Preview A Complete Resource Book in Mathematics for JEE Main 2019 by Dinesh Khattar (2018) Preview A Complete Resource Book in Mathematics for JEE Main 2019 by Dinesh Khattar (2018) Preview A Complete Resource Book in Mathematics for JEE Main 2019 by Dinesh Khattar (2018) Preview A Complete Resource Book in Mathematics for JEE Main 2019 by Dinesh Khattar (2018) Preview A Complete Resource Book in Mathematics for JEE Main 2019 by Dinesh Khattar (2018)
About Pearson Pearson is the world’s learning company, with presence across 70 countries worldwide Our unique insights and world-class expertise comes from a long history of working closely with renowned teachers, authors and thought leaders, as a result of which, we have emerged as the preferred choice for millions of teachers and learners across the world We believe learning opens up opportunities, creates fulfilling careers and hence better lives We hence collaborate with the best of minds to deliver you classleading products, spread across the Higher Education and K12 spectrum Superior learning experience and improved outcomes are at the heart of everything we This product is the result of one such effort Your feedback plays a critical role in the evolution of our products and you can contact us - reachus@pearson.com We look forward to it Objective_Maths_JEE Main 2019_Prelims.indd 5/18/2018 3:11:34 PM This page is intentionally left blank Objective_Maths_JEE Main 2019_Prelims.indd 5/18/2018 3:11:34 PM A Complete Resource Book in Mathematics for Objective_Maths_JEE Main 2019_Prelims.indd JEE Main 2019 5/18/2018 3:11:35 PM This page is intentionally left blank Objective_Maths_JEE Main 2019_Prelims.indd 5/18/2018 3:11:35 PM A Complete Resource Book in Mathematics for JEE Main 2019 Dr Dinesh Khattar Kirori Mal College, University of Delhi Objective_Maths_JEE Main 2019_Prelims.indd 5/18/2018 3:11:35 PM 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 © 2018 Pearson India Education Services 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 publisher of this book No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent This eBook may or may not include all assets that were part of the print version The publisher reserves the right to remove any material in this eBook at any time ISBN 978-93-530-6217-0 eISBN 9789353063436 First Impression Published by Pearson India Education Services Pvt Ltd, CIN: U72200TN2005PTC057128 Head Office: 15th Floor, Tower-B, World Trade Tower, Plot No 1, Block-C, Sector 16, Noida 201 301, Uttar Pradesh, India Registered Office: 4th floor, Software Block, Elnet Software City, TS-140, Block & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India Fax: 080-30461003, Phone: 080-30461060 website: in.pearson.com, Email: companysecretary.india@pearson.com Compositor: SRS Global, Puducherry Printed in India by Objective_Maths_JEE Main 2019_Prelims.indd 5/18/2018 5:08:27 PM Contents Preface ix JEE Mains 2018 Paper xiii JEE Mains 2017 Paper xxv Chapter Set Theory 11–118 Chapter Functions 21–260 Chapter Complex Numbers 31–368 Chapter Quadratic Equations and Expressions 41–452 Chapter Matrices 51–530 Chapter Determinants 61–656 Chapter Permutations and Combinations 71–746 Chapter Mathematical Induction 81–84 Chapter Binomial Theorem 91–946 Chapter 10 Sequence and Series 101–1062 Chapter 11 Limits 111–1148 Chapter 12 Continuity and Differentiability 121–1250 Chapter 13 Differentiation 131–1344 Chapter 14 Applications of Derivatives 141–1462 Chapter 15 Indefinite Integration 151–1552 Chapter 16 Definite Integral and Area 161–1686 Chapter 17 Differential Equations 171–1748 Chapter 18 Coordinates and Straight Lines 181–1854 Chapter 19 Circles 191–1956 Chapter 20 Conic Sections (Parabola, Ellipse and Hyperbola) 201–2062 Chapter 21 Vector Algebra 211–2146 Chapter 22 Three Dimensional Geometry 221–2236 Chapter 23 Measures of Central Tendency and Dispersion 231–2324 Objective_Maths_JEE Main 2019_Prelims.indd 5/18/2018 3:11:35 PM viii Contents Chapter 24 Probability����������������������������������������������������������������������������������������������������������������������������������� 24.1–24.60 Chapter 25 Trigonometric Ratios and Identities�������������������������������������������������������������������������������������������� 25.1–25.30 Chapter 26 Trigonometric Equations������������������������������������������������������������������������������������������������������������� 26.1–26.22 Chapter 27 Inverse Trigonometric Functions������������������������������������������������������������������������������������������������� 27.1–27.36 Chapter 28 Heights and Distances���������������������������������������������������������������������������������������������������������������� 28.1–28.36 Chapter 29 Mathematical Reasoning���������������������������������������������������������������������������������������������������������������� 29.1–29.6 Mock Test - I �������������������������������������������������������������������������������������������������������������������������������������������������������� M1.1–M1.8 Mock Test - II ������������������������������������������������������������������������������������������������������������������������������������������������������� M2.1–M2.6 Mock Test - III ������������������������������������������������������������������������������������������������������������������������������������������������������� M3.1–M3.6 Mock Test - IV ������������������������������������������������������������������������������������������������������������������������������������������������������� M4.1–M4.8 Mock Test - V ������������������������������������������������������������������������������������������������������������������������������������������������������� M5.1–M5.8 Objective_Maths_JEE Main 2019_Prelims.indd 5/18/2018 3:11:36 PM Preface About the Series A Complete Resource Book for JEE Main series is a must-have resource for students preparing for JEE Main examination There are three separate books on Physics, Chemistry, and Mathematics; the main objective of this series is to strengthen the fundamental concepts and prepare students for various engineering entrance examinations It provides class-tested course material and numerical applications that will supplement any ready material available as student resource To ensure high level of accuracy and practicality, this series has been authored by highly qualified and experienced faculties for all three titles About the Book This book, A Complete Resource Book in Mathematics for JEE Main 2019, covers both the text and various types of problems required as per the syllabus of JEE Main examination It also explains various short-cut methods and techniques to solve objective questions in lesser time Salient Features • • • • • • Complete coverage of topics along with ample number of solved examples Large variety of practice problems with complete solutions Chapter-wise Previous 15 years’ AIEEE/JEE Main questions Fully solved JEE Main 2018 and 2017 questions 5 Mock Tests based on JEE Main pattern in the book 5 Free Online Mock Tests as per the recent JEE Main pattern It would have been difficult to prepare this book without aid and support from a number of different quarters I shall be grateful to the readers for their regular feedback I am deeply indebted to my parents without whose encouragement this dream could not have been