Objective mathematics: for engineering entrance examinations [3rd ed] 9788131723630, 8131723631


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Table of contents :
Cover......Page 1
OBJECTIVE MATHEMATICS......Page 4
Copyright......Page 5
Preface to the Third edition......Page 6
Preface......Page 7
Contents......Page 8
Intervals in R......Page 10
Types Of Functions......Page 11
Operations On Functions......Page 13
Even and odd extensions......Page 14
Multiple-Choice Questions......Page 15
Solutions......Page 29
Exercises for Self-Practice......Page 50
Answers......Page 51
Algebraic Limits......Page 52
Exponential and Logarithmic Limits......Page 53
Evaluation of limits using L’Hospital's rule ......Page 54
Multiple-choice questions......Page 55
Solutions......Page 63
Exercises for self-practice......Page 81
Answers......Page 83
Geometrical meaning of continuity......Page 84
Relation between continuity and Differentiability......Page 85
Multiple-choice questions......Page 86
Solutions......Page 95
Exercises for Self-Practice......Page 114
Answers......Page 115
Rules for Differentiation......Page 116
Differentiation of inverse trigonometric functions......Page 117
Differentiation of a function given in the form of a Determinant......Page 118
Multiple-choice questions......Page 119
Solutions......Page 127
Exercise for Self-Practice......Page 143
Answers......Page 145
Equation of normal......Page 146
Increasing and Decreasing Functions (Monotonicity)......Page 147
Maxima and Minima of Functions......Page 148
Multiple-ChoiceQuestions......Page 149
Solutions......Page 164
Exercises for self-Practice......Page 196
Answers......Page 197
Standard Formulae of Integration......Page 198
Method of partial fractions for rational functions......Page 199
Integrals of the form......Page 200
Multiple-Choice Questions......Page 204
Solutions......Page 217
Exercises for self-practice......Page 236
Answers......Page 239
Properties of Definite Integrals......Page 240
Some useful reduction formulae......Page 241
Curve Tracing......Page 242
Multiple-Choice Questions......Page 243
Solutions......Page 265
Exercises for Self-Practice......Page 310
Answers......Page 313
Solution of First Order and First Degree Differential equations......Page 314
Multiple-Choice Questions......Page 316
Solutions......Page 322
Exercises for Self-Practice......Page 332
Answers......Page 333
Area of a Triangle......Page 334
Rotation of Axes......Page 335
Slope-Intercept Form......Page 336
Normal Form......Page 337
Length of perpendicular from a point on a line ......Page 338
Incentre of a Triangle......Page 339
Orthocentre......Page 340
Multiple-Choice Questions......Page 341
Solutions......Page 352
Answers......Page 379
General equation of second degree......Page 380
Equations of the lines joining the origin to the points of intersection of a given line and a given curve......Page 381
Multiple-Choice Questions......Page 382
Solutions......Page 386
Exercises for Self-Practice......Page 395
Answers......Page 396
Different forms of the equation of a circle......Page 397
Position of a point with respect to a circle......Page 398
Notations......Page 399
Chord of contact of tangents......Page 400
Common Chord of Two circles......Page 401
Equation of acircle through the intersection of two circles......Page 402
Transverse common tangents......Page 403
Multiple-Choice Questions......Page 404
Solutions......Page 417
Exercises for Self-Practice......Page 449
Answers......Page 451
Section of a right circular cone by diferent planes......Page 452
Focal Chord......Page 453
Point of intersection of tangents......Page 454
Position of a point with respect to a parabola......Page 455
Prepositions on the Parabola......Page 456
Some Terms and Properties Related to an Ellipse......Page 457
Condition for tangency and points of contact......Page 458
Equation of normal in different forms......Page 459
Equation of Polar of a Point......Page 460
Some Terms and Properties Related to a Hyperbola......Page 461
Conjugate Hyperbola......Page 462
Equation of Tangent in Different Forms......Page 463
Asymptotes of hyperbola......Page 464
Multiple-Choice Questions......Page 465
Solutions......Page 481
Exercises for self-practice......Page 521
Answers......Page 523
Conjugate of a complex number......Page 524
Polar form of a complex number......Page 525
Concept of rotation......Page 526
Geometry of complex numbers......Page 527
Multiple-Choice Questions......Page 528
Solutions......Page 543
Exercises for Self-Practice......Page 572
Answers......Page 573
Arithmetic Progression (A.P.)......Page 574
nth Term of a G.P.......Page 575
nth Term of an H.P.......Page 576
Method of Differences......Page 577
Multiple-Choice Questions......Page 578
Solutions......Page 593
Exercises for Self-Practice......Page 627
Answers......Page 629
Common roots......Page 630
Sign of a quadratic expression......Page 631
Relation between roots and coefficients of a polynomial equation......Page 632
To find the values of a rational expressionin x, where x is real......Page 633
Multiple-Choice Questions......Page 634
Solutions......Page 647
Exercises for Self-Practice......Page 672
Answers......Page 673
Permutation......Page 674
Important Results on Combination......Page 675
Exponent of prime p in n!......Page 676
Multiple-Choice Questions......Page 677
Solutions......Page 687
Exercises for Self-Practice......Page 702
Answers......Page 703
Middle term in the binomial expansion......Page 704
General Term in the Expansion of (1 + x)n......Page 705
Multiple-Choice Questions......Page 706
Solutions......Page 716
Exercises for Self-Practice......Page 736
Answers......Page 737
Deductions from logarithmic series......Page 738
Multiple-Choice Questions......Page 739
Solutions......Page 742
Answers......Page 747
Types of Matrices......Page 748
Properties of Scalar Multiplication......Page 749
Transpose of a Matrix......Page 750
Involutory Matrix......Page 751
Properties of the Adjoint of a Matrix......Page 752
Echelon Form of a Matrix......Page 753
Homogeneous equations......Page 754
Multiple-Choice Questions......Page 755
Solutions......Page 760
Exercises for Self-Practice......Page 769
Answers......Page 771
Expansion of a determinant of order three......Page 772
Properties of determinants......Page 773
Evaluation of determinants using elementary operations......Page 774
Homogeneous and Non-Homogeneous System......Page 775
Multiple-Choice Questions......Page 776
Solutions......Page 784
Exercises for Self-Practice......Page 799
Answers......Page 801
Fundamental Identities......Page 802
Addition and subtraction formulae......Page 803
Trigonometric ratios of submultiple angles......Page 804
Multiple-Choice Questions......Page 805
Solutions......Page 816
Answers......Page 837
General solution......Page 838
Multiple-Choice Questions......Page 839
Solutions......Page 844
Exercises for Self-Practice......Page 854
Answers......Page 855
Some Important Formulae......Page 856
Multiple-Choice Questions......Page 857
Solutions......Page 863
Exercises for Self-Practice......Page 873
Answers......Page 874
Circumcircle of a Triangle......Page 875
The distances of the orthocentre from the vertices......Page 876
Solution of a right angled triangle......Page 877
Solution of an Oblique Triangle......Page 878
Multiple-Choice Questions......Page 879
Solutions......Page 885
Answers......Page 900
Some Useful Results......Page 901
Multiple-Choice Questions......Page 902
Solutions......Page 905
Exercises for Self-Practice......Page 911
Answers......Page 912
Types of Events......Page 913
Important symbols......Page 914
Probability distribution of a random variable......Page 915
Multiple-Choice Questions......Page 916
Solutions......Page 931
Exercises for Self-Practice......Page 953
Answers......Page 957
Combined mean......Page 958
Quartiles, deciles and percentiles......Page 959
Quartile Deviation......Page 960
Combined Standard Deviation......Page 961
Coefficient of Regression of y on x......Page 962
Multiple-Choice Questions......Page 963
Solutions......Page 968
Exercises for Self-Practice......Page 975
Answers......Page 977
Types of vectors......Page 978
Section Formula......Page 979
Coplanarity of four points......Page 980
Some useful identities......Page 981
Triple products......Page 982
Properties of reciprocal system of vectors......Page 983
Multiple-Choice Questions......Page 984
Solutions......Page 999
Exercises for Self-Practice......Page 1026
Answers......Page 1029
Direction Cosines......Page 1030
Vector Equation of a Line through a given point and parallel to a given Vector......Page 1031
The Plane......Page 1032
Angle between Two Planes......Page 1033
Two Sides of a Plane......Page 1034
Equation of a Sphere, the Extremities of Diameter Being given......Page 1035
Multiple-Choice Questions......Page 1036
Solutions......Page 1041
Exercises for Self-Practice......Page 1053
Answers......Page 1055
Forces in Statics......Page 1056
Forces acting at a point......Page 1057
Parallel forces......Page 1058
Moments and couples......Page 1059
Friction......Page 1060
Centre of gravity......Page 1061
Multiple-Choice Questions......Page 1062
Solutions......Page 1072
Exercises for Self-Practice......Page 1093
Answers......Page 1094
Velocity and Acceleration......Page 1095
Equations of Motion......Page 1096
Motion Under Gravity......Page 1097
Third Law of Motion......Page 1098
Projectiles......Page 1099
Multiple-Choice Questions......Page 1100
Solutions......Page 1107
Exercises for Self-Practice......Page 1120
Answers......Page 1121
Subset......Page 1122
Algebra of sets......Page 1123
Inverse relation......Page 1124
Multiple-Choice Questions......Page 1125
Solutions......Page 1129
Exercises for Self-Practice......Page 1135
Answers......Page 1136
Numerical Integration......Page 1137
Methods of solving a system of simultaneous linear equations......Page 1138
Multiple-Choice Questions......Page 1139
Solutions......Page 1141
Exercises for Self-Practice......Page 1146
Answers......Page 1147
The Graph of a Linear Inequality......Page 1148
Multiple-Choice Questions......Page 1149
Solutions......Page 1151
Exercises for Self-Practice......Page 1153
Answers......Page 1154
Some Useful Formulae for Hyperbolic Functions......Page 1155
Multiple-Choice Questions......Page 1156
Solutions......Page 1157
Answers......Page 1159
Model Test Paper-I......Page 1160
Model Test Paper-II......Page 1165
Model Test Paper-III......Page 1169
Model Test Paper-IV......Page 1171
Answers......Page 1173
Model Test Paper-V......Page 1174
Model Test Paper-VI......Page 1176
Answers......Page 1178
Model Test Paper-VII......Page 1179
Model Test Paper-VIII......Page 1185
Model Test Paper-IX......Page 1190
Model Test Paper-X......Page 1192
Answers......Page 1195
Solutions To Model Test Paper-I......Page 1196
Solutions To Model Test Paper-II......Page 1206
Solutions To Model Test Paper-III......Page 1212
Solutions To Model Test Paper-IV......Page 1215
Solutions To Model Test Paper-V......Page 1218
Solutions To Model Test Paper-VI......Page 1221
Solutions To Model Test Paper-VII......Page 1225
Solutions To Model Test Paper-VIII......Page 1229
Solutions To Model Test Paper-IX......Page 1237
Solutions To Model Test Paper-X......Page 1239
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The Pearson Guide to

OBJECTIVE MATHEMATICS For Engineering Entrance Examinations

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The Pearson Guide to

OBJECTIVE MATHEMATICS For Engineering Entrance Examinations Third Edition

IIT (Screening Test), AIEEE (CBSE), CEE (Delhi), UPSEAT (UP), CEET (Haryana), PET (MP), GGS Indraprastha University, Jamia Millia Islamia University, AMU, PET (Rajasthan), EAMCET (Andhra Pradesh), BCA/BSc (Hons) Computer Science and other Common Engineering Entrance Examinations in Orissa, Bihar, Tamil Nadu (TNPCEE), Karnataka (CET), Assam and West Bengal (WBJEE), MBA and MCA

J K Sharma

Anita Khattar

Dinesh Khattar

Professor Faculty of Management Studies University of Delhi Delhi

Department of Mathematics Sarvodaya Vidyalaya Delhi

Head Department of Mathematics Kirori Mal College University of Delhi Delhi

Chandigarh • Delhi • Chennai

The aim of this publication is to supply information taken from sources believed to be valid and reliable. This is not an attempt to render any type of professional advice or analysis, nor is it to be treated as such. While much care has been taken to ensure the veracity and currency of the information presented within, neither the publisher nor its authors bear any responsibility for any damage arising from inadvertent omissions, negligence or inaccuracies (typographical or factual) that may have found their way into this book. Copyright © 2009 Dorling Kindersley (India) Pvt. Ltd. This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out, or otherwise circulated without the publisher’s prior written consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying, recording or otherwise), without the prior written permission of both the copyright owner and the above-mentioned publisher of this book. ISBN: 978-81-317-2363-0 First Impression Published by 2009 Dorling Kindersley (India) Pvt. Ltd., licensees of Pearson Education in South Asia. Head Office: Knowledge Boulevard, a Floor 7th-8(A) Sector 62, Noida, India. Registered Office: 14 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India. Laser typeset by Tantla Compositions Services Pvt. Ltd., Chandigarh Printed in India by Anand Sons

PREFACE to the Third edition I take great pleasure in presenting to the readers, the third revised edition of the book. The third edition of objective Mathematics includes all the basic features of the earlier editions. However, some typical problems have been added in each chapter. These are designed to test the mental, academic and creative skills of the students. Questions from recent examination papers of various engineering entrance examinations have been included in every chapter. Chapters on Mathematical Reasoning and Mathematical Induction have been included in this new edition. In preparing this third edition, I am greatly indebted to many teachers as well as students throughout the country who made constructive criticism and extended valuable suggestions for the improvement of the book. Any suggestions to ensure further improvement of the book will be greatly acknowledged.

Dinesh Khattar

PREFACE The last twenty years of preparing/guiding students to success in different engineering entrance examinations has made us realize that what students really need is a book that imparts the necessary skills to solve questions in the shortest possible time without compromising on the theoretical aspect of the subject. This book introduces the concept briefly and then goes straight into solving real-life problems. The graded problems, techniques to solve them, and huge number of exercises and test papers with solutions will help the student gain the necessary skills and confidence to take the examinations successfully. The book is noteworthy in the following aspects: •

Each chapter contains concise definitions and explanations of basic/fundamental principles and illustrative examples to enable the students to recall the subject matter of the chapter before attempting to answer the questions.



Worked out solutions to a large number of problems have been provided. However, we urge students to attempt the same on their own and only if they fail should they refer to the solutions provided.



Problems have been categorised into various types, and working rules to help the students in solving them have also been provided.



Large number of problems that have been asked in the competitive examinations in the recent past have been included in the text.



To enable the students make a self assessment, practice exercises covering all the topics in each chapter have been provided at the end.



In order to help the students decide how much emphasis they should give to various topics in different competitive examinations, a smart table has been given in the beginning. This table provides the information about how many questions have been asked from various topics during 2001–04 in IIT(Screening test), AIEEE (CBSE), CEE (Delhi) and UPSEAT.



Nine Model Test Papers based on different competitive examinations have been provided at the end to facilitate students understand the pattern and the type of questions asked in different examinations.

The answers to almost all unsolved problems have been thoroughly checked. While every effort has been made to weed out typos, it is possible that a few might have crept in. We will be grateful to the readers for bringing these errors to our notice. It is earnestly hoped that this book will build a strong foundation for success in any competitive examination. We wish to place on record our sincere thanks to our friends and colleagues for their help and suggestions in planning and preparing the manuscript of this book. We would like to thank Mr. Dinesh Kaushik and Mr. Pawan Tyagi for their cooperation in typesetting the book. Suggestions and comments from our esteemed reader to improve the book in content and style are always welcome and will be greatly appreciated and acknowledged. Thank you for choosing our book. May you find it stimulating and rewarding!

Authors

CONTENTS





Preface to Third Edition

v

Preface

vi

  1. Functions   2. Limits   3. Continuity and Differentiability

1 43 75

  4. Differentiation

107

  5. Applications of Derivatives

137



  6. Indefinite Integration

189



  7. Definite Integral and Area

231



  8. Differential Equations

305



  9. Coordinates and Straight Lines

325



10. Pair of Straight Lines

371



11. Circles

388



12. Conic Sections

443



13. Complex Numbers

515



14. Sequences and Series

565



15. Quadratic Equations and Inequations

621



16. Permutations and Combinations

665



17. Binomial Theorem

695



18. Exponential and Logarithmic Series

729



19. Matrices

739



20. Determinants

763



21. Trigonometric Ratios and Identities

793



22. Trigonometric Equations

829



23. Inverse Trigonometric Functions

847



24. Properties and Solutions Triangles

866



25. Heights and Distances

892



26. Probability

904



27. Statistics

949



28. Vectors

969



29. Three Dimensional Geometry

1021



30. Statics

1047



31. Dynamics

1086



32. Set Theory

1113



33. Numerical Methods

1128



34. Linear Programming

1139



35. Hyperbolic Functions

1146





Model Test Papers

1151





Solutions to Model Test Papers

1187

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1

Functions

CHAPTER

Summary of concepts Function or Mapping Let X and Y be any two non-empty sets and there be correspondence or association between the elements of X and Y such that for every element x ∈ X, there exists a unique element y ∈ Y, written as y = f (x). Then we say that f is a mapping or function from X to Y, and is written as



= set of all image points in Y under the map f.



= f (X) = { f (x) : f (x) ∈ Y; x ∈ X}

The set Y is also called the co-domain of f. Clearly f (X) ⊆ Y.

f : X → Y such that y = f (x), x ∈ X, y ∈ Y.

Intervals in R Real Function If f : X → Y be a function from a non-empty set X to another non-empty set Y, where X, Y ⊆ R (set of all real numbers), then we say that f is a real valued function or in short a real function.

Features of a Mapping f : X → Y (i) For each element x ∈ X, there exist is unique element y ∈ Y. (ii) The element y ∈ Y is called the image of x under the mapping f. (iii) If there is an element in X which has more than one image in Y, then f : X ∈ Y is not a function. But distinct elements of X may be associated to the same element of Y. (iv) If there is an element in X which does not have an image in Y, then f : X → Y is not a function. Throughout this chapter a ‘function’ will mean a ‘real function’.

Value of a Function The value of a function y = f (x) at x = a is denoted by f (a). It is obtained by putting x = a in f (x).

Domain and Range of a Function If f : X → Y be a function, then the set X is said to be the domain of f and range of f

The set of all real numbers lying between two given real numbers is called an interval in R. Let a and b be any two real numbers such that a < b, then we define the following types of intervals. Closed Interval  The set of all real numbers x such that a ≤ x ≤ b is called a closed interval and is denoted by [a, b]. That is [a, b] = {x : x ∈ R; a ≤ x ≤ b} Open Interval  The set of all real numbers x such that a < x < b is called an open interval and is denoted by (a, b) or ]a, b[. That is (a, b) = {x : x ∈ R; a < x < b} Semi-Closed or Semi-Open Interval  The intervals [a, b) = {x : x ∈ R; a ≤ x < b} and (a, b] = {x : x ∈  R; a < x ≤ b} are called semi-closed or semi-open intervals. They are also denoted by [a, b[ and ]a, b] respectively. Note that the set R can be thought of as the open interval (– ∞, ∞), so that R = (– ∞, ∞) = {x : x ∈ R; – ∞ < x < ∞}. Also, the infinite intervals in R can be given by (– ∞, a), (a, ∞), (– ∞, a], [a, ∞).

2

types of functions

Objective Mathematics

one-one or Injective function A function f : X → Y is said to be one-one or injective if distinct elements of X have distinct images in Y.

A function f : X → Y is an into function if it is not an onto function.

many-one function A function f : X → Y is said to be many-one if there exists atleast two distinct elements in X whose images are same

1. If X and Y are any two finite sets having m and n elements respectively, where 1 ≤ n ≤ m, then the number of onto functions from X to Y is given by n

∑ (−1)

n−r n

r =1

Rule for checking whether the function f : X → Y is one-one or many-one 1. (a) Consider any two points x, y ∈ X.

Cr r m

2. Any polynomial function f is onto if degree is odd and into if degree of f is even. Rule for checking whether the function f : X → Y is onto or into

(b) Put f (x) = f (y) and solve the equation.

(i) Find the range of the function f.

(c) If we get x = y only, then f is one-one, otherwise it is many-one.

(ii) If range of f = Y, then f is onto, otherwise it is into.

2. If a function is either strictly increasing or strictly decreasing in the whole domain (or equivalently, f ′ (x) > 0 or f ′ (x) < 0, ∨ x ∈ X), then it is one-one, otherwise it is many-one. 3. If any straight line parallel to x-axis intersects the graph of the function atmost at one point, then the function is one-one, otherwise it is many-one (i.e. it intersects the graph of the function in atleast two points).

Bijective function A function f : X → Y is said to be bijective, if f is both one-one and onto.

4. Any continuous function f (x) which has atleast one local maxima or local minima is many-one. 5. All even functions are many-one. 6. All polynomials of even degree defined on R have atleast one local maxima or minima and hence are many one on the domain R. Polynomials of odd degree can be one-one or many-one. Note: If X and Y are any two finite sets having m and n elements respectively, then the number of one-one functions from X to Y would be

 n Pm , if n ≥ m    0, if n < m

= 

onto or Surjective function A function f : X → Y is said to be onto or surjective if every element of Y is the image of some element of X under the map f.

If X and Y are any two finite sets having the same number of elements, say n, then the number of bijective functions from X to Y is n!.

Some Important functions constant function A function f : R → R defined as f (x) = c, ∨ x ∈ R, where c is a constant, is called a constant function. Its domain is R and range is singleton set {c}. The graph of a constant function is a straight line parallel to x-axis when x is the independent variable.

[ x] + [ y ] , if {x} + { y} < 1 (b) [x + y] =  [ x] + [ y ] + 1, if {x} + { y} ≥ 1

Identity Function  The function f : R → R defined as f (x) = x, ∨ x ∈ R, is called the identity function. Its domain is R and range is also R. The graph of the identity function is a straight line passing through origin and inclined at an angle of 45º with x-axis. Modulus Function or Absolute Value Function  The function f : R → R, defined as

where {x} denotes the fractional part of x. (c) n ≤ x < n + 1 ⇔ [x] = n (d) n1 ≤ [x] ≤ n2 ⇒ n1 ≤ x < n2 + 1 (e) x – 1 < [x] ≤ x ( f ) [[x]] = [x] (g) [n + x] = n + [x], where n is any integer

 0 (h) [x] + [– x] =  −1

if x ∈ I . if x ∉ I

Fractional-Part Function  The function f : R →  R defined as f (x) = x – [x] or f (x) = {x}, where {x} denotes the fractional part of x, is called the fractional-part function. Its domain is R and range is [0, 1). The graph of the fractional part function is as shown below:

 x, if x > 0  f (x) = | x | =  0, if x = 0 − x, if x < 0  is called the absolute value function or modulus function. Its domain is R and its range is [0, ∞). The graph of the modulus function is as shown in the adjoining figure. Greatest Integer (Step or Integral) Function  The function f : R → R defined as f (x) = [x] is called the greatest integer function, where  [x] = integral part of x or greatest integer not greater than x or greatest integer less than or equal to x. i.e.   f (x) = n, where n ≤ x < n + 1, n ∈ I (the set of integers).

Note: (a) If x is an integer, then x = [x] ⇒ {x} = 0 ⇒ {[x]} = 0 (b) [{x}] = 0 (c) 0 ≤ {x} < 1

0, if x ∈ integer (d) {x} + {– x} =  1, if x ∉ integer Signum  Function  The function f : R → R defined as,

| x | for x ≠ 0  f (x) =  x  0 for x = 0

Its domain is R and range is I. The graph of the greatest is called the signum function. integer function is as shown below: Its domain is R and range is the set {– 1, 0, 1}. The graph of the signum function is as shown below:

Reciprocal Function  The function f : R\{0} → R defined 1 by f (x) = , is called the reciprocal function. Its domain as x well as range is R\{0}. The graph of the reciprocal function is as shown below:

3

(a)  [x] ≤ x < [x] + 1

Functions



4

Polynomial Function  A function f : R → R, defined by f (x) = a0 + a1 x + a2 x2 + ... + anxn, where n ∈ N and a0, a1, a2, ..., an ∈ R, is called a polynomial function. If an ≠ 0, then n is called the degree of the polynomial. The domain of a polynomial function is R.

Objective Mathematics

p( x) , q( x) where p (x) and q (x) are polynomials over the set of real numbers and q (x) ≠ 0, is called a rational function. Its domain is R\ {x | q (x) = 0}.

Rational  Function  A function of the form f (x) =

Trigonometric Functions Exponential Function  Let a (≠ 1) be a positive real number. Then the function f : R → R, defined by f (x) = ax, is called the exponential function. Its domain is R and range is (0, ∞). The graph of the exponential function is as shown below:



Function • y = sin x

Domain R

Range [– 1, 1]

• y = cos x

R

[– 1, 1]

• y = tan x

  π R\ (2n + 1) n ∈ I 2  

R

• y = cot x

R\{nπ | n ∈ I}

R

• y = sec x • y = cosec x

  π R\ (2n + 1) n ∈ I (– ∞, – 1] ∪ [1, ∞) 2   R\{nπ | n ∈ I}

(– ∞, – 1] ∪ [1, ∞)

Inverse Trigonometric Functions x log a (i)  a = e e , (a > 0). (ii)  a loga x = x, (a > 0, a ≠ 1). x

(iii)  log ab =

log bc , c > 0 and c ≠ 1. log ca

1 (iv)  log b = , provided a ≠ 1 and b ≠ 1. log ba a

Logarithmic Function  Let a (≠ 1) be a positive real number. Then the function f : (0, ∞) → R, defined by f (x) = loga x, is called the logarithmic function. Its domain is (0, ∞) and range is R. The graph of the logarithmic function is as shown below:

Function

Domain

Range  π π  − 2 , 2  [0, π]

• y = sin–1x

– 1 ≤ x ≤ 1

• y = cos–1x

– 1 ≤ x ≤ 1

• y = tan–1x

– ∞ < x < ∞

 π π  − ,  2 2

• y = cot–1x

– ∞ < x < ∞

(0, π)

• y = cosec–1x

(– ∞, – 1] ∪ [1, ∞)

 π   π  − 2 , 0  ∪  0,   2 

• y = sec–1x

(– ∞, – 1] ∪ [1, ∞)

 π π  0, 2  ∪  , π  2  

Operations on Functions Let f and g be two real functions with domain D1 and D2 respectively. Then, (i) The sum function (f + g) is defined by ( f + g) (x) = f (x) + g (x), ∨ x ∈ D1 ∩ D2 The domain of f + g is D1 ∩ D2 (ii) The difference function ( f – g) is defined by (i)  loga a = 1, loga 1 = 0

− ∞, if a > 1 (ii)  loga 0 =  + ∞, if 0 < a < 1 (iii)  loge x is also denoted as: ln x.

( f – g) (x) = f (x) – g (x), ∨ x ∈ D1 ∩ D2 The domain of f – g is D1 ∩ D2 (iii) The product function fg is defined by ( fg) (x) = f (x) ⋅ g (x), ∨ x ∈ D1 ∩ D2 The domain of fg is D1 ∩ D2

f The domain of is D1 ∩ D2 \ {x : g (x) = 0} g (v) The scalar multiple function cf is defined by (cf ) (x) = c ⋅ f (x), ∨ x ∈ D1 The domain of cf is D1

composition of functions Let f and g be two real functions with domain D1 and D2 respectively. If range of f ⊆ domain of g, then composite function is defined (gof ) by (gof ) (x) = g ( f (x)), ∨ x ∈ D1 Also, if range of g ⊆ domain of f, then composite function is defined ( fog) by ( fog) (x) = f (g (x)), ∨ x ∈ D2.

odd function A function f (x) is said to be odd if f (– x) = – f (x) for every real number x in the domain of f. even function A function f (x) is said to be even if f (– x) = f (x) for every real number x in the domain of f. Some Important Results 1. The graph of an odd function is symmetric about origin and it is placed either in the first and third quadrant or in the second and fourth quadrant. 2. The graph of an even function is symmetric about the y-axis. 3. To express a given function f (x) as the sum of an even and odd function, we write f (x) = where

1 1 [ f (x) + f (– x)] + [ f (x) – f (– x)]. 2 2 1 [ f (x) + f (– x)] is an even function and 2

1 [ f (x) – f (– x)] is an odd function. 2 4. f (x) = 0 is the only function which is both even and odd.

Some useful results

5. If f (x) is an odd function, then f ‘ (x) is an even function provided f (x) is differentiable on R.

Let f : X → Y and g : Y → Z.

6. If f (x) is an even function, then f ‘ (x) is an odd function provided f (x) is differentiable on R.

(a) If both f and g are one-one, then so is gof. (b) If both f and g are onto, then is also onto gof. (c) If gof is one-one, then f is one-one but g may not be oneone. (d) If gof is onto, then g is onto but f may not be onto. (e) If f and g are bijective, then is also bijective gof. (f) It may happen that gof may exist and fog may not exist. Moreover, even if both gof and fog exist, they may not be equal.

7. If f and g are even functions, then fog is also an even function, provided fog is defined. 8. If f and g are odd functions, then fog is also an odd function, provided fog is defined. 9. If f is an even function and g is an odd function, then fog is an even function. 10. If f is an odd function and g is an even function, then fog is an even function. 11. For a real domain, even functions are not one-one.

Inverse functions If the function f : X → Y is both one-one and onto, then we define inverse function f –1 : Y → X by the rule y = f (x) ⇔ f –1 (y) = x, ∨ x ∈ X, ∨ y ∈ Y as shown in the figure below:

even and odd extensions A function f (x) defined on the interval [0, a] can be extended to [– a, a], so that f (x) becomes an even or odd function on the interval [– a, a]. If this extension is an even function, it is called even extension and if this extension is an odd function, it is called odd extension. Let g be the extension. Then for even extension, we define

 f ( x) , if x ∈[0, a ] g (x) =   f (− x), if x ∈ [− a, 0] and for odd extension, we define Rule to find the inverse of a function Let f : X → Y be a bijective function. • Put f (x) = y. • Solve the equation y = f (x) to obtain x in terms of y. Interchange x and y to obtain the inverse of f.

 f ( x) , if x ∈[0, a ] g (x) =  − f (− x), if x ∈ [− a, 0].

periodic function A function f (x) is said to be a periodic function of x, provided there exists a real number T > 0 such that f (x + T) = f (x), ∨ x ∈ R.

5

odd and even functions

Functions

f (iv) The quotient function   is defined by g f f ( x)  g  (x) = g ( x) , ∨ x ∈ D1 ∩ D2 \ {x : g (x) = 0}

6

The smallest positive real number T, satisfying the above condition is known as the period or the fundamental period of f (x).

Trigonometric Functions (i) sin x and cos x are defined for all real values of x.

Objective Mathematics

(ii) tan x and sec x are defined for all real values of x except

Rule for testing the periodicity of a function 1. Put f (T + x) = f (x) (x) and solve this equation to find the positive (x values of T independent of x. 2. If no positive value of T independent of x is obtained, then f (x ((x) x) is a non-periodic function. 3. If positive values of T independent of x are obtained, then f (x) (x) is (x a periodic function and the least positive value of T is the period of the function f (x). (x). (x

x = (2n + 1)

π , where n ∈ I. 2

(iii) cot x and cosec x are defined for all real values of x except x = nπ, where n ∈ I. Inverse Trigonometric Functions (i) sin–1x and cos–1x are defined for – 1 ≤ x ≤ 1. (ii) tan–1x and cot–1x are defined for all real values of x. (iii) sec–1x and cosec–1x are defined for x ≤ – 1 or x ≥ 1. Logarithmic Functions logb a is defined when a > 0, b > 0 and b ≠ 1.

Hints for Solving Problems on Periodic Functions 1. Constant function is periodic with no fundamental period. 2. If f (x ((x)) is periodic with period T, then also periodic with same period T.

1 and f ( x)

ax is defined for all real values of x, where a > 0.

f ( x) are

3. If f (x) (x) is periodic with period T1 and g (x (x ((x)) is periodic with period T2, then f (x ((x) x) + g (x ((x)) is periodic with period equal to l.c.m of T1 and T2, provided there is no positive k such that f ((kk + x) = g (x) ( ) (x and g (kk + x) = f (x ((x). x). 4. If f (x ((x) x) is periodic with period T, then kf ((ax ax + b), is also periodic with period

Exponential Functions

T , where a, b, k ∈ R and a, k ≠ 0. | a|

5. sin x, cos x, sec x and cosec x are periodic functions with period 2π.

Rules for solving problems on domain of a function 1. (x ( – a)) ((x – b) > 0 ⇒ x < a or x > b, for a < b 2. (x ( – a)) ((x – b) < 0 ⇒ a < x < b, for a < b 3. | x | < a ⇒ – a < x < a 4. | x | > a ⇒ x < – a or x > a

a > b k , if b > 1 k a < b , if b < 1

5. logb a > k ⇒ 

6. tan x and cot x are periodic functions with period π.

6.

7. | sin x |, | cos x |, | tan x |, | cot x |, | sec x | and | cosec x | are periodic functions with period π.

7.

8. sinn x, cosn x, secn x and cosecn x are periodic functions with period 2π when n is odd or π when n is even. 9. tann x and cotn x are periodic functions with period π. 10. If f (x) (x) is a periodic function with period T and g (x) (x ( ) is any func(x tion such that domain of f ⊂ domain g, then gof is also periodic with period T.

Rules for Finding the Domain of a Function Algebraic Functions (i) Denominator should be non-zero. (ii) Expression under the square root should be non-negative.

x2 = | x | n

x n = | x |, if n is even and

n

x n = x, if n is odd.

Rules for finding the range of a function y = f (x) 1. Find the domain of the function y = f (x ((x). x). 2. If the domain is an infinite interval, solve the equation y = f (x ((x) x) and find x in terms of y to get x = g (y ((y). ). Find the real values of y for which x is real. The set of values of y so obtained constitutes the range of ff. Note that if finite number of values of x are excluded from the domain, find the values of y for these values of x and exclude these values of y from the range of f found earlier. 3. If the domain is a finite interval, find the least and greatest value of y for values of x in the domain. If a is the least value and b the greatest value of y, then range ( f ) = [[a, a, b].

muLtIpLe-cHoIce QueStIonS choose the correct alternative in each of the following problems: 1. Domain of f (x) = 1 (a)  , 4 1 (c)  , 4

1 2  1 3 

sin −1 ( 2 x ) + π/6 is  −1 1  (b)  ,   4 2  1 1 (d)  − ,   4 3

2. The domain of definition of the function y=

 5x − x2  is log10   4 

(a) [1, 4] (c) [1, 4)

(b) (1, 4) (d) (1, 4]

(a) [– 2, – 1) ∪ [1, 2] (c) [– 2, – 1] ∪ [1, 2]

x − 1− x

is

(a) [1, ∞) (c) (1, 5)

9 − x2 is sin (3 − x)

1− x

(b) [2, 3) (d) None of these

15. If f (x) =

1− x

(d) None of these

7. The domain of the function f (x) = (a) (– ∞, – 2) ∪ [4, ∞) (c) (– ∞, – 2) ∪ (4, ∞)

1 [ x]2 − [ x] − 6

(b) (– ∞, – 2] ∪ [4, ∞) (d) None of these

(a) [– 1, 3] (c) [– 1, 3)

 3 − 2x  3 − x + cos   5  −1

(b) (– 1, 3] (d) None of these

x2 + 1 , ([.] denotes the greatest integer func[ x] tion), 1 ≤ x < 4, then

10. If  f (x) =

 17  (a) range of f is  2,   3 (b) f is monotonically increasing in [1, 4] 17 (c) the maximum value of f (x) is 3 17 (d) the maximum value of f (x) is 4 11. The domain of the function 1 is f (x) = 12 9 x − x + x4 − x + 1

7

π  (a) 2nπ +  2 n ∈1  

∪  2nπ +

7π 11π  , 2 nπ +  6 6 

7π   (c) 2nπ +  6 n ∈1  7π 11 π   , 2 nπ + (d)  2nπ +  ∪  6 6  n,m ∈I

π   2mπ +  2

16. The domain of the function x−5 f (x) = log10 2 − 3 x + 5 is x − 10 x + 24

(b) [– 2, 1[ (d) None of these

9. The domain of the function f (x) = is

3| x | − x − 2 and g (x) = sin x, then domain of

n ∈1

is

(b) R – {1, – 1} (d) None of these

definition of f o g (x) is

(b)

8. The domain of definition of the function 1 y = log (1 − x) + x + 2 is 10 (a) [– 2, 1] (c) [– 2, 0[ ∪ ]0, 1[

(b) (– ∞, 5) (d) [1, 5]

(a) {1} (c) x > 3, x ∉ 1

−1

6. Let f : (–∞, 1] → (–∞, 1] such that f (x) = x (2 – x). Then f –1(x) is

(c)

x − 1 + 5 − x is

 1 + x3  2 f (x) = sin − 1  3 / 2  + sin(sin x) + log (3 {x} + 1)(x + 1), 2 x   where { } represents fractional part function, is

5. The domain of the function f (x) =

(b) 1 –

(b) (– 1, 2) ∪ [3, ∞) (d) None of these

14. The domain of the function,

 1  , 1 (d)   2 

1− x

(a) [– 1, 2) ∪ [3, ∞) (c) [– 1, 2] ∪ [3, ∞)

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

 1  1  , + ∞ (c)  −∞, −  ∪   2  2

(a) 1 +

( x + 1)( x − 3) is ( x − 2)

given by 2

1   1   ,1 ∪ (a)  −1, − 2   2   (b) [– 1, 1]

(a) (2, 3) (c) (2, 3]

(b) (1, ∞) (d) (– ∞, ∞)

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

(b) (– 2, – 1] ∪ [1, 2] (d) (– 2, – 1) ∪ (1, 2)

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

(a) (– ∞, – 1) (c) (– 1, 1)

(a) (4, 5) (c) (4, 5) ∪ (6, ∞)

(b) (6, ∞) (d) (4, 5] ∪ (6, ∞)

17. The domain of the function f (x) = (a) (0, ∞) (c) (– ∞, ∞)

1 | x | − x is

(b) (– ∞, 0) (d) None of these

18. The domain of the function f (x) =  3 (a)  0,  2

(b) (0, 3)

 3 (c)  −∞,  2

 3 (d)  0,  2

log10

3− x is x

19. The domain of the function f (x) = log10 | 4 – x2 | is (a) (– ∞, ∞)\{– 2, 2} (b) (0, ∞) (c) (– ∞, 0) (d) None of these

Functions

  1  3. The domain of the function f (x) = sin –1 log 2  x 2    2   is

8

20. The domain of the function f (x) = is

Objective Mathematics

(a) (– ∞, 1) (c) [0, 1]

1 − 1 − 1 − x2

(a) (4, 6) (c) [4, 6)

(b) (– 1, ∞) (d) [– 1, 1]

1 −| x| is 2 −| x|

21. The domain of definition of f (x) = (a) (– ∞, ∞)\[– 1, 1] (b) (– ∞, ∞)\[– 2, 2] (c) [– 1, 1] ∪ (– ∞, – 2) ∪ (2, ∞) (d) None of these

(b) (2, ∞) (d) φ

2 −| x|  f (x) = cos  + [log (3 − x)]−1 is  4  (b) [– 6, 2) ∪ (2, 3] (d) [– 6, 3)

24. The domain of the function f (x) = log 2 log3 log4 x is (b) (4, ∞) (d) None of these

25. The domain of the function f (x) = sin

–1

 x − 3   – log10 (4 – x) is 2 

(a) (1, 4) (c) [1, 4)

(b) [1, 4] (d) (1, 4]

26. The domain of the derivative of the function f (x) = tan–1x, |x| ≤ 1 1 (|x| – 1), |x| > 1 is 2 (a) R – {0} (c) R –­{–1}

(b) R – {1} (d) R – {–1, 1}

27. The domain of the function f (x) = (a) (– ∞, – 3] ∪ (2, 5) (c) (– ∞, – 3] ∪ [2, 5]

x+3 is (2 − x)( x − 5)

(b) (– ∞, – 3) ∪ (2, 5) (d) None of these

3   28. The domain of the function f (x) = cos –1   4 + 2 sin x  is π  π  (a)  − + 2nπ, + 2nπ  6  6  π  π  (b)  − + 2nπ, + 2nπ  6 6

 π π  (c)  − + 2nπ, + 2nπ  6 6   π π  (d)  − + 2nπ, + 2nπ  6  6

f (x) = log10 [1 – log10 (x2 – 5x + 16)] is (a) (2, 3) (c) (2, 3]

1 log [1 − | x |]

(c) x ! {x}

−1

(a) [4, ∞) (c) (– ∞, 4)

30. The domain of the function

(a)

23. The domain of the function

(a) [– 6, 3)\{2} (c) [– 6, 3]

(b) [4, 6] (d) None of these

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

31. Which of the following is a function ([.] denotes the greatest integer function, {.} denotes the fractional part function]?

22. The domain of the function −1 1 1 + 2sin x + f (x) = is 1− x x−2 (a) (– ∞, ∞)\{1} (c) [– 1, 1)

29. The domain of the function f (x) = log10 ( x − 4 + 6 − x ) is

x! {x}

(b)

log ( x − 1)

(d)

1 − x2

32. The domain of the function f (x) = log10 sin (x – 3) + (a) (3, 4) (c) (3, π + 3)

16 − x 2 is

(b) (– 4, 4) (d) None of these

33. The domain of the function f (x) = log x cos x is  π π (a)  − ,  \{1} 2 2

 π π (b)  − ,  \{1}  2 2

 π π (c)  − ,  2 2

(d) None of these

4   34. The domain of the function f (x) = sin –1    3 + 2 cos x  is

π π ≤ x ≤ 2nπ + , n ∈ I 6 6 π (b) 0 ≤ x ≤ 2nπ + , n ∈ I 6 π π < x < 2nπ + , n ∈ I (c) 2nπ – 6 6 π ≤ x ≤ 0, n ∈ I (d) 2nπ – 6

(a) 2nπ –

 1   (where 35. Range of the function f defined by f (x) =   sin{x}  [.] and {.} respectively denote the greatest integer and the fractional part functions) is (a) I, the set of integers (b) N, the set of natural numbers (c) W, the set of whole numbers (d) {2, 3, 4, ...} 36. The domain of the function f (x) = tan–1 x ( x + 1) + sin −1 x 2 + x + 1 is (a) [– 1, 0] (b) {– 1, 0} (c) (– ∞, – 1] ∪ [0, ∞) (d) (– ∞, ∞)

3

46. The domain of the function f (x) = cos –1 (x + [x] ), where [⋅] denotes the greatest integer function, is

38. The domain of the function

(a) [– 1, 1] (c) (– 1, 0)

1 f (x) = log 1  x −  + log 2 4 x 2 − 4 x + 5 is 2  2

n times

(d) None of these

39. The domain of the function f (x) = x − [ x] , where [x] denotes the greatest integer less than or equal to x, is 2

(a) (0, ∞) (c) (– ∞, ∞)

2

(b) (– ∞, 0) (d) None of these

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

1 is | sin x | + sin x

(b) (2nπ, (2n + 1) π) π π (c)  (4n − 1) , (4n + 1)   2 2 (d) None of these 2 f ( n) + 1 , n = 1, 2, .... and f (1) = 2, then 2

(a) 52 (c) 48

(b) 49 (d) 51

f (x) =

C3x – 1 +

40 – 6x

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

C8x – 10 is,

(a) (– ∞, – 3) ∪ (3, ∞) (c) (– ∞, – 3] ∪ [3, ∞)

1 −| x |  is cos −1   2 

(b) [– 3, 3] (d) φ

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

e

sin −1 (log16 x 2 )

1 (a)  , 4  4 

1 1   (b)  −4, −  ∪  , 4 4 4  

1  (c)  −4, −  4 

(d) None of these

45. The domain of the function f (x) = log 

1 x+ 2  

x 2 − 5 x + 6 is

 (a)  3 , 2  ∪ (2, 3) ∪ (3, ∞) 2  3  (b)  , ∞  2 

48. The domain of the function

x − 3 − 2 x − 4 − x − 3 + 2 x − 4 is

f (x) =

(b) (– ∞, 4] (d) (– ∞, 4)

49. The number of solutions of the equation a f (x) + g (x) = 1 0, a > 0, g (x) ≠ 0 and has minimum value is 2 (a) One (b) Two (c) Zero (d) Intinitely many

  16 − x 2 50. The domain of the function f (x) = cos log    3− x is (a) (– 4, 4)

(b) (– 4, 3)

(c) (– ∞, – 4) ∪ (3, ∞)

(d) None of these

f (x) =

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

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

(b) [2n, ∞) (d) None of these

   

51. The domain of the function

42. The domain of the function 24 – x

(a) (2n – 1, ∞) (c) (2n – 2, ∞)

(a) [4, ∞) (c) (4, ∞)

(a) (– 2nπ, 2nπ)

41. If f (n + 1) = f (101) equals

(b) [0, 1) (d) None of these

47. The domain of the function log 2 log 2 log 2 ...log 2 x  is f (x) = 

1  (b)  , ∞  2 

1  (a)  , ∞  2  (c) (– ∞, ∞)

9

1  (c)  , ∞  2  (d) None of these

 2 x − 1  3 tan x is +e 1 − 3 x + 3 cos −1   3  (a) [– 1, 2] (b) (– 1, 2) (c) (– ∞, ∞) (d) None of these

f (x) =

3 + 2 ( x + 25) − 0.5 + (x – 5)0.5 + 5 (x – 3)0.5 is 5 − ( x + 25)0.5

(a) [5, ∞) (c) (– 25, 5)

(b) [3, ∞) (d) None of these

52. Let f : (4, 6) → (6, 8) be a function defined by f (x) =  x x +   (where [ . ] denotes the greatest integer func2 tion), then f– 1 (x) is equal to

is

 x (a) x −   2

(b) – x – 2

(c) x – 2

(d)

1  x x+  2

53. The domain of the function f (x) = (a) (– ∞, ∞) (b) (– ∞, ∞)\{nπ | n ∈ I}

  π (c) (– ∞, ∞)\ (2n + 1) n ∈ I 2   (d) None of these

4

 1  log 3   is  cos x 

Functions

37. The domain of the function

10

54. The domain of the function

63. The range of the function f (x) = loge (3x2 – 4x + 5) is

Objective Mathematics

 1    f (x) = log3   − log 1 1 + 1 5  − 1 is    2 x 

 11 (a)  −∞, log e   3

 11  (b) log e , ∞   3 

11 11 (c)  − log e , log e  (d) None of these  3 3  π2 55. If f : R → R, g : R → R be two given functions then − x 2 is 64. The range of the function y = 3 sin f (x) = 2 min { f (x) – g (x), 0} equals 16 (a) (– ∞, 1) (c) (1, ∞)

(b) (0, 1) (d) None of these

(a) f (x) + g (x) – | g (x) – f (x) | (b) f (x) + g (x) + | g (x) – f (x) | (c) f (x) – g (x) + |g (x) – f (x) | (d) f (x) – g (x) – | g (x) – f (x) | 56. Let f (x) be a function defined on [0, 1] such that

x f (x) =  1 − x,

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

57. The domain of the function

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

1 log x

is

(b) (0, ∞) (d) [0, ∞)\{1}

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

4 − x2 [ x] + 2 , where [x]

denotes the greatest integer less than or equal to x, is (a) [– 1, 2] (b) (– ∞, – 2) (c) (– ∞, – 2) ∪ [– 1, 2] (d) None of these 60. The range of the function y =  (a) 0, 

1 2 

 1  (c)  − , 0  2 

x is 1 + x2

 1 (b)  − ,  2

1 2 

(d) None of these

x2 is 1 + x2 (b) [0, 1] (d) None of these

61. The range of the function y = (a) [0, 1[ (c) ]0, 1[

1 is 62. The range of the function y = 2 − sin 3 x 1  (a)  , 1 3 1  (c)  , 1 3 

(d) None of these

1  (b)  , 1 3

 11  (a)  −∞, 3   

 11  (b)  −∞, 3  

 11  (c)  , ∞   3

 11  (d)  , ∞  3 

1  (a) (– ∞, ∞)\  , 1 5 

(b) (– ∞, ∞)

(c) (– ∞, ∞)\{1}

(d) None of these

 x2  67. The range of the function y = sin    is  1 + x 2  –1

 π (a)  0,  2

 π (b) 0,   2

 π (c) 0,   2

(d) None of these

 π 68. If A =  x : ≤ x ≤  6 f (A) is equal to

π  and f (x) = cos x – x (1 + x), then 3

 1 π π2 3 π π2  (a)  2 − 3 − 9 , 2 − 6 − 36     1 π π2 3 π π2  (b)  2 + 3 − 9 , 2 + 6 − 36     1 π π2 3 π π2  (c)  2 − 3 − 9 , 2 − 6 − 36  (d) None of these 69. The range of the function f (x) = tan (a) [0, (c) [0,

3 ] 3 )

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

π2 − x 2 is 9

3)

3] 70. Period of the function log (sin (x – [x]) ([ . ] denotes the greatest integer function) is –1 

(d) None of these

3 x 2 − 4 x + 5 is

x 2 − 3x + 2 66. The value of the function f (x) = 2 lies in the x + x−6 interval

f (x) = log3 [– (log3 x)2 + 5 log3 x – 6] is (a) (0, 9) ∪ (27, ∞) (b) [9, 27] (c) (9, 27) (d) None of these

(a) (0, ∞)\{1} (c) [0, ∞)

3 3  (b)  − , 2   2

65. The range of the function f (x) =

x ∈Q x ∉Q

Then for all x ∈ [0, 1], fof (x) is (a) a constant (c) x

3  (a) 0, 2   3  (c)  − , 0  2 

71. The range of the function f (x) = sin x – cos x is

2 , 2 ) (c) [0, 2 ] (a) (–

(b) [–

2, 2] (d) None of these

72. The range of the function f (x) = (a) R\{0} (c) {– 1, 1}

x is | x|

(b) R\{– 1, 1} (d) None of these

81. The period of the function f (x) = sin4 2x + cos 4 2x is π 2 π (c) 4

(a)

(b)

π 8

(d) None of these

82. The period of the function f (x) =

tan x is

(a) π π (c) 2

(b) 2π

(a) 1 ≤ n < 2 (c) 1 ≤ n ≤ 2

(b) 1 < n < 2 (d) None of these

(d) None of these

 8x + 5  is 73. If f : R → R is a function satisfying the property 83. The period of the function f (x) = cos   4 π  f (2x + 3) + f (2x + 7) = 2, x ∈ R, then the period (a) 2π (b) π of f (x) is (d) None of these (c) π2 (a) 2 (b) 4 84. The period of the function f (x) = x – [x], where [x] de(c) 8 (d) 12 notes the greatest integer less than or equal to x, is   4 − x2  (a) 2 (b) 1 74. The range of the function f (x) = sin log   (c) 4 (d) None of these 1 − x     is   85. If T1 is the period of the function y = e 3(x – [x]) and T2 is (a) [0, 1] (b) (– 1, 0) the period of the function y = e 3x – [3x] ([.] denotes the (c) [– 1, 1] (d) (– 1, 1) greatest integer function), then 75. Suppose f (x) = (x + 1)2 for x ≥ –1, If g(x) is the function T whose graph is the reflection of the graph of f (x) with (b) T1 = 2 (a) T1 = T2 3 respect to the line y = x, then g(x) equals (c) T = 3T (d) None of these 1 2 1 x f (x) (b) x + 1 2 , x > −1 (a) − x − 1 , x ≥ 0 86. If e + e = e, then range of the function of f is ( ) (a) (– ∞, 1] (b) (– ∞, 1) (c) x + 1, x ≥ –1 (d) x − 1 , x ≥ 0 (c) (1, ∞) (d) [1, ∞) 76. Let the function f : R → R be defined by 87. If the period of the function f (x) = sin ( [n] x), where f (x) = 2x + sin x, [n] denotes the greatest integer less than or equal to x ∈ R. Then f  is n, is 2π, then (a) one-to-one and onto (b) one-to-one but not onto (c) onto but not one-to-one (d) neither one-to-one nor onto 77. The period of the function f (x) = a sin kx + b cos kx is

2π (a) k π (c) | k |

2π (b) | k | (d) None of these

88. Let f (x) = (– 1)[x] (where [ . ] denotes the greatest integer function), then (a) Range of f is {– 1, 1} (b) f is an even function (c) f is an odd function (d) lim f (x) exists, for every integer n x→n

78. The period of the function f (x) = cos x2 is

89. The period of the function

(a) 2π (b) π π (c) (d) None of these 2 79. The period of the function f (x) = | sin 4x | + | cos 4x | is π π (b) (a) 8 2 π (c) (d) None of these 4 80. The period of the function f (x) = sin x is

 πx   πx  is f (x) = cos   − sin   n!   (n + 1)! 

(a) π π (c) 2

(b) 2π (d) None of these

(a) 2 (n + 1)! (c) (n + 1)

(b) 2 (n!) (d) not periodic

90. The function f (x) = k | cos x | + k2 | sin x | + φ (k) has period π if k is equal to 2 (a) 1 (c) 3

(b) 2 (d) None of these

91. The period of the function 3(sin

2 πx + x − [ x ] + sin 4 πx )

denotes the greatest integer function, is

, where [⋅]

11

(b) 2π (d) None of these

Functions

(a) 1 (c) π/2

12

Objective Mathematics

1 2 (c) 2

(a)

101. The period of the function f (x) = sin4 x + cos 4 x is π (a) π (b) 2 (c) 2π (d) None of these

(b) 1 (d) non periodic.

92. The period sin θ is 2

(a) π2 (c) 2π

102. Which of the following functions has period π ?

(b) π (c) π/2

|sin x | − |cos x | 93. The period of the function f (x) = |sin x + cos x | is π 2 (c) π

(a)

(b) 2π (d) None of these

94. The domain of the function f (x) = log 

1 x+ 2  

2 πx   πx  (a) 2 cos  + 3 sin    3   3  (b) | tan x | + cos 2x π π (c) 4 cos  2 πx +  + 2 sin  πx +    2 4 (d) None of these

| x – x – 6 |, 2

103. The function f (x) =

where [⋅] denotes the greatest integer function, is 3  (a)  , 3  ∪ (3, ∞) 2

3  (b)  , 3  ∪ (3, ∞) 2

3  (c)  , ∞  2

(d) None of these

95. If f (x) is defined on (0, 1), then the domain of definition of f (e x) + f (ln | x | ) is (a) (– e, – 1) (c) (– ∞, – 1) ∪ (1, ∞)

(b) (– e, – 1) ∪ (1, e) (d) (– e, e)

(a) [1, 4] (c) [0, 5]

 5x − x2  exists for log10   4  (b) [1, 0] (d) [5, 0]

104. If f (x) is defined on (0, 1), then the domain of definition of f (sin x) is (a) (2nπ, (2n + 1) π), n ∈ I π π  (b)  (2n + 1) , (2n + 3)  , n ∈ I 2 2 (c) ((n – 1) π, (n + 1) π), n ∈ I

(d) None of these 96. The period of the function f (x) = 2 sin x + 3 cos 2x 105. The period of the function f (x) = x [x] is is (a) 1 (b) 2 (a) π (b) 2π (c) non periodic (d) None of these. π (c) (d) None of these 106. If the period of the function f (x) = tan ( [k ] x), where 2 [⋅] denotes the greatest integer function, is π, then 97. The period of the function

1, when x is a rational f (x) = 0, when x is irrational is   (a) 1 (b) 2 (c) non-periodic (d) None of these 98. The domain of sin –1 [log3 (x/3)] is (a) [1, 9] (c) [– 9, 1]

(b) [– 1, 9] (d) [– 9, – 1]

99. The period of the function 2 πx πx + sin f (x) = cos is 5 4 (a) 5 (b) 8 (c) 12 (d) 40

(a) 1 < k < 2 (c) k = 1, 2

(b) 1 ≤ k < 2 (d) None of these

107. The period of the function f (x) = sin 5x + cos 3 x is (a) 3 π (c) non periodic

(b) π (d) None of these

108. The period of the function

πx , 2 where [x] denotes the greatest integer ≤ x, is (a) 4 (b) 1 (c) 2 (d) non-periodic f (x) = 3x + 3 – [3x + 3] + sin

109. The value of n ∈ I for which the function sin nx f (x) = has 4π as its period is x f (x + k) = 1 + [2 – 5 f (x) + 10 { f (x)}2 – 10 { f (x)}3 sin   n + 5 { f (x)}4 – { f (x)}5]1/5 (a) 2 (b) 3 for all real x and some positive constant k, then the period (c) 4 (d) 5 of the function f (x) is 110. π is the period of the function (a) k (a) | sin x | + | sin x | (b) sin4x + cos4x (b) 2k 1 + 2 cos x (c) non periodic (c) sin (sin x) + sin (cos x) (d) sin x (2 + sec x) (d) None of these 100. Let f be a real valued function with domain R satisfying

cos 2 x sin 2 x − 1 + tan x 1 + cot x 112. The period of the function (d) f (x) = 1 –

f (x) =

π 2 (d) None of these

sin x + sin 2 x + sin 4 x + sin 5 x is cos x + cos 2 x + cos 4 x + cos 5 x

π 3 (c) π

(a)

π 4 (d) None of these

115. A function f from the set of natural numbers to integers defined by n − 1  2 , where n is odd is f (n) =  − n , where n is even  2 (a) one-one but not onto (b) onto but not one-one (c) one-one and onto both (d) neither one-one nor onto

(a) an even function (b) an odd function (c) a periodic function (d) neither an even nor an odd function 122. A function whose graph is symmetrical about the y-axis is given by (a) f (x) = sin [log (x +

sec 4 x + cosec 4 x x 3 + x 4 cot x (c) f (x + y) = f (x) + f (y) ∨ x, y ∈ R (d) None of these (b) f (x) =

(a) f (x) = (3x + 3– x) (b) f (x) = cos [log (x + 1 + x 2 )] (c) f (x + y) = f (x) + f (y) ∨ x, y ∈ R (d) None of these

n

∑ f (r )

is

r =1

7n (a) 2

7(n + 1) (b) 2 7 n(n + 1) (c) 7n (n + 1) (d) 2 117. If f is an even function defined on the interval [– 5, 5], then the real values of x satisfying the equation  x +1  f (x) = f   are x + 2

124. Let f (x) = sin x + cos x, g(x) = x2 – 1. Then g( f (x)) is invertible in the domain  π  (a)  − , 0  2 

 π  (b)  − , π   2 

 π π (c)  − ,   4 4

 π (d) 0,   2

125. If f : R → S, defined by f (x) = sin x – onto, then the interval of S is (a) [0, 1] (c) [0, 3]

(b)

x 2 + 1 )]

123. A function whose graph is symmetrical about the origin is given by

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

−1 ± 5 2 −2 ± 5 (c) 2

3 + log10 ( x 3 − x) , is 4 − x2

121. The function f (x) = log( x + x 2 + 1) , is

114. Let f be a real valued function with domain R satisfy1 ing 0 ≤ f (x) ≤ and for some fixed a > 0, 2 1 f (x + a) = − f ( x) − ( f ( x)) 2 ∨ x ∈ R, 2 then the period of the function f (x) is (a) a (b) 2a (c) non-periodic. (d) None of these

(a)

119. Domain of definition of the function:

(a) 3x2 + 4x + 8 log (1 + | x | ) (b) 3x2 – 4x + 8 log (1 + | x | ) (c) 3x2 + 4x – 8 log (1 + | x | ) (d) None of these

(b)

for all x, y, ∈ R and f (1) = 7, then

(b) odd (d) None of these

120. Let the function f (x) = 3x2 – 4x + 8 log (1 + | x | ) be defined on the interval [0, 1]. The even extension of f (x) to the interval [– 1, 1] is

113. The period of the function f (x) =

1 + x 2 )] is

(a) (1, 2) (b) (– 1, 0) ∪ (1, 2) (c) (1, 2) ∪ (2, ∞) (d) (– 1, 0) ∪ (1, 2) ∪ (2, ∞)

(b)

(c) 2π

(a) even (c) constant

f (x) =

sin 8 x cos x − sin 6 x cos 3 x is cos 2 x cos x − sin 3 x sin 4 x

(a) π

118. The function f (x) = sec [log (x +

−3 ± 5 2

(d) None of these

3 cos x + 1 , is

(b) [– 1, 1] (d) [– 1, 3]

126. The graph of the function y = f (x) is symmetrical about the line x = 2, then (a) f (x) = f (–x) (c) f (x + 2) = f (x – 2)

(b) f (2 + x) = f (2 – x) (d) f (x) = – f (– x)

13

(a) f (x) = tan (3x – 2) (b) f (x) = {x}, the fractional part of the number x (c) f (x) = x + cos x

Functions

111. Which of the following functions is non-periodic ?

14

127. The function f : R → R, defined by f (x) = [x], ∨ x ∈ R, is

Objective Mathematics

(a) one-one (b) onto (c) Both one-one and onto (d) neither one-one nor onto

138. The function f : R → R, defined by

128. Let f : R → R be a function defined by f (x) = sin (2x – 3), then f is (a) injective (c) bijective

(b) surjective (d) None of these

129. Which of the following functions from I to itself are bijections? (a) f (x) = x + 3 (c) f (x) = 3x + 2

(b) f (x) = x5 (d) f (x) = x2 + x

130. Which of the following functions (is) are injective map(s)? (a) f (x) = x2 + 2, x ∈ (– ∞, ∞) (b) f (x) = | x + 2 |, x ∈ [– 2, ∞) (c) f (x) = (x – 4) (x – 5), x ∈ (– ∞, ∞) 4 x 2 + 3x − 5 , x ∈ (– ∞, ∞) (d) f (x) = 4 + 3x − 5 x 2 131. Let A = {x ∈ R | – 1 ≤ x ≤ 1} = B. Then the mapping f : A → B given by f (x) = x | x | is (a) Injective but not surjective (b) Surjective but not injective (c) Bijective (d) None of these 132. If the function f : (– ∞, ∞) → B defined by f (x) = – x2 + 6x – 8 is bijective, then B = (a) [1, ∞) (b) (– ∞, 1] (c) (– ∞, ∞) (d) None of these 133. Let f : R → R be defined by f (x) = Then f is

134. The number of bijective functions from a set A to itself when A contains 106 elements is (b) (106)2 (d) 2106

135. The number of surjections from A = {1, 2, ..., n}, n ≥ 2 onto B = {a, b} is (a) nP2 (c) 2n – 1

(b) 2n – 2 (d) None of these

136. Set A has 3 elements and set B has 4 elements. The number of injections that can be defined from A to B is (a) 144 (c) 24

f (x) = x – [x], ∨ x ∈ R is (a) one-one (b) onto (c) Both one-one and onto (d) neither one-one nor onto 139. Let f (x) = [x]2 + [x + 1] – 3, where [⋅] denotes the greatest integer function, then (a) f (x) ≠ 0 for all real x (b) f (x) = 0 for only two real values (c) f (x) = 0 for infinite number of values of x (d) None of these 140. The function f : R → R defined by f (x) = 4x + 4| x | is (a) one-one and into (c) one-one and onto

(b) many-one and into (d) many-one and onto

141. Let f : R → R be a function defined by x2 − 8 f (x) = 2 , then f is x +2 (a) one-one but not onto (b) one-one and onto (c) onto but not one-one (d) neither one-one nor onto 142. Let f : R → R be a function defined by f (x) = x + x 2 , then f is (a) injective (c) bijective

(b) surjective (d) None of these

x−a , where a ≠ b. 143. Let f : (– ∞, 2] → (– ∞, 2] be a function defined by x−b f (x) = 4x – x2. Then f –1 (x) is

(a) Injective but not surjective (b) Surjective but not injective (c) Bijective (d) None of these

(a) 106 (c) (106)!

(a) one-one but not onto (b) onto but not one-one (c) Both one-one and onto (d) neither one-one nor onto

(b) 12 (d) 64

2 π − 2 137. If A =  x : − ≤ x ≤  , B = {y : – 1 ≤ y ≤ 1} and 5 5   f (x) = cos (5x + 2), then the mapping f : A → B is

(a) 2 – (c)

4− x

4− x

(b) 2 +

4− x

(d) None of these

144. Let f : [4, ∞) → [4, ∞) be a function defined by f (x) = 5x (x – 4), then f –1 (x) is (a) 2 –

1 (c)   5

4 + log 5 x

(b) 2 +

4 + log 5 x

x ( x − 4)



(d) None of these

145. If f : R → R is given by f (x) = 3x – 5, then f –1 (x)

1 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 (a) is given by

146. Let f : R → R be given by f (x) = (x + 1)2 – 1, x ≥ – 1. Then f –1 (x)

x +1 (c) does not exist because f is not one-one . (d) does not exist because f is not onto (b) = – 1 –

147. Let f : R → R be given by f (x) = (x + 1)2 – 1, x ≥ – 1. Then, the set of values of x for which f (x) = f –1 (x) is given by (a) {0} (c) {– 1}

(b) {0, – 1} (d) None of these

148. Let f be an injective map with domain {x, y, z} and range {1, 2, 3} such that exactly one of the following statements is correct and the remaining are false : f (x) = 1, f (y) ≠ 1, f (z) ≠ 2. The value of f –1 (1) is (a) y (b) x (c) z (d) None of these 149. The value of the parameter α, for which the function f (x) = 1 + αx, α ≠ 0 is the inverse of itself, is (a) – 2 (c) 1

(b) – 1 (d) 2 [Based on IIT 1992]

150. The inverse of the function f (x) = (a)

1 1 − x  log a   1 + x  2

1+ x  (c) log a   1 − x 

(b)

a x − a− x is a x + a− x

1 1 + x  log a   1 − x  2

(d) None of these

151. If the function f : [1, ∞) → [1, ∞) is defined by f (x) = 2x (x – 1), then f –1 (x) is

1 (a)   2

x ( x −1)

1 (c)   1 − 1 + 4 log 2 x   2 (d) not defined 152. The inverse of the function y = [1 – (x – 3) 4]1/7 is (b) 3 – (1 – x7)1/4 (d) None of these

153. If f (x) = 1 – x, x ∈ [– 3, 3], then the domain of f [ f (x)] is (a) [– 2, 3] (c) [– 2, 3)

x < −1 | x |,  (b) 2 − | x |, − 1 ≤ x ≤ 1 | 2 − x |, x > 1 

| 2 − x |, x < −1  (c) | x |, − 1 ≤ x ≤ 1 2 − | x |, x > 1 

(d) None of these

156. If f (x) = 2 − x and g (x) = of f [g (x)] is

(b) (– 2, 3) (d) (– 2, 3]

1− x   1  ; x ≠ 0 then f [ f (x)] + f  f    154. If f (x) = 1+ x   x  (a) < 2 (b) ≥ 2 (c) = 2 (d) None of these

| x |, x ≤ 1 , then f [ f (x)] is equal to 155. If f (x) =   2 − x, x > 1

1 − 2 x , then the domain

 1 (a)  −∞,   2

1  (b)  , ∞  2 

 3 (c)  −∞, −   2

(d) None of these

157. Let f be a function with domain [– 3, 5] and let g (x) = | 3x + 4 |. Then the domain of ( fog) (x) is 1  (a)  −3,   3

1  (b)  −3,  3 

 (c)  −3, 1  3 

(d) None of these

1 , x ≠ 0, 1, then the graph of the function 1− x y = f [ f { f (x)}], x > 1 is

158. If f (x) =

(a) a straight line (c) an ellipse

(b) a circle (d) a pair of straight lines

159. Let f : R → R be a function defined by f (x) = x – [x], where [⋅] denotes the greatest integer function, then f –1 (x) is (a) [x] – x

1 (b)   1 + 1 + 4 log 2 x  2

(a) 3 + (1 – x7)1/4 (c) 3 – (1 + x7)1/4

2 − | x |, x < −1 (a) | x |, − 1 ≤ x ≤ 1 | 2 − x |, x > 1 

(c) not defined

1 x − [ x] (d) None of these (b)

160. If f (x + y, x – y) = xy, then the arithmetic mean of f (x, y) and f ( y, x) is (a) y (c) 0 161. If f (x) = 64x3 +

(b) x (d) None of these

1 1 3 and a, b are the roots of 4x + x x

= 3, then (a) f (a) = 12 (c) f (a) = f (b)

(b) f (b) = 11 (d) None of these

162. If f, g, h are functions from R to R such that f (x) = x2 – 1, 0, x ≤ 0 g (x) = x 2 + 1 ∨ x ∈ R and h (x) =   x, x ≥ 0 then the composition function ho ( fog) is given by (a) x2 (c) x

(b) 0 (d) None of these

15

x +1

Functions

(a) = – 1 +

16

163. If f (x) = cos (log x), then f (x) ⋅ f (y) –

Objective Mathematics

 1   x  + f ( xy ) is equal to f 2   y  

(a) 2 (c) 0

(b) 1 (d) None of these

171. If f (x) = sin [π2] x + sin [– π2] x, where [⋅] denotes the greatest integer function, then π (a) f    = 1 2

(b) f (π) = 2

π (c) f    = – 1 4

(d) None of these 164. If R denotes the set of all real numbers, then the function f : R → R defined f (x) = | x | is 172. Let f (x) = max. {(1 – x), (1 + x), 2}, ∨ x ∈ R. Then

1 + x, x ≤ −1  (a) f (x) = 2, − 1 < x < 1 1 − x, x ≥ 1 

(a) one-one only (b) onto only (c) both one-one and onto (d) neither one-one nor onto

1 1 – 3, ∨ x (≠ 0) ∈ R, then f (x) 165. If 3 f (x) + 5 f    = x x = (a)

1 3   + 5 x − 6  14  x

1  3  (c)  − + 5 x + 6  14  x

(b)

1  3   − + 5 x − 6  14  x

(d) None of these

1 − x, x ≤ −1  (b) f (x) = 1, − 1 < x < 1 1 + x, x ≥ 1  1 − x, x ≤ −1  (c) f (x) = 2, − 1 < x < 1 1 + x, x ≥ 1 

166. If f (x + y) = f (x) + f (y) – xy – 1 for all x, y ∈ R and (d) None of these f (1) = 1, then the number of solutions of f (n) = n, n 173. The distinct linear function (s) which map (s) [– 1, 1] ∈ N is onto [0, 2] is (are) (a) one (b) two (a) x + 1, – x + 1 (c) no solution (d) None of these (b) x – 1, x + 1 1 (c) – x + 1 , g (x) = f [ f (x)] and h (x) = f [ f { f (x)}], 167. If f (x) = 1− x (d) None of these then the value of f (x) ⋅ g (x) ⋅ h (x) is 174. Let f (x) be a function defined on [– 1, 1]. If the area (a) 1 (b) – 1 of the equilateral triangle with two of its vertices at (c) 0 (d) None of these 3 (0, 0) and (x, f (x)) is , then the function f (x) is 168. If g [ f (x)] = | sin x | and f [g (x)] = (sin )2, then 4 (a) f (x) = sin2x, g (x) = (a) x 2 − 1 (b) – 1 + x 2 (b) f (x) = sin x, g (x) = | x | 2 (c) f (x) = x , g (x) = sin (d) 1 + x 2 = (c) ± 1 − x 2 (d) f and g cannot be determined

4x , then 4x + 2

169. Let f be a function defined on [– 2, 2] and is given by

175. If f (x) =

−1, − 2 ≤ x ≤ 0 f (x) =   x − 1, 0 < x ≤ 2

 1   2   1996  f   + f   + ... + f   is equal to 1997 1997 1997 

and g (x) = f (| x |) + | f (x) |. Then g (x) is equal to

 − x, − 2 ≤ x < 0  (a) 0, 0 ≤ x 20, x 2 x3 g′ (x) = + + ... is 189. Domain of e x = 1 + x + 2! 3! (a) – 1 (b) 1 (a) (1, ∞) (b) (– ∞, ∞) (c) 2 (d) None of these (c) (0, ∞) (d) None of these 182. If S is the set of all real x and such that 190. The range of the function f (x) = | x – 1 | is 2x − 1 is positive, then S contains (a) (– ∞, 0) (b) [0, ∞) 2 x3 + 3x 2 + x (c) (0, ∞) (d) (– ∞, ∞) 3 1   3 2 (b)  − , −  (a)  −∞, −  191. Domain of 4 x − x is   2 2 4 (a) R\[0, 4] (b) R\(0, 4) (c) (0, 4) (d) [0, 4] 1  1 1   (c)  − ,  (d)  , 3   4 2 2  192. The domain of the function f (x) = 2 − 2 x − x 2 is 183. The range of the real function

x+2 is x2 − 8x − 4

1  1   (a)  −∞, −  ∪  − , ∞   4   20  1 1 (b)  −∞, −  ∪  − , ∞   4   20 

 1  1  (c)  −∞, −  ∪  − , ∞   4   20  (d) None of these 184. If the function f, g, h are defined from the set of real numbers R to R such that f (x) = x2 – 1, g (x) =

0, if x ≤ 0 x 2 + 1 , h (x) =   x, if x ≥ 0

then the composite function (hofog) (x) = x=0 0,  (a)  x 2 , x > 0 − x 2 , x < 0 

0, x = 0 (b)  2 x , x ≠ 0

0, x ≤ 0 (c)  2 x , x > 0

(d) None of these

(a) – 2 – 3 ≤ x ≤ – 2 + (b) – 2 ≤ x ≤ 2

3

3 ≤x≤–1+ 3 (d) – 3 ≤ x ≤ 3 193. If f (x) = 2x3 + mx2 – 13x + n and 2, 3 are roots of the equation f (x) = 0, then the values of m and n are (c) – 1 –

(a) 5, 30 (c) – 5, 30 194. If x ∈ R and P =

(b) – 5, – 30 (d) None of these

x2 , then P lies in the inx − 2x2 + 4 4

terval 1 (a) 0,   2 

3 4 (b)  ,  4 5

(c) 0, 

(d) 0, 

1 3 

1 4 

1 + x  3x + x3 195. If f (x) = log   , then f [g (x)]  and g (x) = 1− x 1 + 3x 2 is equal to (a) f (3x) (c) 3 f (x)

(b) [ f (x)]3 (d) – f (x)

17

x2 − x is x2 + 2x

Functions

178. The domain of definition of the function y (x) given 185. The range of the function f (x) = by the equation 2 x + 2y = 2 is

18

2x − 1 is 2 x + 3x 2 + x

196. If S is the set of all real x such that

3

Objective Mathematics

positive, then S contains 1 (a)  − ,  4

1  2

1 (b)  , 3  2 

3 1 (c)  − ,   2 4

206. If e x = y +

3 1 (d)  − ,   2 2

x

(b) 8 (d) 3

is equal to

201. The domain of the function f (x) = log (1 – x) + is

x2 − 1

(b) (– ∞, – 1] (d) [– 1, 1]

(a) R\{nπ : n ∈ I} (c) R\{2nπ : n ∈ I}

log

1 is | sin x |

(b) one-one but not onto (d) many-one onto

1  1 210. Let f   x +  = x2 + 2 (x ≠ 0) then f (x) is equal to x x (a) x2 – 1 (b) x2 – 2 (d) None of these (c) x2

(a) {– 3, 0} (c) [0, 3]

(b) R \ (– π, π) (d) (– ∞, ∞)

(b) [– 3, 0] (d) (– 3, 0)

212. Let f : R → R be a mapping defined by, f (x) = x3 + 5, then f (a) (5 – x) (c) 5 – x 1/3

(b) (sin x) (d) sin x/x2 2

204. Let f : R → R be defined by f (x) = 3x – 4, then f  (x) is

1 (x + 4) 3 (c) 3x + 4

x defined as f : (0, ∞) to (0, ∞), f (x) 209. Let f (x) = 1 + x is

is defined on the set S, where S is equal to

–1

(a)

, n ∈ N}

f (x) = cot–1 [ ( x + 3) x ] + cos–1 ( x 2 + 3 x + 1 )

f (x) = sin x, g : R → R, g (x) = x2 is (a) sin x + x (c) sin x2

(b) x < 0 (d) – ∞ < x < ∞

211. The function

203. The composite mapping fog of the maps f : R → R, 2

2x + 1 is given by x − 10 x − 11 2

(a) one-one onto (c) many-one into

5 2 (d) None of these

(b)

202. The domain of the function y =

3

, n ≥ 0, n ∈ I} (b) R – { (c) R (d) R – {0}

1 200. If 2 f (x) – 3 f    = x2, x is not equal to zero, then f (2) x

(a) (0, 1) (c) (1, ∞)

e x + e− x 2

207. The domain of definition of the function

(a) R – {±

(b) (1, 7/3) (d) (1, 11/7)

7 4 (c) – 1

(d)

  x f (x) = cot–1   , x ∈ R is 2 2  x − [x ] 

x2 + x + 2 is x2 + x + 1

(a) –



208. The domain of the function

(b) {1, cos 1} (d) {0}

(a) (1, ∞) (c) (1, 7/5)

e −e 2

(b) ex – e–x

−x

(a) x > 0 (c) x ≠ – 1, x ≠ 11

π π 1} (d) {y : y < 1}

229. If the function

220. R is the set of real numbers and f : R → R and g : R → R are defined by f (x) = 3x2 + 2 and g (x) = 3x–1 for all x ∈ R. Then (a) ( fog) (x) = 27x2 – 18x + 5 (b) ( fog) (x) = 27x2 + 18x – 5 (c) (gof ) (x) = 9x2 – 5 (d) (gof ) (x) = 9x2 + 5. 221. If x is real, then the expression

x 2 + 34 x − 71 x2 + 2x − 7

4 3

(b) 1

(c) 0

(d)

(a) 0 < x ≤ 1 (c) – ∞ < x ≤ 0

(b) 0 ≤ x ≤ 1 (d) – ∞ < x < 1

1+ x  231. The function f ( x) = log   satisfies the equation 1− x 

x − x +1 x −1 2

(a) cannot lie between – 1 and 3 (b) always lies between – 1 and 3

(a)

3 4 230. The domain of definition of the function y (x) given by the equation 2 x + 2y = 2 is

(a) cannot lie between 5 and 9 (b) always lies between 5 and 9 (c) is not real (d) None of these. 222. If x is real, then the expression

π  π  f (x) = cos2x + cos2   + x  – cos x . cos   + x  3  3  is constant (independent of x), then the value of this constant is

(a)  f(x1) f(x2) = f(x1 + x2) (b)  f(x + 2) – 2f(x + 1) + f(x) = 0 (c)  f(x) + f(x + 1) = f(x2 + x)  x +x  (d)   f ( x1 ) + f ( x2 ) = f  1 2   1 + x1 x2 

Functions

−2

 x − 1 (a) loge    x + 1 

20

Objective Mathematics

232. The largest interval lying in  − π , π  for which the  2 2   function x  – x2 –1  f (x ) = 4 + cos  − 1 + log (cos x) is defined, is 2   π (b)  [0, π] (a)   0,   2  π π (c)    − ,   4 2

 π π (d)    − ,   2 2

233. If A = {1, 2, 3, 4}, B = {1, 2, 3}, then number of mappings from A to B is (a)  34 (c)  43

(b)  12 (d)  27

x π 234. If f : 0,  → [0, ∞) be a function defined by y = sin   , 2  2 then f is (a)  injective (c)  bijective

(b)  surjective (d)  none of these

235. Let f(x) = |x| has an inverse, its domain is (a)  R (c)  {0, 1}

(b)  {–1, 1} (d)  none of these

238. If f(x) = e x and g(x) = loge x, then which of the following is true? (a)  f{g(x)} ≠ g {f(x)} (b)  f{g(x)} = g {f(x)} (c)  f{g(x)} + g {f(x)} = 0 (d)  f{g(x)} – g {f(x)} = 1   x  239.   The domain of sin −1 log 3    is  3   (a)  [1, 9] (c)  [– 9, 1]

240. Let f : N → N defined by f(x) = x2 + x + 1, x ∈ N, then f is (a)  one-one onto (b)  many-one onto (c)  one-one but not onto (d)  None of the above 241. The domain of the function f(x) = log 2x – 1 (x – 1) is

(b)  one-one into (d)  many-one into

237. Let the functions f, g, h be defined from the set of real numbers R to R such that f(x) = x2 – 1, g(x) =

( x 2 + 1)

0, if x < 0 and h( x) =  ,  x, if x ≥ 0 then ho (fog) (x) is defined by (b)  x2 (d)  none of these

(a)  x (c)  0

1  (b)    , ∞  2  (d)  None of these

(a)  (1, ∞) (c)  (0, ∞) 242. Let f ( x) =

236. The function f : R → R defined by f(x) = |(x – 1) (x – 2)|, is (a)  one-one onto (c)  many-one onto

(b)  [–1, 9] (d)  [– 9, – 1]

1  πx  − tan   , 1 < x < 1 and 2  2 

g ( x) = 3 + 4 x − 4 x 2 , then dom (f + g) is given by 1  (a)    , 1 2 

1  (b)    , − 1 2  

 1  (c)    − , 1  2 

 1  (d)    − , − 1  2 

243. Let f : N → Y be a function defined as f(x) = 4x + 3, where Y = {y ∈ N : y = 4x + 3 for some x ∈ N}. Then, the inverse of f(x) is 3y + 4 3 y+3 (c)   g ( y ) = 4

(a)   g ( y ) =

y+3 4 y −3 (d)   g ( y ) = 4

(b)   g ( y ) = 4 +

solutions 1. (b) For f (x) to exist, we must have π π sin –1(2x) + ≥ 0  or,  sin –1 2x ≥ –  6 6 π π –1 Since –  ≤ sin (2x) ≤ 2 2 π π So, – ≤ sin –1 (2x) ≤ 6 2  π π or, sin   −  ≤ 2x ≤ sin   2 6

⇒ –

1 ≤ 2x ≤ 1 2

1 1 ≤ x≤ 4 2



⇒ −



 1 1 ∴  Domain =  − ,   4 2

2. (a) For y to be defined, we must have

 5x − x2  5x − x2 ≥ 0 ⇒ ≥ 100 (a) log10     4  4 ⇒ 5x – x2 ≥ 4 ⇒ x2 – 5x + 4 ≤ 0 ⇒ (x – 1) (x – 4) ≤ 0 ⇒ 1 ≤ x ≤ 4.



1 2 x ≤ 21 ⇒2 ≤ 2 –1

⇒ [x] < – 2 or [x] > 3 But [x] < – 2 ⇒ [x] = – 3, – 4, – 5, ...∴x < – 2. Also, [x] > 3 ⇒ [x] = 4, 5, 6, ... ∴ x≥4 Domain of f = (– ∞, – 2) ∪ [4, ∞). 8. (c) log10 (1 – x) is defined for 1 – x > 0 ⇒ x < 1 ...(1) log10 (1 – x) ≠ 0 ⇒ 1 – x ≠ 100 ⇒ 1 – x ≠ 1 ⇒ x ≠ 0  ...(2)

x + 2 is defined for x + 2 ≥ 0 ⇒ x ≥ – 2 From (1), (2) and (3), we get domain of y = ([– 2, ∞) ∩ (– ∞, 1))\{0} = [– 2, 1) \ {0} i.e. [– 2, 0[ ∪ ] 0, 1[.

...(3)

[∵ the base = 2 > 1] ...(1) 9. (a) 3 − x is defined for 3 – x ≥ 0 ⇒ 1 ≤ x2 ≤ 4 Now, 1 ≤ x2 ⇒ x2 – 1 ≥ 0 i.e. (x – 1) (x + 1) ≥ 0 ⇒ x ≤ 3 ...(1) ⇒ x ≤ – 1 or x ≥ 1  ...(2)  3 − 2x  cos–1    is defined for Also, x2 ≤ 4 ⇒ x2 – 4 ≤ 0 i.e. (x – 2) (x + 2) ≤ 0 5  ⇒ – 2 ≤ x ≤ 2  ...(3) 3 − 2x From (2) and (3), we get the domain of f ≤1 –1 ≤ 5 = ((– ∞, – 1] ∪ [1, ∞)) ∩ [– 2, 2] = [– 2, – 1] ∪ [1, 2]. ⇒ – 5 ≤ 3 – 2x ≤ 5 ⇒ – 8 ≤ – 2x ≤ 2 4. (d) For f (x) to be defined, we must have ⇒ – 1 ≤ x ≤ 4 ...(2) From (1) and (2), we get domain of f x – 1 − x 2 ≥ 0  or  x ≥ 1 − x 2 > 0 (– ∞, 3] ∩ [– 1, 4] = [– 1, 3]. 1 ∴ x2 ≥ 1 – x2 or x2 ≥ . x2 + 1 2 . 10. (a) We have, f (x) = [ x] Also, 1 – x2 ≥ 0 or x2 ≤ 1. When x ∈ [1, 2) then f (x) = x2 + 1 1  1  1  Now, x2 ≥ ⇒ x − ≥0 x+    ⇒ Rf = [2, 5).  2 2  2 When x ∈ [2, 3) then 1 1 5  x2 + 1 or x ≥ . ⇒ x ≤ – ⇒ Rf =  , 5  . f (x) = 2 2 2 2 Also, x2 ≤ 1 ⇒ (x – 1) (x + 1) ≤ 0 When x ∈ [3, 4) then ⇒ – 1 ≤ x ≤ 1 10 17  x2 + 1  1  1 f (x) = ⇒ Rf =  ,  . 2 2 , 1 and x ≤ 1 ⇒ x ∈  . Thus, x > 0, x ≥  3 3 3  2  2 ∴ Rf = [2, 17/3). 5. (b) 9 − x 2 is defined for 11. (d) f (x) is defined for x12 – x9 + x4 – x + 1 > 0 2 9 – x ≥ 0 ⇒ (3 – x) (3 + x) ≥ 0 ⇒ x4 (x8 + 1) – x (x8 + 1) + 1 > 0 ⇒ (x – 3) (x + 3) ≤ 0 ⇒ (x8 + 1) x (x3 – 1) + 1 > 0. ⇒ – 3 ≤ x ≤ 3 ...(1) If x ≥ 1 or x ≤ – 1, then the above expression is sin–1 (3 – x) is defined for positive. – 1 ≤ 3 – x ≤ 1 ⇒ – 4 ≤ – x ≤ – 2 If – 1 < x ≤ 0, the above inequality still holds. ⇒ 2 ≤ x ≤ 4  ...(2) Also, sin–1 (3 – x) = 0 ⇒ 3 – x = 0 or  If 0 < x < 1, then x = 3 ...(3) x12 – xp (x8 + 1) + (x4 + 1) > 0 From (1), (2) and (3), we get the domain of f : [∵ x4 + 1 > x8 + 1 and so x4 + 1 > x (x8 + 1)]. ([– 3, 3] ∩ [2, 4]) \ {3} = [2, 3). The domain of f = (– ∞, ∞). 6. (b) We have, y = x (2 – x) 2 12. (a) We have, ⇒ x – 2x + y = 0



2 ± 2 1− y = 1 ± 1− y 2 But x ≤ 1. So, x = 1 – 1 − y



Therefore, f –1(x) = 1 –



⇒ x =

7. (a) f (x) is defined for [x]2 – [x] – 6 > 0

1− x .

⇒ ([x] – 3) ([x] + 2) > 0

( x + 1)( x − 2)( x − 3) ( x + 1)( x − 3) . = ( x − 2) ( x − 2) f (x) is defined for (x + 1) (x – 2) (x – 3) ≥ 0 or [x – (– 1)] (x – 2) (x – 3) ≥ 0 ⇒ – 1 ≤ x ≤ 2 or x ≥ 3. Also, x – 2 ≠ 0 i.e. x ≠ 2. Domain of f = ([– 1, 2] ∪ [3, ∞))\{2} = [– 1, 2) ∪ [3, ∞).

f (x) =

21

(b)

Functions

5x − x2 > 0 ⇒ 5x – x2 > 0 ⇒ x (x – 5) < 0 4 ⇒ 0 < x < 5 From (a) and (b), we get the domain of f = [1, 4] ∩ (0, 5) = [1, 4]. 3. (c) For f (x) to be defined, we must have 1 – 1 ≤ log2  x 2  ≤ 1 2 

22

13. (d)  x − 1 is defined for x – 1 ≥ 0 i.e. x ≥ 1   5 − x is defined for 5 – x ≥ 0 or x ≤ 5 

...(1) ...(2)

Objective Mathematics

From (1) and (2), we get domain of f (– ∞, 5] ∩ [1, ∞) = [1, 5]. 14. (d) We have, 2  1 + x3  f (x) = sin − 1  3 / 2  + sin (sin x) + log (( 3x{ x+} 1+)1) .  2x 

2

sin (sin x) to be defined sin (sin x) ≥

and for



0, 0 ≤ sin x < 1, x ∈ [2nπ, (2n + 1) π ], n ∈ I



 x − 3/ 2 + x3/ 2  For sin − 1   to be defined 2  −1 ≤



f (x) =

∴ fog (x) =

⇒ ⇒



i.e. x (x – 3) < 0 ⇒ 0 < x < 3



 From (1) and (2), we get domain of f =  0,

3| x | − x − 2 and g (x) = sin x 3|sin x | − sin x − 2

which is defined if 3 | sin x | – sin x – 2 ≥ 0 If sin x > 0 then 2 sin x – 2 ≥ 0 ⇒ sin x ≥ 1 sin x = 1 ⇒ x = 2nπ + π/2. If sin x < 0 then – 4 sin x – 2 ≥ 0 – 1 ≤ sin x ≤ – 1/2

x−5 is defined if x 2 − 10 x + 24 ( x − 4)( x − 5)( x − 6) x−5 > 0 i.e. >0 2 ( x − 4) 2 ( x − 6) 2 x − 10 x + 24

16. (c) log10



⇒ ⇒

(x – 4) (x – 5) (x – 6) > 0 4 < x < 5 or x > 6 

...(2) 3 . 2 

1 – 1 − 1 − x 2 ≥ 0, 1 – 1 − x ≥ 0 and 1 – x2 ≥ 0 1 – x2 ≥ 0 ⇒ (x + 1) (x – 1) ≤ 0 ⇒ – 1 ≤ x ≤ 1. Clearly for these values, the other two inequalities hold. Thus domain of f = [– 1, 1]. 1 −| x| 21. (c) f (x) is defined if 2 − | x | ≥ 0 and 2 – | x | ≠ 0 (1 − | x |)(2 − | x |) ⇒ ≥ 0 and x ≠ – 2, 2 (2 − | x |) 2 ⇒ (| x | – 1) (| x | – 2) ≥ 0 and x ≠ – 2, 2 ⇒ | x | ≤ 1 or | x | > 2 ⇒ – 1 ≤ x ≤ 1 or (x < – 2 or x > 2) Domain of f = [– 1, 1] ∪ (– ∞, – 2) ∪ (2, ∞). 22. (d)

−1 x

is defined if – 1 ≤ x ≤ 1 

1 is defined if x – 2 > 0 i.e. x > 2 x−2 From (1), (2) and (3), domain of f = φ (empty set).

...(1) ...(2) ...(3)

 2 −| x|  23. (a) cos–1   is defined for 4 

...(1)

3 ...(2) x + 5 is defined for all real x. From (1) and (2), we get domain of f = (4, 5) ∪ (6, ∞). 17. (b) f (x) is defined for | x | – x > 0 i.e. | x | > x, which is true for x < 0 only. Thus domain of f = (– ∞, 0).

1 is defined if 1 – x ≠ 0 i.e. x ≠ 1 1− x 2sin

7π 11π   π   ∴ x ∈  2nπ + , 2 nπ +  ∪ 2nπ +  , n,m ∈ I.  6 6   2



x (3 − x) >0 x2

2

x − 3/ 2 + x3/ 2 x − 3/ 2 + x3/ 2 ≤ 1 . But ≥1, 2 2 x − 3/ 2 + x3/ 2 =1 ∴ 2 ⇒ x–3/2 = x3/2 = 1 ⇒ x3 = 1 ⇒ x = 1. Thus Domain = φ. 15. (d) We have,

3− x > 0 or x

19. (a) log10 | 4 – x2 | is defined for all x except when 4 – x2 = 0 i.e., x = ± 2. Hence domain of f = (– ∞, ∞)\{– 2, 2}. 20. (d) f (x) is defined if

For log (( 3x{ x+} 1+)1) to be defined {x} ≠ 0 ⇒ x ∉ I







2 −| x | ≤1 4 ⇒ – 4 ≤ 2 – | x | ≤ 4 ⇒ – 6 ≤ – | x | ≤ 2 ⇒ – 2 ≤ | x | ≤ 6 ⇒ | x | ≤ 6 ⇒ – 6 ≤ x ≤ 6  ...(1) –1 ≤

1 log (3 − x) is defined if log (3 – x) ≠ 0 and

3 – x > 0 ⇒ 3 – x ≠ e0 = 1 and x < 3 ⇒ x ≠ 2 and x < 3  From (1) and (2), we get domain of f 3− x 3− x is defined for log10  ≥0 18. (d) log10  = [– 6, 6] ∩ ((– ∞, 3) \ {2})  x   x  = [– 6, 3) \ {2}. 3− x ≥ 100 = 1 ⇒ 3 – x ≥ x ⇒ 24. (b) f (x) is defined if log3 log4 x > 0, log4 x > 0 x and x > 0 ⇒ 2x ≤ 3 ⇒ x ≤  ...(1) ⇒ log4 x > 3º = 1, x > 4º and x > 0 ⇒ x > 41, x > 1 and x > 0 ⇒ x > 4 3− x Also, log10  Domain of f = (4, ∞).  is defined for x 

...(2)



26. (c) f (x) = tan–1x, –1 ≤ x ≤ 1

1 (– x – 1), x < –1 2 1 (x + 1), x > 1. 2 1 , −1 < x < 1 ∴ f ′(x) = 1 + x2 1 1 ,x>1 –  , x < –1 2 2 1 1 , f ′(–1 + 0) = 1 + −1 + 0 2 = ( ) 2 1 f ′(–1 + 0) = – 2 1 1 1 f ′(1 – 0) = 1 + 1 − 0 2 = , f ′(1 + 0) = ( ) 2 2

∴ f ′(–1) does not exist. ∴ domain of f ′(x) = R – {–1}. 27. (a) We have, f (x) =

=

x+3 (2 − x)( x − 5)

( x + 3)(2 − x)( x − 5) (2 − x)( x − 5)



f (x) is defined if (x + 3) (2 – x) (x – 5) ≥ 0 and (2 – x) (x – 5) ≠ 0 ⇒ (x + 3) (x – 2) (x – 5) ≤ 0 and x ≠ 2, 5 ⇒ (x ≤ – 3 or 2 ≤ x ≤ 5) and x ≠ 2, 5 ⇒ x ≤ – 3 or 2 < x < 5 Domain of  f = (– ∞, – 3] ∪ (2, 5). 3 28. (a) f (x) is defined if – 1 ≤ ≤1 4 + 2 sin x Since 4 + 2 sin x > 0 for all real x, therefore 3 ≤ 1 ⇒ 3 ≤ 4 + 2 sin x 4 + 2 sin x 1 ⇒ sin x ≥ – 2 π π ⇒ – + 2nπ ≤ x ≤ + 2nπ, n ∈ I 6 6 π  π  Domain of f =  − + 2nπ, + 2nπ  . 6  6  29. (b) f (x) is defined if x – 4 ≥ 0 and 6 – x ≥ 0 ⇒ x ≥ 4 and x ≤ 6 Domain of  f = [4, ∞) ∩ (– ∞, 6] = [4, 6].



23

30. (a) f (x) is defined if 1 – log10 (x2 – 5x + 16) > 0 and x2 – 5x + 16 > 0 ⇒ log10 (x2 – 5x + 16) < 1 and 2 5  39  x −  + >0  2 4

Functions

 x − 3 x−3 25. (c) sin–1  ≤1  is defined if – 1 ≤ 2  2 ⇒–2 ≤x–3 ≤2⇒1 ≤x≤5  ...(1) log10 (4 – x) is defined if 4 – x > 0 i.e. x < 4  ...(2) From (1) and (2), we get domain f = [1, 5] ∩ (– ∞, 4) = [1, 4).

⇒ x2 – 5x + 16 < 101 = 10 2    5  39 ∵ x − + > 0 for all real x     2 4   

⇒ x2 – 5x + 6 < 0 ⇒ (x – 3) (x – 2) < 0 ⇒ 2 < x < 3. Domain of f = (2, 3). 31. (c) Since only x ! {x} is defined so x! {x} represents a function. 32. (d) log10 sin (x – 3) is defined if sin (x – 3) > 0

⇒ 2nπ < x – 3 < 2nπ + π, n ∈ I



⇒ 2nπ + 3 < x < 2nπ + π + 3, n ∈ I

...(1)

16 − x is defined for 16 – x ≥ 0 ⇒ (4 – x) (4 + x) ≥ 0 ⇒ (x – 4) (x + 4) ≤ 0 ⇒–4 ≤x≤4  ...(2) The domain of f is the common part of (1) and (2) i.e. 3 < x ≤ 4.  Thus Domain of f = (3, 4]. 33. (d) f (x) is defined if cos x > 0, x > 0 and x ≠ 1 π π 0 and x ≠ 1 ⇒ – 2 2 π  ⇒  0 < x <  \{1} 2 2



 π Domain ( f ) =  0,  \{1}. 2

34. (a) f (x) is defined if – 1 ≤

2

4 ≤1 3 + 2 cos x

Since 3 + 2 cos x > 0 for 4 ≤ 1 ⇒ cos x ≥ 3 + 2 cos x π ≤ x ≤ 2nπ + ⇒ 2nπ – 6

all real x, this gives 1 2 π , n ∈ I. 6

35. (b) ∵  {x} ∈ [0, 1) ∴  sin {x} ∈ [0, sin 1) but f (x) is defined if sin {x} ≠ 0

 1   1  1 ∈ , ∞ . ∴   ∈{1, 2, 3.....} . sin{x}  sin 1   sin{x} 



36. (b) tan–1 x ( x + 1) is defined if



x (x + 1) ≥ 0  or  x ≤ – 1 or x ≥ 0  sin

–1

...(1)

x + x + 1 is defined if 0 ≤ x2 + x + 1 ≤ 1. 2

2

1 3  Now x 2 + x + 1 =  x +  + ≥ 0  2 4 for all real x  Also, x2 + x + 1 ≤ 1 ⇒ x (x + 1) ≤ 0

...(2)

24

⇒–1 ≤x≤0  From (1), (2) and (3), we get domain of  f = {– 1, 0}.

...(3)

Objective Mathematics



| x | ≤ 3 ⇒ – 3 ≤ x ≤ 3.



Also



1 − | x|  cos–1   ≥0 2 

37. (a) 3 1 − 3 x and e3 tan x is defined for all real x.  2x − 1 2x − 1 ≤1   is defined if – 1 ≤ 3  3



3 cos–1



⇒ – 3 ≤ 2x – 1 ≤ 3 ⇒ – 1 ≤ x ≤ 2 Domain of f  = [– 1, 2].

1  1 1 > 0 i.e. x > 38. (b) log1/ 2  x −  is defined if x – 2 2 2

log 2 4 x 2 − 4 x + 5 is defined if 4x2 – 4x + 5 > 0 2   1 x − + 1 > 0    ⇒ 4   2   which is true for all real x.

40. (b) f (x) is defined if | sin x | + sin x > 0 ⇒ sin x > 0 ⇒ 2nπ < x < 2nπ + π Domain of f = (2nπ, (2n + 1) π). 2 f ( n) + 1 and f (1) = 2 2 ∴  f (101) = f (1) + 100 × 1/2 = 2 + 50 ∴ f (101) = 52.

41. (a) We have, f (n + 1) =

42. (a) 24 – xC3x – 1 is defined if, 24 – x > 0, 3x – 1 ≥ 0 and 24 – x ≥ 3x – 1 1 25 and x ≤ ⇒ x < 24, x ≥ 3 4 1 25 ⇒ ≤x≤  ...(1) 3 4 40 – 6x C8x – 10 is defined if 40 – 6x > 0, 8x – 10 ≥ 0 and 40 – 6x ≥ 8x – 10



20 25 5 ,x≥ and x ≤ 3 7 4 25 5 ⇒ ≤x≤  7 4 25 5 ≤x≤ From (1) and (2) we get 7 4 But 24 – x ∈ N, ∴ x must be an integer,



∴ x = 2, 3.



Hence domain ( f ) = {2, 3}.





x
0,  x +  ≠ 1 2 2  

x2 – 5x + 6 ≠ 0 ⇒ (x – 2) (x – 3) ≠ 0 ⇒ x ≠ 2, 3 ...(1)



1   x + 2  > 0 ⇒ x ≥



1 3  1   x + 2  ≠ 1 ⇒ x ∉  2 , 2   



From (1), (2) and (3), we get domain of f



3  =  , 2  ∪ (2, 3) ∪ (3, ∞). 2



...(2) ...(3)

46. (b) f (x) is defined if – 1 ≤ x + [x] ≤ 1 Case I. x is an integer. Let x = n, n ∈ I. Then x + [x] = n + n = 2n. 1 1 So, we get, – 1 ≤ 2n ≤ 1 ⇒ – ≤n≤ 2 2 ...(2) ⇒ n = 0 ...(1) Case II. x is not an integer. Let x = n + k, n ∈ I, 0 < k < 1 Then, x + [x] = n + k + [n + k] = 2n + k So, we get, – 1 ≤ 2n + k ≤ 1 ⇒ n = 0 ∴ x = k, 0 < k < 1 ...(2) From (1) and (2), we get Domain of ( f ) = [0, 1). 47. (d) f (x) is defined if (1) x > 0 (2) log2 x > 0 ⇒ x > 20 = 1. (3) log2 log2 x > 0 ⇒ log2 x > 20 = 1 ⇒ x > 21 = 2.



⇒ log2 x > 22 ⇒ x >



Continuing like this, we get x > 22



∴ Domain of f = ( 22





2...( n −1) times

2...( n −1) times

, ∞).

48. (a) We have,

f (x) =



=



f (x) is defined if x – 4 ≥ 0  or  x ≥ 4. Domain of  f = [4, ∞).

x −3− 2 x − 4 − x −3+ 2 x − 4 x − 4 +1

49. (c) We have, a f (x) + g (x) = 0, a > 0 Since minimum values of g (x) is 1/2 ∴ g (x) > 0 and a f (x) > 0 x. ∴ a f (x) + g (x) > 0, Hence number of solutions is zero.

16 − x 2 >0 50. (b) f (x) is defined if 3− x ⇒ 16 – x2 > 0 and 3 – x > 0 ⇒ (x – 4) (x + 4) < 0 and x < 3 ⇒ – 4 < x < 4 and x < 3  or  – 4 < x < 3 Domain of f = (– 4, 3). 51. (a) f (x) is defined if x + 25 ≥ 0, x ≠ 0, x – 5 ≥ 0, x – 3 ≥ 0 ⇒ x ≥ – 25, x ≠ 0, x ≥ 5, x ≥ 3 ⇒ x ≥ 5. Domain of f = [5, ∞). 52. (c) Since f : (4, 6) → (6, 8) ⇒ f (x) = x + 2 ∴ f – 1 (x) = x – 2. 53. (c)   f (x) is defined if  1  1 log3  ≥ 0, | cos x | > 0 and | cos x | ≠ 0  | cos x |  1 0 | cos x | ≥ 3 = 1 and | cos x | ≠ 0







  1 ∵ | cos x | > 0 for all real x 



  1 ≥ 1 for all real x  ⇒ | cos x | ≠ 0 ∵  | cos x | 



π ,n∈I 2 π   Domain of f = R\ (2n + 1) : n ∈ I . 2   ⇒ x ≠ (2n + 1)

54. (b) f (x) is defined if 1   1 – log 1 2 1 + 1 5  – 1 > 0, 1 + 1 5 > 0, x ≠ 0 x x

1   ⇒ log 1 2 1 + 1 5  < – 1, x1/5 + 1 > 0, x ≠ 0 x −1





x − 4 −1 −



25

(4) log2 log2 log2 x > 0 ⇒ log2 log2 x > 20 = 1 ⇒ log2 x > 21 = 2 ⇒ x > 22 (5) log2 log2 log2 log2 x > 0 ⇒ log2 log2 log2 x > 20 = 1 ⇒ log2 log2 x > 21



1 1 ⇒ 1 + 1 5 >   , x > (– 1)5, x ≠ 0 2 x 1 > 1, x > – 1 and x ≠ 0 ⇒ x1 5 ⇒ 0 < x < 1 and x > – 1 ∴ Domain ( f ) = (0, 1).

Functions



⇒ 0 < x < 1.

55. (d) We have, f (x) = 2 min {| f (x) – g (x)|, 0}

f ( x) > g ( x) 0 = 2( f ( x) − g ( x)), f ( x) ≤ g ( x)   f ( x) − g ( x) − | f ( x) − g ( x) |, =  f ( x) − g ( x) − | f ( x) − g ( x) |, 

f ( x) > g ( x) f ( x) ≤ g ( x)

∴ f (x) = f (x) – g (x) – | g (x) – f (x) |. 56. (c) We have, x f (x) = 1 − x, 

x ∈Q x ∉Q

 f ( x), ∴  f 0 f (x) = 1 − f ( x)   x,  1 − x,  = 1 − x,  1 − (1 − x),

x ∈ Q, x ∉ Q, 1 − x ∈ Q, 1 − x ∉ Q,

f ( x) ∈ Q , f ( x) ∉ Q x ∈Q x ∈Q x ∉Q x ∉Q

 x, x ∈ Q =  x, x ∉ Q  ∴  f (x) = x, x ∈ [0, 1]. 57. (c) f (x) is defined if – (log3 x)2 + 5 log3 x – 6 > 0 and x > 0 ⇒ (log3 x – 3) (2 – log3 x) > 0 and x > 0 ⇒ (log3 x – 2) (log3 x – 3) < 0 and x > 0 ⇒ 2 < log3 x < 3 and x > 0 ⇒ 32 < x < 33 ⇒ 9 < x < 27 Domain of f = (9, 27). 58. (a) f (x) is defined if log x ≠ 0 and x > 0 ⇒ x ≠ 100 = 1 and x > 0 Domain of f  = (0, 1) ∪ (1, ∞) = (0, ∞) \ {1}. 59. (c) f (x) is defined if

4 − x2 ≥ 0 and [x] + 2 ≠ 0 [ x] + 2

Case I. 4 – x2 ≥ 0 and [x] + 2 > 0 ⇒ (x – 2) (x + 2) ≤ 0 and [x] > – 2 ⇒ – 2 ≤ x ≤ 2 and x ≥ – 1 ⇒ – 1 ≤ x ≤ 2  Case II. 4 – x2 ≤ 0, [x] + 2 < 0 ⇒ (x – 2) (x + 2) ≥ 0 and [x] < – 2 ⇒ (x ≤ – 2 or x ≥ 2) and x < – 2

...(1)

26

⇒ x < – 2 

Objective Mathematics



...(2)

From (1) and (2), we get domain ( f ) = (– ∞, – 2) ∪ [– 1, 2].

60. (b) Clearly, y is defined for all real x. ∴ Domain of y = (– ∞, ∞). x ⇒ y + x 2y = x We have, y = 1 + x2 or x2y – x + y = 0 For x to be real, 1 – 4y2 ≥ 0, y ≠ 0⇒ (1 – 2y) (1 + 2y) ≥ 0, y ≠ 0 ⇒ (2y – 1) (2y + 1) ≤ 0, y ≠ 0 1 1 ,y≠0 ⇒ – ≤ y ≤ 2 2 Also, if y = 0, we get x = 0 so that y = 0 is also in the range.  1 1 ∴ Range of y =  − ,  .  2 2 61. (a) Clearly, y is defined for all real x. ∴ Domain of y = (– ∞, ∞) x2 We have, y = ⇒ x 2y + y = x 2 1 + x2 y y i.e. x = 1 − y = 1− y

y (1 − y ) (1 − y )



⇒ x2 =



For x to be real, 1 – y ≠ 0 i.e. y ≠ 1 and y (1 – y) ≥ 0 ⇒ y (y – 1) ≤ 0 and y ≠ 1 ⇒ 0 ≤ y < 1 Range of y = [0, 1).

62. (c) We have, y =

1  2 − sin 3 x



Domain of y = (– ∞, ∞)



From (1), 2 – sin 3x =



...(1)

1 ⇒ sin 3x = y 2y −1 1   2 y − 1 y ⇒x= sin–1  y  3 1   ∵ y = 2 − sin 3 x ∴ y > 0     as − 1 ≤ sin 3 x ≤ 1



For x to be real, – 1 ≤



⇒ – y ≤ 2y – 1 ≤ y





 11  Range of f = log e , ∞  . 3  π2 – x2 ≥ 0 64. (a) For y to be defined, 16



2  π  π π π − x 2 ∈ 0,  . Clearly, for x ∈  − ,  ,  4  4 4  16



 π Since sin x is an increasing function on 0,  .  4



Therefore, sin 0 ≤ sin



⇒ 0 ≤ 3 sin



⇒0 ≤y≤



3   ∴ Range of y = 0, . 2  

⇒ 2y – 1 ≥ – y and 2y – 1 ≤ y 1 ⇒ y ≥ and y ≤ 1 3 1  ∴ Range of y =  , 1 . 3 

2

3 2

which is true for all real x ∴ Domain ( f ) = (– ∞, ∞)



Let y = 3 x 2 − 4 x + 5 ⇒ y2 = 3x2 – 4x + 5 i.e. 3x2 – 4x + (5 – y2) = 0 For x to be real, 16 – 12 (5 – y2) ≥ 0



⇒y≥

11 3

 11  , ∞ ∴ Range of y =  3 .  2 66. (b) f (x) is defined if x + x – 6 ≠ 0 i.e. (x + 3) (x – 2) ≠ 0 i.e. x ≠ – 3, 2 ∴ Domain ( f ) = (– ∞, ∞) \ {– 3, 2}. x 2 − 3x + 2 Let y = x2 + x − 6 2 ⇒ x y + xy – 6y = x2 – 3x + 2

 2  11 4 5 ⇒ 3   x 2 − x +  > 0 ⇒ 3    x −  +  > 0,    3 9  3 3 

π2 − x2 16

π2 − x 2 ≤ sin π 16 4 3 ≤ 2

65. (c) f (x) is defined if 3x2 – 4x + 5 ≥ 0 2  2  11 5  2 4 x − +  ≥0    x x − + ≥ 0 ⇒ 3  ⇒ 3 3 9  3 3   



2y −1 ≤1 y ( ∵ y > 0)

π π  π  π  ⇒  − x   + x  ≥ 0 ⇒  x −   x +  ≤ 0 4 4 4 4 π π ≤x≤ ⇒– 4 4  π π ∴ Domain of y =  − ,   4 4





63. (b) f (x) is defined if 3x2 – 4x + 5 > 0



which is true for all real x. ∴ Domain ( f ) = (– ∞, ∞) Let y = loge (3x2 – 4x + 5) ⇒ ey = 3x2 – 4x + 5. ⇒ 3x2 – 4x + (5 – ey ) = 0 For x to be real, 11 16 – 12 (5 – ey) ≥ 0 ⇒ 12 ey ≥ 44 ⇒ ey ≥ 3 11 ⇒ y ≥ loge 3

 x2  67. (b) Clearly, for y to be defined,  ≤1  1 + x 2  which is true for all x ∈ R. So, the domain = (– ∞, ∞).



 x2  x 2 = sin y Now, y = sin–1  2  ⇒ 1 + x  1 + x2 sin y ⇒x= 1 − sin y For x to be real, sin y ≥ 0 and 1 – sin y > 0

 π ⇒ 0 ≤ sin y < 1 ⇒ y ∈ 0,   2  π ∴ Range = 0,  .  2 68. (a) We have f (x) = cos x – x (1 + x)

π π Since, in the interval  ,  , cos x decreases and 6 3 x (1 + x) increases,

π π ∴ f (x) decreases in  ,  . 6 3



π π π π ∴ f   ≤ f (x) ≤ f   , x ∈  ,  3 6 6 3



 1 π  π  3 π  π  Hence f (A) =  − 1 +  , − 1 +    2 3  3  2 6  6 



 1 π π2 3 π π2  =  − − , − − .  2 3 9 2 6 36 



 π π ∴ Domain of f =  − ,  .  3 3



The greatest value of f (x) = tan



y y π + sin −1 ⇒x= . 4 2 2 y ≤1 For x to be real, – 1 ≤ 2 x –

π = sin–1 4



⇒ – 2≤ y ≤ 2.



∴ Range of y = [– 2 , 2 ].

72. (c) Clearly, f (x) is defined for all real x except at x = 0. ∴ Domain of f = (– ∞, ∞) \ {0} x Now, for x > 0, f (x) = = 1 and for x < 0, f (x) x x = – 1. = −x 0 At x = 0, f (x) = = not defined. 0 Thus, f (x) has only two values, namely 1 and – 1. ∴ Range of f = {– 1, 1}. 73. (b) We have, f (2x + 3) + f (2x + 7) = 2 Replace x by x + 1, f (2x + 5) + f (2x + 9) = 2 Now replace x by x + 2, f (2x + 7) + f (2x + 11) = 2  From (1) – (3),

Since

...(1) ...(2) ...(3)

4 − x2

0, ∴ we have

1 – x > 0 and 4 – x2 > 0

π2 − 0 , when 9 π2 π2 − 9 9

π . 3

∴ The greatest value of f (x) = 3 and the least value of f (x) = 0 ∴ Range of f = [0, 3 ]. 70. (a) We have,  f (x) = log (sin {x}) ⇒   f (x + 1) = f (x) ⇒ period of f = 1. –1

⇒ x < 1 and (x – 2) (x + 2) < 0 ⇒ x < 1 and – 2 < x < 2 ⇒ – 2 < x < 1 ∴ Domain of f = (– 2, 1).  4 − x2  Since – ∞ < log  < ∞  1− x  ⇒

  4 − x2 – 1 ≤ sin log    1− x

  ≤ 1  

∴ Range of f = [– 1, 1]. 75. (d) The graph of f (x) has the equation y = (x + 1)2, x ≥ –1. The reflection of the graph of f (x) is obtained by

27



4 − x2 > 0, 4 – x2 > 0 and 1 – x ≠ 0 1− x

x = 0 and the least value of f (x) = tan , when x =



π  2 sin  x −   4

74. (c) For f (x) to be defined,

π π ≤x≤ . – 3 3



Let y =

we get f (2x + 3) – f (2x + 11) = 0

π – x2 ≥ 0 9





i.e.,  f (2x + 3) = f (2x + 11) ⇒ T = 4.

2

69. (a) For f (x) to be defined,



π  2 sin  x −  .  4 Clearly, f (x) is defined for all real x. ∴ Domain of f = (– ∞, ∞).

71. (b) We have, f (x) = sin x – cos x =

Functions

⇒ x2 (y – 1) + x (y + 3) – (6y + 2) = 0 For x to be real, (y + 3)2 + 4 (y – 1)  (6y + 2) ≥ 0 ⇒ 25y2 – 10y + 1 ≥ 0 i.e. (5y – 1)2 ≥ 0 which is true for all real y. Range of f = (– ∞, ∞).

28

interchanging x, y. So, the graph of g(x) has the equation x = (y + 1)2, y ≥ –1

Objective Mathematics

∴ f (x) = sin4 2x + cos4 2x is a periodic function 1 π with period ⋅ i.e. π . 2 2 4 82. (a) Let f (T + x) = f (x) ⇒ tan (T + x) = tan x ⇒ tan (T + x) = tan x ⇒ T + x = nπ + x, n ∈ I Clearly, from here, the least positive value of T independent of x is π. Therefore, f (x) is a periodic function of period π. 83. (c) cos x is a periodic function with period 2π, therefore  8x + 5  f (x) = cos    will be a periodic function with 4π 

∴ y = x − 1 because y ≥ –1 ∴ φ(x) = x − 1 , x ≥ 0. 76. (a) f ′(x) = 2 + cos x > 0. So, f (x) is strictly m.i. so, f (x) is one-to-one and only. 77. (b) We have, f (x) = a sin kx + b cos kx

2π = π 2. 8 4π 84. (b) Let  f (T + x) = f (x) ⇒  (T + x) – [T + x] = x – [x] ⇒  T = [T + x] – [x] = an integer. Clearly, from here, the least positive value of T independent of x is 1. Hence, f (x) is a periodic function of period 1. period

=

  a b sin kx + cos kx  a 2 + b2  2 2 2 2  a +b  a +b

85. (c) Let g (x) = e3{x} ⇒ T1 = 1

=

a 2 + b 2 (cos θ sin kx + sin θ cos kx),



and f (x) = e{3x} ⇒ T2 = 1/3



∴  T1 = 3T2.

where cos θ = =

a +b 2

2

a a 2 + b2

86. (b) We have, ex + ef (x) = e ⇒ ef (x) = e – ex

sin (kx + θ),

⇒ f (x) = log (e – ex). For f (x) to be defined, e – ex > 0 ⇒ e1 > ex ⇒x 0 ⇒ e1 > ey ⇒ y < 1 ∴ Range of f = (– ∞, 1).

2π which is a periodic function of period | k | . 78. (d) Let f (T + x) = f (x) ⇒ cos (T + x)2 = cos x2 ⇒ (T + x)2 = 2nπ ± x2 Clearly, from here, no positive value of T independent of x is possible because x2 on R.H.S. can be cancelled only when T = 0. ∴ f (x) is a non periodic function.

87. (a) sin x is a periodic function with period 2π, therefore

79. (b) Since, | sin x | + | cos x | is a periodic function with π , ∴ f (x) = | sin 4x | + | cos 4x | is a periodic 2 π 1 π i.e. . function with period 8 4 2

period

80. (d) Let f (T + x) = f (x) ⇒ sin

T + x = sin x

sin ( [n] x) is a periodic function with period

2π . [ n]

But the period of f (x) is 2π (given).

2π = 2π ⇒ [n] = 1 [ n] ⇒ [n] = 1 ⇒ 1 ≤ n < 2.





88. (a) f (x) = (– 1)[x] = {– 1, 1}, since [x] ∈ I. ⇒ T + x = nπ + (– 1)n x Clearly, from here, no positive value of T independ- 89. (a) Since sin x and cos x are periodic functions with period 2π. ent of x is possible because x on RHS can be 2π cancelled only when T = 0.  πx  = 2 (n!) ∴ Period of cos   = π   n! ∴ f (x) is a non-periodic function. n! and period of sin 81. (c) Since sin4 x + cos4 x is a periodic function with π  πx  2π period , = 2 (n + 1)!  (n + 1)!  = π 2 (n + 1)!



2 (n + 1)!} = 2 (n + 1)!.

90. (a) Since | sin x | + | cos x | is a periodic function with π π period , therefore period of f (x) will be when 2 2 k = 1. 91. (b) sin2 πx =

1 − cos 2 πx . 2

96. (b) 2 sin x is a periodic function of period 2π and 3 cos 2x is a periodic function of period π.

∴  f (x) is a periodic function of period = L.C.M. Since cos 2πx is a periodic function will pe{2π, π} = 2π. 2π riod = 1, therefore sin 2 πx is periodic with 97. (d) For every rational number T, we have 2π period 1. ...(1) 1, when x is a rational f (T + x) = 0, when x is irrational , x – [x] is a periodic function with period 1. ...(2)   1 sin4 πx = (sin2 πx)2 = (1 – cos 2πx)2 but there is no least positive value of T for which 4 f (T + x) = f (x) because there are infinite number of 1 = (1 + cos2 2πx – 2 cos 2πx) rational numbers between any two rational numbers. 4 Therefore, f (x) is a periodic function having no 1 fundamental period. = (3 + cos 4πx – 4 cos 2πx) 8  x  −1  S ince cos 4πx is a periodic function with period 98. (a) f (x) = sin  log 3    exists if  3  2π 1 = and cos 2πx is a periodic function with period 4π 2 2π = 1, therefore, period of sin4 πx is equal to 2π  1  L.C.M. (1, 1) 1 ...(3) L.C.M. 1,  = H.C.F. (1, 2) = 1 = 1 2

x x −1≤ log 3   ≤ 1 ⇔ 3−1 ≤ ≤ 31 ⇔ 1 ≤ x ≤ 9. 3 3 99. (d) Since cos

From (1), (2) and (3), we get Period of 3(sin

2 πx + x − [ x ] + sin 4 πx )

= 1.

1 − cos 2θ 2π , ∴  period = = π. 2 2 93. (c) We have, |sin( π + x)| − |cos ( π + x)| f (π + x) = |sin( π + x) + cos( π + x)|

92. (b) sin2θ =





f (x + k) = 1 + [1 + {1 – f (x)}5]1/5 ⇒ f (x + k) – 1 = [1 – ( f (x) – 1)5]1/5 ⇒ φ  ( x + k ) = [ 1 – { φ  ( x ) } 5 ] 1 / 5 ,  where φ (x) = f (x) – 1 ⇒ φ (x + 2k) = [1 – {φ (x + k)}5]1/5 ⇒ φ (x + 2k) = [1 – {1 – (φ (x)) 5 }] 1/5  = φ (x), ∨ x ∈ R ⇒ f (x + 2k) – 1 = f (x) – 1 ⇒ f (x + 2k) = f (x), ∨ x ∈ R ∴ f (x) is periodic with period 2k.

∴ f (x) is periodic with period π. 94. (b) For f (x) to be defined,

⇒  (x – 3) (x + 2) ≠ 0, x + ⇒ x ≠ 3, – 2, x ≥

1 1 ≥ 1, x + ∉ [1, 2) 2 2

1 3  1 , x ∉  ,  2 2 2

3  ⇒ x ∈  , 3  ∪ (3, ∞). 2 95. (a) Since the domain of f is (0, 1), ∴ 0 < ex < 1 and 0 < ln | x | < 1

2π πx 2 π = 5 and sin 4 is a periodic function with 5 2π period π = 8. 4 ∴ f (x) is periodic with period = L.C.M. (5, 8) = 40.

100. (b) We have,

|sin x | − |cos x | = |sin x + cos x | = f (x) for all x.

1 1   x2 – x – 6 ≠ 0,  x +  > 0,  x +  ≠ 1 2 2  

2 πx is a periodic function with period 5

101. (b) f (x) = sin4x + cos4x = (sin2x + cos2x)2 – 2 sin2x cos2x

1 1 1 − cos 4 x  (sin 2 x) 2 = 1 −   2 2 2 3 1 = + cos 4 x 4 4 Since, cos x is periodic with period 2π = 1−

∴  period of f (x) =

2π π = . 4 2

29

⇒ log 0 < x < log 1 and e0 < | x | < e1 ⇒ – ∞ < x < 0 and 1 < | x | < e ⇒ x ∈ (– ∞, 0) and x ∈ ((– ∞, – 1) ∪ (1, ∞)) ∩ (– e, e) ⇒ x ∈ (– ∞, 0) and x ∈ (– e, – 1) ∪ (1, e) ⇒ x ∈ (– e, – 1).

∴ Period of f (x) = L.C.M. of {2 (n!),

Functions



30

2 πx   πx  102. (b) Period of 2 cos  + 3 sin    3   3 

Objective Mathematics



 2π 2π  = L.C.M.  ,  = 6.  2 π π  3 3



π π   Period of 4 cos   2 πx +  + 2 sin  πx +    2 4



2π 2π  = L.C.M.  , =2  2 π π 

 2π  Period of | tan x | + cos 2x = L.C.M.  π,  = π.  2  103. (a) The function f (x) =

110. (d) The period of | sin x | + | cos x | and sin4x + cos4x π . sin (sin x) + sin (cos x) has period 2π. The 2 1 + 2 cos x can be written in a simplifunction sin x (2 + sec x) cos x = cot x, so it has period π. fied form as sin x π 111. (c) tan (3x – 2) is a periodic function with period . 3 The function f (x) = {x} is periodic with period 1. The function in (d) can be written as is

f (x) = 1 –

 5x − x2  exists for log10   4 



5x − x2 ≥ 1 ⇒ 5x – x2 – 4 ≥ 0 ⇒ x ∈ [1, 4]. 4

=

104. (a) Since the domain of f is (0, 1), ∴ 0 < sin x < 1 ⇒ 2nπ < x < (2n + 1) π, n ∈ I. 105. (c) Let n ≤ x < n + 1 Then, f (x) = x ⋅ n, where n changes with x. Clearly no constant k > 0 is possible for which f (x) = f (x + k) corresponding to all x. ∴ f (x) is a non periodic function. 106. (b) The period of f (x) =

π [k ] .

But the period of f (x) = π (Given)

π [k ] = π =

=

cos3 x sin 3 x − sin x + cos x sin x + cos x =1–

sin 3 x + cos3 x sin x + cos x

2 2 1 – (sin x + cos x)(sin x + cos x − sin x cos x) (sin x + cos x) 1  1  1 – 1 − sin 2 x  = sin 2x  2  2

which is periodic with period π. The function x + cos x is non periodic as x is non periodic. 112. (b) We have, 2 sin 8 x cos x − 2 sin 6 x cos 3x f (x) = 2 cos 2 x cos x − 2 sin 3x sin 4 x

=

(sin 9 x + sin 7 x) − (sin 9 x + sin 3x) (cos 3 x + cos x) + (cos 7 x − cos x)

sin 7 x − sin 3 x 2 cos 5 x sin 2 x = = tan 2x, cos 7 x + cos 3 x 2 cos 5 x cos 2 x π ⇒ [k] = 1 ⇒ 1 ≤ k < 2. . which is periodic with period 2 2π 107. (c) The period of sin 5x is and that of cos 3x 113. (a) We have 5 (sin 5 x + sin x) + (sin 4 x + sin 2 x) 2π 2π f (x) = 2π (cos 5 x + cos x) + (cos 4 x + cos 2 x) is . As and do not have a common 3 3 5 2 sin 3x cos 2 x + 2 sin 3x cos x = multiple, f (x) is non periodic. 2 cos 3x cos 2 x + 2 cos 3x cos x πx has 108. (a) 3x + 3 – [3x + 3] has the period 1 and sin 2 sin 3 x (cos 2 x + cos x) 2 = 2 cos 3 x (cos 2 x + cos x) = tan 3x, 2π i.e., 4. Therefore, the period of f (x) the period π π 2 . which is periodic with period 3 is L.C.M. (1, 4) = 4. 114. (b) We have, f (x + 2a) = f ((x + a) + a) 109. (a) For n = 2, we have 1 2 = – f ( x + a ) − ( f ( x + a )) 4 sin x 2 cos x 2 cos x sin 2 x 2 sin x cos x 2 = = f (x) = x x x 1 sin   sin   sin   = 2 2 2 2 x 2 = 4 cos cos x. 1 1  2 − f ( x) − ( f ( x)) 2 −  − f ( x) − ( f ( x)) 2  – x 2  2 The period of cos x is 2π and that of cos is 2 1 1 1 1 2π − f ( x) + ( f ( x)) 2 = −  − f ( x)  = f (x). = – = 4π. 4  2 2 2 1/ 2 ∴

[k ] = 1

Hence the period of f (x) is 4π.



