take 1 u= (3i - 7 j - k ). Hence the maximum value is 59 . 59 Example 30: A vector c perpendicular to the vectors 2i + 3j - k and i - 2 j + 3k satisfying c. ( 2i - j + k ) = - 6 is (a) -2i + j - k (c) -3i + 3j + 3k Ans. (c)

4 (b) 2i - j - k 3 (d) 3i - 3j + 3k

Solution: Vector c perpendicular to a = 2i + 3j - k and

(c) a ◊ b = b ◊ c = c ◊ a

b = i - 2 j + 3k is of the form a (a ¥ b ) = a (7i – 7j – 7k).

(d) a ¥ b = b ¥ c = c ¥ a = 0 Ans. (b)

Since c ◊ ( 2i - j + k ) = – 6 so a [14 + 7 – 7] = – 6 fi a = – 3/7. Hence c = -3i + 3j + 3k .

22.8

Complete Mathematics—JEE Main

Example 31: If a, b, c be three units vectors such that a × (b × c) = (1/2) b; b and c being non-parallel then (a) the angle between a and c is p/3 (b) the angle between a and c is p/2 (c) the angle between a and b is p/3 (d) the angle between a and b is p/6 Ans. (a) Solution: (1/2) b = a ¥ (b ¥ c) = (a ◊ c) b - (a ◊ b ) c Ê a ◊ c - 1 ˆ b = a ◊ b c. ( ) ˜ ÁË 2¯ The last relation is possible only if a ◊ c = 1/2 and a ◊ b = 0 as b and c are non-parallel. Hence the angle between a and c is p/3 and between a and b it is p/2.

K ÈÎ d

15 k, is 2 2 Ê 15 ˆ ± Á 6 i - 8 j - k ˜¯ , so a vector of length 50 along b is ± 25 Ë 2

Example 33: Let there be two points A, B on the curve y = x2 in the plane OXY satisfying OA ◊ i = 1 and OB ◊ i = – 2 then the length of the vector 2OA – 3OB is (a)

(b) 2 51

14

(c) 3 41 Ans. (d)

(d) none of these

Solution: Let OA = x1 i + y1 j and OB = x2 i + y2 j. Since 1 = OA. i = x1 and -2 = OB. i = x2. Moreover, y1 = x21 = 1 and y2 = x22 = 4, so OA = i + j and OB = – 2 i + 4 j. Hence 2OA - 3OB = 8 i - 10 j =

164 = 2 41 .

Example 34: If A, B, C, D are four points in space satisfying AB◊CD = K ÈÎ AD 2 + BC 2 - AC 2 - BD 2 ˘˚ then the value of K is (a) 2 (c) 1/2 Ans. (c)

(b) 1/3 (d) 1

Solution: Let A be the origin of reference and the position vector of B, C, D be b, c, d, w.r.t. A. So AB = b, CD = d – c, AD = d, BC = c – b, AC = c and BD = d – b. The L.H.S. is equal to b ◊(d – c) .The R.H.S. is

2

- c

2

- d - b 2 ˘˚

= 2K [b ◊ (d - c]] . Hence K = 1/2. Example 35: The distance of the point B with position vector i + 2j + 3k from the line passing through the point A with position vector 4i + 2j + 2k and parallel to the vector 2i + 3j + 6k is (a) (c) Ans. (a)

10

(b)

6

(d) none of these

5

Solution: AB = – 3i + k. Since AB◊(2i + 3j + 6k) = – 6 + 6 = 0. Hence AB is perpendicular to the given line. Thus required distance is equal to |AB| = 9 + 1 = 10 . Example 36: If a, b and c are unit vectors then a - b 2 + b - c 2 + c - a 2 does not exceed. (a) 4 (b) 9 (c) 8 (d) 6 Ans. (b)

Solution: A unit vector along b = 6i – 8j –

15 4 ÊÁ 6 i - 8 j - k ˆ˜ . Since a makes obtuse angle with z-axis Ë 2 ¯ so we must have a◊k < 0. Thus a = 24i – 32j – 30k.

+ c-b

= K [ d ◊d + c ◊c + b ◊b - 2 c ◊b - c ◊c - d ◊d - b ◊b + 2 d ◊b]

fi

Example 32: A vector a of magnitude 50 is collinear with the vector 6i – 8j – (15/2)k making on obtuse angle with the z-axis is (a) 24i – 32j – 30k (b) – 24i + 32j + 30k (c) 24i + 32j – 30k (d) none of these Ans. (a)

2

Solution: Since a + b + c ≥ 0 fi fi

0 £ a 2 + b 2 + c 2 + 2 (b ◊ c + c ◊ a + a ◊ b ) = 3 + 2 (b.c + c.a + a.b) b.c + c.a + a.b ≥ –3/2 a-b2 + b-c2 + c-a2

= 2 ( a 2 + b 2 + c 2 - b .c - c. a - a. b ) £ 2 (1 + 1 + 1) + 3 = 9 The maximum value is attained e.g. when 1 1 a = i, b = ( -i + 3j) and c = ( - i - 3j) . 2 2 Example 37: The value of a so that the volume of the parallelepiped formed by i + aj+ k, j + ak and a i +k becomes minimum is (a) –3 (b) 3 (d) 3 (c) 1 3 Ans. (c) Solution: Volume of the 1 V(a) = 0 a

parallelepiped is a 1 1 a = a3 - a + 1 0 1

V¢(a) = 3a2 – 1 = 0 if a = a ± V¢¢(a) = 6a > 0 if a =

Thus V (a) is minimum when a =

1 3

.

1 3

1 3

Vectors

Example 38: If a = i – j – k, a ◊ b = 1 and a ¥ b = –j + k, then k is equal to (a) i + j – k (b) –2j + k (c) i (d) –2i + k Ans (c) Solution: let b = a i +b j + g k then a ◊ b = 1 a – b – g = 1. i j k 1 -1 -1 = j + k And a ¥ b = j + k fi a b g

fi

(1)

fi (b – g) i – (g + a)j + ( b + g) k = – j + k fi b – g = 0, g + a = 1, b + a = 1 fi b = g, a = 1– g Putting there values in (1) 1 – g – g – g =1 fi g = 0 so a = 1; b = 0 Thus b = i. Example 39: The unit vector which is orthogonal to the vector 5 i +2j +6k and is coplanar with the vectors 2i + j + k and i – j + k is 1 1 (a) (2i – 6j + k) (b) (2i – 5j) 41 29 (c)

1 10

(3j – k)

(d)

1 69

(a + b ) + (2d - b ) - (d + b )

fi

1+1+1 This is satisfied when a = 1, b = 1.

(c)

Solution: A vector in the plane of a and b is given by d = a a +b b = (a + b) i + (2a – b)j + (a + b)k Length of projection of d on c is d◊c 1 = c 3

3

19 2

77 2

(d)

Ans (d) Solution: 16 = |a – 2b|2 = |a|2 + 4 |b|2 – 4 a · b = 1 + 16 – 4 a · b 1 4 a·b = 1 fi a·b = 4 |a + 3b|2 = |a|2 + 9 |b|2 + 6 a · b 6 77 = 1 + 36 + = . 4 2 Example 42: If |a|2 = 8 and a × (i + j + 2k) = 0 then the value of a·(–i + j + 4k) is (a) (c)

Ans (c)

Example 40: Let a = i + 2j + k, b = i – j + k and c = i + j – k. A vector in the plane of a and b whose projection on c is 1 3 is (a) 4i – j + 4k (b) 3i + j + 3k (c) 2i + j + 2k (d) 4i + j – 4k Ans. (c)

1

Example 41: If |a| = 1, |b| = 2 and |a – 2b| = 4 then |a + 3b| is equal to 51 (a) 8 (b) 2

(2i – 8j +k)

Solution: A vector coplanar with 2i + j +k and i – j +k is of the form a = a (2i + j + k) + b (i – j + k) = (2a + b) i + (a – b) j + (a + b)k This vector will be orthogonal to 5i + 2j + 6k if 5(2a + b) + 2(a – b) + 6(a + b) = 0 fi 18a + 9b = 0 fi b = –2a So a is of the form a (3j – k) Thus, a required unit vector is 1 (3j – k). 10

=

22.9

4 3 8 3

(b) (d)

16 3 1 3

Ans (b) Solution: Since a × (i + j + 2k) = 0 so a is parallel to i + j + 2k fi a = l(i + j + 2k) 4 but 8 = |a|2 = l2· 6 fi l2 = . 3 a.(–i + j + 4k) = l(i + j + 2k) · (–i + j + 4k) 8.2 16 = l (–1 + 1 + 8) = 8l = = 3 3 Example 43: If a, b, c are unit vectors, then the maximum value of |a + 2b|2 + |b + 3c|2 + |c + 4a|2 is (a) 28 (b) 21 (c) 48 (d) 58 Ans (b) Solution: |a + 2b|2 + |b + 3c|2 + |c + 4a|2 = |a|2 + 4|b|2 + 2 × a·b + |b|2 + 9 |c|2 + 6 b·c + |c|2 + 16 |a|2 + 8 a·c £ 1 + 4 + 2 + 1 + 9 + 6 +1 + 16 + 8 = 48 Example 44: Let a = 2 i + j – 2k, b = i + j. If c is a vector such that a ◊ c = |c| and c - a = 2 2 and angle between a ¥ b and c is 30°, then (a ¥ b) ¥ c equals: 3 2 (a) (b) 2 3

Complete Mathematics—JEE Main

22.10

(c) 2

(d)

3 2

Ans. (a) Solution: a 2 = 4 + 1 + 4 = 9, b 2 = 1 + 1 = 2. c - a 2 = 8 fi c 2 + a 2 - 2 c.a = 8

Also,

2

fi

c +9-2 c =8

fi

c 2 - 2 c + 1 = 0 fi ( c - 1)2 = 0 fi c =1.

Also,

i j k a ¥ b = 2 1 -2 = 2i + 2 j + k 1 1 0

fi

|a ¥ b| = 3

Also, (a ¥ b) ¥ c = a ¥ b c sin30° = (3) (1) (1/2) = 3/2. Example 45: The non-zero vectors a, b and c are related by a = 8b and c= – 7b. Then the angle between a and c is (a) 0 (b) p/4 (c) p/2 (d) p Ans. (d) 8 c so a and c are parallel and a 7 and c have opposite direction so angle between a and c is p. Solution: a = 8b = –

Example 46: If u, v, w are non-coplanar vectors and p, q are real numbers, then the equality [3u pv pw] – [pv w qu] – [2w qv qu] = 0 holds for (a) (b) (c) (d) Ans (c)

more than two but not all values of (p, q) all values of (p, q) exactly one values of (p, q) exactly two values of (p, q)

Solution: 0 =[3u pv pw] – [pv w qu] – [2w qv qu] =(3p2– pq + 2q2) [u v w] As u, v, w are non – coplanar so [u v w ] π 0, so 3p2 – pq +2q2 = 0

(

1 fi 2 q- p 4

)

2

23 2 + p =0 8

1 p=0 4 Thus there is exactly one value of (p, q).

fi

p = 0, q =

Example 47: If the vectors a = i – j + 2k, b = 2i + 4j +k and c = li + j + mk are mutually orthogonal then (l, m) = (a) (–2, 3) (b) (3, –2) (c) (–3, 2) (d) ( 2, –3) Ans. (c)

Solution: As a, b and c are mutually orthogonal, a ◊ c = 0 and b ◊ c = 0 fi l - 1 + 2m = 0 and 2l + 4 + m = 0 fi l + 2m = 1 and 2l + m = –4 fi l = –3, m = 2. Example 48: Let a = j – k and c = i – j – k. Then the vector b satisfying (a ¥ b) + c = 0 and a ◊ b = 3 is (a) i – j – 2k (b) i + j – 2k (c) – i + j – 2k (d) 2i – j + 2k Ans (c) Solution: Let b = ai +bj + gk. We have i j k a ¥ b = 0 1 -1 = (b + g)i – aj – ak a b g As (a ¥ b) + c = 0, we get (b + g + 1)i – (a + 1)j – (a + 1) k = 0 fi b + g + 1 = 0, a = – 1 Also, as a ◊ b = 3, we get b – g = 3 Thus a = – 1, b = 1, g = –2 Hence, b = – i + j – 2k. 1 (2i + 3j – 6k), 7 10 then the value of (2a – b). [(a ¥ b) ¥ (a + 2b)] is: (a) 3 (b) –5 (c) –3 (d) 5 Ans. (b) Example 49: If a =

1

(3i + k) and b =

Solution: we have a . a = 1, b. b = 1 and 1 = ((3) (2) + (1) (–6)) = 0 = b . a a. b = 7 10 Now, (a ¥ b) ¥ (a + 2b) = (a . (a + 2b)) b – (b . (a + 2b)) a = (a . a) b – 2 (b . b)a (a . b = 0) = b – 2a. Thus (2a – b) . [(a ¥ b) ¥ (a + 2b)] = (2a – b) . (b – 2a) = – 2a - b 2 = – ÈÎ4 a 2 + b 2 - 4a . b ˘˚= - 5. Example 50: The vectors a and b are not perpendicular and c and d are the vectors satisfying b ¥ c = b ¥ d and a . d = 0. Then the vector d is equal to : Ê a. c ˆ Ê b. c ˆ b (b) b – Á c (a) c – Á ˜ Ë a. b ¯ Ë a. b ˜¯ Ê a. c ˆ (c) c + Á b Ë a. b ˜¯ And. (a)

Ê b. c ˆ (d) b + Á c Ë a. b ˜¯

Vectors

Solution: b ¥ c = b ¥ d fi b ¥ (c – d) = 0 fi c – d || b fi c – d = ab for some a Œ R fi d=c–ab Also, a · d = a . c – a a . b fi 0 = a · c – aa . b fi

a=

Thus,

a◊c a◊b

d=c–

Example 51: If vectors pi + qj + k, i + qj + k and i + j + rk (p π q π r π 1) are coplanar, then the value of p q r – (p + q + r) is (a) 2 (b) 0 (c) –1 (d) –2 Ans. (d) Solution: Since the given vectors are coplanar so p 1 1 1 q 1 =0 1 1 r fi

q 1 1 1 1 q p + =0 1 r 1 r 1 1

fi

p (qr - 1) - ( r - 1) + (1 - q ) = 0

fi

pqr - p - q - r = - 2

Example 52: Let a, b, c be three non-zero vectors which are pairwise non-collinear. If a + 3b is collinear with c and b + 2c is collinear with a, then a + 3b + 6c is (a) a (b) b (c) 0 (d) a + c Ans. (c) Solution: We are given that a + 3b = ac and

(a) i – 3j + 3k (c) 3i – j + k Ans. (c)

a + 3b + 6c = ac + 6c = (6 + a)c a + 3b + 6c = a + 3(b + 2c)

1 c◊v = c 3

(a + b ) - (a + b ) - (a + b )

fi

3

(6 + a)c = (1 + 3b)a

1 + 3b a, so c is collinear with a. 6 +a Thus 6 + a = 0, therefore, a + 3b + 6c = 0. If 6 + a π 0, we get c =

=

1 3

fi a - b = 1fi a - b = ± 1. The only possible answer is (c). Example 54: Suppose that a and b are two unit vectors. If the vectors c = a + 2b and d = 5a – 4b are perpendicular to each other, then the angle between a and b is (b) p 3 (a) p 2 (c) p 4 Ans. (b)

(d) p 6

Solution: c . d = 0 fi (a + 2b) . (5a – 4b) = 0 fi

5 a 2 + 10b . a – 4 a . b – 8 b 2 = 0

fi

5+6a.b–8=0

fi

a . b = 1/2

fi

a b cosq = 1 2

fi

cos q = 1/2 fi q = p/3.

Example 55: Let ABCD be a parallelogram such that AB = q, AD = p and BAD be an acute angle. If r is the vector that coincides with the altitude directed from the vertex B to the side AD, then r is given by : (a) r = – q +

(p ◊ q ) p◊p

(p ◊ q ) p◊p

(c) r = – 3q +

= a + 3ba = (1 + 3b)b Thus,

3, so

The projection of v on c is 1

(b) r = q –

We have

(b) –3i – 3j – k (d) i + 3j – 3k

Solution: v = a a + bb, where a, b ΠR v = (a + b)i + (a Рb)j + (a + b) k

b + 2c = ba

for some a, b ΠR.

and

Example 53: Let a = i + j + k, b = i – j + k and c = i – j – k be three vectors. A vector v in the plane of a and b, whose projection on c is 1 3, , is given by

fi a◊c b. a◊b

22.11

(d) r = 3q –

p

p

3 (p ◊ q ) p p◊p

3 (p ◊ q ) p p◊p

Ans. (a) Solution: Let M be the foot of perpendicular from B to AD.

22.12

Complete Mathematics—JEE Main B( q)

C

fi

r

A

AM = ap AM = AB + BM = q + r q +r = ap, we get (q + r).p = a p.p p◊q [\ r.p = 0] q.p + r.p = a p.p fi a = p◊p

Suppose Also, As

D( p)

M

Thus,

r=–q+

q◊p p. p◊p

Fig. 22.5

Assertion-Reason Type Questions

Example 56: Statement-1: If vectors a and c are noncollinear then the lines r = 6a – c + l (2c – a) r = a – c + m (a + 3c) are coplanar Statement-2: There exist l and m such that the two values of r become same Ans. (a) Solution: If lines have a common point then there exists l and m such that 6 – l = 1 + m and – 1 + 2l = – 1 + 3m fi l = 3, m = 2. Example 57: Given that a, b, c are the position vectors of the vertices of a D ABC 1 [a ¥ b + b ¥ c + c ¥ a ] 2 Statement-2: Cross product is distributive over addition of vectors Ans. (b) Statement-1: The area of D ABC is

Solution: Required area =

1 ( AB ¥ AC) 2

1 ((b - a ) ¥ (c - a )) 2 1 = [a ¥ b + b ¥ c + c ¥ a ] . 2 =

Example 58: Let A, B, C be three points with position vectors i + 2j – k, 2i + 3k, 3i – j + 2k Statement-1: The angle between AB and AC is acute Statement-2: If q is the angle between AB and AC then cos q = Ans. (c)

17 21 22

Solution: AB = OB – OA = i – 2j + 4k AC = 2i – 3j + 3k AB.AC 20 = AB AC 21 22 Example 59: Statement-1: ((a ¥ b) ¥ (a ¥ c)). d = b.d [abc] Statement-2: (a ¥ b).c = a.(b ¥ c) Ans. (a) Solution: ((a ¥ b) ¥ (a ¥ c)) . d = (a ¥ b) . ((a ¥ c) ¥ d) = (a ¥ b) . [(a . d) c – (c . d) a] = (a . d) [a b c] (the last scalar product is zero) Example 60: If a, b, c be three unit vectors such that 1 a ¥ (b ¥ c) = b, b, c being non-parallel. 2 Statement-1: The angle between a and b is p /2. p Statement-2: The angle between a and c is . 3 Ans. (b) 1 Solution: a ¥ (b ¥ c) = b 2 1 fi (a . c) b – (a . b) c = b 2 fi (a . c – 1/2) b – (a . b) c = 0 fi a . c = 1/2 and a . b = 0 Cosine of angle between a and c = 1/2 fi angle between a and c is p/3 and angle between a and b is p/2. cos q =

Example 61: Let u, v, w be three non-coplanar vectors, then (v – w). [(w – u) × (u – v)] = 0 Statement-1: v – w = l(w – u) + m(u – v) for some l, m Œ R. Statement-2: The vectors v – w, w – u, u – v are coplanar. Ans. (a)

Vectors

Solution: The given condition implies v – w, w – u, and u – v are coplanar. So one vector is linear combination of other two. Example 62: Let r be a vector such that r × b = a × b. Statement-1: r is linear combination of a and b. Statement-2: r = a Ans. (c) Solution: r × b = a × b fi (r – a) × b = 0 fi r – a is parallel to b so r – a = l b, for some l Œ R. fi r = a + l b.

22.13

Example 63: If a, b, c are non-coplanar vectors then Statement-1: b × c, c × a, a × b are non-coplanar. Statement-2: [b × c c × a a × b] = 2[a b c]2 Ans. (c) Solution: [b × c c × a a × b] = ((b × c) × (c × a)) · (a × b) = ((b.(c × a)) c – (c.(c × a)b) · (a × b) = [a. b. c] c.(a × b) (c.(c × a) = 0) = [a b c] [a b c] = [a b c]2 Since R.H.S. is non-zero so is L.H.S. Hence b × c, c × a, a × b are non-coplanar.

LEVEL 2 Straight Objective Type Questions Example 64: The vector c, directed along the internal bisector of the angle between the vectors a = 7i – 4j – 4k and b = – 2i – j + 2k, with |c| = 5 6 is (a) ± (5/3) (i – 7j + 2k) (b) (5/3) (5i + 5j + 2k)

Solution: Let a = (1, 0, 3), b = (- 1, 3, 4), c = (1, 2, 1) and d = (k, 2, 5). Since A, B, C and D are coplanar, we have [d b c] + [d c a] + [d a b] = [a b c]

(c) ± (5/3) (i + 7j + 2k) (d) (5/3) (– 5i + 5j + 2k) Ans. (a) Solution: The required vector c is given by b ˆ Ê a + l Á Ë | a | | b | ˜¯ Now fi

Example 66: The value of k for which the points A(1, 0, 3), B(– 1, 3, 4), C(1, 2, 1) and D(k, 2, 5) are coplanar, are (a) 1 (b) 2 (c) 0 (d) –1 Ans. (d)

fi

a 1 b 1 = (7i - 4j - 4k) and = (- 2i - j + 2k) |a| 9 |b| 3 1 7 2 c = l ÊÁ i - j + k ˆ˜ Ë9 9 9 ¯

fi

|c|2 = l2 ¥

54 81

fi fi

k 2 5 k 2 5 k 2 5 -1 3 4 + 1 2 1 + 1 0 3 1 2 1 1 0 3 -1 3 4 1 0 3 = -1 3 4 1 2 1 - 5k - 15 + 6k - 14 - 9k + 1 = - 20 - 8k - 28 = - 20

fi l2 = 225 fi l = ± 15. fi c = ± (5/3) (i – 7j + 2k)

Alternate Solution

Example 65: Let a, b and c be three non-zero vectors, no two of which are collinear. If the vectors a + 2b is collinear with c, and b + 3c is collinear with a, then a + 2b + 6c is equal to, (l being some non-zero scalar) (a) l a (b) l b (c) l c (d) 0 Ans. (d)

A, B, C, D are coplanar ¤

Solution: Let a + 2b = xc and b + 3c = ya. Then a + 2b + 6c = (x + 6)c and also, a + 2b + 6c = (1 + 2y)a. So (x + 6) c = (1 + 2y)a. Since a and c are non-zero and non-collinear, we have x + 6 = 0 and 1 + 2y = 0, i.e., x = - 6 and y = –1/2. In either case, we have a + 2b + 6c = 0.

fi

k=-1

[AB AC AD] = 0 -2 3 1 0 2 -2 = 0 ¤ k = – 1 k -1 2 2 Example 67: Let a, b, c be distinct non-negative numbers. If the vectors ai + aj + ck, i + k and ci + cj + bk lie in a plane, then c is (a) the arithmetic mean of a and b

22.14

Complete Mathematics—JEE Main

= 2(|a|2 + |b|2 + |c|2 + |d|2 – (a◊b + c◊d + d◊a + a◊c + b◊d + b◊c) < 2 (4 + 2) = 12.

(b) the geometric mean of a and b (c) the harmonic mean of a and b (d) equal to zero Ans. (b) Solution: The three vectors are coplanar if and only if a a c 0 a c 1 0 1 = 0 or 1 0 1 = 0. (C1 Æ C1 - C2) c c b 0 c b - 1 (ab - c2) = 0

¤

¤

ab = c2.

Example 68: Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies the equation p ¥ ((x – q) ¥ p) + q ¥ ((x – r) ¥ q) + r ¥ ((x – p) ¥ r) = 0 then x is given by

Example 70: L e t a = i - k , b = xi + j + (1 - x)k and c = yi + xj (a) (b) (c) (d)

+ (1 + x – y)k. Then [a b c] depends on only x only y neither x nor y both x and y

Ans. (c) 1 0 -1 1- x Solution: [a b c] = x 1 y x 1+ x - y 1 0 0 1 = x 1 (using C3 Æ C3 + C1) y x 1+ x

(a) (1/ 2) (p + q - 2r ) (b) (1/ 2) (p + q + r) (c) (1/ 3) (p + q + r ) Ans. (b)

(d) (1/ 3) ( 2p + q - r )

Solution: Let | p| = | q| = | r| = K. Let a, b, c be unit vectors along p, q, r respectively. Clearly p, q, r are mutually perpendicular vectors, so any vector x can be written as a1 a + a2 b + a3 c. p ¥ ((x – q) ¥ p) = (p◊p) (x – q) – (p◊(x – q)) p = K2 (x – q) – (p◊x) p (p◊q = 0) = K2 ( x - q ) - K a ◊ (a1 a + a2 b + a3 c) K a = K2 ( x - q - a1 a ) Similarly q ¥ ((x – r) ¥ q) = K2(x – r – a2b) and r ¥ ((x – p) ¥ r) = K2 (x – p – a3c) According to the given condition K2 (x – q – a1 a + x – r – a2 b + x – p – a3 c) = 0 fi {3x – (p + q + r) – (a1 a + a2 b + a3 c)} = 0 fi [2x – (p + q + r)] = 0 fi x = (1/2) (p + q + r). Example 69: If a, b, c and d are unit vectors, then |a – b|2 + |b – c|2 + |c - d| 2 + |d – a|2 + |c - a| 2 + |b – d|2 does not exceed (a) 4 (b) 12 (c) 8 (d) 16 Ans. (b) Solution: We have 0 £ a + b + c + d 2 = a 2 + b 2 + c 2 + | d |2 + 2a ◊ b + b ◊ c + c ◊ a

+ a◊d + c◊d + b◊d = 4 + 2 a ◊b + b ◊c + c◊a + a ◊b + c◊d+b ◊d So Now

a◊b + b◊c + c◊a + a◊d + c◊d + b◊d ≥- 2.

a - b 2 + b - c 2 + c ◊ d 2 + | d - a |2 + | c - a |2 + | b - d |2

= (1 + x) – x = 1, which depends neither on x nor on y. Example 71: Let a = 2i + j - 2k and b = i + j. If c is a vector such that a ◊ c = c , c - a = 2 2 and the angle between a ¥ b and c is 30º, then (a ¥ b ) ¥ c = (a) 2/3 (b) 3/2 (c) 2 (d) 3 Ans. (b) i j k Solution: a ¥ b = 2 1 -2 = 2i + 2 j + k 1 1 0 fi

a ¥ b = 4 + 4 + 1 = 3 . Also |c – a|2 = 8

so

c 2 + a 2 - 2a.c = 8

fi

c 2 + 9 - 2 c = 8 fi c 2 - 2 c +1= 0

fi

( c - 1)2

fi

fi

c 2 + a2 -2 c =8

c =1

1 3 2 2 Example 72: A tangent is drawn to the curve y = 8/x2 in xy-plane at the point A (x0, y0), where x0 = 2, and the tangent cuts the x-axis at a point B. Then AB ◊ OB is equal to (a) 2 (b) 1 (c) 0 (d) 3 Ans. (d) Now

( a ¥ b ) ¥ c = a ¥ b c sin 30º = 3.1. = .

Solution: Since x0 = 2 so y0 = 8/4 = 2 8(-2) Equation of tangent is Y – 2 = 3 ( X - 2) i.e. Y = –2X + 6. 2

Vectors

This tangent cuts x-axis at (3, 0) Hence B = (3, 0). Therefore AB = (3 – 2)i + (0 – 2)j and OB = 3i, thus AB ◊ OB = 3 . Example 73: Let P, Q, R be points with position vectors r1 = 3i – 2j – k, r2 = i + 3j + 4k and r3 = 2i + j – 2k relative to an origin 0. The distance of P from the plane OQR is (magnitude) (a) 2 (b) 3 (c) 1 (d) 11/ 3 Ans. (b) Solution: Equation of the plane OQR is r = l r2 + m r3 , i.e. r ◊ (r2 ¥ r3 ) = 0 So the distance of P from the plane OQR is Since

r1 ◊ (r2 ¥ r3 ) . r2 ¥ r3

r2 ¥ r3 = – 10 i + 10 j – 5k so r2 ¥ r3 = 15.

3 - 2 -1 1 |1 3 4 |=3. Required distance = 15 2 1 -2 Example 74: For unit vectors b and c and any non zero vector a, the value of {{(a + b ) ¥ (a + c)} ¥ (b ¥ c)} ◊ (b + c) is 2

2

(a) |a| (c) 3 |a|2 Ans. (d)

(b) 2 |a| (d) none of these

Solution: The given expression = { {a ¥ c + b ¥ a + b ¥ c} ¥ (b ¥ c) } . (b + c) = { (a ¥ c) ¥ (b ¥ c) + (b ¥ a) ¥ (b ¥ c) } . (b + c) = [(a ◊ (b ¥ c)) c - (c ◊ (b ¥ c)) a + (b ◊ (b ¥ c)) a - (a ◊ (b ¥ c) b )] . (b + c) =

[ (a ◊ ( b ¥ c ) ) ( c - b ) ◊ ( b + c ) ]

= (a ◊ (b ¥ c)) ÈÎ c

2

- b 2 ˘˚ = 0 [∵ | b| = | c | = 1].

Example 75: Three non-coplanar vector a , b and c are drawn from a common initial point. The angle between the plane passing through the terminal points of these vectors and the vector a ¥ b + b ¥ c + c ¥ a is (a) p /4 (b) p /2 (c) p /3 (d) none of these Ans. (b) Solution: Let the terminal points be A, B, C and the common initial point be the origin of reference so that AB = b – a and AC = c – a. The vector AB × AC is perpendicular to the plane ABC. AB × AC = (b – a) ¥ (c – a) = b ¥ c + c ¥ a + a ¥ b. Hence the angle between the plane and the given vector is p/2.

22.15

Example 76: A unit tangent vector at t = 2 on the curve x = t2 + 2, y = 4t3 – 5, z = 2t2 – 6t is 1 1 (b) (a) (i + j + k ) (2i + 2 j + k ) 3 3 1 (c) (2i + j + k ) (d) none of these 6 Ans. (d) Solution: The position vector of any point at t is r = (2 + t2 ) i + (4t3 – 5) j + (2t2 – 6t) k dr dr \ = 2t i + 12t2 j + (4t – 6)k, dt dt t = 2 = 4i + 48j + 2k dr = 16 + 2304 + 4 = 2320 . dt t = 2 Hence the required unit tangent vector at t = 2 is (1/ 580) (2i + 24j + k). Example 77: A particle moves along a curve so that 1 2 1 3 t,z= t . The its coordinates at time t are x = t, y = 2 3 acceleration at t = 1 is (a) j + 2k (c) 2j + k Ans. (a)

(b) j + k (d) none of these

1 2 1 t j + t3 k then the velocity 2 3 dr dv = i + t j + t2 k and the acceleration a = = v= dt dt j + 2t k. Hence a |(t =1) = j + 2 k. Example 78: Consider the parallelopipped with sides a = 3i + 2j + k, b = i + j + 2k and c = i + 3j + 3k then the angle between a and the plane containing the face determined by b and c is (b) cos–1 (9/14) (a) sin– 1 (1/3) –1 (d) sin–1 (2/3) (c) sin (9/14) Ans. (c) Solution: b ¥ c = – 3i – j + 2k. If q is the angle between a and the plane containing b and c a ◊ (b ¥ c ) | then. cos (90° – q) = | a b¥c Solution: Let r = ti +

=

1

1

|(– 9 – 2 + 2) | =

9 . 14

14 14 Hence q = sin–1 (9/14). Example 79: A unit vector n perpendicular to the plane determined by the points A (0, –2, 1), B (1, – 1, –2) and C (–1, 1, 0) 1 (a) (b) 1/ 4 6 (8i + 4 j + 4k ) (2i + j + 2k ) 3 (c) Ans. (b)

1 3

(i - j + k )

(d)

1 14

(3i + j + 2k )

22.16

Complete Mathematics—JEE Main

Solution: AB = i + j – 3 k and AC = – i + 3 j - k and A B ¥ A C = 8i + 4j + 4k. Hence n = (1/ 4 6 ) (8 i + 4 j + 4 k ) . Example 80: If the vectors AB = – 3i + 4k and AC = 5i – 2j + 4k are the sides of a triangle ABC. Then the length of the median through A is (a)

14

(b)

(c)

29

(d) none of these

18

Example 84: If a and b are unit vectors and q is the angle between a and b then sin (q/2) is equal to 1 (b) 1 (a) a-b 2 (c)

Solution:

Solution: Let A be the origin of reference so that the position vectors of B and C are – 3i + 4j and – i + 3j – k respectively. The position vector of mid point of BC is i - j + 4 k . Thus the length of the median is

= Hence

18 .

Example 81: If a + b + c = 0 and |a| = 3, | b| = 5 and | c | = 7 then the angle between a and b is (a) p /6 (b) 2p/3 (c) p /3 (d) 5p /3 Ans. (c)

equal to (a) 5i - 4 j - k (c) 4i - 5 j - k Ans. (a)

(b) 3i - 2 j + 5k (d) 5i + 4 j - k

Solution: The given expression

((i - j) ◊ (i + 5 k )) ( j - k ) - (( j - k ) ◊ (i + 5k )) (i - j) = 1 ( j - k ) + 5 (i - j) = 5i - 4 j - k . Example 83: The position vector of a point P is r = xi + yj + zk , where x, y, z Œ N and u = i + j + k. If r ◊ u = 10 , then the number of possible positions of P is (a) 72 (c) 60 Ans. (b)

(b) 36 (d) 108

Solution: r ◊u = 10 fi x + y + z = 10. The number of positive integral solution of this equation is 9C2 = 36 (see Chapter 6).

q 1 1 [1- | a || b | cos q ] = [1 - cos q ] = sin 2 2 2 2

q 1 sin ÊÁ ˆ˜ = ÊÁ ˆ˜ | a - b | . Ë 2¯ Ë 2¯

Solution: a, b, c are coplanar if 1 1 m 1 1 m +1 = 0 1 -1 m

+ b) = | c|2.

Example 82: The vector ((i - j) ¥ ( j - k )) ¥ (i + 5 k ) is

a-b 2 1 = ÈÎ a 2 + b 2 - 2a.b ˘˚ 2 4

Example 85: The vectors a = i + j + mk, b = i + j + (m + 1) k and c = i – j + mk are coplanar if m is equal to (a) 1 (b) 4 (c) 3 (d) none of these Ans. (d)

Solution: a + b + c = 0 fi a + b = – c fi (a + b) ◊(a Thus | a|2 + | b|2 + 2 | a|| b| cos q = | c|2, where q is the 49 - 9 - 25 angle between a and b. Therefore, cos q = = 2 ◊ 3◊ 5 1 fi q = p/3. 2

(d) 0

Ans. (a)

Ans. (b)

1 + 1 + 16 =

1 a+b 2

1 0 m 1 0 m +1 = 0 fi 1 -2 m fi 2 = 0. So there no value of m for which the vectors are coplanar. Example 86: If a, b, c are non-coplanar unit vectors b+c , then the angle between a and such that a ¥ (b ¥ c) = 2 b is (a) 3p /4 (b) p /4 (c) p /2 (d) p Ans. (a) Solution: The given equality implies b+c (a ◊c) b – (a ◊b) c = 2 1 1 ˆ b= Ê Ê a ◊c + a ◊ bˆ˜ c fi ˜ ÁË ÁË ¯ 2 2¯ fi

b and c are collinear thus a, b, c are coplanar if 1 1 a ◊c π and a ◊ b π . Hence a◊b = –1/ 2 2 2

fi

cos q = –1/ 2 , where q is the angle between a and b.

So q = 3p/4.

Vectors

22.17

EXERCISE Concept Based Straight Objective Type Questions 1. If M is the mid point of AB and O is any point, then (a) OM = OA + MA (b) OM = OA – MA (c) OM =

1 (OA – OB) 2

(a)

2. The angle between 3i + 4j and 2j – 5k is p 2

(b) cos-1

8 5 29

p 1 (d) cos-1 3 6 3. A unit vector c perpendicular to a and coplanar with a and b, a = i + j + k, b = i + 2j is given by (c)

(a) (c)

1 2 1 2

(i + k )

(b)

(j + k)

(d)

1 2

(b)

3 2

3 1 (d) 4 3 7. The area of the triangle formed by A (1, 0, 0), B (0, 1, 0), C (1, 1, 1) is (c)

1 (d) OM = (OB + OA) 2

(a)

6. If a and b are non-collinear vectors, then the value of l for which u = (l + 2) a + b and v = (1 + 4l) a – 2b are collinear is

1 2 1 2

(i - j) (- j + k )

4. A vector b, which is collinear with vector a = 2i + j – k and satisfies a·b = 2 is given by (a)

1 (2 i + j - k ) 2

(b)

1 (2 i + j - k ) 3

(c)

1 (2 i + j - k ) 4

(d)

1 (-2i - j + k ) 2

5. If u = i + j – k, v = 2i + j + k and w = i + j + 2k then the magnitude of projection of u × v on w is given by (a)

1 2

(b)

1 3

(c)

3 4

(d)

3 2

(a)

1 2

(b)

3 4

3 1 (d) 2 4 8. A unit vector perpendicular to 3i + 4j and i – j + k is 1 (i + j + k ) (a) 3 (c)

(b) (c) (d)

1 14 1 74 1 74

(i - 2 j + 3k ) (4i + 3 j - 7k ) (4i - 3 j - 7k )

9. The value of scalar triple product i – 2j + 3k, 2i + j – k and j + k is (a) 12 (c) 14

(b) 10 (d) 16

10. The vector [(i – j + k) × (2i – 3j – k)] × [(–3i + j + k) × (2j + k)] is given by (a) (b) (c) (d)

3i + 5j – 3k –5(3i – 5j – 3k) 5(3i + 5j – 3k) (15i – 25j + 15k)

22.18

Complete Mathematics—JEE Main

LEVEL 1 Straight Objective Type Questions 11. Let | a | = 3 and | b | = 4. The value of m for which the vectors a + mb and a – mb will be perpendicular is (a) 3 4

(b) 2 3

(c) – 5/2

(d) – 2 3

12. The value of a for which the vectors 2i – j + k, i + 2j + a k and 3i – 4j + 5k are coplanar is (a) 3 (c) 2

(b) –3 (d) none of these

13. The area of a parallelogram having diagonals a = 3i+ j – 2k and b = i – 3j + 4k is (a) 5 3

(b) 2 3

(c) 4 3

(d) none of these

14. If r satisfies the equation r ¥ (i + 2j + k) = i – k, then for any scalar t, r is equal to (a) (b) (c) (d)

p (a) ÈÍ0, ˆ˜ Î 6¯

5p (b) ÊÁ , 2p ˙˘ Ë 6 ˚

p p (c) ÈÍ , ˙˘ Î6 2˚

p 5p (d) ÈÍ , ˙˘ Î2 6 ˚

19. For non-coplanar vectors a, b and c, | (a ¥ b) ◊ c| = | a | | b | | c | holds if and only if (a) (b) (c) (d)

a◊b = b◊c = c◊a = 0 a◊b = 0 = b◊c a◊b = 0 = c◊a b◊c = 0 = c◊a

20. The volume of the tetrahedron whose vertices are the points with position vectors i – 6j + 10k, –i – 3j + 7k, 5i – j + lk and 7i – 4j + 7k is 11 cubic units if the value of l is (a) –1 (c) –7

(b) 1 (d) 5

21. The vectors (x, x + 1, x + 2), (x + 3, x + 4, x + 5) and (x + 6, x + 7, x+ 8) are coplanar for

i + t(i + 2j + k) j + t(i + 2j + k) k + t(i + 2j + k) i – k + t(i + 2j + k)

15. The vectors a, b and c are equal in length and taken pairwise, make equal angles. If a = i + j, b = j + k and c make an obtuse angle with the base vector i, then c is equal to (a) i + k

(b) –i + 4j – k

1 4 1 (c) - i + j - k 3 3 3

(d)

1 4 1 i + j- k 3 3 3

16. The vectors AB = 3i – 2j + 2k and BC = – i – 2k are the adjacent sides of a parallelogram. An angle between its diagonals is (a) p 4

(b) p 3

(c) p 2

(d) 2p 3

17. Let the unit vectors a and b be perpendicular and the unit vector c be inclined at an angle q to both a and b. If c = aa + b b + g (a ¥ b) then (a) a = 2b (b) g 2 = 1 + 2a 2 1 + cos 2 q (c) g 2 = cos 2q (d) b 2 = 2 18. If the unit vectors a and b are inclined at an angle 2q and |a – b| < 1, then if 0 £ q £ p, q lies in the interval

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

only finite number of values of x x 0. The area of the quadrilateral ABCD is 1 (m + p) |b × d| (b) (m + p) |b × d| 2 1 (c) 2(m + p) |b × d| (d) | m – p| |b × d| 2 (a)

80. If u = a + b and v = a – b and | a| = | b| = k then |u ¥ v | is equal to (a) 2(k2 – (a . b)2)

(b) 2(k4 – (a . b)2)1/2

(c) (k4 + (a . b)2)1/2

(d) (k4 + (a . b)2)1/2

Previous Years' AIEEE/JEE Main Questions 1. If | a | = 4, | b | = 2 and the angle between a and b p then (a × b)2 is equal to is 6 (a) 48 (b) 16 (c) 9 (d) none of these [2002] 2. If a, b, c are vectors such that [a b c] = 4 then [a × b b × c c × a] =

(a) 16 (b) 64 (c) 4 (d) 8 3. If a, b, c are vectors such that a + b + c | a | = 7, | b | = 5, | c | = 3 then the angle vector b and c is (a) 60° (b) 30° (c) 45° (d) 90°

[2002] = 0 and between

[2002]

Vectors

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

4. If | a | = 5, | b | = 4, | c | = 3 then the value of ( a · b + b · c + c · a ) given that a + b + c = 0 (a) 25 (c) –25

(b) 50 (d) –50

[2002]

5. a = 3i – 5j and b = 6i + 3j are two vectors and c is a vector such that c = a × b then | a | : | b | : | c | (a)

34 : 45 : 39

(c) 34 : 39 : 45

(b)

[2002]

13. For any vector a, the value of (a ¥ i) + (a ¥ j)2 + (a ¥ k)2 is equal to

14.

6. Let u = i + j, v = i – j and w = i + 2j + 3k. If n is a unit vector such that u.n = 0 then |w.n| is equal to (a) 1 (b) 2 (c) 3 (d) 0 [2003] 7. A particle acted by constant forces 4i + j – 3k and 3i + j – k is displaced from the point i + 2j + 3k to the point 5i + 4j + k. The total work done by the forces is (a) 30 units (b) 40 units (c) 50 units (d) 20 units [2003, 04] 8. The vector AB = 3i + 4k and AC = 5i – 2j + 4k are the sides of a triangle ABC. The length of the median through A is (a)

72

(b)

33

(c)

288

(d)

18

[2003]

9. If a, b, c are non-coplanar vectors and l is a real number, then the vectors a + 2b + 3c, lb + 4c and (2l – 1)c are non-coplanar for (a) (b) (c) (d)

all except two values of l all except one value of l for all values of l no value of l

[2004]

10. Let u, v, w be such that | u | = 1, | v | = 2, |w| = 3. If the projection v along u is equal to that of w along u and v, w are perpendicular to each other than | u – v + w | equals (a) 14 (c) 2

(b) 7 (d) 14

[2004]

11. Let a, b and c be non-zero vectors such that 1 | b | | c | a. If q is acute angle (a ¥ b) ¥ c = 3 between the vectors b and c, then sin q equals (a) 2/3

(b)

(c) 1/3

(d) 2 2 / 3

2 /3 [2004]

12. If C is the mid point of AB and P is any point outside AB, then

[2005] 2

34 : 45 : 39

(d) 39 : 35 : 34

PA + PB + 2PC = 0 PA + PB + PC = 0 PA + PB = 2PC PA + PB = PC

22.23

15.

16.

17.

(a) 2a2 (b) 4a2 2 (c) 3a (d) a2 [2005] Let a, b and c be distinct non-negative numbers. If the vectors ai + aj + ck, i + k and ci + cj + bk lie in a plane, then c is (a) equal to zero (b) the harmonic mean of a and b (c) the geometric mean of a and b (d) the arithmetic mean of a and b. [2005] If a, b, c are non-coplanar vectors and l is a real number then [l(a + b) l2b lc] = [ a b + c b] for (a) exactly three values of l (b) exactly two values of l (c) exactly one value of l (d) no value of l [2005] Let a = i – k, b = xi + j + (1 – x)k, c = yi + xj + (1 + x – y)k then [a b c] depends on (a) both x and y (b) neither on x nor on y (c) only y (d) only x [2005] If (a ¥ b) ¥ c = a ¥ (b ¥ c), where a, b and c are any three vectors such that a, b π 0, b . c π 0, then a and c are (a) parallel (b) inclined at an angle of p/3 between then (c) inclined at an angle of p/6 between them (d) perpendicular [2006]

18. The value of a, for which the points A, B, C with position vectors 2i – j + k, i – 3j – 5k and ai – 3j + k respectively are vertices of a right-angled triangle with C = p/2 are (a) 2 and – 1 (b) 2 and 1 (c) – 2 and – 1 (d) – 2 and 1 [2007] 19. 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 equals (a) 0 (b) 1 [2007] (c) – 4 (d) – 2 20. If u and v are unit vectors and q is the angle between them, then 2u ¥ 3v is a unit vector for (a) exactly two values of q (b) more than two value of q (c) no value of q (d) Exactly one value of q [2007]

22.24

Complete Mathematics—JEE Main

21. The vector a = a i + 2j + b k lies in the plane of the vectors b = i + j and c = j + k and bisects the angle between b and c. Then which one of the following gives possible values of a and b (a) a = 2, b = 2 (b) a = 1, b = 2 (c) a = 2, b = 1 (d) a = 1, b = 1 [2008] 22. The non-zero vectors a, b and c are related by a = 8b and c = – 7b. Then the angle between a and c is (a) 0 (b) p/4 (c) p/2 (d) p [2008] 23. If u, v, w are non-coplanar vectors and p, q are real numbers, then the equality [3u pv pw] – [pv w qu] – [2w qv qu] holds for (a) more than two but not all values of (p, q) (b) all value of (p, q) (c) exactly one value of (p, q) (d) exactly two values of (p, q) [2009] 24. If the vectors a = i – j + 2k, b = 2i + 4j + k and c = li + j + mk are mutually orthogonal, then (l, m) = (a) (– 2, 3) (c) (– 3, 2)

(b) (3, – 2 ) (d) (2, – 3)

[2010]

25. Let a = j – k and c = i – j – k. Then the vector b satisfying a ¥ b + c = 0 and a . b = 3 is (a) i – j – 2k (c) – i + j – 2k

(b) i + j – 2k (d) 2i – j + 2k

[2010]

1

1 (3i + k) and b = (2i + 3j – 6k), 26. If a = 7 10 then the value of (2a – b) . [(a ¥ b) ¥ (a + 2b)] is (a) 3 (c) –3

(b) –5 (d) 5

Êb ◊ c ˆ (b) b – Ë c a◊b¯

Êa ◊ c ˆ (c) c + Ë b a ◊b¯

Êb ◊ c ˆ (d) b + Ë c a◊b¯

[2011]

[2011]

28. If the vectors pi + j + k, i + qj + k and i + j + rk (p π q π r π 1) are coplanar, then the value of pqr – (p + q + r) is (a) 2 (c) –1

(b) 0 (d) – 2

(a) a (c) 0

[2011]

29. Let a, b, c be three non-zero vectors which are pairwise non-collinear. If a + 3b is collinear with

(b) c (d) a + c

[2011]

30. Let a and b be two unit vectors. If the vectors c = a + 2b and d = 5a – 4b are perpendicular to each other, then the angle between a and b is (a) p/2 (c) p/4

(b) p/3 (d) p/6

[2012]

31. Let ABCD be a parallelogram such that AB = q, AD = p and –BAD be an acute. If r is the vector that coincides with the attitude directed from the vertex B to the side AD, then r is given by p ◊ qˆ p (a) r = – q + ÊÁ Ë p ◊ p ˜¯ p ◊ qˆ p (b) r = q – ÊÁ Ë p ◊ p ˜¯ p ◊ qˆ p (c) r = – 3q + 3 ÊÁ Ë p ◊ p ˜¯ p ◊ qˆ p (d) r = 3q – 3 ÊÁ Ë p ◊ p ˜¯

[2012]

32. If the vectors AB = 3i + 4k and AC = 5i – 2j + 4k are the sides of a triangle ABC, then the length of the median through A is (a)

72

(b)

33

(c)

45

(d)

18

[2013]

33. If a and b are non-collinear vectors, then the value of a for which the vectors u = (a – 2) a + b and v = (2 + 3a) a – 3b are collinear is: (a)

27. The vectors a and b are not perpendicular and c and d are vectors satisfying b ¥ c = b ¥ d and a . d = 0. Then the vector d is equal to Êa ◊ c ˆ b (a) c – Ë a◊b¯

c and b + 2c is collinear with a, then a + 3b + 6c is

3 2

(c) -

(b) 3 2

2 3

(d) -

2 3

[2013, online]

34. Let a = 2i – j + k, b = i + 2j – k and c = i + j – 2k be three vectors. A vector of the type b + lc for some scalar l, whose projection on a is of 2 is magnitude 3 (a) 2i + j + 5k (b) 2i + 3j – 3k (c) 2i – j + 5k (d) 2i + 3j + 5k [2013, online] 35. If [a × b to (a) 2 (c) 0

b×c

c × a] = l [a b c]2 then l is equal (b) 3 (d) 1

[2014]

Vectors

36. If | a | = 2, | b | = 3 and | 2a – b | = 5, then | 2a + b | equals (a) 17 (b) 7 (c) 5 (d) 1 [2014, online] 37. If | c |2 = 60 and c × (i + j + 5k) = 0, then a value of c·(–7i + 2j + 3k) is: (b) 12 (a) 4 2 [2014, online] (c) 24 (d) 12 2 38. If x, y and z are three unit vectors in three dimensional space, then the minimum value of | x + y |2 + | y + z |2 + | z + x |2 is 3 (a) (b) 3 2 (d) 6 [2014, online] (c) 3 3 39. If x = 3i – 6j – k, y = i + 4j – 3k and z = 3i – 4j – 12k, then the magnitude of the projection of x × y on z is (a) 12 (c) 14

(b) 15 (d) 13

[2014, online]

40. Let a, b and c be three non zero vectors such that 1 no two of them are collinear and (a ¥ b) ¥ c = 3 |b| |c| a. If q is the angle between vectors b and c, then a value of sin q is: 2 2 - 2 (b) 3 3 2 -2 3 (c) (d) [2015] 3 3 41. In a parallelogram ABCD, |AB| = a, |AD| = b and |AC| = c, then DB. AB has the value: (choices (a)

1 3a 2 + b 2 - c 2 2 1 2 (c) b + c2 - a2 3 (a)

(

(

)

)

1 2 a + b2 - c2 4 1 2 (d) a + b2 + c2 2 [2015, online] (b)

(

)

(

)

22.25

42. Let a and b be two unit vectors such that |a + b| =

3 . If

c = a + 2b + 3(a ¥ b), then 2|c| is equal to: (a)

55

(b)

51

(c)

43

(d)

37

[2015, online]

43. Let a, b and c be three unit vectors such that a ¥ (b ¥ c) =

3 (b ¥ c). 2

If b is not parallel to c then the angle between a and b is: 3p p (b) (a) 4 2 (c)

2p 3

(d)

5p 6

[2016]

44. In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3i + j – k, – i + 3j + pk and 5i + q j – 4k, then the point (p, q) lies on a line: (a) making an obtuse angle with the positive direction of x-axis. (b) parallel to x-axis. (c) parallel to y-axis. (d) making an acute angle with the positive direction of x-axis. [2016, online] 45. Let ABC be a triangle whose circumcentre is at P. If the positive vectors of A, B, C and P are a, b, 1 c and (a + b + c) respectively, then the position 4 vector of the orthocentre of this triangle, is: 1 (a) - (a + b + c) (b) a + b + c 2 1 (a + b + c) (d) 0 [2016, online] (c) 2

Previous Years' B-Architecture Entrance Examination Questions 1. Let u, v, w be vectors such that u+v+w=0 If |u| = 3, |v| = 4 and |w| = 5, then u·v + v·w + w·u is (a) –25 (b) 0 (c) 25 (d) 47

2. If a and b are two non-parallel vectors having equal magnitude, then the vector (a – b) × (a × b) is parallel to (a) b (b) a – b (c) a + b (d) a [2007] [2006]

22.26

Complete Mathematics—JEE Main

3. If numbers a, b, c are distinct and non-negative. If three vectors ai + aj + ck, ci + cj + bk and i + k are coplanar, then c is (a) geometric mean of a, b (b) harmonic mean of a, b (c) equal to zero (d) arithmetic mean of a, b [2007] 4. Let x, y and z be unit vectors such that | x – y |2 + | y – z |2 + | z – x |2 = 9 Then | x + y – z |2 – 4 x . y = (a) 1 (b) 4 (c) 6 (d) 8 [2008] 5. If a, b and c are three unit vectors satisfying 2a × (a × b) + c = 0 then the acute angle between a and b is p p (b) (a) 5 4 p p (d) [2009] 3 6 6. If b = i – j + 3k, c = j + 2k and a is a unit vector, then the maximum value of the scalar triple product [a b c] is (c)

(a)

30

(b)

29

(c)

26

(d)

60

[2009]

Statement-2: For any two vectors c and d, | c – d | = |c + d| [2010] 8. If a, b and c are non-zero vectors such that a × b = c, b × c = a and c × a = b then (a) [a b c] = 0 (b) a = b = c (c) | a | = | b | = | c | (d) | a | + | b | – | c | = 0 [2010] 9. Let OA = a, OB = 2b + 10a and OC = b where O is the origin. If p is the area of the quadrilateral OABC and q is the area of the parallelogram with OA and OC as adjacent sides then p is equal to (b) 6q (a) q6 (c) q/6 (d) 6 – q [2011] 10. If a and b are two vectors such that 2a + b = e1 and a + 2b = e2, where e1 = (1, 1, 1) and e2 = (1, 1, –1), then the angle between a and b is Ê7ˆ Ê 7ˆ (b) cos-1 Ë ¯ (a) cos-1 Ë ¯ 9 11 Ê 7ˆ (d) cos-1 Ë - ¯ 9

(a)

6

10 + 6

(b)

(c) 59 (d) 60 [2013] 13. Unit vectors a, b, c are coplanar. A unit vector d is perpendicular to them. If 1 1 1 (a × b) × (c × d) = i - j + k 6 3 3 and the angle between a and b is 30°, then c is/are 1 1 (a) ± (-i - 2 j + 2k ) (b) (2i + j - k ) 3 3 1 1 (c) ± (-i + 2 j - 2k ) (d) (-2i - 2 j + k ) [2014] 3 3 14. Let x = 2i + j – 2k and y = i + j. If z is a vector

7. Statement-1: If a and b are two vectors such that | a | = 2, | b | = 3, | 2a – b | = 5 then | 2a + b | = 5

Ê 7ˆ (c) cos-1 Ë- ¯ 11

11. If u, v, w are unit vectors satisfying 2u + 2v + 3w = 0, then | u – v | equals 7 5 (b) (a) 4 2 7 5 (c) (d) [2012] 2 4 12. Let v = 2i + j – k and w = i + 3k. If u is a unit vector, then the maximum value of the scalar triple product [u v w] is

[2012]

such that x.z = |z|, |z – x| = 2 2 and the angle between x ¥ y and z is 30°, then the magnitude of the vector (x ¥ y) ¥ z is: (a)

3 2

(b)

3 2

1 3 3 (d) [2015] 2 2 15. From a point A with position vector p(i + j + k), AB and AC are drawn perpendicular to the lines r = k + l(i + j) and r = – k + m(i – j) respectively. A value of p is equal to (a) –1 (b) 2 (c) 2 (d) –2 [2016] (c)

Answers Concept Based 1. (d) 5. (d) 9. (a)

2. (b) 6. (b) 10. (b)

3. (d) 7. (c)

4. (b) 8. (d)

12. (b) 16. (a) 20. (b)

13. (a) 17. (d) 21. (b)

14. (b) 18. (a) 22. (b)

Level 1 11. (a) 15. (c) 19. (a)

Vectors

23. 27. 31. 35. 39. 43. 47.

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

24. 28. 32. 36. 40. 44. 48.

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

25. 29. 33. 37. 41. 45.

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

26. 30. 34. 38. 42. 46.

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

49. 53. 57. 61. 65. 69. 73. 77.

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

2. (3i + 4j) · (2j – 5k) = 8, |3i + 4j| = 5, 29 . Hence the required angle is

| 2 j – 5k | = cos-1

a◊b 8 = cos-1 . a b 5 29

3. c = la + mb = (l + m)i + (l + 2m)j + lk 0 = a.c = l |a|2 + m a.b = 3l + 3m

Level 2

fi

l+m=0

2

50. 54. 58. 62. 66. 70. 74. 78.

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

51. 55. 59. 63. 67. 71. 75. 79.

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

52. 56. 60. 64. 68. 72. 76. 80.

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

Previous Years‘ AIEEE/JEE Main Questions 1. (b)

2. (a)

3. (a)

4. (c)

5. (b)

6. (c)

7. (b)

8. (b)

9. (a)

10. (a)

11. (d)

12. (c)

13. (a)

14. (c)

15. (d)

16. (b)

17. (a)

18. (b)

19. (d)

20. (d)

21. (d)

22. (d)

23. (c)

24. (c)

25. (c)

26. (b)

27. (a)

28. (d)

29. (c)

30. (b)

31. (a)

32. (b)

33. (b)

34. (b)

35. (d)

36. (c)

37. (d)

38. (b)

39. (c)

40. (a)

41. (a)

42. (a)

43. (d)

44. (d)

45. (c)

Previous Years' B-Architecture Entrance Examination Questions 1. (a)

2. (c)

3. (a)

4. (c)

5. (d)

6. (a)

7. (c)

8. (c)

9. (b)

10. (c)

11. (c)

12. (c)

13. (c)

14. (b)

15. (a) (b) (c) (d)

Hints and Solutions Concept Based 1. Let O be the origin of reference. Then the position vector of A, B, M are OA, OB and OM respectively 1 so OM = (OA + OB) 2

22.27

Also (l + m) + (l + 2m)2 + l2 = 1 fi

2l2 = 1

Hence

fi

l= ±

1

c = –lj + lk = ±

2 1 2

(- j + k )

4. b = la = l(2i + j – k). so l (2i + j – k) · (2i + j – k) = 2 1 fi 6l = 2 fi l = 3 1 Hence b = (2i + j - k ) 3 i j k 5. u × v = 1 1 -1 = 2i – 3j – k 2 1 1 so Projw u × v =

(u ¥ v) ◊ w

w w2 2-3-2 ( i + j + 2k ) = 6 1 = - ( i + j + 2k ) 2 1 3 | Projw u × v |2 = ◊ 6 = 4 2 6. u = t v fi (l + 2)a + b = t(1 + 4l)a – 2t b fi (l + 2 – t – 4t l)a = (–2t – 1)b Since a and b are non-collinear, so 1 t = - and l + 2 – t – 4 + l = 0 2 1 fi l + 2 - – 2l = 0 2 3 3 =0 fi l= . 2 2 7. AB = – i + j, BC = i + k 1 Area of triangle ABC = | AB × BC | 2 i j k 1 1 3 = | -1 1 0 | = i + j - k = 2 2 2 1 0 1 fi

-l +

22.28

Complete Mathematics—JEE Main

i j k 8. (3i + 4j) × (i – j + k) = 3 4 0 1 -1 1 = 4i – 3j – 7k The length of this vector

16 + 9 + 49 = 74 . A unit vector perpendicular to given vectors is 1 74

i j k 10. u = (i – j + k) × (2i – 3j – k) = 1 -1 1 2 -3 -1 = 4i + 3j – k v = (–3i + j + k) × (2j + k) = –i + 3j – 6k i j k u × v = 4 3 -1 = –15i + 25j + 15k -1 3 -6

Level 1 11. 0 = (a + mb) . (a – mb) = | a |2 – m2 | b |2 fi 9 – 16 m2 = 0 fi m = ± 3/4. -1 2 -4

2 1 3

1 a 5

fi

15 + 5a = 0

fi a = – 3. 13. If p and q are adjacent sides of the parallelogram, their sum gives one of the diagonals and their difference gives the other that is p + q = 3i + j – 2k and p – q = i – 3j + 4k fip = 2i – j + k and q = i + 2k – 3k Required area = | p ¥ q | = | i(3 – 2) – j(– 6 – 1) + k(4 + 1) | =

But a. b = 1, a . c = c1 + c2 and b . c = c2 + c3 This gives c1 + c2 = 1 and c2 + c3 = 1 fi c3 = c1

(4i – 3j – 7k)

1 -2 3 9. 2 1 -1 = 1 × 2 – 2 × (–2) + 3 × 2 0 1 1 = 2 + 4 + 6 = 12

12. 0 =

15. | a | = 2 = | b | = | c |. Let c = c1i + c2j + c3k, since c makes an obtuse angle with i, we must have c . i = c1 < 0. It is also given that the angles between the vectors are equal i.e., a◊b 1 a◊c b◊c = cos -1 = cos–1 = cos -1 cos–1 2 a b a c b c

12 + 7 2 + 52 = 5 3 .

14. Put r = xi + yj + zk. The given equation implies (y – 2z)i + (z – x)j + (2x – y)k = i – k That is y – 2z = 1, z – x = 0 and 2x – y = – 1, from which we get x = z = t (say), so that y = 1 + 2t. Substituting these, we get r = ti + (1 + 2t)j + tk = j + t(i + 2j + k)

and c2 = 1 – c1. Putting in c12 + c22 + c32 = 2, we get c1 = 1, – 1/3. Since c1 < 0 so c1 = – 1/3 = c3 and c2 = 4/3. Hence 1 4 1 c= - i + j- k . 3 3 3 16. The two diagonals are AB – BC = 4i – 2j + 4k, AB + BC = 2i – 2j These vectors have magnitudes 6 and 2 2 respectively, and their dot product is 12. Therefore, the angle between them is 12 1 p = . = cos–1 cos–1 (6) (2 2 ) 2 4 17. | a | = | b | = 1 = | c |. It is also given that the angle q between a and c equals that between b and c a◊c

= a . c = cos q =

b◊c

=b.c a c b c Since a . b = 0, we get from the given value of c, a . c = a a . a + b a . b + va . (a ¥ b) = a i.e., a = cos q, and similarly b.c = cos q = b 1 = c . c = 2a2 + g 2 | a ¥ b |2 = 2a2 + g2[| a |2 | b |2 – (a.b)2] = 2a2 + g2 fi g 2 = 1 – 2a2 = 1 – 2cos2 q = – cos2q fi a2 = b2 =

1 - g 2 1 + cos 2 q . = 2 2

18. | a – b |2 = a2 + b2 – 2a.b = 4 sin2 q fi | a – b | = 2 | sin q |. | a – b| < 1 fi | sin q | < 1/2 5p fi q Œ [0, p/6] or Ê , p ˆ . Ë6 ¯ 19. If q is angle between a and b, and f the angle between c and a ¥ b. Then | (a ¥ b) . c | = | a | | b | | c | sin q cos f So we must have sin q cos f = 1 fi sin q = 1, cos f = 1 fi q = p/2, f = 0 fi a and b are perpendicular so a . b = 0. f = 0 fi c must be perpendicular to both a and b, so that b . c = 0 = a . c. 20. Let A, B, C and D be the points with the given position vectors. Then AB = – 2i + 3j – 3k, AC = 4i + 5j + (l – 10)k and AD = 6i + 2j – 3k.

Vectors

The volume of the tetrahedron is 1 | (AB ¥ AC) . AD | 6 -2 3 -3 1 | 4 5 l - 10 | = 6 6 2 -3

22.29

29. The given expression = (i . i)a – (i . a)i + (j . j) a – (j . a)j + (k . k)a – (k . a)k = 3a – [(i . a)i + (j . a)j + (k . a)k] = 3a – a = 2a 30. [ a ¥ b

b¥c

c ¥ a]

= ((a ¥ b) ¥ (b ¥ c)). (c ¥ a)

1 | – 88 + 22 l | = 11 (given) 6 fi l = 1 or 7 x x +1 x + 2 x x +1 x + 2 3 3 3 =0 21. x + 3 x + 4 x + 5 = x+6 x+7 x+8 6 6 6 =

for all x so for x < 0. 22. Required vector = K(2j – k) ¥ (– i + 2j – 3k)

= ((a ¥ b . c)b – ((a ¥ b) . b) c). (c ¥ a) = [a b c] (b . c ¥ a) = [a b c]2. 31. A vector c perpendicular to z-axis is of the form c1 i + c2 j. According to given conditions 3c1 – c2 = 9 and c1 + 2c2 = – 4 fi c2 = 2 and c2 = – 3. Thus required point is (2, – 3, 0). 32. Required area = | (a + 3b) ¥ (3a + b) | = | 9(b ¥ a) + a ¥ b | = 8 | a ¥ b | = 8 | a | | b | sin p /6 = 4.

= K(– 4i + j + 2k). The length of this vector is 21 | K | so | K | = 1 / 3 . Thus K = ± 1 / 3 . But it will make obtuse angle with y-axis if K = – 1 / 3.

33.

x 1 1

5 1 2

7 - 1 = 0 fi x = 2. 2

34. a . (b ¥ c) = – a . (c ¥ b) so a . (c ¥ b) = – 3.

Thus the required vector is 1 / 3 (4i – j – 2k)

35. If b = b1 i + b2 j + b3 k, we have the following

23. | p ¥ q | = | (a + b) ¥ (a – b) | = 2 | a ¥ b | 4 | a |2 | b |2 cos2 q = 4 (K – | a |2 | b |2 cos2 q) fi | a |2 | b |2 = K fi K = 16. 24. The given conditions mean that r is perpendicular to all three vectors a, b and c. This is possible only if they are coplanar which means[a b c] = 0 25. Let A, B and C be the three given vertices and D the fourth vertex, with the position vector xi + yj + zk. Since ABCD is a parallelogram, the diagonals AC and BD bisect each other so (1 + 7)i + (1 + 9) j + (1 + 11)k 2 (1 + x)i + (3 + y ) j + (5 + z )k 2 fi x = 7, y = 7 and z = 7, so that D is the point 7 (i + j + k). =

relations 2b1 + 2b2 + b3 = 14 and a ¥ b = (2b3 – b2)i + (b1 – 2b3)d + 2(b2 – b1)k. So 2b3 – b2 = 3, 1 = b1 – 2b3, b2 – b1 = – 4. Solving b1 = 5, b2 = 1, b3 = 2. 36. Let AB = a, BC = b, then FC = 2a, AD = 2b. Also, DC = AC – AD = a + b – 2b =a–b fi EB = 2 DC = 2(a – b) Now, AD + EB+ FC = 2b + 2(a – b) + 2a = 4a = 4 AB = 4 ED \ l=4 D

E

26. Required volume = (OA ¥ OB) . OC =

2 1 3

-3 0 1 -1 = 4 . 0 -1

F

C ®

b 2

2

27. If a = a1i + a2j + a3k then | a ¥ i | = a2 + a3

2

Therefore | a ¥ i |2 + | a ¥ j |2 + | a ¥ k |2 = 2(a12 + a22 + a32) = 2a2. 28. a ¥ (b ¥ c) = (a . c)b – (a . b)c and (a ¥ b) ¥ c = (a . c)b – (b . c)a, so the given equality implies (a . b) c = (b . c) a which means that a and c are collinear.

A

a®

B

Fig. 22.6

37. Let N1 be a vector normal to plane determined by vectors i, i + k and N2 be a vector normal to the plane determined by vectors i – j, i + k we have N1 = i ¥ (i + j) = k

22.30

Complete Mathematics—JEE Main

and N2 = (i – j) ¥ ( i +k) = – i – j +k Note that a is parallel to N1 ¥ N2 = i – j If q is the angle between a and i – 2j + 2k, then cosq = ±

1+ 2 2 1+ 4 + 4

=±

So the angle between a and c is p/3 fi the angle between b and d is p/3 \ b and d are nonparallel. n1 = n2

1 2

d

Thus, q may be taken as p/4. 38. Let M be the mid point of PR, then position vector of M is 1 1 (-2i - j + 3i + 3j) = i + j 2 2 Let N be the mid point of QS, then position vector of N is

/3 b

1 1 (4i - 3i + 2 j) = i + j 2 2 Thus, PQRS is a parallelogram. PQ = PQ = 4i - ( - 2i - j)

c

Next

= |6i + j| = and

QR = |QR| = |3i + 3j – 4i| = |– i + 3j| =

a

37

1 + 9 = 10

Thus, PQRS is not a rhombus Also, PR = |PR| = |3i + 3j – (– 2i – j)|

Fig. 22.7

40. If V is volume of the parallelopiped, then a◊a a◊b a◊c 2 2 V = [abc] = b ◊ a b ◊ b b ◊ c c◊a c◊b c◊c

= 5i + 4 j = 25 + 16 = 41

and

OS =

1 1/ 2 1/ 2 1 1/ 2 1/ 2 = 1/ 2 1 1/ 2 = 2 1 1 1/ 2 1/ 2 1/ 2 1 1 1/ 2 1

|OS| = |– 3i + 2j – (– 4i)| = i + 2j = 1 + 4 = 5 .

\ PQRS is not a rectangle. Thus, PQRS is a parallelogram which is neither a rhombus nor a rectangle. 39. Let q1 be the angle between a and b and q2 be the angle between c and d. Now, a ¥ b= |a| |b| sin q1 n1 = sin q1 n1 and c ¥ b = |c| |d| sin q2 n2 = sin q2 n2 where n1 is a unit vector parallelogram to the plane of a and b; and n2 is a unit vector perpendicular to the plane of c and d. As (a ¥ b) ◊ (c ¥ d) = 1, we get (sin q1) (sin q2) n1 ◊ n2 = 1 (sin q1) (sin q2) |n1| |n2 | cos f = 1 where f is angle between n1 and n2. fi (sin q1) (sin q2) = cos f = 1 fi q1 = p/2, q2 = p/2 and f = 0 As f = 0 we get n1 and n2 are parallel. Therefore, a, b, c and d are coplanar. Also a. c = 1/2 fi |a| |c| cosq = 1/2,

[C1 Æ C1 + C2 + C3] 1 1/ 2 1/ 2 1 = 2 0 1/ 2 0 = 2 0 0 1/ 2 fi V = 1/ 2 a 2a -3a 41. 2a + 1 2a + 3 a + 1 3a + 5 a + 5 a + 2 1 2 -3 = a 2a + 1 2a + 3 4 + 1 3a + 5 a + 5 a + 2 =a

1 0 0 2a + 1 1 - 2a 7a + 4 3a + 5 -5a - 5 10a + 17

[C2 Æ C2 – 2C1, C3 Æ C3 + 3C1] = a (15a +31a +37) = 0 if and only if a = 0 as 15a2 + 31a + 37 > 0 for any a. 2

Vectors

42. (a ¥ b) ¥ (c ¥ d) = (a ◊ (c ¥ d))b – (b ◊ (c ¥ d)) a = pa + qb So p = – (b ◊ (c ¥ d)) = b ◊ (d ¥ c) = [b d c] = [c b d]. 43. |a| = |b| = |c| = 1. Also cos q = a ◊ c = b ◊ c c = aa + bb + g (a ¥ b) so a ◊ c = a + b(a ◊ b) = a b ◊ c = a(a ◊ b) + b = b Thus a = a ◊ c = cos q = b ◊ c = b. 44. (a ¥ b) . (c ¥ d) = ((a ¥ b) ¥ c) . d = ((a . c) b – (b . c) a) . d = (a . c) (b . d) – (b . c) (a . d) 45. [a ¥ b, b ¥ c, c ¥ a] = [a b c]2 46. | a + b |2 – | a – b |2 = | a |2 + | b |2 + 2a . b – (| a |2 + | b |2 – 2a . b) = 4a . b 47. (i – j + k) ¥ (i + 2j – k) = – i + 2j + 3k 48. [a + b b + c c + a] = 2[a b c].

Level 2 49. Suppose that l, m, n are direction cosine of a line and the cube be a unit cube. The direction ratios of the diagonals will 1,1,1; 1, –1; 1, 1,1; –1,1,1. Hence

=

1◊ 3

+

(l - m + n)2 1◊ 3

+

8 +1+ 6

(

16 + 1 + 4

2

)

(4i – j + 2k)

1 [(14i – 7i + 21k) – 20i + 5j – 10k] 7 1 = (– 6i – 2 j + 11k) 7

=

=–

1 (6i – 2 j – 11k). 7

54. (a ¥ b) ¥ (c ¥ d) = [a b d]c – [a b c]d a, b, c, d lie in the same plane so [a b d] = 0 and [a b c] = O so (a ¥ b) ¥ (c ¥ d) = 0. a◊c b◊c 55. (a ¥ b). (c ¥ d) = a ◊ d b ◊d = (a · c) (b · d) – (a · d) (b · c) so k = –1. 56. Any point on the given line is of the form B(3, 7 + t, 1 + t). The given line is parallel to the vector j + k. The direction ratio of AB will be 3 – 5, 6 + t, –2 + t where A is (5, 1, 3). If AB is perpendicular to the given line then – 2. 0 + (6 + t). 1 + (– 2 + t) 1 = 0 fit=–2

cos2a + cos2b + cos2g + cos2d

(l + m + n)2

= (2i – j + 3k) –

22.31

(-l + m + n)2

(3 - 5)2 + (5 - 1)2 + (-1 - 3)2

|AB| =

1◊ 3

=

4 + 16 + 16 = 6.

2

+

(l + m - n )

57. The vector equation of two lines are

1◊ 3

1 = [4(l2 + m2 + n2) + 2(lm + mn + ln – lm + 3 ln – mn + lm – ln – mn – lm + mn – ln)] 4 = . 3 50. (a ¥ b) ¥ (c ¥ d) = ((a ¥ b). d) c – ((a ¥ b)·c) d

r = i + 2j + 3k + t (–4 i + j + 2k) and r = i + 4j – k + s (i – 2j + k) so distance between them is (Ch. 18, P 6, formula 13) 1 4 -1 1 2 3 -4 1 2 - -4 1 2 1 -2 1 1 -2 1 i j k -4 1 2 1 1 -2 1

= [a b d]c – [a b c]d so, k = – [a b c] = [b a c]. 51.

1 p c◊d = = cos = 2 3 |c||d |

x + x +1 2

2

x +2 x +2 2 fi x + 2 = 4x + 2 fi x = 0, 4.

=

2x + 1 x2 + 2

52. For t = 3, the line intersect xy plane at (–2, 10, 0). 53. The vector component of u orthogonal to a is u.a a u– | a |2

=

=

22 - 38 5i + 6 j + 7k 16 110

.

58. The vector AC is given by r = i – 2j + 2k + t (–5i + 3j – k)

22.32

Complete Mathematics—JEE Main

Any point M on AC is of the form

a·v = 0 fi a·b + t a·a = 0

(1 – 5t, –2 + 3t, 2 – t)

fit=–

d.r. of BM are – 5t, – 6 + 3t, 2 – t. Since BM is an altitude so –5 (– 5t) + 3(– 6 + 3t) – 1(2 – t) = 0 4 fit= 7 Ê - 13 - 2 10 ˆ , , ˜ The point M is given by Á Ë 7 7 7¯ BM is given by

- 20 - 30 10 i j+ k . 7 7 7

=–

a◊b a

2

7 14 7

2+2+3

=-

=-

1+ 4 + 9 4 +1+1

1 2

.

66. For a = a1 i + a2 j + a3 k, we have a ¥ (i ¥ j) = (a · i)j – (a ·j)i |a ¥ (i ¥ j)|2 = (a · i)2 + (a · j)2 = a12 + a22 |a ¥ (j ¥ k)|2 = a22 + a32 |a ¥ (k ¥ i)|2 = a32 + a12 so the required value = 2(a12 + a22 + a32) = 2|a|2.

59. h = 2h + k, 3 – 4h + l = k, 1 + h + k = l fi k = – h so 3 – 3h + l = 0, l = 1

locus passes through

fi h = 4/3, k = – 4/3, l = 1. 60. For a unit cube d.r. of diagonal are 1, 1, 1 and 1, – 1, 1 so the angle between two diagonals will be 1-1+1 1 = cos-1 . cos–1 3 3 3 61. Since (a + b) ¥ a = b ¥ a and (a + b) ¥ b = a ¥ b so the intersection is a + b. 62. (a ¥ b) ¥ (c ¥ d) = [a b d] c –[a b c]d a1 = a2 a3

b1 b2 b3

d1 a1 d2 c - a2 d3 a3

b1 b2 b3

c1 c2 d c3

Also (a ¥ b) ¥ (c ¥ d) = [a c d]b – [b c d]a a1 = a2 a3

c1 c2 c3

d1 b1 d2 b - b2 d3 b3

c1 c2 c3

d1 d2 a d3

Adding, we have k = 2. 63. (b ¥ c). (a ¥ b). b◊a = b◊d

67. The mid point is given by

(a ¥ d) + (c ¥ a). (b ¥ d) + (c ¥ d). c◊a c◊b a◊b a◊c b◊c + + c◊d c◊d a◊d a◊d b◊d

= (b · a) (c · d) – (b · d) (c · a) + (c · b) (a · d) – (c · d) (a · b) + (a · c) (b · d) – (a · d) (b · c) = 0. 64. Equating the two equations we get l = 0 and m = –1 so the point of intersection is b – 2c. 65. a ¥ v = a ¥ b fi a ¥ (v – b) = 0 fiv–b=tafiv=b+ta

1 (a + b). The required 2

1 (a + b) so required locus 2

is of the form 1 (a + b) + t c 2 where c is perpendicular to b – a. Thus

r=

(i)

c . (b – a) = 0 1 Taking dot product in (i) by a – b, we get (r – 2 (a + b)). (a – b) = 0. 68. Since a is perpendicular to d, so x – y + 2z = 0

(i)

Moreover, |b| = |c| so a · b = a · c as a makes equal angles with b and c. Thus xy – 2yz + 3xz = 2xz + 3xy – yz fi xz – 2xy – yz = 0

(ii)

Also x2 + y2 + z2 = 12

(iii)

and y < 0. Substituting the value of y from (i) in (ii) we get x2 + 2xz + z2 = 0 So x = – z and y = z Again substituting these values in (iii) we get z2 = 4 i.e. z = ± 2 but y < 0 and y = z so z = – 2 = y and x = 2. 69. a ¥ b = c fi c is perpendicular to a and b. b ¥ c = a fi a is perpendicular to b and c Thus a, b, c are orthogonal in pairs. Since

Vectors

|c| = |a ¥ b| = |a| |b| and similarly |a| = |b| |c| so |c| = |c| |b|2 Hence |b| = 1 and |c| = |a|.

r = – 2i + 3j – 4k + t (2i – 4i + k) r = – 2i + 3j – 4k + t (– i + 2j + 3k) -2-8+3 4 + 16 + 1 1 + 4 + 9

-7

=

21 14

=-

1 6

Ê 1 ˆ –1 1 Acute angle = p – cos–1 Á . ˜¯ = cos Ë 6 6 71. According to the given conditions, c12 + c32 + c32 = 1, a · c = 0, b · c = 0 and a1b1 + a2 b2 + a3 b3

p 3 = cos = 2 6

c ◊ a = 9 fi 3c1 - 3c2 = 9 ˘ fi c1 = 2, c2 = - 3 c ◊ b = - 4 fi c1 + 2c2 = - 4 ˙˚

 ai2  bi2

76. A unit vector in XZ plane is of the form c = c1 i + c3 k where c12 + c32 = 1 1 p 2c - c 3 Also = cos = 1 3 fi = 2c1 - c3 4 2 2 9 1 p -c = cos = 3 fi c3 = 2 3 2 1 so c1 = . Hence c = 2

3 = 2

a1 a2 a3

b1 b2 b3

2

c1 c2 c3

=

 ai2  ai bi  ai ci  ai bi  bi2  bi ci  ai ci  bi ci  ci2

  ai bi  ai bi  bi2 ai2

=

0 =

(

=

(Â ) (Â )

 ai2

)(

0

)

0 0 1

bi2

3 4

(Â ) (Â ) ai2

ai2

1 = Â ai2 Â bi2 . 4 72. Since the rotation of axes does not affect the distance between the origin and the point, we have

(

)(

2

(i – k).

a 2 + a ◊ b a ◊ b + | b |2 = |b| a

fi |a| +

a◊b a◊b +|b| = |b| |a|

Ê | a | - | b |ˆ fi |a| – |b| = a · b Á Ë | a | | b | ˜¯ If |a| π |b| then |a| |b| = a · b fi q = 0 not possible so |a| = |b|. 78. Treating A as the origin of reference. The position vector M is given by

2

 bi2 - ( ai bi )

ai2

2 1

a. (a + b) b. (a + b) = a a+b b a+b fi

 ai2  bi2 =  ai bi

1

77. According to the given condition

Thus a1c1 + a2c2 + a3c3 = 0 = b1c1 + b2c2 + b3c3 and

c · k = 0 fi c3 = 0

75. Let c = c1 i + c2 j + c3 k.

Thus c = 2i – 3j.

70. Given lines has vector form as

cosq =

22.33

)

4p2 + 1 = (p + 1)2 + 1 fi p + 1= ± 2p fi p = 1 or – 1/3. 73. a. ((b ¥ c) ¥ (a + (b ¥ c)) = a. ((b ¥ c) ¥ a) = a. ((a · b)c – (a · c)b) = 0. 74. X · A = 0, X · B = X · C = 0 fi A, B, C are perpendicular to X. Thus A, B, C are coplanar fi [A, B, C] = 0.

5-3 -2 + 4 8 i+ j + k = i + j + 4k. 2 2 2 The |AM| =

18 .

79. Area of the triangle ABC =

1 |b ¥ AC| 2

1 |b ¥ (mb + pd)| 2 1 = |p (b ¥ d)| 2 Similarly area of the triangle ACD 1 1 = d ¥ (mb + pd) = |m| |b ¥ d| 2 2 Area of the quadrilateral ABCD =

=

1 (|p| + |m| ) |b ¥ d|. 2

22.34

Complete Mathematics—JEE Main

80. |u ¥ v |2 = |(a ¥ b) ¥ (a – b)|2 = |b ¥ a – a ¥ b|2

= (2l – 1) [a + 2b + 3c lb + 4c c]

= 4|a ¥ b|2 = 4|a|2 |b|2 sin2 q

= (2l – 1) [a + 2b lb c]

Ê (ab)2 ˆ = 4k4 (1 – cos2q) = 4k4 Á 1 Ë | a |2 | b |2 ˜¯

= (2l – 1)l [a + 2b b c]

Ê (a ◊ b)2 ˆ = 4k4 Á 1 Ë k 4 ˜¯

if l π 0, 1/2

= (2l – 1)l [a b c] π 0

10. We are given v.u w.u = |u| |u| fi v. u = w. u

|u ¥ v| = 2(k4 – (a·b)2)1/2.

Previous Years’ AIEEE/JEE Main Questions 1. (a ¥ b)2 = |a|2 |b|2 sin2 (p/6)

Also, v. w = 0

= (42) (22) (1/2)2 = 16 2

Now, |u – v + w|2

2

2. [a ¥ b b ¥ c c ¥ a] = [a b c] = 4 = 16 2

3. |b + c| = |–a|

= |u|2 + |v|2 + |w|2 – 2u.v + 2u.w – 2v.w

2

= 1 + 4 + 9 = 14

fi |b|2 + |c|2 + 2b.c = |a|2

fi |u – v + w| =

fi 25 + 9 + 2|b| |c| cos q = 49

1 |b| |c| a 3 1 |b| |c| a fi (a.c)b – (b.c)a = 3

11. (a ¥ b) ¥ c =

fi 2(5) (3) cos q = 15 fi cos q = 1/2 fi q = 60° 4. 0 = |a + b + c|2 2

2

A possible solution is a.c = 0, b.c = -

2

= |a| + |b| + |c| + 2(a.b + b.c + c.a)

fi |b| |c| cos q = -

1 fi a.b + b.c + c.a = - 52 + 42 + 32 = – 25 2 5. c = a ¥ b = 39k

(

34 , |b| =

45 , |c| = 39

\ |a| : |b| : |c| =

34 : 45 : 39

|a| =

)

12. Take P as origin 1 (PA + PB) \ PC = 2 fi PA + PB = 2PC 13. Let a = a1i + a2j + a3k

Thus, w. n = 3 fi |w. n| = 3

fi a ¥ i = –a2k + a3j

7. F = F1 + F2 = 7i + 2j – 4k

fi |a ¥ i|2 = a22 + a32

d = 5i + 4j + k – (i + 2j + 3k)

Similarly, |a ¥ j|2 = a12 + a32

= 4i + 2j – 2k

and |a ¥ k|2 = a12 + a22

Thus, work done = w = F.d = 28 + 4 + 8 = 40

Thus, |a ¥ i|2 + |a ¥ j|2 + |a ¥ k|2

A

8. Let M be mid point of BC,

= 2(a12 + a22 + a32) = 2|a|2

1 (AB + AC) 2

= 4i – j + 4k fi AM = |AM| =

33

B

M

Fig. 22.8

9. As a, b, c are non-coplanar, [a b c] π 0 Now [a + 2b + 3c lb + 4c (2l – 1)c]

C

a a c 14. 1 0 1 = 0 c c b 0 fi 1 0 fi a,

a 0 c b, c

1 |b| |c| 3

1 |b| |c| fi cos q = –1/3 3

\ sin q = 2 2 / 3

6. u ¥ v = –2k fi n = k.

AM =

14

c 2 1 = 0 fi –ab + c = 0 b are in G.P.

Vectors

15. [l(a + b) l2b lc] = [a b + c b] fi (l) (l2) (l) [a + b b c] = [a c b] fi l4 [a b c] = – [a b c] As [a b c] π 0, l4 = –1. Not possible for any real value of l. 1 0 -1 16. [a b c] = x 1 1- x y x 1+ x - y 1 0 0 = x 1 1 [C3 Æ C3 + C1] y x 1+ x

22.35

1

(i + 2j + k) 2 2 Now, a = ld =

l

(i + 2j + k) 2 2 Equating coefficients of j, we get fi ai + 2j + bk =

l= 2 2 Thus, a = 1, b = 1 8 8 22. a = 8b = - (7b) = - c 7 7 fi angle between a and c is p. 23. [3u pv pw] – [pv w qu]

=1

– [2w qv qu] = 0

fi [a b c] is independent of x, y.

fi 3p2 [u v w] – pq[u v w] + 2q2 [u v w] = 0

17. (a ¥ b) ¥ c = a ¥ (b ¥ c)

fi (3p2 – pq + 2q2) [u v w] = 0

fi (a.c) b – (b.c) a = (a.c) b – (a.b) c

As u, v, w are non-coplanar,

a.b c b.c \ a is parallel to c

3p2 – pq + 2q2 = 0

fia=

18. CA = (a – 2)i – 2j + 0k CB = (a – 1)i + 0j + 6k As –C = p/2, CA.CB = 0 fi (a – 2) (a – 1) = 0 fi a = 1, 2 19. As c are lies in the plane of a and b, [a b c] = 0 1 fi 1 x

1 1 -1 2 =0 x - 2 -1

0 0 1 -3 2 =0 fi -1 x + 1 x - 1 -1 [using C1 Æ C1 – C3, C2 Æ C2 – C3] fi – (x – 1) + 3(x + 1) = 0 fi x = –2 1 20. |2u ¥ 3v| = 1 fi |u ¥ v| = 6 1 1 fi |u| |v| sin q = fi sin q = 6 6 –1 –1 fi q = sin (1/6), p – sin (1/6) 21. Unit vector along angle bisector of b and c is 1Ê b c ˆ d= Á + ˜ 2 Ë | b | | c |¯

fi p2 -

1 2 pq + q 2 = 0 3 3 2

1 ˆ 23 2 Ê fi ÁË p - q˜¯ + q = 0 6 36 1 q , q = 0. 6 \ p = 0, q = 0

fip=

That is, there is exactly one value of (p, q). 24. a.c = 0, b.c = 0 fi l – 1 + 2m = 0, 2l + 4 + m = 0 fi l = –3, m = 2 25. Let b = b1i + b2 j + b3k i j a¥b= 0 1 b1 b2

k -1 b3

= (b2 + b3)i – b1 j – b1k Now, a ¥ b + c = 0 fi (b2 + b3 + 1)i + (–b1 – 1)j + (–b1 – 1)k = 0 fi b2 + b3 + 1 = 0, b1 = –1 Also, a.b = 3 fi b2 – b3 = 3

22.36

Complete Mathematics—JEE Main

Hence, b = – i + j – 2k

C

for same a ΠR.

26. (a ¥ b) ¥ (a + 2b)

q

BE = AE – AB

= (a.(a + 2b)) b – (b.(a + 2b)) a 2

B

31. Let AE = a p,

Thus, b2 = 1, b3 = –2

2

fir=ap–q

A

= (|a| + 2a.b) b – (b.a + 2|b| ) a

As, BE.AD = 0

But a.b = 0, |a| = 1, |b| = 1

fi 0 = r.p = ap.q – p.q

Thus, (a ¥ b) ¥ (a + 2b) = b – 2a fi (2a – b). [(a ¥ b) ¥ (a + 2b)] = – |2a – b|2 = –(4|a|2 + |b|2) = – 5. 27. b ¥ c = b ¥ d

fia=

Thus, r = -q +

p.q p p.p

fi d = c – ab

fi |AD| =

Ê a.c ˆ Thus, d = c - Á b Ë a.b ˜¯ 28. As pi + j + k, i + qj + k, i + j + rk are coplanar, so p 1 1 1 q 1 =0 1 1 r fi pqr + 2 – p – q – r = 0 fi pqr – p – q – r = –2 29. a + 3b = ac and b + 2c = ba

32. AD =

33. As u and v are collinear,

D

C

Fig. 22.10

fi a – 2 = k(2 + 3a), 1 = – 3k 1 fi a – 2 = - (2 + 3a ) fi a = 2/3 3 34. Projection of b + lc on a =

(b + l c).a |a|

fi ±

b.a + l c.a 2 = 3 6

fi ±2 = – 1 + l(–1) fi l = –3, 1 For l = 1, – b + lc = 2i + 3j – 3k 35. l = 1 [see Theory]

and a + 3b + 6c = (1 + 3b)a

= 2(4|a|2 + |b|2)

\ (a + 6) c = (1 + 3b) a

fi 25 + |2a + b|2 = 2(4(22) + 32)

As a and c are non-collinear,

fi |2a + b| = 5

30. As c.d = 0

B

u = kv for some k ΠR.

36. |2a – b|2 + |2a + b|2

\ a + 3b + 6c = 0

A

33

Now, a + 3b + 6c = (a + 6)c

We get a + 6 = 0, 1 + 3b = 0.

D

p

p.q p.p

fi c Рd = ab for same a ΠR.

