JEE Main
CHAPTERWISE
SOLUTIONS
2019-2002
Mathematics
All the 16 Question Papers
of JEE Main Online 2019 (Jan & Apr Attempt)
JEE Main
CHAPTERWISE
SOLUTIONS
2019-2002
Mathematics
All the 16 Question Papers
of JEE Main Online 2019 (Jan & Apr Attempt)
ARIHANT PRAKASHAN (Series), MEERUT
Arihant Prakashan (Series), Meerut
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PREFACE
JEE Main is a gateway examination for candidates expecting to seek
admission in Bachelor in Engineering (BE), Bachelor of Technology
(B.Tech) and Bachelor of Architecture (B.Arch) at Indian Institutes of
Information Technology (IIITs), National Institutes of Technology (NITs),
Delhi Technological University and other Centrally Funded Technical
Institutes (CFTIs).
JEE Main is also an examination which is like screening examination for
JEE Advanced (The gateway examination to India's most reputed
Technical Institutes, Indian Institutes of Technology IITs).
To make the students well-versed with the pattern as well as the level
of the questions asked in the exam, this book contains Chapterwise
Solutions of Questions asked in Last 18 Years’ Examinations of
JEE Main (formerly known as AIEEE). Solutions to all the questions
have been kept very detailed and accurate. Along with the indication
of level of the exam, this book will also teach you to solve the
questions objectively in the examination.
To give the students a complete practice, along with Chapterwise
Solutions, this book contains 5 Practice Sets, based exactly on JEE Main
Syllabus and Pattern. By practicing these sets, students can attain
efficiency in Time Management during the examination.
We hope this book would be highly beneficial for the students. We
would be grateful if any discrepancy or mistake in the questions or
answers is brought to our notice so that these could be rectified in
subsequent editions.
Publisher
CONTENTS
1. Sets, Relations and Functions
1-17
2. Complex Numbers and Quadratic Equations
18-42
3. Matrices and Determinants
43-68
4. Permutations and Combinations
69-76
5. Mathematical Induction
77-78
6. Binomial Theorem and Its Simple Applications
79-91
7. Sequences and Series
92-112
8. Limits, Continuity and Differentiability
113-161
9. Integral Calculus
162-201
10. Differential Equations
202-214
11. Coordinate Geometry
215-276
12. Three Dimensional Geometry
277-301
13. Vector Algebra
302-317
14. Statistics and Probability
318-340
15. Trigonometry
341-369
16. Mathematical Reasoning
370-376
PRACTICE SETS for JEE MAIN
Practice Set 1
379-386
Practice Set 2
387-395
Practice Set 3
396-404
Practice Set 4
405-412
Practice Set 5
413-421
SYLLABUS
UNIT 1 Sets, Relations and Functions
Sets and their representation; Union,
intersection and complement of sets and their
algebraic properties; Power set; Relation, Types of
relations, equivalence relations, functions;. oneone, into and onto functions, composition of
functions.
UNIT 2 Complex Numbers and
Quadratic Equations
Complex numbers as ordered pairs of reals,
Representation of complex numbers in the form a
+ib and their representation in a plane, Argand
diagram, algebra of complex numbers, modulus
and argument (or amplitude) of a complex
number, square root of a complex number,
triangle inequality, Quadratic equations in real
and complex number system and their solutions.
Relation between roots and co-efficients, nature of
roots, formation of quadratic equations with given
roots.
UNIT 6 Binomial Theorem and
its Simple Applications
Binomial theorem for a positive integral index,
general term and middle term, properties of
Binomial coefficients and simple applications.
UNIT 7 Sequences and Series
Arithmetic and Geometric progressions,
insertion of arithmetic, geometric means
between two given numbers. Relation between
A.M. and G.M. Sum upto n terms of special
series: ∑ n, ∑ n2, ∑ n3. Arithmetico - Geometric
progression.
UNIT 8 Limit, Continuity
and Differentiability
Real valued functions, algebra of functions,
polynomials, rational, trigonometric,
logarithmic and exponential functions, inverse
functions. Graphs of simple functions. Limits,
continuity and differentiability.
Matrices, algebra of matrices, types of matrices,
determinants and matrices of order two and three.
Properties of determinants, evaluation of determinants, area of triangles using determinants.
Adjoint and evaluation of inverse of a square
matrix using determinants and elementary
transformations, Test of consistency and solution
of simultaneous linear equations in two or three
variables using determinants and matrices.
Differentiation of the sum, difference, product
and quotient of two functions. Differentiation of
trigonometric, inverse trigonometric,
logarithmic exponential, composite and implicit
functions derivatives of order upto two. Rolle's
and Lagrange's Mean Value Theorems.
Applications of derivatives: Rate of change of
quantities, monotonic - increasing and
decreasing functions, Maxima and minima of
functions of one variable, tangents and normals.
UNIT 4 Permutations and Combinations
UNIT 9 Integral Calculus
Fundamental principle of counting, permutation
as an arrangement and combination as selection,
Meaning of P (n,r) and C (n,r), simple applications.
Integral as an anti - derivative. Fundamental
integrals involving algebraic, trigonometric,
exponential and logarithmic functions.
Integration by substitution, by parts and by
partial fractions. Integration using trigonometric
identities.
Evaluation of simple integrals of the type
UNIT 3 Matrices and Determinants
UNIT 5 Mathematical Induction
Principle of Mathematical Induction and its simple
applications.
dx ,
x2 ± a2
dx
Öx 2 ± a 2
dx
,
2
ax + bx + c
(px + q) dx ,
Öax 2 + bx + c
,
dx ,
a2 – x2
dx
Öax 2 + bx + c
dx
2
Öa – x
2
,
(px + q) dx ,
ax 2 + bx + c
,
Öa 2 ± x 2 dx and
Öx 2 – a 2 dx
Integral as limit of a sum. Fundamental Theorem of
Calculus. Properties of definite integrals. Evaluation of
definite integrals, determining areas of the regions
bounded by simple curves in standard form.
UNIT 10 Differential Equations
Ordinary differential equations, their order and degree.
Formation of differential equations. Solution of
differential equations by the method of separation of
variables, solution of homogeneous and linear
differential equations of the type dy +p (x) y = q(x)
dx
UNIT 11 Coordinate Geometry
Cartesian system of rectangular coordinates in a plane,
distance formula, section formula, locus and its
equation, translation of axes, slope of a line, parallel
and perpendicular lines, intercepts of a line on the
coordinate axes.
Ÿ
Ÿ
Straight lines
Various forms of equations of a line, intersection of
lines, angles between two lines, conditions for
concurrence of three lines, distance of a point from
a line, equations of internal and external bisectors
of angles between two lines, coordinates of
centroid, orthocentre and circumcentre of a
triangle, equation of family of lines passing
through the point of intersection of two lines.
Circles, conic sections
Standard form of equation of a circle, general form
of the equation of a circle, its radius and centre,
equation of a circle when the end points of a
diameter are given, points of intersection of a line
and a circle with the centre at the origin and
condition for a line to be tangent to a circle,
equation of the tangent. Sections of cones,
equations of conic sections (parabola, ellipse and
hyperbola) in standard forms, condition for y =
mx + c to be a tangent and point (s) of tangency.
UNIT 12 Three Dimensional Geometry
Coordinates of a point in space, distance between
two points, section formula, direction ratios and
direction cosines, angle between two intersecting
lines. Skew lines, the shortest distance between
them and its equation. Equations of a line and a
plane in different forms, intersection of a line and a
plane, coplanar lines.
UNIT 13 Vector Algebra
Vectors and scalars, addition of vectors, components
of a vector in two dimensions and three dimensional
space, scalar and vector products, scalar and vector
triple product.
UNIT 14 Statistics and Probability
Measures of Dispersion: Calculation of mean,
median, mode of grouped and ungrouped data.
Calculation of standard deviation, variance and
mean deviation for grouped and ungrouped data.
Probability: Probability of an event, addition and
multiplication theorems of probability, Baye's
theorem, probability distribution of a random
variate, Bernoulli trials and Binomial distribution.
UNIT 15 Trigonometry
Trigonometrical identities and equations.
Trigonometrical functions. Inverse trigonometrical
functions and their properties. Heights and
Distances.
UNIT 16 Mathematical Reasoning
Statements, logical operations and implies, implied
by, if and only if. Understanding of tautology,
contradiction, converse and contra positive.
1
Sets, Relations and
Functions
 2x 
1 − x
 is
 ,| x | < 1, then f 
1 + x
1 + x2
(b) 2 f ( x 2 )
written
as
f ( x ) = a x (a > 0) be
f ( x ) = f1( x ) + f 2( x ), where f1( x ) is an even
function and f 2( x ) is an odd function. Then
f1( x + y ) + f 1( x − y ) equals
(d) −2 f ( x )
(a) 2 f 1 ( x + y ) ⋅ f 2 ( x − y )
1. If f ( x ) = loge 
equal to
(a) 2 f ( x )
(c) ( f ( x ))2
[JEE Main 2019, 8 April Shift-I]
2. Let
(b) 2 f 1 ( x + y ) ⋅ f 1 ( x − y )
(c) 2 f 1 ( x ) ⋅ f 2 ( y )
Exp. (a)
1 − x
Given, f( x) = loge 
 ,| x| < 1, then
 1 + x
1− 2x 


  2x  
 2x 
1 + x2 


=
f 
log
< 1
Q
e
2
2
2
x


1 + x 
 1 + x  
1 +
2 
1+ x 

 1 + x2 − 2 x 


 1 + x2 
= loge 

2
1+ x + 2x
 1 + x2 


 (1 − x)2 
1 − x
 = loge 
= loge 

2
1
x
+
(
)
 1 + x


2
1 − x
m
= 2 loge 
 [Qloge| A| = mloge| A|]
 1 + x
= 2f( x)

1 − x

Q f( x) = loge 
 1 + x 

(d) 2 f 1 ( x ) ⋅ f 1 ( y )
[JEE Main 2019, 8 April Shift-II]
Exp. (d)
Given, function f( x) = a x , a > 0 is written as sum of
an even and odd functions f1( x) and f2 ( x)
respectively.
a x + a− x
a x − a− x
Clearly, f1( x) =
and f2 ( x) =
2
2
So, f1( x + y) + f1( x − y)
1
1
= [a x + y + a − ( x + y ) ] + [a x − y + a − ( x − y ) ]
2
2
ax
ay 
1
1
=  a x ay + x y + y + x 
2
a a
a
a 
1 x y
1
1  1

=
a  a + y  + x  y + ay 
 
2  
a  a a
1 x
1  y
1
a + x  a + y 
2
a  
a 
 a x + a− x   ay + a− y 
= 2
 = 2 f1( x) ⋅ f1( y)
 
2
2

 

=
2
JEE Main Chapterwise Mathematics
3. If the function f : R − {1, − 1} → A defined by
f (x ) =
x2
1 − x2
, is surjective, then A is equal to
(a) R − { −1}
(c) R − [−1, 0)
(b) [0, ∞ )
(d) R − ( −1, 0)
[JEE Main 2019, 9 April Shift-I]
Exp. (c)
Given, function f : R – {1, − 1} → A defined as
x2
= y (let)
f ( x) =
1 − x2
⇒
x2 = y(1 − x2 )
[Q x2 ≠ 1]
⇒
x (1 + y) = y
2
x2 =
⇒
Q
⇒
y
1+ y
[provided y ≠ −1]
x2 ≥ 0
y
≥0
1+ y
4. Two newspapers A and B are published in a
city. It is known that 25% of the city
population reads A and 20% reads B while
8% reads both A and B. Further, 30% of those
who read A but not B look into
advertisements and 40% of those who read B
but not A also look into advertisements,
while 50% of those who read both A and B
look into advertisements. Then, the
percentage of the population who look into
advertisements is
(b) 13
(c) 12.8
(d) 13.9
[JEE Main 2019, 9 April Shift-II]
Exp. (d)
Let the population of city is 100.
Then, n( A) = 25, n(B) = 20 and n( A ∩ B) = 8
A
= 5 .1 + 4 .8 + 4 = 13 .9
5. The domain of the definition of the function
f (x ) =
1
4 − x2
+ log 10( x 3 − x ) is
(a) ( −1, 0) ∪ (1, 2 ) ∪ ( 3, ∞ )
(b) ( −2 , − 1) ∪ ( −1, 0) ∪ ( 2 , ∞ )
(c) ( −1, 0) ∪ (1, 2 ) ∪ ( 2 , ∞ )
(d) (1, 2 ) ∪ ( 2 , ∞ ) [JEE Main 2019, 9 April Shift-II]
Exp. (c)
⇒
y ∈ (−∞, − 1) ∪ [0, ∞ )
Since, for surjective function, range of
f = codomain
∴Set A should be R − [−1, 0).
(a) 13.5
So, n( A ∩ B ) = 17 and n( A ∩ B) = 12
According to the question, Percentage of the
population who look into advertisement is
30
40
× n( A ∩ B)
=
× n( A ∩ B ) + 

 100
  100
50
+ 
× n( A ∩ B)
 100

50
30
40
× 8
× 12  + 
= 
× 17  + 

  100
 100
  100
B
U
Given function f( x) =
1
4 − x2
+ log10 ( x3 − x)
For domain of f( x)
4 − x2 ≠ 0 ⇒ x ≠ ± 2
and
…(i)
x − x> 0
3
⇒ x( x − 1)( x + 1) > 0
From Wavy curve method,
+
–∞
–
–1
+
0
–
+1
From Eqs. (i) and (ii), we get the domain of f( x) as
(−1, 0) ∪ (1, 2 ) ∪ (2, ∞ ).
6. Let f ( x ) = x 2 , x ∈R . For any A ⊆ R , define
g ( A ) = { x ∈R : f ( x ) ∈ A }. If S =[0, 4], then
which one of the following statements is not
true?
(a) f ( g (S )) = S
(c) g ( f (S )) = g (S )
(b) g ( f (S )) ≠ S
(d) f(g(S)) ≠ f (S )
[JEE Main 2019, 10 April Shift-I]
Exp. (c)
17
8
12
Venn diagram
n(U)=100
+∞
Given, functions f( x) = x2 , x ∈ R
and g ( A) = { x ∈ R : f( x) ∈ A}; A ⊆ R
Now, for S = [0, 4]
g (S ) = { x ∈ R : f( x) ∈ S = [0, 4]}
3
Sets, Relations and Functions
= { x ∈ R : x2 ∈ [0, 4]}
8. Let A ,B andC be sets such that
= { x ∈ R: x ∈ [−2, 2 ]}
⇒
g (S ) = [−2, 2 ]
So, f(g (S )) = [0, 4] = S
Now, f(S ) = { x2 : x ∈ S = [0, 4]} = [0, 16]
φ ≠ A ∩ B ⊆ C . Then, which of the following
statements is not true?
(a) B ∩ C ≠ φ
(b) If ( A − B ) ⊆ C , then A ⊆ C
(c) (C ∪ A ) ∩ (C ∪ B ) = C
(d) If ( A − C ) ⊆ B , then A ⊆ B
and g (f(S )) = { x ∈ R : f( x) ∈ f(S ) = [0, 16]}
= { x ∈ R : f( x) ∈ [0, 16]}
= { x ∈ R: x2 ∈ [0, 16]}
[JEE Main 2019, 12 April Shift-II]
= { x ∈ R : x ∈ [−4, 4]}
= [−4 ,4]
From above, it is clear that g (f(S )) = g (S ).
7.
Exp. (d)
Key Idea Use Venn diagram for operations of sets.
 3
For x ∈ 0,  , let f ( x ) = x , g ( x ) = tan x and
 2
1 − x2
. If φ( x ) = ((hof )og )( x ), then
h( x ) =
1 + x2
 π
φ   is equal to
 3
π
(a) tan
12
7π
(c) tan
12
According to the question, we have the
following Venn diagram.
Here, A ∩ B ⊆ C
and A ∩ B ≠ φ
C
A
B
11 π
(b) tan
12
5π
(d) tan
12
A∩B
[JEE Main 2019, 12 April Shift-I]
Exp. (b)
Given, for x ∈(0, 3 / 2 ), functions
f ( x) = x
g ( x) = tan x
1 − x2
and
h( x) =
1 + x2
Now, from the Venn diagram, it is clear that
B ∩ C ≠ φ, is true
… (i)
… (ii)
… (iii)
Also given, φ( x) = ((hof )og )( x) = (hof ) (g ( x))
= h(f(g ( x))) = h(f(tan x))
1 − ( tan x )2
= h( tan x ) =
1 + ( tan x )2
1 − tan x
π
= tan − x
=
4

1 + tan x
Now,
π
π π
φ   = tan  − 
 4 3
 3
3π − 4π 
= tan 

 12 
π
= tan  −

 12 
π
π
= − tan   = tan π −

 12 

12 
11π 
= tan 

 12 
Also, (C ∪ A) ∩ (C ∪ B) = C ∪ ( A ∩ B) = C is true.
If ( A − B) ⊆ C, for this statement the Venn diagram
is
A
B
C
From the Venn diagram, it is clear that
if A − B ⊆ C, then A ⊆ C.
Now, if ( A − C ) ⊆ B, for this statement the Venn
diagram.
A
B
C
From the Venn diagram, it is clear that
A ∩ B ≠ φ, A ∩ B ⊆ C and A – C = φ ⊆ B but
A ⊆ B.
4
JEE Main Chapterwise Mathematics
1
x
9. For x ∈R − {0, 1}, let f1( x ) = , f 2( x ) = 1 − x
1
and f 3( x ) =
be three given functions. If
1−x
a function, J (x) satisfies( f 2 ° J ° f1 )( x ) = f 3( x ),
then J ( x ) is equal to
[JEE Main 2019, 9 Jan Shift-I]
(a) f 2 ( x )
(b) f 3 ( x )
(c) f 1 ( x )
(d)
1
f 3( x )
x
Exp. (b)
We have,
1
1
f1( x) = , f2 ( x) = 1 − x and f3 ( x) =
1− x
x
Also, we have (f2 o J o f1 )( x) = f3 ( x)
⇒
f2 ((Jo f1 )( x)) = f3 ( x)
⇒
f2 (J (f1( x)) = f3 ( x)
1
1 − J(f1( x)) =
1− x
⇒
1
]
1− x
1
1
1
[Q f1( x) = ]
1 − J   =
⇒
 x 1 − x
x
1− x − 1
1
1
−x

J  = 1−
⇒
=
=
 x
1− x
1− x
1− x
1
Now, put = X, then
x
−1
Q x = 1 
J( X ) = X

1
X 
1−
X
−1
1
=
=
X − 1 1− X
[Qf2 ( x) = 1 − x and f3 ( x) =
⇒
J( X ) = f3 ( X ) or J( x) = f3 ( x)
10. Let A = { x ∈R : x is not a positive integer}.
2x
,
Define a function f : A → R as f ( x ) =
x −1
then f is
[JEE Main 2019, 9 Jan Shift-II]
One-one Let x1, x2 ∈ A such that
⇒
f( x1 ) = f( x2 )
2 x1
2 x2
=
x1 − 1 x2 − 1
⇒
⇒
2 x1 x2 − 2 x1 = 2 x1 x2 − 2 x2
x1 = x2
Thus, f( x1 ) = f( x2 ) has only one solution, x1 = x2
∴ f( x) is one-one (injective)
2 ×2
Onto Let x = 2, then f(2 ) =
=4
2 −1
But x = 2 is not in the domain, and f( x) is one-one
function
∴ f( x) can never be 4.
Similarly, f( x) can not take many values.
Hence, f( x) is into (not surjective).
∴ f( x) is injective but not surjective.
11. Let f : R → R be a function such that
f ( x ) = x 3 + x 2 f ′ (1) + xf ′ ′ (2 ) + f ′′′( 3), x ∈R .
Then, f (2 ) equals [JEE Main 2019, 10 Jan Shift-I]
(b) − 4
(d) 8
(a) 30
(c) − 2
Exp. (c)
We have,
f( x) = x3 + x2 f ′(1) + xf ′ ′(2 ) + f ′ ′ ′(3)
⇒
f ′( x) = 3 x2 + 2 xf ′(1) + f ′ ′(2 )
… (i)
⇒
f ′ ′( x) = 6 x + 2 f ′(1)
⇒
f ′ ′ ′ ( x) = 6
⇒
f′ ′ ′(3) = 6
Putting x = 1in Eq. (i), we get
… (ii)
… (iii)
f ′(1) = 3 + 2 f ′(1) + f ′ ′(2 )
and putting x = 2 in Eq. (ii), we get
… (iv)
f ′ ′(2 ) = 12 + 2 f ′(1)
…(v)
From Eqs. (iv) and (v), we get
f ′(1) = 3 + 2 f ′(1) + (12 + 2 f ′(1))
⇒
3f′(1) = − 15
(a) injective but not surjective
(b) not injective
(c) surjective but not injective
(d) neither injective nor surjective
⇒
f′ ′(2 ) = 12 + 2 (− 5) = 2 [using Eq. (v)]
∴
f( x) = x3 + x2 f ′(1) + xf ′ ′(2 ) + f ′ ′ ′(3)
Exp. (a)
⇒
f ( x) = x 3 − 5 x 2 + 2 x + 6
⇒
f(2 ) = 2 3 − 5(2 )2 + 2(2 ) + 6
We have a function f : A → R defined as
2x
f ( x) =
x−1
⇒
f′(1) = − 5
= 8 − 20 + 4 + 6 = − 2
5
Sets, Relations and Functions
12. In a class of 140 students numbered 1 to 140,
all even numbered students opted
Mathematics course, those whose number
is divisible by 3 opted Physics course and
those whose number is divisible by 5 opted
Chemistry course. Then, the number of
students who did not opt for any of the three
courses is
[JEE Main 2019, 10 Jan Shift-I]
(a) 42
(b) 102
(c) 38
(d) 1
Exp. (c)
A
B
C
Let A be the set of even numbered students
140 
then n( A) = 
= 70
 2 
([.] denotes greatest integer function)
Let B be the set of those students whose number
is divisible by 3,
140 
then n(B) = 
= 46
 3 
([.] denotes greatest integer function)
Let C be the set of those students whose number
is divisible by 5,
140 
then
= 28
n(C ) = 
 5 
Now,
([.] denotes greatest integer function)
140 
= 23
n( A ∩ B) = 
 6 
(numbers divisible by both 2 and 3)
140 
=9
n(B ∩C ) = 
 15 
(numbers divisible by both 3 and 5)
140 
= 14
n(C ∩ A) = 
 10 
(numbers divisible by both 2 and 5)
140 
=4
n( A ∩ B ∩ C ) = 
 30 
(numbers divisible by 2, 3 and 5)
and n( A ∪ B ∪C )
= Σn( A) − Σn( A ∩ B) + n( A ∩ B ∩C )
= (70 + 46 + 28 ) − (23 + 9 + 14) + 4 = 102
∴ Number of students who did not opt any of the
three courses
= Total students − n( A ∪ B ∪C ) = 140 − 102 = 38
13. Let N be the set of natural numbers and two
functions f
f , g : N → N
n + 1
;

f (n ) =  2
n

;
 2
and g be
such that
defined
as
if n is odd
if n is even
and g (n ) = n − ( −1)n . Then, fog is
[JEE Main 2019, 10 Jan Shift-II]
(a) one-one but not onto
(b) onto but not one-one
(c) both one-one and onto
(d) neither one-one nor onto
Exp. (b)
 n + 1, if n is odd

Given, f(n) =  2
 n , if n is even,
2
n + 1 , if n is odd
andg (n) = n − (−1)n = 
n − 1, if n is even
f(n + 1), if n is odd
Now, f(g (n)) = 
f(n − 1), if n is even
 n + 1, if n is odd

= f ( x)
= 2
 n − 1 + 1 = n , if n is even
 2
2
[Qif n is odd, then (n + 1) is even and
if n is even, then (n − 1) is odd]
Clearly, function is not one-one as f(2 ) = f(1) = 1
But it is onto function.
[Qif m ∈ N (codomain) is odd, then 2m ∈ N
(domain) such that f(2 m) = m and
if m ∈ N codomain is even, then
2 m − 1 ∈ N (domain) such that f(2 m − 1) = m]
∴Function is onto but not one-one.
14. Let f : R → R be defined by f ( x ) =
x ∈R. Then, the range of f is
x
1 + x2
,
[JEE Main 2019, 11 Jan Shift-I]
1 1
(a)  − , 
 2 2 
1 1
(c) R −  − , 
 2 2 
(b) ( −1,1) − {0}
(d) R − [−1,1]
Exp. (a)
We have, f( x) =
x
1 + x2
, x∈R
6
JEE Main Chapterwise Mathematics
Ist Method f( x) is an odd function and maximum
occur at x = 1
Y
(1, 1/2)
y=
–1
1
2
(a)
X
O1
y= –
(–1, 1/2)
1
k
Then, for all x ∈R , the value of f 4 ( x ) − f 6 ( x ) is
equal to
[JEE Main 2019, 11 Jan Shift I]
15. Let fk ( x ) = (sink x + cosk x ) for k = 1, 2 , 3 ... .
1
2
1
12
5
12
(b)
(c)
−1
12
(d)
1
4
Exp. (a)
We have,
From the graph it is clear that range of f( x) is
− 1 , 1 
 2 2 
1
IInd Method f( x) =
1
x+
x
1
If x > 0, then by AM ≥ GM , we get x + ≥ 2
x
1
1
≤
⇒
1 2
x+
x
1
0 < f ( x) ≤
⇒
2
1
If x < 0, then by AM ≥ GM , we get x + ≤ −2
x
1
1
≥−
⇒
1
2
x+
x
1
⇒
– ≤ f ( x) < 0
2
0
If x = 0, then f( x) =
=0
1+ 0
1
1
Thus, − ≤ f( x) ≤
2
2
1 1
Hence, f( x) ∈  − , 
 2 2 
IIIrd Method
x
Let y =
⇒ yx2 − x + y = 0
1 + x2
Q x ∈ R, so
D≥ 0
⇒
1 − 4 y2 ≥ 0
⇒
(1− 2 y) (1 + 2 y) ≥ 0
1 1
y ∈ − , 
 2 2 
⇒
+
–
–1/2
1 1
So, range is  − ,  .
 2 2 
–
1/2
fk ( x) =
1
(sink x + cos k x), k = 1, 2, 3, …
k
1
(sin4 x + cos 4 x)
4
1
= ((sin2 x + cos 2 x)2 − 2 sin2 xcos 2 x)
4
1
1
=  1 − (sin2 x)2 

2
4
1 1 2
= − sin 2 x
4 8
1
and f6 ( x) = (sin6 x + cos 6 x)
6
1
= {(sin2 x + cos 2 x)3 − 3sin2 xcos 2 x
6
(sin2 x + cos 2 x)}
3
1
= 1 − (2 sin xcos x)2 
4
6

1 1 2
= − sin 2 x
6 8
1 1 3−2
1
Now, f4 ( x) − f6 ( x) = − =
=
4 6
12
12
∴
f4 ( x) =
16. Let a function f :(0, ∞ ) → (0, ∞ ) be defined
by f ( x ) = 1 −
1
. Then, f is
x
[JEE Main 2019, 11 Jan Shift-II]
(a)
(b)
(c)
(d)
injective only
both injective as well as surjective
not injective but it is surjective
neither injective nor surjective
Exp. (d)
− ( x − 1), if 0 < x ≤ 1
| x − 1| 
x
=
x−1
x
if x > 1
,


x
 1 − 1, if 0 < x ≤ 1

= x
1
if x > 1
1 − ,

x
We have, f( x) =
7
Sets, Relations and Functions
Now, let us draw the graph of y = f( x)
Note that when x → 0, then f( x) → ∞, when x = 1,
then f( x) = 0, and when x → ∞, then f( x) → 1
Y
occurs = 2 50 − 1
So, the required number of non-empty subsets of
‘S’ such that product of elements is even.
= (2 100 − 1) − (2 50 − 1)
= 2 100 − 2 50 = 2 50(2 50 − 1).
19. Let Z
x=0
y=1
O
1
y=0
X
Clearly, f( x)is not injective because if f( x) < 1, then f
is many one, as shown in figure.
Also, f( x)is not surjective because range of f( x)is [
0, ∞[ and but in problem co-domain is (0, ∞ ), which
is wrong.
∴f ( x) is neither injective nor surjective
17. The number of functions f from {1, 2, 3, … ,
20} onto {1, 2, 3, … , 20} such that f (k ) is a
multiple of 3, wheneverk is a multiple of 4, is
[JEE Main 2019, 11 Jan Shift-II]
(a) (15)! × 6!
(b) 56 × 15
(c) 5 ! × 6 !
(d) 65 × (15)!
Exp. (a)
According to given information, we have if
k ∈{4, 8, 12, 16, 20}
Then, f(k ) ∈ {3, 6, 9, 12, 15, 18}
[Qcodomain (f ) = {1, 2, 3, …, 20}]
Now, we need to assign the value of f(k ) for
k ∈ {4, 8, 12, 16, 20} this can be done in 6 C 5 ⋅ 5!
ways = 6 ⋅ 5! = 6! and remaining 15 element can
be associated by 15! ways.
∴Total number of onto functions = 15 16 = 15! 6!
18. Let S = {1, 2 , 3,... , 100}. The number of
non-empty subsets A of S such that the
product of elements in A is even, is
[JEE Main 2019, 12 Jan Shift-I]
(a) 2 50 ( 2 50 − 1)
(c) 2 50 + 1
(b) 2 50 − 1
(d) 2100 − 1
Exp. (a)
Given, set S = {1, 2, 3, ... ,100}. Total number of
non-empty subsets of ‘S’ = 2 100 − 1
Now, numbers of non-empty subsets of ‘S’ in
which only odd numbers {1, 3, 5, … , 99 }
be the set of integers. If
2
and
A = { x ∈ Z : 2( x + 2 ) ( x − 5x + 6 ) = 1}
B = { x ∈ Z : − 3 < 2 x − 1 < 9}, then the number
of subsets of the set A × B , is
[JEE Main 2019, 12 Jan Shift-II]
(a) 212
(b) 218
(c) 215
(d) 210
Exp. (c)
Given, set A = { x ∈ Z : 2( x + 2 )( x
2
−5 x+ 6)
= 1}
2
( x+ 2) ( x −5 x+ 6)
Consider, 2
= 1 = 2º
⇒ ( x + 2 ) ( x − 3) ( x − 2 ) = 0
⇒
x = −2, 2, 3
⇒
A = {−2, 2, 3}
Also, we have set B = { x ∈ Z : − 3 < 2 x − 1 < 9}
Consider, −3 < 2 x − 1 < 9, x ∈ Z
⇒
−2 < 2 x < 10, x ∈ Z
⇒
−1 < x < 5, x ∈ Z
⇒
B = {0, 1, 2, 3, 4}
So, A × B has 15 elements.
∴ Number of subsets of A × B = 215 .
[Qif n( A) = m, the number
of possible subsets = 2 m ]
20. If the function f given by
f ( x ) = x 3 − 3(a − 2 ) x 2 + 3ax + 7, for some
a ∈R is increasing in (0, 1] and decreasing
in [1, 5), then a root of the equation,
f ( x ) − 14
= 0 ( x ≠ 1) is
( x − 1)2
[JEE Main 2019, 12 Jan Shift-I]
(a) − 7
(b) 6
(c) 7
(d) 5
Exp. (c)
Given that function,
f( x) = x3 − 3 (a − 2 ) x2 + 3ax + 7, for some a ∈ R
is increasing in (0, 1] and decreasing in [1, 5).
f′(1) = 0 [Qtangent at x = 1will be parallel to X-axis]
⇒ (3 x2 − 6(a − 2 ) x + 3a) x = 1 = 0
⇒
⇒
⇒
3 − 6(a − 2 ) + 3a = 0
3 − 6a + 12 + 3a = 0
15 − 3a = 0
8
JEE Main Chapterwise Mathematics
⇒
So,
a=5
f( x) = x3 − 9 x2 + 15 x + 7
⇒
f( x) − 14 = x3 − 9 x2 + 15 x − 7
⇒
f( x) − 14 = ( x − 1) ( x2 − 8 x + 7 )
f( x) − 14
⇒
( x − 1)
f( x) − 14
2
Now,
( x − 1)2
⇒
⇒
Exp. (c)
We have, 2 | x − 3| + x( x − 6) + 6 = 0
Let
x −3= y ⇒ x = y+ 3
∴ 2 | y| + ( y + 3)( y − 3) + 6 = 0
⇒
2 | y| + y2 − 3 = 0
= ( x − 1) ( x − 1)( x − 7 )
…(i)
⇒
[from Eq. (i)]
⇒
⇒
⇒
⇒
= ( x − 7)
= 0, ( x ≠ 1)
x−7 = 0
x=7
| y|2 + 2 | y| − 3 = 0
(| y| + 3)(| y| − 1) = 0
| y| ≠ − 3 ⇒ | y| = 1⇒ y = ± 1
x − 3 = ± 1 ⇒ x = 4, 2
x = 16, 4
23. The function f : R → − ,  defined as
 2 2
x
1 1
21. 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,
[JEE Main 2018]
(a) B ⊂ A
(b) A ⊂ B
(c) A ∩ B = φ (an empty set)
(d) neither A ⊂ B nor B ⊂ A
f (x ) =
1 + x2
is
Exp. (c)
We have, f( x) =
Exp. (b)
We have, |a − 5| < 1 and |b − 5| < 1
∴
−1 < a − 5 < 1 and −1 < b − 5 < 1
⇒
4 < a < 6 and 4 < b < 6
Now,
4(a − 6)2 + 9(b − 5)2 ≤ 36
(a − 6)2 (b − 5)2
+
≤1
9
4
Taking axes as a-axis and b-axis
⇒
(6, 7)
b
P
(0, 5) (3, 5) (4, 5)
S
Q (6, 6)
(6, 5)
R (6, 4)
[JEE Main 2017 (offline)]
(a) invertible
(b) injective but not surjective
(c) surjective but not injective
(d) neither injective nor surjective
∴
∴
So, f( x) is many-one function.
x
Again, let y = f( x) ⇒ y =
1 + x2
⇒
(9, 5)
(6, 3)
a
The set A represents square PQRS inside set B
representing ellipse and hence A ⊂ B.
22. Let S = { x ∈ R : x ≥ 0 and
2| x − 3| + x ( x − 6) + 6 = 0 . Then, S
(a) is an empty set
[JEE Main 2018]
(b) contains exactly one element
(c) contains exactly two elements
(d) contains exactly four elements
x
1 + x2
1
1
x

x =
= f ( x)
f  =
 x 1 + 1
1 + x2
2
x
1
1


f   = f(2 ) or f   = f(3) and so on.
2
 3
y + x2 y = x ⇒ yx2 − x + y = 0
x∈R
(− 1)2 − 4 ( y)( y) ≥ 0 ⇒ 1 − 4 y2 ≥ 0
−1 1
y∈ , 
⇒
 2 2 
−1 1
∴ Range = Codomain =  , 
 2 2 
So, f( x) is surjective.
Hence, f( x) is surjective but not injective.
As,
∴
24. Let a ,b , c ∈R . If f ( x ) = ax 2+ bx + c be such
that
and
a +b + c = 3
f ( x + y ) = f ( x ) + f ( y ) + xy , ∀ x , y ∈R , then
10
∑ f (n ) is
equal to
(a) 330
(b) 165
n =1
[JEE Main 2017 (offline)]
(c) 190
(d) 255
9
Sets, Relations and Functions
Exp. (a)
−
We have, f( x) = ax + bx + c
2
Now, f( x + y) = f( x) + f( y) + xy
Put y = 0
⇒ f( x) = f( x) + f(0) + 0
⇒ f(0) = 0 ⇒c = 0
Again, put y = − x
∴
f(0) = f( x) + f(− x) − x2
⇒
0 = ax2 + bx + ax2 − bx − x2
1
⇒2 ax2 − x2 = 0 ⇒ a =
2
Also,
a+ b+c=3
5
1
+ b+ 0=3 ⇒ b=
⇒
2
2
x2 + 5 x
f ( x) =
∴
2
2
n + 5n 1 2 5
Now,
f(n) =
= n + n
2
2
2
10
1 10 2 5 10
∴
∑ f(n) = 2 ∑ n + 2 ∑ n
n =1
n =1
n =1
1 10 × 11 × 21 5 10 × 11
= ⋅
+ ×
2
6
2
2
385 275 660
=
+
=
= 330
2
2
2
1
25. If f ( x ) + 2 f   = 3x ,
 x
x ≠ 0 and S = { x ∈R : f ( x ) = f ( − x )}; then S
(a) is an empty set
[JEE Main 2016 (offline)]
(b) contains exactly one element
(c) contains exactly two elements
(d) contains more than two elements
Exp. (c)
1
...(i)
We have, f( x) + 2 f   = 3 x, x ≠ 0
 x
1
On replacing x by in the above equation, we
x
get
1
3
f   + 2 f( x) =
 x
x
1 3

...(ii)
⇒
2 f ( x) + f   =
 x x
On multiplying Eq. (ii) by 2 and subtracting
Eq. (i) from Eq. (ii), we get
1
6
4f( x) + 2 f   =
 x x
⇒
1
f( x) + 2 f   = 3 x
−  x −
6
3f( x) = − 3 x
x
2
f ( x) = − x
x
Now, consider
⇒
⇒
f ( x) = f ( − x)
2
2
− x=− + x
x
x
4
= 2x
x
⇒
2 x2 = 4
⇒
x2 = 2
x=± 2
⇒
Hence, S contains exactly two elements.
26. If X = ( 4n − 3n − 1 : n ∈ N ) and
Y = {9 (n − 1): n ∈ N }; where N is the set of
natural numbers, then X ∪ Y is equal to
(b) Y − X
(d) Y
[JEE Main 2014]
(a) N
(c) X
Exp. (d)
Q
X = {4n − 3n − 1 : n ∈ N}
X = {0, 9, 54, 243, ...}
[put n = 1, 2, 3, ... ]
Y = {9 (n − 1) : n ∈ N}
Y = {0, 1, 18, 27, K}
It is clear that X ⊂ Y .
∴
X∪Y =Y
27. If g is the inverse of a function f and
f ′ (x ) =
1
1 + x5
(a) 1 + x 5
1
(c)
1 + { g ( x )}5
, then g ′ ( x ) is equal to
(b) 5x 4 [JEE Main 2014]
(d) 1 + { g ( x )}5
Exp. (d)
Here, g is the inverse of f ( x).
⇒
fog ( x) = x
On differentiating w.r.t. x, we get
f ′ {g ( x)} × g ′ ( x) = 1

1
1 
g ′ ( x) =
= 1 + {g ( x)}5 Q f ′ ( x) =

f ′ {g ( x)}
1 + x5 

5
⇒ g ′ ( x) = 1 + {g ( x)}
10
JEE Main Chapterwise Mathematics
28. Let X = {1, 2, 3, 4, 5}. The number of different
(b) a ∈ Y , i.e., ‘ a’ is present in Y and a ∉ Z, i.e., ‘ a’ is
not present in Z.
(c) a ∉ Y , i.e., ‘ a’ is not present in Y and a ∈ Z,
i.e., ‘ a’ is present in Z.
(d) a ∉ Y , i.e., ‘ a’ is not present in Y and a ∉ Z,
i.e., ‘ a’ is not present in Z.
ordered pairs (Y , Z ) that can formed such
thatY ⊆ X , Z ⊆ X andY ∩ Z is empty, is
[AIEEE 2012]
(a) 52
(b) 35
(c) 2 5
(d) 53
Exp. (b)
Analysis of the above 4 cases
Given A set X = {1, 2, 3, 4, 5}
To find The number of different ordered pairs
(Y , Z ) such that Y ⊆ X, Z ⊆ X and Y ∩ Z = φ.
Since, Y ⊆ X, Z ⊆ X, hence we can only use the
elements of X to construct sets Y and Z.
Method 1
n (Y )
Number of ways to
make Y
Number of ways
to make Z such
that Y ∩ Z = φ
0
5
1
5
C1
24
2
5
C2
2
3
3
5
C3
22
4
5
C4
21
5
5
C0
C5
25
20
Let us explain anyone of the above 6 rows say
third row. In third row,
Number of elements in Y = 2
∴Number of ways to select Y = 5C 2 ways
Because any 2 elements of X can be part of Y.
Now, if Y contains any 2 elements, then these
2 elements cannot be used in any way to
construct Z, because we want Y ∩ Z = φ.
And from the remaining 3 elements which are not
present in Y,2 3 subsets can be made each of which
can be equal to Z and still Y ∩ Z = φ will be true.
Hence, total number of ways to construct sets Y
and Z such that Y ∩ Z = φ
= 5C 0 × 2 5 + 5C1 × 2 5 − 1 + K + 5C 5 × 2 5 − 5
= (2 + 1)5 = 35
Method 2
Since, Y ⊆ X, Z ⊆ X, hence we can only use the
elements of X to construct sets Y and Z.
Every elements in X (say a) has four options (as far
as going to Y and Z is concerned).
(a) a ∈ Y , i.e., ‘ a’ is present in Y and a ∈ Z, i.e., ‘ a’ is
present in Z.
(a) If ‘ a’ is present in Y and also in Z, then it will be
certainly present in Y ∩ Z.
(b) If ‘ a’ is present in Y but not present in Z, then it
will not be present in Y ∩ Z.
(c) If ‘ a’ is not present in Y but present in Z, then
also it will not be present in Y ∩ Z.
(d) If ‘ a’ is not present in both of Y and Z, then it will
not be present in Y ∩ Z.
We want Y ∩ Z = φ to which only case (a) is not
favourable and remaining cases, i.e., cases (b),
(c) and (d) are favourable.
Hence, for every elements ‘ a’ in X, there are
3 favourable ways such that Y ∩ Z = φ.
⇒ Total number of ways
= 3 × 3 × 3 × 3 × 3 = 35
[as there are 5 elements in X and each of them
have 3 options to go or not to go to Y and Z]
29. Let R be the set of real numbers.
Statement I A = {( x , y ) ∈R × R : y − x is an
integer} is an equivalence relation on R.
[AIEEE 2011]
Statement II B = {( x , y ) ∈R × R : x = αy for
some rational number α} is an equivalence
relation on R.
(a) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(b) Statement I is true, Statement II is false
(c) Statement I is false, Statement II is true
(d) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
Condition for equivalence relation A
relation which is symmetric, reflexive and
transitive is equivalence relation.
Exp. (b)
Statement I
A = {( x, y) ∈ R × R : y − x is an integer}
(a) Reflexive
xRx : ( x − x) is an integer. i.e., true
∴ Reflexive
11
Sets, Relations and Functions
(b) Symmetric
xRy : ( x − y) is an integer.
⇒ − ( y − x) is an integer.
⇒ ( y − x) is an integer.
⇒ yRx
∴ Symmetric
(c) Transitive
xRy and yRz
⇒( x − y) is an integer and ( y − z) is an
integer.
⇒( x − y) + ( y − z) is an integer.
⇒( x − z) is an integer.
⇒ xRz
∴ Transitive
Hence, A is an equivalence relation.
Statement II
B = {( x, y) ∈ R × R : x = αy for some
rational number α}
1
If α = , then for reflexive, we have
2
1
x Rx ⇒ x = x, which is not true, ∀ x ∈ R − {0}.
2
∴ B is not reflexive on R.
Hence, B is not an equivalence relation on R.
Hence, statement I is true, statement II is false.
30. The domain of the function
f (x ) =
(a)
(b)
(c)
(d)
1
|x | − x
Statement II For any two invertible 3 × 3
matrices M and N ,( MN )−1 = N −1M −1.
(a) Statement I is false, Statement II is true
[AIEEE 2011]
(b) Statement I is true, Statement II is true;
Statement II is correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Condition for equivalence relation A
relation which is symmetric, reflexive and
transitive is equivalence relation.
Exp. (c)
Given,
R = {( A, B) : A = P −1 BP for some invertible
matrix P}
For Statement I
(i) Reflexive
ARA ⇒ A = P −1 AP, which is true only, if P = I.
Since, A = P −1BP for some invertible matrix P.
∴ We can assume P = I
⇒
ARA ⇒ A = I −1 AI
⇒
A= A
⇒ R is Reflexive
Note Here, due to some invertible matrix, P is used
[AIEEE 2011]
is
(0, ∞ )
( − ∞ , 0)
( − ∞ , ∞ ) − { 0}
(− ∞, ∞)
(reflexive) but if for all invertible matrix is used,
then R is not reflexive.
(ii) Symmetric
ARB ⇒ A = P −1BP
Exp. (b)
y=
1
| x| − x
For domain, | x| − x > 0
⇒
| x| > x
i .e., only possible, if x < 0.
∴
x ∈ (− ∞, 0)
31. Consider the following relation R on the set
of real square matrices of order 3.
R = {( A , B ): A = P −1BP for some invertible
matrix P}
Statement I R is an equivalence relation.
⇒
PAP −1 = P(P −1BP) P −1
⇒
PAP −1 = (PP −1 ) B(PP −1 )
∴
B = PAP −1
Since, for some invertible matrix P, we can let
Q = P −1
∴
B = (P − 1)−1 AP −1
⇒
B = Q −1 AQ ⇒ BRA
⇒ R is symmetric.
(iii) Transitive
ARB and BRC
⇒
A = P −1BP
and
B = P −1CP
⇒
A = P −1 (P −1CP) P
= (P −1 )2 C (P)2
12
JEE Main Chapterwise Mathematics
So, ARC, for some P 2 = P
⇒ R is transitive
So, R is an equivalence relation.
For Statement II It is always true that
(MN)−1 = N−1M −1
33. Let f be a function defined by
f ( x ) = ( x − 1 )2 + 1, ( x ≥ 1 ).
Statement I The set{ x : f ( x ) = f
g ( x ) = a1x 2 + b1 x + c1 and p ( x ) = f ( x ) − g ( x ).
If p ( x ) = 0 only for x = − 1 and p ( − 2 ) = 2 ,
then the value of p (2 ) is
[AIEEE 2011]
(a) 18
(b) 3
(c) 9
(d) 6
Exp. (a)
Given that p ( x) = f ( x) − g ( x) has only one root −1.
⇒ p ( x) = (a − a1 ) x2 + (b − b1 ) x + (c − c1 ) has
one root, − 1only,
⇒ p′ ( x) will also has root as − 1.
⇒ p′ ( x) = 0 at x = − 1
⇒ 2 (a − a1 ) x + (b − b1 ) = 0 at x = − 1.
⇒ − 2 (a − a1 ) + (b − b1 ) = 0
− (b − b1 )
…(i)
⇒
= −2
(a − a1 )
Now, p ( x) = (a − a1 ) x2 + (b − b1 ) x + (c − c1 )
b − b1
(c − c1 )
p ( x)
⇒
x+
= x2 +
a − a1
a − a1
a − a1
Q
p (− 1) = 0
(b − b1 ) (c − c1 )
+
a − a1
a − a1
(c − c1 )
[using Eq. (i)]
0 = 1− 2 +
⇒
a − a1
c − c1
…(ii)
=1
⇒
a − a1
∴
0 = (− 1)2 −
Also, given that p (− 2 ) = 2
…(iii)
⇒ 4 (a − a 1) − 2 (b − b 1) + (c − c 1 ) = 2
From Eqs. (i), (ii) and (iii), we have
4 (a − a 1) − 4 (a − a 1) + (a − a 1) = 2
⇒
a − a1 = 2
On substituting a − a1 = 2 in Eq. (ii), we get
c − c1 = 2
On substituting a − a 1 = 2 in Eq. (i), we get
b − b1 = 4
Now, p (2 ) = 4 (a − a 1) + 2 (b − b 1) + (c − c 1)
= 4×2 + 2 × 4+ 2
= 8 + 8 + 2 = 18
( x )} = {1, 2}
Statement II f is bijection and
Hence, both statements are true but second is
not the correct explanation of first.
32. Let for a ≠ a1 ≠ 0, f ( x ) = ax 2 + bx + c ,
−1
f
−1
( x ) = 1 + x − 1, x ≥ 1 [AIEEE 2011]
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Exp. (b)
f( x) = ( x − 1)2 + 1,
f ( x) = f ( x)
When
⇒
⇒
f ( x) = x
( x − 1)2 + 1 = x
⇒
⇒
as x ≥ 1
−1
( x − 1)2 = x − 1
( x − 1)2 − ( x − 1) = 0
⇒ ( x − 1) { x − 1 − 1} = 0
⇒
∴
x = 1, 2
{ x : f( x) = f −1( x)} = {1, 2}
f ( x) = y
Also, let
∴
⇒
y = ( x − 1)2 + 1
( x − 1) = ±
y−1
x = 1±
y−1
∴
x≥1
Q
Neglecting 1 −
y−1
⇒
x = 1+
y−1
f ( y) = 1 +
y−1
or
∴
−1
−1
f ( x) = 1 +
x−1
So, both statements are correct and Statement II
is correct explanation of Statement I.
34. Consider the following relations
R = {( x , y ) | x , y are real numbers and x = wy
for some rational number w} ;
 m p 
S =   ,  m , n , p and q are integers such
 n q 
that n, q ≠ 0 and qm = pn}. Then, [AIEEE 2010]
13
Sets, Relations and Functions
(a) R is an equivalence relation but S is not an
equivalence relation
(b) Neither R nor S is an equivalence relation
(c) S is an equivalence relation but R is not an
equivalence relation
(d) R and S both are equivalence relations
Exp. (c)
(i) Reflexive xRx ⇒ x = wx
∴ w = 1 ∈ rational number
The relation R is reflexive.
36. For real x, let f ( x ) = x 3 + 5x + 1, then
f is one-one but not onto R
[AIEEE 2009]
f is onto R but not one-one
f is one-one and onto R
f is neither one-one nor onto R
Exp. (c)
f ( x) = x 3 + 5 x + 1
Given
(ii) Symmetric xRy ⇒
/ yRx as 0R1
But 1R0 ⇒1 = w ⋅ (0)
which is not true for any rational number.
The relation R is not symmetric.
Thus, R is not equivalence relation.
Now, for relation S which is defined as
 m p
S =  ,  m, n, p and q ∈ integers such that
 n q 
n, q ≠ 0 and qm = pn}
m m
(i) Reflexive R
⇒ mn = mn
n n
The relation S is reflexive.
m p
(ii) Symmetric
R ⇒ mq = np
n q
p m
np = mq ⇒ R
⇒
q n
[true]
Now, f ′( x) = 3 x2 + 5 > 0, ∀ x ∈ R
Thus, f( x) is strictly increasing function.
So, f( x) is one-one function.
Clearly, f( x) is a continuous function and also
increasing on R.
∴
lim
x→ −∞
f( x) = − ∞ and lim = ∞
x→ ∞
Hence, f( x)takes every value between − ∞ and ∞.
Thus, f( x) is onto function.
37. Let f ( x ) = ( x + 1)2 − 1, x ≥ − 1
Statement I The set{ x : f ( x )= f
−1
Statement II f is a bijection.
The relation S is symmetric.
m p
p r
(iii) Transitive
R and R
n q
q s
⇒
and
⇒
⇒
mq = np
ps = rq
mq ⋅ ps = np ⋅ rq
ms = nr
m r
m r
= ⇒ R
⇒
n s
n s
The relation S is transitive.
Hence, the relation S is equivalence relation.
35. If A , B andC are three sets such that
A ∩ B = A ∩ C and A ∪ B = A ∪ C , then
A =C
B =C
A∩B=φ
A=B
Given, A ∩ B = A ∩ C
and A ∪ B = A ∪ C
∴
B=C
(a)
(b)
(c)
(d)
Given, relation R is defined as
R = {( x, y)| x, y are real numbers and
x = wy for some rational number w}
(a)
(b)
(c)
(d)
Exp. (b)
[AIEEE 2009]
( x )} = {0,−1}
[AIEEE 2009]
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Exp. (d)
Given,
f( x) = ( x + 1)2 − 1, x ≥ − 1
⇒
f ′( x) = 2 ( x + 1) ≥ 0, for x ≥ − 1
⇒ f( x) is one-one.
Since, codomain of the given function is not given,
hence it can be considered as R, the set of real
and consequently f is not onto.
Hence, f is not bijective. Statement II is false.
Also,
f( x) = ( x + 1)2 − 1 ≥ − 1 for x ≥ − 1
⇒
R f = [− 1, ∞ )
Clearly, f( x) = f −1( x) at x = 0 and x = − 1
∴Statement I is true.
14
JEE Main Chapterwise Mathematics
38. Let R be the real line. Consider the following
subsets of the planeR × R
S = {( x , y ): y = x + 1 and 0 < x < 2 }
[AIEEE 2008]
T = {( x , y ): x − y is an integer}
Which one of the following is true?
(a) T is an equivalence relation on R but S is not
(b) NeitherS norT is an equivalence relation on R
(c) Both S and T are equivalence relations on R
(d)S is an equivalence relation on R but T is not
x∈N
y = 4x + 3
y−3
⇒
x=
4
y−3
−1
.
∴ Inverse of f is f ( y) =
4
⇒
π π
40. The largest interval lying in  − ,  for
which the function
 2 2
Since, (1, 2 ) ∈S but (2 , 1) ∉S
2

x
f ( x ) = 4− x + cos −1  − 1 + log (cos x ) is

2
defined, is
[AIEEE 2007]
So, S is not symmetric.
(a) [0 , π ]
Hence, S is not an equivalence relation.
π π
(c)  − , 
 4 2 
Exp. (a)
Given,
T = {( x, y) : ( x − y) ∈ l }
Now, x − x = 0 ∈ I, it is reflexive relation.
Again now, ( x − y) ∈ I
y − x ∈ I, it is symmetric relation.
Let
x − y = I1 and y − z = I2
Now, x − z = ( x − y) + ( y − z) = I1 + I2 ∈ I
So, T is also transitive.
Hence, T is an equivalence relation.
39. Let f : N → Y be a function defined as
f ( x ) = 4x + 3, where
for some x ∈ N }.
Y = { y ∈ N : y = 4x + 3
[AIEEE 2008]
Show that f is invertible and its inverse is
y −3
4
y+3
−1
(c) f ( y ) = 4 +
4
(a) f
−1
(y)=
3y + 4
3
y+3
−1
(d) f ( y ) =
4
(b) f
−1
(y)=
Exp. (a)
The given function is f : N → y defined as
f ( x) = 4 x + 3
where y = { y ∈ N : y = 4 x + 3 for some x ∈ N}
Now,
f ( x) = 4 x + 3
⇒
f ′ ( x) = 4 > 0
⇒ f is a strictly increasing function.
⇒ f is one-one.
Also, here y = {7, 8, 9, ... , ∞}
and range of f = {7, 8, 9, ... , ∞}
QRange of f = y = Codomain of f
⇒ f is onto.
⇒ f is invertible.
Now, let y ∈ Y such that f ( x) = y
⇒
x = f −1 ( y)
π π
(b)  − , 
 2 2
π
(d) 0, 
 2 
Exp. (d)
Given that,
2
x
f( x) = 4− x + cos −1  − 1 + log (cos x)
2

π π

− x2
Here, 4
is defined for  − ,  ,
 2 2
x

−1  x
cos  − 1 is defined, if −1 ≤ − 1 ≤ 1.
2

2
x
0≤ ≤ 2 ⇒0≤ x≤ 4
⇒
2
And log (cos x) is defined, if cos x > 0.
π
π
− < x<
⇒
2
2
2
x
Hence, f( x) = 4− x + cos −1  − 1 + log (cos x)
2

π

is defined, if x ∈ 0,  .
 2 
41. Let W denotes the words in the English
dictionary define the relation R by
R = {( x , y ) ∈W × W : the words x and y have
atleast one letter in common}. Then, R is
(a)
(b)
(c)
(d)
reflexive, symmetric and not transitive
reflexive, symmetric and transitive
reflexive, not symmetric and transitive
not reflexive, symmetric and transitive
[AIEEE 2006]
Exp. (a)
Let W = {CAT, TOY , YOU, . . .}
Clearly, R is reflexive and symmetric but not
transitive.
[QCAT RTOY, TOY R YOU ⇒
/ CAT R YOU ]
15
Sets, Relations and Functions
42. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12),
(3, 9), (3, 12), (3, 6)} be a relation on the set
A = {3, 6, 9, 12}. The relation is [AIEEE 2005]
(a)
(b)
(c)
(d)
reflexive and symmetric only
an equivalence relation
reflexive only
reflexive and transitive only
Exp. (d)
Given, f( x − y) = f( x)f( y) − f(a − x)f(a + y)
Let
x=0= y
⇒
f(0) = [ f(0)]2 − [ f(a)]2
⇒
⇒
∴
Exp. (d)
Since, for every elements of A, there exists
elements (3, 3), (6, 6), (9, 9), (12, 12) ∈ R ⇒ R is
reflexive relation.
Now, (6, 12 ) ∈ R but (12 , 6) ∉R, so it is not a
symmetric relation.
Also, (3, 6), (6, 12 ) ∈ R ⇒(3, 12 ) ∈ R
∴ R is transitive relation.
43. Let f :( −1, 1) → B be a function defined by
 2x 
f ( x ) = tan 
 , then f is both one-one
1 − x2
and onto when B is in the interval
−1
[AIEEE 2005]
π π
(a)  − , 
 2 2
π
(c) 0, 
 2 
π π
(b)  − , 
 2 2 
π
(d)  0, 
 2
45. Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a
relation on the set A = {1, 2, 3, 4}. The
relation R is
[AIEEE 2004]
(a) a function
(c) not symmetric
(b) transitive
(d) reflexive
Exp. (c)
Given, R = {(1, 3), (4, 2 ), (2 , 4), (2 , 3), (3, 1)} is a
relation on the set A = {1, 2 , 3, 4}.
(a) Since, (2 , 4) ∈ R and (2 , 3) ∈ R. So, R is not a
function.
(b) Since, (1, 3) ∈ R and (3, 1) ∈ R but (1, 1) ∉R. So,
R is not transitive.
(c) Since, (2 , 3) ∈ R but (3, 2 ) ∉R. So, R is not
symmetric.
(d) Since, (1, 1), (2, 2), (3, 3), (4, 4) ∉R. So, R is
not reflexive.
symmetrical about the line x = 2 , then
x ∈ (−1, 1)
π π
−1
⇒
tan x ∈  − , 
 4 4
π π
2 tan−1 x ∈  − , 
⇒
 2 2
 2x 

Given that, f( x) = tan−1 
2
1 − x 
Given,
= 2 tan−1 x
π π
f( x) ∈  − , 
 2 2
[AIEEE 2004]
(a)
(b)
(c)
(d)
f (x + 2) = f (x − 2)
f (2 + x) = f (2 − x)
f ( x ) = f ( −x )
f ( x ) = − f ( −x )
Exp. (b)
[Q x2 < 1]
Hence, function is one-one onto.
Given, graph is symmetrical about the line x = 2.
∴
f(2 + x) = f(2 − x)
47. The domain of the function,
f (x ) =
sin −1 ( x − 3)
44. A real valued function f ( x ) satisfies the
functional equation
f ( x − y ) = f ( x ) f ( y ) − f (a − x ) f (a + y ),
wherea is a given constant and f (0) = 1, then
[AIEEE 2005]
f (2a − x ) is equal to
(a) f ( − x )
(c) f ( x )
[given, f(0) = 1]
f ( a) = 0
f(2 a − x) = f{a − ( x − a)}
= f(a)f( x − a) − f(a − a)f( x)
= 0 − f ( x) ⋅ 1 = − f ( x)
46. The graph of the function y = f ( x ) is
Exp. (a)
So,
1 = 1 − [ f(a)]2
(b) f (a ) + f (a − x )
(d) − f ( x )
9 − x2
(a) [2, 3]
(c) [1, 2]
is
[AIEEE 2004]
(b) [2, 3)
(d) [1, 2)
Exp. (b)
Given function f( x) =
sin−1 ( x − 3)
9 − x2
will be defined, if
16
JEE Main Chapterwise Mathematics
⇒
and
−1 ≤ ( x − 3) ≤ 1
2 ≤ x≤ 4
9 − x2 > 0
n
…(i)
∑ f(r ) = f(1) + f(2 ) + f(3) + . . . + f(n)
r =1
= f(1) + 2 f(1) + 3f(1) + . . . + nf(1)
…(ii)
⇒
−3 < x < 3
From Eqs. (i) and (ii), we get
2 ≤ x< 3
Hence, domain of the given function is [2, 3).
48. A function f from the set of natural numbers
to integers defined by
n − 1
, when n is odd

is
f (n ) =  2
n
 − , when n is even
 2
(a)
(b)
(c)
(d)
one-one but not onto
onto but not one-one
one-one and onto both
neither one-one nor onto
[AIEEE 2003]
[Qf( x + y) = f( x) + f( y)]
= (1 + 2 + 3 + . . . + n)f(1) = f(1)Σn
7 n(n + 1)
[Qf(1) = 7, given]
=
2
50. Domain of definition of the function
f (x ) =
3
4− x
[AIEEE 2003]
(b) ( −1, 0) ∪ (1, 2 )
(d) ( −1, 0) ∪ (1, 2 ) ∪ ( 2 , ∞ )
(a) (1, 2)
(c) (1, 2 ) ∪ ( 2 , ∞ )
Exp. (d)
Given, f( x) =
3
4 − x2
+ log10 ( x3 − x)
For domain of f( x),
x3 − x > 0 ⇒ x( x − 1)( x + 1) > 0
Exp. (c)
 n − 1, when n is odd

Given that, f(n) =  2
n
 − , when n is even
 2
and f : N → I, where N is the set of natural numbers
and I is the set of integers.
Let x, y ∈ N and both are even.
Then, f( x) = f( y)
x
y
⇒
⇒ x= y
− =−
2
2
Again, x, y ∈ N and both are odd.
x−1 y−1
Then, f( x) = f( y) ⇒
⇒x= y
=
2
2
So, mapping is one-one.
Since, each negative integer is an image of even
natural number and positive integer is an image of
odd natural number. So, mapping is onto. Hence,
mapping is one-one onto.
49. If f : R → R satisfies f ( x + y ) = f ( x ) + f ( y ),
for all x , y ∈R and f (1) = 7, then
+ log 10 ( x 3 − x ), is
2
n
∑ f (r ) is
–
+
–1
–
0
⇒
x ∈ (−1, 0) ∪ (1, ∞ ) and
⇒
x≠±2
⇒
x ∈ (−∞, − 2 ) ∪ (−2 , 2 ) ∪ (2 , ∞ )
(a)
7n
2
(c) 7n(n + 1 )
Exp. (d)
7(n + 1 )
2
7n(n + 1 )
(d)
2
(b)
4 − x2 ≠ 0
So, common region is (−1, 0) ∪ (1, 2 ) ∪ (2 , ∞ ).
51. The function f ( x ) = log ( x + x 2 + 1 ), is
(a)
(b)
(c)
(d)
an even function
[AIEEE 2003]
an odd function
a periodic function
neither an even nor an odd function
Exp. (b)
Given that, f( x) = log ( x +
Now,
f(− x) = log (− x +
∴ f( x) + f(− x) = log ( x +
r =1
[AIEEE 2003]
+
1
x2 + 1)
x2 + 1)
x2 + 1)
+ log (− x +
x2 + 1)
= log (1) = 0
Hence, f( x) is an odd function.
52. The period of sin 2 θ is
(a) π
2
Exp. (b)
(b) π
1[AIEEE 2002]
(c) 2 π
(d)
π
2
17
Sets, Relations and Functions
We know that,
1 − cos 2 θ 1 1
= − cos 2 θ
2
2 2
2π
2
= π
∴Period of sin θ =
2
sin2 θ =

x 
53. The domain of sin −1 log 3    is
 3 

[AIEEE 2002]
(a) [1, 9]
(b) [–1, 9] (c) [–9, 1]
(d) [–9, –1]
55. The domain of definition of the function
Exp. (a)
−1
Since, domain of sin
x is [–1, 1].
x

∴
−1 ≤ log 3   ≤ 1
 3
x
3−1 ≤ ≤ 3 ⇒ 1 ≤ x ≤ 9
⇒
3
x 

Hence, domain of sin−1 log 3   is [1, 9].

 3  
54. The period of the function
f ( x ) = sin 4 x + cos 4 x is
[AIEEE 2002]
π
2
(d) None of these
(a) π
(b)
(c) 2 π
Exp. (b)
Given that, f( x) = sin4 x + cos 4 x
∴
f( x) = (sin2 x + cos 2 x)2 − 2 sin2 x cos 2 x
= 1−
1
(sin 2 x)2
2
1 1 − cos 4 x 
= 1 − 


2 
2
3 1
= + cos 4 x
4 4
2π π
=
∴The period of f( x) =
4
2
[Q cos x is periodic with period 2 π]
= 1−
1
(2 sin x cos x)2
2
 5x − x 2 
f ( x ) = log 10 
 is
 4 
(a) [1, 4]
(c) [0, 5]
[AIEEE 2002]
(b) [1, 0]
(d) [5, 0]
Exp. (a)
 5 x − x2 

Given that, f( x) = log10 
4 

For domain of f( x),
 5 x − x2 
≥0
log10 
4 

⇒
⇒
⇒
5 x − x2
≥1
4
x2 − 5 x + 4 ≤ 0
( x − 1)( x − 4) ≤ 0 ⇒ x ∈[1, 4]
2
Complex Numbers and
Quadratic Equations
1. If α and β are the roots of the equation
2. The sum of the solutions of the equation
x − 2 x + 2 = 0, then the least value of n for
2
| x − 2 | + x ( x − 4 ) + 2 = 0 ( x > 0 ) is equal to
[JEE Main 2019, 8 April Shift-I]
n
α
which   = 1 is
β
[JEE Main 2019, 8 April Shift-I]
(a) 2
(b) 5
(c) 4
(d) 3
Given, α and β are the roots of the quadratic
equation,
x2 − 2 x + 2 = 0
⇒
( x − 1) + 1 = 0 ⇒( x − 1) = − 1
2
[where i =
⇒
x − 1= ± i
⇒
x = (1 + i )or (1 − i )
Clearly, if α = 1 + i, then β = 1 − i
−1]
n
α
According to the question   = 1
β
1 + i 

 =1
1− i 
⇒
 (1 + i )(1 + i )

 =1
 (1 − i )(1 + i ) 
n
[by rationalization]
n
⇒
⇒
 1 + i 2 + 2i 

 = 1
2
 1− i

n
 2i  = 1⇒ i n = 1
 
2
So, minimum value of nis 4.
Key Idea Reduce the given equation into
quadratic equation.
Given equation is
| x − 2| + x( x − 4) + 2 = 0
⇒
| x − 2| + x − 4 x + 4 = 2
⇒
⇒
| x − 2| + ( x − 2 )2 = 2
(| x − 2|)2 + | x − 2| − 2 = 0
Let| x − 2| = y, then above equation reduced to
y2 + y − 2 = 0
n
⇒
(b) 12
(d) 10
Exp. (d)
Exp. (c)
2
(a) 9
(c) 4
[Qi 4 = 1]
⇒
y2 + 2 y − y − 2 = 0
⇒
y( y + 2 ) − 1( y + 2 ) = 0
⇒
( y + 2 )( y − 1) = 0
⇒
y = 1, − 2
∴
y=1
[Qy = | x − 2| ≥ 0]
| x − 2| = 1
⇒
x −2 = ±1
⇒
x = 3 or 1
⇒
⇒
x = 9 or 1
∴ Sum of roots = 9 + 1 = 10
19
Complex Numbers and Quadratic Equations
3 i
+ (i = −1 ), then (1 + iz + z 5 + iz 8 )9
2
2
is equal to
[JEE Main 2019, 8 April Shift-II]
3. If z =
(a) 1
(b) ( −1 + 2i )9
(c) −1
(d) 0
Given quadratic equation is
(1 + m2 )x2 − 2(1 + 3m)x + (1 + 8m) = 0
= 4[(1 + 3m)2 − (1 + m2 )(1 + 8m)]
= 4[1 + 9m2 + 6m − (1 + 8m + m2 + 8m3 )]
Exp. (c)
= 4[−8m3 + 8m2 − 2 m]
Key Idea Use, e i θ = cos θ + i sin θ
= − 8m(4m2 − 4m + 1) = − 8m(2 m − 1)2
π
i
3  1
π
π
Given, z =
+   i = cos + i sin = e 6
2
2
6
6
According to the question there is no solution of
the quadratic Eq. (i), then
D< 0
∴ −8m(2 m − 1)2 < 0 ⇒ m > 0
so, (1 + iz + z5 + iz8 )9
π
5π
8π 

i
i
i
=  1 + ie 6 + e 6 + ie 6 




9
π
π
π
4π 
5π

i
i
i
i
i
= 1 + e 2 ⋅ e 6 + e 6 + e 2 ⋅ e 3 




2π
5π
11π 

i
i
i
= 1 + e 3 + e 6 + e 6 





= 1+

9
π

i 
Q i = e 2 


9
11π
11π  
+  cos
+ i sin


6
6  

1 i 3
3 1
3 i
= 1 − +
−
+ i +
− 
2
2
2
2
2
2

So, there are infinitely many values of ‘m’ for which,
there is no solution of the given quadratic
equation.
5. All the points in the set
α + i

S =
: α ∈ R (i = −1 ) lie on a
−
i
α


[JEE Main 2019, 9 April Shift-I]
 cos 2 π + i sin 2 π  +  cos 5 π + i sin 5 π 


 

3
3  
6
6 
9
9
9
9
1
π
π
3i 

= +
 =  cos + i sin 

2 
3
3
2
= cos 3 π + i sin 3 π [Qfor any natural number ‘n’
(cos θ + i sinθ)n = cos(nθ) + i sin(nθ)]
= −1
4 The number of integral values ofm for which
the
equation
(1 + m 2 )x 2 − 2(1 + 3m )x + (1 + 8m ) = 0, has no
real root is
[JEE Main 2019, 8 April Shift-II]
(a) 3
(c) 1
…(i)
Now, discriminant
D = [−2(1 + 3m)]2 − 4(1 + m2 )(1 + 8m)
(a)
(b)
(c)
(d)
circle whose radius is 2.
straight line whose slope is −1.
circle whose radius is 1.
straight line whose slope is 1.
Exp. (c)
⇒ x + iy =
=
(α 2 − 1) + (2α )i
α +1
α2 + 1
2
α − 1  2α 
+
= 2
i
α + 1  α 2 + 1
2
On comparing real and imaginary parts, we get
x=
α2 − 1
α2 + 1
and y =
2α
α2 + 1
2
2
 α 2 − 1
2α 
 +  2
Now, x2 + y2 =  2

 α + 1
 α + 1
=
(b) infinitely many
(d) 2
=
Exp. (b)
Key Idea
(i) First convert the given equation in quadratic
equation.
(ii) Use, Discriminant, D = b 2 − 4ac < 0
α+ i
α−i
(α + i )2
Let x + iy =
⇒
α 4 + 1 − 2α 2 + 4α 2
(α 2 + 1)2
2
(α + 1)2
(α 2 + 1)2
=1
x 2 + y2 = 1
α+ i
So, S = 
;α ∈ R lies on a circle with

α − i
radius 1.
20
JEE Main Chapterwise Mathematics
6 Let p, q∈R. If2 − 3 is a root of the quadratic
equation, x + px + q = 0, then
2
[JEE Main 2019, 9 April Shift-I]
(a) q 2 − 4p − 16 = 0
(m 2 + 1)x 2 − 3x + (m 2 + 1)2 = 0 such that the
sum of its roots is greatest, then the absolute
difference of the cubes of its roots is
[JEE Main 2019, 9 April Shift-II]
(b) p 2 − 4q − 12 = 0
(a) 10 5
(c) 8 3
(c) p 2 − 4q + 12 = 0
(d) q 2 + 4p + 14 = 0
Given quadratic equation is
x2 + px + q = 0, where p, q ∈R having one root
2 − 3 , then other root is 2 + 3 (conjugate of
2 − 3)
[Qirrational roots of a quadratic
equation always occurs in pairs]
So, sum of roots = − p = 4 ⇒ p = −4
and product of roots = q = 4 − 3 ⇒ q = 1
Now, from options p2 − 4q − 12 = 16 − 4 − 12 = 0
7 Let z ∈C be such that | z|< 1. If ω =
5 + 3z
,
5(1 − z )
Given quadratic equation is
(m2 + 1)x2 − 3 x + (m2 + 1)2 = 0
According to the question, the sum of roots is
greatest and it is possible only when ‘‘(m2 + 1) is
minimum’’ and ‘‘minimum value of m2 + 1 = 1,
when m = 0’’.
∴α + β = 3 and αβ = 1, as m = 0
Now, the absolute difference of the cubes of roots
= |α 3 − β 3|
= |α − β||α 2 + β 2 + αβ|
(b) 5 Re (ω) > 1
(d) 5 Re(ω) > 4
= (α + β )2 − 4αβ |(α + β )2 − αβ|
= 9 − 4 |9 − 1|
=8 5
Exp. (b)
Given complex number
5 + 3z
ω=
5(1 − z)
⇒
⇒
⇒
Q
∴
⇒
…(i)
Let the roots of quadratic Eq. (i) are α and β, so
3
and αβ = m2 + 1
α+β= 2
m +1
[JEE Main 2019, 9 April Shift-II]
(a) 4 Im(ω) > 5
(c) 5 Im(ω) < 1
(b) 8 5
(d) 4 3
Exp. (b)
Exp. (b)
then
8 If m is chosen in the quadratic equation
9 If α and β are the roots of the quadratic
5ω − 5ω z = 5 + 3 z
(3 + 5ω)z = 5ω − 5
…(i)
|3 + 5ω|| z| = |5ω − 5|
[applying modulus both sides and
| z1 z2| = | z1|| z2|]
| z| < 1
[from Eq. (i)]
|3 + 5ω| > |5ω − 5|
3


ω + > |ω − 1|
5

2
3
Let ω = x + iy, then  x +  + y2 > ( x − 1)2 + y2

5
9
6
+ x > x2 + 1 − 2 x
x2 +
⇒
25 5
16 x 16
⇒
>
5
25
1
x > ⇒ 5x > 1
⇒
5
⇒
5 Re( ω) > 1
 π
equation, x 2 + x sin θ − 2 sin θ = 0, θ ∈ 0,  ,
 2
then
α12 + β12
(α −12 + β −12 )(α − β )24
is equal to
[JEE Main 2019, 10 April Shift-I]
(a)
212
(sin θ + 8)
12
(b)
26
(sin θ + 8)12
12
(c)
2
(sin θ − 4)
12
(d)
212
(sin θ − 8)6
Exp. (a)
Given quadratic equation is
π
x2 + xsinθ − 2 sinθ = 0, θ ∈  0, 
 2
and its roots are α and β.
So, sum of roots = α + β = − sinθ
and product of roots = αβ = − 2sinθ
⇒
αβ = 2(α + β )
…(i)
21
Complex Numbers and Quadratic Equations
α12 + β12
Now, the given expression is −12
(α + β −12 )(α − β)24
sin 2 x − 2 sin x + 5
1
⋅
≤ 1 also satisfy the
α12 + β12
=
 1
1 
24
 12 + 12  (α − β)
α
β 
2
α12 + β12
=
12
 β + α12 
 12 12  (α − β)24
 α β

(a) 2 | sin x| = 3 sin y
(b) sin x = | sin y |
(c) sin x = 2 sin y
(d) 2 sin x = sin y
12
 αβ 
=
2
 (α − β ) 
4
equation


αβ

= 
2
 (α + β ) − 4αβ 


2
=

 (α + β ) − 8 
2
[from Eq. (i)]


2
=

 − sinθ − 8 
12
212
1 3
− i
5 5
3 1
(d) − − i
5 5
1 3
− i
5 5
1 3
(c) − + i
5 5
a2 + 1
2 i (a + i ) −2 + 2 ai
= 2
=
a +1
a2 + 1
z = 2/5
4 + 4a
4
1 + a2
⋅ 2 −2 sin
(sin x − 1) 2 + 4
≤2
2
y
≤1
2 sin 2 y
2
⇒
(sin x − 1) + 4 ≤ 2 sin2 y
2
QRange of (sin x − 1)2 + 4 is [2, 2 2 ]
(sin x − 1)2 + 4 = 2 = 2 sin2 y
⇒
sin x = 1and sin2 y = 1
⇒
sin x = |sin y|
=
=
[from the options]
[JEE Main 2019, 10 April Shift-II]
(1 + i )2
a−i
(1 − 1 + 2 i ) (a + i )
(a2 + 1)2
(sin x − 1) 2 + 4
⇒
(a) zw = − i
The given complex number z =
⇒
2
π
that | zw | = 1 and arg( z ) − arg(w ) = , then
2
Exp. (b)
⇒
⇒
12 If z and w are two complex numbers such
(b) −
(a)
⇒
≤1
∴The above inequality holds, iff
2
,
5
[JEE Main 2019, 10 April Shift-I]
2
sin 2 y
and range of 2 sin2 y is [0, 2 ].
(sinθ + 8)12
=
1
⋅
[if a > 1and am ≤ an ⇒ m ≤ n]
(1 + i )2
If a > 0 and z =
, has magnitude
a −i
then z is equal to
Q
sin 2 x − 2 sin x + 5
4
[Qα + β = − sinθ]
=
[JEE Main 2019, 10 April Shift-I]
Given, inequality is
12
12
12
sin 2 y
Exp. (b)


2(α + β)
=

2
 (α + β ) − 8(α + β) 
10
11 All the pairs ( x , y ) that satisfy the inequality
(c) zw = i
[Q i 2 = − 1]
[given]
2
2
2
=
⇒
2
5
5
1+ a
2
⇒ a2 + 1 = 10
5
[Qa > 0]
a2 = 9 ⇒a = 3
−2 + 6i
1 3
1 3
So, z =
= − + i ⇒z= − − i
10
5 5
5 5
[Qif z = x + iy, then z = x − iy]
1 −i
2
−1 + i
(d) zw =
2
(b) zw =
Exp. (a)
It is given that, there are two complex numbers z
and
such
that
and
w,
| z w| = 1
arg( z) − arg(w) = π / 2
∴
| z|| w| = 1
[Q| z1 z2 | = | z1 || z2 |]
π
and arg( z) = + arg(w)
2
1
…(i)
Let| z| = r, then| w| =
r
π
… (ii)
and let arg(w) = θ, then arg( z) = + θ
2
So, we can assume
…(iii)
z = re i ( π / 2 + θ )
[Qif z = x + iy is a complex number, then it can
be written as z = re iθ where, r =| z| and θ = arg ( z)]
22
JEE Main Chapterwise Mathematics
w=
and
Now,
1 iθ
e
r
…(iv)
z ⋅ w = re − i( π / 2 +
θ)
=e
− i( π / 2 )
Let the complex number z = x + iy
Also given,| z − i | = | z − 1|
⇒| x + iy − i | = | x + iy − 1|
1
⋅ e iθ
r
= e i( − π / 2 − θ +
θ)
⇒ x2 + ( y − 1)2 = ( x − 1)2 + y2
=−i
[Qe − i θ = cos θ − i sinθ]
1
zw = re i ( π / 2 + θ ) ⋅ e − iθ
r
= e i( π / 2 + θ − θ ) = e i ( π / 2 ) = i
and
13. The number of real roots of the equation
5 + | 2 x − 1 | = 2 x (2 x − 2 ) is
[JEE Main 2019, 10 April Shift-II]
(a) 1
(b) 3
(c) 4
(d) 2
Exp. (a)
Given equation 5 + | 2 x − 1| = 2 x (2 x − 2 )
Case I
If 2 x − 1 ≥ 0 ⇒ x ≥ 0,
then 5 + 2 x − 1 = 2 x (2 x − 2 )
Put 2 x = t , then
5 + t − 1 = t 2 − 2t ⇒ t 2 − 3t − 4 = 0
⇒ t − 4t + t − 4 = 0 ⇒ t (t − 4) + 1(t − 4) = 0
2
⇒
t = 4 or − 1 ⇒ t = 4
⇒
2x = 4⇒ x = 2 > 0
⇒ x = 2 is the solution.
Case II
If 2 x − 1 < 0 ⇒ x < 0,
then 5 + 1 − 2 x = 2 x (2 x − 2 )
Put 2 x = y, then 6 − y = y2 − 2 y
⇒
(Qt = 2 x > 0)
y2 − y − 6 = 0 ⇒ y2 − 3 y + 2 y − 6 = 0
⇒ ( y + 2 ) ( y − 3) = 0 ⇒ y = 3 or − 2
⇒
y= 3(as y = 2 x > 0) ⇒2 x = 3
⇒
x = log 2 3 > 0
So, x = log 2 3 is not a solution.
Therefore, number of real roots is one.
14. The equation | z − i | = | z − 1|, i = −1,
represents
(a)
(b)
(c)
(d)
Exp. (b)
[JEE Main 2019, 12 April Shift-I]
1
a circle of radius
2
the line passing through the origin with
slope 1
a circle of radius 1
the line passing through the origin with
slope −1
[Q| z| = (Re( z))2 + (Im( z))2 ]
On squaring both sides, we get
x 2 + y2 − 2 y + 1 = x 2 + y2 − 2 x + 1
⇒ y = x, which represents a line through the origin
with slope 1.
15. Let z ∈C with Im ( z ) = 10 and it satisfies
2z − n
= 2i − 1 for some natural number n,
2z + n
then
[JEE Main 2019, 12 April Shift-II]
(a)
(b)
(c)
(d)
n = 20 and Re( z ) = −10
n = 40 and Re( z ) =10
n = 40 and Re( z ) = −10
n = 20 and Re( z ) =10
Exp. (c)
Let z = x + 10i , as Im ( z) = 10 given.
Since, z satisfies,
2z − n
= 2 i − 1, n ∈ N,
2z + n
∴ (2 x − 20i − n) = (2 i − 1) (2 x + 20i + n)
⇒ (2 x − n) + 20i = (− 2 x − n − 40)
+ (4 x + 2 n − 20)i
On comparing real and imaginary parts, we get
2 x − n = − 2 x − n − 40 and 20 = 4 x + 2 n − 20
⇒
4 x = − 40 and 4 x + 2 n = 40
⇒
x = − 10 and − 40 + 2 n = 40 ⇒ n = 40
So,
n = 40 and x = Re ( z) = − 10
16. Let α and β be two roots of the equation
x 2 + 2 x + 2 = 0, then α15 + β15 is equal to
[JEE Main 2019, 9 Jan Shift-I]
(a) 256
(c) −256
(b) 512
(d) −512
Exp. (c)
We have, x2 + 2 x + 2 = 0
⇒
x=
−2 ±
4−8
2
[Qroots of ax2 + bx + c = 0 are
given by x =
−b±
b 2 − 4ac
2a
]
23
Complex Numbers and Quadratic Equations
⇒
x = − 1± i
Let α = − 1 + i and β = − 1 − i.

Then, α15 + β15 = (−1 + i )15 + (− 1 − i )15
= − [(1 − i )15 + (1 + i )15 ]
15
15

i 
1
  1 + i  
= −   2 
−
 +  2
 
2  
2 
  2
   2




=−

+

3 + 2i sin θ

− 2i sin θ
1


is purely imaginary . Then, the sum of the

elements in A is
[JEE Main 2019, 9 Jan Shift-I]
π
 2
17. Let A = θ ∈  − , π :



15 
π
π 
 
 2  cos + i sin   
4
4 
 
15
π
π
2  cos − i sin  

4
4
(a)
3π
4
5π
6
2π
(d)
3
(b)
(c) π
Exp. (d)
 3 + 2 i sinθ   1 + 2 i sinθ 
Let z = 
 ×

 1 − 2 i sinθ   1 + 2 i sinθ 
15 π
15 π 

= −( 2 )15  cos
− i sin

 
4
4 
15 π
15 π  
+  cos
+ i sin


4
4  
[using De Moivre’s theorem
(cos θ ± i sinθ) = cos nθ ± i sin nθ, n ∈ Z]
15 π 
= − ( 2 ) 2 cos

4 
1 
= − ( 2 )15 2 ×

2 
(rationalising the denominator)
3 − 4sin2 θ + 8i sinθ
=
1 + 4sin2 θ
[Q a2 − b 2 = (a + b )(a − b ) and i 2 = − 1]
n
 3 − 4sin2 θ   8sinθ 
 + 
i
= 
2 
2 
 1 + 4sin θ   1 + 4sin θ 
15
15 π
π
π
1 

Qcos
= cos  4 π −  = cos =


4
4
4
2 
Since, real part of z = 0
3 − 4sin2 θ
=0
∴
1 + 4sin2 θ
⇒
⇒
= − ( 2 )16 = − 2 8 = − 256.
3 − 4sin2 θ = 0
sin2 θ =
Alternate Method
α15 + β15 = (−1 + i )15 + (−1 − i )15
Y
1
√3/2
= − [(1 − i )15 + (1 + i )15 ]
 (1 − i )16 (1 + i )16 
+
=−

1+ i 
 1− i
 [(1 − i )2 ]8 [(1 + i )2 ]8 
+
=−

1+ i 
 1− i
 [1 + i 2 − 2 i ]8 [1 + i 2 + 2 i ]8 
+
=−

1− i
1+ i


 2 
= − 256 
2
 1 − (i ) 
2
= − 256   = − 256
 2 

i 
y=sin θ
–π/2 –π/3
X′
O π/3
−1
2π/3
π
X
–√3/2
Y′
π π 2π
θ ∈ − , ,

 3 3 3 
2π
.
Sum of values of θ =
3
⇒
 (−2 i )8 (2 i )8 
+
=−

1+ i 
 1− i
 1
1
= − 28 
+
−
+
1
i
1

3
3
⇒ sinθ = ±
4
2
[Qi 4 n = 1, n ∈ Z]
18. Let z0 be a root of the quadratic equation,
x 2 + x + 1 = 0, If z = 3 + 6iz081 − 3iz093, then arg z
is equal to
[JEE Main 2019, 9 Jan Shift-II]
(a)
π
4
(b)
π
6
(c) 0
(d)
π
3
24
JEE Main Chapterwise Mathematics
Exp. (a)
20. If both the roots of the quadratic equation
Given, x2 + x + 1 = 0
− 1 ± 3i
2
[QRoots of quadratic equation ax2 + bx + c = 0
⇒
are given by x =
x=
−b±
b − 4ac
2
2a
]
⇒ z0 = ω, ω2
− 1+ 3 i
− 1− 3 i
[where ω =
and ω2 =
are
2
2
2
the cube roots of unity and ω, ω ≠ 1)
x 2 − mx + 4 = 0 are real and distinct and they
lie in the interval [1, 5] then m lies in the
interval
[JEE Main 2019, 9 Jan Shift-II]
(b) ( −5, − 4) (c) (5, 6)
(a) (4, 5)
(d) (3, 4)
Exp. (a)
According to given information, we have the
following graph
Y
Now consider z = 3 + 6i z081 − 3i z093
= 3 + 6i − 3i
(Qω3 n = (ω2 )3 n = 1)
= 3 + 3i = 3(1 + i )
If ‘θ’ is the argument of z, then
Im( z)
[Qz is in the first quadrant]
tan θ =
Re( z)
3
π
= = 1⇒ θ =
3
4
19. The number of all possible positive integral
values of α for which the roots of the
quadratic equation, 6x 2 − 11x + α = 0 are
rational numbers is
[JEE Main 2019, 9 Jan Shift-II]
(a) 5
(c) 4
(b) 2
(d) 3
Exp. (d)
For the roots of quadratic equation
ax2 + bx + c = 0 to be rational D = (b 2 − 4ac )
should be perfect square.
In the equation 6 x2 − 11x + α = 0
a = 6, b = − 11and c = α
∴For roots to be rational
D = (− 11)2 − 4(6)(α) should be a perfect square.
⇒ D(α) = 121 − 24α should be a perfect square
Now,
D(1) = 121 − 24 = 97 is not a perfect square.
D(2 ) = 121 − 24 × 2 = 73 is not a perfect square.
D(3) = 121 − 24 × 3 = 49 is a perfect square.
D(4) = 121 − 24 × 4 = 25 is a perfect square.
D(5) = 121 − 24 × 5 = 1 is a perfect square.
and for α ≥ 6, D(α) < 0, hence imaginary roots.
∴ For 3 values of α (α = 3, 4, 5), the roots are
rational.
O
X
5
1
Now, the following conditions should satisfy
(i) D> 0 ⇒ b 2 − 4ac > 0
⇒
m2 − 4 × 1 × 4 > 0
⇒
m2 − 16 > 0
⇒
(m − 4) (m + 4) > 0
⇒
m ∈ (− ∞, − 4) ∪ (4, ∞ )
(ii) The vertex of the parabola should lie
between x = 1and x = 5
b
m
∴
−
∈ (1, 5) ⇒1 < < 5
2a
2
⇒
m ∈(2, 10)
(iii) f(1) > 0 ⇒1 − m + 4 > 0
⇒ m < 5 ⇒ m ∈ (−∞, 5)
(iv) f(5) > 0 ⇒ 25 − 5m + 4 > 0 ⇒ 5m < 29
29
⇒ m ∈  − ∞, 

5
From the values of m obtained in (i), (ii), (iii) and
(iv), we get m ∈(4, 5).
–∞
–4
2
4
5 29/5
∞
21. Let z1 and z2 be any two non-zero complex
numbers such that 3| z1| = 4| z2 |. If
3z 2 z
z = 1 + 2 , then
2 z2 3z1
[JEE Main 2019, 10 Jan Shift-I]
(a) | z | =
1 17
2 2
(c) Re(z) = 0
(b) Im( z ) = 0
(d) | z | =
5
2
25
Complex Numbers and Quadratic Equations
Exp. (*)
Given, 3| z1| = 4| z2| ⇒
| z1| 4
=
| z2| 3
[Q z2 ≠ 0 ⇒| z2| ≠ 0]
z1
z1 iθ
z2
z
=
= 2 e − iθ
e and
z2
z2
z1
z1
∴
iθ
⇒
⇒
[Q z =| z|(cos θ + i sinθ) = | z| e ]
z1 4 iθ
z
3
= e and 2 = e − iθ
z2 3
z1 4
3 z1
2 z2 1 − iθ
= e
= 2e iθ and
2 z2
3 z1 2
On adding these two, we get
3 z1 2 z2
1
z=
= 2e iθ + e − iθ
+
2 z2
3 z1
2
1
1
= 2 cos θ + 2 i sinθ + cos θ − i sinθ
2
2
[Qe ± iθ = (cos θ ± i sinθ)]
5
3
= cos θ + i sinθ
2
2
| z| =
⇒
2
 5 +  3
 
 
2
2
2
=
34
17
=
4
2
Note that z is neither purely imaginary and nor
purely real.
‘*’ None of the options is correct.
22. Consider the quadratic equation
(c − 5)x 2 − 2cx + (c − 4) = 0, c ≠ 5. Let S be the
set of all integral values of c for which one
root of the equation lies in the interval
(0, 2) and its other root lies in the interval
(2, 3). Then, the number of elements in S is
[JEE Main 2019, 10 Jan Shift-I]
(a) 11
(b) 10
(c) 12
(d) 18
Exp. (a)
Let f( x) = (c − 5)x2 − 2 cx + (c − 4) = 0. Then,
according to problem, the graph of y = f( x) will be
either of the two ways, shown below.
O
2
3
In both cases f(0). f(2 ) < 0 and f(2 )f(3) < 0
Now, consider
f(0)f(2 ) < 0
⇒ (c − 4) [4(c − 5) − 4c + (c − 4)] < 0
⇒
(c − 4) (c − 24) < 0
⇒
c ∈(4, 24)
Similarly, f(2 ) ⋅ f(3) < 0
⇒
[4(c − 5) − 4c + (c − 4)]
[9(c − 5) − 6c + (c − 4)] < 0
⇒
(c − 24) (4c − 49) < 0
–∞
–4
2
3
2
4
5 29/5
∞
49
c ∈  , 24
 4

⇒
…(ii)
From Eqs. (i) and (ii), we get
49
c ∈  , 24
 4

∴Integral values of c are 13, 14, ……, 23.
Thus, 11 integral values of c are possible.
5
5
 3 i
 3 i
−  . If R ( z ) and
+  +
2
2
 2
 2
I ( z ) respectively denote the real and
imaginary parts of z, then
23. Let z = 
[JEE Main 2019, 10 Jan Shift-II]
(a) R( z )> 0 and I ( z )> 0
(b) I ( z ) = 0
(c) R( z )< 0 and I ( z )> 0
(d) R( z ) = − 3
Exp. (b)
5
 3 i
 3 i
Given, z = 
− 
+  +
2
2
 2
 2
5
Q Euler’s form of
π
π
3 i 
+ =  cos + i sin  = e i( π / 6 )
2
2 
6
6
and
−π
3 i
 π
− iπ / 6
− = cos
 + i sin −  = e
 6
 6 
2
2
So,
z = (e iπ / 6 )5 + (e − iπ / 6 )5
i
O
… (i)
5π
−i
5π
=e 6 +e 6
5π
5π
5π
5π
− i sin 
=  cos
+ i sin  +  cos

6
6  
6
6 
[Q e iθ = cos θ + i sinθ]
26
JEE Main Chapterwise Mathematics
= 2 cos
∴
I( z) = 0
5π
6
π
= − 3< 0
6
5π
π
π

Qcos
= cos π −  = − cos


6
6
6 
and R( z) = −2 cos
On equating real and imaginary part, we get
x = − 198 and y = − 107
⇒
y − x = − 107 + 198 = 91
26. If one real root of the quadratic equation
81x 2 + kx + 256 = 0 is cube of the other root,
then a value of k is
[JEE Main 2019, 11 Jan Shift-I]
24. The value of λ such that sum of the squares
of the roots of the quadratic equation,
x 2 + ( 3 − λ )x + 2 = λ has the least value is
[JEE Main 2019, 10 Jan Shift-II]
4
(a)
9
15
(c)
8
(b)1
(d) 2
Exp. (d)
Given quadratic equation is
x2 + ( 3 − λ ) x + 2 = λ
x2 + (3 − λ )x + (2 − λ ) = 0
… (i)
Let Eq. (i) has roots α and β, then α + β = λ − 3
and αβ = 2 − λ
b
[QFor ax2 + bx + c = 0, sum of roots = −
a
c
and product of roots = ]
a
Now, α 2 + β 2 = (α + β )2 − 2αβ
(a) 100
Given quadratic equation is
81x2 + kx + 256 = 0
Let one root be α, then other is α 3 .
256
k
Now, α + α 3 = −
and α ⋅ α 3 =
81
81
b
[Qfor ax2 + bx + c = 0, sum of roots = − and
a c
product of roots = ]
a
4
4

4
⇒
α = 
 3
4
α=±
⇒
3
∴
k = − 81 (α + α 3 )
= − 81 α (1 + α 2 )
16 
4
= − 81  ±   1 +
 = ± 300
 3 
9
= λ2 − 6λ + 9 − 4 + 2 λ
27. Let z be a complex number such that
= (λ − 4λ + 4) +1
| z | + z = 3 + i (where i = − 1).
= (λ − 2 ) + 1
Then, | z | is equal to
2
2
[JEE Main 2019, 11 Jan Shift-II]
Clearly, a + β will be least when λ = 2.
2
2
3
1
x + iy
(i = −1 ), where x and
3
27
y are real numbers, then y − x equals
25. Let  −2 − i =


[JEE Main 2019, 11 Jan Shift-I]
(a) 91
(c) – 85
(b) 85
(d) – 91
Exp. (a)
(a)
3
x + iy 
1
–1
=  − 2 − i  =  (6 + i )

3   3
27

x + iy
1
2
3
=−
(216 + 108i + 18i + i )
⇒
27
27
1
(198 + 107 i )
=−
27
[Q(a + b )3 = a3 + b 3 + 3a2 b + 3ab 2 and i 2 = − 1]
34
3
5
(b)
3
(c)
41
4
(d)
5
4
Exp. (b)
We have,| z| + z = 3 + i
By inspection it is clearly that imaginary part is 1.
∴
z = x + i for some x ∈ R
⇒
|x + i| + x + i = 3 + i
⇒
3
We have,
(d) −300
Exp. (d)
= (λ − 3)2 − 2(2 − λ )
= λ2 − 4λ + 5
(c) −81
(b) 144
⇒
x2 + 1 = 3 − x
x2 + 1 = 9 − 6 x + x2
4
⇒
6x = 8 ⇒ x =
3
4
z= + i
⇒
3
16
25
5
+ 1=
⇒ | z| =
⇒ | z| =
9
9
3
27
Complex Numbers and Quadratic Equations
28. Let α and β be the roots of the quadratic
equation x 2 sin θ − x (sin θ cos θ + 1) + cos θ = 0
∞ 
( − 1)n 
(0 < θ < 45º ) and α < β. Then, ∑ αn + n 
β 
n = 0
is equal to
Given,
x2 sinθ − xsinθcos θ − x + cos θ = 0,
where 0 < θ < 45°
⇒
xsinθ( x − cos θ) − 1( x − cos θ) = 0
⇒
( x − cos θ) ( xsinθ − 1) = 0
⇒
x = cos θ, x = cosec θ
⇒
α = cos θ and β = cosec θ
1
(QFor 0 < θ < 45° ,
< cos θ < 1 and
2
2 < cosec θ < ∞ ⇒ cosθ < cosec θ)
Now, consider,
∞ 
∞
∞
(−1)n 
(−1)n
n
n
∑  α + β n  = ∑ α + ∑ β n
n=0
n=0
n=0
= (1 + α + α + α + .... ∞ )
2
3


1
1
1
+  1 − + 2 − 3 + .... ∞ 
β


β
β
1
1
1
1
=
+
=
+
 1 1 − α 1 + 1
1− α
1− − 
β
 β
1
1

 1
=
+
Q = sinθ
β
1 − cos θ 1 + sinθ


29. If λ be the ratio of the roots of the quadratic
equation in x, 3m 2x 2 + m (m − 4)x + 2 = 0,
1
then the least value of m for which λ + = 1,
λ
is
[JEE Main 2019, 12 Jan Shift-I]
(a) − 2 + 2
(c) 4 − 3 2
⇒
⇒
1
1
(b)
+
1 − cosθ 1 + sin θ
1
1
(d)
+
1 + cosθ 1 − sin θ
Exp. (b)
(b) 4 − 2 3
(d) 2 − 3
Exp. (c)
Let the given quadratic equation in x,
3m2 x2 + m(m − 4)x + 2 = 0, m ≠ 0 have roots α
and β, then
m(m − 4)
2
and αβ =
α+β = −
2
3m
3m2
λ+
Then,
[JEE Main 2019, 11 Jan Shift-II]
1
1
(a)
−
1 − cosθ 1 + sin θ
1
1
(c)
−
1 + cosθ 1 − sin θ
α
=λ
β
Also, let
1
α β
= 1⇒ + = 1
λ
β α
(given)
α 2 + β 2 = αβ ⇒ (α + β )2 = 3αβ
m (m − 4)2
2
⇒
2
=3 2
3m
9m4
(m − 4)2 = 18
[Qm ≠ 0]
m − 4 = ±3 2
⇒
⇒
m = 4± 3 2
The least value of m = 4 − 3 2
z −α
(α ∈R ) is a purely imaginary number
z +α
and | z| = 2, then a value of α is
30. If
[JEE Main 2019, 12 Jan Shift-I]
1
(b)
2
(a) 2
(c) 1
(d) 2
Exp. (d)
z−α
(α ∈ R ) is purely
z+α
imaginary number, therefore
z−α
z−α
[Qα ∈ R]
=0
+
z+α
z+α
Since, the complex number
⇒
⇒
zz − αz + αz − α 2 + zz − αz + αz − α 2 = 0
2 z
⇒
⇒
2
− 2 α2 = 0
α2 = z
2
=4
2
[Qzz = z ]
[| z| = 2 given]
α = ± 2.
31. The number of integral values ofm for which
the
quadratic
expression,
(1 + 2m ) x 2 − 2(1 + 3m )x + 4(1 + m ), x ∈R , is
always positive, is [JEE Main 2019, 12 Jan Shift-II]
(a) 6
(b) 8
(c) 7
(d) 3
Exp. (c)
The quadratic expression
ax2 + bx + c, x ∈ R is always positive,
if a > 0 and D < 0.
So, the quadratic expression
(1 + 2 m) x2 − 2 (1 + 3m)x + 4(1 + m), x ∈ R will be
always positive, if1 + 2 m > 0
…(i)
and D = 4(1 + 3m)2 − 4(2 m + 1) 4(1 + m) < 0 …(ii)
From inequality Eq. (i), we get
1
m> −
2
…(iii)
28
JEE Main Chapterwise Mathematics
From inequality Eq. (ii), we get
1 + 9m2 + 6m − 4 (2 m2 + 3m + 1) < 0
⇒
m2 − 6m − 3 < 0
⇒ [m − ( 3 +
⇒
34. Let ω be a complex number such that
m=
6±
12 )][m − (3 − 12 )] < 0
[Q m2 − 6m − 3 = 0
36 + 12
=3±
12]
2
…(iv)
3 − 12 < m < 3 + 12
⇒
From inequalities Eqs. (iii) and (iv), the integral
values of m are 0, 1, 2, 3, 4, 5, 6
Hence, the number of integral values of m is 7.
32. Let z1 and z2 be two complex numbers
satisfying | z1 | = 9 and | z2 − 3 − 4i | = 4. Then,
the minimum value of | z1 − z2 | is
[JEE Main 2019, 12 Jan Shift-II]
(a) 1
(b) 2
(c)
(d) 0
2
Exp. (d)
Clearly | z1|= 9, represents a circle having centre
C1(0, 0) and radius r1 = 9.
and | z2 − 3 − 4i|= 4 represents a circle having
centre C 2 (3, 4) and radius r2 = 4.
The minimum value of | z1 − z2| is equals to
minimum distance between circles | z1|= 9 and
| z2 − 3 − 4i|= 4.
Q C1C 2 = (3 − 0) + (4 − 0) =
2
2ω + 1 = z, where z = − 3. If
1
1
1
2
1 −ω − 1 ω 2 = 3k , then k is equal to
1
ω7
[JEE Main 2017 (Offline)]
(a) − z
(c) −1
(b) z
(d)1
Exp. (a)
2ω + 1 = z
[Q z =
2ω + 1 = − 3
− 1 + 3i
⇒
ω=
2
Since, ω is cube root of unity.
− 1 − 3i
and ω3 n = 1
ω2 =
∴
2
1
1
1
Now,
1 − ω2 − 1 ω2 = 3k
Given,
⇒
ω2
1
ω7
[Q1 + ω + ω2 = 0 and ω7 = (ω3 )2 ⋅ ω = ω]
On applying R1 → R1 + R 2 + R 3 , we get
3 1 + ω + ω2 1 + ω + ω2
ω
ω2
= 3k
1
2
1
ω
ω
and |r1 − r2|=|9 − 4|= 5 ⇒ C1C 2 =|r1 − r2|
∴ Circles touches each other internally.
Hence, | z1 − z2|min = 0
⇒
equation x 2 − x + 1 = 0, then α101 + β107 is
equal to
[JEE Main 2018]
3 0
0
1 ω ω2 = 3k
1 ω2 ω
⇒
3(ω2 − ω4 ) = 3k
(a) −1
⇒
(ω2 − ω) = k
33. If α , β ∈C are the distinct roots of the
(b) 0
(c) 1
(d) 2
Exp. (c)
∴
We have, α, β are the roots of x − x + 1 = 0
2
∴ Let α = − ω and β = − ω
⇒
= − (ω
+ω
= − (ω + ω)
2
= − (−1)
=1
214
3i = − z
35. For a positive integer n, if the quadratic
equation, x ( x + 1) + ( x + 1)( x + 2 ) + ...
α101 + β107 = (− ω)101 + (− ω2 )107
101
 − 1 − 3i   − 1 + 3i 
k=
 −

2
2

 

=−
Q Roots of x2 − x + 1 = 0 are −ω,−ω2
2
− 3]
1 1
1
1 ω ω2 = 3k
1 ω2 ω
⇒
9 + 16 = 25 = 5
2
ω2
+ ( x + n − 1)( x + n ) = 10n
)
(Qω = 1)
3
[Q1 + ω + ω = 0]
2
has two consecutive integral solutions, then
[JEE Main 2016 (Offline)]
n is equal to
(a) 12
(c) 10
(b) 9
(d) 11
29
Complex Numbers and Quadratic Equations
Exp. (d)
=
Given, quadratic equation is
x( x + 1) + ( x + 1)( x + 2 ) + ... + ( x + n − 1)
( x + n) = 10n
⇒
( x2 + x2 + ... + x2 ) + [(1 + 3 + 5 + ... + (2 n − 1)]x
+ [(1⋅ 2 + 2 ⋅ 3 + ... + (n − 1)n] = 10n
n(n2 − 1)
⇒ nx + n x +
− 10n = 0
3
2
n −1
− 10 = 0
x2 + nx +
⇒
3
2
2
⇒
3 x + 3nx + n − 31 = 0
2
Q
∴
⇒
2
Let α and β be the roots.
|α − β| = 1
(α − β )2 = 1
Again,
(α − β )2 = (α + β )2 − 4αβ
⇒
⇒
n = ± 11
n = 11
π
(a)
3
 3
(c) sin −1 

 4 
2 + 3i sin θ
is purely
1 − 2i sin θ
[JEE Main 2016 (Offline)]
π
6
1 
(d) sin −1 

 3
(b)
Exp. (d)
2 + 3i sin θ
is purely imaginary. Then, we
1 − 2 i sin θ
have Re( z) = 0
2 + 3i sin θ
Now, consider z =
1 − 2 i sin θ
(2 + 3i sin θ) (1 + 2 i sin θ)
=
(1 − 2 i sin θ) (1 + 2 i sin θ)
Let z =
=
=
37. The sum of all real values of x satisfying the
(b) − 4
2 + 4i sin θ + 3i sin θ + 6i 2 sin2 θ
12 − (2 i sin θ) 2
2 + 7 i sin θ − 6 sin2 θ
1 + 4 sin2 θ
2
+ 4 x − 60
= 1 is
(c) 6
Given, ( x2 − 5 x + 5) x
(d) 5
2
+ 4 x − 60
=1
Clearly, this is possible when
I. x2 + 4 x − 60 = 0and x2 − 5 x + 5 ≠ 0
or
II. x2 − 5 x + 5 = 1
or
III. x2 − 5 x + 5 = − 1 and x2 + 4 x − 60
= Even integer
[Qn > 0]
36. A value of θ for which
imaginary, is
1 
−1  1 
⇒ θ = sin− 1  ±
 = ± sin 


 3
3
Exp. (a)
n2 = 121
⇒
∴
Re( z) = 0
2 − 6 sin2 θ
= 0 ⇒ 2 = 6 sin2 θ
1 + 4 sin2 θ
1
1
⇒ sin θ = ±
sin2 θ =
3
3
(a) 3
 n2 − 31
− 3n 

1 = 
 − 4
 3 
3 

4
1 = n2 − (n2 − 31)
3
3 = 3n2 − 4n2 + 124
⇒
7 sin θ
1 + 4 sin2 θ
[JEE Main 2016 (Offline)]
2
⇒
+ i
1 + 4 sin2 θ
equation ( x 2 − 5x + 5)x
Since, α and β are consecutive.
∴
⇒
2 − 6 sin2 θ
Case I When x2 + 4 x − 60 = 0
⇒
x2 + 10 x − 6 x − 60 = 0
⇒
x( x + 10) − 6( x + 10) = 0
⇒
( x + 10) ( x − 6) = 0
⇒
x = − 10 or x = 6
Note that, for these two values of x,
x2 − 5 x + 5 ≠ 0
Case II When
x2 − 5 x + 5 = 1
⇒
x2 − 5 x + 4 = 0
⇒
x2 − 4 x − x + 4 = 0
⇒
x( x − 4) − 1 ( x − 4) = 0
⇒
( x − 4) ( x − 1) = 0
⇒
x = 4 or x = 1
Case III When x2 − 5 x + 5 = − 1
⇒
x2 − 5 x + 6 = 0
⇒
x2 − 2 x − 3 x + 6 = 0
⇒
⇒
⇒
x( x − 2 ) − 3( x − 2 ) = 0
( x − 2 ) ( x − 3) = 0
x = 2 or x = 3
30
JEE Main Chapterwise Mathematics
Now, when x = 2, x2 + 4 x − 60
Exp. (c)
= 4 + 8 − 60 = − 48, which is an even
integer.
When x = 3, x2 + 4 x − 60 = 9 + 12 − 60
= − 39,
which is not an even integer.
Thus, in this case, we get x = 2.
Hence, the sum of all real values of
x = − 10 + 6 + 4 + 1 + 2 = 3
38. A complex number z is said to be
unimodular, if | z | = 1. Suppose z1 and z2
z − 2 z2
are complex numbers such that 1
2 − z1z2
is unimodular and z2 is not unimodular.
Then, the point z1 lies on a [JEE Main 2015]
(a) straight line parallel to X-axis
(b) straight line parallel to Y -axis
(c) circle of radius 2
(d) circle of radius 2
Given, z2 is not unimodular i.e.| z2| ≠ 1
z − 2 z2
is unimodular
and 1
2 − z1 z2
z1 − 2 z2
=1
2 − z1 z2
=
=
=
2(α − β )
9
x2 − 6 x − 2 = 0
2(α 9 − β 9 )
6α 9 − 6β 9
2(α 9 − β 9 )
=
or
x2 = 6 x + 2
⇒
α 2 = 6α + 2
⇒ α 2 − 2 = 6α
6
=3
2
β 2 = 6β + 2
and
∴
⇒
α 2 = 6α + 2
α10 = 6α 9 + 2α 8
…(i)
Similarly,
β
…(ii)
⇒
10
= 6β + 2β
9
8
a10
a10 = 6a9 + 2 a8 (Q an = α n − β n )
a − 2 a8
=3
− 2 a8 = 6a9 ⇒ 10
2 a9
40. If z is a complex number such that | z | ≥ 2 , then
= 4 + | z1|2| z2|2 − 2 z1 z2 − 2 z1 z2
⇒ (| z2|2 − 1)(| z1|2 − 4) = 0
| z2| ≠ 1
| z1| = 2
z1 = x + iy ⇒ x2 + y2 = (2 )2
∴Point z1 lies on a circle of radius 2.
the minimum value of z +
1
2
5
[JEE Main 2014]
2
(b) lies in the interval (1, 2 )
5
(c) is strictly greater than
2
3
5
(d) is strictly greater than but less than
2
2
(a) is equal to
Exp. (b)
39. Let α and β be the roots of equation
x 2 − 6x − 2 = 0. If an = αn − βn , for n ≥ 1,
a − 2a 8
then the value of 10
is equal to
2a 9
[JEE Main 2015]
(c) 3
9
α 8 ⋅ 6α − β 8 ⋅ 6β
2
⇒ | z1|2 + 4| z2|2 − 2 z1 z2 − 2 z1 z2
(b) −6
Qα and β are the
roots of
α 8 (α 2 − 2 ) − β 8 (β 2 − 2 )
| z1 − 2 z2| = |2 − z1 z2|
2
(Q zz = | z| )
(a) 6
a10 − 2 a8 α10 − β10 − 2(α 8 − β 8 )
=
2 a9
2(α 9 − β 9 )
⇒
2
Let
Now, consider
On subtracting Eq. (ii) from Eq. (i), we get
α10 − β10 = 6(α 9 − β 9 ) + 2(α 8 − β 8 )
⇒( z1 − 2 z2 )( z1 − 2 z2 ) = (2 − z1 z2 )(2 − z1 z2 )
Q
∴
a8 = α 8 − β 8 ⇒ a9 = α 9 − β 9
⇒ β − 2 = 6β
Alter
Since, α and β are the roots of the equation
x2 − 6 x − 2 = 0 or x2 = 6 x + 2
Central Idea If z is unimodular, then| z| = 1.
Also, use property of modulus i.e. z z =| z|2 .
⇒
an = α n − β n for n ≥ 1
a10 = α10 − β10
Q
∴
2
Exp. (c)
⇒
Given, α and β are the roots of the equation
x2 − 6 x − 2 = 0.
(d) −3
| z|≥ 2 is the region on or outside circle whose centre is
(0, 0) and radius is 2.
1
Minimum z + is distance of z, which lie on the
2
−1
circle| z|= 2 from  , 0 .
2 
31
Complex Numbers and Quadratic Equations
∴Minimum z +
1
1
= Distance of  − , 0 from
 2 
2
(−2, 0)
=
1 + 3 a2 < 4 ⇒ a2 − 1 < 0
⇒
(a + 1)(a − 1) < 0
+
2
 −2 + 1  + 0 = 3



2
2
2
 −1 + 2  + 0 = 3


2

2
1
= AD =
2
−
–1
Geometrically
Min z +
⇒
+
1
a ∈ (−11
, )
For no integral solution of a, we consider the
interval (−1, 0) ∪ (0,1.)
Note Here, when we figure out the non-integral
solution, we get a ≠ 0. This implies any interval
excluding zero should be correct answer as it
give either no solution or no integral solution.
Y
42. Let α and β be the roots of equation
A
D
X′
(–2, 0)
(– 21 , 0)
(0, 0)
(2, 0)
X
px 2 + qx + r = 0,p ≠ 0. If p ,q and r are in AP
1 1
and + = 4, then the value of |α − β | is
α β
[JEE Main 2014]
(a)
Y′
41. If a ∈R and the equation
− 3( x − [ x ])2 + 2( x − [ x ]) + a 2 = 0 (where,[ x ]
denotes the greatest integer ≤ x) has no
integral solution, then all possible values of
a lie in the interval
[JEE Main 2014]
(a) ( −1,0) ∪ (0,1)
(c) ( −2 , − 1 )
(b) (1, 2 )
(d) ( − ∞ , − 2 ) ∪( 2 , ∞ )
/ x − [x ] = [X ] which is a fractional part function lie
between 0 ≤ {X } < 1.
Exp. (a)
Here, a ∈ R and equation is
−3{ x − [ x]}2 + 2{ x − [ x]} + a2 = 0
Let
t = x − [ x]
∴−3t 2 + 2 t + a2 = 0
⇒
Q
∴
1± 1 + 3a2
3
t = x − [ x] = { X }
0≤ t ≤ 1
t =
1 ± 1 + 3a2
≤1
3
Taking positive sign
[{ x} > 0]
Q
⇒
∴
⇒
0≤
0≤
1 ± 1 + 3a2
<1
3
1 + 3a2 < 2
[fractional part]
(c)
61
9
34
9
2 17
(b)
9
2 13
(d)
9
/ If ax 2 + bx + c
α+β =
= 0 has roots α and β, then
−b
c
and αβ = .
a
a
Exp. (d)
Since, α and β are roots of px2 + qx + r = 0, p ≠ 0.
−q
r
,αβ =
α+β =
∴
p
p
Since, p, q and r are in AP.
∴
2q = p + r
1 1
α+β
Also,
+ =4 ⇒
=4
αβ
α β
− q 4r
=
⇒
α + β = 4αβ ⇒
p
p
Q
2q = p + r
⇒
2(−4r ) = p + r ⇒ p = −9 r
− q 4r
4r
4
Q
α+β =
=
=
=−
P
p −9r
9
r
r
1
and
=
αβ = =
p −9r −9
∴
⇒
⇒
(α − β)2 = (α + β )2 − 4αβ
16 4 16 + 36
=
+ =
81 9
81
52
2
(α − β ) =
81
2
|α − β|=
13
3
32
JEE Main Chapterwise Mathematics
43. The real number k for which the equation,
2 x 3 + 3x + k = 0 has two distinct real roots in
[0, 1]
[JEE Main 2013]
(a)
(b)
(c)
(d)
lies between 1 and 2
lies between 2 and 3
lies between −1 and 0
does not exist
(a)
(b)
(c)
(d)
infinite number of real roots
no real root
exactly one real root
exactly four real roots
Given equation is
esin
f ( x) = 2 x + 3 x + k
3
On differentiating w.r.t. x, we get
f ′( x) = 6 x + 3 > 0, ∀ x ∈ R
Thus, f( x) is strictly increasing function.
Hence, f( x) = 0 has only one real root, so two roots
are not possible.
the equations x + 2 x + 3 = 0 and
ax 2 + bx + c = 0; a ,b , c ∈R , have a common
root, then a :b : c is equal to
[JEE Main 2013]
2
44. If
(a) 1 : 2 : 3 (b) 3 : 2 : 1 (c) 1 : 3 : 2 (d) 3 : 1 : 2
Exp. (a)
Given equations are
x2 + 2 x + 3 = 0
…(i)
ax + bx + c = 0
…(ii)
2
Since, Eq. (i) has imaginary roots.
So, Eq. (ii) will also have both roots same as
Eq. (i).
a b c
Thus,
= =
1 2 3
Hence, a : b : c is 1 : 2 : 3.
45. If z is a complex number of unit modulus
1 + z
and argument θ, then arg 
 is equal to
1 + z 
(a) − θ
(c) θ
π
(b)
−θ
2
(d) π − θ [JEE Main 2013]
Exp. (c)
Given, | z| = 1, arg ( z) = θ
∴
z = e iθ
1
But
z=
z


 1+ z
arg 
∴
 = arg ( z) = θ
1 + 1

z
x
− e − sin
x
= 4 ⇒ esin
y = esin
Now, let
2
and
[AIEEE 2012]
Exp. (b)
Exp. (d)
Let
46. The equation e sin x − e − sin x − 4 = 0 has
x
−
1
esin
x
=4
x
Hence, we get
1
y − = 4 ⇒ y2 − 4 y − 1 = 0
y
16 + 4
⇒ y=2 ±
2
On substituting the value of y, we get
⇒
y=
4±
esin
x
=2 ±
5
5
Now, since sine is a bounded function, i.e.,
−1 ≤ sin x ≤ 1. Hence, we get
1
e −1 ≤ esin x ≤ e ⇒ esin x ∈  , e 
 e 
Also, it is obvious that 2 + 5 > e
1
1
and 2 − 5 < ⋅ ⇒ 2 ± 5 ∉  , e 
e
 e 
So, esin x = 2 + 5 is not possible for any x ∈ R
and esin x = 2 − 5 is also not possible for any
x ∈ R.
Hence, we can say that the given equation has no
solution.
z2
is real, then the point
z −1
represented by the complex number z lies
47. If z ≠ 1 and
(a) either on the real axis or on a circle passing
through the origin
[AIEEE 2012]
(b) on a circle with centre at the origin
(c) either on the real axis or on a circle not passing
through the origin
(d) on the imaginary axis
Exp. (a)
z2
Given A complex number
, ( z ≠ 1) is purely
z−1
real.
To find The locus of the complex number z.
Method 1 Since,
z2
, ( z ≠ 1) is purely real.
z−1
33
Complex Numbers and Quadratic Equations
z2
z2
=
z−1 z−1
Then,
⇒
z2 ( z − 1) = z 2 ( z − 1)
⇒
z2 z − z2 = z 2 z − z 2
⇒
zzz − z2 = zz ⋅ z − z 2
⇒
z| z|2 − z2 = z | z|2 − z 2
Rearranging the terms, we get
z| z|2 − z | z|2 = z2 − z 2
⇒
| z|2 ( z − z) = ( z − z)( z + z)
⇒
| z|2 ( z − z) − ( z − z)( z + z) = 0
⇒
( z − z) [| z|2 − ( z + z)] = 0
Either ( z − z) = 0 or [| z|2 − ( z + z)] = 0
Now, z = z ⇒ Locus of ‘ z ’ is real axis and
{| z|2 − ( z + z)} = 0
⇒
zz − ( z + z) = 0
Hence, locus of ‘ z ’ is a circle passing through
origin.
Method 2 Put z = x + iy, then
( x + iy)2
( x2 − y2 ) + i (2 xy)
z2
=
=
z − 1 ( x + iy) − 1
( x − 1) + iy
=
( x2 − y2 ) + i (2 xy) ( x − 1) − iy
×
( x − 1) + iy
( x − 1) − iy
z2
Since,
, ( z ≠ 1) is purely real, hence its
z−1
imaginary part should be equal to zero.
⇒
( x2 − y2 )(− y) + (2 xy)( x − 1) = 0
⇒
y( x2 − y2 + 2 x − 2 x2 ) = 0
⇒
y( x2 + y2 − 2 x) = 0
Either
y = 0 or x + y − 2 x = 0
2
2
Now,
y=0
⇒ Locus of ‘ z ’ is real axis and x2 + y2 − 2 x = 0.
⇒ Locus of ‘ z ’ is a circle passing through origin.
Hence, locus of ‘ z ’ is either real axis or a circle
passing through origin.
48. Letα , β be real and z be a complex number. If
z 2 + αz + β = 0 has two distinct roots on the
line Re z = 1, then it is necessary that
(a) β ∈( −1, 0)
(c) β ∈[1, ∞ )
(b) | β | =1
[AIEEE 2011]
(d) β ∈(0,1 )
Exp. (c)
Let z = x + iy, given Re( z) = 1
∴
x = 1 ⇒ z = 1 + iy
Since, the complex roots are conjugate of each
other.
∴ z = 1 + iy and 1 − iy are two roots of
z2 + αz + β = 0
⇒
∴
Product of roots = β
(1 + iy) (1 − iy) = β
β = 1 + y2 ≥ 1 ⇒ β ∈ [1, ∞ )
49. If ω ( ≠ 1) is a cube root of unity and
(1 + ω ) 7 = A + Bω. Then, ( A , B ) is equal to
(a) (1,1 )
(c) ( −1,1 )
(b) (1, 0)
(d) (0,1 )
[AIEEE 2011]
Exp. (a)
(1 + ω)7 = A + Bω, we know 1 + ω + ω2 = 0
∴
1 + ω = − ω2
⇒
(− ω2 )7 = A + Bω
⇒
− ω14 = A + Bω
⇒
− ω2 = A + Bω
[Qω14 = ω12 ⋅ ω2 = ω2 ]
⇒
1 + ω = A + Bω
On comparing, we get
A = 1, B = 1
50. Sachin and Rahul attempted to solve a
quadratic equation. Sachin made a mistake
in writing down the constant term and
ended up in roots (4, 3). Rahul made a
mistake in writing down coefficient of x to
get roots (3, 2). The correct roots of equation
are
[AIEEE 2011]
(a) − 4, − 3 (b) 6,1
(c) 4, 3
(d) − 6, − 1
Exp. (b)
Let the quadratic equation be
ax2 + bx + c = 0
Sachin made a mistake in writing down constant
terms.
So, sum of roots is correct.
i.e.,
α + β =7
Rahul made mistake in writing down coefficient of x.
So, product of roots is correct.
i.e.,
αβ = 6
Correct quadratic equation is
x2 − (α + β )x + αβ = 0
x2 − 7 x + 6 = 0 having roots 1 and 6.
34
JEE Main Chapterwise Mathematics
51. The number of complex numbers z such
that | z − 1| = | z + 1| = | z − i | is equal to
(a) 0
(c) 2
(b) 1
(d) ∞
[AIEEE 2010]
Exp. (b)
We have, | z − 1| = | z + 1| = | z − i|
Clearly, z is the circumcentre of the triangle
formed by the vertices (1, 0) and (0, 1) and (−1, 0 )
which is unique.
Y
(0, 1)
(–1, 0)
(1, 0)
O
=
=
X′
X
53. Let f : R → R be a continuous function
.
e + 2e− x
1
Statement I f (c ) = , for some c ∈R .
3
1
Statement II 0 < f ( x ) ≤
,∀ x ∈R .
2 2
[AIEEE 2010]
Exp. (b)
f ( x) =
|z+i|
Y′
52. If α and β are the roots of the equation
x 2 − x + 1 = 0 , then α 2009 + β 2009 is equal to
(b) – 1
(d) 2
[AIEEE 2010]
Exp. (c)
Since, α and β are roots of the equation
x2 − x + 1 = 0.
α + β = 1, αβ = 1
⇒
x=
1±
3i
2
⇒
3i
1− 3 i
or
2
2
or −ω2
x = −ω
Thus,
α = −ω
Then,
β = −ω
or
α = − ω, then β = − ω2
Hence, α
2009
⇒
x=
1+
+β
= − [(ω )
3 669
= − [ω2 + ω]
= − (−1) = 1
e +
2
ex
AM ≥ GM
2
e + x
1/ 2
e ≥  e x ⋅ 2  , as e x > 0
x

2
e 
2
x
e + x ≥2 2
⇒
e
1
1
0<
≤
⇒
2
x
2
2
e + x
e
1
, ∀ x∈R
0 < f ( x) ≤
∴
2 2
Statement II is true and Statement I is also true as
for some ‘c’.
1
[for c = 0]
⇒
f (c ) =
3
1
.
which lies betwen 0 and
2 2
So, statement II is correct explanation of statement I
x
Alternate Solution
2
2009
1
x
Using
Hence, the number of complex number z is one.
⇒
x
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation for
Statement I
(d) Statement I is true, Statement II is false
Q
(a) – 2
(c) 1
1
defined by f ( x ) =
f ( x) =
= ( − ω)
2009
⋅ ω + (ω )
2
[where, ω3 = 1]
+ (− ω )
3 1337
2 2009
⋅ ω]
1
ex
= 2x
−x
e + 2e
e +2
x
(e 2 x + 2 )e x − 2 e 2 x ⋅ e x
⇒
f ′( x) =
⇒
f ′( x) = 0 ⇒ e 2 x + 2 = 2 e 2 x
⇒
e2 x = 2 ⇒ e x = 2
(e 2 x + 2 )2
Maximum value of f ( x) =
2
1
=
4
2 2
35
Complex Numbers and Quadratic Equations
1
, ∀ x∈R
2 2
1
1
Since,
0< <
3 2 2
⇒ For some c ∈ R, f (c ) = 1 / 3
0 < f ( x) ≤
Exp. (d)
z=
Then,
 1 
1
1
z=
=−
 =
i +1
 i − 1 − i − 1
54. If the roots of the equationbx 2 + cx + a = 0 is
imaginary, then for all real values of x, the
expression 3b 2x 2 + 6bcx + 2 c 2 is [AIEEE 2009]
(a) greater than 4ab
(c) greater than − 4ab
Given bx2 + cx + a = 0 has imaginary roots.
⇒
c 2 − 4ab < 0 ⇒ c 2 < 4ab
− c 2 > − 4ab
f( x) = 3b 2 x2 + 6bcx + 2 c 2
…(i)
3b 2 > 0
So, the given expression has a minimum value.
−D
∴ Minimum value =
4a
4ac − b 2 4(3b 2 ) (2 c 2 ) − 36b 2c 2
=
=
4a
4(3b 2 )
2 2
12 b c
[from Eq. (i)]
=−
= − c 2 > − 4ab
12 b 2
4
Ifz − = 2, then the maximum value of| z | is
z

equal to
[AIEEE 2009]
3 +1
(a)
(b)
5 +1
(d) 2 +
(c) 2
2
⇒
 z − 4  + 4

| z| = 
z
z

 z − 4
+ 4
| z| ≤ 
z | z|

4
| z| ≤ 2 +
| z|
⇒
⇒
1
i −1
(b) −
1
i −1
(c)
1
i +1
∴
and
⇒
∴
α + 4β = 6 and 4αβ = a
α + 3β = c and 3αβ = 6
a 4
=
⇒ a=8
6 3
x2 − 6 x + 8 = 0
⇒
⇒
and
( x − 4)( x − 2 ) = 0
x = 2, 4
x2 − cx + 6 = 0
⇒
22 − 2c + 6 = 0 ⇒ c = 5
∴
x2 − 5 x + 6 = 0
[AIEEE 2007]
(b) 10
(c) 6
(d) 0
Exp. (c)
| z + 4| ≤ 3 represents the interior and boundary of
the circle with centre at (− 4, 0) and radius = 3.
5 )] ≤ 0
5 ≤ | z| ≤
1
.
i −1
[AIEEE 2008]
(d) −
Y
5+1
56. The conjugate of a complex number is
(a)
Let the roots of x2 − 6 x + a = 0 be α, 4β and that
of x2 − cx + 6 = 0 be α and 3β.
(a) 4
[| z| − ( 5 + 1)] [| z| − (1 −
Then, the complex number is
(d) 3
| z + 1| is
| z| − 2| z| − 4 ≤ 0
1−
(c) 4
[AIEEE 2008]
58. If | z + 4 | ≤ 3 , then the maximum value of
2
⇒
(b) 1
⇒
x = 2, 3
Hence, common root is 2.
Exp. (b)
⇒
(a) 2
Exp. (a)
Here,
55.
x 2 − 6x + a = 0
x2 −c x + 6 =0
have one root in common. The other roots of
the first and second equations are integers
in the ratio 4 : 3. Then, the common root is
Exp. (c)
⇒
Let
57. The quadratic equations
and
(b) less than 4ab
(d) less than − 4ab
1
i −1
Let
X′
(–7, 0)
X
(–1, 0)
(–4, 0)
1
i +1
Y′
36
JEE Main Chapterwise Mathematics
As −1 is an end point of a diameter of the circle,
hence maximum possible value of| z + 1| is 6.
Alternate Solution
equation x 2 + ax + 1 = 0 is less than 5 , then
the set of possible values of a is [AIEEE 2007]
(b) ( − 3, ∞ )
(d) ( − ∞ , − 3)
Exp. (a)
⇒
|α − β | =
a −4
a − 4 < 5 ⇒ a − 4< 5
a2 < 9 ⇒ |a| < 3
a ∈ (− 3, 3)
2
60. If the roots of the quadratic equation
x 2 + px + q = 0 are
tan 30°
and
(b) 0
(d) 2
11
i



k = 1
(b) −1 < m < 3
(d) −2 < m < 0
Since, both roots of equation
x2 − 2 mx + m2 − 1 = 0 are greater than –2 but
less than 4.
b
∴
D ≥ 0, −2 < −
< 4,
2a
f(4) > 0 and
D ≥ 0;
Now,
–2
f(−2 ) > 0
4m − 4m2 + 4 ≥ 0
2
4
or
-2
⇒
4 > 0, ∀ m ∈ R
b
<4
2a
2m
−2 < 
<4
 2 ⋅ 1
∴
tan 30° + tan 15° = − p
and
tan 30° tan 15° = q
Now, 2 + q − p = 2 + tan 30° tan15°
+ (tan 30° + tan 15° )
= 2 + tan 30° tan 15° + 1 − tan 30° tan 15°

tan 30° + tan 15° 
Q tan 45° = 1 − tan 30° tan 15° 


⇒
−2 < m < 4
f(4) > 0
16 − 8m + m2 − 1 > 0
⇒
⇒
m2 − 8m + 15 > 0
2 + q − p= 3
2 kπ
2 kπ 

The value of ∑  sin
+ i cos
 is

11
11 
k =1
10
(b) –1
(d) i
2 kπ
–b/2a
Since, tan 30° and tan 15° are the roots of
equation x2 + px + q = 0.
(a) 1
(c) −i
−
[AIEEE 2006]
Exp. (a)
61.


 10  − 2 kπ i 


= i  ∑  e 11  − 1 = − i



k = 0 

2 kπ


10
−
i
∴sum of roots of unity is zero ⇒ ∑ e 11 = 0


k =0


tan 15°
respectively, then the value of 2 + q − p is
(a) 3
(c) 1
10
∑  e
Exp. (b)
According to given condition,
⇒
⇒
k =1
2 kπ
2 kπ 
− i sin
 =i
11
11 
(a) m > 3
(c) 1 < m < 4
2
2

∑  cos
the equation x 2 − 2mx + m 2 − 1 = 0 are
greater than –2 but less than 4 lie in the
interval
[AIEEE 2006]
αβ = 1
and
|α − β | = (α + β )2 − 4αβ
10
2 kπ
2 kπ 
+ i cos

11
11 
62. All the values of m for which both roots of
Let α and β be the roots of equation
x2 + ax + 1 = 0, then
α + β = −a

=i
59. If the difference between the roots of the
Now,
10
∑  sin
k =1
| z + 1| = | z + 4 − 3| ≤ | z + 4| + |− 3| ≤ 6
Hence, maximum value of| z + 1| is 6.
(a) ( − 3, 3)
(c) (3, ∞)
Exp. (c)
[AIEEE 2006]
4
…(i)
−2 < −
⇒
⇒
⇒
⇒
and
⇒
⇒
(m − 3)(m − 5) > 0
−∞ < m < 3 and 5 < m < ∞
f(−2 ) > 0
4 + 4m + m2 − 1 > 0
m2 + 4m + 3 > 0
…(ii)
…(iii)
37
Complex Numbers and Quadratic Equations
⇒
⇒
(m + 3)(m + 1) > 0
−∞ < m < − 3 and −1 < m < ∞
…(iv)
From Eqs. (i), (ii), (iii) and (iv), we get
(c) −1, − 1, − 1
(d) −1, − 1 + 2 ω , − 1 − 2 ω2
Exp. (b)
( x − m)2 = 1
Given that, ( x − 1)3 + 8 = 0
x − m = ±1
⇒
x = m± 1
Since, it is given that the roots of the equation
are greater than −2 but less than 4.
∴
−2 < x < 4
⇒ −2 < m ± 1 < 4
⇒
−2 < m + 1 < 4 and −2 < m − 1 < 4
⇒
−3 < m < 3
and
−1 < m < 5
⇒
−1 < m < 3
63. If z 2 + z + 1 = 0, where z is complex number,
then the value of

z +

[AIEEE 2006]
2
2
1
 2 1
 3 1
 + z + 2 + z + 3


z
z 
z 
⇒
3
 x − 1 = 1


 −2 
 x − 1 = (1)1/ 3
⇒


 −2 
x − 1
2
∴ Cube roots of 
 are 1, ω and ω .
 −2 
⇒
Cube roots of ( x − 1) are –2, −2 ω and − 2 ω2 .
Cube roots of x are –1, 1 − 2 ω and 1 − 2 ω2 .
65. If z1 and z2 are two non-zero complex
(a) −
2
π
2
(c) − π
(b) 0
(d)
π
2
Exp. (b)
(b) 6
(d) 18
Let
z1 = x1 + iy1 and z2 = x2 + iy2
Given that,
Exp. (c)
| z1 + z2| = | z1| + | z2|
Given equation is z2 + z + 1 = 0
−1 ±
∴ ( x1 + x2 )2 + ( y1 + y2 )2
1− 4 × 1× 1
⇒
z=
⇒
−1 ± −3
z=
2
⇒
z = ω, ω2
2 ×1
=
 z + 1  +  z2 + 1  +  z3 + 1 









z
z2 
z3 
2
x12 + y12 +
x22 + y22
On squaring both sides, we get
x12 + x22 + 2 x1 x2 + y12 + y22 + 2 y1 y2
2
2
Now,
( x − 1)3 = (−2 )3
numbers such that | z1 + z2| = | z1| + | z2|, then
[AIEEE 2005]
arg ( z1 ) − arg ( z2 ) is equal to
2
1

+ ... +  z 6 + 6  is

z 
(a) 54
(c) 12
[AIEEE 2005]
(b) −1,1 − 2 ω ,1 − 2 ω2
Alternate Solution
The given equation is
x2 − 2 mx + m2 − 1 = 0
⇒
x2 − 2 mx + m2 = 1
⇒
the roots of the equation ( x − 1)3 + 8 = 0, are
(a) −1,1 + 2 ω ,1 + 2 ω2
m lie between –1 and 3.
⇒
64. If the cube roots of unity are1,ω andω 2 ,then
2
= x12 + y12 + x22 + y22 + 2 ( x12 + y12 )( x22 + y22 )
2
1
1
1
+  z4 + 4  +  z5 + 5  +  z6 + 6 



z 
z 
z 
⇒
2
Again squaring, we get
x12 x22 + y12 y22 + 2 x1 x2 y1 y2
= x12 x22 + y12 y22 + x12 y22 + y12 x22
= (−1)2 + (−1)2 + (1 + 1)2 + (−1)2 + (−1)2 + (1 + 1)2
When we put either z = ω or z = ω2 ,
we get the same result
= 1 + 1 + 4 + 1 + 1 + 4 = 12
x1 x2 + y1 y2 = ( x12 + y12 )( x22 + y22 )
⇒
⇒
( x1 y2 − y1 x2 )2 = 0
y1 y2
=
x1 x2
38
JEE Main Chapterwise Mathematics
y 
y 
tan−1  1  = tan−1  2 
 x1 
 x2 
⇒
⇒
⇒
arg ( z1 ) = arg ( z2 )
arg ( z1 ) − arg ( z2 ) = 0
Alternate Solution
Given that, | z1 + z2| = | z1| + | z2|
On squaring both sides, we get
67. Ifw =
z
z−
i
3
(b) a straight line
(c) a circle
(d) an ellipse
Exp. (b)
Given that, w =
2
= | z1|2 + | z2|2 + 2| z1|| z2|
⇒
Re ( z1 z2 ) = | z1|| z2|
z
⇒
z−
⇒ | z1|| z2| cos (θ1 − θ2 ) = | z1|| z2|
⇒
θ1 − θ2 = 0
⇒
arg ( z1 ) − arg ( z2 ) = 0
66. The value of a for which the sum of the
squares of the roots of the equation
x 2 − (a − 2 )x − a − 1 = 0 assume the least
value is
[AIEEE 2005]
(a) 2
(c) 0
(b) 3
(d) 1
Exp. (d)
Let α and β be the roots of equation
x2 − ( a − 2 ) x − a − 1 = 0
Then,
Now,
α + β = a − 2 and αβ = − a − 1
α 2 + β 2 = (α + β )2 − 2 αβ
⇒
α 2 + β 2 = (a − 2 )2 + 2(a + 1)
⇒
α 2 + β 2 = a2 − 2 a + 6
⇒
α 2 + β 2 = (a − 1)2 + 5
The value of α 2 + β 2 will be least, if a − 1 = 0.
⇒
a=1
Alternate Solution
Since, α + β = (a − 2 ) and αβ = − a − 1
Let
f(a) = α 2 + β 2 = (α + β )2 − 2 αβ
= (a − 2 )2 + 2 (a + 1)
= a2 − 2 a + 6
⇒
f ′ ( a) = 2 a − 2
For maxima or minima, put f ′(a) = 0.
∴
2a − 2 = 0 ⇒ a = 1
Now,
f ′ ′ ( a) = 2
⇒
f′ ′(1) = 2 > 0
So, f(a) is minimum at a = 1.
[AIEEE 2005]
(a) a parabola
| z1| + | z2| + 2 Re ( z1 z2 )
2
and |w | = 1, then z lies on
z
z−
i
3
and|w| = 1
z − i 

i = 1 ⇒ | z| = 
3

3
So, z lies on perpendicular bisector of (0, 0) and
 0, 1  .


 3
Hence, z lies on a straight line.
68. If the roots of the equation x 2 − bx + c = 0 is
two consecutive integers, thenb 2 − 4c equal
to
[AIEEE 2005]
(a) 1
(b) 2
(c) 3
(d) –2
Exp. (a)
Let n and (n + 1) be the two consecutive roots of
x2 − bx + c = 0. Then, n + (n + 1) = b
and n(n + 1) = c.
∴
b 2 − 4 c = (2 n + 1)2 − 4n(n + 1)
= 4n2 + 4n + 1 − 4n2 − 4n = 1
69. If both the roots of the quadratic equation
x 2 − 2 kx + k 2 + k − 5 = 0 are less than 5, then
[AIEEE 2005]
k lies in the interval
(a) [4, 5]
(b) ( −∞ , 4) (c) (6, ∞ )
(d) (5, 6]
Exp. (b)
Let f( x) = x2 − 2 kx + k 2 + k − 5
Since, both roots are less than 5. Then,
b
< 5 and f(5) > 0
D ≥ 0, −
2a
Now,
D = 4k 2 − 4(k 2 + k − 5)
⇒
and
⇒
= − 4k + 20 ≥ 0
k≤ 5
b
−
< 5 ⇒ k< 5
2a
f(5) > 0
25 − 10k + k 2 + k − 5 > 0
…(i)
…(ii)
39
Complex Numbers and Quadratic Equations
⇒
k 2 − 9k + 20 > 0
⇒
⇒
(k − 5)(k − 4) > 0
k < 4 and k > 5
Exp. (d)
z1/ 3 = p + iq
Q
…(iii)
∴
z = ( p + iq )3
From Eqs. (i), (ii) and (iii), we get
= p3 − iq 3 + 3 ip2q − 3 pq 2
k< 4
Given that, z = x − iy
∴
x − iy = p3 − 3 pq 2 + i (3 p2q − q 3 )
70. If the equation
an x n + an − 1x n − 1 + ... + a1x = 0, a1 ≠ 0, n ≥ 2 ,
has a positive root x = α, then the equation
nan x n − 1 + (n − 1)an − 1x n − 2 + ... + a1 = 0 has a
positive root, which is
…(i)
f(0) = 0 and f(α ) = 0
According to the Rolle’s theorem,
f ′ ( x) = 0
has atleast one root between (0, α ).
So, f ′( x) = 0 has a positive root less than α.
71. Let z ,w be complex numbers such that
z + iw = 0 and arg ( zw ) = π. Then, arg ( z ) is
equal to
[AIEEE 2004]
π
2
(c)
3π
4
(d)
5π
4
Exp. (c)
Given that, z + i w = 0 ⇒ z = − i w
⇒
z = iw ⇒ w = − i z
and
arg ( zw) = π
⇒
arg (− iz2 ) = π
⇒
⇒
⇒
arg (− i ) + 2 arg ( z) = π
π
π
− + 2 arg ( z) = π Q arg (− i ) = − 

2 
2
3π
arg ( z) =
4
and
z = x − iy
z = p + iq , then
x y
2
2
 +  (p + q ) is equal to
p q 
[AIEEE 2004]
1/3
72. If
(a) 1
(b) –1
and
⇒
Let f( x) = an xn + an − 1 xn − 1 + ... + a1 x = 0
(b)
⇒
∴
Exp. (c)
π
4
x = p3 − 3 pq 2 and − y = 3 p2q − q 3
x
= p2 − 3q 2
p
y
= − 3 p2 + q 2
q
x
y
+ = − 2 p2 − 2q 2
p q
[AIEEE 2005]
(a) equal to α
(b) greater than or equal to α
(c) smaller than α
(d) greater than α
(a)
⇒
(c) 2
(d) –2
x
1
y
 +  = −2
( p2 + q 2 )  p q 
73. If | z 2 − 1| = | z |2 + 1, then z lies on
(a) the real axis
(c) a circle
[AIEEE 2004]
(b) the imaginary axis
(d) an ellipse
Exp. (b)
Using the relation, if
| z1 + z2| = | z1| + | z2|
Then,
arg ( z1 ) = arg ( z2 )
Since,
| z2 + (−1)| = | z2| + |− 1|
Then,
arg ( z2 ) = arg (−1)
⇒
2 arg ( z) = π
[Qarg (−1) = π ]
π
⇒
arg ( z) =
2
So, z lies on Y-axis (imaginary axis).
74. If (1 − p ) is a root of quadratic equation
x 2 + px + (1 − p ) = 0, then its roots are
(a) 0, 1
(c) 0, –1
(b) –1, 1
(d) –1, 2
[AIEEE 2004]
Exp. (c)
Since, (1 − p) is a root of quadratic equation
…(i)
x2 + px + (1 − p) = 0
So, (1 − p) satisfied the above equation.
∴
(1 − p)2 + p (1 − p) + (1 − p) = 0
⇒
(1 − p)(1 − p + p + 1) = 0
⇒
(1 − p) 2 = 0 ⇒ p = 1
On putting this value in Eq. (i), we get
x2 + x = 0 ⇒ x = 0, − 1
40
JEE Main Chapterwise Mathematics
75. If one root of the equation x 2 + px + 12 = 0 is
4, while the equation x 2 + px + q = 0 has
equal roots, then the value of q is
49
4
(c) 3
(a)
(b) 12
[AIEEE 2004]
(d) 4
Exp. (a)
∴
16 + 4 p + 12 = 0
⇒
4 p = − 28 ⇒ p = − 7
So, the other equation is x2 − 7 x + q = 0 whose
roots are equal. Let the roots be α and α.
7
7
∴ Sum of roots = α + α = ⇒ α =
1
2
And product of roots = α ⋅ α = q
∴
9 and geometric mean 4. Then, these
numbers are the roots of the quadratic
equation
[AIEEE 2004]
(a) x 2 + 18x + 16 = 0
(b) x 2 − 18x + 16 = 0
(c) x 2 + 18x − 16 = 0
(d) x 2 − 18x − 16 = 0
Exp. (b)
Since, one of the roots of equation
x2 + px + 12 = 0 is 4.
⇒
77. Let two numbers have arithmetic mean
2
7 = q
 
2
49
q=
4
Let α and β be two numbers whose arithmetic
mean is 9 and geometric mean is 4.
∴
α + β = 18 and αβ = 16
∴ Required equation
x2 − (α + β )x + (αβ ) = 0
⇒
x2 − 18 x + 16 = 0
78. Let z1 and z2 be two roots of the equation
z 2 + az + b = 0, z being complex. Further,
assume that the origin, z1 and z2 form an
equilateral triangle. Then,
[AIEEE 2003]
(a) a 2 = b
(b) a 2 = 2 b
(c) a = 3b
(d) a 2 = 4b
2
76. If 2a + 3b + 6c = 0,then atleast one root of the
equation ax 2 + bx + c = 0 lies in the interval
[AIEEE 2004]
(a) (0, 1)
(c) (2, 3)
(b) (1, 2)
(d) (1, 3)
Let f ′( x) = ax2 + bx + c
On integrating both sides, we get
f ( x) =
ax3
bx2
+
+ cx + d
3
2
2 ax + 3bx + 6 cx + 6 d
6
Since, f( x) is a polynomial function and is
continuous as well as differentiable in its entire real
set.
2 a + 3b + 6 c + 6 d
⇒ f(1) =
6
6d
=
=d
[Q2 a + 3b + 6 c = 0, given]
6
6d
and
=d
f(0) =
6
∴
Since, origin z1 and z2 are the vertices of an
equilateral triangle, then
z12 + z22 = z1 z2
⇒
Exp. (a)
⇒
Exp. (c)
f ( x) =
3
( z1 + z2 )2 = 3 z1 z2
Again, z1, z2 are the roots of the equation
z2 + az + b = 0
Then,
z1 + z2 = − a and z1 z2 = b
On putting these values in Eq. (i), we get
(− a)2 = 3b ⇒ a2 = 3b
2
f(0) = f(1)
Hence, according to Rolle’s theorem, atleast one
root of ax2 + bx + c = 0 lies between 0 and 1.
…(i)
79. If z and w are two non-zero complex
numbers
such
that
and
| zw | = 1
π
, then z (w ) is equal to
2
arg ( z ) − arg (w ) =
(a) 1
(c) i
(b) –1
(d) −i
[AIEEE 2003]
Exp. (d)
Let
z = r1 e iθ and w = r2 e iφ
Given,
⇒ z = r1 e − iθ
| zw| = 1
⇒
|r1 e iθ ⋅ r2 e iφ| = 1
⇒
r1r2 = 1
…(i)
41
Complex Numbers and Quadratic Equations
π
2
π
θ−φ=
2
and arg ( z) − arg (w) =
⇒
Also given,
…(ii)
= 1 ⋅ e iπ / 2
= cos
zw = − i
⇒
[from Eqs. (i) and (ii)]
π
π
− i sin
2
2
[AIEEE 2003]
x = 4n , where n is any positive integer
x = 2 n , where n is any positive integer
x = 4n + 1, where n is any positive integer
x = 2 n + 1, where n is any positive integer
Exp. (a)
1 +
Now, 
1−
x
 (1 +
i
 =
i
 (1 −
 (1 +
=
 1−
x
i )(1 + i )
i )(1 + i ) 
x
x
i )2 
1 − 1 + 2 i 
=



2
i2 
x
⇒
1 + i 
x

 = (i) = 1
1− i 
⇒
+
1
β
2
=
α2 + β2
α 2β 2
α + β
2
α+β=
 −
 αβ 
αβ
⇒
 − b  =  − b/a  − 2


 
 a   c /a 
c/a
2
−
2
2a
b  b
=  −
a c
c
2a b  b c
=  + 
c
c c
a
2a b c
⇒
= +
b
c
a
c a
b
Since, , and are in AP.
a b
c
a b
c
Hence, , and are in HP.
c a
b
⇒
1 + i
 = 1, then
1 − i
80. If 
(a)
(b)
(c)
(d)
α
2
⇒
⇒
x
1
2
zw = r1 e − iθ ⋅ r2 e iφ = r1r2e − i ( θ − φ )
Now,
α+β=
[given]
( i ) x = ( i )4 n
82. The number of the real solutions of the
equation x 2 − 3| x | + 2 = 0 is
[AIEEE 2003]
(a) 2
(d) 3
(b) 4
(c) 1
Exp. (b)
Given equation is
x2 − 3| x| + 2 = 0
Case I
When x > 0, then| x| = x
where, n is any positive integer.
∴
x2 − 3 x + 2 = 0
∴
⇒
⇒
( x − 1)( x − 2 ) = 0
x = 1, 2
x = 4n
81. If the sum of the roots of the quadratic
equation ax 2 + bx + c = 0 is equal to the sum
of the squares of their reciprocals, then
a b
c
, and are in
c a
b
[AIEEE 2003]
(a)
(b)
(c)
(d)
arithmetic progression
geometric progression
harmonic progression
arithmetic-geometric progression
Exp. (c)
Given equation is ax2 + bx + c = 0.
Let α and β be the roots of the equation.
b
Then,
α+β=−
a
c
and
αβ =
a
Case II
∴
When x < 0, then| x| = − x
x2 + 3 x + 2 = 0
⇒
( x + 1)( x + 2 ) = 0
⇒
x = − 1, − 2
Hence, four solutions are possible.
83. The value of ‘a’ for which one root of the
quadratic equation
(a 2 − 5a + 3)x 2 + ( 3a − 1)x + 2 = 0
is twice as large as the other, is [AIEEE 2003]
(a) 2/3
(c) 1/3
(b) –2/3
(d) –1/3
Exp. (a)
Since, one root of the quadratic equation
(a2 − 5a + 3)x2 + (3a − 1)x + 2 = 0 is twice as
large as the other, then let their roots beα and 2 α.
42
JEE Main Chapterwise Mathematics
∴
(3a − 1)
α + 2α = −
3α = −
⇒
α ⋅2α =
and
2α2 =
⇒
(a2 − 5a + 3)
2
Now, (1 + ω − ω2 )7 = (−ω2 − ω2 )7
[Q 1 + ω + ω2 = 0]
1
= (−2 ω2 )7 = − 2 7 ⋅ ω14
(a2 − 5a + 3)
= − 128(ω3 )4 ω2
(3 a − 1)2 = 9(a2 − 5 a + 3)
⇒
= − 128ω2
9a2 − 6a + 1 = 9a2 − 45 a + 27
⇒
45a − 6a = 27 − 1
⇒
a=
86.
26 2
=
39 3
2
equation having α/β and β/α as its roots, is
[AIEEE 2002]
(a) 3x + 19x + 3 = 0
(b) 3x − 19x + 3 = 0
(c) 3x 2 − 19x − 3 = 0
(d) x 2 − 16x + 1 = 0
2
2
Given α 2 = 5α − 3 ⇒
α 2 − 5α + 3 = 0
β2 = 5 β − 3 ⇒
β2 − 5 β + 3 = 0
These two equations shows that α and β are the
roots of the equation x2 − 5 x + 3 = 0.
∴
α + β = 5 and αβ = 3
α β α +β
+ =
β α
αβ
2
Now,
2
(α + β )2 − 2 αβ
αβ
25 − 6 19
=
=
3
3
=
α β
⋅ =1
β α
Thus, the equation having roots
α
β
and is given
β
α
by
α β 
α β
x2 −  +  x + ⋅ = 0
β α
β α
19
x + 1= 0
3
⇒
x2 −
⇒
3 x − 19 x + 3 = 0
2
[Q ω3 = 1]
−3 i 1
3i −1 = x + iy , then
3
i
[AIEEE 2002]
(a) x = 3, y = 1
(c) x = 0, y = 3
(b) x = 1, y = 3
(d) x = 0, y = 0
Exp. (d)
6i − 3i
Given that, 4
3i
20
Exp. (b)
and
6i
If 4
20
84. If α ≠ β and α = 5α − 3 , β = 5β − 3 , then the
2
and
(b) −128 ω
(d) −128 ω2
Exp. (d)
(a2 − 5a + 3)
=
[AIEEE 2002]
(a) 128 ω
(c) 128 ω2
(a − 5 a + 3)
2
9(a2 − 5 a + 3)2
⇒
(1 + ω − ω 2 ) 7 equal to
2
(3a − 1)2
⇒
85. If ω is an imaginary cube root of unity, then
(a2 − 5a + 3)
(3a − 1)
1
−1 = x + iy
3
i
Applying R1 → R1 + R 2
⇒
6i + 4 0
4
3i
0
−1 = x + iy
20
3
i
⇒
(6 i + 4)
3i
−1
3
i
⇒
(6 i + 4)(3 i 2 + 3) = x + iy
⇒
∴
= x + iy
0 + 0i = x + iy
x = 0 and
87. The number of real roots of 32x
(a) 0
(c) 1
(b) 2
(d) 4
y=0
2
− 7x + 7
Exp. (b)
Given that,
32 x
2
− 7x + 7
= 32
⇒
2x − 7x + 7 = 2
⇒
2 x2 − 7 x + 5 = 0
2
Now, D = b 2 − 4ac = (−7 )2 − 4 × 2 × 5
= 49 − 40 = 9 > 0
Hence, it has two real roots.
= 9 is
[AIEEE 2002]
3
Matrices and
Determinants
1. The greatest value of c ∈R for which the
system of linear equations
x −cy − cz =0,cx −y + cz =0,cx + cy − z =0
has a non-trivial solution, is
[JEE Main 2019, 8 April Shift-I]
1
(b)
2
(d) 0
(a) −1
(c) 2
Key Idea A homogeneous system of linear
equations have non-trivial solutions iff ∆ = 0
Given system of linear equations is
x − cy − cz = 0,
cx − y + cz = 0
and
cx + cy − z = 0
We know that a homogeneous system of linear
equations have non-trivial solutions iff
∆=0
1 − c − c
c − 1
c = 0
⇒


c
c
−
1

⇒ 11
( − c 2 ) + c(−c − c 2 ) − c(c 2 + c ) = 0
⇒
1− c − c − c − c − c = 0
⇒
−2c 3 − 3c 2 + 1 = 0
⇒
2c 3 + 3c 2 − 1 = 0
⇒
(c + 1)[2c 2 + c − 1] = 0
2
3
(c + 1)[2c 2 + 2c − c − 1] = 0
⇒
(c + 1)(2c − 1)(c + 1) = 0
⇒
c = − 1or
1
2
1
Clearly, the greatest value of c is .
2
cos α − sin α
 ,(α ∈R ) such that
 sin α cos α 
0 −1
A 32 = 
 . Then, a value of α is
1 0 
2. Let A = 
Exp. (b)
2
⇒
3
2
π
(a)
32
[JEE Main 2019, 8 April Shift-I]
(b) 0
(c)
π
64
(d)
π
16
Exp. (c)
cos α − sinα 
Given, matrix A = 

 sinα cos α 
cos α − sinα  cos α − sinα 
∴ A2 = 


 sinα cos α   sinα cos α 

cos 2 α − sin2 α
=
sinα cos α + cos α sinα
− cos α sinα − sinα cos α 

− sin2 α + cos 2 α

cos 2 α − sin2 α 
=

 sin2 α cos 2 α 
Similarly,
cos(nα ) − sin(nα )
An = 
, n ∈ N
 sin(nα ) cos(nα ) 
44
JEE Main Chapterwise Mathematics
cos(32 α ) − sin(32 α )
⇒ A 32 = 

 sin(32 α ) cos(32 α ) 
 0 −1
=

1 0 
So, cos(32 α ) = 0 and sin(32 α ) = 1
π
π
⇒
32 α = ⇒α =
2
64
On applying, C 2 → C 2 − C1 and C 3 → C 3 − C1,
(given)
3. If the system of linear equations
x − 2 y + kz = 1 , 2 x + y + z = 2 ,
3x − y − kz = 3
has a solution( x , y , z ), z ≠ 0, then( x , y ) lies on
the straight line whose equation is
[JEE Main 2019, 8 April Shift-II]
(a) 3x − 4y − 4 = 0
(c) 4x − 3y − 4 = 0
(b) 3x − 4y − 1 = 0
(d) 4x − 3y − 1 = 0
Exp. (c)
Given system of linear equations
x − 2 y + kz = 1
2x + y + z = 2
and
3 x − y − kz = 3
has a solution ( x, y, z,) z ≠ 0.
x − 2 y + kz + 3 x − y − kz = 1 + 3
4x − 3y = 4
4x − 3y − 4 = 0
This is the required equation of the straight line in
which point ( x, y) lies.
4. Let the numbers 2 ,b , c be in an AP and
1 1 1 
A = 2 b c . If det( A ) ∈[2 , 16], then c