translated into reality The cherubic smiles of my daughters, Nikita and Nishita, have inspired me to treat my work as worship Anuj Agarwal from IIT-Delhi, Ankit Katial from National Institute of Technology (Kurukshetra) and Raudrashish Chakraborty from Kirori Mal College, University of Delhi, with whom I have had fruitful discussions, deserve special mention I earnestly hope that the book will help the students grasp the subject well and respond with a commendable score in the JEE Main examination There are a plethora of options available to students for Mathematics, however, ever grateful to them and to the readers for their candid feedback Despite of our best eff orts, some errors may have crept into the book Constructive comments and suggestions to further improve the book are welcome and shall be acknowledged gratefully Best of luck! Dinesh Khattar Objective_Maths_JEE Main 2019_Prelims.indd 5/18/2018 3:11:36 PM 2.46 Chapter ⎧ x, x ∈ Q =⎨ ⎩ x, x ∉ Q \ fof (x) = x, x ∈ [0, 1] The correct option is (C) 77 We have, f (x – f (y)) = f (f (y)) + xf (y) + f (x) – 1 (1) Put x = f (y) = then f (0) = f (0) + + f (0) – \ f (0) = 1 (2) Again, put x = f (y) = l in (1) Then, f (o) = f (l) + l2 + f (l) – ⇒ 1 = 2f (l) + l2 – − l2 l2 \ f (l) = =1− 2 x2 Hence, f (x) = − The correct option is (C) 78 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 + + x ) is an odd function, \ cos (log (x + + x )) is an even function If f (x + y) = f (x) + f (y) ∀ x, y ∈ R, then f (x) must be odd The correct option is (C) 79 The function f (x) = x2 + 2, x ∈ (–∞, ∞) is not injective as f (1) = f (–1) but ≠ –1 The function f (x) = (x – 4) (x – 5), x ∈ (– ∞, ∞) is not one-one as f (4) = f (5) but ≠ x + 3x − The function, f (x) = , x ∈ (–∞, ∞) is also not + 3x − x injective as f (1) = f (–1) but ≠ –1 For the function, f (x) = |x + 2|, x ∈ [– 2, ∞) Let f (x) = f (y), x, y ∈ [–2, ∞) ⇒ |x + 2| = |y + 2| ⇒ x + = y + ⇒ x = y So, f is an injection The correct option is (B) 80 ( fog) (x) = f [g (x)] = f (|3x + 4|) Since the domain of f is [–3, 5], \ –3 ≤ |3x + 4| ≤ ⇒ |3x + 4| ≤ ⇒ –5 ≤ 3x + ≤ ⇒ –9 ≤ 3x ≤ ⇒ –3 ≤ x ≤ 1⎤ ⎡ \ Domain of fog is ⎢ −3, ⎥ 3⎦ ⎣ The correct option is (B) 81 For any x ∈ R, we have ⎡ x ⎤ ⎡ x + 1⎤ [x] = ⎢ ⎥ + ⎢ ⎥ (1) ⎣2⎦ ⎣ ⎦ Objective_Maths_JEE Main 2017_Ch 2.indd 46 ⎡ n + 1⎤ ⎡ n ⎤ ⇒ [n] = ⎢ ⎥+⎢ ⎥ ⎣ ⎦ ⎣2⎦ ⎡ n + 1⎤ ⎡ n ⎤ ⇒ n = ⎢ ⎥+⎢ ⎥ ⎣ ⎦ ⎣2⎦ ⎡n ⎤ + 1⎥ ⎡ n + 1⎤ ⎡ n ⎤ ⎢ = ⎢ ⎥ [Using (1)] ⎥+⎢ ⎥+⎢ ⎣ ⎦ ⎣4⎦ ⎣ ⎦ ⎡ n + 1⎤ ⎡ n + ⎤ ⎡ n ⎤ = ⎢ ⎥+⎢ ⎥+⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣4⎦ ⎡n ⎤ + 1⎥ ⎡ n + 1⎤ ⎡ n + ⎤ ⎡ n ⎤ ⎢ = ⎢ ⎥ ⎥+⎢ ⎥+⎢ ⎥+⎢ ⎣ ⎦ ⎣ ⎦ ⎣8⎦ ⎣ ⎦ [Using (1)] ⎡ n + 1⎤ ⎡ n + ⎤ ⎡ n + ⎤ ⎡ n ⎤ = ⎢ ⎥+⎢ ⎥ ⎥+⎢ ⎥+⎢ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣8⎦ Continuing in this manner, we have, ⎡ n + 1⎤ ⎡ n + ⎤ ⎡ n + ⎤ ⎢ ⎥ + ⎢ ⎥ + ⎢⎣ ⎥⎦ + = n ⎣ ⎦ ⎣ ⎦ The correct option is (A) 82 We have, ⎧1 + n − n = 1, x = n ∈ I g (x) = ⎨ ⎩1 + n + k − n = + k , x = n + k , where n ∈ I, < k < ⎧ −1, g ( x ) < ⎪ Now, f [g (x)] = ⎨ 0, g ( x ) = ⎪ 1, g ( x) > ⎩ Clearly, g (x) > for all x So, f [g (x)] = 1, for all x The correct option is (B) 83 We have, g (x) = + x and f [g (x)] = + x + x(1) Also, f [g (x)] = f (1 + x )(2) From (1) and (2), we get f (1 + x ) = + x + x Let 1 + x = y or x = ( y – 1)2 \ f ( y) = + ( y – 1) + ( y – 1)2 = + 2y – + y2 – 2y + = + y2 \ f (x) = + x2 The correct option is (B) 84 For f to be defined, we must have ⎞ ⎛ log1/ ⎜1 + ⎟ < – ⇒ + > (2– 1)– = which is pos⎝ ⎠ x x sible only if > 1, i.e., < x < x Hence, the domain of the given function is {x: < x < 1} The correct option is (A) 01/01/2008 03:20:09 Functions 2.47 85 We have, f (x + 2a) = f ((x + a) + a) = − f ( x + a) − ( f ( x + a)) ⎛1 f ( x ) − ( f ( x )) − ⎜ − ⎝2 = 1 − − 2 = − = ⎛1 ⎞ − ⎜ − f ( x )⎟ = f (x) ⎠ ⎝2 ⎞ f ( x ) − ( f ( x )) ⎟ ⎠ − f ( x ) + ( f ( x )) Hence, f (x) is periodic with period 2a The correct option is (B) 86 Since the domain of f is (0, 1), \ 0 < ex < and < ln |x| < ⇒ log < x < log and e0 < |x| < e1 ⇒ –∞ < x < and < |x| < e ⇒ x ∈ (–∞, 0) and x ∈ ((–∞, –1) ∪ (1, ∞)) ∩ (–e, e) ⇒ x ∈ (–∞, 0) and x ∈ (–e, –1) ∪ (1, e) ⇒ x ∈ (–e, –1) The correct option is (A) 87 sin– (x – x2) is defined when – ≤ x – x2 ≤ ⇒ 1− 1+ ≤x≤ (1) 2 1 is defined when − ≥0 |x| |x| ⇒ x ≤ – or x ≥ 1 (2) 2 and is defined when x – < or x – ≥ [ x − 1] i.e., x ∈ (–∞, − ) ∪ (–1, 1) ∪ ( − , ∞)(3) From (1), (2) and (3), we get 1− ⎛ 1+ 5⎞ x ∈ ⎜ 2, ⎟⎠ ⎝ 91 We have, f (x) = f (x) = x3 + x2f ′(1) + xf ′′(2) + f ′(3)(1) ⇒ f ′(x) = 3x2 + 2xf ′(1) + f ′′(2)(2) ⇒ f ′′(x) = 6x + 2f ′(1)(3) sin101 ( − x ) ⎡ x⎤ ⎢− p ⎥ + ⎣ ⎦ Case I: when x = np (n ∈ Z ) ⇒ f (–x) = sin101 ( − np ) =0 ⎡ − np ⎤ + ⎢ p ⎥ ⎣ ⎦ Case II: When x ≠ np, n ∈ Z f (–x) = ⇒ f ′′′(x) = i.e., a constant function Hence, f ′(3) = ⇒ f (–x) = Using equation (3), we have f ′′(2) = 12 + 2f ′(1)(5) Substituting x = 1, in equation (2), we have ⇒ f (–x) = f ′(1) = + 2f ′(1) + f ′′(2) ⇒ f ′(1) = + 2f ′(1) + 12 + 2f ′(1) {using (5)} ⇒ 3f ′(1) = –15 \ f ′(1) = –5 Objective_Maths_JEE Main 2017_Ch 2.indd 47 sin101 x ⎡x⎤ ⎢p ⎥ + ⎣ ⎦ f (–x) = The correct option is (A) 88 Since, Substituting respective value in (5), we have f ′′(2) = Hence, the polynomial f (x) can be written as f (x) = x3 – 5x2 + 2x + Therefore, f (2) – f (1) = (8 – 20 + + 6) – (1 – + + 6) = –2 – = – \ f (2) – f (1) = –6 = – f (0) The correct option is (B) 89 Given, q2 – 4pr = and p > For f (x) to be defined px3 + (p + q)x2 + (q + r)x + r > ⇒ px2(x + 1) + qx(x + 1) + r(x + 1) > ⇒ (x + 1)(px2 + qx + r) > q [Since q2 – 4pr = ⇒ x > – and x ≠ − 2p q q \ at x = − , px2 + qx + r = and at x ≠ − , px2 + qx + p p r > 0] ⎡ ⎧ q ⎫⎤ \ Domain = R − ⎢( − ∞, −1] ∪ ⎨ − ⎬ ⎥ ⎩ p ⎭⎦ ⎣ The correct option is (B) 90 Since the function is periodic, a2 – and b2 + should be perfect squares, which is possible only if a = 2, b = in which case f (x) = sin x + cos 4x, whose period is 2p The correct option is (B) (6) − sin101 ( x ) ⎡ x⎤ ⎢− p ⎥ + ⎣ ⎦ ⎡ ⎡x⎤ ⎡ x⎤ ⎤ − sin101 ( x ) ⎢Q ⎢ p ⎥ + ⎢ − p ⎥ = − 1⎥ ⎡x⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −1 − ⎢ ⎥ + ⎣ p ⎦ sin101 ( x ) = f (x) ⎡x⎤ + ⎢p ⎥ ⎣ ⎦ \ f (x) is an even function The correct option is (B) 01/01/2008 03:20:12 2.48 Chapter 92 Graph is symmetric about x = k if f (k – x) = f (k + x) ⇒ a(k – x)3 + b(k – x)2 + c(k – x) + d = a(k + x)3 + b(k + x) + c(k + x) + d ⇒ 2ax3 – (6ak2 + 4bk + 2c)x = It is true for all x if a = and 6ak2 + 4bk + 2c = c c ⇒a+k= − i.e., a = and k = – 2b 2b The correct option is (A) 93 For the ln operation to be defined, we have {x} > Plotting the curves y = {x} and y = 0, we can see that the values of x satisfying the above inequality, are x ∈ R – I(1) For the square root operation to be defined, we have {x} ≤ x/2 The domain of definition is the darkened portion on the X-axis p⎞ ⎛p p⎤ ⎡ p i.e x ∈ ⎢ − , − ⎟ ∪ ⎜ , ⎥ 3⎠ ⎝ 2⎦ ⎣ The correct option is (C) 95 For the sgn operation to be defined, we have – x2 > i.e., –3 < x < 3 (1) For the square root operation to be defined, we have [x]3 – 4[x] ≥ ⇒ [x] ([x] + 2) ([x] – 2) ≥ ⇒ –2 ≤ [x] ≤ or [x] ≥ i.e., –2 ≤ x < or x ≥ 2 (2) The union of the intervals (1) and (2) gives the domain as [–2, 1) ∪ [2, 3) The correct option is (B) 96 we have f (x) = (log0.2 x )3 + (log0.2 x ) (log0.2 0.0016 x ) + 36 = (log0.2 x )3 + (3 log0.2 x ) {log0.2 x + log0.2 (0.2)4 } + 36 = l + 3l ( l + 4) + 36 [putting log0.2 x = l] = l + 3l + 12l + 36 = Plotting the curves y = {x} and y = x/2 as shown above, we can see that the values of x satisfying the above inequality, are x ∈ {0} ∪ [1, ∞)(2) The