As a.d = 0, we get a.c – a a.b = 0 a.c fia= a.b

E

Fig. 22.9

1 (AB + AC) 2 = 4i – j + 4k

fi b ¥ (c – d) = 0

r

37. As c ¥ (i + 2j + 5k) = 0, is parallel to i + 2j + 5k. Let c = a(i + 2j + 5k) for same a Œ R.

fi (a + 2b).(5a – 4b) = 0

fi 60 = |c|2 = a 2(1 + 4 + 25) fi a2 = 2

fi 5|a|2 + 10b.a – 4a.b – 8|b|2 = 0

Now, c. (–7i + 2j + 3k)

fi 5 + 6|a| |b| cos q – 8 = 0

= a(–7 + 4 + 15)

fi cos q = 1/2 fi q = p/3

= 12a = 12 2

Vectors

38. |x + y|2 + |y + z|2 + |z + x|2 – |x + y + z|2 2

2

2

2

Thus, 2

= |x| + |y| + 2x. y + |y| + |z| + 2y.z + |z| + |x| + 2z.x – (|x|2 + |y|2 + |z|2 + 2x.y + 2x.z + 2y.z)

2

=

Thus, u = |x + y|2 + |y + z|2 + |z + x|2 42.

= 3 + |x + y + z|2 ≥ 3. \ minimum value u is 3 and its attained when x + y + z = 0.

1 3a 2 - b 2 2

(

1 2 c - a 2 - b2 2

( -c )

DB.AB = a 2 -

= |x|2 + |y|2 + |z|2 = 3

2

3 = |a + b| fi 3 = |a + b|2 = |a|2 + |b|2 + 2a.b fi a.b = 1/2

= 1 – 1/4 = 3/4 Now,

= 22i + 8j + 18k

|c|2 = |a|2 + 9|b|2 + 9|a ¥ b|2 + 4a.b + 6a.(a ¥ b) + 12 b.(a ¥ b)

Projection of x ¥ y on z | (x ¥ y ).z | 182 | 66 - 32 - 216 | = = = 14 |z| 13 9 + 16 + 144 1 40. (a ¥ b) ¥ c = |b| |c| a 3 1 fi (a.c) b – (b.c) a = |b| |c| a 3 1ˆ Ê fi |a| |c| cos jb = |b| |c| Á cosq + ˜ a Ë 3¯ As a and b are non-collinear, 1 = 0 and cos j = 0 cos q + 3 fi sin2 q = 1 – 1/9 = 8/9 =

fi sin q = ± 2 2 / 3 .

= 1 + 4 + 9(3/4) + 4(1/2) + 0 + 0 fi 4|c|2 = 55 fi 2|c| =

55 .

3 (b + c) 2 As b and c are not parallel

43. (a.c) b – (a.b) c =

a.c =

3 3 , a.b = 2 2

Now, a.b = - 3 /2 fi |a| |b| cos q = - 3 /2 fi cos q = - 3 /2 fi q = 5p/6 44. AB = –4i + 2j + (p + 1)k

C

and AC = 2i + (q – 1)j – 3k As –CAB = p/2, we get

Thus, a value of sin q is 2 2 / 3

(–4) (2) + 2(q – 1) + (p + 1) (–3) = 0

41. AC = AB + BC fic=a+b

fi –3p + 2q – 13 = 0

and DB = AB – AD = a – b. 2

Now, DB.AB = (a – b).a = a – b.a D

C

A

B

Fig. 22.12

Thus, (p, q) lies on 3x – 2y + 13 = 0 which makes an acute angle with the x-axis. 45. Let p be the position vectors of the circumcentre and orthocentre be p and h. As G divides HP in the ratio 2:1,

c 2 a

A

B

Fig. 22.11

Also c2 = |c|2 = |a + b|2 = |a|2 + |b|2 + 2a.b fi a.b =

)

Also, |a ¥ b|2 = |a|2 |b|2 – (a.b)2

i j k 39. x ¥ y = 3 -6 -1 1 4 -3

b

1 2 c - a 2 - b2 2

(

22.37

)

H (orthocentre)

1 G (centroid)

Fig. 22.13

1 1 (a + b + c) = (2p + h) 3 3 1 fih= (a + b + c) 2

P (circumcentre)

22.38

Complete Mathematics—JEE Main

Previous Years’ B-Architecture Entrance Examination Questions 1. 0 = |u + v + w|2 2

2

7. Statement-2 is false as |i – j| =

2 = |i + j|

Also, |2a – b|2 + |2a + b|2

2

= |u| + |v| + |w| + 2u.v + 2v.w + 2w.u

= 2[|2a|2 + |b|2] = 2[4(4) + 9] = 50

= 9 + 16 + 25 + 2(u.v + v.w + w.u)

fi |2a + b|2 = 50 – 52 = 25

fi u.v + v.w + w.u = –25

fi |2a + b| = 5

2. Let c = (a – b) ¥ (a ¥ b)

\ Statement-1 is true.

= ((a – b).b)a – ((a – b).a) b

8. c = a ¥ b

fi (a.b – |b|2)a – (|a|2 – b.a)b

fi |c|2 = c.c = (a ¥ b).c = [a b c]

2

= (a.b – |a| ) (a – b) [\ |a| = |b|] fi c is parallel to a – b

Similarly, |a|2 = |b|2 = [a b c] Thus, |a|2 = |b|2 = |c|2

3. See solution to Question 14 in Previous Years’ AIEEE/JEE Main Questions. 4. |x – y|2 + |y – z|2 + |z – x|2 = 9

fi |a| = |b| = |c| 9. p = Area of quadrilateral C

fi |x|2 + |y|2 – 2x.y + |y|2 + |z|2 – 2y.z + |z|2 + |x|2 – 2z.x = 9

B

b

fi 2x.y + 2y.z + 2z.x = –3 [|x| = |y| = |z| = 1] Now, |x + y – z|2 – 4x.y O

= |x|2 + |y|2 + |z|2 + 2x.y – 2x.z – 2y.z – 4x.y

A

Fig. 22.14

= 3 – (–3) = 6

= area (D OAB) + area (D OBC) 1 1 = |a ¥ (2b + 10a)| + |(2b + 10a) ¥ b| 2 2 = |a ¥ b| + 5|a ¥ b| = 6|a ¥ b| = 6q

5. 2a ¥ (a ¥ b) + c = 0 fi 2(a.b)a – 2(a.a)b + c = 0 fi c = 2b – 2(a.b)a

10. Solving two equations, we get

[∵ |a| = 1] 2

a

2

2

2

fi 1 = |c| = 4|b| + 4(a.b) |a| – 8(a.b) (b.a) fi 1 = 4 – 4(a.b)2 fi (a.b)2 = 3/4 fi |a|2 |b| cos2q = 3/4 fi cos q = ± 3 / 2 \ a value of q is p/6 i j k 6. b ¥ c = 1 -1 3 0 1 2 = – 5i – 2j + k Now, [a b c] = a.(b ¥ c) = |a| |b ¥ c| cos q where q = angle between a and b ¥ c fi [a b c] £ (1) (29) Max [a b c] = 29 when cos q = 1.

2 1 1 2 e1 - e2 , b = - e1 + e2 3 3 3 3 1 fi a = (1, 1, 3) 3 1 b = (1, 1, –3) 3 \ a.b = |a| |b| cos q a=

1 1 (1 + 1 – 9) = (11) cos q 9 9 fi q = cos–1 (–7/11)

fi

11. 2u + 2v = – 3w fi 9 = 9|w|2 = 4|u|2 + 4|v|2 + 8u.v fi u.v = 1/8 Now, |u – v|2 = |u|2 + |v|2 – 2u.v = 1 + 1 – 1/4 = 7/4 fi |u – v| =

7 /2

Vectors

Also,

i j k 12. v ¥ w = 2 1 -1 1 0 3

i j k x ¥ y = 2 1 -2 1 1 0

= 3i – 7j – k

= 2i + 2j + k

[u v w] = u.(v ¥ w) = |u| |v ¥ w| cos q

fi |x ¥ y| = 3

where q = angle between u and v ¥ w fi [u v w] = (1) \ Max [u v w] =

59 cosq £ 59 59

for cos q = 1 13.

22.39

1 (i – 2j + 2k) = (a ¥ b) ¥ (c ¥ d) 6 = {(a ¥ b).d}c – {(a ¥ b).c}d

Now, |(x ¥ y) ¥ z|2 = |x ¥ y|2 |z|2 sin2 (30°) Ê 1ˆ = 9(1) Á ˜ Ë 4¯ 3 2 15. AB = k + l(i + j) – p(i + j + k)and AC = – k + m(i – j) – p(i + j + k) fi |(x ¥ y) ¥ z| =

= [a b d]c – [a b c]d

As AB is perpendicular to r = k + l(i + j)

As a, b, c are coplanar, [a b c] = 0

k ◊ (i + j) + l(i + j).(i + j) – p(i + j + k).(i + k) = 0

We have [a b d] = (a ¥ b).d

fi 0 + 2l – p(2) fi l = p

= (± |a| |b| sin q d).d

Similarly m(1 + 1) – p(1 – 1) fi m = 0

= ±

1 2

1 1 Thus, ± d = (i – 2j + 2k) 6 2 1 fi d = ± (i – 2j + 2k) 3 1 = ± (–i + 2j – 2k) 3 14. |z – x|2 = 8 fi |z|2 – 2z.x + |x|2 = 8 2

fi |z| – 2|z| + 9 – 8 = 0 fi |z| = 1.

Thus for any given value of p, B(p, p, 1) lying on the line r = k + l(i + j)

(1)

is such that AB is perpendicular to (1), and C(0, 0, –1) lying on the line r = –k + m(i – j)

(2)

is such that AC is perpendicular to (2). \ all the four answers are correct.

CHAPTER TWENTY THREE

MEASURES OF CENTRAL TENDENCY One of the most important objectives of statistical analysis is to get one single value that describes the characteristic of the entire mass of unwiedely data. Such a value is called the central value or an average. The following are the important types of averages: 1. Arithmetic Mean 2. Geometric Mean 3. Harmonic Mean 4. Median 5. Mode We consider these measures in three cases (i) individual series (i.e. each individual observation is given) (ii) discrete series (i.e. the observations along with number of times a particular observation called the frequency is given (iii) continuous series (i.e. the class intervals along with their frequencies are given)

therefore, try to obtain an approximate value of the mean. The method of approximation is to replace all the observed values belonging to a class by mid-value of that class. If x1, x2 … xn are the mid values of the class intervals having corresponding frequencies f1, f2 … fn then we apply the same formula as in discrete series x =

i.e.

1 N

n

n

Â

fi xi , N = Â fi

i =1

i =1

PROPERTIES OF ARITHMETIC MEAN 1. Sum of all the deviations from arithmetic mean is zero i.e. n

 ( xi - x ) = 0

(in case of individual series)

i =1

ARITHMETIC MEAN

n

1. Individual Series If x1, x2, …xn are the values of the variable x, then the arithmetic mean usually denoted by x is given by x + x2 º + x n 1 n = Â xi x = 1 n n i =1 2. Discrete Series If a variable takes values x1, x2 … xn with corresponding frequencies f1, f2 … fn then the arithmetic mean x is given by x =

f1 x1 + f2 x2 º + fn xn 1 = f1 + f2 º + fn N

n

 fi xi i =1

 fi ( xi - x ) = 0 series)

2. If each observation is increased or decreased by a given constant, the mean is also increased or decreased. That is, if x1, x2 … xn are given variables and if we take di = xi – A then the mean d of the new variables di is given by d = x-A or

x = A+

N = Â fi i =1

3. Continuous Series In case of a set of data with class intervals, we cannot find the exact value of the mean because we do not know the exact values of the variables. We,

1 Â di n

If we have discrete/continuous series then

n

where

(in case of discrete or continuous

i =1

x = A+

1 N

n

 fi di ,

N = Â fi i =1

This property is also known as effect of change of origin. A can be taken to be any number. However, to simplify the calculations, A should be taken as a value which is in the middle of the table.

23.2

Complete Mathematics—JEE Main

3. Step Deviation Method or change of scale If x1, x2 …, xn are mid values of class intervals with corresponding frequencies f1, f2, … fn, then we may x -A change the scale by taking di = i , in this case h Ê1 x = A+h¥Á ËN

 fi di ˆ˜¯

5. If x1 and x2 be the means of two related groups having n1 and n2 items respectively then the combined mean x of both the group is given by n1 x1 + n2 x2 n1 + n2 If there are more than two groups then n x + n2 x2 + n2 x3 + º x = 1 1 n1 + n2 + n3 + º x =

ˆ  log xi ˜¯

and N =

or

(

)

Ê1 G = Antilog Á ËN

n

, where N =

 fi i =1

 fi log xi ˆ˜¯

HARMONIC MEAN The harmonic mean is based on the reciprocals of the value of the variable H.M. =

or

1 1Ê 1 1 1ˆ + +º+ ˜ n ÁË xi x2 xn ¯

1 = H 1 N

1 n

1

 fi x i =1

i

h ÊN ˆ ÁË - C ˜¯ f 2

 fi . i =1

i =1

In case of discrete or continuous series G.M. =

n

N , where N = Â fi . Find the commulative frequency (c.f.) 2 i =1 just more than N/2. The corresponding value of x is median. In case of continuous distribution, the class corrsponding to c.f. just more than N/2 is called the median class and the median is obtained by

n

n

1/ N º xnfn

It refers to the middle value in a distribution. In case of individual series, in order to find median, arrange the data in ascending or descending order of magnin +1 tude. In case of odd number of values Med = size of th 2 item. In case of even number of values Med = average of n n+2 th and th observation. 2 2 In case of discrete frequency distribution the median is obtained by considering the commulative frequency. Find

where l = the lower limit of the median class; f = the frequency of the median class; h = the width of the median class; C = the c.f. of the class preceding to the median class

In case of individual series x1, x2, …, xn G.M = (x1. x2 … xn)1/n

f f x11 x22

(in case of discrete series or 1 continuous series) Â fi x i i =1

If x1, x2, … xn > 0 then it is known that A.M. > G.M. > H.M.

Median = 1 +

GEOMETRIC MEAN

or

1

n

MEDIAN

A and h can be any numbers but if the lengths of class intervals are equal then h may be taken as width of the class interval. In particular if each observation is multiplied or divided by a constant, the mean is also multiplied or divided by the same constant. 4. The sum of the squared deviation of the variate from their mean is minimum i.e. the quantity S (xi – A)2 or S fi (xi–A)2 is minimum when A= x

Ê1 G = Antilog Á Ën

1 = H 1 N

and

(in case of individual series)

MODE The mode is that value in a series of observations which occurs with greatest frequency. In case of individual series, for determining the mode, count the number of times the various values repeat themselves and the value occurring the maximum number of times in the model value. In case of discrete series, quite often mode can be determined just by inspection i.e. by looking to that value of the variable around which the items are most heavily concentrated. In case of continuous series, Mode = l +

f1 - f0 ¥h 2 f1 - f2 - f0

where l = the lower limit of the modal class i.e. the class having maximum frequency; f1 = frequency of the modal class; f0 = frequency of the class preceding the modal class; f2 = frequency of the class succeeding the modal class and h = width of the modal class.

Statistics 23.3

When mode is ill-defined i.e. there are two or more values which occur with equal maximum frequency, the mode can computed by mode = 3 median – 2 mean.

4. Standard Deviation Variance s2 in case of individual series is given by s2 =

DISPERSION Literally, dispersion means ‘scatteredness’. Dispersion measures the degree of scatteredness of the variable about a central value. Different measures of dispersion are 1. Range 2. Mean-deviation 3. Quartile deviation 4. Standard deviation

1 n 1 n 2 Ê1 n ˆ 2 x x = ( )  i  xi - ÁË n  xi ˜¯ n i =1 n i =1 i =1

If x1, x2 …, xn occur with frequency f1, f2, …fn respectively then s 2 (variance) =

1. Range

2. Mean-deviation If x1, x2, … xn are n observations then mean deviation about a point A is given by 1 M.D. = Â xi - A n If A = x , then it is called M.D. from mean. In case of discrete or continuous series 1 M.D. = N

1 N

= 1 N

The range is the difference between the largest and smallest observations.

n

 fi i =1

n

xi - A ,

N = Â fi i =1

Q - Q1 Q.D. = 3 , where Q3 and Q1, are the third quartile and 2 the first quartile. Q1 and Q3 can be calculated in a similar manner as median. In fact, quartiles divides the data into four parts. So in case of individual series, arrange the data n +1 in ascending or descending order. Q1 = size of th and 4 3 (n + 1) th item Q3 = size of 4 In case of discrete frequency distribution, Q1 is obtained by considering commulative frequency. Find N/4, where N = S fi. Find the commulative frequency (c.f.) just more than N/4. The corresponding value of x is Q1. Similarly for obtaining Q3, find 3N/4 and the c.f. just more than 3N/4. The corresponding value of x is Q3. In case of continuous distribution hÊ N ˆ Qi = l + Á i - C ˜ , i = 1, 2, 3 ¯ f Ë 4 where l = the lower limit of the class whose c.f. is just more than iN/4, f is its frequency and h is its width. C = c.f. of the class preceding to the class whose c.f. is just more than iN/4, i = 1, 2, 3. Note that i = 2 will give us median.

n

 fi ( xi - x )2 i =1

1 Â fi xi2 - ÊÁË N

 fi xi ˆ˜¯

2

and standard deviation = + variance. There is no effect of change of origin on standard deviax -A then tion i.e. if di = xi – A then s x = s d. If di = i h sx = |h|sd equivalently È1 s 2x = h2 Í ÎN

1 Â fi di2 - ÊÁË N

2 ˆ ˘ f d  i i ˜¯ ˙ ˚

If there are two samples of sizes n1 and n2 with x1 , and x2 as their means and s1 and s2 their standard deviations respectively, then the combined variance is given by

M.D. is least when taken from the median i.e. When A is median. M.D. from mean has no effect of change of orgin.

3. Quartile Deviation

2

s2 = where

(

)

(

n1 s 12 + d12 + n2 s 22 + d22

)

n1 + n2

d1 = x1 - x and d2 = x2 - x ,

x being the combined mean. i.e x =

n1 x1 + n2 x2 n1 + n2

COEFFICIENT OF DISPERSION (C.D.) A- B , where A and B are the A+ B greatest and smallest items in the series.

1. C.D. based on range =

2. C.D. based on quartile deviation =

Q3 - Q1 Q3 + Q1

3. C.D. based on M.D. = M.D. Average from which it is calculated 4. C.D. based on S.D. =

S.D. x

5. Coefficient of variation = 100 ¥

S.D. x

23.4

Complete Mathematics—JEE Main

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Example 1: The mean deviation of an ungrouped data is 50. If each observation is increased by 2%, then the new mean deviation is (a) 50 (b) 51 (c) 49 (d) 50.5 Ans. (b) Solution: If x1 … xn are the observations then the new observations are (1.02) x1, (1.02)x2 .., (1.02)xn Therefore the new mean = (1.02) x New mean deviation =

1 Â |(1.02)xi – (1.02) x | n

1 = (1.02) Â xi - x n

n +1 2 n ( n + 1) (c) 2

Ans. (d) Solution: Mean =

Solution: Combined mean =

22 ¥ 71 + 18 ¥ 78 + 10 ¥ 89 22 + 18 + 10

1562 ¥ 1404 + 890 50 3856 = = 77 (approximately) 50

=

Example 3: The mean marks obtained by 300 students in Mathematics are 45. The mean of top 100 Students was 70 and the mean of last 100 was known to be 20. The mean of remaining 100 students is (a) 40 (b) 50 (c) 45 (d) 43 Ans. (c) Solution: 45 = fi

70 ¥ 100 + x2 ¥ 100 + 100 ¥ 20 300

135 = 90 + x 2 fi x2 = 45

Example 4: The mean of variable 1, 2, … n whose corresponding frequencies are 1, 2,… n is given by

1.1 + 2.2 + . . . . n . n 1+ 2... + n

n

 k2

=

K =1 n

=

Âk

n ( n + 1) (2 n + 1) 2 ¥ n ( n + 1) 6

K =1

=

= (1.02) ¥ 50 = 51 Example 2: In a nuclear engineering class there are 22 juniors, 18 seniors and 10 graduate students. If the juniors averaged 71 in the midterm examination, the senior averaged 78 and the graduate students averaged 89, the mean of the centre class is approximately (a) 77 (b) 80 (c) 74 (d) 81 Ans. (a)

2n + 1 6 2n + 1 (d) 3

(b)

(a)

2n + 1 . 3

Example 5: The mean of 5 observations is 4.4 and the variance is 8.24. If three of the five observations are 1, 2 and 6, the two values are (a) 4 and 9 (b) 3 and 5 (c) 2 and 6 (d) 4 and 6 Ans. (a) Solution: S xi = n x = 5 ¥ 4.4 = 22 So 1 + 2 + 6 + x4 + x5 = 22 x4 + x5 = 13 s2 =

1 1 Â x12 - ÊË n n

 xi ˆ¯

2

fi

(1)

 xi2 = n (s 2 + x 2 )

= 5(8.24 + 19.36) = 138. fi 1 + 4 + 36 + x 42 + x52 = 138 fi x42 + x52 = 97 2 2 So x4 +(13 – x4) = 97 fi x42 + 169 + x42 – 26x4 = 97 fi 2x42 – 26x4 + 72 = 0 fi x42 – 13x4 + 36 = 0 13 ± 169 - 144

13 ± 5

x4 = = = 9,4 2 2 So x5 = 13 – x4 = 4, 9 Hence, the other two nembers are 4 and 9. Example 6: In a set of 20 observations, each of the observation below the median of all observations is increased by 6 and each of the remaining observation is decreased by 2.Then the mean of the new set of observations

Statistics 23.5

(a) increase by 1 (c) increase by 4 Ans. (d)

(b) decrease by 1 (d) increase by 2

will be x1 + 6, . . . x10 + 6, x11 – 2, . . . , x20 – 2. Their mean is equal to ˆ 1 Ê 20 xi + 6 ¥ 10 - 2 ¥ 10˜ Â Á 20 Ë i = 1 ¯

Solution: Arrange all observations in increasing or decreasing order. Let the observations be x1, x2 …. x20 so that median. The new set of observations

=

x10 + x1 is the 2

1 20 Â xi + 2 20 i = 1

LEVEL 1 Straight Objective Type Questions Example 7: If the mean of a set of observations x1, x2, …, x10 is 20 then the mean of x1 + 4, x2 + 8, x3 + 12, …, x10 + 40 is (a) 34 (b) 42 (c) 38 (d) 40 Ans. (b) 1 Solution: Required mean = [(x1 + x2 +… x10 ) + 10 (4 + 8 +… 40) =

1 4 ( x + º + x10 ) + [1 + 2 + º + 10] 10 1 10

= 20 +

4 ¥ 10 ¥ 11 = 42. 10 ¥ 2

Example 8: Let G1, G2 be the geometric means of two series x1, x2, … xn; y1, y2 … yn. If G is the geometric mean of xi/yi, i = 1, 2, … n, then G is equal to (b) log G1/ log G2 (a) G1 – G2 (c) log (G1/G2 ) (d) none of these Ans. (d)

Solution: Let Rs. x be the sale in the last month. Using the formula of combined mean, we have 12, 000 ¥ 11 + x = 11,375 12 fi x = 11,375 × 12 – 12,000 × 11 = 1,36,500 – 1,32,000 = 4500. Example 10: The A.M. of the observations 1.3.5, 3.5.7, 5.7.9, …, (2n – 1) (2n + 1) (2n + 3) is (a) 2n3 + 6n2 + 7n – 2 (b) n3 + 8n2 + 7n – 2 (c) 2n3 + 5n2 + 6n – 1 (d) 2n3 + 8n2 + 7n – 2 Ans. (d) 1 Solution: A.M. = [1.3.5 + 3.5.7 +…+ (2n – 1) (2n n + 1) (2n + 3)] ˘ 1Èn = ÍÂ (2r - 1) (2r + 1) (2r + 3)˙ n Î r =1 ˚ =

Solution: G1 = (x1, x2 … xn)1/n and G2 = (y1, y2 … yn)1/n x ˆ Êx x G =Á 1 ◊ 2º n˜ Ë y1 y2 yn ¯

1/ n

1/ n

=

( x1. x2 º xn ) ( y1. y2 º yn )1 / n

=

G1 G2

Example 9: The monthly sales for the first 11 months of the year of a certain salesman were Rs. 12,000 but due to his illness during the last month the average sales for the whole year came down to Rs. 11,375. The value of the sale during the last month was (a) Rs 4,500 (b) Rs 6,000 (c) Rs 10,000 (d) Rs 8,000 Ans. (a)

=

1 n

1 n

n

 (8r 3 + 12r 2 - 2r - 3) r =1

È Ê n (n + 1) ˆ 2 ˘ n(n + 1)(2n + 1) n(n + 1) -2 - 3n ˙ Í8 ÁË ˜¯ + 12 2 6 2 Î ˚ = 2n(n + 1)2 + 2(n + 1) (2n + 1) – (n + 1) – 3 = 2n3 + 8n2 + 7n – 2. 18

Example 11: If

 ( xi - 8) = 9 i =1

18

and

 ( xi - 8)2 = 45 i =1

then the standard deviation of x1, x2 … x18 is (a) 4/9 (b) 9/4 (c) 3/2 (d) none of these Ans. (c)

Complete Mathematics—JEE Main

23.6

Solution: Let di = xi – 8 but s x2 = s2d =

1 1  di2 - ÊÁË 18  di ˆ˜¯ 18

2 1 5 1 9 Ê 9ˆ ¥ 45 - Á ˜ = - = Ë 18 ¯ 18 2 4 4

= Therefore

Example 14: The variance of first 20-natural numbers 2

n(n + 1) 4

n 2

(b)

n(n - 1) (c) 2

Solution: N = S fi = k [nC0 + nC1 +…+ nCn] = k (1 + 1)n = k2n S fi xi = k [1.nC1 + 2 . nC2 + … + n nCn] (n - 1)(n - 2) È ˘ + º + 1˙ = kn (1 + 1)n -1 = kn Í1 + (n - 1) + 2! Î ˚ = kn2n – 1 x =

1 2

n

(n 2n-1 ) = 2n .

Example 13: The mean of two samples of sizes 200 and 300 were found to be 25, 10 respectively. Their standard deviations were 3 and 4 respectively. The variance of combined sample of size 500 is (a) 64 (b) 65.2 (c) 67.2 (d) 64.2 Ans. (c) n x + n2 x2 Solution: Combined mean x = 1 1 n1 + n2 =

200 ¥ 25 + 300 ¥ 10 = 16 500

d1 = x1 - x = 25 – 16 = 9 and

Let

d2 = x2 - x = 10 – 16 = – 6 Now we know that 2

s = =

(b) 379/12 (d) 399/4

Solution: s2 =

(

)

(

n1 s 12 + d12 + n2 s 22 + d22

2 1 2 2 1 [1 + 2 +…+ 202]– ÈÍ (1 + 2 + º + n )˙˘ 20 Î 20 ˚

=

1 20 ¥ 21 ¥ (2 ¥ 20 + 1) È 1 20 ¥ 21 ˘ 2 -Í 20 6 2 ˙˚ Î 20

=

7 ¥ 41 441 133 = . 2 4 4

In fact, the variance of first n-natural numbers is

n(n + 1) (d) 2

Ans. (b)

Thus

(a) 133/4 (c) 133/2 Ans. (a)

sx = 3/2.

Example 12: If a variable takes the values 0, 1, 2 … n with frequencies proportional to the binomial coefficients n C0, nC1, …, nCn then the mean of the distribution is (a)

is

)

n1 + n2

200 (9 + 81) + 300(16 + 36) 33600 = = 67.2 500 500

n2 - 1 . 12

Example 15: If the standard deviation of x1, x2, …, xn is 3.5, then the standard deviation of –2x, – 3, – 2x2,– 3, …, – 2xn – 3 is (a) – 7 (b) – 4 (c) 7 (d) 1.75 Ans. (c) x -A Solution: We know that if di = i then sx = |h|s d. h xi + 3 / 2 In this case – 2xi – 3 = so h = – 1/2. Thus sd = -1/ 2 1 s = 2 × 3.5 = 7. h x Example 16: The geometric mean of 3, 32 …, 3n is (a) 3n/2 (c) 3n(n + 1)/2 Ans. (b)

(b) 3(n + 1)/2 (d) none of these

Solution: G.M. = (3.32.33 … 3n)1/n = (3

1+ 2 + … + n 1/n

)

Ê n ( n +1) ˆ = Á3 2 ˜ Ë ¯

1/ n

= 3(n + 1)/ 2 Example 17: If the median of 21 observations is 40 and if the observations greater than the median are increased by 6 then the median of the new data will be (a) 40 (b) 46 (c) 46 + 40/21 (d) 46 – 40/21 Ans. (a) Solution: If the observations greater than the median are increased by 6, the median will not be affected.

Statistics 23.7

Example 18: The median from the following data Wages/week (Rs.) 50-59 60-69 70-79 80-89

No. of workers Wages/week No. of workers 15 40 50 60

90–99 100–109 110–119

45 40 15

(b) 84.08

(c) 82.17

(d) 85.67

Solution: Wages/week (Rs.) No. of workers (f )

Median Class

(b) 10, 14 (d) 11,13

Solution: Mean of the given series is 5. If we have to transfer the given series into one having 12 as mean, we must take d = x + 7. The transformed terms are 8, 9, 11, 13, 15, 16. The two middle terms are 11, 13.

is (a) 83.17 Ans. (b)

(a) 7, 8 (c) 9, 15 Ans (d)

c.f.

50–59

15

15

60–69

40

55

70–79

50

105

80–89 90–99 100–109 110–119

60 45 40 15

165 210 250 265

265 Ê Nˆ Median = size of Á ˜ item = = 132.5th item Ë 2¯ 2

Example 22: If a variable takes the values 0, 1, 2 ----, n with frequencies proportional to the binomial coefficients, C (n, 0) , C (n,1) ---- , C (n, n) respectively, then the variance of the distribution is: (a) n n (c) 2 Ans. (d)

(d)

Solution: s2 =

h f

Example 20: Assuming the variance of four numbers a, b, c and d as 9, the variance of 5a, 5b, 5c and 5d is (a) 45 (b) 5/9 (c) 9/5 (d) 225 Ans. (d) x-A Solution: sx2 = h2 su2 if u = h Here h = 1/5 so su2 = 9 ¥ 25 = 225. Example 21: 1, 2, 4, 6, 8, 9 transformed into another series whose mean is 12. Two middle terms of the transformed series are

2

Âk

C (n,k) = n(2n-1)

k=0

n

and Âfx2 =

 k2

C (n,k)

k=0 n

=

 [k (k - 1) + k ]C (n, k )

k =0

= 79.5 + 10 (132.5 - 105) = 84.08. 60

Solution: The standard deviation is independent of the change of origin.

 fx 2 - Ê Â fx ˆ  f ÁË Â f ˜¯

n

Âfx =

Next,

ÊN ˆ ÁË - C ˜¯ 2

Example 19: If the standard deviation of numbers 2, 4, 5 and 6 is a constant a, then the standard deviation of the numbers 4, 6, 7 and 8 is (a) a + 2 (b) 2 a (c) a (d) 2a Ans. (c)

n 4

We have Âf = Â C (n,k) = 2n

Median lies in the class 80–89. But the true limit of the class is 79.5 – 89.5. Here, N = 265, C = 105, f = 60, h = 10 and l = 79.5 Median = l +

n /2

(b)

= n(n - 1)(2 n - 2 ) + n(2 n - 1 ) 2

s2 =

Thus,

n (n - 1) n Ê n ˆ n + -Ë ¯ = . 4 2 2 4

Example 23: If the mean deviation of the numbers 1, 1 + d, ...., 1 + 100d from their mean is 255, then d is equal to (a) 10.1 (b) 20.2 (c) 10.0 (d) 20.0 Ans. (a) 1 Solution: x = [1 + (1 + d ) + (1 + 2d ) + ..... + (1 + 100d )] 101 1 101 ¥ = [1 + (1 + 100d )] = 1 + 50d. 101 2 \ Mean deviation from mean =

1 [| 1 - (1 + 50d ) | + | 1 + d 101 (1 + 50 d ) | + ..... + | 1 + 100 d - (1 + 50 d ) |]

=

2d {1 + 2 + .... + 50] 101

Complete Mathematics—JEE Main

23.8

Ans. (d)

2 d 50(51) 2550 = d 101 2 101 2550 Now, d = 255 fi d = 10.1 101 Thus, we may take d = 10.1 =

Solution: Median of a, 2a , 50a is 1 (25a + 26a) = (25.5) a. 2 M.D. of given numbers from median

Example 24: For two data sets, each of size 5, the variances are given to be 4 and 5 the corresponding means are given to be 2 and 4 respectively. The variance of the combined data set is (a) 6 (b) 13/2 (c) 5/2 (d) 11/2 Ans. (d) Solution: We have n1 = 5, n2 = 5, s 12 = 4, s 22 = 5, We have x– = We know that

where

)+ ( + ) n1 + n2 d1 = x–1 – x– = –1 and d2 = x–2 – x– = 1

Thus,

s2 =

s2 =

(

+

d12

=

n2 s 22

d22

5(4 + 1) + 5(5 + 1) 11 = 5+5 2

Example 25: If the mean deviation about the median of the numbers a, 2a, , 50a is 50 then |a| equal (a) 5 (b) 2 (c) 3 (d) 4

50

1 50

 ka - (25 ◊ 5) a

a 50

Â

k =1 50

k - 25 ◊ 5

k =1

a (2 (24 ◊ 5 + 23 ◊ 5 + ... + 0 ◊ 5)) 50 a = (1 + 3 + ... + 49) 50 a 25 25 = ¥ (1 + 49) = a = 50 50 2 2 |a| = 4 =

x–1 = 2, x–2 = 4

n1 x1 + n2 x2 (5)(2) + (5)(4) = =3 n1 + n2 5+5 n1 s 12

=

fi

Example 26: A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 gm and a standard deviation of 2 gm. Later, it was found that the measuring scale was misaligned and always under reported every fish weight by 2 gm. The correct mean and standard deviation (in gm) of fishes are respectively: (a) 32, 2 (b) 32, 4 (c) 28, 2 (d) 28, 4 Ans. (a) Solution: Since mean (X + b) = mean X + b and Var (X + b) = Var X, so we get correct mean as 30 + 2 = 32 gm and s.d, is 2 gm

Assertion-Reason Type Questions

n

Example 27: If x– = 3 and A =

Â

(xi – 3)2,

i =1 n

B=

Â

n

(xi – 4)2, C =

i =1

Â

(xi – 2)2

i =1

Statement-1: Min (A, B, C) = C. Statement-2: Sum of the squares of deviations is least when taken from mean Ans. (d) Solution: Among all the squares of the deviations, the least is when taken from mean. This can be proved by conn

sidering, F(x) =

Â

i =1

(xi – X)2, F is least when X = x–.

Example 28: The mean monthly salary paid to all employees of a factory was Rs. 6000. The mean salary paid to male and female employees was Rs. 6200 and Rs. 5200. Statement-1: The percentage of males is 80. Statement-2: The exact percentage cannot be determined as the data is insufficient. Ans. (c) Solution: Let there be N1 percentage male and female be N2 percent so N1 + N2 = 100 6200 N1 + 5200 N 2 6000 = N1 + N 2 fi fi

6000 = 62 N1 + 52 (100 – N1) 800 N1 = = 80 and N2 = 20. 10

Statistics 23.9

Example 29: The arithmetic mean of two observations is 127.5 and geometric mean is 60. Statement-1: The harmonic mean is approximately 28. Statement-2: A.M. ¥ H.M. = (G.M.)2. Ans. (a) 2ab a+b Solution: A.M. = H.M. = a +b 2 So A.M ¥ H.M. = ab = (G.M )2 3600 Thus H.M. = = 28.24 127.5

Example 30: Let x1, x2  xn be n observations, Let x– be their arithmetic mean and s 2 be their variance Statement 1: Variance of 2x1, 2x2  2xn is Statement 2: Arithmetic mean of 2x1, 2x2  3xn is 4x– Ans. (c) x-A Solution: If d = then x– = A + h d . In this case h 1 – x = 2x. Also s x2 = h 2 s d2 so s d2 A = 0, h = 1/2 so d = h 1 = 2 s x2 = 4s 2 . h

LEVEL 2 Straight Objective Type Questions Example 31: Given the following frequency distribution with some missing frequencies Class 10–20

Frequency 180

20–30

--

30–40 40–50 50–60 60–70 70–80

34 180 136 -50

fi f1 + f2 = 105 We have 580 + f1 + f2 = 685 As median 42.6 lies in the class 40 – 50, median class is 40 – 50. We have that l = 40, f = 180, C = 214 + f1 and h = 10 Median = l +

Now,

42.6 = 40 +

fi

If the total frequency is 685 and median is 42.6 then the missing frequencies are (a) 81, 24 (b) 80, 25 (c) 82, 23 (d) 832, 22 Ans. (c) Solution: Let the missing frequencies be f1 and f2, respectively. Class

Frequency (f )

Cum. Freq. (c.f.)

10–20

180

180

20–30

f1

180 + f1

30–40

34

214 + f1

40–50

180

394 + f1

50–60

136

530 + f1

60–70

f2

530 + f1 + f2

70–80

50

580 + f1 + f2

Total

685

hÊN ˆ Á - C ˜¯ fË2 10 Ê 685 ˆ - 214 - f1 ˜ Á ¯ 180 Ë 2

fi 2.6 × 18 = 342.5 – 214 – f1 fi f1 = 81.7 As frequency of an item is always a whole number, we take f1 = 82. \ f2 = 105 – 82 = 23. Example 32: The mean deviation from mean of the observations a, a + d, a + 2d, …, a + 2nd is (a)

n (n + 1)d 2 3

(c) a +

(b)

n(n + 1) d 2 2

n (n + 1) 2 d 2

(d) none of these

Ans. (d) Solution: x =

1 [a + (a + d) + … + (a + 2nd)] 2n + 1

1 [(2n + 1)a + d (1 + 2 + … + 2n)] 2n + 1 2n (1 + 2n) = a + nd =a+d 2 2n + 1 M.D. from mean n(n + 1) | d | 1 2 | d | (1 + 2 + º + n) = = 2n + 1 (2n + 1) =

23.10

Complete Mathematics—JEE Main

Example 33: Sum of absolute deviations about median is (a) least (b) greatest (c) zero (d) equal to median Ans. (a)

i =1

2 Ê1 ˆ < 1 a ÁË Â i ˜¯ n n

fi

Solution: See theory. Example 34: In any discrete series (when all the values are not same) the relationship between M.D. about mean and S.D. is (a) M.D. = S.D. (b) M.D. > S.D. (c) M.D. < S.D. (d) M.D. £ S.D. Ans. (c) Solution: M.D. =

2

n Ê n ˆ a < n  ai2 ÁË Â i ˜¯

1 n 1 n xi - x and S.D. = ( xi - x )2 Â Â n i =1 n i =1

Ê1 n ˆ Let ai = |xi – x | so (M.D.) = Á Â ai ˜ Ë n i =1 ¯ 1 n 2 Â ai . By Cauchy Schwarz inequality n i =1 2

i =1

 ai2

fi

M.D. < S.D.

Example 35: The median and standard deviation (S.D.) of a distribution are 20 and 4 respectively. If each item is increased by 2 then (a) median will increase and S.D. will also increase (b) median will will go up by 2 but S.D. will remain same (c) median will increase but S.D. will decrease (d) none of these Ans. (b)

2

and (S.D.)2 =

Solution: Median will go up by 2 and S.D. will remain the same.

EXERCISE Concept-based Straight Objective Type Questions 1. The mean of 5 observations is 3 and variance is 2. If three of the five observations are 1, 3, 5 the other two are (a) 2, 6 (b) 3, 3 (c) 1, 5 (d) 2, 4 2. The mean and standard deviation of a sample of size 20 are found 10 and 2 respectively. If one item 8 is replaced by 12 then (a) standard deviation increases (b) standard deviation decreases (c) standard deviation will remain same (d) mean increases by 4% 3. The mean weight of 150 students in 60 Kgs. The mean weight of boys is 70 and that of girls in 55 with standard deviations 10 and 15 respectively. The combined variance is (a)

700 3

(b)

500 3

(c)

400 7

(d)

300 7

4. If n = 10, x = 12 and S x2 = 1530, then coefficient of variation is (a) 16 (b) 9 (c) 25 (d) 36 5. The first of two samples has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and variance 13.44, the standard deviation of the second sample is (a) 3 (b) 2 (c) 4 (d) 5 6. If the variables with values 10, 20, 30, 40 have frequencies x, x + 2, x – 2, x + 4 and the mean is 2.7 then the value of x is (a) 2 (b) 4 (c) 6 (d) 5

Statistics 23.11

LEVEL 1 Straight Objective Type Questions 7. The arithmetic mean of the first n natural numbers is (a) (n + 1)/2 (b) (n – 1)/2 (c) n/2 (d) none of these 8. The arithmetic mean of the first n odd natural numbers is (a) n (b) (n + 1)/2 (c) (n – 1) (d) none of these 9. The arithmetic mean of the squares of the first n natural numbers is (a) (n + 1)/6 (b) (n + 1) (2n + 1)/6 2

(c) (n – 1)/6

(d) none of these

10. The arithmetic mean of the series 1, 2, 22, …, 2n – 1 is (b) (2n –1)/n (a) 2n/n (d) none of these (c) (2n +1 )/n n

n

n

11. The arithmetic mean of the series C0, C1, C2, …, n Cn is (b) 2n/n (a) 2n/(n + 1) n –1 (d) none of these (c) 2 /(n + 1) 12. The arithmetic mean of a, a + d, a + 2d, …, a + 2nd is (a) a + (n – 1)d (b) a + (n + 1)d (c) a + (n + 2)d (d) none of these 13. The standard deviation of the data given by 0

Variate (x) Frequency (f) (a)

(n + 1) / 2

(c) 2n/n

n

C0

1 n

C1 (b)

2 n

3

C2

n

17. For a certain frequency table which has been partly reproduced here, the arithmetic mean was found to be Rs 28.07. Income (in Rs):

15

n

20

25

30

35

40

No. of workers: 8 12 ? 16 ? 10 If the total number of workers is 75, then the missing frequencies are (a) 14, 15 (b) 15, 14 (c) 13, 16 (d) 12, 17 18. The value of median for the data Income (in Rs.) 1000 1100 1200 1300 1400 1500 No. of persons: 14 26 21 18 28 15 is (a) 1300 (b) 1200 (c) 1250 (d) 1150 19. The median from the following data: Mid-value Frequency

…

C3 …

16. The mean weight of 9 items is 15. If one more item is added to the series the mean becomes 16. The value of 10th item is (a) 35 (b) 30 (c) 25 (d) 20

Mid-value Frequency

n

115

6

165

60

Cn

125

25

175

38

135

48

185

22

145

72

195

3

155

116

n/2

(d) none of these

14. The mean height of 25 male workers in a factory is 161 cms, and the mean height of 35 female workers in the same factory is 158 cms. The combined mean height of 60 workers in the factory is (a) 159.25 (b) 159.5 (c) 159.75 (d) 158.75 15. The average salary of male employees in a firm was Rs 5200 and that of females was 4200. The mean salary of all the employees was Rs 5000. The percentage of male employees is (a) 80 (b) 60 (c) 40 (d) 20

is (a) 153.79 (b) 153.91 (c) 165.18 (d) 165.93 20. In a discrete series, the relation between s.d.s and range r is (a) s > r (b) s > r (c) r = s (d) s < r 21. If A.M., G.M. and H.M. in any series are equal then (a) the distribution is symmetric (b) all the values are same (c) the distribution is unimodal (d) none of these

23.12

Complete Mathematics—JEE Main

22. For a group containing 100 observation, the mean and standard deviation are 8 and 10.5 . For 50 observations selected from these 100 observations, the mean and standard deviation are 10 and 2 respectively . The standard deviation of the other half is (a) 2 (b) 3 (d) (4) (c) 6.5

23. A sample of 90 values has mean 55 and standard deviation 3. A second sample of 110 values has mean 60 and standard deviation 2. The combined variance is equal to (upto two decimal) (a) 12.43 (b) 11. 24 (c) 10.43 (d) 13.25

Assertion-Reason Type Questions 24. Suppose that the mean and standard deviation of 100 observations are 50 and 4. Statement-1: Sum of the square of the observations is 2,51,600.

Statement-2: Sum of the observations is 5000. 25. If Âxi = 440, Â(xi – 44)2 = 1710, N = 10, then. Statement-1: C.V. is approximately 30. Statement-2: Variance is 171.

LEVEL 2 Straight Objective Type Questions 26. If the s.d. of a set of observations is 4 and if each observation is divided by 4, the s.d. of the new set of observations will be (a) 4 (b) 3 (c) 2 (d) 1

30. If a variable takes values 0, 1, 2 … n with frequencies proportional to qn, nC1 pqn –1, nC2 p2 qn–2, …, pn where p + q = 1 then the mean is (a) np (b) nq (c) npq (d) np2

27. If s.d. of a set of observations is 8 and if each observation is divided by – 2, the s.d. of the new set of observations will be (a) – 4 (b) – 8 (c) 8 (d) 4

31. If a variable takes values 0, 1, 2, 3 … with frequen-

28. The mean square deviation of a set of n observations x1, x2, …, xn about a point c is defined as 1 n  ( x - c )2 n i =1 i The mean square deviation about –2 and 2 are 18 and 10 respectively, then standard deviation of this set of observation is (a) 3 (b) 2 (c) 1 (d) none of these 29. If the s.d. of n observations x1, x2, …, xn is 4 and another set of n observations y1, y2, …, yn is 3, the s.d. of n observations x1 – y1, x2 – y2, …, xn – yn is (a) 1

(b) 2 / 3

(c) 5

(d) none of these

cies proportional to e–l, e–ll,

e- l l 2 e- l l 3 , ,º then 2! 3!

the mean of the distribution is (b) l (a) e–l (c) e–ll

Ê 1ˆ (d) Á ˜ e–ll2 Ë 2¯

32. The standard deviation is not affected by the change of (a) origin (b) scale (c) origin and scale (d) none of these 33. If x1, x2, x3 are three non-zero real numbers such that (x21 + x22) (x22 + x23) < (x1 x2 + x2 x3)2 then the G.M. of x1, x2, x3 is (a) x1 (b) x2 (c) x3

(d)

x1 x2 x3 3

Statistics 23.13

34. The mean and s.d. of a sample of size 10 were found to be 9.5 and 2.5 respectively. Later on, an additional observation 15 was added to the original data. The s.d. of 11 observation is (a) 2.6 (b) 2.8 (c) 2.86 (d) 3.24

36. The mean marks of 120 students is 20. It was later discovered that two marks were wrongly taken as 50 and 80 instead of 15 and 18. The correct mean of marks is (a) 19.19 (b) 19.17 (c) 19.21 (d) 19.14

35. For a group of 50 male workers, the mean and s.d. of their daily wages are Rs 630 and Rs 90 respectively. For a group of 40 female workers these Rs 540 and Rs 60 respectively. The s.d. of these 90 workers is (a) 60 (b) 70 (c) 80 (d) 90

37. Let r be the range and S2 =

1 Â ( xi - x )2 . If n -1

S2 < r2 k then k is equal to 1 n -1 n +1 (c) 2 (n - 1) (a)

(b)

n n -1

(d) none of these

Previous Years' AIEEE/JEE Main Questions

1. In a class of 100 students, there are 70 boys whose average marks in a subject is 75. If the average marks of the complete class is 72, then the average of the girls is (a) 73 (b) 65 (c) 68 (d) 74 [2002] 2. 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 (a) is decreased by 2 (b) is two times the original median (c) remains the same as that of the original set (d) is increased by 2 [2003] 3. In an experiment with 15 observations on x, the following results were available S x2 = 2830, S x =170. One observation, 20, was found to be wrong and was replaced by the correct value 30. Then the corrected variance is (a) 78 (b) 188.66 (c) 177.33 (d) 833 [2003] 4. Consider the following statements: (1) Mode can be computed from histogram (2) Median is not independent of change of scale (3) Variance is independent of change of origin and scale Which of these is/are correct? (a) only (1) and (3) (b) only (2) (c) only (1) (d) (1), (2), (3) [2004]

5. In a series of 2n observations, half of them equal a and remaining half equal – a. If the standard deviation of the observations is 2, then |a| equals 1 n

(b)

2

(c) 2

(d)

2 n

(a)

[2004]

6. Let x1, x2 ….xn be n observations such that S x2 = 400 and S x = 80. Then a possible value of n among the following is (a) 15 (b) 18 (c) 9 (d) 12 [2005] 7. If in a frequency distribution, the mean and median are 21 and 22 respectively, then the mode is approximately (a) 25.5 (b) 24.0 (c) 22.0 (d) 20.5 [2005] 8. Suppose a population A has 100 observations 101, 102, º 200 and another population B has 100 observations 151, 152, º 250. If VA and VB represent the variances of the two populations, respectively, then VA /VB is (a) 2/3 (b) 1 (c) 9/4 (d) 4/9 [2006] 9. The average marks of boys in a class is 42 and that of girls is 52. 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 [2007]

23.14

Complete Mathematics—JEE Main

10. 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 = 0, b = 7 (b) a = 5, b = 2 (c) a = 1, b = 6 (d) a = 3, b = 4 [2008]

19. The variance of first 50 even natural numbers is.

11. If the mean deviation of the numbers 1, 1 + d, 1 + 2d, ..., 1 + 100d form their mean is 255 then the d is equal to (a) 10.1 (b) 20.2 (c) 10.0 (d) 20.0 [2009]

20. In a set of 2n distinct observations, each of the observations below the median of all the observations is increased by 5 and each of the remaining observation is decreased by 3. Then the mean of the new set of observations (a) increase by 1 (b) decrease by 1 (c) decrease by 2 (d) increase by 2 [2014, online]

12. 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 combined data set is (a) 6 (b) 13/2 (c) 5/2 (d) 11/2 [2010] 13. If the mean deviation about the median of the numbers a, 2a, ..., 50a is 50 then |a| equals (a) 5 (b) 2 (c) 3 (d) 4 [2011] 14. A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 gm and a standard deviation of 2 gm. Later, it was found that the measuring scale was misaligned and always reported every fish weight by 2 gm. The correct mean and standard deviation (in gm) of fishes are respectively (a) 32, 2 (b) 32, 4 (c) 28, 2 (d) 28, 4 [2011] 15. Let x1, x2 ...xn be n observations, and let x– be their arithmetic mean and s 2 their variance. Statement-1: Variance of 2x1, 2x2 ... 2xn is 4s 2 Statement-2: Arithmetic mean of 2x1, 2x2 ... 2x4 is 4x [2012] 16. All the students of a class performed poorly in mathematics. The teacher decided to give marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given? (a) mean (b) median (c) mode (d) variance [2013]

(a)

833 4

(c) 437

(b) 833 (d)

437 4

[2014]

21. Let x and M.D be the mean and the mean deviation about x of n observations xi, i = 1, 2 …., n. If each of the observation is increased by 5, then the new mean and mean deviation about the new mean, respectively are: (b) x + 5, M.D. (a) x , M.D (d) x – 15, m. D+5 (c) x , M.D. + 5 [2014, online] 22. Let x , M and s 2 be respectively the mean, mode and variance of n observations x1, x2 – xn and di = – xi – a, i = 1, 2 . . . n where a is any number Statement 1: Variance of d1, d2 . . . dn is s 2 Statement 2: Mean and Mode of d1, d2 – dn are – x – a and M – a respectively. (a) Statement 1 and Statement 2 are both false. (b) Statement 1 and Statement 2 are both true. (c) Statement 1 is true and Statement 2 is false. (d) Statement 1 is false and Statement 2 is true. [2014, online] 23. A factory is operating in two shifts, day and night, with 70 and 30 workers respectively. If per day mean wage of the day shift workers is `54 and per day mean wage of all the workers is `60, then per day mean wage of the night shift (in `) is: (a) 66 (b) 69 (c) 74 (d) 75 [2015, online]

17. If the median and the range of four numbers {x, y, 2x + y, x – y}, where 0 < y < x < 2y are 10 and 28 respectively, then the mean of numbers is (a) 18 (b) 10 (c) 5 (d) 14 [2013, online]

24. The mean of the data set comprising of 16 observations is 16. If one of the observations 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 (a) 16.8 (b) 16.0 (c) 15.8 (d) 14.0 [2015]

18. The mean of a data set consisting of 20 observation is 40. If one observation 53 was wrongly recorded as 33, then the correct mean will be (a) 41 (b) 49 (c) 40.5 (d) 42.5 [2013, online]

25. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true? (a) 3a2 – 26a + 55 = 0 (b) 3a2 – 32a + 84 = 0 (c) 3a2 – 34a + 91 = 0 (d) 3a2 – 23a + 44 = 0 [2016]

Statistics 23.15

26. If the mean deviation of the numbers 1, 1 + d, … 14 + 100d from their means is 255, then a value of d is (a) 10.1 (b) 5.05 (c) 20.2 (d) 10 [2016, online]

27. The mean of 5 observations is 5 and their variance is 124. If three of the observations are 1, 2 and 6; then the mean deviation from the mean of the data is: (a) 2.5 (b) 2.6 (c) 2.8 [2016, online] (d) (1.6) (1 + 3 2 ) (modified)

Previous Years' B-Architecture Entrance Examination Questions 1. The mean deviation of an ungrouped data is 10. If each observation is increased by 4%, the revised mean deviation is (a) 10.0 (b) 10.4 (c) 10.04 (d) 9.6 [2006]

6. If the mean and standard deviation of 20 observations

2. There are 30 boys and 20 girls in a class. The mean and variance of their marks in maths are, respectively, 65 and 100. If 5 grace marks are added to the score of each student, then the revised mean and variance will be respectively (a) 70, 100 (b) 70, variance > 100 (c) 65, variance < 100 (d) 65, 100 [2007]

[2011]

3. The mean and the median of 100 observations have been computed to be 60 and 70 respectively. Later it was discovered that three observations which have been recorded 18, 28 and 98 are actually 80, 26 and 38 respectively. If the mean and median are recalculated with actual observations, then (a) median will change but mean will not change (b) neither mean nor median will change (c) both mean and median will change (d) mean will change but median will not change [2009] 4. Let a distribution be made by combining three distributions, each having mean zero, standard deviations 3, 4 and 5 respectively and frequencies 200, 250 and 300 respectively. Then the variance of the combined distribution is equal to 266 (a) 15

(c)

50 3

(b) 17 (d)

62 15

[2010]

5. In a class of 20 students, each student can score either 10 or 0 marks in a certain examination. The maximum possible variance in the marks of the students in the class is (a) 24 (b) 22 (c) 20 (d) 25 [2010]

10

 xi2

x1, x2 . . . x20 are 50 and 10 respectively, then

i =1

is equal to (a) 2510 (c) 52000

(b) 50200 (d) 2600

7. If the mean of a set of observations x1, x2 . . . , x10 is 20 then the mean of x1 + 4, x2 + 8, . . . , x10 + 40 is (a) 34 (b) 42 (c) 38 (d) 40 [2012] 8. Suppose a population A has 100 observation 101, 102 . . . 200 and another population B has 100 observations 151, 152, . . . 250. If VA and VB represents the variances of two populations respectively, then the ratio VA : VB is (a) 1 : 1 (b) 2 : 3 (c) 1 : 2 (d) 3 : 2 [2013] 9. If the mean and the standard deviations of 10 observations x1, x2 . . . xn are 2 and 3 respectively, then the mean of (x1 + 1)2, (x2 + 1)2. . . (x10 + 1)2 is equal to (a) 13.5 (b) 14.4 (c) 16.0 (d) 18.0 [2014]

Answers Concept-based 1. (d)

2. (b)

5. (c)

6. (10)

3. (a)

4. (c)

Level 1 7. 11. 15. 19. 23.

(a) (a) (a) (a) (a)

8. 12. 16. 20. 24.

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

9. 13. 17. 21. 25.

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

10. 14. 18. 22.

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

23.16

Complete Mathematics—JEE Main

Level 2

60 =

26. (d)

27. (d)

28. (a)

29. (d)

30. (a)

31. (b)

32. (a)

33. (b)

34. (c)

35. (d)

36. (a)

37. (b)

fi fi fi

Previous Years' AIEEE/JEE Main Questions 1. 5. 9. 13. 17. 21. 25.

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

2. 6. 10. 14. 18. 22. 26.

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

3. 7. 11. 15. 19. 23. 27.

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

4. 8. 12. 16. 20. 24.

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

Previous Years' B-Architecture Entrance Examination Questions 1. (a)

2. (a)

3. (a)

4. (a)

5. (d)

6. (c)

7. (b)

8. (a)

n1 x1 + n2 x2 fi 70n1 + 55n2 = 150 ¥ 60 = 9000 n1 + n2

14n1 14n1 n1 = d1 =

+ 11n2 = 1800 + 11(150 – n1) = 1800 50 so n2 = 100 x 1 – x = 70 – 60 = 10

d2 = x 2 – x = 55 – 60 = – 5 1 (n1 (s 12 + d12) + n2 (s 22 + d22)) s2 = 150 1 (50 (100 + 100) + 100 (225 + 25)) = 150 700 = . 3 1 1530 4. s 2 = Sx2 – x 2 = – 122 = 9 n 10 s 3 = 100 ¥ = 25. C.V = 100 ¥ x 12 5. n1 = 100 and n1 + n2 = 250 so n2 = 150. Also x 1 = 15, x = 15.6 so 15.6 = x =

9. (d)

n1 x1 + n2 x2 1500 + 150 x2 = n1 + n2 250

fi x 2 = 16

Hints and Solutions

d1 = x 1 – x = 15 – 15.6 = – 0.6 d2 = x 2 – x = 16 – 15.6 = 0.4

Concept-based 1.

(n1 + n2) s 2 = n1 (s12 + d12) + n2 (s 22 + d22)

1 S xi = 3 fi 1 + 3 + 5 + x4 + x5 = 15 5

fi x4 + x5 = 6 1 1 Sxi2 – ÊË Â xi ˆ¯ 5 5

2

=4fi

(i) Sxi2

= 65

So x42 + x52 = 65 – (1 + 9 + 25) = 30

(ii)

Solving (i) and (ii) x4, x5 = 2, 4 2. S xi = 20 ¥ 10 = 200 so new sum = 200 – 8 + 12 = 204. New mean = 10.2 so increase of 2%. Also

1 Sxi2 – 102 = 4. so S xi2 = 2080 20

New sum of squares = 2080 – 64 + 144 = 2160 204 ˆ 2 1 ¥ 2160 – ÊË New variance = 20 ¯ 20

= 108 – 104.04 = 3.96 < 4 3. n = n1 + n2 = 150 (n1 is number of boys, n2 is number of girls)

250 ¥ 13.44 = 100 (9 + 0.36) + 150 (s22 + 0.16) fi s 22 = 16. È x + 2 x + 4 + 3 x - 6 + 4 x + 16 ˘ 6. 10 Í ˙˚ = 27 4x + 4 Î fi 10[10x + 14] = 108 x + 108 fi 8 x = 32 fi x = 4.

Level 1

n (n + 1) 7. Sum n natural numbers is so the mean is 2 n +1 . 2 8. Sum of first n odd numbers n 2 so the mean is n.

9. 12 + 22 º + n2 = n ( n + 1) (2 n + 1) so the mean = 6 (n + 1) (2n + 1) . 6 10. 1 + 2 + º 2 n – 1 = 2 n – 1. 11. nC0 + nC1 º + nCn = 2 n 12. a + (a + d) + º + (a + 2nd) = (2n + 1) a + d n(2n + 1) = (2n + 1) (a + dn)

Statistics 23.17

1 13. (S.D.) = N 2

1 N

 fx

2

=

= =

1 2n

1 Â f x – ÊË N 2

 f xˆ¯

ÈÎ1 . C1 +  n 2

n

2 n

10.5 = s2 =

Cn ˘˚

n È 3 ˘ 1 + 2 (n – 1) + (n – 1)(n – 2 ) +  + n ˙ ˚ 2 2 n ÍÎ

{

}

n È (n – 1) (n – 2 ) ++1 1 + ( n – 1) + n Í 21 2 Î

+ {( n – 1) + ( n – 1)(n – 2 ) +  + (n – 1)}] =

n

È n – 1 C0 + 2n Î

n (n + 1) = 4 s2 =

C1  +

n –1

n–2

24. Â xi = 100 ¥ 50 = 5000, 2

x + s2 =

fi

m = 4f.

4 ¥ 100 = 80% . 5 16. Sum of 9 items = 15 ¥ 9 = 135. If x is added then 135 + x = 16 10 fi x = 25. 17. 8 + 12 + f1 + 16 + f2 + 10 = 75 fi f1 + f2 = 29 120 + 240 + 25 f1 + 480 + 35 f2 + 400 75 fi f1 = 15 f2 = 14. Compute c.f. find median. Turn the data into class intervals e.g. 110 – 120, 120 – 130, º See Theory. See Theory. n1= n2= 50, x–1= 10 , x– = 8, n = 100, s = 10.5 We find x 2 and then s2 Using the combined mean formula, we have 50 ¥ 10 + 50 ¥ x2 8= 100

1 N

 xi2

(

2

)

fi  xi2 = N x + s 2 .

6 ¥ 100. x

25. C.V. =

=

20. 21. 22.

90 ¥ 55 + 110 ¥ 60 = 57.75 200 d1 = 55 – 57.75 = – 2.75 d2 = 60 –57.75 = 2.25 90(9 + 7.5625) + 110(4 + 5.0625) s2 = 200 = 12.43 (App).

x– 2 =

Cn – 2 ˘˚ ˘˚

m So percentage of mole = ¥ 100 m+ f

18. 19.

23. n1 = 90, x–1 = 55, s1= 3, n2 = 110, x–2 = 60 and s2 = 2

25 ¥ 161 + 35 ¥ 158 = 159.25 60

m ¥ 5200 + f ¥ 4200 15. 5000 = m+ f

s 22 = 9, s2 = 3.

fi

Cn – 1 + ( n – 1)

ÈÎ n – 2 C0 +  +

n1 (s 12 + d12 ) + n2 (s 22 + d22 ) n1 + n2

50(4 + 4) + 50(s 22 + 4) 100

=

n –1

n (n + 1) n2 n – = . 4 4 4

14. Mean height =

28. 07 =

x–2 = 6 d1 = 10 – 8 = 2 and d2= 6 – 8 = –2

fi

2

Level 2 26. sX = 4, d = So sd =

4 = 1. 4

27. sX = 8, d = So sd =

X -0 Since sX = |4|sd 4

x-0 . Since sx = |–2| sd -2

8 = 4. 2

28. Given that 18 =

1 n

 (xi + 2)2, 10 =

Subtracting, 8 = 1 ( Â (xi + 2)2 – n

=

8 n

 xi fi

Also 18 = = =

x =1

1 Â ( xi2 + 4 + 4 xi ) n

1 1 xi2 + 4 + 4  xi  n n 1  xi2 + 8 n

So 10 =

1 Â xi2 n

1 (xi – 2)2 n

Â(xi – 2)2)

23.18

Complete Mathematics—JEE Main 2

fi x24 + x12 x32 - 2 x1 x22 x3 £ 0

1 Ê1 ˆ xi2 - Ë Â xi ¯ = 10 – 1 = 9 Â n n

s 2X =

2

fi ( x22 - x1 x3 ) £ 0 fi x22 = x1 x3

Thus sX = 3. 1 2 29. s 2X -Y = S ( xi - yi - ( x - y )) n

Hence G.M. of x1, x2, x3 is x2 34.

1 2 = S (( xi - x ) - ( yi - y )) n =

1 1 2 2 S ( xi - x ) + S ( yi - y ) n n –

= s 2X + s Y2 -

2 n  ( x - x ) ( yi - y ) n i =1 i

2 n  ( x - x ) ( yi - y ) n i =1 i

=

Ê1 n ˆ = 7 - 2 Á Â ( xi - x ) ( yi - y )˜ . Ë n i =1 ¯ n

n

30. x = [0·q + 1· C1 pq K ¥ K [ p + q]n = npqn–1 + 2.

= np[q

n–1

n–1

s=

n

2 n–2

+ 2· C2 p q

+  + n·p ] n

n (n - 1) 2 n–2 p q +  + npn 2

+ (n–1) pq

35. s2 =

1190 - 1100 90 = 11 11

90 30 = 2.86. = 11 110 n1 (s 12 + d12 ) + n2 (s 22 + d22 ) n1 + n2

d1 = x1 - x and d2 = x2 - x

n–2

+

( n - 1) ( n - 2 )

2

x =

2 n–3

pq

È ˘ e-l l 2 e-l l 3 31. x = Í0.e - l + e - l l + 2. + 3. +  +˙ Î ˚ 2! 3! K È Ê ˆ˘ l2 + ˜¯ ˙ K Íe - l ÁË 1 + l + Î ˚ 2! È ˘ 1 l2 + l + + ˙ - l l 1 ÍÎ ˚e e 2!

= le–l el = l. 32. If di = xi – A then sx = sd. 33. ( x12 + x22 ) ( x22 + x32 ) £ (x1 x2 + x2 x3)2 fi x12 x22 + x24 + x12 x32 + x22 + x32 £ x12 x22 + x22 x32 + 2 x1 x22 x2

n1 x1 + n2 x2 1 = [50 ¥ 630 + 40 + 540 ] n1 + n2 90 =

+ 

+ pn–1] = np [q + p]n–1 = np (K is constant of proportionality)

= le-l

1 1  xi = 9.5 and n  xi2 - x 2 = 6.25 10 1 fi Sxi = 95 and  xi2 = 96.5 10 95 + 15 = 10 Corrected Mean = 11 1 Corrected variance = (965 + 152) – 102 11

10 [315 + 216 ] = 590 9

d1 = 590 – 630 = –40 d2 = 590 – 540 = 50 ( ) ( ) So s 2 = 50 8100 + 1600 + 40 3600 + 2500 90

= 48500 + 24400 = 8100 9 s = 90. 36. 1 Â x1 = 20 so Sxi = 120 × 120 120 Corrected sum of the marks = 2400 – 50 – 80 + 15 + 18 = 2303 2303 Corrected mean = = 19.19 . 120 37. S2 =

nr 2 1 1 r2 = , ( xi - x )2 £ Â Â n -1 n -1 n -1

so K =

n . n -1

Statistics 23.19

Previous Years’ AIEEE/JEE Main Questions 1. If the average of girls is x, then 70 ¥ 75 + 30 ¥ x 72 = 100 fi x = 65 2. Median refers to middle value so remain unchanged. 3. Corrected Sx = 170 – 20 + 30 = 180 180 Corrected mean = = 12 15 Corrected Sx2 = 2830 – 400 + 900 = 3330 1 ¥ 3330 – 122 = 78. Corrected variance = 15 4. See theory. 1 ◊ 2n a2 – 0 fi a2 = 4 fi |a| = 2 5. 4 = 2n Ê 80 ˆ 1 ¥ 400 – ÁË ˜¯ n n So n = 15

2

6. s 2 =

≥ 0 fi n ≥ 16

7. Mode = 3 Median – 2 mean = 3 ¥ 22 – 2 ¥ 21 = 24 8. Since variables in B can be obtained by adding 50 to variables in A so VA = VB i.e. VA/VB = 1. 9. Let there be x number of boys and y number of girls. 50 =

42 ¥ x + 52 y x+ y

fi 50x + 50y = 42x + 52y fi 8x = 2y fi y = 4x Percentage of boys =

x x ¥ 100 = ¥ 100 = 20% x+ y 5x

a + b + 8 + 5 + 10 10. =6fia+b=7 5 1 2 [a + b2 + 64 + 25 + 100] – 36 = 6.8 5 fi a2 + b2 = 25 So a2 + (7 – a)2 = 25 fi a = 4, 3 a = 4, b = 3 or a = 3, b = 4 11. Same as Example 23. 2¥5+4¥5 =3 5+5 d1 = x1 - x = – 1, d2 = x2 - x = 1

12. Combined mean =

s 2 (combined variance) =

n1 (s12 + d12 ) + n2 (s 22 + d 22 ) n1 + n2

1 11 [4 + 1 + 5 + 1] = . 2 2 25a + 26a = 25.5a 13. Median = 2 1 [24.5 + 23.5 + … + .5 + .5 + … + 24.5] |a| 50 = 50 1 = [0.5 + 1.5 + .. + 24.5] |a| 25 1 25 = [0.5 + 24.5] |a| ¥ 25 2 100 = 25|a| fi |a| = 4 14. Correct mean = 32 and s = 2 (see properties of mean and standard deviation). 15. Mean of 2x1, 2x2, …, 2xn is equal to 2 x . =

s2

= 4s 2. (1/ 2)2 16. Clearly variance (properties of mean, median, variance). Variance of 2x, 2x2 …, 2xn is equal to

17. Arranging the four numbers, we have x – y, y, x, 2x x+ y = 10 fi x + y = 20 and + y. So the median is 2 the range is 2x + y – (x – y) = x + 2y = 28. Solving x = 12, y = 8. 1 The mean of the four numbers = [4x + y] = 4 1 ¥ 56 = 14. 4 1 [correct sum of observation] 18. Correct mean = 20 1 = [20 ¥ 40 – 33 + 53] 20 = 41. 1 [2 + 4 + … + 100] 50 1 50 = ¥ 102 = 51 ¥ 50 2 1 s2= [22 + 42 + … + 1002] – 512 50 4 = [12 + 22 + … + 502] – 512 50 2 È 50 ¥ 51 ¥ 101 ˘ 2 = ˙˚ – 51 25 ÍÎ 6 = 34 ¥ 101 – 512 = 17(202 – 153) = 833. 20. There are exactly n observations below the median and exactly n observations above the median. Thus, exactly n observations are increased by 5 and n observations are decreased by 3. =

23.20

Complete Mathematics—JEE Main

5n + ( -3n) =1 25 Thus, mean increases by 1. \ Change in mean =

21. Mean will become x + 5 , and there will be no change in M.D. (see theory). 22. Both the statements are true (see theory). 23. Let x be the mean wages of workers in the night shift, then 70 ¥ 54 + 30 x 60 = 70 + 30

Mean deviation from Mean =

1 [|1 – 5| + |2 – 5| + |6 – 5| + 5

|12 2 + 8 – 5| + |8 – 12 2 – 5|] 1 [4 + 3 + 1 + 12 2 + 3 + 12 2 – 3] 5 = 1.6 + 4.8 2

=

Previous Years’ B–Architecture Entrance Examination Questions 1. M.D. remains same with change of origin (see theory).

fi 600 = 7 ¥ 54 + 3x fi x = 200 – 7 ¥ 18 = 74.

2. The new mean will be x + 5 = 70 and variance will be same i.e. 100 (see theory).

24. Required mean

3. Corrected mean.

16 ¥ 16 - 16 + 3 + 4 + 5 16 - 1 + 3 256 - 16 + 12 252 = = = 14 18 18 1 1 (2 + 3 + a + 11) = 4 + a 25. x = 4 4 2 1 ˆ 1 Ê Variance = (22 + 32 + a2 + 112) – Á 4 + a ˜ Ë 4 ¯ 4 =

2

1 1Ê 1 ˆ Ê 7ˆ 2 ÁË ˜¯ = (134 + a ) – ÁË 8 + a ˜¯ 2 2 4 4

2

1 Ê ˆ fi 49 = 134 + a – Á 64 + a 2 + 8a ˜ Ë ¯ 4 3 2 fi a – 8a + 21 = 0 4 fi 3a2 – 32a + 84 = 0 2

26. Same as Q. 11 above. 27. Let remaining two observations be a, b, where a ≥ b. Then 1 (1 + 2 + 6 + a + b) fi a + b = 16 5 1 2 (1 + 22 + 62 + a2 + b2) – 52 Also, 124 = 5 fi 745 = 41 + a2 + b2 fi a2 + b2 = 704 Now, (a – b)2 = 2(a2 + b2) – (a + b)2 = 2(704) – 256 = 1152

5=

fi a – b = 24 2 Also, a + b = 16 fi a = 12 2 + 8, b = 8 – 12 2

1 [60 ¥ 100 – 18 – 28 – 98 + 80 + 26 + 38] 100 = 60 (same as the original mean)

=

Median will change (see theory). 4. Combined mean = 0, d1 = d2 = d3 = 0 s2 = =

n1 (s12 + d12 ) + n2 (s 22 + d 22 ) + n3s 32 n1 + n2 + n3

200 ¥ 9 + 250 ¥ 16 + 300 ¥ 25 750

1330 266 = 75 15 5. Max difference from mean can be 5 so the maximum possible variance is 25. =

1 20 7. (x1 + 4) + = (x1 + x2 = (x1 + x2 6. 100 =

Sx2i – 2500 fi Sx2i = 52000

(x2 + 8) + …, x10 + 40 … + x10) + (4 + 8 + … + 40) … + x10) + 4(1 + 2 + … + 10) 4 ¥ 10 ¥ 11 = (x1 + x2 … + x10) + 2 = 10 ¥ 20 + 220 = 420 So their mean =

420 = 42 10

8. Same as Q.8 in AIEEE [2006] above. 9. Mean of (x1 + 1)2, …, (x10 + 1)2 =

1 10 1 10 2 10 Â ( x j + 1)2 = Â x 2j + Â x j + 1 10 j =1 10 j =1 10 j =1

= s2 + x2 + 2 ¥ 2 + 1 =9+4+4+1 = 18

CHAPTER TWENTY FOUR

Probability

DEFINITIONS 1. A random experiment is an experiment in which (i) all the outcomes of the experiment are known in advance, and (ii) the exact outcome of any specific performance of the experiment is unpredictable, i.e. not known in advance. 2. A sample space, denoted by S, associated with a random experiment is a set of points such that (i) each element of S denotes an outcome of the experiment, and (ii) any performance of the experiment results in an outcome that corresponds to exactly one element of S. In other words, sample space consists of all possible outcomes associated with the random experiment. The element of the sample space associated with a random experiment are called the elementary or simple or indecomposable events of that experiment. 3. An event (associated with a random experiment) is a subset of the sample space S (associated with that experiment). We say that an event E (Õ S) has occurred provided the outcome w of the experiment is an element of E. If the outcome w of the experiment is such that w œ E, we say that the event E has not occurred. The empty set Ø, which is a subset of every set, also represents an event and is said to be an impossible event. The set S is itself a subset of S and it represents the sure event. 4. A set of events is said to be mutually exclusive if the occurrence of one of them precludes the occurrence of any of the remaining events. In set-theoretic notation, events E1, E2, , Em are mutually exclusive if Ei « Ej = Ø for i π j and 1 £ i, j £ m. See Fig. 24.1 E4 E1

E3 E2

Fig. 24.1

5. A set of events is said to be equally likely if, taking into consideration all the relevant factors, there is no reason to expect one of them in preference to the others. 6. A set of events is said to be exhaustive if the performance of the experiment always results in the occurrence of at least one of them.

CLASSICAL DEFINITION OF PROBABILITY If a random experiment results in N mutually exclusive, equally likely and exhaustive outcomes, out of which n are favourable to the occurrence of an event A, then the probability of occurrence of A, usually denoted by P(A), is given by n P(A) = N Since the number of favourable outcomes can never exceed the total number of outcomes, we have 0 £ P(A) £ 1. Note that P(Ø) = 0 and P(S ) = 1.

NOTATIONS Let A and B be two events. Then (i) A¢ or A or Ac stands for the non-occurrence or negation of A. (ii) A » B stands for the occurrence of at least one of A and B. (iii) A « B stands for the simultaneous occurrence of A and B. (iv) A¢ « B¢ stands for the non-occurrence of both A and B. (v) A Õ B stands for “the occurrence of A implies occurrence of B”.

A FEW THEOREMS ON PROBABILITY 1. If A and B are two mutually exclusive events, then P(A » B) = P(A) + P(B). 2. If A is any event, then P(A¢) = 1 – P(A)

24.2

Complete Mathematics—JEE Main

3. If A and B are two events, then P(A « B¢) = P(A) – P(A « B) 4. If A and B are two events, then P(A » B) = P(A) + P(B) – P(A « B) 5. If A and B are two events, then P(exactly one of A, B occurs) = P[(A « B¢) » (A¢ « B)] [Fig. 24.2] A

B

CONDITIONAL PROBABILITY

Fig. 24.2

= P(A) – P(A « B) + P(B) – P(A « B) = P(A) + P(B) – 2P(A « B) = P(A » B) – P(A « B) Also, P(exactly one of A, B occurs) = P(A « B¢) + P (A¢ « B) = P(B¢) – P(A¢ « B¢) + P(A¢) – P(A¢ « B¢) = P(A¢) + P(B¢) – 2P(A¢ « B¢) = P(A¢ » B¢) – P(A¢« B¢) 6. If A and B are two events P(A¢ » B¢) = 1 – P(A « B) and P(A¢ « B¢) = 1 – P(A » B) 7. If A1, A2, , An are n events, then P(A1 » A2 »  » An) n

=

Â

i =1

P (Ai) –

Â

P (Ai « Aj)

1£ i < j £ n

+

Â

(ii) P(A1 « A2 « .... « An) > 1 – P(A¢1) – P (A¢2) – .... – P(A¢n) 10. If A1, A2, , An are n events, then P(A1 « A2 « .... « An) ≥ P(A1) + P(A2) +  + P(An) – (n – 1) 11. If A and B are two events such that A Õ B, then P(A) £ P(B). Odds in Favour and against an Event If x is the number of cases favourable to the occurrence of an event A and y that for the event A¢, then the odds in favour of A are x : y and the odds against A are y : x. In this case x y and P(A¢) = P(A) = x+y x+y

P(Ai « Aj « Ak) –  +

1£ i < j < k £ n

(–1)n – 1 P(A1 « A2  « An) 8. If A, B and C are three events, then (i) P (A » B » C) = P(A) + P(B) + P(C) – P(B « C) – P(C « A) – P(A « B) + P(A « B « C) (ii) P (at least two of A, B, C occur) = P(B « C) + P(C « A) + P(A « B) – 2P (A « B « C) (iii) P (exactly two of A, B, C occur) = P(B « C) + P(C « A) + P(A « B) – 3P(A « B « C) (iv) P(exactly one of A, B, C occurs) = P(A) + P(B) + P(C) – 2P(B « C) – 2P(C « A) – 2P(A « B) + 3P(A « B « C) 9. If A1, A2, , An are n events, then (i) P(A1 » A2 »  » An) < P(A1) + P(A2) +  + P(An)

The probability of occurrence of an event A, given that B has already occurred is called the conditional probability of occurrence of A. It is denoted by P(A | B). If the event B has already occurred, then the sample space reduces to B. Now the outcomes favourable to the occurrence of A (given that B has already occurred) are A those that are common to both A and B, that is, those B A«B which belong to A « B. Thus, Fig. 24.3

P(A | B) =

N A« B

NB where NA « B is the number of elements in A « B and NB π 0 is the number of elements in B and N the total number of elements in S. N A « B / N P ( A « B) fi P(A | B) = = NB / N P ( B) Hence Ï P( B) P( A | B) if P( B) π 0 P(A « B) = Ì Ó P( A) P( B | A) if P( A) π 0 Important Note The function P (| A) is a probability function and hence all theorems on probability apply to this function. For instance, P (B » C | A) = P(B | A) + P(C | A) – P(B « C | A).

MULTIPLICATION THEOREM Let A1 A2, , An, An + 1 be (n + 1) events such that P(A1 « A2 «  « An) > 0.

Probability 24.3

Then

INDEPENDENT EVENTS Ê n +1 ˆ P Á ∩ A j ˜ = P(A1) P(A2 | A1) P(A3 | A1 « A2)  Ë j =1 ¯ P(An + 1 | A1 «  « An)

TOTAL PROBABILITY THEOREM Let {Hi}, i = 1, 2, , n, be n events such that Hi « Hj = Ø n

for i π j and

∪

Hi = S. Suppose that P(Hi) > 0 for 1 £ i £ n.

i =1

Then for any event A n

P(A) =

Â

i =1

P(Hi) P(A | Hi)

H1

Hn

H2

H3

H4

Fig. 24.4

BAYES’ RULE Let S be a sample space and H1, , Hn be n mutually n

exclusive events such that ∪ Hj = S and P(Hj) > 0 for j =1

j = 1, 2, , n. We can think of the Hj’s as the ‘causes’ (or the hypotheses) that lead to the outcome of an experiment. The probabilities P(Hj), j = 1, 2,  , n are called prior probabilities. Suppose the experiment results in an outcome of event A, where P(A) > 0. What is the probability that the observed event was due to cause Hj? In other words, we seek the conditional probability P(Hj | A). These probabilities are frequently called posterior probabilities. The information that A has occurred makes us reassess the probability P(Hj) assigned to Hj.

THEOREM (BAYES’ RULE) Let {Hj} be mutually exclusive events such that P(Hj) > 0 for j = 1, 2, , n and S =

n

∪

P( B « A) = P(B) fi P(A « B) = P(A) P(B) P( A) Thus, two events A and B are independent if and only if P(A « B) = P(A) P(B). If two events are not independent, they are said to be dependent. We note that if P(A) = 0 then for any event B, 0 £ P (A « B) £ P(A) = 0, so that P(A « B) = P(A) P(B) and A is independent of any other event. Difference between Mutually Exclusiveness and Indepen dence fi

Important Note We advise the reader to distinguish between independent and mutually exclusive events. Independence is a property of probability, whereas mutual exclusion is a set-theoretic property. If A and B are mutually exclusive events with P(A) > 0 and P(B) > 0, then P (A « B) = 0 π P(A) P(B), so that A and B cannot be independent. In fact, P(A | B) = 0 = P(B | A), so that if A occurs, B cannot occur and vice-versa. Thus, A and B are strongly dependent when A « B = Ø. We also note the following fact; two events A and B are independent if and only if (i) A and B¢ are independent, or (ii) A¢ and B are independent, or (iii) A¢ and B¢ are independent. Three events A, B and C are said to be pairwise independent if P(A « B) = P(A) P(B), P(B « C) = P(B) P(C) and P(C « A) = P(C)P(A) Three events A, B and C are said to be mutually indepen dent if P(A « B) = P(A) P(B), P(B « C) = P(B) P(C), P(C « A) = P(C) P(A) and P(A « B « C) = P(A) P(B) P(C)

Hj. Let A be an event with

j =1

P(A) > 0. Then for j = 1, 2,  , n P( H j ) P( A | H j ) P(Hj | A) = n

 P ( H k ) P( A | H k )

k =1

We say that two events A and B are independent if the occurrence or non-occurrence of A(B) does not affect the probability of occurrence or non-occurrence of B(A), that is, if P(B | A) = P(B) provided P(A) π 0

RANDOM VARIABLES Let S be a sample space. A random variable X is a function from the set S to R, the set of real numbers. If X is a random variable defined on the sample space S and r is a real number, then {X = r} = {wŒS | X (w) = r } is an event. If the random variable X takes n distinct values x1, x2, , xn, then

24.4

Complete Mathematics—JEE Main

{X = x1}, {X = x2}, , {X = xn} are mutually exclusive and exhaustive events (Fig. 24.5). S X = x1

X = x2 X = x3

X = x4

X = xn

Fig. 24.5

Now, since (X = xi) is an event, we can talk of P(X = xi). If P(X = xi) = pi (1 £ i £ n), then the system of numbers Ê x1 ÁË p 1

x2  x n ˆ p2  pn ˜¯

is said to be the probability distribution of the random variable X.

Mean and Variance of a Random Variable The expectation (mean) of the random variable X is defined as n

E (X ) =

Â

pi xi

i =1

and the variance of X is defined as n

var (X) =

Â

i =1

n

pi (xi – E(X))2 =

Â

pi xi2 – (E (X ))2

i =1

BINOMIAL DISTRIBUTIONS Consider a random experiment with sample space S and an event E (a subset of S) associated with it. Then the event “not E ” may be denoted by E¢, the complement (in S) of the subset E. Let P(E) = p and P(E¢) = q, so that p, q > 0 and p + q = 1. If the experiment results in the event E, we say a success, denoted by S, has occurred. If on the other hand, the event E does not occur (i.e., E¢ occurs) the experiment is said to have resulted in a failure, denoted by F. The probability of a success is equal to p and that of a failure is q = 1 – p. Suppose the above experiment is carried out n times under identical conditions. If the success (occurrence of E) and failure (non-occurrence of E) are recorded successively as the experiment is repeated, we will get a result of the type SFFSSFFFSF

There are 2n such outcomes, and these constitute the sample space of the combined experiment. Since the successive experiments are independent, the probability of the outcome above is p q q p p q q  q p q = pr qn – r where r is the number of successes in the outcome. If X denotes the random variable “number of successes”, then the possible values of X are 0, 1, 2, , n. We shall now calculate the probability of the event (X = r). Out of the 2n outcomes, r successes and (n – r) failures can occur in nCr ways. Also, the probability of r successes and (n – r) failures is pr qn – r. Thus, P(X = r) = nCr pr qn – r. The probability distribution of the random variable X is given by r

0 n

P(x = r) C 0 q

1 n n

C1 p q

2 n–1 n

2 n–2

C 2p q



r n

r n–r

 Cr p q



n n

 Cn p n

Note that the probabilities that the random variable takes the values 0, 1, 2, , n are given by the terms in the binomial expansion of (q + p)n. Because of this, the probability distribution of the random variable X is said to be a binomial distribution and X is said to be a binomial random variable. A binomial distribution B(n, p) is the probability distribution of a random variable X which takes values 0, 1, 2, , n with probabilities n

C0 p0q n, nC1p qn – 1,  , nCr p rqn – r, , nC n pnq0

where p, q > 0 and p + q = 1. The numbers n and p are said to be parameters of the binomial distribution B(n, p). We usually write, X ~ B (n, p).

Mode of Binomial Distribution For the value of r so that P(X = r) is maximum, we have two cases. Case 1: (n + 1)p is not an integer then, P(X = r) is maximum when r = m, where m is the greatest integer £ (n + 1)p, that is, m = [(n + 1)p]. Case 2: (n + 1)p is an integer then, P(X = r) is maximum when r = m – 1 or m, where (n + 1) p = m and m is an integer. If X ~ B (n, p), then E(X) = np and var (X) = npq.

Probability 24.5

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Example 1: If 15 boys of different ages are distributed into 3 groups of 4, 5, and 6 boys randomly then the probability that three youngest boys are in different groups is 24 (a) 91 67 (c) 91 Ans. (a)

71 (b) 91 20 (d) 91

Solution: Total number of ways of distributing 15 boys in three groups is = (15C4) (11C5) (6C6) =

15! 4! 5! 6!

The number of ways of distributing 12 remaining boys (ex12! cluding three youngest) is (12C3) (9C4) (5C5) = 3! 4! 5! and the number of ways of distributing three youngest boys is 3P3 = 3! \ the required probability 3! (12!) 4! 5! 6! 24 = ¥ = 3! 4! 5! 15! 91 Example 2: A bag contains 6 white and 6 black socks. A man randomly takes out two socks from the bag. The probability that socks are the same colour is: (a)

2 11

(b)

5 11

(c)

7 11

(d)

4 11

Ans. (b) Solution: Total number of ways of taking out two socks is 12C2 = 66. The number of favourable ways = Number of ways both the socks are white or both are black = 6C2 + 6C2 = 30. \ probability of required event 30 5 = = 66 11 Example 3: Two numbers are randomly selected from the first 100 natural numbers. The probability that the product of the numbers is divisible by 7 is

1 14 91 (d) 4950

(a) 0 (c)

(b)

4859 4950

Ans. (c) Solution: Product of two natural numbers is divisible by 7 if and only if at least one of the number is divisible by 7. \ required probability = 1 – P(none of the numbers is divisible by 7) 14

= 1-

C2

100

=

4859 4950

C2 Example 4: Let A = {1, 2, 3, 4} and B = {a, b}. A function f : A Æ B is selected randomly. Probability that function is an onto function is 1 5 (a) (b) 8 8 7 3 (c) (d) 8 8 Ans. (c) Solution: Total number of functions from A to B is 24 = 16. There are exactly two functions which are not onto viz. one in which all the elements of A are mapped to a and other in which all the elements of A are mapped to b. Thus, there are 14 onto function. Therefore, probability of required event is 14 7 = . 16 8 Example 5: Sets A, B, C, A « B, A « C, B « C and A « B « C have 35, 40, 45, 13, 12, 14 and 5 elements respectively. An element is selected at random from the set A » B » C. The probability that the selected element belongs to only set A is: 13 35 (b) (a) 86 86 5 15 (c) (d) 86 86 Ans. (d) Solution: By the principle of inclusion and exclusion, n(A » B » C) = n(A) + n(B) + n(C) – n(B « C) – n(A « C) – n(A « B) + n(A « B « C) = 35 + 40 + 45 – 13 – 12 – 14 + 5 = 86

24.6

Complete Mathematics—JEE Main

We have A

C B

Also, n(A « B¢ « C¢) = n(A) – n[(A « B) » (A « C)] = n(A) – [n(A « B) + n(A « C) – n(A « B « C)] = 35 – [13 + 12 – 5] = 15 \ Probability of required event 15 = 86 Example 6: If P(A) = 0.4, and P(A « B) = 0.15, then P (A|A¢ » B¢) is equal to (a)

1 17

(b)

2 17

(c)

5 17

(d)

9 17

Ans. (c) Solution: P (A|A¢ » B¢) =

P ( A « ( A¢» B ¢ )) P ( A¢ » B ¢)

P (( A « A¢ ) » ( A « B ¢ )) = P ( A¢ » B ¢) =

P (f » ( A« B ¢ )) [De Margon’s Law] P (( A « B)¢ )

=

P ( A « B¢) P( A) - P ( A « B) = 1 - P ( A « B) 1 - P ( A « B)

0.4 - 0.15 0.25 5 = = 1 - 0.15 0.85 17 Example 7: A class consists of 80 students, 25 of them are girls. If 10 of the students are rich and 20 of the students are fair complexioned, then the probability of selecting a fair complexioned rich girl from the class (assuming three traits as independent) is (b) 1 (a) 1 10 32 =

(c)

5 512

(d)

25 10 20 , P(R) = and P(F) = 80 80 80 As G, R and F are independent events, P(G « R « F) = P(G) P(R) P(F) 25 10 20 5 = ÊÁ ˆ˜ ÊÁ ˆ˜ ÊÁ ˆ˜ = Ë 80 ¯ Ë 80 ¯ Ë 80 ¯ 512 Example 8: Suppose A and B are two mutually exclusive 1 1 events such that P(A) = (3p + 1) and P(B) = (1 – p), then 3 4 the set of possible values of p lies in the interval: È1 2 ˘ (a) [0, 1] (b) Í , ˙ Î3 9 ˚ P(G) =

7 512

Ans. (c) Solution: Let G, R and F denote the following events G: a girl student is selected R: a rich student is selected F: a fair complexioned student is selected.

