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 do. 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 - We look forward to it.

Objective_Maths_JEE Main 2019_Prelims.indd 1

5/18/2018 3:11:34 PM


This page is intentionally left blank.

Objective_Maths_JEE Main 2019_Prelims.indd 2

5/18/2018 3:11:34 PM


A Complete Resource Book in

Mathematics
for

Objective_Maths_JEE Main 2019_Prelims.indd 3

JEE Main 2019

5/18/2018 3:11:35 PM


This page is intentionally left blank.

Objective_Maths_JEE Main 2019_Prelims.indd 4

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

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 2 & 9, Rajiv Gandhi Salai,
Taramani, Chennai 600 113, Tamil Nadu, India.
Fax:  080-30461003, Phone: 080-30461060
website: in.pearson.com, Email: 
Compositor: SRS Global, Puducherry
Printed in India by

Objective_Maths_JEE Main 2019_Prelims.indd 6

5/18/2018 5:08:27 PM


Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix
JEE Mains 2018 Paper                                                   xiii
JEE Mains 2017 Paper                                                   xxv
Chapter 1

Set Theory                                              11–118

Chapter 2

Functions                                               21–260

Chapter 3

Complex Numbers                                          31–368

Chapter 4

Quadratic Equations and Expressions                                41–452

Chapter 5

Matrices                                               51–530

Chapter 6

Determinants                                             61–656

Chapter 7

Permutations and Combinations                                   71–746

Chapter 8

Mathematical Induction                                        81–84

Chapter 9

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 7

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 8

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
•
•
•
•
•
•

Completecoverageoftopicsalongwithamplenumberofsolvedexamples.
Largevarietyofpracticeproblemswithcompletesolutions.
Chapter-wisePrevious15years’AIEEE/JEEMainquestions.
FullysolvedJEEMain2018and2017questions.
5MockTestsbasedonJEEMainpatterninthebook.
5FreeOnlineMockTestsaspertherecentJEEMainpattern.

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.
AnujAgarwalfromIIT-Delhi,AnkitKatialfromNationalInstituteofTechnology(Kurukshetra)andRaudrashish
ChakrabortyfromKiroriMalCollege,UniversityofDelhi,withwhomIhavehadfruitfuldiscussions,deservespecial
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.
Despiteofourbestefforts,someerrorsmayhavecreptintothebook.Constructivecommentsandsuggestionsto
further improve the book are welcome and shall be acknowledged gratefully.
Best of luck!
Dinesh Khattar

Objective_Maths_JEE Main 2019_Prelims.indd 9

5/18/2018 3:11:36 PM


This page is intentionally left blank.

Objective_Maths_JEE Main 2019_Prelims.indd 10

5/18/2018 3:11:36 PM


Mathematics Trend Analysis
(2007 to 2018)
S. No Chapters

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

Set Relation and
Function
Complex Numbers and
Quadratic Equation
Permutation and
Combination
Mathematical
Induction
Binomial Theorem and
ItsApplication

Sequence and Series
Exponential and
Logarithms Series
Matrices and
Determinant
Statistics and
Probability
Trigonometry
Mathematical
Reasoning
Coordinate Geometry
Limit, Continuity and
Differentiability
Integral Calculus
DifferentialEquation/
Calculus
ThreeDimensionalGeometry
VectorAlgebra
Dynamics
Applicationof
Derivative
Total No. of Questions

Objective_Maths_JEE Main 2019_Prelims.indd 11

07

08

09

10

11

12

13

14

15

16

17

18

2

2

3

2

2

3

2

1

1

1

1

1

2

1

3

3

2

2

1

2

3

3

2

2

3

1

1

2

1

2

1

1

1

1

–

1

1

1

1

–

1

–

–

–

1

–

–

–

–

–

–

0

1

1

1

1

1

–

1

1

1

1

1

1

1

2

1

1

1

1

1

2

2

2

2

2

1

2

1

–

–

–

–

–

–

–

–

–

–

–

–

3

3

2

3

2

3

2

2

2

2

2

2

2

3

3

4

2

3

2

2

2

2

2

2

3

2

2

2

1

2

1

1

1

2

2

2

2

2

2

–

2

1

–

1

1

1

1

1

1

1

1

1

7

5

4

5

4

5

5

5

5

5

5

5

7

4

2

3

3

4

4

2

1

4

2

2

1

2

4

3

2

2

2

2

3

5

3

3

3

3

3

1

1

1

2

2

1

1

1

1

1

1

2

1

3

2

1

1

2

2

2

2

2

2

2

2

1

2
2

2
–

2
–

2
–

2
–

2
–

2
–

1
–

1
–

1
–

1
–

1
–

1
0

1

–

–

–

–

–

1

–

–

2

2

1

1

40

35

30

30

30

30

30

30

30

30

30

30

30

5/18/2018 3:11:37 PM


This page is intentionally left blank.

Objective_Maths_JEE Main 2019_Prelims.indd 12

5/18/2018 3:11:37 PM


JEE MAINs 2018 paper
1.Two sets A and B are as under:

A = {(a, b) ∈ R × R: |a – 5| < 1 and |b – 5| < 1};

B = {(a, b) ∈ R × R: 4(a – 6)2 + 9(b – 5)2 ≤ 36}.