2
2
4 b c 
lies in the interval
[JEE Main 2019, 8 April Shift-II]
(a) [3, 2 + 2 3/ 4 ]
(c) [4, 6]
Exp. (c)
(b) ( 2 + 2 3/ 4 , 4)
(d) [2 , 3)
1 1 1 
Given, matrix A = 2 b c , so


2
2
 4 b c 
1 1 1 
det( A) = 2 b c 

2
2
4 b c 
b−2 c −2

= 2
2
b − 4 c − 4
b−2
c −2
=
(b − 2 )(b + 2 ) (c − 2 )(c + 2 )
1 
 1

= (b − 2 )(c − 2 )
2
2
b
c
+
+

[taking common (b − 2 ) from C1
and (c − 2 ) from C 2 ]
= (b − 2 )(c − 2 )(c − b )
Since, 2, b and c are in AP, if assume common
difference of AP is d, then
b = 2 + d and c = 2 + 2d
So, | A| = d (2d )d = 2d 3 ∈ [2, 16]
[given]
⇒
…(i)
…(ii)
…(iii)
On adding Eqs. (i) and (iii), we get
⇒
0
0 
1
we get det( A) = 2 b − 2 c − 2 


2
2
4 b − 4 c − 4
∴
d 3 ∈[1, 8] ⇒ d ∈[1, 2 ]
2 + 2d ∈ [2 + 2, 2 + 4]
= [4, 6] ⇒ c ∈[4, 6]
1 1 1 2 1 3 1 n − 1 1 78
=
,
1  0 1 
5. If   . 
.
 ... 
0 1 0 1 0 1 0
1 n
then the inverse of 
 is
0 1
[JEE Main 2019, 9 April Shift-I]
1
(a) 
12
1
(c) 
13
0
1 
0
1 
1 −13
(b) 

0 1 
1 −12 
(d) 

0 1 
Exp. (b)
Given
 1 1  1 2   1 3 ...  1 n − 1  1 78
=
 0 1  0 1  0 1  0
1   0 1 



 
Q
 1 1  1 2   1 2 + 1
,
 0 1  0 1 =  0
1 


 
 1 2 + 1
0
1 

 1 3  1 3 + 2 + 1
,
 0 1 =  0
1 

 
:
:
:
:
:
:
1
1
1
2
1
3
1



  n − 1
∴ 




 ... 
1 
 0 1  0 1  0 1  0
45
Matrices and Determinants
 1 (n − 1) + (n − 2 )+ ...+3 + 2 + 1
=

1
0

 1 n (n − 1)  1 78
=
=
2

 0
1   0 1 
Since, both matrices are equal, so equating
corresponding element, we get
n(n − 1)
= 78 ⇒ n(n − 1) = 156
2
= 13 × 12 = 13(13 − 1)
⇒
n = 13
 1 −13
 1 13
−1
So, A = 
 = A = 0 1 
0
1




 a b
−1
[Qif| A|= 1and A = 
 , then A =
c d 
 d − b 
 −c a  


x 2 + x + 1 = 0. Then, for y ≠ 0 in R,
α
α
β
y +β
1
[JEE Main 2019, 9 April Shift-I]
(a) y ( y 2 − 1)
(b) y ( y 2 − 3)
(c) y 3 − 1
(d) y 3
Exp. (d)
Given, quadratic equation is x2 + x + 1 = 0 having
roots α, β.
Then, α + β = −1and αβ = 1
Now, given determinant
∆=
β
α
β
y+ β
1
1
y+ α
On applying R1 → R1 + R 2 + R 3 , we get
y + 1+ α + β
∆=
α
y + 1+ α + β
y+ β
y + 1+ α + β
1
1
y+ α
β
y
= α
β
y
y+ β
y
1
1
y+ α
1− β
y+ α −β
β
= y[( y + (β − α )) ( y − (β − α )) − (1 − α ) (1 − β )]
[expanding along R1]
= y[ y2 − (β − α )2 − (1 − α − β + αβ )]
= y[ y2 − β 2 − α 2 + 2αβ − 1 + (α + β ) − αβ ]
= y[ y2 − (α + β )2 + 2αβ + 2αβ − 1 + (α + β ) − αβ ]
= y[ y2 − 1 + 3 − 1 − 1] = y3 [Qα + β = −1and αβ = 1]
x + ky − 2 z = 0 and 2 x − y + z = 0 has a
non-trivial
solution
then
( x , y , z ),
x y z
+ + + k is equal to
y z x
[JEE Main 2019, 9 April Shift-II]
1
(b)
2
(c) −
1
4
(d)
3
4
Exp. (b)
1 is equal to
y +α
α
0
1− α
(a) −4
β
y+ 1
0
y+ β −α
7. If the system of equations 2 x + 3y − z = 0,
6. Let α and β be the roots of the equation
y +1
y
∆= α
[Qα + β = −1]
On applying C 2 → C 2 − C1 and C 3 → C 3 − C1,
we get
Given system of linear equations
2 x + 3 y − z = 0,
x + ky − 2 z = 0
and 2 x − y + z = 0 has a non-trivial solution
( x, y, z).
2 3 −1 
∴
∆ = 0 ⇒ 1 k − 2  = 0


2 −1 1 
2(k − 2 ) − 3(1 + 4) − 1(−1 − 2 k ) = 0
⇒
2 k − 4 − 15 + 1 + 2 k = 0
9
⇒
4 k = 18 ⇒ k =
2
So, system of linear equations is
2 x + 3y − z = 0
2 x + 9y − 4z = 0
and
2x − y + z = 0
From Eqs. (i) and (ii), we get
y 1
6 y − 3 z = 0, =
z 2
From Eqs. (i) and (iii), we get
x
1
4x + 2 y = 0 ⇒ = −
y
2
1 z
x x y
= × =− ⇒ =− 4
4 x
z y z
x y z
1 1
+ + + k= − +
∴
y z x
2 2
So,
…(i)
…(ii)
…(iii)
 y 1
x
1
Q z = 2 and y = − 2 


9 1
−4+ = .
2 2
46
JEE Main Chapterwise Mathematics
8. The
total
number
of
matrices
 0 2y 1 
A = 2 x y −1 , ( x , y ∈R , x ≠ y ) for which


2 x − y 1 
A T A = 3I 3 is
[JEE Main 2019, 9 April Shift-II]
(a) 2
(c) 3
(b) 4
(d) 6
Exp. (b)
Given matrix
 0 2y 1 
A = 2 x y −1 ,( x, y ∈ R, x ≠ y)


2 x − y 1 
for which
AT A = 3I3
⇒
⇒
 0 2 x 2 x  0
2 y y − y 2 x


 1 −1 1  2 x
 8 x2
0 0


2
 0 6 y 0 =
 0
0 3


2 y 1   3 0 0
y −1 =  0 3 0

 
− y 1   0 0 3
 3 0 0
 0 3 0


 0 0 3
Here, two matrices are equal, therefore equating
the corresponding elements, we get
3
8 x2 = 3 and 6 y2 = 3 ⇒ x = ±
8
1
and
y=±
2
QThere are 2 different values of x and y each.
So, 4 matrices are possible such that AT A = 3I3 .
x
sin θ cos θ
9. If ∆1 = − sin θ
−x
1 and
1
x
sin 2θ cos 2θ
, x ≠ 0, then
∆ 2 = − sin 2θ
−x
1
cos 2θ
1
x
 π
for all θ ∈ 0, 
 2  [JEE Main 2019, 10 April Shift-I]
cos θ
x
(a) ∆1 + ∆ 2 = − 2( x 3 + x − 1)
(b) ∆1 − ∆ 2 = − 2 x 3
(c) ∆1 + ∆ 2 = − 2 x 3
(d) ∆1 − ∆ 2 = x(cos 2θ − cos 4θ)
Exp. (c)
Given determinants are
x
sinθ cos θ
∆1 = − sinθ − x
1
cos θ
1
x
= − x + sinθcos θ − sinθcos θ
3
+ xcos 2 θ − x + x sin2 θ
=− x
3
x
sin2θ cos 2θ
and ∆ 2 = − sin2θ − x
1 ,x≠0
cos 2θ
1
x
= − x (similarly as ∆1)
3
So, according to options, we get ∆1 + ∆ 2 = − 2 x3
10. If the system of linear equations
x + y + z = 5, x + 2 y + 2 z = 6
x + 3y + λz = µ,(λ ,µ ∈R ), has infinitely many
solutions, then the value of λ + µ is
[JEE Main 2019, 10 April Shift-I]
(a) 7
(b) 12
(c) 10
(d) 9
Exp. (c)
Given system of linear equations
…(i)
x+ y+ z=5
…(ii)
x + 2y + 2z = 6
…(ii)
x + 3 y + λz = µ
(λ,µ ∈ R )
The above given system has infinitely many
solutions, then the plane represented by these
equations intersect each other at a line, means
( x + 3 y + λz − µ )
= p( x + y + z − 5) + q ( x + 2 y + 2 z − 6)
= ( p + q )x + ( p + 2q )y + ( p + 2q )z − (5 p + 6q )
On comparing, we get
p + q =1, p + 2q = 3, p + 2q = λ and 5 p + 6q = µ
So, ( p, q ) = (−1, 2 )
⇒
λ = 3 and µ = 7 ⇒ λ + µ = 3 + 7 = 10
11. Let λ be a real number for which the system
of linear equations
x + y + z = 6, 4x + λy − λz = λ − 2 and
3x + 2 y − 4z = − 5
has infinitely many solutions. Then λ is a
root of the quadratic equation
[JEE Main 2019, 10 April Shift-II]
(a) λ2 − 3λ − 4 = 0
(b) λ2 + 3λ − 4 = 0
(c) λ − λ − 6 = 0
(d) λ2 + λ − 6 = 0
2
47
Matrices and Determinants
Exp. (c)
Exp. (b)
Given, system of linear equations
x+ y+ z=6
4 x + λy − λz = λ − 2
and
3x + 2 y − 4z = − 5
has infinitely many solutions, then ∆ = 0
1 1 1
⇒
4 λ −λ =0
Given matrix A is a symmetric and matrix B is a
skew-symmetric.
∴
AT = A and BT = − B
2 3 
(given)… (i)
Since, A + B = 

 5 − 1
On taking transpose both sides, we get
… (i)
…(ii)
…(iii)
2
( A + B)T = 
5
2
⇒
AT + BT = 
3
T
T
Given, A = A and B = −
2 5 
⇒ A− B= 

 3 − 1
−4
3 2
⇒1(− 4λ + 2 λ ) − 1(− 16 + 3λ) + 1(8 − 3λ) = 0
⇒
− 8λ + 24 = 0 ⇒ λ = 3
From, the option λ = 3, satisfy the quadratic
equation λ2 − λ − 6 = 0.
12. The sum of the real roots of the equation
x
2
−6
− 3x
−3
2x
−1
x − 3 = 0, is equal to
[JEE Main 2019, 10 April Shift-II]
(c) 6
Exp. (a)
x
Given equation
−6
2 − 3x
− 3 2x
(d) 1
−1
14.
x−3 =0
x+2
On expansion of determinant along R1, we get
x[(− 3 x)( x + 2 ) − 2 x( x − 3)]+ 6[2( x + 2 ) + 3( x − 3)]
− 1[2(2 x) − (− 3 x) (− 3)] = 0
⇒ x[− 3 x2 − 6 x − 2 x2 + 6 x] + 6[2 x + 4 + 3 x − 9]
⇒
− 1[4 x − 9 x] = 0
x(− 5 x2 ) + 6(5 x − 5) − 1(− 5 x) = 0
⇒
−5 x3 + 30 x − 30 + 5 x = 0
3
⇒ 5 x − 35 x + 30 = 0 ⇒ x3 − 7 x + 6 = 0.
Since all roots are real
coefficient of x2
∴ Sum of roots = −
=0
coefficient of x3
13. If A is a symmetric matrix and B is a
skew-symmetric
matrix
such
2 3 
A +B = 
, then AB is equal to
5 −1
that
[JEE Main 2019, 12 April Shift-I]
 −4 −2 
(a) 

 −1 4 
 4 −2 
(c) 

1 −4 
4
(b) 
 −1
 −4
(d) 
1
−2 
−4 
2
4 
5
− 1
… (ii)
B
On solving Eqs. (i) and (ii), we get
2 4 
 0 − 1
A=
 and B =  1 0 
4
1
−




2 4   0 − 1  4 − 2 
So, AB = 


=
 4 − 1  1 0   − 1 − 4
x +2
(b) − 4
(a) 0
T
3
− 1
5 2 α
If B = 0 2

α 3
1
1  is the inverse of a 3 × 3

−1
matrix A, then the sum of all values of α for
which det ( A ) + 1 = 0, is
[JEE Main 2019, 12 April Shift-I]
(a) 0
(b) −1
(c) 1
(d) 2
Exp. (c)
Given matrix B is the inverse matrix of 3 × 3
matrix A,
 5 2α 1 
where B =  0 2
1


−
α
3
1

We know that,

1
1 
det( A) =
Qdet( A − 1 ) =
det(B) 
det( A)
Since, det( A) + 1 = 0 (given)
1
+ 1 = 0 ⇒ det(B) = − 1
det(B)
⇒ 5(− 2 − 3) − 2α(0 − α ) + 1 (0 − 2α ) = − 1
⇒ − 25 + 2α 2 − 2α = − 1 ⇒ 2α 2 − 2α − 24 = 0
⇒
α 2 − α − 12 = 0 ⇒ (α − 4) (α + 3) = 0
⇒
α = − 3, 4
So, required sum of all values of α is 4 − 3 = 1
48
JEE Main Chapterwise Mathematics
15. A value of θ ∈(0, π / 3), for which
Exp. (c)
sin θ
1 + cos θ
4cos 6θ
cos 2 θ
1 + sin 2 θ
4cos 6θ = 0, is
cos 2 θ
sin 2 θ
1 + 4cos6θ
2
2
[JEE Main 2019, 12 April Shift-II]
π
9
7π
(c)
24
π
18
7π
(d)
36
| A| = cos 2 θ + sin2 θ = 1
 cos θ sinθ 
adj A = 

 − sinθ cos θ
and
 d − b
 a b
[QIf A = 
, then adj A =  −c a  ]

c d 

 cos θ sinθ 
A −1 = 

 − sinθ cos θ
⇒
Exp. (a)
1 + cos 2 θ
sin2 θ
Let ∆ =
cos 2 θ
1 + sin2 θ
cos 2 θ
sin2 θ
4cos 6 θ
=0
4cos 6 θ
1 + 4 cos 6 θ
Applying C1 → C1 + C 2 , we get
2
sin2 θ
4cos 6 θ
∆ = 2 1 + sin2 θ
4cos 6 θ
=0
1
sin2 θ
1 + 4 cos 6 θ
Applying R1 → R1 − 2 R 3 and R 2 → R 2 − 2 R 3 , we get
0 − sin2 θ − 2 − 4cos 6 θ
∆ = 0 1 − sin2 θ − 2 − 4cos 6 θ = 0
1
sin2 θ
1 + 4 cos 6 θ
On expanding w.r.t. C1, we get
⇒sin2 θ (2 + 4 cos 6 θ) + (2 + 4 cos 6 θ)
(1 − sin2 θ) = 0
⇒ 2 + 4 cos 6 θ = 0
⇒
∴
(b)
(a)
⇒
cos θ − sinθ
We have, A = 

 sinθ cos θ 
1
2π
= cos
2
3
2π
π
⇒θ =
6θ =
3
9
cos 6 θ = −
π 

Qθ ∈  0, 

 3  
cos θ − sin θ
 , then the matrix
sin θ cos θ 
π
A −50 when θ = , is equal to
12
16. If A = 
[JEE Main 2019, 9 Jan Shift-I]
 1
3

 2
2
(a) 

− 3 1 
 2
2 
 3 1 

2 
(c)  2
1
3
−
 2 2 


(b) 




(d) 


3
2
1
2
1
2
3
2
1
− 
2
3
2 
3
−
2 
1 
2 

adj A 
−1
Q A =

| A| 

Note that, A −50 = ( A −1 )50
Now,
A −2 = ( A −1 )( A −1 )
⇒
 cos θ sinθ   cos θ sinθ 
A −2 = 


 − sinθ cos θ  − sinθ cos θ

cos 2 θ − sin2 θ
cos θsinθ + sinθcos θ
=

− sin2 θ + cos 2 θ 
 − cos θsinθ − cos θsinθ
 cos 2 θ sin2 θ 
=

 − sin2 θ cos 2 θ
Also, A −3 = ( A −2 )( A −1 )
 cos 2 θ sin2 θ 
A −3 = 

 − sin2 θ cos 2 θ
 cos θ sinθ 


 − sinθ cos θ
 cos 3θ sin 3θ 
=

 − sin 3θ cos 3θ
 cos 50θ sin 50θ 
Similarly, A −50 = 

 − sin 50θ cos 50θ
 cos 25 π sin 25 π 

6
6 
=
25
25 
π
π
sin
cos
−


6
6 
 when θ = π 



12 
 cos π sin π 

6
6
=
π
π
cos 
 − sin

6
6
π
π
25 π 



Qcos 
 = cos 4 π +  = cos 

 6 

6
6

25 π 
π
π

and sin 
 = sin 4 π +  = sin 





6
6
6 
 3

= 2
 −1
 2
1 
2 
3
2 
49
Matrices and Determinants
17. The system of linear equations
18.
x + y + z =2
2 x + 3y + 2 z = 5
2 x + 3y + (a 2 − 1)z = a + 1
[JEE Main 2019, 9 Jan Shift-I]
(a)
(b)
(c)
(d)
has infinitely many solutions for a = 4
is inconsistent when a = 4
has a unique solution for | a | = 3
is inconsistent when | a | = 3
Exp. (d)
According to Cramer’s rule, here
1 1
1
1 0
0
D= 2 3
2
= 2 1
0
2 3 a2 − 1
2 1 a2 − 3
D1 =
2
5
1
3
1
2
2
5
=
a + 1 3 a2 − 1
1
3
0
−1
a + 1 3 a2 − 1 − 3
(ApplyingC 3 → C 3 − C 2 )
=
2
5
0
5
−1
2
a + 1 3 − (a + 1) a2 − 1 − 3
2
0
3−
1
(ApplyingC 2 → C 2 − C1)
2
2
=
5
0
1
2
5 a
a+1
−
2 2
1
= 2 (a2 − 4) +
 2


− e sint + e cost  then A is

−2e −t cost

−t
−t
[JEE Main 2019, 9 Jan Shift-II]
(a) invertible only when t = π
(b) invertible for every t ∈ R
(c) not invertible for any t ∈ R
π
(d) invertible only when t =
2
Exp. (b)
(Applying C 2 → C 2 − C1 and C 3 → C 3 − C1)
= a2 − 3 (Expanding along R1)
and
et
e −t cost
 t
If A = e −e −t cost − e −t sint
et
2e −t sint

e −t sint
0
−1
a −4
2
 5 − a   (Expanding along R )


1
 2 2  
 a2
5 a
= 2
−2 + − 
2 2
2
= a2 − 4 + 5 − a
= a2 − a + 1
Clearly, when a = 4, then D = 13 ≠ 0 ⇒ unique
solution and
when|a| = 3, then D = 0 and D1 ≠ 0.
∴When|a| = 3, then the system has no solution
i.e. system is inconsistent.
et
| A | = et
et
e − t cos t
e − t sint
− e − t cos t − e − t sin t
2e − t sin t
− e − t sint + e −t cos t
− 2e − t cos t
1
cos t
= (et ) (e − t ) (e − t ) 1 − cos t − sin t
1
2 sin t
sint
− sin t + cos t
− 2 cos t
(taking common from each column)
Aplying R 2 → R 2 − R1 and R 3 → R 3 − R1, we get
cos t
sin t
1
−t
=e
0 − 2 cos t − sint − 2 sin t + cos t
0 2 sint − cos t
− 2 cos t − sin t
[Qet
=e
−t
−t
= e 0 = 1]
((2 cos t + sin t ) + (2 sin t − cos t ) )
2
2
(expanding along column 1)
= e − t (5 cos 2 t + 5sin2 t )
= 5e − t
(Qcos 2 t + sin2 t = 1)
−t
for allt ∈ R
⇒
| A | = 5e ≠ 0
∴ A is invertible for all t ∈ R.
[Qif| A | ≠ 0, then A is invertible]
19. If the system of linear equations
x − 4y + 7z = g
3y − 5z = h
− 2 x + 5y − 9z = k
is consistent, then
(a) 2 g + h + k = 0
(b) g + 2h + k = 0
(c) g + h + k = 0
(d) g + h + 2k = 0
[JEE Main 2019, 9 Jan Shift-II]
50
JEE Main Chapterwise Mathematics
Exp. (a)
Here, D =
1
0
21. Let d ∈R , and
−4 7
3 −5
−2
5
−9
= 1(− 27 + 25) + 4(0 − 10) + 7(0 + 6)
[expanding along R1]
= − 2 − 40 + 42 = 0
∴ The system of linear equations have infinite
many solutions.
[Qsystem is consistent and does not have
unique solution as D = 0]
⇒
D1 = D2 = D3 = 0
g −4 7
Now, D1 = 0 ⇒ h 3 − 5 = 0
k
5
4+d
(sin θ ) − 2 
−2

,
A= 1
(sin θ ) + 2
d


 5 (2 sin θ ) − d ( − sin θ ) + 2 + 2d 
θ ∈[θ , 2 π]. If the minimum value of det(A) is
8, then a value of d is
[JEE Main 2019, 10 Jan
Shift-I]
(a)
(b)
(c)
(d)
−5
−7
2( 2 + 1)
2( 2 + 2 )
Exp. (a)
Given,
4+d
(sinθ) − 2 
 −2

A =  1 (sinθ) + 2
d


 5 (2 sinθ) − d (− sinθ) + 2 + 2d 
−9
⇒ g (− 27 + 25) + 4(− 9h + 5k ) + 7(5h − 3k ) = 0
⇒
− 2 g − 36h + 20k + 35h − 21k = 0
⇒
− 2g − h − k = 0
⇒
2g + h + k = 0
20. If the system of equations
−2
4+d
(sinθ) − 2
∴| A| = 1 (sinθ) + 2
d
5 (2 sinθ) − d (− sinθ) + 2 + 2d
−2
4+d
(sinθ) − 2
= 1 (sinθ) + 2
d
x + y + z = 5, x + 2 y + 3z = 9, x + 3y + αz = β
has infinitely many solutions, then β − α
equals
[JEE Main 2019, 10 Jan Shift-I]
(a) 8
(c) 21
1
Exp. (a)
Since, the system of equations has infinitely many
solution, therefore D = D1 = D2 = D3 = 0
1 1 1
D= 1 2
3 = 1(2α − 9) − 1 (α − 3) + 1(3 − 2 )
1 3 α
=α − 5
1 1 5
and D3 = 1 2 9
1 3 β
= 1 (2 β − 27 ) − 1(β − 9) + 5(3 − 2 )
= β − 13
Now, D = 0
⇒
α−5=0 ⇒ α=5
and
D3 = 0 ⇒ β − 13 = 0
⇒
∴
β = 13
β − α = 13 − 5 = 8
0
(R 3 → R 3 − 2 R 2 + R1 )
= 1 [(4 + d )d − (sinθ + 2 ) (sinθ − 2 )]
(expanding along R 3 )
= (d 2 + 4d − sin2 θ + 4)
= (d 2 + 4d + 4) − sin2 θ
= (d + 2 )2 − sin2 θ
(b) 18
(d) 5
Here,
0
Note that| A| will be minimum if sin2 θ is maximum
i.e. if sin2 θ takes value 1.
Q
| A|min = 8,
therefore
(d + 2 )2 − 1 = 8
⇒
(d + 2 )2 = 9
⇒
⇒
d+2=± 3
d = 1, − 5
2
b

1
b
1
22. Let A = b b 2 + 1 b  , whereb > 0. Then, the

2
det ( A )
minimum value of
is
b
[JEE Main 2019, 10 Jan Shift-II]
(a) − 3
(c) 2 3
(b) −2 3
(d) 3
51
Matrices and Determinants
and (sin 3θ)x + (cos 2 θ)y + 2 z = 0
has non-trivial solution, then
1
3
7
−1
4
7 =0
Exp. (c)
Given matrix
1
b
b 2 + 1 b , b > 0

2 
b
2
b
1
2
A = b

 1
det ( A) =| A| = b
1
So,
b2 + 1
b
sin 3θ cos 2 θ 2
b
2
= 2 [2(b 2 + 1) − b 2 ] − b(2 b − b ) +1(b 2 − b 2 − 1)
= 2[2 b 2 + 2 − b 2 ] − b 2 − 1
= 2b + 4 − b − 1= b + 3
det( A) b 2 + 3
3
=
= b+
⇒
b
b
b
Now, by AM ≥ GM, we get
3
b+
1/ 2
b ≥  b × 3 


b
2
3
b+ ≥2 3
⇒
b
det ( A)
So, minimum value of
=2 3
b
2
2
⇒ − 7 (3 sin θ − 4sin3 θ) − 14(1 − 2 sin2 θ) +14 = 0
2
[ Qsin 3 A = 3 sin A − 4 sin3 A and
cos 2 A = 1 − 2 sin2 A]
⇒ 28 sin3 θ + 28 sin2 θ − 21 sinθ − 14 + 14 = 0
{Q b > 0}
23. The number of values of θ ∈(0, π ) for which
the system of linear equations
x + 3y + 7z = 0,
− x + 4y + 7z = 0,
(sin 3θ )x + (cos 2θ )y + 2 z = 0
has a non-trivial solution, is
[JEE Main 2019, 10 Jan Shift-II]
(a) two
(c) four
⇒ 1(8 − 7 cos 2 θ) − 3 (− 2 − 7 sin 3θ)
+ 7 (− cos 2 θ − 4 sin 3θ) = 0
⇒ 8 − 7 cos 2 θ + 6 + 21 sin 3θ
− 7 cos 2 θ − 28 sin 3θ = 0
⇒ − 7 sin 3θ − 14 cos 2 θ + 14 = 0
(b) three
(d) one
Exp. (a)
We know that,
the system of linear equations
a1 x + b1 y + c1 z = 0
⇒
7 sin θ [4 sin2 θ + 4 sinθ − 3] = 0
⇒
sin θ [4sin2 θ + 6sinθ − 2 sinθ − 3] = 0
⇒ sin θ [2 sin θ (2 sin θ + 3) − 1 (2 sin θ + 3)] = 0
⇒
(sin θ) (2 sin θ − 1) (2 sin θ + 3) = 0
1
Now, either sin θ = 0 or
2
Qsin θ ≠ − 3 as − 1 ≤ sinθ ≤ 1


2
In given interval (0, π ),
1
π 5π
sin θ = ⇒ θ = ,
2
6 6
[Qsin θ ≠ 0, θ ∈ (0, π )]
Hence, 2 solutions in (0, π )
24. Let a1 , a 2 , a 3.....,a10 be in GP with ai > 0 for
i = 1, 2 ,.....,10 and S be the set of pairs (r , k ),
r , k ∈ N (the set of natural numbers) for
which
log e a1r a k2 log e a r2a k3 log e a r3 a k4
log e a r4 a k5 log e a r5 a k6 log e a r6 a k7 = 0
log e a r7 a k8
a2 x + b2 y + c 2 z = 0
a3 x + b3 y + c 3 z = 0
has a non-trivial solution, if
a1 b1 c1
a2
a3
b2 c 2 = 0
b3 c 3
log e a r8 a k9
k
log e a r9 a10
Then, the number of elements in S, is
[JEE Main 2019, 10 Jan Shift-II]
(a) 4
(c) 10
(b) 2
(d) infinitely many
Exp. (d)
x + 3y + 7 z = 0
loge a1r a2k
Given, loge a4r a5k
loge a2r a3k
loge a5r a6k
loge a3r a4k
loge a6r a7k = 0
− x + 4 y + 7 z = 0,
loge a7r a8k
loge a8r a9k
k
loge a9r a10
Now, if the given system of linear equations
52
JEE Main Chapterwise Mathematics
On applying elementary operations
C 2 → C 2 − C1 and C 3 → C 3 − C1, we get
loge a1r a2k
loge a4r a5k
loge a7r a8k
⇒
25. If the system of linear equations
− loge a1r a2k
− loge a4r a5k
− loge a7r a8k
loge a3r a4k − loge a1r a2k
loge a6r a7k − loge a4r a5k
k
loge a9r a10
− loge a7r a8k
2 x + 2 y + 3z = a
3x − y + 5z = b
x − 3y + 2 z = c
where a ,b , c are non-zero real numbers, has
more than one solution, then
loge a2r a3k
loge a5r a6k
loge a8r a9k
=0
[JEE Main 2019, 11 Jan Shift-I]
(a) b − c − a = 0
(c) b − c + a = 0
 ar ak 
 ar ak 
loge a1r a2k loge  2 3  loge  3 4 
 ar ak 
 ar ak 
 1 2
 1 2
r
k


 ar ak 
a
a
loge a4r a5k loge  5r 6k  loge  6r 7k  = 0
 a4 a5 
 a4 a5 
k 
 a8r a9k 
 a9r a10
loge a7r a8k loge  r k  loge  r k 
 a7a8 
 a7a8 
Exp. (a)
We know that, if the system of equations
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c 2 z = d 2
a3 x + b3 y + c 3 z = d 3
has more than one solution, then D = 0 and
D1 = D2 = D3 = 0. In the given problem,
a 2 3
D1 = 0 ⇒ b − 1 5 = 0
m 

Qloge m − loge n = loge  

 n  
[Qa1, a2 , a3 ......., a10 are in GP, therefore put
a1 = a, a2 = aR, a3 = aR 2 , ..., a10 = aR 9 ]
loge ar
+ k
Rk
⇒ loge ar
+ k
R3r
+ 4k
loge ar
+ k
R6r
+ 7k
26.

a
R

loge 
r + k k
R

 a
 ar + k R 5 r + 6 k 
loge  r + k 3 r + 4 k  = 0
R

a
 ar + k R 8 r + 9 k 
loge  r + k 6 r + 7k 
R

a
r + k
k
r + k
R ) loge R
loge (a
r + k 3r + 4k
loge R r
⇒ loge a R
r + k 6 r + 7k
loge a R
loge R r
+ k
+ k
loge (ar + k R k ) loge R r
⇒ loge (ar + k R 3 r + 4 k ) loge R r
loge (ar + k R 6 r + 7k ) loge R r
loge R
2r + 2k
2r + 3k
2r + 2k
loge R
loge R 2 r
loge R 2 r
+ 2k
+ k
+
+ k
+
=0
+ 2k
+ k
2 loge R r
k
2 loge R r
k
2 loge R r
c
−3 2
⇒ a (− 2 + 15) − 2(2 b − 5c ) + 3(− 3b + c ) = 0
⇒
13a − 4b + 10c − 9b + 3c = 0
⇒
13a − 13b + 13c = 0
⇒
a − b + c = 0⇒b − a − c = 0
 ar + k R r + 2k 

loge 
r + k k
R

 a
 ar + k R 4 r + 5 k 
loge  r + k 3 r + 4 k 
R

a
 ar + k R 7r + 8 k 
loge  r + k 6 r + 7k 
R

a
r + k
(b) a + b + c = 0
(d) b + c − a = 0
=0
+ k
[Qlog mn = n log m and here
= loge R 2( r + k ) = 2 loge R r + k ]
Q Column C 2 and C 3 are proportional,
So, value of determinant will be zero for any value
of (r, k ), r, k ∈ N.
∴Set ‘S’ has infinitely many elements.
 0 2q

Let A = p q
 p −q

r

−r  . If AA T = I 3, then |p | is
r 
[JEE Main 2019, 11 Jan Shift-I]
1
(a)
5
1
(b)
2
(c)
1
3
(d)
1
6
Exp. (b)
Given,
0
⇒p

 p
AAT
2q
q
−q
=I
r
−r

r 
p p   1 0 0
q −q  =  0 1 0

 
− r r   0 0 1
0
2q

 r
 0 + 4q 2 + r 2

⇒  0 + 2q 2 − r 2
 0 − 2q 2 + r 2

0 + 2q 2 − r 2
p +q + r
p2 − q 2 − r 2
2
2
2
0 − 2q 2 + r 2 

p2 − q 2 − r 2 
p2 + q 2 + r 2 

 1 0 0
=  0 1 0


 0 0 1
We know that, if two matrices are equal, then
corresponding elements are also equal, so
… (i)
4q 2 + r 2 = 1 = p2 + q 2 + r 2 ,
53
Matrices and Determinants
2q 2 − r 2 = 0 ⇒ r 2 = 2q 2
… (ii)
p2 − q 2 − r 2 = 0
and
2c
… (iv)
∆=
[using Eq. (iv)]
and
order
If
3 × 3.
det( ABA T ) = 8
−1
−1 T
det( AB ) = 8, then det(BA B ) is equal to
[JEE Main 2019, 11 Jan Shift-II]
1
4
(b)
1
(c)
16
(d) 16
⇒
[Q| XY| = | X||Y|]
| A|2 |B| = 8
…(i)
[Q| AT| = | A|]
⇒
| AB−1| = 8 ⇒| A||B−1| = 8
| A|
=8
|B|
…(ii)

1
−1
−1
Q| A |=| A| = | A|


On multiplying Eqs. (i) and (ii), we get
| A|3 = 8 ⋅ 8 = 43
⇒
| A| = 4
| A| 4 1
|B| =
= =
⇒
8
8 2
1 1 1
1
1
−1 T
Now, |BA B | = |B| |B| =     =
 2  4  2  16
| A|
28.
a −b − c
2a
If
2b
b −c −a
2c
2c
2a
2b
c − a −b
= (a + b + c )( x + a + b + c )2, x ≠ 0 and
a + b + c ≠ 0, then x is equal to
[JEE Main 2019, 11 Jan Shift-II]
(a) − (a + b + c )
(c) 2(a + b + c )
2b
c−a−b
2c
1
2b
c−a−b
2c
(taking common (a + b + c ) from R1)
Applying C 2 → C 2 − C1 and C 3 → C 3 − C1, we get
∆
0
0
−(a + b + c )
0
Now, expanding along R1, we get
∆ = (a + b + c ) 1. {(a + b + c )2 − 0 }
| A||B|| AT| = 8
Also, we have
b−c − a
2c
2c
| ABAT| = 8
∴
2b
2c
1
0
= (a + b + c ) 2 b −(a + b + c )
Exp. (c)
Given,
c−a−b
2c
1
1
= (a + b + c ) 2 b b − c − a
27. Let A and B be two invertible matrices of
(a) 1
2a
2b
Applying R1 → R1 + R 2 + R 3 , we get
a+ b+c a+ b+c a+ b+c
Using Eqs. (ii) and (iv) in Eq. (i), we get
4q 2 + 2q 2 = 1
6q 2 = 1 ⇒ 2 p2 = 1
1
1
p2 = ⇒ | p| =
2
2
a− b−c
2a
2b
b−c − a
Let ∆ =
Using Eqs. (ii) and (iii), we get
p2 = 3q 2
⇒
Exp. (b)
… (iii)
(b) − 2(a + b + c )
(d) abc
= (a + b + c )3 = (a + b + c )( x + a + b + c )2
(given)
⇒
( x + a + b + c )2 = (a + b + c )2
⇒
⇒
x + a + b + c = ± (a + b + c )
x = − 2(a + b + c )
[Q x ≠ 0]
29. An ordered pair(α , β ) for which the system of
linear equations
(1 + α )x + βy + z = 2
αx + (1 + β )y + z = 3
ax + βy + 2 z = 2
has a unique solution, is
[JEE Main 2019, 12 Jan Shift-I]
(b) ( − 4, 2 )
(d) ( −3,1)
(a) (2, 4)
(c) (1, − 3)
Exp. (a)
Given system of linear equations,
(1 + α )x + βy + z = 2
αx + (1 + β )y + z = 3
αx + βy + 2 z = 2
has a unique solution, if
1+ α
β
α
α
1
(1 + β ) 1 ≠ 0
β
2
54
JEE Main Chapterwise Mathematics
Apply R1 → R1 − R 3 and R 2 → R 2 − R 3
1 0 −1
0 1 −1 ≠ 0
α β
and
2
⇒
⇒ 12
( + β ) − 0(0 + α ) − 1(0 − α ) ≠ 0
⇒
α+β+2≠0
Note that, only (2, 4) satisfy the Eq. (i)
30.
Hence,
… (i)
1 0 0
Let P = 3 1 0 and Q =[qij ] be two 3 × 3


9 3 1
matrices such that Q − P 5 = I 3. Then,
q 21 + q 31
is equal to
q 32
of
linear
equations
x − 2 y − 2 z = λx ,
x + 2 y + z = λy and − x − y = λz
has a non-trivial solution
[JEE Main 2019, 12 Jan Shift-II]
(a)
(b)
(c)
(d)
(b) 135
(d) 15
The given system of linear equations is
x − 2 y − 2 z = λx
x + 2 y + z = λy
− x − y − λz = 0,
which can be rewritten as
(1 − λ )x − 2 y − 2 z = 0
⇒
x + (2 − λ ) y + z = 0
x + y + λz = 0
Now, for non-trivial solution, we should have
1− λ − 2 − 2
1
2−λ 1 =0
Given matrix
 1 0 0  0 0 0  1 0 0
P =  3 1 0 =  3 0 0 +  0 1 0

 
 

 9 3 1  9 3 0  0 0 1
⇒
P = X + I (let)
Now, P 5 = (I + X )5
= I + 5C1( X ) + 5C 2 ( X 2 ) + 5C 3 ( X 3 ) + …
[QI n = I, I ⋅ A = A and (a + x)n = nC 0 an +
n
C1an − 1 x + ...+T nC n xn ]
1
 0 0 0  0 0 0  0 0 0
X =  3 0 0  3 0 0 =  0 0 0

 


 9 3 0  9 3 0  9 0 0
λ
b1 y + c1 z = 0; a2 x + b2 y + c 2 z = 0
a3 x + b3 y + c 3 z = 0]
a1 b1 c1

has a non-trivial solution, then a2 b2 c 2 = 0

a3 b3 c 3

and
 0 0 0  0 0 0  0 0 0
X 3 = X 2 ⋅ X =  0 0 0  3 0 0 =  0 0 0

 


 9 0 0  9 3 0  0 0 0
 0 0 0
⇒ X 4 = X 5 =  0 0 0


 0 0 0
So,
1
[Q If a1 x +
2
 0 0 0
 0 0 0


P = I + 5 3 0 0 + 10  0 0 0




 9 0 0
 9 3 0
0 0
 1

= 15 1 0


135 15 1
contains exactly two elements
contains more than two elements
is a singleton
is an empty set
Exp. (c)
Exp. (a)
5
q 21 + q 31 15 + 135 150
=
=
= 10
q 32
15
15
31. The set of all values of λ for which the system
[JEE Main 2019, 12 Jan Shift-I]
(a) 10
(c) 9
Here,
0 0
 2

Q = I + P = 15 2 0 = [q ij ]


135 15 2 
q 21 = 15, q 31 = 135 and q 32 = 15
5
⇒ (1 − λ ) [(2 − λ )λ − 1] + 2 [λ − 1]
− 2 [1 − 2 + λ ] = 0
⇒ (λ − 1)[λ2 − 2 λ + 1 + 2 − 2 ] = 0
⇒
(λ −1)3 = 0 ⇒ λ = 1
32.
sin θ
1 
 1

If A = − sin θ
1
sin θ; then for all


− sin θ
1 
 − 1
 3π 5π 
,  , det( A ) lies in the interval
θ ∈
 4 4
[JEE Main 2019, 12 Jan Shift-II]
55
Matrices and Determinants
3
(a)  , 3
 2 
3
(c)  0, 
 2 
5
(b)  , 4
 2 
5
(d) 1, 
 2 
Apply R 2 → R 2 − R1 and R 3 → R 3 − R1
2x
0 
1

0 = ( A + Bx)( x − A)2
∴(5 x − 4) 0 − x − 4


0
− x − 4
0
Expanding along C1, we get
(5 x − 4)( x + 4)2 = ( A + Bx)( x − A)2
Exp. (a)
sinθ
1 
 1
Given matrix A =  − sinθ
1 sinθ


− sinθ 1 
 −1
1
sinθ
1
1
sinθ
⇒ det( A) =| A|= − sinθ
−1
− sinθ
Equating, we get
A = − 4 and B = 5
34. If the system of linear equations
x + ky + 3z = 0
3x + ky − 2 z = 0
2 x + 4y − 3z = 0
1
= 11
( + sin2 θ) − sinθ(− sinθ + sinθ) + 1(sin2 θ + 1)
⇒ | A| = 2 (1 + sin2 θ)
…(i)
3π 5π 
As we know that, for θ ∈ 
,

 4 4 
1 1 
sinθ ∈  −
,


2 2
1
1
⇒ sin2 θ ∈  0,  ⇒ 1 + sin2 θ ∈  0 + 1, + 1
 2 


2
3

2
⇒ 1 + sin θ ∈ 1, 
 2 
3 
⇒ 2(1 + sin2θ) ∈ [2, 3) ⇒| A| ∈ [2, 3) ⊂  , 3
 2 
33.
2x 
x − 4 2 x
If 2 x
x − 4 2 x = ( A + Bx )( x − A )2,


2x
x − 4
 2x
then the ordered pair ( A , B ) is equal to
[JEE Main 2018]
(a) ( −4, − 5)
(c) ( −4, 5)
(b) ( −4, 3)
(d) ( 4, 5)
Exp. (c)
Given,
2x 
x − 4 2 x
 2x
x − 4 2 x  = ( A + Bx)( x − A)2


x − 4
2x
 2x
⇒ Apply C1 → C1 + C 2 + C 3
2x 
 5x − 4 2 x
 5 x − 4 x − 4 2 x  = ( A + Bx)( x − A)2


x − 4
 5x − 4 2 x
Taking common (5 x − 4) from C1, we get
2x 
1 2 x
(5 x − 4
) 1 x − 4 2 x  = ( A + Bx)( x − A)2


x − 4
1 2 x
has a non-zero solution ( x , y , z ), then
equal to
(a) −10
xz
y2
is
[JEE Main 2018]
(b) 10
(c) −30
(d) 30
Exp. (b)
We have, x + ky + 3 z = 0; 3 x + ky − 2 z = 0;
2 x + 4y − 3z = 0
System of equation has non-zero solution, if
1 k 3 
 3 k −2  = 0


 2 4 −3 
⇒(−3k + 8) − k(−9 + 4) + 3(12 − 2 k ) = 0
⇒
−3k + 8 + 9k − 4k + 36 − 6k = 0
⇒
−4k + 44 = 0 ⇒ k = 11
Let
z = λ , then we get
x + 11y + 3λ = 0
3 x + 11y − 2 λ = 0
and
2 x + 4 y − 3λ = 0
Solving Eqs. (i) and (ii), we get
5λ
−λ
, z=λ
, y=
x=
2
2
5λ2
xz
= 10
=
⇒
2
2
y
λ
2 ×  − 
 2
…(i)
…(ii)
…(iii)
35. If S is the set of distinct values ofb for which
the following system of linear equations
x + y + z = 1,
x + ay + z = 1
and
ax + by + z = 0
has no solution, then S is
[JEE Main 2017 (Offline)]
56
JEE Main Chapterwise Mathematics
(a) an infinite set
(b) a finite set containing two or more elements
(c) singleton set
(d) an empty set
Exp. (d)
1 1 1
Q ∆ = 1 a 1 = 1 (a − b ) − 1 (1 − a) + 1 (b − a2 )
a b 1
⇒
⇒
⇒ 5k 2 + 13k − 66 = 0 or 5k 2 + 13k − 46 = 0
[Qk ∈ I]
⇒
k =2
Thus, the coordinates of vertices of triangle are
A(2, − 6), B(5, 2 ) and C(− 2, 2 ).
Y
C (–2, 2)
= − (a − 1)2
1 1 1
= 1 (a − b ) − 1 (1) + 1 (b )
∆1 = 1 a 1
0 b 1
= − (a − 1)
1 1 1
∆ 2 = 1 1 1 = 1 (1) − 1 (1 − a) + 1 (0 − a) = 0
a 0 1
= − a(a − 1)
a=1
∆ = ∆1 = ∆ 2 = ∆ 3 = 0
∆ for b = 1only
x + y + z = 1, x + y + z = 1 and x + y + z = 0
i.e. no solution
(QRHS is not equal)
Hence, for no solution b = 1only
36. Let k be an integer such that the triangle
with vertices (k , − 3k ), (5, k ) and ( − k , 2 ) has
area 28 sq units. Then, the orthocentre of
this triangle is at the point
[JEE Main 2017 (Offline)]
1
3
(a)  2 , −  (b) 1, 

 4
2
3
1
(c) 1, −  (d)  2 , 

 2
4
Exp. (d)
Given, vertices of triangle are (k, − 3k ), (5, k ) and
(− k, 2 ).
k − 3k 1
1
5
k
1 = ± 28
∴
2
−k
2
1
k − 3k 1
⇒
5
k
1 = ± 56
−k
2
1
⇒
k(k − 2 ) + 3k(5 + k ) + 110
( + k ) = ± 56
2
D
B (5, 2)
(2, 1/2)
E
X′
X
O
A (2, –6)
Y′
Now, equation of altitude from vertex A is
1 1 1
and ∆ 3 = 1 a 1 = 1 ( − b ) − 1 (− a) + 1 (b − a2 )
a b 0
For
k 2 − 2 k + 15k + 3k 2 + 10 + k 2 = ± 56
5k 2 + 13k + 10 = ± 56
y − (− 6) =
−1
( x − 2) ⇒ x = 2
 2 −2 


 − 2 − 5
…(i)
Equation of altitude from vertex C is
−1
y−2 =
[ x − (− 2 )]
 2 − (− 6)
 5−2 


⇒
…(ii)
3 x + 8 y − 10 = 0
1
On solving Eqs. (i) and (ii), we get x = 2 and y = .
2
1
∴ Orthocentre =  2, 
 2
 2 −3
2
 , then adj( 3A + 12 A ) is equal
−4 1 
37. If A = 
to
[JEE Main 2016 (Offline)]
 72 − 84
(a) 

 − 63 51 
 51 84 
(c) 

63 72 
 51 63 
(b) 

84 72 
 72 − 63
(d) 

 − 84 51 
Exp. (b)
 2 − 3
We have, A = 

− 4 1 
∴
 2 − 3  2 − 3
A2 = A ⋅ A = 


− 4 1  − 4 1 
 4 + 12 − 6 − 3  16 − 9
=

= 
 − 8 − 4 12 + 1   − 12 13 
57
Matrices and Determinants
 2 − 3
 16 − 9
Now, 3 A 2 + 12 A = 3 
+ 12 


− 4 1 
 − 12 13 
 48 − 27   24 − 36  72 − 63
=

= 
+ 
 − 36 39   − 48 12   − 84 51 
 51 63
∴adj (3 A 2 + 12 A) = 

 84 72 
5a − b 
T
 and A adj A = AA , then
3
2


[JEE Main 2016 (Offline)]
5a + b is equal to
38. If A = 
(a) −1
(c) 4
(b) 5
(d) 13
Exp. (b)
 5a − b 
and A adj A = AAT
Given, A = 
2 
3
Clearly, A(adj A) = A I2
[Q if A is square matrix of
order n, then A(adj A) = (adj A) ⋅ A = A In ]
5a − b
=
I2 = (10a + 3b ) I2
3
2
 1 0
= (10a + 3b ) 

 0 1
0
10a + 3b

=
0
10
a
+ 3b 

 5a − b   5a 3
and AAT = 
2   − b 2 
3
25a2 + b 2 15a − 2 b 
=

13 
 15a − 2 b
39. The system of linear equations
x + λy − z = 0; λx − y − z = 0; x + y − λz = 0
has a non-trivial solution for
[JEE Main 2016 (Offline)]
(a) infinitely many values of λ
(b) exactly one value of λ
(c) exactly two values of λ
(d) exactly three values of λ
Exp. (d)
Given, system of linear equations is
x + λy − z = 0; λx − y − z = 0; x + y − λz = 0
Note that, given system will have a non-trivial
solution only if determinant of coefficient matrix is
1 λ −1
zero, i.e.
λ −1 −1 = 0
1
1
−λ
⇒ 1 (λ + 1) − λ(− λ + 1) − 1(λ + 1) = 0
2
...(i)
...(ii)
Q A(adj A) = AAT
0
10a + 3b

∴ 

0
10
a
+
3
b


25a2 + b 2 15a − 2 b 
=

13 
 15a − 2 b
[using Eqs. (i) and (ii)]
15a − 2 b = 0
2b
...(iii)
a=
⇒
15
and
...(iv)
10a + 3b = 13
On substituting the value of ‘a’ from Eq. (iii) in Eq.
(iv), we get
2b
10 ⋅   + 3b = 13
 15 
20b + 45b
= 13
⇒
15
⇒
65b
= 13
15
⇒
b=3
Now, substituting the value of b in Eq. (iii), we get
5a = 2
Hence,
5a + b = 2 + 3 = 5
⇒
⇒
λ + 1 + λ3 − λ − λ − 1 = 0
⇒
λ3 − λ = 0
⇒
λ(λ2 − 1) = 0
⇒
λ = 0 or λ = ± 1
Hence, given system of linear equation has a
non-trivial solution for exactly three values of λ.
1 2
2
40. If A = 2 1 −2 is a matrix satisfying the


a 2 b 
equation AA T = 9I , where I is 3 × 3 identity
matrix, then the ordered pair(a ,b ) is equal to
(a) ( 2 , − 1)
(b) ( −2 ,1)
(c) ( 2 ,1)
(d) ( −2 , − 1)
[JEE Main 2015]
Exp. (d)
Given,
1
A = 2

 a
1
AT = 2

2
2 2 
1 −2 

2 b 
2 a
1 2

−2 b 
58
JEE Main Chapterwise Mathematics
1 2 2  1 2 a
AA = 2 1 −2  2 1 2 



 a 2 b  2 −2 b 
a + 4 + 2b 
9
0

0
9
2a + 2 − 2b
=


2
2
 a + 4 + 2 b 2 a + 2 − 2 b a + 4 + b 
T
It is given that
AAT = 9I
9
0

⇒ 
0
9

 a + 4 + 2 b 2 a + 2 − 2 b
1 0
= 9 0 1

 0 0
⇒
a + 4 + 2b 
2a + 2 − 2b

a2 + 4 + b 2 
0
0

1
a + 4 + 2b 
9
0


0
9
2a + 2 − 2b


2
2
 a + 4 + 2 b 2 a + 2 − 2 b a + 4 + b 
 9 0 0
=  0 9 0


 0 0 9
On comparing, we get
a + 4 + 2b = 0 ⇒ a + 2b = − 4
2a + 2 − 2b = 0 ⇒ a − b = − 1
and
a2 + 4 + b 2 = 9
Since, the system has non-trivial solution.
2−λ
−2
1
∴
2
− (3 + λ ) 2 = 0
−1
⇒ (2 − λ )(3λ + λ − 4) + 2(−2 λ + 2 )
+ 1(4 − 3 − λ) = 0
⇒ (2 − λ )(λ2 + 3λ − 4) + 4(1 − λ ) + (1 − λ) = 0
⇒
⇒
⇒
(2 − λ )(λ + 4)(λ − 1) + 5(1 − λ ) = 0
(λ − 1)[(2 − λ )(λ + 4) − 5] = 0
(λ − 1)(λ2 + 2 λ − 3) = 0
⇒
⇒
(λ − 1)[(λ − 1)(λ + 3)] = 0
(λ − 1)2 (λ + 3) = 0
⇒
42. If α , β ≠ 0, f (n ) = αn + βn and
3
1 + f (1) 1 + f (2 )
1 + f (1) 1 + f (2 ) 1 + f ( 3)
1 + f (2 ) 1 + f ( 3) 1 + f ( 4)
= K (1 − α )2 (1 − β )2 (α − β )2, thenK is equal to
[JEE Main 2014]
1
αβ
(d) −1
(b)
(c) 1
/ Two determinants can be multiplied row-to-row
or row-to-column.
f (n) = α n + β n
f (1) = α + β,
f (2) = α 2 + β 2 , f (3) = α 3 + β 3 , f (4) = α 4 + β 4
On solving Eqs. (i) and (ii), we get
a = − 2, b = − 1
This satisfies Eq. (iii)
Hence, (a, b ) ≡ (−2, − 1)
41. The set of all values of λ for which the system
of linear equations 2 x1 − 2 x 2 + x 3 = λx1,
2 x1 − 3x 2 + 2 x 3 = λx 2 and − x1 + 2 x 2 = λx 3 has
a non-trivial solution,
[JEE Main 2015]
(a) is an empty set
(b) is a singleton set
(c) contains two elements
(d) contains more than two elements
Exp. (c)
1 + f(1) 1 + f(2 )
3
Let ∆ = 1 + f(1) 1 + f(2 ) 1 + f(3)
1 + f(2 ) 1 + f(3) 1 + f(4)
3
⇒ ∆=
=
…(i)
…(ii)
…(iii)
1+ α + β
1 + α2 + β2
1 + α + β 1 + α2 + β2 1 + α3 + β3
1 + α2 + β2 1 + α3 + β3 1 + α4 + β4
1⋅ 1 + 1⋅ 1 + 1⋅ 1
Exp. (c)
Given system of linear equations
2 x1 − 2 x2 + x3 = λx1
⇒
(2 − λ )x1 − 2 x2 + x3 = 0
2 x1 − 3 x2 + 2 x3 = λx2
⇒
2 x1 − (3 + λ )x2 + 2 x3 = 0
− x1 + 2 x2 = λx3
⇒
− x1 + 2 x2 − λx3 = 0
λ = 1, 1, − 3
(a) αβ
…(i)
…(ii)
…(iii)
−λ
2
2
1⋅ 1 + 1⋅ α + 1⋅ β
1⋅ 1 + α ⋅ 1 + β ⋅ 1
1⋅ 1 + α ⋅ α + α ⋅ β
2
2
1⋅ 1 + 1⋅ α + 1⋅ β 1⋅ 1 + α 2 ⋅ α + β 2 ⋅ β
1⋅ 1 + 1⋅ α 2 + 1⋅ β 2
1⋅ 1 + α ⋅ α 2 + β ⋅ β 2
1⋅ 1 + α 2 ⋅ α 2 + β 2 ⋅ β 2
59
Matrices and Determinants
1
1
1
1
1
1
1
1
1
2
Now,
= 1 α β 1 α β = 1 α β
1 α2 β2 1 α2 β2
1 α2 β2
On expanding, we get
∆ = (1 − α )2 (1 − β )2 (α − β 2 )
= (1 − α ) (1 − β ) (α − β )
∴
2
2
K =1
43. If A an 3 × 3 non-singular matrix such that
AA ′ = A ′ A and B = A −1A ′, then BB ′ is equal
to
[JEE Main 2014]
(a) I + B
(c) B −1
/ (AB )T
= B T AT and A −1A = l
(where, AT = A ′ = Transpose of A)
Exp. (b)
−1 T
−1
[Q AAT = AT A ]
−1 T
⇒
k 2 − 4k + 3 = 0
(k − 1) (k − 3) = 0
k = 1 or = 3
k + 3 − 8 
Now,
adj A = 

 − k k + 1
 k + 3 − 8   4k 
Now, (adj A) B = 


 − k k + 1  3k − 1
(k + 3) (4k ) − 8 (3k − 1)
=
2

 − 4k + (k + 1) (3k − 1)
 4k 2 − 12 k + 8
=

2
 − k + 2k − 1
Put k = 1
T
T
(not true)
 36 − 36 + 8  8 
(adj A) B = 
= ≠0
 − 9 + 6 − 1   – 4
(true)
Hence, required value of k is 1.
Alternate Solution
[Q( AB) = B A ]
= ( A A)
T
T
= IT = I
44. The number of values of k, for which the
system of equations
(k + 1)x + 8y = 4k
kx + (k + 3)y = 3k − 1
has no solution, is
[JEE Main 2013]
(a) ∞
(c) 2
(k + 1) (k + 3) − 8k = 0
k 2 + 4k + 3 − 8k = 0
[Q( A′ )′ = A]
= IA ( A ) = A ( A )
T
=0
Put k = 3
BB′ = ( A −1 AT ) ( A −1 AT )T
T
k+ 3
 4 − 12 + 8  0
(adj A) B = 
= 
 − 1 + 2 − 1  0
If A is non-singular matrix, then| A| ≠ 0.
AAT = AT A and B = A −1 AT
= A −1 AAT ( A −1 )T
8
k
⇒
⇒
(b) I
(d) ( B −1 )′
= A −1 AT A ( A −1 )T
k+1
⇒
⇒
Hence, K (1 − α )2 (1 − β )2 (α − β )2
2
| A| =
(b) 1
(d) 3
Exp. (b)
Condition for the system of equations has no
solution.
a1
b
c
= 1 ≠ 1
a2 b2 c 2
∴
k+1
8
4k
=
≠
k
k + 3 3k − 1
Take
k+1
8
=
k
k+ 3
⇒
k 2 + 4k + 3 = 8k
⇒
k 2 − 4k + 3 = 0
⇒
(k − 1) (k − 3) = 0
Given equations can be written in matrix form
AX = B
8 
 x
k + 1
where, A = 
, X =  y
3
+
k
k
 


4 k

and
B= 

3 k − 1 
For no solution, | A | = 0 and (adj A) B ≠ 0
If k = 1, then
If k = 3, then
k = 1, 3
8
4⋅1
≠
1+ 3
2
8
4⋅ 3
≠
6 9−1
k=3
Hence, only one value of k exists.
(not true)
(true)
60
JEE Main Chapterwise Mathematics
1 α
3
45. If P 1 3 3 is the adjoint of a 3 × 3 matrix


2 4 4
A and | A | = 4 , then α is equal to
[JEE Main 2013]
(a) 4
(b) 11
(c) 5
(d) 0
Exp. (b)
If A is matrix of order n , then|adj A| = | A|n − 1.
 1 α 3
Given, P =  1 3 3


2 4 4
∴
Q
∴
| P | = 112
( − 12 ) − α (4 − 6) + 3 (4 − 6)
= 2α − 6
[given]
P = adj ( A)
|P| = |adj A| = | A|2 = 16
⇒ 2 α − 6 = 16 ⇒ 2 α = 22
∴
α = 11
1 0 0
46. Let A = 2 1 0 . If u1 and u 2 are column


3 2 1
0
1


matrices such that Au 1 = 0 and Au 2 = 1 ,
 
 
0
0
thenu 1 + u 2 is equal to
[AIEEE 2012]
 −1 
(a)  1 
 
 0 
 −1 
(b)  1 
 
 −1 
 −1 
(c)  −1 
 
 0 
 1
(d)  −1 
 
 −1 
Exp. (d)
Given Matrices are
 0
 1
 1 0 0




A = 2 1 0 , Au1 = 0 and Au 2 =  1
 
 


 0
 0
 3 2 1
To Find Matrix u1 + u 2
Since, both Au1 and Au 2 are given, hence adding
them, we get
 1  0  1 + 0 
Au1 + Au 2 =  0 +  1 =  0 + 1

    
 0  0  0 + 0
 1
⇒
A (u1 + u 2 ) =  1
 
 0
Since, A is a non-singular matrix, i.e., | A | ≠ 0,
hence multiplying both sides by A −1(from RHS),
we get
 1
A −1 A(u1 + u 2 ) = A −1  1
 
 0
1
⇒ u1 + u 2 = 2

 3
1
Now, | A |= 2
3
−1
 1
×  1
 
 0
2
0
0

1
0
0
1
2
0 = 1×
0
1
1
2
1
…(i)
0
−0+ 0
1
[by expanding the determinant along row 1]
⇒
| A| = 1
Now, cofactor matrix of A (i.e., the matrix in which
every element is replaced by corresponding
cofactor)
2 1 
2 0
 1 0
−
 2 1
3 2 
3 1


 0 0
1 0 
1 0
= −
−

3 2 
3 1
 2 1
 0 0
1 0 
1 0


−
2 1 
2 0
 1 0
1
 1 −2
1 −2
= 0


0
1
 0
T
1 − 2
1
∴ adj ( A) =  0

0
 0
adj ( A)
⇒ A −1 =
| A|
1
 1

− 2 = − 2


1
 1
 1
⇒ A −1 =  − 2

 1
0
0

1
0
1
−2
0
1
−2
[Q | A | = 1]
From Eq. (i), we get
1
u1 + u 2 = 2

 3
 1
=  −2

 1
0
0

1
0
0

1
0
1
2
0
1
−2
−1
 1
×  1
 
 0
0  1
0 ×  1
  
1  0
61
Matrices and Determinants
∴
 1+ 0 +
u1 + u 2 =  −2 + 1 +

 1 − 2 +
0  1
0 =  −1
  
0  −1
49. Let A and B be two symmetric matrices of
47. Let P and Q be 3 × 3 matrices P ≠ Q .If P 3 = Q 3
and P 2Q = Q 2P , then
(P 2 + Q 2 ) is equal to
(a) –2
(c) 0
determinant
of
[AIEEE 2012]
(b) 1
(d) –1
Exp. (c)
Given
(i) Two matrices P and Q of order 3 × 3 such that
P ≠ Q.
(ii) P 3 = Q 3 and P 2Q = Q 2 P
To find The value of determinant of P 2 + Q 2 .
On subtracting the given equations, we get
P 3 − P 2Q = Q 3 − Q 2 P
⇒
⇒
48. The number of values of k for which the
linear
equations
4x + ky + 2 z = 0,
kx + 4y + z = 0 and 2 x + 2 y + z = 0 posses a
non-zero solution, is
[AIEEE 2011]
(b) 1
(d) 3
Exp. (a)
Since, equation has non-zero solution.
⇒
∆=0
4 k 2
∴
k 4 1 =0
2 2 1
⇒
⇒
4 (4 − 2 ) − k (k − 2 ) + 2 (2 k − 8) = 0
8 − k 2 + 2 k + 4k − 16 = 0
⇒
k 2 − 6k + 8 = 0
Exp. (a)
Since, A and B are symmetric matrices.
∴
AT = A
[given]
P2 + Q 2 = 0
(a) 2
(c) 0
(a) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(b) Statement I is true, Statement II is false
(c) Statement I is false, Statement II is true
(d) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
BT = B
Now, to check A (BA) is symmetric.
Consider [ A (BA)]T = (BA)T ⋅ AT = ( AT BT ) AT
(P − Q )(P 2 + Q 2 ) = 0
∴
are
Statement II AB is symmetric matrix, if
matrix multiplication of A with B is
commutative.
[AIEEE 2011]
and
P 2 (P − Q ) = Q 2 (Q − P)
Now, since
P≠Q
⇒
P−Q≠ 0
⇒
| P2 + Q 2 | = 0
order 3.
Statement I A (BA ) and ( AB ) A
symmetric matrices.
⇒
(k − 2 ) (k − 4) = 0
⇒
k = 2, 4
Hence, number of values of k is two.
So,
= ( AB) A = A (BA)
[ A (BA)] = A (BA)
T
⇒ A (BA) is symmetric.
Similarly, ( AB) A is symmetric.
So, Statement I is true. Also, Statement II is true,
as if A and B are symmetric.
⇒( AB) is symmetric, iff AB = BA.
i.e., AB commutative.
Hence, both the statements are true but
statement II is not a correct explanation of
statement I.
50. If the trivial solution is the only solution of
the system of equations
x − ky + z = 0,
kx + 3y − kz = 0
and
3x + y − z = 0
Then, the set of all values of k is [AIEEE 2011]
(a) { 2 , − 3 }
(c) R − { 2 }
(b) R − { 2 , − 3}
(d) R − { − 3}
Exp. (b)
x − ky + z = 0, kx + 3 y − kz = 0 and
3 x + y − z = 0 has trivial solution.
62
JEE Main Chapterwise Mathematics
1 −k
⇒
⇒
1
3 −k ≠0
1 −1
k
3
1 (− 3 + k ) + k (− k + 3k ) + 1 (k − 9) ≠ 0
⇒
2 k + 2 k − 12 ≠ 0
⇒
k2 + k − 6 ≠ 0
⇒
(k + 3) (k − 2 ) ≠ 0
53. Consider the system of linear equations
x1 + 2 x 2 + x 3 = 3,
2 x1 + 3x 2 + x 3 = 3
k ≠ 2, − 3
k ∈ R − {2 , − 3}
51. Statement I Determinant of a skewsymmetric matrix of order 3 is zero.
Statement II For any matrix A,
det( A T ) = det( A )
det( − A ) = − det( A ).
Then,
and 3x1 + 5x 2 + 2 x 3 = 1
The system has
(a)
(b)
(c)
(d)
[AIEEE 2011]
Exp. (a)
Determinant of skew-symmetric matrix of odd
order is zero and of even order is perfect square.
So, Statement I is true.
The given system of linear equations can be put in
the matrix form as
 1 2 1  x1   3
2 3 1  x  =  3
  2  

 3 5 2   x3   1
1  x1   3 
1 2
by R 2 → R 2 − 2 R1,
~  0 − 1 − 1  x2  =  − 3 
      R 3 → R 3 − 3R1 

 0 − 1 − 1  x3   − 8
 1 2 1  x1   3
~  0 1 1  x2  =  3
   

 0 0 0  x3   5
det ( AT ) = det ( A)
det (− A) = (− 1) det ( A)
n
So, Statement II is false.
52. If ω ≠ 1 is the complex cube root of unity and
ω 0 
matrix H = 
, then H
 0 ω
(a) H
(c) −H
Exp. (a)
Here,
70
(b) 0
(d) H 2
is equal to
[AIEEE 2011]
ω 0 
H=

 0 ω
[by R 3 → R 3 − R 2 ]
Clearly, the given system of equations has no
solution.
Alternate Solution
Subtracting the addition of first two equations from
third equation, we get
0 = − 5, which is an absurd result.
Hence, the given system of equations has no
solution.
54. The number of 3 × 3 non-singular matrices,
∴
ω 0  ω 0  ω
H2 = 

= 
 0 ω  0 ω  0
⇒
ωK
HK = 
0
2
0

ωK 
[AIEEE 2010]
infinite number of solutions
exactly 3 solutions
a unique solution
no solution
Exp. (d)
(a) Statement I is true and Statement II is false
(b) Both statements are true
(c) Both statements are false
(d) Statement I is false and Statement II is true
and
ω 0 
 0 ω = H


2
∴
Now,
ω70
0 
=
H70 = 
70 
ω
0


∴
k − 3 + 2 k2 + k − 9 ≠ 0
⇒
and
ωK + 1
0 
HK + 1 = 
K + 1
0
ω


Then,
0

ω2 
with four entries as 1 and all other entries as
0, is
[AIEEE 2010]
(a)
(b)
(c)
(d)
less than 4
5
6
atleast 7
63
Matrices and Determinants
Exp. (d)
1 * *


Consider  * 1 *  . By placing 1 in anyone of


 * * 1
the 6 * position and 0 elsewhere, we get
6 non-singular matrices.
 * * 1


Similarly,  * 1 *  gives atleast one

1
 * *
non-singular matrix.
Hence, we get atleast 7 non-singular matrix.
55. Let A be 2 × 2 matrix with non-zero entries
and A 2 = I , where I is 2 × 2 identity matrix.
Define tr ( A ) = Sum of diagonal elements of
A and | A | = Determinant of matrix A.
Statement I tr ( A ) = 0
Statement II | A | = 1
[AIEEE 2010]
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true,
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
56. Let a, b and c be such that (b + c ) ≠ 0.
a a +1 a −1
If −b b + 1 b − 1
c c −1 c +1
a +1
b +1
c −1
+
b −1
c + 1 = 0,
then the value of ‘n’ is
(a) zero
(c) any odd integer
On comparing with A 2 − I = 0, we get
tr ( A) = 0,| A| = − 1
Therefore, Statement I is true but Statement II is
false.
Alternate Solution
 a b
Let
A=
; a, b,c,d ≠ 0
c d 
 a b  a b
Now
A2 = 
⋅ 

c d  c d 
 a2 + bc ab + bd 
A2 = 
2
 ac + cd bc + d 
Q
A2 = I
⇒
a2 + bc = 1 ⇒ bc + d 2 = 1
and ab + bd = ac + cd = 0
Also,
c ≠ 0 and b ≠ 0
⇒
a + d = 0, tr(A) = a + d = 0
and
| A| = ad − bc = − a2 − bc = − 1
⇒
Hence, Statement I is true but Statement II is false.
[AIEEE 2009]
(b) any even integer
(d) any integer
Exp. (c)
a a+ 1 a−1
a + 1 b + 1 c −1
− b b + 1 b − 1 + (− 1)n a − 1 b − 1 c + 1
c −1 c + 1
c
−b
a
1
c
a a+ 1 a−1
a+1 a −1 a
= − b b + 1 b − 1 + (− 1)n b + 1 b − 1 − b
c
c −1 c + 1
c −1 c + 1
a a+ 1 a−1
= − b b + 1 b − 1 + (− 1)n + 1
c
c −1 c + 1
Exp. (d)
A satisfies A 2 − tr( A) ⋅ A + (det A) I = 0
a −1
( − 1)n + 2a ( − 1)n + 1b ( − 1)n c
a
c
[Q| A| = | AT|]
a+ 1 a a−1
b + 1 −b b − 1
c −1
a+ 1 a−1
c
c+1
[C 2 ↔ C 3 ]
= [1 + (− 1)n + 2 ] − b b + 1 b − 1
c c −1 c + 1
This is equal to zero only, if n + 2 is odd i.e., n is an
odd integer.
57. Let A be 2 × 2 matrix.
Statement I adj (adj A) = A
Statement II |adj A | = A
[AIEEE 2009]
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Exp. (c)
|adj A| = | A|n − 1 = | A|2 − 1 = | A|
adj (adj A) = | A|n − 2 A = | A| 0 A = A
64
JEE Main Chapterwise Mathematics
Hence, both the statements are true but
Statement II is not a correct explanation of
Statement I.
Exp. (a)
Given equations are
x − cy − bz = 0,
cx − y + az = 0
and
bx + ay − z = 0
58. Let A be 2 × 2 matrix with real entries. Let I
be the 2 × 2 identity matrix. Denote by tr( A ),
the sum of diagonal entries of A. Assume
that A 2 = I .
Statement I If A ≠ I and A ≠ − I , then
det ( A ) = − 1.
For non-zero solution,
1 −c − b
c
b
Statement II If A ≠ I and A ≠ − I , then
[AIEEE 2008]
tr( A ) ≠ 0.
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Exp. (d)
 a b
A=

c d 
Let
 a2 + bc
⇒ 
 ac + cd
⇒
and
⇒
 a b
 1 0
c d  =  0 1





1
0


ab + bd
=
bc + d 2   0 1
b(a + d ) = 0
c(a + d ) = 0
a2 + bc = 1
bc + d 2 = 1
a=1
d = −1
b=c = 0
[Q A 2 = I]
⇒
1 − a2 − c 2 − abc − abc − b 2 = 0
⇒
a2 + b 2 + c 2 + 2 abc = 1
60. Let A be a square matrix all of whose entries
are integers. Then, which one of the
following is true ?
[AIEEE 2008]
(a) If det ( A ) = ± 1, then A −1 need not exist
(b) If det ( A ) = ± 1, then A −1 exists but all its entries
are not necessarily integers
(d) If det( A ) = ± 1, then A −1 exists and all its entries
are integers
A −1 =
And
1
(adj A) = ± (adj A)
det ( A)
All entries in adj ( A) are integers.
Hence, A −1 has integer entries.
that there are real numbers x , y , z not all
zero such that x = cy + bz , y = az + cx and
z = bx + ay . Then, a 2 + b 2 + c 2 + 2 abc is equal
to
[AIEEE 2008]
(c) –1
1(1 − a2 ) + c(− c − ab ) − b(ac + b ) = 0
As det ( A) = ± 1, A −1 exists.
59. Let a ,b and c be any real numbers. Suppose
(b) 2
⇒
Exp. (d)
1 0 
If A = 
 , then
 0 −1
 1 0   1 0   1 0
A2 = 
= 
 = I A ≠ I, A ≠ − I

 0 −1  0 −1  0 1
det ( A) = − 1 (Statement I is true)
Statement II, tr( A) = 1 − 1 = 0, Statement II is false.
(a) 1
a =0
−1
(c) If det( A ) ≠ ± 1, then A −1 exists and all its entries
are non-integers
2
∴
−1
a
(d) 0
1
1
61. If D = 1 1 + x
1
1
1
1 for x ≠ 0, y ≠ 0, then D
1+ y
is
[AIEEE 2007]
(a)
(b)
(c)
(d)
divisible by neither x nor y
divisible by both x and y
divisible by x but not y
divisible by y but not x
Exp. (b)
1
1
Given that, D = 1 1 + x
1
1
1
1
1+ y
65
Matrices and Determinants
Applying C 2 → C 2 − C1 and C 3 → C 3 − C1,
1 0 0
= 1 x 0 = xy
1 0 y
[AIEEE 2006]
Hence, D is divisible by both x and y.
5 5 α
62. Let A = 0 α

0
equal to
0
α
5 α. If | A 2 | = 25, then |α| is

5 
[AIEEE 2007]
(a) 52
1
(c)
5
(b) 1
(d) 5
Exp. (c)
 5 5α α 
Since, A =  0 α 5α 


5 
 0 0
 5 5α α   5 5α α 
2
∴
A =  0 α 5α   0 α 5α 



5 
5   0 0
 0 0
25 25α + 5α 2 10α + 25α 2 


α2
5α 2 + 25α 
=0

0
0
25


| A 2|= 0
0
= 25
α2
0
2
4
0
b 
Now,
 1 2   a 0  a 2 b
AB = 

= 

 3 4  0 b   3a 4b 
and
 a 0   1 2   a 2 a
BA = 


= 
 0 b   3 4  3 b 4b 
[AIEEE 2005]
α 2 25α + 5α 2
= 625α 2
0
25
I −A
A−I
A
A+I
Exp. (a)
[given]
such that A 2 − B 2 = ( A − B )( A + B ), then
which of the following will be always true?
[AIEEE 2006]
AB = BA
Either of A or B is a zero matrix
Either of A or B is an identity matrix
A=B
Exp. (a)
1
Given that, A = 
3
a
and
B= 
0
(a)
(b)
(c)
(d)
63. If A andB are square matrices of size n × n
(a)
(b)
(c)
(d)
Exp. (c)
65. If A 2 − A + I = O , then the inverse of A is
5α 2 + 25α
25
⇒ 625α 2 = 25
1
∴
|α| =
5
(a) there exists more than one but finite number
of B’s such that AB = BA
(b) there exists exactly one B such that AB = BA
(c) there exist infnitely many B’s such that
AB = BA
(d) there cannot exist any B such that AB = BA
If AB = BA, then a = b.
Hence, AB = BA is possible for infinitely many
values of B’s.
25 25α + 5α 2 10α + 25α 2
⇒
a 0
1 2
 andB = 0 b ;a ,b ∈ N . Then,
3
4




64. Let A = 
Q
A2 − A + I = O
⇒
A −1 A 2 − A −1 A + A −1I = O
⇒
( A −1 A ) A − ( A −1 A ) + A −1 = O
⇒
A − I + A −1 = O
A −1 = I − A
⇒
66. Ifa 2 + b 2 + c 2 = − 2 and
1 + a 2x (1 + b 2 )x (1 + c 2 )x
f ( x ) = (1 + a 2 )x 1 + b 2x (1 + c 2 )x ,
(1 + a 2 )x (1 + b 2 )x
then f ( x ) is a polynomial of degree
Since, A 2 − B2 = ( A − B)( A + B)
∴
⇒
1 + c 2x
A 2 − B2 = A 2 − B2 + AB − BA
AB = BA
[AIEEE 2005]
(a) 2
(c) 0
(b) 3
(d) 1
66
JEE Main Chapterwise Mathematics
Applying C1 → C1 + C 2 + C 3 ,
α+2 1 1
Exp. (a)
Given that,
 1 + a2 x (1 + b 2 )x (1 + c 2 )x


f( x) = (1 + a2 )x 1 + b 2 x (1 + c 2 )x
(1 + a2 )x (1 + b 2 )x 1 + c 2 x 


Applying C1 → C1 + C 2 + C 3 , we get
 1 + a2 x + x + b 2 x + x + c 2 x

f ( x) =  x + a 2 x + 1 + b 2 x + x + c 2 x
 x + a2 x + x + b 2 x + 1 + c 2 x

(1 + b 2 )x (1 + c 2 )x

1 + b 2 x (1 + c 2 )x
(1 + b 2 )x 1 + c 2 x 

1 + (a2 + b 2 + c 2 + 2 )x (1 + b 2 )x (1 + c 2 )x


= 1 + (a2 + b 2 + c 2 + 2 )x 1 + b 2 x (1 + c 2 )x
1 + (a2 + b 2 + c 2 + 2 )x (1 + b 2 )x 1 + c 2 x 


1 (1 + b 2 )x (1 + c 2 )x


=  1 1 + b 2 x (1 + c 2 )x
 1 (1 + b 2 )x 1 + c 2 x 


[Qa2 + b 2 + c 2 = − 2, given]
Applying R1 → R1 − R 3 , R 2 → R 2 − R 3 , we get
x −1 
0
0
0
x −1  =
1− x
= 0


1
−
x
2
2
 1 (1 + b )x 1 + c x
x−1
x−1
α+2 α 1 =0
α+2 1 α
Applying R 2 → R 2 − R1, R 3 → R 3 − R1,
1
1
1
(α + 2 ) 0 α − 1
0 =0
0
0
α −1
⇒
(α + 2 )(α − 1)2 = 0
∴
α = 1, − 2
But α = 1 makes given three equations same.
So, the system of equations has infinite solution.
So, answer is α = − 2 for which the system of
equations has no solution.
68. If a1 , a 2 ,..., an ,... are in GP, then the
determinant
log an
∆ = log an + 3
log an + 6
(a) 2
log an + 5
log an +
log an + 8
[AIEEE 2005, 2004]
(c) 0
log an , log an + 1, log an + 2 , ..., log an + 8 , ... are in
log an
∆ = log an +
log an +
67. The system of equations
1
not –2
either –2 or 1
–2
(d) 1
Since, a1, a2 , . . . , an , . . . are in GP, then
Given that,
[AIEEE 2005]
7
Exp. (c)
Hence, f( x) is of degree 2.
(a)
(b)
(c)
(d)
log an + 4
(b) 4
AP.
αx + y + z = α − 1,
x + αy + z = α − 1
and x + y + αz = α − 1
has no solution, if α is
log an + 2
is equal to
= ( x − 1)
2
log an + 1
∴
3
6
log an + 1 log an + 2
log an + 4 log an + 5
log an + 7 log an + 8
a
∆ = a + 3d
a+d
a + 4d
a + 2d
a + 5d
a + 6d
a + 7d
a + 8d
=0
where a and d are the first term and common
difference of an AP.
Applying C 2 → C 2 − C1,C 3 → C 3 − C1,
Exp. (d)
The system of given equations has no solution, if
α 1 1
1
1
α 1 =0
1 α
a
⇒
d 2d
∆ = a + 3d d 2 d
a + 6d d 2 d
=0
[since, two columns are similar]
67
Matrices and Determinants
0 −1
−1 0 . The only correct


0 
−1 0
statement about the matrix A is [AIEEE 2004]
0
69. Let A =  0
(a)
(b)
(c)
(d)
A is a zero matrix
A = ( −1 ) I , where I is a unit matrix
A −1 does not exist
A2 = I
Exp. (d)
 0 0 −1
The given matrix, A =  0 −1 0 


 −1 0 0 
 1 0 0  −1 0 0 
(b) Now, (−1) I = − 1  0 1 0 =  0 −1 0  ≠ A

 

 0 0 1  0 0 −1
(−1) I ≠ A
(c) Now,| A | = 0 + 0 − 1
Since,| A | ≠ 0, so A
0
(d) Now, A =  0

 −1
1
⇒
A2 = 0

 0
2
70.
−1
0 −1
= − 1(−1) = 1
−1 0
exists.
0 −1  0
−1 0   0
 
0 0   −1
0 0
1 0 ∴ A 2

0 1
(c) 2
⇒
x + 2 ay + az = 0, x + 3by + bz = 0
and
x + 4cy + cz = 0
has a non-zero solution, then a ,b and c
(a) are in AP
(b) are in GP
(c) are in HP
(d) satisfy a + 2 b + 3c = 0
[AIEEE 2003]
Exp. (c)
Since, the system of linear equations has a
non-zero solution, then
1 2a a
1 3b b = 0
=I
(d) 5
Since, B is the inverse of matrix A, i.e., B = A −1.
 4 2 2
10 A −1 =  −5 0 α 


 1 −2 3 
2 2
 4
10 A −1 A =  − 5 0 α  A


 1 − 2 3 
−5 + α = 0 and 5 + α = 10 ⇒ α = 5
⇒
Exp. (d)
∴
⇒
0
0 
 10
 − 5 + α 5 + α −5 + α 


0
10 
 0
Applying R 2 → R 2 − R1, R 3 → R 3 − R1, we get
If B is the inverse of matrix A , then α is equal
to
[AIEEE 2004]
(b) 1
10 0 0 
⇒  0 10 0  =


 0 0 10
0 −1
−1 0 

0 0 
 4 2 2
1 −1 1 
Let A = 2 1 −3 and 10B = −5 0 α




1 
 1 −2 3
1 1
(a) –2
[Q A −1 A = I]
71. If the system of linear equations
(a) It is clear that A is not a zero matrix.
∴
2 2   1 −1 1 
 4
10I =  − 5 0 α  2 1 − 3

 

1 
 1 − 2 3   1 1
⇒
1 4c c
⇒
⇒
a
1
2a
0 3b − 2 a b − a = 0
0 4c − 2 a c − a
(3 b − 2 a)(c − a) − (4c − 2 a)(b − a) = 0
3 bc − 3 ba − 2 ac + 2 a2
= 4bc − 2 ab − 4ac + 2 a2
⇒
2 ac = bc + ab
On dividing by abc both sides, we get
2 1 1
= +
b a c
So, a, b and c are in HP.
α β 
a b 
2
 and A = β α , then
b
a




[AIEEE 2003]
72. If A = 
(a) α = a 2 + b 2 and β = ab
(b) α = a 2 + b 2 and β = 2ab
(c) α = a 2 + b 2 and β = a 2 − b 2
(d) α = 2ab and β = a 2 + b 2
68
JEE Main Chapterwise Mathematics
Exp. (d)
Exp. (b)
Given that,
Since, l , m and n are the pth, qth and rth terms of a
GP whose first term is A and common ratio is R.
 a b
α β 
A=
⇒ A2 = 


b a
β α 
 a b  a b
A2 = A A = 


b a b a
∴
∴
⇒
 a2 + b 2 ab + ba
A2 = 
2
2
 ba + ab b + a 
⇒
α β   a2 + b 2
β α  =  2 ab

 
⇒
⇒
log l = log A + ( p − 1) log R
Similarly,
log m = log A + (q − 1) log R
and
log n = log A + (r − 1) log R
log n
2
1
∆ = ωn
ω 2n
ωn
ω 2n
ω 2n
1
1 is equal to
ωn
(a) 0
(c) ω
(b) 1
log A + ( r − 1) log R
1
r
0 p 1
= 0 q 1 =0
0 r 1
(d) ω2
1 + i + ω2
1
ω2
1−i
−1
ω 2 − 1 is equal to
−i − 1 + ω − i
−1
[AIEEE 2002]
Given that,
ω2 n
(a) 0
(b) 1
(c) i
(d) ω
ωn ω2 n
ω2 n
1
1
ωn
= 1 (ω3 n − 1) − ωn (ω2 n − ω2 n ) + ω2 n (ωn − ω4 n )
= 1 (1 − 1) − 0 + ω2 n (ωn − ωn )
[Qω3 = 1]
=0
Exp. (a)
ω2
1 1 + i + ω2
2
Let ∆ = 1 − i
ω −1
−1
−i
74. Ifl , m and n are the pth, qth and rth terms of
log l p 1
a GP and all positive, then log m q 1 is
log n
equal to
(a) 3
1
75. If ω( ≠ 1) is a cubic root of unity, then
[AIEEE 2003]
Exp. (a)
1
∆ = ωn
r
Applying C1 → C1 − [C 3 log A + (C 2 − C 3 ) log R ],
ω 2 are the cube roots of unity, then
73. If 1,ω
−1
log l p 1
log A + ( p − 1) log R p 1
Now, log m q 1 = log A + (q − 1) log R q 1
2 ab 

2
a + b2 
α = a + b and β = 2 ab
2
l = AR p
r
1
[AIEEE 2002]
(b) 2
(c) 1
(d) 0
−1 + ω − i
−1
Applying R1 → R1 + R 3 , we get
1− i
= 1− i
−i
=0
−1
−1
ω2 − 1
ω2 − 1
−1 + ω − i
−1
[Q ω + ω2 = − 1]
[since, two rows are identical]
4
Permutations and
Combinations
1. All possible numbers are formed using the
2. The number of four-digit numbers strictly
digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time.
The number of such numbers in which the
odd digits occupy even places is
greater than 4321 that can be formed using
the digits 0, 1, 2, 3, 4, 5 (repetition of digits is
allowed) is
[JEE Main 2019, 8 April Shift-II]
[JEE Main 2019, 8 April Shift-I]
(a) 180
(c) 160
(b) 175
(d) 162
Exp. (a)
Given digits are 1, 1, 2, 2, 2, 2, 3, 4, 4.
According to the question, odd numbers 1, 1, 3
should occur at even places only.
(a) 306
(b) 310
(c) 360
(d) 288
Exp. (b)
Following are the cases in which the 4-digit
numbers strictly greater than 4321 can be formed
using digits 0, 1, 2, 3, 4, 5 (repetition of digits is
allowed)
Case-I
4
3
2
2/3/4/5
4 ways
4 numbers
Case-II
4
3
3/4/5 0/1/2/3/4/5
3 ways
Even places
3×6=18 numbers
6 ways
Case-III
∴The number of ways to arrange odd numbers at
3!
even places are 4 C 3 ×
2!
and the number of ways to arrange remaining even
6!
.
numbers are
4!2 !
So, total number of 9-digit numbers, that can be
formed using the given digits are
3!
6!
4
= 4 × 3 × 15 = 180
C3 ×
×
2 ! 4!2 !
4
4/5
0/1/2/3/4/5
2 ways
6 ways
2×6×6=72 numbers
Case-IV
5
6×6×6=216 numbers
0/1/2/3/4/5
6 ways
So, required total numbers
= 4 + 18 + 72 + 216 = 310
70
JEE Main Chapterwise Mathematics
3. A committee of 11 members is to be formed
from 8 males and 5 females. If m is the
number of ways the committee is formed
with at least 6 males and n is the number of
ways the committee is formed with atleast 3
females, then
[JEE Main 2019, 9 April Shift-I]
(a) m = n = 68
(c) m = n = 78
(b) m + n = 68
(d) n = m − 8
Exp. (c)
Since there are 8 males and 5 females. Out of
these 13 members committee of 11 members is to
be formed.
According to the question, m = number of ways
when there is at least 6 males
= ( 8 C 6 × 5C 5 ) + ( 8 C 7 × 5 C 4 ) + ( 8 C 8 × 5 C 3 )
= (28 × 1) + (8 × 5)+ (1 × 10)
= 28 + 40 + 10 = 78
and n = number of ways when there is at least 3
females
= ( 5C 3 × 8 C 8 ) + ( 5C 4 × 8 C 7 ) + ( 5C 5 × 8 C 6 )
= 10 × 1 + 5 × 8 + 1 × 28 = 78
So, m = n = 78
4. The number of 6 digits numbers that can be
formed using the digits 0, 1, 2,5, 7 and 9
which are divisible by 11 and no digit is
repeated, is
[JEE Main 2019, 10 April Shift-I]
(a) 60
(b) 72
(c) 48
(d) 36
Exp. (a)
Key Idea Use divisibility test of 11 and consider
different situation according to given condition.
Since, the sum of given digits
0 + 1 + 2 + 5 + 7 + 9 = 24
Let the six-digit number be abcdef and to be
divisible by 11, so the difference of sum of odd
placed digits and sum of even placed digits
should be either 0 or a multiple of 11 means
|(a + c + e ) − (b + d + f )| should be either 0 or a
multiple of 11.
Hence, possible case is
a + c + e = 12 = b + d + f (only)
Now, Case I
set {a, c, e} = {0, 5, 7} and set {b, d , f} = {1, 2, 9}
So, number of 6-digits numbers
= (2 × 2 !) × (3!) = 24
[Qa can be selected in ways only either 5 or 7].
Case II
Set {a, c, e} = {1, 2, 9} and set {b, d , f} = {0, 5, 7}
So, number of 6-digits numbers = 3! × 3! = 36
So, total number of 6-digits numbers
= 24 + 36 = 60
5. Suppose that 20 pillars of the same height
have been erected along the boundary of a
circular stadium. If the top of each pillar has
been connected by beams with the top of all
its non-adjacent pillars, then the total
number of beams is
[JEE Main 2019, 10 April Shift-II]
(a) 180
(b) 210
(c) 170
(d) 190
Exp. (c)
It is given that, there are 20 pillars of the same
height have been erected along the boundary of a
circular stadium.
Now, the top of each pillar has been connected by
beams with the top of all its non-adjacent pillars,
then total number of beams = number of diagonals
of 20-sided polygon.
Q 20 C 2 is selection of any two vertices of 20-sided
polygon which included the sides as well.
So, required number of total beams = 20C 2 − 20
[Qthe number of diagonals in a n-sided closed
polygon = nC 2 − n]
20 × 19
=
− 20
2
= 190 − 20 = 170
6. The number of ways of choosing 10 objects
out of 31 objects of which 10 are identical and
the remaining 21 are distinct, is
[JEE Main 2019, 12 April Shift-I]
(a) 2 20 − 1 (b) 2 21
(c) 2 20
(d) 2 20 + 1
Exp. (c)
Given that, out of 31 objects 10 are identical and
remaining 21 are distinct, so in following ways, we
can choose 10 objects.
0 identical + 10 distincts, number of ways
= 1×
21
C10
1 identical + 9 distincts, number of ways
= 1×
21
C9
2 identicals + 8 distincts, number of ways
= 1×
21
C8
71
Permutations and Combinations
10 identicals + 0distinct, number of ways
= 1 × 21C 0
Exp. (d)
So, total number of ways in which we can choose
10 objects is
21
C10 + 21C 9 + 21C 8 + K + 21C 0 = x (let) … (i)
⇒ 21C11 +
C12 +
21
C13 + K +
C 21 = x
21
21
… (ii)
Number of girls in the class = 5 and number of
boys in the class = 7
Now, total ways of forming a team of 3 boys and 2
girls = 7C 3 ⋅5 C 2 = 350
n
But, if two specific boys are in team, then number
of ways = 5C1 ⋅5 C 2 = 50
C12 + K +
C 21
Required ways, i.e. the ways in which two specific
boys are not in the same team = 350 − 50 = 300.
Alternate Method
Number of ways when A is selected and B is not
= 5C 2 ⋅5 C 2 = 100
7. A group of students comprises of 5 boys and
Number of ways when B is selected and A is not
= 5C 2 ⋅5 C 2 = 100
[Q C r = C n − r ]
On adding both Eqs. (i) and (ii), we get
2 x = 21C 0 + 21C1 + 21C 2 + K + 21C10
n
+
⇒
2x = 2
21
⇒x=2
C11 +
21
21
21
20
n girls. If the number of ways, in which a
team of 3 students can randomly be selected
from this group such that there is at least
one boy and at least one girl in each team, is
1750, then n is equal to
[JEE Main 2019, 12 April Shift-II]
(a) 28
(b) 27
(c) 25
(d) 24
Exp. (c)
It is given that a group of students comprises of 5
boys and n girls. The number of ways, in which a
team of 3 students can be selected from this group
such that each team consists of at least one boy
and at least one girls, is = (number of ways
selecting one boy and 2 girls) + (number of ways
selecting two boys and 1 girl)
5
5
[given]
= ( C1 × nC 2 ) ( C 2 × nC1 ) = 1750
⇒
n(n − 1)  5 × 4
× n = 1750
 + 

2   2
2
n (n − 1) + 4n = × 1750
5
n2 + 3n = 2 × 350
⇒
n2 + 3n − 700 = 0
⇒  5 ×

⇒
⇒
n2 + 28n − 25n − 700 = 0
⇒
⇒
n(n + 28) − 25(n + 28) = 0
n = 25
(c) 200
9. The number of natural numbers less than
7,000 which can be formed by using the
digits 0, 1, 3, 7, 9 (repitition of digits allowed)
is equal to
[JEE Main 2019, 9 Jan Shift-II]
(a) 374
(b) 375
(c) 372
(d) 250
Exp. (a)
Using the digits 0, 1, 3, 7, 9
number of one digit natural numbers that can be
formed = 4,
number of two digit natural numbers that can be
formed = 20,
(Q0 can not come in Ist box)
number of three digit natural numbers that can be
formed = 100
[Q n ∈ N]
number of different teams consisting of 2
girls and 3 boys that can be formed from this
class, if there are two specific boys A and B,
who refuse to be the members of the same
team, is
[JEE Main 2019, 9 Jan Shift-I]
(b) 500
∴Required ways = 100 + 100 + 100 = 300.
4×5
8. Consider a class of 5 girls and 7 boys. The
(a) 350
Number of ways when both A and B are not
selected = 5C 3 ⋅5 C 2 = 100
(d) 300
4×5× 5
and number of four digit natural numbers less than
7000, that can be formed = 250
2×5× 5×5
(Qonly 1 or 3 can come in Ist box)
∴Total number of natural numbers formed
= 4 + 20 + 100 + 250 = 374
72
JEE Main Chapterwise Mathematics
10. Consider three boxes, each containing 10
12. From 6 different novels and 3 different
balls labelled 1, 2, …, 10. Suppose one ball is
randomly drawn from each of the boxes.
Denote by ni , the label of the ball drawn
from the ith box, (i = 1, 2 , 3). Then, the
number of ways in which the balls can be
chosen such that n1 < n 2 < n 3 is
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.
[JEE Main 2018]
The number of such arrangements is
[JEE Main 2019, 12 Jan Shift-I]
(a) 82
(b) 120
(c) 240
(d) 164
Exp. (b)
Given there are three boxes, each containing 10
balls labelled 1, 2, 3, … , 10.
Now, one ball is randomly drawn from each boxes,
and ni denote the label of the ball drawn from the
ith box, (i = 1, 2, 3).
Then, the number of ways in which the balls can be
chosen such that n1 < n2 < n3 is same as selection
of 3 different numbers from numbers {1, 2, 3, … ,
10} = 10C 3 = 120.
11. There are m men and two women
participating in a chess tournament. Each
participant plays two games with every
other participant. If the number of games
played by the men between themselves
exceeds the number of games played
between the men and the women by 84,
then the value of m is
[JEE Main 2019, 12 Jan
Shift-II]
(a) 12
(b) 11
(c) 9
(d) 7
Exp. (a)
Since, there are m-men and 2-women and each
participant plays two games with every other
participant.
∴ Number of games played by the men between
themselves = 2 × mC 2
and the number of games played between the
men and the women = 2 × mC1 × 2C1
Exp. (a)
Given 6 different novels and 3 different
dictionaries.
Number of ways of selecting 4 novels from 6
6!
novels is 6 C 4 =
= 15
2 ! 4!
Number of ways of selecting 1 dictionary is from
3!
3 dictionaries is 3 C1 =
=3
1!2 !
∴Total number of arrangement of 4 novels and 1
dictionary where dictionary is always in the middle,
is
15 × 3 × 4! = 45 × 24 = 1080
13. A man X has 7 friends, 4 of them are ladies
and 3 are men. His wifeY also has 7 friends, 3
of them are ladies and 4 are men. Assume X
andY have no common friends.
Then, the total number of ways in which X
and Y together can throw a party inviting 3
ladies and 3 men, so that 3 friends of each of
X andY are in this party, is
[JEE Main 2017 (Offline)]
(a) 485
(c) 469
(b) 468
(d) 484
Exp. (a)
Given, X has 7 friends, 4 of them are ladies and 3
are men while Y has 7 friends, 3 of them are
ladies and 4 are men.
∴Total number of required ways
= 3C 3 × 4C 0 × 4C 0 × 3C 3
Now, according to the question,
2 mC 2 = 2 mC1 2C1 + 84
m!
⇒
= m × 2 + 42
2 !(m − 2 )!
⇒
m(m − 1) = 4m + 84
⇒
m2 − m = 4m + 84
2
⇒ m − 5m − 84 = 0
⇒ m2 − 12 m + 7 m − 84 = 0
⇒ m(m − 12 ) + 7 (m − 12 ) = 0
⇒
m = 12
(a) atleast 1000
(b) less than 500
(c) atleast 500 but less than 750
(d) atleast 750 but less than 1000
+ 3C 2 × 4C1 × 4C1 × 3C 2
+ 3C1 × 4C 2 × 4C 2 × 3C1
+ 3C 0 × 4C 3 × 4C 3 × 3C 0
[Q m > 0]
= 1 + 144 + 324 + 16 = 485
73
Permutations and Combinations
14. If all the words (with or without meaning)
having five letters, formed using the letters
of the word SMALL and arranged as in a
dictionary, then the position of the word
SMALL is
[JEE Main 2016 (Offline)]
(a) 46th
(b) 59th
(c) 52nd
Exp. (b)
The integer greater than 6000 may be of 4 digit or
5 digit. So, here two cases arise.
Case I When number is of 4 digit.
Four digit number can starts from 6, 7 or 8
(d) 58th
6,7 or 8
Exp. (d)
4!
= 12
2!
Number of words start with L = 4! = 24
4!
Number of words start with M =
= 12
2!
3!
Number of words start with SA =
=3
2!
Number of words start with SL = 3! = 6
Note that, next word will be “SMALL”.
Hence, the position of word “SMALL” is 58th.
Clearly, number of words start with A =
15. Let A and B be two sets containing four and
two elements respectively. Then, the
number of subsets of the set A × B , each
having atleast three elements are
[JEE Main 2015]
(a) 219
(b) 256
(c) 275
(d) 510
4
3
2
Thus, total number of 4 digit number, which are
greater than 6000 = 3 × 4 × 3 × 2 = 72
Case II When number is of 5 digit.
Total number of five digit number which are greater
than 6000 = 5! = 120
∴Total number of integers = 72 + 120 = 192
17. Let A and B be two sets containing
2 elements and 4 elements, respectively.
The number of subsets of A × B having 3 or
more elements is
[JEE Main 2013]
(a) 256
(c) 219
(b) 220
(d) 211
Exp. (c)
Exp. (a)
Given,
n( A) = 4, n(B) = 2
⇒
n( A × B) = 8
Total number of subsets of set ( A × B) = 2 8
Number of subsets of set A × B having no
element
(i.e. φ) = 1
Number of subsets of set A × B having one
element
= 8C1
Number of subsets of set A × B having two
elements
= 8C 2
∴Number of subsets having atleast three
elements
= 2 8 − (1 + 8C1 + 8C 2 )
= 2 8 − 1 − 8 − 28
= 2 8 − 37
= 256 − 37 = 219
16. The number of integers greater than 6000
that can be formed, using the digits 3, 5, 6, 7
and 8 without repetition, is [JEE Main 2015]
(a) 216
(c) 120
3
(b) 192
(d) 72
Given, n( A) = 2 and n(B) = 4
∴
n( A × B) = 8
The number of subsets of A × B having 3 or more
elements = 8C 3 + 8C 4 + … + 8C 8
= 2 8 − 8C 0 − 8C1 − 8C 2
= 256 − 1 − 8 − 28 = 219
[Q2 n = nC 0 + nC1 + K + nC n ]
18. Let Tn be the number of all possible triangles
formed by joining vertices of an n-sided
regular polygon. If Tn + 1 − Tn = 10, then the
value of n is
[JEE Main 2013]
(a) 7
(b) 5
(c) 10
(d) 8
Exp. (b)
Given,
∴
Tn = nC 3
Tn + 1 =
n+1
Tn + 1 − Tn =
n+1
C3
⇒ n C 2 + nC 3 − nC 3 = 10
n
C 2 = 10
⇒
∴
n=5
C 3 − nC 3 = 10 [given]
74
JEE Main Chapterwise Mathematics
19. Assuming the balls to be identical except for
difference in colours, the number of ways in
which one or more balls can be selected
from 10 white, 9 green and 7 black balls is
[AIEEE 2012]
(a) 880
(c) 630
(b) 629
(d) 879
Exp. (d)
Given 10 identical white balls, 9 identical green
balls and 7 identical black balls.
To find The number of ways for selecting atleast
one ball.
Number of ways to choose zero or more white
balls
= (10 + 1) [since, all white balls are mutually
identical]
Number of ways to choose zero or more green
balls
= (9 + 1) [since, all green balls are mutually
identical]
Exp. (a)
Let the number of ways of distributing n identical
objects among r persons such that each person
gets atleast one object is same as the number of
ways of selecting (r − 1) places out of (n − 1)
different places, i.e., n − 1C r − 1.
Statement I Here, n = 10 and r = 4
∴ Required number of ways = 10 − 1C 4 − 1 = 9C 3
Statement II
Required number of ways = 9C 3
Hence, both the statements are true but Statement
II is not a correct explanation of Statement I.
21. There are 10 points in a plane, out of these 6
are collinear. If N is the number of triangles
formed by joining these points, then
(a) N >190
(c)100 < N ≤ 140
(b) N ≤100
[AIEEE 2011]
(d)140 < N ≤ 190
Exp. (b)
If out of n points, m are collinear, then
Number of triangles = nC 3 − mC 3
Number of ways to choose zero or more black
balls = (7 + 1) [since, all black balls are mutually
identical]
∴ Number of triangles =
Hence, number of ways to choose zero or more
balls of any colour = (10 + 1)(9 + 1)(7 + 1)
⇒
Also, number of ways to choose a total of zero balls
=1
Hence, the number, if ways to choose atleast
one ball (irrespective of any colour)
= (10 + 1)(9 + 1)(7 + 1) − 1 = 879
[10 × 9 × 7 − 1 = number of ways to select
atleast one ball of each colour
(in case there is some confusion]
20. Statement I The number of ways of
distributing 10 identical balls in 4 distinct
boxes such that no box is empty, is 9C 3.
Statement II The number of ways of
choosing any 3 places from 9 different
places is 9C 3.
(a) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(b) Statement I is true, Statement II is false
(c) Statement I is false, Statement II is true
(d) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
[AIEEE 2011]
C 3 − 6C 3
10
= 120 − 20
N = 100
22. From 6 different novels and 3 different
dictionaries, 4 novels and 1 dictionary are to
be selected and arranged in a row on the
shelf so that the dictionary is always in the
middle. Then, the number of such
arrangements is
[AIEEE 2009]
(a)
(b)
(c)
(d)
atleast 500 but less than 750
atleast 750 but less than 1000
atleast 1000
less than 500
Exp. (c)
The number of ways in which 4 novels can be
selected
= 6C 4 = 15
The number of ways in which 1 dictionary can be
selected
= 3C1 = 3
Now, we have 5 places in which middle place is
fixed.
∴4 novels can be arranged in 4! ways.
∴The total number of ways = 15 × 4! × 3
= 15 × 24 × 3 = 1080
75
Permutations and Combinations
23. In a shop, there are five types of ice-creams
25. The set S = {1, 2 , 3 ,... , 12 } is to be partitioned
available. A child buy six ice-creams.
into three sets A , B andC of equal size.
Statement I The number of different ways
the child can buy the six ice-creams is 10C 5.
Thus, A ∪ B ∪ C = S ,
[AIEEE 2007]
A ∩ B = B ∩C = A ∩C = φ
The number of ways to partition S is
[AIEEE 2008]
Statement II The number of different ways
the child can buy the six ice-creams is equal
to the number of different ways of arranging
6 A’s and 4 B’s in a row.
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Exp. (a)
Since, the number of ways that child can buy the
six ice-creams is equal to the number of different
ways of arranging 6 A’s and 4 B’s in a row.
So, number of ways to arrange 6 A’s and 4 B’s in a
row
10! 10
=
= C4
6! 4!
And number of integral solutions of the equation
x1 + x2 + x3 + x4 + x5 = 6
=
6 + 5 −1
C5 − 1 =
C4 ≠
10
jumbling the letters in the word
‘MISSISSIPPI’ in which no two S are
adjacent?
[AIEEE 2008]
(a) 7 ⋅ 6C 4 ⋅ 8C 4
(b) 8 ⋅ 6C 4 ⋅ 7C 4
(c) 6 ⋅ 7 ⋅ C 4
(d) 6 ⋅ 8 ⋅ 7C 4
=
C 4 × 8C 4 × 4C 4
12 !
8!
12 !
=
×
× 1=
8! × 4! 4! × 4!
(4!)3
12
26. At an election, a voter may vote for any
number of candidates not greater than the
number to be elected. There are
10 candidates and 4 are to be elected. If a
voter votes for atleast one candidate, then
the number of ways in which he can vote, is
(a) 6210
(b) 385
(c) 1110
(d) 5040
Exp. (b)
Total number of ways
= 10C1 +
C2 +
10
C3 +
10
10
C4
= 10 + 45 + 120 + 210 = 385
27. If the letters of the word ‘SACHIN’ are
arranged in all possible ways and these
words are written out as in dictionary, then
the word ‘SACHIN’ appears at serial number
(a) 602
(c) 600
(b) 603
(d) 601
[AIEEE 2005]
In the word ‘SACHIN’, order of alphabets is A, C,
H, I, N and S. Number of words start with A = 5!, so
with C, H, I, N.
Now, words start with S and after that ACHIN are in
ascending order of position, so 5 ⋅ 5! = 600 words
are in dictionary before words with S start and
position of this word is 601.
Given word is MISSISSIPPI.
Here,
I = 4 times,
S = 4 times,
P = 2 times
M = 1time
_ M_ I_ I_ I_ I_ P_ P_
∴ Required number of words = 8C 4 ×
[AIEEE 2006]
Exp. (d)
Exp. (a)
= 8C 4 ×
Required number of ways
C5
24. How many different words can be formed by
(b) 12!/3!(3!) 4
(d) 12!/(3!) 4
Exp. (c)
10
Hence, Statement I is false and Statement II is true.
8
(a) 12!/3!(4!) 3
(c) 12!/(4!) 3
7!
4! 2 !
7 × 6!
= 7 ⋅ 8C 4 ⋅ 6C 4
4! 2 !
28. The range of the function f ( x ) =
7− x
Px − 3 is
[AIEEE 2004]
(a) {1, 2, 3}
(c) {1, 2, 3, 4}
(b) {1, 2, 3, 4, 5, 6}
(d) {1, 2, 3, 4, 5}
76
JEE Main Chapterwise Mathematics
Exp. (a)
Given that, f( x) =
7− x
=
Px − 3 . The above function is
Now,
f(3) = P0 = 1
f(5) = P2 = 2
∴
R f = {1, 2 , 3}
5 women can dine at a round table, if no
two women are to sit together, is given by
2
(a) 6! × 5!
(c) 5! × 4!
29. How many ways are there to arrange the
letters in the word ‘GARDEN’ with the
vowels in alphabetical order?
[AIEEE 2004]
(a) 120
(b) 240
= 5 × 4 × 7 + 8 × 7 = 140 + 56 = 196
32. The number of ways in which 6 men and
f(4) = 3 P1 = 3
and
8×7
8×7 × 6
+ 1×
2
3×2
=5×
defined, if7 − x ≥ 0 and x − 3 ≥ 0and 7 − x ≥ x − 3.
⇒
x ≤ 7, x ≥ 3 and x ≤ 5
∴
Df = {3, 4, 5}
4
5!
8!
5!
8!
×
+
×
4!1! 6!2 ! 5! 0! 5! 3!
(c) 360
(d) 480
(b) 30
(d) 7! × 5!
Exp. (a)
First, we fix the position of men, the number of
ways to sit men = 5! and the number of ways to sit
women = 6 P5
M
Exp. (c)
Total number of ways in which all letters can be
arranged in alphabetical order = 6!.
There are two vowels (A, E) in the word ‘GARDEN’.
Total number of ways in which these two vowels
can be arranged = 2!
6!
∴ Total number of required ways = = 360
2!
30. The number of ways of distributing
8 identical balls in 3 distinct boxes, so that
none of the boxes is empty, is
[AIEEE 2004]
(a) 5
(b) 21
8
(d)
(c) 3
8
C3
Exp. (b)
The required number of ways =
= 7C 2 =
8 −1
C3 − 1
7!
7⋅6
=
= 21
2 ! 5! 2 ⋅ 1
in an examination such that he must choose
atleast 4 from the first five questions. The
number of choices available to him is
(b) 196
(d) 346
[AIEEE 2003]
Exp. (b)
The number of choices available to him
= 5C 4 × 8C 6 + 5C 5 × 8C 5
M
M
M
M
M
∴ Total number of ways = 5! × 6 P5 = 5! × 6!
33. If n C r denotes the number of combinations
of n things taken r at a time, then the
expression n C r + 1 + nC r − 1 + 2 × nC r equal to
(a)
(b)
31. A student is to answer 10 out of 13 questions
(a) 140
(c) 280
[AIEEE 2003]
(c)
(d)
n+2
Cr
n+2
n+1
[AIEEE 2003]
Cr + 1
Cr
Cr + 1
n+1
Exp. (b)
n
Here, we use
Now,
n
Cr
+1
Cr
−1
+ nC r =
=
=
n+ 2
Cr
Cr
+1
+1
Cr
…(i)
+ C r −1 + 2 ⋅ C r
n
n
= nC r + 1 + nC r −1 + nC r + nC r
= nC r + 1 + n + 1C r + nC r
n+1
n+1
+
n+1
Cr
[using Eq. (i)]
[using Eq. (i)]
[using Eq. (i)]
5
Mathematical Induction
1. Consider the statement : ‘‘P (n ): n 2 − n + 41 is
prime.’’ Then, which one of the following is
true?
[JEE Main 2019, 10 Jan Shift I]
(a)
(b)
(c)
(d)
Both P( 3) and P( 5) are true.
P( 3) is false but P( 5) is true.
Both P( 3) and P( 5) are false.
P( 5) is false but P( 3) is true.
For n = k,
P(k ) = k 7 − k
Let P(k ) be divisible by 7.
∴
k 7 − k = 7 λ, for some λ ∈ N
…(i)
For n = K + 1,
P(k + 1) = (k + 1)7 − (k + 1)
= ( 7C 0 k 7 + 7C1 k 6 + 7C 2 k 5 + K + 7C 6 ⋅ k + 7C 7 )
− (k + 1)
Exp. (a)
Given statement is “P(n) : n2 − n + 41 is prime”.
= (k 7 − k ) + 7 {k 6 + 3k 5 + K + k}
Clearly P(3) : 32 − 3 + 41 = 9 − 3 + 41
= 7 λ + 7 {k 6 + 3k 5 + K + k}
= 47 which is a prime number.
and P(5) : 5 − 5 + 41 = 25 − 5 + 41 = 61,
⇒ Divisible by 7.
So, both statements are true and Statement II is
correct explanation of Statement I.
2
which is also a prime number.
∴ Both P(3) and P(5) are true.
2. Statement I For each natural number
[using Eq.(i)]
3. Statement I For every natural number
n ≥ 2,
1
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Exp. (b)
Exp. (c)
P(n) = (n)7 − n
By mathematical induction,
For n = 1,
P(1) = 0, which is divisible by 7.
Let
∴
+ ... +
1
Statement II For each natural number n,
n 7 − n is divisible by 7.
[AIEEE 2011]
Let
+
1
> n
1
2
n
Statement II For every natural number
n ≥ 2, n(n + 1) < n + 1.
[AIEEE 2008]
n ,(n + 1) 7 − n 7 − 1 is dizvisible by 7.
1
1
1
+
+ K+
1
2
n
1
1
+
= 1707
> 2
P( 2 ) =
.
1
2
P(n) =
78
JEE Main Chapterwise Mathematics
Let us assume that
1
1
1
P(k ) =
+
+ K+
> k is true. …(i)
1
2
k
Now, for n = k + 1,
1
1
1
1
LHS =
+
+ K+
+
1
2
k
k+1
>
k+
>
k+1
k+1
1
=
k+1
k(k + 1) + 1
(k + 1)
[from Eq. (i)]
[Q k(k + 1) > k, ∀ k ≥ 0]
= (k + 1)
∴ P (k + 1) > (k + 1)
By mathematical induction, statement I is true for
all n ≥ 2.
Now, let
α(n) = n(n + 1)
∴
α (2 ) = 2 (2 + 1) = 6 < 3
Let us assume that
α (k ) = k(k + 1) < (k + 1) is true.
For n = k + 1,
LHS = (k + 1) (k + 2 ) < (k + 2 )
[Q(k + 1) < (k + 2 )]
∴ α (k + 1) < (k + 2 )
By mathematical induction, Statement II is true but
Statement II is not a correct explanation of
Statement I.
1 0
1 0
 andI = 0 1 ,then which one of
1
1




the following holds for all n ≥ 1, by the
principle of mathematical induction?
4. If A = 
(a)
(b)
(c)
(d)
An = 2n − 1 A + (n − 1 ) I
An = n A + (n − 1 ) I
An = 2n − 1 A − (n − 1 ) I
An = n A − (n − 1 ) I
[AIEEE 2005]
Exp. (d)
Given that,
∴
1 0
A=

1 1
1 0 1 0  1 0
A2 = 

= 

1 1 1 1 2 1
 1 0 1 0  1 0
A3 = 

= 

2 1 1 1  3 1
MMMMMMMMMMM
 1 0
A =
 can be verified by induction.
 n 1
Now, taking options
0 
 1 0  n 0  n − 1
(b)
 n 1 =  n n +  0
n − 1

 
 
0 
 1 0 2 n − 1
⇒ 
≠ n
− 1
n
1
2
n

 
n
n
(d) nA − (n − 1)I = 
n
1
=
n
0  n − 1 0 
−
n  0
n − 1
0
1
= An
5. Let S (k ) = 1 + 3 + 5 + ... + (2 k − 1) = 3 + k 2.
Then, which of the following is true?
(a) S (1 ) is correct
[AIEEE 2004]
(b) S (k ) ⇒S (k + 1 )
(c) S (k ) ⇒S
/ (k + 1 )
(d) Principle of mathematical induction can be
used to prove the formula
Exp. (b)
S (k ) = 1 + 3 + 5 + . . . + (2 k − 1) = 3 + k 2
Put k = 1in both sides, we get
LHS = 1
and
RHS = 3 + 1 = 4
⇒
LHS ≠ RHS
Put (k + 1) in both sides in the place of k, we get
LHS = 1 + 3 + 5 + . . . + (2 k − 1) + (2 k + 1)
RHS = 3 + (k + 1)2 = 3 + k 2 + 2 k + 1
Let LHS = RHS
1 + 3 + 5 + . . . + (2 k − 1) + (2 k + 1)
= 3 + k2 + 2 k + 1
⇒ 1 + 3 + 5 + . . . + (2 k − 1) = 3 + k 2
If S (k ) is true, then S (k + 1) is also true.
Hence, S (k ) ⇒S (k + 1)
6
Binomial Theorem and
Its Simple Applications
(x +
1. The sum of the coefficients of all even
x3 − 1)6 + ( x −
x3 − 1)6
degree terms is x in the expansion of
= 2 [6 C 0 − 6C 2 + 6 C 4 + 6C 4 − 1 − 3]
( x + x 3 − 1 )6 + ( x − x 3 − 1 )6 , ( x > 1) is equal
to
[JEE Main 2019, 8 April Shift-I]
= 2 [1 − 15 + 15 + 15 − 1 − 3]
(a) 29
(b) 32
(c) 26
(d) 24
2. If the fourth term in the binomial expansion
6
  1 
1
  1+ log 10 x 

12
of  x
+ x  is equal to 200, and




Exp. (d)
Key Idea Use formula :
(a + b )n + (a − b )n =
2 [n C 0 an + nC 2 an − 2 b 2 + nC 4 an − 4 b 4 + ......]
x > 1, then the value of x is
[JEE Main 2019, 8 April Shift-II]
Given expression is
(x +
x − 1) + ( x −
3
6
x − 1)
3
6
6
4
6
3
Exp. (c)
+ C 4 x ( x − 1) + C 6 ( x − 1) ]
6
2
3
(b) 104
(d) 103
(a) 100
(c) 10
= 2 [ C 0 x + C 2 x ( x − 1)2
6
= 2(15 − 3) = 24
4
6
3
6
Given binomial is
{Q(a + b )n + (a − b )n



 
1

1

  1 + log x 
10
12
+ x 
 x


= 2 [n C 0 an + nC 2 an − 2 b 2 + nC 4 an − 4 b 4 + …]}
= 2 [6 C 0 x6 + 6C 2 x4 ( x3 − 1) + 6C 4 x2 ( x3 − 1)2
+ 6C 6 ( x3 − 1)3 ]
The sum of the terms with even power of x
= 2 [6 C 0 x6 + 6C 2 (− x4 ) + 6C 4 x8 + 6C 4 x2
+ 6C 6 (−1 − 3 x6 )]
= 2 [6 C 0 x6 − 6C 2 x4 + 6C 4 x8 + 6C 4 x2 − 1 − 3 x6 ]
Now, the required sum of the coefficients of even
powers of x in
6
Since, the fourth term in the given expansion is
200.
3
1

 2  1 3

1 + log 10 x   12 
6
∴ C3  x
  x  = 200



 
⇒
3

 2 ( 1+ log
20 × x 
10 x )
1
+ 
4
= 200
80
JEE Main Chapterwise Mathematics
3
x 2 ( 1 + log 10 x )
⇒
+
1
4
4. If some three consecutive coefficients in the
= 10

3
1
+  log10 x = 1
⇒ 
 2(1 + log10 x) 4 
[applying log10 both sides]
⇒ [6 + (1 + log10 x)]log10 x = 4(1 + log10 x)
⇒
(7 + log10 x)log10 x = 4 + 4log10 x
[letlog10 x = t ]
⇒
t 2 + 7t = 4 + 4t
⇒
t 2 + 3t − 4 = 0
binomial expansion of( x + 1)n in powers of x
are in the ratio 2 : 15 : 70, then the average of
these three coefficients is
[JEE Main 2019, 9 April Shift-II]
(a) 964
(c) 232
(b) 227
(d) 625
Exp. (c)
⇒
⇒
t = 1, −4 = log10 x
x = 10, 10−4
Key Idea Use general term of Binomial
expansion ( x + a)n i.e. Tr + 1 = nC r 1 xn − r ar
Since,
∴
x>1
x = 10
Given binomial is ( x + 1)n , whose general term, is
Tr + 1 = nC r xr
According to the question, we have
n
C r − 1 : nC r : nC r + 1 = 2 : 15 : 70
n
Cr − 1
2
Now,
=
n
15
Cr
n!
(r − 1)!(n − r + 1)! 2
=
⇒
n!
15
r !(n − r )!
r
2
=
⇒
n − r + 1 15
⇒
15r = 2 n − 2 r + 2
…(i)
⇒
2 n − 17 r + 2 = 0
n!
n
r !(n − r )!
C
15
3
Similarly, n r =
=
⇒
n!
14
C r + 1 70
(r + 1)!(n − r − 1)!
r+1 3
⇒
⇒14r + 14 = 3n − 3r
=
n − r 14
3. If the fourth term in the binomial expansion
log x 
2
of  + x 8 

x
value of x is
(a) 8−2
6
( x > 0) is 20 × 8 7 , then the
[JEE Main 2019, 9 April Shift-I]
(b) 83
(d) 82
(c) 8
Exp. (d)
log x
2
Given binomial is  + x 8 
x

6
Since, general term in the expansion of ( x + a)n is
Tr + 1 = n C r xn − r ar
6−3
2
( xlog 8 x )3 = 20 × 87
∴T4 = T3 + 1 = 6 C 3  
 x
(given)
3
2
⇒ 20   x3 log 8 x = 20 × 87
 x
[Q 6 C 3 = 20]
3

 log 2 x − 3 

= (2 3 )6
⇒ 2 3 x[3(log 8 x ) − 3 ] = (2 3 )7 ⇒ x 3
Qlog ( x) = 1 log x for x > 0; a > 0, ≠ 1
a
an


n
x− 3)
x( 2
= 218
⇒
On taking log 2 x both sides, we get
log
(log 2 x − 3)log 2 x = 18
⇒
(log 2 x)2 − 3log 2 x − 18 = 0
⇒(log 2 x)2 − 6log 2 x + 3log 2 x − 18 = 0
⇒ log 2 x(log 2 x − 6) + 3 (log 2 x − 6) = 0
⇒
⇒
⇒
⇒
(log 2 x − 6) (log 2 x + 3) = 0
log 2 x = −3, 6
x = 2 −3 , 2 6
1
x = , 82
8
…(ii)
⇒
3n − 17 r − 14 = 0
On solving Eqs. (i) and (ii), we get
n − 16 = 0
⇒
n = 16 and r = 2
16
C1 + 16C 2 + 16C 3
Now, the average =
3
16 + 120 + 560 696
=
=
= 232
3
3
5. If the coefficients of x 2 and x 3 are both zero,
in the expansion of the expression
(1 + ax + bx 2 ) (1 − 3x )15 in powers of x, then
the ordered pair (a ,b ) is equal to
[JEE Main 2019, 10 April Shift-I]
(a) (28, 315)
(b) ( − 21, 714)
(c) ( 28, 861)
(d) ( − 54, 315)
81
Binomial Theorem and Its Simple Applications
Exp. (a)
7. The coefficient of x 18 in the product
Given expression is
(1 + ax + bx2 )(1 − 3 x)15 in the expansion of
binomial (1 − 3 x)15 the (r + 1th
) term is
Tr
+1
=
C r (−3 x)r =
15
C r (−3)r xr
15
Now, coefficient of x2 , in the expansion of
(1 + ax + bx2 )(1 − 3 x)15 is
15
C 2 (−3)2 + a15C1 (−3)1 + b
⇒
⇒
C 0 (−3)0 = 0 (given)
15
(1 + x )(1 − x )10(1 + x + x 2 )9 is
[JEE Main 2019, 12 April Shift-I]
Exp. (a)
Given expression is
(1 + x) (1 − x)10 (1 + x + x2 )9
(105 × 9) − 45 a + b = 0
45a − b = 945 …(i)
3
(b) −126
(d) 126
(a) 84
(c) − 84
= (1 + x) (1 − x) [(1 − x) (1 + x + x2 )]9
= (1 − x2 ) (1 − x3 )9
Similarly, the coefficient of x , in the expansion of
(1 + ax + bx2 )(1 − 3 x)15 is
Now, coefficient of x18 in the product
(1 + x) (1 − x)10 (1 + x + x2 )9
C 3 (−3)3 + a 15C 2 (−3)2 + b15C1(−3)1 = 0 (given)
⇒
− 12285 + 945a − 45b = 0
⇒
63a − 3b = 819
⇒
21a − b = 273 …(ii)
From Eqs. (i) and (ii), we get
24a = 672 ⇒ a = 28
So,
b = 315
⇒
(a, b ) = (28, 315)
= coefficient of x18 in the product
(1 − x2 ) (1 − x3 )9
15
6. The smallest natural number n, such that
the coefficient of x in the expansion of
n
 2 1
n
 x + 3  is C 23, is

x 
= coefficient of x18 in (1 − x3 )9
− coefficient of x16 in (1 − x3 )9
Since, (r + 1)th term in the expansion of
(1 − x3 )9 is 9 C r (− x3 )r = 9C r (− 1)r x3 r
Now, for x18 , 3r = 18 ⇒ r = 6
and for x16 , 3r = 16
r=
⇒
9!
6! 3!
9× 8×7
= 84
=
3×2
∴Required coefficient is 9 C 6 =
[JEE Main 2019, 10 April Shift-II]
(a) 35
(b) 23
(c) 58
(d) 38
Exp. (d)
n
1
Given binomial is  x2 + 3  , its (r + 1)th term, is

x 
r
Tr
+1
1
1
= nC r ( x2 )n − r  3  = nC r x2 n − 2 r 3 r
x 
x
= Cr x
n
2n − 2r − 3r
= Cr x
n
2n − 5r
For the coefficient of x ,
…(i)
2 n − 5r = 1 ⇒ 2 n = 5r + 1
As coefficient of x is given as n C 23 , then either
r = 23 or n − r = 23.
If r = 23, then from Eq. (i), we get
2 n = 5(23) + 1
⇒ 2 n = 115 + 1 ⇒2 n = 116 ⇒ n = 58.
If n − r = 23, then from Eq. (i) on replacing the
value of ‘r’, we get 2 n = 5(n − 23) + 1
⇒ 2 n = 5n − 115 + 1 ⇒ 3n = 114 ⇒ n = 38
So, the required smallest natural number n = 38.
16
∉ N.
3
8. The term independent of x in the expansion
6
 1 x8 
3
of  −  . 2 x 2 − 2  is equal to
x 
 60 81  
[JEE Main 2019, 12 April Shift-II]
(a) − 72
(c) − 36
(b) 36
(d) −108
Exp. (c)
Key Idea Use the general term (or (r + 1th
) term)
in the expansion of binomial (a + b )n
i.e.
Tr + 1 = nC r an − r b r
6
3
Let a binomial  2 x2 − 2  , it’s (r + 1) th term

x 
r
3
= Tr + 1 = 6C r (2 x2 )6 − r  − 2 
 x 
= 6C r (− 3)r (2 )6 − r x12 − 2 r
6−r
= C r (−3) (2 )
6
r
12 − 4 r
x
− 2r
…(i)
82
JEE Main Chapterwise Mathematics
Now, the term independent of x in the expansion of
6
 1
x8   2
3
  2 x − 2 

−
x 
 60 81 
= the term independent of x in the expansion of
6
3
1  2
 2 x − 2  + the term independent of x in the
60 
x 
3
x8  2
2 x − 2 
81 
x 
expansion of −
6
and when 8(16)100 is divided by 15, gives
remainder 8.
2 403  8
∴

= .
 15  15
(where {}
⋅ is the fractional part function)
⇒
k=8
10. The coefficient of t 4 in the expansion of
3
C3
[put r = 3]
(− 3)3 (2 )6 − 3 x12 − 4 ( 3 )
60
1
+  −  6C 5 (−3)5 (2 )6 − 5 x12 − 4 ( 5 ) x8 [put r = 5]
 81
1 −t 6 
 is

 1 −t 
3 × 2(6)
1
(− 3)3 2 3 +
3
81
= 36 − 72 = − 36
Exp. (c)
=
6
5
=
(a) 14
2 403 k
is ,
15 15
[JEE Main 2019, 9 Jan Shift-I]
(b) 6
(c) 4
(d) 8
(d) 14
∴ Coefficient of t 4 in (1 − t 6 )3 (1 − t )−3
= Coefficient of t 4 in (1 − t 18 − 3t 6 + 3t 12 )(1 − t )− 3
= Coefficient of t 4 in (1 − t )− 3
3 + 4 −1
C 4 = 6C 4 = 15
(Qcoefficient of xr in (1 − x)− n =
3
= 8⋅2
400
= 8 ⋅ (2 )
= 8 (16)
4 100
100
C1(15) +
100
C 2 (15)2 + …
100
C100 (15)100 )
[By binomial theorem,
(1 + x)n = nC 0 + nC1 x + nC 2 x2 + … nC n xn , n ∈ N]
= 8 + 8 (100 C1(15) +
C 2 (15)2 + …
100
+
100
Cr )
of (1 + x log 2 x )5 equals 2560, then a possible
value of x is
[JEE Main 2019, 10 Jan Shift-I]
100
+
n + r −1
11. If the third term in the binomial expansion
= 8(1 + 15)
100
= 8(1 +
(c) 15
1 − t 6 
 = (1 − t 6 )3 (1 − t )− 3
Clearly, 
 1− t 
=
Exp. (d)
Consider,
2 403 = 2 400 +
[JEE Main 2019, 9 Jan Shift-II]
(b) 10
3
9. If the fractional part of the number
then k is equal to
(a) 12
100
C100 (15)
)
= 8 + 8 × 15λ
where λ =100 C1 +......+ 100 C100 (15)99 ∈ N
2 403 8 + 8 × 15λ
8
∴
=
= 8λ +
15
15
15
2 403  8
⇒ 
=
 15  15
(where {}
⋅ is the fractional part function)
∴
k=8
Alternate Method
2 403 = 8 ⋅ 2 400 = 8(16)100
Note that, when 16 is divided by 15, gives
remainder 1.
∴ When (16)100 is divided by 15, gives remainder
1100 = 1
(a) 4 2
(b)
1
4
(c)
1
8
(d) 2 2
Exp. (b)
The (r + 1) th term in the expansion of (a + x)n is
given by Tr + 1 = nC r an − r xr
∴3rd term in the expansion of (1 + xlog 2 x )5 is
5
C 2 (1)5 − 2 ( xlog 2 x )2
⇒
5
C 2 (1)5 − 2 ( xlog 2 x )2 = 2560 (given)
⇒
10( xlog 2 x )2 = 2560
⇒
x( 2 log 2 x ) = 256
⇒
log 2 x2 log 2 x = log 2 256
⇒
(taking log 2 on both sides)
2(log 2 x)(log 2 x) = 8
(Qlog 2 256 = log 2 2 8 = 8)
(log 2 x)2 = 4
⇒
⇒
⇒
log 2 x = ± 2
log 2 x = 2 or log 2 x = − 2
x = 4 or
x = 2 −2 =
1
4
83
Binomial Theorem and Its Simple Applications
12.
10 − r
2
3
20

Ci − 1 
k
 = , then k equals
If ∑  20
20

21
i = 1  Ci + Ci − 1 
20
[JEE Main 2019, 10 Jan Shift-I]
(a) 100
(b) 400
(c) 200
(d) 50
Exp. (a)
3
20
∑
Given,
i =1
20


Ci − 1

 = k
 20 C + 20C 
21

i
i −1
3
20
∑
⇒
i =1
 20 C i − 1 
 = k

 21C 
21

i 
(Q nC r + nC r
−1
=
n+1
Cr )
3
20
∑
⇒
i =1
20
∑
⇒
i =1
⇒

 20

Ci − 1 
k
 =

21
 21 20C i − 1 

 i
Q n C = n

r

r
3
1
 i  = k ⇒
 
 21
21
(21)3
Cr


− 1
20
k
∑ i = 21
i =1
3
k=
2
21  20 × 21

 = 100
(21)3  2 
∴
45 λ2 = 720
λ2 = 16 ⇒ λ = ± 4
∑{ 50C r ⋅ 50 − rC 25 − r } = K (50C 25 ),
r=0
then, K is equal to
[JEE Main 2019, 10 Jan Shift-II]
24
(b) 2
the
(b) 5
(c) 2 2
25
25
Σ 
r =0
C 25
50!
25! 
×
 =K
 25!25! r !(25 − r )!
50
C 25
50
C 25
[on multiplying 25! in
numerator and denominator.]
expression
25
⇒
The general term in the expansion of binomial
expression (a + b )n is
Tr + 1 = n C r an − r b r ,
so the general term in the expansion of binomial
10
λ
expression x2  x + 2  is

x 
50
50!
(50 − r )! 
×
 =K
 r !(50 − r )! (25 − r )!25!
⇒
(d) 4
(d) ( 25)2
Σ 
r =0
50
C 25
Σ 25C r
r =0
=K

50! 
C 25 Q 50C 25 =
25! 25!

50
25
Σ 25C r
r =0
⇒
K=
⇒
[Q nC 0 + nC1 + n C 2 + ....+ nC n = 2 n ]
K = 2 25
Exp. (d)
r

λ 
Tr + 1 = x2  10 C r ( x )10 − r  2  

 x  

(c) 2 25
25
[JEE Main 2019, 10 Jan Shift-II]
(a) 3
−1
Σ {50 C r .50 − r C 25 − r } = K
r =0
Given,
13. The positive value of λ for which the
coefficient of x 2 in
10
λ

x 2  x + 2  is 720, is

x 
25
Exp. (c)
⇒
k = 100
[given]
25
14. If
(a) 2
2
2
 3
 n(n + 1) 
3
3

Q1 + 2 + K + n = 
 2  

⇒
Now, for the coefficient of x2 ,
10 − r
put
− 2r = 2
2+
2
10 − r
− 2r = 0
⇒
2
⇒
10 − r = 4r ⇒ r = 2
So, the coefficient of x2 is 10 C 2 λ2 = 720
10! 2
λ = 720
⇒
2 ! 8!
10 ⋅ 9 ⋅ 8! 2
⇒
λ = 720
2 ⋅ 8!
⇒
k
 n(n + 1)
=
 n = 20 21
(21)3  2
1
10 − r
2+
−2r
2
λr x
=10 C r
⇒
n −1
λr x−2 r
=10C r x2 ⋅ x
= 2 25
15. The sum of the real values of x for which the
middle term in the binomial expansion of
8
 x 3 3
+  equals 5670 is

 3 x
[JEE Main 2019, 11 Jan Shift-I]
(a) 4
(b) 0
(c) 6
(d) 8
84
JEE Main Chapterwise Mathematics
Exp. (b)
17. Let ( x + 10)50 + ( x − 10)50
8
 x3
3
In the expansion of 
+  , the middle term is
x
 3
T4 + 1.
= a 0 + a1x + a 2x 2 + K + a 50x 50, for all x ∈R ;
a
then 2 is equal to
a0
[Qhere, n = 8, which is even, therefore middle term
n + 2
= 
 th term]
 2 
(a) 12.25
4
4
 x3 
3
8⋅7 ⋅ 6⋅ 5 8
x
∴ 5670 = C 4     =


3
1⋅ 2 ⋅ 3 ⋅ 4
x
 
⇒
x8 = 34 ⇒ x = ±
 x3 
= 8C r  
 3
+1
(c) 12.00
r
 3 
 
 x 

8−r
Exp. (a)
+ … + a50 x50
∴ a0 + a1 x + a2 x + … + a50 x
2
= [( 50 C 0 x50 +
3
So, sum of all values of x i.e. + 3 and −
C1 x49 10 +
Cr C 0 +
20
C r −1 C 1 +
20
50
+ …+
3=0
C r − 2 C 2 + ....
20
20
20
20
+ 20C 0 C r
is maximum, is
[JEE Main 2019, 11 Jan Shift-I]
(a) 15
(c) 11
(b) 10
(d) 20
+ ( 50 C 0 x50 −
= 2 [50 C 0 x50 +
20
Cr
− 1x
C1 x +
C 2 x2 + ... +
20
r −1
+
20
C r xr + ... +
20
∴ (1 + x)20 ⋅ (1 + x)20 = ( 20 C 0 +
20
C 2 x + ... +
2
20
Cr
20
C 20 x20
C1 x +
r −1
20
50
−…+
50
C1 x + ...+
20
− 1x
+
C r xr + ...
20
Cr
− 1x
r −1
20
c 20 x20 )
+
C r xr
20
+ ....+ 20 C 20 x20 )
⇒ (1 + x)
40
= ( C0 .
20
Cr +
20
C 2 x48 ⋅ 102 +
50
C 2 x48 102
C 50 1050 )]
C 4 x46 ⋅ 104
50
C 50 ⋅ 1050 ]
50
C 48 (10)48 ; a0 = 2
50
50
48
a2 2( C 2 )(10)
=
a0
2 (10)50
=2
C 50 (10)50 = 2(10)50
50
50 ⋅ 49 (10)48
1⋅ 2 2 ⋅ (10)50
[Q 50C 48 = 50C 2 ]
50 × 49
5 × 49 245
=
=
=
= 12 . 25
2 ⋅ (10 × 10)
20
20
20
+
× ( 20 C 0 +
C 50 1050 )
By comparing coefficients, we get
∴
We know that,
(1 + x)20 = 20C 0 +
50
C1 x49 10 +
50
+ …+
a2 = 2
Exp. (d)
50
C 2 x48 ⋅ 102
50
16. The value of r for which
20
(d) 12.75
We have,
( x + 10)50 + ( x − 10)50 = a0 + a1 x + a2 x2
8

QTr


[JEE Main 2019, 11 Jan Shift-II]
(b) 12.50
20
20
C1 C r
20
−1
...
On comparing the coefficient of xr of both sides,
we get
20
C 0 20C r + 20C120C r − 1 + ... + 20C r
C0 =
the 5th term from the end in the binomial
10
 1


1 
expansion of 2 3 +
is
1


3

2( 3) 
[JEE Main 2019, 12 Jan Shift-I]
C r C 0 ) xr + ...
20
20
18. A ratio of the 5th term from the beginning to
40
Cr
The maximum value of 40 C r is possible only when
r = 20
[Q n C n / 2 is maximum when n is even]
Thus, required value of r is 20.
1
(a)1 : 2(6) 3
1
(c) 4( 36) 3 :1
1
(b)1 : 4(16) 3
1
(d) 2( 36) 3 :1
Exp. (c)
Since, rth term from the end in the expansion of a
binomial ( x + a)n is same as the (n − r + 2 )th term
from the beginning in the expansion of same
binomial.
T
T5
T
∴Required ratio =
= 5 = 4+1
T10 − 5 + 2 T7 T6 + 1
85
Binomial Theorem and Its Simple Applications
10
⇒
T5
T10 − 5 +
=
2
10
 1 
C 4 (21/ 3 )10 − 4 

 2(3)1/ 3 
4
 1 
C 6 (21/ 3 )10 − 6 

 2(3)1/ 3 
6
[QTr
=
2
6/ 3
1/ 3 6
(2(3)
)
2 4 / 3 (2(3)1/ 3 )4
+1
So, the general term in the binomial expansion of
(71/ 5 − 31/ 10 )60 is
Tr + 1 = 60C r (71/ 5 )60 − r (−31/ 10 )r
60 − r
= nC r xn − r ar ]
[Q C 4 =
10
10
C6 ]
= 2 6 / 3 − 4 / 3 (2(3)1/ 3 )6 − 4
= 2 2 / 3 ⋅ 2 2 ⋅ 32 / 3 = 4(6)2 / 3 = 4(36)1/ 3
So, the required ratio is 4(36)1/ 3 : 1 .
19. If n C 4 , n C 5 and n C 6 are in AP, then n can be
[JEE Main 2019, 12 Jan Shift-II]
(a) 9
(b) 11
(c) 14
(d) 12
Exp. (c)
If n C 4 , n C 5 and n C 6 are in AP, then
2 ⋅n C 5 = n C 4 + n C 6
[if a, b, c are in AP , then 2b = a + c]
n!
n!
n!
⇒2
=
+
5!(n − 5)! 4!(n − 4)! 6!(n − 6)!
 n
n! 
Q C r = r !(n − r )!
2


⇒
5 ⋅ 4!(n − 5) (n − 6)!
1
1
=
+
4!(n − 4) (n − 5) (n − 6)! 6 ⋅ 5 ⋅ 4! (n − 6)!
2
1
1
=
+
⇒
5(n − 5) (n − 4) (n − 5) 30
30 + (n − 4) (n − 5)
2
=
⇒
5(n − 5)
30 (n − 4) (n − 5)
⇒ 12 (n − 4) = 30 + n2 − 9n + 20
⇒
⇒
⇒
⇒
n − 21n + 98 = 0
n2 − 14n − 7 n + 98 = 0
n(n − 14) − 7(n − 14) = 0
(n − 7 ) (n − 14) = 0 ⇒ n = 7 or 14
20. The total number of irrational terms in the
binomial expansion of ( 71/5 − 31/10 )60 is
[JEE Main 2019, 12 Jan Shift-II]
(b) 48
(c) 54
60
Cr 7
r
(−1)r 310
5
12 −
r
r
5 310
= (−1)r 60C r 7
The possible non-negative integral values of ‘r’ for
r
r
are integer, where r ≤ 60, are
which and
5
10
r = 0, 10, 20, 30, 40, 50, 60.
∴ There are 7 rational terms in the binomial
expansion and remaining 61 − 7 = 54 terms are
irrational terms.
21. The sum of the coefficients of all odd
degree terms in the expansion of
5
5
 x + x 3 − 1 +  x − x 3 − 1 , ( x > 1) is




[JEE Main 2018]
(a) −1
(b) 0
(c) 1
(d) 2
Exp. (d)
Key idea = (a + b )n + (a − b )n
= 2( n C 0 an + nC 2 an − 2 b 2 + nC 4 an − 4 b 4 … )
We have,
(x +
x3 − 1)5 + ( x −
x3 − 1)5 , x > 1
= 2( 5 C 0 x5 + 5C 2 x3 ( x3 − 1)2 + 5C 4 x( x3 − 1)4 )
= 2( x5 + 10 x3 ( x3 − 1) + 5 x( x3 − 1)2 )
= 2( x5 + 10 x6 − 10 x3 + 5 x7 − 10 x4 + 5 x)
Sum of coefficients of all odd degree terms is
2 (1 − 10 + 5 + 5) = 2
22. The value of ( 21C 1 − 10C 1 ) + ( 21C 2 − 10C 2 )
2
(a) 49
=
+ ( 21C 3 − 10C 3 ) + ( 21C 4 − 10C 4 ) + ... +
( 21C 10 − 10C 10 ) is
(a) 2 21 − 211
(c) 2 20 − 2 9
[JEE Main 2017 (Offline)]
(b) 2 21 − 210
(d) 2 20 − 210
Exp. (d)
(
C1 −
21
C1 ) + ( 21C 2 −
10
C 2 ) + ( 21C 3 −
10
The general term in the binomial expansion of
(a + b )n is Tr + 1 = nC r an − r b r .
C3 )
+ ... + ( C10 −
21
(d) 55
Exp. (c)
10
=(
C1 +
21
C 2 + ... +
21
C 2 + ... +
21
C10 )
10
+
1
= ( 21C1 +
2
10
C10 ) − ( C1
21
C 2 + ... +
10
C 20 ) − (2
21
10
− 1)
10
C10 )
86
JEE Main Chapterwise Mathematics
1 21
( C1 + 21C 2 + ... + 21C 21 − 1) − (210 − 1)
2
1
= (2 21 − 2 ) − (210 − 1) = 2 20 − 1 − 210 + 1
2
= 2 20 − 210
=
Aliter
We have,
(1 − 2 x )50 = C 0 − C12 x + C 2 (2 x )2
+ ...+ C 50 (2 x )50 …(i)
(1 + 2 x )
50
= C 0 + C12 x + C 2 (2 x )2
23. If the number of terms in the expansion of
n
2
4

1 − + 2  , x ≠ 0, is 28, then the sum of the

x x 
coefficients of all the terms in this
expansion, is
[JEE Main 2016 (offline)]
(a) 64
(b) 2187
(c) 243
(d) 729
Exp. (d)
Clearly, number of terms in the expansion of
n
 1 − 2 + 4  is (n + 2 ) (n + 1) or n + 2 C .


2

2
x x2 
1
1
[assuming and 2 distinct]
x
x
(n + 2 ) (n + 1)
∴
= 28
2
⇒ (n + 2 ) (n + 1) = 56 = (6 + 1) (6 + 2 ) ⇒ n = 6
Hence, sum of coefficients = (1 − 2 + 4)6 = 36
= 729
1
1
and 2 are functions of same variables,
x
x
therefore number of dissimilar terms will be
2 n + 1, i.e. odd, which is not possible. Hence, it
contains error.
Note As
24. The sum of coefficients of integral powers of
x in the binomial expansion of (1 − 2 x )50 is
[JEE Main 2015]
1
(a) ( 350 + 1)
2
1 50
(c) ( 3 − 1)
2
1
(b) ( 350 )
2
1 50
(d) ( 2 + 1)
2
+ ... + C 50 (2 x )50 …(ii)
On adding Eqs. (i) and (ii), we get
(1 − 2 x )50 + (1 + 2 x )50 = 2[C 0 + C 2 (2 x )2
+ ... + C 50 (2 x )50 ]
(1 − 2 x )50 + (1 + 2 x )50
= C 0 + C 2 (2 x )2
2
+ ... + C 50 (2 x )50
On putting x = 1, we get
(1 − 2 1)50 + (1 + 2 1)50
= C 0 + C 2 (2 )2
2
+ ... + C 50 (2 )50
⇒
(−1)50 + (3)50
= C 0 + C 2 (2 )2 + ... + C 50 (2 )50
2
1 + 350
= C 0 + C 2 (2 )2 + ... + C 50 (2 )50
⇒
2
⇒
25. If the coefficients of x 3 and x 4 in the
expansion of (1 + ax + bx 2 )(1 − 2 x )18 in
powers of x are both zero, then (a ,b ) is equal
to
[JEE Main 2014]
251 
(a) 16,


3 
272 
(c) 14,


3 
251 
(b) 14,


3 
272 
(d) 16,


3 
Exp. (d)
In the expansion of (1 + ax + bx2 ) (1 − 2 x)18 ,
Coefficient of x3 in (1 + ax + bx2 ) (1 − 2 x)18
= Coefficient of x3 in (1 − 2 x)18
+ Coefficient of x2 in a (1 − 2 x)18
Exp. (a)
+Coefficient of x in b (1 − 2 x)18
Let Tr +1 be the general term in the expansion of
=−
(1 − 2 x )50
∴
Q − 18C 3 ⋅ 2 3 + a18C 2 ⋅ 2 2 − b18C1 ⋅ 2 = 0
18 × 17 2
18 × 17 × 16
⋅ 8 + a⋅
⋅ 2 − b ⋅ 18 ⋅ 2 = 0
⇒
2
3×2
34 × 16
…(i)
⇒ 17 a − b =
3
Similarly, coefficient of x4
50 − r
Tr + 1 =
50
=
50
C r (1)
( −2 x
1/ 2 r
)
C r 2 r xr / 2 (−1)r
For the integral power of x, r should be even
integer.
∴Sum of coefficients =
25
∑
50
C 2 r (2 )2 r
r =0
1
1
= [(1 + 2 )50 + (1 − 2 )50 ] = [350 + 1]
2
2
18
∴
C 3 ⋅ 2 + a C 2 ⋅ 2 2 − b18C1 ⋅ 2
18
3
18
C 4 ⋅ 2 4 − a ⋅ 18C 3 2 3 + b ⋅ 18C 2 ⋅ 2 2 = 0
32 a − 32 b = 240
…(ii)
87
Binomial Theorem and Its Simple Applications
On solving Eqs. (i) and (ii), we get
Adding both the binomial expansions above, we
get
( 3 + 1)2 n − ( 3 − 1)2 n = 2 [2 n C1( 3 )2 n − 1
272
a = 16 and b =
3
+
26. The term independent of x in the expansion
(a) 4
(c) 210
+ K+
10

x +1
x −1 
of  2/3
−

1/3
 x − x + 1 x − x 1/2 
[JEE Main 2013]
28. The coefficient of x 7 in the expansion of
(a) −132
10

( x − 1) 
x+1
Consider  2 / 3
−

1/ 3
− x1/ 2 
1
x
−
x
+
x

= (1 − x)6 ⋅ (1 − x2 )6
 6
  6

=  ∑ (− 1)r 6C r ⋅ xr   ∑ (− 1) s 6C s ⋅ x2 s 
r = 0
  s = 0

10
10
− x
)
C r ( x1/ 3 )10 − r (− x− 1/ 2 )r
=
10
=
10
10 − r
C r (− 1)r x
−
3
r
2
( 3 + 1)
(a)
(b)
(c)
(d)
− ( 3 − 1)
is
Exp. (a)
C 0 ( 3 )2 n +
2n
2n
=
( 3 − 1)
C 0 ( 3 ) (−1) +
2n
2n
+
+
C1( 3 )2 n − 1
2n
C 2 ( 3 )2 n − 2 + K +
+
2n
2n
0
C2n( 3)
C 2 n ( 3 )2 n − 2 n
2n
C1( 3 )2 n − 1(−1)1
2n
C 2 ( 3 )2 n − 2 (−1)2 + K
(−1)2 n
2n
2n − 2n
r + s
⋅ 6C r ⋅ 6C s ⋅ xr
+ 2s
r =0 s=0
{(− 1)5 + 1 ⋅ 6C 5 ⋅ 6C1} + {(− 1)3 + 2 ⋅ 6C 3 ⋅ 6C 2 }
+ {(− 1)1 + 3 ⋅ 6C1 ⋅ 6C 3 }
⇒
⇒
29. Let
(36) − (20) (15) + 6(20)
36 − 300 + 120 = − 144
S1 =
[AIEEE 2012]
an irrational number
an odd positive integer
an even positive integer
a rational number other than positive integers
( 3 + 1)2 n =
6
For coefficient of x7 , we have
r + 2 s = 7 i .e., s = 1, r = 5
or
s = 2, r = 3 or s = 3, r = 1
and S 3 =
27. If n is a positive integer, then
2n
6
∑ ∑ (−1)
∴ Coefficient of x7 is
For independent of x, put
10 − r r
− =0
3
2
⇒
20 − 2 r − 3 r = 0
⇒
20 = 5 r ⇒ r = 4
10 × 9 × 8 × 7
∴
= 210
T5 = 10C 4 =
4× 3×2 ×1
2n
=
− 1/ 2 10
∴The general term is
+1
(d) 144
= {(1 − x) − x2 (1 − x)}6 = {(1 − x) (1 − x2 )}6
 ( x1/ 3 + 1) ( x 2/ 3 + 1 − x1/ 3 ) ( x − 1) ( x + 1)
=
−
x ( x − 1) 
x 2/ 3 − x1/ 3 + 1

= (x
(c) 132
Here, (1 − x − x2 + x3 )6
 ( x1/ 3 )3 + 13
{( x )2 − 1} 
=  2/ 3
−

1/ 3
x ( x − 1)
x − x + 1
1/ 3
[AIEEE 2011]
(b) −144
Exp. (b)
10
Tr
C 2 n − 1 ( 3 )2 n − ( 2 n − 1) ]
2n
(1 − x − x 2 + x 3 )6 is
Exp. (c)

( x + 1)
= ( x1/ 3 + 1) −
x 

C 5 ( 3 )2 n − 5
2n
which is most certainly an irrational number
because of odd powers of 3 in each of the terms.
is
(b) 120
(d) 310
C 3 ( 3 )2 n − 3 +
2n
10
10
j =1
10
j =1
∑ j ( j − 1)10C j , S 2 = ∑ j 10C j
∑ j 2 10C j .
j =1
Statement I
Statement II
S3 = 55 × 2 9
[AIEEE 2010]
S1 = 90 × 2 8 and S2 = 10 × 2 8
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true,
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Exp. (d)
88
JEE Main Chapterwise Mathematics
n
∑
Here, we use
Cr = 2 n
n
r =0
S1 =
Q
10
10!
∑ j ( j − 1) j ( j − 1)( j − 2 )!(10 − j )!
j =1
10
8!
= 90 ⋅ 2 8
= 90 ∑
(
j
)![
2
8
−
− ( j − 2 )]!
j=2
and S 2 =
10
10!
∑ j j ( j − 1)![9 − ( j − 1)]!
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Exp. (b)
n
∑
Since,
On multiplying by x, we get
n
j =1
∑
10
9!
= 10 ∑
= 10 ⋅ 2 9
j = 1 ( j − 1)![9 − ( j − 1)]!
Also, S 3 =
10
n
∑ (r + 1)⋅
j =1
=
∑ j ( j − 1)
Cj =
j =1
10
∑j
Cj
n
∑ (r + 1)⋅
j =1
9
30. The remainder left out when 82n − (62 )2n + 1
is divided by 9, is
[AIEEE 2009]
(b) 2
(c) 7
(d) 8
82 n − (62 )2 n + 1 = (1 + 63)n − (63 − 1)2 n + 1
r =0
32. In the binomial expansion of (a − b )n , n ≥ 5,
the sum of 5th and 6th terms is zero, then
[AIEEE 2007]
a /b is equal to
= (1 + 63) + (1 − 63)
= (1 + nC1 63 + nC 2 (63)2 + … + (63)n
( 2 n + 1)
C 2 (63)2
+ … + (− 1) (63)( 2 n + 1) ]
= 2 + 63 [n C1 + nC 2 (63) + … + (63)n − 1
+
5
n−4
(b)
( 2 n + 1)
C 2 (63) − … − (63)( 2 n ) ]
n
∑(r + 1)⋅nC r = (n + 2 )2n − 1
r=0
n
Statement II
(c)
n− 5
6
(d)
n−4
5
Since, in a binomial expansion of (a − b )n , n ≥ 5,
the sum of 5th and 6th terms is equal to zero.
n
∴
C 4 an − 4 (− b )4 + nC 5 an − 5 (− b )5 = 0
n!
n!
⇒
an − 5 b 5 = 0
an − 4 ⋅ b 4 −
(n − 4)! 4!
(n − 5)! 5!
 a
b
n!
−  =0
an − 5 ⋅ b 4 
(n − 5)! 4!
 n − 4 5
a n−4
=
b
5
⇒
⇒
Hence, the remainder is 2.
31. Statement I
6
n−5
Exp. (d)
2n + 1
+ [1 −( 2 n + 1) C1 63 +
C r = 2 n + n(2 )n − 1 = (n + 2 )2 n − 1
n
Hence, Statements I and II are true and
Statement II is a correct explanation of Statement I.
(a)
Exp. (b)
n
C r ⋅ xr = (1 + x)n + n x(1 + x)n − 1
n
r =0
Hence, Statement I is true and Statement II is false.
(a) 0
= x(1 + x)n
Hence, Statement II is true. If x = 1, then
10
= 90 ⋅ 2 + 10 ⋅ 2
= 90.2 8 + 20.2 8 = 110 ⋅ 2 8 = 55 ⋅ 2 9
8
+1
r =0
10!
10
C r ⋅ xr
n
On differentiating w.r.t. x, we get
∑ [ j ( j − 1) + j ] j !(10 − j )!
10
C r ⋅ xr = (1 + x)n
n
r =0
33. The sum of the series
20
C0 −
C1 +
20
C2 −
20
C 3 + ... +
20
20
C10 is
[AIEEE 2007]
∑ (r + 1) C r ⋅ x
n
r
r=0
n
= (1 + x ) + nx (1 + x )n − 1 [AIEEE 2008]
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(a) −
20
C10
1
(b)
2
(d)
20
C10 x10 + . . . +
20
20
C10 (c) 0
C10
Exp. (b)
We know that,
(1 + x)20 = 20C 0 +
C1 x + . . . +
20
20
C 20 x20
89
Binomial Theorem and Its Simple Applications
On putting x = − 1in the above expansion, we get
0 = 20C 0 − 20C1 + . . . − 20C 9
+ 20 C10 −
⇒
0=
⇒
⇒
⇒
20
C0 −
20
C1 + . . . −
20
−
20
20
0 = 2 ( 20 C 0 −
20
C10 = 2 ( 20 C 0 −
20
20
C0 −
C11 + . . . +
20
C9 +
C10
20
C10 =
C10
C9 ) +
C1 + . . . +
20
20
20
20
C1 + . . . +
C 20
C9 + . . . +
C1 + . . . −
20
20
1
2
C0
+ n [− m(1 − y)m − 1(1 + y)n − 1
+ (1 − y)m (n − 1)(1 + y)n − 2 ]
20
C10
a 0 + a1x + a 2x 2 + a 3x 3 + ..., then an is equal
to
[AIEEE 2006]
(c)
a n + 1 − bn + 1
(b)
a −b
bn + 1 − a n + 1
b −a
(d)
bn − a n
b −a
Now, (1 − ax)−1(1 − bx)−1
= (1 + ax + a2 x2 + . . .)(1 + bx + b 2 x2 + . . .)
Hence, an = Coefficient of xn in (1 − ax)−1(1 − bx)−1
= a b + ab
n −1
…(iii)
= 2 a2 + 6 a3 y + . . .
On putting y = 0 in Eq. (iii), we get
− m[−(m − 1) + n] + n[− m + (n − 1)] = 2 a2 = 20
⇒ m(m − 1) − mn − mn + n(n − 1) = 20
⇒
m2 + n2 − m − n − 2 mn = 20
⇒
(m − n)2 − (m + n) = 20
⇒
100 − (m + n) = 20
∴
m + n = 80
On solving Eqs. (ii) and (iv), we get
m = 35 and n = 45
36. The value of 50C 4 +
Exp. (b)
0 n
+ (1 − y)m − 1 n(1 + y)n − 1 ]
C10 )
1
is
(1 − ax )(1 − bx )
a n − bn
(a)
b −a
+ ... + a b
n 0
 an + 1 
−1 
 

 b
 a −1 

 b


sum of nth terms of a GP with common ratio a 

b 
2


a
a
= a0 b n  1 + +   + ... = a0 b n




b
b


(a)
56
∴
50
C4 +
=
C3 +
=
C4 +
51
54
=
55
(1 − y) (1 + y) = 1 + a1 y + a2 y + a3 y + . . .
2
55
C4
n+1
C3 +
54
C3 +
53
Cr .
52
C3
C4 +
50
C3 +
51
C3 +
52
C3 +
51
C3 +
52
C3 +
53
C3 +
51
C4 +
52
C3 +
53
C4 +
53
C4 +
54
C4 +
55
C3 +
54
C3 +
54
C3 +
55
C3 +
55
C3 =
56
C3 +
50
C3
53
C3
C3 +
54
C3 +
54
55
C3
55
C3
n+1
Cr ]
55
C3
C3
C3
C4
37. If the coefficients of rth,(r + 1)th and(r + 2 )th
terms in the binomial expansion of (1 + y )m
are in AP, then m and r satisfy the equation
Exp. (c)
n
[AIEEE 2005]
(d)
C3
[Q nC r + nC r −1 =
=
m
55
(c)
C3
+
35. For natural numbers m and n, if
(b) (45, 35)
(d) (20, 45)
C3 +
55
50
53
(a) (35, 20)
(c) (35, 45)
r =1
+
=
(1 − y ) (1 + y ) = 1 + a1y + a 2y + ... and
[AIEEE 2006]
a1 = a 2 = 10, then (m , n ) is
6
∑ 56 − rC 3 is
Since, we know that n C r + nC r −1 =
=
2
…(iv)
Exp. (a)
52
n
56
(b)
C4
b n ( an + 1 − b n + 1 )
an + 1 − b n + 1
b
⋅ n+1 =
=
a−b
a−b
b
m
…(i)
On putting y = 0 in Eq. (i), we get
…(ii)
− m + n = a 1 = 10
[Q a 1 = 10, given]
On differentiating Eq. (i), we get
− m[−(m − 1)(1 − y)m − 2 (1 + y)n
20
34. In the expansion, powers of x in the
function
= a1 + 2 a2 y + 3 a3 y2 + . . .
3
On differentiating w.r.t. y, we get
− m(1 − y)m − 1(1 + y)n + (1 − y)m n(1 + y)n − 1
(a) m 2 − m ( 4r − 1 ) + 4r 2 + 2 = 0
(b) m 2 − m ( 4r + 1 ) + 4r 2 − 2 = 0
(c) m 2 − m ( 4r + 1 ) + 4r 2 + 2 = 0
(d) m 2 − m ( 4r − 1 ) + 4r 2 − 2 = 0
[AIEEE 2005]
90
JEE Main Chapterwise Mathematics
Exp. (b)
m
Since, the coefficient of given terms are C r − 1,
m
C r , mC r + 1, respectively and they also in AP.
m
∴
C r − 1 + mC r + 1 = 2 ⋅ mC r
m!
m!
+
⇒
(r − 1)!(m − r + 1)! (r + 1)!(m − r − 1)!
m!
=2
r !(m − r )!
1
1
2
+
=
⇒
(m − r + 1)(m − r ) (r + 1)r r(m − r )
⇒
r(r + 1) + (m − r + 1)(m − r )
2
=
r(r + 1)(m − r + 1)(m − r )
r(m − r )
⇒
r + r + m + r − 2 mr + m − r
2
2
39. If x is so small that x 3 and higher powers of x
may be neglected, then
⇒
m2 − m(4 r + 1) + 4 r 2 − 2 = 0
11
1
 equal

bx 
11
1 

to the coefficient of x −7 in ax − 2  , then

bx 
[AIEEE 2005]
a and b satisfy the relation
38. If the coefficient of x 7 in ax 2 +
(a) ab =1
(b)
a
=1
b
(c) a + b =1 (d) a − b =1
Exp. (a)
Exp. (b)
1
(1 + x)3 / 2 −  1 + x

2 
=
a6
binomial expansion in powers of x of
(1 + α x )4 and of (1 − α x )6 is the same, if α is
equal to
[AIEEE 2004]
(a) −
C5
a6
b5
= C6
11
a5
b6
⇒
a6
b5
=
a5
b6
5
3
(b)
10
3
(c) −
3
10
(d)
3
5
Exp. (c)
⋅ x7
b5
So, the coefficient of x7 in the expansion of
11
6
 ax2 + 1  is 11C a ⋅


5

bx 
b5
Similarly, coefficient of x−7 in the expansion of
11
5
 ax − 1  is 11C a ⋅


6
2

bx
b6
According to the given condition,
11
(1 − x)1/ 2
40. The coefficient of the middle term in the
⇒ r=5
T6 = 11C 5
∴
(1 − x)1/ 2
3 1 

⋅
 

3
3 x 3 ⋅ 2 x2 
2
+
⋅ 
 1 + x + 2 2 x  −  1 +
2
2
2
4
2
 



[since, higher powers of x can be neglected]
11
22 − 3 r = 7
3
[neglecting higher powers of x]
3 x2
− 1/ 2
(1 − x)
=−
8
1 3


⋅

3 x2 
1
3 x2
2
2
2
=−
⋅x =−
1 + x +
8 
2
2
8



Let x7 be contained in (r + 1)th term in the
1
expansion of  ax2 +
 .

bx 
r
1
∴ Tr + 1 = 11C r (ax2 )11 − r  
 bx 
a11 − r
11
= Cr
⋅ x22 − 3 r
br
For coefficient of x7, put
[AIEEE 2005]
3
(b) − x 2
8
3
(d) 1 − x 2
8
x 3
(a) − x 2
2 8
3
(c) 3x + x 2
8
= 2 (mr − r 2 + r + m − r + 1)
4r 2 − 4 mr − m − 2 + m2 = 0
3
(1 − x )1/2
may be approximated as
2
⇒
1 

(1 + x )3/2 − 1 + x 

2 
∴ ab = 1
The coefficient of the middle term in powers of x of
(1 + α x)4 = 4C 2 α 2 .
The coefficient of the middle term in powers of x of
(1 − α x)6 = 6C 3 (− α )3 .
According to given condition, 4 C 2 α 2 = 6C 3 (− α )3
⇒
⇒
4! 2
6! 3
α =−
α ⇒ 6α 2 = − 20α 3
2 !2 !
3! 3!
6
3
∴α = −
α=−
20
10
91
Binomial Theorem and Its Simple Applications
41. The coefficient of x n in the expansion of
(1 + x )(1 − x ) is
n
[AIEEE 2004]
n −1
(c) ( −1 )
(n − 1 )
n −1
(d) ( −1 )
2
n
Exp. (b)
The coefficient of xn in the expansion of
(1 + x)(1 − x)n = Coefficient of xn in (1 − x)n
+ Coefficient of xn − 1 in (1 − x)n
[Q(1 + x) (1 − x)n = (1 + x)n + x (1 − x)n ]
n n
C n + (−1)n − 1 nC n − 1
= (−1)
n
∑
1
r=0
n
Cr
and t n =
n
∑
r=0
r
n
Cr
n
(a)
2
tn
is
Sn
[AIEEE 2004]
n
(b) − 1
2
(c) n −1
2 n −1
2
(d)
Exp. (a)
Given that, S n =
n
∑
n
r =0
Sn =
n
∑
r =0
⇒
nS n =
n
∑
n
r =0
⇒
n
∑
nS n =
r =0
⇒
nS n =
n
∑
r =0
⇒
1
Cr
n
Cn − r
 n−r

+
 nC
 n−r
n−r
+
n
Cn − r
n
r
Cn − r
n
∑
r =0

Q t n =

tn
n
=
⇒
Sn 2
nS n = t n + t n
expansion of ( 3 + 5 )
(a) 32
(b) 33
256
27  27
27
− r + 1
− 1 . . . 