È 1 5˘ È 7 4˘ (c) Í- , ˙ (d) Í- , ˙ Î 3 9˚ Î 9 9˚ Ans. (c) 1 1 Solution: (3p + 1) ≥ 0, (1 – p) ≥ 0 3 4 1 1 (1 – p) £ 1. and (3p + 1) + 3 4 1 5 È 1 5˘ fi p ≥ – , p £ 1, p £ fi p Œ Í- , ˙ 3 Î 3 9˚ 9 Example 9: A number x is chosen at random from the set {1, 2, 3,…, 100}. Define event : A = the chosen number x - 20 x satisfies ≥ 2 . Then P(A) is x - 40 1 1 (b) (a) 4 5 1 1 (c) (d) 8 10 Ans. (b) x - 20 x - 20 ≥2¤ Solution: -2≥0 x - 40 x - 40 60 - x x - 60 ≥0¤ £0 ¤ x - 40 x - 40 ¤ 41 £ x £ 60 \ A contains 20 elements. 20 1 Thus, P(A) = = 100 5 Example 10: A number x is chosen at random from the set S = {1, 2, 3,…,100}.Then the probability that the x - 15 is a positive real number is expression ( x - 10)( x - 20) (a) 11/25 (c) 21/25 Ans. (c)

(b) 7/25 (d) 17/25

Probability 24.7

Solution:

¤

y=

x - 15 is a positive real number ( x - 10)( x - 20) x - 15 >0 ( x - 10)( x - 20)

Following figure shows sign of y in different intervals:

Thus, y > 0 ¤ 11 £ x £ 14 or x ≥ 21 Therefore, there are 84 favourable cases. 84 21 Hence, probability of required event is . = 100 25

LEVEL 1 Straight Objective Type Questions Example 11: Suppose A and B are two events such that 0 < P(B) < 1. If P( A) P(A | B¢) = 1 - P ( B) then which one of the following is not true? (a) P (A « B) = 0 (b) P(A « B) = P(A » B) (c) P(A » B) = P(A) + P(B) (d) P(A | B) = 0 Ans. (b) P( A) P( A « B ¢) Solution: = P(A|B¢) = 1 - P ( B) P( B ¢) fi P(A) = P(A « B¢) = P(A) – P(A « B) fi P(A « B) = 0, Also, P(A » B) = P(A) + P(B) – P(A « B) = P(A) + P(B) P ( A « B) and P(A|B) = = 0. P ( B) Example 12: A man is known to speak truth 3 out of 4 times. He takes out a card at random from a well shuffled pack of 52 playing cards, and reports it is a king. The probability that its actually a king is 1 3 (a) (b) 4 4 4 1 (c) (d) 5 5 Ans. (d) Solution: E1 : card drawn is a king E2 : card drawn is not a king A : man reports it is a king. We have 4 1 12 = , P ( E2 ) = P(E1) = 52 13 13 P(A|E1) =

3 1 and P( A | E2 ) = 4 4

By the Bayes’ rule P(E1|A) =

P( E1 )P( A | E1 ) P( E1 )P( A | E1 ) + P( E2 )P( A | E2 )

Ê 1 ˆ Ê 3ˆ ÁË ˜¯ ÁË ˜¯ 1 13 4 = = Ê 1 ˆ Ê 3 ˆ + Ê 12 ˆ Ê 1 ˆ 5 ÁË ˜¯ ÁË ˜¯ ÁË ˜¯ ÁË ˜¯ 13 4 13 4 Example 13: Suppose A has 7 fair coins and B has 6 fair coins. If A and B toss these coins simultaneously, then the probability that A and B get equal number of heads is (a)

13

(c)

13

Ê 1ˆ C6 Á ˜ Ë 2¯

13

(b)

Ê 1ˆ C12 Á ˜ Ë 2¯

13

13

Ê 1ˆ C5 Á ˜ Ë 2¯

13

13 Ê 1ˆ (d) Á ˜ Ë 2¯

Ans. (a) Solution: Let X = number of heads obtained by A. and Y = number of heads obtained by B. Note that both X and Y follow binomial distribution and X~B (7, 1/2) and Y~B (6, 1/2) 6

Now, P(X = Y ) =

 P( x = k , y = k )

k =0 6

=

Â(

k =0

=

1 213

7

Ê 1ˆ Ck Á ˜ Ë 2¯

)

7

(

6

Ê 1ˆ Ck Á ˜ Ë 2¯

)

6

6

 ( 7 Ck )( 6 C6- k )

k =0

6

But

 ( 7 Ck )( 6 C6- k )

k =0

= Number of ways of choosing 6 persons out of 7 men and 6 women = 13C6. Thus, P(X = Y ) = 13 C ÊÁ 1 ˆ˜ 6 Ë 13 ¯ 2

24.8

Complete Mathematics—JEE Main

Example 14: A box X contains 1 white ball, 3 red balls and 2 black balls. Another box Y contains 2 white balls, 3 red balls and 4 black balls. If one ball is drawn from each of the two boxes, then the probability the balls are of different colours is 19 35 (b) (a) 54 54 17 37 (c) (d) 54 54 Ans. (b) Solution: P(Balls are of different colours) = 1 – P(Balls are of the same colour) = 1 – [P(W1 W2) + P(R1 R2) + P(B1 B2)] ÈÊ 1 ˆ Ê 2 ˆ Ê 3 ˆ Ê 3 ˆ Ê 2 ˆ Ê 4 ˆ ˘ = 1 - ÍÁ ˜ Á ˜ + Á ˜ Á ˜ + Á ˜ Á ˜ ˙ ÎË 6 ¯ Ë 9 ¯ Ë 6 ¯ Ë 9 ¯ Ë 6 ¯ Ë 9 ¯ ˚ 19 35 = = 154 54 Example 15: A sample consists of six points S = {(i, j)| i = 1, 2 , j = 1, 2, 3}. Suppose k . Let A = {(i, j) | i + j = 4} i+ j B = {(i, j) | j = 2}, then P(B|A) is equal to 1 (b) 3 1 (d) 6

P ((i, j)) = and 1 2 1 (c) 4 Ans. (a) (a)

Solution: TIP It is unnecessary to calculate k.

We have P(B|A) =

P( B « A) P(2, 2) = P( A) P {(1, 3), (2, 2)} k 4

1 = . k k 2 + 4 4 Example 16: Let A be a set containing n elements. The probability of randomly choosing two subsets P and Q of A so that P = Q¢, the complement of Q, is. =

(a)

1 2

n

2

n-1

1

(c) Ans. (b)

(b) (d)

1 n

2 -1 2 n

2 -1

Solution: Number of subsets of A is 2n = m (say). The 1 number of ways of choosing P and Q is mC2 = m (m – 1) 2 = 2n–1 (2n– 1). For P = Q¢, we just have to choose Q and put the remaining elements of A in P. But the number of ways of choosing P and Q is 2n. But every selection of P and Q appears twice. 1 Therefore, the number of choosing P and Q is (2 n ) = 2n–1. 2 Thus, probability of the required event is 2 n -1 n

=

1

. 2 -1 2 (2 - 1) Example 17: Four natural numbers are selected at random and are multiplied. The probability that the product is divisible by 5 or 10 is 49 369 (b) (a) 625 625 64 256 (c) (d) 625 625 Ans. (b) Solution: The product will be divisible by 5 or 10 if at least one of the number has last digit as 0 or 5. Thus, required probability = 1 – P(none of the number has last digit 0 or 5.) 8 4 4 4 369 = 1 - ÊÁ ˆ˜ = 1 - ÊÁ ˆ˜ = Ë 10 ¯ Ë 5¯ 625 n -1

n

Example 18: Six boys and six girls sit in a row randomly. The probability that boys and girls sit alternatively is 1 5 (b) (a) 231 462 1 7 (c) (d) 462 101 Ans. (c) Solution: Total number of ways in which six boys and six girls can sit in a row is 12P12 = 12!. The first position can be occupied in two ways either by a boy or by a girl. BG BG BG BG BG BG or GB GB GB GB GB GB Six boys can take their seats in 6P6 = 6! ways and six girls can take their seats 6P6 = 6! ways. Thus, the number of favourable ways = 2(6!) (6!) 2(6!) (6!) 1 = . \ probability of the required event = 12! 462 Example 19: An unbiased die is rolled twice. Let A denote the event that an even number appears on the first throw and B denote the event that an odd number appears on the second throw. Then A and B (a) are mutually exclusive

Probability 24.9

(b) are independent and mutually exclusive (c) are independent (d) none of these Ans. (c) Solution: The sample space is given by S = {11, 12, 13, 14, 15, 16, 21, 22, 23, 24, 25, 26, 31, 32, 33, 34, 35, 36, 41, 42, 43, 44, 45, 46, 51, 52, 53, 54, 55, 56, 61, 62, 63, 64, 65, 66} We have A = {21, 22, 23, 24, 25, 26, 41, 42, 43, 44, 45, 46, 61, 62, 63, 64, 65, 66} B = {11, 13, 15, 21, 23, 25, 31, 33, 35, 41, 43, 45, 51, 53, 55, 61, 63, 65} Note that A«B = {21, 23, 25, 41, 43, 45, 61, 63, 65} As A « B π f, A and B cannot be mutually exclusive. Next,

P(A) =

18 1 18 1 = , P(B) = = , 36 2 36 2

9 1 = = P(A) P(B) 36 4 A and B are independent. P(A« B) =

\

Example 20: Two events A and B have probabilities 0.25 and 0.50 respectively. The probability that both A and B occur simultaneously is 0.14. The probability that neither A nor B occurs is (a) 0.39 (b) 0.25 (c) 0.11 (d) none of these Ans. (a) Solution: We are given P(A) = 0.25, P(B) = 0.50, P(A« B) = 0.14 Now, P (neither A nor B) = P (A¢« B¢ ) = 1 – P (A» B) = 1 – [P(A) + P(B) – P(A « B)] = 1 – [0.25 + 0.50 – 0.14] = 1 – 0.61 = 0.39. Example 21: An anti-aircraft gun can take a maximum of four shots at an enemy plane moving away from it. The probabilities of hitting the plane in the first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively. The probability that gun hits the plane is (a) 0.6976 (b) 0.866 (c) 0.922 (d) 0.934 Ans. (a) Solution: Let Ei denote the event that ith shot hits the plane. Then P(E1) = 0.4, P(E2) = 0.3, P(E3) = 0.2, P(E4) = 0.1 We have P(E1» E2 » E3 » E4)= 1 – P(E¢1« E¢2 « E¢3 « E¢4) = 1 – P(E¢1 ) P(E¢2) P(E¢3) P(E¢4)

[∵ E1, E2, E3, E4 are independent] = 1 – (0.6) (0.5) (0.8) (0.9) = 1 – 0.3024 = 0.6976. Example 22: A and B are two candidates seeking admission in IIT. The probability that A is selected is 0.5 and the probability that both A and B are selected is at most 0.3. The probability of B getting selected cannot exceed (a) 0.6 (b) 0.7 (c) 0.8 (d) 0.9 Ans. (c) Solution: We are given P(A) = 0.5 and P(A « B ) £ 0.3 We have P(A) + P(B) – P(A « B) = P (A » B ) £ 1 fi P (A) + P (B) £ 1 + P (A « B) £ 1 + 0.3 = 1.3 fi 0.5 + P (B) £ 1.3 fi P (B) £ 0.8 Thus, probability of B getting selected cannot exceed 0.8 Example 23: If letters of the word “ASSASSIN” are written down at random in a row, the probability that no two S’s occur together is (a)

1 7

(b)

1 14

(c)

1 28

(d)

1 35

Ans. (b) Solution: The number of ways of permuting the letters of the word “ASSASSIN” is 8! 8¥7¥6¥5 = 840 = 2! 4! 2 Let us now find the number of ways in which two S are not together. 4! = 12 ways. We first arrange A, A, I, N. This can be done 2! If in the following figure X’s shows one such arrangement, then 4 S’s can be arranged at four boxes out of the five boxes. ¥ ¥ ¥ ¥ Thus, corresponding to each arrangement of A, A, I, N, there are 5C4 = 5 ways of arranging S’s. \ the number of favourable ways = (12) (5) = 60 Thus, probability of the required event is

60 1 = . 840 14

Example 24: Fifteen coupons are numbered 1, 2, …, 15 respectively. Three coupons are selected at random without replacement. The probability that maximum number on the selected coupon is 9 is 1 3 (b) (a) 65 65

24.10

Complete Mathematics—JEE Main

(c)

1 13

(d)

4 65

Ans. (d) Solution: Total number of ways of selecting 3 coupons out of 15 is 15C3. Maximum number on the selected ticket will be 9 if coupon bearing number 9 is selected and two coupons from 1 to 8 are selected. This can be done in (8C2) (1C1) ways. Thus, probability of the required event is 8 C2 8¥7 3¥2 4 ¥ = = . 15 2 15 ¥ 14 ¥ 13 65 C3 Example 25: A student appears for tests I, II, and III. The student is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in tests I, II, III are p, q and 1/2, respectively. If the probability that the student is successful is 1/2, then possible values of p and q are (a) p = q = 1 (b) p = q = 1/2 (c) p = 1, q = 0 (d) p = 1, q = 1/2 Ans. (c) Solution: Let A, B and C be the events that the student is successful in tests I, II and III, respectively. Then P(the student is successful) = P [(A « B « C¢) » (A « B¢ « C) » (A « B « C)] = P(A « B « C¢) + P(A « B¢ « C) + P(A « B « C) = P(A) P(B) P(C¢) + P(A) P(B¢) P(C) + P(A) P(B) P(C) [∵ A, B and C are independent] = pq (1 - 1 2 ) + p (1 - q) (1 2 ) + ( pq) (1 2 ) 1 1 [pq + p(1 – q) + pq] = p(1 + q) 2 2 1 1 = p (1 + q) fi p (1 + q) = 1 \ 2 2 This equation is satisfied for the pair of values given in (c). =

Example 26: A bag contains a white and b black balls. Two players, A and B alternately draw a ball from the bag, replacing the ball each time after the draw till one of them draws a white ball and wins the game. A begins the game. If the probability of A winning the game is three times that of B, the ratio a : b is (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) none of these Ans. (c) Solution: Let W denote the event of drawing a white ball at any draw and B that for a black ball. Then P(W ) =

a b and P(B) = a+b a+b

P(A wins the game) = P(W or BBW or BBBBW or  ) = P(W) + P(BBW) + P(BBBBW) +  = P(W) + P(B) P(B) P(W) + P(B) P(B) P(B) P(B) P(W) +  = P(W) + P(W) ◊ P(B)2 + P(W) ◊ P(B)4 +  = Also

P(W ) 1 - P ( B)

2

=

a ( a + b) 2

a + 2ab

P(B wins the game) = 1 –

=

a+b a + 2b

a+b b = a + 2b a + 2b

According to the given condition, fi

a+b b = 3◊ a + 2b a + 2b

fi

a = 2b fi a : b = 2 : 1.

Example 27: A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the determinant chosen is non-zero is 3 3 (a) (b) 16 8 (c)

1 4

(d)

5 16

Ans. (b) Solution: A determinant of order 2 is of the form a b D= c d It is equal to ad – bc. The total number of ways of choosing a, b, c and d is 2 ¥ 2 ¥ 2 ¥ 2 = 16. Now D π 0 if and only if either ad = 1, bc = 0 or ad = 0, bc = 1. But ad = 1, bc = 0 if and only if a = d = 1 and at least one of b, c is zero. Thus, ad = 1, bc = 0 in three cases. Similarly, ad = 0, bc = 1 in three cases. Thus, the probability of the required event is 6/16 = 3/8. Example 28: A fair coin is tossed 100 times. The probability of getting tails an odd number of times is (a) 1/2 (b) 1/8 (c) 3/8 (d) none of these Ans. (a) Solution: Let p = probability of getting a tail in a single trial = 1/2. n = number of trials = 100 and X = number of tails in 100 trials. Note that q = 1/2 We have P(X = r) = 100Cr pr q n – r r 100 - r 100 Ê 1ˆ Ê 1ˆ Ê 1ˆ = 100 Cr Á ˜ Á ˜ = 100 Cr Á ˜ Ë 2¯ Ë 2¯ Ë 2¯

Probability 24.11

= P(A) + P(B) – P(A) P(B) [∵ A and B are independent] = 0.3 + P(B) – (0.3) P(B) 0.5 = (0.7) P(B) fi P(B) = 5/7.

Now, P(X = odd) = P(X = 1) + P(X = 3) + … + P(X = 99) Ê 1ˆ = 100C1 Á ˜ Ë 2¯

100

Ê 1ˆ + 100C99 Á ˜ Ë 2¯

Ê 1ˆ + 100C3 Á ˜ Ë 2¯

100

+…

100

Ê 1ˆ = (100C1 + 100C3 + … + 100C99) Á ˜ Ë 2¯ 100

But Thus,

100

C1 + 100C3 + … + 100C99 = 299

P(X = odd) =

299 100

=

1 . 2

2 Example 29: The probability that at least one of A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, then P(A¢) + P(B¢) is (a) 0.9 (b) 1.15 (c) 1.1 (d) 1.2 Ans. (c)

fi

Example 32: A pair of unbiased dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is 2 3 (b) (a) 5 5 4 1 (c) (d) 5 5 Ans. (a) Solution: Let A denote the event that a sum of 5 occurs, B the event that a sum of 7 occurs and C the event that neither a sum of 5 nor a sum of 7 occurs. We have 4 1 6 1 26 13 = , P(B) = = and P(C) = = P(A) = . Thus, 36 9 36 6 36 18 P(A occurs before B) = P [A or (C « A) or (C « C « A) or ]

Solution: We have P(A » B) = 0.6 and P(A « B) = 0.3. We know that

= P(A) + P(C « A) + P(C « C « A) +  = P(A) + P(C) P(A) + P(C)2 P(A) + 

P(A) + P(B) = P(A » B) + P(A « B) = 0.6 + 0.3 = 0.9 \

P(A¢) + P(B¢) = 1 – P(A) + 1 – P(B) = 2 – 0.9 = 1.1.

Example 30: An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of the four faces obtained, the probability that the minimum value is not less than 2 and the maximum value is not greater than 5 is 16 1 (a) (b) 81 81 (c)

80 81

(d)

65 81

Ans. (a) Solution: The probability of getting a number not less than 2 and not greater than 5 in a single throw is 4/6 = 2/3. As die is rolled four times, the probability of required event = (2/3)4 = 16/81. Example 31: Let A and B be two events such that P(A) = 0.3 and P(A » B) = 0.8. If A and B are independent events, then P(B) is 3 4 (b) (a) 7 7 5 6 (c) (d) 7 7 Ans. (c) Solution: We have, 0.8 = P(A » B) = P(A) + P(B) – P(A « B)

=

1 Ê 13 ˆ 1 Ê 13 ˆ 2 1 +Á ˜ +Á ˜ +º 9 Ë 18 ¯ 9 Ë 18 ¯ 9

=

1/ 9 2 = 1 - 13 /18 5

[sum of an infinite G.P.]

Example 33: E and F be two independent events such that P(E) < P(F). The probability that both E and F happen is 1/12 and the probability that neither E nor F happen is 1/2. Then (a) P(E) = 1/3, P(F) = 1/2 (b) P(E) = 1/2, P(F) = 2/3 (c) P(E) = 2/3, P(F) = 3/4 (d) P(E) = 1/4, P(F) = 1/3 Ans. (d) Solution: We are given P(E « F) = 1/12 and

P(E¢ « F¢ ) = 1/2

As E and F are independent, we get P(E) P(F) = 1/12 and

P(E¢ ) P(F¢ ) = 1/2

fi

(1 – P(E)) (1– P(F)) = 1/2

fi

1 – (P(E) + P(F)) + P(E) P (F) = 1/2

fi

P(E) + P(F) = 1 + 1/12 – 1/2 = 7/12

\

or

Equation whose roots are P(E) and P(F) is x2 – (P(E) + P(F)) x + P(E) P(F) = 0 7 1 x+ fi 12x2 – 7x + 1 = 0 x2 – 12 12

24.12

fi

Complete Mathematics—JEE Main

(3x – 1) (4x – 1) = 0 fi x = 1/3, 1/4

As P(E) < P(F), we take P(E) = 1/4 and P(F) = 1/3. Example 34: A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six is (a) 3/8 (b) 1/5 (c) 3/4 (d) none of these Ans. (a) Solution: Let E denote the event that a six occurs and A the event that the man reports that it is a six. We have P(E) = 1/6 , P(E¢) = 5/6, P(A | E) = 3/4 and P( A | E¢) = 1/4. By Bayes’ theorem P(E ) P( A | E ) P(E | A) = P( E ) P( A | E ) + P( E ¢) P( A | E ¢) 3 (1 6) (3 4) = = (1 6) (3 4) + (5 6) (1 4) 8 Example 35: The event A is independent of itself if and only if P(A) is (a) 0 or 1 (b) 1/2 (c) 0 (d) 0, 1/2 or 1 Ans. (a) Solution: A is independent of itself if and only if P(A « A) = P(A) P(A) ¤ P(A) = P(A)2 ¤ P(A) = 0 or P(A) = 1. Example 36: If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all the three numbers are divisible by both 2 and 3 is 4 7 (a) (b) 57 99 4 1 (c) (d) 1155 1100 Ans. (c) Solution: A number is divisible by both 2 and 3 if and only if it is divisible by 6. In the first 100 natural numbers, there are 16 numbers which are divisible by 6. Thus, probability of the required event is 4 16 C3 100 C3 = 1155 Example 37: One ticket is selected at random from 100 tickets numbered 00, 01, 02, º, 98, 99. If X and Y denote respectively the sum and the product of the digits on the tickets, then P(X = 9 | Y = 0) = 2 2 (a) (b) 17 19 2 2 (c) (d) 21 11

Ans. (b) Solution: (Y = 0) is {00, 01, , 09, 10, 20, , 90}. Also, (X = 9) « (Y = 0) = {09, 90}. We have P(Y = 0) = 19/100 and P[(X = 9) « (Y = 0)] = 2/100. P(X = 9 | Y = 0) =

P[( X = 9) « (Y = 0)] 2 = . 19 P(Y = 0)

Example 38: If three six-faced fair dice are thrown together, the probability that the sum of the numbers appearing on the dice is 16 is (a)

1 36

(b)

1 11

(c)

1 12

(d)

5 36

Ans. (a) Solution: The total number of cases is 6 × 6 × 6 = 6 3. If the first die shows 6, sum of the faces on the remaining two dice must be 10 which can happen in 3 ways viz. outcome should be (6, 4), (5, 5) or (4, 6). If the first die shows 5, then other two must show (5, 6) or (6, 5).If the first die shows 4, then other two must show (6, 6). \ number of favourable ways = 3 + 2 + 1 = 6. 6 1 . Thus, probability of the required event = 3 = 36 6 Example 39: Four person are selected at random from a group of 3 men, 2 women and 4 children. The probability that exactly two of them are children is (a)

9 21

(b)

10 23

(c)

1 2

(d)

10 21

Ans. (d) Solution: Total number of ways of selecting 4 persons out of 9 is 9C4. The number of selecting exactly two children is (4C2) (5C2). \ probability of the required event 4 C2 ) ( 5C2 ) 10 ( = = 9

C4

21

Example 40: 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 2 (b) 0 (a) 5 3 (c) 1 (d) 5

Probability 24.13

Ans. (c) Solution: A = {4, 5, 6} and B = {1, 2, 3, 4} then A » B = {1, 2, 3, 4, 5, 6} \ P(A » B) = 1.

Ans. (d)

Example 41: If A and B are two events. The probability that exactly one of them occurs is equal to (a) P(A) + P(B) – 2P(A « B) (b) P(A) + P(B) + P(A « B) (c) P(A) + P(B) (d) P(A) + P(B) – P(A « B) Ans. (a) Solution: See Theory. Example 42: If A and B are two events such that P(A) > 0 and P(B) < 1, then P(A| B ) is equal to (a) 1 – P( A |B)

(b) 1 – P(A|B)

P ( A) P ( B)

(d) 1– P ( A | B )

(c) Ans. (d)

Solution: P(A| B ) =

P ( A « B) P (B) – P ( A « B) = P (B) P (B)

= 1 – P ( A | B) Example 43: For a biased die, the probabilities for the different faces to turn up are given by the table Face Probability

1

2

3

4

5

6

0.1

0.32

0.21

0.15

0.05

0.17

The die is thrown and you are told that either the face 1 or the face 2 has turned up, then the probability that it is face 1, is 16 1 (a) (b) 21 10 5 5 (c) (d) 16 21 Ans. (d) Solution: P (face 1 | face 1 or face 2) P (face 1) = P (face 1) + P(face 2) 0.1 0.1 5 = = . 0.1 + 0.32 0.42 21 Example 44: A six faced die is so biased that it is twice likely to show an even number as an odd number when thrown. If the die is thrown twice the probability that sum of the numbers is even is 1 2 (a) (b) 3 3 4 5 (c) (d) 9 9

Solution: Let p denote the probability of showing an odd number, then the probability that die will show an even number is 2p. We have p + 2p = 1 fi 3p = 1 fi p = 1/3 Thus, probability of getting an even sum when dice is thrown twice is P (odd, odd) + P (even, even) = P (odd)2 + P (even)2 = p2 + (2p)2 = 5p2 = 5/9 Example 45: While shuffling a pack of playing cards, four are accidently dropped. The probability that the cards that are dropped are one from each suit is 1 2197 (a) (b) 20825 20825 1 301 (c) (d) 256 20825 Ans. (b) Solution: Total number of ways of choosing four cards out of 52 is 52C4. The number of ways of choosing one card from each suit is (13C1) (13C1) (13C1) (13C1) = 134. Thus, probability of the required event is 134 52

=

2197 13 ¥ 13 ¥ 13 ¥ 13 ¥ 4 ¥ 3 ¥ 2 ¥ 1 = . 52 ¥ 51 ¥ 50 ¥ 49 20825

C4 Example 46: The probability that a man who is 85 years old will die before attaining the age of 90 is 1/3. Four persons A1, A2, A3 and A4 are 85 years old. The probability that A1 will die before attaining the age of 90 and will be the first to die is 65 13 (a) (b) 81 81 65 13 (c) (d) 324 108 Ans. (c) Solution: Required probability = (probability at least one of A1, A2, A3 and A4 dies) × (probability that A1 is first to die) Now, probability that at least one of A1, A2, A3 and A4 dies 4 16 65 Ê 2ˆ = 1 – P(none of them dies)= 1 - Á ˜ = 1 = Ë 3¯ 81 81

=

65 1 65 ¥ = . 81 4 324 Example 47: Three persons A, B and C are to speak at a function along with 5 others. If the persons speaks in a random order, the probability that A speaks before B and B speaks before C is 3 1 (a) (b) 8 12

\

Required probability =

24.14

Complete Mathematics—JEE Main

(c)

1 8

(d)

\

1 6

Ans. (d) Solution: The number of ways in which 8 persons can speak is 8P8 = 8!. We can choose 3 places for A, B, C (out of 8) in 8C3 ways and arrange A, B, C in that order in just one way. We can arrange the remaining 5 persons in 5! ways. Thus, the number of favourable ways is (8C3) (5!).

8 C3 ) (5!) 1 1 ( = = . \ probability of the required event =

8!

3!

6

Example 48: The probability that a man will live for 10 more years is 1/4 and the probability that his wife will live for 10 more years is 1/3. The probability that neither the husband nor the wife will be alive after 10 years is 5 1 (b) (a) 12 2 7 11 (c) (d) 12 12 Ans. (b) Solution: We have P(H) = 1/4 and P(W) = 1/3. Now, P(neither husband nor wife will be alive) = P(H¢ « W¢ ) = P(H¢ ) P(W¢ ) = (1 – P(H)) (1 – P(W)) = (1 – 1/4) (1 – 1/3) = 1/2. Example 49: Two numbers x and y are chosen at random without replacement from the first 30 natural numbers. The probability that x2 – y2 is divisible by 3 is 3 3 (a) (b) 29 55 5 47 (c) (d) 29 87 Ans. (d) Solution: Total number of ways of choosing three numbers out of 30 is 30C3 . We rewrite the first 30 natural numbers in three rows as follows: Row I: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28 Row II: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29 Row III: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 For x2 – y2 to be divisible by 3, either both x and y must be chosen from the same row or exactly one of x, y from Row I and the other from Row II. Thus, the number of favourable ways = 3(10C2) + 10 × 10 = 235

probability of the required event 235 235 ¥ 2 47 = = 30 = . C2 30 ¥ 29 87

Example 50: If m is a natural such that m £ 5, then the 1 m probability that the quadratic equation x2 + mx + + =0 2 2 has real roots is (a)

1 5

(b)

2 3

(c)

3 5

(d)

1 5

Ans. (c) Solution: Discriminant D of the given quadratic equation is Ê 1 mˆ D = m2 – 4 Á + ˜ = m2 – 2m – 2 = (m – 1)2 – 3 Ë2 2¯ Now, D ≥ 0 ¤ (m – 1)2 ≥ 3 This is possible for m = 3, 4 and 5. Also, the total number of ways of choosing m is 5. \ probability of the required event = 3/5. Example 51: If two events A and B such that P(A¢) = 0.3, P(B) = 0.5 and P(A « B) = 0.3, then P(B|A » B¢) is (a) 0.6 (b) 0.32 (c) 0.31 (d) 0.28 Ans. (a) Solution: We have P(A» B¢) = P(A) + P(B¢) – P(A « B¢) = [1 – P(A¢ )] + [1 – P(B)] – [P(A) – P(A « B)] = (1 – 0.3) + (1 – 0.5) – (1 – 0.3) = 0.5. Now, P (B|A » B¢) =

P [ B « ( A » B¢ )] P[( B « A) » ( B « B¢ )] = P ( A » B¢ ) P ( A » B¢ )

P ( A « B) 0.3 = = 0.6 P ( A » B¢ ) 0.5 Example 52: Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals 1 7 (b) (a) 2 15 =

(c)

2 15

(d)

1 3

Ans. (b) Solution: The number of ways of placing 3 black balls at 10 places is 10C3. The number of ways in which two black balls are not together is equal to the number of ways of choosing 3 places marked with X out of the eight places XWXWXWXWXWXWXWX

Probability 24.15

This can be done in 8C3 ways. Thus, probability of the required event is 8

C3

10

C3

(a)

13 32

(b)

1 4

(c)

1 32

(d)

3 16

Ans. (a) Solution: P (2 white and 1 black) = P(W1 W2 B3 or W1 B2 W3 or B1 W2 W3) = P(W1) P(W2) P(B3) + P(W1) P(B2) P(W3) + P(B1) P(W2)P(W3)

=

Ê 3ˆ Ê 3ˆ ÁË ˜¯ + ÁË ˜¯ 4 4

Ê 2ˆ ÁË ˜¯ 4

Ê 1ˆ Ê 1ˆ ÁË ˜¯ + ÁË ˜¯ 4 4

Ê 2ˆ ÁË ˜¯ 4

Ê 1ˆ ÁË ˜¯ 4

13 32

Example 54: There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is 1 1 (b) (a) 3 6 (c)

We have

P(E | F ) + P ( E | F )

8¥7¥6 7 = = . 10 ¥ 9 ¥ 8 15

Example 53: If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that two white and one black ball will be drawn is

Ê 3ˆ Ê 2ˆ = Á ˜Á ˜ Ë 4¯ Ë 4¯

Alternate Solution

1 2

(d)

1 4

Ans. (a) Solution: Either both the machines tested should be faulty or both should be good. \ required probability =

2 ¥1 2 ¥1 1 = . ¥ 4¥3 4¥3 3

Example 55: If E and F are two events such that 0 < P(F) < 1, then (a) P(E| F ) + P ( E | F ) = 1 (b) P(E|F) + P(E| F ) = 1 (c) P( E |F) + P(E| F ) = 1 (d) none of these Ans. (a) Solution: As P(| F ) is a probability function, (a) is clearly true.

=

P( E « F ) P ( E « F ) + P (F ) P (F )

=

P (E « F ) + P (E « F ) P (F ) = =1 P (F ) P (F )

Example 56: The probability that an event A occurs in a single trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that A occurs at least once is (a) 0.936 (b) 0.784 (c) 0.904 (d) 0.788 Ans. (b) Solution: Let p= probability that event A occurs in a single trial = 0.4 n = number of independent trials = 3 and X= number of times A occurs. Note that X ~ B (n, p) P (A occurs at least once) = P (X ≥ 1) = 1 – P(X = 0) = 1 – q3 = 1 – (0.6)3 = 1 – 0.216 = 0.784. Example 57: Fifteen coupons are numbered 1, 2, … , 15 respectively. Seven coupons are selected at random one at a time with replacement. The probability that the largest number on a selected coupon as 9 is 6 Ê 9ˆ (a) Á ˜ Ë 15 ¯

7 Ê 8ˆ (b) Á ˜ Ë 15 ¯

7 Ê 3ˆ (c) Á ˜ (d) none of these Ë 5¯ Ans. (d) Solution: Let p = the probability that a selected coupon bears number £ 9. 9 3 fi p = = 15 5 n = number of coupons drawn with replacement and X = the number of coupons bearing number £ 9. Note that X ~ B (n, p). Probability that the largest number on the selected coupons does not exceed 9 = probability that all the coupons bear number £ 9 7 Ê 3ˆ = P(X = 7) = 7C7 p7 = Á ˜ . Ë 5¯ Similarly, probability that largest number on the selected

8 7 coupon is £ 8 is Ê ˆ . Hence, probability of the required Ë 15 ¯ 3 7 8 7 event = Ê ˆ – Ê ˆ . Ë 5¯ Ë 15 ¯

24.16

Complete Mathematics—JEE Main

Example 58: Each of two persons A and B toss three fair coins. The probability that both get the same number of heads is (a) 3/8 (b) 1/9 (c) 5/16 (d) none of these Ans. (c) Solution: Let X be the number of heads obtained by A and Y be the number of heads obtained by B. Note that both X and Y are binomial variate with parameters n = 3 and p = 1/2. Probability that both A and B obtain the same number of heads is \ P(X = 0) P(Y = 0) + P(X = 1) P(Y = 1) + P(X = 2) P(Y = 2) + P(X = 3) P(Y = 3) 2

3 3 È Ê 1ˆ ˘ Ê 1ˆ ˘ È = Í 3 C0 Á ˜ ˙ + Í 3 C1 Á ˜ ˙ Ë 2¯ ˚ Ë 2¯ ˚ Î Î 2

2

3 3 È Ê 1ˆ ˘ Ê 1ˆ ˘ È + Í 3 C2 Á ˜ ˙ + Í 3 C3 Á ˜ ˙ Ë 2¯ ˚ Ë 2¯ ˚ Î Î

2

Solution: Let X be the number of coins showing heads. Then X ~ B (100, p). We have P (X = 51) = P (X = 50) fi

100

C51 p51 q49 = 100C50 p50 q50

p 100! 51! 49! p 51 = fi = q 50! 50! 100! 1 - p 50

fi fi fi

50p = 51 – 51p p = 51/101.

Ê 4ˆ (a) Á ˜ Ë 5¯

Ê 4ˆ (c) 5C4 Á ˜ Ë 5¯ Ans. (b)

4

Ê 4ˆ (b) Á ˜ Ë 5¯

4

Ê 3ˆ (d) Á ˜ Ë 4¯

3

Ê 1ˆ = 1 – qn = 1 – Á ˜ Ë 2¯

4 4 Ê 4ˆ Ê 1ˆ Ê 4ˆ = 5C 4 p 4q = 5 Á ˜ Á ˜ = Á ˜ . Ë 5 ¯ Ë 5¯ Ë 5 ¯ Example 60: One hundred identical coins, each with probability p of showing up heads, are tossed once. If 0 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is 1 49 (b) (a) 2 101

(c) Ans. (d)

50 101

(d)

51 101

n

n Ê 1ˆ P (X ≥ 1) ≥ 0.95 fi 1 – Á ˜ ≥ 0.95 Ë 2¯

Now, n

\

Solution: Let p = probability that a student selected at random is a swimmer = 1 – 1/5 = 4/5. n = number of students selected = 5 and X = number of swimmers Then X ~ B (n, p) where n = 5 and p = 4/5 Probability that exactly 4 students are swimmer is

101p = 51

Solution: Let n be the required number of tosses, and X be the number of heads obtained in n tosses. Then X ~ B(n, 1/2) P (at least one head) = P(X ≥ 1) = 1 – P(X = 0)

20 5 Ê 1ˆ = = Á ˜ [1 + 9 + 9 + 1] = . Ë 2¯ 64 16

3

fi

Example 61: The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.95 (a) 5 (b) 6 (c) 7 (d) 12 Ans. (a)

6

Example 59: The probability that a student is not a swimmer is 1/5. The probability that out of 5 students exactly 4 are swimmer is

(where q = 1 – p)

Ê 1ˆ ÁË ˜¯ £ 0.05 fi 2 least value of n is 5.

2 n ≥ 20

fi

n≥5

Example 62: A box contain N coins, m of which are fair and rest are biased. The probability of getting a head when a fair coin is tossed is 1/2, while it is 2/3 when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. The probability that the coin drawn is fair is 8m m (b) (a) 8N + m 8N + m (c)

9m 8N + m

(d)

9N 8N + m

Ans. (c) Solution: Let E1, E2 and A denote the following events: E1 : coin selected is fair E2 : coin selected is biased A : the first toss results in a head and the second toss results in a tail. m N -m , P(E1) = , P(E2) = N N P(A|E1) =

1 1 1 2 1 2 ¥ = , P(A| E2) = ¥ = 2 2 4 3 3 9

Probability 24.17

By Bayes’ rule P( E1 ) P( A | E1 ) P (E1| A) = P( E1 ) P ( A | E1 ) + P( E2 ) P( A | E2 ) 9m . 8N + m Example 63: The probability of India winning a test match against West Indies is 1/2. Assuming independence from match to match, the probability that in a 5 match series India’s second win occurs at the third test is (a) 1/8 (b) 1/4 (c) 1/2 (d) 2/3 Ans. (b) =

Solution: Required probability = P(winning exactly one match in the first two matches) ¥ P(winning third march) È Ê 1ˆ Ê 1ˆ ˘ È1 ˘ 1 = Í 2 C1 Á ˜ Á ˜ ˙ Í ˙ = Ë 2¯ Ë 2¯ ˚ Î2 ˚ 4 Î Example 64: Let 0 < P(A) < 1, 0 < P(B) < 1 and P(A » B) = P(A) + P(B) – P(A) P(B), then (a) P(A |B) = 0 (b) P(B |A) = 0 (c) P(A¢ « B¢ ) = P(A¢ ) P(B ¢) (d) P(A|B) + P(B|A) = 1 Ans. (c) Solution: We know that P(A » B) = P (A) + P(B) – P(A « B) fi P(A) + P(B) – P(A) P(B) = P(A) + P(B) – P (A « B) fi P(A) P(B) = P(A « B) fi A and B are independent. Thus, P(A¢ « B¢) = P(A¢) P(B ¢). Example 65: If A1, A2, …, An are n independent events 1 , i = 1, 2, …, n. The probability that i +1 none of A1, A2, … An occurs is

such that P(Ai) =

(a) (c)

n n +1

(b)

1 n!

(d)

1 n +1 1 n+2

Ans. (b)

Example 66: A four digit number (numbered from 0000 to 9999) is said to be lucky if the sum of its first two digits is equal to the sum of its last two digits. If a four digit number is picked up at random, then the probability that it is lucky is (a) 0.065 (b) 0.064 (c) 0.066 (d) 0.067 Ans. (d) Solution: The total number of ways of choosing the ticket is 10000. Let the four digits number on the ticket be x1 x2 x3 x4. Note that 0 £ x1 + x2 £ 18 and 0 £ x3 + x4 £ 18. Also, the number of non-negative integral solutions of x + y = m (with 0 £ x, y £ 9) is m + 1 if 0 £ m £ 9 and is 19 – m if 10 £ m £ 18. Thus, the number of favourable ways = 1 ¥ 1 + 2 ¥ 2 + … + 10 ¥ 10 + 9 ¥ 9 + 8 ¥ 8 + … +1¥1 9 ¥ 10 ¥ 19 + 100 = 670 = 2 6

{

\

}

probability of required event =

Example 67: Three numbers are chosen at random without replacement from {1, 2, 3, …, 10}. The probability that minimum of the chosen number is 3 or their maximum is 7, is 11 11 (a) (b) 30 40 (c)

1 7

(d)

1 ˆ Ê 1 ˆ Ê 1ˆ Ê = Á1 - ˜ Á1 - ˜ º Á1 ˜ Ë 2 ¯ Ë 3¯ Ë n + 1¯ =

1 2 3 n -1 n 1 ¥ = . ¥ ¥ ¥º¥ n +1 n +1 2 3 4 n

1 8

Ans. (b) Solution: Let A and B denote the following events: A: minimum of the chosen number is 3 B: maximum of the chosen number is 7 We have P(A) = P (choosing 3 and two other numbers from 4 to 10) 7

=

C2

10

C3

=

7¥6 3¥2 7 ¥ = 2 10 ¥ 9 ¥ 8 40

P(B) = P (choosing 7 and two other numbers from 1 to 6) 6

Solution: P(A¢ « A¢ 2 … « A¢n) = P(A¢1) P (A¢2 ) … P(A¢n) [∵ A1, A2, …, An are independent]

670 = 0.067 . 1000

=

C2

10

C3

=

6¥5 3¥2 1 ¥ = 2 10 ¥ 9 ¥ 8 8

P(A « B) = P (choosing 3 and 7 and one other number from 4 to 6). =

3 10

C3

=

3¥3¥2 1 = . 10 ¥ 9 ¥ 8 40

24.18

Complete Mathematics—JEE Main

Now, P(A » B) = P(A) + P(B) – P(A « B) 7 1 1 11 = + = 40 8 40 40 Example 68: One ticket is selected at random from 50 tickets numbered 00, 01, 02, ..., 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals 5 1 (b) (a) 14 50 (c)

1 14

(d)

1 7

Ans. (c) Solution: Let A and B denote the following events. A : sum of digits is 8 and B: product of digits is 0. Then A « B = {08} and B = {00, 01, 02, ..., 09, 10, ... 30, 40} Now,

P(A | B) =

P ( A « B) n ( A « B) 1 = = 14 P ( B) n ( B)

Example 69: An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is (a) (c)

1 21

(b)

1 3

(d)

Example 71: A signal which can be green or red with 4 1 and respectively, is received by station A 5 5 and then transmitted to station B. The probability of each 3 station receiving the signal correctly is . If the signal 4 received at station B is green, then the probability that the original signal was green is 3 6 (b) (a) 5 7

probability

(c)

20 23

(d)

2 7

Solution: Total number of ways of drawing three balls out of 9 is 9C3 = 84. The number of ways of drawing the balls so that balls are of different colours is (3C1) (4C1) (2C1) = (3) (4) (2) = 24 24 2 probability of the required event = = 84 7

Example 70: Let w be a complex cube root of unity with w π 1. A fair die is thrown three times. If r1, r2 and r3 the numbers obtained on the die, then the probability that wr1 + wr2 + wr3 = 0 is (a)

1 18

(b)

1 9

(c)

2 9

(d)

1 36

Ans. (c) Solution: Total number of ways of throwing a fair die three times is 6 ¥ 6 ¥ 6.

9 20

Ans. (c) Solution:

Station A

2 23

Ans. (d)

\

Note that wr1 + wr2 + wr3 = 0 if r1, r2, r3 is a permutation of the numbers of the form {3k, 3m + 1, 3n + 2} from the set {1, 2, 3, 4, 5, 6} Thus, r1, r2, r3 can be chosen in (2) (2) (2) (3!) = (8) (6) ways. 2 (8)(6) = Hence, probability of the required event = (6)(6)(6) 9

3/4 4/5

G

Station B G

3/4 1/4

G

3/4 1/4

1/4 R

1/ 5

3/4 R

3/4 1/4

R

R R G R G

1/ 4

G

3/4 G

1/4

R

Fig. 24.6

Let G, E1, E2, and E denote the following events : G : Original signal is green E1 : A receives the signal correctly E2 : B receives the signal correctly E : B receives the green signal We have E = G E1 E2 » G E¢1 E¢2 » G¢ E1 E¢2 » G¢ E¢1 E2 fi P (E) = P (G E1 E2) + P (G E¢1 E¢2) + P(G¢ E1 E¢2) + P (G¢ E¢1 E¢2) Ê 4ˆ Ê 3ˆ Ê 3ˆ Ê 4ˆ Ê 1ˆ Ê 1ˆ = Á ˜Á ˜Á ˜ +Á ˜Á ˜Á ˜ Ë 5¯ Ë 4¯ Ë 4¯ Ë 5¯ Ë 4¯ Ë 4¯

Probability 24.19

Ê 1ˆ Ê 3ˆ Ê 1 ˆ Ê 1ˆ Ê 1 ˆ Ê 3ˆ +Á ˜ Á ˜ Á ˜ +Á ˜ Á ˜ Á ˜ Ë 5¯ Ë 4 ¯ Ë 4 ¯ Ë 5¯ Ë 4 ¯ Ë 4 ¯ 36 + 4 + 3 + 3 23 = 80 40 Also, P(G « E) = P(G E1 E2) + P(G E¢1 E¢2) 40 1 = = 80 2 =

\

P(G | E) =

P (G « E ) 1/ 2 20 = = 23 / 40 23 P (E )

Example 72: Three numbers are chosen at random from the numbers 1, 2, ... 20. The probability that the arithmetic mean of these numbers is 4 is (a) 7/2280 (b) 3/4 (c) 2/7281 (d) 0 Ans. (a) Solution: Total number of ways of choosing three numbers out of 20 is 20C3= 1140 Let these numbers be x1, x2, x3. 1 We have ( x1 + x2 + x3 ) = 4, 3 we get x1 + x2 +x3 =12. The possible choices of (x1, x2, x3 ) are (1, 2, 9), (1, 3, 8), (1, 4, 7), (1, 5, 6) (2, 3,7), (2, 4, 6), (3, 4, 5) 7 \ probability of the required event = . 1140 Example 73: An unbiased cubical die is thrown 5 times. The probability that the maximum number appearing on the die is 4 is (b) 1023/65 (a) 7/65 5 (c) 3781/6 (d) 1781/65 Ans. (b) Solution: Let X = maximum number appearing on the cubical die. P(X = 4) = P(X < 4) – P(X < 3)

=

Example 75 Suppose A, B, C are three independent events such that P(A) = P (B) = P(C) = p, then P (at least two of A, B, C) equals (b) 3p2 – 2p3 (a) 2p2 –3p3 (c) p2 – p3 (d) 2p2 – p3 Ans. (b) Solution: P(at least two of A, B, C) = P [( A « B « C¢ ) » (A « B¢« C) » (A¢ « B « C) » (A « B « C)] = P ( A « B « C¢) + P(A « B¢ « C) + P(A¢ « B « C) + P(A « B « C) = p2 (1 – p) + p2 (1 – p) + p2(1 – p) + p3 [A, B, C are independent] 2 3 = 3p –2p Example 76: A and B are two students. Their probabilities of solving a problem correctly are 1/4 and 1/5 respectively. If the probability of their making a common error is 1/40, and they obtain the same answer, then the probability of their answer is correct is (a) 1/12 (b) 1/20 (c) 10/13 (d) 13/200 Ans. (c) Solution: Let A, B, C and S denote the following events. A : A solves the problem correctly; B : B solves the problem correctly; C : A and B make common errors; S : A and B reach the same answer. We have S = (A « B) » (A¢ « B¢ « C) fi P(S) = P (A « B) + P (A¢ « B¢ « C) =P(A) P(B) + P(A¢) P(B¢) P(C) 13 Ê 1 ˆ Ê 1ˆ Ê 3ˆ Ê 4ˆ Ê 1 ˆ = Á ˜Á ˜ +Á ˜Á ˜Á ˜ = Ë 4 ¯ Ë 5 ¯ Ë 4 ¯ Ë 5 ¯ Ë 40 ¯ 200 Now P (answer is correct | S)

5 5 5 5 25 1 1023 Ê 4ˆ Ê 3ˆ Ê 2ˆ Ê 1ˆ = Á ˜ -Á ˜ = Á ˜ -Á ˜ = 5 - 5 = 5 Ë 6¯ Ë 6¯ Ë 3¯ Ë 2¯ 6 3 2

=

Example 74: Let A and B be two events such that P(A) = 3/7, P(B) = 4/7 and P(A » B) = 5/7, P ( A | B) Then equals P ( A ¢ | B) (a) 1/3 (b) 2/3 (c) 1/2 (d) 3/8 Ans. (c)

=

Solution: P(A « B) = P(A) + P(B) – P(A » B) = 2/7 Now,

P ( A | B) P ( A « B) P ( B ) = P ( A ¢ | B) P ( B) P ( A ¢ « B)

P ( A « B) 27 1 = = P ( B) - P ( A « B) 4 7-2 7 2

P( A « B « S ) P ( A « B) = P (S ) P (S )

(1 4)(1 5) 13 200

=

10 13

Example 77: Suppose A and B are two events such that P(A) π 0, P(B) π 0, then (a) P(A|B) = P(A)/P(B) (b) P(A|B) = P(A « B)/P(B) (c) P(A|B).P(B/A) = 1 (d) P(A|B) = P(A)P(B) Ans. (b) Solution: See theory.

24.20

Complete Mathematics—JEE Main

Example 78: Fig. 24.7 shows three events A, B and C. Probabilities of different events are shown in the figure. For instance, P(A « B¢ « C) = 0.18, P(A¢ « B « C¢) = 0.06 etc. A

B

0.24

0.18 0.14

0.06

0.24 0.06 0.06

= pq3 + pq4 + pq5 + ... =

pq3 pq3 = = q3 1- q p

p (X > 6) = q5

and Thus,

P(X > 6 | X > 3) =

q5 q

3

= q2 =

25 . 36

Example 80: If P(A) = 0.4, P(B ¢) = 0.6 and P(A « B) = 0.15 then the value of P(A/A¢ » B¢) is 4 5 (a) (b) 17 17 (c)

10 17

(d)

1 17

Ans.(b) C

Solution: P(A/A¢ » B¢) = Fig. 24.7

Which of the following is not true ? (a) A and B are independent (b) B and C are independent (c) A and C are independent (d) A and B « C are independent Ans. (c) Solution: Note that P(A) = 0.18 + 0.24 + 0.24 + 0.14 = 0.8 P(B) = 0.06 + 0.06 + 0.24 + 0.24 = 0.6 P(C) = 0.06 + 0.14 + 0.24 + 0.06 = 0.5 P(B « C) = 0.24 + 0.06 = 0.3 = P(B) P(C) P(C « A) = 0.14 + 0.24 = 0.38 π P(C) P(A) P(A « B) = 0.24 + 0.24 = 0.48 = P(A) P(B) P(A « B « C) = 0.24 = P(A) P(B « C) Example 79: A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. The probability that X > 3 equals 125 25 (a) (b) 216 216 5 25 (c) (d) 36 36 Ans.(d) 1 5 Solution: Let p = , q = 6 6 We have P (X = r) = pqr – 1 " r > 1. Now, P [( X ≥ 6 ) « ( X > 3 )] P ( X ≥ 6) P(X > 6 | X > 3) = = P ( X > 3) P ( X ≥ 4) But P(X > 4) = P (X = 4) + P(X = 5) + P (X = 6) + ...

P [ A « ( A ¢ » B ¢ )] P( A¢ » B ¢)

=

P ( A « B ¢) 1 - P ( A « B)

=

P( A) - P ( A « B) 0.4 - 0.15 = 1 - P ( A « B) 1 - 0.15

=

0.25 5 = 0.85 17

Example 81: Consider 5 independent Bernoulli‘s trials each with probability of success p. If the probability of at least one failure is greater than or equal to 31/32, then p lies in the interval. (a) [11/12, 1] (b) [1/2, 3/4] (c) [3/4, 11/12] (d) [0, 1/2] Ans.(d) Solution: Let X = number of failure, then X ~ B(5, 1 –p) P (at least one failure) = P (X ≥ 1) = 1– p (X = 0) = 1–5C0 [1– (1–p)]5 = 1 p5 We are given = 1 – p5 ≥ 31/32 fip5 £ 1/32 fi 0 £ p £ 1/2 Example 82: If C and D are two events such that C Õ D and P(D) π 0, then the correct statement among the following is P (C ) (b) P(C/D) = P(C) (a) P(C/D) = P (D) (c) P(C/D) ≥ P(C) (d) P(C/D) < P(C) Ans.(a) ( ) ( ) Solution: P(C/D) = P C « D = P C P (D) P (D)

[∵ C Õ D]

Probability 24.21

Example 83: Three numbers are chosen at random without replacement from {1, 2, 3, .... 8}.The probability that their minimum is 3, given that their maximum is 6, is (a) 1/5 (b) 1/4 (c) 2/5 (d) 3/8 Ans.(a) Solution: Total number of ways of choosing 3 numbers out of 8 is 8C3 = 56 Let A denote the event that minimum is 3 and B denote the event that maximum number is 6. Then P ( A « B) P(A|B) = P ( B) 5

But

P(B) =

\

8

C2 C3

=

10 and P(A « B) = 56

2 8

C1 C3

=

2 56

2 1 = 10 5

P(A|B) =

Example 84: Ten unbiased coins are thrown simultaneously. The probability of getting at least seven heads is 3 5 (b) (a) 64 64 (c)

7 64

(d)

11 64

Ans. (d) Solution: Let X = number of heads when 10 unbiased coins are thrown, then X ~ B (10, 1/2) We have 7 3 8 2 Ê1 ˆ Ê1 ˆ Ê1 ˆ Ê1 ˆ P(X ≥ 7) = 10C7 Ë ¯ Ë ¯ +10 C8 Ë ¯ Ë ¯ 2 2 2 2 9

Ê1 ˆ Ê1 ˆ Ê1 ˆ +10 C9 Ë ¯ Ë ¯ +10 C10 Ë ¯ 2 2 2

10

1

ÈÎ 10 C7 +10 C8 +10 C9 +10 C10 ˘˚ 210 1 11 = 10 (176 ) = 64 2 =

Example 85: Out of 13 applicants for a job there are 5 women and 8 men. It is desired to select 2 persons for this job.The probability that at least one of the selected person will be a woman is (a) 5/13 (b) 10/13 (c) 14/39 (d) 25/39 Ans.(d) Solution: P (at least one woman) = 1 – P (no woman) = 1 – 8 C2 13 C2 = 25 39

Example 86: Four persons A, B, C, D are to speak at a function along with 6 others. If they all speak in random order, the probability that A speaks before B, B before C, C before D is (a) 1/4 (b) 1/6 (c) 1/24 (d) 1/16 Ans.(c) Solution: 10 persons can speak at the function in 10! ways. We can choose 4 places out of 10 in 10C4 ways and arrange A, B, C, D at the selected places in only one way. The remaining 6 persons can be arranged in 6! ways. Thus, probability of the required event is

( 10 C4 ) (6!) =

1 1 = 10! 4! 24 Example 87: Let A, B, C be pairwise independent events with P(C) > 0 and P (A « B « C) = 0. Then P(A¢ « B¢|C) is equal to (a) P (A) – P (B¢) (b) P (A¢) + P (B¢) (c) P (A¢) – P (B¢) (d) P (A¢) – P (B) Ans.(d) Solution: P(A¢ « B¢|C) = 1 – P(A » B|C) = 1– [P (A|C) + P (B|C) – P (A « B|C)] =

P (A « B « C)˘ = 1 – È P ( A) + P ( B ) ÍÎ ˙˚ P (C ) [∵ A, B, C are pairwise independent] = 1 – P (A) – P (B) = P (A¢) – P (B) Example 88: Four fair dice D1, D2, D3 and D4, each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability D4 shows a number appearing on one of D1, D2 and D3 is 91 108 (b) 216 216 125 127 (c) (d) 216 216 Ans.(a) Solution: P (D4 shows number on one of D1, D2 and D3) = 1 – P (D4 shows a number different from D1, D2 and D3 ) (a)

6

C1 (53 )

125 91 = 216 216 6 Example 89: Let A and B be two events such that P (A|B) = 1/2, P (B|A) = 1/3, P(A « B) = 1/6, then (a) P (A » B) = 1/2 (b) A and B are independent (c) P (A¢ » B) = 1/3 (d) none of these Ans.(b) =1–

4

=1-

24.22

Complete Mathematics—JEE Main

P ( A « B) 1 1 1 = fi P ( B) = fi P ( B) 2 2 3 1 1 and P (B/A) = fi P ( A) = 3 2 As P (A « B) = P (A) P (B), we get A and B are independent. Example 90: A ship is fitted with three engines E1, E2 and E3.The engines function independently of each other with respective probabilities 1/2, 1/4, and 1/4. For the ship to be operative at least two of its engines must function. Let X denote the event that ship is operational and let X1, X2 and X3 respectively the events that the engines E1, E2 and E3 are functioning. Let (a) P ( X1¢ / X ) = 3 / 8 (b) P(X/X2) =7/8 (c) P (Exactly two engines are functioning) = 7/8 (d) P(X/X1) =7/16 Ans.(d) Solution: We have Solution: P (A/B) =

X= and

( X1 X2 X3¢ ) » ( X1 X2¢ X3 ) » ( X1¢X2 X3 ) » ( X1 X2 X3 ) ,

Now,

P ( X1¢ / X ) =

P ( X1¢ « X ) P ( X1¢ X 2 X3 ) = P (X ) P (X )

We have P ( X1¢ X 2 X3 ) = P ( X1¢ ) P ( X 2 ) P ( X3 ) and

= (1/2) (1/4) (1/4) = 1/32 P (X) = (1/2) (1/4) (3/4) +(1/2) (3/4) (1/4) + (1/2) (1/4) (1/4) + (1/2) (1/4) (1/4) = 1/4 P ( X1¢ / X ) = 1 / 8

\ Next,

X « X2 = X - X1 X 2¢ X3 , P ( X « X 2 ) = P ( X ) - P ( X1 X 2¢ X3 ) = 5 / 32 P ( X / X2 ) =

\

P ( X « X 2 ) 5 / 32 5 = = P ( X2 ) 1/ 4 8

and P (Exactly two engines are functioning) = P (X – X1 X2 X3) =7/32 Next,

X « X1 = X - X1¢ X 2 X3

fi

P (X « X1) = 7/32

\

P ( X / X1 ) =

P ( X « X1 ) 7 / 32 7 = = P ( X1 ) 1 / 2 16

X1¢ « X = X1¢ X 2 X3

Assertion-Reason Type Questions

Example 91: Four numbers are chosen at random (without replacement) from the set {1, 2, 3.... 20}. Statement-1: The probability that the chosen numbers when arranged in some order will form an A.P. is 1/85. Statement-2: If the four numbers form an A.P., then the set of all possible values of common difference is {+ 1, + 2, + 3, + 4, + 5}. Ans. (c)

Example 92: Let A and B be two events such that P(A » B) = P(A « B).

Solution: Statement-2 is false. For d = 6, we have 1, 7, 13, 19 forms an A.P. We, now check the truth of Statement-1. For d = 1, there are 17 A.P.’s. viz. (1, 2, 3, 4), (2, 3, 4, 5), ..... (17, 18, 19, 20) For d = 2, there are 14 A.P.’s viz. (1, 3, 5, 7), (2, 4, 6, 8), ..... (14, 16, 18, 20) For d = 3, there are 11 A.P.’s For d = 4, there are 8 A.P.’s For d = 5, there are 5 and for d = 6, there are 2 A.P.’s. Thus, there are 17 + 14 + 11 + 8 + 5 + 2 = 57 A.P.’s Also, total number of ways of choosing 4 numbers out of 20 is 20C4. 57 1 = . \ The required probability = 20 85 C4

¤

Statement-1: P(A « B¢) = P(A¢ « B) = 0 Statement-2: P(A) + P(B) = 1. Ans. (c) Solution: P(A » B) = P(A « B) ¤ P(A) + P(B) – P(A « B) = P(A « B) [P(A) – P(A « B)] + [P(B) – P(A « B)] = 0

(1)

But P(A), P(B) ≥ P(A « B). Thus, (1) is possible if and only if P(A) – P(A « B) = 0 and P(B) – P(A « B) = 0 ¤

P(A) = P(B) = P(A « B)

\

P(A) + P(B) = 1 need not hold. Next, P(A « B¢ ) = P(A) – P(A « B) = 0 Similarly P(A¢ « B) = P(B) – P(A « B) = 0 Example 93: Statement-1: If

1 1 (1 + 5p), (1 + 2p), 5 3

1 1 (1 – p) and (1 – 3p) are probabilities of four mutually 3 5 exclusive events, then p can take infinite number of values.

Probability 24.23

Statement-2: If A, B, C and D are three mutually exclusive events, then P(A), P(B), P(C), P(D) ≥ 0 and P(A) + P(B) + P(C) + P(D) £ 1. Ans. (a)

If m = 2n, x1 and x3 can be chosen in nC2 + nC2 = n(n – 1) ways. n ( n – 1) In this case probability of the required event is 2 n C3

Solution: Statement-2 is true. See theory. Now,

=

1 1 (1 + 5p), (1 + 2p), 5 3

If m = 2n + 1, x1 and x3 can be chosen in n + 1C2 + nC2 = n2 ways. In this case probability of the required event is

1 1 (1 – p), (1 – 3p) ≥ 0 3 5 fi fi

n2

p ≥ – 1/5, p ≥ – 1/2, p £ 1, p £ 1/3 – 1/5 £ p £ 1/3

2n + 1

(1)

fi fi fi fi

1 (1 – 3p) £ 1 5

1 1 (2 + 2p) + (2 + p) £ 1 5 3 6 + 6p + 10 + 5p £ 15 11p £ – 1 fi p £ – 1/11. From (1) and (2) we get – 1/5 £ p £ – 1/11 there are infinite values of p.

C3

=

3n 4n2 – 1

.

Example 96: Statement-1: Let A, B, C be three mutually independent events. Statement-1: A and B » C are independent. Statement-2: A and B « C are independent.

1 1 1 and (1 + 5p) + (1 + 2p) + (1 – p) 5 3 3 +

6n ( n – 1) 3 = . 2n (2n – 1)(2n – 2 ) 2 (2 n – 1)

Ans. (b)

(2)

Example 94: Statement-1: A natural number is chosen at random. The probability that sum of the digits of its square is 93 is 0. Statement-2: A number is divisible by 31 if and only if sum of its digits is divisible by 31. Ans. (c) Solution: Let n Œ N, be such that sum of digits of its square is 93. Then 3 | n2 fi 3 | n fi 9 | n2. A contradiction. Statement-2 is clearly false. Example 95: Let m Œ N, and suppose three numbers are chosen at random from the number 1, 2, 3 º m. Statement-1: If m = 2n for some n Œ N, then the chosen 3 numbers are in A.P. with probability . 2 (2n – 1) Statement-2: If m = 2n + 1, then the chosen numbers are in 3n A.P. with probability 4n2 – 1 Ans. (b) Solution: We can choose three numbers out of m in mC3 ways. Let numbers be x1, x2, x3. Now, x1, x2, x3 are in A.P. if and only if x1 + x3 = 2x2, that is, if and only if other both x1, x3 are odd or both x1, x3 are even.

Solution: As A, B, C are mutually independent P(B « C) = P(B) P(C), P(C « A) = P(C) P(A), P(A « B) = P(A) P(B) and P(A « B « C) = P(A) P(B) P(C) Now, P[A « (B » 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) + P(C) – P(B « C)] = P(A) P(B » C) and P(A « B « C) = P(A) P(B) P(C) = P(A) P(B « C) Example 97: Let A and B be two events such that 3 1 3 and £ P(A « B) £ . P(A » B) ≥ 4 8 8 Statement-1: P(A) + P(B) ≥ 7/8 Statement-2: P(A) + P(B) £ 11/8 Ans. (b) 3 Solution: £ P(A » B) £ 1 4 3 fi £ P(A) + P(B) – P(A « B) £ 1 4 3 3 1 + £ P(A) + P(B) £ 1 + fi 4 8 8 7 11 fi £ P(A) + P(B) £ 8 8 Thus, both the statements are correct but Statement-2 is not a correct explanation for Statement-1.

24.24

Complete Mathematics—JEE Main

Example 98: An urn contains four balls bearing numbers 1, 2, 3 and 123 respectively. A ball is drawn at random from the urn. Let Ei, i =1, 2, 3 denote the event that digit i appears on the ball drawn. Statement-1: P(E1 « E2) = P(E2 « E3) 1 = P(E3 « E1) = 4 1 Statement-2: P(E1) = P(E2) = P(E3) = 2

Ans. (b) Solution: We have 2 = P(E2) = P(E3) P(E1) = 4 1 P(E2 « E3) = P(E3 « E1) and P(E1 « E2) = 4 Thus, both the statements are correct but Statement-2 is not a correct explanation of Statement-1.

LEVEL 2 Straight Objective Type Questions Example 99: If p and q are chosen from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, with replacement, the probability that the roots of the equation x2 + 2px + q = 0 are real is (a) 0.84 (b) 0.16 (c) 0.62 (d) 0.38 Ans. (a) Solution: Required probability = 1 – P (the roots of x2 + 2px + q = 0 are non-real) The roots of x2 + 2px + q = 0 will be non-real if (2p)2 – 4q < 0, i.e. if p2 < q. We enumerate the possible values of p and q. When q = 1, there is no value of p When q = 2, 3, 4, possible value of p is 1 When q = 5, 6, 7, 8, 9 possible values of p are 1 and 2 When q = 10, possible values of p are 1, 2 and 3. Thus, the number of pairs for which x2 + 2px + q = 0 have non-real roots is 0 + 3 × 1 + 5 × 2 + 1 × 3 = 16 Also, total number of possible pairs is 10 × 10 = 100. 16 Thus, probability of the required event = 1 – = 0.84. 100 Example 100: A person writes 4 letters and addresses 4 envelopes. If the letters are placed in the envelopes at random, the probability that not all letters are placed in correct envelopes is 1 11 (a) (b) 24 24 (c)

5 8

(d)

23 24

Ans. (d) Solution: Required probability = 1 – P (all the letters are put in correct envelopes)

The number of ways of putting the letters in the envelopes = 4P4 = 4! The number of ways of putting all the letters in correct envelopes = 1 1 23 \ Required probability = 1 = . 24 24 Example 101: A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters, TA, are visible. The probability that the letter has come from CALCUTTA is 4 1 (a) (b) 11 3 (c)

5 12

(d)

7 19

Ans. (a) Solution: Let E1 denote the event that the letter has come from TATANAGAR and E2 the event that the letter has come from CALCUTTA. Let B denote the event that the two consecutive alphabets visible on the envelope are TA. We have P(E1) = 1/2, P(E2) = 1/2, P (B|E1) = 1/7, P(B|E2) = 2/8. Therefore, by the Bayes' theorem we have P(E2 | B) =

P ( E2 ) P ( B | E2 ) 4 = . P ( E1 ) P( B | E1 ) + P( E2 ) P( B | E2 ) 11

Example 102: A fair coin is tossed n times. If the probability that heads occurs 6 times is equal to the probability that heads occurs 8 times, then value of n is (a) 24 (b) 48 (c) 14 (d) 16 Ans. (c) Solution: Let X be the number of times heads occurs. Then X follows a binomial distribution with parameter n and p=1/2. We are given that P(X = 6) = P(X = 8).

Probability 24.25

fi

n

n

fi fi

6 n -6 8 Ê 1ˆ Ê 1ˆ Ê 1ˆ C6 Á ˜ Á ˜ = n C8 Á ˜ Ë 2¯ Ë 2¯ Ë 2¯

n

Ê 1ˆ ÁË ˜¯ 2

n -8

n n Ê 1ˆ Ê 1ˆ C6 Á ˜ = n C8 Á ˜ Ë 2¯ Ë 2¯

Example 106: The mean and variance of random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is 1 1 (b) (a) 16 8

C6 = nC8 = nCn–8

(c)

fi 6 = n–8 or n = 14 Example 103: If the standard deviation of the binomial distribution (q + p)16 is 2, then mean of the distribution is (a) 6 (b) 8 (c) 10 (d) 12

np = 4, npq = 2 fi

Solution: We have n = 16, npq = 2 2 = 4 \ p(1 – p) = 1/4 fi 4p2 – 4p + 1 = 0

Now

(2p – 1)2 = 0 fi p = 1/2 mean = np = 16 (1/2) = 8.

Example 104: If A and B are two events such that P(A » B) = 3/4, P(A « B) = 1/4, P(A¢) = 2/3, then P(A¢ » B) is (a)

5 12

(b)

3 8

(c)

5 8

(d)

1 4

Ans. (a) Solution: P(A) + P(B) =P(A » B) + P(A « B) 3 1 = + =1 4 4 Also P(A¢) = 2/3 fi P(A) = 1/3 \ P(B) = 1 – 1/3 = 2/3 Now, P(A¢ « B) = P(B) – P(A « B) 2 1 5 = - = 3 4 12 Example 105: A die is thrown 5 times. Getting an odd number is consider a success. The variance of the distribution is 8 3 (b) (a) 3 8 (c)

4 5

(d)

5 4

Ans. (d) Solution: Let X = number of successes. Then X ~ B(n, p) where n = 5, p = 1/2 Ê 1ˆ Var(X) = npq = (5) Á ˜ Ë 2¯

5 Ê 1ˆ ÁË ˜¯ = 2 4

(d)

1 32

Ans. (d) Solution: We are given

Ans. (b)

fi Thus,

1 4

p=

fi

q=

2 1 = 4 2

1 2

Ê 1ˆ nÁ ˜ = 4 Ë 2¯

fi

n=8

8 1 Ê 1ˆ P(X = 1) = nC1 pqn – 1 = 8pq7 = 8 Á ˜ = Ë 2¯ 32

\

Example 107: If (1 + 3p)/3, (1 – p)/4 and (1 – 2p)/2 are the probabilities of three mutually exclusive events, then the set of all values of p is (a) 1/3 £ p £ 1/2 (b) 1/4 £ p £ 1/3 (c) – 1 £ p £ 1/5 (d) – 2 £ p £ 1/3 Ans. (a) Solution: As (1 + 3p)/3, (1 – p)/4 and (1 – 2p)/2 are the probabilities of three mutually exclusive events 1 + 3p 1– p 1 – 2p ≥ 0, ≥ 0, ≥0 4 3 2

fi

1 + 3p 1 - p 1 - 2 p £1 + + 2 3 4 p ≥ 1/3, p £ 1, p £ 1/2, p ≥ 1/3.

Thus,

1 3£ p £1 2 .

and

Example 108: 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 letter is (a) 1/45 (b) 13/90 (c) 19/90 (d) 5/18 Ans. (c) Solution: Letters of the word STATISTICS are AIICSSSTTT Letters of the word ASSISSTANT are AAINSSSTT Common letters are A, I, S and T 1 2 2 Probability of choosing A is ¥ = 10 9 90 Probability of choosing I is

2 1 2 = ¥ 10 9 90

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Complete Mathematics—JEE Main

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

Example 109: For three events A, B and C, P (exactly one of the events A or B occurs) = P (exactly one of the events B and C occur) = P (exactly one of the events C or A occurs) = p and P (all the three events occur simultaneously) = p2, where 0 < p < 1/2. Then the probability of at least one of the three events A, B and C occurring is (a)

3 p + 2 p2 2

(b)

p + 3 p2 4

(c)

p + 3 p2 2

(d)

3 p + 2 p2 4

Ans. (a) Solution: We know that P (exactly one of A or B occurs) = P(A) + P(B) – 2P(A « B) Therefore, P(A) + P(B) – 2P(A « B) = p (1) Similarly, P(B) + P(C) – 2P(B « C) = p (2) and P(C) + P(A) – 2P(C « A) = p (3) Adding (1), (2) and (3) we get 2[P(A) + P(B) + P(C) – P(A « B) – P(B « C) – P(C « A)] = 3p fi P(A) + P(B) + P(C) – P(A « B) – P(B « C) – P(C « A) = 3p/2 (4) 2 We are also given that P(A « B « C) = p (5) Now, P(at least one of A, B and C) = P(A) + P(B) + P(C) – P (A « B) – P(B « C) – P(C « A) + P(A « B « C) =

3 p + 2 p2 3p + p2 = 2 2

[using (4) and (5)]

Example 110: Let A, B, C, be the three mutually independent events. Consider the two statements S1 and S2. S1 : A and B » C are independent S2 : A and B « C are independent Then (a) (b) (c) (d) Ans. (a)

Both S1 and S2 are true Only S1 is true Only S2 is true Neither S1 nor S2 is true.

Solution: We are given that P(A « B) = P(A) P(B), P(B « C) = P(B) P(C), P(C « A) = P(C) P(A), and P(A « B « C) = P(A) P(B) P(C) We have P(A « (B « C)) = P(A « B « C) = P(A) P(B) P(C) = P(A) P(B « C). fi A and B « C are independent. Therefore, S2 is true Also 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) + P(C) – P(B « C)] = P(A) P(B »C) \ A and B » C are independent. Example 111: Three identical dice are rolled. The probability that the same number appears on each of them is (a) 1/6 (b) 1/36 (c) 1/18 (d) 3/28 Ans. (b) Solution: Total number of cases = 63. The number of ways in which the same number appears on each of them is 6C1 = 6 6 1 Thus, probability of the required event = 3 = . 36 6 Example 112: A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is 10. If three balls are drawn at random without replacement and all of them are found to be black, the probability that the bag contains 1 white and 9 black balls is (a) 14/55 (b) 12/55 (c) 2/11 (d) 8/55 Ans. (a) Solution: Let Ei denote the event that the bag contains i black and (10 – i) white balls (i = 0, 1, 2, …, 10). Let A denote the event that the three balls drawn at random from the bag are black. We have 1 11 P(A|Ei) = 0 for i = 0, 1, 2 P(Ei) =

i

and P(A|Ei ) =

C3

10

C3

for i ≥ 3

(i = 0, 1, 2, …, 10)

Probability 24.27

= 5C4 + 5C3 + 6C3 + … + 10C3

Now, by the total probability rule

= 6C4 + 6C3 + … + 10C3 = … = 11C4

10

P(A) =

 P(Ei) P( A | Ei)

11

i =0

=

1 1 ¥ [3C3 + 4C3 + … + 10C3] 11 10 C3

Thus,

4

5

C4 10

11 ¥ C3

=

1 4

By the Bayes’ rule

But 3C3 + 4C3 + 5C3 + … + 10C3 4

P(A) =

P (E9|A) =

10

= C4 + C3 + C3 + … + C3

P ( E9 ) P ( A | E9 ) 14 = . P ( A) 55

[ 3C 3 = 1 = 4 C 4]

EXERCISES Concept-based Straight Objective Type Questions 1. If 2n boys are randomly divided into 2 teams containing n boys each, the probability, that the 2 tallest boys are in different teams, is n 1 (b) (a) 2n - 1 2 2n (c) 2n + 1

(d)

n2 + 1 n2 + n

2. Two events A and B are such that P(B) = 0.55 and P(A « B¢) = 0.15, then probability that at least one of A, B occurs is (a) 0.70 (b) 0.20 (c) 0.35 (d) 0.30 3. Two numbers are randomly selected from the first 100 natural numbers. The probability the product is divisible by 17 is 931 97 (b) (a) 990 990 101 879 (c) (d) 990 990 4. Out of 30 consecutive numbers, two are chosen at random, then the probability that sum of the numbers is odd is (a)

14 29

(b)

15 29

1 3 (d) 29 29 5. Suppose A and B are two events such that P(A) = 0.5 and P(B) = 0.8, then which one of the following is not true? (c)

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

P(A « B) £ 0.5 P(A « B) ≥ 0.3 P(A¢ « B) £ 0.5 P(A « B¢) £ 0.1

6. Let A, B and C be three events and suppose that simultaneous occurrence of A and B implies the occurrence of C, then (a) (b) (c) (d)

P(C) ≥ P(A) + P(B) P(C) ≥ P(A) + P(B) – 1 P(C) < P(A) + P(B) – P(A « B) none of these

7. Let A and B be two events such that P(A|B) =

1 , 2

1 1 and P(A « B) = , then which one of 3 6 the following is not true? P(B|A) =

2 3 (b) A and B are independent (c) A and B are not independent (a) P(A » B) =

1 6 8. An urn contains four balls bearing numbers 1, 2, 3 and 123 respectively. A ball is drawn at random from the urn. Let Ei, i = 1, 2, 3 denote the event that digit i appears on the ball drawn. Which one of the following is not true? (d) P(A¢ « B) =

(a) E1 and E2 are independent (b) E2 and E3 are independent

24.28

Complete Mathematics—JEE Main

10. Consider the circuit

(c) E3 and E1 are independent (d) E1, E2, E3 are independent

a

9. In a game called “odd man out” each person in a group of n ≥ 2 persons tosses a coin to determine who will buy refreshment for the entire group. The odd man out is the one whose outcome is different from the rest of the persons. The probability that there is a loser in the game is (a) (c)

n 2

(b)

n -1

n! 2

(d)

n

n +1 2n 1 2

n-1

A

B b

c

Fig. 24.8

If the probability that a circuit is closed is p and the probability that current flows from A to B is 4/9, then value of p is 1 1 (b) (a) 2 9 1 3 (c) (d) 3 4

LEVEL 1 Straight Objective Type Questions 11. A bag contains 100 tickets numbered 1 to 100. If one ticket is picked up at random, then the probability that the number is divisible by 3 or 5 is (a) 0.47 (b) 0.53 (c) 0.52 (d) 0.61 12. Suppose n students appear in an examination. Let X = the number of students who pass the examination. Suppose P(X = k) = lk2, then value of l is 2 6 (b) (a) n(n + 1) n(n + 1)(2 n + 1) (c)

4 2

n (n + 1)

2

(d)

2n + 1 2

n (n + 1)2

13. Suppose two numbers a and b are chosen randomly from the set {1, 2, 3, 4, 5, 6}. The probability that function f (x) = x3 + ax2 + bx is strictly increasing function on R is: 1 2 (b) (a) 15 15 13 14 (d) 15 15 14. Two dice are rolled together and the numbers x, y on them form complex number z = x + iy. The probability that | z | £ 3 is 1 2 (b) (a) 3 3 (c)

(c)

1 12

(d)

5 36

15. Suppose A is an event which is independent of itself and B is any other event, then (a) P(A « B) £ P(A) P(B) (b) P(A « B) < P(B) (c) P(A « B) = 0 (d) P(A¢ » B¢ ) = 1 16. Six boys and six girls sit in a row randomly. The probability that all the six girls sit together is 1 7 (b) (a) 132 462 5 1 (c) (d) 462 924 17. If P(A « B) = 1/2, P ( A¢ « B ¢ ) = 1/ 3 , P(A) = p and P(B) = 2p then value of p is 1 7 (b) (a) 3 18 4 1 (c) (d) 9 2 18. Out of 20 consecutive integers, two are chosen at random. The probability that their product is odd is 1 9 (b) (a) 9 38 2 7 (c) (d) 19 19 19. An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment. If A consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent, is

Probability 24.29

(a) 2, 4, or 8 (c) 4 or 8

(b) 3, 6 or 9 (d) 5 or 10

20. A box contains 24 balls of which 12 are white and 12 are black. The balls are drawn at random from the box one at a time with the replacement. The probability that a white ball is drawn for the 4th time on the 7th draw is 5 27 (b) (a) 64 32 5 1 (c) (d) 32 2 21. India plays two matches each with West Indies and Australia. In any match, the probabilities of India getting 0, 1 and 2 points are 0.45, 0.05 and 0.50, respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is (a) 0.875 (b) 0.0875 (c) 0.0625 (d) 0.0250 22. If E and F are two events associated with a random experiment for which P(E) = 0.6, P (E » F) = 0.85 and P (E « F) = 0.42, then P(F) is (a) 0.6 (b) 0.67 (c) 0.7 (d) 0.73 23. If E and F are events with P(E) £ P(F) and P(E « F) > 0, then (a) occurrence of E fi occurrence of F (b) occurrence of F fi occurrence of E (c) non-occurrence of E fi non-occurrence of F (d) none of the above implication hold 24. If the integers m and n are chosen at random from 1 and 100, then the probability that a number of the form 7 m + 7 n is divisible by 5 equals is 1 1 (a) (b) 4 7 1 1 (c) (d) 8 49 25. The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Then 19 27 (b) p + m + c = (a) p + m + c = 20 20 1 1 (c) pmc = (d) pmc = 3 7 26. One ticket is selected at random from 100 tickets numbered 00, 01, 02, ..., 99. Suppose X and Y are the sum and product of the digit found on the ticket then P(X = 7|Y = 0) is given by

(a)

2 13

(b)

2 19

1 3 (d) 50 19 27. A mapping is selected at random from the set of all the mappings of the set A = {1, 2, …, n} into itself. The probability that the mapping selected is one-toone is 1 1 (b) (a) n n n! (c)

(c)

(n - 1)!

(d)

n -1

n!

n n n -1 28. A natural number x is choosen at random from the first 100 natural numbers. The probability that x+

100 > 50 x

is (a)

1 10

(b)

11 20

1 13 (d) 20 50 29. A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. A coin is picked up at random from the bag and tossed. If the probability 31 that the toss results in a head is , then n is equal 42 to (c)

(a) 10 (c) 12

(b) 11 (d) 13

30. If X and Y are independent binomial variate B (5, 1/2) and B (7, 1/2), then P (X + Y = 3) is 55 55 (b) 1024 4098 55 55 (c) (d) 2048 128 31. Suppose n (≥ 3) persons are arranged in a row. The probability that two particular persons are not together is 2 2 (b) (a) 1 – n n -1 1 2 (c) 1 – (d) n n –1 32. A number x is chosen at random from the set {1, 2, 3, ..., 100}. Let p = probability that x is divisible by 19 and q = probability that x is divisible by 31, then p + q is equal to (a)

24.30

Complete Mathematics—JEE Main

(a) 0.5 (c) 0.7

(b) 0.6 (d) 0.8

33. If the letters of the word PROBABILITY are written down at random in a row, the probability that two Bs are together is 2 10 (b) (a) 11 11 3 6 (d) 11 11 34. If four positive integers are taken at random and multiplied together, then the probability that the last digit is 1, 3, 7 or 9 is 1 2 (b) (a) 8 7 (c)

1 16 (d) 625 625 35. Two contestants play a game as follows: each is asked to select a digit from 1 to 9. If the two digits match they both win a prize. The probability that they will win a prize in a single trial is 1 7 (b) (a) 81 81 1 3 (c) (d) 9 11 36. If A and B are two events, then which of the following does not represent the probability that exactly one of A, B occurs is (a) P(A) + P (B) – P(A « B) (b) P(A« B¢ ) + P(A¢ « B) (c) P(A¢ ) + P (B¢ ) – 2P(A¢ « B¢ ) (d) P(A) + P(B) – 2P(A « B) (c)

37. If A and B are two events, then which one of the following is not always true (a) P(A « B) ≥ P(A) + P(B) – 1 (b) P(A « B) £ P(A) (c) P(A¢ « B¢ ) ≥ P(A¢ ) + P(B¢ ) – 1 (d) P(A « B) = P(A) P(B) 38. A sum of money is rounded off to the nearest rupee. The probability that the round off error is at most 10 paise is 10 11 (b) (a) 101 101 12 21 (d) 101 100 39. The probability that an electric bulb will last 150 days or more is 0.7 and that it will last at most 160 days is 0.8. The probability that the bulb will last 150 to 160 days is

(a) 0.5 (c) 0.56

(b) 0.3 (d) 0.28

40. A bag contains three tickets numbered 1, 2 and 3. A ticket is drawn at random and put back in the bag, and this is done four times. The probability that the sum of the numbers drawn is even is 40 41 (b) (a) 81 81 14 13 (c) (d) 27 81 41. Two numbers x and y are selected at random from the set {1, 2, 3,…, 3N}. The probability that x2 – y2 is divisible by 3 is (3N – 1) ( N – 1) (b) (a) N 3N N ( 5 N – 3) (c) (d) 3N – 5 ( 9 N – 3) 42. Three persons A, B and C are to speak at a function along with 7 other persons. The probability that A, B, and C speak together 3 1 (b) (a) 70 15 3 1 (c) (d) 7 7! 43. A positive integer is chosen at random. The probability that the sum of the digits of its square is 33 is 1 2 (b) (a) 33 33 1 (c) (d) 0 11 44. A lottery sells n2 tickets and declares n prizes. If a man purchases n tickets, the probability of his winning exactly one prize is (a)

(c)

(n2 – n)! (n2 )! (n – 1)!2 (n2 )!

(b)

1 2n

(d) none of these

45. Let x be a non-zero real number. A determinant is chosen from the set of all determinants of order 2 with entries x or – x only. The probability that the value of the determinant is non-zero is

(c)

(a)

3 16

(b)

1 4

(c)

1 2

(d)

1 8

Probability 24.31

46. A die is rolled three times. The probability of getting a number larger than the previous number each time is 5 5 (b) (a) 72 54 (c)

13 216

(d)

1 18

1+ 4p 1- p 1- 2p and are the probabilities of , 2 4 4 three mutually exclusive events, then 1 1 1 2 (a) £ p £ £ p£ (b) 3 2 2 3 1 1 1 (c) (d) 0 £ p £ £ p£ 2 6 2 48. Two non-negative integers are chosen at random. The probability that the sum of the square is divisible by 5 is 9 9 (b) (a) 16 25 9 (c) (d) 0 17 49. A is a set containing n elements. Two subsets P and Q of A are chosen at random. (P and Q may have elements in common). The probability that P « Q = f is n 1 n (b) (a) Ê 3 ˆ Ë4 ¯ 4

47. If

n

(c)

C2

(d) none of these 2n 50. Suppose A is a set containing n elements. Two subsets P and Q of A are chosen at random. If the probability that P is a subset of Q is 243/1024, then n is equal to (a) 3 (b) 4 (c) 5 (d) 6 51. Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertics is equilateral is 1 1 (b) (a) 2 5 1 1 (c) (d) 10 20 52. Each of the n urns contains 4 white and 6 black balls. The (n +1)th urn contains 5 white and 5 black balls. One of the (n +1) urns is chosen at random and two balls are drawn from it without replacement. Both the balls turn out to be black. If the probability that the (n +1)th urn was chosen to draw the balls is 1/16, then value of n is

(a) 10 (c) 12

(b) 11 (d) 13

53. Two numbers x and y are chosen at random from the set {0, 1, 2,º, 10}. The probability that | x – y | £ 5, is (a)

8 11

(b)

7 11

1 3 (d) 2 11 54. In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is 1/3 and the probability that he copies the answer is p. The probability that his answer is correct, given that he copied it, is 1/8. If the probability that he knew the answer to question, given that he correctly answered is 8/11, then value of p is 1 1 (b) (a) 2 3 (c)

1 1 (d) 6 12 55. A fair coin is tossed n times. Let X = the number of times head occurs. If P(X = 4), P(X = 5) and P(X = 6) are in AP, then the value of n is (a) 7 (b) 10 (c) 12 (d) 24 (c)

56. A fair coin is tossed a fixed number of times. If the probability of getting exactly 3 heads equals the probability of getting exactly 5 heads, then probability of getting exactly one head is (a)

1 64

(b)

1 32

1 1 (d) 16 8 57. Let E, F, G be pairwise independent events with P(G) > 0 and P(E « F « G) = 0. Then P(E ¢ « F¢| G) equals (a) P(E¢) + P(F¢) (b) P(E¢) – P(F¢) (c) P(E¢) – P(F) (d) P(E) – P(F¢) (c)

58. If P (E) = 0.40, P (F) = 0.35, P (E » F) = 0.55, then P (F/E) is (a) 0.5 (b) 0.25 (c) 0.6 (d) 0.7 59. If the mean and variance of a binomial distribution are 4 and 3 respectively, then P (X = 1) is

24.32

Complete Mathematics—JEE Main

(a) Ê 3 ˆ Ë4 ¯

16

15

(b) 4 Ê 3 ˆ Ë4 ¯

1 1 (d) 4 16 60. A box contains 20 cards of these 10 have letter I printed on them and the remaining 10 have T printed on them. 3 cards are drawn from the box, the probability that we can write IIT with these cards is 9 1 (b) (a) 80 8 4 15 (c) (d) 27 38 61. A and B toss a coin alternatively till one of them gets a head and wins the game. If A begins the game, the probability B wins the game is 1 1 (b) (a) 2 3 1 2 (c) (d) 4 3 62. A fair coin is tossed repeatedly. If head and tail appear alternatively on first 7 tosses; then the probability that head appears on the eighth toss is 1 1 (b) (a) 2 128 1 7 (c) (d) 256 256 63. If P(A) > 0, then the event A is independent of itself if and only if P(A) is 1 1 (b) (a) 3 2 1 (c) 1 (d) 4 64. If A and B are two events, then P(A » B) = P(A « B) if and only if (a) P(A) + P(B) = 1 (b) P(A) = P(B) (c) P(A) + P(B) > 1 (d) P(A) + P(B) < 1 (c)

65. A fair die is rolled repeatedly. The probability that 1 occurs for the first time at the even throw is 1 5 (b) (a) 6 11 6 3 (c) (d) 11 11 66. Suppose A and B are two events such that P(A) > 0 and P(B) π 1, then P (A¢ | B¢) equals (a) [1–P(A » B)]/P(B¢) (b) P(A¢)/ P(B¢) (c) 1–P(A|B) (d) 1–P(A¢|B)

67. Two unbiased die are rolled together. Let A = {(a, b): a + b = 11} and B = {(a, b): a π 5}, then P(A|B) equals (a) 1/30 (b) 1/15 (c) 2/15 (d) 5/6 68. A bag contains w white and b black balls. A ball is drawn at random and put back in the bag with k additional balls of the same colour. A ball is again drawn at random from the bag. The probability that it is black is (a) (b + k)/(w + b + k) (b) b/(w + b + k) (c) b/(w + b) (d) b/(w + b + 2k) 69. The probability distribution of a random variable X is given as follows: x

1

2

3

2k

3k

5k

P(x = x)

1/2

1/5

3/25

1/10

1/25

1/25

If E(X) = 2.3, then value of k is (a) 1/2 (b) 5/4 (c) 3 (d) 2 70. Let A and B be two events such that P(A|B) = P(A¢|B¢) = p and P(B) = 0.05. The value of p so that P(B|A) = 0.9 is (a) 21/430 (b) 173/430 (c) 171/430 (d) 95/430 71. In the Fig. 24.9 the number represents the probability of the event. Which one of following is not true ? S

A

B 1 6

1 6 1 12

1 12

1 6 1 12

1 12 C

Fig. 24.9

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

A and B are independent A and C are independent B and C are independent A and B « C are dependent.

72. Three numbers are chosen at random from the first 20 numbers 1, 2, 3, … 20. The probability the selected numbers forms a G.P., is

Probability 24.33

(a) 1/20 (c) 1/45

(b) 1/30 (d) none of these

73. From a bag containing 30 tickets numbered 1, 2, 3, … 30, three tickets are selected at random without replacement. Suppose these tickets are arranged as x1 < x2 < x3. The probability that x2 = 15 and x3 > 20, is (a) 1/29 (b) 1/7 (c) 1/19 (d) 1/11 74. Suppose an unbiased die has n sides numbered i = 1, 2, .... n. The die is tossed n times (assume independence) and a “match” is defined to be the occurrence of side i on the its roll. The probability of at least one match is

1ˆ n Ê (a) Á 1 - ˜ Ë n¯

(b)

1 nn

1ˆ n 1 Ê (c) 1– Á 1 - ˜ (d) 1– n Ë n¯ n 75. Three numbers are chosen at random (without replacement) from the numbers 1, 2,... 20. The probability that the numbers are not consecutive are 3 7 (b) (a) 190 190 171 187 (c) (d) 190 190

Assertion-Reason Type Questions 76. Let A, B and C be three events such that P(C) = 0. Statement-1: P(A « B « C) = 0 Statement-2: P(A » B » C) = P(A » B) 77. Let A, B and C be three events such that A is independent of both B and C. Statement-1: A is independent of B » C. Statement-2: A is independent of B « C. 78. A set P contains n elements. Two boys Rakshit and Gitansh independently pick up two subsets Q and R of P. n Ê 3ˆ Statement-1: Probability Q « R = f is Á ˜ Ë 4¯ n Ê 1ˆ Statement-2: Probability Q » R = P is Á ˜ . Ë 2¯

79. A fair die is thrown twice. Let (a, b) denote the outcome in which the first throw shows a and the second shows b. Let A and B be the following two events. A = {(a, b) | a is odd}, B = {(a, b) | b is odd} Statement-1: If C = {(a, b) | a + b is odd}, then P(A « B « C) = 1/8 Statement-2: If D = {(a, b) | a + b is even}, then P(A « B « D | A » B) = 1/3 80. A man P speaks truth with probability p and another man Q speaks truth with probability 2p. Statement-1: If P and Q contradict each other with probability 1/2, then there are two values of p. Statement-2: A quadratic equation with real coefficients has two real roots.