=

Hence, f (x) is periodic with period 2a.

r =1

= f (1) + 2f (1) + 3f (1) + ... + n f (1) [since f (x + y) = f (x) + f ( y)] = (1 + 2 + 3 + ... + n) f (1) =

n (n + 1) 7 n (n + 1) . ⋅7 = 2 2

−1 ± 5  ...(1) 2 Since f (x) is an even function defined on [– 5, 5], ⇒x=



f (– x) = f (x), ∨ x ∈ [– 5, 5]

 x +1  ⇒ x = –   x + 2

therefore sin (log (x +

x 2 + 1 )) must be odd.

123. (c) A function whose graph is symmetrical about the origin must be odd. (3x + 3– x) is an even function. Since cos x is an even function and log (x +

1 + x2 ) 1 + x 2 )) is

is an odd function, ∴ cos (log (x + an even function. If f (x + y) = f (x) + f (y) ∨ x, y ∈ R, then f (x) must be odd (See Q. 122).

⇒ x2 + 3x + 1 = 0

−3 ± 5 . 2 From (1) and (2), the values of x are

⇒ x=

x 2 + 1 ) are odd functions,

4 4 Also, sec x + cosec x is an odd function. 3 4 x + x cot x Now, let f (x + y) = f (x) + f (y) ∨ x, y ∈ R ∴ f (0 + 0) = f (0) + f (0) ∴ f (0) = 0. f (x – x) = f (x) + f (– x) or 0 = f (x) + f (– x) i.e. f (– x) = – f (x) ∴ f (x) is odd.

 x +1  x +1 117. (a), (b) Since f (x) = f    ⇒ x = x + 2 x+2 ⇒ x2 + x – 1 = 0

Since sin x and log (x +

..(2)

124. (c) g( f (x)) = g(sin x + cos x)  π π It will be invertible when x ∈  − ,  .  4 4 125. (d) –2 ≤ sin x – 3 cos x ≤ 2

−1 ± 5 −3 ± 5 . and 2 2 ⇒ –1 ≤ sin x – 3 cos x +1 ≤ 3 118. (a) Since the function sec x is an even function and 2 ∴ range of f = [–1, 3] log (x + 1 + x ) is an odd function, therefore For f to be onto S = [–1, 3] 2 the function sec [log (x + 1 + x )] is an even function. 126. (b) Graph is symmetrical about the line x = 2 ⇒ f (2 – x) = f (2 + x) 3 + log10 ( x 3 − x) 119. (d) f (x) = 2 [Graph is symmetrical about y-axis 4− x For domain of f (x) i.e., about x = 0 iff f (–x) = f (x) x3 – x > 0 i.e., f (0 – x) = f (0 + x)]. ⇒ x (x – 1) (x + 1) > 0. x1 = x2 127. (d) Let f (x1) = f (x2) ⇒ [x1] = [x2] ⇒  the region is (– 1, 0) ∪ (1, ∞) 2 Also,  4 – x ≠ 0 ⇒ x ≠ ± 2 [For example, if x1 = 1.4 and x2 = 1.5, then [1.4] = [1.5] = 1] ⇒  the region is (– ∞, 2) ∪ (2, ∞) ∴ f is not one-one. ∴  Common region is (– 1, 0) ∪ (1, 2) ∪ (2, ∞).

31

Functions

120. (a) To make f (x) an even function in the interval n − 1 n , where is odd [– 1, 1], we re-define f (x) as follows:  2 115. (c) f (n) =   f ( x), 0 ≤ x ≤ 1   f (x) =  − n , where n is even  f (− x), − 1 ≤ x < 0  2 3 x 2 − 4 x + 8 log (1 + | x |), 0 ≤ x ≤ 1  and f : N → I, where N is the set of natural numbers =  2  and I is the set of integers. 3 x + 4 x + 8 log (1 + | x |), − 1 ≤ x < 0 Let x, y, ∈ N and both are even, then f (x) = f ( y) Thus, the even extension of f (x) to the interval x y = –  ⇒ x = y. Again x, y ∈ N and both ⇒– [– 1, 1] is 3x2 + 4x + 8 log (1 + | x | ). 2 2 121. (b) f (x) = log( x + x 2 + 1) are odd, then f (x) = f ( y) 2 ⇒ f (– x) = log(− x + x + 1) x −1 y −1 ⇒ ⇒x=y = ⇒ f (x) + f (– x) = 2 2 i.e., mapping is one-one. log( x + x 2 + 1) + log(− x + x 2 + 1) Since each negative integer is an image of even = log (1) = 0 natural number and positive integer is an image of Hence f (x) is an odd function. odd natural number. So mapping is onto. Hence, mapping is one-one onto. 122. (d) A function whose graph is symmetrical about the n y-axis must be even. 116. (d) ∑ f (r ) = f (1) + f (2) + f (3) + ... + f (n)

32

Objective Mathematics

Also, f is not onto as its range I (set of integers) is a proper subset of its co-domain R.

134. (c) The total number of bijective functions from a set A containing 106 elements to itself is (106)!.

128. (d) Since sin (2x – 3) is a periodic function with period π, therefore f is not injective. Also, f is not surjective as its range [– 1, 1] is a proper subset of its co-domain R.

135. (b) We know that if X and Y are any two finite sets having m and n elements respectively, where 1 ≤ n ≤ m, then the number of onto functions from X to Y is given by

129. (a) The function f (x) = x5 is not a surjective map from I to itself as 3 ∈ I does not have any preimage in I. The function f (x) = 3x + 2 is not a surjection from I to itself as 4 ∈ I does not have any pre-image in I. The function f (x) = x2 + x is not one-one as f (– 4) = f (3) but – 4 ≠ 3. The function f (x) = x + 3 is a bijection from I to itself. 130. (b) The function f (x) = x2 + 2, x ∈ (– ∞, ∞) is not injective as f (1) = f (– 1) but 1 ≠ – 1. The function f (x) = (x – 4) (x – 5), x ∈ (– ∞, ∞) is not one-one as f (4) = f (5) but 4 ≠ 5.



n

∑ (−1)

n−r n

r =1

Cr r m .

Thus, the number of surjective mappings is 2



∑ (−1) r =1

2− r 2

C r r n = (2n – 2).

136. (c) We know that if X and Y are any two finite sets having m and n elements respectively, then the number of one-one functions from X to Y is  n Pm if n ≥ m  given by =   . Thus, the number of  0 if n < m 4 injective mappings is P3 = 24.

4 x 2 + 3x − 5 , x ∈ (– ∞, ∞) is 137. (c) Let t = 5x + 2, then A = {t : 0 ≤ t ≤ π} 4 + 3x − 5 x 2 ∴  f (t) = cos t which is bijective in [0, π]. also not injective as f (1) = f (– 1) but 1 ≠ – 1. Hence, f (x) is bijective. For the function, f (x) = | x + 2 |, x ∈ [– 2, ∞). Let f (x) = f (y), x, y ∈ [– 2, ∞) ⇒ | x + 2 | = 138. (b) Let f (x1) = f (x2) ⇒ x1 – [x1] = x2 – [x2] x1 = x2 for if x1 = 2.7 and x2 = 3.7 | y + 2 | then f (x1) = f (x2) = 0.7 but x1 ≠ x2 ⇒x+2= y+2 ∴ f is not one-one. Clearly, f is onto. ⇒ x = y. So, f is an injection. 139. (c) We have, f (x) = [x]2 + [x + 1] – 3 = [x]2 + [x] + 1 – 3 131. (c) Let x, y be any two elements in A such that = [x]2 + [x] – 2 = ([x] + 2) ([x] – 1) f (x) = f (y) ⇒ x | x | = y | y | ⇒ x = y ∴ f is injective. f (x) = 0 ⇒ [x] + 2 = 0 or [x] – 1 = 0 i.e., [x] = – 2 or [x] = 1 Also, range of f = f (A) = B, ∴ f is surjective. Hence, f is a bijection. [x] = – 2 ⇒ – 2 ≤ x < – 1 [x] = 1 ⇒ 1 ≤ x < 2. 132. (b) Since the function f is bijective, therefore f is onto. Therefore range of f = B. 140. (a) Since, for different x, 4x and 4| x | are different Let y = – x2 + 6x – 8 ⇒ x2 – 6x + (8 + y) = 0 positive numbers, ∴ f is one-one. Also, f is not onto as its range 6 ± 36 − 4 (8 + y ) 6 ± 4 − 4y (0, ∞) is a proper subset of its co-domain R. ⇒ x = = 2 2 141. (d) Since f (x) = f (– x) ∴ f is not one-one. For x to be real, 4 – 4y ≥ 0 ⇒ y ≤ 1 Let y ∈ R. Then f (x) = y ∴ B = range of f = (– ∞, 1]. x2 − 8 ⇒y = 2 133. (a) Let x, y ∈ R be such that x +2 x−a y−a = f (x) = f (y) ⇒ 8 + 2y x−b y−b ⇒ x2 = . 1− y ⇒ (x – a) (y – b) = (x – b) (y – a) For x to be real, (8 + 2y) (1 – y) ≥ 0 ⇒ xy – bx – ay + ab = xy – ax – by + ab and 1 – y ≠ 0 ⇒ (y + 4) (y – 1) ≤ 0 and y ≠ 1 ⇒ (a – b) x = (a – b) y ⇒ x = y ⇒–4 ≤y 1 

Objective Mathematics

161. (c) We have,

x ≤ 1 and | x | ≤ 1 | x |, f (| x |) = 2 − | x |, x ≤ 1 and | x |> 1 =  | x |, − 1 ≤ x ≤ 1  2 − | x |, x < −1



= | 2 – x |, x > 1 ∴

⇒ –

3 1 ≤x≤ 2 2

 3 1 Hence, the domain of f [g (x)] is  − ,  .  2 2 157. (b) ( fog) (x) = f [g (x)] = f ( | 3x + 4 | ). Since the domain of f is [– 3, 5], ∴ – 3 ≤ | 3x + 4 | ≤ 5 ⇒ | 3x + 4 | ≤ 5 ⇒ – 5 ≤ 3x + 4 ≤ 5 ⇒ – 9 ≤ 3x ≤ 1 1 ⇒–3 ≤x≤ . 3 1  ∴ Domain of fog is  −3,  . 3 

=

1− x x −1 = . −x x

=



= (3)3 – 12 ⋅ 3 = 27 – 36 = – 9.



156. (d) f [g (x)] = f ( 1 − 2 x ) = 2 − 1 − 2 x For f [g (x)] to be defined, 1 – 2x ≥ 0 and 2 –

1 158. (a) We have f [ f (x)] = = 1 − f ( x)





2 − | x |, x < −1  f [ f (x)] = | x |, − 1 ≤ x ≤ 1 . | 2 − x |, x > 1 

1 − 2x ≥ 0 1 3 ⇒ x≤ and x ≥ –  2 2

f (a) = 64 a3 + 3

x > 1 and 2 − x ≤ 1 | 2 − x |, f (2 – x) = 2 − (2 − x), x > 1 and 2 − x > 1 (impossible) 

1 1 3 3 = (4a) + a a3



1 1 1   4a +  − 3 ⋅ 4a ⋅  4a +  a a a

1   Since a, b are roots of 4 x + x = 3    ∴ 4a + 1 = 3   a Similarly, f (b) = – 9 ∴ f (a) = f (b) = – 9.

162. (a) Since f (x) = x2 – 1 and g (x) =

1−

1 1− x

2

163. (c) We have f (x) ⋅ f (y) –

 1   x   f   + f ( xy ) 2   y  

= cos (log x) ⋅ cos (log y) –

 1    x   cos log    + cos {log ( xy )} 2  y     

= cos (log x) ⋅ cos (log y) – + cos (log x + log y)}

1

x2 + 1 ∨ x ∈ R

∴ f [g (x)] = [g (x)] – 1 = x + 1 – 1 = x 2, ∨ x ∈ R ∴ [ho ( fog)] (x) = h [( fog) (x)] = h [ f {g (x)}] = h (x2) = x2 [∵ x2 ≥ 0] 2

1 {cos (log x – log y) 2

1 = cos (log x) ⋅ cos (log y) – ⋅ 2 cos (log x) 2 cos (log y) = cos (log x) ⋅ cos (log y) – cos (log x) ⋅ cos (log y) = 0.

164. (d) 1 = x. 165. (b) We have, x −1 1− 1 1 x 3 f (x) + 5 f    = – 3, ∨ x (≠ 0) ∈ R  ...(1)   x x Hence, the graph of the function y = f [ f { f (x)}], x > 1 is a straight line passing through origin. 1 ⇒ 3 f    + 5 f (x) = x – 3  ...(2) 159. (c) Since x – [x] = 0 for all integral values of x, therex –1 fore, the function is not one-one. Hence f  (x) does 1  not exist.  Replacing x by x  160. (c) Let x + y = a and x – y = b Multiplying (1) by 3 and (2) by 5 and subtracting, a+b a−b and y = ⇒ x = we get 2 2 3  9 f (x) – 25 f (x) =  − 9  – (5x – 15) ∴ f (x + y, x – y) = xy x  a+b a−b a 2 − b2 3 ⇒ f (a, b) = = ⋅ ⇒ – 14 f (x) = – 5x + 6 2 2 4 x 2 2 2 2 x −y y −x 1  3  ∴ f (x, y) = and f ( y, x) = ⇒ f (x) =    − + 5 x − 6  , ∨ x (≠ 0) ∈ R. 4 4  14  x ∴

1 f [f {f (x)}] = 1 − f ( f ( x)) =



35

∴ f   π  = sin 9π – sin 5π = 1 – 0 = 1 2 2 f (π) = sin 9π – sin 10π = 0

9π 1 π f    = sin – sin 10π = – 1. 4 2 4 4 172. (c) For x ≤ – 1, 1 – x ≥ 2 and 1 – x ≥ 1 + x ∴ max {(1 – x), 2, (1 + x)} = 1 – x For – 1 < x < 1, 0 < 1 – x < 2 and 0 < 1 + x < 2. ∴ max {(1 + x), 2, (1 + x)} = 2. For x ≥ 1, 1 + x ≥ 2, 1 + x > 1 – x ∴ max {(1 – x), 2, (1 + x)} = 1 + x

167. (b) We have,

 1  = g (x) = f [ f (x)] = f   1 − x 



Also,



 x − 1 = h (x) = f [ f { f (x)}] = f   x 





1 1 1− 1− x

=

x −1 . x

1 = x. x −1 1− x 1 x −1 ⋅ ⋅ x = – 1. f (x) ⋅ g (x) ⋅ h (x) = 1− x x

168. (a) When f (x) = sin2x and g (x) =

x,

( fog) (x) = f [g (x)] = f ( x ) = (sin x )2 and (gof ) (x) = g [ f (x)] = g (sin2x) = | sin x | When f (x) = sin x and g (x) = | x | ( fog) (x) = f (g (x)) = f (| x |) = sin | x | ≠ (sin x )2 When f (x) = x2 and g (x) = sin x ( fog) (x) = f [g (x)] = f (sin x ) = (sin x )2 and (gof ) (x) = g [ f (x)] = g (x2) = sin x 2 = sin | x | ≠ | sin x |. 169. (b) We have,

171. (a) We have, f (x) = sin [π2] x + sin [– π2] x = sin 9x + sin (– 10) x = sin 9x – sin 10x

−1, − 2 ≤ x ≤ 0 f (x) =   x − 1, 0 < x ≤ 2 Since x ∈ [– 2, 2], therefore | x | ∈ [0, 2]. Therefore, f (| x | ) = | x | – 1, ∨ x ∈ [– 2, 2] − x − 1, x ∈[−2, 0]  =   x − 1, x ∈[0, 2] Also, −2≤ x 0, then f (– 1) = 0 and f (1) = 2. ⇒ – a + b = 0 and a + b = 2 ⇒ a = 1 and b = 1. If a < 0, then f (– 1) = 2 and f (1) = 0 ⇒ – a + b = 2 and a + b = 0 ⇒ a = – 1 and b = 1. Hence, f (x) = x + 1 or f (x) = – x + 1. 174. (c) The length of the side of the equilateral triangle is

( x − 0) 2 + ( f ( x) − 0) 2 i.e.

x 2 + ( f ( x)) 2 .

∴ Area of the equilateral triangle =

3 4

[x2 + { f (x)}2] But the area = ...(1)

3 (given) 4

3 2 [x + { f (x)}2] = 4 ⇒ x2 + { f (x)}2 = 1



...(2)

− x − 1 + 1, − 2 ≤ x < 0  g (x) = f (| x |) + | f (x) | =  x − 1 + 1 − x, 0 ≤ x < 1  x − 1 + x − 1, 1 ≤ x ≤ 2   − x, − 2 ≤ x < 0  0 ≤ x 0 =

1 1 ⇒ x ∈ (– ∞, – 1) ∪  − , 0  ∪  , ∞  . 2    2 2  x2 + 1 Thus (a) and (d) are the correct answers. is maximum, i.e. when x2 + 1 is minimum i.e. at x = 0. x+2 183. (a) Let f (x) = 2 ∴ minimum value of f (x) is f (0) = – 1. x − 8x − 4 177. (b) The maximum value of f (x) = cos x + cos ( 2 For f (x) to be defined, x2 – 8x – 4 ≠ 0 ⇒ x ≠ 4 x) is 2 which occurs at x = 0. Also, there is no ±2 value of x for which this value will be attained ∴ Domain of f = R\{4 + 2 ,4–2 } again. x+2 For range, put y = 2 178. (d) We have, 2x + 2y = 2 ⇒ 2y = 2 – 2x x − 8x − 4 log (2 − 2 x ) ⇒ x2y – (8y + 1) x – (4y + 2) = 0 ⇒ y = log 2 For x to be real, (8y + 1)2 + 4y (4y + 2) ≥ 0 and x x y ≠0 For y to be defined, 2 – 2 > 0 ⇒ 2 < 2 x ⇒ 80y2 + 24y + 1 ≥ 0 and y ≠ 0 Since, 2 is an increasing function, therefore we ⇒ (20y + 1) (4y + 1) ≥ 0 and y ≠ 0 get x < 1. 1 ⇒ y ≤ – 179. (a) Since f (x + y) = f (x) ⋅ f ( y) for all natural numbers 4 x and y and f (1) = 2, 1 so, f (2) = f (1 + 1) = [ f (1)]2 = 22  or y ≥ – and y ≠ 0 20 and f (3) = f (2 + 1) = f (2) ⋅ f (1) = 23. Also, if y = 0, we get x + 2 = 0 Continuing this process, we find that f (n) = 2n, ⇒x=–2 k since if f (k) = 2 , then so that y = 0 is also in the range. k k + 1 f (k + 1) = f (k) ⋅ f (1) = 2 ⋅ 2 = 2 . n

n

∑ f (a + k ) = ∑ f (a) f (k )

Now,

k =1

= f (a)

n

∑2

= f (a)

k =1

k

n

∑ f (k ) k =1

n

= 2 f (a) (2 – 1)

...(1)

k =1

n

It is given that

∑ f (a + k ) = 16 (2

n

– 1)

...(2)

k =1

From (1) and (2), we get f (a) = 8. So, f (a) = 23 ⇒ 2a = 23. ∴ a = 3. 180. (a), (c) [π ] = 9 and [– π ] = – 10 Thus, f (x) = cos [π2] x + cos [– π2] x = cos 9x + cos (– 10) x = cos 9x + cos 10x 2

2



π 9π So, f   = cos + cos 5π = – 1 2 2



f (π) = c os 9π + cos 10π = – 1 + 1 = 0 f (– π) = cos (– 9π) + cos (– 10π) = – 1 + 1 = 0



π 9π 10 π f   = cos + cos = 4 4 4

181. (b) g (x) = f [f (x)] = f [ | x – 2 | ] = f (x – 2) as x > 20

1 1 − = 0. 2 2

1  1   ∴ Range of f =  −∞, −  ∪  − , ∞  .  4   20 

184. (b) We have, ( fog) (x) = f [g (x)] =f

(

) (

x2 + 1 =

x2 + 1

)

2

– 1 = x2

∴ (hofog) (x) = h [( fog) (x)] = h (x2) 0 if x = 0 =  2 .  x if x ≠ 0  2 185. (a) For f (x) to be defined, x + 2x ≠ 0 ⇒ x (x + 2) ≠ 0 i.e. x ≠ 0, – 2. ∴  Domain of f = R\{0, – 2} x2 − x , x ≠ 0, – 2 For range, put y = 2 x + 2x 2y +1 x −1 ⇒ yx + 2y = x – 1 or x = , ⇒ y = 1− y x+2 which is not defined for y = 1. 1 and x ≠ – 2 for any real Also, x ≠ 0 ⇒ y ≠ – 2 value of y. 

1 ∴  Range of f = R\ 1, −  . 2 

 x + 2 x > −2   x + 2 = 1,  186. (a) Since, f (x) =  . 2 x +  = −1, x < −2   − ( x + 2)

196. (b) Let f (x) =

∴ Range of f = {1, – 1}. ( x 2 − 3)3 + 27

187. (a) 2 is minimum when (x2 – 3)3 + 27 is minimum. Since (x2 – 3)3 + 27 = x6 – 9x4 + 27x2



1  which clearly contains  , 3  . 2  197. (b) The number of functions = 23 = 8. π 0 i.e. x > 1 ∴ Domain of f = (1, ∞).

= x ⋅

x2 − x + 1 x −1 xy – y = x2 – x + 1 x2 – (y + 1) x + (y + 1) = 0 x to be real, (y + 1)2 – 4 (y + 1) ≥ 0 (y + 1) (y – 3) ≥ 0 ⇒ y ≤ – 1 or y ≥ 3 y cannot lie between – 1 and 3.

222. (a) Let y =

216. (d) Since f (x) is an odd periodic function with period  2 ∴ f (– x) = – f (x) and f (x + 2) = f (x) ∴ f (2) = f (0 + 2) = f (0) and f (– 2) = f (– 2 + 2) = f (0) Now, f (0) = f (– 2) = – f (2) = – f (0) ⇒ 2 f (0) = 0 i.e. f (0) = 0 ∴ f (4) = f (2 + 2) = f (2) = f (0) = 0 Thus, f (4) = 0.



y2 + 289 – 34y ≥ – 7y2 – 71 + 78y 8y2 – 112y + 360 ≥ 0 y2 – 14y + 45 ≥ 0 (y – 9) (y – 5) ≥ 0 ⇒ y ≤ 5 or y ≥ 9 y cannot lie between 5 and 9.

39

⇒ ⇒ ⇒ ⇒ ∴

 y +1 1  log    3 − y  2

Functions

∴ x =

[∵ Discriminent ≥ 0]



2x   1 + x2 − 2x  1 − 1 + x 2   2x  f   = log    = log  1 + x 2 + 2 x   1 + x 2  1 + 2 x  2  1 + x 



= log 

1 − x  1 − x  = 2 log   = 2 f (x).  1 + x   1 + x  2

224. (d) Let x1, x2 ∈ R, then f (x1) = f (x2) ⇒ x12 + x1 = x22 + x2 ⇒ (x1 – x2) (x1 + x2 + 1) = 0. ∴ x2 = x1 or x2 = – (x1 + 1). ∴ f is many one. Again, let y ∈ R and x be such that 1  [– 1 ± 1 − 4 y ]. f (x) = y ⇒ x2 + x = y ⇒ x = 2 1 x ∉ R, if y > , ∴ f is into mapping. 4 Hence, f is many one into mapping. 225. (a) We have, f (x) =

2 x + 2− x 2

∴ f (x + y) ⋅ f (x – y)

1 x + y 1  (2 + 2– x – y) ⋅   (2x – y + 2– x + y) 2 2 1 [(22x + 2– 2x) + (22y + 2–2y)] = 4 1  [ f (2x) + f (2y)].  = 2 226. (b) We have, g (x) = 1 + x =

and f [g (x)] = 3 + 2 x + x

...(1)

Also, f [g (x)] = f (1 +

...(2)

x ) From (1) and (2), we get

f (1 +

x ) = 3 + 2 x + x.

x = y or x = ( y – 1)2. ∴ f ( y) = 3 + 2 ( y – 1) + ( y – 1)2 = 3 + 2y – 2 + y2 – 2y + 1 = 2 + y2 ∴ f (x) = 2 + x2.

Let 1 +

227. (d) We have, f (x) = (a – xn)1/n = y ∴ f ( y) = (a – yn)1/n = [a – {(a – xn)1/n}n]1/n

= [a – (a – xn)]1/n = (xn)1/n = x.

40

Objective Mathematics

228. (d) For y to be defined, we must have x2 – 1 ≥ 0 and x – 1 > 0 ⇒ | x | ≥ 1 and x > 1 ⇒ x > 1. ∴ Domain of y is (1, ∞).

237. (b) We have, ho(fog) (x) = hof{g(x)} = hof { ( x 2 + 1)}

229. (d) Since the function is constant ∨ x ∈ R, therefore, the value of the constant is obtained by assigning any suitable value to x. Choose x = 0, the desired value of the constant π π = cos20 + cos2 – cos 0 ⋅ cos  3 3 1 1 5 1 3 =1+ − = − = . 4 2 4 2 4 230. (d) We have, 2x > 0, ∨ x ∈ R Thus, 2x = 2 – 2y < 2, ∨ x ∈ R. As 2x is increasing function, we get – ∞ < x < 1. 231. (d) We have,



 1 + x1   1 + x2  f ( x1 ) + f ( x2 ) = log   + log   1 x − 1    1 − x2   1 + x1   1 + x2   = log      1 − x1   1 − x2  

x1 + x2  1 + 1 + x x 1 2 = log   1 − x1 + x2  1 + x1 x2

   x1 + x2   = f   1 + x1 x2   

= h{( x 2 + 1) 2 − 1} = h{x2 + 1 – 1} = x2 238. (b) We have, f {g ( x)} = eloge x = x g{f(x)} = loge ex = x f{g(x)} = g{f(x)}

and ∴ 239. (a) Clearly,    ⇒  ⇒  ⇒  ∴ 

x −1 ≤ log 3   ≤ 1 3 1 x ≤ ≤3 3 3 1≤x≤9 x ∈ [1, 9]   x  Domain of sin −1 log 3    is [1, 9].  3  

240. (a) Let x, y ∈ N be such that f(x) = f(y) ⇒  ⇒  ⇒  ∴ 

x2 + x + 1 = y2 + y + 1 (x – y) (x + y + 1) = 0 x = y or x = (– y – 1) ∉ N f is one-one

  Also, f is onto. 241. (a) We have,

232. (a) Clearly, the domain of f is π π = R ∩ [0, 4] ∩  x : − < x < and cos x > 0  2 2   π π π = R ∩ [0, 4] ∩  ,  = 0,  2 2  2 233. (a) Since every element of set A can be associated to any element in set B, therefore, total number of mappings from A to B is 34. 234. (a) Since, 0 ≤ x ≤ π ⇒ 0 ≤ x ≤ π 2 2 4 x 1 ∴  0 ≤ sin ≤ 2 2  1  ⇒   0,  ⊂ [0, ∞) 2  Hence, function is injective.

x – 1 > 0 and 2x – 1 > 0 and 2x – 1 ≠ 1 1 ⇒  x > 1, x > and x ≠ 1 2 ⇒  x > 1 ∴Domain of f(x) is (1, ∞) 242. (c) We have, domain of f is D1 = (–1, 1). For the function g(x) – (2x – 3) (2x + 1) ≥ 0 ⇒  (2x – 3) (2x + 1) ≤ 0 1 3 ⇒  − ≤ x ≤ 2 2  1 ∴  Domain of g(x) is D2 =  − ,  2

3 2 

235. (c) For (a)   and (b)   f is not 1–1, but for (c) f is 1–1.

Therefore, domain of (f + g) = D1 ∩ D2

236. (d) Clearly, from the graph, f is many-one into function.

1 =  − , 1  2 

y

243. (d) We have, y = 4x + 3 ⇒ 

O

x

x=

y −3 4

∴  The inverse of f(x) is g ( x) =

y −3 . 4

1. Let A be a set of n distinct elements. Then the total number of distinct functions from A to A is (a) 2n (b) n2 (c) nn (d) None of these 2. Given f (x) = 4x8, then 1  1  1 (a) f'   = f'  −  (b) f′(x) = f'  −  2  2  2  1 1 (c)  f  −  = f    2 2 3. Let f (x) =

1  1 (d) f   = f'  −  2  2

sin x . If D is the domain of f, then D 1 + sin x

contains. (a) (– 2π, – π) (b) (4π, 6π) (c) (0, π) (d) (2π, 4π) 4. Let the function f : R → R be defined by f (x) = 2x + sin x, x ∈ R. Then f is (a) one-to-one and onto (b) one-to-one but not onto (c) onto but not one-to-one (d) neither one-to-one nor onto 5. Suppose f (x) = (x + 1)2 for x ≥ – 1, If g (x) is the function whose graph is the reflection of the graph of f (x) with respect to the line y = x, then g (x) equals (a) – x – 1, x ≥ 0

(b)

x + 1 , x ≥ – 1

(d)

(c) 6. 7. 8.

1 ,x>–1 ( x + 1) 2

x, x ≥ 0

(a) f (– a)

1 (b) f    a

(c) f (a2)

 −a  (d) f    a − 1 

12. Let f (x) = x, g (x) = = 1, if

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

is is is is

1 and h (x) = f (x) ⋅ g (x). Then, h (x) x

an irrational number a real number ≠ 0 a real number a rational number

13. If the real-valued function f (x) = n equals (b)

1 (c) –  3

(d) 2

1 + 2 x − x 2 is [ x]

(a) (1, 2] (b) (0, 2) (c) (0, 1) (d) [1, 2] 15. The range of the function π −π 0

41

Functions

EXERCISES FOR SELF-PRACTICE

42

Objective Mathematics

 x, when x is an even integer 19. If f (x) =  , then f (x) is 0, when x is an odd integer (a) even (b) odd (c) both even and odd (d) neither even nor odd 20. If f (x) = 3x + 10, g (x) = x2 – 1, then ( fog)– 1 is equal to: 1/ 2 1/ 2  x + 7  x − 7 (a)  (b)     3   3 

x − 3 (c)   7 

x + 3 (d)   7 

1/ 2



1/ 2

23. The value of 24.

2 + 2 + 2 + ..... ∞ is:

(a) 5 (c) 2 If 3x – 3x – 1 = 6, then xx (a) 2 (c) 9

(b) 3 (d) None of these is equal to: (b) 4 (d) None of these

25. Inverse of the function y = 2x – 3 is: x+3 2 1 (c) 2x − 3

(a)

(b)

x−3 2

(d) None of these

26. If x is real, then value of the expression

21. Domain of the function

lies between:

1/ 2

  5x − x2  is: f (x) = log10  4     (a) – ∞ ≤ x ≤ ∞ (b) 1 ≤ x ≤ 4 (c) 4 ≤ x ≤ 16 (d) – 1 ≤ x ≤ 1

(a) 5 and 4 (c) – 5 and 4

x 2 + 14 x + 9 x2 + 2x + 3

(b) 5 and – 4 (d) None of these

27. Which of the following is one-one function: (a) ex (c) sin x

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

2

(b) ex (d) None of these

28. The value of ex – x is always: (a) greater than 1 for all real values (b) less than 1 for all real values (c) greater than 1 for some real values (d) None of these

Answers

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

2. (c) 12. (b) 22. (b)

3. (c) 13. (c) 23. (c)

4. (a) 14. (a) 24. (b)

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

6. (a) 16. (b) 26. (c)

7. (b) 17. (a) 27. (a)

8. (a) 18. (d) 28. (c)

9. (a) 19. (b)

10. (d) 20. (a)

2

Limits

CHAPTER

Summary of conceptS Limit defined Let a function f be defined at every point in the neighbourhood of a (an open interval about a) except possibly at a. If as x approaches closer and closer to a, but not equal to a, then the value of the function f (x) approaches a real number l. The number l is referred to as the limit of f (x) as x tends to a and we write it as

1.

lim [ f ( x) + g ( x)] = lim f ( x) + lim g ( x) = l + m. x→a x→a x→a

2.

lim [ f ( x) − g ( x)] = lim f ( x) – lim g ( x) = l – m. x→a x→a

3.

lim k ⋅ f ( x) = k ⋅ lim g ( x) = kl, x→a x→a

x→a

where k is a constant.

lim f ( x) = l

4.

f ( x) ⋅ lim g ( x) = lm. lim [ f ( x) ⋅ g ( x)] = lim x→a

Note that f (x) approaches l means the absolute difference between f (x) and l, i.e., | f (x) – l | can be made as small as we please. When the values of f (x) do not approach a single finite value as x approaches a, we say that the limit does not exist.

5.

lim f ( x) l x→a  f ( x)  = = lim   lim g ( x ) m x → a g ( x)   x→a

6.

f [ g ( x)] = f lim g ( x) = f (m). lim ( fog ) ( x) = lim x→a x→a

x→a

righthand Limit We say that right hand limit of f (x) as x tends to ‘a’ exists and is equal to l1 if as x approaches ‘a’ through values greater than ‘a’, the values of f (x) approach a definite unique real number l1 and we write

x→a

x→a

(provided m ≠ 0).

(

x→a

In particular,

(

)

)

log g ( x) = log lim g ( x) = log m. (i) lim x→a x→a (ii) lim e g ( x ) = e

lim f ( x) = l1 or f (a + 0) = l1.

lim g ( x ) x→a

x→a

x → a+

= em. n

7.

Working Rule To evaluate lim f ( x)

8.

x → a+

1. Put x = a + h in f (x) to get lim f ( a + h) 2. Take the limit as h → 0.

h→0

lim[ f ( x)]n = lim f ( x)  = ln, for all n ∈ N.  x→a  x → a Sandwich Theorem (or Squeeze Principle). If f, g and h are functions such that f (x) ≤ g (x) ≤ h (x) for all x in some neighbourhood of the point a (except possibly at h ( x) , then lim g ( x) = l. x = a) and if lim f ( x) = l = lim x→a x→a

Lefthand Limit We say that left hand limit of f (x) as x tends to ‘a’ exists and is equal to l2 if as x approaches ‘a’ through values less than ‘a’, the values of f (x) approach a definite unique real number l2 and we write lim f ( x) = l2 or f (a – 0) = l2.

x → a−

x→a

evaLuation of LimitS The problems on limits can be divided into the following categories:

Working Rule To evaluate lim f ( x) . x → a−

1. Put x = a – h in f (x) to get lim f ( a − h) . h→0 2. Take the limit as h → 0.

aLgebra of LimitS If lim f ( x) = l and lim g ( x) = m, then following results are x→a

true:

x→a

algebraic Limits The following methods are useful for evaluating limits of algebraic functions:

44

method of factorisation and g (a) ≠ 0, then we have

If f (x) and g (x) are polynomials

Objective Mathematics

f ( x) f (a) f ( x) lim lim = x→a = . x → a g ( x) lim g ( x) g (a ) x→a

Now, if f (a) = 0 = g (a), then (x – a) is a factor of both f (x) and g (x). We cancel this common factor (x – a) from both the numerator and denominator and again put x = a in the given expression. If we get a meaningful number then that number is the limit of the given expression, otherwise we repeat this process till we get a meaningful number.

Some Useful Summations n (n + 1) 2 n (n + 1) (2n + 1) (ii) Σ n2 = 12 + 22 + 32 + ... + n2 = 6 (i) Σ n = 1 + 2 + 3 + ... + n =

2

 n (n + 1)    2  a (1 − r n ) (iv) Σ arn – 1 = a + ar + ar2 + ... + arn – 1 = ; 1− r

(iii) Σ n3 = 13 + 23 + 33 + ... + n3 = 

provided r < 1.

method of rationalisation This method is useful where radical signs (i.e., expressions of the form a ± b ) are in- trigonometric Limits volved either in the numerator or in the denominator or both. The numerator or (and) the denominator (as required) is For finding the limits of trigonometric functions, we use trigo(are) rationalised and limit taken after cancelling out the com- nometric transformations and simplify. The following results are quite useful. mon factors. sin x 1. (i) lim =1 (ii) lim cos x = 1 xn − an x →0 x →0 x n – 1 Standard formula lim = na , where n ∈ Q, the x→a x − a tan x sin −1 x (iii) lim =1 (iv) lim =1 set of rational numbers. x →0 x → 0 x x π tan −1 x sin x 0 =1 (vi) lim = . (v) lim Limit of an algebraic function when x → ∞ x→0 x →0 180 x x f ( x) as x → ∞, g ( x) where f (x) and g (x) are algebraic functions of x, it is convenient to divide all the terms of f (x) and g (x) by the highest power of x in numerator and denominator both and use the following standard limits:

In order to find the limit of a function of the type

(i) (ii)

lim

x →∞

(iv)

if a > 1 if a = 1 if − 1 < a < 1 if a ≤ −1

a0 x p + a1 x p −1 + ... ... + a p −1 x + a p b0 x q + b1 x q −1 + ... ... + bq −1 x + bq  a0 b ,  0 = 0, ∞,  

if p = q if p < q if p > q

x3 x5 + − ... to ∞ 3! 5! 2 4 x x + − ... to ∞ cos x = 1 – 2! 4! 3 2 5 x x + ... to ∞ + tan x = x + 3 15 2 2 2 12 ⋅ x 3 12 ⋅ 32 5 1 ⋅ 3 ⋅ 5 7 x + x + ... to ∞ sin–1x = x + + 7 ! 3! 5! 2 2 2 ⋅ 22 4 2 ⋅ 22 ⋅ 42 6 x + x + x + ... to ∞ (sin–1 x)2 = 2! 4! 6!

(i) sin x = x –

(v)

∞,   1, (i) lim a n =   n →∞ 0 ,  does not exist,

x →∞

Some Useful Expansions

(iii)

1 = 0, if p > 0. lim x →∞ x p

lim

x→a

(ii)

1 =0 x

Some Useful Limits

(ii)

f (a + h) , where a ≠ 0, on taking x = a + h. 2. lim f ( x) = lim h→0

(vi) tan–1 x = x –

x3 x5 + − ... to ∞ . 3 5

exponential and Logarithmic Limits For finding the limits of exponential and logarithmic functions, the following results are useful: (i) lim x→0

ex − 1 =1 x

(ii) lim

ax − 1 = loge a, a > 0 x

(iii) lim x→0

ax − bx a = log e   ; a, b > 0 x b

(iv) lim x→0

(1 + x) n − 1 =n x

x→0

h→0

log x = 0, (m > 0) xm log a (1 + x) = loga e, (a > 0, a ≠ 1) (viii) lim x→0 x x a  (ix) lim 1 +  = ea x →∞ x  (vii) lim

x →∞

f ( x)

(x) (xi)

 1  = e, where f (x) → ∞ as x → ∞. lim 1 +  x →∞ f ( x)   lim (1 + f ( x)1/ f ( x ) = e.

x→a

Some Useful Expansions (i) ex = 1 +

x x 2 x3 + + + ... to ∞ 1! 2! 3!

(ii) e–x = 1 –

x x 2 x3 + − + ... to ∞ 1! 2! 3!

x 2 x3 (iv) loge (1 – x) = – x – − − ... to ∞, – 1 ≤ x < 1 2 3 ( x log a ) 2 + ... to ∞ 2!

n (n − 1) 2 x + ... to ∞, – 1 < x < 1, 2!

(vi) (1 + x)n = 1 + nx +

n being any negative integer or fraction.

The expansion formulae mentioned above can be used with advantage in simplification and evaluation of limits. For example, lim

cos x − 1 +

x→0

x

4

x2 2

4

exists. ∞ (ii)   form : If lim f ( x) = ∞ and lim g ( x) = ∞, then x→a x→a ∞ ′ f ( x) f ( x) lim = lim , provided the limit on the R.H.S. x → a g ( x) x → a g ′ ( x) exists.

Working Rule (i) L’Hospital’s Rule is applicable only when

6

f ( x) becomes of the g ( x)

0 ∞ or . 0 ∞ 0 ∞ (ii) If the form is not or , simplify the given expression till it 0 ∞ ∞ 0 reduces to the form or and then use L’Hospital’s rule. ∞ 0 form is

(iii) For applying L’Hospital’s rule differentiate the numerator and denominator separately.

Note: L’ Hospital’s rule cannot be applied in every problem. Consider the example, x→0

3 x + sin 2 x  ∞  form  . 3 x − sin 2 x  ∞

Here, if we apply L’ Hospital’s rule, we get 3 x + sin 2 x 3 + 2 cos 2 x = lim 3 x − sin 2 x x → ∞ 3 − 2 cos 2 x . Now, both the numerator and denominator are undefined because lim cos 2 x does not exist. lim

x→∞

x→∞

We can find the above limit as:

Some Useful Limits (i) If lim f ( x) = A > 0 and lim g ( x) = B, then x→a

x→a

lim [ f ( x)]g ( x ) = AB. x→a

(ii) If lim f ( x) = 1 and lim g ( x) = ∞, then x→a

0 (i)   form : If lim f ( x) = 0 and lim g ( x) = 0, then x→a x→a 0 ′ f ( x) f ( x) = lim , provided the limit on the R.H.S. lim x → a g ′ ( x) x → a g ( x)

lim

  x x x x2 1 − 2! + 4! − 6! + ... − 1 + 2 = lim x→0 x4 1 1  1 = lim  + terms containing x and its powers  = = . x → 0 4!   4! 24 2

Besides the methods given above to evaluate limits, there is yet another method for finding limits, usually known as L’Hospital’s Rule as given below for in deter minant forms:

Note that sometimes we have to repeat the process if the form is 0 ∞ or again. 0 ∞

x 2 x3 (iii) loge (1 + x) = x – + − ... to ∞, – 1 < x ≤ 1 2 3

(v) ax = ex log a = 1 + x log a +

evaLuation of LimitS uSing L’HoSpitaL’S ruLe

45

n

Limits

1  (v) lim (1 + x)1/ x = lim 1 +  = e x→0 n →∞ n  1/ h a lim ( 1 + a h ) (vi) =e

x→a

lim [ f ( x)]g ( x ) = e x→a

lim g ( x ) [ f ( x ) − 1] x→a

 sin 2 x  3 + 2  2 x  3 + 2(0) 3 x + sin 2 x = lim = lim x→∞ 2 x sin   3 − 2(0) x → ∞ 3 x − sin 2 x 3 − 2  2 x  = 1, since lim

x→∞

.

sin 2 x = 0. 2x

46

multiple-choice questions

Objective Mathematics

Choose the correct alternative in each of the following problems: 1. The integer n for which lim

( cos x − 1) ( cos x − e x )

x →0

x

n

finite nonzero number is (a) 1 (c) 3 2. lim

(b) 2 (d) 4 is equal to

n 2 + 1 + 4n 2 − 1

1 3 1 (c) –  5

(b) – 

(a)

(1 + x)1/ 3 − (1 − x)1/ 3 is x

x→0

2 (a) 3 (c) 1

1 (b) 3 (d) None of these

1 − cos 2 x 2x

x→0

1 3

(d) None of these

3. The value of lim

4. lim

11. In a circle of radius r, an isosceles triangle ABC is inscribed with AB = AC. If the ∆ ABC has perimeter

1 − 2 + 3 − 4 + 5 − 6 + ... − 2n

n →∞

is (b) – 1 (d) Does not exist

 x 4 sin(1/x) + x 2  5. The value of lim   is x→∞ 1+ | x |3   (a) 1 (b) – 1 (c) 0 (d) ∞  3x 2 + 2 x + 1  6. lim  2  x →∞  x +x+2  (a) 3 (c) 1

is equal to (b) 9 (d) None of these

  7. The value of lim  x + x + x − x  is x →∞   1 (a) 2 (b) 1 (c) 0 (d) None of these x

 x + 5x + 3  8. lim  2  equals x→0  x +x+2  2

P = 2  2 hr − h 2 + 2 hr  and area A = h 2 hr − h 2 ,   A where h is the altitude from A to BC, then lim 3 is equal h → 0+ P to 1 (a) 128 r (b) 128r 1 (c) (d) None of these 64 r

2 x + 23 − x − 6 is equal to 2− x / 2 − 21− x (a) 8 (b) 4 (c) 2 (d) None of these

12. lim x→2

n k sin 2 (n!) , 0 < k < 1, is equal to n+2 (a) ∞ (b) 1 (c) 0 (d) None of these

13. lim

(a) λ (c) zero

6x +1 3x + 2

x + cos x is is a 10. The value of xlim →∞ x + sin x (a) – 1 (b) 0 (c) 1 (d) None of these

(a) e4 (c) e3

(b) e2 (d) e

n →∞

14. Let f (2) = 4 and f' (2) = xf (2) − 2 f ( x) Then lim x→2 x−2 (a) 2 (c) – 4

is given by (b) – 2 (d) 3

2 x +3 3 x +5 5 x is x →∞ 3x − 2 + 3 2 x − 3

15. The value of lim

2 3 1 (c) 3 (a)

16. lim x→0

(a)

(

(b)

x 3 z 2 − ( z − x) 2 3

8 xz − 4 x 2 + 3 8 xz z

11/ 3

3

(d) None of these



)

4

is equal to

1 223/ 3 ⋅ z (d) None of these

(b)

2 (c) 221/3 z

n! is equal to (n + 1)!− n! (a) 0 (b) ∞ (c) 1 (d) None of these

17. lim

n →∞

x

 x−3 9. For x ∈ R, lim   x→∞ x + 2   (a) e (b) e–1 (c) e–5 (d) e5

4.

4

n5 + 2 − 3 n 2 + 1

n →∞ 5

n 4 + 2 − 2 n3 + 1 (b) 0 (d) ∞

18. The value of lim (a) 1 (c) – 1

is

x →∞ 4

(a) 1 (c) 0

x2 + 1 − 3 x2 + 1 4

5

4

x +1 − x −1

is equal to (b) – 1 (d) None of these

28. lim x →0

x2 x4 x2 x2  8  1 − cos − cos + cos cos  is equal to 8  2 4 2 4 x 

1 16 1 (c) 32

1 16 1 (d) – 32 (b) –

(a)

1 − cos 2( x − 1) x −1 (a) exists and it equals 2 ln(1 + 2h) − 2 ln(1 + h) 29. The value of lim is h →0 (b) exists and it equals – 2 h2 (c) does not exist because (x – 1) → 0 (a) 1 (b) – 1 (d) does not exist because left hand limit is not equal to (c) 0 (d) None of these right hand limit

20. lim x →1

1 1 1 1 + + 2 + ... + n 2 2 2 is equal to 21. lim n →∞ 1 1 1 1 + + 2 + ... + n 3 3 3 4 3 1 (c) 3

(b)

(a)

2 3

22. The value of lim (a) 1 (c) 0

(b) – 1 (d) None of these

  x  1 − tan  2   (1 − sin x) is 23. lim π  x   x→ 3 2 1 + tan    ( π − 2 x)  2  1 8 1 (c) 32

(a)

x→0

(c)

1 3 2 (d) –  3

(b) – 

−1

x →0

2 2 − (cos x + sin x)3 is π 1 − sin 2 x x→

27. The value of lim

3 (a) 2 1 (c) 2

4

2 (b) 3 (d)

(b) – 1 (d) None of these

(b) n (d) None of these 1/ x 2

tan x − sin x is equal to x3 1 1 (a) (b) – 2 2 (c) 1 (d) – 1

26. lim

x →0

(a) 1 (c) 0

is equal to (a) ∞ (c) k

log (3 + x) − log (3 − x) = k, the value of k is x

−1

32. The value of lim log e (sin x) tan x is

34. lim log n −1 (n) ⋅ log n (n + 1) ⋅ log n +1 (n + 2)...log k (n k )  n −1  n →∞ 

(d) ∞

2 3

x − sin x is equal to x + cos 2 x (a) – 1 (b) 1 (c) 0 (d) None of these x →∞

1− x

(b) 0

(a) 0

(b) etan a (d) ecos a

 1 + 3 x  1− x is 33. The value of lim   x →∞ 2 + 3 x   (a) 0 (b) – 1 (c) e (d) 1

3n + 2n 24. The value of lim n is n →∞ 3 − 2 n (a) – 1 (b) 1 (c) 0 (d) ∞ 25. If lim

(a) ecot a (c) esin a 31. lim

(d) None of these

x5 is x →∞ 5 x

1

 sin x  x − a 30. lim   , a ≠ nπ, n is an integer, equals x→a  sin a 

2

 1 + 5x2  35. The value of lim   x →0 1 + 3 x 2   (a) e2 (b) e (c) e–1 (d) None of these

is

1/sin x

 1 + tan x  36. lim   x →0 1 + sin x   (a) 0 (c) – 1

is equal to (b) 1 (d) None of these

 1 e1/ n e 2 / n e( n −1)/ n  37. The value of lim  + + + ... +  is n →∞ n n n n   (a) 1 (c) e – 1

 π  38. lim  tan  + x   x →0 4    (a) e (c) e3

(b) 0 (d) e + 1 1/ x

is equal to (b) e2 (d) e–1

47

2

Limits

19. lim

48

39. The value of lim x →1

x n + x n −1 + x n − 2 + ... + x 2 + x − n is x −1

Objective Mathematics

n (n +1) 2 (c) 1

(b) 0

(a)

(d) n

40. Let f (a) = g (a) = k and their nth derivatives f n (a), gn (a) exist and are not equal for some n. Further if f (a) g ( x) − f (a) − g (a) f ( x) + g (a) lim = 4, then the x→a g ( x) − f ( x) value of k is (a) 4 (b) 2 (c) 1 (d) 0 x2

41. The value of lim

2 ∫ cos t dt

0

x →0

(a) 3 2 (c) – 1 42. If lim

x→0

x sin x

is

(b) 1 (d) None of these

x − sin x is nonzero finite, then n may be x − sin n x n

n

equal to (a) 1 (c) 3

(b) 2 (d) None of these

43. lim(sin x + 1 − sin x ) is equal to x →∞

(a) 1 (c) 0

(b) – 1 (d) None of these

 x  44. lim sec −1   = x→∞  x + 1 (a) 0 (c) π/2

(b) π (d) does not exist

xn 45. lim x = 0, (n integer), for x →∞ e (a) no value of n (b) all values of n (c) only negative values of n (d) only positive values of n x sin ( x − [ x]) , where [⋅] denotes the greatest integer x −1 function, is equal to

46. lim x →1

(a) 1 (c) ∞

(b) – 1 (d) does not exist

47. lim (1 + x) (1 + x2) (1 + x4) ... (1 + x2n), | x | < 1, is n→∞

equal to

1 (a) x −1 (c) 1 – x 48. lim

x →∞

x →0

2 3 3 (c) 2

(a) – 1 (c) 0

1 + sin x − 3 1 − sin x is x 2 3 3 (d) – 2 (b) –

(a)

50. If f (x) =

3

x2n − 1 , then lim f(x) is n→∞ x2n + 1 (b) 1 (d) ∞

(1 + x)(1 − x 2 )(1 + x 3 )(1 − x 4 )...(1 − x 4 n ) is equal to x →−1 [(1 + x )(1 − x 2 )(1 + x 3 ) (1 − x 4 )...(1 − x 2 n )]2

51. lim

(a) 4nC2n (c) 2 ⋅ 4nC2n

(b) 2nCn (d) 2 ⋅ 2nCn.

1 is equal to 1 + n sin 2 nx (a) – 1 (b) 0 (c) 1 (d) ∞

52. lim

n →∞

53. lim log tan x (tan 2 x) is equal to x → 0+

(a) 1 (c) 0 54. If f (x) = (a) 0 (c) – 1

(b) – 1 (d) None of these



2 sin x − sin 2 x dx , x ≠ 0, then lim f ′ ( x) is x→0 x3 (b) ∞ (d) 1

11/ x + 21/ x + 31/ x + ... + n1/ x  55. The value of lim   x →∞ n   (a) n! (b) n (c) (n – 1)! (d) 0 56. lim

π x→ 2

nx

is

sin x − (sin x)sin x equals 1 − sin x + ln sin x

(a) 1 (c) 3

(b) 2 (d) 4

x  2  57. lim (where [ . ] denotes the greatest integer x → π / 2 ln(sin x ) function) (a) does not exist (b) equals 1 (c) equals 0 (d) equals – 1 58. lim lim(1 + cos 2 m n!πx) is equal to m →∞ n →∞

1 (b) 1− x (d) x – 1

x + sin x = x − cos x

(a) 0 (c) – 1

49. The value of lim

(a) 2 (c) 0

59. If f (a) = 2, f ' (a) = 1, g (a) = – 1, g′ (a) = 2, then

g ( x) f (a) − g (a) f ( x) is equal to x−a (a) 3 (b) 5 (c) – 3 (d) 0 lim x→a

(b) 1 (d) None of these

(b) 1 (d) None of these

(a) 0 (c) does not exist

(b) 1 (d) sin 1

1 (1 − cos 2 x) 61. The value of lim 2 is x →0 x (a) 1 (b) – 1 (c) 0 (d) None of these

e x − e− x − 2 x is equal to x →0 x − sin x (a) 1 (b) – 1 (c) 2 (d) 0

62. lim

63. lim

3

x →1

x2 − 2 3 x + 1 is equal to ( x − 1) 2

1 (a) 9 (c)

1 (b) 6

1 3

1 + ax + bx + c) lim( x→ α

is

(a + h) 2 sin( a + h) − a 2 sin a is h →0 h (b) 2a sin a – a2 cos a (a) 2a sin a + a2cos a 2 (c) 2a cos a + a sin a (d) None of these

65. The value of lim

66. lim tan x log sin x is equal to x →0

(a) 1 (c) 0

(b) – 1 (d) None of these

67. If g (x) = – (a) (c) – 

g ( x) − g (1) is equal to 25 − x 2 , then lim x →1 x −1

3 24

1 24

(b)

1 24

(d) None of these

68. The values of constants a and b so that

 x2 + 1  − ax − b  = 0 are lim  x →∞ x + 1   (a) a = 1, b = – 1 (c) a = 0, b = 0 69. The value of lim x →0

(b) a = – 1, b = 1 (d) a = 2, b = – 1

27 − 9 − 3 + 1 is 2 − 1 + cos x x

x

πx 2a

is equal to (b) e2/π (d) e– π/2

72. The value of lim 11 cos π x→ 

2x

2

+ 21 cos

2x

cos 2 x

2 + ... + n1 cos x  

(a) 0

(b) n

(c) ∞

(d)

is

n (n +1) 2

x →0

(b) ea (α – β) (d) None of these

(a) log | a (α – β) | (c) ea (β – α)

tan

a

64. If α and β be the roots of ax2 + bx + c = 0, then 1 ( x −α )

71. lim  2 − x  x→a a  (a) eπ/2 (c) e– 2/π

73. lim (cos x + a sin bx) x is equal to

(d) None of these

2

x

(a) 4 2 (log 3)2

(b) 8 2 (log 3)2

(c) 2 2 (log 3)2

(d) None of these

2

(a) e − a b 2

(c) e a b

(b) e ab

2

(d) e − b

2a

sin x n , m, n ∈ N, is equal to x →0 (sin x ) m (a) 1 (b) – 1 (c) 0 (d) ∞

74. lim

sin x

 sin x  x −sin x 75. The value of lim  is  x →0  x  (a) 1 (b) – 1 (c) 0 (d) None of these 76. The value of lim  3 n 2 − n3 + n  is n →∞   (a)

1 3

(b) –

1 3

(c)

2 3

(d) –

2 3

 n2 + 1 + n  77. lim   is equal to n →∞ 4 3  n + n − 4 n  (a) 0 (b) 1 (c) – 1 (d) ∞ 78. The value of lim  3 (n + 1) 2 − 3 (n − 1) 2  is n →∞   (a) 1 (c) 0

(b) – 1 (d) ∞

 1 1 1 1  is equal to 79. lim  + + + ... + n →∞ 1 ⋅ 2 ( 2 ⋅ 3 3 ⋅ 4 n n + 1)   (a) 1 (c) 0

(b) – 1 (d) None of these

49

π − cos −1 x is equal to x →−1 x +1 1 1 (b) (a) 2π π 1 (c) (d) None of these 2

70. lim

Limits

 sin([ x − 3])  60. lim   , where [ . ] represents greatest integer x→0  [ x − 3)  function, is

50

Objective Mathematics

 2n n +1 n n (−1) n  80. The value of lim  2 cos − ⋅ 2  is n →∞ 2 n − 1 2n − 1 1 − 2n n + 1   (a) 1 (b) – 1 (c) 0 (d) None of these 8 x2 +3

 2x2 + 3  81. The value of lim  2 is  x →∞ 2 x + 5   (a) e8 (b) e– 8 4 (c) e (d) e– 4

90. lim sin( a + 3h) − 3 sin(a + 23h) + 3 sin( a + h) − sin a is equal h →0 h to (a) sin a (b) – sin a (c) cos a (d) – cos a x

 x2 − 2x + 2  91. The value of lim  2  is x →∞ x − 4 x + 1   (a) e–2 (c) 1

(b) e2 (d) 0

sin 2 x + a sin x be finite, then the value of a and 92. If f (9) = 9 and f ' (9) = 1, then lim 3 − f ( x) is equal x3 x →9 3− x to the limit are given by (a) 0 (b) 1 (a) – 2, 1 (b) – 2, – 1 (c) – 1 (d) None of these (c) 2, 1 (d) 2, – 1

82. If lim x →0

93. Let f (x) be a twice differentiable function and f ′′ (0) 3 f ( x) − 4 f (3 x) + f (9 x) = 5, then lim is equal to x →0 x2 (a) 30 (b) 120 (c) 40 (d) None of these

1/sin x

 1 + tan x  83. The value of lim  is  x →0 1 + sin x   (a) 1 (b) – 1 (c) 0 (d) ∞ 84. The values of a, b and c such that

x3

 2 1+ x 94. The value of lim  3 x 2 + 1  is x →∞ 4 x − 1  

−x

a e − b cos x + ce = 2 are x sin x (a) a = 1, b = – 2, c = 1 (b) a = 1, b = 2, c = – 1 (c) a = 1, b = 2, c = 1 (d) a = – 1, b = 2, c = 1 lim

x

x →0

 x+6 85. lim   x →∞ x + 1  

(a) 0 (c) 1

log (1 + x + x 2 ) + log (1 − x + x 2 ) is equal to x →0 sec x − cos x (a) 1 (b) – 1 (c) 0 (d) ∞

95. lim

x+4

is equal to

(a) e– 5 (c) 0

(b) e5 (d) None of these

86. If f (5) = 7 and f ' (5) = 7, then lim x →5

given by

96. The value of lim(sin x) tan x is x→

x f (5) − 5 f ( x) is x−5

(b) 1 (d) ∞

x →0

(b) 28 (d) – 35

1   87. If a, b, c, d are positive, then lim 1 +  x →∞  a + bx  d/b c/a (a) e (b) e (c) e(c + d)/a + b (d) e 88. If a = min {x2 + 4x + 5, x ∈ R} and b = lim θ→0

n

∑a

r

⋅ b n − r is

r =0

n +1

(a) 0 (c) – 1

π 2

97. lim x x is equal to

(a) – 28 (c) 35

then the value of

(b) ∞ (d) – 1

(a)

2 −1 4 ⋅ 2n

(b) 2n + 1 – 1

(c)

2n +1 − 1 3 ⋅ 2n

(d) None of these

3 1x 89. If lim (1 + a ) +3 8e1 x = 2, then x →0 1 + (1 − b ) e (a) a = 1, b = (– 3)1/3 (b) a = 1, b = 31/3 1/3 (c) a = – 1, b = – (3) (d) None of these

(a) 0 (c) – 1

c + dx

=

1 − cos 2θ , θ2

(b) 1 (d) None of these

98. The value of lim a 2 − x 2 cot x→a

2a π 4a (c) π

(a)

 99. The value of lim log a x →3  (a) loga 6 (c) loga 2 100. lim x→2

3

2a π 4a (d) – π (b) –

x−3  is x + 6 − 3  (b) loga 3 (d) None of these

10 − x − 2 is equal to x−2

1 12 1 (c) 6

(a)

π a−x is 2 a+x

1 12 1 (d) – 6 (b) –

3 2 (c) 3

3 2 (d) does not exist

(a)

to (a) 4 (c) 2

(b) –

(a) eb (c) e

(a) 4 (c) 0

x →−∞

x →0

(a) e (c) e–1

(b) a = 1, b =

1 2

x → 0+

(a) 1 (c) 0

(d) None of these

x

is equal to

(a) 1 (c) ∞

(a) 1 (c) 0

n

∑ (−1) r =1

r

⋅ tr , then

x →∞

(a) – 1 (c) – 1

n →∞

2 3 1 (d) – 3 (b) –

107. If f (x), g (x) be differentiable functions and f (1) = g (1) = 2 then f (1) g ( x) − f ( x) g (1) − f (1) + g (1) lim is equal to x →1 g ( x) − f ( x) (a) 0 (b) 1 (c) 2 (d) None of these

a0 x n + a1 x n −1 + a2 x n − 2 + ... + an ; m, n > 0, is equal to x →∞ b x m + b x m −1 + b x m − 2 + ... + b 0 1 2 m

108. lim

(c)

(d) None of these

109. lim (1 + cos π x)cot 2 π x is equal to

x →0

is

(a) a (c)

a 2 − ax + x 2 − a 2 + ax + x 2 a+x − a−x (b) – a

a

(d) –

ln x − 1 is equal to | x−e| 1 (a) e (c) e

a

118. lim x →e

1 e (d) Does not exist

(b) –

119. The value of lim [ x 2 ] is x→ 2

(a) 1 (c) 0

(b) 2 (d) Does not exist

sin (2k − 3) x ,x 0 is x →∞  x 

105. The value of lim 1 − cos (14− cos x) is x →0 x 1 1 (b) (a) 4 8

(a)

(b) – 4 (d) Does not exist

113. lim log tan x (sin x) is equal to

(b) e2 (d) 1

1 (c) 16

(b) eb/2 (d) None of these

x →3

and b are given by

104. lim (cos x + sin x)

is

integer function, is equal to

103. If lim ( x − x + 1 − ax − b) = 0, then the values of a

1 x

n

112. lim ([ x − 3] + [3 − x] − x) , where [⋅] denotes the greatest

2

1 2 1 (c) a = 1, b = – 2

(b) – 4 (d) – 2

b  111. The value of lim 1 + tan  n →∞ n 

 ln cos x  102. lim   is equal to 2 x →0 4  1 + x − 1 (a) 2 (b) – 2 (c) 1 (d) – 1

(a) a = – 1, b =

tan (3k − 4) x , x > 0 2x and lim f ( x) exists, then the value of k is given by

(b) – 1

=

(d) None of these



x →0

51

2 x 2 − 4 f ( x) is equal x→2 x−2

110. If f (2) = 2 and f ′ (2) = 1 then lim

Limits

101. If f (x) is differentiable and f ' (2) = 6, f '(1) = 4, then f ( 2 + 2h + h 2 ) − f ( 2 ) lim is equal to h →0 f (1 + h + h 2 ) − f (1)

52

Objective Mathematics

5 4 4 (c) 5

5 4 4 (d) – 5

(a)

(b) –

1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + ... + n (n + 1) is equal to n3 (a) 1 (b) – 1 1 (d) None of these (c) 3 x +1 x +1 3 −5 is 122. The value of lim x x →∞ 3 − 5 x 1 (a) 5 (b) 5 (c) – 5 (d) None of these 121. lim

n →∞

n −1 1 2  1 123. lim 1 + e n + e n + ... + e n  is equal to n →∞ n   (a) e (b) – e (c) e – 1 (d) 1 – e

130. lim π x→ 4

(a) 5 2 (c) 2

(b)

1 2

(c) –

(a) 0 (c) – 1

x →1

(a) 0 (c) – 1

x

x →0

4− x − x (a) 0 (c) – 1

x7 − 2 x5 + 1 is x3 − 3x 2 + 2 (b) 1 (d) None of these

(a)

e x + e − x + 2 cos x − 4 is equal to x →0 x4 (a) 0 (b) 1 1 1 (d) – (c) 6 6

133. lim

3 24 3 3 (c) 3 4 2 (a)

3 24 3 3 (d) – 3 4 2 (b) –

y3 as (x, y) → (1, 0) along the line x →1 x − y 2 − 1 y →0

129. lim

3

y = x – 1 is given by (a) 1 (c) 0

(b) ∞ (d) None of these

(b) –

1 2

(d) – 1

x − 2a + x − 2a x 2 − 4a 2

x→2 a

1 a 1 (c) – a

(a)

is equal to

1 2 a 1 (d) – 2 a (b)

134. lim (tan x cot β)1/( x −β ) is equal to x →β

1 sin β cos β 1 (c) – sin β cos β

(a)

(b) sin β cos β (d) None of these

135. lim(−1)[ x ] , where [x] denotes the greatest integer less x→n

than or equal to x, is equal to (a) (– 1)n (c) 0

(b) (– 1)n – 1 (d) Does not exist

log e [ x] , where [x] denotes the greatest integer less x than or equal to x, is

136. lim

x →∞

127. lim

2− 2+ x 128. lim 3 is equal to x→2 2 − 3 4− x

1 2

(c) 1

is equal to (b) 1 (d) Does not exist

(b) 1 (d) None of these

x +1 π   is 132. The value of lim x  tan −1 − x →∞ + 2 4  x 

(d) None of these

125. The value of lim

126. lim

1 2

(b) 3 2 (d) None of these

131. lim cos  π n 2 + n  , n ∈ Z is equal to n →∞  

 1  1 1 1 124. lim  + + + ... +  is equal to n →∞ 1 ⋅ 3 ( )( ) 3 ⋅ 5 5 ⋅ 7 2 n + 1 2 n + 3   (a) 1

4 2 − (cos x + sin x)5 is equal to 1 − sin 2 x

(a) 1 (c) 0

(b) – 1 (d) Does not exist

137. If α is a repeated root of ax2 + bx + c = 0, then tan (ax 2 + bx + c) lim is x →α ( x − α) 2 (a) a (b) b (c) c (d) 0 138. Let f (x)  = sin x,  x ≠ nπ, where n ∈ Z and = 2,    g (x)

x  = nπ = x2 + 1, x ≠ 2, then lim g [ f ( x)] is



x = 2.

= 3, (a) 1 (c) 3

x →0

(b) 0 (d) Does not exist

139. If the rth term, tr , of a series is given by n r . Then lim ∑ tr is tr = 4 2 n →∞ r =1 r + r +1

(c)

1 3

1 2

x→2

(d) None of these

 x4 + x2 + x + 1  140. lim   2 x →−1  x − x +1 

1− cos ( x +1) ( x +1)2

is equal to 1/ 2

2 (b)   3

(a) 1 1/ 2

3 (c)   2



(d) e1/2

2

(log x) , n > 0 is equal to xn (a) 1 (b) 0 (c) – 1 (d) ∞

141. lim

x →∞

4 142. The value of lim 3x sin  x  is x →∞ 3  (a) 4 log 3 (b) 3 log 4 (c) 4 (d) None of these

  tan[ x] , [ x] ≠ 0    where 143. If f (x) =  [ x]  0, 0 [ ] x =  

[x] denotes the greatest integer less than or equal to x, then lim f ( x) equals (a) 1 (c) 0

x →0

(b) – 1 (d) Does not exist

2x − x2 is equal to x→2 x x − 22

144. lim

log 2 − 1 log 2 + 1 (c) 1 (a)

log 2 + 1 log 2 − 1 (d) – 1

(b)

n

x  145. lim  cos  is equal to n →∞ n  (a) e1 (c) 1

(b) e–1 (d) None of these

 tan −1 ([ x] + x)  , [ x] ≠ 0   2 [ ] − x x 146. If f (x) =   0 , [ x] = 0 

1+ 2 + x − 3 is equal to x−2 1 1 (a) (b) 8 3 3

148. lim

53

(b)

(c) 8 3

(d)

3

149. Let f (x) = x – [x] where [x] denotes the greatest integer { f ( x)}2 n − 1 , then g (x) = ≤ x and g (x) = lim n →∞ { f ( x )}2 n + 1 (a) 0 (b) 1 (c) – 1 (d) None of these 150. lim  1 2 + 2 2 + ... + n 2  is equal to n →∞ 1 − n 1− n 1 − n   1 (a) 0 (b) – 2 1 (d) None of these (c) 2 151. lim (1 − x) tan  π x  is equal to x →1  2  π 2 (a) (b) 2 π π 2 (c) – (d) – 2 π 152. Let f : R → R be such that f (1) = 3 and f ′ (1) = 6. 1

 f (1 + x )  x Then lim   equals x →0  f (1)  (a) 1 (b) e1/2 2 (c) e (d) e3  π  π  2  3 sin  + h  − cos  + h   6 6      is equal to 153. lim  h →0 3h ( 3 cos h − sin h) 4 4 (b) – (a) 3 3 2 3 (c) (d) 3 4 154. lim x →0

(1 − cos 2 x)sin 5 x is equal to x 2 sin 3 x

6 5 10 (c) 3 (a)

(b)

3 10

(d) –

3 10

where [x] denotes the greatest integer less than or equal to  e x − esin x  155. lim  x, then lim f ( x) is equal to  is equal to x → 0 x − sin x x →0   (a) – 1 (b) 0 1 (b) 1 (a) – (c) 1 (d) None of these 2 x2 π e − cos x (d) Does not exist (c) is 156. lim x →0 4 x2 x 1 2  x−3 (b) (a) 147. For x ∈ R, lim  =  x →∞ x + 2 2 3   3 (a) e (b) e–1 (c) (d) 2 (c) e–5 (d) e5 2

Limits

(a) 1

54

157. If lim

( sin nx ) ( a − n ) nx − tan x  x2

x →0

= 0, then find the

Objective Mathematics

value of a (a)

1 n

(c) n +

(b) n −

1 n

1 n

(d) None of these

1   x sin , x ≠ 0 x  158. If f (x) = , then lim f ( x) equals x →0 0 ,x = 0 (a) 1 (c) – 1

(b) 0 (d) None of these

159. If f ( x) = ∫

x + sin x f ′ ( x) is equal to dx, then xlim →∞ x + cos x

(a)  0 (c)  ∞

(b)  1 (d)  –1

1 + sin x − cos x + log(1 − x) is 160. The value of lim x→0 x3 (a)  –1 (b)  1/2 (c)  –1/2 (d)  1

161. If S1 = Σ n, S2 = Σ n2, S3 = Σ n3, then the value of S   S1 1 + 3  8  is equal to  lim n→∞ S 22 (a)  3/32 (c)  9/32 162. lim

x→∞

(b)  0 (d)  

3 10

 x + bx + 4  163. The value of lim  2  is x → ∞ x + ax + 5   2

(a)  

b a

(c)  1

(b)  3 (d)  None of the above

x+2 165. lim   x →∞  x +1  (a)  1 (c)  e2

(b)  0 (d)  

4 5

x +3

is (b)  e (d)  e3

166. lim x loge(sin x) is equal to x→ 0

(a)  –1 (c)  1

(b)  loge 1 (d)  None of these

1 + tan x    1 + sin x 

cos ec x

167. lim  x →0 (a)  

is equal to

1 e

(b)  1

(c)  e

(d)  e2

168. lim (–1)[x] is equal to x→n

(a)  (–1)n (c)  (–1)n – 1

(b)  0 (d)  does not exist.

ax 2 + bx + c = 2 then (a, b, c) is ( x − 1) 2

x →1

(2 x − 3) (3 x − 4) is equal to (4 x − 5) (5 x − 6)

1 10 1 (c)   5

(a)  9 (c)  1

169. If lim

(b)  3/64 (d)  9/64

(a)  

164. If f be a function such that f(9) = 9 and f ′ (9) = 3, f ( x) − 3 then lim is equal to x →9 x −3

(a)  (2, – 4, 2) (c)  (2, 4, – 2) sec2 x

170. lim x→



x2 −

8 π (c)   2 π

(a)  

f (t ) dt

2

π 4

(b)  (2, 4, 2) (d)  (2, – 4, – 2)

π2 16

equals 2 f (2) π

f (2)

(b)  

1 f   2

(d)  4f(2)

( x − 1) n ; 0 < x < 2, m and n are intelog cos m ( x − 1) gers, m ≠ 0, n > 0, and let p be the left hand derivative of |x – 1| at x = 1. If lim g(x) = p, then x →1

171. Let g(x) =

(a)  n = 1, m = 1 (c)  n = 2, m = 2

(b)  n = 1, m = – 1 (d)  n > 2, m = n

SOLUTIONS 1. (c) n cannot be negative integer for then the limit = 0 x 2 sin 2 x x 2 . e − cos x = 1 lim e − cos x , Limit = lim 2 n − 2 x →0 x 2 x →0 x n − 2  x 22.  2 (n ≠ 1 for then the limit = 0)

1 e x + sin x = lim . x → 0 2 ( n − 2 ) x n −3 So, if n = 3, the limit is

1 which is finite. 2 ( n − 2)

If n = 4, the limit is infinite.

2

2

n + 1 + 4n − 1

[1 + 3 + 5 + 7 + ... + (2n − 1)] − (2 + 4 + 6 + ... + 2n) 1 1 n 1+ 2 + n 4 − 2 n n n n [2 ⋅1 + (n − 1) ⋅ 2] − [2 ⋅ 2 + (n − 1) ⋅ 2] 2 2 = lim n →∞  1 1  n  1+ 2 + 4 − 2  n n   n n ⋅ 2n − 2 (n + 1) 2 = lim 2 n →∞  1 1  n  1+ 2 + 4 − 2  n n   = lim

n →∞

n2 − n2 − n

= nlim →∞

 1 n  1+ 2 + 4 − n  −n = lim n →∞  1 n  1+ 2 + 4 − n  −1 = nlim →∞ 1 1 1+ 2 + 4 − 2 n n

1   n2 

−1 −1 = . 1+ 2 3

 a 3 − b3   Using a − b = 2  a + ab + b 2   2 = lim x → 0 (1 + x ) 2 / 3 + (1 + x )1/ 3 (1 − x )1/ 3 + (1 − x ) 2 / 3 2 2 = . = 1+1+1 3 

x → −∞

= lim

x→0

2x

1 − (1 − 2 sin 2 x)

|sin x | 2 sin 2 x = lim . x→0 x 2x

Let f (x) =

|sin x | x

hh |sin (0 + h) | limsinsin =lim =1 h → 0 h→0 h→0 0+h h− h

Then, f (0 + 0) = lim

sin h |sin (0 − h) | = hlim = – 1. →0 −h −h ∵  f (0 + 0) ≠ f (0 – 0)

and f (0 – 0) = lim

h→0

∴  the limit does not exist.

 x 4hsin(1/x) + x 2  sin 5. (b) hlim   =  lim 3 → 0 − h 1+ | x | y →∞  

 3 =   1

6/3

= 9.

7. (a) lim  x + x + x − x   x →∞   

x+ x+ x −x x+ x+ x + x

x+ x

= xlim →∞

1   x 1 +   x

 4x + 1  = xlim 1 + 2  →∞ x + x + 2 

x+ x+ x + x

1/ 2

1 − y 4 sin + y 2 y 1+ y 3

[Putting x = – y; as x → – ∞, y → ∞]

1 1 = 1+1 = 2 .

x

[(1 + α)1/ α ]αx = xlim →∞

1 4+ 4x + 1 x →0 where α = 2 = 1 2 x + x +2  x x + + 2  x x    as x → ∞ 4+

1 x

→ 4 as x → ∞ 1 2 + 2 x x ∴  Given limit = e4. and αx =

1+

x

2x

x→0

    

1/ 2   1 1  1+ x  1 + + 1  x x    8. (a) Given limit

− (1 − x)1/ 3 x→0 x 1 (1 + x) − (1 − x) ⋅ = lim x → 0 (1 + x ) 2 / 3 + (1 + x )1/ 3 (1 − x )1/ 3 + (1 − x ) 2 / 3 x

= lim

2 1   3 + x + x2  = xlim →∞  1 + 1 + 22 x x 

= xlim →∞

1/ 3

1 − cos 2 x

1 x 2 3+ x

6+

x →∞

3. (a) lim (1 + x)

4. (d) lim

6x +1

 2 3x + 2 6. (b) lim  3 x2 + 2 x + 1  x →∞ x + x + 2  

= lim

1   n2  =

1  sin  y 1 + − 1  y    y  −1 + 0 = = – 1. = ylim →∞ 1 1+ 0 1+ 3 y

55

1 − 2 + 3 − 4 + 5 − 6 + ... − 2n

n →∞

Limits

2. (b) lim

x

    9. (c) lim  x − 3  = lim  1 − (3 / x)  x→∞ x + 2 x → ∞ 1 + (2 / x)     x 3  1 −  e −3 x = lim = 2 = e–5. x x→∞ e 2  1+    x 1 1 + cos   h x + cos x h lim 10. (c) xlim →∞ x + sin x = h → 0 1 1 + sin   h h

1    Putting x = h ; as x → ∞, h → 0    1 1 + h cos h = 1+ 0 = lim h→0 1 1+ 0 1 + h sin h

56

Objective Mathematics

1 1  , ∵−1 ≤ sin h ≤ 1 and −1 ≤ cos h ≤ 1 



∴ when h → 0, h cos 1 → 0 and h sin 1 → 0.  h h  = 1. 11. (b) lim

h → 0+

h 2 hr − h 2 A = lim P 3 h → 0+ 8  2 hr − h 2 + 2 hr  3  

= lim h→0 = lim h→0

=

(

h ⋅ h 2r − h

8 2 2r

12. (a) lim x→2

)

3

= lim x→2

(2

1 = . 128 r

x/2

+ 2) (2 − 2) (2 − 2) (2 x / 2 − 2)

x→2

n k sin 2 (n!) n k sin 2 (n!) = nlim →∞ 2  n →∞ n+2 n 1 +  n  

13. (c) lim

z= nlim →∞

sin 2 (n!) a finite quantity = 2   ∞ n1− k 1 +  n 

[ ∵ sin2 (n !) always lies between 0 and 1. Also, since 1 – k > 0, ∴ n1 – k → ∞ as n → ∞] = 0. 14. (c) lim xf (2) − 2 f ( x) x→2 x−2 xf (2) − 2 f (2) + 2 f (2) − 2 f ( x) = lim x→2 x−2 = xlim →2

f ( x) − f (2) ( x − 2) f (2) − 2 lim x→2 x−2 x−2

= f (2) – 2f ′ (2) = 4 – 2 × 4 = – 4. 3 5 15. (a) lim 2 x + 3 x + 5 x 3 x →∞ 3x − 2 + 2 x − 3

= xlim →∞

2 x +3 x +5 x 2 3 x 3− + 3 x 3 2− x x 3

3

z 2 − ( z − x) 2

( 3 8 xz − 4 x 2 + 3 8 xz ) 4

= lim x →0

x 3 2 xz − x 2 ( x 8z − 4x + 3 8z 3

3

x4 3

3

3

x )4

4

=

2z − x

3

2z

 3 8z − 4x + 3 8z  2 3 8z      n! n! = lim 17. (a) lim n →∞ ( n + 1)!− n ! n →∞ ( n + 1) n !− n ! = lim n →∞

x

43

=

1 . 223 3 ⋅ z

1 1 = lim = 0. n + 1 − 1 n→∞ n

4

n5 + 2 − 3 n 2 + 1

n →∞ 5

n 4 + 2 − 2 n3 + 1

18. (b) lim

4

x

= lim(2 x / 2 + 2) (2 x − 2) = (2 + 2) ⋅ (4 – 2) = 8.



x

= lim x →0

22 x − 6 ⋅ 2 x + 8 ( 2 x − 4) ( 2 x − 2) = lim x/2 x→2 2 −2 ( 2 x / 2 − 2) x/2

2 . 3

3

( 2 2 x + 23 − 6 ⋅ 2 x ) / 2 x 2 x + 23 − x − 6 lim − x/2 1− x = x → 2 1 2 2 −2 − 2x/2 2x

= lim x→2

16. (b) lim

3

2r − h

2r

=

x →0

8 ⋅ h ⋅ h  2r − h + 2r  8  2r − h + 2r 

3 5 + 3/10 1/ 6 x x = xlim →∞ 2 1 3 3 − + 1/ 6 3 2 − x x x [Dividing the numerator and denominator by the highest power x1/2] 2+

5

2 1 − n2 3 3 1 + 2 5 n n = lim n →∞ 2 1 n 4 5 5 1 + 4 − n3 2 2 1 + 3 n n n5 4 4 1 +

n5 4 4 2 n2 3 1 1+ 5 − 3 2 3 1+ 2 32 n n n n = lim n →∞ n 4 5 2 n3 2 2 1 5 1+ − 1+ 3 32 n n 4 n3 2 n [Dividing the numerator and denominator by the highest power n3/2]

1 4 2 1 1 1+ 5 − 5 6 3 1+ 2 14 0−0 n n n n = = 0. = lim n →∞ 0 −1 1 5 2 2 1 + − + 1 1 n 7 10 n4 n3 2

x2 + 1 − 3 x2 + 1

x →∞ 4

x4 + 1 − 5 x4 − 1

19. (a) lim

1 1 − x2 3 ⋅ 3 1 + 2 2 x x = lim x →∞ 1 1 x 4 1 + 4 − x4 5 5 1 − 4 x x x 1+

1 1 1 − 1 3 3 1+ 2 2 x x x = lim x →∞ 1 1 5 1 4 1+ − 1− 4 x 4 x1 5 x 1+

h = lim tan h→0 2

2

x →1

2

=–

2 ⋅1=–

2.

RHL = lim+ x →1

h  h  tan   sin  1  1 2 2 ⋅ = hlim .  h  = → 0 32 h 32     2  2 

2 sin( x − 1) 2 sin(1 − h − 1) = lim − h →0 ( x − 1) (1 − h − 1)

2 − sin h −h

= lim h →0 =–

tan

2 sin( x − 1) ( x − 1)

LHL = lim−

8h 3

h 2

h  h  sin 2  1 1 2 ⋅ × × = hlim →0 4 h h  4 ×2  2  2 

1 − cos 2( x − 1) 2 sin 2 ( x − 1) = lim 20. (d) lim x →1 x →1 x −1 x −1 = lim x →1

2 sin 2

57

1− 0 = = 1. 1− 0

Limits

[Dividing the numerator and denominator by x]

2 lim h →0

sin h h

24. (b) lim

n→ ∞

2 sin (1 + h − 1) 2 sin( x − 1) = lim h →0 (1 + h − 1) ( x − 1)

2 sin h sin h = 2 lim = h →0 h h Since LHL ≠ RHL,

= lim h →0

2⋅ 1 =

3 +2 = lim 3n − 2n n → ∞ n

n

2 1+   3

n

2 1−   3

n

[Dividing by 3n]

2 1+ 0 2 [ ∵ < 1, ∴   → 0 as n → ∞] 1− 0 3 3 = 1. n

=

2.

log (3 + x) − log (3 − x) = k x Applying L’Hospital rule

25. (c) lim

x →∞

∴ lim x →1

1 − cos 2( x − 1) does not exist. x −1

 1 1  3 + x + 3 − x    ⇒ xlim =k →∞ 1

1 1 1 1 1 1 − n +1 1− 1 + + 2 + ... + n 3 2 2 2 2 = lim × 21. (a) lim n →∞ 1 1 n →∞ 1 1 1 1− 1 − n +1 1 + + 2 + ... + n 2 3 3 3 3

1 1 + =k 3 3 2 . ⇒k= 3



n +1

1 1−   4 1− 0 4 2 ⋅  n +1 = ⋅ = lim n →∞ 3 3 1+ 0 1 1−   3

[ ∵ xn + 1 → 0 as n → ∞ if 0 < x < 1] =

22. (c) lim

x →∞

tan −1 x − sin −1 x  x →0 x3

4 . 3

1 1 − 1 + x2 − x2 1 = lim 2 x →0 3x

x5 x5 x5 = lim x log 5 = lim kx , where k = log 5 x x x →∞ →∞ 5 e e

= lim x →∞



= lim x →0



= lim x →0

1  1 1 k2 1 k3 1 k4 1   5 + k ⋅ 4 + ⋅ 3 + ⋅ 2 + ⋅  2! x 3! x 4! x  x  x



= lim x →0

 k5  k6 +  x + ...   5!  6! 



5

x   k 2 x 2 k 3 x3 k 4 x 4 x5k 5 k 6 x6 + + + + + ...  1 + kx + ! ! ! ! ! 2 3 4 5 6  

= lim x →∞

+

0   form  0 

26. (b) lim

1 = = 0. ∞ π π 23. (c) Put x = − h as x → , h → 0 2 2 π h 1 − tan  −   4 2  (1 − cos h) ∴  Given limit = lim ⋅ h→0 ( 2h)3 π h 1 + tan  +   4 2

[Using L’Hospital’s Rule]

1 − x 2 − (1 + x 2 ) 3 x 2 (1 + x 2 ) 1 − x 2 (1 − x 2 ) − (1 + x 2 ) 2 3 x 2 (1 + x 2 ) 1 − x 2  1 − x 2 + (1 + x 2 )   

−3 − x 2 3 (1 + x ) 1 − x 2  1 − x 2 + (1 + x 2 )    3 1 = – =− . 6 2

27. (a) lim π x→ 4

2

 2 2 − (cos x + sin x)3  0  form  1 − sin 2 x 0 

= limπ x→

4

−3 (cos x + sin x) 2 (− sin x + cos x) −2 cos 2 x [Using L’Hospital’s Rule]

58

= limπ x→

4

Objective Mathematics

= limπ x→

4

−3 (cos x + sin x)(cos 2 x − sin 2 x) −2 cos 2 x



= lim x→ a

−3 (cos x + sin x)cos 2 x −2 cos 2 x



x+a cos  = lim  2  ex→ a



= e sin a = ecot a.

3 (cos x + sin x) 3  1 1  3 + = ⋅ . = 2 2  2 2 2 x→ 4

= limπ 28. (c) lim x →0

x2 x2 x2 x2  8  1 − cos − cos + cos cos  8  2 4 2 4 x 



= lim x →0

x2  x4  x2  8  1 − cos  − cos 1 − cos   8  2 4 2  x 



= lim x →0

x2   x2  8 1 − cos  1 − cos  8  2  4 x 



2 2 8 2 x 2 x 2 2 ⋅ sin ⋅ sin = lim x →0 x 8 4 8





 x2 sin 32   24 = lim x →0 x 8  x   4 =

2

  x2   x 2  2  sin    ⋅  28   4  x     8

sin x x = 1− 0 cos 2 x 1+ 0 1+ x 1−

 sin x  cos 2 x ∵ x → 0, x → 0 as x → ∞   



2

   x 2 2  ⋅    8   



x→0

log e sin x ∞    form  ∞ cot x cot x = lim  [Using L’Hospital’s Rule] x → 0 − cosec 2 x − cos x ⋅ sin x) = 0. = lim( x→0

= lim x →0

1

1 + 3x  33. (d) lim   x →∞  2 + 3 x 

ln(1 + h) 2 − ln(1 + 2h) h →0 h2

3 =   3

= – lim



  (1 + h) 2  h2  ln  ln 1 +   1 + 2h   1 + 2h  = – lim = – lim 2 h→0  h 2 h→0 h   1 + 2h  (1 + 2h)



 h2  ln 1 +  1 + 2h  1 ⋅ = – lim =–1 h→0 1 + 2h  h2   1 + 2h 

1

  x 1 +3   = lim  x  x →∞ 2  + 3  x



1  1    −1 x x 

0 −1

= 10 = 1.

34. (c) lim log n −1 (n) ⋅ log n (n + 1) ⋅ log n +1 (n + 2)...log k (n k )  n −1 n →∞   log n log(n + 1) log(n + 2) log(n k )  ⋅ ⋅ ... = nlim  →∞ log( n − 1) log n log(n + 1) log(n k − 1)  

= nlim →∞

1

 sin x  x − a  sin x − sin a  x − a 30. (a) lim   = lim 1 +  x → a  sin a  x→ a  sin a sin x − sin a

sin a  ( x − a )sin a  sin x − sin a   x a sin − sin     1 lim + = x→ a        sin a   

sin x − sin a ( x − a )sin a

1− x 1− x

 log m   Using log n m = log n   

log(1 + x)   = 1  Using lim x→0 x 

e

x − sin x 31. (b) lim = lim x →∞ x + cos 2 x x →∞

x →0

1 . 32

= lim x→ a

cos a

32. (c) lim log e (sin x) tan x = lim tan x ⋅ log e sin x (0 ⋅ ∞ form)





  x − a   x − a  1 sin  2   2   sin a  

= 1.

ln(1 + 2h) − 2 ln(1 + h) 29. (b) lim h →0 h2



2 x+a x−a 1 ⋅ cos  sin  2   2  sin a x−a

e

log n k log n  ∞ = k lim  form  n → ∞ log( n − 1) ∞  log(n − 1)

= k lim

n →∞

1 1

n

[Using L’Hospital’s Rule]

n −1

 1 = k lim 1 −  = k. n →∞  n 1 + 5x2  35. (a) lim  x→0  1 + 3x 2  

1

x2

= e

1 1+ 5 x 2   lim −1 x → 0 x 2 1+ 3 x 2 

lim g ( x )[ f ( x ) −1]   [ f ( x ) ]g ( x ) = e x → a   Using lim x→ a  provided f ( x) → 1 and g ( x) → ∞ as x → a   

36. (b) Let f (x) =

2

1 + tan x 1 + sin x

⇒ lim

= e 2.

x→a

and g (x) =

1 sin x

 1 + tan x  ∴ lim  x →0 1 + sin x   

=e

⇒ k = 4.

1 . sin x

Clearly f (x) → 1 and g (x) → ∞ as x → 0.

x2

41. (b) lim x →0

1  1+ tan x  −1 lim  x →0 sin x  1+ sin x 

lim g ( x )  f ( x ) −1 g ( x)    f ( x) = e x →a  Using lim  x→a    lim

1  tan x − sin x    1+ sin x 

= e x →0 sin x 

lim

1− cos x

∫ cos t dt 0

2

= lim x →0

2 x cos x 4 x cos x + sin x

= lim x →0

2 cos x 4 − 8 x 4 sin x 4 2 cos x − x sin x

2−0 = 1. = 2−0



1



x

 1 + tan x  = lim  x →0 1 − tan x    = e

1

x

= e



=

39. (a) lim x →1

2 tan x lim x →0 x (1− tan x ) 1  tan x  lim 2  ⋅  x  1− tan x

e x →0

= e 2.

x n + x n −1 + x n − 2 + ... + x 2 + x − n x −1

0   0 form   

nx n −1 + (n − 1) x n − 2 + ... + 2 x + 1 1 [Using L’Hospital’s Rule] n (n +1) = n + (n – 1) + ... + 2 + 1 = . 2

= lim x →1



40. (a) lim

x→a

f (a ) g ( x) − f (a) − g (a) f ( x) + g (a) =4 g ( x) − f ( x)

Applying L’ Hospital rule ⇒ lim

f (a ) g' ( x) − g (a ) f ' ( x) =4 g' ( x) − f ' ( x)

⇒ lim

kg' ( x) − kf ' ( x) =4 g' ( x) − f ' ( x)

x→a

x→a

x − sin x = 1. x − sin x

x →∞



2 cos = lim x →∞

x +1 + x x +1 − x ⋅ sin 2 2



2 cos = lim x →∞

x +1 + x 1 ⋅ sin 2 2( x +1 + x )



= 2 × (some number between – 1 and 1) × sin 0 = 0.

x 44. (d) For x > 0,  ∵ x < x + 1,  ∴  x + 1 < 1

1  1+ tan x  lim  −1 x →0 x  1− tan x 

1  1 + tan x  ∵ 1 − tan x → 1 and x → ∞ as x → 0   

[Using L’Hospital’s Rule again]

43. (c) lim(sin x + 1 − sin x )

1 ⋅ (e ) − 1 1 = (e − 1) lim 1 n = lim n →∞ n →∞  e n (e1 n − 1) −1    n 1   = (e – 1) × 1 = (e – 1).

  π 38. (b) lim  tan  + x   x →0   4

[Using L’Hospital’s Rule]

42. (a) Clearly, for n = 1, lim x→0

 1 + e1 n + (e1 n ) 2 + ... + (e1 n ) n −1  = lim   n →∞ n   1n n



0   0 form   



x sin x

= e x →0 cos x (1+sin x ) = e0 = 1.

 1 e1 n e 2 n e( n −1) n  37. (c) lim  + + + ... +  n →∞ n n n n  

k[ g' ( x) − f ' ( x)] =4 g' ( x) − f ' ( x)

59

2

lim

= e x→ 0 1+3 x

Limits

= e

2 x2 1 lim ⋅ x → 01+ 3 x 2 x 2

x  −1  ∴  sec  x + 1  is not defined    x  sec −1   does not exist. Hence xLt →∞  x +1 45. (b) Case I. n is a positive integer xn nx n −1 lim x = lim x x →∞ e x →∞ e = lim x →∞

n (n − 1) x n − 2 n! = ... = lim x x →∞ e ex [Using L’Hospital’s Rule repeatedly]

= 0. Case II. n is a negative integer. xn x−m lim x = lim x x →∞ e x →∞ e [Putting n = – m, where m is a positive integer] 1 1 = = lim = 0. x →∞ x m e x ∞ Case III. n = 0 1 1 xn lim x = lim x = = 0. x →∞ e x →∞ e ∞ Hence lim

x →∞

xn = 0 for all values of n. ex

60

(1 + h)sin (1 + h − [1 + h]) 1 + h −1 (1 + h)sin (1 + h − 1) =  lim h →0 h sin h 1 + h) =  lim(  = 1. h →0 h (1 − h)sin (1 − h − [1 − h]) LHL = lim h →0 1− h −1 (1 − h)sin (1 − h) = lim = – ∞. h →0 −h Since LHL ≠ RHL,

Case III. | x | = 1

46. (d) RHL = lim h →0

Objective Mathematics

x sin ( x − [ x]) does not exist. x −1 47. (b) lim (1 + x) (1 + x2) (1 + x4) ... (1 + x2n)

lim f ( x) = lim n →∞

(1 + x)(1 − x 2 )(1 + x 3 )(1 − x 4 )...(1 − x 4 n ) x →−1 [(1 + x )(1 − x 2 )(1 + x 3 ) (1 − x 4 )...(1 − x 2 n )]2



(1 − x 2 )(1 + x 2 )(1 + x 4 )...(1 + x 2 n ) 1− x . . . . . . . . . 1 − x4n + 2 1 for | x | < 1. = lim = n →∞ 1− x 1− x = lim

n →∞

sin x 1+ x + sin x x = 1 + 0 = 1. 48. (b) lim = lim x →∞ x − cos x x →∞ 1 − cos x 1− 0 x   1 sin x = lim y sin   = O × (a finite quantity)  ∵lim x →∞ y → 0 x  y     cos x   =0. Similarly lim = 0. x →∞ x   49. (a) lim x →0

 = lim x →0

3

1 + sin x − 3 1 − sin x x

1 (1 + sin x) − (1 − sin x) × (1 + sin x) 2 / 3 + (1 + sin x)1/ 3 (1 − sin x) 2 / 3 + (1 − sin x) 2 / 3 x

2⋅ = lim x →0 ×

sin x x

(1 + sin x)

2/3

1 + (1 + sin x) (1 − sin x)1/ 3 + (1 − sin x) 2 / 3

= xlim →−1

1 + x 2 n +1 1 − x 2 n + 2 1 − x4n ... × × × 1+ x 1 − x2 1 − x2n

= xlim →−1

x 2 n +1 − (−1) 2 n +1 x 2 n + 2 − (−1) 2 n + 2 × x − (−1) x 2 − (−1) 2 ×...×

x →1

(1 − x)(1 + x)(1 + x 2 )(1 + x 4 )...(1 + x 2 n ) = lim n →∞ 1− x

1 −1 = 0. 1+1

51. (a) lim

∴ lim n→∞

n →∞

=

2n + 1 2n + 2 2n + 3 2n + 4 4n ⋅ ⋅ ⋅ ... = 4nC2n. 1 2 3 4 2n

52. (b), (c) Case I. x ≠ mπ (m is an integer) 1 1 lim = = 0. n →∞ 1 + n sin 2 nx ∞

Case II. x = mπ (m is an integer) lim

n →∞

1 1 = = 1. 1 + n sin 2 nx 1

log (tan 2 x)  ∞  log tan x (tan 2 x) = lim  form  53. (a) xlim x →0+ log (tan x )  ∞ → 0+   log m   Using log n m =  log n   1 (2 sec 2 2 x) x 2 tan = lim [Using L’Hospital’s Rule] x →0 1 (sec 2 x) tan x

50. (a), (b), (c) Case I. | x | < 1 x2n − 1 0 − 1 lim f ( x) = lim 2 n = = – 1. n →∞ n →∞ x +1 0 +1 Case II. | x | > 1 1 1 − 2n x = 1 − 0 = 1. lim f ( x) = lim n →∞ n →∞ 1 1 + 2n 1 + 0 x

0   0 form   

= lim x →0

2 sin x cos x 2 sin 2 x = lim  sin 2 x cos 2 x x →0 sin 4 x

= lim x →0

4 cos 2 x [Using L’Hospital’s Rule] 4 cos 4 x

= 1. 2 sin x − sin 2 x dx x3 d 2 sin x − sin 2 x 2 sin x − sin 2 x dx = ⇒ f ′ (x) = dx ∫ x3 x3

54. (d) f (x) =



1/ 3

1 2 =2×1× = . 3 3

x 4 n − (−1) 4 n x 2 n − (−1) 2 n

∴ lim f' ( x) = lim x →0

= lim x →0

x →0

2 sin x − sin 2 x  x3

2 cos x − 2 cos 2 x  3x 2

0   0 form    0   0 form   

[Using L’Hospital’s Rule] = lim x →0

−2 sin x + 4 sin 2 x  6x

0   0 form    [Using L’Hospital’s Rule]

[Using L’Hospital’s Rule] 6 = = 1. 6  11 x + 2 1 x + 3 1 x + ... + n 1 x  55. (a) lim   x →∞  n    1 + 2 + 3 + ... + n    = lim y →0 n   y





= e = e

y

y

y

nx

 sin([ x − 3])   sin( − 4)  60. (c) L.H.L. = lim−   =   x→0  [ x − 3]   −4 

 sin 4  =   =–1  4 



∵ π 20.

13. (a) When t ≥ 0, | t | = t ∴ x = 2t – t = t and y = t2 + t.t = 2t2 ⇒ y = 2x2, when x ≥ 0 [As t ≥ 0 ⇒ x ≥ 0]

87

⇒ (a + 1) + 1 = xlim →0

Continuity and Differentiability



88



Also, for t ≤ 0, | t | = – t



∴ x = 2t + t = 3t and y = t2 + t (– t) = 0

Objective Mathematics



sin ( x − 1), if x ≥ 1 =  sin ( 1 − x ), if x < 1



⇒ y = 0 when x ≤ 0 [As t ≤ 0 ⇒ x ≤ 0]



Hence the function is defined as :



Now, L (gof ) ' (1) = lim



2 x 2 y = f (x) =  0



= lim



∴ (gof ) (x) is not differentiable at x = 1.

,x≥0 ,x≤0



Now, L f ′ (0) = hlim →0

f (0 − h) − f (0) 0 = lim =0 −h h→0 −h



and R f ′ (0) = hlim →0

2h 2 f (0 + h) − f (0) = hlim =0 →0 h h



f (1 − h) − f (1) 31− h − 3 = hlim →0 −h −h

 3− h − 1    = 3 log 3 = 3 hlim →0  −h  f (1 + h) − f (1) R f ′ (1) = hlim →0 h 4 − (1 + h) − 3 = lim h→0 h −h = hlim = – 1. →0 h

Since L f ′ (1) ≠ R f ′ (1), therefore, f (x) is not differentiable at x = 1 but f (x) is continuous at x = 1 (as L f ′ (1) and R f ′ (1) are finite). 15. (c) The function f (x) can be rewritten as

 x3 , − 1 < x < 1 f (x) =   x, x ≤ − 1 or x ≥ 1 f (1 − h) − f (1) Now, L f ′ (1) = hlim →0 −h

( gof ) (1 − h) − ( gof ) (1) −h

sin  1 − (1 − h)  − sin 0

h→0

−h

= lim

h→0

sin h →– ∞ −h

17. (c) We have,

Since L f ′ (0) = R f ′ (0), therefore, f (x) is differentiable and hence continuous at x = 0. 14. (b) L f ′ (1) = hlim →0

h→0



 x 1 − x , x ≤ − 1   x 1 + x , − 1 < x < 0 f (x) =  .  x , 0 ≤ x 0 f (x) =  =  xe , x > 0 If x = – 1, then x = x . So, f (x) = x.  0 0 , x = 0 ,x=0 If – 1 < x < 0, then x < x3. So, f (x) = x3.   3 3  If x = 0, then x = x . So, f (x) = x .  If 0 < x < 1, then x > x3. So, f (x) = x. f (x) is differentiable as well as continuous 3 If x = 1, then x = x . So, f (x) = x. everywhere except possibly at x = 0. If x > 1, then x < x3. So, f (x) = x3. Now, Thus, f (x) = x, x ≤ – 1 f ( 0 + h ) − f ( 0) R f ′ (0) = hlim x3, – 1 < x ≤ 0   x, 0 < x ≤ 1   x3, x > 1. →0 h Clearly, f (x) is not differentiable at x = – 1, 0, 1. − 2/ h he − 0 = hlim = hlim e– 2/h = 0 →0 →0 1  2 h x≠0  x sin , x 24. (a), (b)  We have, g (x) =  f (0 − h) − f (0) lim − h − 0 L f ′ (0) = lim = h→0 = 1. 0, x = 0. h→0 −h −h For x ≠ 0, 1  1 1 1 Since L f ' (0) ≠ R f '(0), therefore f (x) is not dif g′ (x) = x2 cos     − 2  + 2x sin  = – cos  +   x x x x ferentiable at x = 0 but f (x) is continuous at x = 0 1 2x sin  . [as L f ′ (0) and R f ′ (0) are finite]. x For x = 0

27. 1 x sin − 0 g ( x ) − g ( 0) x lim = g′ (0) = lim x→0 x→0 x−0 x 1 28. = lim x sin = 0. x→0 x 1 1  x≠0 2 x sin − cos , ∴ g′ (x) =  x x 0, x=0 29. 1 g′ is not continuous at x = 0 as cos  is not continux ous at x = 0. Also, f is not differentiable at x = 0.

2

(d) Since | x – 1 |, | x – 2 | and cos x are continuous functions and the sum of continuous functions is also continuous, so the given function is continuous every where, i.e.,, there is no point of discontinuity. (c) Since logax is continuous where x > 0 and a > 0, a ≠ 1, therefore log (1 + x) is continuous when 1 + x > 0 i.e., x > – 1. ∴ interval = (– 1, + ∞). 1 (a) Let f (x) = log | x | . The points of discontinuity of f (x) are those points where f (x) is undefined or infinite. It is undefined

89



For f to be continuous at x = 0, we must have

Continuity and Differentiability



90

Objective Mathematics

when x = 0 and is infinite when log | x | = 0, | x | RHL = lim {x – [x]} = lim {x – n} = p – n x→ p x→ p = 1, i.e., x = ± 1. x> p ∴ Set of points of discontinuity = {– 1, 0, 1}. f ( p) = p – [ p] = p – n 1 30. (b) We have, f (x) = 1 − x . ∴ f (x) is continuous at all non-integral points p. 35. (c) Let x = n, n ∈ Z As at x = 1, f (x) is not defined, x = 1 is a point of discontinuity of f (x). Then, LHL = xlim (x) = n; →n If x ≠ 1, f [ f (x)] xn

Since, LHL ≠ RHL, therefore f (x) is discontinuous at all integers n. Now, let x = p, n < p < n + 1, where n is an integer. (x) = n + 1, Then, LHL = xlim →p x< p

x> p

36.

 e[ x ] + | x | − 2 , x≠0  32. (d) f (x) =  [ x] + | x | − 1 , x=0 

37.

e[ x ] + | x| − 2 e−1 − 2 lim f ( x) = lim− = − x → 0 [ x] + | x | x→0 −1 e[ x ] + | x| − 2 ex − 2 lim lim f ( x) = lim+ →−∞. = + + x → 0 [ x] + | x | x→0 x→0 x 33. (b) We have,

38.

39.



f (x) = lim (sin x)



π π  0, if |sin x | < 1 i.e. − 2 + nπ < x < 2 + nπ =  1, if |sin x | = 1 i.e. x = nπ + π , n ∈ I  2

f ( p) = ( p) = n + 1. Since LHL = RHL = f ( p), therefore f (x) is continuous at all non-integral points p. (a) As the function log | x | is not defined at x = 0, therefore the set of points of discontinuity is {0}. (c) As the function | sin x | is defined for all real x, therefore the function f (x) is continuous at all real x. Hence, the required set is φ (empty set).  |sin x | (b) The function is not defined for x = nπ, n ∈ I.  sin x Hence the set of points of discontinuity is {nπ : n ∈ I}. (b) f is continuous at x = π/4, if lim f ( x) = f (π/4).

2n

x → π/ 4

x → π/ 4



lim

{x – [x]} = lim {x – (n – 1)} x→n

= n – (n – 1) = 1. RHL = xlim {x – [x]} = lim {x – n} = n – n = 0. →n x→n

xn

x → π/ 4

Now, L = lim (sin 2 x)

n→∞

Then LHL =

(x) = n + 1 RHL = xlim →p





f (π/4) = e–1/2 = 1/ e .

x+ y f ( x) + f ( y ) f , f (0) = 0 and f ′ (0) = 3 = 3 3   f ( x + h) − f ( x ) f ′ (x) = hlim →0 h  3 x + 3h  f  − f ( x)  3  = hlim →0 h f (3 x) + f (3h) f (3 x) + f (0) − 3 3 = hlim →0 h f (3h) − f (0) = hlim =3 →0 3h

RHL = lim

h→0



= lim [6 – 5 (1 + h)] = 6 – 5 = 1 h→0



 ∴ f (x) continuous at x = 1. At x = 3 : LHL = lim (3 – h) = hlim [6 – 5 (3 – h)] = – 9, →0



f (3) = 3 – 3 = 0.



0   0 form   

lim  f (x) = lim 1 − tan x  x → π/ 4 4x − π

x → π/ 4

lim

x → π/ 4

2 1 −sec 2 x = − = − 4 4 2

∴  For f (x) to be continuous at x =

π 4

π 1 f  =– 2 4 43. (a) Differentiability at x = 0 :



f (0 − h) − f (0) −h − h (e − 1/ h − e1/ h ) − h (e − 1/ h + e1/ h )

L f ′ (0) = hlim →0 =  lim

h→0

e− 2 / h − 1  = – 1. e− 2 / h + 1 f ( 0 + h ) − f ( 0) R f ′ (0) = hlim →0 h

 =  hlim →0

1/ h

− 1/ h

h (e − e ) = hlim →0 h (e1/ h + e − 1/ h )

f (x) =

xn



∑ n!

(log a)n =

n=0

= ex log a =



( x log a ) n n! n=0 ∞



= a x. f (0 − h) − f (0) a− h − 1 lim = −h −h h→0

= logea

f ( 0 + h ) − f ( 0) h

R f ′ (0) = lim

h→0



= lim

h→0

ah − 1 = logea h

Since L f ′ (0) = R f ′ (0), ∴ f (x) is differentiable at x = 0. Since every differentiable function is continuous, therefore, f (x) is continuous at x = 0. 46. (b) We have, for x ≠ 0,





h→0

L  f ′ (0) = hlim →0

Therefore, the only point of discontinuity is x = 3. 42. (a) We have,

= lim (1 + h)2 [(1 + h)1/h ]– 2

= 1 × e– 2 = e– 2. Since LHL ≠ RHL,  ∴ f (x) is not continuous at x = 0. 45. (d) We have

h→0

Since LHL ≠ f (3),  ∴ f (x) is discontinuous at x = 3.

1 1 2− +  h h

h→0

2 2− h

h→0

1 (2 + 3) = 1. and f (1) = 5 Since LHL = RHL = f (1),



f (0 + h) = lim (h + 1)

= lim (1 + h)

RHL = lim f (1 + h) h→0

h→0

h→0

1 [2 (1 – h)2 + 3] = 1 = lim h→0 5



1 1 2− −  h h

= lim (1 – h)2 = 1

LHL = lim f (1 – h) h→0

f (0 – h) = lim (− h + 1)

LHL = hlim →0

1 − e− 2 / h = lim 1 + e − 2 / h = 1. h→0 Since L f ′ (0) ≠ R f  ‘ (0), ∴ f (x) is not differentiable at x = 0. But since L f ′ (0) and R f ′ (0) are finite, therefore f (x) is continuous at x = 0. Hence, f (x) is continuous every where but not differentiable at x = 0.

f (x) = 

x x x + + 1 + x ( x + 1) (2 x + 1) (2 x + 1) (3 x + 1) + ... to ∞

=

x ∑ [( − 1 ) + 1] [nx + 1] n x n =1

=

∑  (n − 1) x + 1 − nx + 1







1

1

  1  = nlim 1 −  = 1. →∞ nx + 1  n =1

 

1, x ≠ 0  For x = 0, f (x) = 0. Thus, f (x) = 0, x = 0    Clearly, lim f (x) = lim f (x) = 1 ≠ f (0). x → 0−

x → 0+

So, f (x) is discontinuous at x = 0. 47. (b) For x ≠ 0, we have x /1+ x x /1+ x f (x) = x + = x + x / 1 + x = x + 1. 1 1− 1+ x For x = 0, f (x) = 0  x + 1, x ≠ 0 Thus, f (x) =  0 , x = 0  Clearly, lim f (x) = lim+ f (x) = 1 ≠ f (0). x → 0−

x→0

91

  f (0) = 0 ⇒ c = 0 ∴  f (x) = 3x. 41. (b) At x = 1 :

Continuity and Differentiability

44. (a) The only doubtful point is x = 0.

∴  f (x) = 3x + c,

92

So, f (x) is discontinuous and hence not differentiable at x = 0.

Objective Mathematics

48. (b) Continuity at x = 2 : LHL = lim f (2 – h) h→0

h→0

RHL = lim f (2 + h) h→0

= lim [(2 + h)2] + [– 2 – h]2 h→0

= lim {4 + (– 3)2} = 13. h→0

Since LHL ≠ RHL,  ∴ f (x) is not continuous at x = 2. Differentiability at x = 2 : f ( 2 − h ) − f ( 2) LHD = hlim →0 ( 2 − h) − 2 7−8 → ∞. = lim h→0 −h So, f (x) is not differentiable at x = 2. Hence, the function f (x) is neither continuous nor derivable at x = 2. 49. (b) For x ≠ – 1, we have

f (x) = 1 – 2x + 3x2 – 4x3 + ...∞ 1 = (1 + x)– 1 = 1 + x . 1 lim f (– 1 – h) = lim 1 − 1 − h → – ∞ h→0 h→0 So, f (x) is not continuous at x = – 1. Also, lim

h→0

f (− 1 − h) − f (−1) = hlim (−1 − h) − (−1) →0

−1 −1 h −h

1+ h → ∞. = lim h→0 h2 So, f (x) is not derivable at x = – 1. Hence, f (x) is neither continuous nor derivable at x = – 1. 1 50. (a) Since f (x) = , ∴  sin x ∉ [0, 1] [sin x] ⇒  x ∉ [2nπ, (2n + 1) π] – (4n + 1) π/2, n ∈ I. ∴ f (x) is not continuous if x ∈ (2nπ, 2nπ + π), n ∈ I. 51. (b) Since f (x) is continuous at x =

Thus the required value is f ′ (0). f (h) − f (0) = k (say) h f (0) − f (0 − h) ∴  f ′ (0 – ) = hlim →0 h f (0) − f (h) = hlim =– k →0 h + – f ′ (0 ) ≠ f ′ (0 ), but both are finite so f (x) is continuous at x = 0 but not differentiable at x = 0. 54. (c) Since f (x) is continuous at x = 0, ∴ lim f (x) = f (0). 53. (b) Let f ′ (0 + ) = hlim →0

x→0







b (1 − sin x) (π − 2 x) 2 = a

lim

 π  b 1 − sin  + h   cos x 2   = lim 2 = a h→0 3 cos 2 x  π  π − 2  2 + h     2

x → π/ 2

lim

h→0



= f (a) + lim f (h) = f (a) + f (0)

h→0

h→0

= f (a + 0) = f (a). ∴  f (x) is continuous at x = a. Since x = a is any arbitrary point, therefore f (x) is continuous for all x. 55. (c) We have,

f (x) = nlim →∞



= nlim →∞

(2 sin x) 2 n 3 − (2 cos x) 2 n n

(2 sin x) 2 n ( 3 ) 2 n − (2 cos x) 2 n

f (x) is discontinuous when ( 3 ) 2 n − (2 cos x) 2 n = 0 i.e.,, cos x =

1 − sin x = 3 cos 2 x

x → π/ 2 x > π/ 2

lim f (x) = lim f (a + h)

x→a

= lim [ f (a) + f (h)] [ ∵ f (x + y) = f (x) + f ( y)]

π , 2

lim

x → π/ 2 x < π/ 2

Take any point x = a, then at x = a



π ∴ LHL = RHL = f   2 2



b  sin h / 2  1 = hlim   =a → 0 8  h/2  3 1 b = = a. ⇒ 3 8 ∴ a = 1/3 and b = 8/3. f ( x) 52. (d) We have, f (0) = 0, g (x) = , x f ( x) = lim f' ( x) = f ′ (0) lim g ( x) = lim x→0 x→0 x→0 x 2

{3 + (– 2)2} = 7. = hlim →0



1 b (1 − cos h) = hlim =a →0 3 4h 2



= lim [(2 – h)2] + [– 2 + h]2





⇒x =

π . 6

3 2

56. (c) Since f (x) is continuous in [0, 1], therefore

  n  n   lim f  2 n + 1  = f  nlim →∞ 2 n + 1  n→∞   



1 = f   = 2. 2

L f ′ (0) = hlim →0

f (h) − f (0) h2 = hlim =0 →0 h h



log cos h = hlim → 0 log (1 + h 2 ) 

f (− h) − f (0) −h



− tan h = hlim → 0 2 h / (1 + h 2 ) = 1/2.

− ( − h) − 0 = – 1. −h Since R f ′ (0) ≠ L f ′ (0), ∴ f ′ (0) does not exist.

= hlim →0

At x = 1: f (1 + h) − f (1) h 2 [(1 + h) − (1 + h) + 1] − 1 lim h→0 h

R f ′ (1) = hlim →0

= =

h2 + h lim = lim (h + 1) = 1 h→0 h h→0

L f ′ (1) = hlim →0 =

lim

h→0

(1 − h) 2 − 1 f (1 − h) − f (1) = hlim →0 −h (1 − h) − 1

h 2 − 2h = lim (2 – h) = 2. −h h→0

Since R f ′ (1) ≠ L f ′ (1), ∴ f ′ (1) does not exist. ∴ f (x) is not differentiable at x = 0 and x = 1. x x 58. (b) f (x) = 1 + x , x > 0; f (x) = 1 − x , x < 0;

f (x) = 0, x = 0.

If x > 0 or x < 0, f (x) is a rational function of x, having non-zero denominator. Hence f (x) is continuous and differentiable for x > 0 or x < 0. So the only doubtful point is x = 0. At x = 0 : h −0 f (h) − f (0) 1+ h lim lim = h→0 =1 R f ′ (0) = h → 0 h h −h −0 f (− h) − f (0) 1 − ( − h) lim lim L f ′ (0) = h → 0 = h→0 =1 −h −h

0   form  0 

Since L f ′ (0) = R f ′ (0), therefore f (x) is differentiable at x = 0. Since differentiability ⇒ continuity, therefore f (x) is continuous at x = 0. 60. (b) We have, f (0 − h) − f (0) L f ′ (0) = lim −h h→0 − h (1 + h sin 1 / h) −h 1 1  = hlim + h sin   →0 h  h

=  hlim →0

= ∞ + 0 × a finite quantity = ∞ ∴ f (x) is not differentiable at x = 0. Clearly, f (x) is continuous at x = 0. 61. (c) We have f (2 ⋅ 5 − h) − f (2 ⋅ 5) L f ′ (2 ⋅ 5) = lim −h h→0



[2 ⋅ 5 − h − 2] − [2 ⋅ 5 − 2] = hlim →0 −h



= hlim →0



and R f ′ (2 ⋅ 5) = hlim →0



= lim



∴  f ′ (2 ⋅ 5) = 0.



Also, L f ′ (5) = lim



2−3 → ∞. = lim [5 − h − 2] − [5 − 2] = hlim →0 − h h→0 −h

h→0

0 =0 −h

f (2 ⋅ 5 + h) − f (2 ⋅ 5) h

[ 2 ⋅ 5 + h − 2 ] − [ 2 ⋅ 5 − 2] 0 = lim =0 h→0 h h

h→0

f (5 − h) − f (5) −h

Since R  f ′ (0) = L f ′ (0) = 1, ∴ f ′ (0) = 1 i.e., f (x) is differentiable at x = 0. Since differentiability ⇒ continuity,  Hence, f ′ (2 ⋅ 5) = 0 while f ′ (5) does not exist. ∴ f (x) is also continuous at x = 0. Hence f (x) is continuous as well as differentiable π  π for all real x. f  − h − f   π 4  4 59. (a) We have, 62. (c) L f ′   = lim  h→0 −h 4 f (0 − h) − f (0) lim L f ′ (0)= hlim = h → 0 − h log cos h  π π   →0 −h  tan  4 − h   −  tan 4  − h log (1 + h 2 )     = lim 0 − 1 → ∞   = lim h→0 h→0 −h −h log cos h 0  = lim  π  form  log (1 + h 2 ) h→0 0  ∴ f ′    does not exist. 4 − tan h f (0 − h) − f (0) = lim 2h / (1 + h 2 ) = – 1/2. h→0 63. (c) L f ′ (0) = lim   −h h→0 h log cos h f (0 + h) − f (0) lim R f ′ (0) = hlim = h → 0 h log (1 + h 2 ) 1 − 1 − h2 →0 h = hlim   →0 −h

93



R f ′ (0) = hlim →0

Continuity and Differentiability

57. (d) At x = 0 :

94



Objective Mathematics





= lim   h→0

= lim

h→0

1 − (1 − h 2 ) x −h −1 1 + 1 − h2 h→0

= lim



= lim

1− 1− h

h→0

h→0

h→0

1 + 1 − h2

h p cos

−1 = . 2

= lim

h→0

1+ 1− h



2

=

1 . 2



Therefore, f (x) is not differentiable at x = 0. Since L f ′ (0) and R f ′ (0) are finite, therefore, f (x) is continuous at x = 0. Hence, f (x) is continuous but not differentiable at x = 0. f (0 − h) − f (0) 64. (b) L f ′ (0) = lim h→0 −h



= lim

h→0

= hlim →0

1− e −h

 h2  h ⋅ 1 − + ... 2!   = lim = – 1. h→0 −h f ( 0 + h ) − f ( 0) R f ′ (0) = lim h→0 h 2

1 − e− h = lim = 1. h→0 h Hence, f (x) is not differentiable at x = 0. Since L f ′ (0) and R f ′ (0) are finite, therefore, f (x) is continuous at x = 0. Hence, f (x) is continuous but not differentiable at x = 0. 65. (a) Continuity at x = 0: 1 LHL = lim f (0 – h) = lim (– h) p cos = 0 if p h→0 h→0 h >0 RHL = lim f (0 + h) = lim hp cos h→0

h→0



=  lim

(− h) p cos

h→0

= lim (– h) h→0



−h p – 1

cos

= lim h p – 1 cos h→0

1 = 0 if p > 1 h

1 −0 h 1 = 0 if p – 1 > 0, i.e., p > 1; h

(| 2 − h − 1| + | 2 − h − 3|) − (| 2 − 1| + | 2 − 3|) = hlim →0 −h | 1 − h | + | 1 + h | − 2 1− h +1+ h − 2 = hlim = hlim →0 →0 −h −h 0 = hlim = 0. →0 −h f ( 2 + h ) − f ( 2) R f ′  (2) = hlim →0 h (| 2 + h − 1 | + | 2 + h − 3|) − 2 = lim h→0 h



= lim |1 + h | + | −1 + h | − 2 h→0 h



= lim 1 + h + 1 − h − 2 = lim 0 = 0. h→0 h→0 h h

Since L f ′ (2) = R f ′ (2), therefore f (x) is differentiable at x = 2 and f ′ (x) = 0 at x = 2. 67. (b) We have,  0, 0 ≤ x ≤ b  2b3  −8 f ′ (x) =  x + 2 , b < x ≤ a 9x 9  −2 3 3  2 (a − b ), x > a 9x

∴ f ′ (a+) = lim f ′ (a + h)



 − 2 (a 3 − b3 )  ⋅ = lim  2  h→0  9 ( a + h) 



=

1 = 0 if p > 0 h

and f (0) = 0. ∴ f (x) is continuous at x = 0 if p > 0 Differentiability at x = 0 : f (0 − h) − f (0) L f ′ (0) = lim −h h→0



1/ 2

1/ 2





− h2

   h4 2 − ...   1 − 1 − h + ! 2    −h

h

1 −0 h

∴ f (x) is differentiable at x = 0 if p > 1. f (2 − h) − f (2) 66. (b) L f ′ (2) = lim −h h→0

2

h 1

f ( 0 + h ) − f ( 0) h

R f ′ (0) = lim

f ( 0 + h ) − f ( 0) h

R f ′ (0) = lim



1

h→0

− 2 a 3 − b3 ⋅ . 9 a2

and f ′ (a– ) = lim f ′ (a – h) h→0



 −8 2b3  ( a − h) + = lim   h→0 9 ( a − h) 2   9



=

− 8a 2b3 − 8a 3 + 2b3 + 2 = 9 9a 9a 2

∴ f ′ (a+ ) ≠ f ′ (a– ). 68. (a) We have, f (1 − h) − f (1) L f ′ (1) = lim −h h→0



0 −1 = hlim →∞ →0 −h ∴ f ′ (1) does not exist. f (2 − h) − f (2) Also, L f ′ (2) = lim −h h→0



(2 − h) [2 − h] − (2 − 1) [2] −h 2−h−2 = lim =1 −h h→0

= lim

h→0

f (2 + h) − f (2) h (1 + h) [2 + h] − (2 − 1) [2] = lim h→0 h 2 + 2 h − 2 = lim = 2. h→0 h ∴ f ′ (2) also does not exist. 69. (c) We have, and R  f ′ (2) = lim

h→0

f (0 − h) − f (0) −h

L f ′ (0) = lim

h→0

= lim 3 − 1 → – ∞ h→0 −h ∴ f (x) is not differentiable at x = 0 Also, if x < 0 or x ≥ 0 then | x | ≥ 0 ∴ f ( | x | ) = 2 | x | + 1 for all x. f ( 0 + h ) − f ( 0) ∴ R f ′ (0) = lim h→0 h 2h + 1 − 1 =2 = lim h→0 h f (0 − h) − f (0) and L f ′ (0) = lim −h h→0 = lim

h→0

2 (− h) + 1 − 1  = 2 −h

h 2 + 5h = hlim = 5. →0 h ∴ f ( | x | ) is not differentiable at x = 2 but f ( | x | ) is continuous at x = 2 (as L f ′ (2) and R f ′ (2) are finite). Hence, f ( | x | ) is continuous but not differentiable in (– 3, 3). f ( 4 + h ) − f ( 4) 71. (b) f ′ (4) = hlim →0 h f ( 4 + h ) − f ( 4 + 0) = hlim →0 h 2 f (4) ⋅ f (h) − 2 f (4) f (0) = hlim →0 h



[Using f (x + y) = 2f  (x) ⋅ f ( y) for all x, y] f (h) − f (0) = 2f (4) hlim = 2f (4) × f ′ (0) →0 h = 2 × 2 × 3 = 12. f ( 4 + h ) − f ( 4) h f ( x ) f ( h) − f ( x ) = hlim →0 h 1 + h φ ( h) − 1 = f (x) hlim = f (x) hlim φ (h) →0 →0 h = f (x) ⋅ 1 = f (x) f ( x − h) − f ( x ) and L f ′ (x) = hlim →0 −h

72. (c) R f ′ (x) = hlim →0

= hlim →0



1 − h φ ( − h) − 1 = f (x) ⋅ 1 = f (x). −h Hence f ′ (x) exists and is equal to f (x). = f (x) hlim →0

73. (a) Let f (x) be a periodic function with period T. i.e., f (x + T) = f (x). Now, f ′ (x) = lim

h→0

∴ f ( | x | ) is differentiable at x = 0. 70. (b) If – 3 < x < – 1 then 1 < | x | < 3

The nature of f ( | x | ) changes only at x = 2. f (2 − h) − f (2) L f ′ (2) = lim −h h→0

= lim 5 − (2 − h) − (4 + 2 − 3) = – 1 h→0 −h f (2 + h) − f (2) R f ′ (2) = hlim →0 h [(2 + h) + (2 + h) − 3] − (4 + 2 − 3) = hlim →0 h 2



f ( x + h) − f ( x ) h

f ( x + T + h) − f ( x + T ) h f ( x + h) − f ( x ) = f ′ (x). h

∴ f ′ (x + T) = lim

h→0

and if – 1 ≤ x < 0 then 0 < | x | ≤ 1. 5 − | x |, 0 < | x | ≤ 1  ∴ f ( | x | ) = 5 − | x |, 1 < | x | < 2  2 | x | + | x | − 3, 2 ≤ | x | < 3

f ( x ) f ( − h) − f ( x ) f ( − h) − 1 = f (x) hlim →0 −h −h





= hlim →0

Similarly, f “ (x + T) = f “ (x). ∴ f ′ and f “ are also periodic functions with the same period. 74. (d) We have,

  | x|  sgn   sgn [sgn (x)] =   x  sgn (0) 



 | xx |  | x| =  x   0

,x≠0 ,x=0

,x≠0 ,x=0

95

(1 − h) [1 − h] − 1 [1] = hlim →0 −h

Continuity and Differentiability



96

x  ,x≠0 ∴ f (x) =  | x | 0 , x = 0 

Objective Mathematics



Continuity at x = 0 :

LHL = hlim f (0 – h) = lim →0

h→0

−h = – 1 and f (0) = 0 |− h|

∴ f (x) is not continuous and hence not differentiable at x = 0. 75. (a) When t < 0, x = 2t – (– t) = 3t < 0 and y = t2 + t (– t) = 0 ∴ y = 0 when x < 0. When t ≥ 0, x = 2t – t = t ≥ 0 and y = t2 + t (t) = 2t2 = 2x2 ∴ y = 2x2 when x ≥ 0. Thus, the function f is given by : 0 , x < 0 f (x) =  2 2 x , x ≥ 0

Since, L f ′ (0) = R f ′ (0), therefore f (x) is differentiable at x = 0. Since differentiability ⇒ continuity, therefore f (x) is continuous at x = 0. Hence, f (x) is continuous as well as differentiable in the interval [– 1, 1]. 76. (b) If | 2sin x | < 1, f (x) x x = nlim = = x. →∞ (2 sin x) 2 n + 1 0 +1 If | 2 sin x | = 1, f (x) = nlim →∞

x (2 sin x) 2 n + 1

x 1 = = x. 1+1 2 If | 2 sin x | > 1, f (x) = nlim →∞





1 or 2

−1 2

sin x


1 2

Clearly, f (x) is not continuous at x = nπ ±

The only doubtful point is x = 0. f (0 − h) − f (0) 0−0 L f ′ (0) = lim = hlim = 0. →0 −h −h h→0 R f ′ (0) = hlim →0

⇒ | sin x | >

and | 2 sin x | > 1

78. (a) We have, f (x) = 2x + | x – x2 |, – 1 ≤ x ≤ 1

= 2x + | x | | 1 – x |, ­– 1 ≤ x ≤ 1



2 x + (− x) (1 − x), − 1 ≤ x < 0 =  2 x + x (1 − x), 0 ≤ x ≤ 1



2  x + x , − 1 ≤ x < 0  = 2 3 x − x , 0 ≤ x ≤ 1

Since the polynomial functions are continuous as well as differentiable everywhere, so the only doubtful point is the point x = 0.