domain of definition is the intersection of (1) and (2) which gives x ∈ (1, ∞) – I + The correct option is (C) 94 For the ln operation to be defined, we have – |cos x| > i.e., |cos x| < (1) For the inverse cos operation to be defined, we have 2x ≤1 –1≤ p p p i.e., − ≤ x ≤ (2) 2 To find the values of x satisfying both (1) and (2), let us plot in the interval [–p/2, p/2] as the curves y = |cos x| and y = shown below ( l + 3) ( l + 12) For the square root operation to be defined, we have (l + 3) (l2 + 12) ≥ i.e., l + ≥ 0 (Q l2 + 12 is a positive quantity) i.e., log0.2 x ≥ – i.e., x ≤ (0.2)–3 = 53 = 125 Also, x > for log to be defined Hence, the required domain of definition is (0, 125] The correct option is (C) 97 We have, y = [x2] – [x]2, x ∈ [0, 2] i.e., y = [x ], 0≤x – 9/8 i.e., 2 ≤ a ≤ 14 and a > – 9/8 i.e., 2 ≤ a ≤ 14 Hence, f : R → R is onto for ≤ a ≤ 14 The correct option is (D) 101 Let y = ⎛ 2| x|⎞ 102 The function f (x) = sin −1 ⎜ is defined for ⎝ + x ⎟⎠ 2| x| 2| x| ≤ i.e., ≤1 + x2 + x2 ⎡ 2| x| ⎤ is a positive quantity⎥ ⎢Q + x ⎣ ⎦ i.e., x2 – |x| + ≥ i.e., (|x| – 1)2 ≥ 0, which is true ∀ x ∈ R(1) The function sin x is defined for, sin x ≥ i.e., 2np ≤ x ≤ (2n + 1)p, n ∈ I(2) Intersection of inequalities (1) and (2) gives the domain as [2np, (2n + 1)p], n ∈ I The correct option is (B) 103 [sin x] is always defined ⎛ p ⎞ cos ⎜ is also defined everywhere except when ⎝ [ x − 1] ⎟⎠ [x – 1] = ⇒ 0 ≤ x – < ⇒ 1 ≤ x < Hence, domain ∈ R – [1, 2) The correct option is (B) x − [ x] {x} = + x − [ x ] + {x} Thus, domain = (–∞, ∞) 104 Here, y = {x} we have, + {x} y + y {x} = {x} y ⇒ {x} = 1− y y Here, ≤ {x} < so, ≤ ⎧7 − x if x < ⎪ if ≤ x ≤ ⇒ | f (x)| + |f (x)| = ⎨ ⎪2 x − if x > ⎩ We need those points for which the L.H.S is greater than Clearly, we can exclude values of x between and Now, for values of x less than 3, – 2x is greater than 1, and for values of x greater than 4, 2x – is greater than Therefore, the given inequality is true for values of x given by (–∞, ∞) – [3, 4] The correct option is (C) More Than One Option Correct Type x +1 ⎛ x + 1⎞ 118 Since f (x) = f ⎜ ⇒x= ⎝ x + ⎠⎟ x+2 ⇒ x2 + x – = −1 ± (1) Since f (x) is an even function defined on [– 5, 5], ⇒ x = Objective_Maths_JEE Main 2017_Ch 2.indd 51 \ f (– x) = f (x), ∀ x ∈ [–5, 5] ⎛ x + 1⎞ ⇒ x = – ⎜ ⇒ x2 + 3x + = ⎝ x + ⎟⎠ ⇒ x = −3 ± (2) 01/01/2008 03:20:24 2.52 Chapter From (1) and (2), the values of x are −1 ± −3 ± and 2 The correct option is (A) and (B) 119 Let the required function be f (x) = ax + b If a > 0, then f (–1) = and f (1) = ⇒ –a + b = and a + b = ⇒ a = and b = If a < 0, then f (–1) = and f (1) = ⇒ –a + b = and a + b = ⇒ a = –1 and b = Hence, f (x) = x + 1 or f (x) = –x + The correct option is (B) and (C) 120 Taking f (x) = log x, we see that ⎛ x⎞ f ⎜ ⎟ = f (x) – f ( y) ⎝ y⎠ Clearly, f (x) is not bounded ⎛ 1⎞ and f ⎜ ⎟ = – log x → ∞ as x → ⎝ x⎠ Also, x f (x) = x log x → as x → The correct option is (C) and (D) 121 Since, f (– x) = f (x) ⇒ f (x) is an even function, its graph will be symmetrical about y-axis ⎛ | x |3 + | x | ⎞ Also, f (x) = − ⎜ ⎝ + x ⎟⎠ ⇒ f (x) = –(positive) = negative i.e., the graph of f (x) completely lies below the x-axis, and is also symmetric about y-axis (as discussed above) \ The graph of f (x) lies in III and IV quadrants The correct option is (C) and (D) 122 When x1 = –1 and x2 = 1, ⎛ −1 − ⎞ then f (– 1) – f (1) = f ⎜ = f (– 1) ⎝ + 1(1) ⎟⎠ ⇒ f (1) = 0, ⎛1 − x⎞ which is satisfied when f (x) = tan − ⎜ ⎝ + x ⎟⎠ When x1 = x2 = 0, then ⎛ − 0⎞ f (0) – f (0) = f ⎜ = f (0) ⇒ f (0) = ⎝ − ⎟⎠ When x1 = –1 and x2 = 0, then ⎛ −1 − ⎞ = f (– 1) ⇒ f (0) = 0, f (–1) – f (0) = f ⎜ ⎝ − ⎟⎠ ⎛1 − x⎞ which is satisfied when f (x) = log ⎜ ⎝ + x ⎟⎠ ⎛1 + x⎞ and, f (x) = log ⎜ ⎝ − x ⎟⎠ The