Then:
(A)  B ⊂ A
(B) A ⊂ B
(C) A ∩ B = ϕ (an empty set)
(D)  neither A ⊂ B nor B ⊂ A
2.Let S = {x ∈ R: x ≥ 0 and 2| x – 3| +
+ 6 = 0}. Then S
(A)  is an empty set
(B)  contains exactly one element
(C)  contains exactly two elements
(D)  contains exactly four elements

x ( x – 6)

3.If α, β ∈ C are the distinct roots, of the equation x2 –
x + 1 = 0, then α101 + β107 is equal to
(A)  –1
(B)  0
(C)  1
(D)  2
x − 4 2x
2x
4.If 2 x x − 4 2 x = (A + Bx)(x – A)2, then the
2x
2x x − 4
ordered pair (A, B) is equal to:

(A)  (–4, –5)
(C)  (4, 5)

(B)  (–4, 3)
(D)  (–4, 5)

5.If the system of linear equations, x + ky + 3z = 0,
3x + ky – 2z = 0, 2x + 4y – 3z = 0 has a non-zero soluxz
tion (x, y, z), then 2 is equal to
y


(A)  –10

(B)  10

(C)  –30

(D)  30

6.From 6 different novels and 3 different dictionaries, 4  novels and 1 dictionary are to be selected and
arranged in a row on a shelf so that the dictionary is
always in the middle. The number of such arrangements is
(A)  at least 1000
(B)  less than 500
(C)  at least 500 but less than 750
(D)  at least 750 but less than 1000
7.The sum of the coefficients of all odd degree terms in
the expansion of

Objective_Maths_JEE Main 2019_Prelims.indd 13



(x +




(A)  –1

x3 − 1

) +(x −

x 3 − 1 , ( x > 1)

(B)  0

(C)  1

5

)

5

(D)  2
12

8.Let a1, a2, a3, …, a49 be in A.P. such that



∑ a4 k +1 = 416

k =0

2
and a9 + a43 = 66. If a12 + a22 +…+ a17
= 140 m , then
m is equal to
(A)  66
(B)  68
(C)  34
(D)  33

9.Let A be the sum of the first 20 terms and B be the sum
of the first 40 terms of the series 12 + 2 · 22 + 32 + 2 ·
42 + 52 + 2 · 62 + …. If B – 2A = 100λ, then λ is equal
to
(A)  232
(B)  248
(C)  464
(D)  496
10. For each t ∈ R, let [t] be the greatest integer less than
or equal to t. Then

⎛⎡1⎤ ⎡2⎤
⎡15 ⎤ ⎞
lim x ⎜ ⎢ ⎥ + ⎢ ⎥ +…+ ⎢ ⎥ ⎟
x →0 + ⎝ ⎣ x ⎦ ⎣ x ⎦
⎣ x ⎦⎠




(A)  is equal to 0.
(B)  is equal to 15.
(C)  is equal to 120. (D)  does not exist (in R).

11. Let S = {t ∈ R: f(x) = |x – π|⋅(e|x| – 1) sin |x| is not differentiable at t}. Then the set S is equal to
(A)  ϕ (an empty set)
(B)  {0}
(C) {π}


(D)  {0 , π}
12. If the curves y2 = 6x, 9x2 + by2 = 16 intersect each
other at right angles, then the value of b is
7
9
(A)  6
(B) 
(C)  4
(D) 
2
2

13. Let f(x) = x2 +



1
2

and g(x) = x –

1
, x ∈ R – {–1,
x

x
f ( x)
0, 1}. If h(x) =
, then the local minimum value
g( x)
of h(x) is
(A)  3
(B)  –3
(C)  −2 2 (D)  2 2

14. The integral

∫

sin 2 x cos 2 x

(sin 5 x + cos3 x sin 2 x + sin 3 x cos 2 x + cos5 x ) 2

is equal to

dx

5/18/2018 3:11:40 PM


xiv  JEE Mains 2018 Paper


(A) 

(C) 

1
3(1 + tan 3 x )
1
3

1 + cot x

+ C

+C

(B) 

(D) 

−1
3(1 + tan 3 x )
−1
1 + cot 3 x

+C

+C

(where C is a constant of integration)
π
2

15. The value of

sin 2 x

∫ 1 + 2 x dx

is

π
−
2



π

(A) 
8

π
(B) 
2

π
(D) 
4

(C) 4π

16. Let g(x) = cos x2, f(x) = x , and α, β(α < β) be the
roots of the quadratic equation 18x2 – 9πx + π2 = 0.
Then the area (in sq. units) bounded by the curve y =
(gof) (x) and the lines x = α, x = β and y = 0, is


1
2
1
(C) 
2

(A) 

(
(

)

3 −1

1
2
1
(D) 
2

(B) 

)

(

)

3 +1

(

)

3− 2
2 −1

17. Let y = y(x) be the solution of the differential equation
dy
π 

sin x
+ y cos = 4x, x ∈ (0, π). If y   = 0, then
dx
2
⎛π ⎞
y ⎜ ⎟ is equal to
⎝6⎠
4 2
−8 2
(A) 
π
(B) 
π
9 3
9 3
8
4
(C)  − π 2
(D)  − π 2
9
9

18. A straight line through a fixed point (2, 3) intersects
the coordinate axes at distinct points P and Q. If O is
the origin and the rectangle OPRQ is completed, then
the locus of R is
(A)  3x + 2y = 6
(B)  2x + 3y = xy
(C) 3x + 2y = xy
(D)  3x + 2y = 6xy

19. Let the orthocentre and centroid of a triangle be
A(–3,  5) and B(3, 3), respectively. If C is the circumcentre of this triangle, then the radius of the circle
having line segment AC as diameter, is



(A) 

3 5

2

(C)  3

5

2

Objective_Maths_JEE Main 2019_Prelims.indd 14

(B)  2 10

20. If the tangent at (1, 7) to the curve x2 = y – 6 touches the
circle x2 + y2 + 16x + 12y + c = 0 then the value of c is
(A)  195
(B)  185
(C)  85
(D)  95
21. Tangent and normal are drawn at P(16, 16) on the
parabola y2 = 16x, which intersect the axis of the parabola at A and B, respectively. If C is the centre of the

circle through the points P, A and B and ∠CPB = θ,
then a value of tan θ is
1
4
(A) 
(B) 2
(C)  3
(D) 
2
3
22. Tangents are drawn to the hyperbola 4x2 – y2 = 36 at
the points P and Q. If these tangents intersect at the
point T(0, 3) then the area (in sq. units) of ΔPTQ is