 5
 r
 5

5
x
=
r!
Now, this term will be negative, if the last factor in
numerator is the only one negative factor.
27
32
⇒
− r + 1< 0 ⇒
<r
5
5
6.4 < r ⇒ Least value of r is 7.
Thus, first negative term will be 8th.




r =0
n
is
(c) 34
(b) 25
(d) None of these
Exp. (d)
(1 + 2 x + 3 x2 + . . .)−3 / 2 = [(1 − x)−2 ]−3 / 2
r
n
Cr
n
∑
[AIEEE 2002]
(a) 21
(c) 26
n
= (1 − x)3
So, coefficient of x in (1 + 2 x + 3 x2 + . . .)−3 / 2
5
1 
C1 
= Coefficient of x5 in (1 − x)3 = 0
n
∑
n
r
Cr
r
, given
Cr



46. If | x | < 1, then the coefficient of x n in the
expansion of (1 + x + x 2 + x 3 + ...)2 is
(b) n −1
(d) n + 1
(a) n
(c) n + 2
[AIEEE 2002]
Exp. (d)
(1 + x + x2 + x3 + . . .)2 = [(1 − x)−1 ]2 = (1 − x)−2
43. The number of integral terms in the
8
Since,(r + 1)th term in the expansion of(1 + x)27/ 5 .
is
 n
n−1
nS n =  n
+ n
+ ... +
 C
Cn − 1
 n
nS n = 2 t n
[AIEEE 2003]
(b) 5th term
(d) 6th term
45. The coefficient of x 5 in (1 + 2 x + 3x 2 + ...)−3/2
r =0
⇒
256
expansion of (1 + x )27/5 is
[Q nC r = nC n − r ]
+
⇒
=
44. If x is positive, the first negative term in the
⇒
1
n
Cn−r
5 )256 is
Exp. (c)
, then
equal to
8
C r (3)( 256 − r )/ 2 (5)r / 8
256 − r
r
For integral terms,
and are both positive
2
8
integers.
i.e., r = 0, 8, 16, 24, 32 , . . . , 256
Hence, total number of terms are 33.
+1
(a) 7th term
(c) 8th term

n! 
n
= (−1)n  1 −
 = (−1) (1 − n)
1!(n − 1)!

42. If Sn =
The general term of ( 3 +
Tr
(b) ( −1 )n (1 − n )
(a) (n −1 )
Exp. (b)
[AIEEE 2003]
(d) 35
Coefficient of xn in (1 + x + x2 + . . .)2
= Coefficient of xn in (1 − x)−2
=
n + 2 −1
C2 − 1 =
n+1
C1 = n + 1
7
Sequences and Series
1. The sum of the series
2 ⋅ C0+ 5⋅
20
20
= (60 × 219 ) + (2 × 2 20 )
C1+ 8⋅ C2+ 11⋅ C3
+ ..... + 62 ⋅20 C20 is equal to
20
20
= (15 × 2 21 ) + 2 21
[JEE Main 2019, 8 April Shift-I]
(a) 2 26
(c) 2 23
 n n

n
Q ∑ C r = 2 
 r = 0

(b) 2 25
(d) 2 24
= 16 × 2 21 = 2 25
2. The sum of all natural numbers ‘n’ such that
100 < n < 200 and HCF (91, n)>1 is
Exp. (b)
[JEE Main 2019, 8 April Shift-I]
Given series is
2 ⋅ 20C 0 + 5 ⋅ 20C1 + 8 ⋅ 20C 2 + … + 62 ⋅ 20C 20
=
20
∑ (3r + 2 )⋅
20
(a) 3203
(c) 3221
Cr
Exp. (d)
r =0
[Qgeneral term of the sequence 2, 5, 8, …, which
forms an AP, is 2 + (n − 1)3 = 3n − 1, where
n = 1, 2, 3 ... and it can be written as 3n + 2,
where n = 0, 1, 2, 3]
= 3⋅
20
∑r
r =0
20
C r + 2 ∑ 20C r
20
r =0
20
20
= 3 ∑ r   19C r

r 
r =1
−1+2
20
∑
20
Cr
r =0
Q nC = n n − 1C
r
r

r
20
= 3 × 20 ∑ C r
19
r =1
19
20
−1
+2∑
20


− 1
Cr
r =0
20
r =0
 20 19
Q ∑ C r
 r = 1
−1
=
The natural numbers between 100 and 200 are
101, 102, 103, …, 199.
Since, 91 = 13 × 7, so the natural numbers
between 100 and 200 whose HCF with 91 is more
than 1 are the numbers which are either divisible
by 7 or 13.
So, the required sum of numbers between 100 and
200 = (sum of numbers divisible by 7) + (sum of
numbers divisible by 13) − (sum of numbers
divisible by 91)
=
14
8
r =1
r =1
∑ (98 + 7 r ) + ∑ (91 + 13r ) − (182 )
14 × 15 
= (98 × 14) + 7 
 + (91 × 8)
 2 
8 × 9
+ 13 
 − (182 )
 2 
= 60 ∑ 19C r + 2 ∑ 20C r
r =0
(b) 3303
(d) 3121
19
∑
19
r =0

Cr 

= 1372 + 735 + 728 + 468 − 182
= 3303 − 182
= 3121
93
Sequences and Series
20
3. The sum
1
∑k 2k is equal to
k =1
[JEE Main 2019, 8 April Shift-II]
(a) 2 −
(c) 2 −
11
(b) 1 −
219
3
(d) 2 −
217
⇒
11
2 20
21
⇒
2 20
⇒
Exp. (a)
1
Let S = ∑ k  k 
 
k =1 2
1
2
3
4
20
…(i)
+
+
+
+ … + 20
2 22 23 24
2
1
On multiplying by   both sides, we get
2
S
1
2
3
19
20
…(ii)
=
+
+
+ … + 20 + 21
2 22 23 24
2
2
On subtracting Eq. (ii) from Eq. (i), we get
S 1
1
1
1
20
S − = + 2 + 3 + … + 20 − 21
2 2 2
2
2
2
1
1 
 1 − 20 

S 2
20
2
=
− 21
⇒
1
2
2
1−
2


a(1 − r n )
, r < 1
Qsum of GP =
1− r


S=
⇒
S =2 −
1
20
2
11
−
20
2
21
= 1−
1
2
20
−
10
2
20
= 1−
11
2
20
219
4. If the lengths of the sides of a triangle are in
AP and the greatest angle is double the
smallest, then a ratio of lengths of the sides
of this triangle is [JEE Main 2019, 8 April Shift-II]
(a) 3 : 4 : 5
(c) 5 : 9 : 13
6 − 8sin2 θ = 1 + 2 cos θ
[Qsinθ can not be zero]
6 − 8(1 − cos 2 θ) = 1 + 2 cos θ
8 cos 2 θ − 2 cos θ − 3 = 0
⇒ (2 cos θ + 1)(4cos θ − 3) = 0
3
cosθ =
⇒
4
20
= 1−
⇒
2 sin( π − 3θ) = sinθ + sin2 θ [From Eq. (ii)]
⇒
2 sin 3 θ = sinθ + sin2 θ
⇒ 2 [3sinθ − 4sin3 θ] = sinθ + 2 sinθcos θ
(b) 4 : 5 : 6
(d) 5 : 6 : 7
Exp. (b)
Let a, b and c be the lengths of sides of a ∆ABC
such that a < b < c.
Since, sides are in AP.
…(i)
∴
2b = a + c
Let
∠A =θ
Then,
C = 2θ [according to the question]
So,
…(ii)
B = π − 3θ
On applying sine rule in Eq. (i), we get
2sin B = sin A + sinC
1
(rejected).
2
Clearly, the ratio of sides is a : b : c
= sinθ : sin 3θ : sin2 θ
= sinθ : (3sinθ − 4sin3 θ) : 2 sinθcos θ
or
cosθ = −
= 1 : (3 − 4sin2 θ) : 2 cos θ
= 1 : (4cos 2 θ − 1) : 2 cos θ
5 6
= 1: : = 4 : 5 : 6
4 4
5. If three distinct numbers a ,b and c are in GP
and the equations ax 2 + 2bx + c = 0 and
dx 2 + 2ex + f = 0 have a common root, then
which one of the following statements is
correct?
[JEE Main 2019, 8 April Shift-II]
(a) d , e and f are in GP
f
d e
(b) , and are in AP
a b
c
(c) d , e and f are in AP
f
d e
(d) , and are in GP
a b
c
Exp. (b)
Given, three distinct numbers a, b and c are in GP.
…(i)
∴
b 2 = ac
and the given quadratic equations
ax2 + 2 bx + c = 0
dx2 + 2ex + f = 0
…(ii)
…(iii)
For quadratic Eq. (ii),
the discriminant D = (2 b )2 − 4ac
= 4(b 2 − ac ) = 0
[from Eq. (i)]
⇒ Quadratic Eq. (ii) have equal roots, and it is
b
equal to x = − , and it is given that quadratic
a
Eqs. (ii) and (iii) have a common root, so
94
JEE Main Chapterwise Mathematics
10
10
 10 
So, Σ f(a + k ) = Σ λa + k = λa  Σ λk 
k =1
k =1
 k =1 
2
b
b
d  −  + 2e  −  + f = 0
 a
 a
⇒
db 2 − 2eba + a2 f = 0
⇒
d (ac ) − 2eab + a2 f = 0
⇒
⇒
= 2 a [21 + 2 2 + 2 3 + ......+210 ]
[Qa ≠ 0]
dc − 2eb + af = 0
2eb = dc + af
e dc
af
2 = 2 + 2
b b
b
[dividing each term by b 2 ]
e d
f

[Qb 2 = ac]
2  = +
 b a c
⇒
⇒
 2(210 − 1)
= 2a 

 2 −1 
[by using formula of sum of n-terms of a GP
having first term ‘a’ and common ratio ‘r’, is

a(r n − 1)
, where r > 1
Sn =
r −1

[Qb 2 = ac]
d e f
, , are in AP.
a b c
Alternate Solution
Given, three distinct numbers a, b and c are in GP.
Let a = a, b = ar, c = ar 2 are in GP, which satisfies
ax2 + 2 bx + c = 0
⇒ 2 a + 1 (210 − 1) = 16 (210 − 1) (given)
⇒
2 a + 1 = 16 = 2 4 ⇒ a + 1 = 4 ⇒ a = 3
So,
∴
ax2 + 2(ar )x + ar 2 = 0
⇒
x2 + 2 rx + r 2 = 0
⇒
[Qa ≠ 0]
( x + r )2 = 0 ⇒ x = − r.
According to the question, ax2 + 2 bx + c = 0 and
dx2 + 2ex + f = 0 have a common root.
So, x = − r satisfies dx + 2ex + f = 0
2
∴
d (− r ) + 2e(− r ) + f = 0
⇒
dr 2 − 2er + f = 0
c
c
d   − 2e   + f = 0
 b
 a
d 2e
f
−
+ =0
a
b
c
d
f 2e
+ =
a c
b
⇒
⇒
[Qc ≠ 0]
10
6. Let
∑ f (a + k ) = 16(210 − 1), where the
k =1
function f satisfies f ( x + y ) = f ( x ) f ( y ) for
all natural numbers x , y and f (1) = 2. Then,
the natural number ‘a’ is
[JEE Main 2019, 9 April Shift-I]
(a) 2
(c) 3
(b) 4
(d) 16
Exp. (c)
Given,
Let
Q
∴
f( x + y) = f( x) ⋅ f( y)
f ( x) = λ x
f(1) = 2
λ =2
non-constant AP a1 , a 2 , a 3.....be
n(n − 7)
A , where A is a constant. If d
50n +
2
is the common difference of this AP, then
the ordered pair (d , a 50 ) is equal to
[JEE Main 2019, 9 April Shift-I]
(a) (A, 50 + 46A)
(c) (50, 50 + 46A)
(b) (50, 50 + 45A)
(d) (A, 50 + 45A)
Exp. (a)
Key Idea Use the formula of sum of first n terms
n
of AP, i.e Sn = [2 a + (n − 1)d ]
2
2
⇒
7. Let the sum of the first n terms of a
[where λ > 0]
(given)
Given AP, is a1, a2 , a3 ,… having sum of first n-terms
n
= [2 a1 + (n − 1)d ]
2
[where, d is the common difference of AP]
n(n − 7 )
(given)
= 50n +
A
2
1
n−7
[2 a1 + (n − 1)d ] = 50 +
A
⇒
2
2
n
7
1
⇒
[2 a1 + nd − d ] =  50 − A + A


2
2
2
d  nd 
7  n

⇒
=  50 − A + A
 a1 −  +


2  2
2
2
On comparing corresponding term, we get
d
7
d = A and a1 − = 50 − A
2
2
A
7
[Q d = A]
a1 − = 50 − A
⇒
2
2
⇒
a1 = 50 − 3 A
So
a50 = a1 + 49d
= (50 − 3 A) + 49 A [Q d = A]
= 50 + 46A
Therefore, (d , a50 ) = ( A, 50 + 46 A)
95
Sequences and Series
8. The sum of the series
1 + 2 × 3 + 3 × 5 + 4 × 7 +... upto 11th term is
[JEE Main 2019, 9 April Shift-II]
(a) 915
(b) 946
(c) 916
(d) 945
Exp. (b)
Given series is
1 + (2 × 3) + (3 × 5) + (4 × 7 ) + …upto
11 terms.
Now, the rth term of the series is ar = r(2 r − 1)
∴Sum of first 11-terms is
S11 =
11
11
r =1
r =1
∑ r(2 r − 1) = ∑ (2 r
=2
2
11
− r) = 2 ∑ r −
2
r =1
11
∑r
r =1
11 × (11 + 1)(2 × 11 + 1) 11 × (11 + 1)
−
6
2
n
 n 2 n(n + 1)(2 n + 1)
n(n + 1)
and ∑ r =
Q ∑ r =

6
2 
 r = 1
r =1
11 × 12 × 23   11 × 12 
= 
 −


  2 
3
= (11 × 4 × 23) − (11 × 6)
= 11(92 − 6) = 11 × 86 = 946
9. Some identical balls are arranged in rows to
form an equilateral triangle. The first row
consists of one ball, the second row consists
of two balls and so on. If 99 more identical
balls are added to the total number of balls
used in forming the equilateral triangle,
then all these balls can be arranged in a
square whose each side contains exactly 2
balls less than the number of balls each side
of the triangle contains. Then, the number
of balls used to form the equilateral triangle
is
[JEE Main 2019, 9 April Shift-II]
(a) 262
(b) 190
(c) 225
(d) 157
Exp. (b)
Let there are n balls used to form the sides of
equilateral triangle.
According to the question, we have
n(n + 1)
+ 99 = (n − 2 )2
2
⇒
n2 + n + 198 = 2 [n2 − 4n + 4]
⇒
n = 19, − 10
⇒
n = 19
[Qnumber of balls n > 0]
Now, number of balls used to form an equilateral
n(n + 1) 19 × 20
triangle is
=
= 190.
2
2
10. If the sum and product of the first three
terms in an AP are 33 and 1155, respectively,
then a value of its 11th term is
[JEE Main 2019, 9 April Shift-II]
(a) 25
(b) –36
(c) –25
(d) –35
Exp. (c)
Let first three terms of an AP as a − d , a, a + d .
So,
3a = 33 ⇒ a = 11
[given sum of three terms = 33
and product of terms = 1155]
[given]
⇒ (11 − d )11(11 + d ) = 1155
⇒
112 − d 2 = 105
⇒
d 2 = 121 − 105 = 16
⇒
d=±4
So the first three terms of the AP are either 7, 11,
15 or 15, 11, 7.
So, the 11th term is either 7 + (10 × 4) = 47
or 15 + (10 × (−4)) = − 25.
11. If a1 , a 2 , a 3 ,...,an are in AP and
a1 + a 4 + a 7 + ... + a16 = 114 , then
a1 + a 6 + a11 + a16 is equal to
[JEE Main 2019, 10 April Shift-I]
(a) 64
(b) 76
(c) 98
(d) 38
Exp. (b)
Key Idea Use nth term of an AP i.e.
an = a + (n − 1)d, simplify the given equation and
use result.
Given AP is a1, a2 , a3 , … , an
Let the above AP has common difference ‘d’, then
a1 + a4 + a7 + … + a16
= a1 + (a1 + 3d ) + (a1 + 6d ) + … + (a1 + 15d )
= 6a1 + (3 + 6 + 9 + 12 + 15)d
= 6a1 + 45d = 114
(given)
⇒
2 a1 + 15d = 38
Now, a1 + a6 + a11 + a16
…(i)
⇒
n2 − 9n − 190 = 0
= a1 + (a1 + 5d ) + (a1 + 10d ) + (a1 + 15d )
⇒
n − 19n + 10n − 190 = 0
⇒
(n − 19)(n + 10) = 0
= 4a1 + 30d = 2(2 a1 + 15d ) = 2 × 38 = 76
[from Eq. (i)]
2
96
JEE Main Chapterwise Mathematics
12. The sum of series
7 × (13 + 2 3 + 33 )
+
1 +2 + 3
2
2
2
is
3 × 13
1
2
+
5 × (13 + 2 3 )
12 + 2 2
Exp. (a)
Given series,
13 + 2 3 13 + 2 3 + 33
+ ... +
+
1+ 2
1+ 2 + 3
13 + 2 3 + 33 + K + 153
1 + 2 + 3 + K + 15
1
− (1 + 2 + 3 + K + 15)
2
= S1 − S 2 (let)
S = 1+
+ ....... + upto 10th term,
[JEE Main 2019, 10 April Shift-I]
(a) 680
(c) 660
(b) 600
(d) 620
Exp. (c)
Given series is
7 × (13 + 2 3 + 33 )
3 × 13 5 × (13 + 2 3 )
+
+
+ ...
2
2
2
1
1 +2
12 + 2 2 + 33
where,
S1 = 1+
13 + 2 3 13 + 2 3 + 33
+ K+
+
1+ 2
1+ 2 + 3
13 + 2 3 + 33 + K + 153
1 + 2 + 3 + K + 15
So, nth term
Tn =
(3 + (n − 12
) )(13 + 2 3 + 33 ... + n3 )
2
 n(n + 1)


1 + 2 + K+ n
 2 
= ∑
= ∑
n(n + 1)
n =1 1+ 2 + K + n
n =1
2
2
n
 n
n(n + 1)
 n(n + 1) and
r=
Q ∑ r 3 = 


∑


2
2 
 r = 1
r =1
12 + 2 2 + 32 + K + n2
15
2
n (n + 1)
(2 n + 1) × 

 2

=
n(n + 1)(2 n + 1)
6
2
n
n
n(n + 1)(2 n + 1)
n(n + 1)

3
and Σ r 2 =
[Qr Σ= 1 r = 


r
=1
6
 2 

3n(n + 1) 3 2
So,
Tn =
= (n + n)
2
2
Now, sum of the given series upto n terms
3
S n = ΣTn = [Σn2 + Σn]
2
3 n(n + 1)(2 n + 1) n(n + 1)
= 
+
2 
6
2 
∴
3  10 × 11 × 21 10 × 11
+
2 
6
2 
3
3
= [(5 × 11 × 7 ) + (5 × 11)] = × 55(7 + 1)
2
2
3
= × 55 × 8 = 3 × 55 × 4
2
S10 =
= 12 × 55 = 660
13. The sum of series
=
3
3
3
n(n + 1) 1
=
2
2
n =1
15
∑
15
∑ (n
2
15
+ n)
n =1
1  15 × 16 × 31 15 × 16 
+
2 
6
2 
 n 2 n(n + 1) (2 n + 1)

Q ∑ r =
6

 r = 1
1
= [(5 × 8 × 31) + (15 × 8)]
2
= (5 × 4 × 31) + (15 × 4)
= 620 + 60 = 680
1
and S 2 = (1 + 2 + 3 + K + 15)
2
1 15 × 16
= ×
= 60
2
2
Therefore, S = S1 − S 2 = 680 − 60 = 620.
=
14. The angles A , B and C of a ∆ABC are in AP
1 +2
1 +2 + 3
+ ...
+
1+2
1+2 + 3
13 + 2 3 + 33 + K + 153 1
− (1 + 2 + 3 + K + 15)
+
1 + 2 + 3 + K + 15
2
is equal to
[JEE Main 2019, 10 April Shift-II]
and a :b = 1 : 3. If c = 4 cm, then the area (in
sq cm) of this triangle is
(a) 620
(c) 1240
Exp. (c)
1+
3
3
3
3
3
(b) 660
(d) 1860
[JEE Main 2019, 10 April Shift-II]
(a)
2
3
(b) 4 3
(c) 2 3
(d)
4
3
It is given that angles of a ∆ABC are in AP.
So,
∠A + ∠B + ∠C = 180º
97
Sequences and Series
⇒
∠B − d + ∠B + ∠B + d = 180º
[if ∠A, ∠B and ∠C are in AP, then it taken as
∠B − d, ∠B, ∠B + d respectively, where d is
common difference of AP]
…(i)
⇒ 3∠B = 180º ⇒ ∠B = 60º
1
a
[given]
and
=
3
b
1
sin A
=
⇒
3
sin B
bu sine rule sin A = sin B = sinC 

a
b
c 

3
1
sin A
⇒
=
Qsin B = sin 60° =

3 
3
2 
2
1
⇒ sin A = ⇒ ∠A = 30º
2
So,
∠C = 90º
∴ From sine rule,
a
b
c
=
=
sin A sin B sinC
4
a
b
⇒
=
=
[Qc = 4 cm]
1
3 1
2
2
⇒
a = 2 cm, b = 2 3 cm
1
1
∴ Area of ∆ABC = ab sinC = × 2 × 2 3 × 1
2
2
= 2 3 sq. cm
15. Let a1 , a 2 , a 3 ,K be an AP with a 6 = 2. Then,
the common difference of this AP, which
maximises the product a1 , a 4 , a 5, is
[JEE Main 2019, 10 April Shift-II]
8
(a)
5
2
(b)
3
(c)
3
2
(d)
6
5
Exp. (a)
Given, the terms a1, a2 , a3 , K , are an AP.
Let the common difference of this AP is ‘d’ and first
term a1 = a, then
(given) … (i)
a6 = a + 5d = 2
Now, a1, a4 , a5
= a(a + 3d ) (a + 4d )
[from Eq. (i)]
= (2 − 5d ) (2 − 2d ) (2 − d )
= (4 − 14d + 10d 2 ) (2 − d )
= 8 − 4d − 28d + 14d 2 + 20d 2 − 10d 3
= − 10d 3 + 34d 2 − 32d + 8
(let)
= f(d )
On differentiating f(d ) w.r.t.d, we get
f ′(d ) = − 30d 2 + 68d − 32
For maxima or minima, f ′(d ) = 0
⇒
− 30d 2 + 68d − 32 = 0
⇒
15d 2 − 34d + 16 = 0
⇒
15d − 24d − 10d + 16 = 0
2
⇒
3d (5d − 8) − 2(5d − 8) = 0
2 8
d= ,
⇒
3 5
and f ′ ′(d ) = − 60d + 68
2
120
2
at d = , f ′ ′  d =  = −
+ 68 = 28 > 0, and at

3
3
3
8
8
480
d = , f ′ ′  d =  = −
+ 68 = − 28 < 0

5
5
5
So, at d = 8 / 5, product a1a4 a5 is maximum.
16. Let a ,b and c be in GP with common ratio r,
1
where a ≠ 0 and 0 < r ≤ . If 3a, 7b and 15c are
2
the first three terms of an AP, then the 4th
term of this AP is
[JEE Main 2019, 10 April Shift-II]
2
(b) a
3
(a) 5a
(c) a
(d)
7
a
3
Exp. (c)
Key Idea Use nth term of AP i.e.,
an = a + (n − 1) d, If a, A, b are in AP, then
2A = a + b and nth term of G.P. i.e., an = ar n − 1.
It is given that, the terms a, b, c are in GP with
1
common ratio r, where a ≠ 0 and 0 < r ≤ .
2
So, let, b = ar and c = ar 2
Now, the terms 3a, 7b and 15c are the first three
terms of an AP, then
2(7 b ) = 3a + 15c
⇒
14ar = 3a + 15ar 2
⇒
14r = 3 + 15r 2
⇒
15r − 14r + 3 = 0
⇒
15r − 5r − 9r + 3 = 0
⇒
5r(3r − 1) − 3(3r − 1) = 0
⇒
[as b = ar, c = ar 2 ]
[as a ≠ 0]
2
2
(3r − 1) (5r − 3) = 0
1 3
r = or
3 5
1
1
as, r ∈  0, , so r =
 2 
3
Now, the common difference of AP = 7 b − 3a
7
2a
= 7 ar − 3a = a − 3 = −
3 
3
−2 a 

So, 4th term of AP = 3a + 3 
 =a
 3 
⇒
98
JEE Main Chapterwise Mathematics
17. For x ∈R , let [ x ] denote the greatest integer
So given series
− 1  + − 1 − 1  + − 1 − 2  + …
 3   3 100   3 100 
− 1 99 
K+ 
−
100 
 3
1 
  1
  1

= −
− 1 +  −
+
− 1
  3 
   3 100 

1
99 
1
2  



− 1
− 1 + K +  −  +
+ −  +
  3 100 

  3 100  
1


= (− 1) × 100 −
× 100 = − 100 − 33 = − 133.
 3

≤ x, then the sum of the series
1   1
2 
 1  1
− 3 + − 3 − 100 + − 3 − 100 +
 1 99 
is
… + − −
 3 100
[JEE Main 2019, 12 April Shift-I]
(a) −153
(c) −131
(b) −133
(d) −135
Exp. (b)
Given series is
− 1  + − 1 − 1  + − 1 − 2  + K
 3   3 100   3 100 
1
99 
... +  − −
 3 100 
[where, [ x] denotes the greatest integer ≤ x]
Now,
 − 1 ,  − 1 − 1  +  − 1 − 2 ,…+  − 1 − 66 
 3   3 100   3 100 
 3 100 
all the term have value − 1
1 67   1
68 
1
99 
, − −
, …,  − −
all
and  − −
 3 100   3 100 
 3 100 
the term have value − 2.
1
1
1   1
2 
So,  −  +  − −
+ ... +
+ − −
 3   3 100   3 100 
 − 1 − 66 
 3 100 
= − 1 − 1 − 1 − 1 K 67 times.
= (− 1) × 67 = − 67
1 67   1
68 
and  − −
+ K+
+ − −
 3 100   3 100 
 − 1 − 99 
 3 100 
= − 2 − 2 − 2 − 2 K 33 times = (−2 ) × 33 = −66
1
1
1   1
2 
∴ −  + − −
+ K+
+ − −
 3   3 100   3 100 
 − 1 − 99 
 3 100 
= (− 67 ) + (− 66) = − 133.
Alternate Solution
Q [− x] = − [ x] − 1, if x ∉Integer,
n − 1
1
2
and [ x] +  x +  +  x +  + K +  x +

n 
n  
n 

= [nx], n ∈ N.
18. Let Sn denote the sum of the first n terms of
an AP. If S 4 = 16 and S 6 = − 48, then S10 is
equal to
[JEE Main 2019, 12 April Shift-I]
(a) − 260
(b) − 410
(c) − 320
(d) − 380
Exp. (c)
Given S n denote the sum of the first n terms of an
AP.
Let first term and common difference of the AP be ‘
a’ and ‘d’, respectively.
(given)
∴
S 4 = 2[2 a + 3d ] = 16
QS = n [2 a + (n − 1)d ]
 n 2

… (i)
⇒
2 a + 3d = 8
and
[given]
S 6 = 3[2 a + 5d ] = − 48
… (ii)
⇒
2 a + 5d = − 16
On subtracting Eq. (i) from Eq. (ii), we get
2d = − 24
⇒
d = − 12
So,
[putd = −12 in Eq. (i)]
2 a = 44
Now,
S10 = 5[2 a + 9d ]
5[44 + 9(−12 )] = 5[44 − 108]
= 5 × (− 64) = − 320
19. If a1 , a 2 , a 3 ,... are in AP such that
a1 + a 7 + a16 = 40, then the sum of the first 15
terms of this AP is
[JEE Main 2019, 12 April Shift-II]
(a) 200
(b) 280
(c) 120
(d) 150
Exp. (a)
Let the common difference of given AP is ‘d’.
Since,
a1 + a7 + a16 = 40
∴ a1 + a1 + 6d + a1 + 15d = 40
[Q an = a1 + (n − 1) d ]
99
Sequences and Series
…(i)
⇒
3a1 + 21d = 40
Now, sum of first 15 terms is given by
15
[2 a1 + (15 − 1) d ]
S15 =
2
15
= [2 a1 + 14d ] = 15 [a1 + 7d ]
2
From Eq. (i), we have
40
a1 + 7d =
3
40
So, S15 = 15 ×
= 5 × 40 = 200
3
20. If α , β and γ are three consecutive terms of a
non-constant GP such that the equations
αx 2 + 2βx + γ = 0 andx 2 + x − 1 = 0 have a
common root, then, α(β + γ) is equal to
[JEE Main 2019, 12 April Shift-II]
(a) 0
(b) αβ
(c) αγ
(d) βγ
Exp. (d)
Given α, β and γ are three consecutive terms of a
non-constant GP.
Let
α = α, β = αr, γ = αr 2, {r ≠ 0, 1}
and given quadratic equation is
αx2 + 2 βx + γ = 0
…(i)
Exp. (b)
Let b = ar and c = ar 2 , where r is the common ratio.
Then,
⇒
a + b + c = xb
a + ar + ar 2 = xar
⇒
1 + r + r 2 = xr
1+ r + r
1
= 1+ r +
r
r
1
We know that, r + ≥ 2 (for r > 0)
r
1
and
r + ≤ − 2 (for r < 0)
r
[using AM ≥ GM]
1
1
1 + r + ≥ 3 or 1 + r + ≤ −1
∴
r
r
⇒
x ≥ 3 or x ≤ −1
⇒
x ∈ (− ∞,−1] ∪ [3, ∞ )
Hence, x cannot be 2.
Alternate Method
From Eq. (i), we have
1 + r + r 2 = xr
⇒
x=
⇒
r 2 + (1 − x)r + 1 = 0
For real solution of r, D ≥ 0.
⇒
(1 − x)2 − 4 ≥ 0
On putting the values of α,β, γ in Eq. (i), we get
αx2 + 2αrx + αr 2 = 0
⇒
x2 − 2 x − 3 ≥ 0
⇒
( x − 3)( x + 1) ≥ 0
⇒
x + 2 rx + r = 0
⇒
( x + r) = 0
⇒
x ∈ (−∞, − 1] ∪ [3, ∞ )
∴ x cannot be 2.
2
2
2
⇒
x=−r
QThe quadratic equations αx2 + 2 βx + γ = 0 and
x2 + x − 1 = 0 have a common root, so x = − r
must be root of equation x2 + x −1 = 0, so
r2 − r − 1 = 0
Now,
…(ii)
α (β + γ ) = α (αr + αr )
2
= α 2 (r + r 2 )
From the options,
βγ = αr ⋅ αr 2 = α 2 r 3 = α 2 (r + r 2 )
[Q r 2 − r − 1 = 0 ⇒ r 3 = r + r 2 ]
21. If a ,b and c be three distinct real numbers in
GP and a + b + c = xb , then x cannot be
[JEE Main 2019, 9 Jan Shift-I]
(a) 4
(c) −2
(b) 2
(d) −3
… (i) [Qa ≠ 0]
2
30
22. Let a1 , a 2 ,.....a 30 be an AP, S = ∑ai and
i =1
T =
15
∑ a(2i − 1). If a5 = 27 and S − 2T = 75,
i=1
then a10 is equal to
(a) 42
(b) 57
[JEE Main 2019, 9 Jan Shift-I]
(c) 52
(d) 47
Exp. (c)
We have, S = a1 + a2 + … + a30
…(i)
= 15[2 a1 + 29d ]
(where d is the common difference)
QS = n [2 a + (n − 1)d ]
 n 2

and
T = a1 + a3 + … + a29
15
= [2 a1 + 14 × 2 d )]
2
(Qcommon difference is 2d)
…(ii)
⇒
2T = 15[2 a1 + 28d ]
100
JEE Main Chapterwise Mathematics
From Eqs. (i) and (ii), we get
[QS − 2T = 75]
S − 2T = 15d = 75
⇒
d=5
Now,
a10 = a5 + 5d = 27 + 25 = 52
Now, required sum =
respectively of a non-constant AP. If these
are also the three consecutive terms of a GP,
a
then is equal to
[JEE Main 2019, 9 Jan Shift-II]
c
(b)
7
13
(c) 4
(d)
1
2
Exp. (c)
( A + 10d )2 = ( A + 6d ) ( A + 12 d )
⇒ A 2 + 20 Ad + 100d 2 = A 2 + 18 Ad + 72d 2
⇒
2 Ad + 28d 2 = 0
⇒
⇒
But
⇒
∴
and
⇒
2d ( A + 14d ) = 0
d = 0 or A + 14d = 0
[Qthe series is non constant AP]
d≠0
A = − 14d
a = A + 6d = − 14d + 6d = − 8d
c = A + 12 d = − 14d + 12 d = − 2d
a − 8d
=4
=
c − 2d
24. The sum of the following series
9(12 + 2 2 + 32 ) 12 (12 + 2 2 + 32 + 42 )
+
7
9
15 (12 + 2 2 + ... + 52 )
+
+ ... up to 15 terms is
11
1+6+
[JEE Main 2019, 9 Jan Shift-II]
(a) 7510
1
2
15
∑ (r
3
+ r2 )
r =1
=
1
2
 n(n + 1) 2 n(n + 1) (2 n + 1)
+


6
 n = 15
 2 
=
1
2
=
1
2
=
1
2
 n(n + 1)  n2 + n 2 n + 1 
+



3 
 2
 2
n = 15
 n(n + 1) (3n2 + 7 n + 2 )


6
 n = 15
 2
15 × 16 (3 × 225 + 105 + 2 )
×
×
= 7820
2
6
25. The sum of all two digit positive numbers
Let A be the Ist term of AP and d be the common
difference.
∴ 7th term = a = A + 6d
[Qnth term = A + (n − 1)d ]
11th term = b = A + 10d
13th term = c = A + 12d
Qa, b, c are also in GP
∴
b 2 = ac
⇒
=
r =1
23. Leta ,b andc be the 7th, 11th and13th terms
(a) 2
15
∑ Tr
(b) 7820
(c) 7830
(d) 7520
Exp. (b)
General term of the given series is
3r(12 + 2 2 + K + r 2 ) 3r[r(r + 1) (2 r + 1)]
Tr =
=
2r + 1
6(2 r + 1)
1 3
2
= (r + r )
2
which when divided by 7 yield 2 or 5 as
remainder is
[JEE Main 2019, 10 Jan Shift-I]
(a) 1256
(b) 1465
(c) 1356
(d) 1365
Exp. (c)
Clearly, the two digit number which leaves
remainder 2 when divided by 7 is of the form
[by Division Algorithm]
N = 7k + 2
For,
k = 2, N = 16
k = 3, N = 23
M
M
k = 13, N = 93
∴ 12 such numbers are possible and these
numbers forms an AP.
12
QS = n (a + l )
Now, S =
[16 + 93] = 654


n


2
2
Similarly, the two digit number which leaves
remainder 5 when divided by 7 is of the form
N = 7k + 5
For k = 1,
N = 12
k = 2,
N = 19
M
k = 13,
N = 96
∴ 13 such numbers are possible and these
numbers also forms an AP.
13
n
Now, S ′ = [12 + 96] = 702 QS n = (a + l )


2
2
Total sum = S + S ′ = 654 + 702 = 1356
26. Let a1 , a 2 ,...., a10 be a GP. If
a9
equals
a5
a3
= 25, then
a1
[JEE Main 2019, 11 Jan Shift-I]
(a) 53
(b) 2( 52 )
(c) 4( 52 )
(d) 54
101
Sequences and Series
Exp. (d)
Exp. (c)
Let r be the common ratio of given GP, then we
have the following sequence
a1, a2 = a1r, a3 = a1r 2 , ..., a10 = a1r 9
a3 = 25 a1
a1r 2 = 25 a1 ⇒ r 2 = 25
Now,
⇒
a
a r8
Consider, 9 = 1 4 = r 4 = (25)2 = 54
a5 a1r
27. The sum of an infinite geometric series with
positive terms is 3 and the sum of the cubes
27
of its terms is . Then, the common ratio of
19
this series is
[JEE Main 2019, 11 Jan Shift-I]
(a)
4
9
(b)
2
3
(c)
2
9
(d)
1
3
Let the GP be a, ar, ar , ar , .... ∞; where a > 0 and
0 < r < 1.
Then, according the problem, we have
a
3=
1− r
27
and
= a3 + (ar )3 + (ar 2 )3 + (ar 3 )3 + ...
19

a3
27
a 
=
⇒
QS ∞ = 1 − r 
19 1 − r 3



a
27 (3(1 − r ))3 
=
⇒
Q 3 = 1 − r ⇒ a = 3(1 − r )
19
1 − r3


3
27 27 (1 − r ) (1 + r 2 − 2 r )
=
19
(1 − r ) (1 + r + r 2 )
⇒
r + r + 1 = 19(r 2 − 2 r + 1)
2
⇒
18r 2 − 39r + 18 = 0
⇒
6r 2 − 13r + 6 = 0
⇒
(3r − 2 )(2 r − 3) = 0
2
3
r = or r = (reject)
3
2
By using AM ≥ GM (because x, y ∈ R + ), we get
( xm + x− m ) ≥ 2 and ( yn + y− n ) ≥ 2
1
[Qif x > 0, then x + ≥ 2]
x
⇒
( xm + x− m )( yn + y− n ) ≥ 4
1
1
≤
⇒
( xm + x− m )( yn + y− n ) 4
1
∴Maximum value = .
4
29. If 19th term of a non-zero AP is zero, then its
[Q0 < r < 1]
[JEE Main 2019, 11 Jan Shift-II]
(c)
1
4
(b) 4 : 1
(c) 2 : 1
(d) 3 : 1
Exp. (d)
Let t n be the nth term of given AP. Then, we have
t 19 = 0
[Qt n = a + (n − 1)d ]
⇒
a + (19 − 1)d = 0
…(i)
⇒
a + 18d = 0
t 49 a + 48d − 18d + 48d
Now,
[using Eq. (i)]
=
=
t 29
a + 28d − 18d + 28d
30d
=
= 3:1
10d
30. Let Sn = 1 + q + q 2 + K + q n and
n
 q + 1
 q + 1  q + 1
Tn = 1 + 
 ,
 +K + 
 +
 2 
 2   2 
where q is a real number and q ≠ 1. If
101
C 1 + 101C 2 ⋅ S1 + K + 101C 101 ⋅ S100 = αT100,
then α is equal to
(a) 2100
(c) 200
(b) 202
(d) 2 99
Exp. (a)
positive integers. The maximum value of the
xm y n
is
expression
(1 + x 2m )(1 + y 2n )
(b) 1
(a) 1 : 3
[JEE Main 2019, 11 Jan Shift-II]
28. Let x , y be positive real numbers and m, n
1
2
)(1 + y2 n )
1
= m
( x + x− m )( yn + y− n )
2
[Q(1 − r )3 = (1 − r ) (1 − r )2 ]
(a)
(1 + x
[JEE Main 2019, 11 Jan Shift-II]
2
∴
x m yn
2m
(49th term) : (29th term) is
Exp. (b)
⇒
Consider,
(d)
m+n
6mn
We have,
Sn = 1 + q + q 2 + … + q n
2
q + 1  q + 1
 q + 1
and Tn = 1 + 

 + …+ 
 + 
 2 
 2   2 
n
Also, we have
C1 + 101C 2S1 + 101C 3S 2 + … + 101C101S100 = αT100
101
102
JEE Main Chapterwise Mathematics
⇒
C1 +
101
C 2 (1 + q ) +
101
+ …+
C 3 (1 + q + q 2 )
101
C101(1 + q + q + … + q
101
2
100
)
= α ⋅ T100
(1 − q ) 101  1 − q 3 

+ C 3 
⇒ 101C1 + 101C 2
1− q
 1− q 
2
+
⇒
1 − q4 
 + … +
C 4 
 1− q 
101
 1 − q 101 

C101 
 1− q 
101
 1 − rn 
= α ⋅ T100 [Qfor a GP, S n = a
 , r ≠ 1]
 1− r 
1
[{101C1 + 101C 2 + … + 101C101}
1− q
− {101C1q +
C 2q 2 + … + 101 C101 q 101}]
101
= α ⋅ T100
1
⇒
[(2101 − 1) − ((1 + q )101 − 1)] = αT100
(1 − q )
[Q n C 0 + nC1 + … + nC n = 2 n ]
2101 − (q + 1)101
=α
1− q
2

q + 1  q + 1
q +
+ 
 + …+ 
1 +
 2 
 2
2

101

 q + 1 
 
 1− 
101
101
 2  
2 − (q + 1)
= α 1⋅
⇒
q+1 

1− q
1−


2


q+
[Qq ≠ 1⇒q + 1 ≠ 2 ⇒
2
α[2101 − (q + 1)101]
=
(1 − q ) ⋅ 2100
⇒
⇒
1


100 


1
≠ 1]
GP is 512. If 4 is added to each of the first and
the second of these terms, the three terms
now form an AP. Then, the sum of the
original three terms of the given GP is
[JEE Main 2019, 11 Jan Shift-I]
Let the three consecutive terms of a GP are
a
, a and ar.
r
⇒
⇒
2 r(r − 2 ) − 1(r − 2 ) = 0
(r − 2 ) (2 r − 1) = 0
1
2
Thus, the terms are either 16, 8, 4 or 4, 8, 16.
Hence, required sum = 28.
⇒
r = 2,
1 + 2 + 3 + ... + k
. If
k
5
2
S12 + S 22 + ... + S10
A , then A is equal to
=
12
[JEE Main 2019, 12 Jan Shift-I]
(b) 301
(d) 303
Exp. (d)
α =2
Exp. (b)
2 r 2 − 4r − r + 2 = 0
(a) 156
(c) 283
100
(b) 28
(d) 24
⇒
32. Let Sk =
31. The product of three consecutive terms of a
(a) 36
(c) 32
Now, according to the question, we have
a
⋅ a ⋅ ar = 512 ⇒ a3 = 512
r
... (i)
⇒
a=8
Also, after adding 4 to first two terms, we get
8
+ 4, 8 + 4, 8r are in AP
r
8
2 (12 ) = + 4 + 8r
⇒
r
8
2
⇒
24 = + 8r + 4 ⇒ 20 = 4  + 2 r 
r

r
2
5 = + 2r
⇒
r
⇒
2 r 2 − 5r + 2 = 0
1 + 2 + 3 + ... + k
k
k (k + 1) k + 1
=
=
2k
2
Since,
Sk =
So,
k + 1
1
2
S 2k = 
 = (k + 1)
 2 
4
2
Now,
⇒
5
2
A = S 12 + S 22 + S 23 + ... S 10
=
12
5
1
A=
12
4
10
… (i)
10
∑S
2
k
k =1
∑ (k + 1)
2
k =1
1
= [2 2 + 32 + 42 + ... 112 ]
4
1 11 × (11 + 1) (2 × 11 + 1) 2 
= 
−1
4 
6

n (n + 1) (2 n + 1)
2
[Q∑ n =
]
6
1 11 × 12 × 23
= 
− 1

4 
6
103
Sequences and Series
1
[(22 × 23) − 1]
4
1
1
= [506 − 1] = [505]
4
4
=
A
= 101 ⇒ A = 303
3
⇒
33. If the sum of the first 15 terms of the series
3
3
3
3
 3
 1
 1
 3
3
  + 1  +  2  + 3 +  3  + . . .
 4
 4
 2
 4
is equal to 225 k, then k is equal to
[JEE Main 2019, 12 Jan Shift-II]
(a) 108
(c) 54
(b) 27
(d) 9
Let a1 = a and d = common difference
Q
a1 + a5 + a9 + L + a49 = 416
∴ a + (a + 4d ) + (a + 8d ) + …(a + 48d ) = 416
13
⇒
(2 a + 48d ) = 416
2
…(i)
⇒
a + 24d = 32
Also,
a9 + a43 = 66
∴
a + 8d + a + 42d = 66
⇒
2 a + 50d = 66
…(ii)
⇒
a + 25d = 33
Solving Eqs. (i) and (ii), we get
a = 8 and d = 1
2
Now, a12 + a22 + a32 + L + a17
= 140m
82 + 92 + 102 + … + 242 = 140m
Exp. (b)
⇒ (12 + 2 2 + 32 + … + 242 ) − (12 + 2 2
Given series is
3
3
3
3
 3  +  1 1  +  2 1  + 33 +  3 3  + ...
 
 
 
 
 4
 2
 4
 4
3
3
3
3
6
9
12
Let S =   +   +   +  
 4
 4
 4
 4
⇒
3
3
15
+   + … + upto 15 terms
 4
3
=   [13 + 2 3 + 33 + 43 + 53 + ... + 153 ]
 4
3
15 × 16 
3
=   

 4  2 
be the sum of the first 40 terms of the series
 3
 n (n + 1) , n ∈ N
3
3
3


Q1 + 2 + 3 + ... + n = 


2


27 225 × 256
= 27 × 225
=
×
64
4
[given]
⇒ S = 27 × 225 = 225 k
⇒ k = 27
34. Let a1 , a 2 , a 3, …, a 49 be in AP such that
12
∑a 4k + 1 = 416 and a 9 + a 43 = 66. If
2
+ … + a17
= 140 m, then m is equal to
[JEE Main 2018]
(a) 66
(c) 34
(b) 68
(d) 33
Exp. (c)
We have, a1, a2 , a3 , … a49 are in AP.
12
∑ a4 k + 1 = 416 and a9 + a43 = 66
k =0
⇒
⇒
⇒
35. Let A be the sum of the first 20 terms and B
2
2
k=0
a12 + a 22
⇒
3
+ 32 + … + 7 2 ) = 140m
24 × 25 × 49 7 × 8 × 15
−
= 140m
6
6
3×7 × 8× 5
(7 × 5 − 1) = 140m
6
7 × 4 × 5 × 34 = 140m
140 × 34 = 140m
m = 34
12 + 2 ⋅ 2 2 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + …
If B − 2 A = 100λ, then λ is equal to
[JEE Main 2018]
(a) 232
(b) 248
(c) 464
(d) 496
Exp. (b)
We have,
12 + 2 ⋅ 2 2 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + …
A = sum of first 20 terms
B = sum of first 40 terms
∴ A = 12 + 2 ⋅ 2 2 + 32 + 2 ⋅ 42 + 52
+ 2 ⋅ 62 + … + 2 ⋅ 202
A = (12 + 2 2 + 32 + … + 202 ) + (2 2 + 42
+ 62 + … + 202 )
2
2
2
2
A = (1 + 2 + 3 + … + 20 )
+ 4 (12 + 2 2 + 32 + …+102 )
20 × 21 × 41 4 × 10 × 11 × 21
A=
+
6
6
20 × 21
20 × 41 × 63
(41 + 22 ) =
A=
6
6
104
JEE Main Chapterwise Mathematics
Similarly
B = (12 + 2 2 + 32 + … + 402 )
+ 4(12 + 2 2 +…+ 202 )
40 × 41 × 81 4 × 20 × 21 × 41
B=
+
6
6
40 × 41
40 × 41 × 123
(81 + 42 ) =
B=
6
6
Now, B − 2 A = 100λ
40 × 41 × 123 2 × 20 × 21 × 63
−
= 100λ
∴
6
6
40
⇒
(5043 − 1323) = 100λ
6
40
× 3720 = 100λ
⇒
6
⇒
40 × 620 = 100λ
40 × 620
⇒
λ=
= 248
100
36. For any three positive real numbers a ,b and
c, if 9 (25a 2 + b 2 ) + 25 (c 2 − 3ac ) = 15b ( 3a + c ),
then
[JEE Main 2017 (Offline)]
(a) b , c and a are in GP
(c) a , b and c are in AP
(b) b , c and a are in AP
(d) a , b and c are in GP
Exp. (c)
We have,
225a2 + 9b 2 + 25c 2 − 75ac − 45ab − 15bc = 0
⇒
a 2 + 16 d
− (3b )(5c ) = 0
1
[(15a − 3b )2 + (3b − 5c )2 + (5c − 15a)2 ] = 0
2
⇒
15a = 3b, 3b = 5c and 5c = 15a
∴
15a = 3b = 5c
a b c
= = = λ (say)
⇒
1 5 3
⇒
a = λ, b = 5λ, c = 3λ
Hence, a, b and c are in AP.
37. If the 2nd, 5th and 9th terms of a nonconstant AP are in GP, then the common
ratio of this GP is
[JEE Main 2016 (Offline)]
(a)
8
5
(b)
4
3
(c) 1
(d)
7
4
Exp. (b)
Let a be the first term and d be the common
difference. Then, we have a + d , a + 4d , a + 8 d
in GP,
i.e.
(a + 4d ) 2 = (a + d ) (a + 8 d )
+ 8ad = a2 + 8ad + ad + 8 d
2
⇒
8 d = ad
[Q d ≠ 0]
⇒
8d = a
Now, common ratio,
a + 4d 8 d + 4d 12 d 4
=
=
=
r=
8d + d
9d
3
a+d
2
38. If the sum of the first ten terms of the series
2
2
2
2
 4
 1
 2
 3
2
1  + 2  +  3  + 4 +  4  + K , is
 5
 5
 5
 5
16
m, then m is equal to
[JEE Main 2016 (Offline)]
5
(a) 102
(b) 101
(c) 100
(d) 99
Exp. (b)
Let S10 be the sum of first ten terms of the series.
Then, we have
2
2
2
4
1
2
3
S10 =  1  +  2  +  3  + 42 +  4 
 5
 5
 5
 5
2
+ ... to 10 terms
2
2
2
8
12
16
24
=   +   +   + 42 +  
 5
 5
 5
 5
2
+ ... to 10 terms
=
⇒ (15a)2 + (3b )2 + (5c )2 − (15a)(5c ) − (15a)(3b )
⇒
2
1
52
(82 + 12 2 + 162 + 202 + 242
+ ... to 10 terms)
=
=
42
52
42
(2 2 + 32 + 42 + 52 + ... to 10 terms)
(2 2 + 32 + 42 + 52 + ... + 112 )
52
16 2
=
((1 + 2 2 + ... + 112 ) − 12 )
25
16  11 ⋅ (11 + 1) (2 ⋅ 11 + 1)
=
− 1


25 
6
16
16
(506 − 1) =
=
× 505
25
25
16
16
⇒
m=
× 505
5
25
⇒
m = 101
39. If m is the AM of two distinct real numbers l
and n(l , n > 1) and G 1, G 2 and G 3 are three
geometric means between l and n, then
G 14 + 2G 24 + G 34 equals
[JEE Main 2015]
(a) 4 l 2mn
(b) 4 lm 2n
(c) 4 lmn 2
(d) 4l 2m 2n 2
105
Sequences and Series
Exp. (b)
41. If (10)9 + 2 (11)1 (10)8 + 3 (11)2 (10) 7
Given, m is the AM of l and n
…(i)
∴
l + n = 2m
and G1, G2 , G3 are geometric means between l
and n
∴ l, G1, G2 , G3 , n are in GP.
Let r be the common ratio of this GP.
∴
G1 = lr
G2 = lr 2
G3 = lr
3
n = lr
4
[JEE Main 2014]
(a)
121
10
(b)
441
100
(c) 100
Exp. (c)
k ⋅ 109 = 109 + 2 (11)1 (10)8 + 3 (11)2 (10)7 + ...
+ 10 (11)9
2
9
2
1
 11 k = 1  11 + 2  11 + ... + 9  11
 
 
 
 
 10 
 10 
 10 
 10 
= l 4 × r 4 (1 + 2 r 4 + r 8 )
= l 4 × r 4 (r 4 + 1)2
= l4 ×
n n + l


l l 
…(ii)
2
11  11
11
11
+   + ... +  
k  1 −  = 1 +
 10 

10  10 
10 
= ln × 4m2 = 4 lm2 n
1
1 +2
1 +2 + 3
+ ... is
+
+
1
1+ 3
1+ 3+5
3
3
3
3
3
[JEE Main 2015]
(a) 71
(b) 96
(c) 142
1 +2
1 +2 + 3
1
+ ... ∞
+
+
1
1+ 3
1+ 3 + 5
3
3
3
3
3
3
Let Tn be the nth term of the given series.
∴ Tn =
13 + 2 3 + 33 + ... + n3
1 + 3 + 5 + ... + to n terms
2
 n(n + 1)


(n + 1)2
=  22  =
4
n
S9 =
=
10
10

11 
11
− k = 10 10   − 10 − 10   
 10  
  10 

⇒
k = 100
42. Three positive numbers form an increasing
GP. If the middle term in this GP is doubled,
then new numbers are in AP. Then, the
common ratio of the GP is
[JEE Main 2014]
(a) 2 + 3
(c) 2 − 3
(b) 3 + 2
(d) 2 + 3
Exp. (d)
Let a, ar, ar 2 are in GP (r > 1.)
According to the question, a, 2 ar, ar 2 in AP.
⇒
4ar = a + ar 2
(n + 1)2 1 2
= [(2 + 32 + K + 102 ) + 12 − 12 ]
4
4
n =1
⇒
⇒
r − 4r + 1 = 0
4±
r=
1  10 (10 + 1)(20 + 1)  384
−1 =
= 96

4
4 
6
⇒
r =2 +
9
∑
10
 11 10

1    − 1
10


 10

10 − 11
11
− 10  
⇒ k 
 =
 10 
 10 
 11 − 1


 10

⇒
Central Idea Write the nth term of the given
series and simplify it to get its lowest form. Then,
apply, S n = ΣTn .
Given series is
9
11
− 10  
 10 
(d) 192
Exp. (b)
10
On subtracting Eq. (ii) from Eq. (i), we get
2
40. The sum of first 9 terms of the series
9
11
+ 10  
 10 
Now, G14 + 2G24 + G34 = (lr )4 + 2(lr 2 )4 + (lr 3 )4
3
(d) 110
11
11
11
k = 1 + 2   + 3   + ... 10   …(i)
 10 
 10 
 10 
n 4
r =  
 l
⇒
+ ... + 10 (11)9 = K (10)9, then k is equal to
2
16 − 4
=2 ± 3
2
3
[QAP is increasing]
106
JEE Main Chapterwise Mathematics
43. The sum of first 20 terms of the sequence
Also, tan− 1 x, tan− 1 y and tan− 1 z are in AP.
0.7, 0.77, 0.777,…, is
7
(a)
(179 − 10− 20 )
81
7
(c)
(179 + 10− 20 )
81
[JEE Main 2013]
7
(b) ( 99 − 10− 20 )
9
7
(d) ( 99 + 10− 20 )
9
⇒
=
∴
x= y= z
n
Statement II
∑ [k 3 − (k − 1)3] = n 3 , for any
k =1
natural number n.
[AIEEE 2012]
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Exp. (b)

r

20
7  179 1  1  
+   

9  9
9  10  
7
(179 + 10− 20 )
=
81
=
Statement I
S = (1) + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16)
+ K + (361 + 380 + 400)
S = (0 + 0 + 1) + (1 + 2 + 4) + (4 + 6 + 9)
+ (9 + 12 + 16) + K + (361 + 380 + 400)
Now, we can clearly observe the first elements in
each bracket.
In second bracket, the first element is 1 = 12
In third bracket, the first element is 4 = 2 2
44. If x , y and z are in AP and tan − 1 x, tan − 1 y
and tan − 1 z are also in AP, then
[JEE Main 2013]
(b) 2 x = 3y = 6z
(d) 6x = 4y = 3z
Since, x, y and z are in AP.
∴
2y = x + z
y2 = xz
1 + (1 + 2 + 4) + ( 4 + 6 + 9) + (9 + 12 + 16) + K
+ ( 361 + 380 + 400) is 8000.
7
(1 + 1+…+ upto 20 terms)
9
1
1
1

− 
+
+
+…+ upto 20 terms 
 10 102 103

Exp. (a)
x+ z
1 − xz
45. Statement I The sum of the series
1 
1  
1 

  1 − 10  +  1 − 102  +  1 − 103 

+…+ upto 20 terms]
(a) x = y = z
(c) 6x = 3y = 2 z
=
Since x, y and z are in AP as well as in GP.
7 9
99
999
+
+
+ … + upto 20 terms
9  10 100 1000

20

1 
 1   

1 −    
 10   
10 
7

= 20 −

1
9
1−

10





a(1 − r n )
, where 1 >
QS n =
1− r

20 
1
1 
7

= 20 − 1 −   
 10  
9 
9


x+ z
1 − y2
⇒
0.7 + 0.77 + 0.777 + … + upto 20 terms
7
77
777
=
+
+
+ … + upto 20 terms
10 102
103
1
11
111
= 7  + 2 + 3 + … + upto 20 terms
 10 10

10
7
=
9
 2y 
 x + z
 = tan− 1 
tan−1 

2
 1 − xz 
1 − y 
⇒
Exp. (c)
=
2 tan− 1 y = tan− 1 x + tan− 1 z
∴
In fourth bracket, the first element is 9 = 32
… … … … … … … … … …
In last bracket, the first element is 361 = 192
Hence, we can conclude that there are 20 brackets
in all.
Also, in each of the brackets, there are 3 terms out
of which the first and last terms are perfect squares
of consecutive integers and the middle term is
their product.
⇒ The general term of the series is
Tr = (r − 1)2 + (r − 1) r + (r )2
107
Sequences and Series
⇒ The sum of the n terms of the series is
n
∑ [(r − 1)
Sn =
Since, the number of terms is 20, hence
substituting n = 20, we get
20 (21)( 41)
20(21)
S20 = 3
−3
+ 20 = 8000
6
2
Whenever we are solving a question of this
Statement I, Statement II type, then we should
try our level best to exhaust every possibility that
Statement II cannot be used to explain
Statement I in any way before marking (b) as the
answer, otherwise we may end up finding the
wrong answer.
Since, these kind of questions are very tricky
and ambiguous, that is the reason why they are
no more asked in IIT-JEE from the last 2 yr.
+ (r − 1) r + (r ) ]
2
2
r =1
 r 3 − (r − 1)3 

= 1  r − ( r − 1) 
n
∑
Sn =
r
[Q a3 − b 3 = (a − b )(a2 + ab + b 2 )]
n
∑ [r
Sn =
⇒
3
− (r − 1)3 ]
r =1
n
Now, let S n =
∑ [k
3
− (k − 1)3 ]
k =1
On substituting the value of k, we get
Sn =
n
∑
k =1
= (13 − 03 ) + (2 3 − 13 )
+ (33 − 2 3 ) + K+ [n3 − (n − 1)3 ]
On rearranging the terms, we get
S n = − 03 + (13 − 13 ) + (2 3 − 2 3 ) + (33 − 33 )
+ K + [(n − 1)3 − (n − 1)3 ] + n3
(a) –150
(b) 150 times its 50th term
(c) 150
(d) zero
Exp. (d)
Given 100 times the 100th term of an AP
= 50 times its 50th term.
⇒ S n = n3
Since, the number of terms is 20, hence
substituting n = 20, we get
S 20 = 8000
Hence, Statement I is correct.
Statement II We have already proved in the
Statement I, that
Sn =
46. If 100 times the 100th term of an AP with
non-zero common difference equal to the
50 times its 50th term, then the 150th term of
this AP is
[AIEEE 2012]
[k 3 − (k − 1)3 ]
n
∑ [k
Now, according to the given data,
− (k − 1)3 ] = n3
3
To find The 150th term of the same AP.
Let a be the first term and d (d ≠ 0) be the common
difference of the given AP, then
T100 = a + (100 − 1) d = a + 99d ,
T50 = a + (50 − 1) d = a + 49d ,
T150 = a + (150 − 1) d = a + 149d
100 × T100 = 50 × T50
k =1
Hence, Statement II is also correct and is a correct
explanation of Statement I.
Caution If you have solved the series in Statement I, in
the following way, then you may get option (b)
as the correct answer.
The sum of the n terms of the series is
Sn =
n
∑ [( r − 1)
2
+ ( r − 1) r + ( r )2 ]
r=1
⇒
Sn =
n
2
∑ (3 r − 3 r + 1)
r=0
n
∑r
n
n
r=1
r=1
Sn = 3
⇒
n( n + 1)(2 n + 1)
n( n + 1)
Sn = 3
−3
+ n
6
2
r=1
2
−3
∑ r + ∑1
⇒
⇒
100(a + 99d ) = 50(a + 49d )
⇒
2 (a + 99d ) = (a + 49d )
⇒
2 a + 198d = a + 49d
⇒
a + 149d = 0
∴
T150 = 0
47. A man saves ` 200 in each of the first three
months of his service. In each of the
subsequent months, his saving increases by
` 40 more than the saving of immediately
previous month. His total saving from the
start of service will be ` 11040 after
(a) 19 months
(c) 21 months
(b) 20 months
(d) 18 months
[AIEEE 2011]
108
JEE Main Chapterwise Mathematics
Exp. (b)
Exp. (c)
Let the time taken to save ` 11040 be (n + 3)
months.
For first three months, he saves ` 200 each month.
In (n + 3) months,
n
3 × 200 + {2 (240) + (n − 1) × 40} = 11040
2
n
⇒ 600 + {40 (12 + n − 1)} = 11040
2
⇒
600 + 20n (n + 11) = 11040
⇒
30 + n2 + 11 n = 552
⇒
n2 + 11n − 522 = 0
2
⇒
n + 29n − 18n − 522 = 0
⇒
n (n + 29) − 18 (n + 29) = 0
⇒
(n − 18) (n + 29) = 0
∴
n = 18,
n = − 29, neglecting
∴ Total time = (n + 3) = 21months
48. Let an be the nth term of an AP. If
100
100
∑ a 2r = α and ∑ a 2r − 1 = β,
r =1
then
the
r =1
common difference of the AP is [AIEEE 2011]
α −β
200
α −β
(c)
100
(a)
(b) α − β
…(i)
…(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(a 2 − a 1 ) + (a 4 − a 3 ) + (a 6 − a 5 )
+ K + (a 200 − a 199 ) = α − β
⇒ d + d + d + K + 100 times = (α − β )
⇒
100 d = (α − β )
α −β
∴
d=
100
49. A person is to count 4500 currency notes. Let
an denotes the number of notes he counts in
the nth min. If a1 = a 2 = .... = a10 = 150 and
a 10 , a 11 ,... are in AP with common difference
– 2, then the time taken by him to count all
notes is
[AIEEE 2010]
(b) 34 min
(d) 135 min
50. The sum to the infinity of the series
1+
2 6 10 14
+
+
+
+…is
3 32 33 34
(a) 3
(b) 4
(c) 6
[AIEEE 2009]
(d) 2
2
6
10 14
+
+ 3 + 4 +…
3 32
3
3
2
6
10 14
⇒ S − 1= + 2 + 3 + 4 + …
3 3
3
3
S −1 2
6
10 14
= 2 + 3 + 4 + 5 +…
⇒
3
3
3
3
3
On subtracting Eq. (ii) from Eq. (i), we get
2
4
4
4
2
(S − 1) = + 2 + 3 + 4 + …
3 3
3
3
3
2
2
2
S − 1= 1+ + 2 + 3 + …
⇒
3 3
3
2
⇒
S = 2 + 3 = 2 + 1= 3
1
1−
3
Let
Exp. (c)
(a) 24 min
(c) 125 min
⇒
(n − 24)(n − 125) = 0
∴
n = 24, 125
Then, the total time taken by the person to count all
notes
= 10 + 24
= 34 min
[neglecting n = 125 because for this value of n,
a 125 will be negative, which is not possible as
currency notes cannot be negative]
Exp. (a)
(d) β − α
Given, a2 + a 4 + a 6 + K + a 200 = α
and
a 1 + a 3 + a 5 + K + a 199 = β
Number of notes that the person counts in 10 min
= 10 × 150 = 1500
Since, a 10 , a 11, a 12 ,.... are in AP with common
difference −2.
Let n be the time taken to count remaining
3000 notes, then
n
[2 × 148 + (n − 1) × −2 ] = 3000
2
⇒
n2 − 149 n + 3000 = 0
S = 1+
…(i)
…(ii)
51. The first two terms of a geometric
progression add upto 12. The sum of the
third and the fourth terms is 48. If the terms
of the geometric progression are alternately
positive and negative, then the first term is
(a) 4
(c) –12
(b) − 4
(d) 12
[AIEEE 2008]
109
Sequences and Series
⇒
( p + q )2 = 1 + 2 pq
…(i)
⇒
( p + q )2 ≤ 1 + 1
…(ii)
⇒
Exp. (c)
a + ar = a(1 + r ) = 12
Since,
ar 2 + ar 3 = ar 2 (1 + r ) = 48
and
From Eqs. (i) and (ii), we get
[since, the series is alternately sign, so
we take negative values]
On putting the value of r in Eq. (i), we get
a = − 12
52. In a geometric progression consisting of
1
(1 − 5 )
2
(c)
1
5
2
1
(d) ( 5 − 1 )
2
(c) e
ar n − 1 = ar n + ar n + 1
1
= 1+ r
r
r=

− 5 − 1
Q r ≠

2


5 −1
2
53. If p and q are positive real numbers such
that p 2 + q 2 = 1, then the maximum value of
[AIEEE 2007]
(p + q ) is
(a) 2
(c)
(b)
1
2
(d)
1
2
2
1
1
1
−
+
−K
2 ! 3! 4!
a1 + a 2 + ... + ap
a1 + a 2 + ... + aq
p2 + q 2
≥
2
pq ≤
p2q 2 = pq
1
2
1
1
1
−
+
− ... = e −1
2 ! 3! 4!
=
p2
q
2
, p ≠ q , then
equal to
7
(a)
2
11
(c)
41
a6
is
a 21
[AIEEE 2006]
2
(b)
7
41
(d)
11
Given that,
a1 + a 2 + . . . + a p
a1 + a 2 + . . . + aq
[Q AM ≥ GM]
[Q p2 + q 2 = 1] …(i)
( p + q )2 = p2 + q 2 + 2 pq
=
p2
q2
p
[2 a 1 + ( p − 1)d ]
p2
2
= 2
∴
q
[2 a 1 + (q − 1)d ] q
2
where, d is a common difference of an AP.
⇒
Applying AM-GM inequality in the positive real
numbers p2 and q 2 ,
Now,
= 1− 1+
⇒
Exp. (d)
⇒
(d) e 2
Exp. (c)
r2 + r − 1 = 0
∴
1
1
2
Exp. (b)
If
Since, each term is equal to the sum of two
preceding terms.
⇒
−
(b) e −1
55. Let a1 , a 2 , a 3 ,... be terms of an AP.
Exp. (d)
⇒
[AIEEE 2007]
(a) e −2
(b)
5
∴
1 1 1
− + − ... upto
2 ! 3! 4!
infinity is
positive terms, each term equals the sum of
the next two terms. Then, the common ratio
of this progression is equal to
[AIEEE 2007]
(a)
p+ q ≤ 2
54. The sum of the series
r 2 = 4 ⇒ r = −2
[from Eq. (i)]
∴
Now,
(2 a 1 − d ) + pd
(2 a 1 − d ) + qd
=
p
q
(2 a 1 − d )( p − q ) = 0
d
a1 =
2
a6
a + 5d
= 1
a 21 a 1 + 20d
d
+ 5d
11
= 2
=
d
+ 20d 41
2
110
JEE Main Chapterwise Mathematics
56. If a1 , a 2 ,..., an are in HP, then the expression
a1a 2 + a 2a 3 + ... + an − 1an is equal to
(a) (n − 1 )(a1 − an )
(b) na 1an
(c) (n −1 )a1an
(d) n (a1 − an )
Similarly, y =
⇒
⇒
Similarly,
a 1 − a 2 = a 1a 2d
a 2 − a 3 = a 2 a 3d
⇒
x=
1
,
1− a
a=
z−1
y−1
x−1
, b=
, c=
z
y
x
On adding all of these, we get
a 1 − a n = d (a 1a 2 + a 2 a 3 + . . . + a n − 1a n ) …(i)
a1 − an
a 1an (n − 1)
On putting the value of d in Eq. (i), we get
a 1 − an
a1 − a n =
(a 1a 2 + a 2 a 3 + . . .
a 1a n (n − 1)
+ a n − 1a n )
⇒ a 1a 2 + a 2 a 3 + . . . + a n − 1a n = a 1a n (n − 1)
∞
n =0
n =0
n =0
∑ a n , y = ∑ b n , z = ∑ cn , where a , b
and c are in AP and |a | < 1,|b | < 1,|c | < 1, then
[AIEEE 2005]
x , y and z are in
HP
Arithmetic geometric progression
AP
GP
58. The sum of the series
1+
1
1
1
+
+
+ ... ∞ is
4 ⋅ 2 ! 16 ⋅ 4! 64 ⋅ 6!
e +1
(a)
2 e
e +1
(c)
e
∞
∑a ,
n=0
n
y=
∞
∑b
n=0
n
,
z=
∞
∑c
n=0
n
e −1
(b)
2 e
e −1
(d)
e
[AIEEE 2005]
Exp. (a)
We know that,
e x + e− x
x2
x4
x6
= 1+
+
+
+ ...
2
2!
4!
6!
1
On putting x = both sides, we get
2
e1/ 2 + e −1/ 2
= 1+
2
x=
1
1− c
2
1
1
= 1− + 1−
y
x
z
2 1 1
= +
⇒
y x z
Hence, x, y and z are in HP.
Exp. (a)
Given that,
z=
2−
⇒
1
1
=
+ (n − 1)d
a n a1
∞
1
,
1− b
 y − 1 x − 1 z − 1
2
+
 =
 y 
x
z
⇒
∞
y=
Since, a, b and c are in AP.
∴
2b = a + c
a n − 1 − a n = a n − 1a nd
d=
1
1
1
and
are in HP.
,
1− a 1− b
1− c
Hence, x, y and z are in HP.
… … … … …
… … … … …
(a)
(b)
(c)
(d)
…(ii)
Alternate Solution
From Eqs. (i) and (ii), we get
Let d be the common difference of AP.
1
1
−
=d
∴
a 2 a1
57. If x =
1
1
and z =
1− b
1− c
⇒ 1 − a, 1 − b and 1 − c are also in AP.
1 1 1
1
are in AP.
,
,
,...,
a1 a 2 a 3
an
⇒
…(i)
Now, a, b and c are in AP.
⇒ − a, − b and − c are in AP.
Since, a 1, a 2 , a 3 , . . . , an are in HP.
Also,
1
1− a
x = 1 + a + a2 + . . . =
[AIEEE 2006]
Exp. (c)
∴
⇒
⇒
2
 1 1 +
 
 2  2!
 1
 
2
4
1
+ ...
4!
e+1
1
1
1
= 1+
+
+
+ ... ∞
4 ⋅ 2 ! 16 ⋅ 4! 64 ⋅ 6!
2 e
111
Sequences and Series
59. Let Tr be the rth term of an AP whose first
term is a and common difference is d. If for
1
some positive integers m , n , m ≠ n , Tm = ,
n
1
and Tn = , then a − d is the equal to
m
(a) 0
(b) 1
1
(c)
mn
1 1
(d)
+
m n
[AIEEE 2004]
61. The sum of the series
[AIEEE 2004]
(e 2 − 1 )
(a)
2
(e 2 − 1 )
(c)
2e
We know that,
1
+
1!
1
= 1− +
1!
e = 1+
1
Given that,
Tm =
n
1
a + (m − 1)d =
⇒
n
1
and
Tn =
m
1
a + (n − 1)d =
⇒
m
On solving Eqs. (i) and (ii), we get
1
a=d =
mn
∴
a−d = 0
and
…(i)
…(ii)
60. The sum of the first n terms of the series
1 + 2 ⋅ 2 + 3 + 2 ⋅ 4 + 5 + 2 ⋅ 6 + ... is
n (n + 1)2
, when n is even. When n is odd, the
2
sum is
[AIEEE 2004]
2
(e −1 )2
(b)
2e
(e 2 − 2 )
(d)
e
Exp. (b)
Exp. (a)
2
2
3n(n + 1 )
(a)
2
n(n + 1 )2
(c)
4
2
2
1 1 1
+ + + ... is
2 ! 4! 6!
2
n 2 (n + 1 )
(b)
2
2
n(n + 1 ) 

(d)


2

e −1
1
+
2!
1
−
2!
1
+
3!
1
+
3!
1
+ ... ∞
4!
1
− ... ∞
4!
…(i)
…(ii)
On adding Eqs. (i) and (ii), we get
2
2
e + e −1 = 2 +
+
+ ... ∞
2 ! 4!
2
e +1
2
2
⇒
−2 =
+
+ ... ∞
2 ! 4!
e
e2 + 1 − 2e
1
1
⇒
= 2  +
+ . . . ∞ 
 2 ! 4!

e
(e − 1)2
1
1
=
+
+ ... ∞
2e
2 ! 4!
⇒
62. The sum of the series
upto ∞ is equal to
1
1
1
−
+
− ...
1 ⋅2 2 ⋅ 3 3 ⋅ 4
[AIEEE 2003]
(a) 2 loge 2
(b) loge 2 − 1
(c) loge 2
4
(d) loge  
e 
Exp. (d)
Exp. (b)
Given that, the sum of n terms of given series is
n(n + 1)2
, if n is even.
2
Let n be odd i.e, n = 2 m + 1
Then, S 2 m + 1 = S 2 m + (2 m + 1)th term
=
Q n = 2 m + 1 
(n − 1)n2
+ nth term 

2
 ⇒ 2 m = n − 1
=
(n − 1)n2
n − 1+ 2
+ n2 = n2 



2
2
=
(n + 1)n2
2
Now,
1
1
1
−
+
− ...
1⋅ 2 2 ⋅ 3 3 ⋅ 4
1
1 1
1 1
=  1 −  −  −  +  −  − . . .

2   2 3  3 4
1
1
1
= 1− 2⋅ + 2⋅ − 2⋅ + …
2
3
4
1 1 1

= 2  1 − + − + . . . − 1


2 3 4
= 2 log (1 + 1) − 1
= log 2 2 − log e
= loge
4
e
112
JEE Main Chapterwise Mathematics
1
2
2
1 
1
1−
1
−


2

2
=2 + 2 = 4
63. If 1, log 3 ( 31 − x + 2 ), log 3 ( 4 ⋅ 3 x − 1) are in
AP. Then, x is equal to
[AIEEE 2002]
(a) log 3 4
(b) 1 − log 3 4
(c) 1 − log 4 3
(d) log 4 3
1
Exp. (b)
∴
31 −
Since, 1, log 3
x
+ 2 , log 3 (4 ⋅ 3 x − 1) are in
AP.
∴ 2 log 3 (31 − x + 2 )1/ 2 = log 3 3 + log 3 (4 ⋅ 3 x − 1)
⇒
1
=
log 3 (31 −
31 −
⇒
3
+ 2 = 12 t − 3
t
12 t 2 − 5t − 3 = 0
∴
⇒
⇒
(3t + 1)(4t − 3) = 0
⇒
t =−
⇒
∴
Since, 5th term of a GP = 2
∴
64. The value of 2
1/4
⋅4
1/8
(a) 1
3
(c)
2
⋅8
1/16
... ∞ is [AIEEE 2002]
(b) 2
ar 4 = 2
…(i)
where, a and r are the first term and common ratio
of a GP.
Now, required product
1 3
,
3 4
x = 1 − log 3 4
(b) 512
(d) None of these
Exp. (b)
[let 3 x = t ]
3
4
[since, 3 x cannot be negative]
3
log 3   = x
 4
[AIEEE 2002]
(a) 256
(c) 1024
= a × ar × ar 2 × ar 3 × ar 4 × ar 5 × ar 6 × ar 7 × ar 8
3x =
⇒
=2
terms is
+ 2 = 12 ⋅ 3 x − 3
x
( 4)
65. 5th term of a GP is 2, then the product of its 9
+ 2 ) = log 3 3(4 ⋅ 3 x − 1)
x
S = 24
1⋅
+
= a9 r 36 = (ar 4 )9
= 2 9 = 512
66.
[from Eq. (i)]
∞
(log e x )n
is equal to
n!
n =0
∑
(a) loge x
(c) log x e
[AIEEE 2002]
(b) x
(d) None of these
Exp. (b)
(d) 4
∞
(loge x)n
loge x (loge x)2
= 1+
+
+ ...
n!
1!
2!
n=0
∑
Exp. (b)
= elog e
Let S = 21/ 4 ⋅ 41/ 8 ⋅ 81/ 16 . . .
=2
1/ 4
⋅2
2/ 8
⋅2
3 / 16
1 
2
3

1 + 2 + 2 + . . . 

2
= 24 
...
1
= 24
67. e
( S1 )
2
3
+
+ ... ∞
2 22
It is an infinite arithmetic geometric progression.
d⋅r
a
∴
S1 =
+
1 − r (1 − r )2
where, S1 = 1 +
( x − 1) −
x
= x
( x − 1 )3 ( x − 1 )4
1
( x − 1 )2 +
−
+...
2
3
4
to
is equal
[AIEEE 2002]
(a) log ( x −1 )
(c) x
(b) log x
(d) None of these
Exp. (c)
( x − 1) −
e
1
1
( x − 1) 2 + ( x − 1) 3 − . . .
2
3
= elog ( 1 +
x − 1)
= elog
x
= x
8
Limits, Continuity
and Differentiability
1. lim
x→ 0
sin 2 x
2 − 1 + cos x
equals
[JEE Main 2019, 8 April Shift-I]
(b) 2
(d) 4
(a) 4 2
(c) 2 2
2
x→ 0
sin x
2 − 1 + cos x
 0 form
 0

sin2 x
Q1 + cos x = 2 cos 2 x 

x
2 
2 − 2 cos
2
sin2 x
sin2 x
= lim
= lim
x→ 0
x→ 0
x
x
2 × 2 sin2  
2  1 − cos 

 4
2
Q1 − cos x = 2 sin2 x 
2
4 

= lim
x→ 0
= lim
x→ 0
=
x2
x
2 2  
 4
2
[ lim sin x = lim x]
x→ 0
x→ 0
Given function is
f( x) = 9 x4 + 12 x3 − 36 x2 + 25 = y (let)
dy
For maxima or minima put
=0
dx
dy
= 36 x3 + 36 x2 − 72 x = 0
⇒
dx
⇒
x3 + x2 − 2 x = 0
⇒
x[ x2 + x − 2 ] = 0
⇒
x[ x + 2 x − x − 2 ] = 0
2
⇒
x[ x( x + 2 ) − 1( x + 2 )] = 0
⇒
x( x − 1)( x + 2 ) = 0
⇒
x = − 2, 0, 1
By sign method, we have following
–
+
–2
16
=4 2
2 2
minimum and local maximum points of the
function, f ( x ) = 9x 4 + 12 x 3 − 36x 2 + 25, x ∈R ,
[JEE Main 2019, 8 April Shift-I]
–
0
+
1
dy
changes it’s sign from negative to
dx
positive at x = ‘−2 ’ and ‘1’, so x = − 2, 1 are points
dy
of local minima. Also,
changes it’s sign from
dx
positive to negative at x = 0, so x = 0 is point of
local maxima.
∴
S1 = {−2, 1} and S 2 = {0}.
Since,
2. If S1 and S 2 are respectively the sets of local
then
S 1 = { −2 } ; S 2 = {01
,}
S 1 = { −2 , 0} ; S 2 = {1}
S 1 = { −2 ,1} ; S 2 = {0}
S 1 = { −1} ; S 2 = {0, 2 }
Exp. (c)
Exp. (a)
Given limit is lim
(a)
(b)
(c)
(d)
114
JEE Main Chapterwise Mathematics
2
3.

 3 cos x + sin x  
 π
If 2 y = cot −1 
  , x ∈ 0 , 
 2
 cos x − 3 sin x  

dy
then
is equal to
dx
(a)
π
−x
6
[JEE Main 2019, 8 April Shift-I]
(b) x −
π
6
π
−x
3
(c)
(d) 2 x −
π
3
Exp. (b)
function such that f ′ ′ ( x ) > 0, for all x ∈(0,2 ).
If φ( x ) = f ( x ) + f (2 − x ) , then φ is
[JEE Main 2019, 8 April Shift-I]
(a) increasing on (0, 1) and decreasing on (1, 2)
(b) decreasing on (0, 2)
(c) decreasing on (0, 1) and increasing on (1, 2)
(d) increasing on (0, 2)
Exp. (c)
Given expression is

 3 cos x + sin x  
2 y =  cot −1 

 cos x − 3 sin x  


 3 cot x + 1 
=  cot −1 

 cot x − 3  

2
2
[dividing each term of numerator and
denominator by sin x]

 cot π cot x + 1 



−1
6
=  cot 

π

 cot x − cot  


6 

π


=  cot −1  cot  − x  

 6

2
Qcot π =

6
3

2

cot A cot B + 1
Qcot( A − B) = cot B − cot A 





=
 π

4. Let f :[0, 2] → R be a twice differentiable
2
 π − x ,


6

π
6
π
π
< x<
6
2
0< x<
2
π

+  − x  ,
6


Qcot −1(cot θ) =


 π + θ, − π < θ < 0 

0< θ< π 
 θ,

θ − π, π < θ < 2 π 


π
π

  − x , 0 < x <

6
6
⇒ 2y = 
2
  7 π − x , π < x < π

 6

6
2
 2  π − x (−1), 0 < x <


dy   6

=
⇒2
dx 2  7 π − x (−1), π < x <



6
  6
π
π
 x − , 0< x<
dy 
6
6
⇒ =
dx  x − 7 π , π < x < π

6
6
2
Given, φ( x) = f( x) + f(2 − x), ∀ x ∈ (0, 2 )
…(i)
⇒
φ′( x) = f ′( x) − f ′(2 − x)
Also, we have f ′ ′( x) > 0 ∀ x ∈ (0, 2 )
⇒ f ′( x)is a strictly increasing function
∀ x ∈ (0, 2 ).
Now, for φ( x) to be increasing,
φ′( x) ≥ 0
[using Eq. (i)]
⇒
f ′( x) − f ′(2 − x) ≥ 0
⇒ f ′( x) ≥ f ′(2 − x) ⇒ x > 2 − x
[Qf′ is a strictly increasing function]
⇒
2x> 2
⇒
x>1
Thus, φ( x) is increasing on (1, 2).
Similarly, for φ( x) to be decreasing,
φ′( x) ≤ 0
[using Eq. (i)]
⇒ f ′( x) − f ′(2 − x) ≤ 0
⇒
f ′( x) ≤ f ′(2 − x)
⇒
x<2 − x
[Qf′ is a strictly increasing function]
⇒
2x< 2
⇒
x<1
Thus, φ( x) is decreasing on (0, 1).
5. If f (1) = 1, f ′ (1) = 3, then the derivative of
f ( f ( f ( x ))) + ( f ( x ))2 at x = 1 is
2
[JEE Main 2019, 8 April Shift-II]
(a) 12
(c) 15
π
6
π
2
(b) 9
(d) 33
Exp. (d)
Let y = f(f(f( x))) + (f( x))2
On differentiating both sides w.r.t. x, we get
dy
= f ′(f(f( x))) ⋅ f ′(f( x)) ⋅ f ′( x) + 2 f( x)f ′( x)
dx
[by chain rule]
So,
dy
dx at
= f ′ ( f ( f (1))) ⋅ f ′ ( f (1)) ⋅ f ′ (1) + 2 f (1)f ′ (1)
x =1
115
Limits, Continuity and Differentiability
∴
dy
= f ′(f(1)) ⋅ f ′(1) ⋅ (3) + 2(1)(3)
dx x = 1
Exp. (b)
[Qf(1) = 1 and f′(1) = 3]
= f ′(1) ⋅ (3) ⋅ (3) + 6
= (3 × 9) + 6
= 27 + 6 = 33
(ii) For maximum or minimum, put first
derivative of V equal to zero
6. Let f :[ −1, 3] → R be defined as
| x | + [ x ], −1 ≤ x < 1

f ( x ) =  x + | x |, 1 ≤ x < 2
 x + [ x ], 2 ≤ x ≤ 3 ,

where, [t ] denotes the greatest integer less
than or equal to t . Then, f is discontinuous
at
[JEE Main 2019, 8 April Shift-II]
(a)
(b)
(c)
(d)
Key Idea
(i) Use formula of volume of cylinder, V = πr 2 h
where, r = radius and h = height
Let a sphere of radius 3, which inscribed a right
circular cylinder having radius r and height is h, so
h
From the figure, = 3cosθ
2
⇒
h = 6cosθ
and
…(i)
r = 3sinθ
r
four or more points
only two points
only three points
only one point
h
Exp. (c)
h/2
Given function f : [−1, 3] → R is defined as
| x| + [ x], −1 ≤ x < 1

f( x) =  x + | x|, 1 ≤ x < 2
 x + [ x ], 2 ≤ x ≤ 3

− x − 1, −1 ≤ x < 0
 x,
0≤ x< 1

=  2 x,
1≤ x < 2
 x + 2, 2 ≤ x < 3

 6,
x=3
[Qif n ≤ x < n + 1, ∀ n ∈ Integer, [ x] = n]
Q
lim f( x) = − 1 ≠ f(0)
[Q f(0) = 0]
lim f( x) = 1 ≠ f(1)
[Q f(1) = 2]
x→ 0 −
Q
x→1−
Q lim f( x) = 4 = f(2 ) = lim f( x) = 4
x→ 2 −
and
x→ 2 +
lim f( x) = 5 ≠ f(3)
x→ 3 −
[Q f(2 ) = 4]
[Q f(3) = 6]
∴ Function f( x) is discontinuous at points 0, 1
and 3.
7. The height of a right circular cylinder of
maximum volume inscribed in a sphere of
radius 3 is
[JEE Main 2019, 8 April Shift-II]
(a)
6
(c)
3
(b) 2 3
2
(d)
3
3
3
θ
r
Q Volume of cylinder V = πr 2 h
= π(3sinθ)2 (6cos θ) = 54 π sin2 θcos θ .
dV
For maxima or minima,
=0
dθ
⇒ 54 π[2 sinθcos 2 θ − sin3 θ] = 0
⇒
⇒
sinθ[2 cos 2 θ − sin2 θ] = 0
tan2 θ = 2
π 

Qθ ∈  0, 

 2  
tanθ = 2
2
1
and cosθ =
⇒ sinθ =
3
3
From Eqs. (i) and (ii), we get
1
h=6
=2 3
3
⇒
…(ii)
8. Let f : R → R be a differentiable function
satisfying f ′ ( 3) + f ′ (2 ) = 0. Then
1
 1 + f ( 3 + x ) − f ( 3) x
lim 
 is equal to
x → 0  1 + f (2 − x ) − f (2 )
[JEE Main 2019, 8 April Shift-II]
(a) e
(c) e 2
(b) e −1
(d) 1
116
JEE Main Chapterwise Mathematics
π π
6 3
10. If the function f defined on  ,  by


Exp. (d)
Let
1
f(3) x
 1 + f ( 3 + x) −
l = lim 

x→ 0  1 + f(2 − x) − f(2 ) 
l =e
⇒
=e
=e
[1∞ form]
lim
1 + f( 3 + x ) − f( 3 ) 
1
1 −

1 + f( 2 − x ) − f( 2 ) 
lim
 1 + f( 2 − x ) − f( 2 ) − 1 − f( 3 + x ) + f( 3 ) 

x( 1 + f( 2 − x ) − f( 2 ))

x→ 0 x 
x→ 0 
 f( 2 − x ) − f( 3 + x ) + f( 3 ) − f( 2 ) 
lim

x→ 0 
x( 1 + f( 2 − x ) − f( 2 ))

On applying L’Hopital rule, we get
l =e
− f ′( 2 − x ) − f ′( 3 + x )


lim
x→ 0  1 − xf ′( 2 − x ) + f( 2 − x ) − f( 2 ) 
On applying limit, we get
l =e
 − f ′( 2 ) − f ′( 3 ) 


 1 − 0 + f( 2 ) − f( 2 ) 
1
f(3) x
 1 + f ( 3 + x) −
So, lim 

x→ 0  1 + f(2 − x) − f(2 ) 
= e0 = 1
=1
9. If f ( x ) is a non-zero polynomial of degree
four, having local extreme points at
x = − 1, 0, 1, then the set
 2 cos x − 1
π
 cot x − 1 , x ≠ 4
is continuous,
f (x ) = 
π

k,
x=

4
then k is equal to [JEE Main 2019, 9 April Shift-I]
(a)
1
2
(b) 2
[JEE Main 2019, 9 April Shift-I]
four rational numbers
two irrational and two rational numbers
four irrational numbers
two irrational and one rational number
Exp. (d)
The non-zero four degree polynomial f( x) has
extremum points at x = −1, 0,1, so we can
assume f ′( x) = a( x + 1)( x − 0) ( x − 1) = ax( x2 − 1)
where, a is non-zero constant.
f ′( x) = ax3 − ax
a
a
⇒
f ( x) = x 4 − x 2 + C
4
2
[integrating both sides]
where, C is constant of integration.
Now, since f( x) = f(0)
a 4 a 2
x4
x2
⇒
x − x +C =C ⇒
=
4
2
4
2
⇒
x2 ( x2 − 2 ) = 0 ⇒ x = − 2 , 0, 2
Thus, f( x) = f(0) has one rational and two irrational
roots.
(d)
1
2
Exp. (a)
Given function is
 2 cos x − 1
π
,x≠

4
f( x) =  cot x − 1
π

, x=
k

4
QFunction f( x) is continuous, so it is continuous at
π
x= .
4
π
f   = lim f ( x)
∴
 4  x→ π
4
lim
k = x→ π
4
π
x = + h,
4
⇒
S = { x ∈ R : f ( x ) = f (0)} contains exactly
(a)
(b)
(c)
(d)
(c) 1
Put
2 cos x − 1
cot x − 1
π
, then h → 0
4
π
2 cos  + h − 1
lim
4

k=
π
h→ 0


cot  + h − 1
4

1
1

2
cos h −
sin h − 1
 2

lim
2
=
1
cot
h
−
h→ 0
−1
cot h + 1
when x →

Q cos ( x + y) = cos x cos y − sin x sin y and

cot x cot y − 1
cot ( x + y) =
cot y + cot x 
lim cos h − sin h − 1
=
−2
h→ 0
1 + cot h
=

lim  (1 − cos h) + sin h
(sin h + cos h)
h → 0 
2 sin h

117
Limits, Continuity and Differentiability
h
h


2h
lim  2 sin 2 + 2 sin 2 cos 2

+
=
(sin
h
cos
h
)

h
h
h→ 0 
sin
cos
4




2
2
h

 h
lim  sin 2 + cos 2

×
+
(sin
h
cos
h
)
=

h → 0  2 cos h




2
1
⇒ k=
2
11. If the tangent to the curve, y = x 3 + ax − b at
the point (1, − 5) is perpendicular to the line,
− x + y + 4 = 0, then which one of the
following points lies on the curve ?
[JEE Main 2019, 9 April Shift-I]
(a) ( −2 , 2 )
(c) ( −2 ,1)
(b) ( 2 , − 2 )
(d) ( 2 , − 1)
Exp. (b)
Given curve is y = x3 + ax − b
…(i)
passes through point P(1, − 5).
…(ii)
∴
− 5 = 1 + a − b ⇒b − a = 6
and slope of tangent at point P(1, − 5) to the curve
(i), is
dy
= [3 x2 + a]( 1, − 5 ) = a + 3
m1 =
dx ( 1, − 5 )
Q The tangent having slope m1 = a + 3 at point
P(1, − 5) is perpendicular to line − x + y + 4 = 0 ,
whose slope is m2 = 1.
∴
a + 3 = −1 ⇒a = −4 [Q m1m2 = −1]
Now, on substituting a = −4in Eq. (ii), we get b = 2
On putting a = −2 and b = 2 in Eq. (i), we get
y = x3 − 4 x − 2
Now, from option (2, − 2 ) is the required point
which lie on it.
12. Let f ( x ) = 15 − x − 10 ; x ∈ R. Then, the set of
all values of x , at which the function,
g ( x ) = f ( f ( x )) is not differentiable, is
[JEE Main 2019, 9 April Shift-I]
(a) { 5,10,15, 20}
(b) { 5,10,15}
(c) {10}
(d) {10,15}
Exp. (b)
Given function is f( x) = 15 − | x − 10|, x ∈ R
and g ( x) = f(f( x))
= f(15 −| x − 10|)
= 15 − | 15 − | x − 10| − 10|
= 15 − | 5 − | x − 10||
15 − | 5 − ( x − 10)| , x ≥ 10
=
15 − | 5 + ( x − 10)| , x < 10
15 − | 15 − x| , x ≥ 10
=
 15 − | x − 5| , x < 10
x< 5
 15 + ( x − 5) = 10 + x ,
 15 − ( x − 5) = 20 − x , 5 ≤ x < 10

=
 15 + ( x − 15) = x , 10 ≤ x < 15
15 − ( x − 15) = 30 − x ,
x ≥ 15
From the above definition it is clear that g ( x) is not
differentiable at x = 5, 10, 15.
13. Let S be the set of all values of x for which the
tangent to the curve y = f ( x ) = x 3 − x 2 − 2 x
at( x , y ) is parallel to the line segment joining
the points (1, f (1)) and ( −1, f ( −1)), then S is
equal to
[JEE Main 2019, 9 April Shift-I]
1
(a)  , − 1

3
1 

(c) − ,1
 3 
1
(b)  ,1
3 
1
(d) − , − 1
 3

Exp. (c)
Given curve is y = f( x) = x3 − x2 − 2 x
...(i)
So,
f(1) = 1 − 1 − 2 = −2
and f(−1) = −1 − 1 + 2 = 0
Since, slope of a line passing through ( x1, y1 ) and
( x2 , y2 ) is given by
y − y1
m = tanθ = 2
x2 − x1
∴Slope of line joining points (1, f(1)) and (−1, f(−1))
is
f(1) − f(−1) −2 − 0
= −1
=
m=
1 − (−1)
1+ 1
dy
Now,
= 3 x2 − 2 x − 2
dx
[differentiating Eq. (i), w.r.t. ‘x’]
According to the question,
dy
=m
dx
⇒
3 x 2 − 2 x − 2 = −1
⇒
3 x2 − 2 x − 1 = 0
⇒
( x − 1) (3 x + 1) = 0
1
⇒
x = 1,−
3
1 

Therefore, set S = − , 1.
 3 
118
JEE Main Chapterwise Mathematics
14. A water tank has the shape of an inverted
right circular cone, whose semi-vertical
 1
angle is tan −1   . Water is poured into it at a
2
constant rate of
5 cu m/min. Then, the rate (in m/min) at
which the level of water is rising at the
instant when the depth of water in the tank
is 10 m is
[JEE Main 2019, 9 April Shift-II]
2
(a)
π
1
(b)
5π
1
(c)
15π
1
(d)
10π
Key Idea Use formula :
1
Volume of cone = πr 2 h, where r = radius
3
and h = height of the cone.
Given, semi-vertical angle of right circular cone
1
= tan−1  
2
1
Let
α = tan−1  
2
1
tanα =
⇒
2
r 1
r
[from fig. tanα = ]
=
⇒
h 2
h
1
…(i)
r= h
⇒
2
r
 π
y 
 3
and
[JEE Main 2019, 9 April Shift-II]
π
2 3
2
(a)
(b) −
π2
2 3
(c) −
π2
4 3
(d) −
π2
2
α
QVolume of cone is (V ) =
1 2
πr h
3
2
1 1 
1
π  h (h)=
πh3
3 2 
12
dV
1
dh
=
π(3h2 )
dt 12
dt
dh
4 dV
=
dt
πh2 dt
⇒
4
dh
=
×5
dt
πh2
[Q given
Given differential equation
dy
cos x − (sin x)y = 6 x
dx
6x
dy
, which is the
− (tan x)y =
⇒
cos x
dx
linear differential equation of the form
dy
+ Px = Q,
dx
6x
where P = − tan x and Q =
cos x
− tan xdx
So,
IF = e ∫
= e − log(sec x ) = cos x
2
[from Eq. (i)]
On differentiating both sides w.r.t. ‘t’, we get
⇒
Key Idea
(i) First convert the given differential equation
into linear differential equation of the form
dy
+ Py = Q
dx
(ii) Find IF
(iii) Apply formula, y(IF) = ∫ Q (IF )dx + C
∴Required solution of differential equation is
cos x
dx + C
y(cos x) = ∫ (6 x)
cos x
x2
=6
+ C = 3 x2 + C
2
π
Given, y  = 0
 3
h
l
V=
x
dy

− y sin x = 6x , 0 < x < 

2
dx
π
 
= 0, then y   is equal to
 6
15. If cos x
Exp. (b)
Exp. (b)
∴
Now, at h = 10 m, the rate at which height of water
4
1
dh
m/min
level is rising =
=
× 5=
5π
dt h = 10 π(10)2
dV
= 5m 3 /min]
dt
π
π2
0 = 3   + C ⇒C = −
 3
3
π2
2
y(cos x) = 3 x −
∴
3
π
Now, at x =
6
 3
π2
π2
π2
−
=−
y
 =3
36
3
4
 2 
So,
⇒
y=−
π2
2 3
119
Limits, Continuity and Differentiability
x
16. If f ( x ) = [ x ] −   , x ∈R where[ x ] denotes the
 4
greatest integer function, then
x→ 5 +
[JEE Main 2019, 9 April Shift-II]
(a) lim f ( x ) exists but lim f ( x ) does not exist
x→ 4−
x→ 4+
(b) f is continuous at x = 4
(c) Both lim f ( x ) and lim f ( x ) exist but are
x→ 4−
lim f( x) = lim [a| π − (5 − h)| + 1] = a(5 − π ) + 1
h→ 0
…(ii)
and lim f( x) = lim[b| (5 + h) − π| + 3]
x→ 5 −
x→ 4+
= b(5 − π ) + 3
QFunction f( x) is continuous at x = 5.
∴
⇒
⇒
⇒
not equal
h→ 0
…(iii)
f(5) = lim f( x) = lim f( x)
x→ 5 +
x→ 5 −
a(5 − π ) + 1 = b(5 − π ) + 3
(a − b )(5 − π ) = 2
2
a−b=
5− π
(d) lim f ( x ) exists but lim f ( x ) does not exist
x→ 4−
x→ 4+
18. Let f ( x ) = e x − x and g ( x ) = x 2 − x , ∀ x ∈R.
Then, the set of all x ∈R , where the function
h( x ) = ( fog )( x ) is increasing, is
Exp. (b)
x
Given function f( x) = [ x] −  , x ∈ R
 4 
[JEE Main 2019, 10 April Shift-I]
4 + h

Now, lim f( x) = lim  [4 + h] − 
 4  
h→ 0 
x→ 4 +
[Qput x = 4 + h, when x → 4+ , then h → 0]
= lim(4 − 1) = 3
h→ 0
4 − h

and lim f( x) = lim  [4 − h] − 
 4  
h→ 0 
x→ 4 −
−
[Q put x = 4 − h, when x → 4 then h → 0]
= lim(3 − 0) = 3
h→ 0
Q
lim f( x) = f(4) = lim f( x) = 3
x→ 4 −
x→ 4 +
a | π − x | + 1 , x ≤ 5
is
b | x − π |+3, x > 5
continuous at x = 5, then the value ofa − b is
[JEE Main 2019, 9 April Shift-II]
2
π+ 5
2
(d)
5− π
(b)
Exp. (d)
Given function
 a| π − x| + 1, x ≤ 5
f ( x) = 
b| x − π| + 3, x > 5
and it is also given that f( x)is continuous at
Clearly, f(5) = a(5 − π ) + 1
h( x) = (fog )( x) = f(g ( x))
h′( x) = f ′(g ( x)) ⋅ g ′( x)
= (e g( x ) − 1) ⋅ (2 x − 1)
= (e
17. If the function f ( x ) = 
−2
π+ 5
2
(c)
π−5
The given functions are
f( x) = e x − x,
and
g ( x) = x2 − x, ∀ x ∈ R
x = 5.
…(i)
2
− x)
− 1) (2 x − 1)
x( x − 1)
− 1) (2 x − 1)
= (e( x
So, function f( x) is continuous at x = 4.
(a)
Exp. (a)
Then,
Now,
4
and f(4) = [4] −   = 4 − 1 = 3
 4 
−1
1
(b)  −1,  ∪  , ∞ 



2  2 
−1 

(d)
, 0 ∪ [1, ∞ )
 2 
1
(a) 0,  ∪ [1, ∞ )
 2 
(c) [0, ∞ )
QIt is given that h( x) is an increasing function, so
h′( x) ≥ 0
⇒
(e x( x − 1) − 1)(2 x − 1) ≥ 0
Case I (2 x − 1) ≥ 0 and (e x( x − 1) − 1) ≥ 0
1
⇒
x ≥ and x( x − 1) ≥ 0
2
⇒ x ∈ [1 / 2, ∞ ) and x∈(− ∞, 0] ∪ [1, ∞ ), so x∈ [1, ∞ )
Case II (2 x − 1) ≤ 0 and [e x( x − 1) − 1] ≤ 0
1
x ≤ and x( x − 1) ≤ 0
⇒
2
1
⇒
x ∈  −∞,  and x ∈[0, 1]

2 
1
So,
x ∈  0, 
 2 
1
From, the above cases, x ∈  0,  ∪ [1, ∞ ).
 2 
120
JEE Main Chapterwise Mathematics
x4 − 1
x3 − k 3
, then k is
= lim 2
x→1 x −1
x→k x −k2
19. If lim
[JEE Main 2019, 10 April Shift-I]
4
(a)
3
3
(c)
2
3
8
8
(d)
3
(b)
Exp. (d)
x4 − 1
x3 − k 3
Given, lim
= lim 2
x→ 1 x − 1
x → k x − k2
⇒
⇒
( x − 1)( x + 1)( x2 + 1)
lim
x→ 1
x−1
( x − k )( x2 + k 2 + xk )
= lim
x→ k
( x − k )( x + k )
3k 2
2k
8
k=
3
2 ×2 =
⇒
20. Let f : R → R be differentiable at c ∈R and
f (c ) = 0. If g ( x ) = | f ( x )|, then at x = c , g is
[JEE Main 2019, 10 April Shift-I]
(a)
(b)
(c)
(d)
not differentiable
differentiable if f ′(c ) ≠ 0
not differentiable if f ′(c ) = 0
differentiable if f ′(c ) = 0
21.
 sin (p + 1) x + sin x
,x<0

x

If f ( x ) = 
q,
x =0

x>0
x + x2 − x

,
3/2

x
is continuous at x = 0 , then the ordered pair
(p , q ) is equal to [JEE Main 2019, 10 April Shift-I]
3 1
(a)  − , − 
 2
2
5 1
(c)  , 
 2 2
Given function, g ( x) = |f( x)|
where f : R → R be differentiable at c ∈ R and
f(c ) = 0, then for function ‘g’ at x = c
g (c + h) − g (c )
[where h > 0]
g ′(c ) = lim
h→ 0
h
|f(c + h)| − |f(c )|
= lim
h→ 0
h
|f(c + h)|
[as f(c ) = 0(given)]
= lim
h→ 0
h
f(c + h) − f(c )
[Qh > 0]
= lim
h→ 0
h
f(c + h) − f(c )
= lim
h→ 0
h
= f ′(c )
[Qf is differentiable at x = c]
Now, if f ′(c ) = 0, then g ( x) is differentiable at x = c,
otherwise LHD (at x = c) and RHD
(at x = c ) is different.
3 1
(d)  − , 
 2 2
Exp. (d)
Given function
sin( p + 1)x + sin x
, x< 0

x

q
f ( x) = 
, x =0

2
x+x − x

, x> 0

x3 / 2
is continuous at x =`0, then
f(0) = lim f( x) = lim f( x)
x→ 0 −
x→ 0 +
…(i)
sin( p + 1)x + sin x
x
sin(ax)
=
= p + 1 + 1 = p + 2 Q lim
 x→ 0 x
lim f( x) = lim
x→ 0 −
x→ 0 −
and lim f( x) = lim
x→ 0 +
Exp. (d)
1 3
(b)  − , 
 2 2
a

x + x2 − x
x→ 0 +
x3 / 2
x[(1 + x) − 1]
= lim
x→ 0 +
x x
11 


 − 1


 1 + 1 x + 2  2  x2 + .... − 1


2!
2




= lim
x→ 0 +
x
1/ 2
[Q(1+ x)n
n(n − 1) 2 n(n − 1(n − 2 )) 3
x +
x + ...,| x|< 1]
1⋅ 2
1⋅ 2 ⋅ 3
11 


 − 1


1
1 2 2 

= lim
+
x + ... =
 2
x→ 0 +  2
2!




From Eq. (i), we get
1
1
3
f(0) = q = and lim f( x) = p + 2 = ⇒ p = −
x→ 0 −
2
2
2
3 1
So, ( p, q ) =  − , 
 2 2
= 1 + nx +
121
Limits, Continuity and Differentiability
x
, x ∈R ,
x2 − 3
( x ≠ ± 3 ), at a point (α , β ) ≠ (0, 0) on it is
parallel to the line 2 x + 6y − 11 = 0, then
22. If the tangent to the curve y =
[JEE Main 2019, 10 April Shift-II]
(a) | 6α + 2β | = 19
(c) | 2α + 6β | = 19
(b) | 6α + 2β | = 9
(d) | 2α + 6β | = 11
…(i)
3)
On differentiating Eq. (i) w.r.t. x, we get
dy ( x2 − 3) − x(2 x) (− x2 − 3)
= 2
=
dx
( x − 3)2
( x2 − 3)2
It is given that tangent at a point (α, β ) ≠ (0, 0) on it
is parallel to the line
2 x + 6 y − 11 = 0.
2 dy
∴ Slope of this line = − =
6 dx ( α, β )
⇒
⇒
α + 3
2
(α 2 − 3)2
=−
1
3
According to the options,| 6α + 2 β | = 19
1
at (α, β ) =  ± 3, ± 

2
x 2 − ax + b
= 5, then a + b is equal to
x→1
x −1
[JEE Main 2019, 10 April Shift-II]
(d) 5
Exp. (c)
x2 − ax + b
=5
x→ 1
x−1
Exp. (b)
Given functions, f( x) = loge (sin x), (0 < x < π ) and
g ( x) = sin− 1(e − x ), x ≥ 0.
Now, fog ( x) = f(g ( x)) = f(sin− 1(e − x ))
= loge (sin(sin− 1(e − x )))
= loge (e − x )
{Qsin(sin− 1 x) = x, if x ∈ [− 1, 1}
]
=− x
23. If lim
It is given that lim
[JEE Main 2019, 10 April Shift-II]
(d) aα 2 + bα + a = 0
α − 9 α = 0 ⇒ α = 0, − 3, 3
(c) − 7
f ( x ) = log e (sin x ),
(0 < x < π ) and
−1 − x
g ( x ) = sin (e ), ( x ≥ 0). If α is a positive real
number such that a = ( fog )′ (α ) and
b = ( fog )(α ), then
24. Let
(c) aα 2 + bα − a = − 2α 2
2
(b) 1
…(iii)
⇒
1− b = 5 ⇒ b = − 4
On putting value of ‘b’ from Eq. (iii) to Eq. (ii), we
get
a=−3
So,
a+ b= −7
(b) aα 2 − bα − a = 1
3α 2 + 9 = α 4 − 6α 2 + 9
4
lim ( x − b ) = 5
x→ 1
(a) aα 2 − bα − a = 0
⇒
α = 3or − 3,
[Qα ≠ 0]
Now, from Eq. (i),
α
β= 2
α −3
−3
1
1
3
or
β=
= or −
⇒
9−3 2
2
9−3
(a) − 4
( x2 − x) − b( x − 1)
=5
x→ 1
x−1
( x − 1) ( x − b )
=5
lim
x→ 1
x−1
lim
⇒
Equation of given curve is
x
, x ∈ R, ( x ≠ ±
y= 2
x −3
−
⇒
⇒
Exp. (a)
⇒
On putting the value of ‘a’ from Eq. (ii) in
Eq. (i), we get
x2 − (1 + b ) x + b
lim
=5
x→ 1
x−1
…(i)
Since, limit exist and equal to 5 and denominator
is zero at x = 1, so numerator x2 − ax + b should
be zero at x = 1,
So
…(ii)
1− a + b = 0 ⇒ a = 1+ b
…(i)
d
and
…(ii)
(fog )′( x) = (− x) = − 1
dx
According to the question,
[from Eq. (ii)]
Q
a = (fog )′(α ) = − 1
and
[from Eq. (i)]
b = (fog ) (α ) = − (α )
for a positive real value ‘α’.
Since, the value of a = − 1and b = − α, satisfy the
quadratic equation (from the given options)
aα 2 − bα − a = 1.
25. A spherical iron ball of radius 10 cm is
coated with a layer of ice of uniform
thickness that melts at a rate of 50 cm 3/min.
When the thickness of the ice is 5 cm, then
122
JEE Main Chapterwise Mathematics
the rate at which the thickness (in cm/min)
of the ice decreases, is
[JEE Main 2019, 10 April Shift-II]
1
(a)
9π
(c)
1
36 π
(b)
1
18 π
(d)
5
6π
Exp. (b)
Let the thickness of layer of ice is x cm, the volume
of spherical ball (only ice layer) is
4
…(i)
V = π[(10 + x)3 − 103 ]
3
On differentiating Eq. (i) w.r.t. ‘t’, we get
dx
dV 4
[given]
= π(3(10 + x)2 ) = − 50
3
dt
dt
[− ve sign indicate that volume is
decreasing as time passes].
dx
4 π(10 + x)2
= − 50
⇒
dt
At x = 5 cm
dx
[4 π(10 + 5)2 ] = − 50
dt
1
1
50
dx
cm / min
=−
=−
=−
⇒
9(2 π )
18 π
225(4 π )
dt
So, the thickness of the ice decreases at the rate
1
of
cm / min.
18π
26.
 dy d 2y 
If e + xy = e, the ordered pair  , 2  at
 dx dx 
x = 0 is equal to [JEE Main 2019, 12 April Shift-I]
y
1
1
(a)  , − 2 
e
e 
1 1
(c)  , 2 
e e 
1 1
(b)  − , 2 
 e e 
1
1
(d)  − , − 2 
 e
e 
2
dy
d 2 y dy dy
+ e y   + x 2 +
+
= 0 …(iv)
 dx 
dx dx
dx
dx
Now, on putting x = 0 in Eq. (i), we get
e y = e1 ⇒ y = 1
ey
d2y
2
On putting x = 0, y = 1in Eq. (iii), we get
1
1
dy
=−
=−
dx
e+ 0
e
1
dy
Now, on putting x = 0, y = 1and
= − in
dx
e
Eq. (iv), we get
2
 d 2 y
1
1
1
d2y
e1 2 + e1  −  + 0 2  +  −  +  −  = 0


 e


e
e
dx
dx


d2y
⇒
dx2
Given equation is
…(i)
e y + xy = e
On differentiating both sides w.r.t. x, we get
dy
dy
…(ii)
+ x + y=0
ey
dx
dx
 y 
dy
⇒
=− y
 …(iii)
dx
 e + x
Again differentiating Eq. (ii) w.r.t. ‘x’, we get
( 0, 1)
1
e2
 dy d 2 y 
1 1
 , 2  at (0, 1) is  − , 2  .

dx
e
dx 
e 

So,
27. Let f : R → R be a continuously differentiable
1
. If
48
f(x) 3
∫ 4t dt = (x − 2 )g (x ), then lim g (x ) is
function such that f (2 ) = 6 and f ′ (2 ) =
x→ 2
6
equal to
[JEE Main 2019, 12 April Shift-I]
(a) 18
(b) 24
(c) 12
(d) 36
Exp. (a)
Given ∫
f( x )
6
4t 3dt = ( x − 2 ) g ( x)
f( x )
∫
g ( x) = 6
⇒
4t 3 dt
( x − 2)
f( x )
lim g ( x) = lim
So,
x→ 2
∫6
x→ 2
[provided x ≠ 2]
4t 3dt
x−2
Q 0 form as x → 2 ⇒ f(2 ) = 6
 0

Exp. (b)
Key Idea Differentiating the given equation
twice w.r.t. ‘x’.
=
lim g ( x) = lim
x→ 2
Q d
 dx
x→ 2
φ 2( x )
∫φ
1 ( x)
4(f( x))3 f ′( x)
1
f(t ) dt = f(φ2 ( x)), φ2′ ( x) − f(φ1( x)) ⋅ φ1′ ( x)

On applying limit, we get
lim g ( x) = 4(f(2 ))3 f ′(2 ) = 4 × (6)3
x→ 2
1
,
48
Q f(2 ) = 6 and f ′(2 ) = 1 
48 

4 × 216
=
= 18
48
123
Limits, Continuity and Differentiability
28. If α and β are the roots of the equation
Given function,
375x 2 − 25x − 2 = 0, then
n
lim
n→∞
n→∞
∑βr is equal to
r =1
[JEE Main 2019, 12 April Shift-I]
21
(a)
346
1
(c)
12
29
358
7
(d)
116
(b)
Exp. (c)
Given α and β are roots of quadratic equation
375 x2 − 25 x − 2 = 0
25
1
… (i)
α+β=
=
∴
375 15
2
… (ii)
and
αβ = −
375
n
Now, lim
n→ ∞
∑α
r
+ lim
r =1
n
n→ ∞
∑β
(β + β 2 + β 3 + K + upto infinite terms)


a
QS ∞ = 1 − r for GP


α(1 − β) + β(1 − α ) α − αβ + β − αβ
=
=
(1 − α ) (1 − β )
1 − α − β + αβ
=
α
β
+
1− α 1− β
=
(α + β ) − 2αβ
1 − (α + β ) + αβ
−2
1
and αβ =
375
15
from Eqs. (i) and (ii) respectively, we get
1
4
+
15
375
=
1
2
1−
−
15 375
29
29
1
=
=
=
375 − 25 − 2 348 12
On substituting the value α + β =
29. If m is the minimum value of k for which the
function f ( x ) = x kx − x 2 is increasing in the
interval [0, 3] and M is the maximum value of
f in the interval [0, 3] when k = m , then the
ordered pair (m , M ) is equal to
[JEE Main 2019, 12 April Shift-I]
(b) ( 4, 3 3 )
(d) ( 5, 3 6 )
… (i)
the function f( x) is defined if
kx − x2 ≥ 0
⇒
x2 − kx ≤ 0
… (ii)
⇒
x ∈[0, k ]
because it is given that f( x)is increasing in interval
x ∈[0, 3], so k should be positive.
Now, on differentiating the function f( x) w.r.t. x, we
get
x
f ′( x) = kx − x2 +
× ( k − 2 x)
2 kx − x2
=
=
r
r =1
= (α + α 2 + α 3 + K + upto infinite terms)+
(a) ( 4, 3 2 )
(c) ( 3, 3 3 )
f( x) = x kx − x2
n
∑αr + lim
r =1
Exp. (b)
2(kx − x2 ) + kx − 2 x2
2 kx − x2
3kx − 4 x2
2 kx − x2
as f( x) is increasing in interval x ∈[0, 3], so
f ′( x) ≥ 0 ∀ x ∈ (0, 3)
⇒
3kx − 4 x2 ≥ 0
⇒
4 x2 − 3kx ≤ 0
3k 
4 x  x −
⇒
≤0

4
3k
(as k is positive)
x ∈  0, 
⇒
 4 
3k
So,
3≤
4
⇒
k≥ 4
⇒ Minimum value of k = m = 4
and the maximum value of f in [0, 3] is f(3).
Qf is increasing function in interval x ∈[0, 3]
QM = f(3) = 3 4 × 3 − 32 = 3 3
Therefore, ordered pair (m, M ) = (4, 3 3 )
Hence, option (b) is correct.
 sin x − cos x 
 , with
 sin x + cos x 
x

 π
respect to , where  x ∈ 0,   is
 2 

2
30. The derivative of tan −1 
[JEE Main 2019, 12 April Shift-II]
(a) 1
(c)
1
2
2
3
(d) 2
(b)
124
JEE Main Chapterwise Mathematics
= lim ( x2 + 2 sin x + 1 +
Exp. (d)
x→ 0
 sin x − cos x 
Let f( x) = tan−1 

 sin x + cos x 
x2 − sin2 x + 2 sin x + x
x + 2 sin x
 0 form
= 2 × lim 2
 0

x → 0 x − sin2 x + 2 sin x + x
x→ 0
[dividing numerator and
π 
denominator by cos x > 0, x ∈  0, 
 2  

π 
 tan x − tan

−1 
4 
= tan
 1 +  tan π  (tan x)







4
π 

= tan−1 tan x − 
 
4  
 tan A − tan B

Q 1 + tan A tan B = tan ( A − B)


π

Since, it is given that x ∈  0,  , so
 2
π
π π
x − ∈  − , 
4  4 4
π
π π
Also, for  x −  ∈  − ,  ,

4  4 4
π 
π

Then, f( x) = tan−1  tan  x −   = x −


4
4
π π 

Q tan−1 tanθ = θ, for θ ∈  − , 

 2 2  
x
Now, derivative of f( x) w.r.t. is
2
d 
π
d (f( x))
df( x)
=2 ×
=2
x−  =2
dx 
4
d( x / 2 )
d ( x)
x→ 0
x + 2 sin x
is
x + 2 sin x + 1 − sin 2 x − x + 1
2
[JEE Main 2019, 12 April Shift-II]
(a) 6
(b) 2
(c) 3
(d) 1
Now applying the L′ Hopital’s rule, we get
1 + 2 cos x
P = 2 × lim
x → 0 2 x − sin 2 x +2 cos x + 1
(1 + 2 )
[on applying limit]
=2
0−0+2 +1
3
=2 × =2
3
x + 2 sin x
⇒ lim
=2
x→ 0
x2 + 2 sin x + 1 − sin2 x − x + 1
32. If 20C 1 + (2 2 ) 20C 2 + ( 32 ) 20C 3 + .....
+ (202 )20C 20 = A (2β ) , then the ordered pair
( A , β) is equal to [JEE Main 2019, 12 April Shift-II]
(a) (420, 19)
(c) (380, 18)
x→ 0
We know,
(1 + x)n = nC 0 + nC1 x + nC 2 x2 + ... + nC n xn
On differentiating both sides w.r.t. x, we get
n(1 + x)n − 1 = nC1 + 2 nC 2 x + ... + n nC n xn − 1
On multiplying both sides by x, we get
n x(1 + x)n − 1 = nC1 x + 2 n C 2 x2 + ... + nnC n xn
Again on differentiating both sides w.r.t. x,
we get
n [(1 + x)n − 1 + (n − 1) x (1 + x)n − 2 ]
= nC1 + 2 2 nC 2 x + ... + n2 nC n xn − 1
Now putting x = 1in both sides, we get
n
C1 + (2 2 ) nC 2 + (32 ) nC 3 + ... + (n2 ) nC n
= n(2 n − 1 + (n − 1) 2 n − 2 )
x + 2 sin x
x2 + 2 sin x + 1 − sin2 x − x + 1
 0 form
 0

On rationalization, we get
( x + 2 sin x)
P = lim 2
x → 0 x + 2 sin x + 1 − sin2 x + x − 1
( x2 + 2 sin x + 1 +
(b) (420, 18)
(d) (380, 19)
Exp. (b)
Exp. (b)
Let P = lim
x + 2 sin x
× lim
 tan x − 1
= tan−1 

 tan x + 1
31. lim
sin2 x − x + 1)
sin2 x − x + 1)
For n = 20, we get
20
C1 + (2 2 )
C 2 + (32 )
20
= 20(2
19
C 3 + ... + (20)2
20
+ (19) 2 )
18
= 20 (2 + 19) 218 = 420 (218 )
= A(2 B ) (given)
On comparing, we get
( A, B) = (420, 18)
20
C 20
125
Limits, Continuity and Differentiability
33. Let f ( x ) = 5 − | x − 2 | and g ( x ) = | x + 1|, x ∈R . If
f ( x ) attains maximum value at α and g ( x )
attains minimum value of β, then
( x − 1)( x 2 − 5x + 6)
is equal to
lim
x → − αβ
x 2 − 6x + 8
Clearly, for f( x) to be continuous, it has to be
continuous at x = 1, x = 3 and x = 5
[QIn rest portion it is continuous everywhere]
…(i)
∴ lim (a + bx) = a + b = 5
x → 1+
[Q lim f( x) = lim f( x) = f(1)]
x→1−
[JEE Main 2019, 12 April Shift-II]
(b) − 3 / 2
(d) 3/2
Exp. (a)
Given functions are f( x) = 5 − | x − 2|
and
g ( x) = | x + 1|, where x ∈ R.
Clearly, maximum of f( x) occurred at x =2, so
α = 2.
and minimum of g ( x) occurred at x = − 1, so
β = − 1.
⇒
αβ = − 2
( x − 1) ( x2 − 5 x + 6)
Now,
lim
x → − αβ
x2 − 6 x + 8
( x − 1) ( x − 3) ( x − 2 )
[Qαβ = − 2]
= lim
x→ 2
( x − 4) ( x −2 )
( x − 1) ( x − 3)
= lim
x→ 2
( x − 4)
(2 − 1) (2 − 3) 1 × (− 1) 1
=
=
=
(− 2 )
2
(2 − 4)
34. Let f :R → R be a function defined as
 5,
 a + bx,

f (x) = 
b + 5 x ,
 30,
Then, f is
(a)
(b)
(c)
(d)
x → 5−
x → 5−
and
x → 3−
lim (b + 5 x) = b + 15 = 20
x→ 3+
Hence, for a = 0 and b = 5, f( x)is not continuous at
x=3
∴ f( x) cannot be continuous for any values of a
and b.
35. The maximum volume (in cum) of the right
circular cone having slant height 3m is
[JEE Main 2019, 9 Jan Shift-I]
(a)
4
π
3
(c) 3 3π
(b) 2 3π
Let h = height of the cone,
r = radius of circular base
= (3)2 − h2
[JEE Main 2019, 9 Jan Shift-I]
[Ql 2 = h2 + r 2 ]
9 − h2
…(i)
h
r
Now, volume (V ) of cone =
if
x≤1
if 1 < x < 3
if 3 ≤ x < 5
x≥ 5
if
(d) 6π
Exp. (b)
continuous if a = − 5 and b =10
continuous if a = 5 and b = 5
continuous if a =0 and b = 5
not continuous for any values of a and b
 5
 a + bx

We have, f( x) = 
b + 5 x
 30
f( x) = f(5)]
lim (a + bx) = a + 3b = 15
Here,
=
Key Idea A function is said to be continuous
if it is continuous at each point of the domain .
x→ 5+
On solving Eqs. (i) and (ii), we get b = 5and a = 0
Now, let us check the continuity of f( x) at x = 3.
if
x ≤1
if 1 < x < 3
if 3 ≤ x < 5
if
x≥5
Exp. (d)
…(ii)
[Q lim f( x) = lim
l=
3
(a) 1/2
(c) −1 / 2
x→1 +
lim (b + 5 x) = b + 25 = 30
1 2
πr h
3
1
[from Eq. (i)]
π(9 − h2 )h
3
1
…(ii)
= π[9h − h3 ]
3
For maximum volume V ′(h) = 0 and V ′′(h) < 0.
Here,
V ′(h) = 0 ⇒(9 − 3h2 ) = 0
⇒
V(h) =
⇒
h=
3
[Qh </ 0]
126
JEE Main Chapterwise Mathematics
1
π(−6h) < 0for h = 3
3
Thus, volume is maximum when h = 3
Now, maximum volume
1
V( 3 ) = π(9 3 − 3 3 ) [from Eq. (ii)]
3
= 2 3π
V ′′(h) =
and
36. If θ denotes the acute angle between the
curves, y = 10 − x and y = 2 + x at a point of
their intersection, then |tan θ| is equal to
2
(a) exists and equals
(b) does not exist
(c) exists and equals
1
2 2
(d) exists and equals
1
2 2 ( 2 + 1)
Exp. (a)
7
17
(b)
1+
2
Clearly, lim
8
15
(c)
4
9
(d)
8
17
1+
= lim
1 + y4 − 2
y→ 0
y
Given equation of curves are
y = 10 − x2
y = 2 + x2
and
y→ 0
×
1+
1 + y4 +
2
1+
1+ y +
2
4
2 x2 = 8
⇒
x2 = 4
…(i)
y4 ( 1 +
1 + y4 +
2)
At (−2, 6), slopes m1 = 4 and m2 = − 4 and in that
case
m − m1
−4 − 4
8
=
|tanθ| = 2
=
1 + m1m2
1 − 16
15
1+ 1+ y4 − 2
4
[JEE Main 2019, 9 Jan Shift-I]
1 + y4 − 1
= lim
y→ 0
…(ii)
⇒
x=±2
Clearly, when x = 2 , then y = 6 (using Eq. (i)) and
when x = − 2, then y = 6
Thus, the point of intersection are (2, 6) and
(−2, 6).
Let m1 be the slope of tangent to the curve (i) and
m2 be the slope of tangent to the curve (ii)
dy
dy
For curve (i)
= −2 x and for curve (ii)
= 2x
dx
dx
∴ At (2, 6), slopes m1 = − 4 and m2 = 4, and in that
case
m − m1
4+ 4
8
=
|tanθ| = 2
=
1 + m1m2
1 − 16 15
y
1 + y4 ) − 2
[Q(a + b ) (a − b ) = a2 − b 2 ]
For point of intersection, consider
10 − x2 = 2 + x2
⇒
(1 +
= lim
Key Idea Angle between two curves is the
angle between the tangents to the curves at
the point of intersection.
y→ 0
4
[rationalising the numerator]
Exp. (b)
37. lim
1 + y4 − 2
y4
y→ 0
[JEE Main 2019, 9 Jan Shift-I]
(a)
1
4 2
y4 ( 1 +
=
1 + y4 +
1 + y4 + 1
2)
[again, rationalising the numerator]
y4
= lim
y→ 0
1 + y4 + 1
×
y4 ( 1 +
1 + y4 +
2 )( 1 + y4 + 1)
1
2 2 ×2
(by cancelling y4 and then by direct substitution).
1
=
4 2
38. Let f be a differentiable function from R to R
3
such that | f ( x ) − f ( y )| ≤ 2 | x − y | 2 , for all
1
x , y ∈R . If f (0) = 1, then ∫ f 2( x ) dx is equal to
0
[JEE Main 2019, 9 Jan Shift-II]
(a) 2
(b)
1
2
(c) 1
(d) 0
Exp. (c)
3
Given,| f( x) − f( y)| ≤ 2 | x − y|2 , ∀ x, y ∈ R
1
⇒
| f( x) − f( y)|
≤ 2 | x − y|2
| x − y|
(dividing both sides by| x − y| )
127
Limits, Continuity and Differentiability
Put x = x + h and y = x, where h is very close to
zero.
1
⇒
lim
h→ 0
f( x + h) − f( x)
≤ lim 2 | ( x + h) − x|2
h→ 0
( x + h) − x
1
f( x + h) − f( x)
≤ lim 2 | h|2
h→ 0
h→ 0
h
f( x + h) − f( x)
⇒ lim
≤0
h→ 0
h
[substituting limit directly on right hand
⇒ lim
side and using lim | f( x)| = lim f( x) ]
x→ a
x→ a
Q lim f( x + h) − f( x) = f ′( x)


 h→ 0

h
⇒| f ′( x)| ≤ 0
⇒| f ′( x)| = 0 (Q| f ′( x)| can not be less than zero)
⇒ f ′( x) = 0 [Q| x| = 0 ⇔ x = 0]
⇒ f( x) is a constant function.
Since, f(0) = 1, therefore f( x) is always equal
to 1.
Now,
1
∫0
1
(f( x))2 dx = ∫ dx
0
= [ x]10 = (1 − 0) = 1
39. For each x ∈R , let [ x ] be the greatest integer
less than or equal to x. Then,
x ([ x ] + | x |) sin[ x ]
is equal to
lim
−
|x |
x→ 0
(b) sin1
(c) − sin1
(d)1
Exp. (c)
lim
x→ 0
−
d2y
d  dy  d  dy  dt
  =  ⋅
dx  dx  dt  dx  dx
d  dy 
d
(sin t )
 
 dx 
dt
dt
=
=
d
dx
(3 tan t )
dt
dt
cos t
cos 3 t
=
=
2
3
3sec t
3 π
cos
1
1
d2y 
π
4 =
Now,
=
=
at
=
t


3
3(2 2 ) 6 2
4
dx2 
and
dx2
=
2

41. Let f ( x ) = max { | x |, x },
x([ x] − x) sin [ x]
x([ x] + | x|) sin [ x]
= lim
x → 0−
| x|
− x
(Q | x| = − x, if x < 0)
x(− 1 − x) sin(− 1)
(Q lim [ x] = − 1)
x → 0−
x → 0−
− x
− x( x + 1) sin(− 1)
= lim ( x + 1)sin(− 1)
= lim
x → 0−
x → 0−
− x
= lim
2 < |x |≤ 4
 8 − 2 | x |,
LetS be the set of points in the interval( −4, 4)
at which f is not differentiable. Then, S
[JEE Main 2019, 10 Jan Shift-I]
Exp. (a)
Key Idea
This type of problem can be solved graphically
max{| x|, x2 }, | x| ≤ 2
We have, f( x) = 
2 < | x| ≤ 4
 8 − 2| x|,
Let us draw the graph of y = f( x)
For|x|≤ 2 f( x) = max{| x| x2 }
Let us first draw the graph of y =| x| and y = x2 as
shown in the following figure.
y=x2
= (0 + 1) sin (− 1) (by direct substitution)
[Qsin(− θ) = − sinθ]
= − sin 1
2
40. If x = 3 tant and y = 3 sect , then the value of
1
π
att = , is
4
dx
|x |≤ 2
(a) equals {−2 , − 1, 0,1, 2 } (b) equals {−2 , 2}
(c) is an empty set
(d) equals {−2 ,−1,1, 2}
[9 Jan 2019, Shift II]
(a) 0
We have, x = 3tan t and y = 3 sec t
d
dy
(3sec t )
dy dt
dt
Clearly,
=
=
d
dx dx
(3 tant )
dt
dt
3 sec t tant
tant
= sin t
=
=
sec t
3 sec 2 t
y=|x|
2
d y
2
(a)
1
6
Exp. (b)
(b)
1
6 2
[JEE Main 2019, 9 Jan Shift-II]
(c)
1
3 2
(d)
3
2 2
–2
0
–1
1
2
–1
–2
Clearly, y =| x| and y = x2 intersect at x = − 1, 0, 1
128
JEE Main Chapterwise Mathematics
Now, the graph of y = max{| x|, x2 } for| x| ≤ 2 is
4
Given,
π
(1 − | x| + sin|1 − x|) sin  [1 − x]
2

lim
x → 1+
|1 − x| [1 − x]
y=x2
1
–2
–1
Put x = 1 + h, then
1
x → 1+ ⇒ h → 0+
2
π
(1 − | x| + sin|1 − x|) sin  [1 − x]
2

∴ lim
x → 1+
|1 − x| [1 − x]
y =|x|
For| x| ∈ (2, 4]
x ∈ (2, 4]
 8 − 2 x,
f( x) = 8 − 2| x| = 
8
2
x
,
x
[− 4, − 2 )
+
∈

Q2 < | x| ≤ 4


 ⇒| x| > 2 and | x| ≤ 4



 –4 –2 2
4


Y
5
4
3
2
y=8+2x
y=8–2x
1
X′
–5 –4 –3 –2 –1 O
Exp. (a)
1
2
3
4
5
X
Y′
Hence, the graph of y = f( x) is
π
(1 − |h + 1| + sin|− h|) sin  [− h]
2

= lim
h→ 0+
|− h| [− h]
π
(1 − (h + 1) + sin h) sin  [− h]
2

= lim
h→ 0+
h [− h]
(Q|− h| = h and|h + 1| = h + 1as h > 0)
π
( − h + sin h) sin  (− 1)
2

= lim
h→ 0+
h (− 1)
(Q[ x] = − 1for − 1 < x < 0 and h → 0+ ⇒− h → 0− )
(− h + sinh)  − π 
sin 
= lim

 2 
h→ 0+
−h
(− h + sin h)
sin h − h
(−1) = lim
= lim
h→ 0+
h→ 0+
h
−h
sin h 
h


= lim 
 − lim +   = 1 − 1 = 0
h→ 0+  h 
h → 0  h
sin h


= 1
lim
Q h →
0+
h

1
–2
x
–4
8
y=
y=
8+
2x
4
–2
–1
1
2
4
From the graph it is clear that at x = − 2, − 1, 0, 1, 2
the curve has sharp edges and hence at these
points fis not differentiable.
42. For eacht ∈R, let[t ] be the greatest integer
less than or equal tot . Then,

π
(1 − | x | + sin|1 − x |) sin  [1 − x ]

2
lim
x → 1+
|1 − x |[1 − x ]
[JEE Main 2019, 10 Jan Shift-I]
(a) equals 0
(c) equals −1
(b) does not exist
(d) equals 1
43. A helicopter is flying along the curve given
by y − x 3/2 = 7,( x ≥ 0). A soldier positioned at
1 
the point  , 7 wants to shoot down the
2 
helicopter when it is nearest to him. Then,
this nearest distance is
[JEE Main 2019, 10 Jan Shift-II]
1 7
(a)
3 3
5
(b)
6
(c)
1 7
6 3
(d)
1
2
Exp. (c)
The helicopter is nearest to the soldier, if the
tangent to the path, y = x3 / 2 + 7, ( x ≥ 0) of
helicopter at point ( x, y) is perpendicular to the line
1
joining ( x, y) and the position of soldier  , 7  .
2 
129
Limits, Continuity and Differentiability
4
7
(d) exists and equals 4
(a) does not exist
(b) exists and equals
(c) exists and equals 0
Exp. (d)
(x, y)
(1/2, 7)
QSlope of tangent at point ( x, y) is
dy 3 1/ 2
= x = m1(let )
dx 2
…(i)
1
and slope of line joining ( x, y) and  , 7  is
2 
y−7
…(ii)
m2 =
1
x−
2
Now,
m1 ⋅ m2 = −1
 y−7 
3
⇒ x1/ 2 
 = −1 [from Eqs. (i) and (ii)]
2
 x − (1/2 )
3 1/ 2 x3 / 2
[Qy = x3 / 2 + 7]
x
= −1
⇒
1
2
x−
2
3 2
1
⇒
x = −x +
2
2
⇒
3 x2 + 2 x − 1 = 0
2
⇒
3x + 3x − x − 1 = 0
1
⇒
3 x( x + 1) − 1( x + 1) = 0 ⇒ x = , − 1
3
1
Q
x≥ 0 ∴ x =
3
3/ 2
1

and so,
+ 7 [Q y = x3 / 2 + 7 ]
y= 
 3
 1 1 3/ 2

Thus, the nearest point is  ,  
+ 7
 3  3



Now, the nearest distance
=
3/ 2

 1 − 1  +  7 −  1 
− 7








 2 3
3


=
 1 +  1
 
 
 6
 3
=
3+ 4
7
1
=
=
108
108 6
2
2
3
=
2
1
1
+
36 27
7
3
44. Let f be a differentiable function such that
3 f (x )
,( x > 0) and f (1) ≠ 4.
4 x
 1
Then, lim x f  
+
 x
x→ 0
f ′ (x ) = 7 −
3 f ( x)
, ( x > 0)
4 x
dy
On putting f( x) = y and f ′( x) = , then we get
dx
3 y
dy
=7−
4 x
dx
dy 3
…(i)
+
y=7
⇒
dx 4 x
which is a linear differential equation of the form
dy
3
and Q = 7.
+ Py = Q, where P =
dx
4x
Given, f ′( x) = 7 −
y=x3/2+7
3
Now, integrating factor (IF) = e
∫ 4 x dx
= elog
3
= e4
x 3/ 4
log x
= x3 / 4
and solution of differential Eq. (i) is given by
y(IF) = ∫ (Q ⋅ (IF)) dx + C
yx3 / 4 = ∫ 7 x3 / 4dx + C
3
⇒
⇒
+1
7
x4
yx
+ C ⇒ y x3 / 4 = 4 x 4 + C
=7
3
+1
4
y = 4 x + C x −3 / 4
3/ 4
y = f ( x ) = 4 x + C ⋅ x −3 / 4
1
4
Now,
f   = + C ⋅ x3 / 4
 x x
1
4
∴ lim x f   = lim x + Cx3 / 4 

 x  x→ 0 +  x
x→ 0 +
So,
= lim (4 + Cx7/ 4 ) = 4
x→ 0 +
2
45. The tangent to the curve, y = xe x passing
through the point (1, e ) also passes through
the point
[JEE Main 2019, 10 Jan Shift-II]
4
(a)  , 2e 
3

(c) ( 2 , 3e )
(b) ( 3, 6e )
5
(d)  , 2e 
3

Exp. (a)
Given equation of curve is
y = xe x
2
…(i)
Note that (1, e ) lie on the curve, so the point of
contact is (1, e).
130
JEE Main Chapterwise Mathematics
Now, slope of tangent, at point(1e
, ), to the curve (i) is
2
2
dy
=  x(2 x) e x + e x 
 ( 1, e )
dx ( 1,e ) 
= 2e + e = 3e
Now, equation of tangent is given by
( y − y1 ) = m ( x − x1 )
y − e = 3e( x − 1) ⇒ y = 3ex − 2e
4
On checking all the options, the option  , 2e 
3

satisfy the equation of tangent.
46. Let f :( −1, 1) → R be a function defined by
 − 1 − x2 ,



⇒ f( x) =  −| x|,


2
 − 1− x ,

From the figure, it is clear that function have sharp
1
1
edges, at x = −
, 0,
2
2
∴ Function is not differentiable at 3 points.
47. Let [ x ] denote the greatest integer less than
f ( x ) = max { − x , − 1 − x 2 }. If K is the set of
or equal to x.
all points at which f is not differentiable,
then K has exactly [JEE Main 2019, 10 Jan Shift-II]
Then, lim
(a) three elements
(c) two elements
(b) five elements
(d) one element
Exp. (a)
Key Idea This type of questions can be solved
graphically.
Given, f : (−1, 1)→ R, such that
f( x) = max −| x|, − 1 − x2 


On drawing the graph, we get the follwong figure.
–1 , –1 ,
√2 √2
tan( π sin 2 x ) + (| x | − sin( x[ x ]))2
x2
x→ 0
[JEE Main 2019, 11 Jan Shift-I]
(a)
(b)
(c)
(d)
equals π
equals π + 1
equals 0
does not exist
Exp. (d)
Key Idea
lim f( x) exist iff lim f( x) = lim f( x)
x→ a+
x→ a
x → a−
At x = 0,
Y
O
1
2
1
1
−
< x≤
2
2
1
< x<1
2
− 1< x ≤ −
1
y=f(x)
–1
tan( π sin2 x) + (| x| − sin( x[ x]))2
RHL = lim
+
x→ 0
1 , –1 ,
√2 √2
= lim
x→ 0
y=–|x|
x2
tan( π sin2 x) + ( x − sin( x ⋅ 0))2
+
x2
 Q| x| = x for x > 0 
and [ x] = 0 for 0 < x < 1


Q graph of y = −| x| is
Y
X
= lim
x→ 0
tan( π sin2 x) + x2
+
x2

 tan( π sin2 x) π sin2 x
.
= lim 
+ 1
2
2
+
x→ 0 
π sin x
x

and graph of y = − 1 − x2
= π lim
Y
x→ 0
X
[Q x2 + y2 = 1 represent a complete circle]
= π+1
tan ( π sin2 x)
+
π sin x
2
. lim
x→ 0
sin2 x
+
x2
+1
Q lim tan x = 1 


x→ 0
x


sin
x
and lim
=1
x→ 0
x


131
Limits, Continuity and Differentiability
and LHL
= lim
2
x → 0−
x → 0−
= lim
x→ 0
= lim
x2
g ( x) = |f( x)| + f(| x|)
and
tan ( π sin2 x) + (− x − sin( x (− 1))2
= lim
x2
 Q| x| = − x for x < 0
and [ x] = − 1 for − 1 < x <

−2 ≤ x< 0
 − 1,
f ( x) =  2
 x − 1, 0 ≤ x ≤ 2
We have,
tan( π sin x) + (| x| − sin( x[ x])
2

0
−2 ≤ x< 0
 1,
Clearly, |f( x)| =  2
1
0≤ x≤ 2
|
x
−
|,

−2 ≤ x< 0
1,


= − ( x2 − 1), 0 ≤ x < 1
 x2 − 1,
1≤ x ≤ 2

tan( π sin2 x) + ( x + sin(− x))2
−
x2
tan( π sin2 x) + ( x − sin x)2
x → 0−
and f(| x|) = | x|2 − 1, 0 ≤ | x| ≤ 2
[Qf(| x|) = − 1is not possible as| x| </ 0]
[Q| x|2 = x2 ]
= x2 − 1, | x| ≤ 2
2
= x − 1, − 2 ≤ x ≤ 2
g ( x) = |f( x)| + f(| x|)
x2
[Qsin(− θ) = − sinθ]
 tan( π sin x) + x + sin2 x − 2 xsin x 

= lim 
x → 0− 
x2

2
2
∴

1 + x2 − 1,
−2 ≤ x< 0

2
0≤ x< 1
= − ( x − 1) + x2 − 1,
 x2 − 1 + x2 − 1,
1≤ x ≤ 2

 tan( π sin2 x)
sin2 x 2 xsin x 

= lim 
+ 1+
−
2
−
x→ 0 
x
x2
x2 
 tan ( π sin2 x) π sin2 x
= lim 
+1
.
−
x→ 0 
π sin2 x
x2
sin x 

−2
x 
x2
π sin2 x
tan( π sin2 x)
= lim
+
.
lim
x → 0−
x → 0−
π sin2 x
x2
2
sin x
sin x
− 2 lim
1 + lim
2
−
x → 0−
x
→
0
x
x
= π + 1+ 1− 2 = π
RHL ≠ LHL
Q
∴ Limit does not exist.
+
 x2 ,
−2 ≤ x< 0

0,
0≤ x< 1
=
2 ( x2 − 1), 1 ≤ x ≤ 2

sin2 x
Now, let us draw the graph of y = g ( x), as shown in
the figure.
Y
(–2,4)
X′
–2
−2 ≤ x < 0
 −1 ,
2
x
−
1
,
0≤ x ≤ 2

and g ( x ) = | f ( x )| + f (| x |).
(2,6)
y=x2 y=0
y=2 (x2 –1)
–1 O
X
1
2
48. Let f ( x ) = 
Then, in the interval ( −2 , 2 ), g is
[JEE Main 2019, 11 Jan Shift-I]
(a)
(b)
(c)
(d)
not differentiable at one point
not differentiable at two points
differentiable at all points
not continuous
Exp. (a)
Key Idea
This type of problem can be solved graphically.
Y′
1
( y + 2 ) represent a
2
parabola with vertex (0, − 2 ) and it open upward]
Note that there is a sharp edge at x = 1 only, so
g ( x) is not differentiable at x = 1only.
[ Here, y = 2 ( x2 − 1) or x2 =
49. The maximum value of the function
f ( x ) = 3x 3 − 18x 2 + 27x − 40
S = { x ∈R : x 2 + 30 ≤ 11x } is
on
the
set
[JEE Main 2019, 11 Jan Shift-I]
(a) 122
(c) − 222
(b) −122
(d) 222
132
JEE Main Chapterwise Mathematics
Exp. (a)
⇒ f ′( x) = 9 x2 − 36 x + 27
= 9( x2 − 4 x + 3) = 9( x − 1) ( x − 3) …(i)
Also, we have
S = { x ∈ R : x2 + 30 ≤ 11 x}
x2 − 11x + 30 ≤ 0
50. If x loge (loge x ) − x 2 + y 2 = 4( y > 0), then
dy
dx
at x = e is equal to [JEE Main 2019, 11 Jan Shift-I]
(c)
e
(b)
4+e2
(1 + 2e )
(d)
4+e2
( 2e − 1)
2 4+e2
We have, xloge (loge x) − x2 + y2 = 4, which can
be written as
… (i)
y2 = 4 + x2 − xloge (loge x)
Now, differentiating Eq. (i) w.r.t. x, we get
dy
1
1
. − 1⋅loge (loge x)
2y = 2x − x
dx
loge x x
[by using product rule of derivative]
1
2x −
− loge (loge x)
dy 
loge x

⇒   =
 dx 
2y
f ( x) =
x
(a2 + x2 )1/ 2
−
(d − x)
(b 2 + (d − x)2 )1/ 2
Differentiating above w.r.t. x, we get
1
2x
(a2 + x2 )1/ 2 − x
2 (a2 + x2 )1/ 2
f ′ ( x) =
( a2 + x 2 )
( b 2 + (d − x )2 )1/ 2( −1) − (d − x )
−
2(d − x )( −1)
2( b 2 + (d − x )2 )1/ 2
( b 2 + (d − x )2 )
a2
(a + x )
2
2 3/ 2
+
b2
(b + (d − x)2 )3 / 2
2
> 0,
∀ x∈R
Hence, f( x) is an increasing function of x.
52. Let K be the set of all real values of x, where
the function
f ( x ) = sin | x | − | x | + 2( x − π )cos| x | is not
differentiable. Then, the set K is equal to
[JEE Main 2019, 11 Jan Shift-II]
… (ii)
Now, at x = e, y2 = 4 + e 2 − e loge (loge e )
[using Eq. (i)]
= 4 + e 2 − e loge (1) = 4 + e 2 − 0
= e2 + 4
[Q y > 0]
∴ At x = e and y = e + 4,
2e − 1
dy 2e − 1 − 0
=
=
dx 2 e 2 + 4 2 e 2 + 4
2
f is an increasing function of x
f ′ is not a continuous function of x
f is a decreasing function of x
f is neither increasing nor decreasing
function of x
We have,
=
y = e2 + 4
2
[by using quotient rule of derivative]
a2 + x 2 − x 2
b 2 + (d − x)2 − (d − x)2
+
= 2
2 3/ 2
(a + x )
(b 2 + (d − x)2 )3 / 2
2 4+e2
(1 + 2e )
Exp. (b)
⇒
2
Exp. (a)
⇒
( x − 5) ( x − 6) ≤ 0 ⇒ x ∈[5, 6]
So,
S = [5, 6]
Note that f( x) is increasing in [5, 6]
[Qf ′( x) > 0 for x ∈[5, 6]
∴f(6) is maximum, where
f(6) = 3(6)3 − 18(6)2 + 27(6) − 40 = 122
(a)
(a)
(b)
(c)
(d)
x2 + 30 ≤ 11x
Clearly,
d−x
−
, x ∈R ,
a +x
b + (d − x )2
wherea,b andd are non-zero real constants.
Then,
[JEE Main 2019, 11 Jan Shift-II]
We have,
f( x) = 3 x3 − 18 x2 + 27 x − 40
⇒
x
51. Let f ( x ) =
2
[using Eq. (ii)]
(a) {0}
(c) { π }
(b) φ (an empty set)
(d) {0, π }
Exp. (b)
We have,
f( x) = sin| x| − | x| + 2 ( x − π ) cos| x|
− sin x + x + 2( x − π )cos x, if x < 0
f ( x) = 
 sin x − x + 2( x − π )cos x, if x ≥ 0
[Qsin(−θ) = − sinθ and cos(−θ) = cos θ]
∴ f ′ ( x) =
− cos x + 1 + 2 cos x − 2( x − π )sin x; if x < 0

 cos x − 1 + 2 cos x − 2( x − π ) sin x , if x > 0
133
Limits, Continuity and Differentiability
Clearly, f( x) is differentiable everywhere except
possibly at x = 0
[Qf ′( x) exist for x < 0 and x > 0]
Here, Rf ′(0) = lim (3cos x − 1 − 2( x − π )sin x)
x→ 0
+
= Lt
(1 − tan2 x)
2 (1 + tan2 x)
×
cos x − sin x
tan3 x
= Lt
cos 2 x − sin2 x
2 (sec 2 x)
×
π cos x − sin x
cos 2 x tan3 x
π
x→
4
x→
= 3 − 1− 0 = 2
x→ 0
−
= 1+ 1− 0 = 2
Q
Rf ′(0) = Lf ′(0)
So, f( x) is differentiable at all values of x.
⇒
K=φ
x cot( 4x )
53. lim
x → 0 sin 2 x cot 2 (2 x )
(b) 1
(d) 2
Exp. (b)
sin x ⋅ cot 2 x
x→ 0
= lim
x→ 0
= lim
x→ 0
=
54.
2
[Q(a2 − b 2 ) = (a − b ) (a + b )]
2 sec 4 x
= Lt
tan3 x
(cos x + sin x)
2 ( 2 )4  1
1 
+

 [on applying limit]
 2
2
(1)3
2 
= 4 2 
 =8
 2
=
(1 + log e 2 x )2
2
x
1 tan 2 x
.
tan 4 x sin2 x
1
2
2
dy
is equal to
dx
[JEE Main 2019, 12 Jan Shift-I]
2
1 4x
x
tan 2 x
.
2
4 (tan 4 x) sin x
x2
2
1 4 x  x   tan2 x  4

 .
 .
4 (tan 4 x)  sin x   2 x  1
4
1
x
tan x 
=1
. 1 .11. Q lim
= 1 = lim

x→ 0 x 
1  x→ 0 sin x
4
x loge 2 x + loge 2
(a)
x
(c) x loge 2 x
x loge 2 x − loge 2
x
(d) loge 2 x
(b)
Exp. (b)
Given equation is
(2 x)2 y = 4 ⋅ e 2 x − 2 y
... (i)
On applying ‘loge ’ both sides, we get
loge (2 x)2 y = loge 4 + loge e 2 x − 2 y
cot 3 x − tan x
is
lim
π
π

x→
cos
x
+


4

4
2 yloge (2 x) = loge (2 )2 + (2 x − 2 y)
[Qloge nm = mloge n and loge e f( x ) = f( x)]
[JEE Main 2019, 12 Jan Shift-I]
(a) 4 2
(c) 8
(cos x − sin x) (cos x + sin x)
2 sec 4 x
×
(cos x − sin x)
tan3 x
55. For x > 1, if (2 x )2y = 4e 2x − 2y , then
xcot 4 x
= lim
π
x→
4
is equal to
(a) 0
(c) 4
2
= Lt
π
x→
4
[JEE Main 2019, 11 Jan Shift-II]
x→ 0
[Q1 + tan2 x = sec 2 x]
Lf ′(0) = lim (cos x + 1 − 2( x − π )sin x)
and
lim
4
(b) 4
(d) 8 2
⇒ (2 loge (2 x) + 2 )y = 2 x + 2 loge (2 )
x + loge 2
y=
⇒
1 + loge (2 x)
On differentiating ‘y’ w.r.t. ‘x’, we get
Exp. (c)
Given, limit =
=
cot x − tan x
x→ π/4
π
cos x + 

4
3
Lt
Lt
x→ π/4
1 − tan4 x
1
×
3
1
(cos x − sin x) tan x
2
Qcot x = 1 

tan x 
dy
=
dx
(1 + loge (2 x))1 − ( x + loge 2 )
2
2x
(1 + loge (2 x))2
1
1 + loge (2 x) − 1 − loge 2
x
=
(1 + loge (2 x))2
dy  xloge (2 x) − loge 2 
So, (1 + loge (2 x))2
=


dx 
x
134
JEE Main Chapterwise Mathematics
56. Let S be the set of all points in ( − π , π ) at
which the function, f ( x ) = min {sin x ,cos x }
is not differentiable. Then, S is a subset of
which of the following?
=
π
π π π
(b)  − , − , , 
4 4 2
 2
=
3π
π 3π π 
(c)  −
,− ,
, 
4 4 4
 4
Let us draw the graph of y = f( x), as shown below
–3π
4
–
π
2
O
y=cos x y=sin x
π
π/4
–1
X
=
y=min {sin x, cos x}
Clearly, the function f( x) = min {sin x, cos x} is not
− 3π
π
differentiable at x =
and [these are point
4
4
of intersection of graphs of sin x and cos x in
(− π, π ), on which function has sharp edges]. So,
−3 π π 
S = 
, , which is a subset of
 4 4
 −3 π , − π , 3 π , π 


4 4 4
 4
π − 2 sin − 1 x
57. lim−
1− x
x→1
is equal to
[JEE Main 2019, 12 Jan Shift-II]
(a)
π
2
(b)
2
π
(c)
π
(d)
1
2π
Exp. (b)
Let L =
lim
x → 1−
π − 2 sin−1 x
, then
1− x
L=
lim
x → 1−
π − 2 sin−1 x
×
1− x
=
lim
x → 1−
lim 2 cos −1 x
lim
×
x → 1−
x → 1−
1− x
−1
1
π + 2 sin−1 x
lim

π
−1
Q x → 1− sin x = 2 


lim 2 cos x
1
−
1− x
2 π x→1
2θ
θ
2 sin 
2
θ
2 ⋅  
2
1
=
⋅ 2 lim
θ
θ→ 0 +
2 π
sin 
2
=
Exp. (c)
–π
π + 2 sin−1 x
Put x = cosθ, then as x → 1− , therefore θ → 0+
lim
2θ
1
Now, L =
+
1 − cos θ
2 π θ→ 0
3π
π π 3π 
(d)  −
,− , ,

2 2 4 
 4
1
1
Qsin−1 x + cos −1 x = π 

2 
[JEE Main 2019, 12 Jan Shift-I]
π
π
(a)  − , 0, 
4
 4
π
π − 2  − cos −1 x
2

×
1− x
lim
=
x → 1−
π − 2 sin−1 x
×
1− x
π + 2 sin−1 x
π + 2 sin−1 x
[on rationalization]
1
π + 2 sin−1 x
lim
1
+
2 π θ→ 0
1
2 π
2 2 =

58. lim 
n → ∞ n
n
2
+1
2
+
equal to
Q1 − cos θ = 2 sin2 θ 
2 

lim


θ
Q x → 0+ sinθ = 1


2
π
n
n +2
2
2
+
n
n +3
2
2
+ ... +
1
 is
5n 
[JEE Main 2019, 12 Jan Shift-II]
(a) tan −1 ( 3)
(c) π / 4
(b) tan −1 ( 2 )
(d) π / 2
Exp. (b)
Clearly,
1
n
n
n
+ 2
+ 2
+ ...+ 
lim 
n→ ∞  n2 + 12
5n 
n + 22
n + 32
n
n
n
= lim  2
+ 2
+
n→ ∞  n + 12
n + 2 2 n2 + 32

n
+ ....+ 2

2
n + (2 n) 
2n
n
= lim ∑ 2
n → ∞ r =1 n + r 2
2n
2 dx
1
1
⋅ =∫
= lim ∑
2
0
n→ ∞
r =1
n
1 + x2
r
1 +  
 n
pn

1  r
f  =
Q nlim
∑
→∞
 n
n
r =1

−1
= [tan
x]20
−1
= tan 2
p

∫ 0 f( x)dx

135
Limits, Continuity and Differentiability
59. Let f be a differentiable function such that
60. The tangent to the curve y = x 2 − 5x + 5,
f (1) = 2 and f ′ ( x ) = f ( x ) for all x ∈R . If
h( x ) = f ( f ( x )), then h ′(1) is equal to
parallel to the line 2 y = 4x + 1, also passes
through the point
[JEE Main 2019, 12 Jan Shift-II]
[JEE Main 2019, 12 Jan Shift-II]
(a) 4e 2
(c) 2e
(b) 4e
(d) 2e 2
Exp. (b)
f ′ ( x) = f ( x)
f ′ ( x)
=1
f ( x)
f ′ ( x)
∫ f( x) dx = ∫ 1⋅ dx
Given that,
⇒
⇒
[by integrating both sides w.r.t. x]
⇒ Put f( x) = t ⇒ f ′( x)dx = dt
dt
∴
∫ t = ∫ 1 dx
Q dx = ln| x|+ C 
⇒
ln|t|= x + C
 ∫ x

…(i)
⇒
ln|f( x)|= x + C
[Qt = f( x)]
Q
f(1) = 2
So,
[using Eq. (i)]
ln (2 ) = 1 + C
⇒
C = ln2 − lne
[Qlne = 1]
2
A

[Qln A − ln B = ln  ]
C = ln  
⇒
 B
e
From Eq. (i), we get
⇒
⇒
⇒
⇒
2
ln|f( x)|= x + ln  
e
2
ln|f( x)|− ln  = x
e
ef( x)
A
[Qln A − ln B = ln ]
ln
= x
2
B
e
f( x) = e x [Qln a = b ⇒ a = e b, a > 0]
2
e
 e

Q f( x) = | f( x)|
|f( x)|= 2e x −1
 2

2
f( x) = 2e
Now,
⇒
⇒
x −1
or −2e
x −1
h( x) = f(f( x))
h′( x) = f ′(f( x)) ⋅ f ′( x)
[on differentiating both sides w.r.t. ‘x’]
h′(1) = f ′(f(1)) ⋅ f ′(1)
[Qf(1) = 2 (given)]
= f ′(2 ) ⋅ f ′(1)
= 2e 2 −1 ⋅ 2e1−1
[Qf ′( x) = 2e x −1 or −2e x −1]
= 4e
1 7
(a)  , 
 4 2
1
(c)  − , 7
 8 
7 1
(b)  , 
 2 4
1
(d)  , − 7
8

Exp. (d)
The given curve is y = x2 − 5 x + 5
…(i)
Now, slope of tangent at any point ( x, y)
on the curve is
dy
…(ii)
= 2x− 5
dx
[on differentiating Eq. (i) w.r.t. x]
QIt is given that tangent is parallel to line
2 y = 4x + 1
dy
So,
= 2 [Qslope of line 2 y = 4 x + 1 is 2]
dx
7
⇒ 2x− 5 = 2 ⇒ 2x = 7 ⇒ x =
2
7
On putting x = in Eq. (i), we get
2
49 35
69 35
1
y=
−
+5=
−
=−
4
2
4
2
4
Now, equation of tangent to the curve (i) at point
 7 ,− 1  and having slope 2, is


 2 4
7
1
1
y + = 2  x −  ⇒ y + = 2 x − 7

2
4
4
29
…(iii)
y = 2x−
⇒
4
On checking all the options, we get the point
 1 , − 7  satisfy the line (iii).