correct option is (A), (B) and (C) Objective_Maths_JEE Main 2017_Ch 2.indd 52 2p = 2, and period of {x} is p Hence, period of the given function is L.C.M of (1, 2) = ⎛p ⎞ ⎛p ⎞ (B) Solving tan ⎜ [ x + T ]⎟ = tan ⎜ [ x ]⎟ ⎝2 ⎠ ⎝2 ⎠ 123 (A) The period of cos p x is i.e., [x + T] – [x] = 2n gives a value of T independent of x only if T is an integer In that case, the above equation reduces to [x] + T – [x] = 2n i.e., T = 2n Hence, period of f (x), is the smallest positive value of T, i.e., (C) We have period of sin x = 2p and period of {x} = Hence, period of the given functions is L.C.M of (2p, 1) which does not exist since 2p is an irrational number Hence, the function is not periodic (D) Let us solve sin{cos (x + T)} = sin{cos x} i.e., cos (x + T) = np + (– 1) n cos x, n ∈ I Putting n = 0, gives cos (x + T) = cos x, which gives T = 2p as the smallest positive value For no other value of n can a value of T be found independent of x Hence, the required fundamental period is 2p The correct option is (A) and (B) 124 The function is defined for all real values of x except those which satisfy the equation [|x – 1|] + [|7 – x|] – = 0 (1) Case I: (1 < x < 7) Equation (1) reduces to [x – 1] + [7 – x] – = i.e., [x] – + [– x] + – = or [x] + [– x] = which is true ∀ x ∈ I Thus, every integer in (1, 7) satisfies equation (1) Case II: (x ≤ 1) Equation (1) reduces to [1 – x] + [7 – x] – = i.e., 1 + [– x] + + [– x] – = i.e., [– x] = – i.e., – ≤ – x < or < x ≤ Thus, equation (1) is satisfied ∀ < x ≤ Case III: (x ≥ 7) Equation (1) reduces to [x – 1] + [x – 7] – = i.e., [x] – + [x] – – = or [x] = \ 7 ≤ x < Thus, equation (1) is satisfied ∀ ≤ x < The union of the intervals obtained in the above three cases gives the domain of definition as R – (0, 1] – [7, 8) – {2, 3, 4, 5, 6} The correct option is (A), (B) and (C) 01/01/2008 03:20:26 Functions 2.53 125 f (x) = x(sin x + tan x ) x(sin x + tan x ) = ⎡x + p⎤ ⎡x⎤ ⎢ p ⎥−2 ⎢p ⎥ + − ⎣ ⎦ ⎣ ⎦ x(sin x + tan x ) ⎡x⎤ ⎢ p ⎥ + 0.5 ⎣ ⎦ − x(sin( − x ) + tan( − x )) ⇒ f (–x) = ⎡− x⎤ ⎢ p ⎥ + 0.5 ⎣ ⎦ = x(sin x + tan x ) ⎡x⎤ −1 − ⎢ ⎥ + 0.5 ⎣p ⎦ ⎧ ⎪⎪ ⇒ f (–x) = ⎨ ⎪ ⎪⎩ , x ≠ np , x = np ⎞ ⎛ ⎜ x(sin x + tan x ) ⎟ ⇒ f (–x) = – ⎜ ⎟ , when x ≠ np ⎜ ⎡ x ⎤ + 0.5 ⎟ ⎟⎠ ⎜⎝ ⎢ p ⎥ ⎣ ⎦ and f (–x) = 0, when x = np Hence, f (x) is an odd function (if x ≠ np) and f (x) is an even function (if x = np) The correct option is (C) and (D) 126 Let us check for invertibility of f (x) (A) one-one: we have, f (x) = ⇒ f ′(x) = e x + e− x e2 x + , which is strictly increasing as e2x 2e x > for all x Thus, f is one-one (B) Onto; Let y = f (x) e x + e− x , where y is strictly monotonic Hence, range of f (x) = ( f (–∞), f (∞)) ⇒ range of f (x) = (–∞, ∞) So, range of f (x) = co-domain Hence, f (x) is one-one and onto ⇒ y = e2 x − 2e x 2x x ⇒ e – 2e y – = ⇒ ex = ⇒ x = log(y ± ⇒ f –1(y) = log ( y ± (C) To find f –1 : y = Since, e f sign y ± y2 + −1 ( x) y2 + ) y + 1) is always positive, so, neglecting negative Hence, f –1(x) = log(x + x + ) The correct option is (A), (B) and (D) Objective_Maths_JEE Main 2017_Ch 2.indd 53 127 We have, f (x) = sin–1(log [x]) + log(sin–1[x])(1) Let g(x) = sin–1(log [x])(2) and, h(x) = log(sin–1[x])(3) Now for g(x); – ≤ log [x] ≤ {as sin–1 q exists when – ≤ q ≤ 1} and, [x] > {as log [x] exists when [x] > 0} ⇒ ≤ [x] ≤ e and [x] > e ⇒ [x] = 1, ⇒ x ∈ [1, 3) (4) Again, from (3), we have h(x) = log (sin–1[x]) exists when; sin–1 [x] > and – ≤ [x] ≤ ⇒ [x] > and –1 ≤ [x] ≤ ⇒ 0 < [x] ≤ ⇒ [x] = ⇒ x ∈ [1, 2) (5) ⇒ Domain of f (x) is [1, 2) Now, for range, we know, f (x) = sin–1(log[x]) + log(sin–1[x]) where x ∈ [1, 2) ⇒ [x] = \ Range of f (x) = sin–1(log 1) + log(sin–1 1) ⎛p⎞ = sin–1(0) + log ⎜ ⎟ ⎝ 2⎠ = log(p/2) p⎫ ⎧ ⇒ Range of f (x) = ⎨log ⎬ 2⎭ ⎩ The