(A)  45 5 (B) 54 3 (C) 60 3 (D)  36 5

23. If L1 is the line of intersection of the planes 2x – 2y +
3z – 2 = 0, x – y + z + 1 = 0 and L2 is the line of intersection of the planes x + 2y – z – 3 = 0, 3x – y + 2z –
1 = 0, then the distance of the origin from the plane,
containing the lines L1 and L2, is
1
1
(A) 

(B) 
4 2
3 2
(C) 

1
2 2

10

(D) 

1

2

24. The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the plane, x +
y + z = 7 is
2
2
2
1
(A) 

(B) 
(C) 
(D) 
3
3
3
3
25. Let u be a vector co-planar with the vectors
a = 2iˆ + 3 ˆj − kˆ and b = ˆj + kˆ. If u is perpendicular
2
to a and u ⋅ b = 24, then u is equal to

(A)  336
(B)  315
(C)  256
(D)  84
26. A bag contains 4 red and 6 black balls. A ball is drawn
at random from the bag, its color is observed and this
ball along with two additional balls of the same color
are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn
ball is red, is
3
2
1
3
(A) 

(B) 
(C) 
(D) 
10
5
5
4
9

27. If

∑ ( xi − 5) = 9
i =1

(D) 

9

and

∑ ( xi − 5)2 = 45,

then the stan-

i =1

dard deviation of the 9 items x1, x2, …, x9 is
(A)  9
(B)  4
(C)  2
(D)  3

5/18/2018 3:11:48 PM


JEE Mains 2018 Paper  xv
28. If sum of all the solutions of the equation
⎛
⎛π
⎞
⎛π

⎞ 1⎞
8 cos x ⋅ ⎜ cos ⎜ + x ⎟ ⋅ cos ⎜ − x ⎟ − ⎟ = 1 in [0, π]
⎝6
⎠
⎝6
⎠ 2⎠
⎝


is kπ, then k is equal to
2
13
(A) 
(B) 

3
9

(C) 

8

9

(D) 

20
9

29. PQR is a triangular park with PQ = PR = 200 m. A T.V.

tower stands at the midpoint of QR. If the angles of

elevation of the top of the tower at P, Q and R are
respectively 45°, 30° and 30°, then the height of the
tower (in m) is
(A)  100

(B)  50
(C) 100 3
(D)  50 2
30. The Boolean expression: ∼ (p ∨ q) ∨ (∼ p ∧ q) is equivalent to
(A)  p(B) 
p(C) 
q
(D)  ∼p

Answer keys
1. (B)
11.  (A)
21. (B)

2. (C)
12.  (D)
22.  (A)

3. (C)
13.  (D)
23. (B)

4.  (D)

14.  (A)
24.  (D)

5. (B)
15.  (D)
25.  (A)

6.  (A)
16.  (A)
26. (B)

7.  (D)
17. (C)
27. (C)

8. (C)
18. (C)
28. (B)

9.  (B)
19.  (C)
29.  (A)

10. (C)
20.  (D)
30.  (D)

Hints and Solutions
1.The figure is represented as
A

Even through x and y ≠ 4, 6. We will check whether the
boundary lies in the ellipse B
  (i)(x, y) = (4, 4)

 (ii) (x, y) = (4, 6)

(iii)(x, y) = (6, 4)

  (iv) (x, y) = (6, 6)

Objective_Maths_JEE Main 2019_Prelims.indd 15

Result

y

  (i)

4

4

 (ii)

4

6

25
<1
36

Inside

(iii)

6

4

Inside

  (iv)

6

6

1
<1
4
1
<1
4

B

The region A fully falls in the region B, hence A is the subset
of B

Put a = x and b = y
( x − 6) 2 ( y − 5) 2
B represents an ellipse
+
≤1
9
4

A represent a rectangle lines
−1 < x − 5 < 1
4< x<6


−1 < y − 5 < 1
4< y<6

( x − 6 )2 ( y − 5)2
+
9
4
25
<1
36

x

Inside

Inside


Hence check also prove that A lies wholly in B.
2.We know that S = { x ∈ R : x ≥ 0}

2 x − 3 + x ( x − 6) + 6 = 0
Case I: 0 ≤ x < 9 or x ∈[0, 9)

The equation reduces to

⇒ 2(3 − x ) + x ( x − 6) + 6 = 0

⇒ 6 − 2 x + x − 6 x + 6 = 0

⇒ x − 8 x + 12 = 0

⇒ x − 6 x − 2 x + 12





x [ x − 6] − 2[ x − 6] = 0
( x − 6)( x − 2) = 0

5/18/2018 3:11:52 PM

xvi  JEE Mains 2018 Paper

∴ x = 22 = 4


x = 62 = 36

For Case 1


R1 → R1 – R2 and R2 → R2 – R3

x

x

Condition (x)

Result

0
x+4
0
⇒ (5 x − 4) 0 −( x + 4) x + 4

1
2x
x−4

1

4

2

[0, 9)

Accepted

2

36

6

[0, 9)

Rejected


⇒(5x – 4)[(x + 4)2 – 0]

⇒(5x – 4) ⋅ (x + 4)2

On comparing with (A + Bx)(x – A)2, we get A = –4 and
B=5

Case II: 9 ≤ x < ∞ or x ∈[9, ∞)

The equation reduces to

1 k 3
5. Δ = 3 k −2 = 0
2 4 −3


⇒ 2( x − 3) + x ( x − 6) + 6 = 0

⇒ 2 x − 6 + x − 6 x + 6

⇒ x − 4 x = 0

⇒ ( x − 0)( x − 4) = 0

∴ x = 0


x = 16
x

x

Condition (x)

Result

1

0

0

[9, ∞)

Rejected

2

16

4

[9, ∞)

Accepted


Hence, S contains exactly two solution
3.Given x2 – x + 1 = 0

Roots are α and β


 α + β = 1


αβ=1


1 + ω + ω2 = 0

 ω3 = 1


 –ω – ω2 = 1


α = –ω


 β = –ω2

To calculate

α101 + β107

⇒(–ω)101 + (–ω2)107
⇒ −1 × ⎡⎣(ω 33 )3ω 2 + ω 214 ⎤⎦




⇒–[ω2 + (ω3)71 ⋅ ω]

⇒–[ω + ω2] = –1 × –1 = 1
x − 4 2x
2x
4. 2 x x − 4 2 x
2x

2x x − 1

C1 → C1 + C2 + C3
⇒


5x − 4
5x − 4
5x − 4

2x
2x
x − 4 2x
2x x − 4

1 2x

⇒ (5 x − 4) 1 x − 4
1 2x

Objective_Maths_JEE Main 2019_Prelims.indd 16

2x
2x
x−4

[cube roots fully]


Then the equation has infinitely many solution.