8

61. The equation of a tangent to the parabola,
x 2 = 8y , which makes an angle θ with the
positive direction of X -axis, is
[JEE Main 2019, 12 Jan Shift-II]
(a) y = x tan θ − 2 cot θ
(c) y = x tan θ + 2 cot θ
(b) x = y cot θ + 2 tan θ
(d) x = y cot θ − 2 tan θ
Exp. (b)
Given parabola is x2 = 8 y
…(i)
Now, slope of tangent at any point ( x, y) on the
parabola (i) is
136
JEE Main Chapterwise Mathematics
dy x
= = tanθ
dx 4
[Qtangent is making an angle θ with the positive
direction of X-axis]
So,
x = 4tanθ ⇒ 8 y = (4 tanθ)2
[on putting x = 4 tanθ in Eq. (i)]
⇒
y = 2 tan2 θ
Now, equation of required tangent is
y − 2 tan2 θ = tanθ ( x − 4 tanθ)
⇒
y = x tanθ − 2 tan2 θ
⇒
x = ycot θ + 2 tanθ
63. 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
[JEE Main 2018]
(a) φ (an empty set)
(b) {0}
(c) { π }
(d) {0, π }
Exp. (a)
We have,
f( x) = | x − π|(⋅ e| x| − 1)sin| x|
( x − π )(e − x − 1)sin x,
x< 0

x
f( x) = −( x − π )(e − 1)sin x, 0 ≤ x < π
 ( x − π )(e x − 1)sin x,
x≥ π

62. For each t ∈R, let [t ] be the greatest integer
less than or equal tot . Then,
1   2
15
+…+
+
lim x 
 x 
  x   x 
x→ 0 +
[JEE Main 2018]
(a) is equal to 0
(b) is equal to 15
(c) is equal to 120
(d) does not exist (in R)
Exp. (c)
Key idea Use property of greatest integer
function [ x] = x − { x}.
We have,
2
15 
 1
lim x   +   + …+   
 x  
x→ 0 +  
 x   x 
Clearly, lim f ′( x) = 0 = lim f ′( x)
and
x − { x}
1  1
− 
∴
x  x
n  n
Similarly,
− 
x  x
1
2
2
1
∴Given limit = lim x −   + −   + …
x→ 0 +  x
 x x  x
We know,
We check the differentiability at x = 0 and π.
We have,
( x − π )(e − x − 1)cos x + (e − x − 1) sin x

+ ( x − π ) sin xe − x (−1), x < 0


x
x
− [( x − π )(e − 1)cos x + (e − 1) sin x
f ′ ( x) = 
+ ( x − π ) sin xe x ],0 < x < π


x
x
( x − π )(e − 1)cos x + (e − 1) sin x

+ ( x − π ) sin xe x , x > π

[ x] =
 1 =
 x 
 n =
 x 
15 15 
−  
x  x 
= lim (1 + 2 + 3+ ...+15) − x
x→ 0 +
  1  2 
15 
   +   + ... +  
  x  x
 x 
= 120 − 0 = 120
n


Q 0 ≤   < 1, therefore


 x


 0 ≤ x n  < x ⇒ lim x n  = 0






x→ 0 +  x 
 x
x→ 0 −
x→ 0 +
x→ π −
x→ π +
lim f ′( x) = 0 = lim f ′( x)
∴ f is differentiable at x = 0 and x = π
Hence, f is differentiable for all x.
64. If the curves y 2 = 6x , 9x 2 + by 2 = 16 intersect
each other at right angles, then the value ofb is
[JEE Main 2018]
(a) 6
7
(b)
2
(c) 4
(d)
Exp. (d)
We have,
⇒
⇒
y2 = 6 x
dy
2y = 6
dx
dy 3
=
dx y
Slope of tangent at ( x1, y1 ) is m1 =
Also,
⇒
9 x2 + by2 = 16
dy
18 x + 2 by = 0
dx
3
y1
9
2
137
Limits, Continuity and Differentiability
dy −9 x
=
dx
by
−9 x1
Slope of tangent at ( x1, y1 ) is m2 =
by1
⇒
Since, these are intersection at right angle.
27 x1
=1
∴
m1m2 = − 1 ⇒
by12
27 x1
[Q y12 = 6 x1]
=1
⇒
6bx1
9
b=
⇒
2
1
and g ( x ) = x − ,
x
x
f (x )
, then the
x ∈R − { −1, 0, 1}. If h( x ) =
g (x )
local minimum value of h( x ) is
65. Let f ( x ) = x 2 +
1
2
[JEE Main 2018]
(b) −3
(d) 2 2
(a) 3
(c) −2 2
cot x − cos x
67. lim
( π − 2 x )3
1
x2
and g ( x) = x −
1
f ( x)
⇒ h( x) =
g ( x)
x
2
1

x + 2 x−  + 2

x
x =
∴ h( x) =
1
1
x−
x−
x
x
1
2

⇒ h( x) =  x −  +
1

x
x−
x
1
2
1

∈ [2 2 , ∞ )
x − > 0,  x −  +
1

x
x
x−
x
1
2
1

∈ (−∞, 2 2 ]
x − < 0,  x −  +
1

x
x
x−
x
∴Local minimum value is 2 2.
2
1
66. If 20 m of wire is available for fencing off a
flower-bed in the form of a circular sector,
then the maximum area (in sq m) of the
flower-bed is
[JEE Main 2017 (Offline)]
(b) 10
(c) 25
Exp. (c)
Total length = 2 r + rθ = 20
20 − 2r
θ=
r
θ
r
rθ
equals
[JEE Main 2017 (Offline)]
We have,
⇒
A = 10r − r 2
dA
= 10 − 2 r
∴
dr
dA
For maxima or minima, put
= 0.
dr
⇒
10 − 2 r = 0
⇒
r=5
20 − 2 (5)
1
∴
Amax = (5)2 


5
2
1
= × 25 × 2 = 25 sq m
2
1
24
1
(c)
8
1
16
1
(d)
4
(a)
f ( x) = x 2 +
r
⇒
x → π /2
Exp. (d)
(a) 12.5
Now, area of flower-bed,
1
A = r 2θ
2
20 − 2 r 
1
⇒
A = r 2 

2 
r 
(d) 30
(b)
Exp. (b)
lim
x→ π / 2
cot x − cos x
( π − 2 x)
3
1
= lim ⋅
h→ 0 8
=
=
lim
x→ π/2
1 cos x(1 − sin x)
⋅
3
8
π
sin x − x
2

π
π


cos − h 1 − sin − h
2
 
2
 
3
π π
π
sin − h  − + h

2
 2 2
sin h (1 − cos h)
1
lim
8 h → 0 cos h ⋅ h3
1
= lim
8h→ 0
1
= lim
4h→ 0
h
sin h 2 sin2 

2
cos h ⋅ h3
h
sin h ⋅ sin2  
2
h3 cos h
2
 sin h 
sin h   2 
1
1
1

= lim 
⋅
  h  ⋅
h
0
→


h
4

 cos h 4
 2 
1 1
1
= × =
4 4 16
138
JEE Main Chapterwise Mathematics
68. The normal to the curve
y ( x − 2 )( x − 3) = x + 6 at the point, where the
curve intersects theY -axis passes through
the point
[JEE Main 2017 (Offline)]
1
1
(a)  − , − 
 2
2
1 1

(c)  , − 
 2 3
1 1
(b)  , 
 2 2
1 1
(d)  , 
 2 3
∴
Given curve is
…(i)
y( x − 2 )( x − 3) = x + 6
Put x = 0 in Eq. (i), we get
y(− 2 ) (− 3) = 6
⇒
y=1
So, point of intersection is (0, 1).
x+ 6
Now,
y=
( x − 2 )( x − 3)
dy 1( x − 2 )( x − 3) − ( x + 6)( x − 3 + x − 2 )
⇒
=
dx
( x − 2 )2 ( x − 3)2
6 + 30 36
 dy 
=
=1
=
 
 dx  ( 0, 1)
36
4×9
69. The radius of a circle having minimum area,
which touches the curve y = 4 − x 2 and the
[JEE Main 2017 (Offline)]
(a) 2 ( 2 + 1)
(c) 4 ( 2 − 1)
(b) 2 ( 2 − 1)
(d) 4 ( 2 + 1)
Let the radius of circle with least area be r.
Then, coordinates of centre = (0, 4 − r ).
y=|x|
(0
,4
)
–r
r
X′
r
2
Y′
equal to
(a) 2
[JEE Main 2016 (Offline)]
(b) 1
(c)
1
2
(d)
1
4
Exp. (c)
1
Given, p = lim (1 + tan2
x→ 0+
tan 2
lim
=e
x→ 0+
∴ log p = log e 2 =
2x
x
=e
x) 2 x
(1∞ form)
1
 tan x 
lim


2 x→ 0+ 
x 
2
1
= e2
1
2
71. For x ∈R , f ( x ) = |log 2 − sin x | and
g ( x ) = f ( f ( x )), then
[JEE Main 2016 (Offline)]
(a) g is not differentiable at x = 0
(b) g ′(0) = cos(log 2 )
(c) g ′(0) = − cos(log 2 )
(d) g is differentiable at x = 0 and
g ′(0) = − sin (log 2 )
We have, f( x) = log 2 − sin x
and
g ( x) = f(f( x)), x ∈ R
Note that, for x → 0, log 2 > sin x
∴
f( x) = log 2 − sin x
⇒
g ( x) = log 2 − sin (f( x))
= log 2 − sin (log 2 − sin x)
Clearly, g ( x) is differentiable at x = 0 as sin x is
differentiable.
Now,
g ′( x) = − cos (log 2 − sin x) (− cos x)
⇒
g ′(0) = 1 ⋅ cos (log 2 )
= cos x ⋅ cos (log 2 − sin x)
X
O
x→ 0
Exp. (b)
Exp. (c)
Y
70. Let p = lim+ (1 + tan 2 x )1/2x , then log p is
1
∴ Equation of normal at (0, 1) is given by
−1
( x − 0) ⇒ x + y − 1 = 0
y − 1=
1
1 1
which passes through the point  ,  .
2 2
lines y = | x |, is
4
Q 4 < 0

1− 2
 1 − 2
4
r=
= 4 ( 2 − 1)
2 +1
r≠
But
Exp. (b)
⇒
Since, circle touches the line y = x in first
quadrant.
0 − (4 − r )
=r ⇒ r−4=± r 2
∴
2
4
4
or
r=
⇒
2 + 1 1− 2
y=4 – x
139
Limits, Continuity and Differentiability
 1 + sin x 
 π
 , x ∈ 0 ,  .
 2
 1 − sin x 
72. Consider f ( x ) = tan −1 
π
A normal to y = f ( x ) at x = also passes
6
through the point
[JEE Main 2016 (Offline)]
(a) (0, 0)
2π
π 
(b)  0,
 (c)  , 0
 3 
6 
π
(d)  , 0
4 
Exp. (b)
We have, f( x) = tan− 1
⇒
f( x) = tan− 1
1 + sin x
π
, x ∈  0, 
 2
1 − sin x
 cos x + sin x 



2
2
According to given information, we have
Perimeter of square + Perimeter of circle = 2 units
⇒
4 x + 2 πr = 2
1− 2x
...(i)
⇒
r=
π
Now, let A be the sum of the areas of the square
and the circle. Then,
A = x 2 + πr 2
= x2 + π
2
2
 cos x − sin x 



2
2
 cos x + sin x 


−1
2
2
= tan 
x
x
 cos − sin 

2
2
Q cos x > sin x for 0 < x <


2
2
2
 1 + tan x 


2
= tan− 1 
x
 1 − tan 

2
x
π
x  π

−1 
= tan
tan  +  = +

 4 2   4 2
π
1
1
f ′( x) = ⇒ f′   =
⇒
 6 2
2
π
Now, equation of normal at x = is given by
6
π
π



y− f   = −2 x− 


 6  
6
π
π


⇒
y−  = −2 x− 


3
6
π π
π
4π


=
=
Qf   = +

 6  4 12 12
2π
which passes through  0,
.

3 
Exp. (c)
π

4
(1 − 2 x) 2
π2
(1 − 2 x)2
⇒
A( x) = x2 +
π
dA
Now, for minimum value of A( x),
=0
dx
2 (1 − 2 x)
⇒
2x +
⋅ (− 2 ) = 0
π
2 − 4x
x=
⇒
π
⇒
πx + 4 x = 2
2
⇒
x=
π+ 4
...(ii)
Now, from Eq. (i), we get
2
1− 2 ⋅
π+ 4
r=
π
π+ 4−4
1
=
=
π(π + 4)
π+ 4
...(iii)
From Eqs. (ii) and (iii), we get
x = 2r
74. lim 
n→∞
(n + 1)(n + 2 )K 3n 


n 2n
1/n
is equal to
[JEE Main 2016 (Offline)]
(a)
π
3 
(c)
18
e4
9
(b)
27
e2
(d) 3 log 3 − 2
e2
Exp. (b)
1
73. A wire of length 2 units is cut into two parts
which are bent respectively to form a square
of side = x units and a circle of radius = r
units. If the sum of the areas of the square
and the circle so formed is minimum, then
[JEE Main 2016 (Offline)]
(a) 2 x = ( π + 4)r
(c) x = 2r
(b) ( 4 − π )x = πr
(d) 2x = r
(n + 1) ⋅ (n + 2 ) K (3n) n
Let l = lim 

n→ ∞ 

n2 n
1
(n + 1) ⋅(n + 2 ) ... (n + 2 n) n
= lim 

n→ ∞ 

n2 n
 n+
= lim  
n→ ∞  n
1
1  n + 2 
 n + 2n  n
 
 K

  n 
 n 
140
JEE Main Chapterwise Mathematics
Taking log on both sides, we get
1
n→ ∞ n

1 
2 
2 n  

log  1 + n   1 + n  ...  1 + n  



1
⇒ log l = lim
n→ ∞ n
1
2
2n 

log  1 +  + log  1 +  + ... + log  1 +





n
n
n  
log l = lim
1
n→ ∞ n
⇒ log l = lim
2
∫0
⇒ log l =
2n
r

∑ log  1 + n 
r =1
2

1 
 dx
x
∫0  1 − 1 +
⇒ log l = 2 ⋅ log 3 − [ x − log 1 + x ] 20
⇒ log l = 2 ⋅ log 3 − [2 − log 3]
⇒ log l = 3 ⋅ log 3 − 2
⇒ log l = log 27 − 2
⇒
l = elog 27 − 2
27
= 27 ⋅ e − 2 = 2
e
75. lim
x→ 0
(c) 2
(d)
1
2
Exp. (c)
We have,
(1 − cos 2 x)(3 + cos x)
2 sin2 x(3 + cos x)
lim
= lim
tan4 x
x→ 0
x→ 0
x tan 4 x
× 4x
x×
4x
2
(3 + cos x)
1
2 sin x
= lim
× lim
×
2
4x
tan
0
x→ 0
x
→
4
x
lim
x→ 0 4 x
=2 ×
=2
4
×1
4
(d) 4
Exp. (a)
Since, g ( x) is differentiable ⇒ g ( x) must be
k x + 1 , 0 ≤ x ≤ 3
g ( x) = 
 mx + 2 , 3 < x ≤ 5
At x = 3,
RHL = 3m + 2
and at x = 3, LHL = 2k
∴
…(i)
2 k = 3m + 2
k

, 0≤ x< 3

Also,
g ′ ( x) =  2 x + 1
 m
, 3< x≤ 5
k
and R{g ′(3)} = m
L {g ′(3)} =
∴
4
k
…(ii)
⇒
= m i.e. k = 4m
4
77. The normal to the curve x 2 + 2 xy − 3y 2 = 0
[JEE Main 2015]
(b) 3
(a) 2
10
(c)
3
On solving Eqs. (i) and (ii), we get
8
2
k = ,m=
5
5
⇒
k + m=2
(1 − cos 2 x )( 3 + cos x )
is equal to
x tan 4x
(a) 4
[JEE Main 2015]
16
(b)
5
∴


1
⋅ x dx
⇒ log l = log (1 + x) ⋅ x − ∫
1+ x

0
2
1
1
+
−
x
⇒ log l = [log (1 + x) ⋅ x]20 − ∫
dx
0
1+ x
2
 mx + 2 , 3 < x ≤ 5
differentiable, then the value of k + m is
continuous.
log (1 + x) dx
⇒ log l = 2 ⋅ log 3 −

76. If the function g ( x ) = k x + 1 , 0 ≤ x ≤ 3 is
Q lim sinθ = 1 
 θ →0

θ


 and lim tanθ = 1




θ→ 0 θ
at (1, 1)
[JEE Main 2015]
(a) does not meet the curve again
(b) meets the curve again in the second
quadrant
(c) meets the curve again in the third quadrant
(d) meets the curve again in the fourth quadrant
Exp. (d)
Given equation of curve is
x2 + 2 xy − 3 y2 = 0
On differentiating w.r.t. x, we get
⇒
2 x + 2 xy′ + 2 y − 6 yy′ = 0
x+ y
y′ =
3y − x
At x = 1, y = 1, y′ = 1
 dy 
i.e.
=1
 
 dx  ( 1, 1)
…(i)
141
Limits, Continuity and Differentiability
Equation of normal at (1, 1) is
1
y − 1 = − ( x − 1)
1
⇒
y − 1 = − ( x − 1)
…(ii)
⇒
x+ y=2
On solving Eqs. (i) and (ii) simultaneously, we
get
x2 + 2 x(2 − x) − 3(2 − x)2 = 0
⇒
x2 + 4 x − 2 x2 − 3(4 + x2 − 4 x) = 0
⇒
− x2 + 4 x − 12 − 3 x2 + 12 x = 0
⇒
−4 x2 + 16 x − 12 = 0
⇒
4 x2 − 16 x + 12 = 0
⇒
x2 − 4 x + 3 = 0
⇒
( x − 1)( x − 3) = 0
⇒
x = 1, 3
Now, when x = 1, then y = 1
and when x = 3, then y = − 1
∴
P = (1, 1) and Q = (3, − 1)
Hence, normal meets the curve again at (3, − 1) in
fourth quadrant.
Aliter
Given,
x2 + 2 xy − 3 y2 = 0
⇒
⇒
( x − y)( x + 3 y) = 0
x − y = 0 or x + 3 y = 0
Equation of normal at (1, 1) is
y − 1 = − 1( x − 1)
⇒
x + y− 2= 0
It intersects x + 3 y = 0 at (3, − 1)
and hence normal meet the curve in fourth
quadrant.
x+y=2
y=x
Central Idea Any function have extreme values
(maximum or minimum) at its critical points, where
f ′( x) = 0.
Since, the function have extreme values at x = 1
and x = 2.
∴
f ′( x) = 0 at x = 1 and x = 2
⇒
f′(1) = 0 and f′(2 ) = 0
Also it is given that
f ( x)
lim 1 + 2  = 3
x→ 0 

x 
f ( x)
f ( x)
⇒ 1 + lim 2 = 3 ⇒ lim 2 = 2
x→ 0 x
x→ 0 x
⇒ f( x) will be of the form
ax4 + bx3 + 2 x2
[Qf( x) is of four degree polynomial]
f( x) = ax4 + bx3 + 2 x2
Let
⇒
f ′( x) = 4ax3 + 3bx2 + 4 x
⇒
and
f ′(1) = 4a + 3b + 4 = 0
f ′(2 ) = 32 a + 12 b + 8 = 0
…(i)
⇒
8a + 3b + 2 = 0
On solving Eqs. (i) and (ii), we get
1
a = , b = −2
2
x4
f ( x) =
− 2 x3 + 2 x2
∴
2
⇒
f(2 ) = 8 − 16 + 8 = 0
79. lim
sin( π cos 2 x )
x2
x→ 0
π
(a)
2
(b) 1
is equal to
…(ii)
[JEE Main 2014]
(c) − π
(d) π
Y
Exp. (d)
x+3y=0
(1,1)
X′
lim
X
O
78. Let f ( x ) be a polynomial of degree four
having extreme values at x = 1 and x = 2. If
f ( x )

lim 1 + 2  = 3, then f (2 ) is equal to
x→ 0 
x 
[JEE Main 2015]
(c) 0
x2
x→ 0
Y′
(b) −4
x→ 0
sin( π cos 2 x)
= lim
(3, –1)
(a) −8
Exp. (c)
(d) 4
= lim
x→ 0
= lim
x→ 0
= lim
x→ 0
= π
sin π (1 − sin2 x)
x2
sin( π − π sin2 x)
x2
sin ( π sin2 x)
x2
sin( π sin2 x)
π sin x
2
 sin2
× ( π )  2
 x
x


Q lim sin x = 1

 x → 0 x
142
JEE Main Chapterwise Mathematics
80. If f and g are differentiable functions in
(0,1) satisfying f (0) = 2 = g (1), g (0) = 0 and
f (1) = 6, then for some c ∈]0,1[
(a) 2 f ′(c ) = g ′(c )
(c) f ′(c ) = g ′(c )
Exp. (d)
(1 − cos 2 x) (3 + cos x)
x
⋅
⋅
1
tan4 x
x2
2 sin2 x 3 + cos x
x
= lim
⋅
⋅
x→ 0
1
tan 4 x
x2
Let I = lim
x→ 0
(b) 2 f ′(c ) = 3g ′(c )
(d) f ′(c ) = 2 g ′(c )
2
sin x 
= 2 lim 
 ⋅ lim (3 + cos x)
x→ 0  x 
x→ 0
[JEE Main 2014]
Exp. (d)
⋅ lim
Here, f (0) = 2 = g (1), g (0) = 0 and f (1) = 6
Q f and g are differentiable in (1, 0).
Let
h( x) = f ( x) − 2 g ( x)
h(0) = f (0) − 2 g (0)
h(0) = 2 − 0
h(0) = 2
Now,
h(1) = f (1) − 2 g (1) = 6 − 2 (2 )
h(1) = 2, h(0) = h(1) = 2
Hence, using Rolle's theorem,
There exists c ∈]0,1[, such that
h′(c ) = 0
f ′(c ) − 2 g ′(c ) = 0, for some c ∈]0,1[
⇒
f ′(c ) = 2 g ′(c )
⇒
81. If x = −1 and x = 2 are extreme points of
f ( x ) = α log | x | + βx + x , then [JEE Main 2014]
2
1
2
1
(c) α = 2 , β = −
2
(a) α = −6, β =
(b) α = −6, β = −
(d) α = 2 , β =
1
2
1
2
Exp. (c)
Here, x = −1and x = 2 are extreme points of
f ( x) = α log| x|+ β x2 + x , then
α
f ′ ( x) = + 2 β x + 1
x
…(i)
∴ f ′(−1) = − α − 2 β + 1 = 0
[at extreme point, f ′( x) = 0]
α
and f ′(−2 ) = + 4β + 1 = 0
…(ii)
2
On solving Eqs. (i) and (ii), we get
1
α = 2, β = −
2
82.
(1 − cos 2 x )( 3 + cos x )
is equal to
lim
x→ 0
x tan 4x
[JEE Main 2013]
(a) −
1
4
(b)
1
2
(c) 1
(d) 2
x→ 0
Q
lim
θ→ 0
⇒
4x
4 tan 4 x
θ
sinθ
=1
= 1 and lim
θ → 0 tanθ
θ
1
I = 2 ⋅ (1)2 ⋅ (3 + cos 0° ) ⋅ (1)
4
1
1
= 2 ⋅ 1⋅ (3 + 1) ⋅ = 2 ⋅ 4 ⋅ = 2
4
4
83. The intercepts on X -axis made by tangents
x
to the curve, y = ∫ |t | dt , x ∈R , which are
0
parallel to the line y = 2 x , are equal to
(a) ± 1
(c) ± 3
(b) ± 2
(d) ± 4
[JEE Main 2013]
Exp. (a)
Given,
y=
x
∫0 |t| dt
dy
=| x|
dx
Since, tangent to the curve is parallel to the line
y = 2 x.
dy
⇒
=2
dx
∴
x=±2
∴
∴ Points,
y=
± 2
∫0
| t | dt = ± 2
∴ Equation of tangents are
y − 2 = 2 ( x − 2)
y + 2 = 2 ( x + 2)
For x-intercept, put y = 0, we get
0 − 2 = 2 ( x − 2)
0 + 2 = 2 ( x + 2)
⇒
x=±1
84. If y = sec (tan − 1 x ), then
dy
at x = 1 is equal to
dx
[JEE Main 2013]
1
(a)
2
(c) 1
1
(b)
2
(d) 2
143
Limits, Continuity and Differentiability
Exp. (a)
−1
Given, y = sec (tan
dr dV /dt
=
dt
4 πr 2
⇒
x)
…(i)
dr
dV
and
at t = 49 min, we require
dt
dt
the radius (r ) at that stage,
dV
= − 72 π m3 / min
dt
Now, to find
2
√
1+
x
x
θ
∴ Amount of volume lost in 49 min
= 72 π × 49 m3
1
Let tan− 1 x = θ ⇒ x = tan θ
∴
y = sec θ = 1 + x
∴ Final volume at the end of 49 min
= (4500 π − 3528 π ) m3
2
= 972 π m3
On differentiating w.r.t. x, we get
1
dy
=
⋅2 x
dx 2 1 + x2
At x = 1,
If r is the radius at the end of 49 min, then
4 3
πr = 972 π ⇒ r 3 = 729 ⇒ r = 9
3
∴ Radius of the balloon at the end of 49 min
= 9m
dy
1
=
dx
2
85. A spherical balloon is filled with 4500π cu m
of helium gas. If a leak in the balloon causes
the gas to escape at the rate of
72 π cu m/min, then the rate (in m/min) at
which the radius of the balloon decreases 49
min after the leakage began is [AIEEE 2012]
(a)
9
7
(b)
7
9
(c)
2
9
(d)
9
2
Exp. (c)
Given
(i) Volume (V = 4500 π m3 /min) of the helium
gas filled in a spherical balloon.
(ii) Due to a leak, the gas escapes the balloon at
the rate of 72 π m3 /min.
∴ Rate of decrease of volume of the balloon is
dV
= − 72 π m3 /min
dt
To find The rate of decrease of the radius of the
balloon 49 min after the leakage started.
dr
i.e.,
at t = 49 min
dt
[assuming that the leakage started at time t = 0]
Now, the balloon is spherical in shape, hence the
4
volume of the balloon is V = πr 3 ⋅
3
On differentiating both sides w.r.t. t, we get
dV 4  2 dr 
= π  3r × 
dt 
dt
3 
Hence, from Eq. (i), we get
dr d V /dt
=
dt
4 πr 2
dr
⇒  
 dt  t
=
= 49
(dV /dt )t
4 π(r 2 )t
= 49
= 49
=
72 π
4 π(92 )
=
2
m / min
9
86. Let a ,b ∈R be such that the function f given
by f ( x ) = log | x | + bx 2 + ax , x ≠ 0 has extreme
values at x = − 1 and x = 2.
Statement I f has local maximum at x = − 1
and at x = 2.
1
−1
Statement II a = andb =
2
4
[AIEEE 2012]
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Exp. (c)
Given
(i) A function f, such that
f( x) = log| x| + bx2 + ax, x ≠ 0
(ii) The function f has extrema at x = − 1and x = 2 ,
i.e., f ′(−1) = f ′(2 ) = 0 and f ′ ′(−1) ≠ 0 ≠ f ′ ′(2 ).
144
JEE Main Chapterwise Mathematics
Now, given function f is given by
f( x) = log| x| + bx2 + ax
1
f ′( x) = + 2 bx + a
⇒
x
−1
f ′ ′ ( x) = 2 + 2 b
⇒
x
Since, f has extrema at x = − 1and x = 2 .
Hence,
f ′(−1) = 0 = f ′(2 )
f′(−1) = 0
…(i)
⇒
a − 2b = 1
and
f′(2 ) = 0
−1
…(ii)
a + 4b =
⇒
2
On solving Eqs. (i) and (ii), we get
1
−1
and b =
a=
2
4
 x2 + 2 
−1 −1

= − 
⇒ f ′ ′ ( x) = 2 +
2 
2
x
 2x 
⇒
f′ ′(−1) < 0 and f′ ′(2 ) < 0
Hence, f has local maxima at both x = − 1and
x = 2.
Hence, Statement I is correct.
Also, while solving for Statement I, we found
the values of a and b, which justify that
Statement II is also correct.
However, Statement II does not explain
Statement I in any way.
87. If f : R → R is a function defined by
 2 x − 1
f ( x ) = [ x ]cos 
 π , where [ x ] denotes
 2 
the greatest integer function, then f is
(a) continuous for every real x
[AIEEE 2012]
(b) discontinuous only at x = 0
(c) discontinuous only at non-zero integral
values of x
(d) continuous only at x = 0
Exp. (a)
A function f : R → R defined by
1
f( x) = [ x] cos π  x −  , where [] denotes the

2
greatest integer function.
Given
To discuss The continuity of function f.
Now, cos x is continuous, ∀x ∈ R.
1
⇒cos π  x −  is also continuous, ∀x ∈ R.

2
Hence, the continuity of f depends upon the
continuity of [ x]. Now, [ x] is discontinuous, ∀x ∈ I.
So, we should check the conitnuity of f at x = n,
∀n ∈ I.
LHL at x = n is given by
1
f(n− ) = lim f( x) = lim [ x] cos π  x − 

x → n−
x → n−
2
(2 n − 1) π
= (n − 1) cos
=0
2
RHL at x = n is given by
f(n+ ) = lim f( x)
x→ n+
1
= lim [ x] cos π  x − 

x→ n+
2
(2 n − 1) π
= (n) cos
=0
2
Also, value of the function at x = n is
1
f(n) = [n] cos π  n − 

2
(2 n − 1) π
= (n) cos
=0
2
∴ f(n+ ) = f(n− ) = f(n)
Hence, f is continuous at x = n, ∀n ∈ I.
88. Consider the function,
f ( x ) = | x − 2 | + | x − 5 |, x ∈R .
Statement I f ′ ( 4) = 0
Statement II f is continuous in [2 , 5],
differentiable in (2 , 5) and f (2 ) = f (5).
(a) Statement I is false, Statement II is true
(b) Statement II is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
[AIEEE 2012]
Exp. (c)
Given A function f such that
f( x) = | x − 2 | + | x − 5|.
To discuss Continuity and differentiability of f in
interval (2, 5).
We know that,
x≥2
x − 2
| x − 2| = 
2 − x, x < 2
and
 x − 5,
| x − 5| = 
5 − x,
x≥ 5
x< 5
145
Limits, Continuity and Differentiability
2 − x,

⇒ | x − 2 | =  x − 2,
 x − 2,

5 − x,

and| x − 5| = 5 − x,
 x − 5,

x<2
2 ≤ x≤ 5
x> 5
x→ 2
RHL at x = 2,
x<2
2 ≤ x≤ 5
x> 5
lim
h→ 0
f( x) = | x − 2 | + | x − 5|
(2 − x) + (5 − x), x < 2

= ( x − 2 ) + (5 − x), 2 ≤ x ≤ 5
( x − 2 ) + ( x − 5), x > 5

7 − 2 x, x < 2

= 3
, 2 ≤ x≤ 5
2 x − 7, x > 5

Now, we can draw the graph of f very easily.
Y
y = 2x – 7
2
LHL at x = 2,
lim
h→ 0
2 |sin (2 − h − 2 )|
2 |sin (− h)|
= lim
h→ 0
(2 − h) − 2
−h
⇒
5
So, the limit does not exist.
90.
d 2x
dy 2
is equal to
[AIEEE 2011]
From the above graph, we can analyse all the
required things.
Exp. (c)
Statement I f′(4) = 0
It is obviously clear that f is constant around x = 4,
hence f′(4) = 0.
Hence, Statement I is correct.
Here,
Statement II
It can be clearly seen that
(i) f is continuous, ∀x ∈ [2 , 5].
(ii) f is differential, ∀x ∈ (2 , 5).
(iii) f(2 ) = f(5) = 3
Hence, Statement II is also correct but obviously
not a correct explanation of Statement I.
 1 − {cos 2 ( x − 2 )} 
 is equal to
x −2

89. lim 
x→ 2
(b) − 2
(a) 2
1
(c)
2
[AIEEE 2011]
(d) Does not exist
Exp. (d)
lim
x→ 2
1 − cos 2 ( x − 2 )
( x − 2)
= lim
x→ 2
2 sin2 ( x − 2 )
( x − 2)
2 sin h
=− 2
−h
lim
h→ 0
d2 y 

(c) − 
2
 dx 
X
2 sin h
= 2
h
lim
h→ 0
d2 y 
(a) −  2 
 dx 
y=3
O
2 |sin (2 + h − 2 )|
2 |sin h|
= lim
h→ 0
(2 + h) − 2
h
⇒
⇒
y = 7 – 2x
2 |sin ( x − 2 )|
( x − 2)
= lim
−1
 dy 
 
 dx 
 dy 
 
 dx 
−3
d2 y 

(b) 
2
 dx 
−3
d2 y 

(d) 
2
 dx 
dx  dy 
= 
dy  dx 
 dy 
 
 dx 
−2
−1
−1
On differentiating both sides w.r.t. y, we get
d2x
dy
= −  
 dx 
dy
−2
2
dy
= −  
 dx 
dy
= −  
 dx 
dy
= −  
 dx 
−2
−2
−2
⋅
d  dy 
 
dy  dx 
⋅
d
dy
⋅
d  dy  dx
⋅ ⋅
dx  dx  dy
⋅
d 2 y  dy 
dy
⋅   = −  
 dx 
dx2  dx 
 dy  ⋅ dx
 
 dx  dx
−1
−3
 d 2 y
⋅  2 
 dx 
91. The values ofp andq for which the function
 sin (p + 1) x + sin x
,x<0

x

f (x ) = 
q,
x =0

2
 x+x − x ,
x>0

x 3/2
is continuous for all x in R, are [AIEEE 2011]
146
JEE Main Chapterwise Mathematics
(a) p =
5
1
,q =
2
2
(b) p = −
1
3
(c) p = , q =
2
2
3
1
,q =
2
2
Exp. (c)
Here, f( x) =
1
3
(d) p = , q = −
2
2
5 π
t sin t dt , where x ∈  0,


2 
x
∫0
f ′( x) = { x sin x − 0}
Exp. (b)
∴

 sin ( p + 1) x + sin x
, x< 0

x

Here, f ( x) = 
, x=0
q

2
 x+ x − x
, x> 0
3


x2
Since, f( x) is continuous for x ∈ R.
h→ 0
h
3
h2
= lim
h→ 0
x cos x +
1
sin x
2 x
So, f( x) has local minimum at x = 2 π.
93. Let f : R → [0, ∞ ) be such that lim f ( x ) exists
x→ 5
1
1
=
h+ 1+ 1 2
and lim
1

1
| x − 5|
x→ 5
= 0. Then, lim f ( x ) is
x→ 5
equal to
[AIEEE 2011]
(a) 3
(b) 0
(c) 1
(d) 2
Exp. (a)
…(i)
LHL at x = 0,
sin ( p + 1) (− h) + sin (− h)
lim
h→ 0
−h
sin ( p + 1) h sin h
= lim
+
h→ 0
h
h
⇒ ( p + 1) + 1 = ( p + 2 )
f(0) = q
[ f ( x )] − 9
2
h h
 h+ 1−1
h+ 1+
= lim 
×
h→ 0 
h
h+ 1+
(h + 1) − 1
= lim
h → 0 h { h + 1 + 1}
= lim
x = π, 2 π
f ′′( π ) = − π < 0
So, f( x) has local maximum at x = π.
f ′′(2 π ) = π > 0
h { h + 1 − 1}
h→ 0
x sin x = 0 ⇒ sin x = 0
f ′′( x) =
RHL at x = 0,
h + h2 −
f ′ ( x) =
∴
So, f( x) is continuous at x = 0.
lim
…(i)
[using Newton-Leibnitz formula]
Given, lim f( x) exists and lim
x→ 5
⇒
x→ 5
lim [f( x)] − 9 = 0
2
x→ 5
2
 lim [f( x)] = 9


 x→ 5

⇒
∴
…(ii)
…(iii)
From Eqs. (i), (ii) and (iii), we get
1
=q = p+ 2
2
3
1
∴
p= − ,q =
2
2
x
5π 
 , define f ( x ) = ∫ 0 t sint dt .
 2
Then, f has
[AIEEE 2011]
92. For x ∈ 0,
(a) local minimum at π and 2 π
(b) local minimum at π and local maximum at 2 π
(c) local maximum at π and local minimum at 2 π
(d) local maximum at π and 2 π
[f( x)]2 − 9
=0
| x − 5|
lim f( x) = 3, − 3
x→ 5
But f : R → [0, ∞ )
∴
Range of f( x) ≥ 0 ⇒
lim f( x) = 3
x→ 5
94. Let f be a function defined by
 tan x
,x ≠0

f (x ) =  x
1 , x = 0
Statement I
Statement II
x = 0 is point of minima of f .
f ′ (0) = 0
[AIEEE 2011]
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is correct explanation of
Statement I
147
Limits, Continuity and Differentiability
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Exp. (c)
tan x
, x≠0

f ( x) =  x
 1 , x = 0
tan x
> 1, ∀ x ≠ 0
x
As,
∴
f(0 + h) > f(0)
and
f(0 − h) > f(0)
= lim
h→ 0
= lim
h→ 0
Statement II f1( x ) and f 2( x ) are
continuous on IR.
[AIEEE 2011]
= lim
h→ 0
tan h
−1
h
h
tan h − h
h2
sec 2 h − 1
[using L’ Hospital rule]
2h
tan2 h
2 h2
[Qtan θ = sec θ − 1]
⋅h
2
2
1
Q lim tanθ = 1 
⋅0 = 0
2
 θ → 0 θ

So, Statement II is true.
Hence, both statements are true but Statement II
is not the correct explanation of Statement I.
=
95. If function f ( x ) is differentiable at x = a ,
then lim
x→a
x 2 f (a ) − a 2 f ( x )
is equal to
x −a
[AIEEE 2011]
(a) 2a f (a ) + a 2 f ′(a )
(b) − a 2 f ′(a )
(c) a f (a ) − a 2 f ′(a )
(d) 2af (a ) − a 2 f ′(a )
x→ a
x 2 f ( a) − a 2 f ( x )
x−a
= lim
x→ a
2 xf(a) − a f ′( x)
1− 0
[using L’ Hospital rule]
= 2 af(a) − a f ′(a)
Exp. (d)
 x ⋅ sin  1  , x ≠ 0

 
Here, f( x) = 
 x
 0
, x=0
To check continuity at x = 0,
1
LHL = lim (− h) sin  −   = 0
h→ 0 
 h 
1
RHL = lim h sin    = 0
h→ 0 
 h 
f(0) = 0
So, f( x) is continuous at x = 0.
Hence, Statement I is correct.
sin  1  , x ≠ 0

f2 ( x) =   x 
 0
, x=0
1
Here, lim f2 ( x) = lim sin  
x→ 0
x→ 0
 x
97. If f : ( − 1 , 1) → R be a differentiable function
2
2
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is correct explanation of
Statement I.
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
which does not exist.
So, f2 ( x) is not continuous at x = 0.
Hence, Statement II is false.
Exp. (d)
Here, lim
functions f1( x ) = x , x ∈IR
 sin 1 , if x ≠ 0
as follows
and f 2( x ) = 
x
 0 , if x = 0
 f ( x ) ⋅ f 2( x ), if x ≠ 0
f (x ) =  1
0
, if x = 0

Statement I F ( x ) is continuous on IR.
At x = 0, f( x) attains minima.
f(h) − f(0)
= lim
∴ f ′(0) = lim
h→ 0
h→ 0
h
96. Define f ( x ) as the product of two real
Let
with
and
f ′(0) = 1.
f (0) = − 1
g ( x ) = [ f (2 f ( x ) + 2 )]2. Then, g ′ (0) is equal to
(a) 4
(c) 0
(b) – 4
(d) – 2
[AIEEE 2010]
148
JEE Main Chapterwise Mathematics
Exp. (b)
100. Let f : R → R be defined by
We have, f : (− 1, 1) → R
f ′(0) = 1
f (0) = −1,
g ( x) = [f (2 f ( x) + 2 )]2
⇒
g ′( x) = 2 [f (2 f ( x) + 2 )] × f ′(2 f ( x) +2 ) × 2 f ′( x)
⇒
g ′(0) = 2 [f {2 f (0) + 2}] × f ′{2 f (0) +2} × 2 f ′(0)
= 2 [f (0)] × f ′(0) × 2 f ′(0)
= 2 × (− 1) × 1 × 2 × 1 = − 4
98. Let f : R → R be a positive increasing
function with lim
x→ ∞
lim
x→ ∞
f (2 x )
is equal to
f (x )
(b) 0
1
(c) −
2
(d) −1
Exp. (d)
O
2
3
k+2
(d) 3
Since, f ( x) is a positive increasing function.
⇒
0 < f ( x) < f (2 x) < f (3 x)
0 < 1<
By Sandwich theorem,
f (2 x)
=1
lim
x → ∞ f ( x)
99. The equation of the tangent to the curve
4
x2
,i.e., parallel to the X -axis, is
(a) y = 0
(c) y = 2
Exp. (d)
We have, y = x +
(b) y =1
(d) y = 3
1
Q
∴
k+2≤1
k ≤ −1
101. Let y be an implicit function of x defined by
x 2x − 2 x x cot y − 1 = 0. Then, y ′(1) is equal to
f (2 x) f (3 x)
<
f ( x)
f ( x)
f (2 x)
f ( 3 x)
lim 1 ≤ lim
≤ lim
x→ ∞
x → ∞ f ( x)
x → ∞ f ( x)
⇒
{
–1
Exp. (a)
⇒
2x + 3
k – 2x
[AIEEE 2010]
(b)
3
2
y =x+
(a) 1
f ( 3x )
= 1. Then,
f (x )
(a) 1
(c)
k − 2 x , if x ≤ − 1
.
f (x ) = 
2 x + 3, if x > − 1
If f has a local minimum at x = −1 , then a
possible value of k is
[AIEEE 2010]
[AIEEE 2010]
4
x2
On differentiating w.r.t. x, we get
8
dy
= 1− 3
dx
x
Since, the tangent is parallel to X-axis, therefore
dy
= 0 ⇒ x3 = 8
dx
⇒
x = 2 and y = 3
(a) −1
(c) log 2
(b) 1
[AIEEE 2009]
(d) − log 2
Exp. (a)
x2 x − 2 x x cot y − 1 = 0
Now,
x=1
∴
1 − 2 cot y − 1 = 0
π
⇒
cot y = 0 ⇒ y =
2
On differentiating Eq. (i) w.r.t. x, we get
dy
2 x2 x (1 + log x) − 2 [ x x (− cosec 2 y)
dx
+ cot y x x (1 + log x)] = 0
π
At  1,  ,
 2


dy


+ 0 = 0
2 (1 + log 1) − 2  1 (− 1)  
 dx   1, π 




 2
dy
=0
2 + 2  
⇒
 dx   1, π 

∴
2
 dy 
 
 dx   1,

π

2
= −1
…(i)
149
Limits, Continuity and Differentiability
102. Given, P ( x ) = x 4 + ax 3 + bx 2 + cx + d such
103. Let f ( x ) = x | x |and g ( x ) = sin x
that x = 0 is the only real root of P ′ ( x ) = 0. If
P ( − 1) < P (1), then in the interval[ − 1, 1],
Statement I gof is differentiable at x = 0 and
its derivative is continuous at that point.
[AIEEE 2009]
Statement II gof is twice differentiable at
[AIEEE 2009]
x = 0.
(a) P( −1 ) is the minimum and P(1 ) is the
maximum of P
(b) P( −1 ) is not minimum but P(1 ) is the
maximum of P
(c) P( −1 ) is the minimum and P(1 ) is not the
maximum of P
(d) Neither P( −1 ) is the minimum nor P(1 ) is the
maximum of P
Exp. (b)
Given,
P( x) = x4 + a x3 + bx2 + cx + d
⇒
P′( x) = 4 x3 + 3 ax2 + 2 bx + c
Since, x = 0 is a solution for P′( x) = 0, then
c=0
∴
P( x) = x4 + ax3 + bx2 + d
Exp. (d)
f( x) = x| x| and g ( x) = sin x
…(i)
Also, we have P(− 1) < P(1)
⇒
1− a + b + d < 1+ a + b + d
⇒
a> 0
Since, P′( x) = 0, only when x = 0 and P( x) is
differentiable in (− 1, 1), we should have the
maximum and minimum at the points x = − 1, 0
and 1.
Also, we have P (− 1) < P(1)
∴ Maximum of P( x) = Max {P(0), P(1)}
and minimum of P ( x) = Min {P (− 1), P(0)}
In the interval [0, 1],
P′( x) = 4 x3 + 3ax2 + 2 bx
= x (4 x + 3ax + 2 b )
2
Since, P′( x) has only one root x = 0, then
4 x2 + 3 ax + 2 b = 0 has no real roots.
∴
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
(3a)2 − 32 b < 0
3 a2
<b
32
∴
b> 0
Thus, we have a > 0 and b > 0.
∴ P′( x) = 4 x3 + 4 ax2 + 2 bx > 0, ∀ x ∈ (0, 1)
⇒
Hence, P( x) is increasing in [0, 1].
∴ Maximum of P( x) = P(1)
Similarly, P( x) is decreasing in [−1 , 0].
Therefore, minimum P( x)does not occur at x = − 1.
− sin x2 , x < 0
gof ( x) = sin ( x| x|) = 
 sin x2 , x ≥ 0
2
− 2 x cos x , x < 0
∴ (gof )′( x) = 
2
 2 x cos x , x ≥ 0
Clearly, L ( gof )′ (0) = 0 = R ( gof )′(0)
So, gof is differentiable at x = 0 and also its
derivative is continuous at x = 0.
− 2 cos x2 + 4 x2 sin x2 , x < 0
Now, ( gof )′ ′( x) = 
2 cos x2 − 4 x2 sin x2 , x ≥ 0
∴
L ( gof )′ ′(0) = − 2 and R ( gof )′′(0) = 2
∴
L ( gof )′′(0) ≠ R ( gof )′′(0)
Hence, gof( x) is not twice differentiable at x = 0.
Therefore, Statement I is true, Statement II is false.

( x − 1) sin
1
, if x ≠ 1
x −1
0,
if x = 1
104. Let f ( x ) = 

Then, which one of the following is true?
[AIEEE 2008]
(a) f
(b) f
(c) f
(d) f
is differentiable at x =1 but not at x = 0
is neither differentiable at x = 0 nor at x =1
is differentiable at x = 0 and at x =1
is differentiable at x = 0 but not at x =1
Exp. (d)
f(1 − h) − f(1)
−h
1 
(1 − h − 1) ⋅ sin 
−0
 1 − h − 1
= lim
h→ 0
−h
Now, f ′(1− ) = lim
h→ 0
150
JEE Main Chapterwise Mathematics
1
1
= lim sin  −  = − lim sin
h→ 0
h→ 0
 h
h
f(1 + h) − f(1)
and f ′(1+ ) = lim
h→ 0
h
1 

(1 + h − 1) sin 
−0
 1 + h − 1
1
= lim
= lim sin
h→ 0
h→ 0
h
h
∴ f ′(1− ) ≠ f ′(1+ )
Exp. (b)
Let
f( x) = x3 − px + q
Then,
f ′ ( x) = 3 x 2 − p
Maxima
– p/3
Minima
Hence, 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   
 (h − 1) cos 


 h − 1  (h − 1)2  




1


+ sin 

 h − 1 

= lim 
h→ 0
−1
[using L’ Hospital rule]
= cos 1 − sin 1 and


1
(0 + h − 1) sin 
 − sin 1
 0 + h − 1
f′(0+ ) = lim
h→ 0
h
1   −1 

 1 
(h − 1) cos 
 + sin


 h − 1  (h − 1)2 
 h − 1
= lim
h→ 0
1
[using L’Hospital rule]
= cos 1 − sin 1
⇒
f ′(0− ) = f ′(0+ )
Hence, f is differentiable at x = 0.
105. If the cubic x 3 − px + q has three distinct
real roots, where p > 0 and q > 0. Then, which
one of the following holds?
[AIEEE 2008]
p
p
(a) The cubic has maxima at both
and −
3
3
p
(b) The cubic has minima at
and maxima at
3
p
−
3
p
and maxima at
(c) The cubic has minima at −
3
p
3
p
p
and −
(d) The cubic has minima at both
3
3
p/3
Put
f ′ ( x) = 0
⇒
x=
p
,−
3
p
3
Now, f ′ ′( x) = 6 x
∴ At x =
p
,
3
f ′ ′ ( x) = 6
and at
x=−
p
>0
3
p
,
3
[minima]
f ′ ′( x ) < 0
[maxima]
106. How many real solutions does the equation
x 7 + 14x 5 + 16x 3 + 30x − 560 = 0 have?
(a) 5
(c) 1
(b) 7
(d) 3
[AIEEE 2008]
Exp. (c)
Let f( x) = x7 + 14 x5 + 16 x3 + 30 x − 560
∴ f ′( x) = 7 x6 + 70 x4 + 48 x2 + 30 > 0, ∀ x ∈ R
So, f( x) is increasing.
Hence, f( x) = 0 has only one solution.
107. The normal to a curve at P ( x , y ) meets the
X -axis at G . If the distance of G from the
origin is twice the abscissa of P , then the
curve is a
[AIEEE 2007]
(a) ellipse
(c) circle
(b) parabola
(d) hyperbola
Exp. (d)
Let the equation of normal be
dx
( X − x)
Y − y=−
dy
It meets the X-axis at G. Therefore, coordinates of
dy
G are  x + y , 0 .

dx 
151
Limits, Continuity and Differentiability
π
π π
3π
π
< x+ <
⇒ −
< x<
2
4 2
4
4
Hence, option (b) is correct, which lies in the
above interval.
According to given condition,
dy
dy
= 2x ⇒ y
= x
x+ y
dx
dx
⇒
−
y dy = x dx
110. Let f : R → R be a function defined by
On integrating, we get
2
f ( x ) = min { x + 1,| x | + 1}. Then, which one of
the following is true?
[AIEEE 2007]
2
y
x
=
+C
2
2
⇒
(a)
(b)
(c)
(d)
x 2 − y2 = − 2 C
which shows a hyperbola.
108. A value of C for which the conclusion of
mean value theorem holds for the function
f ( x ) = log e x on the interval [1, 3] is
1
(b) loge 3
2
(d) loge 3 [AIEEE 2007]
(a) 2 log 3 e
(c) log 3 e
f ( x )≥1 for all x ∈ R
f ( x ) is not differentiable at x =1
f ( x ) is differentiable everywhere
f ( x ) is not differentiable at x = 0
Exp. (c)
f( x) = min { x + 1,| x| + 1}
f( x) = x + 1, ∀x ∈ R
y=– x+1
Exp. (a)
⇒
∴
1 loge 3 − loge 1
=
c
2
2
c=
= 2 log 3 e
loge 3
increasing function in
O
X
Y′
It is clear from the figure that f( x) is differentiable
everywhere.
109.The function f ( x ) = tan −1 (sin x + cos x )is an
π π
(a)  , 
 4 2
π
(c)  0, 
 2
X′

f ( b ) − f ( a) 
Q f ′(c ) = b − a 


f(3) − f(1)
3−1
y=1
(0, 1)
Using mean value theorem,
f ′(c ) =
y=x+1
Y
111. The function f : R/{0} → R given by
f (x ) =
[AIEEE 2007]
π π
(b)  − , 
 2 4
π π
(d)  − , 
 2 2
1
2
−
x e 2x − 1
can be made continuous at x = 0 by defining
[AIEEE 2007]
f (0) as
(a) 2
(b) –1
(c) 0
(d) 1
Exp. (d)
Exp. (b)
Since, f( x) = tan−1 (sin x + cos x)
On differentiating w.r.t. x, we get
1
f ′ ( x) =
(cos x − sin x)
1 + (sin x + cos x)2
π
π
2 cos x ⋅ cos − sin x ⋅ sin 
4
4

=
1 + (sin x + cos x)2
=
π
2 cos  x + 

4
1 + (sin x + cos x)2
For f( x) to be increasing,
1
e2 x − 1 − 2 x
2 
 = lim
Now, lim  − 2 x
x→ 0  x
e − 1 x → 0 x(e 2 x − 1)
= lim
2e2 x − 2
(e 2 x − 1) + 2 xe 2 x
[using L’Hospital rule]
4e 2 x
= lim
=1
x → 0 4 e 2 x + 4 xe 2 x
[using L’Hospital rule]
So, f( x) is continuous at x = 0, then
lim f( x) = f(0)
x→ 0
x→ 0
⇒
1 = f(0)
152
JEE Main Chapterwise Mathematics
112. The set of points, where f ( x ) =
differentiable, is
(a)
(b)
(c)
(d)
x
is
1 + |x |
π
2
π
(c)
4
(b)
Exp. (a)
x
1 + | x|
g ( x)
x
=
f ( x) =
h( x) 1 + | x|
∴
Now,
Hence, f( x) is differentiable on (− ∞, ∞ ).
x 2
113. The function f ( x ) = + has a local
2 x
minimum at
[AIEEE 2006]
(a) x = − 2
(c) x =1
y = x2 − 5 x + 6
dy
= 2x − 5
dx
dy
m1 =  
= 4 − 5 = −1
 dx  ( 2 , 0 )
Q
f ( x) =
It is clear that g ( x) = x and h( x) = 1 + | x| are
differentiable on (− ∞, ∞ ) and (− ∞, 0) ∪ (0, ∞ ),
respectively.
Thus, f( x) is differentiable on (− ∞, 0) ∪ (0, ∞ ).
Now, we have to check the differentiability at
x = 0.
x
−0
1 + | x|
f( x) − f(0)
∴ lim
= lim
x→ 0
x→ 0
x
x−0
1
= lim
=1
x → 0 1 + | x|
(b) x = 0
(d) x = 2
dy
m2 =  
 dx  ( 3, 0 )
and
= 6− 5=1
m1m2 = − 1 × 1 = − 1
Now,
Hence, angle between the tangents is
∴
x
+
2
1
f ′ ( x) = −
2
f ( x) =
3x 2 + 9x + 17
3x + 9x + 7
2
(a) 41
[AIEEE 2006]
is
(b) 1
(c)
Let f( x) =
2
x
2
3 x2 + 9 x + 17
3 x2 + 9 x + 7
= 1+
x2
Now,
⇒
and
x= ±2
4
f ′ ′ ( x) = 3
x
4 1
f′ ′(2 ) = = > 0
8 2
4
1
f′ ′(− 2 ) = − = − < 0
8
2
Hence, f( x) is minimum at x = 2.
17
7
(d)
1
4
Exp. (a)
10
7

2
3  x + 3 x + 

3
[by division algorithm]
= 1+
10
2

3
1
3   x +  +



2
12


3
⋅
2
10
So, the maximum value of f( x) = 1 +
1

3  
 12 
= 1 + 40 = 41
Hence, f( x) will be maximum at x = −
For maxima or minima, put f ′( x) = 0
1 2
−
= 0 ⇒ x2 = 4
∴
2 x2
⇒
π
.
2
115. If x is real, then maximum value of
Exp. (d)
Q
[AIEEE 2006]
π
6
π
(d)
3
(a)
Exp. (b)
Let
y = x 2 − 5x + 6 at the points (2, 0) and (3, 0) is
[AIEEE 2006]
( − ∞ , − 1) ∪ ( −1, ∞ )
(− ∞, ∞)
(0, ∞ )
( − ∞ , 0) ∪ (0, ∞ )
Since,
114. Angle between the tangents to the curve
[minima]
116. If x m y n = ( x + y )m + n , then
x+y
xy
x
(c)
y
(a)
[maxima]
dy
is equal to
dx
(b) xy
(d)
y
x
[AIEEE 2006]
153
Limits, Continuity and Differentiability
Exp. (d)
Given that, xm yn = ( x + y)m + n
Taking log on both sides, we get
m log x + n log y = (m + n) log ( x + y)
On differentiating w.r.t. x, we get
m n dy (m + n) 
dy 
+
=
1 + 
x
y dx ( x + y) 
dx 
dy  m + n n  m m + n
−  =
−
⇒

dx  x + y
y
x
x+ y
⇒
∴
dy  my + ny − nx − ny 


dx 
y( x + y)

mx + my − mx − nx
=
x( x + y)
dy y
=
dx x
117. A function is matched below against an
interval, where it is supposed to be
increasing. Which of the following pair is
incorrectly matched?
[AIEEE 2005]
Interval
(a) ( − ∞ , − 4)
1
(b)  − ∞ , 

3 
(c) [2 , ∞ )
Function
x 3 + 6x 2 + 6
(d) ( − ∞ , ∞ )
x 3 − 3x 2 + 3x + 3
3x 2 − 2 x + 1
2 x 3 − 3x 2 − 12 x + 6
Exp. (b)
On differentiating w.r.t. x, we get
f ′( x) = 3 x2 + 12 x = 3 x( x + 4)
For f( x) to be increasing,
f ′ ( x) > 0
+
–4
⇒
(b) Let
+
–
lim
f ( x) = 3 x 2 − 2 x + 1
∴
f ′ ( x) = 6 x − 2
⇒
1
f ′( x) > 0, ∀ x ∈  , ∞ 
3 
Since, this is wrong.
Hence, option (b) is the required answer.
f (1 + h ) = 5, then f ′(1) is equal to
(a) 6
(b) 5
(c) 4
(d) 3
[AIEEE 2005]
Exp. (b)
f(1 + h) − f(1)
h
f(1 + h)
f(1)
= lim
− lim
→
h→ 0
h
0
h
h
f(1 + h)
f(1)
Since, lim
must be finite
= 5, so lim
h→ 0
h→ 0 h
h
f(1)
can be finite only, if
as f′(1) exists and lim
h→ 0 h
f(1)
= 0.
f(1) = 0 and lim
h→ 0 h
f(1 + h)
∴
=5
f ′(1) = lim
h→ 0
h
f ′(1) = lim
h→ 0
119. Let f be differentiable for all x. If f (1) = − 2
and f ′ ( x ) ≥ 2 for x ∈[1, 6], then
(a) f (6) = 5
(c) f (6) < 8
[AIEEE 2005]
(b) f (6) < 5
(d) f (6) ≥ 8
Exp. (d)
f(1) = − 2 and f ′( x) ≥ 2
dy
≥ 2 ⇒ d y ≥ 2 dx
dx
Given that,
⇒
f( 6 )
∫ f(1)
dy ≥ 2
6
∫1
dx
⇒
[ y]ff(( 16)) ≥ 2 [ x]16
⇒
⇒
⇒
f(6) − f(1) ≥ 10
f(6) ≥ 10 + f(1)
f(6) ≥ 8
[Q f(1) = − 2 ]
Alternate Solution
0
x ∈ (−∞, − 4) ∪ (0, ∞ )
1
h→ 0h
⇒
f ( x) = x 3 + 6 x 2 + 6
(a) Let
118. If f ( x ) is differentiable at x = 1 and
⇒
⇒
f(6) − f(1)
≥2
6−1
f(6) − f(1) ≥ 10
f(6) + 2 ≥ 10 ⇒ f(6) ≥ 8
120. If f is a real-valued differentiable function
satisfying| f ( x ) − f ( y )| ≤ ( x − y )2 ; x , y ∈R and
[AIEEE 2005]
f (0) = 0, then f (1) is equal to
(a) 1
(c) 0
(b) 2
(d) –1
154
JEE Main Chapterwise Mathematics
Exp. (c)
122. If α and β are the distinct roots of
|f( x) − f( y)| ≤ ( x − y)2
Q
∴ lim
x→ y
ax 2 + bx + c = 0, then
|f( x) − f( y)|
≤ lim | x − y|
x→ y
| x − y|
⇒
lim
1 − cos (ax 2 + bx + c )
( x − α )2
x→ α
|f ′( y)| ≤ 0 ⇒ f ′( y) = 0
⇒
f( y) = Constant
⇒
f(1) = 0
⇒ f( y) = 0
[Q f(0) = 0, given]
[AIEEE 2005]
x = a (cos θ + θ sin θ ), y = a (sin θ − θ cos θ ) at
any point θ is such that
[AIEEE 2005]
Exp. (d)
Now, lim
= lim
dy
= tan θ
dx
( x − α )2
a
2 sin2  ( x − α )( x − β )
2

= lim
2
x→ α
 a  ( x − α )2 ( x − β )2
 
2
= lim
=
x→ α
2
a2
( x − β )2
2
So, equation of normal is
y − a sin θ + aθ cos θ
cos θ
=−
( x − a cos θ − a θ sin θ)
sin θ
⇒ ysin θ − asin2 θ + aθ cos θ sin θ
= − x cos θ + acos 2 θ + aθsin θ cos θ
⇒ x cos θ + y sin θ = a
So, it is always at a constant distance ‘a’ from
origin.
2
 a  ( x − β )2
 
2
Q lim sin x = 1
 x → 0 x

a
(α − β )2
2
123. A spherical iron ball 10 cm in radius is
coated with a layer of ice of uniform
thickness that melts at a rate of 50 cm 3/min.
When the thickness of ice is 15 cm, then the
rate at which the thickness of ice decreases,
is
[AIEEE 2005]
5
cm/min
6π
1
cm/min
(c)
18π
(a)
dx
= − cot θ
dy
( x − α )2
 ax2 + bx + c 

2 sin2 
2


x→ α
Exp. (a)
Given that, x = a(cos θ + θ sin θ)
and
y = a (sin θ − θ cos θ)
On differentiating w.r.t. θ respectively, we get
dx
= a(− sin θ + sin θ + θ cos θ)
dθ
dx
…(i)
= aθ cos θ
⇒
dθ
dy
and
= a(cos θ − cos θ + θ sin θ)
dθ
dy
…(ii)
= aθ sin θ
⇒
dθ
On dividing Eq. (ii) by Eq. (i), we get
1 − cos (ax2 + bx + c )
x→ α
(a) it is at a constant distance from the origin
aπ
(b) it passes through  , − a 
 2

π
(c) it makes angle − θ with the X-axis
2
(d) it passes through the origin
Since, slope of normal = −
a2
(b) −
(α − β )2
2
a2
(d)
(α − β )2
2
1
(a) (α − β )2
2
(c) 0
121. The normal to the curve
is equal to
1
cm/min
54π
1
(d)
cm/min
36π
(b)
Exp. (c)
dV
= 50 cm3 /min
dt
d  4 3
∴
 πr  = 50

dt  3
50 × 3
dr
dr
50
=
⇒
⇒
=
3r 2
dt
4π
dt 4 πr 2
1
dr
50
cm/min
=
=
⇒  
 dt  r = 15 4 π × 225 18π
Given that,
155
Limits, Continuity and Differentiability
124. A lizard, at an initial distance of 21 cm
behind an insect, moves from rest with an
acceleration of 2 cm/s 2 and pursues the
insect which is crawling uniformly along a
straight line at a speed of 20 cm/s. Then, the
lizard will catch the insect after [AIEEE 2005]
(a) 24 s
(b) 21 s
(c) 1 s
⇒ u 2 = ( a2 + b 2 ) + 2 ( a2 + b 2 ) x − x 2
On differentiating w.r.t. θ, we get
dx
du 2
2
( a 2 + b 2 − 2 x) ×
=
2
2
2
θ
d
dθ
2 (a + b )x − x
and
(d) 20 s
[Q x = a2 cos 2 θ + b 2 sin2 θ]
Exp. (b)
Let lizard catch the insect C.
And distance covered by insect = S
S
Time taken by insect, t =
…(i)
20
Distance covered by lizard = 21 + S
1
…(ii)
21 + S = (2 ) ⋅ t 2
∴
2
1 2
2
[QS = ut + at ; here u = 0, a = 2 cm/s and
2
S = 20 t]
[from Eq. (i)]
⇒
21 + 20t = t 2
⇒
t 2 − 20t − 21 = 0
⇒
t 2 − 21t + t − 21 = 0
⇒ t (t − 21) + 1( t − 21) = 0
⇒
( t + 1)( t − 21) = 0
⇒
t = −1, 21
∴
t = 21s [Qneglecting t = −1]
125. Ifu = a 2 cos 2 θ + b 2 sin 2 θ
+ a 2 sin 2 θ + b 2 cos 2 θ,
then the difference between the maximum
and minimum values ofu 2 is given by
(b) 2 a + b
(d) (a − b )2 [AIEEE 2004]
(a) 2 (a + b )
(c) (a + b )2
2
2
2
a2 cos 2 θ + b 2 sin2 θ
+
a2 sin2 θ + b 2 cos 2 θ
2
2
2
+ b cos θ + 2 (a cos θ + b sin θ)
2
2
2
2
2
2
× (a2 sin2 θ + b 2 cos 2 θ)
⇒ u = a + b + 2 x(a + b − x)
2
du 2
=0
dθ
a2 + b 2 = 2 (a2 cos 2 θ + b 2 sin2 θ)
For maxima and minima, put
∴
sin 2 θ = 0
and
⇒
cos 2 θ(b 2 − a2 ) = 0
sin 2 θ = 0
cos 2 θ = 0
θ=0
π
2θ =
⇒
2
and
θ=0
π
θ=
⇒
4
and
θ=0
So, u 2 will be minimum at θ = 0 and will be
π
maximum at θ = ⋅
4
2
∴
umin
= (a + b )2
and
⇒
and
2
umax
= 2 ( a2 + b 2 )
and
2
2
Hence, umax
− umin
= 2 (a2 + b 2 ) − (a + b )2
= (a − b )2
a b
126. If lim 1 + + 2 
x→ ∞

2
2
2
[where, x = a cos θ + b sin θ)
2
2
(a) a ∈ R , b ∈ R
(c) a ∈ R , b = 2
2x
= e 2 , then the values
x
[AIEEE 2004]
(b) a = 1, b ∈ R
(d) a = 1, b = 2
Exp. (b)
∴ u = a cos θ + b sin θ + a2 sin2 θ
2
2
( a 2 + b 2 − 2 x)
du 2
=
× (b 2 − a2 ) sin 2 θ
dθ
( a2 + b 2 ) x − x 2
x
of a and b are
Given that,
2
⇒
2
Exp. (d)
u=
dx
= (b 2 − a2 ) sin 2 θ
dθ
2
2
a
b
Now, lim  1 + + 2 
x→ ∞ 
x x 
2x


2x 
b 

2
 
a
= lim  1 + +
x→ ∞ 
x x
a
+
x
a
+
x
b 

x2 
b 

x2 
156
JEE Main Chapterwise Mathematics
=e
b 
a
lim 2 x  +

 x
x2 
Q lim (1 + x)1/ x = e 

 x → ∞
x→ ∞
= e 2a
a
b
lim  1 + + 2 
x→ ∞ 
x x 
But
⇒
2x
129. A point on the parabola y 2 = 18x at which
= e2
the ordinate increases at twice the rate of the
abscissa, is
[AIEEE 2004]
e 2a = e 2
⇒
and
(a) (2, 4)
9 9
(c)  − , 
 8 2
a=1
b∈R
1 − tan x
π
 π
, x ≠ , x ∈ 0,  ⋅ If f ( x )
4x − π
4
 2
π


 π
is continuous in 0,  , then f   is equal
 4
 2
to
127. Let f ( x ) =
(a) 1
(c) –1/2
(b) 1/2
(d) –1
[AIEEE 2004]
Exp. (c)
f ( x) =
1 − tan x
4x − π
lim
f ( x) =
lim
lim
 − sec 2 x 


4


Q
∴
x→ π/4
=
x→ π/4
x→ π/4
(a)
(c)
y + ... ∞
x
1+ x
(b)
1− x
x
(d)
dy
is equal to
dx
1
x
1+ x
x
Exp. (c)
Given that, x = e y
Equation of parabola is y2 = 18 x.
On differentiating w.r.t. t, we get
dy
dx
= 18
2y
dt
dt
Q dy = 2 dx , given
⇒
2 ⋅ 2 y = 18
 dt

dt
9
⇒
y=
2
From equation of parabola,
2
[using L’Hospital rule]
, x > 0, then
+ e y +... ∞
∴
x = ey + x
Taking log on both sides, we get
log x = ( y + x)
(b) (2, – 4)
9 9
(d)  , 
8 2
Exp. (d)
 9  = 18 x
 
2
81
= 18 x
4
9
x=
8
 1 − tan x 


 4x − π 
− sec 2 ( π / 4)
2
=
=−
4
4
1
lim f( x) = −
⇒
x→ π/4
2
Also, f( x) is continuous in [0, π/2 ], so f( x) will be
π
continuous at ⋅
4
π
1

∴
f   = lim f( x) = −
 4 x→ π/4
2
128. If x = e y + e
On differentiating w.r.t. x, we get
1 dy
dy 1 − x
=
=
+1 ⇒
dx
x
x dx
⇒
⇒
9 9
Hence, required point is  ,  .
 8 2
130. 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
[AIEEE 2004]
(a) ( x −1 )2
(b) ( x −1 )3
(c) ( x + 1 )
(d) ( x + 1 )2
3
Exp. (b)
[AIEEE 2004]
Since,
f ′ ′( x) = 6( x − 1)
On integrating, we get
f ′( x) = 3( x − 1)2 + C
…(i)
Also, at the point (2, 1), the tangent to graph is
y = 3x − 5
Slope of tangent = 3
⇒
f′(2 ) = 3
∴
f ′(2 ) = 3(2 − 1)2 + C = 3 [from Eq. (i)]
157
Limits, Continuity and Differentiability
⇒
∴
⇒
3+C =3
⇒
C=0
From Eq. (i),
f ′( x) = 3( x − 1)2
⇒
∴
On integrating, we get
f( x) = ( x − 1)3 + k
…(ii)
Since, graph passes through (2, 1).
∴
1 = (2 − 1)2 + k
⇒
k=0
Hence, equation of function is f( x) = ( x − 1)3 .
131. The normal to the curve x = a (1 + cos θ),
y = a sin θ at θ always passes through the
fixed point
[AIEEE 2004]
(a) (a , 0)
(b) (0, a )
(c) (0, 0)
2B = 0 ⇒ B = 0
f( x) = Ax2 + C
(d) (a , a )
f ′( x) = 2 Ax
f ′(a) = 2 Aa,
f ′(b ) = 2 Ab
and
f ′(c ) = 2 Ac
Also, a, b, c are in AP.
So, 2 Aa, 2 Ab and 2 Ac are in AP.
Hence, f ′(a), f ′(b ) and f ′(c ) are also in AP.
133. The real number x when added to its
inverse gives the minimum value of the
sum at x equal to
[AIEEE 2003]
(a) 2
(b) 1
f ( x) = x +
f ′ ( x) = 1 −
Given that,
x = a(1 + cos θ), y = a sin θ
On differentiating w.r.t. θ, we get
dx
dy
= a(− sin θ) and
= a cos θ
dθ
dθ
dy −cos θ
=
∴
dx
sin θ
−1
∴Slope of normal =
(− cos θ /sinθ)
Equation of normal at the given points is
sin θ
[ x − a(1 + cos θ)]
y − a sin θ =
cos θ
It is clear that in the given options, normal passes
through the point (a, 0).
132. Let f ( x )be a polynomial function of second
134. If f ( x ) = x n , then the value of
f (1) −
f ′ (1) f ′ ′ (1) f ′ ′ ′ (1)
+
−
+ ... +
1!
2!
3!
( −1)n f n (1)
is
n!
(a) 2n
(c) 0
(a)
(b)
(c)
(d)
Exp. (c)
A+ B+ C
A− B+ C
f(−1)
A− B+ C
⇒
⇒
[AIEEE 2003]
f(1) = 1
f ′(1) = n
f ′ ′( x) = n(n − 1)xn − 2 ⇒ f ′ ′(1) = n(n − 1)
… … … … … … …
… … … … … … …
f n ( x) = n(n − 1)(n − 2 ) . . . 2 ⋅ 1
f( x) = Ax2 + Bx + C
∴
f(1) =
and
f(−1) =
Q
f(1) =
⇒ A+ B+ C =
(b) 2n − 1
(d) 1
f ( x) = x n
f ′( x) = nxn − 1
Exp. (a)
Let
1
x
1
x2
For maxima and minima, put f ′( x) = 0
1
1− 2 = 0 ⇒ x = ± 1
⇒
x
2
Now,
f ′ ′ ( x) = 3
x
At
[minima]
x = 1, f ′ ′( x) = + ve
and at x = − 1, f ′ ′( x) = −ve
[maxima]
Thus, f( x) attains minimum value at x = 1.
degree. If f (1) = f ( −1) and a ,b , c are in AP,
then f ′ (a ), f ′ (b )and f ′ (c ) are in [AIEEE 2003]
AP
GP
HP
Arithmetic-Geometric progression
(d) –2
Exp. (b)
Let
Exp. (a)
(c) –1
[given]
⇒
f n (1) = n(n − 1)(n − 2 ) . . . 2 ⋅ 1
Now, f(1) −
(−1)n f n (1)
f ′(1) f ′ ′(1) f ′ ′ ′(1)
+
−
+ ...+
n!
1!
2!
3!
158
JEE Main Chapterwise Mathematics
= 1−
n n(n − 1) n(n − 1)(n − 2 )
+
−
1!
2!
3!
(−1)n n(n − 1)(n − 2 ) . . . 2 ⋅ 1
+. . . +
n!
n
n
[Q(1 − x) = 1 − C1 x + nC 2 x2
− nC 3 x3 + ... + (− 1)n nC n ]
= (1 − 1) = 0
Exp. (c)
log (3 + x) − log (3 − x)
=k
x
[using L’Hospital rule]
 1
1 
+


 3 + x 3 − x
1 1
lim
+ =k
=k ⇒
x→ 0
1
3 3
2
k=
3
Q lim
x→ 0
⇒
n

 x
1 − tan  2   (1 − sin x )

135. lim 
is equal to
π 
x
 
3
x→
2 1 + tan    ( π − 2 x )
 2 

[AIEEE 2003]
1
(a)
8
(b) 0
1
(c)
32
(d) ∞
Exp. (c)
137. Let f (a ) = g (a ) = k and their nth derivatives
f n (a ), g n (a )exist and are not equal for some
n.Further, if
f (a )g ( x ) − f (a ) − g (a ) f ( x ) + g (a )
lim
= 4,
x→a
g (x ) − f (x )
then the value of k is
 1 − tan x  (1 − sin x)



2
lim
π 
x
3
x→
2  1 + tan  ( π − 2 x)

2
π
π
Let
x = − h as x → , h → 0
2
2
π h
1 − tan  − 
 4 2  (1 − cos h)
⋅
∴ lim
h→ 0
π h
(2 h)3
1 + tan  − 
 4 2
h
2 sin2
 π  π h 
2
⋅
= lim tan
−
−
 4  4 2  
h→ 0
8 h3

π
 1 − tan x 
Q tan  4 − x = 1 + tan x 


2
h 
h
tan
 sin 
1
2 × 1
2 ×
= lim ⋅
h
h→ 0 4 h
4

×2 
 2 
2
2
 sin h 
 tan h 




1
2
2 = 1
lim 
=
 lim 
h
h
h
h
0
0
→
→
32
32




 2 
 2 
Q lim sinθ = 1 and lim tanθ = 1

θ→ 0 θ
 θ → 0 θ
log ( 3 + x ) − log ( 3 − x )
= k , then the
x→ 0
x
value of k is
[AIEEE 2003]
136. If lim
(a) 0
(c) 2/3
∴
(b) –1/3
(d) –2/3
(a) 4
(c) 1
[AIEEE 2003]
(b) 2
(d) 0
Exp. (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, we get
f(a) g ′( x) − g (a)f ′( x)
=4
lim
x→ a
g ′ ( x) − f ′ ( x)
kg ′( x) − kf ′( x)
=4
⇒
lim
x → a g ′ ( x) − f ′ ( x)
∴
k=4

138. If f ( x ) =  xe
 1
1
−
+ 
|x | x 


0,
, x ≠ 0, then f ( x ) is
x =0
[AIEEE 2003]
(a) continuous as well as differentiable for all x
(b) continuous for all x but not differentiable at
x =0
(c) Neither differentiable nor continuous at x = 0
(d) discontinuous everywhere
Exp. (b)
Continuity at x = 0,
LHL = lim f( x) = lim (0 − h) e
x → 0−
h→ 0
= lim (− h) e
1 1
−  − 
 h h
x→ 0+
= lim (− h) = 0
h→ 0
h→ 0
RHL = lim
1 
 1
−
+

 |− h | ( − h ) 
f( x) = lim (0 + h) e
h→ 0
1
 1
−
+ 
 |h | h 
159
Limits, Continuity and Differentiability
= lim he
1
1
−  + 
h
h
⇒
= lim
h→ 0
h→ 0
h
e
2/ h
=0
lim f( x) = lim f( x) = f(0)
x → 0−
x→ 0+
Therefore, f( x) is continuous for all x.
Differentiability at x = 0,
Lf ′(0) = lim
h→ 0
1 1
−  − 
 h h
(− h)e
−0
(− h) − 0
= lim e 0 = 1
Rf ′(0) = lim
h→ 0
1
1
−  + 
h
h
a=2
140. If y = ( x + 1 + x 2 )n , then
d 2y
(1 + x 2 )
dx 2
+x
dy
is equal to
dx
[AIEEE 2002]
(c) − y
(b) −n 2 y
(a) n 2 y
y = (x +
Q
−0
h−0
1
=0
e2/ h
⇒
Rf ′(0) ≠ Lf ′(0)
Therefore, f( x ) is not differentiable at x = 0.
= lim
h→ 0
139. If the function
∴
(d) 2 x 2 y
Exp. (a)
h→ 0
he
So, f( x) will be minimum at x = 2 a.
i.e.,
q = 2a
Given,
p2 = q ⇒ a2 = 2 a
1 + x2 )n
…(i)
On differentiating w.r.t. x, we get


2x
dy

= n( x + 1 + x2 )n − 1 ⋅  1 +


2
dx
+
x
2
1


n( x +
=
⇒
1 + x2 )n
1 + x2
2
dy
(1 + x2 )   = n2 y2
 dx 
[from Eq. (i)]
f ( x ) = 2 x − 9a x + 12 a x + 1, where a > 0,
attains its maximum and minimum at
p and q respectively, such that p 2 = q , then a
is equal to
[AIEEE 2003]
Again differentiating w.r.t. x, we get
(a) 3
(c) 2
⇒
3
2
2
(b) 1
(d) 1/2
Exp. (c)
Q
f( x) = 2 x3 − 9ax2 + 12 a2 x + 1
∴
f ′( x) = 6 x2 − 18ax + 12 a2
(1 + x2 ) ⋅ 2
141. lim
(1 + x2 )
1 − cos 2 x
x→ 0
2x
For maxima or minima, put f ′( x) = 0
(a) λ
(c) zero
∴
Exp. (d)
6( x2 − 3ax + 2 a2 ) = 0
⇒
x − 3ax + 2 a = 0
2
2
⇒
x2 − 2 ax − ax + 2 a2 = 0
⇒
x( x − 2 a) − a( x − 2 a) = 0
⇒
( x − a)( x − 2 a) = 0
⇒
Now,
At
x = a, x = 2 a
f ′ ′( x) = 12 x − 18 a
x=a
f ′ ′( x) = 12 a − 18 a = − 6 a
So, f( x) will be maximum at x = a.
i.e.,
p= a
Again, at x = 2 a
f ′ ′( x) = 24 a − 18 a = 6 a
2
dy
dy d 2 y
dy
⋅
+ 2 x   = n2 2 y
 dx 
dx
dx dx2
Now, lim
d2y
dx2
+ x
dy
= n2 y
dx
is equal to
[AIEEE 2002]
(b) –1
(d) Does not exist
1 − cos 2 x
x→ 0
2x
2 |sin x|
|sin x|
= lim
= lim
x→ 0
→
x
0
2x
x
|sin x|
Let
f ( x) =
x
sin h
|sin (0 − h)|
Now, LHL = lim
= lim
= −1
h → 0 −h
h→ 0
0−h
and RHL = lim
h→ 0
Q
sin h
|sin (0 + h)|
= lim
=1
→
h
0
h
0+ h
LHL ≠ RHL
|sin x|
does not exist.
x→ 0
x
Hence, lim
160
JEE Main Chapterwise Mathematics
x
Exp. (d)
 x 2 + 5x + 3
 is equal to
x → ∞  x2 + x + 2 
142. lim 
(a) e 4
(b) e 2
(c) e
[AIEEE 2002]
3
(d) e
Exp. (a)
 x2 + 5 x + 3 

Now, lim  2
x→ ∞  x + x + 2 

⇒
x

1
4x + 1 


= lim   1 + 2
x→ ∞ 
x + x + 2


=e
( 4 x + 1)
143. If sin y = x sin (a + y ),then
x
 x2 +
x
1 + 1  = e 




x

dy
is equal to
dx
[AIEEE 2002]
(a)
sin a
sin 2 (a + y )
(c) sin a sin 2 (a + y )
x=
⇒
(a)
[AIEEE 2002]
(b) touch each other
π
(d) cut at an angle
4
Exp. (a)
The equations of two curves are
x3 − 3 xy2 + 2 = 0
…(i)
On differentiating Eqs. (i) and (ii) w.r.t. x, we get
2
2
 dy  = x − y
 
 dx  C1
2 xy
(d)
…(ii)
 dy  = −2 xy
 
 dx  C 2
x 2 − y2
 x2 − y2   −2 xy 
dy
dy


Now,   ×   = 
 dx  C1  dx  C 2  2 xy   x2 − y2 
dy
is equal to
dx
[AIEEE 2002]
(b)
3 x 2 y − y3 − 2 = 0
and
dy sin2 (a + y)
=
sin a
dx
(c) not defined
(a) cut at right angle
π
(c) cut at an angle
3
sin 2 (a − y )
sin a
dx sin (a + y) cos y − sin y cos (a + y)
=
dy
sin2 (a + y)
sin a
dx
=
dy sin2 (a + y)
1+ x
1 + log x
3x 2y − y 3 − 2 = 0
(d)
sin y
sin (a + y)
144. If x y = e x − y , then
145. The two curves x 3 − 3xy 2 + 2 = 0 and
and
On differentiating w.r.t. y, we get
⇒
(1 + log x )2
sin 2 (a + y )
sin a
sin y = x sin (a + y)
⇒
(1 + log x )2
log x
1
x
(b)
Exp. (b)
Q
=
x+ 2



(1 + log x ) − x ⋅
dy
=
dx
( 4 x + 1) x
x2 + x + 2 

Q xlim
→ 0

= e4
y log x = ( x − y) loge e
x
y=
1 + log x
On differentiating w.r.t. x, we get

4x + 1 

= lim  1 + 2
x→ ∞ 
x + x + 2 
1

4 + 

x
lim
1
2
x→ ∞
1+ +
x
x2
Since,
xy = e x − y
Taking log on both sides, we get
= −1
Hence, the two curves cut at right angle.
146. The function f ( x ) = cot −1 x + x increases in
the interval
(a) (1, ∞ )
(c) ( − ∞ , ∞ )
[AIEEE 2002]
(b) ( −1, ∞ )
(d) (0, ∞ )
Exp. (c)
Since,
f( x) = cot −1 x + x
1 − log x
1 + log x
log x
On differentiating w.r.t. x, we get
1
x2
f ′ ( x) = −
+ 1=
≥0
2
1 + x2
1+ x
(1 + log x )2
Hence, f( x) is increasing function for all
x ∈ (− ∞, ∞ ).
161
Limits, Continuity and Differentiability
Exp. (c)
147. The greatest value of
f ( x ) = ( x + 1)1/3 − ( x − 1)1/3 on [0, 1] is
[AIEEE 2002]
(a) 1
(c) 3
(b) 2
(d) 1/3

1
 −5  

= lim  1 + 

x→ ∞
 x + 2


Exp. (b)
Given that,
f( x) = ( x + 1)1/ 3 − ( x − 1)1/ 3
=e
On differentiating w.r.t. x, we get

1
1
1
f ′ ( x) = 
−
2/ 3
2/ 3 
3  ( x + 1)
( x − 1) 
=
( x − 1)2 / 3 − ( x + 1)2 / 3
3( x2 − 1)2 / 3
( x − 1)2 / 3 = ( x + 1)2 / 3
5
1 + 2/


x
 −5 x 


x + 2






Q xlim
→ ∞

x
1 + 1  = e 




x

149. Let f (2 ) = 4 and f ′ (2 ) = 4. Then,
x→ 2
x f (2 ) − 2 f ( x )
is given by
x −2
(a) 2
Now, put f ′( x) = 0, then
(b) –2
(c) – 4
[AIEEE 2002]
(d) 3
Exp. (c)
x=0
xf(2 ) − 2 f( x)
x−2
xf(2 ) − 2 f(2 ) + 2 f(2 ) − 2 f( x)
= lim
x→ 2
x−2
f(2 )( x − 2 ) − 2 {f( x) − f(2 )}
= lim
x→ 2
x−2
Now, lim
x→ 2
At x = 0
f( x) = (0 + 1)1/ 3 − (0 − 1)1/ 3 = 2
Hence, the greatest value of f( x) is 2.
x
 x − 3
 is equal to
 x + 2
148. For x ∈R , lim 
x→ ∞
(a) e
(c) e −5

lim  −
x→ ∞ 
 −5 


 x + 2
x
= e −5
lim
Clearly, f ′( x) does not exist at x = ± 1.
⇒
x
 x − 3

5 
Now, lim 
 = lim 1 −
x→ ∞  x + 2
x→ ∞
2 
+
x

−1
(b) e
(d) e 5
[AIEEE 2002]
= f(2 ) − 2 lim
x→ 2
f( x) − f(2 ) 
f ( x ) − f ( a) 
Q f ′( x) = xlim
→ a
x−2
x − a 

= f(2 ) − 2 f ′(2 )
= 4−2 × 4= −4
9
Integral Calculus
5x
∫ 2x dx is equal to
sin
2
(where,C is a constant of integration )
sin
1.
[JEE Main 2019, 8 April Shift-I]
(a)
(b)
(c)
(d)
2 x + sin x + 2 sin 2 x + C
x + 2 sin x + 2 sin 2 x + C
x + 2 sin x + sin 2 x + C
2 x + sin x + sin 2 x + C
A = {( x , y ) ∈R × R |0 ≤ x ≤ 3,
0 ≤ y ≤ 4, y ≤ x 2 + 3x } is
[JEE Main 2019, 8 April Shift-I]
53
(a)
6
59
(c)
6
(b) 8
(d)
x
5x
5x
2 sin cos
2 dx =
2
2 dx
∫ x
∫
x
x
sin
2 sin cos
2
2
2
x
[multiplying by 2 cos in numerator and
2
denominator]
sin 3 x + sin2 x
dx
=∫
sin x
[Q 2sin A cos B = sin( A + B) + sin( A − B) and
sin 2 A = 2 sin A cos A]
(3sin x − 4sin3 x) + 2 sin xcos x
dx
=∫
sin x
[Qsin 3 x = 3sin x − 4 sin3 x]
sin
= ∫ (3 − 4sin2 x + 2 cos x) dx
Given,
y ≤ x2 + 3 x
⇒
3
9
y ≤  x +  −

2
4
2
2
 x + 3 ≥  y + 9






2
4
⇒
Since,
0≤ y≤ 4
and
0≤ x≤ 3
∴The diagram for the given inequalities is
Y
y=x2+3x
y=4
–3/2
O
9
–
4
(–3, 0)
= ∫ [3 − 2(1 − cos 2 x) + 2 cos x] dx
[Q 2 sin x = 1 − cos 2 x]
2
= ∫ [3 − 2 + 2 cos 2 x + 2 cos x] dx
= ∫ [1 + 2 cos 2 x + 2 cos x] dx
= x + 2 sin x + sin 2 x + C
26
3
Exp. (c)
Exp. (c)
Let I =
2. The area (in sq units) of the region
1
X
3
x=3
and points of intersection of curves y = x2 + 3 x
and y = 4 are (1, 4) and (−4, 4)
1
Now required area = ∫ ( x2 + 3 x)dx +
0
3
∫ 4 dx
1
163
Integral Calculus
1
 x3
1 3
3 x2 
3
=
+
 + [4 x]1 = + + 4(3 − 1)
3 2
2 0
3
2+ 9
11
59
sq units
=
+ 8=
+ 8=
6
6
6
at x = 1
Now,
 tan−1(1)

a 
 1+ 1 
π
aπ
π
(given)
= a 4 =
=
8
32
2
1
1
a = ⇒a =
4
16
a y (1) =
3. Let y = y ( x )be the solution of the differential
dy
+ 2 x ( x 2 + 1)y = 1 such
dx
π
that y(0) = 0. If a y(1) = , then the value of
32
‘a’ is
[JEE Main 2019, 8 April Shift-I]
∴
equation, ( x 2 + 1)2
1
(a)
4
1
(b)
2
1
(d)
16
(c) 1
4. If f ( x ) =
then the value of the integral ∫
Given differential equation is
dy
+ 2 x( x2 + 1)y = 1
( x2 + 1)2
dx
dy  2 x 
1
y=
⇒
+ 
dx  1 + x2 
(1 + x2 )2
is
= eln( 1 +
x2 )
Let
∫ Q (IF) dx + C
1
⇒
y(1 + x2 ) =
∫1+
⇒
y(1 + x2 ) = tan−1( x) + C
Q
∴
∴
y(0) = 0
C=0
y(1 + x2 ) = tan−1 x
⇒
y=
⇒
dx
x2
π /4
∫− π / 4 g(f( x))dx
π /4
∫− π / 4 loge
 2 − xcos x 

 dx
 2 + xcos x 
…(i)
b
b
a
a
∫ f( x) dx = ∫ f(a + b − x) dx, we get
= (1 + x2 )
∫ (1 + x2 )2 (1 + x
2 − xcos x
2 + xcos x
Now, by using the property
x2
y(1 + x2 ) =
I=
Then, I =
dx
⇒
(b) loge e
(d) loge 1
g ( x) = loge x, x > 0 and f( x) =
and required solution of differential Eq. (i) is given
by
y ⋅ (IF) =
[JEE Main 2019, 8 April Shift-I]
The given functions are
This is a linear differential equation of the form
dy
+ P⋅ y = Q
dx
2x
1
Here, P =
and Q =
2
(1 + x )
(1 + x2 )2
∴ Integrating Factor (IF) = e
g ( f ( x ))dx
Exp. (d)
[dividing each term by(1 + x2 )2 ] …(i)
2x
π /4
− π /4
(a) loge 3
(c) loge 2
Exp. (d)
∫1+
2 − x cos x
and g ( x ) = log e x , ( x > 0)
2 + x cos x
2
) dx + C
+C
I=
π /4
∫− π / 4 loge
2I =
⇒
tan−1 x
1+ x
 tan−1 x 

ay = a 
2 
 1+ x 
π /4
[Qloge A + loge B = loge AB]
∫− π / 4 loge
(1)dx = 0
I = 0 = loge (1)
x
5. Let f ( x ) = ∫ g (t )dt , where g is a non-zero
0
2
[multiplying both sides by a]
…(ii)
On adding Eqs. (i) and (ii), we get
π /4 
 2 − xcos x 
 2 + cos x  
2I = ∫
loge 
 + loge 
 dx
−π / 4 
+
x
x
2
cos


 2 xcos x  

π /4
 2 − xcos x 2 + xcos x 
×
=∫
loge 
 dx
−π / 4
 2 + xcos x 2 − xcos x 
⇒
[QC = 0]
 2 + xcos x 

 dx
 2 − xcos x 
even
x
function.
∫ f (t )dt equals
0
If
f ( x + 5) = g ( x ),
then
[JEE Main 2019, 8 April Shift-II]
164
JEE Main Chapterwise Mathematics
x+5
5
∫ g (t )dt
(a) 5
(b)
x+5
5
x+5
(c) −
5
∫ g (t )dt
(c) 2
(a) −
∫ g (t )dt
(d)
∫ g (t )dt
(b) −
Let I =
Exp. (d)
x
Given, f( x) = ∫ g (t ) dt
∫
=
∫
On replacing x by (− x), we get
−x
∫ g(t )dt
0
Now, put t = − u, so
x
x
0
0
x2
dx
∫ x3 (1 +
=
0
f ( − x) =
(d)
2x 2
1
2x 3
3
Exp. (b)
x+5
5
1
6x 3
1
f(− x) = − ∫ g (−u )du = − ∫ g (u )du = − f( x)
[Qg is an even function]
⇒ f(− x) = − f( x) ⇒ f is an odd function.
Now, it is given that f( x + 5) = g ( x)
∴f(5 − x) = g (− x) = g ( x) = f( x + 5)
[Qg is an even function]
…(i)
⇒
f(5 − x) = f( x + 5)
x
x6 )2 / 3
dx
1
x3 ⋅ x4  6 + 1
x

dx
2/3
2/3
1
x7  6 + 1
x

1
Now, put
+ 1= t3
x6
6
⇒
− 7 dx = 3t 2dt
x
dx
t2
= − dt
⇒
7
2
x
1 2
− t dt
1
So, I = ∫ 2 2
= − ∫ dt
2
t
1 1
1
= − t + C = −  6 + 1

2x
2
Let I = ∫ f(t )dt
0
1/ 3
+C
Qt 3 = 1 + 1


x6
Put t = u + 5 ⇒ t − 5 = u ⇒ dt = du
∴
x−5
x−5
−5
−5
∫ f (u + 5)du = ∫ g(u )du
I=
1 1
(1 + x6 )1/ 3 + C
2 x2
= x ⋅ f( x) ⋅ (1 + x6 )1/ 3 + C
=−
Put u = − t ⇒du = − dt , we get
5− x
I=−
5
∫ g(−t )dt = ∫ g(t )dt
5− x
5
b
a
a
b
[Q− ∫ f( x)dx = ∫ f( x)dx and g is an even function]
5
∫ f ′(t )dt
I=
[by Leibnitz rule f ′( x) = g ( x)]
5− x
= f(5) − f(5 − x) = f(5) − f(5 + x)
5
5
5+ x
5+ x
6. If ∫
dx
x (1 + x 6 )2/3
3
= xf ( x )(1 +
7. Let S (α ) = {( x , y ): y 2 ≤ x , 0 ≤ x ≤ α} and A(α ) is
area of the region S(α ). If for λ, 0 < λ < 4,
A ( λ ): A ( 4) = 2 : 5, then λ equals
[JEE Main 2019, 8 April Shift-II]
[from Eq. (i)]
1
3
4
(a) 2  
 25 
∫ f ′(t )dt = ∫ g(t )dt
=
1
1
x 6 )3
+C
where, C is a constant of integration, then
the function f ( x ) is equal to
[JEE Main 2019, 8 April Shift-II]
[given]
On comparing both sides, we get
1
f ( x) = − 3
2x
4 3
(c) 4 
 25 
1
2 3
(b) 4 
 5
1
2 3
(d) 2  
 5
Exp. (c)
Given,
S(α ) = {( x, y) : y2 ≤ x, 0 ≤ x ≤ α} and
165
Integral Calculus
A(α ) is area of the region S(α )
=
Y
y2 = x


1
π/2
1
=  x + cos 2 x

 0
4
1
π
=  − 0 + (−1 − 1)
2
 4
π 1
= −
2 2
π 1 π −1
I= − =
4 4
4
X
O
π
∫02  1 − 2 sin2 x dx
A(λ)
⇒
x=λ
9. The area (in sq units) of the region
λ
λ
 x3 / 2 
4 3/ 2
Clearly, A(λ ) = 2 ∫ xdx = 2 
 = λ
/
3
2
0 3

0
A(λ ) 2
Since,
= , (0 < λ < 4)
A(4) 5
λ3 / 2
⇒
43 / 2
λ  4
= 
4  25 
⇒
=
3
2
λ
2
⇒   =  
 5
 4
5
1/ 3
4
⇒ λ = 4 
 25 
π /2
8. The value of ∫0
A = {( x , y ): x 2 ≤ y ≤ x + 2} is
[JEE Main 2019, 9 April Shift-I]
13
(a)
6
31
(c)
6
2
1/ 3
9
2
10
(d)
3
(b)
Exp. (b)
Given region is A = {( x, y) : x2 ≤ y ≤ x + 2}
sin 3 x
dx is
sin x + cos x
Now, the region is shown in the following graph
Y
[JEE Main 2019, 9 April Shift-I]
π −1
2
π −1
(c)
4
π−2
8
π−2
(d)
4
(b)
(a)
B(2,4)
A
Exp. (c)
X'
(–2,0)
–1
b
sin3 x
dx
sin x + cos x
On applying the property,
∫
b
…(i)
b
∫a f( x)dx = ∫a f(a + b − x) dx, we get
cos 3 x
dx
cos x + sin x
On adding integrals (i) and (ii), we get
π / 2 sin3 x + cos 3 x
dx
2I = ∫
0
sin x + cos x
I=
π
=
∫02
=
∫02
π
π /2
∫0
…(ii)
(sin x + cos x) (sin2 x + cos 2 x − sin xcos x)
dx
sin x + cos x
1 − 1 (2 sin xcos x) dx
 2

2
X'
Y'
b
∫a f( x) dx = ∫a f(a + b − x) dx
π
2
0
(0,2)
O
Key Idea Use property of definite integral.
Let I =
y=x+2
x2=y
For intersecting points A and B
Taking, x2 = x + 2 ⇒ x2 − x − 2 = 0
⇒
x2 − 2 x + x − 2 = 0
⇒
x( x − 2 ) + 1( x − 2 ) = 0
⇒
x = −1,2 ⇒ y = 1, 4
So, A(−1, 1) and B (2, 4).
2
Now, shaded area = ∫ [( x + 2 ) − x2 ] dx
−1
2
x
x 
1
8
1
4
=  + 2 x −  =  + 4 −  −  − 2 + 



3
3
2
2
2
3
 −1

1 9
= 8− −
2 3
1 9
1
= 8 − − 3 = 5 − = sq units
2 2
2
2
3
166
JEE Main Chapterwise Mathematics
10. The integral ∫ sec 2/3 x cosec4/3x dx is equal to
Graphical representation of A is
Y
(hereC is a constant of integration)
[JEE Main 2019, 9 April Shift-I]
Q
y2
=x
2
(a) 3 tan −1/ 3 x + C
(b) −3 tan −1/ 3 x + C
−1/ 3
(c) −3cot
X'
x+C
x=
Exp. (b)
Let I =
∫
4
x cosec 3 x dx =
∫
2
4
⇒
dx
4
4
∫
4
[dividing and multiplying by cos 4 / 3 x
in denominator]
sec 2 xdx
dx
=∫
4
tan 3 xcos 2 x
(tan x)3
Now, put tan x = t ⇒sec x dx = dt
2
−4
∴I=
dt
∫ t 4/ 3
1
= −3
t
1
3
+1
−
−3
1
(tan x)3
+ C = −3 tan
1
3
x+C
11. The area (in sq units) of the region


y2
A = ( x , y ): ≤ x ≤ y + 4 is
2


x2 − 10 x + 16 = 0
⇒
( x − 2 )( x − 8) = 0
⇒
x = 2, 8
[from Eq. (ii)]
∴
y = − 2, 4
So, the point of intersection of Eqs. (i) and (ii) are
P(2, − 2 ) and Q(8, 4).
Now, the area enclosed by the region A
4
 y2

y3 
y2 
∫ ( y + 4) − 2 dy =  2 + 4 y − 6 
−2
−2
8
64   4
16

=
+ 16 −
 − −8+ 
2
6
6  2
4
32
= 8 + 16 −
−2 + 8−
3
3
= 30 − 12
= 18 sq units
=
t 3
+C
=
−4
+1
3
+C =
x2 − 8 x + 16 = 2 x
⇒
2
 sin x  3

 cos 3 xcos 3 x
 cos x 
=
4
12. If f : R → R is a differentiable function and
f (2 ) = 6, then
[JEE Main 2019, 9 April Shift-II]
53
(b)
3
(d) 18
(a) 30
(c) 16
⇒
and
∫
6
2t dt
is
(x − 2 )
(b) 0
(d) 2 f ′( 2 )
Key Idea
(i) First use L’ Hopital rule
(ii) Now, use formula
2
x = y + 4⇒ y = x − 4
x→2
Exp. (a)


y2
Given region A = ( x, y) :
≤ x ≤ y + 4
2


y
= x
2
y2 = 2 x
f(x)
lim
[JEE Main 2019, 9 April Shift-II]
(a) 12 f ′( 2 )
(c) 24 f ′( 2 )
Exp. (d)
∴
Y'
On substituting y = x − 4 from Eq. (ii) to Eq. (i), we
get
( x − 4)2 = 2 x
dx
cos 3 xsin 3 x
∫
P
y+
4
3
(d) − tan −4 / 3 x + C
4
2
sec 3
X
O
…(i)
…(ii)
φ ( x)
d 2
f (t )dt = f[φ2 ( x)]⋅ φ′2 ( x) − f[φ1( x)]⋅ φ′1 ( x)
dx φ ∫( x )
1
167
Integral Calculus
f( x )
Let l = lim
x→ 2
f( x )
∫
6
∫ 2tdt
2tdt
= lim 6
( x − 2 ) x→ 2 ( x − 2 )
14. The value of the integral
1
∫ x cot
 0 form, as f(2 ) = 6
 0

On applying the L’ Hopital rule, we get
2 f( x)f ′( x)
l = lim
x→ 2
1
 d φ 2( x )
Q
∫ f (t )dt = f(φ 2 ( x))⋅ φ2 ′( x)
 dx φ ( x )

1
− f(φ1( x)) ⋅ φ1 ′( x)]
[Qf(2 ) = 6]
l = 2 f(2 ) ⋅ f ′(2 )= 12 f ′(2 )
So,
∴ lim
f( x )
x→ 2
2tdt
= 12 f ′(2 ), if f(2 ) = 6
x−2
∫
6
(1 − x 2 + x 4 )dx is
0
[JEE Main 2019, 9 April Shift-II]
π 1
(a)
− loge 2
4 2
π
(c)
− loge 2
4
(sec x tan x f ( x ) + (sec x tan x + sec 2 x ))
Exp. (a)
Let I =
1
∫ xcot
∴I=
1
2
1
2
1
(c) sec x + x tan x −
2
1
(d) sec x − tan x −
2
1
1
cot −1(1 − t + t 2 ) dt
2 ∫0
=
1


1
1
1
 dt Qcot −1 x = tan−1 
tan−1 
∫
2

20
x 
1 − t + t 
=
 t − (t − 1) 
1
tan−1 
 dt
2 ∫0
 1 + t (t − 1)
=
1

1
−1
−1
 ∫ ( tan t − tan (t − 1) dt 
2  0

1

−1 x − y
−1
−1 
Q tan 1 + xy = tan x − tan y


Exp. (b)
Given,
sec x
2
∫ e [(sec x tan x)f( x) + (sec x tan x + sec x)]dx
1
1
0
0
Q ∫ tan−1(t − 1)dt = ∫ tan−1(1 − t − 1) dt
1
⋅ f ( x) + C
= − ∫ tan−1(t ) dt
On differentiating both sides w.r.t. x, we get
sec x
e
=e
[(sec x tan x)f( x) + (sec x tan x + sec x)]
2
sec x
f ′( x) + esec x (sec x tan x)f( x)
⇒ esec x (sec x tan x + sec 2 x) = esec x f ′( x)
⇒
f ′( x) = sec x tan x + sec 2 x
So,
f( x) = ∫ f ′( x)dx = ∫ (sec x tan x + sec 2 x)dx
= sec x + tan x + C
So, possible value of f( x)from options, is
1
f( x) = sec x + tan x + .
2
(1 − x2 + x4 ) dx
0
(b) sec x + tan x +
=e
−1
Now, put x2 = t
dx = e sec x f ( x ) + C , then a possible choice of
[JEE Main 2019, 9 April Shift-II]
f ( x ) is
sec x
π 1
− loge 2
2 2
π
(d)
− loge 2
2
(b)
⇒
2 xdx = dt
Lower limit at x = 0, t = 0
Upper limit at x = 1, t = 1
13. If ∫ e sec x
(a) x sec x + tan x +
−1
0
a
a
0
0
because ∫ f( x)dx = ∫ f(a − x) dx
1
1
So, I = ∫ ( tan−1 t + tan−1 t ) dt
20
1
= ∫ tan−1 tdt = [t tan−1 t ]10 −
0
1
t
∫ 1 + t 2 dt
0
[by integration by parts method]
=
π 1
π 1
− [loge (1 + t 2 )]10 = − loge 2
4 2
4 2
168
JEE Main Chapterwise Mathematics
dx
15. If ∫
( x 2 − 2 x + 10)2


f (x )
 x − 1
= A  tan −1 
 +C ,
+
 3  x 2 − 2 x + 10

where,C is a constant of integration, then
[JEE Main 2019, 10 April Shift-I]
1
(a) A =
and f ( x ) = 9 ( x − 1)
27
1
(b) A =
and f ( x ) = 3 ( x − 1)
81
1
(c) A =
and f ( x ) = 3 ( x − 1)
54
1
(d) A =
and f ( x ) = 9 ( x − 1)2
54
 (n + 1)1/3
16. lim 
 n
equal to
n→∞
n
4 /3
+ ..... +
(2n )1/3 

n 4/3 
is
[JEE Main 2019, 10 April Shift-I]
3 4/ 3 4
(2) −
3
4
4
3/ 4
(d) ( 2 )
3
(b)
Exp. (c)
Let
dx
∫ ( x2 − 2 x + 10)2
=
dx
∫ (( x − 1)2 + 32 )2
 (n + 1)1/ 3 (n + 2 )1/ 3
(2 n)1/ 3 
p = lim 
+
+ … + 4 / 3 
4
3
4
3
/
/
n→ ∞ 
n
n
n

Now, put x − 1 = 3tanθ ⇒dx = 3 sec2θ dθ
So, I =
(n + 2 )1/3
+
4 /3
4
(a) ( 2 )4/ 3
3
3
3
(c) ( 2 )4/ 3 −
4
4
Exp. (c)
Let I =
It is given, that


x − 1
f ( x)
I = A  tan−1 
 + 2
+C


3
x
−
x
+
2
10


1
On comparing, we get A =
and f( x) = 3( x − 1.)
54
3sec θ dθ
2
∫ (32 tan2 θ + 32 )2 = ∫
3sec θ dθ
3 sec θ
+
1
cos
2θ
1
1
cos 2θ dθ =
=
dθ
27 ∫
27 ∫
2
Qcos 2 θ = 1 + cos 2θ 


2
1
=
(1 + cos 2θ) dθ
54 ∫
4
n→ ∞
4
=
sin2θ 
1 
θ +
 +C
54 
2 
=
x − 1
1
1  2 tanθ 

 +C
tan−1 
 +


54
3
108  1 + tan2 θ 

2 tanθ 
Qsin2θ =

1 + tan2 θ 

x − 1
1
1
tan−1 
=
 +
 3  54
54
= lim
2
 x − 1


 3 
x − 1
1 + 

 3 
2

x − 1
x−1
1
1 

 +C
tan−1 
 +
 3  18  ( x − 1)2 + 32 
54
=
x − 1
1
1
tan−1 
 +
 3  18
54
=
3( x − 1) 
1  −1  x − 1
 + 2
 tan 
+C
54 
3 
x − 2 x + 10 


x−1
 2
 + C
 x − 2 x + 10 
(n + r )1/ 3
n4 / 3
r =1
1/ 3
1 + r 
n1/ 3




n
= lim ∑
n→ ∞
n4 / 3
r =1
n
= lim
n→ ∞
1
n
n

∑  1 +
r =1
r

n
1/ 3
Now, as per integration as limit of sum.
r
1
Let = x and = dx
[Q n → ∞]
n
n
Then, upper limit of integral is 1 and lower limit of
integral is 0.
1
So,
p = ∫ (1 + x)1/ 3 dx
0

1
Q lim
 n → ∞ n
+C
=
n
∑
n
 r
1

∑ f  n  = ∫0 f( x) dx
r =1

1
3
3
3
3
=  (1 + x)4 / 3  = (2 4 / 3 − 1) = (2 )4 / 3 −
4
4
 4
 0 4
17. The region represented by | x − y | ≤ 2 and
| x + y | ≤ 2 is bounded by a
[JEE Main 2019, 10 April Shift-I]
(a)
(b)
(c)
(d)
rhombus of side length 2 units
rhombus of area 8 2 sq units
square of side length 2 2 units
square of area 16 sq units
169
Integral Calculus
Exp. (c)
The given inequalities are
| x − y| ≤ 2 and | x + y| ≤ 2 .
On drawing, the above inequalities, we get a
square
Y
equal to
(2, 0)
O
[JEE Main 2019, 10 April Shift-II]
(b) 37/ 6 − 35/ 6
(c) 35/ 3 − 31/ 3
(d) 34/ 3 − 31/ 3
π /3
∫ sec
2/3
x cosec 4 / 3 x dx
π /6
X
=
π /3
∫
π /6
1
cos 2 / 3 x sin4 / 3 x
Y′
Now, the area of shaded region is equal to the
area of a square having side length
(2 − 0)2 + (0 − 2 )2 = 2 2 units.
2π
18. The value of ∫ [sin 2 x (1 + cos 3x )] dx , where
So,
[sin2 x ⋅ (1 + cos 3 x)]dx
π
= I1 + I2 (let)
∴
I=
0
∫π [− sin2 x ⋅ (1 + cos 3 x)]dx
∫0 [− sin2 x ⋅ (1 + cos 3 x)]dx
3
1
−
2 loge 2
Given, equations of curves
 x + 1 ,x≥ − 1
y = 2 x and y = | x + 1| = 
− x − 1 , x < − 1
Q The figure of above given curves is
…(ii)
Y
y=x+1
π
∫0 [sin2 x ⋅ (1 + cos 3 x)]dx
+
=
(d)
(1,2)
y=–x–1
π
∫0 [− sin2 x ⋅ (1 + cos 3 x)]dx
π
∫0 (−1) dx]
= −π
3
2
Exp. (d)
∫π [sin2 x ⋅ (1 + cos 3 x)]dx
let 2 π − x = t, upper limit t = 0 and lower limit t = π
and
dx = −dt
=
3
... (i)
2π
π
3
(b) loge 2 +
(a)
2π
I2 = −
t 4/ 3
 t − 1/ 3 
=

 − 1 / 3 1/
3
2
1
(c)
2
∫0 [sin2 x ⋅ (1 + cos 3 x)]dx
π
So,
∫
dt
by the curves y = 2 x and y = | x + 1 |, in the first
quadrant is
[JEE Main 2019, 10 April Shift-II]
2π
I2 =
3
3
20. The area (in sq units) of the region bounded
+ ∫ [sin2 x ⋅ (1 + cos 3 x)]dx
Now,
dx
= 37/ 6 − 35 / 6
Given integral
=
(tan x)4 / 3
= 3 ⋅ 31/ 6 − 3 ⋅ 3− 1/ 6
(d) π
Exp. (a)
∫0
∫
π/6
sec 2 x
1
= − 3  1/ 6 − 31/ 6 
3

[JEE Main 2019, 10 April Shift-I]
I=
I=
1/ 3
[t ] denotes the greatest integer function, is
(c) − 2 π
π /3
Put, tan x = t , upper limit, at x = π / 3 ⇒t =
and lower limit, at x = π / 6 ⇒ t = 1 / 3
and sec 2 x dx = dt
0
(b) 2 π
dx =
[multiplying and dividing the
denominator bycos 4/ 3 x]
(0, –2)
(a) − π
x cosec4/3x dx is
(a) 35/ 6 − 32 / 3
Let I =
(–2, 0)
2/ 3
Exp. (b)
(0, 2)
X′
π /3
∫π /6 sec
19. The integral
(0,1)
[from Eqs. (i) and (ii)]
[Q[ x] + [− x] = − 1, x ∉Integer]
y=2x
X′
(–1,0)
O
X
170
JEE Main Chapterwise Mathematics
In first quadrant, the above given curves intersect
each other at (1, 2).
So, the required area =
1
∫0
(( x + 1) − 2 x ) dx
Exp. (b)
Given region is {( x, y) : y2 ≤ 4 x, x + y ≤ 1, x ≥ 0,
y ≥ 0}
1
x


2 
ax
x
+ C
=
+ x−

Q ∫ a dx =
2
log
2
a
log
e
e 0



1
2
1  3
1
=  + 1−
+
= −
loge 2 loge 2  2 loge 2
2
2
x
2
B(0,1)
P
x+y=1
2
21. If ∫ x 5e − x dx = g ( x )e − x + C , where C is a
constant of integration, then g ( − 1) is equal
to
[JEE Main 2019, 10 April Shift-II]
(a) −1
1
(c) −
2
(b) 1
(d) −
5
2
5 − x
∫xe
dx
x = t ⇒2 xdx = dt
1
So,
I = ∫ t 2e − t dt
2
1
= [(− t 2e − t ) + ∫ e − t (2t ) dt ]
2
[integration by parts]
1
= [− t 2e − t + 2t (− e − t ) + ∫ 2e − t dt ]
2
1
= [− t 2e − t − 2te − t − 2e − t ] + C
2
e− t 2
=−
(t + 2t + 2 ) + C
2
2
2
e− x
( x4 + 2 x2 + 2 ) + C [Qt = x2 ]
2
Q It is given that,
=−
I=
5 − x2
∫xe
dx = g ( x) ⋅ e −
x2
…(i)
+C
36 − 4
= 3 ± 2 2.
2
Since, x-coordinate of P less than x-coordinate of
point A(1, 0).
∴ x= 3−2 2
Now, required area
x=
x ≥ 0, y ≥ 0}
{( x , y ): y 2 ≤ 4x , x + y ≤ 1,
a 2 + b , then a − b is equal to
is
[JEE Main 2019, 12 April Shift-I]
(c)
8
3
3−2 2
∫0
=2
x3 / 2
3/2
2 x dx +
3−2 2
0
1
∫3 − 2
2
(1 − x) dx
1

x2 
+ x −

2 3 − 2

2
4
1
= ( 3 − 2 2 )3/ 2 +  1 −  − ( 3 − 2 2 )

3
2
+
(3 − 2 2 )2
2
4
1
1
[( 2 − 1)2 ]3 / 2 + − 3 + 2 2 +
3
2
2
(9 + 8 − 12 2 )
=
17
5
4
( 2 − 1)3 − + 2 2 +
−6 2
2
2
3
4
= (2 2 − 3(2 ) + 3( 2 ) − 1) − 4 2 + 6
3
4
= (5 2 − 7 ) − 4 2 + 6
3
22. If the area (in sq units) of the region
(b) 6
=
6±
=
By Eq. (i), comparing both sides, we get
1
g ( x) = − ( x 4 + 2 x 2 + 2 )
2
5
1
So, g(− 1) = − (1 + 2 + 2 ) = −
2
2
10
(a)
3
X
A(1,0)
Now, for point P, put value of y = 1 − x to y2 = 4 x,
we get
(1 − x)2 = 4 x ⇒ x2 + 1 − 2 x = 4 x
⇒
Let given integral, I =
Put
O
⇒ x2 − 6 x + 1 = 0
Exp. (d)
2
y2=4x
(d) −
2
3
=
8 2 10
−
3
3
=a 2 + b
8
10
So, on comparing a = and b = −
3
3
8 10
a−b= +
=6
∴
3
3
(given)
171
Integral Calculus
23. The integral ∫
2x3 − 1
x4 + x
dx is equal to
(hereC is a constant of integration)
| x 3 + 1|
1
loge
+C
2
x2
(c) loge
x3 + 1
+C
x
(b)
( x 3 + 1)2
1
+C
loge
2
| x 3|
(d) loge
| x 3 + 1|
x2
2 x3 − 1
2 x − 1 / x2
dx = ∫
dx
4
1
x + x
x2 +
x
[dividing each term of numerator and
denominator by x2 ]
1
1 

Put x2 + = t ⇒  2 x +  − 2   dx = dt
 x 

x
dt
= loge| (t )| + C
I=∫
∴
t
1
= loge  x2 +  + C

x
= loge
(cosec x cot x − cosec 2 x + 1) dx
cos x − 1
= x +

sin x  0
π /2
x
=  x − tan 
2  0

=


= x +


∫
x3 + 1
+C
x
1
2
(b) 1
(c)
1
2
(d) −1
Exp. (d)
π /2
cot x
dx
cot x + cosec x
cos x
π /2
π / 2 cos x
sin x
dx = ∫
=∫
dx
0
0
1
cos x
1 + cos x
+
sin x
sin x
π / 2 cos x (1 − cos x)
dx
=∫
0
1 − cos 2 x
∫0
π / 2 cos
∫0
x − cos 2 x
sin2 x
dx
[given]
1
and n = − 2
2
∴ m⋅ n = − 1
Alternate Solution
π /2
cot x
Let I = ∫
dx
0
cot x + cosec x
cos x
π /2
π / 2 cos x
sin x
dx = ∫
=∫
dx
0
0
1
cos x
cox + 1
+
sin x
sin x
2 x
2 cos
−1
π /2
2
dx
=∫
0
x
2 cos 2
2
θ
2θ
[Qcos θ = 2 cos
− 1 and cos θ + 1 = 2 cos 2 ]
2
2
π /2
1
2 x
=∫
 1 − sec  dx
0

2
2
π /2
cot x
If ∫
dx = m ( π + n ), then m ⋅ n
0
cot x + cosec x
is equal to
[JEE Main 2019, 12 April Shift-I]
π /2
 − 2 sin2 x  



2 
x
x
2 sin cos 
2
2  0
π
1
− 1 = [ π − 2]
2
2
On comparing, we get m =
x
=  x − tan 

2  0
π /2
=
π /2
= m [ π + n]
Let integral is I =
Let I =
∫0
+C
Key Idea
(i) Divide each term of numerator and
denominator by x2 .
1
(ii) Let x2 +
= t
x
(a) −
=
(cosec x cot x − cot 2 x) dx
π /2
Exp. (c)
24.
∫0
= [− cosec x + cot x + x]π0 / 2
[JEE Main 2019, 12 April Shift-I]
(a)
π /2
=
=
π
1
− 1 = ( π − 2)
2
2
I = m( π − n)
1
∴m( π − n) = ( π − 2 )
2
On comparing both sides, we get
1
m = and n = − 2
2
1
Now, mn = × (− 2 ) = − 1
2
Since,
25. A value of α such that
α +1
∫
α
dx
 9
= log e   is
 8
( x + α )( x + α + 1)
[JEE Main 2019, 12 April Shift-II]
(a) − 2
(b)
1
2
(c) −
1
2
(d) 2
172
JEE Main Chapterwise Mathematics
sin α
sin x
+
cos x cos α
=∫
dx
sin x
sinα
−
cos x cos α
sin x cos α + sin α cos x
=∫
dx
sin x cos α − sin α cos x
sin ( x + α )
=∫
dx
sin ( x − α )
Exp. (a)
α+1
∫
Let I =
α
α+1
=
∫
α
α+1
=
∫
α
dx
( x + α ) ( x + α + 1)
( x + α + 1) − ( x + α )
dx
( x + α ) ( x + α + 1)
 1

1
−

 dx
+
+
+
x
α
x
α
1


Now, put x − α = t ⇒dx = dt , so
sin (t + 2 α )
I=∫
dt
sin t
sin t cos 2 α + sin 2α cos t
=∫
dt
sint
cos t 
= ∫  cos 2 α + sin 2 α
 dt

sint 
α +1
α
= [loge ( x + α ) − loge ( x + α + 1)]
α+1

 x + α 
= loge 

 x + α + 1  α

2α + 1
2α
= loge
− loge
2α + 2
2α + 1
 2α + 1 2α + 1
= loge 
×

2α 
 2α + 2
9
= loge  
 8
(given)
(2 α + 1)2
9
=
4α (α + 1) 8
⇒
27. If the area (in sq units) bounded by the
⇒ 8 [4α + 4α + 1] = 36 (α + α )
2
2
⇒ 8α 2 + 8α + 2 = 9α 2 + 9α
⇒
= t (cos 2 α ) + (sin 2 α ) loge | sint | + C
= ( x − α ) cos 2 α + (sin 2 α ) loge |sin ( x − α )| + C
= A( x) cos 2 α + B( x) sin 2 α + C (given)
Now on comparing, we get
A( x) = x − α and B( x) = loge |sin ( x − α )|
α2 + α − 2 = 0
⇒ (α + 2 ) (α − 1) = 0
⇒
α = 1, − 2
From the options we get α = − 2
26. Let α ∈(0, π / 2) be fixed. If the integral
tan x + tan α
∫ tan x − tan α dx = A (x )cos2α + B (x )
parabola y 2 = 4λx and the line y = λx , λ > 0, is
1
, then λ is equal to
9
[JEE Main 2019, 12 April Shift-II]
(a) 2 6
(b) 48
(c) 24
(d) 4 3
Exp. (c)
Given, equation of curves are
y2 = 4λx
…(i)
…(ii)
y = λx
λ> 0
Area bounded by above two curve is, as per figure
and
Y
sin2 α +C , where C is a constant of
integration, then the functions A ( x ) and
B ( x ) are respectively
A
y2=4λx
O
[JEE Main 2019, 12 April Shift-II]
X
(a) x + α and loge | sin( x + α )|
(b) x − α and loge | sin( x − α )|
y=λx
(c) x − α and loge |cos( x − α )|
(d) x + α and loge | sin( x − α )|
Exp. (b)
tan x + tanα
π
Let I = ∫
dx, α ∈  0, 
 2
tan x − tanα
the intersection point A we will get on the solving
Eqs. (i) and (ii), we get
⇒
λ2 x2 = 4λx
4
x = ,so y = 4
λ
173
Integral Calculus
4
A  , 4
λ 
So,
Now, as cos 3 x = 4cos 3 x − 3cos x
1
∴ cos 3 x = (cos 3 x + 3cos x)
4
Now, required area
=
4/ λ
∫ (2
λx − λx) dx
∴
I=
0
π
4/ λ


 x3 / 2 
=2 λ 
3 


 2 0
1  sin 3 x
2
+ 3sin x
 0
2  3
π
1 1 3π
1
=  sin
+ 3sin  −  sin 0 + 3sin 0 
 
2  3
2
2   3
1 1
=  (−1) + 3 − [0 + 0]
2  3


=
4/ λ
 x2 
−λ 
 2 0
2
4
4 4 λ  4
32 8
λ
−   =
−
3
λ λ 2  λ
3λ λ
32 − 24
8
=
=
3λ
3λ
1
It is given that area =
9
8
1
=
⇒
3λ 9
⇒
λ = 24
=
π
π
π
3π


= sin π +  = − sin = − 1
Qsin



2
2
2
1 1
4
=  − + 3 =
 3
2  3
29. For x 2 ≠ nπ + 1, n ∈ N (the set of natural
numbers), the integral
28. The value of ∫0 |cos x | dx is
3
2
3
(b) −
4
3
(c) 0
(d)
4
3
1
loge | sec( x 2 − 1)| + C
2
 x 2 −1
 + C
(b) loge sec 
 2 
(a)
We know, graph of y = cos x is
Y
π
π/2
O
1
sec 2 ( x 2 − 1) + C
2
 x 2 −1
1
 + C
(d) loge sec 2 
2
 2 
(c) loge
X
Y′
∴ The graph of y =|cos x| is
Y
dx
2 sin( x 2 − 1) + sin 2( x 2 − 1)
is equal to (where C is a constant of
integration )
[JEE Main 2019, 9 Jan Shift-I]
Exp. (d)
X′
2 sin( x 2 − 1) − sin 2( x 2 − 1)
∫x
[JEE Main 2019, 9 Jan Shift-I]
(a)
π
2 2
(cos 3 x + 3cos x)dx
4 ∫0
Exp. (b)
y=|cos x|
Let I =
2 sin( x2 − 1) − sin2( x2 − 1)
∫x
2 sin( x2 − 1) + sin2( x2 − 1)
dx
x2 − 1
=θ
2
⇒
x2 − 1 = 2θ ⇒ 2 x dx = 2 dθ
Put
X′
∴ I=
π
π
π/2
O
Y′
∫0 |cos x|
3
=2
π
2|cos
0
∫
X
⇒
3
x| dx
(Q y =|cos x| is symmetric about x =
π
2 cos 3 x dx
0
= 2∫
π
)
2
π 

Qcos x ≥ 0 for x ∈  0, 

 2  
x dx = dθ
Now, I =
∫
2 sin2 θ − sin 4θ
dθ
2 sin2 θ + sin 4θ
=
∫
2 sin2 θ − 2 sin2 θcos 2 θ
dθ
2 sin2 θ + 2 sin2 θ cos 2 θ
(Qsin2 A = 2 sin A cos A)
174
JEE Main Chapterwise Mathematics
=
=
∫
2 sin2 θ(1 − cos 2 θ)
dθ
2 sin2 θ(1 + cos 2 θ)
∫
1 − cos 2 θ
dθ =
1 + cos 2 θ
∫
2 sin2 θ
2 cos 2 θ
[Q1 − cos 2 A = 2 sin A and 1 + cos 2 A = 2 cos A]
=
∫
2
tan θ d θ =
2
∫ tanθd θ
 x − 1
= loge|sec θ| + C = loge sec 
 + C
 2 

x2 − 1

Qθ =
2 

∫0 ( x
=
( x − 2 )3
3
2
2
− 4 x + 4) dx =
2
=
0
(a) 2
(2 − 2 )3 (0 − 2 )3 8
−
= sq units
3
3
3
32
(c)
3
(b)
4
3
(c)
14
(d)
3
2
2
y=–x +1
⇒ Vertex of parabola is (0, − 1) and it is open
upward.
Equation of tangent at (2, 3) is given by T = 0
y + y1
= x x1 − 1, where, x1 = 2
⇒
2
and
y1 = 3.
x
1
parabola with vertex (0,1) but open downward]
We need to calculate the shaded area, which is
equal to
0
∫−1(− x
2
1
+ 1)dx + ∫ ( x2 + 1)dx
0
1
0
 x3

 x3

= −
+ x + 
+ x
 3
− 1  3
0
(2, 3)

 (− 1)3
  1

=  0 − −
+ (− 1)  +   + 1 − 0
   3



3



1
4 2 4
= −  − 1 + = + = 2
3
 3 3 3
y=4x–5
32. If
∫0 (y(parabola ) − y(tangent )) dx
y=0
[Q y = x2 + 1 ⇒ x2 = ( y − 1), parabola with vertex
(0, 1) and y = − x2 + 1⇒ x2 = − ( y − 1) ,
2
=
y=x2+1
1
–1
y+ 3
= 2x − 1
2
y = 4x − 5
2
2
3
y
Given, equation of parabola is y = x2 − 1, which
can be rewritten as x2 = y + 1 or x2 = ( y − (−1)).
Now, required area = area of shaded region
(d)
Now, the required region is the shaded region.
Exp. (a)
(0, –1)
1
3
and when x < 0, then 0 ≤ y ≤ − x2 + 1
[JEE Main 2019, 9 Jan Shift-I]
y=x2–1
dx
We have,
A = {( x, y) : 0 ≤ y≤ x| x| + 1and − 1 ≤ x ≤ 1}
When x ≥ 0, then 0 ≤ y ≤ x2 + 1
parabola y = x − 1, the tangent at the point
(2, 3) to it and theY -axis is
⇒
2
Exp. (a)
2
⇒
2
∫0 ( x − 2 )
A = {(x , y ); 0 ≤ y ≤ x | x | + 1 and − 1 ≤ x ≤ 1} in
sq. units, is
[JEE Main 2019, 9 Jan Shift-II]
30. The area (in sq units) bounded by the
56
(b)
3
=
− 1) − (4 x − 5)] dx
2
31. The area of the region
2
8
(a)
3
∫0 [( x
dθ
2
2
=
π /3
∫0
tan θ
2k sec θ
value of k is
(a) 1
1
(b)
2
dθ =1 −
1
2
,(k > 0), then the
[JEE Main 2019, 9 Jan Shift-II]
(c) 2
(d) 4
175
Integral Calculus
Exp. (c)
=
tan θ
1
, (k > 0)
dθ = 1 −
0
2 k sec θ
2
π/3
tan θ
Let I = ∫
dθ
0
2 k sec θ
1 π / 3 tanθ
dθ
=
sec θ
2 k ∫0
π/3
(sin θ)
1
dθ
=
2 k ∫0
1
(cos θ)
cos θ
π / 3 sin θ
1
dθ
=
cos θ
2 k ∫0
We have, ∫
π/3
⇒ I=
−1
1 1/ 2 − dt
=
t
2 k ∫1
2 k ∫1
t
(5 x− 6 + 7 x− 8 )dx = − dt
dt
f( x) = ∫ − 2 = − ∫ t −2dt
∴
t
t−2 +1
t−1
1
=−
+ C =−
+C= +C
t
−2 + 1
−1
⇒
1
2 dt
2 = 2k ⇒ 2k = 4 ⇒ k = 2
5x + 7x
f (x ) = ∫
1
2
(b) −
1
4
(c)
and
1
4
(d)
1
2
Exp. (c)
We have, f( x) =
=
f(0) = 0
∴
0=
∴
f( x) =
⇒
f(1) =
∫ ( x2 + 1 + 2 x7 )2 dx
2 x7 
 7 + 7 + 7 
x
x 
x
1
+2
+C=
x7
2 x + x2 + 1
7
+C
0
+ C ⇒C = 0
0+ 0+1
x7
2 x + x2 + 1
1
7
2(1)7 + 12 + 1
=
1
4
π
2
34. Let n ≥ 2 be a natural number and 0 < θ < .
1
Then, ∫
(sinn θ − sin θ )n cos θ
dθ is equal to
sinn + 1 θ
(whereC is a constant of integration)
[JEE Main 2019, 10 Jan Shift-I]
(a)
1
1 −



sinn + 1 θ 
n2 −1 
n
1
1 +

(b) 2

n −1 

sin
θ
n −1
(c)
1
1 −



sinn − 1 θ 
n2 −1 
(d)
1
1 −



sinn − 1 θ 
n2 + 1 
n
n
Let I =
 x6 
 x8 
5 14  + 7  14 
x 
x 
x2
−7
n+1
n
+C
n+1
n
+C
n+1
n
+C
n+1
n
+C
Exp. (c)
5 x8 + 7 x6
∫
+ x
Q
[JEE Main 2019, 9 Jan Shift-II]
(a) −
x
−5
n
6
dx ,( x ≥ 0),
( x 2 + 1 + 2 x 7 )2
f (0) = 0, then the value of f (1) is
33. If
1
=

1
1 
1
[2 t ]12
=−
 =−

2 k  − 1 + 1
2k
1
 2

1 
2 
2  1
=−
− 1 =
1 −


2
2k 
2k  2

1
(given)
Q I = 1−
2
1
1 
2 
2
⇒
=1
∴
1 −
 = 1−
2
2
2k 
2k
8
dx
⇒ (− 5 x− 6 − 7 x− 8 )dx = dt
1
1
− +1 2

t 2
⇒
x− 7 + 2 )2
x− 5 + x− 7 + 2 = t
Let
Let cos θ = t ⇒ − sin θ dθ = dt ⇒sin θ dθ = − dt
for lower limit, θ = 0 ⇒t = cos 0 = 1
π
π 1
for upper limit, θ = ⇒t = cos =
3 2
3
1/ 2 −
5 x− 6 + 7 x− 8
∫ ( x− 5 +
2
dx
(dividing both numerator and denominator by x14 )
∫
(sinn θ − sinθ)1/ n cos θ
dθ
sinn + 1 θ
Put sinθ = t ⇒ cosθ dθ = dt
1/ n
t 
 n
t  1 − n 
n
1/ n
 
(t − t )
t   dt
∴
I=∫
dt = ∫ 
n+1
n+1
t
t
t (1 − 1 / t n −1 )1/ n
(1 − 1 / t n − 1 )1/ n
=∫
dt
dt = ∫
tn
tn +1
176
JEE Main Chapterwise Mathematics
1−
Put
1
or
1 − t −( n − 1) = u ⇒
⇒
dt
du
=
n
n
−1
t
(n − 1)
dt = du
tn
1
⇒
Also, as long as f( x) lie below the X-axis, the value
of definite integral will be minimum.
∴ (a, b ) = (− 2 , 2 ) for minimum of I.
=u
tn −1
36. If the area enclosed between the curves
y = kx 2 and x = ky 2 ,(k > 0), is 1 square unit.
Then, k is
[JEE Main 2019, 10 Jan Shift-I]
+1
un
u1/ ndu
+C
I=∫
=
n − 1 (n − 1)  1 + 1


n

(a)
n+1
n+1
n
+C
Qu = 1 − 1 and t = sinθ


tn −1
b
35. Let I = ∫ (x 4 − 2 x 2 )dx . If I is minimum, then
a
the ordered pair (a ,b ) is [JEE Main 2019, 10 Jan
Shift-I]
(a) ( − 2 , 0)
(c) ( 2 , − 2 )
(b) (0, 2 )
(d) ( − 2 , 2 )
Exp. (d)
We have, I =
b
∫a ( x
4
− 2 x2 )dx
f ( x) = x 4 − 2 x 2 = x 2 ( x 2 − 2 )
Let
= x2 ( x − 2 ) ( x +
16  1   1 
1
   =
3  4k   4k  3k 2
1
[given, area = 1 sq.unit]
⇒
=1
3k 2
1
1
k2 = ⇒ k = ±
⇒
3
3
1
[Qk > 0]
⇒
k=
3
π /2
dx
∫− π /2[ x] + [sin x] + 4 , where [t ]
denotes the greatest integer less than or
equal tot , is
[JEE Main 2019, 10 Jan Shift-II]
y=f(x)
1
( 7 π − 5)
12
3
(c) ( 4 π − 3)
10
1
( 7 π + 5)
12
3
(d)
( 4 π − 3)
20
(b)
(a)
X
√2 ∴
f(x) < 0 for – √2 < x < √2
–
–
+
+
– √2
3
2
∴ Area enclosed between y = kx2 and x = ky2 is
37. The value of
Y
O
(d)
We know that, area of region bounded by the
parabolas x2 = 4ay and y2 = 4bx is
16
(ab ) sq units.
3
On comparing y = kx2 and x = ky2 with above
equations, we get
1
1
4a = and 4b =
k
k
1
1
and b =
a=
⇒
4k
4k
2)
Graph of y = f( x) = x4 − 2 x2 is
– √2
2
3
(c)
Exp. (b)
1
n
n  1 − n −1 


t
=
+C
(n − 1) (n + 1)
1 
n  1 −
n −1 

sin
θ
=
n2 − 1
1
3
(b)
3
Note that the definite integral
Exp. (d)
√2
0
b
∫a ( x
4
− 2 x )dx
2
represent the area bounded byy = f( x) , x = a, b
and the X -axis.
But between x = − 2 and x = 2, f( x) lies below
the X-axis and so value definite integral will be
negative.
π
dx
2
−π [ x] + [sin x] + 4
2
−1
dx
+
−π
[ x] + [sin x] + 4
2
Let I =
∫
=
∫
dx
+
0 [ x] + [sin x] + 4
+∫
1
dx
0
∫−1 [ x] + [sin x] + 4
π
dx
∫ 12 [ x] + [sin x] + 4
177
Integral Calculus
−2,
 − 1,

Q [ x] = 
 0,
 1,
− π / 2 < x < −1
− 1≤ x < 0
=−
3
=−
∴
So, I =
∫
dx
+
−2 − 1 + 4
1
x
1
39. If ∫0 f (t ) dt = x 2 + ∫xt 2 f (t )dt , then f ′   is
2
dx
∫−1 −1 − 1 + 4 +
0
dx
∫0 0 + 0 + 4
1
π
+∫ 2
1
=
∫
−1
−π
2
dx
+
1
dx
∫−1 2 +
0
1 dx
π
2
1
∫0 4 + ∫
dx
1+ 0 + 4
dx
5
π
1 π
1
1
=  −1 +  + (0 + 1) + (1 − 0) +  − 1



52
4
2
2
1
1
1
π
π
=  −1 + + −  +  + 

2 4 5   2 10 
−20 + 10 + 5 − 4 5 π + π
=
+
20
10
9
3π
=−
+
20
5
3
(4 π − 3)
=
20
[JEE Main 2019, 10 Jan Shift-II]
24
25
6
(c)
25
[using integration by parts]
1  te −4t e −4t 
= 
+
+C
3  −4
−16 
x

d
Q
 dx



 ψ( x )
d
d
 ∫ f(t )dt  = f(ψ( x)) ψ( x) − f(φ( x)) φ( x)
dx
dx


φ ( x)



⇒
(1 + x2 ) f( x) = 2 x
2x
f ( x) =
⇒
1 + x2
On differentiating w.r.t. ‘x’ we get
f ′ ( x) =
=
(1 + x2 )(2 ) − (2 x) (0 + 2 x)
(1 + x2 )2
2 + 2 x2 − 4 x2
(1 + x )
2 2
=
2 − 2 x2
(1 + x2 )2
2
∴
1 −4 x 3
e
f ( x) + C
48
In LHS, put x3 = t
⇒
3 x2dx = dt
3
1
So, ∫ x5e −4 x dx = ∫ t e −4t dt
3
e −4t 
1  e −4t
dt 
= t
−∫
3  −4
−4 
1
On differentiating both sides, w.r.t. ‘x’, we get
f ( x) = 2 x + 0 − x 2 f ( x)
Exp. (a)
3
x
0
Given, ∫ f(t ) dt = x2 + ∫ t 2 f(t )dt
1 −4 x 3
e
f (x ) + C ,
48
where C is a constant of integration, then
f ( x ) is equal to [JEE Main 2019, 10 Jan Shift-II]
Given, ∫ x5e −4 x dx =
(b)
Exp. (a)
3
(b) 4x 3 + 1
(d) − 2 x 3 + 1
18
25
4
(d)
5
(a)
38. If ∫ x 5e −4 x dx =
(a) − 4x 3 − 1
(c) − 2 x 3 − 1
[Qt = x3 ]
(comparing with given equation)
[QFor x < 0, − 1 ≤ sin x < 0 and for x > 0, 0 < sin x ≤ 1 ]
−1
−π
2
e −4 x
[4 x3 + 1] + C
48
f ( x) = − 1 − 4 x 3
0≤ x< 1
1≤ x < π / 2
 − 1, − π / 2 < x < −1
−1, − 1 < x < 0

[sin x]= 
0< x< 1
 0,
1< x < π / 2
 0,
and
1 −4t
e [4t + 1] + C
48
1
1
2 − 2  
2 − 2  
 4
2
1

f′   =
=
2
2
2 
2
1 + 1 
1




1 +   

4

 2  

3
1
2−
24
2
2
=
=
=
2
25 25
 5
 
16
 4
40. The area (in sq units) of the region bounded
by the curve x 2 = 4y and the straight line
[JEE Main 2019, 11 Jan Shift-I]
x = 4y − 2 is
7
8
5
(c)
4
(a)
9
8
3
(d)
4
(b)
178
JEE Main Chapterwise Mathematics
Exp. (c)
Exp. (b)
Given equation of curve is x = 4 y, which
represent a parabola with vertex (0, 0) and it open
upward.
2
x2
4
x+2
y=
4
Y
∫
Let I =
Y′
Now, let us find the points of intersection of
x2 = 4 y and 4 y = x + 2
For this consider, x2 = x + 2
x2 − x − 2 = 0
⇒
⇒
( x − 2 ) ( x + 1) = 0
x = − 1, x = 2
1
When x = − 1, then y =
4
and when x = 2, then y = 1
1
Thus, the points of intersection are A − 1,  and

4
B (2, 1.)
Now, required area = area of shaded region
2
∫−1{ y (line ) − y (parabola )} dx
2
 x + 2 x2 
x3 
1  x2
∫−1  4 − 4  dx = 4  2 + 2 x − 3 
−1
1 
8  1
1 
=
2 + 4 −  −  − 2 +  
4  
3  2
3 
=
2
1
4
1
=
4
=
41. If
∫
8 −

5 −

1
− 3

2
1 9
= sq units
2  8
1 − x2
4
dx = A ( x )( 1 − x 2 )m + C , for a
[JEE Main 2019, 11 Jan Shift-I]
(c)
1
9x 4
−1
27x 9
∫
(b)
(d)
x4
1
x2
x4
−1
3x 3
1
27x 6
−1
∫
1
−1 = t 2 ⇒
x
2
∫
dx =
=
dx =
∴I = − ∫ t 2dt = −
t
−2
x
3
… (i)
1
x2  2 − 1
x

dx
4
x
1
∫ x3
1
x2
− 1dx
dx = 2t dt ⇒
3
1
x3
dx = − t dt
+C
3
3/ 2
1/ 2

1  1 − x2 
 1
 
= − .  2 
+C
Qt =  2 − 1 
x
 
3  x 


1 1
2 3
…(ii)
=−
( 1− x ) + C
3 x3
On comparing Eqs. (i) and (ii), we get
1
A( x) = − 3 and m = 3
3x
1
∴( A( x))m = ( A( x))3 = −
27 x9
sin 2 x
dx
x 1
+
 π  2
(where, [x] denotes the greatest integer less
than or equal to x) is
2
42. The value of the integral ∫−2
[JEE Main 2019, 11 Jan Shift-I]
(a) 4 − sin 4 (b) 4
(c) sin 4
(d) 0
Exp. (d)
Let I =
2
sin2 x
dx
x
+  
2  π 
∫−2 1
Also, let f( x) =
x
suitable chosen integer m and a function
A ( x ), where C is a constant of integration,
then ( A ( x ))m equals
(a)
dx = A( x) ( 1 − x2 )m + C
1 − x2
x
X
2
Put
=
x4
B
–1 O
⇒
1 − x2
y=
A
X′
We have,
Then, f(− x) =
=
sin2 x
1  x
+
2  π 
sin2 (− x)
(replacing x by − x)
1  x
+ −
2  π 
sin2 x
x 
1 
+  − 1−   
 π  
2 

 − [ x], if
Q[− x] = − 1 − [ x], if


x ∈ I
x ∉ I
179
Integral Calculus
sin2 x
= − f ( x)
1  x
+
2  π 
i.e. f( x) is odd function
∴
I=0
0, if f( x) is odd function

 a


Q ∫ f( x) dx =  a
−a
2
if
x)
is
even
function
(
)
,
(
f
x
dx
f
 ∫0


f ( − x) = −
⇒
43. The
π /4
dx
∫π / 6 sin 2 x(tan 5 x + cot 5 x )
integral
equals
[JEE Main 2019, 11 Jan Shift-II]
1π
1 
(a)  − tan − 1 

 3 3
5 4
1 π
−1  1  
 − tan 

 9 3
10  4
π
(d)
40
(c)
Exp. (c)
=
bounded by the parabola, y = x 2 + 1, the
tangent to it at the point (2, 5) and the
coordinate axes is
[JEE Main 2019, 11 Jan Shift-II]
14
(a)
3
8
(c)
3
187
24
37
(d)
24
(b)
Exp. (d)
Given, equation of parabola is y = x2 + 1, which
can be written as x2 = ( y − 1).
Clearly, vertex of parabola is (0, 1) and it will open
upward.
Now, equation of tangent at (2, 5) is
y+ 5
= 2x + 1
2
[Qequation of the tangent at ( x1, y1 ) is given by
1
T = 0. Here, ( y + y1 ) = xx1 + 1]
2
⇒
y = 4x − 3
1
1 
(b)
tan − 1 

 9 3
20
Let I =
44. The area (in sq units) in the first quadrant
y= 4x–3
π /4
dx
Y
∫π / 6 sin2 x(tan5 x + cot 5 x)
π / 4 (1 +
P (2, 5)
tan2 x) tan5 x
∫π / 6 2 tan x (tan10 x + 1) dx

2 tan x 
Qsin2 x =

1 + tan2 x 

4
π / 4 tan
1
=
2
∫π / 6
x sec x
(tan10 x + 1)
dx
π
6
x
∴ I=
1 1
⋅
2 5
 1 


 3
1
∫(1/
R
Q (2, 0)
3,
0
4
X
[Qsec 2 x = 1 + tan2 x]
5 tan4 x sec 2 x dx = dt
t
O
2
Put tan5 x = t
⇒
(0, 1)
+1
=
∫0 y(parabola ) dx − (Area of ∆PQR )
=
1
∫0 ( x
 x3

3
1
= 
+ x −  2 −  ⋅ 5
4
 3
0 2 
1
(tan−1(t ))1
( 1/
10
=
1  −1
−1  1  
 tan (1) − tan 

 9 3
10 
=
1π
−1  1  
 − tan 

 9 3
10  4
2
=
5
dt
3 )5 t 2
Required area = Area of shaded region
π
4
2
2
+ 1) dx − (Area of ∆PQR)
2
3 )5
[Qarea of a triangle =
8
1 5
=  + 2  − 0 −   5
3

2  4
14 25 112 − 75 37
=
−
=
=
24
24
3
8
1
× base × height]
2
180
JEE Main Chapterwise Mathematics
45. If ∫
x +1
dx = f ( x ) 2 x − 1 + C , where C is
2x − 1
a constant of integration, then f ( x ) is equal
to
[JEE Main 2019, 11 Jan Shift-II]
2
(x + 2)
3
2
(c) ( x − 4)
3
1
( x + 4)
3
1
(d) ( x + 1)
3
Put 2 x − 1 = t
...(i)
2dx = 2tdt ⇒ dx = tdt
t +1
+1
1
tdt = ∫ (t 2 + 3) dt
I=∫ 2
2
t

t 2 + 1
2

Q2 x − 1 = t ⇒ x =
2 


t
1 t 3
=  + 3 t  + C = (t 2 + 9) + C
2 3
6

2
=
2x − 1
6
2x − 1
(2 x − 1 + 9) + C
47. The area (in sq units) of the region bounded
by the parabola, y = x 2 + 2 and the lines,
y = x + 1, x = 0 and x = 3, is
15
(a)
2
17
(b)
4
46. The integral ∫ cos (loge x ) dx is equal to
(where C is a constant of integration)
[JEE Main 2019, 12 Jan Shift-I]
(b) x [cos(loge x ) + sin(loge x )] + C
(d)
15
4
Given equation of parabola is y = x2 + 2, and the
line is y = x + 1
y
y=x2 +2
y=x+1
(0,2)
[Qt = 2 x − 1 ]
6
x+ 4
=
2x−1+ C
3
On comparing it with Eq. (i), we get
x+ 4
f ( x) =
3
21
2
(c)
Exp. (a)
1
1
(2 x + 8) + C
x
(a) [cos(loge x ) + sin(loge x )] + C
2
O
x
(3,0)
The required area = area of shaded region
=
3
∫0
(( x2 + 2 ) − ( x + 1)) dx =
3
∫0
( x2 − x + 1)dx
3
 x3 x2

27 9
= −
+ x = 
− + 3 − 0


3 2
2
3
0
9
9
= 9 − + 3 = 12 −
2
2
15
sq units
=
2
48. Let f and g be continuous functions on
[0, a] such that f ( x ) = f (a − x ) and
a
g ( x ) + g (a − x ) = 4, then ∫ f ( x ) g ( x ) dx is
(c) x [cos(loge x ) − sin(loge x )] + C
x
(d) [sin(loge x ) − cos(loge x )] + C
2
equal to
Exp. (a)
(a) 4∫ f ( x ) dx
(b) ∫ f ( x ) dx
(c) 2 ∫ f ( x ) dx
(d) − 3∫ f ( x ) dx
0
[JEE Main 2019, 12 Jan Shift-I]
a
Let I = ∫ cos(loge x)dx
1
dx
x
[JEE Main 2019, 12 Jan Shift-I]
2
⇒
=
[using integration by parts]
= xcos(loge x) + ∫ sin(loge x) dx
[again, using integration by parts]
⇒ I = xcos(loge x) + xsin((loge x) − I
x
⇒ I = [cos(loge x) + sin(loge x)] + C
2
Exp. (b)
We have,
x+1
∫ 2 x − 1 dx = f( x) 2 x − 1 + C
x+1
Let I = ∫
dx
2x − 1
1
∫ x(− sin(loge x)) x ⋅ dx
= x cos(loge x) + x sin(loge x) − ∫ x(cos(loge x))
(b)
(a)
= xcos(loge x) −
0
a
0
a
0
a
0
181
Integral Calculus
1
  t 2 1 
I =   +  
 2 t   1
Exp. (c)
a
∫0 f( x) g( x) dx
I=
Let
=
a
∫0
… (i)
e
f(a − x) g (a − x) dx
Q af( x) dx =
 ∫0
⇒
a

∫0 f(a − x) dx
a
∫0 f( x) [4 − g( x)] dx
I=
[Qf( x) = f(a − x) and g ( x) + g (a − x) = 4]
=
⇒
1
1
=  + 1 −  2 + e  
  2e

 2
3
1
= −e − 2
2
2e
a
a
∫0 4f( x) dx − ∫0 f( x) g( x) dx
a
I = 4 ∫ f( x) dx − I
dx is equal to
(2 x 4 + 3x 2 + 1)4
(whereC is a constant of integration)
[JEE Main 2019, 12 Jan Shift-II]
[from Eq. (i)]
0
(a)
a
⇒ 2 I = 4 ∫ f( x) dx
x
I=2
(b)
a
∫0 f ( x) dx
e
49. The integral ∫1
equal to
  x  2x  e  x 
   −    log e x dx is
 x 
e


[JEE Main 2019, 12 Jan Shift-II]
3
1
−e − 2
2
2e
1
1
(c) − e − 2
2
e
1 1
1
+ −
2 e 2e 2
3 1
1
(d) − − 2
2 e 2e
(b) −
(a)
Exp. (a)
Let I =
e
 x 
∫1  e 
2x

e
−  
 x
x

 loge x dx

x
x
Now, put   = t
e
x
x loge   = log t
e
⇒
⇒
x (loge x − loge e ) = log t
1
  1

⇒ x   + (loge x − loge e ) dx = dt
  x 

t
1
⇒
(1 + loge x − 1) dx = dt
t
1
(loge x) dx = dt
⇒
t
Also, upper limit x = e ⇒ t = 1 and lower limit
1
x = 1 ⇒t =
e
1
1 1
∴ I = ∫  t 2 −  ⋅ dt
1/ e 
t t
⇒ I=
1
∫1/ e (t − t
−2
)dt
(c)
(d)
4
+C
6( 2 x 4 + 3x 2 + 1)3
x12
0
⇒
3x 13 + 2 x 11
50. The integral ∫
6( 2 x + 3x 2 + 1)3
4
x4
( 2 x + 3x 2 + 1)3
4
x12
( 2 x + 3x 2 + 1)3
4
+C
+C
+C
Exp. (b)
Let I =
3 x13 + 2 x11
∫ (2 x4 + 3 x2 + 1)4 dx
3
2
+ 5
x3
x
∫  3 1  4 dx
2 + 2 + 4 

x
x 
[on dividing numerator and denominator by x16 ]
3
1
Now, put 2 + 2 + 4 = t
x
x
=
⇒
 − 6 − 4  dx = dt
 3

 x
x5 
⇒
 3 + 2  dx = − dt
 3

x
2
x5 
So, I =
∫ 2t 4
=
=
− dt
=−
1
1 t−4 +1
×
+C= 3 +C
2 −4+1
6t
1
3
1
6  2 + 2 + 4 

x
x 
3
x12
6(2 x + 3 x2 + 1)3
4
3
1
+ C Qt = 2 + 2 + 4 

x
x 
+C
182
JEE Main Chapterwise Mathematics
51. The integral
sin 2 x cos 2 x
∫ (sin 5 x + cos3 x sin 2 x + sin 3 x cos2 x
⇒
dx
π /2
∫
I=
−π / 2
π
π
sin2  − +
− x
 2

2
dx
π
π
−
1+ 2
[JEE Main 2018]
1
(a)
3(1 + tan x )
1
(c)
+C
1 + cot 3 x
3
+C
(b)
−1
3(1 + tan x )
−1
(d)
+C
1 + cot 3 x
3
+C
(whereC is a constant of integration)
Exp. (b)
We have,
I=
=
=
=
sin2 x ⋅ cos 2 x
∫ (sin5 x + cos3 x ⋅ sin2 x + sin3 x ⋅ cos2 x + cos5 x)2
dx
2
2
2
sin2 x
dx
1 + 2− x
−π / 2
∫
I=
⇒
I=
⇒
2I =
⇒
2I =
π/2
2 x sin2 x
dx
2x + 1
−π / 2
∫
sin xcos x
tan2 xsec 2 x
∫ (1 + tan3 x)2
∴
⇒
⇒
I=
⇒
⇒
tan x = t
3
∫ sin
2
x dx ⇒ 2 I = 2
π/2
∫ sin
2
x dx
0
I=
π/2
∫ sin
xdx
∫ cos
2
0
π/2
2I =
π/2
∫ dx
2
xdx
a

 a
Q ∫ f( x)dx = ∫ f(a − x)dx

 0
0
⇒ 2 I = [ x]π0 / 2 ⇒ I =
0
3 tan xsec xdx = dt
1
dt
I= ∫
3 (1 + t )2
−1
+C
I=
3 (1 + t )
−1
+C
I=
3 ( 1 + tan3 x)
2
π
4
2
52. The value of
sin 2 x
∫
π
−
2
1 + 2x
π
(b)
2
dx is
[JEE Main 2018]
(c) 4π
π
(d)
4
53. Let g ( x ) = cos x 2, f ( x ) = x and α , β (α < β ) be
the roots of the quadratic equation
18x 2 − 9πx + π 2 = 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
[JEE Main 2018]
1
(a) ( 3 − 1)
2
1
(c) ( 3 − 2 )
2
Key idea
b
b
a
a
Use property = ∫ f( x)dx = ∫ f(a + b − x) dx
π /2
sin2 x
dx
1+ 2x
−π / 2
∫
1
(b) ( 3 + 1)
2
1
(d) ( 2 − 1)
2
Exp. (a)
We have,
⇒
⇒
Exp. (d)
I=
π/2
0
π
2
Let
⇒
dx
π
(a)
8
 2 x + 1
sin2 x x
 dx
 2 + 1
−π / 2
∫
[Q sin2 x is an even function]
sin xcos x
Put
⇒
π/2
−π / 2
2
∫ (sin3 x + cos3 x)2 dx = ∫ cos6 x(1 + tan3 x)2 dx
− x
π/2
⇒
sin2 xcos 2 x
∫ {sin3 x(sin2 x + cos2 x) + cos3 x(sin2 x + cos2 x)}2 dx
2
b
 b

Q ∫ f( x)dx = ∫ f(a + b − x)dx
 a

a
+ cos 5 x )2
is equal to
+
2
18 x2 − 9 πx + π 2 = 0
18 x − 6 πx − 3 πx + π 2 = 0
⇒
Now,
∴
Given,
2
(6 x − π )(3 x − π ) = 0
π π
x= ,
6 3
α<β
π
π
α = ,β =
6
3
g ( x) = cos x2 and f( x) =
x
183
Integral Calculus
y = gOf( x)
∴
y = g (f( x)) = cos x
Area of region bounded by x = α,x = β, y = 0
and curve y = g (f( x)) is
A=
Exp. (b)
We have, In = ∫ tann x dx
∴ In + In +
π/3
2
∫ cos xdx
π
π
3 1
− sin =
−
3
6
2
2
 3 − 1
A=

 2 
tan5 x
+C
5
1
a = and b = 0
5
∴
{( x , y ): x ≥ 0, x + y ≤ 3, x ≤ 4y and y ≤ 1+ x }
is
[JEE Main 2017 (offline)]
2
7
(c)
3
5
(d)
2
56.
3 π /4
∫π /4
dx
is equal to
1 + cos x
[JEE Main 2017 (offline)]
(a) − 2
(b) 2
x ) dx +
2
2
∫1 (3 − x) dx − ∫0
∫π / 4
=
∫π / 4
=
∫π / 4
x2
dx
4
Y
y=1+√x
(0, 3) (1, 2)
4y=x 2
(0, 1)
57. The integral ∫
X
(a)
1
2
2
x 


x 
x 
= x +
 − 
 + 3 x −
3
/
2
2
 1  12  0
0 

2
3
1
8
2
=  1 +  +  6 − 2 − 3 +  −  

2   12 
3 
5 3 2
3 5
= + − = 1 + = sq units
3 2 3
2 2
(c)
2x
12
dx
2 )] = 2
+ 5x
9
( x + x 3 + 1)3
5
+C
x5
2( x + x + 1)
dx is equal to
5
3
2
+C
(b)
(d)
x10
2( x 5 + x 3 + 1)2
− x10
2( x + x 3 + 1)2
5
where, C is an arbitrary constant.
Exp. (b)
Let
5
1
(b)  , 0
5 
1
(d)  − , 0
 5 
2 ) − (− 1 +
( x 5 + x 3 + 1)2
I 4 + I 6 = a tan x + bx + C , where C is a
constant of integration, then the ordered
pair (a ,b ) is equal to
[JEE Main 2017 (offline)]
1
(a)  − ,1
 5 
1
(c)  , − 1
5

1 − cos 2 x
(cosec2 x − cosec x cot x)dx
− x5
55. Let I n = ∫ tann x dx (n > 1). If
5
∫π / 4
[JEE Main 2016 (offline)]
Y′
3/ 2
3π / 4
= [(1 +
(2, 1)
(0, 0) (1, 0)(2, 0) (3, 0)
3π / 4
1 − cos x
3π / 4
dx
=
1 + cos x
1 − cos x
dx
sin2 x
= [− cot x + cosec x]3π π/ 4/ 4
x+y=3
X′
3π / 4
Let I =
Required area
(1 +
(d) −1
(c) 4
Exp. (b)
Exp. (d)
1
tann + 1 x
+C
n+1
Put n = 4, we get I4 + I6 =
54. The area (in sq units) of the region
3
(b)
2
2
= ∫ tann x sec 2 x dx =
A = sin
∫0
x dx
= ∫ tan x(1 + tan x) dx
A = [sin x]ππ // 36
=
∫ tan
n
π /6
59
(a)
12
n+ 2
= ∫ tann x dx +
2 x12 + 5 x9
I=
∫ ( x5 +
=
∫ x15 (1 +
=
∫ (1 +
x3 + 1) 3
dx
2 x12 + 5 x9
x− 2 + x− 5 ) 3
2 x− 3 + 5 x− 6
x− 2 + x− 5 ) 3
dx
Now, put 1 + x− 2 + x− 5 = t
⇒
⇒
(− 2 x− 3 − 5 x− 6 ) dx = dt
(2 x− 3 + 5 x− 6 ) dx = − dt
dx
+C
+C
184
JEE Main Chapterwise Mathematics
I=−
∴
=−
dt
∫t3
= − ∫ t − 3 dt
−3+1
x
=
2 ( x5 + x3 + 1) 2
+C
{( x , y ): y 2 ≥ 2 x and x 2 + y 2 ≤ 4x , x ≥ 0, y ≥ 0} is
[JEE Main 2016 (offline)]
+c
⇒
x2 + 2 x = 4 x
⇒
x = 0 or x = 2
⇒
y = 0 or y = ± 2 [using Eq. (i)]
Now, the required area is the area of shaded
region, i.e.
1
x4
dx
x5
log x 2
∫2 log x 2 + log(36 − 12 x + x 2 )dx
4
is equal to
(a) 2
[JEE Main 2015]
(b) 4
(c) 1
(d) 6
Central Idea Apply the property
b
b
∫a f( x)dx = ∫a f(a +
Let
I=
b − x) dx and then add.
log x2
4
∫2 log x2 + log(36 − 12 x +
=∫
x2 )
dx
2 log x
4
dx
x + log(6 − x)2
4
2 log xdx
=∫
2 2 [log x + log( 6 − x )]
4
log xdx
I=∫
2 [log x + log( 6 − x)]
4
log(6 − x)
dx
I=∫
2 log( 6 − x) + log x
2 2 log
2
2
X
⇒
y2=2x
⇒
Y′
Area of circle
−
4
2
∫0
Q bf( x)dx =
 ∫a
2 x dx
2
2
 x3 / 2 
π(2 )2
− 2 ∫ x1/ 2dx = π − 2 

0
4
3/2  0
= π−
3
1 4
x5  1 + 4 

x 
4
= t 4 ⇒ − 5 dx = 4t 3dt
x
−t 3dt
= − t 3dt = ∫ 3 = − ∫ dt = − t + c
t
60. The integral
x +y =4x
=
dx
∫
1 4
= −  1 + 4  + c

x 
Y
Required area =
+c
Exp. (c)
x2 = 2 x
B (2,0)
=
1/ 4
1
which is a circle with centre (2, 0) and radius
...(ii)
=2
On substituting y2 = 2 x in Eq. (ii), we get
(0, 0)
 x 4 + 1
(d) − 

4
 x 
3
1)4
x2 ( x4 +
which is a parabola with vertex (0, 0) and axis
parallel to X-axis.
...(i)
And x2 + y2 = 4 x
X′
(b) ( x 4 + 1)1/ 4 + c
dx
Put 1 +
Given equations of curves are
y2 = 2 x
1/ 4
Exp. (d)
Exp. (b)
A (2,2)
equals
3
+ 1)4
(c) −( x 4 + 1)1/ 4 + c
∫
8
(b) π −
3
π 2 2
(d) −
2
3
⇒
4
[JEE Main 2015]
 x 4 + 1
(a) 

4
 x 
58. The area (in sq units) of the region
4
(a) π −
3
4 2
(c) π −
3
2
x (x
t
1
+C= 2 +C
−3+1
2t
10
dx
59. The integral ∫
2 2
8
[2 2 − 0] =  π −  sq units

3
3
b
⇒
4
2 I = ∫ dx = [ x]42
2
2I = 2 ⇒ I = 1
…(ii)

∫a f(a + b − x)dx
On adding Eqs. (i) and (ii), we get
4log x + log( 6 − x)
dx
2I = ∫
2 log x + log( 6 − x)
⇒
…(i)
185
Integral Calculus
61. The area (in sq units) of the region described
by {( x , y ): y ≤ 2 x and y ≥ 4x − 1} is
2
[JEE Main 2015]
7
(a)
32
15
(c)
64
5
(b)
64
9
(d)
32
Exp. (d)
Given region is {( x, y) : y2 ≤ 2 x and y ≥ 4 x − 1}
y2 ≤ 2 x represents a region inside the parabola
y = 2x
2
…(i)
and y ≥ 4 x − 1represents a region to the left of the
line
…(ii)
y = 4x − 1
The point of intersection of the curve (i) and (ii) is
(4 x − 1)2 = 2 x
2
⇒
16 x + 1 − 8 x = 2 x
⇒
16 x2 − 10 x + 1 = 0
1 1
x= ,
⇒
2 8
∴ The points where these curves intersect, are
 1 ,1 and  1 , − 1  .