= hlim →0

 1  (cos x + sin x)  83. (b) We have, f (x) =   2 

(− h + h 2 ) − 0 −h



= hlim (1 – h) = 1. →0



R f ′ (0) = hlim →0



= hlim (3 – h) = 3. →0



f ( 0 + h ) − f ( 0) (3h − h 2 ) − 0 = hlim →0 h h

Since L f ′ (0) ≠ R  f ′ (0), therefore f (x) is not differentiable at x = 0 but f (x) is continuous at x = 0 (as L  f ′ (0) and R  f ′ (0) are finite). Hence f (x) is continuous but not differentiable in the interval [– 1, 1]. 79. (c) Since [x] and e| x | are not differentiable at x = 0, therefore, for f (x) to be differentiable at x = 0, we must have a = 0, b = 0 and c can be any real number. 80. (a) Since f (x) is continuous at x = 0, therefore lim f (0 – h) = lim f (0 + h) = f (0) h→0



h→0

log (1 + 3h) − log (1 − 2h) h log (1 + 3h) log (1 − 2h) +2 = hlim 3 →0 3h − 2h

⇒ a = hlim →0

= 3 × 1 + 2 × 1 = 5.   81. (b) Since f (x) is differentiable at x = 0, therefore L f ′ (0) = R f ′ (0) ⇒ hlim →0 ⇒ hlim →0

f (0 − h) − f (0) f ( 0 + h ) − f ( 0) = hlim →0 −h h (a + b | − h | + c | − h |4 ) − a −h = hlim →0



f (1 + h) = hlim →0



 [1 + cos (π + π h)]n + 1   nlim  n = hlim →0  → ∞ [1 + cos (π + π h)] − 1 



 (1 − cos π h) n + 1  lim   = hlim n →0  n → ∞ (1 − cos π h) − 1 





 0 + 1   = – 1. = hlim →0  0 −1 f (1 – 0) = hlim f (1 – h) →0  [1 + cos (π − π h)]n + 1  lim lim  = h → 0    n → ∞ [1 + cos (π − π h)]n − 1   (1 − cos π h) n + 1  lim   = hlim   n →0  n → ∞ (1 − cos π h) − 1   0 + 1 = hlim  =–1 → 0    0 −1 0 +1 (1 + cos π) n + 1 Also, f (1) = nlim   = = – 1. → 0 (1 + cos π) n − 1 0 −1 Since, f (1 + 0) = f (1 – 0) = f (1), therefore f (x) is continuous at x = 1. x

1

∫ t cos t

dt

0

3

⇒ – b = b i.e., b = 0. , 0 < x 0  sgn x = 0, x = 0 −1, x < 0 

⇒ k = hlim →0



= hlim →0



= hlim →0



=



= hlim →0



= hlim →0

( fog ) (− 2 − h) − ( fog ) (− 2) −h |[− 2 − h]| − |[ − 2]| −h

sin 2 (π − h) log [1 + π − 2π (π − h) + (π − h) 2 ] 2

1 − cos h sin 2 h ⋅ 2 h log (1 + h 2 ) 1  sin h / 2  h2  sin h  ⋅  ⋅  ⋅  2  h / 2  log (1 + h 2 )  h  2

2

1 . ∴ k = 1/2. 2

94. (a) Since f (x) is continuous at x = π/2 ∴ x lim f (x) = f (π/2) → π/ 2  π  π  sin cos  − h   − cos  − h  2  2   ⇒ k = hlim 2 →0  π  π − 2  2 − h    

Clearly, f (x) is continuous as well as differentiable at x = 0. 90. (b) ( fog) (x) = f (g (x)) = f ([x]) = | [x] |. Now, L ( fog) ‘ (– 2)

1 + cos (π − h) ( π − π + h) 2 ·

 x5 ,x>0  5 0 ,x=0 Therefore, f (x) = x sgn x =   5 − x , x < 0



= hlim →0

 sin (sin h) − sin h  0 = hlim  form  →0 4h 2 0 

= hlim →0



= hlim →0



=

cos (sin h) cos h − cos h 8h

0   form  0  2 − sin (sin h) cos h − sin h cos (sin h) + sin h 8

0 = 0.     ∴ k = 0. 8



1 −3 ⋅ 4 ⋅ (− 8) 8 1 = = . = 4 −4 −2 64 8 ⋅ 4 ⋅ (3) 3



lim f (1 – h) = lim a (1 – h)2 + b = a + b, h→0

h→0

lim f (1 + h) = lim (1 + h) + 3 = 4 and h→0

h→0

f (1) = 4.

∴ f (x) will not be continuous at x = 1 if a + b ≠ 4.

97. (b) We have, 4x + 3  | y | = 5y





Taking limit as h → 0, we get



96. (d) We have,

⇒ 4x + 3y = 5y if y ≥ 0

 2 x, x ≥ 0  and 4x – 3y = 5y if y < 0 ⇒ y =  1  2 x, x < 0

Clearly, y is continuous at x = 0 but not differentiable at x = 0. 2, x ≥ 0  dy = 1 . Also, dx  2 , x < 0 98. (d) We have hlim f (3 – h) →0 = hlim | 3 – (3 – h) | + (3 + 3 – h) →0 = hlim (h + 6) = 6, hlim f (3 + h) →0 →0

f ( x + h) − f ( x ) = – 10x + 3x2h h



lim

h→0

f ( x + h) − f ( x ) = – 10x ⇒ f ′ (x) = – 10x. h

101. (c) Since | x – 1 |, | x – 1 | 2, etc, are continuous at x = 1 ∴ f (x) is continuous at x =1 for all ak ∈ R. Also, | x – 1 |2, | x – 1 |4, etc, are all differentiable at x = 1, whereas | x – 1 |, | x – 1 |3, etc, are not differentiable at x = 1. Therefore, f (x) is differentiable at x = 1 for all a2k + 1 = 0. 102. (b) We have,



f ( x + h) − f ( x ) h f ( x ) ⋅ f ( h ) − f ( x) = hlim →0 h [Using f (x + y) = f (x) ⋅ f ( y)]



= f (x) hlim →0



f ′  (x) = hlim →0

f ( h) − 1 h



1 + h φ ( h) + h 2 φ ( h) ψ ( h) − 1 = f (x) hlim →0 h lim = f (x)  [ φ (h) + h φ (h) ψ (h)]



= f (x) (a + 0 ⋅ a ⋅ b) = a ⋅ f (x).



h→0

103. (c) For f (x) to be continuous at x = 0, we must have  (1 + x)cot x f (0) = lim x→0

= hlim | 3 – (3 + h) | + (3 + 3 + h) →0

⇒ log  f (0) = lim  log (1 + x)cot x x→0

= hlim → 0 (h + 7) = 7



= lim  cot x log (1 + x) x→0

Since hlim f (3 – h) ≠ hlim f (3 + h), therefore f (x) →0 →0



log (1 + x)   = lim x→0 tan x



= lim   x→0

is not continuous and hence not differentiable at x = 3.

log (1 + x) x ⋅ = 1 ⋅ 1 = 1. x tan x ∴ f (0) = e1 = e.

99. (a) Clearly π [x – π] is an integral multiple of π for all x as [x – π] is an integer for all x. Therefore, sin (π [x – π]) = 0, for all x 104. (a) Since f (x) is continuous for all x, therefore, it is Also, 4 + [x]2 ≠ 0 for all x. continuous at x = 1 also. Therefore, f (x) = 0 for all x. ∴ f (1) = hlim f (1 – h) →0 Hence, f (x) is continuous as well as differentiable lim for all x. ⇒ 1 = h → 0 [a (1 – h)2 + b]

99

Continuity and Differentiability

95. (b) For f (x) to be continuous everywhere, we must 100. (b) We have, have f (s + t) = f (s) + as t + 3s 2 t2 for all s, t ∈ R  ...(1) f (0) = lim f (x) Putting s = 3 and t = 2 in (1), we get x→0 f (5) = f (3) + 6a + 108 (256 − 8 x)1/ 4 − 4 0  ⇒ 52 = 4 + 6a + 108 form = lim   x → 0 16 − 4 (64 + 3 x )1/ 3 0  [Putting f (3) = 4 and f (5) = 52] ⇒ 6a = – 60 i.e., a = – 10. 1 (256 − 8 x) − 3 / 4 ⋅ (− 8) Therefore, from (1), we have 4 = lim x→0 f (s + t) = f (s) – 10st + 3s2 t2 −4 − 2/3 (64 + 3 x) ⋅ (3) Putting s = x and t = h, we get 3 f (x + h) = f (x) – 10xh + 3x2 h2 [Using L' Hospital’s Rule]

100

⇒ a + b = 1

...(1)

Also, f (x) is differentiable at x = 1

Objective Mathematics

⇒ hlim →0

f (1 − h) − f (1) f (1 + h) − f (1) = hlim →0 −h h

108. (b), (c) Clearly, x = 1 is a point of discontinuity of the 1 function f (x) = . 1− x  1   = If x ≠ 1, then ( fof ) (x) = f [ f (x)] = f  1 − x  x −1 , which is discontinuous at x = 0. x If x ≠ 0 and x ≠ 1, then

1 −1 a (1 − h) 2 + b − 1 | 1 + h| lim ⇒ hlim = →0 h→0 −h h ⇒ hlim →0

(a + b − 1) + (h 2 − 2h) a 1−1− h = hlim →0 −h h (1 + h)

 x − 1 ( fo fof ) (x) = f [( fof ) (x)] = f   = x,  x  which is continuous everywhere.

⇒ 2a = – 1 (Using a + b = 1) −1 ∴ a = . 2 3 . Hence, a + b = 1 ⇒ b = 1 – a = 2

Hence f 3n (x) = ( fofof )n(x) = x, which is continuous everywhere.

105. (c) Since the function f (x) is continuous at x = 0, therefore, f (0 + h) lim f (0 – h) = f (0) = hlim →0 h→0

⇒ lim   (1 − |tan h |)

a | tan h |

h→0

e = b = hlim →0

sin 3 h sin 2 h

e ⇒ hlim [(1 – | tan h | )– 1/|tan h |]– a = b = hlim →0 →0

So, the only points of discontinuity are x = 0 and x = 1. e[ h ] + h − 1 f (0 + h) = hlim 109. (a), (d) lim+   f (x) = lim x→0 →0 x→0 [ h] + h



and t = 3 ⇒

1 = 3 ⇒ x = 7/3. x−2

1+ x − 4 1+ x x

(1 /3)(−2 /3) 2 (1 /4) (−3 /4) 2  1   1  x + ... − 1 + x + x + ... 1 + 3 x + 2! 4 2!      = lim x→0 x



 1  −1 3  + + x + terms containing x 2 and x 12  9 32    higher powers lim = x→0  x 1 1 = . ∴ f (0) = . 12 12

1 . e Since lim+ f (x) ≠ lim− f (x), therefore, f (x) is not

=1–

x→0

continuous at x = 0.

x→a

107. (a) For f (x) to be continuous at x = 0, we must have 3

e−1− h − 1 e−1 − 1 = ( − 1 − h) −1

110. (c) Let a be any real number. lim f (x) = lim x = a

Therefore, the values of x which make the function y discontinuous are x = 2, 3 and 7 . 2 3

f (0) = lim x→0

= hlim →0

x→0

1 is discontinuous t2 − t − 6

at the points where t2 – t – 6 = 0 i.e., (t + 2) (t – 3) = 0 ⇒ t = – 2, 3. 1 3 =–2 ⇒x= But t = – 2 ⇒ x−2 2

eh − 1 =1 h

e[ − h ] − h − 1 lim f (0 – h) =0 hlim f (x) = and xlim − →0 →0 h→0 [ − h] − h

sin 3 h / 3 h 3 ⋅ sin 2 h / 2 h 2

−3 ⇒ e– a = b = e3/2 ⇒ a = and b = e3/2. 2 1 106. (b) Clearly, the function t = is discontinuous at x x−2 = 2 and the function y =

= hlim →0

   

x→a



(when x → a through rational values) lim f (x) = lim (1 – x) = 1 – a x→a x→a



(when x → a through irrational values) lim f (x) will exist only when a = 1 – a i.e., when

x→a

a =

1 . 2

1 , then xlim f (x) will not exist and →a 2 hence f (x) will be discontinuous at x = a where a 1 . We have shown that ≠ 2 lim f (x) = 1 and f  1  = 1 1   x→ 2 2 2 2 Thus if x ≠

1 . 2 −π π  111. (a) 3 ≤ 3 + 2 cos x ≤ 5 for x ∈  ,   2 2 Hence f (x) is continuous at x =

f (x) = [3 + 2 cos x] is discontinuous at those points where 3 + 2 cos x is an integer.



 2x  1−  2  1 + x 

(not possible) 1 . 2



3 + 2 cos x = 4 if cos x =



So x has two values π and − π . 3 3 3 + 2 cos x = 5 if cos x = 1. So, x = 0. ∴ The number of values of x = 2 + 1 = 3. 

h 3 φ ( h) = hlim →0 h



 h2 φ (h) = 0 × φ (0) = hlim →0 h→0



∴ f (x) = (– 1)n = ± 1



Hence f (x) is discontinuous for x = n1/3, n ∈ I.

114. (a), (c) The function f (x) is defined when – 1 ≤ sin x < 0 or sin x = 1. ⇒ x ∈ ((2n + 1) π, 

(2n + 2) π) ∪ 2nπ + π  , n ∈ I. 2 



When x ∈ (2nπ + π, 2nπ + 2π), f (x) = – 1 ∴ f (x) is a constant function. Hence f (x) is continuous when x ∈ (2nπ + π, 2nπ + 2π).

π  π f  − h − f   2   2 Now, L f ′ (π/2) = hlim →0 −h 1 −1  π  sin − h   2    = hlim →0 −h 1 −1 [cos h] lim = h→0 = does not exist. −h

Hence, f (x) is not differentiable at x = π/2.

115. (c) Let g (x) = x3 – 3, then g (x) is an increasing function on the interval (1, 2). Since g (1) = – 2 and g (2) = 5, therefore between – 2 and 5 there are 6 points where f (x) is discontinuous (as [x3 – 3] is discontinuous at the points where x3 – 3 is an integer).

d  2x    dx  1 + x 2 

2 (1 − x 2 ) (1 + x 2 ) 2



 −2 1 + x 2 , if | x | < 1 − 2 1 − x2  ⋅ = =  1 + x 2 |1 − x 2 |  2 , if | x | > 1 1 + x 2

(1 + x 2 ) 2 − 4 x 2

x

Clearly, f (x) is differentiable everywhere except at the points where | x | = 1 i.e., x = ± 1. Hence, f (x) is differentiable on  (– ∞, ∞) \ { – 1, 1}.



= 0. 113. (a) Let x3 = n, n ∈ I ⇒ x = n1/3

x

=

[ f (x) = x3 φ (x)] 117. (a) f ′ (x) = lim h→0

[ φ is continuous at x = 0, ∴ lim φ (h) = φ (0)]



− (1 + x 2 )

2



f ( x + h) − f ( x ) 112. (c) f ′ (x) = hlim →0 h f ( x ) + f ( h) − f ( x ) = hlim →0 h [ f (x + y) = f (x) + f ( y)]

−1

116. (c) f ′ (x) =

101

−π π , 2 2

= hlim →0

f ( x + h) − f ( x ) h

f ( x ) ⋅ f ( h) − f ( x ) h



[ f (x + y) = f (x) ⋅ f ( y)] 

= f (x) lim  f (h) − 1   h→0 h  



= f (x) lim 1 + h φ (h) log 2 − 1 h→0 h [ f (x) = 1 + x φ (x) log 2]



= f (x) log 2 hlim φ (h) →0



= f (x) ⋅ log 2 ⋅ 1



= log 2



f (x)

∵ lim φ (h) = 1  h → 0 

.

118. (b) We have, f (x + y2n + 1) = f (x) + { f ( y)}2n + 1 ...(1) ⇒ f (0) = f (0) + { f (0)}2n + 1 [Putting x = y = 0] ⇒ f (0) = 0. Also, from (1) f (1) = f (0) + { f (1)}2n + 1 [Putting x = 0, y = 1] = 0 + { f (1)}2n + 1 [ f (0) = 0] ⇒ f (1) [{ f (1)}2n – 1] = 0 ⇒ f (1) = 0 or 1 or – 1. Now, f ′ (0) = lim x→0

f ( x ) − f ( 0) f ( x) = lim x→0 x x

[ f (0) = 0]

Since f ′ (0) ≥ 0 [Given] f ( x) ≥0 x ⇒ f (x) ≥ 0 for x > 0. ∴ f (1) ≠ – 1. Also, f (1) = 0 ⇒ f (x + 1) = f (x) + { f (1)}2n + 1 [Putting y = 1 in (1)] = f (x)  ⇒ f (2) = f (3) = ... = 0, ⇒ lim

x→0

Continuity and Differentiability

Now, 3 + 2 cos x = 3 if cos x = 0. So, x =

102



which is not true. Thus f (1) = 1.

Objective Mathematics



∴ f (x + 1) = f (x ) + 1



⇒  f (x) is a function whose value is increased by 1 when x is increased by 1, ∴ it is a linear function whose graph is a straight line passing through origin ( f (0) = 0) having slope 1. Hence f (x) = x. ∴ f ′ (6) = 1.

f  ′ (– 1+) = 0. Similarly, f ′ (1–) = 0 and f ′ (1+) = 1, so f is differentiable everywhere except at x = – 1, 1.

119. (c) Let x0 be any arbitrary real number. Case I. x0 is rational Then f (x0) = 1. 122. (c) The function f (x) is continuous at all points where In any vicinity of a rational point there are irrational points, where f (x) = 0. Hence, in any vicinity of x0 there are points x for which 1 – cos2x ≥ 0 ⇒ | cos x | ≤ 1 | ∆ y | = | f (x0) – f (x) | = 1. 2 2 Case II. x0 is irrational ⇒ π + 2nπ ≤ x ≤ 3π + 2nπ Then f (x0) = 0. 4 4 In any vicinity of an irrational point there are rational or 5π + 2nπ ≤ x ≤ 7 π + 2nπ, n ∈ I. points at which f (x) = 1. Hence, it is possible to 4 4 find the values of x for which 1 | ∆ y | = | f (x0) – f (x) | = 1. suffers a discontinuity at 123. (b) The function u = x −1 Thus, in both cases, the difference ∆ y does not tend the point x = 1. to zero as ∆ x → 0. Therefore, x0 is a point of discontinuity. Since x0 is an arbitrary point, the Dirichlet 1 function f (x) is discontinuous at each point. The function f (x) = 2 suffers a discontinuity u +u−2 120. (a) We have, at the points where u2 + u – 2 = 0 i.e., u = – 2 and u = 1. Using these values of u, the corresponding 1 values of x are obtained by solving the equations lim f (5 – h) = lim tan– 1 h→0 h→0 (5 − h) − 5 1 1  −1  –2= and 1 = i.e., x = 1/2 and x = 2. = hlim tan– 1   →0 x −1 x − 1  h 

= tan– 1 (– ∞) = − π 2

Hence, the composite function is discontinuous at three points x = 1/2, x = 1 and x = 2.

and lim f (5 + h) = hlim tan– 1 →0 h→0



1 (5 + h) − 5

π 1 = hlim tan– 1   = tan– 1 (∞) = . →0 2 h f (5 – h) ≠ hlim f (5 + h), therefore, f (x) Since hlim →0 →0

has discontinuity of the first kind at x = 5. 1 − x, 121. (a), (c)  f (x) = 2, 1 + x, 

x ≤ −1 −1 < x ≤ 1 x >1

lim f ( x) = lim (1 − x) −

x → −1−

x → −1

= 2 = lim+ f ( x)

124. (a), (b), (d)  We have  x, if x is an integer  f (x) = n, if x is not an integer i.e.  n < x < n +1  0, if x is an integer and  g (x) =  2  x , otherwise Now, ( gof ) (x) = g[ f (x)] = g (integer)  [ f (x) is an integer ∨ x ∈ R] = 0. ∴  (gof ) (x) is a constant function, so it is continuous and differentiable at all x ∈ R. Also, ( fog) (x) = f [g (x)]

x → −1

a nd 

lim f ( x) = 2, so f is continuous at all x →1

points. f ′ (– 1–) = lim− h→0



= lim− h→0

f (−1 + h) − f (−1) h

1+1− h − 2 = – 1. h

0, if x ∈ I  = n, if n < x < n + 1   or − n + 1 < x < − n , n = 0, 1, 2, ... Thus ( fog) (x) is continuous and differentiable at all x except when x = ± n , n = 1, 2, 3,... .

h→0

f (0 + h) − f (0) f (0 − h) − f (0) = lim h → 0 h −h f (− h) − f (0) = – hlim →0 h ( ) f h − f (0) = – lim h→0 h [ f (– h) = f (h)]

⇒ lim

= lim

h→0



h→0

f ( h) − 1  = f (x) ⋅ lim    h→0 h   = f (x) lim

h→0

= f (x) hlim g (h) lim G (h) = ab f (x). ∴ k = ab →0 h→0

126. (b) L f ′ (1) = lim f (1 − h) − f (1) h→0 −h

129. (c) f (x) will be a continuous function at x = π f   = limπ f (x) x→ 2 2 1 − sin x ⇒ λ = limπ π − 2 x) 2 ( x→

 (1 − h) 2 3 13  − (1 − h) +  − 2  4 2 4  = lim h→0 −h



h→0

= lim

h→0

sin x 1 8 = 8. ∴ λ = 1/8.

|1 + h − 3| − 2 f (1 + h) − f (1) = lim h→0 h h

2−h−2 = ­– 1. h

Since L f ′ (1) = R f ′ (1), therefore f (x) is derivable at x = 1. Hence f (x) is continuous at x = 1. π , we must have 127. (b) For f to be continuous at x = 4 lim f (x) = f  π  π   x→ 4 4 π  tan  − x  4   lim But π f (x) = lim x→ π cot 2 x x→ 4

− cos x = xlim π → 2 ( π − 2 x ) ( − 2) 2

h+4 h 2 + 4h = lim = hlim = – 1. →0 −4 h→0 − 4h

π if 2

0   form  0  0   form  0 

2

(1 + h 2 − 2h) − 6 (1 − h) + 13 − 8 − 4h

R f ′ (1) = lim

1 + h g (h) G (h) − 1 h

= f (x) hlim g (h) G (h) →0

⇒ f ′ (0) = 0.

h→0

f ( x ) ⋅ f ( h) − f ( x ) h [ f (x + y) = f (x) ⋅ f (x)]

f (h) − f (0) = 0 ⇒ 2 f ′ (0) = 0 ⇒ 2 lim h→0 h [ f ′ (0) exists]

= lim

f ( x + h) − f ( x ) h

103

128. (d) f ′ (x) = lim

=

lim

π x→ 2

130. (c) Since, x is not defined for negative values of x, therefore, f is neither continuous nor differentiable at x = 0. 131. (d)

y

3 2 x'

1 -3

-2 -1 O

1

2

3

4

x

4



 1 − tan x  = xlim π tan 2x   → 4  1 + tan x 



=



=



2 (1) 2 1 = = = . (1 + 1) 2 4 2

lim x→

π 4

lim x→

π 4

2 tan x 1 − tan x ⋅ 1 − tan 2 x 1 + tan x 2 tan x (1 + tan x) 2

π 1 ∴ f   = . 4 2

y' Clearly, from the figure, f is continuous and derivable for all x ∈ (R – I).

132. (a) We have, h(x) = min {x, x2}  x, for x ≤ 0  =  x 2 , for 0 < x < 1  x, for x ≥ 1  Clearly from the figure, h(x) is continuous everywhere, but not derivable at 0 and 1.

Continuity and Differentiability

125. (a) Since f ′ (0) exists, ∴ R f ′ (0) = L f ′ (0)

104

y

Objective Mathematics

1   ⇒ k = lim (2 + h) 2 + e 2 − (2 + h )  h→0  



−1



−1

−1

⇒ k = lim  4 + h 2 + 4h + e h  h→0 x



x

O



⇒ k = [4 + 0 + 0 + ε–∞]–1 ⇒ k = 1

4

138. (d) Given that, f(x) = ex sin x, x ∈ [0, π] y

133. (a) Since |x| and sin x are continuous for all x, therefore f(x) = sin |x| is continuous for all x ∈ R. 134. (c) f(x) being a polynomial function, is continuous for all real values of x except at x = 2. At x = 2, LHL = lim− x – 1 = lim 2 – h – 1 = 1 h→0

x→2

RHL = lim+ 2x – 3 = lim 2(2 + h) – 3 = 1 h→0

x→2

f(2) = 2(2) – 3 = 1

and

∴ LHL = RHL = f(2) Thus, f(x) is continuous for all real values of x. 135. (a)

y

At x = 0, f(0) = 0 and at x = π, f (π) = 0 Also, f(x) is continuous and differentiable in the interval. [0, π]. ∴ f(x) satisfies conditions of Rolle’s theorem in [0, π]. Hence, option (d) is the required answer. 139. (c) We have, at x = a 3 3 3 3 LHL = lim x − a = lim ( a − h) − a − x→a

= lim h→0

x−a

h→0

a−h−a

[(a − h) + a + a (a − h)] 1 2

2

= (a2 + a2 + a2) = 3a2 Since, f(x) is continuous at x = a. ∴  LHL = f(a) ⇒  3a2 = b y=−x+1

y = x +1 (0,1)

140. (c) Continuity at x = 0: LHL = lim−

tan x − tan h = lim− =1 h→0 −h x

RHL = lim

tan x tan h = lim =1 h → 0 x h

x →0

x

O

x →0

Clearly, from the figure, f(x) ≥ 1, for every x ∈ R.

Since LHL = RHL = f(0) = 1, therefore f(x) is continuous at x = 0. Differentiability at x = 0:

136. (d) We have,

2  1 e2 x − 1 − 2 x lim  − 2 x  = lim x →0 x → 0 x (e 2 x − 1)  x e −1 2x

= lim

2e − 2  (e 2 x − 1) + 2 x e 2 x

= lim

4e 2 x =1 4e 2 x + 4 xe 2 x

x →0

x →0

(using L’Hospital’s rule)

137. (b) f(x) is continuous from right at x = 2, x→2



1



−1

⇒ lim  x 2 + e 2 − x  = k x → 2+





f (0 − h) − f (0) −h

tan (−h) h 2 2h 4 −1 + + + ... 15 = lim 3 =0 = lim −h h→0 h→0 −h −h h→0

x→ 0

if lim+ f(x) = f(2) = k

h→0

RHD = lim

f(x) is continuous at x = 0, if lim f(x) = f(0) i.e.,, 1 = f(0)

LHD = lim

f (0 + h) − f (0) h

tan h h 2 2h 4 −1 + + ... 15 = lim h = lim 3 =0 h→0 h→0 h h  ince LHD = RHD, therefore f(x) is differentiable at S x=0 Hence, Option (c) is the required answer. 141. (c) At no point, function is continuous.

x →0

h→ 0

and lim f(x) = lim 51/x = lim 5–1/h = 0 − − h→ 0 x →0

∴ f is differentiable at x = 0.

x →0

Also, f(0) = λ[0] = 0 ∴ f(x) is continuous for x = 0, whatever λ may be. f (1 − h) − f (1) 143. (c) Now, f ′ (1 ) = lim h →0 −h –

 1  (1 − h − 1) sin  −0  1− h −1  = lim h →0 −h

h →0

x →0

x →0

x →0

x →0

Since L.H.L ≠ R.H.L ∴ f (x) is discontinuous at x = 0 log e (1 + x 2 tan x) sin x 3 This function is continuous at x = 0, then

145. (a) Given, f (x) =

f (1 + h) − f (1) h

log e (1 + x 2 tan x) = f (0) x→0 sin x 3

lim

 1  (1 + h − 1) sin  −0 1  1+ h −1  = lim = lim sin h →0 h →0 h h Since f ′ (1–) ≠ f´ (1+) ∴ f is not differentiable at x = 1. Again,   1   − sin 1 (0 − h − 1) sin    0 − h −1  , f ′ (0− ) = lim  h →0 −h   1   1    1  − (h + 1) cos   ×  (h + 1) 2   + sin  h + 1  + ! h         = lim  h →0 1 = cos 1 – sin 1 1   (0 + h − 1) sin  − sin 1 0 + h − 1   + and f ′ (0 ) = lim h →0 h   1   −1   1  (h − 1) cos    (h − 1) 2  + sin  h − 1   − 1 h        = lim  h →0 1 

2 x, x < 0 f (x) =   2 x + 1, x ≥ 0 L.H.L = lim f ( x) = lim 2 x = 0 − − R.H.L = lim f ( x) = lim 2 x + 1 = 1 + +

1  1 = lim sin  −  = − lim sin h →0 h → 0 h  h and f′ (1+) = lim

144. (c) We have,

(using L’ Hospital’s rule)

   x3 log e 1 + x 2  x + + ...  3    ⇒ lim = f (0) 9 15 x→0 x x − ... x3 − + 3! 5! log e (1 + x 3 ) ⇒ lim = f (0) x→0 x 9 x15 x3 − + − ... 3! 5! [neglecting higher power of x in x2 tan x] x6 x9 x3 − + − 2 3 ⇒ lim = f (0) ⇒ 1 = f (0) x→0 x 9 x15 3 − ... x − + 3! 5! 146. (a) We have lim t→x

t 2 f ( x) − x 2 f (t ) =1 t−x

⇒ x2f´(x) – 2xf(x) + 1 = 0 ⇒ f(x) = cx2 + 1 also f(1) = 1 3x ⇒c= 2 3 2 1 Hence, f(x) = x 2 + 3 3x

EXERCISEs FOR SELF-PRACTICE 1. If f (x) = x3Sgn, then 

(a) f is continuous but not derivable at x = 0 (b) f ′ (0 +) = 2 (c) f ′ (0 – ) = 1 (d) f is derivable at x = 0.

2. In order that the function f (x) = (x + 1)cot x is continuous at x = 0, f (0) must be defined as

1 e (c) f (0) = e

(a) f (0) =

(b) f (0) = 0 (d) None of these

3. The domain of the derivative of the function

 tan −1 x, | x | ≤ 1  is f (x) =  1  (| x | − 1), | x | > 1 2

Continuity and Differentiability

x →0

105

= cos 1 – sin 1 Since, f ′ (0–) = f′ (0+),

142. (c) lim+ f(x) = lim+ λ[x] = lim λ[h] = 0

106

(a) R – [0] (c) R – {– 1}

Objective Mathematics

4. If f (x) = x

(

(b) R – {1} (d) R – {– 1, 1}

(a) 3

(c) 9 (d) 12 9. The value of the constants a, b and c for which the function

)

x − x + 1 , then

(a) f (x) is continuous but not differentiable at x = 0 (b) f (x) is differentiable at x = 0 (c) f (x) is not differentiable at x = 0 (d) none of these. 2 x − sin −1 x 5. If the function f (x) = is continuous at each 2 x + tan −1 x point of its domain, then the value of f (0) is

(a) 2

1 (c) –  3

(b) (d)

(b) 6

2 3



 (1 + ax)1/ x ,x0  ( x + 1)1/ 2 − 1

may be continuous at x = 0, are −2 2 (a) a = log , b= , c = 1 3 3

2 2 , b= , c = – 1 3 3 2 2 (c) a = log , b = , c = 1 3 3

(b) a = log

2 3

1  x≠0  x sin , x 6. If f (x) =  , then, at x = 0, the function 0, x=0 f (x) is: (a) Differentiable (b) Not differentiable (c) Continuous but not differentiable (d) None of these

(d) None of these

10.

7. If f (x) = | x – 3 |, then f ′ (3) is: (a) – 1 (b) 1 (c) 0 (d) Does not exist



sin 3 x , x≠0 x 8. The function f (x) = k , x = 0, 2 is continuous at x = 0, then k =



 −1/ x2 1 sin   , x ≠ 0 e x then graph of the function If f (x) =   0 ,x=0  f (x) (a) is broken at the point x = 0 (b) is continuous at the point x = 0 (c) has a definite tangent at the point x = 0 (d) does not have a definite tangent at the point x = 0.

11. The function f (x) = a [x + 1] + b [x – 1], where [x] is the greatest integer function is continuous at x = 1, if: (a) a + b = 0 (b) a = b (c) 2a – b = 0 (d) None of these

Answers

1. (d) 11. (a)

2. (c)

3. (c)

4. (b)

5. (b)

6. (c)

7. (d)

8. (b)

9. (c)

10. (b)

4

Differentiation

CHAPTER

Summary of concepts Derivative of a Function Let y = f (x) be a function defined on the interval [a, b]. Let for a small increment δx in x, the corresponding increment in the value of y be δy. Then y = f (x) and y + δy = f (x + δx) On subtraction, we get δy = f (x + δx) – f (x) or

δy f ( x + δx ) − f ( x ) = δx δx

Taking limit on both sides when δx → 0 we have, δy f ( x + δx ) − f ( x ) = lim . δ x δx → 0 δx if this limit exists, is called the derivative or differential coeffidy or f ' (x). Thus cient of y with respect to x and is written as dx lim

δx → 0



dy δy f ( x + δx ) − f ( x ) = lim = δlim . x → 0 δx δx → 0 dx δx

Derivative at a Point The value of f′ (x) obtained by putting x = a, is called the deriva dy  tive of f (x) at x = a and it is denoted by f ' (a) or   .  dx  x = a

Standard Derivatives The following formulae can be applied directly for finding the derivative of a function: 1.

d (sin x) = cos x dx

2.

d (cos x) = – sin x dx

3.

d (tan x) = sec2x dx

4.

d (cot x) = – cosec2x dx

d 5.   (sec x) = sec x tan x dx

d dx d 7. dx d 8. dx d 9. dx 6.

(cosec x) = – cosec x cot x (ex ) = ex (ax ) = ax logea, a > 1 (logex) =

1 , x>0 x

d (xn) = nxn – 1 dx d 1 11. (sin–1x) = ,–1< x 0 and x2 + y2 ≤ 1 14. cos–1x ± cos–1y = π – cos–1 ( xy  1 − x 2 1 − y 2 ) ,  2x  15. sin–1   = 2 tan–1x 1 + x2  1 − x2  16. cos–1  = 2 tan–1x  1 + x 2  17. tan–1

1 1− x = cos–1x 2 1+ x

SucceSSive Differentiation dy is again a function of x dx and is called the first derivative of y w.r.t. x. If the first derivative is differentiable, its derivative is called second derivative of d2y the original function and is denoted by or y2. If the second dx 2 derivative is differentiable, its derivative is called the third ded3y or y3 and rivative of the original function and is denoted by dx 3 so on. This process of differentiating a function more than once is called successive differentiation. Let y = f (x) be a function of x, then

Differentiation of a function given in the form of a Determinant

π 1 − tan x 6. tan  − x  = 4  1 + tan x –1

–1

 x+ y   1 − xy  ,

provided x, y > 0 and xy > 1 9. tan–1x – tan–1y = tan–1  x − y  ; if x, y > 0  1 + xy  π = tan–1x + cot–1x = sec–1x + cosec–1x 10. sin–1x + cos–1x = 2

f ( x ) g ( x ) h( x ) If y = p ( x) q ( x) r ( x) , then u ( x) v( x) w( x) f' ( x) g' ( x) h' ( x) dy = p( x) q( x) r ( x) dx u ( x) v( x) w( x) f ( x ) g ( x ) h( x ) f ( x ) g ( x ) h( x ) + p' ( x) q' ( x) r' ( x) + p ( x) q ( x) r ( x) u' ( x) v' ( x) w' ( x) u ( x) v( x) w( x) Note that differentiation of a determinant can be done in columns also.

109

provided x, y ≥ 0 and x2 + y2 ≤ 1

if x, y > 0 and x2 + y2 > 1

a+x a−x

Some useful trigonometric and inverse trigonometric transformations

–1

11. sin–1x ± sin–1y = sin–1 ( x 1 − y 2 ± y 1 − x 2 ) ,

Differentiation

Important Substitutions to Reduce the Function to a Simpler Form

110

multiple-choice questions

Objective Mathematics

Choose the correct alternative in each of the following problems:  xa  1. If f (x) =  b  x  to

a +b

 xb  ⋅ c  x 

(a) 1 (c) xa + b + c  sin x  2. If f (x) =  n   sin x  is equal to

m+n

 sin x  ⋅ p   sin x 

(a) 0 (c) cosm + n + p x

then

 xc  ⋅ a  x 

n

n+ p

(a)

dy 1 = dx a + b cos x

(c)

dy 1 = dx a − b cos x

c+a

, then f ′ (x) is equal

(b) 0 (d) None of these m

3. If y =

b+c

 sin x  ⋅ m   sin x  p

b sin x d2y = (a + b cos x) 2 dx 2 −b sin x d2y (d) = (a − b cos x) 2 dx 2 (b)

9. If sin y = a sin (x + y), then

p+m

, then f ′ (x)

(b) 1 (d) None of these

1 1 1 + + 1 + xβ − α + x γ − α 1 + x α − β + x γ − β 1 + x α − γ + xβ − γ

dy = dx

(a)

sin a sin a sin 2 (a + y )

(c) sin a sin2 (a + y) 10. If f (x) =

4. If y = ( x + 1 + x 2 ) n , then (1 + x 2 ) (a) n2y (c) – y

(b) – n2y (d) 2x2y

d2y dy +x is dx 2 dx

5. If φ (x) = log5 log3 x, then φ′ (e) is equal to (a) e log 5 1 (c) elog 5

(b) – e log 5

1+ x 1 + log x

dy is dx

x − 1 − x2

12. If y = tan –1

(d) None of these

(d)

sin 2 (a − y ) sin a

1 − log x 1 + log x log x (d) (1 + log x) 2

(b)

(c) not defined

(a)

sin 2 (a + y ) sin a

(b) 0 (d) Does not exist

11. If x y = e x – y then (a)

(b)

1 − cos x π , then f ′   is equal to 1 − sin x 2

(a) 1 (c) ∞

(a) 0 (b) 1 (c) (α + β + γ ) xα + β + γ – 1 (d) None of these

dy is dx

−1 1− x −x

2

x + 1− x

2

dy is equal to dx

, then



(b)

1 1 − x2



(c) (d) None of these 1 − x2  2x − 1 dy 2 and f ′ (x) = sin x , then is equal 6. If y = f  2 1  x + 1  , then g′ (x) is 13. If g is the inverse of f and f ′ (x) = dx 1 + x3 to equal to 2x − 1   2 + 2x + x2  (a) sin  2 ⋅  x + 1   ( x 2 + 1) 2  2

2x − 1   2 + 2 x − 2 x2  (b) sin  2  x + 1   ( x 2 + 1) 2  2

2x − 1  2 + 2x − x  (c) sin  2 ⋅  x + 1   ( x 2 + 1) 2  2

7. If f (x) = log x (log x), then f ′(x) at x = e is (a) e (c)

(b)

2 e

8. If y =

1 e

(d) 0  a−b x tan −1  tan  , a > b > 0, then 2  a+b a −b 2

2

2

(b)

(c) [g (x)]3 14. If y = cos –1

2

(d) None of these

1 1 + [ g ( x)]3 (d) None of these

(a) 1 + [g (x)]3

1 1 + x2 1 (c) 2 (1 + x 2 ) (a)

1 + x2 + 1 2 1 + x2

, then (b)

dy = dx

−1 2 (1 + x 2 )

(d) None of these

15. Let f (x) be a polynomial function of second degree. If f (1) = f (– 1) and a, b, c are in AP, then f ′ (a), f ′ (b) and f ′ (c) are in (a) AP (b) GP (c) HP (d) Arithmetico-Geometric progression

1 1 + tan −1 2 x + x +1 x + 3x + 3 1 dy −1 + tan 2 + ... to n terms, then = x + 5x + 7 dx

17. If y = tan –1

2

1 1 1 1 + − (b) 2 2 2 1 + ( x + n) 1 + x 1 + ( x + n) 1 + x 2 2 1 − (c) (d) None of these 1 + ( x + n) 2 1 + x 2

(a)

18. If y = etan x , then cos2 x dy (a) (1 – sin2x) dx dy (c) (1 + sin2x) dx 19. If y = x (log x ) (a)

log log x

, then

d y = dx 2 2

dy (b) – (1 + sin2x) dx (d) None of these dy is equal to dx

y log y (2 log log x + 1) x log x

x2

x

∑ tan r =1

(a)

−1

(c) 0 23. If y = eax sin bx, then (a) – (a2 + b2) y (c) – y

26. The derivative of tan –1

1 + x2 − 1 w.r.t. x

cos–1

1 + 1 + x2 2 1 + x2

is

(a) 1 1 (c) 2

(b) – 1 (d) None of these

2u 1 1− u 1 2u  + sin −1 27. If y = tan  cos −1 and x = , 2 2  1 − u2 2 1 + 2 1 + u u   dy = then dx (b) 0 (d) None of these π , where 4

1 2 (c) 1 (a)

(b)

2

(d) None of these

d2y is equal to dx 2 dy (b) x –2 dx (d) None of these

30. If x = t t and y = t t , then

(b) y (d) None of these

dy 1 then is equal to dx 1+ r + r2

1 1 + x2

(b) P (x) ⋅ P''' (x) (d) None of these.

29. If x = a (cos θ + θ sin θ) and y = a (sin θ – θ cos θ), π d2y π where 0 < θ < , then at x = is equal to 2 2 dx 4 4 2 8 2 (a) (b) aπ aπ 4 (c) (d) None of these aπ 2

21. If y = (sin –1 x)2, then (1 – x2)

22. If y =

(a) – P (x) ⋅ P''' (x) (c) P (x) ⋅ P′′ (x)

f ′ (1) = 2 and g′ ( 2 ) = 4, is

20. If y = a cos (log x) + b sin (log x), then

dy + 2 dx dy (c) – x + 2 dx

2 d  3d y y 2  = dx  dx 

28. The derivative of f (tan x) w.r.t. g (sec x) at x =

(d) None of these

(a) x

2

(a) – 1 (c) 1

2 y log y (log log x + 1) x log x

d2y dy +x = dx 2 dx (a) 0 (c) – y

25. If y 2 = P (x), a polynomial of degree n ≥ 3, then

2

x log x (b) (2 log log x + 1) y log y (c)

(b)

1 1 + (1 + x) 2

(d) None of these d2y dy − 2a is equal to dx 2 dx (b) (a2 + b2) y (d) None of these

t

t  1 t t ⋅ (1 + log t )log t +  t  (a) (1 + log t ) t  1 t t 1 + log t +  t  (b) (1 + log t ) t  1 t t (1 + log t )log t +  t  (c) 2 (1 + log t )

(d) None of these

dy is equal to dx

111

2x 1 − x2 24. The differential coefficient of tan –1 w.r.t. 1 − 2x2 1 1 sec–1 at x = is equal to 2 2x2 − 1 1 1 (a) (b) – 2 2 (c) – 1 (d) None of these.

Differentiation

16. If S n denotes the sum of n terms of a G.P. whose comdSn mon ratio is r, then (r – 1) is equal to dr (a) (n – 1) Sn + n Sn – 1 (b) (n – 1) Sn – n Sn – 1 (c) (n – 1) Sn (d) None of these

112

Objective Mathematics

dy 31. Let y = x3 – 8x + 7 and x = f (t). If = 2 and x = 3 dt dx at t = 0, then at t = 0 is given by dt 19 (a) 1 (b) 2 2 (c) (d) None of these 19 32. If ax2 + 2hxy + by 2 = 1, then

d2y is equal to dx 2

ab − h 2 (a) (hx + by ) 2

h 2 − ab (b) (hx + by ) 2

h 2 + ab (c) (hx + by ) 2

(d) None of these

33. If y = sin x, then

d2 (cos7x) is equal to dy 2

sin x x x x ⋅ cos 2 cos 3 ... to ∞ = , then x 2 2 2 x 1 x x 1 1 1 tan + 2 tan 2 + 3 tan 3 + ... to ∞ = – cot x +   2 2 2 2 2 2 x x 1 x x 1 1 1 tan + 2 tan 2 + 3 tan 3 + ... to ∞ = cot x – 2 2 2 2 2 2 x 1 1 1 2 x 2 x 2 sec + 4 sec 2 + ... + ∞ = cosec x – 2 x 22 2 2 2 1 x x 1 1 sec 2 + 4 sec 2 2 + ... + ∞ = – cosec2x + 2 2 x 2 2 2 2

34. If cos

(b) (c) (d)

 3π  35. If f (x) = | cos x |, then f ′   is equal to  4  1 1 (a) – (b) 2 2 (c) 1 (d) None of these. a+ x 36. If f (x) =   b+x

(a) m2y (c) – m2y

(b) my (d) None of these

39. Let f be a twice differentiable function such that f ′′(x) = – f (x) and f ′ (x) = g (x). If h (x) = [ f (x)]2 + [g (x)]2 and h (5) = 11 then h (10) = (a) 11 (c) – 1

(b) 0 (d) None of these

40. A function f (x) is so defined that for all x, [ f (x)]n = f (nx). If f ′ (x) denotes derivative of f (x) w.r.t. x, then f ′ (x) ⋅ f (nx) = (a) f (x)

(b) 0

(c) f (x) ⋅ f ′ (nx)

(d) None of these

41. If y = x – x2, then the derivative of y 2 w.r.t. x2 is (a) 2x2 + 3x – 1 (c) 2x2 + 3x + 1

(b) 2x2 – 3x + 1 (d) None of these

ax + b , where a, b, c are constants then x2 + c (2xy′ + y) y′′′ is equal to

42. If y =

(a) 35 cos3x – 42 cos5x (b) 35 cos3x + 42 cos5x (c) 42 cos3x – 35 cos5x (d) None of these.

(a)

38. If y = sin (m sin –1 x), then (1 – x2) y′′– xy′ is equal to

a +b+ 2 x

, then f ′ (0) is equal to

 a a 2 − b2   a  (a)  2 log +   b ab   b  

a +b

 a b2 − a 2   a  (b)  2 log +   b ab   b  

a +b

 a a 2 + b2   a  (c)  2 log +   b ab   b   (d) None of these

a +b

(a) 3 (xy′′+ y′) y′′

(b) 3 (xy′ + y′′ ) y′′

(c) 3 (xy′′+ y′) y’

(d) None of these

43. If f (x) = (ax + b) sin x + (cx + d) cos x, then the values of a, b, c and d such that f ′ (x) = x cos x for all x are (a) b = c = 0, a = d = 1 (c) c = d = 0, a = b = 1

(b) b = d = 0, a = c = 1 (d) None of these d 2P is equal to d θ2 (b) 2a2 + 2b2 – 3P (d) None of these

44. If P = a2 cos2θ + b 2 sin 2θ, then P + (a) 2a2 + 2b2 + 3P (c) 2a2 – 2b2 – 3P 45. If y = then (a)

ax 2 bx c + + +1 , ( x − a )( x − b)( x − c) ( x − b)( x − c) x − c

y' = y 1 a b c  + +   x a− x b− x c− x

a b c  (b)  + +  a − x b − x c − x  (c)

1 a b c  + +   x x−a x−b x−c

(d) None of these

π 37. If f (x) = | cos x – sin x |, then f ′    is equal to 2 (a) 1 (b) – 1 (c) 0 (d) None of these

46. A triangle has two of its vertices at P (a, 0), Q (0, b) and the third vertex R (x, y) is moving along the straight line dA y = x. If A be the area of the triangle, then = dx a−b a−b (a) (b) 2 4 a+b a+b (c) (d) 2 4

(a)

f1φ2 − φ1 f 2 f12

f φ −φ f (b) 1 2 3 1 2 f1

(c)

φ1 f 2 − f1φ2 f13

(d) None of these,

where the suffixes denote diff. w.r.t. t.  5 x + 12 1 − x 2 48. If y = sin –1   13  1

(a)

1 − x2 3

(c)

1 − x2

 dy  , then is equal to  dx  1



(b) −



(d) None of these

54. If f, g, h are differentiable functions of x and f ( xf )' ( x 2 f )"

f' (a) f ( x 3 f" )'

g' g ( x 3 g" )'

h' h ( x 3h" )'

(b)

f f' ( x 2 f" )'

g g' ( x 2 g" )'

h h' ( x 2 h" )'

(c)

f f' ( x 3 f" )'

g g' ( x 3 g" )'

h h' ( x 3h" )'

1 − x2

(d) None of these

 dx   dy  then   +   is equal to  dt   dt  2

(a) f (t) – f ′′(t)

(b) [ f (t) – f ′′(t)]2

(c) [ f (t) + f ′′(t)]2

(d) None of these

50. If x = sec θ – cos θ, y = sec nθ – cos nθ, then dy (x2 + 4)   is equal to  dx  2

(b) n (4 – y ) (d) None of these

(a) n (y – 4) (c) n2 (y2 + 4) 2

2

2

f1 ( x) f 2 ( x) f 3 ( x) F (x) = g1 ( x) g 2 ( x) g 3 ( x) , then F′ (a) is equal to h1 ( x) h2 ( x) h3 ( x) (b) – a (d) None of these

52. If y = k sin px, then the value of the determinant y y3 y6

y1 y4 y7

55. If 2x3 – 3x2y 2 + 4x – y + 7 = 0 and y (1) = 1 then the value of y′′(1) is equal to (a) – (c) 56. If

343 474

(b) –

474 343

(d)

474 343

343 474

1 − x 6 + 1 − y 6 = a3 (x3 – y 3), then

2

51. If f r (x), g r (x), hr (x), r = 1, 2, 3 are polynomials in x such that f r (a) = g r (a) = hr (a), r = 1, 2, 3 and

(a) a (c) 0

h ( xh)' ( x 2 h)"

then ∆′ (the derivative of ∆ w.r.t. x) is given by

49. If x = f (t) cos t – f ′ (t) sin t, y = f (t) sin t + f ′ (t) cos t 2

g ( xg )' ( x 2 g )"

∆=

y2 y5 is equal to y8

dy is equal to dx

(a)

x2 1 − y6 y 2 1 − x6

(b)

y2 1 − y6 x2 1 − x6

(c)

x2 1 − x6 y2 1 − y6

(d) None of these

f ( x1 ) + f ( x2 ) + ... + f ( xn )  x + x + ... + xn  57. Let f  1 2 ,  = n n   where all xi ∈ R are independent to each other and n ∈ N. If f (x) is differentiable and f ′ (0) = a, f (0) = b, then f ′ (x) is equal to (a) a (c) b

(b) 0 (d) None of these

58. If f (x) = (1 – x) n , then the value of

(a) 1

(b) 0

(c) – 1

(d) None of these,

where yn denotes nth derivative of y w.r.t. x.

f (0) + f ′ (0) +

f"(0) f n (0) + ... + is equal to 2! n!

(a) 2n (c) 2n – 1

(b) 0 (d) None of these

53. If f (x), g (x), h (x) are polynomials in x of degree 2 and 59. If y = f (x3), z = g (x5), f ′ (x) = tan x and g′ (x) = sec x, dy f g h then the value of is dz F (x) = f' g' h' , then F′ (x) is equal to f" (a) 1 (c) – 1

g"

h" (b) 0 (d) None of these

(a)

3 tan x 3 ⋅ 5 x 2 sec x 5

(b)

(c)

3 x 2 tan x 3 ⋅ 5 sec x 5

(d) None of these

5 x 2 sec x 5 ⋅ 3 tan x 3

113

d2y is equal to dx 2

Differentiation

47. If x = f (t), y = φ (t), then

114

60. If f ′ (x) = φ (x) and φ′ (x) = f (x) for all x. Also, f (3) = 5 and f ′ (3) = 4. Then the value of [ f (10)]2 – [ φ (10)]2 is

Objective Mathematics

(a) 0 (b) 9 (c) 41 (d) None of these

(a) 0 (c) – 1

1 dy 61. If 8 f (x) + 6 f   = x + 5 and y = x2 f (x), then at dx  x x = – 1 is equal to (a) 0 (c) –

1 14

1 (b) 14 (d) None of these

62. If cos y = x cos (a + y) and the value of k is (a) sin a (c) 1

k dy = then 2 1 + x − 2 x cos a dx

(b) cos a (d) – sin a

63. If e x + e y = e x + y, then the value of (a) 0 (c) 1

dy at (1, 1) is dx

(b) – 1 (d) None of these.

(b) 45 (d) None of these

(b) 1 (d) None of these xn

n! nπ 70. If f (x) = cos x cos 2 nπ sin x sin 2 n d [f (x)]x = 0 is dx n (a) 0 (c) – 1 71. If

64. Let f be a function defined for all x ∈ R. If f is differentiable and f (x3) = x5 for all x ∈ R (x ≠ 0), then the value of f ′ (27) is (a) 15 (c) 0

 m ( x)  69. If f (x) = log    , m (1) = n (1) = 1 and m′ (1) = n′ (1)  n ( x)  = 2, then f ′ (1) is equal to

2 4

then the value of

8

(b) 1 (d) None of these

d  1 + x 4 + x8  3   = ax + bx then dx  1 + x 2 + x 4 

(a) a = 4, b = 2 (c) a = – 2, b = 4

(b) a = 4, b = – 2 (d) None of these

72. Let f (x) = α (x) β (x) γ (x) for all real x, where α (x), β (x) and γ (x) are differentiable functions of x. If f ′ (2) = 18 f (2),  α′ (2) = 3α (2), β' (2) = – 4β (2) and γ′ (2) = k γ (2), then the value of k is (a) 14 (c) 19

(b) 16 (d) None of these

π π  2 2  1 dy 1 and x 4 + y4 = t2 + 2 , then is 73. If f (x) = cos x + cos  x + 3  + sin x sin  x + 3  and   t dx t 5 equal to g    = 3 then (gof ) (x) is equal to 4 y y (a) 1 (b) 2 (a) (b) – x x (c) 3 (d) None of these 3 30 n x x 74. Let f (x) = (x + 2) . If f  (x) is a polynomial of degree (c) (d) – y y 20, where f n (x) denotes the nth derivative of f (x) w.r.t. x, then the value of n is 66. If f (x) = x 2 − 10 x + 25 , then the derivative of f (x) on (a) 60 (b) 40 the interval [0, 7] is (c) 70 (d) None of these.

65. If x2 + y 2 = t +

(a) 1 (c) 0

(b) – 1 (d) Does not exist

67. Let f (x) = 22x – 1 and φ (x) = – 2 x + 2x log 2. If f ′ (x) > φ′ (x), then (a) 0 < x < 1 (c) x > 0

(b) 0 ≤ x < 1 (d) x ≥ 0

1 68. Let f (x) be a polynomial function satisfying f (x) ⋅ f     x 1 = f (x) + f    . If f (4) = 65 and l1, l2, l3 are in G.P.  x then f ′ (l1), f ′ (l2), f ′ (l3) are in (a) A.P. (b) G.P. (c) H.P. (d) None of these

75. Let f (x) = 3x2 + 4x g′ (1) + g′′ (2) and g (x) = 2x2 + 3x f ′ (2) + f ′′(3) then (a) f ′ (1) = 22 + 12 f ′ (2) (b) g′ (2) = 44 + 12 g′ (1) (c) f ′′(3) + g′ ′ (2) = 10 (d) None of these 76. If y = xn – 1 log x, then x2y2 + (3 – 2n) xy1 is equal to (a) – (n – 1)2 y (b) (n – 1)2y (c) – n2y (d) n2y 77. If 2x = y1/5 + y–1/5, then (x2 – 1) y2 + xy1 = ky where k is equal to (a) – 25 (b) 25 (c) 16 (d) – 16 dy π at x = is equal to dx 4 (b) 2 (d) None of these

78. If y = [(tan x)tan x]tan x , then (a) 1 (c) 0

(a) – 1 (c) 0

(b) 1 (d) None of these

π 80. If f (x) = cos x cos 2x cos 4x cos 8x, then f ′    is 4 (a) – 1 (b) 2 (d) None of these. (c) 2 π  81. If f (x) = sin  [ x] − x 5  , 1 < x < 2 and [x] denotes the 2   π greatest integer less than or equal to x, then f ′  5   2 is equal to π (a) 5   2 (c) 0

4/5

π (b) – 5   2 (d) None of these 4/5



1 x 1 (c) |x|

(a)

(b) –

1 x

(d) None of these

89. If the parametric equation of a curve is given by θ x = cos θ + log tan and y = sin θ, 2 d2y = 0 are given by then the points for which dx 2 π ,n∈I 2 (c) θ = (2n + 1) π, n ∈ I (d) θ = 2nπ, n ∈ I.

(a) θ = nπ, n ∈ I

(b) θ = (2n + 1)

90. If f (x) = | x – 3 | and φ (x) = ( fof ) (x), then for x > 10,  φ′ (x) is equal to (a) 1 (c) – 1

(b) 0 (d) None of these

82. If f (x) + f (y) + f (z) + f (x) ⋅ f (y) ⋅ f (z) = 14 for all x, y, z dy  2x  ∈ R, then 91. If y = sin–1  is equal to  , then 1 + x2  dx  (a) f (0) = 2 2 (b) f ′ (x) = 0, for all x ∈ R (a) , when – 1 < x < 1 (c) f ′ (x) > 0, for all x ∈ R 1 + x2 (d) None of these 2 , when x < – 1 or x > 1 (b) + x2 1 83. Let f (x) be a polynomial of degree 3 such that f (3) = 1, 2 f ′ (3) = – 1, f ′′(3) = 0 and f ′′′ (3) = 12. Then the value (c) ­–  , when – 1 < x < 1 of f ′ (1) is 1 + x2 (a) 12 (c) – 13

(b) 23 (d) None of these

sin x cos x sin x dy 84. If y = cos x − sin x cos x , then is equal to dx x 1 1 (a) 1 (c) 0

(b) – 1 (d) None of these

85. If f (x) = xn; then the value of f' (1) f" (1) f"' (1) (−1) n f n (1) + − + ... + is 1! 2! 3! n! (a) 2n (b) 0 (c) 2n – 1 (d) None of these f (1) –

86. If f (x) = (cos x + i sin x) (cos 2x + i sin 2x) (cos 3x + i sin 3x) ... (cos nx + i sin nx) and f (1) = 1 then f ′′ (1) is equal to (a)

n (n + 1) 2

 n (n + 1)  (b)   2   2

 n (n + 1)  (c) –   2   2

(d) None of these

87. Let f (x) = sin x, g (x) = 2x and h (x) = cos x. If φ (x) = π [go ( f h)] (x), then φ′′   is equal to 4 (a) 4 (b) 0 (c) – 4 (d) None of these

(d) None of these 92. If f (x) =

x 2 + 6 x + 9 , then f ′ (x) is equal to (a) 1 for x < – 3 (b) – 1 for x < – 3 (c) 1 for all x ∈ R (d) None of these

93. If f (x) = | (x – 4) (x – 5) |, then f ′ (x) is equal to (a) – 2x + 9, for all x ∈ R (b) 2x – 9 if 4 < x < 5 (c) – 2x + 9 if 4 < x < 5 (d) None of these 94. If f (x) = | x – 1 | and g (x) = f [ f { f (x)}], then for x > 2, g′ (x) is equal to (a) – 1 if 2 ≤ x < 3 (c) 1 for all x > 2

(b) 1 if 2 ≤ x < 3 (d) None of these

95. If x2 + y2 = a2, then a is equal to (a)

(1 + y' 2 )3 | y" |

(b)

(c)

(1 + y' 2 )3 2 | y" |

(d) None of these

96. If y = tan

–1

( )  + tan

 log e  x3  log (ex3 )  

  

| y" | (1 + y' 2 )3

−1

  log (e 4 x3 )   log e 12 x 

( )

  d2y  , then dx 2  

is equal to (a) 1 (c) – 1

(b) 0 (d) None of these

115

88. If f (x) = log | 2x |, x ≠ 0, then f ′ (x) is equal to

Differentiation

dy 79. If y = (1 + x) (1 + x2) (1 + x 4) ... (1 + x2n), then at dx x = 0 is

116

97. If f (x) = xm, m being a non-negative integer, then the value of m for which f ′ (α + β) = f ′ (α) + f ′ (β), for all α, β > 0, is

Objective Mathematics

(a) 1 (c) 0

(b) 2 (d) None of these

98. If xexy = y + sin2x, then at x = 0, (a) – 1 (c) 0

dy is equal to dx

(b) 1 (d) None of these

99. If x2 + y2 = 1, then (a) yy′′– 2 (y′)2 + 1 = 0 (c) yy′′+ (y′)2 – 1 = 0 100. Let f (x) =

(b) yy′′+ (y′)2 + 1 = 0 (d) yy′′+ 2 (y′)2 + 1 = 0

x 3 sin x cos x 6 −1 0 , where p is a constant. p p2 p3

d3 [ f (x)] at x = 0 is dx 3 (a) p (b) p + p2 3 (d) independent of p (c) p + p Then

101. Let F (x) = f (x) g (x) h (x) for all real x, where f (x), g (x) and h (x) are differentiable functions. At some point x0, if F′ (x0) = 21 F (x0), f ′ (x0) = 4 f (x0), g′ (x0) = – 7 g (x0) and h′ (x0) = kh (x0) then k is equal to (a) 24 (c) – 12

(b) 12 (d) – 24

  dy  (a > 0); then 1 +     dx   is given by a 3 2 (c) 3a

2

  

2

d2y dx 2

at t =

cos 2t π 6

(b) a 2

(a)

(d)

2a 3

103. If f (x) is a polynomial of degree n (> 2) and f (x) = f (k – x), (where k is a fixed real number), then degree of f ′ (x) is (a) n (c) n – 2 104. If x + y y′′(0) = 2

(b) n – 1 (d) None of these 2

2 −π (a) e 2 a 2 −π (c) – e 2 a

= ae

( x ) , a > 0, then, assuming y > 0,

tan −1 y

2 π (b) – e 2 a (d) None of these

x sin x cos x f' ( x) 105. If f (x) = x 2 tan x − x 3  , then lim    is equal to x→0 x 2 x sin 2 x 5 x (a) – 4 (c) 0

(a) A (x) (c) C (x)

(b) B (x) (d) f (x)

107. The function y defined by the equation xy – log y = 1 satisfies x ( yy′′+ y′ 2) – y′′+ kyy′ = 0. The value k is (a) – 3 (c) 1

(b) 3 (d) None of these

108. If the function y (x) represented by x = sin t, y = aet

2

+ bet

2

 π π , t ∈  − ,  satisfies the equation  2 2

(1 – x2) y′′– xy′ = ky, then k is equal to (a) 1 (c) 2

(b) – 2 (d) None of these

109. The derivative of sec–1

1

w.r.t.

2x − 1 2

1 − x 2 at x =

1 2

is (a) – 4 (c) 2

(b) 4 (d) – 2

110. If f (x) = logx (lnx) then f ′ (x) at x = e is

102. Given : x = a cos t cos 2t and y = a sin t 3

106. If α is a repeated root of a quadratic equation f (x) = 0 and A (x), B (x), C (x) be polynomials of degree > 2, then the determinant A( x) B( x) C( x) A(α) B(α) C(α) is divisible by A' (α) B' (α) C' (α)

(b) 4 (d) None of these

1 e 1 (c) –  e

(a)

(b) e

111. If f (x) =

x −x df −1 ( x) , then is equal to 2 x + 2x dx

(d) 0 2

3 (1 − x) 2 1 (c) (1 − x) 2 (a) – 

112. If y = x + ex, then (a)

1 (1 + e x ) 2

(c) –

ex (1 + e x )3

113. If x = e y + e 1− x x x (c) 1+ x

(a)

y + ... to ∝

(b)

3 (1 − x) 2

(d) None of these d 2x is dy 2 (b) –

ex (1 + e x ) 2

(d) ex. , x > 0 then

dy is dx

(b)

1 x

(d)

1+ x x

114. If y = f (x) is an odd differentiable function defined on (– ∞, ∞) such that f ′ (3) = – 2, then f ′ (– 3) equals (a) 4 (b) 2 (c) – 2 (d) 0

(a) tan x

(b) cot x

(c) f (cos x)

(d)

116. If y =

sin x + y , then

1 x

dy equals dx

(a)

cos x 2y −1

(b)

cos x 1− 2y

(c)

sin x 1− 2y

(d)

sin x 2y −1

 1 + cos d  −1  tan 125. dx  1 − cos  

x 2 x 2

   is equal to   

(a)  –1/4 (c)  –1/2

 1 − x2  dy 117. If y = log  , then = 2  dx 1+ x  4 x3 (a) 1 − x4 1 (c) 4 − x4

(b)  1/4 (d)  1/2

126. If f ′′ (x) = – f(x), where f(x) is a continuous double differentiable function and g(x) = f ‘ (x). If F(x) =

4x (b) – 1 − x4 4 x3 (d) – 1 − x4

2

2

  x    x   f    +  g    and F(5) = 5, then F(10) is   2    2 

(a)  0 (b)  5 (c)  10 (d)  25 118. Let f and g be differentiable functions satisfying g′ (a) = 2, g (a) = b and fog = I (identity function). x −1 −1 x + 1 + sin −1 , then dy is 127. If y = sec Then f ′ (b) is equal to x −1 x +1 dx 2 (a)  1 (b)  0 (a) 2 (b) 3 1 (c)   x − 1 (d)   x + 1 (c) (d) None x +1 x −1 2 dx 128. Let φ(x) be the inverse of the function f(x) and 119. If x + y = 4, then at y = 1, is dy 1 d φ(x) is f ' ( x) = , then (a) – 1 (b) – 3 1 + x5 dx (c) 3 (d) None of these 1 1 (b)  (a)   2 5 d y dy 1 + [φ( x)] 1 + [ f ( x)]5 120. If y = log ( x + 1 + x 2 ), then is equal to 2 dx dx (c)  1 + [φ(x)]5 (d)  1 + f(x) 2x ( x 2 + 1)3/ 2 x (c)   2 ( x + 1)3/ 2

(a)   −

129. If f(x) is differentiable function and f ′′ (0) = a, then 2 f ( x) − 3 f (2 x) + f (4 x) is equal to lim x→0 x2 (a)  3a (b)  2a (c)  5a (d)  4a

1

(b)  

x2 + 1 x (d)   − 2 ( x + 1)3/ 2

121. Derivative of log | x | w.r.t. | x | is (a)   1 x (c)   ± 1 x

(b)  

 dy  1+    dx 

2

is equal

(b)  sec2 θ (d)  | sec θ |

π 123. The derivative of the function cot −1 ( cos 2 x ) at x = 6 is 2 1 (b)   (a)   3 3

(c)   3

(d)   6

sec x is

1

sec x sin x 4 x 1 (b)   (sec x )3/ 2 ⋅ sin x 4 x (c)   1 x sec x sin x 2 (d)   1 x (sec x )3/ 2 ⋅ sin x 2 (a)  

(d)  none of these

122. If x = a cos3 θ, y = a sin 3 θ, then to (a)  tan2 θ (c)  sec θ

130. Differential coefficient of

1 |x|

131. If x = y 1 − y 2 , then dy is equal to dx 1 − y2 (a)  x (b)   1 + 2 y2 2 1 − y (c)   (d)  0 1 − 2 y2

117

d  −1 1 + x 2 − 1   is equal to 124. dx  tan  x   x2 (a)   1 (b)   2 1 + x 2 ( 1 + x 2 − 1) 1 + x2 1 (c)   2 (d)   2(1 + x 2 ) 1 + x2

Differentiation

115. The differential coefficient of f (sin x) with respect to x where f (x) = log x is

118

132.

Objective Mathematics

d 2x is equal to dy 2 1 (a)   2 dy    dx   

2 (b)    d y   dy    2  dx   dx 

(c)   1 (–1 + loge 2) 3e (d)   − 1 (1 + loge 2) 3e

2

 d y   dy  (d)    − 2     dx   dx  2

2 (c)   d y dx 2

2

133. If f(x) = mx2 + nx + p, then f ' (1) + f ' (4) – f ' (5) is equal to (a)  m (c)  n 134. If y =

(b)  –m (d)  –n dy 1− x + y is equal to , then (1 – x2) dx 1+ x

(a)  1 (c)  2

(b)  – 1 (d)  0

2 135. y = eax sin bx, d y − 2a dy + a 2 y + is dx 2 dx

(a)  –a2y. (c)  –ay

d 2x equals dy 2

137.

−1

−1

 d2y  (a)    2   dx 

 d2y  dy (b)   −  2     dx   dx  −2

 d 2 y  dy (c)    2     dx   dx 

 d 2 y  dy (d)   −  2     dx   dx 

−3

−3

138. Let g(x) = log f(x) where f(x) is a twice differentiable positive function on (0, ∞) such that f(x + 1) = x f(x). Then, for N = 1, 2, 3,…,  1 1  1 + ... + (a)   −4 1 + +  (2 N − 1) 2   9 25

….

 1 1  1 (b)   4 1 + + + ... + 2 9 25 (2 N − 1)  

(b)  –b2y (d)  –by

136. If f(x) = log x3 (loge x2), then f´ (x) at x = e is (a)   1 (1 – loge 2) 3e (b)   1 (1 + loge 2) 3e

 1 1  1 (c)   −4 1 + + + ... + 2 9 25 (2 N + 1)    1 1  1 (d)   4 1 + + + ... + 2 9 25 (2 N + 1)  

SOLUTIONS 1. (b) We have



a = x

2 − b2



a = x

2 − b2 + b2 − c2 + c2 − a 2



4. (a) We have, y = ( x + 1 + x 2 ) n 

f (x) = (xa – b)a + b ⋅ (xb – c)b + c ⋅ (xc – a)c + a ⋅ xb

2 − c2

⋅ xc



2 − a2



= x0 = 1.

∴ f ′ (x) = 0.

2. (a) We have, f (x) = (sinm – nx)m + n ⋅ (sinn – px)n + p ⋅ (sinp – mx)p + m 2 − n2



m = sin



m = (sin x)



x ⋅ sin n

2 − p2

x ⋅ sin p

2 − n2 + n2 − p 2 + p 2 − m2

2 − m2

x



= (sin x)0 = 1

∴ f ′ (x) = 0.

3. (a) We have,

1

y=



γ

+

1 α

γ

+

1 α

x α + xβ + x γ = α = 1 ∴ x + xβ + x γ

dy = 0. dx

= or

n[ x + 1 + x2 ] 1 + x2 dy = dx

ny 1 + x2

⇒ y12 (1 + x2) = n2 y2.



Again differentiating, we get



2y1 y2 (1 + x2) + 2xy12 = 2n2yy1



Dividing by 2y1, we get

y2 (1 + x2) + xy1 = n2y d2y dy or (1 + x 2 ) + x = n2y. dx 2 dx 5. (c) We have,

β

x x x x x x + 1+ β + β 1+ γ + γ xα xα x x x x xα xβ xγ + + = α x + xβ + x γ x α + xβ + x γ x α + xβ + x γ 1+



β



Differentiating Eq. (1), we get  dy x  = n [ x + 1 + x 2 ]n − 1 1 +  2 dx  x + 1  



 log x  φ (x) = log5 log3 x = log5    log 3 



= log5 (log x) – log5 (log 3)

...(1)



φ′ (x) =



∴ φ′ (e) =

1 1 1 ⋅ ⋅ –0 log 5 log x x

=

1 1 1 ⋅ ⋅ log 5 log e e



=

1 . a + b cos x





−1 b sin x d2y (−b sin x) = . 2 2 = (a + b cos x) (a + b cos x) 2 dx

9. (b) sin y = x sin (a + y) sin y x = sin (a + y )

1 . elog 5



 2x − 1  6. (b) We have, y = f  2   x +1



dy  2 x − 1   ( x + 1) 2 − (2 x − 1) ⋅ 2 x  = f ′  2  ⋅ dx ( x 2 + 1) 2  x +1  

dx sin (a + y ) cos y − sin y cos (a + y ) = dy sin 2 (a + y ) sin (a + y − y ) = sin 2 (a + y ) ⇒

2







 2x − 1  = sin  2   x +1



2

 2 + 2x − 2x2  ⋅  2 2  ( x + 1) 

2   2x − 1   2x − 1   2 sin =  f' ( x) = sin x , ∴ f'  2   2    x +1  x + 1   

7. (b) Given f (x) = logx (log x)

log e (log e x) = loge (loge x) × logx e = log e x log e x ×



∴ f ′ (x) =



1 [1 − log e (log e x)] = x [log e x]2



1 1 1 × − log e (log e x) × log e x x x (log e x) 2

[   loge e = 1 and loge 1 = 0]

dy = dx





a−b x 1 ⋅ × sec 2 ⋅ 2 2 2 2 2 a + b a −b  a−b x 1+  tan  2  a+b 2

1 ⋅ a−b a+b

=

=

1

1 x a − b sin 2 1+ ⋅ a + b cos 2 x 2 2



a−b x sec 2 2 a+b

x ( a + b)cos 2 x 1 2 ⋅ ⋅ sec 2 a + b ( a + b)cos 2 x + ( a − b)sin 2 x 2 2 2 1 x x x x    a  cos 2 + sin 2  + b  cos 2 − sin 2     2 2 2 2

sin 2 (a + y ) dy = . sin a dx



10. (d) f ′ (x) =

=

(1 − sin x)(sin x) − (1 − cos x)(− cos x) (1 − sin x) 2

sin x − sin 2 x + cos x − cos 2 x sin x + cos x − 1 = (1 − sin x) 2 (1 − sin x) 2



1+ 0 −1 π 0 ∴ f ′   = = (1 − 1) 2 0 2



π Hence, f ′   does not exist. 2

11. (d) xy = ex– y

Taking log on both sides, we get



y log x = x – y x ⇒ y = 1 + log x



8. (a), (b)

=



1 [1 − log e (log e e)] 1 ∴ f ′ (x)at x = e = e = [log e]2 e



119

=

dy ∴ = dx

y (log x + 1) = x

(1 + log x) − x ⋅ (1 + log x)

2

1 x =

log x . (1 + log x) 2

12. (b) Put x = cos θ ⇒ θ = cos–1 x

 cos θ − sin θ   1 − tan θ  –1 ∴ y = tan–1   = tan   cos θ + sin θ   1 + tan θ  



π  π π = tan–1 tan  − θ  = – θ= – cos–1 x 4 4 4 





 1  dy  = = 0 – − 2  1− x  dx

13. (a) We have, g = inverse of f = f

1 1 − x2 –1



⇒ g (x) = f –1 (x) ⇒ f [g (x)] = x.



Differentiating w.r.t. x, we get



f ′ [g (x)] ⋅ g′ (x) = 1



1 ∴ g′ (x) = f' g ( x) = 1 + [g (x)]3 [ ]



.

  1 1 , ∴ f'[ g ( x)] =  f' ( x) = 3 3 + x + [ g ( x )] 1 1  

Differentiation

log (log x) – log5 (log 3) log 5



120

14. (c) Put x = tan θ ⇒ θ = tan–1x 1 + tan 2 θ + 1

Objective Mathematics



∴ y = cos–1



= cos–1



θ  θ 1 = cos–1  cos  = = tan–1x 2 2 2 

2 1 + tan 2 θ

1 + cos θ = cos–1 2

sec θ + 1 2 sec θ

= cos–1 2 cos 2 θ

18. (c) We have, y = etan x ⇒ log y = tan x Differentiating w.r.t. x, we get 1 dy dy = sec2x or = y sec2x y dx dx dy =y ⇒ cos2x dx Differentiating again w.r.t. x, we get d2y dy dy cos2x – 2 cos x sin x = dx 2 dx dx

2

2

1 dy = . ( + x2 ) 2 1 dx 15. (a) Let f (x) = ax2 + bx + c







Then,  f (1) = a + b + c and f (– 1) = a – b + c Since f (1) = f (– 1) ⇒ a + b + c = a – b + c ⇒ 2b = 0  or  b = 0 i.e.,, f (x) = ax2 + c ∴  f ′ (x) = 2ax. f ′ (a) = 2a2, f ′ (b) = 2ab, f ′ (c) = 2ac Now 2f ′ (b) = f ′ (a) + f ′ (c) if 2 ⋅ 2ab = 2a + 2ac if 2b = a + c if  a, b, c are in AP, which is given. ∴  f ′ (a), f ′ (b), f ′ (c) are in AP. 2

a (r n − 1) ⇒ (r – 1) Sn = arn – a r −1 Differentiating both sides w.r.t. r, we get

16. (b) We have, Sn =

dS (r – 1) n   + Sn = narn – 1 – 0 dr dSn ⇒ (r – 1) = narn – 1 – Sn dr



= n [nth term of G.P.] – Sn = n (Sn – Sn – 1) – Sn



= (n – 1) Sn – n Sn – 1.

17. (b) We have,

y = tan–1







⇒ log log y = log log x ⋅ log log x + log log x



Differentiating w.r.t. x, we get y log y dy = [2 log log x + 1]. x log x dx



20. (c) We have,



y = a cos (log x) + b sin (log x)  ...(1) dy 1 1 ⇒ = – a sin (log x) ⋅   + b cos (log x) ⋅   dx x x  [Differentiating w.r.t. x] dy ⇒ x  = – a sin (log x) + b cos (log x) dx d2y dy + dx 2 dx



⇒ x



= – a cos (log x) ⋅



⇒ x2 



= – y. 

1 1 – b sin (log x) ⋅ x x

d2y dy +x = – [a cos (log x) + b sin (log x)] dx 2 dx [From (1)]

21. (a) We have, y = (sin x)  –1

2

dy = 2 sin–1x ⋅ dx

...(1) 1

dy dx

1 − x2

1 + ... to n terms 1 + ( x + 5 x + 6)



= 2 sin–1x



 dy  ⇒ (1 – x2)   = 4 (sin–1x)2 = 4y  [Using (1)]  dx 

+ [tan–1 (x + 3) – tan–1 (x + 2)] + ... + [tan  (x + n) – tan (x + n – 1)] –1 

= tan–1 (x + n) – tan–1x 1 1 dy − ∴ = . 1 + ( x + n) 2 1 + x 2 dx

1 − x2



i.e.,

2

⇒ (1 – x2) ⋅ 2 ⋅

−1  ( x + 3) − ( x + 2)  + tan   + ... to n terms  1 + ( x + 3)( x + 2) 





⇒ log y = (log x)log log x ⋅ log x

2

= [tan–1 (x + 1) – tan–1x] + [tan–1 (x + 2) – tan–1 (x + 1)]





⇒ 

−1 + tan

–1

log log x





(log x ) 19. (a) We have, y = x

1 1 + tan −1 2 2 1 + ( x + x) 1 + ( x + 3 x + 2)

−1  ( x + 1) − x  −1  ( x + 2) − ( x + 1)    + tan  = tan   1 + ( x + 1) x   1 + ( x + 2)( x + 1) 



d2y dy = (1 + sin 2x) . dx 2 dx

i.e., cos2x

=

dy dx

4

⇒ (1 – x2) 

 dy  dy d2y ⋅ + (– 2x) ⋅   dx dx 2  dx 

d2y dy =x +2 2 dx dx [Dividing both sides by 2

22. (b) We have,

y=

x

∑ tan r =1

−1

1 = 1+ r + r2

x

∑ tan r =1

−1

 (r + 1) − r     1 + (r + 1) r 

2

dy ] dx

(r + 1) − tan −1 r 





= [tan–12 – tan–11 + tan–13 – tan–12 + ...

r =1



+ tan–1x – tan–1 (x – 1) + tan–1 (x + 1) – tan–1x]

= [tan–1 (x + 1) – tan–11] 1 dy ∴ = . 1 + ( x + 1) 2 dx



23. (a) We have,

y = eax sin bx

...(1)



dy = eax ⋅ b cos bx + a eax sin bx ⇒ dx



= b eax cos bx + ay



d2y dy ⇒ = ba eax cos bx + beax ⋅ b (– sin bx) + a   2 dx dx

[From (1)]

...(2)



 dy  dy = a  − ay    – b2y + a [From (1) and (2)] dx dx  



dy = 2a – (a2 + b2) y dx



d2y dy ∴ – 2a = – (a2 + b2) y. dx 2 dx

24. (c) Let y = tan–1

2x 1 − x 1 − 2x2

2

and t = sec–1

dy = – 2. dθ







1   Also, t = sec–1  2 cos 2 θ − 1    = sec–1 (sec 2θ) = 2θ













= P (x) P′′′ (x). 

 1 + x2 − 1   26. (a) Let y = tan–1    x  

dt dy  = – 1.  ∴ = – 1. dθ dt  x = 1



2 1 + x2

Put x = tan θ  sec θ − 1   1 − cos θ  –1 ∴ y = tan–1   = tan    tan θ   sin θ  = tan–1 tan

θ θ = 2 2

1 + sec θ 1 + cos θ = cos–1 2 sec θ 2 θ θ = cos–1 cos = 2 2 θ dy ∴ y = = t; ∴ = 1. 2 dt 1 1  −1 −1 27. (c) y = tan  (2 tan u ) + (2 tan u )  2 2  and  t = cos–1

 −1 2u  2u = tan  tan = x.  = 1 − u2  1 − u2 

y2 = P (x) dy = P′ (x) ⇒ 2y  dx

28. (a) Let y = f (tan x) and u = g (sec x)



dy = f ′ (tan x) sec2x dx



and

du = g′ (sec x) ⋅ sec x tan x dx





dy dy = dx du





...(1) ...(2)

dy dy d2y ⋅ + 2y ⋅ = P′′ (x) dx dx dx 2



d2y  dy  = y2 P′′ (x) ⇒ 2y2   + 2y3 ⋅ dx 2  dx  2

 dy  d2y ⇒ 2y3 = y2 P′′ (x) – 2y2   dx dx 2 1 = y2 P′′ (x) – [P′′ (x)]2 2

dy = 1. dx



2

⇒2



1 + 1 + x2

and t = cos–1







[ y2 = P (x)]

= tan (tan–1u + tan–1u) = tan (2 tan–1u)

25. (b) We have,

= y2 P′′′ (x)



1 2x2 − 1

dy 1 P′′ (x) + y2 P′′′ (x) – 2P′ (x) ⋅ P′′ (x) dx 2 = P′ (x) P′′ (x) + y2 P′′ (x) – P′ (x) P′′ (x) dy    2 y dx = P' (x) = 2y





Put x = cos θ Then,  2 cos θ sin θ  –1 y = tan–1   = tan (– tan 2θ) = – 2θ 2  1 − 2 cos θ 

dt =2 dθ dy dy = dθ dt



2 d  y3 d y   dx 2  dx

⇒2

121

−1

=

dy  du  x = π 4

2

= [from 2]

f ′(tan x) sec 2 x du = g ′(sec x) sec x tan x dx π  f'  tan  4  = π π  g'  sec  sin 4 4 

f' (1) 1 g' ( 2 ) ⋅ 2

=

2×2 = 4

1 . 2

Differentiation

x

∑  tan



122

29. (b)

Objective Mathematics





dx = a (– sin θ + 1 ⋅ sin θ + θ ⋅ cos θ) = aθ cos θ dθ dy = a (cos θ – 1 ⋅ cos θ + θ sin θ) = aθ sin θ dθ a θ sin θ dy dy dx = = = tan θ. ∴ a θ cos θ dθ dθ dx ⇒



1 sec3 θ d2y dθ 2 2 = sec θ ⋅ = sec θ ⋅ = a θ cos θ aθ dx 2 dx sec3 π d2y 4 = 8 2 . =  dx 2  θ = π aπ π a  4 4  

30. (a) We have,

x = t = e dx ⇒ = et log t (1 + log t) = tt (1 + log t) dt t

t log t

t

Also,  y = t t ⇒ log y = tt log t = et log t ⋅ log t 1 dy 1 ⇒ = et log t (1 + log t) log t + et log t ⋅ y dt t 1 dy tt t  ⇒ = t ⋅ t (1 + log t )log t +  dt t 



dy dt dy = = dx dt dx

t  1 t t (1 + log t )log t +  t  . (1 + log t )

dy 31. (c) We have, y = x3 – 8x + 7 ⇒ = 3x2 – 8 dx It is given that when t = 0, x = 3. dy ∴ When t = 0, = 3 ⋅ 32 – 8 = 19. dx dy dy dt =  ...(1) Also, dx dx dt dy dy = 19 and = 2, Since, when t = 0, dx dt 2 dx 2 ∴  from (1), 19 = ⇒ = . dx dt dt 19 32. (b) Differentiating the given equation w.r.t. x, we get dy dy + 2hy + 2by =0 2ax + 2hx dx dx dy ax + hy =– ⇒ dx hx + by d2y ⇒  dx 2  dy  dy      (hx + by )  a + h dx  − (ax + hy )  h + b dx        =–  (hx + by ) 2     dy 2   2  y (ab − h ) + dx (h x − abx)  =– (hx + by ) 2

dy   (h 2 − ab)  y − x  dx   = 2 (hx + by ) (h 2 − ab)  ax + hy  h 2 − ab ⋅ y + x ⋅ = .   (hx + by ) 2  hx + by  (hx + by ) 2 dy 33. (a) We have, y = sin x ⇒ = cos x. Now, dx d2 d  d 7  7  cos x  2  (cos x) = dy dy  dy  =



d dy d = dy

=

 dx  6  7 cos x ⋅ (− sin x)  dy   (– 7sin x cos5x)



= – 7 [cos x ⋅ cos5x – 5cos4x sin2x] 



= – 42 cos5x + 35 cos3x

dx dy

34. (a), (c) Taking log on both sides of the given equation, we get, x x x log cos + log cos 2 + log cos 3 + ...∞ 2 2 2  = log sin x – log x Differentiating w.r.t. x 1 x 1 x 1 − tan − 2 tan 2 − ...∞ = cot x – 2 2 2 2 x ⇒

x 1 x 1 1 tan + 2 tan 2 + ...∞ = – cot x + 2 2 2 2 x

Differentiating again w.r.t. x, we get x 1 x 1 1 2 sec 2 + 4 sec 2 2 + ...∞ = cosec2x – 2 . x 2 2 2 2 π 35. (b) When < x < π, cos x < 0, ∴ | cos x | = – cos x 2 ∴ f (x) = – cos x ⇒ f ′ (x) = sin x  3π  3π ∴ f ′   = sin = 4 4  

1 . 2

36. (b) We have, log f (x) = (a + b + 2x) [log (a + x) – log (b + x)] Differentiating both sides w.r.t. x, we get f' ( x) = 2 [log (a + x) – log (b + x)] f ( x) 1   1 − + (a + b + 2x)   a + x b + x ⇒ f ′ (0) = f (0)   1 1   2 (log a − log b) + (a + b)  a − b     



=

a   b

a +b

 a b2 − a 2   2 log +  b ab  

π , cos x > sin x ∴ cos x – sin x > 0. 4 π π – 2x + 2



22x + 2x – 2 > 0

⇒ ⇒

(2x – 1) (2x + 2) > 0 2x – 1 > 0 [  2x + 2 > 0 for all x]



2x > 1 ∴ x > 0.

68. (b) Since f (x) is a polynomial function satisfying 1 1 f (x) ⋅ f   = f (x) + f    ,  x  x

⇒ cot y (1 – x cos a) = – x sin a  x sin a  ⇒ y = cot–1    x cos a − 1 

– e– y

f (x3) = x5

= x4 + y4 + 2x2y2 – 2 ⇒ x2y2 = 1

1 8 f (x) + 6 f   = x + 5 for all x  x Therefore, 1 1 8 f   + 6 f (x) = + 5 x x  

dy  ∴ = dx



∴ f ′ (27) = f ′ (33) =

∴ [ f (10)]2 – [φ (10]2 = [ f (3)]2 – [φ (3)]2 = [ f (3)]2 – [ f ′ (3)]2

64. (a) We have, for all x (x ≠ 0)

–x

=1

∴ f (x) = xn + 1 or f (x) = – xn + 1 If f (x) = – xn + 1, then f (4) = – 4n + 1 ≠ 65 So, f (x) = xn + 1. Since f (4) = 65 ∴ 4n + 1 = 65 ⇒ n = 3 ∴ f (x) = x3 + 1 ⇒ f ′ (x) = 3x2 ∴ f ′ (l1) = 3l12 , f ′ (l2) = 3l22 , f ′ (l3) = 3l32 Since l1, l2, l3 are in G.P., ∴ f ′ (l1), f ′ (l2), f ′ (l3) are also in G.P. 69. (a) f ′ (x) =

n( x)  n( x)m' ( x) − m( x)n' ( x)    m( x )  [n( x)]2 

⇒ f ′ (1) =

n(1)  n(1)m' (1) − m(1)n' (1)    = 0. m(1)  [n(1)]2 



n

d [ f (x)]x = 0 dx n

n! n! nπ nπ cos = cos 2 2 nπ nπ sin sin 2 2



Thus,

2 4 =0 8

[ C1 and C2 are identical]

71. (b) We have, d  (1 + x 2 + x 4 )(1 − x 2 + x 4 )    = ax3 + bx dx  (1 + x 2 + x 4 ) 



75. (a), (b), (c)  We have,

f ′ (x) = 6x + 4g′ (1)

...(1)



f ′′(x) = 6

...(2)



g′ (x) = 4x + 3 f ′ (2)

...(3)



g′′ (x) = 4

...(4)

From (1), f ′ (1) = 6 + 4g′ (1) = 6 + 4 [4 + 3 f ′ (2)] [  g′ (1) = 4 + 3 f ′ (2)] = 22 + 12 f ′ (2)

⇒ – 2x + 4x3 = ax3 + bx ⇒ a = 4 and b = – 2.

From (3), g′ (2) = 8 + 3 f ′ (2) = 8 + 3 [12 + 4g′ (1)]

72. (c) We have, f (x) = α (x) β (x) γ (x), for all real x

[ f ′ (2) = 12 + 4g′ (1)] = 44 + 12 g′ (1)

⇒ f ′ (x) = α′ (x) β (x) γ (x) + α (x) β′ (x) γ (x) + α (x) β (x) γ′ (x)

Also, from (2) and (4), f ′′(3) + g′′ (2) = 6 + 4 = 10. 76. (a) We have, y1 = (n – 1)xn – 2 log x + xn – 1 ⋅

⇒ 18 f (2) = 3α (2) β (2) γ (2) – 4α (2) β (2) γ (2)

1 x



+ kα (2) β (2) γ (2)

⇒ xy1 = (n – 1) xn – 1 log x + xn – 1



[ f ′ (2) = 18 f (2), α′ (2) = 3α (2), β′ (2) = – 4β (2)

Differentiating again w.r.t. x, we get



= (n – 1) y + xn – 1 

⇒ x2y2 + xy1 = (n – 1) xy1 + (n – 1) xn – 1

⇒ 18 f (2) = (– 1 + k) α (2) β (2) γ (2) = (k – 1) f (2) ⇒  k – 1 = 18  ∴ k = 19.

π π   73. (c) f ′ (x) = – 2 cos x sin x – 2 cos   x +   sin   x +  3 3   π π   + cos x sin  x +    + sin x cos   x +  3 3   2π  π   = – sin 2x – sin   2 x +  + sin   x + x +  3  3   π π   π = – 2 sin  2 x +  cos  + sin   2 x +  3 3 3   π π   = – sin   2 x +  + sin   2 x +  = 0. 3 3   ⇒

= (n – 1) xy1 + (n – 1) (xy1 – (n – 1) y)

[ f (2) = α (2) β (2) γ (2)]



f (x) = constant for all x.

π π 5 But, f (0) = cos2 0 + cos2 + sin 0 ⋅ sin = 3 3 4

...(1)

y1 + xy2 = (n – 1) y1 + (n – 1) xn – 2

and γ′ (2) = kγ (2)] 

n = 70.

d (1 – x2 + x4) = ax3 + bx ⇒ dx

⇒ f ′ (2) = α′ (2) β (2) γ (2) + α (2) β′ (2) γ (2) + α (2) β (2) γ′ (2)

127

74. (c) f (x) is a polynomial of degree 90. f ′ (x) reduces the degree of f (x) by one. Thus, in order to get a polynomial of degree 20, we must reduce the degree of f (x) by 70. Hence, f (x) should be differentiated 70 times to get a polynomial of degree 20.



[using (1)]

⇒ x2y2 + (3 – 2n) xy1 + (n – 1)2 y = 0. 77. (b) We have,

4x2 = ( y1/5 + y–1/5)2 = y2/5 + y–2/5 + 2 = ( y1/5 – y–1/5)2 + 4 ⇒ y1/5 – y–1/5 = 2 x 2 − 1 . Adding with the given relation, we get 2y1/5 = 2  x + x 2 − 1    ⇒ y =  x + x 2 − 1      2 4 ⇒ y1 = 5 ( x + x − 1) 1 + 

⇒ y1 =

5 ( x + x 2 − 1) 5 x −1 2

⇒ (x2 – 1) y12 = 25y2

=

x   x2 − 1  5y x2 − 1

5

Differentiation

2 n! n! dn nπ  nπ  70. (a) n [ f (x)] = cos  x + 4  cos dx 2  2  nπ  nπ  8 sin  x +  sin 2  2 

5 for all x. 4 5 (gof ) (x) = g [ f (x)] = g    = 3. 4

f (x) =



128

Differentiating again w.r.t. x, we get

⇒ 3 f ′ (x) [1 + { f (x)}2] = 0

(x – 1) 2y1 y2 + y1 ⋅ 2x = 50 yy1

⇒ f ′ (x) = 0, for all x.

2

2

Objective Mathematics



[Dividing by 2y1] 83. (b) Let f (x) = a (x – 3)3 + b (x – 3)2 + c (x – 3) + d 2 ⇒ (x – 1) y2 + xy1 = 25y. ∴ k = 25. Then, f (3) = 1 = d ⇒ d = 1

78. (b) We have, log y = tan2x ⋅ log tan x ⇒ ⇒ ∴

1 dy 1 = 2 tan x sec2x ⋅ log tan x + tan2x ⋅ sec2x y dx tan x

∴ f (x) = 2 (x – 3)3 – x + 4 ⇒ f ′ (x) = 6 (x – 3)2 – 1

dy  = 1 ⋅ 1 ⋅ 2 (2 ⋅ 0 + 1) = 2. dx  x = π

∴ f ′ (1) = 6 (4) – 1 = 23. 84. (a)

79. (b) We have, 2n

y = (1 + x) (1 + x2) (1 + x4) ... (1 + x ) n +1

(1 − x)(1 + x)(1 + x 2 )(1 + x 4 )...(1 + x 2 ) 1 − x2 = 1− x 1− x n +1

(1 − x) ⋅ −2n +1 ⋅ x 2 −1 + (1 − x 2 dy ⇒ = (1 − x) 2 dx

n +1

sin x cos x sin x cos x sin x = 0 – sin x cos x sin x + 1 − sin x cos x x 1 1

dy  = 1. dx  x =0

1  sin x ⋅ cos16 x ⋅ 16 − sin 16 x ⋅ cos x  ⇒ f ′ (x) = a 16  sin 2 x   1  1 ⋅ 1 ⋅ 16 − ⋅0  π 1  2 2  = ∴ f ′   = 2  16  4  1         2

= 0 + (cos2x + sin2x) = 1. 85. (b) We have, f (x) = xn ⇒ f (1) = 1 = nC0

82. (a), (b) We have, f (x) + f (y) + f (z) + f (x) ⋅ f (y) ⋅ f (z) = 14 ...(1) Putting x = y = z = 0, we get 3 f (0) + [ f (0)]3 = 14 ⇒ [ f (0)]3 + 3 f (0) – 14 = 0

3 f ′ (x) + 3 [ f (x)]2 ⋅ f ′ (x) = 0

f"(1) n (n −1) = = nC 2 2! 2! f'"(1) n (n − 1)(n − 2) n = = C3 3! 3! f n (1) n! = = nC n n! n! f' (1) f" (1) f"' (1) (−1) n f n (1) + − + ... + 1! 2! 3! n! = nC0 – nC1 + nC2 – nC3 + ... + (– 1)n nCn ∴ f (1) –

= (1 – 1)n = 0. 86. (c) f (x) = (cos x + i sin x) (cos 2x + i sin 2x) (cos 3x + i sin 3x)

for all x, y, z ∈ R

Differentiating w.r.t. x, we get





4/5 4/5  π π π π ⇒ f ′  5  = – 5   ⋅ sin   = – 5   . 2 2  2 2

3 f (x) + [ f (x)]3 = 14

f'(1) n = = nC 1 1! 1!



π 5 ∴ f (x) = sin  − x  = cos x5 2  ⇒ f ′ (x) = – sin x5 ⋅ 5x4

⇒ f (0) = 2.



2.

81. (b) Since, 1 < x < 2. ∴ [x] = 1

Now, putting y = z = x in (1), we get

sin x cos x sin x + cos x − sin x cos x 1 0 0



)

sin 16 x 80. (c) f (x) = 2 sin x cos x cos 2 x cos 4 x cos 8 x = 4 2 sin x 2 sin x



cos x − sin x cos x sin x cos x sin x dy = cos x − sin x cos x + − sin x − cos x − sin x dx 1 1 x x 1 1



n



f ′′(3) = 0 = 2b ⇒ b = 0 f ′′′ (3) = 12 = 6a ⇒ a = 2.

dy = [(tan x)tan x]tan x ⋅ tan x sec2x (2 log tan x + 1) dx

4

=

f ′ (3) = – 1 = c ⇒ c = – 1



... (cos nx + i sin nx) = cos (x + 2x + 3x + ... + nx) + i sin (x + 2x + 3x + ... + nx) n (n + 1) n (n + 1) = cos x + i sin x 2 2 ⇒ f ′ (x) = ⇒  f ′′(x)

n (n + 1)  n (n + 1) n (n + 1)  − sin x + i cos x 2  2 2 

 n (n + 1)  = –   f ( x)   2   2

 n (n + 1)   n (n + 1)  ∴ f ′′ (1) = –   f (1) = –   . 2  2    2

2

92. (b) f (x) =

∴  [go ( f h)] (x) = g [( f h) (x)] = g [ f (x) ⋅ h (x)]

1 if x ≥ −3  . ∴ f ′ (x) =  −1 if x < −3

= g (sin x cos x) = 2sin x cos x = sin 2x i.e., φ (x) = sin 2x

93. (c) f (x) = | (x – 4) (x – 5) | ( x − 4)( x − 5) if ( x − 4)( x − 5) ≥ 0   =  −( x − 4)( x − 5) if ( x − 4)( x − 5) < 0 

π π ∴ φ′′    = – 4 sin = – 4. 4 2   log 2 x, 2 x > 0 i.e. x > 0   88. (a) f (x) = log | 2x | =  log(−2 x), 2 x < 0 i.e. x < 0  1  1   2 x × 2, x > 0   x , x > 0   =   ⇒ f ′ (x) =   1 × −2, x < 0   1 , x < 0  −2 x   x 

∴ f ′ (x) =

2 if x ≤ 4 or x ≥ 5  x − 9 x + 20  =  2 − ( x − 9 x + 20 ) if 4< x 2)

1 , x ≠ 0. x



89. (a) We have, dx 1 sec θ / 2 = – sin θ + = – sin θ + θ θ dθ 2 tan θ / 2 2 sin cos 2 2 1 1 − sin 2 θ cos 2 θ = – sin θ + = = sin θ sin θ sin θ dy = cos θ dθ dy d θ dy cos θ sin θ = = = tan θ ∴ dx d θ dx cos 2 θ

and



d2y dθ sin θ = sec2θ ⋅  = dx 2 dx cos 4 θ

d2y = 0 ⇒ sin θ = 0 dx 2



90. (a) Since x > 10, therefore, f (x) = x – 3 ∴ φ (x) = ( fof ) (x) = f [ f (x)] = f (x – 3) = x – 3 – 3 = x – 6

 ⇒ φ′ (x) = 1.



dy = dx = 2⋅

1 1−

4x2 (1 + x 2 ) 2

1 + x2 (1 − x 2 ) 2



 (1 + x 2 ) ⋅ 2 − 2 x ⋅ 2 x  ×  (1 + x 2 ) 2  

1 − x2 1 1 − x2 ⋅ = 2 ⋅   2 2 2 | 1 − x | 1 + x2 (1 + x )

 x − 3, if x ≥ 3  = | (x – 2) – 1 | = | x – 3 | =   − x + 3, if 2 ≤ x < 3 1, if x ≥ 3  ∴ g′ (x) =  . −1, if 2 ≤ x < 3

95. (a) We have, x2 + y2 = a2 ⇒ 2x + 2yy’ = 0 x ⇒ y’ = – . y Differentiating again w.r.t. x, we get 1 + y' 2 y′2 + yy′′+ 1 = 0 ⇒ y = – y" 2   x ∴ a = x2 + y 2 = y 1 +    y

⇒ θ = nπ, n ∈ I

91. (a)

f [ f (x)] = f (x – 1) = | (x – 1) – 1 | = | x – 2 | = (x – 2) (  x > 2) ∴ g (x) = f [ f { f (x)}] = f (x – 2)

2

Also,

( x + 3) 2 = | x + 3 |

 x + 3, x + 3 ≥ 0 i.e. x ≥ −3  =  − x − 3, x + 3 < 0 i.e. x < −3 

87. (c) We have, ( f h) (x) = f (x) ⋅ h (x) = sin x cos x

⇒ φ′ (x) = 2 cos 2x and φ′′ (x) = – 4 sin 2x

x2 + 6x + 9 =



= −

(1 + y' 2 ) ⋅ 1 + y' 2 y"

=

(1 + y' 2 )3 . | y" |

96. (b) We have,

−1  1 − 3 log x  −1  4 + 3 log x   + tan   y = tan   1 + 3 log x   1 − 12 log x 

= tan–11 – tan–1 (3 log x) + tan–14 + tan–1 (3 log x)  −1 −1 −1  A − B    Using tan A − tan B = tan  1+AB       A + B   −1  −1 −1  and tan A + tan B = tan     1 − AB   

129

2

Differentiation

 2  1 + x 2 , − 1 < x < 1  dy  =  −2 ∴   dx , x or x < − > 1 1 1 + x 2 

n (n + 1) n (n + 1)   n (n + 1)   x + i sin x = –   cos 2   2 2  

130



Objective Mathematics

= tan–11 + tan–1 (4) dy d2y ∴ = 0 and hence = 0. dx dx 2

102. (d) We have, −a sin 3t cos t sin 2t  dx  = a  − sin t cos 2t − =  cos 2t dt cos 2t  

97. (b), (c) We have, f ′ (α + β) = f ′ (α) + f ′ (β) ⇒ (α + β)m – 1 = αm – 1 + βm – 1

and

Since, for m > 2, the above equality is not valid ∴ we must have m = 2. Also, for m = 0, f ′ (x) = 0 for all x. So the equality is trivially true. 98. (b) When x = 0, we get y = 0. Differentiating both sides of the given equation w.r.t. x, we get  dy  dy exy + xexy  x + y  = + 2 sin x cos x dx dx



dy dy dt = = – cot 3t. dx dx dt



dt −3cosec 2 3t ⋅ cos 2t d2y 2 = 3 cosec 3t ⋅ = dx a sin 3t dx 2

2   ∴ 1 +  dy     dx  



99. (b) Differentiating x2 + y2 = 1 w.r.t. x, we get 2x + 2yy’ = 0 or x + yy’ = 0. Again differentiating w.r.t. x, we get 1 + y′y’ + yy′′= 0 or 1 + (y′)2 + yy′′= 0.



3 = –   cosec33t cos 2t a

Putting x = 0, y = 0, we get  dy   dy  e0 + 0 e0  0 ⋅  + 2 sin 0 cos 0 + 0 =   dx  ( 0, 0 )  dx  ( 0, 0 ) dy  = 1. dx  ( 0, 0 )

sin t sin 2t  a cos 3t dy  = a cos t cos 2t −  = cos 2t dt cos 2t  



= (1 + cot 2 3t )



3

3

d2y dx 2

2

2

  dy  2  1 +      dx  

 3 3  −  cosec 3t cos 2t  a 3

2

d2y dx 2

at

t =



π is 6

100. (d) We have,





3x 2 f ′ (x) = 6 p

cos x − sin x −1 0 p2 p3

6 x − sin x − cos x f ′′(x) = 6 −1 0 p p2 p3 and f ′′′ (x) =

6 ∴ f ′′′ (0) = 6 p

6 − cos x sin x 6 −1 0 p p2 p3 −1 −1 p2



[ R1 and R2 are identical]

101. (a) We have,

3 cos



+ ... + an – 1 [x + (k – x)] + an.

It may be noted that n must be even for otherwise f (x) will become a polynomial of degree n – 1. Clearly, f ′ (x) is a polynomial of degree n – 1.

104. (c) When x = 0, y > 0 ⇒ y = aeπ/2.  aking log on both sides of the given equation, T we get  y 1 log (x2 + y2) = log a + tan–1   . 2 x

Differentiating both the sides w.r.t. x, we get 1 2 x + 2 yy' ⋅ = 2 x2 + y 2

1

( x)

1+ y

2



xy' − y x2

⇒ x + yy′ = xy′ – y Differentiating again w.r.t. x, we get



⇒ log F (x) = log f (x) + log g (x) + log h (x) Differentiating both the sides w.r.t. x, we get



F' ( x) f' ( x) g' ( x) h' ( x) = + + F(x) f (x) g (x) h(x)







⇒ 21 = 4 – 7 + k ⇒ k = 24.

F' ( x0 ) f' ( x0 ) g' ( x0 ) h' ( x0 ) = + + F(x0 ) f (x0 ) g (x0 ) h(x0 )

2a . 3

f (x) = a0 [xn + (k – x)n] + a1 [xn – 1 + (k – x)n – 1]

F (x) = f (x) g (x) h (x)



π 3

=

103. (b) Clearly, f (x) must be of the form

0 0 =0 p3

a

...(1)

1 + (y′)2 + yy′′= xy′′+ y’ – y’ ⇒ 1 + (y′)2 = (x – y) y′′ 

⇒ y′′=

When x = 0, we get, from (1), y’ = – 1 2 2 −π ∴ y′′(0) = = – e 2. π/ 2 −ae a

1 + ( y' ) 2 . x− y



x sin x − sin x + x 2 tan x −3 x 2 . 2 x sin 2 x 5



∴ For x ≠ 0, 1 sin x cos x 1 cos x cos x f' ( x) 2 2 tan x − x + x sec x − x3 = 2 x x 2 2 cos 2 x 5 x 2 sin 2 x 5x

⇒ f ′ (x) =

= – 2 – 2 = – 4.

∆′ (x) =

and ⇒

dz =– dx

=

1 − x2

dy  = 4. dz  x = 1 2

log (log x) log x

1 − log (log x)  x (log x) 2

∴ f ′ (e) =

1 − log (log e) 1 = . e (log e) 2 e

x2 − x x2 + 2x



Clearly, f (0) and f (– 2) are not defined.



So domain of f = R\{0, – 2}.



Then, in this domain, we have



y = f (x) =



A ′ ( x) B′ ( x) C′ ( x) A(α) B(α) C(α) A' (α) B' (α) C' (α)

x

1 1 1 ⋅ − log (log x) ⋅ log x x x (log x) 2

111. (b) We have, f (x) =

A( x) B( x) C( x) 106. (d) Let ∆ (x) = A(α) B(α) C(α) , then A' (α) B' (α) C' (α)



dy dy dx 2 = = dz dz dx x



log x ⋅

1 0 1 1 1 1 1 0 0 f' ( x) ⇒ lim = 2 1 0 + 0 1 0 + 0 0 0 x→0 x 2 0 0 2 2 0 2 0 5



and z = 1 − x 2 dy 2 ⇒ =– dx 1 − x2

110. (a) We have, f (x) = logx (lnx) =

1 sin x − sin x + x tan x −3 x 2 2 sin 2 x 5





1 = cos–1 (2x2 – 1) = 2cos–1x 2x2 − 1



or x =

x −1  x+2

⇒ yx + 2y = x – 1

2y +1 2x + 1 , i.e., f –1 (x) = 1− y 1− x

df −1 ( x) 2 (1 − x) + 2 x + 1 3 = = . (1 − x) 2 dx (1 − x) 2



dy Since, ∆ (α) = 0 = ∆’ (α), therefore α is a repeated root of  = 1 + ex 112. (c) y = x + ex ⇒ ∆ (x) and α is a repeated root of the quadratic equation dx f (x) = 0, so ∆ is divisible by f (x). dx dx d 2x −1 1 ex ⋅ ∴ =  ⇒ = (1 + e x ) 2 dy dy dy 2 107. (b) Differentiating the given equation w.r.t. x, we get 1 + ex



xy′ + y ⋅ 1 –

1 y′ = 0 y





⇒ xyy′ – y′ + y2 = 0 i.e., (xy – 1) y′ + y2 = 0



Differentiating again w.r.t. x, we get



⇒ x (yy′′+ y′ ) – y′′+ 3yy’ = 0

108. (c) We have, ∴

113. (a) x = ey + x ⇒ log x = x + y

(xy – 1) y′′+ y′ (xy′ + y ⋅ 1) + 2yy’ = 0 dx dy = cos t and = dt dt

dy dt dy = = dx dt dx

2 ( a + b) et cos t

2 (a + b) et 2

=

2y 1 − x2

 dy  ⇒ (1 – x2)   = 2y2  dx  Differentiating w.r.t. x, we get (1 – x2) 2y′ y′′– 2x (y′ )2 = 4yy’ ⇒ (1 – x2) y′′– xy′ = 2y [dividing by 2y′] 2

∴ k = 2.



∴ k = 3.

2

1 −e x ex ⋅ . x 2 x = – (1 + e ) 1 + e (1 + e x )3

=

2

dy 1 =1+  dx x

114. (c) Since f (x) is odd

∴ 

dy 1− x = . dx x

∴ f (– x) = – f (x)

⇒ f ′ (– x) (– 1) = – f ′ (x) i.e., f ′ (– x) = f ′ (x) ∴ f ′ (– 3) = f ′ (3) = – 2. 115. (b) We have,

y = f (sin x)  or  y = log (sin x)



1 dy x cos x = cot x. = sin x dx

116. (a) y =

sin x + y

⇒ 2y

⇒ y2 = sin x + y

dy dy = cos x + dx dx

131

109. (b) Let y = sec–1

1 sin x cos x cos x cos x x f ′ (x) = 2 x tan x − x 3 + x 2 sec 2 x − x3 2 sin 2 x 5 x 2 x 2 cos 2 x 5 x

Differentiation

105. (a) We have,

132

Objective Mathematics

dy = cos x dx cos x dy = . ( 2 y − 1) dx

⇒ (2y – 1) ∴

π 3 sin   3 2 =   π   π  1 + 1  1 + 1 cos cos   3  3  2  2      

117. (b) We have, y = log (1 – x2) – log (1 + x2) dy −2 x 2x (1 + x + 1 − x ) = = – 2x ⋅   − 2 2 dx 1− x 1+ x 1 − x4 4x . =– 1 − x4 118. (c) We have, 2





fog = I ⇒ ( fog) (x) = x for all x



⇒ f ′ [g (x)] ⋅ g′ (x) = 1, for all x



⇒ f ′ [g (a)] ⋅ g′ (a) = 1



⇒ f ′ [g (a)] =

1 1 = g' (a ) 2

( g′ (a) = 2) 1 ⇒ f ′ (b) =  (  g (a) = b) 2 119. (b) Differentiating the given equation w.r.t. y, we get 1 dx 1 + =0 2 x dy 2 y dx =– dy

x = y

y −4 y

 ∴

dx  1− 4 = – 3.  = dy  y =1 1

d  −1 1 + tan 2 θ − 1   = dx  tan  tan θ  

1 dy d = (1 + x 2 ) −1/ 2 = log ( x + 1 + x 2 ) = 1 + x2 dx dx d2y 1 x = − (1 + x 2 ) −3/ 2 ⋅ 2 x = − dx 2 2 (1 + x 2 )3/ 2



121. (b) Let |x| = t, 

∴ 

d 1 1 (log t ) = = dt t |x|

122. (d) We have, dy = 3a sin2 θ cos θ dθ  and dx = –3a cos2 θ sin θ dθ 3 a sin 2 θ cos θ ⇒  dy = dy dx  = = − tan θ 3a cos 2 θ(− sin θ) dx d θ d θ 2

∴ 

 dy  2 1 +   = 1 + tan 2 θ = sec θ =| sec θ |  dx 

123. (a) We have, d 1 1 (− sin 2 x) [cot −1 ( cos 2 x )] = − ⋅ ⋅2 dx 1 + cos 2 x 2 cos 2 x =

sin 2 x (1 + cos 2 x) cos 2 x

Therefore, the value of the derivative at x = π is 6

(put x = tan θ)

=

 1 − cos θ  d d θ d θ tan −1  tan −1 tan =  = θ dx sin dx 2 dx  2   

=

1 1 d tan −1 x = 2(1 + x 2 ) 2 dx

 1 + cos d  −1  tan 125. (a) We have, dx  1 − cos    x  2 cos 2  d  −1 4 = d  tan = dx  dx 2 x  2sin  4   =

120. (d) We have,

2 3

2   124. (d) We have, d  tan −1 1 + x − 1   dx  x 





3 2= 3

=

2

x 2 x 2

     

x  −1  tan cot  4 

1 − d  −1  π x   −d  π x  + =−  tan tan  +   =  dx  4  2 4   dx  2 4 

126. (b) We have, f ′′ (x) = – f(x) ⇒ d (f ′ (x)) = – f(x) ⇒ g ′ (x) = – f(x) dx 2 2   x    x  Now, F ( x) =  f    +  g      2    2    x  x 1 ⇒ F '( x) = 2  f    ⋅ f '   ⋅ 2   2 2     x  x 1 x x  +2  g    ⋅ g '   ⋅ = f   ⋅ f '   − 2 2 2 2   2 

x x f   f '  =0 2 2 ∴ F(x) is constant  Since F(5) = 5, ∴ F (10) = 5.

 x +1  x −1  127. (b) We have, y = sec −1  + sin −1     x −1   x +1  x −1   x −1  = cos −1  + sin −1     x +1  x +1 ⇒ 

y=

π 2

∴ 

dy =0 dx

128. (c) We have, φ(x) = f –1(x) ⇒  x = f[φ (x)] On differentiating both sides w.r.t. x, we get 1 = f ′ [φ(x)]. φ ' (x)

f ' [φ( x)] =

∴ 

1  1 + [φ( x)]5

=−

∴  φ ' (x) = 1 + [φ(x)]5

129. (a) lim 2 f ( x) − 3 f (2 x) + f (4 x) x→0 x2 2 f ' ( x) − 6 f ' (2 x) + 4 f ' (4 x) = lim x→0 2x  (by L’ Hospital’s rule) = lim

x→0

 =

2 f '' ( x) − 12 f '' (2 x) + 16 f '' (4 x) 2 (by L’ Hospital’s rule)

2 f '' (0) − 12 f '' (0) + 16 f '' (0) 6a = = 3a 2 2

130. (b) Let y =

sec x

⇒ dy = 1 (sec x ) −1/ 2 ⋅ d (sec x ) dx 2 dx 1 1 = ⋅ sec x ⋅ tan x ⋅ 2 x 2 sec x =

dy = aeax sinbx + beax cosbx dx = ay + beax cosbx d2y dy =a + abe ax cos bx − b 2e ax sin bx of dx 2 dx

….(ii)

dy + abeax cosbx – b2y [using (i)] dx log (2 log x) 136. (a) Since, f (x) = log x3 (log x2) = 3 log x On differentiating w.r.t. x, we get 1 2 1 log x ⋅ ⋅ − log (2 log x) 2 log x x x f ′ (x) = 3 (log x) 2

= a

1 1 − log 2  1   = (1 − log 2) 3e  (1) 2  3e

dx 1  dy  = = dy dy / dx  dx 

−1

−1

 dx  d  dy  dx  =    dy  dx  dx  dy −2 −3  d 2 y   dy   dx   d 2 y   dy  d 2x = − 2      = −  2    ⇒ 2 dy  dx   dx   dy   dx   dx  138. (a) Given: g(x) = log f(x)

131. (c) We have, x = y 1 − y 2 dy y (− y ) dy 1 − y2 + dx 1 − y 2 dx  dy  1 − y 2 − y 2   = 1 or dx  1 − y 2  

….(i)

⇒ 

d ⇒ dy

1

dy  y2  1 − y2 − dx  1 − y2 

1− x 1− x 1 =− ⋅ 1− x 1 + x 1 − x2

dy + y=0 dx 135. (b) We have, y = eax sin bx

137. (d) Since,

(sec x )1/ 2 ⋅ sin x ⋅ sec x 4 x 1 = (sec x )3/ 2 ⋅ sin x 4 x

1=

×

(1 – x2)

⇒ f ′ (e) =

1 sin x (sec x )1/ 2 4 x cos x

=

1 (1 + x) 1 − x 2

133

1 1 Since f ' ( x) = 1 + x5 f ' [φ( x)]

Differentiation

⇒  φ ' (x) = =

  =1  

1− y dy = dx 1 − 2 y 2

2

−d 2 y d 2 x d  dx  d  1  dx 2 132. (d) We have, =  =  = dy 2 dy  dy  dy  dy / dx   dy  2  dx    133. (c) We have, f(x) = mx2 + nx + p

⇒ g (x + N) = log f (x + N) f (x + 1) = x f(x) ∴ f (x + N) = (x + N – 1) f (x + N – 1) = (x + N – 1)(x + N – 2) f (x + N – 2)… = (x + N – 1)(x + N – 2)… … …(x + 1) xf (x) ∴ log f(x + N) = log(x + N – 1) + log(x + N – 2)  + … log (x + 1)+log x log f(x) ⇒ log f(x + N) – log f(x) = logx + log(x + 1)  + …log(x + N – 1) ⇒ g(x + N) – g(x) = log x+ log(x + 1)  + … + log (x + N – 1) 1 and g ′ (x + N) – g ′ (x) = 1 + 1 + ... + x x −1 x + N −1

∴ g ′ (x + N) – g′ (x) = 1  1 1 1 puting x = we get − 2 + + ... 2 2 x ( x + 1) ( x + N − 1) 2   1− x   134. (d) We have, y =   1+ x 1 1 1 1  1  = −4 1 + 1 + ... g ′′  N +  = g ''   = −  ... + +  2   1 2  3 2  (−1) 1 2 2   2 N 1 −    9 − 1− x × 1+ x ×         dy − + x x 2 1 2 1 2 2 2        =   dx ( 1 + x )2    1 1 1  1  1 + 1 + ...  = −4 1 + 1 + ... 1 1 + x + 1 − x    N +  = g ''   = −  2 2 2 2  −  2 (2n − 1)   2  1 3  2N − 1   9 2  1 − x 2         2 2 2   =      (1 + x) ⇒  f ' (x) = 2mx + n  ∴f ' (1) + f ' (4) – f ' (5) = (2m + n) + (8m + n) – (10m + n) = n

134

EXERCISEs FOR SELF-PRACTICE

Objective Mathematics

1. Differential co-efficient of sec (tan – 1 x) is x 1 (a) (b) 2 1+ x 1 + x2 x (c) x 1 + x 2 (d) 1 + x2 2. If f (0) = 1, f ′ (0) = – 1, f (x) > 0 for all x then there exists a function f (x) such that (a) f ′ (x) < 0 for all x (b) – 1 < f ′′(x) < 0 for all x (c) – 2 ≤ f ′′(x) ≤ – 1 for all x (d) f ′′(x) ≤ – 2 for all x. 3. If 2x + 2y = 2x + y, then the value of (a) 0 (c) 1

dy at x = y = 1 is dx

(b) – 1 (d) 2.

1 − x2   2x  –1 4. The derivative of sin–1   is 2  w.r.t. sin    1 + x   1 + x 2  (a) – 1

(b) 1 (c) 2 (d) 1 2 5. Let f be a polynomial. Then, the second derivative of f (ex) is

(a) f ′′(ex) e2x + f ′ (ex) ex (b) f ′′(ex) ex + f ′ (ex) (c) f ′′(ex) e2x + f ′′(ex) ex (d) f ′′(ex).

6. If y = log (sin x), then

 x x − x− x  9. If f (x) = cot–1   , then f ′ (1) is equal to 2   (a) – 1 (c) log 2

(b) 1 (d) – log 2

2t 1− t2 dy , then = 2 , y = 1+ t 1+ t2 dx

10. If x = (a)

2t t2 −1

(b)

2t t2 +1

(c)

2t 1− t2

(d) None of these

 1 + x2 − 1   w.r.t. tan–1x is 11. Derivative of tan–1  x   (a) 1 1 (c) 2

(b) 2 (d) None of these

 3a 2 x − x 3  dy = 12. If y = tan–1   2 2  , then  a (a − 3x )  dx 3a 2 a + x2 a (c) 2 a + x2

(a)

2

3a a2 + x2 3 (d) 2 a + x2 (b)

13. Differential coefficient of esin d y equals dx 2 2

(a) sec x tan x (b) – cosec x cot x (c) sec2x (d) – cosec2x 7. If y = cos 2x sin 3x, then yn is equal to nπ  nπ    (a) 6n sin   2 x +  cos  3 x +   2  2  nπ  nπ    (b) 6  cos   2 x +  cos  3 x +   2 2 

(c) e

w.r.t. sin–1x is −1

(a) sin–1x cos −1 x

−1 x

(b) esin x (d) cos–1x



dy  cos x  14. If y = tan–1   , then is equal to dx  1 + sin x  1 2 (c) 1

1 2 (d) None of these

(b) – 

(a)

15. Differential coefficient of sin–1x w.r.t. cos–1 1 − x 2 is

n

 n nπ  nπ     5 sin  5 x + 2  + sin  x + 2     (d) None of these dy If y = (sin x)tan x, then is equal to dx (a) (sin x)tan x ⋅ (1 + sec2x ⋅ log sin x) (b) tan x ⋅ (sin x)tan x – 1 ⋅ cos x (c) (sin x)tan x ⋅ sec2x ⋅ log sin x (d) tan x ⋅ (sin x)tan x ­– 1

(c) 8.