On solving, we get

⇒1[kx –3 –(–2)x 4] –k[3x –3 –2x –2] + 3[3x 4 –2x k] = 0

⇒ –3k + 8 – kx – 5 + [12 – 2k] = 0

⇒–3k + 8 + 5k + 36 – 6k = 0

⇒–4k = –44


k = 11

The equation are


x + 11y + 3z = 0
(i)

3x + 11y – 2z = 0
(ii)

2x + 4y – 3z = 0
(iii)

Put z = k and taking equation (i) and (iii)


x + 11y = –3k

2x + 4y = 3k

∴ Δ =

1 11
= 4 − 22 = −18
2 4




Δx =

−3k 11
= −12k − 33k = −45k
3k 4




Δy =

1 −3k
= 3k + 6 k = 9k
2 3k


By Cramer Rule,
Δx −45k 5k

=
x=
=
Δ
−18
2
Δy 9k
−k


y=
=
=
Δ −18 2
5k
× k 5k 2 x 4
xz
2

∴
=
=
= 10
y2
k2
2 × k2
4
6.Refer to the position where N denotes Novel and D denotes
Dictionary

N

N D N

N


Number of ways of selection 4 novel and one dictionary
form 6 different novel and three different dictionary

⇒

6

C4 × 3C1

5/18/2018 3:11:56 PM


JEE Mains 2018 Paper  xvii

∴ Total number of arrangement are

⇒ 6 C4 × 3C1 × 4! 
[Position of dictionary is fixed]
6!
3!

⇒
×

× 4!
4! × 2! 2! × 1!
6! × 3!
= 1080

⇒
( 2!) 2
7.( x + x 3 − 1)5 + ( x − x 3 − 1)5



+ [5C0x5t0 + 5C2x3t2 + 5C4xt4 + other term]

Remove other terms

⇒2[5C0x5 + 5C2x3t2 + 5C4 xt4]

⇒2[x5 + 10x3 (x3 – 1) + 5x(x3 – 1)2]

⇒2[x5 + 10x6 – 10x3 + 5x(x6 – 2x3 + 1)]

⇒2[x5 + 10x6 – 10x3 + 5x7 – 10x4 + 5x]

Taking coefficient of odd degree term

⇒2[5x7 + x5 – 10x3 + 5x]

Sum of coefficient is

⇒ 2[5 + 1 – 10 + 5] = 2 × 1 = 2

8.Let the first term be ‘a’ and common difference be ‘d’


a9 + a43 = 66

(a + 8d) + (a + 42d) = 66

2a + 50d = 66


a + 25d = 33

(i)

∑ a4 k +1 = 416



a1 + a5 + a9 + … + a49 = 416

Number of terms of the above series


1 + (n – 1)4 = 49

(n – 1)4 = 48


n = 13

Common difference of series A is 4d
a1 = a

(A)

n
[2a + ( n − 1)4 d ] = 416
2

13
[2a + 48d ] = 416
2

13[a + 24d] = 416


a + 24d = 32

Solving equation (i) and (ii), we get


d = 1 and a = 8




In the series

+

th

+…+

2
a17

= 140 n


The n term is (8 + (n – 1) × 1)2

Tn = (n + 7)2

Objective_Maths_JEE Main 2019_Prelims.indd 17

17

17

17

n =1

n =1

n =1

= ∑ n2 + 14∑ n + 49∑1

n( n + 1)( 2n + 1) 14 n( n + 1)
⎫

⇒
+
+ 49n ⎬ n = 17
6
2
⎭

( 2n1 + 1)
6
( 2n + 1)
( 2n + 1)

S = n( n + 1)
+ 4 n1( n1 + 1) 1
6
6

A = S where n = 20 and n1 = 10

Solving, we get,
41
21


A = 20 × 21 × + 4 × 10 × 11 × = 4410

6
6


B=S

Solving, we get
81
41


B = 40 × 41 × + 4 × 20 × 21 × = 22140 + 11480
6
6


= 33620

B – 2A ⇒ 33620 – 2 × 4410


= 100λ

⇒24800 = 100λ


λ = 248
10. x = [x] + {x} [Integral + FRACTIONAL PART]
= 4 n1 ( n1 + 1)

k =0

a22

n =1

1

12

a12

n =1

17 × 18 × 35
18
+ 14 × 17 × + 49 × 17 = 140 m
6
2

⇒ 1785 + 2142 + 833 = 140m

⇒140m = 4760


m = 34
9.We are required to find the sum of 20 terms and 40 terms,

therefore the number of terms are even


S = 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + …


S = S1 + S2
S1 = 12 + 22 + 32 + 42 + …




S2 = 22 + 42 + 62+ …

Sum of n terms of series S1
( 2n + 1)
S1n = n( n + 1)


6

Sum of n1 terms of series S2 where
n
n1 =


2


S2 n = ∑ ( 2n1 ) 2 = 4∑ n12

⇒[5C0 x5t0 + 5C2x3t2 + 5C4 xt4 + other term]



17


⇒


( x 3 − 1)1 2 = t

⇒(x + t)5 + (x – t)5

Root needs to be removed

Hence, we need t 0 , t 2 , t 4




17

∑ Tn = ∑ ( n + 7)2

(ii)