2 
 8 2
62.
1
x
1 x +
e
x

The integral ∫ 1 + x −

is equal to
[JEE Main 2014]
(a) ( x −1)e
(c) ( x + 1)e
x+
1
x
x+
1
x
+C
(b) x e
+C
x+
(d) − xe
1
x
x+
+C
1
x
+C
1 + x − 1  e x +



x
= ∫e
=
∫
x+
1
x
1
x+
e x
= ∫e
=
π
∫0
1 + 4 sin 2
x
x
− 4 sin dx is
2
2
equal to
[JEE Main 2014]
(a) π − 4
2π
(b)
− 4− 4 3
3
(c) 4 3 − 4
π
(d) 4 3 − 4 −
3
Exp. (d)
π
∫0
=
2
 1 − 2 sin x  dx = π |1 − 2 sin x|dx


∫0

2
2
π

x
π
∫03  1 − 2 sin 2  dx = ∫π
3
π
 1 − 2 sin x  dx



2

 x − a, x ≥ a 
Q| x − a|= −( x − a), x < a



x
x 3
x
=  x + 4cos  −  x + 4cos 
2  0 
2  π

= 4 3 − 4−
3
π
3
64. The area of the region described by
A = {( x , y ): x 2 + y 2 ≤ 1and y 2 ≤ 1 − x } is
π 4
+
2 3
π 2
(c)
−
2 3
π 4
−
2 3
π 2
(d)
+
2 3
(a)
(b)
[JEE Main 2014]
Exp. (a)
A = {( x, y): x2 + y2 ≤ 1 and y2 ≤ 1 − x}
Exp. (b)
∫
63. The integral
x+
1
x
1
x+
xe x
1
x
dx +
dx +
Y
dx

1
∫ x  1 − x2  e
1
x+
xe x
dx + xe
x+
−∫
1
x
d
dx
− ∫e
x+
1
x
dx
x+
(–1, 0)
1
x
1
1

x+ 
1  x+

Q∫  x − 2  e x = e x 
x 

 
+C
X′
(0, 1)
X
1
x+
( x)e x
Y′
Required area =
1
1 2
πr + 2 ∫ (1 − y2 )dy
0
2
1
=

y3 
π 4
1
π(1)2 + 2  y −  = +
2
3
2
3

0
186
JEE Main Chapterwise Mathematics
65. If ∫ f ( x ) dx = ψ( x ), then ∫ x 5 f ( x 3 )dx is equal
⇒ I=
to
[JEE Main 2013]
π/3
tan x dx
∫π / 6 1 +
…(ii)
tan x
1 3
[x ψ( x 3 ) − ∫ x 2 ψ( x 3 )dx ] + C
3
1
(b) x 3 ψ( x 3 ) − 3 ∫ x 3 ψ( x 3 ) dx + C
3
1 3
(c) x ψ( x 3 ) − ∫ x 2 ψ( x 3 ) dx + C
3
1
(d) [x 3 ψ( x 3 ) − ∫ x 3 ψ( x 3 ) dx ] + C
3
∴
Exp. (c)
statement by property of definite integrals.
On adding Eqs. (i) and (ii), we get
(a)
2I =
∫
I=
Put
x3 = t
x2dx =
⇒
But
…(i)
Statement II
b
∫a
Exp. (a)
Given curves are
y= x
and
2y − x + 3 = 0
On solving Eqs. (i) and (ii), we get
2 x − ( x )2 + 3 = 0
Y
x
y=√
b
f ( x ) dx = ∫ f (a + b − x ) dx
a
–
2y
X′
∴
Exp. (d)
∴ Required area =
∴
I=
π/3
π/3
∫π / 6
dx
tan x
dx
1+
…(i)
x+
3=
0
X
3
(a) Statement I is true Statement II is true;
Statement II is a correct explanation of
Statement I
(b) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
∫π / 6 1 +
…(i)
…(ii)
( x )2 − 2 x − 3 = 0
⇒
[JEE Main 2013]
Let I =
is a true
(b) 36
27
(d)
4
(c) 18
is equal to π/6 .
tan x
b
(a) 9
dt
3
66. Statement I The value of the integral
dx
b
∫a f( x) dx = ∫a f(a + b − x) dx
y = x , 2 y − x + 3 = 0, X -axis and lying in the
first quadrant is
[JEE Main 2013]
1
1
t f( t ) dt = [t ψ( t ) − ∫ ψ(t ) dt ]
3∫
3
[using integration by parts]
1
= [ x3 ψ( x3 ) – 3 ∫ x2 ψ( x3 ) dx] + C
3
[from Eq. (i)]
1 3
3
2
3
= x ψ( x ) − ∫ x ψ( x ) dx + C
3
π /3
 π − π = π
 3 6  12
67. The area (in sq units) bounded by the curves
x5 f( x3 ) dx
∴I=
∫π /6 1 +
1
2
2 I = [ x]ππ // 36
⇒
Hence, Statement I is false.
Given, ∫ f( x) dx = ψ( x)
Let
I=
π/3
∫π / 6 dx
–3
2
Y′
⇒
⇒
( x − 3) ( x + 1) = 0
x=3
[Q x = − 1is not possible]
y=3
=
3
∫0 ( x2 − x1 ) dy
3
∫0 {(2 y + 3) − y } dy
2
3
π/3
dx
=∫
π
/
6
cot x
1
+
π
tan  − x
2


y3 
=  y2 + 3 y −

3 0

=9+ 9−9=9
187
Integral Calculus
Now, by comparing the value of I in Eqs. (i) and (ii),
we get
a=2
68. If the integral
5 tan x
∫ tan x − 2 dx = x + a log
| sin x − 2 cos x | + k ,
then a is equal to
(a) –1
(b) –2
[AIEEE 2012]
(c) 1
(d) 2
69. The area bounded between the parabolas
x2 =
y
and x 2 = 9y and the straight line y = 2,
4
is
Exp. (d)
5 tan x
Given Integral ∫
dx
tan x − 2
To find The value of ‘ a’, if
[AIEEE 2012]
10 2
(b)
3
(a) 20 2
5 tan x
∫ tan x − 2 dx
= x + a log| sin x − 2 cos x| + k
20 2
(c)
3
Exp. (c)
…(i)
Now, let us assume that
5 tan x
dx
I=∫
tan x − 2
On multiplying by cos x in numerator and
denominator, we get
5 sin x
dx
I=∫
sin x − 2 cos x
y
and x2 = 9 y
4
To find The area bounded between the parabolas
and the straight line y = 2 .
The required area is equal to the shaded region in
the drawn figure.
Given Two parabolas x2 =
Y
y = 4x 2
1
y = x2
9
y=2
This special integration requires special
substitution of type
 dDr 
Nr = A(Dr ) + B 
 , A and B are constants.
 dx 
Let
5 sin x = A (sin x − 2 cos x)
+ B (cos x + 2 sin x)
⇒ 0 cos x + 5 sin x = ( A + 2 B) sin x
+ (B − 2 A) cos x
On comparing the coefficients of sin x and cos x,
we get
A + 2 B = 5 and B − 2 A = 0
On solving the above two equations in A and B, we
get
A = 1 and B = 2
⇒ 5 sin x = (sin x − 2 cos x) + 2 (cos x + 2 sin x)
5 sin x
⇒ I=∫
dx
sin x − 2 cos x
(sin x − 2 cos x) + 2 (cos x + 2 sin x)
dx
=∫
(sin x − 2 cos x)
sin x − 2 cos x
(cos x + 2 sin x)
dx + 2 ∫
dx
=∫
sin x − 2 cos x
(sin x − 2 cos x)
d (sin x − 2 cos x)
= ∫ 1 dx + 2 ∫
(sin x − 2 cos x)
= x + 2 log| (sin x − 2 cos x)| + k
where, k is the constant of integration.
(d) 10 2
…(ii)
X
The area of the shaded region (which can be very
easily found by using integration) is twice the area
shaded in first quadrant.
∴Required area
25
2
y
y  dy
= 2 ∫ 3 y −
 dy = 2 ∫ 
0 
0 2

2 
2
 y3 / 2 
20 2
10 3 / 2
 =
= 5 
(2
− 0) =
3
3
 3 /2  0
x
70. If g ( x ) = ∫ 0 cos 4t dt , then g ( x + π ) is equal to
[AIEEE 2012]
(a)
g (x )
g ( π)
(b) g ( x ) + g ( π )
(c) g ( x ) − g ( π )
(d) g ( x ) ⋅ g ( π )
Exp. (b, c)
Given Integral g ( x) =
x
∫ 0 cos 4t dt
To find g ( x + π ) in terms of g ( x) and g( π ).
g ( x) =
x
∫ 0 cos 4t dt
188
JEE Main Chapterwise Mathematics
⇒
t = x+ π
∫t = 0
g( x + π ) =
=
On adding Eqs. (i) and (ii), we get
cos 4t dt
x+ π
x
∫ 0 cos 4t dt + ∫ x
cos 4t dt
= g ( x) + I1
I1 =
where,
=
x+ π
∫x
[say]
2I = 8 ∫
⇒
cos 4t dt
∫ 0 cos 4t dt
π
sin 4t 
 sin 4 π − sin 0  = 0
I1 = 
 =

 4 0  4
4 
i.e., the value of I1 is zero.
Since, g ( π ) = I1 = 0, so value of g ( x + π ) does not
depend on g( π ) and we can add or subtract g ( π )
to or from g ( x).
Hence, g ( x + π ) = g ( x) + g ( π )
or
g ( x) − g ( π )
1 8 log (1 +
71. The value of ∫ 0
1+ x
2
x)
dx is [AIEEE 2011]
π
π
(a) log 2 (b) log 2 (c) log 2
8
2
(d) π log 2
Exp. (d)
Let I =
1
∫0
8 log (1 + x)
(1 + x2 )
I=4∫

log {1 + tan θ} + log

π/4
0
log 2 d θ
π
= 4 log 2 ⋅  − 0
4

= π log2
72. The area of the region enclosed by the
curves y = x , x = e , y =
1
and the positive
x
X-axis is
[AIEEE 2011]
3
(b) sq units
2
1
(d) sq unit
2
(a) 1 sq unit
(c)
5
sq units
2
Exp. (b)
1
Given, y = x, x = e and y = , x ≥ 0
x
Since, y = x and x ≥ 0 ⇒ y ≥ 0
∴ Area to be calculated in I quadrant shown as
dx
Y
y=x
Put x = tan θ ⇒ dx = sec 2 θd θ
When x = 0 ⇒ tan θ = 0
∴
θ=0
When
x = 1 = tanθ
π
θ=
⇒
4
π / 4 8log (1 + tanθ)
⋅ sec 2 θd θ
I=∫
∴
0
1 + tan2 θ
=8∫
I=8∫
a
π/4
0
π/4
0
Using ∫ f( x) dx =
0
∴
A
B
log (1 + tan θ) d θ
D (1, 0)
…(i)
=8∫
0
π/4
1
dx
x
1
(1 × 1) +
2
π
log 1 + tan  − θ  d θ


4

=
1
+ [log| x|]1e
2

1 − tan θ 
log 1 +
 dθ
1
+ tan θ 

=
1
+ {log|e| − log 1}
2
=
1
3
+ 1 = sq units
2
2
π/4
0
∴ Required area = Area of ∆ ODA + Area of
ABCD
=
a
=8∫
X
C (e, 0)
x=e
∫ 0 f(a − x) dx
0
y = 1/x
O
log (1 + tan θ) d θ
I=8∫
 

2
 d θ

1
+
tan
θ
 

= 4 ⋅ log 2(θ)π0 / 4
π
[using definite integral property]
= g( π )
Also,
π/4
0
π/4


2
log 
 dθ
1 + tan θ 
…(ii)
e
∫1
189
Integral Calculus
73. The area bounded by the curves y 2 = 4x and
x = 4y is
2
[AIEEE 2011]
(a) 0
(b)
32
3
(c)
16
3
(d)
f ( 3x )
= 1 , p ′ ( x ) = p ′ (1 − x ), for all
f (x )
x ∈[0 , 1], p (0) = 1 and p (1) = 41. Then,
1
∫ p (x )dx is equal to
that lim
x→ ∞
8
3
Exp. (c)
[AIEEE 2010]
0
For the point of intersection, solve y2 = 4 x and
x2 = 4 y.
2
 x2 
  = 4 x ⇒ x4 = 43 x ⇒ x = 0, 4
 4
⇒
75. Let p ( x ) be a function defined on R such
(b) 21
(d) 42
(a) 41
(c) 41
Exp. (b)
We have,
p′( x) = p′(1 − x), ∀ x ∈ [0, 1],
Y
x 2 = 4y
y 2 = 4x
p(0) = 1, p(1) = 41
[given]
On integrating, we get
(4, 4)
∫ p′( x) dx = ∫ p′(1 − x) dx
X′
⇒
X
(0, 0) D
(4, 0)
p( x) = − p(1 − x) + C
Put x = 1;
⇒
41 = − 1 + C
⇒
C = 42
∴
Y′
∴ Area bounded between curves
p(1) = − p(0) + C
p( x) + p(1 − x) = 42
I=
∫0 p( x)dx
⇒
I=
∫0 p(1 − x)dx Q ∫ 0 f( x) dx = ∫0f(a − x) dx
⇒
2I =
4
=
4
∫0

x2 
 dx =
 4 x −
4 



 x3 / 2
x3 
2
⋅
−

3
12 

0

2
3
(4)
4
32 16 16
= ⋅ (4)3 / 2 −
=
−
=
12
3
3
3
3
=
74. Let [ ] denotes the greatest integer function,
then the value of ∫
(a)
5
4
1. 5
x[ x 2] dx is
0
(b) 0
(c)
3
2
[AIEEE 2011]
(d)
3
4
Exp. (d)
Here, ∫
1. 5
0
∴
1
1
∫ 0 x ⋅ 0 dx + ∫ 1
2
x ⋅ 1 dx +
∫
1. 5
2
x ⋅ 2 dx

a
a
1
∫0 [( p( x) + p(1 − x)]dx
[on adding]
1
I = 21
76. The area bounded by the curves y = cos x
and y = sin x between the ordinates x = 0 and
3π
x=
, is
2
[AIEEE 2009]
(b) ( 4 2 + 2 ) sq units
(d) ( 4 2 + 1 ) sq units
Exp. (a)
2
 x2 
= 0 +   + [ x2 ]1. 52
 2 1
1
. )2 − 2}
= {2 − 1} + {(15
2
1
1 1 3
= + 2 .25 − 2 = + =
2
2 4 4

∫0 42 dx = 42
(a) ( 4 2 − 2 ) sq units
(c) ( 4 2 − 1 ) sq units
x [ x2 ] dx
I=
1
Now,
π
O
π/2
3π/2
190
JEE Main Chapterwise Mathematics
∴Required area
=
π /4
∫0
5π / 4
∫π / 4
(cos x − sin x) dx +
+∫
Exp. (b)
(sin x − cos x)dx
3π / 2
5π / 4
(cos x − sin x)dx
The equation of tangent at (2, 3) to the given
parabola is
x = 2y − 4
Y
= [sin x + cos x]π0 / 4 + [− cos x − sin x[5π π/ 4/ 4
(2, 3)
+ [sin x + cos x]35 ππ // 24
 1
1   −1
1 
1
1
= 
+
− 1 +  
+
−
 −

 2
  2
2  2
2  
2
X′
−1

1 
+ (− 1 + 0) − 
−


2
2  

Y′
=
= (4 2 − 2 ) sq units
77.
π
∫0
π
2
[AIEEE 2009]
(b) 1
(d) −
(c) −1
1 cos x
dx and J = ∫
dx . Then,
0
x
x
which one of the following is true?
1
79. If I = ∫ 0
π
2
π
Let
I=
∫0
⇒
I=
∫0
=
π
π
∫0
…(i)
[cot x] dx
[cot ( π − x)] dx
[− cot x] dx
…(ii)
On adding Eqs. (i) and (ii), we get
π
∫0
π
∫0
[cot x] dx +
(− 1) dx
π
∫0
[− cot x] dx

− 1 if x ∉ Z 
Q[ x] + [− x] =  0,if x ∈ Z 



= [− x] π0
∴
sin x
[AIEEE 2008]
Exp. (d)
=
{( y − 2 )2 + 1 − 2 y + 4}dy
3
integer function, is equal to
2I =
3
∫0
 ( y − 2 )3

=
− y2 + 5 y
3

0
8
1
= − 9 + 15 + = 9 sq units
3
3
[cot x ] dx , where [] denotes the greatest
(a)
(y – 2)2 = (x – 1)
∴ Required area
8
−2
2
=
X
(– 4, 0)
=− π
π
I=−
2
78. The area of the region bounded by the
2
(a) I > and J < 2
3
2
(b) I > and J > 2
3
2
(c) I < and J < 2
3
2
(d) I < and J > 2
3
Exp. (c)
Since,
1
∫0
sin x
dx <
x
x
dx,
x
1
∫0
because in x ∈ (0, 1), x > sin x
1
2
⇒
I< ∫
x dx = [ x3 / 2 ]10
0
3
2
⇒
I<
3
parabola ( y − 2 )2 = x − 1, the tangent to
the parabola at the point(2 , 3) and the X-axis
is
[AIEEE 2009]
and
(a) 6 sq units
(c) 12 sq units
∴
(b) 9 sq units
(d) 3 sq units
I=
J=
1
∫0
cos x
dx <
x
= 2 [ x1/ 2 ]10 = 2
J<2
1
∫0
x
−
1
2
dx
191
Integral Calculus
80. The area of the plane region bounded by
the curves x + 2 y = 0 and x + 3y = 1 is equal
to
[AIEEE 2008]
2
2
4
sq units
3
1
(c) sq unit
3
5
sq units
3
2
(d) sq unit
3
(a)
(b)
Let
I= 2
∴
Given curves are
x + 3 y2 = 1
…(i)
x + 2 y2 = 0
…(ii)
sin x
dx
π

sin  x − 

4
 1 cot t + 1  dt
2
2 
∫ 
= 2
= log| sin t | + t + C
π
sin  x −  
+ C
= x + log 

4 

On solving Eqs. (i) and (ii), we get
y = ± 1 and x = − 2
1
82. If F ( x ) = f ( x ) + f   ,
 
Y
(–2, 1)
2
x + 3y = 1
x + 2y 2 = 0
X′
X
where f ( x ) = ∫
x
log t
dt . Then, F (e ) is equal
1 +t
x
1
to
(–2, –1)
[AIEEE 2007]
1
(a)
2
Y′


∴Required area = 
( x1 − x2 ) dy
∫−1

1
2
2



=
(
1
−
3
y
+
2
y
)
dy
 ∫−1

1
1
2







= ∫ (1 − y ) dy = 2 ∫ (1 − y2 ) dy
 −1
  0

[since, integral is an even]
1

1 
y3 
 
= 2 y −
 =  2  1 −  
3 
3


1
0
4
= sq units
3
81. The value of 2 ∫
∫
π
= t ⇒ dx = dt
4
π
sin  + t  dt
4

I= 2 ∫
sin t
Put x −
Exp. (a)
and
Exp. (d)
sin x dx
is
π

sin  x − 

4
[AIEEE 2008]
π
(a) x − log cos  x −  + C

4
π
(b) x + log cos  x −  + C

4
π
(c) x − log sin  x −  + C

4
π
(d) x + log sin  x −  + C

4
(b) 0
(c) 1
(d) 2
Exp. (a)
x
log t
1
dt and F(e ) = f(e ) + f  
e
1+ t
e
log t
dt +
1+ t
Since,
f ( x) =
∫1
⇒
F(e ) =
∫1
1/ e
∫1
log t
dt
1+ t
1
in second integration
t
e log t
e − log t
1
d 
dt + ∫
∴ F(e ) = ∫
1 1+ t
1
1  t 
1+
t
e log t
e t log t
− dt
=∫
×  2 
dt − ∫
1 1+ t
1 (1 + t )
 t 
Put t =
=
∫1
e
log t
dt +
1+ t
∫ 1 t (1 + t ) dt
=
∫1
e
log t
dt +
1+ t
∫1
=
∫1
e
log t
dt
t
e
e
e
log t
e log t
log t
dt
dt − ∫
1 (1 + t )
t

1
1 1 
Qt (1 + t ) = t − t + 1


 (log t )2 
=

 2 1
1
1
= [(log e )2 − (log 1)2 ] =
2
2
192
JEE Main Chapterwise Mathematics
Exp. (c)
83. The solution for x of the equation
∫
x
2
dt
t t −1
2
=
π
is
2
(a) − 2
3
(c)
2
Required area,
[AIEEE 2007]
1
∫0 (
A=
x − x) dx
(b) π
Y
y=|x|
(d) 2 2
y2 = x
Exp. (a)
π
2
t t −1
π
[sec −1 t ] x =
2
2
π
sec −1 x − sec −1 2 =
2
π
π 3π
sec −1 x = +
=
2
4
4
3π
x = sec
=− 2
4
∫
Given,
⇒
⇒
⇒
∴
84.
∫
dt
x
2
2
dx
cos x + 3 sin x
=
X′
Y′
is equal to
Let
⇒
dx
85. The area enclosed between the curves y 2 = x
2
sq unit
3
1
(c) sq unit
6
[AIEEE 2007]
(b) 1 sq unit
(d)
(b) 2
(d) 1/2
[AIEEE 2006]
x
9− x +
6
I=
∫3
=
∫3
I=
∫3
x
…(i)
dx
9− x
6
9−9+ x +
9− x
6
x+
9− x
9− x
dx
…(ii)
dx
On adding Eqs. (i) and (ii), we get
dx
cos x + 3 sin x

1
3
sin x
2  cos x +
2

2
π
1

= ∫ sec  x −  dx

2
3
x π
π
1

= log tan  − +  + C
2 6
2
4
1
x
π
= log tan  +
 +C
 2 12 
2
(a)
dx is
Exp. (a)
Exp. (a)
and y = | x | is
9− x + x
(a) 3/2
(c) 1
[AIEEE 2007]
∫
x
6
86. The value of ∫ 3
π
1
x
log tan  +  + C
 2 12 
2
x π
1
(b) log tan  −  + C
 2 12 
2
π
x
(c) log tan  +  + C
 2 12 
x π
(d) log tan  −  + C
 2 12 
=
1
2
x 
2 1 1
=  x3 / 2 −
 = − = sq unit
3
2
0 3 2 6

2
(a)
Now, ∫
X
(1, 0)
O
1
sq unit
3
∴
87.
6
2I =
∫3
I=
3
2
− π /2
∫− 3π /2 [(x + π )
3
x+
9− x
x+
9− x
dx = [ x]63 = 6 − 3
+ cos 2 ( x + 3π )] dx is equal to
 π4   π 
(a) 
+ 
 32   2 
(b)
π
(c)   − 1
 4
π4
(d)
32
π
2
[AIEEE 2006]
Exp. (b)
−π / 2
Let I =
∫ − 3 π / 2 [( x +
⇒ I=
∫ − 3 π / 2 [( x +
−π / 2
π )3 + cos 2 ( x + 3 π )] dx
π )3 + cos 2 x] dx
…(i)
193
Integral Calculus
−π / 2

π
∫ − 3 π / 2   − 2
⇒ I=
3π
− x − π 

2
−

3
π 3π

+ cos 2  − −
− x dx
 2
 
2
Q b f( x) dx = b f(a + b − x) dx
∫a

 ∫ a
−π / 2
∫ − 3 π / 2 [−( x +
⇒ I=
π ) + cos x] dx
3
2
∫ − 3 π / 2 2 cos
2I =
=
2
88.
π
∫0
∫ − 3 π / 2 (1 + cos 2 x) dx
 − 3 π + sin (−3 π ) 


 2

2
π
0
(c) π ∫
π /2
I=
∫0
⇒
I=
∫0
I=
π
π
2
[AIEEE 2006]
π /2
∫0
π
0
f (sin x ) dx
⇒
2
∫1
f ′( x) dx +
I=
π
∫0
π
2
…(ii)
I= π
π f(sin x) dx
π
∫0
f(sin x) dx
…(iii)
2 af( x) dx, if f(2 a − x) = f( x)

 ∫0
 0
,if f(2 a − x) = − f( x) 
π /2
∫0
π
− x=t
2
π
x= −t
2
3
∫ 2 2 f ′( x) dx
a
∫ [a ] [a]f ′( x) dx
= [f( x)]12 + 2 [f( x)]32 + . . . + [a][f( x)][aa ]
= f(2 ) − f(1) + 2 f(3) − 2 f(2 ) + . . . + [a]f(a) − [a]f([a])
= [a]f(a) − {f(1) + f(2 ) + . . . + f([a])}
1
…(i)
( π − x) f(sin x) dx
 2a
Q ∫ f( x) dx =
 0
Put
Exp. (a)
90. lim  2 sec2 2 + 2 sec2 2 +...+ 2 sec21 is
n → ∞ n
n
n
n
n

f (cos x ) dx
( π − x)f[sin ( π − x)] dx
2I =
⇒
[a ]f (a ) − { f (1 ) + f ( 2 ) + ... + f ([a ])}
[a ]f ([a ]) − { f (1 ) + f ( 2 ) + ... + f (a )}
af ([a ]) − { f (1 ) + f ( 2 ) + ... + f (a )}
af (a ) − { f (1 ) + f ( 2 ) + ... + f ([a ])}
+ ... +
On adding Eqs. (i) and (ii), we get
⇒
(a)
(b)
(c)
(d)
a
x f(sin x) dx
π
∫0
a
∫ 1 [ x]f ′ (x )dx , a > 1, where [ x]
1
2
4
n
equal to
π
Let
f(cos x) dx
[Qf(cos x) is an even function]
1
(d) π ∫
f (cos x ) dx
0
Exp. (c)
⇒
(b)
f (sin x ) dx
π/2
∫0
Since, ∫ [ x]f ′( x) dx =
x f (sin x ) dx is equal to
(a) π ∫
π/2
∫ − π / 2 f(cos x) dx
denotes the greatest integer not exceeding x ,
is
[AIEEE 2006]
−π / 2
−π / 2
∴
π
2
= π
89. The value of
x dx
sin 2 x 
= x +

2  −3 π / 2
 π sin (− π )
= − +
−
 2
2
π 3π
=− +
= π
2
2
π
I=
2
=
…(ii)
On adding Eqs. (i) and (ii), we get
−π / 2
Put dx = − dt in Eq. (iii), we get
π π/2
I= ∫
f(cos t ) dt
2 −π / 2
(a)
[AIEEE 2005]
1
tan 1 (b) tan1
2
(c)
1
1
cosec 1 (d) sec 1
2
2
Exp. (a)
1
1
2
4
Let A = lim  2 sec 2 2 + 2 sec 2 2
n→ ∞  n
n
n
n
n
+ . . . + 2 sec 2 1

n
2
2
1 1
2
2  1
2 2
= lim
 sec   + sec  
n→ ∞ n n
 n


n
n

2
n
n 
+ . . . + sec 2   
 n 
n

f(sin x) dx
= lim
n→ ∞
⇒
A=
1
n
1
n
 r
∑  n  sec
2
r =1
∫ 0 x sec
2
( x2 ) dx
 r
 
 n
2
194
JEE Main Chapterwise Mathematics
⇒ xdx =
x2 = t
Put
∴
A=
1
2
1
∫ 0 sec
2
dt
2
Exp. (a)
t dt =
f( x )
1
1
[tan t ]10 = tan 1
2
2
lim
x→ 2
2
 (log x − 1) 
dx is equal to
2
 1 + (log x ) 
91. ∫ 
x
(a)
(log x )2 + 1
x
(c) 2
+C
x +1
+C
(b)
(d)
[AIEEE 2005]
+C
1 + x2
log x
(log x )2 + 1
+C
d 

x
x
∫ dx (log x)2 + 1 dx = (log x)2 + 1 + C
Alternate Solution
2
Let
 (log x − 1) 
dx
I = ∫
2
1 + (log x) 
⇒
x = et
⇒
t = log x
and
2
Using by parts,
−1
2 t et
1
dt
⋅ 2 t ⋅ et dt − ∫
I=
⋅ et − ∫
2
2 2
(1 + t 2 )2
1+ t
(1 + t )
1+ t2
I=
∴
+
2 t et
2 t et
∫ (1 + t 2 )2 dt − ∫ (1 + t 2 )2 dt
t
e
1+ t2
lim
x→ 2
f(x)
∫6
(a) 18
(c) 36
 form 0 

0 
( x − 2)
f ′( x) = 4{ f(2 )} f ′(2 )
3
Q f(2 ) = 6 and f ′(2 ) = 1 , given


48
= 18
+C=
f (2 ) = 6,
x
(log x)2 + 1
+C
π
(a) 1 − + 2 


4
π


(c)  − 2 + 1
4

π
(b) 1 − − 2 


4
π

(d)  + 2 − 1
4

4t 3
dt is equal to
x −2
According to the given condition,
β
π
∫ π / 4 f( x) dx = β sin β + 4 cos β + 2 β
On differentiating both sides w.r.t. β, we get
π
f( β ) = sin β + β cos β − sin β + 2
4
π
π

∴ f   = 1+ 0 − + 2
2
4
π
= 1− + 2
4
1
1
2
I 1 = ∫ 2 x dx ,
2
2
1
Then,
3
I 2 = ∫ 2 x dx ,
0
0
I 3 = ∫ 2 x dx and I 4 = ∫
1
f ′ (2 ) =   .
 48
(b) 12
(d) 24
function such that the area bounded by the
curve y = f ( x ), X-axis and the ordinates
π
π
and
is
x=
x =β>
4
4
π


 π
β sin β + cos β + 2 β . Then, f   is


2
4
equal to
[AIEEE 2005]
94. If
92. Let f : R → R be a differentiable function
having
4t 3 dt
Exp. (a)
dx = et dt
 t −1 t
(1 + t 2 ) − 2 t t
e dt = ∫
⋅ e dt
I = ∫
2
(1 + t 2 )2
1 + t 
2 tet
et
=∫
⋅ dt − ∫
⋅ dt
2
1+ t
(1 + t 2 )2
et
∫6
93. If f ( x ) is a non-negative continuous
2
=
4{f( x)}3
= lim
x→ 2
1
1
3
= 4 × (6) ×
48
xe x
 (log x − 1) 
(log x)2 + 1 − 2 log x
∫ 1 + (log x)2  dx = ∫ [(log x)2 + 1]2 dx
1
(log x)2 + 1 − 2 x  log x ⋅ 

x  dx
=∫
[(log x)2 + 1]2
∴
∫6
4t 3
dt = lim
x→ 2
x−2
[by Leibnitz’s rule]
Exp. (a)
=
f( x )
2
1
3
2 x dx , then
(a) I 3 > I 4
(b) I 3 = I 4 [AIEEE 2005]
(c) I 1 > I 2
(d) I 2 > I 1
Exp. (c)
Given that,
[AIEEE 2005]
1
I1 =
∫0 2
I3 =
∫1
2
2
x2
dx
x2
dx
and
1
I2 =
∫0 2
x3
I4 =
∫1
2
x3
2
dx,
dx
195
Integral Calculus
Since,
2
and
2
1
∴
∫0 2
x3
and
∫1
2
x3
⇒
2
x3
<2
x2
for 0 < x < 1
x3
>2
x2
for x > 1
1
dx <
∫0 2
x2
dx >
∫1
2
x2
2
y=4
S1
S2
dx
X′
dx
O
y2 = 4x
S3
x=4
X
I2 < I1 and I4 > I3
95. The area enclosed between the curve
y = log e ( x + e ) and the coordinate axes is
[AIEEE 2005]
(a) 4 sq units
(c) 2 sq units
(b) 3 sq units
(d) 1 sq unit
Y′
1
16
sq units
× 64 =
S1 = S 3 =
12
3
⇒
S2 + S3 =
and
Exp. (d)
Required area, A =
0
∫ 1 − e loge ( x + e ) dx
Put x + e = t ⇒ dx = dt
∴
x2 = 4y
Y
A=
e
∫1
loge t dt
= [t loge t − t ]e1
= (e − e − 0 + 1) = 1sq unit
the square region bounded by the lines
x = 4, y = 4 and the coordinate axes. If S1 , S 2
and S 3 are respectively the areas of these
parts numbered from top to bottom, then
S1 : S 2 : S 3 is equal to
x = –e
Y
32 16
[from Eq. (i)]
−
3
3
16
sq units
S2 =
⇒
3
16 16 16
:
:
= 1: 1: 1
∴ S1 : S 2 : S 3 =
3 3 3
97. The value of ∫ − π
X
Y′
(a) 1 : 1 : 1
(c) 1 : 2 : 3
[AIEEE 2005]
(b) 2 : 1 : 2
(d) 1 : 2 : 1
Exp. (a)
(a) 2 π
=
4
∫ 0 y dx
4
∫0
dx , where a > 0, is
π
(c)
2
[AIEEE 2005]
(d) aπ
Exp. (c)
Let I =
π
cos 2 x
∫ − π 1 + a x dx,
a> 0
…(i)
4
1  x3 
x2
dx =  
4  3 0
4
π
cos 2 x
∫ − π 1 + a− x dx
…(ii)
On adding Eqs. (i) and (ii), we get
π (1 + a x ) cos 2 x
π
2I = ∫
dx = ∫ cos 2 x dx
−π
−π
(1 + a x )
π cos 2 x + 1
= ∫ 
 dx
−π 

2
=
1
2
π

  sin 2 x
+ x
  2
  − π
sin (−2 π )
1  sin 2 π

+ π − 
− π 

 
2  2
2
1
⇒ 2 I = ( π + π)
2
π
∴ I=
2
=
It is clear from the figure that
S1 = S 3 =
1+ax
π
(b)
a
I=
1–e O
cos 2 x
Put x = − x, we get
(0, 1)
X′
∫0
4
 x3 / 2 
4
4 x dx = 2 
 = ×8
3
2
/
0 3

S2 =
⇒
π
96. The parabolas y 2 = 4x and x 2 = 4y divide
4
…(i)
196
JEE Main Chapterwise Mathematics
sin x
∫ sin (x − α ) dx = Ax + B log sin(x − α ) + C ,
98. If
then the value of ( A , B ) is
(a)
(b)
(c)
(d)
[AIEEE 2004]
(sin α ,cos α )
(cos α , sin α )
( − sin α ,cos α )
( − cos α , sin α )
dx
∫ cos x − sin x is equal to
1
2
1
2
1
2
1
2
(a)
(b)
(c)
Exp. (b)
Let
99.
I=
∫
sin x
dx
sin ( x − α )
(d)
Put x − α = t ⇒ dx = dt
sin (t + α )
dt
I=∫
sin t
sin t cos α + cos t sin α
dt
I=∫
⇒
sin t
∴
[AIEEE 2004]
x π
log tan  −  + C
2 8
x
log cot   + C
 2
x 3π 
log tan  −
 +C
2
8 
x 3π 
log tan  +
 +C
2
8 
Exp. (d)
Let I =
dx
cos x − sin x
∫
=
1
2
∫
dx
 1 cos x − 1 sin x


 2

2
=
1
2
∫
dx
1
=
π

2
cos  x + 

4
[let C = − α cos α + C1]
But I = Ax + B log sin ( x − α ) + C
=
1
tan  π + x + π  
+ C
log 
4 2
2
8 

∴ x cos α + sin α log sin ( x − α ) + C
= Ax + B log sin ( x − α ) + C
On comparing both sides, we get
A = cos α and B = sin α
=
1
x 3 π 
tan  +
log 
 + C
2
8 
 2
∫
∫
cos t
dt
sin t
⇒
I = cos α
⇒
⇒
I = cos α(t ) + sin α log sin t + C1
I = cos α( x − α ) + sin α log sin ( x − α ) + C1
⇒
I = x cos α + sin α log sin ( x − α ) + C
1 dt + sin α
3
100. The value of ∫ − 2 |1 − x 2| dx is
Alternate Solution
Given that,
sin x
∫ sin ( x − α ) dx = Ax + B log sin ( x − α ) + C
28
(a)
3
7
(c)
3
On differentiating both sides w.r.t. x, we get
sin x
cos ( x − α )
= A+ B
sin ( x − α )
sin ( x − α )
Exp. (a)
⇒
⇒
⇒
3
sin x = sin x ( A cos α + B sin α )
+ cos x (B cos α − A sin α )
On comparing the coefficients of sin x and cos x
both sides, we get
…(i)
A cos α + B sin α = 1
and B cos α − A sin α = 0
…(ii)
On solving Eqs. (i) and (ii), we get
A = cos α,
B = sin α
[AIEEE 2004]
14
(b)
3
1
(d)
3
−1
∫ − 2 |1 − x | dx = ∫ − 2 ( x
sin x = A sin ( x − α ) + B cos ( x − α )
sin x = A(sin x cos α − cos x sin α )
+ B(cos x cos α + sin x sin α )
π
sec  x +  dx

4
∫
2
+
−1
2
− 1) dx
1
∫ −1 (1 − x
1
 x3


x3 
=
− x +  x −
 +
3  −1
 3
 −2 
2
) dx +
3
∫1
( x2 − 1) dx
3
 x3

− x

 3
1
1
1
8
1
=  − + 1 + − 2  +  1 − + 1 − 
 3
 
3
3
3
1
+  9 − 3 − + 1


3
=
4 4 20 28
+ +
=
3 3
3
3
197
Integral Calculus
π / 2 (sin
101. The value of ∫ 0
x + cos x )2
Exp. (a)
dx is
1 + sin 2 x
[AIEEE 2004]
(a) 0
(b) 1
(c) 2
(d) 3
Given that,
f ( x) =
ex
1+ ex
∴
f ( a) =
ea
1 + ea
…(i)
e −a
1
=
1 + e −a 1 + e a
…(ii)
Exp. (c)
Now, ∫
π / 2 (sin
0
x + cos x)2
dx
1 + sin 2 x
π/2
∫0
=
(sin x + cos x)2
(sin x + cos x)
2
On adding Eqs. (i) and (ii), we get
dx
f ( a) + f ( − a) = 1
⇒
π/2
∫0
=
(sin x + cos x) dx
⇒
Now, I1 =
A is equal to
[AIEEE 2004]
(b) π
(a) 0
Exp. (b)
π
Let
I=
∫0
∴
I=
∫0
⇒
π
(d) 2 π
2I =
∫0
π
∫0
…(i)
xf(sin x) dx
( π − x)f(sin x) dx
2I = π
π
∫0
2I = 2π
⇒
I= π
⇒ A
∴
π /2
∫0
∫0
π /2
∫0
103. If f ( x ) =
and I 2 = ∫
1 + ex
f (a )
f ( −a )
(1 − x)g [ x(1 − x)] dx
…(iv)
On adding Eqs. (iii) and (iv), we get
1−t
∫t
g [ x(1 − x)] dx = I2
, I1 = ∫
[given]
I2
=2
I1
∴
the curves y = | x − 2 |, x = 1, x = 3 and the
X-axis is
[AIEEE 2004]
(b) 2 sq units
(d) 4 sq units
Required area =
3
∫1
ydx =
3
∫ 1 | x − 2| dx
Y
π /2
∫0
f(sin x) dx
π /2
∫0
f (a )
f ( −a )
f(sin x) dx, given

I2
is
I1
[AIEEE 2004]
(c) –1
X′
(d) 1
X
O
x=1
x g [ x (1 − x )] dx
g [ x (1 − x )] dx , then the value
(b) – 3
1−t
∫t

f(sin x) dx
Q I = A

ex
I1 =
b
b
Exp. (a)
f(sin x) dx
f(sin x) dx = π
A= π
⇒
…(iii)
∫a f( x) dx = ∫a f(a + b − x) dx
(a) 1 sq unit
(c) 3 sq units
f(sin x) dx
π /2
xg [ x(1 − x)] dx
104. The area of the region bounded by
…(ii)
( x + π − x) f(sin x) dx
⇒
(a) 2
∫t
f ( a) = 1 − t
Q I =

2 I1 =
( π − x)f[sin ( π − x)] dx
π
I=
(c) π/4
On adding Eqs. (i) and (ii), we get
of
1−t
π /2
102. If ∫ 0 x f (sin x ) dx = A ∫ 0 f (sin x ) dx , then
⇒
f ( a) = 1 − f ( − a)
f ( − a) = t
Let
= [− cos x + sin x]π0 / 2
π
π
= − cos + sin + cos 0 − sin 0
2
2
= − 0 + 1+ 1− 0= 2
π
f ( − a) =
and
x=3
Y′
2
=
∫1
=
∫1
2
− ( x − 2 ) dx +
(2 − x) dx +
3
∫ 2 ( x − 2 ) dx
3
∫ 2 ( x − 2 ) dx
198
JEE Main Chapterwise Mathematics
3
2

 x2

x2 
= 2 x −
− 2 x
 + 
2 1  2

2
1
9


= 4 − 2 −  2 −  +  − 6 − (2 − 4)



2    2
3 3
= 2 − − + 2 = 1sq unit
2 2
n
r =1
(b) e −1
(a) e
n
∑
n→ ∞
r =1
1 r/n
e =
n
1
∫0
e x dx = [e x ]10 = e − 1
t
F (t ) = ∫ f (t − y )g ( y ) dy , then
0
(a) F (t ) = 1 − e
(c) F (t ) = te t
(1 + t )
[AIEEE 2003]
(b) F (t ) = e − (1 + t )
(d) F (t ) = te − t
∫0
−y
⋅ y dy = et
= et (− ye − y ) t0 −

t
∫0
t
∫ 0 1 (− e
−y
) dy

107. If f (a + b − x ) = f ( x ), then
a+b
f ( b − x ) dx
2 ∫a
a+b b
(b)
f ( x ) dx
2 ∫a
b −a b
(c)
f ( x ) dx
2 ∫a
b
a+b
(d)
f (a + b + x ) dx
2 ∫a
b
∫a
x f ( x )dx is
[AIEEE 2003]
b
Let
⇒
x→ 0
sec 2t dt
x sin x
is
[AIEEE
2003]
(b) 2
(d) –1
Exp. (c)
x2
∫0
sec 2 t dt
 form 0 

0 
x sin x
∫ a xf( x) dx
b
I = ∫ (a + b − x) f(a + b − x) dx
a
x2
∫0

sec 2 t dt = sec 2 x2 ⋅ 2 x (Leibnitz’s rule)

0 + 2 sec 2 0
=1
0 + 2 cos 0
1
109. The value of I = ∫ 0 x (1 − x )n dx is
1
(a)
n +1
1
1
(c)
−
n +1 n + 2
I=
Given,
Put
I=
∫0
108. The value of lim
[AIEEE 2003]
1
(b)
n+2
1
1
(d)
+
n +1 n + 2
Exp. (c)
Exp. (b)
b
b
∫ a f( x) dx
x2
=
F( t ) = et − (1 + t )
(a)
a
Again using L’Hospital rule,
2 x ⋅ 2 sec 2 x2 ⋅ tan x2 ⋅ 2 x + 2 sec 2 x2
= lim
x→ 0
− x sin x + cos x + cos x
= et [− te −t − e − t + 1]
equal to
2 I = (a + b ) ∫ f( x) dx
∴
 d
Q
 dx
e − y y dy
= et [(− te − t − 0) − ( e − y )t0 ]
∴
[from Eq. (i)]
b
Using L’Hospital rule,
sec 2 x2 ⋅ 2 x
= lim
x → 0 x cos x + sin x
t
et
b
x→ 0
∫ 0 f(t − y)g( y) dy
t
b
∫ a x f( x) dx
a
a + b
I = 

 2 
lim
Given that, f( y) = e y , g ( y) = y
=
I = (a + b ) ∫ f( x) dx − I
t
Exp. (b)
and F( t ) =
⇒
a
(a) 3
(c) 1
106. If f ( y ) = e y , g ( y ) = y ; y > 0 and
−t
b
I = (a + b ) ∫ f( x) dx −
(d) e + 1
Exp. (b)
Now, lim
b
∫ a (a + b − x)f( x) dx
⇒
⇒
[AIEEE 2004]
(c) 1 − e
I=
[Qf(a + b − x) = f( x), given]
1
∑ n er /n is equal to
n→∞
105. lim
⇒
…(i)
∴
1
∫ 0 x(1 − x)
n
dx
1 − x = z ⇒ − dx = dz
I=
=
0
∫1
1
(1 − z)zn (− dz)
∫ 0 (1 − z) z
n
b
dz Q ∫ f( x) dx = −
 a
a

∫b f( x) dx
199
Integral Calculus
1
∫0 (z
n
− zn + 1 ) dz
⇒
[F( z)]164 = F(k ) − F(1)
1
⇒
∴
F(64) − F(1) = F(k ) − F(1)
k = 64
 zn + 1 zn + 2 
1
1
=
−
−
 =
1
2
+
+
n
n
n
+
1
n
+
2

0
1 + 2 4 + 34 


4
 + ... + n 
110. lim
n→∞
n5
equal to
curves y = | x − 1| and y = 3 − | x | is [AIEEE 2003]
is
[AIEEE 2003]
(a) 1/30
(c) 1/4
Exp. (d)
1 + 2 3 + 33 


3
 + . . . + n 
n5
4
1
∫0
y
–x
X′
(a) 15
(c) 63
Now,
⇒
Exp. (d)
∫1
x3
dx =
e
x
4
∫1
3 x2
x3
⋅ esin
∴
3 x dx = dz
2
64
∫1
sin z
e
z
y
=
X
3
–
x
AB = 2 and BC 2 = (0 − 2 )2 + (3 − 1)2
=4+ 4=8
BC = 2 2
= 4 sq units
…(i)
dx
x3 = z
⇒
A
(1, 0)
= 2 ×2 2
sin x
= F(k ) − F(1)
Let
O
∴Area of rectangle ABCD = AB × BC
esin x
d
, x> 0
F( x ) =
x
dx
On integrating both sides, we get
∫
1
AB2 = (2 − 1)2 + (1 − 0)2 = 1 + 1 = 2
Given,
3 sin
e
x
=
–
On solving, we get
y = x − 1 and y = 3 − x
⇒
x − 1= 3 − x ⇒ x = 2
and
y= 3−2 ⇒ y=1
⇒
(b) 16
(d) 64
4
y
x
Y′
4
Also,
C (0, 3)
+
B (2, 1)
x dx
∫1
F( x ) =
x> 0
r =1
3
3 sin x 3
e
dx = F (k ) − F (1 ), then one of
x
the possible value of k , is
[AIEEE 2003]
If
x≤ 0
1
(–1, 2) D
3
 e sin x 
d
F (x ) = 
 , x > 0.
dx
 x 
111. Let
=
+
4
 r
x>1
x≤1
Y
3
1
= ∫ x dx − lim
×
0
n→ ∞ n
1
 x5 
1
=  −0=
5
5
 0
1
n
∑  n 
Exp. (c)
=
n→ ∞
1 n  r
1
1
× lim
∑   − nlim
→ ∞ n n→ ∞ n
n r = 1 n 
(b) 3 sq units
(d) 6 sq units
y
= lim
(a) 2 sq units
(c) 4 sq units
 x − 1,
Since, y = | x − 1| = 
− x + 1,
3 + x,
and
y = 3 − | x| = 
 3 − x,
(b) 0
(d) 1/5
1 + 2 4 + 34 


4
 + . . . + n 
lim
− lim
n→ ∞
n→ ∞
n5
[from Eq.(i)]
112. The area of the region bounded by the
1 + 2 3 + 33


3
 + ... + n 
− lim
n→∞
n5
x
=
dz = F(k ) − F(1)
x3
dx
113. If f ( x ) is a function satisfying f ′ ( x ) = f ( x )
with f (0) = 1 and g ( x ) is a function that
satisfies f ( x ) + g ( x ) = x 2. Then, the value of
1
∫ 0 f (x )g (x )dx , is
e2 5
−
2 2
e2 3
(c) e −
−
2 2
(a) e −
[AIEEE 2003]
e2 3
−
2 2
e2 5
(d) e +
+
2
2
(b) e +
200
JEE Main Chapterwise Mathematics
Exp. (c)
f ′( x) = f( x) and f(0) = 1
Given,
f ( x) = e x
Let
…(i)
f ( x) + g ( x) = x 2
Also,
⇒
1
1
∫ 0 f( x)g( x) dx = ∫ 0 e
Now,
x
x
…(ii)
( x2 − e x ) dx
=
∫0 x e
2
x
dx −
1
∫0e
2x
dx
1 2x1
[e ]0
2
1
= [ x2e x − 2 xe x + 2 e x ]10 − (e 2 − 1)
2
1
1
= [( x2 − 2 x + 2 )e x ]10 − e 2 +
2
2
1
1
= [(1 − 2 + 2 )e1 − (0 − 0 + 2 )e 0 ] − e 2 +
2
2
e2
e2 3
1
=e −2 −
+ =e −
−
2
2
2
2
= [ x2e
−
x
∫ 2 xe
x
dx]10 −
(a) 20
(b) 8
10 π
2
∫ 0 [x
π
117. ∫ − π
[AIEEE 2002]
(d) 0
π/4
=
∫0
Now, In + In +
2
=
∫0
π/4
[ x2 ] dx +
∫
∫ 0 0 dx + ∫ 1
1
2
= [ x]1 2 + [2 x]
3
2
1 dx +
∫
3
2
∫
2 dx +
tann x dx
tann +
2
2
[ x2 ] dx
3
∫
2
3
3 dx
+ [3 x]2 3
1 + cos 2 x
(b) π
dx is equal to
[AIEEE 2002]
2
(c) 0
(d)
π
2
Exp. (b)
Let I =
π
∫ −π
2 x(1 + sin x)

2x
is an odd function
Q
2
+
x
1
cos

x dx
tann x dx
+
[ x2 ] dx
3− 2
2 x (1 + sin x )
π2
(a)
4
3
2
dx
1 + cos 2 x
π
π
2x
2 x sin x
dx
dx + ∫
=∫
− π 1 + cos 2 x
− π 1 + cos 2 x
π
x sin x
…(i)
⇒ I=0+ 4∫
dx
0 1 + cos 2 x
(b) 1
2
2
=5−
n→∞
In +
∫1
] dx +
= 2 − 1+ 2 3 − 2 2 + 6 − 3 3
is equal to
∴
2
π
tann x dx , then lim n[I n + I n + 2]
π/4
1
∫ 0 [x
0
= 10[1 + 1] = 20
∫0
] dx =
=
= 10[− cos π + cos 0]
In =
2
+
(d) 18
0
Since,
[AIEEE 2002]
(b) 2 + 2
(d) − 2 − 3 + 5
Exp. (d)
π
Exp. (b)
1
1 n
=
[AIEEE 2002]
(c) 10
1
(a)
2
(c) ∞
2
(a) 2 − 2
(c) 2 − 1
= 10 ∫ sin x dx = 10[− cos x]π0
115. If I n = ∫ 0
sec 2 x tann x dx
2
|sin x| dx = 10 ∫ |sin x| dx
π /4
π/4
t n + 1 
1
∫ 0 t dt =  n + 1 = n + 1
0
n
⇒ lim n[In + In + 2 ] = lim
n→ ∞
n→ ∞ n + 1
1
= lim
=1
1
n→ ∞
1+
n
In + In +
Since,|sin x| is a periodic function with period π.
∫0
∫0
sec 2 x dx = dt
Exp. (a)
∴
=
tann x(1 + tan2 x) dx
116. ∫ 0 [ x 2] dx is equal to
10x
114. ∫ 0 |sin x | dx is equal to
⇒
∴
[from Eqs. (i) and (ii)]
1
∫0
tan x = t
Put
g ( x) = x 2 − e
π/4
=
π/4
∫0
and
tann +
2
x dx

is an even function
1 + cos x

2 x sin x
2
201
Integral Calculus
⇒ I=4∫
⇒ I=4∫
π
( π − x) sin ( π − x)
0
1 + cos 2 ( π − x)
π
π sin x
0
1 + cos 2 x
π
⇒ I = 4π
∫0
⇒I = 2 π
∫0
dx − 4 ∫
sin x
1 + cos 2 x
π
sin x
1 + cos 2 x
Exp. (a)
dx
π
x sin x
0
1 + cos 2 x
dx − I
dx
[from Eq. (i)]
∴
π /2
118. ∫ 0
sin x
sin x + cos x
π
4
(c) 0
π
2
(d) 1
(b)
[AIEEE 2002]
∴
I=
1
n
=
t − 1
1
log 
 +C
 t 
n
=
 xn 
1
log  n
 +C
n
 x + 1
I=
I=
I=
dt
1
=
t (t − 1) n
∫
119. ∫
π/2
∫0
sin x
cos x +
sin x
π
cos  − x +
2

π/2
∫0
…(i)
dx
1
 dt
t
and the straight line y = − x is given by
9
(a) sq units
2
35
sq units
(c)
6
43
(b)
sq units
6
(d) None of these
cos x
sin x +
cos x
The equations of given curve and a line are
…(i)
y = 2 x − x2
y=−x
and
π
sin  − x
2

π/2
∫0
π/2
∫0
dx
x ( x n + 1)
dx
y=
…(ii)
dx
X′
–x
y = 2x – x 2
(0, 0)
Y′
π
4
X
(3, –3)
On solving Eqs. (i) and (ii), we get the points of
intersection of curves which are (0, 0) and (3, –3).
is equal to
[AIEEE 2002]
(b)
…(ii)
Y
π
sin  − x
2

1 dx = [ x]π0 / 2 ⇒ I =
 xn 
1
log  n
 +C
n
 x + 1
 xn 
(c) log  n
 +C
 x + 1
(a)
 1
−

t − 1
120. The area bounded by the curve y = 2 x − x 2
On adding Eqs. (i) and (ii), we get
2I =
∫
Exp. (a)
Exp. (a)
Let
n
[AIEEE 2002]
dx is equal to
(a)
xn − 1
dx
x ( xn + 1)
∫
xn + 1 = t ⇒ nxn − 1 dx = dt
dx
π
π
= 2 π[tan−1 t ]1−1 = 2 π  +  = π 2
4 
 4
dx
=
x( xn + 1)
∫
Put
cos x = t ⇒ − sin x dx = dt
−1
1
I = −2π ∫
dt
1 1+ t2
Put
I=
Let
 x n + 1
1
log 
 +C
n
n

 x
(d) None of these
∴ Required area =
3
∫ 0 [(2 x − x
2
) − (− x)] dx
3
 3 x2
x3 
2
∫ 0 (3 x − x ) dx =  2 − 3 
0
27 27 9
=
−
= sq units
2
3 2
=
3
10
Differential Equations
1. Given that the slope of the tangent to a curve
y = y ( x ) at any point ( x , y ) is
2y
. If the curve
x2
passes through the centre of the circle
x 2 + y 2 − 2 x − 2 y = 0, then its equation is
[JEE Main 2019, 8 April Shift-II]
(a) x loge | y | = − 2( x − 1)
2
(b) x loge | y | = x −1
(c) x loge | y | = 2( x − 1)
(d) x loge | y | = − 2( x − 1)
Given,
[integrating both sides]
2
…(i)
+C
x
Since, curve (i) passes through centre (1, 1) of the
circle
x 2 + y2 − 2 x − 2 y = 0
2
∴loge (1) = − + C ⇒C = 2
1
∴ Equation required curve is
2
[putC = 2 in Eq. (i)]
loge| y| = − + 2
x
⇒
xloge| y| = 2( x − 1)
⇒
loge| y| = −
2. The solution of the differential equation
x
1
x3
+
5
5x 2
4 3
1
(d) y = x +
5
5x 2
(b) y =
Exp. (a)
Given differential equation is
dy
x + 2 y = x2 , ( x ≠ 0)
dx
dy  2 
+   y = x,
⇒
dx  x 
which is a linear differential equation of the form
dy
+ Py = Q
dx
2
Here, P = and Q = x
x
Exp. (c)
dy 2 y
=
dx x2
2
dy
=
dx
⇒ ∫
y ∫ x2
3
x2
+
4
4x 2
3 2
1
(c) y = x +
4
4x 2
(a) y =
dy
+ 2 y = x 2( x ≠ 0) with y(1) = 1, is
dx
[JEE Main 2019, 9 April Shift-I]
∴
IF = e
∫
2
dx
x
= e 2 log
x
= x2
Since, solution of the given differential equation is
y × IF = ∫ (Q × IF) dx + C
∴ y( x2 ) = ∫ ( x × x2 ) dx + C ⇒ yx2 =
Q y(1) = 1, so 1 =
1
+C
4
⇒C =
x4
+C
4
3
4
x4
3
+
4
4
3
x2
+
y=
4 4 x2
∴ yx2 =
⇒
3. If y = y ( x ) is the solution of the differential
equation
dy
= (tan x − y ) sec 2 x ,
dx
203
Differential Equations
 π π
 π
x ∈  − ,  , such that y (0) = 0, then y  − 
 2 2
 4
is equal to
[JEE Main 2019, 10 April Shift-I]
1
(a) − 2
e
1
(c) 2 +
e
(b)
1
−e
2
(d) e − 2
Exp. (d)
Given differential equation
dy
= (tan x − y)sec 2 x
dx
dy
+ (sec 2 x)y = sec2 x tan x,
⇒
dx
which is linear differential equation of the form
dy
+ Py = Q,
dx
where P = sec 2 x and Q = sec 2 x tan x
= e∫
IF
sec 2 xdx
= e tan x
So, solution of given differential equation is
y × IF = ∫ (Q × IF)dx + C
y(e tan x ) = ∫ e tan x ⋅ sec2 x tan x dx + C
Let
tan x = t ⇒ sec xdx = dt
2
ye tan x = ∫ et ⋅ t dt + C = tet − ∫ et dt + C
[using integration by parts method]
= et (t − 1) + C
tan x
⇒
y⋅e
= e tan x (tan x − 1) + C
[Qt = tan x]
Q
y(0) = 0
⇒
0 = 1(0 − 1) + C
⇒
C =1
∴
y ⋅ e tan x = e tan x (tan x − 1) + 1
π
Now, at x = −
4
ye −1 = e −1(−1 − 1) + 1
⇒
ye −1 = − 2e −1 + 1⇒ y = e −2
4. Let y = y ( x ) be the solution of the
differential equation,
dy
 π π
+ y tan x = 2 x + x 2 tan x , x ∈  − ,  ,
 2 2
dx
such that y(0) = 1. Then
[JEE Main 2019, 10 April Shift-II]
π
π
(a) y ′   − y ′  −  = π − 2
 4
 4
π
π
(b) y ′   + y ′  −  = − 2
 4
 4
π
(c) y   + y  −

 4
π

(d) y   − y  −

 4
π  π2
+2
=
4
2
π
= 2
4
Exp. (a)
Given differential equation is
dy
which
is
+ y tan x = 2 x + x2 tan x ,
dx
linear differential equation in the form of
dy
+ Py = Q .
dx
Here, P = tan x and Q = 2 x + x2 tan x
∴IF = e ∫ tan x dx = elog e (sec
x)
= sec x
Now, solution of linear differential equation is
given as
y × IF = ∫ (Q × IF )dx + C
∴ y(sec x) = ∫ (2 x + x2 tan x) sec x dx + C
= ∫ (2 x sec x) dx +
∫x
2
sec x tan x dx + C
Q∫ x sec x tan x dx = x2 sec x − ∫ (2 xsec x) dx
2
Therefore, solution is
ysec x = 2 ∫ xsec x dx + x2 sec x − 2 ∫ xsec x dx + C
⇒ ysec x = x2 sec x + C
…(i)
Q y(0) = 1 ⇒11
( ) = 0(1) + C ⇒C = 1
Now, y = x2 + cos x
[from Eq. (i)]
and y′ = 2 x − sin x
According to options,
− π   π
π
1 
y′   − y′ 
 = 2   −

 4
 4    4
2
π
1 

−  2  −  +
 = π− 2
  4
2
π
π
1 
 π
and y′   + y′  −  =  2   −

 4
 4   4
2
π
1 

+  2  −  +
 =0
  4
2
π
π
π2
1
and y  + y −  =
+
 4  16
 4
2
+
π2
π2
1
+
=
+ 2
16
2
4
204
JEE Main Chapterwise Mathematics
π
π
π2
1
π2
1
and y  − y −  =
+
−
−
=0
 4  16
 4
2
16
2
5. Consider
the
differential
equation,

1
y dx +  x −  dy = 0. If value of y is 1 when
y

x = 1, then the value of x for which y = 2, is
2
So, at y = 2, the value of
1
1
e1/ 2 3
= −
x = + 1−
2
2
e
e
6. The general solution of the differential
equation ( y 2 − x 3 )dx − xydy = 0 ( x ≠ 0) is
(where,C is a constant of integration)
[JEE Main 2019, 12 April Shift-II]
[JEE Main 2019, 12 April Shift-I]
5
(a) +
2
1
(c) +
2
1
e
1
e
(a) y 2 − 2 x 2 + Cx 3 = 0
3 1
(b) −
2
e
3
(d) − e
2
(b) y 2 + 2 x 3 + Cx 2 = 0
(c) y 2 + 2 x 2 + Cx 3 = 0
(d) y 2 − 2 x 3 + Cx 2 = 0
Exp. (b)
Exp. (b)
Given differential equation is

1
y2dx +  x −  dy = 0

y
dx
1
1
+
x= 3
⇒
dy y2
y
which is the linear differential equation of the
dx
form
+ Px = Q.
dy
1
1
Here, P = 2 and Q = 3
y
y
Now, IF = e
∫
1
y2
dy
=e
−
1
y
∴The solution of linear differential equation is
x ⋅ (IF ) ∫ Q (IF ) dy + C
⇒ x e − 1/ y =
1
∫ y3 e
− 1/ y
dy + C
∴ x e − 1/ y = ∫ (− t ) et dt + C
Now,
1
1
[Qlet − = t ⇒ + 2 dy = dt]
y
y
= − tet + ∫ et dt + C
[integration by parts]
= − te + e + C
1
⇒ x e − 1/ y = e − 1/ y + e − 1/ y + C
y
Now, at y = 1, the value of x = 1, so
t
Given differential equation is
( y2 − x3 ) dx − xy dy = 0, ( x ≠ 0)
dy
xy
− y2 = − x 3
⇒
dx
dy dt
Now, put y2 = t ⇒ 2 y
=
dx dx
dy 1 dt
y =
⇒
dx 2 dx
x dt
∴
− t = − x3
2 dx
dt 2
− t = − 2 x2
⇒
dx x
which is the linear differential equation of the form
dt
+ Pt = Q.
dx
2
Here,
P = − and Q = − 2 x2 .
x
1⋅ e − 1 = e − 1 + e − 1 + C
1
C=−
⇒
e
On putting the value of C, in Eq. (i), we get
x=
1/ y
1
e
+ 1−
y
e
−∫
2
dx
x
=
1
x2
QSolution of the linear differential equation is
(IF) t = ∫ Q(IF )dx + λ
[where λ is integrating constant]
t
… (i)
IF = e
∴
⇒
⇒
1
1
t  2  = − 2 ∫  x2 × 2  dx + λ
x 

x 
t
= − 2x + λ
x2
y2
x2
+ 2x = λ
⇒
y2 + 2 x 3 − λ x 2 = 0
or
y2 + 2 x3 + Cx2 = 0
[Qt = y2 ]
[let C = − λ]
205
Differential Equations
7. If y = y ( x ) is the solution of the differential
equation, x
Exp. (b)
Given, f( xy) = f( x) ⋅ f( y), ∀ x, y ∈ [0, 1]
Putting x = y = 0 in Eq. (i), we get
f(0) = f(0) ⋅ f(0)
⇒ f(0)[f(0) − 1] = 0
⇒
f(0) = 1 as f(0) ≠ 0
Now, put y = 0 in Eq. (i), we get
f(0) = f( x) ⋅ f(0)
⇒
f ( x) = 1
dy
dy
So,
= f ( x) ⇒
=1
dx
dx
⇒
∫ dy = ∫ dx ⇒ y = x + C
dy
+ 2 y = x 2 satisfying y(1) = 1,
dx
 1
then y   is equal to
2
[JEE Main 2019, 9 Jan Shift-I]
13
(a)
16
1
(b)
4
(c)
49
16
(d)
7
64
Exp. (c)
Given differential equation can be rewritten as
dy  2 
+   ⋅ y = x, which is a linear differential
dx  x 
2
dy
equation of the form
+ Py = Q, where P =
x
dx
and Q = x.
Now, integrating factor
(IF) = e
2
dx
x =
∫
e 2 log
x
= elog
x2
y(0) = 1
1= 0 + C
C= 1
y= x+1
5
1 1

Now, y   = + 1 = and
 4 4
4
1
3 5 7


⇒ y  + y  = + =
 4 4 4
 4
Q
∴
⇒
∴
= x2
[Qelog f( x ) = f( x) ]
and the solution is given by
y(IF ) =
⇒
yx2 =
∫
(Q × IF ) dx + C
∫x
3
⇒
Now,
yx2 =
y=
dx + C
4
x4 + 3
1
+ 3
49
1  16

=
y  =
1
2
16
4×
4
3
1
dy
 −π π
+
y=
,  and
, x ∈
2
2
 3 3
dx cos x
cos x
 π
 π 4
y   = , then y  −  equals
 4
 4 3
[JEE Main 2019, 10 Jan Shift-I]
1
+ e6
3
1
(c) + e 3
3
(d)
1
3
4
3
Exp. (a)
where P =
is
(d) 4
3
and Q =
1
a
linear
dy
+ Py = Q,
dx
.
cos x
cos 2 x
Now, Integrating factor
3
∫ 2 dx
3 sec 2 xdx
IF = e cos x = e ∫
= e 3 tan x and the
solution of differential equation is given by
y(IF ) = ∫ (Q.(IF ))dx
⇒
[JEE Main 2019, 9 Jan Shift-II]
(c) 2
(b) −
(a)
differential equation of the form
f ( xy ) = f ( x ). f ( y ), for all x , y ∈[0, 1] and
f (0) ≠ 0. If y = y ( x ) satisfies the differential
dy
equation,
= f ( x ) with y(0) = 1, then
dx
 3
 1
y   + y   is equal to
 4
 4
(b) 3
3
Given, differential equation is
dy  3 
1
, which
+ 
y=
dx  cos 2 x 
cos 2 x
4 x2
8. Let f :[0, 1] → R be such that
(a) 5
7
3
3
y   = + 1 =
 4 4
4
9. If
x
…(i)
+C
4
Since, it is given that y = 1when x = 1
∴ From Eq. (i), we get
1
3
…(ii)
1= + C ⇒C =
4
4
∴
4 x2 y = x4 + 3 [using Eqs. (i) and (ii)]
⇒
...(i)
Let
2
e 3 tan x . y = ∫ e 3 tan x sec 2 xdx
I = ∫ e 3 tan x sec 2 xdx
… (i)
206
JEE Main Chapterwise Mathematics
Put
⇒
∴
⇒ ln|(1 + v 2 )Cx|= 0
3tan x = t
3sec 2 x dx = dt
I=
∫
et
et
e 3 tan x
dt =
+C=
+C
3
3
3
From Eq. (i)
e 3 tan x
+C
3
It is given that when,
π
4 e3
4
+C
x = , y is ⇒ e 3 =
4
3
3
3
⇒
C =e3
e 3 tan x . y =
e 3 tan x
+ e3
3
π
e −3
Now, when x = − , e −3 y =
+ e3
4
3
π
1


y = e6 +
Q tan −  = − 1
⇒


 4
3
Thus,
e 3 tan x y =
10. The curve amongst the family of curves
represented by the differential equation,
which
passes
( x 2 − y 2 )dx + 2 xydy = 0,
through (1, 1), is [JEE Main 2019, 10 Jan Shift-II]
(a) a circle with centre on the Y-axis
(b) a circle with centre on the X-axis
(c) an ellipse with major axis along the Y-axis
(d) a hyperbola with transverse axis along the
X-axis.
(1 + v 2 )Cx = 1
[loge x = 0 ⇒ x = e 0 = 1]
y
Now, putting v = , we get
x
2

y
 1 + 2  Cx = 1
x 

⇒
C( x2 + y2 ) = x
Q The curve passes through (1, 1), so
1
C(1 + 1) = 1 ⇒ C =
2
Thus, required curve is x2 + y2 − 2 x = 0, which
represent a circle having centre (1, 0)
∴ The solution of given differential equation
represents a circle with centre on the X-axis.
11. If y ( x ) is the solution of the differential
equation
dy  2 x + 1
−2 x
+
 y = e , x > 0,
dx  x 
where y (1) =
1 −2
e , then
2
[JEE Main 2019, 11 Jan Shift-I]
1
(a) y ( x ) is decreasing in  ,1
2 
(b) y ( x ) is decreasing in (0, 1)
(c) y (loge 2 ) = loge 4
loge 2
(d) y (loge 2 ) =
4
Exp. (a)
Exp. (b)
Given differential equation is
( x2 − y2 )dx + 2 xy dy = 0, which can be written as
dy y2 − x2
=
dx
2 xy
Put y = vx
[Qit is in homogeneous form]
dv
dy
⇒
=v+ x
dx
dx
Now, differential equation becomes
dv v 2 x2 − x2
dv (v 2 − 1)x2
=
=
v+ x
⇒ v+ x
dx
dx
2 x(vx)
2 vx2
dv v 2 − 1
v2 − 1− 2v2
=
−v=
2v
2v
dx
2 v dv
dx
1+ v2
dv
⇒
⇒ ∫
= −∫
=−
x
x
2v
dx
1+ v2
⇒ ln (1 + v 2 ) = − ln x − lnC
⇒
[Qln A + ln B = ln AB]
⇒
x


f ′ ( x)
Q ∫ f( x) dx ⇒ ln|f( x)| + C 


dy  2 x + 1
−2x
+ 
y=e
dx  x 
dy
which is of the form
+ Py = Q, where
dx
2x + 1
and Q = e −2 x
P=
x
We have,
Now, IF = e ∫
= eln x +
Pdx
2x
=e
 1 + 2 x  dx

x 
∫ 
=e
 1 + 2  dx


x
∫ 
= eln x . e 2 x = x.e 2 x
and the solution of the given equation is
y ⋅ (IF ) = ∫ (IF )Q dx + C
⇒
y( xe 2 x ) = ∫ ( x e 2 x . e −2 x ) dx + C
=
∫ xdx + C =
x2
+C
2
1 −2
e when x = 1
2
1 −2 2 1
e .e = + C ⇒ C = 0
2
2
… (i)
Since, y =
∴
(using Eq. (i))
207
Differential Equations
x2
x
⇒ y = e −2 x
2
2
x
dy 1 −2 x
1
Now,
= e
+ e −2 x (− 2 ) = e −2 x  − x
dx 2
2
2

< 0,
1
[by using product rule of derivative]
if < x < 1
2
−2
loge 2 −2 log e 2 1
and y(loge 2 ) =
e
= loge 2 elog e 2
2
2
1
1
= . loge 2 ⋅ 2 −2 = loge 2
2
8
∴ y ( xe 2 x ) =
12. The solution of the differential equation,
dy
= ( x − y )2, when y(1) = 1, is
dx
[JEE Main 2019, 11 Jan Shift-II]
(a) loge
2−y
= 2( y − 1)
2−x
(b) − loge
(c) loge
1+ x − y
=x+ y −2
1− x + y
2−x
=x−y
2−y
(d) − loge
1− x + y
= 2( x − 1)
1+ x − y
Exp. (d)
dy
= ( x − y)2 which is a differential
dx
dy
equation of the form
= f(ax + by + c )
dx
Put x − y = t
dy dt
dy
dt
⇒ = 1−
1−
=
⇒
dx dx dx
dx
dy
dt
2
[Q = ( x − y)2 ]
⇒
1−
=t
dx
dx
dt
dt
= 1− t 2 ⇒ ∫
=
dx
⇒
dx
1− t2 ∫
We have,
[separating the variables]
⇒
⇒
1 + t 
1
loge 
 = x+C
2
1− t 


a+ x
dx
1
+ C
=
loge
∫ 2
2
a− x
2a
 a − x



1+ x − y
1
loge 
 = x + C [Qt = x − y]
2
 1 − x + y
Since, y = 1when x = 1, therefore
 1 + 0
1
loge 
 = 1+ C
2
 1 + 0
⇒
C = −1
[Qlog 1 = 0]
 1+ x − y
1
loge 
∴
 = x−1
 1− x + y
2
1− x + y
= 2( x − 1)
⇒ − loge
1+ x − y
1
[Qlog = log x−1 = − log x]
x
13. If a curve passes through the point (1, − 2 )
and has slope of the tangent at any point
x2 − 2y
, then the curve also
( x , y ) on it as
x
passes through the point
[JEE Main 2019, 12 Jan Shift-II]
(b) ( −1, 2 )
(d) (3, 0)
(a) ( 3 , 0)
(c) ( − 2 ,1)
Exp. (a)
We know that, slope of the tangent at any point
( x, y) on the curve is
dy x2 − 2 y
(given)
=
dx
x
dy 2
…(i)
⇒
+ y= x
dx x
which is a linear differential equation of the form
dy
+ P( x) ⋅ y = Q( x),
dx
2
where
P( x) = and Q( x) = x
x
Now, integrating factor
(IF) = e ∫
P( x )dx
= elog e
x
=e
∫
2
dx
x
= e 2log e
2
x
[Q m log a = log am ]
= x2
[Qelog e f( x ) = f( x)]
and the solution of differential Eq. (i) is
y(IF ) = ∫ Q( x)(IF )dx + C
⇒
y( x2 ) =
∫ x⋅ x
2
dx + C
4
x
+C
4
Q The curve (ii) passes through the point
(1, − 2 ), therefore
1
9
− 2 = + C ⇒C = −
4
4
∴ Equation of required curve is
4 yx2 = x4 − 9.
⇒
yx2 =
…(ii)
Now, checking all the option, we get only ( 3, 0)
satisfy the above equation.
208
JEE Main Chapterwise Mathematics
 π2  π2
1
π2
8π2
 −
⇒ y=
y  = 2 
− π2 = −
2
9
9
 36  2
14. Let y = y ( x )be the solution of the differential
dy
+ y cos x = 4x , x ∈(0, π ).
dx
 π
 π
If y   = 0, then y   is equal to
2
 6
equation sin x
(a)
4
−8 2
8
π 2 (b)
π (c) − π 2
9 3
9 3
9
[JEE Main 2018]
(d) −
4 2
π
9
Exp. (c)
We have,
dy
dy
+ ycot x = 4 xcosec x
sin x + ycos x = 4 x ⇒
dx
dx
This is a linear differential equation of form
dy
+ Py = Q
dx
where P = cot x, Q = 4 x cosec x
cot xdx
Now, IF = e ∫ Pdx = e ∫
= elogsin x = sin x
Solution of the differential equation is
y ⋅ sin x = ∫ 4x cosec xsin xdx + C
⇒
ysin x = ∫ 4 x dx + C = 2 x2 + C
Put x =
Put
∴
⇒
π
, y = 0, we get
2
π2
π2
C=−
⇒ ysin x = 2 x2 −
2
2
π
x=
6
 π2  π2
1

 −
y  = 2 
2
2
 36 
π2
8π2
y=
− π2 ⇒ y = −
9
9
Alternate Method
dy
We have, sin x + ycos x = 4 x, which can be
dx
d
written as
(sin x ⋅ y) = 4 x
dx
On integrating both sides, we get
d
∫ dx (sin x ⋅ y)⋅ dx = ∫ 4 x ⋅ dx
4 x2
y ⋅ sin x =
+ C ⇒ y ⋅ sin x = 2 x2 + C
⇒
2
π
Now, as y = 0 when x =
2
π2
π2
C=−
⇒ y ⋅ sin x = 2 x2 −
∴
2
2
π
Now, putting x = , we get
6
15. If (2 + sin x)
dy
+ ( y + 1)cos x = 0 and y(0) = 1,
dx
 π
then y   is equal to
2
[JEE Main 2017 (offline)]
(a)
1
3
(b) −
2
3
(c) −
1
3
(d)
4
3
Exp. (a)
dy
+ ( y + 1)cos x = 0
dx
− cos x
cos x
dy
+
y=
2 + sin x
dx 2 + sin x
We have, (2 + sin x)
⇒
which is a linear differential equation.
∴
IF = e
∫
cos x
dx
2 + sin x
= elog ( 2 + sin x ) = 2 + sin x
∴Required solution is given by
− cos x
y ⋅ (2 + sin x) = ∫
⋅ (2 + sin x)dx + C
2 + sin x
⇒
y(2 + sin x) = − sin x + C
Also,
y(0) = 1
∴
12
( + sin 0) = − sin 0 + C ⇒ C = 2
π
2 − sin
2 − sin x
π
2 = 1
∴ y=
⇒ y  =
 2  2 + sin π 3
2 + sin x
2
16. If a curve y = f ( x ) passes through the point
(1,
− 1)
and
(a) −
2
5
(b) −
satisfies
differential
 1
equation, y (1 + xy )dx = x dy , then f  −  is
 2
equal to
[JEE Main 2016 (offline)]
4
5
(c)
the
2
5
(d)
4
5
Exp. (d)
Given differential equation is
y(1 + xy)dx = x dy ⇒ y dx + xy2 dx = x dy
x dy − y dx
( y dx − x dy)
⇒
= x dx
= x dx ⇒ −
y2
y2
⇒
 x
− d   = x dx
 y
On integrating both sides, we get
x x2
− =
+C
y 2
Q It passes through (1, − 1).
...(i)
209
Differential Equations
1
1
+C ⇒C=
2
2
x x2
1
Now, from Eq. (i) − =
+
y 2
2
2x
2x
⇒ y=− 2
x2 + 1 = −
⇒
y
x +1
1
4
∴
f  −  =
 2 5
17. Let y ( x ) be the solution of the differential
dy
+ y = 2 x log x , ( x ≥ 1).
dx
Then, y (e ) is equal to
[JEE Main 2015]
equation ( x log x )
(a) e
(b) 0
(c) 2
(d) 2e
Q
∴
⇒
p (0) = 100 ⇒ K = −300
p(t )e − t / 2 = 400e − t / 2 − 300
p(t ) = 400 − 300et / 2
19. At present, a firm is manufacturing
2000 items. It is estimated that the rate of
change of production P w.r.t. additional
number of workers x is given by
dP
= 100 − 12 x . If the firm employees
dx
25 more workers, then the new level of
production of items is
[JEE Main 2013]
(a) 2500
(b) 3000
(c) 3500
(d) 4500
Exp. (c)
Exp. (c)
Given differential equation is
dy
( xlog x) + y = 2 xlog x,
dx
dy
y
+
=2
⇒
dx xlog x
dP
= (100 − 12 x )
dx
dP = (100 − 12 x ) dx
⇒
On integrating both sides, we get
∫ dP = ∫ (100 − 12 x ) dx
Given,
( x ≥ 1)
This is a linear differential equation.
1
dx
∫
IF = e x log x = elog(log x ) = log x
∴
Now, the solution of given differential equation is
given by
y ⋅ log x = ∫ log x ⋅ 2dx
⇒
y ⋅ log x = 2[ xlog x − x] + c
x = 1, c = 2
y ⋅ log x = 2[ xlog x − x] + 2
x = e,
y = 2(e − e ) + 2 ⇒ y = 2
18. Let the population of rabbits surviving at a
time t be governed by the differential
dp (t ) 1
equation
= p (t ) − 200. If p (0) = 100,
2
dt
then p (t ) is equal to
[JEE Main 2014]
(b) 300 − 200e − t / 2
(d) 400 − 300e − t / 2
(a) 400 − 300e t / 2
(c) 600 − 500e t / 2
Exp. (a)
dp 1
− p(t ) = −200
dt 2
IF = e
 1
∫ −   dt
 2
= e −t / 2
Hence, p(t )⋅ e − t / 2 = ∫ −200 ⋅ e − t / 2dt
P = 100 x − 8 x3 / 2 + C
When x = 0, then P = 2000 ⇒ C = 2000
Now, when x = 25, then
P = 100 × 25 − 8 × (25)3 / 2 + 2000
= 2500 − 8 × 125 + 2000
= 4500 − 1000 = 3500
y ⋅ log x = 2 ∫ log x dx
⇒
⇒
At
⇒
At
p(t )⋅ e − t / 2 = 400 e − t / 2 + K
1=
∴
20. The population p (t ) at time t of a certain
mouse species satisfies the differential
dp (t )
equation
= 0.5(t ) − 450. If p(0) = 850,
dt
then the time at which the population
becomes zero is
[AIEEE 2012]
(a) 2 log 18 (b) log 9
(c)
1
log 18 (d) log18
2
Exp. (a)
Given
(i) The population of mouse at time ‘t ’ satisfies
the differential equation
dp(t )
p′(t ) =
= 0.5p(t ) − 450
dt
(ii) Population of mouse at time
t = 0 is p(0) = 850
To find The time at which the population of
the mouse will become zero, i.e., to find the
value of ‘t ’ at which p(t ) = 0.
210
JEE Main Chapterwise Mathematics
Let’s solve the differential equation first.
dp(t )
= 0.5p(t ) − 450
p′(t ) =
dt
2dp(t )
2dp(t )
⇒
= dt ⇒ ∫
= dt
p(t ) − 900
p(t ) − 900 ∫
⇒ 2 log| p( t ) − 900| = t + C
where, C is the constant of integration.
To find the value of ‘C’, let’s substitute t = 0.
⇒ 2 log| p(0) − 900| = 0 + C
⇒
∴
C = 2 log|850 − 900|
C = 2 log 50
[Q P (θ) = 850]
Now, substituting the value of C back in the
solution, we get
2 log| p(t ) − 900| = t + 2 log 50
Here, since we want to find the value of t at
which p(t ) = 0, hence substituting p(t ) = 0, we
get
2 log| 0 − 900| = t + 2 log 50
t = 2 log
⇒
900
50
∴ t = 2 log 18
dy
= y + 3 > 0 and y (0) = 2 , then y (log 2 ) is
dx
equal to
[AIEEE 2011]
21. If
(a) 5
(c) −2
(b) 13
(d) 7
Exp. (d)
Here,
and
⇒
⇒
Since,
⇒
∴
⇒
∫
dy
= y + 3> 0
dx
y(0) = 2
dy
= dx
y+ 3 ∫
log| y + 3| = x + C
y(0) = 2
log|2 + 3| = 0 + C
(a) I −
log| y + 3| = x + loge 5
and V (t ) be the value after it has been used
for t years. The value V (t ) depreciates at a
rate given by differential equation
(b) I −
Exp. (a)
Given,
d {V(t )}
= − k (T − t )
dt
∴
d {V(t )} = − k (T − t ) dt
t = 0, V( t ) = I
When
T
…(i)
T
∫ 0 d {V(t )} = ∫ 0 − k (T − t ) dt
⇒
T
( t − T )2 
V(T ) − V(0) = k 

 2 0
k
V(T ) − I = {(T − T )2 − (0 − T )2 }
2
k
V(T ) = I − T 2
2
⇒
⇒
∴
23. Consider the differential equation

1
y 2dx +  x −  dy = 0. If y(1) = 1, then x is

y
given by
[AIEEE 2011]
(a)1 −
1 e 1/ y
+
y
e
(b) 4 −
2 e 1/ y
−
y
e
(c) 3 −
1 e 1/ y
+
y
e
(d)1 +
1 e 1/ y
−
y
e
Exp. (d)
Here,
∴
dx
1
1
+
⋅x= 3
dy y2
y
[linear differential equation in x]
IF = e
∫
1
y2
dy
=e
−
1
y
Complete solution is
y + 3 = 10
y=7
22. Let I be the purchase value of an equipment
k (T − t )2
2
1
(d) T 2 −
k
k 2
T
2
(c) e −kT
C = loge 5
When x = loge 2
⇒
log| y + 3| = loge 2 + loge 5 = loge 10
⇒
∴
dV (t )
= − k (T − t ), wherek > 0 is a constant
dt
and T is the total life in years of the
equipment. Then, the scrap value V (T ) of
the equipment is
[AIEEE 2011]
x⋅e
⇒
Put
−
1
y
=
1
∫ y3
⋅e
−
1
y
dy
1
1 1 −y
∫ y ⋅ y2 ⋅ e dy
1
1
− = t ⇒ 2 dy = dt
y
y
x⋅e
−
1
y
=
211
Differential Equations
−
⇒
xe
⇒
xe
⇒
= ∫ − t ⋅ et dt
1
−
y
1
−
y
xe
−
⇒
1
y
xe
1
y
= − {t ⋅ et − ∫ 1⋅ et dt } + C
= − tet + et + C
=
1
⋅e
y
−
1
y
+e
−
1
y
+C
y(1) = 1
e −1 = e −1 + e −1 + C
1
C=−
e
Q
∴
⇒
−
∴
xe
1
y
=
1
⋅e
y
−
1
y
+e
−
1
y
−
1
e
[AIEEE 2010]
(a) sec x = (tan x + C ) y (b) y sec x = tan x + C
(c) y tan x = sec x + C (d) tan x = (sec x + C ) y
Exp. (a)
Since, cos x dy = ysin x dx − y2 dx
1 dy 1
⇒
− tan x = − sec x
y2 dx y
which is the Bernaulli’s form.
⇒
y = c1ec 2 x
y ′ = c2 y
y ′ ′ = c2 y ′
( y′ )2
y ′′ =
y
yy ′′ = ( y ′ )2
⇒ y ′ = c 2 c1 ec 2 x
…(i)
…(ii)

y′
 from Eq. (i), c 2 = y 


circles with fixed radius 5 units and centre
on the line y = 2 is
[AIEEE 2008]
(a) ( x − 2 )2 y ′2 = 25 − ( y − 2 )2
(b) ( x − 2 )y ′2 = 25 − ( y − 2 )2
Exp. (d)
The equation of family of circles with centre on
y = 2 and radius 5 is
…(i)
( x − α )2 + ( y − 2 )2 = 52
⇒
x2 + α 2 − 2 α x + y2 + 4 − 4 y = 25
On differentiating w.r.t. x, we get
dy
dy
2 x − 2α + 2 y
−4
=0
dx
dx
dy
( y − 2)
α= x+
⇒
dx
On putting the value of α in Eq. (i), we get
2
1
1 dy dz
= z ⇒ 2
=
y
y dx dx
dz
⇒
+ (tan x) z = − sec x
dx
This is a linear differential equation. Therefore,
−
= sec x
x
z ⋅ (sec x) = ∫ − sec x ⋅ sec x dx + C1
= − ∫ sec 2 x dx + C1
⇒
Given,
⇒
⇒
(d) ( y − 2 )2 y ′2 = 25 − ( y − 2 )2
π
cos x dy = y (sin x − y )dx , 0 < x < , is
2
⇒
Exp. (d)
(c) ( y − 2 )y ′2 = 25 − ( y − 2 )2
24. Solution of the differential equation
IF = e ∫ tan xdx = elog sec
Hence, the solution is
(d) yy ′′ = ( y ′ )2
26. The differential equation of the family of
1
Put
(b) y ′′ = y ′ y
(c) yy ′′ = y ′
⇒
1
1
x = + 1 − ⋅ey
y
e
⇒
(a) y ′ = y 2
1
− sec x = − tan x + C1
y
sec x = y(tan x + C )
2
⇒
 dy  ( y − 2 )2 = 25 − ( y − 2 )2
 
 dx 
or
y ′2 ( y − 2 )2 = 25 − ( y − 2 )2
27. The solution of the differential equation
dy x + y
satisfying the condition y(1) = 1 is
=
dx
x
(a) y = x log x + x
(b) y = log x + x
(c) y = x log x + x 2
(d) y = xe ( x − 1 )
[AIEEE 2008]
[put C = − C1 ]
25. The differential equation which represents
the family of curves y = c1e , where c1 and
[AIEEE 2009]
c 2 are arbitrary constants, is
c 2x
 x − x − dy ( y − 2 ) + ( y − 2 )2 = 52
dx


Exp. (a)
Given equation can be rewritten as
dy 1
− ⋅ y=1
dx x
212
JEE Main Chapterwise Mathematics
Now, IF = e
−∫
1
dx
x
= e − log
x
=
dy
=0
dx
Again differentiating, we get
1
x
2 Ax + 2 By
∴ Required solution,
1
y   =
 x
1
∫ x dx = log x + C
y(1) = 1 ⇒ 1 = log 1 + C
Q
∴
⇒ C =1
y = x log x + x
 dy 2
d 2 y 
…(ii)
2 A + 2 B   + y 2  = 0
dx 
 dx 
Eliminating A and B from Eqs. (i) and (ii), we get
passing through the origin and having their
centres on the X-axis is
[AIEEE 2007]
dy
dx
dy
2
2
(c) y = x + 2 xy
dx
(a) x 2 = y 2 + xy
30. The differential equation representing the
dy
dx
dy
2
2
(d) y = x − 2 xy
dx
(b) x 2 = y 2 + 3xy
family of curves y 2 = 2 c ( x + c ), where c > 0,
is a parameter, is of order and degree as
follows
[AIEEE 2005]
Exp. (c)
General equation of all such circles which pass
through the origin and whose centre lie on X-axis,
is
x2 + y2 + 2 gx = 0
2
d2y
dy
y dy
=0
+   − ⋅
2


dx
x dx
dx
This is the required differential equation whose
order is two and degree is one.
y
28. The differential equation of all circles
…(i)
…(i)
On differentiating w.r.t. x, we get
dy
2x+2y
+ 2g = 0
dx
dy
⇒ 2 g = −  2 x + 2 y 

dx 
(a) order 2, degree 2
(c) order 1, degree 1
(b) order 1, degree 3
(d) order 1, degree 2
Exp. (b)
Given equation of family of curves is
y2 = 2 c ( x +
c)
…(i)
On differentiating Eq. (i) w.r.t. x, we get
2 y y1 = 2 c ⇒ c = y y1
On putting the value of c in Eq. (i), we get
y2 = 2 y y1( x + y y1 )
On putting the value of 2g in Eq. (i), we get
∴ ( y2 − 2 y y1 x)2 = 4( y y1 )3
dy
x2 + y2 +  −2 x − 2 y  x = 0

dx 
dy
2
2
2
=0
x + y − 2 x − 2 xy
⇒
dx
dy
y2 = x2 + 2 xy
⇒
dx
which is the required equation.
Hence, the degree and order of above equation
are three and one, respectively.
29. The differential equation whose solution is
Ax + By = 1, where A and B are arbitrary
constant, is of
[AIEEE 2006]
2
(a)
(b)
(c)
(d)
2
first order and second degree
first order and first degree
second order and first degree
second order and second degree
Exp. (c)
The given equation is Ax2 + By2 = 1.
On differentiating w.r.t. x, we get
dy
= y (log y − log x + 1), then the
dx
solution of the equation is
[AIEEE 2005]
31. If x
x
(a) log   = C y
y
y
(c) x log   = C y
x
y
(b) log   = C x
x
x
(d) y log   = C x
y
Exp. (b)
dy
= y(log y − log x + 1)
dx
dy  y  
y

…(i)
=    log   + 1
∴
 x

dx  x  
which is the homogeneous equation.
dt
dy
Put y = t x ⇒
=t + x
dx
dx
dt
[from Eq. (i)]
∴
t + x
= t log t + t
dx
Given that, x
213
Differential Equations
⇒
t log t dx = x dt
dt
dx
∫ t log t = ∫ x
⇒
34. The degree and order of the differential
equation of the family of all parabolas whose
axis is X-axis, are respectively
[AIEEE 2003]
On integrating, we get
log t = log x + log C
y
log   = C x
⇒
 x
(a) 2, 1
(c) 3, 2
Exp. (b)
32. The differential equation for the family of
curves x + y − 2 ay = 0, where a is an
arbitrary constant, is
[AIEEE 2004]
2
2
(a) 2 ( x 2 − y 2 ) y ′ = xy
(b) 2 ( x 2 + y 2 ) y ′ = xy
(c) ( x − y ) y ′ = 2 xy
(d) ( x + y ) y ′ = 2 xy
2
2
2
(b) 1, 2
(d) 2, 3
2
Exp. (c)
The equation of the family of curves is
x2 + y2 − 2 ay = 0
…(i)
General equation of parabola whose axis is X-axis,
is
y2 = 4 a( x + h)
On differentiating w.r.t. x, we get
dy
= 4a
2y
dx
dy
= 2a
y
⇒
dx
Again differentiating w.r.t. x, we get
2
On differentiating w.r.t. x, we get
⇒
d2y
dy
y   + y 2 = 0
 dx 
dx
2 x + 2 yy ′ − 2 ay ′ = 0
2 x + 2 yy ′
= 2a
y′
…(ii)
From Eq. (i), we get
2a =
35. The solution of the differential equation
x 2 + y2
y
On putting the value of 2a in Eq. (ii), we get
2 x + 2 yy ′ x2 + y2
=
y′
y
⇒
2 xy + 2 y2 y ′ = x2 y ′ + y2 y ′
∴
This is a differential equation whose degree and
order are one and two, respectively.
(1 + y 2 ) + ( x − e tan
(a) ( x − 2 ) = Ce
(b) 2 xe tan
(c) xe
( x2 − y2 )y ′ = 2 xy
(d) xe
33. The solution of the differential equation
y dx + ( x + x 2y ) dy = 0 is
(a) −
(c)
1
=C
xy
1
+ log y = C
xy
(b) −
[AIEEE 2004]
1
+ log y = C
xy
(d) log y = Cx
 1
∫ d  − xy  = −
= e 2 tan
=e
−1
−1
+C
y +C
tan −1 y
+C
−1
1
dx
e tan y
+
x=
2
dy 1 + y
1 + y2
1
∫1+
∴
Pdy
IF = e ∫
=e
∴
Required solution,
1
∫ y dy
On integrating both sides, we get
1
1
−
= − log y + C ⇒ −
+ log y = C
xy
xy
y
[AIEEE 2003]
−2 tan −1 y
= tan
2 tan −1 y
dy
= 0, is
dx
Given differential equation can be rewritten as
−1
dx
+ x = e tan y
(1 + y2 )
dy
Exp. (b)
⇒
tan −1 y
y
)
y
Exp. (b)
or
Given that, y dx + ( x + x2 y) dy = 0
y dx + x dy
1
= − dy
∴
y
x 2 y2
−1
−1
xe tan
−1
y
=
∫
e 2 tan
dt =
dy
dy
1 + y2
= e tan
−1
1+ y
Put t = tan−1 y
⇒
y2
2
y
−1
y
dy + C1
214
JEE Main Chapterwise Mathematics
−1
y
−1
y
∴
xe tan
⇒
x e tan
⇒
xe tan
⇒
2 xe tan
−1
y
−1
y
= ∫ e 2 t dt + C1
Exp. (b)
1 2t
e + C1
2
−1
1
= e 2 tan y + C1
2
=
= e 2 tan
−1
y
+C
d2y
= e −2 x
dx2
On integrating both sides, we get
dy e −2 x
=
+C
dx
−2
Again integrating, we get
Q
[Qput 2C1 = C]
36. The order and degree of the differential
2
3
dy 
d y

equation 1 + 3  = 4 3 are


dx
dx
[AIEEE 2002]
2
(a) 1, 
 3
(b) (3, 1)
(c) (3, 3)
(d) (1, 2)
3
d 2y
dx 2
d2 y
=0
dx 2
dy
(c)
=0
dx
[AIEEE 2002]
d 2x
=0
dy 2
dx
(d)
=0
dy
(b)
Exp. (a)
This shows that the order and degree of given
equation are 3 and 3, respectively.
37. The solution of the equation
38. The differential equation of all non-vertical
(a)
The given differential equation can be rewritten as
2
3
 1 + 3 dy  = 4  d y 



3 


dx
 dx 
e −2 x
+ Cx + D
4
lines in a plane is
Exp. (c)
= e −2 x is
[AIEEE 2002]
e −2 x
(a)
4
1
(c) e −2 x + Cx 2 + D
4
y=
3
e −2 x
(b)
+ Cx + D
4
1
(d) e −2 x + C + D
4
The general equation of all non-vertical lines in a
plane is ax + by = 1, where b ≠ 0.
On differentiating, we get
dy
=0
a+ b
dx
Again differentiating, we get
d2y
b 2 =0
dx
d2y
[Q b ≠ 0]
⇒
=0
dx2
11
Coordinate Geometry
1. The shortest distance between the line y = x
and the curve y 2 = x − 2 is
[JEE Main 2019, 8 April Shift-I]
7
(b)
8
11
(d)
4 2
(a) 2
7
(c)
4 2
1
1
= 1⇒t =
2
2t
9 1
So, the point P is  ,  .
 4 2
Given equation of curve is
y2 = x − 2
…(i)
and the equation of line is
y= x
…(ii)
y=x
y2=x–2
M
(2, 0)
dy
= 1]
dx
[Q P( x, y) = P(t 2 + 2, t )]

9 − 1

Now, minimum distance = PM = 4 2
2
[Qdistance of a point P( x1, y1 ) from a line
|ax + by1 + c |
ax + by + c = 0 is 1


a2 + b 2
7
units
=
4 2
2. The sum of the squares of the lengths of the
P(t2+2, t)
O
[differentiating the curve (i), we get 2 y
⇒
Exp. (c)
Y
1
=1
2yP
⇒
X
chords intercepted on the circle, x 2 + y 2 = 16,
by the lines, x + y = n, n ∈ N , where N is the
set of all natural numbers, is
[JEE Main 2019, 8 April Shift-I]
(a) 320
(b) 105
(c) 160
(d) 210
Exp. (d)
Consider a point P(t 2 + 2, t ) on parabola (i).
For the shortest distance between curve (i) and
line (ii), the line PM should be perpendicular to line
(ii) and parabola (i), i.e. tangent at P should be
parallel to y = x.
dy
∴
= Slope of tangent at point P to
dx at point P
curve (i)
=1
[Qtangent is parallel to line y = x]
Given equation of line is x + y = n,n∈N
and equation of circle is x2 + y2 = 16
…(i)
…(ii)
Now, for intercept, made by circle (ii) with line (i)
x2+y2=16
(0, 0)
4
A
d
B
x+y=n, n 0 N
216
JEE Main Chapterwise Mathematics
⇒
d<4
n
<4
2
[Qd = perpendicular distance from (0, 0) to the
|0 + 0 − n|
n 
line x + y = n and it equal to
=

2
2
2
1 +1
n< 4 2
⇒
Q n ∈ N, so n = 1, 2, 3, 4, 5
…(iii)
n2
2
Qd = n 
2 

∴Sum of square of all possible lengths of chords
(for n = 1, 2, 3, 4, 5)
1
= 4 (16 × 5) − (12 + 2 2 + 32 + 42 + 52 )


2
5(6)(11)
= 320 − 2
= 320 − 110 = 210
6
3. If the tangents on the ellipse 4x 2 + y 2 = 8 at
the points (1, 2) and (a ,b ) are perpendicular
to each other, then a 2 is equal to
[JEE Main 2019, 8 April Shift-I]
128
(a)
17
64
(b)
17
(c)
4
17
(d)
⇒
4a2 + 64a2 = 8
[from Eq.(iv)]
8
68a2 = 8 ⇒ a2 =
68
2
2
a =
17
⇒
⇒
4. A point on the straight line, 3x + 5y = 15
Clearly, length of chord AB = 2 42 − d 2
= 2 16 −
Also, the point (a, b ) lies on the ellipse (i), so
4a2 + b 2 = 8
2
17
which is equidistant from the coordinate
axes will lie only in
[JEE Main 2019, 8 April Shift-I]
(a)
(b)
(c)
(d)
IV quadrant
I quadrant
I and II quadrants
I, II and IV quadrants
Exp. (c)
Given equation of line is 3 x + 5 y = 15
…(i)
Clearly, a point on the line (i), which is
equidistance from X and Y-axes will lie on the line
either y = x or y = − x.
Y
B
y=x
Exp. (d)
(0, 3)
Equation of given ellipse is
4 x 2 + y2 = 8
⇒
y=–x
x 2 y2
x2
y2
+
=1
+
= 1⇒
2
2
8
( 2)
(2 2 )2
Now, equation of tangent at point (1, 2) is
...(ii)
2x+ y = 4
x2
y2
[Qequation of tangent to the ellipse 2 + 2 = 1
a
b
xx
yy
at ( x1, y1 ) is 21 + 21 = 1]
a
b
and equation of another tangent at point (a, b ) is
…(iii)
4ax + by = 8
Since, lines (ii) and (iii) are perpendicular to each
other.
 − 2  ×  − 4a  = − 1
∴


 
 1  b 
[if lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c 2 = 0
⇒
A
…(i)
 a  a 
are perpendicular, then  − 1   − 2  = − 1]
 b1   b2 
…(iv)
b = − 8a
(5, 0)
0
X
3x+5y=15
In the above figure, points A and B are
on the line (i) and are equidistance from the
coordinate axes.
15 15
On solving line (i) and y = x, we get A ,  .
 8 8
Similarly, on solving line (i) and y = − x, we get
15 15
B  − ,  .
 2 2
So, the required points lie only in I and II
quadrants.
5. Let O(0, 0) and A(0, 1) be two fixed points,
then the locus of a point P such that the
perimeter of ∆AOP is 4, is
[JEE Main 2019, 8 April Shift-I]
(a) 8x2 − 9 y2 + 9 y = 18 (b) 9x2 − 8 y2 + 8 y = 16
(c) 9x2 + 8 y2 − 8 y = 16 (d) 8x2 + 9 y2 − 9 y = 18
217
Coordinate Geometry
Exp. (c)
Given vertices of ∆AOP are O(0, 0) and A(0, 1)
Let the coordinates of point P are ( x, y).
Clearly, perimeter = OA + AP + OP = 4 (given)
⇒ (0 − 0)2 + (0 − 1)2 +
(0 − x)2 + (1 − y)2
+
⇒1 +
x + ( y − 1) +
x + y =4
⇒ x + y − 2y + 1 +
x 2 + y2 = 3
2
2
2
2
2
⇒ x 2 + y2 − 2 y + 1 = 3 −
x 2 + y2 = 4
2
x 2 + y2
⇒ x 2 + y2 − 2 y + 1 = 9 + x 2 + y2 − 6 x 2 + y2
[squaring both sides]
1 − 2 y = 9 − 6 x 2 + y2
⇒
⇒
6 x 2 + y2 = 2 y + 8
⇒
3 x 2 + y2 = y + 4
⇒
9( x2 + y2 ) = ( y + 4)2
Since, lines L1 and L2 are perpendicular to each
other.
∴
m1m2 = − 1

− k
3
1
− 
⇒
 = − 1 [from Eqs. (i) and (iii)]

 
 2   4 − h
⇒
3 − k = 8 − 2h
⇒
2h − k = 5
On solving Eqs. (ii) and (iv), we get
(h, k ) = (3, 1)
k 3
So,
= =3
h 1
…(iv)
7. In an ellipse, with centre at the origin, if the
difference of the lengths of major axis and
minor axis is 10 and one of the foci is at
(0, 5 3 ), then the length of its latus rectum is
[JEE Main 2019, 8 April Shift-II]
(a) 5
[squaring both sides]
(b) 10
(c) 8
(d) 6
Exp. (a)
x2
+
y2
= 1is on Y-axis
⇒
9 x2 + 9 y2 = y2 + 8 y + 16
One of the focus of ellipse
⇒
9 x2 + 8 y2 − 8 y = 16
(0, 5 3 )
∴
be = 5 3
[where e is eccentricity of ellipse]
According to the question,
2 b − 2 a = 10 ⇒ b − a = 5
On squaring Eq. (i) both sides, we get
b 2e 2 = 75
Thus, the locus of point P( x, y) is
9 x2 + 8 y2 − 8 y = 16
6. Suppose that the points (h , k ), (1, 2 ) and
( −3, 4) lie on the line L1. If a line L 2 passing
through the points (h , k ) and
(4, 3) is perpendicular to L1, thenk /h equals
[JEE Main 2019, 8 April Shift-II]
(a) −
1
7
(c) 3
1
3
(d) 0
(b)
Exp. (c)
Given, points (1, 2), (−3, 4) and (h, k ) are lies on line
L1, so slope of line L1 is
4−2
k −2
=
m1 =
−3 − 1 h − 1
−1 k − 2
…(i)
⇒
=
m1 =
2
h−1
⇒
2(k − 2 ) = − 1(h − 1)
⇒
2k − 4 = − h + 1
…(ii)
⇒
h + 2k = 5
and slope of line L2 joining points (h, k ) and
(4, 3), is m2 =
3−k
4−h
…(iii)
a
2
⇒

a2 
b 2  1 − 2  = 75
b 

⇒
b 2 − a2 = 75
b2
…(i)
…(ii)
 2
a2 
Qe = 1 − 2 
b 

⇒
(b + a)(b − a) = 75
⇒
b + a = 15 [from Eq. (ii)] …(iii)
On solving Eqs. (ii) and (iii), we get
b = 10 and a = 5
2 a2 2 × 25
So, length of latusrectum is
=
=5
b
10
8. The tangent to the parabola y 2 = 4x at the
point where it intersects the circle x 2 + y 2 = 5
in the first quadrant, passes through the
point
[JEE Main 2019, 8 April Shift-II]
 1 3
(a)  , 
 4 4
 1 4
(c)  − , 
 3 3
 3 7
(b)  , 
 4 4
 1 1
(d)  − , 
 4 2
218
JEE Main Chapterwise Mathematics
Exp. (b)
Given equations of the parabola y2 = 4 x
and circle
x 2 + y2 = 5
10. If a point R ( 4, y , z ) lies on the line segment
…(i)
…(ii)
So, for point of intersection of curves (i) and (ii),
put y2 = 4 x in Eq. (ii), we get
x + 4x − 5 = 0 ⇒ x + 5x − x − 5 = 0
2
2
⇒
( x − 1)( x + 5) = 0 ⇒ x = 1, − 5
For first quadrant x = 1, so y = 2.
Now, equation of tangent of parabola (i) at point
(1, 2) is T = 0
⇒
2 y = 2( x + 1)
⇒
x − y + 1= 0
3 7

The point  ,  satisfies, the equation of line
 4 4
x − y + 1= 0
9. If the eccentricity of the standard hyperbola
passing through the point ( 4, 6) is 2, then the
equation of the tangent to the hyperbola at
[JEE Main 2019, 8 April Shift-II]
( 4, 6) is
(a) 3x − 2 y = 0
(c) 2x − y − 2 = 0
(b) x − 2 y + 8 = 0
(d) 2x − 3 y + 10 = 0
Exp. (c)
Let the equation of standard hyperbola is
x2
y2
− 2 =1
2
a
b
Now, eccentricity of hyperbola is
1+
⇒
⇒
b2
a2
=2
…(i)
(given)
b = 3a
2
[JEE Main 2019, 8 April Shift-II]
(a) 2 21
(b)
53
(c) 2 14
(d) 6
Exp. (c)
Given points are P(2, − 3, 4), Q(8, 0, 10) and
R(4, y, z).
Now, equation of line passing through points P
x − 8 y − 0 z − 10
and Q is
=
=
6
3
6
[since equation of a line passing through two
points A( x1, y1, z1 ) and B( x2 , y2 , z2 ) is given by
x − x1
y − y1
z − z1 
=
=

x2 − x1 y2 − y1 z2 − z1 
x − 8 y z − 10
…(i)
= =
⇒
2
1
2
Q Points P, Q and R are collinear, so
z − 10
4 − 8 y z − 10
= =
⇒ −2 = y =
2
1
2
2
⇒
y = − 2 and z = 6
So, point R is (4, − 2, 6), therefore the distance of
point R from origin is
OR = 16 + 4 + 36 = 56 = 2 14
11. The tangent and the normal lines at the
point ( 3 , 1) to the circle x 2 + y 2 = 4 and the
a2 + b 2 = 4a2
2
joining the points P(2 , − 3, 4) and Q(8, 0, 10),
then the distance of R from the origin is
X-axis form a triangle. The area of this
triangle (in square units) is
[JEE Main 2019, 8 April Shift-II]
…(ii)
Since, hyperbola (i) passes through the point
(4, 6)
16 36
…(iii)
∴
−
=1
a2 b 2
On solving Eqs. (ii) and (iii), we get
…(iv)
a2 = 4 and b 2 = 12
Now, equation of tangent to hyperbola (i) at point
(4, 6), is
4x 6y
−
=1
a2 b 2
4x 6y
[from Eq. (iv)]
−
=1
⇒
4 12
y
⇒
x − = 1 ⇒ 2x − y − 2 = 0
2
1
(a)
3
4
(b)
3
(c)
2
3
(d)
1
3
Exp. (c)
Let T = 0 and N = 0 represents the tangent and
normal lines at the point P( 3, 1) to the circle
x2 + y2 = 4.
Y
N=0
P
T=0
O
M
x2+y2=4
A
X
219
Coordinate Geometry
So, equation of tangent (T = 0) is
…(i)
3x + y = 4
For point A, put y = 0, we get
4
x=
3
1
Q Area of required ∆OPA = (OA)(PM )
2
1
4
= ×
×1
2
3
[QPM = y-coordinate of P]
2
sq unit
=
3
12. Slope of a line passing through P (2, 3) and
intersecting the line, x + y = 7 at a distance
of 4 units from P, is
[JEE Main 2019, 9 April Shift-I]
(a)
1− 5
1+ 5
(b)
1− 7
(c)
1+ 7
(d)
7 −1
7+1
5 −1
5+1
Y
8
7
6
5
3
P (2 ,
2
R
3)
4
1
X′
1
Y′
2
3
4
θ
5
Q
6
7
8
⇒
⇒
⇒
m=
m2 − m1 

1 + m1m2 
1
7
1− 7
−1 − 7
or m =
1+ 7
7 −1
13. If the line y = mx + 7 3 is normal to the
hyperbola
x2 y 2
−
= 1, then a value of m is
24 18
[JEE Main 2019, 9 April Shift-I]
Let the slope of line is m, which is passing through
P(2, 3).
d
∴
QR = 16 − d 2 = 16 − 2 = 14
d
2
1
m+ 1
tanθ =
=
=
=
QR
14
7
1− m

Q tan θ =

m+ 1
1
=±
1− m
7
1
m+ 1
m+ 1
or
=
=−
1− m
7
1− m
3
(a)
5
2
(c)
5
Exp. (c)
4
Now, in ∆PRQ,
9
X
x+y=7
Since, the distance of a point ( x1, y1 ) from the line
ax + by1 + c
.
ax + by + c = 0 is d = 1
a2 + b 2
∴ The distance of a point P(2, 3) from the line
x + y − 7 = 0, is
|2 + 3 − 7| 2
d=
=
= 2
1+ 1
2
15
2
5
(d)
2
(b)
Exp. (c)
Given equation of hyperbola, is
x 2 y2
…(i)
−
=1
24 18
Since, the equation of the normals of slope m to
x 2 y2
the hyperbola 2 − 2 = 1, are given by
a
b
m(a2 + b 2 )
y = mx m
a2 − b 2 m2
∴ Equation of normals of slope m, to the hyperbola
(i), are
m(24 + 18)
…(ii)
y = mx ±
24 − m2 (18)
QLine y = mx + 7 3 is normal to hyperbola (i)
∴On comparing with Eq. (ii), we get
m(42 )
6m
±
= 7 3 ⇒±
= 3
24 − 18m2
24 − 18m2
⇒
⇒
⇒
⇒
36m2
[squaring both sides]
=3
24 − 18m2
12 m2 = 24 − 18m2
30m2 = 24
5m2 = 4 ⇒ m = ±
2
5
220
JEE Main Chapterwise Mathematics
14. If a tangent to the circle x 2 + y 2 = 1 intersects
the coordinate axes at distinct points P and
Q, then the locus of the mid-point of PQ is
[JEE Main 2019, 9 April Shift-I]
(a) x2 + y2 − 2x2y2 = 0
∴ Slope of focal chord having one end point is (1, 4)
4− 0
4
is m = tanα =
=−
1− 4
3
[where, ‘α’ is the inclination of focal
chord with X-axis.]
Since, the length of focal chord = 4a cosec 2α
(b) x2 + y2 − 2xy = 0
(c) x2 + y2 − 4x2y2 = 0
(d) x2 + y2 − 16x2y2 = 0
Exp. (c)
Equation of given circle is x2 + y2 = 1, then
equation of tangent at the point (cos θ,sinθ) on the
given circle is
…(i)
x cos θ + ysinθ = 1
[QEquation of tangent at the point
P(cos θ, sinθ) to the circle x2 + y2 = r 2
is x cos θ + ysinθ = r]
Now, the point of intersection with coordinate
axes are P(sec θ, 0) and Q(0, cos ec θ).
QMid-point of line joining points P and Q is
sec θ cos ecθ 
M 
,
 = (h, k ) (let)
 2
2 
1
1
and sinθ =
So, cosθ =
2h
2k
Qsin2 θ + cos 2 θ = 1
1
1
1
1
+ 2 = 1⇒ 2 + 2 = 4
∴
2
4h
4k
h
k
Now, locus of mid-point M is
1
1
+
=4
x 2 y2
⇒
x 2 + y2 − 4 x 2 y2 = 0
So, correct option is (c).
15. If one end of a focal chord of the parabola,
y 2 = 16x is at (1, 4), then the length of this
focal chord is
[JEE Main 2019, 9 April Shift-I]
(a) 22
(b) 25
(c) 24
Equation of given parabola is y2 = 16 x, its focus is
(4, 0).
Since, slope of line passing through ( x1, y1 ) and
y − y1
.
( x2 , y2 ) is given by m = tanθ = 2
x2 − x1
(d) 20
Exp. (b)
Key Idea
(i) First find the focus of the given parabola
(ii) Then, find the slope of the focal chord by
y − y1
using m = 2
x2 − x1
(iii) Now, find the length of the focal chord by
using the formula 4a cosec 2 α.
∴ The required length of the focal chord
= 16 [1 + cot 2 α ]
[Q a = 4 and cosec 2 α = 1 + cot 2 α ]
9
1
3
= 16 1 +  = 25units Qcot α =
=− 
 16 

4 
tanα
16. If the tangent to the parabola y 2 = x at a
point (α , β ),(β > 0) is also a tangent to the
ellipse, x 2 + 2 y 2 = 1, then α is equal to
[JEE Main 2019, 9 April Shift-II]
(a) 2 + 1
(c) 2 2 + 1
(b) 2 − 1
(d) 2 2 − 1
Exp. (a)
Since the point(α, β ) is on the parabola y2 = x, so
α = β2
…(i)
Now, equation of tangent at point (α, β ) to the
parabola y2 = x, is T = 0
1
yβ = ( x + α )
⇒
2
[Qequation of the tangent to the parabola
y2 = 4ax at a point ( x1 , y1 ) is given by
yy1 = 2 a( x + x1 )]
[from Eq. (i)]
⇒
2 yβ = x + β 2
x
β
…(ii)
y=
+
⇒
2β 2
Since, line (ii) is also a tangent of the ellipse
x 2 + 2 y2 = 1
2
2
 β  = (1)2  1  + 1
∴
 
 
2
2
 2β 
[Qcondition of tangency of line y = mx + c to
x2
y2
ellipse 2 + 2 = 1 is c 2 = a2 m2 + b 2 ,
a
b
β
1
1
here m =
, a = 1, b =
and c = 
2 
2β
2
221
Coordinate Geometry
⇒
1
1
β2
=
+
4
2
4β 2
⇒
β4 = 1 + 2β2
18. If the two lines x + (a − 1)y = 1 and
are
2 x + a 2y = 1, (a ∈R − {0, 1})
perpendicular, then the distance of their
⇒ β4 − 2β2 − 1 = 0
β2 =
⇒
2±
point of intersection from the origin is
4+ 4
[JEE Main 2019, 9 April Shift-II]
2
2±2 2
=
= 1±
2
β2 = 1 + 2
⇒
α = β = 1+
2
Q
2
(a)
5
2
[Qβ 2 > 0]
2
17. A rectangle is inscribed in a circle with a
diameter lying along the line 3y = x + 7. If the
two adjacent vertices of the rectangle are
(–8, 5) and (6, 5), then the area of the
rectangle (in sq units) is
[JEE Main 2019, 9 April Shift-II]
(a) 72
(c) 98
(b) 84
(d) 56
2
(b)
5
(c)
2
5
(d)
2
5
Exp. (d)
Key Idea
(i) If lines are perpendicular to each other, then
product of their slopes is −1, i.e. m1m2 = − 1
(ii) Distance between two points ( x1, y1 )
and ( x2 , y2 ) =
( x2 − x1 )2 + ( y2 − y1 )2
Given, lines x + (a − 1)y = 1
and 2 x + a2 y = 1, where a ∈ R − {0, 1}
are perpendicular to each other

1   2 
∴
−
 × −  = −1
 a − 1  a2 
Exp. (b)
Given points are (−8, 5) and (6, 5) in which
y-coordinate is same, i.e. these points lie on
horizontal line y = 5.
[QIf lines are perpendicular,
then product of their slopes is −1]
a2 (a − 1) = − 2
⇒
⇒
3y=x+7
a3 − a2 + 2 = 0
⇒ (a + 1)(a − 2 a + 2 ) = 0 ⇒ a = − 1
2
(–8, β)
(6, β)
(–8, 5)
(6, 5)
Let (−8, β) and (6, β ) are the coordinates of the
other vertices of rectangle as shown in the figure.
Since, the mid-point of line joining points (−8, 5)
and (6, β ) lies on the line 3 y = x + 7.
5 + β  −8 + 6
3 
+7
∴
 =
 2 
2
⇒
15 + 3β = − 2 + 14
⇒
3β = − 3
⇒
β = −1
Now, area of rectangle = |−8 − 6| × |β − 5|
= 14 × 6 = 84
∴Equation of lines are
…(i)
x − 2y = 1
and
…(ii)
2x + y = 1
On solving Eq. (i) and Eq. (ii), we get
3
1
x = and y = −
5
5
∴Point of intersection of the lines (i) and (ii)
3 1
is P , −  .
 5 5
3 1
Now, required distance of the point P , − 
 5 5
9
1
10
2
from origin =
+
=
=
25 25
25
5
19. The area (in sq units) of the smaller of the
two circles that touch the parabola, y 2 = 4x
at the point (1, 2) and the X -axis is
[JEE Main 2019, 9 April Shift-II]
(a) 8π(3 − 2 2 )
(c) 8π (2 − 2 )
(b) 4π (3 + 2 )
(d) 4π (2 − 2 )
222
JEE Main Chapterwise Mathematics
Exp. (a)
Given parabola y = 4 x
2
…(i)
So, equation of tangent to parabola (i) at point(1, 2 )
is 2 y = 2( x + 1)
[Qequation of the tangent to the parabola
y2 = 4ax at a point ( x1, y1 ) is given by
yy1 = 2 a( x + x1 )]
…(ii)
⇒
y= x+1
Now, equation of circle, touch the parabola at
point (1, 2) is
( x − 1)2 + ( y − 2 )2 + λ( x − y + 1) = 0
⇒ x + y + ( λ − 2 ) x + ( −4 − λ ) y + ( 5 + λ ) = 0
2
2
…(iii)
Also, Circle (iii) touches the x-axis, so g 2 = c
c1c 2 = 9 + 16 = 5 and|r1 − r2| = 5
Q
Q
c1c 2 = |r1 − r2| = 5
∴circle x2 + y2 = 4 touches the circle
x2 + y2 + 6 x + 8 y − 24 = 0 internally.
So, equation of common tangent is
S1 − S 2 = 0
⇒
6 x + 8 y − 20 = 0
…(i)
⇒
3 x + 4 y = 10
The common tangent passes through the point
(6, − 2 ), from the given options.
21. If the circles x 2 + y 2 + 5Kx + 2 y + K = 0 and
2 ( x 2 + y 2 ) + 2Kx + 3y −1 = 0,(K ∈R ), intersect
at the points P and Q, then the line
4x + 5y − K = 0 passes through P andQ, for
2
[JEE Main 2019, 10 April Shift-I]
⇒
 λ − 2 = 5 + λ


 2 
⇒
λ2 − 4λ + 4 = 4λ + 20
⇒
(a) no values of K
(b) exactly one value of K
(c) exactly two values of K
(d) infinitely many values of K
λ2 − 8λ − 16 = 0
8±
⇒
λ=
⇒
λ=4±
64 + 64
Now, radius of circle is r =
⇒
Exp. (a)
2
32 = 4 ± 4 2
g + f −c
2
2
r =|f|
[Qg 2 = c]
= 8 + 4 2
λ + 4
=
2
 2 
8−4 2
or
2
8−4 2
For least area r =
= 4 − 2 2 units
2
So, area = πr 2 = π(16 + 8 − 16 2 )
= 8 π(3 − 2 2 ) sq unit
20. The common tangent to the circles
x 2 + y 2 = 4 and x 2 + y 2 + 6x + 8y − 24 = 0 also
passes through the point
[JEE Main 2019, 9 April Shift-II]
(a) (6, − 2)
(c) (−6, 4)
(b) (4, − 2)
(d) (−4, 6)
Exp. (a)
Given circles are
x2 + y2 = 4, centre c1(0, 0) and radius r1 = 2
and x2 + y2 + 6 x + 8 y − 24 = 0, centre c 2 (−3, − 4)
and radius r2 = 7
Equation of given circles
x2 + y2 + 5Kx + 2 y + K = 0
…(i)
2( x2 + y2 ) + 2 Kx + 3 y − 1 = 0
3
1
…(ii)
x2 + y2 + Kx + y − = 0
⇒
2
2
On subtracting Eq. (ii) from Eq. (i), we get
1
1
4Kx + y + K + = 0
2
2
…(iii)
⇒
8Kx + y + (2 K + 1) = 0
[Qif S1 = 0 and S 2 = 0 be two circles, then their
common chord is given by S1 − S 2 = 0.]
Eq. (iii) represents equation of common chord as
it is given that circles (i) and (ii) intersects each
other at points P and Q.
Since, line 4 x + 5 y − K = 0 passes through point P
and Q.
8K 1 2 K + 1
1
∴
⇒K =
= =
−K
4
5
10
[equating first and second terms]
and − K = 10K + 5
[equating second and third terms]
5
⇒
11K + 5 = 0 ⇒ K = −
11
1
5
≠ − , so there is no such value of K, for
Q
10
11
which line 4 x + 5 y − K = 0 passes through points
P and Q.
and
223
Coordinate Geometry
22. If a directrix of a hyperbola centred at the
origin and passing through the point
( 4, − 2 3 ) is 5x = 4 5 and its eccentricity is e,
then
[JEE Main 2019, 10 April Shift-I]
(a)
(b)
(c)
(d)
4e4 − 12e2 − 27 = 0
4e4 − 24e2 + 27 = 0
4e4 + 8e2 − 35 = 0
4e4 − 24e2 + 35 = 0
23. If the line x − 2 y = 12 is tangent to the
 −9 
= 1 at the point  3,  , then
 2
a
b2
the length of the latusrectum of the ellipse
is
[JEE Main 2019, 10 April Shift-I]
ellipse
x2
+
y2
(a) 8 3
(c) 5
(b) 9
(d) 12 2
Exp. (b)
Exp. (d)
Let the equation of hyperbola is
x2
y2
…(i)
− 2 =1
2
a
b
Since, equation of given directrix is 5 x = 4 5
a
a
so 5  = 4 5
[Qequation of directrix is x = ]
e
e
a
4
…(ii)
=
⇒
e
5
and hyperbola (i) passes through point (4, − 2 3 )
16
so,
a2
−
12
b2
=1
e2 = 1 +
⇒
⇒
b
…(iii)
a
Key Idea Write equation of the tangent to the
ellipse at any point and use formula for
latusrectum of ellipse.
Equation of given ellipse is
x2
y2
+
=1
a2
b2
9
Now, equation of tangent at the point  3, −  on

2
the ellipse (i) is
3x
9y
…(ii)
−
=1
⇒
2
a
2 b2
x2
2
a
2
2
+
y2
b2
…(iv)
From Eqs. (ii) and (iv), we get
16 4 16 2
e − e = b2
5
5
From Eqs. (ii) and (iii), we get
…(v)
⇒
b2 =
b2
=
5
e2
−1 ⇒
12
b2
xx1
a2
+
yy1
b2
= 1]
=
⇒
1
2
12
=
=
3
9
1
a2
2 b2
a2 = 36and b 2 = 27
2 b 2 2 × 27
=
a
6
= 9units
Now, Length of latusrectum =
16
12
5
12
−
=1 ⇒ 2 − 2 =1
16 2 b 2
e
b
e
5
12
= 1 at the point ( x1, y1 ) is
QTangent (ii) represent the line x − 2 y = 12, so
a2
a2e 2 − a2 = b 2
⇒
…(i)
[Q the equation of the tangent to the ellipse
b2
The eccentricity e = 1 +
24. The line x = y touches a circle at the point
5−e
12e 2
⇒
16(e 2 − 1)(5 − e 2 ) = 60
⇒
4(5e 2 − e 4 − 5 + e 2 ) = 15
4e 4 − 24e 2 + 35 = 0
2
(1, 1). If the circle also passes through the
point (1, − 3), then its radius is
e2
5 − e2
From Eqs. (v) and (vi), we get
 12e 2 

16e 4 − 16e 2 = 5
2
5−e 
⇒
2
[JEE Main 2019, 10 April Shift-I]
…(vi)
(a) 3 2
(b) 2 2
(c) 2
(d) 3
Exp. (b)
Since, the equation of a family of circles touching
line L = 0 at their point of contact( x1, y1 ) is
( x − x1 )2 + ( y − y1 )2 + λ L = 0, where λ ∈ R.
∴ Equation of circle, touches the x = y at point (1, 1)
is
( x − 1)2 + ( y − 1)2 + λ( x − y) = 0
224
JEE Main Chapterwise Mathematics
⇒ x 2 + y2 + ( λ − 2 ) x + ( − λ − 2 ) y + 2 = 0
…(i)
QCircle (i) passes through point (1, − 3).
∴
1 + 9 + (λ − 2 ) + 3(λ + 2 ) + 2 = 0
⇒
4λ + 16 = 0
⇒
Exp. (a)
Let (h, k ) be the centre of the circle and radius
r = h, as circle touch the Y-axis and other circle
x2 + y2 = 1 whose centre (0, 0) and radius is 1.
Y
λ = −4
So, equation of circle (i) at λ = − 4, is
(h,k)
C
r
x 2 + y2 − 6 x + 2 y + 2 = 0
Now, radius of the circle =
9 + 1 − 2 = 2 2.
r=h>O
k>O
for first
quadrant
25. If 5x + 9 = 0 is the directrix of the hyperbola
 5 
(a)  − , 0
 3 
(b) (− 5, 0)
5 
(c)  , 0
3 
(d) (5, 0)
x2+y2=1
∴
OC = r + 1
[Qif circles touch each other externally,
then C1C 2 = r1 + r2 ]
h2 + k 2 = h + 1, h > 0
⇒
and k > 0, for first quadrant.
⇒
h2 + k 2 = h2 + 2 h + 1
Exp. (b)
Equation of given hyperbola is
⇒
16 x2 − 9 y2 = 144
x2
y2
−
=1
9
16
So, the eccentricity of Eq. (i)
⇒
e = 1+
X
O
16x 2 − 9y 2 = 144, then its corresponding
focus is
[JEE Main 2019, 10 April Shift-II]
…(i)
16 5
=
9
3
3x 2 + 5y 2 = 32 at the point P(2 , 2 ) meets the
x2
a2
−
y2
b2
= 1 is 1 + (b / a) ]
2
and given directrix is 5 x + 9 = 0 ⇒ x = − 9 / 5

So, corresponding focus is  −

5

3   , 0 = (− 5, 0)
 3 
26. The locus of the centres of the circles, which
touch the circle, x 2 + y 2 = 1 externally, also
touch the Y-axis and lie in the first quadrant,
is
[JEE Main 2019, 10 April Shift-II]
(b) y = 1 + 4x, x ≥ 0
(c) x = 1 + 2 y, y ≥ 0
(d) x = 1 + 4 y, y ≥ 0
k = 1 + 2 h, as k > 0
⇒
Now, on taking locus of centre (h, k ), we get
y = 1 + 2 x, x ≥ 0
27. The tangent and normal to the ellipse
[Qthe eccentricity (e) of the hyperbola
(a) y = 1 + 2x, x ≥ 0
k2 = 2 h + 1
X-axis at Q and R, respectively. Then, the
area (in sq units) of the ∆PQR is
[JEE Main 2019, 10 April Shift-II]
16
(a)
3
14
(b)
3
(c)
34
15
(d)
68
15
Exp. (d)
Equation of given ellipse is
3 x2 + 5 y2 = 32
…(i)
Now, the slope of tangent and normal at point
P(2, 2 ) to the ellipse (i) are respectively
dy
dx
and mN = −
mT =
dx ( 2, 2 )
dy ( 2, 2 )
On differentiating ellipse (i), w.r.t. x, we get
dy
dy
3x
6 x + 10 y
=0 ⇒
=−
dx
dx
5y
So, mT = −
3x
3
5y
5
= − and mN =
=
5 y ( 2, 2 )
5
3 y ( 2, 2 ) 3
225
Coordinate Geometry
Now, equation of tangent and normal to the given
ellipse (i) at point P(2, 2 ) are
3
( y − 2) = − ( x − 2)
5
5
and ( y − 2 ) = ( x − 2 ) respectively.
3
It is given that point of intersection of tangent and
normal are Q and R at X-axis respectively.
16
4
So,
Q  , 0 and R  , 0
 3 
5 
1
∴ Area of ∆PQR = (QR ) × height
2
1 68
= ×
×2
2 15
68
sq units
=
15
[QQR =
2
2
 16 − 4  =  68  = 68

 

 3
 15 
5
15
and height = 2]
29. If the line ax + y = c, touches both the curves
x 2 + y 2 = 1 and y 2 = 4 2 x , then| c | is equal to
[JEE Main 2019, 10 April Shift-II]
1
(a)
2
(b) 2
(c)
2019, 10 April Shift-II]
2
 1
(a)  − , − 
 4
3
1
1
(c)  , − 
4
3
 1 2
(b)  − , 
 4 3
 1 1
(d)  , 
 4 3
Key Idea Use the equation of tangent of slope
a
‘m’ to the parabola y 2 = 4ax is y = mx +
and
m
a line ax + by + c = 0 touches the circle
| c|
x 2 + y 2 = r 2 , if
= r.
a2 + b2
Since, equation of given parabola is y2 = 4 2 x
and equation of tangent line is ax + y = c or
y = − ax + c,
2
2
[Qm = slope of line = − a]
thenc =
=
m
−a
⇒
∴Equation of line parallel to line
4 x − 3 y + 2 = 0 is 4 x − 3 y + k = 0
…(i)
Now, distance of line (i) from the origin is
| k|
3
=
2
2
5
4 + 3
[as per question’s requirement]
⇒
Then, equation of tangent line becomes
2
…(i)
y = − ax −
a
QLine (i) is also tangent to the circle x2 + y2 = 1.
∴ Radius = 1 =
Since, equation of a line parallel to line
ax + by + c = 0 is ax + by + k = 0
⇒
[Qline y = mx + c touches the
parabola y2 = 4ax iff c = a / m].
2
a
−
Exp. (b)
| k| = 3
k=± 3
So, possible lines having equation, either
4 x − 3 y + 3 = 0 or 4 x − 3 y − 3 = 0
1 2
Now, from the given options the point  − , 
 4 3
lies on the line 4 x − 3 y + 3 = 0.
1
2
Exp. (c)
28. Lines are drawn parallel to the line
3
from the
4x − 3y + 2 = 0, at a distance
5
origin. Then which one of the following
points lies on any of these lines?
[JEE Main
(d)
2
1 + a2
1 + a2 = −
2
a
2
[squaring both sides]
a2
⇒ a4 + a2 − 2 = 0 ⇒ (a2 + 2 ) (a2 − 1) = 0
⇒
⇒
∴
1 + a2 =
a2 = 1
| c| =
[Q a2 > 0, ∀ a ∈ R ]
2
= 2
| a|
30. If the angle of intersection at a point where
the two circles with radii 5 cm and 12 cm
intersect is 90°, then the length (in cm) of
their common chord is
[JEE Main 2019, 12 April Shift-I]
13
(a)
5
60
(c)
13
120
13
13
(d)
2
(b)
226
JEE Main Chapterwise Mathematics
In ∆C1 AC 2 ,
Exp. (b)
Let,
the
length
= AB = 2 AM = 2 x
of
common
chord
[Q∠C1 AC 2 = 90º, because circles
intersects each other at 90º]
A
12
C1
12
= (12 )2 + (5)2
5
= 144 + 25
= 169 = 13 cm
C2
M
5
and
C1C 2 =
AC12 + AC 22
… (i)
[Qcircles intersect each other at 90º]
C1C 2 = C1M + MC 2
⇒ C1C 2 = 12 2 − AM 2 +
52 − AM 2
… (ii)
From Eqs. (i) and (ii), we get
AC12 + AC 22 = 144 − AM 2 +
⇒
144 + 25 = 144 − x2 +
13 = 144 − x2 +
⇒
25 − AM 2
25 − x2
31. If the normal to the ellipse 3x 2 + 4y 2 = 12 at a
25 − x2
point P on it is parallel to the line, 2 x + y = 4
and the tangent to the ellipse at P passes
throughQ( 4, 4) then PQ is equal to
On squaring both sides, we get
169 = 144 − x2 + 25 − x2
+ 2 144 − x2
⇒
x = 144 − x
2
25 − x
2
25 − x2
2
Again, on squaring both sides, we get
= (144 × 25) − (25 + 144)x2 + x4
⇒
144 × 25
x =
169
12 × 5 60
cm
x=
=
13
13
2
Now, length of common chord 2 x =
120
cm
13
Alternate Solution
Given, AC1 = 12 cm and AC 2 = 5 cm
A
12
C1
5
M
C2
B
Common
chord
[JEE Main 2019, 12 April Shift-I]
5 5
(a)
2
(b)
61
2
(c)
221
2
(d)
157
2
Exp. (a)
x4 = (144 − x2 ) (25 − x2 )
⇒
1
AC1 × AC 2
2
1
= × 12 × 5 = 30 cm2
2
1
Also, area of ∆C1 AC 2 = C1C 2 × AM
2
1
AB
Q AM = AB 
= × 13 ×

2 
2
2
1
∴ × 13 × AM = 30 cm
4
120
AM =
cm
13
Now, area of ∆C1 AC 2 =
B
Now,
C1C 2 = (C1 A)2 + ( AC 2 )2
Key Idea Equation of tangent and normal to the
x2
y2
ellipse 2 + 2 = 1 at point p(x1, y1 ) is
a
b
xx
yy
T = 0 ⇒ 21 + 21 = 1
a
b
a2 x
b2 y
and
−
= a2 − b 2 respectively.
x1
y1
Equation of given ellipse is 3 x2 + 4 y2 = 12
x2
y2
… (i)
+
=1
⇒
4
3
Now, let point P(2 cos θ, 3 sinθ) , so equation of
tangent to ellipse (i) at point P is
xcos θ ysinθ
… (ii)
+
=1
2
3
Since, tangent (ii) passes through point Q(4, 4)
4
… (iii)
∴
sinθ = 1
2 cos θ +
3
227
Coordinate Geometry
and equation of normal to ellipse (i) at point P is
4x
3y
−
=4−3
2 cos θ
3 sinθ
… (iv)
2 xsinθ − 3 cos θy = sinθ cos θ
⇒
Since, normal (iv) is parallel to line, 2 x + y = 4
∴ Slope of normal (iv) = slope of line, 2 x + y = 4
2
⇒
tanθ = − 2
3
tanθ = − 3 ⇒θ = 120º
⇒
 3
1
,− 
⇒ (sinθ, cos θ) = 
2
 2
3
Hence, point P − 1, 

2
3
Now, PQ = (4 + 1)2 +  4 − 

2
2
25 5 5
=
4
2
common tangents to the parabola y 2 = 12 x
and the hyperbola 8x 2 − y 2 = 8. If S and S ′
denotes the foci of the hyperbola where S lies
on the positive X-axis then P divides SS ′ in a
ratio
[JEE Main 2019, 12 April Shift-I]
(b) 14 : 13
(d) 2 : 1
Exp. (c)
Equation of given parabola y2 = 12 x
… (i)
and hyperbola 8 x2 − y2 = 8
… (ii)
Now, equation of tangent to parabola y = 12 x
3
having slope ‘m’ is y = mx +
… (iii)
m
and equation of tangent to hyperbola
x2
y2
−
= 1having slope ‘m’ is
1
8
2
y = mx ±
12 m2 − 8
…(iv)
Since, tangents (iii) and (iv) represent the same
line
∴
⇒
⇒
⇒
3
m − 8 =  
 m
2
minor axis of length 4, passes through
which of the following points?
(a) ( 2 , 2)
(c) (2, 2 2 )
2
Let the equation of ellipse be
x2
y2
…(i)
+ 2 =1
2
a
b
Since, foci are at (0, 2) and (0, − 2 ), major axis is
along the Y-axis.
So,
…(ii)
be = 2
[where e is the eccentricity of ellipse]
and 2a = length of minor axis = 4
[given]
…(iii)
⇒
a=2
a2
2
Q
e = 1− 2
b
m = ± 3.
2
2 = 1− 4
Qe = 2 
 

 b
b 
b2
8
= 1 ⇒ b2 = 8
⇒
b2
Thus, equation of required ellipse is
x2
y2
+
=1
4
8
x2
y2
Now, from the option the ellipse
+
=1
4
8
passes through the point ( 2 , 2 ).
∴
34. A circle touching the X -axis at (3, 0) and
making a intercept of length 8 on the
Y -axis passes through the point
[JEE Main 2019, 12 April Shift-II]
m4 − 8m2 − 9 = 0
(m2 − 9) (m2 + 1) = 0
(b) (2, 2 )
(d) (1, 2 2 )
Exp. (a)
32. Let P be the point of intersection of the
(a) 13 : 11
(c) 5 : 4
33. An ellipse, with foci at (0, 2) and (0, − 2 ) and
[JEE Main 2019, 12 April Shift-II]
[given cordinates of Q ≡≡ (4, 4)]
= 25 +
Now, equation of common tangents to the
parabola (i) and hyperbola (ii) are
y = 3 x + 1and y = − 3 x − 1
Q Point ‘P’ is point of intersection of above
common tangents,
∴
P(− 1 / 3, 0)
and focus of hyperbola S(3, 0) and S ′(− 3, 0).
Thus, the required ratio
3 + 1 / 3 10 5
PS
=
=
=
=
PS ′ 3 − 1 / 3
8
4
(a) (3, 10)
(c) (2, 3)
(b) (3, 5)
(d) (1, 5)
228
JEE Main Chapterwise Mathematics
C
which will give the points of intersection of tangent
and rectangular hyperbola.
4
Since, line y = mx +
is also a tangent to the
m
rectangular hyperbola.
∴Discriminant of quadratic equation
4
mx2 + x + 4 = 0, should be zero.
m
[Qthere will be only one point of intersection]
r
⇒
Exp. (a)
It is given that the circle touches the X-axis at (3, 0)
and making an intercept of 8 on the Y-axis.
Y
B
M
8
r
A
O
X
(3,0)
2
4
D =   − 4 (m) (4) = 0
 m
⇒
Let the radius of the circle is ‘r’, then the
coordinates of centre of circle are (3, r ).
From the figure, we have
CM = 3, CA = radius = r
AB
and
AM = BM =
=4
2
Then, r 2 = CM 2 + AM 2 = 9 + 16 = 25
⇒
r=± 5
Now, the equation of circle having centre (3, ± 5)
and radius = 5 is
( x − 3)2 + ( y ± 5)2 = 25
Now, from the options (3, 10) satisfy the equation
of circle ( x − 3)2 + ( y − 5)2 = 25
m3 = 1 ⇒ m = 1
So, equation of required tangent is y = x + 4.
36. A triangle has a vertex at (1, 2) and the
mid-points of the two sides through it are
( −1, 1) and (2 , 3). Then, the centroid of this
triangle is
[JEE Main 2019, 12 April Shift-II]
 7
(a) 1, 
 3
1 
(c)  , 1
3 
1 
(b)  , 2
3 
 1 5
(d)  , 
 3 3
Exp. (b)
Let a ∆ABC is such that vertices
A(1, 2 ), B( x1 y1 ) and C( x2 , y2 ).
35. The equation of a common tangent to the
A(1,2)
curves, y 2 = 16x and xy = − 4, is
[JEE Main 2019, 12 April Shift-II]
(a) x − y + 4 = 0
(c) x − 2 y + 16 = 0
(b) x + y + 4 = 0
(d) 2x − y + 2 = 0
B(x1,y1)
Exp. (a)
Key Idea An equation of tangent having slope
a
‘m’ to parabola y2 = 4ax is y = mx + .
m
Given equation of curves are
y2 = 16 x (parabola)
…(i)
and
xy = − 4 (rectangular hyperbola)…(ii)
Clearly, equation of tangent having slope ‘m’ to
4
parabola (i) is y = mx +
…(iii)
m
Now, eliminating y from Eqs. (ii) and (iii), we get
4
x  mx +  = − 4

m
4
mx2 +
x + 4 = 0,
⇒
m
C(x2,y2)
It is given that mid-point of side AB is (− 1, 1.)
x1 + 1
So,
= −1
2
y1 + 2
and
=1
2
⇒
x1 = − 3 and y1 = 0
So, point B is (− 3, 0)
Also, it is given that mid-point of side AC is
(2, 3), so
x2 + 1
y +2
= 2 and 2
=3
2
2
⇒
x2 = 3and y2 = 4
So, point C is (3, 4).
Now, centroid of ∆ABC is
1 + (− 3) + 3 2 + 0 + 4 
1 
G 
,
 = G  , 2
3 


3
3
229
Coordinate Geometry
37. A straight line L at a distance of 4 units from
the origin makes positive intercepts on the
coordinate axes and the perpendicular from
the origin to this line makes an angle of 60°
with the linex + y = 0. Then, an equation of
the line L is
[JEE Main 2019, 12 April Shift-II]
(a) x + 3 y = 8
(b) ( 3 + 1) x + ( 3 − 1) y = 8 2
(c) 3x + y = 8
(d) ( 3 − 1)x + ( 3 + 1) y = 8 2
Exp. (d)
According to the question, we have the following
figure.
(0,b)
M
60°
O
(a,0)
α
x+
y=
0
x
y
— + — =1
a b
Letθ be the inclination of the line x + y = 0.Then,
tanθ = − 1 = tan (180° − 45° )
⇒
tanθ = tan 135°
⇒
θ = 135°
⇒
α + 60° = 135°
⇒
α = 75°
Since, line L having perpendicular distance
OM = 4.
So, equation of the line ‘L’ is
x cos α + y sinα = 4
⇒
x cos 75° + y sin 75° = 4
⇒ x cos (45° + 30° ) + y sin (45° + 30° ) = 4
 3
 3
1 
1 
⇒ x
−
+
+ y
=4
2
2
2
2
2
2
2
2



⇒
( 3 − 1) x + y ( 3 + 1) = 8 2
Exp. (c)
Given equation of parabola is
y = ( x − 2 )2 − 1
⇒
y = x2 − 4 x + 3
…(i)
Now, let (x1, y1) be the point of intersection of
tangents of parabola (i) and line.
x − y = 3, then
Equation of chord of contact of point ( x1, y1 ) w.r.t.
parabola (i) is
T=0
1
⇒
( y + y1 ) = xx1 − 2( x + x1 ) + 3
2
⇒
y + y1 = 2 x ( x1 −2 ) − 4 x1 + 6
⇒ 2 x( x1 − 2 ) − y = 4 x1 + y1 − 6, this equation
represent the line x − y = 3 only, so on comparing,
we get
2( x1 − 2 ) − 1 4 x1 + y1 − 6
=
=
−1
1
3
5
x1 = and y1 = − 1
⇒
2
5
So, the required point is  , − 1 .
2

39. Axis of a parabola lies along X -axis. If its
vertex and focus are at distances 2 and 4
respectively from the origin, on the positive
X-axis, then which of the following points
does not lie on it? [JEE Main 2019, 9 Jan Shift-I]
(a) (4, −4)
(c) (8, 6)
(b) (6, 4 2)
(d) (5, 2 6)
Exp. (c)
According to given information, we have the
following figure.
Y
(2, 0)
(4, 0)
X
38. The tangents to the curve y = ( x − 2 )2 − 1 at
its points of intersection with the line
x − y = 3, intersect at the point
[JEE Main 2019, 12 April Shift-II]
5 
(a)  , 1
2 

 5
(b)  − , − 1

 2
5

(c)  , − 1
2

 5 
(d)  − , 1
 2 
Now, if the origin is shifted to (2, 0) and ( X, Y ) are
the coordinates with respect to new origin, then
equation of parabola is Y 2 = 4aX,
where, X = x − 2 and Y = y and a = 4 − 2 = 2
∴ y2 = 8( x − 2 )
Note that (8, 6) is the only point which does not
satisfy the equation.
230
JEE Main Chapterwise Mathematics
40. Consider the set of all lines px + qy + r = 0
such that 3p + 2q + 4r = 0. Which one of the
following statements is true?
[JEE Main 2019, 9 Jan Shift-I]
(a) Each line passes through the origin.
3 1
(b) The lines are concurrent at the point  , 
 4 2
(c) The lines are all parallel
(d) The lines are not concurrent
Given, px + qy + r = 0 is the equation of line such
that
3 p + 2q + 4r = 0
Consider,
3 p + 2q + 4r = 0
3 p 2q
⇒
+
+ r=0
4
4
(dividing the equation by 4)
3
1


p  + q   + r = 0
⇒
2
 4
3 1
⇒  ,  satisfy px + qy + r = 0
 4 2
So, the lines always passes through the point
 3, 1 .