correct option is (A) and (C) 128 (A) one-one: f (x) = x − x ⇒ f ′(x) = x − x (2x – 1) · log2 For f (x) to be one-one, it should be strictly increasing or strictly decreasing So, f ′(x) > ⇒ x − x (2x – 1) > 0, where 2x2 – x > for all x ⇒ 2x – > or x > Thus, for given domain [1, ∞), f (x) is always increasing Hence, f is one-one (B) onto: As f (x) is strictly increasing ⇒ Range f (x) ∈ [ f (1), f (∞)) ⇒ Range f (x) ∈ [1, ∞) ⇒ Range of f (x) = Co-domain of f (x), thus, f is onto (C) Inverse: As f is one-one and onto, f –1 can be obtained Let y = f (x) ⇒ y = 2x2 – x ⇒ x2 – x = log2 y ⇒ x2 – x – log2 y = 01/01/2008 03:20:29 2.54 Chapter ⇒ x = ± + log y ⇒ f –1(y) = ⇒ + + log y [as y > 0, ∀ x ∈ D] The correct option is (A), (B) and (C) 9x 129 f (x) = x (1) +3 91 − x and, f (1 – x) = − x +3 x 9 ⇒ f (1 – x) = = 9 + 3.9 x +3 x 9 f (1 – x) = (2) 3(3 + x ) Adding (1) and (2), we get f (x) + f (1 – x) = = 9x + x + 3(3 + x ) x + 3(9 x + 3) = x 3(9 + 3) 3(9 x + 3) \ f (x) + f (1 – x) = 1 (3) 998 , , , , in (3), we get Now, putting x = 1996 1996 1996 1996 ⎛ ⎞ ⎛ 1995 ⎞ + f⎜ =1 f ⎜ ⎝ 1996 ⎟⎠ ⎝ 1996 ⎟⎠ ⎛ ⎞ ⎛ 1994 ⎞ + f⎜ =1 f⎜ ⎝ 1996 ⎟⎠ ⎝ 1996 ⎟⎠ ⎛ ⎞ ⎛ 1993 ⎞ f⎜ + f⎜ =1 ⎝ 1996 ⎟⎠ ⎝ 1996 ⎟⎠ … … … … … … ⎛ 997 ⎞ ⎛ 999 ⎞ ⇒ f ⎜ + f⎜ =1 ⎝ 1996 ⎟⎠ ⎝ 1996 ⎟⎠ ⇒ ⇒ ⎛ 998 ⎞ ⎛ 998 ⎞ ⎛ 998 ⎞ f⎜ = + f⎜ = or f ⎜ ⎝ 1996 ⎟⎠ ⎝ 1996 ⎟⎠ ⎝ 1996 ⎟⎠ Adding all the above expressions, we get ⎛ ⎞ ⎛ ⎞ ⎛ 1995 ⎞ f ⎜ + f⎜ + + f ⎜ ⎝ 1996 ⎟⎠ ⎝ 1996 ⎟⎠ ⎝ 1996 ⎟⎠ = (1 + + + + 997) + 1 = 997 + = 997 2 The correct option is (A) and (D) 130 We have f (n + 2) – f (n + 1) = (n + 2) ! = (n + 2) (n + 1) ! = (n + 2) [ f (n + 1) – f (n)] ⇒ f (n + 2) = (n + 3) f (n + 1) – (n + 2) f (n) \ P (x) = x + and Q(x) = –x – The correct option is (A) and (B) Passage Based Questions 131 sin x is a periodic function with period 2p, therefore 2p sin ( [n] x) is a periodic function with period [n] But the period of f (x) is 2p (given) 2p = 2p ⇒ [n] = ⇒ [n] = ⇒ ≤ n < [n] The correct option is (A) 132 f (x + T) = f (x) ⇒ cos (sin (x + T) + cos (cos (x + T) = cos (sin x) + cos (cos x) If x = 0, then cos (sin T) + cos (cos T) \ p⎞ ⎛ ⎛ p⎞ = cos(0) + cos (1) = cos ⎜ cos ⎟ + cos ⎜ sin ⎟ ⎝ ⎠ ⎝ 2⎠ p On comparing, we get T = The correct option is (A) Objective_Maths_JEE Main 2017_Ch 2.indd 54 p 133 Since |sin x| + |cos x| is a periodic function with period , p when k = therefore period of f (x) will be The correct option is (A) px 134 3x + – [3x + 3] has the period and sin has the 2p i.e., Therefore, the period of f (x) is L.C.M period p /2 (1, 4) = The correct option is (A) p 135 The period of |sin x| + |cos x| and sin4x + cos4x is sin (sin x) + cos x + sin (cos x) has period 2p The function sin x ( + sec x ) cos x = cot x, so it can be written in a simplified form as sin x has period p The correct option is (D) 01/01/2008 03:20:34 Functions 2.55 2p 2p 136 The period of sin 5x is and that of cos 3x is 2p 2p As and not have a common multiple, f (x) is non-periodic The correct option is (C) 137 We have, ex + ef (x) = e ⇒ ef (x) = e – ex ⇒ f (x) = log (e – ex) For f (x) to be defined, e – ex > ⇒ e1 > ex ⇒ x < \ Domain of f = (– ∞, 1) Let y = log (e – ex) ⇒ ey = e – ex x y ⇒ e = e – e ⇒ x = log (e – ey) For x to be real, e – ey > ⇒ e1 > ey ⇒ y < \ Range of f = (– ∞, 1) The correct option is (B) 138 Since [x2 + 1] is an integer, \ sin (p[x2 + 1]) = sin (p [ x + 1]) ⇒ f (x) = =0 x4 + Hence, Range of f = Rf = {0} The correct option is (C) 139 We have, f (x) = + sin x cos x cos x (cos x ) + sin x ( cos x sin x ) cos x 2 cos x (cos x + sin x ) + sin x = = cos x cos3 x ⇒ f ′(x) > \ f (x) is increasing function ⇒ f ′(x) = sin x ⎞ ⎛ =–∞ lim ⎜1 + ⎟ −p ⎝ cos x⎠ x→ sin x ⎞ ⎛ =∞ and, lim ⎜1 + ⎟ p ⎝ cos x⎠ x→ \ Range = (–∞, ∞) The correct option is (B) 140 