⎛ 1 ⎧1 ⎫ 2 ⎧2 ⎫

⎞
lim x ⎜ − ⎨ ⎬ + − ⎨ ⎬ + …⎟
x →0
x
x
x
x
⎩ ⎭
⎩ ⎭
⎝
⎠
+

⎛ ⎧1⎫ ⎧2⎫ ⎧3⎫ ⎞
⎛1 2 3 ⎞
lim x ⎜ + + …⎟ − x ⎜ ⎨ ⎬ + ⎨ ⎬ + ⎨ ⎬ …⎟ ,
⎝x x x ⎠
x→0
⎝⎩x⎭ ⎩x⎭ ⎩x⎭ ⎠
+


⇒

⎡ n( n + 1) ⎤
n( n + 1) 15 × 16
lim x ⎢ 2 ⎥ − 0 ⇒ lim
=
= 120
x→0

x →0
2
2
⎢
⎥
⎣ x ⎦
 
+

+

5/18/2018 3:12:01 PM


xviii  JEE Mains 2018 Paper

(

12. y2 = 6x, 9x2 + by2 = 16

Set the parametric point of the parabola y2 = 6x
x = 6t2 and y = 6t
dy
dx
= 6 and
= 12t
dt
dt

)

x

11. f ( x ) = x − π e − 1 sin x

x<0

f ( x ) = −1( x − π )(e − x − 1)sin( − x )
f ( x ) = ( x − π )(e − x − 1)sin x when x < 0

dy 1
=
dx 2t
1

∴ The normal slope is
× m = −1
2t
m = –2t

Finding the slope of the ellipse 9x2 + by2 = 16

18x + 2byy′ = 0

[x = 6t2 and y = 6t because of intersection point]

y′ = –2t

0≤ x <π

f ( x ) = −1( x − π )(e x − 1)sin x



= (π − x )(e x − 1)sin x when 0 ≤ x < π


For x ≥ π
f ( x ) = ( x − π )(e x − 1)sin x
f (0) = (π − 0)(e 0 − 1)sin 0 = 0
f (π ) = (π − π )(eπ − 1) sin π = 0

Check for derivative or differentiability

At x = 0
f (0 + h) − f (0)
f ( 0 + h) − f ( 0 )
R.H.D = lim
= lim
h→0
h→0
( 0 + h) − 0
h
L.H.D = lim

h→0

f ( 0 − h) − f ( 0 )
f ( 0 − h) − f ( 0 )
= lim

h
→
0
(0 − h) − 0
−h

(π − (0 + h))(e( 0 + h) − 1) sin(0 + h)) − f (0)
h→0
(0 + h) − 0

R.H.D = lim

(π − h)(e h − 1) sin h − 0
=0
= lim
h→0
h

sin h


=1 
 hlim
→
0
h



((0 − h) − π )(e −( 0 − h) − 1) sin(0 − h)) − f (0)

h→0
(0 − h) − 0

L.H.D = lim

−(π + h)(e − h − 1) sin h − 0
=0
h→0
−h

= lim

sin h


=1 
 hlim
→
0
h




At x = 0 R.H.D = L.H.D hence, it is differentiable function

At x = π
R.H.D = lim

h→0

L.H.D = lim

h→0

R.H.D = lim

f (π + h) − f (0)
f (π + h) − f (π )
= lim
h→0
(π + h) − π
h
f (π − h) − f (0)
f (π − h) − f (0)
= lim
h→0
(π − h) − 0
−h
((π + h) − π )(e

(π + h )

− 1) sin(0 + h)) − f (π )
(π + h) − π

h(e(π + h) − 1) sin h − 0
=0
h→0
h

= lim

sin h 

= 1
 hlim
→
0
h



(π − (π + h))(e(π − h) − 1) sin(π + h)) − f (π )
h→0
(π − h) − π

L.H.D = lim

h→0


⇒ 108t 2 − 24bt 2 = 0

b=

108 9
=
24 2

13. f ( x ) = x 2 +

− h(e

(π − h )

− 1)( −sin h) − 0
=0
−h

sin h 

= 1
 hlim
 →0 h



At x = 0 R.H.D = L.H.D hence it is differentiable function.

Hence, the function is differentiable at x = π

Therefore, set S is an empty set of ϕ

Objective_Maths_JEE Main 2019_Prelims.indd 18

1
x2

1

x

g( x) = x −


1
x2 + 2
f ( x)
x

h( x ) =
=
1
g( x)
x−
x

x−

1
=t
x


∴

x2 +

x2 +

h→0

= lim


⇒ 18 × 6t 2 + 2b × 6t × ( −2t ) = 0

1
x2


h(t ) =

1
1

2
 f ( x ) = x + 2 and g ( x ) = x − x 
x


2

1⎞
⎛
=⎜x− ⎟ +2
x⎠
x2 ⎝
1

= t2 + 2

t2 + 2
2
=t+
t
t

h′(t ) = 1 −


2
t2

=0


t2 = 2
t=± 2

+ve

–ve

− 2

2

+ve

Hence, at t = 2 we have minimum value as the sign
changes from –ve to +ve.
2

h(t ) = t +
t
2
h( 2 ) = 2 +
=2 2

2

5/18/2018 3:12:09 PM


JEE Mains 2018 Paper  xix

14.

sin 2 x cos 2 x dx
∫ (sin5 x + cos3 x sin 2 x + sin3 x cos2 x + cos5 x)2
sin 2 x cos 2 x dx

∫ (sin 2 x(sin3 x + cos3 x)cos2 x(sin3 x + cos3 x))2


⇒

sin 2 x cos 2 x dx

∫ (sin3 x + cos3 x)2 (sin 2 x + cos2 x)2


⇒

2

2

sin x cos x dx

∫ (sin3 x + cos3 x)2  [sin x + cos x = 1]


⇒

2

2

∫ cos6 x(tan3 x + 1)2
2

αβ =



( β − α ) 2 = (α + β ) 2 − 4αβ

β=


π
and
3


α=

π
6

π 3

−1

−1
−1
1t
=
⇒
3 x − 1 3t
3(tan 3 x + 1)
sin 2 x

π 2

−π 2 1 +

2x

−π 2 sin

2

t ( − dt )
−t

1+ 2

π 2

2 sin 2t dt
t

t

−π 2

            (ii)