 4 2
41. Three circles of radii a ,b ,c(a < b < c ) touch
each other externally. If they have X -axis as
a common tangent, then [JEE Main 2019, 9 Jan
Shift-I]
(a) a , b , c are in AP
1
1
1
(b)
=
+
a
b
c
∴In ∆ACE, AE = (a + c )2 − (c − a)2 = 2 ac
Similarly, BC = b + c and CF = c − b
Q
AD + AE = BF
∴ 2 ab + 2 ac = 2 bc
1
1
1
⇒
+
=
c
b
a
42. Equation of a common tangent to the circle,
x 2 + y 2 − 6x = 0 and the parabola, y 2 = 4x , is
[JEE Main 2019, 9 Jan Shift-I]
(a)
(c)
3y = 3x + 1
3y = x + 3
(b) 2 3y = 12 x + 1
(d) 2 3y = − x − 12
Exp. (c)
We know that, equation of tangent to parabola
y2 = 4ax is
a
y = mx +
m
∴Equation of tangent to the parabola y2 = 4 x is
1
(Qa = 1)
y = mx +
m
…(i)
⇒
m2 x − my + 1 = 0
Now, let line (i) is also a tangent to the circle.
Equation of circle x2 + y2 − 6 x = 0
Clearly, centre of given circle is (3, 0) and radius
=3
[Qfor the circle x2 + y2 + 2 gx + 2 fy + c = 0,
a , b , c are in AP
1
1
1
(d)
=
+
b
a
c
(c)
centre = (− g , − f ) and radius =
Exp. (b)
According to given information, we have the
following figure.
C
B
D
(using Pythagoras theorem in ∆ABD)
= 2 ab
Similarly, AC = a + c and CE = c − a
∴In ∆BCF, BF = (b + c )2 − (c − b )2 = 2 bc
Exp. (b)
b
Clearly, AB = a + b (sum of radii) and BD = b − a
∴ AD = (a + b )2 − (b − a)2
∴ The perpendicular distance of (3, 0) from the line
(i) is 3.
[Qradius is perpendicularto the
tangent of circle]
|m2 ⋅ 3 − m ⋅ 0 + 1|
=3
⇒
(m2 )2 + (− m)2
The length of perpendicular from a point ( x1, y1 ) to
F
c
the line ax + by + c = 0 is
E
A
g 2 + f 2 − c]
a
where A, B, C are the centres of the circles
⇒
3m2 + 1
m4 + m2
=3
ax1 + by1 + c
a2 + b 2
.
231
Coordinate Geometry
⇒
9m4 + 6m2 + 1 = 9(m4 + m2 )
1
⇒
m ≈ ∞or m = ±
3


1
3+ 2
2


3m + 1
m
= lim
= 3
Q lim

 m → ∞ m4 + m2 m → ∞ 1 + 1


m2
∴ Equation of common tangents are x = 0,
x
−x
y=
+ 3 and y =
− 3
3
3
 using y = mx + 1 



m
x = 0, 3 y = x + 3
3y = − x − 3
i.e.
and
π
2
x2
43. Let 0 < θ < . If the eccentricity of the
hyperbola
y2
−
= 1 is greater than 2,
Since, both sinθ and tanθ are increasing functions
π π
in  , 
3 2
∴ Least value of latus rectum is
π
π
3
π
> 2 sin ⋅ tan = 2 ⋅
⋅ 3 = 3  at θ = 

3
2
3
3
and greatest value of latusrectum is < ∞
Hence, latusrectum length ∈ (3, ∞ ).
44. Let the equations of two sides of a triangle be
3x − 2 y + 6 = 0 and 4x + 5y − 20 = 0. If the
orthocentre of this triangle is at (1, 1) then
the equation of its third side is
[JEE Main 2019, 9 Jan Shift-I]
(a)122 y − 26x − 1675 = 0 (b) 26x − 122 y − 1675 = 0
(c)122 y + 26x + 1675 = 0 (d) 26x + 61y + 1675 = 0
Exp. (b)
cos θ sin θ
then the length of its latus rectum lies in the
interval
[JEE Main 2019, 9 Jan Shift-I]
2
2
3
(a) (1, ]
2
3
(c) ( ,2]
2
Exp. (b)
For the hyperbola
e = 1+
x2
a
2
−
y2
b2
= 1,
b2
a2
∴ For the given hyperbola,
e = 1+
sin2 θ
cos 2 θ
>2
(Qa2 = cos 2 θ and b 2 = sin2 θ)
⇒
1 + tan2 θ > 4
⇒
tan2 θ > 3
⇒
E
tanθ ∈ (− ∞, − 3 ) ∪ ( 3, ∞ )
[x > 3 ⇒| x| > 3 ⇒ x ∈ (−∞, − 3 ) ∪ ( 3, ∞ )]
π
But θ ∈  0,  ⇒ tanθ ∈ ( 3, ∞ )
 2
π π
θ ∈  , 
⇒
3 2
2
Now, length of latusrectum
2 b2
sin2 θ
=
=2
= 2sinθ tanθ
a
cos θ
H
F
(1, 1)
(b) (3,∞)
(d) (2, 3]
A
C
3x–2y+6=0
B
4x+5y–20=0
Let equation of AB be 4 x + 5 y − 20 = 0 and AC be
3x − 2 y + 6 = 0
3
Clearly, slope of AC =
2
a
[Qslope of ax + by + c = 0 is − ]
b
∴ Slope of altitude BH, which is perpendicular to

1 
2
AC = − .
Q mBH = −

mAC 
3

Equation of BH is given by y − y1 = m( x − x1 )
2
m = − , x1 = 1and y1 = 1
3
2
y − 1 = − ( x − 1) ⇒ 2 x + 3 y − 5 = 0
∴
3
Now, equation of AB is 4 x + 5 y − 20 = 0 and
Here,
equation of BH is 2 x + 3 y − 5 = 0
Solving these, we get point of intersection
(i.e. coordinates of B).
4 x + 5 y − 20 = 0 
 ⇒ y = − 10
4 x + 6 y − 10 = 0
On substituting y = − 10 in 2 x + 3 y − 5 = 0, we get
35
x=
2
232
JEE Main Chapterwise Mathematics
35
B  , − 10
2

∴
Solving 4 x + 5 y − 20 = 0 and 3 x − 2 y + 6 = 0, we
get coordinate of A.
12 x + 15 y − 60 = 0
 ⇒23 y = 84
12 x − 8 y + 24 = 0 
84
10
⇒
⇒x=
y=
23
23
10 84 

∴A , 
 23 23 

 84
 y − y1   23 − 1
61
.
Now, slope of AH =  2
=
=
10
−
x
x
−
13
 2

1
− 1

 23
QBC is perpendicular to AH.
13
∴Slope of BC is
61

1 
Q mBC = −

m

AH 
Now, equation of line BC is given by
y − y1 = m( x − x1 ), where ( x1, y1 ) are coordinates of
B.
13 
35 
∴
y − (− 10) =
x−

61 
2 
13
y + 10 =
(2 x − 35)
⇒
61 × 2
⇒
122 y + 1220 = 26 x − 455
⇒ 26 x − 122 y − 1675 = 0
For y2 = 4ax, parametric coordinates of a point is
(at 2 , 2at).
∴For y2 = 4 x, let coordinates of C be (t 2 , 2t).
1
Then, area of ∆ABC =
2
[JEE Main 2019, 9 Jan Shift-II]
(b) 32
(c) 31
3
4
(d) 30
1
2
Exp. (a)
C (t2, 2t)
B (9, 6)
Clearly, A(4, − 4) ≡
⇒
and
A(t 12 , 2t 1 ) ⇒2t 1
...(i)
=−4
t1 = − 2
B(9, 6) ≡ B(t 22 ,2t 2 )
⇒
2t 2 = 6 ⇒ t 2 = 3
Since, C is on the arc AOB, the parameter ‘t’ for
point C∈ (− 2, 3).
Let f(t ) = t 2 − t − 6 ⇒ f ′(t ) = 2t − 1
1
Now,
f ′(t ) = 0 ⇒t =
2
1
Thus, for A(t ), critical point is at t =
2
2
1
1
1


Now, A  = 5   − − 6
2
2
2
=
125
1
= 31
4
4
[using Eq. (i)]
46. Let S be the set of all triangles in the
xy-plane, each having one vertex at the
origin and the other two vertices lie on
coordinate axes with integral coordinates.
If each triangle in S has area 50 sq. units,
then the number of elements in the set S is
[JEE Main 2019, 9 Jan Shift-II]
(b) 32
(c) 18
(d) 9
Exp. (a)
According to given information, we have the
following figure.
Y
B
A (4, –4)
−4 1
A(t ) = 5| t 2 − t − 6|
Let
(a) 36
According to given information, we have the
following figure.
4
1
1
1 2
|t (6 − (− 4)) − 2t (9 − 4) + 1(− 36 − 24)|
2
1
10
= |10t 2 − 10t − 60| = |t 2 − t − 6|
2
2
= 5| t 2 − t − 6|
2
1
(a) 31
4
2t
6
=
45. Let A ( 4, − 4) and B(9, 6) be points on the
parabola, y = 4x . LetC be chosen on the arc
AOB of the parabola, where O is the origin,
such that the area of ∆ACB is maximum.
Then, the area (in sq. units) of ∆ACB , is
t2
9
O
(0, b)
A (a, 0)
X
233
Coordinate Geometry
(Note that as a and b are integers so they can be
negative also)
Here O(0, 0), A(a, 0) and B(0, b ) are the three
vertices of the triangle.
Clearly, OA =| a| and OB =| b|.
1
∴Area of ∆OAB = | a||b |.
2
But area of such triangles is given as 50 sq units.
1
|a|| b| = 50
∴
2
⇒
|a|| b | = 100 = 2 2 ⋅ 52
Number of ways of distributing two 2’s in|a|
and| b| = 3
| a|
| b|
0
2
1
1
2
0
⇒ | r − 6|< (8 − 4)2 + (10 − 7 )2 < r + 6
⇒
| r − 6| < 16 + 9 < r + 6
⇒
| r − 6| < 5 < r + 6
Now as, 5 < r + 6 always, we have to solve only
| r − 6| < 5
⇒
− 5< r − 6< 5
⇒
6 − 5 < r < 5 + 6 ⇒1 < r < 11
48. A hyperbola has its centre at the origin,
passes through the point (4, 2) and has
transverse axis of length 4 along the X -axis.
Then the eccentricity of the hyperbola is
[JEE Main 2019, 9 Jan Shift-II]
(a) 2
2
(b)
3
(c)
3
2
(d) 3
Exp. (b)
⇒ 3 ways
Similarly, number of ways of distributing two 5’s in
| a| and|b | = 3 ways.
∴Total number of ways of distributing 2’s and 5’s
= 3 × 3 = 9 ways
Note that for one value of| a| , there are 2 possible
values of a and for one value of |b |, there are 2
possible values of b.
∴Number of such triangles possible
= 2 × 2 × 9 = 36.
So, number of elements in S is 36.
47. If the circles x 2 + y 2 −16x − 20y + 164 = r 2
and ( x − 4)2 + ( y − 7)2 = 36 intersect at two
distinct points, then
[JEE Main 2019, 9 Jan
Equation of hyperbola is given by
x2
y2
− 2 =1
2
a
b
QLength of transverse axis = 2 a = 4
∴
a= 2
x2
y2
Thus,
− 2 = 1 is the equation of hyperbola
4
b
QIt passes through (4, 2 ).
16
4
4
− 2 = 1⇒ 4 − 2 = 1
∴
4
b
b
4
2
b2 = ⇒ b =
⇒
3
3
Now, eccentricity,
4
b2
1
2
e = 1+ 2 = 1+ 3 = 1+ =
4
3
3
a
Shift-II]
(a) 0 < r < 1
(c) 1 < r < 11
(b) r > 11
(d) r =11
4x 2 − 5y 2 = 20 parallel to the line x − y = 2 is
[JEE Main 2019, 10 Jan Shift-I]
Exp. (c)
Circle I is
x2 + y2 − 16 x − 20 y + 164 = r 2
⇒
49. The equation of a tangent to the hyperbola
( x − 8)2 + ( y − 10)2 = r 2
⇒ C1(8, 10)is the centre of Ist circle and r1 = r is its
radius
Circle II is ( x − 4)2 + ( y − 7 )2 = 36
⇒ C 2 (4, 7 ) is the centre of 2nd circle and r2 = 6 is
its radius.
Two circles intersect if|r1 − r2 | < C1C 2 < r1 + r2
(a) x − y − 3 = 0
(c) x − y + 1 = 0
(b) x − y + 9 = 0
(d) x − y + 7 = 0
Exp. (c)
Given equation of hyperbola is
4 x2 − 5 y2 = 20
which can be rewritten as
x2
y2
⇒
−
=1
5
4
The line x − y = 2 has slope, m = 1
∴ Slope of tangent parallel to this line = 1
234
JEE Main Chapterwise Mathematics
We know equation of tangent to hyperbola
x2
y2
−
= 1 having slope m is given by
a2 b 2
y = mx ±
a2 m2 − b 2
[JEE Main 2019, 10 Jan Shift-I]
2
∴Required equation of tangent is
y= x± 5−4
⇒
⇒
y = x± 1 ⇒ x − y ± 1 = 0
have a common normal, then which one of
the following is a valid choice for the ordered
triad (a ,b , c ) ?
[JEE Main 2019, 10 Jan Shift-I]
(b) (1, 1, 0)
(c) (1, 1, 3)
1
(d)  , 2 , 3
2

Exp. (c)
Normal to parabola y2 = 4ax is given by
y = mx − 2 am − am3
∴ Normal to parabola y2 = 4b( x − c ) is
y = m( x − c ) − 2 bm − bm3
[replacing a by b and x by x − c]
… (i)
= mx − (2 b + c )m − bm3
and normal to parabola y2 = 8 ax is
y = mx − 4am − 2 am3
…(ii)
[replacing a by 2a]
For common normal, we should have
mx − 4am − 2 am3 = mx − (2 b + c )m − bm3
[using Eqs. (i) and (ii)]
4am + 2 am3 = (2 b + c )m + bm3
⇒
(2 a − b )m3 + (4a − 2 b − c )m = 0
⇒
m((2 a − b )m2 + (4a − 2 b − c )) = 0
3
2
(d) parallel to X-axis
(b) with slope
Exp. (a)
50. If the parabolas y 2 = 4b ( x − c ) and y 2 = 8ax
1
(a)  , 2 , 0
2

Q(1, 4) and R( 3, − 2 ) are fixed points, then the
locus of the centroid of ∆PQR is a line
2
(a) with slope
3
(c) parallel to Y -axis
Here, a = 5, b = 4 and m = 1
2
51. A point P moves on the line 2 x − 3y + 4 = 0. If
⇒
m= 0
2 b + c − 4a
c
or
m2 =
=
−2
2a − b
2a − b
c
As, m2 > 0, therefore
>2
2a − b
Note that if m = 0, then all options satisfy
(Q y = 0 is a common normal) and if common
normal is other than the axis, then only option (c)
satisfies.


c
3
Q for option (c), 2 a − b = 2 − 1 = 3 > 2 


Let the coordinates of point P be ( x1, y1 )
QP lies on the line 2 x − 3 y + 4 = 0
∴
2 x1 − 3 y1 + 4 = 0
2x + 4
…(i)
y1 = 1
⇒
3
Now, let the centroid of ∆PQR be G(h, k ), then
x + 1+ 3
h= 1
3
…(ii)
⇒
x1 = 3h − 4
y1 + 4 − 2
and
k=
3
2 x1 + 4
+2
3
[from Eq. (i)]
⇒
k=
3
2x + 4 + 6
3k = 1
⇒
3
…(iii)
⇒
9k − 10 = 2 x1
Now, from Eqs. (ii) and (iii), we get
2(3h − 4) = 9k − 10
⇒
6h − 8 = 9k − 10
⇒ 6h − 9k + 2 = 0
Now, replace h by x and k by y.
⇒6 x − 9 y + 2 = 0, which is the required locus and
2
slope of this line is
3
Qslope of ax + by + c = 0 is − a 

b 
52. If a circle C passing through the point (4, 0)
touches the circle x 2 + y 2 + 4x − 6y = 12
externally at the point(1, − 1), then the radius
ofC is
[JEE Main 2019, 10 Jan Shift-I]
(a) 5
(b) 2 5
(c)
57
(d) 4
Exp. (a)
Equation of tangent to the circle
x2 + y2 + 4 x − 6 y − 12 = 0 at (1, − 1) is given by
xx1 + yy1 + 2 ( x + x1 ) − 3( y + y1 ) − 12 = 0,
where x1 = 1 and y1 = − 1
235
Coordinate Geometry
⇒ x − y + 2 ( x + 1) − 3( y − 1) − 12 = 0
⇒
3x − 4y − 7 = 0
This will also a tangent to the required circle.
Now, equation of family of circles touching the line
3 x − 4 y − 7 = 0 at point (1, − 1) is given by
( x − 1)2 + ( y + 1)2 + λ (3 x − 4 y − 7 ) = 0
Also, let incentre is (h k ), then
ax + bx2 + cx3
h= 1
a+ b+c
So, the equation of required circle will be
( x − 1)2 + ( y + 1)2 + λ(3 x − 4 y − 7 ) = 0, for some
…(i)
λ ∈R
QThe required circle passes through (4, 0)
∴(4 − 1)2 + (0 + 1)2 + λ (3 × 4 − 4 × 0 − 7 ) = 0
and
⇒ 9 + 1 + λ (5) = 0 ⇒ λ = − 2
Substituting λ = − 2 in Eq. (i), we get
( x − 1)2 + ( y + 1)2 − 2 (3 x − 4 y − 7 ) = 0
∴ incentre is (2, 2)
⇒
x2 + y2 − 8 x + 10 y + 16 = 0
On comparing it with
x2 + y2 + 2 gx + 2 fy + c = 0, we get
g = − 4, f = 5, c = 16
∴
Radius =
g2 + f2 − c
53. If the line 3x + 4y − 24 = 0 intersects the
X -axis at the point A and theY -axis at the
point B, then the incentre of the triangle
OAB, whereO is the origin, is
[JEE Main 2019, 10 Jan Shift-I]
(b) (3, 4)
(c) (4, 4)
(here, y1 = 0, y2 = 6, y3 = 0)
6 × 0 + 8 × 6 + 10 × 0 48
=
=2
=
24
6 + 8 + 10
54. The shortest distance between the point
3 
 , 0 and the curve y = x ,( x > 0), is
2 
[JEE Main 2019, 10 Jan Shift-I]
3
(a)
2
5
(b)
4
(c)
(d) (2, 2)
Exp. (d)
(d)
5
2
Let P( x1, y1 ) be any point on the curve y = x .
Clearly,y1 = x1 ⇒ x1 = y12 [Q( x1, y1 ) lies on y = x]
∴ The point is P( y12 , y1 )
3
Now, let the given point be A  , 0 , then
2 
PA =
2
 y2 − 3  + y2
 1

1

2
9
+ y12
4
9
5
= y14 − 2 y12 + = ( y12 − 1)2 +
4
4
Clearly, PA will be least when y12 − 1 = 0.
=
Given equation of line is
3 x + 4 y − 24 = 0
For intersection with X-axis put y = 0
⇒
3 x − 24 = 0 ⇒ x = 8
For intersection with Y-axis, put x = 0
⇒
4 y − 24 = 0 ⇒ y = 6
∴ A(8, 0) and B (0, 6)
⇒ PA min =
y14 − 3 y12 +
0+

5
5
=
4
2
55. Let S = ( x , y ) ∈R 2 :
B(0,6)
3
2
Exp. (d)
= 16 + 25 − 16 = 5
(a) (4, 3)
(here, x1 = 8, x2 = 0, x3 = 0)
6 × 8 + 8 × 0 + 10 × 0
48
=
=
=2
6 + 8 + 10
24
ay + by2 + cy3
k= 1
a+ b+c


y2
x2
−
= 1 ,
1+r 1−r

where r ≠ ± 1. Then, S represents
[JEE Main 2019, 10 Jan Shift-II]
(a) a hyperbola whose eccentricity is
O
Let AB = c =
A(8,0)
8 + 6 = 10
2
2
OB = a = 6 and OA = b = 8
when 0 < r < 1.
(b) a hyperbola whose eccentricity is
when 0 < r < 1.
2
,
1−r
2
,
r +1
236
JEE Main Chapterwise Mathematics
2
, when
r +1
(c) an ellipse whose eccentricity is
r >1.
1
, when
r +1
(d) an ellipse whose eccentricity is
r >1.
Exp. (c)


y2
x2
Given, S = ( x, y) ∈ R 2 :
−
= 1
1+ r 1− r




y2
x2
= ( x, y) ∈ R 2 :
+
= 1
r
r
1
1
+
−


y2
x2
+
= 1, represents a vertical
1+ r r − 1
ellipse.
[Qfor r > 1, r − 1 < r + 1and r − 1 > 0]
r −1
Now, eccentricity (e ) = 1 −
r+1

x2
y2
a2 
Q for 2 + 2 = 1, a < b, e = 1 − 2 
a
b
b 

(r + 1) − (r − 1)
2
=
=
r+1
r+1
For r > 1,
56. Two sides of a parallelogram are along the
lines, x + y = 3 and x − y + 3 = 0. If its
diagonals intersect at (2, 4), then one of its
vertex is
[JEE Main 2019, 10 Jan Shift-II]
(a) (3, 6)
(b) (2, 6)
(c) (2, 1)
(d) (3, 5)
Now, as point M (2, 4) is mid-point of line joining
the points A and C, so
0 + x2 3 + y2 
(2, 4) = 
,

 2
2 

 x1 + x2 y1 + y2  
Qmid-point =  2 , 2  


3 + y2
0 + x2
;4=
2=
⇒
2
2
⇒
x2 = 4 and y2 = 5
∴Thus,
C ≡ (4, 5)
Now, equation of line BC is given by
( y − y1 ) = m ( x − x1 )
y − 5 = 1( x − 4)
[line BC is parallel to x − y + 3 = 0 and slope
(−1)
of x − y + 3 = 0 is
= 1]
(−1)
…(iii)
⇒
y = x+1
and equation of line DC is
y − 5 = −1 ( x − 4)
[line DC is parallel to x + y = 3and
−1
slope of x + y = 3is
= −1]
1
…(iv)
⇒
x+ y = 9
On solving Eqs. (i) and (iii), we get B (1, 2 ) and on
solving Eqs. (ii) and (iv), we get D (3, 6)
57. Two vertices of a triangle are (0, 2) and
(4, 3). If its orthocentre is at the origin, then its
third vertex lies in which quadrant?
Exp. (a)
[JEE Main 2019, 10 Jan Shift-II]
According to given information, we have the
following figure
(a) Fourth
(c) Second
(b) Third
(d) First
Exp. (c)
C (x2 , y2 )
D
x–y+3=0
M
Let ABC be a given triangle with vertices
B(0, 2 ), C(4, 3) and let third vertex be A(a, b )
A (a, b)
(2,4)
B
A
E
F
x+y=3
[Note that given lines are perpendicular to each
other as m1 × m2 = −1]
Clearly, point A is point of intersection of lines
…(i)
x+ y = 3
and
…(ii)
x − y = −3
So,
A = (0, 3)
[solving Eqs. (i) and (ii)]
(0,0)
(0, 2) B
D
C (4,3)
Also, let D, E and F are the foot of perpendiculars
drawn from A, B and C respectively.
b − 0 3−2
Then, AD⊥ BC ⇒
= −1
×
a− 0 4− 0
[if two lines having slopes m1 and m2 , are
perpendicular then m1m2 = −1]
237
Coordinate Geometry
…(i)
⇒
b + 4a = 0
and CF⊥ AB
b −2 3− 0
⇒
= −1
×
a− 0 4− 0
⇒
3b − 6 = −4a
…(ii)
⇒
4a + 3b = 6
From Eqs. (i) and (ii), we get
− b + 3b = 6 ⇒ 2 b = 6
⇒
b=3
3
and
[from Eq. (i)]
a=−
4
So, the third vertex
3
(a, b ) ≡  − , 3 , which lies in II quadrant.
 4 
58. The length of the chord of the parabola
x 2 = 4y having equation x − 2 y + 4 2 = 0 is
[JEE Main 2019, 10 Jan Shift-II]
(a) 8 2
(b) 2 11
(c) 3 2
(d) 6 3
Given, equation of parabola is x2 = 4 y
… (i)
and the chord is x − 2 y + 4 2 = 0
From Eqs. (i) and (ii), we have
[ 2 ( y − 4)]2 = 4 y
… (ii)
2( y − 4)2 = 4 y
⇒
( y − 4) = 2 y
⇒
y − 8 y + 16 = 2 y
⇒
y2 − 10 y +16 = 0
= ( x1 + x2 )2 − 4 x1 x2 + ( y1 + y2 )2 − 4 y1 y2
[Q(a − b )2 = (a + b )2 − 4ab ]
= 8 + 64 + 100 − 64
= 108
[from Eqs. (iv) and (vi)]
=6 3
59. The straight line x + 2 y = 1 meets the
coordinate axes at A and B. A circle is drawn
through A , B and the origin. Then, the sum
of perpendicular distances from A and B on
the tangent to the circle at the origin is
[JEE Main 2019, 11 Jan Shift-I]
(a) 2 5
5
(b)
4
(c) 4 5
B (x2 y2 )
O
Y
(0,1/2)
B
A
O
…(iii)
x– √ 2 y+4 √ 2 =0
(x1 , y1 )
X
Let the roots of Eq. (iii) be y1 and y2
Then, y1 + y2 = 10 and y1 y2 = 16
Again from Eqs. (i) and (ii), we have
x
x2 = 4 
+ 4
 2

x2 − 2 2 x − 16 = 0
Let the roots of Eq. (v) be x1 and x2
Then,
x1 + x2 = 2 2
and
x1 x2 = −16
5
2
N
2
A
(d)
Exp. (d)
2
Y
⇒
= ( x1 − x2 )2 + ( y1 − y2 )2
According to given information, we have the
following figure.
Exp. (d)
⇒
Clearly, length of the chord AB
… (iv)
… (v)
… (vi)
x
(1, 0) x+2y=1
M
From figure, equation of circle (diameter form) is
1
( x − 1) ( x − 0) + ( y − 0) y −  = 0

2
y
x 2 + y2 − x − = 0
⇒
2
y
Equation of tangent at (0, 0) is x + = 0
2
[Qequation of tangent at ( x1, y1 ) is given by T = 0.
Here, T = 0
1
1
⇒ xx1 + yy1 − ( x + x1 ) − ( y + y1 ) = 0]
2
4
⇒
2x + y = 0
|2 ⋅ 1 + 1⋅ 0|
2
Now,
AM =
=
5
5
[Qdistance of a point P( x1, y1 ) from a line
|ax + by1 + c |
]
ax + by + c = 0 is 1
a2 + b 2
238
JEE Main Chapterwise Mathematics
1
2 ⋅ 0 + 1 
2
1
and
BN =
=
5
2 5
4+1
2
1
5
+
=
=
∴ AM + BN =
5 2 5 2 5
2
60. A square is inscribed in the circle
x 2 + y 2 − 6x + 8y − 103 = 0 with its sides
parallel to the coordinate axes. Then, the
distance of the vertex of this square which is
nearest to the origin is
[JEE Main 2019, 11 Jan Shift-I]
(a) 6
(c) 41
(b) 13
(d) 137
Now,
and
Given equation of circle is
x2 + y2 − 6 x + 8 y − 103 = 0, which can be written
as ( x − 3)2 + ( y + 4)2 = 128 = (8 2 )2
61. Two circles with equal radii are intersecting
at the points (0, 1) and (0, −1). The tangent at
the point (0,1) to one of the circles passes
through the centre of the other circle. Then,
the distance between the centres of these
circles is
[JEE Main 2019, 11 Jan Shift-I]
(a)
2
(c) 1
(d) 2
Y
S2
(0, 1)
r
r
(–α, 0)
Y
S1
(α, 0)
X
(0,–1)
C
X
O
45°
Let C1(α, 0) and C 2 (− α, 0) are the centres.
Then,
S1 ≡ ( x − α )2 + y2 = α 2 + 1
⇒
(3,–4)
A
(b) 2 2
Clearly, circles are orthogonal because tangent at
one point of intersection is passing through centre
of the other.
∴ Centre = (3, − 4) and radius = 8 2
Now, according to given information, we have the
following figure.
G
25 + 144 = 169 = 13;
121 + 144 = 265
121 + 16 = 137
25 + 16 = 41
Exp. (d)
Exp. (c)
D
OA =
OB =
OC =
OD =
S1 ≡ x2 + y2 − 2 αx − 1 = 0
[Qradius, r = (α − 0)2 + (0 − 1)2 ]
B
For the coordinates of A and C.
x−3 y+ 4
Consider,
=
=± 8 2
1
1
2
2
[using distance (parametric) form of line,
x − x1 y − y1
=
= r]
cos θ
sinθ
⇒
x = 3 ± 8, y = − 4 ± 8
∴ A(− 5, − 12 ) and C(11, 4)
Similarly, for the coordinates of Band D, consider
x−3 y+ 4
=
=±8 2
1
1
−
2
2
[in this case, θ = 135°]
⇒
x = 3 m 8, y = − 4 ± 8
∴ B (11, − 12 ) and D(− 5, 4)
and
S 2 ≡ ( x + α )2 + y2 = α 2 + 1
⇒
S 2 ≡ x2 + y2 + 2αx − 1 = 0
Now, 2(α ) (− α ) + 2 ⋅ 0 ⋅ 0 = (− 1) + (− 1) ⇒α = ± 1
[Qcondition of orthogonality is
2 g1g 2 + 2 f1f2 = c1 + c 2 ]
∴C1(1, 0) and C 2 (− 1, 0) ⇒C1C 2 = 2
62. Equation of a common tangent to the
parabola y 2 = 4x and the hyperbola xy = 2 is
[JEE Main 2019, 11 Jan Shift-I]
(a) x + 2 y + 4 = 0
(c) 4x + 2 y + 1 = 0
(b) x − 2 y + 4 = 0
(d) x + y + 1 = 0
Exp. (a)
a
is the equation of
m
2
tangent to the parabola y = 4ax.
1
is a tangent to the parabola
∴ y = mx +
m
[Q a = 1]
y2 = 4 x
We know that, y = mx +
239
Coordinate Geometry
Let, this tangent is also a tangent to the hyperbola
xy = 2
1
Now, on substituting y = mx + in xy = 2, we get
m
1
x  mx +  = 2.

m
⇒
m2 x2 + x − 2 m = 0
Note that tangent touch the curve exactly at one
point, therefore both roots of above equations are
equal.
⇒
D = 0 ⇒1 = 4(m2 ) (− 2 m)
3
1
m3 =  − 
 2
1
m= −
⇒
2
∴Required equation of tangent is
x
y= − −2
2
⇒
2y = − x − 4
⇒
x + 2y + 4 = 0
⇒
63. If tangents are drawn to the ellipse
x 2 + 2 y 2 = 2 at all points on the ellipse other
than its four vertices, then the mid-points of
the tangents intercepted between the
coordinate axes lie on the curve
Now, the equation of tangent at P is
x 2 cos θ ysinθ
+
=1
2
1
[Qequation of tangent at ( x1, y1 ) is given by T = 0
xx
yy
⇒ 21 + 21 = 1

a
b
x
y
+
=1
⇒
2 sec θ cosec θ
∴ A ( 2 sec θ, 0) and B (0, cosec θ)
Let mid-point of AB be R(h, k ), then
cosec θ
2 sec θ
and k =
h=
2
2
2 h = 2 sec θand 2k = cosec θ
1
1
and sinθ =
⇒
cosθ =
2h
2k
We know that, cos 2 θ + sin2 θ = 1
1
1
∴
+
=1
2 h2
4k 2
1
1
So, locus of (h, k ) is 2 +
=1
2x
4 y2
64. A circle cuts a chord of length 4a on the
X-axis and passes through a point on the
Y-axis, distant 2b from the origin. Then, the
locus of the centre of this circle, is
[JEE Main 2019, 11 Jan Shift-II]
[JEE Main 2019, 11 Jan Shift-I]
x2 y 2
(a)
+
=1
4
2
(b)
x2 y 2
+
=1
2
4
(d)
(c)
1
4x 2
1
2x
2
+
+
1
2y 2
1
4y 2
=1
(a) a parabola
(c) a straight line
=1
Exp. (a)
(b) an ellipse
(d) a hyperbola
According to given information, we have the
following figure.
Exp. (d)
Given equation of ellipse is x2 + 2 y2 = 2 , which
can be written as
x2
y2
+
=1
2
1
Let P be a point on the ellipse, other than its four
vertices. Then, the parametric
coordinates of P be ( 2 cos θ, sinθ)
y
(0, 2b)
P
4a
A
B
O
Let the equation of circle be
x2 + y2 + 2 gx + 2 fy + c = 0
…(i)
According the problem,
B
4a = 2 g 2 − c
P (√2 cos θ, sin θ)
A
x
…(ii)
[Qthe length of intercepts made by the
circle x2 + y2 + 2 gx + 2 fy + c = 0
with X-axis is 2 g 2 − c ]
240
JEE Main Chapterwise Mathematics
Also, as the circle is passing through P(0, 2 b )
∴
0 + 4b + 0 + 4bf + c = 0
[using Eq. (i)]
4b + 4bf + c = 0
…(iii)
2
⇒
2
Eliminating ‘c’ from Eqs. (ii) and (iii), we get
4b 2 + 4bf + g 2 − 4a2 = 0
66. If the area of the triangle whose one vertex is
[Q4a = 2 g 2 − c ⇒c = g 2 − 4a2 ]
So, locus of (− g , − f ) is
4b 2 − 4by + x2 − 4a2 = 0
⇒
x2 = 4by + 4a2 − 4b 2
which is a parabola.
65. Let the length of the latus rectum of an
ellipse with its major axis along X-axis and
centre at the origin, be 8. If the distance
between the foci of this ellipse is equal to the
length of its minor axis, then which one of
the following points lies on it?
[JEE Main 2019, 11 Jan Shift-II]
(a) ( 4 2 , 2 3 )
(c) ( 4 2 , 2 2 )
x2
y2
+
=1
64 32
Now, check all the options.
Only (4 3, 2 2 ), satisfy the above equation.
∴Equation of ellipse be
(b) ( 4 3 , 2 2 )
(d) ( 4 3 , 2 3 )
at
the
vertex
of
the
parabola,
y 2 + 4( x − a 2 ) = 0 and the other two vertices
are the points of intersection of the parabola
andY -axis, is 250 sq units, then a value of ‘a’
is
[JEE Main 2019, 11 Jan Shift-II]
Exp. (b)
Vertex of parabola
y2 = − 4( x − a2 )is (a2 , 0).
For point of intersection with Y-axis, put x = 0 in the
given equation of parabola.
This gives,
y2 = 4a2
⇒
y = ± 2a
Thus, the point of intersection are (0, 2a) and
(0, − 2 a).
Exp. (b)
Let the equation of ellipse be
x2
y2
+ 2 =1
2
a
b
Then, according the problem, we have
2 b2
= 8 and 2 ae = 2 b
a
2 b2
and
[length of latus rectum =
a
length of minor axis = 2b]
b
b
b   = 4and = e
 a
a
⇒
b(e ) = 4
1
b = 4.
⇒
e
Also, we know that b 2 = a2 (1 − e 2 )
⇒
⇒
⇒
b2
a2
1
2
From Eqs. (i) and (ii), we get
b=4 2
32
b2
Now,
a2 =
=
= 64
1 − e2 1 − 1
2
⇒
…(i)
= 1 − e2
e2 = 1 − e2
2e 2 = 1 ⇒ e =
(b) 5
(d) (10)2 / 3
(a) 5 5
(c) 5( 21/ 3 )
Q b = e 
 a

…(ii)
Y
B (0, 2a)
O
A
(a2, 0)
X
C(0, –2a)
From the given condition, we have
Area of ∆ABC = 250
1
∴
(BC )(OA) = 250
2
1
[Qarea = × base × height]
2
1
(4a)a2 = 250
⇒
2
⇒
a3 = 125 = 53
∴
a=5
67. If a hyperbola has length of its conjugate axis
equal to 5 and the distance between its foci
is 13, then the eccentricity of the hyperbola
is
[JEE Main 2019, 11 Jan Shift-II]
(a)
13
12
(b) 2
(c)
13
8
(d)
13
6
241
Coordinate Geometry
Exp. (a)
2
2
x
y
−
= 1, where
a2 b 2
2
2 2
b = a (e − 1), the length of conjugate axis is 2b
and distance between the foci is 2ae.
∴According the problem, 2 b = 5and 2 ae = 13
Now, b 2 = a2 (e 2 − 1)
We
⇒
know
that
in
2
 5  = a2e 2 − a2
 
2
3y − 6 = 5x − 5
5x − 3y + 1 = 0
69. Let P( 4, − 4) andQ(9, 6) be two points on the
parabola, y 2 = 4x and let X be any point on
the arc POQ of this parabola, whereO is the
vertex of this parabola, such that the area of
∆PXQ is maximum. Then, this maximum
area (in sq units) is
[JEE Main 2019, 12 Jan Shift-I]
125
(a)
2
2
25 (2 ae )
=
− a2
4
4
25 169
=
− a2
4
4
169 − 25 144
⇒
=
= 36
a2 =
4
4
⇒
a=6
Now, 2 ae = 13
⇒
⇒
⇒
[Q2 ae = 13]
2 × 6 × e = 13
13
e=
12
75
(b)
2
(c)
625
4
(d)
125
4
Exp. (d)
Given parabola is y2 = 4 x,
Since, X lies on the parabola, so let the
coordinates of X be (t 2 , 2t ). Thus, the coordinates
of the vertices of the triangle PXQ are P(4,–4), X
(t 2 ,2t ) and Q(9, 6).
Y
Q(9,6)
X (t 2,2t)
y 2=4x
68. If in a parallelogram ABDC , the coordinates
of A , B andC are respectively (1, 2), (3, 4) and
(2, 5), then the equation of the diagonal AD
is
[JEE Main 2019, 11 Jan Shift-II]
(a) 3x + 5y − 13 = 0
(c) 5x − 3y + 1 = 0
(b) 3x − 5y + 7 = 0
(d) 5x + 3y − 11 = 0
Exp. (c)
According to given information, we have the
following figure
C(2, 5)
D
P
A(1, 2)
X′
B(3, 4)
We know that, diagonals of a parallelogram
intersect at mid-point.
5 9
∴ P = Mid-point of BC and so, P ≡  , 
2 2
Now, equation of AD is.
9
−2
( y − 2) = 2
( x − 1)
5
−1
2
5
y − 2 = ( x − 1)
⇒
3
X
O
P(4,–4)
Y′
∴Area of ∆PXQ =
1
2
4
−4 1
t2
9
2t
6
1
1
1
[4(2t − 6) + 4(t 2 − 9) + 1(6t 2 − 18t ]
2
1
= | [8t − 24 + 4t 2 − 36 + 6t 2 − 18t ]|
2
= | 5t 2 − 5t − 30| = | 5(t + 2 ) (t − 3)|
=
Now, as X is any point on the arc POQ of the
parabola, therefore ordinate of point X, 2t ∈ (− 4, 6)
⇒t ∈ (− 2, 3).
∴ Area of ∆PXQ = − 5(t + 2 ) (t − 3)
= − 5t 2 + 5t + 30
[Q| x − a| = − ( x − a), if x < a]
The maximum area (in square units)
 25 − 4(− 5) (30) 125
=−
= 4
4(− 5)


[Qmaximum value of quadratic expression
D
ax2 + bx + c, when a < 0 is − ]
4a
242
JEE Main Chapterwise Mathematics
70. Let C 1 and C 2 be the centres of the circles
72. The maximum area (in sq. units) of a
and
x + y − 2x − 2y − 2 = 0
x 2 + y 2 − 6x − 6y + 14 = 0 respectively. If P
andQ are the points of intersection of these
circles, then the area (in sq units) of the
quadrilateral PC 1QC 2 is [JEE Main 2019, 12 Jan
rectangle having its base on the X -axis and
its other two vertices on the parabola,
y = 12 − x 2 such that the rectangle lies inside
the parabola, is [JEE Main 2019, 12 Jan Shift-I]
2
2
Shift-I]
(a) 8
(c) 6
(b) 4
(d) 9
… (i)
x2 + y2 − 6 x − 6 y + 14 = 0
… (ii)
are intersecting each other orthogonally, because
2(1)(3) + 2(1)(3) = 14 − 2
[Qtwo circles are intersected
orthogonally if 2 g1g 2 + 2 f1f2 = c1 + c 2 ]
(d)18 3
Exp. (c)
x2 = − ( y − 12 ).
Note that vertex of parabola is (0, 12) and its open
downward.
Let Q be one of the vertices of rectangle which lies
on parabola. Then, the coordinates of Q be
(a, 12− a2 )
Y
(0, 12)
R
P
2
C1
(1,1)
2
S
O
P(a,0)
X
y=12–x2
Y′
Q
Then, area of rectangle PQRS
= 2 × (Area of rectangle PQMO)
[due to symmetry about Y-axis]
= 2 × [a(12 − a2 )] = 24a − 2 a3 = ∆ (let).
71. If the straight line, 2 x − 3y + 17 = 0 is
perpendicular to the line passing through
the points (7, 17) and (15, β), then β equals
[JEE Main 2019, 12 Jan Shift-I]
(b) − 5
Q(a, 12–a2)
M
X′
(3,3)
C2
So, area of quadrilateral
PC1QC 2 = 2 × ar (∆PC1C 2 ).
1
= 2 ×  × 2 × 2  = 4 sq units
2

35
3
(c) 32
or
Given circles,
x 2 + y2 − 2 x − 2 y − 2 = 0
(a)
(b) 20 2
Equation of parabola is given, y = 12 − x2
Exp. (b)
and
(a) 36
(c) −
35
3
(d) 5
Exp. (d)
Slope of the line 2 x − 3 y + 17 = 0 is
2
= m1, (let) and the slope of line joining the points
3
(7, 17) and
β − 17 β − 17
(15, β ) is
=
= m2 (let)
15 − 7
8
According to the question, m1m2 = − 1
2 β − 17
×
= −1
⇒
3
8
⇒
β − 17 = − 12 ⇒ β = 5.
The area function ∆ a will be maximum, when
d∆
=0
da
⇒
24 − 6a2 = 0
⇒
a2 = 4 ⇒a = 2
[Q a > 0]
So, maximum area of rectangle
PQRS = (24 × 2 ) − 2 (2 )3
= 48 − 16 = 32 sq units
73. If a variable line, 3x + 4y − λ = 0 is such that
the two circles
x 2 + y 2 − 2 x − 2 y + 1 = 0 and
x 2 + y 2 − 18x − 2 y + 78 = 0
are on its opposite sides, then the set of all
values of λ is the interval
[JEE Main 2019, 12 Jan Shift-I]
(a) [13, 23]
(c) [12, 21]
(b) (2, 17)
(d) (23, 31)
243
Coordinate Geometry
Exp. (c)
The given circles,
x 2 + y2 − 2 x − 2 y + 1 = 0
... (i)
x + y − 18 x − 2 y + 78 = 0,
... (ii)
and
2
2
are on the opposite sides of the variable line
3 x + 4 y − λ = 0. So, their centres also lie on the
opposite sides of the variable line.
⇒
[3(1) + 4(1) − λ ] [3(9) + 4(1) − λ ] < 0
[QThe points P( x1, y1 ) and Q( x2 , y2 ) lie on the
opposite sides of the line ax + by + c = 0,
if (ax1 + by1 + c )(ax2 + by2 + c ) < 0]
⇒
(λ − 7 )(λ − 31) < 0
⇒
Also, we have
λ ∈(7, 31)
3(1) + 4(1) − λ
≥ 1+ 1− 1
5
⇒ |7 − λ|≥ 5 ⇒λ ∈ (− ∞, 2 ] ∪ [12 , ∞ )
3(9) + 4(1) − λ
and
≥ 81 + 1 − 78
5
... (iii)
(d) ( x 2 + y 2 )2 = 4Rx 2 y 2
Let the foot of perpendicular be P(h, k ). Then, the
k
slope of line OP =
h
Y
B
... (v)
A
X
O
Shift-I]
(b) (6, 5 2 )
(d) ( − 6, 2 10 )
Exp. (b)
The vertices of hyperbola are given as (± 2, 0) and
one of its foci is at (− 3, 0).
∴ (a, 0) = (2, 0) and (− ae, 0), = (− 3, 0)
On comparing x-coordinates both sides, we get
a = 2 and − ae = − 3
QLine AB is perpendicular to line OP, so slope of
h
[Qproduct of slopes of two
line AB = −
k
perpendicular lines is(−1)]
Now, the equation of line AB is
h
y − k = − ( x − h)
k
⇒
hx + ky = h2 + k 2
x
y
or
+
=1
 h2 + k 2   h2 + k 2 

 

h  
k 

 h2 + k 2 
 h2 + k 2 

So, point A
, 0 and B 0,
h
k 



Q∆AOB is a right angled triangle, so AB is one of
the diameter of the circle having radius R (given).
⇒ AB = 2 R
2e = 3
3
e=
2
b2
9
Also, = 1 +
⇒ b2 = 5
4
4
(b) ( x 2 + y 2 )3 = 4R 2 x 2 y 2
... (iv)
(2 ,0) and one of its foci be at ( −3, 0), then
which one of the following points does not
lie on this hyperbola? [JEE Main 2019, 12 Jan
⇒
[JEE Main 2019, 12 Jan Shift-II]
(a) ( x 2 + y 2 )2 = 4R 2 x 2 y 2
Exp. (b)
74. If the vertices of a hyperbola be at ( −2 , 0) and
⇒
origin O and intersects the coordinate axes
at A and B, then the locus of the foot of
perpendicular fromO on AB is
P(h, k)
⇒
|λ − 31| ≥ 10
⇒
λ ∈ (− ∞, 21] ∪ [41, ∞ )
From Eqs. (iii), (iv) and (v), we get
λ ∈[12, 21]
⇒
75. If a circle of radius R passes through the
(c) ( x 2 + y 2 )( x + y ) = R 2 xy
Qdistance of centre from the given line is 


ax + by1 + c
≥ r
greater than the radius,i.e. 1


a2 + b 2
(a) ( 2 6 , 5)
(c) ( 4, 15 )
So, equation of the hyperbola is
x2
y2
… (i)
−
=1
4
5
The point (6, 5 2 ) from the given options does not
satisfy the above equation of hyperbola.
 2
b2 
Qe = 1 + 2 
a 

2
⇒
 h2 + k 2 
 h2 + k 2 

 + 

h 
k 


2
= 2R
244
JEE Main Chapterwise Mathematics
⇒
1
1
(h2 + k 2 )2  2 + 2  = 4R 2
h
k 
⇒
(h2 + k 2 )3 = 4R 2 h2 k 2
On replacing h by x and k by y, we get
( x2 + y2 )3 = 4R 2 x2 y2 ,
⇒
2e 2 = 1
1
e2 =
2
⇒
From Eqs. (i) and (iii), we get
1
1
b 2 = a2   = 16 
2
2
which is the required locus.
76. Let S and S ′ be the foci of an ellipse and B be
any one of the extremities of its minor axis. If
∆S ′ BS is a right angled triangle with right
angle atB and area( ∆S ′ BS ) = 8 sq units, then
the length of a latus rectum of the ellipse is
[JEE Main 2019, 12 Jan Shift-II]
(a) 2 2
(c) 2
(b) 4 2
(d) 4
x2
2
+
y2
2
Y
S′(–ae,o)O
b2 = 8
Now, length of latus rectum
=
2 b2
a
=
2×8
= 4 units
4
76. If a straight line passing through the point
[JEE Main 2019, 12 Jan Shift-II]
= 1.
a
b
Then, according to given information, we have the
following figure.
B
⇒
[using Eq. (ii)]
P( − 3, 4) is such that its intercepted portion
between the coordinate axes is bisected atP,
then its equation is
Exp. (d)
Let the ellipse be
…(iii)
(a)
(b)
(c)
(d)
x − y + 7=0
4x − 3y + 24 = 0
3x − 4y + 25 = 0
4x + 3y = 0
Exp. (b)
(0,b)
S(ae,0)
X
Let the equation of required line having intercepts
a and b with the axes is
x y
…(i)
+ =1
a b
Y
B (0,b)
Clearly, slope of line SB =
b
− ae
b
b
and slope of line S ′B =
ae
Q Lines SB and S ′ B are perpendicular, so
 b  .  b  = −1

  
 − ae   ae 
[Qproduct of slopes of two
perpendicular lines is (−1]
)
…(i)
⇒
b 2 = a2e 2
Also, it is given that area of ∆S ′ BS = 8
1 2
a =8
∴
2
[QS ′ B = SB = a because S ′ B + SB = 2 a and
S ′ B = SB]
…(ii)
⇒
a2 = 16 ⇒ a = 4
b2
2
2
[from Eq. (i)]
e = 1− 2 = 1− e
Q
a
P (–3,4)
A (a,0)
O
a
X
Now, according to given information,
P is the mid-point of AB
a b
P =  ,  = (−3, 4)
∴
2 2 
⇒
[given]
(a, b ) = (−6, 8)
On putting the value of a and b in Eq. (i), we get
x
y
+ =1
−6 8
⇒
8 x − 6 y = −48
⇒
4 x − 3 y + 24 = 0
245
Coordinate Geometry
78. A straight line through a fixed point (2, 3)
80. If the tangent at (1, 7) to the curve x 2 = y − 6
intersects the coordinate axes at distinct
points P and Q. If O is the origin and the
rectangleOPRQ is completed, then the locus
of R is
[JEE Main 2018]
touches the circle x 2 + y 2 + 16x + 12 y + c = 0,
then the value of c is
[JEE Main 2018]
(a) 3x + 2 y = 6
(c) 3x + 2 y = xy
(b) 2 x + 3y = xy
(d) 3x + 2 y = 6xy
Exp. (c)
R(α, β)
Q(0, β)
(2, 3)
P(α, 0)
x
y
+ =1
α β
Since this line is passes through fixed point (2, 3).
2
3
+ =1
∴
α β
∴Locus of R is 2β + 3α = αβ
i.e. 2 y + 3 x = xy ⇒ 3 x + 2 y = xy
Equation of line PQ is
79. Let the orthocentre and centroid of a
triangle be A( −3, 5) and B( 3, 3), respectively.
IfC is the circumcentre of this triangle, then
the radius of the circle having line segment
[JEE Main 2018]
AC as diameter, is
(a) 10
(b) 2 10
(c) 3
5
2
(d)
3 5
2
Exp. (c)
Key Idea Otrhocentre, centroid and circumcentre
are collinear and centroid divide orthocentre and
circumcentre in 2 : 1 ratio.
We have orthocentre and centroid of a triangle
be A(−3, 5) and B(3, 3) respectively and C
circumcentre.
A(–3, 5)
Clearly,
B(3,3)
(a) 195
(b) 185
(d) 95
Exp. (d)
Key Idea Equation of tangent to the curve
y + y1 
x2 = 4ay at ( x1, y1 ) is xx1 = 4a

 2 
Tangent to the curve x2 = y − 6 at (1, 7 ) is
y+7
x=
−6
2
…(i)
⇒ 2x − y + 5 = 0
Equation of circle is x2 + y2 + 16 x + 12 y + c = 0
Centre (−8, − 6)
r=
82 + 62 − c = 100 − c
Since, line 2 x − y + 5 = 0 also touches the circle.
2(−8) − (−6) + 5
100 − c = 
∴



2 2 + 12 

−16 + 6 + 5

⇒
100 − c = 
5


100 − c = |− 5| ⇒100 − c = 5 ⇒ c = 95
⇒
81. Tangent and normal are drawn at P(16, 16)
on the parabola y 2 = 16x , which intersect the
axis of the parabola at A and B, respectively.
If C is the centre of the circle through the
pointsP, A and B and ∠CPB = θ, then a value
of tanθ is
[JEE Main 2018]
(a)
1
2
(b) 2
(c) 3
(d)
4
3
Exp. (b)
Equation of tangent and normal to the curve
y2 = 16 x at (16, 16) is x − 2 y + 16 = 0 and
2 x + y − 48 = 0, respectively.
Y
C
P
AB = (3 + 3)2 + (3 − 5)2
= 36 + 4 = 2 10
We know that, AB : BC = 2 : 1
BC = 10
⇒
Now, AC = AB + BC = 2 10 + 10 = 3 10
Since, AC is a diameter of circle.
5
3 10
AC
r=
=3
∴
⇒ r=
2
2
2
(c) 85
y
x–2
X′
+1
θ
0
6=
C(4, 0)
A(–16, 0)
Y′
(16, 16)
2x
+
y–
48
=
0
B(24, 0)
X
246
JEE Main Chapterwise Mathematics
A = (−16, 0) ; B = (24, 0)
C is the centre of circle passing through PAB
i.e. C = (4, 0)
16 − 0 16 4
Slope of PC =
=
= = m1
16 − 4 12 3
16 − 0 16
Slope of PB =
=
= − 2 = m2
16 − 24 −8
 m − m2 
tanθ =  1

1 + m1m2
⇒

 4
 3+2 
tanθ = 
⇒ tanθ = 2
1 −  4  (2 )
  3 
4x 2 − y 2 = 36 at the points P and Q. If these
tangents intersect at the point T (0, 3), then
the area (in sq units) of ∆PTQ is
[JEE Main 2018]
(b) 54 3
(c) 60 3
(d) 36 5
Exp. (a)
Tangents are drawn to the hyperbola
4 x2 − y2 = 36 at the point P and Q.
Tangent intersects at point T(0, 3)
Y
T (0, 3)
(–3√5, –12)Q
∴

Now, b 2 = a2 (1 − e 2 ) = (2 )2 1 −

=3
b= 3
⇒
x2
∴Equation of the ellipse is 2
(2 )
2
 1   = 4 1 − 1 
  


2 

4

S(0, –12)
P(3√5, –12)
we get x = ± 3 5
1
1
× PQ × ST = (6 5 × 15)
2
2
= 45 5
Area of ∆PQT =
=1
( 3 )2
x2
y2
+
=1
4
3
3
Now, the equation of normal at  1,  is
 2
a2 x b 2 y
−
= a2 − b 2
x1
y1
4x
3y
−
=4−3
1 ( 3 /2 )
⇒
⇒
4x − 2 y = 1
84. If a hyperbola passes through the point
P( 2 , 3 ) and has foci at ( ± 2 , 0), then the
tangent to this hyperbola at P also passes
through the point
[JEE Main 2017 (offline)]
(b) ( 2 2 , 3 3 )
(d) ( − 2 , − 3 )
Exp. (b)
∴
⇒
ae = 2
a2e 2 = 4
⇒
a2 + b 2 = 4
⇒
∴
x2
a
2
−
b 2 = 4 − a2
x
2
a2
−
y
2
4 − a2
=1
Since, ( 2 , 3 ) lie on hyperbola.
2
3
−
=1
a2 4 − a2
∴
83. The eccentricity of an ellipse whose centre is
⇒
at the origin is 1/2. If one of its directrices is
x = − 4, then the equation of the normal to it
 3
at 1,  is
 2
[JEE Main 2017 (offline)]
⇒
(b) 4x − 2 y = 1
(d) x + 2 y = 4
y2
+
Let the equation of hyperbola be
Clearly, P Q is chord of contact.
∴Equation of PQ is −3 y = 36
⇒
y = − 12
Solving the curve 4 x2 − y2 = 36 and y = − 12 ,
(a) 2 y − x = 2
(c) 4x + 2 y = 7
a
1
and = 4
e
2
a=2
(a) ( 3 2 , 2 3 )
(c) ( 3 , 2 )
X
O
We have, e =
⇒
82. Tangents are drawn to the hyperbola
(a) 45 5
Exp. (b)
8 − 2 a2 − 3a2 = a2 (4 − a2 )
8 − 5a2 = 4a2 − a4
⇒
a − 9a2 + 8 = 0
⇒
(a4 − 8)(a4 − 1) = 0
⇒
a4 = 8, a4 = 1
∴
4
a=1
y2
b2
= 1.
247
Coordinate Geometry
x2
y2
−
= 1.
1
3
∴ Equation of tangent at ( 2 , 3 ) is given by
3y
=1
2x −
3
y
2x −
=1
⇒
3
which passes through the point (2 2 , 3 3 ).
Now, equation of hyperbola is
85. Two sides of a rhombus are along the lines,
x − y + 1 = 0 and 7x − y − 5 = 0. If its diagonals
intersect at (− 1, − 2), then which one of the
following is a vertex of this rhombus?
[JEE Main 2016 (offline)]
(a) ( − 3, − 9)
1
8
(c)  , − 
 3 3
(b) ( − 3, − 8)
10 7
(d)  − , − 
 3
3
Exp. (c)
Exp. (d)
Given equation of circle is
x2 + y2 − 8 x − 8 y − 4 = 0,
whose centre is C(4, 4) and radius
= 42 + 42 + 4 =
Let the centre of required circle be C1( x, y). Now,
as it touch the X-axis, therefore its radius = y .
Also, it touch the circle
x2 + y2 − 8 x − 8 y − 4 = 0,
therefore CC1 = 6 + y
( x − 4)2 + ( y − 4)2 = 6 + y
⇒
⇒
x2 + 16 − 8 x + y2 + 16 − 8 y
= 36 + y2 + 12 y
⇒
x2 − 8 x − 8 y + 32 = 36 + 12 y
⇒
D
C (x, y)
x2 − 8 x − 8 y − 4 = 12 y
Case I If y > 0, then we have
x2 − 8 x − 8 y − 4 = 12 y
⇒
x − 8 x − 20 y − 4 = 0
2
⇒
x2 − 8 x − 4 = 20 y
⇒
( x − 4)2 − 20 = 20 y
⇒
(–1, –2)
7x –
y–5
=0
As the given lines x − y + 1 = 0 and
7 x − y − 5 = 0 are not parallel, therefore they
represent the adjacent sides of the rhombus.
On solving x − y + 1 = 0 and 7 x − y − 5 = 0, we
get x = 1and y = 2. Thus, one of the vertex is
A(1, 2 ).
A x – y+1=0
(1, 2)
( x − 4) 2 = 20 ( y + 1),
which is a parabola.
Case II If y< 0, then we have
x2 − 8 x − 8 y − 4 = − 12 y
B
Let the coordinate of point C be ( x, y).
x+1
y+2
Then,
and − 2 =
− 1=
2
2
⇒
x + 1 = − 2 and y = − 4 − 2
⇒
x = − 3 and y = − 6
Hence, coordinates of C = (− 3, − 6)
Note that, vertices B and D will satisfy
x − y + 1 = 0 and 7 x − y − 5 = 0, respectively.
Since, option (c) satisfies 7 x − y − 5 = 0,
1 − 8
therefore coordinate of vertex D is  ,
.
3 3 
86. The centres of those circles which touch the
circle, x 2 + y 2 − 8x − 8y − 4 = 0, externally
and also touch the X -axis, lie on
(a) a circle
[JEE Main 2016 (offline)]
(b) an ellipse which is not a circle
(c) a hyperbola
(d) a parabola
36 = 6
⇒
x − 8 x − 8 y − 4 + 12 y = 0
⇒
x2 − 8 x + 4 y − 4 = 0
2
⇒
x2 − 8 x − 4 = − 4 y
⇒
( x − 4) 2 = 20 − 4 y
⇒
( x − 4)2 = − 4( y − 5)
which is again a parabola.
87. If one of the diameters of the circle, given by
the equation, x 2 + y 2 − 4x + 6y − 12 = 0, is a
chord of a circle S, whose centre is at ( −3, 2 ),
then the radius of S is [JEE Main 2016 (offline)]
(a) 5 2
(b) 5 3
(c) 5
(d) 10
Exp. (b)
Given equation of circle is x2 + y2 − 4 x
+ 6 y − 12 = 0, whose centre is(2, − 3) and radius
= 2 2 + (− 3) 2 + 12 =
4 + 9 + 12 = 5
248
JEE Main Chapterwise Mathematics
Now, according to given information, we have
the following figure.
S
A (–3, 2)
Thus, coordinate of point P are (2, − 4).
CP = 2 2 + (− 4 + 6)2 =
Now ,
4+ 4=2 2
Hence, required equation of circle is
( x − 2 )2 + ( y + 4)2 = (2 2 )2
C
⇒ x2 + 4 − 4 x + y2 + 16 + 8 y = 8
⇒
O (2,–3)
B
x2 + y2 − 4 x + 8 y + 12 = 0
89. The eccentricity of the hyperbola whose
Clearly, AO ⊥ BC, as O is mid-point of the chord.
length of the latusrectum is equal to 8 and
the length of its conjugate axis is equal to
half of the distance between its foci, is
Now, in ∆AOB, we have
(a)
x2 + y2 – 4x + 6y – 12 = 0
OA = (− 3 − 2 ) + (2 + 3)
2
2
= 25 + 25 =
50 = 5 2
and
OB= 5
∴
AB = OA 2 + OB2
=
50 + 25 = 75 = 5 3
88. Let P be the point on the parabola, y 2 = 8x ,
which is at a minimum distance from the
centre C of the circle, x 2 + ( y + 6)2 = 1. Then,
the equation of the circle, passing throughC
and having its centre at P is
[JEE Main 2016 (offline)]
4
3
(b)
4
3
(c)
2
3
(d) 3
Exp. (c)
We have,
2 b2
= 8 and 2b = ae ⇒ b 2 = 4a and 2b = ae
a
Consider, 2b = ae
⇒
4b 2 = a2e 2
⇒
4a2 (e 2 − 1) = a2e 2
⇒
4e 2 − 4 = e 2
[Q a ≠ 0]
2
3e = 4 ⇒ e =
3
⇒
2
[Q e > 0]
[JEE Main 2016 (offline)]
(a) x + y − 4x + 8y + 12 = 0
90. The number of points having both
(b) x + y − x + 4y − 12 = 0
x
(c) x 2 + y 2 − + 2 y − 24 = 0
4
(d) x 2 + y 2 − 4x + 9y + 18 = 0
coordinates as integers that lie in the
interior of the triangle with vertices (0, 0), (0,
41) and (41, 0) is
[JEE Main 2015]
2
2
2
2
(a) 901
(c) 820
Exp. (a)
Centre of circle x2 + ( y + 6)2 = 1 is C(0, − 6).
Let the coordinates of point P be (2t 2 , 4 t ).
Now, let
D = CP = (2t 2 )2 + (4 t + 6)2
⇒
D=
4 t 4 + 16 t 2 + 36 + 48 t
Squaring on both side
⇒
D2 (t ) = 4 t 4 + 16 t 2 + 48 t + 36
Let
F(t ) = 4 t + 16 t + 48 t + 36
4
For minimum,
Exp. (d)
Required points (x, y) are such that it satisfy
x + y < 41
and
x > 0, y > 0
Number of positive integral solution of the
equationx + y + k = 41will be number of integral
coordinates in the bounded region.
(0,41)
2
F ′(t ) = 0
⇒
16 t + 32t + 48 = 0
⇒
t 3 + 2t + 3 = 0
⇒
(b) 861
(d) 780
3
(t + 1) (t 2 − t + 3) = 0 ⇒ t = − 1
(0,0)
(41,0)
249
Coordinate Geometry
4=
y+
2x
–3
Aliter
Consider the following figure :
(1, 2)
M
A
2
x=
y+
0
∴
1 point
⇒
(40,1)
(40,1)
(0,0)
x=1 x=2
P′ (h, k)
3=
(0,41)
(1,40)
(2,39)
P (2, 3)
0
∴Total number of integral coordinates
40!
= 780
= 41−1C 3 −1 = 40C 2 =
2 ! 38!
39 points
x=40 (41,0)
Clearly, the number of required points
= 1 + 2 + 3 + ... + 39
39
(39 + 1) = 780
=
2
91. Locus of the image of the point (2, 3) in the
line(2 x − 3y + 4) + k ( x − 2 y + 3) = 0,k ∈R , is a
[JEE Main 2015]
(a) straight line parallel to X-axis
(b) straight line parallel to Y -axis
(c) circle of radius 2
(d) circle of radius 3
Exp. (c)
Central Idea First of all find the point of
intersection of the lines 2 x − 3 y + 4 = 0 and
x − 2 y + 3 = 0 (say A). Now, the line
(2 x − 3 y + 4) + k( x − 2 y + 3) = 0 is the
perpendicular bisector of the line joining points
P(2, 3) and image P′(h, k ). Now, AP = AP′ and
simplify.
Given line is
(2 − 1)2 + (3 − 2 )2 = (h − 1)2 + (k − 2 )2
⇒
2 =
h2 + k 2 − 2 h − 4k + 1 + 4
⇒
2 =
h2 + k 2 − 2 h − 4k + 5
⇒
h2 + k 2 − 2 h − 4k + 5 = 2
⇒
h2 + k 2 − 2 h − 4k + 3 = 0
Thus, the required locus is
x 2 + y2 − 2 x − 4 y + 3 = 0
which is a equation of circle with
radius = 1 + 4 − 3 = 2
Aliter
(2 x − 3 y + 4) + k( x − 2 y + 3) = 0 is family of lines
passing through (1, 2). By congruency of
triangles, we can prove that mirror image (h, k ) and
the point (2, 3) will be equidistant from (1, 2).
∴Locus of (h, k ) is PR = PQ
⇒ (h − 1)2 + (k − 2 )2 = (2 − 1)2 + (3 − 2 )2
or ( x − 1)2 + ( y − 2 )2 = 2
∴Locus is a circle of radius = 2
2 x – 3y + 4 = 0
Q
(2, 3)
P
(2 x − 3 y + 4) + k( x − 2 y + 3) = 0, k ∈ R …(i)
This line will pass through the point of
intersection of the lines
…(ii)
2 x − 3y + 4 = 0
and
…(iii)
x − 2y + 3 = 0
On solving Eqs. (ii) and (iii), we get
x = 1, y = 2
∴Point of intersection of lines (ii) and (iii) is (1, 2 ).
Let M be the mid-point of PP′, then AM is
perpendicular bisector of PP′ (where, A is the
point of intersection of given lines).
AP = AP′
(1, 2)
R
(h, k)
x – 2y + 3 = 0
92. The number of common tangents to the
circles
and
x 2 + y 2 − 4x − 6y −12 = 0
2
2
x + y + 6x +18y + 26 = 0 is
[JEE Main 2015]
(a) 1
(c) 3
(b) 2
(d) 4
250
JEE Main Chapterwise Mathematics
Y
Exp. (c)
Central Idea Number of common tangents
depend on the position of the circle with respect to
each other.
(i) If circles touch externally
⇒C1C 2 = r1 + r2 , 3 common tangents
(ii) If circles touch internally
⇒ C1C 2 = r2 − r1, 1 common tangent
(iii) If circles do not touch each other, 4 common
tangents
Given equations of circles are
…(i)
x2 + y2 − 4 x − 6 y − 12 = 0
x + y + 6 x + 18 y + 26 = 0
2
2
…(ii)
Centre of circle (i) is C1(2, 3) and radius
= 4 + 9 + 12 = 5(r1 )
(say)
Centre of circle (ii) is C 2 (−3, − 9) and radius
= 9 + 81 − 26
(say)
= 8(r2 )
Now,
C1C 2 = (2 + 3)2 + (3 + 9)2
⇒
C1C 2 =
⇒
C1C 2 = 25 + 144 = 13
∴
r1 + r2 = 5 + 8 = 13
Also,
C1C 2 = r1 + r2
Thus, both circles touch each other externally.
Hence, there are three common tangents.
93. The area (in sq units) of the quadrilateral
formed by the tangents at the end points of
2
x2 y
the latera recta to the ellipse
+
= 1 is
9
5
[JEE Main 2015]
27
(a)
4
27
(c)
2
(b) 18
(d) 27
(–2,0)
O
(2,0)
Q
X
(9/2, 0)
L′
(2, –5/3)
M′
(–2, –5/3)
Y′
∴Extremities of one of latusrectum are
 2, 5  and  2, −5  .




 3
 3
5
∴Equation of tangent at  2,  is,
 3
x(2 ) y(5 / 3)
+
=1
9
5
2 x + 3y = 9
…(ii)
9 

Eq.(ii) intersects X and Y-axes at  , 0 and (0, 3),
2 
respectively.
∴ Area of quadrilateral
= 4 ×Area of ∆POQ
1 9
= 4 ×  × × 3 = 27 sq units
2 2

94. LetO be the vertex andQ be any point on the
parabola x 2 = 8y . If the point P divides the
line segment OQ internally in the ratio 1 : 3,
then the locus of P is
[JEE Main 2015]
(b) y 2 = x
(c) y 2 = 2 x
Given equation of ellipse is
x2
y2
+
=1
9
5
…(i)
∴
a2 = 9, b 2 = 5
⇒
a = 3, b =
and
X′
L (2, 5/3)
(a) x 2 = y
Exp. (d)
Now,
(–2, 5/3)
M
or
52 + 12 2
(0,3)
P
Exp. (d)
Central Idea Any point on the parabola x2 = 8 y is
5
e = 1−
(d) x 2 = 2 y
b2
a2
5 2
= 1− =
9 3
foci = (± ae, 0) = (± 2, 0)
b2 5
=
3
a
(4t , 2t 2 ). Point P divides the line segment joining of
O(0, 0) and Q(4t , 2t 2 ) in the ratio 1 : 3. Apply the
section formula for internal division.
Equation of parabola is
x2 = 8 y
…(i)
251
Coordinate Geometry
Let any point Q on the parabola (i) is (4t , 2t 2 ).
Exp. (c)
Let coordinate of the point be (α, − α ).
Y
Since, (α, − α ) lie on 4ax + 2 ay + c = 0
)
,k
1: P (
h
3
X′
5bx + 2 by + d = 0.
−c
∴
4aα − 2 aα + c = 0 ⇒ α =
2a
−d
Also, 5bα − 2 bα + d = 0 ⇒ α =
3b
−c −d
From Eqs. (i) and (ii),
=
2a
3b
∴
3bc = 2 ad
and
Q (4t,2t2)
X
(0, 0) O
Y′
Let P(h, k ) be the point which divides the line
segment joining (0,0) and (4t , 2t 2 ) in the ratio 1 : 3.
1 × 4t + 3 × 0
∴
⇒ h=t
h=
4
1 × 2t 2 + 3 × 0
t2
and
⇒ k=
k=
2
4
1
k = h2
(Qt = h)
⇒
2
⇒
2 k = h2 ⇒ 2 y = x2 , which is required locus.
…(i)
…(ii)
97. The locus of the foot of perpendicular drawn
from the centre of the ellipse x 2 + 3y 2 = 6 on
any tangent to it, is
[JEE Main 2014]
(a) ( x 2 − y 2 )2 = 6x 2 + 2 y 2
(b) ( x 2 − y 2 )2 = 6x 2 − 2 y 2
(c) ( x 2 + y 2 )2 = 6x 2 + 2 y 2
(d) ( x 2 + y 2 )2 = 6x 2 − 2 y 2
Exp. (c)
95. If PS is the median of the triangle with
vertices P (2 , 2 ), Q (6, − 1) and R ( 7, 3), then
equation of the line passing through (1, – 1)
and parallel to PS is
[JEE Main 2014]
(a) 4x − 7y − 11 = 0
(c) 4x + 7y + 3 = 0
(b) 2 x + 9y + 7 = 0
(d) 2 x − 9y − 11 = 0
/ Product of slope of two perpendicular lines is − 1,
is the key concept used.
Exp. (b)
P (2, 2)
Coordinate of
7 + 6 3 − 1
S = 
,

 2
2 
13
=  , 1
2 
R
Q
S
(7, 3)
−2 (6, –1)
Slope of the line PS is
.
9
Required equation passes through (1, − 1) is
−2
( x − 1)
y+ 1=
9
⇒ 2 x + 9y + 7 = 0
96. Let a ,b , c and d be non-zero numbers. If the
point of intersection of the lines
4ax + 2 ay + c = 0 and 5bx + 2by + d = 0 lies in
the fourth quadrant and is equidistant from
the two axes, then
[JEE Main 2014]
(a) 2 bc − 3ad = 0
(c) 2ad − 3bc = 0
(b) 2 bc + 3ad = 0
(d) 3bc + 2ad = 0
Equation of ellipse is x2 + 3 y2 = 6.
xcos θ ysinθ
Equation of the tangent is
+
= 1.
a
b
Let (h, k ) be any point on the locus, then
h
k
…(i)
cos θ + sinθ = 1
a
b
−b
Slope of the tangent line is
cotθ.
a
Slope of perpendicular drawn from centre (0, 0) to
k
(h, k ) is .
h
Since, both the lines are perpendicular.
 k  ×  − b cotθ = −1
∴

  

 h  a
cos θ sinθ
[say]
=
=α
⇒
ha
kb
⇒ cos θ = αha, sinθ = αkb
From Eq. (i),
h
k
(αha) + (αkb ) = 1
a
b
⇒
h2α + k 2α = 1
1
⇒
α= 2
h + k2
2
2
Also,
sin θ + cos θ = 1
⇒
(αkb )2 + (αha)2 = 1
⇒
α k b + α 2 h2 a2 = 1
⇒
2 2 2
k 2 b2
(h + k )
2
2 2
+
h2 a2
(h + k 2 )2
2
=1
252
JEE Main Chapterwise Mathematics
2 k2
⇒
(h + k )
2
⇒
2 2
+
6h2
=1
(h + k )
[Q a2 = 6 and b 2 = 2 ]
2
2 2
6 x2 + 2 y2 = ( x2 + y2 )2
[replacing k by y and h by x]
98. Let C be the circle with centre at (1, 1) and
radius 1. If T is the circle centred at (0, y )
passing through origin and touching the
circle C externally, then the radius of T is
equal to
[JEE Main 2014]
3
2
(a)
(b)
3
2
(c)
1
2
(d)
1
4
1
is tangent line and it touches the
m
parabola x2 = −32 y.
1
x2 = −32  mx + 
∴

m
32
⇒ x2 + 32 mx +
=0
m
Q
D=0
32
1
1
∴ (32 m)2 − 4 ⋅   = 0 ⇒ m3 = ⇒ m =
 m
8
2
Let y = mx +
100. A ray of light along x + 3y = 3 gets
reflected upon reaching X -axis,
equation of the reflected ray is
[JEE Main 2013]
Exp. (d)
Let the coordinate of the centre of T be (0, k ).
Y
C
(a) y = x + 3
(c) y = 3x − 3
3y = x − 3
3y = x − 1
(b)
(d)
Exp. (b)
(1, 1)
Take any point B (0, 1) on given line.
(1– k)
T
(0, k)
X′
B (0, 1)
O
3y = x – 3
X
(1, 0)
A ( 3, 0)
Y′
B′(0, –1)
Distance between their centre
Equation of AB′,
− 1− 0
(x −
y−0=
0− 3
k + 1 = 1 + (k − 1)2
⇒
k + 1 = 1 + k2 + 1 − 2 k
⇒
k + 1=
⇒
⇒
⇒
k2 + 2 − 2 k
k2 + 1 + 2 k = k2 + 2 − 2 k
1
k=
4
−
3y = − x +
⇒ x−
3y =
⇒
3y = x −
3)
3
3
3
Alternate Solution
1
So, the radius of circle T is k i .e., .
4
Slope of given line is −
1
⋅
3
Y
99. The slope of the line touching both the
parabolas y 2 = 4x and x 2 = −32 is
3y =x – 3
[JEE Main 2014]
(a)
1
2
the
(b)
3
2
(c)
1
8
(d)
2
3
= mx + a / m, if it touches
the other curve, then D = 0.
/ Tangent to parabola is y
Exp. (a)
For parabola, y2 = 4 x
30°
X′
30°
( 3, 0)
X
Y′
From figure, equation of reflected rays is
1
( x − 3)
y=
3
⇒
3y = x − 3
253
Coordinate Geometry
101. The circle passing through (1, − 2 ) and
touching the axis of x at ( 3, 0) also passes
through the point
[JEE Main 2013]
(a) ( − 5, 2 )
(c) ( 5, − 2 )
(b) ( 2 , − 5)
(d) ( − 2 , 5)

7 
, 0 = (±
∴ Foci is (± ae, 0) =  ± 4 ×
4


∴ Radius of the circle, r = (ae )2 + b 2
= 7 + 9 = 16 = 4
Now, equation of circle is
( x − 0)2 + ( y − 3)2 = 16
Exp. (c)
∴
Let the equation of circle be
( x − 3)2 + ( y − 0)2 + λy = 0
x 2 + y2 − 6 y − 7 = 0
103. The x-coordinate of the incentre of the
Y
triangle that has the coordinates of
mid-points of its sides as (0, 1), (1, 1) and
(1, 0) is
[JEE Main 2013]
A (3, 0)
X′
(a) 2 + 2
(c) 1 + 2
X
P
(1, –2)
(b) 2 − 2
(d) 1 − 2
Exp. (b)
Y′
Given, mid-points of a triangle are (0, 1), (1, 1) and
(1, 0). Plotting these points on a graph paper and
make a triangle.
As it passes through (1, − 2 ).
∴
(1 − 3)2 + (− 2 )2 + λ(− 2 ) = 0
⇒
4 + 4 − 2λ = 0
⇒
λ=4
∴ Equation of circle is ( x − 3)2 + y2 + 4 y = 0.
So, the sides of a triangle will be 2, 2 and
22 + 22 = 2 2.
Y
By hit and trial method, we see that point (5, − 2 )
satisfies equation of circle.
C (0,2)
102. The equation of the circle passing through
x2 y 2
the foci of the ellipse
+
= 1 and
16 9
having centre at (0, 3) is
[JEE Main 2013]
2 (0, 1)
X′
B
(0, 0)
(a) x 2 + y 2 − 6y − 7 = 0 (b) x 2 + y 2 − 6y + 7 = 0
(c) x + y − 6y − 5 = 0 (d) x + y − 6y + 5 = 0
2
2
2
Given equation of ellipse is
x2
y2
+
=1
16
9
X
x-coordinate of incentre
2 × 0 + 2 2 ⋅ 0 + 2 ⋅2
=
2+2+2 2
2− 2
2
×
=2 − 2
2+ 2 2− 2
parabola, y 2 = 4 5x .
S
(–ae, 0)
A (2, 0)
104. Given A circle, 2 x 2 + 2 y 2 = 5 and a
(0, b)
S′
(1, 0)
2
Y′
=
Y
r
(1, 1)
2
Exp. (a)
X′
7 ,0)
X
(ae, 0)
Y′
Here, a = 4, b = 3, e = 1 −
9
⇒
16
[JEE Main 2013]
Statement I An equation of a common
tangent to these curves is y = x + 5.
7
4
Statement II If the line,
5
y = mx +
, (m ≠ 0) is the common
m
tangent, then m satisfies m 4 − 3m 2 + 2 = 0.
254
JEE Main Chapterwise Mathematics
(a) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(b) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
Exp. (b)
Equation of circle can be rewritten as x2 + y2 =
Here, centre → (0, 0) and radius →
5
2
5
⋅
2
Let common tangent be
5
m
y = mx +
The perpendicular from centre to the tangent is
equal to radius.
5
5
m
=
2
1 + m2
∴
⇒
m 1+ m =
2
⇒
2
Statement I
Analysis given, a parabola y2 = 16 3 x and an
ellipse 2 x2 + y2 = 4.
To find The equation of common tangent to the
given parabola and the ellipse. This can be very
easily done by comparing the standard equation
of tangents. Standard equation of tangent with
slope ‘ m’ to the parabola y2 = 16 3 x is
4 3
…(i)
y = mx +
m
Standard equation of tangent with slope ‘ m’ to the
x2
y2
ellipse
+
= 1 is
2
4
…(ii)
y = mx ± 2 m2 + 4
If a line L is a common tangent to both parabola
and ellipse, then L should be tangent to parabola,
i.e., its equation should be like Eq. (i).
L should be tangent to ellipse i.e., its equation
should be like Eq. (ii).
i.e., L must be like both of the Eqs. (i) and (ii).
2 ⇒ m (1 + m ) = 2
2
2
⇒ m + m − 2 = 0 ⇒(m + 2 ) (m − 1) = 0
4
Exp. (c)
2
2
m = ± 1 [Qm2 + 2 ≠ 0, as m ∈ R ]
y = ± ( x + 5)
∴
Both statements are correct as m = ± 1 satisfies
the given equation of Statement II but statement II is
not a correct explanation of statement I.
105. Statement I An equation of a common
tangent to the parabola y 2 = 16 3x and the
ellipse 2 x 2 + y 2 = 4 is y = 2 x + 2 3.
4 3
Statement II If the line y = mx +
,
m
(m ≠ 0) is a common tangent to the
parabola y 2 = 16 3x and the ellipse
2 x 2 + y 2 = 4,then m satisfiesm 4 + 2m 2 = 24.
[AIEEE 2012]
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
Hence, comparing Eqs. (i) and (ii), we get
4 3
= ± 2 m2 + 4
m
On squaring both sides we get
⇒
m2 (2 m2 + 4) = 48
⇒
m4 + 2 m2 − 24 = 0
⇒
(m2 + 6)(m2 − 4) = 0
⇒
m2 = 4
∴
[Q m2 ≠ − 6]
m= ± 2
Substituting m = ± 2 in the Eq. (i), we get the
required equation of the common tangents as
y = 2x + 2 3
and
y = −2x − 2 3
Hence, Statement I is correct.
Statement II
In Statement II, we have already seen that, if the
4 3
is a common tangent to the
line y = mx +
m
x2
y2
parabola y2 = 16 3 x and the ellipse
+
= 1,
2
4
then it satisfies the equation m4 + 2 m2 − 24 = 0.
Hence, Statement II is also correct but is not able
to explain the Statement I. It is an intermediate
step in the final answer.
255
Coordinate Geometry
106. If the line 2 x + y = k passes through the
point which divides the line segment
joining the points(1, 1) and( 2, 4) in the ratio
[AIEEE 2012]
3 : 2, then k is equal to
(a)
29
5
(b) 5
(c) 6
(d)
11
5
Hence, the equation of the circle is
( x − h)2 + ( y − k )2 = k 2
Also, given that the circle passes through points
(1, 0) and (2 , 3). Hence, substituting them in the
equation of the circle, we get
(1 − h)2 + (0 − k )2 = k 2
…(i)
(2 − h) + (3 − k ) = k
…(ii)
2
Exp. (c)
Given line L : 2 x + y = k passes through point (say
P) which divides a line segment (say AB) in ratio
3 : 2, where A(1, 1) and B(2, 4).
Using section formula, the coordinates of the
point P which divides AB internally in the ratio 3 : 2
are
 3 × 2 + 2 × 1 3 × 4 + 2 × 1
 8 14 
,
P
 ≡P , 
5 5 
3+2
3+2


Also, since the line L passes through P, hence
8 14
substituting the coordinates of P  ,  in the
5 5 
equation of L : 2 x + y = k, we get
8
14
2   +   = k
 5  5 
∴
k=6
107. The length of the diameter of the circle
which touches the X -axis at the point (1, 0)
and passes through the point ( 2, 3) is
[AIEEE 2012]
(a)
10
3
(b)
3
5
(c)
6
5
(d)
5
3
Exp. (a)
Given
(i) A circle which touches X-axis at the point(1, 0).
(ii) The circle also passes through the point(2, 3).
To find The length of the diameter of the circle.
Y
(h, k)
k
O
(1, 0)
k
(2, 3)
X
Let us assume that the coordinates of the centre
of the circle are C(h, k ) and its radius is r.
Now, since the circle touches X-axis at(1, 0), hence
its radius should be equal to ordinate of centre.
⇒
r=k
2
2
From Eq. (i), we get h = 1
On substituting in Eq. (ii), we get
(2 − 1)2 + (3 − k )2 = k 2
k=
⇒
5
3
The diameter of the circle is 2 k =
10
3
108. An ellipse is drawn by taking a diameter of
the circle ( x − 1)2 + y 2 = 1 as its semi-minor
axis and a diameter of the circle
x 2 + ( y − 2 )2 = 4 is semi-major axis. If the
centre of the ellipse is at the origin and its
axes are the coordinate axes, then the
equation of the ellipse is
[AIEEE 2012]
(a) 4x 2 + y 2 = 4
(b) x 2 + 4y 2 = 8
(c) 4x + y = 8
(d) x 2 + 4y 2 = 16
2
2
Exp. (d)
Given
(i) An ellipse whose semi-minor axis coincides
with one of the diameters of the circle
( x − 1)2 + y2 = 1.
(ii) The semi-major axis of the ellipse coincides
with one of the diameters of circle
x2 + ( y − 2 )2 = 4.
(iii) The centre of the ellipse is at origin.
(iv) The axes of the ellipse are coordinate axes.
To find The equation of the ellipse.
Diameter of circle ( x − 1)2 + y2 = 1 is 2 units and
that of circle x2 + ( y − 2 )2 = 4 is 4 units.
⇒ Semi-minor axis of ellipse, b = 2 units and
semi-major axis of ellipse, a = 4 units.
Hence, the equation of the ellipse is
x2
y2
+ 2 =1
2
a
b
x2
y2
⇒
+
=1
16
4
∴
x2 + 4 y2 = 16
256
JEE Main Chapterwise Mathematics
109. A line is drawn through the point (1, 2 ) to
meet the coordinate axes at P and Q such
that it forms a ∆OPQ, where O is the origin,
if the area of the ∆OPQ is least, then the
slope of the line PQ is
[AIEEE 2012]
(a) −
1
4
(c) − 2
(b) – 4
(d) −
1
2
Exp. (c)
Given
(i) A line through (1, 2 ) meets the coordinate axes
at P and Q.
(ii) The area of ∆OPQ is minimum.
To find The slope of line PQ.
Let mbe the slope of the line PQ, then the equation
of PQ is
y − 2 = m( x − 1)
2
Now, PQ meets X-axis at P  1 − , 0 and Y-axis

m 
at Q(0, 2 − m).
2
⇒
OP = 1 −
m
and
OQ = 2 − m
110. The lines L1 : y − x = 0 and L 2 : 2 x + y = 0
intersect the line L 3 : y + 2 = 0 at P and Q ,
respectively. The bisector of the acute
angle between L1 and L 2 intersects L 3 atR.
Statement I The
2 2 : 5.
ratio
PR : RQ
Statement II In any triangle, bisector of an
angle divides the triangle into two similar
triangles.
(a) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(b) Statement I is true, Statement II is false
(c) Statement I is false, Statement II is true
(d) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
[AIEEE 2011]
Exp. (b)
Here, L1 : y − x = 0, L2 : 2 x + y = 0
and L3 : y + 2 = 0 shown as
Y
L1
Q
y=x
(1, 2)
X
O (0, 0)
O
1
(OP)(OQ )
2
2
1 
=
 1 −  (2 − m)
2 
m
1
4
=
2 − m−
+2
2
m
4
1
4 −  m + 
=

2
m
⇒
Now,
4
f(m) = 4 −  m + 

m
4
f ′(m) = − 1 + 2
m
f ′(m) = 0
⇒
m2 = 4
⇒ m= ± 2
⇒
f(2 ) = 0 and f(− 2 ) = 8
Since, the area cannot be zero, hence the
required value of m is −2.
R
(– 2, –2)
L3
Q
P
P
Also, area of ∆OPQ =
Let
equals
–2
(1, –2)
L2
Angle bisector
|PO| =
y = –2
y = – 2x
4+ 4 =2 2
|OQ| = 1 + 4 =
5
Since, OR is angle bisector.
OP PR
=
∴
OQ RQ
In a triangle, angle bisector divides the opposite
sides in the ratio of side containing the angle.
⇒
PR 2 2
=
RQ
5
So, Statement I is true.
But, it does not divide the triangle in two similar
triangles.
So, Statement II is false.
Hence, option (b) is correct.
257
Coordinate Geometry
111. Equation of the ellipse whose axes are the
axes of coordinates and which passes
through the point ( −3, 1) and has
2
eccentricity
is
5
[AIEEE 2011]
(a) 5x 2 + 3y 2 − 48 = 0
(b) 3x 2 + 5y 2 − 15 = 0
(c) 5x + 3y − 32 = 0
(d) 3x + 5y − 32 = 0
2
2
2
(i) If circles touch internally,
a
a
a a
c−
=
⇒ c− =
2
2
2 2
⇒
c = a, c > 0
∴
|a| = c
(ii) If circles touch externally,
2
Y
Exp. (d)
x2
Given,
a
2
y2
+
b2
=1
a/2
c
X′
C
(0, 0)
Y
X
a/2
(– 3, 1) P+
X′
Y′
(ii) Externally
X
O
c+
Y′
Passes through P(− 3, 1) and e =
9
∴
a2
9
+
1
b2
5
=1
and e = 1 −
two
circles
x 2 + y 2 = ax
2
2
2
x + y = c , (c > 0) touch each other, if
112. The
113. The shortest distance between line y − x = 1
and curve x = y 2 is
3 2
(a)
8
(b)
8
[AIEEE 2011, 2009]
4
(c)
3
3 2
(d)
3
4
Exp. (a)
To find The shortest distance between y − x = 1
and x = y2 along the common normal.
∴ Tangent at P is parallel to
y= x+1
and
Y
[AIEEE 2011]
(b) a = 2 c
a a
=
2 2
Hence, the circles should touch internally and
|a| = c.
b2
a2
b2
2
= 1− 2
+
= 1 and
⇒
5
a
a2
3a2
2
27 + 5
3
b
and
⇒
=
=
1
5
a2
3 a2
32
32
2
2
and b =
⇒
a =
3
5
∴ Equation on ellipse
3 x2
5 y2
⇒
+
= 1 ⇒ 3 x2 + 5 y2 = 32
32
32
(a) | a | = c
⇒ c+
∴ c = 0, i .e., not possible as c > 0.
2
.
5
2
a
a
=
2
2
(c) | a | = 2 c (d) 2 | a | = c
…(i)
y=1+x
1 Q
P (t 2, t)
Exp. (a)
x2 + y2 − ax = 0 and x2 + y2 = c 2 touch each other.
Y
X′
–1
X
O
c
X′
a/2
O
a/2
C
Y′
X
∴ Slope of tangent at P (t 2 , t ).
⇒
Y′
(i) Internally
y2 = x
dy  1 
1
= 
=
dx  2 y  ( t 2, t ) 2 t
…(ii)
258
JEE Main Chapterwise Mathematics
⇒
⇒
∴
1
=1
2t
1
t =
2
1 1
P  , 
 4 2
[Eqs. (i) and (ii) are parallel]
x = y2 ⇒ 1 = 2 y
and
⇒
⇒
As P is mid-point of AB.
∴
…(i)
dy
dx
1
dy
=
= Slope of the line (i)
dx 2 y
1
1
=1 ⇒ y=
2
2y
2
1
1
x =   =
2
4
⇒
 dx 
2 x = x − y⋅  
 dy 
dx dy
+
=0
x
y
∴
3 2
⋅
8
Alternate Solution
Given,
x− y+ 1= 0
dx
=− x
dy
are ( − 2 , 0) and (2 , 0) and eccentricity is 2, is
given by
[AIEEE 2011]
(a) − 3x 2 + y 2 = 3
(b) x 2 − 3y 2 = 3
(c) 3x − y = 3
(d) − x 2 + 3y 2 = 3
2
Exp. (c)
Y
X′
(–2, 0)
114. The curve that passes through the point
(2, 3) and has the property that the
segment of any tangent to it lying between
the coordinate axes is bisected by the point
of contact, is given by
[AIEEE 2011]
2
y
115. The equation of the hyperbola whose foci
∴The shortest distance is
1 1
− +1
3
3 2
4 2
=
=
1+ 1
4 2
8
2
⇒
Integrating both sides, we get
log x + log y = log c
⇒
xy = c, as it passes through (2, 3).
∴
c = 6 ⇒ xy = 6
2
1 1
⇒ The point is ( x, y) =  , 
 4 2
y
x
(a)   +   = 2
 2
 3
6
(c) y =
x
…(i)

 dx  
x-intercept is  x − y ⋅   , 0
 dy  

dy
y-intercept is  0, y − x 

dx 
1 1
− +1
3
Shortest distance =|PQ| = 4 2
=
1+ 1
4 2
Hence, shortest distance is
Y − y  dy 
= 
X − x  dx  ( x, y )
Equation of AB is
(b) 2 y − 3x = 0
(d) x 2 + y 2 = 13
(2, 0)
O
X
Y′
Let equation of hyperbola be
x2
y2
− 2 =1
2
a
b
where, 2 ae = 4 and e = 2 ⇒ a = 1
Q
a2e 2 = a2 + b 2 ⇒ 4 = 1 + b 2
∴
b2 = 3
Thus, equation of hyperbola is
x2
y2
−
= 1 or 3 x2 − y2 = 3
1
3
Exp. (c)
Y
116. The equation of the circle passing through
B
(x, y)
the point (1, 0) and (0, 1) and having the
smallest radius is
[AIEEE 2011]
y = f (x)
(a) x 2 + y 2 + x + y − 2 = 0
P
O
(b) x 2 + y 2 − 2 x − 2 y + 1 = 0
A
X
(c) x 2 + y 2 − x − y = 0
(d) x 2 + y 2 + 2 x + 2 y − 7 = 0
259
Coordinate Geometry
Exp. (c)
Circle whose diametric end points are (1, 0) and
(0, 1) will be of smallest radius.
⇒ ( x − 1) ( x − 0) + ( y − 0) ( y − 1) = 0
⇒
x 2 + y2 − x − y = 0
117. The lines x + y = |a | and ax − y = 1 intersect
each other in the first quadrant. Then, the
set of all possible values of a in the interval
(a) ( −1,1]
(c) [1, ∞)
(b) (0, ∞ )
[AIEEE 2011]
(d) ( −1, ∞ )
Exp. (c)
As, x + y =|a| and ax − y = 1, intersect in
I quadrant.
Therefore, x and y intercept are positive.
1 + |a|
a|a| − 1
∴
≥ 0 and y =
≥0
x=
1+ a
a+1
⇒
and
⇒
and
Case I If
⇒
1+ a≥ 0
a|a| − 1 ≥ 0
a≥ −1
a|a| ≥ 1
− 1< a < 0
− a2 > 1
…(i)
[not possible]
Case II If
a ≥ 0 ⇒ a2 ≥ 1
⇒
∴
a≥1
a ≥ 1or a ∈[1, ∞)
…(ii)
118. If A(2 , − 3) and B( −2 , 1) are two vertices of a
triangle and third vertex moves on the line
2 x + 3y = 9, then the locus of the centroid of
the triangle is
[AIEEE 2011]
(a) 2 x − 3y = 1
(c) 2 x + 3y = 1
(b) x − y =1
(d) 2 x + 3y = 3
Exp. (c)
The third vertex lies on 2 x + 3 y = 9 .
 x, 9 − 2 x 
i .e.,



3 
A (2, –3)
B (–2, 1)
∴ Locus of centroid is
(
C x,
9 – 2x
3
)
9 − 2x

+ 1
2 − 2 + x − 3 +
3
,
 = (h, k )

3
3




3 − 2x
x
h= ,k=
∴
9
3
⇒
9 k = 3 − 2 (3 h)
⇒
9k = 3 − 6h
⇒
2 h + 3 k = 1 or 2 x + 3 y = 1
119. If two tangents drawn from a point P to the
parabola y 2 = 4x are at right angles, then
the locus of P is
[AIEEE 2010]
(a) x =1
(c) x = −1
(b) 2 x + 1 = 0
(d) 2 x − 1 = 0
Exp. (c)
We know that, the locus of point P from which two
perpendicular tangents are drawn to the parabola,
is the directrix of the parabola.
Hence, the required locus is x = −1.
x y
+ = 1 passes
5 b
through the point (13, 32). The line K is
x y
parallel to L and has the equation + = 1.
c 3
Then, the distance between L and K is
120. The line L given by
[AIEEE 2010]
23
(a)
15
(b)
17
17
(c)
15
(d)
23
17
Exp. (d)
Since, the line L is passing through the point
(13, 32 ). Therefore,
13 32
32
8
= − ⇒ b = −20
+
=1 ⇒
5
b
5
b
The line K is parallel to the line L, its equation must
be
x y
x
y
−
=1
−
= a or
5 a 20a
5 20
x y
On comparing with + = 1, we get
c 3
3
20a = − 3, c = 5 a = −
4
Hence, the distance between lines
−3
−1
|a − 1|
23
= 20
=
=
17
17
1
1
+
400
25 400
260
JEE Main Chapterwise Mathematics
121. The circle x 2 + y 2 = 4x + 8y + 5 intersects
124. Three distinct points A , B and C given in
the line 3x − 4y = m at two distinct points, if
the 2-dimensional coordinate plane such
that the ratio of the distance of any one of
them from the point (1, 0) to the distance
1
from the point ( − 1, 0) is equal to . Then,
3
the circumcentre of the ∆ABC is at the
point
(a) − 85 < m < − 35
(c) 15 < m < 65
(b) − 35 < m < 15
(d) 35 < m < 85
[AIEEE 2010]
Exp. (b)
Since, the coordinates of the centre of the circle
are (2 , 4).
Also, r 2 = 4 + 16 + 5 = 25
The line will intersect the circle at two distinct
points, if the distance of (2 , 4) from 3 x − 4 y = m is
less than radius of the circle.
|6 − 16 − m|
∴
< 5 ⇒ − 25 < 10 + m < 25
5
∴
− 35 < m < 15
122. The linesp (p 2 + 1) x − y + q = 0
and (p 2 + 1)2 x + (p 2 + 1) y + 2q = 0
are perpendicular to a common line for
(a)
(b)
(c)
(d)
exactly one value of p
exactly two values of p
more than two values of p
no value of p
[AIEEE 2009]
Exp. (a)
Lines perpendicular to same line are parallel to
each other.
∴ − p( p2 + 1) = p2 + 1 ⇒ p = − 1
Hence, there is exactly one value of p.
123. If P and Q are the points of intersection of
the circles x 2 + y 2 + 3x + 7y + 2p − 5 = 0 and
x 2 + y 2 + 2 x + 2 y − p 2 = 0, then there is a
circle passing through P ,Q and (1, 1) and
(a)
(b)
(c)
(d)
all values of p
all except one value of p
all except two values of p
exactly one value of p
[AIEEE 2009]
5
(a)  , 0
4 
5
(c)  , 0
3 
5
(b)  , 0
2 
(d) (0, 0)
Exp. (a)
Let ( x, y) denotes the coordinates in A, B and C
plane.
( x − 1)2 + y2
1
Then,
=
2
2
9
( x + 1) + y
⇒
9 x2 + 9 y2 − 18 x + 9 = x2 + y2 + 2 x + 1
⇒
8 x2 + 8 y2 − 20 x + 8 = 0
5
⇒
x 2 + y2 − x + 1 = 0
2
5
Hence, A, B and C lie on a circle with C  , 0 .
4 
125. The ellipse x 2 + 4y 2 = 4 is inscribed in a
rectangle aligned with the coordinate axes,
which in turn is inscribed in another ellipse
that passes through the point (4, 0). Then,
the equation of the ellipse is
[AIEEE 2009]
(a) x 2 + 12 y 2 = 16
(b) 4x 2 + 48y 2 = 48
(c) 4x 2 + 64y 2 = 48
(d) x 2 + 16y 2 = 16
Exp. (a)
Let the equation of the required ellipse be
x2
y2
+ 2 = 1.
16 b
But the ellipse passes through the point (2, 1).
2
x2 + y = 1
1
4
A(2, 1)
Y
Exp. (c)
Let
S = x 2 + y2 + 3 x + 7 y + 2 p − 5 = 0
and
S′ = x + y + 2 x + 2 y − p = 0
2
2
[AIEEE 2009]
2
Equation of the required circle is S + λS ′ = 0.
As it passes through (1, 1), the value of
λ = − (7 + 2 p)/(6 − p2 )
Here, λ is not defined at p = ± 6
Hence, it is true for all except two values of p.
(0, 1)
X′
O
(2, 0)
Y′
⇒
1
1
1
3
+
=1 ⇒ 2 =
4 b2
4
b
X
(4, 0)
Coordinate Geometry
261
4
3
128. The point diametrically opposite to the
b2 =
∴
Hence, equation is
x2
3 y2
+
= 1 ⇒ x2 + 12 y2 = 16
16
4
126. The perpendicular bisector of the line
segment joining P(1, 4) and Q (k , 3) has y
intercept − 4. Then, a possible value of k is
(a) – 4
(b) 1
(c) 2
(d) – 2
[AIEEE 2008]
(a) (3, 4)
(c) (– 3, 4)
the
circle
[AIEEE 2008]
(b) (3, – 4)
(d) (–3, – 4)
Exp. (d)
Given equation can be rewritten as
( x + 1)2 + ( y + 2 )2 = (2 2 )2
Let required point be Q(α, β ).
Then, mid-point of P(1, 0) and Q(α, β ) is the centre
of the circle.
α+1
= −1
i .e.,
2
β+ 0
and
= −2
2
Exp. (a)
Since, slope of PQ =
point
on
P(1, 0)
x 2 + y 2 + 2 x + 4y − 3 = 0 is
4−3
1
=
1− k 1− k
Slope of AM = (k − 1)
A
⇒
and
α = −3
β = −4
So, required point is (– 3, – 4).
P
(1, 4)
M
Q
(k, 3)
k + 1, 7
2 2
∴ Equation of AM is
7

y − = (k − 1) x −

2
k +

 2
1 


For y intercept, x = 0, y = − 4
k + 1
7
− 4 − = − (k − 1) 

 2 
2
⇒
⇒
15 k 2 − 1
=
2
2
the line x = 2 as the directrix. Then, the
vertex of the parabola is at
[AIEEE 2008]
(c) (1, 0)
8
3
4
(d)
3
∴
Y
(d) (0, 1)
Exp. (c)
Since, the vertex is the mid-point of the focus and
foot of the directrix vertex of the parabola at (1, 0).
X′
X
Y
Y′
(1, 0)
X′
X
(0, 0)
⇒
(2, 0)
⇒
Y′
x=2
[AIEEE 2008]
(b)
1
a
− ae = 4 and e =
2
e
a
2a − = 4
2
Q
127. A parabola has the origin as its focus and
(b) (0, 2)
5
3
2
(c)
3
is
Exp. (b)
k=±4
(a) (2, 0)
the line x = 4 and the
1
eccentricity is
, then the length of
2
semi-major axis is
directrix
(a)
k 2 − 1 = 15 ⇒ k 2 = 16
∴
129. A focus of an ellipse is at the origin. The
3a
=4
2
8
a=
3
x=4
262
JEE Main Chapterwise Mathematics
130. Consider a family of circles which are
passing through the point (–1, 1) and are
tangent to X -axis. If (h , k ) are the
coordinates of the centre of the circles,
then the set of values of k is given by the
interval
(a) 0 < k <
(c) −
1
2
1
2
1
(d) k ≤
2
(b) k ≥
1
1
≤k ≤
2
2
[AIEEE 2007]
( x − h) + ( y − k ) = k
2
Since it is passing through (–1, 1), then
(−1 − h)2 + (1 − k )2 = k 2
⇒
(a) (–1, 1)
(c) (2, 4)
h2 + 2 h − 2 k + 2 = 0
133. Let A (h , k ), B (1, 1)andC (2 , 1) be the vertices
of a right angled triangle with AC as its
hypotenuse. If the area of the triangle is 1,
then the set of values which k can take is
given by
[AIEEE 2007]
(a) {1, 3}
(c) {–1, 3}
For real circles,
D≥ 0
⇒
(2 )2 − 4(−2 k + 2 ) ≥ 0
1
k≥
∴
2
(a)
(b)
(c)
(d)
(b) {0, 2}
(d) {–3, –2}
Exp. (c)
Since,A(h, k ), B(1, 1) and C(2 , 1) are the vertices of a
right angled ∆ABC.
x2
−
y2
= 1,
cos 2 α sin 2 α
which of the following remains constant
when α varies?
[AIEEE 2007]
131. For the hyperbola
(b) (0, 2) [AIEEE 2007]
(d) (–2, 0)
Since, perpendicular tangents intersect on the
directrix, then point must lie on the directrix
x = − 2.
Hence, the required point is (– 2, 0).
Equation of circle which touches X-axis and
coordinates of centre are (h, k ), is
2
y 2 = 8x is y = x + 2. The point on this line
from which the other tangent to the parabola
is perpendicular to the given tangent is
Exp. (d)
Exp. (b)
2
132. The equation of a tangent to the parabola
Y
A (h, k)
Eccentricity
Directrix
Abscissae of vertices
Abscissae of foci
B (1, 1)
C (2, 1)
O
Exp. (d)
The given equation of hyperbola is
x2
y2
−
=1
2
cos α sin2 α
Here,
a = cos α
and
b 2 = sin2 α
Now,
e = 1+
⇒
e = 1+
2
Now, AB = (1 − h)2 + (1 − k )2
BC = (2 − 1)2 + (1 − 1)2
2
=1
and CA = (h − 2 )2 + (k − 1)2
b2
From Pythagoras theorem,
AC 2 = AB2 + BC 2
a2
sin2 α
cos 2 α
X
= 1 + tan2 α
⇒
e = sec α
Coordinates of foci are (± ae, 0) i.e., (±1, 0).
Hence, abscissae of foci remains constant when α
varies.
⇒
4 + h2 − 4 h + k 2 + 1 − 2 k
= h2 + 1 − 2 h + k 2 + 1 − 2 k + 1
⇒
5 − 4h = 3 − 2 h
∴
h=1
Now, given that area of triangle is 1.
…(i)
263
Coordinate Geometry
1
× AB × BC
2
1
1 = × (1 − h)2 + (1 − k )2 × 1
2
Then, area (∆ABC ) =
⇒
⇒
2 = (1 − h)2 + (1 − k )2
⇒
2 = (k − 1)2
⇒
4 = k2 + 1 − 2 k
⇒
…(ii)
[from Eq. (i)]
⇒
(k − 3)(k + 1) = 0
∴
k = − 1, 3
Thus, the set of values of k is {–1, 3}.
134. Let P = ( −1, 0),Q = (0, 0)and R = ( 3 , 3 3 ) be
three points. The equation of the bisector
of the ∠PQR is
[AIEEE 2007]
(a)
3x + y = 0
(b) x +
3
y =0
2
(c)
3
x + y =0
2
(d) x +
3y = 0
1
2
(c) ±1
(a) −
[AIEEE 2007]
(b) –2
(d) 2
Equation of bisectors of line xy = 0 are
y = ± x.
Put y = ± x in my2 + (1 − m2 )xy − mx2 = 0, we get
mx2 ± (1 − m2 )x2 − mx2 = 0
⇒
(1 − m2 )x2 = 0
∴
m= ±1
136. In an ellipse, the distances between its foci
is 6 and minor axis is 8. Then, its
eccentricity is
[AIEEE 2007]
(a)
1
2
(b)
4
5
(c)
1
5
(d)
3
5
Exp. (d)
Exp. (a)
Now, slope of QR =
3 3−0
=
3−0
3 = tan θ
Y
R (3, 3√3)
M
π/3
P
(0, 0)Q
(–1, 0)
X
2π/3
Y′
⇒
xy − mx 2 = 0 is a bisector of the angle
between the lines xy = 0, then m is
Exp. (c)
k2 − 2 k − 3 = 0
X′
135. If one of the line of my 2 + (1 − m 2 )
θ=
π
3
2π
, so the line QM
3
2π
makes an angle
from positive direction of
3
X-axis.
2π
Slope of the line QM = tan
=− 3
3
Hence, equation of line QM is y = − 3 x
or
3x + y = 0
Given that,
⇒
2 ae = 6 and 2 b = 8
ae = 3
and
b=4
ae 3
b 2 16e 2
⇒
=
=
⇒
4
b
9
a2
We know that,
16e 2
b2
2
=
1
−
⇒
= 1 − e2
e
9
a2
9
 16 + 9  e 2 = 1
⇒
⇒
e2 =


 9 
25
3
e=
∴
5
137. The locus of the vertices of the family of
parabolas y =
The angle between PQR is
a 3x 2 a 2x
+
− 2 a is
3
2
[AIEEE 2006]
3
4
64
(c) xy =
105
(a) xy =
35
16
105
(d) xy =
64
(b) xy =
Exp. (d)
The given equation of parabola is
a3 x 2
a2 x
y=
+
− 2a
3
2
264
JEE Main Chapterwise Mathematics
3 
a3  2
x
x +
3 
2a 
a3  2
3
9
9 
x +

x+
−
⇒ y + 2a =
2
3 
2a
16 a
16 a2 
⇒ y + 2a =
⇒ y + 2a =
⇒
a3 
a3
3
9
×
x+
 −
3 
4 a  16 a2
3
y + 2a +
3 a a3 
3
=
x+

16
3 
4a
2
3
 y + 35 a  = a  x + 3 





16 
3 
4a
⇒
2
 3
35 a 
Thus, the vertices of parabola is  − , −
⋅
16 
 4a
35 a
3
and k = −
Let h = −
4a
16
105
Now,
hk =
64
Thus, the locus of vertices of a parabola is
105
xy =
64
138. A straight line through the point A( 3 , 4) is
139. If (a , a 2 ) falls inside the angle made by the
x
, x > 0 and y = 3x , x > 0, then a
2
belongs to
[AIEEE 2006]
lines y =
1
(c)  −3 , − 

2
Exp. (b)
The graph of equations x − 2 y = 0 and 3 x − y = 0
is as shown in the figure. Since, given point (a, a2 )
lies in the shaded region.
a
Then,
a2 − > 0 and a2 − 3 a < 0
2
⇒
a(2 a − 1) > 0 and a (a − 3) < 0
+ – +
0
+ – +
1/2
0
3
⇒
1
a ∈ (− ∞, 0) ∪  , ∞ 
2 
and
0< a< 3
Y
such that its intercept between the axes is
bisected at A. Its equation is
[AIEEE 2006]
(a) 3x − 4y + 7 = 0
(c) 3x + 4y = 25
1
(b)  , 3
2 
1
(d)  0 , 
 2
(a) ( 3 , ∞ )
y = 3x
(b) 4x + 3y = 24
(d) x + y = 7
y=
Exp. (b)
Since, A is the mid-point of line PQ.
X′
x
2
X
O
Y
Y′
P (0, b)
i .e.,
A (3, 4)
a ∈(0,3)
1
⇒ a ∈  , 3
2 
140. If the lines 3x − 4y − 7 = 0 and2 x − 3y − 5 = 0
X′
O
Q
(a, 0)
Y′
a+ 0
⇒ a=6
2
0+ b
and 4 =
⇒ b=8
2
Hence, the equation of line is
x y
+ = 1 or 4 x + 3 y = 24
6 8
∴
3=
X
are two diameters of a circle of area 49π sq
units, then the equation of the circle is
(a) x 2 + y 2 + 2 x − 2 y − 62 = 0
[AIEEE 2006]
(b) x 2 + y 2 − 2 x + 2 y − 62 = 0
(c) x 2 + y 2 − 2 x + 2 y − 47 = 0
(d) x 2 + y 2 + 2 x − 2 y − 47 = 0
Exp. (c)
The given equations of diameters are
3x − 4y − 7 = 0
and
2 x − 3y − 5 = 0
…(i)
…(ii)
265
Coordinate Geometry
On solving Eqs. (i) and (ii), we get
x = 1 and y = − 1
So, the intersection of two diameters is the centre
of circle, is (1, –1).
Let r be the radius of circle, then
Area of circle = 49 π
⇒
πr 2 = 49 π ⇒ r = 7 units
∴ Equation of required circle is
( x − 1)2 + ( y + 1)2 = 49
⇒
x2 − 2 x + 1 + y2 + 2 y + 1 = 49
⇒
x2 + y2 − 2 x + 2 y − 47 = 0
141. Let C be the circle with centre (0, 0) and
radius 3 units. The equation of the locus of
the mid-points of the chords of the circleC
2π
that subtend an angle of
at its centre, is
3
[AIEEE 2006]
27
4
3
(d) x 2 + y 2 =
2
(a) x + y = 1
2
(b) x + y =
2
2
9
(c) x + y =
4
2
2
2
Exp. (c)
OP = (h − 0)2 + (k − 0)2
=
h + k
2
3
A
⇒
(b) x 2 + 4y + 2 = 0
(c) y + 4x + 2 = 0
(d) y 2 − 4x + 2 = 0
2
[AIEEE 2005]
Exp. (d)
Since, the coordinates of P are (1, 0).
Let any point Q on y2 = 8 x be (2 t 2 , 4t ).
Again, let mid-point of PQ is (h, k ), so
2t 2 + 1
…(i)
⇒ 2 h = 2t 2 + 1
h=
2
4t + 0
k
and k =
…(ii)
⇒
t =
2
2
On putting the value of t from Eq. (ii) in Eq. (i), we
get
2 k2
+ 1 ⇒ 4 h = k2 + 2
2h =
4
Hence, locus of (h, k ) is y2 − 4 x + 2 = 0.
143. Area of the greatest rectangle that can be
h2 + k 2 =
(b)
y2
b2
= 1 is
ab sq unit
(d) 2ab sq unit
B
Let the coordinates of the vertices of rectangle
ABCD be
A(a cos θ, b sin θ),
B(− a cos θ, b sin θ),
C(− a cos θ, − b sin θ) and D(a cos θ, − b sin θ)
Then length of rectangle, AB = 2 a cos θ
and breadth of rectangle, AD = 2 b sin θ
∴ Area of rectangle, A = AB × AD
= 2 a cos θ × 2 b sin θ
⇒
π OP
=
3 OA
1
=
2
a2
+
[AIEEE 2005]
π/3 π/3
P (h, k)
cos
a
sq unit
b
(c) ab sq unit
(a)
x2
Exp. (d)
A = 2 ab sin 2 θ
…(i)
Y
h2 + k 2
(– a cos θ, b sin θ)
A (a cos θ, b sin θ)
B
3
9
4
Hence, the required locus is
9
x 2 + y2 =
4
⇒
(a) x 2 − 4y + 2 = 0
2
O (0, 0)
In ∆AOP,
on y 2 = 8x . The locus of mid-point ofPQ is
inscribed in the ellipse
Let the coordinates of a point P be (h, k ) which is
mid-point of the chord AB.
Now,
142. Let P be the point (1, 0) andQ be the point
X′
X
O
C
(– a cos θ, – b sin θ)
D (a cos θ, – b sin θ)
Y′
266
JEE Main Chapterwise Mathematics
On differentiating Eq. (i) w.r.t. θ, we get
dA
= 2 × 2 ab cos 2 θ
dθ
dA
For maxima or minima, put
=0
dθ
⇒
4ab cos 2 θ = 0
π
2θ =
⇒
2
π
θ=
∴
4
d2A
Now,
= − 8 ab sin 2 θ
dθ2
At θ =
The negative sign shows that the line is below
3
X-axis at a distance from it.
2
Alternate Solution
Equation of given lines are
…(i)
ax + 2 by + 3 b = 0
and
…(ii)
bx − 2 ay − 3 a = 0
On solving Eqs. (i) and (ii), we get the point of
intersection is
3
x = 0, y = −
2
Also, required line is parallel to X-axis.
∴
m= 0
d2 A
 2  < 0
 dθ 
π
,
4
π
.
4
⇒ Maximum area of rectangle = 2 ab sq units
[from Eq. (i)]
Alternate Solution
From Eq. (i),
Area of rectangle, A = 2 ab sin 2 θ
∴ A is maximum when sin 2 θ = 1
⇒ Maximum area of rectangle = 2 ab sq units
∴ Area is maximum at θ =
144. The line parallel to the X -axis and passing
through the intersection of the lines
a x + 2by + 3b = 0 and bx − 2 ay − 3a = 0,
where (a ,b ) ≠ (0, 0) is
[AIEEE 2005]
(a) above the X-axis at a distance of (2/3) from it
(b) above the X-axis at a distance of (3/2) from it
(c) below the X-axis at a distance of (2/3) from it
(d) below the X-axis at a distance of (3/2) from it
Exp. (d)
Equation of a line passing through the intersection
of lines ax + 2 by + 3b = 0 and bx − 2 a y − 3 a = 0
is
(ax + 2 by + 3 b ) + λ(bx − 2 ay − 3 a) = 0 …(i)
Now, this line is parallel to X-axis, so coefficient of
x should be zero.
a
i.e.,
a + λb = 0 ⇒ λ = −
b
On putting this value in Eq. (i), we get
b(ax + 2 by + 3 b ) − a(bx − 2 ay − 3 a) = 0
⇒
2 b y + 3 b + 2 a y + 3a
⇒
2 (b 2 + a2 ) y + 3(b 2 + a2 ) = 0
∴
2
2
2
=0
2
y=−
3
2
∴ Equation of line which is passing through
 0, − 3  having slope m = 0 is



2
 y + 3  = 0( x − 0)



2
3
2
Thus, the required line is below X-axis at a
3
distance from X-axis.
2
⇒
y=−
145. If non-zero numbers a ,b and c are in HP,
x y 1
+ + = 0 always
a b c
passes through a fixed point. That point is
then the straight line
1
(a) 1, − 

2
(b) (1, –2) [AIEEE 2005]
(c) (–1, –2)
(d) (–1, 2)
Exp. (b)
1 1
1
Since, a, b and c are in HP. Then , and are in
a b
c
AP.
2 1 1
1 2
1
∴
= +
⇒ − + =0
b a c
a b c
x
y 1
Hence, straight line
+ + = 0 is always
a b c
passes through a fixed point (1, – 2).
146. If a vertex of a triangle is (1, 1) and the
mid-points of two sides through this vertex
are (–1, 2) and (3, 2), then the centroid of
the triangle is
[AIEEE 2005]
7
1 7
(a)  ,  (b) 1, 
 3
 3 3
7
1 7
(c)  − ,  (d)  −1, 
 3 3

3
267
Coordinate Geometry
On comparing it with 5 x + by − a = 0, we get
Exp. (b)
Let D and E be the mid-points of AB and AC. The
coordinates of B are
1 + x2
−1 =
2
1 + y2
and
2=
2
i.e.,
x2 = − 3
and
y2 = 3
⇒
⇒
which gives no real value of a.
Hence, the line will passes through P and Q for no
value of a.
148. A circle touches the X -axis and also
and coordinates of C are
1 + x3
3=
2
1 + y3
and
2=
2
i.e.,
x3 = 5 and y3 = 3
touches the circle with centre at (0, 3) and
radius 2. The locus of the centre of the
circle is
[AIEEE 2005]
(a) a parabola
(c) a circle
Since, circle touches the X-axis and also touches
circle with the centre at (0, 3) and radius 2, then
C1C 2 = r1 + r2
E (3, 2)
∴
B (–3, 3)
147. If the circles x 2 + y 2 + 2 a x + cy + a = 0 and
x + y − 3a x + dy − 1 = 0 intersect in two
distinct points P and Q , then the line
5x + by − a = 0 passes through P and Q for
(a)
(b)
(c)
(d)
Y
(h, k)
C2 |k|
X′
Let equation of circles be
S1 ≡ x2 + y2 + 2 a x + c y + a = 0
and
S 2 ≡ x2 + y2 − 3 a x + dy − 1 = 0
Chord through intersection points P and Q of the
given circles is S1 − S 2 = 0.
∴ ( x2 + y2 + 2 ax + cy + a)
− ( x2 + y2 − 3 a x + dy − 1) = 0
5 a x + (c − d )y + a + 1 = 0
X
O
Y′
∴ Locus of centre of circle is
x2 = − 5 + 6 y + 4| y|
[AIEEE 2005]
Exp. (c)
⇒
(0,3)
C1 2
2
exactly two values of a
infinitely many values of a
no value of a
exactly one value of a
h2 + (k − 3)2 = (|k| + 2 )2
h2 + k 2 + 9 − 6 k = k 2 + 4 + 4|k|
C (5, 3)
∴ Centroid of triangle
x + x2 + x3 y1 + y2 + y3 
=  1
,



3
3
1 − 3 + 5 1 + 3 + 3  7 

,
=
 =  1, 

  3
3
3
2
(b) a hyperbola
(d) an ellipse
Exp. (a)
A (1, 1)
(–1, 2) D
5a c − d a + 1
=
=
−a
5
b
a ( − a) = a + 1
a2 + a + 1 = 0
⇒
x2 = 10 y − 5
[Q y > 0]
This equation represents a parabola. Thus, locus
of the centre of the circle is a parabola.
149. If a circle passes through the point (a ,b )
and
cuts
the
circle
x2 + y 2 = p2
orthogonally, then the equation of the
locus of its centre is
(a) 2 ax + 2 by − (a 2 + b 2 + p 2 ) = 0
[AIEEE 2005]
(b) x 2 + y 2 − 2 ax − 3 by + (a 2 − b 2 − p 2 ) = 0
(c) 2 ax + 2 by − (a 2 − b 2 + p 2 ) = 0
(d) x 2 + y 2 − 3ax − 4 by + (a 2 + b 2 − p 2 ) = 0
268
JEE Main Chapterwise Mathematics
Exp. (a)
Let the equation of circle be
x2 + y2 + 2 gx + 2 fy + c = 0. It cuts the circle
x2 + y2 = p2 orthogonally.
2 g1g 2 + 2 f1f2 = c1 + c 2
2 g (0) + 2 f(0) = c − p2
Q
∴
⇒
and slope of BF ′ =
c = p2
Also, it passes through (a, b ).
∴
a2 + b 2 + 2 ga + 2 fb + p2 = 0
So, locus of (− g , − f ) is
a2 + b 2 − 2 a x − 2 by + p2 = 0
⇒
2 a x + 2 by − (a2 + b 2 + p2 ) = 0
150. An ellipse has OB as semi-minor axis,
F and F ′ its foci and the ∠FBF is a right
angle. Then, the eccentricity of the ellipse
is
[AIEEE 2005]
1
(a)
3
1
(c)
2
1
(b)
4
1
(d)
2
∠FBF ′ = 90°
∴
b  b
−
⋅   = −1
ae  ae 
⇒
∴
⇒
151. The locus of a point P(α , β ) moving under
Since, the line y = αx + β is tangent to the
hyperbola
∴
45°
(ae,0)
(b) a parabola
(d) an ellipse
Exp. (a)
B(0, b)
O F
b 2 = a2e 2
b2
a2e 2
e2 = 1 − 2 = 1 − 2
a
a
1
2
2e = 1 ∴ e =
2
[AIEEE 2005]
Y
(–ae,0)
Q
(a) a hyperbola
(c) a circle
Since, ∠FBF ′ = 90°, then
∠OBF ′ = 45°
and ∠BF ′O = 45°
F'
b
ae
the condition that the line y = α x + β is a
x2 y 2
tangent to the hyperbola 2 − 2 = 1 , is
a
b
Exp. (d)
X′
Alternate Solution
Since, F and F ′ are foci of an ellipse, whose
coordinates are (ae, 0) and (− ae, 0) respectively
and coordiantes of B are (0, b ).
b
∴
Slope of BF =
− ae
A(a, 0)
X
x2
a
2
−
y2
b2
= 1.
β 2 = a2α 2 − b 2. .
So, locus of (α, β ) is
y2 = a 2 x 2 − b 2
⇒ a 2 x 2 − y2 − b 2 = 0
Y′
⇒
and
⇒
⇒
⇒
∴
ae = b
[since, ∆BOF′ is an isosceles triangle]
b2
e2 = 1 − 2
a
2 2
a
e
e2 = 1 − 2
a
e2 = 1 − e2
2e2 = 1
1
[since, e cannot be negative]
e=
2
Since, this equation represents a hyperbola, so
locus of a point P(α, β ) is a hyperbola.
152. If the pair of lines ax 2 + 2 (a + b )xy + by 2 = 0
lie along diameters of a circle and divide
the circle into four sectors such that the
area of one of the sectors is thrice the area
of another sector, then
(a) 3a 2 + 2ab + 3 b 2 = 0
(b) 3a 2 + 10ab + 3 b 2 = 0
(c) 3a 2 − 2ab + 3 b 2 = 0
(d) 3a 2 − 10ab + 3 b 2 = 0
[AIEEE 2005]
269
Coordinate Geometry
Exp. (a)
154. The equation of the straight line passing
Given equation of pair of lines is
ax2 + 2 (a + b ) xy + by2 = 0
Here,
H = a + b, A = a, B = b
π
Since,
4θ = π ⇒ θ =
4
3θ
θ
O
(0, 0)
through the point (4, 3) and making
intercepts on the coordinate axes whose
sum is –1, is
[AIEEE 2004]
x
+
2
x
(b) −
2
x
(c) +
2
x
(d) −
2
(a)
y
y
x
= − 1 and
+ = −1
3
−2 1
y
y
x
= − 1 and
+ = −1
3
−2 1
y
y
x
= 1 and
+ =1
3
−2 1
y
y
x
= 1 and
+ =1
3
−2 1
Exp. (d)
Angle between lines is given by
tan θ =
⇒
⇒
tan
2 H − AB
2
Let the intercepts on the coordinate axes be
a and b.
Y
A+ B
2 (a + b )2 − ab
π
= 1=
4
a+ b
b
a2 + b 2 + 2 ab = 4(a2 + b 2 + ab )
⇒ 3 a2 + 3 b 2 + 2 ab = 0
X′
153. Let A (2 , − 3)and B ( −2 , 1) be the vertices of a
∆ABC . If the centroid of this triangle moves
on the line 2 x + 3y = 1,then the locus of the
vertexC is the line
[AIEEE 2004]
(a)
(b)
(c)
(d)
2 x + 3y = 9
2 x − 3y = 7
3x + 2 y = 5
3x − 2 y = 3
Exp. (a)
Let ( x, y) be the coordinates of vertex C and ( x1, y1 )
be the coordinates of centroid of the triangle.
x+ 2 −2
y−3+1
and y1 =
∴ x1 =
3
3
y−2
x
…(i)
and y1 =
⇒ x1 =
3
3
Since, the centroid lies on the line 2 x + 3 y = 1.
So, point ( x1, y1 ) satisfied the equation of line.
∴
2 x1 + 3 y1 = 1
2 x 3( y − 2 )
[from Eq. (i)]
+
=1
3
3
⇒
2 x + 3y − 6 = 3
⇒
2 x + 3y = 9
Hence, this is the required equation of locus of the
vertex C.
⇒
O
X
a
Y′
According to the given condition,
a + b = − 1 ⇒ b = − a − 1 = − (a + 1)
x
y
Let equation of line be + = 1.
a b
x
y
−
=1
⇒
a a+1
…(i)
Since, this line passes through a point (4, 3).
4a + 4 − 3a
4
3
=1
−
=1 ⇒
∴
a(a + 1)
a a+1
⇒
a + 4 = a2 + a ⇒ a2 = 4
∴
a=±2
On putting the values of a in Eq. (i), we get
x y
x
y
− = 1 and
+ =1
2 3
−2 1
155. If the sum of the slopes of the lines given by
x 2 − 2 cxy − 7y 2 = 0 is four times their
product, then c has the value [AIEEE 2004]
(a) 1
(c) 2
(b) –1
(d) –2
Exp. (c)
The given pair of lines is x2 − 2cxy − 7 y2 = 0.
270
JEE Main Chapterwise Mathematics
On comparing the standard equation
ax2 + 2 hxy + by2 = 0, we get
a = 1, 2 h = − 2 c, b = − 7
2h
2c
Now, m1 + m2 = −
=−
b
7
1
a
and
m1m2 = = −
7
b
According to the given condition,
m1 + m2 = 4m1m2
2c
4
−
=−
⇒ c =2
7
7
∴
156. If
one of the lines given by
6x − xy + 4cy 2 = 0 is 3x + 4y = 0, then c is
equal to
[AIEEE 2004]
(b) –1
(c) 3
(d) –3
Exp. (d)
Since, one of the two lines is 3 x + 4 y = 0. Then,
3x
will satisfy the equation
y=−
4
6 x2 − xy + 4cy2 = 0
2
3x
3x
∴ 6 x2 − x  −  + 4c  −  = 0
 4
 4
⇒
6 x2 +
9 x2
3 x2
+ 4c ⋅
=0
16
4
⇒
a2 + b 2 + 2 ag + 2 f b + 4 = 0
Locus of centre (− g , − f ) will be
a2 + b 2 − 2 xa − 2 yb + 4 = 0
⇒
2 ax + 2 by − (a2 + b 2 + 4) = 0
Alternate Solution
Let the centre of required circle be (− g , − f ). This
circle cuts the circle x2 + y2 = 4 orthogonally. The
centre and radius of circle x2 + y2 = 4 are (0, 0)
and 2, respectively.
2
(a) 1
Since, it passes through the point (a, b ).
∴
∴
g 2 + f 2 = 4 + (a + g )2 + (b + f )2
⇒
g 2 + f 2 = 4 + a2 + g 2 + 2 ag
+ 2 bf + b 2 + f 2
⇒
4 + a + b + 2 ag + 2 bf = 0
2
2
Hence, the locus of centre is
2 ax + 2 by − (a2 + b 2 + 4) = 0
158. A variable circle passes through the fixed
point A (p , q )and touches X -axis. The locus
of the other end of the diameter through A
is
(a) ( x − p )2 = 4qy
(b) ( x − q )2 = 4py
(c) ( y − p ) = 4qx
(d) ( y − q )2 = 4px
2
[AIEEE 2004]
x2 (27 + 9c ) = 0
∴
c = −3
157. If a circle passes through the point (a ,b )
and cuts the circle x 2 + y 2 = 4 orthogonally,
then the locus of its centre is [AIEEE 2004]
(a) 2 ax + 2 by + (a 2 + b 2 + 4) = 0
(b) 2 ax + 2 by − (a 2 + b 2 + 4) = 0
(c) 2 ax − 2 by + (a 2 + b 2 + 4) = 0
(d) 2 ax − 2 by − (a + b + 4) = 0
2
2
Exp. (b)
Exp. (a)
Since, the coordinates of one end of a diameter of
a circle A are ( p, q ) and let the coordinates of other
end B be ( x1, y1 ).
Equation of circle in diameter form is
( x − p)( x − x1 ) + ( y − q )( y − y1 ) = 0
⇒ x2 − ( p + x1 )x + px1 + y2
⇒
x2 − ( p + x1 )x + y2
− (q + y1 )y + q y1 = 0
− ( y1 + q )y + px1 + q y1 = 0
Let the equation of circle be
x2 + y2 + 2 gx + 2 fy + c = 0
Since, this circle touches X-axis.
i.e.,
y=0
∴
x2 − ( p + x1 ) x + px1 + q y1 = 0
It cuts the circle x2 + y2 = 4 orthogonally, if
Also, the discriminant of above equation will be
2 g1g 2 + 2 f1f2 = c1 + c 2
∴
2 g ⋅ 0 + 2f ⋅ 0 = c − 4
⇒
c=4
∴ Equation of circle is
x2 + y2 + 2 gx + 2 fy + 4 = 0
( p + x1 )2 = 4( px1 + q y1 )
⇒
p2 + x12 + 2 px1 = 4 px1 + 4q y1
⇒
x12 − 2 px1 + p2 = 4q y1
Hence, locus of point B is ( x − p)2 = 4q y.
271
Coordinate Geometry
159. If the lines 2 x + 3y + 1 = 0 and 3x − y − 4 = 0
lie along diameters of a circle of
circumference 10π, then the equation of
the circle is
(a) x 2 + y 2 − 2 x + 2 y − 23 = 0
[AIEEE 2004]
(b) x 2 + y 2 − 2 x − 2 y − 23 = 0
Given lines 2 x + 3 y + 1 = 0 and 3 x − y − 4 = 0 are
the diameters of circle.
The intersection of two lines is the centre of circle
(1, –1).
Circumference of circle = 10π
[given]
2 πr = 10 π
⇒
r=5
∴ Equation of circle having centre (1, –1) and
radius 5 is
( x − 1)2 + ( y + 1)2 = 52
⇒
x2 + 1 − 2 x + y2 + 2 y + 1 = 25
⇒
x2 + y2 − 2 x + 2 y − 23 = 0
(d) d 2 + ( 3 b − 2 c )2 = 0
[AIEEE 2004]
x 2 + y 2 − 2 x = 0 is AB. Equation of the circle
on AB as a diameter is
[AIEEE 2004]
(a) x 2 + y 2 − x − y = 0
(b) x 2 + y 2 − x + y = 0
(c) x + y + x + y = 0
(d) x + y + x − y = 0
2
2
Exp. (a)
Given, equation of line is y = x.
And equation of circle is
x 2 + y2 − 2 x = 0
…(i)
…(ii)
From Eqs. (i) and (ii), we get
x2 + x2 − 2 x = 0 ⇒ 2 x( x − 1) = 0
⇒
x = 0, x = 1
On putting the values of x in Eq. (i) respectively, we
get
y = 0, y = 1
Let coordinates of A be (0, 0) and coordinates of B
be (1, 1).
∴ Equation of circle when AB as a diameter, is
( x − 0)( x − 1) + ( y − 0)( y − 1) = 0
x 2 − x + y2 − y = 0
x 2 + y2 − x − y = 0
and
x2 = 4 ay
The point of intersection of parabolas are
A(0, 0) and B(4 a, 4 a).
Also, given line 2 bx + 3cy + 4d = 0 passes
through the points A and B, respectively.
…(i)
∴
d=0
and 2 b ⋅ 4 a + 3c ⋅ 4 a + 4d = 0
⇒
2 ab + 3 ac + d = 0
⇒
a( 2 b + 3c ) = 0
[Qd = 0]
…(ii)
⇒
2 b + 3c = 0
On squaring and adding Eqs. (i) and (ii), we get
d 2 + ( 2 b + 3c )2 = 0
162. The eccentricity of an ellipse with its centre
160. The intercept on the line y = x by the circle
⇒
⇒
(c) d 2 + ( 2 b − 3c )2 = 0
y2 = 4 ax
Exp. (a)
2
(a) d 2 + ( 2 b + 3c )2 = 0 (b) d 2 + ( 3 b + 2 c )2 = 0
Given, equation of parabolas are
(d) x 2 + y 2 + 2 x − 2 y − 23 = 0
2
through the points of intersection of the
parabolas y 2 = 4ax and x 2 = 4ay , then
Exp. (a)
(c) x 2 + y 2 + 2 x + 2 y − 23 = 0
⇒
161. Ifa ≠ 0 and the line2bx + 3cy + 4d = 0 passes
1
at the origin, is .If one of the directrices is
2
x = 4, then the equation of the ellipse is
(a) 3x 2 + 4y 2 = 1
(b) 3x 2 + 4y 2 = 12
(c) 4x + 3y = 12
(d) 4x 2 + 3y 2 = 1
2
2
[AIEEE 2004]
Exp. (b)
Since, equation of directrix is x = 4, then major axis
of an ellipse is along X-axis.
1
a
Qe = 1 
∴
=4 ⇒ a=4×
2 
2
e

⇒
a=2
Now,
b 2 = a2 (1 − e 2 )
1
3
b 2 = 4  1 −  = 4 × ⇒ b 2 = 3
∴

4
4
Hence, equation of ellipse is
x2
y2
+
= 1 or 3 x2 + 4 y2 = 12
4
3
163. If x1 , x 2 , x 3 and y1 , y 2 , y 3 are both in GP with
the same common ratio, then the points
[AIEEE 2003]
( x1 , y1 ),( x 2 , y 2 )and ( x 3 , y 3 )
(a)
(b)
(c)
(d)
lie on a straight line
lie on an ellipse
lie on a circle
are vertices of a triangle
272
JEE Main Chapterwise Mathematics
Exp. (a)
Exp. (a)
Since, x1, x2 , x3 and y1, y2 , y3 are in GP.
Then,
x2 = rx1,
x3 = r 2 x1
and
y2 = ry1,
y3 = r 2 y1,
where, r is a common ratio.
The points become ( x1, y1 ), (r x1, r y1 ) and
(r 2 x1, r 2 y1 ).
x1
x2
Now,
x3
y1 1
x1
y1 1
y2 1 = rx1
ry1 1
y3 1
r 2 x1 r 2 y1 1
Taking x1 common from C1 and y1 from C 2
1
1 1
r 1 = x1 y1(0) = 0
r2 1
= x1 y1 r
r2
Let P(α, β ) be the point which is equidistant from
A(a1, b1 ) and B(a2 , b2 ).
∴
PA = PB
⇒ (α − a1 )2 + ( β − b1 )2 = (α − a2 )2 + ( β − b2 )2
⇒ α 2 + a12 − 2 α a1 + β 2 + b12 − 2 βb1
= α 2 + a22 − 2 αa2 + β 2 + b22 − 2 βb2
⇒ 2 (a2 − a1 )α + 2 (b2 − b1 )β
+ (a12 + b12 − a22 − b22 ) = 0
Thus, the equation of locus (α, β ) is
1
(a2 − a1 )x + (b2 − b1 )y + (a12 + b12 − a22 − b22 ) = 0
2
But the given equation is
(a2 − a1 ) x + (b2 − b1 ) y − c = 0
c=−
∴
[since, two columns are identical]
Hence, these points lie on a straight line.
Alternate Solution
Let
x1 = a
⇒
x2 = ar and x3 = ar 2 and y1 = b
⇒
y2 = br and y3 = br 2
where, r is a common ratio.
The given points will be A(a, b ), B(ar, br ) and
C(ar 2 , br 2 ).
b(r − 1) b
Now,
slope of AB =
=
a(r − 1) a
b(r 2 − r )
b
slope of BC =
=
2
a(r − r ) a
and
Slope of AB = Slope of BC ⇒ AB|| BC
∴
But B is a common point.
So, A , B and C are collinear.
i.e., the points ( x1, y1 ), ( x2 , y2 ) and ( x3 , y3 ) lie on a
straight line.
164. If the equation of the locus of a point
equidistant from the points
(a1 ,b1 )and (a 2 ,b 2 ) is
(a1 − a 2 )x + (b1 − b 2 )y + c = 0, then the value
of c is
[AIEEE 2003]
1
(a) (a 22 + b22 − a12 − b12 )
2
(b) a12 − a 22 + b12 − b22
1
(c) (a12 + a 22 + b12 + b22 )
2
(d)
a12
+
b12
− a 22
−
b22
=
1 2
(a1 + b12 − a22 − b22 )
2
1 2
(a2 + b22 − a12 − b12 )
2
165. Locus of centroid of the triangle whose
vertices
are
(a cos t , a sin t ),(b sin t , − b cos t ) and (1, 0),
wheret is a parameter, is
[AIEEE 2003]
(a) ( 3x − 1 ) + ( 3y ) = a − b
2
2
2
2
(b) ( 3x − 1 )2 + ( 3y )2 = a 2 + b 2
(c) ( 3x + 1 )2 + ( 3y )2 = a 2 + b 2
(d) ( 3x + 1 )2 + ( 3y )2 = a 2 − b 2
Exp. (b)
Since, the triangle, whose vertices are
(a cos t , a sin t ), (b sin t , − b cos t ) and (1, 0).
Let the coordinates of centroid be
a cos t + b sin t + 1
x=
3
⇒
3 x − 1 = a cos t + b sin t
a sin t − b cos t + 0
and
y=
3
…(i)
⇒
…(ii)
3y = a sin t − b cos t
On squaring and adding Eqs. (i) and (ii), we get
(3 x − 1)2 + (3 y)2
= a2 (cos 2 t + sin2 t ) + b 2 (sin2 t + cos 2 t )
⇒ (3 x − 1)2 + (3 y)2 = a2 + b 2
[Qsin2 θ + cos 2 θ = 1]
273
Coordinate Geometry
166. If the pair of straight lines x 2 − 2pxy − y 2 = 0
and x 2 − 2qxy − y 2 = 0 be such that each
pair bisects the angle between the other
pair, then
[AIEEE 2003]
(a) p = q
(b) p = − q
(c) pq =1
(d) pq = −1
The given equation is
x2 − 2 pxy − y2 = 0
a = 1,
b = − 1,
h=− p
∴
Equation of the bisector of angles is
x 2 − y2
xy
=
a−b
h
∴
⇒
⇒
x − y
xy
=
1+ 1
−p
2 xy
x 2 − y2 = −
p
2 xy
2
2
− y =0
x +
p
O
…(i)
…(ii)
167. A square of side a lies above the X -axis and
has one vertex at the origin. The side
passing through the origin makes an angle
π

α 0 < α <  with the positive direction of

4
X -axis. The equation of its diagonal not
passing through the origin is [AIEEE 2003]
(b) y (cos α + sin α ) + x (sin α − cos α ) = a
(c) y (cosα + sin α ) + x (sin α + cos α ) = a
(d) y (cosα + sin α ) + x (cosα − sin α ) = a
A (a cos α, a sin α)
90°
α
On comparing Eqs. (i) and (ii), we get
2
= − 2 q ⇒ pq = − 1
p
(a) y (cos α − sin α ) − x (sin α − cos α ) = a
y(sin α + cos α ) + x(cos α − sin α ) = a
[a cos (90° + α),
a sin (90° + α)] B
2
But given bisector equation of angles is
x2 − 2 qxy − y2 = 0
Also, OB ⊥ OA, then OB makes an angle (90° + α )
with X-axis, then coordinates of B are
[a cos (90° + α ), a sin (90° + α )].
i.e., (− a sin α, a cos α )
⇒ (sin α + cos α )( y − asin α )
= (sin α − cos α )( x − acos α )
⇒ y(sin α + cos α ) + x(cos α − sin α )
= a sin α(sin α + cos α ) − acos α(sin α − cos α )
= a(sin2 α + sin α cos α − cos α sin α + cos 2 α )
On comparing with standard equation
ax2 + 2 h x y + by2 = 0
2
Since, line OA makes an angle α with X-axis and
given OA = a, then coordinates of A are
(a cos α, a sin α ).
Equation of the diagonal AB not passing through
the origin is
acos α − asin α
( x − acos α )
( y − a sin α ) =
− a sin α − a cos α
Exp. (d)
we get
Exp. (d)
168. If the two circles ( x − 1)2 + ( y − 3)2 = r 2 and
x 2 + y 2 − 8x + 2 y + 8 = 0 intersect in two
distinct points, then
[AIEEE 2003]
(a) 2 < r < 8 (b) r < 2
(c) r = 2
(d) r > 2
Exp. (a)
The equation of first circle
( x − 1)2 + ( y − 3)2 = r 2
whose centre is C1(1, 3) and radius r1 = r .
and equation of second circle
x 2 + y2 − 8 x + 2 y + 8 = 0
whose centre is C 2 (4, − 1) and radius
r2 =
42 + 12 − 8 = 17 − 8 =
9=3
Two circles intersect in two distinct points, then
r1 − r2 < C1C 2 < r1 + r2
⇒
r − 3 < (4 − 1)2 + (−1 − 3)2 < r + 3
⇒
r − 3<
⇒
r − 3< 5< r + 3
9 + 16 < r + 3
274
JEE Main Chapterwise Mathematics
⇒
⇒
∴
r − 3 < 5 and 5 < r + 3
r < 8 and 2 < r
2< r< 8
169. The lines 2 x − 3y = 5 and 3x − 4y = 7 are
diameters of a circle having area as 154 sq
units. Then, the equation of the circle is
(a) x + y + 2 x − 2 y = 62
2
2
[AIEEE 2003]
(c) x + y − 2 x + 2 y = 47
2
2
Exp. (c)
Given equation of diameters of a circle are
…(i)
2 x − 3y = 5
and
…(ii)
3x − 4y = 7
The intersection of Eqs. (i) and (ii), we get the
coordinate of centre (1, –1).
Given area of circle = πr 2
22 2
⇒
r ∴r =7
154 =
7
Equation of the circle whose centre (1, –1) and
radius 7 is
( x − 1)2 + ( y + 1)2 = 7 2
⇒
x2 − 2 x + 1 + y2 + 2 y + 1 = 49
⇒
x2 + y2 − 2 x + 2 y = 47
170. The normal at the point (bt 12 , 2bt 1 ) on a
parabola meets the parabola again in the
point (bt 22 , 2bt 2 ), then
[AIEEE 2003]
(a) t 2 = − t 1 −
(c) t 2 = t 1 −
2
t1
2
t1
(b) t 2 = − t 1 +
(d) t 2 = t 1 +
(b) 5
(d) 9
Given equation of hyperbola can be rewritten as
x2
y2
−
=1
2
2
 12 
 9
 
 
 5
 5
(d) x + y − 2 x + 2 y = 62
2
(a) 1
(c) 7
Exp. (c)
(b) x 2 + y 2 + 2 x − 2 y = 47
2
x2 y 2
+
= 1 and the
16 b 2
2
2
x
y
1
hyperbola
coincide. Then,
−
=
144 81 25
the value ofb 2 is
[AIEEE 2003]
171. The foci of the ellipse
2
t1
2
t1
Exp. (a)
∴ Eccentricity, e ′2 = 1 +
a′ 2
9 25
e′2 = 1 +
=
16 16
5
e′ =
4
⇒
∴
The foci of a hyperbola are
12 5 
(± a ′e ′, 0) =  ±
× , 0 = (± 3, 0)
 5
4 
Given equation of ellipse is
x2
y2
+ 2 =1
16 b
Foci of an ellipse are (± ae, 0) = (± 4e, 0) but given
focus of ellipse and hyperbola coincide, then
3
4e = 3 ⇒ e =
4
Also, b 2 = a2 (1 − e 2 )
9
= 16  1 −  = 169 − 9 = 7
 16 
172. A triangle with vertices (4, 0), (–1, –1), (3, 5)
is
Equation of the normal at point (bt 12 , 2 bt 1 ) on
parabola is
y = − t 1 x + 2 bt 1 + bt 13
(a)
(b)
(c)
(d)
It also passes through (bt 22 , 2 bt 2 ) , then
Exp. (a)
2 bt 2 = − t 1 ⋅ bt 22 + 2 bt 1 + bt 13
⇒
2 t 2 − 2 t 1 = − t 1(t 22 − t 12 )
= − t 1(t 2 + t 1 )(t 2 − t 1 )
⇒
∴
2 = − t 1(t 2 + t 1 )
2
t 2 = − t1 −
t1
b ′2
isosceles and right angled
[AIEEE 2002]
isosceles but not right angled
right angled but not isosceles
neither right angled nor isosceles
Let the vertices of ∆ABC be A(4, 0), B(−1, − 1) and
C(3, 5).
Now, AB = (−1 − 4)2 + (−1 − 0)2 = 25 + 1 = 26
BC = (3 + 1)2 + (5 + 1)2 =
= 16 + 36 =
52
42 + 62
275
Coordinate Geometry
and CA = (4 − 3)2 + (0 − 5)2
175. The radius of the circle passing through the
= 1 + 25 = 26
foci of the ellipse
∴ CA 2 + AB2 = ( 26 )2 + ( 26 )2
centre at (0, 3), is
= 26 + 26 = 52 = BC 2
[AIEEE 2002]
(a) 4 units
(c)
⇒ CA 2 + AB2 = BC 2
Thus, the triangle is isosceles and right angled
triangle.
173. The equation of the directrix of the
parabola y 2 + 4y + 4x + 2 = 0 is [AIEEE 2002]
(a) x = −1
(c) x = − 3/2
x2 y 2
+
= 1 and having its
16 9
(b) 3 units
7
(d) units
2
12 units
Exp. (a)
The equation of an ellipse is
x2
y2
+
=1
16
9
Y
(b) x =1
(d) x = 3/ 2
(0,3)
Exp. (d)
Given equation of parabola can be rewritten as
1
( y + 2 )2 = − 4  x − 