1 2

2

(a)

1− x

2



1

(c) – 

1 + x2

(b)

1 1 − x2

(d) None of these

16. Differential coefficient of tan–1  (a) –  (c)

1 2

1 2

1 − x2 w.r.t. cos–1x2 is 1 + x2

(b) 1 (d) None of these

1 (1 + x 2 ) 1 (c) 1 + x2 (a)

(d) None of these

(a) 2 x (loge x − 1) ⋅ log e x (b) x (loge x − 1) 2 (c) log e x x (d) x

dy = dx

(a) 1 (b) – 1 (c) 2 (d) None of these

28.

sec x is

(sec x )3 / 2 sin x

2 1− x

29. If y = (a) – (c)

(b) sec xº tan xº (d)

180 sec x º tan x º π

22. Differential coefficient of cos–1 x w.r.t. (a)

x

(b) –  x

(c)

1 x

(d) – 

1 x

(b)

1 − x is

π 2

(d) None of these

2

(c) cos 2x

dy = dx

π sec x º tan x º 180



d (sin–12x 1 − x 2 ) is equal to dx

(a) – 

4 x 1 (b) sec x sin x 4 x 1 (c) x (sec x )3 / 2 sin x 2 1 (d) x sec x sin x 2

(a) sec x tan x

1 − x2

(c) 0

(d) e y

21. If y = sec xº, then

2

(a)

1 (b) x

20. Differential coefficient of

(c)



dP at V = 9 equals dV

(a) 0 (b) e (c) 2e (d) 1/e 27. d (cos–1x + sin–1x) is dx

⋅ log e x

1 (a) x log x 1 (c) log x

1

3 2

26. If f (x) = logx (ln x) then f ′ (x) at x = e is

19. If y = log log x, then ey 

(a)

(c) ± 



2

(b)

1 − x2

(d) None of these.

x + x + x + ...∞ , then 1 2y −1

1 xy

(b)

dy is equal to dx

1 2y −1

(d) 1

30. The differential coefficient of x6 w.r.t. x3 is (a) 6x5 (b) 3x2 (d) x3 (c) 2x3  sin x + cos x  dy 31. If y = tan– 1   is  , then  cos x − sin x  dx (a) 1/2 (b) π/4 (c) 0 (d) 1. 32. If xp yq = (x + y)p + q, then dy/dx is equal to

 1+ x   + xx 23. The first derivative of the function cos −1  sin 2   (a) y/x (b) py/qx w.r.t. x at x = 1 is (c) x/y (d) qy/px 3 (a) (b) 0 4 x .... ∞ dy , then is: 33. If e x + e x + e 1 1 dx (c) (d) –  2 2 x y (a) (b) 24. The values of x, at which the first derivative of the func1− x 1+ y 1   tion  x +   x

2

w.r.t. x is

3 are 4

(c)

y 1− y

(d)

1− y y

135

(b) ± 

25. If PV = 81, then

18. The differential coefficient of the function x loge x w.r.t. x is

(loge x − 1)

1 2 2 (d) ±  3

(a) ± 2

Differentiation

dy is equal to dx 1 (b) –  (1 + x 2 )

17. If x 1 + y + y 1 + x = 0, then

136

34. If x = a sin θ and y = b cos θ, then

Objective Mathematics

(a)

a sec 2 θ b2 −b (c) 2 sec3 θ a

(a) 4 (c) 1/2

d2y is  dx 2

(b) 1 (d) 6.

(b)

−b sec 2 θ a

36. If y = sin x + ex, then

(d)

−b sec3 θ a2

(a) (– sin x + ex)– 1

35. Let f (x + y) = f (x) f ( y) 3, then f ′ (5) equals:

x, y ∈ R, f (5) = 2, f ′ (0) = (c)

sin x − e x (cos x + e x )3

d 2x equals dy 2 (b)

sin x + e x (cos x + e x ) 2

(d)

sin x + e x (cos x + e x )3

Answers

1. 11. 21. 31.

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

2. 12. 22. 32.

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

3. 13. 23. 33.

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

4. 14. 24. 34.

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

5. 15. 25. 35.

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

6. 16. 26. 36.

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

7. (c) 17. (b) 27. (c)

8. (a) 18. (a) 28. (b)

9. (a) 19. (b) 29. (b)

10. (a) 20. (a) 30. (c)

5

Applications of Derivatives

CHAPTER

Summary of conceptS tangentS and normalS geometrical meaning of derivative at a point The derivative of a function f (x) at a point x = a is the slope of the tangent to the curve y = f (x) at the point [a, f (a)]. Consider a curve y = f (x) and a point P(x, y) on this curve. If tangent to the curve at P(x, y) makes an angle θ with the positive dy direction of x-axis, then, at the point P(x, y): = tan θ = m = dx gradient or slope of tangent to the curve at P (x, y).

[ the normal to the curve at the point P(x1, y1) is a line perpendicular to the tangent at the point P (x1, y1) and passing through it. Therefore, slope of normal at P (x1, y1) =

−1 −1 . = Slope of tangent at P ( x1 , y1 )  dy    dx ( x1 , y1 )

Important Results

equation of tangent The equation of a tangent to a curve y = f (x) at a given point P (x1, y1) is given by  dy  (x – x1) y – y1 =    dx  ( x , y ) 1

1. If

dy > 0, the tangent makes an acute angle with the x-axis. dx

2. If

dy < 0, the tangent makes an obtuse angle with the x-axis. dx

3. If

dy = 0, the tangent is parallel to x-axis. dx

1

[Using point slope form of equation of the straight line]

4. If the tangent is perpendicular to x-axis, then

equation of normal The equation of normal to a curve y = f (x) at a given point P (x1, y1) is given by −1 (x – x1) y – y1 =  dy    dx ( x , y ) 1

dy dy = ∞, i.e. = 0. dx dx 5. If the tangent is equally inclined to the axes, then

dy = tan 45º or tan 135º = ± 1. dx

1

angle of Intersection of two curves Let y = f (x) and y = g (x) be two curves intersecting at a point P (x1, y1). Then, the angle of intersection of these two curves is defined as the angle between the tangents to the two curves at their point of intersection.

138

(iii) Subtangent = TM = | y cot θ | = Objective Mathematics

y  dy   dx   

 dy  (iv) Subnormal = MN = | y tan θ | = y   .  dx  If θ is the required angle of intersection, then, θ = ± (θ1 – θ2), where θ1 and θ2 are the inclination of tangent to the curves y = f (x) and y = g (x) respectively at the point P.

IncreaSIng and decreaSIng functIonS (monotonIcIty) Increasing function A function f(x) is said to be an increasing function on an interval I, if x1 < x2 ⇒ f (x1) ≤ f (x2), ∨ x1, x2 ∈ I.

Working rule for finding the angle of intersection 1. Find f ′ (x) and g′ (x). 2. If f ′ (x) × g′(x) = – 1, then the two curves are said to cut each other orthogonally, wherever they cut. 3. If the product is not – 1, solve the equation of the two curves to get their point of intersection. If (α, β) be their point of intersection, then find f ′ (α) and g′ (α). Let m1 = f ′ (α) and m2 = g′ (α). 4. If θ is the angle between the tangents, then tan θ = ±

f ' (α) − g' (α) m1 − m2 =± . 1 + f ' (α) g' (α) 1 + m1m2

Increasing Function Strictly Increasing function A function f (x) is said to be a strictly increasing function on an interval I, if

Repeat this process for other points of intersection. Note: The two curves are said to touch each other at their point of intersection (α, β), if the slope of their tangents at (α, β) are equal.

x1 < x2 ⇒ f (x1) < f (x2), ∨ x1, x2 ∈ I.

length of tangent, length of normal, Sub-tangent and Subnormal Let the tangent and normal at the point P (x, y) on the curve meet the axis of x at the points T and N respectively. Let M be the foot of the ordinate at P. Then, Strictly Increasing Function decreasing function A function f (x) is said to be a decreasing function on an interval I, if x1 < x2 ⇒ f (x1) ≥ f (x2), ∨ x1, x2 ∈ I Decreasing Function Strictly decreasing function A function f (x) is said to be a strictly decreasing function on an interval I, if

(i) Length of the tangent = PT = | y cosec θ |

=

 dy  y 1+    dx  y 1 + cot θ = dy dx

2

x1 < x2 ⇒ f (x1) > f (x2) ∨ x1, x2 ∈ I.

2

(ii) Length of the normal = PN = | y sec θ | =

 dy  y 1 + tan 2 θ = y 1 +    dx 

2

Strictly Decreasing Function

test for monotonicity of functions (i) (ii) (iii) (iv)

f (x) is increasing in [a, b] if f ′ (x) ≥ 0, ∨ x ∈ [a, b]. f (x) is strictly increasing in [a, b] if f ′ (x) > 0, ∨ x ∈ [a, b]. f (x) is decreasing in [a, b] if f ′ (x) ≤ 0, ∨ x ∈ [a, b]. f (x) is strictly decreasing in [a, b] if f ′ (x) < 0, ∨ x ∈ [a, b].

Remarks 1. If a function f (x) is strictly increasing (strictly decreasing) on an interval I, then f –1 exists and is also strictly increasing (strictly decreasing). 2. If f (x) is monotonic on an interval I, then f (x) has at the most one zero in the interval I. 3. If the functions f (x) and g (x), both are increasing or decreasing on an interval I, then the composite function (gof ) (x) is an increasing function on I. 4. If the function f (x) is increasing and g (x) decreasing on an interval I, then the composite function (gof ) (x) is decreasing on the interval I.

(i) The points at which a function attains either the local maximum value or local minimum value are called the extreme points and both local maximum and local minimum values are called the extreme values of the function f (x). (ii) The local maximum and local minimum values are also known as relative maximum and relative minimum values respectively.

Working rule to determine the points of local maxima and local minima method I (first derivative test) 1. For the function y = f (x), find f ′ (x). 2. Put f ′ (x) = 0 and solve this equation for x. Let its roots be a, b, c etc. These points are called stationary points or critical points. 3. At x = a, determine the sign of f ′ (x) for values of x slightly less than a and that for values of x slightly greater than a. (i) If f ′ (x) changes sign from positive to negative as x increases through a, then x = a is a point of maximum.

5. A function may be increasing in some interval I1 and decreasing in some other interval I2.

maxIma and mInIma of functIonS local maximum A function y = f (x) is said to have a local maximum value at a point x = a, if f (x) ≤ f (a), ∨ x ∈ (a – h, a + h), for small h > 0, i.e., f (a) is the greatest of all the values of f (x) in the interval (a – h, a + h).

(ii) If f ′ (x) changes sign from negative to positive as x increases through a, then x = a is a point of minimum.

The point x = a is called a point of local maximum of the function f (x). local minimum A function y = f (x) is said to have a local minimum value at a point x = a, if f (x) ≥ f (a), ∨ x ∈ (a – h, a + h), for small h > 0, i.e., f (a) is the smallest of all the values of f (x) in the interval (a – h, a + h).

(iii)f ′ (x) does not change sign as x increases through a, then x = a is neither a point of maximum nor a point of minimum. Such a point is called a point of inflexion. We repeat this process for other values of x and examine them for maxima or minima.

Method II (Second Derivative Test) 1. For the function y = f (x), find f ′ (x) and f ′′ (x). 2. Put f ′ (x) = 0 and solve this equation for x. Let its roots be a, b, c etc.

3. At x = a (i) if f ′′ (a) < 0, then x = a is a point of local maxima;

The point x = a is called a point of local minimum of the function f (x).

(ii) if f ′′ (a) > 0, then x = a is a point of local minima; (iii) if f ′′ (a) = 0, we cannot say any thing.

139

Remarks

Applications of Derivatives

monotonic function A function f (x) is said to be monotonic on an interval I if it is either increasing or decreasing on I.

140

Objective Mathematics

General Test  In the cases where the second derivative vanishes, the method discussed above fails to give any result. In those cases, we make use of still higher derivatives, and the following working rule proves very useful. If f (x) has a derivative at x = a such that (i) f ′ (a) = f ′′ (a) = f ′′′ (a) = ... = f  n – 1 (a) = 0, and (ii) f n (a) exists and is not zero, then for n odd, f (a) is not an extreme value, while for n even f (a) The theorem states that between two points with equal oris a maximum or minimum value according as f n (a) is negative dinates on the graph of f, there exists atleast one point where the or positive. tangent is parallel to x-axis.

Greatest and Least Values of a Function in a Closed Interval (Absolute Maximum and Absolute Minimum)

Algebraic Interpretation Between two zeros a and b of f (x) (i.e., between two roots a and b of f (x) = 0) there exists atleast one zero of f′ (x).

If f (x) is continuous in an interval [a, b], then greatest or absolute maximum value of f (x) = max. { f (a), f (b), values of f (x) at all Lagrange’s Mean Value Theorem critical points in (a, b)}. Also, least or absolute minimum value of f (x) = min.  If a function f defined on the closed interval [a, b], is {f (a), f (b), values of f (x) at all critical points in (a, b)} If a function is defined and continuous on an interval which (i) continuous on [a, b] and is not a closed interval, then it cannot have any greatest or least (ii) derivable on (a, b), then there exists atleast one real number c between a and b (a < c < b) such that value other than local maximum or local minimum values. f (b) − f ( a ) f ′ (c) = b−a Rolle’s and Lagrange’s Mean Geometrical Interpretation  The theorem states that Value Theorem between two points A and B on the graph of f there exists atleast Rolle’s Theorem one point where the tangent is parallel to the chord AB. If a function f defined on the closed interval [a, b], is (i) continuous on [a, b], (ii) derivable on (a, b) and (iii) f (a) = f (b), then there exists atleast one real number c between a and b (a < c < b) such that f ′ (c) = 0. Geometrical Interpretation  Let the curve y = f (x), which is continuous on [a, b] and derivable on (a, b), be drawn.

MULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1. The tangent to the curve y = x3 – 2x2 + x – 2 is parallel to x-axis at the point (a) (1, – 2)  1 50  (c)  , −   3 27  2. The tangent to the curve

(b) (– 1, 2)  1 50  (d)  ,   3 27  x + y = 4 is equally inclined

to the axes at the point (a) (1, – 2) (c) (4, – 4)

(b) (4, 4) (d) (– 4, 4)

3. The tangent to the curve x2 + y2 = 25 is parallel to the line 3x – 4y = 7 at the point

(a) (– 3, – 4) (c) (– 3, 4)

(b) (3, – 4) (d) (3, 4)

4. The minimum value of a tan2x + b cot2x equals the maximum value of a sin2 θ + b cos2 θ where a > b > 0, when (a) a = b (c) a = 3b

(b) a = 2b (d) a = 4b

x  5. The points on the curve y2 = 4a  x + a sin  at which a  the tangent is parallel to x-axis, lie on (a) a straight line (c) a circle

(b) a parabola (d) an ellipse

(a) m < 1 (c) | m | > 1

(b) | m | ≤ 1 (d) None of these

ax 7. If the slope of the curve y = b − x at the point (1, 1) is 2, then the values of a and b are (a) 1, – 2 (c) 1, 2

(b) – 1, 2 (d) None of these

8. If the tangent at each point of the curve 2 3 x – 2ax2 + 2x + 5 y= 3 makes an acute angle with the positive direction of x-axis, then (a) a ≥ 1 (c) a ≤ – 1

(b) – 1 ≤ a ≤ 1 (d) None of these

9. The angle between the tangents to the curve y2 = 2ax at a the points where x = is 2 π π (b) (a) 6 4 π π (c) (d) 3 2 10. The equation of the tangent to the curve y = 9 − 2 x 2 at the point where the ordinate and the abscissa are equal, is (a) 2x + y – 3 3 = 0

(b) 2x + y + 3 3 = 0 (d) None of these

(c) 2x – y – 3 3 = 0 11. The angle between the tangents at those points on the curve y = (x + 1) (x – 3) where it meets x-axis is (a) ± tan–1

15 8

8 (b) ± tan–1    15 

π (d) None of these 4 12. The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point t = 2 is (c) ±

7 1 (b) 6 2 6 (c) (d) 1 7 13. The equation of the tangent to the curve y = 2x2 + 5x at the point where the line y = 3 cuts the curve in the first quadrant is (a)

(a) 14x – 2y – 1 = 0 (c) 14x + 2y – 1 = 0

(b) 14x – 2y + 13 = 0 (d) None of these

2 2 14. The angle between the tangents to the curve x + y 2 2 a b = 1 at the points (a, 0) and (0, b) is

π 4 π (c) 3

(a)

(b)

π 2

(d) None of these

(a) at an angle π 3 (c) orthogonally

π 4 (d) None of these

(b) at an angle

16. The angle of intersection of the parabolas y2 = 4ax and x2 = 4ay at origin is π π (b) 3 2 π (c) (d) None of these 4 17. If a < 0, f (x) = eax + e–ax and S = {x : f (x) is monotonically decreasing}, then S equals (a)

(a) {x : x > 0} (c) {x : x > 1 }

(b) {x : x < 0} (d) {x : x < 1}

18. The two curves y2 = 4x and x2 + y2 – 6x + 1 = 0 at the point (1, 2) (a) intersect orthogonally π (b) intersect at an angle 3 (c) touch each other (d) None of these 19. The critical points of the function f (x) = (x – 2)2/3 (2x + 1) are (a) –1 and 2 (c) 1 and (–1/2)

(b) 1 (d) 1 and 2

20. The angle of intersection of the curves y = 2 sin2x and y π = cos 2x at x = is 6 π (a) (b) π 4 3 (c) π (d) None of these 2 21. An extreme value of the function f (x) = (sin–1 x)3 + (cos–1 x)3, (–1 < x < 1) is (a)

7 π3 8

3 (c) π 32

(b)

π3 8

(d)

π3 16

22. The angle at which the curves y = sin x and y = cos x intersect in [0, π] is (a) ± tan–1 2 2

(b) ± tan–1 2

 1  (c) ± tan–1    2

(d) None of these

α  x log x, x > 0 23. If f (x) =  , find the value of α so that the x=0  0, function obeys the Rolle’s theorem in [0, 1]

(a) – 2 (c) 0

(b) – 1 1 (d) 2

141

15. The curves x3 – 3xy2 = a and 3x2y – y3 = b, where a and b are constants, cut each other

Applications of Derivatives

6. If m be the slope of a tangent to the curve e2y = 1 + 4x2, then

142

n

n

Objective Mathematics

x  y x y 24. The line a + b = 2 touches the curve  a  +  b  = 2 at the point (a, b) for (a) n = 2 only (b) n = – 3 only (c) n is any real number (d) None of these

a 2

a (c) (d) None of these 3 26. Let f (x) and g(x) be defined and differentiable for x ≥ x0 and f (x0) = g(x0), f ′(x) > g′(x) for x > x0, then (b) f (x) = g(x), x > x0 (d) None of these

27. A function f such that f ′(a) = f ′′(a) = f ′′′ (a) = ... = f (2n) (a) = 0 and f has a local maximum value b at x = a, if f (x) is (a) (x – a)2n+2 (c) b – (x – a)2n+2

(b) b – 1 – (x + 1 – a)2n–1 (d) (x – a)2n+2 – b

28. f (x) = x3 + ax2 + bx + 5 sin2 x is an increasing function in the set of real numbers if a and b satisfy the condition (a) a – 3b – 15 > 0 (c) a2 – 3b + 15 < 0 2

(b) a – 3b + 15 > 0 (d) a > 0 b > 0 2

29. The curve y – exy + x = 0 has a vertical tangent at the point (a) (1, 1) (c) (0, 1)

(b) at no point (d) (1, 0)

30. If f (x) satisfies the conditions for Rolle’s theorem in [3,5] then



5

3

f ( x)dx equals to

(a) 2 (c) 0

(b) –1 (d) – 4/3

31. Let f (x) = x3 + bx2 + cx + d, 0 < b2 < c. Then, f (a) is bounded (c) has a local minima

(b) has a local maxima (d) is strictly increasing

32. Tangents are drawn from the origin to the curve y = sin x. Their points of contact lie on the curve (a) x2y2 = x2 + y2 (c) x2y2 = y2 – x2

π 4

35. The points where the normal to the curve makes equal intercepts on the axes are

(b)

(a) f (x) < g(x), x > x0 (c) f (x) > g(x), x > x0

(c) inclined at an angle (d) None of these

25. The length of the perpendicular from the origin to the curve x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ) at any point θ is (a) a

(a) parallel (b) perpendicular

(b) x2y2 = x2 – y2 (d) None of these

33. If the line y = 2x touches the curve y = ax2 + bx + c at the point where x = 1 and the curve passes through the point (– 1, 0), then the values of a, b and c are 1 1 1 1 (a) a = 2 , b = 1, c = 2 (b) a = 1, b = 2 , c = 2 1 1 (c) a = 2 , c = 2 , b = 1 (d) None of these 34. The two tangents to the curve ax2 + 2hxy + by2 = 1, a > 0 at the points where it crosses x-axis, are

 (a) ±    (b) ±  

xy = a + x

a a  , 2a − ± 2a  2 2  a a  , − 2a + ± 2a  2 2 

a  a  , 2a + ± 2a  (c) ±  2  2  (d) None of these 36. The maximum value of (x – p)2 + (x – q)2 + (x – r)2 will be at x equal to p+q+r 3 (c) qpr (a)

(b) 3 qpr 2 2 2 (d) p + q + r

37. The normals to the curve x = a (θ + sin θ), y = a (1 – cos θ) at the points θ = (2n + 1) π, n ∈ I are all (a) parallel to x-axis (b) parallel to y-axis (c) parallel to the line y = x

(d) None of these 38. The value of θ, θ ∈ [0, π/2] for which the sum of intercepts on co-ordinate axes by tangent at point ( 3 3 cos θ, sin θ) x2 + y 2 = 1 is minimum, is: of ellipse 27 (a) π/6 (b) π/4 (c) π/3 (d) π/2 39. The equation of the tangent to the curve 1  2 x≠0  x sin , x at the origin is y=  0, x=0 (a) x = 0 (c) y = 0

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

40. Number of possible tangents to the curve y = cos (x + y), –3π ≤ x ≤ 3π, that are parallel to the line x + 2y = 0, is (a) 1 (c) 3

(b) 2 (d) 4

41. The condition that the line lx + my = 1 may be normal to the curve y2 = 4ax is (a) al3 – 2alm2 = m2 (b) al2 + 2alm3 = m2 (c) al3 + 2alm2 = m3 (d) al3 + 2alm2 = m2. 42. The point in the interval [0, 2π], where f (x) = ex sin x has maximum slope, is (a) π/4 (c) π

(b) π/2 (d) 3π/2

44. If the normal at the point “t1” on the curve xy = c2 meets the curve again at “t2”, then 3 (a) t1 t2 = 1

3 (b) t1 t2 = – 1

(c) t t = – 1

(d) t t = 1.

3 1 2

3 1 2

45. The tangent to the curve 3xy2 – 2x2y = 1 at (1, 1) meets the curve again at 16 1 (a)  − , −  20   5

16 1 (b)  , −  20   5

 16 1  (c)  − ,   5 20 

(d) None of these

3  1  2 (1 − x ) 2 − 8 3   3   1  2 2 (b) f (x) = (1 − x ) + 8 3   1 (c) f (x) = – 3 {sin x + 7}

(a) f (x) = –

x+ y

(b) a (d) None of these 2/3

 y +  b

2/3

= 1 meets

the coordinate axes in A and B, where p = OA and q = OB, O being the origin, then the locus of ( p, q) is p q p q = 1 (b) 2 + 2 = 2 + a b b2 a 2 p2 q2 (c) 2 + 2 = 1 (d) None of these a b 49. The part of the tangent to the curve xy = c2 included between the coordinate axes, is divided by the point of tengency in the ratio 2

2

(a)

(a) 1 : 1 (b) 1 : 2 (c) 1 : 3 (d) None of these 50. The portion of the tangent to the curve a − a2 − y2 a x = a − y + log 2 a + a2 − y2 intercepted between the curve and x-axis, is of length. 2

(b) 2y + x = 0 (d) None of these

52. The equation of the normal to the curve y = e–2 | x | at the 1 point where the curve cuts the line x = is 2 (a) 2e (ex + 2y) = e2 – 4 (b) 2e (ex – 2y) = e2 – 4 (c) 2e (ey – 2x) = e2 – 4 (d) None of these

(a) 9 5 (c)

9 5 4

9 5 2 (d) None of these

(b)

(b) a > 0, b < 0 (d) a < 0, b < 0

55. If y = f (x) be the equation of an ellipse to which the line y = 2x + 3 is a tangent at the point where x = 2, then

x 48. If the tangent to the curve   a

2

(a) 2y – x log 2 = 0 (c) 2y + x log 2 = 0

(a) a > 0, b > 0 (c) a < 0, b > 0

47. The sum of intercepts of the tangent to the curve = a upon the coordinates axes is

2

51. The equation of the normal to the curve y = 1 – 2x/2 at the point of intersection with the y-axis is

54. If the line ax + by + c = 0 is a tangent to the curve xy = 4, then

(d) None of these

2

(d) None of these

53. The area of the triangle formed by the positive x-axis, the tangent and the normal to the circle x2 + y2 = 9 at (2, 5 ) is

46. The function f whose graph passes through the point  7  and whose derivative is x 1 − x 2 , is given by 0 ,   3

(a) 2a (c) 2 2 a

(b) a

143

(a) 3 (b) a (c) no maximum value (d) None of these

a 2 (c) 2a

(a)

(a) f ′ (2) = 2 (b) f (2) = 2 f ′ (2) (c) f (2) + f ′ (2) + f ′′ (2) = 2 (d) None of these 56. The length of the portion of the tangent, at any point on the curve x = a cos3θ, y = a sin3θ, intercepted between the coordinate axes, is (a) a (c) 3a

(b) 2a (d) None of these

57. If the curve y = px2 + qx + r passes through the point (1, 2) and the line y = x touches it at the origin, then the values of p, q and r are (a) p = 1, q = – 1, r = 0 (b) p = 1, q = 1, r = 0 (c) p = – 1, q = 1, r = 0 (d) None of these 58. The length of the perpendicular from the origin to the tangent to the curve y = e4x + 2 drawn at the point x = 0 is (a)

4 17

(b)

(c)

2 17

(d) None of these

3 17

59. The area of the triangle formed by a tangent to the curve 2xy = a2 and the coordinate axes is

Applications of Derivatives

43. If f (x) = a – (x – 3)89, then greatest value of f (x) is

144

(a) 2a2 (c) 3a2

(b) a2 (d) None of these

Objective Mathematics

60. Tangents are drawn to the parabola x2 = 4y at its points of intersection with another parabola y2 = 4x. The point of intersection of the tangents drawn, is given by (a) (2, 0) (c) (– 2, 0)

(b) (0, 2) (d) None of these

π 4 with the straight line y = 3x + 5, then the point of contact is

61. If a tangent to the parabola y2 = 8x makes an angle

1  (a)  , 2  2  1 (c)  , − 2  2 

(b) (0, 0)

62. The set of critical points of (a) {0} (c) {2, –2}

x

π t + 2 2

(d) None of these

(a) decreases on (0, 1) ∪ (2, ∞) (b) increase on (– ∞, 0) ∪ (1, 2) (c) has a local maximum value 0 (d) has a local maximum value 1. (b) k2 = 4 (d) None of these

71. For the curve, x = a cos3θ, y = a sin3θ, at any point θ, is

(b) {0, 2, – 2} (d) {2}

(b) is – 2 3 (d) None of these

64. The area of the triangle, formed by the x-axis and the tangent and the normal to the curve y = 6x – x2 at the point (5, – 5), is 85 (a) 425 (b) 8 8 5 (c) (d) None of these 8 65. If the curve y = x2 + bx + c touches the line y = x at the point (1, 1), then the values of x for which the curve has a negative gradient are 1 (a) x < 2 1 (c) x < – 2

(b)

69. The function f (x) = x2(x – 2)2

(a) k2 = 8 (c) k2 = 2

1 63. If P = x3 – 1 and Q = x – , x ∈ (0, x) then minimum 3 x x value of P/Q2 (a) is 2 3 (c) does not exist

(a) π 4 π (c) + t 4 2

70. The curves y2 = 2x and 2xy = k cut at right angles if

1 (d)  − , 2   2  x2 − 4

68. The angle made by the tangent at any point on the curve x = a (t + sin t cos t), y = a (1 + sin t)2, with x-axis, is

1 (b) x > 2

(d) x > – 1 2 66. The angle subtended by x2 + y2 = 25 at the point (10, 0) is

(a) the length of the tangent is asin2θ (b) the length of the normal is a tan θ sin2θ (c) the length of sub-tangent = a sin2θ cos θ (d) satisfies all the three conditions. 72. The sub-normal at any point of the curve x2y2 = a2 (x2 – a2) varies as (a) (abscissa)– 3 (c) (ordinate)– 3

(b) (abscissa)3 (d) None of these

73. The sub-tangent at any point of the curve xmyn = am+n varies as (a) (abscissa)2 (c) abscissa

(b) (ordinate)2 (d) ordinate

74. The sum of tangent and sub-tangent at any point of the curve y = a log (x2 – a2) varies as (a) abscissa (c) ordinate

(b) product of the coordinates (d) None of these

75. For the curve xm + n = am – n y2n, where a is a positive constant and m, n are positive integers, (a) (sub-tangent)m ∝ (sub-normal)n (b) (sub-normal)m ∝ (sub-tangent)n (c) the ratio of subtangent and subnormal is constant (d) None of these.

76. The length of the normal at any point on the ellipse x2 y 2 + a 2 b 2 = 1 varies as (a) π/6 (b) π/4 (a) (abscissa)3 (c) π/2 (d) π/3 (b) (ordinate)3 (c) the ⊥ r from origin on tangent 67. The intercepts made on the coordinate axes by the tangent (d) None of these m m –1 m at any point on the curve y = ax + x are 2 2 ay 77. The angle between the ellipse x + y = 1 and the circle − ax 2 2 (a) , a b (m − 1) a + mx m (a + y ) x2 + y2 = ab at their point of intersection is ay − ax a−b a+b (b) , (a) tan–1  (b) tan–1    m ( a + x) (m − 1) a + mx  ab   ab  ay ax (c) , (c) tan–1  a − b  (d) None of these (m − 1) a + mx m (a + x)    2 ab  (d) None of these

79. The points where the curves x2 + y2 = 2a2 and 2y2 – x2 = a2 cut orthogonally are  2  4 4  2  a  a (b)  ± a, ± (a)  ± a, ± 3 3 3 3      a 2  ,± a (c)  ± 3  3 

(d) None of these

80. The two curves y = 3x and y = 5x intersect at an angle  log 3 − log 5  (a) tan–1    1 + log 3 ⋅ log 5   log 3 + log 5  (b) tan–1    1 − log 3 ⋅ log 5 

(d) None of these

2 ordinate

(b) ordinate

(c)

2 ordinate

(d) None of these

82. For the parabola y2 = 4ax, the ratio of the sub-tangent to the abscissa is (b) 2 : 1 (d) x2 : y.

83. Let f (x) = cos 2πx + x – [x] ([ . ] denotes the greatest integer function). Then number of points in [0, 10] at which f (x) assumes its local maximum value, is (b) 10 (d) infinite

84. The subtangent, ordinate and subnormal to the parabola y2 = 4ax at a point (different from the origin) are in (b) AP (d) None of these

85. The curve y = ax3 + bx2 + cx + 5 touches the x-axis at A (– 2, 0) and cuts the y-axis at a point B where its slope is 3. The values of a, b and c are 1 3 (a) a = , b = – , c = 3 2 4 1 3 (b) a = – , b = – , c = 3 2 4 1 3 (c) a = , b = , c = 3 2 4 (d) None of these 86. The normal to the curve 5x5 – 10x3 + x + 2y + 6 = 0 at P (0, – 3) meets the curve again at the point (a) (– 1, 1) (c) (– 1, – 5)

88. The tangent to the graph of the function y = f (x) at the π point with abscissa x = 1 forms an angle of and at 6 π and at the point x = 3 the point x = 2 an angle of 3 π an angle of with the positive direction of x-axis. The 4 value of



3

1

f ' ( x) f '' ( x) dx +

continuous) 4 3 −1 3 3



3

2

f '' ( x) dx is f ′′(x) being

(b) 3 3 − 1 2

4− 3 (d) None of these 3 89. The abscissa of the point on the curve xy = (c + x)2, the normal at which cuts off numerically equal intercepts from the axes of coordinates, is (a)

(a)

(a) GP (c) HP

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

(c)

81. If at any point on a curve the sub-tangent and sub-normal are equal, then the length of the normal is equal to

(a) 0 (c) 9

(a) 1 ≤ b ≤ 2 (c) b ∈ (–∞, –1)

(a)

 log 3 + log 5  (c) tan–1    1 + log 3 ⋅ log 5 

(a) 1 : 1 (c) x : y

 x 3 − x 2 + 10 x − 5, x ≤ 1 87. Let f (x) =  2 −2 x + log 2 (b − 2) x > 1 The set of values of b for which f (x) has greatest value at x = 1, is given by

(b) (1, – 1) (d) (– 1, 5)

c 2

(b) c

2 c

(d) –

c 2 90. The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 – 2 = 0 (c)

(a) cut at right angles (b) touch each other π (c) cut at an angle 3 π (d) cut at an angle 4 91. The intercept made by the tangent to the curve y =

x

∫ | t | dt ,

which is parallel to the line y = 2x, on

0

x-axis is equal to (a) 1 (c) 2

(b) – 1 (d) – 2.

x2 y 2 = 1 where the tangent is + 4 16 equally inclined to the axes is

92. A point on the curve

 2 −8  , (a)    5 5

 2 8  , (b)    5 5

 −2 8  (c)  (d) all the above ,   5 5 93. The area bounded by the tangent to the curve y = loge x at the point (2, 0) and the coordinate axes is 1 sq. unit 2 (c) 1 sq. unit

(a)

(b) 2 sq. unit (d) None of these

145

(a) set of critical points is {–1/2, 0, 1/2} (b) f (x) is increasing in (–∞, –1/2] ∪ (0, 1/2] (c) f (x) is decreasing in [–1/2, 0) ∪ [1/2, ∞) (d) None of these

Applications of Derivatives

78. For the function f (x) = 2x2 – ln | x |

146

94. Any tangent to the curve y = 3x7 + 5x + 3

Objective Mathematics

(a) is parallel to x-axis (b) is parallel to y-axis (c) makes an acute angle with the x-axis (d) makes an obtuse angle with the x-axis 95. The function f (x) = cot–1 x + x increases in the interval (a) (1, ∞) (c) (–∞, ∞)

(b) (–1, ∞) (d) (0, ∞)

8 (c)  4, −   3

(d) None of these

104. The line

8  3

x y + = 1 touches the curve y = be–x/a at a b

(a) (– a, ba)

f (x) = (x + 1)1/3 – (x – 1)1/3 on [0, 1] is

b (c)  a,   a

(b) 2 (d) 1/3

97. The equation of the normal to the curve y = (1 + x) y + sin–1 (sin2x) at x = 0 is (a) x + y = 2 (c) x – y = 1

(b)  −4, 

the point

96. The greatest value of (a) 1 (c) 3

(a)  4, 8   3

(b) x + y = 1 (d) None of these

105. For the curve x = t2 – 1, y = t2 – t, the tangent is parallel to x-axis where (a) t =

1 3

(c) t = 0

98. If the function

 a (b)  a,   b (d) None of these

(b) t = –

1 3

(d) t = 1

2 f (x) = 2x3 – 9ax2 + 12a2x + 1, 106. For the curve x = 3 cos θ, y = 3 sin θ, 0 ≤ θ ≤ π, the where a > 0, attains its maximum and minimum at p and q tangent is parallel to the x-axis when respectively such that p2 = q, then a equals (a) θ = π (b) θ = 0 (a) 3 (b) 1 π π (c) θ = (d) θ = 1 (c) 2 (d) 3 2 2 107. The length of the subtangent to the curve x2 + xy + y2 99. On the ellipse 4x2 + 9y2 = 1, the points at which the = 7 at (1, 3) is tangents are parallel to the line 8x = 9y are (a) 3 (b) 5 2 1 (c) 15 (d) 3/5 (a)  2 , 1  (b)  − ,   5 5 5 5 108. The curve y – exy + x = 0 has a vertical tangent at the point 2 1 2 1 (c)  − , −  (d)  , −  .  5 5 5 (a) (1, 1) (b) at no point 5 (c) (1, 0) (d) (0, 0) 100. If the normal to the curve y = f (x) at the point (3, 4) 3 2 109. The curve y = ax + bx + cx is inclined at 45º to x-axis makes an angle 3π with the positive x-axis, then f ′(3) = at (0, 0) but it touches x-axis at (1, 0), then the values 4 of a, b, c are given by 3 (a) – 1 (b) – (a) a = 1, b = – 2, c = 1 4 4 (b) a = 1, b = 1, c = – 2 (c) 3 (d) 1. (c) a = – 2, b = 1, c = 1 101. A curve y = f (x) passes through the point P (1, 1). The normal to the curve at P is a ( y – 1) + (x – 1) = 0. If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, then the equation of the curve is (a) y = ea (x–1) (c) y = e

a ( x −1) 2

(b) y = ea (1–x)

(d) y = e

a ( x +1) 2

.

102. If the line ax + by + c = 0 is a normal to the curve xy = 1, then (a) a > 0, b > 0 (b) a > 0, b < 0 (c) a < 0, b > 0 (d) a < 0, b < 0 103. The point on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes is

(d) a = – 1, b = 2, c = 1 π   x sin , x > 0 x 110. If f (x) =  , then in the interval (0, 1), f ′ (x) 0, x=0 vanishes at (a) exactly one point (b) exactly two points (c) at no point (d) infinite number of points 111. If f (x) and g (x) are differentiable functions for 0 ≤ x ≤ 1 such that f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2, then in the interval (0, 1), (a) f ′ (x) = 0 for all x (b) f ′ (x) = 2g′ (x) for atleast one x (c) f ′ (x) = 2g′ (x) for atmost one x (d) None of these

n – 1

(a) exactly one root (c) atleast one root

2

(b) atmost one root (d) no root

113. If a + b + c = 0, then the equation 3ax2 + 2bx + c = 0 has, in the interval (0, 1) (a) atleast one root (c) no root

(b) atmost one root (d) None of these

114. If α and β (α < β) be two different real roots of the equation ax2 + bx + c = 0, then (a) α > –

b 2a

(b) β < –

b 2a

b < β 2a

(d) β < – b < α 2a n n – 1 + ... + a1x = 0 has a 115. If the equation anx + an – 1x positive root x = α, then the equation (c) α < –

nanxn – 1 + (n – 1) an – 1xn – 2 + ... + a1 = 0 has a positive root, which is (a) smaller than α (c) equal to α

(b) greater than α (d) greater than or equal to α.

sin x sin a sin b π 116. If f (x) = cos x cos a cos b , where 0 < a < b < , 2 tan x tan a tan b then the equation f ′ (x) = 0 has, in the interval (a, b) (a) atleast one root (c) no root

(b) atmost one root (d) None of these

121. If the function f (x) and g (x) are continuous in [a, b] and differentiable in (a, b), then the equation f (a ) f (b) = (b – a) g (a ) g (b) interval [a, b],

f (a) g (a)

f ' ( x) g' ( x)

has, in the

(a) atleast one root (b) exactly one root (c) atmost one root (d) no root 1 for all x and f (0) = 0, then 1 + x2

122. If f ′ (x) =

(a) f (2) < 0.4 (c) 0.4 < f (2) < 2

(b) f (2) > 2 (d) f (2) = 2

123. If a, b, c be non-zero real numbers such that 1

∫ (1 + cos 0

=

8

x)(ax 2 + bx + c) dx

2

∫ (1 + cos

8

x)(ax 2 + bx + c) dx = 0,

0

then the equation ax2 + bx + c = 0 will have (a) one root between 0 and 1 and other root between 1 and 2 (b) both the roots between 0 and 1 (c) both the roots between 1 and 2 (d) None of these 124. The value of c in Lagrange’s theorem for the function f (x) = log sin x in the interval  π , 5π  is 6 6    (a)

π 4

(b)

π 2

2π (c) (d) None of these a0 a a a + 1 + 2 + ... + n −1 + an = 0, then the equation 3 2 n +1 n n −1 a0xn + a1xn – 1 + ... + an – 1x + an = 0 has, in the interval 125. The value of c in Lagrange’s theorem for the function  1 (0, 1),  x cos   , x ≠ 0 f (x) = in the interval [– 1, 1] is  x  (a) exactly one root (b) atleast one root  x=0 (c) atmost one root (d) no root 0, (a) 0 (b) 1 118. The equation x log x = 3 – x has, in the interval (1, 3), 2 (a) exactly one root (b) atmost one root 1 (c) – (d) non existent in the interval (c) atleast one root (d) no root. 2 119. Between any two real roots of the equation ex sin x = 1, 126. If 27a + 9b + 3c + d = 0, then the equation the equation ex cos x = – 1 has 4ax3 + 3bx2 + 2cx + d = 0 has atleast one real root lying (a) atleast one root (b) exactly one root between (c) atmost one root (d) no root (a) 0 and 1 (b) 1 and 3 120. If f (x) is differentiable in the interval [2, 5], where (c) 0 and 3 (d) None of these 1 1 f (2) = and f (5) = , then there exists a number c, 127. Let f be a function which is continuous and differenti5 2 able for all real x. If f (2) = – 4 and f ′ (x) ≥ 6 for all 2 < c < 5 for which f ′ (c) = x ∈ [2, 4], then 1 1 (a) f (4) < 8 (b) f (4) ≥ 8 (b) (a) (c) f (4) ≥ 12 (d) None of these 2 5 2 1 1 28. The function f (x) = 2x – log | x |, x ≠ 0 is increasing (c) (d) None of these in the interval 10

117. If

147

anx + an – 1x + ... + a2x + a1x + a0 = 0, n positive integer, has two different real real roots α and β, then between α and β, the equation nanxn – 1 + (n – 1) an – 1xn – 2 + ... + a1 = 0 has n

Applications of Derivatives

112. If the polynomial equation

148

1   (a)  − 1 , 0  ∪  1 , ∞  (b)  −∞, −  ∪  0,  2   2  2 

Objective Mathematics

1 1 (c)  − ,   2 2

1  2

(d) None of these

129. The set of values of x for which log (1 + x) < x, is (a) x < 0 (c) 0 < x < 1

(b) x > 0 (d) None of these

130. The values of x for which 1 + x ln (x + (a) x ≤ 0 (c) x ≥ 0

x2 + 1 ) ≥

1 + x 2 , are (b) 0 ≤ x ≤ 1 (d) None of these

131. If 0 < x < π , then 2 x2 (a) cos x < 1 – 2

x2 (b) cos x > 1 – 2

x 2

x 2

2

(c) cos x = 1 –

(d) cos x ≥ 1 –

2

132. The function f (x) = x – log (1 + x), x > – 1 is increasing in the interval (a) (0, ∞) (c) (– ∞, 0)

(b) (– 1, 0) (d) None of these

133. The range of values of x for which the function x , x > 0 and x ≠ 1, may be decreasing, is log x (a) (0, e) (b) (e, ∞) (c) (0, e)\{1} (d) None of these f (x) =

134. The function f (x) = x1/x is increasing in the interval (a) (e, ∞) (b) (– ∞, e) (c) (– e, e) (d) None of these 135. log x – tan–1x increases in the interval (a) (– ∞, 0) (c) (– ∞, ∞)

(b) (0, ∞) (d) None of these

136. The real number x when added to its inverse gives the minimum value of the sum at x equal to (a) 2 (c) –1 137. The function f (x) = val π (a)  − , 0   2  (c) (0, π) π , then 2 2 (a) > sin x π x sin x (c) < 1 x

138. If 0 < x
cos x (b) cos (sin x) < cos x (c) cos (sin x) > sin (cos x) (d) cos (sin x) < sin (cos x) 140. (1 + x)p ≤ 1 + xp, where (a) p > 1 (c) x > 0

(b) 0 ≤ p ≤ 1 (d) x < 0

141. If ax2 + b ≥ c for all positive x, where a, b > 0, x then (a) 27ab2 ≥ 4c3 (c) 4ab2 ≥ 27c3

(b) 27ab2 < 4c3 (d) None of these

b ≥ c for all positive x, where a, b > 0, x

142. If ax + then

c2 c2 (b) ab ≥ 4 4 c (d) None of these (c) ab ≥ 4 2x 143. The function f (x) = log x – is increasing in the 2+ x interval (a) ab
2 5 145. The function f (x) = – 2x3 + 21x2 – 60x + 41, in the interval (– ∞, 1), is (b) ≤ 0 (d) ≥ 0

(a) < 0 (c) > 0 146. If 0 < α < β < α (a) tan β < β tan α tan α α (c) < tan β β

π , then 2 tan β α > tan α β tan α α (d) > tan β β (b)

sin x is decreasing in the interx 147. A point on the parabola y 2 = 18x at which the ordinate increases at twice the rate of the abscissa is  π 0 , (b)   (a)  −9 , 9  (b) (2, – 4)  2  8 2 (d) None of these 9 9 (c) (2, 4) (d)  ,  8 2 2 –x 148. Let y = x e , then the interval in which y increases 2 sin x (b) < w.r.t. x, is : π x (d) sin x > 1 x

(a) (– ∞, ∞) (c) (2, ∞)

(b) (– 2, 0) (d) (0, 2)

(a) x > 0 (c) x > 1

(b) x < 0 (d) x < 1

150. If y = 2x + arc cot x + ln [ 1 + x 2 – x], then y (a) increases in [0, ∞[ (b) decreases in [0, ∞[ (c) neither increases nor decreases in [0, ∞[ (d) increases in ] – ∞, 0] 151. The function f defined by f (x) = (x + 2) e–x is (a) decreasing for all x (b) decreasing in (– ∞, – 1) and increasing (– 1, ∞) (c) increasing for all x (d) decreasing in (– 1, ∞) and increasing in (– ∞, – 1) 152. The function f (x) =

ln ( π + x) is ln (e + x)

(a) increasing on (0, ∞) (b) decreasing on (0, ∞) (c) increasing on (0, π/e), decreasing on (π/e, ∞) (d) decreasing on (0, π/e), increasing on (π/e, ∞) 153. The function f (x) = tan x – x (a) somtimes increases and sometimes decreases (b) never increases (c) never decreases (d) can’t say. 154. Let the function f (x) = sin x + cos x, be defined in [0, 2π], then f (x) π π (a) increases in  ,  4 2  π 5π  (b) decreases in  ,  4 4 

1 (b) g (x) increases in  0,   2 1 (c) g (x) decreases in  , 1 2  1 (d) g (x) decreases in  , ∞  2  158. The function f (x) =

4 sin x − 2 x − x cos x , 0 < x < 2π 2 + cos x

 π  3π  (a) increases in  0,  ∪  , 2π   2  2   π 3π  (b) increases in  ,  2 2   3π  π (c) decreases in  0,  ∪  , 2π  2 2     π 3π (d) decreases in  ,  2 2  159. A function y = f (x) 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 (d) (x + 1)2

| x −1| x2 (a) increases in (– ∞, 0) ∪ (1, 2) (b) increases in (0, 1) ∪ (2, ∞) (c) decreases in (0, 1) ∪ (2, ∞) (d) decreases in (– ∞, ∞) ∪ (1, 2)

160. The function f (x) =

 π π  (d) decreases in 0,  ∪  , 2π  4 2     –1

(sin x + cos x), be defined

 π  5π  (a) increases in 0,  ∪  , 2π    4  4 π 5 π   (b) decreases in  ,  4 4   5π  π (c) increases in  0,  ∪  , 2π  4 4      π 5π  (d) decreases in  4 , 4  156. The function f (x) = | x + 2 | + | x – 1 | is (a) increasing in (1, ∞) (b) increasing in [1, ∞) (c) decreasing in (– ∞, – 2] (d) decreasing in (– ∞, – 2)

1 (a) g (x) increases in  −∞,  2 

(a) (x + 1)3 (c) (x – 1)2

 5π 7 π   π (c) increases in 0,  ∪  ,  4  4 4  

155. Let the function f (x) = tan in [0, 2π], then f (x)

161. The function f (x) = 3cos4x + 10cos3x + 6cos2x – 3, 0 ≤ x ≤ π is (a) increasing in  π , 2π  2 3  (b) increasing in  0, π  ∪  2 π , π      2  3 (c) decreasing in  0, π  ∪  2 π , π   2  3  π 2π  (d) decreasing in  ,  2 3  162. The values of k for which the function f (x) = kx3 – 9x2 + 9x + 3 may be increasing on R are (a) k > 3 (b) k < 3 (c) k ≤ 3 (d) None of these 163. The least possible value of k for which the function f (x) = x2 + kx + 1 may be increasing on [1, 2] is

149

157. If g (x) = f (x) + f (1 – x) and f ′′ (x) < 0 for 0 ≤ x ≤ 1, then

Applications of Derivatives

149. If a < 0, the function (eax + e–ax) is a monotonic decreasing function for all values of x, where

150

(a) 2 (c) 0

(b) – 2 (d) None of these

Objective Mathematics

164. If f (x) = 2x3 + 9x2 + λx + 20 is a decreasing function of x in the largest possible interval (– 2, – 1) then λ is equal to (a) 12 (c) 6

(b) – 12 (d) None of these

165. The function f (x) = sin4x + cos4x increases in the interval π 3π  (b)  ,  4 8 

(a)  0, π   8

 5π 3π  3π 5π  (c)  , (d)  ,  8 4   8 8  166. Let f (x) = cot–1 [g (x)], where g (x) is an increasing function for 0 < x < π. Then f (x) is (a) increasing in (0, π) (b) decreasing in (0, π) (c) increasing in  0, π  and decreasing in  π , π   2 2  (d) None of these 167. Let f ′ (x) > 0 and g′ (x) < 0 for all x ∈ R. Then, (a) f  [ g (x)] > f  [g (x + 1)] (b) f [g (x)] > f [g (x – 1)] (c) g [  f (x)] > g [ f (x + 1)] (d) g [  f (x)] > g [  f (x – 1)] 168. If the function f (x) = 3 cos | x | – 6ax + b increases for all x ∈ R, then the range of values of a is given by (a) a > – 1 2 (c) a ≤ b

(b) a < –

1 2

(d) a ≥ b

169. The function f (x) = 2 log (x – 1) – x2 + 2x + 3 increases in the interval (a) (– ∞, 0) ∪ (1, 2) (c) (0, 1) ∪ (2, ∞)

(b) (– ∞, 0) ∪ (2, ∞) (d) None of these

170. The equation x + ex = 0 has (a) only one real root (c) no real root

(b) only two real roots (d) None of these

171. The interval in which the function 2x3 + 15 increases less rapidly than the function 9x2 – 12x, is (a) (– ∞, 1) (c) (2, ∞)

(b) (1, 2) (d) None of these

172. Let h (x) = f (x) – [ f (x)] + [ f (x)] for every real number x. Then 2

3

(a) h is increasing whenever f is increasing (b) h is increasing whenever f is decreasing (c) h is decreasing whenever f is decreasing (d) nothing can be said in general 173. Consider the following statements S and R:

Which of the following is true? (a) Both S and R are wrong (b) Both S and R are correct, but R is not the correct explanation for S (c) S is correct and R is the correct explanation for S (d) S is correct and R is wrong 174. Let f (x) = interval

∫ e ( x − 1)( x − 2) dx . Then f decreases in the x

(a) (– ∞, – 2) (c) (1, 2)

(b) (– 2, – 1) (d) (2, + ∞)

175. The value of a in order that f (x) = sin x – cos x – ax + b decreases for all real values is given by (a) a ≥ 2 (c) a ≥ 1

(b) a < 2 (d) a < 1

3 x 2 + 12 x − 1, − 1 ≤ x ≤ 2 176. If f (x) =  , then 2< x≤3 37 − x, (a) f (x) is increasing on [– 1, 2] (b) f (x) is continuous on [– 1, 3] (c) f ′ (2) does not exist (d) all of these 177. Let f and g be increasing and decreasing functions respectively from [0, ∞) to [0, ∞). Let h (x) = f (g (x)). If h (0) = 0, then h (x) is (a) always zero (c) always positive

(b) always negative (d) strictly increasing

178. The normal to the curve x = a(1 + cosθ), y = a sin θ at θ always passes through the fixed point (a) (0, 0) (c) (a, 0)

(b) (0, a) (d) (a, a)

179. Given that f ′ (x) > g′ (x) for all real x and f (0) = g (0), then (a) f (x) > g (x) ∨ x ∈ (0, ∞) (b) f (x) < g (x) ∨ x ∈ (– ∞, 0) (c) f (x) < g (x) ∨ x ∈ (0, ∞) (d) f (x) > g (x) ∨ x ∈ (– ∞, 0) 180. If f ′′ (x) < 0 ∨ x ∈ (a, b), then f ′ (x) = 0 (a) exactly once in (a, b) (b) atmost once in (a, b) (c) atleast once in (a, b) (d) None of these 181. The maximum value of x1/x, x > 0 is (a) e1/e (c) 1 182. The minimum value of (a) e

e

(b)  1  e (d) None of these x is log x 1 e (d) None of these

(b)

S: Both sin x and cos x are decreasing functions in the (c) 1 interval  π , π    2  183. The maximum value of sin x + 1 sin 2x + 1 sin 3x, 2 3 R: If a differentiable function decreases in an interval π 0≤x≤ , is (a, b), then its derivative also decreases in (a, b). 2

(c) 1 – 2 2 (d) None of these 2 4 184. If P (x) = a0 + a1x + a2x + ... + anx2n be a polynomial in x ∈ R with 0 < a1 < a2 ... < an , then P (x) has (a) no point of minimum (b) only one point of minimum (c) only two points of minimum (d) None of these 185. If y = a loge x + bx2 + x has its extreme values (i.e., maximum or minimum value) at x = 1 and x = 2, then the values of a and b are

1 4 ,b= 6 3 4 1 (c) a = , b = – 3 6

(a) a = –

(b) a = –

4 ,b= 1 3 6

(d) None of these

P (2, – 1), then the values of a and b are (b) a = 0, b = – 1 (d) a = – 1, b = 0

sin ( x + a ) 187. If y = ; a ≠ b, then y has sin ( x + b) (a) maximum at x = 0 (b) minimum at x = 0 (c) neither maximum nor minimum (d) None of these 1 , x (a) x = 1 is a point of maximum (b) x = – 1 is a point of minimum (c) maximum value > minimum value (d) maximum value < minimum value.

188. For the function y = x +

(a) y must be maximum at x = a (b) y must be minimum at x = a (c) y may not have a maximum or minimum at x = a (d) it is a constant function 195. The fraction exceeding its pth power by the greatest number possible, where p ≥ 2, is 1 (a)    p

1

p −1



(c) p1/p – 1

ax + b 186. If the function y = has an extremum at ( x − 4)( x − 1) (a) a = 0, b = 1 (c) a = 1, b = 0

(a) maximum at x =1 7 (b) minimum at x = 5 (c) neither maximum nor minimum at x = 2 (d) all the above d2y dy 194. For a function y = f (x), if = 0 and = 0 at a dx 2 dx point x = a, then

1 (b)    p

p −1



(d) None of these

196. If y = x2 + ax + b has a minimum at x = 3 and the minimum value is 5, then the values of a and b are (a) a = 6, b = – 14 (c) a = 14, b = – 6

(b) a = – 6, b = 14 (d) a = – 14, b = 6

197. The greatest value of the function f (x) = xe–x in [0, ∞) is (a) 0 (b) 1 e (c) – e (d) None of these 198. If 2a + 3b + 6c = 0, then at least one root of the equation ax2 + bx + c = 0 lies in the interval (a) (2, 3) (c) (0, 1)

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

199. The greatest value of the function 1  1  f (x) = tan–1x – log x in  , 3  is 2  3  π 1 (a) π + 1 log 3 (b) − log 3 3 4 6 4 189. If xz = 1, where x > 0, then the least value of x + z π 1 π 1 is (c) − log 3 (d) + log 3 3 4 6 4 (a) 1 (b) 2 200. The maximum value of the function (c) – 2 (d) None of these y = x (x – 1)2, 0 ≤ x ≤ 2 is 190. The minimum value of loga x + logx a, 0 < x < a, is 4 (a) 0 (b) (a) 1 (b) 2 27 (c) – 2 (d) None of these (c) – 4 (d) None of these 191. The minimum value of 2 log10 x – logx .01, x > 1, is n2 (a) 1 (b) – 1 201. The largest term in the sequence xn = n3 + 200 , n ∈ N, (c) 2 (d) None of these is 49 8 192. For a differentiable curve y = f (x) having atleast two (a) (b) extremum in the interval [a, b], 543 89 1 (a) two of its maximum values occurs successively (c) (d) None of these 52 (b) two of its minimum values occur successively 202. The shortest distance of the point (0, 0) from the curve (c) maximum and minimum values occur alternatively 1 (d) None of the above (ex + e–x) is y = 2 193. The function x (a) 2 (b) 1 f (x) = ∫  2 (t − 1)(t − 2)3 + 3 (t − 1) 2 (t − 2) 2  dt has (c) 3 (d) None of these 1

151

(b) 2 2 – 1

Applications of Derivatives

(a) 1 + 2 2

152

203. The greatest height of the graph of the curve y = 6 cos x – 8 sin x above the x-axis is

Objective Mathematics

(a) 4 (c) 10

(b) 8 (d) None of these

204. The minimum value of a2sec2x + b2cosec2x, 0 < a < b, is (a) a + b (c) (a + b)4

(b) (a + b)2 (d) None of these

205. The points on the curve xy2 = 1 which are nearest to the origin are 1/ 3 −1/ 6  1 1/ 3    (a)  1  , ±  1   (b)   , 2−1/ 6   2    2    2   1/ 3  1  (c)  2 , ±   2 

−1/ 6

   

(d) None of these.

206. N characters of information are held on magnetic tape, in batches of x characters each; the batch processing time is α + βx2 seconds; α, β are constants. The optimum value of x for fast processing is β α (a) (b) α β (c)

α β

(d)

β α

207. AB is a diameter of a circle and C is any point on the circumference of the circle, then (a) area of ∆ABC is maximum when it is an isosceles (b) area of ∆ABC is minimum when it is an isosceles (c) the perimeter of ∆ABC is minimum when it is isosceles (d) the perimeter of ∆ABC is maximum when it is isosceles. 208. If xy = k and z = lx + my, where k, l, m are positive constants, then the minimum value of z is (a) mlk (c) 2mlk

(b) 2 mlk (d) None of these

209. The number of values of x where the function f (x) = 2 (cos 3x + cos 3 x) attains its maximum is (a) 1 (c) 0

(b) 2 (d) infinite

210. If the slope of the tangent to the curve y = excos x is minimum at x = a, 0 ≤ a ≤ 2π, then the value of a is (a) 0 (c) 2π

(b) π (d) None of these

π , then cos α ⋅ cos β has a maximum value 2 at β equal to π π (a) 4 (b) 2 π (c) 6 (d) None of these

211. If α + β =

212. Let f (x) = 1 + 3x2 + 32x4 + ... + 330 ⋅ x60. Then f (x) has

(a) atleast one maximum (b) exactly one maximum (c) atleast one minimum (d) exactly one minimum. 213. Let the function f (x) be defined as: −1 2  tan α − 3 x , 0 < x < 1 f (x) =  x ≥1 − 6 x, f (x) can have a maximum at x = 1 if the value of α is (a) 0 (b) 2 (c) 1 (d) None of these

214. If the roots of the equation x3 – ax2 + 4x – 8 = 0 are real and positive, then the minimum value of a is (a) 2 (b) 6 (d) None of these (c) 3 3 4 215. Let f (x) = (x – 2)2 xn, n ∈ N. Then f (x) has a (a) minimum at x = 2, ∨ n ∈ N (b) minimum at x = 2 if n is even (c) minimum at x = 0 if n is even (d) minimum at x = 0 if n is odd. 216. For the function f (x) =

x

sin t dt , where x > 0, t 0



(a) maximum occurs at x = nπ, n even (b) minimum occurs at x = nπ, n odd (c) maximum occurs at x = nπ, n odd (d) minimum occurs at x = nπ, n even. 217. For the function f (x) =

x

∫e

−t 4

4

(4 − t 2 ) dt ,

2

(a) maximum occurs at x = 2 (b) minimum occurs at x = – 2 (c) maximum occurs at x = – 2 (d) minimum occurs at x = 2 218. A function f is such that f ′ (4) = f ′′ (4) = 0 and f has minimum value 10 at x = 4. Then f (x) = (a) 4 + (x – 4)4 (c) (x – 4)4

(b) 10 + (x – 4)4 (d) None of these

219. The difference between the greatest and the least value of the function f (x) =

x

∫ (6t

2

− 24) dt on [1, 3] is

0

(a) 14 (c) 4

(b) 10 (d) None of these

220. The range of values of k for which the function f (x) = (k2 – 7k + 12) cos x + 2 (k – 4) x + log 2 does not possess critical points, is (a) (1, 5) (c) (1, 4)

(b) (1, 5) – {4} (d) None of these

221. The range of values of k for which the function f (x) = (2k – 3) (x + tan 2) + (k – 1) (sin4x + cos4x) does not possess critical points, is 4 (a)  −∞,  ∪ (2, ∞) 3 

4 (b)  , 2  3 

4 (c)  , ∞  3 

(d) (2, ∞)

223. The minimum value of the function f (x) = 2 | x – 2 | + 5 | x – 3 |, ∨ x ∈ R is (a) 3 (c) 5

(b) 2 (d) 7

224. The maximum value of x2/3 + (x – 2)2/3 is (a) 0 (c) 22/3

(b) 2 (d) None of these

225. The minimum value of the function x p x−q f (x) = , where 1 + 1 = 1, p > 1 is + p q p q (a) 1 (c) 2

(b) 0 (d) None of these

226. If (x – a)2n (x – b)2m + 1, where m and n are positive integers and a > b, is the derivative of a function f, then (a) x = a gives neither a maximum nor a minimum (b) x = a gives a maximum (c) x = b gives a minimum (d) x = b gives neither a maximum nor a minimum 227. On the interval [0, 1], the function x25 ⋅ (1 – x)75 takes its maximum value at the point

(b)  –6 (d)  0

232. If the function f(x) = kx3 – 9x2 + 9x + 3 is monotonically increasing in every interval, then (a)  k < 3 (c)  k > 3

(b)  k ≤ 3 (d)  k ≥ 3

233. The function f(x) = 2 + 4x2 + 6x4 + 8x6 has (a)  only one maxima (b)  only one minima (c)  no maxima and minima (d)  many maxima and minima 234. The length of the largest interval in which the function 3 sin x – 4 sin3 x is increasing, is (b)   π (a)   π 3 2 3 π (c)   (d)  π 2 235. The maximum value of x1/x is (a)  1/ee (c)  e1/e

(b)  e (d)  1/e

236. The function f defined by f(x) = 4x4 – 2x + 1 is increasing for (a)  x < 1 (c)  x < 1/2

(b)  x > 0 (d)  x > 1/2

237. The function f ( x) = x + 2 has a local minimum at 2 x (a)  x = – 2 (b)  x = 0 (c)  x = 1 (d)  x = 2

1 4 1 (c) 1 (d) 238. Angle between the tangents to the curve y = x2 – 5x + 3 2 6 at the points (2, 0) and (3, 0) is 228. The value of k so that the sum of the cubes of the (a)  π/2 (b)  π/6 roots of the equation x2 – kx + (2k – 3) = 0 assumes (c)  π/4 (d)  π/3 the minimum value, is x 239. The function x is increasing, when (a) k = 1 (b) k = 3 (b)   x < 1 (a)   x > 1 (c) k = 0 (d) None of these e e 1 (c)  x < 0 (d)  for all real x 229. Let f (x) = sin x + cos 2x. Then 2 240. For what value of a, f(x) = – x3 + 4ax2 + 2x – 5 is 4 3 decreasing ∀ x. (a) min f (x)  π  π 3 4 x∈ 0 ,  x∈ 0 ,  (a)  (1, 2) (b)  (3, 4)  2  2 (c)  R (d)  no value of a 2 x2 + 1 (c) min f (x) >   (d) min f (x) < 3 −t 2  π  π 241. If f ( x) = ∫ 2 e dt , then f(x) increases in 3 x∈ 0 ,  x∈ 0 ,  2 x  2  2 (a) 0

(b)

230. A private telephone company serving a small community makes a profit of Rs. 12 per subscriber, if it has 725 subscribers. It decides to reduce the rate by a fixed sum for each subscriber over 725, thereby reducing the profit by 1 paise per subscriber. Thus, there will be profit of Rs. 11.99 on each of the 726 subscribers, Rs. 11.98 on each 727 subscribers etc. The number of subscribers which will give the company the maximum profit, is (a) 961 (c) 963

(b) 962 (d) None of these

231. If the curres ay + x2 = 7 and x3 = y cut orthogonally at (1,1), then a is equal to

(a)  (–2, 2) (c)  (0, ∞)

(b)  no value of x (d)  (–∞, 0)

242. The function x5 – 5x4 + 5x3 – 1 is (a)  neither maximum nor minimum at x = 0 (b)  maximum at x = 0 (c)  maximum at x = 1 and minimum at x = 3 (d)  minimum at x = 0 x 243. The maximum value of f ( x) = on [–1, 1] is 4 + x + x2 (a)   − 1 (b)   − 1 3 4 (c)   1 (d)   1 5 6

153

(a)  1 (c)  6

Applications of Derivatives

222. The minimum value of e( x4 − x3 + x2 ) is (a) e (b) e2 (c) 1 (d) None of these

154

Objective Mathematics

244. A value of c for which the conclusion of mean value theorem holds for the function f(x) = loge x on the interval [1, 3] is (b)   1 log e 3 (a)  2log3 e 2 (c)  log3 e (d)  loge 3 245. The function f(x) = tan–1 (sin x + cos x) is an increasing function in (a)  (π/4, π/2) (c)  (0, π/2)

(b)  (–π/2, π/4) (d)  (–π/2, π/2)

246. The tangent to the curve y = e drawn at the point (c, ec) intersects the line joining the points (c –1, ec –1) and (c + 1, ec + 1) x

(a)  on the left of x = c (b)  on the right of x = c (c)  at no point (d)  at all points 247. If the Mean Value theorem is f(b) – f(a)  Then for the function x2 – 2x + 3 in 1,  c is (a)  6/5 (b)  5/4 (c)  4/3 (d)  7/6

= (b – a) f ′ (c). 3 , the value of 2 

p and maxima at – 3

p 3

p and maxima at 3 p p (c)  The cubic has minima at both and – 3 3 p p (d)  The cubic has maxima at both and – 3 3

p 3

(a)  The cubic has minima at (b)  The cubic has minima at –

256. The minimum value of 2log10 x – logx 0.01, x > 1 is…. (a)  2 (c)  6

(b)  4 (d)  8

257. In between any two real roots of an ex sinx = 1 there exists how many roots satisfying equation ex cosx = – 1 (a)  at least one root (c)  at most one root

(b)  no root (d)  none of these.

258. If the curves x2 = 9A(9 – y) and x2 = A(y + 1) intersect orthogonally, then the value of A is

(a)  3 (b)  4 (c)  5 (d)  7 248. The equation of the curve which passes through the point (0, 1) and has the slope 3x2 + 2x + 5 at any point 259. The total number of local maxima and local minima of (x, y) is (2 + x)3 , −3 < x ≤ −1 the function f(x) =  2/3 is (a)  y = 3x3 + 2x2 + 5x + 1 (b)  y = 2x3 + 3x2 + 5x + 1 −1 < x < 2  x , 3 2 3 2 (c)  y = x + x + 5x + 1 (d)  y = x + x + 5x – 1 (a)  0 (b)  1 249. The abscissa of the point on the curve y = a (ex/a + (c)  2 (d)  3 –x/a e ), where the tangent is parallel to the x-axis, is 260. Let f(x) be a non-constant twice differentiable function (a)  0 (b)  a 1 defined on (– ∞, ∞) such that f(x) = f(1 – x) and f '   (c)  2a (d)  – 2a 4  = 0.  e2 x − 1  This section contains 4 multiple correct answer(s) type 250. Function f ( x) =  2 x  is questions. Each question has 4 choices (A), (B), (C) and (D),  e +1 out of which ONE OR MORE is/are correct. Then (a)  increasing (b)  decreasing (a)  f” (x) vanishes at least twice on [0, 1] (c)  even (d)  None of the above 1 −  x

251. The function f ( x) = x( x + 3) e  2  satisfies all the conditions of Rolle’s theorem in [–3, 0]. The value of c is (a)  0 (c)  –2

(b)  –1 (d)  –3

x −x 252. If f ( x) = cot  2  (a)  –1 (c)  log 2 −1

x

−x

  , then f ′ (1) is equal to  (b)  1 (d)  – log 2

253. If a + b = 8, then ab is greatest when (a)  a = 4, b = 4 (c)  a = 6, b = 2

(b)  a = 3, b = 5 (d)  None of these

254. How many real solutions does the equation x7 + 14x5 + 16x3 + 30x – 560 = 0 have? (a)  7 (c)  3

(b)  1 (d)  5

255. Suppose the cubic x3 – px + q has three distinct real roots where p > 0 and q > 0. Then which one of the following holds?

1 (b)   f '   = 0 2 1/ 2

(c)  



−1/ 2 1/ 2

(d)  

∫ 0

1  f  x +  sin x dx = 0 2 

f (t )esin πt dt =

1



f (1 − t )esin πt dt

1/ 2

261. A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then 1 1 2 + < PS ST QS × SR 1 1 2 + > (b)   PS ST QS × SR 1 1 4 (c)   + < PS ST QR 1 1 4 (d)   + > PS ST QR

(a)  

1. (a), (c) We have, x  ...(1) y = x3 – 2x2 + x – 2  ...(1) 5. (b) We have, y2 = 4a  x + a sin   a  dy = 3x2 – 4x + 1 dy x 1 x dx   dy =4a 1 + a ⋅ cos ⋅  = 4a 1 + cos  ⇒ 2y Since the tangent is parallel to x-axis, = 0. dx a a a   dx That is 2 dy 3x – 4x + 1 = 0   or  (x – 1) (3x – 1) = 0 Since the tangent is parallel to x-axis, =0 1 dx ⇒  x = 1, . 3 x x ⇒ 4a 1 + cos  = 0 ⇒ cos = – 1 = cos π 50 1 a a  and if x ∴ If x = , from (1) we get, y = – 27 3 x ∴ = π i.e., x = aπ. a = 1, from (1) we get, y = – 2. Putting this value of x in equation (1), we get y2 = 4a (x + a sin π) = 4a (x + 0) = 4ax.  1 50  The points are (1, – 2) and  , −  . ∴ y2 = 4ax, which is a parabola.  3 27  2. (b) We have,  1



2 x

x+ y = 4  +

...(1)

1 dy = 0,  or  2 y dx

dy =– dx

y x

.

Since the tangent is equally inclined to the axes, dy = tan 45º or tan 135º i.e., 1 or – 1. Thus dx –

y x

= ± 1. This gives y = x.

From (1), x + x = 4 ⇒ x = 4. Also, y = x = 4. The point is (4, 4). 3. (b), (c)   We have, x2 + y2 = 25  ⇒

2x + 2y

dy = 0,  or  dx

...(1)

x dy =– . y dx

Now, slope of the line 3x – 4y = 7 is m =

3 . 4

Since the tangent is parallel to the given line, x dy 3 3 4 = ⇒– = i.e., y = – x. ...(2) y dx 4 4 3 From (1) : x2 +

16 2 x = 25 ⇒ x = ± 3. 9

If x = 3, from (2), y = –

4 (3) = – 4. 3

If x = – 3, from (2), y = –

4 (– 3) = 4. 3

The points are (3, – 4) and (– 3, 4). 4. (d) Minimum value of a tan2 x + b cot2 x is 2 ab and maximum value of a sin2 θ + b cos2 θ is a ( a sin2 θ + b cos2 θ = (a – b) sin2 θ + b) Given: a = 2 ab .  ∴ a = 4b.

6. (b) We have, e2y = 1 + 4x2 ⇒

dy 4x 4x = 2y = . dx e 1 + 4x2

∴  Slope of tangent = m = ⇒

| m | =

4| x | ≤1 1 + 4 | x |2

 (1 − 2 | x |) 2 ≥ 0    7. (c) We have,

⇒ e2y ⋅ 2

dy = 8x dx

4x 1 + 4x2

⇒ 1 + 4 | x |2 −4 | x |≥ 0   4| x |  ⇒ ≤1 2  1 + 4| x |

ax y = b−x



(b − x) a − ax ⋅ (−1) ab dy = = . 2 (b − x) (b − x) 2 dx



dy  ab dx  (1, 1) = (b −1) 2 = 2 (given) 

...(1)

Since the curve passes through the point (1, 1), therefore, a 1= i.e., a = b – 1 ...(2) b −1 Putting a = b – 1 in (1), we get (b − 1) b = 2 ⇒ b = 2. ∴ a = 2 – 1 = 1. (b − 1) 2 Hence, a = 1, b = 2. 8. (b) We have, y = ⇒

2 3 x – 2ax2 + 2x + 5 3

dy = 2x2 – 4ax + 2. dx

Since, the tangent makes an acute angle with the positive direction of x-axis, therefore dy ≥ 0 ⇒ 2x2 – 4ax + 2 ≥ 0 for all x dx

Applications of Derivatives

155

Solutions

156

Objective Mathematics

⇒ 16a2 – 16 ≤ 0 ( Disc. = (4a)2 – 4 (2) (2) ≤ 0) ⇒ a2 – 1 ≤ 0 i.e., (a – 1) (a + 1) ≤ 0 ⇒ – 1 ≤ a ≤ 1.

dx dy = 2t + 3 and = 4t – 2 dt dt dy dt dy 4t − 2 ∴ = = . dx dt dx 2t + 3

12. (c) We have,

9. (d) We have, y2 = 2ax ...(1) a a 2 Put x = ; y = 2a     ⇒ y = ± a 2 2

Thus, slope of the tangent to the curve at the point t = 2 is

a  a  ∴ The points are  , − a  and  , − a  2   2  Differentiating (1) w.r.t x, we get a dy dy 2y = 2a ⇒ = y dx dx a dy a At  a , a  : = = = 1 = m1 (say).   y dx a 2  a a  dy a At  , − a  : = = = – 1 = m2 (say). y −a 2  dx Since m1m2 = – 1, the two tangents are at right angles. 10. (a) Putting y = x in y = 9 − 2 x 2 , we get x = 9 − 2 x 2 ⇒ x2 = 9 – 2x2 ⇒ x = 3 , – 3 . Since y > 0, therefore, the point is ( 3 , 3 ). Now, we have, y2 = 9 – 2x2 Differentiating w.r.t. x, we get dy 2y = – 4x dx ⇒

2x dy =– y dx dy  dx  (



3)

3 ) = – 2 (x –

3 ) is

3) i.e., 2x + y – 3 3 = 0.



11. (b) We have, y = (x + 1) (x – 3) = x2 – 2x – 3 ...(1) For points on x-axis, y = 0. This gives (x + 1) (x – 3) = 0 ⇒ x = – 1, 3. Therefore, the curve meets x-axis at (– 1, 0) and (3, 0). Differentiating the given equation w.r.t. x, we get dy = 2x – 2. dx dy  ∴ dx  = – 4 = tan θ1 i.e., θ1 = tan–1 (– 4)  ( −1, 0 ) and

dy  = 4 = tan θ2 i.e., θ2 = tan–14 dx  ( 3, 0 )

Hence, the angle between these tangents = ± (θ1 – θ2) = ± [tan–1 (– 4) – tan–1 (4)] =

...(1) ...(2)

Therefore, the point of intersection in the first 1  quadrant is  , 3  . 2  Now, differentiating equation (1) w.r.t. x, we get 1 dy dy  = 4x + 5 ⇒ = 4   + 5 = 7. dx dx  ( 1 , 3) 2 2 1  Therefore, the equation of tangent at  , 3  is 2  1  ( y – 3) = 7  x −  i.e., 14x – 2y – 1 = 0. 2  x2 y 2 + =1 a 2 b2 −b 2 x 2 x 2 y dy dy =0 ⇒ = ⇒ 2 + 2 a2 y a b dx dx

2 3 = – 2. 3

So, the equation of tangent at ( 3 , ( y –

13. (a) y = 2x2 + 5x  y =3  Solving equations (1) and (2), we get 2x2 + 5x – 3 = 0 i.e., (2x – 1) (x + 3) = 0 1 ⇒ x = , x = – 3. 2

14. (b) We have, =–

3,

4 (2) − 2 dy  6 = = . 2 (2) + 3 dx  t =2 7

 −4 − 4  8  = ± tan–1 ± tan–1   . 15  1 + (−4)(4) 

∴ tan θ1 =

dy  π = ∞ ⇒ θ1 = dx  ( a , 0 ) 2

and  tan θ2 =

dy  = 0 ⇒ θ2 = 0 dx  ( 0, b )

Hence, the angle between the two tangents is π π –0= . θ = θ1 – θ2 = 2 2 15. (c) The two curves are x3 – 3xy2 = a  ...(1) and 3x2y – y3 = b  ...(2) Differentiating (1) w.r.t. x, we get x2 − y 2 dy dy 3x2 – 3y2 – 6xy =0 ⇒ = 2 xy dx dx Differentiating (2) w.r.t. x, we get 6xy + 3x2

2 xy dy dy dy – 3y2 =0 ⇒ =– 2 x − y2 dx dx dx

The product of

dy for the two curves dx

∴ The curves cut each other orthogonally.

⇒ 16. (b) Differentiating y = 4ax w.r.t. x, we get 2a dy dy 2y = 4a ⇒ = . y dx dx 2



dy  2a π dx  ( 0, 0 ) = 0 = ∞ = tan 2 = tan θ1 (say)

∴ θ1 =

π . 2

dy  0 = = 0 = tan 0 = tan θ2 (say) dx  ( 0, 0 ) 2a

∴ θ2 = 0. ∴ Angle between the two curves = (θ1 – θ2) =

π . 2

17. (b) Since f (x) is monotonically decreasing ∴ f ′ (x) = a eax – a e–ax < 0 ⇒ eax > e–ax ( a < 0) ⇒ e2ax > 1 ⇒ 2ax > 0 ⇒ 2x < 0 ( a < 0) ∴ x < 0. 18. (c) We have, y2 = 4x and x2 + y2 – 6x + 1 = 0  Differentiating (1) w.r.t. x, we get 2 dy dy 2y =4 ⇒ = y dx dx ∴

...(1) ...(2)

157

Differentiating (2) w.r.t. x, we get dy  dy π = – 2 sin 2x ⇒  π = – 2 sin dx  dx 3 x= 6 = – 3 = m2 (say). Hence, angle between the two curves is

=

3π 3π2 π3 (sin −1 x) 2 − sin −1 x + 2 4 8

=

3π  π π2  3π3 π3 −1 2 −1 + (sin x) − 2 ⋅ ⋅ sin x +  − 2  3 16  32 8

=

3π  −1 π  π3  π3 7 π3  ⋅  sin x −  + ∈  ,  2  4  32  32 32 

22. (a) We have, y = sin x  and y = cos x  The two curves intersect at x =

π in [0, π]. 4

...(1) ...(2)

Differentiating (1) w.r.t. x, we get dy  1 dy = cos x. ∴ = = m1 (say). dx  x = π 2 dx

dy  2 = = 1 = m1 (say). dx  (1, 2 ) 2

4

Differentiating (2), w.r.t. x, we get dy  1 dy = – sin x. ∴ = m2 (say)  π =– dx  2 dx x= 4

dy  3 −1 = = 1 = m2 (say) dx  (1, 2 ) 2

Hence, the angle between the two curves is

Since m1 = m2, therefore the two curves touch each other at (1, 2). 19. (b) We have, f (x) = (x – 2)2/3 (2x + 1) ⇒ f ′(x) = 2/3 (x – 2)–1/3 (2x + 1) + (x – 2)2/3.2 10 ( x − 1) = 3 ( x − 2)1/ 3 Now, f ′(x) = 0  ⇒ x = 1 and f ′(x) is not defined at x = 2. 20. (b) We have, y = 2sin2x  and y = cos 2x

3 = m1 (say)

2

Differentiating (2) w.r.t. x, we get 3− x dy dy 2x + 2y –6=0 ⇒ = y dx dx ∴

3 = 2

 m − m2  π 2π θ = ± tan–1  1 or .  = ± tan–1 3 = 1 + m m 3 3 1 2   21. (c) f (x) = (sin –1 x + cos –1 x) 3  – 3sin –1 x cos –1 x(sin –1  x + cos–1 x) f (x) = (sin– 1x + cos– 1x)3 – 3 sin– 1 ⋅ cos– 1  (sin– 1x + cos– 1x)  3 π 3π  π  −  − sin −1 x  sin −1 x = 8 2 2 

Differentiating x2 = 4ay w.r.t. x, we get dy dy x 2x = 4a ⇒ = dx dx 2a ∴

dy  1 =4 ⋅   ⋅ dx  x = π 2 6

...(1) ...(2)

 m1 − m2  θ = ± tan–1  1 + m m  = ± tan–12 2 . 1 2   23. (d) f (x) = xα log x For Rolle’s theorem to be applicable, f (x) is continuous in [0, 1] f (x) is differentiable in (0, 1); f ′(x) = 0 ∴ f (0) = f (1) given f ′(x) = xα–1[1 + α log x] = 0 As xα–1 ≠ 0 −1 ∴ 1 + α log x = 0  or  α = log x log x < 0 for x ∈ [0, 1] 1 . ∴ α must be + ve ∴ α = 2

Applications of Derivatives

=

Differentiating (1) w.r.t. x, we get dy = 4 sin x cos x dx

 x 2 − y 2   − 2 xy   × 2 2  = – 1.  2 xy   x − y 

158

n

n

Objective Mathematics

x  y 24. (c) We have,   +   = 2 a b nx n −1 ny n −1 dy =0 ⇒ + n an b dx n

Now,

dx xe xy − 1 = = 0 for vertical tangents dy 1 − ye xy

∴ xexy – 1 = 0 ⇒ exy = 1/x, y = 0 ⇒ x = 1. 30. (d) Let f (x) = (x – 3) (x – 5) = x2 – 8x + 15

n −1

dy  b ⋅a b = − n n −1 = − . ⇒ dx  a b a (a, b)

5

∴ ∫ f ( x)dx = 3

∴  The equation of tangent at (a, b) is b x y y–b =– (x – a) ⇒ + = 2. a a b

x y = 2 touches the curve at + a b (a, b), for all n.

5

∫ (x 3

2

− 8 x + 15)dx

3

 x3 8 x 2  =  3 − 2 + 15 x   5 125  − 100 + 75 − [9 − 36 + 45] =   3 

∴ The line

50 − 54 4 =– . 3 3 31. (d) f (x) = x3 + bx2 + cx + d = [50/3 – 18] =

25. (a) We have, x = a (cos θ + θ sin θ) and y = a (sin θ – θ cos θ) dx = a (– sin θ + θ cos θ + sin θ) = aθ cos θ ⇒ dθ dy = a (cos θ + θ sin θ – cos θ) = aθ sin θ and dθ dy d θ a θ sin θ dy = = = tan θ. ∴ dx d θ a θ cos θ dx

32. (b) The given curve is

 ence, the equation of the normal at any point θ H on the curve is given by [ y – a (sin θ – θ cos θ)]

y = sin x dy = cos x. ⇒ dx

cos θ [x – a (cos θ + θ sin θ)] sin θ ⇒ x cos θ + y sin θ = a (cos2θ + sin2θ) = a. Therefore, the length of the perpendicular from the origin to the normal

 o, the equation of tangent through the origin (0, S 0) is dy y (x – 0) = x cos x, ∴ = cos x ...(2) y–0= dx x



f ′(x) = 3x2 + 2bx + c; D = 4 (b2 – 3c) As, 0 < b2 < c ∴ b2 – 3c < 0 ⇒ f ′(x) > 0 Hence, function is strictly increasing.

=–



=

0 ⋅ cos θ + 0 ⋅ sin θ − a cos 2 θ + sin 2 θ

= a.

26. (c) Since f (x0) = g(x0)  and  f ′(x) > g′(x), ∀x > x0

Squaring and adding (1) and (2), we get y2 y2 + 2 = 1, ∴ x2y2 = x2 – y2. x 33. (a) The given curve is y = ax2 + bx + c

⇒  f (x) – g(x) is an increasing function ∀x > x0. ∴ f (x) – g(x) > f (x0) – g(x0) = 0 ∴ f (x) > g(x), ∀x > x0. 27. (c) Since f 2n + 1 (a) = 0  f 2n + 2(a) = –ve and f (a) = b, ∴  f (x) = – (x – a)2n + 2 + b = b – (x – a)2n + 2. 28. (c) We have, f (x) = x3 + ax2 + bx + 5 sin2x ⇒  f ′(x) = 3x2 + 2ax + b + 5 sin 2x Since f (x) is an increasing function ∴ 3x2 + 2ax + b – 5 > 0, ∀x ∈ R ⇒ 4a2 – 4 · 3 · (b – 5) < 0 ∴ a2 – 3b + 15 < 0. 29. (d) We have, y – exy + x = 0 ⇒

 dx  dx + = 0. 1 – exy  x + y dy   dy

...(1)

Since the point (– 1, 0) lie on it ∴ a – b + c = 0 ...(2) Also, y = 2x is a tangent to (1) at x = 1, so that y = 2. Since the point (1, 2) lies on (1), ∴ a + b + c = 2  ...(3)

∴  f ′(x) – g′(x) > 0



...(1)

dy  = (2ax + b)](1, 2 ) dx  (1, 2 ) = 2, ∴ 2a + b = 2 ...(4) 1 1 , b = 1, c = . Solving (2), (3) and (4) : a = 2 2

Also 

34. (a) When the curve meets x-axis, then 1 y = 0 ⇒ ax2 = 1 ⇒ x = ± a  1   1  , 0 . , 0  and  − Hence the points are  a a     Differentiating the given equation w.r.t. x, we get



dy   dy 2ax + 2h  y + x  + 2by =0 dx  dx 

The point (a cos θ, b sin θ) is (3 3 cos θ, sin θ). Tangent at the above point is

159



ax + hy dy =– . hx + by dx



dy  dx  

x3 3 cos θ y sin θ =1 + 9 1 ∴  Sum of intercepts =

Applications of Derivatives

1

 a

 , 0 

dy  a =– and dx    − h 

1 a

 , 0 

a =– . h

Hence the tangents at these points are parallel. 35. (c) We have, xy = (a + x)2 ⇒

a2 dy a2 + 2a + x ⇒ =1– 2 . y= x dx x

I f the normal makes equal intercepts then its slope is ± 1. x2 = ± 1 ⇒ x2 = x2 – a2 or ⇒ – 2 x − a2 x2 = – (x2 – a2) a a2 i.e., x = ± . ⇒ a2 = 0 or x2 = 2 2 a a2 a + 2a ± = ± 2a ± Then, y = + 2a . a 2 2 ± 2 a   a , 2a + ± 2a  . Hence the points are ±  2   2 36. (a) We have,

9 1 + 3 cos θ sin θ

or s = 3 3 sec θ + cosec θ ds = 3 3 sec θ tan θ – cosec θ cot ⇒ dθ 1 1   ∴  tan θ = ⇒θ ⇒ tan3 θ = 3 3 3 d 2s π is positive at θ = . d θ2 6 π Therefore, sum is minimum at θ = . 6 39. (c) We have, 1 h 2 sin − 0 dy  h lim = h→0 = lim h sin h→0 dx  ( 0, 0 ) h

θ=0

π 6

=

1 = 0. h

∴ Slope of the tangent at origin is 0. Hence, the equation of the tangent at (0, 0) is y – 0 = 0 (x – 0) ⇒ y = 0.

40. (c) We have, y = cos (x + y) dy dy  = sin (x + y) 1 + ⇒  ...(1) z = (x – p)2 + (x – q)2 + (x – r)2 dx dx   Differentiating equation (1) with respect to x, we get Since, the tangents are parallel to the line x + 2y = 0 dz = 2(x – p) + 2(x – q) + 2(x – r) ...(2) 1 1 = sin (x + y) 1 −  ⇒ sin (x + y) = –1 ∴ – dx 2 2 dz  For minima or maxima, put =0 dx ⇒ x + y = –π/2, 3π/2, –5π/2 Form equation (2), we get

2(x – p) + 2(x – q) + 2(x – r) = 0 ⇒ 3x – ( p + q + r) = 0 1 or, x = ( p + q + r) 3 ⇒ the given equation is maximum at 1 x = ( p + q + r). 3 37. (a) We have, x = a (θ + sin θ) and y = a (1 – cos θ) dx dy = a (1 + cos θ) and = a sin θ ⇒ dθ dθ 2 sin θ cos θ dy dy d θ sin θ 2 2 = tan θ = = = 2 cos 2 θ dx dx d θ 1 + cos θ 2 2 dy  π = tan (2n + 1) = ∞. ∴ dx   θ = ( 2 n +1) π 2



∴ Slope of the normal is 0. Hence, the normal is parallel to x-axis at θ = (2n + 1) π. 38. (a) Here a2 = 27, b2 = 1,

a=3 3,b=1



–1 ≤ y ≤ 1.

41. (d) Let P (x1, y1) be a point on the curve y2 = 4ax

...(1)

Differentiating w.r.t. x, we get 2a dy dy    = 4a ; ∴ = . 2y  y1 dx dx  ( x1 , y1 ) Hence, the equation of normal at (x1, y1) is y y – y1 = – 1 (x – x1) 2a ⇒ xy1 + 2ay = y1 (x1 + 2a) But lx + my = 1 is also normal; hence coefficients must be proportional. y 2a y1 ( x1 + 2a ) ∴ 1 = = l m 1 2al 1 , x1 = – 2a. m l Hence, from (1),  1 − 2al  4 a 2l 2 = 4a   2 m  l  ⇒ y1 =

⇒ al3 + 2alm2 = m2.

160

42. (b) We have, f ′(x) = ex cos x + sin x.ex and

f ′′(x) = –sin x ex + cos x ex + cos x ex

Objective Mathematics



+ sin x ex.

Now, f ′′(x) = 2 cos x e = 0 ⇒ cos x = 0 x

⇒ x = π/2.



7 1 8 = –­ +c ⇒c = 3 3 3 / 3 2 1 ∴ f (x) = – (1 − x 2 ) − 8 . 3 47. (b) Let P (x1, y1) be a point on the curve ∴

{

Also, f ′′’(x) = –2 sin x e + 2 cos x e = –ve x

x

x + y = a  ...(1) Differentiating w.r.t. x, we get

∴  slope is maximum at x = π/2.

1 1 dy dy  + =0 ⇒ =– 2 x 2 y dx dx  ( x1 , y1 )

43. (b) f (x) = a – (x – 3)89 ≤ a (equality holds at x = 3). 44. (b) We have, xy = c2 dy dy y ⇒ x +y⋅1=0 ⇒ =– ; dx dx x ∴

3 ⇒ t1 t2 = – 1 (as t1 – t2 ≠ 0).



a [from (1)]



a ⋅ a = a.

=

48. (c) Let P (x1, y1) be a point on the curve

x   a

2/3

 y +  b

2/3

= 1 ...(1)

Differentiating w.r.t. x, we get 2

2

−1

−1



2  x  3 1 2  y  3 1 dy   ⋅ +   ⋅ 3 a  a 3 b  b dx



2/3 dy   b   y1  = –     dx  ( x1 , y1 )  a   x1 

=0

1/ 3

The equation of the tangent at (1, 1) is 1 ( y – 1) = (x – 1) 4 ⇒ x = 4y – 3.

b y – y1 = –   a ⇒

x x a

13 23 1



46. (a) We have, f′(x) = x 1 − x 2

1/ 3

 y1     x1 

y y b

13 23 1

(x – x1)

x  =  1 a

2/3

y  + 1 b

2/3

=1

p2 q2 x 2 / 3a 4 / 3 y 2 / 3b 4 / 3 + 2 = 1 2 + 1 2 2 a b a b



2 3/2

+c

+

2/3

 [from (1)] Hence, p = x11 3a 2 3 and q = y11 3b 2 3

 ence, the tangent meets the curve again at H 1   16 − , −  . 20   5 1 (1 − x ) 1 − x 2 dx = – 2 3/2

.

The equation of the tangent at (x1, y1) is

When this tangent meets the curve again, then 3 ⋅ (4y – 3) y2 – 2 (4y – 3)2 y = 1 ⇒ 20y3 – 39y2 + 18y + 1 = 0 1 16 ⇒ x = 1, 1, – . ⇒ y = 1, 1, – 20 5

7 Since it passes through  0 ,   3

.

Hence, the intercepts on the axes are a x1 and a y1 . ∴  Sum of the intercepts = a ( x1 + y1 )

Differentiating w.r.t. x, we get dy dy – 4xy – 2x2 =0 3y2 + 6xy dx dx 1 dy  = . ⇒  4 dx  (1, 1)

3/2 1 1 − x2 ) + c ( 3

x y + = x1 + y1 = x1 y1 x y + = 1. a x1 a y1



45. (a) We have, 3xy2 – 2x2y = 1

⇒ f (x) = –

x1

y1 (x – x1) x1

y – y1 = –

 c The equation of the normal at  ct1 ,  is t 1  c 2 y – = t1 (x – ct1). t1  c Since, this normal passes through  ct2 ,  , theret2   fore, c c − = t12 (ct2 – ct1) t2 t1

∫x

y1

Equation of tangent at (x1, y1) is

1 dy  =– 2 .  t dx  ( ct1 , c t1 ) 1

⇒ f (x) =

}

x  =  1 a

2/3

y  + 1 b

2/3

= 1.

[from (1)]

49. (a) Let P (x1, y1) be a point on the curve xy = c2 ...(1)

 Differentiating w.r.t. x, we get

x

dy  y1 dy + y ⋅ 1 = 0 ⇒ dx  =– . x1  ( x1 , y1 ) dx

 herefore, the tangent at P(x1, y1) meets x-axis at T A (2x 1 , 0) and y-axis at B (0, 2y 1 ). Clearly P (x1, y1) is the mid point of AB.

53. (c) We have, dy =0 x2 + y2 = 9 ⇒ 2x + 2y dx dy  2 =– . ∴ dx  ( 2, 5 ) 5 The equation of tangent is 2 (x – 2) ( y – 5 ) = – 5 and the equation of normal is

a − a2 − y2 a a 2 − y 2 + log  ...(1) 2 a + a2 − y2 Differentiating w.r.t. x, we get



x=

dx = dy

a2 − y2 dy  ⇒ = y dx  ( x1 , y1 )

y1 a 2 − y12

i.e., 2e (ex – 2y) = e2 – 4.



50. (b) Let P(x1, y1) be a point on the curves

1 1  e  y−  = x−  e 2 2  

161



.

(y –

Equation of the tangent at (x1, y1) is y1 (x – x1). y – y1 = 2 a − y12

5)=

5 (x – 2) 2

...(1)

...(2)

The tangent and the normal intersect x-axis at the 9  points A  , 0  and O (0, 0) respectively. 2  Therefore, the area of the triangle OAP 1 9 9 5 = × × 5 = . 2 2 4

This passes through A. ∴ α = x1 – a 2 − y12 . ∴ AP2 = (x1 – x1 + a 2 − y12 )2 + (y1 – 0)2 = a 2 – y 12 + y 12 = a 2. Hence, the length of the portion of the tangent to the given curve, intercepted between the curve and x-axis is a. 51. (d) We have, y = 1 – 2 x/2

54. (a), (d)  We have, xy = 4

...(1)

For y-axis, x = 0 ∴ y = 1 – 20 = 1 – 1 = 0. Differentiating (1) w.r.t. x, we get dy  x dy 1 1 = − 2 2 ⋅ log 2 ⇒ =– log 2. dx  ( 0, 0 ) dx 2 2 Therefore, equation of normal is 2 ( y – 0) =  (x – 0) ⇒ 2x – y log 2 = 0. log 2 1 , y = e–1 52. (b) At the point x = 2 1 Since, x = > 0, ∴ y = e–2x 2 Differentiating w.r.t. x, we get dy  dy −2 = – 2e–2x ⇒ = – 2e–1 = . dx   1 , 1  dx e 2 e

Thus, the equation of normal is



⇒ x ⋅ 

dy + y ⋅ 1 = 0 dx

dy y 4 =– = – 2  ( xy = 4) x dx x 4 ∴ Slope of tangent = –  2 . x a Slope of the line ax + by + c = 0 is = –  . b Since the given line is a tangent to the curve 4 a a ∴ – 2 = – ⇒ >0 x b b It is possible only when a > 0, b > 0 or a < 0, b < 0. i.e.,

dy = f ′ (x). dx ∴ Slope of the tangent to the ellipse at the point where x = 2, is = f ′ (2) = 2 [ y = 2x + 3 is a tangent to the ellipse at x = 2] Hence, f ′ (2) = 2.

55. (a) We have, y = f (x) ⇒

Applications of Derivatives

Equation of tangent at (x1, y1) is y – y1 = – y1 (x – x1) ⇒ xy1 + yx1 = 2x1 y1 x x y 1 + ⇒ = 1. 2 x1 2 y1

162

56. (a) We have,

Objective Mathematics

x = a cos θ and y = a sin θ dy dy = – 3acos2θ sin θ and = 3a sin2θ cos θ ⇒ dθ dθ 3

3

dy d θ dy 3a sin 2 θ cos θ sin θ = = =– . dx d θ dx −3a cos 2 θ sin θ cos θ The equation of tangent, at any point (a cos3θ, a sin3θ) on the curve, is sin θ (x – a cos3θ) (y – a sin3θ) = – cos θ This meets the coordinates axes at A (a cos θ, 0) and B (0, a sin θ). ∴ AB = (a cos θ − 0) 2 + (0 − a sin θ) 2 = a. ∴

57. (b) Since the curve passes through the point (1, 2) ∴ 2 = p + q + r Also, the curve passes through the origin, ∴ r = 0. The equation of the tangent at (0, 0) is dy  (x – 0) ⇒ y = qx. y – 0 = dx   ( 0, 0)

...(1)

61. (b), (c)  Let P (x1, y1) be a point on the parabola y2 = 8x ...(1) Differentiating equation (1) w.r.t. x, we get 4 dy  dy =8 ⇒ = = m1 (say). 2y  y dx  ( x1 , y1 ) dx 1

But y = x is the tangent at origin, ∴ q = 1. ∴ (1) ⇒ 2 = p + 1 + 0 i.e., p = 1. Hence, p = 1, q = 1, r = 0. 58. (b) We have, y = e4x + 2  ...(1) Putting x = 0, we get y = 3. So, the given point is (0, 3). Differentiating equation (1) w.r.t. x, we get dy  dy = 4e4x ⇒ = 4. dx  ( 0, 3) dx The equation of the tangent at (0, 3) is

(y – 3) = 4 (x – 0) i.e., 4x – y + 3 = 0.

∴ Length of the ⊥ r from origin to the tangent

| 4(0) − 0 + 3| = = 16 + 1

3 . 17

59. (b) Let P (x1, y1) be a point on the curve

2xy = a2 

...(1)

Differentiating equation (1) w.r.t. x, we get

∴  The area of the required triangle is 1 (2x1) (2y1) = 2 (x1 y1) = a2. = 2 60. (a) We have, ...(1) x2 = 4y  ...(2) and y2 = 4x Solving the equation (1) and (2), we get x = 0 and x = 4. Thus, the two parabolas intersect at the points (0, 0) and (4, 4). Now, differentiating equation (1) w.r.t x, we get dy dy x ⇒ = . 2x = 4 dx dx 2 dy  dy  = 0 and = 2. ⇒ dx  ( 4, 4 ) dx  ( 0, 0 ) Thus, the two tangents to the parabola x2 = 4y are (y – 0) = 0 (x – 0) and (y – 4) = 2 (x – 4) ⇒ y = 0 and 2x – y – 4 = 0 Clearly, the two tangents intersect at (2, 0).

dy  y dy 2x + 2y ⋅ 1 = 0 ⇒ =– 1.  dx x dx  ( x1 , y1 ) 1

Equation of the tangent at P (x1, y1) is y y – y1 = – 1 (x – x1) x1 ⇒ xy1 + yx1 = 2x1 y1 x y + ⇒ = 1. 2 x1 2 y1 Hence, intercepts on x-axis and y-axis are 2x1 and 2y1 respectively.

Also, the slope of the given line is = 3 = m2 (say) 4 −3 m1 − m2 y1 π = ⇒1= ∴ tan 12 1 + m1m2 4 1+ y1 ⇒ y1 = 0 or y1 = – 2. 1 The corresponding values of x1 are 0, . 2 1  Hence, the points of contact are (0, 0) and  , − 2  . 2  62. (c) f (x) =

x2 − 4 x

, f (x) = 0 for x = ±2.

4  ∴ f (x) = ±  x −  = f ′(x) x  4  ⇒ ±  x + 2  ≠ 0. x   63. (a) P = x3 –

P ∴ Q2

1 ,Q x3   x − =

= x –

1 x

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

1   x −  + 3 x 1 3  = = x −  + 1  1 x    x −  x −  x x  Clearly, the minimum does not exist.

y = 6x – x2 ⇒



dy  = – 4. dx  ( 5, 5)

dy = 6 – 2x dx

Y –y =



x-intercept, put Y = 0 mx m −1 (a + x) − ax ∴ X = x – m − 2 = x (am − a + mx) (m − 1) a + mx ay Similarly, intercept on y-axis = . m (a + x)

Hence, the equation of the tangent at (5, 5) is y – 5 = – 4 (x – 5) ⇒ 4x + y = 25. 25 ∴ Co-ordinates of A are  , 0  .  4  Also, the equation of the normal at (5, 5) is

x m − 2 (am − a + mx) (X – x) my m −1

68. (c) We have, x = a (t + sin t cos t) and y = a (1 + sin t)2 dx ⇒ = a (1 + cos 2t) dt dy and = 2a cos t (1 + sin t) dt = a (2 cos t + sin 2t) dy dt dy 2 cos t + sin 2t 1 + sin t ∴ = = = dx dt dx 1 + cos 2t cos t

1 y–5= (x – 5) ⇒ x – 4y = – 15. 4 ∴ Co-ordinates of B are (– 15, 0). 1 85 425 ×5= . Thus, the area of the ∆APB = × 2 4 8 2 65. (a) We have, y = x + bx + c dy = 2x + b. ⇒ dx Since the curve touches the line y = x at the point (1, 1) ∴ (2 x + b)](1, 1) = 1 i.e., 2 + b = 1 ⇒ b = – 1. Also, the curve passes through the point (1, 1) ∴ 1 = 1 + b + c i.e., c = – b = 1. dy = 2x – 1. ∴ y = x2 – x + 1 ⇒ dx dy 1 < 0 ⇒ 2x – 1 < 0 ⇒ x < . Now, dx 2 66. (d) sin θ =

5 1 π = , ∴θ = 10 2 6

∴ 2θ =

π 3



t t t cos + sin 1 + tan π t  2 2 2 + = t t = t = tan  4 2  . cos − sin 1 − tan 2 2 2

π t Therefore, the tangent makes an angle + with 4 2 x-axis. 69. (d)

f (x) = x2 (x – 2)2 f ′(x) = 2x (x – 2)2 + 2x2(x – 2) = 4x (x – 2) (x – 1)



f ′(x) ≥ 0 for x ∈ [0, 1] ∪ [2, ∞)

and

f ′(x) ≤ 0 for x ∈ (–∞, 0] ∪ [1, 2]

70. (a) Let P (x1, y1) be the point of intersection of the two curves. We have,

y2 – 2x ⇒ 2y

dy =2 dx 1 = . y1

 dy   ⇒ m1 =     dx   ( x1 , y1 )

and 2xy = k ⇒ x 67. (b) We have, ym = axm – 1 + xm ⇒ mym – 1

dy = a (m – 1) xm – 2 + mxm – 1 dx



dy +y =0 dx

y1  dy   ⇒ m2 =    =– . x 1  dx   ( x1 , y1 )

163



x m − 2 (am − a + mx) dy = . my m −1 dx Equation of tangent: ⇒

Applications of Derivatives

64. (a) We have,

164

Objective Mathematics

Since the two curves intersect at right angles,  1  y  ∴ m1m2 = – 1 ⇒    −  = – 1  y1   x1 

⇒ x1 = 1 2 and hence from y12 = 2x1, we get y1 = 2. Since (x1, y1) also lies on 2xy = k ∴ k2 = 4 x12 y12 = 4 × 1 × 2 = 8.

71. (d) We have, x = acos3θ and y = asin3θ dx dy = – 3acos2θ sin θ and = 3a sin2θ cos θ ⇒ dθ dθ dy d θ dy = = – tan θ. ∴ dx d θ dx Length of the tangent =



2

dy dx

= a sin 3 θ 1 + tan 2 θ tan θ = asin2θ.

Length of the normal =

 dy  y 1+    dx 

 dy  y 1+    dx 

2

= asin3θ 1 + tan 2 θ = atanθ sin2θ dy Length of sub-tangent = y dx =

a sin 3 θ = asin2θ cos θ. tan θ

72. (a) We have, x2y2 = a2 (x2 – a2) ...(1) dy + y2 ⋅ 2x = a2 ⋅ 2x ⇒ x2 ⋅ 2y dx

75. (a) We have xm + n = am – n ⋅ y2n ⇒ (m + n) log x = (m – n) log a + 2n log y Differentiating w.r.t. x, we get 2n dy m+n dy y m+n = ⇒ = ⋅ . y dx x dx x 2n y 2n Sub-tangent = dy =x⋅ m +n dx m

 2n  m ∴ (sub-tangent)m =   ⋅x  m+n  dy  m + n y2 ⋅ Sub-normal =  y  = 2n x  dx  n

2n m+n y ∴ (Sub-normal)n =   ⋅ n  2n  x n

xm+n m+n =   ⋅ m−n n  2n  a ⋅ x = constant xm  ...(2) From (1) and (2), we have (Sub-tangent)m ∝ (sub-normal)n.

2 2 76. (c) We have, x + y = 1 ⇒ 2 x + 2 y dy = 0, 2 2 a b a 2 b 2 dx 2 dy b x =– 2 ∴ dx a y



p = length of the ⊥r from (0, 0) on the tangent dy x −y dx =



a2 − y2 dy ⇒ = . xy dx dy a 2 − y 2 x 2 (a 2 − y 2 ) a 4 = = = 3 dx x x3 x [ from (1) x2 (a2 – y2) = a4] ∴ Sub-normal = y

⇒ The sub-normal varies inversely as the cube of its abscissa.

 dy  1+    dx 

∴ Sub-tangent =

y

dx dy

=

nx ∝ x. m

74. (b) We have, y = a log (x2 – a2) dx dy 2x x2 − a2 =a⋅ 2 ⇒ = . ⇒ 2 dy 2ax dx x −a ∴ Sum of tangent and sub-tangent dx  dy  dx x2 + a2 x2 − a2 +y =y⋅ = y 1 +   ⋅ + y dy  dx  dy 2ax 2ax 2

=

xy ∝ product of the co-ordinates a

2

p′ = length of normal = y

 dy  1+    dx 

2

 b2 x2  dy   − y = y − 2 − y ∴ p × p′ = y  x − a y dx    



73. (c) We have, xmyn = am + n ⇒ m log x + n log y = (m + n) log a m n dy dx + ∴ =0 ⇒ = – nx x y dx dy my

...(1)

− (b 2 x 2 + a 2 y 2 ) b2a 2 =– 2 a a2



=



= – b2 = constant.

∴ p′ ∝

1 . p

77. (a) Let (x1, y1) be the point of intersection. Solving the two equations, we get 2 2 2 2 x12 = a (ab − b ) and y12 = b (a − ab)  2 2 2 a − b2 (a − b )

...(1)

Also, x2b2 + y2a2 = a2b2 ⇒

b 2 x1 dy  = – = m1 (say) a 2 y1 dx  ( x1 , y1 )

and x2 + y2 = ab ⇒

x dy  = – 1 = m2 (say)  y1 dx  ( x1 , y1 )

= tan

–1

 (a 2 − b 2 ) x1 y1   2 2 2 2   a y1 + b x1 

a 2b 2 (ab − b 2 )(a 2 − ab) = tan–1  a − b  .    ab  a 2b 2 78. (d) f (x) = 2x2 – ln | x |

165

 m − m2  θ = tan–1   1   1 + m1m2   log 3 − log 5  = tan–1    .  1 + log 3 log 5 



81. (a) We are given, sub-tangent = sub-normal ⇒

y dy dx

=

y

dy dx



dy = ± 1. dx

= tan–1

1 =0 x ⇒ 4x | x | = 1 ⇒ x = 1/2. f ′(x) = 4x –

79. (b) Let (x1, y1) be the point of intersection where the curves cut orthogonally. Now, x2 + y2 = 2a2 x dy   ⇒ = – 1 = m1 (say)  y1 dx  ( x1 , y1 ) and 2y2 – x2 = a2 

x1 dy  ⇒ = = m2 (say)  2 y1 dx  ( x1 , y1 )

 dy  1+    dx 

So, length of the normal = y

=y ⋅

2 =

2

2

(ordinate).

82. (b) We have, y = 4ax 2

dy dy = 4a i.e., = dx dx y y Sub-tangent = dy = 2a y dx ∴ Sub-tangent : Abscissa = ⇒ 2y

2a . y =

y2 4ax = = 2x. 2a 2a

2x : x = 2 : 1.

83. (b) f (x) = cos (2πx) + {x} f (x) is periodic with period 1 and f (x) has one local maximum in x ∈ [0, 1].

Now, the product of the slopes at (x1, y1) is 84. (a) We have, y2 = 4ax x12 2a dy dy = – 1 ⇒ x12 = 2 y12  ...(1) = m1 × m2 = – = 4a i.e., = . ⇒ 2y 2 y12 y dx dx Also, (x1, y1) lies on the curve, x2 + y2 = 2a2 y y y2 ∴ x12 + y12 = 2a2  ...(2) Sub-tangent = dy = 2a = 2a y dx From (1) and (2) 2a dy Sub-normal = y =y × = 2a. 2 2 2 y ⇒ 3 y1 = 2a ⇒ y1 = ± a and hence dx 3 y2 4 , y, 2a are in G.P. Clearly, x1 = ± a. 2a 3 ...(1) 85. (b) We have, y = ax3 + bx2 + cx + 5  Thus, the points of intersection of the two curves dy are = 3ax2 + 2bx + c  ...(2) ⇒ dx  4 2  Since the curve touches x-axis at A (– 2, 0) a  .  ± a, ± 3   3 dy  ∴ = 0 ⇒ 12a – 4b + c = 0 ...(3) 80. (a) We have, dx  ( −2, 0 ) ...(1) y = 3x Also, the curve passes through the point A (– 2, 0) x and y = 5 ...(2) ∴ – 8a + 4b – 2c + 5 = 0  ...(4) Clearly, the two curves intersect at the point (0, 1). dy  =3 Also, it is given that Differentiating (1) w.r.t. x, we get dx  x =0 dy  dy ⇒ 3a (0)2 + 2b (0) + c = 3 i.e., c = 3. = 3x log 3, ∴ dx  = log 3 = m1 (say)  ( 0, 1) dx Solving (3) and (4), we get Differentiating (2) w.r.t. x, we get dy  dy = 5x log 5, ∴ dx  = log 5 = m2 (say).  ( 0, 1) dx ∴ The angle between the two curves is given by

a=–

1 3 and b = – . 2 4

Therefore, a = –

1 3 , b=– and c = 3. 2 4

Applications of Derivatives

Hence, the angle between the two curves is  x1 b 2 x1   − 2  y a y θ = tan–1 m1 − m2 = tan–1  1 2 2 1   b x1  1 + m1m2  1+ 2 2  a y1  

166

86. (b), (c)  We have,

Objective Mathematics

5x – 10x + x + 2y + 6 = 0 dy =0 ⇒ 25x4 – 30x2 + 1 + 2 dx dy 1 ⇒ =– (25x4 – 30x2 + 1). dx 2 dy  1 ∴ dx  =– .  ( 0 , − 3) 2 5

3

...(1)

Equation of the normal to the curve at P (0, – 3) is (y + 3) = 2 (x – 0) i.e., y = 2x – 3 ...(2) Substituting the value of y from (2) in (1), we get 5x5 – 10x3 + 5x = 0 ⇒ 5x (x2 – 1)2 = 0 ⇒ x = 0, 1, – 1. When x = 0, y = – 3; when x = 1, y = – 1 and when x = – 1, y = – 5. ∴ The normal at P (0, – 3) meets the curve again at the points (1, – 1) and (– 1, – 5). 87. (d) For x £ 1 2  1  29  f ′ (x) = 3x2 – + 10 = 3  x −  +    > 0 3 3  

=

( f' (3) )

2

− ( f' (1) )

2

2  1 − 1/3  =  + 1 − 3  2 

(

+ f' (3) − f' (2)

)

=

1 4 +1− 3 = − 3 3 3

4−3 3 . 3 89. (a), (d)  We have, xy = (c + x)2 dy dy 2(c + x) − y ⇒ x + y = 2 (c + x) ⇒ = dx dx x =



dy = dx

(c + x ) 2 ( x2 − c2 ) x = . x2 x

2 (c + x ) −

At the point, the normal cuts off numerically, equal  dx  intercepts on axes, slope of normal = –   = ± 1  dy  − x2 ⇒ 2 =±1 x − c2 ⇒ – x2 = x2 – c2 or – x2 = – (x2 – c2) c2 or c2 = 0 2 c ,  c2 = 0 is not possible. ⇒ x = ± 2

⇒ x2 =

 o f (x) is an increasing function for x £ 1, for x > 1, S f′ (x) = – 1 So, f (x) is decreasing functon for x > 1. 90. (a) We have  x3 – 3xy2 + 2 = 0 Now f (x) will have greatest value at x = 1. If and  3x2y – y3 – 2 = 0

...(1) ...(2)

Diff. (1) and (2) w.r.t. x, we obtain  dy   dy  x2 − y 2 −2 xy   = and  dx  = 2 dx 2 xy x − y2  C1  C2 Since m1 × m2 = –1, therefore the two curves cut at right angles. 91. (a), (b)  We have, y =

x

∫ | t | dt .

dy = | x |. dx Since the tangent is parallel to the line y = 2x ∴ | x | = 2 ⇒ x = ± 2. 0

lim f ( x) ≤ f (1) ⇒ lim f (1 + h) ≤ 5



x → 1+

h→0

− 1(1 + h) + log 2 (b 2 − 2) ≤ 5 ⇒ lim h→0 ⇒ ⇒ ⇒ ⇒

– 2 + log2 (b2 – 2) ≤ 5 ⇒ log2(b2 – 2) ≤ 7 b2 – 2 ≤ 128 b2 ≤ 130 but b2 > 2 2 < b2 ≤ 130

∴ b ∈ [− 130 , − 2 ) ∪ ( 2 , 130 ] . 88. (d) We have, f ′(1) = tan π/6 =

3 and f ′(3) = tan π/4 = 1.

f ′ (2) = tan π/3 =



3

1

f' ( x) f'' ( x) dx + 3



1 , 3



3

2

f'' ( x) dx

 ( f' ( x) )2  3  + [ f' ( x)]2 =  2  1

Differentiating w.r.t. x, we get

When x = 2, y =

2

∫ | t | dt 0

When x = – 2, y =

−2

2

=

∫ | t | dt 0

∫ t dt

= 2.

0

0

=–

∫ −t dt

=–2

−2

The tangent at (2, 2) is y – 2 = 2 (x – 2) i.e., 2x – y – 2 = 0 ∴ x-intercept = 1. (Putting y = 0) Also, the tangent at (– 2, – 2) is y + 2 = 2 (x + 2) i.e., 2x – y + 2 = 0 ∴ x-intercept = – 1 (Putting y = 0) 92. (d) We have, ⇒

x2 y 2 + = 1 4 16

4x 2 x 2 y dy dy =0 ⇒ =– . + y 4 16 dx dx

...(1)

 2 −8   2 8   − 2 8   − 2 −8  , , , ,  ,  ,  ,  .  5 5  5 5  5 5  5 5 93. (c) We have, y = loge x



dy   y (0) dy = 1 ⋅  + dx  x =0 dx  1

 log1 + 0 = y (0) = 1. x =0 

∴ The equation of the normal at (0, 1) is ( y – 1) = – 1 (x – 0) i.e., x + y = 1. 98. (c) We have, f (x) = 2x3 – 9ax2 + 12a2x + 1 ∴ f ′(x) = 6x2 – 18ax + 12a2 = 0 ⇒ 6[x2 – 3ax + 2a2] = 0 ⇒ x2 – 3ax + 2a2 = 0 ⇒ x2 – 2ax – ax + 2a2 = 0 ⇒ x (x – 2a) – a (x – 2a) = 0 ⇒ (x – a) (x – 2a) = 0 ⇒ x = a,  x = 2a Now, f ′′(x) = 12x – 18a

dy 1 1 dy  ⇒ = ⇒ = .  dx x 2 dx  ( 2, 0 ) The equation of the tangent at (2, 0) is 1 (x – 2) i.e., 2y = x – 2. (y – 0) = 2 1 ∴ Area of the ∆OAB = (2) (1) = 1 square 2 unit. dy = 21x6 + 5, dx which is positive ∨ x ∈ R. Therefore, the tangent makes an acute angle with x-axis.

94. (c) We have,

95. (c) We have, f (x) = cot–1 x + x ⇒ f ′(x) = – 

1 +1 1 + x2

x2 . Clearly, f ′(x) > 0 for all x 1 + x2 therefore f (x) increases in (–∞, ∞).

=

96. (b) We have, f (x) = (x + 1)1/3 – (x – 1)1/3  1 1 1 − ∴ f ′(x) =  . 3  ( x + 1) 2/3 ( x − 1) 2/3  Clearly, f ′ (x) does not exists at x = ± 1. Now, f ′ (x) = 0 ⇒ (x – 1)2/3 = (x + 1)2/3 ⇒ x = 0 Clearly f ′ (x) ≠ 0 for any other value of x ∈ [0, 1]. The value of f (x) at x = 0 is 2. Hence, the greatest value of f (x) is 2.

∴ f ′′(a) = 12a – 18a = –6a < 0 ∴  f (x) will be maximum at x = a i.e.,,

p =a

Also, f ′′(2a) = 24a – 18a = 6a ∴  f (x) will be minimum at x = 2a i.e.,,



q = 2a

Given: p2 = q ⇒ a2 = 2a ⇒ a = 2. 99. (b), (d)  We have, 4x2 + 9y2 = 1

...(1)

Differentiating w.r.t. x, we get

4x dy dy =0 ⇒ =– . 9y dx dx The tangent at point (x, y) will be parallel to the the line 4x 8 = i.e., x = – 2y. 8x = 9y if – 9y 9 Subsituting x = – 2y in (1), we get 8x + 18y

4 (– 2y)2 + 9y2 = 1 or 25y2 = 1 ⇒ y = ±

1 . 5

Thus, the points where the tangents are parallel to  2 1 2 1 8x = 9y are  − ,  and  , −  .  5 5 5 5 100. (d) Slope of normal to y = f (x) at (3, 4) is Thus,

−1  3π  = tan    4  f'(3)

 ⇒ f ′ (3) = 1.

−1 . f'(3)

π π π = – 1. = tan  +  = – cot 4 2 4

167

4x dy =– = ± ⇒ y = 4x, – 4x. y dx 2 4 Subtituting in (1), we get x2 = i.e., x = ± 5 5 ∴ The points on the curve, where the tangent is equally inclined to the axes, are ∴

97. (b) We have, y = (1 + x)y + sin–1 (sin2x) dy ⇒ dx dy  y  2 sin x cos x + log (1 + x)  +  = (1 + x)y  1 1 + x dx   (1 − sin 4 x) 2

Applications of Derivatives

Since the tangent is equally inclined to the axes,

168

1 . a ⇒ Slope of the tangent at (1, 1) is a

101. (a) Slope of the normal at (1, 1) is –

Objective Mathematics

i.e.,

dy  = a dx  (1, 1)

⇒ 1 = e–x/a ⇒ –

...(1)

dy ∝y We are given that dx dy = ky, where k is some constant dx

dy = k dx y log | y | = kx + c, where c is a constant | y | = ekx + c y = ± ec ekx = Aekx, where A is a constant. Since the curve passes through (1, 1), therefore 1 = Aek ⇒ A = e– k Therefore, y = e– k ⋅ ekx = ek (x – 1) dy dy  = kek (x – 1) ⇒ =k dx dx  (1, 1) ⇒ a = k [Using (1)] Thus, the required curve is y = ea (x – 1). ⇒

...(1)

6y = ± 1 x2

...(2)

( any line making equal intercepts on axes will have its slope as 1 or – 1) Now from (2), we have x2 x2 or y = y =– 6 6 Solving these with the equation (1), we get the 8 8 points  4,  ,  4, −  .  3  3 104. (d) We have, y = be–x/a dy b –x/a =– e dx a x y Since the line + = 1 touches (1) a b b −1 a b –x/a b –x/a ∴ =– e ⇒– a =– e 1b a a ⇒

105. (d) We have, x = t2 – 1, y = t2 – t dy dy = 2t and = 2t – 1 ⇒ dt dx dy dt dy 2t − 1 ∴ = = . dx dt dx 2t Since the tangent is parallel to x-axis, dy 1 ∴ =0 ⇒t= . dx 2 106. (d) We have, x = 3 cos θ and y = 3 sin θ dx dy = – 3 sin θ and = 3 cos θ ⇒ dθ dθ dy d θ dy 3 cos θ cos θ = = =– . ∴ dx d θ dx −3 sin θ sin θ

On differentiating (1) w.r.t. ‘x’, we get 2x + x

(2 x + y ) dy dy dy =0⇒ =– + y + 2y ( x + 2 y) dx dx dx

y − y ( x + 2 y) Length of subtangent = dy = 2x + y dx ∴  Length of subtangent at (1, –3) = 15. 108. (c) We have, y – ex y + x = 0. Taking log on both sides, we get log (x + y) = log ex y = xy log e = xy 1  dy  dy ⇒ +y  + 1 = x ( x + y )  dx  dx

Differentiating w.r.t. x, we get x2 dy dy 18y = 3x2 ⇒ = . 6y dx dx Slope of the normal = –

∴ y = be0 = b. Hence, the required point is (0, b).

Since the tangent is parallel to x-axis, dy π = 0 ⇒ cos θ = 0 ⇒ θ = . ∴ dx 2 107. (c) We have, x2 + xy + y2 = 7 ...(1)

1 102. (b), (c) We have xy = 1 ⇒ y = x dy 1 ∴ =– 2 . x dx ∴  The slope of the normal = x2. If ax + by + c = 0 is normal to the curve xy = 1 a a     ∴  – >0 then x2 = – b b ⇒ a > 0, b < 0 or a < 0, b > 0. 103. (a), (c) We have, 9y2 = x3 

x = 0 i.e., x = 0. a

...(1)



 dy  1 1 − x = y –  dx  y + x y+x 



 dy  1 1 − x = y –  dx  y + x y +x 

y ( y + x) − 1 dy = 1 − x( y + x) dx Since the curve has a vertical tangent, dy = ∞ ⇒ 1 – x(y + x) = 0 ∴ dx which is satisfied by the point (1, 0). ⇒

109. (a) We have, y = ax3 + bx2 + cx dy = 3ax2 + 2bx + c dx dy  = c = tan 45º = 1 (Given) ⇒ c = 1. ∴ dx  ( 0, 0 ) ⇒

dy  = 3a + 2b + c = 0 dx  (1, 0 )

 ( x-axis is tangent at (1, 0)) ⇒ 3a + 2b + 1 = 0 which is true if a = 1, b = – 2. Hence, a = 1, b = – 2, c = 1. π π π – cos , x > 0, which x x x remains finite and unique for (0, 1]. Hence f (x) is differentiable and continuous in (0, 1]. π f (0 + h) = lim h sin = 0 = f (0), Also lim h→0 h→0 h ∴ f (x) is continuous at x = 0 also. Also, f (0) = 0 = f (1). Hence, Rolle’s theorem is applicable in [0, 1]. Consider the intervals  1 1   r + 1 , r  , (r = 1, 2, 3, ...).  

110. (d) We have, f ′ (x) = sin

 learly, Rolle’s theorem is applicable in each these C intervals. Hence, by Rolle’s theorem f ′ (cr) = 0,  1 1 , . where cr ∈   r +1 r   1 , As   r +1

1  ⊂ [0, 1], cr ∈ [0, 1]. r

Hence, f ′ (x) = 0 at cr ; r = 1, 2, 3, ... Thus, f ′ (x) = 0 at infinite number of points. 111. (b) Let h (x) = f (x) – 2g (x), x ∈ [0, 1]. ⇒ h′ (x) = f ′ (x) – 2g′ (x) Also, h (0) = f (0) – 2g (0) = 2 – 2 ⋅ 0 = 2. h (1) = f (1) – 2g (1) = 6 – 2 ⋅ 2 = 2. ∴ h (0) = h (1). Since f (x) and g (x) are differentiable in [0, 1], h (x) is also differentiable in [0, 1]. Hence h (x) is also continuous in [0, 1]. So, all the conditions of Rolle’s theorem are satisfied. Hence, there exists a point c, 0 < c < 1 for which h′ (c) = 0. ∴ f ′ (c) – 2g′ (c) = 0 i.e., f ′ (c) = 2g′ (c). 112. (c) Let f (x) = anxn + an – 1xn–1 + ... + a2x2 + a1x + a0, which is a polynomail function in x of degree n. Hence f (x) is continuous and differentiable for all x. Let α < β. We are given, f (α) = 0 = f (β). By Rolle’s theorem, f ′ (c) = 0 for some value c, α < c < β. Hence the equation f ′ (x) = nanxn–1 + (n – 1) an – 1xn–2 + ... + a1 = 0 has atleast one root between α and β. 113. (a) Let f (x) = ax3 + bx2 + cx, x ∈ [0, 1]. ∴ f ′ (x) = 3ax2 + 2bx + c.

 ince f (x) is a polynomial function of x, it is conS tinuous and differentiable for all x ∈ [0, 1]. Also, f (0) = 0; f (1) = a + b + c = 0. ∴ f (0) = f (1). Applying Rolle’s theorem, f ′ (k) = 0 for atleast one value k, 0 < k < 1. Hence k is a root of the equation 3ax2 + 2bx + c = 0, where 0 < k < 1. 114. (c) Let f (x) = ax2 + bx + c. Then f (α) = 0 = f (β). Also, f (x) is continuous and differentiable in [α, β] as it is a polynomial function of x. Hence, by Rolle’s theorem, there exists a k in (α, β), such that b f ′ (k) = 0 ⇒ 2ak + b = 0 ⇒ k = – . 2a b ∴ α < – < β. 2a n 115. (a) Let f (x) = anx + an – 1xn – 1 + ... + a1x. Then f (α) = 0 (Given). Also f (0) = 0. Moreover, f (x) is continuous and differentiable in [0, α] as it is a polynomial function of x. Hence, by Rolle’s theorem, there exists a ‘c’ in (0, α) such that f ′ (x) = 0 for x = c i.e., nanxn – 1 + (n – 1) an – 1xn – 2 + ... + 2a2x + a1 = 0. sin a sin a sin b 116. (a) Here f (a) = cos a cos a cos b = 0. tan a tan a tan b Also f (b) = 0. Moreover, as sin x, cos x and tan x are continuous and π differentiable in (a, b) for 0 < a < b < 2 , therefore f (x) is also continuous and differentiable in [a, b]. Hence, by Rolle’s theorem, there exists a real number c in (a, b) such that f ′ (c) = 0. 117. (b) Let

x n −1 x2 x n +1 xn + a1 + a2 + ... + an −1 + an x . 2 n +1 n n −1 Then f (x) is continuous and differentiable in [0, 1], as it is a polynomial function of x. Also, f (0) = 0 a a a a and f (1) = 0 + 1 + 2 + ... + n −1 + an = 0. 2 n +1 n n −1 (Given) Hence, by Rolle’s theorem, there exists atleast one real number c ∈ (0, 1) such that f ′ (c) = 0 i.e., c is a root of the equation a0xn + a1xn – 1 + ... + an – 1x + an = 0.

f (x) = a0

118. (c) Let f (x) = (x – 3) log x Then, f (1) = – 2 log 1 = 0 and f (3) = (3 – 3) log 3 = 0. As, (x – 3) and log x are continuous and differentiable in [1, 3], therefore (x – 3) log x = f (x) is also continuous and differentiable in [1, 3]. Hence, by Rolle’s theorem, there exists a value of x in (1, 3) such that

169

Also,

Applications of Derivatives



170



f ′ (x) = 0 ⇒ log x + (x – 3)

Objective Mathematics

⇒ x log x = 3 – x.