LEVEL 2 Straight Objective Type Questions 81. The probability that an event A occurrs in a single trial of an experiment is 0.6. Three independent trials of the experiment are performed. The probability that the event A occurs at least once is (a) 0.936 (b) 0.932 (c) 0.948 (d) 0.946 82. The probability of occurrence of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive, then the probability of occurrence of neither A nor B is (a) 0.6 (b) 0.5 (c) 0.4 (d) 0.2

83. Suppose X ~ B(n, p) and P(X = 3) = P(X = 5). If p > 1/2, then (a) n £ 7 (b) n > 8 (c) n ≥ 9 (d) none of these 84. In a three throw of two dice, the probability throwing doublets not more than twice is 1 5 (b) (a) 6 72 215 (c) (d) none of these 216 85. A bag contains 4 brown and 5 white socks. A man pulls two socks at random without replacement. The

24.34

Complete Mathematics—JEE Main

probability that the man gets both the socks of the same colour is 5 1 (a) (b) 108 6 5 4 (c) (d) 18 9 86. Given a throw of three dice shows different face, the probability that at least one face shows 6 is 1 1 (b) (a) 2 6 1 5 (d) 3 19 87. The probability that a student is not a car driver is 1/5. The probability that out of 5 students, exactly 4 are car drivers is (c)

4 Ê 4ˆ (a) Á ˜ Ë 5¯

4 Ê 4ˆ (b) (5C4) Á ˜ Ë 5¯

4

Ê 4ˆ Ê 1ˆ (c) Á ˜ Á ˜ (d) none of these Ë 5 ¯ Ë 5¯ 88. Four persons are selected from a group of 4 men, 2 women and 3 children. The probability that exactly two of them are men is (a)

9 11

(b)

10 23

(c)

11 24

(d)

10 21

89. Fifteen persons, among whom are A and B, sit down at random around a round table. The probability that there are exactly 4 persons between A and B is (a)

1 11

(b)

2 7

1 1 (d) 7 9 90. There are 3 works. One is of 3 volumes and one is of 4 volumes and one is of exactly one volume. If the books are placed in a random order on a shelf, the probability that all the volumes of the same work are together is 1 3 (b) (a) 70 140 (c)

1 3 (d) 65 130 91. If the mean of a binomial distribution with 9 trials is 6, then its variance is (a) 2 (b) 3 (c)

(c) 4

2

(d)

92. For a binomial variable X if n = 5 and P (X = 1) = 8P(X = 3), then p is given by 4 1 (b) (a) 5 5 (c)

1 3

(d)

2 3

Previous Years’ AIEEE/JEE Main Questions

1. Events A, B, C are mutually exclusive events such 1 1 that P(A) = (3x + 1), P(B) = (1 – x) and P(C) = 3 4 1 (1 – 2x). Then set of possible values of x are in 2 the interval 1 2 (a) ÍÈ , ˘˙ Î3 3 ˚

1 13 (b) ÍÈ , ˘˙ Î3 3 ˚

1 1 (d) ÍÈ , ˘˙ [2003] Î3 2 ˚ 2. Five horses are in a race. Mr. A selected two of the horses at random and bets on them. The probability than Mr. A selected the wining horse is (c) [0, 1]

3 (a) 5

1 (b) 5

2 4 (d) [2003] 5 5 3. The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is (c)

(a)

1 16

(b)

1 8

1 1 (d) [2003] 4 32 4. A random variable X has the probability distribution: (c)

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 prime number} and F = {X < 4}, the probability P(E » F) is

Probability 24.35

(a) 0.35 (c) 0.87

(b) 0.77 (d) 0.50

5. The probability A speaks truth is

[2004] 4 , while the proba 5

3 . The probability that they contradict 4 each other when asked to speak on a fact is 7 1 (a) (b) 20 5 3 4 (c) (d) [2004] 20 5

bility for B is

6. The mean and the variance of a binomial distribution are 4 and 2 respectively. The probability of 2 successes is 128 219 (b) (a) 256 256 37 28 (c) (d) [2004] 256 256 1 7. If A and B are two events such that P[(A » B)¢] = , 6 1 1 P(A » B) = and P(A¢) = , then events A and B 4 4 are (a) (b) (c) (d)

independent but not equally likely mutually exclusive and independent equally likely and mutually exclusive equally likely but not independent

[2005]

8. Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house. 8 7 (b) (a) 9 9 2 1 (c) (d) [2005] 9 9 9. A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is 1 8 (a) (b) 729 9 8 8 (c) (d) [2007] 729 243 10*. 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) 0.2 (d) 0.7 [2007] * None of the given options is correct.

11. 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 3 (b) 0 (a) 5 2 (c) 1 (d) [2008] 5 12. One ticket is selected at random from 50 tickets numbered 00, 01, 02, .., 49. Then the probability that the sum of the digits on the selected ticket is 8, given that product of these digits is zero, equals 5 1 (b) (a) 14 50 1 1 (c) (d) [2009] 14 7 1 13. In a binomial distribution B ÊÁ n, p = ˆ˜ , if the probË 4¯ ability of at least one success is greater than or equal 9 to , then n is greater than 10 (a)

9 log 4 - log 3

(b)

4 log 4 - log 3

(c)

1 log 4 - log 3

(d)

1 log 4 - log 3

[2009]

14. An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that three balls have different colours is 1 2 (b) (a) 21 23 1 2 (c) (d) [2010] 3 7 15. Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ..., 20}. Statement-1: The probability that the chosen numbers when arranged in some order will form an AP is 1/85. Statement-2: If the four chosen numbers form and AP, then the set of all possible values of common difference is {±1, ±2, ±3, ±4, ±5} [2010] 16. Consider 5 independent Bernoulli‘s trials each with probability of success p. If the probability of at least one failure is greater than or equal to 31/32, then p lies in the interval (a) [11/12, 1] (b) [1/2, 3/4] (c) [3/4, 11/12] (d) [0, 1/2] [2011]

24.36

Complete Mathematics—JEE Main

17. If C and D are two events such that C Õ D and P(D) π 0 then the correct statement among the following is P (C ) (a) P (C | D ) = (b) P (C | D) = P (C) P (D) (c) P (C | D) ≥ P (C) (d) P (C | D) < P (C) [2011] 18. Let A, B, C be pairwise independent events such that P(C) > 0 and P (A«B«C) = 0, then P(A¢«B¢|C) is equal to (a) P(A) –P (B¢) (b) P(A¢) + P (B¢) [2011] (c) P(A¢) – P (B¢) (d) P(A¢) – P (B) 19. Three numbers are chosen at random without replacement from {1, 2, 3, ...., 8}. The probability that their minimum is 3, given that their maximum is 6 is (a) 1/5 (b) 1/4 (c) 2/5 (d) 3/8 [2012] 20. 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 10 17 (b) 5 (a) 5 3 3 13 11 (c) 5 (d) 5 [2013] 3 3 21. Given two independent events, if the probability that exactly one of them occurs is 26/49 and the probability that none of them occurs is 15/49, then the probability of more probable of two events is 4 6 (b) (a) 7 7 3 5 (c) (d) [2013, online] 7 7 22. If the events A and B are mutually exclusive events such that 1 1 P(A) = (3x + 1) and P(B) = (1 – x), then the set 3 4 of possible values of x lies in the interval: (a) [0, 1] È 1 5˘ (c) Í- , ˙ Î 3 9˚

È1 2 ˘ (b) Í , ˙ Î3 9 ˚ È 7 4˘ (d) Í- , ˙ Î 9 9˚ [2013, online]

2 . He 5 fires at the target k times (k, a given number). Then the minimum k, so that the probability of hitting the 7 , is target at least once is more than 10

23. The probability of a man hitting a target is

(a) 3 (c) 2

(b) 5 (d) 4

[2013, online]

24. A, B, C try to hit a target simultaneously but independently. Their respective probabilities of hitting the 3 1 5 , , . The probability that target is hit 4 2 8 by A or B but not by C is 21 7 (b) (a) 64 8 7 9 (c) (d) [2013, online] 32 64 1 25. Let A and B be two events such that P ( A » B )¢ = , 6 1 1 P ( A « B ) = and P ( A¢ ) = . Then the events A and 4 4 B are (a) equally likely but not independent (b) independent but not equally likely (c) independent and equally likely (d) mutually exclusive and independent [2014] 26. If A and B are two events such that P(A » B) = P(A « B), then the incorrect statement amongst the following is (a) A and B are equally likely (b) P(A « B¢) = 0 (c) P(A¢ « B) = 0 (d) P(A) + P(B) = 1. [2014, online] target are

(

)

27. If X has a binomial distribution, B(n, p) with parameters n and p such that P(X = 2) = P(X = 3), then E(X), the mean of variable X, is (a) 2 – p (b) 3 – p (c) p/2 (d) p/3 [2014, online] 28. A number x is chosen at random from the set {1, 2, 3, 4, …,100}. Define the event: A = the chosen number ( x - 10)( x - 50) ≥ 0 . Then P(A) is x satisfies ( x - 30) (a) 0.71 (b) 0.70 (c) 0.51 (d) 0.20 [2014, online] 29. Let A and E be two events with positive probabilities: Statement-1: P(E|A) ≥ P(A|E) P(E) Statement-2: P(A|E) ≥ P(A « E) (a) Both statements are true (b) Both statements are false (c) Statement-1 is true, Statement-2 is false (d) Statement-1 is false, Statement-2 is true [2014, online]

Probability 24.37

30. A set S contains 7 elements. A non-empty subset A of S and an element x of S are chosen at random. Then the probability that x ΠA is (a)

1 2

(b)

64 127

(a)

63 31 (d) [2014, online] 128 128 31*. If 12 distinct baalls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is (c)

(a)

55 Ê 2 ˆ 11 Á ˜ 3 Ë 3¯

1 (c) 220 ÊÁ ˆ˜ Ë 3¯

12

2 10 (b) 55 ÊÁ ˆ˜ Ë 3¯ 1 (d) 22 ÊÁ ˆ˜ Ë 3¯

(a) (c)

C10

210

(210 - 1)

(b)

220 20

C10

[2015, online] 220 2 33. If the lengths of the sides of a triangle are decided by the three throws of a single fair die, then the probability that the triangle is of maximum area given that it is an isosceles triangle, is 1 1 (b) (a) 26 27 1 1 (c) (d) [2015, online] 21 15 10

(d)

(b)

9 16

3 15 (d) [2015, online] 4 16 35. Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four and E2 be the event die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements in NOT true? (a) (b) (c) (d)

[2014]

(210 - 1)

1 16

(c)

11

32. Let X be a set containing 10 elements and P(X) be its power set. If A and B are picked up at random form P(X), with replacement, then the probability that A and B have equal number of elements, is 20

34. If the mean and the variance of a bionomial variate are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is

E1 and E2 are independent E2 and E3 are independent E1 and E3 are independent E1, E2 and E3 are independent

[2016]

36. If A and B are any two events such that P(A) = 2/5 and P(A « B) = 3/20, then the conditional probability, P(A | A¢ » B¢), where A¢ denotes the complement of A, is equal to (a) 11/20 (b) 5/17 (c) 8/17 (d) 1/4 [2016, online] 37. An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is (a)

496 729

(b)

192 729

(c)

240 729

(d)

256 729

[2016, online]

Previous Years' B-Architecture Entrance Examination Questions 1. Two events A and B are such that P(B) = 0.55 and P(AB¢) = 0.15. The probability of occurrence of at least one of event is (a) 0.70 (b) 0.20 (c) 0.35 (d) 0.30 [2006] 2. A certain water-supply system consists of a source, three pumping stations, and a destination. Each pumping station has a probability p (0 < p < 1) of being operable at a specified time to and station functions independently of one another. The stations are connected as shown in figure: * Word ‘identical’ in the question should be replaced by ‘distinct’

2 Source

Destination

1 3

The probability that water is available to the destination at time to is (a) 2p2

(b) p2 (2 – p)

(c) p3

(d) p2

[2007]

24.38

Complete Mathematics—JEE Main

3. An urn contains four balls bearing numbers 1, 2, 3 and 123 respectively. A ball is drawn at random from the urn. Let Ei, i = 1, 2, 3 donote the event that digit i appears on the ball drawn. 1 Statement-1: P(E1 « E2) = P(E1 « E3) = P(E2 « E3) = 4 1 [2008] 2 4. Three dice, red, blue and green in colour are rolled together. Let B be the event that sum of the numbers shown up is 7. Let A be the event that the red die shows 1. The conditional probability of the event A given B, P(A|B) is 2 1 (b) (a) 7 6 Statement-2: P(E1) = P(E2) = P(E3) =

(c)

1 7

(d)

1 3

[2009]

5. Sets A, B, C A « B, A « C, B « C and A « B « C have 35, 40, 45, 13, 12, 14 and 5 elements respectively. An element is selected at random from the set A » B » C. The probability that the selected element belongs to only set A is (a)

13 86

(b)

35 86

5 15 (d) [2009] 86 86 6. A man is known to speak the truth on an average 3 out of 4 times. He throws a fair die and reports that it is a six. The probability that it is actually a six is (c)

(a)

3 5

(b)

3 8

3 1 (d) [2010] 4 5 7. If P(A) = 0.4, P(B¢) = 0.6, and P(A « B) = 0.15, then the value of P(A| A¢ » B¢) is (c)

(a)

1 17

(b)

4 17

5 10 (d) [2011] 17 17 8. A class consists of 80 students, 25 of them are girls. If 10 of the students are rich and 20 of the students are fair complexioned, then the probability of selecting a fair complexioned rich girl from the class (assuming three traits as independent) is (c)

(a)

1 10

(b)

1 32

5 7 (d) [2012] 512 512 9. Let A and B be two events such that P(A » B) ≥ 3/4 and 1/8 £ P(A « B) £ 3/8 Statement-1: P(A) + P(B) ≥ 7/8 Statement-2: P(A) + P(B) £ 11/8 [2013] 10. A biased coin with probability p, 0 < p < 1 of showing head is tossed until a head appears for the first time. If the probability that the number of tosses required is even is 2/5, then p is equal to 2 1 (a) (b) 3 2 1 1 (c) (d) [2014] 3 4 11. If p and q are chosen at random from the set {1, 2, 3, ..., 10}, with replacement, then the probability that the roots of equation x2 + px + q = 0 are real, is 31 9 (a) (b) 50 25 29 13 (c) (d) [2015] 50 25 1 12. In a binomial distribution B ÊÁ n, p = ˆ˜ , if the probË 4¯ ability of at least one success is greater than or equal 9 to , then n is an integer greater than 10 1 - log10 9 9 (a) (b) log10 4 - log10 3 log10 4 - log10 3 (c)

(c)

1 log10 4 - log10 3

(d)

1 [2015] log10 4 + log10 3

13. A box contains 5 black and 4 white balls. A ball is drawn at random and its colour is noted. The ball is then put back in the box along with two additional balls of its opposite colour. If a ball is drawn again from the box, then the probability that the ball drawn now is black, is 5 53 (b) (a) 11 99 48 7 (c) (d) [2016] 99 11

Answers Concept-based 1. (a)

2. (a)

3. (b)

4. (b)

5. (d)

6. (b)

7. (c)

8. (d)

9. (a)

10. (c)

Probability 24.39

Level 1

Hints and Solutions

11. (a)

12. (b)

13. (c)

14. (c)

15. (a)

16. (a)

17. (b)

18. (b)

19. (d)

20. (c)

21. (b)

22. (b)

23. (d)

24. (a)

25. (b)

26. (b)

27. (c)

28. (b)

29. (a)

30. (a)

31. (a)

32. (d)

33. (a)

34. (d)

35. (c)

36. (a)

37. (d)

38. (d)

39. (a)

40. (b)

41. (c)

42. (b)

43. (d)

44. (d)

45. (c)

46. (b)

47. (d)

48. (b)

49. (a)

50. (c)

51. (c)

52. (a)

53. (a)

54. (b)

55. (a)

56. (b)

57. (c)

58. (a)

59. (b)

60. (d)

61. (b)

62. (a)

63. (c)

64. (b)

65. (b)

66. (a)

67. (a)

68. (c)

69. (d)

70. (c)

71. (d)

72. (d)

73. (a)

74. (c)

75. (d)

76. (a)

77. (a)

78. (c)

79. (d)

80. (c)

Level 2 81. (a)

82. (d)

83. (a)

84. (c)

85. (d)

86. (a)

87. (a)

88. (d)

89. (c)

90. (b)

91. (a)

92. (b)

Previous Years’ AIEEE/JEE Main Questions 1. 5. 9. 11. 15. 19. 23. 27. 31. 35.

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

2. 6. 10. 12. 16. 20. 24. 28. 32. 36.

(c) 3. (d) 4. (b) 7. (a) 8. None of the answer matches. (c) 13. (c) 14. (d) 17. (a) 18. (d) 21. (a) 22. (a) 25. (b) 26. (a) 29. (a) 30. (a) 33. (b) 34. (b) 37. (d)

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

Previous Years’ B-Architecture Entrance Examination Questions

Concept-based 1. Total number of ways of dividing 2n boys is (2n)! (2nCn) (nCn) = n! n! Dividing (2n– 2) boys (excluding 2 tallest boys) is

(2n - 2)! ((n - 1)!)2

. Two tallest boys can be divided in

2

P2 = 2! ways. Thus, required probability is n! n! n (2!)(2 n - 2)! ¥ = (n - 1)!(n - 1)! (2n)! 2n - 1

2. Use P(A » B) = P(A « B¢) + P(B) 3. Use 17 | ab fi 17 | a or 17 | b. 4. Use : one of them must be even and other must be odd. 5. P(A « B) £ min {P(A), P(B)} = 0.5, P(A « B) ≥ P(A) + P(B) –1 = 0.3, P(A¢ « B) = P(B) – P(A « B) £ 0.8 – 0.3 = 0.5 and P(A « B¢) = P(A) – P(A « B) ≥ 0.2 6. A « B Õ C fi P(C) ≥ P(A « B) ≥ P(A) + P(B) –1 7. P(A) P(B|A) = P(A « B) fi P(A) = 1/2. Similarly, P(B) = 1/3 Now, P(A » B) = P(A) + P(B) – P(A « B) = 2/3 As P(A « B) = P(A) P(B), A and B are independent. \ (c) cannot be correct. Also, P(A¢ « B) = P(B) – P(A « B) = 1/3 – 1/6 = 1/6 2 1 8. P(E1) = P(E2) = P(E3) = = 4 2 1 P(E1 « E2) = P(E2 « E3) = P(E3 « E1) = 4 1 and P(E1 « E2 « E3) = 4 9. Total number of possible outcomes is 2n. We can choose a loser in n ways and he can lose in 2 ways when he gets a head and rest of them tails or when he gets a tail and rest of them get head. Thus, probability of required event is 2n/2n = n/2n – 1. 10. Current flows from A to B with a probability of p + p2. Now, p + p2 = 4/9 fi p = 1/3.

Level 1

1. (a)

2. (a)

3. (b)

4. (b)

5. (d)

6. (b)

7. (c)

8. (c)

9. (b)

10. (c)

11. Let A, B denote the following events. A : number is divisible by 3 B : number is divisible by 5, then A « B : number is divisible by 15.

24.40

Complete Mathematics—JEE Main

33 20 6 , P(B) = , P(A « B) = 100 100 100 Now, use formula for P(A » B). 12. Note that X can take values from 0 to n. Now, P(A) =

20. Required probability = (probability of drawing 3 white balls in the first six draws) ¥ (probability of getting a white ball at the 7th draw)

n

 P( X = k )

= 1

k=0

21.

n

fi

lÂk

2

k=0 2

6 = 1 fi =l n(n + 1)(2n + 1)

13. f ¢(x) = 3x + 2ax + b. Note that f ¢(x) > 0 ¤ 4a2 – 12b < 0 fi a2 < 3b. Possible values of (a, b) are (1, 2), (1, 3), (2, 3), (1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5), (1, 6), (2, 6), (3, 6), (4, 6). Total number of ways of choosing (a, b) is 6C2 = 15. 13 Thus, required probability = . 15 2 14. | z | £ 3 ¤ | z | £ 9 ¤ x2 + y2 £ 9. Possible ordered pairs (x, y) are : (1, 1), (1, 2), (2, 1). Thus, probability of required event is 3/36 = 1/12 15. As A is independent of itself, P(A) = P(A « A) = P(A) P(A) fi P(A) = 0 or P(A) = 1. If P(A) = 0, then f Õ A « B Õ A fi P(f) £ P(A « B) £ P(A) = 0 fi P(A « B) = 0 £ P(A) P(B) If P(A) = 1, use A « B Õ B 16. Total number of ways = 12! The number of ways when all the six girls sit together is (7!) (6!) 7! 6 ! 1 = \ probability of the required event = 12! 132 17. P(A¢ « B¢) = P(A¢) + P(B¢) – P(A¢ » B¢) = (1 – p) + (1 – 2p) – {1 – (A « B)} 1 3 fi = – 3p 3 2 3 1 7 7 fi 3p = – = fi p= 2 3 6 18 18. Use product is odd if and only if both numbers are odd. 19. As A and B are independent P (A « B) = P(A) P(B) fi fi fi fi

nA « B 10

=

4 nB ¥ 10 10

2nB = 5n A « B 5/2nB fi 5 nB = 5 or 10.

22. 23. 24.

25.

26.

3 3 5 ˘ 1 È = Í 6 C3 Ê 1 ˆ Ê 1 ˆ ˙ Ê ˆ = Ë ¯ Ë ¯ Ë ¯ Î 2 2 ˚ 2 32 Probability of getting at least 7 points = (probability of getting 2 points each from 3 matches) ¥ (probability of getting 1 point from one match) + probability of getting 2 points from each of the four matches 4 3 4 = C3(0.5) (0.05) + (0.5) = (0.5)3 (0.2) + (0.5)4 = (0.125) (0.7) = 0.0875 P(F) = P(E » F) + P(E « F) – P(E) See Theory. Last digit of 7 m is 7, 9, 3, 1. If m is of the form 4r + 1, then n must be of the form 4s + 3 etc. P(M » P » C) = 0.75, P(at least two of M, P, C) = 0.5, and P(exactly two) = 0.4 Now, P(exactly two) = P(at least two) – P(M « P « C) fi 0.4 = 0.5 – P(M « P « C) fi P(M « P « C) = 0.1 As the events are independent P(exactly two) = pm(1 – c) + pc(1 – m) + mc(1 – p) 0.4 = pm + pc + mc – 3pmc Also, P(M « P « C) = pmc = 0.1 \ pm + pc + mc = 0.7 Now, P(M » P » C) = p + m + c – (pm + mc + cp) + pmc fi 0.75 = p + m + c – 0.7 + 0.1 fi 1.35 = p + m + c {Y = 0} = {00, 01, 02, º, 09, 10, 20, 30 º 90} and {(X = 7) « (Y = 0)} = {07, 70}

2 19 27. Total number of mappings = n n and number of one-to-one mappings = n! 100 > 50. 28. Use x = 1, 2 or x ≥ 48 for x + x 29. E1 : a biased coin is picked up; E2 : a fair coin is picked up. A : the toss results in a head. Thus, P(X = 7ΩY = 0) =

Probability 24.41

40. P(even) = P(all four odd) + P(two evens and two odds) + P(all four even)

We have n n +1 , P(E2) = P(E1) = 2n + 1 2n + 1

=

1 . 2 Now, use the total probability

P(AΩE1) = 1, P(AΩE2) =

4 2 2 4 41 Ê 2ˆ Ê 2 ˆ Ê 1ˆ Ê 1ˆ C0 Á ˜ + 4 C2 Á ˜ Á ˜ + 4 C 4 Á ˜ = Ë 3¯ Ë 3 ¯ Ë 3¯ Ë 3¯ 81

41. Split the set {1, 2, º 3N} into three sets as follows: A = {1, 4, 7, º 3N – 2} B = {2, 5, 8, º 3N – 1} C = {3, 6, º 3N} Choose x, y from A » B or from C. \ required probability

P(A) = P(E1) P(AΩE1) + P(E2) P(AΩE2) 3

30. P(X + Y = 3) =

4

 P ( X = k, Y = 3 - k )

k=0

2N

3

=

=

 P ( X = k ) . P (Y = 3 - k )

k=0

[∵ X and Y are independent] 3

=

Â(

k=0

5

Ê 1ˆ Ck Á ˜ Ë 2¯

)

5

12 3 Ê 1ˆ = Á ˜ Â 5 Ck Ë 2¯ k = 0

(

(

7

Ê 1ˆ C3 – k Á ˜ Ë 2¯

7

)

)( 7 C3 – k )

12 55 Ê 1ˆ = Á ˜ 12 C3 = Ë 2¯ 1024 31. Use P(they are not together) = 1 – P (they are together)

(

= 1–

n!

5 3 32. P = ,q= , p + q = 0.08 100 100 33. Total number of ways of arranging the letters of the 11! word PROBABILITY is and the number of 2! 2! ways in which two B’s are together is

10! 2!

34. Last digit of the product will be 1, 3, 7 or 9 if the last digit of each of the four numbers is 1, 3, 7 or 9. 35. They can win prize in 9 ways out of 9 ¥ 9 ways. 36. See Theory. 37. See Theory. 38. Sample space is {– 0.50, – 0.49, º, 0, 0.01, º, 0,49} and the round off error is at most 10 paise if A = {– 0.10, – 0.09, º 0.10} 39. Let A : the bulb will last at least 150 days and B : the bulb will last at most 160 days. Use P(A) = 0.7, P(B) = 0.8 and P(A » B) = 1.

C2

=

2 N (2 N – 1) + N ( N – 1) 3N (3N – 1)

5N – 3 9N – 3 42. The number of ways in which 10 persons can speak is 10!. The number of ways in A, B and C speak together (8!) (3!) 1 is = 10! 15 43. Use if 3Ωn 2 fi 3Ωn and hence 9Ωn2 44. Required probability =

45.

2 =1– n

3N

=

)

(n – 1)!(2!)

C2 + N C2

( n C1 ) ( n n

2

2

–n

Cn – 1

)

Cn

a b = ad – bc. c d

Each of the element can take two values, therefore total number of ways = 16. But ad – bc π 0 if and only ad = x 2 and bc = – x2 or ad = – x2 and bc = x2 But ad = x2 and bc = – x2 ¤ (a = d = x or a = – d = x) and (b = x, c = – x or b = – x, c = x) That is, there are four ways Similarly, ad = – x 2 and bc = x2 in four ways. 4+4 1 Thus, required probability = = 16 2 46. Total number of ways = 6 ¥ 6 ¥ 6. The number of favourable ways = 6C3 47. We have 0 £ and 0 £

1 + 4p 1 – p 1 – 2p , , £1 4 2 4

1 + 4p 1 – p 1 – 2p + + £1 4 4 2

1 3 1 1 £ p £ , – 3 £ p £ 1, – £ p £ 4 4 2 2 and 0 £ p £ 4 1 1 3 1 fi max – , – 3, – , 0 £ p £ min , 1, , 4 4 2 4 2 fi –

{

}

{

}

24.42

Complete Mathematics—JEE Main

fi 0 £ p £ 1/2. 48. Let x = 5m + r, y = 5n + s where m, n Œ I, 0 £ r, s £ 4 fi x 2 + y2 = 25(m2 + n2) + 10(mr + ns) + r 2 + s2 The ordered pair (r, s) can take 5 ¥ 5 = 25 values. But r 2 + s2 is divisible by 5 if (r, s) = (0, 0), (1, 2), (2, 1), (1, 3), (3, 1), (2, 4), (4, 2), (3, 4), (4, 3) 9 . Thus, probability of the required event = 25 49. Let A = {a1, a2, º, an}. For each ai, we have four choices: (i) ai Œ P, ai Œ Q (ii) ai œ P, ai Œ Q (iii) ai Œ P, ai œ Q (iv) ai œ P, ai œ Q For P « Q = f, just last three cases are favourable. 50. Let A = {x1, x2, ..., xn}. For xi (1 £ 1 £ n), we have four choice (i) xi Œ P, xi : Œ Q (ii) xi Œ P, xi : œ Q (iii) xi œ P, xi : Œ Q (iv) xi œ P, xi œ Q There are 3 cases out of 4 which are favourable to event P Õ Q. Thus, probability of required event is (3/4)n. \ (3/4)n = 243/1024 = (3/4)5 fi n = 5 51. Total number of cases = 6C3 = 20. There are just two favourable cases viz. D A1A3A5 or DA2 A4 A6. Required probability = A5

2 1 . See Fig. 24.10. = 20 10

5

P(A | E2) =

C2

10

=

C2 By the Bayes’ rule P(E2 | A) =

P ( E 2 ) P ( A | E2 ) P ( E1 ) P ( A | E1 ) + P ( E2 ) P ( A | E2 )

2 fi n = 10. 9 53. The number of ways of choosing x and y is 11C2 = 55. We now count the number of ways in which | x – y | > 5 or | x – y | ≥ 6. For x = 0, y = 6, 7, 8, 9, 10 For x = 1, y = 7, 8, 9, 10 For x = 2, y = 8, 9, 10 For x = 3, y = 9, 10 For x = 4, y = 10 \ the number of favourable ways for | x – y | > 5 is 15. Thus, probability of the required event 40/55 = 8/11. 1 2 54. P(G) = , P(C) = p, P(K) = – p 3 3 1 1 P(R | G) = , P(R | C) = , P(R | K) = 1. 4 8 Now, use Bayes’ rule. 55. Use 2P(X = 5) = P(X = 4) + P(X = 6) Now, use

P(E2 | A) =

56. P(X = 3) = P(X = 5) 8

fi

n

Ê 1ˆ C3 Á ˜ = Ë 2¯

57. P(E¢ « F ¢ | G) = A3

A1

A2

Fig. 24.10

52. E1 : one of the n urns is chosen E2 : (n + 1)th urn is chosen A : both the balls drawn are black. P(E1) =

n 1 , P(E2) = , n +1 n +1 6

P(A | E1) =

C2

10

C2

=

1 , 3

n

Ê 1ˆ C5 Á ˜ Ë 2¯

8

fi

n = 8.

8 1 Ê 1ˆ Thus, P(X = 1) = 8 C1 Á ˜ = Ë 2¯ 32

A4

A6

2 9

P (E ¢ « F ¢ « G ) P (G )

=

P (G ) – P (G « E ) – P ( G « F ) P (G )

=

P (G ) – P (G ) P ( E ) – P (G ) P ( F ) P (G )

= 1 – P(E) – P(F) = P(E¢) – P(F) 58. P(E « F) = P(E) + P(F) – P(E » F) = 0.2. Now, P(F | E) = 59. np = 4, npq = 3 Thus, n = 16. P(1) =

16

P (F « E ) 0.2 = = 0.5 0.4 P (E ) fi

q = 3/4

15 15 Ê 1ˆ Ê 3ˆ Ê 3ˆ C1 Á ˜ Á ˜ = 4 Á ˜ Ë 4¯ Ë 4¯ Ë 4¯

fi

p = 1/4

Probability 24.43 10

60. 61.

C2 ¥ 10C1 20

C3

P(B wins) = P(TH or T T T H or T T T T T H or º) 14 1 1 Ê 1ˆ 2 Ê 1ˆ3 = + Á ˜ + Á ˜ + = Ë 4¯ 1–1 4 3 4 Ë 4¯ Use result of eighth toss is independent of the first seven tosses. P(A « A) = P(A) P(A) fi P(A) = P(A)2 fi P(A) = 1 as P(A) > 0. P(A) + P(B) = P(A » B) + P(A « B) = 2P(A « B) fi (P(A) – P(A « B)) + (P(B) – P(A « B)) = 0 fi P(A) = P(B) = P(A « B) Let E : 1 occurs; F : 1 does not occur. P(1 occurs at even throw) = P(F E or F F F E or F F F F F E or º) =

62. 63.

64.

65.

=

5 5 Ê 25 ˆ 5 Ê 25 ˆ 2 5 36 5 + Á ˜ + Á ˜ + = = Ë ¯ Ë ¯ 36 36 36 36 36 1 – 25 36 11

P ( A¢ « B¢ ) 1 - P ( A » B ) 66. P(A¢|B¢) = = P ( B¢ ) P ( B¢ ) 67. P(B) = 1– P(B¢) = 1 – 6/36 = 5/6 P(A|B) =

P ( A « B) P {(6, 5)} 1 36 1 = = = 30 56 P ( B) P ( B)

68. Let W and B denote the events of drawing a white and black balls in the first draw. Let A denote the event of drawing a black ball in the second draw. P(W) =

w b and P(B) w+b w+b

b b+k and P(A|B) = w+b+k w+b+k By the total probability rule, P(A) = P(W) P(A|W) + P(B) P(A|B) b(w + b + k ) wb + b(b + k ) = = ( w + b) ( w + b + k ) ( w + b) ( w + b + k ) P(A| W) =

=

b w+b

Ê 1ˆ Ê 1ˆ Ê 3ˆ 69. E(X) = 1Á ˜ + 2 Á ˜ + 3 Á ˜ Ë 2¯ Ë 5¯ Ë 25 ¯ Ê 1ˆ Ê 1ˆ Ê 1ˆ + (2k ) Á ˜ + (3k ) Á ˜ + (5k ) Á ˜ Ë 10 ¯ Ë 25 ¯ Ë 25 ¯ =

1 (63 + 26k) fi 115 = 63 + 26k 50 fi k=2 P ( A « B) 70. p = P(A|B) = P ( B) fi P(A « B) = 0.05p Similarly, P(A¢ « B¢) = 0.95p P( B « A) Next, 0.9 = P(B/A) = P( A) 0.05 p p fi P(A) = = 0.9 18 We have 0.95p = 1– P(A » B) fi P(A) + P(B) – P(A « B) = 1 – 0.95p p 5 fi + - 0.05 p = 1 – 0.95p 18 100 fi

1 [25 + 20 + 18 + 10k + 6k + 10k] 50

2.3 =

9ˆ Ê 1 p Á + ˜ = 0.95 Ë 18 10 ¯

fi fi

p=

0.95 ¥ 90 171 = 86 430

71. We have 1 1 1 1 1 = + + + 6 6 12 12 2 1 1 1 1 P(B) = + + + = 1 6 6 12 12 2 1 1 1 1 1 + + + P(C) = = 12 12 12 12 3 1 1 3 1 P(A « B) = + = = = P(A) P(B) 6 12 12 4 1 1 1 + = = P(B) P(C) P(B « C) = 12 12 6 1 1 1 P(C « A) = + = = P(C) P(A) 12 12 6 1 P(A « B « C) = = P(A) P(B « C) 12 72. Three numbers can be chosen out of 20 in 20C3 ways. For different values of common ratios, G.P.’s are listed below. P(A) =

r 2

G.P. (1, 2, 4), (2, 4, 8), (3, 6, 12), (4, 8, 16), (5, 10, 20) 3 (1, 3, 9), (2, 6, 18) 4 (1, 4, 16) 3/2 (4, 6, 9), (8, 12, 18) 4/3 (9, 12, 16) Total

Number of G.P.’s 5 2 1 2 1 11

24.44

Complete Mathematics—JEE Main

\

probability of required event =

11 20

C3

73. Total number of ways of choosing 3 tickets out of 30 is 30C3. We can choose x1 is 14 ways, x3 in 10 ways. Therefore, the probability of the required

(14 )(10 )

14 ¥ 10 ¥ 3 ¥ 2 1 = 29 30 ¥ 29 ¥ 28 C3 74. P(Match does not occur at ith trial) n –1 1 = =1– n n \ P(at least one match) 1ˆ n Ê = 1 – P(no match) = 1 – Á 1 – ˜ . Ë n¯ 75. Required probability = 1 – P (three numbers chosen are consecutive) 18 3 187 = 1 - 20 = 1 = 190 190 C3 events =

30

Thus, probability Q » R = P is (3/4) n. 79. If a and b are both odd, then a + b is even, therefore, P(A « B « C) = 0.

Now, P[(A « B « D) | A » B]

=

76. A « B « C Õ C fi P(A « B « C) £ P(C) = 0 \ P(A « B « C) = 0 Similarly, P(A « C) = 0 P(B « C) = 0 Next, P(A » B » C) = P(A » B) + P(C) – P[(A » B) « C] But P[(A » B) « C] = P[(A « C) » (B « C)] = P(A « C) + P(B « C) – P(A « B « C) = 0 Thus, P(A » B » C) = P(A » B) 77. P[A « (B » 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 « B « C) Now, A and B » C are independent ¤ P[A « (B » C)] = P(A) P(B » C) ¤ P(A) P(B) + P(A) P(C) – P(A « B « C) = P(A) [P(B) + P(C) – P(B « C)] ¤ P(A « B « C) = P(A) P(B « C) ¤ A and B « C are independent 78. Let P = {a1, a2, º an}. For ai Œ P (1 £ i £ n), we have four choices. (i) ai Œ Q and ai Œ R (ii) ai Œ Q and ai œ R (iii) ai œ Q and ai Œ R (iv) ai œ Q and ai œ R Thus, total number of choices for picking up Q and R is 4 n. For ai not to belong to Q « R, we have 3 choices. Therefore, Q « R = f in 3n ways. Thus, probability Q « R = f is (3/4)n. Next, ai Œ Q » R in 3 ways. Therefore, Q » R = P in 3n ways.

1 1 1 , P(B) = , P(D) = . 2 2 2

We have P(A) =

=

P [( A « B « D ) « ( A » B )] P ( A « B ) = P ( A » B) P ( A » B)

=

14 1 = 1 2 +1 2 –1 4 3

80. Statement-2 is false. P(P and Q contradict each other) = p(1 – 2p) + 2p(1 – p) = 1/2 fi 8p2 – 6p + 1 = 0. fi (2p – 1) (4p – 1) = 0 fi p = 1/2, 1/4.

Level 2 81. Let X = number of times the event occurs, then X ~ B (n, p) where n = 3, p = 0.6 P (the event occur at least once) = P (X ≥ 1) = 1 – P (X = 0) = 1 – (0.4)3 = 0.936 82. P (neither A nor B) = P (A¢ « B ¢) = 1 – P (A » B) = 1 – [P (A) + P (B)] = 1 – 0.8 = 0.2 83. P (X = 3) = P (X = 5) fi nC3 p3 (1 – p)n– 3 = nC5 p5 (1 – p)n – 5 fi

2 (n - 3) (n - 4) = (1 - p) < 1 p2 ( 5) ( 4 )

fi n2 – 7n – 8 < 0 fi (n – 8) (n + 1) < 0 fi n – 8 < 0 fi n £ 7. 84. Let X = number of times a doublet is thrown, then X ~ B (n, p) where n = 3, p = 1/6. \ P (doublet not more than twice) 215 216 85. Probability of the required event = P (both brown or both white) = P (both brown) + P (both white) = P (X £ 2) = 1 – P (X = 3) =

4

=

9

C2 C2

5

+

9

C2 C2

=

4 9

86. Let A denote the event that three dice show different faces and B denote the event that at least one dice shows 6.

Probability 24.45

We have 5

1 1 £x£ 3 2 2. Winning horse can be selected in 5C1 = 5 ways. The number of ways in which winning horse is amongst the horses selected by man is 2. \

3

C /6 P ( B « A) 1 = 6 2 3 = P ( A) 2 C3 / 6 87. P (exactly 4 are car drivers) P (B | A) =

4 4 = 5C4 ÊÁ 4 ˆ˜ ÊÁ 1 ˆ˜ = ÊÁ 4 ˆ˜ Ë 5 ¯ Ë 5¯ Ë 5 ¯ 88. Probability of the required event

=

(

4

C2 9

)(

5

C2

C4

3. np = 4, np(1 – p) = 2

) = 10

21

fi

(1 - p )2 p

2

=

(8) (10) 5

2 1 1 = fi p= 4 2 2 Thus. n = 8

fi 1- p = A

89. Suppose A occupy any seat at the round table. Then there are 14 seats available B1 B2 for B. If there are four persons between A and B, then B has only two ways to sit, as shown in figure. Thus, probability of the required event = 2/14 = 1/7. 90. Total number of ways of arranging (3 + 4 + 1) books = 8! The number of favourable ways = (3P3) (3P3) (4P4) (1P1) = (3!) (3!) (4!) Thus, probability of the required event (3!) (3!)(4!) 3 = = 8! 140 91. As n = 9 and np = 6, we get p = 2/3. Thus, variance = np (1 – p) = 2 92. P (X = 1) = 8P (X = 3) fi nC1 p (1 – p)n – 1 = 8 (nC3) p3 (1 – p)n – 3 fi

1 8 1 Now, P ( X = 1) = 8C1 ÊÁ ˆ˜ = Ë 2¯ 32 4. P(E) = P(2, 3, 5, 7) = 0.23+ 0.12 + 0.20 + 0.07 = 0.62 P(F) = P(1, 2, 3) = 0.15 + 0.23 + 0.12 = 0.5 P(E « F) = P(2, 3) = 0.23 + 0.12 = 0.35 Thus, P(E » F) = P(E) + P(F) – P(E « F) = 0.62 + 0.5 – 0.35 = 0.77 5. Probability they contradict each other = P(A¢B or AB¢) = P(A¢B) + P(AB¢) = P(A¢) P(B) + P(A) P(B¢) 4 3 4 3 7 = ÊÁ1 - ˆ˜ ÊÁ ˆ˜ + ÊÁ ˆ˜ ÊÁ1 - ˆ˜ = Ë 5 ¯ Ë 4 ¯ Ë 5 ¯ Ë 4 ¯ 20 6. From solution to Question 3, 1 n = 8, p = 2

= 16

1- p = 4 fi p = 1/ 5 p

1 Thus, P(X = 2) = 8C2 ÊÁ ˆ˜ Ë 2¯ 7.

Previous Years’ AIEEE/JEE Main Questions 1.

1 1 1 (3x + 1) ≥ 0, (1 - x ) ≥ 0, (1 - 2 x ) ≥ 0 3 4 2 and

2 5

\ Required probability =

1 1 1 (3x + 1) + (1 - x ) + (1 - 2 x ) £ 1 3 4 2

1 1 fi x ≥ - , x £ 1, x £ 3 2 and 4(3x + 1) + 3(1 – x) + 6(1 – 2x) £ 12 Last equation gives 13 – 3x £ 12 1 fi x≥ 3

fi

8

=

7 28 = 64 256

1 = P( A » B) 6 = 1 – P(A » B) 5 P(A » B) = 6

fi P(A) + P(B) – P(A « B) =

5 6

fi

5 1 13 + = 6 4 12

P(A) + P(B) =

fi

3 13 + P( B) = 4 12

fi

P(B) =

13 3 1 - = 12 4 3

24.46

Complete Mathematics—JEE Main

3 1 1 P(A) P(B) = ÊÁ ˆ˜ ÊÁ ˆ˜ = Ë 4 ¯ Ë 3¯ 4

We have

= P(A « B) 8. The number of ways in which any person can choose a house is 3. \ total number of ways = 33 = 27 Number of favourable ways = 3 3 1 = \ Required probability = 27 9 9. Let S denote the event of getting a score of 9 in a single throw and F = S¢, then 4 1 8 = and q = P( F ) = p = P( S ) = 36 9 9 Let X = Number of times a score of 9 occurs. Then X ~ B(3, p) 1 2 8 8 P(X = 2) = 3C2 p 2 q = 3 ÊÁ ˆ˜ ÊÁ ˆ˜ = Ë 9¯ Ë 9¯ 9 10. Probability of the required event is (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 0.14 7 = = = 1 - 0.56 0.44 22 None of the Answer Matches. 11. Let X denote the nummber on the die 1 P(A) = P(X > 3) = P(4, 5, 6) = 2 4 P(B) = P(X < 5) = 6 \

1 6 \ P(A » B) = P(A) + P(B) – P(A « B) 1 4 1 = + - =1 2 6 6 12. Let A and B denote the following events A : sum of digits is 8 and B : product of digits is 0. Then A « B = {08} and B = {00,01, 02, ..., 09, 10, 20, 30, 40} Now, P(A « B) = P(4) =

P ( A | B) = 13. P ( X ≥ 1) ≥

P ( A « B) n ( A « B) 1 = = P( B) n( B ) 14 9 10

fi 1 - P ( X = 0) ≥

9 10

3 n 9 fi 1 - ÊÁ ˆ˜ ≥ Ë 4¯ 10 3 n 1 fi ÊÁ ˆ˜ £ Ë 4¯ 10 4 n fi ÊÁ ˆ˜ ≥ 10 Ë 3¯ fi n [log 4 - log 3] ≥ log10 10 1 log 4 - log 3 14. Total number of ways of drawing three balls out of 9 is 9C3. The number of ways of drawing the balls so that balls are of different colours is (3C1) (4C1) (2C1) = (3) (4) (2) \ probability of the required event (3)(4)(2) (3)(4)(2)(3)(2) 2 = = = 9 9¥8¥7 7 C3 15. Statement-2 is false. For d = 6, we have 1, 7, 13, 19 form an A.P. Thus, the only choice of answer is (c). However, let us count the number of AP¢s. For d = 1, there are 17 such AP¢s, viz. (1, 2, 3, 4), (2, 3, 4, 5), ..., (17, 18, 19, 20). For d = 2, there are 14 such AP¢s, viz. (1, 3, 5, 7), (2, 4, 6, 8), ... (14, 16, 18, 20). Similarly, for d = 3, there are 11 AP¢s, for d = 4, there are 8 AP¢s. d = 5, there are 5 and for d = 6, there are 2 AP¢s. Thus, the number of AP¢s is 17 + 14 + 11 + 8 + 5 + 2 = (3) (19) Also, the total number of ways of choosing 4 numbers out of 20 is 20C4. Hence, probability of the required event is (3)(19) (3)(19)(4)(3)(2) 1 = = 20 (20)(19)(18)(17) 85 C4 \ Statement-1 is true. 16. Let X = number of failures, then X ~ B(5, 1 – p) P(at least one failure) = P(X ≥ 1) = 1 – P(X = 0) = 1 – (1 – (1 – p))5 = 1 – p5 We are given 31 1 fi p5 £ 1 - p5 ≥ 32 32 fi n≥

fi 0£ p£

1 2

Probability 24.47

17. P (C | D ) =

P (C « D ) P(C ) = P( D) P( D)

18. We have P(A¢ « B¢ | C) P ( A¢ « B ¢ « C ) = P(C )

(1 - a )(1 - b ) =

[∵ C Õ D]

fi a + b - 2ab =

A

C

B

P(C ) - P[( A « C ) » ( B « C )] P(C )

=

P(C ) - P( A « C ) - P( B « C ) + P( A « B « C ) P(C )

=

P(C ) - P( A) P(C ) - P( B) P (C ) P(C )

34 49 42 6 8 = , ab = Thus, a + b = 49 7 49 2 4 fi a = ,b= 7 7 4 \ most probable = 7

[∵ A, B, C are pairwise independent and P(A « B « C) = 0] = 1 – P(A) – P(B) = P(A¢) – P(B) 19. Total number of ways of choosing 3 numbers out of 8 is 8C3 = 56 Let A denote the event that minimum is 3 and B denote the event that maximum number is 6. Then P( A « B) P( A | B) = P( B) C2

8

C3

=

10 and 56

=

2 56

2

P( A « B) =

C1

8

C3

2 1 = 10 5 20. Let X = number of correct answers obtained by just guessing, then X ~ B(n, p) where n = 5, p = 1/3. P(getting 4 or more correct answers) = P(X ≥ 4) = P(X = 4) + P(X = 5) \ P( A | B) =

1 4 2 1 5 11 = 5C4 ÊÁ ˆ˜ ÊÁ ˆ˜ + 5C5 ÊÁ ˆ˜ = 5 Ë 3¯ Ë 3 ¯ Ë 3¯ 3 21. P((A¢ « B) » (A « B¢)) =

26 49

(1)

15 (2) 49 Let P(A) = a, P(B) = b, then (1) and (2) give 26 (1 - a ) b + a (1 - b ) = 49 and P(A¢ « B¢) =

15 49

fi a + b - ab =

=

5

26 49

and 1 - (a + b ) + ab =

P[( A » B) ¢ « C ] = P(C )

But P( B) =

15 49

22.

1 1 (3x + 1) ≥ 0, (1 - x ) ≥ 0 3 4 1 fi x ≥ - , x £1 3 Also, P(A) + P(B) £ 1 fi 4(3x + 1) + 3(1 – x) £ 12 5 fi 9x £ 5 fi x £ 9 1 5 Thus, x Œ ÈÍ- , ˘˙ Î 3 9˚

23. Let X = number of times the man hits the target 2 Then X ~ B (k, p) where p = 5 7 Now, P ( X ≥ 1) ≥ 10 fi 1 - P ( X = 0) ≥

7 10

k

3 3 fi k C0 ÊÁ ˆ˜ £ Ë 5¯ 10 fi 2(3k – 1) £ 5k – 1 Least value of k is 3. 24. P((A » B) « C¢) = P[(A « C¢) » (B « C¢)] = P(A « C¢) + P(B « C¢) – P(A « B « C¢) = P(A) P(C¢) + P(B) P(C¢) – P(A) P(B) P(C¢) 3 1 3 3 21 = ÊÁ + - ˆ˜ = Ë 4 2 8 ¯ 8 64 25.

1 = P(( A » B )¢ ) 6 = 1 – P(A » B) = 1 – (P(A) + P(B) – P(A « B))

24.48

Complete Mathematics—JEE Main

3 1 = 1 - ÊÁ + P( B) - ˆ˜ Ë4 4¯ 1 - P( B) 2 1 fi P( B) = 3 =

1 Ê 3ˆ Ê 1ˆ = Á ˜ Á ˜ = P( A) P( B) 4 Ë 4 ¯ Ë 3¯ we get A and B are independent. 26. P(A » B) = P(A « B) fi [P(A) – P(A « B)] + [P(B) – P(A « B)] = 0 As P(A « B) £ P(A), P(B), we get P(A) – P(A « B) = 0, P(B) – P(A « B) = 0 fi P(A) = P(B) = P(A « B) Thus, A and B are equally likely. Now, P(A « B¢) = P(A) – P(A « B) = 0 Smilarly, P(A¢ « B) = 0 However, we cannot say P(A) + P(B) = 1 27. P(X = 2) = P(X = 3) fi nC 2 p 2q n – 2 = nC 3 p 3q n – 3 p n (n - 1) 3¥ 2 3 = ◊ = fi q 2 n (n - 1)(n - 2) n - 2 fi np – 2p = 3q Now, E(X) = np = 2p + 3q = 3(p + q) – p fi E(X) = 3 – p. ( x - 10)( x - 50) ≥0 28. ( x - 30) fi 10 £ x < 30 or x ≥ 50 \ x can take (20 + 51) values. Thus, probability of required event is 71 = 0.71 100 As P( A « B) =

29. P ( E | A) =

P ( A « E ) P (E ) P ( A | E ) = P (E ) P ( A)

≥ P(E) P(A|E) as 0 < P(A) £ 1 Also, P ( A | E ) =

P (A | E) ≥ P (A « E) P (E )

as 0 < P(E) £ 1 30. Number of ways of choosing A is 27 – 1 = 127. Now x Œ A if and only if A = {x} » B where B is a subset of S – {x}. The number of ways of choosing B is 64. 64 . \ Probability of required event is 127

31. The question is incorrect We solve it by taking balls to be distinct. For each ball we have three choice. Thus, total number of choice is 312. We can choose one box three balls out of 12 in 12 C3 ways and distribute the remaining 9 balls in two boxes in 29 ways. Thus, number of favourable ways is 12C3 (29) \ probability of the required event is 12 ¥ 11 ¥ 10 29 55 Ê 2 ˆ 11 ◊ 12 = Á ˜ 3¥ 2 3 Ë 3¯ 312 3 32. Number of elements in P(X) is 210. Sets A and B can be chosen in (210) (210) = 220 ways We now count number of favourable ways. A and B each can have k (0 £ k £ 10) elements in (10C k) (10Ck) ways. \ Number of favourable ways 12

C3 (29 )

10

=

=

2

10

 ( 10Ck ) =  ( 10Ck )( 10C10- k )

k =10

k =0

= Number of ways of selecting 10 persons from 10 men and 10 women = 20C10 20

\ probability of required event =

C10

220

33*. Possible isosceless triangle are (1, 1, 1), (2, 2, a) where a = 1, 2, 3, (3, 3, b), where b = 1, 2, 3, 4, 5 and (c, c, d) where c = 4, 5, 6 and d = 1, 2, 3, 4, 5, 6 Thus, there are 27 isosceles triangle and maximum area occurs when the triangle is (6, 6, 6). Hence, probability of the required event s 1/27. 34. np = 2, np(1 – p) = 1 fi 1 – p = 1/2 fi p = 1/2 and n = 4 Thus, P(x ≥ 1) = 1 – P (X = 0) 1 4 15 = 1 - 4C0 ÊÁ ˆ˜ = Ë 2¯ 16 35. 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)} E3 = {(x, y) : 1 £ x, y £ 6, x + y is odd} We have E1 « E2 = {(4, 2)} E2 « E3 = {(1, 2), (3, 2), (5, 2)} E1 « E3 = {(4, 1), (4, 3), (4, 5)} E1 « E2 « E3 = f We have 1 1 1 P ( E1 ) = , P ( E2 ) = , P ( E3 ) = 6 6 2

Probability 24.49

P ( E1 « E2 ) =

1 1 1 , P ( E2 « E3 ) = , P ( E3 « E1 ) = 36 12 12

P(E1 « E2 « E3) = 0 As P(E1 « E2) = P(E1) P(E2), P(E2 « E3) = P(E2) P(E3), P(E3 « E1) = P(E3) P(E1) We get (a), (b), (c) are true. But P(E1 « E2 « E3) = 0 π P(E1) P(E2) P(E3) fi (d) is not true. 2 3 36. P ( A) = , P ( A « B ) = 5 20 Now, P [ A « ( A¢ » B ¢ )] P ( A | ( A¢ » B¢ )) = P ( A¢ » B ¢ ) =

P ( A « B¢) 1 - P ( A « B)

P ( A) - P ( A « B ) = 1 - P ( A « B) =

2 5 - 3 20 1 - 3 20

=

5 20 5 = 17 20 17

37. P(S) = 2P(F) Also, P(S) + P(F) = 1 P(F) = 1/3, P(S) = 2/3 Let X = No. of successes, then P(X ≥ 5) = P(X = 5) + P(X = 6) 2 5 1 2 6 = C5 ÊÁ ˆ˜ ÊÁ ˆ˜ + 6C6 ÊÁ ˆ˜ Ë 3 ¯ Ë 3¯ Ë 3¯ 6

2 5 6 2 256 = ÊÁ ˆ˜ ÈÍ + ˘˙ = Ë 3 ¯ Î 3 3 ˚ 729

Previous Years’ B-Architecture Entrance Examination Questions 1. P(A » B) = P(AB¢) + P(B) = 0.15 + 0.55 = 0.7 2. Required probability = P(1 and 2 are operational) + P(1 and 3 are operational) = P(1) P(2) + P(1) P(3) = (p) (p) + (p) (p) = 2p2 3. P(Ei) = P(digit i appears on the ball)

2 1 = , i = 1, 2, 3 4 2 P(Ei « Ej) = P(digits i and j appear on the ball) 1 = for i π j 4 Statement 1 and 2 both are true but statement-2 is not a correct explanation for statement-1. 4. Total number of cases = 63 Number of favourable cases = Coefficient of x7 in (x + x2 + ... + x6)3 = Coefficient of x7 in x3(1 – x6)3 (1 – x)–3 = Coefficient of x4 in (1 + 3x + 4C2 x2 + 5C3 x3 + ...) = 6C4 = 15 Also, A « B = {(1, 1, 5), (1, 2, 4), (1, 3, 3), (1, 4, 2), (1, 5, 1)} P ( A « B) 5 1 P ( A | B) = = = P( B) 15 3 =

5. P (Element belongs to A only) = P(A « B¢ « C¢)

C

n ( A « B¢ « C ¢) A B n(A » B » C) We have n(A » B » C) = 35 + 40 + 45 – 13 – 12 – 14 +5 = 86 and n(A « B¢ « C¢) = n(A) – [n(A « B) + n(A « C) – n(A « B « C)] =

= 35 – 13 – 12 + 5 = 15 Thus, P(A « B¢ « C¢) =

15 86

6. E1 : die shows a six E2 : die doesn’t show a six A : man reports it is a six We have 1 5 P ( E1 ) = , P ( E2 ) = 6 6 3 1 P ( A | E1 ) = , P ( A | E2 ) = 4 4 By the Bayes’ rule P ( E1 | A) = =

P ( E1 ) P ( A | E1 ) P ( E1 ) P ( A | E1 ) + P ( E2 ) P ( A | E2 )

(1 6)(3 4) (1 6)(3 4) + (5 6)(1 4)

= 3/8 7. P ( A | A¢ » B ¢ ) =

P ( A « ( A¢ » B ¢ )) P ( A¢ » B ¢ )

24.50

Complete Mathematics—JEE Main

=

P ( A « B¢) 1 - ( P ( A) + P ( B ) - P ( A » B ))

=

P ( A) - P ( A « B ) 1 - P ( A « B)

=

0.4 - 0.15 1 - 0.15

q = 2, p = 1, 2 q = 3, 4, p = 1, 2, 3 q = 5, 6, p = 1, 2, 3, 4 q = 7, 8, 9 p = 1, 2, 3, 4, 5 q = 10, p = 1, 2, 3, 4, 5, 6 Thus, (1) is satisfied by 1 + 2 + 2 ¥ 3 + 2 ¥ 4 + 3 ¥ 5 + 6 = 38 ordered pairs (p, q) Also, total number of ways of choosig (p, q) is 10 ¥ 10 = 100 \ probability of required event 38 31 = 1= . 100 50

0.25 0.85 5 = 17 =

8. P(F « R « G) = P(F) P(R) P(G) [∵ traits are independent] 20 10 25 5 = ÊÁ ˆ˜ ÊÁ ˆ˜ ÊÁ ˆ˜ = Ë 80 ¯ Ë 80 ¯ Ë 80 ¯ 512 3 £ P ( A » B) £ 1 4 3 fi £ P ( A) + P ( B ) - P ( A « B ) £ 1 4 3 1 3 fi + £ P ( A) + P ( B ) £ 1 + 4 8 8 7 11 fi £ P ( A) + P ( B ) £ 8 8 Thus both the statements are corret but statement-2 is not a correct explanation for statement-2. 10. P(Even number of tosses are required) = P(TH or TTTH or TTTTTH or ---) = P(T) P(H) + P(T)3 P(H) + P(T)5 P(H) + --9.

= =

P (T ) P ( H ) 2

1 - P (T )

=

P (T ) P ( H ) (1 - P (T )) (1 + P (T ))

P (T ) P ( H ) 1- p = P ( H ) (1 + P (T )) 2 - p

2 1- p Now, = 5 2- p

fi

4 – 2p = 5 – 5p

fi p = 1/3 11. P(root are real) = 1 – P(roots are non-real) Root are non-real if and only if p2 – 4q < 0 fi 4q < p2 When q = 1, p = 1

9 10 9 1 – P(X = 0) ≥ 10 1 P(X = 0) £ 10 P (x ≥ 1) ≥

12. fi fi

Ê 3ˆ ÁË ˜¯ 4

n

fi

Ê 4ˆ ÁË ˜¯ 3

n

fi fi

n log10(4/3) ≥ 1

fi

£

1 10

≥ 10

n≥

1 log10 4 - log10 3

13. Let E1 : black ball is drawn in the first draw E2 : white ball is drawn in the first draw A : a black ball is drawn in the second draw 5 4 P ( E1 ) = , P ( E2 ) = 9 9 P ( A | E1 ) =

5 5+2 , P ( A | E2 ) = 9+2 9+2

By the total probability rule P(A) = P(E1)P(A | E1) + P(E2)P(A | E2) 5 5 4 7 53 = ÁÊ ˜ˆ ÁÊ ˜ˆ + ÁÊ ˜ˆ ÁÊ ˜ˆ = Ë 9 ¯ Ë 11¯ Ë 9 ¯ Ë 11¯ 99

CHAPTER TWENTY FIVE

INTRODUCTION An angle is the amount of revolution which a line OP revolving about the point O has undergone in passing from its initial position OA into its final position OB. B

The trigonometrical ratios of an angle are numerical quantities. Each one of them represents the ratio of the length of one side to another of a right angled triangle.

DOMAIN AND RANGE OF TRIGONOMETRICAL FUNCTIONS Domain

Range

sin A

R

[– 1, 1]

cos A

R

[– 1, 1]

tan A

R Р{(2n + 1) p/2, n ΠI}

R = (– •, •)

R Р{np, n ΠI}

(– •, – 1] » [1, •)

P A

O

Fig. 25.1

If the rotation is in the clockwise sense, the angle measured is negative and it is positive if the rotation is in the anticlockwise sense. The two commonly used systems of measuring an angle are 1. Sexagesimal system in which 1 right angle = 90 degrees (90∞) 1 degree = 60 minutes (60¢) 1 minute = 60 seconds (60≤) 2. Circular systems in which the unit of measurement is the angle subtended at the centre of a circle by an arc whose length is equal to the radius and is called a radian. Relation between degree and radian p radian = 180 degrees 180 degrees 1 radian = p = 57∞ 17¢ 45≤ (approximately)

cosec A sec A cot A

R – {(2n + 1) p/2, n Œ I} (– •, – 1] » [1, •) R – {np, n Œ I}

R = (– •, •)

We find, |sin A| £ 1, |cos A| £ 1 and

sec A ≥ 1 or sec A £ – 1 cosec A ≥ 1 or cosec A £ – 1

SOME BASIC FORMULAE 1. sin2 A + cos2 A = 1 or cos2 A = 1 – sin2 A or sin2 A = 1 – cos2 A. 2. 1 + tan2 A = sec2 A or sec2 A – tan2 A = 1. 3. 1 + cot2 A = cosec2 A or cosec2 A – cot2 A = 1. 4. sin A cosec A = tan A cot A = cos A sec A = 1. A system of rectangular coordinate axes divides a plane into four quadrants. An angle q lies in one and only one of these quadrants. The values of the trigonometric ratios in the various quadrants are shown in Fig. 25.2.

Complete Mathematics—JEE Main

25.2

2. If n is even, the result contains the same trigonometric function as the given function. But if n is odd, the result contains the corresponding co-function, i.e., sine becomes cosine, tangent becomes cotangent, secant becomes cosecant and vice versa.

Y II quadrant only sin q cosec q

I quadrant All t-ratios are + ve

are + ve

X¢

X III quadrant only tan q cot q

IV quadrant only cos q

are + ve

sec q

are + ve

Y¢

Fig. 25.2

Sine, cosine and tangent of some angles less than 90° 0° sin

0

cos

1

tan

0

15°

6- 2 4 6+ 2 4 2- 3

18°

5 -1 4

30°

1 2

10 + 2 5 4

3 2

25 - 10 5 5

1 3

36°

45°

60°

90°

10 - 2 5 4

1

3 2

1

cos

5 +1 4

1

tan

5-2 5

1

sin

2

2

1 2 3

Illustration

1

(a) To determine sin (540∞ – q), we note that 540∞ – q = 6 ¥ 90° – q is a second quadrant angle if 0∞ < q < 90∞. In this quadrant, sine is positive and, since the given angle contains an even multiple of p / 2, the sine function is retained. Hence, sin (540° – q) = sin q. (b) To determine cos (630° – q), we note 630° – q = 7 ¥ 90° – q is a third quadrant angle if 0° < q < 90°. In this quadrant, cosine is negative and, since the given angle contains an odd multiple of p / 2, cosine is replaced by sine. Hence, cos (630° – q) = – sin q.

COMPOUND ANGLES An angle made up of the algebraic sum of two or more angles is called a compound angle. Some formulae and results regarding compound angles: 1. sin (A + B) = sin A cos B + cos A sin B. 2. sin (A – B) = sin A cos B – cos A sin B. 3. cos (A + B) = cos A cos B – sin A sin B. 4. cos (A – B) = cos A cos B + sin A sin B. tan A + tan B 5. tan (A + B) = . 1 - tan A tan B 6. tan (A – B) =

0

not defined

ALLIED OR RELATED ANGLES np np The angles + q and – q, where n is any integer, 2 2 are known as allied or related angles. The trigonometric functions of these angles can be expressed as trigonometric functions of q, with either a plus or a minus sign. The following working rules can be used in determining these functions : 1. Assuming that 0 < q < 90°, note the quadrant in which the given angle lies. The result has a plus or minus sign according as the given function is positive or negative in that quadrant.

tan A - tan B . 1 + tan A tan B

7. sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A. 8. cos (A + B) cos (A – B) = cos2 A – sin2 B = cos2 B – sin2 A. 2 tan A 9. sin 2A = 2 sin A cos A = . 1 + tan 2 A 10. cos 2A = cos2 A – sin2 A = 1 – 2 sin2 A = 2 cos2 A – 1 =

1 - tan 2 A 1 + tan 2 A

11. 1 + cos 2A = 2 cos2 A, 12. tan 2A = 13. (a) (c)

2 tan A 1 - tan 2 A

.

1 – cos2A = 2 sin2 A.

.

1 - cos q q = tan sin q 2 1 - cos q q = tan 2 . 1 + cos q 2

(b)

1 + cos q q = cot sin q 2

Trigonometrical Ratios, Identities and Equations 25.3

14. (a) sin 3x = 3 sin x – 4 sin3 x (b) cos 3x = 4 cos3 x – 3 cos x (c) tan 3x =

3 tan x - tan3 x 1 - 3 tan 2 x

.

15. tan (A + B + C) tan A + tan B + tan C - tan A tan B tan C = . 1 - tan B tan C - tan C tan A - tan A tan B 16. (a) sin C + sin D = 2 sin

C+D C-D cos 2 2

C+D C-D sin 2 2 C+D C-D (c) cos C + cos D = 2 cos cos 2 2 C+D D-C (d) cos C – cos D = 2 sin sin . 2 2 (Note) sin ( A + B) 17. (a) tan A + tan B = cos A cos B (b) sin C – sin D = 2 cos

sin ( A - B) (b) tan A – tan B = . cos A cos B 18. (a) (b) (c) (d)

2 2 2 2

sin A cos B = sin (A + B) + sin (A – B) cos A sin B = sin (A + B) – sin (A – B) cos A cos B = cos (A + B) + cos (A – B) sin A sin B = cos (A – B) – cos (A + B) (Note) 19. sin nA = cosn A (nC1 tan A – nC3 tan3 A + n C5 tan5 A – º) 20. cos nA = cosn A (1 – nC2 tan2 A + nC4 tan4 A º) C1 tan A - nC3 tan3 A + nC5 tan 5 A º

n

21. tan nA =

1 - nC2 tan 2 A + nC4 tan 4 A º 22. sin a + sin (a + b) + sin (a + 2b) + º + sin (a + (n – 1)b) sin (a + ( n - 1) b 2 ) sin (nb /2) = sin ( b 2 ) 23. cos a + cos (a + b) + cos (a + 2b) + º + cos (a + (n – 1)b) =

cos (a + ( n - 1) b 2 ) sin (nb/2) sin ( b 2 )

S - S3 + S5 - S7 + º 24. tan (A1 + A2 + º + An) = 1 1 - S2 + S4 - S6 + º where S1 = tan A1 + tan A2 + º + tan An (sum of the tangents taken one at a time) S2 = tan A1 tan A2 + tan A2 tan A3 + º (sum of the tangents taken two at a time, there are nC2 such terms)

S3 = tan A1 tan A2 tan A3 + tan A2 tan A3 tan A4 + º (sum of the tangents taken three at a time, there are n C3 such terms and so on.

SOME IMPORTANT RESULTS 1. sin q sin (60∞ – q) sin (60∞ + q) = (1/4) sin 3q. 2. cos q cos (60∞ – q) cos (60∞ + q) = (1/4) cos 3q. 3. tan q tan (60∞ – q) tan (60∞ + q) = tan 3q 4. sin 9∞ = (1 / 4 ) È 3 + 5 - 5 - 5 ˘ = cos 81∞ Î ˚ 5. cos 9∞ = (1 / 4 ) È 3 + 5 + 5 - 5 ˘ = sin 81∞ Î ˚ 6. cos 36∞ – cos 72∞ = 1/2 7. cos 36∞ cos 72∞ = 1/4 1 Ê 1ˆ 8. sin 22 ∞ = Á ˜ È 2 - 2 ˘ Ë 2¯ Î ˚ 2 9. cos 22

1 ∞ Ê 1ˆ È = Á ˜ 2+ 2˘ Ë 2¯ Î ˚ 2

10. tan 22

1∞ = 2

2 –1

11. cot 22

1∞ = 2

2 +1

12. - a 2 + b2 £ a sin x + b cos x £ x Œ R.

a 2 + b2 for all

IDENTITIES A trigonometric equation is an identity if it is true for all values of the angle or angles involved. A given identity may be established by (i) reducing either side to the other one, or (ii) reducing each side to the same expression, or (iii) any convenient modifications of the methods given in (i) and (ii).

CONDITIONAL IDENTITIES When the angles A, B and C satisfy a given relation, many interesting identities can be established connecting the trigonometric functions of these angles. In proving these identities, we require the properties of complementary and supplementary angles. For example, if A + B + C = p, then 1. sin (B + C) = sin A,

cos B = – cos (C + A)

2. cos (A + B) = – cos C,

sin C = sin (A + B)

3. tan (C + A) = – tan B,

cot A = – cot (B + C).

4. cos

A+ B C = sin , 2 2

cos

5. sin

C+A B = cos , 2 2

sin

C A+B = sin . 2 2 A B+C = cos . 2 2

25.4

Complete Mathematics—JEE Main

B+C A B C+A = cot , tan = cot . 2 2 2 2 Some Important Identities If A + B + C = p, then 1. tan A + tan B + tan C = tan A tan B tan C. 2. cot B cot C + cot C cot A + cot A cot B = 1. B C C A A B tan + tan tan + tan tan = 1. 3. tan 2 2 2 2 2 2 A B C A B C 4. cot + cot + cot = cot cot cot . 2 2 2 2 2 2 6. tan

5. sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C. 6. cos 2A + cos 2B + cos 2C = – 1 – 4 cos A cos B cos C. 2 7. cos A + cos2 B + cos2 C = 1 – 2 cos A cos B cos C. A B C cos cos . 8. sin A + sin B + sin C = 4 cos 2 2 2 9. cos A + cos B + cos C = 1 + 4 sin

A B C sin sin . 2 2 2

GENERAL SOLUTIONS OF TRIGONOMETRICAL EQUATIONS The following formulae are used in solving trigonometrical equations: 1. If sin q = sin a, then q = np + (– 1)n a (n Œ I) 2. If cos q = cos a, then q = 2np ± a (n Œ I) 3. If tan q = tan a, then q = np + a (n Œ I) 4. If sin q = sin a, cos q = cos a, then q = 2np + a (n Œ I) Illustration

2

If sin q = 1/2, then sin q = sin (p/6) so that q = np + (Р1)n p/6 (n ΠI) In solving trigonometrical equations, the general values of the angle should be given, unless the solution is required in any specified interval or range.

Principal Value Since all the trigonometrical ratios are periodic functions, an equation of the form sin q = k, cos q = k or tan q = k etc. can have an infinite number of angles satisfying it. The set of all such values is called general value of q. Numerically least value in this set is called Principal Value. Illustration

3

If sin q = 1/ 2 , the general value of q is np + (– 1)n p/4 and the principal value is p/4.

Method for Finding Principal Value (i) First note the quadrant in which the angle lies. (ii) For 1st and 2nd quadrants consider anticlockwise direction and for 3rd and 4th clockwise direction. (iii) Find the angles in the Ist rotation. (iv) Select the numerically least angle among these two values. The angle thus found will be the principal value. Illustration

4

If tan q = – 1, q lies in 2nd or 4th quadrant. For 2nd quadrant we will select anticlockwise and for 4th quadrant we will select clockwise direction. So we get two values 3p/4 and – p/4, of which – p/4 is numerically least angle. Hence principal value is – p/4.

Illustration

5

For sin q = – 1/2, the principal value of q is – p/6 and for cos q = 1/2, the principal value of q is p/3. [Note cos q is positive in Ist and 4th quadrant, so we get two values – p/3 and p/3 of q satisfying the relation. In such case. We select the angle with positive sign as principal value].

SOLVED EXAMPLES Concept-based Straight Objective Type Questions Example 1: If 0 < q < p/2, and f (q) = (1 + cot q – cosec q) (1 + tan q + sec q), then f (p/4) is equal to (a) 0 (b) 1 (c) 2 (d) 2 + 2 Ans. (c)

cos q 1 ˆÊ sin q 1 ˆ Solution: f (q) = ÊÁ 1 + 1+ + ˜ Á ˜ Ë sin q sin q ¯ Ë cos q cos q ¯ sin q + cos q - 1ˆ Ê sin q + cos q + 1ˆ = ÊÁ ˜¯ ÁË ˜¯ Ë sin q cos q

Trigonometrical Ratios, Identities and Equations 25.5 2 = (sin q + cos q ) - 1 = 1 + 2 sin q + cos q - 1 = 2 sin q cos q sin q cos q \ f (p/4) = 2 1 Example 2: If 2 cos q = x + for some x > 0, then x which one of the following is not equal to cos(5q) 1Ê 1 ˆ 1Ê 1 ˆ (b) Á x 3 2 + 3 2 ˜ (a) Á x 5 2 + 5 2 ˜ Ë ¯ Ë 2 2 x x ¯ (c) 1 (d) – 1 Ans. (d) 1 Solution: For x > 0, x + ≥ 2 and x 1 x+ = 2 ¤ x = 1 or x = 1. x 1 We have 2cos q £ 2 and x + ≥ 2. x 1 ¤ x = 1, cos q = 1 Thus, 2cos q = x + x \ q = 2np, where n Œ Z. fi cos(5q) = cos(10n p) = 1 π – 1

Also, for x = 1,

1Ê 5 2 1 ˆ 1Ê 3 2 1 ˆ Á x + 5 2 ˜¯ = ÁË x + 3 2 ˜¯ = 1 2Ë 2 x x

Example 3: For 0 < a, b < p /2, let E = (tan a + tan b) (1 – cot a cot b)+ (cota + cotb) (1– tana tanb) then E is (a) independent of a only (b) independent of b only (c) independent of a and b both (d) equal to 2 (tana + tanb) Ans. (c) 1 ˆ Ê Solution: E = (tana + tanb) Á 1 Ë tan a tan b ˜¯ (tan a + tan b ) (1 - tan a tan b ) tan a tan b = (tana + tanb) (1 – tana tanb)p +

where p= \

1 1 + =0 tan a tan b tan a tan b

E=0

Example 4: If cot A = tan ((n–1)A), where n Œ N, then a value of A can be (a) p/2n (b) p/n (c) 2p/n (d) p/4n Ans. (a) Solution: tan (p /2 – A) = tan ((n–1)A) One possible solution is p /2 – A = (n – 1) A fi A = p /2n

1 ( 3 cos 23° – sin 23°) is not equal to 2 (a) cos 53° (b) sin 53° (c) sin 37° (d) sin 143° Ans. (b) 1 Solution: ( 3 cos 23° – sin 23°) 2 = cos 30° cos 23° – sin 30° sin 23° = cos (30° + 23°) = cos 53° = cos (90° – 37°) = sin 37° = sin (180° – 37°) = sin (143°) Example 5:

Example 6: Suppose a, b ŒR, and let E = sin2 (a + b) + sin2a + sin2b – 2cosa cosb cos (a + b) then E is equal to: (a) 0 (b) 1 (c) 2 (d) 2cos a cosb Ans. (c) Solution: E = 1+ sin2a + sin2b – cos2 (a + b) – 2 cosa cosb cos (a + b) 2 = 1 + sin a + sin2b – cos (a + b) [cos (a +b) – 2 cosa cosb] 2 2 = 1 + sin a + sin b + cos(a + b) cos (a – b) = 1 + sin2a + sin2b + cos2a – sin2b =2 Example 7: If a, b, g are three distinct real numbers such that 0 < a, b, g < p/2, then tan (a – b) + tan (b – g ) + tan (g – a) is equal to: (a) 0 (b) – 1 (c) 1 (d) none of these Ans. (d) Solution: Let A = a – b, B = b – g, C = g – a, then A + B + C = 0. We have tan (A + B) = tan (–C) tan A + tan B = – tan C fi 1 - tan A tan B fi tan A + tan B + tan C = tan A tan B tan C 1 1 Example 8: If tan A – tan B = and cot A – cot B = , 2 3 then cot(A – B) is equal to: (a) 1 (c) 5 Ans. (b)

(b) –1 (d) –5

cot A - cot B 1 1 1 = =cot A cot B 2 cot A cot B fi cot A cot B = – 2/3. Now, cot B - cot A -1 / 3 = = -1 cot(A – B) = 1 + cot A cot B 1 - 2 / 3 Solution:

25.6

Complete Mathematics—JEE Main

Example 9: If tan b =

3 sin a cos a 2

1 - 3 sin a

is equal to: (a) 3 (c) – 2 Ans. (c)

, then

tan(a - b ) tan a

(b) 2 (d) none of these

Solution: tan (a – b) =

tan a - tan b 1 + tan a tan b

sin a 3 sin a cos a cos a 1 - 3 sin 2 a = sin a 3 sin a cos a 1+ ◊ cos a 1 - 3 sin 2 a = – 2tan a p p Example 10: Let f (q) = tan ÊÁ + q ˆ˜ tan ÊÁ - q ˆ˜ , Ë3 ¯ Ë3 ¯ p p p Ê ˆ - < q < , then f Á ˜ is equal to: Ë 12 ¯ 3 3 (a) 2 + 3 (c) Ans. (a)

(b) 2 - 3

3 +1

(d)

3 -1

p p sin ÊÁ + q ˆ˜ sin ÊÁ - q ˆ˜ ¯ Ë3 ¯ Ë3 Solution: f (q) = p p cos ÊÁ + q ˆ˜ cos ÊÁ - q ˆ˜ ¯ Ë3 ¯ Ë3 p sin 2 ÊÁ ˆ˜ - sin 2 q Ë 3¯ = p cos2 ÊÁ ˆ˜ - sin 2 q Ë 3¯ 2

=

Êpˆ fÁ ˜ = Ë 12 ¯

\

3 - 4 sin q 2

1 - 4 sin q

=

2 cos(2q ) + 1 2 cos(2q ) - 1

p 2 cos ÊÁ ˆ˜ + 1 Ë 6¯ 3 +1 = p 3 -1 2 cos ÊÁ ˆ˜ - 1 Ë 6¯

Êp qˆ cot Á + ˜ is equal to: Ë 4 2¯

p+q where p > q > 0, then p-q

(a)

p q

(b)

(c)

pq

(d) pq

Ans. (a)

q p

p-q Êp ˆ p-q fi cos Á + q ˜ = Ë2 ¯ p+q p+q

p q 1 - tan 2 ÊÁ + ˆ˜ Ë 4 2¯ 1- q / p = q q 1+ q / p 1 + tan 2 ÊÁ + ˆ˜ Ë 4 2¯ q Êp qˆ tan 2 Á + ˜ = Ë 4 2¯ p

fi

fi

Êp qˆ cot Á + ˜ = Ë 4 2¯

fi

p q

Example 12: If 2 cos q + sin q = 1, (q π (4k + 1) p /2, k Œ I) then 7 cos q + 6 sinq is equal to: (a) 1/2 (b) 2 (c) 11/2 (d) 46/5 Ans. (b) Solution: 2cos q + sin q = 1 fi 4 cos2 q = (1– sin q)2 fi 4(1–sin2 q) = (1– sin q)2 fi 4(1+ sin q) = 1 – sin q fi sin q = –3/5 As 2cos q + sin q = 1, we get cos q = 4/5. Ê 4ˆ Ê -3 ˆ Thus, 7cos q + 6 sin q = 7 Á ˜ + 6 Á ˜ = 2 Ë 5¯ Ë 5¯ Example 13: The number of solutions of the equation tan x + sec x = 2cos x, x Œ[0, 2p] is (a) 1 (b) 2 (c) 3 (d) 0 Ans.(b) Solution: tan x + sec x = 2 cos x fi 1 + sin x = 2 cos2 x = 2(1 – sin2 x) fi (1 + sin x) (1– 2 + 2 sin x) = 0 fi sin x = – 1 or sin x = 1/2 But sin x = – 1 fi cos x = 0. This is not possible. Thus, sin x = 1/2 fi x = p /6, 5p /6, Œ[0, 2p ] Example 14: Let A = sin4x + cos4x, x Œ R, then 1 1 £ A £1 £ A £1 (b) 2 2 1 1 £ A £1 £ A £1 (c) (d) 4 2 2 Ans. (a) Solution: A = (sin2x + cos2x)2 – 2 sin2x cos2x 1 = 1 - sin 2 2x 2 2 Now, 0 £ sin 2x £ 1 1 1 1 fi = 1 - £ 1 - sin 2 2x = A £ 1 2 2 2 (a)

= 2+ 3 Example 11: If cosec q =

Solution: sin q =

Trigonometrical Ratios, Identities and Equations 25.7

Example 15: If 0 < x < p, and sin x + cos x = 1/3, then a possible value of tan x is 1 1 -9 + 17 9 - 17 (b) (a) 8 8 1 1 9- 5 (c) - 9 + 5 (d) 5 5 Ans. (a)

(

(

)

(

)

(

)

)

Solution: 3(tan x + 1) = sec x fi 9 (1 + tan2 x + 2 tan x) = 1 + tan2x fi 4 tan2x + 9 tan x + 4 = 0 - 9 ± 81 - 64 - 9 ± 17 = 8 8 Example 16: Let f(q) = (1 + cos q) (1 + cos 3q) Ê pˆ (1 + cos 5q) (1 + cos 7q) then f Á - ˜ is equal to: Ë 8¯ (a) – 1/8 (b) 1/4 (c) – 1/4 (d) 1/8 Ans. (d) fi

tan x =

Solution: As f (–q) = f (q), we evaluate f (p /8). For q = p /8, 3q = p /2 – p /8, 5q = p /2 + p /8 and 7q = p – p /8, so that p p p p Êpˆ f Á ˜ = ÊÁ 1 + cos ˆ˜ ÊÁ 1 + sin ˆ˜ ÊÁ 1 - sin ˆ˜ ÊÁ 1 - cos ˆ˜ Ë 8¯ Ë 8¯ Ë 8¯ Ë 8¯ Ë 8¯ p p = cos2 ÊÁ ˆ˜ sin 2 ÊÁ ˆ˜ Ë 8¯ Ë 8¯ 2 1È Êpˆ Êp ˆ˘ sin cos 2 ÁË ˜¯ ÁË ˜¯ ˙ 8 8 ˚ 4 ÍÎ 1 Êpˆ 1 = sin 2 Á ˜ = Ë 4¯ 8 4

=

Example 17: If q = p /8, and f (q) = cos6 q + cos6 3q + cos 5q + cos6 7q then f (q) is equal to: (a) 3/4 (b) 1 (c) 5/4 (d) 3/2 Ans. (c) 6

Solution: As in example 16, using 3q = p/2 – p/8, 5q = p/2 + p/8, 7q = p – p/8, we get f (q) = 2(cos6 q + sin6 q) 2

2

3

2

2

2

= 2[(cos q + sin q) –3cos q sin q (cos q + sin2 q)] È 3 Ê 1 ˆ2˘ È 3 ˘ = 2 Í1 - (2 cos q sin q )2 ˙ = 2 Í1 - Á ˜ ˙ Î 4 ˚ Î 4 Ë 2¯ ˚ = 5/4 Example 18: If a, b, g are three real numbers, then cos2 (b – g ) + cos2(g – a) + cos2 (a – b) – 2cos(b – g ) cos(g – a) cos (a – b) is equal to

(a) 0 (c) –1 Ans. (b)

(b) 1 (d) none of these

Solution: Put A = b – g, B = g – a, C = a – b, so that A + B + C = 0. Now cos2 A + cos2 B + cos2 C – 2 cos A cos B cos C = 1 + cos2 A – sin2 B + cos C [cos C – 2cos A cos B] = 1 + cos (A + B) cos (A – B) + cos C [cos (– A – B) –2cos A cos B] = 1 + cos (– C) cos (A – B) – cos C [sin A sin B + cos A cos B] = 1 + cos C cos (A – B) – cos C cos (A – B) =1 Example 19: Number of values of x Œ[0, 2p] and satisfying the equation cos x cos 2x cos 3x = 1/4 is: (a) 6 (b) 8 (c) 20 (d) 24 Ans. (b) Solution: 2 (2cos 3x cos x) cos 2x = 1 fi 2 (cos 4x + cos 2x) cos 2x = 1 fi 2 cos 4x cos 2x = 1 – 2 cos2 2x = – cos 4x fi cos 4x (2cos 2x + 1) = 0 fi cos 4x = 0 or cos 2x = – 1/2 Now cos 4x = 0 fi 4x = (2n + 1)p /2, n Œ I fi x = (2n + 1) p /8, n Œ I. and 0 £ (2n + 1)p /4 £ 2 p fi 0 £ 2n + 1 £ 8 fi n = 0, 1, 2, 3. Also, cos (2x) = – 1/2 = cos (2p /3) fi 2x = 2np ± 2p /3, n Œ I fi x = np ± p /3, n Œ I. As 0 £ x £ 2p, x = p /3, 2p /3, 4p /3, 5p /3. Thus, there are 8 values of x. Example 20: Number of values of x Œ [0, 4p] and satisfying 2 secx + tanx = 1 is (a) 0 (b) 1 (c) 2 (d) 4 Ans. (c) Solution: We can write the given equation as 2 = cosx – sinx 1 1 cos x sin x = 1 fi 2 2 pˆ Ê fi cos Á x + ˜ = 1 Ë 4¯ fi fi

p = 2np, n ΠI. 4 7p 15p x= , 4 4 x+

25.8

Complete Mathematics—JEE Main

LEVEL 1 Straight Objective Type Questions Example 21: cos (540∞ – q) – sin (630∞ – q) is equal to (a) 0 (b) 2 cos q (c) 2 sin q (d) sin q – cos q Ans. (a) Solution: cos (540∞ – q) = cos (6(p/2) – q) = – cos q sin (630∞ – q) = sin (7(p/2) – q) = – cos q Example 22: 2 sec2 a – sec4 a – 2 cosec2 a + cosec4 a = 15/4 if tan a is equal to (a) 1/ 2

(b) 1/2

(d) 1/4 (c) 1/2 2 Ans. (a) Solution: L.H.S. = 2(1 + tan2 a – 1 – cot2 a) – [(1 + tan2 a)2 – (1 + cot2 a)2] = cot4 a – tan4 a = 15/4 tan4 a = 1/4 fi tan a = ±1/ 2 . Example 23: If the roots of the quadratic equation x2 + px + q = 0 are tan 30° and tan 15°, then the values of 2 + q – p is (a) 1 (b) 2 (c) 3 (d) 0 Ans. (c) Solution: tan 30° + tan 15° = – p and tan 30° tan 15° = q tan 30∞ + tan 15∞ Now 1 = tan 45° = 1 - tan 30∞ tan 15∞

=

1=

fi

–p fi1–q+p=0 1– q

2 + q – p = 2 + 1 = 3.

tan 155∞ - tan 115∞ Example 24: If tan 25° = x, then 1 + tan 155∞ tan 115∞ is equal to (a) (c)

1 - x2 2x 1+ x

2

1- x

2

(b) (d)

1 + x2 2x 1- x

2

1 + x2

Ans. (a) tan 155∞ - tan 115∞ Solution: 1 + tan 155∞ tan 115∞ =

tan (180∞ - 25∞) - tan (90∞ + 25∞) 1 + tan (180∞ - 25∞) tan (90∞ + 25∞)

1 - x2 1ˆ ˜¯ = 2 x . x

Example 25: The value of the determinant sin 2 13∞ sin 2 77∞ tan 135∞ sin 2 77∞ tan 135∞ sin 2 13∞ is equal to tan 135∞ sin 2 13∞ sin 2 77∞ (a) – 1 (c) 1 Ans. (b)

(b) 0 (d) 2

Solution: The given determinant is equal to sin 2 13∞ cos2 13∞ cos2 13∞

fi

fi

- tan 25∞ + cot 25∞ 1 = ÊÁ - x + 1 + tan 25∞ cot 25∞ 2Ë

sin 2 13∞

-1

sin 2 13∞ cos2 13∞

-1

0 cos2 13∞ = 0

-1

-1

-1 sin 2 13∞ = 0

0 sin 2 13∞ cos2 13∞ (Applying C1 Æ C1 + C2 + C3) Example 26: If A = 130∞ and x = sin A + cos A, then (a) x > 0 (b) x < 0 (c) x = 0 (d) x ≥ 0 Ans. (a) Solution: x = sin 130∞ + cos 130∞ = sin 50∞ – sin 40∞ > 0 (Q sin x is increasing for 0 < x < p/2) Example 27: If tan2 36∞ + k(sin 18∞ + cos 36∞) = 5, then the value of k is (a) 2

(b) 2 5

(c) 4 Ans. (c)

(d) 4 5

Solution: From the tables, we have Ê 5 -1 5 + 1ˆ + L.H.S. = 5 - 2 5 + k Á 4 ˜¯ Ë 4 = 5-2 5 +

5 k = 5 (given) 2

fi

k=4

Trigonometrical Ratios, Identities and Equations 25.9

Example 28: If bc +

sin x cos x tan x = = = k, then a b c

1 ak + is equal to ck 1 + bk 1ˆ Ê (a) k Á a + ˜ Ë a¯ (c)

(b)

1

(d)

k2

1Ê 1ˆ ÁË a + ˜¯ k a a k

Solution: The given expression is equal to k = =

sin x k2

2

+

+

1 sin x + tan x 1 + cos x

cos x (1 + cos x ) + sin 2 x sin x (1 + cos x )

a 1 a 1 1 1 + = + = ÊÁ a + ˆ˜ . k sin x k ak kË a¯

Example 29: sin2 a + cos2 (a + b) + 2 sin a sin b cos (a + b) is independent of (a) a (b) b (c) both a and b (d) none Ans. (a) Solution: The given expression is equal to sin2 a + cos (a + b) [cos (a + b) + 2 sin a sin b] = sin2 a + cos (a + b) [cos a cos b + sin a sin b] = sin2 a + cos (a + b) cos (a – b) = sin2 a + cos2 a – sin2 b = 1– sin2 b = cos2 b which is independent of a only. Example 30: tan 203∞ + tan 22∞ + tan 203∞ tan 22∞ = (a) – 1 (b) 0 (c) 1 (d) 2 Ans. (c) tan 203∞ + tan 22∞ Solution: tan (203∞ + 22∞) = 1 - tan 203∞ tan 22∞ tan 203∞ + tan 22∞ fi 1 = tan (180∞ + 45∞) = 1 - tan 203∞ tan 22∞ fi tan 203∞ + tan 22∞ + tan 203∞ tan 22∞ = 1. Example 31: If sin 32∞ = k and cos x = 1 – 2k2; a, b are the values of x between 0∞ and 360∞ with a < b, then (a) a + b = 180∞ (c) b = 4a + 40∞ Ans. (c)

(b) b – a = 200∞ (d) b = 5a – 20∞ 2

is (a) 9 (c) 27 Ans. (b) Solution: A.M. ≥ G.M.