2
1
Let
y + 2 = Y and x − = X
2
∴
Y 2 = − 4X
2 1 
(b)  ,

 3 3
 2 3
(c)  ,

3 2 
1 
(d) 1,


3
[AIEEE 2002]
X
Here, a = 4,
b=3
b2
= ( 7 − 0)2 + (0 − 3)2
= 7 + 9 = 16 = 4 units
176. Three straight lines 2 x + 11y − 5 = 0,
24x + 7y − 20 = 0 and 4x − 3y − 2 = 0
Exp. (d)
Let the vertices of ∆ABC be A(1, 3 ), B(0, 0) and
C(2 , 0).
Again, let a = BC = (2 − 0)2 + (0 − 0)2 = 2
and
A(a, 0)
Y′
Eccentricity, e = 1 −
174. The incentre of the triangle with vertices

3
(a) 1,

 2 
O S
S′
(–ae,0) (ae,0)
9
7
= 1−
=
16
4
a2
Foci of an ellipse are (± ae, 0) i.e., (± 7 , 0).
∴ Radius of required circle
Here,
a=1
∴ Equation of directrix is X = a.
1
3
x− =1 ⇒ x=
∴
2
2
(1, 3 ),(0, 0)and (2 , 0) is
X′
b = AC = (2 − 1)2 + (0 −
3 )2 = 2
c = AB = (0 − 1)2 + (0 −
3 )2 = 2
Since, all sides of a triangle are equal, then the
triangle is an equilateral triangle.
Also, incentre is same as centroid of the triangle.
Hence, coordinates of incentre are
 1 + 0 + 2 3 + 0 + 0
 1 .
,

 i.e.,  1,


3
3
3


(a) form a triangle
[AIEEE 2002]
(b) are only concurrent
(c) are concurrent with one line bisecting the
angle between the other two
(d) None of the above
Exp. (c)
The angle bisector for the given two lines
24 x + 7 y − 20 = 0 and 4 x − 3 y − 2 = 0, are
24 x + 7 y − 20
4x − 3y − 2
=±
25
5
Taking positive sign, we get
2 x + 11y − 5 = 0
This equation of line is already given.
Therefore, the given three lines are concurrent
with one line bisecting the angle between the other
two.
276
JEE Main Chapterwise Mathematics
177. A straight line through the point (2, 2)
intersects the lines
3 x + y = 0 and
3x − y = 0 at the points A and B . The
equation to the line AB so that the ∆OAB is
equilateral, is
[AIEEE 2002]
(a) x − 2 = 0
(c) x + y − 4 = 0
(b) y − 2 = 0
(d) None of these
178. The greatest distance of the point P(10, 7)
from the circle x 2 + y 2 − 4x − 2 y − 20 = 0 is
10 units
15 units
5 units
None of the above
[AIEEE 2002]
x 2 y2
(a)
+
=1
12 16
2
2
y
x
(c)
+
=1
16
8
(d) None of the above
1
, ae = 2 ⇒ a = 4
2
1
b 2 = a2 (1 − e 2 ) = 16  1 −  = 12

4
Given that, e =
Hence, equation of an ellipse is
Given equation of circle
x2 + y2 − 4 x − 2 y − 20 = 0
= 75 > 0
So, P lies outside the circle.
Now, PC = (2 − 10) + (1 − 7 )
2
=
x2
y2
+
= 1.
16 12
181. The equation of the chord joining two
whose centre is C(2 , 1) and radius is 5.
At point (10, 7),
S1 = 102 + 7 2 − 4 × 10 − 2 × 7 − 20
2
82 + 62 = 102 = 10
∴Greatest distance between circle and the point is
P = 10 + 5 = 15 units.
179. The equation of the tangent to the circle
x + y + 4x − 4y + 4 = 0 which make equal
intercepts on the positive coordinate axes,
is
[AIEEE 2002]
2
(b) x + y = 2 2
(d) x + y = 8
points
on
the
( x1 , y1 )and ( x 2 , y 2 )
rectangular hyperbola xy = c 2 is [AIEEE 2002]
x
x1 + x 2
x
(b)
x1 − x 2
x
(c)
y1 + y 2
x
(d)
y1 − y 2
(a)
Given equation of circle is
x2 + y2 + 4 x − 4 y + 4 = 0 whose centre (– 2, 2)
y
y1 + y 2
y
+
y1 − y 2
y
+
x1 + x 2
y
+
x1 − x 2
+
=1
=1
=1
=1
Exp. (a)
x + x2 y1 + y2 
The mid-point of the chord is  1
,
.
 2
2 
The equation of the chord T = S1.
y + y2 
 x1 + x2 
∴ x  1

 + y
 2 
 2 
x + x2   y1 + y2 
= 2  1

 
 2   2 
Exp. (b)
and radius = 2
x 2 y2
(b)
+
=1
16 12
Exp. (b)
∴
Exp. (b)
(a) x + y = 2
(c) x + y = 4
180. The equation of the ellipse whose foci are
[AIEEE 2002]
Given lines 3 x + y = 0 makes an angle of 120°
with OX and 3 x − y = 0 makes an angle of 60°
with OX. So, the required line is y − 2 = 0.
2
Hence, the equation of tangent is x + y = 2 2.
( ± 2 , 0) and eccentricity is 1/2, is
Exp. (b)
(a)
(b)
(c)
(d)
Let the equation of required tangent be x + y = a.
The perpendicular distance from centre to the
circle is equal to radius of circle.
−2 + 2 − a
= 2 ⇒ a = 2 2

∴
2


⇒ x( y1 + y2 ) + y( x1 + x2 ) = ( x1 + x2 )( y1 + y2 )
⇒
x
y
+
=1
x1 + x2
y1 + y2
12
Three Dimensional
Geometry
1. The equation of a plane containing the line
of intersection of the planes 2 x − y − 4 = 0
and y + 2 z − 4 = 0 and passing through the
point (1, 1, 0) is [JEE Main 2019, 8 April Shift-I]
(a) x − 3y − 2 z = − 2
(c) x − y − z =0
(b) 2 x − z = 2
(d) x + 3y + z = 4
Exp. (c)
Equations of given planes are
…(i)
2x − y − 4 = 0
and
…(ii)
y + 2z − 4 = 0
Now, equation of family of planes passes through
the line of intersection of given planes (i) and (ii) is
…(iii)
(2 x − y − 4) + λ( y + 2 z − 4) = 0
According to the question,
Plane (iii) passes through the point (1, 1, 0), so
(2 − 1 − 4) + λ(1 + 0 − 4) = 0
⇒
−3 − 3λ = 0
⇒
λ = −1
Now, equation of required plane can be obtained
by putting λ = − 1in the equation of plane (iii).
⇒
(2 x − y − 4) − 1( y + 2 z − 4) = 0
⇒
2x − y − 4 − y − 2z + 4 = 0
⇒
2x − 2y − 2z = 0
⇒
x− y− z=0
2. The magnitude of the projection of the
$
vector
on
the
vector
2 i$ + 3$j + k
perpendicular to the plane containing the
$ and $i + 2 $j + 3k
$ , is
vectors $i + $j + k
[JEE Main 2019, 8 April Shift-I]
(a)
3
2
(c) 3 6
(b)
6
(d)
3
2
Exp. (d)
The normal vector to the plane containing the
vectors ($i + $j + k$ ) and ($i + 2 $j + 3k$ ) is
n = ($i + $j + k$ ) × ($i + 2 $j + 3k$ )
$i $j k$ 
= 1 1 1




1 2 3
= $i(3 − 2 ) − $j(3 − 1) + k$ (2 − 1)
= $i − 2 $j + k$
Now, magnitude of the projection of vector
2 $i + 3$j + k$ on normal vector n is
|(2 $i + 3$j + k$ ) ⋅ n| |(2 $i + 3$j + k$ ) ⋅ ($i − 2 $j + k$ )|
=
|n|
1+ 4 + 1
=
|2 − 6 + 1|
3
3
units
=
=
6
6
2
3. The length of the perpendicular from the
point (2 , − 1, 4) on
x +3 y −2 z
= is
=
10
1
−7
the
straight
line,
[JEE Main 2019, 8 April Shift-I]
(a)
(b)
(c)
(d)
greater than 3 but less than 4
less than 2
greater than 2 but less than 3
greater than 4
278
JEE Main Chapterwise Mathematics
Exp. (a)
Equation of given line is
x+ 3 y−2 z
…(i)
= = r (let)
=
10
1
−7
Coordinates of a point on line (i) is
A(10r − 3, − 7 r + 2, r )
Now, let the line joining the points P(2, − 1, 4) and
A(10r − 3, − 7 r + 2, r ) is perpendicular to line (i).
Then,
PA ⋅ (10$i − 7 $j + k$ ) = 0
[Qvector along line (i) is(10$i − 7 $j + k$ )]
⇒ [(10r − 5)$i + (−7 r + 3)$j + (r − 4)k$ ]
⋅[10$i − 7 $j + k$ ] = 0
⇒ 10(10r − 5) − 7(3 − 7 r ) + (r − 4) = 0
⇒ 100r − 50 − 21 + 49r + r − 4 = 0
1
⇒
150r = 75 ⇒ r =
2
3 1
So, the foot of perpendicular is A 2, − , 

2 2
1
[put r = in the coordinates of point A]
2
Now, perpendicular distance of point P(2, − 1, 4)
from the line (i) is
2
1
3
PA = (2 − 2 )2 +  − + 1 +  − 4
2

 2

1 49
+
=
4
4
which lies in (3, 4).
=
2
50
5
=
4
2
4. The vector equation of the plane through
the line of intersection of the planes
x + y + z = 1 and 2 x + 3y + 4z = 5, which is
perpendicular to the plane x − y + z = 0 is
[JEE Main 2019, 8 April Shift-II]
$ )− 2 =0
(a) r ⋅ ( i$ − k
$ )+ 2 =0
(c) r × ( $i − k
$ )+ 2 =0
(b) r × ( $i + k
$ )+ 2 =0
(d) r ⋅ ( $i − k
Exp. (d)
Since, equation of planes passes through the line
of intersection of the planes
x+ y+ z=1
and 2 x + 3 y + 4 z = 5, is
( x + y + z − 1) + λ(2 x + 3 y + 4 z − 5) = 0
⇒(1 + 2 λ )x + (1 + 3λ )y + (1 + 4λ )z − (1 + 5λ ) = 0
…(i)
Q The plane (i) is perpendicular to the plane
x − y + z = 0.
∴(1 + 2 λ ) − (1 + 3λ ) + (1 + 4λ ) = 0
[Q if plane a1 x + b1 y + c1 z + d1 = 0 is
perpendicular to plane a2 x + b2 y + c 2 z + d 2 = 0,
then a1a2 + b1b2 + c1c 2 = 0]
⇒
3λ + 1 = 0
1
…(ii)
⇒
λ=−
3
So, the equation of required plane, is
1 − 2  x + 1 − 3  y + 1 − 4  z − 1 − 5  = 0












3
3
3
3
1
1
2
x− z+ =0 ⇒ x− z+2 =0
⇒
3
3
3
$)+ 2 = 0
Now, vector form, is r ⋅ (i$ − k
x −1 y +1 z −2
meets the
=
=
2
3
4
plane, x + 2 y + 3z = 15 at a point P, then the
distance of P from the origin is
5. If the line
[JEE Main 2019, 9 April Shift-I]
(a) 7 / 2
(b) 9 / 2
(c)
5/2
(d) 2 5
Exp. (b)
Equation of given plane is
…(i)
x + 2 y + 3 z = 15
x−1 y+ 1 z−2
and line is,
…(ii)
=
=
= r (let)
2
3
4
So, the coordinates of any point on line (ii) is
P(1 + 2 r, − 1 + 3r, 2 + 4r ).
QPoint P is intersecting point of plane (i) and line
(ii)
∴(1 + 2 r ) + 2 (–1 + 3r ) + 3 (2 + 4r ) = 15
⇒1 + 2 r − 2 + 6r + 6 + 12 r = 15 ⇒20r = 10
1
r=
⇒
2
∴Coordinates of P
3
1
=  1 + 1, − 1 + , 2 + 2  =  2, , 4

  2 
2
Now, distance of the point P from the origin
1
1
81 9
= 4 + + 16 = 20 + =
= units
4
4
4 2
6. A plane passing through the points (0, − 1, 0)
π
with the
4
plane y − z + 5 = 0, also passes through the
point
[JEE Main 2019, 9 April Shift-I]
and (0, 0, 1) and making an angle
(a) ( 2 ,1, 4)
(c) ( − 2 , − 1, − 4)
(b) ( − 2 ,1, − 4)
(d) ( 2 , − 1, 4)
279
Three Dimensional Geometry
Exp. (a)
Let the equation of plane is
…(i)
ax + by + cz = d
Since plane (i) passes through the points (0,−1, 0)
and (0, 0, 1,) then
− b = d and c = d
∴Equation of plane becomes
…(ii)
ax − dy + dz = d
π
QThe plane (ii) makes an angle of with the plane
4
y − z + 5 = 0.
cos
⇒
π
=
4
1
=
2
⇒
[Qdistance of ( x1, y1, z1 ) from the plane

ax + by1 + cz1 − d

ax + by + cz − d = 0, is 1
a2 + b 2 + c 2 
−d − d
a + d2 + d2
2
1+ 1
[Q The angle between the two planes
a1 x + b1 y + c1 z + d = 0 and
a2 x + b2 y + c 2 z + d = 0 is

a1 a2 + b1 b2 + c1 c 2

cosθ =
a12 + b12 + c12 a22 + b22 + c 22 
|−2d|
⇒ a2 + 2d 2 = |−d|
2
2
a + 2d 2
a2 + 2d 2 = 4d 2
⇒
On substituting λ = − 1in Eq. (i),
we get
− x − 2 y − 11 = 0
…(ii)
⇒
x + 2 y + 11 = 0
which is the required equation of the plane.
Now, the distance of the point (0, 0, 256) from
plane P is
0 + 0 + 11 11
=
1+ 4
5
[squaring both sides]
a = 2d ⇒ a = ± 2d
2
2
8. The vertices B and C of a ∆ABC lie on the
x +2 y −1 z
=
= such that BC = 5 units.
3
0
4
Then, the area (in sq units) of this triangle,
given that the point A(1, − 1, 2 ) is
line,
[JEE Main 2019, 9 April Shift-II]
(a) 34
(c) 5 17
Exp. (a)
Given line is
x+ 2 y−1 z
=
=
3
0
4
$
Vector along line is, a = 3i$ + 4k
So, the Eq. (ii) becomes
…(iii)
± 2x− y+ z = 1
Now, from options ( 2 , 1, 4) satisfy the plane
− 2x− y+ z = 1
7. Let P be the plane, which contains the line
of intersection of the planes, x + y + z − 6 = 0
and2 x + 3y + z + 5 = 0 and it is perpendicular
to the XY -plane. Then, the distance of the
point (0, 0, 256) from P is equal to
[JEE Main 2019, 9 April Shift-II]
(a) 63 5
(b) 205 5 (c)
11
5
(b) 2 34
(d) 6
(d)
17
5
Exp. (c)
Equation of plane, which contains the line of
intersection of the planes
x + y + z − 6 = 0 and 2 x + 3 y + z + 5 = 0, is
( x + y + z − 6) + λ(2 x + 3 y + z + 5) = 0
⇒(1 + 2 λ )x + (1 + 3λ )y + (1 + λ )z + (5λ − 6) = 0 …(i)
QThe plane (i) is perpendicular to XY-plane
(as DR’s of normal to XY-plane is (0, 0, 1)).
∴ 0(1 + 2 λ ) + 0(1 + 3λ ) + 11
( + λ) = 0
⇒
λ = −1
and vector joining the points (1, − 1, 2 ) to (−2, 1, 0) is
$
b = (1 + 2 )i$ + (−1 − 1)$j + (2 − 0)k
$
= 3$j − 2 $j + 2 k
and |BC| = 5 units
Now, area of required ∆ABC
1
…(ii)
= |BC|| b| |sinθ|
2
[where θ is angle between vectors a and b]
|a × b|
,
Q | b|sinθ =
|a|
$i $j k
$
$
Q |a × b|= 3 0 4 = 8i$ + 6$j − 6k
3 –2 2
⇒ | a × b| =
64 + 36 + 36
= 136 = 2 34
and
∴
|a| =
9 + 16 = 5
2 34
| b|sinθ =
5
280
JEE Main Chapterwise Mathematics
On substituting these values in Eq. (i), we get
2 34
1
Required area = × 5 ×
= 34 sq. units
5
2
Alternate Method
Given line is
x+ 2 y−1 z
…(i)
=
= = λ (let)
3
0
4
A(1, –1, 2)
B
C
D
x+2 y–1 z
=
=
0
4
3
Since, point D lies on the line BC.
∴Coordinates of D = (3λ − 2, 1, 4λ )
Now,
DR of BC ⇒ a1 = 3, b1 = 0, c1 = 4
and DR of AD ⇒ a2 = 3λ − 3, b2 = 2, c 2 = 4λ − 2
Since, AD ⊥ BC, a1a2 + b1b2 + c1c 2 = 0
3 × (3λ − 3) + 0(2 ) + 4(4λ − 2 ) = 0
⇒
9λ − 9 + 0 + 16λ − 8 = 0
⇒ 25λ − 17 = 0
17
λ=
⇒
25
68
1
∴Coordinates of D =  , 1,  .
 25 25 
Now, AD =
=
2
 1 − 1  + (−1 − 1)2 +  2 − 68 






25 
25 
2
 24  + (−2 )2 +  −18 
 


 25 
 25 
2
2
9. If Q(0, − 1, − 3) is the image of the point P in
the plane 3x − y + 4z = 2 and R is the point
( 3, − 1, − 2 ) , then the area (in sq units) of
[JEE Main 2019, 10 April Shift-I]
∆PQR is
91
2
(b) 2 13
(c)
91
4
(d)
=
91
1
sq units
1 + 81 + 9 =
2
2
10. If the length of the perpendicular from the
point (β, 0, β ) (β ≠ 0) to the line,
x y −1 z +1
3
is
, then β is equal to
=
=
−1
1
0
2
[JEE Main 2019, 10 April Shift-I]
324 2
576
34
=
+ 4+
=
625 5
625
2
1
1
34
∴Area of ∆ABC = BC × AD = × 5 ×
5
2
2
= 34 sq units
(a)
Let the coordinates of point P is ( x1, y1, z1 ), then
x1 − 0 y1 + 1 z1 + 3
=
=
3
−1
4
[3(0) − 1(−1) + 4(−3) − 2 ]
= −2
32 + (−1)2 + 42
[∴image of the point ( x1, y1, z1 ) in the plane
ax + by + cz + d = 0 is ( x, y, z), where
x − x1 y − y1 z − z1 − 2 (ax1 + by1 + cz1 + d )
]
=
=
=
a
b
c
a2 + b 2 + c 2
x1 − 0 y1 + 1 z1 + 3
=
=
⇒
3
−1
4
(1 − 12 − 2 ) 26
=2
=
=1
26
26
⇒ P( x1, y1, z1 ) = (3, − 2,1)
Now, area of ∆PQR, where point R(3, − 1, − 2 )
1 → →
= PQ × PR
2
1
= (−3i$ + $j − 4k$ ) × (0i$ + $j − 3k$ )
2
i$ $j
k$
1
1
=
−3 1 −4 = $i − 9 $j − 3k$
2
2
0 1 −3
(a) 2
(b) − 2
Given, equation of plane 3 x − y + 4 z = 2 ...(i) and
the point Q(0, − 1, − 3) is the image of point P in the
plane (i), so point P is also image of point Q w.r.t.
plane (i).
(d) 1
Exp. (c)
Equation of given line is
x y−1 z+ 1
…(i)
=
=
−1
1
0
Now, one of the point on line is P(0, 1, − 1) and the
given point is Q(β, 0, β ).
Q(β,0, β)
l
65
2
Exp. (a)
(c) −1
P(0, 1, –1)
M
x y–1 z+1
=
=
1
0
–1
From the figure, the length of the perpendicular
3
(given)
QM = l =
2
281
Three Dimensional Geometry
PQ × PM
⇒
PM
=
3
2
12. A perpendicular is drawn from a point on
$
and PM = a vector along given line (i) = i$ − k
$i $j
$
k
So, PQ × PM = β −1 β + 1
1
0
[JEE Main 2019, 10 April Shift-II]
−1
(a) ( −1, 0, 4)
(c) (2, 0, 1)
$
= $i − $j(− β − β − 1) + k
$
$
$
= i + (2 β + 1)j + k
So,
PQ × PM
⇒
PM
=
1 + (2 β + 1)2 + 1
1+ 1
4β 2 + 4β + 3 =
⇒
⇒
x −1 y +1 z
to the plane
=
=
2
−1 1
x + y + z = 3 such that the foot of the
perpendicular Q also lies on the plane
x − y + z = 3. Then, the coordinates ofQ are
the line
$
PQ = βi$ − $j + (β + 1)k
Q
Exp. (c)
3
=
2
Key Idea Use the foot of perpendicular
Q(x2 , y 2 , z2 ) drawn from point P(x1, y1, z1 ) to the
plane ax + by + cz + d = 0, is given by
3 ⇒ 4 β (β + 1) = 0
β = 0, − 1
β = −1
P(x1,y1,z1 )
[as β ≠ 0]
11. If the plane 2 x − y + 2 z + 3 = 0 has the
1
2
and
units from the planes
3
3
4x − 2 y + 4z + λ = 0 and 2 x − y + 2 z + µ = 0,
respectively, then the maximum value of
λ + µ is equal to
Q(x2,y2,z2)
ax+by+z+d=0
distances
[JEE Main 2019, 10 April Shift-II]
(a) 13
(b) 15
(c) 5
(b) ( 4, 0, − 1)
(d) (1, 0, 2)
(d) 9
Exp. (a)
Equation of given planes are
2x − y + 2z + 3 = 0
4x − 2 y + 4z + λ = 0
and
2x − y + 2z + µ = 0
QDistance between two parallel planes
ax + by + cz + d1 = 0,
and
ax + by + cz + d 2 = 0is
| d1 − d 2 |
distance =
a2 + b 2 + c 2
∴Distance between planes (i) and (ii) is
| λ − 2(3)|
1
=
16 + 4 + 16 3
…(i)
…(ii)
…(iii)
x2 − x1 y 2 − y1 z2 − z1
=
=
a
b
c
ax1 + by1 + cz1 + d
=−
a2 + b2 + c 2
Let a general point on the given line
x−1 y+ 1 z
=
= =r
−1
2
1
(say)
is P(2 r + 1, − 1 − r, r ).
Now, let foot of perpendicular Q( x1, y1, z1 ) to be
drawn from point P(2 r + 1, − 1 − r, r ) to the plane
x + y + z = 3, then
x1 − 2 r − 1 y + 1 + r z − r
=
=
1
1
1
 2r + 1 − 1 − r + r − 3
=−

1+ 1+ 1


⇒
[given]
⇒ | λ − 6| = 2 ⇒ λ − 6 = ± 2 ⇒ λ = 8 or 4
and distance between planes (i) and (iii) is
| µ − 3|
2
= [given]
4 + 1+ 4 3
⇒ |µ − 3| = 2 ⇒µ − 3 = ± 2 ⇒ µ = 5 or 1
So, maximum value of (λ + µ ) at λ = 8 and µ = 5
and it is equal to 13.
x1 − 2 r − 1 = y + r + 1
1
2
= z − r = (3 − 2 r ) = 1 − r
3
3
4r
5r
r
x1 = 2 + , y = − and z = 1+
⇒
3
3
3
4r
5r
r
So, point Q ≡  2 + , − , 1 +  , lies on the

3
3
3
plane x − y + z = 3 also.
r
4r 5r
So, 2 +
+
+ 1+ = 3 ⇒ r = 0
3
3
3
Therefore, the coordinates of point Q are (2, 0, 1)
282
JEE Main Chapterwise Mathematics
x −2 y +1 z −1
intersects the
=
=
3
2
−1
plane 2 x + 3y − z + 13 = 0 at a point P and the
plane 3x + y + 4z = 16 at a pointQ, then PQ is
equal to
[JEE Main 2019, 12 April Shift-I]
13. If the line
(a) 14
(b)
14
(c) 2 7
(d) 2 14
Exp. (d)
Equation of given line is
x−2 y+ 1 z−1
(let) …(i)
=r
=
=
3
2
−1
Now, coordinates of a general point over given
line is
R(3r + 2, 2 r − 1, − r + 1)
Let the coordinates of point P are
(3r1 + 2, 2 r1 − 1, r1 + 1) and Q are
(3r2 + 2, 2 r2 − 1, − r2 + 1).
Since, P is the point of intersection of line (i) and
the plane 2 x + 3 y − z + 13 = 0, so
2(3r1 + 2 ) + 3(2 r1 − 1) − (− r1 + 1) + 13 = 0
⇒
6r1 + 4 + 6r1 − 3 + r1 − 1 + 13 = 0
⇒
13r1 + 13 = 0 ⇒ r1 = − 1
So, point P(− 1, − 3, 2 )
And, similarly for point ‘Q’, we get
3(3r2 + 2 ) + (2 r2 − 1) + 4(− r2 + 1) = 16
⇒
7 r2 = 7 ⇒ r2 = 1
So, point is Q (5, 1, 0)
Now,
PQ = (5 + 1)2 + (1 + 3)2 + 2 2
=
36 + 16 + 4 =
56 = 2 14
14. A plane which bisects the angle between the
two given planes 2 x − y + 2 z − 4 = 0 and
x + 2 y + 2 z − 2 = 0, passes through the point
[JEE Main 2019, 12 April Shift-II]
(a) (1, − 4,1)
(c) ( 2 , 4,1)
(b) (1, 4, − 1)
(d) ( 2 , − 4,1)
Now, equation of planes bisecting the angles
between the planes (i) and (ii) are
2x − y + 2z − 4
x + 2y + 2z − 2
=±
4 + 1+ 4
1+ 4 + 4
⇒ 2 x − y + 2 z − 4 = ± ( x + 2 y + 2 z − 2)
On taking (+ ve) sign, we get a plane
…(iii)
x − 3y = 2
On taking (− ve) sign, we get a plane
…(iv)
3x + y + 4z = 6
Now from the given options, the point (2, − 4, 1)
satisfy the plane of angle bisector
3x + y + 4z = 6
15. The length of the perpendicular drawn from
the point (2, 1, 4) to the plane containing the
$)
lines r = ( i$ + $j) + λ( i$ + 2 $j − k
$ ) is
and r = (i$ + $j) + µ (− i$ + $j − 2k
[JEE Main 2019, 12 April Shift-II]
1
(b)
3
(a) 3
(c) 3
(d)
1
3
Exp. (c)
Key Idea QLength of the perpendicular drawn
from point (x1, y1, z1 ) to the plane
ax + by + cz + d = 0 is
| ax1 + by1 + cz1 + d |
d1 =
a2 + b 2 + c 2
Given line vectors
$ ) and
r = ($i + $j) + λ (i$ + 2 $j − k
$
$
$
$
$)
r = (i + j) + µ (− i + j − 2 k
…(i)
…(ii)
Now, a vector perpendicular to the given vectors
(i) and (ii) is
$i $j k
$
n = 1 2 −1
−1 1 − 2
Exp. (d)
Key Idea Equation of planes bisecting the
angles between the planes
a1 x + b1 y + c1 z + d1 = 0 and
a2 x + b2 y + c 2 z + d 2 = 0, are
a1 x + b1 y + c1 z + d1
=±
a12 + b12 + c12
a2 x + b2 y + c 2 z + d 2
a22 + b22 + c 22
Equation of given planes are
2x − y + 2z − 4 = 0
and
x + 2y + 2z − 2 = 0
…(i)
…(ii)
$ (1 + 2 )
= $i (−4 + 1) − $j(− 2 − 1) + k
$
= − 3$i + 3$j + 3k
∴The equation of plane containing given vectors
(i) and (ii) is
− 3 ( x − 1) + 3( y − 1) + 3 ( z − 0) = 0
⇒
−3 x + 3 y + 3 z = 0
…(iii)
⇒
x− y− z=0
Now, the length of perpendicular drawn from the
point (2, 1, 4) to the plane x − y − z = 0, is
| 2 − 1 − 4|
3
d1 =
=
= 3
1+ 1+ 1
3
283
Three Dimensional Geometry
16. The equation of the line passing through
or
(−4, 3, 1), parallel to the plane
x + 2 y − z − 5 = 0 and intersecting the line
x +1 y − 3 z − 2
is
=
=
−3
2
−1
[JEE Main 2019, 9 Jan Shift-I]
x + 4 y − 3 z −1
x + 4 y − 3 z −1
(b)
=
=
=
=
−1
−1
3
1
1
1
x + 4 y − 3 z −1
x −4 y + 3 z +1
(c)
(d)
=
=
=
=
1
1
3
2
1
4
(a)
x+ 4 y− 3 z−1
=
=
−1
3
1
[multiplying by 2]
17. The plane through the intersection of the
planes x + y + z = 1 and 2 x + 3y − z + 4 = 0
and parallel to Y -axis also passes through
the point
[JEE Main 2019, 9 Jan Shift-I]
(a) (3, 3, −1)
(c) (3, 2, 1)
(b) (−3, 1, 1)
(d) (−3, 0, −1)
Exp. (c)
Exp. (a)
x + 1 y −3 z − 2
is of the
=
=
2
−1
−3
form (−3λ − 1, 2 λ + 3, − λ + 2 )
 take x + 1 = y − 3 = z − 2 = λ ⇒ x = − 3λ − 1,
2
−1
−3

Any point on the line
y = 2 λ + 3 and z = − λ + 2]
So, the coordinates of point of intersection of two
lines will be (− 3λ − 1, 2 λ + 3, − λ + 2 ) for some
λ ∈ R.
Let the point A ≡ (− 3λ − 1, 2 λ + 3, − λ + 2 ) and
B ≡ (−4, 3, 1)
Then,
AB = OB − OA
= (−4$i + 3$j + k$ ) − {(−3λ − 1)$i + (2 λ + 3)$j
$
+ (− λ + 2 ) k}
= (3λ − 3)i$ − 2 λ$j + (λ − 1)k$
Now, as the line is parallel to the given plane,
therefore AB will be parallel to the given plane and
so AB will be perpendicular to the normal of plane.
$ is normal to
⇒ AB ⋅ λ = 0, where n = $i + 2 $j − k
the plane.
$ ⋅ (i$ + 2 $j − k$ ) = 0
⇒ ((3λ − 3)i$ − 2 λ$j + (λ − 1)k)
⇒ 3(λ − 1) − 4λ + (−1)(λ − 1) = 0
[QIf a = a1i$ + a2 $j + a3 k$ and
b = b1i$ + b2 $j + b3 k$ ,
then a ⋅ b = a1b1 + a2 b2 + a3 b3 ]
⇒
3λ − 3 − 4λ − λ + 1 = 0
⇒
− 2λ = 2
⇒
λ = −1
Now, the required equation is the equation of line
joining A(2, 1, 3) and B(−4, 3, 1,) which is
x − (− 4) y − 3 z − 1
=
=
2 − (−4) 1 − 3 3 − 1
[Qequation of line joining ( x1, y1, z1 ) and ( x2 , y2 , z2 )
x − x1
y − y1
z − z1 
is
=
=

x2 − x1 y2 − y1 z2 − z1 
⇒
x+ 4 y− 3 z−1
=
=
6
2
−2
Key Idea Equation of plane through the
intersection of two planes P1 and P2 is given by
P1 + λP2 = 0
The plane through the intersection of the planes
x + y + z − 1 = 0 and 2 x + 3 y − z + 4 = 0 is given
by
( x + y + z − 1) + λ(2 x + 3 y − z + 4) = 0,
where λ ∈ R
⇒(1 + 2 λ )x + (1 + 3λ )y + (1 − λ )z + (4λ − 1) = 0,
where λ ∈ R
…(i)
Since, this plane is parallel to Y-axis, therefore its
normal is perpendicular to Y-axis.
⇒
{(1 + 2 λ )i$ + (1 + 3λ )$j + (1 − λ )k$ } ⋅ $j = 0
⇒
1 + 3λ = 0
1
⇒
λ=−
3
Now, required equation of plane is
 1 − 2  x +  1 − 3  y +  1 + 1  z +  − 4 − 1 = 0








 3 



3
3
3
−1
[substituting λ =
in Eq. (i)]
3
⇒
x + 4z − 7 = 0
Here, only (3, 2, 1) satisfy the above equation.
18. If lines x = ay + b , z = cy + d and x = a ′ z + b ′,
y = c ′ z + d ′ are perpendicular, then
[JEE Main 2019, 9 Jan Shift-II]
(a) ab ′+ bc ′+1 = 0
(c) aa ′+ c + c ′ = 0
(b) bb ′+ cc ′+1 = 0
(d) cc ′+ a + a ′ = 0
Exp. (c)
Let 1st line is x = ay + b, z = cy + d .
z−d
x −b
z−d
x−b
= y,
= y⇒
= y=
⇒
c
a
c
a
The direction vector of this line is
b1 = ai$ + $j + ck$ .
Let 2nd line is x = a′ z + b ′, y = c ′ z + d ′.
284
JEE Main Chapterwise Mathematics
x − b′
y− d′
x − b′ y − d ′
= z,
= z⇒
=
= z
a′
c′
a′
c′
The direction vector of this line is
b 2 = a′ $i + c ′ $j + k$ .
QThe two lines are perpendicular, therefore,
b1 ⋅ b 2 = 0.
⇒ (ai$ + $j + ck$ ) ⋅(a′ i$ + c ′ $j + k$ ) = 0
⇒ aa′ + c ′ + c = 0 ⇒ aa′ + c + c ′ = 0
⇒
19. The equation of the plane containing the
x y z
straight line = = and perpendicular to
2 3 4
the plane containing the straight lines
x y z
x y z
= = and = = is
3 4 2
4 2 3
[JEE Main 2019, 9 Jan Shift-II]
(a) 5x + 2 y − 4z = 0
(b) x + 2 y − 2 z = 0
(c) 3x + 2 y − 3z = 0
(d) x − 2 y + z = 0
Exp. (d)
Let P1 be the plane containing the lines
x y z
x y z
= = and = = .
3 4 2
4 2 3
For these two lines, direction vectors are
b1 = 3i$ + 4$j + 2 k$ and b 2 = 4i$ + 2 $j + 3k$ .
A vector along the normal to the plane P1 is given by
$i $j k$
n1 = b1 × b 2 = 3 4 2
4 2 3
= $i(12 − 4) − $j(9 − 8) + k$ (6 − 16) = 8i$ − $j − 10k$
x y z
Let P2 be the plane containing the line = =
2 3 4
and perpendicular to plane P1.
x y z
For the line
= = , the direction vector is
2 3 4
b = 2 i$ + 3$j + 4k$ and it passes through the point
with position vector a = 0$i + 0$j + 0k$ .
QP2 is perpendicular to P1, therefore n1 and b lies
along the plane.
Also, P2 also passes through the point with
position vector a.
∴ Equation of plane P2 is given by
x−0 y−0 z−0
− 1 − 10 = 0
(r − a) ⋅(n1 × b ) = 0 ⇒ 8
2
3
4
⇒ x(− 4 + 30) − y(32 + 20) + z(24 + 2 ) = 0
⇒
26 x − 52 y + 26 z = 0
⇒
x − 2y + z = 0
20. Let
A be a point on the line
$ and B( 3, 2 , 6)
r = (1 − 3µ )$i + (µ − 1)$j + (2 + 5µ )k
be a point in the space. Then, the value of µ
for which the vector AB is parallel to the
plane x − 4y + 3z = 1 is
[JEE Main 2019, 10 Jan Shift-I]
1
(a)
4
1
(c)
8
(b) −
(d)
1
4
1
2
Exp. (a)
Given equation of line is
$
r = (1 − 3 µ )$i + (µ − 1) $j + ( 2 + 5 µ ) k
Clearly, any point on the above line is of the form
(1 − 3µ ,µ − 1, 2 + 5µ)
Let A be (− 3 µ + 1, µ − 1, 5µ + 2 ) for some µ ∈ R.
Then, AB = (3 − (− 3 µ + 1)) i$ + (2 − (µ − 1)) $j
$ [Q AB = OB − OA]
+ (6 − (5µ + 2 ))k
$ … (i)
= (3 µ + 2 ) $i + (3 − µ )$j + (4 − 5µ ) k
Normal vector (n ) of the plane x − 4 y + 3 z = 1is
$
…(ii)
n = $i − 4$j + 3k
QAB is parallel to the plane
∴n is perpendicular to the AB
⇒
AB ⋅ n = 0
$ ]⋅
⇒ [(3 µ + 2 )$i + (3 − µ )$j + (4 − 5 µ )k
$
$
$]= 0
[i − 4 j + 3k
[from Eqs. (i) and (ii)]
⇒ (3 µ + 2 ) − 4(3 − µ ) + 3 (4 − 5 µ ) = 0
⇒
⇒
− 8µ + 2 = 0
1
µ=
4
21. The plane passing through the point
( 4, − 1, 2 ) and parallel to the lines
x +2 y −2 z +1
x −2 y − 3 z − 4
and
=
=
=
=
3
2
−1
1
2
3
also passes through the point
[JEE Main 2019, 10 Jan Shift-I]
(a) ( −1, − 1, − 1)
(c) (1, 1, 1)
(b) (1,1, − 1)
(d) ( −1, − 1,1)
Exp. (c)
Let a be the position vector of the given point
(4, − 1, 2 ).
$
Then,
a = 4$i − $j + 2 k
285
Three Dimensional Geometry
The direction vector of the lines
Exp. (d)
x+ 2 y−2 z+ 1
=
=
3
2
−1
and
x − 2 y − 3 z− 4
=
=
1
2
3
are respectively
$
b1 = 3$i − $j + 2 k
$
and
b = i$ + 2 $j + 3k
2
Now, as the plane is parallel to both b1 and b2
[Q plane is parallel to the given lines]
So, normal vector (n ) of the plane is perpendicular
to both b1 and b2 .
⇒
n = b1 × b2 and
Required equation of plane is
(r − a ) ⋅ n = 0
⇒
(r − a ) ⋅ ( b1 × b2 ) = 0
x− 4 y+ 1 z−2
⇒
3
1
−1
2
2
3
=0
$ ) − (4$i − $j + 2 k
$ )
Q r − a = ( x$i + y$j + zk


$i + ( y + 1)$j + ( z − 2 ) k
$
=
(
x
−
4
)


 and we know that, [a b c] = a ⋅ ( b × c)


a1 a2 a3




= b1 b2 b3


c1 c 2 c 3


⇒ ( x − 4) (− 3 − 4) − ( y + 1) (9 − 2 )
+ ( z − 2 ) (6 + 1) = 0
⇒ − 7 ( x − 4) − 7 ( y + 1) + 7 ( z − 2 ) = 0
⇒
( x − 4) + ( y + 1) − ( z − 2 ) = 0
⇒
x + y − z − 1= 0
(1, 1, 1) is the only point that satisfies.
22. On which of the following lines lies the
point of intersection of the line,
x − 4 y −5 z − 3
and the plane,
=
=
2
2
1
[JEE Main 2019, 10 Jan Shift-II]
x + y + z =2 ?
x−4 y −5 z−5
=
=
1
1
−1
x + 3 4− y z +1
(b)
=
=
3
3
−2
x−2 y −3 z+ 3
(c)
=
=
2
2
3
x −1 y − 3 z + 4
(d)
=
=
1
2
−5
(a)
Given equation of line is
x− 4 y− 5 z− 3
…(i)
=
=
= r (let)
2
2
1
⇒
x = 2 r + 4; y = 2 r + 5 and z = r + 3
∴ General point on the line (i) is
P(2 r + 4, 2 r + 5, r + 3)
So, the point of intersection of line (i) and plane
x + y + z = 2 will be of the form
P (2 r + 4, 2 r + 5, r + 3) for some r ∈ R.
⇒(2 r + 4) + (2 r + 5) + (r + 3) = 2
[Qthe point will lie on the plane]
⇒
5r = −10 ⇒ r = −2
So, the point of intersection is P(0, 1, 1)
[putting r = −2 in (2 r + 4, 2 r + 5, r + 3)]
Now, on checking the options, we get
x−1 y− 3 z+ 4
contain the point (0, 1, 1)
=
=
1
2
−5
23. The plane which bisects the line segment
joining the points ( −3, − 3, 4) and ( 3, 7, 6) at
right angles, passes through which one of
the following points ?
[JEE Main 2019, 10 Jan Shift-II]
(a) ( 4, − 1, 7)
(c) ( −2 , 3, 5)
(b) ( 2 ,1, 3)
(d) ( 4,1, − 2 )
Exp. (d)
Let the given points be A(− 3, − 3, 4) and B (3, 7 6).
Then, mid-point of line joining A, B is
P 

− 3 + 3 − 3 + 7 4 + 6
,
,
 = P (0, 2, 5)
2
2
2 
QThe required plane is perpendicular
bisector of line joining A, B, so direction ratios of
normal to the plane is proportional to the direction
ratios of line joining A, B.
So, the direction ratios of normal to the plane are
6, 10, 2.
[QDR’s of AB are 3 + 3, 7 + 3, 6 − 4, i.e. 6, 10, 2]
Now, equation of plane is given by
a( x − x1 ) + b ( y − y1 ) + c ( z − z1 ) = 0
6( x − 0) + 10( y − 2 ) + 2( z − 5) = 0
[QP(0,2, 5) line on the plane]
⇒
3 x + 5 y − 10 + z − 5 = 0
… (i)
⇒
3 x + 5 y + z = 15
On checking all the options, the option (4, 1, − 2)
satisfy the equation of plane (i).
286
JEE Main Chapterwise Mathematics
Exp. (c)
24. The direction ratios of normal to the plane
through the points (0, − 1, 0) and (0, 0, 1) and
π
making an angle with the plane
4
y − z + 5 = 0 are
Let the direction vector of the line
x− 3 y+ 2 z−1
is
=
=
2
3
−1
b = 2 $i − $j + 3k$ .
Since, the required plane contains this line and its
projection along the plane 2 x + 3 y − z = 5, it will
also contain the normal of the plane
2 x + 3 y − z = 5.
[JEE Main 2019, 11 Jan Shift-I]
(a) 2 , − 1,1
(c) 2 , 2 , − 2
(b) 2 ,1, − 1
(d) 2 3 ,1, − 1
Exp. (b, c)
=
|b − c|
a2 + b 2 + c 2 2
⇒ a + b 2 + c 2 = |b − c|2 = (b − c )2
2
= b 2 + c 2 − 2 bc
⇒
a = −2 bc
[using Eq. (ii)]
⇒
a2 = 2 b 2
⇒
a = ± 2b
⇒ Direction ratios (a, b,c ) = (± 2 , 1, − 1)
So, options (b) and (c) are correct because 2,
2 , − 2 and
2 , 1, − 1. are multiple of each
other.
b×n
plane
containing
the
line
x −3 y +2 z −1
and also containing its
=
=
2
3
−1
projection on the plane 2 x + 3y − z = 5,
contains which one of the following points?
[JEE Main 2019, 11 Jan Shift-I]
(a) ( −2 , 2 , 2 )
(c) ( 2 , 0, − 2 )
(b) ( 2 , 2 , 0)
(d) (0, − 2 , 2 )
B
3 y–
z=
5
C
z–1
y+2
x–3
=
=
–1
3
2
L
required plane (ABCD)
b
P
D
Projection of line L
Normal vector of the plane 2 x + 3 y − z = 5 is
n = 2 i$ + 3$j − k$ .
Now, the required plane contains b = 2 $i − $j + 3k$
and n = 2 i$ + 3$j − k$ .
∴Normal of the required plane is b × n.
Since, the plane contains the line
x − 3 y + 2 z −1
, therefore it also contains
=
=
2
3
−1
the point a = 3$i − 2 $j + k$ .
Now, the equation of required plane is
(r − a) ⋅ (b × n ) = 0
x− 3 y+ 2 z−1
2
−1
3 =0
2
25. The
n
A
2x+
Let the equation of plane be
a( x − 0) + b( y + 1) + c ( z − 0) = 0
[Qequation of plane passing through a point
( x1, y1, z1 ) is given by
a( x − x1 ) + b( y − y1 ) + c( z − z1 ) = 0]
... (i)
⇒
ax + by + cz + b = 0
Since, it also passes through (0, 0, 1), therefore,
we get
...(ii)
c+ b= 0
Now, as angle between the planes
ax + by + cz + b = 0
π
and y − z + 5 = 0 is .
4
π  |n1 ⋅ n 2|

; where n1 = ai$ + b$j + c k$
∴ cos   =
 4  |n1||n 2|
and n 2 = 0i$ + $j − k$
1 |(ai$ + b$j + ck$ ) ⋅ (0i$ + $j − k$ )|
⇒
=
2
a2 + b 2 + c 2 0 + 1 + 1
2
−1
3
⇒( x − 3) [1 − 9] − ( y + 2 ) [ − 2 − 6] +
( z − 1) [6 + 2 ] = 0
⇒
− 8 x + 8 y + 8 z + 32 = 0
⇒
x− y− z=4
Note that (2, 0, − 2) is the only point which satisfy
above equation.
26. If the point (2 , α , β ) lies on the plane which
passes through the points (3, 4, 2) and
(7, 0, 6) and is perpendicular to the plane
2 x − 5y = 15, then 2α − 3β is equal to
[JEE Main 2019, 11 Jan Shift-II]
(a) 17
(c) 5
(b) 7
(d) 12
287
Three Dimensional Geometry
Exp. (b)
According to given information, we have the
following figure.
2, –5, 0
n
2x–5y=15
p
A
(3, 4, 2)
B (7, 0, 6)
C (2, a, b)
From figure, it is clear that
(AB × BC) = p and n = 2 i$ − 5$j + 0k$
$i
$j
k$
∴ p = 4 −4
4
−5 α β − 6
28. The perpendicular distance from the origin
to the plane containing the two lines,
[QAB = (7 − 3)i$ + (0 − 4)$j + (6 − 2 )k$
= 4i$ − 4$j + 4k$ and
BC = (2 − 7 )i$ + (α − 0)$j + (β − 6)k$
= − 5$i + α$j + (β − 6)k$ ]
$
$
= i(−4β + 24 − 4α ) − j(4β − 24 + 20) + k$ (4α − 20)
⇒ p = (24 − 4α − 4β )$i + $j(4 − 4β ) + k$ (4α − 20)
Now,as the planes are perpendicular, therefore
p⋅ n = 0
⇒ ((24 − 4α − 4β )i$ + (4 − 4β )$j + (4α − 20)k$ )
⋅ (2 $i − 5$j + 0k$ )) = 0
⇒ 2(24 − 4α − 4β ) − 5(4 − 4β ) + 0 = 0
⇒ 8(6 − α − β ) − 4(5 − 5β ) = 0
⇒ 12 − 2α − 2 β − 5 + 5β = 0
⇒
2 α − 3β = 7
x − 3 y +1 z −6
and
=
=
1
3
−1
x +5 y −2 z − 3
intersect at the point R.
=
=
7
4
−6
The reflection of R in the xy-plane has
coordinates
[JEE Main 2019, 11 Jan Shift-II]
27. Two
lines
(a) ( 2 , − 4, − 7)
(c) ( − 2 , 4, 7)
(b) ( 2 , − 4, 7)
(d) (2, 4, 7)
Exp. (a)
x−3 y+1 z−6
= a (say). Then, any
=
=
1
3
−1
point on this line is of the form
P(a + 3, 3a − 1, − a + 6)
Let
Similarly, any point on the line.
x+ 5 y−2 z− 3
=
=
= b(say), is of the form
−6
7
4
Q (7 b − 5, − 6b + 2, 4b + 3)
Now, if the lines are intersect, then P = Q for some
a and b.
⇒
a + 3 = 7b − 5
3a − 1 = − 6b + 2
and
− a + 6 = 4b + 3
⇒ a − 7 b = − 8, a + 2 b = 1and a + 4b = 3
On solving a − 7 b = − 8 and a + 2 b = 1, we get
b = 1 and a = − 1 ,
which also satisfy a + 4b = 3
∴ P = Q = (2, − 4, 7 ) for a = − 1and b = 1
Thus, coordinates of point R are (2, − 4, 7 )
and reflection of R in xy-plane is (2, − 4, − 7 ).
x+2 y−2 z+5
x −1 y − 4 z + 4
and
,
=
=
=
=
3
5
7
1
4
7
is
[JEE Main 2019, 12 Jan Shift-I]
(a)11 6
(b)
11
6
(c)11
(d) 6 11
Exp. (b)
Let the equation of plane, containing the two lines
x+ 2 y−2 z+ 5
=
=
3
5
7
x−1 y− 4 z+ 4
and
is
=
=
1
4
7
… (i)
a( x + 2 ) + b( y − 2 ) + c( z + 5) = 0
Q Plane (i) contain lines, so
... (ii)
3a + 5b + 7c = 0
and
... (iii)
a + 4b + 7c = 0
From Eqs. (ii) and (iii), we get
a
b
c
=
=
35 − 28 7 − 21 12 − 5
a
b
c
a
b
c
=
= ⇒ =
=
⇒
7 − 14 7
1 −2 1
So, equation of plane will be
1( x + 2 ) − 2( y − 2 ) + 1( z + 5) = 0
... (iv)
⇒
x − 2 y + z + 11 = 0
Now, perpendicular distance from origin to plane
11
11
is =
=
1+ 4 + 1
6
[Qperpendicular distance from origin to the
|d|
]
plane ax + by + cz + d = 0, is
a2 + b 2 + c 2
288
JEE Main Chapterwise Mathematics
29. A tetrahedron has vertices P(1, 2 , 1),
1 − λ2
2
2
1 − λ2
0
2
2
− λ2 − 1
Q (2 , 1, 3), R ( − 1, 1, 2 ) and O(0, 0, 0). The angle
between the facesOPQ and PQR is
[JEE Main 2019, 12 Jan Shift-I]
 7
 
 31 
9
(b) cos−1  
 35 
19
(c) cos−1  
 35 
17
(d) cos−1  
 31 
(a) cos
−1
Exp. (c)
=0
0
Qcondition of coplanarity is
 x −x
y2 − y1 z2 − z1
1
 2
 x3 − x1 y3 − y1 z3 − z1 =

 x4 − x1 y4 − y1 z4 − z1
⇒



0


(−1 − λ2 ) [(1 − λ2 )2 − 4] = 0
⇒(1 + λ ) [(1 − λ2 − 2 ) (1 − λ2 + 2 )] = 0
2
The given vertices of tetrahedron PQRO are
P(1, 2, 1,) Q (2, 1, 3), R (−1, 1, 2 ) and O (0, 0, 0).
The normal vector to the face OPQ
= OP × OQ
= (i$ + 2 i$ + k$ ) × (2 i$ + $j + 3k$ )
$i $j k$
= 1 2 1 = 5$i − $j − 3k$
2 1 3
and the normal vector to the face PQR
= PQ × PR
= (i$ − $j + 2 k$ ) × (−2 $i − $j + k$ )
$j k$
i$
=
1
−1 2
−2 −1 1
= $i (− 1 + 2 ) − $j (1 + 4) + k$ (− 1 − 2 )
= $i − 5$j − 3k$
Now, the angle between the faces OPQ and PQR
is the angle between their normals
5+ 5+ 9
= cos − 1
25 + 1 + 9 1 + 25 + 9
19
= cos − 1  
 35 
(1 + λ2 )2 (3 − λ2 ) = 0
⇒
λ2 = 3
that a plane passing through the points
( − λ 2 , 1, 1), (1, − λ 2 ,1) and (1, 1, − λ 2 ) also
passes through the point ( − 1, − 1, 1). Then, S
is equal to
[JEE Main 2019, 12 Jan Shift-II]
(b) { 3, − 3}
(d) { 3 }
Exp. (a)
According
to
the
question
points
(− λ2 , 1, 1), (1, − λ2 , 1) and (1, 1, − λ2 ) are coplanar
with the point (−1, − 1, 1,) so
[Q1 + λ2 ≠ 0 ∀ λ ∈ R]
λ=± 3
⇒
31. If an angle between the line,
x +1 y −2 z − 3
and the plane,
=
=
2
1
−2
2 2 
x − 2 y − kz = 3 is cos − 1 
 , then a value
 3 
of k is
[JEE Main 2019, 12 Jan Shift-II]
(a)
5
3
(b)
3
5
(c) −
3
5
(d) −
5
3
Exp. (a)
Clearly, direction ratios of given line are
2, 1, –2 and direction ratios of normal to the given
plane are 1, –2, –k.
As we know angle ‘θ’ between line and plane can
be obtained by
|a1 a2 + b1 b2 + c1c 2|
sinθ =
2
a1 + b12 + c12 a22 + b22 + c 22
So, sinθ =
30. Let S be the set of all real values of λ such
(a) { 3 , − 3 }
(c) {1, − 1}
⇒
⇒
⇒
|2 − 2 + 2 k|
4 + 1 + 4 1 + 4 + k2

|2 k|
2 2
sin cos −1
 =
3  3 5 + k2


−1 2 2 
given θ = cos
3 

2 |k|
1
=
3
3 5 + k2

2 2
1
−1 1
Qcosθ = 3 ⇒sinθ = 3 ⇒ θ = sin 3

2 2
1
= sin−1 
⇒ cos −1
3
3
289
Three Dimensional Geometry
(a)
3
√32–(2√2)2 =1
4k = 5 + k ⇒ 3k = 5
5
k=±
3
2
2
32. If L1 is the line of intersection of the planes
2 x − 2 y + 3z − 2 = 0, x − y + z + 1 = 0 and L 2 is
the line of intersection of the planes
x + 2 y − z − 3 = 0, 3x − y + 2 z − 1 = 0, then the
distance of the origin from the plane,
containing the lines L1 and L 2 is
[JEE Main 2018]
(a)
1
4 2
2
3
(c)
1
3
(d)
2
3
Key idea Length of projection of the line
segment joining a 1 and a 2 on the plane r ⋅ n = d
(a − a 1 ) × n
is 2
| n|
2√2
2
⇒
(b)
Exp. (d)
θ
⇒
2
3
(b)
1
3 2
(c)
1
2 2
(d)
1
2
Exp. (b)
L1 is the line of intersection of the plane
2 x − 2 y + 3 z − 2 = 0 and x − y + z + 1 = 0 and
L2 is the line of intersection of the plane
x + 2 y − z − 3 = 0 and 3 x − y + 2 z − 1 = 0
$i $j k$
Since Li is parallel to2 −2 3 = $i + $j




1 −1 1
k$ 
$i $j

L2 is parallel to 1 2 −1 = 3$i − 5$j − 7 k$




3 −1 2 
5 8
Also, L2 passes through  , , 0 .
7 7 
[put z = 0 in last two planes]
x − 5 y − 8 z 

 7
7
So, equation of plane is 1
1
0 = 0
 3
−5
−7




⇒ 7 x − 7 y + 8z + 3 = 0
Now, perpendicular distance from origin is


3
1
3
=
=

 2
2
2
162 3 2
7
+
7
+
8


33. 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
[JEE Main 2018]
Length of projection the line segment joining the
points (5, −1, 4) and (4, −1, 3) on the plane
x + y + z = 7 is
B (4, –1, 3)
(a2)
^
^
^
n=i+j+k
(5, –1, 4)
C
A(a1)
(a − a1 ) × n |(− $i − k$ ) × ($i + $j + k$ )|
AC =  2
=
| $i + $j + k$|
|n |


|^i − ^
k|
2
2
=
⇒ AC =
3
3
3
Alternative Method
Clearly, DR’s of AB are 4 − 5, − 1 + 1, 3 − 4,
i.e. −1, 0, −1
and DR’s of normal to plane are 1, 1, 1.
Now, let θ be the angle between the line and plane,
then θ is given by
− 1+ 0 − 1
2
2
sinθ =
=
=
2
2
2
2
2
2
3
3
(−1) + (−1) 1 + 1 + 1
AC =
B (4, –1, 3)
A (5, –1, 4)
θ
C
θ
Plane : x + y + z = 7
2
3
⇒
sinθ =
⇒
cos θ = 1 − sin2 θ = 1 −
2
1
=
3
3
1
3
[Q AB = 2 ]
Clearly, length of projection = ABcosθ = 2 ⋅
=
2
3
290
JEE Main Chapterwise Mathematics
34. If the image of the point P(1, − 2 , 3) in the
plane2 x + 3y − 4z + 22 = 0 measured parallel
x y z
to the line = = isQ, then PQ is equal to
1 4 5
[JEE Main 2017 (offline)]
(a) 3 5
(c) 42
(b) 2 42
(d) 6 5
Exp. (b)
2
x y z
= = and passing
1 4 5
through P(1, − 2, 3) is
Any line parallel to
−1 −1
∴Equation of a plane passing through (1, − 1, − 1)
and perpendicular to n is given by
5( x − 1) + 7( y + 1) + 3( z + 1) = 0
P (1, –2, 3)
<1, 4, 5>
x y z
= =
1 4 5
x−2 y+ 1 z+ 7
=
=
−1
2
−1
$
$
$
Let n1 = i − 2 j + 3k and n 2 = 2 $i − $j − k$
∴Any vector n perpendicular to both n1, n 2 is given
by
n = n1 × n2
$i
$j
k$
n = 1 − 2 3 = 5$i + 7 $j + 3k$
⇒
and
⇒
5x + 7 y + 3z + 5 = 0
∴Required distance =
R
2x + 3y – 4z + 22 = 0
=
Q
x−1 y+ 2 z− 3
=
=
=λ
1
4
5
(say)
42
35. The distance of the point (1, 3, −7) from the
plane passing through the point (1, − 1, − 1)
having normal perpendicular to both the
lines
x −1 y +2 z − 4
x −2 y +1
and
=
=
=
=
1
3
2
−2
−1
z+7
, is
−1
[JEE Main 2017 (offline)]
10
(b)
units
83
10
(d)
units
74
Exp. (b)
Given, equations of lines are
x−1 y+ 2 z− 4
=
=
1
3
−2
(b)10 3
(c)
10
3
(d)
20
3
Exp. (b)
PQ = 2 PR = 2 42
20
(a)
units
74
5
(c)
units
83
10
units
83
36. The distance of the point (1, − 5, 9) from the
(a) 3 10
∴Coordinates of R are (λ + 1, 4λ − 2, 5λ + 3).
∴
52 + 7 2 + 32
plane x − y + z = 5 measured along the line
[JEE Main 2016 (offline)]
x = y = z is
Any point on above line can be written as
(λ + 1, 4λ − 2, 5λ + 3).
Since, point R lies on the above plane.
∴2(λ + 1) + 3(4λ − 2 ) − 4(5λ + 3) + 22 = 0
⇒
λ =1
So, point R is (2, 2, 8).
Now, PR = (2 − 1)2 + (2 + 2 )2 + (8 − 3)2 =
5 + 21 − 21 + 5
Equation of line passing through (1, − 5, 9) and
parallel to x = y = z is
x−1 y+ 5 z− 9
(say)
=
=
=λ
1
1
1
Thus, any point on this line is of the form
(λ + 1, λ − 5, λ + 9).
Now, if P(λ + 1, λ − 5, λ + 9) is the point of
intersection of line and plane, then
(λ + 1) − (λ − 5) + λ + 9 = 5
⇒
λ + 15 = 5
⇒
λ = − 10
∴Coordinates of point P are (− 9, − 15, − 1.)
Hence, the required distance
= (1 + 9)2 + (− 5 + 15)2 + (9 + 1)2
= 102 + 102 + 102 = 10 3
x − 3 y +2 z + 4
lies in the
=
=
2
3
−1
plane,lx + my − z = 9, thenl 2 + m 2 is equal to
37. If the line,
[JEE Main 2016 (offline)]
(a) 26
(b) 18
(c) 5
(d) 2
291
Three Dimensional Geometry
Exp. (c)
Exp. (d)
x−3 y+2 z+ 4
lies in the
=
=
2
3
−1
plane lx + my − z = 9, therefore we have
2l − m − 3 = 0
[Q normal will be perpendicular to the line]
...(i)
⇒
2l − m = 3
and 3l − 2 m + 4 = 9
[Q point (3, − 2, − 4) lies on the plane]
...(ii)
⇒
3l − 2 m = 5
On solving Eqs. (i) and (ii), we get
l = 1and m = − 1
∴
l 2 + m2 = 2
Since, the line
38. The distance of the point (1, 0, 2) from the
point of intersection of the line
x −2 y +1 z −2
and
the
plane
=
=
3
4
12
[JEE Main 2015]
x − y + z = 16 is
(a) 2 14
(b) 8
(c) 3 21
(d) 13
Exp. (d)
Given equation of line is
x−2 y+ 1 z−2
=
=
=λ
3
4
12
and equation of plane is
(say)…(i)
x − y + z = 16
…(ii)
Any point on the line (i) is
(3λ + 2, 4λ − 1, 12 λ + 2)
Let this point be point of intersection of the line
and plane.
(3∴
λ + 2 ) − (4λ − 1) + (12 λ + 2 ) = 16
⇒
11λ + 5 = 16
⇒
11λ = 11
⇒
λ =1
∴Point of intersection is (5, 3, 14).
Now, distance between the points (1, 0, 2) and
(5, 3, 14)
= (5 − 1)2 + (3 − 0)2 + (14 − 2 )2
= 16 + 9 + 144
= 169 = 13
39. The equation of the plane containing the
line 2 x − 5y + z = 3, x + y + 4z = 5 and parallel
to the plane x + 3y + 6z = 1 is [JEE Main 2015]
(a) 2 x + 6y + 12 z = 13
(c) x + 3y + 6z = 7
(b) x + 3y + 6z = − 7
(d) 2 x + 6y + 12 z = − 13
Let equation of plane containing the lines
2 x − 5 y + z = 3 and x + y + 4 z = 5 be
(2 x − 5 y + z − 3) + λ( x + y + 4 z − 5) = 0
⇒ (2 + λ )x + (λ − 5)y + (4λ + 1)z − 3 − 5λ = 0
…(i)
This plane is parallel to the plane x + 3 y + 6 z = 1.
2 + λ λ − 5 4λ + 1
∴
=
=
1
3
6
On taking first two equalities, we get
6 + 3λ = λ − 5
⇒
2 λ = − 11
11
⇒
λ=−
2
On taking last two equalities, we get
6λ − 30 = 3 + 12 λ
11
⇒
−6λ = 33 ⇒ λ = −
2
So, the equation of required plane is
 2 − 11 x +  −11 − 5 y +  − 44 + 1 z − 3






 2

 2


2
11
+ 5× =0
2
7
21
42
49
y−
z+
− x−
=0
⇒
2
2
2
2
⇒
x + 3y + 6z − 7 = 0
40. The angle between the lines whose direction
cosines satisfy the equation l + m + n = 0 and
l 2 = m 2 + n 2 is
[JEE Main 2014]
(a)
π
3
(b)
/ Use formula
cos θ =
a12
π
4
(c)
π
6
(d)
π
2
a1a2 + b1b 2 + c1c2
+ b12 + c12
a22 + b 22 + c22
Exp. (a)
Given,
⇒
⇒
⇒
l + m+ n= 0
l = − (m + n)
(m + n)2 = l 2
m2 + n2 + 2 mn = m2 + n2
[Ql 2 = m2 + n2 , given]
⇒
2 mn = 0
Case I When m = 0, then
l=−n
Hence, (l, m, n) is (1, 0, − 1).
292
JEE Main Chapterwise Mathematics
Case II
Given planes are
2x + y + 2z − 8 = 0
5
and
2x + y + 2z + = 0
2
Distance between two planes
5
21
−8−
| c1 − c 2 |
7
2
=
=
= 2 =
2
2
2
2
2
2
3
2
a + b +c
2 +1 +2
When n = 0, then
l=−m
Hence, (l, m, n) is (1, − 1, 0).
∴
cos θ =
⇒
θ=
1+ 0 + 0 1
=
2 × 2 2
π
3
x −1 y − 3 z − 4
=
=
3
1
−5
in the plane 2 x − y + z + 3 = 0 is the line
41. The image of the line
[JEE Main 2014]
x+ 3 y −5 z−2
x+ 3 y −5 z+ 2
(a)
(b)
=
=
=
=
5
−1
3
1
−5
−3
x−3 y + 5 z−2
x−3 y + 5 z−2
(d)
(c)
+
=
=
=
3
1
−5
−3
−1
5
/ Distance of mirror from the object is equal to
distance of image from the mirror.
Exp. (a)
Plane and line are parallel to each other. Equation
of normal to the plane through the point(1, 3, 4) is
x−1 y− 3 z− 4
[say]
=
=k
=
2
1
−1
Any point in this normal is
(2 k + 1, − k + 3, 4 + k )
2k + 1 + 1 3 − k + 3 4 + k + 4
,
,
⇒ 
 lies on


2
2
2
plane.
6 − k  8 + k
3
0
+
+
=
⇒ 2(k + 1) − 
 

 2   2 
⇒
k = −2
Hence, point through which this image pass is
(2 k + 1, 3 − k, 4 + k )
i.e., [2 (− 2 ) + 1, 3 + 2, 4, − 2 ]
= (− 3, 5, 2 )
Hence, equation of image line is
x+ 3 y− 5 z−2
=
=
3
1
−5
42. Distance between two parallel planes
2 x + y + 2 z = 8 and 4x + 2 y + 4z + 5 = 0 is
[JEE Main 2013]
(a)
3
2
Exp. (c)
(b)
5
2
(c)
7
2
(d)
9
2
x −2 y − 3 z − 4
and
=
=
1
1
−k
x −1 y − 4 z −5
are coplanar, then k can
=
=
k
2
1
have
[JEE Main 2013]
43. If the lines
(a) any value
(c) exactly two values
(b) exactly one value
(d) exactly three values
Exp. (c)
The given lines are
x−2 y− 3 z− 4
=
=
1
1
−k
x−1 y− 4 z− 5
and
=
=
k
2
1
Condition for two lines are coplanar.
x1 − x2 y1 − y2 z1 − z2
l1
l2
m1
m2
…(i)
…(ii)
=0
n1
n2
where, ( x1, y1, z1 ) and ( x2 , y2 , z2 ) are any points on
the lines (i) and (ii), respectively and < l1, m1, n1 >
and < l2 , m2 , n2 > are direction cosines of lines (i)
and (ii), respectively.
2 −1 3− 4 4− 5
1
1
−k =0
∴
k
⇒
2
1
−1 −1
1 −k =0
1
1
k
2
1
⇒
1(1 + 2 k ) + (1 + k ) − (2 − k ) = 0
⇒
k2 + 2 k + k = 0
⇒
⇒
2
k 2 + 3k = 0
k = 0, − 3
Note If 0 appears in the denominator, then the
correct way of representing the equation of
straight line is
y −4 z −5
x −2 y −3
; z = 4 and x = 1;
=
=
2
1
1
1
293
Three Dimensional Geometry
44. An equation of a plane parallel to the plane
x − 2 y + 2 z − 5 = 0 and at a unit distance from
the origin is
[AIEEE 2012]
(a) x − 2 y + 2 z − 3 = 0
(c) x − 2 y + 2 z − 1 = 0
(b) x − 2 y + 2 z + 1 = 0
(d) x − 2 y + 2 z + 5 = 0
Exp. (a)
Given A plane P : x − 2 y + 2 z − 5 = 0
To find The equation of a plane parallel to given
plane P and at a distance of 1 unit from origin.
Equation of family of plane parallel to the given
plane P is
Q : x − 2y + 2z + d = 0
Also, perpendicular distance of Q from origin is
1 unit.
0 − 2(0) + 2(0) + d
⇒
12 + 2 2 + 2 2
=1
d
=1
3
⇒
the
x −1 y +1 z −1
=
=
2
3
4
line
[AIEEE 2012]
(b)
2
9
(c)
9
2
On solving 2 p + 1 = q + 3 and 4 p + 1 = q , we get
the values of p and q as
−3
and q = − 5
p=
2
On substituting the values of p and q in the third
equation 3 p − 1 = 2q + k, we get
− 3
9
3 
 − 1 = 2(− 5) + k ∴ k =
 2 
2
y −1
=
2

and the plane x + 2 y + 3z = 4 is cos −1 

46. If the angle between the line x =
3
(a)
2
Exp. (d)
(d) 0
z−3
λ
5
,
14 
[AIEEE 2011]
2
(b)
5
5
(c)
3
(d)
2
3
Angle between straight line
^
r = a + λb and plane r ⋅ n = d is
^
b⋅n
sin θ =
and
x − 3 y −k z
=
= intersect, then k is equal to
1
2
1
(a) –1
2 p + 1 = q + 3, 3 p − 1 = 2q + k and 4 p + 1 = q
then λ is equal to
⇒
d=± 3
Hence, the required equation of the plane
parallel to P and at unit distance from origin is
x − 2y + 2z ± 3 = 0
Hence, out of the given equations, option (a) is
the only correct.
45. If
Since, L1 and L2 are intersecting each other,
hence both points P and Q should coincide at the
point of intersection, i.e., corresponding
coordinates of P and Q should be same.
∴
sin θ =
sin θ =
⇒
^
|b||n|
($i + 2 $j + λ k$ ) ⋅ ($i + 2 $j + 3 k$ )
1 + 4 + λ2
1+ 4 + 9
5 + 3λ
λ2 + 5 ⋅ 14
5
14
3
⇒
sin θ =
14
Given, cos θ =
Exp. (c)
x−1 y+ 1 z−1
=
=
2
3
4
x−3 y−k z−0
=
=
L2 :
1
2
1
3
=
14
Given Two lines L1 :
∴
and
⇒ 9 (λ + 5) = 9λ + 30λ + 25
To find The value of k of the given lines L1 and L2
are intersecting each other.
x−1 y+ 1 z−1
Let
=
=
= p
L1 :
2
3
4
x−3 y−k z−0
and
=
=
=q
L2 :
1
2
1
⇒ Any point P on line L1 is of type
P(2 p + 1, 3 p − 1, 4 p + 1) and any point Q on line L2 is
of type Q(q + 3, 2q + k, q ).
2
5 + 3λ
λ + 5 ⋅ 14
2
2
⇒ 9 λ2 + 45 = 9λ2 + 30λ + 25 ⇒ 30λ = 20
2
∴
λ=
3
47. Statement I The point A (1, 0, 7) is the
mirror image of the pointB (1, 6, 3) in the line
x y −1 z −2
=
=
.
1
2
3
294
JEE Main Chapterwise Mathematics
x y −1
z −2
=
=
1
2
3
bisects the line segment joining A (1, 0 , 7)
and B (1, 6, 3).
[AIEEE 2011]
Statement II The
(a) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(b) Statement I is true, Statement II is false
(c) Statement I is false, Statement II is true
(d) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
Exp. (d)
Mid-point of AB is M (1, 3, 5)
A (1, 0, 7)
M
x
1
B
Q
∴ DR’s of PQ is (2 λ − 3, 3λ + 3, 4λ − 8).
Also, perpendicular to straight line AB
x y−2 z− 3
=
=
=λ
2
3
4
having DR’s (2, 3, 4).
Thus, 2 (2 λ − 3) + 3 (3λ + 3) + 4 (4λ − 8) = 0
⇒
4λ − 6 + 9λ + 9 + 16λ − 32 = 0
⇒
29λ − 29 = 0
∴
λ =1
Hence, coordinates of Q are (2, 5, 7).
|PQ| = (3 − 2 )2 + (− 1 − 5)2 + (7 − 11)2
= 1 + 36 + 16 =
53
49. The distance of the point (1, − 5, 9) from the
which lies on
x y−1 z−2
=
=
1
2
3
1 3−1 5−2
as
=
=
⇒ 1= 1= 1
1
2
3
Hence, Statement II is true.
Also, direction ratios of AB is
(1 − 1, 6 − 0, 3 − 7 )
…(i)
i .e.,
(0, 6, − 4)
And direction ratios of straight line is
…(ii)
(1, 2 , 3)
The two lines are perpendicular, if
0 (1) + 6 (2 ) − 4 (3) = 12 − 12 = 0
Hence, Statement I is true and Statement II is a
correct explanation of Statement I.
48. The length of the perpendicular drawn from
the
point ( 3, − 1, 11)
x y −2 z − 3
is
=
=
2
3
4
A
∴
y–1 z–2
2
3
B (1, 6, 3)
(a) 66
(c) 33
P (3, _1,11)
line
to
the
line
[AIEEE 2011]
(b) 29
(d) 53
Exp. (d)
Let the coordinates of Q be (2 λ , 3λ + 2 , 4λ + 3)
which is any point on the straight line AB.
plane x − y + z = 5 measured along a straight
line x = y = z , is
[AIEEE 2011]
(b)10 3
(d) 3 10
(a) 3 5
(c) 5 3
Exp. (b)
Let Q be any point on the plane.
Then equation of PQ is
x−1 y+ 5 z− 9
=
= λ,
=
1
1
1
where P = (1, − 5, 9)
Q
∴ x = λ + 1, y = λ − 5, z = λ + 9 lies on the
plane
x− y+ z=5
⇒ λ + 1− λ + 5 + λ + 9 = 5
∴
λ = − 10
Hence, coordinate of Q is Q (− 9, − 15, − 1.)
∴
|PQ| = (10)2 + (10)2 + (10)2
= 10 3
295
Three Dimensional Geometry
50. A line AB in three-dimensional space makes
angles 45° and 120° with the positive X-axis
and the positive Y-axis, respectively. If AB
makes an acute angle θ with the positive
Z-axis, then θ is equal to
[AIEEE 2010]
(a) 30°
(b) 45°
(c) 60°
(d) 75°
Exp. (c)
We know that,
cos 2 45°+ cos 2 120°+ cos 2 θ = 1
1 1
1
⇒
+ + cos 2 θ = 1 ⇒ cos 2 θ =
2 4
4
1
cosθ = ±
⇒
2
∴
θ = 60° or 120°
51. Statement I The point A(3, 1, 6) is the
mirror image of the point B(1, 3, 4) in the
plane x − y + z = 5.
Statement II The plane x − y + z = 5 bisects
the line segment joining A(3, 1, 6) and
[AIEEE 2010]
B(1, 3, 4)
(a) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(b) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
The image of the point (3, 1, 6) with respect to
the plane x − y + z = 5 is
x − 3 y − 1 z − 6 − 2 (3 − 1 + 6 − 5)
=
=
=
1
1
1+ 1+ 1
−1
Q x − x1 = y − y1 = z − z1

a
b
c
− 2(ax1 + by1 + cz1 + d )

a2 + b 2 + c 2

x− 3 y−1 z− 6
=
= −2
=
1
1
−1
=
⇒
x −2 y −1 z +2
lies in the
=
=
3
2
−5
plane x + 3y − αz + β = 0. Then, (α , β ) is equal
to
[AIEEE 2009]
52. Let the line
(a) (6, − 17)
(c) ( 5, − 15)
x = 3−2 = 1
y = 1+ 2 = 3
and z = 6 − 2 = 4
which shows that Statement I is true.
We observe that the line segment joining the
points A(3, 1, 6) and B(1, 3, 4) has direction ratios
(b) ( − 6, 7)
(d) ( − 5,15)
Exp. (b)
DR’s of given line are (3, − 5, 2 ) .
DR’s of normal to the plane = (1, 3, − α )
∴ Line is perpendicular to the normal
3(1) − 5(3) + 2 (− α ) = 0
⇒
3 − 15 − 2 α = 0
⇒
2 α = − 12
⇒
α=−6
Also, point (2 , 1, − 2 ) lies on the plane.
∴ 2 + 3 + 6 (− 2 ) + β = 0 ⇒ β = 7
⇒
(α, β ) = (− 6, 7 )
53. The projections of a vector on the three
coordinate axes are6, − 3 and 2, respectively.
The direction cosines of the vector are
6
3 2
,− ,
5 5 5
6
3 2
(d) − , − ,
7 7 7
(a) 6, − 3, 2
(c)
Exp. (a)
⇒
2, – 2, 2 which are proportional to 1, − 1, 1, the
direction ratios of the normal to the plane.
Hence, Statement II is true. Thus, both statements
are true and Statement II is a correct explanation
of Statement I.
(b)
6
3 2
,− ,
7 7 7
[AIEEE 2009]
Exp. (c)
Projection of a vector on coordinate axes are
x2 − x1, y2 − y1, z2 − z1
⇒
x2 − x1 = 6, y2 − y1 = − 3, z2 − z1 = 2
Now, ( x2 − x1 )2 + ( y2 − y1 )2 + ( z2 − z1 )2
=
36 + 9 + 4 = 7
6
3
2
So, the DC’s of the vector are , − and ⋅
7
7
7
54. The line passing through the points (5, 1, a )
and ( 3,b , 1) crosses the yz-plane at the point
 17 13
0, , −  . Then,
 2
2
[AIEEE 2008]
(a) a = 8, b = 2
(c) a = 4, b = 6
(b) a = 2 , b = 8
(d) a = 6, b = 4
296
JEE Main Chapterwise Mathematics
Exp. (d)
56. Let L be the line of intersection of the planes
Equation of line passing through (5, 1, a) and
(3, b, 1) is
x− 3 y− b z−1
…(i)
=
=
5 − 3 1− b a − 1
 x − x1
y − y1
z − z1 
=
=
Q

 x2 − x1 y2 − y1 z2 − z1 
17
13
Point  0, , −  satisfies Eq. (i), we get
 2
2
13
17
−1
−b −
3
2
=
− = 2
a−1
1− b
2
 − 15 


 2
=5 ⇒ a=6
a − 1=
⇒
 − 3


 2
17
Also, − 3(1 − b ) = 2 
− b 
2

⇒
⇒
⇒
and
(b) – 5
(d) 2
…(i)
…(ii)
1 1 −2
k (− 2 k − 2 ) − 2 (− 6 − 2 ) + 3(3 − k ) = 0
− 2 k 2 − 5k + 25 = 0
⇒
2 k 2 + 5k − 25 = 0
⇒
2 k 2 + 10k − 5k − 25 = 0
⇒
(c) 1
(d)
1
2
and l + 3 m + 2 n = 0
…(ii)
From Eqs. (i) and (ii), we get
l
m n
=
= =k
3 −3 3
[say]
(3k )2 + (− 3k )2 + (3k )2 = 1
27 k 2 = 1
∴
2 k(k + 5) − 5(k + 5) = 0
5
k= ,−5
⇒
2
Hence, integer value of k is − 5.
l=
1
3
1
3 3
1
⇒ cos α =
3
⇒
k=
π
4
directions of each of X-axis and Y-axis, then
the angle that the line makes with the
positive direction of the Z-axis is [AIEEE 2007]
57. If a line makes an angle of with the positive
π
6
π
(c)
4
(a)
Since, lines intersect at a point. Then, shortest
distance between them is zero.
k 2 3
∴
3 k 2 =0
⇒
⇒
1
2
Exp. (a)
⇒
Exp. (b)
Given,
(b)
∴
x −1 y −2 z − 3
and
=
=
k
2
3
x−1 y−2 z− 3
=
=
k
2
3
x−2 y− 3 z−1
=
=
3
k
2
1
3
We know that, l 2 + m2 + n2 = 1
x −2 y − 3 z −1
intersect at a point, then
=
=
3
k
2
the integer k is equal to
[AIEEE 2008]
(a) – 2
(c) 5
(a)
Let the direction cosines of line L be l , m and n.
Since, the line intersect the given planes, then the
normal to the planes are perpendicular to the line
L.
…(i)
∴
2 l + 3m + n = 0
3 b − 3 = 17 − 2 b
5b = 20
b=4
55. If the straight lines
2 x + 3y + z = 1 and x + 3y + 2 z = 2 .If L makes
an angle α with the positive X-axis, then
[AIEEE 2007]
cos α is equal to
π
3
π
(d)
2
(b)
Exp. (d)
π
with positive
4
directions of each of X-axis and Y-axis, therefore
π
π
α= , β=
4
4
We know that,
cos 2 α + cos 2 β + cos 2 γ = 1
π
π
cos 2
+ cos 2
+ cos 2 γ = 1
∴
4
4
1 1
⇒
+ + cos 2 γ = 1
2 2
⇒
cos 2 γ = 0
π
γ=
∴
2
Since, a line makes an angle of
297
Three Dimensional Geometry
x − x1 y − y1 z − z1
=
=
a
b
c
− 2 (ax1 + by1 + cz1 + d )
=
a2 + b 2 + c 2
58. If (2, 3, 5) is one end of a diameter of the
sphere x 2 + y 2 + z 2 − 6x − 12 y − 2 z + 20 = 0,
then the coordinates of the other end of the
diameter are
[AIEEE 2007]
(a) (4, 9, –3)
(c) (4, 3, 5)
(b) (4, –3, 3)
(d) (4, 3, –3)
Exp. (a)
Given equation of sphere
x2 + y2 + z2 − 6 x − 12 y − 2 z + 20 = 0
whose coordinates of centre are (3, 6, 1).
Since, one end of diameter are (2, 3, 5) and let the
other end of diameter be (α, β, γ ), then
α+2
β+ 3
γ+ 5
= 3,
= 6,
=1
2
2
2
⇒
α = 4, β = 9 and γ = − 3
Hence, the coordinates of other points be
(4, 9, − 3).
59. The two lines x = ay + b , z = cy + d and
x = a ′ y + b ′ , z = c ′ y + d ′ are perpendicular
to each other, if
[AIEEE 2006, 2003]
(a) aa ′ + cc ′ =1
(c)
a
c
+
=1
a′ c′
(b)
a
c
+
= −1
a′ c′
(d) aa ′ + cc ′ = −1
− 2 [1 × (−1) + (− 2 ) × 3 + 0 × 4]
1+ 4
x + 1 y − 3 z − 4 − 2 ( −7 )
=
=
=
⇒
1
0
5
−2
9
14
⇒ x=
− 1= ,
5
5
13
28
and z = 4
y=−
+ 3=−
5
5
Hence, the image of point (–1, 3, 4) is
 9 , − 13 , 4 .


5
5 
=
61. If the plane 2 ax − 3ay + 4az + 6 = 0 passes
through the mid-point of the line joining the
centres of the spheres
x 2 + y 2 + z 2 + 6x − 8y − 2 z = 13 and
x 2 + y 2 + z 2 − 10x + 4y − 2 z = 8, then a is
equal to
[AIEEE 2005]
(b) − 2
(a) 2
Exp. (d)
(d) −1
(c) 1
Exp. (b)
Given equations can be rewritten as
x− b y− 0 z−d
=
=
a
1
c
x − b′ y − 0 z − d ′
and
=
=
a′
c′
1
Equation of given spheres are
x2 + y2 + z2 + 6 x − 8 y − 2 z = 13
x + y + z − 10 x + 4 y − 2 z = 8
2
and
These lines will be perpendicular, if
aa ′ + cc ′ + 1 = 0[Q l1l2 + m1m2 + n1n2 = 0]
60. The image of the point (–1, 3, 4) in the plane
x − 2 y = 0 is
Thus, the image of point (–1, 3, 4) in a plane
x − 2 y = 0 is given by
x+1 y−3 z−4
=
=
1
0
−2
[AIEEE 2006]
(a) (15, 11, 4)
17 19
(b)  − , − ,1
 3
3 
(c) (8, 4, 4)
9 13
(d)  , − , 4
5
5 
Exp. (d)
We know that, the image ( x, y, z) of a point
( x1, y1, z1 ) in a plane ax + by + cz + d = 0 is given
by
2
2
…(i)
…(ii)
whose centres are (– 3, 4, 1) and (5, – 2, 1).
Mid-point of (– 3, 4, 1) and (5, – 2, 1) is (1, 1, 1).
Since, the given plane 2 ax − 3ay + 4az + 6 = 0
passes through (1, 1, 1).
∴
2 a − 3a + 4a + 6 = 0
⇒
3a = − 6 ⇒ a = − 2
62. If the angle θ between the line
x +1 y −1 z −2
and the plane
=
=
1
2
2
1
2 x − y + λ z + 4 = 0 is such that sin θ = .
3
The value of λ is
[AIEEE 2005]
(a) −
4
3
(b)
3
4
(c) −
3
5
(d)
5
3
298
JEE Main Chapterwise Mathematics
Exp. (d)
and radius =
Direction ratios of line are
(a1, b1, c1 ) = (1, 2 , 2 )
And direction ratios of a plane are
(a2, b2 , c 2 ) = (2 , − 1, λ )
a1a2 + b1b2 + c1c 2
∴ sin θ =
2
a1 + b12 + c12 a22 + b22 + c 22
=
1 × 2 + 2 (−1) + 2 ×
12 + 2 2 + 2 2
63. The angle between the lines 2 x = 3y = − z
and 6x = − y = − 4z is
(a) 30°
(c) 90°
[AIEEE 2005]
(b) 45°
(d) 0°
Exp. (c)
The given equations of lines can be rewritten as
x y
z
= =
3 2 −6
x
y
z
and
=
=
2 −12 − 3
∴ Angle between the lines is
θ = cos −1


3 × 2 + 2 (−12 ) − 6(−3)


 32 + 2 2 + (−6)2 (2 )2 + (−12 )2 + (−3)2 




a1 ⋅ a2 + b2 ⋅ b2 + c1 ⋅ c 2
Qcosθ =

2
2
2
2
2
2 

a
+
b
+
c
a
+
b
+
c
1
1
1
2
2
2


= cos −1 (0) = 90°
64. The plane x + 2 y − z = 4 cuts the sphere
x 2 + y 2 + z 2 − x + z − 2 = 0 in a circle of radius
(a) 2
(c) 1
(b) 2
(d) 3
O
B
A
λ
2 2 + 12 + ( λ )2
1
2 λ
5
=
= 5 + λ = 4λ ⇒ λ =
3 3 5+ λ
3
⇒
1 1
10
+ +2=
4 4
2
[AIEEE 2005]
Distance from centre of sphere to the given plane
 1 + 1 − 4
 3

OA = 2 2
=


+
+
1
4
1
6




So, radius of circle,
10 9
AB = OB2 − OA 2 =
−
4
6
30 − 18
12
=
=
=1
12
12
65. A line makes the same angle θ with each of
the X and Z -axes. If the angle β, which it
makes with Y-axis, is such that
sin 2 β = 3 sin 2 θ , then cos 2 θ is equal to
(a) 2/3
(c) 3/5
(b) 1/5
(d) 2/5
Exp. (c)
A line makes angle θ with X-axis and Z-axis and β
with Y-axis.
∴
l = cos θ, m = cos β, n = cos θ
Q
l 2 + m2 + n2 = 1
∴
cos 2 θ + cos 2 β + cos 2 θ = 1
⇒
2 cos 2 θ = 1 − cos 2 β
⇒
2 cos 2 θ = sin2 β
But it is given that,
sin2 β = 3 sin2 θ
From Eqs. (i) and (ii), we get
3 sin2 θ = 2 cos 2 θ
⇒
3(1 − cos 2 θ) = 2 cos 2 θ
Since, the centre of sphere
⇒
1
1
x2 + y2 + z2 − x + z − 2 = 0 is  , 0, − 
2
2
∴
3 = 5cos 2 θ
3
cos 2 θ =
5
Exp. (c)
[AIEEE 2004]
…(i)
…(ii)
299
Three Dimensional Geometry
66. Distance between two parallel planes
2 x + y + 2 z = 8 and 4x + 2 y + 4z + 5 = 0 is
[AIEEE 2004]
3
(a)
2
5
(b)
2
7
(c)
2
(d)
9
2
Exp. (c)
We know that the distance between two parallel
planes ax + by + cz + d1 = 0
and ax + by + cz + d 2 = 0 is
d 2 − d1
On solving second and third fractions, we get
2k − 2λ − 2a = k − λ
⇒
k − λ = 2a
⇒
λ = 3 a − 2 a [from Eq. (i)]
∴
λ=a
Hence, coordinates of E are (3 a, 2 a, 3 a) and
coordinates of F are (a, a, a).
68. If the straight lines x = 1 + s , y = − 3 − λ s ,
a + b2 + c 2
2
Therefore, the distance between
4 x + 2 y + 4 z − 16 = 0 and 4 x + 2 y + 4 z + 5 = 0
is
 5 + 16   21  21 7
=
=

= 
 16 + 4 + 16  36 6 2
67. A line with direction cosines proportional to
2, 1, 2 meets each of the lines x = y + a = z
and x + a = 2 y = 2 z. The coordinates of each
of the points of intersection are given by
(a)
(b)
(c)
(d)
On solving first and second fractions, we get
k − 2λ + a = 2k − 2λ − 2a
…(i)
⇒
k = 3a
( 3a , 3a , 3a ),(a , a , a )
( 3a , 2a , 3a ),(a , a , a )
( 3a , 2a , 3a ),(a , a , 2a )
( 2a , 3a , 3a ),( 2a , a , a )
[AIEEE 2004]
t
, y = 1 + t , z = 2 − t , with
2
parameterss and t respectively are coplanar,
then λ is equal to
[AIEEE 2004]
z = 1 + λ s and x =
(a) – 2
The given straight line can be rewritten as
x−1 y+ 3 z−1
=
=
=s
−λ
1
λ
x− 0 y−1 z−2
and
=t
=
=
1
2
−2
l1
l2
[say]
m1
m2
F
B
D
1
1
−λ
2
=0
λ =0
−2
1 −4
1 −λ
1
2
−1
λ =0
−2
⇒ 1(2 λ − 2 λ ) + 4(−2 − λ ) − 1(2 + λ ) = 0
⇒
− 8 − 4λ − 2 − λ = 0
⇒
−10 = 5λ ⇒ λ = − 2
Any point on the line is F (k, k − a, k ).
Also, the equation of other line CD is
x+ a y−0 z−0
=
=
=λ
2
1
1
Any point on the line is E (2 λ − a, λ, λ ).
Direction ratios of EF are
[(k − 2 λ + a), (k − a − λ ), (k − λ )].
n1
n2
1− 0 − 3 − 1 1− 2
⇒
⇒
C
(d) 0
These two lines are coplanar, if
x1 − x2 y1 − y2 z1 − z2
Let the equation of line AB be
x−0 y+ a z=0
=
=
=k
1
1
1
E
1
2
Exp. (a)
Exp. (b)
A
(c) −
(b) –1
[say]
Since, it is given that direction ratios of EF are
proportional to 2, 1, 2.
k − 2λ + a k − λ − a k − λ
=
=
∴
2
1
2
69. The
intersection
of
the
x 2 + y 2 + z 2 + 7x − 2 y − z = 13
spheres
and
x 2 + y 2 + z 2 − 3x + 3y + 4z = 8 is the same as
the intersection of one of the sphere and the
plane
[AIEEE 2004]
(a) x − y − z =1
(c) x − y − 2 z = 1
(b) x − 2 y − z = 1
(d) 2 x − y − z = 1
300
JEE Main Chapterwise Mathematics
Exp. (d)
Given equation of two spheres are
S ≡ x2 + y2 + z2 + 7 x − 2 y − z − 13 = 0
and S ′ ≡ x2 + y2 + z2 − 3 x + 3 y + 4 z − 8 = 0
(a) k = 0 or –1
(c) k = 0 or – 3
(b) k =1 or –1
(d) k = 3 or – 3
[AIEEE 2003]
Exp. (c)
If these spheres intersect, then S − S ′ = 0
represents the equation of common plane of
intersection.
∴ ( x2 + y2 + z2 + 7 x − 2 y − z − 13)
− ( x2 + y2 + z2 − 3 x + 3 y + 4 z − 8) = 0
Given lines
x−2 y− 3 z− 4
=
=
1
1
−k
x−1 y− 4 z− 5
and
=
=
k
2
1
⇒
⇒
are coplanar.
10 x − 5 y − 5 z − 5 = 0
2x− y− z=1
∴
1
k
70. The radius of the circle in which the sphere
x 2 + y 2 + z 2 + 2 x − 2 y − 4z − 19 = 0 is cut by
the plane x + 2 y + 2 z + 7 = 0, is [AIEEE 2003]
(a) 1
(b) 2
(c) 3
−1 1
…(ii)
1
1 −k = 0
2 1
 x2 − x1
Q
l1

l2

(d) 4
…(i)
y2 − y1
m1
m2
z2 − z1

n1
= 0

n2

⇒ −11
( + 2 k ) − 11
( + k 2 ) + 12
( − k) = 0
Exp. (c)
Given equation of sphere is
x2 + y2 + z2 + 2 x − 2 y − 4 z − 19 = 0
⇒
−2 k − 1 − 1 − k 2 + 2 − k = 0
⇒
− k 2 − 3k = 0
whose centre (–1, 1, 2)
⇒
k = 0 or − 3
and
radius = (−1)2 + (1)2 + (2 )2 + 19
= 25 = 5
72. The shortest distance from the plane
12 x + 4y + 3z = 327 to the sphere
x 2 + y 2 + z 2 + 4x − 2 y − 6z = 155 is
(a) 26
O
B
N
(c) 13
A
Also, equation of plane is
x+2y+2z+7 =0
Length of the perpendicular from centre O on the
plane is
−1 × 1 + 1 × 2 + 2 × 2 + 7 12
ON =
=
=4
3
12 + 2 2 + 2 2
In ∆OAN,
52 = 42 + AN2 [by Pythagoras theorem]
⇒
∴
AN2 = 25 − 16 ⇒ AN2 = 9
AN = 3
x −2 y − 3 z − 4
and
=
=
1
1
−k
x −1 y − 4 z −5
are coplanar, if
=
=
k
2
1
71. The lines
4
13
(d) 39
(b) 11
[AIEEE 2003]
Exp. (c)
Given equations of sphere and plane are
x2 + y2 + z2 + 4 x − 2 y − 6 z − 155 = 0
and
12 x + 4 y + 3 z − 327 = 0
Centre of the sphere is (– 2, 1, 3) and radius
= 4 + 1 + 9 + 155 = 169 = 13
Length of the perpendicular from centre to the
plane
− 2 × 12 + 1 × 4 + 3 × 3 − 327
=



(12 )2 + (4)2 + 32




ax + by1 + cz1 + d

Q P = 1
2
2
2


a + b +c


= 338 = 26
−11 − 327
=
169  13

301
Three Dimensional Geometry
∴Shortest distance between the plane and sphere
= 26 − Radius of sphere
= 26 − 13 = 13
Let a, b and c be the lengths of edges, then
a = 5 − 2 = 3, b = 9 − 3 = 6 and c = 7 − 5 = 2
So, the length of diagonal of a parallelopiped
73. Two systems of rectangular axes have the
same origin. If a plane cuts them at
distances a ,b , c and a ′ ,b ′ , c ′ from the origin,
then
[AIEEE 2003]
(a)
(b)
(c)
(d)
1
a2
1
a2
1
2
a
1
a2
1
+
b2
1
+
−
+
b2
1
2
b
1
1
+
c2
1
−
−
b2
c2
1
2
c
+
+
+
+
1
+
a ′2
1
a ′2
1
a′
1
2
−
c2
1
+
−
a ′2
1
b ′2
1
b ′2
1
b′
1
2
−
+
−
−
=0
c ′2
1
c ′2
1
c′
1
2
−
b ′2
1
=0
1
a
a2
2
+
+
1
b2
1
b
2
+
+
1
c2
1
c
2
−
=
1
a′
1
a ′2
2
−
+
1
b′
1
b ′2
2
−
+
1
c ′2
1
c′
2
=0
74. A parallelopiped is formed by planes drawn
through the points (2, 3, 5) and (5, 9, 7),
parallel to the coordinate planes. The length
of a diagonal of the parallelopiped is
[AIEEE 2002]
(a) 7 units
(c)
x − x1 y − y1 z − z1
is
=
=
l
m
n
[AIEEE 2002]
Exp. (b)
Length of perpendicular from origin to Eqs. (i)
and (ii) must be same.
1
1
=
∴
1
1
1
1
1
1
+ 2 + 2
+ 2 + 2
a2
b
c
a ′2
b′
c′
⇒
49 = 7 units
(b) al + bm + cn = 0
a b c
(c) = =
l m n
(d) lx1 + my1 + nz1 = 0
Consider OX, OY , OZ and Ox, Oy, Oz are two
systems of rectangular axes.
Let their corresponding equations of plane be
x
y
z
…(i)
+ + =1
a b c
x
y
z
and
…(ii)
+
+
=1
a′ b ′ c ′
1
9 + 36 + 4
=
(a) ax1 + by1 + cz1 = 0
Exp. (d)
⇒
=
a ( x − x1 ) + b ( y − y1 ) + c ( z − z1 ) = 0, where
=0
c ′2
a2 + b 2 + c 2
75. The equation of the plane containing the
line
=0
=
155 units
(b)
38 units
(d) None of these
Exp. (a)
A parallelopiped is formed by planes drawn
through the points (2, 3, 5) and (5, 9, 7), parallel to
the coordinate planes.
The equation of plane containing the line
x − x1 y − y1 z − z1
is
=
=
l
m
n
a( x − x1 ) + b ( y − y1 ) + c( z − z1 ) = 0
Since, the normal to the plane is perpendicular to
the given line.
∴
al + bm + cn = 0
76. The centre of the circle given by
r ⋅ ( i + 2 j + 2 k ) = 15 and | r − ( j + 2 k )| = 4, is
(a) (0, 1, 2)
[AIEEE 2002]
(b) (1, 3, 4)
(c) (–1, 3, 4)
(d) None of the above
Exp. (b)
The equation of a line through the centre j + 2k
and normal to the given plane is
…(i)
r = j + 2 k + λ(i + 2 j + 2 k )
This meets the plane at a point for which we
must have
[(j + 2 k ) + λ(i + 2 j + 2 k )] ⋅ (i + 2 j + 2 k ) = 15
⇒
2 (1 + 2 λ ) + λ + 2 (2 + 2 λ ) = 15
⇒
6 + 9 λ = 15
∴
λ =1
On putting λ = 1in Eq. (i), we get
r = i + 3j + 4k
Hence, centre of circle is (1, 3, 4).
13
Vector Algebra
1. Let a = 3$i + 2 $j + xk$ and b = $i − $j + k$ , for some
real x. Then |a × b| = r is possible if
[JEE Main 2019, 8 April Shift-II]
3
(a) 0 < r ≤
2
3
3
(c) 3
<r < 5
2
2
3
3
<r ≤ 3
2
2
3
(d) r ≥ 5
2
(b)
Exp. (d)
r
r
r
r
r
r
where β1 is parallel to α and β 2 is
r r
r
perpendicular to α, then β1 × β 2 is equal to
[JEE Main 2019, 9 April Shift-I]
1
$)
(a) ( 3i$ − 9$j + 5k
2
$
(c) −3$i + 9$j + 5k
1
$)
( −3i$ + 9$j + 5k
2
$)
(d) 3i$ − 9$j − 5k
(b)
Exp. (b)
$
Given vectors are a = 3$i + 2 $j + xk
$
$
$
and
b= i − j+ k
$
 $i $j k


∴
a × b= 3 2 x


1 −1 1


$
$ ( −3 − 2 )
= i(2 + x) − $j(3 − x) + k
$
= ( x + 2 )$i + ( x − 3)$j − 5k
⇒ |a × b| = ( x + 2 )2 + ( x − 3)2 + 25
= 2 x − 2 x + 4 + 9 + 25
2
1
1
= 2  x2 − x +  − + 38

4 2
2
1
75
= 2  x −  +

2
2
So,|a × b| ≥
⇒
r
r
$ If β = β − β ,
2. Let α = 3i$ + $j and β = 2 i$ − $j + 3k.
1
2
r≥ 5
75
1
[at x = ,|a × b| is minimum]
2
2
3
2
→
→
$
Given vectors α = 3$i + $j and β = 2 $i − $j + 3k
→
→
→
and β = β1 − β 2
→
→
→
such that β1 is parallel to α and β 2 is
perpendicular to α
→
So,
β = λα = λ(3i$ + $j)
1
→
→
→
$)
Now, β 2 = β1 − β = λ(3$i + $j) − (2 $i − $j + 3k
$
= (3λ − 2 )$i + (λ + 1)$j – 3k
→
→
Q β 2 is perpendicular to α, so β 2 ⋅ α = 0
[since if non-zero vectors a and b are
perpendicular to each other, then a ⋅ b = 0]
∴ (3λ − 2 )(3) + (λ + 1)(1) = 0
⇒
9λ − 6 + λ + 1 = 0
1
⇒
10λ = 5 ⇒ λ =
2
→
3
1
So, β1 = $i + $j
2
2
→
3
1
$
and β 2 =  − 2  i$ +  + 1 $j − 3k
2

2

1
3
$
= − $i + $j − 3k
2
2
303
Vector Algebra
$j k
$i
$
→
3 1
0
∴ β1× β 2 =
2 2
−1 3 −3
2 2
9
3

$
$  9 + 1 
= i  − − 0 − $j − − 0 + k
 2

 2
 4 4

3$ 9$ 5 $
=− i+ j+ k
2
2
2
1
$)
= (−3i$ + 9$j + 5k
2
→
π
3
3. If a unit vector a makes angles with $i,
π
$ then a value
with $j and θ ∈(0, π ) with k,
4
of θ is
[JEE Main 2019, 9 April Shift-II]
(a)
5π
6
(b)
π
4
(c)
5π
12
(d)
2π
3
Exp. (d)
Given unit vector a makes an angle
$
with $j and θ ∈(0, π ) with k.
π
π
with $i,
3
4
Now, we know that cos 2 α + cos 2 β + cos 2 γ = 1,
where α, β, γ are angles made by the vectors
$
with respectively $i, $j and k.
π
π
∴ cos 2   + cos 2   + cos 2 θ = 1
 4
 3
1 1
⇒
+ + cos 2 θ = 1
4 2
1
cos 2 θ =
⇒
4
1
cosθ = ±
⇒
2
π
2π
cos θ = cos   or cos 
⇒

 3
 3 
π
2π
or
θ=
⇒
3
3
2π
, according to options.
So, θ is
3
Exp. (a)
Key Idea Use the angle between two non-zero
a⋅b
and
vectors a and b is given by cos θ =
|a|| b|
coordinates of the centroid i.e.
 x1 + x2 + x3 , y1 + y 2 + y 3 , z1 + z2 + z3  of a




3
3
3
triangle formed with vertices; (x1, y1, z1 ),
(x2 , y 2 , z2 ) and (x3 , y 3 , z3 ).
Given vertices of a ∆ABC are A(3, 0, − 1),
B(2, 10, 6) and C(1, 2, 1) and a point M is mid-point
of AC. An another point G divides BM in ratio
2 : 1, so G is the centroid of ∆ABC.
3 + 2 + 1 0 + 10 + 2 −1 + 6 + 1
∴ G 
,
,
 = (2, 4, 2 ).


3
3
3
OG ⋅ OA
, where O is the
Now, cos(∠GOA) =
OG OA
origin.
Q
⇒
and
⇒
and
∴
$
OG =`2 $i + 4$j + 2 k
OG = 4 + 16 + 4 = 24
$
OA = 3$i − k
OA = 9 + 1 = 10
OG ⋅ OA = 6 − 2 = 4
4
1
cos(∠GOA) =
=
24 10
15
5. The distance of the point having position
$ from the straight line
vector − $i + 2 $j + 6k
passing through the point (2 , 3, − 4) and
$ is
parallel to the vector, 6i$ + 3$j − 4k
[JEE Main 2019, 10 April Shift-II]
(a) 2 13
(b) 4 3
[JEE Main 2019, 10 April Shift-I]
(a)
1
15
(b)
1
2 15
(c)
1
30
(d)
1
6 10
(d) 7
Exp. (d)
Let point P whose position vector is
$ ) and a straight line passing through
(− i$ + 2 $j + 6k
$.
Q(2, 3, − 4) parallel to the vector n = 6$i + 3$j − 4k
P(–1,2,6)
4. Let A( 3, 0, −1), B (2 , 10, 6) and C (1, 2 , 1) be the
vertices of a triangle and M be the mid-point
of AC . If G divides BM in the ratio 2 : 1, then
cos ( ∠GOA ) (O being the origin) is equal to
(c) 6
d
Q(2,3,–4)
n=6^
i+3^
j–4^
k
Q Required distance d = Projection of line
segment PQ perpendicular to vector n.
304
JEE Main Chapterwise Mathematics
|PQ × n |
| n|
$
$ , so
Now, PQ = 3i + $j − 10k
$i $j
$
k
Exp. (b)
=
PQ × n = 3 1 − 10
6 3
−4
$
= 26i$ − 48$j + 3k
d=
So,
(26)2 + (48)2 + (3)2
(6)2 + (3)2 + (4)2
676 + 2304 + 9
2989
=
36 + 9 + 16
61
=
=
49 = 7 units
6. Let a = 3$i + 2 $j + 2 k$ and b = $i + 2 $j − 2 k$ be two
vectors. If a vector perpendicular to both the
vectors a + b and a − b has the magnitude 12,
then one such vector is
[JEE Main 2019, 12 April Shift-I]
$)
(a) 4( 2 $i + 2 $j + k
$
$
$
(c) 4( 2 i + 2 j − k)
$)
(b) 4( 2 i$ − 2 $j − k
$
$
$)
(d) 4( − 2 i − 2 j + k
Exp. (b)
Given vectors are
$ and b = $i + 2 $j − 2 k
$
a = 3i$ + 2 $j + 2 k
$
$
Now, vectors a + b = 4i + 4 j
$
and
a − b = 2 i$ + 4k
∴ A vector which is perpendicular to both the
vectors a + b and a − b is
$i $j k
$
Key Idea Volume of parallelopiped formed by
the vectors a, b and c is V = [ a b c].
$ and λ$i + k
$,
$ , $j + λk
Given vectors are $i + λ$j + k
which forms a parallelopiped.
∴Volume of the parallelopiped is
1 λ 1
V = 0 1 λ = 1 + λ3 − λ
λ 0 1
⇒
V = λ3 − λ + 1
On differentiating w.r.t. λ, we get
dV
= 3 λ2 − 1
dλ
dV
For maxima or minima,
=0
dλ
1
λ=±
⇒
3
2 3 > 0 , for λ = 1

d 2V
3
and
= 6λ = 
1
dλ2
2 3 < 0 , for λ = −
3

d 2V
1
, so volume ‘V’ is
Q 2 is positive for λ =
3
dλ
1
minimum for λ =
3
8. Let α ∈R and the three vectors
$ , b = 2 i$ + $j − αk
$
a = α$i + $j + 3k
$
and c = α$i − 2 $j + 3k. Then, the set
S = {α : a, b and c are coplanar}
(a + b) × (a − b) = 4 4 0
[JEE Main 2019, 12 April Shift-II]
(a)is singleton
(b) is empty
(c) contains exactly two positive numbers
2 0 4
$ (− 8)
= i$(16) − $j(16) + k
$
$
$
= 8(2 i − 2 j − k)
Then, the required vector along (a + b) × (a − b)
having magnitude 12 is
$)
8(2 i$ − 2 $j − k
$)
= ± 4(2 i$ − 2 $j − k
± 12 ×
8× 4+ 4+1
7. If the volume of parallelopiped formed by the
$ , $j + λk
$ and
vectors $i + λ$j + k
minimum, then λ is equal to
$ is
λ$i + k
[JEE Main 2019, 12 April Shift-I]
(a) −
(c)
1
3
3
1
3
(d) − 3
(b)
(d) contains exactly two numbers only one of
which is positive
Exp. (b)
Given three vectors are
$
a = α$i + $j + 3k
$
$
$
b = 2 i + j − αk
$
and
c = αi$ − 2 $j + 3k
3
α 1
Clearly, [a b c] = 2 1 −α
α −2 3
= α(3 − 2α ) − 1(6 + α 2 ) + 3 (− 4 − α )
= − 3 α 2 − 18 = − 3 (α 2 + 6)
305
Vector Algebra
Q There is no value of α for which − 3(α 2 + 6)
α 1
3
becomes zero, so = 2 1 −α [a b c] ≠ 0
α −2 3
⇒ vectors a, b and c are not coplanar for any
value α ∈ R.
So, the set S = {α : a, b and c are coplanar} is
empty set.
9. Let a = i$ − $j, b = $i + $j + k$ and c be a vector
such that a × c + b = 0 and a ⋅ c = 4, then | c|2 is
equal to
[JEE Main 2019, 9 Jan Shift-I]
19
(b)
2
(a) 8
17
(d)
2
(c) 9
Exp. (b)
We have,
(a × c ) + b = 0
⇒
a × (a × c ) + a × b = 0
(taking cross product with a on both sides)
$i $j k$
⇒(a ⋅ c )a − (a ⋅ a )c + 1 −1 0 = 0
1
1
1
(Qa × (b × c ) = (a . c )b − (a ⋅ b)c)
$
$
⇒ 4( i − j ) − 2c + (− $i − $j + 2k$ ) = 0
[Qa ⋅ a = ($i − $j )($i − $j ) = 1 + 1 = 2 and a ⋅ c = 4]
⇒
⇒
⇒
2c = 4$i − 4$j − $i − $j + 2k$
3$i − 5$j + 2k$
c=
2
9 + 25 + 4 19
2
|c| =
=
4
2
=
12 + 12 + ( 2 )2
b + b2 + 2 b1 + b2 + 2
= 1
=
4
2
But projection of b on a =| a|
b + b2 + 2
= 12 + 12 + ( 2 )2
∴ 1
2
b1 + b2 + 2
⇒
= 2 ⇒ b1 + b2 = 2
2
Now, a + b = ($i + $j + 2 k$ ) + (b1$i + b2 $j +
= (b1 + 1)$i + (b2 + 1)$j + 2 2 k$
[JEE Main 2019, 9 Jan Shift-II]
(b) 4
(d) 32
⇒ | b | = (− 3)2 + (5)2 + ( 2 )2 =
a+b
a
...(ii)
36 = 6
11. Let a = 2 i$ + λ 1$j + 3k$ , b = 4i$ + ( 3 − λ 2 )$j + 6k$
$ be three vectors
and c = 3$i + 6$j + ( λ 3 − 1)k
such that b = 2a and a is perpendicular to c.
Then a possible value of ( λ 1 , λ 2 , λ 3 ) is
(b) (1, 5, 1)
1
(d)  , 4, − 2 
2

Exp. (c)
$;
We have, a = 2 i$ + λ1$j + 3k
$
$
$
b = 4i + (3 − λ 2 )j + 6k
$
$
$,
and
c = 3i + 6 j + (λ 3 − 1)k
such that b = 2 a
Now,
b = 2a
$ = 2 (2 $i + λ $j + 3k
$)
⇒ 4$i + (3 − λ 2 )$j + 6k
1
$
$
$
$
$
$
⇒ 4i + (3 − λ )j + 6k = 4i + 2 λ j + 6k
2
According to given information, we have the
following figure.
θ
2 k$ )
Q (a + b ) ⊥ c, therefore (a + b ) ⋅ c = 0
$
⇒ { (b1 + 1)i$ + (b2 + 1)$j + 2 2 k}
(5i$ + $j + 2 k$ ) = 0
⇒ 5(b1 + 1) + 1(b2 + 1) + 2 2 ( 2 ) = 0
⇒
5b1 + b2 = − 10
From Eqs. (i) and (ii), b1 = − 3 and b2 = 5
⇒
b = − 3i$ + 5$j + 2 k$
⇒
Exp. (a)
b
...(i)
[JEE Main 2019, 10 Jan Shift-I]
$ be three vectors such that
c = 5 $i + $j + 2 k
the projection vector of b on a is a. If a + b is
perpendicular to c, then | b | is equal to
c
(b1$i + b2 $j +
(a) (1, 3, 1)
1
(c)  − , 4, 0
 2

10. Let a = i$ + $j + 2 k$ , b = b1 i$ + b 2 $j + 2 k$ and
(a) 6
(c) 22
b⋅ a
|a|
2k$ ) ($i + $j + 2k$ )
Clearly, projection of b on a =
1
(3 − 2 λ1 − λ 2 )$j = 0
⇒
3 − 2 λ1 − λ 2 = 0
...(i)
⇒
2 λ1 + λ 2 = 3
Also, as a is perpendicular to c, therefore a . c = 0
$ ) ⋅ (3$i + 6$j + (λ − 1)k
$)= 0
⇒ (2 $i + λ1$j + 3k
3
⇒
⇒
6 + 6λ1 + 3(λ 3 − 1) = 0
6λ1 + 3λ 3 + 3 = 0
306
JEE Main Chapterwise Mathematics
... (ii)
⇒
2 λ1 + λ 3 = − 1
Now, from Eq. (i), λ 2 = 3 − 2 λ1 and from Eq. (ii)
λ 3 = − 2 λ1 − 1
∴ (λ1, λ 2 , λ 3 ) ≡ (λ1, 3 − 2 λ1, − 2 λ1 − 1)
1
If λ1 = − , then λ 2 = 4, and λ 3 = 0
2
1
Thus, a possible value of (λ1, λ 2 , λ 3 ) =  − , 4, 0
 2

12. Let α = ( λ − 2 ) a + b and β = ( 4λ − 2 ) a + 3b be
two given vectors where vectors a and b are
non-collinear. The value of λ for which
vectors α and β are collinear, is
⇒ 1 {λ(λ2 − 1) − 16} − 2((λ2 − 1) − 8) + 4
(4 − 2 λ ) = 0
⇒ λ3 − λ − 16 − 2 λ2 + 18 + 16 − 8λ = 0
⇒
λ3 − 2 λ2 − 9λ + 18 = 0
⇒
λ2 (λ − 2 ) − 9 (λ − 2 ) = 0
⇒
(λ − 2 )(λ2 − 9) = 0
⇒
(λ − 2 ) (λ + 3) (λ − 3) = 0
∴
λ = 2, 3 or − 3
If λ = 2 , then
$i $j k$
a×c= 1 2
2 4 3
[JEE Main 2019, 10 Jan Shift-II]
(b) −3
(a) 4
(c) 3
(d) −4
= i$(6 − 16) −
= − 10$i + 5$j
$i
Exp. (d)
Two vectors c and d are said to be collinear if we
can write c = λb for some non-zero scalar λ.
Let the vectors α = (λ − 2 ) a + b
and
β = (4λ − 2 )a + 3 b are
collinear, where a and b are non-collinear.
∴ We can write
α = kβ, for some k ∈ R − {0}
⇒
(λ − 2 )a + b = k[(4λ − 2 )a + 3 b]
⇒ [(λ − 2 ) − k (4λ − 2 )]a+(1 − 3k )b = 0
Now, as a and b are non-collinear, therefore they
are linearly independent and hence
(λ − 2 ) − k (4λ − 2 ) = 0 and 1 − 3k=0
⇒
λ − 2 = k(4λ − 2 ) and 3k = 1
1
Q 3k = 1 ⇒ k = 1 
⇒
λ − 2 = (4λ − 2 )

3
3 
⇒
13.
3λ − 6 = 4λ − 2 ⇒ λ = −4
$ , b = $i + λ$j + 4k
$ and
Let a = i$ + 2 $j + 4k
$ be coplanar vectors.
c = 2 $i + 4$j + ( λ 2 − 1) k
$j(3 − 8) + k$ (4 − 4)
$j k$
If λ = ± 3, then a × c = 1 2 4 = 0
2 4 8
(because last two rows are proportional).
14. Let 3 $i + $j, $i + 3$j and β$i + (1 − β )$j
respectively be the position vectors of the
points A , B andC with respect to the origin
O. If the distance ofC from the bisector of
3
the acute angle betweenOA andOB is
,
2
then the sum of all possible values of β is
[JEE Main 2019, 11 Jan Shift-II]
(a) 1
(b) 3
(c) 4
(d) 2
Exp. (a)
According to given information, we have the
following figure.
A(Ö3, 1)
+1
Ö3
+1 , 2
3
Ö
2
M
Then, the non-zero vector a × c is
[JEE Main 2019, 11 Jan Shift-I]
(a) − 10 i$ + 5$j
(c) − 14 $i − 5$j
4
(b) − 10 i$ − 5$j
(d) − 14 $i + 5$j
D
2
C (b, (1–b))
Exp. (a)
We know that, if a, b, c are coplanar vectors, then
[a b c] = 0
1 2
4
=0
1 λ
4
∴
2
4 λ −1
2
O
2
B(1, Ö3)
Clearly, angle bisector divides the sides AB in
OA : OB,
i.e., 2 : 2 = 1 : 1 [using angle bisector theorem]
307
Vector Algebra
So, D is the mid-point of AB and hence
 3 + 1 3 + 1
coordinates of D are 
,

2 
 2
Now, equation of bisector OD is
 3+1

− 0

 ( x − 0) ⇒ y = x
( y − 0) =  2
 3+1

0
−


 2

⇒
x− y = 0
According to the problem,
β − (1 − β )
3
= CM =
2
2
[Distance of a point P( x1, y1 ) from the line
ax + by1 + c 
ax + by + c = 0 is 1

a2 + b 2 
⇒
|2 β − 1| = 3 ⇒ 2 β = ± 3 + 1
⇒
2 β = 4, − 2 ⇒ β = 2, − 1
Sum of 2 and −1is 1.
15. The sum of the distinct real values of µ, for
$ i$ + µ$j + k
$,
which the vectors, µi$ + $j + k,
$ are coplanar, is
$i + $j + µk
[JEE Main 2019, 12 Jan Shift-I]
(a) 2
(b) 0
(c) 1
(d) −1
Exp. (d)
Given vectors, µ$i + $j + k$ , $i + µ$j + k$ , $i + $j + µk$
will be coplanar, if
µ 1 1
1 µ 1 =0
1 1 µ
µ (µ 2 − 1) − 1(µ − 1) + 1(1 − µ ) = 0
⇒
⇒
(µ − 1) [µ (µ + 1) − 1 − 1] = 0
(µ − 1) [µ 2 + µ − 2 ] = 0
⇒
(µ − 1) [(µ + 2 ) (µ − 1)] = 0
⇒
µ = 1or − 2
So, sum of the distinct real values of
µ = 1 − 2 = − 1.
16. Let a, b and c be three unit vectors, out of
which vectors b and c are non-parallel. If α
and β are the angles which vector a makes
with vectors b and c respectively and
1
a × ( b × c) = b, then | α − β | is equal to
2
[JEE Main 2019, 12 Jan Shift-II]
(b) 45º
(c) 90º
1
b
2
1
(a ⋅ c)b − (a ⋅ b)c = b
⇒
2
[Qa × ( b × c) = (a ⋅ c)b − (a ⋅ b)c]
On comparing both sides, we get
1
…(i)
a⋅ c =
2
and
…(ii)
a⋅ b = 0
Qa, b and c are unit vectors, and angle between a
and b is α and angle between a and c is β, so
1
[from Eq. (i)]
|a||c| cos β =
2
1
[Q|a|= 1 =|c|]
cosβ =
⇒
2
π
π 1
…(iii) Qcos = 
⇒
β=

3 2 
3
and|a|| b| cos α = 0
[from Eq. (ii)]
π
…(iv)
⇒
α=
2
From Eqs. (iii) and (iv), we get
π π π
|α − β| = − = = 30º
2
3 6
Given,
a × ( b × c) =
17. Let u be a vector coplanar with the vectors
$
If u is
a = 2 $i + 3$j − k$ and b = $j + k.
perpendicular to a and u ⋅ b = 24, then | u|2 is
equal to
[JEE Main 2018]
(a) 336
(b) 315
(c) 256
(d) 84
Exp. (a)
⇒
(a) 30º
Exp. (a)
(d) 60º
Key idea If any vector x is coplanar with the
vector y and z, then x = λy + µz
Here, u is coplanar with a and b.
∴
u = λa + µb
Dot product with a, we get
u ⋅ a = λ(a ⋅ a ) + µ (b ⋅ a ) ⇒ 0 = 14λ + 2µ…(i)
[Q a = 2^i + 3^j − ^
k, b = ^j + ^
k, u ⋅ a = 0]
Dot product with b, we get
u ⋅ b = λ(a ⋅ b ) + µ (b ⋅ b )
…(ii) [Qu ⋅ b = 24]
24 = 2 λ + 2µ
Solving Eqs. (i) and (ii), we get
λ = − 2, µ = 14
Dot product with u, we get
|u |2 = λ(u ⋅ a ) + µ (u ⋅ b )
|u |2 = − 2(0) + 14(24) ⇒ |u |2 = 336
308
JEE Main Chapterwise Mathematics
$ b = $i + $j and c be a vector
18. Let a = 2 $i + $j − 2 k,
⇒
such that | c − a | = 3, |( a × b ) × c | = 3 and the
angle between c and a × b is 30°. Then, a ⋅ c is
equal to
[JEE Main 2017 (offline)]
⇒
(a)
25
8
(b) 2
(c) 5
(d)
1
8
( c − a) ⋅ ( c − a) = 9
|c| + |a|2 − 2 c ⋅ a = 9
⇒ |a × b|| c| sin 30° = 3 ⇒|c| =
$i
$j
…(i)
6
|a × b|
|c| =
6
=2
4+ 4+1
…(ii)
c⋅ a = 2
19. Let a, b and c be three unit vectors such that
a × (b × c ) =
3
( b + c ). If b is not parallel to c,
2
then the angle between a and b is
3π
(a)
4
π
(b)
2
[JEE Main 2016 (offline)]
(c)
2π
3
(d)
5π
6
Exp. (d)
Given,
|a$| = |b$| = |c$| = 1
3 $ $
(b+ c )
2
3 $ $
Now, consider
(b+ c )
a$ × ( b$ × c$ ) =
2
3
3
(a$ ⋅ c$ )b$ − (a$ ⋅ b$)c$ = b$ + c$
⇒
2
2
On comparing, we get
and
a$ × ( b$ × c$ ) =
3
3
a$ ⋅ b$ = −
⇒|a$||b$| cos θ = −
2
2
− 2
3
−2 3
(d)
3
2 2
(a)
3
2
(c)
3
(b)
1
(a × b) × c = |b||c|a
3
1
− c × (a × b) = |b||c|a
⇒
3
1
⇒ − (c ⋅ b) ⋅ a + (c ⋅ a )b = |b||c|a
3
 1|b||c| + (c ⋅ b) a = (c ⋅ a )b
 3

Given,
From Eqs. (i) and (ii), we get
(2 )2 + (3)2 − 2 c ⋅ a = 9 ⇒ 4 + 9 − 2 c ⋅ a = 9
⇒
⇒
Exp. (a)
$
k
$
But a × b = 2 1 − 2 = 2 $i − 2 $j + k
1 1 0
∴
π
cos θ = cos  π − 

6
5π
θ=
6
[JEE Main 2015 (offline)]
|(a × b ) × c| = 3
Again,
[Q|a$| = |b$| = 1]
that no two of them are collinear and
1
( a × b) × c = | b|| c| a. If θ is the angle between
3
vectors band c, then a value of sinθ is
$
We have, a = 2 $i + $j − 2 k
⇒
| a| = 4 + 1 + 4 = 3
and
b = $i + $j ⇒|b| = 1 + 1 = 2
Now, |c − a| = 3 ⇒|c − a|2 = 9
2
3
2
20. Let a, band c be three non-zero vectors such
Exp. (b)
⇒
⇒
cos θ = −
Since, a and b are not collinear.
1
c ⋅ b + |b||c| = 0 and c ⋅ a = 0
3
1
⇒ |c||b|cos θ + |b||c| = 0
3
1

|b||c| cos θ +  = 0
⇒

3
1
⇒
(Q|b| ≠ 0,|c| ≠ 0)
cosθ + = 0
3
1
⇒
cosθ = −
3
8 2 2
sinθ=
=
⇒
3
3
21. If[a × b b × cc × a] = λ [a b c]2, then λ is equal
to
[JEE Main 2014]
(a) 0
(b) 1
(c) 2
Exp. (b)
/ Use the formulae
a × (b × c) = (a ⋅ c)b − (a ⋅ b) c,
[a b c ] = [b c a ] = [c a b]
(d) 3
309
Vector Algebra
and [a a b] = [a b b] = [ a c c ] = 0
23. Let a and b be two unit vectors. If the vectors
c = a + 2 b and d = 5a − 4a are perpendicular
to each other, then the angle between a and
[AIEEE 2012]
b is
Further simplify it and get the result.
Now, [a × b b × c c × a ]
= a × b ⋅ ((b × c ) × (c × a ))
= a × b ⋅ ((k × c × a ))
[here, k = b × c]
π
6
(a)
= a × b ⋅ [(k ⋅ a ) c − (k ⋅ c ) a ]
= (a × b) ⋅ ((b × c ⋅ a ) c − (b × c ⋅ c )a )
= (a × b) ⋅ ([b c a ] c ) − 0
[Q[b × c ⋅ c ] = 0]
= a × b ⋅ c [b c a ]
= [a b c ] [b c a ]
= [a b c ] 2
{Q[a b c ] = [b c a ]}
Hence, [a × b b × c c × a ] = λ [a b c] 2
⇒
[a b c ] 2 = λ [a b c ] 2
⇒
λ =1
(b)
π
2
(c)
π
3
(d)
π
4
Exp. (c)
Given that,
(i) a and b are unit vectors,
i.e.,| a | = | b| = 1
(ii) c = a + 2b and d = 5a − 4b
(iii) c and d are perpendicular to each other.
i.e., c⋅ d = 0
To find Angle between a and b.
Now,
c ⋅ d = 0 ⇒ (a + 2 b) ⋅ (5 a − 4 b) = 0
the
vectors
and
AB = 3i + 4k
$
$
$
AC = 5i − 2 j + 4k are the sides of a ∆ABC ,
⇒
5 a ⋅ a − 4 a ⋅ b + 10 b ⋅ a − 8 b ⋅ b = 0
⇒
6 a ⋅b = 3
then the length of the median through A is
⇒
a ⋅b =
So, the angle between a and b is
π
.
3
22. If
(a)
(b)
18
(c)
72
(d)
33
45
[JEE Main 2013, 2003]
Exp. (c)
We know that, the sum of three vectors of a
triangle is zero.
A
24. Let ABCD be a parallelogram such that
AB = q ,AD = p and ∠BAD be an acute angle.
If r is the vector that coincides with the
altitude directed from the vertex B to the
side AD, then r is given by
[AIEEE 2012]
(a) r = 3q
B
∴
⇒
⇒
M
C
AB + BC + CA = 0
BC = AC − AB
[QAC = − CA ]
AC − AB
[QM is a mid-point of BC]
BM =
2
Also, AB + BM + MA = 0
[by properties of a triangle]
⇒
⇒
⇒
AC − AB
= AM
[QAM = − MA ]
2
AB + AC 3$i + 4k$ + 5$i − 2 $j + 4k$
AM =
=
2
2
= 4 $i − $j + 4k$
AB +
|AM| =
42 + 12 + 42 =
33
1
2
3( p ⋅ q )
p
(p ⋅ p )
p⋅q
(c) r = q − 
p
 p⋅p
p⋅q
(b) r = − q + 
p
 p⋅p
(d) r = − 3q +
3( p ⋅ q )
p
(p ⋅ p)
Exp. (b)
Given
(i) A parallelogram ABCD such that AB = q and
AD = p.
(ii) The altitude from vertex B to side AD coincides
with a vector r.
To find The vector r in terms of p and q.
Let E be the foot of perpendicular from B to
side AD.
q⋅ p
AE = Projection of vector q on p = q ⋅ p =
| p|
310
JEE Main Chapterwise Mathematics
D
C
E
p
r
A
B
q
AE = Vector along AE of length AE
 q ⋅ p
(q ⋅ p)p
= | AE | AE = 
 p=
| p|2
 | p| 
Now, applying triangles law in ∆ABE, we get
⇒
AB + BE = AE
(q ⋅ p)p
(q ⋅ p)p
⇒ r=
−q
q+r=
| p|2
| p|2
 q ⋅ p
r = −q+ 
p
 p ⋅ p
⇒
1
(a) −3
(b) 5
(c) 3
(d) −5
[AIEEE 2011]
Exp. (d)
a =
Given, a ⋅ b ≠ 0, a ⋅ d = 0
and
b×c =b×d
⇒
b × (c − d) = 0
∴
b|| (c − d)
⇒
c − d = λb
⇒
d = c − λb
Taking dot product with a, we get
a ⋅ d = a ⋅ c − λa ⋅ b
⇒
0 = a ⋅ c − λ (a ⋅ b)
a ⋅c
λ=
∴
a ⋅b
(a ⋅ c )
∴
b
d=c −
(a ⋅ b)
…(i)
…(ii)
…(iii)
27. If the vectors p $i + $j + k$ , i$ + q $j + k$ and
i$ + $j + r k$ ( where, p ≠ q ≠ r ≠ 1) are coplanar,
then the value of pqr − (p + q + r ) is
1
( 3i$ + k$ ) and b = (2 i$ + 3$j − 6k$ ),
7
10
then the value of(2 − b) ⋅[(a × b) × (a + 2 b)] is
25. If a =
Exp. (c)
1
(3$i + k$ )
10
(a) − 2
(c) 0
(b) 2
(d) −1
[AIEEE 2011]
Exp. (a)
$ b = $i + q$j + k$ and
Given, a = p $i + $j + k,
c = $i + $j + rk$ are coplanar and p ≠ q ≠ r ≠ 1.
Since, a , b and c are coplanar.
⇒
[a b c ] = 0
p 1 1
1 q 1 =0
⇒
1 $
(2 i + 3$j − 6 k$ )
7
∴
(2a − b) ⋅ {(a × b) × (a + 2b)}
= (2a − b) ⋅ {(a × b) × a + (a × b) × 2b}
= (2a − b) ⋅ {(a ⋅ a ) b − (b ⋅ a ) a
+ 2 (a ⋅ b) b − 2 (b ⋅ b) a }
= (2a − b) ⋅ {1 (b) − (0) a + 2 (0) b − 2 (1) a }
28. Let a , b and c be three non-zero vectors
[as a ⋅ b = 0 and a ⋅ a = b ⋅ b = 1]
= (2a − b) (b − 2a )
= − (4|a|2 − 4 a ⋅ b + |b|2 )
which are pairwise non-collinear. If a + 3b is
collinear with c and b + 2 c is collinear witha,
then a + 3b + 6 c is
[AIEEE 2011]
= − {4 − 0 + 1} = − 5
(a) a + c
and b =
26. The vectors a and b are not perpendicular
and c and d are two vectors satisfying
b × c = b × d anda ⋅ b = 0. Then, the vector d is
equal to
[AIEEE 2011]
 a⋅ c 
(a) c + 
b
 a⋅ b
 b⋅ c 
(b) b + 
c
 a⋅ b
 a⋅ c 
(c) c − 
b
 a⋅ b
 b⋅ c 
(d) b − 
c
 a⋅ b
1 1 r
⇒
⇒
∴
p (q r − 1) − 1 (r − 1) + 1 (1 − q ) = 0
pqr − p − r + 1 + 1 − q = 0
pqr − ( p + q + r ) = − 2
(b) a
(c) c
(d) 0
Exp. (d)
As, a + 3 b is collinear with c.
∴
a + 3 b = λc
Also, b + 2c is collinear with a.
⇒
b + 2c = µa
From Eq. (i), we get
a + 3 b + 6 c = (λ + 6) c
…(iii)
From Eq. (ii), we get
a + 3 b + 6 c = (1 + 3µ ) a
…(iv)
…(i)
…(ii)
311
Vector Algebra
From Eqs. (iii) and (iv), we get
∴
(λ + 6) c = (1 + 3 µ ) a
Exp. (d)
Since, [3u pv pw ] − [ p v w q u] − [2 w qv qu] = 0
Since, a is not collinear with c.
⇒
λ + 6 = 1 + 3µ = 0
∴ 3 p2 [u ⋅ (v × w )] − pq [v ⋅ (w × u)]
− 2q 2 [w ⋅ (v × u)] = 0
From Eq. (iv), we get
a + 3b + 6c = 0
⇒
29. Leta = $j − k$ anda = $i − $j − k$ . Then, the vector
bsatisfying a × b + c = 0 and a ⋅ b = 3, is
[AIEEE 2010]
(a) − $i + $j − 2 k$
(c) $i − $j − 2 k$
(b) 2 $i − $j + 2 k$
(d) $i + $j − 2 k$
Exp. (a)
We have,
a ×b+c = 0
⇒
a × (a × b) + a × c = 0
⇒
(a ⋅ b)a − (a ⋅ a )b + a × c = 0
⇒
3a − 2 b + a × c = 0
⇒
2 b = 3a + a × c
⇒
2 b = 3$j − 3k$ − 2 $i − $j − k$
= − 2 $i + 2 $j − 4k$
∴
b = − $i + $j − 2 k$
30. If the vectors a = $i − $j + 2 k$ , b = 2 $i + 4 $j + k$
and c = λ i$ + $j + µ k$ are mutually orthogonal,
then ( λ , µ ) is equal to
[AIEEE 2010]
(a) (– 3, 2)
(c) (– 2, 3)
(b) (2, – 3)
(d) (3, – 2)
Exp. (a)
Since, the given vectors are mutually orthogonal,
therefore
a ⋅b = 2 − 4 + 2 = 0
...(i)
a ⋅ c = λ − 1 + 2µ = 0
and
...(ii)
b⋅ c = 2 λ + 4 + µ = 0
On solving Eqs. (i) and (ii), we get
µ =2
and
λ = −3
Hence,
(λ , µ ) = (−3, 2 )
31. If u , v and w are non-coplanar vectors and
p , q are real numbers, then the equality
[ 3 u p v p w ] − [p v w q u] − [2 w q v q u] = 0
holds for
[AIEEE 2009]
(a)
(b)
(c)
(d)
exactly two values of (p , q )
more than two but not all values of (p , q )
all values of (p , q )
exactly one value of (p , q )
(3 p2 − pq + 2q 2 ) [u ⋅ (v × w )] = 0
But
⇒
∴
[u v w ] ≠ 0
3 p2 − pq + 2q 2 = 0
p=q = 0
32. The vector a = αi$ + 2 $j + βk$ lies in the plane of
the vectors b = i$ + $j and c = $j + k$ and bisects
the angle between b and c. Then, which one
of the following gives possible values of
[AIEEE 2008]
α and β?
(a) α = 1, β = 1
(c) α = 1, β = 2
(b) α = 2 , β = 2
(d) α = 2 , β = 1
Exp. (a)
Given that, b = $i + $j and c = $j + k$ .
The equation of bisector of b and c is
r = λ(b + c )
 $i + $j $j + k$ 
=λ
+

2 
 2
λ $
…(i)
=
( i + 2 $j + k$ )
2
Since, vector a lies in plane of b and c.
∴
a = b + µc
λ $
( i + 2 $j + k$ ) = ($i + $j ) + µ( $j + k$ )
⇒
2
On equating the coefficient of i both sides, we
get
λ
=1 ⇒ λ = 2
2
On putting λ = 2 in Eq. (i), we get
r = $i + 2 $j + k$
Since, the given vector a represents the same
bisector equation r.
∴
α = 1 and β = 1
Alternate Solution
Since, a , b and c are coplanar.
α 2 β
⇒
1 1 0 =0
0 1 1
⇒ α(1 − 0) − 2 (1 − 0) + β(1 − 0) = 0
⇒ α + β = 2, which is possible for α = 1, β = 1.
312
JEE Main Chapterwise Mathematics
33. The non-zero vectors a , b and c are related
by a = 8b and c = − 7 b. Then, the angle
between a and c is
[AIEEE 2008]
(a) π
(b) 0
(c)
π
4
(d)
π
2
Exp. (a)
Since, a = 8 b and c = −7 b
So, a is parallel to b and c is anti-parallel to b.
⇒ a and c are anti-parallel.
So, the angle between a and c is π.
34. If u and v are unit vectors and θ is the acute
angle between them, then 2 u × 3 v is a unit
vector for
[AIEEE 2007]
(a)
(b)
(c)
(d)
exactly two values of θ
more than two values of θ
no value of θ
exactly one value of θ
Exp. (c)
Since,
(a × b) × c = a × (b × c )
∴
(a ⋅ c )b − (b⋅ c )a = (a ⋅ c )b − (a ⋅ b)c
⇒
(b⋅ c )a = (a ⋅ b)c
(a ⋅ b)
⇒
⋅c
a =
(b ⋅ c )
Hence, a is parallel to c.
37. The value of a , for which the points, A , B ,C
with position vectors 2 $i − $j + k$ , $i − 3 $j − 5 k$
and a $i − 3 $j + k$ respectively are the
π
vertices of a right angled triangle withC =
2
are
[AIEEE 2006]
(a) –2 and –1
(c) 2 and –1
Exp. (d)
Exp. (d)
Since, (2u × 3v ) is a unit vector.
⇒
|2u × 3v| = 1
⇒
6|u || v|| sin θ|= 1
1
sin θ =
⇒
[Q|u| = |v| = 1]
6
Since, θ is an acute angle, then there is exactly one
value of θ for which (2u × 3v ) is a unit vector.
35. Let a = $i + $j + k$ , b = $i − $j + 2 k$ and
Since, position vectors of A, B, C are 2 $i − $j + k$ ,
$i − 3 $j − 5k$ and a $i − 3 $j + k$ , respectively.
Now,
AC = (a $j − 3 $j + k$ ) − (2 $i − $j + k$ )
Since, the ∆ABC is right angled at C, then
AC ⋅ BC = 0
⇒
(a) 0
∴
(c) – 4
= (a − 2 ) $i − 2 $j
BC = (a $i − 3 $j + k$ ) − ($i − 3 $j − 5k$ )
= (a − 1) $i + 6k$
and
c = x$i + ( x − 2 )$j − k$ . If the vector c lies in the
plane of a and b, then x equal to [AIEEE 2007]
(b) 1
(b) –2 and 1
(d) 2 and 1
(d) –2
Exp. (d)
Since, given vectors a , b and c are coplanar.
1
1
1
1 −1 2 = 0
∴
x x − 2 −1
⇒ 1{1 − 2 ( x − 2 )} − 1(−1 − 2 x) + 1( x − 2 + x) = 0
⇒
1− 2 x + 4 + 1+ 2 x + 2 x − 2 = 0
⇒
2 x = − 4 ⇒ x = −2
36. If (a × b) × c = a × ( b × c ), where a, b and c are
any three vectors such that a ⋅ b ≠ 0, b ⋅ c ≠ 0,
then a and c are
[AIEEE 2006]
π
(a) inclined at an angle of between them
6
(b) perpendicular
(c) parallel
π
(d) inclined at an angle of between them
3
{(a − 2 ) $i − 2 $j } ⋅ {(a − 1) $i + 6k$ } = 0
⇒
(a − 2 )(a − 1) = 0
a=1
a=2
and
38. IfC is the mid-point of AB and P is any point
outside AB , then
[AIEEE 2005]
(a) PA + PB + PC = 0
(b) PA + PB + 2 PC = 0
(c) PA + PB = PC
(d) PA + PB = 2 PC
Exp. (d)
Let P be the origin outside of ABand C is mid-point
of AB, then
C
A
B
P
PA + PB
PC =
2
⇒ 2PC = PA + PB
313
Vector Algebra
r = 2 $i − 2 $j + 3k$ + λ ( $i − $j + 4k$ ) and the plane
r ⋅ ( i$ + 5 $j + k$ ) = 5 is
1 0
−1
1− x
[a b c ] = x 1
y x 1+ x − y
∴
[AIEEE 2005]
(a)
10
3
(b)
3
10
(c)
10
3 3
c = y $i + x $j + (1 + x − y)k$
and
39. The distance between the line
(d)
10
9
Applying C 3 → C 3 + C1, we get
1 0
= x 1
Exp. (c)
Line is parallel to plane as
($i − $j + 4k$ ) ⋅ ($i + 5 $j + k$ ) = 1 − 5 + 4 = 0
y
(λ + 2 , − λ − 2 , 4λ + 3).
42. Let a , b and c be distinct non-negative
For λ = 0, a point on this line is (2, –2, 3) and
distance from
r ⋅ ($i + 5$j + k$ ) = 5 or x + 5 y + z = 5 is
numbers. If the vectors a $i + a $j + c k$ , i + k
and c $i + c $j + b k$ lie in a plane, then c is
[AIEEE 2005]
2 + 5(−2 ) + 3 − 5
d =

1 + 25 + 1 

(a)
(b)
(c)
(d)
 = 10
 −10 
d =
 3 3 3 3
40. For
any vector a, the value
2
$
( a × i ) + (a × $j )2 + ( a × k$ )2 is equal to
(a) 4 a
2
(b) 2 a
(c) a 2
2
of
[AIEEE 2005]
(d) 3 a 2
1
3
2
= a 22 + a 23 + a 12 + a 23 + a 12 + a 22
=2
(a 12
+
a 22
+
a 23 ) =
2a
2
41. Let a = $i − k$ , b = xi$ + $j + (1 − x ) k$ and
c = y $i + x $j + (1 + x − y )k$ . Then,[a b c]
depends on
[AIEEE 2005]
(a) Neither x nor y
(c) Only x
a a c
1 0 1 =0
∴
−1 (ab − c 2 ) = 0
⇒
∴ (a × i) + (a × j) + (a × k )
2
Since, the given vectors lie in a plane.
⇒
a × k = − a1 j + a2 i
2
Exp. (d)
Applying C1 → C1 − C 2 , we get
0 a c
1 0 1 =0
0 c b
a = a1 $i + a2 $j + a3 k$
a × $i = − a2k$ + a3 $j
a × $j = a k$ − a $i
Then,
the harmonic mean of a and b
equal to zero
the arithmetic mean of a and b
the geometric mean of a and b
c c b
Exp. (b)
Let
= 11
( + x) − x = 1
x 1+ x
Thus, [a b c ] depends upon neither x nor y.
General point on the line is
⇒
0
1
(b) Both x and y
(d) Only y
Exp. (a)
Given, vectors are
a = $i − k$ , b = x $i + $j + (1 − x) k$
c 2 = ab
Hence, c is GM of a and b.
43. If a , b, c are non-coplanar vectors and λ is a
real number, then
[ λ (a + b) λ 2b λc] = [a b + c b] for
(a)
(b)
(c)
(d)
exactly two values of λ
exactly three values of λ
no value of λ
exactly one value of λ
[AIEEE 2005]
Exp. (c)
Given that,
[λ(a + b) λ2 b
λ c ] = [a
b + c b]
314
JEE Main Chapterwise Mathematics
∴
λ(a1 + b1 ) λ(a2 + b2 ) λ(a3 + b3 )
λ2 b1
λ2 b2
λ2 b3
λc1
λc 2
λc 3
a1
⇒
a2
a3
= b1 + c1 b2 + c 2
b1
b2
a3
a1 a2 a3
b3 = − b1 b2 b3
a1 a2
λ4 b1 b2
c1 c 2
c3
⇒
The particle is displaced from
A($i + 2 $j + 3k$ ) to B(5 $i + 4 $j + k$ ).
c1 c 2
b3 + c 3
b3
c3
λ4 = − 1
44. If a , b and c are three non-zero vectors such
that no two of these are collinear. If the
vector a + 2 b is collinear with c and b + 3c is
collinear with a ( λ being some non-zero
scalar), then a + 2 b + 6c equal to [AIEEE 2004]
(b) λb
(c) λc
(d) 0
Exp. (d)
If a + 2b is collinear with c, then a + 2b = tc …(i)
Also, b + 3 c is collinear with a, then
b + 3 c = λa
…(ii)
⇒
b = λa − 3 c
From Eqs. (i) and (ii), we get
a + 2 (λa − 3 c ) = tc
⇒
(a − 6c ) = tc − 2 λa
On comparing the coefficients of a and c, we get
1
1= − 2λ ⇒ λ = −
2
and
−6 = t
⇒ t = −6
From Eq. (i), we get
⇒
a + 2 b = −6c
a + 2b + 6c = 0
45. A particle is acted upon by constant forces
4i$ + $j − 3k$ and 3i$ + $j − k$ which displace it
from a point $i + 2 $j + 3k$ to the point
5$i + 4$j + k$ . The work done in standard units
by the forces is given by
(a) 40 units
(c) 25 units
∴Work done = F ⋅ AB
= (7 $i + 2 $j − 4k$ ) ⋅ (4 $i + 2 $j − 2 k$ )
= 28 + 4 + 8
= 40 units
46. If a, b, c are non-coplanar vectors and λ is a
real number, then the vectors a + 2 b + 3c ,
λb + 4c and (2 λ − 1) c are non-coplanar for
So, no real value of λ exists.
(a) λa
Now, displacement,
AB = ( 5$i + 4 $j + k ) − ( $i + 2 $j + 3 k$) = 4 $i + 2 $j − 2 k$
[AIEEE 2004]
(b) 30 units
(d) 15 units
Exp. (a)
Total force, F = (4 $i + $j − 3k$ ) + (3 $i + $j − k$ )
∴
F = 7 $i + 2 $j − 4k$
(a)
(b)
(c)
(d)
all values of λ
all except one value of λ
all except two values of λ
no value of λ
[AIEEE 2004]
Exp. (c)
The three vectors (a + 2b + 3c ),(λb + 4c ) and
(2 λ − 1)c are non-coplanar, if
1 2
3
0 λ
4
≠0
0 0 2λ − 1
⇒
(2 λ − 1)(λ ) ≠ 0 ⇒ λ ≠ 0,
1
2
So, these three vectors are non-coplanar for all
except two values of λ.
47. Let u , v , w be such that | u | = 1,| v | = 2 ,| w | = 3.
If the projection v along u is equal to that of
w along u and v , w are perpendicular to
each other, then | u − v + w | equal to
(a) 2
(c) 14
(b) 7
(d) 14
[AIEEE 2004]
Exp. (c)
|u |= 1,|v|= 2 ,|w|= 3
v ⋅u
The projection of v along u =
|u |
Since,
and the projection of w along u =
w ⋅u
|u |
According to given condition,
v ⋅u w ⋅u
=
|u|
|u|
⇒
v ⋅u = w ⋅u
…(i)
315
Vector Algebra
Since, v, w are perpendicular to each other.
…(ii)
∴
v ⋅w = 0
2
2
2
2
Now,
|u − v + w| = |u| + |v| + |w|
Exp. (a)
Vector perpendicular to face OAB is n1.
Y
−2 u ⋅ v − 2 v ⋅ w + 2 u ⋅ w
⇒ |u − v + w|2 = 1 + 4 + 9 − 2u ⋅ v + 2 v ⋅ u
B(2, 1, 3)
[from Eqs. (i) and (ii)]
⇒ |u − v + w|2 = 1 + 4 + 9
⇒
|u − v + w| = 14
O
48. Let a , b and c be non-zero vectors such that
1
(a × b) × c = | b|| c |a. If θ is an acute angle
3
between the vectors b and c, then sin θ is
equal to
[AIEEE 2004]
1
3
2
(c)
3
(a)
2
3
2 2
(d)
3
1
|b||c|a = (a × b) × c
3
We know that,
(a × b) × c = (a ⋅ c ) b − (b ⋅ c )a
C (–1, 1, 2)
$i
1
|b|| c|a = (a ⋅ c ) b − (b ⋅ c )a
3
On comparing the coefficients of a and b,
we get
1
|b|| c|= −b ⋅ c and a ⋅ c = 0
3
1
⇒
|b||c|= −|b||c| cos θ
3
1
1
cos θ = − ⇒ 1 − sin2 θ =
⇒
3
9
8
2
sin θ =
⇒
9
2 2
Q 0 ≤ θ ≤ π 
∴
sin θ =
3
2 

49. A tetrahedron has vertices at O(0, 0, 0),
A (1, 2 , 1), B (2 , 1, 3) and C ( −1, 1, 2 ). Then, the
angle between the faces OAB and ABC will
be
[AIEEE 2003]
19
(a) cos−1  
 35 
17
(b) cos−1  
 31 
(c) 30°
(d) 90°
$j k$
= OA × OB = 1 2 1
2 1 3
= 5 $i − $j − 3k$
Vector perpendicular to face ABC is n 2
$i
$j k$
= AB × AC = 1
Given that,
∴
X
Z
(b)
Exp. (d)
A (1, 2, 1)
−1 2
−2 −1 1
= $i − 5 $j − 3k$
Since, angle between faces is equal to the angle
between their normals.
n ⋅n
∴ cos θ = 1 2
|n1||n 2|
5 × 1 + (−1) × (−5) + (−3) × (−3)
=
=
⇒
5 + (−1)2 + (−3)2
2
12 + (−5)2 + (−3)2
5 + 5 + 9 19
=
35 35 35
19
θ = cos −1  
 35 
50. a , b, c are
three vectors, such that
a + b + c = 0 , |a | = 1, | b| = 2 , | c | = 3, then
[AIEEE 2003]
a ⋅ b + b ⋅ c + c ⋅ a is equal to
(a) 0
(c) 7
(b) –7
(d) 1
Exp. (b)
Given that, |a|= 1,|b|= 2 ,|c|= 3
and
a +b+c =0
Now, (a + b + c )2 = |a|2 + |b|2 + |c|2
+ 2(a ⋅ b + b ⋅ c + c ⋅ a )
316
JEE Main Chapterwise Mathematics
⇒
⇒
0 = 12 + 2 2 + 32 + 2 (a ⋅ b + b ⋅ c + c ⋅ a )
2 (a ⋅ b + b ⋅ c + c ⋅ a ) = − 14
⇒
53.
a a2 1+a3
If b b 2 1 + b 3 = 0 and vectors
a ⋅ b + b ⋅ c + c ⋅ a = −7
c
51. If u , v and w are three non-coplanar vectors,
then ( u + v − w ) ⋅[( u − v ) × ( v − w )] equal to
[AIEEE 2003]
(a) 0
(c) u ⋅ w × v
(b) u ⋅ v × w
(d) 3u ⋅ v × w
Exp. (b)
(u + v − w ) ⋅ [(u − v ) × (v − w )]
= (u + v − w ) ⋅ [u × v − u × w − v × v + v × w ]
= u ⋅ (u × v ) − u ⋅ (u × w ) + u ⋅ (v × w ) + v ⋅ (u × v )
− v ⋅ (u × w) + v ⋅ (v × w) − w ⋅ (u × v)
+ w ⋅ (u × w)− w ⋅ (v × w)
{[a ,a b] = 0}
= u⋅ v × w − v ⋅ u × w − w ⋅ u × v
= u⋅ v × w + w ⋅ u × v − w ⋅ u × v
= u⋅ v × w
52. Consider points A , B ,C and D with position
vectors 7$i − 4$j + 7k$ , $i − 6 $j + 10 k$ , − $i − 3$j + 4k$
and5 i$ − $j + 5 k$ , respectively. Then, ABCD is a
[AIEEE 2003]
(a) square
(c) rectangle
(b) rhombus
(d) None of these
Exp. (d)
Given that,
and
OA = 7 $i − 4 $j + 7 k$ ,
OB = $i − 6 $j + 10 k$ ,
OC = − $i − 3 $j + 4 k$
OD = 5$i − $j + 5 k$
Now, AB = ( 7 − 1)2 + (−4 + 6)2 + ( 7 − 10)2
=
=
36 + 4 + 9
49 = 7
BC = (1 + 1)2 + (−6 + 3)2 + (10 − 4)2
=
=
4 + 9 + 36
49 = 7
c2
1 + c3
(1, a , a 2 ),(1, b , b 2 ) and (1, c , c 2 ) are noncoplanar, then the product abc equal to
(a) 2
(c) 1
(b) –1
(d) 0
Exp. (b)
Since,
a a2 1 + a3
a a2 1
a a2
2
3
2
b b 1 + b = b b 1 + b b2
c c2 1 + c3
c c2 1
c c2
and
36 + 4 + 1
41
DA = (5 − 7 )2 + (−1 + 4)2 + (5 − 7 )2
= 4 + 9 + 4 = 17
Hence, option (d) is correct.
a3
b3 = 0
c3
a a2 1
a a2 1
2
b b 1 + abc b b 2 1 = 0
c c2 1
c c2 1
⇒
⇒
a a2 1
(1 + abc ) b b 2 1 = 0
c c2 1
⇒
1 + abc = 0
⇒

 a a2 1


2
Q b b 1 ≠ 0
2

 c c 1


abc = − 1
54. Let u = $i + $j , v = $i − $j and w = i$ + 2 $j + 3k$ . If n
is a unit vector such that u ⋅ n = 0 and v ⋅ n = 0,
then | w ⋅ n | is equal to
[AIEEE 2003]
(a) 0
(c) 2
(b) 1
(d) 3
Exp. (d)
Given that, u = $i + $j, v = $i − $j , w = $i + 2 $j + 3k$ ,
u ⋅ n = 0 and v ⋅ n = 0
u×v
i .e.,
n=
|u × v |
$i
Now,
u×v= 1
CD = (−1 − 5)2 + (−3 + 1)2 + (4 − 5)2
=
=
[AIEEE 2003]
$j
k$
1 0
1 −1 0
= 0 $i − 0 $j − 2 k$ = − 2 k$
|w ⋅ u × v| |− 6k|
=
=3
|− 2 k|
|u × v|
[Qw ⋅ (u × w ) = ($i + 2 $j + 3k$ ) ⋅ (− 2k$ ) = − 6k$ ]
Hence,
|w ⋅ n|= 3
∴
|w ⋅ n|=
317
Vector Algebra
55. Given, two vectors are $i − $j and $i + 2 $j , the
unit vector coplanar with the two vectors
and perpendicular to first is
[AIEEE 2002]
1 $ $
(i + j)
2
1 $ $
(c) ±
( i + j)
2
(b)
(a)
1
( 2 $i + $j )
5
(d) None of these
Exp. (a)
2|$i + x $j + 3k$|= |4 $i + (4 x − 2 ) $j + 2 k$|
⇒
2 1 + x2 + 9 = 16 + (4 x − 2 )2 + 4
⇒
40 + 4 x2 = 20 + (4 x − 2 )2
⇒
3 x2 − 4 x − 4 = 0
⇒ ( x − 2 )(3 x + 2 ) = 0
x = 2, −
⇒
2
3
57. If the vectors a , b and c form the sides
Given two vectors lie in xy-plane. So, a vector
coplanar with them is
a = x $i + y $j
Since,
a ⊥ ($i − $j )
⇒
∴
( x $i + y $j ) ⋅ ($i − $j ) = 0
⇒
⇒
∴
and
x− y=0
x= y
a = x $i + x $j
|a|=
x + x = x 2
2
2
∴ Required unit vector
x($i + $j )
a
1 $ $
=
=
=
(i + j )
|a|
x 2
2
BC ,CA and AB respectively of a ∆ABC , then
(a)
(b)
(c)
(d)
a ⋅ b = b⋅ c = c⋅ b = 0
a ×b=b×c=c×a
a ⋅ b = b⋅ c = c⋅ a = 0
a × a + a × c + a × a =0
[AIEEE 2002]
Exp. (b)
Since,
⇒
⇒
⇒
Similarly,
Hence,
a + b+ c =0
a +b= −c
(a + b) × c = − c × c
b×c =c×a
a ×b = b× c
a ×b = b× c = c × a
56. The vector $i + x $j + 3 k$ is rotated through an
58. If the vectors c , a = x i$ + y $j + z k$ and b = $j are
angle θ and doubled in magnitude, then it
becomes 4 $i + ( 4x − 2 ) $j + 2 k$ . The value of x
are
[AIEEE 2002]
such that a , c and b form a right handed
system, then c is
[AIEEE 2002]
2
(a) − , 2 
 3 
2
(c)  , 0
3 
1
(b)  , 2 
3 
(d) {2, 7}
Exp. (a)
Since, the vector $i + x $j + 3k$ is doubled in
magnitude, then it becomes
4 $i + (4 x − 2 ) $j + 2 k$
(a) z $i − x k$
(c) y $j
(b) 0
(d) − z $i + x k$
Exp. (a)
Since, the vectors a = x $i + y $j + z k$ and b = $j
are such that a , c and b form a right handed
system.
$i $j k$
∴
c = b × a = 0 1 0 = z $i − xk$
x
y
z
14
Statistics and Probability
1. The
mean and variance of seven
observations are 8 and 16, respectively. If 5
of the observations are 2, 4, 10, 12, 14, then
the product of the remaining two
observations is
[JEE Main 2019, 8 April Shift-I]
(a) 45
(c) 48
(b) 49
(d) 40
Let the remaining two observations are a and b.
According to the question,
2 + 4 + 10 + 12 + 14 + a + b
Mean =
=8
7
⇒
42 + a + b = 56
…(i)
⇒
a + b = 14
and variance
a2 + b 2 + 4 + 16 + 100 + 144 + 196
=
− 82 = 16
7
a2 + b 2 + 460
⇒
− 64 = 16
7
a2 + b 2 + 460
⇒
= 80
7
⇒
a2 + b 2 + 460 = 560
a2 + b 2 = 100
A ⊂ B . Then, which of the following
statements is always correct.
[JEE Main 2019, 8 April Shift-I]
(a) P( A /B ) = P( B ) − P( A ) (b) P( A/B ) ≥ P( A )
(c) P( A/B ) ≤ P( A )
(d) P( A/B ) =1
Exp. (b)
Exp. (c)
⇒
2. Let A andB be two non-null events such that
…(ii)
We know that,
(a + b )2 = (a2 + b 2 ) + 2 ab
⇒
(14)2 = 100 + 2 ab
⇒
⇒
⇒
[from Eqs. (i) and (ii)]
196 = 100 + 2 ab
2 ab = 96
ab = 48
We know that, P ( A / B) =
Q
⇒
∴
P( A ∩ B)
P(B)
[by the definition of conditional probability]
A⊂B
A∩B= A
P( A)
…(i)
P( A / B) =
P(B)
As we know that, 0 ≤ P(B) ≤ 1
1
1≤
<∞
∴
P(B)
P( A)
<∞
P(B)
⇒
P( A) ≤
⇒
P( A)
≥ P( A)
P(B)
…(ii)
Now, from Eqs (i) and (ii), we get
P ( A/B) ≥ P(A)
3. The minimum number of times one has to
toss a fair coin so that the probability of
observing atleast one head is atleast 90% is
[JEE Main 2019, 8 April Shift-II]
(a) 2
(c) 5
(b) 3
(d) 4
319
Statistics and Probability
Exp. (d)
The required probability of observing atleast one
head = 1 − P (no head)
1
[let number of toss are n]
= 1− n
2
Q P(Head) = P(Tail ) = 1 