Now, 0 < c < 2,

1 =0 x

119. (a) Let α, β (α < β) be any two real roots of f (x) = e – sin x. Then, f (α) = 0 = f (β) Moreover, f (x) is continuous and differentiable for x ∈ [α, β]. Hence, from Rolle’s theorem, there exists atleast one x in (α, β) such that f ′ (x) = 0 ⇒ – e– x – cos x = 0 ⇒ – e–x (1 + ex cos x) = 0 ⇒ ex cos x = – 1.

and

–x

120. (c) As f (x) is differentiable in [2, 5], therefore, it is also continuous in [2, 5]. Hence, by mean value theorem, there exists a real number c in (2, 5) such that 1 1 − f (5) − f (2) 1 ⇒ f ′ (c) = 2 5 = . f ′ (c) = 5−2 10 3 f (a) g (a)

121. (a) Let h (x) =

f ( x) = f (a) g (x) – g (a) f (x). g ( x)

Then, h' (x) = f (a) g′ (x) – g (a) f ′ (x) =

f (a) g (a)

f ' ( x) . g' ( x)

Since f (x) and g (x) are continuous in [a, b] and differentiable in (a, b), therefore, h (x) is also continuous in [a, b] and differentiable in (a, b). So, by Mean Value Theorem, there exists atleast one real number c, a < c < b for which h (b) − h (a ) , h′ (c) = b−a ∴ h (b) – h (a) = (b – a) h′ (c) Here h (a) =

f (a) g (a)

∴ From (1) :

= (b – a)



f (a) f (a) = 0, h (b) = g (a) g (a)

f (a) g (a)

...(1) f (b) . g (b)

f (b) = (b – a) h′ (c) g (b)

1 for all x.  1 + x2

 rom (1), it follows that f (x) is differentiable at all F x, therefore f (x) is also continuous at all x. ∴ By mean value theorem in [0, 2] f (2) − f (0) 1 = f ′ (c) = where 0 < c < 2 2−0 1 + c2 f ( 2) − 0 1 2 = or f (2) =  2 1 + c2 1 + c2

2 > 0.4 1 + c2

...(4)

From (2), (3) and (4) it follows that 0.4 < f (2) < 2. 123. (a) Let f ( y) =

y

∫ (1 + cos

8

x)(ax 2 + bx + c) dx

0

⇒ f ′ (y) = (1 + cos8y) (ay2 + by + c)

...(1)

1

∫ (1 + cos

Now, f (1) =

8

x)(ax 2 + bx + c) dx = 0

0

and  f (2) =

2

∫ (1 + cos

8

x)(ax 2 + bx + c) dx = 0

0

Also, f (0) = 0. ∴ f (0) = f (1) = f (2). Now by Rolle’s theorem for f (x) in [0, 1]. f ′ (α) = 0, for atleast one α, 0 < α < 1 and by Rolle’s theorem for f (x) in [1, 2], f ′ (β) = 0, for atleast one β, 1 < β < 2. From (1),  f ′ (α) = 0 ⇒ (1 + cos8α) (aα2 + bα + 2) = 0. But 1 + cos8α ≠ 0, ∴ aα2 + bα + c = 0, i.e., α is a root of the equation ax2 + bx + c = 0. Similarly f ′ (β) = 0 ⇒ aβ2 + bβ + c = 0, i.e., β is a root of the equation ax2 + bx + c = 0. But the equation ax2 + bx + c = 0, being a quadratic equation, cannot have more than two roots. ∴ The equation ax2 + bx + c = 0 has one root α between 0 and 1 and other root β between 1 and 2. 124. (b) We have, f (x) = log sin x 1 cos x = cot x ⇒ f ′ (x) = sin x

applicable. ...(1)

⇒ f ′ (x) > 0, for all x ( 1 + x2 > 0)



2 2 2 > = = 0.4 1 + c2 1 + 22 5

Clearly f (x) is continuous and differentiable  π 5π  in  ,  . Hence mean value theorem is 6 6 

f (a ) f' (c) . g (a ) g' (c)

122. (c) We have, f ′ (x) =

or

2 2 2 < or < 2 ...(3) 1 + c2 1 + 02 1 + c2





...(2)

 π 5π  ∴ There exists a real number c in  ,  such 6 6  that



π  5π  f  − f   6  6  f ′ (c) = 5π π − 6 6 log sin

⇒ cot c =

5π π − log sin 6 6 2 π 3

of the polynomial f ′ (x) = 4ax3 + 3bx2 + 2cx + d lying between 0 and 3.

127. (b) By mean value theorem, there exists a real number c ∈ (2, 4) such that f (4) − f (2) f (4) + 4 f ′ (c) = ⇒ f ′ (c) = 4−2 2 Since f ′ (x) ≥ 6 ∨ x ∈ [2, 4] f (4) + 4 ≥6 ∴ f ′ (c) ≥ 6 ⇒ 2 ⇒ f (4) + 4 ≥ 12

⇒ f (4) ≥ 8.

128. (a) We have, f (x) = 2x – log | x |, x ≠ 0 Case I. When x < 0, f (x) = 2x – log | x | = 2x – log (– x) 1 1 4x2 − 1 ∴ f ′ (x) = 4x – (– 1) = 4x – = . (−x) x x Case II. When  x > 0, f (x) = 2x2 – log | x | = 2x2 – log x 1 ∴ f ′ (x) = 4x – x Thus, when 1 4x2 − 1  x < 0 or x > 0, f ′ (x) = 4x – = x x 4x2 − 1  : Sign scheme for x 

2

Thus, f (x) is an increasing function in the interval

 1  1   − , 0  ∪  , ∞  . 2 2

129. (b) Let f (x) = log (1 + x) – x 1 x –1=– ⇒ f ′ (x) = 1+ x 1+ x ⇒ f ′ (x) < 0, for x > 0 ⇒ f (x) is decreasing for x > 0 ⇒ f (x) < f (0), for x > 0

⇒ 1 + x log (x +

x2 + 1 ) ≥

131. (b) Let f (x) = cos x – 1 +

⇒ f ′ (x) = – sin x + x Let Then

1 + x2

≥ 0,

1 + x 2 , for x ≥ 0.

x2 2

g (x) = x – sin x. g′ (x) = 1 – cos x.

Clearly, g′ (x) > 0, for 0 < x
0, for 0 < x < 2

π 2

g (x) > g (0), for 0 < x
f (0), for 0 < x
0, for 0 < x < 2 2 x2 π i.e., cos x > 1 – , for 0 < x < . 2 2 132. (a) We have, f (x) = x – log (1 + x), x > – 1 x (1 + x) 1 x = = ⇒ f ′ (x) = 1 – (1 + x) 2 1+ x 1+ x cos x – 1 +

Sign scheme for

x (1 + x) : (1 + x) 2

f ′ (x) > 0 if x < – 1 or x > 0. But x > – 1. ∴ f ′ (x) > 0 if x > 0. Thus, f (x) is increasing in the interval (0, ∞).

171

Applications of Derivatives

⇒ log (1 + x) – x < 0, for x > 0 1 1 log − log i.e., log (1 + x) < x, for x > 0. 2 2 ⇒ cot c = =0 2π 130. (c) Let f (x) = 1 + x log (x + x 2 + 1 ) – 1 + x 2 3 1  π 5π  π ⇒ f ′ (x) = 1 ⋅ log (x + x 2 + 1 ) + x ⋅ ⇒c= ∈  , . x + x2 + 1 2 6 6  125. (d) f (x) is continuous in the interval [– 1, 1], but f (x)  x  x  × 1 + − is not differentiable at x = 0. Hence mean value  x2 + 1  1 + x2 theorem is not applicable. So, no c can be found. x x − 126. (c) Let f (x) = ax4 + bx3 + cx2 + dx. = log (x + x 2 + 1 ) + x2 + 1 x2 + 1 Then, f (0) = 0 and f (3) = 81a + 27b + 9c + 3d = log (x + x 2 + 1 ). = 3 (27a + 9b + 3c + d) Clearly, f ′ (x) ≥ 0, for x ≥ 0 = 0 ( 27a + 9b + 3c + d = 0) ⇒ f (x) is increasing for x ≥ 0 Therefore 0 and 3 are roots of the polynomial f (x). ⇒ f (x) ≥ f (0), for x ≥ 0 So, by Rolle’s theorem, there exists atleast one root

172

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

Objective Mathematics

⇒ f ′ (x) =

x , x > 0, x ≠ 1. log x

log x −1 (log x) 2

134. (b) We have, f (x) = x 1 ⇒ f ′ (x) = 2 (1 – log x) x1/x. x f ′ (x) > 0 if 1 – log x > 0, i.e., log x < 1 ⇒ x < e.  ∴  f (x) is increasing in the interval (–∞, e). 1/x

135. (b) Let f (x) = log x – tan–1x 1 + x2 − x 1 1 − = x (1 + x 2 ) x 1 + x2 2



1 3   x −  + 2 4 > 0 for all x > 0. = x (1 + x 2 )

 ence, f (x) is an increasing function in the interval H (0, ∞). 136. (a) We have, f (x) = x +

1 x

∴ f ′(x) = 1 –

1 =0 x2

Now, f ′′(x) =

2 .  ∴  f ′′ (1) = 2 > 0 x3

⇒ x=±1

 herefore, f (x) will be minimum at x = 1 and the T minimum value is 1 = 1 + 1 = 2. f (1) = 1 + 1 137. (b) We have, f (x) = sin x x x cos x − sin x cos x ( x − tan x) = 2 x x2 π But tan x > x and cos x > 0, for 0 < x < 2 π  ∴ f ′ (x) < 0 in the interval  0,  .  2 π Thus, f (x) is decreasing in  0,  .  2 ⇒ f ′ (x) =

 sin x ,  138. (b), (c)  Let f (x) =  x 1, Then f ′ (x) =

cos x ( x − tan x) π < 0 if x ∈  0,  . x2  2

π   tan x > x and cos x > 0 when 0 < x <  2 

f ′ (x) < 0 if log x – 1 < 0, i.e., log x < 1 ⇒ x < e. But x > 0 and x ≠ 1 ∴ f ′ (x) < 0 if x > 0, x ≠ 1, x < e. Thus, f (x) is decreasing for (0, e) \ {1}.

⇒ f ′ (x) =

=

x≠0 x=0

x cos x − sin x x2

 π ∴ f (x) is decreasing in  0,  . Since  2 π ,  0< x < 2 π 2 sin x < < 1. ∴ f   < f (x) < f (0) ⇒ 2 π x   139. (a), (c)  We know that π  ...(1) 2  π Since cos x is decreasing in  0,  , cos (sin x) >  2 cos x. π π , ∴ 0 < cos x < 1 < Also, since 0 < x < 2 2 sin x < x if 0 < x
cos x > sin (cos x) if 0 < x < 140. (b), (c)  Let f (x) = 1 + xp – (1 + x)p ⇒ f ′ (x) = pxp – 1 – p (1 + x)p – 1

 1 = p  1 − 1− p 1− p  ( ) x x + 1  

≥ 0 if p ≥ 0, 1 – p ≥ 0, x > 0 ∴ f (x) increases when x > 0 and 0 ≤ p ≤ 1. Since x > 0, f (x) > f (0) ∴ 1 + xp – (1 + x)p > 0    ( f (0) = 0) i.e., 1 + xp > (1 + x)p if 0 ≤ p ≤ 1 and x > 0. 141. (a) Let f (x) = ax2 +

b – c; x > 0; a, b > 0 x

⇒ f ′ (x) = 2ax –

b 2a  b  = 2  x3 −  . 2  x x 2a 

 b  b ,∴x=    2a  2a

f ′ (x) = 0 gives x3 = Since ax2 +

1/ 3

b ≥ c, ∴ f (x) ≥ 0 for all x > 0. x

  b 1 / 3  ∴ f     ≥ 0   2a   ⇒ a  b   2a   

2/3

1/ 3

 2a  + b   b 

– c≥0

⇒ a ⋅  b  + b − c  b      2a 2a

1/ 3

3b  b  ≥ c⋅    2a  2 2 ∴ 27ab ≥ 4c3.

27b3 b ≥ c3 ⋅ 8 2a



> 0.

1/ 3



≥0

π . 2

b ax − b = . x2 x2 b b =0 ⇒x =   2 x a

⇒ y = 9/2 ∴ x =

1/ 2



f ′ (x) = 0 ⇒ a –

.

b ≥ c; ∴ f (x) ≥ 0 for all x > 0 But ax + x

1/ 2

dy = x2e–x (– 1) + 2xe–x = e–x x (2 – x) dx y increases in the interval where ⇒

a + b    b

1/ 2

c2 . 4 2x 143. (b) We have, f (x) = log x – 2+ x



=



– c≥0

⇒ 2 ab ≥ c ⇒ ab ≥

⇒ f ′ (x) =

x2 + 4 ( x 2 + 4) x = > 0, for x > 0 x (2 + x) 2 x 2 ( x + 2) 2

∴ f (x) is increasing for x > 0. 144. (c) We have, y = x3 (x – 2)2 ⇒

dy = 3x2 (x – 2)2 + 2x3 (x – 2) dx



= x2 (x – 2) (5x – 6)

dy 6 > 0 if (x – 2) (5x – 6) > 0 i.e., x < or x > 2. dx 5 6 or x > 2. 5 145. (c) We have, f (x) = – 2x3 + 21x2 – 60x + 41 ∴ f (x) is increasing for x
f (1) as x < 1. Since f (1) = 0. ∴ f (x) > 0 for x ∈ (– ∞, 1).

dy >0 dx

⇒ e–x ⋅ x (2 – x) > 0

⇒ x (2 – x) > 0 ⇒ x (x – 2) < 0 ⇒ 0 < x < 2. 149. (a) Let f (x) = eax + e–ax a (e 2 ax − 1) . e ax f (x) is a decreasing function if f ′ (x) < 0 a (e 2 ax − 1) ⇒ < 0 ⇒ e2ax – 1 < 0 ⇒ e2ax < 1 e ax ⇒ 2ax < 0 ⇒ x > 0 ( a < 0) Thus, f (x) is a decreasing function for x > 0. ⇒ f ′ (x) = a (eax – e–ax) =

1  2 (2 + x) − 2 x  ( x + 2) 2 − 4 x −  = 2 x  (2 + x) x (2 + x) 2 

146. (b) Let f (x) = x tan x

y2 81 9 = = 18 4 × 18 8

9 9 ∴ Required point is  ,  8 2 148. (d) We have, y = x2e–x

 b 1/ 2  ∴ f  a   ≥ 0   b ⇒ a    a

dx   dy dy dx = 18 ⇒ 2y⋅ 2 = 18  = 2  dt  dt dt  dt

150. (a), (d) We have, y = 2x + cot–1x + log [ 1 + x 2 – x]

 ⇒

dy  x  1 1 =2– + × − 1 2 2 2 dx 1+ x 1+ x − x  1+ x 



=



Now,

2x2 + 1 1 (2 x 2 + 1) − 1 + x 2 − = . 2 2 1+ x 1+ x 1 + x2 dy ≥ 0 dx

⇒ (2x2 + 1) – 1 + x 2 ≥ 0 ⇒ (2x2 + 1)2 ≥ 1 + x2 ⇒ 4x4 + 3x2 ≥ 0, which is true for all real values of x. ∴ y increases for all real values of x. 151. (d) We have, f (x) = (x + 2) e–x ⇒ f ′ (x) = – (x + 2) e–x + e–x = – e–x (1 + x).

f (x) is increasing if f ′ (x) > 0.

⇒ – e–x (1 + x) > 0 ⇒ 1 + x < 0 i.e., x < – 1.

Also, f (x) is decreasing if f ′ (x) < 0 π   ⇒ – e–x (1 + x) < 0 ⇒ 1 + x > 0 i.e., x > – 1. ⇒ f ′ (x) = tan x ⋅ 1 + x sec2x > 0, for x ∈  0,  .  2 Hence, f (x) decreases in (– 1, ∞) and increases in  π (– ∞, – 1). 0 , So, f (x) is increasing for x ∈  .  2 ln ( π + x) 152. (b) We have, f (x) = π . ∴ f (β) > f (α) Since, 0 < α < β < ln (e + x) 2 1 1 α tan β ln (e + x) ⋅ − ln ( π + x) ⇒ β tan β > α tan α i.e., > . π+ x e+ x ⇒ f ′ (x) = β tan α 2 [ln (e + x)]

173

∴ 2y

2

Applications of Derivatives

⇒ f ′ (x) = a –

147. (d) Given curve is y2 = 18x

b – c; x > 0; a, b > 0 x

142. (b) Let f (x) = ax +

174

Objective Mathematics

(e + x)ln (e + x) − ( π + x)ln ( π + x)

=

 < 0 on (0, ∞) since 1 < e < π ∴ f (x) decreases on (0, ∞).

( π + x)(e + x) [ ln (e + x)]

2

153. (c) We have, f (x) = tan x – x ⇒ f ′ (x) = sec2x – 1 = tan2x ≥ 0, ∨ x ∈ R ⇒ f (x) never decreases. 154. (b), (c)  We have, f (x) = sin x + cos x ⇒ f ′ (x) = cos x – sin x = f ′ (x) > 0 if 0 ≤ x + i.e., –

π 2 cos   x +  4 

π π 3π π < or 1 2,   hus, f (x) is increasing in (1, ∞) and decreasing in T (– ∞, – 2). 157. (b), (c)  We have, f ′′ (x) < 0, for 0 ≤ x ≤ 1 ⇒ f ′ (x) is a decreasing function in [0, 1]. Now, g′ (x) = f ′ (x) – f ′ (1 – x). g (x) is increasing if g′ (x) > 0 ⇒ f ′ (x) > f ′ (1 – x) ⇒ x < 1 – x ( f ′ (x) is decreasing) 1 . i.e., x < 2 1 ∴ g (x) is an increasing function for 0 < x < . 2 Also, g (x) is decreasing if g′ (x) < 0 ⇒ f ′ (x) < f ′ (1 – x) ⇒ x > 1 – x  ( f ′ (x) is decreasing) 1 . i.e., x > 2 1 < x < 1. ∴ g (x) is a decreasing function for 2

π π 3π 0 for all x, therefore π 3π 0 if cos x > sin x 2 2 π 5π π ⇒ 0 ≤ x < or < x ≤ 2π Hence, f (x) is an increasing function in  0,  ∪ 4 4  2 π 5π and f ′ (x) < 0 if cos x < sin x ⇒ < x < .  π 3π   3π  4 4  , 2π  and a decreasing function in  2 , 2  .    2   5π  π  , 2 π Thus, f (x) is increasing in 0,  ∪    4  159. (b) f ′′(x) = 6(x – 1)  4  π 5π  and decreasing in  , ⇒ f ′(x) = 3(x – 1)2 + c ...(1) . 4 4  But at (2, 1), y = 3x – 5 is tangent to y = f (x) 156. (a), (d)  We have, ∴ f ′ (2) = 3 ∴ from (1), 3 = 3 + c ⇒ c = 0 −2 x − 1, x < −2  ∴ f ′(x) = 3(x – 1)2 ⇒ f (x) = (x – 1)3 + k   − 2 ≤ x ≤ 1 f (x) = | x + 2 | + | x – 1 | = 3, But the curve passes through (2, 1) 2 x + 1, x > 1    ∴ 1 = (2 – 1)3 + k ⇒ k = 0 ∴ f (x) = (x – 1)3. Also, f ′ (x) < 0 if

x − 2 x < 1, x ≠ 0  x3 ,  ∴ f ′ (x) = does not exist, x = 1 2 − x  3 , x >1  x Clearly, f ′ (x) > 0 for x < 0 or 1 < x < 2 and f ′ (x) < 0 for 0 < x < 1 or x > 2. Thus, f (x) is increasing for (– ∞, 0) ∪ (1, 2) and decreasing for (0, 1) ∪ (2, ∞). 161. (a), (c) f (x) = 3 cos4x + 10 cos3x + 6 cos2x – 3, 0 ≤x≤π ⇒ f ′ (x) = – 12 cos3x sin x – 30 cos2x sin x – 12 cos x sin x = – 6 sin x cos x (cos x + 2) (2 cos x + 1). π 2π , , π. 2 3 2π π 0, for 3 2 2π π and f ′ (x) < 0, for 0 < x < or < x < π. 3 2 Thus, f (x) is increasing in the interval  π , 2 π  2 3 

f ′ (x) = 0, for x = 0,

π 2π and decreasing in the interval  0,  ∪  , π  .  2  3  162. (a) We have, f (x) = kx3 – 9x2 + 9x + 3 ⇒ f ′ (x) = 3kx2 – 18x + 9. Since f (x) is increasing on R, therefore, f ′ (x) > 0 ∨ x ∈ R ⇒ 3kx2 – 18x + 9 > 0 ∨ x ∈ R ⇒ kx2 – 6x + 3 > 0 ∨ x ∈ R ⇒ k > 0 and 36 – 12k < 0 [ ax2 + bx + c > 0 ∨ x ∈ R ⇒ a > 0 and discrininant < 0] ⇒ k > 3. Hence, f (x) is increasing on R, if k > 3. 163. (b) We have, f (x) = x + kx + 1 2

⇒ f ′ (x) = 2x + k. Also, f ′′ (x) = 2. Now, f ′′ (x) = 2, ∨ x ∈ [1, 2] ⇒ f ′′ (x) > 0, ∨ x ∈ [1, 2] ⇒ f ′ (x) is an increasing function in the interval [1, 2]. ⇒ f ′ (1) is the least value of f ′ (x) on [1, 2] But f ′ (x) > 0 ∨ x ∈ [1, 2]  [ f (x) is increasing on [1, 2]] ∴ f ′ (1) > 0, ∨ x ∈ [1, 2] ⇒ k > – 2. Thus, the least value of k is – 2.

f ′ (x) < 0 ⇒ 6x2 + 18x + λ < 0. The value of λ should be such that the equation 6x2 + 18x + λ = 0 has roots – 2 and – 1. λ ⇒ λ = 12. 6 165. (b) We have, f (x) = sin4x + cos4x Therefore, (– 2) (– 1) =

⇒ f ′ (x) = 4 sin3x cos x – 4 cos3x sin x = – 4 sin x cos x (cos2x – sin2x) = – 2 (2 sin x cos x) cos 2x = – 2 sin 2x cos 2x = – sin 4x. f (x) is an increasing function if f ′ (x) > 0 ⇒ – sin 4x > 0 ⇒ sin 4x < 0 π π 0] Thus, f (x) is decreasing in (0, π). 167. (a), (c)  f ′ (x) > 0 ∨ x ∈ R ⇒ f (x) is increasing ∨ x ∈ R ∴ f (x) < f (x + 1) and f (x) > f (x – 1) ∨ x ∈ R ⇒ g ( f (x)) > g ( f (x + 1))  and g ( f (x)) < g ( f (x – 1)) as g (x) is decreasing ∨ x ∈ R. Also, g′ (x) < 0 ∨ x ∈ R ⇒ g (x) is decreasing ∨ x ∈ R ⇒ g (x) > g (x + 1) and  g (x) < g (x – 1) ∨ x ∈ R ∴ f (g (x)) > f (g (x + 1)) and f (g (x)) < f (g (x – 1)) ∨ x ∈ R, as f (x) is increasing ∨ x ∈ R. 168. (b) We have, f (x)  = 3 cos | x | – 6ax + b   = 3cos x – 6ax + b ( cos (– x) = cos x) ⇒ f ′ (x) = – 3sin x – 6a. Since f (x) is an increasing function ∨ x ∈ R ⇒ f ′ (x) > 0 ∨ x ∈ R ⇒ – 3sin x – 6a > 0 ∨ x ∈ R Inparticular, at x =

π 2

1 . 2 169. (a) We have, f (x) = 2 log (x – 1) – x2 + 2x + 3 – 3 – 6a > 0 ⇒ a < –

⇒ f ′ (x) =

1 − ( x − 1) 2  2 − 2x + 2 = 2   x −1  x −1 

175

164. (a) Since f (x) is decreasing in the interval (– 2, – 1), therefore,

Applications of Derivatives

 x −1  x 2 , x ≥ 1 | x −1| 160. (a), (c)  f (x) = = 1 − x  x2 , x < 1, x ≠ 0  x 2

176

Objective Mathematics

− 2 x ( x − 1)( x − 2) − 2 x ( x − 2) = ( x − 1) 2 x −1 − 2 x ( x − 1)( x − 2) Sign scheme for : ( x − 1) 2 =

f ′ (x) > 0 if x ∈ (– ∞, 0) or x ∈ (1, 2) ∴ f (x) is increasing in the interval (– ∞, 0) ∪ (1, 2). 170. (a) Let f (x) = x + ex = 0. Since f (– ∞) = – ∞ and f (+ ∞) = ∞, ∴ f (x) = 0 has a real root. Let the real root be α. Then f (α) = 0. Now, f ′ (x) = 1 + ex > 0, ∨ x ∈ R ∴ f (x) is an increasing function ∨ x ∈ R. ∴ for any other real number β, f (β) > f (α) or f (β) < f (α). But f (α) = 0; so, f (β) ≠ 0. ∴ f (x) = 0 has no other real root. Hence, the equation has only one real root. 171. (b) Let f (x) = 2x3 + 15 and g (x) = 9x2 – 12x. Then, f ′ (x) = 6x2 > 0 ∨ x ∈ R ∴ f (x) is an increasing function ∨ x ∈ R. 2 . Also, g′ (x) > 0 ⇒ 18x – 12 > 0 i.e., x > 3 2 Thus, f (x) and g (x) both increase for x > . 3 Let F (x) = f (x) – g (x). Since f (x) increases less rapidly than the function g (x), therefore, F (x) is a decreasing function. ⇒ F′ (x) < 0 ⇒ 6x2 – 18x + 12 < 0 i.e., 6 (x2 – 3x + 2) < 0 ⇒ (x – 1) (x – 2) < 0 ⇒ 1 < x < 2. Hence, the function 2x3 + 15 increases less rapidly than the function 9x2 – 12x on the interval (1, 2). 172. (a), (c) We have, h′ (x) = f ′ (x) [1 – 2 f (x) + 3 ( f (x))2] 2 1  = 3 f ′ (x) ( f ( x)) 2 − f ( x) +  3 3 

174. (c) We have, f ′ (x) = ex (x – 1) (x – 2). As ex > 0, f ′ (x) < 0 if and only if (x – 1) (x – 2) < 0 i.e., if 1 < x < 2. 175. (a) We have, f (x) = sin x – cos x – ax + b ⇒ f ′ (x) = cos x + sin x – a. f (x) is a decreasing function for all real values of x, if f ′ (x) < 0 ∨ x ∈ R ⇒ cos x + sin x < a ∨ x ∈ R.  s the maximum value of cos x + sin x is 2 , the A above is possible when a ≥ 2 . ∴ f (x) decreases for each x ∈ R if a ≥ 2 . 176. (d) In the interval [– 1, 2], f ′ (x) = 6x + 12 > 0. Hence f (x) is increasing in [– 1, 2]. Now, f (x) being a polynomial in x is continuous in – 1 ≤ x < 2 and in 2 < x ≤ 3. We check at x = 2.

lim f (2 – h)= lim 3 (2 – h)2 + 12 (2 – h) – 1 h→0 h→0 = 12 + 24 – 1 = 35

lim f (2 + h) = lim 37 – (2 + h) = 35. h→0 h→0

Also, f (2) = 3 (2)2 + 12 (2) – 1 = 35. ∴ f (x) is continuous at x = 2 and hence in the interval [– 1, 3]. f (2 − h) − f (2) Now, L f ′ (2) = lim h→0 −h 2 3 ( 2 − h ) + 12 ( 2 − h ) − 1 − 35 = lim h→0 −h 2 = lim 3h − 24h = 24. h→0 −h f (2 + h) − f (2) R f ′ (2) = lim h→0 h 37 − ( 2 + h ) − 35 = – 1. = lim h→0 h  ince, L f ′ (2) ≠ R f ′ (2), therefore, f ′ (2) does not S exist. 177. (a) Since g (x) is decreasing, ∴ g (x2) ≤ g (x1) when x2 ≥ x1. Since f (x) is increasing, ∴ f [g (x1)] ≥ f [g (x2)] ⇒ h (x1) ≥ h (x2) when x2 ≥ x1. ⇒ h (x) is a decreasing function of x and h (0) = 0. Also, domain of h = [0, ∞) and range of h = [0, ∞). ∴ h (x) = 0, ∨ x ∈ [0, ∞).

2  1  2 = 3 f ′ (x)   f ( x) −  +  3  9    Note that h′ (x) < 0 whenever f ′ (x) < 0 and h′ (x) dx dx/d θ −a sin θ > 0 whenever f ′ (x) > 0. Thus, h (x) increases (de- 178. (c) = = = – tan θ dy dy/d θ a cos θ creases) whenever f (x) increases (decreases). Equation of normal at θ is 173. (d) S is clearly correct. R is incorrect, can be seen by sin θ 3π  (y – a sinθ) = ( x − a (1 + cos θ )) taking f (x) = sin x, x ∈  π, . cos θ   2 

I n this interval f (x) decreases from 0 to – 1 but f ′ (x) = cos x increases from – 1 to 0.

Clearly this line passes through (a, 0).

180. (b) Suppose, there are two points x1 and x2 in (a, b) such that f ′ (x1) = f ′ (x2) = 0. By Rolle’s theorem applied to f ′ on [x1, x2], there must then be a c ∈ (x1, x2) such that f ′′ (c) = 0. This contradicts the given condition f ′′ (x) < 0 ∨ x ∈ (a, b). Hence our assumption is wrong. Therefore, there can be atmost one point in (a, b) at which f ′ (x) is zero. 181. (a) Let y = x1/x, x > 0  ...(1) 1 log x ⇒ log y = x log x Let z = log y = x 1 x ⋅ − log x ⋅ 1 dz 1 − log x x  ⇒ = = 2 x dx x2 dz 1 − log x =0 ⇒ = 0 ⇒ log x = 1 ⇒ x = e. dx x2  1 x 2 ⋅  −  − (1 − log x) ⋅ 2 x d 2z − 3 + 2 log x  x = = x4 dx 2 x3 ∴

d 2z  − 3 + 2 log e −3 + 2 1 = = – 3 < 0.  = dx 2  x = e e3 e e3

∴ x = e gives maximum value of z and hence, for y also  ( z = log y). Also, from (1), Max. (y) = e1/e. x log x −1 ⇒ f ′ (x) = 182. (a) Let f (x) = log x (log x) 2 For maximum or minimum, f ′ (x) = 0 ⇒ log x = 1 ⇒ x = e. 1 2 log x ⋅ (log x) 2 − (log x − 1) ⋅ x x Now, f ′′ (x) = (log x) 4 1 −0 1 ⇒ f ′′ (x)]x = e = e = > 0. 1 e ∴ f (x) is minimum at x = e. ∴ Minimum value of f (x) = f (e) = 1 1 183. (d) Let y = sin x + sin 2x + sin 3x 2 3 dy = cos x + cos 2x + cos 3x ⇒ dx

e = e. 1

For maximum or minimum,

177

Since, f ′ (x) > g′ (x) ∨ x ∈ R, therefore, h′ (x) = f ′ (x) – g′ (x) > 0 ∨ x ∈ R. ⇒ h (x) is an increasing function ∨ x ∈ R But h (0) = f (0) – g (0) = 0, So, for x > 0, we must have h (x) > h (0) = 0 and for x < 0, we have h (x) < h (0) = 0. ⇒ f (x) > g (x) ∨ x ∈ (0, ∞) and f (x) < g (x) ∨ x ∈ (– ∞, 0).

= cos 2x + 2 cos 2x cos x = cos 2x (1 + 2 cos x). dy =0 dx

⇒ cos 2x (1 + 2 cos x) = 0 ⇒ cos 2x = 0 or cos x = –

1 2

π 2π or x = 2 3 π 2π ∴ x = as x = is not required. 4 3 ⇒ 2x =

Now, ⇒

d2y = – sin x – 2 sin 2x – 3 sin 3x dx 2

d2y =–  dx 2  x = π

1 1 − 2 − 3⋅ < 0. 2 2

4

π  y is maximum, when x = 4 and maximum ∴ value is 4  1 1 1 1 1 8+3 2 y ]x = π = + + ⋅ = =  + . 2 3 2 4 2 2 3 2 6 2 184. (b) We have, P (x) = a0 + a1x2 + a2x4 + ... + anx2n ⇒ P′ (x) = 2a1x + 4a2x3 + ... + 2nan x2n – 1. For maximum or minimum, P′ (x) = 0 ⇒ x (2a1 + 4a2x2 + ... + 2nanx2n – 2) = 0 ⇒ x = 0  [ each ai > 0 and powers of x are even] Now, P′′(x) = 2a1 + 12a2x2 + ... + 2n (2n – 1) anx2n – 2 ∴ P′′ (x)]x = 0 = 2a1 > 0 i.e., P (x)  has a minimum at x = 0 only. 185. (d) We have, y = a loge x + bx2 + x dy a = + 2bx + 1 ⇒ dx x Since y has its extreme values at x = 1 and x = 2. dy  dy  = 0 and =0 ∴ dx  x =1 dx  x = 2 ⇒ a + 2b + 1 = 0 ...(1) and a + 8b + 2 = 0 ...(2) Solving (1) and (2), we get −2 1 a= and b = – . 3 6 ax + b  ...(1) 186. (c) We have, y = ( x − 4)( x − 1) dy a ( x − 4)( x − 1) − (ax + b)(2 x − 5) = dx ( x − 4) 2 ( x − 1) 2 For an extremum (maximum or minimum), dy =0 dx



Applications of Derivatives

179. (a), (b)   Let h (x) = f (x) – g (x)

178

Objective Mathematics

⇒ a (x – 4) (x – 1) – (ax + b) (2x – 5) = 0...(2) Since, y has an extremum at P (2, – 1), therefore, x = 2 satisfies the equation (2) and y = – 1 when x = 2. ∴ a (– 2) (1) – (2a + b) (4 – 5) = 0 i.e., b = 0 2a + b and – 1 = (2 − 4)(2 − 1) [Putting x = 2, y = – 1 in (1)]



⇒ 2 = 2a + b ⇒ 2 = 2a i.e., a = 1. sin ( x + a ) sin ( x + b)

dy dx sin ( x + b) ⋅ cos ( x + a ) ⋅ 1 − sin ( x + a )cos ( x + b) ⋅ 1  = sin 2 ( x + b) sin (b − a ) = ≠ 0 for any x as a ≠ b. sin 2 ( x + b) ⇒

Hence, y has neither maximum nor minimum.

1 dy 1 ⇒ =1– 2 . x x dx dy =0 For maximum or minimum, dx 1 ⇒ 1 – 2 = 0 ⇒ x2 = 1, i.e., x = ± 1. x

188. (d) We have, y = x +

Now,

d2y =2>0  dx 2  x =1

2 d2y = 3. ∴ x dx 2

x+ 1  x  ≥ 2  x ⋅ 1  ( A.M. ≥ G.M.) = 2    x  2     ∴ y ≥ 2 ∴ Min. (y) = 2. 2

2

2 ≥ 2.

 ∴ y ≥ 2, ∴ Min. (y) = 2.

1 1 = = logx a z log a x 2



1  z− 1  =  + 2, z z 

whose minimum value is 2.

∴ 2 log10 x – logx .01 = z +

4  z− 2  =  + 4, z z 

2

whose minimum value is 4.

193. (d) We have, f (x) =

x

∫ 2 (t − 1)(t − 2) 1

3

+ 3 (t − 1) 2 (t − 2) 2  dt

⇒ f ′ (x) = 2 (x – 1) (x – 2)3 + 3 (x – 1)2 (x – 2)2 = (x – 1) (x – 2)2 [2 (x – 2) + 3 (x – 1)] = (x – 1) (x – 2)2 (5x – 7). For maximum or minimum, f ′ (x) = 0 ⇒ (x – 1) (x – 2)2 (5x – 7) = 0 7 . ⇒ x = 1, 2, 5 Sign scheme for f ′ (x):

195. (a) Let y = x – xp, where x is the fraction ⇒

dy = 1 – pxp – 1. dx

For maximum or minimum,

1  1   1  = x−  liter: y = x + A = ( x )2 +   +   x  x x

∴ loga x + logx a = z +

−2 4 =– . log10 x z

d2y = 0 at x = a, then we have to obtain dx 2 d3y d4y , , ... and so on to ascertain the existence dx 3 dx 4 of points of extremum.

1 1    xz = 1; ∴ z = , x > 0 x x  



= logx 10–2 = – 2 logx 10 =

194. (c) When

⇒ y is maximum when x = – 1. Also, Max. (y) = – 2 and Min. (y) = 2 ∴ Max. (y) < Min. (y).

190. (b) Let z = loga x

1 100

Clearly, f (x) has maximum value at x = 1, minimum 7 and neither maximum nor minimum value at x = 5 at x = 2.

⇒ y is minimum when x = 1. d2y =–2 f (8).

6k = k ∴ f (0) = f (1) ⇒ f ′(x) = 0 6 i.e., equation ax2 + bx + c = 0 has at least one root between 0 and 1. Also, f (0) =

199. (b) We have, 1 log x 2 1 1 − ( x − 1) 2 = . ⇒ f ′ (x) = − 2 1+ x 2x 2 x (1 + x 2 )

f (x) = tan–1x –

Now, f (1) = tan–1 1 –

1 π log 1 = 2 4

 1   1  1 π 1 f   log 3 = + log 3  = tan–1  + 3    3 4 6 4 1 π 1 and f ( 3 ) = tan–1 ( 3 ) – log 3 = − log 3 . 4 3 4 π 1 Hence, the least value of f (x) is − log 3 . 3 4 2 200. (b) We have, y = x (x – 1)

dy ⇒ = (x – 1)2 ⋅ 1 + 2x (x – 1) dx

= 3x2 – 4x + 1 = (x – 1) (3x – 1). dy For maximum or minimum, =0 dx

3

3

f (x) =

∴ f ′ (x) = 0 ⇒ x = 1.

d2y  dx 2  x = 1 = – 2 < 0.

1 ∴ y is maximum when x = and maximum value 3 is 4 . y ]x = 1 = 27 3 d2y Also, = 2 > 0.  dx 2  x =1  ∴ y is minimum when x = 1.

⇒ 6 + a = 0 ∴ a = – 6. Also, y]x = 3 = 5 ⇒ 9 + (– 6) (3) + b = 5



d2y = 6x – 4. ∴ dx 2

1 , 1. 3

179

dy = 2x + a dx

49 8 and f (8) = . 543 89

49 . 543 202. (b) Let P (x, y) be the point on the curve which is nearest to O (0, 0). 1 (e2x + e–2x + 2). Let z = OP2 = x2 + y2 = x2 + 4 dz 1 ⇒ = 2x + (e2x – e–2x) dx 2 dz =0 For maximum or minimum, dx 1 ⇒ 2x + (e2x – e–2x ) = 0 2 e −2 x − e 2 x ⇒ = 2x 2 ⇒ x = 0 is a solution and then 1 (e0 + e0) = 1.  y = 2 d2y Also, = 2 + e2x + e–2x > 0, 2 dx  hence z is minimum. Hence the largest value is

∴ The shortest distance OP =

02 + 12 = 1.

203. (c) We have, y = 6 cos x – 8 sin x ⇒

dy = – 6 sin x – 8 cos x. dx

⇒ – 6 sin x = 8 cos x ⇒ sin x = ±



dy =0 dx

⇒ tan x = –

4 3 , cos x =  . 5 5

4 3

Applications of Derivatives

196. (b) We have, y = x2 + ax + b ⇒

180

When sin x =

4 3 , cos x = – 5 5



Objective Mathematics

 3 4 y = 6   −  – 8   = – 10 and  5 5 4 3 , cos x = when sin x = – 5 5 then

 4 3 y = 6   – 8  −  = 10.  5 5 Hence, the greatest height above the x-axis is 10.

then

204. (b) Let f (x) = a2sec2x + b2cosec2x ⇒ f ′ (x) = 2a2sec2x tan x – 2b2cosec2x cot x 2a 2 sin x 2b 2 cos x =0 − ∴ f ′ (x) = 0 ⇒ cos3 x sin 3 x 2 b ⇒ tan4x = 2 a

α dT = 0 ⇒ x2 = ⇒x= β dx

α . β

α  2α  d 2T = N  3  > 0 at x = . 2 β dx x  α ∴ T is minimum for x = . β 207. (a) Let the equation of the circle be x2 + y2 = a2 Also,

Let A (a, 0), B (– a, 0) be the ends of the diameter and C (x, y) be any point on the circle. Then, area of ∆ABC 1 × AB × y =A = 2 2 = ay = a a − x 2 .

b a i.e., cot2x = . a b b a and cosec2x = 1 + . Hence sec2x = 1 + a b ⇒ tan2x =

Also, f ′′ (x) = 2a2 (sec4x + 2sec2x tan2x)  + 2b2 (cosec4x + 2cosec2x cot2x) > 0,  for all x b2 . a2 Hence the minimum value of f (x) is a+b b+a + b2 ⋅ = (a + b)2. = a2 ⋅ a b 205. (a) Let P (x, y) be such a point then OP2 = x2 + y2. ⇒ f (x) is minimum at tan4x =

1 dS 1 ⇒ = 2x – 2 . x x dx dS 1 1 = 0 ⇒ 2x – 2 = 0 ⇒ x3 = . ∴ x dx 2

Let S = OP2 = x2 +

1/ 3

1 ∴ x =   2



d 2S 2 Also, =2+ 3 x dx 2

1/ 3

1 ∴ S is minimum at x =   2

z

= lx + my = lx + dz mk ⇒ =l– 2 . dx x ∴

( xy = k)

dz mk =0 ⇒l– 2 =0 dx x

⇒ x = ± Now,

mk . l

d 2z 2mk = dx 2 x3 ⇒

d 2z   dx 2  x =

mk l

=

2mk  mk     l 

3

2

mk . l

Also,

1/ 3

α  N (α + βx2) = N  + βx  x x 

mk  x

∴ z is minimum at x =

.

1 Thus, for nearest point x =   2 −1/ 6 1 y =±   . 2

 α  dT = N  − 2 + β . ⇒ x dx  

208. (b) We have,



d 2S 2 ⇒ =2+ >0  1 dx 2  x =  1 1/ 3   2 2  

206. (c) T =

∴  A is maximum if x = 0 i .e., C lies on y-axis and then CAB is an isosceles triangle.

and then

min. z = l ⋅

mk   + m ⋅ l

= 2 lmk .

k  = mk l

lmk + lmk

209. (a) We have, f (x) = 2 (cos 3x + cos 3 x) 3− 3  3+ 3  = 4 cos  ≤4  2  x ⋅ cos  2  x    

>0

This is possible only when x = 0.

dS = ex (– sin x + cos x – cos x – sin x) = – 2ex dx sin x

Now, 

dS = 0 ⇒ – 2ex sin x = 0 ⇒ x = 0, π, 2π dx ( 0 ≤ x ≤ 2π)

d 2S Also,  2 = – 2ex (cos x + sin x) > 0 at x = π dx only ∴ S is minimum at x = π. Thus, a = π. 1 (2 cos α cos β) 211. (a) Let y = cos α ⋅ cos β = 2 1 [cos (α + β) + cos (α – β)] = 2 1 cos π + cos  π − 2β   1 sin 2β.   =  2 2  2  2 π π i.e., β = Clearly, y is maximum when 2β = 2 4

=

f (x) = 1 + 3x2 + 32 ⋅ x4 + ... + 330 ⋅ x60 ⇒ f ′ (x) = x (6 + 4 ⋅ 32 ⋅ x2 + ... + 60 ⋅ 330 ⋅ x58) f ′ (x) = 0 ⇒ x = 0. Also, f ′′ (x) = 1 ⋅ (6 + 4 ⋅ 32 x2 + ... + 60 ⋅ 330 ⋅ x58)  + x (4 ⋅ 32 ⋅ 2x + ... + 60 ⋅ 330 ⋅ 58x57) ⇒ f ′′ (0) = 6 > 0.  ∴ f (x) has minimum at x = 0 only. 213. (d) We have,  − 6 x, 0 < x < 1 f ′ (x) =  − 6, x ≥ 1 ∴ f ′ (1 – h) = – 6 (1 – h) < 0

⇒ f ′ (x) = 2 (x – 2) ⋅ xn + (x – 2)2 ⋅ nxn – 1

= xn – 1 ⋅ (x – 2) [(n + 2) x – 2n] 2n ∴ f ′ (x) = 0 ⇒ x = 0, 2, . n+2 Now,



f ′ (0 – h) = (– h)n – 1 (– h – 2) ((n + 2) (– h) – 2n)

= (– ve)n – 1 ⋅ (– ve) ⋅ (– ve) = (– ve)n – 1

and

f ′ (0 + h) = hn – 1 (h – 2) ((n + 2) h – 2n)

= (+ ve) (– ve) (– ve) = +ve

∴ f (x) has a minimum at x = 0 if n is even. Also, f ′ (2 – h) = (2 – h)n – 1 ⋅ (– h) ⋅ (2 (2 – h) – nh) = (2 – h)n – 1 ⋅ (– h) ⋅ (4 – 2h – nh) = (+ ve)n – 1 ⋅ (– ve) ⋅ (+ ve) = – ve and f ′ (2 + h) = (2 + h)n – 1 ⋅ h ⋅ (2 (2 + h) + nh) = +ve. ∴ f (x) has a minimum at x = 2, ∨ n ∈ N. 216. (c), (d)  We have, sin x x cos x − sin x and f ′′ (x) = f ′ (x) = x x2 For maximum or minimum, f ′ (x) = 0 sin x = 0 ⇒ sin x = 0; x ≠ 0. x ∴ x = nπ; n = 1, 2, 3, ... ( x > 0). nπ cos nπ − sin nπ (nπ) 2 cos nπ (−1) n = .  = nπ nπ ∴ Extreme points are x = nπ, n = 1, 2, 3, ..., where the maximum occurs at x = π, 3π, 5π, ... and the minimum occurs at x = 2π, 4π, 6π, ... At x = nπ, f ′′ (x) =

− x4

4

(4 – x2).

For maximum or minimum, f ′ (x) = 0

f ′ (1 + h) = – 6 < 0.

⇒ e

= 4, αβγ = 8.

− x4

4

(4 – x2) = 0 ⇒ x = – 2, 2.

Also, − x4



f ′′ (x) = e



< 0 if x = 2  and > 0 if x = −2 

214. (b) Let α, β, γ be the roots of the given equation. Then, α + β + γ = a, αβ + αγ + βγ

= (x – 2) ⋅ xn – 1 [2x + n (x – 2)]

217. (a), (b)  We have, f ′ (x) = e

 ince f ′ (x) does not change sign as x passes through S 1, therefore, f (x) does not have a maximum or minimum at x = 1, whatever be the value of α.



4

 (– 2x) + e

− x4

4

 ⋅ (4 – x2) ⋅ (– x3)

181

αβγ



212. (d) We have,

and

3

215. (a), (c)  We have, f (x) = (x – 2)2 ⋅ xn

210. (b) The slope of the tangent to the curve y = ex cos x is dy = ex (– sin x + cos x) S= dx

.

α+β+ γ Since A.M. ≥ G.M. ⇒ ≥ 3 a 3 ⇒ ≥ 8 ⇒ a ≥ 6. 3 ∴ The minimum value of a = 6.

Applications of Derivatives

3+ 3  and it is equal to 4 when both cos  x and  2    3− 3  cos  x are equal to 1 for a value of x.  2   

182

∴ f (x) has a maximum at x

=2

Objective Mathematics

= (2k – 3) – (k – 1) 2 sin 2x cos 2x = (2k – 3) – (k – 1) sin 4x. Since f (x) does not have critical points, therefore, f ′ (x) = 0 does not have any solution in R. ⇒ (2k – 3) – (k – 1) sin 4x = 0 has no solution in R 2k − 3 is not solvable in R ⇒ sin 4x = k −1 2k − 3 2k − 3 2k − 3 >1 ⇒ < – 1 or >1 ⇒ k −1 k −1 k −1

and minimum at x = – 2. 218. (b) Since f ′ (4) = f ′′ (4) = 0, therefore, f (x) = (x – 4)n + k, where n ≥ 3 But f has minimum at x = 4, so n = 4. ∴ f (x) = (x – 4)4 + k. Since f (4) = 10, therefore, k = 10. Thus, f (x) = (x – 4)4 + 10. 219. (a) We have, f (x) =

x

∫ (6t

2

− 24) dt

⇒ 2k – 3 < – k + 1 or 2k – 3 > k – 1

0

⇒ f ′ (x) = (6x2 – 24) ⋅ 1

4  4 or k > 2 ⇒ k ∈  −∞,  ∪ (2, ∞). 3 3 

⇒ k
3 f (x) = (k – 7k + 12) cos x + 2 (k – 4) x + log 2 Thus, we find that f (x) has a minimum value = (k – 3) (k – 4) cos x + 2 (k – 4) x + log 2 2 at x = 3. ⇒ f ′ (x) = – (k – 3) (k – 4) sin x + 2 (k – 4) = (k – 4) [– (k – 3) sin x + 2] 224. (b) Let f (x) = x2/3 + (x – 2)2/3 Since f (x) does not have critical points, therefore 2 –1/3 2 ⇒ f ′ (x) = x + (x – 2)–1/3 f ′ (x) = 0 does not have any solution in R. 3 3 –1/3 Now, f ′ (x) = 0 ⇒ x + (x – 2)–1/3 = 0 ⇒ k ≠ 4 and 2 – (k – 3) sin x = 0 is not solvable ⇒ (x – 2)–1/3 = – x–1/3 ⇒ x – 2 = – x in R 2 ⇒ x = 1 is the critical point. is not solvable in R ⇒ k ≠ 4 and sin x = 2  1 −4 / 3 1 k −3  − x − ( x − 2) −4 / 3  Also, f ′′ (x) = 2 3  3 3  >1 ⇒ k ≠ 4 and k −3 2 −4 / 3  x + ( x − 2) −4 / 3  =– ⇒ k ≠ 4 and | k – 3 | < 2 9 2 −4 / 3 4 ⇒ k ≠ 4 and – 2 < k – 3 < 2 1 + (−1) −4 / 3  = – ∴ f ′′ (1) = – b, (x – b)2m + 1 is positive ( 2m + 1 is odd). Thus, f ′ changes sign from negative to positive as x passes through b and so, f has a minimum at x = b. Since 2n is an even integer, (x – a)2n does not change sign as x passes through a i.e., f ′ (x) does not change sign as x passes through a. Hence f has neither a maximum nor a minimum at x = a. 227. (b) Let f (x) = x25 ⋅ (1 – x)75

=

475 = 0 i.e., x = 237.5. 2 But x is a positive integer, so x can be taken 237 or 238. P is maximum when x –

 ince P (237) = P(238), for maximum profit, total S number of subscribers should be 725 + 237 = 962 or 725 + 238 = 963. Since there is no profit from the 963rd subscriber, therefore number of subscribers should be 962.

231. (c) ⇒ f ′ (x) = 25x24 (1 – x)75 – 75x25 (1 – x)74 = 25x24 (1 – x)74 (1 – 4x) 1 are the critical points. f ′ (x) = 0 ⇒ x = 0, 1, 4 25 75 1 1 3 Now, f (0) = 0, f (1) = 0 and f   =   ⋅   . 4 4 4 1 . ∴ f (x) takes its maximum value at x = 4 228. (b) Let α and β be the roots of the given equation. Then α + β = k and αβ = (2k – 3) Let z = α3 + β3 = (α + β)3 – 3αβ (α + β) = k3 – 3k (2k – 3) = k3 – 6k2 + 9k dz = 3k2 – 12k + 9 = 3 (k2 – 4k + 3) ∴ dk  = 3 (k – 1) (k –3) dz = 0 ⇒ 3 (k – 1) (k – 3) = 0 Now, dk  ⇒ k = 1, 3. d 2z  Also,  = (6k − 12) k = 3 = 6 (3) – 12 > 0. dk 2  k = 3 Hence, z is minimum when k = 3.

π  π π or x = 0 ≤ x ≤  2 2 6 

⇒ x =



Now, f (0) =

On differentiating w.r.t. x, respectively, we get and 3 x 2 = dy dx

2  dy  m1 =   = − a  dx (1,1)

 dy  and  m2 =   = 3(1)  dx (1,1) Since, the two curves cut each other orthogonally

∴  m1m2 = – 1 ⇒  − 2 × 3 = −1 a 232. (d) We have,

or

a=6

f ′ (x) = 3kx2 – 18x + 9 f(x) is increasing on every interval, if  f ′ (x) ≥ 0, ∀ x ∈ R ∴  3kx2 – 18x + 9 ≥ 0, ∀ x ∈ R ⇒  If k > 0 and (–18)2 – 4(3k) (9) ≤ 0. ⇒  If k > 0 and 324 – 108k ≤ 0 ⇒  If k > 0 and k ≥ 3 ∴ k ≥ 3.

and f ′′ (x) = 8 + 72x2 + 240x4 ≥ 8, ∀ x ∈ R For maxima and minima, put 1 2

3 π π 1 1 , f    = and f    = . 4 2 2 6 2

Of these values, the minimum value is

ay + x2 = 7 and x3 = y

⇒ f ′ (x) = 8x + 24x3 + 48x5, ∀ x ∈ R

⇒ f ′ (x) = cos x – sin 2x = cos x (1 – 2 sin x). ∴ f ′ (x) = 0



We have,

233. (b) We have f(x) = 2 + 4x2 + 6x4 + 8x6

1 cos 2x 229. (a), (d)  We have, f (x) = sin x + 2

⇒ cos x (1 – 2 sin x) = 0 ⇒ cos x = 0 or sin x =

2 2 1  475    475   − x − 870000 −  .    2   100  2  

1 4 3 < < . 2 3 2



f ′ (x) = 0

⇒ x(8 + 24x2 + 48x4) = 0 ⇒ x = 0 Since, f ′′ (x) > 0, therefore, f has a local minima at x = 0. Moreover, f has no other critical point and f ′ (x) exists at all x ∈ R. Therefore, there can be only one minima.

183

230. (b) Let the additional number of subscribes be x, so the number of subscribers becomes 725 + x and then x   the profit per subscriber is Rs. 12 − .  100 

Applications of Derivatives

f ′ (x) = 0 gives xp + q = 1 ⇒ x = 1. When x < 1, f ′ (x) < 0 and when x > 1, f ′ (x) > 0

184

234. (a) Let f(x) = 3 sin x – 4 sin3 x = sin 3x

Objective Mathematics

 π π Since, sin x is increasing in the interval  − ,   2 2 π π − ≤ 3x ≤ 2 2 π π − ≤x≤ 6 6 Therefore, the length of interval = 235. (c) Let y = x1/x

π  π π − − = 6  6  3

⇒ log y = 1 log x x log x Let f(x) = x f ′ (x) = 1 − log x x2 Put f ′ (x) = 0 for maxima and minima ⇒ 1 − log x = 0 x2 ⇒ 1 – log x = 0 ⇒ x = e Now, f '' ( x) =

− x − 2(1 − log x) x < 0 at x = e ( x 2 )2

∴ at x = e, f(x) is maximum and maximum value of x1/x = e1/e. 236. (d) We have, f(x) = 4x4 – 2x + 1 ⇒ f ′ (x) = 16x3 – 2 Put f ′ (x) = 0 for maxima and minima ⇒ 16x3 – 2 = 0 ⇒ x = 1/2 Since for x > 1 , f ′ (x) > 0 2 ∴ f(x) is increasing for x > 1 . 2 237. (d) We have, f ( x) =

x 2 + 2 x

1 2 − 2 x2 Put f ′ (x) = 0 for maxima or minima, we get



f ' ( x) =

1 2 − =0 2 x2 ⇒ x2 = 4 ⇒ x = ± 2

f ′′ (x) = 4 ⇒ f ′′ (2) = 4 = 1 > 0 x3 8 2 4 1 and f ′′ (–2) = − = − < 0 8 2 ∴ f(x) is minimum at x = 2. Now, 

238. (a) We have, y = x – 5x + 6 2

⇒ dy = 2 x − 5 dx

 dy  = 4 − 5 = −1 and ∴ m1 =    dx (2, 0)  dy  = 6 −5 =1  m2 =    dx (3, 0) Since, m1m2 = – 1 ∴ Angle between the tangents is π . 2 239. (a) Let y = xx ⇒ dy = x x (1 + log x) dx For y to be increasing, dy > 0 dx ⇒ xx (1 + log x) > 0 or 1 + log x > 0

1 e

x>

or

240. (d) We have, f(x) = – x3 + 4ax2 + 2x – 5 ⇒f ′ (x) = –3x2 + 8ax + 2

Since, f(x) is decreasing ∀x, therefore f ′ (x) < 0 ⇒ –3x2 + 8ax + 2 < 0  rom above, it is clear that for no value of a, f(x) is F decreasing. x2 + 1

241. (d) We have, f ( x) = ∫x2 ⇒ f ′ (x) = 2xe –( x = 2x(e − ( x

2

+1) 2

+1) 2

2

− 2 xe − x – 2xe–x4 4

− e− x )

 ince, e − x > e − ( x S be x < 0. 4

2

e − t dt

4

2

+ 1) 2

, therefore for f ′ (x) > 0, it should

242. (c) Let f(x) = x5 – 5x4 + 5x3 – 1 ⇒ f ′ (x) = 5x4 – 20x3 + 15x2 For maximum or minimum, put f ′ (x) = 0 ⇒ 5x4 – 20x3 + 15x2 = 0 ⇒ x2(5x2 – 20x + 15) = 0 ⇒ 5x2(x – 1) (x – 3) = 0 ⇒ x = 0, 1, 3 Now,  f ′′ (x) = 5(4x3 – 12x2 + 6x) At x = 1, f ′′ (1) = 5(4 – 12 + 6) = – 10 < 0, maximum. At x = 3, f ′′ (3) = 5(4 × 27 – 12 × 9 + 6 × 3)  = 90 > 0, minimum. At x = 0, f ′′ (0) = 5(0 – 12 × 0 + 6 × 0) = 0, f ′′′ (x) = 5(12x2 – 24x + 6) ⇒ f ′′′ (0) = 30 ≠ 0 ∴ f (x) is maximum at x = 1, minimum at x = 3 and x = 0 is point of inflection. 243. (d) We have, x 4 − x2 ⇒ f '( x) = 2 4+x+x (4 + x + x 2 ) 2 On the interval [–1, 1], f ′ (x) > 0 f ( x) =

Therefore, f(x) is an increasing function on [–1, 1]



244. (a) Using mean value theorem, we have f (3) − f (1) ⇒ 1 log 3 − log 1 = 3 −1 c 2 2 ⇒ c = = 2log 3 e log e 3

e2 x − 1 e −2 x − 1 1 − e 2 x ⇒ f (− x) = −2 x = 2x e +1 e + 1 1 + e2 x e2 x − 1 = − 2x = − f ( x) ∴  f(x) is an odd function. e +1 f ( x) =

e2 x − 1 e2 x + 1 4e 2 x ⇒ f ' ( x) = > 0, ∀ x ∈ R (1 + e 2 x ) 2 ⇒ f(x) is an increasing function.

1 (cos x − sin x) 1 + (sin x + cos x) 2

Again, f ( x) =

π  2 cos  x +  4  = 1 + (sin x + cos x) 2 3π π 1

190

metHoDS of InteGratIon

Thus,

Objective Mathematics

method of transformation When the integrand is a trigonometric function, we transform the given function into standard integrals or their algebraic sum by using trigonometric formulae:

∫ [ f ( x)] f ′ (x) dx =

n ∫ t dt =

=

[ f ( x)]n + 1 n +1

n

(iv) When the integrand is of the form

1 − cos 2mx 2 1 + cos 2mx cos2mx = 2 sin mx mx cos sin mx = 2 sin 2 2 3 sin mx − sin 3 mx sin3mx = 4 3 cos mx + cos 3 mx cos3mx = 4

(ii) (iii) (iv) (v)

Thus,

(i) Choose the first and second function in such a way that the derivative of the first function and the integral of the second function can be easily found. (ii) In case of integrals of the form

By suitable substitution, the variable x in



f ( x ) dx is changed

into another variable t so that the integrand f (x) is changed into F (t) which is some standard integral or algebraic sum of standard integrals. There is no general rule for finding a proper substitution and the best guide in this matter is experience. However, the following suggestions will prove useful. (i) If the integrand is of the form f ′ (ax + b), then we put ax + b = t and dx =

1 dt. a

f ' (t )

dt 1 = a a

dt

=



1 n

∫ f ' (t ) dt

1 1 = f (t) = f (xn). n n (iii) When the integrand is of the form [ f (x)] n. f ′ (x), we put f (x) = t and f ′ (x) dx = dt.

n

function and f (x ((x) x) as the second function. (iii) In case of integrals of the form

log x ) ∫ (log

n

· dx, take 1 as the

second function and (log (logx)) as the first function. (logx n

(iv) Rule of integration by parts may be used repeatedly, if required. (v) If the two functions are of different type, we can choose the first function as the one whose initial comes first in the word ‘‘ILATE’’, where I

— Inverse Trigonometric function

L — Logarithmic function

E — Exponential function.

f ' (t ) dt

xn = t and nxn – 1 dx = dt.

∫ f ' (t ) n

n

T — Trigonometric function

f (t ) f ( ax + b ) ⋅ = a a (ii) When the integrand is of the form xn – 1 f ′ (xn), we put

f ' ( x n ) dx =

∫ f ( x) . x dx, take x as the first

A — Algebraic function

=

n −1

dx.

Working Hints

method of Substitution

∫x



In words, integral of the product of two functions = first function × integral of the second – integral of (differential of first × integral of the second function).

(x) 2 sin A sin B = cos (A – B) – cos (A + B).

Thus,

dt = log t = log f (x). t

 du

(ix) 2 sin A cos B = sin (A + B) + sin (A – B)





∫ (uv) dx = u. ∫ vdx − ∫  dx ⋅ ∫ vdx 

(viii) 2 cos A cos B = cos (A + B) + cos (A – B)



dx =

The process of integration of the product of two functions is known as integration by parts. For example, if u and v are two functions of x, then

(vii) cot2mx = cosec2mx – 1

f ' ( ax + b ) dx =

f ' ( x)

∫ f ( x)

method of Integration by parts

(vi) tan2mx = sec2mx – 1

Thus,

f' ( x ) , we put f ( x)

f (x) = t and f ′ (x) dx = dt.

The following trigonometric identities are important: (i) sin2mx =

tn +1 n +1

(vi) In case, both the functions are trigonometric, take that function as second function whose integral is simpler. If both the functions are algebraic, take that function as first function whose derivative is simpler. (vii) If the integral consists of an inverse trigonometric function of an algebraic expression in x, first simplify the integrand by a suitable trigonometric substitution and then integrate the new integrand.

method of partial fractions for rational functions Integrals of the type

p ( x)

∫ g ( x)

dx can be integrated by resolving

the integrand into partial fractions. We proceed as follows:

caSe 1: When the denominator contains non-repeated linear factors. That is g (x) = (x – α1) (x – α2) ... (x – αn).

In such a case write f (x) and g (x) as:

f ( x) A1 A2 An + + ... + = x − α1 x − α 2 x − αn g ( x) where A1, A2, ... An are constants to be determined by comparing the coefficients of various powers of x on both sides after taking L.C.M. caSe 2: When the denominator contains repeated as well as non-repeated linear factors. That is

Some SpecIaL InteGraLS 1.



dx x 1 = tan − 1 + C x + a2 a a

2.



dx 1 x−a log = +C x − a2 2a x+a

3.

∫a

4.



a −x

5.



x + a2

6.



x − a2



a 2 − x 2 dx =



x 2 + a 2 dx =

g (x) = (x – α1)2 (x – α3) ... (x – αn). In such a case express f (x) and g (x) as:

f ( x) A1 A2 A3 An + + ... + + = g ( x) x − α1 ( x − α1 )2 x − α 3 x ( − αn ) where A1, A2, ... An are constants to be determined by comparing the coefficients of various powers of x on both sides after taking L.C.M. caSe 3: When the denominator contains a non repeated quadratic factor which cannot be factorised further:

7.

8.

g (x) = (ax2 + bx + c) (x – α3) (x – α4) ... (x – αn).

2

2

2

dx 1 a+x log = +C − x2 2a a−x

dx 2

2

dx 2

dx 2

= sin– 1

= log x + x 2 + a 2 + C = log x + x 2 − a 2 + C

x x a2 a2 − x2 + sin– 1 +C a 2 2

x 2

where A1, A2, ... An are constants to be determined by comparing the coefficients of various powers of x on both sides after taking L.C.M.

9.



x 2 − a 2 dx =

(a)

In such a case write f (x) and g (x) as:

Note:

where A1, A2, ... An are constants to be determined by comparing the coefficients of various powers of x on both sides after taking L.C.M. Corresponding to repeated linear factor (x – a)r in the denominator, a sum of r partial fractions of the type

A1 A2 Ar + ... + + is taken. x − a ( x − a )2 ( x − a )r

x 2

x2 − a2 −

a2 2

Integrals of the form

g (x) = (ax2 + bx + c)2 (x – α5) (x – α6) ... (x – αn)

An + ... + x − α ( n)

a2 2

log x + x 2 − a 2 +C

caSe 4: When the denominator contains a repeated quadratic factor which cannot be factorised further. That is

f ( x) A1 x + A 2 A3 x + A 4 A5 = + + g ( x) ax 2 + bx + c ( ax 2 + bx + c )2 x − α 5

x2 + a2 +

2 2 log x + x + a + C

In such a case express f (x) and g (x) as:

f ( x) A1 x + A 2 A3 An + + ... + = 2 ax + bx + c x − α 3 x − αn g ( x)

x +C a

(c)

− x 2 ) dx ,

(b)

∫ f (a

f ( x 2 − a 2 ) dx ,

(d)



∫ f (a ∫

2

2

+ x 2 ) dx ,

a − x f  dx,  a + x 

Working Rule Integral

∫ f (a ∫ f (x

Substitution

) dx + a ) dx dx ∫ f (x − a ) ∫

2

x = a sin θ or x = a cos θ

2

2

x = a tan θ or x = a cot θ

2

2

x = a sec θ or x = a cosec θ

2

−x

a − x f dx or  a + x 



a + x dx f  a − x 

x = a cos 2θ

191

Indefinite Integration

• Check degree of p (x) and g (x). caSe 5: If the integrand contains only even powers of x • If degree of p (x) > degree of g (x), then divide p (x) by g (x) (i) Put x2 = z in the integrand. till its degree is less, i.e., put in the form (ii) Resolve the resulting rational expression in z into partial p ( x) f ( x) fractions = r (x) + (iii) Put z = x2 again in the partial fractions and then integrate g ( x) g ( x) both sides. where degree of f (x) < degree of g (x).

192

∫ sin

Integrals of the form

m

x cos n x dx

Working Rule (i) Divide the numerator and denominator by cos2x.

Objective Mathematics

Working Rule

(ii) In the denominator, replace sec2x, if any, by 1 + tan2x.

(i) If the power of sin x is an odd positive integer, put cos x = z.

(iii) Put tan x = z ⇒ sec2x dx = dz.

(ii) If the power of cos x is an odd positive integer, put sin x = z.

(iv) Integrate the resulting rational algebraic function of z.

(iii) If the power of sin x and cos x are both odd positive integers, put sin x = z or cos x = z.

(iv) In the answer, put z = tan x.

(iv) If the power of sin x and cos x are both even positive integers, use De’ Moivre’s theorem as follows :

a cos x + b sin x

∫ c cos x + d sin x dx

Integrals of the form:

Let cos x + i sin x = z. Then cos x – isin x = z–1 Adding these, we get z+



Working Rule

1 1 = 2 cos x and z – = 2i sin x z z

(i) Put Numerator = λ (denominator) + µ (derivative of denominator)

By De’ Moivre’s theorem, we have

1 1 zn + n = 2cos nx and zn – n = 2i sinnx ...(1) z z n m 1 1  1  1 ⋅ n z +  z −  m ∴ sinmx cosnx = (2 i ) 2  z   z  1 =

2m + n

1 ⋅ m i

n

m

1  1   z +   z −  . z z

Now expand each of the factors on the R.H.S. using Binomial theorem. Then group the terms equidistant from the beginning and the end. Thus express all such pairs as the sines or cosines of multiple angles. Further integrate term by term. (v) If the sum of powers of sin x and cos x is an even negative integer, put tan x = z.

a cos x + b sin x = λ (c cos x + d sin x) + µ (– c sin x + d cos x). (ii) Equate coefficients of sin x and cos x on both sides and find the values of λ and µ. (iii) Split the given integral into two integrals and evaluate each integral separately, i.e.,

a cos x + b sin x

∫ c cos x + d sin x dx = λ

∫ 1 dx + µ ∫ = λx + µ log | a cos x + b sin x |.

(iv) Substitute the values of λ and µ found in step 2.

Integral of the form dx , (a) ∫ a + b cosx dx (c) ∫ a + b cos x + c sin x

dx (b) ∫ , a + b sinx

− c sin x + d cos x dx a cos x + b sin x

Integrals of the form

a + b cos x + c sin x

∫ e + f cos x + g sin x dx

Working Rule (i) Put Numerator = l (denominator) + m (derivative of denominator) + n

Working Rule

x x 1 − tan 2 tan 2 and sin x = 2 so that the given (i) Put cos x = x x 1 + tan 2 1 + tan 2 2 2 x integrand becomes a function of tan . 2 x 1 x (ii) Put tan =z⇒ sec2 dx = dz 2 2 2 2

(iii) Integrate the resulting rational algebraic function of z

x (iv) In the answer, put z = tan ⋅ 2

dx ∫ a + b cos2 x ,

(c)

∫ a cos

2

(b)

+ m (– f sin x + g cos x) + n (ii) Equate coefficients of sin x, cos x and constant term on both sides and find the values of l, m, n. (iii) Split the given integral into three integrals and evaluate each integral separately, i.e.,

a + b cos x + c sin x

∫ e + f cos x + g sin x dx ∫

= l 1dx + m

− f sin x + g cos x

∫ e + f cos x + g sin x dx + n dx

∫ e + f cos x + g sin x

Integrals of the form (a)

a + b cos x + c sin x = l (e + f cos x + g sin x)

dx

∫ a + b sin

dx x + b sin x cos x + c sin 2 x

2

x

,

= lx + m log | e + f cos x + g sin x | +n

dx

∫ e + f cos x + g sin x dx

(iv) Substitute the values of l, m, n found in Step (ii).



(b)



ax + bx + c

(c)



ax + bx + c dx a

dx

Working Rule (i) Make the coefficient of x2 unity by taking the coefficient of x2 outside the quadratic. (ii) Complete the square in the terms involving x, i.e., write ax2 + bx + c in the form a [x ± α)2 ± β2].

,

2

(iii) The integrand is converted to one of the nine special integrals. (iv) Integrate the function.

2

px + q

Integrals of the form (a) ∫ ax 2 + bx + c dx, (b) Integral

∫ ax

193

(a)

dx , ax 2 + bx + c



px + q

dx,

ax 2 + bx + c

(c)

∫ ( px + q )

ax 2 + bx + c dx

Working rule

px + q dx + bx + c

Put px + q = λ (2ax + b) + µ or px + q = λ (derivative of quadratic) + µ. Comparing

2

the coefficient of x and constant term on both sides, we get p = 2aλ and q = bλ + µ p bp   ⇒λ= and µ =  q −  Then the integral becomes 2a  2a 

∫ ax

px + q dx + bx + c

2

px + q



ax 2 + bx + c

=

p 2a

=

bp  dx p  ⋅ log | ax2 + bx + c | +  q −  ∫ 2  2a 2a ax + bx + c

In this case the integral becomes

dx

px + q

∫ =

∫ ( px + q )

2ax + b dx dx +  q − bp  2   + bx + c 2a  ∫ ax 2 + bx + c

∫ ax

ax 2 + bx + c dx

ax 2 + bx + c p a

dx =

p 2a



bp   dx +  q − ∫ 2a  ax 2 + bx + c

dx

2ax + b

bp   ax 2 + bx + c +  q −   2a  ∫

ax + bx + c 2

dx ax 2 + bx + c

The integral in this case is converted to p 2 ∫ ( px + q ) ax + bx + c dx = 2a ∫ (2ax + b) ax 2 + bx + c dx

bp   +  q −  2a  ∫ =

ax 2 + bx + c dx

p (ax2 + bx + c)3/2 + 3a

bp    q −  2a  ∫

ax 2 + bx + c dx

P ( x)

x2 + 1 4 Integrals of the form: dx, where P (x) is a ∫ x + kx 2 + 1 dx or ax 2 + bx + c x2 − 1 polynomial in x of degree n. ∫ x 4 + kx 2 + 1 dx, where k is a constant positive, negative

Integrals of the form:



or zero. Working Rule: Write



Working Rule

P( x)

2 n–1 ax + bx + c dx = (a0 + a1x + a2x + ... + an– 1 x )

2

ax + bx + c 2

+k



dx

ax + bx + c 2

where k, a0, a1, ... an – 1 are constants to be determined by differentiating the above relation and equating the coefficients of various powers of x on both sides.

(i) Divide the numerator and denominator by x2. (ii) Put x –

1 1 = z or x + =z x x

whichever substitution, on differentiation gives, the numerator of the resulting integrand. (iii) Evaluate the resulting integral in z (iv) Express the result in terms of x.

Indefinite Integration

Integrals of the form

194

∫P

Integrals of the form: quadratic functions of x.

dx , where P, Q are linear or Q

Objective Mathematics

Integral

Substitution

1

∫ (ax + b)

∫ (ax

2

dx + bx + c )

px + q

ax 2 + bx + c

dx

2



dx

x m ( a + bx )

p

Substitution Put a + bx = zx

,

cx + d = z2

where either (m and p are positive integers) or (m and p are fractions, but m + p = integers > 1)

px + q = z2

∫ x (a + bx )

1 px + q = z

(i) p is a + ve integer

Apply Binomial theorem to (a + bxn)p.

(ii) p is a – ve integer

Put x = zk where k = common denominator of m and n.

n p

m

dx,

where m, n, p are rationals.

dx

∫ ( px + q )

∫ (ax

dx

cx + d

Integral

x=

+ b ) cx 2 + d



Integrals of the form: n are integers.

1 ⋅ z

(

f x, ( ax + b )

α/n

) dx, where α and

Working Rule

(iii)

m +1 is an integer n

(iv)

m +1 + P is an integer n

Put ax + b = zn.

α/n

, ( ax + b )

β/ m

) where α, β, m, n are integers.

(ii) Integrate only the first integral by parts, i.e.,

Put ax + b = zk, where k = l.c.m. (n, m).

x ∫ e  f ( x ) + f ' ( x ) dx =

Integrals of the form (a) (b) (c)

x =  f ( x ). e −



m

∫ (a + bx)

p

p

m

p

Integral

∫ (a + bx)

m

n p

dx.

Substitution Put a + bx = z p

dx,

m is a + ve integer

∫ f ' ( x).e

x

x

dx  +

∫e

x

x

f ' ( x ) dx

f ' ( x ) dx

Integrals of the form

dx

m

∫ e f ( x) dx + ∫ e

= ex f (x) + C.

dx

∫ x (a + bx) dx ∫ x (a + bx) a ∫ x (a + bx ) xm

Working Rule (i) Split the integral into two integrals.

Working Rule

x

Put a + bxn = xn zk where k = denominator of fraction p.

x Integrals of the form: ∫ e  f ( x ) + f ' ( x ) dx

Integrals of the form

∫ f ( x, (ax + b)

Put a + bx n = z k , where k = denominator of p.

where the initial integrand reappears after integrating by parts. Working Rule (i) Apply the method of integration by parts twice. (ii) On integrating by parts second time, we will obtain the given integrand again. Put it equal to I. (iii) Transpose and collect terms involving I on one side and evaluate I.

Choose the correct alternative in each of the following problems: 1.

1 + x4

∫ (1 − x )

4 3/2

(a)

6.

dx =

1

+c

(b)

1 x − 2 x 1 +c 1 2 +x x2 2

(c)

2. If

1 +c 1 2 − x x2

1 sin (2x – c) + a, then 2

π , a = arbitrary constant 2 π (b) c = , a = arbitrary constant 6 π (c) c = , a = arbitrary constant 4 (d) None of these

(b) cosec α log

sin ( x + α ) +C sin x

(c) cosec α log

sec ( x + α ) +C sec x

(d) cosec α log

sec x + C. sec ( x + α )

(a) c =

3. If



4. If

dx

∫ 1 + sin x

1 1 (b) A = , B = 9 5

∫x

dx

 a+x − a−x

∫ 

(a) – (b)

a−x dx is equal to a + x 

a2 − x2 + C a2 − x2 + C

(c) –

x2 − a2 + C

(d) None of these 9.

∫ [1 + tan x. tan ( x + α)

dx is equal to

sin x (a) cot α . log sin ( x + α ) + C sin x (b) tan α. log sin ( x + α ) + C sin ( x + α) + C sin x (d) None of these (c) cot α. log

=

(a)

1 log 3

1 − x3 − 1

(b)

1 log 3

1 − x2 + 1

1 log 3

8.

x  = tan  + a  + b, then 2 

1 − x3

(c)

cos3 x + cos5 x dx is 2 x + sin 4 x

∫ sin

(a) sin x – 6 tan–1 (sin x) + c (b) sin x – 2 (sin x)–1 + c (c) sin x – 2 (sin x)–1 – 6 tan–1 (sin x) + c (d) None of these

(d) None of these

π (a) a = – , b = arbitrary constant 4 π , b = arbitrary constant (b) a = 4 π (c) a = , b = arbitrary constant 2 (d) None of these 5.

7. The value of

−9 −5   cos3 x dx = − 2  A tan 2 x + B tan 2 x  + C, then 11 sin x  

−1 1 (a) A = ,B= 5 9 1 1 (c) A = –  , B = 9 5

is equal to

sin x (a) cosec α log sin ( x + α ) + C

(d) None of these

∫ (sin 2 x + cos 2 x) dx =

dx

∫ sin x. sin ( x + α)

1 − x3 + 1 1 − x2 − 1 1 1− x

3

+c

1 (d) log 1 − x 3 + c 3

+c +c

10.



ex − 1 dx is equal to ex + 1

( (b) log (e (c) log (e

) − 1) + sec − 1) – sec

(a) log e x + e 2 x − 1 – sec– 1 (ex) + C x

2x

+ e

x

− e2 x

(d) None of these

– 1

(ex) + C

– 1

(ex) + C

195

Indefinite Integration

mULTIPLE-CHOICE QUESTIONS

196

11.

∫ tan ( x − α) tan ( x + α)

tan 2x dx is equal to

17. If

Objective Mathematics

4 4 8 , B = –  , C = 15 35 25 4 4 8 , B = –  , C = –  (b) A = 15 35 25 4 4 8 (c) A = , B = –  , C = 15 35 25 (d) None of these

sec 2x . sec ( x − α ) +C sec ( x + α )

(c) log

(d) None of these 12. If



dx A B x = 2 + + log + C, then x 4 + x3 x x x +1

1 , B = 1 2 1 (c) A = –  , B = 1 2

(a) A =

13.



dx

x 2 (1 + x 4 )

3/ 4

(a) –

(1 + x )

(c) –

(1 + x )

4 1/ 4

x

4 3/ 4

14.

x

∫ ( x + 1)

∫ sin

(b)

1 2

19.

(b) (1 + x x

)

4 1/ 4

(d) None of these

4

x dx is equal to x+3x



(a) x –

6 5/ 6 3 2/3 x − x – 2x1/2 + 3x1/3 + 6x1/6 5 2 + 6 log (1 + x1/6) + C

(b) x –

6 5/ 6 3 2/3 x + x – 2x1/2 + 3x1/3 – 6x1/6 5 2 + 6 log (1 + x1/6) + C

6 5/ 6 3 2/3 x + x – 2x1/2 + 3x1/3 – 6x1/6 5 2 + 6 log (1 + x 1/6) + C (d) None of these (c) x +

 +C

20. If

dx is equal to x + cos 4 x

(a) (– sin α, cos α) (c) (sin α, cos α) 21.

 1  2 tan– 1  tan 2 x  + C  2 

 1  1 cot 2 x  + C tan– 1    2 2 (d) None of these

then f (x) =

cos x dx =

1 log f (x) + c, 2 (b − a 2 )

= Ax + B log sin(x – a) + C, then value

of (A, B) is

1  1  tan– 1  tan 2 x  + C   2 2

∫ f ( x) sin x

sin x

∫ sin( x − α) dx

∫ cos x

(b) (cos α, sin α) (d) (– cos α, sin α)

dx is equal to cos 2 x

(a)

5   1 2  tan x + tan 2 x  + C 5  

(b)

5   1 2  cot x + tan 2 x  + C 5  

(c)

5   1 2  tan x − tan 2 x  + C 5  

(c)

16. If

3 3  = −  f ( x ) + g ( x )  + C, then   8 2 sin x cos x 11



is equal to

( x + 1)2

dx

3



+C

x2 + 2x + 1



(a) f (x) = tan–8/3 x, g (x) = tan–2/3x (b) f (x) = tan8/3x, g (x) = tan–2/3x (c) f (x) = tan–8/3x, g (x) = tan2/3x (d) None of these

x2 + 2x + 2 + C (d) None of these x +1

(c) –

(a)

x + 2x + 2 2

18. If

(d) None of these

x 2 + 2 x + 2 + C (b) – x +1

(a)

15.

+ C

dx 2

(b) A = 1, B = – 

is equal to

+ C

⋅ (1 + x 5/ 2 )1/ 2 dx

(a) A = – 

sec 2x +C sec ( x − α ). sec ( x + α )

(b) log

13/ 2

= A (1 + x5/2)7/2 + B (1 + x5/2)5/2 + C (1 + x5/2)3/2, then

sec 2x . sec ( x + α ) +C sec ( x − α )

(a) log

∫x

(d) None of these

2

22.

dx

∫ cos x − sin x

is equal to

(a)

1 1 (b) 2 2 a 2 sin 2 x + b 2 cos 2 x a sin x − b 2 cos 2 x

(a)

1  x 3π  log tan  −  + C 2 8  2

(c)

1 1 (b) 2 a 2 cos 2 x + b 2 sin 2 x a cos 2 x − b 2 sin 2 x

(b)

1 x log cot   + C 2 2

1  x π log tan  −  + C 2 8 2

(d)

1  x 3π  log tan  +  + C 2 8  2

(a) log

x  + C x + cos x

(b) log

dx is equal to x sin 2 x

(c) log

1  + C x + cos x

(d) log| x + cos x | + C

∫ cos

3

1  5/ 2  2    cot x + tan x  + C   5

(a)

30.



x2 − 2 x3

x2

(a) (c) 25.



(b) –

x2 − 1 x2

(d) –

x2 x2 − 1 x2 − 1 x2

(a) 2

1 − x + cos − 1

x + x − x2 + C

(b) 2

1 − x − cos − 1

x + x − x2 + C

1 − x + cos − 1

x + x − x2 + C

(d) None of these 26.



d (cos θ)

32.

1 − cos θ

dx

∫ sec x + cos ec x

(b) θ + c (d) sin–1 (cos θ) + c =

1 1 g (x), then f ( x) − 2 2 2

 x π (a) f (x) = sin x – cos x, g (x) = log cot  +  2 8  x π (b) f (x) = sin x – cos x, g (x) = log tan  +  2 8  x π (c) f (x) = cos x – sin x, g (x) = log tan  +  2 8  x π (d) f (x) = cos x – sin x, g (x) = log cot  +  . 2 8 28. If



2 1 + sin x dx = – 4 cos (ax + b) + c, then the value

of (a, b) is equal to 1 π (a) , 2 4 (c) 1, 1

x2 x2

2

2

2

2

2

2

2

2

2

x x ⋅ tan 2 2 dx is equal to 31. ∫ cos x x x 1 − 2 sin 1 + sin 1 2 2 − log log +C (a) x x 2 1 + 2 sin 1 − sin 2 2 x x 1 + 2 sin 1 + sin 1 2 − log 2 +C log (b) x x 2 1 − 2 sin 1 − sin 2 2 x x 1 + 2 sin 1 − sin 1 2 2 log − log +C (c) x x 2 1 − 2 sin 1 + sin 2 2 (d) None of these

2

(a) cos–1 θ + c (c) sin–1 θ + c 27. If

) dx is equal to +a )+ x +a +C +a )−2 x +a +C +a )− x +a +C

x2 + a2

(d) None of these

1− x dx is equal to 1+ x

(c) – 2

x + cos x +C x

sin



x2 − 1

dx is equal to

(a) x log x + x 2

dx is equal to

x2 − 1

∫ log ( x +

( (b) x log ( x + (c) x log ( x +

1   2  tan x + tan 5 / 2 x  + C   5 1 (c) 2  tan x − tan 5 / 2 x  + C   5 (d) None of these (b)

24.

∫ x ( x + cos x)

197

29.

π (b) 1, 2 (d) none of these

∫ sin

−1

x dx is equal to a+x

(a) (x + a) tan– 1

x − ax + C a

(b) (x + a) tan– 1

x + ax + C a

(c) (x + a) cot– 1

x − ax + C a

(d) None of these 33.



tan x a + b tan 2 x

dx, a > b > 0, is equal to

(a)

 1 b  +C log  cos x + cos 2 x + a − b  a−b 

(b)

 1 b  +C log  sin x + sin 2 x + a − b  a−b 

(c) –

 1 b  +C log  cos x + cos 2 x + a − b  a−b 

(d) None of these

Indefinite Integration

23.

cos x + x sin x

(c)

198

34. For the function f (x) = 1 + 3x log 3, the antiderivative F assumes the value 7 for x = 2. At what value of x does the curve y = F(x) cut the abscissa?

Objective Mathematics

(a) x = 3 (c) x = 0 35.

(b) x = 1 (d) None of these

tan x



41.

(

)

(a) log tan 2 x + 1 + tan 4 x + C

( 1 (c) log ( tan x + 4

42.

) 1 + tan x ) + C

1 log tan 2 x + 1 + tan 4 x + C 2

(b)

2



4

43.

then x 1+ x 2x

(c) f (x) = 37.

2

1 + x2

−x



(b) f (x) =



(d) None of these

dx



1 + x2 44.

sin x sin ( x + α )

cos α + sin α cot x + C

(a) 2 cosec α

cos α + sin α cot x + C

(b) – 2 cosec α

45.

x

∫ log (1 + cos x) − x tan 2 

dx is equal to

x 2 (b) log (1 + cos x) (c) x log (1 + cos x) (d) None of these

(a) x tan

39. Let f (x) =

40.

(a)

−1 2

(c)

2





(c)

2 3/ 2

and f (0) = 0, then f (1) = (b)

)

x −1 1− x

2

+ C

+ C

sin 3 x dx 1 + cos 2 x + cos 4 x

2

x

∫ cos x. log tan 2

(a) 1 –

1 2 47.

is equal to

(b)

(

2 1− x 1− x

π 4

(c) tan 1 –

(d) None of these.

1− x

(

∫ (1 + cos x)

is equal to

dx is equal to



(x

2

+ sin 2 x ) sec 2 x dx and f (0) = 0, 1 + x2

then f  (1) =

(1 + x )

(1 + x ) x − x 2 ( x − 1)

(a)

1 − sin x − x / 2 ⋅e dx is equal to 1 + cos x x x (a) sec ⋅ e − x / 2 + C (b) – sec ⋅ e − x / 2 + C 2 2 x (c) tan ⋅ e − x / 2 + C (d) None of these 2



46. If f (x) =

dx

dx

− sin 2 x – sin x + C 2 − sin 2 x (d) + sin x + C. 2 (b)

(a) sin x. log tan

(d) None of these  38.  

cos 5 x + cos 4 x dx is equal to 1 − 2 cos 3 x

x + x + C 2 x (b) sin x log tan –x+C 2 x (c) – sin x log tan – x + C 2 (d) None of these

cos α + sin α cot x + C

(c) cosec α

(d) None of these

(a) sec–1(sec x + cos x) + C (b) sec–1(sec x – cos x) + C (c) sec–1(cos x – tan x) + C (d) None of these

is equal to

3



 x −1  (b) ex  +C  1 + x 2 

sin 2 x  + sinx + C 2 sin 2 x – sinx + C (c) 2

1 1 − tan −1 x arc tan x − 2 − log f (x) + C, dx = 4 3x3 6x 3 x

(a) f (x) =

dx is equal to

(a)

(d) None of these 36. If

2

 1− x  (a) ex  + C  1 + x 2  1 + C (c) ex. 1 + x2

dx is equal to

sin 4 x + cos 4 x



1 − x  e   1 + x  x

)



(d) None of these

π 4

(

log x + 1 + x 2

(a) +C

(b)

1+ x

2

π – 1 4

(d) None of these

)

dx is equal to

(

)

2 1 log x + 1 + x 2  + C   2

(

)

2

(b) log x + 1 + x 2  + C   1 (c) log x + 1 + x 2  + C  2  (d) None of these

(

)

49.

3

x − sin x )

cos 2 x (a) esin x (sec x – x) + C (c) esin x (tan x – x) + C

dx is equal to (b) esin x (x – sec x) + C (d) None of these



x 2 + 1  log ( x 2 + 1) − 2 log x  dx is equal to x4

(a)

( x 2 + 1)3 / 2 x3

2  x2 + 1   − log  2   + C  x  3

(b)

( x + 1) 3x3

  x + 1 log  2  −  x  

2

3/ 2

2

 x2 + 1  ( x 2 + 1)3 / 2  2  − log  2   + C 3 3x  x   3 (d) None of these

50. If







54. If

2  +C 3

(c)

55.

1 log x 56.

(d) None of these 51.

3x + 1

∫ ( x − 1) ( x + 1) dx is equal to 1 1 1 + − log (a) ( x − 1) 2 2 ( x − 1) 4

x +1 x −1 + C

−1 1 1 − + (b) log ( x − 1) 2 2 ( x − 1) 4

x +1 x −1 + C

−1 1 1 + + log ( x − 1) 2 2 ( x − 1) 4

x +1 x −1 + C

57.



 x + 2 ex   x + 4 

x2

∫ ( x sin x + cos x)

52. If

2

dx =

x cos x −x (c) f (x) = cos x

f ( x) + tan x + C, x sin x + cos x



(b) f (x) =

cos x x

(d) None of these

x4 dx is equal to ( x − 1) ( x 2 + 1)

(a)

x2 1 1 + x +  log (x – 1) –  log (x2 + 1) 2 2 4 1  tan– 1x + C – 2

dx is equal to

(a)

xe x + C x+4

 x + 2 (b) ex  +C  x + 4 

(c)

ex + C x+4

(d) None of these

dx

∫ sin x + sin 2 x

is equal to

1 1 log (1 – cos x) – log (1 + cos x) 6 2 2 + log (2 cos x + 1) + C 3 1 1 (b) log (1 – cos x) + log (1 + cos x) 6 2 2 log (2 cos x + 1) + C – 3 1 1 (c) log (1 – cos x) + log (1 + cos x) 6 2 2  log (2 cos x + 1) + C + 3 (d) None of these

then (a) f (x) =

2

(a)

(d) None of these



x3 − 1 dx is equal to x3 + x

3

(c)

53.



(b) 2 (d) –2

1 log (x2 + 1) – tan– 1x + C 2 1 log (x2 + 1) – tan– 1x + C (b) x – log x + 2 1 log (x2 + 1) – tan– 1x + C (c) x – log x – 2 (d) None of these

1 log x

1 , g (x) = log (log x) (c) f (x) = log x

cot x + Q, then P equals

(a) x + log x +

then

(b) f (x) = log x, g (x) =

cot x

∫ sin x cos x dx = P

(a) 1 (c) –1

 1   log (log x ) +  dx = x [ f (x) – g (x)] + C, (log x )2  

(a) f (x) = log (log x), g (x) =

x2 1 1 +x+ log (x – 1) +  log (x2 + 1) 2 2 4 1  tan–1 x + C – 2 x2 1 1 – x +  log (x – 1) +  log (x2 + 1) (c) 2 2 4 1 +  tan–1x + C 2 (d) None of these (b)

−1

58.



e tan x (1 + x + x2) dx is equal to (1 + x 2 ) −1

(a)

e tan x 1 + x2

(c) xe tan

−1 x

(b) e tan

−1 x

 · (1 + x2)

(d) None of these

199

∫e

( x cos .

Indefinite Integration

48.

sin x

200

59.

∫ sin x.

(c) f (x) = tan

1 1 + 2 sin x 1 1 + sin x log +C − log 2 2 1 − sin x 1 − 2 sin x 4

(a)

Objective Mathematics

α π x  tan   −   2 4 2 (d) None of these

cosec 4x dx is equal to

1 + 2 sin x 1 1 + sin x log +C + log (b) 2 2 1 − sin x 1 − 2 sin x 4 1

64. If

(a) f (x) =

1 − 2 sin x 1 1 + sin x − log log +C (c) 1 − sin x 2 2 1 + 2 sin x 4 65.

 cos x + sin x  1 sin 2 x log  − log sec 2x + C 2  cos x − sin x  2

x



∫ cos  b log a  (a) –

66.

x  x x    cos  b log  + b sin  b log   + C   1 + b 2  a a 

x  x x    cos  b log  − b sin  b log   + C   1 + b 2  a a 

(c)

x  x x    cos  b log  + b sin  b log   + C   1 + b 2  a a 

∫ (x (a)

2

67.

2

2

1 − a2 )

x  −1 x − a tan − 1  + C  b tan a b

(b)

x 1  −1 x + a tan − 1   + C  b tan b2 − a 2  b a

(c)

1 b2 − a 2

x  −1 x + a tan − 1   + C  − b tan b a

(d) None of these. dx π 63. If ∫ (0 < α < and α is a constant) 1 + sin α. cos x 2 = 2 sec α tan– 1 [f (x)] + C, then (a) f (x) = cot

x π α tan  −  4 2 2

(b) f (x) = tan

x π α tan  −  4 2 2

1 2

1+ x dx is equal to 3 x

∫ 1+

sin 3 x dx ∫ (cos4 x + 3 cos2 x + 1) tan −1 (sec x + cos x) =

∫x

x1/ 2 dx is equal to − x1/ 3

1/ 2

5 2 1 1   1 1 6 3 3 2 x x x x x (a) 6  + + + + + x 6 + log( x 6 − 1) + C 6 5  4 3 2   5 2 1 1   1 1 x x6 x3 x2 x3 (b) 6  − + − + + x 6 − log( x 6 − 1) + C 6 5  4 3 2   5 2 1 1   1 1 x x6 x3 x2 x3 (c) 6  + − + − + x 6 + log( x 6 − 1) + C 6 5  4 3 2  

x2 dx is equal to + a ) ( x2 + b2 )

(b

(d) L =

(d) None of these

(d) None of these 62.

(b) g(x) = log x

(a) tan–1 (sec x + cos x) + c (b) log tan–1 (sec x + cos x) + c 1 +c (c) (sec x + cos x )2

dx is equal to

(b)

+ Lx + c,

3 5/3 3 4/3 x − x + x + C 5 4 3 5/3 3 4/3 (b) x + x + x + C 5 4 3 5/3 3 4/3 (c) x + x – x + C 5 4 (d) None of these

(c)

61.

2

(a)

 cos x + sin x  1 sin 2 x (a) log  + log sec 2x + C 2  cos x − sin x  2

 cos x − sin x  1 sin 2 x log  − log sec 2x + C 2  cos x + sin x  2 (d) None of these

x2 2

(c) L = 1

(d) None of these

(b)

1

then

1

 cos x + sin x  dx is equal to 60. ∫ cos 2x log   cos x − sin x 



∫ x log 1 + x  dx = f (x) log (x + 1) + g(x) log x

(d) None of these 68.

dx

∫ ( x − α) (β − x) , where β > α, is equal to  2x + α + β  (a) sin– 1  + C  β − α   x − α − β (b) sin– 1  +C  β − α   2x − α − β  (c) sin– 1  +C  β − α  (d) None of these

dx 5

(a)

3 4

4

x −1 + C x+2

(b)

(c)

4 3

4

x+2 + C x −1

(d) None of these

70. If

dx

∫ (1 + x )

4 3

x −1 +C x+2

4

(a) f (x) =

1− x 1 + x2

(b) f (x) =

(c) f (x) =

x2 − 1 x2 + 1

(d) None of these

2

71.



 3x − 4  = x + 2, then 74. If   3 x + 4  (a) ex – 2 ln

1 sin– 1 [  f (x)] + C, then 2

=–

1 − x2

2

4  1 + x 3 / 4 − log (1 + x 3 / 4 ) + C 3 4 (c)  1 − x 3 / 4 + log (1 + x 3 / 4 ) + C 3 (d) None of these (b)

is equal to

( x − 1) ( x + 2) 3

4

1+ x 1 − x2

75.



5

 +C 

4

 (1 + x )3 / 2 (1 + x )4 / 3 (1 + x )7 / 6 (b) 6  − + 9 8 7  −



76.

6

5



4

 

+

+

5

(d) None of these 72.

(x

∫x

2

− 1)

x 4 + 3x 2 + 1

(a) log x + (b) log x −

(c)

(c) log x + x + 3

+C

2

∫ 1+ (a)

x 4

x3

dx is equal to

4  1 + x 3 / 4 + log (1 + x 3 / 4 ) + C 3

dx is equal to

)

tan x + cot x

)

dx is equal to

2 cos– 1 (sin x – cos x) + C

4

+ C 

 3  x x + x +1  e ∫  1 + x 2 3/2  dx is equal to )  (

(c)

(d) None of these 73.



77.

dx is equal to

1 1 + x2 + 2 − 3 + C x x 2

∫(

(a)

1 1 + x2 + 2 + 3 + C x x

(

x 1+ 3 x

(d) None of these

(1 + x ) (1 + x )5 / 6 (1 + x )2 / 3  6

x + 3 x2 + 6 x

(b) 2 sin– 1 (sin x – cos x) + C

 (1 + x )3 / 2 (1 + x )4 / 3 (1 + x )7 / 6 (c) 6  + − 9 8 7  



(a) sin– 1 (sin x – cos x) + C

(1 + x ) (1 + x )5 / 6 (1 + x )2 / 3  + C +

3x − 4 +c 3x + 4

(a) 3 x2/3 + 6 tan– 1 x1/6 + C 2 (b) 3 x2/3 – 6 tan– 1 x1/6 + C 2 − 3 2/3 x + 6 tan– 1 x1/6 + C (c) 2 (d) None of these

(1 + x ) (1 + x )5 / 6 (1 + x )2 / 3  +

is equal to

(d) None of these

 (1 + x )3 / 2 (1 + x )4 / 3 (1 + x )7 / 6 + + (a) 6  9 8 7 

6

∫ f ( x) dx

(b) – 8 ln x − 1 + 2 x + c 3 3 (c) 8 ln x − 1 + x + c 3 3

2

x dx is equal to 3 1+ x + 2 1+ x



201



x 2e x

+c

(b)

+c

(d)

(1 + x )

2 1/ 2

ex

(1 + x )

2 1/2

78. The value of



1+ x −1 1+ x +1

 1 + x − 1 (b) ln  +c  1 + x + 1 (c) 2 1 + x + c 1+ x −1 1+ x +1

x (1 + x 2 )

1/2

xe x

(1 + x )

2 1/2

1− x dx is x

(a) 2 1 + x + ln

(d)

ex

+c

+c

+c

+c

Indefinite Integration

69.

202

79.

2

∫ (2 − x )

2

3

2− x dx is equal to 2+ x

Objective Mathematics

4 2 + x 3  2 − x 

2/3

3 2 − x (c) 4  2 + x 

2/3

(a)

80. The value of

∫x

(b)

 + C

(d) None of these

dx

n

(1 + x n )

1− n

1 n

1−

1 n

1−

1  1 1 − n  x  1+ n 

1  1 1 − n  x  1− n 

81.

1 n

1 1− n

1  1 1 + n  n −1  x 

(d) –

1− x

∫1+x

2

2



dx 1 + x4

85.

+c 86.

is equal to

(b)

x 2 1 sin − 1  2  +C 2  x + 1

87.

(a) f (θ) = sin 2θ + 1

(b) f (θ) = 1 – sin 2θ

(c) f (θ) = sin 2θ – 1

(d) None of these



x3

(1 + x )

2 1/ 3



88.



dx

( 1+ x

)

(n ≠ ± 1) =

n

1  zn +1 zn −1  + + C, 2  n + 1 n − 1 

1 + x2

(b) z =

(c) z = x +

1 + x2

(d) None of these

(1 + x ) 4

∫ (1 − x )

4 3/ 2

(a)

 xn  1 log  n +c n  x + 1 

(c)

(d) None of these

−x

2

(a) z = x –

dx dx is equal to n + 1)

∫ x (x

 xn  +c (c) log  n  x + 1 

dx is equal to

where

= cosec– 1 [ f (θ)] + C, then

 xn + 1  1 (b) log  n  + c n  x 

x 2

(a) 20 (1 + x2)2/3 (2x2 – 3) + C 3 (b) 3 (1 + x2)2/3 (2x2 – 3) + C 20 (c) 3 (1 + x2)2/3 (2x2 + 3) + C 20 (d) None of these

x 2 1 sin − 1  2   + C 2  x + 1

(sin θ − cos θ) ∫ (sin θ + cos θ) sin θ cos θ + sin 2 θ cos2 θ

(a)

dx is equal to

(1 + xe )

 1 + xe x  1 (c) log  +C +  xe x  1 + xe x (d) None of these

(d) None of these

83.

( x + 1)

 x  1 (b) log  +C +  1 + xe x  1 + xe x

+c

x 2 2 sin − 1  2 +C  x + 1

82. If

∫x

 xe x  1 +C (a) log  +  1 + xe x  1 + xe x

(a)

(c)

, n ∈ N, is

+c

1−

(c) –

(a) tan– 1x +

+C

+c

x4 + 1 ∫ x6 + 1 dx is equal to 1 tan– 1 x3 + C 3 1 (b) tan– 1x – tan– 1 x3 + C 3 1 (c) – tan x – tan– 1 x3 + C 3 (d) None of these

2/3

 + C

1  1 (a) 1 + n  x  1− n  (b)

3 2 + x 4  2 − x 

84.

89.



x 1− x 2x

1 − x4

(x x

(a) (c)

4

2

4

1 + x2 – x

dx is equal to + C

(b)

+ C

(d)

− 1)

x + x2 + 1 4

x x4 + x2 + 1 2x x4 + x2 + 1

−x 1 − x4 − 2x 1 − x4

+C +C

dx is equal to

 + C

(b)

x4 + x2 + 1 +C x

 + C

(d)

x4 + x2 + 1 +C 2x

x −1 x +x +x

(a) tan– 1

3

2

dx is equal to

x + x + 1 + C x

(b) 2 tan– 1

x + x +1 + C x

(c) 3 tan– 1

x 2 + x + 1 + C x

(1 + x 2 )

 (a) – log   x −    (b) log   x −  

98.

1   +  x − x

1   +  x − x

2  1  − 2  + C x  

2  1  + 2  + C x  

2   1 1  (c) log   x −  +  x −  − 2  + C  x x    

dx

∫ x (a + bx)

2

=

a + bx a a + bx (c) z = x

(b) z =

(a)

2sec α (tan x + tan α) + C

(b)

2sec α (tan x − tan α) + C

(c)

2sec α (tan α − tan x) + C



3

x2

1 12 (2 + x2/3)5/4 + C (2 + x2/3)9/4 – 3 5

dx is equal to

(a) (1 + x1/3)3/2 + C (c) 2 (1 + x1/3)3/2 + C

(b) – (1 + x1/3)3/2 + C (d) None of these

dx

∫x

1 − x3

1 − x3 − 1

= a log

1 − x3 + 1

(a) 1/3

(b) 2/3

(c) – 1/3

(d) – 2/3

+ C, then a =

1 + x dx is equal to x



(a) 2

 1 + x − 1  +C 1 + x – 2 log   1 + x + 1 

(b) 4

 1 + x + 1  +C 1 + x + 2 log   1 + x − 1 

4e x + 6e − x dx = Ax + Bln (9e2x – 4) + C, then x − 4e − x

∫ 9e

35 3  , B = , C = a constant 36 2 3 − 35 , C = a constant (b) A = , B = 2 36 3 − 35 , C = A constant (c) A = – , B =   2 36 (d) None of these

(d) None of these

(d) None of these 95.

(c)

(a) A = –

a + bx b

sec x dx is equal to sin (2 x + α) + sin α

1+ 3 x

2 12 (2 + x2/3)9/4 – (2 + x2/3)5/4 + C 3 5

99. If

− 1  z2 b3  − 3bz + 3b 2 log z +  + C, 4  a 2 z

(a) z =



(b)

 1 + x − 1 1 + x + 2 log   +C  1 + x + 1  (d) None of these

where

94.

2 12 (2 + x2/3)9/4 + (2 + x2/3)5/4 + C 3 5

(c) 4

(d) None of these 3

(a)

97. If

dx is equal to

1 + x4

(2 + x 2 / 3 )1/ 4 dx is equal to

(d) None of these

(a) tan– 1 (tan x + cot x) + C (b) tan– 1 (cot x – tan x) + C (c) tan– 1 (tan x – cot x) + C (d) None of these

93.

1/ 3

2

dx 91. ∫ 6 is equal to sin x + cos 6 x

∫x

∫x

2

(d) None of these

92.

96.

203

∫ ( x + 1)

100.



3

1+ 4 x x

dx is equal to

(

 1+ 4 x (a) 12    7 

(

 1+ 4 x (b) 12    7 

(

 1+ 4 x (c) 6    7 

)

7/3

(1 + x ) + 4

4/3

  +C  

4/3

  +C  

4

)

7/3

)

7/3

(d) None of these

(1 + x ) − 4

4

(1 + x ) − 4

4

4/3

  +C  

Indefinite Integration

90.

204

∫ f ( x) sin x cos x

101. If

dx =

Objective Mathematics

then f (x) is equal to

1 log [ f (x)] + C, 2 (b − a 2 ) 2

106.

(c)

2

(d) None of these 102.



3

x

7

107.

1 + 3 x 4 dx is equal to

21 1 + 3 x4 32 3

4

3

4

8/ 7



(x − x )

5 1/ 5

8/ 7

+C

8/ 7

+C

5 1   − 1 24  x 4

(b)

5  1 1 − 4  x 24

(c) –

∫e

104.

∫x

(1 + x ) 3

2

109.

2/3

+C

+C

1/ 2

(a)

+C

(c)

110.

is equal to



2 5/ 4

1+ x

1 + x2

1+ x

(a) z =

 1  x + (a) 3  log +C 1 + x1/ 3 1 + 3 x  

2

− x 4

dx 4

(d) A = 1

(1 + x )

−2 x 4

6/5

4

4

=

is equal to

+ C

(b)

+ C

(d)

111.

 1+ x 1  − (c) 3  log 3  +C 1+ 3 x  x  3

∫ (a)

(d) None of these dx = a cos 8x + C, then

(b)

2 x 4

1 + x2

4

1 + x2

x

1 + x4 x

(b) z =

1 + x4

(d) None of these

(

− 1 + x2 + log x + 1 + x 2 x

( + log ( x +

) +C

) 1+ x ) + C

1 + x2 + log x + 1 + x 2 + C x

(a) a =

−1 16

(b) a =

1 8

(c)

(c) a =

1 16

(d) a =

−1 ⋅ 8

(d) None of these

x

+C

x 4

x 2 + 1 dx is equal to x2

1 + x2

+C

 1 1 1+ z log − tan −1 z  + C, where 2  2 1− z 

4 4 (c) z = − 1 + x x

 1+ 3 x 1  + (b) 3  log 3  +C 1+ 3 x  x 

∫ tan 2 x − cot 2 x

. cos ec3 x (cos2x + sin x + cos x + sin x cos x) dx

dx

∫x

6/5

cos 8 x + 1

+C

(c) A = – 1

1/ 3

105. If

1  x2  2a  a + bx 2 

cos ec x

(d) None of these dx

3/ 2

sin 8 x − cos8 x dx = A sin 2x + B, then 2 x cos 2 x 1 1 (a) A = –  (b) A = 2 2

+C

5 1   − 1 24  x 4

+ C

∫ 1 − 2 sin

108. If

6/5

(a)

3/ 2

(a) e cosec x (cosec x + cot x) + C (b) e cosec x + C (c) – e cosec x (cosec x + cot x) + C (d) None of these

dx is equal to

x6

1  x2  3a  a + bx 2 

is equal to

+C

(d) None of these 103.

dx is equal to

(d) None of these

( ) 32 (d) (1 + x ) 21 7 (c) (1 + x ) 32

(a)

2 5/ 2

1  x2  (b) 3a  a + bx 2 

1 a cos x − b 2 sin 2 x 2

(a + bx )

(a) –

1 (a) 2 2 a sin x − b 2 cos 2 x 1 (b) 2 2 a sin x + b 2 cos 2 x (c)



x2

2

(a) tan

−1

(log x) + C x

120.

x 2 + 4 + C

(c) 2 114.

(b)

∫ x. (x ) . x x

1 x2 + 4

(a) x + C (c) xx. log x + C

(b) (xx)x + C (d) None of these

(d) θ = 2 sin– 1 (x + 1)

∫ u v" dx = uv′ – vu′ + a, then a = (a) ∫ u" .v dx (b) ∫ u' .v dx (c) ∫ uv' dx (d) None of these

121. If

122. If

115. The equation of a curve passing through origin is given by y =

written in the form x = g ( y), then (a) g (y) = (b) g (y) = (c) g (y) =

sin sin

4

−1

(4 y )

−1

(4 y )

123.

(a) 3 sin x – (3x + 4) cos x (b) 3 sin x + (3x + 4) cos x (c) – 3 sin x + (3x + 4) cos x (d) None of these

[ f ( x) g' ( x) + g ( x) f ( x).g ( x)

124.

f' ( x)] [log f (x) + log g (x)] dx is

(a) f (x) g (x) log (f (x) g (x)) + C

118. If f ′′ (x) = sec2x and f (0) = f ′ (0) = 0 then

dx

(a) A =

2 3

5 (c) B = 3

(d) None of these

∫ sin x d (cos x)

is equal to

(b) f (x) = sec2x (d) None of these

x 4  = A tan– 1  B tan +  + C, then  2 3 (b) A =

d2

∫ dx

2

(b)

1  sin 2 x  − x + C   2  2

(d) None of these

(tan–1x) dx is equal to

1 +C 1 + x2 (b) tan– 1 x + C 1 log (1 + x2) + C (c) x tan–1 x – 2 (d) None of these



dx 2ax − x 2

= ( f o g) (x) + C, then

x+a a x−a –1 (b) f (x) = sin x, g (x) = a x−a –1 (c) f (x) = cos x, g (x) = a x−a –1 (d) f (x) = tan x, g (x) = a sec 2x 126. If ∫ dx = f [g (x)] + C, then (a) f (x) = sin– 1x, g (x) =

(d) log f (x).g (x) + C

∫ 5 + 4 sin x

(b) ag (ax + b) + C

1 (c) [g(ax + b) + C] a

125. If

(b) 1 [ log f ( x) g ( x)]2 + C 2 (c) [ log f ( x) g ( x)]2 + C

119. If

dx is equal

(a)

equal to

(a) f (x) = log sec x (c) f (x) = log secx + x

∫ f (ax + b)

(a) g (ax + b) + C

(a)

116. The anti-derivative of the function (3x + 4) | sin x |, when 0 < x < π, is given by



dx = g (x) + C, then

sin 2 x – x + C 2 1  sin 2 x  + x   + C (c)   2 2

sin − 1 ( 4 y )

(d) None of these

117.

∫ f ( x)

to

3 4 ∫ x cos x dx. If the equation of the curve is

3

sin 4 θ  −1   3θ − 2 sin 2θ +  + C, 8 4 

 1  (c) θ = 2 sin– 1    1 + x 

(2 log x + 1) dx is equal to

( xx )

=

 1  (b) θ = sin– 1    1 + x 

+C

(d) None of these

x + 2x 2

(a) θ = sin–1(x + 1)

(d) None of these (c) tan x + C x 1 w.r.t (x2 + 3) is equal to 113. Integral of x2 + 4 x 2 + 4 + C

dx

5

where

(b) tan– 1 (log x) + C

−1

(a)

∫ (1 + x)

1 3

5 (d) B = –  3

(a) dom ·  f (x) = R – {0} (c) f ′ (x) =

205

1 w.r.t. log x is 1 + (log x) 2

Indefinite Integration

112. Integral of

(b) range of g (x) = R

1 π , ∨ x ∈ R+ (d) g ′ (x) = – cosec2   − x  . 2x 4 

206

x +1 ⋅ If 127. Let f (x) = x+2

Objective Mathematics

= 

 f ( x)  ∫  x 2 

 2 f ( x) − 1   1 g   −h 2  2 f ( x) + 1  

1/ 2

dx

132.

1+ x dx is equal to 1− x

(a)   − sin

f ( x) − 1   , then f ( x) + 1 

(a) g (x) = log | x |, h (x) = log | x | (b) g (x) = log | x |, h (x) = tan– 1x (c) g (x) = tan– 1x, h (x) = log | x | (d) None of these



−1

x − 1 − x2 + c

(b)   sin

−1

x + 1 − x2 + c

(c)   sin

−1

x − 1 − x2 + c

(d)   − sin

−1

x − x2 − 1 + c

133. ∫ 32x3(log x)2 dx is equal to

(a)  8x4(log x)2 + c 128. Let f (x) be a polynomial of degree three satisfying (b)  x4 {8(log x)2 – 4(log x) + 1} + c f (0) = – 1 and f (1) = 0. Also, 0 is a stationary point (c)  x4 {8(log x)2 – 4 log x} + c of f (x). If f (x) does not have an extremum at x = 0, (d)  x3 {(log x)2 – 2 log x} + c f ( x) dx is equal to then ∫ 3 cos x − 1 x x −1 e dx is equal to 134. ∫ sin x + 1 2 x e x sin x (a) + C (b) x + C. (b)   c − (a)  ex cos x + c 2 1 + sin x x3 + C (d) None of these (c) ex e x cos x (c)   c − (d)   c + 6 1 + sin x 1 + sin x 129. If In = ∫ tan n x dx , then I0 + I1 + 2 (I2 + ... + I8) + I9 + −1 e m tan x I10 is equal to 135. ∫ dx equals 1 + x2 2 9  tan x tan x tan x  m tan –1 x + + ... + −1 1 (a)  + c (b)   e (a)   e tan x + c 2 9   1 m −1 1  tan x tan 2 x tan 9 x  (c)   e m tan x + c (d)  none of these + + ... + (b) –  m 2 9   1 dx is 136. The value of ∫ 2 sin x cos 2 x  cot x cot 2 x cot 9 x  + + ... + (c)  (a)  tan x – cot x + c (b)  tan x + cot x + c 2 9   1 (c)  sec x – tan x + c (d)  none of these  cot x cot 2 x cot 9 x  + + ... + (d) –  2 9   1 130.

∫x

( x + 1)

dx is eqaul to

(1 + x e )

x 2

1 xe x +C (a) log − 1 + xe x 1 + xe x 1 xe x (b) log +C + x 1 + xe 1 + xe x (c) log

137.

∫x

2

dx is equal to + 4 x + 13

(a)  log (x2 + 4x + 13) + c (b)  

1  x+2 tan −1  +c 3  3 

(c)  log (2x + 4) + c

2x + 4 +c ( x 2 + 4 x + 13) 2

(d)   1

∫ [( x − 1) ( x + 2) ]

138. The value of

3

5 1/ 4

1/ 4

x

1 + xe 1 +C + xe x 1 + xe x

(a)  

4  x −1  3  x + 2 

(c)  

4  x +1  3  x − 2 

1/ 4

+c

(b)  

4  x +1  3  x + 2 

+c

(d)  

4  x −1  3  x − 2 

1/ 4

(d) None of these ex 131. ∫ dx is equal to x (2 + e ) (e x +1)  e x +1  (a)   log  x  + c  e +2 

 ex + 2  (b)   log  x +c  e +1 

 e x +1  (c)    x  + c  e +2 

 ex + 2  (d)    x +c  e +1 

139.



ax/2 a− x − a x

(a)  

dx is

+c

1/ 4

+c

dx is equal to

1 1 sin −1 (a x ) + c (b)   tan −1 (a x ) + c log a log a

(c)   2 a − x − a x + c

(d)  log (ax – 1) + c

ex ∫ (2 + e x ) (e x + 1) dx is equal to

π (c)   x + log sin  x −  + c 4 

207

 ex + 1  (a)   log  x +c e +2  ex + 1  (c)    x +c e +2

π  (d)   x − log cos  x −  + c 4 

Indefinite Integration

140.

 ex + 2  (b)   log  x +c  e +1   ex + 2  (d)    x +c  e +1 

141. ∫ [sin (log x) + cos (log x)] dx is equal to (a)  x cos (log x) + c (c)  cos (log x) + c

142.

(b)  sin (log x) + c (d)  x sin (log x) + c

1 + x + x + x2 ∫ x + 1 + x dx is equal to 1 1+ x + c 2 2 (b)   (1 + x)3/ 2 + c 3

(a)  

(c)   1 + x + c (d)  2 (1 + x)3/2 + c 143.

dx

∫ cos x +

equals

3 sin x

1 x π (a)   log tan  +  + c 2  2 12  1 x π (b)   log tan  −  + c 2  2 12  x π (c)   log tan  +  + c  2 12  x π (d)   log tan  −  + c  2 12  x f o f o .. o f . 144. Let f ( x) = for n ≥ 2 and g ( x) =    (1 + x n )1/ n f occurs n times Then ∫ xn – 2g(x) dx equals

(a)  

1 1− 1 (1 + nx n ) n + c n(n − 1) 1−

1 n

(b)  

1 (1 + nx n ) n −1

(c)  

1 1+ 1 (1 + nx n ) n + c n(n + 1)

(d)  

1 1+ 1 (1 + nx n ) n + c n +1

145. The value of

2∫

+c

sin x dx is π  sin  x −  4 

146.

∫ x( x )

x x

(2log x + 1) dx = ….

(a)  (xx)x + c

(b)  log (x)x + c

(c)  xx + c

(d)  none of these.

147. Let f(x) = f(x). Then

x fofo ... of ) for n ≥ 2 and g(x) = (  (1 + x n )1/ n f occurs n times

∫x

n– 2

g(x)dx equals

(a)  

1 1− 1 (1 + nx n ) n + K n(n − 1)

(b)  

1 1− 1 (1 + nx n ) n + K n −1

(c)  

1 1+ 1 (1 + nx n ) n + K n(n + 1)

(d)  

1 1+ 1 (1 + nx n ) n + K n +1

148. Let l =

∫e

4x

ex e− x dx, J = ∫ −4 x dx. 2x + e +1 e + e −2 x + 1

The, for an arbitrary constant C, the value of J – l equals  e4 x − e2 x + 1  1 (a)   log  4 x +C 2x 2  e + e +1  e2 x + e x + 1  1 (b)   log  2 x +C x 2  e − e +1  e2 x − e x + 1  1 (c)   log  2 x +C x 2  e − e +1  e4 x + e2 x + 1  1 (d)   log  4 x +C 2x 2  e − e +1 1 + sin x 149. The area of the region between the curves y = cos x 1 − sin x y = and bounded by the lines x = 0 and cos x π x= is 4 2 −1

(a)  

∫ 0

t (1 + t 2 ) 1 − t 2

2 −1

(b)  

∫ 0

π (a)   x + log cos  x −  + c 4 

(c)  

π (b)   x − log sin  x −  + c 4 

(d)  

4t (1 + t 2 ) 1 − t 2

2 +1

∫ 0

4t (1 + t ) 1 − t 2 2

2 +1

∫ 0

t (1 + t ) 1 − t 2 2

dt

dt dt dt

208

solutions

Objective Mathematics

1. (b) I =



=

Let



1 + x4



dx =

(1 − x ) ( x + 1 x ) dx 4 3/2

∫1

x 3 ( x + 1/ x 3 ) dx

4. (a)

(1 − x )

4 3/2

dx

=





=

1 2



π x = – tan  −  + k, 4 2

3

2  2 − x  x

3/2

 −2  ⇒  3 − 2 x  dx = dt x

1 – x2 = t x2

5. (a) I =

∫ x (1 − x )

3 −1/2

−1

dt 2 3 ∫ 1 − t2 = 2 dt = 2 ⋅ 1 log t − 1 + c 3 ∫ t2 −1 3 2 t +1

π  2 ∫ cos  2 x −  dx  4



1 π  = 2 . sin  2 x −  + K,  2 4 where K is arbitrary constant

−1

1  ln  3 

=

1 π  sin  2 x −  + K.  4 2 π and a = arbitrary constant. ∴ C = 4 3. (b)  Since, the sum of powers of

6. (a)

3 11 sin x and cos x = = – 4 = negative even − 2 2 integer, ∴ put tan x = z ⇒ sec2x dx = dz





=

=

∫ ∫



dx tan x. cos x 11

(1 + z ) dz

4



2

tan

11/ 2

x

dx

2

x

=

∫ (z

− 11/ 2

+z

− 7/2

)

1 1 and B = 5 9

=

1 sin α

sin (( x + α ) − x ) ∫ sin x. sin ( x + α)

∫ cot x − cot ( x + α)

sin x. sin ( x + α ) dx

= cosec α [log sin x – log sin (x + α)] + C

7. (c) I = dx

=





(cos

2

sin x + C. sin ( x + α )

x + cos 4 x ) cos xdx

sin 2 x + sin 4 x

1 − t 2 + (1 − t 2 ) t2 + t4

2

dt,  where sin x = t

(1 − t ) (2 − t ) dt t (1 + t ) (2 − t ) dt 2−t 2 dt − 2



2

2

∫t

2

2

2

∫ t2 + t4 dt 3dt = 2⋅ ∫ 2 − 4 ∫ 2 dt + ∫ dt 2 t t (1 + t ) =

2

dx

(sin ( x + α) cos x − cos ( x + α) sin x)

= cosec α

= –

∴ A =

=−



dz

2 − 9/ 2 2 − 5/ 2 z − z +C 9 5 1 1  − 9/ 2 x + tan − 5 / 2 x  + C. = – 2  tan  9 5

2

= cosec α

=

2

z11/ 2

=

(1 + tan x) sec

tdt

1 − x3 − 1  +c. 1 − x 3 + 1 

∫ sin x. sin ( x + α)

= cosec α log

cos3 x dx tan11 x. cos11 x

−2

  ∫ (t )  3  1 − t

I =

=



2 tdt 2 tdt = − 3 x3 3 1 − t2

⇒ x–1dx = –

cos3 x dx = sin11 x

dx

Let 1 – x3 = t2 ⇒ –­ 3x2dx = 2t dt





π x  −  dx 4 2

π and b = arbitrary constant. 4

∴ a = –

2. (c) We have, dx =

2

 x π = tan  −  + k. 2 4



3 − +1

∫ (sin 2 x + cos 2 x)

∫ sec

where k is any arbitrary constant.

1 1  ⇒  x + 3  dx = – dt  x  2 1 dt 1 t 2 1 ∴ I = – = ⋅ +c = +c 3/2 ∫ 2 t 2 −3/2 + 1 t 1 +c. = 1 2 −x x2

dx π  1 + cos x  − x  2 

∫ 1 + sin x

dx

−2 –6 tan–1 (t) + t + c t = sin x – 2(sin x)–1 – 6 tan–1 (sin x) + c.



= tan 2x – tan (x – α) – tan (x + α).



∫ tan ( x − α) tan ( x + α)



=

∫ [ tan 2x − tan ( x − α) − tan ( x + α)]  dx



=

1 log | sec 2x |  – log  | sec (x – α) |  2

=

8. (a)

 a+x ∫  a − x −



dx

z dz dx = − ∫ 2 2 z a −x 2 [Put a – x2 = z2 ⇒ – 2x dx = 2 zdz]

12. (c)

cos α

∫ cos ( x + α) cos x

sin [ ( x + α) − x ] = cot α. ∫ cos ( x + α) cos x dx.