(b) 18 (d) 30

27 tan 2 q + 3 cot 2 q 2 2 ≥ 27 tan q ◊ 3 cot q 2 fi 27 tan2 q + 3 cot2 q ≥ 18. Example 33: If 3 sin b = sin (2a + b), then tan (a + b) – 2 tan a is (a) independent of a (b) independent of b (c) independent of both a and b (d) independent of none of them Ans. (c) Solution: sin (2a + b ) = 3 sin b sin (2a + b ) + sin b 3 +1 = fi 3 -1 sin (2a + b ) - sin b fi

Ans. (b) cos x ◊ tan x

Example 32: The minimum value of 27 tan2 q + 3 cot2 q

2 sin (a + b ) cos a = 2 fi tan(a + b) – 2tan a = 0 2 cos (a + b ) sin a Example 34: If A = sin2 q + cos4 q, then for all values of q (a) 1 £ A £ 2 (b) 3/4 £ A £ 1 (c) 13/16 £ A £ 1 (d) 3/4 £ A £ 13/16 Ans. (b) Solution: A = sin2 q + (1– sin2 q)2 = 1 + sin2 q (sin2 q –1) = 1– sin2 q cos2 q £ 1 Also A = 1– (1/4) sin2 2q ≥ 1– (1/4) = (3/4). Hence 3/4 £ A £ 1 fi

Example 35: If u =

a 2 sin 2 q + b2 cos2 q where a, b > 0. Then the difference between the maximum and minimum values of u2 is given by (a) (a + b)2 (c) 2(a2 + b2) Ans. (d) Solution: We have

+ =

(b) 2 a 2 + b2 (d) (a – b)2

a2 b2 (1 + cos 2q ) + (1 - cos 2q ) 2 2

u=

2

Solution: cos x = 1 – 2k = 1 – 2 sin 32∞ = cos 64∞ fi x = 64∞ or 296∞ \ a = 64∞ and b = 296∞ which satisfy (c).

a 2 cos2 q + b2 sin 2 q +

a2 b2 (1 - cos 2 q ) + (1 + cos 2 q ) 2 2 a 2 + b2 a 2 – b2 + cos 2 q . 2 2

25.10

Complete Mathematics—JEE Main

+

a 2 + b2 a 2 - b2 cos 2 q 2 2

Squaring we get 2

2

2 2 Ê 2 ˆ Ê 2 ˆ u = a + b + 2 Á a + b ˜ – Á a – b ˜ cos2 2 q Ë 2 ¯ Ë 2 ¯

2

2

2

Thus max (u2) = a2 + b2 + a2 + b2 = 2(a2 + b2) 2

2 2 2 2 and min (u2) = a2 + b2 + 2 ÊÁ a + b ˆ˜ - ÊÁ a - b ˆ˜ Ë 2 ¯ Ë 2 ¯

2

= a2 + b2 + 2ab = (a + b)2 So the required difference = 2(a2 + b2) – (a + b)2 = (a – b)2. sin 2 x cos2 x Example 36: cos2 x sin 2 x - 10 12 (a) (b) (c) (d) Ans. (a)

1 1 2

is equal to

0 12 cos2 x – 10 sin2 x 12 sin2 x – 10 cos2 x – 2 10 sin2 x

Solution: Apply C1 Æ C1 + C2 – C3, then the given determinant is equal to 0 1 cos2 x = 0. 0 1 sin 2 x 0 12 2 cos A cos B 1 = = , - p / 2 < A, B < 0, 3 4 5 then value of 2 sin A + 4 sin B is equal to (a) 4 (b) 2 (c) – 4 (d) 0 Ans. (c) Example 37: If

Solution: cos A = 3/5 fi sin A = – (4/5) cos B = (4/5) fi sin B = – 3/5. So that 2 sin A + 4 sin B = – (8/5) – (12/5) = – 4. Example 38: If sin A = 3 sin (A + 2B), angle B is acute and A is obtuse; then (a) tan B = 1 / 2 (c) tan B < 1 / 2 Ans. (d)

(b) tan B > 1 / 2 (d) 0 < tan B < 1 / 2

sin (A + 2 B) 1 = , 3 sin A By componendo and dividendo, tan(A + B) = – 2 tanB Solution:

fi

tan A + tan B = - 2 tan B 1 - tan A tan B

3 tan B

fi

tan A =

fi

2 tan2 B – 1 < 0 (as tan B > 0)

fi

0 < tan B < 1 / 2

2 tan 2 B - 1

< 0 (as A is obtuse)

3p 4p Example 39: The value of cos2 Ê ˆ + cos2 Ê ˆ is Ë 5¯ Ë 5¯ equal to (a) 3/4 (b) 5/4 (c) 5/2 (d) 4/5 Ans. (a) 3p 4p Solution: cos2 Ê ˆ + cos2 Ê ˆ Ë 5¯ Ë 5¯ = cos2 (108°) + cos2 (144°) = (cos (90° + 18°))2 + (cos (180° – 36°))2 = sin2 18° + cos2 36° 2

2

2¥6 3 ˆ ˆ Ê Ê = . = Á 5 – 1 ˜ + Á 5 + 1˜ = Ë 4 ¯ Ë 4 ¯ 16 4 Example 40: The value of cos 12° + cos 84° + cos 156° + cos 132° is (a) 1/8 (b) – 1/2 (c) 1 (d) 1/2 Ans. (b) Solution: (cos 12° + cos 132°) + (cos 84° + cos 156°) = 2 cos

12∞ + 132∞ 12∞ - 132∞ cos + 2 2

84∞ + 156∞ 84∞ - 156∞ cos . 2 2 = 2 cos 72° cos 60° + 2 cos 120° cos 36° 2 cos

= 2¥

5 –1 1 1 5 +1 1 ¥ + 2 ¥ Ê– ˆ ¥ =– . Ë 2¯ 4 4 2 2

1 Example 41: Let D = - sin q -1 D lies in the interval. (a) [2, 4] (c) [1, 4] Ans. (a)

0 Solution: D = - sin q -1

sin q 1 - sin q

1 sin q , then 1

(b) [3, 4] (d) none of these 0 1 - sin q

2 sin q [R1Æ R1 + R3] 1

= 2(1 + sin2 q) fi 2 £ D £ 4 as 0 £ sin2 q £ 1 Example 42: If sin (120° – a ) = sin (120° – b) and 0 < a, b < p, then all values of a, b are given by (a) a + b = p /3 (b) a = b or a + b = p /3 (c) a = b (d) a + b = 0 Ans. (b)

Trigonometrical Ratios, Identities and Equations 25.11

Solution: sin (120° – a) = sin (120° – b) fi either 120° – a = 120° – b fi a = b or 120° – a = 180° – (120° – b) fi a + b = 60° = p /3 Example 43: If sin q + cosec q = 2, then sinn q + cosecn q is equal to (a) 2 (b) 2n n (c) 4 (d) none of these Ans. (a)

Example 46: If cos q = cos a cos b , then q +a q -a tan is equal to tan 2 2 (b) tan2 (b/2) (a) tan2 (a/2) 2 (c) tan (q/2) (d) cot2 (b/2) Ans. (b) q +a q -a Solution: tan tan 2 2 =

Solution: We can write sin2 q + 1 = 2 sin q fi sin2 q – 2 sinq + 1 = 0 fi fi sin q = 1 fi n and thus sin q + cosecn q = 2.

(sin q – 1)2 = 0 cosec q = 1

b = cos f sin y - sin f cosy cos d

Solution: a2 + b2 + c2 = cos2 f cos2 y + sin2 f sin2 y cos2 d + cos2 f sin2 y + sin2 f cos2 y cos2 d + sin2 f sin2 d = cos2 f + sin 2 f cos2 d + sin 2 f sin 2 d

=

(AB, AC are the radii of the circle on CP as diametre) BP =2 = 2 sin C. CP

Fig. 25.3

2 (cos a - cos q ) cos a (1 - cos b ) b = = tan2 . 2 2 (cos a + cos q ) cos a (1 + cos b )

Example 47: The equation a sin x + b cos x = c, where |c| >

a 2 + b2 has

(a) (b) (c) (d) Ans. (c)

a unique solution infinite number of solutions no solution none of these

Solution: Let a = r cos a, b = r sin a so that

= cos2 f + sin 2 f = 1. Example 45: In a triangle ABC, BP is drawn perpendicBP = ular to BC to meet CA in P, such that CA = AP, then AB (a) 2 sin A (b) 2 sin B (c) 2 sin C (d) none of these Ans. (c) BP BP Solution: We have = AB AC

1 - tan 2 (q 2 ) tan 2 (a 2 )

1 - cos q 1 - cos a 1 + cos q 1 + cos a = 1 - cos q 1 - cos a ◊ 11 + cos q 1 + cos a

Example 44: If a = cos f cosy + sin f sin y cos d and c = sin f sin d . Then a2 + b2 + c2 = (a) – 1 (b) 0 (c) 1 (d) none of these Ans. (c)

tan 2 (q 2 ) - tan 2 (a 2 )

r = a 2 + b2 . The given equation can be written as r sin (x + a) = c fi sin (x + a) = fi

|sin (x + a)| =

c

c 2

a + b2

> 1 as |c | > a 2 + b2 which is not possible for any value of x.

a 2 + b2

Example 48: If cot a equals the integral solution of the inequality 4x2 – 16x + 15 < 0 and sin b equals to the slope of the bisector of the first quadrant, then sin (a + b) sin (a – b) is equal to (a) – 3/5 (b) – 4/5 (d) 3 (c) 2/ 5 Ans. (b) Solution: We have 4x2 – 16x + 15 < 0 fi 3/2 < x < 5/2 fi cot a = 2, the integral solution of the given inequality and sin b = tan 45∞ = 1 \ sin (a + b) sin (a – b) = sin2 a – sin2 b 1 1 4 = –1= –1= - . 2 1+ 4 1 + cot a 5

25.12

Complete Mathematics—JEE Main

Example 49: The value of cos 6p 7p + cos is cos 7 7 (a) 1 (c) 1/2 Ans. (d) 2p Solution: cos + cos 7

2p 4p + cos + 7 7

5p 3p p È 3p sin - sin + sin - sin + = p ÍÎ 7 7 7 7 2 sin 7 7p 5p ˘ sin - sin ˙ –1 7 7 ˚ 1

1 pˆ 3 Ê - sin ˜ –1 = - - 1 = - . p ÁË 7¯ 2 2 2 sin 7 1

Example 50: The greatest value of f (x) = 2 sin x + sin 2x on [0, 3p/2], is given by (a) 9/2 (b) 5/2 (c) 3 3 /2 Ans. (c)

(d) 3/2

Solution: f ¢(x) = 2 cos x + 2 cos 2x and f ≤(x) = – 2 sin x – 4 sin 2x For extreme value f ¢(x) = 0 fi fi fi

=

2 3 3 3 3 = + 2 2 2

Also f(0) = 0 and f (3p/2) = –2 so the greatest value of f(x) is 3 3 2 Example 51: If x = a cos3 q sin2 q, y = a sin3 q cos2 q

( x 2 + y2 ) and q

( xy )

p

(p, q, ΠN) is independent of q, then

(a) 4p = 5q (c) p + q = 9 Ans. (a)

a 2 p (sin q cos q )

4p 5q

a 2 q (sin q cos q )

Example 52: If cos a + cos b = a, sin a + sin b = b and cos 3q = a – b = 2q, then cos q (a) a2 + b2 – 2 (c) 3 – a2 – b2 Ans. (b)

(b) 4q = 5p (d) pq = 20

(b) a2 + b2 – 3 (d) (a2 + b2)/4

Solution: From the given relations we have a2 + b2 = cos2 a + cos2 b + 2 cos a cos b + sin2 a + sin2 b + 2 sin a sin b = 2 + 2 cos (a – b) = 2 + 2 cos2 q = 4 cos2 q Now

cos 3q 4 cos3 q - 3 cos q = cos q cos q = 4 cos2 q – 3 = a 2 + b2 – 3.

Example 53: If cos A = 3/4, then the value of 16 cos2 (A/2) – 32 sin (A/2) sin (5A/2) is (a) – 4 (b) – 3 (c) 3 (d) 4 Ans. (c) Solution: The given expression is equal to 8 (1 + cos A) – 16 (cos 2A – cos 3A) = 8(1 + cos A) – 16 [2 cos2 A – 1 – cos A (4 cos2 A – 3)] 3ˆ 9 3Ê 9 Ê È ˆ˘ = 8 Á 1 + ˜ –16 Í2 ¥ - 1 - Á 4 ¥ - 3˜ ˙ Ë 4¯ 4Ë 16 ¯ ˚ Î 16

cos x + 2 cos2 x Р1 = 0 cos x = Р1 or 1/2 x = p or p/3 as x Π[0, 3p/2]

Now f(p) = 0 and f(p/3) =

2p

Èa sin 2 q cos2 q ˘˚ = Î q ÈÎa 2 sin 5 q cos5 q ˘˚

which is independent of q if 4p = 5q. 4p 6p 7p + cos + cos 7 7 7

2p 4p p p È 2 sin cos + 2 sin cos + Í p 7 7 7 7 2 sin Î 7 6p ˘ p 2 sin cos ˙ –1 7 7 ˚

=

p

( xy )q

(b) – 1 (d) – 3/2

1

=

Solution:

( x 2 + y2 )

= 14 – (18 – 16 – 27 + 36) = 3. 1 cos q 1 1 - cos q Example 54: If D = - sin q -1 sin q 1 then D lies in the interval (a) [0, 4] (b) [2, 4] (c) [2 - 2 , 2 + 2 ]

(d) [–2, 2]

Ans. (c) Solution: Expanding D, we get D = 1 + sinq cosq – cosq (–sinq – cosq) + (– sin2q + 1) = 2 + sin2q + cos2q = 2 +

2 cos (2q – p/4)

As –1 £ cos (2q – p/4) £ 1, 2 –

2£ D£2+

2

Trigonometrical Ratios, Identities and Equations 25.13

Example 55: A value of q lying between q = 0 and q = p/2 and satisfying the equation 1 + sin 2 q 2

cos2 q

4 sin 4q

2

sin q

1 + cos q

4 sin 4q

sin 2 q

cos2 q

1 + 4 sin 4q

(a) 3p/24 (c) 11p/24 Ans. (c)

= sin p/4 fi

(b) 5p/24 (d) p/24

0 1

sin 2 q fi fi fi

-1 -1

=0

1 + 4 sin 4q + cos2 q + sin2 q = 0 sin 4q = –1/2 fi 4q = p + p /6 or 2p – p/6 [Q 0 < 4q < 2p] q = 7p/24 or 11p/24.

Example 56: Let q Π(p/4, p/2), which of the following statements is true? (a) (b) (c) (d) Ans. (b)

cos q

x + p/4 = np + (– 1)n p /4

fi x = np + (– 1)n p/4 – p/4 which includes x = 2np and x = 2np + p/2.

cos2 q 1 + 4 sin 4q

cos q

Solution: sin x + cos x = 1 fi sin (x + p/4) = 1 / 2

= 0 is

Solution: Applying R1 Æ R1 – R3, R2 Æ R2 – R3 to the given determinant we get 1 0

(c) x = np + (– 1)n p/4 – p/4, nŒI (d) none of these Ans. (c)

sin q

< (sin q) < (cos q) (cos q) sin q cos q (cos q) < (cos q) < (sin q) cos q (sin q)cos q < (cos q) cos q < (cos q) sin q (cos q)cos q < (cos q) sin q < (sin q) cos q

Example 59: The values of x between 0 and 2p which satisfy the equation sin x

8 cos2 x = 1 are in A.P. The com-

mon difference of the A.P. is (a) p /8 (b) p/4 (c) 3p/8 (d) 5p/8 Ans. (b) Solution: From the given equation we have 2 sin x |cos x| = 1/ 2 fi sin 2x = 1/ 2 if cos x > 0 and sin 2x = – 1/2 if cos x < 0 \

when cos x > 0, sin 2x = 1/ 2 fi x = p/8, 3p/8

when cos x < 0, sin 2x = – 1/ 2 fi x = 5p/8, 7p/8. So the required values of x are p/8, 3p/8, 5p/8, 7p/8 which form an A.P. with common difference p/4.

Solution: For q Π(p/4, p/2), 0 < cos q < 1 / 2 < sin q (cos q) sin q (as 0 < cos q < 1 and cos q < sin q) Showing that (b) is correct.

Solution: We have tan 2q + sin 2q = – 8/15 2 tan q 2 tan q -8 + = fi 2 2 1 - tan q 1 + tan q 15

Example 57: If a cos A – b sin A = c, then a sin A + b cos A is equal to

fi

15 ¥ 4 tan q + 8(1 – tan4 q) = 0

fi

8tan4 q + 60 tan q – 8 = 0

(a) ± a 2 + b2 - c 2

(b) ± b2 + c 2 - a 2

which is satisfied if tan q = 2

(c) ± c 2 + a 2 - b2 Ans. (a)

(d) ± a 2 + b2 + c 2

Example 61: The numbers of solutions of the pair of equations

Solution: Put a sin A + b cos A = x, then c2 + x2 = (a cos A – b sinA)2 + (a sin A + b cosA)2 = a2 + b2 fi

x = ± a 2 + b2 - c 2

Example 58: The general solution of the trigonometrical equation sin x + cos x = 1 is (a) x = 2np, nŒI (b) x = 2np + p/2, nŒI

2 sin2q – cos2q = 0 2 cos2q – 3 sin q = 0 in the interval [0, 2p] is (a) zero (c) two Ans. (c)

(b) one (d) four

Solution: 2 sin2q – cos2q = 0 fi 1 – cos2q – cos2q = 0 fi cos2q = 1/2

25.14

Complete Mathematics—JEE Main

fi 2 cos2q Р1 = 1/2 fi 2 cos2 q = 3/2 So that from 2 cos2q Р3 sin q = 0, we have sin q = 1/2 fi q = p /6, 5p/6 as q Π[0, 2p].

(a) 1/2 (c) 2/3 Ans. (b)

Solution: cos B + cos C = 4 sin2(A/2) B+C B - C = 4 sin A sin Ê p - B + C ˆ fi 2 cos cos ÁË ˜ 2 2 2 ¯ 2 2

Example 62: The solution set of the equation tan (p tan x) = cot (p cot x) is (a) {0} (c) f Ans. (c)

(b) {p/4} (d) none of these

A B-C A B+C = 4 sin cos cos 2 2 2 2 B C B C fi cos cos + sin sin = 2 2 2 2 B C B Cˆ Ê 2 Á cos cos - sin sin ˜ Ë 2 2¯ 2 2 fi

Solution: tan (p tan x) = tan (p 2 – p cot x ) fi fi

p tan x = p /2 – p cot x fi tan x + cot x = 1/2 2 tan2 x – tan x + 2 = 0

1 ± 1 - 16 4 which does not give real values of tan x. fi

Ans. (a) p 22 8 Solution: sin x = –1= –1= 2 14 14 4 = so the least positive root lies in [0, p/2] 7 Example 64: The number of solutions of the equation

(a) 1 (c) more than two Ans. (a)

x ) = 1 is

(b) 2 (d) 0

Solution: Given equation is possible if cos (p = 1 and cos (p x ) = 1. Since So

B C B C sin = cos cos 2 2 2 2 1 B C tan tan = . 3 2 2

3 sin

fi

Example 63: The least positive root of the function sin x – p/2 + 1 = 0 lies in the interval (a) [0, p/2] (b) [p/2, p] (c) [p/2, 3p/2] (d) none of these

x - 4 ) cos (p

2 sin

fi

tan x =

cos (p

(b) 1/3 (d) 3/2

Example 67: If 15 sin4x + 10cos4x = 6, Then tan2x = (a) 1/5 (b) 2/5 (c) 2/3 (d) 1/3 Ans. (c) Solution: 15 sin4x + 10cos4 x = 6 (sin2x + cos2x)2 fi 9 sin4x + 4 cos4x – 12 sin2x cos2x = 0 fi (3 sin2x – 2 cos2x)2 = 0 fi tan2x = 2/3. Example 68: Sum of the root of the equation 2 sin2q + sin22q = 2; 0 £ q £ p / 2 is (a) p/2 (b) 3p/4 (c) 7p/2 (d) 5p/12 Ans. (b) Solution: 4 sin2 q cos2 q = 2(1 – sin2q)

x - 4)

x – 4 ≥ 0 fi x ≥ 4 and x ≥ 0 x = 4 is the only solution.

Example 65: The number of values of x in the interval [0, 3p] satisfying the equation 2 sin2 x + 5 sin x Р3 = 0 is (a) 2 (b) 4 (c) 6 (d) 1 Ans. (b) Solution: We have (sin x + 3) (2 sin x Р1) = 0 fi sin x = Р3 which is not possible or sin x = 1/2 fi x = p/6, 5p/6, 13p/6, 17p/6 as x Π[0, 3p]. Example 66: If A, B, C are the angles of a triangle and cos B + cos C = 4 sin2(A/2), then tan B/2 tan C/2 is equal to

(2 sin2q – 1) cos2q = 0

fi fi

sin2q = 1/2

or

cos2 q = 0

fi

q = p/4

or

q = p/2

Example 69: If tan x/2 = cosec x – sin x, then sec (x/2) = 2

(a)

5 +1

(b)

5 -1

(c)

5-2

(d)

5+2.

Ans. (b) Solution: tan (x/2) =

1 + tan 2 ( x / 2 ) 2 tan ( x / 2 ) 2 tan ( x / 2 ) 1 + tan 2 ( x / 2 )

fi 2 tan2 (x/2) (1 + tan2 (x/2) = [1 + tan2(x/2)]2 –4 tan2(x/2) fi 2 tan4 (x/2) + 2 tan2 (x/2) = 1 + tan4 (x/2) – 2 tan2(x/2) fi tan4 (x/2) + 4 tan2 (x/2) – 1 = 0 fi tan2 (x/2) =

5-2

fi sec2 (x/2) =

5 -1

Trigonometrical Ratios, Identities and Equations 25.15

Example 70: Let A and B denote the statements A : cosa + cosb + cos g = 0 B : sina + sinb + sin g = 0 If cos(b – g) + cos(g – a) + cos (a – b) = – 3/2 Then (a) both A and B are true (b) both A and B are false (c) A is true B is false (d) A is false B is true. Ans. (a)

r 1 p p fi n=3 = , cos = cos R 2 n 3 r 1 p p , cos = cos fi n=4 and when = n 4 R 2 r 2 p 2 But when = , cos = R 3 n 3 Which does not give a positive integral value of n. when

O

R

Solution: cos (b – g) + cos (g – a) + cos(a – b) = – 3/2 fi

p/n r

2[cos b cos g + cos g cos a + cos a cos b + sin b sin g + sin g sin a + sin a sin b] +

(sin2 a + cos2 a ) + (sin2 b + cos2 b ) + (sin 2 g + cos2 g ) = 0 2

2

fi (sin a + sin b + sin g ) + (cos a + cos b + cos g ) = 0 fi and so both A and B are true.

cos a + cosb + cosg = 0 sina + sinb + sing = 0

Example 71: cos2u + cos2(u + x) – 2 cosu cosx cos(u + x) = 1/2 if (a) x = p/4 (b) u = p/4 (c) x = p/2 (d) u = p/2 Ans. (a) 2

A1

Example 73: If cos (a + b) = 4/5 and sin (a – b) = 5/13 where 0 £ a, b £ p/4, then tan 2a = (a) 19/12 (c) 25/16 Ans. (d)

so

sin x = 1/2

fi

sin x = ± 1/ 2

(c)

r 2 = R 3

(b)

r 1 = R 2

(d)

=

tan (a + b ) + tan (a - b ) 1 - tan (a + b ) tan (a - b )

=

(3 / 4) + (5 /12) 14 /12 56 = = . 33 1 - (3 / 4 )(5 /12 ) 33 / 48

Example 74: The possible values of q Π(0, p) such that sin(q) + sin(4q) + sin(7q) = 0 are (a)

p 5p p 2p 3p 8p , , , , , 4 12 12 3 4 9

(b)

3 r = R 2

2p p p 2p 3p 35p , , , , , 9 4 2 3 4 36

(c)

r 1 . = R 2

2p p p 2p 3p 8p , , , , , 9 4 2 3 4 9

(d)

2p p 4p p 3p 8p , , , , , 9 4 9 2 4 9

Example 72: For a regular polygon, let r, R be the radii of the inscribed and circumscribed circles. There is no regular polygon with (a)

(b) 20/7 (d) 56/33

Solution: 0 £ a, b £ p/4 fi 0 £ a + b £ p/2 fi – p/4 £ a – b £ p/4 Now cos (a + b) = 4/5 fi tan (a + b) = 3/4 and sin (a – b) = 5/13 fi tan (a – b) = 5/12 we have tan2a = tan [(a + b) + (a – b)]

Solution: L.H.S = cos u + cos (u + x) – [cos(u + x) + cos(u – x)] cos (u + x)

2

A2

Fig. 25.4

2

= cos2u – cos(u – x) cos(u + x) = cos2u – (cos2u – sin2x) = sin2x.

M

Ans. (a) r p Solution: We have = cos R n p p 3 r When , cos = cos . fi n = 6 = n 6 R 2

Ans. (d) Solution: sinq + sin 4q + sin 7q = 0 fi 2sin 4q cos 3q + sin 4q = 0 fi sin 4q (2 cos 3q + 1) = 0 fi sin 4q = 0 or cos 3q = –1/2 = cos (2p/3)

25.16

Complete Mathematics—JEE Main

fi

q=

p p 3p , , or 3q = 2p 3 , 2p ± 2p 3 4 2 4

fi

q=

2p p p 4p 3p 8p , , , , , . 9 4 2 9 4 9

16 + 9 + 24 sin (P + Q) = 37

Example 75: In a D PQR, if 3 sin P + 4 cosQ = 6 and 4 sin Q + 3 cos P = 1, Then the angle R is equal to (a) p 4 (c) 5p 6 Ans. (d)

Solution: Squaring and adding the given relations we get fi

sin (P + Q) = 1 2 fi sin R = 1 2

fi

R = p /6

or

5p /6

If R = 5p /6, then P < p /6

fi 3sin P < 3/2.

(b) 3p 4

fi 3 sin P + 4 cos Q < 3/2 + 4 < 6

(d) p 6.

So R π 5p /6

Assertion-Reason Type Questions

Example 76: Statement-1: 2

2

1/2 £ sin q + 3 sin q cos q + 5cos q £ 11/2

fi

Gives 6 values of q, ±

Ans. (a) Solution: Take a = r cosa 2

and

b = r sina

2

a + b and a sinq + b cosq =r sin(q + a) – r £ r sin (q + a) £ r.

fi - a 2 + b2 £ a sin q + b cos q £ a 2 + b2 fi Statement-2 is true. Now sin2q + 3 sinq cosq + 5 cos2q 2

= 1 + (3 / 2 ) sin 2q + 4 cos q = 1 + (3 / 2 ) sin 2q + 2 (1 + cos 2 q )

3-

(9 / 4 ) + 4

5p - p p , and 12 12 4 satisfy sin 2q = cos4q. so statement -1: is true and statement-2 is false. Example 78: Statement-1: sin 52° + sin 78° + sin 50° = 4 cos 26° cos 39° cos 25° Statement-2: If A + B + C = p, then sin A + sin B + sin C = 4 cos (A/2) cos (B/2) cos (C/2) Ans. (a) fi

£ A£ 3+

9/4 + 4

fi 1/2 £ A £ 11 / 2 and the statement-1 is also true. Example 77: Statement-1: The number of values of q in the interval (– p/2, p/2) such that q π np/5 for n = 0, ± 1, ± 2 and tan q = cot 5q is 6. Statement-2: No value of q in statement-1 satisfies sin2q = cos4q. Ans. (c) fi fi fi

Solution: tan q = cot 5q 2 sinq sin5q = 2 cosq cos5q cos 4q – cos6q = cos6q + cos4q cos6q = 0

5p 3p p ,± ,± 12 12 12

out of which

= 3 + (3 / 2 ) sin 2q + 2 cos 2 q = A (say ) From statement-2

p 5p 3p ,± ,± 2 2 2

as – 3p < 6q < 3p

Statement-2: - a 2 + b2 £ a sin q + b cos q £ a 2 + b2 .

fi r= Now

6q = ±

Solution: Statement-2 is True from conditional identities statement-1, is also True.

Example 79: Statement-1: If A, B, C are the angles of a triangle such that angle A is obtuse, then tan B tan C < 1 Statement-2: In a triangle ABC tan A =

tan B + tan C 1 - tan B tan C

Ans. (c) Solution: Statement-2 is false because A = tan (p – (B + C)) tan B + tan C tan A = – tan (B + C) = . tan B tan C - 1 In statement-1 If A is obtuse, tan A < 0 1 and the statement-1 is True.

fi

tan B tan C
0 and cos2a < 0 Solution: cos 2 a or sin 3 a < 0 and cos2a > 0 i.e.

if 3a Œ(0, p) and 2a Œ(p/2, 3p/2) or 3a Œ(p, 2p) and 2a Œ(–p/2, p/2)

Trigonometrical Ratios, Identities and Equations 25.19

i.e. i.e.

if aŒ(0, p/3) and a Œ(p/4, 3p/4)

= cos2 x + cos2 y – (1/2)[cos 2x + cos2 y]

or aŒ(p/3, 2p/3) and aŒ(–p/4, p/4)

= (1/2) [2cos2 x + 2 cos2 y – cos2 x – cos 2 y]

if aŒ(p/4, p/3)

= (1/2) [2 cos2 x + 2 cos2 y – 2cos2 x + 1 – 2 cos2 y + 1] = 1

since (13p/48, 14p/48) Ã (p/4, p/3), (a) is correct

Example 94: If sin 2q = k, then the value of

Example 91: If cos a + cos b = a, sin a + sin b = b and q is the arithmetic mean between a and b then sin 2q + cos 2q is equal to (a) (a + b)2/(a2 + b2)

(a)

(d) none of these

a +b a -b a +b a -b cos = a and 2sin cos =b 2 2 2 2

a +b b By dividing we get tan = fi 2 a 1 - b2 / a 2 1 + b2 / a 2

\ sin 2 q + cos2q = Example 92: If

=

a 2 - b2 a 2 + b2

a +b

2ab a 2 + b2

.

1 - k2 k

2 - k2 k

(d) 2 – k2 tan3 q 1 + tan 2 q

+

cot 3 q 1 + cot 2 q

1 + tan a tan b = tan g, cos a cos b

2

)

2 1 - k 2 /2 = 2-k , k /2 k

k˘ È . ÍÎQsin q cos q = 2 ˙˚

Example 95: If sin2 A = x, then sin A sin 2A sin 3A sin 4A is a polynomial in x, the sum of whose coefficients is (a) 0 (b) 40 (c) 168 (d) 336 Ans. (a) Solution: We have sin A sin 2A sin 3A sin 4A = sin A (2 sin A cos A) (3 sin A – 4 sin3 A) ¥ 2 sin 2A cos2A = 2 sin2 A cos A ¥ sin A (3 – 4 sin2 A) ¥ 2 ¥ 2 sin A cos A (1 – 2 sin2 A)

2

cos2 a cos2 b - (1 + sin a sin b ) cos2 a cos2 b

= 8sin4 A cos2 A (3 – 4 sin2 A) (1 – 2 sin2 A) = 8x2 (1 – x) (3 – 4x) (1 – 2x)

(1 - sin2 a ) (1 - sin2 b ) - (1 + 2 sin a sin b + sin2 a sin2 b ) = cos2 a cos2 b

2

cos2 a cos2 b

(b)

(

Solution: We have 1 – tan2 g =

- (sin a + sin b )

is equal to

sin 2 q + cos2 q - 2 sin 2 q cos2 q sin 4 q + cos4 q = = sin q cos q sin q cos q =

2

0 < a, b < p then 1 – tan2 g < 0 for (a) all values of a and b (b) no values of a and b (c) finite number of values of a and b (d) infinite number of values of a and b Ans. (a)

=

1 + cot 2 q

Solution: We have

a 2 - b2 + 2 ab 2

cot 3 q

3 2 3 sin3 q cos3 q = sin q ◊ cos q + cos q ◊ sin 2 q = + cos q sin q cos3 q sin3 q

b tan q = a

and sin 2q =

+

(c) k 2 + 1 Ans. (b)

Solution: From the given relations we have

so that cos 2 q =

1 + tan 2 q

(b) (a – b)2/(a2 + b2)

(c) (a2 – b2)/(a2 + b2) Ans. (d)

2 cos

tan3 q

< 0 (sin a + sin b π 0, as 0 < a, b < p)

Example 93: If x + y = z, then 2 cos x + cos2 y + cos2z –2 cos xcos ycos z is equal to (b) sin2 z (a) cos2 z (c) 0 (d) 1 Ans. (d) Solution: The given expression can be written as cos2 x + cos2 y + cos2 z – cos z [cos (x + y) + cos (x – y)] = cos2 x + cos2 y + cos2 z – cos2 z – cos (x + y) cos (x – y)

= 24x2 – 104x3 + 144x4 – 64x5. The required sum = 24 – 104 + 144 – 64 = 0. Example 96: If

sin A 3 5 cos A = and = , sin B 2 2 cos B

0 < A, B < p/2, then tan A + tan B is equal to (a) (c) 1 Ans. (d)

3

5

(b) (d)

5

(

3

3+ 5

)

5

Complete Mathematics—JEE Main

25.20

Solution: From the given relation we have tan A

fi

3

=

5

2 tan A 1 + tan A

2

fi

1 + tan B 2 fi 4(1 + 5k ) = 5(1 + 3k2) fi k2 = 1/5 fi k = 1 / 5 3

so that tan A =

5

2 3k 1 + 3k

2

=

3 ¥ 5k 1 + 5k 2

fi fi

, tan B = 1 fi tan A + tan B =

1- k

3+ 5 5

.

(b) 2 1 + k

(c) 2 k

Solution: a < b < g < d and sin a = sin b = sin g = sin d = k fi b = p - a, g = 2p + a, d = 3p - a. so that the given expression is equal to a p -a 2p +a 3p -a + 3 sin + 2 sin + sin 2 2 2 2 a a a a = 4 sin + 3 cos – 2 sin – cos 2 2 2 2 a aˆ Ê = 2 Á sin + cos ˜ Ë 2 2¯ a a = 2 1+ k . cos 2 2

Example 98: If (a – b) sin (q + f) = (a + b) sin (q – f) and a tan (q/2) – b tan (f/2) = c, then the value of sin f is equal to 2

2

2

(a) 2ab/(a – b – c ) (b) 2bc/(a2 – b2 – c2) (c) 2bc/(a2 – b2 + c2) (d) 2ab/(a2 – b2 + c2) Ans. (b) Solution: From the first relation we have a [sin (q + f) – sin (q – f)] = b[sin (q – f) + sin (q + f)] fi 2a sin f cos q = 2 b sin q cos f fi a tan f = b tan q

1 - tan 2 (q 2 )

sin 2 a cos b cos x - cos a = sin 2 b cos a cos x - cos b

then cos x is equal to (a)

cos a - cos b 1 + cos a cos b

(b)

cos a - cos b 1 - cos a cos b

(c)

cos a + cos b 1 + cos a cos b

(d) none of these

Ans. (c) Solution: The given relation can be written as cos x (sin2 b cos a – sin2 a cos b) = cos2 a sin2 b – sin2 a cos2 b

(d) none of these

= 2 1 + 2 sin

2b tan (q 2 )

tan (f/2) (a2 – b2 – c2) = bc (1 + tan2 (f/2)) 2 tan (f 2 ) 2bc = sinf = 2 1 + tan 2 (f 2 ) a - b2 - c 2 Example 99: If

Ans. (b)

4 sin

1 - tan (f 2 )

=

From the second relation we have tan (q/2) = (b tan (f/2) + c)/a b (b tan (f 2 ) + c ) a a tan (f 2 ) = so that 2 1 - (b tan (f 2 ) + c ) a 2 1 - tan 2 (f 2 )

Example 97: If a, b, g, d are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity k, then the values of a b g d + 3 sin + 2 sin + sin is equal to 4 sin 2 2 2 2 (a) 2

2

= k (say), (clearly k > 0)

3 tan B

=

2

2a tan (f 2 )

fi

3 sin B.

Also 2 sin A = fi

tan B

fi

( ) ( ) 2 cos a (1 - cos b ) - cos b (1 - cos2 a )

cos2 a 1 - cos2 b - 1 - cos2 a cos2 b

cos x =

=

cos2 a - cos2 b (cos a - cos b ) (1 + cos a cos b )

=

cos a + cos b . 1 + cos a cos b

Example 100: If 0 < a, b < p and cos a + cos b – cos (a + b) = 3/2 then sin a + cos b is equal to (a) 0 (b) 1 (c) Ans. (c)

(

)

3 +1 2

(d)

3

Solution: From the given equation we have a+b 3 a+b a-b cos - 2 cos2 +1= 2 2 2 2 a+b a-b a+b fi 4 cos2 - 4 cos cos +1=0 2 2 2 2 cos

fi

a +b a - bˆ2 Ê 2a -b - cos -1 ÁË 2 cos ˜¯ = cos 2 2 2

so the only possibility is cos2

(i)

a-b -1 = 0 2 Êa - bˆ since cos2 Á £1 Ë 2 ˜¯

Trigonometrical Ratios, Identities and Equations 25.21

As 0 < a, b < p, we have a = b From (i) we get cos a = 1/2 = cos b and

sin b = sin a =

Example 103: sin x + 2 sin 2x = 3 + sin 3x, 0 £ x £ 2p has

3 . so that sin a + cos b = 2

3 +1 2

Example 101: If sin a, sin b, sin g are in A.P.and cos a, cos b, cos g are in G.P. then

cos2 a + cos2 g - 4 cos a cos g = 1 - sin a sin g

(a) – 2 (c) 0 Ans. (a)

(b) – 1 (d) 2

cos2 a + cos2 g - 4 cos a cos g = – 2. 1 - sin a sin g Example 102: If x =

sin3 p

cos2 p sin p + cos p = 1/2, then x + y = (a) 75/18 (c) 79/18 Ans. (c)

=

cos3 p sin 2 p

; and

sin 5 p + cos5 p sin 2 p cos2 p

(

)

(

sin3 p 1 - cos2 p + cos3 p 1 - sin 2 p 2

Solution: From the given relation we have fi fi

2 sin x cos 2x – 2 sin 2x + 3 = 0 (sin x + cos 2x)2 + (sin 2x – 1)2 + 3 = sin2 x + cos2 2x + sin2 2x + 1 fi (sin x + cos 2x)2 + (sin 2x – 1)2 + cos2 x = 0 which is possible only if sin x + cos 2x = 0, sin 2x = 1 and cos x = 0 which is not possible for any value of x. Example 104: The equation (cos p – 1)x2 + (cos p)x + sin p = 0 where x is a variable, has real roots if p lies in the interval (a) (0, 2p) (b) (– p, 0) (c) (– p/2, p/2) (d) (0, p) Ans. (d) Solution: The equation has real roots if cos2 p – 4(cos p – 1) sin p ≥ 0 or cos2 p – 4 cos p sin p + 4 sin p ≥ 0 or (cos p – 2 sin p)2 – 4 sin2 p + 4 sin p ≥ 0 or (cos p – 2 sin p)2 + 4 sin p (1 – sin p) ≥ 0 since (cos p – 2 sin p)2 ≥ 0, 1 – sin p ≥ 0 for all values of p, and for p Œ (0, p), sin p ≥ 0 so that the discriminant is non-negative.

(b) 44/9 (d) 48/9

Solution: x + y =

=

,y=

2 solutions in I quadrant one solution in II quadrant no solution in any quadrant one solution in each quadrant

sin x – sin 3x + 2 sin 2x = 3

Solution: From the given conditions we have 2 sin b = sin a + sin g (i) 2 cos b = cos a cos g (ii) Squaring (i), 4 sin2 b = sin2 a + sin2 g + 2 sin a sin g Using (ii), 4(1 – cos a cos g) = 1 – cos2 a + 1 – cos2 g + 2 sin a sin g fi cos2 a + cos2 g – 4 cos a cos g = 2(sin a sin g – 1) fi

(a) (b) (c) (d) Ans. (c)

)

Example 105: The general solution of the equation

2

sin p cos p

n

1 - sin x + º + ( -1) sin n x + º 1 + sin x + º + sin n x + º

sin3 p + cos3 p - sin 2 p cos2 p (sin p + cos p ) sin 2 p cos2 p =

=

(sin p + cos p)3 - 3 sin p cos p (sin p + cos p) sin 2 p cos2 p (sin p + cos p ) sin 2 p cos2 p

Now 2 sin p cos p = (sin p + cos p)2– 1 = – 3/4

\

3 2 Ê 1 ˆ - 3 Ê - 3ˆ Ê 1 ˆ - Ê - 3ˆ Ê 1 ˆ ÁË ˜¯ ÁË ˜¯ ÁË ˜¯ ÁË ˜¯ ÁË ˜¯ 79 2 8 2 8 2 x+y= = . 2 18 Ê - 3ˆ ÁË ˜¯ 8

(a) (b) (c) (d) Ans. (b)

1 - cos 2 x , x π (2n + 1) p/2, nŒI is 1 + cos 2 x

(Р1)n (p/3) + np (Р1)n (p/6) + np (Р1)n + 1 (p/6) + np (Р1)n Р1 (p/3) + np, (n ΠI)

Solution: The equation n

1 - sin x +º + ( -1) sin n x + º n

1 + sin x + º + sin x + º

=

1 - cos 2 x 1 + cos 2 x

25.22

Complete Mathematics—JEE Main

1 1 - sin x 2 sin 2 x ¥ = 1 + sin x 1 2 cos2 x

fi

x-y x-y x+y + 2 sin cos +1=0 2 2 2

fi

sin2

fi

x-y = sin 2

as –1 < sin x < 1 2

fi

1 – sin x =

sin x (1 + sin x ) 1 - sin 2 x

fi fi

(1 – sin x)2 = sin2x fi 1 – 2sin x = 0 sin x = 1/2 = sin (p/6)

fi

x = np + (–1)n p/6.

Solution: The given equation can be written as sin4 x + cos4 y + 2 – 4 sin x cos y = 0 fi (sin2 x – 1)2 + (cos2 y – 1)2 + 2 sin2 x + 2 cos2 y – 4 sin x cos y = 0 fi (sin2 x – 1)2 + (cos2 y – 1)2 + 2 (sin x – cos y)2 = 0 which is true if sin2 x = 1, cos2 y = 1 and sin x = cos y, so sin x + cos y = 2 as 0 £ x, y £ p/2. Example 107: A solution (x, y) of the system of equation x – y = 1/3 and cos2 (px) – sin2 (py) = 1/2 is given by (a) (2/3, 1/3) (b) (7/6, 1/6) (c) (13/6, 11/6) (d) (1/6, 5/6) Ans. (c) Solution: cos2 (px) – sin2 (py) = 1/2 fi cos p (x + y) cos p (x – y) = 1/2 cos p (x + y) cos (p/3) = 1/2

fi

cos p (x + y) = 1

fi

p (x + y) = 2np fi x + y = 2n

Now x + y = 2n and x – y = 1/3 fi x = n + 1/6, y = n – 1/6,

[Q x – y = 1/3]

(n ŒI)

\ (x, y) = ÊÁ n + 1 , n - 1 ˆ˜ which is satisfied by (c) for n = 2 Ë 6 6¯ Example 108: cos (x – y) – 2 sin x + 2 sin y = 3 if (a) (b) (c) (d) Ans. (c)

sin x = sin y x + y = 2np, x – y = (2k – 1)p/2 x = 2kp – p/2, y = 2np + p/2 cos (x – y) = – 1

x-y x+y , we have cos2 =1 2 2 x+y x+y fi sin2 = 0 or sin =0 2 2 fi x + y = 2np p x-y x-y and then sin = (2k - 1) = ±1 fi 2 2 2 fi x – y = (2k – 1) p so (b) is not correct. Also if sin x = sin y then the given equation becomes cos (x – y) = 3 which is not correct; (a) is not correct. Next if x = 2kp – p /2 and y = 2n p + p /2. Then cos (x – y) –2 sin x + 2 sin y = – 1 + 2 + 2 = 3, so (c) is correct. Finally if cos (x – y) = – 1, the given equation becomes sin x – sin y = – 2 which is not true for any real values of x and y so (d) is not correct. Example 109: If 6 cos 2q + 2 cos2 (q/2) + 2 sin2 q = 0, – p < q < p, then q is equal to (a) p/3 (b) p/3, cos–1 (3/5) –1 (c) cos (3/5) (d) p/3, p – cos–1 (3/5) Ans. (d)

fi fi fi fi fi

Solution: The given equation can be written as 6(2 cos2 q – 1) + (1 + cos q) + 2(1 – cos2 q) = 0 10 cos2 q + cos q – 3 = 0 (5 cos q + 3) (2 cos q – 1) = 0 cos q = 1/2 or cos q = – 3/5 –1 q = p/3 or q = p – cos (3/5) as – p < q < p q = p/3, p – cos–1 (3/5).

Example 110: The number of integral values of a for which the equation cos 2x + a sin x = 2a –7 possesses solution is (a) 2 (b) 3 (c) 4 (d) 5 Ans. (d)

(n, k ΠI)

Solution: cos (x – y) – 2 sin x + 2 sin y = 3 fi

x+y x+y ± 4 cos2 -4 2 2 2

For real values of sin

Example 106: If sin4 x + cos4 y + 2 = 4 sin x cos y, 0 £ x, y £ p/2 then sin x + cos y = (a) – 2 (b) 0 (c) 2 (d) none of these Ans. (c)

fi

-2 cos

x-y x-y x+y – 4 sin cos –3=0 1 – 2 sin2 2 2 2

Solution: The given equation can be written as 1 – 2 sin2 x + a sin x = 2a – 7 fi 2 sin2 x – a sin x + 2a – 8 = 0 fi

sin x =

a ± a 2 - 8 (2 a - 8) a ± ( a - 8) = 4 4

Trigonometrical Ratios, Identities and Equations 25.23

fi

sin x =

a-4 a-4 which is possible if – 1 £ £ 1 or 2 2

2 £ a £ 6. so the required values of a are 2, 3, 4, 5, 6 and hence the required number is 5. Example 111: The least difference between the roots of the equation 4 cos x (2 – 3 sin2 x) + (cos 2x + 1) = 0 (0 £ x £ p/2) is (a) p/6 (b) p/4 (c) p/3 (d) p/2 Ans. (a) Solution: The given equation can be written as 4 cos x (3 cos2 x – 1) + 2 cos2 x = 0 fi 2 cos x (6 cos2 x + cos x – 2) = 0 fi 2 cos x (3 cos x + 2) (2 cos x – 1) = 0 fi either cos x = 0 which gives x = p/2 or cos x = – 2/3, which gives no value of x for 0 £ x £ p/2 or cos x = 1/2, which gives x = p/3. so the required difference = p/2 – p/3 = p/6 Example 112: The solution of |cos x| = cos x – 2 sin x is (a) x = np (b) x = np + p/4 (c) x = np + (– 1)n (p/4) (n Œ I) (d) x = (2n + 1)p + p/4 Ans. (d) Solution: |cos x| = cos x – 2 sin x fi cos x = cos x – 2 sin x if cos x ≥ 0 fi sin x = 0 fi x = 2np (as cos x ≥ 0) n Œ I and x π np (as x = p fi cos x < 0, for n = 1) Next |cos x| = cos x – 2 sin x fi – cos x = cos x – 2 sin x if cos x < 0 fi cos x – sin x = 0 fi tan x = 1 Now cos x < 0 and tan x = 1 fi tan x = tan (5p/4) fi x = 2np + (5p/4) = (2n + 1)p + p/4 and x π 2np + p/4 (as x = p/4 fi cos x > 0, for n = 0).

Example 113: If 0 £ a, b £ 3 and the equation x2 + 4 + 3cos (ax + b) = 2x has at least one solution, then the value of a + b is (a) p/4 (b) p/2 (c) p (d) 2p Ans. (c) Solution: We have x2 – 2x + 4 = – 3cos (ax + b) fi (x – 1)2 + 3 = – 3 cos (ax + b) As (x – 1) ≥ 0 and –1 £ cos (ax + b) £ 1 Equation (1) s possible if cos (ax + b) = –1 and x – 1 = 0 fi a + b = p, 3p, 5p, ...... Since a + b £ 6 and 3p > 6, a + b = p

Example 114: If 0

2

(d) none of these

53. If 2 tan (p/3) cos (2px) = 3 , the general solution of the equation is (a) 2np ± p/3 (b) n ± 1/3 (c) n ± 1/6 (d) n ± 1/2 (n Œ I) 54. 2 cos2 x + 4 cos x = 3 sin2 x if (a) cos x =

-2 + 14 5

(b) cos x =

-2 + 19 5

-2 + 14 -2 + 19 (d) sin x = 5 5 2 2 55. 6 tan x – 2 cos x = cos 2x if (a) cos 2x = – 1 (b) cos 2x = 1 (c) cos 3x = – 1/2 (d) cos 2x = 1/2 56. The greatest value of cos q for which cos 5q = 0 is (b) (1 +

5 )/4

5+ 5 5 +1 (d) 8 4 57. If tan pq = tan qq, then the values of q form an A.P. with common difference (a) p/(p + q) (b) p/p (c) p/q (d) p/(p – q) (c)

48. If cos x – sin x = 1/2, then tan 2x = (b)

51. If 4na = p, then the value of tan a tan 2a tan 3a º tan (2n – 1)a is (a) – 1 (b) 0 (c) 1 (d) none of these x sin q cos q x 1 = 0 then 52. If x > 0 and - sin q cos q 1 x

(a) 0

(b) 1 (d) 4

11∞ – cos 2∞ is a positive integer a negative integer a positive rational number a negative rational number

7 /3

n Êpˆ Êpˆ , then a 50. If n Œ N and sin Á ˜ + cos Á ˜ = Ë 2n ¯ Ë 2n ¯ 2 possible value of n is (a) 4 (b) 6 (c) 8 (d) 12

(c) sin x =

47. If sin A, cos A and tan A are in G.P., then cot6 A – cot2 A = (a) – 1 (b) 0 (c) 1 (d) none of these (a)

49. Which of the following gives the least value of A (a) cos 2A = sin 3A (b) cos 3A = sin 7A (c) tan A = cot 3A (d) cot A = tan 2A

7 /4

(d) 2/ 7

58. The equation

x

Ú0

(t2 – 8t + 13) dt = x sin (a/x) has a

solution if sin (a/x) = (a) 0 (b) 1 (c) 3 (d) 6 59. The smallest positive root of the equation sin (1 - x ) = (a) 1/2 + p/4 (c) 1/2 + 5p/4

cos x is (b) 1/2 + 3p/4 (d) 1/2 + 7p/4

Trigonometrical Ratios, Identities and Equations 25.27

60. The sum of the roots of the equation 4 cos3 x – 4 cos2 x – cos (p + x) – 1 = 0 in the interval [0, 315] is pp, where p is equal to (a) 2500 (b) 2550 (c) 2600 (d) 2651 61. A solution (x, y) of x2 + 2x sin xy + 1 = 0 is (a) (1, 0) (b) (1, 7p/2) (c) (– 1, –7p/2) (d) (– 1, 0) 62. esin x – e–sin x = 4 for (a) all real values of x (b) some x Œ[0, p/2] (c) some x Œ(– p/2, p/2) (d) no real value of x 63. The number of values of x satisfying the condition sin x + sin 5x = sin 3x in the interval [0, p] is (a) 6 (b) 2 (c) 10 (d) 0 64. sin x, sin 2x, sin 3x are in A.P. if (a) x = 165° (b) x = 195° (c) x = 15° (d) x = 2np 2

(a) 32 (c) 50

(b) 16 (d) 64

69. If 15 sin4a + 10 cos4a = 6 then the value of 8 cosec6 a – 27 sec6 a is (a) 0 (b) 1 (c) –1 (d) 5 •

70. If

1

 tan - 1 2r 2 = t , then tan t is equal to

r =1

(a) 0 (c)

(b) 1 3

(d) 1/ 3 .

71. If q Œ [– 2011p, 2011p], then the number of solutions of the equation. q [sec q – 1] + {tan q tan } 2 q }] 2 (where {◊} denotes the fractional part function and [◊] denotes the greatest integer function) is (a) 0 (b) 2010 (c) 2011 (d) 2012. = [{secq – 1 + tanq tan

n ΠI.

2

65. 2 sin ((p/2) cos x) = 1 Рcos (p sin 2x) if (a) x = (2n + 1)p/2 (b) x = np/2 (c) x = 2np (d) none of these, n ΠI 66. Number of solutions of the equation 4 sin2 x + tan2x + cot2 x + cosec2 x = 6 in [0, p] is (a) 0 (b) 2 (c) 4 (d) 8. 67. Sum of integral values of n such that sinx (2 sin x + cos x) = n, has at least one real solution is (a) 1 (b) 2 (c) 3 (d) 0. 68. If 0 < x < p/2 and sin (2 sinx) = cos (2 cosx) then a where a + b + c = tanx + cotx = C p -b

2p 4p 8p + sin + sin 7 7 7 2p 4p 8p + cos + cos and y = cos , then x2 + y2 is 7 7 7 (a) 1 (b) 2 (c) 3 (d) 4.

72. If x = sin

73. Number of solutions of the equation Ê p x ˆ = x 2 - 2 3 x + 4 is sin Á Ë 2 3 ˜¯ (a) 0 (b) 1 (c) 2 (d) 4

Assertion-Reason Type Questions

p 2p 4p 1 =cos cos 7 7 7 8 Statement-2: cosq cos2q cos4q … cos2n – 1q = 1 p - n if q = . n 2 2 -1

74. Statement-1: cos

75. Statement-1: Let f(x) = 1 + sin x - cos x 1 + sin x + cos x

sin 2 x and g(x) = 1 + cos 2 x

If b, g, Π(0, p) such that cos a + cos (a + b) + cos (a + b + g ) = 0 and sin a + sin (a + b) + sin (a + b + g ) = 0 Then f(b) + g(g ) = 0 Statement-2: For all values of x, f(x) + g(x) = 0, f(x) and g(x) are functions defined in statement-1. 76. Statement-1: If x and y are real number such that x2 + y2 = 27, then the maximum possible value of x Рy is 3 6 .

25.28

Complete Mathematics—JEE Main

Statement-2: – 1 £ cos (q + p /4) £ 1 77. Statement-1: If p = 7 + tan a tan b, q = 5 + tan b tan g and r = 3 + tan g tan a, then the maximum p + q + r is 4 3 ; a, b, g > 0 and

value of

a + b + g = p/2 Statement-2: If a + b + g = p/2, a, b, g > 0, then cot a cot b + cot b cot g + cot g cota = 1. 78. Statement-1: If x sin3 q + y cos3 q = sin q cos q.

and x sin q = y cos q, then x2 + y2 = 1 Statement-2: 4 sin3 q = 3 sin q – sin 3q 4 cos3 q = 3 cos q + cos 3q 79. Statement-1: 2 sin2 x + 3 sin x – 2 > 0 and x2 – x – 2 < 0, if – 1 < x < 2. Statement-2: x 2 – x – 2 < 0 if – 1 < x < 2 80. Statement-1: cos7 x + sin4 x = 1 has only two nonzero solution in the interval – p < x < p. Statement-2: cos5 x + cos2 x – 2 = 0 is possible only when cos x = 1

LEVEL 2 Straight Objective Type Questions 81. The acute angle of a rhombus whose side is a mean proportional between its diagonals is (a) 15° (b) 20° (c) 30° (d) 80° 82. Given the height h and the angle bisector l drawn from the vertex of the right angle of a triangle, then cosine of an acute angle of the triangle is given by (a)

(c)

h + l 2 - h2

(b)

2h h l

(d)

83. If 2 sin2 (x + p /4) +

•

 cos2 n q ,

n=0

2h h - l 2 - h2 2l

3 cos 2x > 0, then

(a) cos (2x – p /6) > – 1/2 (c) sin (2x – p /6) > – 1/2 84. x =

h - l 2 - h2

(b) sin (2x – p /6) < – 1/2 (d) cos (2x – p /6) < – 1/2

•

y=

 sin2 n q ,

n=0

sin 2 2 x + 4 sin 4 x - 4 sin 2 x cos2 x = 1 is 9 4 - sin 2 2 x - 4 sin 2 x

86. cos 7.5∞ =

2 2

x+y xy

(c)

n=0

85. The least positive value of x satisfying

(b) p/6 (d) 5p/6

2 2 + 3 +1

(b)

4+ 2 + 6 8

(d)

4+ 2 + 6 4

87. If x = X cos q – Y sin q, y = X sin q + Y cos q; and ax2 + 2bxy + cy2 = AX2 + 2HXY + BY2, then (a) H = 0 If q = 0 (b) H = b if q = p/2 (c) A + B = a + c (d) H = c – a if q = p/4 88. If sin 5q = a sin5 q + b sin3 q + c sin q + d, then (a) a + b + c = 0 (b) a + b + c + d = 0 (c) 5a + 3b – 4c = 0 (d) a – 3c + d = 0 89. If tan A – tan B = x and cot B – cot A = y, then the value of cot (A – B) is 1 1 x-y (a) (b) 2 + 2 x y xy

 cos2 n q sin2 n q ,

|cosq| < 1, |sinq | < 1 then x + y + z is equal to (a) xy (b) yz (c) zx (d) xyz

(a) p/3 (c) 2p/3

(c)

•

z=

2+ 2 + 6 8

(a)

90.

cos 3q

+

sin 3q

(d) xy

is equal to cos q sin3 q (a) 3 cos 2q cosec 2q (b) 3 cot 2q cosec 2q (c) 12 cot 2q cosec 2q (d) 12 tan 2q sec 2q 91. (x tan a + y cot a) (x cot a + y tan a) – 4xy cot2 2a = (b) 4xy (a) x2 + y2 (d) none of these (c) (x + y)2 92. If tan A, tan B, tan C satisfy the equation 3 tan3 q – 4 tan2 q + 3 tan q + 1 = 0, then A+B+C= (a) 0 (b) p/2 (c) 3p/4 (d) 2p 3

Trigonometrical Ratios, Identities and Equations 25.29

93. If x sin q + y sin 2q + z sin 3q = sin 4q, (q π np) then 8 cos3 q – 4z cos2 q – 2(y + 2) cos q is equal to (a) x – y (b) x – z (c) y – z (d) none of these 94. The number of values of sin x satisfying sin 5x = 5 sin x is (a) 0 (b) 1 (c) 2 (d) 3 95. If sin a, sin b are the roots of the equation a sin2 q + b sin q + c = 0 and sin a + 2 sin b = 1 then a2 + 2b2 + 3ab + ac = (a) – 1 (b) 0 (c) 1 (d) a + b + c 96. If sin (q/2) = a, cos (q/2) = b, then (1 + sin q) (3 sin q + 4 cos q + 5) = (b) (a + b)2 (3a + b)2 (a) (a + b)2 (a + 3b)2 (c) (a – b)2 (a – 3b)2 (d) (a – b)2 (3a – b)2 97. If sin A = sin B and cos A = cos B; A π B, then A- B (a) tan =0 (b) cos (A + B) = 1 2 A+ B (c) tan =0 (d) sin (A – B) = 1/2 2 98. cos 22∞ + cos 78∞ + cos 80∞ = (a) 4 sin 11∞ sin 39∞ sin 40∞ (b) 1 + 4 cos 11∞ cos 39∞ cos 40∞ (c) 1 + 4 sin 11∞ sin 39∞ sin 40∞ (d) 4 cos 11∞ cos 39∞cos 40∞ x 1 x x 1 1 99. tan x + tan + 2 tan 2 + º + n -1 tan n -1 is 2 2 2 2 2 2 equal to (a) (b)

1 2

n

cot

1 2 n-1

x 2n

– 2 cot 2x

Ê x ˆ cot Á n -1 ˜ – 2 cot 2x Ë2 ¯ n

(c) tan

2 -1 2 n -1

(d) 2 cot 2x – 100. The value of (a) cot 44° (c) tan 2∞

(a) 1/ 3 (c)

(b) 1/3

3

(d) 3

103. If tan q, 2 tan q + 2, 3 tan q + 3 are in G.P., then the value of

7 - 5 cot q 9 - 4 sec 2 q - 1

1 n -1

cot

x 2

n -1

3 + cot 76∞ cot 16∞ is cot 76∞ + cot 16∞ (b) cot 46° (d) cot 92∞

101. If x cos q = y cos (q + 2p/3) = z cos (q + 4p/3) then xy + yz + zx = (a) cos2 q (b) sin2 q (c) 1 (d) 0

is

(a) 12/5

(b) – 33/28

(c) 33/100

(d) 12/13

104. If sin q + cos q = a and cos q – sin q = b, then sin q (sin q – cos q) + sin2 q (sin2 q – cos2 q) + sin3 q (sin3 q – cos3 q) + º is equal to (a)

1 - ab 1 + ab

(b)

(c)

1 - ab 1 - a 2 + 1 + ab 3 - a 2

(d)

(

105. If 1 + 1 + x

)

(

1 - a2 3 - a2 1 + ab a 2 - 1 1 - ab 3 - a 2

tan a = 1 - 1 - x

(a) sin a (c) sin 4a

)

then x =

(b) sin 2a (d) cos 4a

106. If f (q) = sin q (sin q + sin 3q), then f(q) (a) ≥ 0 only when q ≥ 0 (b) £ 0 for all real q (c) ≥ 0 for all real q (d) £ 0 only when q £ 0 107. In a right angled triangle, the hypotenuse is 2 2 times the length of the perpendicular drawn from the opposite vertex on its hypotenuse then the other two angles are (a) p/3, p/6 (b) p/4, p/4 (c) p/8, 3p/8

(d) p/12, 5p/12

108. cos 2 x + 1 + sin 2 x = 2 sin x + cos x if (a) sin x + cos x = 1 (c) x = np + p/4

x

2

102. If A > 0, B > 0 and A + B = p/3 then the maximum value of tan A tan B is

109. sin (a) (b) (c) (d)

(b) x = 2np (d) sin x – cos x = 0

x + 2 sin 2x = 3 + sin 3x if sin x + cos 2x = 0 if sin 2x – 1 = 0 if cos x = 0 for no real value of x

110. The number of pairs (x, y) satisfying the equation sin x + sin y = sin (x + y) and |x| + |y| = 1 is (a) 2 (b) 4 (c) 6 (d) infinite

25.30

Complete Mathematics—JEE Main

Previous Years’ AIEEE/JEE Main Questions

1. The numbers of solutions of tan x + sec x = 2 cos x in [0, 2p] is (a) 2 (b) 3 (c) 0 (d) 1 [2003] 2. Let a, b be such that p < a – b < 3p. If sin a + sin b = – 21/65 and cos a + cos b = – 27/65 then the a-b is values of cos 2 (a) 6/65

(b) 3 / 130

(c) - 3 / 130

(d) – 6/65

3. If u =

[2004]

a 2 sin 2 q + b2 cos2 q , then the difference between the maximum and minimum values of u2 is given by

(b) 2 a 2 + b2

(c) 2(a2 + b2)

(d) (a – b)2

4. If f : R Æ S is defined by f(x) = sin x – is onto, then the interval of S is (a) [0, 1] (b) [– 1, 1] (c) [0, 3] (d) [– 1, 3]

(b) There is a regular polygon with r / R = 3 / 2 (c) There is a regular polygon with r /R = 1/2 (d) There is a regular polygon with r / R = 1 / 2 [2010]

a 2 cos2 q + b2 sin 2 q +

(a) (a + b)2

9. For a regular polygon, let r, R be the radii of the inscribed and the circumscribed circles. A false statement among the following is (a) There is a regular polygon with r /R = 3/3

[2004] 3 cos x + 1

[2004]

5. The number of values of x in the interval [0, 3p] satisfying the equation 2 sin2 x + 5 sin x – 3 = 0 is (a) 2 (b) 4 (c) 6 (d) 1 [2006] 6. If the roots of the quadratic equation x2 + px + q = 0 are tan 30° and tan 15°, then the value of 2 + q – p is (a) 1 (b) 2 (c) 3 (d) 0. [2006] 7. If 0 < x < p, and cos x + sin x = 1/2, then tan x is (a) (1 + 7 ) / 4

(b) (1 - 7 ) / 4

(c) (4 - 7 ) / 3

(d) - (4 + 7 ) / 3

[2006]

8. Let A and B denote the statements A : cosa + cosb + cosg = 0 B : sina + sinb + sing = 0 If cos (b – g) + cos (g – a) + cos (a – b) = – 3/2 then (a) both A and B are true (b) both A and B are false (c) A is ture and B is false (d) A is false and B is true. [2009]

10. Let cos(a + b) = 4/5 and let sin (a – b) = 5/13 where 0 £ a, b £ p/4. Then tan2a = (a) 19/12 (b) 20/7 (c) 25/16 (d) 56/33 [2010] 11. If A = sin2x + cos4x, then for all real x (a) 3 4 £ A £ 13 16

(b) 3 4 £ A £ 1

(c) 13 16 £ A £ 1

(d) 1 £ A £ 2

[2011]

12. The possible values of q Π(0, p) such that sin (q) + sin (4q) + sin (7q) = 0 are: p 5p p 2p 3p 8p , , , , , (a) 4 9 4 12 12 3 (b)

2p p p 2p 3p 35p , , , , , 9 4 2 3 4 36

2p p p 2p 3p 8p , , , , , 9 4 2 3 4 9 2p p 4p p 3p 8p , , , , , (d) [2011] 9 4 9 2 4 9 13. In a D PQR, if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then the angle R is equal to (c)

(a) p 4

(b) 3p 4

(c) 5 p 6

(d) p 6

[2012]

14. ABCD is a trapezium such that AB and CD are parallel and BC ^ CD. If ADB = q, BC = p and CD = q, then AB is equal to (a)

(c)

p2 + q 2 cos q p cos q + q sin q ( p2 + q 2 )cos q ( p cos q + q sin q )2

(b)

(d)

p2 + q 2 p2 cos q + q 2 sin q ( p2 + q 2 )sin q [2013] p cos q + q sin q

Trigonometrical Ratios, Identities and Equations 25.31

tan A cot A can be written as + 1 - cot A 1 - tan A (a) sec A cosec A + 1 (b) tan A + cot A (c) sec A + cosec A (d) sin A cos A + 1 [2013]

15. The expression

16. Statement-1: The number of common solutions of the trigonometric equations 2 sin2q – cos2q = 0 and 2 cos2q – 3 sin q = 0 in the interval [0, 2p] is two. Statement-2: The number of solutions of the equation, 2cos2q – 3 sin q = 0 in the interval [0, p] is two [2013, online] 17. The number of solutions of the equation sin2x – 2cos x + 4 sin x = 4 in the interval [0, 5p] is: (a) 3 (b) 4 (c) 5 (d) 6 [2013, online]

(

)

18. Let fk(x) = (1 / k ) sin k x + cosk x where x ŒR and k ≥ 1. Then f4(x) – f6(x) equals: (a) 1/6 (b) 1/3 (c) 1/4 (d) 1/12

[2014]

19. The number of values of a in [0, 2p] for which 2sin3a – 7sin2a + 7 sin a = 2, is: (a) 6 (b) 4 (c) 3 (d) 1 [2014, online] p+q 20. If cosec q = (p π q π 0), then p-q Êp qˆ cot Á + ˜ is equal to: Ë 4 2¯ (a)

p q

(b)

q p

(c)

pq

(d) pq

[2014, online]

21. If 2 cos q + sin q = 1, (q π p /2), then 7 cos q + 6 sin q is equal to: (a) 1/2 (b) 2 (c) 11/2 (d) 46/5 22. If f (q) =

1 - sin q -1

cos q 1 sin q

[2014, online]

1 - cos q and A and B are 1

respectively the maximum and the minimum values of f (q), then (A, B) is equal to (a) (3, – 1) (b) 4, 2 - 2

(

(c) 2 + 2 , 2 - 2

)

(d)

( (2 +

)

)

2, - 1

[2014, online] 3 1 23. If cos a + cos b = and sin a + sin b = and q is the 2 2 arithmetic mean of a and b, then sin 2q is equal to:

3 5 7 (c) 5

(a)

4 5 8 (d) 5

(b)

[2015, online]

a = 2 + 3 and c = 60∞. Then the b ordered pair ( A, B) is equal to: (a) (15°, 105°) (b) (105°, 15°) (c) (45°, 75°) (d) (75°, 45°) [2015, online]

24. In a DABC,

25. If 0 £ x < 2p, then the number of real values of x which satisfy the equation cos x + cos 2x + cos 3x + cos 4x = 0, is: (a) 3 (b) 5 (c) 7 (d) 9 [2016] 26. If m and M are the minimum and the maximum values 1 of 4 + sin 2 2 x - 2 cos 4 x, x Œ R, then M – m is equal 2 to: 9 15 (b) (a) 4 4 7 1 (c) (d) [2016, online] 4 4 27. The number of x Œ [0, 2p] for which 2sin 4 x + 18cos 2 x - 2 cos 4 x + 18sin 2 x = 1 is: (a) 2 (b) 6 (c) 4 (d) 8 [2016, online] 28. The number of distinct real roots of the equation, cos x sin x sin x D = sin x cos x sin x = 0 sin x sin x cos x p p in the interval ÈÍ- , ˘˙ is: Î 4 4˚ (a) 1 (b) 4 (c) 2 (d) 3 29. If A > 0, B > 0 and A + B =

[2016, online]

p , then the minimum 6

value of tan A + tan B is: (a)

3- 2 2

(b) ü -

(d) 2 - 3 [2016, online] 3 30. Let P = {q : sin q – cos q = 2 cos q} and Q = {q sin q + cos q = 2 sin q} be two sets. Then: (a) P Ã Q and Q – P π f (b) Q À P (c) P = Q (d) P À Q [2016, online] (c)

25.32

Complete Mathematics—JEE Main

Previous Years' B-Architecture Entrance Examination Questions 1. The number of solutions of the equation tan x + sec x = 2cos x lying in the interval [0, 2p] is (a) 1 (b) 2 (c) 3 (d) 0 [2006] 2. For q π 0, if cos q + sec q = 2, then cosnq + secnq equals (b) (–2)n (a) 2n n+1 (c) 2 (d) 2 [2006] 3. If for real p the equation sin6x + cos6x = p always has a real solutions, then p lies in the interval: È1 È (a) ]1, 2] (b) Í , 1 Í Î4 Î È1 (c) Í , Î4

˘ 1˙ ˚

(d) [1, 2]

4. The value of cos4 (a) 3/2 (c) 3/4

[2007]

p 3p 5p 7p + cos4 + cos4 + cos4 is 8 8 8 8 (b) 1/2 (d) 1/4 [2008]

p 3p 5p is 5. The value of sin ◊ sin ◊ sin 14 14 14 (a) 1/4 (c) 1/2

(a) (c)

(d)

2 sin q

cos 2 q tan 27° – tan 63° + tan 81° is (b) 3 (d) 0 [2012] b a = 0, then c

sin2A + sin2B + sin2C is: (a) 9/4 (b) 5/4 (c) 2 (d) 3 3 2 3p < q < p and 4 then K equals (a) – 1 (c) 1/2

13. Let

2 cot q +

1 sin 2 q

(b) 0 (d) 1

[2014] = K – cot q,

[2014]

3p f (x) = 2 sin x + sin 2x, in the interval ÈÍ0, ˘˙ is: Î 2˚ (a)

6. Let 0 < a, b £ p /4. If cos (a + b) = 4/5 and sin (a – b) = 5/13 then the value of tan 2a is: (a) 19/12 (b) 56/33 (c) 20/7 (d) 25/16 [2009]

(c)

7. The sides of a triangle are sin b, cos b and 1 + a sin 2b , p for some a > 0, 0 < b < . If the greatest angle of 2 this triangle is 120°, then a equals (a) 1 (b) 3/2 (c) 2/3 (d) 1/2 [2010] 8. If cos 25° + sin 25° = K, then cos 50° equals (b) K 1 - K

2 cos q

cos 2 q 11. The value of tan 9° – (a) 4 (c) 2 1 a 12. In a DABC, if 1 c 1 b

[2008]

(a) - K 2 - K

(b) 2 cos q

cos 2 q

14. The maximum value of

(b) 1/8 (d) 1/16

2

2

2

(c) - K 1 - K 2 (d) K 2 - K 2 [2010] 9. tan 9° – tan 27° – tan 63° + tan 81° is equal to: (a) 4 (b) 0 (c) 1 (d) –1 [2011, 12] 10. If a and b are such that a sin q = b cos q for p a-b a+b 0£q< , then equals + 4 a+b a-b

2 +1

(b) 2 3

3 3 2

(d)

[2015]

3

p 15. For all values of q ŒÊÁ 0, ˆ˜ , the determinant of the Ë 2¯ È -2 Í matrix D = Í- sin q Í -3 Î

tan q + sec2 q cos q -4

3 ˘ ˙ sin q ˙ always lies 3 ˙˚

in the interval: (a) [3, 5]

(b) (4, 6)

5 19 (c) ÊÁ , ˆ˜ Ë2 4¯

7 21 (d) ÈÍ , ˘˙ Î2 4 ˚

[2016]

Answers Concept-based 1. (b) 5. (a) 9. (a)

2. (c) 6. (a) 10. (b)

3. (b) 7. (a) 11. (b)

4. (d) 8. (c) 12. (a)

Trigonometrical Ratios, Identities and Equations 25.33

13. (c)

14. (a)

15. (b)

16. (d)

9. (a)

10. (c)

11. (a)

17. (b)

18. (a)

19. (b)

20. (c)

13. (a)

14. (c)

15. (a)

Level 1 21. 25. 29. 33. 37. 41. 45. 49. 53. 57. 61. 65. 69. 73. 77.

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

12. (a)

Hints and Solutions 22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62. 66. 70. 74. 78.

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

23. 27. 31. 35. 39. 43. 47. 51. 55. 59. 63. 67. 71. 75. 79.

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

24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 64. 68. 72. 76. 80.

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

Concept-based 1. f(q) = =

82. (d)

83. (a)

84. (d)

85. (b)

86. (b)

87. (c)

88. (c)

89. (c)

90. (c)

91. (c)

92. (b)

93. (b)

94. (b)

95. (b)

96. (a)

97. (a)

98. (c)

99. (b)

100. (a)

101. (d)

102. (b)

103. (c)

104. (c)

105. (c)

106. (c)

107. (c)

108. (b)

109. (d)

110. (c)

1 - tan3 q 1 + tan q + tan 2 q = tan q (1 - tan q ) tan q

= 1+

2 sin 2q

Ê 9p ˆ \ f Á ˜ = 1+ Ë 8 ¯

2 =1+ 2 2 9p ˆ Ê sin Á ˜ Ë 4¯

3 3 2. f (q) = sin q - cos q cos q sin q

sin 4 q - cos4 q - (cos2 q - sin 2 q )(1) = cos q sin q cos q sin q = – 2 cot (2q)

=

Level 2 81. (c)

tan q ◊ tan q 1 1 + tan q (1 - tan q ) tan q - 1

tan q + sec q - (sec 2 q - tan 2 q ) tan q - sec q + 1 (sec q + tan q )(1 - sec q + tan q ) = tan q - sec q + 1 1 + sin q = cos q 4. tan 15° + tan 75° = tan 15° + cot 15° 3.

tan 2 15∞ + 1 tan 15∞ 1 ˆ = 2 ÊÁ =4 Ë sin 30∞ ˜¯ 5. Use : sin2A – sin2B = sin (A – B) sin (A + B) cos 11∞ - sin 11∞ cos 11∞ + cos (90∞ + 11∞) = 6. cos 11∞ + sin 11∞ cos 11∞ - cos (90∞ + 11∞) =

Previous Years’ AIEEE/JEE Main Questions 1. (a)

2. (c)

3. (d)

4. (d)

5. (b)

6. (c)

7. (d).

8. (d)

9. (a)

10. (d)

11. (b)

12. (d)

13. (d)

14. (d)

15. (a)

16. (b)

17. (a)

18. (a)

19. (c)

20. (a)

21. (b)

22. (c)

23. (a)

24. (b)

25. (c)

26. (a)

27. (d)

28. (c)

29. (b)

30. (c)

2 cos 56∞ cos 45∞ = cot 56° 2 sin 56∞ sin 45∞ 7. tan (a + b) = tan (p/2 – b) tan a + tan b = cot b fi 1 - tan a tan b =

fi fi

Previous Years' B-Architecture Entrance Examination Questions 1. (b)

2. (d)

3. (c)

4. (a)

5. (b)

6. (b)

7. (d)

8. (d)

tan a + tan b = cot b – tan a 2 tan a + tan b = cot b = tan (a + b)

8. sin q + cosec q = 2 fi sin q +

1 =2 sinq

25.34

Complete Mathematics—JEE Main

fi sin q = 1 = cosec q \ cos q = 0. Thus, cos2015q + cosec2015q = 1 9. Multiply both the sides by (1 – sin A) (1 – sin B) (1 –sin C) to obtain (1 – sin A)2 (1 – sin B)2 (1 – sin C)2 = cos2A cos2B cos2C fi (1 – sin A) (1 – sin B) (1 – sin C) = ± cos A cos B cos C p 9p 3p 5p 10. 2 cos cos + cos + cos 13 13 13 13 9p 4p p p + 2 cos cos = 2 cos cos 13 13 13 13 p È Ê 9p ˆ Ê 4p ˆ ˘ = 2 cos cos Á ˜ + cos Á ˜ ˙ Ë 13 ¯ Ë 13 ¯ ˚ 13 ÍÎ p = 2 cos ÊÁ ˆ˜ Ë 13 ¯

p 5p ˘ È ÍÎ2 cos 2 cos 26 ˙˚ = 0

11. sec A – sec B = cosec B – cosec A cos B - cos A sin A - sin B = fi cos A cos B sin A sin B fi

sin A sin B sin A - sin B = cos A cos B cos B - cos A

fi

A + Bˆ A - Bˆ 2 cos ÊÁ sin ÊÁ Ë 2 ˜¯ Ë 2 ˜¯ tan A tan B = A + Bˆ A - Bˆ sin ÊÁ 2 sin ÊÁ Ë 2 ˜¯ Ë 2 ˜¯ Ê A + Bˆ = cot Á Ë 2 ˜¯

12. Use 1 + cos2A = 2 cos2A. 13. f (q) =

1 (2 sinq cosq) cos2q cos4q cos7q 2 sinq

=

1 (2 sin2q cos2q) cos4q cos7q 2 ¥ 2 sinq

=

1 (2 sin4q cos4q) cos7q 2 ¥ 4 sinq

1 = [2 sin8q cos7q] 2 ¥ 8 sinq =

1 [sin(15q) + sinq] 16 sinq

When q = 14. As and

p 3p 2p Êpˆ f Á ˜ = cos cos cos . Ë 14 ¯ 7 7 7 2p 2p 3p p Êpˆ 2 sin f Á ˜ = sin cos cos 7 Ë 14 ¯ 7 7 7 p Êpˆ 4p 3p 4 sin f Á ˜ = sin cos Ë ¯ 7 14 7 7 p Ê 7p ˆ Êpˆ Êpˆ Ê p ˆ 8 sin Á ˜ f Á ˜ = sin Á ˜ + sin Á ˜ = sin Ë 7¯ Ë 7¯ Ë 7 ¯ Ë 14 ¯ 7

fi fi fi

1 Êpˆ fÁ ˜ = Ë 14 ¯ 8

fi

15. sin2A + sin2B + sin2C + 2sin A sin B sin C = 1 – (cos2A – sin2B) + sin C [sin C + 2sin A sin B] = 1– cos(A + B) cos (A – B) + sin C [cos(A + B) + 2sin A sin B] [Q C = p /2 – A – B] = 1 – cos (p /2 – C) cos (A – B) + sin C cos (A – B) =1 16. tan 5x = – cot 3x = tan (p /2 + 3x) p + 3x, n Œ I fi 5x = np + 2 p fi 2x = (2n + 1) ,nŒI 2 p fi x = (2n + 1) , n Œ I. 4 Now, x Œ [0, 4p] p fi 0 £ (2n + 1) £ 4p 4 1 15 £n£ fi0£n£7 2 2 \ There are 8 values of x. 17. sin x = tan x fi sin x (cos x –1) = 0 fi sin x = 0 or cos x = 1. However, when cos x = 1, then sin x = 0 \ sin x = 0 fi x = 0, p, 2p, 3p, 4p, 5p fi

pˆ 3 Ê 2 sin Á x + ˜ £ 2 < Ë ¯ 4 2 Thus, no values of x. 19. tan2x = 1 fi tan x = ±1 18. sin x + cos x =

tan x = 1 fi x =

p 1 , f (q) = . 15 16

p 5p 9p 13p 17p , , , , 4 4 4 4 4

3p 7p 11p 15p 19p , , , , 4 4 4 4 4 Thus, there are 10 values of x. and tan x = – 1 fi x =

p p 3p 3p p 2p , = = 14 2 7 14 2 7 5p p p - , we can write = 14 2 7

-

20.

sec (8 q ) - 1 tan (2 q ) ◊ sec (4 q ) - 1 tan (8 q )

Trigonometrical Ratios, Identities and Equations 25.35

=

1 - cos (8 q ) cos (4 q ) sin (2 q ) cos (8 q ) ◊ ◊ ◊ 1 - cos (4 q ) cos (8 q ) cos (2 q ) sin (8q )

=

2 sin 2 (4 q ) cos (4 q ) sin (2 q ) =1 2 sin 2 (2 q ) cos (2 q ) sin (8 q )

fi i3 – pi2 – r = (i – tan a) (i – tan b) (i – tan g) and – i3 – pi2 – r = (– i – tan a) (– i – tan b) (– i – tan g) Multiplying we get (p – r)2 – (i)2 = – [((i)2 – tan2 a) (i2 – tan2 b) (i2 – tan2 g) fi (p – r)2 + 1 = (1 + tan2 a) (1 + tan2 b) (1 + tan2 g)

Level 1 21. tan (b – a) = =

tan b - tan a tan a + cot b = 1 + tan b tan a 1 + tan b tan a tan a tan b + 1 tan b (1 + tan a tan b )

30. A > p/2 B + C < p/2 fi tan (B + C) > 0 tan B + tan C fi >0 1 - tan B tan C fi

= cot b. 22.

cos ( x - y) =a cos ( x + y)

31.

fi cot x cot y = 2

a +1 a -1

2

23. sin A + sin (A – B) + [sin (A + B) + sin (A – B)] sin (B – A) = sin2 A + sin2 (A – B) + sin (A + B) sin (B – A) – sin2 (A – B) = sin2 A + sin2 B – sin2 A = sin2 B. 24.

6 sin q cos q 2

8 cos q + 1

=

6 tan q 9 + tan 2 q

=1

x sin q y sin3 q y a ay = , = fi sin 2 q = ¥ = 4 4 a cos q b cos q b x bx

cos q sin 3 q - sin q cos 3 q 26. cotq – cot 3q = sin q sin 3 q =

sin 2 q sin q sin 3 q

= 2 cos q cosec 3q. 27. cos (x – y) = 1 fi x – y = 0 (Note x – y π 2p) fi x = y. cos (x + y) = – 1/2 fi x + y = 120° or 240° fi x = y = 60° or x = y = 120°. 28. sin 2B = (3/2) sin 2A, cos 2B = 3 sin2 A. fi tan 2B = cot A fi tan A tan 2B = 1 fi A + 2B = p/2 fi B = p/4 – A/2 29. x3 – px2 – r = (x – tan a) (x – tan b) (x – tan g)

fi cos q =

-1± 1+ 4 -1± 5 = 2 2

But cos q π

-1- 5 so cos q = 2

fi 2 cos2(q/2) = fi cos2 (q/2) =

5 -1 +1= 2

5 -1 2 5 +1 2

5 +1 = cos 36°. 4

32. a tan2 q + b = 1 + tan2 q fi (a – 1) tan2 q = 1 – b

fi tan2 q – 6 tan q + 9 = 0 fi tan q = 3 25.

2 sin (q / 2) = 1 - sin q fi 2 sin2 (q/2) = cos2 q cos (q / 2) sin q

fi cos2 q + cos q – 1 = 0

cos ( x - y) + cos ( x + y) a + 1 = cos ( x - y) - cos ( x + y) a - 1

fi

tan B tan C < 1 (as tan B, tan C > 0)

and (b – 1) tan2 f = 1 – a fi

33.

34. 35.

36.

37.

tan 2 q

=

(1 - b)2

=

b2

tan 2 f (1 - a)2 a 2 fi (a – b) (a + b – 2ab) = 0 fi a + b = 2ab [as a π b] a = (sin2 x + cos2 x)2 – 2 sin2 x cos2 x fi sin2 2x = – 2(a – 1) fi 1 – cos 4x = – 4(a – 1) fi cos 4x = 4a – 3 so – 1 £ 4a – 3 £ 1 fi 1/2 £ a £ 1 For x = np + p/2, tan x is infinite and all other graphs have finite values. x + y + z = p fi tan x + tan y + tan z = tan x tan y tan z fi 2k + 3k + 5k = 2k ¥ 3k ¥ 5k fi k2 = 1/3. tan2 x + tan2 y + tan2 z = (1/3) (4 + 9 + 25) = 38/3. B = 60°, sin (2A + B) = 1/2 fi 2A + B = 150° fi A = 45°, C = 75° so that B + 2C = 210° = 180° + 30° fi sin (B + 2C) = – sin 30° = – 1/2. a = tan q tan f, tan2 a = tan2 (q – j) b 2 Ê tan q - tan f ˆ = Á Ë 1 + tan q tan f ˜¯

25.36

Complete Mathematics—JEE Main

=

a 2 - 4( a / b )

=

(b) fi 10A = p/2 fi A = p/20, (c) fi A = p/8, (d) fi A = p/6 Least value of A is p /20 given by (b).

ab (ab - 4)

(1 + (a / b)2 ( a + b )2 Since tan2a > 0, ab > 4. x y x tan A + y tan B = so 38. cos A cos B x+y =

cos A tan A + cos B tan B cos A + cos B

A+B A-B cos sin A + sin B 2 2 = = A+B A-B cos A + cos B cos 2 cos 2 2 2 sin

= tan

A+B . 2

39. x2 + y2 = a2[cos2 q + sin2 q + q 2 sin2 q + q 2 cos2 q] = a2(1 + q 2) fi

aq =

x 2 + y2 - a2

40. We can write (1 + cos (p/8)) (1 + cos (3p/8)) (1 – cos (3p/8)) (1 – cos (p/8)) = (1 – cos2 (p/8)) (1 – cos2 (3p/8)) = sin2 p /8 sin2 3p/8 = sin2 p/8 cos2 p/8 = (1/4) sin2 p/4) = 1/8. 41. cos y = 2 ¥ 2 cos 2x cos x sin y = 2 ¥ 2 sin 2 x cos x fi 1 = cos2 y + sin2 y = 16 cos2 x So cos 2x = 2 cos2x – 1 = – 7/8. 42. cos A = 3/5 fi sin A = – 4/5, cos B = 4/5 fi sin B = 3/5. 43. log3 tan 1° tan 2° ... tan 44° tan 45° cot 44° ... cot 1° = log31 = 0 44. cotn q =

cos5 q sin3 q

¥

cos2 q cos q = cot8 q ¥ sin q sin 4 q

fin=8 45. Given equation is possible if sin q = sin 2q = sin 3q = sin 4q = 1 for some q, 0 < q < p which is not possible. 46. cos 11° < cos 2° fi cos 11° – cos 2° < 0, cos 11° and cos 2° are rational numbers. cos 11° – cos 2° is a negative rational number. 47. cos2 A = sin A tan A = sin2 A . sec A fi cot2 A = sec A. cot6 A – cot2 A = cot2 A(cot4 A – 1) = cot2 A(sec2 A – 1) = cot2 A ¥ tan2 A = 1 2 48. (cos x – sin x) = 1/4 fi 1 – sin 2x = 1/4 fi sin 2x = 3/4 fi cos 2x = 7 / 4 = tan 2x = 3 / 7 . \

49. (a) cos 2A = sin 3A = cos (p/2– 3A) fi 5A = p/2 = A = p/10

50.

n Êpˆ Êpˆ Ê 2p ˆ = sin 2 Á ˜ + cos2 Á ˜ + sin Á ˜ Ë ¯ Ë ¯ Ë 2n ¯ 4 2n 2n fi

Êpˆ n-4 sin Á ˜ = Ë n¯ 4

Êpˆ For n Œ N, sin Á ˜ > 0 Ë n¯ \

n-4 >0 4

or

n>4

p p Êpˆ < fi sin Á ˜ < 1 Ë n¯ n 2

For n > 4,

1 ( n - 4) < 1 fi 4 < n < 8 4 Thus, a possible value of n = 6. In fact, for n = 6 fi

0
0 and Ë2 4 ¯ cos x = cos Ê 1 + 2p - p ˆ > 0. Ë2 4¯ 1 7p is the smallest positive root of + 2 4 the equation. 60. 4 cos3 x – 4 cos2 x + cos x – 1 = 0 fi (4 cos2 x + 1) (cos x – 1) = 0 fi cos x = 1 fi x = 2np 100p < 315 < 101p Hence x =

Required sum= 2p + 4p + ... 100p = 2(1 + 2 + ... 50)p =

2 ¥ 50 ¥ 51 p = 2550p 2

61. When x = 1, 2 + 2 sin y = 0 fi sin y = – 1 7p p = sin Ê3p + ˆ = – 1 and sin Ë 2 2¯ So (1, 7p/2) is a solution. For x = – 1, sin y = 1 which is not satisfied by y = 7p/2 or 0. 1 (1) 62. esinx = 4 + sin x e – 1 £ sin x £ 1 and e > 1 e

–1

£ esinx £ e < 3.

esin x < 3, also esin x > 0. 1 \ + 4 > 4. sin x e So two sides of (1) cannot be equal for any real value of x. 63. 2 sin 3x cos 2x = sin 3x fi sin 3x (2 cos 2x – 1) = 0 fi sin 3x = 0 or 2 cos 2x = 1, fi x = 0, p/3, 2p/3, p or x = p/6, 5p/6 as x Œ [0, p]. 64. sin x, sin 2x, sin 3x are in A.P. If 2 sin 2x = sin x + sin 3x = 2 sin 2x cos x or if 2 sin 2x(1 – cos x) = 0 or if either sin 2x = 0 or cos x = 1 which is satisfied by x = 2np for which cos x = 1. 65. 2 sin2 ((p/2) cos2 x) = 2 sin2 ((p/2) sin 2x) or if (p/2) cos2x = ± (p/2) sin 2x or if either cos x = 0 or cos x = ± 2 sin x. and cos x = 0 if x = (2n + 1)p/2. fi

fi

4t 2

+ t2 +

2

=5 1+ t t2 fi t6 – 3t2 + 2 = 0 fi (t2 – 1) (t4 + t2 – 2) = 0 fi t2 = 1, t2 = – 2 so t = ±1, x = p/4, 3p/4

66. tan x = t

2

2 sin x cos x =n 2 fi sin2x + 2(1 – cos2x) = 2n fi sin2x – 2 cos 2x = 2(n – 1)

67. 2 sin2x +

fi

- 5 £ 2n - 2 £ 5

5 5 £ n £ 1+ 2 2 Integral values of n are 0, 1, 2. fi

1-

68. sin (2 sinx) = sin (p/2 fi 2 sinx = fi sinx + cosx = fi 1 + sin2x =

– 2 cosx) p/2 – 2 cosx p/4 p2/16

25.38

Complete Mathematics—JEE Main

2 32 tanx + cotx = sin 2 x = 2 p - 16 a = 32, b = 16, c = 2 4 69. 15 tan a + 10 = 6(1 + tan2a)2 fi 9 tan4a – 12tan2a + 4 = 0 fi tan2a = 2/3. so 8 cosec6 a – 27 sec6 a = 8(1 + 3/2)3 – 27(1 + 2/3)3 = 0 •

70.

 tan - 1

r =1

from statement-2 77. If a + b + g = p/2 tan (a + b) = tan (p/2 – g) =

•

1 2r 2

=

 ÈÎtan - 1 (2r + 1) - tan - 1 (2r - 1)˘˚ 1

= p/2 – p/4 = p/4 71. [x] + {x} = x and [{x}] = 0. 2 sin 2 (q / 2 ) ¥ sin q 1 - cos q sec q – 1 = = cos q cos q ¥ 2 sin (q / 2 ) cos (q / 2 ) = tan q tan (q/2) Given equation reduces to cosq = 1 fi q = 2np, n Œ I Number of solutions in [– 2011p, 2011p] is 2011. 2p 4p 6p ˆ Ê + cos + cos ˜ 72. x2 + y2 = 3 + 2 Á cos Ë 7 7 7¯ = 3 + 2 (–1/2) = 2 (Ref. Ex 49,) px

(

= =

= x- 3

2

)

+1 fi 2 3 74. cosq cos2 q … cos 2n – 1q =

x=

3.