1+ 2

Adding equation (i) and (ii)

2I = ∫

π 2

−π 2

π 2

I =∫

0

∫

π 2

0


I=

[sin x ]ππ 36 ⇒

17. sin x

dy
+ y cos x = 4 x   x ∈ (0, π)
dx


⇒

dy

+ y cot x = 4 x cosec x
dx


⇒

dy
+ Py = Q where P and Q are function f(x)
dx
Pdx

a

a

∫ −a f ( x) dx = 2∫ 0


⇒ e ∫
f ( x ) dx

Pdx

sin 2 x dx

sin n x dx =

n −1 n − 3 1 π
⋅
… ⋅ when n is even

2 2
n
n

1 π π
× =
2 2 4

16. g(x) = cos x2


f ( x) = x


g[ f ( x )] = cos( x ) 2 = cos x

Given β > α

Objective_Maths_JEE Main 2019_Prelims.indd 19

Pdx

dx + C

cos x

=e

∫ sin x dx

= e ln sin x = sin x


⇒ yx sin x = ∫ 4 x cosec x sin x dx + C

2

sin x dx

3 1
−
2 2

= ∫ Q ⋅ e∫

2


If f (x) = f (–x) then
2 I = 2∫

0


⇒


⇒ ye ∫

sin x dx

π 2

cos x dx = AreaBounded

dx               
(i)

π 2


I =∫


∫

π 6


x = –t

I =∫

π
6


Solving, we get

(tan 3 x + 1) 2

1 dt
1
⇒ ∫ t −2 dt
3 ∫ t2
3

15. I = ∫

π 2 4π 2
−
4
18

=




3 tan 2 x sec 2 x dx = dt



⇒

π2
18


β −α =

2


tan3x + 1 = t (substituting)


⇒

π
2


α +β =

tan x sec x dx

∫


⇒

18π 2 − 9π x + π 2 = 0


π 2/36


sin 2 x cos 2 x dx

⇒


The equation


⇒ y sin x = 4 ⋅

x2
+C
2


⇒ y sin x = 2x2 + C
⎛π ⎞

Given y ⎜ ⎟ = 0
⎝2⎠

⇒ yx sin

π
π2
= 2⋅
+C
2
4

⇒ 0 × 1 =

π2
π2
+C ⇒C = −
2
2


⇒ y sin x = 2 x 2 −

π2
π
x=
6
2

5/18/2018 3:12:17 PM


xx  JEE Mains 2018 Paper


2


⇒ y sin

⇒

2
π
⎛π ⎞ π
= 2⎜ ⎟ −
6
2
⎝6⎠


∴

y π2 π2
y −8π 2
=
−
⇒ =
2 18
2
2
18


⇒ y = −

8π 2
9

x
y

+
=1
2m − 3 3 − 2m
m

R

Q

2m − 3
,0
m

Q = (0, 3 –2m)

P=

P

2m − 3
y = 3 – 2m
m

3− y −3

x=
(3 − y )
2
−2 y
3− y

⇒3x – xy = –2y

or3x + 2y = xy
19. Ortho Centre, A = (–3, 5)

Centroid, B = (3, 3)

Circumcentre, C = (x, y)

Diameter, AC = ?
Note: Centroid divides the orthocenter and circumcentre in
the ratio 2 : 1
C




3=




2

B
(3, 3)

1

(x, 6y)

2× x − 3
(i)
2 +1
[calculating by ratio method]

2× y + 5
(ii)
3

∴ x = 6


y=2




3=

Objective_Maths_JEE Main 2019_Prelims.indd 20

20. Equation of tangent at (1, 7) to the curve x2 = y – 6
1

= x1 ⋅ x = ( y + y1 ) − 6 where x1 = 1 and y1 = 7
2
1

⇒ x = ( y + 7) − 6
2

⇒2x = y + 7 – 12

⇒2x – y + 5 = 0 → Tangent equation

Consider the equation of circle


x2 + y2 + 16x + 12y + c = 0


x2 + 16x + 64 + y2 + 12y + 36 = 100 – c

⇒ ( x + 8) 2 + ( y + 6) 2 = ( 100 − c ) 2


r=


2m = 3 – y

Substituting the value of m in x

A
(–3, 5)

AC 3

=
10
2
2


Radius is 100 − c and centre (–8, –6)

We can calculate the radius of tangent 2x – y + 5 = 0 (Normal
Distance of Tangent From Centre)

⎛ 2m − 3
⎞
R=⎜
, 3 − 2m ⎟

⎝ m
⎠


⇒ x =

AC = ( −3 − 6) 2 + (5 − 2) 2 = 81 + 9 = 3 10


Radius =

y −3
18. Any line passing through (2, 3) with slope in m =
x−2

mx – 2m = y – 3

mx – y = 2m – 3


x=

C = (6, 2)

2 × ( −8) − ( −6) + 5
2

2 +1

=

−16 + 11
5

= 5


Given, 100 − c = 5

100 – c = 5

c = 95
21. Let the parabola be y2 = 16x; y2 = 4ax a = 4

Equation of tangent
a


y = mx +
m
4
m

It passes through (16, 16)
4

⇒ 16 ± 16 m +
m

(Proved)


⇒ y = mx +

P (16, 16)


⇒ 16 m = 16 m 2 + 4

⇒ m 2 − m +

1
=0
4

1
2

Equation of tangent is
x
y = +8


2


x – 2y + 16 = 0

Equation of normal

2x + y = k
x = 16


y = 16


k = 48

A(–16, 0)

B
(24, 0)

Solving, we get m =

(i)

5/18/2018 3:12:25 PM


JEE Mains 2018 Paper  xxi

Normal equation

2x + y = 48

Axis of Parabola is y = 0


x – 2y + 16 = 0


y=0


x = –16


A = (–16, 0)