∫ [ tan ( x + α) − tan x]

13. (a)





θ 2 θ cos 2 sin

(1 − cos θ) ∫ cos θ dθ

=



=

∫ (sec θ − 1) d θ = log (sec θ + tan θ) – θ + C

(

)

x 2x = log e + e − 1 – sec– 1 (ex ) + C.

=

tan ( x − α) + tan ( x + α) 1 − tan ( x − α). tan ( x + α)

 1  dx x ( x + 1) 



 1 1 +  x2 x ( x + 1)  dx



1 1 1  + − x 2 x x + 1  dx

2

∫  x

=

−1 1 + + log | x | – log | x + 1 | + C 2x2 x

3

3

1 and B = 1. 2

∫x

−1 4

dt

∫t

3/ 4

1 dx −1    Putting 1 + x 4 = t ⇒ x 5 = 4 dt 



1/ 4

+C

dx x2 + 2x + 2

2

∫ ( x +1)

dx 2

( x +1) 2 + 1

=

sec 2 θ d θ ∫ tan 2 θ sec θ

[Putting x + 1 = tan θ ⇒ dx = sec2θ dθ]

 =

dx 1  x 5 1 + 4   x 

(1 + x 4 )1/ 4 + C. x

∫ ( x +1) =

3/ 4



1  −1 t1/ 4 ⋅ + C = – 1 + 4  x 4 1/ 4

= –  14. (c)

dx = (1 + x 4 )3 / 4

2

−1 cos θ dθ = +C = − 2 sin θ θ

∫ sin

( x + 1) 2 + 1 +C x +1 x + 1  ∵ tan θ = 1 

11. (b) tan 2x = tan [(x – α) + (x + α)]



=

=

sec θ − 1 ⋅ tan θ d θ   sec θ + 1

θ θ  2 sin cos  2 2  dθ =   cos θ  

3

∫  x

=

[Put ex = sec θ ⇒ dx tan θ d θ]

1

∫  x

=

dx

sin ( x + α ) + C. sin x

ex − 1 dx = ex + 1

( x + 1) − x dx = x 3 ( x + 1)

x −1 1 = 2 x 2 + x + log x + 1 + C.

= cot α. log sin ( x + α) − log| sin x | + C  

10. (a)



∴ A = –

sin ( x + α) cos x − cos ( x + α) sin x dx = cot α. ∫ cos ( x + α) cos x

= cot α. log

dx = + x3

dx

sin α = cot α. ∫ dx cos ( x + α) cos x

= cot α.

4

sec 2x + C. sec ( x − α). sec ( x + α)

1

dx



∫x

= log

1

a 2 − x 2 + C.

∫ 1 + tan x. tan ( x + α) =



2x

=–z+c=–

9. (c)

(a + x ) − (a − x )

=

– log | sec (x + α) | + C

a − x  dx a + x 

∫ (a − x) (a + x)

=

tan 2x dx



=–

x2 + 2x + 2 + C. x +1

Indefinite Integration

dt dt −6∫ + dt 1 + t2 ∫ t2

= 2∫

209

⇒ tan (x – α). tan (x + α) tan 2x

1 dt 1  = 6 ∫ 2 − dt − 4 ∫ 2 + ∫ dt t 1 + t 2  t

210

15. (a)

∫ sin

4

dx dx = x + cos 4 x ∫ (sin 2 x + cos 2 x) 2 − 2 sin 2 x cos 2 x

Objective Mathematics

2dx = ∫ 2 − sin 2 2 x =

sec 2 2 x dx = 2 2x



∴ put tan x = z ⇒ sec2x dx = dz ∴

dt

( 2)

2

+t2



dx 3

[Putting tan 2x = t ⇒ 2sec2 2x dx = dt] 1  t  tan − 1   2  + C 2

=



=

16. (c) f (x) sin x cos x = f' ( x)

∫ ( f ( x))



⇒ –

2

1  1  tan 2 x  + C. tan– 1   2  2

1 1 ⋅ ⋅ f' ( x) 2 (b 2 − a 2 ) f ( x )

dx =

=

(b

2

2 − a 2 ) (cos 2 x − sin 2 x )

2 b 2 ( 2 cos 2 x − 1) + a 2 ( 2 sin 2 x − 1)

17. (c)

f (x) =

∫x

13 / 2

1/ 2

dx

20. (b)

5 3/ 2 5 / 2 1/ 2 = ∫ x . x . (1 + x )

5 4 dx = ∫ x . z. z dz 5

5 3/ 2  5/ 2 2  Putting 1 + x = z ⇒ 2 x dx = 2 z dz  4  i.e. x 3 / 2 dx = z dz  5

 = 4 5

a 2 − b2 2

−1 . b 2 cos 2 x + a 2 sin 2 x + c

. (1 + x 5 / 2 )

2 2 ∫ z ⋅ (z – 1) dz = 2

4 5

    

2 4 2 ∫ z ( z − 2 z + 1) dz

4  z7 2z5 z3  = +  +C  − 5 7 5 3 4 8 = (1 + x5/2)7/2 – (1 + x5/2)5/2 35 25  ∴ A =

4 8 , B = –  35 25

dx

∫ 3

(1 + z ) 2



=





=

∫ (z



3 3  − 8/3 x + tan − 2 / 3 x  + C = –  tan 8  2

3

tan11 x − 11/ 3

=



sin x ⋅ cos12 x cos11 x 11

z

11/ 3

dz

+ z − 5 / 3 ) dz

x dx = x+3x





=6

∫ 

=6





z 3 . 6 z 5 dz z 6 dz =6∫ 3 2 z +z z +1

 z6 − 1 1  dz + z +1 z + 1  ( z 3 − 1) ( z +1) ( z 2 − z +1) dz dz + 6 ∫ z +1 z +1

5 4 3 2 = 6  ∫ ( z − z + z − z + z − 1)  dz + 6 log (z + 1)

1 b 2 cos 2 x + a 2 sin 2 x −

sec 4 xdx

=

19. (b) Put x = z6 ⇒ dx = 6z5 dz.

2 2 ∫ (b − a ) sin 2x.dx

=

sin x cos x 11

∴ f (x) = tan–8/3x and g (x) = tan–2/3x.

1  − cos 2 x  = (b 2 − a 2 )    f ( x) 2

⇒ f (x) =

−11 1 − = – 4 = – ve even 3 3

integer,

2 sec 2 2 x ∫ 2 sec2 2 x − tan 2 2 x dx

∫ 2 + tan

=2

18. (a) ∵ Sum of powers =

4 +  (1 + x5/2)3/2 + C. 15 4 ⋅ and C = 15

= z6 –

6 5 3 4 z + z – 2z3 + 3z2 – 6z + 6 log (z + 1) + C. z 2

=x–

6 5 / 6 3 2/3 x + x – 2x1/2 + 3x1/3 5 2 – 6x1/6 + 6 log (1 + x1/6) + C.

sin x

∫ sin( x − α) dx = Ax + B log sin (x – α) + c sin x B = A+ cos( x − α) sin( x − α) sin( x − α) ⇒ sin x = A sin (x – α) + B cos (x – α)



= (A cos α + B sin α)sin x



+ (– A sin α + B cos α) cos x

∴ A cos α + B sin α = 1 and – A sin α + B cos α = 0 ∴ B = sin α, A = cos α. 21. (d)   ∫

=

dx = cos x cos 2 x



sec 2 x 1 − tan x 2



dx =

dx cos x



1 − tan 2 x 1 + tan 2 x dx

1 − z2

[Putting tan x = z ⇒ sec2x dx = dz] = sin– 1 z + C = sin– 1 (tan x) + C.

1 π x π log tan  + +  + c 4 2 8 2

=

1  x 3π  log tan  +  + c 2 8  2 dx dx 23. (b) ∫ = ∫ 3 cos x sin 2 x 2 tan x cos3 x 1 + tan 2 x =

=



sec 4 x 1 dx = 2 tan x 2



24. (d)

∫x

dx

x −1

= sec– 1x – 2

∫x

x −1 2

−2 ∫

∫ sec x + cos ec x

x

x2 − 1

sec θ tan θ 3 θ tan θ dθ

∫ sec

= sec– 1 x – 2



= sec x –



sin 2θ  = sec– 1 x –  θ +  +C  2 



= sec– 1 x – sec– 1x –

∫ (1 + cos 2θ)

∴ 







=

1 2sin x cos x dx 2 ∫ sin x + cos x

=

1 2



=

1 2

∫ (sin x + cos x) dx

=

1 1 (– cos x + sin x) – 2 2 2

=

1 1 (sin x – cos x) – log tan  x + π  + C. 2 2 8 2 2

∫ 



(sin x + cos x )2 − 1 sin x + cos x

dx −

1 dx ∫ 2 sin x + cos x



= – 4  ∫ cos 2θ. 2 sin 2 θ d θ



= – 4  ∫ cos 2θ (1 − cos 2θ) d θ



= – 4  ∫ cos 2θ d θ + 2 ∫ (1 + cos 4θ) d θ

dx π  sin  x +   4

 x π f (x) = sin x – cos x and g (x) = log tan  +  . 2 8 2 1 + sin x dx = – 4 cos (ax + b) + c



2 (sin x / 2 + cos x / 2) dx = – 4 cos (ax + b) + c = – 4 cos (ax + b) + c



π x ⇒ − 4 cos  +  = – 4 cos (ax + b) 4 2 ∴ a = 1/2, b = π/4. 29. (a)

1 − cos 2θ 1 + cos 2θ (– 4 sin 2θ cos 2θ) dθ

= – 4  ∫ tan θ. cos 2θ. sin 2θ d θ



⇒ 2 2 (sin x / 2 − cos x / 2) + c

x2 − 1 +C x2



211

dx 1 1 + cos x sin x

=



x2 − 1 = – + C. x2 25. (c) Put x = cos2 2θ ⇒ dx = – 4 cos 2θ sin 2θ dθ 1− x dx = 1+ x

dx

∴ 28. (a)

2 ∫ cos θ  dθ



+ C.

= sin– 1 (cos θ) + c

1 − cos 2 θ

dx 3

[Putting x = sec θ ⇒ dx = sec θ tan θ dθ]

–1

1  sin 4θ + C 2

1 − x + cos– 1 x + x . 1 − x

d (cos θ)

z

x2 − 2

dx =

=–2

27. (b)

4

1   2  tan x + tan 5 / 2 x  + C.   5

2





(1 + z ). 2 z dz

 z5  2 z + +C = 5 

3

= – 2sin 2θ + 2θ +

26. (d)

[Putting tan x = z2 ⇒ sec2x dx = 2z dz] =



30. (c)

cos x + x sin x

∫ x ( x + cos x) 1



dx =

( x + cos x ) − x (1 − sin x ) x ( x + cos x )

1 − sin x 

∫  x − x + cos x 

dx



=



= log | x | – log | x + cos x | + C



= log

∫ log ( x +

dx

x + C. x + cos x x2 + a2

(

)

dx

= x  ∫ log x + x 2 + a 2 –

)

∫ x⋅ x+

1 x2 + a2

 1 + 

  dx x2 + a2  x

Indefinite Integration

dx 1 π 1  = sec  x +  dx ∫ ∫  π 4 2 2 cos  + x    4

22. (d) I =

Objective Mathematics

)

212

(

= x log x + x 2 + a 2 − ∫ = x log x + x + a

(

31. (b)



2

x x ⋅ tan 2 2 dx = cos x

sin

2

)− sin



x

dx

=–

x + a + C.

=–

x +a 2

2

x  ⋅ 2 

2

x sin 2 cos

1 − 2 sin 2

x x cos dx 2 2 = ∫  2 x  2 x 1 − sin  1 − 2 sin  2 2

2

x 2

x 2

dz

∫ (a − b) z 1 a−b

  dx

z dz

=–

 1 b  + C. log  cos x + cos 2 x + a − b  a−b 

2

=

dz −2∫ 1 − z2

2

1  2   −z 2 1 +z 1 1+ z 1 log 2 − 2 ⋅ log = +C 1 1 2 1− z −z 2. 2 2 x 1 + 2 sin 1 + sin 1 2 = ⋅ log − log x 2 1 − 2 sin 1 − sin 2

33. (c)

x = a+x

∫ θ.

x 2 + C. x 2

∫ sin



 tan 2 θ tan 2 θ  = 2a θ. −∫ d θ 2 2  



= a θ tan2θ – a



= a θ tan2θ – a (tan θ – θ) + C



= a θ (1 + tan2θ) – a tan θ + C



= (x + a) tan– 1

∫ =–

a + b tan 2 x



∫ (sec

dx =

dz

az 2 + b (1 − z 2

x log 3) dx = x + 3 + c

2

)

a cos 2 x + b sin 2 x

1 2



dx =



tan x. sec 2 x 1 + tan 4 x

dx

dz 1 + z2

(

)

=

1 log z + 1 + z 2 2

=

1 log tan 2 x + 1 + tan 4 x 2

+C

(



=

tan − 1 x dx = x4

∫ θ cot

=–

2

) + C.

=–  dx

dz

[Putting cos x = z ⇒ sin x dx = – dz]

=–

θ

∫ tan

4

θ

sec2θ d θ

θ cos ec 2θ d θ

θ cot 3 θ 1 + 3 3

3 = – θ cot θ + 1 3 3

θ − 1) d θ

sin x

sin 4 x + cos 4 x

∴ 

x − ax + C. a



tan x

36. (a) Put x = tan θ ⇒ dx = sec2θ d θ

2a tan θ sec2θ dθ



tan x

x

1   2 2  Putting tan x = z ⇒ tan x. sec x dx = dz  2

32. (a) Put x = a tan2θ ⇒ dx = 2a tan θ sec2θ dθ −1

∫ (1 + 3

∴ F(x) = x + 3x – 4 = 0 ⇒ x = 1. 35. (b)   ∫

 1 1  = 2 ∫ dz − 2 1 − z 2  1 − 2z

∫

+C

Since F (2) = 2 + 9 + c = 7 ⇒ c = –4

x 1 x    Putting sin 2 = z ⇒ 2 cos 2 dx = dz 

=

2

 1 b  log  z + z 2 + a − b  a−b 

∴ F(x) =

∫ (1 − z ) (1 − 2 z )

dz

 b  z2 +    a−b

34. (b) f (x) = 1 + 3x log 3.

2

2

dz

=–

sin 2

=2



+b

2

∫ cot

3

θ dθ⋅

∫ cot θ

1 1 θ cot3θ + 3 3

(cosec2θ – 1) dθ

∫ cot θ

cosec2θ dθ –

∫ cot θ 1 1 cot 2 θ 1 θ cot3θ – ⋅ − log sin θ + C 3 3 2 3

 x  1 1 tan − 1 x − 2 − log   + C. 3 2 3x 6x 3  1+ x  x ⋅ ∴ f (x) = 1 + x2 =–

1 3 dθ

=

=

3

dx x sin ( x + α )

42. (b) dx

∫ ∫

z

∫ z dz

[Putting cos α + sin α cot x = z2 ⇒ – sin α cosec2x dx = 2 zdz] = – 2 cosec α .  cos α + sin α cot x  + C. 38. (c)

=

x

∫ log (1 + cos x) dx − ∫ x tan 2

dx

x sin x x ∫ 1 + cos x dx − ∫ x tan 2 dx

= x log (1 + cos x) +

∫ x tan 2 dx − ∫ x tan 2

x

 sin 2 x  + sin x  + C. =–   2 



x

=–

dx

3/ 2

=

sec 2 θ d θ ∫ sec3 θ

44. (a)

=

Thus, f (x) =



x

dθ = sin θ + C =

∫ cos θ

Since f (0) = 0,

∫ (1 + cos x) 2

x + C. 2 sin 3 x

1+ x therefore C = 0. x . Hence, f (1) = 1 + x2

dx

(1 + x )

x−x

2

=

2

+ C.

1 2

2 sin θ cos θ d θ

∫ (1 + sin θ)

sin 3 x

∫ cos x (sec x + cos x) cos x

=



=

∫z

sin θ − sin θ 2

(

∫e

x

 1− x   1 + x 2  dx =

)

2 x −1  + C. +C = x 1− x

2

∫e . x

1 + x − 2x 2

(1 + x )

2 2

sec 2 x + 1 + cos 2 x

 dx

sin 3 x dx cos 2 x (sec x + cos x )

(sec x + cos x )2 − 1

dz z2 − 1

4

dθ 1 − sin θ = 2∫ = 2∫ d θ = 2 (tan θ – sec θ) cos 2 θ 1 + sin θ  x 1 − =2  1 − x 1 − 

dx

1 + cos 2 x + cos 4 x

=

  sin 3 x dx = dz   Putting sec x + cos x = z ⇒ cos 2 x  

[Putting x = sin2θ ⇒ dx = 2 sin θ cos θ d θ]

41. (c)

 ez dz = – ez sec z + C

[Putting x = tan θ ⇒ dx = sec2θ dθ]



40. (a)

d



∫ sec z + dz (sec z )

= – e– x/2 · sec

(1 + x 2 )

cos

−x    Put 2 = z ⇒ dx = − 2 dz 

= x log (1 + cos x). dx

sin 3 x (cos 5 x + cos 4 x ) dx sin 3 x − sin 6 x

x x − sin 2 2 43. (b) ∫ e– x/2 dx ∫ 2 x 2 cos 2 z = – ∫ (sec z + sec z tan z ) e dz

= x log (1 + cos x) +

39. (b) We have, f (x) =



1 − sin x ⋅ e– x/2 dx = 1 + cos x

x  dx



∫ log (1 + cos x) − x tan 2 

cos 5 x + cos 4 x dx = 1 − 2 cos 3 x

3x 3x   9x x  cos   2 cos cos   2 sin    dx 2 2 2 2 = ∫ 9x 3x − 2 cos sin 2 2 3x x = – ∫ 2 cos cos dx = − ∫ (cos 2 x + cos x ) dx 2 2

sin x. (sin x cos α + cos x sin α ) 3

cos ec 2 x −2 dx = sin α cos α + sin α cot x



dx

 1 2 x  x −  dx  =  ∫ e  2 2 1 + x (1 + x 2 )    1 d  1  ex =  ∫ e ⋅  + dx = + C. 2 2   dx  1 + x   1 + x2 1 + x

= sec– 1 z + c = sec– 1 (sec x + cos x) + C. 45. (b) ∫ cos x. log tan = log tan

x dx 2

x. sin x – 2

∫ sin x.

x – ∫ 1. dx 2 x – x + C. = sin x. log tan 2

= sin x ⋅ log tan

46. (c) We have, f (x) =

x

=





(x

x 2 + (1 − cos 2 x ) 1 + x2

2

x 1 ⋅ 2 2 dx x tan 2

sec 2

+ sin 2 x )

1 + x2

⋅ sec2x dx

sec2x dx

213

∫ sin

Indefinite Integration

37. (b)

214

Objective Mathematics

 2 1  ∫  sec x − 1 + x 2  dx

=

= tan x – tan– 1 x + C. ∵ f (0) = 0, ∴ C = 0.

 x2 + 1  ( x 2 +1)3 / 2  2  − log  2   + C. 3 3x  x  3 50. (a) Put log x = z ⇒ x = ez ⇒ dx = ez dz =

Thus, f (x) = tan x – tan– 1x. Hence, f (1) = tan 1 – tan– 1 1 = tan 1 –

π . 4

 1  1 z  ∴ ∫  log (log x ) +  dx = ∫  log z + 2  e dz z (log x )2    1   1 1  = ∫   log z −  +  + 2   ez dz z   z z  

47. (a) Put x = tan θ ⇒ dx = sec2θ dθ ∴



(

log x + 1 + x 2 1+ x

)

dx

2



=

∫ sec θ.

log (sec θ + tan θ) dθ d

=

∫ log (sec θ + tan θ). d θ

=

2 1 log (sec θ + tan θ) + C 2 

=

2 1  2  log x + + x 1  + C. 2 

(

48. (b) ∫ esin x

3

x − sin x )

∫e

=

∫ x (e

sin x

cos x ) dx –

d = ∫x (e sin x) dx – dx = xesin x –

∫e

sin x

∫e

=

=

∫e

sin x

sin x

∴ ∴

x + 1 log ( x 2 + 1) − 2 log x  dx 49. (c) ∫ x4 1 1 1  = ∫ 1 + 2 ⋅ 3 ⋅ log 1 + 2  dx  x x x 

( x − 1)

 −1  2z 1 2 − ∫ ⋅ z 3 / 2 dz  log z. z 3 2  3  −1 3/2 2 3/2 = z log z + z +C 3 9



2

=

53. (a)

1 2 ( x − 1)

2



 1 1 +  dx 4 ( x − 1 ) 4 ( x + 1)  x +1 x −1

1 1 + log 2 ( x − 1) 4

+ C.

1 =z x sin x + cos x −1

(sinx + x cosx – sinx) dx = dz

( x sin x + cos x )2 x cos x

 dx = – dz,

( x sin x + cos x )2 x cos x

∫ ( x sin x + cos x) x2

∫ ( x sin x + cos x)

2

2

 dx = – z.

dx =

x

x cos x

∫ cos x ⋅ ( x sin x + cos x)

 cos x. 1 + x sin x  ∫ (− z )  cos x 

x ⋅ (– z) – cos x

=

−x + sec 2 x  dx cos x ( x sin x + cos x ) ∫

=

−x + tan x + C. cos x ( x sin x + cos x )

x4

∫ ( x − 1) ( x

2

−x ⋅ cos x

∴ f (x) =

3/ 2

+

=

z log z dz 1 1 −1    Putting 1 + x 2 = z ⇒ x 3 dx = 2 dz 

2

−1



.1 dx

3

3



d (sec x) dx dx

2





(sec x tan x) dx

= esin x (x – sec x) + C.

−1 = 2

3x + 1

∫  ( x − 1)

52. (c) Put

sin x sin x – e . sec x − ∫ e . sec x. cos x dx 



1   dz z  

∫ ( x − 1) ( x + 1)

we have dx

(x cos x – sec x tan x) dx

sin x

1 d  +  log z − z  dz 

 1 1   ez  log z −  + C = x log (log x ) − + C  z log x   1 ∴ f (x) = log (log x) and g (x) = ⋅ log x 51. (b) Using partial fractions,

dx

cos 2 x

=

(log (sec θ + tan θ)) dθ

   log z − 

x

=

)

( x cos .

∫e

=

log ( tan θ + sec θ)  · sec2θ d θ sec θ

=

z 3/ 2  2   − log z  + C 3 3

=

2

+ 1)

dx =

(x

4

− 1) + 1

∫ ( x − 1) ( x

2

+ 1)

dx

dx

2

dx

=

∫ ( x + 1)  dx

  dx +1) 

=

1

1

∫  2 ⋅ x − 1 − 2 ⋅ x

+

2

x 1 1  − ⋅  + 1 2 x 2 + 1 dx

[Using partial fractions] 1 x2 + x + log (x – 1) 2 2 1 1  – log ( x 2 + 1) − tan–1x + C. 4 2

=

54. (d)

=–

cosec xdx , let cotx = t, cosec2 x dx = –dt cot x



dt = – 2 t

x −1 3 +x

∫x =

t + c = – 2 cot x + c

x + x − ( x + 1) dx = ∫ x3 + x

=

∫ e ( tan z + sec z ) dz

=

∫e



( x + 1)



2

  dx = + 1) 



z

z

= x – log x +

dx

1 − x   dx 2 + 1  

1

∫ 1 −  x + x

1 log (x2 + 1) – tan– 1 x + C. 2

 2  56. (a) e x  x + 2  dx = e x  x + 4 x + 4  dx ∫  x + 4  ∫  ( x + 4 )2  4  x x  ∫ e  x + 4 + ( x + 4)2  dx

 x d  x  +   x + 4   dx + 4 x dx   x = e x⋅ + C. x+4

57. (b)

dz ∫ (1 − z 2 ) (1 + 2 z )

[Putting cos x = z ⇒ sin x dx = – dz]

dx

1 dx 4 ∫ cos x (1 − 2 sin 2 x )



=

1 4

∫ (1 − sin x) (1 − 2 sin x)



=

1 4

∫ (1 − z ) (1 −2 z )

=

cos x

2

2

dx

dz

2

2

[Putting sin x = z ⇒ cos x dx = dz]

=



=



=

= 

1 2

dz

∫ 1 − 2z 1

2 2 1 2 2

2



1 4

dz

∫ (1 − z ) 2

log

1+ 2 z 1 1+ z +C − log 1− z 1− 2 z 4

log

1 + 2 sin x 1 1 + sin x log + C. − 4 1 − sin x 1 − 2 sin x

 cos x + sin x  sin 2 x log  2  cos x − sin x  –



sin 2 x cos x − sin x 2 dx ⋅ ⋅ 2 cos x + sin x (cos x − sin x )2

=

 cos x + sin x  sin 2 x sin 2 x log  cos x − sin x  − ∫ cos 2 x dx 2

=

 cos x + sin x  1 sin 2 x log  − log sec2x + C. 2  cos x − sin x  2

dz

∫ ( z + 1) ( z − 1) (2 z + 1)

sin x

∫ sin 4 x

 cos x + sin x  60. (b) ∫ cos 2x log  dx  cos x − sin x 

dx ∫ sin x (1 + 2 cos x)

sin x dx = − = ∫ 2 − 1 cos x ) (1 + 2 cos x ) (

=

dz

+ C.

∫ 4 cos x cos 2 x

x

dx ∫ sin x + sin 2 x =

−1 x

=

2

∫e



d  tan z + ( tan z )  dz  dz



1 dx x + dx = ∫ 1 dx − ∫ dx − ∫ x 1 + x2 ∫ x2 + 1

=

1

2

= ez tan z + C = xe tan

dx

[Using partial fractions]

=

4

e tan x z 2 2 ∫ 1 + x 2 (1 + x + x ) dx = ∫ e (1 + tan z + tan z ) dz

59. (a) ∫ sin x. cosec 4x dx =

3

∫ 1 − x ( x

1

1 1 2 log (1 – z) + log ( z + 1) − log (2z + 1) + C 6 2 3 1 1 log (1 + cos x) = log (1 − cos x ) + 6 2 2 log (2 cos x + 1) + C. – 3 58. (c) Put tan– 1 x = z i.e., x = tan z

2

∴ P = – 2 55. (b)

1

=





3

1

−1

cot x ∫ 2 cos x dx = P cot x + Q sin x sin x I=

1

∫  6 ⋅ z − 1 + 2 ⋅ z + 1 − 3 ⋅ 2 z + 1

[Using partial fractions] 1



2

215

1

∫  x − 1 + ( x −1)( x

Indefinite Integration

 x2 − 1

=

216

x   dx =



∫ cos  b log a

61. (c) Let I =

z  ∫ cosbz ⋅ ae dz

64. (d) x log 1 + ∫ 

Objective Mathematics

x   z z  Put log a = z ⇒ x = ae ⇒ dx = ae dz 

∫e

= a

z

where L =

z ∫ e cos bz dz

∫e

z

sin bz dz



= ez cos bz + b e sin bz − b ∫ e cos bz dz 



= ez cos bz + bez sin bz – b2 L

z

z



62. (a)

63. (b)

=

∫ (x

2

=

x  x x    cos  b log  + b sin  b log   + C. 2     1+ b  a a  x2

+ a 2 ) ( x2 + b2 )

dx

b2 ( x2 + a 2 ) − a 2 ( x2 + b2 )

=

1 b − a2





=

1 b − a2

∫  x



x x 1  b tan − 1 − a tan − 1  + C. = 2 2   b −a b a

(x



2

dx ∫ 1 + sin α. cos x =

2



2

=

+ a 2 ) ( x2 + b2 )

− 1)

log (1 + x) −

2

∴  f (x) =

x2 x log x 2 + + c 4 2

x2 − 1 x2 1 ,L= . , g ( x) = – 4 2 2

1+ x (1 + z 3 ) 3 z 2 2 2 dx = ∫ 1 + z dz = 3 ∫ z ( z − z + 1)dz 1+ 3 x  z5 z 4 z3  = 3 ∫ ( z 4 − z 3 + z 2 ) dz = 3 ∫  − +  +C 4 3 5 3 5/3 3 4/3 = x − x +x+C 5 4

dx

b2 a2  − 2 dx 2 +b x + a 2 

dx

66. (b) I =

∫ (cos

4

sin 3 x dx x + 3 cos x + 1) tan −1 (sec x + cos x ) 2

Let tan–1 (sec x + cos x) = t ⇒

1 − tan 2

1 1 + (sec x + cos x )

2

(sec x tan x – sin x) dx = dt

sin 3 xdx = dt cos 4 x + 3 cos 2 x + I dt ∴  I = ∫ = log | t | + c = log | tan–1 t  (sec x + cos x) | + c. ⇒

67. (a) Put x = z6 ⇒ dx = 6z5 dz

dz 2 ⋅ 1 − sin α ∫  1 + sin α  2 2  1 − sin α  + z  

2 ⋅ 1 − sin α

2

∴∫

x 2 1 + sin α. x 1 + tan 2 2 x sec 2 2 dx = ∫ x (1 + sin α ) + (1 − sin α ) tan 2 2 =

(x

1 dx x2 x2 − + +c log x 2 ∫1+ x 2 4

65. (a) Put x = z3 ⇒ dx = 3z2 dz



2

∫ ( x − 1) dx −

ae z (cos bz + b sin bz) ∴ I = 1 + b2

dx

x2 1 x2 x2 x − ∫ log x + ∫ dx dx − 2 2 1+ x 2 2

1 x2 log (1 + x) − 2 2

=

ez (cos bz + sin bz) 1 + b2

∴ L=

∫ ( x log (1 + x) − x log x)

= log (1 + x)

= ez cos bz + b



= f (x) log (x + 1) + g(x) log x2 + Lx + c I=

⋅ cos bz dz = a L,

1  dx x



x1/ 2 z6 ( z 6 − 1) + 1 dz dx = 6 dz = 6 ∫ x1/ 2 − x1/ 3 ∫ z −1 ∫ ( z − 1)

x   2 x  Putting tan 2 = z ⇒ sec 2 dx = 2dz 

=6

 1 − sin α 1 − sin α tan– 1  1 + sin α  1 + sin α

 z6 z5 z 4 z3 z 2  + + + z + log ( z − 1) + C =6  + + 5 4 3 2 6 

 x  π α  = 2 sec α tan– 1  tan tan  −   + C  4 2  2  ∴ f (x) = tan

x π α ⋅ tan  −  . 4 2 2

 z + C 



∫  z

5

+ z 4 + z3 + z 2 z + 1 +

1  dz z − 1 

 x x 5/ 6 x 2 / 3 x1/ 2 =6  + + + 5 4 3 6 1/ 3  x 1/ 6 1/ 6 + 2 + x + log ( x − 1)  + C 

α + β  β − α     −  x − 2  2 



69. (b)



=

dx

=

( x − 1) ( x + 2) 3

4

∫  x −1

5

dx 3/ 4

 x + 2 

( x + 2 )2

− 1) ( z 3 + 1)

3

z 2 ( z + 1)

12 z 3

(1 − z )

4 2

(1 − z ) ×

4 2

9



1 dz z3 72. (a)

∫x =



(x

− 1)

2

x + 3x + 1 4

2

x −1 4 4 dz = ⋅ 4 + C. x+2 3∫ 3

1 − x2 1 − x2 = t  ⇒ = t2 2 1+ x 1 + x2 1− t2 ⇒ x2 = 1+ t2 −4t ⇒ 2x dx = dt. (1 + t 2 ) 2 dx

∫ (1 + x ) 2

=



=–

= ∴

=

1 − x2

−2t

(1 + t )

2 2

1 2





−2t

∫ x 1+ t

(

)

2 2

2

1   x +  + 1 x

dt 1− t2

=

1 − x2 . 1 + x2

dx =

(x

x2



2

− 1)

x2 + 3 +

1 x2

217

dx

dz z2 + 1

  1 1   Putting x + x = z ⇒ 1 − x 2  dx = dz   

73. (b) Put x = z4 ⇒ dx = 4z3 dz. ∴ 1

 1− t2  1− t2 + − 1 1 2 1+ t2  1 + t 

−1 sin– 1 t + C 2



= log x + 1 + x 2 + 1 + 3 + C. x x2

1+ t2 1+ t2 1+ t2 ⋅ ⋅ dt 2 1− t2 2t

−1 1 − x2 sin −1 + C. 1 + x2 2 f (x) =



z 2 + z3

= log z + z 2 + 1 + C

70. (a) Put

 ∴

− 1) 6 z 5 dz

dz

dx =

1  1 − 2  x





6

 (1 + x )3 / 2 (1 + x )4 / 3 (1 + x )7 / 6 − + = 6   9 8 7  (1 + x ) (1 + x )5 / 6 (1 + x )2 / 3   − + −  + C. 6 5 4 

5/ 4

  x −1 4 1+2z 4 = z ⇒ x=   Putting 4 x 2 1 z + −   3 12z dz   and dx =  (1 − z 4 ) 2      3 2 and = x +  1 − z 4  

=

(z

 z9 z8 z 7 z 6 z5 z 4  = 6   − + − + −  + C 6 5 4 9 8 7

dx

3/ 4



(z



8 7 6 5 4 3 = 6  ∫ ( z − z + z − z + z − z ) dz

∫ ( x − 1) ( x + 2) =

x dx = 1+ x + 2 1+ x

= 6  ∫ z 3 ( z 3 − 1) ( z 2 − z + 1) dz

2

 2x − α − β  = sin–1  + C.  β − α 



3

= 6  ∫

dx 2





2 2  2 α + β  α + β − − + . + x x α β ( )  − αβ   2    2  



=

− x 2 + x (α + β) − αβ dx



=

71. (b) Put (1 + x) = z6 ⇒ dx = 6z5 dz

dx



=

Indefinite Integration

dx

∫ ( x − α) (β − x)

68. (c)

 dt

2

∫1+

x 4

x3

dx =

z 2 .4 z 3 ∫ 1 + z 3 dz = 4

z 3 .z 2 ∫ z 3 + 1 dz

1   3 2 4 ( y − 1) dy  Putting z + 1=y ⇒ z dz = dy  ∫ 3   3 y 4 = (y – log y) + C 3 =

=

4 [1 + x3/4 – log (1 + x3/4)] + C. 3

 3x − 4  74. (b) We have f  =x +2  3 x + 4  Let

3x − 4 = t ⇒ 3x – 4 = 3xt + 4t 3x + 4

⇒ x =

4t + 4 3 (1 − t )

4t + 4 +2 3 (1 − t )

218

∴ f (t) =

Objective Mathematics

∴ f (x) =

 1 1   t − 1 − dt = 2t + ln  +c = 2 ∫ dt + 2 ∫   t − 1 t + 1   t + 1 

4x + 4 + 2 = 4 ( x − 1) + 8 + 2 3 (1 − x ) 3 (1 − x ) 4 8 − 3 3 ( x − 1)



=2–



∫ f ( x) dx

 1 + x − 1 2 1 + x + ln   +c.  1 + x + 1 2− x 2 − 2z3 = z3 ⇒ x = 2+ x 1 + z3 2 −12 z dz 2 ⇒ dx = ∴ 2 . ∫ ( 2 − x )2 (1 + z 3 )

79. (b) Put

2 8 ln | x – 1 | + c. x− 3 3

=

75. (a) Put x = z6 ⇒ dx = 6z5 dz

76. (b)

x + 3 x2 + 6 x









= 6  ∫



3 4 =  z + 6 tan– 1 z + C 2



=

∫(

tan x + cot x  dx =

=

(

x 1+ 3 x

)

dx =

= =

2

+ z 4 + z ) 6 z 5 dz

∫ ∫

80. (d) I =

 sin x cos x  +  dx cos x sin x 



sin x + cos x

sin x + cos x 1 − (sin x − cos x )

2

sin x + cos x dx 2 sin x cos x

dx =

2



dz 1 − z2

[Putting sin x – cos x = z ⇒ (sin x + cos x) dx = dz] = =

2 sin– 1 z + C 2 sin– 1 (sin x – cos x) + C.

 x  1 x  e + / 3 2 ∫  1 + x 2 1 + x 2  dx ( )   x Let f (x) = 1 + x2

77. (d) I =

⇒ f ′ (x) = =

1 + x2 −

x2

1 + x2 2 1 + x ( ) 1

(1 + x )

2 3/2

∴  I = ex f (x) + c =

ex x 1 + x2



⋅ z⋅

3 2

∫x

2/3

+ C.

dx

n

dz

(1 + x n )

1/n

=



dx 1  x n + 1 1 + n   x 

1/n

1 −n = t ⇒ n + 1 dx = dt xn x

Let 1 +

  1 dt 1  t − 1/n + 1  +c ∴ I = – =– n ∫ t1/n n  − 1 + 1     n 1−

1

1 n  1 + n  x = – +c n −1 1 − x2 dx 81. (b) ∫ ⋅ 1 + x2 1 + x4 1  1 − 2  dx x =–





z

=



+c.

= 

2

1  1 1   x +   x +  − 2.x. x x x dz

= –  ∫

=

1− x dx . Let 1 + x = t2, dx = 2t dt x t2 −1 + 1 t ⋅ 2t dt ∴ I = ∫ 2 =2 ∫ 2 dt t −1 t −1

78. (a) I =



dx

1 − (1 − 2 sin x cos x )

2

3 2 + x = 4  2 − x 

∫  2

∫

(− 12 z ) (1 + z )

2 − x dx 2+ x

2

2

2 − 2z3   2 − 1 + z 3  −3 dz 3 + C. = = 2 ∫ z3 4z2

z 6 (1 + z 2 )

3 2/3  x + 6 tan– 1 x1/6 + C. 2

sin x + cos x  dx = sin x cos x

2

6

z5 + z3 + 1 1  dz = 6  ∫  z 3 + 2  dz  z + 1 z2 + 1

)



(z

=

3

  1 1   Putting x + x = z ⇒ 1 − x 2  dx = dz    z −2 2

dt 1 − 2t 2



1 −1    Putting z = t ⇒ dz = t 2 dt 

1 dt = 2 ∫  1 2 2   −t 2  2 1 sin– 1  z  + C 2 =

 t  1 sin– 1  +C  1 / 2  2

 2x 1 sin– 1  x 2 + 1  + C. 2



=

(sin

2

θ − cos 2 θ) d θ   sin θ cos θ +

(sin θ + cos θ)2

(sin 2θ + 1) (sin 2θ + 1)2 − 1

= −∫

dt t t2 − 1

f (θ) = sin 2θ + 1. dx n + 1)

1 dx 1  1 1 I = n ∫ t (t − 1) = n ∫  t − 1 − t  dt

=

∫x

∫ (t

(x

∫ (x

2

4

− x 2 + 1) + x 2

+ 1) ( x − x + 1) 4

3 2

∫ (t

3

− 1) tdt

3  t5 t2  +C − 2  5 2 

− t ) dt =

=

3 2  t (2t3 – 5) + C 20

=

3  (1 + x2)2/3 (2x2 – 3) + C. 20

4

 x  ⇒ 1 +  dx = dz  1 + x2 

1 + x 2 = z z dx. z−x

2

z2 − 1 ⋅ 2z

⇒ x =

 x  1 1  t − 1 = n log  t  + c = n log  x n + 1  + c . x4 + 1 dx = 6 +1

2 1/ 3

Also, 1 + x2 = (z – x)2 = z2 – 2 xz + x2

n

∫x

(1 + x )

3 2

⇒ dz =

Put xn + 1 = t,  nxn–1dx = dt

84. (a)



x2 ⋅ x

= 

87. (c) Put x +

∫ x (x



dx =

dx =

= cosec– 1 t + C = cosec– 1 (sin 2θ + 1) + C.

83. (a) I =

(1 + x )

2 1/ 3



[Putting sin 2θ + 1 = t ⇒ 2 cos 2θ dθ = dt] ∴

x3

2

1 1  − 2 4

2 cos 2θ d θ

= –  ∫





219

86. (b) Put 1 + x2 = t3 ⇒ 2x dx = 3t2 dt

(sin θ − cos θ) d θ ∫ (sin θ + cos θ) sin θ cos θ + sin 2 θ cos2 θ





dx

dx x dx + + 1 ∫ ( x 3 )2 + 1

1 + x2 − x n

=



= 1 2

1 d (x ) dx ∫ x 2 + 1 + 3 ∫ x3 2 + 1 dx ( )

)

n

=

∫  1+ x (

z − x  dz = z 

∫ z . 



2

2

(

( 1 + x + x) 2

dx

∫ (z

n

∫z . n

2

n

) − x  2

n

dx

2 z 2 − ( z 2 − 1) 2z2

dz

1  z n +1 z n −1  + + C,  2  n + 1 n − 1 

+ z n − 2 ) dz =

3

=

= tan– 1x +

where z = x +

1 tan– 1 x3 + C. 3

88. (a)

85. (a) Put 1 + xex = t

`



1 + x4

(1 − x )

4 3/ 2

1   3 + x  x

⇒ (x + 1) exdx = dt ∴

∫x

( x + 1)

(1 + xe )

x 2

dx =

( x + 1) e x dx

∫ xe

x

(1 + xe )

 = =

x 2

=



1

∫  − t

2

1 – log t + log (t – 1) + C t

1  xe x  + log  + C. 1+ xe x  1 + xe x 

∫ 1

2  2 − x  x

dt

∫ (t − 1) t

3/ 2

1  x  2 − x2  x 

3/ 2

dx

3

dx =

−1 dz 2 ∫ z 3/ 2

 1 − dz  1  2  Put x 2 − x = z ⇒  x 3 + x  dx = 2   

2

1 1  − +  dt (by partial fractions) t t − 1

 t −1 1 = + log   +C t  t  =

=



dx =

1  x2  2 + x2  x 

1 + x2 .

= z–1/2 + C =

89. (b)

∫x

=



1 − x4

+ C.

1  x2  x2 − 2    x dx = dx ∫ 2 2 x4 + x2 + 1 1 x .x x + 1 + 2 x 1   x − 3  x 1 dz dx = ∫ 1 2 z +1 x2 + 1 + 2 x x4 − 1

2

x

Indefinite Integration

82. (a)

220

 dz  1 1  2  Putting x + x 2 = z ⇒  x − x 3  dx = 2   

Objective Mathematics

z +1 + C =

=

x 4 + x 2 + 1 + C. x x −1

= 90. (b)

∫ ( x + 1) =

x2 + 1 +



x3 + x 2 + x

(x

1 + C. x2

2 = log  z + z + 2  + C

 1 = log   x −  +  x  

2  1   x −  + 2  + C. x   a 93. (c) Put a + bx = zx ⇒ x = z−b

dx

− 1)

2

( x 2 + 2 x + 1).x x + 1x + 1

dx



=

=



2

1 + tan 6 x

dx =



(1 + t )

94. (a)

2 2



dt

1 + t6

=

)

∫ (1 + t ) (1 − t 2

2

1  1 + 2  dt t = = ∫ 1 2 − + t 1 t2

+t

4

)

dt =

a

∫ ( z − b)

2

1



a3



dz

a2 z 2

( z − b)3 ( z − b)2

1  3b 2 b3  z − 3b + − 2  dz 4 ∫ a  z z 

b3  − 1  z2 2  + C, bz b log z − + + 3 3 a 4  2 z 

sec x

sin ( 2 x + α ) + sin α

=



=



[Putting tan x = t ⇒ sec x dx = dt]

(1 + t

= −

dz

−1 ( z − b ) −1  z 3 − 3 z 2b + 3 zb 2 − b3  dz = 4 ∫  4 ∫ 2  dz a z a  z2

sec x dx

2 sin ( x + α ) cos x

2

2 2

2

( z − b )2

where z =

sec6 x dx ∫ 1 + tan 6 x

(1 + tan x) . sec x 2

=

=–

 x2 + x + 1   + C. – 1  = 2 tan    x 

2

3

−a

3



dz = 2 tan– 1 z + C 1 + z2

dx 91. (c) ∫ sin 6 x + cos6 x =

dx

∫ x (a + bx)



2 z dz ∫ ( z 2 + 1).z

  1 1  2  Putting x + x + 1 = z ⇒ 1 − x 2  dx = 2 z dz   

= 2∫

⇒ dx =



1  1 − 2  x dx = = ∫ 1 1   + + + + x x 2 1   x x 

  1 1   Putting x − x = z ⇒ 1 + x 2  dx = dz   



1+ t dt 2 + t4

∫1− t

1  1 + 2  dt t ∫  12 =  t −  + 1 t

2

=

∫z

dz +1

 dx

2 sin x ( x + α ) / cos x 1 2



 dx

sec 2 x dx = tan x cos α + sin α

1 2 cos α



dt t

⇒ sec2x dx = dt/cos α]



  1 1   Putting t − t = z ⇒ 1 + t 2  dt = dz     1 = tan – 1 z + C = tan– 1   t −  t = tan– 1 (tan x – cot x) + C.

( )

dx

[Putting tan x cos α + sin α = t

2

1  1 + 2  dx 1 + x2 ) ( x dx = 92. (b) ∫ ∫ 2 1 x 1 + x4 x + 2 x 1  1 + 2  dz x = ∫ dx = ∫ 2 2 z + 2 1   x −  + 2 x

sec 2 x

a + bx ⋅ x

95. (c)

=

1 ⋅2 t + C 2 cos α

=

2 sec α tan x cos α + sin α + C

=

2sec α ( tan x + tan α ) + C.



1+ 3 x 3

x2

Put 1 + x

dx =

1/3

∫x

=z 2

2

⇒ x–2/3 dx = 6z dz

−2 3

1

1 2   3 1 + x  dx

1 1 −2   Here m = 3 , n = 3 , p = 2    ∴ m + 1 = 1 (an integer )    n



= 2 (1 + x1/3)3/2 + C.

x2

=–

35 −3 , C = a constant. , B= 36 2 100. (b) Put 1 + x1/4 = z3 ⇒ dx = 12 z2 x3/4 dz ⇒ A =

96. (b) Put 2 + x2/3 = z4 ⇒ dx = 6z3. x1/3 dz

∫ x (2 + x )



2 / 3 1/ 4

1/ 3

dx =

∫x

1/ 3

3 1/3 . z (6z . x ) dz

 z9 2z5  4 4 = 6 ∫ z ( z − 2) dz = 6 ∫  − +C 5  9 2 12 = (2 + x2/3)9/4 – (2 + x2/3)5/4 + C. 3 5 97. (a) Put 1 – x3 = t2 ⇒ – 3x2 dx = 2t dt dx



∫x

=

1 log 3

1 − x3 t −1 t +1

∫x

=

x2

dx =

1 − x3

3

+C=

dt ∫ t2 − 1

2 3

1 − x3 − 1

1 log 3

1 − x3 + 1



= 

∫ 1 + z

99. (a)

−x

2

z ∫ x 4z

(

x

)

dz = 4

2x

4e + 6 dx = 2x −4

∫ 9e

 2x  Put 9e − 4=t       





 z7 z4  = 12  −  + C 4 7



 (1 + x1/ 4 )7 / 3 (1 + x1/ 4 ) 4 / 3  − = 12   + C. 7 4 

∫z .x 3

1/ 4

1/ 2

⋅ (12 z2 x3/4) dz

∫ z (z

dz = 12

3

3

− 1) dz

1 f ( x) sin x cos x dx = 2 (b 2 − a 2 log [ f (x)] + C )

∫z

z2 − 1 dz



2

102. (a) Put 1 + x4/3 = z7 ⇒ x1/3 dx = ∴

=

35 3  log (9e2x – 4) –  (log 9 + 2x) + C 36 4

3  log (9e2x) + C 4





2

dx



3

x

7

1 + 3 x 4 dx =

21 6 z dz 4

 21

∫ z  4

 z 6  dz 

 =

 35 3  = ∫ −  dt (by partial fractions)  36t 4 (t + 4) 

35  log (9e2x – 4) – 36

f ' ( x)

∫  f ( x)

=

1 ⇒ f (x) = a 2 sin 2 x + b 2 cos 2 x ⋅

 1 (4 + t ) 9  35 1  2x ⇒ 2e dx = dt  dt 36 9  dt  ⇒ dx = 2 (t + 4) 

=

sin x cos x − 2a 2 sin x cos x )  dx

2

−1 ⇒ – b2 cos2x – a2 sin2x = f ( x )

⇒ e2 x =

35 3  log t –  log (t + 4) + C = 36 4

∫ (2b



35 + 2t

∫ 9t (t + 4) dt

1 1 ⋅ ⋅ f ′ (x) 2 (b 2 − a 2 ) f ( x )

f ' ( x) ⇒ 2 (b2 – a2) sin x cos x =  f ( x ) 2  

1   1 z − 1 dz = 4  z + log +C − 1  2 z + 1  

4e + 6e dx = x − 4e − x

∫ 9e

= 12

x

f (x) sin x cos x =

 1 + x − 1 = 4  1 + x + 2 log    + C.  1 + x + 1  x



dx =

∫x

z





x = z2 ⇒ dx = 4 z x dz

 =4

1+ 4 x

Differentiating both sides w.r.t x, we get

1 + x dx x



3



101. (b) We have, + C.

1 ∴ a = . 3 98. (c) Put 1 +

3   3 35 x+  log (9e2x – 4) +  C − log 3  2 2 36

221



3

∫ z. 6 z dz = 2z3 + C.

dx =

Indefinite Integration

1+ 3 x



103. (c)



=

(

21 1 + 3 x4 32

(x − x )

5 1/ 5

x6

)

8/ 7

21 z 8 ⋅ +C 4 8

+ C.

1   4 − 1 x dx = ∫ x5

1/ 5

dx =

−1 1/ 5 dt t 4 ∫

dx −1  1   Putting x 4 − 1 = t ⇒ x 5 = 4 dt  1  = −5 t6/5 + C = −5  4 − 1  24 24  x

6/5

+ C.

222

104. (a) Put x = z3 ⇒ dx = 3z2 dz

∫x

Objective Mathematics



=3

(

dx

1+ x 3

1

)

2

3 z dz

∫ z (1 + z )

=

3

1

1

∫  z − 1 + z − (1 + z ) 

2

= 3∫

2

dz

z ( z + 1)

  dz 

 1  x = 3  log 1 + x1/ 3 + 1 + x1/ 3  + C.

=

sin 8 x − cos8 x dx 2 x cos 2 x

=

x2

(a + bx 2 )

=– a

5/ 2

dz

(z

2



107. (c)

∫e =

x

∫ (x z)

dx =

− b)

2

5

 − az    dz 2  x ( z 2 − b ) 

(1 − 2 sin



(1 − 2 sin

−1 sin 2x + B. 2

x ( z 4 − 1)

1 z −1

+ C.

dx

(1 + x )

2 5/ 4





=–2



=

∫x

(z

4

2 x

dz 4

− 1) z 2

cosec x cot x (cot x + cosec x) dx

dx

= – (cosec x + cot x) e

1 



=–



=

∫ (1 + z ) (1 − z )



=

1 2

4

∫ xz  x

=

1 + x4

∫x

z2

4

(z

4

− 1)

)

= −2

dz

∫z

2

=

2

1

∫  1 − z

1 z4 − 1 − z3

x 3 ( z 4 − 1)

2

2



− z3

2

 dz = –

2

z2

2



z

 − 2z3    dz 4  x ( z − 1) 

  dz 3 4  ( z − 1) 



(cosec2x + cosec x cot x) dx

cosec x

2

2

⇒ x4 =



+ (cosec x cot x + cosec x)] dx

cosec x

(x

1/ 2

4 5/ 4

2 + C. z

+ C.

1 + x2

2

∫e

1

∫x

=

⇒ dx =

cosec x [cosec x cot x (cot x + cosec x) = ∫e

+

4

dz

2

110. (a) Put 1 + x4 = x4z4

3/ 2

dx

x cos 2 x )

2

−2 z 3

1/ 2

4

cosec x ∫ e (cot2x cosec x + cosec2x + cot x cosec2x

∫e

x cos 2 x )

−1 . 2

∫x

+ cosec x cot x) dx

=

2

cos 2 x. (1 − 2 sin 2 x cos 2 x )

cosec3x (cos2x + sin x + cos x + sin x cos x) dx

cosec x

dx

x cos 2 x )

2

x − cos 2 x ) (sin 2 x + cos 2 x )

2

2

dz 1 ∫ a 2 z 4 = 3az 3  + C

1  x2  = 3a  a + bx 2  cosec x

(sin

⇒ dx =

⇒ dx = x ( z 2 − b )2 dz

4 4

(1 − 2 sin

109. (b) Put 1 + x2 = x2z4 ⇒ x2 =

− az

∫x z

x − cos 4 x ) (sin 4 x + cos 4 x )

4

∴ A =

a 106. (b) Put a + bx2 = x2z2 ⇒ x2 = z 2 − b

=– a



(sin

= –  cos 2x dx = ∫

−1 sin 8 x dx 2 ∫

1 ∴ a = . 16





=–

1 cos8x + C. 16



(cosec2x + cosec x cot x) dx + C

cosec x

× [(sin2x + cos2x)2 – 2 sin2x cos2x] dx

cos 8 x + 1 ∫ tan 2 x − cot 2 x dx 2 cos 2 4 x = ∫ sin 2 2 x − cos 2 2 x ⋅ sin 2x cos 2x dx

∫ sin 4 x cos 4 x dx =

 (cosec2x + cosec xcot x) dx

∫ 1 − 2 sin

=

1/ 3

∫e

+

cosec x

– ecosec x (cosec x + cot x) + C. 108. (a)

 1  = 3  log z − log (1 + z ) + 1 + z  + C.

=–



2

(by partial fractions)

105. (c)

∫e

– 2

∫z

dz

1  dz 1 + z 2 

z2 dz −1

4

dz

where z =

=

Since the curve passes through (0, 0), ∴ C = 0.

1 4 1 + x4 . x

∴  y =

111. (a) Put x = tan θ ⇒ dx = sec2θ dθ x2 + 1  dx = x2







=



sec θ. sec 2 θ dθ tan 2 θ

sec θ (1 + tan 2 θ)

tan 2 θ sec θ = ∫ dθ + tan 2 θ



=

cos θ dθ + log (sec θ + tan θ) + C 2 θ



(

)

= − 1 + x + log x + 1 + x + C x  x x 1  and cos θ = ∵ tan θ = , ∴ sin θ =  2 1 1+ x 1 + x2   1

∫ 1 + (log x)

2

d (log x)

2

dt

∫1+ t

=

[Putting log x = t ⇒ d (log x) = dt]

= tan– 1 t + C = tan– 1 (log x) + C.



d ( x 2 + 3) x +4 2

∫ (3 x + 4 )

sin x dx = – (3x + 4) cos x  +  ∫ 3 cos x dx

= – (3x + 4) cos x + 3 sin x⋅ 117. (b)





 f ( x ) g' ( x ) + g ( x ) f' ( x ) f ( x) ⋅ g ( x)

=

d ( x 2 + 3)



(x

2

+ 3) + 1

=



dt t +1

t +1 + C = 2

x 2 + 4 + C.

2 2 d  x x 114. (b) ( x )  = dxd  x x  = dxd e x . log x  dx  2 = e x . log x . d (x2. log x) dx 1  2 = e x . log x .  log x. 2 x + x 2 .   x

=



 f ( x ) g' ( x ) + g ( x ) f' ( x ) log f (x) g (x) dx f ( x) ⋅ g ( x)

=



log t dt t ⇒ [f (x) g ′ (x) + g (x) f ′ (x)] dx = dt]



1 1 (log t)2 + C = [log f (x)⋅ g (x)]2 + C. 2 2 118. (a) We have, f ′′ (x) = sec2x ⇒ f ′ (x) =

∫ sec

x x



=

∫ tan x

dx + C2 = log sec x + C2

Since f (0) = 0, ∴ C2 = 0 ∴ f (x) = log sec x. 119. (a), (c) 

115. (c) We have, y =



⋅ (2 log x + 1) dx = ( x x )x + C⋅

∫x

3

cos x 4

1 dx = cost dt 4∫

1   4 3  Putting x = t ⇒ x dx = 4 dt 

x dx + C1 = tan x + C1.

∴ f ′ (x) = tan x ⇒ f (x)

dx

∫ 5 + 4 sin x

x

∫ x ⋅ (x )

2

Since f ′ (0) = 0, ∴ C1 = 0

= ( x x ) ⋅ x (2 log x + 1)⋅ ∴

[log f (x) + log g (x)] dx

=

[Putting x2 + 3 = t ⇒ d (x2 + 3) = dt] =2

sin − 1 ( 4 y ) ⋅

[Putting f (x). g (x) = t

2



113. (c)

4

∴ The antiderivative of (3x + 4) | sin x | is

−1 = + log (sec θ + tan θ) + C sin θ

112. (b)

⇒x=

116. (a) In the interval (0, π), sin x is positive, therefore, (3x + 4) | sin x | = (3x + 4) sin x.

∫ secθ d θ

2

1 sin x4 ⇒ x4 = sin–1 (4 y) 4



∫ sin

=

1 1 sin t + C = sin x4 + C. 4 4

223

 1 1 1+ z log − tan −1 z  + C,  2 2 1− z 

=

=



sec 2

x 2

=



dx x   2 tan   2 5+4  2 x   1 + tan 2

x x 5 tan 2 + 8 tan + 5 2 2

=2

∫ 5z

2

dz + 8z + 5

x   2 x  Putting tan 2 = z ⇒ sec 2 dx = 2dz 

Indefinite Integration



224

=

2 5

dz



Objective Mathematics

8 z +1+ z 5 2

=

2 dz 2 2 ∫ 5  4 3 + z +     5 5



2 5 ⋅ tan–1 5 3

=

4  z+  5 + C  3     5 

x   5 tan + 4 2   2 –1 tan  =  +C 3 3  



2 5 ∴ A = and B = ⋅ 3 3

1 ⋅  z2

z 5 dz

(1 − z )2 z

=–

z



1 − z2

dz = –

∫ sin

4

2

+

=–

1 4 1 8



∫ 1 − 2 cos 2θ +

θ dθ

1 + cos 4θ  dθ  2

sin 4θ    3θ − 2 sin 2θ +  + C, 4 

 1  ⋅ where θ = sin– 1z = sin– 1    1 + x 

= uv′ – vu′ +



∴a=



∫ u'' v dx ⋅

∫ u'' v dx ⋅ 1 dz a

122. (c) Put ax + b = z ⇒ dx =





∫ f (ax + b)

dx =

1 a

∫ f (z)

dz =

1 [g (ax + b) + c]. a 123. (b) ∫ sin xd (cos x ) = ∫ sin x ( − sin x ) dx





=

1 − cos 2 x dx ∫ 2 = −1  x − sin 2 x  + C   2  2 

=–

1  sin 2 x  − x  + C. =   2  2



=







d d

∫ dx  dx tan

1 [g (z) + c] a

−1

dx 2ax − x 2

=

dx



a − ( x − 2ax + a 2 ) 2

dx



a − ( x − a) 2

2

f (x) = sin–1x

2

x − a = sin– 1  + C.  a 

1 d 2 2 sec 2 x (sec 2 x + tan 2 x ) dx ∫ (sec 2 x + tan 2 x)

∫ sec 2x dx =

=

1 log | sec 2x + tan 2x | + C. 2



=

1 log 1 + sin 2 x + C 2 cos 2 x



1 log = 2









x−a ⋅ a

and g (x) =





 x  dx 

1 d (tan–1x) = + C. 1 + x2 dx

125. (b)   ∫



121. (a) ∫ uv'' dx = uv′ – ∫ u' v' dx = uv′ – vu' − ∫ vu'' dx   

=

2 (1 − z ) z

[Putting z = sin θ ⇒ dz = cos θ dθ] =–



(tan–1x) dx=

2



= –∫ 4

∫ dx

126. (a), (b), (c) 

1 120. (b) Put 1 + x = z 1 ⇒ dx = –  2 dz. z dx ∴ ∫ 5 (1 + x ) x 2 + 2 x 

d2

124. (a)

π  1 + cos  − 2 x  2  +C π  sin  − 2 x  2 

π 1 log cot  − 4 2 1 log ∴ f (x) = 2 =

 x  + C.  π  | x | and g (x) = cot  − x  ⋅ 4 

∴ dom⋅ f (x) = (– ∞, ∞) – {0} and range g (x) = (– ∞, ∞)⋅ Also, f ′ (x) =

π  1 , ∨ x ∈ R+ and g ′ (x) = cosec2  − x  ⋅ 4  2x

127. (a) Put f (x) = y2 =



⇒x=







2 y2 − 1 1 − y2

 f ( x)  ∫  x 2 

x +1 x+2 and  dx =

1/ 2

dx =

∫ y⋅

2y

(1 − y )

2 2

dy.

1 − y2 2y ⋅ dy 2 2 y − 1 (1 − y 2 )2

y2  1 1  − 2 2  dy = 2 ∫    dy 2 2 − 1 1 − y y ( )( )  2 y − 1 y − 1



=2∫



=

1  log 2



=

1  log 2



Thus, g (x) = log | x | and h (x) = log | x |.

2

2 y −1 y −1 − log y +1 2 y +1 2 f ( x) − 1 2 f ( x) + 1

− log

f ( x) − 1 f ( x) + 1





Since f (0) = – 1 and f (1) = 0 ⇒ d = – 1 and a + b + c + d = 0 ⇒ d = – 1 and a + b + c = 1. ...(1) Since 0 is a stationary point of f (x), ∴ f ‘ (0) = 0 ⇒ 3a (0)2 + 2b (0) + c = 0 ⇒ c = 0. Since f (x) does not have an extremum at x = 0, ∴ f ′′ (0) = 0 ⇒ b = 0. and ∴ from (1), a = 1. So, f (x) = x3 – 1. f ( x) ∴ ∫ 3 dx = ∫ 1 dx = x + C. x −1

129. (a) We have,

I n



=



=









∫ tan ∫ tan

n

∫ tan

x dx =

n−2

= 8x4(log x)2 – 16 ∫ x3 log x dx  x4 1 x 4  = 8 x 4 (log x) 2 − 16 log x ⋅ − ∫ ⋅ dx  4 x 4   4 2 4 3 = 8x (log x) – 4x log x + 4 ∫ x dx = 8x4 (log x)2 – 4x4 log x +x4 + c = x4 [8(log x)2 – 4 log x + 1] + c

n−2

x (sec2x – 1) dx

n−2

x sec2 x dx –

∫ tan

n−2

= (I2 + I0) + (I3 + I1) + (I4 + I2) + (I5 + I3)



=  tan x + tan x + ... + tan x  ⋅  1 2 9 

130. (b)

∫x



=

∫ (t − 1) t



=

∫  t − 1 − t − t



1 t −1 1 = log | t – 1 | – log | t | + + C = log + +C t t t



= log



( x + 1)

(1 + xe ) dt

 1

9

dx =

x 2

x

(1 + xe )

x 2

dx

[Putting 1 + xex = t ⇒ (x + 1) ex dx = dt]

2

1

1 dt 2  

1 xe x + C. + 1 + xe x 1 + xe x dt

 1

1 

∫ (2 + t ) (1 + t ) = ∫ 1 + t − 2 + t  dt

= log (1 + t) – log (2 + t) + c  1 + ex = log  x 2+e 132. (c)

 + c 

1+ x 1+ x dx = ∫ dx 1− x 1 − x2 dx x =∫ +∫ dx = sin −1 x − 1 − x 2 + c 1 − x2 1 − x2



(cos x)e x −(1 + sin x) sin x − cos 2 x x −∫ e dx 1 + sin x (1 + sin x) 2 ex dx 1 + sin x

e x cos x ex ex +∫ dx − ∫ dx + c 1 +sin x 1 + sin x 1 + sin x e x cos x = +c 1 + sin x

=

−1

135. (c) Let I = ∫

e m tan x dx 1 + x2

Put m tan–1 x = t dx 1 = dt 2 1+ x m −1 1 1 1 ∴ I = ∫ et dt = et + c = e m tan x + c m m m sin 2 x + cos 2 x dx 136. (a) Let I = ∫ sin 2 x cos 2 x = ∫ (sec2x + cosec2x)dx = tan x – cot x + c

(by partial fractions)

131. (a) Put ex = t ⇒ ex dx = dt ∴

=



( x + 1) e x

∫ xe

(cos x − 1)e x e x cos x ex dx = ∫ dx − ∫ dx sin x + 1 1 + sin x 1 + sin x

−∫

x dx

+ (I6 + I4)+ (I7 + I5) + (I8 + I6) + (I9 + I7) + (I10 + I8) 2



x ⋅ tan2x dx

n −1 = tan x – In – 2 n −1 n −1 ∴ In + In – 2 = tan x , n ≥ 2 n −1 ∴ I0 + I1 + 2 (I2 + I3 + ... + I8) + I9 + I10



 x4 1 x 4  = 32 (log x) 2 − ∫ 2log x ⋅ ⋅ dx  4 x 4  

134. (d)

∫ tan

=

133. (b) ∫ 32x3(log x)2 dx

137. (b)

∫x

2

dx + 4 x + 13

dx 1  x+2 = tan −1  +c ( x + 2) 2 + 32 3  3  x −1 1 1 =t ⇒ dx = dt 138. (a) Let x+2 ( x + 2) 2 3 =∫

∫ [( x − 1)



=

4 3

1 1 dx = ∫ t −3/ 4 dt 5 1/ 4 ( x + 2) ] 3

1/ 4

 x −1   x+2  

139. (a) Let I = ∫ =∫

3

+c ax/2

a

−x

− ax

ax/2 a− x/ 2 1 − a2x

dx

dx = ∫

ax 1 − a2x

Let ax = t ⇒ ax log a dx = dt

dx

225



Indefinite Integration

128. (b) Let f (x) = ax3 + bx2 + cx + d.

226

∴ I =



Objective Mathematics

=

d (1 + nx n ) n 2 x n −1 1 1 dx dx = 2 ∫ dx = 2∫ n (1 + nx n )1/ n n (1 + nx n )1/ n

1 1 dt log a ∫ 1 − t 2

1 sin −1 t + c log a

1 sin −1 (a x ) + c log a ex dx 140. (a) Let I = ∫ x (2 + e ) (e x + 1) Put ex = t =

⇒ ex dx = dt ∴ I = ∫



dt (2 + t ) (t + 1)

 1 1  = ∫ −  dt (1 + t ) (2 + t)   = log (1 + t) – log(2 + t) + c = log (1 + ex) – log (2 + ex) + c  1+ e  = log  x  + c  2+e  141. (d) ∫ [sin (log x) + cos (log x)] dx x

= ∫ sin (log x) dx + ∫ cos (log x) dx x cos (log x) dx + ∫ cos (log x) dx + c x = x sin (log x) + c

= x sin (log x) − ∫

142. (b)



1 + x + x + x2 dx x + 1+ x

=∫ =∫

(1 + x) 2 + x x + 1 x + 1+ x

dx

1+ x( 1+ x + x) dx ( x + 1+ x)

2 (1 + x)3/ 2 + c 3 dx 1 π  = sec  x −  dx 143. (a) ∫ 3 cos x + 3 sin x 2 ∫ 

1  x π π log tan  − +  + c 2 2 6 4

1 x π = log tan  +  + c 2  2 12  f ( x) x = [1 + f ( x) n ]1/ n (1 + 2 x n )1/ n x and f ( f ( f ( x))) = (1 + 3 x n )1/ n x g ( x) = ( f o f o . . . o f )( x) = n 1/ n  (1 ) + nx n times

144. (a) We have, f ( f ( x)) =





∫x

n−2

g ( x) dx = ∫

 1 1  = 2∫  cot t +  dt 2  2 = t + log | sin t | + c π  = x + log sin  x −  + c 4  146. (a) ∫ x( x x ) x (2log x + 1) dx = ∫1. dt Let (xx)x = t = t + c ⇒ x log xx = log t = (xx)x + c ⇒ x2 log x = log t 1 dt ⇒ 2x log x + x = t dx ⇒ (2log x + 1) x · (xx)x dx = dt

= ∫ 1 + x dx =

=

1 1− 1 (1 + nx n ) n + c n(n − 1) sin x 145. (c) Let I = 2 ∫ dx π  sin  x −  4  π Put x − = t 4 ⇒ dx = dt π  sin  + t  dt 4  ∴ I = 2∫ sin t

=

x n −1 dx (1 + nx n )1/ n

147. (a) We have f ( x) x = [1 + f ( x) n ]1/ n (1 + 2 x n )1/ x x Also, fofof(x) = (1 + 3 x n )1/ n

fof(x) =

∴ g(x) = ( fofo...of ) ( x) = n terms

Thus,

∫x

n−2

g ( x)dx = ∫

n 2 x n −1dx 1 1 = 2∫ = n (1 + nx n)1/ n n 2 =

x n −1dx (1 + nx n )1/ n



d (1 + nx n ) dx dx (1 + nx n )1/ n

1 1− 1 (1 + nx n ) n + k . n(n − 1)

148. (c) J – l =



∫  e

−4 x

 e− x ex − 4x dx −2 x 2x + e +1 e + e +1

 e3 x − e x  = ∫  4x dx 2x  e + e +1 =

x (1 + nx n )1/ n

e 2 x (e x − e − x ) dx 4x + e 2 x + 1)

∫ (e

π/4

0

dt dt Put e t ⇒ dx = x or e t (t − 1 / t ) dt ⋅ ∴J–1= ∫ 2 (t + 1 + 1 / t ) t x

π/4

=

0

1  1 − 2  t   =∫ dt  2 1  t + + 1   t2  

π/4

=

 | (tan x / 2 + 1) |

1 − tan 2 x / 2 0 put tan x/2 = t ⇒ dx =



1 as u = t + t

2t 1− t

4t



1+ t2 1− t2

0

1 t + −1 1 t = log 1 2 t + +1 t

2

0

2 −1

1 u −1 log 2 u +1

[as 0 < × < π/4]

(1 + t ) 2 −1

=

dx

2dt

∴ Area =

1 du 1 du − 2 ∫ u −1 2 ∫ u +1

| (1 − tan x / 2) |  dx 1 − tan 2 x / 2 

2

2 tan x / 2



2 tan x / 2 1 + tan 2 x / 2 1 − tan 2 x / 2 1 + tan 2 x / 2

1−



 1 − tan x / 2

π/4

=

1 1  Put t + = u ⇒ 1 − 2  dt = du t t   du ∴J–l= ∫ 2 u −1

=

 2 tan x / 2  1+ 1 + tan 2 x / 2   1 − tan 2 x / 2  1 + tan 2 x / 2 

∫  0

1  1 − 2  t   = ∫ dt 2  1 t 1  +  −  t

=



 1 + sin x 1 − sin x − cos x cos x 

∫ 

149. (b) Required Area =

2



2dt (1 + t 2 )

dt

=

 t2 − t +1  1 log  2  2  t + t +1

=

 e2 x − e x + 1  1 log  2 x  as t = ex x 2  e + e −1 

Exercises for self-practice 1. ∫ x log x dx is equal to 2

(b) x log x − 2 2 x (c) log x − 2 x2 (d) log x + 2 2

2.

tan x

∫ log sec x dx

x log x + 4 x log x − 4 x log x − 4



(

(d) None of these

x +c 4 2 x +c 4 x2 +c 4 2

4.

1

∫ sin x + cos x

dx =

1 π  log tan  x +  + c  4 2 1 x π  (b) log tan  +  + c 2 8 2  x π (c) log tan  +  + c 2 8

(a)

is equal to

5.

sin x − cos x dx is equal to 2 + sin 2 x

(

)

)

(c) log sin x + cos x − 2 + sin 2 x + C

(a) log (log (log x)) + c (b) log (log tan x) + c (c) log (log sec x) + c (d) None of these 3.

(

(b) – log sin x + cos x + 2 + sin 2 x + C

x x (a) log x − +c 2 4 2

)

(a) log sin x + cos x + 2 + sin 2 x + C

(d)

1  x π log tan  +  + c 2 8 2

∫e

(1 + tan x + tan 2 x) dx =

x

(a) ex cos x + c (b) ex tan x + c (c) ex sec x + c (d) ex sin x + c

 dx 

  dx   

227

(e x − e − x ) ∫ e2 x + 1 + e−2 x )dx

Indefinite Integration

=

228

6. If

∫ g ( x) dx

= g (x), then

∫ g ( x)[ f ( x) + f '( x)] dx is equal

to Objective Mathematics

5 3 −5 (d) 4

5 4 5 (c) –  6

(b) – 

(a)

(a) g (x) f (x) (b) g (x) f ′ (x) (c) g (x) f (x) – g (x) f ′ (x) (d) g (x) f 2(x)

1 dx = f (x) + A, where A is any arbitrary 4 2 15. If ∫ 3 sec x cos ec x dx = K tan x + L tan x + M cot  ( 1 + ) x x ∫ constant, then the function f (x) is x + const., then (b) 2 tan– 1 x (a) 2 tan–1x 1 (a) K = – 1, L = 0, M = 1 (b) K = , L = 1, M = 2 (c) 2 cot–1 x (d) loge (1 + x) 3 1 16. ∫ e x dx is equal to: (c) K = , L = 2, M = – 1 (d) None of these 3

7. If

8.

∫e

x

(a) e

(1 − cot x + cot 2 x) dx =

(a) e cot x + c (c) ex cosec x + c x

(b) – e cot x + c (d) – ex cosec x + c

1 (b) e 2

x

9. If ∫ x sin x dx = – x cos x + α, then α = (a) sin x + C (c) C 10.

(b) cos x + C (d) None of these

x2 + 1 dx is equal to 2 − 1)

∫ x (x

x2 − 1 + C x x + C (c) log 2 x +1 (a) log

11.

17.

∫e

18.

1 x e (sin x – cos x) + C 2 1 (d) (tan– 1x2)3 + C 2

t 3t 2

1 (c) –  (tan–1x3)2 + C 2 13.

∫ x cos x

2

dx is equal to :

1 sin2x + C 2 1 (c) – sin x2 + C 2

(a) –

14. If

2x + 3

∫ ( x − 1) ( x

2

1 = loge ( x − 1)5/ 2 ( x 2 + 1) a  − tan– 1x + A, 2 where A is any arbitrary constant, then the value of ‘a’ is

+ A , A is an arbitrary constant.

dt =

(c)

1 −3t 2 e + C 6

1 2 (d) –  e −3t + C 6

xe x

∫ ( x + 1)

2

dx =

(a)

ex  + C 1+ x

(b)

xe x  + C (1 + x)

(c)

ex (1 + x)3

(d)

ex +C (1 + x) 2

1 dx = x − x3

1 (1 − x 2 ) log +C 2 x2

(c) log x (1 – x2) + C

∫ x (1 + log x) dx x

(1 + log x) 2 (c) x2x 21.

∫ (a

2

(d)

(1 − x) +C x(1 + x)

1 x2 log +C 2 1 − x2

(b) xx log x (d) xx

dx = + x 2 )3 / 2

x ( a 2 + x 2 )3/ 2 x (c) 2 2 a (a + x 2 )1/ 2

(a)

(b) log

=

2

(a)

(b)

+ 1) dx

x

1 3t 2 (b) –  e + C 6

(a)

20. 1 sin2x + C 2 1 (d) sin x2 + C 2

+A

1 3t 2 e + C 6

19. ∫

1 (b) (tan– 1x3)2 + C 6 1 (d) (tan– 1x2)3 + C 2

x

(a)

(c)

(a) tan x + C

+A

(d) 2 ( x + 1) e

(a) ex (sin x – cos x) + C (b) ex (sin x + cos x) + C

–1 3

x



x2 − 1 +C x x (d) – log 2 +C x +1

x 2 tan − 1 x 3 dx is equal to 12. ∫ 1 + x6

+A

(c) 2 ( x − 1) e

(b) – log

x ∫ e sin x dx is equal to

x

(b)

1 a 2 (a 2 + x 2 )1/ 2

(d) None of these

e2 x − 1

(a) sin  (e ) + C (c) tan–1 (ex) + C 23. The value of

1 (a) sin −1 2

(



3

3

3

30.

)

x  (b) – 2 sin–1   2 sin  + C 2 (c) 2 sin–1 

(

)

1  [sin (log x) + cos (log x)] + C 2 x (b)  [sin (log x) + cos (log x)] + C 2 x (c)  [sin (log x) – cos (log x)] + C 2 1 (d)  [sin (log x) – cos (log x)] + C 2

33.

(1 + sin x) dx is 25. ∫ e (1 + cos x) (a)

1 x x  e  sec  + C 2 2

(b) ex sec 

x +C 2

(c)

1 x x  e  tan  + C 2 2

(d) ex tan 

x +C 2

dx 26. ∫ is equal to x( x 4 − 1)

27.

∫ sin

4

(a) a

1 x4 − 1 log 4 + C 4 x

(d) log

x4  + C x −1 4

sin 2 x dx is equal to x + cos 4 x

28. The value of x



a

(b) tan–1 (x tan2x) + C (d) None of these

x

x

loge a + C

(b) 2a

x

loge a + C

(c) 2a

x

log10a + C

(d) 2a

x

loga e + C

log x

dx is equal to

x2 2 (d) ex (b)

dx



x2 − 1

is

(a) sin–1x

(b) log  x + x 2 − 1  

(c) log  x − x 2 − 1  

(d) – cos–1x

∫ sin

−1

x dx is equal to

(a) x sin– 1x –

1 − x2 + C

(b) x sin– 1x +

1 − x2 + C

(c) – x sin– 1x +

1 − x2 + C

(d) None of these 34.

∫ x log (1 + x)

dx is equal to

1 [2 (x2 – 1) log (1 + x) – x2 + 2x] + C 4 1 (b) [2 (x2 – 1) log (1 + x) + x2 + 2x] + C 4 1 [2 (x2 – 1) log (1 + x) + x2 – 2x] + C (c) 4

(a) (b)

(a) tan–1 (tan2x) + C (c) 2 tan–1 (tan2x) + C

∫e

1 log (log x) + C 2

(d) (log x)2 + C

2

32. The value of

x

x4 − 1  + C x4

(b)

x + log x 2 (c) a log x

(a)

(c) log 

1 (log x)2 + C 2

(a)

3 1 sin (e x ) 3

is equal to

31. One value of

∫ cos (log x) dx is

1 x4 log 4 +C 4 x −1

(d)

(c) log (log x) + C

x  (d) 2 sin–1   2  + C 2

(a)

dx

∫ x log x (a)

2 sin x + C

24. The value of

3

(b) sin (e x )

(c) 3sin (e x )

1 + sec x dx is

2 sin x + C

3

(a) ex sin (e x )

(b) cos–1 (ex) + C (d) sec–1 (ex) + C

x

–1

29. ∫ x 2 ⋅ e x ⋅ cos (e x ) dx is equal to

is:

229

dx



dx is

(d) None of these 35. If

4e x + 6e − x dx = Ax + B log (9e2x – 4) + C, then x − 4e − x

∫ 9e

(a) A =

3 2

(c) 4

(b) B =

35 36

(d) None of these

x dx 36. ∫ 1 + x4 (a) log (1 + x2) + k (c)

1 tan–1 x2 + k 2

(b) tan–1 x2 + k (d) None of these

Indefinite Integration

22. The value of

230

37.

xe x

∫ (1 + x)

Objective Mathematics

(a)

ex +c x +1

(b) ex (x + 1) + c

−e x

(c) x + 1 2 + c ( ) 38. If

∫xe

2x

(a) (3x – 1)/4 (b) (2x + 1)/2 (c) (2x – 1)/4 (d) (x – 4)/6

dx is equal to

2

39. If

ex +c (d) 1 + x2



2 1 + sin x dx = 2 cos (ax + b) + c then the value

of (a, b)

π 2 (d) None of these

1 π , 2 4 (c) 1, 1

(b) 1,

(a)

dx is equal to e f (x) + c, where c is constant 2x

of integration, then f (x) is

Answers

1. 11. 21. 31.

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

2. 12. 22. 32.

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

3. 13. 23. 33.

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

4. 14. 24. 34.

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

5. 15. 25. 35.

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

6. 16. 26. 36.

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

7. 17. 27. 37.

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

8. 18. 28. 38.

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

9. 19. 29. 39.

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

10. (a) 20. (d) 30. (c)

7

Definite Integral and Area

CHAPTER

Summary of concepts The Definite Integral Let F (x) be any antiderivative of f (x), then for any two values of the independent variable x, say a and b, the difference F (b) – F (a) is called the definite integral of f (x) from a to b and is b

denoted by

∫ f ( x) dx . Thus

6.

2a

∫ o

7.

2a

∫ o

a

b

∫ f ( x) dx

8.

= F (b) – F (a),

a

a

The reason for using the term ‘‘definite integral’’ follows from the fact that the value of the definite integral in independent of the particular choice of the antiderivative of f (x). For, if F (x) + C is any other antiderivative of f (x), then b





f ( x) dx = F (x) + C

b



= [F (b) + C] – [F (a) + C]



= F (b) – F (a),



a

f ( x) dx + ∫ f (2a − x) dx o

0  a dx = f ( x)  2 ∫ f ( x) dx  o

if f (2a − x) = − f ( x)   if f (2a − x) = f ( x)  

 a  2 ∫ f ( x)dx if f (− x)=f ( x) i.e. f ( x) is even  f ( x) dx =  o  0 if f (− x) = f ( x) i.e. f ( x) iss odd   b

∫ f ( x) dx

∫ f (a + b − x)

=

a

10.

a

o

b

9.

b

dx

a

f ( x)

∫ f ( x) + f (a + b − x)

b−a 2

=

a

11. If f (x) ≥ 0 on the interval [a, b], then

12. If f(x) ≤ g(x) on the interval [a, b], then

a

13.

b

∫ f ( x) dx a

b

∫ f ( x) dx , it is sufficient to find an antiderivaa

tive of f (x), say F (x) and subtract the value of F (x) at the lower limit a from its value at the upper limit b.

b



1.

∫ f ( x) dx

a

=–

a

∫ f ( x) dx a

4. 5.

c

=

∫ a

a

a

b

a

14. If f (x) is continuous on [a, b], m is the least and M is the greatest value of f (x) on [a, b], then m (b – a) ≤

b

∫ f ( x) dx

≤ M (b – a)

15. If f (x) is a periodic function of period a, i.e.,

o

o

∫ o

∫ f ( x) dx a

b

=

∫ f ( y) dy

na

(a)

a

na

(b)

c

dx

f ( x) a dx = 2 f ( x) + f (a − x)

∫ f ( x)

dx = n

o

b

f ( x) dx + ∫ f ( x) dx, where a < c < b

∫ f ( x) dx = ∫ f (a − x) a

2. 

b

b

3.

∫ f ( x)

b

∫ g ( x)

≤ ∫ | f ( x)| dx

f (a + x) = f (x), then b

b

a

a

Properties of Definite Integrals

≥ 0.

f ( x) dx ≤

a

dx

which is same as before.

b

∫ f ( x) dx a

a

Thus to find

a

−a

where F (x) is any antiderivative of f (x). The numbers a and b are called the limits of integration; a is the lower limit and b is the upper limit. Usually F (b) – F (a) is abbreviated by writing b F (x) .

f ( x) dx =



a

∫ f ( x) dx o

f ( x) dx = (n – 1)

o

∫ f ( x) dx o

b + na

(c)

a



na

f ( x) dx =

b

∫ f ( x) dx, where b ∈ R

+

0

b+a

(d)

∫ b

f ( x) dx independent of b.

232

b + na

(e)

∫ b

b

a

∫ f ( x) dx

f ( x) dx = n ∫ f ( x) dx, where n ∈ I

Objective Mathematics

where nh = b – a.

In particlular, (i) if b = 0, (ii) if n = 1,



b+a

a

Put a = 0 and b = 1 ⇒ nh = 1 ⇒ h =

o

Substitute a = 0, b = 1 and h =



h→0

∫ f ( x) dx

f ( x) dx =

b

∫ a

a

= f (c) (b – a), where a < c < b.

(i) Find the rth term of the given series and express it in the form 1  r  Then the given series can be written as f  ⋅ n n

lim 1 n

n→∞

In the functions g (x) and h (x) are defined on [a, b] and differentiable at all points x ∈ (a, b) and f (t) is continuous on [a, b], then d dx

b

∫ | f ( x )| dx Working Rule

1

lim

n→∞

Note: Here each term tends to zero when n → ∞, so addition or omission of finite number of terms does not affect the required limit.



π/2

a

β

2.

with proper sign of f (x) in each (a, α), (α, β), (β, b).

Summation of SerieS with the help of Definite integral aS the limit of a Sum If f (x) is a continuous and single valued function defined on the

follows:

∫ cos

n

x dx

0

∫ sin

n

x cosmx dx

0

dx

interval [a, b], then the definite integral

π/2

2  n −1 n − 3 n − 5   n ⋅ n − 2 ⋅ n − 4 ⋅ ⋅ ⋅ 3 , when n is odd  =  1 π  n −1 n − 3 n − 5 , when n is even ⋅ ⋅ ⋅ ⋅ ⋅ ⋅   n n−2 n−4 2 2  

b

α

sin n x dx =

0

∫ f ( x) dx

a

If n is a positive integer, then π/2

(iii) Check the sign of f (x) by taking any point in each of these subintervals.

∫ f ( x) dx + ∫ f ( x) dx + ∫ f ( x)

r1 n

r2 , n where r1 and r2 are the least and greatest values of r respectively.

(ii) Divide the interval (a, b) into subintervals (a, α), (α, β) and (β, b).

=

∫.

and upper limit of integration = nlim →∞

Let the roots be α and β such that a < α < β < b.

b

by

n→∞

(i) Solve the equation f (x) = 0.

β



Also, lower limit of integration = lim

1.

α

r =1

Some uSeful reDuction formulae

a

(iv) Use the property

r

n

∑ f  n 

(ii) Write the above series equal to a definite integral by replacing n r by x and by dx, n

 h ( x)   ∫ f (t ) dt  = d [h (x)]⋅ f [h (x)] – d [g (x)]⋅ f [g (x)]  g ( x)  dx dx

integration of moD functionS

1 . n

Working Rule

a

Differentiation unDer integral Sign

...(1)

1 in (1), we get n 1 1 n r lim f   = ∫ f ( x) dx. ∑ n→∞ n n 0 r =1

b

f 2 ( x) dx ⋅ ∫ g 2 ( x) dx

If a function f (x) is continuous on the interval [a, b], then there exists a point c ∈ (a, b) such that

∫ f ( x) dx

a

o

f ( x) ⋅ g ( x) dx ≤

b

r =1

o

h or nlim →∞

∫ f ( x) dx = n ∫ f ( x) dx

a

17.

b

a

For any two functions f (x) and g (x), integrable on the interval [a, b], the Schwarz – Bunyakovsky inequality holds b

n

∑ f (a + rh) = ∫ f ( x) dx

na

b

16.

= hlim h [ f (a + h) + f (a + 2h) + ⋅ ⋅ ⋅ + f (a + nh)] →0

a

o

b

∫ f ( x) dx a

is defined as

m−3 2 1   m −1   m + n ⋅ m + n − 2 ⋅ ⋅ ⋅ 3 + n ⋅ 1 + n ; if m is odd   and n may be even or odd     m−3 n −1 n − 3 m −1 1 2  ⋅ ⋅⋅⋅ ⋅ ⋅ ⋅⋅⋅ =   m+n m+n−2 2+n n n−2 3   m n if is even and is odd   m −1 m−3 1 n −1 n − 3 1 π  ⋅ ⋅⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ,  2+ n n n − 2 2 2  m + n m + n − 2  if m is even and n is evenn 

∫ sin

m

x ⋅ cos n x dx

0

=

[(m − 1) (m − 3) ...][(n − 1) (n − 3) ⋅ ⋅⋅] ( m + n ) ( m + n − 2) ⋅ ⋅ ⋅

to be multiplied by

4. If we have two curve y = f (x) and y = g (x), such that y = f (x) lies above the curve y = g (x) and both are above the axis of x then the area bounded between them and the ordinates x = a and x = b (b > a), is given by

π when m and n are both even integers. 2

Area of Plane Regions 1. The are bounded by the curve y = f (x), x-axis and the ordinates x = a. and x = b (where b > a) is given by

b

A=

∫ a

b

f ( x) dx − ∫ g ( x) dx a

i.e., upper curve area – lower curve area. 5. The area bounded by the curves y = f (x) and y = g (x) between the ordinates x = a and x = b is given by b

A=

b

∫ y dx =

∫ f ( x) dx

a

a

2. The area bounded by the curve x = g (y), y – axis and the abscissae y = c and y = d (where d > c) is given by

c

A= d

A=

∫ x dy = c

∫ a

d

∫ g ( y) dy .

b

f ( x) dx + ∫ g ( x) dx , c

where x = c is the point of intersection of the two curves.

c

3. The area bounded by the curve y = h (x), x-axis and the two ordinates x = a and x = b is given by

Curve Tracing In order to find the area bounded by several curves, it is important to have rough sketch of the required portion. The following steps are very useful in tracing a cartesian curve f (x, y) = 0. Step 1: Symmetry



A=

c

b

a

c

∫ y dx + ∫ y dx

where c is a point in between a and b.

(i) The curve is symmetrical about x-axis if all powers of y in the equation of the given curve are even. (ii) The curve is symmetrical about y – axis if all powers of x in the equation of the given curve are even. (iii) The curve is symmetrical about the line y = x, if the equation of the given curve remains unchanged on interchanging x and y. (iv) The curve is symmetrical in opposite quadrants, if the equation of the given curve remains unchanged when x and y are replaced by – x and – y respectively.

233

π/2

Definite Integral and Area

These formulae can be expressed as a single formula :

234

Step 2: Origin

Objective Mathematics

If there is no constant term in the equation of the given curve, then the curve passes through the origin. In that case, the tangents at the origin are given by equating to zero the lowest degree terms in the equation of the given curve. For example, the curve y3 = x3 + axy passes through the origin and the tangents at the origin are given by axy = 0 i.e., x = 0 and y = 0.

the coefficient of the highest power of y in the equation of the given curve. (ii) The horizontal asymptotes or the asymptotes parallel to x-axis of the given curve are obtained by equating to zero the coefficient of the highest power of x in the equation of the given curve. Step 5: Region

Find out the regions of the plane in which no part of the curve lies. To determine such regions we solve the given equation for (i) To find the points of intersection of the curve with y in terms of x or vice-versa. Suppose that y becomes imaginary X-axis, put y = 0 in the equation of the given curve and get for x > a, the curve does not lie in the region x > a. the corresponding values of x. Step 6: Critical Points (ii) To find the points of intersection of the curve with dy Y-axis, put x = 0 in the equation of the given curve and get Find out the values of x at which = 0. dx the corresponding values of y. At such points y generally changes its character from an increasing function of x to a decreasing function of x or viceStep 4: Asymptotes versa. Find out the asymptotes of the curve. Step 7: (i) The vertical asymptotes or the asymptotes parallel to y-axis of the given curve are obtained by equating to zero Trace the curve with the help of the above points. Step 3: Intersection with the Co-ordinate Axes

mULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1. The value of the integral

π

∫e

cos 2 x

⋅ cos3 (2n + 1) x dx, n

2

5.

2.

π/ 2

∫ 0

3.

6.

 x cos   ⋅ sin x dx is equal to 2 5

−4  1  (b) 1 −  7  8 2 

4 1  (c) 1 −  7 8 2 

(d) None of these



2 ∫0  e x  dx (where [ . ] denotes the greatest integer function) equals (b) e2 (d) 2 e

(c) 0 −4

4. If



f ( x) dx = 4 and

−1

∫ f ( − x ) dx

1 6 1 (c) 4

(a)

(b)

(d) None of these π

∫ x (sin

7. The value of the integral 3 π2 (a) 64 2 3π (c) 256

1 12

0

(b)

4

x cos 4 x) dx is

3 π2 128

(d) None of these r

0

= 7, then the value

is

(a)

(b)

r3 3

r3 (d) None of these 2 9. The area of the figure bounded by f (x) = sin x, g(x) = cos x in the first quadrant is (c)

(b) –3 (d) None of these

r 6 3

2

−2

(a) 2 (c) 5

sin 2 x ⋅ cos 2 x dx is equal to 3 x + cos3 x) 2

∫ (sin

−4

∫ ( 3 − f ( x) ) dx

(b) 1 (d) 101/2

8. The value of the integral ∫ xy dx, where y = r 2 − x 2 , is

1

of

π/ 4

0

2 1  (a) 1 −  7 8 2 

1   x −  dx = x 

(a) 1/4 (c) 0

(b) π (d) None of these

(a) loge 2

101

1/2

0

integer, is (a) 0 (c) 2π

1

∫ x cos ec

(b)

3 +1

(a) –

(d) None of these π

∫e

10. The value of the integral

x

π/ 2

(a) e– π/2 (c) eπ/2



1 − sin x dx is 1 − cos x

a

19.



∫e

12. The value of the integral

x

0

2 (e2π + 1) 5 (c) – 2 (e2π – 1) 5 (a)

(b)

π/6

π 4 π (c) 12

(d) None of these

π/ 2

∫ (tan x + cot x)

dx is equal to

(a)

π log 2 2

1

(c) π log 2

(d) None of these ∞

x

∫ (1 + x ) (1 + x )

21. The value of the integral

dx is

2

0

π 2 π (c) 6

(b)

π 4

(d) None of these π

∫ log (1 + cos x )

22. The value of the integral

dx is

0

(d) None of these

∫ e ( x − α ) dx x2

π 6

π log 2 2

(b) –

(d) None of these

(b)

(a)

π –1 2

0

sin x dx is equal to sin x + cos x



14. If

is equal to

a2 − x2

(a)

π/3

13.

dx

π + 1 2 π (c) 1 – 2

20.

π2 log 2 2

(d) None of these

(a)

π x sin  +  dx is  4 2

(b) – 2 (e2π + 1) 5

∫a+ 0

11. The area bounded by the curves y = ln, y = ln | x |, y = | ln x | and y = | ln | x | | is (b) 2 sq. units (d) None of these

(b)

(c) π2 log 2

(b) eπ/4 (d) None of these

(a) 5 sq. units (c) 4 sq. units

π2 log 2 2

235

) 3 − 1)

2 −1

Definite Integral and Area

( (c) 2 ( (a) 2

(a)

π log 2 2

(b) π log 2

(c) – π log 2

= 0, then

(d) None of these

0

(a) 1 < α < 2 (c) 0 < α < 1 15.

π/ 2



(b) α < 0 (d) α = 0

23. The value of

(b) –

(c) π log 2 16. If



r2

0

xf ( x) dx =

17.



1

0

(a)

π log 2 2

(d) None of these

24. The value of c for which the area of the figure bounded by the curve y = 8x2 – x5, the straight lines x = 1 and x = c and the x-axis is equal to 16/3, is

2 5 , t > 0 then f  4  = t  25  5

2 (a) 5 2 (c) – 5

5 (b) 2

(c) π 18. The value of the integral

25.

(b)

(c) 3

(d) –1

1



log (1 + x )

(d) 1

(b)

π –1 2

(d) 1 π

∫ x log sin x dx is 0

8 − 17

(a) 2

0

1− x dx = 1+ x π + 1 2

is

(a) –π (b) –2π (c) –­3π (d) –4π [ . ] representing the greatest integer function.

0

π log 2 2

∫ [ 2 sin x] dx 0

log sin x dx is equal to

(a)



1 + x2

dx is equal to

(a)

π log 2 8

(b)

(c)

π log 2 4

(d) None of these

26. The value of the integral π 2 (c) 0

(a)

π 1 log 8 2

π

sin 2kx  dx, where k ∈I, is 0 sin x



(b) π (d) None of these

236

27.

π/ 2

∫ 0

cos 2 θ d θ is equal to cos 2 θ + 4 sin 2 θ

Objective Mathematics

π 2 π (c) 6

π log 2 2

29.

π





1  dx  ∫0 log  x + x  1 + x 2 is

37.

x 3 cos 4 x sin 2 x dx is equal to 2 − 3πx + 3 x 2

38. If

π2 8 π2 (d) . 32

π2 16 π2 (c) 4

π

π2 4

x 2 sin x

∫ ( 2 x − π ) (1 + cos x ) 2

(c)

π 6

(b)

n

dx is

n n2 − 1 −n (d) 2 n +1 (b)

et dt ∫0 t + 1 = a, then 1

b

e − t dt is equal to ∫ b −1 t − b − 1 (b) ­–a e–b (d) None of these

π2 2

dx = sin x +1 e − π /2



(a) 0 (c) –π/2

(b) 1 (d) π/2

40. The value of the integral

3

∫ | ( x − 1) ( x − 2) ( x − 3) | 1

(d) None of these

(a) 1 3 (c) 9 4

 x2  31. Area bounded by the curves y =  + 2  ([ . ] denotes 64   the greatest integer function), y = x – 1 and x = 0 above the x-axis is (a) 2 (c) 4

π/3

41.

π

x 2 cos x

∫ (1 + sin x )

2

dx is

(b) – π (2 + π) (d) π (2 – π)

33. The area bounded by y = x e y = 0 is (a) 4 (c) 1

| x |

a

and the lines | x | = 1,

0

0

34. If ∫ f (2a − x)dx = µ and ∫ f ( x) dx = λ, then



2a

0

f ( x)dx

equals to: (a) 2λ – µ (c) µ – λ 35. The value of the integral

(b) λ + µ (d) λ – 2µ π/ 4

sin x ⋅ cos x dx is 2 x + sin 4 x

∫ cos 0

(b)

π 6 3

(d) None of these

(d) None of these

sin x + cos x dx is equal to sin 2 x (b) sin − 1 1 ( 3 − 1) 2 (d) None of these

3/ 2

42.

(b) 6 (d) 2

a

(b) 9 2

(a) 2 sin − 1 1 ( 3 − 1) 2 (c) 2 sin − 1 1 ( 3 + 1) 2

0

(a) π (2 + π) (c) π (π – 2)



π/6

(b) 3 (d) None of these

32. The value of the integral

π 3 π (c) 4 3

ln ( x) dx is

where n is an integer > 1, is

2

(a)



π /2

39.

0

(a)

)

(a) a e–b (c) b e–b

(b)

30. The value of the integral

1 + x2

n 1 − n2 n (c) 2 n +1

(d) π log 2.

(a)

(x +

dx

e4 e

(a)

∫π 0

∫ 0

π log 2 2

(b) –

(c) – π log 2

dx = a, the value of

(a) e4 – e (b) e4 – a (c) 2e4 – a (d) 2e4 – e – a

(d) None of these

28. The value of the integral

∫e

x2

1

(b) 0

(a)

(a)

36. Given

2

∫ | x sin πx |

dx is equal to  

−1

(a)

1 1 1 +  π π

(b)

1 1 1 −  π π

(c)

1 1   − 1 ππ 

(d) None of these

43. For x ∈ R and a continuous function f, 1 + cos 2 t

let  I1 =



x f { x (2 − x)} dx

sin 2 t 1 + cos 2 t

and  I2 =



f { x (2 − x)} dx . Then I1/I2 is

sin 2 t

(a) 0 (c) 2

(b) 1 (d) 3

dx is

0

2 (a) π −2 π

(c)

0

0

∫ log (1 + tan x ) dx

(b) I3 > I1 > I2

(c) I1 > I2 > I3

(d) I3 > I2 > I1

(b) k 4 (d) k 8

1 2 1 (c) 16

2

∫| x

(c) − 3 x 2 + 2 x | dx is

3

(b) n – 1

1 n −1

(b) 4 log 4 3

56.

n +1



f ( x) dx = n2. The value



10π

0

(b) 8 (d) 18

(b) 14 (d) None of these

1 − x   −1 50. If I = ∫ cos 2 cot  dx , then 1 + x  0  1

51. If A =

∫ 1

A A B of e e 1

(b) 1 = –1/2 (d) None of these

2

A B2 2 A + B2

cosec θ

B −1 is −1

∫ 1

dt , then the value t (1 + t 2 )

∫ cos

p

x cos qx dx, then

(a) f (p, q) =

q f ( p – 1, q – 1) p+q

(b) f (p, q) =

p f ( p – 1, q – 1) p+q

(c) f (p, q) =

−p f ( p – 1, q – 1) p+q

(d) f (p, q) =

−q f ( p – 1, q – 1) p+q

−2

t dt and B = 1 + t2

π/ 2

0

n

sin θ

dx is

|sin x | dx is

57. If f (p, q) =

f ( x) dx is

(a) I > 1/2 (c) 0 < 1 < 1/2

2

(b)

(a) 20 (c) 10

(d) None of these

49. Suppose for every integer n,

1 + 2 cos x

∫ ( 2 + cos x )

1 2 −1 (d) 4

1 4 −1 (c) 2

2

(a) 16 (c) 19

(d) None of these

(a)

 x + 1  x − 1   +  − 2 dx is  x − 1  x + 1



θ dθ, where n is a positive integer, then

0

48. The value of the integral

4

n

55. The value of the integral

(b) 1 (d) 0

(a) 2 log 4 3 4 (c) log 3

∫ tan

π/ 2

−1



π/ 4

(d) None of these

(a) 1

2−x

− 1/ 2

1 n −1

n (In – 1 + In + 1) =

1 4

(c)

2

x dx, then Un + Un – 2 = (b) n – 1

54. If In =

(d) None of these

(a) – 1 (c) 2

n

(a) 1

∫ log 2 + x dx is equal to

1/ 2

∫ tan

0

(b)

(a)

1

π/ 4

0

0

of

(a) I1 > I3 > I2

53. If Un =

is

46. The value of the integral

0

0

π/4

k 4 k (c) – 8

∫ sin ( cos x ) dx

=

∫ cos x dx , then

and  I3 =

π

(a) –

π/2

2

π/2

−1 π

(d)

∫ cos ( sin x ) dx ; I 0

45. If ∫ log sin x dx = k, then the value of

47.

π/2

52. If I1 =

1 (b) π

(b) cosec θ (d) 1

237

∫ | x cos πx | dx is

58.

π

∫ x sin

6

x cos 4 x dx is equal to

0

(a)

3π2 512

(b)

3π2 256

(c)

3π2 1024

(d) None of these

Definite Integral and Area

(a) sin θ (c) 0

1

44. The value of the integral

238

59. In =

Objective Mathematics

(a)



π/4

0

tan n x dx , then nlim n [In + In + 2] equals →∞

1 2

60.

π/3

∫ cos

4

68.

(d) zero 3φ sin2 6φ dφ is equal to

π

2 x (1 + sin x )

−π

1 + cos 2 x



π 32 5π (c) 96

61.

(b)

7π 96 69.

9/ 2

− 1/ 2

dx is equal to

(a) 63π  a5 4

(b) 63π  a5 2 63 π a5 (d) 8

(c) 63π a5 2

∫x

3

1

∫x

7π (b) 4 7π (d) 16

5



(1 + x )

6 7

0

72.

73.

dx is equal to 8 (b) 45

74.

(a) 0 (c) – 1

67.

(d) None of these

π [ x] dx is equal to 2

2

∫ sin





(b) – 1 (d) None of these

1 − cos 2 x dx is equal to 2 (b) – 2 (d) – 4

∫ [2 x]

dx is equal to

(a)

1 2

(b)

−1 2

(c)

3 2

(d)

−3 2

π

∫ | cos θ − sin θ | (a)

(b) 1 (d) None of these

dθ is equal to

2

(c) 3 3 2

[ x 2 ] dx is

75.

∫ [x ] 2

dx is equal to

−2

(a) 2 – (c)

π2 2 3

0

−1

2

(c)

(d) None of these

is equal to

0

(b)

−1

∫ [ x] dx, where [.] denotes the greatest integer function,



π2 6 3

(a) 2 (c) 4

(d) None of these

π2 4 3

(a)

0

1

66.

x+π/4 dx is equal to 2 − cos 2 x − π/ 4



(a) 1 (c) 0

π+2 (b) 8

4 (a) 45 2 (c) 45 65.

(b) 2 (d) – 2

2

x2



π 2

x  dx is equal to

0

1+ x dx is equal to 1 − x2

3π + 8 (a) 24 3π + 8 (c) 16 64.

71.

2

0

(d)

π/ 4

70.

0

63.

∫ 

2 x − x 2 dx is equal to

7π (a) 2 7π (c) 8

(b) π2

(a) 1 (c) – 1

0

62.

3

0

∫ x ( 2a − x )

dx is

(c) zero

(d) None of these

2a

(d) None of these

π2 4

(a)

0

(a)

(b) 2 − 2

(c) 2 + 2

(b) 1

(c) ∞

2 −2

(a)

2

2 – 1

1.5

(b) 2 + (d) –

2 2 –

3 +5

2 ∫ [ x ] dx, where [.] denotes the greatest in integer func0

tion, is equal to

(a) 10 − 2 3 − 2 2 (b) 10 + 2 3 − 2 2 (c) 10 − 2 3 + 2 2 (d) None of these

(b) 2 2 (d) None of these

dx is equal to

x

∫ f (t ) dt

85. If f (x) is an odd function, then

0

(a)

2

(d) None of these

1

(a) odd (b) even (c) neither even nor odd (d) None of these 86. If f (x) is an even function, then

77. ∫ [ x + 1] dx is equal to −2

(a) 0 (c) – 1 78.

100 π



(b) 1 (d) None of these

∫ [ x − 1]

(d) 400 2

| cos x | dx is equal to (b) 20 (d) None of these

n2

0

(a)

n (n + 1) (4n + 1) 6

(b)

(c)

n (n − 1) (4n − 1) 6

(d) None of these

∫ |sin x |

n (n − 1) (4n + 1) 6

(b) 20 (d) None of these

] dt, where n is any positive integer, is equal to

( ) (b) n n − (1 + 2 + 3 + ⋅ ⋅ ⋅ + n ) (c) − n n + (1 + 2 + 3 + ⋅ ⋅ ⋅ + n )

(a) n n + 1 + 2 + 3 + ⋅ ⋅ ⋅ + n

(d) None of these 84.

∫ ( x − [ x ])

−1

dx is equal to

t

∫ n

2

sin z 2 z

dz, then dy is dx

equal to tan t 2t 2 tan t (c) 2 t

(a)

(b)

2t 2 tan t

(d) None of these



sin x

π/2

sin x + cos x

0

π 4

(b)

dx π 2

(d) 1

x2

∫ sin

t dt is equal to

0

1 3 −1 (c) 3

2 3 −2 (d) 3 (b)

91. If f (x) be a periodic function of period a, then

a

(a)



a

(b) n  ∫ f ( x)  dx

f ( x)  dx

0

0

a

(c) (n – 1) ∫ f ( x)  dx

(d) None of these

0

92. lim

h→0

1 h

x+h

∫ x

dz z + z2 + 1

(a) 0 (c)

is equal to

(b) 1 x +1 2



na

∫ f ( x) a

dx is equal to

0

(a) 50 (b) 100 (c) 200 (d) None of these

(d) None of these

z dz and y =

(a)

0

100

∫ sin

1 90. lim 3 x→0 x

dx is equal to

(a) 18 (c) 40 2

0

(c) zero

π

n

sin t

`88. If x =

0

10 π

(b) n ∫ f ( x) dx

a

89. Evaluate

∫  x  dx is equal to

∫ [t

dx

a

∫ f ( x)  dx

(a)

− 20 π

83.

∫ f ( x)

0

(b) 3 (d) 2

(a) 40 (c) 60

a

(a)

(c) (n – 1) ∫ f ( x)   dx

dx is equal to

(a) – 3 (c) –2

82.

na

is equal to

−1

81.

(a) odd (b) even (c) neither even nor odd (d) None of these

(b) 50 2

2



dt is

0

(c) 200 2

20 π

∫ f (t )

87. If f (x) is a periodic function of period a, then

1 − cos 2 x dx is equal to

(a) 100 2

80.

x

0

0

79.

is

a

(b) 2 2

(c) 3 2

239

π

∫ |sin x + cos x |

1 x + x2 + 1



(d) None of these

Definite Integral and Area

76.

240

x3

dt 93. If f (x) = ∫ , then f ′′ (x) is equal to 1 + t4 1

Objective Mathematics

(a)

6 x (1 − 5 x ) (1 + x12 ) 2

(b)

(c)

− 6 x (1 − 5 x12 ) (1 + x12 ) 2

(d) None of these

12

−1 2 ∫ (tan t ) dt 0

x4

∫ sin

2 π x = 0 and x = a is a + a sin a + π cos a, then f   = 2 2 2 2 1 (a) 1 (b) 2 (c) 1 (d) None of these 3

6 x (1 + 5 x ) (1 + x12 ) 2 12

x2

94. lim x→0

101. Let f (x) be a continuous function such that the area bounded by the curve y = f (x), x-axis and the lines

is equal to

1 102. If 2 f (x) + 3f   = 1 − 2 x ≠ 0, then x  x equal to

t dt

0

(a) 1

−1 1 (d) 2 2 95. The intervals of increase of the function f (x) defined by x

∫ (t

2

−1

(b)

(c)

2 1 log 2 + 5 2

(d) None of these

103. If an =

π/ 2

∫ (|sin x | + |cos x |)

0

(a) G.P. (c) H.P. 104. If an =

(b) A.P. (d) None of these

π/ 4

∫ cot

(b) 40 (d) None of these

∫ ( sin

− 100 π

4

dx. then

x + cos 4 x ) dx is equal to

(a) 100 π (c) 200 π

(b) 150 π (d) None of these

99. If f (x) and φ (x) are continuous functions on the interval [0, 4] satisfying f (x) = f (4 – x), φ (x) + φ (4 – x) = 3 and 4

∫ f ( x)

dx = 2, then

0

4

∫ f ( x)

φ (x) dx =

0

(a) 3 (c) 2

(b) 6 (d) None of these

100. If f (y) = e , g(y) = y; y > 0 and y

F(t) =



t

0

f (t − y ) g ( y ) dy , then

(a) F(t) = 1 – e–t (1 + t) (c) F(t) = tet

(b) F(t) = et – (1 + t) (d) F(t) = te–t

(b) A.P. (d) None of these log ( sin x )

3

3 log ( sin x 3 )

∫1 x x = φ (k) – φ (1) then the possible value of k is

105. If φ ′ (x) =

, x ≠ n π, n ∈ I and

(a) 27 (c) 9

−2

(a) f (x) is continuous in [–1, 1] (b) f (x) is differentiable in [– 1, 1] (c) f ′ (x) is continuous in [– 1, 1] (d) f ′ (x) is differentiable in [– 1, 1].

x dx, then a2 + a4, a3 + a5, a4 + a6 are

(a) G.P. (c) H.P.

x

∫ | x + 1|

n

0

in

dx is equal to

(a) 20 (c) 10

98.

−2 1 log 2 − 5 2

cos 2 nx dx, then a2 – a1, a3 – a2, a4 – a3 are sin x



0

100

dx is

1

in

+ 2t ) ( t − 1) dt are

10π

97. Let f (x) =

∫ f ( x)

2

(a) (– ∞, – 2) ∪ (2, – ∞) (b) (– ∞, – 2) ∪ (– 1, 0) ∪ (1, – ∞) (c) (– 1, 0) ∪ (1, ∞) (d) None of these 96.

−2 1 log 2 + 5 2

(b) – 1

(c)

f (x) =

(a)

2

106.

16

∫ log x

 dx

(b) 18 (d) None of these dz, where z = x4, is equal to

1

(a) 16 log 2 + 15 4

(b) 16 log 2 − 15 4

15 (c) − 16 log 2 +  4 

(d) None of these

107. Let f (x) be a polynomial of degree 2 satisfying f (0) = 1, 2

f ′ (0) = – 2 and f ′′ (0) = 6, then to (a) 6 (c) 9 108. Let In =

∫ f ( x)

dx is equal

−1

(b) 0 (d) None of these π/ 2

∫ sin

n

x dx, n is a positive integer. Then

0

(a) In : In – 2 = n : (n – 1) (b) In > In – 2

(c) n (In – 2 – In) = In – 2 (d) None of these

∫ log

(a) 1 (c) 0

3e x cos (3e ) dx is x

117. The value of the integral

3π / 2

∫ [sin x] dx, where [.] denotes

π/ 2

π 6

the greatest integer function, is π −π (a) (b) 2 2

(b) – 1 (d) None of these

(c) 0 (d) π 110. If (– 1, 2) and (2, 4) are two points on the curve y = f (x) and if g (x) is the gradient of the curve at the 14 + 24 + 34 + ... + n 4 13 + 23 + 33 + ... + n3 118. lim is − lim point 5 n→∞ n → ∞ n n5 2

∫ g ( x) dx

(x, y), then

1 30 1 (c) 4

is

(a)

−1

(a) 2 (c) 0

(b) – 2 (d) 1

111. If f (a + b – x) = f (x), then

112.

(a)

a+b 2



b

(c)

b−a 2



b

5

∫e

f [φ ( x )]

a

a



b

a

x f ( x) dx is equal to

a+b 2



b

f ( x) dx

a+b 2



b

(d)

a

a

f ( x) dx f (a + b − x) dx

x→0



0

l

(a) 3 (c) 1

(b) 2 (d) 1

(b)

a log 2 π

(d)

2a log 2 π

e6

 log x  ∫1  3  dx, where [.] denotes the greatest integer function, is

115. The value of the integral

(a) 0 (c) e6 + e3

(b) e6 – e3 (d) e3 – e6

116. The value of the integral I = (a)

1 n +1

1 1 (c) − n +1 n + 2

(b)



1

0

dx, then

1− l



dx

∫ 1+ e

sin x

(b) 2 I1 = I2 (d) None of these

is equal to

(a) π π (c) 2

where a, b, c are constants, then c =

−a log 2 π

∫ sin ( x (1 − x))

0

a  b tan x + c  dx = 0, 114. If ∫  |tan x | + 1 + sec x  − π/3  3

(c)

dx and

1− l

(a) I1 = 2 I2 (c) I1 = I2

is

π/3

(a) a log 2

∫ x sin ( x (1 − x))

l

I2 =

121.

2

x sin x

dx is

(b) – 8 (d) – 9

120. If I1 =

(b) 0 (d) None of these

113. The value of lim

∫ max {(1 − x), (1 + x), 2}

(a) 8 (c) 9

to

sec t dt

1 5

2

3

x2

(d)

119. The value of

⋅ f ' [φ ( x)] ⋅ φ ′ (x) dx, where φ (3) = φ (5), is equal

(a) 1 (c) 3

(b) zero

−2

f (b − x) dx (b)

x (1 − x) n dx is

1 n+2

1 1 (d) + n +1 n + 2

241

109. The value of the integral

π 3

122.

π/6

∫ sec

2

(b) 2π (d) None of these x d ( x − [ x]) is equal to

0

(a)

3

1 3

(b)

(c) 1

(d) None of these

123. Let f (x) = minimum {x + | x |, x – [x]}, where [.] de1

notes the greatest integer function. Then

∫ f ( x)

dx is

−1

equal to 1 (a) 2

(b)

(c) 1

−1 2

(d) – 1

124. If φ and f are two continuous functions, then the value π/ 4

of the integral



{ f ( x) + f (− x)} . {g (x) – g (– x)} dx

− π/ 4

is π 4 −π (c) 4

(a)

(b) 0 (d) None of these

Definite Integral and Area

log

242

125. If f (a – x) = f (x) and

a/2



19

f ( x) dx = p, then

133.

a

Objective Mathematics

∫ f ( x)

dx is equal to

(a) 2p (c) p

(b) 0 (d) None of these

(a) b – a (c) b + a

3 sin x3 dx = F(k) – F(1), e x then one of the possible values of k, is 4

log 3



  f (0) + 4  1 (b)  f ( 0 ) + 2 6 1 2

(a)

(

log x + 1 + x 2

log 1/ 3

)

dx is

(b) 2 log 3 (d) None of these

a

128. If I =

∫ (α sin

5

x + β tan 3 x + γ cos x) dx, where α,β,γ

−a

b

∫ x f ( x) dx equals

b

∫ f ( x)

dx

a

(c) 0 2π



1 + sin

0

(a) 0 (c) 8 131.

∫e

− ax

b

∫ f ( x) a

(d) None of these

130. The value of



a−b (b) 2

x dx is 2

dx

cos bx dx equals

b (a) 2 a + b2 a (c) 2 a + b2

b (b) 2 a − b2

∫ |(1 − x)|

137. If

x dx is 132. The value of ∫ 2 −4 x 3

15 (b) 2 + log e   7 (c) 2 + 4 loge 3 – 4 loge 7 + 4 loge5 (d) 2 – tan– 1 (15/7)

dx = p log (qe – 1) – r, then x −1

1

(b) p = 1, q = 2, r = 1 (d) None of these

138. The smallest interval [a, b] such that 1 dx ∫0 1 + x 4 ∈ [a, b] is given by  1  (a)  , 1  2 

(b) [0, 1]

1 (c)  , 1 2 

3 (d)  , 1 4 

139. If f (x + y) = f (x). f (y) for all x, y where f ′ (0) = k ≠ 0, then f (x) can be expressed as (a) aekx (c) kx 140. If

2

15 (a) 2 − log e   7

(b) 0 (d) 4

∫ 2e 0

(d) None of these 5

dx equals

(a) p = 1, q = 1, r = –1 (c) p = 1, q = 2, r = –1

(b) 2 (d) 4

0

1 6

(a) – 2 (c) 2

a

a+b (a) 2

 1 f   + f (1)  2   1   f   + f (1)  2 

−1

(b) α, β, γ, a (d) None of these

129. If f (a + b – x) = f (x), then

dx equals

1

136.

are constants, then the value of I depends on (a) γ, a (c) α, β, a

1

∫ f ( x)

  1  f ( 0 ) + 4 f  2  + f (1)      (d) None of these (c)

(a) log 3 (c) 0

| x| dx, a < b is a x (b) a – b (d) | b | – | a |

0

(b) 16 (d) 64

127. The value of the integral

b



135. If f (x) = a + bx + cx2, then

1

(a) 15 (c) 63

(b) 10– 11 (d) 10– 9

134. The value of

 esin x  126. Let d F(x) =  , x > 0 dx  x 



sin x dx is less then 1 + x8

(a) 10– 10 (c) 10– 7

0

If



10

0

1

∫e

x2

(b) a cos kx + b sin kx (d) None of these

( x − α) dx = 0, then

0

(a) 1 < α < 2 (c) 0 < α < 1 141.

(b) α < 0 (d) α = 0

tan − 1 x dx equals x 0

1



(a)

π/ 2

x

∫ sin x

dx

0

π/ 2

(c)

∫ 0

sin x dx x

(b)

1 2

π/ 2

x

∫ sin x

dx

0

(d) None of these

0

1

(a)

(b)

143. The value of

π/ 2



151.

− x2

0

0

(a) – 1 (c) 1 + e–1

(b) 2 (d) None of these

144. If f (x) = ae + be + cx satisfies the conditions x

2x

log 4

∫ [ f ( x) − cx]

f (0) = – 1,  f ′ (log 2) = 31 and

0

dx = 39 , 2

152.

dx equals π π   sin  x −  ⋅ sin  x −  3 6  

(a) 4 log

3

(b) – 4 log

(c) 2 log

3

(d) None of these

∫ ( cox px − sin qx )

2

(a) – π (c) π 1

2

× (ax2 + bx + c) dx =

∫ (1 + cos

∫ (1 + cos 0

8

8

x)

3

dx, where p and q are integers,

x) (ax2 + bx + c) dx,

1

0

(a) 0 (c) 1/π

0

(a) no root in (0, 2) (b) atleast one root in (1, 2) (c) atleast one root in (0, 1) (d) two imaginary roots 1/ 2

146. The value of



cos x log

− 1/ 2

(a) 0 −1 (c) 2

k

(b) – 1/π (d) 2/π

π

∫ x f ( sin

154. If

0 π/ 2

∫ f ( sin

3

3

x + cos 2 x ) dx =

x + cos 2 x ) dx, then k =

0

1+ x dx is 1− x

(b)

(b) 0 (d) 2π

153. ∫ |sin 2π x | dx is equal to

then the quadractic equation ax2 + bx + c = 0 has

(a) π 2 (c) 2 π

1 2

(d) None of these

(b) π (d) None of these

155. The value of

π/2

∫ |sin x − cos x |

(a) π (c) 25

(b) 2 ( 2 − 1)

(c) 2 2 (d) 2 ( 2 + 1) x 156. If f (x) = A. 2 + B, where f ′ (1) = 2 and

(b) 0 (d) None of these

3

∫ f ( x)

π

148. The value of

∫ (1 − x ) 2

(c) 2π – π3

sin x cos2x dx is (b) π – (d)

x < 1  x, 149. If f (x) =  , then x − 1 , x ≥1   (a) 1 5 3

dx = 7, then

0

(a) A =

−π

(a) 0

dx is

0

(a) 0 sin10 x (6x9 – 25x7 + 4x3 – 2x) dx equals

− π/ 2

(c)

10 8 6 4 2 π ⋅ ⋅ ⋅ ⋅ ⋅ 11 9 7 5 3 2

−π

(b) b = – 6 (d) a = 3

145. Let a, b, c be non-zero real numbers such that



(b)

is equal to

(a) a = 5 (c) c = 2

π/ 2

x dx is

(d) 0

π

then

147.

10 8 6 4 2 ⋅ ⋅ ⋅ ⋅ 11 9 7 5 3

(c) 1

∫ (1 + e ) dx is 1

11

−1

1 3 1 (d) 8

1 2 1 (c) 4

(a)

∫ sin

150. The value of the integral

π3 3

(b) B =

7 – 2π3 2 2

∫ x f ( x) 2

dx equals 157.

1 log 2 7 3 ( log )

2

(log 2) 2 − 1

(c) A =

7 (log 2) 2 − 1 3 (log 2) 2 

(d) B =

1 log 2

0

4 (b) 3 5 (d) 2

243



3 + 2 3  cos x dx = k log   , then k is 3   3 + 4 sin x

4

∫ log [ x]

dx equals

1

(a) log 6 (c) log 2

(b) log 3 (d) None of these

Definite Integral and Area

142. If

π / /3

244

x

∫ cos t

158. If f (x) =

2

dt, x > 0, then f ′ (x) is equal to

(a)

π 1 + log 2 4 2

(b)

π 1 − log 2 4 2

1/ x

Objective Mathematics

π 1  (c) –  + log 2  4 2 

1 1 1 sin x (a) 2 cos 2 + x x 2 x 1 1 1 cos x (b) 2 cos 2 − x x 2 2 1 1 1 cos x (c) 2 cos 2 + x x 2 x (d) None of these 159. The area of the region bounded by the curves y = | x – 1 | and y = 3 – | x | is (a) 2 sq. units (b) 3 sq. units (c) 4 sq. units (d) 6 sq. units

 1  + 166. nlim →∞  2n − 12 equal to π (a) 4 π (c) 6 167. The value of

(d) None of these 1

4n − 2

2

1 + ⋅ ⋅ ⋅ +  is n 6n − 3  1

+

2

π 2 π (d) 3 (b)

 n 1 n n lim  2 2 + 2 + 2 + ⋅ ⋅ ⋅ +  is 2 2 + 1 + 2 + 3 2 n n n n 

n→∞

160. 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 (a) e – e − 5 2 2 2



1

0

f ( x) g ( x) dx , is

(b) e + e − 3 2 2 2

2 2 (c) e – e − 3 (d) e + e + 5 2 2 2 2  1 1 1  161. nlim + + ⋅⋅⋅   is equal to →∞ n + 1 n + 2 n + n 

(a) 3 log 2 (c) 2 log 2

(b) log 2 (d) None of these

1 n n 1 + + ⋅ ⋅ ⋅+  is equal to 162. lim  + 3 3 n→∞ n ( n + ) ( n + ) n 1 2 8  3 1 (b) (a) 8 4 1 (c) (d) None of these 8 163. The value of the nlim →∞  1  1 1 n2   is + + + ⋅⋅⋅ +  n 2 n2 − 1 n 2 − 22 n 2 − (n − 1) 2  2

π 4 (c) π

π π (b) 4 3 π (c) (d) None of these 2 164. The area bounded by the curve y = 2x – x2 and the straight line y = – x is given by (a) 9 (b) 43 2 6 35 (c) (d) None of these 6  n +1 n+2 n+3 1 + 2 + + ⋅ ⋅ ⋅ +  is equal to 165.  nlim →∞  2 2 n + 22 n 2 + 32 n n +1

(b)

2 r sec2 r is equal to ⋅ 2 n2 n

n



168. lim

n→∞

r =1

(a) tan 1 (c)

(b)

1 tan 1 2 1 n

169. nlim →∞

n

∑ sin

1 tan 1 3

(d) None of these 2k

r =1

rπ is equal to 2n 2k ! 2 (k !)

(a)

2k ! 2 (k !) 2

(b)

(c)

2k ! 2k (k !) 2

(d) None of these

2

(a)

π 2 (d) None of these

(a)

170. lim

n →∞

2k

k

(n !)1/ n is equal to n

(a) e

(b) 1 e

(c) e – 1

(d) None of these

1 2   n   171. nlim  1 + n   1 + n  ⋅ ⋅ ⋅  1 + n   →∞     

1/ n

(b) e 2 (d) 4 e

(a) 2 e e (c) 4

 12   22   n2  172. nlim  1 + 2   1 + 2  ⋅ ⋅ ⋅  1 + 2   →∞ n  n   n   (a) 2e (c) 2e

π+4 2 π−4 2

is equal to



(b) e



(d) e

π+4 2 π−4 2

1/ n

is equal to

(a)

1 p +1

(b)

1 (c) p+2

1 p

1

(d) None of these 181.

4n 2 − 1

+

1 4n 2 − 4

+ ⋅⋅⋅ +

 , 3n 2 + 2n − 1  1

Sn is equal to then nlim →∞ π 4 π (c) 3

π/2

∫ 0

(a)

1 2

dx is

−1 2

176. If f and g are continuous functions on [0, π] satisfying f (x) + f (π – x) = g (x) + g (π – x) = 1, then π

dx is equal to

0

(a) π π (c) 2

(b) 2π 3π (d) 2

2



f ( x) dx is equal

−2

n →∞

179.