1 sin 2 n q ) = n sin (p + q ) ( 2 sin q sin q 1

2

1

(- sin q ) = -

1

. 2 sin q 2n So statement-2 is true which implies statement-1 is also true. 75. f(x) = tan x, g(x) = tan (x/2) fi f(x) + g(x) π 0 for all x. so statement-2 is False. we have [cosa + cos (a + b)]2 + [sina + sin (a + b)]2 = 1 n

1 . tang

fi

1 tan a + tan b = 1 - tan a tan b tan g

fi fi

tan a tan b + tan b tan g + tan g tan a = 1 statement-2 is False.

In statement-1,

(

p+ q+ r

2

)

=p+q+r+2

(

pq + qr + rp

)

£ p + q + r + 2(p + q + r) = 3(p + q + r) = 48 78. Statement-2 is True. Statement-1 fi y cosq . sin2 q + y cos3 q = sin q cos q fi y cos q = sin q cos q fi y = sin q fi x = cos q fi x2 + y2 = 1 fi statement-1 is also true but does not follow from statement-1.

1 È2 sin q cos q cos 2q K cos 2 n - 1q ˘˚ 2 sin q Î n

cos q, y = 3 3 sin q then x – y = 3 3 (cos q – sin q) = 3 3 ¥ 2 cos (q + p/4) £ 3 6 follows

= tan - 1 3 - tan - 1 1 + tan - 1 5 - tan - 1 3 + K

73. sin

76. Statement-2 is True. In statement-1, x = 3 3

fi 2 + 2 cos (a – a – b) = 1 fi cos b = –1/2 fi b = 2p/3. Similarly g = 2p/3. fi f(b) + g(g ) = tan (2p/3) + tan(p/3) = 0 fi statement-1 is true.

79. x2 – x – 2 = (x – 2) (x + 1) < 0 fi – 1 < x < 2 fi Statement-2 is True 2 sin2 x + 3 sin x – 2 = (2 sin x – 1) (sin x + 2) > 0 fi sin x > 1/2 which is true if p /6 < x £ p /2 11 11 i.e. 0 3 cos 2x > 0

1 3 1 sin 2x + cos 2x > – 2 2 2

fi cos (2x – p /6) > –

1 2

1 9

=

1 2 2

(

1 2 2

)

3 +1

(

)

3 +1 + 1

3 +1+ 2 2

=

6+ 2+4 8

2¥2 2 87. We have A = a cos2q + 2 b sinq cosq + c sin2q B = a sin2q – 2b sinq cosq + c cos2q H = (c – a) sin2q + 2b cos2q So A + B = a + c 88. Sin5q = sin (2q + 3q) = sin 2q cos 3q + cos 2q sin 3q = 2 sinq cosq ¥ cosq (4cos2q – 3) + (1 – 2 sin2q) (3 sinq – 4 sin3q) = 2 sinq (1 – sin2q) (1 – 4 sin2q) + (1 – 2 sin2q) (3 sinq – 4 sin3q) = 16 sin5q – 20 sin3q + 5 sinq So a = 16, b = – 20, c = 5, d = 0 fi 5a + 3b – 4c = 0 cot A cot B + 1 y/ x +1 x + y = = 89. cot (A – B) = cot B - cot A y xy 90.

= cos B.

fi 1 – cos (2x + p /2) +

1 xy = 1 1 xy - 1 1- ◊ x y

cos x fi x = p /6 86. Let q = 7.5° fi 2q = 15° = 45° – 30°

A

D

cos2 x

1 1 1 fi tan4x = fi tan x = 9 9 3

fi 2 cos2q =

h

2

4 cos x - 4 sin x cos x fi

82. Let ACB = q

h - l 2 - h2

=

2

1

fi z(xy – 1) = xy = x + y fi x + y + z = xyz.

x b = fi x2 = ab a x

fi

sin x

,y=

B

x

Fig. 25.5

fi

2

2

q

B

1

1

and z = A

=

2

=

+

sin 3q sin 3 q

4 cos3 q - 3 cos q

=4¥3

cos3 q

+

3 sin q - 4 sin3 q sin3 q

(cos2 q - sin2 q ) 2

2

4 sin q cos q

=

12 cos 2q 1 ¥ sin 2q sin 2q

25.40

Complete Mathematics—JEE Main

91. The given expression is equal to x2 + y2 + xy [(tan2a + cot2a) – 4 cot22a] = x2 + y2 + xy 2 È 2 2 cos2 a - sin 2 a ˘ Í1 - 2 sin a cos a - 4 ˙ Í sin 2 a cos2 a 4 sin 2 a cos2 a ˙ Î ˚ 2 2 = x + y + xy

(

)

2

1 x 1 x tan + 2 tan 2 + L 2 2 2 2 1 x + n - 1 tan n- 1 2 2 1 x = cot x – 2cot 2x + (cot – 2 cot x) 2 2 so tan x +

+

2

È 1 1 - 4 sin a cos a ˘ -2Í 2 ˙ 2 sin 2 a cos2 a ˚ Î sin a cos a 2

2

= x + y + 2xy = (x + y)

+L+

2

92. S1 = tan A + tan B + tan C = 4/3 S2 = Âtan A tan B = 1, S3 = tan A tan B tan C = – 1/3 S - S3 tan (A + B + C) = 1 Æ• 1 - S2

c which satisfies the given equation a+b (Note c π 0) and the required value is 0. fi sina =

2

= (a + b)2 [6 ab + 8 b2 + a2 + b2] = (a + b)2 (a + 3b)2 97. sin A = sin B and cos A = cos B fi A = 2np + B where n is an integer, n π 0 fi

A- B A-B = np fi tan =0 2 2

98. Apply the identity, if A + B + C = 180° then cos A + cos B + cos C = 1 + 4 sin (A/2) sin (B/2) sin (C/2) 99. We can write tan x = = cot x – 2 cot 2x

tan 2 x - 1 + 1 tan x

n -1

1 2

n -1

xˆ ˜ 2¯

Ê x x ˆ ÁË cot 2 n - 1 - 2 cot 2 n - 2 ˜¯

Ê x ˆ cot Á n - 1 ˜ - 2 cot 2 x Ë2 ¯

Note Remember tan x + 2cot 2x = cot x 100. We can write =

4 + cot 76∞ cot 16∞ - 1 cot 76∞ + cot 16∞

4 sin 76∞ sin 16∞ + cot 92∞ sin 92∞

2 (cos 60∞ - cos 92∞) + cos 92∞ 1 - cos 92∞ = sin 92∞ sin 92∞ = cot 46° = tan 44° =

Ê 1 1 1ˆ 101. K Á + + ˜ = cos q + cos (q + 2p /3) Ë x y z¯ + cos (q + 4 p /3) = cos q + 2 cos (q + p) cos p/3 = cos q – cos q = 0 fi xy + yz + zx = 0 102. Let tan A tan B = x so tan (A + B) = tan (p /3) = 3

96. (1 + sinq) (3 sinq + 4 cosq + 5) = (cos(q/2) + sin(q/2))2 [6 sin(q/2) cos(q/2) + 4 (2 cos2 q/2 – 1) + 5]

1

=

fi A + B + C = p /2 93. We have sinq [x + 2 y cos q + z (3 – 4 + 4 cos2q)] = 4 cosq (2 cos2q – 1) sinq fi 8 cos3q – 4 z cos2q – 2( y + 2) cosq = x – z (Q sinq π 0) 5 3 94. We have 16 sin x – 20 sin x + 5 sin x = 5sinx (see Ex 68) fi sin3x (16 sin2x – 20) = 0 fi sin x = 0, only one value 95. sin a + sin b = – b/a, sin a sin b = c/a, sin a + 2 sin b = 1

1 Ê x cot 2 - 2 cot 2 Á 2 Ë 2

fi fi

tan A + tan B = 3 1 - tan A tan B 3 (1 – x) = tan A + tan B £ 2 tan A tan B = 2 x

fi 3(1 – x)2 £ 4x fi 3x2 – 10 x + 3 £ 0 fi (3x – 1) (x – 3) £ 0 1 fi £x£3 3 103. tanq (3tanq + 3) = (2 + 2 tanq)2 fi (1 + tan q) (4 + tan q) = 0 fi tan q = –4 as tan q π – 1 7 - 5 cot q 7 + 5 / 4 33 = So = 9 + 16 100 9 - 4 sec 2 q - 1 104. We have sin2q + sin4q + sin6q + L – (sinq cosq + sin2q cos2q + sin3q + cos3q + L)

Trigonometrical Ratios, Identities and Equations 25.41

sin 2 q

sin q cos q = 2 1 - sin q 1 - sin q cos q =

1 - cos 2q sin 2q 1 + cos 2q (2 - sin 2q )

=

1 - ab 1 - a2 + 1 + ab 3 - a2

(

105. 1 + 1 + x fi

)

tana = 1 - 1 - x

1+ 1+ x

=

1- 1- x

cos a 2 cos a (cos a + sin a ) = sin a 2 sin a (cos a + sin a )

2 cos2 a + 2 sin a cos a

= =

2 sin a cos a + 2 sin 2 a

fi 2 sin

1 + cos 2a + sin 2a sin 2a + 1 - cos 2a

fi sin 2

1 + (cos 2a + sin 2a )

=

fi cos2 x + 4 cos x – 5 = 0 fi cos x = 1 fi x = 2np Note The answer could be obtained by trial method, taking x = 2np 109. sin x + 2 sin 2x – sin 3x = 3 fi sin 3x ≥ 0 fi 0 £ 3x £ p fi 0 £ x £ p /3 fi 0 £ sin x £ 3 / 2 and 0 £ 2x £ 2p /3 fi 0 £ sin 2x £ 1 – 1 £ – sin 3x £ 0 So – 1 £ sin x + sin 2x – sin 3x £ 3 / 2 + 1 < 3 So the given equation is not satisfied for any real value of x 110. sin x + sin y = sin (x + y) x+y x-y x+y x+y cos cos = 2 sin 2 2 2 2

x+y x+ y Ê x- cos ÁË cos 2 2 2

fi either sin

2

1 - (cos 2a - sin 2a )

Now sin

1 + (1 + sin 4a )

=

1 - 1 - sin 4a 3

106. f (q ) = sinq (sinq + 3 sinq – 4 sin q) = 4 sin2q (1 – sin2q) = sin2 2q ≥ 0 for all real q.

and cos

x+ y = 0 and |x| + |y| = 1 2

x+ y x- y = cos 2 2

fi sin

x y sin =0 2 2

fi sin

x y = 0 or sin =0 2 2

h

l

x+ y x+ y x- y = 0 or cos = cos 2 2 2

Ê 1 1ˆ Ê 1 1ˆ (x, y) = Á , - ˜ or Á , - ˜ Ë 2 2¯ Ë 2 2¯

107. l cot B + l cot C = h = 2 2 l C

yˆ ˜¯ = 0

fi (x, y) = (1, 0), (– 1, 0), (0, 1), (0, – 1) A

B

Fig. 25.7

Previous Years' AIEEE/JEE Main Questions

fi cot B + cot C = 2 2 B + C = p /2 fi cot B cot C = 1 fi cot B =

2 - 1 , cot C =

2 +1

The required angles are p /8 and 3p /8 108.

2

2

cos x - sin x + cos x + sin x = 2 sin x + cos x fi

cos x + sin x ÈÎ cos x - sin x + cos x + sin x - 2 ˘˚

=0 fi either sin x + cos x = 0 or

So the required number of pairs is 6.

cos x - sin x + cos x + sin x = 2

fi 2 cos x + 2 cos 2 x = 4 fi 2 cos2 x – 1 = (2 – cos x)2 = 4 – 4 cos x + cos2 x

1. tan x + sec x = 2 cos x fi

1 + sin x = 2 cos2 x = 2 (1 – sin2 x)

fi

(1 + sin x) (1 – 2 + 2 sin x) = 0

fi

sin x = –1 or sin x =

But sin x = –1 Thus sin x =

fi

1 2 cos x = 0. This is not possible

1 p 5p fix= , Œ[0, 2p ] 2 6 6

-21 -27 , cos a + cos b = 65 65 Squaring and adding we get

2. sin a + sin b =

Complete Mathematics—JEE Main

25.42

2 + 2 cos (a – b) = fi

441 + 729 2

(65)

=

1170 2

(65)

=

18 65

fi

p a - b 3p < < 2 2 2 a -b -3 cos = 2 130

=

2

2

2

2

Ê a 2 + b2 ˆ Ê a 2 - b2 ˆ cos 2 2q u =a +b + 2 Á ˜ Á ˜ Ë 2 ¯ Ë 2 ¯

and

q=

2

2

q–p=

1 3 1 3 3 3

+ (2 - 3) (2 - 3) =

2 3

-1

+ 2 - 3 -1 = 1

2 + q – p =3

Thus, cos x < 0 fi

p /2 < x < p

Also, (sin x – cos x)2

2

2

2

2

2

2

and

2

Êa +b ˆ Êa -b ˆ -Á and min (u2) = a 2 + b 2 + 2 Á ˜ Ë 2 ¯ Ë 2 ˜¯ 2

7 2 Since sin x + cos x = 1/2

fi sin x – cos x =

2 sin x = ( 7 + 1) / 2

2

Thus max (u ) = a + b + a + b = 2(a + b ) 2

= 1 + 3/4 = 7/4

We get

2

2

–p =

and

= sin2 x + cos2 x – 2 sin x cos x

Squaring we get

2

\

3

As 0 < x < p, sin x > 0.

a +b a -b cos 2q + 2 2

2

1

7. 2 sin x cos x = (sin x + cos x)2 – 1 = 1/4 – 1 = –3/4

a 2 + b2 a 2 - b2 cos 2q + 2 2

2

x = p/6, 5p/6, 13p/6, 17p/6.

\

a2 b2 + (1 - cos 2q ) + (1 + cos 2q ) 2 2 2

fi

Thus

a2 b2 (1 + cos 2q ) + (1 - cos 2q ) 2 2

2

[Q sin x π –3]

b = 2- 3.

3. We have u=

sin x = 1/2

6. Let a = tan 30°, b = tan 15°, then a =

a - b 18 = 2 ¥ 2 cos 2 65 2

a -b 3 =± fi cos 2 130 p < a – b < 3p fi

fi

2 cos x = (1 - 7) / 2

2

2

\

tan x = -

7 +1 7 -1

=-

( 7 + 1) 2 7 -1

fi

–2 + 1 £ 2 sin (x – p/3) + 1 £ 2 + 1

fi

–1 £ f (x) £ 3

1 = - (4 + 7) 3 8. cos (b – g) + cos (g – a) + cos (a – b) = –3/2 fi 2[cos b cos g + cos g cos a + cos a cos b + sin b sin g + sin g sin a + sin a sin b] + (cos2 a + sin2 a) + (cos2 b + sin2 b) + (cos2 g + sin2 g) = 0 fi (cos a + cos b + cos g)2 + (sin a + sin b + sin g)2 =0 O fi cos a + cos b + cos g = 0 and sin a + sin b + sin g = 0 p/n Thus, both A and B are true.

fi

S = [–1, 3]

9. We have

= a + b + 2ab = (a + b) 2

2

2

So the required difference = 2(a + b ) – (a + b) = (a – b)2. 4. f (x) = sin x - 3 cos x + 1 È1 ˘ 3 = 2 Í sin x cos x ˙ + 1 2 Î2 ˚ = 2 sin (x – p/3) + 1 –2 £ 2 sin (x – p/3) £ 2

p r = cos ÊÁ ˆ˜ Ë n¯ R

5. 2 sin2 x + 5 sin x – 3 = 0 fi

(2 sin x – 1) (sin x + 3) = 0

R

When

r p/2

A1

M

Fig. 25.8

A2

Trigonometrical Ratios, Identities and Equations 25.43

r 3 = , we get 2 R p p cos ÊÁ ˆ˜ = cos ÊÁ ˆ˜ Ë n¯ Ë 6¯ fi

sin(P + Q) = 1/2

fi

sinR = 1/2 R = 5p/6

fi

n =6

fi

p = BD

and cos f =

R = p/6 or 5p/6 P < p/6

fi

3sinP < 3/2

Now, cos (a + b) = 4/5 tan (a – b) = ± 5/12

We have tan (2a) = tan [(a + b) + (a – b)] tan (a + b ) + tan (a - b ) = 1 - tan (a + b ) tan (a - b )

=

3 5 14 + 56 4 12 = = 12 = 3 5 33 33 1- Ê ˆ Ê ˆ Ë 4 ¯ Ë 12 ¯ 48

=

11. A = sin2 x + cos4 x £ sin2 x + cos2 x = 1 4

q

2

[Q 0 £ cos x £ 1 fi cos x £ cos x] Also, A = 1 – cos2 x + cos4 x = (cos2 x – 1/2)2 + 3/4 ≥ 3/4 3 £ A £1 4 12. We have

2

q

D q

C

f p f

A

p 2 + q 2 sin q sin q cos j + cos q sin j

B

Fig. 25.9

( p 2 + q 2 ) sin q p cos q + q sin q

tan A cot A + 1 - cot A 1 - tan A

15.

5 Note tan (a – b) = does not satisty any of the 12 choices. 2

p + q2

sin (p - q - f ) sin q = BD AB sin q BD AB = sin (q + j )

tan (a + b) = 3/4 fi

p 2

p + q2 By the law of sines

fi 0 £ a + b £ p/2 and –p/4 £ a – b £ p/4

=

tan 2 A 1 tan A - 1 tan A ( tan A - 1)

=

tan 3 A - 1 tan 2 A + tan A + 1 = tan A ( tan A - 1) tan A

=

sec2 A + 1 = sec A cosec A + 1 tan A

16. 2 cos2q – 3 sinq = 0

Thus,

fi

2 sin2q + 3 sin q – 2 = 0 p 5p 1 sin q = , fi 6 6 2 Statement-2 is True.

sin q + sin(4q) + sin(7q) = 0

fi

fi

2 sin(4q) cos(3q) + sin(4q) = 0

fi

fi

sin(4q) [2cos(3q) + 1] = 0

Next, 2 sin2q – cos 2q = 0

fi

sin(4q) = 0 or cos(3q) = –1/2 = cos(2p/3)

fi

fi

q = p/4, p/2, 3p/4 or 3q = 2p/3, 2p ± 2p/3

fi

q=

2p p p 4p 3p 8p , , , , , 9 4 2 9 4 9

13. We have (3sinP + 4cosQ)2 + (3cosP + 4sinQ)2 = 62 + 1 fi

sin(p – R) = 1/2

3sinP + 4cosQ < 6, so R = p/6.

sin f =

10. 0 £ a, b £ p/4

and sin (a – b) = 5/13

fi

fi

14. Let –BDC = f, then –DBA = f. We have

r 3 p = = cos when R 3 n there is no value of n

fi

fi

9(sin2P + cos2P) + 16(cos2Q + sin2Q) + 24 (sin P cosQ + cosP sinQ) = 37

1 2 Which gives four values of q in [0, 2p] 4 sin2q = 1

fi

sin q = ±

2 cos2q – 3 sin q = 0 gives two values of q in [0, 2p] So, Statement-1 is also true but does not follow from Statement-2. 17. sin 2x – 2 cos x + 4 sin x = 4 fi

(2 cos x + 4) (sin x – 1) = 0

25.44

fi

Complete Mathematics—JEE Main

sin x = 1

fi

x=

= 2 + cos(2q) + sin(2q)

p 5p 9p , , 2 2 2

1 sin k x + cos k x k 1 fi f4(x) – f6(x) = sin 4 x + cos 4 x 4 1 - sin 6 x + cos6 x 6 1 1 - 2sin 2 x cos 2 x = 4 1 - 1 - 3sin 2 x cos 2 x 6 1 = 12 3 2 19. 2 sin a – 7 sin a + 7 sin a = 2 18. f k ( x) =

(

p = 2 + 2 sin ÊÁ 2q + ˆ˜ Ë 4¯

)

(

A = Max( f (q )) = 2 + 2

)

(

(

B = Min( f (q )) = 2 - 2

)

)

(

)

fi

2(sin3a – 1) – 7 sin a (sin a – 1) = 0

fi

(sin a – 1) [2(sin2a + sin a + 1) – 7 sin a] = 0

fi

(sin a – 1)[2 sin2a – 5 sin a + 2] = 0

fi

(sin a – 1) (2sin a – 1) (sin a – 2) = 0

fi

a = p/6, 5p/6, p/2 Π[0, 2p]

(

\ ( A, B ) = 2 + 2, 2 - 2

23. cos a + cos b = 3/2 a +b a -b 3 cos = fi 2 cos 2 2 2 1 2 a +b a -b 1 cos = fi 2sin 2 2 2 From (1) and (2), we get

24. 2 + 3 =

cos 2 (q 2) - sin 2 (q 2) cos q = = 1 + 2sin (q 2) cos (q 2) 1 + sin q fi

(2)

1 1 tan q = , where q = (a + b ) 3 2

\ There are 3 solutions.

cos (q 2) - sin (q 2) = cos (q 2) + sin (q 2)

(1)

and sin a + sin b =

sin 2q =

p q cot (q 2) - 1 20. cot ÊÁ + ˆ˜ = Ë 4 2 ¯ cot (q 2) + 1

)

2 tan q = 1 + tan 2 q

1 2Ê ˆ Ë 3¯ 1 1+ Ê ˆ Ë 3¯

2

=

2 9 3 ¥ = 3 10 5

a sin A = b sin B sin (120∞ - B ) sin B

fi

2+ 3 =

fi

(2 + 3) sin B = sin 120° cos B – cos 120° sin B

fi

(2 + 3 + cos(180∞ - 60∞))sin B = sin (180° – 60°) cos B

p q 1 - sin q cosec q - 1 q cot 2 ÊÁ + ˆ˜ = = = Ë 4 2 ¯ 1 + sin q cosec q + 1 p

fi

Ê 2 + 3 - 1 ˆ sin B = 3 cos B ÁË ˜ 2¯ 2

p q cot ÊÁ + ˆ˜ = Ë 4 2¯

fi

tan B =

q p

21. 2 cos q + sin q = 1 fi 4 cos2 q = (1 – sin q)2 fi 4(1 – sin2 q) = (1 – sin q)2 fi

4(1 + sin q) = 1 – sin q 3 fi sin q = 5 From (1) and (2) cos q = 4/5

(1)

fi

3

=

1

= 2- 3 3+ 2 3 2+ 3 B = 15° [use tan 15° = tan (60° – 45°)]

\ A = 105° (Q q π p/2) (2)

fi

(A, B) = (105°, 15°)

25. cos x + cos 2x + cos 3x + cos 4x = 0 fi (cos 4x + cos x) + (cos 3x + cos 2x) = 0 fi 2 cos

x 5x 3x 5x cos + 2 cos cos = 0 2 2 2 2

fi 2 cos

x 5x È 3x cos + cos ˘˙ = 0 Í 2 Î 2 2˚

fi 2 cos

x 5x È 2 cos x ◊ cos ˘˙ = 0 Í 2 Î 2˚

Thus, 7 cos q + 6 sin q 4 3 = 7 ÊÁ ˆ˜ + 6 ÊÁ - ˆ˜ = 2 Ë 5¯ Ë 5¯ 22. f (q) = 1 + cos2 q – sin2 q + 1 + sin 2q

Trigonometrical Ratios, Identities and Equations 25.45

x 5x = 0, cos x = 0 or cos = 0 2 2 x p p 3p 5 x p 3p 5p 7p 9p fi = ;x = , ; = , , , , 2 2 2 2 2 2 2 2 2 2 p 3p p 3p 7p 9p fi x = p, , , , , , 2 2 5 5 5 5 1 26. Let E = 4 + sin 2 2 x - 2 cos 4 x 2 1 2 1 = 4 + sin 2 x - (2 cos 2 x)2 2 2 1 1 = 4 + 1 - cos 2 2 x - (1 + cos 2 x )2 2 2 1 = 4 + (1 + cos 2 x )[1 - cos 2 x - 1 - cos 2 x ] 2 1 = 4 - (1 + cos 2 x )(2 cos 2 x ) 2 = 4 – cos 2x – cos2 2x fi cos

(

=

)

2 17 Ê 1 - Á + cos 2 xˆ˜ ¯ 4 Ë2

Note that 17 9 - = 2 when cos 2x = 1 4 4 17 9 when cos 2x = –1/2 fi M – m = and M = 4 4

m=

27. Let a = 2sin 4 x + 18cos 2 x , b = 2 cos 4 x + 18sin 2 x Now, 2a2 = (2sin2 x)2 + 18(2cos2x) 2

= (1 – cos 2x) + 18(1 + cos 2x)

When a – b = –1, we get the same 8 solutions in [0, 2p] 28. Using C1 Æ C1 + C2 + C3, we get 1 sin x sin x D = (2sin x + cos x ) 1 cos x sin x 1 sin x cos x Using C2 Æ C2 – (sin x)C1 and C3 Æ C3 – (sin x)C1, we get ü D = (2sin x + cos x ) 1 cos x - sin x 0 1 0 cos x - sin x = (2sin x + cos x)(cos x – sin x)2 1 \ D = 0 fi tan x = - ,1 2 As –p/4 £ x £ p/4, –1 £ tan x £ 1 and tan x is one-to-one in the interval [–p/4, p/4]. Thus, there are two values of x. p p 29. Let A = +q , B = -q 12 12 where –p/12 < q < p/12 We have S = tan A + tan B sin ( A + B ) 2sin (p 6) = = cos A cos B cos ( A + B ) + cos ( A - B ) =

2

2 (1 2)

(

)

3 2 + cos (2q )

= 19 + 16 cos 2x + cos 2x 2

and 2b = (2cos2x)2 + 18(2sin2x) = (1 + cos 2x)2 + 18(1 – cos 2x) = 19 – 16cos 2x + cos2 2x We have |a – b| = 1 and 2(a2 – b2) = 32cos 2x If a – b = 1, 2(a – b)(a + b) = 32 cos 2x fi a + b = 16 cos 2x \ 2a = 1 + 16 cos 2x fi 4a2 = (1 + 16 cos 2x)2 = 1 + 32 cos 2x + 256 cos2 2x fi 38 + 32 cos 2x + 2 cos2 2x = 1 + 32 cos 2x + 256 cos2 2x 37 254 which gives us 8 solutions in [0, 2p].

fi 37 = 254 cos 2 2 x fi cos 2 x = ±

2

=

3 + cos (2q )

As –p/6 < 2q < p/6 fi

3 /2 < cos (2q) £ 1

fi

3 < 2 cos (2q) £ 2

fi

3 + 3 < 3 + 2 cos (2q ) £ 3 + 2

fi

1 3 + 2 cos (2q )

≥

1 2+ 3

30. q Œ P ¤ sin q - cos q = 2 cos q ¤ sin q = ( 2 + 1) cos q ¤ sin q =

1 2 -1

cos q

¤ ( 2 - 1)sin q = cos q

=2–

3

25.46

Complete Mathematics—JEE Main

fi 0 £ a + b £ p/2

¤ 2 sin q = sin q + cos q ¤ q ŒQ \P=Q

fi

Now cos (a + b) = 4/5 and sin(a – b) = 5/13 we have

Previous Years' B-Architecture Entrance Examination Questions cos2 q – 2 cos q + 1 = 0 fi

=

tan (a + b ) + tan (a - b ) 1 - tan (a + b ) tan (a - b )

=

(3 4) + (5 12) 14 12 56 = = 1 - (3 4)(5 12) 33 48 33

cosn q + secn q = 2

3. (sin2x + cos2x)3 = sin6x + cos6x + 3sin2x cos2x (sin2x + cos2x)

Same as Q10 AIEEE/JEE Main 7. 1 + a sin 2b = sin2b + cos2b – 2 sinb cosb cos A

fi A is obtuse.

4 sin2 2x = (1 - p ) 3

fi

4 0 £ (1 - p ) £ 1 3 1 1 £ p £ 1 fi p Œ ÍÈ ,1˙˘ 4 Î4 ˚

fi fi

p 3p 5p 7p + cos 4 + cos 4 + cos 4 4. cos 8 8 8 8 p p 3 = 2 ÁÊ cos 4 + cos 4 ˜ˆ Ë 8 8¯

A

fi cos A = –a

3 1 = p + sin 2 2 x 4

fi

tan(a – b ) = 5/12

co

b

cos q = sec q = 1

fi

tan (a + b ) = 3/4

sb

sin

fi

fi

fi

tan2a = tan[(a + b) + (a – b)]

1. See Question No. 1 in AIEEE/JEE Main Questions 2. cos q + sec q = 2

–p/4 £ a – b £ p/4

fi A = 120° 1 2

fi a = - cos120∞ =

B

1 + a sin 2b

8. cos 25° + sin 25° = k

C

Fig. 25.10

Squaring we get 1 + sin 50° = k2

4

=

fi

p 3p 2p cos cos 7 7 7 1 2p 2p 3p ◊ sin cos cos = p 7 7 7 2sin 7

2

)

= tan 9° + tan (90° – 9) – (tan 27° + tan(90° – 27°)) = tan 9° + cot 9° – (tan 27° + cot 27°) =

10.

2 2 2¥4 2¥4 = =4 sin18∞ sin 54∞ 5 -1 5 +1 a+b a-b + a-b a+b =

a+b+a-b 2

a -b

2

2a

=

a 1-

= cos

=

=

1 p 4sin 7

sin

4p 3p cos 7 7

Èsin 7p + sin p ˘ = 1 p ÍÎ 7 7 ˙˚ 8 8sin 7 1

6. 0 £ a, b £ p/4

= k 2 - k2

9. tan 9° – tan 27° – tan 63° + tan 81°

p ˆ2 Ê 1 ÈÊ 3p ˆ 2 ˘ + + + 1 cos 1 cos ˜ ÁË ˜ ˙ ÍÁ 2 ÎË 4¯ 4¯ ˚

1È 1 ˆ2 Ê 1 ˆ2˘ 1 Ê 1ˆ 3 + Á1 = ÍÊÁ1 + ˜ ˜¯ ˙ = ¥ 2 ÁË1 + ˜¯ = Ë ¯ Ë 2Î 2 2 2 2 ˚ 2 p 3p 5p sin 5. sin sin 14 14 14 p 3p p 2p p p = sin ÊÁ - ˆ˜ sin ÊÁ - ˆ˜ sin ÊÁ - ˆ˜ Ë2 7 ¯ Ë2 7 ¯ Ë 2 7¯

(

cos 50∞ = 1 - k 2 - 1

=

2 2

1 - tan q

=

b2 a2

2 cos q cos 2q

11. Same as Question No. 9. 1 a b 12. 1 c a = 0 1 b c fi a2 + b2 + c2 – (ab + bc + ca) = 0 fi (a – b)2 + (b – c)2 + (c – a)2 = 0 fia=b=c

fi

A = B = C = 60°

Trigonometrical Ratios, Identities and Equations 25.47 2

Ê 3ˆ 9 fi sin2A + sin2B + sin2C = 3 Á ˜ = 4 Ë 2 ¯ 2 cos q sin q + 1 ± (cos q + sin q ) = sin q sin q = ± [cot q + 1] = k – cot q

13. L.H.S. =

fi k = –1 14. f (x) = 2 sin x + sin 2x = 2 sin x + 2 sin x cos x = 2 sin x (1 + cos x) f ¢(x) = 2 cos x (1 + cos x) – 2 sin x sin x = 2 [cos x + cos2x – (1 – cos2 x)] = 2 [2cos2x + cos x – 1]

3 3. 2 15. Using C1 Æ C1 + C3, we get Thus, maximum value is

1 tan q + sec2 q cos q D= 0 0 -4

3 sin q 3

= 3cos q + 4sin q dD = -3sin q + 4 cos q dq dD sin q cos q 1 =0fi = = dq 4 3 5 fi sin q = 4/5, cos q = 3/5.

= 2 (cos x + 1) (2 cos x – 1) Now, f ¢(x) = 0

fi

x = p/3, p (Q 0 £ x £ 3p/2)

Ê 3p ˆ We have f (0) = 0, of Á ˜ = – 2 Ë 2¯ Ê 3ˆ p 3 3 f ÊÁ ˆ˜ = 2 Á ˜ + = 3, Ë 3¯ 2 2 2 Ë ¯ Ê 3p ˆ

and f (p) = 0, f Á ˜ = - 2 Ë 2¯

Êpˆ

Max D = max {D(0), D(sin–1 (4/5)), D ÁË ˜¯ } 2 24 24 = max 3, , 4 = 5 5 min D = 3

{

\ D Π[3, 5]

}

CHAPTER TWENTY SIX

INVERSE TRIGONOMETRIC FUNCTIONS AND FORMULAE Function

Domain

y = sin–1 x y = cos–1 x y = tan–1 x y = cot–1 x y = sec–1 x

–1 £ x £ 1 –1 £ x £ 1 –• 0, and x2 + y2 >1

2 2 = p – sin–1 ÎÈ x 1 - y - y 1 - x ˘˚

if 0 < x £ 1, –1 £ y < 0, x2 + y2 >1

2 2 = – sin–1 ÎÈ x 1 - y - y 1 - x ˘˚ – p

if –1 £ x < 0, 0 < y £ 1, x2 + y2 >1

(iii) cos–1 x + cos–1 y 2 2˘ È = cos–1 Î xy - 1 - x 1 - y ˚

if –1 £ x, y £ 1, x + y ≥ 0

2 2 = 2p – cos–1 ÈÎ xy - 1 - x 1 - y ˘˚

(iv)

if –1 £ x, y < 0, x + y £ 0

cos–1 x – cos–1 y

(

= cos–1 xy + 1 - x 2 1 - y 2

)

(

= – cos–1 xy + 1 - x 2 1 - y 2

if –1 £ x, y £ 1, x £ y

)

Ê x+ yˆ (v) tan–1 x + tan–1 y = tan–1 Á Ë 1 - xy ˜¯

Illustration

if –1 £ y < 0, 0 < x £ 1, x ≥ y if x y < 1

Ê x+ yˆ = p + tan–1 Á Ë 1 - xy ˜¯

if x y > 1, x > 0, y > 0

Ê x+ yˆ –p = tan–1 Á Ë 1 - xy ˜¯

if xy > 1, x < 0, y < 0

3

Find the value of tan–1 (1/2) – tan–1 (– 3) – tan – 1 (– 7) Solution: –3 < 0, – 7 < 0, (–3)(–7) > 1 So tan–1 (1/2) – tan–1 (–3) – tan–1 (–7) -3-7 ˘ +p=p = tan–1 (1/2) – Ètan -1 ÍÎ 1 - (-3) (-7) ˙˚

Ê x- yˆ (vi) tan–1 x – tan–1 y = tan–1 Á Ë 1 + xy ˜¯

if xy > –1

Inverse Trigonometric Functions 26.3

Ê x- yˆ = tan–1 Á –p Ë 1 + xy ˜¯

if x < 0, y > 0, xy < –1

Ê x- yˆ = p + tan–1 Á Ë 1 + xy ˜¯

if x > 0, y < 0, xy < –1

Ê1 + xˆ p = + tan -1 x (vii) tan–1 Á Ë 1 - x ˜¯ 4 = tan–1 x –

3p 4

if x > 1

Ê1 - xˆ p = - tan -1 x (viii) tan–1 Á Ë 1 + x ˜¯ 4

if x ≥ –1

p 4

if x < –1

= tan–1 x – (f)

if x £ 1

(i) 2sin–1 x = sin–1 (2x

1 - x2 )

= p – sin–1 (2x

= sin–1 (2x –1

–1

1 - x2 )

1 - x2 ) – p 2

(ii) 2 cos x = cos (2x – 1) = p – cos–1 (2x2 – 1)

Illustration

if –

1 2 if

£x£ 1 2

1 2

1

Ê 2x ˆ = tan–1 Á –p Ë 1 - x 2 ˜¯

if x < –1

26.4

Complete Mathematics—JEE Main

Ê 2x ˆ 2 tan–1 x = sin–1 Á Ë 1 + x 2 ˜¯

(iv)

Ê 2x ˆ = p + sin–1 Á Ë 1 + x 2 ˜¯

if x > 1

Ê 2x ˆ = sin–1 Á –p Ë 1 + x 2 ˜¯

if x < –1

Ê 1 - x2 ˆ (v) 2 tan–1 x = cos–1 Á Ë 1 + x 2 ˜¯ Ê 1 - x2 ˆ = - cos - 1 Á Ë 1 + x 2 ˜¯ (g)

if –1 £ x £ 1

if 0 £ x < •

if - • < x £ 0

(i) 3 sin–1 x = sin–1 (3x – 4x3)

if –

(ii) 3 cos–1 x = cos–1 (4x3 – 3x)

if

Ê 3 x - x3 ˆ (iii) 3 tan–1 x = tan–1 Á Ë 1 - 3x 2 ˜¯

if –

Illustration

5

Find the values of 1 2 1 sin Ê 3 sin -1 Ê ˆ ˆ , cos Ê 3 cos-1 Ê ˆ ˆ , tan Ê 3 tan -1 Ê ˆ ˆ Ë Ë 2¯¯ Ë Ë 3¯ ¯ Ë Ë 3¯ ¯ Solution:

23 1 1 1 sin Ê 3 sin -1 Ê ˆ ˆ = sin Ê sin -1 Ê 3 ¥ - 4 ¥ ˆ ˆ = Ë Ë 3¯ ¯ Ë Ë 3 27 ¯ ¯ 27 2 8 2ˆˆ cos Ê 3 cos-1 Ê ˆ ˆ = cos Ê cos-1 Ê 4 ¥ - 3 ¥ ¯¯ Ë Ë 3¯ ¯ Ë Ë 27 3 Ê 22 ˆ = cos cos-1 Ë - ¯ 27 22 = cos Ê p - cos-1 Ê ˆ ˆ Ë Ë 27 ¯ ¯ = -

22 . 27

Ê Ê 3 ¥ 1 - 1 ˆˆ 1 Á Ê Ê ˆ ˆ 2 8 ˜˜ = tan tan -1 Á tan 3 tan -1 Á Á Ë Ë 2¯¯ 1 ˜˜ ÁË ÁË 1 - 3 ¥ ˜¯ ˜¯ 4 11 11 = tan Ê tan -1 Ê ˆ ˆ = Ë Ë 2 ¯¯ 2

1 1 £x£ 2 2

1 £x£1 2 1 3

£x£

1 3

Inverse Trigonometric Functions 26.5

If an expression contains a2 - x2 , a2 + x2 , x2 - a2 simplify them by applying the following substitution.

Expression

Substitution

a2 - x2

x = a sin q

Ê- p 1 fi f (5) = p

Previous Years' B-Architecture Entrance Examination Questions Ê 3ˆ 1. 2 cot–1(7) + cos–1 Á ˜ Ë 5¯ 1 4 + tan–1 7 3 2 4 = tan–1 7 + tan–1 1 3 149 7 4 = tan–1 + tan–1 24 3 = 2 tan–1

7 4 + 24 3 = tan–1 117 = tan 7¥4 44 124 ¥ 3 –1

n

2. 1 + Â (2 p) = 1 + p =1

=

cot–1

2n (n + 1) = 1 + n2 + n 2

44 117

Inverse Trigonometric Functions 26.25 n Ê ˆ \ cot -1 Á1 + Â 2 p˜ = cot -1 (1 + n (n + 1)) Ë p =1 ¯

n +1- n ˆ = tan Á Ë n + (n + 1) n ˜¯ -1 Ê

= tan–1 (n + 1) – tan–1(n) 19 n Ê ˆ p fi S = Â cot -1 Á1 + Â 2 p˜ = tan -1 (20) 4 Ë p =1 ¯ n =1

p Thus, cot( S ) = cot ÊÁ tan -1 (20) - ˆ˜ Ë 4¯ p cot(tan -1 (20)) cot Ê ˆ + 1 Ë 4¯ = p cot Ê ˆ - cot(tan -1 (20)) Ë 4¯ 1 +1 21 = 20 = 1 19 120

CHAPTER TWENTY SEVEN

Heights and Distances

ANGLES OF ELEVATION AND DEPRESSION Let OP be a horizontal line in the vertical plane in which an object R is given and let OR be joined.

P

OP = tan 30∞ = 1 / 3 OA fi OP = (1 / 3 ) OA = 10 / 3

So height of the tower is 10 3

10

O

Fig. 27.1

2

Illustration

Fig. 27(a)

30°

A

A car is moving towards a pole of height a. The angle of depression of the car from the top of the pole at a time is observed to be a. After 10 minutes it is observed to be B. Find the distance travelled by the car in 10 minutes. Solution: Let OP be the pole of height a. A and B be the two positions of the car then OAP = a, OBP = b OA = a cot a, OB = a cot b

AB = OA – OB = a (cot a – cot b ) b

P

a

Fig. 27(b)

In Fig. 27(a), where the object R is above the horizontal line OP, the angle POR is called the angle of elevation of the object R as seen from the point O. In Fig. 27(b), where the object R is below the horizontal line OP, the angle POR is called the angle of depression of the object R as seen from the point O. Remark Unless stated to the contrary, it is assumed that the height of the observer is neglected, and that the angles of elevation are measured from the ground. The following results regarding the sides and angles of a triangle are useful in solving the problems on heights and distances.

a b O

Solution:

A

Fig. 27.2

If the angle of elevation of the top P of a tower OP at a point A on the ground is 30° and OA = 10 m. Find the height of the tower. In a triangle ABC, the angles are denoted by the capital letters A, B, C and the lengths of the sides opposite these angles are denoted by a, b, c, respectively (Fig. 27.3). A c

Illustration

a B

b

1 B

a

Fig. 27.3

C

27.2

Complete Mathematics—JEE Main

1. The law of sines In any triangle ABC a b c = 2R = = sin A sin B sin C where R is the radius of the circumcircle of the triangle ABC. R is also known as the circumradius of the triangle. 2. The law of cosines In any triangle ABC cos A = or

b2 + c 2 - a 2 2bc

a2 = b2 + c2 – 2bc cos A

Similar formulae for cos B, cos C exist. 3. Projection rule In any triangle ABC a = b cos C + c cos B, b = c cos A + a cos C and C = a cos B + b cos A. 4. Area of the triangle If D represents the area of the triangle ABC, then D = (1/2)bc sin A = (1/2)ca sin B = (1/2)ab sin C. abc = rs = s(s - a )(s - b)(s - c) = 4R where 2s = a + b + c, R is the radius of the circum circle of triangle ABC and r is the radius of the circle inscribed in the triangle ABC.

SOLVED EXAMPLES Concept-based Straight Objective Type Questions

Example 1: The angle of elevation of the top of two poles at a point on the line joining the foot of the towers on the ground is 45°. If the distance between the towers is 1 m., the difference between the heights of the tower is (a) 2m (b) 1m (c) 1/2 m (d) 3/2 m Ans. (b)

(a) 45° (c) tan -1 ( 3 / 2) Ans. (c)

(b) 30° (d) tan -1 (2 / 3 )

Solution: Let P and Q be the opposite corners of the park with centre O. If h is the height of the lampost at P h

Solution: Let PQ and LM be the two towers PAQ = LAM = 45°

P 60°

P

O b

L

Q

A

45°

Fig. 27.5 M

Q

Fig. 27.4

fi AQ = PQ and AM = LM. fi PQ – LM = AQ – AM = MQ = 1m. Hence the required difference is 1m. Example 2: Apoorv is standing in the centre of a rectangular park and observes that the angle of elevation of the top of a lamp post at a corner of the park in 60°. He then moves diagonally towards the opposite corner of the park and observes that the angle of elevation is now b, then the value of b is

then OP = h cot 60° = h / 3 . PQ = h cot b = 2 OP = 2h / 3 . fi cot b = (2 / 3 ) fi b = tan -1 ( 3 / 2) . Example 3: A bird is sitting on the top of a vertical pole 20m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (m/s) of the bird is (a) 40

(

)

2 -1

(c) 20 2 Ans. (d)

( 20 (

(b) 40

3- 2

(d)

3 -1

)

)

Heights and Distances 27.3

Solution: Let PQ be the pole of height 20 m POQ = 45°

AOB = 120°

Let P ¢ be the position of the bird after one second

OA = OP cot 60° = 15 / 3

P ¢OQ ¢ = 30°.

OB = OP cot 30° = 15 3 P

P¢

From D AOB (AB)2 = (OA)2 + (OB)2 – 2OA ¥ OB ¥ cos 120° È1 Ê 1ˆ ˘ = 152 Í + 3 - 2 Á - ˜ ˙ Ë 2¯ ˚ Î3

( )

= 152 ¥

30° O

Q

45°

Q¢

Fig. 27.6

PP¢ = QQ¢ = PQ¢ – PQ = 20 cot 30° – 20 cot 45° = 20

(

)

3 -1

Example 4: Two ships A and B are sailing straight away from the foot of a tower OP along routes such that AOB is always 120°. At a certain instance, the angles of depression of the ships A and B from the top P of the towers are 60° and 30° respectively. The distance between the ships when the height of the tower is 15m is

A

60°

120° O

30°

13 15 ¥ 15 ¥ 13 = = 5 39 3 3

Example 5: A vertical pole stands at a point A on the boundary of a circular park of radius 2 km. and subtends an angle 60° at another point B on the boundary. If the chord AB subtends the same angle 60° at the centre of the park, the height of the pole is

B

15

Fig. 27.8

(a) 2 3 km

(b)

(c) 2 / 3 km

(d) 1 km.

3 km

Ans. (a) Solution: Let AP be the pole, PBA = 60°

P

Fig. 27.7

fi AB = AP cot 60° =

h 3

, h being the height of the pole.

(a) 5 39 m

(b) 5 30 m

Let O be the centre of the park.

(c) 5 21 m

(d) 5 3 m

Then AOB = 60° fi DOAB is equilateral and hence AB = OB = 2km.

Ans. (a) Solution: OP = 15m PAO = 60°

fi

h 3

= 2 fi h = 2 3 km .

PBO = 30°

LEVEL 1 Straight Objective Type Questions

Example 6: A flagstaff stands in the centre of a rectangular field whose diagonal is 1200 m, and subtends angles 15° and 45° at the mid points of the sides of the field. The height of the flagstaff is

(a) 200 m (c) 300 Ans. (c)

(b) 300 2- 3 m

(d) 400 m

2+ 3 m

Complete Mathematics—JEE Main

27.4

Solution: Let OP be the flagstaff of height h standing at the centre O of the rectangular field ABCD subtending angles 15° and 45° at E and F the mid points of the sides AD and DC of the field.(Fig. 27.9) F

D

C 45∞

E

15∞

O

A

B

P

Fig. 27.10

Fig. 27.9

fi

then

OE = h cot 15°

and

= h(2 + 3 ) OF = h. cot 45° = h.

(

EF = h 1 + 2 + 3

fi

= 2h fi

h=

(

Similarly

)

= 300

30 = y

fi

(c) 60 - 15 3

(d) 60 + 15 3

Ans. (d) Solution: Let x and y be the heights of the flagstaffs at P and Q respectively. AP = x cot 60° = x/ 3 ,

3 - 1/ 3 3.

(

= (60 + 15

BP – AP = x – x / 3 = AB

3 ) m.

(a) sin-1 (1/ 3 )

(b) cos-1 (1/ 3 )

(c) tan-1 (1/ 3 ) Ans. (a)

(d) cot-1 (1/ 3 )

Solution: Let H be the mid point of BC since –TBH = 90°, TH2 = BT2 + BH2 = 52 + 52 = 50 Also since –THG = 90°, TG2 = TH2 + GH2 = 50 + 25 = 75 Let q be the required angle of elevation of G at T. (Fig. 27.11) then sin q =

5 GH 1 = = TG 5 3 3

BP = x cot 45° = x,

fi

)

Example 8: In a cubical hall A B C D P Q R S with each side 10 m, G is the centre of the wall B C R Q and T is the mid point of the side AB. The angle of elevation of G at the point T is

AQ = y cot 30° = y 3

BQ = y cot 60° = y/ 3 (Fig. 27.10)

)

= 15 3 + 3 + 15

2- 3.

(b) 45 + 15 3

)

PQ = BP + BQ = x + y/ 3

2+ 3

(a) 30 + 15 3

(

y = 15

so that

Example 7: Two flagstaffs stand on a horizontal plane. A and B are two points on the line joining their feet and between them. The angles of elevation of the tops of the flagstaffs as seen from A are 30° and 60° and as seen from B are 60° and 45°. If AB is 30 m, the distance between the flagstaffs in metres is

)

3 -1 x

x = 15 3 + 3

2+ 3

Then

(

2

1200 = AC = 2EF = 4h

fi

3 =

fi

2+ 3

300

30

fi

q = sin

–1

(1/ 3 )

Heights and Distances 27.5

3

(a) 100

(b) 200/ 3

(c) 100 / 3 Ans. (b)

(d) 200

3

Solution: Let OP be the tower and C1, C2 be the positions of the car (Fig. 27.13) P 75 50

100 m

60∞

Fig. 27.11

0

Example 9: Two vertical poles 20 m and 80 m high stand apart on a horizontal plane. The height of the point of intersection of the lines joining the top of each pole to the foot of the other is (a) 15 m (b) 16 m (c) 18 m (d) 50 m Ans. (b) Solution: Let PQ and RS be the poles of height 20 m and 80 m subtending angles a and b at R and P respectively. Let h be the height of the point T, the intersection of QR and PS. (Fig. 27.12) S

Q

80

T 20

h

a

b V

R

P

Fig. 27.12

Then PR = h cot a + h cot b = 20 cot a = 80 cot b fi

cot a =4 cot b

cot a = 4 cot b or

Again h cot a + h cot b = 20 cot a fi (h – 20) cot a = – h cot b fi

h cot a = =4 cot b 20 - h

fi

h = 80 – 4 h

fi

C2

C1

Fig. 27.13

We have

OC1 = cot 30° = OP

fi

OC1 = 100

Similarly Now

3

3

OC2 = 100/ 3 C1 C2 = OC1 – OC2 = 100 ( 3 – 1/ 3 )

= 200/ 3 . Example 11: A pole stands vertically, inside a triangular park ABC. If the angle of elevation of the top of the pole from each corner of the park is same, then in DABC, the foot of the pole is at the (a) centroid (b) circumcentre (c) incentre (d) orthocentre Ans. (b) Solution: Let OP be the pole with O as the foot, and the angle of elevation of the top P of the pole at each of the corner A, B, C be a. (Fig. 27.14) OP = tan a then OA fi OA = OP cot a Similarly OB= OC = OP cot a Since OA = OB = OC, O is the circumcentre of the triangle ABC. A

P a

h = 16 m.

Example 10: A man from the top of a 100 metres high tower observes a car moving towards the tower at an angle of depression of 30°. After some time, the angle of depression becomes 60°. The distance (in metres) travelled by the car during this time is

30∞

O

B

C

Fig. 27.14

Example 12: A man observes that the angle of elevation of the top of a tower from a point P on the ground is q.

Complete Mathematics—JEE Main

27.6

He moves a certain distance towards the foot of the tower and finds that the angle of elevation of the top has doubled. He further moves a distance 3/4 of the previous and finds that the angle of elevation is three times that at P. The angle q is given by (b) cos q = 5 /12 (d) cos q = 3/8

(a) sin q = 5 /12 (c) sin q = 3/4 Ans. (a)

Solution: Let AB be the tower, –APB = q, –AQB = 2q and –ARB = 3q

tan q - tan (q - a ) = = 1 + tan q tan (q - a ) fi a = tan–1 (2/9).

Example 14: An aeroplane flying at a height of 3000 m above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 60° and 45° respectively. The height of the lower plane from the ground is (a) 1000

Then QR = (3/4) PQ and –PBQ = –QBR = q fi BQ is the bisector of –PBR (Fig. 27.15) 4 P B PQ AB cosec q sin 3q 4 = fi = fi = 3 BR QR AB cosec 3q 3 sin q

fi

3 – 4 sin2 q = 4/3 fi 12sin2 q = 5 fi sin q = 5 / 12 .

fi

B

1 1 2 4 =2 1 1 9 1+ ¥ 2 4

(b) 1000 / 3 m

3 m

(c) 500 m

(d) 1500 ( 3 + 1) m

Ans. (a) Solution: Let Q and P be the position of the two planes and A be the point on the ground such that –QAO = 60°, –PAO = 45° (Fig. 27.17) and OP, the height of lower plane be h

q q

2q Q R

q P

3q A

Fig. 27.15

Example 13: A vertical pole subtends an angle tan (1/2) at a point P on the ground. The angle subtended by the upper half of the pole at the point P is (b) tan-1 (2/9) (a) tan-1 (1/4) -1 (d) tan-1 (2/3) (c) tan (1/8) Ans. (b) –1

Solution: Let the pole AB subtend angle q at P and the upper half BC of the pole subtend angle a at P (Fig. 27.16) 1 AB then tan q = = 2 AP B

C a q P

A

Fig. 27.16

tan (q – a) =

AC (1/ 2) A B 1 = . = AP AP 4

Now tan a = tan (q – (q – a))

Fig. 27.17

Also OQ = 3000 m Then OA = OP = h and OQ = h tan 60° = 3000 fi

h = 3000 cot 60° =

3000 3

= 1000 3 m

Example 15: A pole of height h stands at one corner of a park in the shape of an equilateral triangle. If a is the angle which the pole subtends at the midpoint of the opposite side, the length of each side of the park is (a) ( 3 / 2) h cot a

(b) (2 / 3 ) h cot a

(c) ( 3 / 2) h tan a

(d) (2 / 3 ) h tan a

Ans. (b) Solution: Let ABC be the triangular park, AP the pole at A, and D the midpoint of BC. Let each side of the equilateral triangle ABC be a. Then (Fig. 27.18)

Heights and Distances 27.7 P

OA = sec 30° a/2

So

h A

fi

OA =

a 2 a ◊ = . 2 3 3

Now from right angled triangle POA PA2 = OP2 + OA2 a

B

D

C

Fig. 27.18

AD2 = AB2 – BD2 = a2 – a 2 / 4 = 3 a 2 / 4 fi and, since we have Therefore, fi

AD = a 3 / 2

fi

a2 = h2 +

fi

2a2 = 3h2.

Example 17: A pole 50 m high stands on a building 250 m high. To an observer at a height of 300 m, the building and the pole subtend equal angles. The horizontal distance of the observer from the pole is (a) 25 m (b) 50 m (c) 25 6 m

AP = h and –ADP = a, AD = h cot a. a 3 2 = h cot a

(

)

a = 2 / 3 h cot a

(d) 25 3 m

Ans. (c) Solution: Let PQ be the pole on the building QR and O be the observer. Then

PQ = 50, QR = 250 (Fig. 27.20)

Example 16: If each side of length a of an equilateral triangle subtends an angle of 60° at the top of a tower h metre high situated at the centre of the triangle, then (b) 2a2 = 3h2 (a) 3a2 = 2h2 2 2 (d) 3a2 = h2 (c) a = 3h Ans. (b)

x

O

q q

50 m

250 m R

Fig. 27.20

fi

PR = 300 so the observer is at the same height as the top P of the pole. Let OP = x. Then from right angled triangles OPQ and OPR,

tan q = so that

fi

In triangle ABC, OA is the bisector of the angle A,

P

Q

Solution: Let O be the centre of the equilateral triangle ABC and OP be the tower of height h. (Fig. 27.19) Then each of the triangles PAB, PBC and PCA are equilateral and thus PA = PB = AB = a.

Fig. 27.19

a2 2a 2 fi = h2 3 3

50 300 and tan 2q = . x x 2 tan q 2

1 - tan q

2 ¥ (50 / x ) 2

1 - (50 / x )

=

=

300 x

ÏÔ Ê 50 ˆ 2 Ô¸ 300 fi 3 Ì1 - Á ˜ ˝ = 1 x ÔÓ Ë x ¯ Ô˛

2 2 50 3 Ê 50 ˆ fi Á ˜ = fix= = 25 6 . Ë x¯ 3 2

27.8

Complete Mathematics—JEE Main

Example 18: Two vertical poles of height a and b subtend the same angle 45° at a point on the line joining their feet, the square of the distance between their tops is (b) a2 + b2 (a) (1/2) (a2 + b2) (c) 2 (a2 + b2)

(d) (a + b)2

and c = AD = x tan g so that a + b + c = x (tan a + tan b + tan g) = x tan a tan b tan g [∵ a + b + g = 180°] and abc = x3 tan a tan b tan g x2 =

fi

Ans. (c) Solution: Let OP and QR be the poles of height a and b respectively subtending angle 45° at S, a point between O and Q. (Fig. 27.21) then OS = a, QS = b PS2 = 2a2 and RS2 = 2b2 Since –PSR = 180° – (45° + 45°) = 90° PR2 = PS2 + RS2 = 2(a2 + b2)

abc a+b+c

Example 20: A vertical tower CP subtends the same angle q, at point B on the horizontal plane through C, the foot of the tower, and at point A in the vertical plane. If the triangle ABC is equilateral with length of each side equal to 4 m, the height of the tower is (a) 8 3 m

(b) 4 3 /3 m

(c) 4 3 m

(d) 8 / 3 m

Ans. (b)

R

P

Solution: Since –CBP = –CAP = q (angles in the same segment are equal) PCBA is a cyclic quadrilateral. (Fig. 27.23) A

b a

q Q

P

45∞

45∞ S

O

60∞

Fig. 27.21

Example 19: A monument ABCD stands at A on a level ground. At a point P on the ground the portions AB, AC, AD subtend angles a, b, g respectively. If AB = a, AC = b, AD = c, AP = x and a + b + g = 180° then x2 is equal to a b (a) (b) a+b+c a+b+c c (c) a+b+c

abc (d) a+b+c

60∞ q = 30∞ B

4

C

Fig. 27.23

fi fi fi

Ans. (d)

–BAP + –BCP = 180° –BAP = 90° –PBC = q = –BAP – –BAC = 90° – 60° = 30°

ÈÎ∵ –BC P = 90 ˘˚

If h is the height of the tower CP, then

Solution: We have

CP = tan q = CB

tan A fi

Fig. 27.22

a = AB = AP tan a = x tan a (Fig. 27.22) b = AC = x tan b

h 1 = 4 3

fi

h=

4 3

m

Example 21: A man on the ground observes that the angle of elevation of the top of a tower is 68° 11¢, and a flagstaff 24 m high on the summit of the tower subtends an angle of 2° 10 ¢ at the observer’s eye. If tan 70° 21¢ = 2.8 and cot 68° 11¢ = 0.4, the height of the tower is (a) 120 m (b) 168 m (c) 200 m (d) 300 m Ans. (c)

Heights and Distances 27.9 R 24 m Q

Solution: Let PQ be the tower of height h whose elevation at A is 68° 11¢ and QR be the flagstaff of height 24 m subtending an angle of 2° 10¢ at A. (Fig. 27.24) Then = fi

h cot 68° 11¢ = AP (h + 24) cot (68° 11¢ + 2° 10¢)

fi

h

h cot 68°11¢ =

fi

2°10¢

(h + 24 ) tan 70 21¢

È (1 / 8) + (5 / 12 ) ˘ x = 60 Í ˙ – 25 Î1 - (1 / 8) ¥ (5 / 12 ) ˚ = 60 ¥ (52/91) – 25 = 60 ¥ (4/ 7) – 25 = 65/7 = 9.2857 = 9.29 (approximately).

fi

Example 23: A tower BCD surmounted by a spire DE stands on a horizontal plane. At the extremity A of a horizontal line BA it is found that BC and DE subtend equal angles. If BC = 3 m, CD = 28 m and DE = 5 m, then BA is equal to

68°11¢

h (0◊ 4) (2◊8) = h + 24 24 24 h= = 1◊12 - 1 ◊12

A

P

(a)

18 ¥ 93

(b)

36 ¥ 93

Fig. 27.24

(c)

34 ¥ 93

(d)

34 ¥ 36

Ans. (a)

= 200 m

Solution: Let –BAC = –DAE = q

Example 22: A statue, standing on the top of a pillar 25 m high, subtends an angle whose tangent is 0.125 at a point 60 m from the foot of the pillar. The best approximation for the height of the statue is (a) 9.28 m (b) 9.29 m (c) 9.30 m (d) 10 m Ans. (b)

–DAB = a, AB = x. (Fig. 27.26) E 5m D 28 m

Solution: Let PQ be the statue of height x on the top of the pillar OP, 25 m high, subtending an angle a at the point A, 60 m from O, the foot of the tower and –OAP = q. (Fig. 27.25)

q a A

q x

C 3m B

Q

Fig. 27.26 x

a

P

Now BC = 3 m, CD = 28 m, DE = 5 m. \ tan (a + q) = 36 /x tan (a) = 31/x, tan q = 3/x

25 m

so, tan (a + q) =

q A

60 m

O

tan a = ◊125 (given)

and

tan q =

then fi

O P 25 5 = = O A 60 12

(x + 25) cot (a + q) = 60 x + 25 = 60 tan (a + q) È tan a + tan q ˘ = 60 Í ˙ Î1 - tan a tan q ˚

31 3 + x x 31 3 1¥ x x

fi

36 = x

fi fi

36(x2 – 93) = 34 x2 fi 2x2 = 36 ¥ 93 x2 = 18 ¥ 93

Fig. 27.25

Then

tan a + tan q 1 - tan a tan q

fi

x=

18 ¥ 93 .

Example 24: A lamp post standing at a point A on a circular path of radius r subtends an angle a at some point B on the path, and AB subtends an angle of 45° at any other point on the path, then height of the lampost is

Complete Mathematics—JEE Main

27.10

(a)

2 r cot a

(b) (r / 2 ) tan a

(c)

2 r tan a

(d) (r / 2 ) cot a

Ans. (c) Solution: Let AP be the lampost of height h at a point A on a circular path of radius r and centre C. Let B be the point on this path such that –PBA = a. fi AB = h cot a (Fig. 27.27)

Example 26: From the top of a cliff x m high, the angle of depression of the foot of a tower is found to be double the angle of elevation of the tower. If the height of the tower is h, the angle of elevation is (a) sin-1

x (2 - h )

(b) tan - 1 3 - 2 h x

(c) sin-1

2h x

(d) cos-1

2h x

Ans. (b) Solution: Let OQ be the tower of height h with O as its foot. AB be the cliff of height x. (Fig. 27.29) Q h-x q 2q

B

P

x

x

Fig. 27.27

Since AB subtends an angle 45° at another point of the path, it subtends an angle of 90° at the centre C so that –BCA = 90° Also CA = CB = r and then h cot a = fi

h=

fi

AB =

2 r

2 r

2 r tan a.

Example 25: A tree is broken by wind, its upper part touches the ground at a point 10 m from the foot of the tree and makes an angle 45° with the ground. The entire length of the tree was (a) 15m (b) 20m (c) 10 ( 2 + 1)m Ans. (c)

(d) 5( 3 + 2) m

Solution: Let OPQ be the tree whose upper part PQ makes an angle 45° with the ground. Then the length of the tree (Fig. 27.28)

A

O

Fig. 27.29

PQ = h – x, –QBP = q and –PBO = 2q Then (h – x) cot q = x cot 2q fi (h – x) tan 2q = x tan q fi

fi fi

tan q h-x = tan 2 q x h-x 1 - tan 2 q +1= +1 x 2 2h = 3 – tan2 q x

fi q = tan–1

3 - 2h / x .

Example 27: A river flows due North, and a tower stands on its left bank. From a point A upstream and on the same bank as the tower, the elevation of the tower is 60°, and from a point B just opposite A on the other bank the elevation is 45°. If the tower is 360 m high, the breadth of the river is (a) 120 6 m (c) 240

3 m

(b) 240 / 3 m (d) 240

6 m

Ans. (a) Fig. 27.28

= OP + PQ = 10 tan 45° + 10 sec 45° = 10 + 10

2 = 10 ( 2 + 1) m

Solution: Let PQ be the tower of height 360 m standing on the left bank of the river subtending an angle 60° at A on the same bank as the tower. (Fig. 27.30) Then PA = 360 cot 60°

Heights and Distances 27.11

Let B be the point opposite to A on the other bank where the elevation of the tower is 45°. So that PB = 360 cot 45° = 360. If x be the width of the river, then AB = x and from the right angled triangle PAB, (PA)2 + (AB)2 = (PB)2 fi fi fi

fi

Ê tan b + tan a ˆ =h x=h Á Ë tan b - tan a ˜¯ È sin (a + b ) ˘ =h Í ˙, Î sin ( b - a ) ˚

1 ˆ2 Ê 2 2 360 ¥ ˜¯ + x = (360) ÁË 3

L(Cloud)

1ˆ Ê (360)2 Á 1 - ˜ = x2 Ë 3¯ x = 360 ¥

2 3

= 120

Ê cot a + cot b ˆ ÁË cot a - cot b ˜¯

x–h A h

6 m.

a b

Q h P

O Water level

x

L¢

Fig. 27.31

Example 29: A balloon of radius r subtends an angle a at the eye of an observer and the elevation of the centre of the balloon from the eye is b, the height h of the centre of the balloon is given by r sin b (b) r sin b sin a (a) sin a (c) Fig. 27.30

Example 28: The angle of elevation of a cloud at a height h above the level of water in a lake is a, and the angle of depression of its image in the lake is b. The height of the cloud above the surface of the lake is h (cot a + cot b ) h (tan a - tan b ) (a) (b) cot b - cot a tan a + tan b (c)

h sin (a + b ) sin (b - a )

(d)

h sin (a - b ) sin (a + b )

Ans. (c) Solution: Let the height of the cloud above the water level be x. Let A be the point at a height h above the water level from which the cloud’s angles of depression and elevation are measured. Let L and L¢ be the locations of the cloud and its image, respectively. Further, as shown in Fig. 27.31, let P be the point directly below the cloud, midway between L and L¢, and Q the point directly above P at a height h. Then –LAQ = a, –QAL¢ = b, PQ = h, QL = x – h, PL¢ = x, and we have (x – h)cot a = (x + h)cot b = AQ, i.e., x + h tan b 2 x tan b + tan a = fi = x - h tan a 2 h tan b - tan a

r sin b sin (a / 2 )

(d)

r sin a sin ( b / 2 )

Ans. (c) Solution: Let O be the centre of the balloon of radius r which subtends an angle a at E, the eye of the observer. If EA and EB are the tangents to the balloon, then –OEA = –OEB = a /2 (Fig. 27.32) Let OL = h be the height of the centre of the balloon, then from DOLE h = OE sin b = r cosec (a /2) sin b = r sin b/sin (a /2).

r

A

O r B

a/

2

a

/2

b E

L

Fig. 27.32

27.12

Complete Mathematics—JEE Main

Example 30: Two poles of height a and b stand at the centres of two circular plots which touch each other externally at a point and the two poles subtend angles of 30° and 60° respectively at this point, then distance between the centres of these poles is

B b A¢

(b) (3a + b)/ 3

(a) a + b (c) (a + 3b)/ 3 Ans. (b)

P

g

O

a A

S

(d) a 3 + b B¢

Solution: Let A and B be the centres of the two circles where the poles of height a and b respectively stand making angles 30° and 60° respectively at the point O where these circle touch each other externally. (Fig. 27.33)

Fig. 27.34

OA = hcota = a OS = hcotb = ae OB = hcotg = b b2 = a2(1 – e2)

Since

cot2g = cot2a – cot2b

fi O

b

a

30°

60° B

A

Fig. 27.33

sum of the radii of the two circles = a cot 30° + b cot 60° = a 3+

b 3

=

Example 32: A tower of height h stands at a point O on the ground. Two poles of height a and b stand at the points A and B respectively such that O lies on the line joining A and B. If the angle of elevation of the top of the tower at the foot of one pole is same as at the top of the other pole, then h is equal to ab a+b (b) (a) a+b ab

3a + b

(c) a + b

(d)

3

= distance between the centres of the two circles. Example 31: A tower is standing in the centre of an elliptic field. If Adya observes that the angle of elevation of the top of the tower at an extremity of the major axis of the field is a, at its focus is b and an extremity of the minor axis is g, then (a) cot2a = cot2b – cot2g (b) cot2b = cot2g – cot2a (c) cot2g = cot2a – cot2b (d) none of these Ans. (c)

Ans. (b) Solution: Let L and M be the tops of the poles at A and B respectively. Angle of elevation of P at A and M be a, and at B and L be b. P

b L a

x

+

y

b

a

2

A

=1

a 2 b2 OP be the tower of height h at O, the centre of the field, Let A & B be extremities of major and minor axis respectively and S be a focus, then

M b

a

Solution: Let the equation of the elliptic field be 2

a+b |a -b|

O

B

Fig. 27.35

then and fi

OA= hcota = (h – a)cotb OB= hcotb = (h – b)cota ab h h–a = fi h= h – b a +b h

Heights and Distances 27.13 P

Example 33: Rajat observes that the angle of elevation of the first floor of a building at a point A on the ground is 30°. He moves 3 units towards the building to the point B and finds that the angle of elevation of the second floor of the building is 60°. If each floor has the same height, height of the 7th floor from the ground in units is (a) 5 (b) 7 (c) 21 (d) 35 Ans. (c) Solution: Let the height of each floor be x units

h

A

P x A

30° 60° 3 B

O

Fig. 27.36

fi

3 = AB = OA – OB = x cot30° – 2x cot60° 2x =x 3 – 3 x = 3 and hence the required height is 21 units.

Example 34: A pole of height h stands in the centre of a circular platform in the centre of a circular field. Another pole of equal height is at a point on the boundary of the field. The angles of elevation of the top of the first pole from the bottom and top of the second pole are respectively a and b. Height of the platform from the ground is h cot a h cot b (b) (a) cot b - cot a cot a - cot b (c) cota – cotb (d) none of these

O

a

Fig. 27.37

Q x

}Ma

L

Ans. (a) Solution: Let a be the height of the platform then height of the first pole from the ground is h + a and from the top of the second pole is a, so (h + a) cot a = a cot b h cot a fi a = cot b - cot a Example 35: The tangents of the angles subtended by a tower at four points A, B, C and D on the ground are in H.P. If O be the foot of the tower on the ground, then (a) OA + OC = OB + OD (b) OA + OB = OC + OD (c) OA + OD = OB + OC (d) AB + CD = BC + CD Ans. (c) Solution: If a, b, g, d be the four angles such that tana, tanb, tang, tand are in H.P, then cota, cotb, cotg, cotd are in A.P fi hcota, hcotb, hcotg, hcotd are in A.P. fi OA, OB, OC, OD are in A.P fi OA + OD = OB + OC

Assertion-Reason Type Questions

Example 36: Statement-1: A pole standing in the centre of a rectangular field of area 2500 sq. units subtends angle a and b respectively at the mid-points of two adjacent sides of the field such that a + b = p/2, the height of the pole is 25 sq. units. Statement-2: Area of a rectangle is equal to the product of the length of the adjacent sides. Ans. (a) Solution: Statement-2 is True. In statement-1 if the height of the pole is h, then the length of the adjacent

sides of the field are 2h cot a and 2h cot b and the area is 4h2 cot a cot b = 4h2 as a + b = p/2 fi cot a cot b = 1. So 4h2 = 2500 fi h = 25 sq. units and the statement-1 is True using statement-2 Example 37: Statement-1: Apoorv, standing on the ground wants to observe the angle a of elevation of the top of a tower in front of him. He walks half the distance towards the foot of the tower and finds the angle of elevation is p/4. He observes a = tan–1(1/2)

27.14

Complete Mathematics—JEE Main

Statement-2: If the angles of elevation of the top of a tower at three distinct points on the ground is a, the points lie on a circle with centre at the foot of the tower. Ans. (b) Solution: Statement-2 is True, because if h is the height of the tower, the points are at a distance of h cot a from the foot of the tower and hence lie on a circle. In statement-1, if h is the height of the tower then h cot a = 2h cot p/4 = 2h fi a = tan–1 1/2 fi Statement-1 is true but does not follow from statement-2 Example 38: A and B are two points in a line on the horizontal plane through the foot O of a tower lying on opposite sides of the tower. Statement-1: If the angles of elevation of the top of the tower at A and B are a and 2a respectively and the distance between the points is twice the height of the tower, then tan2 a + 4 tan a = 3. Statement-2: If OB = 2(OA), a, b are respectively the angles of elevation of the top of the tower at A and B, then b = 2a. Ans. (c) Solution: If h is the height of the tower, in statement-1 OA = h cot a, OB = h cot 2a so h (cot a + cot 2a) = 2h. fi

1 1 =2 + tan a tan 2a 1 1 - tan a =2 + 2 tan a tan a

fi fi

tan2 a + 4 tan a = 3 statement-1 is True.

In statement 2, h cot b = 2h cot a fi

Statement-1: If AB subtends an angle p /2 at the foot of the tower, then 2h = 3a. Statement-2: If AB subtends an angle p /3 at the foot of the tower then h = a 3 Ans. (d) Solution: OA = OB = h cot p /3 = h 3 . In statement-1 AOB is a right angled triangle. so (AB)2 = (OA)2 + (OB)2 fi a2 = 2h2/3 fi 3a2 = 2h2 and the statement is false. In statement-2 AOB is an equilateral triangle. so AB = OA fi a = h

3

fi h = a 3 and the statement is True. Example 40: ABC is an equilateral triangle on the horizontal ground with length of each side equal to a. Statement-1: If a tower standing at the centre O of the triangle makes an angle a at each corner such that a = tan–1 9, then height of the tower is 3 3 . Statement-2: If a tower of height 2a standing at one corner of the triangle makes an a at any other corner, then a = tan –1 2. Ans. (b) Solution: Since ABC is an equilateral triangle. OA = OB = OC = a 3 . In statement-1 if the height of the

2

fi

the horizontal plane through the foot O of the pole is p/3. AB = a.

cot b = 2cot a fi b = 2a and the statement is false.

Example 39: Mansi observes that the angle of elevation of a vertical pole of height h at two points A and B on

tower is h, h cot a = a / 3 a ¥ 9 = 3 3 a. fi h= 3 fi statement-1 is True. In statement-2, 2a cot a = a fi tan a = 2 fi a = tan–1 2 so the statement-2 is True but does not lead to statement-1.

LEVEL 2 Straight Objective Type Questions Example 41: A and B are two points 30 m apart in a line on the horizontal plane through the foot of a tower lying on opposite sides of the tower. If the distances of the top of the tower from A and B are 20 m and 15 m respectively, the angle of elevation of the top of the tower at A is

(a) (b) (c) (d) Ans. (a)

cos-1 (43/48) sin-1 (43/48) cos-1 (29/36) sin-1 (29/36)

Heights and Distances 27.15 Solution: Let OP be the tower, then AB = 30, AP = 20

fi

and BP = 15. (Fig. 27.38) P 20

15

(c)

(50

(b)

)

3 tan q m

(70

)

3 tan q m

(d) (75 3 ) tan q m

Ans. (b)

If q be the elevation at A of the top P of the tower, then from triangle ABP 202 + 302 - 152 1075 43 = = 2 ¥ 20 ¥ 30 1200 48

q= cos–1 (43/48).

fi

fi a = h( 3 – 1).

sin 30

(a) (70 3 ) tan q m

B

O 30

Fig. 27.38

cos q =

2 h ◊ sin 15

Example 43: The angles of elevation of a vertical tower standing inside a triangular field at the vertices of the field are each equal to q. If the length of the sides of the field are 30 m, 50 m and 70 m, the height of the tower is

q

A

a=

Example 42: The angle of elevation of the top C of a vertical tower CD of height h from a point A in the horizontal plane is 45° and from a point B at a distance a from A on the line making an angle 30° with AD, it is 60°, then (a) a = h ( 3 + 1)

(b) h = a ( 3 + 1)

(c) a = h ( 3 - 1)

(d) h = a( 3 - 1)

Solution: Let OP be the tower of height h at the point O in the triangular field ABC with sides BC = 30, CA = 50 and AB = 70. Then OA = OB = OC = h cot q fi O is the circumcentre of the triangle ABC fi h cot q = R (the radius of the circumcircle) fi h = R tan q We know that R = abc/4D where

D= =

Thus,

h=

Ans. (c)

=

Solution: We have –CAD = 45°, –BAD = 30°, –CBH = 60° (Fig. 27.39) fi –ACD = 45°, –BCH = 30° so that –ACB = 15° and –CAB = 45° – 30° = 15° fi –ABC = 150°

s (s - a ) (s - b) (s - c ) 75 ¥ 45 ¥ 25 ¥ 5 = 25 ¥ 5 ¥ 3 3 30 ¥ 50 ¥ 70

tan q

4 ¥ 25 ¥ 5 ¥ 3 3 70 3

tan q m.

Example 44: In a triangular plot ABC with BC = 7 m, CA = 8 m and AB = 9 m. A lamp post is situated at the middle point E of the side AC and subtends an angle tan–1 3 at the point B, the height of the lamp post is (a) 21 m (b) 24 m (c) 27 m (d) cannot be determined Ans. (a) Solution: In triangle ABC A

9

P

90∞ -1 3

E

8

tan

B

Fig. 27.39 2

2

2

sin 15

=

AC sin 150

fi

C

Fig. 27.40 2

From D ADC, AC = h + h = 2h and from D ABC AB

7

a sin 15

=

a = BC = 7, b = CA = 8 and c = AB = 9

2h sin 150

so that cos C =

a 2 + b2 - c 2 2 49 + 64 - 81 = = 7 2¥7¥8 2a b

27.16

Complete Mathematics—JEE Main

In triangle BEC, BC = 7, CE = 4 and cos C = so that BE2 = 42 + 72 – 2 ¥ 4 ¥ 7 ¥ fi

2 7

2 = 49 7

BE = 7.

If h is the height of the lamp post EP at E PE h then = = tan (tan– 13) = 3 fi h = 21 m. BE 7 Example 45: Two objects at the points P and Q subtend an angle of 30° at a point A. Lengths AR = 20 m and AS = 10 m are measured from A at right angles to AP and AQ respectively. If PQ subtends equal angles of 30°, at R

fi

PQ =

500 - 200 3 .

Example 46: From a ship at sea it is observed that the angle subtended by feet A and B of two light houses, at the ship is 30°. The ship sails 4 km towards A and this angle is then 48°, the distance of B from the ship at the second observation is (a) 6.460 km (b) 6.472 km (c) 6.476 km (d) 6.478 km Ans. (b) Solution: If S1 and S2 are the two positions of the ship then S1S2 = 4. A

and S, then length of PQ is (a)

300 - 200 3

(b)

500 - 200 3

(c)

500 3 - 200

(d)

300

S2

Ans. (b)

48∞

4 30∞

Solution: Since PQ subtends the same angle of 30° at each of the points A, R and S, the points A, P, Q, R and S lie on a circle. (Fig. 27.41)

B

S1

Fig. 27.42

P

–AS1B = 30° and –AS2B = 48°

30∞

A

18∞

Q

fi

–S1BS2 = 18° (Fig. 27.42)

Now from DS1BS2 30∞

S1 S2 

10 m

sin 18 20

fi S 2B =

m

30∞

30∞ S

Fig. 27.41

Also AR ^ AP and AS ^ AQ fi –RAS = 90° – –RAQ = –PAQ = 30° As it is given AR = 20 m, AS = 10 m. Then PQ and RS are chords of the same circle making an angle of 30° at a point A on the circumference of the circle and hence are equal in length. Now from D SAR RS2 = 202 + 102 – 2 ¥ 20 ¥ 10 cos 30° PQ2 = 500 – 200

S2 B sin 30

4 ¥ (1/ 2) sin 18

= 2 ( 5 + 1) = 6.472.

Example 47: From a point on the horizontal plane, the elevation of the top of a hill is 45°. After walking 500 m towards its summit up a slope inclined at an angle of 15° to the horizon the elevation is 75°, the height of the hill is

R

fi

=

3

(a) 500

6 m

(b) 500

3 m

(c) 250

6 m

(d) 250

3 m

Ans. (c) Solution: Let P be the top of the hill CP, A and B be the points where the angles of elevation of the top P of the hill are 45° and 75° respectively. Let BE be the line parallel to AC through B. (Fig. 27.43)

Heights and Distances 27.17

ÈÊ 3 + 1ˆ 2 ˘ ˙ = 4a2 1 fi h ÍÁ ˜ ÍÎË 3 - 1¯ ˙˚ 2

h2 ¥ 4 3 = 4a2

fi

3 -1

fi h=

Then –PAC = 45° and –BAC = 15° fi CA = CP = x (say)

fi and

2x –PBE =75° fi –BPE = 15° –APB = 30° = –BAP –ABP =120° \ From D ABP sin 120



=

500 sin 30

( 3 - 1)

(b) a

2

1 È 1 - ˘ (c) a Í 3 4 - 3 4 ˙ ÍÎ ˙˚

-

1 4

]a.

fi

dow and A and B be the two positions of the pole such that –LAX = a and –MBX = b. LM = x, X being the foot of the vertical wall. (Fig. 27.45) L (Coping)

x = 250

x M (Sill)

6 m.

Example 48: The elevation of a steeple at a place due south of it is 45° and at a place B due west of A the elevation is 15°. If AB = 2a, the height of the steeple is (a) a

1

a = [3 4 - 3

Solution: Let L be the coping and M the sill of the win-

PA =

2x

( 3)

b B

( 3 + 1)

a a

A

1 È 1 - ˘ (d) a Í 3 4 + 3 4 ˙ ÍÎ ˙˚

Solution: Let OP be the steeple of height h. A be due south of OP and –OAP = 45°and B be due west of A and –OBP = 15° then OA = h cot 45° = h and OB = h cot 15° and AB = 2a (given) so that from right angled DOAB, OB2 = OA2 + AB2 fi h2 cot2 15 = h2 + 4a2 (Fig. 27.44) P

Let then

AL = BM = l and AB = a (given) x = LX – MX = l(sin a – sin b) a = BA = BX – AX = l(cos b – cos a).

2a S

h

45∞ A

N

O E

Fig. 27.44

fi

x sin a - sin b = = a cos b - cos a

fi

x = a cot

a+b a-b sin 2 2 a+b a-b sin 2 sin ◊ 2 2 2 cos

a+b . 2

Example 50: OAB is a triangle in the horizontal plane through the foot P of the tower at the middle point of the side OB of the triangle. If OA = 2 m, OB = 6 m, AB = 5 m and –AOB is equal to the angle subtended by the tower at A then the height of the tower is

W

15∞

X

Fig. 27.45

2

Ans. (c)

B

2

)

3 -1

Example 49: The top of a pole, placed against a wall at an angle a with the horizon, just touches the coping, and when its foot is moved a m, away from the wall and its angle of inclination is b, it rests on the sill of a window; the vertical distance of the sill from the coping is (a) a sin ((a + b)/2) (b) a cos ((a + b)/2) (c) a cot ((a + b)/2) (d) a tan ((a + b)/2) Ans. (c)

Fig. 27.43

fi

1/ 2

(

(a)

11 ¥ 39 25 ¥ 3

(b)

(c)

11 ¥ 25 39 ¥ 2

(d) none of these

Ans. (b)

11 ¥ 39 25 ¥ 2

Complete Mathematics—JEE Main

27.18

Solution: Let PQ be the tower of height h at the middle point P of the side OB of the triangle OAB, where (Fig. 27.46)

\ From D SBQ, we have sin (g - b ) SQ a - b sin g ¥ = = sin a SB b sin ( b - a )

2m

A 5m

a

a 6m

O

P

B

h Q

Fig. 27.46

OA = 2, OB = 6, AB = 5 and –AOB = –PAQ = a, then AP = h cot a, OP = 3 From triangle OAB,

cos a =

22 + 62 - 52 5 = 2¥2¥6 8

and from D OAP cos a =

5 = 8

fi fi

h2 =

22 + 32 - h 2 cot 2 a 2¥2¥3

13 - h 2 ¥

25 39

(a) a

11 ¥ 39 25 ¥ 2

b sin a

(a - b) sin g sin g sin a

(c)

Example 52: A person standing on the ground observes the angle of elevation of the top of a tower to be 30°. On walking a distance a in a certain direction, he finds the elevation of the top to be the same as before. He then walks a distance 5a/3 at right angles to his former direction, and finds that the elevation of the top has doubled. The height of the tower is

12

(c)

Example 51: If two vertical towers PQ and RS of lengths a and b (a > b) respectively subtend the same angle a at a point A on the line joining their feet P and R in the horizontal plane and angles b and g at another point B on this line nearer the towers on the same side of the towers as sin ( b - g ) is equal to A, then sin ( b - a ) (a)

Fig. 27.47

(b)

(d)

6/5 a

(b)

85 / 48 a

(d)

48 / 85 a

Ans. (b) Solution: Let PQ be the tower of height h and A, B, C be the three positions of the person from initial to final then (Fig. 27.48) –PAQ =–PBQ = 30° and –PCQ = 60° fi

AQ = BQ = h cot 30° = h 3

and

CQ = h cot 60° = h

3◊

(b - a ) sin g b sin a

(b - a ) sin a b sin g

Ans. (b) Solution: We have

P

A 30∞ a/2 M a/2

h

L Q 30∞ B

60∞

5a/3 C

–PAQ = a = –RAS –PBQ = b, –RBS = g fi

–QBS = g – b

and –BQS = 90° – a – (90° – b ) = b – a.

Fig. 27.48

Let QM be perpendicular to AB and CL be perpendicular to QM Now AB = a and BC = 5a/3 and –ABC = 90°

Heights and Distances 27.19

fi

Also since –PAB = a given, –PQB = a (angles in the same segment.)

ML = BC = 5a/3 and CL = BM = a/2

From right angled triangle BMQ, QM2 = BQ2 – BM2 = 3h2 – (a / 2)2 = 3h2 – (a2/4) and from right angled triangle CLQ. h2 a2 QL2 = QC2 – CL2 = 3 4 Since QM = QL + LM 2

2

fi

3h2 –

2

fi

64h4 –

fi fi fi

576h4 – 1500a2h2 + 850a4 = 0 288h4 – 750a2h2 + 425a4 = 0 (48h2 – 85a2) (6h2 – 5a2) = 0

fi

h = 85 / 48 a or

2

2

h a 3 4

h2 a2 3 4

400 a 2 h 2 625 a 4 100 a 2 h 2 – 25a4 + = 3 9 3

5/6 ◊ a

AB = (1/2) sin a BQ

(d)

ABQ is an isosceles triangle. AB = AQ, and 

Example 53: A tower PQ subtends an angle a at a point A on the same level as the foot Q of the tower. It also subtends the same angle a at a point B where AB subtends the angle a with AP then (a) AB = BQ (b) BQ = 2AQ (c)

fi

AB

25 a 10 a a h a + = + 3 4 9 3 4

10 a 8 h 2 25 a 2 = 3 3 9

Now –PBQ = a = –PQB –ABQ = –AQB = 90° – a

fi

2

fi

fi fi

sin (90 - a )

=

BQ sin 2 a

fi

2AB sin a = BQ.

1 AB 1 = cosec a. = 2 BQ 2 sin a

fi

Example 54: The angle of elevation of the top of a tree at a point B due south of it is 60° and at a point C due north of it is 30°. D is a point due north of C where the angle 8 and BC × CD = of elevation is 15°, then given 3 = 1 11 23 × 32 × 19 × 11, the height of the tree is (a) 33 (b) 38 (c) 57 (d) 88 Ans. (c) Solution: Let A be the top of the tree OA = h (Fig. 27.50)

B, C, D be the three point of observation such then –ABO = 60°, –ACO = 30° and –ADO = 15°

AB = (1/2) cosec a BQ

Ans. (d) Fig. 27.50

Solution: Since –PAQ = –PBQ = a (angles in the

same segment are equal) PQAB is a cyclic quadrilateral (Fig. 27.49)

Then BC = BO + OC = h (cot 60° + cot 30)

a P

So that

a 90∞- a Q

Fig. 27.49

(

)

)

23 ¥ 32 ¥ 19 ¥ 11 4 ¥ 2 ¥ 11 2 4 ¥ 2h 2 = = h 19 3

Ê ÁË∵

3=

19 ˆ ˜ 11 ¯

h2 = 32 ¥ 192 fi h = 57 Example 55: n poles standing at equal distances on a straight road subtend the same angle a 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

fi

a a

) (

= h 2 + 3 - 3 = 2h

90∞- a

A

(

= h 3 + 1/ 3 = 4 / 3 h CD = h cot 15° – h cot 30°

B

27.20

Complete Mathematics—JEE Main

(a)

h sin a - a cos a (n - 1) sin a

(b)

h cos a - a cos a (n - 1) cos a

(c)

h cos a - a sin a (n - 1) sin a

(d)

h sin a - a cos a (n - 1) cos a

Ans. (c) Solution: Let A1, A2 ... An be the feet of the n poles subtending angle a at O, such that OA1 = a, if d be the distance between two consecutive poles OA2 = a + d OA3 = a + 2d  OAn = a + (n - 1)d

Fig. 27.51

Since the height of the pole at An is h; OAn = h cot a. fi

a + (n – 1)d = h cot a

fi

d=

h cot a - a h cos a - a sin a = . n -1 (n - 1) sin a

EXERCISE Concept-based Straight Objective Type Questions 1. A pole is standing at a point O between two milestones at A and B such that the angles of elevation of the top of the pole at A and B are respectively a and b. If the distance between the milestones is half the height of the pole then (a) 2 sin (a + b) = sin a sin b (b) 2 sin (a + b) = cos a cos b (c) sin (a + b) = 2 sin a sin b (d) sin (a + b) = 2 cos a cos b 2. OP is a tower of height 20 m and AB is a pole of height 5 m. The angle of elevation of the top P of the tower from the top B of the pole is 45°. Both pole and tower stand on the same ground. The angle of elevation of the top P of the tower from the base A of the pole is 3 3 (b) sin -1 (a) cos-1 5 5 3 4 (c) tan -1 (d) cot -1 4 3 3. A circular path is 50 m. wide. The angle of elevation of the top of a pole at the centre of the circular park at a point on the inner circle is a and at a point on the outer circle is 45°. The height of the pole is

(a)

50 cos a cos a - sin a

(b)

50 sin a sin a - cos a

(c)

50 cos a - sin a

(d)

50 cos a + sin a

4. A pole 10 m high stands on a tower 30 m. high. The angle of elevation of the top of the pole at a point A on the ground in 45° and the pole subtends an angle a at the same point A then a is equal to (b) cot – 1 7 (a) cot – 1 (1/7) –1 (d) sin – 1 (1/7) (c) cos (1/7) with each side equal to 100 m. 5. ABCD is a square field with each side equal to 100 m. Two poles of equal heights stand at E, the mid point of DC and at the corner B of the field, subtending respectively angles a and 30° at the corner A of the field. The value of a satisfies (a) cos2 a =

11 19

(b) sin2 a =

15 19

(c) tan2 a =

4 19

(d) tan2 a =

19 15

Heights and Distances 27.21

LEVEL 1 Straight Objective Type Questions 6. The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of 100 m from its base is 45°. If the angle of elevation of the top of the pillar after completion at the same point is 60°, then the height to be increased for the completion of the pillar in metres is (b) 100 2 (a) 50 3 (c) 100

3

(d) 100 ( 3 - 1)

7. The angles of elevation of the top of a tower at the top and the foot of a pole 10 m high are 30° and 60° respectively. The height of the tower is (a) 15 m (b) 20 m (d) 25 3 m (c) 10 3 m 8. A tower subtends an angle a at a point A on the ground, and the angle of depression of its foot from a point B just above A and at distance b from A, is b. The height of the tower is (a) b tan a tan b (b) b tan a cot b (c) b cot a cot b (d) b cot a tan b 9. A person walking along a straight road observes that at two points 1 km apart, the angles of elevation of a pole in front of him are 30° and 75°. The height of the pole is (a) 250 ( 3 + 1) m

(b) 250 ( 3 - 1) m

(c) 500 ( 2 + 1) m

(d) 500 ( 2 - 1) m

13. An observer finds that the angular elevation of a tower is A, on advancing 3 m towards the tower the elevation is 45° and on advancing 2 m nearer, the elevation is 90° - A, the height of the tower is (a) 1m (b) 5m (c) 6m (d) 8m 14. ABC is a triangular park with all sides equal. If a pillar at A subtends an angle of 45° at C, the angle of elevation of the pillar at D, the middle point of BC is (b) tan-1 2 / 3 (a) tan-1 3 / 2

(

(c) cot-1

11.