2x + y = 48

y=0


x = 24


B = (24, 0)


Using T1
(ii)


⇒

x2

y = − 5x + 3





⇒

B (24, 0)

2

−

y2

= 1 at point
a
b2
yy
⎡ xx
⎤
(x1, y1) on Hyperbola  ⎢ 21 − 21 = 1⎥
a
b
⎣
⎦

(x1, y1) are point of contact

∴ First point of contact is x1 = −3 5 and y1 = –12

Taking equation T2

θ

C (4, 0)

x
y
+ =1

3
3
−
5


Equation of tangent to hyperbola

P (16, 16)

A (–16, 0)

− 5x + y = 3




5x + y = 3

5x y
⎡x x y y ⎤
+ = 1    ⎢ 22 − 22 = 1⎥
3
3
b
⎣a
⎦


∴ We get (x2, y2) on comparison as x2 = 3 5 and y2 = –12

AB is diameter as slope of AP × slope of PB = –1


C = (x, y)

T (0, 3)

−16 + 24
=4
2


y=0
16 − 0 16 4
=
=

Slope of PC =
16 − 4 12 3



x=


Referring to figure

P
(–3√5, –12)

4
− ( −2)
4+6
10
tan θ = 3
=
=
=2
4
3
−
8
−5
1 + × −2
3

22. 4x2 – y2 = 36 is a hyperbola
⎡ x2 y2
⎤
x2 y2
−
= 1    ⎢ 2 − 2 = 1 form ⎥
9 36
b
⎢⎣ a
⎥⎦

Equation of tangent

⇒

y = mx + a 2 m 2 − h2

1 0, 3

2

y = mx + 9m − 36

It passes through (0, 3)

Hence, 3 = 0 ± 9m 2 − 36
2


⇒9m – 36 = 9


m2 = 5



m=± 5


Equation of Tangents are

T1 : y = 5 x + 3




T2 : y = 5 x + 3

Objective_Maths_JEE Main 2019_Prelims.indd 21

P

Q

Q
(3√5, –12)


∴ PQ = 2 × 3 5 = 6 5

Altitude = 3 – (–12) = 15

Area of triangle
1
PQT = × 6 5 × 15


2

= 45 5 unit
23. Let the plane pass through line of intersection of the planes
2x – 2y + 3z – 2 = 0 and x – y + z + 1 = 0

Let as assume x = 0

Hence the equation reduces to

–2y + 3z = 2

–y + z = –1

Solving simultaneously, we get

–2y + 3z = 2

–2y + 2z = –2


z=4


y=5

Point passing through the plane 2x – 2y + 3z – 2 = 0 and x – y
+ z + 1 = 0 is (0, 5, 4) vector direct of the line is the vector
normal to the plane 2x – 2y + 3z – 2 = 0 and x – y + z + 1 = 0
iˆ ˆj kˆ

⇒

2 −2 3
1 −1 1

5/18/2018 3:12:31 PM


xxii  JEE Mains 2018 Paper

= iˆ[−2 + 3] − ˆj[2 − 3] + kˆ[−2 + 2]

= iˆ + ˆj

L2 is the line of intersection of the plane x + 2y – z – 3 = 0
and 3x – y + 2z – 1 = 0

Hence, the direction normal to x + 2y – z – 3 = 0 and 3x – y
+ 2z – 1 = 0
iˆ ˆj kˆ
1 2 −1
3 −1 2

y −5 z −4
1
0 =0
−5
−7

−7 × 0 + 7 × 0 − 8 × 0 − 3
7 2 + 7 2 + 82

−3
81 2

=

−3
9 2

=

=

−3
162

1
3 2

24. We have to find the projection of AB on the plane, i.e., length
‘PQ’ equation of the plane x + y + z = 7
B (4, –1, 3)
R

(5, –1, 4)
A

n=i+j+k

Q
P

Normal vector = iˆ + ˆj + kˆ ; AB = −iˆ − kˆ

Length BR = |projection of AB on nˆ |





2
3

iˆ ˆj kˆ

a × b = 2 3 −1
0 1 1






⇒(x – 0) (–7 + 0) – (y – 5) (–7 + 0) + (z – 4) (–5 –3) = 0

⇒–7x + 7y – 35 – 8z + 32 = 0

⇒–7x + 7y – 8z – 3 = 0

Hence, distance of plane from origin

=

2

4
⎛ 2 ⎞

= ( 2 )2 − ⎜
⎟ = 2−
3
3
⎝
⎠

u ⋅ (a × b ) = 0


Hence, the direction <x – 0, y – 5, z – 4>, <1, 1, 0> and
<3, –5, –7> are coplanar.

Hence, required plane is


=

3

25. If u is coplanar with a = 2iˆ + 3 ˆj − kˆ and b ⋅ ˆj + kˆ, then

= 3iˆ − 5 ˆj − 7kˆ

x−0
1
3

2


PQ = AR = AB 2 − BR 2


PQ =


= iˆ[2 x + 2 − 1] − ˆj[2 + 3] + kˆ[−1 − 6]


⇒

=




= |projection of
iˆ ( −iˆ − kˆ ) on
iˆ (iˆ + ˆj + kˆ ) |

= iˆ[3 + 1] − ˆj[2 − 0] + kˆ[2 − 0]
= 4iˆ − 2 ˆj + 2kˆ
u ⋅ ( a × b ) = 4u1 − 2u2 + 2u3 = 0 (i)


Where u = u1iˆ + u2 ˆj + u3kˆ

u is perpendicular to a

Hence, u ⋅ a = 0


2u1 + 3u2 – u3 = 0

(ii)

u ⋅b = 0



u2 + u3 = 24

Equation (ii) + equation (iii)


2u1 + 4u2 = 24

Equation (i) – 2(equation (iii))

4u1 – 4u2 = –48


u1 – u2 = –12

Solving equation (iv) and equation (v), we get u2 = 8


u1 = –4


u3 = 16



u = −4iˆ + 8 ˆj + 16 kˆ




u = 4 2 + 82 + 16 2 = 336

(iii)
(iv)