(b) 5 +

9 log 2 4

(d) None of these

 12 22 1 + 3 +⋅ ⋅ ⋅ +  is equal to  3 3 3 2n  1 + n 2 + n

(a) 1 log 2 3

(b) 1 log 2 2

(c) log 2

(d) None of these

π/ 2

 ax + bx + c  log  2 ⋅ ( a + b ) |sin x | dx is equal to  ax − bx + c  −π/ 2



(c)

π log (a + b) 2

dx is

a +b (b) π log    2  (d)

1

∫e

x2

dx is

(a) greater than e (b) less than e (c) greater than 1 (d) less than 1 184. The value of α which satisfies

α

∫ cos x dx

= cos 2α,

0

α ∈ [0, 2π]  is π (a) 6 π (c) 2

(b)

π 3

(d) None of these 100

∫2

x −[ x]

dx, where [.] denotes

the greatest integer function, is (a) log 2 (c) 100 log 2

π a +b log   2  2 

(b) 50 log 2 (d) None of these

186. The value of the integral 3



−2



∫ cot

−1

  x − 1 −1  x + 1    + cot    is  x +1   x − 1 

(a)

5π 2

(b)

(c)

π 2

(d) None of these

187. Let I1 =

2 − tan 2 z



3π 2

x f ( x (3 −x)) dx and

sec2 z

2

(a) π log (a + b)

∫ |log 2 x |

0

to

178. lim

e/2

(b) 1 – e– 1 (d) None of these

185. The value of the integral

177. If f (x) = | 2x – 1 | + | x – 1 |, then

9 (a) 5 − log 2 4 9   (c) −  5 + log 2  4  

(b) 2 log (a + 1) (d) 2 log a

0

(d) 0

∫ [ f ( x) + g ( x)]

182. The value of the integral

183. The value of the integral

sin 8 x log (cot x) cos 2 x

(b)

(c) 1

(a) log (a + 1) (c) 3 log a

(a) 1 + e– 1 (c) e– 1 – 1

(b)

175. The value of the integral

xa − 1 ∫0 log x dx, where a > 0, is equal to 1

1/ 2 e

π 6 π (d) 2

(a)

(b) B = x – [x] (d) B = x + [x]

(a) A = [x] – 1 (c) A = [x] + 1

174. If Sn = 1  +  2n

x

1 180. If x > 0 and ∫ [ x] dx = [x]  A + B  , where [.] denotes 2  0 the greatest integer function, then

2 − tan 2 z

I2 =



f ( x (3 −x)) dx,

sec2 z

where f is a continuous function and z is any real number, then I1/I2 = (a) 3/2 (c) 1

(b) 1/2 (d) None of these

245

n

n→∞

1p + 2 p +⋅ ⋅ ⋅ + n p  , p > – 1, is equal to

p +1

Definite Integral and Area

1

173. lim

246

π/ 2

∫ log sin x

188. If

4

dx = a, then the value of

196. The value of the integral

0

3

Objective Mathematics



log (1 + x 2 ) ∫0 1 + x 2

dx in terms of a is

(a) 2a (c) a/2

(b) – 2a (d) a log 2

b + 2T

197. If I1 =

f ( x) dx is equal to



(a)

b

f ( x) dx



(b)

a+T



f ( x) dx

b

f ( x) dx

∫ f ( x)

(d)

a

dx

198. The value of the integral

(c) 3 [x] 199. The value of the integral

48

(b)

(c)

73

(d) None of these

(a) ae (c) a (1 – e)

66

192. If the variables x and y are cannected by the relation y 2 dz x= ∫ , then d y is proportional to 3 1 1 + 6z dx 2 (a) y (c) y3

2 x − 9 x + 12 x + 4 3

1

2

(a) 1 < I < 1 2 3 1 (c) < I < 1 4 194.

∫{ x} 3

tan θ

201.

, then

α

(d) None of these

(c) π

0

(d) None of these π 2

∫ (|sin x | + |cos x |) (b) 2a π (d) independent of a

(b) sin −1

2+ x dx is equal to 2−x

(a) π + 1 2 (c) π + 1

(b) 7 2

a

(a) aπ aπ (c) 2

∫ 0

is equal to

195. The value of the integral

2

α β α (d) cos −1 β

β−α

(a)

202.

(b) 2 tan θ (d) None of these

dx , β > α > 0, is equal to ( x − α) (β − x)



(b) 1 < I < 1 4 3

a+

dx, a ∈ Z+, is

(b) a (e + 1) (d) a (e – 1)

(a) tan2θ (c) 0

dx, where {x} denotes the fractional part of x,

(a) 5 2 3 (c) 2

[ x]

is

β

dx



ex

1 200. If f (x) is function satisfying f     + x2 f (x) = 0 for cot θ  x all x (x ≠ 0), then the value of the integral ∫ f ( x) dx

(b) y2 (d) None of these 2

193. If I =

a

∫e 0

1

(a)

dx is

(b) 1 [x] 2 (d) 2 [x]

(a) [x]

(b) 0, 4) (d) None of these

(2 x + 3) (3 x 2 + 4) dx cannot exceed

∫ ( x − [ x ]) 0

191. The value of the interval



where [x] and {x}

0

2[ x]

k

 5 3/ 2 1/ 2   x + 2 x − 3 x  dx < 0, lie in the interval ∫ k k 12 

2

∫{x} dx,

(d) I2 = a I1

a

(a) (0, 1) (c) (1, 4)

a

and I2 =

(a) I2 = (a – 1) I1 (b) I1 = (a – 1) I2 (c) I1 = a I2

190. All values of k for which the integral 1

 x 2  dx, − 14 x + 49  +  x 2 

denote, respectively, the integral and fractional parts of x and a is a positive integer, then

a+T

b+T

(c)

a

∫ [ x] dx 0

a + 2T b+T



2

where [.] denotes the greatest integer function, is 3 (a) 1 (b) 2 1 (c) (d) None of these 2

189. If f (x) is a periodic function with period T, then



∫ x

e

203. If dx is

 1

(b) π + 3 2 (d) None of these 1

∫  log x − ( log x ) 2



(a) a = e, b = – 2 (b) a = e, b = 2 (c) a = – e, b = 2 (d) None of these

2

  dx = a + b , then  log 2

(b)

b a dx

1

∫ (1 + x )

205. If

0

1 − x2

(a) a = π, b = (c) a =

= a , then b

π , b = – 2 2 (d) a = b 2 2

206. The value of

∫ ( ax

−2

3

(b) value of c (d) value of a and b

207. If d g (x) = f (x), where f (x) is continuous in [a, b], dx then b



dx = a + b log 2 , then 3

(a) a =

3 3 ,b= 4 2

(b) a =

3 3 ,b= 2 4

(c) a =

−3 3 ,b= 2 2

(d) a =

3 −3 ,b= 2 2

dx is equal to tan x − cot x

∫ 0

π 2

(a)

+ bx + c ) dx depends on the

(a) value of b (c) value of a

∫ x log 1 + 2 

π/ 2

213.

(b) a = π, b = 2 2

2

x



1

0

a b 1 (d) ab

(a) ab (c)

212. If

f ( x) ⋅  g (x) dx equals

(b)

(c) 0

4

214. The value of the integral

∫ |log

(a)

 e9  1 log 4  15  4 4 

(b)

(c)

 e9  1 log 4  15  2 4 

(d) None of these

215. The value of the integral

 415  1 log 4  9  4 e 

1

∫ 0

1 f   dx  x

1

(b)

1

209. If f (x) =

∫ x f ( x) 1

(d)

∫ f ( x)

1 2 log   n 3

2 (b) n log   3

(c)

1 3 log   n 2

3 (d) n log   2

216. The value of the integral 4

(a)

4 3 log 3

dx

+

1 3 12 log 4



42 x + 1 − 32 x − 1 dx is 12 x 0

1



(b)

0

x

sin t f ′ (x) = ∫1 t dt, then xlim →∞

(a) 1 (c) – 1

(a)

dx

0

1 1 (c) ∫ f   dx  x 0 x

dx is n + 1)

∫ x (x 1

1  1 2  n  f   + f   + ⋅ ⋅ ⋅ + f    is equal to n   n  n    n 

(b) 0 (d) ∞

x | dx is

4

1/ 4

31/ n

208. nlim →∞ (a)

π 4

(d) None of these

a

1 (a) [ f (b)]2 − [ f (a )]2  2 (b) f (b) – f (a) 1 (c) [ g (b)]2 − [ g (a )]2  2 (d) g (b) – g (a)

247

dx ∫0 [ax + (1 − x) b]2 is equal to 1

Definite Integral and Area

204.

4

(c)

4 3 log 3

4

+

3 log

3 4



4 4 3 log 3

3

+

4 log

3 4

(d) None of these 41 π / 4

217. The value of the integral



|cos x | dx is

0

1 (e1/n + e2/n + e3/n + ··· + en/n) is equal to 210. nlim →∞ n (a) e (b) 1 – e 211.

π/ 2

∫ sin θ cos θ

(b) e – 1 (d) None of these a 2 sin 2 θ + b 2 cos 2 θ dθ is equal to

0

(a) (c)

(a) 20 −

1 2

(b) 20 +

1 2

(c) 19 +

1 2

(d) 19 −

1 2

2

218. If

∫ f ( x)

dx = 2 and

−3

a 2 + ab + b 2 2 ( a + b)

(b)

a 2 + ab + b 2 ( a + b)

a 2 + ab + b 2 3 ( a + b)

(d) None of these

−3

of the integral

5

∫ [5 + f ( x ) ]

∫ f ( x)

2

dx is

5

(a) 2 (c) 4

(b) 3 (d) 5

dx = 9, then the value

248

y

219. If

∫e

− t2

0

x2

dt + ∫ sin t dt = 0, then dy dx 0 2

Objective Mathematics

(a) 2 x sin 2 x 2 e y

2

∫ cos

2n

x dx is equal to

π/ 2

∫ cos

2n

0 π/ 2

∫ cos

(c) 2k

π

(b) 2k ∫ cos 2 n x dx

x dx

0

2n

x  dx

(d) None of these

0

f (m, n) =

1k + 2k + 3k + ⋅ ⋅ ⋅ + n k  221. lim   is equal to nk +1 n→ ∞   1 (b) (a) 1 k +1 k 1 (c) (d) None of these k+2 3

x−4

∫x

n −1

(log x)m dx, then f (m, n) =

0

(a) m  f (m – 1, n) n (c) n f (m – 1, n) m 230.



∫x

n

(b) − m f (m – 1, n) n (d) None of these

e– x dx (n is a + ve integer) is equal to

(a) n ! (c) (n – 2)!

dx is

−3

(a) 0 (c) – 6

1

0

∫ | x − 4|

222. The value of the integral

dx is equal to

x

2 2 (a) π (b) π 4 8 2 π (c) (d) None of these 12 229. If m and n are positive integers and

0

(a) k

x

∫1 + e

2 2 (a) π (b) π 6 8 π2 (c) (d) None of these 12 1 1 1+ x 228. ∫ log dx is equal to 1− x 0 x

2





0

2

(d) − x sin 2 x 2 e y 220. The integral

is equal to

2

2 2 y (b) − 2 x sin x e

(c) x sin 2 x 2 e y

227.

(b) 6 (d) None of these

231. If In =

π/ 2

∫x

n

(b) (n – 1)! (d) None of these sin x dx, where n > 1, then

0

223. The value of the integral is π ( 2n) !

(a)

22 n ⋅ ( n!)

2



 n   2n + 1  !  2  

2



x dx (n even integer),

π (a) In + n (n – 1) In – 2 = n   2

n

π ⋅ n!

π (b) In + n (n – 1) In – 2 = n   2

n −1

 n   2n ⋅  !  2  

2



π (c) In – n (n – 1) In – 2 = n   2 (d) None of these

(d) None of these

π

3

224. The value of

n

0

(b)

π ⋅ n!

(c)

π

∫ cos

∫ |1 − x

2

232.

| dx is

225. If

(b) 14/3 (d) 1/3 π/ 2

π

∫ x f (sin x) dx 0

(a) π/4 (c) 0

I2 =



f (− a)

(a) –1 (c) 2



f (sin x) dx , then A is

0

233.

(b) π (d) 2π

226. If f (x) = f (a)

= A

ex , I1 = 1 + ex

4

2

dθ is equal to

0

−2

(a) 7/3 (c) 28/3

1 − cos θ

∫ sin θ (1 + cos θ )

n

∫ 0

f (a)



(a) 8 2 15 32 2 (c) 15 a xn

xg {x (1 − x)} dx

and

(b) –3 (d) 1

(d) None of these dx is equal to

(a)

1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅ (2n − 1) (– π an) 2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ 2n

(b)

1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅ (2n − 1) (π an) 2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ 2n

(c)

1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅ (2n − 1) (π an – 1) 2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ ⋅ 2n

f (− a)

I g{x(1 − x)} dx , then the value of 2 is I1

ax − x 2

(b) 64 2 15

(d) None of these

(a) 2 (c) 8

(b) 4 (d) None of these

236. The area bounded by the semi-circle y = its diameter y = 0 is

4 − x 2 and

(a) 2π (b) π π (d) None of these (c) 2 237. The area bounded by y = logex, x – axis and the ordinate x = e is given by (a) 4 (c) 1

(b) 1 2 (d) None of these

238. The area bounded by the curve | x | + y = 1 and axis of x is given by (a) 2 (c) 4

(b) 1 (d) None of these

239. The area of the bounded by y = cos x, y = 0, | x | = 1 is given by (a) sin 1 (c) 4 sin 1

(b) 2 sin 1 (d) None of these

240. The area of the region bounded by x-axis and the curves defined by y = tan x, − π ≤ x ≤ π and y = cot x, π ≤ x ≤ 3π is 3 3 6 2 (a) log 2 (c) 3 log 3

(b) 2 log 2 (d) None of these

241. If the ordinate x = a divides the area bounded by x-axis, part of the curve y = 1 + 8 and the ordinates x = 2, x2 x = 4, into two equal parts, then a is equal to (a)

(b) 2 2 2 (c) 3 2 (d) None of these 242. The area bounded by y = | x – 1|, y = 0 and | x | = 2 is  

246. The area bounded by y = cos x, y = 1 + x and x – axis is (a) 1

(b) 1 2

(c) 3 (d) None of these 2 247. The area bounded by the lines | x |+ | y | = 1, is (a) 1 (b) 2 (c) 4 (d) None of these 248. The area of the portion of the circle x2 + y2 = 64 which is exterior to the parabola y2 = 12x is (b) 8 (8π + 3 ) (a) 16 (8π + 3 ) 3 3 16 (c) (d) None of these (8π − 3 ) 3 249. The area bounded by the curve y = sin– 1x and the lines x = 0 , | y | = π is 2 (a) 2 (b) 4 (c) 8 (d) 16 250. The area of one the curvilinear triangles formed by the curves y = sin x, y = cos x and x-axis is (a) 2 + 2 (c) 2 + 2 2

(b) 2 − 2 (d) None of these

251. The total area enclosed by the lines y = | x |, y = 0 and | x | = 1 is (a) 2 (c) 1

(b) 4 (d) None of these

252. The area bounded by y = tan x, y = cot x, x-axis in 0 ≤ x ≤ π is 2 (a) 3 log 2 (b) log 2 (c) 2 log 2 (d) None of these

(a) 4 (b) 5 (c) 3 (d) None of these 243. The area bounded by the y = | sin x |, x-axis and the lines | x | = π is (a) 2 (c) 4

(b) f (x) = sin (3x + 4) + 3 (x – 1) cos (3x + 4) (c) f (x) = sin (3x + 4) – 3 (x – 1) cos (3x + 4) (d) None of these

(b) 1 (d) None of these

2 2 253. The area of the region bounded by the ellipse x + y a 2 b2 = 1 and the ordinates x = ± ae where b2 = a2 (1 – e2)

249

Definite Integral and Area

234. The area of the smaller part of the circle x2 + y2 = a2, 244. The smaller area enclosed by the circle x2 + y2 = a2 and the line x + y = a is a , is given by cut off by the line x = 2 2 2 (b) a (π + 2) (a) a (π – 2) 4 4 a2  π a2  π   (b) (a) 2  + 1  − 1 a 2 2 2 2   (2 – π) (d) None of these (c) 4 π (c) a 2  − 1 (d) None of these 245. If the area bounded by the curve y = f (x), x-axis and 2  the ordinates x = 1 and x = b is (b – 1) sin (3b + 4), then 235. The area bounded by the curve y = 2cos x and the x-axis from x = 0 to x = 2π is (a) f (x) = cos (3x + 4) + 3 (x – 1) sin (3x + 4)

250

Objective Mathematics

and e < 1, is given by

the coordinate axes in the first quadrant is

( ) (b) 2 ab ( e 1 − e − sin e ) (c) ab e 1 − e − sin e ( )

(a) 1 (c) 3

2 −1 (a) 2 ab e 1 − e + sin e 2

2

263. The area bounded by the lines y = 2, x = 1, x = a and the curve y = f (x), which cuts the last two lines above the first line for all a ≥ 1, is equal to

−1

−1

(d) None of these 254. The area bounded by y = e , y = e and x = 1 is 1 1 (a) e + + 2 (b) e + − 2 e e 1 (d) None of these (c) e − + 2 e 255. The area bounded by the curve x = 2 – y – y2 and y-axis is (b) 7 (a) 9 2 2 5 (c) (d) None of these 2 256. The area of the smaller part bounded by the semi-circle y = 4 − x 2 , y = x 3 and x-axis is x

π 3 4π (c) 3

(a)

(b)

(b) 2 (d) None of these

– x

2π 3

(d) None of these

257. The area of the region bounded by the curve y = and y = | x – 1 | is

5 − x2

(a) 5π + 1 (b) 3π + 1 4 2 4 2 5 π 1 3 (d) π − 1 (c) − 4 2 4 2 258. The area lying above x-axis, bounded by the circle x2 + y2 = 2ax and the parabola y2 = ax is π 2 π 2 (b)  −  a 2 (a)  +  a 2  4 3  4 3 π 1 2  (d) None of these (c)  +  a  4 3 259. The area bounded by the curves x2 + y2 = 25, 4y = | 4 – x2 | and x = 0, above x-axis is (a) 2 + 25 sin − 1 4 (b) 2 + 25 sin − 1 4 2 5 4 5 25 1 − 1 (c) 2 + (d) None of these sin 2 5 260. The area bounded by y = x3 – 4x and x – axis is

2 3/ 2 ( 2a ) − 3a + 3 − 2 2  ⋅ Then f (x) = 3 (a) 2  2x x ≥ 1 (c) 2

x , x ≥ 1

(b)

2x , x ≥ 1

(d) None of these

264. The area of the region in xy – plane enclosed by the curve a2y2 = x2 (a2 – x2) is 4 4 (a)  a (b)  a2 3 3 2 2 (c)  a (d)  a2 3 3 265. The area bounded by the normal at (1, 2) to the parabola y2 = 4x (y ≥ 0), the curve itself and the axis of the parabola is 2 4 (a) (b) 3 3 8 10 (c) (d) 3 3 266. The area bounded by the curve xy2 = 1 and the lines x = 1, x = 2 is (a) 4 ( 2 – 1)

(b) 4 ( 2 + 1)

(d) 2 ( 2 + 1). (c) 2 ( 2 – 1) 267. The area bounded by the curve y = 2x4 – x2, x-axis and the two ordinates corresponding to the minima of the function, is 3 5 (a) (b) 120 120 1 7 (c) (d) 20 120 268. The area contained between the curve y = x-axis is (a) πa2 (c) 3πa2

a2 and x + a2 2

(b) 2πa2 (d) None of these

269. The area bounded by y = [x], x-axis and the two ordinates x = 1 and x = 1.7 is 17 (a) (b) 1 10 17 7 (d) (c) (a) 4 (b) 8 5 10 (c) 16 (d) None of these 270. The area of the loop of the curve y2 = x (1 – x)2 is 261. The area bounded by the curve y = x (3 – x)2, the 7 (a) 8 (b) x-axis and the ordinates of the maximum and minimum 15 15 points of the curve is given by 4 (c) (d) None of these (a) 1 (b) 2 15 (c) 4 (d) None of these et + e − t and y = et − e − t is a point 262. The area bounded by the curve y = sin x + cos x and 271. For any real t, x = 2 2

(a) 1 t1 2 (c) 2 t1

(b) t1 (d) None of these

272. The area bounded by the curve y2 (a2 + x2) = x2 (a2 – x2) is (a) a2 (π – 2) (c) a2 (π – 1)

(b) a2 (π + 2) (d) a2 (π + 1)

273. The area of the region bounded by x = 1 , x = 2, y = 2 logex and y = 2x is (a)

4− 2 5 3 + log 2 + log 2 2 2

(b)

4− 2 5 3 − log 2 − log 2 2 2

(c)

4− 2 5 3 − log 2 + log 2 2 2

(d) None of these 2 274. The area included between the parabola y = x (a > 0) 4a 8a 3 is and the witch of Agnessi y = 2 2 x + 4a

4 (a) a 2  2 π −   3

4 (b) a 2  2 π +   3

4 (c) a 2  π +  (d) None of these  3 275. If the ordinate x = a divides the area bounded by x-axis, part of the curve y = 1 + 8 and the ordinates x = 2, x2 x = 4 into two equal parts, then a = (a)

2

(b) 2 2

(a) 2 (c) 4

(b) 1 (d) None of these

280. The area bounded by the curves x + y = 1 and x + y = 1 is 1 1 (a) (b) 3 6 1 (c) (d) None of these 2 281. The area of the region bounded by the curves y = x2 2 is and y = 1 + x2 1 2 (a) π + (b) π + 3 3 2 (c) π − (d) None of these 3 282. The area above x – axis, bounded by the line x = 4 and the curve y = f (x), where f (x) = x2, 0 ≤ x ≤ 1 and f (x) = x , x ≥ 1, is (a) 1 (b) 2 (c) 4 (d) 5 283. The area of the region bounded by the parabolas x = – 2y2 and x = 1 – 3y2 is 4 2 (a) (b) 3 3 1 (c) (d) None of these 3 284. The area common to the curves y=

9 − x 2 , x2 + y2 = 6x in y ≥ 0 is

(a) 3 π +

9 3 4

(b) 3 π −

9 3 4

(c) 3 2 (d) None of these 9 3 9 3 (c) 3 π − (d) 3 π + 276. The area bounded by the curves x2 + y2 = 4; x2 = – 2 y 2 2 and x = y is 2 285. The area of the loop the curve ay = x2 (a – x) is 1 1 (a) 2 π + (b) π − 2 2 3 3 (b) 4 a (a) 8 a 1 15 15 (c) − π + (d) None of these 2 3 2 a (c) (d) None of these 15 277. The area of the region bounded by the curves y = ex 286. The area of the region bounded by x2 + y2 – 2x ≤ 0, logex and y = log e x is x + y ≤ 1; y ≥ 0 is ex 2 2 π 1 π 1 (a) − (b) + (a) e − 5 (b) e + 5 8 2 8 2 2e 2e 2 2 π 1 π 1 (c) − (d) + (c) e + 5 (d) e − 5 4 2 4 2 4e 4e 278. The ratio of the areas into which the circle x2 + y2 = 287. The curve y = a  x + bx passes through the point (1, 2) and the area enclosed by the curve, the x-axis 64 is divided by the parabola y2 = 12x, is and the line x = 4 is 8 square units. The values of a 4π − 3 4π + 3 and b are (b) (a) 8π + 3 8π − 3 (a) a = 3, b = 1 (b) a = 3, b = – 1 4π − 3 (c) a = – 3, b = 1 (d) a = – 3, b = – 1 (d) None of these (c) 8π − 3

251

279. The area enclosed between the curves y = loge (x + e), 1 and the x – axis is x = loge y

Definite Integral and Area

on the hyperbola x2 – y2 = 1. The area bounded by the hyperbola and the lines joining its centre to the points corresponding to t1 and – t1 is

252

288. The area of the portion of the circle x2 + y2 = 1, which lies inside the parabola y2 = 1 – x, is

Objective Mathematics

(b) π + 2 (a) π − 2 2 3 2 3 π 4 π (c) + (d) − 4 2 3 2 3 289. The area bounded by the parabolas y2 = 5x + 6 and x2 = y is (a) 19 (b) 21 5 5 23 27 (c) (d) 5 5 290. The area bounded by the parabolas y2 = 4a (x + a) and y2 = – 4a (x – a) is 16 2 8 (a) a (b) a 2 3 3 4 2 (c) a (d) None of these 3 291. The area inside the parabola 5x2 – y = 0 but outside the parabola 2x2 – y + 9 = 0 is (a) 8 3 (c) 4  3

(b) 12 3 (d) None of these

2 292. The area bounded by the parabolas y = 4x2, y = x 9 and the line y = 2 is

(a) 20 2 3

(b) 10 2 3

(c) 40 2 3

(d) None of these

293. The area bounded by the circle x2 + y2 = 8, the parabola x2 = 2y and the line y = x in y ≥ 0 is

296. The area bounded by the curves y = sin x and y = cos x between two consecutive points of their intersection is (a) 2 (c) 3 2

(b) 2 2 (d) None of these

297. The area of the region formed by x2 + y2 – 6x – 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is (a)

π 3 +1 − 6 8

(b)

π 3 −1 + 6 8

π 3 −1 (d) None of these − 6 8 298. If An be the area bounded by the curve y = (tan x)n and the lines x = 0, y = 0 and x = π/4, then, for n > 2, (c)

(a) An + An – 2 = (c) An >

1 n −1

1 2n + 2

(b) An + An – 2 < (d) An
0 be a fixed number. Suppose f is a continuous function such that for all x ∈ R, f (x + T) = f (x). If

∫ f ( x) dx T

I=

0

3 I 2 (c) 3I

(a)

316.



∫e

x 2

0

3 + 3T

3

f ( 2 x ) dx , is

(b) 2I (d) 6I

 x π sin  +  dx = 2 4 (b) eπ (d) 2 2

(a) 2π (c) 0

(1 – t2) dt are



then the value of

317. Suppose that f ′′ (x) is continuous for all x and 1 (b) x = – 2 f (0) = f ′ (1) = 1. If ∫ t f '' (t ) dt = 0, then the value 1 (c) x = 0 (d) x = 0 2 of f (1) is 309. The area of the region bounded by the curve y = x – x2, (a) 3 (b) 2 x-axis between x = 0 and x = 1 is 1 (c) 4 (d) None of these 5 1 2 (a) (b) 6 2 a dx 1 1 = π , then a equals 318. If ∫ (c) (d) 1 + 4x2 8 0 3 6 310. The area between the curve y = 1 – | x | and x-axis is (b) 1 (a) π 1 2 2 (a) (b) 2 3 π (c) (d) 1 1 4 (d) 1 (c) e2 2 log x 319. The value of ∫e −1 xe dx is 1 1 1 311. If ∫ f ( x) dx = 1, ∫ x f ( x) dx = a, ∫ x 2 f ( x) dx = a2, 5 3 0 0 0 (b) (a) 1 2 2 2 then ∫ ( a − x ) f ( x) dx equals (c) 3 (d) 5 1

(a) x = 1, – 1

0

π

320.

(a) 4a2 (b) 0 (c) 2a2 (d) None of the above 312. The integral



1/2

−1/2

∫ (1 − x ) 2

π − 2 π3 2 π3 (c) π − 3

321. If

b

∫ (b − 4x )

(b) 0

(c) 1

(d) 2log 1 2

(a) 3 (c) 2

esin x dx is

322. If I1 =

π/ 2

∫ cos x ⋅ 0

(a) 1 (c) 0

(b) 2π – π3

(a)

1 (a) – 2

313. The value of

sin x cos2x dx is equal to

−π

  1 + x  [ x ] + log   dx equals  1 − x   

(d) 0 dx ≥ 6 – 5b, b > 1, then b equals

1

(b) 4 (d) 1 3π

∫ f ( cos x ) 2

0

(b) e – 1 (d) – 1

253

2

314. The value of

(a) I1 = 5I2 (c) I1 = 3I2

dx and I2 =

π

∫ f ( cos x ) 2

0

(b) I1 = I2 (d) None of these

dx, then

Definite Integral and Area

305. The area of the region bounded by y = | [x – 2] |, the x-axis and the lines x = – 1 and x = 2 is

254

323. Area of the region bounded by y = x-axis lying in Ist quadrant is

Objective Mathematics

(a) 18 sq. units (c) 36 sq. units

x , x = 2y + 3 &

log 5

332.

ex

0

(b) 9 sq. units (d) 27/4 sq. units

324. The equation of the common tangent to the curves y2 = 8x and xy = –1 is (b) y = 2x + 1 (d) y = x + 2

(a) 3y = 9x + 2 (c) 2y = x + 8



e x −1 dx = e +3 x

(a) 4 + π (c) 4 – π 333. If u10 =

(b) 2 + π (d) 3 + 2π

π/ 2

∫x

10

sin x dx, then the value of u10 + 90u8 is

0

π (a) 9   2

π (b) 10   2

π (c)   2

π (d) 9   2

9

 n n n n  + 2 + 2 + ⋅⋅⋅ + 2 325. lim  2  2 2 2 n→∞ + + + + 1 n 2 n 3 n n n2   is equal to

9

9

(a) log 2

8

(b) 0 334. Let f(x) be a continuous function in R such that f(x) (c) 1 (d) π 3 3 4 + f(y) = f(x + y). If ∫ f ( x) dx = k , then ∫−3 f ( x) dx is 0 3 326. The area enclosed between the curves y = x and equal to y = x is, (in square units) (a)  2k (b)  0 k (c)   (d)  –2k (a) 5 (b) 5 2 3 4 (c) 5 12

335.

(d) 12 5 x3

dt 327. The derivative of F (x) = ∫ (x > 0) is log t 2 x (a)

1 3log x

(b)



1 1 − 3 log x 2 log x 3x log x

(d)

ecos x ⋅ sin x, | x | ≤ 2 328. If f (x) =  , then 2, otherwise 3

f ( x) dx is equal to

−2

(a) 0 (c) 2

(b) 1 (d) 3

329. nlim   1 + 2 2 + 3 3 +⋅ ⋅ ⋅ + n n is equal to →∞ n5 / 2 1

(a)

∫x

(b) 5 2 (d) 1

x dx

0

(c) 0 π

330. The value of

3 2 ∫ sin x cos x dx is

−π 4 (a) π 2 (c) 0

4 (b) π 4 (d) None of these x

1 − x  331. The function F (x) = ∫ log   dx is 1 + x  0 (a) an odd function (c) an even function

π/ 2

−π / 2

sin 2 x cos 2 x (sin x + cos x) dx is equal to

(a)   2 15 (c)   6 15 336. The value of

(b)   4 15 (d)   8 15



π/ 2 0

log | tan x + cot x | dx is

2

(c) (log x)– 1· x (x – 1)





(b) a periodic function (d) None of these

(a)  π log 2 (b)  –π log 2 π (c)   log 2 (d)   − π log 2 2 2 337. The ratio of the areas between the curves y = cos x and y = cos 2x and x-axis from x = 0 to x = π is 3 (a)  1 : 2 (b)  2 : 1 (d)  1 : 4 (c)   3 :1 338. The value of the integral (a)  π (c)   π 2



b

a

x dx is x + a+b− x

(b)   1 (b − a ) 2 (d)  b – a

339. The area bounded by y = log x, x-axis and ordinates x = 1, x = 2 is (a)   1 (log 2)2 2 (c)  log 4/e 340. The value of the integral (a)   3 2 (c)  1

(b)  log 2/e (d)  log 4



6

3

x dx is 9− x + x

(b)  2 (d)   1 2

0

350.

| sin 3 θ | d θ is (b)   3 8



1000

0

e x −[ x ] dx is

255

π



1000 (b)   e − 1 e −1 (c)   4 (d)  π e −1 (c)  1000 (e – 1) (d)   3 1000 342. The area bounded by the curves y = log x, y = log | x |, 1 sin x 1 cos x 351. Let I = ∫ y = |log x | and y = | log | x || is dx and J = ∫ dx 0 0 x x (a)  4 sq unit (b)  6 sq unit Then, which one of the following is true ? (c)  10 sq unit (d)  none of these 10 2 2 (b)   I < and J < 2 (a)   I > and J > 2 343. ∫ | x( x − 1) ( x − 2) | dx is equal to 0 3 3 (a)  160.05 (b)  1600.5 2 2 (c)   I < and J > 2 (d)   I > and J < 2 (c)  16.005 (d)  none of these 3 3 344. 30 |x3 + x2 + 3x| dx is equal to 3 52. The area of the plane region bounded by the curves ∫ 2 2 x + 2y = 0 and x + 3y = 1 is equal to (a)   171 (b)   171 2 4 (b)   1 sq unit (a)   5 sq unit 3 3 170 170 (c)   (d)   2 4 3 (d)   4 sq unit (c)   sq unit 3 3 3 x +1 345. The value of ∫ 2 dx is 353. The area bounded by the curve y = 2x – x2 and the line 2 x ( x − 1) y = –x is given by (b)   log 16 − 1 (a)   log 16 + 1 (a)  1/2 (b)  9/3 9 6 9 6 (c)  9/2 (d)  none of these.

(a)  0

(c)   2log 2 − 1 6 346.

x) dx + ∫

5π / 4

π/ 4



π/ 4

(sin x − cos x) dx + ∫

(cos x − sin x) dx + ∫

π/ 4



5π / 4

347. The value of

(sin x − cos x) dx + ∫

π/ 4





1

348. The value of the integral

(b)   1 (3π + 4) 6 1 (d)   (3π − 4) 6



a

0

1 π 1 − 16a 3  4 3 

(b)  

1 3 π 1 a − 16  4 3 

(d)  

x 4 dx is (a + x 2 )4 2

1  π 1 + 16a 3  4 3  1 3 π 1 a + 16  4 3 

349. The area bounded by the curves y2 = 4a2(x – 1) and lines x = 1 and y = 4a is (a)  4a2 sq unit (c)   16a 3

(b)   16a sq unit 3

2

sq unit

(d)  None of these

is ..

(cos x − sin x) dx (a)  log 2

x4 + 1 dx is x2 + 1

(a)   1 (3 − 4π) 6 1 (c)   (3 + 4π) 6

∫ log [ x] dx 2

(b)   2 2 − 2 (d)   4 2 − 2 0

(c)  

354.

(cos x − sin x) dxis equal to

(a)   2 − 2 (c)   3 2 − 2

(a)  

4

(d)   log 4 − 1 3 6 π/ 4

0

(a)  e1000 – 1

(b)  log 3 (d)  none of these.

(c)  log 5 355. If I10 =

π/ 2

∫x

10

sin x dx. Then the value of I10 + 90 I8 is

0

(a)  10(π/2)6

(b)  10(π/2)9

(c)  10(π/2)7

(d)  none of these.

356. If I = π/ 2



π/ 2



cos(sin x)dx, j =

0

π/ 2



sin(cos x) dx, K =

0

cos x dx then

0

(a)  I > J > K (c)  K > J > I 357. If g(x) =



x

0

(b)  J > K > I (d)  I > K > J

cos 4 t dt, then g(x + π) is equal to

(a)  g(x) + g(π)

(b)  g(x) – g(π)

(c)  g(x) · g(π)

(d)  

358. Let Sn =

n

∑ k =1

g ( x) g (π)

n and Tn = n + kn + k 2 2

n −1

∑n k =0

n = 1, 2, 3,…  . Then, π 3 3 π (c)   Tn < 3 3

(a)   S n
3 3 (b)   S n >

2

n for + kn + k 2

Definite Integral and Area

341. The value of

256

solutions

Objective Mathematics

π

∫e

1. (a) Let I=

cos 2 x

 Put 1 + tan 3 x = z    1  ⇒  tan 2 x sec 2 x dx = dz   3  2  −1  1 1  1 =    = –  − 1 = ⋅ 32  3 z 1 6 

⋅ cos3  (2n + 1) x dx

0

π

=

∫e

cos 2 ( π − x )

∫e

cos 2 x

0 π

=

⋅ cos3 (2n + 1) (π – x) dx

⋅ cos3 ( 2n + 1) π − ( 2n +1) x  dx

7. (c) Let I =

0

π

∫e

=–

cos 2 x

=

0

π/ 2

2. (c) Let I =

∫ 0

π/ 2



=2

0

0

 x cos5    · sin x dx 2

x  x cos   sin dx = 2 2 2

∴ I =

1/ 2



z (−2) dz

0

 − 4  1  4 1  − 1 = 1 −  =  ·   7  2  7 8 2    7

z  = – 4.    7 1 ∞

ln 2

2 3. (a) ∫  x  dx = 0 e 

∫ 1dx + 0

4. (b)

∫ f ( x ) dx

−1

2

∫ ( 3 − f ( x ) ) dx −4

2

−2

= 4 + 25 = 29.

−1

−1

1 1 101  ∴ I = –  ∫ cos ec  t −  dt t  1/2 t

∫ 0

π/ 4

=

∫ 0

3

tan 2 x ⋅ sec 2 x

(1 + tan x ) 3

2

x + cos3 x )

2

2

 dx. =

1

dx

2

dz

∫ 3z

0

2

−1 z dz 2 r2



= 1 (r3 ) ⋅ 3

π/2

∫ sin xdx + ∫ cos xdx = ( − cos x ) π/4

π

∫e

x



π/4 0

2

1 − sin x dx 1 − cos x

x

=

∫e

x

π/2 π

∫e

π/2

sin x ⋅ cos x

(sin

0

π/2

=

⇒ I = – I ⇒ 2I = 0 ⇒ I = 0. 6. (a) Let I =



π

2

x cos 4 x dx

0

x x  1 − 2 sin 2 cos 2  = ∫e   dx x π/2   2 sin 2   2

Let 1 = t, – 1 dx = dt x2 x

4

0

r2

10. (c) Let I =

π

1 1 101  5. (c) I = ∫ cos ec  x −  dx   x x  1/2 2

2

∫ sin

2 2 ∫ x r − x dx =

 z 3/ 2  z dz =    3 0

r2

2

−4

π/ 2

r

1  1  =  =2– + 1 + 1 − 2  2 

2

∫ f ( − x ) dx = ∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx

π/ 4

1 2

0

= 7,

π 2

−1   2 2  Put r − x = z ⇒ x dx = 2 dz   

π/4 −4

−4 1

I=

∫ xy dx =



9. (d) A =

f ( x ) dx = 7 + 18 = 25





r

0

= ln 2

= 4 and

2 3 ⋅ 1⋅ 3 ⋅ 1 π = 3π ⋅ ⋅ 8⋅ 6⋅ 4⋅ 2 2 256

8. (b) Let I =

=

0

−4

π

π sin 4 x cos 4 x dx 2 ∫0 = 2·

= π · 

ln 2

=

x cos 4 x dx – I



2 2 dx + ∫  x  dx x   ln 2  e 

∫  e 0

4



6

x x    Putting cos 2 = z ⇒ sin 2 dx = − 2 dz    1/ 2

π

∫ sin



6

7

∫ (π − x) sin4 (π – x)cos4 (π – x) dx 0

∴ 2 I = 0 ⇒ I = 0.

x cos 4 x ) dx

4

0

π

cos3 (2n + 1) x dx = – I.

π

∫ x ( sin

x

x 1 2 x − cot  dx  cos ec 2 2 2  x d  x   − cot 2 + dx  − cot 2   dx   π

x  =  − e x cot  2 π/2  π π =  − e π cot + e π / 2 cot  = eπ/2. 2 4 

+ ( sin x )π/4 π/2

∫ ln xdx = 4 ·  εlim →0

Required area = 4

0

 x ( ln x − 1)  ε = 4 ·  εlim →0 1

1

∫ ln xdx

1

∫ e ( x − α ) dx

( x − α)

must be +ve

257

1

and –ve both for x ∈ (0, 1) i.e., ex (x – α) = 0 for one x∈ (0, 1)

Definite Integral and Area

11. (c) y = ln x, y = ln | x |, y = | ln x |, y = | ln | x | |

14. (a) 

x2

= 0, ∴ e x

2

0

ε

= 4 sq. units

∴ α ∈ (0, 1). 15. (b) Let I =

π/ 2

∫ log sin x

dx =

0

=

π/ 2

∫ log cos x

π/ 2

π



∫ log sin  2 − x 

dx

0

dx.

0

π/ 2

∫ ( log cos x + log cos x )

⇒ 2 I =

dx

0

π/ 2

∫ log ( sin x cos x )

=

dx

0

π/ 2

0

π/ 2

∫ ( log sin 2 x − log 2 )

= 12. (b) Let I =



∫e 0

x

 sin 2 x   dx 2 

∫ log 

=

dx

0

π π sin  +  dx 4 2

=

π/ 2

∫ log sin 2 x

π/ 2

∫ dx

dx – log 2

0



 π x  = sin  +  e x  −   4 2  0



π

1

x

∫ 2 cos  4 + 2  ⋅ e

x

dx

0



∫ 0

 1 π x sin  +  e x dx    2 4 2 



π/6

π/3

=



π/6

π/3

=



π/6

cos x dx sin x + cos x π/3



π/6

∴ I =

sin x dx sin x + cos x

π π  sin  + − x  6 3  dx π π  π π  sin  + − x  + cos  + − x  6 3  6 3 

∴ 2I =

π ⋅ 12

π/ 2 0

0

sin x + cos x π π/3 dx = [ x ]π / 6 = ⋅ 6 sin x + cos x

π/ 2

π

∫ log sin z dz − 2 log 2 0

∵log sin z = log sin ( π − z )   π  π/ 2   log sin z dz = log sin z dz ∴ 2 ∫  ∫  0 0   π/ 2

π

∫ log sin x dx − 2 log 2

=

⇒ I = − 2 (e 2 π +1) ⋅ 5 13. (c) Let I =

1 2 2

=

1  −1 2 π 1  −1 2 π 1  1 =  I e − e −  –  −  2 2 2 2 2 4

π/3

∫ log sin z dz − log 2 ⋅ [ x]

[Putting 2x = z in the first integral, ∴ 2 dx = dz]

  π π = sin  π +  e 2 π − sin  4 4    2π 1   π x x  cos  +  e  + 2  4 2 0

0

π

1 = 2

0

∴ I = 16. (a)



=I–

π log 2 2

−π log 2 . 2

t2

xf ( x) dx = 2 t 5 By Leibnitz’s rule, 5 2 2t [t2 f (t2)] = 5t 4 or f (t2) = t 5 2 2. 4 Putting t = , f   = 5 5 25   0

17. (b) l =



1

0

1− x dx 1+ x

Put x = cos 2θ, dx = – 2 sin 2θ dθ 0

∴ l = − 2 ∫π sin 2θ tan θ d θ = 2 ∫

π 4

0

4

2 tan 2 θ dθ 1 + tan 2 θ

258

π  π  = 4  ∫ 4 θ d θ − ∫ 4 cos 2 θ d θ 0 0  

=

Objective Mathematics

sin 2θ  π  = 4   − 2 θ + 2  0 4  π 1 = π − 2  +   4 2

∫ x log sin x



21. (b) Let I =

dx

=

∫ ( π − x ) log sin ( π − x ) ∫ ( π − x ) log sin x

dx

=

1 2x 1 dx 1 dx dx + − ∫ 2 2 ∫ ∫ 4 0 x +1 2 0 x +1 2 0 1 + x



dx



π

∫ log sin x

dx = 2π

0

π/ 2

∫ log sin x

dx

0

−π = 2π· log 2 [see Q. No. 15] 2 − π2 log 2 ⋅ ∴ I = 2 a

dx

∫a+

=

π/ 2

=

a2 − x2

0

∫ 0

a cos θ d θ a + a cos θ

[Putting x = a sin θ = ⇒ dx = a cos θ dθ] π/ 2

∫ 0

π/ 2

cos θ dθ = 1 + cos θ

π/ 2

π/ 2

1 = ∫ 1 dθ − 2 0

∫ 0

∫ 0

  1 1 −  dθ  1 + cos θ 

θ sec dθ 2

 1   1+ 2    1 1 1 + x2  1 π x  − log = log   + ⋅ 2 4 4  (1 + x) 2  x =0 2 2 1     + 1   x  x = ∞  = 0 – 0 + π = π. 4 8 22. (c) Let I = =

0

⇒ 2 I =

0 π

=

π/ 2



dx

0

 sin x cos x  log  +  dx  cos x sin x 

π/ 2

=



2



∫ log  sin 2 x 

dx

π/ 2

π/ 2

0

0

∫ 1 dx −



∫ log sin

x) dx =

0

dx = 2· 2

log sin 2 x dx

π π log 2 – 1 ⋅ log sin z dz = ∫ 2 2 0 1     Putting 2 x = z ⇒ dx = 2 dz   

2

x dx

π/ 2

∫ log sin x

dx

0

[see Q. No. 15]



∫ [ 2 sin x] dx  

23. (a) I =

0

–2 ≤ 2 sin x ≤ 2

2 sin x = 1 ⇒ sin x = 1 ⇒ x = π , 5π 2 6 6



2 sin x = –1 ⇒ sin x = – 1 ⇒ x = 7 π , 11π 2 6 6

0

= log 2

π

∫ log sin x

2

−π  = 4·  log 2    2  ∴ I = – π log 2.

0

=

∫ log (1 − cos

dx

π

0

π  π π =  − 1 · − tan 2 4 2 

dx

0

π

= 2

∫ log ( tan x + cot x )

π

∫ log (1 + cos x) (1 − cos x) 0

π/ 2

dx

∫ log [1 + cos (π − x)] dx =  ∫ log (1 − cos x )

2

θ  = θ − tan  2 0 

20. (c) Let I =

π

∫ log (1 + cos x) 0

π

π/ 2

=



1  1 + x2  ∞ 1 =  log    +  tan − 1 x  0  (1 + x )2   2  4 0 

∫ log sin x dx – I

⇒ 2 I = π







π

0

19. (b) Let I =

(by partial fraction)

1 1 1  =  log (1 + x 2 ) − log (1 + x ) + tan − 1 x  4 2 2  0

0



dx

∞ 1  x +1 1  −   dx 2 ∫0  x 2 +1 1 + x  ∞

0 π



2

=

0

=

x

∫ (1 + x ) (1 + x ) 0

π



0

= π log 2.

π π = π − − 1 = − 1 . 2 2 18. (a) Let I =

∫ log sin z dz

π π = log 2 + log 2 [see Q. No. 15] 2 2

π 4

π

π/ 2

π 1 log 2 – ⋅ 2 2 2

=

+

π/6

π/2

5 π/6

0

π/6

π/2

∫ [ 2 sin x] dx + ∫ [ 2 sin x] dx + ∫ [ 2 sin x] dx π

7 π/6

11π/6

5 π/6

π

7 π/6

∫ [ 2 sin x] dx + ∫ [ 2 sin x] dx + ∫ [ 2 sin x] dx

11π / 6 π/6

=

π/2

5 π/6

π/6

π/2

7 π/6







11π/6

dx − 2

π



dx −

7 π/6



dx

=

11π/6



2π π 4π π = – π 2π = –π. = − − − − 3 6 3 6 3 3

=

24. (d) y′ = 16x – 5x4 = 0 ⇒ x = 0, x3 = 16 5 c

2

1

− x 5 ) dx = 16 3

∫ 8 x dx − ∫ x dx 2

5

−1

= 16 3

[Putting x = tan θ ⇒ dx = sec2θ dθ] π/ 4

=

∫ log (1 + tan θ )

=



=

  dθ

=

∫ 0

0

2   log   dθ  1 + tan θ 

π/ 4

∫ 1 dθ

= log 2

0

–I

0

π/ 2

∫ log ( tan θ + cot θ )

=

π

sin 2 kx ∫0 sin x dx =

sin 2k (π − x) ∫0 sin (π − x) dx

π

sin (2k π − 2kx) dx = – sin x 0



π

sin 2kx dx = – I 0 sin x



∴ 2 I = 0 ⇒ I = 0.

= π log 2 [see Q. No 20] π

=

x 3 cos 4 x sin 2 x dx 2 − 3πx + 3 x 2

∫π ∫

(π − x)

cos 4 ( π − x ) sin 2 ( π − x )

π − 3π ( π − x ) + 3 ( π − x ) 2

0

=  ∫

3



π/ 2

∫ cos 0

π/ 2

=

∫ 0

2

cos 2 θ dθ θ + 4 sin 2 θ

cos θ dθ = 1 + 3 sin 2 θ

= −1 3

2

π/ 2

∫ 0

π/ 2

∫ 0

1 + cos 2θ dθ 2 + 3 (1 − cos 2θ )

( 5 − 3 cos 2θ ) − 8 5 − 3 cos 2θ

dx

π2 − 3πx + 3 x 2

0

π





− 3π2 x + 3πx 2 ) cos 4 x sin 2 x

3

π2 − 3πx + 3 x 2 π

⇒ 2 I = π





2



= π·2

π/ 2

∫ cos

4

2

− 3πx + 3 x 2 )

x sin 2 x dx = 2π·

0

∴ I = π /32. π

x 2 sin x

∫ (2 x − π) (1 + cos

2

π

=

x)

dx

(π − x) 2 sin (π − x) dx 2 (π − x))

∫ (2π − 2 x − π) (1 + cos 0

π

(π2 − 2πx + x 2 ) sin x dx 2 x) 0

=

∫ (π − 2 x) (1 + cos

=

∫ (π − 2 x) (1 + cos 0

(π2 − 2π x) sin x dx – I 2 x)

dx

3 ⋅ 1⋅ 1 π ⋅ 6⋅ 4⋅ 2 2

2

30. (a) Let I =

dx

dx – I

− 3πx + 3 x 2 ) cos 4 x sin 2 x

0

π



2

− 3π2 x + 3πx 2 − x 3 ) cos 4 x sin 2 x

3

0

27. (c) Let I =



0

0

π

2

sec 2 θ dθ 2 θ [Putting x = tan θ ⇒ dx = sec2θ dθ]



=

⇒ 2I = π log 2   ∴  I = π log 2. 4 8

=

1  dx



0

26. (c) Let  I =

∞ −π 2 +  tan − 1 ( 2 z )  0 6 3

=

2

∫ log ( tan θ + cot θ ) sec

π



1 z2 +   2

π/ 2

=

0

π/ 4

dz



∫ log  x + x  1 + x

0

 1 − tan θ  1 +  dθ =  1 + tan θ 





π

∫ log 1+ tan  4 − θ 

π/ 4

0

sec 2 θ d θ θ) − 3 (1 − tan 2 θ)

2

π −π 2 π + ⋅ = . 6 6 3 2

29. (d) Let I =



π

∫ 5 (1 + tan

−π 1 + 6 3

0

π/ 4

π/ 2

0

16 16    ∴   c = –1. = 3 3 1 π/ 4 log (1 + tan θ ) log (1 + x ) dx = ∫ sec2θ dθ 25. (a) Let I = ∫ 2 + 1 x (1 + tan 2 θ) 0 0 

0

[Putting tan θ = z ⇒ sec2θ dθ = dz]

1

−1



∫ 5 − 3 cos 2θ

−π 8 dz + ∫ 2 6 3 0 8z + 2

28. (d) Let I =

1

If c = –1,

0

π/ 2

8 3





= (5π/6 – π/6) – (7π/6 – π) – 2(11π/6 – 7π/6) – (2π – 11π/6)

∫ (8 x

∫ 1⋅ dθ +

= −π + 8 6 3

∫ 0 ⋅ dx + ∫ 1⋅ dx + ∫ 1⋅ dx + 0 0

π/ 2

259

= −1 3

∫ [ 2 sin x] dx

Definite Integral and Area





260

⇒ 2 I = π

Objective Mathematics



π

sin x ∫0 1 + cos2 x dx = – π

= 2 (e – e + 1) = 2

−1

dt ∫1 1 + t 2

[Putting cos x = t ⇒ sin x dx = – dt]

2 1 π π = π  tan − 1 t  = π  +  = π −1 2 4 4 π2 ∴ I = . 4 31. (b) x ∈ (–8, 8)   y = 2

x ∈ (–8 2 , –8] ∪ [8, 8 2 )   y = 3



34. (b)

1 ×2×2 2 = 6 – 2 = 4 sq. units Required area = 2 × 3 –

2a

∫ f ( x) dx 0

0



=λ–

=

a

2a

0

a

∫ f ( x) dx + ∫ f ( x) dx

f (2a − t ) dt

a

[Put x = 2a – t ⇒ dx = –dt]

 0

∫ f (2a − x) dt

=l+

a

= λ + µ. π/ 4

sin x ⋅ cos x dx 2 x + sin 4 x

∫ cos

35. (b) Let I =

0

1 2 32. (d)

π

x 2 cos x

∫ (1 + sin x)

π

=2

 −1   x   + 2 ∫ dx sin sin x x 1 + 1 +  0 0

= – π2 + 2

π

π

x

∫ 1 + sin x x

∫ 1 + sin x

Let I =

dx =

0

(π − x) ∫0 1 + sin x dx = π

⇒ 2 I = π

π

(π − x)

∫ 1 + sin (π − x) 1

∫ 1 + sin x

dx

dx – I.

0

1 − sin x dx = π 2 0 cos x



π

∫ ( sec

= [ π (tan x − sec) ]0

2

x − sec x tan x ) dx

0

x

− ex )

1 0

∫z 1

dz +3

2

0

1 π  1  tan − 1  .  = 3 6 3  3 2

e4

x ∫ e dx = a, let I =

36. (d) Given



2

Put ln (x) = t2 ⇒ 1 dx = 2tdt x

∴  I =

∫e

t2

( ) − ∫ e dt = 2e

⋅ 2t 2 dt = t et

∫ 0

(x +

2

1 + x 2 dz; dx 1 + x2



)

n

2

t2

4

– e – a.

1

 x  dx = dz ⇒ 1 +  1 + x2  

37. (b) Put x +  1 + x 2 = z ⇒ z dx =

2

1

1



ln ( x) dx

e

1



( xe

0

 −1  z  tan − 1  =    3  1  3

33. (d) y = x e| x | = 1, y = 0 x ∫ xe dx = 2

dx

2

−1    Putting cos 2 x = z ⇒ sin 2 x dx = 2 dz   

2

= π [(tan π – sec π) – (tan 0 – sec 0)] = 2π ⇒ I = π. π x 2 cos x dx So, by (1) ∫ 2 0 (1 + sin x )  = – π2 + 2π = π (2 – π).

1

sin 2 x dx = – cos 2 2 x + 3



0

π

Required area = 2

dx

2

sin 2 x



=

π



∫ 0

...(1)

0

π

=

 1 − cos 2 x  +  2  

∫ 2 (1 + cos 2 x) + (1 − cos 2 x)

π/ 4

=2

dx  π

sin 2 x

0

0

π

0

π/ 4

0

 2 = x 

∫ 1 + cos 2 x 2

dx

2

π/ 4

=∫ 1

1 1 z+  1 + x 2 dz ∞ 2  z  dz =∫ n +1 zn ⋅ z z 1





1 2

=



∫ 1

  1 1 2 ,∴2 1 + x = z +  z z 2

z2 + 1 1 dz = zn + 2 2



∫ (z

3 −1  3 − 1 = sin − 1 z  1 −2 3 = 2 sin– 1  .    2    2

42. (a) x sin πx = 0, gives x = 0 or πx = nπ, n integer.

+ z − n − 2 ) dz

−n

∴ x = 0; x = n, n = 0, ± 1, ± 2, ··· ; out of

1

which 0, 1 ∈  −1, 





1  z1 − n z − n −1  n − ⋅   = 2 2 1 − n (n + 1) 1 n −1

=

et dt ∫0 t + 1 = a,  Let I = 1

38. (b) Given

3/ 2

b

e − t dt ∫b − 1 t − b − 1

dt = dZ =

1



e

0

1

=

dZ = Z +1

eZ − b

∫ Z + 1 ⋅ dZ 0

0

1



e

− (1 − Z ) − b + 1

0

dZ 1− Z − 2

=

e Z dZ = –e–b · a 0 Z +1

=

−1

0

1

1

3/ 2

−1

1

∫ x sin πx dx − ∫ x sin πx

1 1 1  1 −1 1 =  −  −  2 −  = 1 +  . π π  π π π π

dx = π ⇒ I = π . 2

1 + cos 2 t

3

1

1 + cos 2 t

3

+  ∫ |( x − 1) ( x − 2) ( x − 3)| dx

∫ ( 2 − x ) f ( x ( 2 − x ) ) dx

=

sin 2 t

2

2

∫(x

3

xf ( x ( 2 − x ) ) dx

sin 2 t

= ∫ |( x − 1) ( x − 2) ( x − 3)| dx 

=



43. (d) I1 =

2



− 6 x 2 + 11 x − 6 ) dx

⇒ 2I1 = 2I2 ⇒

1

3





∫(x

3

− 6 x 2 + 11x − 6 ) dx

2

2

 x4  11 2 3 =   − 2 x + x − 6 x  4 2  1

3

x  11 2 3 –  − 2x + x − 6x 4 2  2 4



 11 1  = (4 − 16 + 22 − 12) −  − 2 + − 6   2 4    81  99  –  − 54 + − 18  − (4 − 16 + 22 − 12)  = 9 . 4 2     2 π/3

π/3 sin x + cos x sin x + cos x dx = ∫ dx 41. (a) Let I = ∫ sin 2 x − (1 − sin 2 x) 1 π/6 π/6

π/3

=



π/6

sin x + cos x 1 − (sin x − cos x)

dx

1  −x  –  cos πx + 2 sin πx  π π 1



40. (b) ∫ |( x − 1) ( x − 2) ( x − 3)| dx 1

dx

3/ 2

b b  esin x dx  = ∫ sin x  ∫ f ( x) dx = ∫ f (a + b − x) dx  + 1 e a π /2 a 



3/ 2

1

π /2

π /2

1

1

1  −x  =   cos πx + 2 sin πx    π π  −1

dx 39. (d) I = ∫ sin x +1 π /2 e

2I =

0

0

∫ x sin πx dx + ∫ x sin πx dx − ∫ x sin πx

1

= –e–b ·  ∫

3/ 2

1

∫ | x sin πx | dx + ∫ | x sin πx | dx + ∫ | x sin πx | dx

=

π /2

π /2

dx

−1

−Z − b + 1

3 . 2

−1

Let t – b + 1 = Z



∫ | x sin πx |

∴ 

2

dx =

3 −1 2



1− 3 2

1 − z2

I1 = 1. I2

1

1/ 2

0

0

∫ | x cos πx | dx =

1/ 2

=



x cos πx dx −

0

∫ | x cos πx | dx +

1

∫ x cos πx

1

∫ | x cos πx |

dx

1/ 2

dx

1/ 2

1 1 x  x  =  sin πx + 2 cos πx  −  sin πx + 2 cos πx  π π π 0  π 1 / 2 1/ 2

1

1 1   −1 1  1 =  − 2 − 2 −  = . 2π  π  2π π   π 45. (c) I =

dz

= 2 · I2 – I1

44. (b) x cos πx = 0 gives x = 0, πx = π , 3π etc. 2 2 ∴ x = 0, 1 ∈ [0, 1] 2 ∴

π/4

π/4

0

0

∫ log (1 + tan x ) dx = ∫ log (1 + tan ( π/4 − x ) ) dx

 sin ( π/4 − x ) + cos ( π/4 − x )   dx = ∫ log  cos ( π/4 − x ) 0   π/4

261

[Putting sin x – cos x = z ⇒ (cos x + sin x) dx = dz]

) ( 1 + x + x) = 1

Definite Integral and Area

(

∵ 1+ x 2 − x   ∴ 1 + x 2 − x = 

262

π/4





2

∫ log  1 + tan x  dx

=

= (log 2) π/4 – I

0

Objective Mathematics

  k = –π log 2  ∴  I = –  k 8

⇒ I = π ⋅log 2 .   8

0

1

2

0

1

3 2 3 2 ∫ | x − 3x + 2 x | dx + ∫ | x − 3x + 2 x | dx 1

∫(x

=

2

− 3 x 2 + 2 x ) dx − ∫ ( x 3 − 3 x 2 + 2 x ) dx

3

0

−1

1 − x   dx . 1 + x 

0   et cos–1 x = 2θ ⇒ x = cos 2θ ⇒ dx = –2 L sin 2θdθ

1 − cos 2θ cot–1 1 + cos 2θ = cot–1 (tan θ)

2

46. (a) ∫ | x 3 − 3 x 2 + 2 x | dx =



1

∫ cos 2 cot

50. (b) I =

1

1

π = cot–1cot  − θ  = π − θ 2 2  π   cos 2  − θ  = cos(π –­ 2θ) = –cos 2θ = –x 2 



2

 x4  x4 3 3 2 2 =  − x +x  −  −x +x  4 0  4 1

sin θ

1 16 1 =  − 1 + 1 −  − 8 + 4  +  − 1 + 1 = 1 . 2 4   4  4 

1

 x2  1 ∫0 − xdx =  − 2  = – 2 1

∴  I = 51. (c) A =

t dt

∫ 1+ t

2

1

cosec θ



=

2 −x 2 +x , then f (– x) = log  47. (d) Let f (x) = log   2 + x   2 − x 

1

. Let t = 1 , dt = – 1 dx x2 x

cosec θ 1 −1 dx dx =– ∫ = –B ⋅ 2⋅ 2 x x x +1 x (1 + x 2 ) 1 2 x

2 −x = – log  = – f (x)·  2 + x  So, the function f (x) is odd.



2−x

1

∫ log 2 + x



dx = 0.

−1

2



− 1/ 2

 x+1 x − 1  −    x − 1 x + 1

1/ 2



=

− 1/ 2

1/ 2

=

2

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

1/ 2

48. (b)



− 1/ 2

2

1/ 2

dx =



− 1/ 2

x + 1 x −1 − dx x − 1 x +1

1/ 2 4x 4x dx = 2 ∫ 2 dx x −1 x −1 0

 4x  ∫0  x 2 − 1  dx

3 4 = – 4  log  = 4 log   . 4  3

4



f ( x)dx =

−2

−1



f ( x) dx +

−2 2



0



−1 3

0

I2 = I3 =

∫ cos ( cos x ) dx 0

∫ sin ( cos x ) dx ∫ cos x dx 0

f1 (x) = cos (sin x) Area under curve f1 is more than f2 where f2 (x) = sin (cos x), x ∈ [0, π/2] ∴  I1 > I2    area under f3 is more than others ∴  I3 > I1 > I2. 53. (c) We have, Un = π/ 4

4

f ( x) dx + ∫ f ( x) dx + ∫ f ( x) dx

+



= 4 + 1 + 0 + 1 + 4 + 9 = 19.

2

π/2

π/2

1

f ( x) dx + ∫ f ( x) dx



1

= 0.

0

= – 4 log 1 − x 2 1/ 2 ( ) 0 

49. (c)

∫ cos ( sin x ) dx =

−A −1 −1

π/2

 4x  1  < 0 in the integral  0,   ∵ 2  2   x −1



1

0

1/ 2



A+B

π/2

(  integrand is even)

= –2

∆ = e

A A2 B 2 −1 = 1 A 1 2A 2 −1

A2 B2 A 2 + B2

A

52. (b) I1 =

2



∴  A + B = 0.

= =

∫ tan

n

x dx =

0

n−2

x ⋅ (sec 2 x − 1) dx

∫ tan

n−2

x ⋅ sec 2 x dx −

0

π/ 4

∫ tan

n−2

x ⋅ tan 2 x dx

0

∫ tan

0 π/ 4

3

π/ 4

π/ 4

∫ tan 0

n−2

x dx

π/ 4

π/ 4

∫ tan

θ dθ

0

∫ tan

Now, In + 1 =

n

n +1

θ dq

0

58. (a) Let I =

π/ 4



=

∫ tan

n −1



=

∫ tan

n −1

θ ⋅ tan 2 θ dθ.



=

∫ tan

n −1

θ sec 2 θ d θ −

0



n −1

θ dθ

i.e., ∫ 0

10π

0

59. (b) In + In

=

∫ t dt 1

∴ In + In

6

x cos 4 x dx

π/3

∫ cos 0

π/2 π = 10  ∫ sin x dx + ∫ sin x dx  π/2  0 





π/2 π = 10 [ − cos x ]0 + [ − cos x ]π/2   

=



= 10 [1 + 1] = 20.  x cos qx dx

=

π/ 2

 61. (d)

tan n x (1 + tan 2 x ) dx

tan n x sec 2 x dx

n

+2

=

where t = tan x 1 n +1

⇒ nlim n [In + In + 2] →∞ n n 1 = = = 1. = lim n ⋅ n→∞ 1  n +1 n +1 n 1 +  n 

=

π/ 2  sin qx  p = cos p x ⋅ + cos p − 1 x sin x sin qx dx  ∫ q 0  0 q π/ 2 p =0+ cos p − 1 x [cos (q – 1) x – cos qx cos x] dx q ∫0 [Since cos (q – 1) x = cos qx cos x + sin qx sin x ∴ cos (q – 1) x – cos qx cos x = sin qx sin x]

π/4



0

π/4

0

=



0



=

+2

0

π/ 2

∫ cos

∫ sin

2 ∴ I = 3π . 512

|sin x | dx

57. (b) f (p, q) =

π/ 2

2 5 ⋅ 3 ⋅ 1⋅ 3 ⋅ 1 π ⋅ = 3π 10 ⋅ 8 ⋅ 6 ⋅ 4 ⋅ 2 2 256

= 2π·

 sin x  2 cos x + 1 dx =   2 (2 + cos x)  2 + cos x  0

p

x cos 4 x dx – I

0

2 cos x + 1 . (2 + cos x) 2

π/ 2

6

⇒ 2 I = π· 2·

1 1 =  − 0  = . 60. (c) 2 2 



x cos 4 x dx

0

– In – 1

 56. (a)

π

∫ sin

= π

I ntegrating both sides w.r.t. x between the limits 0 and π , 2 π/ 2 2 cos x + 1 we get [P]0π / 2 = ∫ dx 2 0 ( 2 + cos x ) π/ 2

6

0

π/ 4

(2 + cos x) ⋅ cos x − sin x ⋅ (− sin x) ⇒ dP = (2 + cos x) 2 dx =

∫ (π − x) sin

=

⇒ In – 1 + In + 1 = 1 or n (In – 1 + In + 1) = 1. n sin x 55. (b) Let P = 2 + cos x



(π − x) cos 4 (π − x) dx

π

0

 tan θ  =    n 0 n

∫ tan

6

0

θ ⋅ (sec 2 θ − 1) dθ π/ 4

x cos 4 x dx

6

∫ (π − x) sin

=

0

π/ 4

π

∫ x sin 0 π

0

π/ 4

=

 p p ⇒ 1 +  f ( p, q) = f ( p – 1, q – 1) q q  p f ( p – 1, q – 1). ⇒ f ( p, q) = p+q

1 . n −1

∴ Un + Un – 2 = 54. (a) We have, In =



– Un – 2

2a

3φ ⋅ sin2 6 φ d φ

π

1 cos 4 θ ⋅ sin2 2 θ dθ 3 ∫0 1    Putting 3φ = θ ⇒ d φ = 3 d θ    π 1 4 2 cos θ ⋅ (2 sin θ cosθ) dθ 3 ∫0 π 4 = ∫ cos 6 θ ⋅ sin2θ dθ 30 π/ 2 4 6 2 x 2 ∫ cos θ ⋅ sin θ dθ 3 0 8 5 ⋅ 3⋅1 π = ⋅ ⋅ = 5π . 3 8⋅ 6⋅ 4⋅ 2 2 96

∫x

9/ 2

0



4

(2a − x) − 1/ 2 dx π/ 2

=

∫ 0

(2a )9 / 2 sin 9 θ ⋅ 4a sin θ cos θ d θ 2a − 2a sin 2 θ

263



Definite Integral and Area

p p f ( p − 1, q − 1) − f ( p, q ) q q

π/ 4

 tan n − 1 x  =    n − 1 0

[Putting x = 2a sin2θ ⇒ dx = 4a sin θ cos θ dθ]

264



π/ 2

∫ sin

= 64a5

10

66. (d)

θ dθ

Objective Mathematics

0

9 ⋅ 7 ⋅ 5 ⋅ 3⋅1 π = 64a5 · ⋅   = 63π a5. 10 ⋅ 8 ⋅ 6 ⋅ 4 ⋅ 2 2 8 62. (c)

2

∫x

2 x − x 2 dx

3

0

π/ 2

=

∫ 0

8 sin 6 θ 4 sin 2 θ − 4 sin 4 θ (4 sin θ cos θ) dθ [Putting x = 2 sin θ ⇒ dx = 4 sin θ cos θ dθ] 2



π/ 2

∫ sin

= 64

8



0

1 2

=–

∫ 0

π/ 2

1 = 2 =

1 2

∫ 0

0

∫ cos

2

θ

π/ 2

1+ cos θ sin θ dθ 1 − cos θ

0

  = cos θ  = − sin θ d θ  − sin θ  = d θ 2 

64. (b)

∫ 0

=

=

x

1 3

1 3

∫ 0

68. (b)

) (

=5–

3− 2. 1

3

1

5

θ dθ =

0

1

dx =

−1

π

2 x (1 + sin x ) 1 + cos 2 x

=



π

−π

1 4⋅ 2 = 8 . ⋅ 3 5⋅3 45

0

1

−1

0

= – ( x) −1 = – 1.

∵[ x] = − 1 if − 1 ≤ x < 0   0 if 0 ≤ x < 1 

2

3/ 2 2

2.

dx

π 2x x sin x dx + 2 ∫ dx 2 − π 1 + cos x 1 + cos 2 x



π

0

x sin x dx 1 + cos 2 x

( π − x ) sin ( π − x ) 1 + cos 2 ( π − x ) π ( π − x ) sin x π

0

1 + cos 2 x

0

⇒ I = 4π



π

2I = 4π



π

0

0

π sin x x sin x dx − 4 ∫ dx 0 1 + cos 2 x 1 + cos 2 x

sin x dx 1 + cos 2 x

 Put cos x = t ⇒ – sin x dx = dt ⇒ sin x dx = –dt 1 1 − dt dt = 2π ∫ = 2π  tan −1 t  −1 −1 1 + t 2 1 + t2 π π = 2π [tan–1 (1) – tan–1 (–1)] = 2π  +  4 4 = 2π × π = π2. 2



−1

1

3

69. (b) ∫  x  dx = 0

∫ (−1) dx + ∫ 0 dx

 0

 Putting x 3 = tan θ    1  ⇒ x 2 dx = sec 2 θ d θ    3

dx = ( x) 1 + 2 ( x)

2

−π

∴ I = 2π

sec θ d θ  sec7 θ

dx

1.5

2

θ dθ

dx

2

2

3  2 −1+ 2  − 2  = 2 – 2 



0

2

∫ cos

1

∫ 0 dx + ∫ 1 ⋅ dx + ∫ 2



2

2

∵[ x 2 ] = 0, 0 ≤ x < 1    1, 1 ≤ x < 2     2, 2 ≤ x < 1.5 

I = 4

∫ cos

)

1.5

2

0



π/ 2

) (

3− 2 +3 2− 3

2

2



1 cos 2 θ d θ + 2

3

∫ [ x ] dx = ∫ [ x ] dx + ∫ [ x ] dx + ∫ [ x ]

I = 4

θ (1 + cos θ) dθ

3

2

2

2 −1 + 2

1.5

2

1⋅ dx + ∫ 2dx + ∫ 3dx 3

(



2

3

2

=0+1

cos θ / 2 θ θ cos 2 θ ⋅ ⋅ 2 sin cos dθ sin θ / 2 2 2

π/ 2

∫ [ x]

65. (c)

1

I = 0 + 4

(1 + x 2 )7 π/ 2

0

2



1 1 π 1 2 = π + 1 = 3π + 8 . = ⋅ ⋅ + ⋅ 2 2 2 2 3⋅1 8 3 24 ∞

1

=

0



∫ 0 dx + ∫

2

1

2 cos 2 θ / 2 cos θ ⋅ ⋅ sin θ dθ 2 sin 2 θ / 2

∫ cos

π/ 2

[ x 2 ] dx + ∫ [ x 2 ] dx + ∫ [ x 2 ] dx + ∫ [ x 2 ] dx

0

2

π/ 2

1 = 2

=

1

0

1

 π/ 2



=

  Putting x 2  ⇒ 2 x dx  ⇒ x dx 

1 = 2

=



1 + x2 dx 1 − x2

5 ∫x

[ x 2 ] dx =

0

7 ⋅ 5 ⋅ 3 ⋅ 1⋅ 1 π ⋅ = 7π . 10 ⋅ 8 ⋅ 6 ⋅ 4 ⋅ 2 2 8

= 64· 1

2

0

θ ⋅ cos2θ dθ

0

63. (a)

67. (b)



1

3

0

1

∫  x  dx + ∫  x  dx  0 if 0 ≤ x < 1  ∵  x  =   1 if 1 ≤ x < 3 

 1

=

3

∫ 0 dx + ∫1 dx = x 0

1

3 1

= 2.

x π dx + = ∫ − cos 2 2 x 4 − π/ 4



π ⋅2 4

π/ 4

∫ 0

π/ 4

dx dx ∫ − cos 2 2x − π/ 4

dx 2 − cos 2 x

π/ 4



=

π2 . 6 3

2

71. (a) ∫ sin 0

π[ x]  dx = 2

1

=

∫ sin 0

∫ sin 0

π/ 4

=

2

− 1/ 2

72. (c)

∫ 0



∫ (cos θ − sin θ)

= (sin θ + cos θ)

75. (a)

π



=



∫ |sin x | dx + ∫ |sin x | dx π

0

∵sin x = 0 gives x = nπ, n = 0, ±1, ± 2, ...; out of which only x = π ∈ (0, 2π) ⋅   

 =

π



0

π

∫ sin x dx − ∫ sin x

dx = – [cos x]0π + [cos x]2ππ

= – (– 1 – 1) + (1 + 1) = 4. −1/ 2

2

∫ [2 x]

73. (c)

−1





 ∵[2 x] =      

dx =

1/ 2

0

∫ [2 x] dx + ∫ [2 x] dx + ∫ [2 x]

−1

−1/ 2

+

0

1

3/ 2

1/ 2

1

∫ [2 x] dx +

−2 if −1 ≤ x < 1 0 if 0 ≤ x < ; 2 3 2 if 1 ≤ x < ; 2

−1 ; 2

dx

∫ [2 x] dx +

2

∫ [2 x] dx

3/ 2

−1  ≤ x < 0; 2  1 1 if ≤ x < 1;   2  3 3 if ≤ x < 2   2

+ 3 ( x)

2 3/ 2

=

3 . 2

π

∫ |cos θ − sin θ|



dθ –

π

∫ (cos θ − sin θ)



π/ 4

π/ 4

π

− (sin θ + cos θ)

0

π/ 4

 1 1    1   1 + + =    − 1 − (−1) −  2   2   2  2

0 if 0 ≤ x < 1 1 if 1 ≤ x < 2 

∫ |sin x | dx

3/ 2 1

π/ 4

2.

=2

0

1/ 2

0



3/ 2

+ 2 ( x)

1

|cos θ − sin θ | d θ +

π/ 4

= 2 – 1 = 1.

1 − cos 2 x dx = 2

+ ( x)

0

2



1

∵cos θ − sin θ = 0 gives cosθ = sin θ    π   ∴ tan θ = 1 = tan   4   π  ∴ θ = n π + , n = 0, ± 1, ± 2, ⋅ ⋅ ⋅ ⋅,  4     π out of which only θ = ∈ (0, π) ⋅  4  

=

2

1

1/ 2 −1

0

π⋅ 0 π ⋅1 dx + ∫ sin dx 2 2 1

=0+ x

2

0

π [ x] π [ x] dx + ∫ sin dx 2 2 1



3/ 2

π

2

∵[ x] =  

1

∫ 1dx + ∫ 2dx + ∫ 3dx

74. (b) ∫ |cos θ − sin θ | dθ

 1

0

− 1/ 2

– ( x)



1

π 2

∫ 0 ⋅ dx +

 tan − 1 3 z  π π π tan − 1 3 = ⋅   = 3 2 3 2 3 3   0

=

1/ 2

− 1/ 2

= – 2 ( x)

π/ 4 dx π sec 2 x = dx 2 ∫ 1 − tan x 2 0 1 + 3 tan 2 x 0 2− 2 1 + tan x 1 π dz =  [Put tan x = z ⇒ sec2x dx = dz] 2 ∫0 1 + 3 z 2

π 2



(−1)dx +



  x 1 is an odd function while is even  ∵ 2 − cos 2 x 2 − cos 2 x   =

0

(−2)dx +

−1

π/ 4

=0+



=

265

x+π/4 dx ∫ − π / 4 2 − cos 2 x

∫ [x ] 2

−2

2

dx = 2

∫[x ] 2

dx 

[ integrand is even]

0

2 3 1 = 2  ∫ [ x 2 ] dx + ∫ [ x 2 ] dx + ∫ [ x 2 ] dx +  0 1 2

2



∫ [ x ] dx  2

3



∵[ x 2 ] = 0 if 0 ≤ x < 1; 1 if 1 ≤ x < 2 ;    2 if 2 ≤ x < 3 ; 3 if 3 ≤ x < 2  



1 = 2  ∫ 0 dx +  0 = 2 ( x)

2 1

2

3

2



2

3



∫ 1 dx + ∫ 2 dx + ∫ 3 dx  1

+ 4 ( x)

(

3 2

+ 6 ( x)

2 3

)

= 10 − 2 3 − 2 2 . π

76. (b) |sin x + cos x | dx ∫ 0

−1 if

3π / 4

=

∫ 0

|sin x + cos x | dx +

π



3π / 4

|sin x + cos x | dx

∵sin x + cos x = 0 ⇒ cos x = − sin x    ⇒ tan x − 1 ⇒ x = 3π ∈ (0, π)    4

Definite Integral and Area

− 1/ 2

π/ 4

70. (a)

266

3π/ 4

=



π



(sin x + cos x) dx −

3π/ 4

0

Objective Mathematics

∵ x  = 0 if 0 ≤ x < 1; 1 if 1 ≤ x < 4;       2 if 4 ≤ x < 9 ⋅ ⋅ ⋅ ; (n −1) if (n − 1) 2 ≤ x < n 2  

(sin x + cos x) dx

= (− cos x + sin x)

3π / 4 0

π

− (− cos x + sin x)

3π / 4

 1 1     1 1  + + =   = 2 2 .  +1 − 1 −  2 2 2 2       1

−1

0

1

−2

−2

−1

0

∫ [ x + 1] dx = ∫ [ x + 1] dx + ∫ [ x +1] dx + ∫ [ x + 1]

77. (a)

−1

0

1

∫ (−1) dx + ∫ 0 ⋅ dx + ∫1

−2

−1

dx = – ( x)

dx



+ ( x)

−2

100 π

1 − cos 2 x dx =

1 0

= 0.

1

4

( n − 1)2

2

∫ |sin x |

π

10 π

∫ |sin x |

dx

0

[ sin x ≥ 0 in [0, π]]

n

∫ [t

83. (b)

1

2

−1

0

1



=

− 2 if −1 ≤ x < 0; −1 if 0 ≤ x < 1  0 if 1 ≤ x < 2 

0

1

2



80. (d) Let I =

0

20 π



1

= – 2 ( x)

|cos x | dx = 2



− 20 π

∫ |cos x |

∴  I = 2⋅ 20

0 −1

− ( x) 0 = – 3.

|cos x | dx



= 40 π/ 2

= 40

∫ 0

dx π/ 2

π

0 π

π/ 2

∫ |cos x | dx + 40 ∫ |cos x |

= 40[sin x]



= 40 (1 – 0) – 40 (0 – 1) = 80.

∫  x  dx = 0

n

] dt + ⋅ ⋅ ⋅



2

∫ 0 dt +

∫ 1 ⋅ dt +

(

) (

n

3

∫ 2 dt + ⋅ ⋅ ⋅

1

2

2 −1 + 2

3− 2



(n − 1) dt

)

+ ⋅ ⋅ ⋅ (n − 1)

(

∫ ( x − [ x ])

).

1

∫ ( x − [ x ])

dx = 100

0

0

1

= 100

1

∫ x dx − 100 ∫ [ x] dx 0

4

9

0

1

4

x

∫ f (t )

85. (b) Let φ (x) =

∫  x  dx + ∫  x  dx + ∫  x  dx ∫

( n − 1)2

 x   dx

n − n −1

84. (a) Since x – [x] is a periodic function of period 1 100

dt, then

a

−x

φ (– x) =

∫ a



f (t ) dt = –

x



    

n −1

1  x2  = 100   − 100 ∫ 0 dx = 50.  20 0

1

[t 2 ] dt

n −1

1

− 40 [sin x]

+ ⋅ ⋅ ⋅ +

2

2

0

n2



3

∫ [t

(

dx

π π/ 2



] dt +

= n n − 1 + 2 + 3 + ⋅⋅⋅ + n



π/ 2

2



cos x dx − 40 ∫ cos x dx π/ 2 0

2

∫ [t 1

0

1

0

n2

] dt +

1

=

 [ integrand is even] Since | cos x | is a periodic function of period π,



2

0

π

sin x > 0 in (0, π)]

∵[t 2 ] = 0 if 0 ≤ t < 1; 1 if 1 ≤ t < 2  2 if 2 ≤ t < 3 ; 3 if 3 ≤ t < 2 ⋅ ⋅⋅;   (n −1) if n − 1 ≤ t < n 

= 1 ⋅ 20π

dx

0

] dt

∫ [t



2

∫ (− 2) dx + ∫ (−1) dx + ∫ (0) dx

−1

∫ sin x

0

=

0

∵[ x − 1] =  

π

= – 9 [cos x] = – 9 (– 1 – 1) = 18.

∫ [ x − 1] dx = ∫ [ x − 1] dx + ∫ [ x − 1] dx + ∫ [ x − 1] dx



0

[

0

−1

∫ |sin x | dx = 9

π 0

1

2 [− cos x]0π = 200 2 .

2

π



2 ∫ sin x  dx

0

79. (a)

dx = 9

π

2 ∫ |sin x | dx = 100



(n − 1) dx



n (n + 1) (2n + 1) n (n − 1) (4n + 1) = . 6 6

π

Since | sin x | is a periodic function of the period π

81. (b)

0

82. (a) Since | sin x | is a periodic function of period π,

0

= 100

n2



0

∴   I = 100

9

= n3 –

0

100 π

78. (c) Let I =

−1

4

= 1 (4 – 1) + 2 (9 – 4) + ⋅ ⋅ ⋅ + (n – 1) [n2 – (n – 1)2] = – 1 – 4 – 9 ⋅ ⋅ ⋅ – (n – 1)2 + (n – 1)⋅ n2 = – (12 + 22 + 32 + ⋅⋅⋅ + n2) + n3

∵[ x + 1] = −1 if − 2 ≤ x < − 1; 0 if −1 ≤ x < 0;   1 if 0 ≤ x 0



π/ 2 π/ 2  π = 400  ∫ sin 4 x + ∫ cos 4  − x  dx  2   0 0 



= 800 ⋅ 

⇒ (x + 2) (x + 1) x (x – 1) > 0 ⇒ x ∈ (– ∞, – 2) ∪ (– 1, 0) ∪ (1, ∞). 96. (b)

−1

− 100 π

2

x

100 π

98. (b)

 tan − 1 x 2    1 x2  lim  =  sin x 2  2 x→0  2   x 



x

Clearly, f (x) is continuous as well as differentiable in the interval [– 1, 1]. Also, f ′ (x) = | x + 1 | ⇒ f ′ (x) is continuous in [– 1, 1] but differentiable everywhere except at x = – 1.

[Using L’ Hospital’s Rule)



− x2 −x ; 2

∫ | x + 1| dx + ∫ | x + 1| dx

 x2  = −  + x   2 

d 2 (tan − 1 t ) 2  2 ⋅ (x ) t=x dx = lim    x→0 sin t  4 ⋅ d ( x 4 )   t = x dx

−1

−1

−2

t ) 2 dt

0

x4

=

−1

0



−2

=

∫ (tan

x→0

∫ − ( x + 1) dx

When – 1 ≤ x ≤ 1

6 x (1 + x12 − 6 x12 ) 6 x (1 − 5 x12 ) = . 12 2 (1 + x ) (1 + x12 ) 2 x2

= 20 (1 + 1) = 40.

97. (a), (b), (c)  When – 2 ≤ x ≤ – 1

 (1 + x12 ) ⋅ 2 x − x 2 ⋅ 12 x11  ∴ f ′′ (x) = 3   (1 + x12 ) 2  

94. (d) lim

   π ∵ in  0, 2  both sin x and cos x are positive     



 1  d 3 1 ⋅ 3x2 ⇒ f ′ (x) = 1 + t 4  3 ⋅ dx ( x ) = 1 + x12  t = x

=

dx

∫ (sin x + cos x)

π/ 2

10π

∫ (|sin x | + |cos x |) dx

=



99. (a)



x dx = 800 ⋅

3⋅1 π ⋅ = 150 π. 4⋅ 2 2

4

∫ f (4 − x) φ (4 − x)

dx

0

4

=

0

∵|sin x | + |cos x | is a periodic    π   function of period   2

f ( x) φ ( x) dx =

0

π/ 2

∫ (|sin x | + |cos x |) dx = 20 ∫ (|sin x | + |cos x |) dx 0

4

0

4

0

 π 20   2

∫ sin

   

∫ f ( x) ⋅ (3 − φ ( x))

dx

0



∵ f ( x) = f (4 − x)  and φ (x) + φ (4 − x) = 3  

0

⇒ 2 I = 3 ⋅ 2 100. (b) F(t) =



t

0



t

∴ a2 – a1, a3 – a2, a4 – a3 are in H⋅P.

f (t − y ) g ( y ) dy

∫e

=

∴ I = 3.

t−y

0

t

⋅ y dy = e  ·  ∫ e t

0

−y

⋅ y dy



= et ( −te − t − 0 ) − ( e 

) 

−y t

0

π a a + sin a + cos a 101. (b) We have, ∫ f ( x) dx = 2 2 2 0 Differentiating w.r.t a, we get

f (a) = a +

1 π (sin a + a cos a ) − sin a 2 2

π π π 1 π 1 + − Put a = ;f   = = . 2 2 2 2 2 2 1 1 102. (a) We have, 2 f (x) + 3 f    = –2 x  x 1 ⇒ 2 f   + 3 f (x) = x – 2  x Solving the above two equations, we get − 2 3x 2 + − 5x 5 5

f (x) = ∴

2

∫ f ( x)

2

 −2

∫  5 x +

dx =

1

1

 −2 3x 2  =  5 log x + 10 − 5 x    2

3x 2  −  dx 5 5 2

6 4  3 2  −2 log 2 + −  −  −  =  5 5   10 5   5 1  −2 log 2+  . =  2  5

103. (c) an – an – 1=

π/ 2

∫ 0

= −1 2 =–

π/ 2

∫ 0

π/ 2

∫ 0

π/ 2

=–

∫ 0

cos 2 nx − cos 2 (n − 1) x dx, n ≥ 2 sin x

cos 2 (n − 1) x − cos 2 nx dx sin x

sin (2n − 1) x sin x dx sin x

x + cot n + 2 x) dx, n ≥ 2

∫ cot

n

x cos ec 2 x dx π/ 4

 − cot n + 1 x  −1 =  =  1 n + n +1  0 ∴ a2 + a4, a3 + a5, a4 + a6 are in H⋅P. 105. (a)

3 log (sin x 3 ) dx = x 1

27

3





1

dz   3  Putting x = z ⇒ 3dx = x 2   

 27

=

∫ φ ( z)

dz = [φ ( z )]1 = φ (27) – φ (1). 27

1

∴ k = 27. 16

106. (b) ∫ log x  dz = 1

2

2

∫ log x d ( x ) 4

=4

1

∫x

3

log x dx

1

2

4 4 = 4 log x ⋅ x − 1 ⋅ x dx  x ∫x x  1

2

2  4 x4   x4 x4  = 4  log x −  =  x log x −  4 1 16 1  4

 = (16 log 2 – 4) –  0 − 

1  4

15   = 16 log 2 −  . 4 

⇒ f ′ (x) = 2ax + b and f ′′ (x) = 2a. We are given f (0) = c = 1, f ′ (0) = b = – 2 and f ′′ (0) = 2a = 6 ⇒ a = 3, b = – 2 and c = 1. ∴ f (x) = 3x2 – 2x + 1. 2



∫ f ( x)

dx =

−1

2

∫ (3x

2

− 2 x + 1) dx

−1

= [ x 3 − x 2 + x]2− 1



= (8 – 4 + 2) – (– 1 – 1 – 1) = 6 + 3 = 9. π/ 2

108. (c) In =

∫ sin 0

n

x dx =

π/ 2

∫ sin

n −1

x sin x dx

0

π/ 2

sin (2n − 1) x dx

log (sin z ) dz z

107. (c) Let f (x) = ax2 + bx + c

1



n

0

= et [–te–t – (e–t – e0)]

2

∫ (cot 0

π/ 4

=

= et [–te–t – e–t + 1] = [1 – e–t (1 + t)] et = et – (1 + t). a

π/ 4

104. (c) an + an + 2 =

t −y t −y   = et ( − ye )0 − ∫0 1 ( −e ) dy   



1 (2n − 1)

=–

269



π/ 2

=  cos (2n − 1) x   (2n − 1)  0

f ( x) dx – I

n −1 =  − sin x cos x  0 +

π/ 2

∫ (n − 1) sin 0

n−2

x cos 2 x dx

Definite Integral and Area

4

=3

270

π/ 2



= (n – 1)

sin n − 2 x (1 − sin 2 x) dx

113. (c)

0

Objective Mathematics

π/ 2

∫ sin

= (n – 1)

n−2

x dx − (n − 1)

0

π/ 2

∫ sin

n

x dx

⇒ n In = (n – 1) In – 2 ⇒ In =

n −1 In – 2 n

⇒ n (In – 2 – In) = In – 2. Also, In : In – 2 = (n – 1) : n and In – 2 > In.



∫ log

x

π 3

3e x cos (3e x ) dx =

π/3

∫ 3 cos 3t  dt

π/6

π 6

= [sin 3t ]π / 6 = sin π – sin π/3

π 2

∫ g ( x)

∫ f '( x) dx

dx =

= lim x→0



Again applying L’Hospital rule



111. (b) I =

a

0 + 2 sec 2 0 = 1. 0 + 2 cos 0

114. (c) We have π/3

a  b tan x + c  dx = 0  |tan x | + + sec 3 1 x  − π/3  π/3

a   |tan x | + c  dx = 0 3   − π/3



−1

[ f ( x)]−1 2

= f (2) – f (– 1)

⇒ ⇒

x f ( x) dx b

= ∫ ( a + b − x ) f ( a + b − x ) dx a

b

= ∫ ( a + b − x ) f ( x) dx a

[Since given f (a + b – x) = f (x)]



2 x ⋅ 2 sec 2 x 2 ⋅ tan x 2 ⋅ 2 x+ 2 sec 2 x 2 − x sin x + cos x + cos x

= lim x→0

= 4 – 2 [As f (– 1) = 2 and f (2) = 4] = 2. b

= (a + b)



b

= (a + b)



b

a

a



f ( x) dx – I



b

a

f ( x) dx

a +b b ∴ I =   ∫ f ( x) dx .  2  a 5

112. (b) e f [ φ ( x )] ⋅ f ' [φ ( x)] ⋅ φ ' ( x) dx ∫

π/3

a 3



− π/3

2a 3

π/3

2a 3

π/3



e f ( t ) ⋅ f ' (t ) dt [Putting φ (x) = t ⇒ φ′ (x) dx = dt]

= 0. [ φ (3) = φ (5)].

∫ tan x dx + 0

2π c=0 3

2a 2π −a (log 2 − 0) + log 2 . c=0 ⇒c= 3 3 π

 log x  115. (b) When 1 < x < e3,   =0  3   log x  and when e3 < x < e6,   = 1.  3  ∴

e6

 log x  ∫1  3  dx =

= 116. (c) I =



1

0

φ (5)

f (t ) = e  φ ( 3) = e f[φ (5)] – e f [φ (3)]

0

2π c = 0 [ | tan x | is even] 3



φ ( 3)



∫ |tan x | dx +

2π =0 3

2a 2π π/3 c=0 [log | sec x |]0 + 3 3

3

=

|tan x | dx + c ⋅



b

f ( x) dx − ∫ x f ( x) dx a

∴    2I = (a + b)

φ(5)

sec 2 x 2 ⋅ 2 x x cos x + sin x

π/3   tan x tan x is an odd function ∴ ∫ dx = 0  ∵  1 + sec x  − π / 3 1 + sec x

2

−1

=

Applying L’Hospital rule



110. (a) Since g (x) = f ′ (x) ∴

x→0



= – 1.

2

sec 2 t dt  0   form  x sin x  0 

0

=

109. (b) Put e = t ⇒ e dx = dt log

x2

0

∴ In = (n – 1) In – 2 – (n – 1) In

x

∫ lim



e3

e6

1

e3

e3

e6

 log x   log x  ∫1  3  dx + ∫3  3  dx e

6 3 ∫ 0 dx + ∫ 1 dx = (e – e ).

x (1 − x) n dx

Let 1 – x = z ⇒ – dx = dz

∴ I =



0

1

(1 − z ) z n (− dz ) =



1

0

(1 − z ) z n dz

n



dx 121. (a) Let I = ∫ = sin x 0 1+ e

− zn + 1) 1

1 1 z  z − =   = n +1 − n + 2 .  n + 1 n + 2 0 n +1

117. (b)

n+2

π

∫ [sin x]  dx = ∫ [sin x] dx + ∫ [sin x]

π/ 2

π/ 2

π

3π / 2

π/ 2

π

∫ 0 dx + ∫

π

( − 1) dx = – [ x]

π/6

14 + 24 + 34 + ... + n 4 13 + 23 + 33 + ... + n3 − lim 5 n→∞ n n5



1 r 1 1 ∑   − lim x lim n r = 1 n  n → ∞ n n → ∞ n

∑  n 

3

π/6

x  1 dx =   – 0 = . 5  5 0

2

∫ max {(1 − x), (1 + x), 2}



1

l

=

1− l

 l

=

dx

∫ x sin [ x (1 − x)]

⇒ 2 I1 = I2.

1− l

0

0

1

1

0

0

∫ x dx − ∫ x dx + ∫ x dx − ∫ 0 dx −1

2 0

x 2

=



−1

0

2 1

x  x  + 2  −1 2 2

= 0

1 . 2

124. (b) S ince f (x) + f (– x) is an even function and g (x) – g (– x) is an odd function, therefore, { f (x) + f (– x)}⋅ {g (x) – g (– x)} is an odd function, therefore π/ 4



{ f ( x) + f (− x)} ⋅ {g ( x) − g (− x)} dx = 0.



f ( x) dx =

a/2



f ( x) dx +

0



a



f ( x) dx

a/2

0

=p–

f (a − z ) dz

a/2

[ Putting x = a – z in the second integral so that dx = – dz]

dx

=p+

dx

∫ sin [(1 − x) x] dx − ∫ x sin [(1 − x) x]

1− l

1

0

a/2

∫ 0

b  b  ∵ ∫ f ( x) dx = ∫ f (a + b − x) dx  a  a  l

( x) dx

∫ ( x + | x |) dx + ∫ ( x − [ x]) dx

−1

a

1− l

∫ (1 −x) sin [(1 − x) x]

0

−1

125. (a)

2 2 =  x − x  + [2 x]1− 1 +  x + x  = 9. 2 − 2 2 1  

120. (b) We have, I1 =

−1

0

dx

2

l

−1

=

1

−1

1

0

2

−1

0

− π/ 4

∫ (1 − x) dx + ∫ 2 dx + ∫ (1 + x)

−2

1

=

−2 −1

1 . 3

=

∫ f ( x) dx = ∫ f ( x) dx + ∫ f

1

and 1 – x > 1 + x ∴ max {(1 – x), (1 + x), 2} = 1 – x. For – 1 < x < 1, we have 0 < 1 – x < 2 and 0 < 1 + x  1– x ∴ max {(1 – x), (1 + x), 2} = 1 + x.

t dt

123. (a) Since | x | ≥ – [x] for all position x and | x | ≤ –­ [x] for all negative x, therefore

r =1 5

1 1 x 4 dx − lim ⋅ × ∫ x 3 0 n→∞ n

1

0

r

n

2

0

= tan t 0

119. (c) For – 2 ≤ x ≤ – 1, we have 1 – x ≥ 2

=

π/6

0

= nlim →∞

= 2π

0

2

n→∞

=

∫ 1 dx

∫ sec x d ( x − [ x]) = ∫ sec



3



 π 122. (b) Let x – [x] = t. In the interval 0,  , [x] = 0  6 ∴ x = t

lim

n

− sinx

0

∴ I = π .

3π / 2 π

−π  3π  − π = . =–  2 2  

118. (d)

dx

∫ 1+ e



1 + esinx ∫o 1 + esin x dx =

∴ 2 I =

dx

=

sin x

0

3π / 2

sin ( 2 π − x )

0



3π / 2

=

∫e

∫ 1+ e



dx

esinx dx +1



=



dx = I2 – I1

a/2

=p+



 f (a − z ) dz  ∵  f (a − x) dx 

0

=p+

a/2



f ( x) dx

0

[

f (a – x) = f (x)]

= p + p = 2p.

b

∫ a

 ∵ 

271

1

0

a  f ( x) dx = − ∫ f ( x) dx  b  b

∫f a

b  ( x) dx = ∫ f ( y ) dy  a 

Definite Integral and Area

∫ (z

=

272

d esin x F(x) = ,x>0 dx x

126. (d)

131. (c) Let I =

Objective Mathematics



64

1

64

Using eq. (1), we get

132. (b)

(

)

2 Then, f (– x) = log − x + 1 + x



(x +

= – log

1 + x2

)

5

)





(

log x + 1 + x 2

log 1/ 3

)

= – f (x)

 15  = 2 + loge   . 7 sin x dx 8 10 1 + x 19

19



dx = 0.

5 3 128. (a) We have, I = ∫ (α sin x + β tan x + γ cos x)  dx −a

∫ sin





5

a

∫ tan

x dx + β

−a

3

x dx + γ

−a

=

−a

=α×0+ β×0+ γ×2

∫ cos x

dx

b

∫ (a + b − x) f (a + b − x)

a

=

b

a

a

∫ [(a + b) − x] f ( x)  dx = (a + b) ∫ f ( x) dx

∴ I = 130. (c)





a+b 2

1 + sin

0



=

a

b

∫ 0

sin 2

b

∫ f ( x)

∴ I = dx



b

−x dx = a – b = – b – (– a) = | b | – | a |. a x



(iii) If a < 0 < b, then 0



I =



= ­–

a

x dx 2

= 4 [(1 + 1)] = 8.

b

| x| | x| dx = dx + ∫ x 0 x 0

b

a

0

∫ dx + ∫ dx

b

−x x dx + ∫ x a 0 x 0



dx

= a + b = b – (– a) = | b |– | a |.

Hence (d) is correct answer. 1 135. (c) f (0) = a, f (1) = a + b + c, f   = a + b + c . 2 4 2

dx

x x  x x  = ∫  sin + cos  dx = 4  − cos + sin  4 4   4 4 0 

−1 − 7 19 1 1 1  x  = 10− 7 − 19− 7 < 10− 7 1 and y = – x + 1,  x < 1

f (sin 3 x + cos 2 x) dx

0

f (sin 3 x + cos 2 x)  dx

275

π ⋅2 2

Definite Integral and Area

=

∴ k = π.

0

155. (b)

π/2

∫ |sin x − cos x |

dx

0

π/4

π/2

0

π/4

∫ |sin x − cos x | dx + ∫ |sin x − cos x |

=

π/4

π/2

0

π/4

dx

∫ (sin x − cos x) dx + ∫ (sin x − cos x)

=–

dx

= [ cos x + sin x ]0 + [ − cos x − sin x ]π / 4 π/4

π/2

Also, y = 3 – | x | ⇒ y = 3 – x,  x > 0

1 1 1 1    =  + + − 1 +  −1 +  = 2 2−2 .  2   2 2 2

(

)

and y = 3 + x,  x < 0 Solving y = x – 1 and y = 3 – x

156. (a), (b)  We have f (x) = A ⋅ 2x + B

⇒ x – 1 = 3 – x ⇒ x = 2

⇒ f ′ (x) = A ⋅ 2x log 2

and y = 3 – 2   i.e., y = 1 1 . log 2

f ′ (1) = 2 ⇒ 2 = 2 A log 2 ⇒ A = 3





0

AB2 = (2 – 1)2 + (1 – 0)2 = 1 + 1 = 2 ∴ AB = 2 BC2= (2 – 0)2 + (1 – 3)2 = 4 + 4

 1  ⋅ 2 x + B  dx = 7 f ( x) dx = 7 ⇒ ∫  log 2  0 3

∴ BC = 2 2 Area of rectangle

3

 1  ⋅ 2 x + Bx  = 7 ⇒  2  (log 2) 0

ABCD = AB × BC =

160. (c) Since f ′(x) = f (x) and f (0) = 1

8 1 + 3B − =7 2 (log 2) (log) 2



therefore let f (x) = ex Also, f (x) + g(x) = x2 Again,

7 (log 2) 2 −1 ⇒ B = 3 (log 2) 2 



4

2

4

2

3

=

1

2

3

4

1

2

3

∫ log 1 dx + ∫ log 2 dx + ∫ log 3 dx

= log 2 + log 3 = log 6. 158. (c) f (x) =

x

∫ cos t

2

dt =

1/ x

1/ x 1/ x

=–

2 ∫ cos t dt + 0



0

∫ cos t

2

dt +

x

∫ cos t

2

dt

0

x

∫ cos t dt . 2

0

d 1 2 ∴  f′(x) = – cos t 2  1   t = dx  x  + cot t  t =   x 1 1 1 = 2 cos 2 + cos x. x x 2 x

x

d dx

( x)

f ( x) g ( x) dx =



1

0

∫ e (x 1

0

x

∴ g(x) = x2 – ex 2

− e x ) dx =

∫ (x e 1

0

2 x

− e2 x )

x 2e x dx − ∫ e 2 x dx 1

0

1 1 2x 1 2 x x =  x e − ∫ 2 xe dx  − e  0 0 2 1 1 2 2 x x x 0 =  x e − 2 ( xe − e )  0 − (e − e ) 2 1 1 =  x 2e x − 2 xe x + 2e x  − (e 2 − e0 ) 0 2 1 2 1 2 x 1   =  ( x − 2 x + 2) e  − e + 0 2 2

∫ log [ x] dx + ∫ log [ x] dx + ∫ log [ x] dx

1

=

3

1

0

1 7 (log 2) 2 − 1 . ∴ A = and B = log 2 3 (log 2) 2  157. (a) ∫ log [ x] dx =

2 × 2 2 = 4 sq. units

= [(1 – 2 + 2) e1 – (0 – 0 + 2) e0] – 1 e 2 + 1 2 2 = [e – 2] – 1 e 2 + 1 = e – 1 e 2 − 3 . 2 2 2 2 n  1 1 1  1 = nlim + +⋅ ⋅ ⋅ + 161. (b) lim  ∑  → ∞ n→ ∞ +r n + 1 n + 2 n + n n r = 1   

= lim

n→∞

n



r =1

1 1 1 1 1 ⋅ =∫ dx = [log (1 + x)]0 = log 2. n 1+ r 01+x n

276

1 n2 n2 1 162. (a) nlim + + ⋅⋅⋅ +  →∞  + 3 3 8n   n (n + 1) (n + 2)

Objective Mathematics



 n2 n2 n2 n2  = nlim + + + ⋅⋅⋅ +   3 3 3 →∞ (n + 1) (n + 2) ( n + n)3   (n + 0) n

n

= nlim →∞

n

2

∑ (n + r )

r=0

= nlim →∞

3



r=0

1 1 ⋅ n  r 3 1 +   n

1

 −1  dx −1  1  3 ∫0 (1 + x)3 =  2 (1 + x)2  0 = 2  4 − 1 = 8 . 163. (c) lim 1

1 1 =  tan − 1 x  + 1 log (1 + x 2 )   0 2  0 = π + 1 log 2 . 4 2  1 1 1  + + + ⋅⋅⋅ + 166. (b) nlim →∞ 2 2  2n − 1 4n − 2 6n − 32

 1  + = nlim →∞  2n − 12

=

= lim

n→∞

 1  1 1 1   + + +⋅ ⋅ ⋅ +  n 2 n2 − 1 n 2 − 22 n 2 − (n −1) 2 

=

 1 1 1 = nlim  + + + ⋅⋅⋅ →∞  n 2 − 02 n 2 − 12 n 2 − 22

=

  2 2 n − (n − 1) 



+ n −1

= nlim →∞

=

1



n −r 2

r=0

dx

1



1 − x2

0

n −1

2

= nlim →∞



r=0

1 1 ⋅ n 1 − r 2 / n2

1 = sin − 1 x  = π . 0 2

164. (a) Required area =



3

0

4n − 2

n

n→∞

1

1

1

2x − x

0

0

2

dx =

n→∞

1 n



r =1

  2n ⋅ n − n  1

2

1 2r r 2 − n n2

[1 − ( x − 2 x + 1)] 2

1

−1 = sin ( x − 1)  0

1 − ( x − 1) 2

 −π  π =0–  .  = 2 2  



 n 1 n n + 2 + ⋅⋅⋅ +  167. (a) nlim  2 2 + 2 2 2 →∞ 2n  n +2 n +3  n +1  n n n n  + 2 + ⋅⋅⋅ + 2 = nlim    2 2 2 2 →∞  2 n +3 n + n2  n +1 n + 2 n n 2 = lim ∑ n→∞ r =1 n + r r =1 n

= nlim ∑ →∞

( y1 − y2 ) dx

+ ⋅⋅⋅ +

dx

∫ 0

dx

1

= lim

2

1

1

∫ ∫

2r n − r

r =1

6n − 3

2

n

1



1

+

2

1  n 

2

1 π −1 =  tan x  0 = . 4

168. (c) lim

n→∞

r

n

∑n r =1

2

r

n

= lim

n→∞

⋅ sec 2

r 1+ 2 n 2



1 = n

1

1

∫1 + x

2

dx

0

r2 n2

∑  n ⋅ sec r =1

1

2



r2  1 =  n2  n

1

∫ x sec

2

x 2 dx

0

1

= =



3

0

(2 x − x 2 ) − (− x) dx =



3

0

(3 x − x 2 ) dx

1 sec 2 t dt [Putting x2 = t ⇒ 2x dx = dt] 2 ∫0

= 1 [tan t ]10 = 1 tan 1. 2 2

3

 3x 2 x3  −  = 27 – 9 = 9 =  3 0 2 2  2  n +1 n +2 n+3 165. (a) lim  2 2 + 2 + 2 +⋅ ⋅ ⋅ + 2 2 n→∞  n +1 n + 2 n + 3

169. (a) lim

n→∞

1  n

 n +1 n+2 n+3 n+n  = nlim + 2 + ⋅⋅⋅ + 2 →∞  2 2 + 2  2 2 n +2 n +3 n + n2   n +1 n

= nlim ∑ →∞

r =1

1+ x ∫0 1 + x 2 dx = 1

=

1 1+ r / n 1 n+r lim ⋅ 2 2 2 2 = n→∞ ∑ n n +r r = 11 + r / n 1

dx

∫1 + x 0

1 2x dx 2 ∫0 1 + x 2 1

2

+

1

=

∫ sin 0



1 n

n

∑ sin r =1

2k

2k

rπ 2n

πx 2 dx = 2 π

π/ 2

∫ sin

2k

t dt

0

πx 2    Putting 2 = ⇒ dx = π dt   

=

2 (2k − 1) (2k − 3) ⋅ ⋅ ⋅ ⋅ 1 π ⋅ ⋅ π 2k (2k − 2) ⋅ ⋅ ⋅ ⋅ 2 2

=

[(2k − 1) (2k − 3) (2k − 5) ⋅ ⋅ ⋅ 1] [2k ⋅ (2k − 2) ⋅ ⋅ ⋅ ⋅ 2] 2k [k (k − 1) (k − 2) ⋅ ⋅ ⋅ 1] [2k ⋅ (2k − 2) ⋅ ⋅ ⋅ 2]

=

(2k ) ! . 22k ⋅ (k !) 2

170. (b) Let y = lim

n→∞



(n !)1/ n n! = lim  n  n→∞ n n  

1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ n  = nlim  →∞  nn  



n→∞

= [ x log x ]0 − ∫ x ⋅ 1

0

∴ y = e– 1 =

n→∞

1 n



∴ y = e

= lim

n→∞

2

  dx 

log 2 +

π −2 2

= elog 2 ⋅ e

π−4 2

= 2e

π−4 2

.

n

1

∑r

n p +1

p

r =1

1 r ⋅  n n

n

= lim ∑ n→∞

r =1

1

1

 1  + 174. (b) lim Sn= nlim →∞ n→∞  4n 2 − 0 1

+

 n −1

= lim

n→∞

dx

p

p +1 x p + 1  1 p  x x a dx =  ∫0  p + 1p + 1 = p + 1 . 0 0

1/ n



r=0

4n − 2 2

2

1 4n 2 − 12 +⋅ ⋅ ⋅ +

  4n − (n − 1)  1

2

2

1 ( 2n) 2 − r 2

0

x dx = [ x log (1 + x)]0 − ∫ 01+ x 1

1

(1 + x) − 1 dx 1+ x 0

=

n −1 1 lim, ∑ 2 n→∞ r=0

=

1 2

1



 1  = log 2 – ∫ 1 −  dx x 1 + 0 1

= log 2 – [ x − log (1 + x)]10 = log 2 – [(1 – log 2) – 0] = 2 log 2 – log e = log 4 ⋅ e ∴ y = 4 . e

1

∫ 0

1  r  1−    2n 

=



1 n

1 π dx= 1 ⋅  2 sin − 1 x  = .   6 x 2 2    0 1−   2 2

175. (d) Let f (x) =

1/ n

2

1

sin 8 x ⋅ log (cot x) cos 2 x

π  ⇒ f  − x  = 2 

2 2 2 172. (c) Let y = lim 1 + 1  1 + 2  ⋅ ⋅ ⋅ 1 + n      n→∞ 2  2  n  n   n2  

⇒ log y = lim 1 n→∞ n

dx

1 p p p 173. (a) nlim → ∞ n p + 1 [1 + 2  + ⋅ ⋅ ⋅ + n  ]

2   1   n   1 + n   1 + n  ⋅ ⋅ ⋅  1 + n         

= log 2 –

2

0

1 −1 π = log 2 – 2  x − tan x  0 = log 2 + – 2. 2



1 ⇒ log y = lim × log  n→∞ n

∫ log (1 + x)

1

0

1

r  log 1 +  = ∑ n  r =1



∫ 1 − 1 + x

= log 2 – 2

=

1

∫ log (1 + x )

0

1 dx = 0 – 1 = – 1. x

n

1

1 2x =  x log (1 + x 2 )  − ∫ x ⋅ dx 0 1 + x2 1

1 . e



 r2  log 1 + 2  = ∑ n  r =1  n

1/ n

1 2   n  171. (d) Let y = nlim  1 + n   1 + n  ⋅ ⋅ ⋅  1 + n   →∞       

1 = lim n→∞ n

= lim

1

1 n

1







n 1 2  log n + log n + ⋅ ⋅ ⋅ + log n    1 n r 1 log = ∫ log x dx = nlim ∑ →∞ n r =1 n 0

= lim

    12  22  n2  log 1 + 2  + log 1 + 2  + ⋅ ⋅ ⋅ log 1 + 2   n  n  n     

1/ n

n 1 1 2 3 ⋅ log  ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅  ⇒ log y = nlim →∞ n n n n n



 π  π  sin 8  − x  ⋅ log cot  − x     2  2  π  cos 2  − x  2 

sin (4 π − 8 x) log tan x − sin 8 x ⋅ log cot x = =– f (x) cos 2 x cos ( π − 2 x) π/2



∫ 0

sin 8 x log cot x dx = 0. cos 2 x

277

2k (2k − 1) (2k − 2) (2k − 3)  2 ⋅ 1 2k ⋅ [(k ( k − 1) (k − 2)1] ⋅ 2k [k ⋅ (k − 1)(k − 2) 1]

Definite Integral and Area

=

278

176. (a)

π



f ( x) ⋅ g ( x) dx =

0

π



π/ 2

f (π − x) g (π − x)

Objective Mathematics

π

∫ [1 − f ( x)] ⋅

=

 −π  = π log (a + b) + 2 ⋅  log 2    2  

[1 – g (x)] dx (given)

0

a +b = π log (a + b) – π log 2 ⋅ = π log  .  2 

π



∫ [1 − f ( x) − g ( x) + f ( x) ⋅ g ( x)]

=

dx

0



π

π

π

0

0

0

∫ [ f ( x) + g ( x)]

∫1 ⋅ dx

dx =

0

2



−2

x

= π. =

0

0

1

−2

0

=

2

x + ∫ (2 − 1 + x − 1) dx



1

0

1

   x x x2  x =  2 x − 2 − 2 log 2  +  2 log 2 − 2   − 2  0 2



0

1

2

l

l+k

0

1

l −1

l

∫ [ x] dx + ∫ [ x] dx + ⋅ ⋅ ⋅ + ∫ [ x] dx + ∫ [ x] dx

= 0 + 1 + 2 + ⋅ ⋅ + (l – 1) + l ⋅ k

x x ∫ (1 − 2 + 1 − x) dx + ∫ (2 − 1 + 1 − x) dx

=

∫ [ x] dx

=

0

f ( x) dx



l+k

∫ [ x] dx



π

π



180. (a), (b)  Let [x] = l then x = l + k, 0 ≤ k < 1.

∫1 ⋅ dx − ∫ [ f ( x) + g ( x)] dx + ∫ f ( x) ⋅ g ( x) dx

=

dx

0

0



177. (b)

∫ log sin x

= π log (a + b) + 2

1 1 (l – 1) l + lk = [x] ([ x] − 1) + [ x] ( x − [ x]) 2 2

1  = [x]  ([ x] − 1) + ( x − [ x])  2  ∴ A = [x] – 1 1

181. (a) Let f (a) =

and B = x – [x].

a

x −1

∫ log x

dx

0

2

  x2 +  2 x log 2 + − 2 x  2  1



9 log2. 4

=5+

n→∞

1

=

∫x

r 1 n  n   r2  = nlim ∑  3 3 3  →∞ n r =1  r  r +n    +1 n

e/2



x2 1 dx = 1 log ( x 3 + 1)    0 +1 3

π/ 2

 ax 2 + bx + c  179. (b) Let I = ∫ log  2 ⋅ (a + b) |sin x | dx  ax − bx + c  − π/ 2 =



− π/ 2 π/ 2

+



−π/ 2

 ax 2 + bx + c  log  2  dx  ax − bx + c  log (a + b) dx +

π/ 2



= 0 + log (a + b) [ x]

+2

π/ 2

∫ log |sin x |

|log 2 x | dx =

dx

  ax 2 + bx + c  ∵log  2  is an odd function and  ax − bx + c   log | sin x | is an even function 

    

e

1 | log t | dt 2 1∫/ e

e 1  1  ∫ |log t | dt + ∫ |log t | dt  2 1/ e 1 



=



e 1  1 − log t dt + log t dt   = ∫ ∫ 2  1/ e 1 



=

1 (− t log t ) 2



=

1 1 1 1  log + 1 − + e − e + 1 2  e e e 



=

1 1  −2   + 2  = 1 −  .=  2 e e  

log | sin x | dx

0





−π/ 2 π/ 2 −π/ 2

= log (a + 1) + C.

1 dt 2

1/ 2 e

1 1 (log 2 – log 1) = log 2. 3 3

π/ 2

1

∫ a + 1 da + C

182. (b) Put 2x = t ⇒ dx =

3

0

=

r =1

1

1 xa + 1 x a log x dx = ∫ x a dx = = a +1 0 log x 0

If a = 0, then f (a) = 0, ∴ C = 0. Hence, f (a) = log (a + 1).

2

n

0

∴ f (a) =

 12 22 1 + 3 +⋅ ⋅ ⋅ ⋅ +  178. (a) lim  3 3 3 n→∞ 1 + n + n n 2 2  = lim ∑

1

⇒ f ′ (a) = ∫

1 1/ e

+t

1 1/ e

e e + (t log t ) 1 − t 1  

183. (b), (c)  We have, 0 < x2 < 1, ⇒ e0 < e x 1

1

0

0

1

1

0

0

2

< e1

x x ⇒ ∫1 dx < ∫ e dx < ∫ e dx ⇒ 1 < ∫ e dx < e . 2

2



=–2

0 π/ 2

0

⇒ sin α = 1 – 2 sin2α

∫ log sin θ

=–2

i.e., 2 sin2α + sin α – 1 = 0

b + 2T



189. (a), (d) 

100



2 x − [ x ] dx = k

100





k =1



k −1

∑2

(1 − k )

∫ f ( z)

100

∑ 2− ( k − 1)

k =1

k



k −1

= 2 x dx

∫ f ( x)

= 100 log 2. 

=

 −1 x − 1 −1 x + 1  ∫− 2  cot x + 1 + cot x − 1  dx 

∫  tan

−1

−2

dx. Hence, (a) and (d) are correct.

k

1 k k

190. (a)

 ⋅ (2k – 2k – 1) log 2

3

=

dz

a



k =1

3

f ( y + T) dy

b

100

186. (a)



a+T

[Again, putting z = y + T so that dz = dy]



2 x − [ x ] dx

2 x − ( k − 1) dx =

∑ (2 − 1)

b+T

b

k −1

= log 2



5

∫  2 x



π π 5π = ∫ dx = [3 − (2)] = . −2 2 2 2

3/ 2

1

1 k k

 + 2 x − 3 x1/ 2  dx  k

( x5 / 2 + x 2 − 2 x3/ 2 ) k

1

5

∫  2 x

k k

3/ 2

1

⇒ k +

x +1 x + 1 + cot − 1  dx x −1 x −1

1

(

= k+ k −2

(

k –20

2 ∫ (2 x + 3) dx ⋅ ∫ (3x + 4) dx



2 − tan 2 z

=

0 be a fixed number. Suppose f is a continuous 2 2 function such that for all x ∈ R, f (x + T) = f (x). If I (a)

=



(a)

T

0

f ( x) dx then the value of

3 I 2

3 + 3T

3

f (2 x) dx , is



1/2

−1/2

 1 + x    [ x] + log   dx equals  1 − x   

1 2

10. The value of

(d) 2log π

−π

(a) π π (c) 2

cos 2 x

∫ 1+ a

x

∫ log (1 + tan x)

dx =

π log 2 4 (c) π log 2 tan x 8

17.

π/2

∫ log (tan x)

π log 2 8 π (d) log 3 8

(b)

dx =

0

(b) 0

(c) 1

π/4

(a)

(d) 6I

9. The integral

16.

0

(b) 2I

(c) 3I

(a) –



1 2

dx, a > 0 is (b) aπ

(a) π 4 π (c) 2 18.

1

∫ sin 0

(d) 2π.

m m m   11. lim 1   1  +  2  +  n − 1   is equal to n→∞ n n  n   n 

−1

(d) 2

 2 x  dx =  1 + x 2 

π + log 2 4 (c) π + log 2 2

(a)

(b) 0

π − log 2 4 π (d) − log 2 2 (b)

∫ 0

(a) π ab π 2ab

(c) 20.

∫ 0

1

∫ |sin 2π x |

(a)

(b) π 4 (d) (a + b) π

0

∫x

−1

2

(b)

24.

1 e 2 e

(d) 1 –

dx = + 2x + 2 (b) π/4 (d) – π/4

(a) 0 (c) π/2 =

(a) 0 (c) 2

x4 + 1 ∫0 x 2 + 1 dx is 1

(c) 1 (3π + 4) 6

(d) 1 (3 + 4π) 6

26. If f (x) is an odd function of x then



f (cos x) dx is equal to

− π/2 π/2

(a) 0

(b)



f (cos x) dx

0

π/2

(c) 2

∫ 0

=



(b)

π 24

(d) 0



1

dx

∫ log  x + x  1 + x

π log 2 2 (c) – π log 2 (a) –

2

= π log 2 2 (d) π log 2 (b)

  1 1 1 1  + + + ... 31. lim  2 2 2 2 n→∞  n n +n n + 2n n + (n − 1) n  (a) 2  2

(b) 2

(c) 2 – 2  2

(d) 2

1

x dx

∫ (1 − x)

5/ 4

2 –2

=

15 16 3 (c) – 16

(b) 1 (3 – 4π) 6

π/2

2

(a)

(b) 1 (d) 4

(a) 1 (3π – 4) 6



∫ 4 + 9x

3 (d) ( π − 8) 2

0

−2

25. The value of

dx

2/3

π3 4

0

32.

2

∫ | x | dx

29.

30.

(d) 0

22. The area bounded by the the curves y = xex, y = xe– x and line x = 1 is

23.

(b)

(a) π 12 (c) π 4

(b) 1 π

1 (a) 1 –  e 2 (c) e

( π3 − 8) 4

0

0

2 π −1 (c) π

(d) 2 : 1

3 (c) ( π − 16) 4

dx is equal to

(a)

(b) 1 : 1

28. The area in the first quadrant between x2 + y2 = π2 and y = sin x is

a sin x + b cos x dx = sin x + cos x

π 4 (c) (a + b) π 2

2 : 1

(c) 1 : 2

(d) πab

(a) (a + b)

21.

(a)

π (b) ab

2

π/2

27. The ratio of the areas bounded by the curves y = cos x and y = cos 2x between x = 0, x = π/3 and x-axis, is

dx = a 2 cos 2 x + b 2 sin 2 x

π/2

f (sin x) dx

(d)



− π/2

f (cos x) dx

3 16 −16 (d) 3

(b)

33. Let f : (0, ∞) → R and F (x2) =

x2



f (t ) dt .

0

If F (x2) = x2 (1 + x), then f (4) equals (b) 7 (a) 5 4 (c) 4 (d) 2.  1 1 1  34. lim  + + ... +  = n → ∞ 2n + 1 2n + 2 2n + n   1 (a) log e   3

2 (b) log e   3

3 (c) log e   2

4 (d) log e   3

303

π/2

Definite Integral and Area

19.

304

Objective Mathematics

35. The area of the region bounded by y = | x – 1 | and y = 1 is (a) 2 (b) 1 (c) 1/2 (d) None of these

42.



π/2

0

sin 2θ d θ is equal to: sin 2θ + cos 2θ

(a) π/2 (c) π/4 36. The value of ∫ [2 sin x] dx, where [.] represents the great-

(b) 2π/3 (d) π/6 



π

43. If g(x) =

est integer function, is 5π 3 5π (c) 3 (a) –

(d) – 2π

(b)

(c) 0 and – 4 π

(d) 4 and 0 π

2

x

1

cos 4 t dt , then g(x + π) is equal to:

(d) g(x)/g(π)

π (a) π and – 2 2

∫e

x

0

(a) g(x) + g(π) (b) g(x) – g(π) (c) g(x) g(π)

(b) – π

1 1 πx 37. If f (x) = A sin   + B, f '   = 2 and f ( x) dx ∫   2  2  0 2A then the constants A and B are, respectively = π

38.



π 3 and 2 π

(b) e (e – 1)

(c) 0

(d) None of these



π/2

0

|sin x − cos x | dx is equal to:

(a) 2 2 – 2

(b)

(c) 1/ 2

(d) 0

45. If I =

 1 1  dx is equal to  − 2  x x

e  (a) e  − 1 2 

44.



cos x

π/2

0

(a)

π 2

(b)

(c)

π 6

(d) None of these

1

π 4 (c) 1

40.



∫ 0

41. If un =



0

(d)

π

0

x tan x dx is: sec x + cos x

3π 2 4

(b)

1 2n + 1

π2 3

2 (d) π 2

48. The value of the integral b are integers), is:

1 (b) n +1

1 2n − 1



2 (c) π 4

tan n x dx , then un + un – 1:

1 (a) n −1 (c)

(a)

(b) 2 (d) 4 π/4

 x + 2 dx where [ ] is the greatest   (b) 22 (d) None of these

47. The value of

x dx is equal to 2

(a) zero (c) 8

9

0

(a) 31 (c) 23

π2 32

(d) None of these

1 + sin



π 4

integer function: (b)

(a)

then the value of I is:

sin x + cos x

46. Find the value of

tan − 1 x 39. ∫ dx is equal to 1 + x2 0

2



π

−π

(cos ax − sin bx) 2 dx (a and

(a) –π

(b) 0

(c) π

(d) 2π 

Answers

1. 11. 21. 31. 41.

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

2. 12. 22. 32. 42.

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

3. 13. 23. 33. 43.

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

4. 14. 24. 34. 44.

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

5. 15. 25. 35. 45.

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

6. 16. 26. 36. 46.

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

7. 17. 27. 37. 47.

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

8. 18. 28. 38. 48.

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

9. 19. 29. 39.

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

10. 20. 30. 40.

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

8

Differential Equations

CHAPTER

Summary of conceptS Differential equation an equation involving an independent variable, a dependent variable and the derivatives of the dependent variable, is called a differential equation. a few examples of differential equations are as follows: dy 1. x + y = x3, dx d3y d2y dy + 2 +6 + 7 y = 0, 2. 3 2 dx dx dx 2

 d2y   dy  3.  2  + 4   + 3 y = 0, dx  dx      dy  2  4. 1 +      dx  

3

3/ 2

=

d3y . dx 3

orDer of a Differential equation The order of the highest derivative appearing in a differential equation is called the order of the differential equation.