From a point on the ground 100 m away from the base of a building, the angle of elevation of the top of the building is 60°. Which of the following is the best approximation for the height of the building? (a) 172 m (b) 173 m (c) 174 m (d) 175 m 12. From the top of a tower 100 m height, the angles of depression of two objects 200 m apart on the horizontal plane and in a line passing through the foot of the tower and on the same side of the tower are 45° - A and 45° + A, then angle A is equal to (a) 15° (b) 22.5° (c) 30° (d) 35°

3

(

(d) tan-1

)

3

15. A kite is flying with the string inclined at 75° to the horizon. If the length of the string is 25 m, the height of the kite is (a) (25 / 2) ( 3 - 1)2

(b) (25 / 4) ( 3 + 1) 2

(c) (25 / 2) ( 3 + 1)2

(d) (25 / 2) ( 6 + 2 )

16. AB is a vertical pole. The end A is on the level ground, C is the middle point of AB. P is a point on the level ground. The portion BC subtends an angle b at P. If AP = n AB, then tan b = n n (a) (b) 2 2 2 n +1 n -1 (c)

10. If a flagstaff subtends the same angle at the points A, B, C and D on the horizontal plane through its foot, then ABCD is a (a) square (b) cyclic quadrilateral (c) rectangle (d) none of these

)

n 2

n +1

(d) none of these

17. A man in a boat rowing away from a cliff 150 metres high observes that it takes 2 minutes to change the angle of elevation of the top of the cliff from 60° to 45°. The speed of the boat is (a) (1/2) (9 - 3 3 ) km/h (b) (1/2) (9 + 3 3 ) km/h (c) (1/2) (9 3 ) km/h (d) none of these 18. A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°, when he retires 40 metres from the banks he finds the angle to 30°. The breadth of the river is (a) 40 m (b) 60 m (c) 20 m (d) 30 m

27.22

Complete Mathematics—JEE Main

19. The elevation of the top of a mountain at each of the three angular points A, B and C of a plane horizontal triangle is a, if BC = a the height of the mountain is (a) (a/2) cosec A tan a (b) (a/2) sec A tan a (c) (a/2) cosec a cot A (d) (a/2) sec a tan A 20. The angles of elevation of the top of a tower standing on a horizontal plane, from two points on a line passing through its foot at distances a and b, respectively, are complementary angles. If the line joining the two points subtends an angle q at the top of the tower, then if a > b, sin q = a-b a+b (b) (a) a+b a-b

(a) a cot a a+b (c) cot a + cot b

(b) b cot b (d) none of these

25. Three poles of height a, b, c stand on the same side of a road and subtend an angle of 45° at a point on the line joining their feet. The pole of height a subtends an angle a at the foot of the pole of height b which subtends an angle b at the foot of the pole with height c, if a > b > c, then cot a – cot b = (a)

ac - b2 ab

(b)

bc - a 2 ab

ab - c 2 ac - b2 (d) bc bc 26. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. If after 10 seconds, the elevation is observed to be 30°, then the uniform speed of the aeroplane per hour is (a) 120 km (b) 240 km (c)

(c)

2 ab a+b

(d)

2 ab a-b

21. The upper three-quarters of a vertical pole subtends an angle tan-1 (3/5) at a point in the horizontal plane through its foot and distant 40 m from it. The height of the pole is (a) 80 m (b) 100 m (c) 160 m (d) 200 m 22. PQ is a vertical tower and A, B, C are three points on a horizontal line through Q, the foot of the tower and on the same side of the tower. If the angles of elevation of the top of the tower from A, B and C are a, b, g respectively, then AB/BC = cot a - cot g cot a - cot b (b) (a) cot b - cot g cot b - cot g (c)

cot a - cot b cot a - cot g

(d)

cot a - cot g cot a - cot b

23. ABCD is a rectangular park with AB = a. A tower standing at C makes angles a and b at A and B respectively, the height of the tower is (a)

a 2

2

(b)

cot a + cot b (c)

a tan a tan b 2

2

tan b + tan a

(d)

a 2

cot a - cot 2 b a cot a cot b cot 2 a - cot 2 b

24. Two circular path of radii a and b intersect at a point O and AB is a line through O meeting the circles at A and B respectively. Chords OA and OB subtend equal angles of 60° at their respective centres. A vertical pole at O subtends angles a and b respectively at A and B then height of the pole is

(c) 240

3 km

(d) 240/ 3 km

27. If a flagstaff 6 metres high placed on the top of a tower throws a shadow of 2 3 metres along the ground then the angle (in degrees) that the sun makes with the ground is (a) 15° (b) 30° (c) 60° (d) tan–1 2 3 28. Three poles whose feet lie on a circle subtend angle a, b, g respectively at the centre of the circle. If the height of the poles are in A.P. then cot a, cot b, cot g are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 29. A, B, C are three points on a vertical pole whose distances from the foot of the pole are in A.P. and whose angles of elevation at a point on the ground are a, b and g respectively. If a + b + g = p, then tan a tan g is equal to (a) 3 (b) 2 (c) 1 (d) – 1 30. A ladder rests against a wall at an angle of 35°. Its foot is pulled away through a distance a, so that it slides a distance b down the wall, finally making an angle of 25° with the horizontal, then a/b = (a) 1 (b) 1/ 3 (c)

3

(d)

3 /2

Heights and Distances 27.23

Assertion-Reason Type Questions 31. Statement-1: Three poles of height a, b, c stand at the points A, B, C respectively and subtend the same angle a at a point O on the horizontal line through the feet of the poles. If a, b, c are in A.P., then AB = BC. Statement-2: O is the centre of a circular field and A is any point on its boundary. Two poles standing at A and O subtend the same angle a at a point B on the other end of the diameter through A. Height of the pole at A is twice the height of the plot at O. 32. Statement-1: Rajat observes that the angle of elevation of the top P of a tower OP at a point A on the ground is a. He travels a distance a in the direction AP and reaches the point B. He then travels a horizontal distance a towards the tower and reaches the point C, where the angle of elevation of the top of the tower is p /4, the height of the tower is

a (cos a + 1 - sin a ) cot a - 1

Statement-2: On the top of building a pole of height equal to 1/3 of the height of the building is placed so that the angles of elevations of the top of the pole and the top of the building at a point on the ground are a and b respectively then a = (3/4)b.

33. Statement-1: ABC is a triangular field with AC = b and AB = c. A pole standing at a point D on BC subtends angles a at B and b at C. If BAD = DAC then b cot a = c cot b. Statement-2: Bisector of an angle of a triangle divides the opposite side in the ratio of the side containing the angle. 34. Statement-1: The angle of elevation of the top P of a tower OP at a point A on the ground is a, the angle of elevation of the mid-point Q of the tower at the mid-point B of OA is also a. Statement-2: The line joining the mid-points of two sides of a triangle is parallel to the third side. 35. Statement-1: A tower standing at the centre of a square field subtends an angle a at each corner. If the height of the tower is twice the length of a side of the square, then a = tan–1 2. Statement-2: A, B, C are three points on the horizontal line through the foot of a tower and the angles of elevation of the top of the tower at these points are AB 30°, 45° and 60° respectively, = 3. BC

LEVEL 2 Straight Objective Type Questions 36. The angle of elevation of the top Q of a tower PQ at a point A on the horizontal plane through P the foot of the tower is a. At a point B on AQ at a vertical height of a, the angle of elevation of the middle point R of the tower PQ is b, then the height of the tower is (a)

2a ( tan a - tan b ) tan a - 2 tan b

(c)

2a ( tan a tan b - 1) 2a ( tan a cot b - 1) (d) 2 tan a cot b - 1 2 tan a cot b - 1

(b)

2a ( tan a - 2 tan b ) tan a - tan b

37. A lamppost stands in the centre of a circular garden and makes angle a at points A and B on the boundary where AB subtends an angle 2b at the foot of the lamppost. If g is the angle which the lamppost subtends at C, the middle point of the line joining A and B, then tan g =

(a) tan a tan b (c) tan a sec b

(b) sec a tan b (d) none of these

38. From a point on the ground, if the angles of elevation of a bird flying at constant speed in a horizontal direction, measured at equal intervals of time are a, b, g and d, then (a) cot2 b – cot2 g = 3(cot2 a – cot2 d) (b) cot2 b – cot2 d = 3(cot2 a – cot2 g) (c) cot2 g – cot2 d = 3(cot2 a – cot2 b) (d) cot2 a – cot2 d = 3(cot2 b – cot2 g) 39. A vertical tower standing at O has marks P, Q, R, S at heights of 1m, 2m, 3m and 4m from the foot O and A is a point on the horizontal plane through O. If PQ and RS subtend angles a and b respectively at A where OA = 2m then cos (a + b) =

27.24

Complete Mathematics—JEE Main

(a) 5/ 26

(b) 24/ 650

(c) 23/ 650

(d) 1/ 26

40. ABCD is a rectangular field with AB = a and BC = b. A lamp post of height h at A subtends an angle a at P, the middle point of CD and another lamp post of equal height at D subtends an angle b at Q, the middle point of BC. If PQ subtends an angle q at A, then cot2 a cot2 b cos2 q = k2, where k = (b) (a2 – b2)/2h2 (a) (a2 + b2)/2h2 2 2 2 (c) 2h /(a + b ) (d) 2(a2 + b2)h2 41. A vertical tower OP of height h subtends angle a, b, g respectively at the points A, B, C on the horizontal plane through the foot O of the tower. A is due west of the tower. B is due east of A and on the same side of the tower as A. C is due south of B, then AC = (a) h(cot a – cot b) (b) h cot 2 g - cot 2 b (c) h cot 2 a + cot 2 g - 2 cot a cot b (d) h cot 2 g + cot 2 b - 2 cot a cot b 42. PQ and RS are two vertical towers of the same height where S is on the ground Q is above the ground. The line joining the top P and the foot S of the two towers meets the horizontal line through Q at a point A where the angles of elevation of the tops P and R of the two towers are a and b respectively. If AS = a, the height of the towers is (a)

a sin ( b + a ) cos b

(b)

a cos ( b + a ) cos b

(c)

a sin ( b + a ) sin b

(d)

a cos ( b + a ) sin b

43. From the top of a building of height h, a tower standing on the ground is observed to make an angle q. If the horizontal distance between the building and the tower is h, the height of the tower is (a)

2h cos q sin q + cos q

2h (c) 1 + tan q

(b)

2h 1 + cot q

2h (d) sin q + cos q

44. A tower stands at the foot of a hill whose inclination to the horizon is 9°; at a point 40 m up the hill the tower subtends an angle of 54°. The height of the tower is

(a) 17.56 m (c) 54.76 m

(b) 45.76 m (d) none of these

45. The angle of elevation of a stationary cloud from a point 2500 metres above a lake is 15° and the angle of depression of its reflection in the lake is 45°. The height of the cloud above the lake level is (a) 2500/ 3 m (c) 2500

3 m

(b) 2500 m (d) 5000

3 m

46. A tower PQ stands at a point P with in the triangular park ABC such that the sides a, b, c of the triangle subtend equal angles at P, the foot of the tower and the tower subtends angles a, b, g at A, B, C respectively, then a2 (cot b –cot g) + b2(cot g – cot a) + c2 (cot a – cot b ) is equal to (a) – 1 (b) 0 (c) 1 (d) a + b + c 47. A spherical balloon subtends an angle 2a at a man’s eye and the elevation of its centre is b. If q is the elevation of the highest point of the balloon at A then tan q is equal to (a)

sin a + cos b sin b

(b)

sin a + sin b cos b

(c)

sin a + cos b sin a

(d)

sin a + sin b cos a

48. A person stands at a point A due south of a tower and observes that its elevation is 60°. He then walks westwards towards B, where the elevation is 45°. At a point C on AB produced, he finds it to be 30°. Then AB/BC is equal to (a) 1/2 (b) 1 (c) 2 (d) 5/2 49. A pole stands at a point A on the boundary of a circular park of radius a and subtends an angle a at another point B on the boundary. If the chord AB subtends an angle a at the centre of the path, the height of the pole is (a) 2a cos (a/2) tan a (b) 2a sin (a/2) cot a (c) 2a sin (a/2) tan a (d) 2a cos (a/2) cot a 50. A, B, C are three points on a horizontal line through the base O of a pillar OP, such that OA, OB, OC are in A.P. If a, b, g the angles of elevation of the top of the pillar at A, B, C respectively are also in A.P. then sin a, sin b, sin g are in (a) A.P. (b) G.P. (c) H.P. (d) none of these

Heights and Distances 27.25

Previous Years' AIEEE/JEE Main Questions

1. The upper 3/4th portion of a vertical pole subtends an angle tan–1 (3/5) at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is (a) 40 m (b) 60 m (c) 80 m (d) 20 m [2003] 2. A person standing on the bank of a river observe that the angle of elevation of the top of a tree on the opposite bank of the river is 60º and when he retires 40 m away from the tree the angle of elevation becomes 30º. The breadth of the river is (a) 40 m (b) 30 m (c) 20 m (d) 60 m [2004] 3. A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB (= a) subtends an angle of 60º at the foot of the tower and the angle of elevation of the top of the tower from A or B is 30º. The height of the tower is (b) 2a 3 (a) 2a / 3 (c) a / 3 (d) a 3 [2007] 4. AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to a point D such that CD = 7m. From D the angle of elevation of the point A is 45°. Then the height of the pole is (a)

7 3 1 . m. 2 3 +1

(b)

7 3 1 . m. 2 3 -1

(c)

7 3 . 2

(d)

7 3 . 2

(

)

3 + 1 m.

(

)

3 - 1 m. [2008]

5. A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is (b) 40 ( 3 - 2 ) (a) 40 ( 2 - 1) (c) 20 2 (d) 20 ( 3 - 1) [2014] 6. The angle of elevation of the top of a vertical tower from a point P on the horizontal ground was observed to be a. After moving a distance 2 meters from P towards the foot of the tower, the angle of elevation

changes to b. Then the height (in meters) of the tower is: 2 sin a sin b sin a sin b (b) (a) sin ( b - a ) cos ( b - a ) (c)

2 sin ( b - a ) sin a sin b

(d)

cos ( b - a ) sin a sin b [2014, online]

7. If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot of the tower, art 30°, 45° and 60° respectively, then the ration AB: BC, is: (a)

3 :1

(b)

3: 2

(c) 1 : 3 (d) 2 : 3 [2015] 8. Let 10 vertical poles standing at equal distances on a straight line, subtend the same angle of elevation a at a point O on this line and all the poles are on the same side of O. If the height of the longest pole is 'h' and the distance of the foot of the smallest pole from O is 'a' then the distance between two consecutive poles, is: h sin a + a cos a h cos a - a sin a (b) (a) 9 sin a 9 cos a h cos a - a sin a 9 sin a

(d) h sin a + a cos a 9 cos a [2014, online] 9. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, from B to reach the pillar is (a) 6 (b) 10 (c) 20 (d) 5 [2016] 10. The angle of elevation of the top of a vertical tower from a point A, due east of it is 45°. The angle of elevation of the top of the same tower from a point B, due south of A is 30°. If the distance between A and B is 54 2 m, then the height of the tower (in meters) is; (a) 108 (b) 36 3 (c) 54 3 (d) 54 [2016, online] (c)

27.26

Complete Mathematics—JEE Main

Previous Years' B-Architecture Entrance Examination Questions 1. A vertical pole stands at a point A on the boundary of a circular park of radius a and subtends an angle a at another point b on the boundary. If the chord AB subtends an angle a at the centre of the park, the height of the pole is a tan a (a) 2a sin 2

a (b) 2a cos tan a 2

a (c) 2a sin cot a 2

a (d) 2a cos cot a. 2 [2014]

2. Two vehicles C1 and C2 start from a point P and travel east of P at the speeds 20 km/hr and 60 km/hr respectively. If an observer, one kilometre north of P, is able to see both the vehicles at the same time, then the maximum angle of sight between the observer's view of C1 and C2, is: (a)

p 3

2 (b) tan–1 ÊÁ ˆ˜ Ë 3¯

(c)

p 6

Ê 3ˆ (d) tan–1 Á ˜ Ë 2 ¯

44. (b)

45. (c)

46. (b)

48. (b)

49. (c)

50. (b)

Previous Years' AIEEE/JEE Main Questions 1. (a)

2. (c)

3. (c)

4. (c)

5. (d)

6. (a)

7. (a)

8. (c)

9. (d)

10. (d)

Previous Years' B-Architecture Entrance Examination Questions 1. (a)

2. (c)

Hints and Solutions Concept-based h 2 fi 2 sin (a + b ) = sin a sin b

1. h cot a + h cot b =

P

[2015] h

Answers

a

2. (a)

b O

A

Concept-based 1. (a)

47. (b)

3. (b)

Fig. 27.52

4. (b)

5. (a)

B

2. OA = BL = 15 cot 45° = 15 P

Level 1 6. (d)

7. (a)

8. (b)

9. (a)

10. (b)

11. (b)

12. (b)

13. (c)

14. (b)

15. (b)

16. (a)

17. (a)

18. (c)

19. (a)

20. (a)

21. (c)

22. (b)

23. (b)

24. (c)

25. (a)

26. (c)

27. (c)

28. (c)

29. (a)

30. (b)

31. (b)

32. (c)

33. (a)

34. (a)

35. (d)

15

B

45°

5 A

5 a

O

Fig. 27.53

Level 2

tan a =

36. (a)

37. (c)

38. (d)

39. (c)

40. (a)

41. (c)

42. (a)

43. (b)

OP 20 4 = = OA 15 3

fi cos a =

3 . 5

L

Heights and Distances 27.27

3. OA = h cot a OB = h cot 45° = h 50 = OB – OA = h [1 – cot a] 50 50 sin a = fih= 1 - cot a sin a - cos a

cos 2a =

15 4 , sin2 a = 19 19

cos 2a =

11 . 19

Level 1 6. PQ = OQ – OP = 100 tan 60º – 100 tan 45º = 100( 3 – 1)

P

45° B A

a

O 60°

Fig. 27.57

Fig. 27.54

7. (h – 10) cot 30º = h cot 60º 10 cot 30∞ fih= cot 30∞ - cot 60∞

4. OA = 40 cot – 45° a = 40 = 30 cot (45° – a) fi 4 tan (45° – a) = 3 fi 4 (1 – tan a) = 3 (1 + tan a)

=

fi tan a = 1/7 fi cot a = 7

10 3 3 - (1 / 3)

= 15

Q

P

10 m P

A

60°

O

A

O

Fig. 27.58

Fig. 27.55

5. Let the height of the poles be h. then h cot a = AE =

h

10

30 m

a 45° -a

45°

30°

B

8. h cot a = b cot b fi h = b tan a cot b

100 5 2

P

h cot 30° = 100 h=

100 3

cot a =

h

b

.

b

b

b

a

A

15 2

O

Fig. 27.59 Q h E

D

9. h cot 30º – h cot 75º = 1000 m. C P

fi

h=

1000 cot 30∞ - cot 75∞

a h A

30° 100 m B

Fig. 27.56

=

1000 3 -1 33 +1

= 250( 3 + 1 )m

27.28

Complete Mathematics—JEE Main P

P

h

A

30° 1 km

75°

q

O B

Fig. 27.60

10. A, B, C, D lie on a circle with centre at O, the foot of the flagstaff as OA = OB = OC = OD = h cot a, h being the height of the flagstaff and a the angle given. 11. Height = 100 tan 60º = 100 × 3 = 100 × 1.73 = 173 m. 12. 100(cot (45º – A) – cot(45º + A)) = 200 fi

45 °

h

1 + tan A 1 - tan A – =2 1 - tan A 1 + tan A

D

C

Fig. 27.63

15. sin 75° = h/25 fi

3 +1

h = 25 ¥

2 2

= (25/4) ( 3 + 1) 2 .

16. AC = AP tan a fi (1/2) AB = n AB tan a B

fi 4 tan A = 2(1 – tan2 A) fi

2 tan A 2

1 - tan A

C

= 1 = tan 45º b a

1 fi tan 2A = 45º fi A = 22 º . 2 P

Fig. 27.64 45°–A

45°+A

fi

100 m

tan a =

tan(a + b) =

200 m

O

Fig. 27.61

fi

13. h cot A = OM + ML = h + 3 h cot (90° – A) = h – 2 h = (h + 3) (h – 2)

AB 1 = . AP n

tan a + tan b 1 = . 1 - tan a tan b n

h= 6 m fi

P h

L

A 45° 90° – A 3 M 2 N

O

2n + 1

.

1 ˆ Ê = 150 Á 1 – ˜ = 50 (3 - 3 ) Ë 3¯

14. AC = h cot 45° = h

fi q = tan–1 (2 / 3 )

2

17. A B = (cot 45° – cot 60°) 150

speed = 3h = h cot q 2

n

tan b =

Fig. 27.62

AD =

1 2n

1 ˆ = 1 – 1 tan b fi nÊ Ë 2n + tan b ¯ 2n

2

fi

A

P

50 (3 - 3 ) m/min. 2

= 25 (3 - 3 ) ¥

=

60 km/hr 1000

9-3 3 km/h. 2

Heights and Distances 27.29 P

fi h tan a + h tan q = a – a tan a tan q fi tan q =

a–b a–b = . h + a tan a 2 ab

fi sin q =

a–b a+b

150 m

45° A

60° B

P

O

Fig. 27.65

q

h

18. OP = x tan 60° = (40 + x) tan 30° 40 tan 30∞ fi x= tan 60∞ - tan 30∞ 1 ˆ 3Ê 3– Ë 3¯

O

Fig. 27.68

21. tanq =

h h , tan (a + q) = . 160 40

tan a + fi 30° 60° x 40 B

B

= 20 m

P

A

90° – a b

a A

a

40

=

a

O

Fig. 27.66

19. The base of the mountain is at the circumcentre of the circle passing through A, B and C if O is the centre and R the radius of the circle, then R = h cot a.

h 160

= h 40

h 1◊ tan a 160

h 3 + h 5 160 fi = h 3 40 ¥ 1– 160 5 fi 3h2 – 600h + 19200 = 0 fi h = 40 or 160. P

A

3

4

h

Q a

O R B

a

A

a a = 2 R 2h cot a

fi

a tan a ◊ cosec A 2 20. tan a = h/a, cot a = h/b. fi h = ab .

P

tan (a + q) = a /h. tan a + tan q 1 - tan a tan q

=

O

AB cot a - cot b . = BC cot b - cot g

fih=

fi

h

22. AB = h[cot a – cot b] BC = h[cot b – cot g]

Fig. 27.67

sin A =

40

4

Fig. 27.69

C

D

2

1

q

a . h Fig. 27.70

27.30

Complete Mathematics—JEE Main

23. BC= h cot b AC= h cot a. (AC)2= a2 + (BC)2 fi h2 cot2 a = a2 + h2 cot2 b fi

a

h=

cot 2 a - cot 2 b

26. Distance travelled in 10 sec = AB = PQ. = OQ – OP = (cot 30° – cot 60°) km. .

3 – 1/ 3 ¥ 60 ¥ 60 km. 10

in 1 hour =

=

2 ¥ 60 ¥ 60 3 = 240 30 A

3 km.

B

Fig. 27.71 1 km

24. OA= a, OB = b. h cot a = a, h cot b = b

O

a+b h= . cot a + cot b

fi

27. AB is the shadow of the tower PQ (= 6 m) on the ground.

h

So 2 3 = (x + 6) cot q – x cot q

O

a

Q

Fig. 27.74

P

A

60°

30° P

b

B

60°

60°

C1

C2

fi cot q = fi

2 3 1 = 6 3

q = 60°.

Fig. 27.72 Q

25. PQ = a – b fi QL = a – b fi OA = a – b fi a cot a = a – b a–b fi cot a = a b–c Similarly cot b = . b

6 P x q A 2 3 B

c b ac – b2 – = b a ab

So that cot a – cot b =

P

45°

L

Q a

N b

M

c 45° C

b B

a A

Fig. 27.73

O

q O

Fig. 27.75

28. If r is the radius of the circle, then the heights of the poles are r tan a, r tan b, r tan g. Since they are in A.P. cot a, cot b, cot g are in H.P. 29. If a is the distance of the point from the foot O, of the poles, then the distances of the points A, B, C from O are a tan a, a tan b, a tan g, since they are in A.P; tan a, tan b, tan g are in A.P. So 2 tan b = tan a + tan g. Also a + b + g = p fi tan a + tan b + tan g = tan a tan b tan g fi tan a tan g = 3. 30. AP and BQ be the two positions of the ladder of length l.

Heights and Distances 27.31 P b Q

fi

h cot a = a cos a + a + h – a sin a. h (cot a – 1) = a(cos a + 1 – sin a) statement-1 is True. Q h/3 P

35° O

25° A a

h

B a

Fig. 27.76

b

A

O

a l (cos 25∞ - cos 35∞) = b l (sin 35∞ - sin 25∞) =

Fig. 27.79

In statement-2 (Fig. 27.79) OA = h cot b = (h + h/3) cot a.

2 sin 30∞ sin 5∞ 2 cos 30∞ sin 5∞

fi = tan 30° = 1 / 3 . AB = OA – OB = (a – b) cot a

31.

BC = OB – OC = (b – c) cot a fi AB = BC

a = (3/4) b fi / fi statement-2 is False. 33. BD = h cot a, CD = h cot b. Statement-2 is True, so using it

a b a C

BD c h cot a = = . CD b h cot b

fi b cot a = c cot b fi statement-1 is True. 34. Statement-2 is True. Using it BQ || AP fi QBO = PAO = a and the statement-1 is also True.

c O

cot a = (3/4) cot b

B

A

P

Fig. 27.77

fi

fi 32. Let OP OA

[∵ a, b, c are in A.P. fi b – a = c – b] Statement-1 is True. In statement-2 if r is the radius of the circular field, then OB = r, AB = 2r Height of the pole at A = 2r tan a = 2 ¥ height of the pole at O. Statement-2 is also correct but does not lead to statement-1. h be the height of the tower, BD ^ OA; BE ^ then (Fig. 27.78) = AD + BC + CE

Q

a A

a O

B

Fig. 27.80

35. Let h be the height of the tower in statement-1 and x be the side of the square, then diagonal of the square is 2h cot a fi (2h cot a)2 = 2x2 fi 4h2 cot2 a = 2x2 [∵ h = 2x] fi 16 cot2 a = 2 fi tan a = 2 2 fi

statement-1 is false.

In statement-2 3 -1 AB cot 30∞ - cot 45∞ = 3 = = BC cot 45∞ - cot 60∞ 1 - 1 3

(

Fig. 27.78

fi

statement-2 is True.

)

27.32

Complete Mathematics—JEE Main

Level 2

POA = a, QOB = b, ROC = g, SOD = d

36. Let the height of the tower be h PAQ = a, RBN = b, BM = a

fi OA = h cota, OB = h cotb, OC = h cot g and OD = h cot d, AB = BC = CD.

fi RN = (h/2) – a AP = h cot a = AM + MP = a cota + (h/2 – a) cotb

AD = h (cot d – cota) = 3 BC = 3h (cot g – cot b) fi cot a – cot d = 3 (cot b – cot g )

Q

Also cot a + cot d = cot b + cot g R b

B

So cot2a – cot2d = 3 (cot2b – cot2g )

h

39. (AP)2 = 4 + 1 = 5

N

a

a A

(AQ)2 = 4 + 4 = 8

M

P

(AR)2 = 4 + 9 = 13

Fig. 27.81

fi h (cot a – (h/2) cot b) = a (cot a – cot b ) fi h = 2a

(AS)2 = 4 + 4 = 16 S

(tan b - tan a ) 2 tan b - tan a

÷20 ÷13 R

37. If h is the height of the tower OP at the centre O of the circular garden. Then AOB = 2 b fi AOC = b

Q ÷8 b P

÷5

P

a q A

1 O

2

Fig. 27.84 b

O

3

cos a = 8 + 5 - 1 = 2 8¥5

a

1

fi sina =

10

10

g A

C

B

cos b =

20 + 13 - 1 2 20 ¥ 13

8

=

65

Fig. 27.82

OA = OB = h cot a OC = h cot g. OA = sec b fi tan g = tan a sec b. fi OC 38. Let P, Q, R, S be the positions of the bird when it makes angles a, b, g, d respectively at a point, O, on the ground. P

Q

R

3¥8

cos (a + b ) =

fi sinb =

650

1¥1

-

650

65 =

P

q

g

AP = h cota =

d A

B

Fig. 27.83

C

D

C

Q

a

Fig. 27.85

O

650

b

S A

b a

23

40. Let the height of the lamppost be h. D

b2 + a2 4

AQ = DQ = hcot b =

1

a2 + b2 4

B

Heights and Distances 27.33

cosq =

=

= RB + BS = AB (tan a + tan b ) = a cosa (tan a + tan b ) sin (a + b ) = a cos b

( AP )2 + ( AQ)2 - (a2 4 + b2 4) 2 AP ◊ AQ ◊

a2 + b2 2 h cot a ¥ h cot b a 2 + b2

fi k = cota cotb cosq =

2h

43. PBO = q, QBO = 45°

2

fi PBQ = q – 45°

41. AB = h (cota – cotb)

PQ = h tan(q – 45°)

OB = h cot b, OC = h cot g

q - 45°

P

B

Q

From right angled triangle OBC (BC)2 = h2 [cot2 g – cot2 b] From right angled triangle ABC

45°

h

h

P A

h

O

N

Fig. 27.89 h

a

O

b B

A

W

Height of the tower is PQ + OQ. È tan q - 1 ˘ 2h + 1˙ = = hÍ Î tan q + 1 ˚ 1 + cotq

90°

E

g

S

44. ROX = 9° = ORQ . ORP = 54° fi QRP = 45°

C

P

Fig. 27.86 C

hc

ot

g

54° 45°

Q B

h cot b

O

O

Fig. 27.87 2

2

2

40 m

R

9° 9°

X

Fig. 27.90 2

2

(AC) = h [(cota – cotb ) + cot g – cot b] = h2 [cot2 a + cot2 g – 2 cota cotb]

fi QPR = 45° fi PQ = QR OR = 40 m fi QR = 40 cos 9°, OQ = 40 sin 9° Height of the tower OP = OQ + QP = 40 (sin 9° + cos 9°)

42. h = PQ = RS

= 40 1 + sin 18∞ = 40 1 +

a

5 -1 4

= 20 3 + 5 = 45.76 m. 45. QPC = 15°, QPR = 45°, CQ = x

Fig. 27.88

27.34

Complete Mathematics—JEE Main R

PQ = QR = 2500 + 2500 + x C (cloud)

O

x

P

15°

A

Q

45°

2500

r

2500 m Lake

A

a

b

P

Q

q

2500 + x

Fig. 27.93

48. Let OP be the tower of height h. R

(Reflection)

Fig. 27.91

Also PQ = x cot 15° 5000 So x = = 2500 cot 15∞ - 1

(

OAP = 60°, OBP = 45°, OCP = 30°, OAC = 90° P

)

3 -1

W

Height of the cloud above the lake = x + 2500 = 2500 3 m.

C

45°

B

46. Let h be the height of the tower at P, then A

15°

60° A

S

N

O E

c

120° P 120°

120°

B

Fig. 27.94

b

a

C

C

h ÷3

B

h

Fig. 27.92

PA = h cota, PB = h cot b, PC = h cot g . From triangle PBC cos 120° = 2

2

A

Fig. 27.95

h2 cot 2 g + h2 cot 2 b - a2 2 h2 cot b cot g 2

O

h / ÷3

OA = h cot 60° = h

3

2

fi a = h [cot b + cot g + cot b cot g ] fi a2 (cot b – cot g ) = h2 [cot3 b – cot3 g ] Similarly for b2 (cot g – cot a) and c2(cot a – cot b )

OB = h, OC = h cot 30° = h (AB)2 = h2 (1 –1/3) = (2/3) h2 (AC)2 = h2 (3 – 1/3) = (8/3) h2

47. Let r be the radius of the sphere with centre O OPA = a, OPQ = b, RPQ = q OP = r cosec a (From D OAP) OQ = OP sin b = r cosec a sin b PQ = OP cos b = r cosec a cos b QR r + r cos ec a sin b sin a + sin b tan q = = = cos b PQ r cos ec a cos b

3

(

)

BC = AC – AB = 2 2 / 3 - 2 / 3 h = 2 / 3 h = AB Hence AB/BC = 1 49. Let AP be the pole of height h. PBA = a fi AB = h cot a

Heights and Distances 27.35 P P

3/4 h O a a a

A

Q a B

B

O

40

Fig. 27.98

Fig. 27.96

OA = OB = a From triangle OAB cos a =

1/4 h

q

fi

a2 + a2 - h2 cot 2 a 2a 2

fi h2 cot2 a = 2a2 (1 – cos a) = 4a2 sin2 (a /2) fi h = 2a sin (a /2) tana 50. Let OP be the pillar of height h OAP = a, OBP = b, OCP = g

h 3 + 5 160 = h h 3 40 1¥ 160 5

fi 3h2 – 600h + 19200 = 0 fi h = 40 or 160. 2. We have h = tan 60° b fi h= 3b T

2 h cotb = h (cota + cot g ) P

h 30° 40 m

B

A

60° b

O

Fig. 27.99 g C

b B

a A

O

h 1 = tan 30∞ = 40 + b 3

and

Fig. 27.97

fi 2

sin (a + g ) cos b = sin a sin g sin b sin 2b = sin a sin g =

2 sin b cos b sin a sin g 2

fi sin a sin g = sin b fi sina, sin b, sin g are in G·P.

Previous Year's AIEEE/JEE Main Questions 1. tan q =

h h , tan(a + q) = . 160 40

h 160 = h h 40 1◊ tan a 160 tan a +

fi

fi

( 3b)

3 = 40 + b fi 2b = 40 fi b = 20

3. OA = OB = AB = a fi

h = tan 30∞ a

fi

h=

a

h O a A

60° a 30° a

3

B

Fig. 27.100

4. h – h cot 60° = 7 fi

h=

1 3

17 3 = 2

(

A

7

)

3 +1

5. RS = PQ = 20 fi OS = 20 Let SP = d = 20 cot 30° – 20 = 20

(

h

3 -1

)

D

45° 7 C

60° b

Fig. 27.101

B

Complete Mathematics—JEE Main

27.36

B

Q

Then OA10 = a + 9d Now, c tan a = a

20 m

h - c h - a tana = 9d 9d 9d tan a = h – a tan a h cos a - a sin a fi d= 9 sin a 9. Refer figure and tan a =

30° 45° O

S

P

Fig. 27.102

Required speed is 20

(

)

3 - 1 m/s A

6. h cot a – h cot b = 2 2 fih= cot a - cot b = 7.

h

2 sin a sin b sin(b - a )

P

b a 2 Q

PQ = tan 60∞ CQ fi

B

h = tan 30∞ a+b and

h = tan 60∞ b

fi

a + b = 3h, b =

Fig. 27.103

h 3 T

CQ = h cot 60° = hl 3 P h h 30°

45°

A

B

30° A

60° C

Q

fi

Similarly 3h

Time taken to cover a = fi

AB AQ - BQ 3h - h 3 = = = 1 BC BQ - CQ h - h / 3

A 1B 1 = C B10 B9

h-c

B1 a

time taken to cover b =

Fig. 27.105

h units is 10 minutes 1 3

h units is 5 minutes.

A9 d A10

T

h

OT 1 = tan 30∞ = OB 3 2

Also, OB = OA + AB fi

45°

A

OB = 3OT = 3h 2

A1 d A2

3

OT = tan 45∞ = 1 OB fi OA = OT = h In right triangle, DBOT

fi

c O

2

10. Let OT = h In right triangle DAOT,

8. Let OA1 = a and A1A2 = A2A3 = … d.

a a

O

1 ˆ 2 a = ÊÁ 3 h= h ˜ Ë 3¯ 3

Now,

B2

60° b

Fig. 27.106

Fig. 27.104

BQ = h and AQ =

B

a

54 ÷2 2

3h 2 = h 2 + (54 2) 2

30°

B

Fig. 27.107

O

Heights and Distances 27.37

fi

2h 2 = (54 2)2

fi

h = 54

O q1 q2 q

Previous Years' B-Architecture Entrance Examination Questions 1. AP = h ABP = a , AOB = a

P

a fi BOQ = . OB = a 2 AB = h cot a Also AB = 2BQ a fi h cot a = 2a sin 2 fi

a h = 2a sin tan a 2

= h O a/2

tan q = tan (q 2 - q1 ) =

sec2 q

a B

Q

(

)

(

= 40

Fig. 27.108

We have and

40t 1 + 1200t 2

1 + 1200t 2 (1) - t (2400t ) dq = 40 2 dt 1 + 1200t 2

A

)

1 - 1200t 2 2

(1 + 1200t 2 )

dq 1 =0fit = dt 20 3

dq 1 > 0 for 0 < t < dt 20 3

dq 1 < 0 for t < dt 20 3 \ q is maximum when 1 1 p t= fi tan q = fiq = 6 20 3 3 and

tan q 2 - tan q1 1 + tan q 2 tan q1

C2

Fig. 27.109

P

2. At time t, PC1 = 20t and PC2 = 60t As OP = 1, PC1 PC2 tan q1 = = 20t , tan q 2 = = 60t OP OP Now

C1

CHAPTER TWENTY EIGHT

Some Logical Equivalences

SOME DEFINITIONS 1. A statement or proposition is a declarative sentence which is either true or false but not both simultaneously. 2. A paradox is a sentence which is both true and false simultaneously. 3. A statement is said to be simple if it cannot be broken further into simple statements otherwise statement is said to be compound. Truth value of a compound statement is completely determined by its constituent statements. 4. Table for basic logical connectives. p

5. 6. 7. 8.

q

pŸq p⁄q pÆq p´p

~p

T T T T T T F T F F T F F F F T F T T F T F F F F T T T Contrapositive of p Æ q is ~ q Æ ~ p Converse of the conditional statement p Æ q is q Æ p Inverse of the conditional statement p Æ q is ~ p Æ ~ q. p ´ q ∫ ( p Æ q) Ÿ (q Æ p) TIP Negation of an if statement does not start with the word if.

Some Definitions 1. A statement is said to be a tautology if it is always true. 2. A statement is a contradiction if it is always false. 3. Dual of a compound statement not involving logical connective other than Ÿ and ⁄ is obtained by replacing each occurrence of Ÿ and ⁄ by ⁄ and Ÿ respectively, and each occurance of T by F and F by T.

Given any statement variables p, q and r, a tautology t and a contradiction c, the following logical equivalances hold: 1. Commutative laws: (i) p Ÿ q ∫ q Ÿ p (ii) p ⁄ q ∫ q ⁄ p 2. Associative laws: (i) (p Ÿ q) Ÿ r = p Ÿ (q Ÿ r) (ii) (p ⁄ q) ⁄ r = p ⁄ (q ⁄ r) 3. Distributive laws: (i) p Ÿ (q ⁄ r) ∫ (p Ÿ q) ⁄ (p Ÿ r) (ii) p ⁄(q Ÿ r) ∫ (p ⁄ q) Ÿ (q ⁄ r) 4. Identity laws: (i) p Ÿ t ∫ p (ii) p ⁄ c ∫ p 5. Negation laws: (i) p ⁄ (~p) ∫ t (ii) p Ÿ (~p) ∫ c 6. Idempotent laws: (i) p Ÿ p ∫ p (ii) p ⁄ p ∫ p 7. De Morgan’s Laws: (i) ~ (p Ÿ q) ∫ (~p) ⁄ (~q) (ii) ~(p ⁄ q) ∫ (~p) Ÿ (~q) 8. Universal bound laws: (i) p ⁄ t ∫ t (ii) p Ÿ c ∫ c 9. Absorption laws: (i) p ⁄ (p Ÿ q) ∫ p (ii) p Ÿ (p ⁄ q) ∫ p 10. Double negative laws: ~ (~p) ∫ p 11. Negation of t and c (i) ~ t ∫ c (ii) ~ c ∫ t

28.2

Complete Mathematics—JEE Main

The principle of Duality Let s and t be statements that contain no logical connectives other than Ÿ and

⁄. If s ´ t, then sd ´ td where sd denotes dual of s etc. Quantifiers are the phrases like “There exists” and “for every”.

SOLVED EXAMPLES Concept-based Straight Objective Type Questions

Example 1: Which of the following is not a statement? (a) 17 is a prime number (b) 22 is an odd number (c) What a beautiful flower ! (d) New Delhi is Capital of India Ans. (c) Solution: (a), (b) and (d) can be assigned values true or false: Example 2: Let p and q be the statements: p : Rakshit gets 100% marks in mathematics q : Rakshit is a good dancer. The verbal form of ~ p Æ q is (a) If Rakshit does not get 100% marks in mathematics, then Rakshit is a good dancer (b) If Rakshit gets 100% marks in mathematics then Rakshit is a good dancer (c) Mathematics is good for dancing (d) Rakshit is good in mathematics and dance Ans. (a) Solution: Verbal interpretation of ~ p Æ q is If Rakshit does not get 100% marks in methematics then Rakshit is a good dancer. Example 3: If truth value of p is T, q is F, then truth values of (p Æ q) and (q Æ p) ⁄ (~p) is are respectively (a) F, F (b) F, T (c) T, F (d) T, T Ans. (b) Solution: If p = T, q = F, then p Æ q is F. and q Æ p is T. \ truth value of (q Æ p) ⁄ (~p) is T Example 4: If p, q, r are three statements and truth value of p Ÿ q Æ r is F, then truth values of p, q, r are respectively (a) T, F, T (b) T, T, F (c) F, T, T (d) F, F, T Ans. (b) Solution: As p Ÿ q Æ r has truth

value F, p Ÿ q is T and r is F \ p = T, q = T, r = F Example 5: If p is any statement, then which of the following is a contradiction? (a) p Ÿ p (b) p Ÿ ~p (c) p ⁄ (~p) (d) (~p) Ÿ (~p) Ans. (b) Solution: p Ÿ (~p) ∫ F for both the truth values of p. Example 6: Let p and q be two statements. If truth value of p Æ (~p Ÿ q) is F, then truth values of p, q are respectively: (a) F, F (b) T, F (c) F, T (d) T, T Ans. (b) Solution: As p Æ (~p ⁄ q) has truth value F, p must be T and ~p ⁄ q must be F. As ~p is F, q must be F. Thus, truth values of p and q are respectively T and F. Example 6: For integers m and n, both greater than 1, consider the following three statements p : m divides n q : m divides n2 r : m is prime Then (a) q Ÿ r Æ p (b) p Ÿ q Æ r (c) q Æ r (d) q Æ p Ans. (a) Solution: We know that if m is prime and m|n2 the nm|n. Thus, q Ÿ r Æ p Example 7: The statement ~ (p ´ ~ q) is (a) equivalent to ~p ´ q (b) a tautology (c) a fallacy (d) equivalent to p ´ q Ans. (d) Solution: ~ (p ´ ~ q) ∫ ~((p Æ ~q) Ÿ (~q Æ p) ∫ ~((~p ⁄ ~ q) Ÿ (q ⁄ p))

Mathematical Reasoning 28.3

∫ ~ (~q ⁄ ~ p) ⁄ (~(q ⁄ p)) ∫ (~(q) Ÿ ~(~p)) ⁄ (~q Ÿ ~p) ∫ (q Ÿ p) ⁄ (~q Ÿ ~p) ∫ [(q Ÿ p) ⁄ (~q)] Ÿ [(q Ÿ p) ⁄ (~p)] ∫ [(q ⁄ (~q)) Ÿ (p ⁄ (~q))] Ÿ [(q ⁄ (~p)) Ÿ (p ⁄ (~p))] ∫ [t Ÿ (~q ⁄ p)] Ÿ [((~p) ⁄ q)) Ÿ t] ∫ (~q ⁄ p) Ÿ (~p ⁄ q) ∫ (q Æ p) Ÿ (p Æ q) ∫p´q Alternative Solution. Use the following table. p

q

~q

p ´ ~q ~(p ´ ~q) p ´ q

T

T

F

F

T

T

T

F

T

T

F

F

F

T

F

T

F

F

F

F

T

F

T

T

Ans. (a) Solution: ~(p Ÿ q) ⁄ q ∫ ((~p) ⁄ (~q)) ⁄ q ∫ (~p) ⁄ ((~q) ⁄ q) ∫ (~p) ⁄ t ∫ t Thus, ~(p Ÿ q) ⁄ q is a tautology. Example 9: If p is any logical statement, then: (a) p Ÿ (~p) is a tautology. (b) p ⁄ (~p) is a contradiction (c) p Ÿ p ∫ p (d) p ⁄ (~p) ∫ p Ans. (c) Solution: Idempotent law p Ÿ p ∫ p is always true.

From the last two colums, we get ~(p ´ (~q)) ∫ p ´ q

Example 10: If p, q are two statements, then ~(~p Ÿ q) Ÿ ( p Ÿ q) is logically equivalent to (a) p (b) q (c) p Ÿ q (d) (~p) ⁄ q Ans. (a) Solution: ~(~p Ÿ q) Ÿ (p ⁄ q) ∫ ((~(~p)) ⁄ (~ q)) Ÿ (p ⁄ q) ∫ (p ⁄(~q)) Ÿ (p ⁄ q) ∫ p ⁄ (~q Ÿ q) ∫ p ⁄ c ∫ p

Example 8: The statement ~(p Ÿ q) ⁄ q: (a) is a tautology (b) is equivalent to (p Ÿ q) ⁄ (~q) (c) is equivalent to p ⁄ q (d) is a contradiction

LEVEL 1 Straight Objective Type Questions Example 11: Which of the following is not a negation of the statement p: 5 is rational? (a) It is not the case that (b)

5 is rational

5 is not rational

(c) 5 is an irrational number (d) none of these Ans. (d) Solution: (a), (b) and (c) are negation of p. Example 12: Negation of tional is

5 is irrational or 3 is ra-

(a)

5 is rational or 3 is irrational

(b)

5 is rational and 3 is rational

(c) 5 is rational and 3 is irrational (d) none of these Ans. (c)

Solution: Use De Morgan’s Law. Example 13: Contrapositive of the statement. If a number is divisible by 9, then it is divisible by 3, is (a) If a number is not divisible by 3, it is not divisible by 9. (b) If a number is not divisible by 3, it is divisible by 9. (c) If a number is not divisible by 9, it is not divisible by 3. (d) none of these Ans. (a) Solution: Contrapositive of p Æ q is ~ q Æ ~ p. Example 14: Converse of the statement: If a number n is even, then n2 is even, is (a) If a number n 2 is even, then n is even (b) If n is not even, then n 2 is not even

Complete Mathematics—JEE Main

28.4

(c) Neither n nor n 2 is even (d) none of these Ans. (a) Solution: Converse of p Æ q is q Æ p. Example 15: Which of the following statement is not equivalent to p ´ q? (a) p if and only if q (b) q if and only if p (c) p is necessary and sufficient for q (d) none of these Ans. (d) Solution: Each of (a), (b), (c) is equivalent to p ´ q. Example 16: Negation of the statement p: for every real number, either x > 1 or x < 1 is (a) There exist a real number x such that neither x > 1 nor x < 1 (b) There exist a real number x such that 0 < x < 1 (c) There exist a real number x such that neither x ≥ 1 nor x £ 1 (d) none of these Ans. (a) Solution: Negation of p ⁄ q is (~ p) Ÿ (~ q). Example 17: Contrapositive of p: “If x and y are integers such that xy is odd then both x and y are odd” is (a) If both x and y are odd, then xy is odd (b) If both x and y are even, then xy is even (c) If x or y is odd, then xy is odd (d) If it is false that both x and y are odd then the product xy is odd Ans. (d) Solution: Contrapositive of p Æ q is ~ q Æ ~ p. Example 18: The statement p Æ (q Æ p) is equivalent to (a) p Æ (p Æ q) (c) p Æ (p Ÿ q)

(b) p Æ (p ⁄ q) (d) p Æ (p ´ q)

Solution: p Æ (q Æ p) ∫ ~ p ⁄ (q Æ p) ∫ (~ p) ⁄ (~ q ⁄ p) ∫ (~ q) ⁄ ( p ⁄ ~ p) ∫ (~ q) ⁄ T = T p Æ (q Æ p) is a tautology.

Also

∫ (~ p ⁄ q) ⁄ (~ p ⁄ r) ∫ (p Æ q) ⁄ (p Æ r) Also,

p Æ (q ⁄ r) ∫ (~ p) ⁄ (q ⁄ r) ∫ (~ p ⁄ q) ⁄ r ∫ ~ (p Ÿ (~ q)) ⁄ r

∫ p Ÿ (~ q) Æ r Interchanging the roles of q and r in the above paragraph, we find p Æ (q ⁄ r) ∫ p Ÿ (~ q) Æ r ∫ p Ÿ (~ r) Æ q For p = T, q = F, r = F, p Æ (q ⁄ r) is F, but (p Ÿ q) Æ (p ⁄ r) ⁄ (q Ÿ r) is T. Therefore, p Æ (q ⁄ r) and p Ÿ q Æ (p Ÿ r) ⁄ (q Ÿ r) are not equivalent. Example 20: Negation of p Æ q is (a) p Ÿ (~ q) (b) ~ p ⁄ q (c) ~ q Æ ~ p (d) p ⁄ (~ q) Ans. (a) Solution: ~ (p Æ q) ∫ ~ (~ p ⁄ q) ∫ ~ (~ p) Ÿ (~ q) [De Morgan’s Laws] \ ~ (p Æ q) ∫ p Ÿ (~ q) Example 21: Contrapositive of p Æ (q Æ r) is logically equivalent to (a) p Æ (q Æ r) (b) (q Æ r) Æ ~ p (c) p ⁄ q Æ r (d) (q Æ r) Æ p Ans. (a) Solution: Contrapositive of p Æ (q Æ r) is ~ (q Æ r) Æ ~ p ∫ ~ [~ (q Æ r)] ⁄ (~ p)

Ans. (b)

\

Example 19: The statement p Æ (q ⁄ r) is not equivalent to (a) (p Æ q) ⁄ (p Æ r) (b) p Ÿ (~ q) Æ r (c) p Ÿ (~ r) Æ q (d) p Ÿ q Æ (p Ÿ r) ⁄ (q Ÿ r) Ans. (d) Solution: p Æ (q ⁄ r) ∫ (~ p) ⁄ (q ⁄ r)

p Æ (p ⁄ q) ∫ ~ p ⁄ (p ⁄ q)

∫ (~ p ⁄ p) ⁄ q ∫ T ⁄ q = T \ p Æ (p ⁄ q) is also a tautology. Thus, p Æ (q Æ p) is equivalent to p Æ (p ⁄ q).

∫ (q Æ r) ⁄ (~ p) ∫ (~ p) ⁄ (q Æ r) ∫ p Æ (q Æ r) Example 22: Let S be a non-empty subset of R. Consider the following statement: P : There is a rational number x Œ S such that x > 0. Which of the following statements is the negation of the statement P ? (a) Every rational number x Œ S satisfies x £ 0. (b) x Œ S and x £ 0 fi x is not a rational number.

Mathematical Reasoning 28.5

(c) There is a rational number x Œ S such that x £ 0. (d) There is no rational number x Œ S such that x < 0. Ans. (a) Solution: The statement P can be written as follows : P : $ x Œ Q « S such that x > 0 Negation of P is ~ P : " x Œ Q « S satisfies x < 0. Example 23: The negation of the following statement : P : Neha lives in Ludhiana or she lives in Gurudaspur. (a) Neha does not live in Ludhiana or she does not live in Gurudaspur. (b) Neha does not live in Ludhiana and she does not live in Gurudaspur (c) Neha does not live in Punjab. (d) None of these. Ans. (b) Solution: Let p and q be the statements defined as follows : p : Neha lives in Ludhiana q : Neha lives in Gurudaspur. The statement P is P:p⁄q Negation of p is ~ P : ~ (p ⁄ q) or ~ P : (~ p) Ÿ (~ q) [Q ~ (p ⁄ q) ∫ (~ p) Ÿ (~q)] Thus, negation of P is given by (b) Example 24: The converse of the statement “If a < b then x + a < x + b”, is (a) If a > b then x + a > x + b (b) If a > b then x + a > x + b (c) If x + a < x + b then a < b (d) If x + a > x + b then a > b Ans. (c) Solution: See theory. Example 25: Which of the following is the conditional statement p Æ q ? (a) p is necessary for q (b) p is sufficient for q (c) p only if q (d) if q then p Ans. (b) Solution: As p Æ q, therefore truth of p is sufficient for truth of q. Example 26: Converse of the statement “if x2 is odd then x is odd” is

(a) (b) (c) (d) Ans. (b)

if x2 is even then x is even if x is odd then x2 is odd if x is odd then x2 is even if x is even then x2 is odd

Solution: See theory. Example 27: Consider the following statements : P: Suman is brilliant Q: Suman is rich R: Suman is honest The negation of the “Suman is brilliant and dishonest if and only if Suman is rich” can be expressed as (a) ~ (P Ÿ ~R ) ´ Q (b) ~ p Ÿ (Q ´ ~ R) (c) ~ (Q ´ (P Ÿ ~ R)) (d) ~ Q ´ ~ p Ÿ R Ans. (c) Solution: P Ÿ ~ R stands for Suman is brilliant and dishonest. Thus, P Ÿ ~ R ´ Q stands for Suman is brilliant and dishonest if and only if Suman is rich. Its negation is ~ (P Ÿ ~ R ´ Q) or ~ (Q ´ P Ÿ ~ R) Example 28: The only statement among the following that is a tautology is: (a) A Ÿ (A ⁄ B) (b) A ⁄ (A Ÿ B) (c) [A Ÿ (A Æ B)] Æ B (d) B Æ [A Ÿ (A Æ B)] Ans. (c) Solution: Note that A Ÿ (A ⁄ B) is F when A = F, A ⁄ (A Ÿ B) is F when A = F, B = F, and B Æ [A Ÿ (A Æ B)] is F when A = F, B = T \ We check only (c) [A Ÿ (A Æ B)] Æ B ∫ [A Ÿ (~ A ⁄ B)] Æ B ∫ [(A Ÿ (~ A))] ⁄ (A Ÿ B)] Æ B ∫ AŸBÆB ∫ ~ (A Ÿ B) ⁄ B ∫ ~ [(A Ÿ B) Ÿ (~ B)] ∫ ~ [A Ÿ (B Ÿ ~ B)] ∫ ~ [A Ÿ F] ∫ ~ F ∫ T Thus, [A Ÿ (A Æ B)] Æ B is a tautology. Example 29: The negation of the statement “If I become teacher, then I will open a school.” (a) Either I will not become a teacher or I will not open a school. (b) Neither I will become a teacher nor I will open a school (c) I will not become a teacher or I will not open a school (d) I will become a teacher and I will not open a school Ans. (d)

28.6

Complete Mathematics—JEE Main

It negation is ~ (~ p) ⁄ q) ∫ p Ÿ (~ q) Thus, negation of the given statement is I will become a teacher and I will not open a school.

Solution: Let p : I become a teacher q : I will open a school. The given statement is p Æ q ∫ (~ p) ⁄ q

Assertion-Reason Type Questions

(p Æ q) Ÿ (q Æ r) Æ (p Æ r) Assume it to be F then p Æ r is F and (p Æ q) Ÿ (q Ÿ r) is T. But p Æ r is F ¤ p = T and r = F. Now, (T Æ q) Ÿ (q Æ F) must be T

Example 30: Statement-1 ~ (p ´ ~ q) is equivalent to p ´ q Statement-2 ~ (p ´ ~ q) is a tautology Ans. (c) Solution: Table for basic logical connectives. p ´ ~ q ~ (p Æ ~q) (p ´ q)

fi

T Æ q is T and q Æ F is T

fi

q is T and q is F.

p

q

~q

T

T

F

F

T

T

A contradiction.

T

F

T

T

F

F

\

F

T

F

T

F

F

cannot be F for any assignment of T and F to p, q, r.

F

F

T

F

T

T

\

Note that ~ (p ´ ~ q) is not tautology. \ Statement-2 is false. From table ~ (p ´ ~ q) is equivalent to (p ´ q) Thus, Statement-1 is true. Example 31: Let p, q and r be the statements. p : X is a rectangle q : X is a square r:pÆq Statement-1: p Æ r is a tautology. Statement-2: A tautology is equivalent to T. Ans. (d) Solution: p Æ r ∫ p Æ (p Æ q) ∫ (~ p) ⁄ (p Æ q) ∫ (~ p) ⁄ [(~ p) ⁄ q] ∫ [(~ p) ⁄ (~ p)] ⁄ q ∫ (~ p) ⁄ q ∫ p Æ q ∫ r Thus, Statement-1 is not a tautology. Example 32: Let p, q, r be three statements. Statement-1: If p Æ q and q Æ r then p Æ r is a tautology. Statement-2: (p Æ q) Ÿ (q Æ r) ´ (p Æ r) Ans. (c) Solution: Statement-2 is not true. It can be checked by taking p = T, q = F, r = T We can write Statement-1 as

(p Æ q) Ÿ (q Æ r) Æ (p Æ r) (p Æ q) Ÿ (q Æ r) Æ (p Æ r) is a tautology.

We may abbreviate the above procedure as follows: 1. (p Æ q) Ÿ (q Æ r) Æ (p Æ r) F

2. (p Æ q) Ÿ (q Æ r) Æ (p Æ r) T

F

F

3. (p Æ q) Ÿ (q Æ r) Æ (p Æ r) T

T

T

T F F

T F F

4. (p Æ q) Ÿ (q Æ r) Æ (p Æ r) T T T T FT F F T F F This assignment is not possible. Example 33: Let p, q, r and s be three statements. Statement-1: (p Æ r) Ÿ (~ p Æ q) Ÿ (q Æ s) Ÿ (~ r Æ s) is a tautology. Statement-2: (p Æ q) Ÿ (~ p Æ q) Æ q is a tautology. Ans. (b) Solution: Assume Statement-1 is false. Then 1. (p Æ r) Ÿ (~ p Æ q) Ÿ (q Æ s) Æ (~ r Æ s) T

T

F

F

2. (p Æ r) Ÿ (~ p Æ q) Ÿ (q Æ s) Æ (~ r Æ s) T T T T T F T FF F 3. (p Æ r) Ÿ (~ p Æ q) Ÿ (q Æ s) Æ (~ r Æ s) F T F T TF T F T F T F F TF F F Look at the encircled part. This means ~ p Æ q is both T and F. A contradiction.

Mathematical Reasoning 28.7

\

∫ ~ [~(A Ÿ B) Ÿ (A ⁄ B)]

Statement-1 is always true.

Next assume Statement-2 is false.

∫ [(A Ÿ B) ⁄ (~ A Ÿ ~ B)]

Then

∫ [(A Ÿ B) ⁄ (~ A)] Ÿ [(A Ÿ B) ⁄ (~B)] ∫ [(A ⁄ ~ A) ⁄ (B ⁄ ~ A)] Ÿ [(A ⁄ ~ B) Ÿ (B ⁄ ~ B)]

( p Æ q) Ÿ (~ p Æ q) Æ q F T F T

TF T T

F F

∫ [T Ÿ (~ A ⁄ B)] Ÿ [(A ⁄ ~ B) Ÿ T] ∫ (A Æ B) Ÿ (B Æ A) ∫ A ¤ B

Not possible

\

Statement-2 is a tautology.

Example 34: Statement-1: ~ (A ¤ ~ B) is equivalent to A ¤ B Statement-2: A ⁄ (~(A Ÿ ~ B)) is a tautology. Ans. (b) Solution: We have A ⁄ [ ~ (A Ÿ ~ B)] ∫ A ⁄ [~ A ⁄ B] ∫ (A ⁄ ~ A) ⁄ B ∫ T ⁄ B ∫ T Thus, Statement-2 is true. We have ∫ ~ [A ¤ ~ B] ∫ ~ [(A Æ ~ B) Ÿ (~ B Æ A)] ∫ ~ [(~A ⁄ ~ B) Ÿ (B ⁄ A)]

Example 35: Statement-1: Consider the following statements p : Delhi is in India q : Mumbai is not in Italy Then negation of p ⁄ q is Delhi is not in India and Mumbai is in Italy. Statement-2: For two statements p and q ~ (p ⁄ q) ∫ ~ p Ÿ ~ q Ans. (a) Solution: Statement-2 is true. [See Theory] As ~ (p ⁄ q) ∫ ~ p Ÿ ~q ∫ ~ (Delhi is in India) and ~(Mumbai is not is Italy) ∫ Delhi is not in India and Mumbai is in Italy.

LEVEL 2 Straight Objective Type Questions Example 36: Let p, q and r be the statements: p: city X is in U.P. q: city X is in India r:pÆq Contrapositive of r is (a) if city X is not in India then X is not in U.P. (b) city X is neither in U.P. nor in India (c) city X is in India but not in U.P. (d) none of these Ans. (a) Solution: Contrapositive of p Æ q is ~ q Æ ~ p. Example 37: If p, q are r are as in Example 31, then which one of the following represents converse of p Æ q. (a) If X is a rectangle then X is a square. (b) If X is a rectangle then X is not a square (c) X is a rectangle but X is not a square (d) none of these Ans. (a) Solution: Converse of p Æ q is q Æ p.

Example 38: Let p, q, r be three statements, then (p Æ (q Æ r)) ´ ((p Ÿ q) Æ r), is a (a) tautology (b) contradiction (c) fallacy (d) none of these Ans. (a) Solution: p Æ (q Æ r) ∫ ~ p ⁄ (q Æ r) ∫ ~ p ⁄ (~ q ⁄ r) ∫ [(~ p) ⁄ (~ q)] ⁄ r ∫ ~ (p Ÿ q) ⁄ r ∫pŸqÆr \ given statement is a tautology. Example 39: Let p and q be two statements, then ~ (~ p Ÿ q) Ÿ (p ⁄ q) is logically equivalent to (a) q (b) p Ÿ q (c) p (d) p ⁄ ~ q Ans. (c) Solution: ~ (~ p Ÿ q) Ÿ (p ⁄ q) ∫ [~ (~ p) ⁄ (~ q)] Ÿ (p ⁄ q) ∫ [p ⁄ (~ q) Ÿ (p ⁄ q] ∫ p ⁄ [(~ q) Ÿ q] ∫p⁄F∫p

28.8

Complete Mathematics—JEE Main

EXERCISES Concept-based Straight Objective Type Questions 1. Which of the following pairs are not logically equivalent? (a) p ⁄ ( p Ÿ q) and p (b) p ⁄ t and p (c) ~( p ⁄ q) and (~p) Ÿ (~q) (d) p ⁄ c and p 2. The statement (p Ÿ q) ⁄ (~p ⁄ (p Ÿ(~q))) is logically equivalent to (a) p Ÿ q (b) p (c) q (d) t 3. If (~p) ⁄ q Æ ~q has value F, then p, q are respectively (a) F, F (b) T, F (d) none of these (c) p is T or F, F 4. The statement p Ÿ ~q Æ r is logically equivalent to (a) ~p ⁄ q ⁄ r (b) (~p Ÿ q) ⁄ r (c) (p Ÿ ~ q) ⁄ r (d) (~p ⁄ q) Ÿ r 5. The statement (p ⁄ ~q) Ÿ (~p ⁄ ~q) is logically equivalent to (a) p (b) q (c) ~p (d) ~q 6. Let p, q, r be the following: p : Rohit is healthy q : Rohit is wealthy r : Rohit is wise. The statement p ⁄ q Æ ~r means

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

If Rohit is healthy or wealthy then Rohit is not wise If Rohit is healthy and wealthy then Rohit is wise Rohit is neither healthy nor wealthy nor wise none of these

7. Let p, q, r be the following three statements: p : n is prime q : n is odd r : n is 2 Then p Æ q ⁄ r means (a) If n is prime then n is odd or 2 (b) If n is prime and odd then n cannot be 2 (c) If n is odd or 2 then n is prime (d) If n is odd then n is prime or 2 8. The contrapositive of the statement "I go to college if its not a holiday" (a) If I do not go to the college then its a holiday (b) If I go the college then its not a holiday (c) I go the college and its not a holiday (d) If I go to college then its a holiday 9. Suppose p and q are two statements such that p Æ q is false, then which one of the following not true? (a) Truth value of ~p ⁄ q is F (b) Truth values of p Ÿ (~q) is T (c) Truth values of (~p) Ÿ (~q) is T (d) Truth values of p ⁄ q is T 10. Which of the following is equivalent to p Æ q ⁄ r? (a) p Ÿ (~q) Æ r (b) p ⁄ (~r) Æ q (c) p Ÿ (~r) Æ ~q (d) p ⁄ (~r) Æ q

LEVEL 1 Straight Objective Type Questions 11. Let p be the proposition: Mathematics is interesting and q be mathematics is difficult, then p Ÿ q means (a) Mathematics is interesting or difficult (b) Mathematics is interesting and difficult (c) Mathematics is interesting implies it is difficult (d) Mathematics is interesting is equivalent to saying that it is difficult 12. If p Æ (~ p ⁄ q) is false, the truth value of p and q are respectively

(a) T, T (c) F, T

(b) T, F (d) F, F

13. The contrapositive of p Æ q is (a) q Æ p (b) ~ p Æ ~ q (c) ~ q Æ ~ p (d) p Æ q 14. The contrapositive of (p ⁄ q) Æ r is (a) r Æ (p ⁄ q) (b) ~ r Æ (~ p) Ÿ (~ q) (c) (~ p) ⁄ (~ q) Æ ~ r (d) none of these

Mathematical Reasoning 28.9

15. ~ p Ÿ q is logically equivalent to (a) p Æ q (b) q Æ p (c) ~ (p Æ q) (d) ~ (q Æ p)

~ (p ⁄ q) ∫ ~ p Ÿ ~ q, is (a) a tautology (b) a contradiction (c) a simple statement (d) none of these

16. Negation of q ⁄ ~ (p Ÿ r) is (a) ~ q ⁄ ~ (p ⁄ r) (b) ~ q Ÿ (p Ÿ r) (c) ~ q Ÿ (p Ÿ r) (d) ~ q ⁄ (p Ÿ r)

19. Which one of the following is a tautology? (a) (p Æ q) Ÿ p Æ q (b) (p Æ q) ⁄ p Æ q (c) (p Æ q) ⁄ p Æ q (d) (p Æ q) Ÿ (~ q) Æ p

17. Negation of the statement “If a number is prime then it odd”, is (a) A number is not prime but odd (b) A number is prime but it is not odd (c) A number is neither prime nor odd (d) none of these

20. Which of the following is not equivalent to ~ p Ÿ q? (a) ~ (q Æ p) (b) ~ (p ⁄ ~ q) (c) ~ p Æ ~ q (d) ~ (p ⁄ q)

18. If p, q are two propositions, then

21. Which of the following is equivalent to p ´ q? (a) (~ p ⁄ q) ⁄ (p ⁄ q) (b) (p Ÿ q) ⁄ (~ p Ÿ ~ q) (c) (p ⁄ q) Ÿ (p ⁄ ~ q) (d) (p Ÿ q) ⁄ (p ⁄ q)

Assertion-Reason Type Questions

22. Let p, q, r be three statements.

24. Let p, q and r be three statements.

Statement-1: p ´ q ∫ (p Æ q) Ÿ (~ q ⁄ p) is a tautology. Statement-2: p ⁄ q Æ r ∫ (p Æ r) Ÿ (q Æ r) is a tautology. 23. Let p and q be two statements. Statement-1: (p ⁄ q) ⁄ ~ (~ p Ÿ q) is logically equivalent to p. Statement-2: p ⁄ T ∫ p

Statement-1: [p ⁄ (q Ÿ r)] ⁄ [~ (p ⁄ q)] ∫ p ⁄ q is a tautology. Statement-2: [p ⁄ q Æ r] ´ [~ r Æ (~ p) Ÿ (~ q)] is a tautology. 25. Let p, q and r be three statements. Statement-1: Negation of p Ÿ (q ⁄ r) is ~ p ⁄ (~ q Ÿ ~ r) Statement-2: Negation of p ⁄ q is (~ p) Ÿ (~ q); and that of p Ÿ q is (~ p) ⁄ (~ q)

LEVEL 2 Straight Objective Type Questions 26. Let p and q be two statements, then q ´ (~ p ⁄ ~ q) is logically equivalent to (a) p (b) q (c) p Æ q (d) ~ p Ÿ q 27. Dual of (p Æ q) Æ r is

(a) (q Æ p) Ÿ r (c) (p ⁄ ~ q) ⁄ r

(b) p Æ (q Æ r) (d) none of these.

28. Statement (p ⁄ q) Æ (p Ÿ q) is equivalent (a) F (b) p ´ q (c) T (d) q Æ p Ÿ q

28.10

Complete Mathematics—JEE Main

Previous Years' AIEEE/JEE Main Questions 1. The statement p Æ (q Æ p) is equivalent to (a) p Æ (q ´ p) (b) p Æ (p Æ q) (c) p Æ (p ⁄ q) (d) p Æ (p Ÿ q) [2008] 2. Let p be the statement “x is an irrational number”, q be the statement “y is a transcendental number”, and r be the statement “x is a rational number iff y is a transcendental number”. Statement-1: r is equivalent to either q or p. Statement-2: r is equivalent to ~ (p ´ ~ q). [2008] 3. Statement-1: ~ (p ´ ~ q) is equivalent to p ´ q. Statement -2: ~ (p ´ ~ q) is tautology. [2009] 4. Let S be a non-empty subset of R. Consider the following statement : P : There is a rational number x ŒS such that x > 0. Which of the following statements is the negation of the statement P ? (a) (b) (c) (d)

Every rational number x Œ S satisfies x £ 0. x ŒS and x < 0 fi x is not rational. There is a rational number x Œ S such that x £ 0. There is no rational number x such that x £ 0. [2010]

5. Consider the following statements P : Suman is brilliant Q : Suman is rich R : Suman is honest The negation of the statement “Suman is brilliant and dishonest if and only if Suman is rich” can be expressed as (a) ~ (P Ÿ ~ R) ´ Q (b) ~ P Ÿ (Q Ÿ ~ R) (c) ~ (Q ´ (P Ÿ ~ R)) (d) ~ Q ´ ~ P Ÿ R [2011] 6. The only statement among the followings that is a tautology is (a) A Ÿ (A ⁄ B) (b) A ⁄ (A Ÿ B) (c) [A Ÿ (A Æ B)] Æ B (d) B Æ [A Ÿ (A Æ B)] [2011]

7. The negation of the statement “If I become a teacher, then I will open a school”, is (a) Either I will not become a teacher or I will not open a school. (b) Neither I will become a teacher nor I will open a school. (c) I will not become a teacher or I will open a school. (d) I will become a teacher and I will not open a school. [2012] 8. Consider: Statement-1: (p Ÿ ~q) Ÿ (~p Ÿ q) is a fallacy Statement-2: (p Æ q) ´ (~q Æ ~p) is a tautology. [2013] 9. The statement p Æ (q Æ p) is equivalent to (a) p Æ q (b) p Æ (p ⁄ q) (c) p Æ (p Æ q) (d) p Æ (p Ÿ q) [2013, online] 10. Let p and q be any two logical statements and r : p Æ (~p ⁄ q). If r has a truth value F, the truth values of p and q are respectively (a) F, F (b) T, T (c) T, F (d) F, T [2013, online] 11. Statement-1: The statement A Æ (B Æ A) is equivalent to A Æ (A ⁄ B). Statement-2: The statement ~[(A Ÿ B) Æ (~A ⁄ B)] is a tautology. [2013, online] 12. For integers m and n, both greater than 1, consider the following three statements: P : m divides n Q : m divides n2 R : m is prime then (a) Q Ÿ R Æ P (c) Q Æ R

(b) P Ÿ Q Æ R (d) Q Æ P [2013, online]

13. The statement ~(p ´ ~q) is (a) equivalent to ~p ´ q (b) a tautology (c) a fallacy (d) equivalent to p ´ q [2014] 14. The contrapositive of the statement " if I am not feeling well, then I will go to the doctor" is

Mathematical Reasoning 28.11

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

If I am feeling well, then I will not go to the doctor If I will go to the doctor, then I am feeling well If I will not go to the doctor, then I am feeling well If I will go to the doctor, then I am not feeling well [2014, online]

15. The proposition ~(p ⁄ ~q) ⁄ ~(p ⁄ q) is logically equivalent to (a) p (b) q (c) ~p (d) ~q [2014, online] 16. Let p, q, r denote three arbitrary statements. The logically equivalent of the statement p Æ (q ⁄ r) is (a) (p Æ ~q) Ÿ (p Æ r) (b) (p Æ q) ⁄ (p Æ r) (c) (p Æ q) Ÿ (p Æ ~r) (d) p ⁄ q Æ r [2014, online] 17. The contrapositive of the statement “I go to school if it does not rain” is (a) if it rains, I do not go to school (b) if I do not go to school, it rains (c) if it rains, I go to school (d) if I go to school, it rains [2014, online] 18. The negation of ~ s ⁄ (~ r Ÿ s) is equivalent to: (a) s Ÿ ~ r (b) s Ÿ (r Ÿ ~) (c) s ⁄ (r ⁄ ~ s) (d) s Ÿ r [2015] 19. Contrapositive of the statement “If it is raining then I will not come”, is (a) if I will come, then it is not raining (b) if I will not come, then it is raining (c) if I will not come, then it is not raining (d) if I will come, then it is raining [2015, online] 20. Consider the following statements: P : Suman is brilliant

O : Suman is rich R : Suman is honest The negation of the statement, "Suman is brilliant and dishonest if and only if Suman is rich" can be equivalently expressed as: (a) ~ Q ´ P Ÿ R (b) ~ Q ´ ~ P ⁄ R (c) ~ Q ´ P ⁄ ~ R (d) ~ Q ´ P Ÿ ~ R [2015, online] 21. The Boolean Expression (p Ÿ –q) ⁄ q ⁄ (~p Ÿ q) is equivalent to (a) ~p Ÿ q (b) p Ÿ q (c) p ⁄ q (d) p ⁄ ~q [2016] 22. Consider the following two statements: P: If 7 is an odd number, then 7 is divisible by 2 Q: If 7 is a prime number, then 7 is an odd number If V1 is the truth value of the contrapositive of P and V2 is the truth value of contrapositive of Q, then the ordered pair (V1, V2) equals: (a) (F, F) (b) (F, T) (c) (T, F) (d) (T, T) [2016, online] 23. The contrapositive of the following statement, "If the side of a square doubles, then its area increases four times", is (a) if the area of a square increases four times, then its side is not doubled (b) if the area of a square increases four times, then its side is doubled (c) if the area of a square does not increase four times, then its side is not doubled (d) if the side of a square is not doubled, then its area does not increase four times [2016, online]

Previous Years' B-Architecture Entrance Examination Questions 1. The statement ~(p Ÿ q) ⁄ q: (a) is a tautology (b) is equivalent to (p Ÿ q) ⁄ ~q (c) is equivalent to p ⁄ q (d) is a contradiction

(c) (~ r Ÿ ~ s) Æ (~ p ⁄ ~ q) (d) (r ⁄ s) Æ (~ p ⁄ ~ q)

[2010]

3. Statement-1: [2009]

2. 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, s, as (a) (~ r ⁄ ~ s) Æ (~ p Ÿ ~ q) (b) (r Ÿ s) Æ (~ p Ÿ ~ q)

~ (A ¤ ~ B) is equivalent to A ¤ B. Statement-2: A ⁄ (~(A Ÿ ~ B)) a tautology. 4. Statement-1: Consider the statements p : Delhi is in India q : Mumbai is not in Italy

[2011]

28.12

Complete Mathematics—JEE Main

Then the negation of the statement p ⁄ q, is Delhi is not in India and Mumbai is in Italy. Statement-2: For any two statements p and q ~(p ⁄ q) = ~p ⁄ ~q [2012] 5. If p is any logical statement, then (a) p Ÿ(~p) is a tautology (b) p ⁄(~p) is a contradiction (c) p Ÿ p = p (d) p ⁄(~p) = p

[2013]

6. Let p and q be any two propositions. Statement-1: ( p Æ q) ´ q ⁄ ~p is a tautology. Statement-2: ~(~p Ÿ q) Ÿ (p ⁄ q) ´ p is a fallacy. (a) Both statements 1 and statements 2 are true (b) Both statements 1 and statement 2 are false (c) Statement 1 is true and statement 2 is false (d) Statement 1 is false and statement 2 is true [2014] 7. The statement [p Ÿ (p Æ q)] Æ q, is (a) a fallacy (b) a tautology (c) neither a fallacy nor a tautology (d) not a compound statement

[2015]

8. The negation of A Æ (A ⁄ ~ B) is (a) a tautology (b) equivalent to (A ⁄ ~ B) Æ A (c) equivalent to A Æ (A Ÿ ~ B) (d) a fallacy

12. (a)

13. (d)

14. (c)

15. (c)

16. (b)

17. (b)

18. (d)

19. (a)

20. (b)

21. (c)

22. (b)

23. (a)

Previous Years' B-Architecture Entrance Examination Questions 1. (a)

2. (c)

3. (b)

4. (c)

5. (c)

6. (c)

7. (b)

8. (d)

Hints and Solutions Concept-based 1. p ⁄ t ∫ t not p. [see Results on Logical equivalences] 2. (p Ÿ q) ⁄ (~p ⁄ (p Ÿ (~q))) ∫ (p Ÿ q) ⁄ ((~ p ⁄ p) Ÿ (~p ⁄ ~ q)) ∫ (p Ÿ q) ⁄ (t Ÿ ~(p Ÿ q)) ∫ (p Ÿ q) ⁄ (~(p Ÿ q)) ∫ t 3. (~p ⁄ q) Æ ~q is F if ~p ⁄ q is T but ~q is F or q is T. But then ~p ⁄ q is T irrespective of value of p. 4. Use p Æ q is logically equivalent to ~p ⁄ q. 5. (p ⁄ ~q) Ÿ (~p ⁄ ~q) ∫ (p Ÿ(~p)) ⁄ (~q) ∫ c ⁄ (~q) ∫ ~q

[2016]

6. Use p ⁄ q mean Rohit is healthy or wealthy 7. If n is prime then n is odd or 2.

Answers

8. Let p : It is a holiday q : I go the college.

Concept-based 1. (b) 5. (d) 9. (c)

2. (d) 6. (a) 10. (a)

3. (c) 7. (a)

4. (a) 8. (a)

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

9. p Æ q is false mean p is T and q is F. ~p ⁄ q = (F) ⁄ (F) = F, p Ÿ (~ q) = T 12. 16. 20. 24.

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

13. 17. 21. 25.

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

14. (b) 18. (a) 22. (b)

27. (a)

(~p) Ÿ (~q) = F, p ⁄ q = F \ (c) is incorrect. 10. p Æ q ⁄ r ∫ ~p ⁄ (q ⁄ r) ∫ (~p ⁄ q) ⁄ r ∫ p Ÿ (~q) Æ r

Level 2 26. (d)

~p Æ q Its contrapositive is ~q Æ p.

Level 1 11. 15. 19. 23.

The given statement is

28. (b)

Level 1

Previous Years' AIEEE/JEE Main Questions 1. (c) 4. (a)

2. None of the answer matches 3. (c) 5. (c) 6. (c) 7. (d)

8. (b)

9. (b)

10. (c)

11. (c)

11. Mathematics is interesting and difficult 12. p Æ (~ p ⁄ q) is F if p = T and ~ p ⁄ q = F or F ⁄ q = F or if q = F.

Mathematical Reasoning 28.13

13. Contrapositive of p Æ q is ~ q Æ ~ p.

∫ q ´ ~ ( p Ÿ q) ∫ [q Æ ~ ( p Ÿ q)] Ÿ [~ ( p Ÿ q) Æ q] ∫ [~ q ⁄ ~ ( p Ÿ q)] Ÿ [( p Ÿ q) ⁄ q] ∫ ~ [q Ÿ ( p Ÿ q)] Ÿ q ∫ ~ ( p Ÿ q) Ÿ q ∫ (~ p ⁄ ~ q) Ÿ q ∫ (~ p Ÿ q) ⁄ (~ q Ÿ q) ∫ (~ p Ÿ q)

14. Contrapositive of p ⁄ q Æ r is ~ r Æ ~ (p ⁄ q) i.e. ~ r Æ (~ p) Ÿ (~ q) 15. See Theory 16. ~ [q ⁄ ~ (p Ÿ r)] ∫ ~ q Ÿ (p Ÿ r) 17. Negation of p Æ q = ~ p ⁄ q is ~ (~ p ⁄ q) = p Ÿ ~ q, that is, a number is prime and but it is not odd. 18. It is De Morgan’s law. 19. (p Æ q) Ÿ p Æ q. FT F T T FF Contradiction.

Thus, (p Æ q) Ÿ p Æ q can never take value F. 20. ~ p Ÿ q ∫ ~ (p ⁄ ~ q) by De Morgan’s Law. 21. Use p ´ q ∫ (p Æ q) Ÿ (q Æ p) ∫ (~ p ⁄ q) Ÿ (~ q ⁄ p) ∫ (p Ÿ q) ⁄ (~ p Ÿ ~ q) 22. p ´ q ∫ (p Æ q) Ÿ (q Æ p) ∫ (p Æ q) Ÿ (~ q ⁄ p) and p ⁄ q Æ r ∫ ~ (p ⁄ q) ⁄ r ∫ (~ p Ÿ ~ q) ⁄ r ∫ (~ p ⁄ r) Ÿ (~ q ⁄ r) ∫ (p Æ r) Ÿ (q Æ r) 23. (p ⁄ q) Ÿ ~ (~ p Ÿ q) ∫ (p ⁄ q) Ÿ [~ (~ p) ⁄ (~ q)] ∫ (p ⁄ q) Ÿ [p ⁄ (~ q)] ∫ p ⁄ [q Ÿ (~ q)] ∫ p ⁄ F ∫ p Note that p ⁄ T = T not p. 24. [p ⁄ (q Ÿ r)] ⁄ [~ (p ⁄ q)] = [(p ⁄ q) Ÿ (p ⁄ r)] ⁄ [~ (p ⁄ q)] = [~ (p ⁄ q)] ⁄ [(p ⁄ q) Ÿ (p ⁄ r)] = T Ÿ (p ⁄ r) = p ⁄ r Next, ~ r Æ (~ p) Ÿ (~ q) ∫ ~ r Æ ~ (p ⁄ q) ∫ p ⁄ q Æ r. 25. ~ [p Ÿ (q ⁄ r)] ∫ ~ p ⁄ [~ (q ⁄ r)] ∫ ~ p ⁄ (~ q Ÿ ~ r)

Level 2 26. q ´ (~ p ⁄ ~ q)

27. ( p Æ q) Æ r ∫ (~ p ⁄ q) Æ r ∫ ~ (~ p ⁄ q) ⁄ r ∫ [ p Ÿ (~ q)] ⁄ r. Dual of ( p Æ q) Æ r is [ p ⁄ (~ q)] Ÿ r ∫ (q Æ p) Ÿ r 28. ( p ⁄ q) Æ (p Ÿ q) ∫ ~ ( p ⁄ q) ⁄ ( p Ÿ q) ∫ (~ p Ÿ ~ q) ⁄ (p Ÿ q) ∫ [~ p ⁄ ( p Ÿ q)] Ÿ [~ q ⁄ ( p Ÿ q)] ∫ [~ p ⁄ q] Ÿ [~ q ⁄ p] ∫ ( p Æ q) Ÿ (q Æ p) ∫p´q

Previous Years’ AIEEE/JEE Main Questions 1. p Æ (q Æ p) ∫ ~ p ⁄ (q Æ p) ∫ (~ p) ⁄ (~ q ⁄ p) ∫ (~ q) ⁄ (p ⁄ ~ p) ∫ (~ q) ⁄ T = T \ p Æ (q Æ p) is a tautology. Also p Æ (p ⁄ q) ∫ ~ p ⁄ (p ⁄ q) ∫ (~ p ⁄ q) ⁄ q ∫ T ⁄ q = T \ p Æ (p ⁄ q) is also an tautology. Thus, p Æ (q Æ p) is equivalent to p Æ (p ⁄ q). 2. Note that the statement r is ~ p ´ q. Now, ~ p ´ q ∫ (~ p Æ q) Ÿ (q Æ ~ p) ∫ [~(~ p) ⁄ q Ÿ (~ q ⁄ ~ p)] ∫ (p ⁄ q) Ÿ [~(p Ÿ q)] πp⁄q Next, p ´ ~ q ∫ ~ q ´ p ∫ (p ⁄ q) Ÿ [~(p Ÿ q)] fi ~(p ´ ~ q) ∫ ~(p ⁄ q) ⁄ (p Ÿ q) Thus, neither Statement-1 nor Statement-2 is true. 3. p T T F F

q T F T F

~q F T F T

p´~q F T T F

~(p ´ ~ q) p ´ q T T F F F F T T

28.14

Complete Mathematics—JEE Main

Note that ~(p ´ ~ q) is not a tautology. \ Statement-2 is false. From table ~(p ´ ~ q) is equivalent to p ´ q. Thus, Statement-1 is true. 4. Negation of P is “for each rational number x Œ S, x £ 0”. 5. P Ÿ ~ R stands for Suman is brilliant and dishonest. Thus P Ÿ ~ R ´ Q stands for Suman is brilliant and dishonest if and only if Suman is rich. Its negation is ~(P Ÿ ~ R ´ Q) or ~(Q ´ P Ÿ ~ R) 6. A Ÿ (A ⁄ B) is F when A = F A ⁄ (A Ÿ B) is F when A = F, B = F We have [A Ÿ (A Æ B)] Æ B ∫ [A Ÿ (~ A ⁄ B)] Æ B ∫ [(A Ÿ (~ A)) ⁄ (A Ÿ B)] Æ B ∫AŸBÆB ∫ ~ (A Ÿ B) ⁄ B ∫ [(~ A) ⁄ (~ B)] ⁄ B ∫ (~ A) ⁄ [(~ B) ⁄ B] ∫ (~ A) ⁄ T ∫ T \ [A Ÿ (A Æ B)] Æ B is a tautology. 7. Let p : I become a teacher q : I will open a school. The given statement is p Æ q ∫ (~ p) ⁄ q Its negation is ~ ((~ p) ⁄ q) ∫ p Ÿ (~ q) Thus, negation of the given Statement is 1 will become a teacher and I will not open a school. 8. As ~ q Æ ~ p ∫ p Æ q, Statement 2 can be written as (p Æ q) ´ (p Æ q) Thus, Statement 2 is a tautology. Also, (p Ÿ ~ q) Ÿ (~ p Ÿ q) ∫ (p Ÿ ~ p) Ÿ (~ q Ÿ q) = F Ÿ F ∫ F, which is a fallacy. However Statement 2 is not a correct reason for Statement 1. 9. See Solution to Question 1. 10. Statement r is p Æ (~ p ⁄ q) If r is false, then p must be T and ~ p ⁄ q must be F fi p is T and F ⁄ q is F fi p is T and q is F. 11. A Ÿ B Æ (~ A ⁄ B) ∫ ~ (A Ÿ B) ⁄ (~ A ⁄ B) ∫ (~ A ⁄ ~ B) ⁄ (~ A ⁄ B)

∫ (~ A) ⁄ (~ B ⁄ B) ∫ (~ A) ⁄ T ∫ ~ A fi ~[(A Ÿ B) Æ (~ A ⁄ B)] ∫ A \ Statement-2 is false. Next, A Æ (B Æ A) ∫ A Æ (~ B ⁄ A) ∫ ~ A ⁄ (~ B ⁄ A) ∫ (~ B) ⁄ (~ A ⁄ A) ∫ (~ B) ⁄ T ∫ T ∫ (~ A ⁄ A) ⁄ B ∫ ~ A ⁄ (A ⁄ B) ∫AÆA⁄B Thus, Statement-1 is true. 12. Q Ÿ R Æ P In words it means m is prime and m|n2 fi m|n. 13. Sec Solution to Questions 3. 14. Let p, q be statements p: I am not feeling well q: I will go to the doctor Contrapositive of p Æ q is ~q Æ ~p i.e. if I do not go to the docor then I am feeling well. 15. ~(p ⁄ ~q) ⁄ ~(p ⁄ q) = ~[(p ⁄ ~q) Ÿ (p ⁄ q)] = ~[p ⁄ (~q Ÿ q)] = ~[p ⁄ F] = ~p 16. p Æ q ⁄ r ∫ ~p ⁄ (q ⁄ r) ∫ (~p ⁄ q) ⁄ (~p ⁄ r) ∫ (p Æ q) ⁄ (p Æ r) 17. Let p and q be the statements p : It does not rain q : I go to school The given statement is p Æ q. Its contrapositive is ~q Æ ~p i.e., if I do not go to school it rains. 18. ~[~ s ⁄ (~r Ÿ s)] ∫ ~(~s) Ÿ ~(~r Ÿ s) ∫ s Ÿ (r ⁄ ~s) ∫ (s Ÿ r) ⁄ (s Ÿ ~s) ∫ (s Ÿ r) ⁄ F ∫ s Ÿ r 19. Let p and q be the statements: p: It is raining q: I will not come. Contrapositive of p Æ q is ~q Æ ~p that is, if I (will) come then it is not raining.