(v)

2

26. Case I: Red ball is drawn in first draw
Ball

1st

2nd

Red

4

6

Black

6

6

Total

10

12

Color

( −iˆ − kˆ ) ⋅ (iˆ + ˆj + kˆ )
−1 + 0 − 1
iˆ =

=
ˆi 2 + ˆj 2 + kˆ 2
3

Objective_Maths_JEE Main 2019_Prelims.indd 22

5/18/2018 3:12:38 PM


JEE Mains 2018 Paper  xxiii
Case II: Black Ball is drawn in 1st draw
Ball


⇒

1st

Color

2nd

Red

4

4

Black

6

8

Total

10

12

st


Let R1 be Red ball drawn in 1 draw

Let B1 be Black ball drawn in 1st draw
R
2 represent red ball drawn in 2nd draw proved black ball is
B1
drawn in 1st draw
R
2 represent red ball drawn in 2nd and 1st draw.
R1
4
6

P ( R1 ) = ; P ( B1 ) =
10
10

4 6
6 4
48
= × + × =
10 12 10 12 120




2
=
5
( xi − x ) 2
n
i =1
n

i =1

9

9

9

9

i =1

i =1

i =1

i =1

i =1

∑ xi2 − 12∑ xi + 36∑1 = 45 − 2∑ xi + 11∑1
9

∑ ( xi − 6)2 = 45 − 2 × 54 + 99
i =1


⇒

9

∑ ( xi − 6)2 = 45 − 108 + 99 = 36
i =1


⇒

( xi − 6)
= 4 = σ 2    [VARIANCE]
9

i =1
9

∑


Hence, standard deviation, σ = 4 = 2
⎡ ⎛π
⎞
⎛π
⎞ 1⎤
28. 8 cos x ⋅ ⎢cos ⎜ + x ⎟ ⋅ cos ⎜ − x ⎟ − ⎥ = 1
6
6
⎠
⎝
⎠ 2⎦
⎣ ⎝


Formula of Standard deviation, σ =



n=9




∑ xi − 5∑1 = 9

9

9

i =1

i =1

∑ ( xi − x )2
i =1

9

∑ xi − 45 = 9
i =1
9

∑ xi = 54
i=1

2

⎤
⎡ π

⇒ 4 cos x ⎢cos + cos 2 x − 1⎥ = 1
3
⎣
⎦

cos 2 x = 2 cos 2 x − 1


n


⇒ 4 cos x ⎡⎣ 4 cos 2 x − 3⎤⎦ = 2

⇒ 4 cos3 x − 3 cos x =

⇒ cos3 x =

3 x = 2nπ ±

π
3

2nπ π
±
3
9

All roots lie in the region [0, π]

x=


n = 0 one root

= 45

n = 1 two roots

i =1

Objective_Maths_JEE Main 2019_Prelims.indd 23

1
2

1
2

54
= 6 [ x = Mean ]
9

x=

∑ ( xi − 5)


2cos A cos B = cos( A + B ) + cos( A − B)

1⎤
⎡

⇒ 4 cos x ⎢ 2 cos 2 x − 2 + ⎥ = 1
2⎦

⎣
n

9

i =1

9

⎡1
⎤

⇒ 4 cos x ⎢ + 2 cos 2 x − 1 − 1⎥ = 1
⎣2
⎦

27. Formula of variance, σ 2 = ∑


Given

i =1


4 cos x[2 cos A cos B − 1] = 1

⎛R ⎞
⎛R ⎞

Required probability = P ( R1 ) × P ⎜ 2 ⎟ + P ( B1 ) × P ⎜ 2 ⎟

⎝ R1 ⎠
⎝ B1 ⎠




9

⎛π
⎞
cos B = cos ⎜ − x ⎟

⎝6
⎠

⎛R ⎞ 6

P⎜ 2 ⎟ =
⎝ R1 ⎠ 12





⇒

9

⎛π
⎞

Let cos A = cos ⎜ + x ⎟
6
⎝
⎠

⎛R ⎞ 4

P⎜ 2 ⎟ =
⎝ B1 ⎠ 12





⇒

9

∑ xi2 − 10∑ xi + 25∑1 = 45

π
9
2π π
±
3
9

5/18/2018 3:12:45 PM

xxiv  JEE Mains 2018 Paper

Summing the three roots


Truth table: 2

π ⎛ 2π π ⎞ ⎛ 2π π ⎞
+ ⎜
+ ⎟+⎜
−
9 ⎝ 3
9 ⎠ ⎝ 3 9 ⎟⎠
π 2π 2π 13π
+
=
+
9
3
3
9
29. Refer to figure:

Let height = h
QM = h cot 30

PM = h cot 45

Using Pythagoras theorem

PQ2 = PM 2 + QM 2
P

R
200 m

30°

L

45°

200 2 = h2 + ( h 3 ) 2

m
30°

200 m
Q

30. Truth table: 1
p

q

p∨q

∼(p ∨ q)

1

T

T

T

F

2

T

F

T

F

3

F

T

T

F

4

F

F

F

T

Objective_Maths_JEE Main 2019_Prelims.indd 24

∼p

q

∼p ∧ q

1

T

F

T

F

2

T

F

F

F

3

F

T

T

T

4

F

T

F

F

∼(p ∨ q)

∼p ∧ q

∼(p ∨ q) ∨ (∼p ∧ q)

1

F

F

F

2

F

F

F

3

F

T

T

4

T

F

T


Truth table: 3

h




⇒4h2 = 200 × 200

h2 = 100 × 100

h = 100m

p


The result is equal to ∼p refer to truth table 2
Alternate solution:

∼(p ∨ q) = ∼p ∧ ∼q (Theorem)

∼(p ∨ q) ∨ (∼p ∧ q) ⇒ (∼p ∧ ∼q) ∨ (∼p ∧ q)

⇒(∼p ∨ ∼p) ∧ (∼q ∨ q) = ∼p

5/18/2018 3:12:46 PM