Degree of a Differential equation The power of the highest order derivative appearing in a differential equation, after it is made free from radicals and fractions, is called the degree of the differential equation. For example in the above examples of differential equations the

• • • •

are also solutions of this differential equation. The solution y = sin x + c, where c is an arbitrary constant, is called the general solution of (1), since every solution of (1) can be obtained from y = sin x + c for a suitable choice of c.

general anD particular SolutionS The solution of a differential equation which contains a number of arbitrary constants equal to the order of the differential equation is called the general solution. Thus, the general solution of a differential equation of the nth order has n arbitrary constants. a solution obtained by giving particular values to arbitrary constants in the general solution is called a particular solution.

formation of a Differential equation Let f (x, y, c1, c2, ... cn) = 0 be the solution of a differential equation, where c1, c2, ..., cn are n arbitrary constants. If we eliminate these n constants, we obtain the differential equation of the nth order satisfied by the given solution values. Any equation taken together with n relations obtained by differentiating it n times helps us to eliminate the n constants. Working Rule 1. Write the given equation. 2. Differentiate the given equation w.r.t. independent variable x as many times as the number of arbitrary constants.

differential equation (1) is of first order and first degree, differential equation (2) is of order 3 and degree 1, differential equation (3) is of order 2 and degree 2, and differential equation (4) is of order 3 and degree 2.

Solution of a Differential equation any relation between the dependent and independent variables (not involving the derivatives) which, when substituted in the differential equation reduces it to an identity is called a solution of the differential equation. For example, the solution to the differential equation: dy = cos x ...(1) dx is y = sin x. However y = sin x + 1, y = sin x – 3

3. Eliminate the arbitrary constants with the help of the given equation and the equations obtained by differentiation to get the required differential equation.

Solution of firSt orDer anD firSt Degree Differential equationS The following methods may be used to solve first order and first degree differential equations. 1. Variable Separable Differentiable equations a differential equation of the form dy =0 dx f (x) dx + g (y) dy = 0, f (x) + g (y)

or

is said to have separated variables.

...(2)

306

Objective Mathematics

We substitute x = X + h and y = Y + k in eqn. (5), where h, k are constants to be determined suitably. dx dy We have = 1 and = 1, so that dX dY

Integrating eqn. (2), we obtain dy

∫ f ( x) dx + ∫ g ( y) dx dx

= c,

where c is an arbitrary constant. (2).

dy dy d Y d X d Y = ⋅ ⋅ = . dx d Y d X dx d X

Hence ∫ f ( x) dx + ∫ g ( y ) dy = c is the solution of eqn.

2. equations reducible to Variable Separable form Sometimes in a given differential equation, the variables are not separable. But, some suitable substitution reduces it to a form in which the variables are separable. For example, the differential dy equations of the type = f (ax + by + c) can be reduced to varidx able separable form by substituting ax + by + c = t. The reduced variable separable form is : dt = dx. b f (t ) + a

3. Homogeneous equations a differential equation of the form ...(3)

...(4)

To solve this equation, we put y = Vx, where V is a function of x. Differentiating y = Vx w.r.t x dy dV . =V+x dx dx Substituting in eqn. (2), we obtain V + x d V . = φ (V) dx ⇒

...(6)

ah + bk + c = 0, ah + Bk + C = 0. These equations give bC −Bc ac − aC ,k= ⋅ aB − a b a B − ab

...(7)

Now equation (6) becomes d Y aX + bY = , d X aX + BY which being a homogeneous equation can be solved by means of the substitution Y = VX. type ii: Consider a differential equation of the form

where f (x, y) and g (x, y) are homogeneous functions of x and y of the same degree, i.e., degree of each term in f (x, y) and g (x, y) is same, is called a homogeneous differential equation. If f (x, y) and g (x, y) are homogeneous functions of degree n each, then we can write f (x, y) = xn f1 (y/x) and g (x, y) = xn g1 (y/x).  y dy Now eqn. (3) takes the form =φ   dx x

a X + b Y + ( ah + bk + c ) dY = d X a X + BY + (a h + B k + C) Choose h and k so that

h=

Integrating both sides to obtain the solution of this differential equation.

dy f ( x, y ) = dx g ( x, y )

Now eqn. (5) becomes

dx dV = , x φ(V) − V

which being a differentiable equation with separated variables, can be solved. 4. equations reducible to the Homogeneous form type i: Consider a differential equation of the form : dy ax + by + c a b , where ...(5) = ≠ dx ax + By + C a B This is clearly non-homogeneous. In order to make it homogeneous, we proceed as follows :

a b dy ax + by + c , where = k (say) = = a B d x ax + By +C Since a B – a b = 0, the above method fails in view of eqn. (7). We have

dy k (a x + B y ) + c = dx ax + By +C

...(8)

Substitute ax + By = z so that a+B Now eqn. (8) becomes

dy dz = ⋅ dx dx

dz kz + c = B⋅ + a, dx z+C which is an equation with variables separable. 5. linear Differential equations a differential equation of the form dy + Py = Q, dx

...(9)

where P and Q are functions of x (or constants), is called a linear differential equation of the first order. Working Rule For Solving

dy + Py = Q. dx

1. Find integrating factor (I. F.) = e ∫

Pdx

.

2. The solution of the differential equation is y (I.F) =

∫ Q (I.F) dx + c,

where c is constant of integration.

where R and S are functions of y alone (or constants). The integrating factor in this case is given by

 x  y dx − x dy (iii) d   = y2  y

I.F. = e ∫ . The solution of this equation is

y x dy − y dx (iv) d   = x x2  

R dy

x ⋅ (I.F.) =

∫ (S x (I.F.) ) dy + C ,

(v) d (log xy) =

where C is the constant of integration. 2. The fact elogt = t will be frequently used in the solution of linear equations. 6. Equations Reducible to the Linear Form Consider a differential equation of the form:

y dx + x dy xy

y x dy − y dx (vi) d  log  = x xy  (vii) d (xm yn) = xm – 1 yn – 1 (my dx + nx dy)

dy n + Py = Q y ,  dx

...(10)

1 x dx + y dy (viii) d  log ( x 2 + y 2 )  = x2 + y 2 2 

where P and Q are functions of x. This equation can be reduced (ix) d  1 log x + y  = x dy − y dx   to the linear form as follows: x− y x2 − y 2 2 n Dividing both sides of eqn (10) by y , we get y x dy − y dx (x) d  tan −1  = dy Py –n+1 = Q  ...(11) y– n x x2 + y 2  dx d [ f ( x, y ) ] 1-n dy dz f ¢ ( x, y ) Put y– n + 1 = z ⇒ (– n + 1) y– n = ⋅ = (xi) n dx dx 1- n f [ ( x, y ) ] Substituting in eqn. (11), we obtain x dx + y dy (xii) d x 2 + y 2 = dz x2 + y 2 + (1 − n) Pz = (1 – n) Q, dx which is now a linear equation with z as the dependent variable.

(

)

mULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1. If φ(x) is a differentiable function then the solution of dy + (yφ′ (x) – φ(x) φ′(x)) dx = 0 is (a) y = (φ(x) – 1) + ce (b) yφ(x) = (φ(x))2 + c (c) yeφ(x) = φ(x) eφ(x) + c (d) (y – φ(x)) = φ(x) e–φ(x)

–φ(x)

(b) 2 (d) None of these

dy + 4x2 tan y = e x sec y satisfying 3. The solution of x dx y(1) = 0, is 3

x

(a)  tan y = (x–2)e logx (c) tan y = (x – 1) ex x–3

y2 = 4a (x + a) is a solution, is (a) 1 (c) 3

2. The order of the differential equation, of which xy = ce x + be – x + x2 is a solution, is (a) 1 (c) 3

4. The degree of the differential equation, of which

(b) sin y = e (x – 1)x (d) sin y = ex(x – 1)x–3 x

–4

(b) 2 (d) None of these

5. The differential equation of the family of curves y = e x (A cos x + B sin x), where A and B are arbitrary constants, is 2 (a) d y + 2 dy + 2 y = 0 dx 2 dx 2 (b) d y − 2 dy − 2 y = 0 dx 2 dx 2 (c) d y − 2 dy + 2 y = 0 dx 2 dx

(d) None of these

307

Solution by Inspection The following derivatives must be remembered as they 1. Sometimes a first order differential equation which is not are very useful in solving some differential equations expressible as eqn (9) becomes a linear equation of the directly. form (i) d (x + y) = dx + dy dx + Ry = S, (ii) d (xy) = ydx + xdy dy

Differential Equations

Remarks

308

6. The differential equation of all circles passing through the origin and having their centres on the x-axis is

Objective Mathematics

(a) y2 = x2 + 2xy (c) x2 = y2 + xy

dy dx

(b) y2 = x2 – 2xy

dy dx

dy dx

(d) None of these

7. The differential equation of family of parabolas with foci at the origin and axis along the x-axis is dy  dy  − y = 0 (a) y   + 2 x  dx  dx 2

(c) ex + 2 = 3

dy  dy  (c) y   + 2 x + y  = 0  dx  dx 2

8. The differential equation that represents all parabolas each of which has a latus rectum 4a and whose axes are parallel to x-axis, is d 2 y  dy  d 2 y  dy  +   = 0 (b) 2a 2 +   = 0 2  dx   dx  dx dx 3

3

d 2 y  dy  −   = 0 (d) None of these dx 2  dx  3

9. The differential equation of all parabolas whose axes are parallel to the axis of y, is 3 (b) d y = – 1 dx 3

(c) d y = 0 dx 3 3

(d) None of these

10. The degree of differential equation dy 1 dy 1 dy x = 1 +   +   +   + ............ is  dx    2! dx 3!  dx  2

(a) three (c) not defined

3

(b) one (d) None of these

x2+ y 2+ 1 11. The solution of dy = , satisfying y(1) = 0, is 2 xy dx given by (a) a hyperbola (b) a circle (c) y2 = x(1 + x) – 10 (d) (x – 2)2 + (y – 3)2 = 5 12. The order and degree of the differential equation dy   1 + 3  dx

2/3

2 cos y (d) None of these

2 15. The solution of the equation d y = e –2x is y = dx 2

(d) None of these

3 (a) d y = 1 dx 3

(a) (x + a) (1 + ay) = – 4a2y (b) (x + a) (1 – ay) = 4a2y (c) (x + a) (1 – ay) = – 4a2y (d) None of these cos y dx + (1 + 2e– x) sin y dy = 0, when x = 0, y = π is 2 (a) ex – 2 = 3 2 cos y (b) ex + 2 = 2 cos y

2

(c) 2a

dy y – x dy = a  y 2 +  is  dx  dx

14. The particular solution of

dy  dy  − y = 0 (b) x   + 2 y  dx  dx

(a) a

13. The equation of the curve passing through the point 1   a, −  and satisfying the differential equation a

3 = 4 d y are dx 3

2 (a) 1,   3

(b) (3, 1)

(c) (3, 3)

(d) (1, 2)

−2 x −2 x (a) e (b) e + cx + d 4 4 (c) 1 e–2x + cx2 + d (d) None of these 4 16. The equation of the family of curves for which subnormal is constant, is

(a) y2 = cx + k (c) x2 = 2cy + k

(b) y2 = 2cx + k (d) None of these

17. The equation of the curve for which the cartesian subtangent varies as the reciprocal of the square of the abscissa, is (a) x = c e y (c) y = c e

3 / 3k

x3 / 3 k



(b) x = c e y



(d) None of these

2 / 3k



18. A curve is such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2). The equation of the curve is (a) xy = 1 (c) xy = 3

(b) xy = 2 (d) None of these

19. The equation of the curve through the point (1, 1) and 2ay , is whose slope is x ( y − a) (a) ya . x2a = ey – 1 (b) ya. x2a = ey 2a a y – 1 (c) y . x = e (d) None of these 20. Solution of the differential equation (x + 2y 3) dy = y is dx (a) x = y2 (c + y2) (b) x = y (c – y2) (c) x = 2y (c – y2) (d) x = y (c + y2) 21. The differential equation of all non-vertical lines in a plane is 2 (a) d y = 0 dx 2 (c) dy = 0 dx

d 2x =0 dy 2 dx (d) =0 dy (b)

23. The degree and order of the differential equation of the family of all parabolas whose axis is x-axis, are respectively (a) 2, 1 (c) 3, 2

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

π 24. The equation of a curve passing through 1,  and  4 sin 2 y at (x, y) is having slope x + tan y (a) x = tan y (c) x = 2 tan y

(b) y = tan x (d) y = 2 tan x

25. The equation of the curve satisfying the equation −1 y

(1 + y2) dx + ( x − e − tan ) dy = 0 and passing through origin is −1 y

= cot– 1y

−1 y

= tan– 1y

(a) x ⋅ e tan (b) x ⋅ ecot (c) y ⋅ e

tan −1 x

= tan x

−1 y

= tan– 1y

(d) x ⋅ e tan

26. The solution of the differential equation (1 + y2) + (x – e

dy ) = 0, is dx

(a) (x – 2) = k (b) 2x e tan (c) x e tan

−1 y

−1 y

(d) x e 2 tan

−1 y

= e 2 tan

−1 y

+k

= tan–1 y + k tan −1 y

x (b) log y = cos   + C  y x (c) log x = cos   + C  y (d) None of these 32. The general solution of the differential equation  2 xy − x  dy + ydx = 0 is   x =C y

(a) log x +

y = C x

(b) log y −

(c) log y +

x = C y

(d) None of these

33. The general solution of the differential equation

– 1

− tan −1 y

y (a) log x = cos   + C x

+k

dy + y g′ (x) = g (x). g′ (x), dx where g (x) is a given function of x, is (a) g (x) + log [1 + y + g (x)] = C (b) g (x) + log [1 + y – g (x)] = C (c) g (x) – log [1 + y – g (x)] = C (d) None of these 34. Solution of the equation (x – y) (2dy – dx) = 3dx – 5dy is (a) 2y – x = log (x – y + 2) + C (b) 2x – y = log (y – x + 2) + C (c) 2y + x = log (x – y + 2) + C (d) None of these

= e 27. The equation of the curve which passes through the point (2a, a) and for which the sum of the cartesian sub tangent and the abscissa is equal to the constant xdy − ydx 35. Solution of the equation xdx + ydy + 2 = 0 is a, is x + y2 (a) y (x – a) = a2 (b) y (x + a) = a2  c + x2 + y 2  (c) x (y – a) = a2 (d) x (y + a) = a2. (a) y = x tan   2  28. The general solution of the differential equation dy = y dx  c + x2 + y 2  tan x – y 2 sec x is (b) x = y tan   2  (a) tan x = (c + sec x) y (b) sec y = (c + tan y) x (c) sec x = (c + tan x) y (d) None of these  c − x2 − y 2  (c) y = x tan   2  29. The equation of the curve satisfying the equation (xy – x2) (d) None of these dy = y 2 and passing through the point (–1, 1) is 36. The equation of the curve satisfying the differential dx equation y (x + y 3) dx = x (y 3 – x) dy and passing through (a) y = (log y – 1) x (b) y = (log y + 1) x the point (1, 1) is (c) x = (log x – 1) y (d) x = (log x + 1) y

309

Differential Equations

22. The equation of a curve passing through (0, 1) and hav- 30. The general solution of the differential equation y (x2y + e x) dx – e x dy = 0 is − ( y + y3 ) at (x, y) is ing gradient 1 + x + xy 2 (a) x3y – 3ex = cy (b) x3y + 3ex = cy (c) y3x – 3ey = cx (d) y3x + 3ey = cx (a) xy + tan– 1y = π (b) xy + tan– 1y = π 31. Solution of the differential equation 2 4 y y   (c) xy – tan– 1y = π (d) xy – tan– 1y = π x sin dy =  y sin − x  dx is   x x 2 4

310

(a) y3 – 2x + 3x2y = 0 (c) y3 + 2x – 3x2y = 0

(b) y3 + 2x + 3x2y = 0 (d) None of these

Objective Mathematics

37. A curve passes through the point (5, 3) and at any point (x, y) on it, the product of its slope and the ordinate is equal to its abscissa. The curve is a (a) parabola (c) hyperbola

(b) ellipse (d) circle

38. Differental equation of y = sec (tan – 1 x) is (a) (1 + x 2 ) dy = y + x + c dx dy 2 =y–x+c (b) (1 + x ) dx (c) (1 + x 2 ) dy = xy + c dx x dy = +c (d) (1 + x 2 ) y dx 1  2 2  xy − e ⋅ 3  dx –x y dy = 0, given y = 0 when x = 1 is x e (a) y = 2 (c) y =

1 2  4 − x  x

e1 2  + x  3  x4

(a) x log (xy) = 2 (x – y) (b) y log (xy) = 2 ( y – x) (c) x log (xy) = 2 (x + y) (d) None of these 45. If y+ d (xy) = x (sin x + log x), then dx (a) y = cos x + 2 sin x + 2 cos x + x log x − x + c x x2 3 9 x2 (b) y = − cos x − 2 sin x + 2 cos x + x log x − x + c x x2 3 9 x2 (c) y = − cos x + 2 sin x + 2 cos x − x log x − x + c x x2 3 9 x2 (d) None of these

39. The solution of the equation

2

y (2 y − x) , 44. The solution of the differential equation dy = x (2 y + x) dx which satisfies y = 1 at x = 1, is

e 1 (b) y =  4 − x 2   3 x 2

(d) None of these

40. The solution of the equation  y2  y sin x dy = cos x  sin x −  , 2  dx given y = 1 when x = π is 2 (a) y2 = sin x (b) y2 = 2 sin x 2 (c) x = sin y (d) x2 = 2 sin y  f ( y / x)  41. Solution of the equation x dy =  y + x dx is f ' ( y / x)   x (a) f   = cy  y

y (b) f   = cx x

y (c) f   = cxy x

(d) None of these

46. The general solution of the differential equation (1 + tan y) (dx – dy) + 2xdy = 0 is (a) x (sin y + cos y) = sin y + cey (b) x (sin y + cos y) = sin y + ce– y (c) y (sin x + cos x) = sin x + cex (d) None of these 47. Solution of the equation dy = e x – y (e x – e y) is dx x

(a) ey = e x − 1 + ce − e y

(c) ex = e y − 1 + ce − e

(b) ey = e x − 1 + cee

x

(d) None of these

dy dy 48. Solution of the equation x   + ( y − x ) − y = 0 is  dx  dx 2

(a) (x – y + c) (xy – c) = 0 (b) (x + y + c) (xy – c) = 0 (c) (x – y + c) (2xy – c) = 0 (d) ( y – x + c) (xy – c) = 0 49. Solution of equation (xy4 + y) dx – xdy = 0 is (a) 4x4y3 + 3x3 = cy3 (c) 3x4y3 + 4x3 = cy3

(b) 3x3y4 + 4y3 = cx3 (d) None of these

50. A curve C has the property that if the tangent drawn 42. The solution of the equation log dy = 9x – xy + 6, given at any point P on C meets the coordinate axis at A and dx B, then P is the mid- point of AB. If the curve passes that y = 1 when x = 0, is through the point (1, 1), then the equation of the cure is (a) 3e6y = 2e9x– 6 + 6e6 (a) xy = 2 (b) xy = 3 (b) 3e6y = 2e9x + 6 – 6e6 (c) xy = 1 (d) None of these (c) 3e6y = 2e9x + 6 + e6 (d) None of these 43. The differential equation of all straight lines passing through origin is (a) y = (d)

x

dy dx

dy = y – x dx

(b) dy = y + x dx (d) None of these

51. A normal is drawn at a point P (x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k, then the differential equation describing such a curve is 2 2 (a) y dy = ± k 2 − y 2 (b) x dy = ± k − x dx dx

(c) y dy = ± dx

y 2 − k 2 (d) x dy = ± x 2 − k 2 dx

(b) x3 = 3y3 sin x (d) None of these

(a) y3 = 3x3 sin x (c) x3 = y3 sin x

(a) 5 (c) 3

53. The differential equation which is satisfied by all the curves, y = Ae 2x + Be –x/2, where A and B are non-zero constants, is

2 (b) 2 d y + 3 dy − 2 y = 0 dx 2 dx

54. The degree of the differential equation of all tangent lines to the parabola y 2 = 4ax is (b) 2 (d) None of these

55. A solution of the differential equation dy  dy  + y = 0 is   − x dx dx 2

(b) y = 2x (d) y = 2x2 – 4

)

ves y 2 = 2c x + c , where c is a positive parameter, is of (a) order 1 (c) degree 3

(b) order 2 (d) degree 4

57. The order of the differential equation whose general solution is given by y = c1 cos (2x + c2) – (c3 + c4)  a x + c5 + c6 sin (x – c7) is (a) 3 (c) 5

(b) 4 (d) 2.

58. The degree of the differential equation d4y   dx 4 

3/ 5

−5

d3y d2y dy +6 2 −8 + 5 = 0 is 3 dx dx dx

(a) 2 (c) 4

(b) 3 (d) 5

59. The order of the differential equation satisfying 1 − x 4 + 1 − y 4 = a (x2 – y2). is (a) 1 (c) 3

(b) 2 (d) None of these

60. The degree of the differential equation d3y  dy  d 4 y is + x   = 4 log 3  dx  dx dx 4 (a) 1 (b) 3 (c) 4 (d) None of these 4

311

63. The solution of the equation, x 2 d y = log x , when x = dx 2 1, y = 0 and dy = −1 is dx (a)   y = 1 (log x) 2 + log x 2 (b)   y = 1 (log x) 2 − log x 2 (c)   y = − 1 (log x) 2 + log x 2 (d)   y = − 1 (log x) 2 + log x 2 64. Which of the following is a solution of the differential 2

56. The differential equation representing the family of cur-

(

62. 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

2

2 (c) 2 d y − 3 dy + 2 y = 0 dx 2 dx (d) None of these

(a) y = 2 (c) y = 2x – 4

(b) 4 (d) 2

(a) 6x + 9y + 2 = 26e3 (x – 1) (b) 6x – 9y + 2 = 26e3 (x – 1) (c) 6x + 9y – 2 = 26e3 (x – 1) (d) None of these

2 (a) 2 d y − 3 dy − 2 y = 0 dx 2 dx

(a) 1 (c) 3

61. The order of the differential equation whose general solution is given by y = (c1 + c 2) cos (x + c 3) – c4 e x + c5 where c1, c 2, c3, c4, c5 are arbitrary constants, is

Differential Equations

52. Solution of the equation x2y – x3 dy = y4 cos x, when dx y (0) = 1 is

dy  dy  equation   − x + y=0 dx dx   (a)  y = 2x – 4 (c)  y = 2

(b)  y = 2x2 – 4 (d)  y = 2x

65. Given that, dy = ye x such that x = 0, y = e. The value dx of y(y > 0), when x = 1 will be (a)  e (c)  ee

(b)   1 e (d)  none of these

dy 2 yx 2 + = 66. The solution of the differential equation dx 1 + x 2 (1 + x 2 ) dy 2 yx 2 + = is dx 1 + x 2 (1 + x 2 ) (a)  y(1 + x2) = c + x y = c + tan −1 x 1 + x2 (c)  y log (1 + x2) = c + tan–1 x

(b)  

(d)  y(1 + x2) = c + sin–1 x 67. The solution of the differential equation x dy − y dx = x 2 + y 2 dx is 2 2 2 (a)   x + x + y = cx

2 2 (b)   y − x + y = cx

(c)   x − x 2 + y 2 = cx

2 2 2 (d)   y + x + y = cx

68. The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants, is of

312

Objective Mathematics

(a)  first order and second degree (b)  first order and first degree (c)  second order and first degree (d)  second order and second degree

76. The differential equation

(a)  variable radii and a fixed centre at (0, 1) (b)  variable radii and a fixed centre at (0, –1) (c)  fixed radius 1 and variable centres along the x-axis (d)  fixed radius 1 and variable centres along the y-axis

(b)  x + y3 = cy (d)  none of these

70. The equation of family of a curve is y 2 = 4a(x + a), then differential equation of the family is (a)  y = y’ + x (c)  y = 2y’ x + yy’2

(b)  y = y” + x (d)  y” + y’ + y2 = 0

71. The solution of the differential equation dy 2x 1 is + y= dx 1 + x 2 (1 + x 2 ) 2 (a)  y(1 – x ) = tan x + c (b)  y(1 + x2) = tan–1 x + c (c)  y(1 + x2)2 = tan–1 x + c (d)  y(1 – x2)2 = tan–1 x + c 2

–1

(c)   log ( y + x 2 + y 2 ) + log y + c = 0 x (d)   sinh −1   + log y + c = 0  y 2 73. The solution of the equation d y = e −2 x is dx 2

(a)   y = 1 e −2 x + cx + d (b)   y = 1 e −2 x + cx + d 4 2 4 (c)   y = 1 e −2 x + cx 2 + d (d)   y = 1 e −2 x + cx 3 + d 4 4

77. The differential equation of all coaxial parabolas y 2 = 4a (x – b), where a and b are arbitrary constants, is 2

2 d 2 y  dy  (b)   y 2 +   = 1 (a)   y d y + dy = 1 dx  dx  dx 2 dx 2 2 d 2 y  dy  (c)   y 2 +   = 0 (d)   y d y + dy = 0 dx  dx  dx 2 dx

78. The solution of the differential equation ( x 2 − yx 2 ) dy + y 2 + xy 2 = 0 dx dy ( x 2 − yx 2 ) + y 2 + xy 2 = 0 is dx x 1 1 x 1 1 (a)   log   = + + c (b)   log   = + + c y x y    y x y

72. The general solution of y 2 dx + (x2 – xy + y 2) dy = 0 is −1  x  (a)   tan   + log y + c = 0  y −1  x  (b)   2 tan   + log x + c = 0  y

determines a

family of circles with

69. Solution of (x + 2y 3) dy = y dx is (a)  x = y3 + cy (c)  y2 – x = cy

1 − y2 dy = dx y

(c)   log( xy ) =

1 1 1 1 + + c (d)   log( xy ) + + = c x y x y

79. The differential equation of all circles which pass through the origin and whose centre lies on y-axis is (a)   ( x 2 − y 2 ) dy − 2 xy = 0 dx (b)   ( x 2 − y 2 ) dy + 2 xy = 0 dx dy (c)   ( x 2 − y 2 ) − xy = 0 dx (d)   ( x 2 − y 2 ) dy + xy = 0 dx 80. The differential equation of the family of circles with fixed radius 5 unit 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 – 2)2y´2 = 25 – (y – 2)2

dy = y tan xx –− 2 sin 81. x y = 2e 2x – e –x is a solution of the differential equation dx sin x, is (a)  y2 + y1 + 2y = 0 (b)  y2 – y1 + 2y = 0 1 (c)  y2 + y1 = 0 (a)  y sin x = c + sin 2x (b)   y cos x = c + sin 2 x 2 (d)  y2 – y1 – 2y = 0 1 (c)  y cos x = c – sin 2x (d)   y cos x = c + cos 2 x 1 − y2 dy 2 82. The differential equation determines a fam= dx y 75. The differential equation of all circles passing through ily of circles with the origin and having their centres on the x-axis is (a)  variable radii and a fixed centre at (0, 1) (b)   x 2 = y 2 + 3 xy dy (a)   x 2 = y 2 + xy dy (b)  variable radii and a fixed centre at (0, –1) dx dx (c)  fixed radius 1 and variable centres along the x-axis (c)   y 2 = x 2 + 2 xy dy (d)   y 2 = x 2 − 2 xy dy (d)  fixed radius 1 and variable centres along the y-axis dx dx

74. The solution of the differential equation

5. (c) We have, y = ex (A cos x + B sin x) ...(1) Differentiating w.r.t. x, we get dy = ex (A cos x + B sin x) + ex (– A sin x + B cos x) dx

∴ The solution is y.e

φ(x)

=

∫e

φ( x)

= y + ex (– A sin x + B cos x)

.φ( x.).φ' ( x) dx = φ(x)e

φ(x)

– e

φ(x)

+c

y = (φ(x) – 1) + c.e–φ(x). 2. (b) We have, xy = ce x + be – x + x2 Differentiating w.r.t. x, we get dy x + y = cex – be– x + 2x dx Differentiating again,

...(1) ...(2)

d y dy = cex + be– x + 2 = xy – x2 + 2 +2 dx 2 dx [Using (1)] Hence, the required differential equation is x

x

dy    x + y  . dx

dy   dy . x2 + y2 – 2x  x + y  = 0, i.e., y2 = x2 + 2xy  dx  dx 7. (a) Let the directrix be x = – 2a and latus rectum be 4a. Then, the equation of the parabola is (distance from focus = distance from directrix), ...(1) x2 + y2 = (2a + x)2 or y2 = 4a (a + x)  Differentiating w.r.t. x, we get dy 1 dy = 2a or a = y y ⋅ dx 2 dx

sin y  x4 = xex – ex + c x = 1, y = 0  ∴ c = 0 ∴ sin y = ex (x – 1) x–4.

...(1)

dy y dy = 4a i.e., a = ⋅ dx 2 dx

Substituting the value of a in (1), we get y dy    x + 2 dx  dy  dy  + y   dx  dx

...(1)

Putting this value of g in (1), we get

4 log x 4 = x4 dx = e e∫ x The solution is ex ∴ tx4 = ∫ x 4 . 3 dx = xex – ex + c x

or y = 2 x

[from (1) and (2)]

d2y dy + 2y = 0, −2 dx 2 dx

where g is an arbitrary constant. Differentiating w.r.t. x, we get dy x +y + g = 0. i.e., g = – dx

I.F. =

dy dx

 dy  dy – y +  − y   dx  dx

x2 + y2 + 2gx = 0,

dy 4 ex 3. (b) We have, cos y  +  sin y = 3 dx x x dy dt Let sin y = t  ⇒  cos y  = dx dx dt 4 ex ∴ +   t= 3 dx x x

y2 = 2 y

=

6. (a) The equation of circles passing through the origin and having their centres on the x-axis is

The order of this differential equation is 2.

2y

d2y dy =  – ex (A cos x + B sin x) dx 2 dx  + ex (– A sin x + B cos x)

which is the required differential equation.

d2y dy +2 − xy + x 2 − 2 = 0. 2 dx dx

4. (b) We have y2 = 4a (x + a) Differentiating w.r.t. x, we get

...(2)

Differentiating again,

or

2

[from (1)] 

2

  dy  2  dy , or y 1 −  dx   = 2x   dx which is the required differential equation. The degree of the differential equation is 2.

Putting this value of a in (1), the differential equation is dy  y dy  + x y2 = 2 y   dx  2 dx  dy   dy  or y   + 2 x   − y = 0.  dx   dx  2

8. (b) Equation of the family of such parabolas is ( y – k)2 = 4a (x – h) where h and k are arbitrary constants. Differentiating w.r.t. x, we get dy = 2a  dx Differentiating again,

( y – k)



( y– k)

...(1)

...(2)

d 2 y  dy  +   = 0 dx 2  dx  2

...(3)

Differential Equations

1. (a) dy + (yφ′(x) – φ(x) φ′(x)) dx = 0 dy + yφ′(x) = φ(x) φ′(x) dx φ' ( x ) dx I.F = e ∫ = eφ(x)

313

solutions

314

Putting value of y – k from (2) in (3), we get

2a

14. (c) We have, cos y dx + (1 + 2e– x) sin y dy = 0

Objective Mathematics

d y  dy  +   = 0, dx 2  dx 

which is the required differential equation. 9. (c) Such parabolas are given by

d y = 0. dx 3 This is the required differential equation.

∴ The solution is ex + 2 = 3 15. (b)

 dy   

dy = ln x dx ∴  order = 1  and degree = 1.

10. (b) x = e  dx  ⇒

x + y +1 dy = ⇒ 2xydy = (x2 + 1) dx + y2dx 2 xy dx 2



2

=



1 dx 2  

∫ 1 + x

Integrating ,

⇒ y dx – x dy = ay2 dx + a dy ⇒ y (1 – ay) dx = (x + a) dy dx dy − =0 ⇒ x + a y (1 − ay )

dy = c (a constant) dx

y2 k = cx + i.e., y2 = 2cx + k. 2 2

17. (c) We have, Cartesian sub-tangent α i.e.,

y k = 2 dy / dx x

Integrating, log y =

3

 2 dy  dy = a y +  13. (c) We have, y – x  dx  dx

e −2 x + cx + d. 4

or y dy = c dx.

x = 1, y = 0,  ∴  0 = 1 – 1 + c ⇒ c = 0 y2 = x2 – 1  ⇒ x2 – y2 = 1. 2

dy e −2 x = +c dx −2

16. (b) Cartesian sub-normal = y

y2 1 =x– +c ⇒ y2 = (x2 – 1 + cx) x x

 d3y  dy   12. (d) 1 + 3  =  4 3  . dx  dx 

2 cos y.

d y = e–2x dx 2

⇒ y =

 x2+ 1  xd ( y 2 ) − y 2 dx =  2  dx 2 x  x 

2.

2



2

∫ d ( y / x)

1 2 i.e., c = 3



Differentiating once again,



− sin y ex ∫ e x + 2 dx − ∫ cos y dy = log c

⇒ ex + 2 = c cos y, where 1 + 2 = c.

2

3





 ex + 2  = log c ⇒ log   cos y 

d y d y 1 i.e. 2 = ⋅ dx 2 dx 2a 2

Differentiating again, 1 = 2a



dx sin y + dy = 0 1 + 2e − x cos y

x ⇒ log (e + 2) − log cos y = log c

(x – h)2 = 4a (y – k), where h, k, a are three arbitrary constants. dy . Differentiating w.r.t. x, (x – h) = 2a dx

11. (a)



3

2

or

1 square of abscissa dy x2 = dx y k

x3 + log c or y = . 3 3k ce x / 3k

18. (b) L et P (x, y) be any point on the curve, PM the perpendicular to x-axis PT the tangent at P meeting the axis of x at T. As given OT = 2. OM = 2x. Equation of the tangent at P (x, y) is

Integrating, we get

log (x + a) – log y + log (1 – ay) = log C

or log 

(a + x ) (1 − ay ) y = log C i.e., (x + a) (1 – ay) = Cy.

1  Since the curve passes through  a, −  ,  a C i.e., C = – 4a2. ∴ 2a × (1 + 1) = − a So, (x + a) (1 – ay) = – 4a2y.

dy (X − x ) dx It intersects the axis of x where Y = 0 Y– y=

i.e., – y = OT.

dx dy (X – x) or X = x – y = dy dx

2ay dy = x ( y − a) dx y−a 2a dy = dx y x

19. (a) We have, slope = ⇒

Integrating both sides, we get a log | y | – y = – 2a log | x | + log c ⇒ ya . x2a = cey.

or xy + tan– 1 y = C. This passes through (0,1), therefore, tan– 11 = C π . i.e., C = 4 Thus, the equation of the curve is π . ny + tan– 1 y = 4 23. (b) Parabola, whose axis is x-axis, is given by y2 = 4ax Differentiating, we get dy dy = 4a ⇒ y = 2a 2y dx dx Again differentiating, we get

d 2 y  dy  y 2 +  = 0  his passes through (1, 1), therefore 1 = ce i.e., c = T  dx  dx 1 . So, the equation of the curve is ya . x2a = ey – 1. ∴  degree = 1, order = 2. e 20. (d) We have, 24. (a) We have, sin 2 y dy x tan y dx dy dy − = = y ⇒ y = x + 2y3 (x + 2y3) = ⇒ x + tan y dx sin 2 y sin 2y dy dx dx dy dx 1 I.F. = − ∫ sin 2 y − x = 2y2, ⇒ e = elog cot y = cot y . dy y Hence the solution is which is a linear equation, if we take x as the detan y pendent variable. x cot y = ∫ ⋅ cot y dy + C. sin 2 y 1 log   2 1 1  y 1 sec y − ∫ dy = e– log y = e = ⋅ I.F. = ∫ pdy = e y e = ∫2 dy + C. = tan y + C y tan y 1 2 1 ∴ The solution is x. = ∫ 2 y ⋅ dy + c  π S ince the curve passes through 1,  , therey y  4 1 2 2 fore, 1 = 1 + C i.e., C = 0. = y + c or x = y (c + y ). ⇒ x ⋅  y Thus, the equation of curve is x = tan y. 21. (a) The general equation of all non vertical lines in a 25. (d) We have (1 + y2) dx + (x – e– tan  – 1y) dy = 0 plane is ax + hy = 1, where h ≠ 0 dx 1 1 − tan −1y e + x = ⇒ 2 2 dy d2y d2y dy 1 + y 1+ y ⇒ a + h =0 ⇒h 2 =0 ⇒ = 0. dx dx dx 2 1 I.F. = ∫ 1 + y 2 dy = tan–1y. e − ( y + y3 ) dy 22. (b) We have, = 1 + x 1 + y2 ⋅ Hence the solution is ( ) dx −1y − 1y −1y 1 x ⋅ e tan = ∫ ⋅ e − tan ⋅ e tan dy + C. 2 2 1 + x (1 + y ) 1+ y dx or =– dy y + y3 dy + C = tan– 1y + C = ∫  x (1 + y 2 )  1 1 + y2 +    = –  y ( y 2 + 1) y (1 + y 2 )  This passes through the origin,  ∴ 0 = 0 + C i.e.,   −1 dx x + ⇒ = y 1 + y2 ⋅ ( ) dy y dy = elog y = y. ∫ e y Hence the solution is, 1 x ⋅ y = –  ∫ y 1 + y 2 ⋅ y dy + C ( ) dy + C = – tan– 1 y + C =– ∫ 1 + y2

I.F. =

2

C=0  ence, the equation of the curve is x.e tan H tan–1y. 26. (b) (1 + y 2) + (x – e tan

−1 y

)

dy =0 dx

dx −1 + x = e tan y dy



(1 + y2)



dx 1 e tan y + 2  x = 1 + y2 dy 1 + y

−1

−1y

=

315

dx dy + =0 x y log C i.e., xy = C. ∴  C = 2. xy = 2.

Differential Equations

dx = 2x or dy Integrating, log x + log y = This passes through (1, 2),  Hence the required curve is Hence x – y

316



I.F. = e

∫ p dy

= e



1 dy 1 + y2

Objective Mathematics

The solution is −1 y



x . e tan



x e tan

−1 y



x e tan

−1 y



2x e tan

tan ∫e

=

−1 y

tan = e

−1 y

−1

.

e tan y . dy + k1 1 + y2

−1

−1 y

=

e 2 tan y ∫ 1 + y 2 . dy + k1

=

1 2 tan −1 y e + k1 2

2 tan = e

−1 y

30. (b) We have y (x2y + ex) dx – ex dy = 0 dy = x2y2 + yex ⇒ ex dx 1 dy 1 − Dividing by y2ex, we get 2 = x2e– x y dx y

+k

27. (a) We have, Cartesian subtangent + abscissa = constant dx y +x = a +x = a ⇒ y ⇒ dy dy / dx dy dx =  ⇒ y a−x Integrating, we get

y = log y + C x or y = x (log y + C). This passes through the point (– 1, 1), ∴ 1 = – 1 (log 1 + C)  i.e.,  C = – 1. Thus, the equation of the curve is  y = x (log  y – 1). or

log y + log (x – a) = log C

dV 1 −1 dy = ⋅ = V so that 2 y dx dx y dV We thus have + V = –x2e– x, which is linear dx Put

1dx ∴ I. F. = e ∫ = e x. Hence the solution is

∴ y (x – a) = C.

V⋅ ex = –

As the curve passes through the point (2a, a), we have

or



C = a 2.

Hence the required curve is y (x – a) = a2. dy = y tan x – y2 sec x dx

28. (c) We have ⇒

1 dy 1 − tan x = – sec x. y 2 dx y

Putting

dV 1 −1 dy = =V⇒ 2 , y dx dx y

we obtain

dV + tan x ⋅ V = sec x which is linear. dx

I. F. = e ∫ = elog sec x = sec x. Hence the solution is 1 2 sec x = tan x + c V sec x = ∫ sec x dx + c or y or sec x = y (c + tan x). tan x dx

29. (a) We have, (xy – x2) ⇒ y2 Putting

dy = y2 dx

−1 1 dx 1 1 dx = xy – x2 ⇒ 2 − ⋅ = 2 x dy x y y dy dV −1 dx 1 = = V so that 2 , x dy dy x

we obtain

dV V 1 + = 2 , which is linear. dy y y



I. F. = e

2 −x

. e x dx +

x3 C 1 x e = − + y 3 3

C 3 or

x3y + 3ex = Cy.

y −x x ⋅ y x sin x dy dV Put y = Vx so that =V+ x . dx dx

dy = 31. (a) We have, dx

Hence V + x

y sin

1 dV V sin V − 1 = = V− sin V dx sin V

1 dV = − ⇒ sin V dx

dx + sin V dV = C. x ∫ y ⇒ log x – cos V = C ⇒ log x = cos  +C x y dy = which is homogeneous. 32. (c) We have, x − 2 xy dx ⇒ x



dV dy = x +V dx dx V dy Hence x +V= 1− 2 V dx Put y = Vx so that

⇒ x ⇒

dV V 2V 3 / 2 = −V = dx 1− 2 V 1− 2 V

dx 1− 2 V 1 1 = dV =  3 / 2 −  dV x 2 V3/ 2  2V V

Integrating, we get – C + log x = – V1/2 – log V = –

1

∫ y dy

∫x e

= elog y = y.

Hence the solution is Vy =

1

∫y

2

⋅ y dy + C

or

log y +

x = C. y

x – log y + log x y



dV

∫1− V

=

∫ g' ( x) dx

 ⇒ – log (1 – V) = g (x) – C ⇒ g (x) + log (1 – V) = C ∴ g (x) + log [1 + y – g (x)] = C. 34. (a) We have, (2x – 2y + 5) dy = (x – y + 3) dx x− y+3 dy ⋅ ⇒ = 2 x − y) + 5 ( dx Put x – y = V so that dy dV dy dV = 1− 1− = or dx dx dx dx



∴ The equation becomes V+3 dV dV V+3 V+2 or = =1– 1− = ⋅ 2 V + 5 dx dx 2V + 5 2V + 5 ⇒ dx =

 1  2V + 5 dV. dV =  2 + V + 2  V+2 

Integrating, x = 2V + log (V + 2) + C, ⇒ x = 2 (x – y) + log (x – y + 2) + C as V = x – y. or 2y – x = log (x – y + 2) + C is the required solution. 35. (c) We have, x dx + y dy + ⇒

x dy − y dx =0 x2 + y 2

 y 1 d (x2 + y2) + d tan– 1   = 0 x 2

y 1 c (x2 + y2) + tan– 1 x = 2 2 y ⇒ x2 + y2 + 2 tan– 1 x = c. Integrating,

 c − x2 − y 2  ∴ y = x tan   is the required solution. 2  36. (c) We have, y (x + y3) dx = x (y3 – x) dy ⇒ y3 ( y dx – x dy) + x ( y dx + x dy) = 0  y dx − x dy  ⇒ x2y3.   + xd (xy) = 0  x2 ⇒ –

 y  d ( xy ) y d  + 2 2 = 0 x x y x

Integrating, we get

 y −  x − 2 1 y2 + ⇒ 2 x 2 xy ⇒ y3 + 2x +

1 xy

317

2

=C

+C=0 2cx2y = 0.

Since it passes through the point (1, 1), therefore 1 + 2 + 2C = 0 −3 or C = ⋅ 2  ∴  The curve is y3 + 2x – 3x2y = 0. 37. (c) We have,

dy ⋅ y = x (Given) dx

⇒ y dy = x dx Integrating,

y2 x2 C i.e., y2 = x2 + C. = + 2 2 2

Since the curve passes through the point (5, 3), ∴ 9 = 25 + C ⇒ C = – 16. So the curve is y2 = x2 – 16 i.e., x2 – y2 = 16 which is clearly a rectangular hyperbola. 38. (c) y = sec (tan– 1x) Let, tan–1x = t 1 dx = dt  or  ⇒ 1 + x2 ∴ y = sec t

dt 1 = dx 1 + x2

dy = sec t tan t dt dy sec t tan t xy = = ∴ 2 dx 1+ x 1 + x2 dy ⇒ (1 + x 2 ) = xy + c. dx ⇒

e dx x3 e −1 2 2  (x . 2y dy – y · 2x dx) = 3  dx ⇒ x 2 x2 d ( y 2 ) − y 2 d ( x2 ) − 2e 1 ⇒ dx = ⋅ x4 x3 x 4  y2   y2  −2e − 2e ⇒ d  2  = 7 dx ⇒ ∫ d  2  = ∫ 7 dx x x x  x 

39. (b) We have, xy 2 dx – x2y dy =

x −6 y2 = – 2e x −7 dx = – 2e. +C 2 −6 ∫ x e 1 ⇒ y2 = ⋅ 4 + Cx2. 3 x e Putting y = 0, x = 1 we get 0 = +C 3 −e ⇒C= ⋅ 3 e1 2 ∴ The solution is y2 =  4 − x  .  3 x 40. (a) Put y2 sin x = V. dy dV sin x + y2 cos x = . Then 2y dx dx ⇒

Differential Equations

dy = (g (x) – y). g’ (x) dx dy dV Put g (x) – y = V ⇒ g’ (x) – = dx dx dV Hence g′(x) – = V · g ′ (x) dx dV dV = (1 – V) g’ (x) ⇒ = g ′ (x) dx ⇒ 1 −V dx

33. (b) We have,

318

So, the given equation becomes

1 dV = sin x cos x 2 dx

⇒ 2 

Objective Mathematics

⇒ d V = 2 sin x cos x dx Integrating, we get V = sin2x + C. i.e., y2 sin x = sin2x + C π Putting y = 1, x = , we get 1 = 1 + C 2  i.e., C = 0 2 2 2 So, the solution is y sin x = sin x i.e., y = sin x.

Integrating, we get 2log x + 2V + log V = C ⇒ log (x2V) = – 2V + C

y f ( y / x) dy + = which is homogeneous. x f' ( y / x ) dx dy dV =V+ x , Put y = Vx so that dx dx we obtain



∴ C = log 1 + 2 = 2. So the solution is log (xy) = 2 – i.e., x log (xy) = 2 (x – y). 45. (d) We have, y + ⇒ x ⇒

f (V) dV dx =V+ dV = V+x f' ( V) dx x

∴ The solution is

e e = 6 9

+

e6 . 18

I.F. = e

e 18 6

y = mx  ...(i) H ence, the required differential equation of all such lines is  dy  [   m = dy/dx]. y =  x  dx  dy dV =V+ x , we get 44. (a) Putting y = Vx so that dx dx

⇒ x

dV − 2V = dx 2V + 1

x2 x2 +c log x − 3 9

⇒ (1 + tan y) dx = (1 + tan y – 2x) dy dx 2 + ⇒ x = 1, which is linear in x. dy 1 + tan y 2

2 cos y

1

∫ 1 + tan y dy 

= e

∫ sin y + cos y dy

cos y − sin y 

∫ 1 + cos y + sin y  dy = e = ey + log (cos y + sin y) = (cos y + sin y) ey. So, the solution is

43. (d) The equation of all the straight lines passing through origin (0, 0) is



(sin x + log x) dx + c

46. (b) We have, (1 + tan y) (dx – dy) + 2x dy = 0

⇒ 3e6y = 2e9x + 6 + e6

V ( 2V − 1) dV = V+x dx (2V + 1)

2

2 2 x i.e., y = – cos x + sin x + 2 cos x + log x – x x 3 x c + 9 x2

e6 y e9 x + 6 +C = 6 9 e6 e6 +C Putting x = 0, y = 1, we get = 6 9

9x + 6

∫x

= – x2cos x + 2x sin x + 2 cos x +

Integrating, we get

6y

d + 2y = x (sin x + log x) dx

dy 2 + · y = sin x + log x, which is linear in y. dx x

y ⋅ x2 =

 y ⇒ log f (V) = log cx ⇒ f   = cx. x dy = 9x – 6y + 6 42. (c) We have, log dx dy = e9x – 6y + 6 = e9x + 6 · e– 6y ⇒ dx ⇒ e6y dy = e9x + 6 dx

i.e., C =

d (xy) = x (sin x + log x) dx

1

log f (V) = log x + log c



2y x

2 ∫ dx I.F. = e x = e2log x = x2. So, the solution is

Integrating, we get

2y . x When x = 1, y = 1,

⇒ log (xy) = C –

 xf ( y / x )  41. (b) We have, x dy =  y + f' y / x  dx ( )  ⇒

dx  1 +  2 +  d V = 0  x V

y xey (sin y + cos y) = ∫ e (sin y + cos y ) dy + c i.e., xey (sin y + cos y) = eysin y + c.

i.e.,

x (sin y + cos y) = sin y + ce– y.

47. (a) We have, 

dy = e x – y (e x – e y) dx dy + e x ⋅ e y = e 2x. ⇒ ey dx

Putting ey = V so that e y

dy dV = , we get dx dx

dV + ex ⋅ V = e2x, which is linear in V. dx



x

∫e

ex

x

∫e

z

V ⋅ ee =

e ⇒ V ⋅e =

.e

2x

dx + c

Since P is the mid point of AB, ∴ 2x = x – y

⋅ z dz + c [Putting ex = z ⇒ ex dx = dz]

 ⇒ V ⋅ee

x

x = (z – 1) ez + c = (e − 1) e

ex

+c

 dy  dy 48. (a) We have, x   + (y – x) – y=0  dx  dx  dy   dy  + y = 0 ⇒  − 1  x  dx   dx  2

dy dy = 1 or x =– y dx dx

dy and solution of x =– y dx 

dy dx + = 0 is y x

i.e., 

log (xy) = log c  i.e.,  xy = c. Hence general solution is (x – y + c) (xy – c) = 0. 49. (c) We have, xy4 dx + y dx – x dy = 0 ⇒ x dx +

y dx − x dy =0 y4 2

 x  y dx − x dy =0 ⇒ x3 dx +   ⋅ y2  y x x ⇒ x3 dx +   d   = 0  y  y 1 x x + 4 3  y  4

∴  Equation of curve is xy = 1. 51. (a) We have, k = PQ = length of normal  dy  ⇒ k = y 1 +    dx 

3

= c′

or 3x4y3 + 4x3 = c y3, which is the required solution. 50. (c) Equation of tangent at P (x, y) is dy (X – x). Y– y= dx

dy = ± dx

2



k2  dy  = 1+   2  dx  y

2

k 2 − y2 ,

which is the required differential equation. dy = y4cos x dx dy – x2y = y4cos x i.e., x3 dx Dividing by, – y4x3, we get

52. (b) We have, x2y – x3

−1 dy 1 1 1 + ⋅ = 3 cos x y 4 dx y 3 x x 1 −1 dy 1 d V = , we get Putting 3 = V so that 4 y dx 3 dx y

1 dV 1 1 dV 3 3 + V = 3 cos x ⇒ + V = 3 cos x, 3 dx x x dx x x which is linear in V. 3



2

Integrating, we get

log y + log x = log C ⇒ log (xy) = log C or xy = C.

∴ y

dy = 1 is y = x + c dx

The solution of

Integrating, we get Since the curve passes through (1, 1), ∴ C = 1.

x

⇒ ey = ex – 1 + ce − e .



dx dy dy dx ⇒ + and 2y = y − x =0 dy dx y x

∫ I.F. = e x

dx

= e3log x = x3.

So the solution is 3 x3 V = ∫ x 3 . 3 cos x dx + C = 3 sin x + C. x x3 ⇒ = 3 sin x + C. y3 Putting x = 0, y = 1, we get C = 0. Hence the solution is x3 = 3y3 sin x. 53. (a) We have, y = Ae2x + Be– x/2 

...(1)

dy 1 = 2ae 2 x − Be − x / 2  dx 2

...(2)



d2y 1 = 4ae 2 x + Be − x / 2  dx 2 4 Eliminating A, between (1) and (2), we get 2 dy  Be– x/2 =  2 y −   5 dx

and

It meets the coordinate axes in A and B

∴ from (1), Ae2x = y – Be– x/2 = y −

...(3)

...(4)

dy  2  2 y −  dx 5

319

 dx  dy   ∴ A ≡  x − y , 0  and B ≡  0, y − x  . dy    dx 

Differential Equations

x

x e dx I.F. = e ∫ = ee . So, the solution is

320

⇒ Ae2x =

58. (b) We have,

1 dy   y + 2   5 dx

...(5)

Objective Mathematics

So, from (3), (4) and (5), we get 4 dy  1 2  dy  d2y =  y + 2  + ⋅  2 y −  5 dx 4 5 dx dx 2 d2y dy or 2 2 − 3 – 2y = 0. dx dx

which is a differential equation of degree 2. 55. (c) D  irect substitution of y = 2x – 4 in the equation shows that it is a solution of the given differential equation. ...(1)

Differentiating w.r.t. x, we get dy = 2c 2y dx dy . ⇒ c = y dx

⇒ y – 2 x

y

dy  dy  =2 y   dx  dx  2

dy dx

5

59. (a) Put x2 = sin α, y 2 = sin β. ∴  Given equation reduces to   cos α + cos β = a (sin α – sin β) α + β α − β α + β α − β = 2a cos  cos  . sin  ⇒ 2 cos   2   2   2   2  α + β α − β α + β α − β = 2a cos  2 cos  cos  . sin   2   2   2   2   α − β ⇒ cot  = a ⇒ α – β = 2 cot– 1 a.  2  ⇒ sin– 1x2 – sin– 1y2 = 2 cot– 1 a. Differentiating w.r.t. x, we get 1 1 dy ⋅ 2x − ⋅ 2y 4 4 dx = 0 1− x 1− y ⇒

x 1 − y4 dy = , y 1 − x4 dx

which is a differential equation of first order and first degree. 60. (d) Since, the given differential equation is not a polynomial in differential coefficients, so its degree is not defined.

Putting this value of c in (1), dy  we get y2 = 2y dx  x + 

d3y d2y dy − 6 +8 −5 3 2 dx dx dx

 hich is a differential equation of order 4 and w degree 3.

2

)

= 5 3

dy  dy  ⇒ x   − y + a = 0,  dx  dx

(

3/ 5

d4y   d3y  d2y dy − 5 ⇒  4  =  5 3 − 6 2 + 8 dx dx  dx   dx 

54. (b) The equation of any tangent to the parabola a y2 = 4ax is y = mx + , m where m is any arbitrary constant. dy = m. Differentiating w.r.t x, we get dx Substituting the value of m in (1), we get dy a y = x + dx dy dx

56. (a), (c)  We have, y 2 = 2c x + c .

d4y   dx 4 

  

3/ 2

61. (c) We can write y = A cos (x + B) – ce x where A = c1 + c2, B = c3 and C = c4 eC5. dy = – A sin (x + B) – Cex dx

3

dy    dy  ⇒  y − 2 x  = 4 y  dx  , dx      hich is a differential equation of order 1 and w degree 3. 57. (c) We have, y = c1 cos (2x + c2) – (c3 + c4)  ac + c5 + c6 sin (x – c7) c5 = c1 cos (2x + c2) – c8. a . ax + c6 sin (x – c7) where c3 + c4 = c8. ⇒ y = c1 cos (2x + c2) – c9 ax + c6 sin (x – c7), where c9 = c8. ac 5. S ince the above relation contains five arbitrary constants, so the order of the differential equation satisfying it is 5.

d2y = – A cos (x + B) – Cex dx 2

d2y = – 2cex dx 2 d 3 y dy d2y ⇒ = – 2cex = + +y 3 dx dx dx 2 ⇒

d 3 y d 2 y dy – y=0 − + dx 3 dx 2 dx which is a differential equation of order 3.



dy = tan θ = 2x + 3y, where θ is the dx angle that the tangent at any point (x, y) on the curve makes with x-axis. dy ⇒ − 3 y = 2x, which is a linear dx − 3 dx = e– 3x. ∴ I. F. = e ∫

62. (a) We have

⇒ y.e– 3x =

− 3x

∴ Solution of differential equation is 1 y (1 + x 2 ) = ∫ (1 + x 2 ) dx + c (1 + x 2 ) ⇒ y(1 + x2) = x + c

dx + C

−2 −3 x 2 −3 x xe − e + C 3 9

Since it passes through (1, 2)  ∴  C = −2 − 3 x 2 − 3 x 26 −3 xe − e + e 3 9 9 or 2 + 6x + 9y = 26 e3(x – 1).

26 −3 e 9

∴ ye– 3x =

63. (d) We have, d y log x = 2 dx 2 x ⇒ dy = − log x + 1 dx + c ∫ x2 dx x log x 1 =− − +c x x At x = 1, y = 0 and dy = −1 ⇒ c = 0 dx dy (log x + 1)  log x 1  ∴ =− ⇒ ∫ dy = − ∫  +  dx dx x x  x 2

⇒ y = − 1 (log x) 2 − log x + c1 2 At x = 1, y = 0 ⇒ c1 = 0 ∴ y = − 1 (log x) 2 − log x 2 64. (a) We have, 2

dy  dy   dx  − x dx + y = 0  

…(i)

From option (a) y = 2x – 4 ⇒ dy = 2 ∴ From Eq. (i), dx (2)2 – x(2) + 2x – 4 = 0 ⇒ 0 = 0 Hence, option (a) is correct answer. 65. (c) We have, dy = ye x dx dy ⇒ ⇒ log y = ex + c = e x dx y When x = 0, y = e ∴ log e = e0 + c ⇒ 1 = 1 + c ⇒ c=0 ∴ log y = ex When x = 1, log y = e1 = e ∴ y = ee

66. (a) We have, dy 2 yx 1 , which is a linear differential =  + 2 dx (1 + x ) (1 + x 2 ) equation. 2 x / (1+ x 2 ) dx IF = e∫ p dx = e ∫ = elog(1+ x ) = (1 + x2) 2

321

∫ 2x. e

Differential Equations

∴  Solution is y.e– 3x =

67. (d) We have, x dy − y dx = x 2 + y 2 dx ⇒ x dy = x 2 + y 2 dx + y dx 2 2 ⇒ x dy = ( x + y + y ) dx

x2 + y 2 + y x Now, put y = vx ⇒ dy = v + x dv dx dx ∴ dy = dx

∴ v + x dv = dx ⇒



dv 1 + v2

x 2 + v 2 x 2 + vx x

=∫

dx x

⇒ log (v + 1 + v 2 ) = log x + log c ⇒ y + x 2 + y 2 = cx 2 68. (c) We have, Ax2 + By2 = 1 ⇒ 2 Ax + 2 By dy = 0  dx

…(i)

 dy  2 d 2 y  ⇒ 2 A + 2 B   + y 2  = 0  dx   dx 

…(ii)

From Eqs. (i) and (ii), 2

y

d 2 y  dy  y  dy  + −  =0 dx 2  dx  x  dx 

 hich is the required differential equation whose orW der is 2 and degree is 1. 69. (a) Given equation can be rewritten as dx x = + 2 y2 dy y ⇒

dx x − = 2 y2 dy y

∴ IF = e ∫ Pdx = e ∫ ∴ Solution is,

1 − dy y

=

1 y

x = 2 y dy + c = y 2 + c y ∫

⇒ x = y3 + cy 70. (c) The given equation is y2 = 4ax + 4a2 On differentiating w.r.t. x, we get 2yy′ = 4a On putting the value of 4a in Eq. (i), we get y 2 y '2 y 2 = 2 yy ' x + 4 4

…(i)

322

⇒ y2 = 2yy’ x + y2y′ 2 ⇒ y = 2y’ x + yy’2

Objective Mathematics

71. (b) The given equation is dy 2 1  + , which is a linear differeny= dx 1 + x 2 (1 + x 2 ) 2 tial equation. IF = e

∫ Pdx

=e

On differentiating w.r.t. x, we get dy 2x + 2 y + 2g = 0 dx dy   ⇒ 2 g = −  2 x + 2 y  dx   From Eq. (i),

log(1+ x 2 )

=1+ x

2

dy   x 2 + y 2 +  −2 x − 2 y  x = 0 dx   ⇒ x 2 + y 2 − 2 x 2 − 2 xy dy = 0 dx ⇒ y 2 = x 2 + 2 xy dy dx 76. (c) We have,

∴ The complete solution is 1 y (1 + x 2 ) = ∫ (1 + x 2 ) ⋅ dx + c (1 + x 2 )2 ⇒ y(1 + x2) = tan–1 x + c 72. (a) We have, y2 dx + (x2 – xy + y2) dy = 0 x 2 − xy + y 2 dy = 0 y2 2 dx  x   x  +   −   +1 = 0 ⇒ dy  y   y 

1 − y2 dy = dx y y dy = ∫ dx ⇒ ∫ 1 − y2

⇒ dx +

x ⇒ x = vy y dx dv ⇒ =v+ y dy dy dv ⇒ y = −(v 2 + 1) dy dv dy ⇒ ∫ 2 +∫ =0 v +1 y Let v =

⇒ − 1 − y 2 = x + c ⇒ (x + c)2 + y2 = 1, which represents a family of circles with centre (–c, 0) and radius = 1. 77. (c) We have,

y2 = 4a (x – b)

⇒ 2 y dy = 4a dx

⇒ tan–1 v + log y + c = 0 −1  x  ⇒ tan   + log y + c = 0  y 2 73. (b) Given equation is d y = e −2 x dx 2

Again differentiating, we get 2

2y

d2y  dy  + 2  = 0 dx 2  dx  2

⇒ y

On integrating both sides, we get

d 2 y  dy  + =0 dx 2  dx 

dy e −2 x = +c 78. (a) We have −2 dx dy ( x 2 − yx 2 ) + y 2 + xy 2 = 0 Again integrating, we get −2 x dx e + cx + d y= ⇒ x2 (1 – y) dy = – (1 + x)y2 dx 4 dy 74. (d) We have, 1− y = y tan x − 2 sin x 1+ x  ⇒ dy = −  2  dx dx 2 y  x  ⇒  dy − y tan x = −2 sin x , which is a linear differdx  1 1  1 1 ⇒  2 −  dy = −  2 +  dx ential equation y y x x    ∫ p dx –∫ tan x dx IF = e =e = cos x On integrating, we get ∴ The solution 1  1  − − log y = −  − + log x  + c y(cos x) = ∫ – 2 sin x cos x dx + c y x   = – ∫ sin 2x dx + c 1 1 ⇒ log x − log y = + + c cos 2 x x y ⇒ y cos x = +c x 1 1 2 ⇒ log   = + + c 75. (c) General equation of all such circles which pass  y x y through the origin and whose centre lie on x-axis, 79. (a) The equation of all circles which passes through is the origin and whose centre lies on y-axis is x2 + y2 + 2gx = 0

…(i)



x2 + y2 – 2ky = 0

…(i)

⇒ 2 x + 2 y dy − 2k dy = 0 dx dx

⇒ y′2 (y – 2)2 = 25 – (y – 2)2 81. (d) Given that, y = 2e2x – e–x On differentiating w.r.t. x, we get y1 = 4e2x + e–x Again differentiating w.r.t x, we get y2 = 8e2x – e–x Now, y2 – y1 – 2y = 8e2x – e–x – 4e2x – e–x – 4e2x + 2e–x =0

On putting this value of k in Eq. (i), we get    x  x + y − 2 + y y = 0  dy   dx  2

323

x +y  dy   dx   

Differential Equations

⇒ k =

2

 dy  ⇒   ( y − 2) 2 = 25 − ( y − 2) 2  dx 

2

80. (c) The equation of family of circles with centre on y = 2 and of radius 5 is (x – α)2 + (y – 2)2 = 52

…(i) 82. (c)

⇒ x2 + α2 – 2αx + y2 + 4 – 4y = 25 ⇒ 2 x − 2α + 2 y dy − 4 dy = 0 dx dx

1 − y2 dy = dx y y dy = ⇒ ∫ 1 − y2

∫ dx

⇒ − 1 − y 2 = x + c

⇒ α = x + dy ( y − 2) dx On putting the value of α in Eq. (i), we get

⇒ (x + c)2 + y2 = 1.

2

dy   2 2  x − x − dx ( y − 2)  + ( y − 2) = 5  

centre (–c, 0) and, which determines a family of circles with radius

c 2 − c 2 + 1 = 1.

EXERCISES FOR SELF-PRACTICE 5. dy = e–2y and y = 0 when x = 5, then the value of x dx for y = 3 is e6 + 9 (b) (a) loge 6 2 (c) e6 + 1 (d) e5 6. 1. If

The solution of the differential equation dy = e y + x + e y – x is dx

 d2y  d3y  2  + 4 = x2 – 1, then + dx 3  d3y   dx   dx 3 

(a) ex (x + 1) + 1 = y (b) ex (x – 1) + 1 = y (c) ex (x + 1) = y (d) None of these The curve for which the slope of the tangent at any point equals the ratio of the obscissa to the ordinate of the point is (a) a circle (b) an ellipse (c) a rectangular hyperbola (d) None of these 7. Solution of the differential equation dy y = sin x is + dx x (a) x ( y + cos x) = cos x + C (b) x ( y – cos x) = sin x + C (c) x ( y + cos x) = sin x + C (d) None of these 8. The second order differential equation is: (b) y′ y′′ + y = sin x (a) y′ 2 + x = y2 (c) y′′′ + y′′ + y = (d) y′ = y x− y 9. If y′′z = , then its solution is: x+ y

(a) m = 3, n = 1 (c) m = 3, n = 2

(a) y2 + 2xy – x2 = C (c) y2 – 2xy – x2 = C

2. If

dI = 3cos y . sin y, then I is equal to dx

3cos y + c log 3 (c) sin y + c

(b) 3cos y + c

(a) –

3. 4.

(d) None of these dy y – x = 2 is The solution of dx (b) 2x – 2 · 2 y = k (a) 2x + 2 y = k 1 1 1 1 (c) x − y = k (d) 2 + y = k 2 2 2 2 If m and n are the order and degree of the differential equation 5

 d2y   dx 2 

3

(b) m = 3, n = 3 (d) m = 3, n = 3

(b) y2 + 2xy + x2 = C (d) y2 – 2xy + x2 = C

324

Objective Mathematics

10. The solution of y′ = 1 + x + y2 + xy2, y (0) = 0 is:  x2  (a) y2 = exp  x +  − 1 2    x2  (b) y2 = 1 + c exp  x +  2  

2 3 (a) y = ae (2a – x)

(b) y = ae

−2 3

(a – x)

x+a x+a

2 3

(c) y = ae (2a + x) x + a −

2

(d) y = ae 3 (2a − x) x + a

(c) y = tan (c + x + x2)  x2  (d) y = tan   x +  2   11. 12. 13. 14. 15. 16. 17.

where A is an arbitrary constant. dy = 4 is 18. The solutuion of x2 + y2 dx Family y = Ax + A3 of curves is represented by the dif (a) x2 + y2 = 12x + C (d) x2 + y2 = 3x + C ferential equation of degree: (c) x3 + y3 = 3x + C (d) x3 + y3 = 12x + C (a) 3 (b) 2 19. The solution of x dx + y dx = x2y dy – xy2 dx is (c) 1 (d) None of these (a) x2 – 1 = c (1 + y2) (b) x2 + 1 = c (1 – y2) Family of curves y = ex (A cos x + B sin x), represents the 3 3 (c) x – 1 = c (1 + y ) (d) x3 + 1 = c (1 – y3) differential equation : 20. The family of curves in which the subtangent at any point 2 d2y dy (b) – 2y (a) d y = 2 dy – y =2 of a curve is double the abscissa, is given by 2 2 dx dx dx dx (a) x = cy2 (b) y = cx2 2 2 2 2 d y dy d y dy (d) x = cy (d) y = cx – 2y (d) +y (c) = =2 dx 2 dx dx 2 dx 21. Consider the differential equation Solution of y dx – x dy = x2y dx is: 3/2   dy  2   d2y  x2 − x2 2 2 ye ye = cx (b) = cx (a) 1 +    = k  2    dx    dx  2 2 − x2 (c) y 2e x = cx2 (d) y e = cx2 Find the degree and order of the equation. dy For solving = (4x + y + 1), suitable substitution (a) 2, 2 (b) 3, 2 dx is: (c) 2, 3 (d) None of these (a) y = Vxm (b) y = 4x + V 22. Find the integral factor of expansion (c) y =4x (d) y + 4x + 1 = V dy If integrating factor of + 2xy = x2 – 1. (x2 + 1) dx x (1 – x2) dy + (2x2y – y – ax3) dx = 0 is ∫ p dx then p = e 2x (b) 2 (a) x2 + 1 2 x 2 − ax 3 x +1 (b) 2x2 – 1 (a) 2 x (1 − x 2 ) x −1 (c) 2 (d) None of these 2x2 − 1 x +1 2x2 − 1 (a) (d) x (1 − x 2 ) ax 3 23. Differential equation of y = sec (tan–1 x) is The order and degree of the differential equation: dy 1/ 3 =y +x+c (a) (1 + x2) d 2 y  dy  dx +   + x1/4 = 0 are respectively: 2 dx  dx  dy (b) (1 + x2) =y – x +c dx (a) 2, 3 (b) 3, 3 (c) 2, 6 (d) 2, 4 dy = xy + c (c) (1 + x2) dx Solution of the differential equation dy dy x a+x + xy = 0 is: = (d) (1 + x2) +c dx dx y

Answers

1. (b) 11. (a) 21. (a)

2. (a) 12. (b) 22. (b)

3. (c) 13. (c) 23. (c)

4. (c) 14. (d)

5. (d) 15. (d)

6. (c) 16. (a)

7. (c) 17. (a)

8. (b) 18. (d)

9. (a) 19. (a)

10. (d) 20. (a)

9

Coordinates and Straight Lines

CHAPTER

Summary of conceptS DiStance formula

 x1 + x2 y1 + y2   2 , 2 .  

The distance between two points A (x1, y1) and B (x2, y2) is given by

AB =

( x2 − x1 ) 2 + ( y2 − y1 ) 2

Note 1. If the distance between two points is given, then use ± sign. 2. The distance of any point A(x, y) from origin is x2 + y 2 .

Note: The coordinates of any point on a line joining the two  λx2 + x1 λy2 + y1  points A and B are given by  λ + 1 , λ + 1  . Such a point   divides the given line in the ratio λ : 1. If λ is positive, then the point divides internally and if λ is negative, then the point divides externally.

Section formula

Important Tips In order to prove that a given figure is a

1. The coordinates of the point P (x, y) dividing the line segment joining the two points A (x1, y1) and B (x2, y2) internally in the ratio m1 : m2, are given by x=

(i) Square: Prove that the four sides are equal and the diagonals are also equal. (ii) Rhombus (but not a square): Prove that the four sides are equal but the diagonals are not equal.

m1 x2 + m2 x1 m y + m2 y1 ,y= 1 2 m1 + m2 m1 + m2

(iii) Rectangle: Prove that the opposite sides are equal and the diagonals are also equal. (iv) Parallelogram (but not a ractangle): Prove that the opposite sides are equal but diagonals are not equal. Note that in each of these cases diagonals bisect each other.

2. The coordinates of the point P (x, y), dividing the line segment joining the two points A (x1, y1) and B (x2, y2) externally in the ratio m1 : m2, are given by x=

m1 x2 − m2 x1 m y − m2 y1 ,y= 1 2 m1 − m2 m1 − m2

area of a triangle The area of ∆ABC with vertices A (x1, y1), B (x2, y2) and C (x3, y3) is given by:

∆=

=

1 [x ( y – y ) + x2 ( y3 – y1) + x3 ( y1 – y2)] 2 1 2 3

1 [(x y + x2 y3 + x3 y1) – (x2 y1 + x3 y2 + x1y2)] 2 1 2

= 3. The coordinates of the mid point of the line segment joining the two points A (x1, y1) and B (x2, y2) are given by

x1 1 x2 2 x3

y1 1 y2 1 . y3 1

326

tranSlation of axeS

Objective Mathematics

Sometimes a problem with a given set of axes can be solved more easily by translation of axes. The translation of axes involves the shifting of the origin to a new point, the new axes remaining parallel to the original axes.

Notes: (i) Area of a triangle is always taken as positive. (ii) If area of a triangle is given, then use ± sign. (iii) If the three points A, B, C are collinear, then area of ∆ABC is zero. (iv) The area of a quadrilateral, whose vertices are A (x1, y1), B (x2, y2), C (x3, y3) and D (x4, y4), is

 x1   x2

x y1 + 2 x3 y2

y2 x + 3 y3 x4

y3 x + 4 y4 x1

y4 y1

  

Let OX, OY be the original axes and O′  be the new origin. Let coodinates of O′  referred to original axes, i.e., OX, OY be (h, k). Let O′ X′  and O′ Y′  be drawn parallel to and in the same If the area of a quadrileteral is zero, then its four vertices direction as OX and OY respectively. Let P be any point in the lie on a stright line i.e., points are collinear. (v) The area of a polygon of n sides with vertices A1 (x1, y1), plane having coordinates (x, y) referred to old axes and (X, Y) referred to new axes. Then, A2 (x2, y2), ..., An (xn, yn) is x = OM = ON + NM = ON + O′ M′  xn −1 yn −1 x2 y2 xn yn  1  x1 y1 = h + X = X + h or X = x – h + + + .... + =  . xn yn x3 y3 x1 y1  2  x2 y2 and y = MP = MM′  + M′ P = NO′  + M′ P =

1 2

= k + Y = Y + k. or Y = y – k (vi) If a1x + b1 y + c1 = 0, a2x + b2 y + c2 = 0 and a3x + b3 y + c3 = 0 are the equations of the sides of a triangle, then the area Thus, the point whose coordinates were (x, y) has now the coorof the triangle is dinates (x – h, y – k).

1 = 2C1C 2C3

a1 a2 a3

b1 b2 b3

2

c1 c2 , c3

rotation of axeS

rotation of axes without changing the origin Let OX, OY be the original axes and OX′ , OY′  be the new axes where C1, C2, C3 are the cofactors of c1, c2, c3 in the de- obtained by rotating OX and OY through an angle θ in the anterminant i.e., C1 = a2b3 – a3b2, C2 = a3b1 – a1b3 and ticlockwise sense. Let P be any point in the plane having coorC3 = a1b2 – a2b1. dinates (x, y) w.r.t. axes OX and OY and (x′ , y′ ) w.r.t. axes OX′  and OY′ . Then,

locuS

The locus of a moving point is the path traced by it under certain geometrical condition or conditions. For example, if a point moves in a plane under the geometrical condition that its distance from a fixed point O in the plane is always equal to a constant quantity a, then the curve traced by the moving point will be a circle with centre O and radius a. Thus, locus of the point is a circle with centre O and radius a. Working rule to find the locus of a point (a) Let the coordinates of the moving point P be (h, k). (b) Using the given geometrical conditions, find the relation between h and k. This relation must contain only h, k and known quantities. (c) Express the given relation in h and k in the simplest form and then put x for h and y for k. The relation, thus obtained, will be the required equation of the locus of (h, k).

x = x′   cos θ – y′  sin θ; y = x′   sin θ + y′  cos θ and  

x′   = x cos θ + y sin θ; y′  = – x sin θ + y cos θ.

Note: 1. The above transformation of coordinates may be displayed by a table

x

cos θ

– sin θ

y

sin θ

cos θ

2. If f (x, y) = 0 is the equation of a curve then its transformed equation is f (x′  cos θ – y′  sin θ, x′ sin θ + y′ cos θ) = 0.



3. If f (x′ , y′ ) = 0 is the transformed equation, then the equation w.r.t. original axes is f (x cos θ + y sin θ, – x sin θ + y cos θ) = 0. 4. The angle of rotation of coordinate axes in order to remove the xy term in the equation

 2h  1 ax + 2hxy + by + 2gx + 2fy + c = 0 is tan −1  . 2  a−b 2

2

5. If the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is translated to the form in which first degree terms are missing, then

 hf − bg gh − af  , the coordinates of the new origin are  2 2 ,  ab − h ab − h  where ab ≠ h2. Change of Origin and Rotation of Axes  If origin is changed to O′  (h, k) and axes are rotated about the new origin O′  by angle θ in the anticlockwise sense such that the new coordinates of P (x, y) become (x′ , y′ ), then the equations of transformation will be x = h + x′  cos θ – y′  sin θ and y = β + x′  sin θ + y′  cos θ.

(b) If three points A, B, C are collinear, then slope of AB = slope of BC = slope of AC. (c) If a line is equally inclined to the axes, then it will make an angle of 45º or 135º with x-axis (i.e., positive direction of x-axis) and hence its slope will be tan 45º or tan 135º, = ± 1. (d) Slope of the line joining two points (x1, y1) and (x2, y2) is given as

y1 − y2 y2 − y1 Difference of ordinates m = x − x = x − x = Difference of abscissae . 1 2 2 1

Intercept of a Line on the Axes (i) Intercept of a line on x-axis  If a line cuts x-axis at a point (a, 0), then a is called the intercept of the line on x-axis. | a | is called the length of the intercept of the line on x-axis. Intercept of a line on x-axis may be positive or negative. (ii) Intercept of a line on y-axis  If a line cuts y-axis at a point (0, b), then b is called the intercept of the line on y-axis and | b | is called the length of the intercept of the line on y-axis. Intercept of a line on y-axis may be positive or negative.

Equations of Lines Parallel to Axes Equation of x-axis  The equation of x-axis is y = 0. Equation of y-axis  The equation of y-axis is x = 0. Equation of a line parallel to y-axis  The equation of the straight line parallel to y-axis at a distance ‘a′  from it on the positive side of x-axis is x = a. If a line is parallel to y-axis, at a distance a from it and is on the negative side of x-axis, then its equation is x = – a. Equation of a line parallel to x-axis  The equation of the straight line parallel to x-axis at a distance b from it on the positive side of y-axis is y = b. If a line is parallel to x-axis, at a distance b from it and is on the negative side of y-axis, then its equation is y = – b.

equation of a Straight line in various forms General Equation of a straight Line An equation of the form: ax + by + c = 0, where a, b, c are any real numbers not all zero, always represents a straight line. Equation of a straight line is always of first degree in x and y. π  Slope of a Line  If a line makes an angle θ  θ ≠  with the 2  positive direction of x-axis, the slope or gradient of that line is usually denoted by m, i.e., tan θ = m. Note:

Slope-Intercept Form The equation of a straight line whose slope is m and which cuts an intercept c on the y-axis is given by y = mx + c.

If the line passes through the origin, then c = 0 and hence (a) The slope of a line parallel to x-axis = 0 and perpendicular the equation of the line will become y = mx. to x-axis is undefined.

327

y′

Coordinates and Straight Lines

x′

328

Point-Slope Form

Objective Mathematics

The equation of a straight line passing through the point (x1, y1) and having slope m is given by ( y – y1) = m (x – x1).

Two-Point Form The equation of a straight line passing through two points (x1, y1) and (x2, y2) is given by (y – y1) =

y2 − y1 (x – x1). x2 − x1

where r is the distance of the point (x, y) from the point (x1, y1). The coordinates (x, y) of any point P on the line at a distance r from the point A (x1, y1) can be taken as (x1 + r cos θ, y1 + r + sin θ) or (x1 – r cos θ, y1 – r sin θ).

Intercept Form The equation of a straight line which cuts off intercepts a and b on x-axis and y-axis respectively is given by

x y + = 1. a b

Reduction of the general equation to different standard forms Slope-Intercept Form: The general form, Ax + By + C = 0, of the straight line can be reduced to the form y = mx + c by expressing y as:

A C x− = mx + c B B A C and c = – . where   m = – B B Thus, slope of the line Ax + By + C = 0 is y=–

m=–

Normal Form (or Perpendicular Form) The equation of a straight line upon which the length of the perpendicular from the origin is p and the perpendicular makes an angle α with the positive direction of x-axis is given by x cos α + y sin α = p.

In normal form of equation of a straight line p is always taken as positive and α is measured from positive direction of x-axis in anticlockwise direction between 0 and 2π.

Parametric Form (or Symmetric Form)

coefficient of x A =− . coefficient of y B

Intercept Form: The equation Ax + By + C = 0 can be reduced to the form

x y + = 1 by exa b

pressing it as: Ax + By = – C A B or –  x − y = 1, where C ≠ 0 C C x y x y or − C/A + − C/B = 1, which is of the form + = 1, a b C C and b = – are intercepts on x-axis and where a = – A B y-axis, respectively. Intercept of a straight line on x-axis can be found by putting y = 0 in the equation of the line and then finding the value of x. Similarly intercept on y-axis can be found by putting x = 0 in the equation of the line and then finding the value of y.

Normal Form To reduce the equation Ax + By + C = 0 to the form x cos α + y sin α = p, first express it as:

Ax + By = – C ...(1) The equation of a straight line passing through the point (x1, y1) and making an angle θ with the positive direction of x-axis is Case 1. If C < 0 or – C > 0, then divide both sides of eqn. (1) by A 2 + B2 , we get given by

x − x1 y − y1 = =r cos θ sin θ

A A +B 2

2

x+

B A +B 2

2

y=−

which is of the form x cos α + y sin α = p,

C A + B2 2

and

p=–

A

, sin α

A 2 + B2

Thus to write the equation of any line parallel to a given line, do not change the coefficient of x and y and change the constant term only.

B A 2 + B2

C

equation of a line perpenDicular to a given line

A + B2 2

caSe 2. If C > 0 or – C < 0, then divide Eqn. (1) by

−A

– A + B , we get 2

x−

2

B

A 2 + B2 A 2 + B2 which is of the form x cos α + y sin α = p, where and

p=

cos α = –

A A 2 + B2

, sin α = –

y=

C A 2 + B2 B

A 2 + B2

The equation of a line perpendicular to a given line ax + by + c = 0 is bx – ay + k = 0, where k is a constant.

point of interSection of two given lineS Let the two given lines be

C A 2 + B2

a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0.

.

angle between two interSecting lineS

Solving these two equations, the point of intersection of the given two lines is given by

 b1c2 − b2c1 c1a2 − c2 a1  ,  .  a1b2 − a2b1 a1b2 − a2b1 

The angle θ between two lines y = m1x + c1 and y = m2x + c2 is given by concurrent lineS m1 − m2 The three given lines are concurrent if they meet in a point. tan θ = ± 1 + m m , 1 2 provided no line is perpendicular to x-axis and the acute angle θ is given by tan θ = Note:

m1 − m2 1 + m1m2 .

(a) If both the lines are perpendicular to x-axis, then the angle between them is 0º. (b) If slope of one line is not defined (one of the lines is perpendicular to x-axis and other makes an angle θ with the positive direction of x-axis), then angle between them = π – θ. (c) The two lines are parallel if and only if m1 = m2. (d) The two lines are perpendicular if and only if m1 × m2 = – 1.

conDition for two lineS to be coinciDent, parallel, perpenDicular or interSecting Two lines a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0 are a b c (i) Coincident, if 1 = 1 = 1 ; a2 b2 c2 (ii) Parallel, if

a1 b1 c1 = ≠ ; a2 b2 c2

(iii) Perpendicular, if a1a2 + b1b2 = 0;

Working rule: Following three methods can be used to prove that the three lines are concurrent 1. Find the point of intersection of any two lines by solving them simultaneously. If this point satisfies the third equation also, then the given lines are concurrent. 2. The three lines P ≡ a1x + b1 y + c1 = 0, Q ≡ a2x + b2 y + c2 = 0, R ≡ a3x + b3 y + c3 = 0 are concurrent if

a1 a2 a3

b1 b2 b3

c1 c2 = 0 c3

3. The three lines P = 0, Q = 0 and R = 0 are concurrent if there exist constants l, m and n, not all zero at the same time, such that lP + mQ + nR = 0. This method is particularly useful in theoretical results.

poSition of two pointS relative to a line Two points (x1, y1) and (x2, y2) are on the same side or on opposite sides of the line ax + by + c = 0 according as the expressions: ax1 + by1 + c and ax2 + by2 + c have same sign or opposite signs.

length of perpenDicular from a

point on a line a b (iv) Intersecting, if 1 ≠ 1 i.e., if they are neither coincident a2 b2 The length of the perpendicular from the point (x1, y1) to the line nor parallel. ax + by + c = 0 is given by equation of a line parallel to a given line The equation of a line parallel to a given line ax + by + c = 0 is ax + by + k = 0, where k is a constant.

p=

a x1 + b x2 + c

. a 2 + b2 The length of perpendicular from origin (0, 0) to the given line is

a 2 + b2 .

329

cos α =

Coordinates and Straight Lines

where,

330

Distance between Two parallel lines

Objective Mathematics

The distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is given by d=

| c1 − c2 | a +b 2

2

.

(a) The distance between two parallel lines can also be obtained by taking a suitable point (take y = 0 and find x or take x = 0 and find y) on one straight line and then finding the length of the perpendicular from this point to the second line. (b) Area of a parallelogram or a rhombus, equations of whose sides are given, can be obtained by using the following formula pp Area = 1 2 , sin θ where p1 = DL = distance between lines AB and CD, p2 = BM = distance between lines AD and BC, θ = angle between adjacent sides AB and AD.

Equation of the Bisector of the Acute and Obtuse Angle between Two Lines  Let the equations of the two lines be

a1x + b1 y + c1 = 0 

...(1)

and

a2x + b2 y + c2 = 0 

...(2)

where c1 > 0 and c2 > 0. Then the equation

a1 x + b1 y + c1 a +b 2 1

2 1

=+

a2 x + b2 y + c2 a22 + b22

is the bisector of the acute or obtuse angle between the lines (1) and (2) according as a1a2 + b1b2 < 0 or > 0. Similarly, the equation

a1 x + b1 y + c1 a +b 2 1

2 1

=−

a2 x + b2 y + c2 a22 + b22

is the bisector of the acute or obtuse angle between the lines (1) and (2) according as a1a2 + b1b2 > 0 or < 0. Note: If a1a2 + b1b2 > 0, then the origin lies in obtuse angle and if a1a2 + b1b2 < 0, then the origin lies in acute angle.

Equations of Lines Passing through the point of intersection of two given lines The equation of any line passing through the point of intersection of the lines a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0 is

In the case of a rhombus, p1 = p2. Thus,

(a1x + b1 y + c1) + k (a2x + b2 y + c2) = 0, p12 where k is a parameter. The value of k can be obtained by using Area of rhombus = . one more conditions which the required line satisfies. sin θ 1 Also, area of rhombus = dd 2 1 2 Standard Points of a Triangle where d1 and d2 are the lengths of two perpendicular diagoCentroid of a Triangle  The point of intersection of nals of a rhombus. the medians of the triangle is called the centroid of the triangle. The centroid divides the medians in the ratio 2 : 1 EquationS of straight lines (2 from the vertex and 1 from the opposite side).

passing through a given point and making a given angle with a given line

The equations of the striaght lines which pass through a given point (x1, y1) and make a given angle α with the given straight line y = mx + c are y – y1 =

m ± tan α (x – x1) 1 m tan α

EquationS of the bisectors of the Angles between two lines The equations of the bisectors of the angles between the lines a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0 are given by

a1 x + b1 y + c1 a12 + b12



a2 x + b2 y + c2 a22 + b22

.

Any point on a bisector is equidistant from the given lines.

The coordinates of the centroid of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) are

 x1 + x2 + x3 y1 + y2 + y3  ,  . 3 3  

Incentre of a Triangle The point of intersection of the internal bisectors of the angles of a triangle is called the incentre of the triangle.

The coordinates of the incentre of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) are

 ax1 + bx2 + cx3 ay1 + by2 + cy3   a+b+c , a + b + c  .  Note: The incentre of he triangle formed by (0, 0), (a, 0) and   ab ab (0, b) is  , . 2 2 2 2  a +b + a +b a+b+ a +b  ex-centres of a triangle A circle touches one side outside the triangle and the other two extended sides then circle is known as excircle.

Solving any two of the above equations, we get the circumcentre ((x,, y). (x 3. (a) If the equations of the three sides of the triangle are given, first of all find the coordinates of the vertices of the triangle by solving the equations of the sides of the triangle taken two at a time. (b) Find the coordinates of the middle points of two sides of the triangle. (c) Find the equations of the perpendicular bisectors of these two sides and solve them. This will give the coordinates of the circumcentre of the triangle. 4. If angles A, B, C and vertices A(x A (x ( 1, y1), B (x ( 2, y2) and C (x ( 3, y3) of a ∆ABC are given, then its circumcentre is given by

sinn 2A+x2 sin sin 2B+x B+x3 sin 2C  x1 si  sinn 2A+ ssin , 2 B+ in 2B+ sin 2C 

Let ABC be a triangle then there are three excircles, with three excentres I1, I2, I3 opposite to vertices A, B and C respectively. If the vertices of triangle are A (x1, y1), B (x2, y2) and C (x3, y3) then

 − ax1 + bx2 + cx3 − ay1 + by2 + cy3   I1 =  − a + b + c , −a + b + c    ax1 − bx2 + cx3 ay1 − by2 + cy3  I2 =  a − b + c , a − b + c    ax1 + bx2 − cx3 ay1 + by2 − cy3  I3 =  a + b − c , a + b − c  .  circumcentre The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle. It is the centre of the circle which passes through the vertices of the triangle and so its distance from the vertices of the triangle is same and this distance is known as the circum-radius of the triangle. Working Rules 1. Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of the ∆ABC and let circumcentre be P (x, y). Then (x, y) can be found by solving (OA)2 = (OB)2 = (OC)2 or

(x – x1)2 + ( y – y1)2 = (x – x2)2 + ( y – y2)2 = (x – x3)2 + ( y – y3)2

y1 sin sin 2A+y2 sin sin 2B+ B+yy3 sin 2C  sinn 2A+ ssin B+ sin 2C  in 22B+ Note: 1. The circumcentre of a right angled triangle is the mid point of its hypotenuse. 2. The circumcentre of the triangle formed by (0, 0), (x1, y1) and (x2, y2) is

 y2 ( x12 + y12 ) − y1 ( x22 + y22 ) ,  2 ( x1 y2 − x2 y1 )  x2 ( x12 + y12 ) − x1 ( x22 + y22 )  . 2 ( x2 y1 − x1 y2 ) 

orthocentre The orthocentre of a triangle is the point of intersection of altitudes. Working Rules 1. Let O be the orthocentre. Since AD ⊥ BC, BE ⊥ CA and CF ⊥ AB, then OA ⊥ BC

and

OB ⊥ CA OC ⊥ AB.

331

Coordinates and Straight Lines

2. Let D, E and F be the mid points of the sides BC, CA and AB of the ∆ABC respectively. Then, OD ⊥ BC, OE ⊥ AC, OF ⊥ AB. slope of OD × slope of BC = – 1 slope of OE × slope of AC = – 1 slope of OF × slope of AB = – 1

332

Objective Mathematics

 x1 tan A+x2 tan B+x3 tan C  tan A+ tan B+ tan C ,  y1 tan A+y2 tan B+y3 tan C  tan A+ tan B+ tan C  Note: Solving any two of these, we can get coordinates of O. 2. (a) Write down the equations of any two sides of the triangle. (b) Find the equations of the lines perpendicular to these two sides and passing through the opposite vertices. (c) Solve these equations to get the coordinates of the orthocentre. 3. If angles A, B and C and vertices A ((x1, y1), B ((x2, y2) and C ((x3, y3) of a ∆ABC are given, then orthocentre of ∆ABC is given by

1. If any two lines out of three lines, i.e., AB, BC and CA are perpendicular, then orthocentre is the point of intersection of two perpendicular lines. 2. The orthocentre of the triangle with vertices (0, 0), (x1, y1) and (x2, y2) is

 x1 x2 + y1 y2    x1 x2 − y1 y2   ( y1 − y2 )   .  , ( x1 − x2 )  x y − x y  x1 y2 − x2 y1    2 1 1 2 

multiple-choice queStionS choose the correct alternative in each of the following problems: 1. The intercepts on the straight line y = mx by the line y = 2 and y = 6 is less than 5, then m belongs to  4 4 (a)  − ,   3 3

 4 3 (b)  ,   3 8

4  4   (c)  −∞,  ∪  , , ∞  3  3  

4  (d)  , ∞  3 

2. 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 (c) 2x + 3y = 9

(b) 2x – 3y = 7 (d) 3x – 2y = 3

3. The ratio in which the line y – x + 2 = 0 divides the line joining (3, – 1) and (8, 9) is (a) 2:3 (c) – 2:3

(b) 3:2 (d) – 3:2

4. The equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is –1 is (a)

x y x y + = 1 and + = 1 2 3 2 1

(b)

x y x y − = – 1 and + =1 2 3 −2 1

x y x y (c) + = –1 and + = –1 2 3 −2 1 (d)

x y x y − = 1 and + =1 2 3 −2 1

5. The extremities of the diagonal of a parallelogram are the points (3, – 4) and (– 6, 5). If the third vertex is the point (– 2, 1), then the coordinates of the fourth vertex are (a) (1, 0) (c) (0, 1)

(b) (– 1, 0) (d) (0, – 1)

6. The length of the median through A of a triangle whose vertices are A (– 1, 3), B (1, – 1) and C (5, 1), is (a) 5 (c) 1

(b) 4 (d) None of these

7. The coordinates of the circumcentre of the triangle having vertices (– 2, – 3), (– 1, 0) and (7, – 6) are (a) (– 3, 3) (c) (3, – 3)

(b) (3, 3) (d) (– 3, – 3)

8. The coordinates of A, B, C are (6, 3), (– 3, 5) and (4, – 2) respectively and P is any point (x, y). The ratio of the areas of triangles PBC and ABC is x+ y−2 | x + y − 2| (b) 7 7 x+ y+2 (c) (d) None of these 7 9. The ends of a rod of length l move on two mutually perpendicular lines. The locus of the point on the rod which divides it in the ratio 1 : 2 is (a)

(a)

x2 y 2 l 2 + = 1 4 9

(b)

x2 y 2 l2 + = 4 1 9

(c)

x2 y 2 l2 − = 1 4 9

(d)

x2 y 2 l 2 − = 4 1 9

2 ⋅ a2 + a + 1 3 1 a2 + a + 1 (c) 3 (a)

2 a2 − a + 1 3 1 a2 − a + 1 (d) 3

(b)

11. What is the equation of the locus of a point which moves such that 4 times its distance from the x-axis is the square or its distance from the origin? (a) x2 + y2 – 4y = 0 (c) x2 + y2 – 4x = 0

(b) x2 + y2 – 4 | y | = 0 (d) x2 + y2 –­4 | x | = 0

12. x-coordinates of two points B and C are the roots of equation x2 + 4x + 3 = 0 and their y-coordinates are the roots of equation x2 – x – 6 = 0. If x-coordinate of B is less than x-coordinate of C and y-coordinate of B is greater than the y-coordinate of C and coordinates of a third point A be (3, – 5), then the length of the bisector of the interior angle at A is (a)

7 2 3

(b)

14 2 3

5 2 (d) None of these 3 13. The point A divides the join of P (– 5, 1) and Q (3, 5) in the ratio k : 1. The two values of k for which the area of ∆ABC, where B ≡ (1, 5) and C ≡ (7, – 2), is equal to 2 units, are 31 (a) 7, 14 (b) – 7, 9 31 31 (c) 7, (d) 17, – 9 9 14. Equation of the straight line making equal intercepts on the axes and passing through the point (2, 4) is (c)

(a) 4x – y – 4 = 0 (c) x + y – 6 = 0

(b) 2x + y – 8 = 0 (d) x + 2y – 10 = 0

15. If A, B, C, D are points whose coordinates are (– 2, 3), (8, 9), (0, 4) and (3, 0) respectively, then the ratio in which AB is divided by CD, is (a) 11 : 23 (c) 23 : 47

(b) 11 : 47 (d) None of these

16. The straight line ax + by + c = 0 divides the join of points A (x1, y1) and B (x2, y2) in the ratio (a)

ax1 + by1 + c ax2 + by2 + c

(b) –

ax1 + by1 + c ax2 + by2 + c

(c)

ax2 + by2 + c ax1 + by1 + c

(d) –

ax2 + by2 + c ax1 + by1 + c

17. The value of k for which the points (k, 2 – 2k), (– k + 1, 2k) and (– 4 – k, 6 – 2k) are collinear, is 1 2 (c) k = 1

(a) k =

1 2 (d) k = – 1

(b) k = –

19. If α, β, γ are the real roots of the equation x3 – 3px2 + 3qx – 1 = 0, then the centroid of the triangle  1  1   1 having vertices  α,  ,  β,  and  γ ,  are  α  β  γ (a) ( p, q) (c) (– p, q)

(b) ( p, – q) (d) (– p, – q)

20. The area of the quadrilateral whose vertices are (– 3, 2), (7, – 6), (– 5, – 4) and (5, 4) is (a) 68 (c) 80

(b) 70 (d) None of these

21. The locus of the point whose distance from x-axis is twice that from y-axis, is (a) y = x (c) x = y

(b) y = 2x (d) x = 2y

22. The locus of the moving point P such that 2PA = 3PB, where A is (0, 0) and B is (4, – 3), is (a) 5x2 + 5y2 + 72x + 54y + 225 = 0 (b) 5x2 + 5y2 – 72x – 54y + 225 = 0 (c) 5x2 + 5y2 – 72x + 54y + 225 = 0 (d) None of these 23. The points (3, 3), (h, 0) and (0, k) are collinear if 1 1 1 + = 3 h k 1 1 1 (c) − = 3 k h

(a)

(b)

1 1 1 − = 3 h k

(d) None of these

24. If (1, 4) be the C.G. of a triangle and the coordiantes of its any two vertices be (4, – 8) and (– 9, 7), then the area of the triangle is(a) 84

(b) 132

333 (d) None of these (c) 2 25. The coordinates of incentre of the triangle whose sides are 3x – 4y = 0, 5x + 12y = 0 and y – 15 = 0, are (a) (1, 8) (c) (– 1, 8)

(b) (1, – 8) (d) (– 1, – 8)

26. The coordiantes of the orthocentre of the triangle, having vertices (0, 0), (2, – 1) and (– 1, 3), are (a) (4, – 3) (c) (4, 3)

(b) (– 4, 3) (d) (– 4, – 3)

27. The coordiantes of the orthocentre of the triangle, formed by lines xy = 0 and x + y = 1, are (a) (0, 0) (c) (– 2, 1)

(b) (2, – 1) (d) None of these

28. The coordiantes of the orthocentre of the triangle, having sides 3x – 2y = 6, 3x + 4y + 12 = 0 and 3x – 8y + 12 = 0, are

333

x defined as f : (0, ∞) to (0, ∞), f (x) is 1+ x (a) one-one onto (b) one-one but not onto (c) many-one into (d) many-one onto

18. Let f (x) =

Coordinates and Straight Lines

10. The distance between two parallel lines is unity. A point P lies between the lines at a distance a from one of them. The length of a side of an equilateral triangle PQR, vertex Q of which lies on one of the parallel lines and vertex R lies on the other line, is

334

Objective Mathematics

 1 23  (a)  − ,  6 9 

23   1 (b)  − , −   6 9

 1 23  (c)  ,  6 9 

(d) None of these

45   36 (c)  , −  7 7 

 36 45  (d)  ,  7 7 

36. If the point A is symmetric to the point B (4, – 1) with respect to the bisector of the first quadrant, then the length of AB is

(a) 5 (b) 5 2 (c) 3 2 (d) 3 29. Two vertices of a triangle are (3, – 1) and (– 2, 3) and 37. The equation of the straight line which makes an angle its orthocentre is origin, the coordinates of the third of 15º with the positive direction of x-axis and cuts an vertex are intercept of length 4 on the negative direction of y-axis, 45  is  36 45   36 (b)  − , −  (a)  − ,   7 7   7  7 (a) (2 – 3 ) x – y – 4 = 0

30. If a triangle has its orthocentre at (1, 1) and circumcentre 3 3 at  ,  , then the coordinates of the centroid of the 2 4 triangle are 5 4 (a)  , −  3 6

4 5 (b)  ,  3 6

 4 5 (c)  − ,   3 6

5  4 (d)  − , −   3 6

31. If G be the centroid and I be the incentre of the triangle with vertices A (– 36, 7), B (20, 7) and C (0, – 8) and 25 205 λ, then the value of λ is GI = 3 1 (a) 25 (b) 25 4 25 32. If a vertex of a triangle of two sides through it centroid of the triangle (c)

(d) None of these be (1, 1) and the middle points be (– 2, 3) and (5, 2) then the is

5  (a)  , 3  3 

5  (b)  , − 3  3 

5 (c)  − , 3   3 

5 (d)  − , − 3   3 

33. The equation of the locus of a point such that the sum of its distances from (0, 2) and (0, – 2) is 6, is (a)

x2 y 2 − = 1 5 9

(b)

x2 y 2 − =1 9 5

(c)

x2 y 2 + = 1 5 9

(d)

x2 y 2 + =1 9 5

34. A ladder of length ‘a’ rests against the floor and a wall of a room. If the ladder begins to slide on the floor, then the locus of its middle point is (a) x2 + y2 = a2 (c) x2 + y2 = 2a2

(b) 2 (x2 + y2) = a2 (d) 4 (x2 + y2) = a2

35. A and B are two fixed points. The locus of a point P such that ∠APB is a right angle, is (a) x2 + y2 = a2 (c) 2x2 + y2 = a2

(b) x2 – y2 = a2 (d) None of these

3 ) x + y – 4 = 0 (c) (2 + 3 ) x – y – 4 = 0 (d) None of these (b) (2 –

38. The equation of the straight line cutting off an intercept 8 from x-axis and making an angle of 60º with the positive direction of y-axis is (a) x + (c) y –

3 y = 8 3 x = 8

(b) x – 3 y = 8 (d) None of these

39. A line through the point A (2, 0), which makes an angle of 30º with the positive direction of x-axis is rotated about A in clockwise direction through an angle 15º. The equation of the straight line in the new position is 3)x–y–4+2 3 =0 (b) (2 – 3 ) x + y – 4 + 2 3 = 0 (c) (2 – 3 ) x – y + 4 + 2 3 = 0 (d) None of these

(a) (2 –

40. The equation of the straight line, the portion of which intercepted between the coordinate axes being divided by the point (– 5, 4) in the ratio 1 : 2, is (a) 8x + 5y = 60 (c) – 8x + 5y = 60

(b) 8x – 5y = 60 (d) None of these

41. The coordinates of a point on the line x + y = 4 that lies at a unit distance from the line 4x + 3y – 10 = 0 are (a) (3, 1) (c) (3, – 1)

(b) (– 7, 11) (d) (7, – 11)

42. The equation of the internal bisector of ∠BAC of ∆ABC with vertices A (5, 2), B (2, 3) and C (6, 5), is (a) 2x + y + 12 = 0 (c) 2x + y – 12 = 0

(b) x + 2y – 12 = 0 (d) None of these

43. The range of values of θ in the interval (0, π) such that the points (3, 5) and (sin θ, cos θ) lie on the same side of the line x + y – 1 = 0, is π (b)  0,  (a)  0, π   4  2 π π (c)  ,  (d) None of these 4 2 44. A rectangle has two opposite vertices at the points (1, 2) and (5, 5). If the other vertices lie on the line x = 3, then the coordinates of the other vertices are (a) (3, – 1), (3, – 6) (c) (3, 2), (3, 6)

(b) (3, 1), (3, 5) (d) (3, 1), (3, 6)

(a) 9x + 20y + 96 = 0 (c) 9x + 20y – 96 = 0

(b) 9x – 20y + 96 = 0 (d) None of these

46. The equation of the straight line which passes through the point (3, 4) and whose intercept on y-axis is twice that on x-axis, is (a) 2x – y = 10 (c) 2x + y = 10

(b) x + 2y = 10 (d) None of these

47. The equation of the straight line whose intercepts on x-axis and y-axis are respectively twice and thrice of those by the line 3x + 4y = 12, is (a) 9x + 8y = 72 (c) 8x + 9y = 72

(b) 9x – 8y = 72 (d) None of these

48. The equation of the straight line passing through the origin and the middle point of the intercept of the line ax + by + c = 0 between the axes, is (a) ax + by = 0 (c) bx + ay = 0

(b) ax – by = 0 (d) bx – ay = 0

 (a)  2, 

3 2 

 3 (b)  2, − 2  

 1 3 (c)  2 + (d) None of these , 2  2  55. The acute angle between the line x + y = 3 and the line joining the points (1, 1) and (– 3, 4) is 3 (a) tan–1   7

3 (b) π – tan–1   7

1 (c) tan–1   7

1 (d) π – tan–1   7

(

56. 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)

3 x+ y =0 2

(b) x + 3 y = 0

3 49. If m1 and m2 are the roots of the equation x2 + ( 3 + 2) x y =0 (d) x + (c) 3x + y = 0 2 + ( 3 – 1) = 0, then the area of the triangle formed 57. The equation of the straight line, passing through by the lines y = m1x, y = m2x and y = 2 is the point (2, – 4) and perpendicular to the line (a) 33 − 11 (b) 33 + 11 8x – 4y + 7 = 0, is (c) 33 + 7 (d) None of these (a) x + 2y + 6 = 0 (b) x – 2y + 6 = 0 50. Through the point P (α, β), where αβ > 0 the straight (c) 2x + y + 6 = 0 (d) 2x – y + 6 = 0 x y = 1 is drawn so as to form with coordinate 58. The equation of the straight line, having x-intercept equal line + a b 4 to – and is perpendicular to the line 2x – 5y + 8 = 0, axes a triangle of area S. If ab > 0, then the least value 5 of S is is (a) αβ (c) 4αβ

(b) 2αβ (d) None of these

(a) 2x + 5y + 4 = 0 (c) 2x – 5y + 4 = 0

(b) 5x – 2y + 4 = 0 (d) 5x + 2y + 4 = 0.

51. The equation of the straight line upon which the length 59. The equation of the perpendicular bisector of the line of perpendicular from origin is 3 2 units and this segment joining the points (1, 4) and (3, 6) is perpendicular makes an angle of 75º with the positive (a) x – y – 7 = 0 (b) x + y – 7 = 0 direction of x-axis, is (c) x + y + 7 = 0 (d) None of these (a) ( 3 – 1) x + ( 3 + 1) y – 12 = 0 60. A line 4x + y = 1 through the point A (2, – 7) meets the (b) ( 3 – 1) x + ( 3 + 1) y + 12 = 0 (c) ( 3 + 1) x + ( 3 – 1) y – 12 = 0 (d) None of these

52. If the straight line drawn through the point P ( 3 , 2) π with the x-axis meets the line and making an angle 6 3 x – 4y + 8 = 0 at Q, then the length of PQ is (a) 4 (c) 6

(b) 5 (d) None of these

53. The coordinates of the points at a distance 4 2 units from the point (– 2, 3) in the direction making an angle of 45º with the positive direction of x-axis, are (a) (2, 7), (– 6, – 1) (c) (2, – 7) , (– 6, – 1)

(b) (2, 7), (6, – 1) (d) None of these

line BC whose equation is 3x – 4y + 1 = 0 at the point B. If AB = AC, then the equation of the line AC is (a) 52x – 89y + 519 = 0 (b) 52x + 89y – 519 = 0 (c) 52x + 89y + 519 = 0 (d) None of these 61. The coordinates of the foot of the perpendicular drawn from the point (2, 3) to the line y = 3x + 4 are  1 37  (a)  − ,   10 10 

 1 37  (b)  , −   10 10 

1 37 (c)  − , −  (d) None of these  10 10  62. The image of the point (– 8, 12) with respect to the line mirror 4x + 7y + 13 = 0 is (a) (16, – 2) (c) (16, 2)

(b) (– 16, 2) (d) (– 16, – 2)

335

54. A line joining two points A (2, 0) and B (3, 1) is rotated about A in anticlockwise direction through an angle 15º. If B goes to C in the new position, then the coordinates of C are

Coordinates and Straight Lines

45. A straight line passes through the point (– 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point. The equation of the line is

336

63. The image of the point (3, – 8) under the transformation (x, y) → (2x + y, 3x – y) is

Objective Mathematics

(a) (– 2, 17) (c) (– 2, – 17)

(b) (2, 17) (d) (2, – 17)

64. The image of the point P (3, 5) with respect to the line y = x is the point Q and the image of Q with respect to the line y = 0 is the point R (a, b), then (a, b) = (a) (5, 3) (c) (– 5, 3)

(b) (5, – 3) (d) (– 5, – 3)

65. 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. Then the point O divides the segment PQ in the ratio (a) 1:2 (c) 2:1

(b) 3:4 (d) 4:3

72. If the lines ax + y + 1 = 0, x + by + 1 = 0 and x + y + c = 0 (a, b and c being distinct and different from 1) are con1 1 1 = current, then + + 1− a 1− b 1− c (a) 1 (c) 0

73. The distance between the lines 5x – 12y + 2 = 0 and 5x – 12y – 3 = 0 is 3 5 (b) 13 13 7 (c) (d) None of these 13 74. The sum of the abscissas of all the points on the line x + y = 4 that lie at a unit distance from the line 4x + 3y – 10 = 0, is (a)

66. The equation of the straight line passing through the point of intersection of the lines x – y = 1 and 2x – 3y + 1 = 0 and parallel to the line 3x + 4y = 14 is 75. (a) 3x + 4y + 24 = 0 (b) 3x + 4y – 24 = 0 (c) 4x + 3y + 24 = 0 (d) 4x + 3y – 24 = 0 67. The equation of the straight line passing through the point of intersection of lines 3x – 4y – 7 = 0 and 12x – 5y – 13 = 0 and perpendicular to the line 2x – 3y + 5 = 0 is

68. The equation of the straight line passing through the point of intersection of the lines x + 3y + 4 = 0 and 3x + y + 4 = 0 and equally inclined to the axes is (a) x – y = 0 (c) x – y + 2 = 0

(b) x + y + 2 = 0 (d) None of these

69. The distance of the point (1, 2) from the straight line with slope 5 and passing through the point of intersection of lines x + 2y = 5 and x – 3y = 7 is (a)

64 650

(b)

(c)

132 650

(d) None of these

16 650

70. If a and b are the intercepts of a straight line on the xaxis and y-axis respectively and p be its perpendicular distance from the origin, then 1 1 1 (a) 2 = 2 + 2 p a b

1 1 1 (b) 2 = 2 − 2 p a b

(c) p2 = a2 + b2

(d) None of these

71. If the family of lines x (a + 2b) + y (a + 3b) = a + b passes through the point for all values of a and b, then the coordinates of the point are (a) (2, 1) (c) (– 2, 1)

(a) 3 (c) 4

(b) (2, – 1) (d) None of these

(b) – 3 (d) – 4

If p1 and p2 are the lengths of the perpendiculars from the origin to the straight lines x sec θ + y cosec θ = a and x cos θ – y sin θ = a cos 2θ respectively, then the value of 4 p12 + p22 is (a) 4a2 (c) a2

(b) 2a2 (d) None of these

76. The equation of the straight line that can be drawn through the point (4, – 5) so that its distance from the point (– 2, 3) is equal to 12, is (a) x – 2y = 3 (c) 2x – y = 3

(a) 33x + 22y + 13 = 0 (b) 33x + 22y – 13 = 0 (c) 33x – 22y + 13 = 0 (d) None of these

(b) – 1 (d) None of these

(b) 2x + y = 3 (d) no such line is possible

77. The equation of the straight line which cuts off intercepts on x-axis twice that on y-axis and is at a unit distance from the origin, is (a) x – 2y + (c) x + 2y +

(b) x + 2y – 5 = 0 5 = 0 (d) x – 2y – 5 = 0 5 = 0 78. A line L has intercepts a and b on the coordinate axes. When the axes are rotated through an angle, keeping the origin fixed, the same line L has intercepts p and q. Then, (a)

1 1 1 1 + = + a 2 b2 p 2 q 2

(b)

1 1 1 1 − = − a 2 b2 p 2 q 2

 1 1 1 1  (c) a 2 + b 2 = 2  p 2 + q 2    (d) None of these 79. A straight road passes through two towns, one 5 km. east 1 and other 2 km. north from a tower. A rest house is 2 to be constructed by the side of the road. The nearest position of the rest house from the tower is (a) 1 km. east and 2 km. north (b) 2 km. east and 1 km. north (c) 1 km. east and 1 km. north (d) None of these

(a) 9x – 7y – 41 = 0 (c) 7x + 9y – 3 = 0

(b) 7x + 9y – 3 = 0 (d) None of these

81. The equation of the bisector of the acute angle between the lines 3x + 4y – 11 = 0 and 12x – 5y – 2 = 0 is (a) 11x + 3y + 17 = 0 (b) 11x + 3y – 17 = 0 (c) 3x + 11y + 17 = 0 (d) None of these

(b) 7x – 9y + 3 = 0 (d) None of these

83. The value of k so that the lines 2x – 3y + k = 0, 3x – 4y – 13 = 0 and 8x – 11y – 33 = 0 are concurrent, is (a) 7 (c) 5

(a) all real values of t

(b) some real values of t

−4 ± 7 (c) t = 8

(d) None of these

91. P (3, 1), Q (6, 5) and R (x, y) are three points such that the angle RPQ is a right angle and the area of ∆RPQ = 7, then the number of such points R is (a) 0 (c) 2

82. For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 =  0, the equation of the bisector of the angle which contains the origin is (a) 7x + 9y + 3 = 0 (c) 7x + 9y – 3 = 0

(b) – 7 (d) – 5

(b) 1 (d) 4

92. If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is (a) square (b) circle (c) straight line (d) two intersecting lines 93. If P (1, 0), Q (– 1, 0), R (2, 0) are three given points, then the locus of point S satisfying the relation SQ2 + SR2 = 2SP2 is (a) a straight line | | to x-axis (b) a straight line | | to y-axis (c) circle through the origin (d) circle with centre at the origin.

84. If the equal sides AB and AC (each equal to a) of a right angled isosceles triangle ABC be produced to P and Q so that BP ⋅ CQ = AB2, then the line PQ always passes 94. A line passes through the point (2, 2) and is perpendicular through the fixed point to the line 3x + y = 3. Its y intercept is (a) (a, 0) (b) (0, a) 1 2 (c) (a, a) (d) None of these (b) (a) 3 3 85. If a straight line cuts intercepts from the axes of coordi4 nates the sum of the reciprocals of which is a constant k, (c) 1 (d) 3 then the line passes through the fixed point (a) (k, k)

1 1 (b)  ,  k k 

(c) (k, – k)

(d) (– k, k)

86. The three lines 3x + 4y + 6 = 0, and 4x + 7y + 8 = 0 are (a) sides of a triangle (c) parallel

2x+

3y+2 2=0

(b) concurrent (d) None of these

87. The diagonals of a parallelogram PQRS are along the lines x + 3y = 4 and 6x – 2y = 7. Then PQRS must be  a (a) rectangle (c) cyclic quadrilateral

(b) square (d) rhombus

88. The equation of the straight line equally inclined to the axes and equidistant from the points (1, – 2) and (3, 4)  is (a) x + y + 1 = 0 (c) x – y – 2 = 0

(b) x + y + 2 = 0 (d) x – y – 1 = 0

(a) (– 1, 2) (c) (1, 2)

(b) (1, – 2) (d) (– 1, – 2)

95. A line L passes through the points (1, 1) and (2, 0). Another line M which is perpendicular to L passes through 1 the point  , 0  . Then the area of the triangle formed 2  by the lines L, M and y-axis is 25 16 25 25 (c) (d) 32 4 96. The coordinates of the foot of the perpendicular from the point (2, 4) on the line x + y = 1 are (a)

25 8

(b)

1 3 (a)  2 , 2   

 1 3 (b)  − 2 , 2   

4 1 (c)  3 , 2   

1 3 (d)  4 , − 2   

97. Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 89. Without changing the direction of coordinates axes, 2x + y = 3, then the equation representing the pair of origin is transferred to (α, β) so that the linear terms in lines PQ and PR is the equation x2 + y2 + 2x – 4y + 6 = 0 are eliminated. (a) 3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0 The point (α, β) is (b) 3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0 (c) 3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0 (d) 3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0

337

90. The point (2t2 + 2t + 4, t2 + t + 1) lies on the line x + 2y = 1 for

Coordinates and Straight Lines

80. For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0, the equation of the bisector of the obtuse angle between them is

338

98. The orthocentre of the triangle formed by (0, 0), (8, 0) and (4, 6) is

Objective Mathematics

8 (a)  4,   3 (c) (4, 3)

(b) (3, 4) (d) (– 3, 4)

104. Let the algebraic sum of the perpendicular distances from the points A (2, 0), B (0, 2), C (1, 1) to a variable line be zero. Then all such lines (a) are concurrent (b) pass through the fixed point (1, 1) (c) touch some fixed circle (d) pass through the centroid of ∆ ABC.

99. P is a point on either of the two lines y – 3 | x | = 2 at a distance of 5 units from their point of intersection. 105. If α + β + γ = 0, the line 3αx + β y + 2γ = 0 passes The coordinates of the foot of the perpendicular from through the fixed point P on the bisector of the angle between them are  2 2  (b)  , 2  (a)  2,   1   1   3 3  (a) 0, 2 (4 + 5 3 )  or 0, 2 (4 − 5 3 )  depending on     2 (c)  −2,  (d) None of these which line the point P is taken  3  1  106. The area of the region enclosed by 4 | x | + 5 | y | ≤ 20 (b) 0, 2 (4 + 5 3 )    is 1   (a) 10 (b) 20 (c) 0, 2 (4 − 5 3 )  (c) 40 (d) None of these  

5 5 3  (d)  2 , 2  .   100. Let P be the image of the point (– 3, 2) with respect to x-axis. Keeping the origin as same, the coordinate axes are rotated through an angle 60º in the clockwise sense. The coordinates of point P with respect to the new axes are 2 3 −3 (3 3 + 2)  (a)  , −  2  2  2 3 − 3 3 3 + 2 (b)  , 2   2  (2 3 − 3) 3 3 + 2  (c)  − , 2 2   (d) None of these

x y 1 + + a b c = 0 always passes through a fixed point, that point is

107. If a, b, c are in H.P. then the straight line

(a) (– 1, – 2) (c) (1, – 2)

(b) (– 1, 2) 1 (d) 1, −   2

108. The distance of the middle point of the line joining the point (a sin θ, 0) and (0, a cos θ) from the origin is a 2 (c) a (sin θ + cos θ)

(a)

a (sin θ + cos θ) 2 (d) a

(b)

109. Two vertices of a triangle are (2, – 1) and (3, 2) and third vertex lies on the line x + y = 5. If the area of the triangle is 4 units then third vertex is: (a) (0, 5) or (4, 1) (c) (5, 0) or (4, 1)

(b) (5, 0) or (1, 4) (d) (0, 5) or (1, 4)

101. A square is constructed on the portion of the line 110. The figure which is designed by the line ax ± by ± c = 0 is: x + y = 5 which is intercepted between the axes, on the side of the line away from origin. The equations to the (a) Rectangle (b) Square diagonals of the square are (c) Rhombus (d) None of these (a) x = 5, y = – 5 (b) x = 5, y = 5 111. The locus of the mid point of the intercept of the line (c) x = – 5, y = 5 (d) x – y = 5, x – y = – 5 x cos α + y sin α = p between coordinate axes is : 102. If one of the diagonals of a square is along the line (b) x–2 + y–2 = p–2 (a) x–2 + y–2 = 4 p–2 x = 2y and one of its vertices is (3, 0), then its sides (c) x2 + y2 = 4 p–2 (d) x2 + y2 = p2 through this vertex are given by the equations 112. If the points (x, y), (x′, y′ ) and (x – x′, y – y′ ) are col(a) y – 3x + 9 = 0, 3y + x – 3 = 0 linear, then, (b) y + 3x + 9 = 0, 3y + x – 3 = 0 (a) xy = x′ y′ (b) xx′ = yy′ (c) y – 3x + 9 = 0, 3y – x + 3 = 0 (c) xy′ = x′ y (d) None of these (d) y – 3x + 3 = 0, 3y + x + 9 = 0 x y 103. The line (p + 2q) x + ( p – 3q) y = p – q for different 113. If for a variable line + = 1, the condition a–2 + b–2 a b values of p and q passes through the fixed point = c–2 (c is a constant) is satisfied, then the locus of foot 2 2 3 5 of the perpendicular drawn from origin to this is: (b)  ,  (a)  ,  5 5 2 2 c2 (b) x2 + y2 = 2c2 (a) x2 + y2 = 2 3 3 2 3   (c)  ,  (d)  ,  5 5 5 5 (c) x2 + y2 = c2 (d) x2 – y2 = c2

(a) 2 (c) 4

(b) 0 (d) 1

115. If the centroid and circumcentre of a triangle are (3, 3) and (6, 2) respectively, then the orthocentre is (a) (– 3, 5) (c) (3, – 1)

(b) (– 3, 1) (d) (9, 5)

116. A stick of length l rests against the floor and a wall of a room. If the stick begins to slide on the floor, then the locus of its middle point is (a) an ellipse (c) a circle

(b) a parabola (d) a straight line

3 and which is 2 concurrent with lines 4x + 3y – 7 = 0 and 8x + 5y – 1 = 0 is

117. The equation of the line with slope –

(a) centroid (c) circumcentre

(b) incentre (d) orthocentre

(a rational point is a point both of whose coordinates are rational numbers) 125. 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 (c) 2x + 9y – 11 = 0

(b) 2x – 9y – 11 = 0 (d) 2x + 9y + 7 = 0

126. The incentre of the triangle with vertices (1, and (2, 0) is

 3 (a) 1,   2  

3 ), (0, 0)

2 1  (b)  ,  3 3

2 3  1  (d) 1, (c)  ,   3 2  3    127. If the lines ax + by + c = 0, bx + cy + a = 0 and 118. The equations ax + by + c = 0 and dx + ey + f = 0 cx + ay + b = 0 are concurrent (a + b + c ≠ 0) then represent the same straight line if and only if (a) a3 + b3 + c3 – 3a bc = 0 a b (b) a + b = 0 (b) c = f (a) = d e (c) a + b + c = 0 a b c (d) a2 + b2 + c2 + ab + bc + ca = 0 (d) a = d, b = e, c = f = = (c) d e f 128. If the lines x – 2y – 6 = 0, 3x + y – 4 = 0 and λ x + 119. The distance between the lines 4x + 3y = 11 and 4y + λ2 = 0 are concurrent, then 8x + 6y = 15 is (a) λ = 2 (b) λ = – 3 (a) 2y – 3x – 2 = 0 (c) 3x + 2y – 63 = 0

(b) 3x + 2y – 2 = 0 (d) None of these

7 10 (c) 4

7 2 (d) None of these

(a)

(b)

120. The point (3, 2) is reflected in the y-axis and then moved a distance 5 units towards the negative side of y-axis. The coordinates of the point thus obtained are (a) (3, – 3) (c) (3, 3)

(b) (– 3, 3) (d) (– 3, – 3)

121. If a, b, c are in A.P., then the straight line ax + by + c = 0 will always pass through a fixed point whose coordinates are (a) (– 1, – 2) (c) (– 1, 2)

(b) (1, 2) (d) (1, – 2)

122. If x1, x2, x3 as well as y1, y2, y3 are in G.P. with the same common ratio, then the points (x1, y1), (x2, y2) and (x3, y3) (a) lie on a straight line (b) lie on an ellipse (c) lie on a circle (d) are vertices of a triangle 123. If P (1, 2), Q (4, 6), R (5, 7) and S (a, b) are the vertices of a parallelogram PQRS, then (a) a = 2, b = 4 (c) a = 2, b = 3

(b) a = 3, b = 4 (d) a = 3, b = 5

(c) λ = 4

(d) None of these

129. If A and B are two variable points on axes of x and y respectively, such that OA + OB = c, the locus of the foot of perpendicular from origin to AB is (a) x2 + y2 = cxy (b) x + y = cxy (c) (x + y) (x2 + y2) = cxy (d) x2 + y2 = c2 130. Area of the triangle formed by the points [(a + 3) (a + 4), a + 3], [(a + 2) (a + 3), a + 2] and [(a + 1) (a + 2), a + 1] is (a) 25a2 (c) 24a2

(b) 5a2 (d) None of these

131. If the lines ax + 12y + 1 = 0, bx + 13y + 1 = 0 and cx + 14y + 1 = 0 are concurrent, then a, b, c are in (a) H.P. (c) A.P.

(b) G.P. (d) None of these

132. Equation of a straight line passing through the point of intersection of x – y + 1 = 0 and 3x + y – 5 = 0 and perpendicular to one of them is (a) x + y + 3 = 0 (c) x – 3y – 5 = 0

(b) x + y – 3 = 0 (d) x + 3y + 5 = 0

133. If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is

339

Coordinates and Straight Lines

114. The number of integer values of m, for which the 124. If the vertices P, Q, R of a ∆PQR are rational points, x-coordinate of the point of intersection of the lines which of the following points of the ∆PQR is (are) 3x + 4y = 9 and y = mx + 1 is also an integer, is always rational point (s) ?

340

(a) square (c) straight line

(b) circle (d) two intersecting lines

Objective Mathematics

134. The vertices of a triangle are (0, 0), (3, 0) and (0, 4). Its orthocentre is at 3  (a)  , 2  2 

(b) (0, 0)

 4 (d) None of these (c) 1,   3 135. The vertices of a triangle are (0, 3), (– 3, 0) and (3, 0). The coordinates of its orthocentre are (a) (0, 2) (c) (0, 3)

(b) (0, – 3) (d) (0, – 2)

136. The lines x cos α + y sin α = p1 and x cos β + y sin β = p2 will be perpendicular if π π (b) α = (a) α ± β = 2 2 π (d) α = β (c) | α – β | = 2 137. The vertices of a ∆OBC are O (0, 0), B (– 3, – 1) and C (– 1, – 3). The equation of a line parallel to BC and intersecting sides OB and OC whose distance from the 1 , is origin is 2 1 1 = 0 (b) x + y – =0 2 2 1 1 = 0 (d) x + y + =0 (c) x + y – 2 2 138. The medians AD and BE of a triangle with vertices A (0, b), B (0, 0) and C (a, 0) are perpendicular to each other if (a) x + y +

(a) a =

b 2

(c) ab = 1

(b) b =

a 2

(d) a = ±

2b

139. The equation of a straight line passing through the point (– 5, 4) and which cuts off an intercept of 2 units between the lines x + y + 1 = 0 and x + y – 1 = 0 is (a) x – y + 10 = 0 (b) x – y + 9 = 0 (c) 2x – y + 14 = 0 (d) x – 2y + 13 = 0 140. The point which divides the join of (1, 2) and (3, 4) externally in the ratio 1 : 1 (a) lies in the IIIrd