Mathematical Reasoning 28.15

20. Suman is brilliant and dishonest if and only if Suman is rich, can be expressed as P Ÿ (~R) ´ Q Its negation is ~Q ´ ~P ⁄ R 21. (p Ÿ ~q) ⁄ q ⁄ (~p Ÿ q) = (p Ÿ ~q) ⁄ [q ⁄ (~p Ÿ q)] = (p Ÿ ~q) ⁄ q [absorption law] = (p ⁄ q) Ÿ (~q ⁄ q) = (p ⁄ q) Ÿ T = p ⁄ q 22. As a statement and its counter positive have the same truth values, truth values of (V1, V2) = truth values of (P, Q) = (F, T) 23. Contrapositive of p Æ q is ~q Æ ~p. Thus, contrapositive of given statement is (c), that is, if the area of a square does not increase four times, then its side is not doubled.

Previous Years’ B-Architecture Entrance Examination Questions 1. ~(p Ÿ q) ⁄ q ∫ (~p ⁄ ~q) ⁄ q ∫ ~p ⁄ (~q ⁄ q) ∫ ~p ⁄ T ∫ T 2. Given statement is pŸqÆr⁄s Its contrapositive is ~(r ⁄ s) Æ ~(p Ÿ q) ¤ (~r Ÿ ~s) Æ (~p ⁄ ~q) 3. A ⁄ (~(A Ÿ ~B)) ∫ A ⁄ (~A ⁄ B) ∫ (A ⁄ ~A) ⁄ B ∫T⁄B∫T

For Statement-1, see to Question 3 in the previous year AIEEE/JEE Questions. Thus, Statement-1 is true. However Statement-2 is not a correct reason for statement-1. 4. Statement-2 is true. [See theory] As ~(p ⁄ q) ∫ ~p Ÿ ~q ∫ ~(Delhi is in India) and ~(Mumbai is not is Italy) ∫ Delhi is not in India and Mumbai is in Italy. 5. p Ÿ (~p) ∫ F; p ⁄ (~p) ∫ T and p Ÿ p ∫ p is true for each logical statement p. 6. ~(~p Ÿ q) Ÿ (p ⁄ q) ∫ (p ⁄ ~q) Ÿ (p ⁄ q) ∫ p ⁄ (~q Ÿ q) ∫ p ⁄ F ∫ p. As, ~(~p Ÿ q) Ÿ (p ⁄ q) ´ p p´p is a tautology, Statement-2 is false. For truth Statement-1 see theory. 7. p Ÿ (p Æ q) Æ q ∫ [p Ÿ (~p ⁄ q)] Æ q ∫ [(p Ÿ (~p)) ⁄ (p Ÿ q)] Æ q ∫ [F ⁄ (p Ÿ q)] Æ q ∫ (p Ÿ q) Æ q ∫ ~(p Ÿ q) ⁄ q ∫ (~p ⁄ ~q) ⁄ q ∫ ~p ⁄ (~q ⁄ q) ∫ (~p) ⁄ T ∫ T 8. A Æ (A ⁄ ~B) ∫ ~A ⁄ (A ⁄ ~B) ∫ (~A ⁄ A) ⁄ (~B) ∫ T ⁄ (~B) ∫ T \ Negative of A Æ A ⁄ ~B is ~T = F.

Architecture Entrance – 2017 1. If A and B be two finite sets that in total number of subsets of A is 960 more than the total number of subsets of B, then n(a) – n(b) (where n(X) denotes the number of elements in set X) is equal to: (a) 6 (b) 2 (c) 3 (d) 4 2. The order and the degree of the differential equation of all ellipses with centre at the origin, major axis 3 are, respectively: along x-axis and eccentricity 2 (a) 2, 2 (b) 1, 1 (c) 2 , 1 (d) 1, 2 3. If A and B are two independent events such that P(A) 3 4 and P(A ∪ B) = , then P(A ∩ B) is equal to: 10 5 (b) 1 (a) 3 35 5 1 (c) (d) 3 10 14

=

4. Let S be the set of all real values of ‘a’ for which the following system of linear equations. ax + 2y + 5z = 1 2x + y + 3z = 1 3y + 7z = 1 Is consistent. Then the set S is: (a) an empty set (b) equal to R (c) equal to R – {1} (d) equal to {1} 5.

If an equilateral triangle, having centroid at the origin has a side along the line, x + y = 2, then the area (in sq. units) of this triangle is: (b) 6 (a) 3 6 9 (d) (c) 6 3 3 2

6. If the shortest distance between the lines x + 2λ = 2y = –12z, x = y + 4λ = 6z – 12λ is 4 2 units, then a value of λ is: (a)

2 2

2

(b) 2

(d) 2 2 7. If the digits at ten’s and hundred’s places in (11)2016 are x and y respectively, then the ordered pair (x, y) is equal to:(a) (1, 8) (b) (1, 6) (c) (6, 1) (d) (8, 1) (c)

8.

Which one of the following points does not lie on x2 y 2 = 1 drawn at the normal to the hyperbola, − 16 9 the point (8, 3 3 )? ⎛ (a) ⎜10, ⎝ ⎛ (c) ⎜12, ⎝

1 ⎞ ⎟ 3⎠

1 ⎞ ⎛ (b) ⎜13, − ⎟ 3⎠ ⎝

1 ⎞ ⎟ 3⎠

(d)

(11, 3 )

9. An urn contains 5 red, 4 black and 3 white marbles, Then the number of ways in which 4 marbles can be drawn from it so that at most 3 of them are red, is: (a) 495 (b) 455 (c) 460 (d) 490 10. If (4 + 3i)2 = 7 + 24i, then a value of

(7 +

−576

1 2

) − (7 −

−576

(a) –6i (c) 2i

)

1 2

is:-

(b) –3i (d) 6

11. The equation of the circle, which is the mirror image of the circle, x2 + y2 – 2x = 0, in the line, y = 3 – x is: (a) x2 + y2 – 6x – 4y + 12 = 0 (b) x2 + y2 – 6x – 8y + 24 = 0 (c) x2 + y2 – 8x – 6y + 24 = 0 (d) x2 + y2 – 4x – 6y + 12 = 0 log (sin 7 x + cos 7 x ) 12. lim equals: x→0 sin3 x (a) 1 log 7

(b) 7

3 (c) 14 3

3 (d) 1 3

13. If the line, y = mx, bisects the area of the region {(x, y) : 0 ≤ x ≤ (a) 39

16 13 (c) 3

3 , 0 ≤ y ≤ 1 + 4x – x2}, then m equals: 2 (b) 9 8 13 (d) 6

14. The product of the perpendiculars drawn from the 2 2 foci of the ellipse, x + y = 1 upon the tangent to it 9 25 ⎛3 5 3⎞

at the point ⎜⎜ , ⎟⎟ , is: ⎝2 2 ⎠

2

Complete Mathematics—JEE Main

(a) 3 13

189 (c) 13

(b) 9

(a)

(d) 18

(

)

16. Two numbers are selected at random (without replacement) from the first six positive integers. If X denotes the smaller of the two numbers, then the expectation of X, is: 5 3 13 (c) 3

(b) 14 3 7 (d) 3

17. The perpendicular distance from the point (3, 1, 1) on the plane passing through the point (1, 2, 3) and containing the line. r = i + j + λ(2i + j + 4k) is 1 3 (a) (b) 11 11 4 (c) (d) 0 41 x+2 18. The integral I = ∫ 2 dx is equal to ( x + 3x + 3) x + 1 (a) (b)

(c)

(d)

⎡ x 3 ⎤ 1 cot −1 ⎢ ⎥+c 3 ⎣ x +1 ⎦ ⎡ ⎤ 1 x ⎥+c tan −1 ⎢ 3 ⎢⎣ 3 ( x + 1) ⎥⎦ ⎡ ⎤ 2 x ⎥+c tan −1 ⎢ 3 ⎢⎣ 3 ( x + 1) ⎥⎦ ⎡ ⎤ 2 x ⎥+c cot −1 ⎢ 3 ⎢⎣ 3 ( x + 1) ⎥⎦

(Where C is a constant of integration) 19. The value of E =

(b) 2 2

4 2 (d) 2 2 3 3 20. Let a1, a2, a3, a4, a5 be a G.P. of positive real numbers such that the A.M., of a2 and a4 is 117 and the G.M. of a2 and a4 is 108. Then the A.M. of a1 and a5 is (a) 145.5 (b) 108 (c) 117 (d) 144.5 5π /24 dx is equal to 21. The integral ∫π /24 3 1 + tan 2 x (a) π (b) π 3 18 π π (c) (d) 12 6

(c)

15. An observer at a point P on the top of a hill near the sea-shore notices that the angle of depression of a ship moving towards the hill in a straight line at a constant speed is 30°. After 45 minutes, this angle becomes 45°. If T (in minutes) is the total time taken by the ship to move to a point in the sea where the angle of depression from P of the ship is 60°, then T is equal to: 1 ⎞ ⎛ (b) 45 1 + 3 (a) 45 ⎜1 + ⎟ 3⎠ ⎝ 1 ⎞ ⎛ 2 ⎞ ⎛ (c) 45 ⎜1 + (d) 45 ⎜ 2 + ⎟ ⎟ 3⎠ ⎝ 3⎠ ⎝

(a)

3− 2

1 1 + is cos 285° 3 sin 255°

22. Three vectors a, b and c are such that |a|=1, |b|=2, |c| = 4 and a + b + c = 0. Then the value of 4a·b + 3b·c + 3c·a is equal to: (a) 27 (b) –68 (c) –26 (d) –34 23. For all real numbers x, y and z, the determinant 2x xy − xz y Δ = 2 x + z + 1 xy − xz + yz − z 2 1 + y is equal to: 3x + 1 2 xy − 2 xz 1+ y (a) (y – xz)(z – x) (c) (x – y)(y – z)(z – x)

(b) zero (d) (x – yz)(y – z)

24. If λ1 and λ2 are the two values of λ such that the roots α and β of the quadratic equation, λ(x2 – x) + x + 5 λ λ α β 4 = 0 satisfy + + = 0, then 21 + 22 is equal to: λ 2 λ1 β α 5 (a) 488 (b) 536 (c) 512 (d) 501 25. If the sum of the first 15 terms of the series 3 + 7 + 14 + 24 + 37 + …. is 15k, then k is equal to: (a) 126 (b) 122 (c) 81 (d) 119 26. Water is running into an underground right circular conical reservoir, which is 10 m deep and radius of its base is 5 m. If the rate of change in the volume of water in the reservoir is

3 πm3/min., then the rate 2

(in m/min) at which water rises in it, when the water level is 4m, is: (b) 3 (a) 3 2 1 (c) 8

8 (d) 1 4

27. A bag contains three coins, one of which has head on both sides, another is a biased coin that shows up heads 90% of the time and the third one is an unbiased

Architecture Entrance – 2017 3 ∴ equation of ellipse is x2 4 y 2 + = 1 or x2 + 4y2 = a2 a2 a2

coin. A coin is taken out from the bag at random and tossed. If it shows up a head, then the probability that it is the unbiased coin, is: (b) 5 (a) 3 8

(c)

Differentiating we get

12 (d) 1 3

5 24

2x + 8 y

or 4 y

28. If the functiuon f : R → R, defined by ax x b > 0, is five times their goemetric mean, then a+b is equal to: a −b

6. If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is: (b) 45th (a) 44th th (c) 46 (d) 47th

(b) 3 2

(c)

7 3 12

(d) 5 6

10. lim x →3

16 . 3 16 (d) an ellipse with length of minor axis . 9 4. The number of real values of λ for which the system

5. Let A be any 3 × 3 invertible matrix. Then which one of the following is not always true? (a) adj (A) =|A|·A–1 (b) adj (adj(A)) = |A|·A (c) adj (adj(A)) = |A|2·(adj(A))–1 (d) adj (adj(A)) = |A|.(adj(A))-1

6 2

4

12

9. If the sum of the first n terms of the series 3 + 75 + 243 + 507 + .... is 435 3 then n equals: (a) 18 (b) 15 (c) 13 (d) 29

(c) an ellipse with length of major axis

of linear equations 2x + 4y – λz = 0 4x + λy + 2z = 0 λx + 2y + 2z = 0 has infinitely many solutions, is: (a) 0 (b) 1 (c) 2 (d) 3

(a)

3x − 3 2x − 4 − 2

is equal to:

(a)

3

(b)

(c)

3 2

(d)

1 2 1 2 2

11. The tangent at the point (2, –2) to the curve, x2y2 – 2x = 4(1 – y) does not pass through the point: (a)

⎛ 1⎞ ⎜ 4, ⎟ ⎝ 3⎠

(c) (–4, –9) 12.

If

y

=

(b) (8, 5) (d) (–2, –7) 15

15

⎡ x + x 2 − 1⎤ + ⎡ x − x 2 − 1⎤ ⎣ ⎦ ⎣ ⎦

then

d2y dy +x is equal to: 2 dx dx (a) 125 y (b) 224 y2 2 (c) 225 y (d) 225 y

(x

2

− 1)

13. If a point P has co-ordinates (0, –2) and Q is any point on the circle, x2 + y2 – 5x – y + 5 = 0, then the maximum value of (PQ)2 is: (a)

25 + 6 2

(b) 14 + 5 3

(c)

47 + 10 6 2

(d) 8 + 5 3

10

Complete Mathematics—JEE Main

14. The integral I =

∫

1 + 2 cot x ( cosec x + cot x )dx

π⎞ ⎛ ⎜ 0 < x < ⎟ is equal to: 2⎠ ⎝

(a)

x⎞ ⎛ 4 log ⎜ sin ⎟ + C ⎝ 2⎠

x (b) 2 log ⎛⎜ sin ⎞⎟ + C

(c)

x⎞ ⎛ 2 log ⎜ cos ⎟ + C ⎝ 2⎠

x (d) 4 log ⎛⎜ cos ⎞⎟ + C ⎝ 2⎠

π

15. the integral I =

8cos 2 x

∫π ( tan x + cot x ) 4

2⎠

15 128

(c)

13 32

(b)

15 64

(d)

13 256

1 2 3

+

2π 3

17. The curve satisfying the differential equation, ydx – (x + 3y2)dy = 0 and passing through the point (1, 1), also passes through the point: (a) (c)

⎛1 1⎞ ⎜ ,− ⎟ ⎝4 2⎠

⎛ 1 1⎞

(b) ⎜ − , ⎟ ⎝ 3 3⎠

⎛1 1⎞ ⎜ ,− ⎟ ⎝3 3⎠

(d) ⎛⎜ 1 , 1 ⎞⎟ 4 2 ⎝

2

(

19. If two parallel chords of a circle, having diameter 4 units, lie on the opposite sides of the centre and subtend angles cos −1 ⎜ ⎟ and sec ( 7 ) at the centre ⎝7⎠ respectively, then the distance between these chords, is: −1

16 7

(

)

)

(d) 3 + 2 2

2 +1

21. Consider an ellipse, whose centre is at the origin and its major axis is along the x-axis. If its eccentricity area (in sq. units) of the quadrilateral inscribed in the ellipse, whith the vertices as the vertices of the ellipse, is: (a) 8 (b) 32 (c) 80 (d) 40 22. The coordinates of the foot of the perpendicular from the point (1, –2, 1) on the plane containing the lines, x +1 y −1 z −3 x −1 y−2 z −3 = = and = = , is: 6 7 8 3 5 7

(a) (2, –4, 2) (c) (0, 0, 0)

(b) (–1, 2, –1) (d) (1, 1, 1)

23. The line of intersection of the planes r ⋅ ( 3i − j + k ) = 1 and r ⋅ ( i + 4 j − 2k ) = 2 , is: y = = 7

(a)

⎠

18. The locus of the point of intersection of the straight lines, tx – 2y – 3t = 0 x – 2ty + 3 = 0 (r ∈ R), is : 2 (a) an ellipse with eccentricity 5 (b) an ellipse with the length of major axis 6 (c) a hyperbola with eccentricity 5 (d) a hyperbola with the length of conjugate axis 3

⎛1⎞

(d)

5

1 4π + 3 3

(d)

8 7

is 3 and the distance between its foci is 6, then the

16. The area (in sq. units) of the smaller portion enclosed between the curves, x2 + y2 = 4 and y2 = 3x, is: 1 1 2π π + + (b) (a) 2 3 3 3 3 (c)

(b)

20. If the common tangents to the parabola, x2 = 4y and the circle, x2 + y2 = 4 intersect at the point P, then the distance of P from the origin, is: (b) 2 3 + 2 2 (a) 2 +1 (c)

dx equals: 3

12

(a)

8 7

(c)

(where C is a constant of integration) ⎝

4 7

(a)

(b)

(c)

(d)

x− 2

4 7=

y = −7

5 7 13

z−

5 7 13

z+

6 5 y− 13 = 13 = z 2 −7 −13

x−

6 5 y− 13 = 13 = z 2 7 −13

x−

24. The area (in sq. units) of whose diagonals are along the 3i + 4 j − 12k , is: (a) 26 (b) (c) 20 (d)

the parallelogram vectors 8i − 6 j and 65 52

25. The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean

JEE (Main) 2017 Questions with Solutions Mathematics (8th April - online) 11

age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is: (a) 25 (b) 30 (c) 35 (d) 40

13. (b)

14. (b)

15. (a)

16. (d)

17. (b)

18. (d)

19. (b)

20. (c)

21. (d)

22. (c)

23. (c)

24. (b)

26. Three persons P,Q and R independentely try to hit a target. If the probabilities of their hitting the target

25. (c)

26. (a)

27. (b)

28. (d)

29. (a)

30. (b)

are

3 1 5 , and respectively, then the probabiity 4 2 8

that the target is hit by P or Q but not by R is: (a) (c)

21 64 15 64

(b)

9 64

(d)

39 64

Hints and Solutions 1.

⇒ f (g(x)) = x ⇒ f (310x – 1) = x ⇒ 210(310x – 1) + 1 = x

27. An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is: (a)

255 256

(b) 127

(c)

63 64

(d)

⇒ 310x – 1 + 2-10 = 2–10 x ⇒ (310 – 2–10)x = 1 – 2–10

2.

1 2

⎧ 0 cos x − sin x ⎫ ⎪ ⎪ 28. If S = ⎨ x ∈ [0, 2π] : sin x 0 cos x = 0 ⎬ , ⎪ ⎪ cos x sin x 0 ⎩ ⎭ ⎛π ⎞ Then ∑ tan ⎜ + x⎟ is equal to: ⎝3 ⎠ x ∈s

(a) 4 + 2 3

(b) −2 + 3

(c)

(d) −4 − 2 3

1 + cos −1 x 2 4 2

(b)

π

(c)

π

(d)

π

1 − cos −1 x 2 4 2

4 4

⇒ ∴ ⇒

⇒ ⇒

⎡ 1 + x2 + 1 − x2 ⎤ ⎥, 29. the value of y = tan ⎢ 2 2 ⎣⎢ 1 + x − 1 − x ⎥⎦ 1 |x|< , x ≠ 0, is equal to: 2

π

⇒

3. Let

−1

(a)

1 − 2−10 310 − 2−10 Let p(x) = (x – 1)(ax + b) + 4 p(0) = (–1)(0 + b) + 4 = 1 –b + 4 = 1 ⇒ b = 3 Also, 6 = p(–1) = (–1 – 1)(–a + b) + 4 1=a–b⇒a=4 p(x) = (x – 1)(4x + 3) + 4 p(2) = (2 – 1)(11) + 4 = 15 p(–2) = (–2 – 1)(–8 + 3) + 4 = 19 z = x + iy 2|x + iy + 3i| = |x + iy – i| 4[x2 + (y + 3)2] = x2 + (y – 1)2 3x2 + 3y2 + 26y + 35 = 0

⇒x=

128

−2 − 3

(fog)(x) = x

It’s a circle with radius 2

=

+ cos −1 x 2

8 ⎛ 13 ⎞ 35 = ⎜ ⎟ − 3 3 ⎝3⎠

4. As the system has infinitely many solutions

− cos −1 x 2

30. The proposition (∼p) ∨ (p∧∼q) is equivalent to: (b) p → ∼q (a) p ∨ ∼q (c) p ∧ ∼q (d) q → p

2 4 −λ 4 λ 2 =0 λ 2 2 ⇒ 2

λ 2 4 2 4 λ −4 −λ =0 2 2 λ 2 λ 2

⇒ 2(2λ – 4) –4(8 – 2λ) – λ(8 – λ2) = 0

Answers 1. (d)

2. (c)

3. (a)

4. (b)

5. (d)

6. (c)

7. (d)

8. (d)

9. (b)

10. (b)

11. (d)

12. (d)

⇒ λ3 – 4λ – 40 = 0

Let g(x) = x3 – 4x – 40 g′(x) = 3x2 – 4

12

Complete Mathematics—JEE Main

7. (27)994 = (28 – 1)999 = (7 × 4 – 1)999

g′(x) = 0 ⇒ x = ± 2 / 3

= 7m – 1 = 7(m –1) + 6

g′′(x) = 6x

∴

⎛ 2 ⎞ ⎛ 2 ⎞ As g′′ ⎜ ⎟ > 0 , g′′ ⎜ − 3 ⎟ < 0 ⎝ ⎠ ⎝ 3⎠

remainder is 6 a+b = 5 ab 2 a+b = 10 ⇒ ab

8.

Graph of y(x) is

⇒ g(x)

2 − 3

a b + + 2 = 100 b a a b + = 98 b a

⇒

2 3

⇒

x

O

a b = 10 + b a

⇒ a2 + b2 + 2ab = 100ab

And a2 + b2 – 2ab = 96ab

(a + b) = 2 (a − b) 2

⇒ ∴ g(x) = 0 has exactly one root.

9.

1 adjA | A| = adjA

We know A–1 = ⇒ ⇒

|A|A–1

5 a+b = = 5 6 2 6 a −b 12 3 + 75 + 243 + 507 + ...

⇒

⇒ there is exactly one value of λ

5.

25 100 = 24 96

Upto n terms = 435 3

adj (adjA) = adj (|A|A–1)

⇒

3 + 5 3 + 9 3 + 13 3 + .... upto n terms

= 435 3

= |A|2 adj(A–1)

Also,

= |A|2(|A–1|(A–1)–1)

⇒ 1 + 5 + 9 + 13 + …. Upto n terms = 435

= |A|A

⇒

(adjA)–1 =

1 adj(adjA) | adjA |

1 adj(adjA) = | A |2

adj(adjA) = |A|2 (adj(A))–1 Thus, adj (adj A) = |A|(adj A)–1 is not necessarily true. ⇒

n {2 (1) + 4 ( n − 1)} = 435 2

⇒ 2n2 – 30n + 29n – 435 = 0 ⇒ 2n(n – 15) + 29(n – 15) = 0 ⇒ (2n + 29)(n – 15) = 0 ⇒ n = 15 as n ∈ N.

10. lim x →3

3x − 3 2x − 4 − 2

6. Letters of the word QUEEN are = lim x →3

E,E,N,Q,U Words beginning with E (4!) ⎛ 4! ⎞ ⎟ ⎝ 2! ⎠

R 1 to 24

Words beginning with N ⎜

25 to 36

Words beginning with QE (3!)

37 to 42

⎛ 3! ⎞ Words beginning with QN ⎜ ⎟ ⎝ 2! ⎠

3x − 3

)(

2x − 4 − 2

3x + 3

)(

)(

=

3 2x − 4 + 2 lim 2 x →3 3x + 3

=

3 2

2+ 2 2

3 +3

=

1 . 2

2x − 4 + 2

2x − 4 + 2

( 3x − 9 ) ( 2 x − 4 + 2 ) x →3 ( 2 x − 4 − 2 ) ( 3 x + 3)

= lim

43 to 45

QUEEN is the next word and has rank 46th.

(

(

)(

)

3x + 3

)

JEE (Main) 2017 Questions with Solutions Mathematics (8th April - online) 13 2 2

11. x y – 2x = 4(1 – y)

25 3 3 +5 cosθ + cos 2 θ 4 2 2

=

x2y2 – 2x + 4y – 4 = 0

⇒

Differentiating w.r.t. x, we get 2

π⎞ ⎛ = 14 + 5 3 cos ⎜⎝ θ − ⎟⎠ 4

When x = 2, y = –2 dy ⎤ dy ⎤ –2 + 4 dx ⎥⎦ =0 16 + 2(22)(–2) dx ⎥⎦ (2, −2) (2, −2)

∴ max (PQ)2 = 14 + 5 3

dy ⎤ 14 7 = = dx ⎥⎦ (2,−2) 12 6

⇒

14. 1 + 2cot x (cosec x + cot x) = 1 + 2cot x cosec x + 2cot2 x

Equation of tangent at (2, –2) is

= (1 + cot2 x) + 2cot x cosec x + cot2 x

7 y + 2 = (x – 2) 6

= cosec2 x + 2cot x cosec x + cot2 x = (cosec x + cot x)2

It does not pass through (–2, –7). 15

15

y = ⎡ x + x 2 − 1⎤ + ⎡ x − x 2 − 1⎤

12.

⎣

⎦

⎣

∴ I =

⎦

14 ⎧ dy x ⎪⎫ ⎪ = 15 ⎡ x + x 2 − 1⎤ ⎨1 + 2 ⎬ ⎣ ⎦ dx x − 1 ⎭⎪ ⎩⎪

= 14

2 + 15 ⎡⎣ x − x − 1⎤⎦

15

⇒

(

⎡ 2 ⎢⎣ x + x − 1 2 x −1

=

) −(x − 15

(

⎡ dy 2 x − 1 dx = 15 ⎢⎣ x + x − 1 2

x ⎪⎫ ⎪⎧ ⎨1 − 2 ⎬ ⎪⎩ x − 1 ⎪⎭ 2

x −1

) −(x − 15

)

15

⎤ ⎥⎦

x2 − 1

15. tan x + cot x =

)

15

⎤ ⎥⎦

∴ I =

=

x2 − 1

(1 +

x −1

)

15

+

π /4

∫π

x⎞

/12

⎠

2 1 + tan 2 x = sin 2x tan x

8cos 2 x

sin 3 2 x dx 23 π /4

225 x2 − 1

(1 +

2

x −1

)

=

15

2 ⇒ (x2 – 1) d y + x dy = 225 y dx 2 dx

4 1 ⎛ ⎛ 1⎞ ⎞ 15 1 − . = ⎜ ⎟ ⎜ ⎟ 8 ⎝ ⎝ 2⎠ ⎠ 128

16.

y (1, √3) E

A

13. Rewrite the circle as 2

O

2

5⎞ ⎛ 1⎞ ⎛ 3 ⎜x− ⎟ +⎜ y− ⎟ = 2 2 2 ⎝ ⎠ ⎝ ⎠ A point on (1) is

x=

(1)

⎛5 ⎞ ⎛5 ⎞ 3 3 cosθ ⎟⎟ + ⎜⎜ + sin θ ⎟⎟ ∴ (PQ)2 = ⎜⎜ + 2 2 ⎝2 ⎠ ⎝2 ⎠

D C

5 3 1 3 + cosθ , y = + sin θ , 0 ≤ θ < 2π. 2 2 2 2 2

x

B

2

Required area is shaded in the figure. Area = area (ODCBAEO) = 2 Area (O B A E O) = 2∫

0

3

4 − y 2 dy − 2∫

0

3

y2 dy 3

⎛x⎞

∫ cot ⎜⎝ 2 ⎟⎠ dx

1 1 ⎤ 4 = × ( sin 2 x ) ⎥ 4 2 ⎦ π/12

d y x dy + 2 dx 2 x − 1 dx 2

1 + cos x dx sin x

2cos 2 ( x / 2 ) ∫ 2cos ( x / 2 ) sin ( x / 2 ) dx =

⎛ ⎝

2

225

∫ (cosec x + cot x ) dx = ∫

= 2log ⎜ sin ⎟ + c. 2

Differentiating w.r.t. x, we get x2 − 1

25 3 3 + 5 sin θ + sin 2 θ 4 2 2

+

dy dy –2+4 =0 2xy + 2x y dx dx 2

14

Complete Mathematics—JEE Main

19. As sec–1 (7) = cos–1(1/7), both the chards are equidistant from the centre.

3

⎡ ⎛ y ⎞⎤ 2 = ⎢ y 4 − y 2 + 4sin −1 ⎜ ⎟ ⎥ – ⎡⎣ y 3 ⎤⎦ 9 ⎝ 2 ⎠⎦0 ⎣

3

Let AB = 2a.

⎛ 3⎞ 2 3 (1) + 4sin −1 ⎜⎜ ⎟⎟ − 3 3 ⎝ 2 ⎠ 9

( )

= =

⎛π ⎞ 2 3 + 4⎜ ⎟ − 3 ⎝3⎠

=

1 4 + π. 3 3

From ΔOAB

I.F. = e

⎛ 1⎞ ⎜⎝ − y ⎟⎠ dy

= e

− ln ( y )

(1)

a2 8 − 4a 2 1 = = 1− 2 8 7

⇒

6 a2 3 1 = 1− = ⇒ a = 2 7 2 7 7

Now, OM2 = OA2 – AM2

8 . 7 20. Equation of any tangent to x2 = 4y is

x = my + 1

m

It will be tangent to x2 + y2 = 4 if

1 = 3 + C ⇒ C = –2

| m (0) − 0 +

∴ x = (3y – 2)y

m2 + 1

⎛ 1 1⎞

It passes through ⎜ − , ⎟ . ⎝ 3 3⎠

2

x – 2ty + 3 = 0

−6 ( t 2 + 1) −2 ( t 2 − 1)

=

6t

⇒

⇒ (2m2 + 1)2 = 2

t 2 −1

⇒ m2 =

3t

2

=

(t

2

Let m1 =

)

2

+ 1 − 4t 2

(

)

t2 −1

2

2

⇒ 2m2 + 1 =

3 ( t 2 + 1)

y = 2 t 2 −1 = 2 ( ) t −1

2

O

⇒ 4m4 + 4m2 + 1 = 2

−y 1 x = 2 = 3t + 3t −2t 2 + 2 −6 − 6t

∴

1 | m = 2

1 ⇒ ⎛⎜ ⎞⎟ = 4 ( m 2 + 1) ⎝m⎠

18. tx – 2y – 3t = 0

2

12 16 = 7 7

= 2(OM) =

As it passes through (1, 1), we get

⎛ x⎞ ⎛ y⎞ ⎜⎝ ⎟⎠ − 4 ⎜⎝ ⎟⎠ 3 3

D

4 7 ∴ distance between porallel chards

x = 3y + C y

⇒ x =

C

⇒ OM =

d ⎛1 ⎞ x = 3 dy ⎜⎝ y ⎟⎠ ⇒

2

⇒

= 22 – a2 = 4 –

1 = y

1 to obtain y

Multiply (1) by

O

2 2 + 2 2 − ( 2a ) = ( 2 )( 2 )( 2 )

dx ⎛ −1 ⎞ + ⎜ ⎟ x = 3y dy ⎝ y ⎠

=

(t (t

2 2

− 1) − 1)

2 2

=1

2

x y − =1 9 ( 3 / 2 )2

It’s a hyperbola with length of conjugate axis as 3

2 −1 2 2 −1 , m2 = – 2

2 −1 2

∴Two common tangents are

x = m 1y +

1 1 , x = m 2y + m m1 2

These two intersect at ⎛

p ⎜ 0, ⎝

1 ⎞ ⎟ m1m2 ⎠

B 2

2

Cos(∠AOB)

17. Write the given differential equation as

M

A

JEE (Main) 2017 Questions with Solutions Mathematics (8th April - online) 15 ∴ OP =

1 = m1m2

2 = 2 2 −1

(

Also (6/13, 5/13, 0) lies on both (1) and (2)

)

2 +1 .

∴ the correct answer is (c)

24. Area of parallelogram is

⎛3⎞ 21. 2ae = 6 ⇒ 2a ⎜ ⎟ = 6 ⎝5⎠ ⇒ 2a = 10 ⇒ a = 5

=

1 | d1 × d 2 | 2

=

i j k 1 | 8 −6 0 | 2 3 4 −12

=

1 72i + 96 j + 50k 2

B

Now, b2 = a2 – a2e2 = 25 – 9 = 16

A′

⇒ b = 4

O

A

B′

Area of quadrilateral = 4[area ΔOAB]

=

⎛1⎞ = (4) ⎜ 2 ⎟ (5)(4) = 40. ⎝ ⎠

362 + 482 + 252 =

4225 = 65.

25. Let age of new teacher be x years, then

22. Equation of plane containing

40 × 25 − 60 + x = 39 25 60 − x ⇒ 40 – = 39 25

Both the lines is x +1 y −1 z − 3 6 7 8 =0 3 5 7

⇒

⇒ (49 – 40)(x + 1) – (42 – 24)(y – 1)

60 − x = 1 ⇒ x = 35. 25

26. P(P or Q but not R hits the target)

+ (30 – 21)(z – 3) = 0

= P((P ∪ Q) ∩ R′)

⇒ (x + 1) – 2(y – 1) + (z – 3) = 0

= P(P ∪ Q) P(R′) = [1 – P (P′ ∩ Q′)] P (R′)

Note that it passes through (1, 2, 3)

= [1 – P(P′) P (Q′)]P (R′)

As points (2, –4, 2), (–1, –2, –1) do not lie on (1), ensures (a), (b) are not possible. Also, (0, 0, 0) lies on (1) and line joining (1, –2, 1) and (0, 0, 0) is perpendicular to (1) 23. Let a, b, c, be direction ratios of the line of intersection of the planes 3x – y + z = 1

Then X ∼ B (n, p) where n = 8 p = ½ P(obtaining at leart one head and at leant one tail)

(1)

= 1 – P(no head) – P(no tail)

(2)

1 ⎛1⎞ ⎛1⎞ = 1 – ⎜ ⎟ −⎜ ⎟ = 1 – 7 2 2 2 ⎝ ⎠ ⎝ ⎠

and

8

x + 4y – 2z = 2 Then 3a – b + c = 0 a + 4b – 2c = 0 Solving we get

⇒

21 ⎡ ⎛ 1 ⎞⎛ 1 ⎞ ⎤ ⎛ 3 ⎞ = ⎢1 − ⎜ ⎟⎜ ⎟ ⎥ ⎜ ⎟ = . 64 4 2 8 ⎣ ⎝ ⎠⎝ ⎠ ⎦ ⎝ ⎠ 27. X = number of heads obtained,

a −b c = = 2−4 −6 − 1 12 + 1 b c a = = 7 13 −2

Note that (4/7, 0, 5/7) doesn’t lie on (1)

8

= 127

128

0 cos x − sin x 0 cos x 28. Let Δ(x) = sin x cos x sin x 0

= cos3x – sin3x = (cos x – sin x)(cos2 x + sin2 x + cos xsin x) = (cos x – sin x) ⎛⎜1 + 1 sin 2 x ⎞⎟ ⎝

2

⎠

16

Complete Mathematics—JEE Main

29. Put x2 = cos2θ, then

Now, Δ(x) = 0 ⇒ tan x = 1 ⇒ x =

π 5π

1 + x2 = 1 + cos2θ = 2cos2θ

4 4

And 1 – x2 = 1 – cos2θ = 2sin2θ

,

⎧ π 5π ⎫ ∴ S = ⎨ , ⎬ ⎩4 4 ⎭

⎛ 2 cosθ + 2 sin θ ⎞ ⎟⎟ ∴ y = tan–1 ⎜⎜ ⎝ 2 cosθ − 2 sin θ ⎠ ⎛ ⎛π ⎞⎞ ⎛ 1 + tan θ ⎞ –1 tan ⎜ + θ ⎟ ⎟ = tan–1 ⎜ ⎟ = tan ⎜⎝ ⎝4 ⎠⎠ ⎝ 1 − tan θ ⎠

⎛π π ⎞ + ⎟ + tan ⎛⎜ 5π + π ⎞⎟ ⎝ 4 3⎠ ⎝4 3⎠

Now, tan ⎜

⎛ π π⎞ ⎛ π π⎞ = tan ⎜ + ⎟ + tan ⎜ + ⎟ ⎝ 4 3⎠ ⎝ 4 3⎠ ⎛π ⎞ ⎛π ⎞ tan ⎜ ⎟ + tan ⎜ ⎟ ⎝4⎠ ⎝3⎠ = 2 ⎛π ⎞ ⎛π ⎞ 1 − tan ⎜ ⎟ tan ⎜ ⎟ ⎝4⎠ ⎝3⎠

=2

1+ 3 = –2 1− 3

(

30.

π 4

+θ =

π 4

1 + cos −1 x 2 2

(∼p) ∨ (p ∧ ∼ q) = (∼p ∨ p) ∧ (∼p ∨ ∼q)

)

3 +1 3 −1

=

2

= –(3 + 1 + 2 3 ) = –4 – 2 3

= T ∧ (∼p ∨ ∼q) = ∼p ∨ ∼q = p → ∼q

JEE (Main) 2017 Questions with Solution (9th April – online) 17

JEE (Main) 2017 Questions with Solution (9th April – online) ⎡ x⎤ 1. The function f : N → N defined by f ( x ) = x − 5 ⎢ ⎥ , ⎣5⎦ where N is the set of natural numbers and [x] denotes the greatest integer less than or equal to x, is: (a) one–one and onto. (b) one–one but not onto (c) onto but not one–one (d) neither one–one or onto

2. The sum of all the real values of x satisfying the 2 equation 2( x −1)( x +5 x −50) = 1 is: (a) 16 (b) 14 (c) –4 (d) –5 ⎛ iz − 2 ⎞

3. The equation Im ⎜ ⎟ + 1 = 0, n ∈ C , z ≠ i ⎝ z −i ⎠ represents a part of a circle having radius equal to: (a) 2 (b) 1 (c) 4. For and and (a) (c)

3 4

(d) 1

2

two 3 × 3 matrices A and B, let A + B = 2B′ 3A + 2B = I3, where B′ is the transpose of B I3 is 3 × 3 identity matrix. Then: (b) 10 A + 5B = 3I3 5A + 10B = 2I3 B + 2A = I3 (d) 3A + 6B = 2I3

5. If x = a, y = b, z = c is a solution of the system of linear equations x + 8y + 7z = 0 9x + 2y + 3z = 0 x+y+z=0 such that the point (a, b, c) lies on the plane x + 2y + z = 6, then 2a + b + c equals: (a) –1 (b) 0 (c) 1 (d) 2 6. The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boy B1 and a particular girl G1 never sit adjacent to each other, is: (a) 5 × 6! (b) 6 × 6! (c) 7! (d) 5 × 7! 7. The coefficient of x–5 in the binomial expansion of 10 ⎛ ⎞ x +1 x −1 ⎜ 2 ⎟ , where x ≠ 0, 1, is: − 1 1 ⎜⎝ 3 ⎟ 3 2 ⎠ x − x +1 x − x

(a) 1 (c) –4

(b) 4 (d) –1

8. If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is: (a) 2 (c)

(b)

1

43 (d) 4

2

43

1 1+ 2 1+ 2 + 3 .......... + + + 13 13 + 23 13 + 23 + 33 1 + 2 + ...... + n . If 100 Sn = n then n is equal to: + 3 1 + 23 + .....n3 Sn =

9. Let

(a) 199 (c) 200 10. The

(b) 99 (d) 19

value

of

k

for

which

the

function

tan 4 x ⎧ tan 5 x π ⎪⎛ 4 ⎞ π ,0 < x < ⎪⎜⎝ 5 ⎟⎠ 2 is continuous at x = , is f ( x) = ⎨ 2 ⎪ 2 π , k x + = ⎪ 5 2 ⎩

2 5

(a)

17 20

(b)

(c)

3 5

(d) − 1

11. If 2x = y 5 + y

−

1 5

and

then λ + k is equal to (a) –23 (c) 26 12. The is (a) (b) (c) (d)

(x

2

− 1)

2 5

d2y dy + λ x + ky = 0 2 dx dx

(b) –24 (d) –26

function f defined by f (x) = x3 – 3x2 + 5x + 7, increasing in R decreasing in R decreasing in (0, ∞) and increasing in (– ∞, 0) increasing in (0, ∞) and decreasing in (–∞, 0)

13. Let f be polynomial function such that f (3x) = f ′(x) · f ′′(x), for all x ∈R , then (a) f (2) + f ′(2) = 28 (b) f ′′(2) – f ′(2) = 0 (c) f ′′(2) – f(2) = 4 (d) f (2) – f ′(2) + f ′′(2) = 10 14. If

4 ⎛ 3x − 4 ⎞ f⎜ ⎟ = x + 2, x ≠ − , 3 ⎝ 3x + 4 ⎠

and

∫ f ( x)dx

=

A log |1 – x| + Bx + C, then the ordered pair

18

Complete Mathematics—JEE Main

(A, B) is equal to (where C is a constant of integration, (a)

⎛8 2⎞ ⎜ , ⎟ ⎝3 3⎠

⎛ 8 2⎞ (b) ⎜ − , ⎟ ⎝ 3 3⎠

(c)

⎛ 8 2⎞ ⎜− ,− ⎟ ⎝ 3 3⎠

(d) ⎜ , − ⎟ ⎝3 3⎠

15. If

2

∫ 1

⎛8

dx

(x

2

− 2x + 4)

(a) 1 (c) 3 16. If lim n →∞

3 2

=

2⎞

k , then k is equal to: k + 5'

( n + 1)

⎣⎡( na + 1) + ( na + 2 ) + ..... + (na + n ⎦⎤

= 1 for 60

some positive real number a, then a is equal to (a) 7 (b) 8 (c)

15 2

(d0 17 2

17. A tangent to the curve, y = f(x) at P(x, y) meets x– axis at A and y–axis at B. If AP : BP = 1 : 3 and f(1) = 1, then the curve also passes through the point: (a)

⎛1 ⎞ ⎜ , 24 ⎟ 3 ⎝ ⎠

(b) ⎜ , 4 ⎟ ⎝2 ⎠

(c)

⎛ 1⎞ ⎜ 2, ⎟ ⎝ 8⎠

⎛ 1 ⎞ (d) ⎜ 3, ⎟ ⎝ 28 ⎠

⎛1

⎞

18. A square, of each side 2, lies above the x–axis and has one vertex at the origin. If one of the sides passing through the origin makes an angle 30° with the positive direction of the x–axis, then the sum of the x–coordinates of the vertices of the square is: (b) 2 3 − 2 (a) 2 3 − 1 (c)

3−2

(d)

3 −1

19. A line drawn through the point P(4, 7) cuts the circle x2 + y2 = 9 at the points A and B. Then PA · PB is equal to: (a) 53 (b) 56 (c) 74 (d) 65 20. The eccentricity of an ellipse having centre at the origin, axes along the co–ordinate axes and passing through the points (4, –1) and (–2, 2) is: 2 (a) 1 (b) 5 2 (c)

3 2

(d)

1 1 1 1 + 2+ 2 = 2 x y z 9

23. If the line,

1a + 2a + ...... + n a

3 4

21. If y = mx + c is the normal at a point on the parabola y2 = 8x whose focal distance is 8 units, then |c| is equal to:

(b) 8 3 (d) 16 3

2 3

(c) 10 3 22. If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at A, B and C, then the locus of the centroid of ΔABC is 1 1 1 1 1 1 + 2 + 2 =1 (b) 2 + 2 + 2 = 3 (a) 2 x y z x y z (c)

(b) 2 (d) 4 a −1

(a)

(d)

1 1 1 + 2 + 2 =9 2 x y z

x−3 y + 2 z +λ lies in the plane, = = 1 −1 −2

2x – 4y + 3z = 2, then the shortest distance between this line and the line x − 1 = y = z is: (a) 2 (c) 0

12

9

4

(b) 1 (d) 3

24. If the vector b = 3j + 4k is written as the sum of a vector b1 parallel to a = i + j and a vector b 2 , perpendicular to a, then b1 × b 2 is equal to: (a)

−3i + 3 j − 9k

9 (b) 6i − 6 j + k 2

(c)

9 −6i + 6 j − k 2

(d) 3i − 3j + 9k

21 220 1 11

(b)

25. From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is: (a) (c)

3 11 2 (d) 23

26. Let E and F be two independent events. The 1 and the 12 1 probability that neither E nor F happens is , then 2 P(E) a value of P F is ( ) 4 (b) (a) 3 1 5 (c) (d) 3 12

probability that both E and F happen is

27. The sum of 100 observations and the sum of their squares are 400 and 2475, respectively. Later on, three observations, 3, 4 and 5, were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is:

JEE (Main) 2017 Questions with Solution (9th April – online) 19

(a) 8.25 (c) 8.00

(b) 8.50 (d) 9.00

28. A value of x satisfying sin[cot–1(1 + x)] = cos[tan–1x] is (a)

−

(c) 0

1 2

Range f = (0, 1, 2, 3, 4) In view of one may answer it as (d). the

equation

2

+ 5x – 50)

=1

(i)

2

⇒ (x – 1)(x + 5x – 50) = 0

(b) –1 (d)

2(x – 1)(x

2.

⇒ (x – 1)(x + 10)(x – 5) = 0

1 2

∴ x = 1, 5, – 10

29. The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60°. If the area of the quadrilateral is 4 3 , then the perimeter of the quadrilateral is: (a) 12.5 (b) 13.2 (c) 12 (d) 13 30. Contra positive of the statement ‘If two numbers are not equal, then their squares are not equal’, is: (a) If the squares of two numbers are equal, then the numbers are equal. (b) If the squares of two numbers are equal, then the numbers are not equal. (c) If the squares of two numbers are not equal, then the numbers are not equal. (d) If the squares of two numbers are not equal, then the numbers are equal.

∴ sum of the real values satisfying (i) is 1 + 5 – 10 = – 4.

3. Let z = x + iy ⎛ iz − 2 ⎞ ⎟ +1 = 0 ⎝ z −1 ⎠

Now, Im ⎜ ⇒ Im

⇒ Im

⇒ ⇒

ix − y − 2 +1 = 0 x + i ( y − 1)

( ix − y − 2 ) ( x − i ( y − 1) ) x 2 + ( y − 1) 2

x 2 + ( y + 2)( y − 1) x 2 + ( y − 1)

+1 = 0

2 x 2 + ( y − 1)( 2 y + 1) = 0

⇒ 2x2 + 2y2 – y – 1 = 0 x2 + y 2 −

Answers

2

+1= 0

1 1 y− = 0 2 2

2

1. (d)

2. (c)

3. (c)

4. (b)

5. (c)

6. (a)

7. (a)

8. (a)

9. (a)

10. (c)

11. (b)

12. (a)

13. (b)

14. (b)

15. (a)

16. (a)

17. (c)

18. (b)

19. (b)

20. (c)

21. (c)

22. (a)

23. (c)

24. (b)

25. (c)

26. (a)

27. (d)

28. (a)

29. (d)

30. (a)

Hints and Solutions

⇒ r2 = ⎛ 1 ⎞ + 1 = 9 ⎜ ⎟ ⎝ 4 ⎠ 2 16 3 ⇒r= 4

4. A + B = 2B′ 3A + 2B = I3

(1) (2)

From (1), (2), we get A’ + B’ = 2B

(3)

and 3A’ + 2B’ = I3

(4)

From (1) and (3) A – A′ + B – B′ = 2(B′ – B)

⎡5⎤ 1. Note that f (5) = 5 − 5 ⎢ ⎥ = 0 = f(10). ⎣5⎦

Thus, function f is not defined as 0 ∉N , the codomain. As, for 0 ≤ r ≤ 4 f (5m + r) = 5m + r – 5m =r

⇒ A – A′ = 3(B′ – B)

(5)

From (2) and (4) 3(A – A′) + 2(B – B′) = 0 From (5) and (6) 9(B′ – B) – 2(B′ – B) = 0 ⇒ 7(B’ – B) = 0 ⇒ B′ = B

(6)

20

Complete Mathematics—JEE Main

⇒ From (1) we get A = B

10

=

Cr x

( −1)r

1 and from (2) A = B = I 3 5

For coefficient of x–5, set

10A + 5B = 15A = 3I3 It is easy to check (c), (d) are not true.

10 − r r − = –5 3 2 ⇒ 20 – 2r – 3r = –30

Now, 5A + 10B = 15A = 3I3 ≠ 2I2

5. x + 8y + 7z = 0

(1)

⇒ r = 10

9x + 2y + 3z = 0

(2)

∴

x+y+z=0

(3)

From (2), (3) x −y z = = 2−3 9−3 9−2

⇒

x y z = k (say) = = −1 −6 7

2 3

⎛ 13 ⎞ ⎜ x ⎟ +1 ⎝ ⎠

=

1 3

2 3

1 3

x −1 x−x

∴

=

1 2

(

)(

x

(

)

10

⎞ ⎟ ⎟ ⎠

10

Sn = 2 ∑ ⎜ −

n = 199

10 − r

π 2

−1 2

⎛ ⎞ ⎜ −x ⎟ ⎝ ⎠

r

, tan 4 x

1

)

= 1+ x

−

1 2

⇒ k =

3 . 5

1

11. 2x = y 5 + y − 5 2

⇒

⇒

⎛ ⎞ Cr ⎜ x ⎟ ⎝ ⎠

n

1 ⎞ 1 ⎞ ⎛ ⎛1 ⎟ ⎟ = 2 ⎜1 − r +1⎠ ⎝ n +1⎠ r =1 ⎝ r 2n ⇒ Sn = n +1 n 2n n Now, Sn = ∴ = 100 n + 1 100

⇒

⇒

10

1 3

1 + 2 + ..... + r 1 + 23 + ..... + r 3 3

r ( r + 1) 2 2 = 2 2 = r r r ( r + 1) ( + 1) 4 1 ⎞ ⎛1 = 2⎜ − ⎝ r r + 1⎟⎠

⇒

1 − ⎞ ⎛ 1 = ⎜ x3 − x 2 ⎟ ⎝ ⎠

Tr + 1 =

tr =

∴ k+2 = 1 5

x +1

x −1

=1

1

tan 4 x

x −1

⎛ x +1 x −1 ⎜ − 1 1 ⎜ 32 ⎝ x − x3 +1 x − x2

10

lim ⎛π ⎞ ⎛ 4 ⎞ tan 5 x = ⎛ 4 ⎞ x→π2 tan 5 x = ⎛ 4 ⎞ ° = 1 ⎜ ⎟ f ⎜ ⎟ = lim ⎜ ⎟ ⎜ ⎟ ⎝ 5⎠ ⎝5⎠ ⎝ 2 ⎠ x → π2 ⎝ 5 ⎠

1 ⎛ 13 ⎞⎛ 32 ⎞ ⎜ x + 1⎟⎜ x − x 3 + 1⎟ ⎝ ⎠⎝ ⎠ = 1 ⎛ 32 ⎞ ⎜ x − x 3 + 1⎟ ⎝ ⎠

Also,

C10 ( −1)

1

10. As f is continuous at x =

x − x +1

x − x +1

10

8. 3b = a + b + c ≥ 3 ( abc ) 3 = 3 ( 8 ) 3 ⇒ b ≥ 2 ∴ minimum value of b is 2.

⇒

3

x +1

coefficient of x–5 is

9.

⇒ x = –k, y = –6k, z = 7k Note (–k, –6k, 7k) satisfies (1) Let a = –k, b = –6k, c = 7k As (a, b, c) lies on x + 2y + z = 6 we get, –k – 12k + 7k = 6 ⇒ k = –1 ∴ a = 1, b = 6, c = –7 Now, 2a + b + c = 2 + 6 – 7 = 1 6. Required number of ways = Total number of ways – Number of ways in which B1 and G1 are together = 7! – (2!)6! = (5)(6!)

7.

10 − r r − 3 2

⎛ 15 ⎞ ⎛ 1⎞ ⎜ y ⎟ − 2x ⎜ y 5 ⎟ + 1 = 0 ⎝ ⎠ ⎝ ⎠ 1

y 5 = x ± x2 − 1

(

y = x ± x2 − 1

)

5

)

(

4⎛ x ⎞ 2 dy = 5 x ± x − 1 ⎜1 ± 2 ⎟ x −1 ⎠ ⎝ dx

= ±

5y x2 − 1

JEE (Main) 2017 Questions with Solution (9th April – online) 21 ⇒

x2 − 1

dy = ±5 y dx

⇒

x2 − 1

d2y x dy dy + = ±5 2 2 dx dx x − 1 dx

⇒

(x

2

− 1)

dy ⎞ ⎛ 2 d2y dy = ±5 ⎜ x − 1 ⎟ +x 2 dx ⎠ ⎝ dx dx

= ±5 ( ±5 y ) = 25 y ∴ λ = 1, k = –25 ⇒ λ + k = –24 12. f (x) = x3 – 3x2 + 5x + 7 f ’(x) = 3x2 – 6x + 5 = 3(x – 1)2 + 2 > 0 ∨ x ∈ R Thus, f is increasing in R. 13. Let deg(f (x)) = n, then f (3x) = f ′(x)f ′′(x) ⇒ deg(f (3x)) = deg(f ′(x)) + deg(f ′′(x)) ⇒ n = (n – 1) + (n – 2) ⇒ n = 3 Let f (x) = ax3 + bx2 + cx + d, then f (3x) = f ′(x)f ′′(x) ⇒ 27ax3 + 9bx2 + 3cx + d = (3ax2 + 2bx + c)(6ax + 2b) ⇒ 27a = 18a2 ⇒

a=

3 2

(∵ a ≠ 0 )

Next, 9b = 6ab + 12ab = 18ab = 27b ⇒ b = 0 and 3c = 6ac + 4b2 = 9c + 0 ⇒ c = 0 and d = 2bc = 0 Thus, f (x) = ⇒ f ′(x) =

3 3 x 2

9 2 x , f ′′(x) = 9x 2

f ′′(2) – f ′(2) = 0. ⎛ 3x − 4 ⎞ ⎟ = x + 2 ⎝ 3x + 4 ⎠

14. f ⎜

Put 3x − 4 = t 3x + 4 8 1–t= 3x + 4

⇒

8 ⇒ 1− =t 3x + 4

8 1− t 1⎛ 8 1 ⎛ 8 − 4 + 4t ⎞ ⎞ − 4⎟ = ⎜ x= ⎜ ⎟ 3 ⎝1− t 3 ⎝ 1− t ⎠ ⎠

⇒ 3x + 4 =

4 ⎛1+ t ⎞ ⎜ ⎟ 3 ⎝ 1− t ⎠ 4⎛ 2 4 ⎛1+ x ⎞ ⎞ − 1⎟ + 2 ∴ f (x) = ⎜ ⎟+2 = ⎜ 3 ⎝ 1− x ⎠ 3 ⎝1− x ⎠ 8 1 2 + = 3 1− x 3 8 2 ⇒ ∫ f ( x)dx = − log |1 − x | + x + C 3 3 8 ∴ A= − 3 2 B= . 3

=

2

15. I =

∫ 1

=

dx

(x

3

− 2x + 4)2

2

2

dx

1

⎡( x − 1)2 + 3⎤ 2 ⎣ ⎦

∫

3

Put x – 1 =

3 tan θ

π

π

3 sec 2 θ 16 dθ = ∫ cosθ dθ I= ∫ 3 30 0 3 3 sec θ 6

π 1 [sin θ ]06 = 1 3 6 k 1 Now, = k +5 6 ⇒ k=1

=

16. L =

1a + 2a + ....... + n a

lim

( n + 1)

n →∞

a −1

⎡⎣( na + 1) + ( na + 2 ) + ....( na + n ) ⎤⎦ a

n ⎛k⎞ na ∑ ⎜ ⎟ k =1 ⎝ n ⎠ = lim a −1 n →∞ 1 k⎞ ⎛ ⎞ n ⎛ n a ⎜1 + ⎟ ∑ ⎜ a + ⎟ n⎠ ⎝ n ⎠ k =1 ⎝ a

1 n ⎛k⎞ ∑⎜ ⎟ n k =1 ⎝ n ⎠ = lim a −1 n →∞ k⎞ ⎛ 1⎞ 1 n ⎛ ∑ ⎜1 + ⎟ ⎜a + ⎟ n n n⎠ ⎝ ⎠ k =1 ⎝

=

∫

1

0

x a dx

1 a +1

= 1 2 ( a + 1) − a 2 2 2 1 = ( a + 1)( 2a + 1) = 60 1

∫ ( a + x ) dx 0

(

⇒ (a + 1)(2a + 1) = 120 ⇒ a=7

(∵ a > 0 )

)

22

Complete Mathematics—JEE Main

17. An equation of tangent at P(x, y) is B

3

=

3 + 2 2 cos ( 45° + 30° ) − 1

=

⎛ 1 3 1 1⎞ ⋅ − ⋅ ⎟ −1 3 + 2 2 ⎜⎜ 2 2 ⎠⎟ ⎝ 2 2

=

3 + 3 −1−1 = 2 3 − 2

P(x, y) 1

19.

T

P (4, 7)

A

O

A B

dy Y–y= ( X − x) dx

(PA)(PB) = PT2 and PT2 = 42 + 72 – 9 = 56 20. Let equation of ellipse be

It meets x – axis in ⎛ ⎞ dx A⎜ x − y,0 ⎟ dy ⎝ ⎠ dy ⎞

⎛ ⎝

and y–axis in B ⎜ 0, y − x ⎟ dx ⎠

Coordinates of the point which divide it in the ratio 1 : 3 is ⎛ 3⎛ dx ⎜ ⎜x− 4 dy ⎝ ⎝

⎞ 1⎛ dy ⎞ ⎞ y ⎟ , ⎜ y − x ⎟ ⎟ = (x, y) dx ⎠ ⎠ ⎠ 4⎝

⇒

3⎛ dx ⎞ 1⎛ dy ⎞ ⎜ x − y ⎟ = x, ⎜ y − x ⎟ = y 4⎝ dy ⎠ 4⎝ dx ⎠

⇒

dx dy 3 + =0 x y

⇒ log(x3y) = const. ⇒ x 3y = c When x = 1, y = 1, therefore c = 1 Thus, required curve x3y = 1

(i)

Note that ⎜ 2, ⎟ satisfies (i) ⎝ 8⎠ 18. x–coordinate of A is 2cos30° = 3 , x–coordinate of B is 2 2 cos 75° and x–coordinate of C is 2cos (120° ) = 2 cos(90° + 30°) = –2sin30° = –1 B C

2 45° 30°

Solving we get b2 = 5, a2 = 20 Now a2(1 – e2) = b2 ⇒ 20(1 – e2) = 5 3 2

21. c = –2am – am3 = –4m – 2m3 ∴ y = mx – 4m – 2m3 is a normal at P(2m2, 4m) Distance of P from focus = distance of P from directrix (x = –2) = 2(m2 + 1) As 2(m2 + 1) = 8 ⇒m= ± 3

(

)

For m = − 3 , c = 4 3 + 2 3 3 = 10 3 22. Let the centroid of ΔABC be G (α , β , γ ) then coordinates of A, B, C are ( 3α ,0,0 ) , ( 0,3β ,0 ) , (0, 0,3γ ) respectively Equation of the plane through A, B and C is x y z + + =1 3α β 3γ

2 45°

x2 y 2 + =1 a 2 b2 As it passes through (4, –1) and (–2, 2) we get 16 1 4 4 + 2 = 1 and 2 + 2 = 1 2 a b a b

⇒e=

1⎞

⎛

O

A

Its distance from (0, 0, 0) is | 0 −3| =3 1 1 1 + 2+ 2 2

O

α

Thus, sum of x–coordinates = 0 + 3 + 2 2 cos 75° − 1

⇒

1

α2

β

+

1

β2

γ

+

1

γ2

=1

JEE (Main) 2017 Questions with Solution (9th April – online) 23

Thus, locus of centroid of ΔABC is 1 1 1 + 2 + 2 =1 2 x y z

Number of ways in which there are more women than men in the committee

x−3 y + 2 z +λ = = lie in the plane 1 −1 −2 2 x − 4 y + 3z = 2

= (1)(5) + (10)(10) = 105 ∴ Probability of the required event

=

23. As

(3, −2, − λ )

lies on the plane. ∴ 2(3) – 4(–2) + 3(–λ) = 2 ⇒ λ = 4 d1 = i − j − 2k and d 2 + 12i + 9 j − 4k

Now

i j k d1 × d 2 = 1 −1 −2 12 9 4

= 14i − 28 j + 21k

3 − 1 −2 − 0 −4 − 0 1 −1 −2 12

9

4

= 14(2) – 28(–2) + 21(–4) = 7(4 + 8 – 12) = 0 ∴ Shortest distance between two lines is 0. 24. We have b = b1 + b 2 where b1 = α a1 , b 2 ⋅ a = 0 ∴

a ⋅ b = αa ⋅ a

⇒ 3 = 2α ⇒ α =

3 2

  3 3 Thus, b1 = a ⇒ b 2 = b = a 2 2 3 3 ⎞ ⎛ 3 ⇒ b1 × b 2 = a × ⎜⎝ b − a⎟⎠ = a × b 2 2 2

=

10

C0 )( 5 C4 ) + ( 10 C1 )( 5 C3 )

105 1 = . 1155 11

P(E ∩ F ) =

26.

1 1 ⇒ P(E) P(F ) = 12 12 1

1 and P ( E '∩ F ') = 2 ⇒ P ( E ') P ( F ') =

2

⇒

(1 − P ( E ) ) (1 − P ( F ) ) = 12

⇒ 1 − ( P ( E ) + P( F ) ) +

= 7 ( 2i − 4 j + 3k ) Also,

=

(

3 (i + j) × (3j + 4k ) 2

i j k 3 = 1 1 0 2 0 3 4 3 9 = ( 4i − 4 j + 3k ) = 6i − 6 j + k . 2 2

25. Number of ways of choosing committee so that there is at least one woman is 15 C4 −10 C4 = 1365 – 210 = 1155

1 1 = 12 2

7 12

⇒

P(E) + P(F ) =

∴

Equation whose roots are P(E), P(F) is 1 ⎛7⎞ x2 − ⎜ ⎟ x + =0 12 ⎝ 12 ⎠

or

12 x 2 − 7 x + 1 = 0

⇒

( 3x − 1)( 4 x − 1)

⇒

1 1 x= , 3 4

∴

1 1 P(E) 3 = or 4 P(F ) 1 1 4 3

⇒

P(E) 4 3 = or . P(F ) 3 4

27. Correct Correct

=0

∑ x = 400 − 3 − 4 − 5 = 388 ∑ x = 2475 − 3 − 4 − 5 = 2425 2

2 ∴ Correct σ =

2

2

1 1 ( 2425) − ⎛⎜ ( 388) ⎞⎟ 97 ⎝ 97 ⎠

= 25 – 42 = 9. 28. Let α = cot −1 (1 + x ) ⇒ cot α = 1 + x ⇒ sin α =

2

1 1 + cot 2 α

2

24

Complete Mathematics—JEE Main

=

1

∴

2 + 2x + x 2

Next, let β = tan −1 x ⇒

tan β = x

= a 2 + b 2 − 2ab cos120°

⇒ 2 + 2x + x2 = 1 + x2

⎛ 1⎞ ⎝ ⎠

2 2 ⇒ 19 = a + b − 2ab ⎜ − ⎟ = a 2 + b 2 + 6 2

1 x=− . 2

29. Area of ΔABC = 1 ( 2 )( 5 ) sin 60° = 2

B A

2 a D

60°

5 C

120°

b

5 3 3 3 = 2 2

⎛ 3⎞ 1 1 Thus, 3 3 = ( ab ) sin120° = ( ab ) ⎜⎜ ⎟⎟ 2 2 2 ⎝ 2 ⎠ ⇒ ab = 6 Also, AC2 = 22 + 52 – (2)(2)(5)cos60°

1 1 = ⇒ cos β = 2 1 + tan β 1 + x2 ∴ sin α = cos β

⇒

area of ΔACD = 4 3 −

5 3 2

⇒ a 2 + b 2 = 13 ⇒ a2 + b2 = 13, ab = 6 ⇒ a = 2, b = 3 Thus, perimeter = 2 + 5 + 2 + 3 = 12. 30. Contrapositive of p → q is ~ q →~ p ∴ required contrapositive statement is If the squares of two numbers are equal, then the numbers are equal.

JEE Main 2017 (Offline) 25

JEE Main 2017 (Offline) 1. Let k be an integer such that the triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point: (a)

⎛ 1⎞ ⎜ 2, ⎟ ⎝ 2⎠

(b) ⎜ 2, − ⎟

(c)

⎛ 3⎞ ⎜1, ⎟ ⎝ 4⎠

3⎞ ⎛ (d) ⎜1, − ⎟ ⎝ 4⎠

⎛ ⎝

1⎞ 2⎠

+ ( x + n − 1) ( x + n) = 10n

has two consecutive intergal solutions, then n is equal to: (a) 11 (b) 12 (c) 9 (d) 10

f ( x) =

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

⎡ 1 1⎤ f : R → ⎢− , ⎥ ⎣ 2 2⎦

defined

as

x , is 1 + x2

neither injective nor surjective. invertible. injective but not surjective. surjective but not injective.

4. The following statement ( p → q) → [(~ p → q )] is: (a) a fallacy (b) a tautology (c) equivalent to ~ p → q (d) equivalent to p → ~ q 5. 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

ax + by + z = 0

has (a) (b) (c) (d) 6. The

59 12

(c)

3 2

(d)

7 3

Then: (a) a, (b) b, (c) b, (d) a,

b c c b

and and and and

c a a c

are are are are

in G.P. in G.P. in A.P. in A.P.

8. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is: (a) 484 (b) 485 (c) 468 (d) 469 9. 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: (a)

⎛1 1⎞ ⎜ , ⎟ ⎝ 2 3⎠

(b) ⎜ − , − ⎟

⎛ 1 ⎝ 2

(c)

⎛1 1⎞ ⎜ , ⎟ ⎝2 2⎠

(d) ⎜ , − ⎟

⎛1 ⎝2

1⎞ 2⎠

1⎞ 3⎠

10. A hyperbola passes through the point P( 2, 3) and has foci at (± 2, 0). Then the tangent to this hyperbola at P also passes through the point: (a)

(−

2, − 3 )

(b)

(3

(c)

(2

2,3 3 )

(d)

(

2, 2 3 ) 3, 2 )

2 11. Let a, b, c ∈ R. If f ( x) = ax + bx + c is such that a

no solution, then S is: A singleton An empty set An infinite set A finite set contaning two or more elements area

(b)

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

x( x + 1) + ( x + 1)( x + 2) + ....

function

5 2

7. For any three positive real numbers a, b and c,

2. If, for a positive integer n, the quadratic equation,

3. The

(a)

(in

sq.

units) of the region 2 {( x, y ) : x ≥ 0, x + y ≤ 3, x ≤ 4 y and y ≤ 1 + x } is:

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

10

∑ f ( n)

is equal to:

n =1

(a) 255

(b) 330

(c) 165

(d) 190

26

Complete Mathematics—JEE Main

12. Let a = 2i + j – 2k and b = i + j Let c be a vector such that | c − a | = 3, (a × b ) × c = 3 and the angle between c and a × b be 30o. Then a · c is equal to: 1 25 (b) (a) 8 8 (c) 2

(d) 5

13. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If ∠BPC = β then tanβ is equal to: (a) (c)

4 9 1 4

(b)

6 7

(d)

2 9

15. The integral

dx

∫π 1 + cos x

is equal to:

4

(a) –1 (c) 2 16. If (2+sin x)

(b) –2 (d) 4 dy + (y + 1) cos x = 0 and y(0) = 1, dx

⎛π ⎞ then y ⎜ ⎟ is equal to: ⎝2⎠ 4 (a) 3

(c)

−

2 3

(b)

1 3

(d) −

1 3 5

17. Let In = ∫ tan x dx, (n > 1). If I4 + I6 = atan x n

+ bx5+ c, where c is a constant of integration, then the ordered pair (a, b) is equal to: ⎛ 1 ⎞ ⎛ 1 ⎞ (b) ⎜ − ,1⎟ (a) ⎜ − ,0 ⎟ ⎝ 5 ⎠ ⎝ 5 ⎠ (c)

⎛1 ⎞ ⎜ ,0 ⎟ ⎝5 ⎠

then k is equal to : (a) 1 (c) z

(b) –z (d) –1

19. The value of

( (

) ( C − C )+ C ) + ( C − C ) + ...... (

21

C 1 − 10C1 +

21

C 3 − 10

21

10

2

21

3

2

⎛1 ⎞ (d) ⎜ , −1⎟ ⎝5 ⎠

18. Let ω be a complex number such that 2ω + 1 = z where z = −3 . If

20. limπ x→

2

cot x − cos x

(π − 2 x )

3

C 10 − 10C10 ) is :

21

10

4

4

(a) 220 – 210 (c) 221 – 210

14. Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is: (a) 30 (b) 12.5 (c) 10 (d) 25 3π 4

1 1 1 1 −ω 2 − 1 ω 2 = 3k 1 ω2 ω7

(b) 221 – 211 (d) 220 – 29 equals:

(a)

1 4

(b)

1 24

(c)

1 16

(d)

1 8

21. If 5(tan2 x – cos2x) =2cos 2x + 9, then the value of cos 4x is: 7 3 (a) − (b) − 9 5 1 2 (c) (d) 3 9 22. If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line, x y 7 = = is Q, then PQ is equal to: 1 4 5 (a)

6 5

(b) 3 5

(c)

2 42

(d)

42

23. 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 and

y+2 z−4 x −1 = = −2 1 −2

y +1 z+7 x−2 = = , is : − 1 −1 2

(a)

10 74

(b)

20 74

(c)

10 83

(d)

5 83

JEE Main 2017 (Offline) 27 ⎛ 6x x ⎞ ⎛ 1⎞ −1 24. If for x ∈ ⎜ 0, ⎟ , the derivative of tan ⎜⎜ 3 ⎟ ⎟ is ⎝ 4⎠ ⎝ 1 − 9x ⎠ x . g(x), then g(x) equal: 9 3 (b) (a) 3 1 + 9x 3 1 + 9x

(c)

3x x 1 − 9 x3

(d)

3x 1 − 9 x3

25. The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = | x | is: (a)

4

(

2 +1

)

(b) 2

(

2 +1

)

(c)

2

(

2 −1

)

(d) 4

(

2 −1

(c)

(d) 4

27. The eccentricity of an ellipse whose centre is at the 1 origin is . If one of its directrices is x = –4, then 2 ⎛ 3⎞ the equation of the normal to it at ⎜1, ⎟ is: ⎝ 2⎠ (a) x + 2y = 4 (b) 2y – x =2 (c) 4x – 2y = 1 (d) 4x +2y = 7 28. If two different numbers are taken from the set {0, 1, 3,….., 10}; then the probability that their sum as well as absolute difference are both multiple of 4, is : 6 7 (b) (a) 55 55 12 14 (c) (d) 55 45 29. For three events A, B and C, P (Exactly one of A or B occurs) = P(Exactly one of B or C occurs) 1 =P(Exactly one of C or A occurs)= and P(All the 4 1 three events occur simultaneously) = . 16

Then the probability that at least one of the events occurs, is:

3 16 7 16

7 32 7 (d) 46

(b)

⎡ 2 −3⎤

2 30. If A = ⎢ ⎥ , then adj (3A +12A) is equal to: − 4 1 ⎣ ⎦

(a)

⎡ 72 −63⎤ ⎢ −84 51 ⎥ ⎣ ⎦

⎡ 72 −84⎤ (b) ⎢ ⎥ ⎣ −63 51 ⎦

(c)

⎡ 51 63⎤ ⎢84 72⎥ ⎣ ⎦

⎡ 51 84 ⎤ (d) ⎢ ⎥ ⎣63 72⎦

)

26. 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: 6 12 (b) (a) 25 5 (c) 6

(a)

Answers 1. (a)

2. (a)

3. (d)

4. (b)

5. (a)

6. (a)

7. (c)

8. (b)

9. (c)

10. (c)

11. (b)

12. (c)

13. (d)

14. (d)

15. (c)

16. (b)

17. (c)

18. (b)

19. (a)

20. (c)

21. (a)

22. (d)

23. (c)

24. (b)

25. (d)

26. (b)

27. (c)

28. (b)

29. (c)

30. (c)

Hints and Solutions k 1 | 5 1. 2 −k

−3k 1 k 1 | = 28 2 1

R 2 → R 2 – R 1, R 3→ R 3 – R 1 To get k 5−k −2k

−3k 1 4k 0 = ± 56 2 + 3k 0

⇒ (5 – k)(2 + 3k) + 8k2 = ±56 ⇒ 10 + 13k – 3k2 + 8k2 = ±56 ⇒ 5k2 + 13k2 + 66 = 0 or 5k2 + 13k – 46 = 0

But 5k2 + 13k + 66 = 0 does not have real roots. Next, 5k2 + 13k – 46 = 0 has two roots, 23 5 As k is an integer, k = 2.

viz. 2 and −

28

Complete Mathematics—JEE Main

Thus, vertices of triangle are

x = α ⇔ α x2 − x + α = 0 1 + x2

(2, –6), (5, 2) and (–2, 2) Equation of altitude AD is x = 2

x=

8 −6 − 2 = 3 2−5 3 C(–2, 2) ∴ Slope os CF is − 8

Slope of AB is

D

y–2= −

3 (x + 2) 8

⎛ 1 ± 1 − 4α 2 ∴f⎜ ⎜ 2α ⎝

B(5, 2)

G

Equation of CF is

Thus, f is surjective.

A (2, –6)

4. Let P be the statement p → q; Q be the statement ~p → q and

Thus, orthocenter H of ΔABC is (2, 2. Equation

R be the statement Q → q.

1 ) 2

x(x + 1) + (x + 1)(x + 2) + ……. + (x + n – 1)(x + n) = 10n Can be written as n −1

nx + (1 + 3 + ….. + 2n – 1)x +

∑ k ( k + 1) = 10n k =1

1 1 ⇒ nx + n x + (n – 1)n (2n – 1) + n(n – 1) 6 2 2

2

p T T F F

q T F T F

~p F F T T

P T F T T

Q T T T F

Q→P T F T T

P→(Q→q) T T T T

Thus, (p → q) → [(~p → q)→ q] is a tautology. 5. Let 1 1 1

= 10n 1 2 (n – 31) = 0 3 If α, β are roots of (i), then ⇒ x2 + nx +

|α – β| = 1 ⇒ (α+ β)2 – 4αβ = 1 ⇒ n2 –

⎞ ⎟ = α ⎟ ⎠

Also, f(0) = 0. F

1 Intersection of x = 2, is y = . 2

2

1 ± 1 − 4α 2 2α

4 2 (n – 31) = 1 3

(i)

D= 1 a 1 a b 1 Applying, C1 → C1 – C3, C2 → C2 – C3, to obtain 0 0 1 D = 0 a − 1 1 = –(a – 1)2 a −1 b −1 1

⇒ –n2 + 121 = 0

If D ≠ 0, the system has unique solution, therefore,

⇒ n = 11.

D=0⇒a=1

3. Note that 1 x 2 = 3 1+ x ⇒ x2 – 3x + 1 = 0

3± 5 ⇒ x = 2 ⎛ 3− 5 ⎞ ⎛ 3+ 5 ⎞ 1 ∴ f ⎜⎜ = f ⎜⎜ ⎟⎟ = ⎟⎟ 3 ⎝ 2 ⎠ ⎝ 2 ⎠

Thus, f is not injective. For α≠ 0,

System of Equations become x+y+z=1 x + by + z = 0 ⇒ (1 – b)y = 1

For system to have no solution, b = 1 6. y = 1 +

x and

y = 3 – x meet at (1, 2)

JEE Main 2017 (Offline) 29

9. The curve meets the y–axis at (0, 1). We have

y

y= x+y=3

x2 = 4y

(1, 2) (0, 1)

(2, 1)

A2

A1 0

y = 1 + √x

1

2

x

Also, x2 = 4y and x + y = 3 meet at (2, 1) Required area

9 8 − x−3 x−2 dy 9 8 + = − 2 2 dx x − 3 x − ( ) ( 2)

=

∴

dy ⎤ 9 8 = − + =1 dx ⎥⎦ ( 0,1) 9 4

Thus, slope of normal at (0, 1) is –1. Equation of normal at (0, 1) is

= A1 + A2

y – 1 = –1(x – 0) or x + y = 1.

1

1 ⎞ ⎛ = ∫ ⎜1 + x − x 2 ⎟ dx + 4 ⎠ 0⎝

2

1 2⎞ ⎛ ∫1 ⎜⎝ 3 − x − 4 x ⎟⎠ dx

⎛1 1⎞ It passes through ⎜ , ⎟ ⎝2 2⎠

1

2 ⎛ 2 3 1 3 ⎞⎤ 1 1 ⎞⎤ ⎛ = ⎜ x + x 2 − x ⎟ ⎥ + ⎜ 3x − x 2 − x3 ⎟ ⎥ 3 12 ⎠ ⎥⎦ 0 ⎝ 2 12 ⎠ ⎦1 ⎝

8 1 1⎞ ⎛ 2 1⎞ ⎛ = ⎜1 + − ⎟ + ⎜ 6 − 2 − − 3 + + ⎟ 12 2 12 ⎠ ⎝ 3 12 ⎠ ⎝

=

x+6 x − ( 2 )( x − 3)

10. Let the equation of the hyperbola be x2 y 2 =1 − a 2 b2

We know b2 = a2(e2 – 1)

5 sq.units 2

⇒ a2 + b2 = (ae)2 = 22 = 4

7. 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c) ⇒ (15a)2 + (3b)2 + (5c)2 – 45ab – 15bc – 75ac =

0 ⇒ ((15a)2 + (3b)2 – 90ab) + ((3b)2 + (5c)2 – 30bc)

+ ((15a)2 + (5c)2 – 150ac) = 0

Also, 2 3 − = 1 a 2 b2 2 3 ⇒ 2− =1 a 4 − a2 ⇒ 8 – 2a2 – 3a2 = a2(4 – a2)

⇒ (15a – 3b)2 + (3b – 5c)2 + (15a – 5c)2 = 0

⇒ a4 – 9a2 + 8 = 0 ⇒ a2 = 1, 8

⇒ 15a – 3b = 0, 3b – 5c = 0 and 15a – 5c = 0

As a2 + b2 = 4, a2 ≠ 8

⇒

Thus, a2 = 1 and b2 = 3

a b c = = = k(say) 1 5 3

∴ equation of hyperbola is

⇒ a = k, b = 5k, c = 3k ∴ b, c, a are in A.P.

8. Different possible arrangements are given in the following table M F

H W 3 0 0 3

H W 2 1 1 2

H W 1 2 2 1

H W 0 3 3 0

∴ The number of ways

= (4C3) (4C3) + (4C2)(3C1)(3C1)(4C2) + (4C1)(3C2)(3C2)(4C1) + (3C3)(3C3) = 485

1 2 y = 1 3 An equation of tangent at

x2 –

2x −

(

)

2, 3 is

3 y= 1 3

(

)

It passes through 2 2,3 3 . 11. f(x + y) = f(x) + f(y) + xy Also, f(1) = 3 f(k + 1) = f(k) + f(1) + (k)(1) = f(k) + (k + 3) Let

30

Complete Mathematics—JEE Main

S = f(1) + f(2) + . . . . . + f(k – 1) + f(k)

PB2 = 16a2 + 4a2 = 20a2

S=

PC2 = 16a2 + a2 = 17a2

f(1) + . . . . + f(k – 2) + f(k – 1) + f(k)

Subtract to obtain 0 = f(1) + (f(2) – f(1)) + (f(3) – f(2))

∴ cosβ =

+ …. + (f(k) – f(k – 1)) – f(k) ⇒ f(k) = 3 + 4 + 5 + …… + (k + 2)

1 1 = k ( 3 + k + 2 ) = k ( k + 5) 2 2 10

Now,

∑ f (n) = n =1

=

1 10 2 ∑ n + 5n 2 n =1

(

)

20a

)(

17 a

1 − cos 2 β = cos β

tanβ =

)

2 85 − 81 = 9 9

Then arc(AB) = θr A

B

20 − 2r r

r

Also, area A of sector is given by A= =

= 2i − 2 j + k ⇒ | a × b |= 3

=

=3

θ

θ (π r 2 ) 2π

O

1 2 rθ 2

1 2 ⎛ 20 − 2r ⎞ r ⎜ ⎟ = r(10 – r) 2 ⎝ r ⎠

Thus, A is maximum when r = 5 and maximum A is 25m2

⎛ 1⎞ ⇒ 3 | c | ⎜ ⎟ = 3 ⇒| c |= 2 ⎝ 2⎠

3π 4

Also,

dx

∫ π 1 + cos x

15. Let I =

3 = |c−a|

(1)

4

2 2 ⇒ 9 =| c | + | a | −2a ⋅ c

3π 4

1 (9 + 4 – 9) = 2 2

=

∫ π 4

13. Let AC = a, then AB = 2a

dx 3 ⎛ π π ⎞ 1 + cos ⎜ + − x⎟ ⎝ 4 4 ⎠ 3π 4

Also, AP = 2(AB) = 4a B

dx

∫ π 1 − cos x

I=

(2)

4

a C

P

We have

r

= 25 – (5 – r)2

⇒ a × b c sin 30° = 3

β α

9 85

14. Let radius of circle be r, and ∠AOB = θ,

⇒ θ =

i j k 12. a × b = 2 1 −2 1 1 0

⇒ a⋅c =

(

36 = 4 85

=

r + r + θr = 20

⎤ 1 ⎡ (10 )(11)( 21) 5 + (10 )(11) ⎥ ⎢ 2⎣ 6 2 ⎦

(a × b ) × c

2

We are given

1 = (10 )(11)( 7 + 5 ) = 330 4

As

20a 2 + 17 a 2 − a 2

Adding (1) and (2), we get 3π 4

2I =

a 4a

A

⎛

1

1

⎞

+ ⎟ dx ∫⎜ 1 − cos x ⎠ π ⎝ 1 + cos x 4

3π 4

2 dx = 2 = ∫ 2 π 1 − cos x 4

3π 4

∫ cos ec xdx

π

4

2

JEE Main 2017 (Offline) 31 3π

⇒

I = ( − cot x ) ⎤⎦ π4 = 4

Adding (1), (2) we get

⎛ 3π ⎞ ⎛π⎞ − cot ⎜ ⎟ + cot ⎜ ⎟ ⎝ 4⎠ ⎝ 4⎠

20

∑(

2s =

⇒ I = 2.

21

k =1

⇒ S = 220-1

16. Rewrite the equation as

Next, let

dy cos x dx = − y +1 2 + sin x

10

T=

C1 +

Integrating, we get

= 210 – 1

ln|y + 1| = –ln|2 + sinx| + ln c

∴ (21C1 – 10

⇒ ln|(y + 1)(2 + sin x) = ln c

C10)

We are given y(0) = 1

20. limπ x→

∴ (1 + 1)(2 + sin 0) = A ⇒ A = 4

4 2 + sin x

∫ ( tan

17. I4 + I6 = =

∫ tan

4

2

x→

2

= lim

x + tan x ) dx 6

θ →0

x sec 2 xdx

(

1 −1 + 3i 2

= lim

)

θ →0

= ⇒ 2w + 1 =

Using R1→ R1 + R2 + R3 in the given determinant, We get 0 w2 w

⎛π ⎞ 8sin x ⎜ − x⎟ ⎝2 ⎠

=

3

⎛π ⎞⎡ ⎛π ⎞⎤ cos ⎜ + θ ⎟ ⎢1 − sin ⎜ + θ ⎟ ⎥ ⎝2 ⎠⎣ ⎝2 ⎠⎦ ⎛ ⎛π 3 ⎞⎞ 8 ⎜ sin ⎜ + θ ⎟ ⎟ ( −θ ) ⎠⎠ ⎝ ⎝2

21

C10

(1)

Using nCr = nCn-r, We get C20 +

21

C19 + …. +

+θ ]

1 (sin θ )(1 − cos θ )(1 + cos θ ) lim 8 θ →0 (cos θ )(1 + cos θ )θ 3 3

1 cos θ 1 ( )( + cos θ )

1 1 1 (1)3 (1)(1 + 1) = . 8 16

21. 5(tan2x – cos2x) = 2cos(2x) + 9

1 3

⇒ cos2x = 2cos2x – 1 = −

21

2

(sin θ )(1 − cos θ ) 8 ( cos θ )θ 3

⇒ cos2x =

19. Let C2 + …….+

π

⇒ (3cos2x – 1)(3cos2x + 5) = 0

= –1 – 2w = –z 21

[x =

⇒ 9cos4x + 12cos2x – 5 = 0

= w2 + w – 2w

S=

(cos x )(1 − sin x )

⇒ 5sec2x = 9cos2x + 12

∴ k = w – w

C1 +

C2) + …. + (21C10 –

⇒ 5tan2x = 9cos2x + 7

2

21

10

⇒ 5tan2x = 5cos2x + 2(2cos2x – 1) + 9

= 3(w2 – w4) = 3(w2 – w)

S=

C10

C1) + (21C2 –

⎛ sin θ ⎞ 1 = lim ⎜ ⎟ θ →0 ⎝ θ ⎠ 8

3i

Also, w7 = w and w2 + 1 = –w

3 0 Δ = 1 w 1 w2

10

(π − 2 x )3

= limπ

1 = tan 5 x + c 5 1 ∴ a = , b = 0 5

18. w =

10

C2 + …. +

cot x − cos x

⎛π ⎞ 4 1 ⇒ y ⎜ ⎟ = −1 + = 2 +1 3 ⎝2⎠ 4

10

= (220 – 1) – (210 – 1) = 220 – 210

⇒ (y + 1) (2 + sin x) = ± c = A (say)

Thus, y = –1 +

Ck ) = 221 – 1 – 1 = 221 – 2

21

C11

(2)

[∵ 3cos2x + 5 > 0] 1 3

7 ∴ cos(4x) = 2cos2(2x) – 1 = − . 9

22. Equation of line through P(1, –2, 3)

32

Complete Mathematics—JEE Main

And parallel to

y

x y z = = is 1 4 5

x −1 y + 2 z − 3 = = = r(say) 1 4 5

4

⇒ x = r + 1, y = 4r – 2, z = 5r + 3.

(0, a)

Let coordinates of point Q be (r + 1, 4r – 2, 5r + 3) As Q lies on 2x + 3y – 4z + 22 = 0, 2(r + 1) + 3(4r – 2) – 4(5r + 3) + 22 = 0

0

⇒ –6r + 6 = 0

(0, a) where a > 0.

⇒ r = 1

Then the radius r = 4 – a

Now,

Also, length of perpendicular from

PQ2 = (r + 1 – 1)2 + (4r – 2 + 2)2 + (5r + 3 – 3)2 42

N = ( i − 2 j + 3k ) × ( 2i − j − k )

= 5i + 7 j + 3k Therefore equation of plane is 5(x – 1) + 7(y + 1) + 3(z + 1) = 0

(1)

Distance of (1, 3, -7) from (1) is 25 + 49 + 9

=

10 83

⎛ 6x x ⎞ 24. y = tan–1 ⎜⎜ 3 ⎟ ⎟ ⎝ 1 − 9x ⎠ ⎛ ⎛ 3⎞ ⎞ ⎜ 2 ⎜ 3x 2 ⎟ ⎟ –1 ⎜ ⎠ ⎟ = tan ⎜ ⎝ ⎟ 3 2 ⎜ 1 − ⎛ 3x 2 ⎞ ⎟ ⎟ ⎟ ⎜ ⎜ ⎠ ⎠ ⎝ ⎝ ⎛ 3⎞ = 2tan–1 ⎜ 3x 2 ⎟ ⎝ ⎠

⎛ 1 ⎞⎛ 3 ⎞ 2.⎜ 3 x 2 ⎟ ⎜ ⎟ 9 dy ⎠⎝ 2 ⎠ = x = ⎝ 3 1 + 9 x3 dx 1 + 9x ∴g(x) =

a 2

Thus, 4 – a =

23. Note that normal to plane is

| 5 (1 − 1) + 7 ( 3 + 1) + 3 ( −7 + 1) |

(0, a) to y = x is r ∴ r =

= r2(1 + 16 + 25) = 42r2 = 42 ⇒ PQ =

x

9 1 + 9x 3

25. As the circle touches y = |x|, that is, y = x, y = –x, y > 0, the centre of circle lies on the positive of y–axis. Let centre of the circle be

a 1 ⎞ ⎛ ⇒ 4 = ⎜1 + ⎟a 2 2⎠ ⎝

⇒ a =

4 2 2 +1

⇒ r =

4 = 4 2 +1

(

)

2 −1 .

26. Let p = probability of drawing a green ball in a single draw. =

15 3 = 25 5

n = number of balls drawn = 10, and x = number of green balls drawn, Then x ~ B(n, p) Var(x) = np(1 – p) ⎛ 3 ⎞⎛ 3 ⎞ 12 = 10 ⎜ ⎟⎜1 − ⎟ = 5 ⎝ 5 ⎠⎝ 5 ⎠

27. We are given e =

1 , and 2

a − = –4 ⇒ 2a = 4 or a = 2 e ⎛ 1⎞ ∴ b2 = a2(1 – e2) = 4 ⎜1 − ⎟ = 3 ⎝ 4⎠ ∴ Equation of ellipse is x2 y 2 =1 + 4 3

⎛ 3⎞ And equation of tangent at ⎜1, ⎟ is ⎝ 2⎠

JEE Main 2017 (Offline) 33

Similarly,

1 3y x+ =1 4 23

P(B) + P(C) – 2P(B ∩ C) =

x + 2y = 4 ⎛ 3⎞ ∴ An equation of normal at ⎜1, ⎟ is ⎝ 2⎠ 3⎞ ⎛ 2(x – 1) – ⎜ y − ⎟ = 0 2⎠ ⎝

And P(C) + P(A) – 2P(C ∩ A) =

11

(2) 1 4

(3)

Adding (1), (2), (3), we get 2[P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) –

1 Or 2x – y – = 0 or 4x – 2y = 1 2

28. Two numbers can be chosen in

1 4

3 4

P(C ∩ A)] =

C2 = 55 ways.

Suppose selected number are a, b with a > b, then

⇒ P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) –

a + b = 4r, a – b = 4s

P(C ∩ A) =

⇒ 2a = 4(r + s), 2b = 4(r – s) ∴ both a, b are even, and possible pairs of (a , b)

are

⎡2

6 . 55

30. A2 = ⎢ ⎣ −4 ⎡ 16

1 4

⇒ P(A ∩ B′) + P(A′ ∩ B) =

1 16

From (4), (5), we get P(A ∪ B ∪ C) =

(4, 0), (8, 0), (6, 2), (10, 2), (8, 4), (10, 6),

29. P(Exactly one of A, B) =

(4)

Also, P(A ∩ B ∩ C) =

⇒ a = 2(r + s), b = 2(r – s)

Thus, probability of required event is

3 8

−3⎤ 1 ⎥⎦

3 1 7 + = . 8 16 16

⎡ 2 −3⎤ ⎢ −4 1 ⎥ ⎣ ⎦

−9 ⎤

= ⎢ ⎥ ⎣ −12 13 ⎦ 1 4

1 ⇒ P(A) + P(B) – 2P(A ∩ B) = 4

∴ 3A2 + 12A ⎡ 72

(1)

= ⎢ ⎣ −84

−63⎤ 51 ⎥⎦

⎡ 51 63⎤ ⇒ adj (3A2 + 12A) = ⎢ ⎥. ⎣84 72 ⎦

(5)