For f (x) to be defined, (1) [2x2 – 3] = –1, 0, ⇒ –1 ≤ 2x2 – < ⇒ ≤ 2x2 < 5 ⇒ 1 ≤ x2 < ⎧ ≤ x ⇒ x ≤ −1 or x ≥ ⎪ ⇒ ⎨ 5 ⇒ − (1) 5− 5+ or x > (2) 2 (3) log1/2 (x2 – 5x + 5) > ⇒ x < ⎛ 1⎞ ⇒ x2 – 5x + < ⎜ ⎟ ⇒ x2 – 5x + < ⎝ 2⎠ ⇒ x2 – 5x + < ⇒ 1 < x < 4 5− From Eqs (1), (2) and (3), ≤ x < The correct option is (D) 141 If f (x) ≥ 0, then x + f (x) = 2f (x) or, f (x) = x \ f –1(x) = x, when f–1(x) ≥ 0 Also, when f (x) ≤ 0, x – f (x) = 2f (x) x or, f (x) = \ f –1(x) = 3x, when f –1(x) ≤ 0 Clearly, option (d) satisfies both (1) and (2) The correct option is (D) (3) (1) (2) Match the Column Type 142 I. f (x) is defined if | sin x | + sin x > ⇒ sin x > ⇒ 2np < x < 2np + p \ Domain of f = (2np, (2n + 1) p) The correct option is (B) II. f (x) is defined if ⎞ ⎛ – log1/2 ⎜1 + ⎟ – > 0, + > 0, x ≠ ⎝ x x ⎠ ⎞ ⎛ ⇒ log1/2 ⎜1 + ⎟ < – 1, x1/5 + > 0, x ≠ ⎝ x ⎠ Objective_Maths_JEE Main 2017_Ch 2.indd 55 −1 ⎛ 1⎞ > ⎜ ⎟ , x > (– 1)5, x ≠ x1 ⎝ ⎠ ⇒ 1 + ⇒ ⇒ 0 < x < and x > – ⇒ < x < \ Domain ( f ) = (0, 1) The correct option is (C) > 1, x > – and x ≠ x1 01/01/2008 03:20:37 2.56 Chapter III. f (x) is defined if –(log3 x)2 + log3 x – > and x > ⇒ (log3 x – 3) (2 – log3 x) > and x > ⇒ (log3 x – 2) (log3 x – 3) < and x > ⇒ 2 < log3 x < and x > ⇒ 32 < x < 33 ⇒ < x < 27 Domain of f = (9, 27) The correct option is (D) x IV. Domain of cot–1x is R and is defined if x − [x2 ] x2 ≠ [x2] (Qx2 ≥ [x2]) ⇒ x2 ≠ or positive integer Hence, domain = R – { n : n ≥ 0, n ∈ I} The correct option is (A) 143 I. f (x) is defined if 3x2 – 4x + ≥ ⎡⎛ 5⎤ ⎞ 11⎤ ⎡ ⇒ 3 ⎢ x − x + ⎥ ≥ ⇒ ⎢⎜ x − ⎟ + ⎥ ≥ 3⎦ 3⎠ ⎥⎦ ⎣ ⎢⎣⎝ which is true for all real x \ Domain ( f ) = (–∞, ∞) Let y = 3x − x + ⇒ y2 = 3x2 – 4x + i.e., 3x2 – 4x + (5 – y2) = For x to be real, 16 – 12 (5 – y2) ≥ ⇒ y ≥ ⎡ 11 ⎞ \ Range of y = ⎢ , ∞⎟ ⎠ ⎢⎣ The correct option is (C) II. f (x) is defined if 3x2 – 4x + > 11 ⎡⎛ 5⎤ ⎞ 11⎤ ⎡ ⇒ 3 ⎢ x − x + ⎥ > ⇒ ⎢⎜ x − ⎟ + ⎥ > 0, 3⎦ 3⎠ ⎥⎦ ⎣ ⎢⎣⎝ which is true for all real x \ Domain ( f ) = (– ∞, ∞) Let, y = loge (3x2 – 4x + 5) ⇒ ey = 3x2 – 4x + ⇒ 3x2 – 4x + (5 – ey ) = For x to be real, 11 16 – 12 (5 – ey) ≥ ⇒ 12 ey ≥ 44 ⇒ ey ≥ ⇒ y ≥ loge 11 ⎡ 11 ⎞ Range of f = ⎢log e , ∞⎟ ⎠ ⎣ The correct option is (A) III. f (x) is defined if x2 + x – ≠ i.e., (x + 3) (x – 2) ≠ i.e., x ≠ – 3, \ Domain ( f ) = (– ∞, ∞)\{– 3, 2} x − 3x + Let y = x + x −6 ⇒ x2y + xy – 6y = x2 – 3x + ⇒ x2 (y – 1) + x (y + 3) – (6y + 2) = Objective_Maths_JEE Main 2017_Ch 2.indd 56 For x to be real, (y + 3)2 + (y – 1) (6y + 2) ≥ ⇒ 25y2 – 10y + ≥ i.e., (5y – 1)2 ≥ which is true for all real y \ Range of f = (– ∞, ∞) The correct option is (B) IV. For f (x) to be defined, − x2 > 0, – x2 > and – x ≠ 1− x Since − x and – x2 > ⇒ x < and (x – 2) (x + 2) < ⇒ x < and – < x < ⇒ – < x < \ Domain of f = (– 2, 1) ⎛ − x2 ⎞ Since – ∞ < log ⎜ ⎟ ⎝ 4⎠ which is true " x ∈ R Now, we have p⎞ ⎛ –2 ≤ sin ⎜ x − ⎟ ≤ ⎝ 4⎠ p⎞ ⎛ i.e., 1 ≤ sin ⎜ x − ⎟ + ≤ ⎝ 4⎠ ⎧ ⎫ p⎞ ⎛ i.e., 0 ≤ log ⎨2 sin ⎜ x − ⎟ + 3⎬ ≤ log 5 ⎝ 4⎠ ⎩ ⎭ i.e., 0 ≤ y ≤ Hence, the range is y ∈ [0, 2] The correct option is (C) II. The function is defined for values of x such that – log (16 sin2x + 1) > Also, we have – log (16 sin2x + 1) ≤ [Q log Together, we have < – log 5 (16 sin2x + 1) ≥ 0] (16 sin2 x + 1) ≤ 01/01/2008 03:20:41 Functions 2.57 i.e., –∞ < log2{2 – log (16 sin2 x + 1)} ≤ log22 i.e., –∞ < y ≤ Hence, the range is y ∈(–∞, 1] The correct option is (B) III. We have, ex − ex y = x = 0, x < e + e− x e x − e− x − e− 2x = ,x≥0 = x e + ex Now, we have " x ≥ 0, < e–2x ≤ i.e., – ≤ – e – 2x < i.e., ≤ – e–2x < 1 − e− 2x 1 i.e., 0 ≤ < i.e., ≤ y < 2 ⎡ 1⎞ Hence, the range is y ∈ ⎢0, ⎟ ⎣ 2⎠ The correct option is (D) IV. We have, e x − e− x e2 x + − = 1− = y= x −x + e2 x e +e e2 x + Now, we have " x ≥ 0, ≤ + e2x < ∞ 1 i.e., ≥ >0 + e2 x −2