2 
1
90
According to the question, 1 − n ≥
100
2
1
1
≤
⇒ 2 n ≥ 10 ⇒ n ≥ 4
⇒
2 n 10
So, minimum number of times one has to toss a
fair coin so that the probability of observing atleast
one head is atleast 90% is 4.
4. A student scores the following marks in five
tests 45, 54, 41, 57, 43. His score is not known
for the sixth test. If the mean score is 48 in
the six tests, then the standard deviation of
the marks in six tests is
[JEE Main 2019, 8 April Shift-II]
10
3
100
(c)
3
10
3
100
(d)
3
(a)
(b)
41 + 45 + 43 + 54 + 57 + 48
− 482
6
Σxi2
[Q standard deviation (SD) =
− ( x )2 ]
6
2
2
2
2
2
(412 − 482 ) + (452 − 482 ) + (432 − 482 )
=
+ (542 − 482 ) + (57 2 − 482 )
6
(−7 × 89) + (−3 × 93) + (−5 × 91)
=
−1, 0, 1, k is 5 where k > 0, then k is equal to
[JEE Main 2019, 9 April Shift-I]
10
(a) 2
3
+ (6 × 102 ) + (9 × 105)
6
(b) 2 6
(c) 4
5
3
(d)
6
Exp. (b)
Given observations are −1, 0, 1 and k.
Also, standard deviation of these
observations = 5
four
2
(−1)2 + (0)2 + (1)2 + k 2  −1 + 0 + 1 + k 
−
 =


4
4
∴
5
[Qif x1, x2 .... xn are n observation, then standard
2
1 n 2 1 n  
deviation =
Σ xi −  Σ xi 
 n i =1  
n i =1

⇒
Let the marks in sixth tests is ‘x’, so
41 + 45 + 43 + 54 + 57 + x
mean =
= 48 (given)
6
240 + x
= 48
⇒
6
x
⇒
40 + = 48
6
x
= 8 ⇒ x = 48
⇒
6
Now, standard deviation of these marks
=
5. If the standard deviation of the numbers
⇒
Exp. (b)
2
945 + 612 − 455 − 279 − 623
6
1557 − 1357
200
100 10
=
=
=
=
6
3
6
3
=
⇒
⇒
⇒
2 + k2 k2
−
= 5 [squaring both sides]
4
16
8 + 4k 2 − k 2
8 + 3k 2
=5 ⇒
=5
16
16
8 + 3k 2 = 80 ⇒ 3k 2 = 72
2
k = 24 ⇒ k = 2 6 or −2 6
[Qk > 0]
k =2 6
6. The mean and the median of the following
ten numbers in increasing order 10, 22, 26,
29, 34, x, 42, 67, 70, y are 42 and 35
y
respectively, then is equal to
x
[JEE Main 2019, 9 April Shift-II]
7
(a)
3
7
(b)
2
(c)
8
3
(d)
9
4
Exp. (a)
Given ten numbers are
10, 22, 26, 29, 34, x, 42, 67, 70, y
and their mean = 42
10 + 22 + 26 + 29 + 34 + x + 42 + 67 + 70 + y
∴
10
= 42
300 + x + y
⇒
= 42
10
…(i)
⇒
x + y = 120
320
JEE Main Chapterwise Mathematics
and their median (arranged numbers are in
increasing order) = 35
34 + x
⇒
= 35
2
⇒
34 + x = 70
⇒
x = 36
On substituting x = 36 in Eq. (i), we get
36 + y = 120 ⇒ y = 84
y 84 7
=
=
∴
x 36 3
7. If for some x ∈R , the frequency distribution
of the marks obtained by 20 students in a
test is
Marks
2
3
5
7
Frequency
( x + 1)2
2x − 5
x2 − 3 x
x
Then, the mean of the marks is
[JEE Main 2019, 10 April Shift-I]
(a) 3.0
(c) 2.5
(b) 2.8
(d) 3.2
8. Assume that each born child is equally likely
to be a boy or a girl. If two families have two
children each, then the conditional
probability that all children are girls given
that at least two are girls; is
[JEE Main 2019, 10 April Shift-I]
1
(a)
17
Key Idea Use ∑ fi = Number of students;
i =1
Σ fi xi
Σ fi
The given frequency distribution, for some x ∈ R,
of marks obtained by 20 students is
Marks
2
Frequency
3
( x + 1)
2
5
7
2 x − 5 x2 − 3 x
 1
 
2
4
Exp. (b)
⇒ 2 x + 2 x − 24 = 0 ⇒ x + x − 12 = 0
[as x > 0]
2( x + 1)2 + 3(2 x − 5) + 5( x2 − 3 x) + 7 x
20
2(4)2 + 3(1) + 5(0) + 7(3) 32 + 3 + 21
=
=
20
20
56
=
= 2.8
20
Hence, option (b) is correct.
=
3
2
2
[JEE Main 2019, 10 April Shift-II]
⇒ ( x2 + 2 x + 1) + (2 x − 5) + ( x2 − 3 x) + x = 20
( x + 4)( x − 3) = 0⇒ x = 3
Σf x
Now, mean ( x ) = i i
Σ fi
4
deviation of 50 observations x1 , x 2 ,K , x 50
are equal to 16, then the mean of ( x1 − 4)2,
( x 2 − 4)2 ,K ,( x 50 − 4)2 is
x
2
1
11
9. If both the mean and the standard
QNumber of students = 20 = Σ fi
⇒ ( x + 1)2 + (2 x − 5) + ( x2 − 3 x) + x = 20
⇒
(d)
Let event B is being boy while event G being girl.
According to the question,
1
P(B) = P(G ) =
2
Now, required conditional probability that all
children are girls given that at least two are girls, is
All 4 girls
=
(All 4 girls ) + (exactly 3 girls +1 boy)
+(exactly 2 girls + 2 boys)
(a) 480
(c) 380
2
1
10
 1  + 4C  1   1  + 4C  1   1 
 
3   
2   
2 2
2 2
2
1
1
=
=
1 + 4 + 6 11
n
and Mean (x ) =
(c)
Exp. (d)
=
Exp. (b)
1
(b)
12
(b) 400
(d) 525
It is given that both mean and standard deviation
of 50 observations x1, x2 , x3 , K , x50 are equal to
16,
Σx
So,
mean = i = 16
…(i)
50
and standard deviation =
Σxi2  Σxi 
−

50  50 
⇒
Σxi2
− (16)2 = (16)2
50
⇒
Σxi2
= 2 × 256 = 512
50
2
= 16
…(ii)
321
Statistics and Probability
Now, mean of ( x1 − 4)2 , ( x2 − 4)2 , K ,( x50 − 4)2
=
Σ( xi − 4)
=
50
2
Σ( xi2
− 8 xi + 16)
50
regular hexagon alternate, here A1, A3 , A5 or A2 ,
A4 , A6 will result in an equilateral triangle.
A6
Σx2
Σx
16
= i − 8 i  +
Σ1
 50  50
50
= 512 − 128 + 16 = 400
10. Minimum number of times a fair coin must
be tossed so that the probability of getting
atleast one head is more than 99% is
[JEE Main 2019, 10 April Shift-II]
(b) 6
(d) 5
Q P(Tail ) = P(Head) = 1 

2 
According to the question,
n
99
1
1
99
1 −   >
⇒   < 1 −
2
100
100  2 
⇒
⇒
12. Let a random variable X have a binomial
distribution with mean 8 and variance 4. If
k
P ( X ≤ 2 ) = 16 , then k is equal to
2
[JEE Main 2019, 12 April Shift-I]
As we know probability of getting a head on a toss
1
of a fair coin is P(H) = = p (let)
2
Now, let n be the minimum numbers of toss
required to get at least one head, then required
probability = 1 − (probability that on all ‘n’ toss we
are getting tail)
n
∴Required probability
2 × 3×2 × 3×2
2
2
1
= 6
=
=
=
6!
6 × 5 × 4 × 3 × 2 × 1 10
C3
3! 3!
(a) 17
(c) 1
Exp. (c)
n
Let for the given random variable ‘X’ the binomial
probability distribution have n-number of
independent trials and probability of success and
failure are p and q respectively. According to the
question,
Mean = np = 8
and variance = npq = 4
1
1
q = ⇒ p = 1− q =
∴
2
2
1
Now, n × = 8 ⇒ n = 16
2
1
P( X = r ) =16 C r  
2
n
[for minimum]
16
are chosen at random, then the probability
that the triangle formed with these chosen
vertices is equilateral is
[JEE Main 2019, 12 April Shift-I]
1
10
3
(c)
10
16
∴P( X ≤ 2 ) = P( X = 0) + P( X = 1) + P( X = 2 )
11. If there of the six vertices of a regular hexagon
(a)
(b) 121
(d) 137
Exp. (d)
 1  < 1 ⇒2 n > 100
 
2
100
n=7
A3
A4
[from Eqs. (i) and (ii)]
1
= 1 −  
2
A2
A5
16
= 512 − (8 × 16) + 
× 50
 50

(a) 8
(c) 7
A1
1
5
3
(d)
20
(b)
Exp. (a)
Since, there is a regular hexagon, then the number
of ways of choosing three vertices is 6 C 3 . And,
there is only two ways i.e. choosing vertices of a
⇒
16
1
1
1
=16 C 0   + 16 C1   + 16 C 2  
2
2
2
k
1 + 16 + 120 137
=
= 16 = 16 (given)
216
2
2
k = 137
16
13. If the data x1 , x 2 ,… , x10 is such that the mean
of first four of these is 11, the mean of the
remaining six is 16 and the sum of squares of
all of these is 2000, then the standard
deviation of this data is
[JEE Main 2019, 12 April Shift-I]
(a) 2 2
(c) 4
(b) 2
(d) 2
322
JEE Main Chapterwise Mathematics
Now, according to binomial probability distribution
concept
50 − r
r
1
4
, r = 0, 1, ..., 50
P( X = r ) = 50C r    
 5  5
Exp. (b)
Key Idea Formula of standard deviation (σ ),
for n observations =
Σxi2  Σxi 
−

 n 
n
2
Given 10 observations are x1, x2 , x3 , K , x10
x1 + x2 + x3 + x4
∴
= 11
4
… (i)
⇒
x1 + x2 + x3 + x4 = 44
x5 + x6 + x7 + x8 + x9 + x10
and
= 16
6
… (ii)
⇒ x5 + x6 + x7 + x8 + x9 + x10 = 96
So, mean of given 10 observations
44 + 96 140
=
=
= 14
10
10
Since, the sum of squares of all the observations
= 2000
2
… (iii)
∴
x12 + x22 + x32 + K + x10
= 2000
Σxi2  Σxi 
−

 10 
10
2000
2
=
− (14) = 200 − 196 = 4
10
σ =2
Now, σ 2 = (standard deviation)2 =
So,
2
14. For an initial screening of an admission test,
a candidate is given fifty problems to solve.
If the probability that the candidate can
4
solve any problem is , then the probability
5
that he is unable to solve less than two
problem is
[JEE Main 2019, 12 April Shift-II]
(a)
(c)
201  1 
 
5  5
54  4 
 
5  5
49
316  4 
 
25  5 
48
(b)
164  1 
 
25  5 
48
(d)
49
Exp. (c)
Given that, there are 50 problems to solve in an
admission test and probability that the candidate
4
can solve any problem is
= q (say). So,
5
probability that the candidate cannot solve a
4 1
problem is p = 1 − q = 1− = .
5 5
Now, let X be a random variable which denotes
the number of problems that the candidate is
unable to solve. Then, X follows binomial
1
distribution with parameters n = 50 and p = .
5
∴Required probability
= P( X < 2 ) = P( X = 0) + P( X = 1)
=
4
C 0  
 5
50
4
=  
 5
49
50
+
449
50
C1
(5)50
 4 + 50  = 54


5 5 
5
 4
 
 5
49
15 A person throws two fair dice. He wins
` 15 for throwing a doublet (same numbers
on the two dice), wins ` 12 when the throw
results in the sum of 9, and loses ` 6 for any
other outcome on the throw. Then, the
expected gain/loss (in `) of the person is
[JEE Main 2019, 12 April Shift-II]
1
(a) gain
2
1
(c) loss
2
(b)
1
loss
4
(d) 2 gain
Exp. (c)
It is given that a person wins
`15 for throwing a doublet (1, 1) (2, 2), (3, 3),
(4, 4), (5, 5), (6, 6) and win `12 when the throw
results in sum of 9, i.e., when (3, 6), (4, 5),
(5, 4), (6, 3) occurs.
Also, losses `6 for throwing any other outcome,
i.e., when any of the rest
36 − 6 − 4 = 26
outcomes occurs.
Now, the expected gain/loss = 15 × P (getting a
doublet) + 12 × P (getting sum 9) − 6 × P
(getting any of rest 26 outcome)
26 
4 
6 

=  15 ×
 +  12 ×
 − 6 ×


36 
36  
36  
5 4 26 15 + 8 − 26
= + −
=
2 3
6
6
23 − 26
3
1
1
=
= − = − , means loss of `
6
6
2
2
16. 5 students of a class have an average height
150 cm and variance 18 cm 2. A new student,
whose height is 156 cm, joined them. The
variance (in cm 2) of the height of these six
students is
[JEE Main 2019, 9 Jan Shift-I]
(a) 16
(c) 20
(b) 22
(d) 18
323
Statistics and Probability
x
1
12
Now, P( X = x) = 2C x    
 13   13 
Exp. (c)
Let x1, x2 , x3 , x4 , x5 be the heights of five students.
Then, we have
∴
x=
Mean,
i =1
5
5
= 150 ⇒ Σ xi = 750
1
5
Σ xi2
i =1
variance =
and
n
5
⇒
Σ xi2
i =1
5
− ( x )2
− (150)2 = 18
Σ xi2 = 112590
…(ii)
i =1
6
Now, new mean =
Σ xi
i =1
6
5
=
Σ xi + 156
i =1
6
xnew
⇒
and new variance
750 + 156
6
= 151
=
=
i =1
[using Eq. (i)]
− ( xnew )2 =
Σ xi2 + (156)2
i =1
6
6
112590 + (156)2
=
− (151)2
6
= 22821 − 22801 = 20
random a ball is drawn from this pot. If a
drawn ball is green then put a red ball in the
pot and if a drawn ball is red, then put a
green ball in the pot, while drawn ball is not
replace in the pot. Now we draw another ball
randomly, the probability of second ball to
be red is
[JEE Main 2019, 9 Jan Shift-II]
(a)
[JEE Main 2019, 9 Jan Shift-I]
49
(c)
169
26
49
(c)
21
49
(d)
32
49
24
(d)
169
Exp. (a)
Let p = probability of getting an ace in a draw =
probability of success
and q = probability of not getting an ace in a draw
= probability of failure
4
1
and q = 1 − p
Then, p =
=
52 13
1 12
= 1−
=
13 13
Here, number of trials, n = 2
Clearly, X follows binomial distribution with
1
parameter n = 2 and p = .
13
2
7
P(R1) = Probability that Ist ball drawn in red
5
=
7
P(R ) = Probability that 2nd ball drawn is red.
4
6
= P(G1) ⋅ + P(R1) ⋅
7
7
2 6 5 4 32
= × + × =
7 7 7 7 49
[using Eq. (ii)]
replacement from a well shuffled deck of 52
cards. Let X denote the random variable of
number of aces obtained in the two drawn
cards. Then, P ( X = 1) + P ( X = 2 ) equals
52
(b)
169
(b)
P(G1) = Probability that Ist ball drawn is green =
− (151)2
17. Two cards are drawn successively with
25
(a)
169
27
49
Exp. (d)
5
6
Σ xi2
0
18. A pot contain 5 red and 2 green balls. At
5
⇒
2
1
12
1
12
= 2C1     + 2C 2    
 13   13 
 13   13 
12 
1
= 2 
 +
 169  169
24
1
25
=
+
=
169 169 169
…(i)
i =1
, x = 0, 1, 2
P( X = 1) + P( X = 2 )
5
Σ xi
2− x
19. In a group of data, there are n observations,
n
x , x 2 ,...., xn . If Σ ( xi + 1)2 = 9n and
i =1
n
Σ ( xi − 1) = 5n, the standard deviation of
2
i =1
the data is
[JEE Main 2019, 9 Jan Shift-II]
(a) 2
(c) 5
(b) 7
(d) 5
Exp. (d)
We have,
n
∑ ( xi + 1)
2
= 9n
...(i)
= 5n
...(ii)
i =1
n
and
∑ ( xi − 1)
i =1
2
324
JEE Main Chapterwise Mathematics
On subtracting Eq. (ii) from Eq. (i) is, we get
n
∑ {( xi + 1)
⇒
2
− ( xi − 1) } = 4n
2
i =1
⇒
n
n
n
i =1
i =1
∑ xi
∑ 4 xi = 4n ⇒ ∑ xi = n ⇒
i =1
n
=1
21. The mean of five observations is 5 and their
∴mean ( x ) = 1
n
Now, standard deviation =
∑ ( xi − x)
2
i =1
=
i =1
n
=
(a) 4 : 9
(c) 10 : 3
5n
= 5
n
20. An unbiased coin is tossed. If the outcome is
a head, then a pair of unbiased dice is rolled
and the sum of the numbers obtained on
them is noted. If the toss of the coin results
in tail, then a card from a well-shuffled pack
of nine cards numbered 1, 2, 3, …, 9 is
randomly picked and the number on the
card is noted. The probability that the noted
number is either 7 or 8 is
[JEE Main 2019, 10 Jan Shift-I]
15
(a)
72
13
(b)
36
19
(c)
72
variance is 9.20. If three of the given five
observations are 1, 3 and 8, then a ratio of
other two observations is
[JEE Main 2019, 10 Jan Shift-I]
n
n
2
∑ ( xi − 1)
= P(H) ⋅ P(E1 ) + P(T ) ⋅ P(E2 )
[Q{H, E1} and {T, E2 } both are sets of
independent events]
1 11 1 2 19
= ×
+ × =
2 36 2 9 72
19
(d)
36
Exp. (c)
Exp. (a)
Let 1, 3, 8, x and y be the five observations.
Σx i
Then, mean x =
n
1+ 3 + 8 + x + y
(given)
⇒
x=
=5
5
⇒
x + y = 25 − 12 = 13
…(i)
⇒
x + y = 13
2
Σ( xi. − x )
and variance = σ 2 =
n
(1 − 5)2 + (3 − 5)2 + (8 − 5)2 


+ ( x − 5)2 + ( y − 5)2 

= 
= 92
. (given)
5
⇒16 + 4 + 9 + ( x2 − 10 x + 25) +
( y2 − 10 y + 25) = 46
Clearly,
P(H) = Probability of getting head =
(b) 6 : 7
(d) 5 : 8
1
2
1
2
Now, let E1 be the event of getting a sum 7 or 8,
when a pair of dice is rolled.
Then, E1 = {(6, 1), (5, 2), (4, 3), (3, 4), (2, 5),
(1, 6), (6, 2), (5, 3), (4, 4), (3, 5), (2, 6)}
⇒ P(E1 ) = Probability of getting 7 or 8 when a pair
11
of dice is thrown =
36
Also, let P(E2 ) = Probability of getting 7 or 8 when a
card is picked from cards numbered
2
1, 2, ...., 9 =
9
∴Probability that the noted number is 7 or 8
= P ((H ∩ E1 ) or (T ∩ E2 ))
= P (H ∩ E1 ) + P (T ∩ E2 )
[Q(H ∩ E1 ) and (T ∩ E2 ) are mutually exclusive]
and P(T ) = Probability of getting tail =
⇒
⇒
x2 + y2 − 10( x + y) = 46 − 79
x2 + y2 − 10 × 13 = − 33 (Qx + y = 13)
⇒
x2 + y2 = 97
…(ii)
y
Let
= t ⇒ y = xt
x
Putting y = xt in Eq. (i), we get
x(1 + t ) = 13
⇒
x (1 + t )2 = 169
2
…(iii)
Putting y = xt in Eq. (ii), we get
x2 (1 + t 2 ) = 97
Dividing Eq (iii) by Eq. (iv), we get
x2 (1 + t )2
x2 (1 + t 2 )
⇒
=
169
97
97(t 2 + 2t + 1) = 169 (1 + t 2 )
⇒ (169 − 97 ) t 2 − 194 t + (169 − 97 ) = 0
… (iv)
325
Statistics and Probability
⇒
36t 2 − 97 t + 36 = 0
⇒
(4t − 9) (9t − 4) = 0
t =
⇒
and standard deviation
5
SD =
9
4
or t =
4
9
5
22. If the probability of hitting a target by a
1
shooter in any shot, is , then the minimum
3
number of independent shots at the target
required by him so that the probability of
hitting the target at least once is greater than
5
, is
[JEE Main 2019, 10 Jan Shift-II]
6
(a) 6
(b) 3
(c) 5
n
n
n
 2  < 1− 5 ⇒  2  < 1
 
 
 3
 3
6
6
Clearly, minimum value of n is 5.
observations x1 , x 2 , x 3 , x 4 , x 5 are 10 and 3,
respectively, then the variance of 6
observations x1 , x 2 ,..... x 5 and − 50 is equal to
[JEE Main 2019, 10 Jan Shift-II]
(a) 507.5
(c) 582.5
(b) 586.5
(d) 509.5
Exp. (a)
The mean of five observation, x = 10
(given)
x1 + x2 + x3 + x4 + x5
= 10
5
x1 + x2 + x3 + x4 + x5 = 50
… (i)
⇒
⇒
(given)
5
∑ xi
2
= 5 × 109
5
∑ xi
2
= 545
… (ii)
Now, variance of 6 observations x1, x2 , x3 , x4 , x5
and − 50, is equal to
5
2
2
∑ xi + (− 50)
i =1
6

 5
 ∑ xi − 50 

 i =1
−

6






=
545 + 2500  50 − 50 
−


6
6 
=
3045
= 507.5
6
2
2
[from Eqs. (i) and (ii)]
24. The outcome of each of 30 items was
observed ; 10 items gave an outcome
1
−d
2
1
each and the
2
1
remaining 10 items gave outcome
+d
2
each. If the variance of this outcome data is
4
, then |d | equals
[JEE Main 2019, 11 Jan Shift-I]
3
each, 10 items gave outcome
(a)
23. If mean and standard deviation of 5
− ( x )2 = 3
i =1
The probability of hitting a target at least once
= 1 − (probability of not hitting the target in any trial)
= 1 − n C 0 p0q n
0
5
i =1
⇒
σ2 =
1
2
5
1 − nC 0     >
 3  3
6
i =1
− 100 = 9 ⇒
5
(d) 4
where n is the number of independent trials and p
and q are the probability of success and failure
respectively.
[by using binomial distribution]
1
Here,
p=
3
1 2
and
q = 1− p = 1− =
3 3
According to the question,
2
2
i =1
⇒
Exp. (c)
⇒
∑ xi
∑ xi
2
3
(b)
5
2
(c)
2
(d) 2
Exp. (c)
Σx2
− µ2
n
2
2
1
1
1
10  – d  + 10 × + 10  + d 
2

2

4
=
30
2
1
 1
1


 10  − d  + 10 × + 10  + d  
2

2

2
−


30




Σx
Qµ = i 

n 
We know, σ 2 =
326
JEE Main Chapterwise Mathematics
=
=
∴
⇒
1
20  + d 2  + 5 / 2
4

−
30
15
+ 20d 2
1 1
2
− = +
30
4 4
2 2 4
d =
3
3
d 2 = 2 ⇒ |d| = 2
 1
 
 4
2d 2 1 2 2
− = d
3
4 3
25. Two integers are selected at random from
the set { 1, 2, …… , 11}. Given that the sum of
selected numbers is even, the conditional
probability that both the numbers are even
is
[JEE Main 2019, 11 Jan Shift-I]
(a)
2
5
(b)
1
2
(c)
7
10
(d)
3
5
Exp. (a)
In {1, 2, 3, ...., 11} there are 5 even numbers and 6
odd numbers. The sum even is possible only
when both are odd or both are even.
Let A be the event that denotes both numbers are
even and B be the event that denotes sum of
numbers is even.
Then, n( A) = 5C 2 and n(B) = 5C 2 + 6C 2
Required probability
5
C / 11C
P( A ∩ B)
P( A / B) =
= 6 2 5 2
P(B)
( C2 + C2 )
11
C2
5
C2
10
2
= 6
=
=
C 2 + 5C 2 15 + 10 5
26. A bag contains 30 white balls and 10 red
balls. 16 balls are drawn one by one
randomly from the bag with replacement. If
X be the number of white balls drawn,


mean of X
then 
 is equal to
 standard deviation of X 
[JEE Main 2019, 11 Jan Shift-II]
4 3
(a)
3
(b) 4
(c) 3 2
(d) 4 3
Exp. (d)
Number of white balls = 30
and number of red balls = 10
Let p = probability of success in a trial = probability
30 3
of getting a white ball in a trial =
= .
40 4
and q = probability of failure in a trial
3 1
= 1− p= 1− =
4 4
Here, n = number of trials = 16.
Clearly, X follows binomial distribution with
3
parameter n = 16 and p = .
4
3
∴Mean of X = np, = 16. = 12
4
3 1
and variance of X = npq = 16 ⋅ ⋅ = 3
4 4
12
mean of X
Now,
=
=4 3
3
standard deviation of X
[Q SD = variance ]
27. Let S = {1, 2 ,K , 20}. A subset B of S is said to
be “nice”, if the sum of the elements of B is
203. Then, the probability that a randomly
chosen subset of S is ‘‘nice’’, is
[JEE Main 2019, 11 Jan Shift-II]
(a)
(c)
6
(b)
2 20
7
(d)
2 20
4
2 20
5
2 20
Exp. (d)
Number of subset of S = 2 20
Sum of elements in S is
20(21)
1 + 2 + .....+20 =
= 210
2
Q1 + 2 + ...... + n = n(n + 1)

2 
Clearly, the sum of elements of a subset would be
203, if we consider it as follows
S − {7}, S − {1, 6} S − {2, 5}, S − {3, 4}
S − {1, 2, 4}
∴Number of favourables cases = 5
5
Hence, required probability = 20
2
28. If the sum of the deviations of 50
observations from 30 is 50, then the mean of
these observations is
[JEE Main 2019, 12 Jan Shift-I]
(a) 50
(b) 30
(c) 51
(d) 31
Exp. (d)
Let the 50 observations are x1, x2 , x3 , ..., x50 .
Now, deviations of these observations from 30 are
( x1 − 30), ( x2 − 30), ( x3 − 30), ... , ( x50 − 30)
327
Statistics and Probability
50
∑ ( xi − 30) = 50
Q
(given)
i =1
50
∑ xi − (30 × 50) = 50
⇒
i =1
⇒
50
∑ xi = 50(30 + 1) = 50 × 31
i =1
∴ Mean of 50 observations = ( x ) =
50
∑ xi
i =1
50
= 31
29. In a random experiment, a fair die is
rolled until two fours are obtained in
succession. The probability that the
experiment will end in the fifth throw of the
die is equal to
[JEE Main 2019, 12 Jan Shift-I]
(a)
175
65
(b)
225
65
(c)
200
65
(d)
(c) 0
[Qwhen a pair of dice is thrown
then (4, 4) occur only once]
5 × 35 175
Hence, the required probability =
= 5
65
6
30. In a class of 60 students, 40 opted for NCC,
30 opted for NSS and 20 opted for both NCC
and NSS. If one of these students is selected
at random, then the probability that the
student selected has opted neither for NCC
nor for NSS is
[JEE Main 2019, 12 Jan Shift-II]
1
3
(c)
2
3
(d)
[JEE Main 2019, 12 Jan Shift-II]
65
So, the third throw can be 1, 2, 3, 5 or 6 (not 4).
Also, throw number (i) and (ii) can not take two
fours in succession, therefore number of
possibililites for throw (i) and (ii) = 62 − 1 = 35
(b)
on a throw of a fair die and loses ` 50 for
getting any other number on the die. If he
decides to throw the die either till he gets a
five or a six or to a maximum of three throws,
then his expected gain/loss (in rupees) is
150
Since, the experiment should be end in the
fifth throw of the die, so total number of outcomes
are 65
Now, as the last two throws should be result in two
4 4
fours
(i) (ii) (iii) (iv) (v)
1
6
31. In a game, a man wins ` 100 if he gets 5 or 6
400
(a)
loss
3
Exp. (a)
(a)
Now, n(C ∪ S ) = n(C ∪ S )
= n(∪) − n(C ∪ S )
= 60 − [n(C ) + n(S ) − n(C ∩ S )]
= 60 − [40 + 30 − 20] = 10
So, required probability
10 1
=
=
60 6
5
6
Exp. (a)
Let C and S represent the set of students who
opted for NCC and NSS respectively.
Then, n(C ) = 40, n(S ) = 30, n(C ∩ S ) = 20
and n(U ) = 60
400
loss
9
400
gain
(d)
3
(b)
Exp. (c)
Let p and q represents the probability of success
and failure in a trial respectively. Then,
2 1
4 2
p = P(5 or 6) = = and q = 1 − p = = .
6 3
6 3
Now, as the man decides to throw the die either till
he gets a five or a six or to a maximum of three
throws, so he can get the success in first, second
and third throw or not get the success in any of the
three throws.
So, the expected gain/loss (in `)
= ( p × 100) + qp(− 50 + 100)
+ q 2 p(− 50 − 50 + 100) + q 3 (− 50 − 50 − 50)
2
2
1
2 1
1
=  × 100 +  ×  (50) +     (0)
 3  3
3
  3 3
3
2
+   (− 150)
 3
100 100
1200
=
+
+ 0−
3
9
27
900 + 300 − 1200 1200 − 1200
=
=
=0
27
27
32. The mean and the variance of five
observations are 4 and 5.20, respectively. If
three of the observations are 3, 4 and 4, then
the absolute value of the difference of the
other two observations, is
[JEE Main 2019, 12 Jan Shift-II]
(a) 1
(c) 5
(b) 7
(d) 3
328
JEE Main Chapterwise Mathematics
A = Event that second ball drawn is red
6
4  A
P(E1 ) = , P  =
10  E1  12
Exp. (b)
Given mean x = 4
variance σ 2 = 5.20
and numbers of observation n = 5
Let x1 = 3, x2 = 4, x3 = 4 and x4 , x5 be the five
observations
⇒
∑ xi = 5 ⋅ x = 5 × 4 = 20
i =1
⇒ x1 + x2 + x3 + x4 + x5 = 20
⇒
3 + 4 + 4 + x4 + x5 = 20
⇒
x4 + x5 = 9
5
Now, variance σ 2 =
∑
…(i)
34. If
xi2
i =1
− ( x )2
5
x12 + x22 + x32 + x42 + x52
⇒
− (4)2 = 520
.
5
9 + 16 + 16 + x42 + x52
= 16 + 520
.
⇒
5
⇒
41 + x42 + x52 = 5 × 21. 20
⇒
x42 + x52 = 106 − 41
⇒
x42 + x52 = 65
Q
( x4 + x5 ) =
2
∴
x42
…(ii)
+
x52
9
9
i =1
i =1
∑(xi − 5) = 9 and ∑(xi − 5)2 = 45, then the
standard deviation of the 9 items x1 , x 2 ,…, x 9
is
[JEE Main 2018]
(a) 9
(b) 4
We have,
+ 2 x4 x5
9
9
i =1
i =1
∑ (x1 − 5) = 9 and ∑ (x1 − 5)
9
SD =
[from Eqs. (ii) and (iii)]
33. A bag contains 4 red and 6 black balls. A ball
is drawn at random from the bag, its colour
is observed and this ball along with two
additional balls of the same colour are
returned to the bag. If now a ball is drawn at
random from the bag, then the probability
that this drawn ball is red, is [JEE Main 2018]
(a)
3
10
(b)
2
5
(c)
1
5
(d)
3
4
Exp. (b)
Key idea Use the theorem of total probability
Let E1 = Event that first ball drawn is red
E2 = Event that first ball drawn is black
(d) 3
Key idea Standard deviation is remain
unchanged, if observations are added or
subtracted by a fixed number.
Now, (| x4 − x5|)2 = x42 + x52 − 2 x4 x5
= 65 − 16
= 49
⇒ | x4 − x5| = 7
(c) 2
Exp. (c)
81 = 65 + 2 x4 x5
[from Eqs. (i) and (ii)]
16 = 2 x4 x5
…(iii)
x4 x5 = 8
⇒
⇒
6  A
4
, P  =
10  E2  12
By law of total probability
 A
 A
P( A) = P(E1 ) × P  + P(E2 ) × P 
 E1 
 E2 
48 2
4
6
6
4 24 + 24
=
×
+
×
=
=
=
120
120 5
10 12 10 12
5
So,
P(E2 ) =
2
∑ (x1 − 5)
i =1
9
2
= 45

 9
 ∑ ( x1 − 5)


−  i =1

9






⇒
SD =
45  9 
− 
9  9
2
⇒
SD =
5−1=
4 =2
2
35. For three events A , B andC , if P (exactly one
of A or B occurs) = P(exactly one of B or C
1
occurs) = P (exactly one ofC or A occurs) =
4
and P (all the three events occur
1
simultaneously) = , then the probability
16
that atleast one of the events occurs, is
[JEE Main 2017]
7
(a)
32
7
(c)
64
7
(b)
16
3
(d)
16
329
Statistics and Probability
37. A box contains 15 green and 10 yellow balls.
Exp. (b)
We have, P (exactly one of A or B occurs)
= P( A ∪ B) − P( A ∩ B)
= P( A) + P(B) − 2 P( A ∩ B)
According to the question,
1
…(i)
P( A) + P(B) − 2 P( A ∩ B) =
4
1
…(ii)
P(B) + P(C ) − 2 P(B ∩ C ) =
4
1
and P(C ) + P( A) − 2 P(C ∩ A) =
…(iii)
4
On adding Eqs. (i), (ii) and (iii), we get
2 [P( A) + P(B) + P(C ) − P( A ∩ B) − P(B ∩ C )
3
− P(C ∩ A)] =
4
⇒ P( A) + P(B) + P(C ) − P( A ∩ B) − P(B ∩ C )
3
− P(C ∩ A) =
8
∴P (atleast one event occurs) = P( A ∪ B ∪ C )
= P( A) + P(B) + P(C ) − P( A ∩ B) − P(B ∩ C )
− P(C ∩ A) + P( A ∩ B ∩ C )
3
1
7
Q P( A ∩ B ∩ C ) = 1 
= +
=

16 
8 16 16
36. If two different numbers are taken from the
set {0, 1, 2, 3, …, 10}, then the probability that
their sum as well as absolute difference are
both multiple of 4, is [JEE Main 2017 (offline)]
(a)
6
55
(b)
12
55
(c)
14
45
(d)
7
55
Exp. (a)
If 10 balls are randomly drawn one-by-one
with replacement, then the variance of the
number of green balls drawn is
[JEE Main 2017 (offline)]
12
(a)
5
(b) 6
(c) 4
(d)
6
25
Exp. (a)
Given box contains 15 green and 10 yellow balls.
∴Total number of balls = 15 + 10 = 25
15 3
P(green balls) =
= = p = Probability of
25 5
success
10 2
P(yellow balls) =
= = q = Probability of
25 5
unsuccess and n = 10 = Number of trials.
3 2 12
∴Variance = npq = 10 × × =
5 5 5
38. If the standard deviation of the numbers 2,
3, a and 11 is 3.5, then which of the following
is true?
[JEE Main 2016 (offline)]
(a) 3a 2 − 26a + 55 = 0
(b) 3a 2 − 32a + 84 = 0
(c) 3a 2 − 34a + 91 = 0
(d) 3a 2 − 23a + 44 = 0
Exp. (b)
Total number of ways of selecting 2 different
numbers from {0, 1, 2, ..., 10} = 11C 2 = 55
Let two numbers selected be x and y.
Then,
…(i)
x + y = 4m
and
…(ii)
x − y = 4n
⇒ 2 x = 4(m + n) and 2 y = 4(m − n)
⇒
x = 2(m + n) and y = 2(m − n)
∴ x and y both are even numbers.
x
y
0
4, 8
2
6, 10
4
0, 8
6
2, 10
8
0, 4
10
2, 6
∴Required probability =
6
55
We know that, if x1, x2 , ..., xn are n observations,
then their standard deviation is given by
1 2  Σxi 
Σxi − 

 n 
n
We have, (3.5)2 =
⇒
⇒
⇒
2
(2 2 + 32 + a2 + 112 )
4
2
2 + 3 + a + 11
− 



4
49 4 + 9 + a2 + 121  16 + a 
=
−

 4 
4
4
49 134 + a2 256 + a2 + 32 a
=
−
4
4
16
49 4a2 + 536 − 256 − a2 − 32 a
=
4
16
⇒ 49 × 4 = 3a2 − 32 a + 280
⇒
3a2 − 32 a + 84 = 0
2
330
JEE Main Chapterwise Mathematics
39. Let two fair six-faced dice A and B be thrown
simultaneously. If E 1 is the event that die A
shows up four, E 2 is the event that die B
shows up two and E 3 is the event that the
sum of numbers on both dice is odd, then
which of the following statements is not
true?
[JEE Main 2016 (offline)]
(a) E 1 and E 2 are independent
(b) E 2 and E 3 are independent
(c) E 1 and E 3 are independent
(d) E 1 , E 2 and E 3 are not independent
identical boxes, then the probability that
one of the boxes contains exactly 3 balls, is
[JEE Main 2015]
10
12
1
1
(c) 220  (d) 22  
 3
 3
12
C3 × 2 9
12
3
=
According to the question,
C1 ×
C 3 2 9 − 3C 212C 3 9C 3 +
12
55  2 
 
3  3
11
12 ! × 3!
3! 3! 6! 3!
41. The mean of the data set comprising of 16
40. If 12 identical balls are to be placed in 3
2
(b) 55 
 3
Then, number of ways =
312
Clearly, E1 = {(4, 1),(4, 2 ),(4, 3),(4, 4),(4, 5),(4, 6)}
E2 = {(1, 2 ),(2, 2 ),(3, 2 ),(4, 2 ),(5, 2 ),(6, 2 )}
and
E3 = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3),
(2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5),
(5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}
6
1
6
1
⇒ P(E1 ) =
= , P(E2 ) =
=
36 6
36 6
18 1
and P(E3 ) =
=
36 2
Now, P(E1 ∩ E2 ) = P
(Probability of getting 4 on die A and 2 on die B)
1
=
= P(E1 ) ⋅ P(E2 )
36
P(E2 ∩ E3 ) = P
(Probability of getting 2 on die B and sum of
numbers on both dice is odd)
3
=
= P(E2 ) ⋅ P(E3 )
36
P(E1 ∩ E3 ) = P
(getting 4 on die A and sum of numbers
on both dice is odd)
3
=
= P(E1 ) ⋅ P(E3 )
36
and P(E1 ∩ E2 ∩ E3 ) = P
(getting 4 on die A, 2 on die B
and sum of numbers is odd)
= P (impossible event) = 0
Hence, E1, E2 and E3 are not independent.
11
There seems to be ambiguity in this question. It
should be mentioned that boxes are different
and one particular box has 3 balls.
3
Exp. (d)
55  2 
(a)
 
3  3
Exp. (a)
11
observations is 16. If one of the observation
valued 16 is deleted and three new
observations valued 3, 4 and 5 are added to
the data, then the mean of the resultant data
is
[JEE Main 2015]
(a) 16.8
(b) 16.0
(c) 15.8
(d) 14.0
Exp. (d)
Given,
x1 + x2 + x3 + ... + x16
= 16
16
16
∑ xi = 16 × 16
⇒
i=1
Sum of new observations
=
18
∑ yi = (16 × 16 − 16) + (3 + 4 + 5) = 252
i =1
Number of observations = 18
18
∴
New mean =
∑ yi
i =1
18
=
252
= 14
18
42. Let A and B be two events such that
1
1
1
and P ( A ) = ,
P (A ∪ B ) = , P (A ∩ B ) =
6
4
4
where A stands for the complement of the
event A. Then, the events A and B are
(a)
(b)
(c)
(d)
independent but not equally likely
independent and equally likely
mutually exclusive and independent
equally likely but not independent
[JEE Main 2014]
Exp. (a)
1
1
1
, P ( A ∩ B) = , P ( A ) =
6
4
4
1 5
∴ P ( A ∪ B) = 1 − P ( A ∪ B) = 1 − =
6 6
1 3
and P ( A) = 1 − P ( A ) = 1 − =
4 4
P ( A ∪ B) =
331
Statistics and Probability
P ( A ∪ B) = P ( A) + P (B) − P ( A ∩ B)
Exp. (d)
1
5 3
1
⇒
= + P (B) − ⇒ P (B) =
3
6 4
4
⇒ A and B are not equally likely.
1
Also, P ( A ∩ B) = P ( A) ⋅ P(B) =
4
So, events are independent.
43. The variance of first 50 even natural
numbers is
833
(a)
4
[JEE Main 2014]
(b) 833
(c) 437
(d)
437
4
Exp. (b)
1
Σ xi2 − ( x )2
n
Σx 2 + 4 + 6 + 8 + ... + 100
X= i=
50
n
50 × 51
= 51
=
50
[QΣ 2 n = n (n + 1), here n = 50]
1 2
2
(2 + 42 + ... + 1002 ) − (51)2
σ =
50
= 833
44. A
multiple choice examination has
5 questions. Each question has three
alternative answers of which exactly one is
correct. The probability that a student will
get 4 or more correct answers just by
guessing is
[JEE Main 2013]
17
35
(b)
13
35
(c)
11
35
(d)
10
35
Exp. (c)
1
3
2
and probability of guessing a wrong answer, q =
3
So, the probability of guessing 4 or more correct
answers
4
5
1
2
1
2
1 11
= 5C 4   ⋅ + 5C 5   = 5 ⋅ 5 + 5 = 5
 3
 3 3
3
3
3
Probability of guessing a correct answer, p =
45. All the students of a class performed poorly
in Mathematics. The teacher decided to give
grace marks of 10 to each of the students.
Which of the following statistical measure
will not change even after the grace marks
were given?
[JEE Main 2013]
(a) Mean
(c) Mode
Σ ( xi − x )2
N
Now, each is increased by 10.
Σ [ xi + 10) − ( x + 10)]2
∴New
= σ 2i
σ i2 =
N
So, variance will not change whereas mean,
median and mode will increase by 10.
σ 2i =
46. Let x1 , x 2 ,..., xn be n observations x be their
arithmetic mean and σ 2 be the variance.
Variance, σ 2 =
(a)
If initially all marks were xi, then
(b) Median
(d) Variance
Statement I
is 4σ 2.
Variance of 2 x1 , 2 x 2 ,..., 2 xn
Statement II Arithmetic mean of
[AIEEE 2012]
2 x1 , 2 x 2 ,..., 2 xn is 4x.
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation for
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation for
Statement I
(d) Statement I is true, Statement II is false
Exp. (d)
Given, x is the AM and σ 2 is the variance of
n observations x1, x2 , x3 , ..., xn . AM of
2 x1, 2 x2 , 2 x3 , ..., 2 xn
2 x + 2 x2 + 2 x3 + K + 2 xn
= 1
n
x1 + x2 + x3 + K + xn 

=2 
 = 2x


n
Hence, it is obvious that Statement II is false.
Variance of 2 x1, 2 x2 , 2 x3 , ..., 2 xn = Variance
(2 xi ) = 2 2 Variance ( xi ) = 4σ 2
Hence, Statement I is correct.
Finally, Statement I is true and Statement II is false.
47. Three numbers are chosen
at random
without replacement from numbers
{1, 2, 3,..., 8}. The probability that their
minimum is 3, given that their maximum is
6, is
[AIEEE 2012]
(a)
3
8
(b)
1
5
(c)
1
4
(d)
2
5
332
JEE Main Chapterwise Mathematics
Exp. (b)
Exp. (b)
Given 3 numbers are drawn without replacement
from the set {1, 2 , 3, 4, 5, 6, 7, 8}.
To find The probability that their minimum is
3 given that their maximum is 6.
Let A denotes the event that the minimum of the
3 selected numbers is 3 and B denotes the event
that the maximum of the 3 selected numbers is 6,
then
A
The required probability is P   ⋅
 B
P ( A ∩ B)
A
Now,
P   =
 B
P(B)
P(B) = The probability that the maximum number
selected is 6
If the maximum number is 6, then
(i) 6 should be one of the 3 selected numbers
(1 way) and
(ii) the remaining two numbers should be less
than 6 i.e., any 2 from 1 to 5 ( 5 C 2 ways).
⇒
P(B) =
1 × 5C 2
8
C3
=
10
8
C3
Similarly, P( A ∩ B) = The probability that the
minimum number is 3 and the maximum number
is 6.
If the minimum number is 3 and the maximum
number is 6, then
(i) 6 should be one of the 3 selected numbers
(1 way).
(ii) 3 should be one of the 3 selected numbers
(1 way) and
(iii) the remaining 1 number should lie between 3
and 6 i.e., any one of 4 and 5 (2 ways).
1 × 1 × 2C1
2
⇒ P( A ∩ B) =
= 8
8
C3
C3
P( A ∩ B) 2
A
1
P   =
=
=
 B
P(B)
10 5
1
Hence, the required probability is .
5
∴
48. Consider 5 independent Bernoulli’s trials
each with probability of success p. If the
probability of atleast one failure is greater
31
than or equal to , thenp lies in the interval
32
3 11
1
(a)  ,  (b) 0, 
 2 
 4 12 
11
(c)  ,1
 12 
1 3
(d)  , 
 2 4 
[AIEEE 2011]
Here, n = 5 and r ≥ 1
∴
P( X = r ) = nC r pn − r q r
P ( X ≥ 1) = 1 − P ( x = 0)
= 1 − 5C 0 ⋅ p5 ⋅ q 0 ≥
⇒
p5 ≤ 1 −
∴
p≤
31
1
=
32 32
31
[given]
32
1
1
and p ≥ 0 ⇒ p ∈  0, 
 2 
2
49. If C and D are two events such that C ⊂ D
and P ( D ) ≠ 0, then the correct statement
among the following is
[AIEEE 2011]
(a) P (C / D ) ≥ P (C )
P(D )
(c) P (C / D ) =
P(C )
(b) P (C / D ) < P (C )
(d) P (C / D ) = P(C )
Exp. (a)
QC ⊂ D

∴ P (C ∩ D) = P(C )


P (C ∩ D)
C
As, P   =
 D
P (D)
P (C )
=
P (D)
Also, as
∴
…(i)
P (D) ≤ 1
1
P(C )
≥ P(C )
≥ 1 and
P(D)
P (D)
…(ii)
From Eqs. (i) and (ii), we get
P (C )
C
≥ P (C )
P   =
 D  P (D)
50. If the mean deviations about the median of
the numbers a , 2a ,... , 5a is 50, then |a | equal
to
[AIEEE 2011]
(a) 3
(b) 4
(c) 5
(d) 2
Exp. (b)
Median of a, 2 a, 3a, 4a, K, 50a is
25a + 26a
= (25.5) a
2

Q if total term is even, then

1  n 


n
Median =
  th term +  + 1 th term 

2
2   2 

50
MD (about median) =
Σ | xi − Median|
i =1
N
333
Statistics and Probability
1
{2|a| ⋅ (0.5 + 15
. + 2 .5 + K + 24.5)}
50
25
⇒ 2500 = 2|a| ⋅
(25) ⇒ |a| = 4
2
⇒
50 =
51. Let A , B and C be pairwise independent
events with P (C )> 0 and P ( A ∩ B ∩ C ) = 0.
Then, P ( A C ∩ B C /C ) is equal to [AIEEE 2011]
(a) P( AC ) − P( B )
(b) P( A ) − P ( BC )
(c) P( AC ) + P( BC )
(d) P( AC ) − P ( BC )
C
C
C
Now,
and
x = 2 and y = 4
Σ xi
=2 ⇒
5
1
σ x2 = 
5
Σ xi
2
Σ xi = 10; Σ yi = 20
Σ xi 2  − ( x)2 , σ y2 = 1 (Σ yi 2 ) − 16
= 40 and
5
Σ yi
2
= 105
…(i)
52. A scientist is weighing each of 30 fishes.
Their mean weight worked out is 30 g and a
standard deviation of 2 g. Later, it was found
that the measuring scale was misaligned
and always under reported every fish weight
by 2 g. The correct mean and standard
deviation (in gram) of fishes are respectively
(a) 28, 4
(c) 32, 4
(b) 32, 2
(d) 28, 2
[AIEEE 2011]
Exp. (b)
Correct mean = Old mean + 2 = 30 + 2 = 32
As standard deviation is independent of change
of origin.
∴ It remains same.
⇒ Standard deviation = 2
53. For two data sets, each of size 5, the
variances are given to be 4 and 5 and the
corresponding means are given to be 2 and
4, respectively. The variance of the
combined data set is
[AIEEE 2010]
(b)
11
2
(c) 6
(d)
13
2
2
1
x + y
(Σ xi 2 + Σ yi 2 ) − 

 2 
10
1
145 − 90 55 11
= (40 + 105) − 9 =
=
=
10
10
10 2
σ z2 =
Given, P( A ∩ B ∩ C ) = 0 and A, B, C are pairwise
independent.
∴
P( A ∩ C ) = P( A) ⋅ P(C )
and
P(B ∩ C ) = P(B) ⋅ P(C )
 AC ∩ B C 

∴ P 
C


P(C ) − P( A) ⋅ P(C ) − P(B) ⋅ P(C ) + 0
=
P(C )
= 1 − P( A) − P(B) = P( AC ) − P(B)
5
2
Also,
∴Variance of the combined data set is
 A ∩ B  P ( A ∩ B ∩ C)
 =
P 
C
P (C )


P(C ) − P( A ∩ C ) − P(B ∩ C ) + P( A ∩ B ∩ C )
=
P(C )
(a)
σ x2 = 4 and σ y2 = 5
Q
⇒
Exp. (a)
C
Exp. (b)
54. An urn contains nine balls of which three
are red, four are blue and two are green.
Three balls are drawn at random
without replacement from the urn. The
probability that the three balls have
different colours, is
[AIEEE 2010]
(a)
1
3
(b)
2
7
(c)
1
21
(d)
2
23
Exp. (b)
Total number of cases = 9C 3 = 84
Number of favourable cases = 3C1 ⋅ 4C1 ⋅ 2C1 = 24
24 2
P=
=
∴
84 7
55. Four numbers are chosen at random
(without replacement) from the set
{1, 2, 3, ..., 20}.
Statement I The probability that the
chosen numbers when arranged in some
1
order will form an AP, is .
85
Statement II If the four chosen numbers
form an AP, then the set of all possible
values of common difference is
[AIEEE 2010]
{ ±1 , ± 2 , ± 3 , ± 4 , ± 5}.
(a) Statement I is true, Statement II is true;
Statement II is a correct explanation for
Statement I
(b) Statement I is true, Statement II is true;
Statement II is not a correct explanation for
Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
334
JEE Main Chapterwise Mathematics
Exp. (a)
Exp. (c)
n(S ) = C 4
20
According to the given condition,
n
3
9
1 −   ≥
 4
10
Statement I
Common difference is 1;
Total number of cases = 17
Common difference is 2;
Total number of cases = 14
Common difference is 3;
Total number of cases = 11
Common difference is 4;
Total number of cases = 8
Common difference is 5;
Total number of cases = 5
Common difference is 6;
Total number of cases = 2
Hence, required probability
17 + 14 + 11 + 8 + 5 + 2
1
=
=
20
85
C4
⇒
58. One ticket is selected at random from
50 tickets numbered 00, 01, 02, …, 49.
Then, the probability that the sum of
the digits on the selected ticket is 8,
given that the product of these digits is zero
equal to
[AIEEE 2009]
(a)
56. If
the mean deviation of number
1, 1 + d , 1 + 2d ,…, 1 + 100d from their mean is
255, then d is equal to
[AIEEE 2009]
(b) 20.0
(c) 10.1
(d) 20.2
Exp. (c)
n
(a + l )
Sum of quantities 2
=
n
n
1
= [1 + 1 + 100 d ] = 1 + 50 d
2
1
∴ MD = Σ| xi − x | ⇒ 255
n
1
[50d + 49d + 48d + … + d + 0 + d
=
101
+ … + 50d ]
2d  50 × 51
=


101  2 
( x) =
∴ d=
57.
255 × 101
= 101
.
50 × 51
1

In a binomial distribution B n , p =  , if the

4
probability of atleast one success is greater
9
than or equal to , then n is greater than
10
[AIEEE 2009]
1
log10 4 − log10 3
9
(c)
log10 4 − log10 3
(a)
1
log10 4 + log10 3
4
(d)
log10 4 − log10 3
(b)
n
 4  ≥ 10
 
 3
Taking log on both sides, we get
n [log 4 − log 3] ≥ log10 10 = 1
1
n≥
⇒
log10 4 − log10 3
⇒
Hence, Statement I is true and Statement II is false.
(a) 10.0
n
 3 ≤ 1 − 9 = 1
 
 4
10 10
1
14
(b)
1
7
(c)
5
14
(d)
1
50
Exp. (a)
S = {00, 01, 02 , …, 49}
Let A be the event that sum of the digits on the
selected ticket is 8, then
A = {08, 17, 26, 35, 44}
Let B be the event that the product of the digits is
zero.
B = {00, 01, 02 , 03, …, 09, 10, 20, 30, 40}
∴ A ∩ B = {8}
A
∴ Required probability = P  
 B
1
P ( A ∩ B) 50
1
=
=
=
14 14
P (B)
50
59. Statement I The variance of first n even
natural numbers is
n2 − 1
.
4
Statement II The sum of first n natural
n (n + 1)
and the sum of squares of
numbers is
2
n (n + 1)(2n + 1)
.
first n natural numbers is
6
[AIEEE 2009]
335
Statistics and Probability
(a) Statement I is true, Statement II is true;
Statement II is a correct explanation for
Statement I
(b) Statement I is true, Statement II is true;
Statement II is not a correct explanation for
Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
Statement II is true because they are standard
formulae.
Statement I Sum of n even natural numbers
= n (n + 1)
n(n + 1)
Mean ( x ) =
⇒
=n+1
n
1
∴ Variance =  Σ ( xi )2  − ( x )2
 n

1 2
[2 + 42 + … + (2 n)2 ] − (n + 1)2
n
1
= 2 2 [12 + 2 2 + … + n2 ] − (n + 1)2
n
4 n (n + 1) (2 n + 1)
=
− (n + 1)2
6
n
(n + 1) [2 (2 n + 1) − 3 (n + 1)]
=
3
(n + 1) (n − 1) n2 − 1
=
=
3
3
=
Hence, Statement I is false.
1
1
2
, P ( A / B ) = and P (B / A ) = .
4
2
3
Then, P (B ) is equal to
[AIEEE 2008]
that P ( A ) =
(b)
1
6
(c)
1
3
(d)
2
3
Exp. (c)
1
2
1
A
B
, P   = and P   =




2
3
4
B
A
A  P( A ∩ B)

…(i)
We know that, P   =
 B
P(B)
P(B ∩ A)
B
…(ii)
and P   =
 A
P( A)
Given that, P( A) =
∴
(a)
2
5
(b)
3
5
(c) 0
(d) 1
Exp. (d)
∴
∴
⇒
B = {1, 2 , 3, 4}
A ∩ B = {4}
[by addition theorem of probability]
P( A ∪ B) = P( A) + P(B) − P( A ∩ B)
3 4 1
P( A ∪ B) = + − = 1
6 6 6
62. The mean of the numbers a ,b , 8, 5, 10 is 6
and the variance is 6.80. Then, which one of
the following gives possible values of
[AIEEE 2008]
a and b ?
(a) a = 3 and b = 4
(c) a = 5 and b = 2
(b) a = 0 and b = 7
(d) a =1 and b = 6
Exp. (a)
According to the given condition,
(6 − a)2 + (6 − b )2 + (6 − 8)2 


+ (6 − 5)2 + (6 − 10)2 

6.80 = 
5
⇒ 34 = (6 − a)2 + (6 − b )2 + 4 + 1 + 16
⇒ (6 − a)2 + (6 − b )2 = 13 = 9 + 4
60. It is given that the events A and B are such
1
2
number obtained is greater than 3. Let B be
the event that the number obtained is less
than 5. Then, P ( A ∪ B ) is
[AIEEE 2008]
Q A = {4, 5, 6} and
Exp. (d)
(a)
61. A die is thrown. Let A be the event that the
2
1
B
P   ⋅ P( A)    
 A
 3  4 1
=
P(B) =
=
A
 1
3
P 
 
 B
2
⇒ (6 − a)2 + (6 − b )2 = 32 + 2 2
⇒
a = 3, b = 4
63. A pair of fair dice is thrown independently
three times. The probability of getting a
score of exactly 9 twice is
[AIEEE 2007]
(a) 1/729
(c) 8/729
(b) 8/9
(d) 8/243
Exp. (d)
Probability of getting score 9 in a single throw
4
1
=
=
36 9
Probability of getting score 9 exactly in double
throw
2
1
8
8
= 3C 2 ×   × =
 9
9 243
336
JEE Main Chapterwise Mathematics
64. Two aeroplanes I and II bomb a target in
succession. The probabilities of I and II
scoring a hit correctly are 0.3 and 0.2,
respectively. The second plane will bomb
only if the first misses the target. The
probability that the target is hit by the
second plane, is
[AIEEE 2007]
(a) 0.06
(b) 0.14
(c) 0.32
(d) 0.7
Exp. (d)
Since, variance is independent of change of
origin. Therefore, variance of observations 101,
102, ... , 200 is same as variance of 151, 152, ...,
250.
V
∴
VA = VB ⇒ A = 1
VB
67. At a telephone enquiry system, the number
Exp. (c)
Let the events, A = Ist aeroplane hit the target
B = IInd aeroplane hit the target
And their corresponding probabilities are
P( A) = 0.3 and P(B) = 0.2
⇒
P( A ) = 07
. and P(B ) = 0. 8
∴ Required probability
of phone calls regarding relevant enquiry
follow. Poisson distribution with an average
of 5 phone calls during 10 min time
intervals. The probability that there is
atmost one phone call during a 10 min time
period, is
[AIEEE 2006]
= P( A ) P(B) + P( A ) P(B ) P( A ) P(B) + . . .
= (07
. )(0.2 ) + (07
. )(0.8)(07
. )(0.2 )
(a)
+ (07
. )(0.8)(07
. )(0.8)(07
. )(02
. )+ ...
= 014
. [1 + (0.56) + (0.56)2 + . . .]
5
6
(b)
6
55
(c)
(d) 60%
e −5 0 e −5 1
⋅5 +
⋅5
0!
1!
6
= e −5 + 5e −5 = 5
e
Σxi2 = 400 and Σxi = 80. Then, a
value of n among the following is
(a) 12
Let the number of boys and girls be x and y,
respectively.
∴
52 x + 42 y = 50( x + y)
⇒
52 x + 42 y = 50 x + 50 y
⇒
2 x = 8y ⇒ x = 4y
∴Total number of students in the class
= x + y = 4y + y = 5y
∴ Required percentage of boys
4y
=
× 100% = 80%
5y
66. Suppose a population A has
100 observations 101, 102, …, 200 and
another population B has 100 observations
151, 152, … , 250. IfV A and V B represent the
variances of the two populations
V
respectively, then A is
VB
[AIEEE 2006]
9
4
68. Let x1 , x 2 ,... , xn be n observations such that
possible
[AIEEE 2005]
Exp. (c)
(a)
6
5e
=
52 and that of girls is 42. The average marks
of boys and girls combined is 50. The
percentage of boys in the class is [AIEEE 2007]
(c) 80%
(d)
Required probability = P( X = 0) + P( X = 1)
65. The average marks of boys in a class is
(b) 20%
e
5
Exp. (c)

.
7
1  014
= 014
=
= 0.32
. 
=
 1 − 0.56  0.44 22
(a) 40%
6
(b)
4
9
(c)
2
3
(d) 1
(b) 9
(c) 18
(d) 15
Exp. (c)
Given that, Σ xi2 = 400 and Σ xi = 80
σ2 ≥ 0
Q
∴
Σ xi2
2
Σx
400 6400
−  i  ≥ 0 ⇒
− 2 ≥0
 n 
n
n
n
∴
n ≥ 16
69. If in a frequency distribution, the mean and
median are 21 and 22 respectively, then its
mode is approximately
[AIEEE 2005]
(a) 24.0
(b) 25.5
(c) 20.5
(d) 22.0
Exp. (a)
Given that, mean = 21and median = 22
Using the Emperical relation,
Mode = 3 Median − 2 Mean
∴
Mode = 3(22 ) − 2 (21) = 66 − 42 = 24
337
Statistics and Probability
70. Let A and B be two events such that
1
1
1
P ( A ∪ B ) = , P ( A ∩ B ) = and P ( A ) = ,
6
4
4
where A stands for complement of event A.
Then, events A and B are
[AIEEE 2005]
(a)
(b)
(c)
(d)
mutually exclusive and independent
independent but not equally likely
equally likely but not independent
equally likely and mutually exclusive
72. A random variable X
has Poisson
distribution with mean 2. Then, P ( X > 1.5)
equal to
[AIEEE 2005]
(a)
3
and
⇒
1
,
4
1
P( A ∪ B) =
6
1
1 − P( A ∪ B) =
6
P( A ∩ B) =
P( A ) =
∴
1
4
[Q P( A) + P( A ) = 1]
1 − P( A) − P(B) + P( A ∩ B) =
71. Three houses are available in a locality.
Three persons apply for the houses. Each
applies for one house without consulting
others. The probability that all the three
apply for the same house, is
[AIEEE 2005]
(a)
7
9
(b)
8
9
(c)
1
9
(d)
(c) 0
(d)
2
e2
P( X > 15
. ) = P(2 ) + P(3) + . . . ∞
= 1 − [P(0) + P(1)]

e −2 × 2 
3
 = 1 − 2
= 1 −  e −2 +
1
e


73. Consider the following statements
1
6
[Q P( A ∪ B) = P( A) + P(B) − P( A ∩ B)]
1 1
P( A ) − P(B) + =
⇒
4 6
1 1 1
P(B) = + −
⇒
4 4 6
1
3
⇒
and P( A) =
P(B) =
3
4
1 3 1
Now, P( A ∩ B) = = × = P( A) P(B)
4 4 3
Hence, the events A and B are independent
events but not equally likely.
⇒
3
e2
Exp. (b)
Exp. (b)
Given that,
(b) 1 −
e2
2
9
Exp. (c)
All the three persons has three options to apply a
house.
∴Total number of cases = 33
Now, favourable cases = 3 (as either all has
applied for house 1 or 2 or 3)
3
1
∴ Required probability = 3 =
9
3
I. Mode can be computed from histogram
II. Median is not independent of change of
scale
III. Variance is independent of change of
origin and scale
Which of these is/are correct?
(a) Only I
(c) I and II
[AIEEE 2004]
(b) Only II
(d) I, II and III
Exp. (c)
It is true that mode can be computed from
histogram and median is not independent of
change of scale.
But variance is independent of change of origin
and not of scale.
Hence, option (c) is correct.
74. In a series of 2 n observations, half of them
equal a and remaining half equal −a . If the
standard deviation of the observations is 2,
then |a | equal to
[AIEEE 2004]
(a)
1
n
(b)
2
(c) 2
(d)
2
n
Exp. (c)
In the 2 n observations, half of them equal to a and
remaining half equal to − a. Then, the mean of total
2 n observations is equal to zero.
∴
Σ( x − x )2
N
SD =
Σ x2
2n
⇒
4=
⇒
a2 = 4
⇒ 2=
⇒ 4=
∴ |a|= 2
Σ x2
2n
2 na2
2n
338
JEE Main Chapterwise Mathematics
75. The probability that A speaks truth is 4/5
77. The mean and the variance of a binomial
while this probability for B is 3/4. The
probability that they contradict each other
when asked to speak on a fact, is [AIEEE 2004]
distribution are 4 and 2, respectively. Then,
the probability of 2 successes is [AIEEE 2004]
3
(a)
20
1
(b)
5
7
(c)
20
4
(d)
5
Given, probabilities of speaking truth are
4
3
and P(B) =
P( A) =
5
4
And their corresponding probabilities of not
speaking truth are
1
P( A ) =
5
1
and
P(B ) =
4
The probability that they contradict each other
= P( A) × P(B ) + P( A ) × P(B)
4 1 1 3 1
3
7
= × + × = +
=
5 4 5 4 5 20 20
76. A random variable X has the probability
distribution
1
2
P( X ) 0.15 0.23
3
4
5
6
7
8
0.12
0.10
0.20
0.08
0.07
0.05
For the events, E = {X is a prime number}
and F = { X < 4}, the probability P ( E ∪ F ) is
(a) 0.87
(c) 0.35
(b) 0.77
(d) 0.50
[AIEEE 2004]
Exp. (b)
Given, E = { X is a prime number} = {2 , 3, 5, 7}
P(E ) = P( X = 2 ) + P( X = 3) + P( X = 5) + P( X = 7 )
and
⇒
= 0.23 + 012
. + 0.20 + 0.07 = 0.62
F = { X < 4} = {1, 2 , 3}
P(F ) = P( X = 1) + P( X = 2 ) + P( X = 3)
= 015
. + 0.23 + 012
. = 0.5
and E ∩ F = {X is prime number as well as < 4}
= {2 , 3}
P(E ∩ F ) = P( X = 2 ) + P( X = 3)
= 0.23 + 012
. = 0.35
∴ Required probability,
P(E ∪ F ) = P(E ) + P(F ) − P(E ∩ F )
= 0.62 + 0.5 − 0.35 = 077
.
37
256
(b)
219
256
(c)
128
256
(d)
28
256
Exp. (d)
Exp. (c)
X
(a)
Given that, mean = 4 ⇒ np = 4 and variance = 2
⇒
npq = 2 ⇒ 4q = 2
1
⇒
q=
2
1 1
p= 1− q = 1− =
∴
2 2
Also,
n=8
Probability of 2 successes = P( X = 2 ) = 8C 2 p2q 6
=
8!
×
2 ! × 6!
2
6
 1  ×  1  = 28 × 1 = 28
 
 
2
2
2 8 256
78. The median of a set of 9 distinct
observations is 20.5. If each of the largest
4 observations of the set is increased by 2,
then the median of the new set [AIEEE 2003]
(a) is increased by 2
(b) is decreased by 2
(c) is two times the original median
(d) remains the same as that of the original set
Exp. (d)
Median of new set remains the same as that of the
original set.
79. In an experiment with 15 observations on x ,
the following results were available
Σ x 2 = 2830, Σ x = 170
One observation that was 20, was found to
be wrong and was replaced by the correct
value 30. Then, the corrected variance is
(a) 78.00
(c) 177.33
(b) 188.66 [AIEEE 2003]
(d) 8.33
Exp. (a)
Given, N = 15, Σ x2 = 2830, Σ x = 170
Since, one observation 20 was replaced by 30,
then
Σ x2 = 2830 − 400 + 900 = 3330
Σ x = 170 − 20 + 30 = 180
2
Σ x2  Σ x 
Variance, σ 2 =
−

 N
N
and
Statistics and Probability
339
2
82. The mean and variance of a random variable
3330 − 12 × 180
3330  180 
−
 =
 15 
15
15
3330 − 2160 1170
=
=
= 78.0
15
15
=
X having a binomial distribution are 4 and
2 respectively, then P ( X = 1) is
[2003]
(a)
80. Five horses are in a race. Mr A selects two of
the horses at random and bets on them. The
probability that Mr A selected the winning
horse, is
[AIEEE 2003]
4
5
1
(c)
5
3
5
2
(d)
5
(a)
(b)
The probability that Mr A selected the loosing
4 3 3
horse = × =
5 4 5
The probability that Mr A selected the winning
3 2
horse = 1 − =
5 5
81. Events A , B andC are mutually exclusive
3x + 1
1− x
, P (B ) =
3
4
1 − 2x
. The set of possible values
and P (C ) =
2
of x are in the interval
[AIEEE 2003]
1 1
1 2
1 13
(a)  ,  (b)  ,  (c)  ,  (d) [0, 1]
 3 2 
 3 3 
 3 3 
3x + 1
1
2
≤1 ⇒ − ≤ x≤
3
3
3
1− x
0≤
≤ 1 ⇒ −3 ≤ x ≤ 1
4
1− 2x
1
1
≤1 ⇒ − ≤ x≤
0≤
2
2
2
3x + 1 1 − x 1 − 2 x
and 0 ≤
+
+
≤1
3
4
2
0≤
0 ≤ 13 − 3 x ≤ 12
1
13
≤ x≤
3
3
From Eqs. (i), (ii), (iii) and (iv), we get
1
1
≤ x≤
3
2
…(ii)
…(iii)
1
4
1
students A , B ,C and their respective
probability of solving the problem is
1 1
1
, and . Probability that the problem is
2 3
4
solved, is
[AIEEE 2002]
(a) 3/4
(b) 1/2
(c) 2/3
(d) 1/3
Exp. (a)
Since, the probabilities of solving the problem by
1 1
1
A, B and C are , and , respectively.
2 3
4
∴ Probability that the problem is not solved
= P( A ) P(B ) P(C )
1
1
1
=  1 −   1 −   1 − 

4
3 
2 
1 2 3 1
× × =
2 3 4 4
Hence, the probability that the problem is solved
1 3
= 1− =
4 4
84. In a class of 100 students, there are 70 boys
whose average marks in a subject are 75. If
the average marks of the complete class is
72, then what is the average of the girls?
(a) 73
(c) 68
…(iv)
(d)
83. A problem in Mathematics is given to three
=
…(i)
1
8
1
1
1
1
1
∴ P( X = 1) = 8C1     = 8 × 8 = 5 =
2 2
32
2
2
and 0 ≤ P( A) + P(B) + P(C ) ≤ 1
⇒
(c)
Given that, for binomial distribution mean, np = 4
and variance, npq = 2.
1
1
∴ q = , but p + q = 1 ⇒ p =
2
2
1
and
n× =4 ⇒ n=8
2
We know that,
P( X = r ) = nC r pr q n − r
Since, 0 ≤ P( A) ≤ 1, 0 ≤ P(B) ≤ 1, 0 ≤ P(C ) ≤ 1
⇒
1
16
Exp. (a)
Exp. (a)
∴
(b)
7
Exp. (d)
events such that P ( A ) =
1
32
(b) 65
(d) 74
[AIEEE 2002]
Exp. (b)
Since, total number of students = 100
and
number of boys = 70
∴ Number of girls = (100 − 70) = 30
340
JEE Main Chapterwise Mathematics
Now, the total marks of 100 students
= 100 × 72 = 7200
And total marks of 70 boys = 70 × 75 = 5250
P( A1 ) = P( A 2 ) = P( A 3 ) =
∴ Required probability
= P( A 1 A′2 A 3 ) + P( A′1 A 2 A 3 )
Total marks of 30 girls = 7200 − 5250 = 1950
1950
= 65
∴ Average marks of 30 girls =
30
= P( A 1 ) P( A′2 ) P( A 3 ) + P( A′1 ) P( A 2 ) P( A 3 )
3
1
25
2
(c)
25
(a)
(b)
24
25
(d) None of these
Exp. (b)
88. A biased coin with probability p , 0 < p < 1, of
heads is tossed until a head appears for the
first time. If the probability that the number
of tosses required is even, is 2/5, then p
equal to
[AIEEE 2002]
(a)
1
3
(b)
2
3
(c)
2
5
(d)
3
5
Exp. (a)
The total number of ways in which numbers can
be choosed = 25 × 25 = 625
The number of ways in which either players can
choose same numbers = 25
25
1
=
∴ Probability that they win a prize =
625 25
Thus, the probability that they will not win a prize
1 24
in a single trial = 1 −
=
25 25
86. If A and B are two mutually exclusive events,
then
[AIEEE 2002]
(a) P( A ) < P( B )
(c) P( A ) < P( B )
(b) P( A ) > P( B )
(d) None of these
Exp. (d)
Since, A and B are two mutually exclusive events.
∴ A ∩ B = φ ⇒ Either A ⊆ B or B ⊆ A
⇒ P( A) ≤ P(B ) or
P(B) ≤ P( A )
match against the West Indies is
1/2 assuming
independence
from
match-to-match. The probability that in a
match series India’s second win occurs at
the third test, is
[AIEEE 2002]
1
(b)
4
1
(c)
2
Let the probability of getting a head be p and not
getting a head be q.
Since, head appears first time in an even throw
2 or 4 or 6. . .
2
= qp + q 3 p + q 5 p + . . .
∴
5
2
qp
⇒
=
5 1 − q2
(1 − p)p
2
=
[Qq = 1 − p]
⇒
5 1 − (1 − p)2
2 1− p
⇒
=
5 2− p
1
⇒
4 − 2 p = 5 − 5p ⇒ p =
3
89. A fair die is tossed eight times. The
probability that a third six is observed on the
eight throw, is
[AIEEE 2002]
7
87. The probability of India winning a test
1
(a)
8
3
1
1
1 1 1
=   +   = + =
2
2
8 8 4
85. A and B play a game, where each is asked to
select a number from 1 to 25. If the two
numbers match, both of them win a prize.
The probability that they will not win a prize
in a single trial, is
[AIEEE 2002]
1
2
2
(d)
3
Exp. (b)
Let A1, A2 and A3 be the events of match winning
in first, second and third matches respectively and
whose probabilities are
(a)
7
(c)
C 2 × 55
7
(b)
67
C 2 × 55
C 2 × 55
68
(d) None of these
66
Exp. (b)
Probability of getting success, p =
and probability of failure, q =
1
2
1
2
∴ Required probability
2
5
1
5
1 7C 2 × 55
= 7C 2     × =
 6  6
6
68
15
Trigonometry
3
5
1. If cos(α + β ) = , sin(α − β) =
Now, tan(2α ) = tan[(α + β ) + (α − β )]
5
and
13
tan(α + β ) + tan(α − β )
1 − tan(α + β )tan(α − β )
4
5
+
= 3 12
5
4
1− ×
3 12
[from Eqs. (i) and (ii)]
48 + 15 63
=
=
36 − 20 16
=
π
0 < α ,β < , then tan(2α ) is equal to
4
[JEE Main 2019, 8 April Shift-I]
63
(a)
52
63
(b)
16
(c)
21
16
(d)
33
52
Exp. (b)
5
13
3
π
and cos(α + β ) = , where α, β ∈  0, 
 4
5
π
π
Since, 0 < α < and 0 < β <
4
4
π π π
∴ 0< α + β < + =
4 4 2
π
⇒ 0< α + β <
2
π
Also, − < −β < 0
4
π
π
∴
0− < α −β < + 0
4
4
π
π
− < α −β <
⇒
4
4
π
π π

∴ α + β ∈  0,  and α − β ∈  − , 
 2
 4 4
π
But sin(α − β ) > 0, therefore α − β ∈  0,  .
 4
5
Now,
sin(α − β ) =
13
5
…(i)
tan(α − β ) =
⇒
12
3
and
cos(α + β ) =
5
4
tan(α + β ) =
…(ii)
⇒
3
Given, sin(α − β ) =
 3
 1
α = cos −1   ,
β = tan −1   ,
 5
 3
π
0 < α ,β < , then α − β is equal to
2
2. If
where
[JEE Main 2019, 8 April Shift-I]
9 
(a) tan −1 

 5 10 
9
(c) tan −1  
 14 
9 
(b) cos−1 

 5 10 
9 
(d) sin −1 

 5 10 
Exp. (d)
3
1
Given, α = cos −1   and β = tan−1  
 5
 3
π
where 0 < α, β <
2
5
52 – 32 = 4
α
3
4
3
4
1
So, α − β = tan−1 − tan−1
3
3
Clearly, α = tan−1
342
JEE Main Chapterwise Mathematics

4 1 
−


−1 
3 3 
= tan
 1 +  4 × 1 



 3 3  


−1
−1
−1 x − y
Q tan x − tan y = tan 1 + xy, if xy > − 1]

9
1
= tan−1
= tan−1
4
13
1+
9
250
2 =
3
2 1
9+
α –β
4. Let S = {θ ∈[ −2 π , 2 π]: 2 cos 2 θ + 3 sin θ = 0},
then the sum of the elements of S is
9
[JEE Main 2019, 9 April Shift-I]
13
9
−1
= sin
−1
92 + 132
= sin
(a) 2 π
9
250
(b) π
(c)
5π
3
(d)
13 π
6
Exp. (a)
9 
= sin−1 

 5 10 
3. Two vertical poles of heights, 20 m and
80 m stand apart on a horizontal plane. The
height (in m) of the point of intersection of
the lines joining the top of each pole to the
foot of the other, from this horizontal plane
is
[JEE Main 2019, 8 April Shift-II]
(a) 15
(c) 12
h
20
=
y x+ y
[in ∆MNQ and ∆ABQ] …(ii)
From Eqs. (i) and (ii), we get
y
…(iii)
= 4 ⇒ y = 4x
x
From Eqs. (i) and (iii), we get
h
80
80
=
= 16m
⇒ h=
x x + 4x
5
tanβ =
and
(b) 16
(d) 18
Exp. (b)
Let a first pole AB having height 20 m and second
pole PQ having height 80 m
and ∠PBQ = α, ∠AQB = β
P
We have, θ ∈ [−2 π, 2 π ]
and
2 cos 2 θ + 3sinθ = 0
⇒
2 (1 − sin2 θ) + 3sinθ = 0
⇒
2 − 2 sin2 θ + 3 sin θ = 0
⇒
2 sin2 θ − 3sinθ − 2 = 0
2
⇒ 2 sin θ − 4sinθ + sinθ − 2 = 0
⇒2 sinθ (sinθ − 2 ) + 1(sinθ − 2 ) = 0
⇒
(sinθ − 2 ) (2 sinθ + 1) = 0
−1
[Q(sinθ − 2 ) ≠ 0]
sinθ =
⇒
2
π
π
π
π
⇒ θ = 2π – , − π + , − , π +
6
6
6
6
[Qθ ∈ [−2 π, 2 π ]]
Now, sum of all solutions
π
π π
π
= 2π − − π + − + π + = 2π
6
6 6
6
5. The value of
cos 2 10° − cos 10° cos 50° + cos 2 50° is
[JEE Main 2019, 9 April Shift-I]
80 m
A
M
20 m
α
B
3
(a) (1 + cos 20° )
2
(c) 3 / 2
x
h
N
β
y
3
+ cos 20°
4
(d) 3 / 4
(b)
Exp. (d)
Q
and MN = hm is the height of intersection point
from the horizontal plane
h
80
Q
tanα = =
x x+ y
[in ∆MNB and ∆PQB] …(i)
We have, cos 2 10º − cos 10º cos 50º + cos 2 50º
1
[2 cos 2 10º −2 cos 10º cos 50º +2 cos 2 50º ]
2
1
= [1 + cos 20º −(cos 60º + cos 40º ) + 1 + cos 100º ]
2
[Q 2 cos 2 A = 1 + cos 2 A and
2cos A cos B = cos( A + B) + cos( A − B)]
=
343
Trigonometry
=
=
1
2
2 + cos 20º + cos 100º − 1 − cos 40º 


2
Qcos 60º = 1 
2 

1 3
+ (cos 20º − cos 40º ) + cos 100º 

2  2
Qcos C − cos D = −2 sin C + D sin C − D 

2
2 
=
1 3
− 2 sin 30º sin(−10º ) + cos(90º +10º )
2  2

=
1 3
+ sin10º − sin 10º 

2  2
[Qcos (90º + θ) = − sinθ]
1 3 3
= × =
2 2 4
6. The value of sin 10º sin 30º sin 50º sin 70º is
[JEE Main 2019, 9 April Shift-II]
1
36
1
(c)
16
Given heights of two poles are 5 m and 10 m.
A
5m
20º + 40º
20°− 40°
1 3
sin
=  − 2 sin
+ cos 100° 

2  2
2
2
1
32
1
(d)
18
(a)
Exp. (d)
15°
B
10 m
E
d
5m
15°
C
i.e. from figure AC = 10 m, DE = 5 m
∴ AB = AC − DE = 10 − 5 = 5 m
Let d be the distance between two poles.
Clearly, ∆ABE ~ ∆ACF
[by AA-similarity criterion]
∴
∠AEB = 15°
In ∆ABE, we have
3 −1 5
AB
tan15° =
⇒
=
BE
3+1 d

Q tan15° =

(b)
Exp. (c)
We have, sin10° sin 30° sin 50° sin70°
d=
⇒
d=5
3+1
×
3 −1
3+1
3+1
=
1
= [sin(10° )sin(60° − 10° )sin(60° + 10° )]
2
1 1
=  sin(3(10° ))
2  4

5(3 + 2 3 + 1 5(2 3 + 4)
=
3−1
2
=
2 × 5( 3 + 2 )
= 5(2 +
2
=
1
sin 3θ]
4
1
1 1 1
sin 30° = × =
8
8 2 16
7. Two poles standing on a horizontal ground
are of heights 5 m and 10 m, respectively.
The line joining their tops makes an angle of
15º with the ground. Then, the distance (in
m) between the poles, is
[JEE Main 2019, 9 April Shift-II]
(a) 5( 3 + 1)
(c) 10( 3 − 1)
5
(2 + 3)
2
(d) 5( 2 + 3 )
(b)
3 − 1
3 + 1
5( 3 + 1)
( 3 − 1)
⇒
= sin(30° )[sin(10° )sin(50° )sin(70° )]
[Qsinθsin(60° − θ)sin(60° + θ) =
F
D
d
3) m
8. ABC is a triangular park with AB = AC = 100
m. A vertical tower is situated at the
mid-point of BC . If the angles of elevation of
the top of the tower at A and B are
cot −1 ( 3 2 ) and cosec−1(2 2 ) respectively,
then the height of the tower (in m) is
[JEE Main 2019, 10 April Shift-I]
(a) 25
(b) 20
(c) 10 5
(d)
100
3 3
Exp. (b)
Given ABC is a triangular park with AB = AC = 100
m. A vertical tower is situated at the mid-point of
BC.
Let the height of the tower is h m.
344
JEE Main Chapterwise Mathematics
Now, according to given information, we have the
following figure.
Q
C
h
100
P
α
l
β
B
100
A
From the figure and given information, we have
β = cot −1(3 2 )
α = cosec −1 (2 2 )
and
l
h
cot α =
…(i)
(100) − l
BP
=
h
h
[JEE Main 2019, 12 April Shift-I]
2
h2 cot 2 α = (100)2 − l 2
⇒
h2 (cosec2α − 1) = (100)2 − (3 2 h)2
[from Eq. (i)]
⇒
h2 (8 − 1) = (100)2 − 18h2
25h2 = (100)2
2
100 
h2 = 
 ⇒ h = 20m
 5 
y
2
9. If cos − 1 x − cos − 1 = α, where − 1 ≤ x ≤ 1,
y
− 2 ≤ y ≤ 2, x ≤ , then for all x , y ,
2
4x 2 − 4xy cosα + y 2 is equal to
[JEE Main 2019, 10 April Shift-II]
(a) 2 sin 2 α
(b) 4cos2 α + 2 x 2 y 2
(c) 4 sin 2 α
(d) 4 sin 2 α − 2 x 2 y 2
y2
− xycos α = 1 − cos 2 α
4
⇒ 4 x2 − 4 xycos α + y2 = 4sin2 α
represents a straight line lying in
2
[Q p is mid-point of isosceles ∆ABC, AP ⊥ BC]
⇒
y2
x 2 y2
x 2 y2
+
= cos 2 α +
− xycos α
4
4
4
10. The equation y = sin x sin( x + 2 ) − sin 2( x + 1)
⇒
⇒
[Qcos − 1 x − cos − 1 y = cos − 1( xy + 1 − x2 1 − y2 ),
| x|,| y| ≤ 1and x + y ≥ 0]
xy
+ 1 − x2 1 − ( y / 2 )2 = cosα
⇒
2
xy
1 − x2 1 − ( y / 2 )2 = cosα −
⇒
2
On squaring both sides, we get

y2 
x 2 y2
xy
 = cos 2 α +
− 2 cos α
(1 − x2 )  1 −
4
4
2

⇒ x2 +
⇒
l = (3 2 )h
and in ∆BPQ,
h
tanα =
BP
⇒
Given equation is
y
cos − 1 x − cos − 1 = α, where − 1 ≤ x ≤ 1,
2
y
− 2 ≤ y ≤ 2 and x ≤
2
y
∴ cos − 1  x + 1 − x2 1 − ( y / 2 )2  = α
 2

⇒1 − x2 −
Now, in ∆QPA,
cotβ =
Exp. (c)
(a)
(b)
(c)
(d)
second and third quadrants only
first, second and fourth quadrants
first, third and fourth quadrants
third and fourth quadrants only
Exp. (d)
Key Idea Use formulae :
2sin A sin B = cos(A − B ) − cos(A + B )
and cos 2 θ = 1 − 2 sin2 θ
Given equation is y = sin x sin( x + 2 ) − sin2 ( x + 1)
1
1
= [cos 2 − cos(2 x + 2 )] − [1 − cos(2 x + 2 )]
2
2
[Q2sin A sin B = cos( A − B) − cos( A + B) and
cos 2 θ = 1 − 2 sin2 θ
⇒2 sin2 θ = 1 − cos 2 θ]
1 1
1
1
= cos 2 − cos(2 x + 2 ) − + cos(2 x + 2 )
2 2
2
2
1
1
= (cos(2 ) − 1) = − (2 sin2 (1))
2
2
= − sin2 (1) < 0 ⇒ y < 0
and as we know that y < 0, is in third and fourth
quadrants only.
345
Trigonometry
12 4 3
5
= sin− 1 
× − × 
 13 5 5 13 
48 − 15 
= sin− 1 

 65 
11. The number of solutions of the equation
 5π 5π 
is
,
1 + sin 4 x = cos 2 3x , x ∈ −
 2 2 
[JEE Main 2019, 12 April Shift-I]
(a) 3
(b) 5
(c) 7
33
= sin− 1   = cos − 1 1 −
 65 
(d) 4
Exp. (b)
3136
[Qsin− 1 x = cos − 1 1 − x2 ]
4225
56 π
56
= cos − 1  = − sin− 1  
 65  2
 65 
Qsin− 1 θ + cot − 1 θ = π 

2 
1 + sin4 x = cos 2 (3 x)
Since, range of (1 + sin4 x) = [1, 2 ]
and range of cos 2 (3 x) = [0, 1]
So, the given equation holds if
1 + sin4 x = 1 = cos 2 (3 x)
13. A 2 m ladder leans against a vertical wall. If
⇒
sin4 x = 0and cos 2 3 x = 1
5π 5π 
Since, x ∈  −
,
 2 2 
∴
x = − 2 π, − π, 0, π, 2 π.
Thus, there are five different values of x is
possible.
 3
 12 
The value of sin −1   − sin −1   is equal to
 5
 13 
[JEE Main 2019, 12 April Shift-I]
π
56
− sin −1  
 65 
2
33
(d) π − cos−1  
 65 
−1  63 
(a) π − sin  
 65 
π
−1  9 
(c)
− cos  
 65 
2
(b)
Exp. (b)
Key Idea Use formulae
(i) sin− 1 x − sin− 1 y
the top of the ladder begins to slide down the
wall at the rate 25 cm/s, then the rate (in
cm/s) at which the bottom of the ladder
slides away from the wall on the horizontal
ground when the top of the ladder is 1 m
above the ground is
[JEE Main 2019, 12 April Shift-I]
(a) 25 3
x = cos
 12
1−
= sin− 1 
 13

 3
 
 5
2
−
3
1−
5
[Qsin− 1 x − sin− 1 y = sin− 1( x 1 − y2 − y 1 − x2 ),
if x2 + y2 ≤ 1 or
if xy > 0 and x2 + y2 > 1 ∀x, y ∈ [− 1, 1]]
(d) 25
Ladder
x
2
 12  
  
 13 

25
3
l
y
2
We have,
12
3
sin− 1   − sin− 1  
 5
 13 
(c)
Wall
1 − x and
π
(iii) sin− 1 θ + cos − 1 θ =
2
(ii) sin
25
3
Given a ladder of length l = 2 m leans against a
vertical wall. Now, the top of ladder begins to slide
down the wall at the rate
25 cm/s.
Let the rate at which bottom of the ladder slides
away from the wall on the horizontal ground is
dx
cm/s.
dt
or if xy > 0 and x 2 + y 2 > 1 ∀x, y ∈ [− 1, 1]
−1
(b)
Exp. (b)
= sin− 1( x 1 − y2 − y 1 − x 2 ) if x 2 + y 2 ≤ 1
−1
2
= cos − 1
Given equation is
12.
 33 
 
 65 
Ground
Q
x 2 + y2 = l 2
⇒
[by Pythagoras theorem]
[Ql = 2m]… (i)
x 2 + y2 = 4
On differentiating both sides of Eq. (i) w.r.t. ‘t’,
we get
dx
dy
2x + 2y = 0
dt
dt
dx
y dy
… (ii)
⇒
= −  
 x  dt
dt
346
JEE Main Chapterwise Mathematics
From Eq. (i), when y = 1 m, then
x2 + 12 = 4 ⇒ x2 = 3 ⇒ x = 3 m
On substituting x =
get
[Q x > 0]
3 mand y = 1min Eq. (ii), we
dx
1  25 
=−
−
 m/s
dt
3  100 
given dy = − 25 cm / sec 


dt
25
cm / s
3
Hence, option (b) is correct.
On the elimination of quantity x from
Eqs. (i) and (ii), we get
1
y=
y + 30
3
1 
⇒ y  1 −
 = 30

3
30 3 ( 3 + 1)
30 3
=
3 −1
3−1
30
=
3 ( 3 + 1) = 15 (3 + 3 )
2
⇒
y=
=
15. Let S be the set of all α ∈R such that the
14. The angle of elevation of the top of a vertical
tower standing on a horizontal plane is
observed to be 45° from a point A on the
plane. Let B be the point 30 m vertically
above the point A. If the angle of elevation of
the top of the tower from B be 30°, then the
distance (in m) of the foot of the tower from
the point A is [JEE Main 2019, 12 April Shift-II]
(a) 15 ( 3 +
(b)15( 5 − 3 )
3)
(c)15( 3 − 3 )
(d)15 (1 +
3)
According to the question, we have the following
figure.
T
xm
B
P
30 m
S
45°
ym
A
Now, let distance of foot of the tower from the
point A is y m.
Draw BP ⊥ ST such that PT = x m.
Then, in ∆TPB, we have
x
tan30° =
y
1
…(i)
x=
y
⇒
3
and in ∆TSA, we have
x + 30
tan45° =
y
⇒
y = x + 30
has
a
[JEE Main 2019, 12 April Shift-II]
(a) R
(b) [1, 4]
(c) [3, 7]
(d) [2, 6]
Exp. (d)
The given trigonometric equation is
cos 2 x + α sin x = 2α − 7
⇒
1 − 2 sin2 x + α sin x = 2α − 7
[Qcos 2 x = 1 − 2 sin2 x]
Exp. (a)
30°
equation, cos 2 x + α sin x = 2α − 7
solution. Then, S is equal to
…(ii)
⇒
2 sin2 x − α sin x + 2α − 8 = 0
⇒
2(sin2 x − 4) − α (sin x − 2 ) = 0
⇒
2 (sin2 x − 2 ) (sin x + 2 ) − α (sin x − 2 ) = 0
⇒ (sin x −2 ) (2 sin x + 4 − α ) = 0
⇒ 2 sin x + 4 − α = 0
[Qsin x + 2 ≠ 0]
α−4
…(i)
⇒
sin x =
2
Now, as we know −1 ≤ sin x ≤ 1
α−4
[from Eq. (i)]
∴
− 1≤
≤1
2
⇒
− 2 ≤ α − 4≤ 2
⇒
2 ≤ α ≤ 6 ⇒ α ∈[2, 6]
16. If [ x ] denotes the greatest integer ≤ x , then
the
system
of
liner
equations
[sin θ]x + [ − cos θ]y = 0,[cot θ]x + y = 0
[JEE Main 2019, 12 April Shift-II]
(a) have infinitely many solutions if
π 2π
θ ∈  ,
 and has a unique solution if
2 3 
7π 
θ ∈  π ,
.

6 
(b) has a unique solution if
π 2 π   7π 
θ ∈  ,

 ∪  π,
2 3  
6 
347
Trigonometry
π 2π
(c) has a unique solution if θ ∈  ,
 and
2 3 
7π 
have infinitely many solutions if θ ∈  π ,


6 
(d) have infinitely many solutions if
π 2 π   7π 
θ ∈  ,

 ∪  π,
2 3  
6 
Given system of linear equations is
…(i)
[sin θ] x + [− cos θ] y = 0
and
…(ii)
[cot θ] x + y = 0
where, [ x] denotes the greatest integer ≤ x.
[sin θ] [− cos θ]
Here, ∆ =
[cot θ]
1
⇒
∆ = [sin θ] − [− cos θ] [cot θ]
π 2π
When θ ∈  ,

2 3 
 3 
sinθ ∈ 
, 1

 2
…(iii)
…(iv)
[sin θ] = − 1
 3 
, 1 ⇒ [cos θ] = 0
− cos θ ∈ 
 2 
π
3
2 
−1  3 
 + cos   =  x >  , then x
 3x 
 4x  2 
4
is equal to
[JEE Main 2019, 9 Jan Shift-I]
17. If cos −1 
(a)
(b)
145
(c)
12
⇒
(d)
1 − y2 )]
 1
9 x2 − 4 16 x2 − 9  π
=
cos −1  2 −
 2
2 x
12 x2


6−
9 x2 − 4 16 x2 − 9
12 x
2
= cos
π
=0
2
9 x2 − 4 16 x2 − 9 = 6
On squaring both sides, we get
(9 x2 − 4)(16 x2 − 9) = 36
⇒ 144 x4 − 81x2 − 64 x2 + 36 = 36
⇒
144 x4 − 145 x2 = 0
⇒
x2 (144 x2 − 145) = 0
But
x>
∴
x=
145
=±
144
145
12
3
4
145
12
 π π
θ∈  ,  , the expression
 4 2
3(sin θ − cos θ )4 + 6(sin θ + cos θ )2 + 4 sin 6 θ
equals
[JEE Main 2019, 9 Jan Shift-I]
18. For
any
(b) 13 − 4cos2 θ + 6cos4 θ
cot θ ∈ ( 3, ∞ ) ⇒ [cot θ] = n , n ∈ N.
∆ = − 1 − (0 × n) = − 1
7π
Thus, for θ ∈  π,
 , the given system has a

6 
unique solution.
146
12
= cos −1( xy − 1 − x2
(a) 13 − 4cos4 θ + 2 sin 2 θcos2 θ
and
So,
145
10
[Qcos −1 x + cos −1 y
⇒ x = 0 or x = ±
…(v)
⇒
[cot θ] = − 1
So,
∆ = [sin θ] − [− cos θ] [cot θ]
− (0 × (− 1)) = 0[from Eqs.(iii), (iv) and (v)]
π 2π
Thus, for θ ∈  ,
 , the given system have
2 3 
infinitely many solutions.
7π
 1 
When
θ ∈  π,
 , sinθ ∈  − , 0

 2 
6 
⇒
1 − y2 )
2
3
π
cos −1   + cos −1   =
 4x 2
 3x
2 3
9  π
4
− 1− 2 1−
⇒cos −1  ⋅
=
16 x2  2
9x
 3x 4x
⇒
[− cos θ] = 0
1
cot θ ∈  −
, 0

3 
and
cos −1 x + cos −1 y = cos −1( xy − 1 − x2
⇒
[sin θ] = 0
1
− cos θ ∈  0, 
 2
⇒
Key Idea Use the formula,
We have,
Exp. (a)
⇒
Exp. (c)
145
11
(c) 13 − 4cos2 θ + 6 sin 2 θcos2 θ
(d) 13 − 4cos6 θ
Exp. (d)
Given expression
= 3(sinθ − cos θ)4 + 6(sinθ + cos θ)2 + 4sin6 θ
= 3((sinθ − cos θ)2 )2 + 6(sinθ + cos θ)2 + 4(sin2 θ)3
= 3(1 − sin2 θ)2 + 6(1 + sin2 θ) + 4(1 − cos 2 θ)3
[Q1 + sin2 θ = (cos θ + sinθ)2
and 1 − sin2 θ = (cos θ − sinθ)2 ]
348
JEE Main Chapterwise Mathematics
= 3(12 + sin2 2 θ − 2 sin2 θ) + 6(1 + sin2 θ)
+ 4(1 − cos θ − 3cos θ + 3cos θ)
6
2
4
[Q(a − b )2 = a2 + b 2 − 2 ab
and (a − b )3 = a3 − b 3 − 3a2 b + 3ab 2 ]
= 3 + 3sin 2 θ − 6sin2 θ + 6 + 6sin2 θ + 4
2
− 4cos 6 θ − 12 cos 2 θ + 12 cos 4 θ
= 13 + 3sin 2 θ − 4cos 6 θ − 12 cos 2 θ + 12 cos 4 θ
2
= 13 + 3(2 sinθcos θ)2 − 4cos 6 θ
− 12 cos 2 θ(1 − cos 2 θ)
= 13 + 12 sin θcos θ − 4cos 6 θ − 12 cos 2 θsin2 θ
2
2
= 13 − 4cos 6 θ
19. If x = sin −1(sin 10) and y = cos −1(cos 10), then
y − x is equal to
[JEE Main 2019, 9 Jan Shift-II]
(a) 0
(c) 7π
The graph of y = sin− 1(sin x) is
x
y=
π/2
y=–2π+x
3π/2
π
21. Consider a triangular plot ABC with sides
[JEE Main 2019, 10 Jan Shift-I]
2
(a)
21
3
(c) 7 3
y
y=π–x

 C + D
 C − D 
 QsinC + sin D = 2 sin  2  cos  2  


⇒2 sin 2 x cos x − sin2 x = 0 [Qcos(− θ) = cos θ]
⇒
sin2 x(2 cos x − 1) = 0
⇒
sin 2 x = 0or 2 cos x − 1 = 0
1
⇒
2 x = 0, π, ... or cos x =
2
π
π
⇒
x = 0, ... or x =
2
3
π

In the interval 0,  only two values satisfy,
 2 
π
namely x = 0 and x = .
3
AB = 7 m, BC = 5 m and CA = 6 m. A vertical
lamp-post at the mid-point D of AC
subtends an angle 30° at B. The height (in m)
of the lamp-post is
(b) 10
(d) π
Exp. (d)
–π/2
x + 3x
 x − 3 x  − sin 2 x = 0
⇒ 2 sin 
 cos 

 2 
 2 
y=–x+3π
(b) 2 21
(d)
x
2π 5π/2 3π
Exp. (a)
∴ x = sin− 1(sin 10) = − 10 + 3π
...(i)
−1
and the graph of y = cos (cos x) is
According to given information, we have the
following figure.
E
3π 10
4π
∴ y = cos − 1(cos 10) = − 10 + 4π
c=7
y=
x
y=
4π
+
–x
2π
X
...(ii)
π
If0 ≤ x < , then the number of values of x for
2
which sin x − sin 2 x + sin 3x = 0, is
[JEE Main 2019, 9 Jan Shift-II]
(a) 2
(b) 3
(c) 1
Exp. (a)
We have, sin x − sin 2 x + sin 3 x = 0
⇒
(sin x + sin 3 x) − sin2 x = 0
(d) 4
D
b=6
30°
Now, from Eqs. (i) and (ii),
y − x = (− 10 + 4 π) − (− 10 + 3 π) = π
20.
A
y=
–x
2π
y=
π
–2
π+
x
Y
O
3
21
2
B
a=5
C
1
Clearly, length of BD =
2 a2 + 2c 2 − b 2 ,
2
(using Appollonius theorem)
where,
and
∴
c = AB = 7, a = BC = 5
b = CA = 6
1
BD =
2 × 25 + 2 × 49 − 36
2
1
1
112 = 4 7 = 2 7
=
2
2
349
Trigonometry
Now, let ED = h be the height of the lamp post.
E
⇒
30°
D
h
Then, in ∆BDE, tan30° =
BD
1
h
2 7 2
=
21
=
⇒h =
⇒
3
3
3 2 7
2
22. If 5, 5r , 5r are the lengths of the sides of a
triangle, then r cannot be equal to
[JEE Main 2019, 10 Jan Shift-I]
5
(a)
4
7
(b)
4
(c)
3
2
(d)
We know that, in a triangle sum of 2 sides is
always greater than the third side.
∴a + b > c; b + c > a and c + a > b
⇒
5 + 5r > 5r 2
⇒
5r 2 − 5r − 5 < 0
⇒
r2 − r − 1 < 0
− b ± b 2 − 4ac
2a
r=
1±
Similarly,
⇒
b+ c> a
5r + 5r 2 > 5
⇒
r2 + r − 1 > 0
5


 r − 1 + 3 > 0



2
4
2
⇒
r ∈R
From (i), (ii) and (iii), we get
 −1 + 5 1 + 5 
r ∈
,

2
2 

and
… (iii)
1– √ 5
2
–1+ √ 5
2
1+ √ 5
2
∞
7
is the only value that does not satisfy.
4
π
23. The sum of all values of θ ∈ 0,  satisfying
3π
(a)
8
and r 2 − r − 1 = 0
1+ 4
2
5π
(b)
4
[JEE Main 2019, 10 Jan Shift-I]
(c)
π
2
(d) π
Exp. (c)
=
1± 5
]
2
1 − 5 1 + 5
⇒ r ∈
,

2 
 2

− 1−
r − 
2


⇒
 2
[Qroots of ax2 + bx + c = 0 are given by
⇒
2
1
1
1
r 2 − 2 ⋅ r +   + 1 −   > 0
2
2
2
3
sin 2 2θ + cos 4 2θ = is
4

1 − 5 
1 + 5
⇒ r − 
  r − 
 < 0
 2  
 2 

x=
2
⇒
–∞ –1– √ 5
2
Let a = 5, b = 5r and c = 5r 2
a+ b>c
r2 − r + 1 > 0
3
4
Exp. (b)
Now,
…(ii)
c + a> b
5r 2 + 5 > 5r
and
⇒
h
B

−1 − 5   −1 + 5 
, ∞
⇒ r ∈  − ∞,
 ∪
2
2

 

...(i)

 −1 + 5  
 > 0
r − 
2



 2
−1± 1 + 4 − 1 ±
=
Q r + r − 1 = 0 ⇒ r =
2
2

5


Given, sin2 2 θ + cos 4 2 θ =
3
4
3
4
(Qsin2 x = 1 − cos 2 x)
⇒ (1 − cos 2 2 θ) + cos 4 2 θ =
⇒ 4cos 4 2 θ − 4cos 2 2 θ + 1 = 0
⇒
(2 cos 2 2 θ − 1)2 = 0 ⇒2 cos 2 2 θ − 1 = 0
1
1
⇒
cos 2 2 θ = ⇒cos2 θ = ±
2
2
π

If θ ∈  0,  , then 2 θ ∈(0, π )
 2
1
∴
cos2 θ = ±
2
π 3π
2θ = , ,
⇒
4 4
3π 
π
π
1 


Qcos 
 = cos  π −  = − cos = −

 4 


4
4
2 
350
JEE Main Chapterwise Mathematics
π 3π
,
8 8
π 3π π
Sum of values of θ = +
=
8
8
2
⇒
θ=
∴ cos
24. With the usual notation, in ∆ABC , if
∠A + ∠B = 120°, a = 3 + 1 and b = 3 − 1,
then the ratio ∠A : ∠B , is
π
π
π
π
…cos 10 ⋅ sin 10
23
2
2
sin π 2 9  
  210  
π
=
 sin 10
π


9
2
2 sin


 210  
22
[Qhere, α =
[JEE Main 2019, 10 Jan Shift-II]
(a) 7 :1
(c) 9 : 7
=
(b) 3 :1
(d) 5 : 3
B
b
C
a

19
21
(b)
19
(c)
19
21
 19

Consider, cot  Σ cot −1  1 +

n =1


 19

= cot  Σ cot −1(1 + n(n + 1))
n =1



=
( 3 + 1) − ( 3 − 1)  60º 
cot 

( 3 + 1) + ( 3 − 1)  2 
 19

= cot  Σ cot −1(1 + n + n2 )
n =1



=
2
1
cot (30º ) =
× 3 =1
2 3
3
 19

1

= cot  Σ tan−1
n =1
1 + n(n + 1)

A − B
⇒ tan
 = 1= tan 45º
 2 
A−B
= 45º ⇒ ∠A − ∠B = 90º
⇒
2
On solving ∠A − ∠B = 90º and ∠A + ∠B = 120º ,
we get
∠A = 105º and ∠B = 15º
So, ∠A : ∠B = 7 : 1
25. The value of
π
π
π
π
⋅ cos 3 .......cos 10 ⋅ sin 10 is
22
2
2
2
1
2
(c)
(d)
22
23
1
512
n

Σ 2 p 
p =1

 n
n(n + 1)

Q Σ p =
2 
 p = 1
1
[Qcot −1 x = tan−1 ,if x > 0 ]
x
 19



+
−
1
n
n
(
)
= cot  Σ tan−1 
  [put1 = (n + 1) − n]
n =1
 1 + n (n + 1) 

19
= cot
Σ (tan−1(n + 1) − tan−1 n)
n =1

−1 x − y
−1
−1 
Q tan 1 + xy = tan x − tan y


= cot [(tan−1 2 − tan−1 1) + (tan−1 3 − tan−1 2 ) +
[JEE Main 2019, 10 Jan Shift-II]
(b)

n
Exp. (b)
Clearly, ∠C = 60º
[Q ∠A + ∠B + ∠C = 180º ]
Now, by tangent law, we have
A−B a−b
C
tan
=
cot
2
a+ b
2
1
1024
9
[JEE Main 2019, 10 Jan Shift-II]
23
(a)
22
A
c
and n = 9 ]
1

For a ∆ABC, it is given that a = 3 + 1,
b = 3 − 1and ∠A + ∠B = 120º
(a)
π
1
1
sin  = 9 =


2
512
2
2
π
210
26. The value of cot  ∑cot −1 1 + ∑2p   is
 p = 1 
n = 1
Exp. (a)
cos
⋅ cos
(d)
1
256
......+ (tan−1 20 − tan− 1 19)]
= cot(tan−1 20 − tan−1 1)
π
 π

= cot  − cot −1 20 −  − cot − 1 1
  2
 2
 
Exp. (c)
[Q tan−1 x + cot −1 x = π / 2 ]
We know that,
cos α ⋅ cos (2α ) cos(2 2 α )…cos (2 n−1α ) =
sin (2 α )
2 n sinα
n
= cot(cot −1 1 − cot −1 20)
351
Trigonometry
=
cot(cot −1 1)cot (cot −1 20) + 1
6 = 25 + 36 − c
[Qin the given equation of circle 2 g = 10
and 2 f = 12 ⇒ g = 5 and f = 6]
36 = 25 + 36 − c
c = 25
⇒
cot (cot −1 20) − cot (cot −1 1)
[Q cot ( A − B)] =
(1 × 20) + 1
20 − 1
21
=
19
cot A cot B + 1
cot B − cot A 
[Qcot (cot −1 x) = x]
=
28. In a triangle, the sum of lengths of two sides
27. If the area of an equilateral triangle
inscribed in the circle,
x 2 + y 2 + 10x + 12 y + c = 0
[JEE Main 2019, 11 Jan Shift-I]
[JEE Main 2019, 10 Jan Shift-II]
(b) −25
(c)13
is x and the product of the lengths of the
same two sides is y. If x 2 − c 2 = y , where c is
the length of the third side of the triangle,
then the circumradius of the triangle is
c
(a)
3
is 27 3 sq units, then c is equal to
(a) 20
⇒
⇒
(d) 25
c
(b)
3
(c)
3
y
2
(d)
y
3
Exp. (b)
a
b
c
=
=
= 2R
sin A sin B sinC
and given that, a + b = x, ab = y and x2 − c 2 = y
We know that,
Exp. (d)
Clearly, centre of the circumscribed circle is the
centroid (G) of the equilateral ∆ABC.
[Qin an equilateral triangle
circumcentre and centroid coincide]
A
C
c
60°
r
(–5,–6)
G
120°
B
r
B
Also, we know that
∆AGB ≅ ∆BGC ≅ ∆CGA
[by SAS congruence rule]
∴ ar(∆ABC ) = 3 ar(∆AGB)
1
= 3  r 2 sin 120° 
2

1
[Qarea of triangle ∆ = ab sin (∠C )]
2
Q ar(∆ABC ) = 27 3
[given]
r2 = 4 × 9 ⇒ r = 6
Now, radius of circle,
r=
g2 + f2 − c
3
]
2
a
C
∴
(a + b )2 − c 2 = ab
⇒
a2 + b 2 − c 2 = − 2 ab + ab
⇒
a2 + b 2 − c 2 = − ab
⇒
∴
a2 + b 2 − c 2 − ab
1
=
=−
2 ab
2 ab
2
1
cosC = − ⇒ C = 120°
2
[using cosine rule, cosC =
Now,
⇒
3 2 3
r
= 27 3
2
2
[sin 120° = sin (180° − 60° ) = sin 60° =
⇒
b
R
A
∴
O
∴
a2 + b 2 − c 2
]
2 ab
c
= 2R
sinC
1
c
c 2
R=
=
2 sin(120° ) 2 3
c
R=
3
29. All x satisfying the inequality
(cot − 1 x )2 − 7(cot − 1 x ) + 10 > 0, lie in the
interval
[JEE Main 2019, 11 Jan Shift-II]
352
JEE Main Chapterwise Mathematics
(a)
(b)
(c)
(d)
( − ∞ ,cot 5) ∪ (cot 2 , ∞ )
(cot 5, cot 4)
(cot 2 , ∞ )
( − ∞ ,cot 5) ∪ (cot 4,cot 2 )
Exp. (c)
Given,
(cot −1 x)2 − 7(cot −1 x) + 10 > 0
⇒(cot −1 x − 2 )(cot −1 x − 5) > 0
(by factorisation)
⇒cot −1 x < 2 or cot −1 x > 5
By wavy curve method,
cot −1 x ∈ (−∞, 2 ) ∪ (5, ∞ )
cot −1 x ∈ (0, 2 )
[Qrange of cot −1 x is (0, π )]
∴
a2 + b 2 − c 2 λ2 [49 + 36 − 25]
=
2 ab
84λ2
60 5
=
=
84 7
1 7
19
25
Thus, cos A = = , cos B = , cosC =
5 35
35
35
cos A cos B cos C
1
=
=
=
7
19
25
35
⇒ (α, β, γ ) = (7, 19, 25)
and
x ∈ (cot 2, ∞ )
b + c c + a a +b
for a ∆ABC with
=
=
11
12
13
cos A cos B cosC
, then
usual notation. If
=
=
α
β
γ
the ordered triad (α , β , γ ) has a value
30. Given,
[JEE Main 2019, 11 Jan Shift-II]
(a) (19, 7, 25)
(b) (3, 4, 5)
(c) (5, 12, 13)
(d) (7, 19, 25)
cos C =
31. Let y = y ( x )be the solution of the differential
dy
+ y = x log e x ,( x > 1).
dx
2 y (2 ) = log e 4 − 1, then y (e ) is equal to
equation,
b+c c+ a a+ b
Given,
=
=
= λ (say)
11
12
13
B
a
C
...(i)
b + c = 11λ,c + a = 12λ and a + b = 13λ
⇒
2(a + b + c ) = 36λ
...(ii)
⇒
a + b + c = 18λ
From Eqs. (i) and (ii), we get
a = 7λ, b = 6λ, c = 5λ
b 2 + c 2 − a2
Now, cos A =
2 bc
λ2 [36 + 25 − 49] 12 1
=
=
=
60 5
60λ2
a2 + c 2 − b 2 λ2 [49 + 25 − 36] 19
=
=
cos B =
2 ac
35
70λ2
e
4
(d)
Given differential equation is
dy
x + y = xloge x, ( x > 1)
dx
dy 1
+ y = loge x
⇒
dx x
Which is a linear differential equation.
e2
4
∫
1
dx
x
= elog e
x
…(i)
= x
Now, solution of differential Eq. (i), is
y × x = ∫ (loge x) x dx + C
x2
x2 1
× dx + C
loge x − ∫
x
2
2
[using integration by parts]
x2
x2
… (ii)
⇒
yx =
+C
loge x −
2
4
Given that, 2 y(2 ) = loge 4 − 1
… (iii)
On substituting, x = 2, in Eq. (ii), we get
4
4
2 y(2 ) = loge 2 − + C,
4
2
[where, y(2 ) represents value of y at x = 2]
⇒
b
(c)
Exp. (c)
A
c
If
[JEE Main 2019, 12 Jan Shift-I]
e2
(b) −
2
e
(a) −
2
So, if = e
Exp. (d)
x
yx =
⇒
… (iv)
2 y(2 ) = loge 4 − 1 + C
[Q mlog a = log am ]
From Eqs. (iii) and (iv), we get
C=0
So, required solution is
yx =
x2
x2
loge x −
2
4
353
Trigonometry
e2
e2
loge e −
2
4
[where, y(e ) represents value of y at x = e]
e
y(e ) =
[Qloge e = 1.]
4
Now, at x = e, ey(e ) =
⇒
32.
π

The maximum value of 3 cos θ + 5 sin θ − 

6
for any real value of θ is
[JEE Main 2019, 12 Jan Shift-I]
(a)
79
2
(c) 31
(b) 34
(d) 19
1
6
⇒
6 x2 + 5 x − 1 = 0, 0 ≤ x <
⇒
6 x2 + 6 x − x − 1 = 0,
0≤ x<
⇒
6 x ( x + 1) − 1 ( x + 1) = 0,
0≤ x<
⇒
(6 x − 1)( x + 1) = 0,
0≤ x<
[Q x ≥ 0]
1
, − 1, 0 ≤ x <
6
1
[Q 0 ≤ x <
x= ,
6
x=
⇒
⇒
1
6
1
6
1
6
1
6
1
]
6
So ‘A’ is a singleton set.
Exp. (d)
π
Given expression 3cos θ + 5sin θ − 

6
π
π

= 3cos θ + 5  sinθcos − sin cos θ


6
6

 3
1
sinθ − cos θ
= 3cos θ + 5 
2

 2
5
5 3
sinθ
= 3cos θ − cos θ +
2
2
1
5 3
sinθ
= cos θ +
2
2
34. If sin 4 α + 4cos 4 β + 2 = 4 2 sin α cos β;
α, β ∈[0, π], then cos(α + β ) − cos(α − β ) is
equal to
[JEE Main 2019, 12 Jan Shift-II]
(a) −1
(b)
(c) − 2
2
(d) 0
Exp. (c)
QThe maximum value of acos θ + b sinθ is a2 + b 2
By applying AM ≥ GM inequality, on the numbers
sin4 α, 4cos 4β , 1 and 1, we get
sin4 α + 4cos 4 β + 2
≥ ((sin4 α ) (4cos 4 β ) ⋅ 1⋅ 1)1/ 4
4
⇒sin4 α + 4cos 4 β + 2 ≥ 4 2 sinα cos β
1
5 3
So, maximum value of cos θ +
sinθ is
2
2
But, it is given that
sin4 α + 4cos 4 β + 2 = 4 2 sinα cos β
2
=
2
 1 +  5 3  =


 
2
 2 
1 75
76
+
=
= 19.
4 4
4
33. Considering only the principal values of
inverse functions, the set
π

A = x ≥ 0 : tan −1 ( 2 x ) + tan −1 ( 3 x ) = 
4

[JEE Main 2019, 12 Jan Shift-I]
(a) is an empty set
(b) is a singleton
(c) contains more than two elements
(d) contains two elements
Exp. (b)
Given equation is
π
tan− 1 (2 x) + tan−1 (3 x) = , x ≥ 0
4
5x
π
−1
tan
= , 6 x2 < 1
⇒
4
1 − 6 x2
 x + y
[Q tan− 1 x + tan− 1 y = tan−1 
 , xy < 1]
 1 − xy 
1
5x
= 1, x2 <
⇒
2
6
1 − 6x
So, sin4 α = 4cos 4 β = 1
[Qin AM ≥ GM , equality holds when all given
positive quantities are equal.]
1
…(i)
⇒
sinα = 1 and sinβ =
2
[Qα, β ∈ [0, π ]]
Now, cos (α + β ) − cos (α − β )
= −2sinα sinβ
Qcos C − cos D = 2 sin C + D sin D − C 

2
2 
1
[from Eq. (i)]
= −2 × 1 ×
2
=− 2
35. If the angle of elevation of a cloud from a
pointP which is 25 m above a lake be 30º and
the angle of depression of reflection of the
cloud in the lake from P be 60º, then the
height of the cloud (in meters) from the
surface of the lake is
[JEE Main 2019, 12 Jan Shift-II]
(a) 50
(b) 60
(c) 45
(d) 42
354
JEE Main Chapterwise Mathematics
Exp. (a)
According to given information, we have the
following figure,
Q Cloud
30°
P 60°
25 m
xm
M
25 m
R Image of cloud
In ∆PRM,
⇒
2(4cos 3 x − 3cos x) = 1
⇒
2 cos 3 x = 1 ⇒ cos 3 x =
x
y
25 + (25 + x)
tan 60° =
y
tan 30° =
x
 = 1

1
2
π 5π 7 π
[0 ≤ 3 x ≤ 3 π]
,
,
3 3 3
π 5 π 7π
⇒
,
x= ,
9 9 9
π 5 π 7 π 13 π
13 π
Sum = +
⇒ kπ =
+
=
9
9
9
9
9
13
Hence, k =
9
(25 +x)m
In ∆PQM,
 −3 + 4cos 2
8cos x
4

⇒
Surface
y
⇒
…(i)
…(ii)
On eliminating ‘y’ from Eqs. (i) and (ii), we get
25 + (25 + x)
3=
3x
⇒
3 x = 50 + x
⇒
2 x = 50 ⇒ x = 25 m.
∴ Height of cloud from surface
= x + 25 = 50 m
3x =
37. PQR is a triangular park with PQ = PR = 200
m. A TV tower stands at the mid-point ofQR.
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
[JEE Main 2018]
(a) 100
(c)100 3
(b) 50
(d) 50 2
Exp. (a)
P
36. If sum of all the solutions of the equation
 π
 1
π

8 cos x ⋅  cos + x ⋅ cos − x − 
 2
6

 6
= 1 in [ 0, π ] is kπ, then k is equal to
2
3
8
(c)
9
13
9
20
(d)
9
(b)
Exp. (b)
Key Idea Apply the identity
cos( x + y)cos( x − y) = cos 2 x − sin2 y
and cos 3 x = 4cos 3 x − 3cos x
We have,
π
π
1

8cos x cos + x cos − x −  = 1
6
 2
6


π
1
8cos x cos 2 − sin2 x −  = 1
⇒

6
2
3
1

2
8cos x − sin x −  = 1
⇒
4
2
3 1

8cos x − − 1 + cos 2 x = 1
⇒
4 2

200 m
T
[JEE Main 2018]
(a)
45°
200 m
Q
30°
90°
30°
M
Let height of tower TM be h.
TM
In ∆PMT, tan45° =
PM
h
1=
⇒
PM
⇒
PM = h
h
;
In ∆TQM, tan30° =
QM
QM = 3h
In ∆PMQ, PM 2 + QM 2 = PQ 2
h2 + ( 3h)2 = (200)2
⇒
4h2 = (200)2
⇒
h = 100 m
R
355
Trigonometry
38. If 5 (tan 2 x − cos 2 x ) = 2 cos 2 x + 9, then the
value of cos 4x is
3
(a) −
5
3
1
dy
= 2⋅
⋅ 3 × ( x)1/ 2
3/ 2 2
2
dx
1 + (3 x )
∴
[JEE Main 2017 (offline)]
1
(b)
3
(c)
2
9
(d) −
∴
Exp. (d)
Given,
5 (tan x − cos x) = 2 cos 2 x + 9
 1 − cos 2 x 1 + cos 2 x 
−
⇒ 5
 = 2 cos 2 x + 9
2
 1 + cos 2 x

2
9
=
7
9
g ( x) =
2
1 + 9 x3
9
⋅
x
1 + 9 x3
40. Let a vertical tower AB have its end A on the
level ground. Let C be the mid-point of AB
and P be a point on the ground such that
Put cos2 x = y, we have
 1 − y 1 + y
−
5
 = 2y + 9
2 
1 + y
AP = 2 AB . If ∠BPC = β, then tanβ is equal to
⇒
5 (2 − 2 y − 1 − y − 2 y) = 2(1 + y)(2 y + 9)
⇒
5(1 − 4 y − y2 ) = 2(2 y + 9 + 2 y2 + 9 y)
6
(a)
7
2
⇒
5 − 20 y − 5 y2 = 22 y + 18 + 4 y2
⇒
9 y2 + 42 y + 13 = 0
⇒
9 y2 + 3 y + 39 y + 13 = 0
⇒
⇒
⇒
3 y(3 y + 1) + 13(3 y + 1) = 0
(3 y + 1)(3 y + 13) = 0
1 13
y=− ,−
3
3
1 13
cos 2 x = − , −
3
3
1
Qcos2 x ≠ − 13 
cos2 x = −

3
3 
2
1
cos 4 x = 2 cos 2 2 x − 1 = 2  −  − 1
 3
7
2
= − 1= −
9
9
∴
⇒
Now,
1
39. For x ∈ 0,  , if the derivative of

tan
−1
4
 6x x 

 is x ⋅ g ( x ), then g ( x )equals
 1 − 9x 3 
[JEE Main 2017 (offline)]
(a)
(c)
9
1 + 9x 3
3x
1 − 9x 3
(b)
(d)
3x x
1 − 9x 3
3
1 + 9x 3
Exp. (a)
3/ 2 
 6x x 
)
− 1  2 ⋅ (3 x

Let y = tan− 1 
=
tan


3/ 2 2 
3
1
3
x
−
x
−
1
9
(
)





2x 
= 2 tan− 1(3 x3 / 2 ) Q2 tan− 1 x = tan− 1

1 − x2 

[JEE Main 2017 (offline)]
1
(b)
4
(c)
2
9
(d)
4
9
Exp. (c)
Let AB = h, then AD = 2 h and AC = BC =
h
2
Again, let ∠CPA = α
B
h/2
C
h
h/2
β
α
A
P
2h
AB
h
1
=
=
AP 2 h 2
h
AC
1
2
Also, in ∆ACP, tanα =
=
=
AP 2 h 4
Now, in ∆ABP, tan (α + β ) =
Now, tanβ = tan[(α + β ) − α ]
1 1
1
−
tan(α + β ) − tanα
2
=
= 2 4 = 4 =
9
1 + tan(α + β )tanα 1 + 1 × 1 9
2 4 8
41. If0 ≤ x < 2 π, then the number of real values of
x, which satisfy the equation
cos x + cos 2 x + cos 3x + cos 4x = 0, is
[JEE Main 2016 (offline)]
(a) 3
(b) 5
(c) 7
(d) 9
Exp. (c)
Given equation is
cos x + cos 2 x + cos 3 x + cos 4 x = 0
⇒ (cos x + cos 3 x) + (cos 2 x + cos 4 x) = 0
356
JEE Main Chapterwise Mathematics
⇒
⇒
2 cos 2 x cos x + 2 cos 3 x cos x = 0
2 cos x (cos 2 x + cos 3 x) = 0
x
5x
cos  = 0
2 cos x  2 cos
⇒

2
2
5x
x
cos x ⋅ cos
⋅ cos = 0
⇒
2
2
5x
x
= 0 or cos = 0
⇒ cos x = 0 or cos
2
2
π 3π
Now, cos x = 0 ⇒ x = ,
[Q 0 ≤ x < 2 π]
2 2
5 x π 3 π 5 π 7 π 9 π 11π
5x
,
= ,
,
,
= 0⇒
cos
2
2 2 2 2 2 2
2
...,
π 3π
7 π 9π
[Q 0 ≤ x < 2 π]
⇒
, π,
,
x= ,
5 5
5 5
x
x π 3π 5π
and cos = 0 ⇒ = ,
,
, ...
2
2 2 2 2
[Q 0 ≤ x < 2 π]
⇒
x= π
π 3π
π 3π 7 π 9π
Hence,
, π, ,
,
,
x= ,
2 2
5 5 5 5
42. A man is walking towards a vertical pillar in a
straight path, at a uniform speed. At a
certain point A on the path, he observes that
the angle of elevation of the top of the pillar
is 30°. After walking for 10 min from A in the
same direction, at a pointB, he observes that
the angle of elevation of the top of the pillar
is 60°. Then, the time taken (in minutes) by
him, from B to reach the pillar, is
and
h= 3 y
From Eqs. (i) and (ii),
x+ y
= 3 y
3
⇒
x + y = 3y
⇒
x − 2y = 0
x
⇒
y=
2
Q Speed is uniform.
∴ Distance y will be cover in 5 min.
Q Distance x covered in 10 min.
x
∴Distance will be cover in 5 min.
2
...(ii)
43. If the angles of elevation of the top of a tower
from three collinear points A, B and C on a
line leading to the foot of the tower are 30°,
45° and 60° respectively, then the ratio
[JEE Main 2015]
AB : BC is
(a) 3 :1
(c)1 : 3
(b) 3 : 2
(d) 2 : 3
Exp. (a)
According to the given information, the figure
should be as follows.
Let the height of tower = h
E
h
[JEE Main 2016 (offline)]
(a) 6
(b) 10
(c) 20
(d) 5
30°
Exp. (d)
A
According to given information, we have the
following figure
A
30º
x
B
60º
y
D
ED
AD
ED
h
1
=
=
3 AD AD
AD = h 3
tan30° =
⇒
In ∆EDB,
tan45° =
C
Now, from Right ∆ACD and Right ∆BCD, we have
h
tan 30° =
x+ y
x+ y
h
...(i)
and
tan 60° = ⇒ h =
3
y
C
In ∆EDA,
D
Pillar h
60°
45°
B
In ∆EDC,
h
h
⇒ CD =
CD
3
AB AD − BD
AB h 3 − h
⇒
=
=
h
BC BD − CD
BC
h−
3
tan60° =
Now ,
h
⇒ BD = h
BD
357
Trigonometry
⇒
AB AD − BD
=
BC BD − CD
∴
⇒
3 −1
AB
×
=
BC ( 3 − 1)
⇒
3
∴
 2x 
 , where
1 − x2
(a)
(c)
1
3
. Then, a value of y is
[JEE Main 2015]
3x − x 3
3x − x
3
1 + 3x
2
3x + x 3
(b)
1 − 3x 2
(d)
1 − 3x 2
3x + x 3
1 + 3x 2
Given,
 2x 
,
tan−1 y = tan−1 x + tan−1 
2
1 − x 
1
where| x| <
3


2x
 x+
2 
1− x 

tan−1 y = tan−1 
⇒


1 − x 2 x  
 1 − x2  




 x + y
[Qtan−1 x + tan−1 y = tan−1 
,
 1 − xy 
x > 0, y > 0, xy < 1]
3

−
+ 2x 
x
x

= tan−1 
2
2
1 − x − 2 x 
−1 
3 x − x3 

tan y = tan 
2 
 1 − 3x 
−1
y=
1
k
k ≥ 1, then f 4 ( x ) − f 6 ( x ) is equal to
1
6
1
(c)
4
1
3
1
(d)
12
(a)
(b)
[JEE Main 2014]
Exp. (d)
1
(sink x + cos k x), where x ∈ R and k ≥ 1
k
Now, f4 ( x) − f6 ( x)
3 x − x3
1 − 3 x2
=
1
1
(sin4 x + cos 4 x) − (sin6 x + cos 6 x)
4
6
1
1
(1 − 2 sin2 x ⋅ cos 2 x) − (1 − 3sin2 x ⋅ cos 2 x)
6
4
1 1
1
= − =
4 6 12
=
46. A bird is sitting on the top of a vertical pole
20 m high and its elevation from a pointO on
the ground is 45°. It flies off horizontally
straight away from the pointO. After 1s, the
elevation of the bird fromO is reduced to 30°.
Then, the speed (in m/s) of the bird is
(a) 40( 2 − 1)
(c) 20 2
(b) 40( 3 − 2 )
(d) 20( 3 − 1)
[JEE Main 2014]
Exp. (d)
In ∆OA1 B1,
tan45° =
1
1
1
< x<
⇒ −
3
3
3
x = tanθ
π
π
− < θ<
6
6
| x| <
⇒
1 − 3 x2
45. If fk ( x ) = (sink x + cosk x ), where x ∈R ,
Aliter
Let
3 x − x3
fk ( x) =
Exp. (a)
⇒
y=
⇒
44. Let tan −1 y = tan −1 x + tan −1 
|x |<
= θ + 2θ = 3θ
y = tan3θ
3 tanθ− tan3 θ
y=
1 − 3 tan2 θ
⇒
3
AB
=
1
BC
AB : BC = 3 : 1
⇒
tan−1 y = θ + tan−1(tan2θ)
⇒
AB1
20
⇒
=1
OB1
OB1
OB = 20
In ∆OA2 B2 , tan 30° =
20
OB2
⇒
OB2 = 20 3
⇒
B1B2 + OB1 = 20 3
358
JEE Main Chapterwise Mathematics
⇒
B1B2 = 20 3 − 20
A1
⇒ x=
A2
q 2 sin θ + pq cos θ − pq cos θ + p2 sin θ
p cos θ + q sin θ
( p2 + q 2 ) sin θ
p cos θ + q sin θ
⇒ AB =
20 m
20 m
45°
30°
O
Alternate Solution
20
a
b
c 
Applying sine rule in ∆ABD, 
=
=

 sin A sin B sinC 
B2
B1
B1B2 = 20 ( 3 − 1) m
Distance 20( 3 − 1)
Speed =
=
Time
1
⇒
∴
q
D
θ
p2
+q
= 20( 3 − 1) m/s
47. ABCD is a trapezium such that AB and CD
are parallel and BC ⊥ CD. If ∠ADB = θ,
BC = p andCD = q , then AB is equal to
[JEE Main 2013]
(p + q ) sin θ
p cos θ + q sin θ
p2 + q2
2
(a)
(c)
2
p + q 2 cos θ
p cos θ + q sin θ
(p 2 + q 2 ) sin θ
(d)
p 2 cos θ + q 2 sin θ
(p cos θ + q sin θ)2
Exp. (a)
Let AB = x
q
D
θ
B
p2 + q 2
p2 + q 2
AB
AB
=
=
⇒
sin θ sin { π − (θ + α )}
sin θ sin (θ + α )
p
p2
+q
AB =
⇒
2
p
p2 + q 2 sin θ
sin θ cos α + cos θ sin α
( p2 + q 2 ) sin θ
q sin θ + p cos θ

q
and sin α =
Qcos α =

p2 + q 2
=
C
α
=
x–q
M
q
B
x
p
x−q
p
tan (θ + α ) =
q− x
In ∆DAM, tan ( π − θ − α ) =
⇒
⇒
⇒
q − x = p cot (θ + α )
x = q − p cot (θ + α )
 cot θ cot α − 1 
q
=q − p
 Qin ∆BDC, cotα = 
p
 cot α + cot θ  
 q cot θ − 1




 = q − p  q cot θ − p 
=q − p p
 q + cot θ 
 q + p cot θ 



 p
 q cos θ − p sin θ 
=q − p

 q sin θ + p cos θ 


p + q 
p
2
2
( p2 + q 2 ) sin θ
p cos θ + q sin θ
48. The expression
π–(θ+α)
A
p
2
π–(θ+α)
A
2
(b)
C
α
tan A
cot A
can be
+
1 − cot A 1 − tan A
written as
[JEE Main 2013]
(a) sin A cos A + 1
(c) tan A + cot A
(b) sec A cosec A + 1
(d) sec A + cosec A
Exp. (b)
Given expression is
tan A
cot A
sin A
sin A
+
=
×
1 − cot A 1 − tan A cos A sin A − cos A
cos A
cos A
+
×
sin A cos A − sin A
=
sin3 A − cos 3 A 
1


sin A − cos A  cos A sin A 
sin2 A + sin A cos A + cos 2 A
sin A cos A
1 + sin A cos A
=
sin A cos A
=
= 1 + sec A cosec A
359
Trigonometry
49. In a ∆PQR , if 3 sin P + 4 cos Q = 6 and
⇒
4 sin Q + 3 cos P = 1, then the angle R is equal
to
[AIEEE 2012]
5π
6
π
(c)
4
π
6
3π
(d)
4
A = 1 − cos 2 x + cos 4 x
1 3
= cos 4 x − cos 2 x + +
4 4
2
1
3
=  cos 2 x −  +

2
4
(b)
(a)
2
1
1
where, 0 ≤  cos 2 x −  ≤

2
4
3
∴
≤ A≤1
4
Exp. (b)
Given A ∆PQR such that
3 sin P + 4 cos Q = 6
4 sin Q + 3 cos P = 1
…(i)
…(ii)
On squaring and adding the Eqs. (i) and (ii), we
get
(3 sin P + 4 cos Q )2 + (4 sin Q + 3 cos P)2 = 36 + 1
⇒ 9 (sin2 P + cos 2 P) + 16(sin2 Q + cos 2 Q )
+ 2 × 3 × 4 (sin P cos Q + sin Q cos P) = 37
⇒
24[sin (P + Q )] = 37 − 25
1
sin (P + Q ) =
2
Since, P and Q are angles of ∆PQR, hence
0° < P, Q < 180°.
⇒
P + Q = 30° or 150°
⇒
R = 150° or 30°
Hence, two cases arise here.
Case I
R = 150°
R = 150°
⇒ P + Q = 30°
⇒
0 < P, Q < 30°
1
sin P < , cos Q < 1
⇒
2
3
3 sin P + 4 cos Q < + 4
⇒
2
11
3 sin P + 4 cos Q <
<6
⇒
2
⇒ 3 sin P + 4 cos Q ⇒ 6 is not possible.
⇒
Case II
R = 30°
Hence, R = 30° is the only possibility.
50. If A = sin 2 x + cos 4 x , then for all real x
13
(a) ≤ A ≤ 1
16
3
13
(c) ≤ A ≤
4
16
(b)1 ≤ A ≤ 2
3
(d) ≤ A ≤ 1
4
Exp. (d)
A = sin2 x + cos 4 x
…(i)
[AIEEE 2011]
…(ii)
51. The possible values of θ ∈(0, π ) such that
sin (θ ) + sin ( 4θ ) + sin ( 7θ ) = 0 are [AIEEE 2011]
2 π π 4π π 3π 8π
, ,
, ,
,
9 4 9 2 4 9
π 5π π 2 π 3π 8π
(b) ,
, ,
,
,
4 12 2 3 4 9
2 π π π 2 π 3 π 35 π
(c)
, , ,
,
,
9 4 2 3 4 36
2 π π π 2 π 3π 8π
(d)
, , ,
,
,
9 4 2 3 4 9
(a)
Exp. (a)
sin θ + sin 4θ + sin 7 θ = 0
⇒
sin 4θ + (sin θ + sin 7 θ) = 0
⇒
sin 4θ + 2 sin 4θ ⋅ cos 3θ = 0
⇒
sin 4θ {1 + 2 cos 3θ} = 0
⇒
sin 4 θ = 0, cos 3 θ = −
As ,
∴
∴
⇒
⇒
0< θ< π
0 < 4θ < 4 π
4 θ = π, 2 π, 3 π
1
cos 3 θ = −
2
0 < 3 θ < 3π
2 π 4π 8π
,
,
3θ =
3 3 3
π π 3 π 2 π 4π 8 π
θ= , ,
,
,
,
4 2 4 9 9 9
4
5
52. Let cos(α + β ) = and sin(α − β ) =
0≤ α ,β ≤
25
16
19
(c)
12
(a)
1
2
5
, where
13
π
. Then, tan2 α is equal to
4
56
33
20
(d)
7
(b)
[AIEEE 2010]
360
JEE Main Chapterwise Mathematics
Exp. (b)
4
⇒ α + β ∈ Ist quadrant
5
5
and sin(α − β ) =
⇒ α − β ∈ Ist quadrant
13
Now, 2α = (α + β ) + (α − β )
tan(α + β ) + tan(α − β )
tan2 α =
∴
1 − tan(α + β )tan(α − β )
3 5
+
56
= 4 12 =
3 5
33
1− ⋅
4 12
cos(α + β ) =
53. For a regular polygon, let r and R be the radii
of the inscribed and the circumscribed
circles. A false statement among the
following is
[AIEEE 2010]
r 1
=
R 2
1
r
(b) there is a regular polygon with =
2
R
r 2
(c) there is a regular polygon with =
R 3
3
r
(d) there is a regular polygon with =
2
R
(a) there is a regular polygon with
Exp. (c)
R
2π
n
π r
n
By formula of regular polygon,
π
a
= sin
2R
n
π
a
and
= tan
2r
n
54. Let A and B denote the statements
A :cos α + cos β + cos γ = 0
B : sin α + sin β + sin γ = 0
3
If cos (β − γ ) + cos ( γ − α ) + cos (α − β ) = − ,
2
then
(a)
(b)
(c)
(d)
∴
[AIEEE 2009]
Exp. (c)
cos (β − γ ) + cos (γ − α ) + cos (α − β ) = −
3
2
⇒ 2 [cos (β − γ ) + cos (γ − α ) + cos (α − β )]
+ 3=0
⇒ 2 [cos (β − γ ) + cos (γ − α ) + cos (α − β )]
+ sin2 α + cos 2 α + sin2 β + cos 2 β + sin2 γ
+ cos 2 γ = 0
⇒(sin α + sin β + sin γ )2
+ (cos α + cos β + cos γ )2 = 0
It is possible when,
sinα + sinβ + sin γ = 0
and
cos α + cos β + cos γ = 0
Hence, both statements A and B are true.
5
2
55. The value of cot cosec −1 + tan −1  is

5
(a)
17
3
(c)
17
Exp. (b)
3
6
(b)
17
4
(d)
17
3
[AIEEE 2008]
5
3
Since, cosec −1   = tan−1  
 4
 3
∴
π
r
= cos
R
n
r
1
=
n = 3 gives
R 2
r
1
=
n = 4 gives
R
2
3
r
n = 6 gives
=
2
R
A is true and B is false
A is false and B is true
Both A and B are true
Both A and B are false
3
2
cot  tan−1 + tan−1 

4
3
3
2
 + 


= cot tan−1  4 3 
1
 1− 

2 
  17  
   
= cot tan−1  12  = cot
  1  
  2  
5
4
6
 −1  17  
tan   =

 6   17
3
361
Trigonometry
56. AB is a vertical pole with B at the ground
level and A at the top. A man finds that the
angle of elevation of the point A from a
certain point C on the ground is 60°. He
moves away from the pole along the line BC
to a point D such thatCD = 7 m. From D, the
angle of elevation of the point A is 45°. Then,
the height of the pole is
[AIEEE 2008]
(a)
7 3 1 

m
2  3 + 1
(b)
7 3 1 

m
2  3 −1
(c)
7 3
( 3 + 1) m
2
(d)
7 3
( 3 −1) m
2
Exp. (c)
In ∆ABC,
BC = hcot 60°
7 3
×
3 −1
=
park. A and B are two points on the
boundary of the park such that AB ( = a )
subtends an angle of 60° at the foot of the
tower and the angle of elevation of the top of
the tower from A orB is 30°. The height of the
tower is
[AIEEE 2007]
(b) 2 a 3
(d)
3
C
h
O
a
B
2
of x is
(b) 3
(d) 5
Exp. (b)
x
5
π
Q sin−1   + cosec −1   =
 4 2
 5
x
4
π
sin−1   + sin−1   =
⇒
 5
 5 2
60°
a
90°
a 30°
A
OB = OA = radii
∠OBA = ∠OAB
= 60°
So, ∆OAB is an equilateral.
∴
OA = OB = AB = a.
In ∆OAC,
h
tan 30° =
a
1
h
⇒
=
3 a
a
h=
⇒
3
Also,
∴
x
5
π
57. If sin −1   + cosec −1   = , then the value
(a) 1
(c) 4
58. A tower stands at the centre of a circular
B
3+1
3+1
 4
x=3
Let h be the height of a tower.
Since, ∠ AOB = 60°
7 3
( 3 + 1) m
2
 5
∴
Exp. (c)
and in ∆ABD,
BD = hcot 45°
Q
BD − BC = DC
⇒ h cot 45° − h cot 60° = 7
7
h=
cot 45° − cot 60°
7
=
1 − 1 



3
=
⇒
2a
3
a
(c)
3
h
60°
45°
7m C
⇒
x
π
4
sin−1   = − sin−1  
 5
 5 2
x
4
sin−1   = cos −1  
 5
 5
x
3
sin−1   = sin−1  
 5
 5
(a)
A
D
⇒
[AIEEE 2007]
59. The number of values of x in the interval
satisfying
the
[0, 3π]
2 sin 2 x + 5 sin x − 3 = 0 is
(a) 6
(c) 2
(b) 1
(d) 4
equation
[AIEEE 2006]
362
JEE Main Chapterwise Mathematics
Exp. (d)
Exp. (b)
Given equation is 2 sin x + 5sin x − 3 = 0.
⇒
(2 sin x − 1)(sin x + 3) = 0
1
⇒ sin x =
[Qsin x ≠ − 3]
2
2
Y
y = sin x
π
O π/6
3π
2π
y = 1/2
X
60. A triangular park is enclosed on two sides by
a fence and on the third side by a straight
river bank. The two sides having fence are of
same length x. The maximum area enclosed
by the park is
[AIEEE 2006]
x3
8
(b)
1 2
x
2
(c) πx 2
(d)
3 2
x
2
Exp. (b)
1
× Base × Altitude
2
1
= × (2 x cos θ) × ( x sin θ)
2
1
= x2 sin 2 θ
2
[(since, maximum value of sin 2θ is 1)]
1
Maximum area = x2
2
Area =
∴
1
2
x
x
2 tan
1
2
2
+
=
∴
2 x
2 x
2
1 + tan
1 + tan
2
2
x
Let
tan = t
2
1− t 2
2t
1
+
=
⇒
1+ t2 1+ t2 2
1 − tan2
⇒
It is clear from figure that the curve intersect the
line at four points in the given interval.
Hence, number of solutions are 4.
(a)
cos x + sin x =
Given,
2(1 − t 2 + 2t ) = 1 + t 2
⇒
3t 2 − 4t − 1 = 0
2± 7
3
x π
As
0< x< π ⇒ 0< <
2 2
x
So, tan is positive.
2
x 2+ 7
t = tan =
∴
2
3
x
2 tan
2 = 2t
Now,
tan x =
x 1− t2
1 − tan2
2
2 + 7 
2

 3 
⇒
tan x =
2
2 + 7 
1− 

 3 
⇒
t =
−3 (2 + 7 ) 1 − 2 7
×
1+ 2 7
1− 2 7
⇒
tan x =
⇒
4+ 7
tan x = − 

 3 
π
P
Q
62. In a ∆PQR , ∠R = . If tan   and tan  
2
2
are the roots of ax + bx + c = 0,a ≠ 0, then
x sin θ
2
x
2
x
θ
x cos θ
(b) b = c
(c) c = a + b
(d) a = b + c
[AIEEE 2005]
x cos θ
1
2
61. If 0 < x < π and cos x + sin x = , then tan x is
(4 − 7)
(a)
3
(1 + 7 )
(c)
4
(a) b = a + c
(4 + 7)
(b) −
3
(1 − 7 )
(d)
4
[AIEEE 2006]
Exp. (c)
Since, tan
∴
P
Q
and tan are the roots of equation
2
2
ax2 + bx + c = 0.
tan
P
Q
b
+ tan = −
2
2
a
…(i)
363
Trigonometry
P
Q c
tan =
2
2 a
P Q R π
+
+
=
[Q P + Q + R = π ]
2
2
2 2
P+Q π R
= −
2
2 2
P+Q π
π
=
[Q ∠R = (given)]
2
4
2
P Q
tan  +  = 1
2
2
P
Q
tan + tan
2
2 =1
P
Q
1 − tan tan
2
2
b
−
a =1
c
1−
a
b
c
[from Eq. (i)]
− = 1−
a
a
and
tan
Also,
⇒
⇒
⇒
⇒
⇒
⇒
⇒
−b = a − c
⇒
c=a+ b
Alternate Solution
Since,
⇒
⇒
∴
⇒
⇒
⇒
⇒
π
2
π
∠P + ∠Q =
2
∠P π ∠Q
= −
2
4
2
π Q
P

tan = tan  − 
4 2
2
Q
π
tan − tan
4
2
=
Q
π
1 + tan tan
4
2
P
P
Q
Q
tan + tan tan = 1 − tan
2
2
2
2
P
Q
P
Q
tan + tan = 1 − tan tan
2
2
2
2
b
c
⇒ −b = a − c
− = 1−
a
a
c=a+ b
∠R =
π
2
R is the circumradius of the ∆ABC , then
[AIEEE 2005]
2 (r + R ) equal to
63. In a ∆ABC , let ∠C = ,if r is the inradius and
(a) c + a
(c) a + b
(b) a + b + c
(d) b + c
Exp. (c)
We know that,
c
= 2R
sin C
⇒
c = 2R
…(i)
[QC = 90° ]
C
r
=
2 s−c
π
r
tan =
4 s−c
and
tan
⇒
∴
r = s−c
a+ b+c
r=
−c
2
⇒
a + b − c = 2r
On adding Eqs. (i) and (ii), we get
2(r + R ) = a + b
⇒
y
2
4x 2 − 4xy cos α + y 2 is equal to
…(ii)
64. If cos −1 x − cos −1 = α , then
(a) − 4 sin 2 α
(b) 4 sin 2 α
(c) 4
(d) 2 sin 2 α
[AIEEE 2005]
Exp. (b)
y
=α
2
y2 
=α
1−
4 
cos −1 x − cos −1
Given that,
⇒
 xy
cos −1 
+
2

1 − x2
⇒
xy
+
2
1 − x2
1−
y2
= cos α
4
2 1 − x2
1−
y2
= 2 cos α − xy
4
⇒
On squaring both sides, we get
4(1 − x2 )(4 − y2 )
= 4cos 2 α + x2 y2 − 4xy cos α
4
⇒ 4 − 4 x2 − y2 + x2 y2 = 4cos 2 α + x2 y2
− 4xy cos α
∴
4 x − 4 xy cos α + y = 4sin α
2
2
2
65. If in a ∆ABC , the altitudes from the vertices
A , B andC on opposite sides are in HP, then
[AIEEE 2005]
sin A , sin B and sinC are in
(a) HP
(c) AP
Exp. (c)
(b) AGP
(d) GP
364
JEE Main Chapterwise Mathematics
In ∆ BAD,
cos (90° − B) =
AD
c
sin2 α + sin2 β + 2 sin α sin β
+ cos 2 α + cos 2 β + 2 cos α cos β
A
°–B
E
F
b
⇒
90
c
2
90°–A
B
90°–C
D
C
⇒
a
⇒
AD = c sin B
Similarly,
and
BE = asin C
CF = b sin A
⇒
⇒
Since, AD, BE and CF are in HP.
So, c sin B, asin C and b sin A are in HP.
1
1
1
and
,
⇒
sin C sin B sin A sin C
sin B sin A
are in AP. Hence, sin A, sin B and sin C are in AP.
67. The sides of a triangle are sin α ,cos α and
Alternate Solution
1
× BC × AD
2
1
∆ = × a × AD
⇒
2
2∆
AD =
⇒
a
2∆
2∆
and CF =
Similarly, BE =
b
c
Since, AD, BE and CF are in HP.
1 1
1
So, , and are in HP.
a b
c
ar (∆ABC ) =
Here, we see that the greatest side is c.
a2 + b 2 − c 2
∴ cos C =
2 ab
21
27
and cos α + cos β = − ,
65
65
α − β
then the value of cos 
 is
 2 
[AIEEE 2004]
6
65
(d) −
6
65
Exp. (a)
21
…(i)
65
27
and
…(ii)
cos α + cos β = −
65
On squaring and adding Eqs. (i) and (ii), we get
Given that, sin α + sin β = −
[AIEEE 2004]
(b) 90°
(d) 150°
and c = 1 + sin α cos α
sin α + sin β = −
(c)
greatest angle of the triangle is
Let a = sin α, b = cos α
66. Let α and β be such that π < α − β < 3π. If
3
130
π
. Then, the
2
Exp. (c)
Hence, a, b and c are in AP.
3
(b)
130
1 + sin α cos α for some 0 < α <
(a) 60°
(c) 120°
∴ sin A, sin B and sin C are in AP.
(a) −
∴
2
21
27 
=  −  +  −

 65 
 65 
2 + 2 (cos α cos β + sin α sin β )
441
729
=
+
4225 4225
1170
2 [1 + cos (α − β )] =
4225
α − β
1170
cos 2 
=

 2  4 × 4225
α − β
9
cos 2 
 =
 2  130
α − β
3
cos 
 =−
 2 
130
Q π < α − β < 3 π ⇒ π < α − β < 3 π 
2
2
2 

sin2 α + cos 2 α − 1 − sin α cos α
2 sin α cos α
sin α cos α
⇒ cos C = −
2 sin α cos α
1
⇒ cos C = − = cos 120°
2
⇒
∠C = 120°
⇒ cos C =
68. A person standing on the bank of a river,
observes that the angle of elevation of the
top of a tree on the opposite bank of the river
is 60° and when he retires 40 m away from
the tree the angle of elevation becomes 30°.
The breadth of the river is
(a) 20 m
(c) 40 m
(b) 30 m
(d) 60 m
[AIEEE 2004]
365
Trigonometry
Exp. (a)
Exp. (b)
Let CD(= h) be the height of the tree
and BC (= x) be the width of the river.
Let
AB = a, ON ⊥ AB
tan
h
B
40
In ∆CBD, tan 60° =
3=
⇒
⇒
60°
30°
and in ∆CAD,
x
C
CD
BC
h
⇒ h= x 3
x
…(i)
CD
AC
1
h
=
3 40 + x
h 3 = 40 + x
3 x = 40 + x
2 x = 40
x = 20 m
⇒
⇒
⇒
∴
[from Eq. (i)]
69. If f : R → S , defined by
f ( x ) = sin x − 3 cos x + 1, is onto, then the
interval of S is
[AIEEE 2004]
(a) [0, 3]
(b) [–1, 1] (c) [0, 1]
(d) [–1, 3]
Exp. (d)
Given that, f( x) = sin x − 3 cos x + 1
Q
−2 ≤ sin x − 3 cos x ≤ 2
[Q a2 + b 2 ≤ a sin x + b cos x ≤
⇒
−1 ≤ sin x −
∴ Range of f( x) = [−1, 3 ]
A
π AN
=
n ON
ON = AN cot
B
a/2 N a/2
π a
π
= cot
n 2
n
…(i)
π AN
=
n OA
π a
π
…(ii)
⇒ OA = AN cosec = cosec
n 2
n
Now, sum of the radii = ON + OA
π a
π
a
[from Eqs. (i) and (ii)]
= cot + cosec
2
n 2
n
 cos π

1 
a
n
= 
+
π
2  sin π
sin 

n
n

 2 cos 2 π
 1 + cos π 


2n 
a
a
n
= 
= 

2  sin π  2  2 sin π cos π 


2n
2 n 
n 
a
π
= cot
2
2n
and sin
tan 30° =
⇒
π/n
In ∆AON,
D
A
O
and AN = BN
a2 + b 2 ]
70. The sum of the radii of inscribed and
circumscribed circles for an n sided regular
polygon of side a , is
[AIEEE 2003]
π
(a) a cot  
n
 π 
(c) a cot 

 2n 
 π 
a
(d) cot 

4
 2n 
2
then the sides a ,b and c
(a) are in AP
(c) are in HP
2
2
[AIEEE 2003]
(b) are in GP
(d) satisfy a + b = c
Exp. (a)
3 cos x + 1 ≤ 3
 π 
a
(b) cot 

2
 2n 
3b
C
A
71. If in a ∆ABC , a cos 2   + c cos 2   = ,
⇒
C
A 3b
+ c cos 2
=
2
2
2
s(s − a) 3 b
s(s − c )


+c
=
a
 ab 
 bc 
2
s(s − c + s − a) 3 b
=
b
2
2 s (2 s − c − a) = 3b 2
⇒
2 s (a + b + c − c − a) = 3b 2
⇒
(a + b + c )b = 3b 2
Given,
⇒
⇒
acos 2
⇒
a + b + c = 3b
⇒
2b = a + c
Hence, a, b and c are in AP.
366
JEE Main Chapterwise Mathematics
72. In a ∆ABC ,medians AD and BE are drawn. If
π
π
and ∠ABE = , then the
3
6
area of the ∆ABC is
[AIEEE 2003]
AD = 4, ∠DAB =
8
sq units
3
32
(c)
sq units
3 3
(a)
16
sq units
3
64
(d)
sq units
3
(b)
AD = 4
π/6 8/3
E
and BD = DC
G
Since, the centroid G
π/3
4/3
divides the line AD
B
D
in the ratio 2 :1.
8
4
and DG =
AG =
∴
3
3
π AG
π
In ∆ ABG, tan =
⇒ BG = AG cot
3 BG
3
8
1
8
BG = ×
=
⇒
3
3 3 3
1
Area of ∆ADB = × AD × BG
2
8
1
16
= ×4×
=
3 3 3 3
2
C
∴ Area of ∆ABC = 2 × Area of ∆ ADB
16
32
sq units
=2 ×
=
3 3 3 3
73. The trigonometric equation
sin −1 x = 2 sin −1 a , has a solution for
(b) all real values of a
(d) | a | ≥
1
2 [AIEEE 2003]
Exp. (c)
Given that, sin−1 x = 2 sin−1 a
⇒
⇒
π
≤ sin−1 x ≤
2
π
− ≤ sin−1 a ≤
4
1
−
≤ a≤
2
−
(b) 40 m
(c) 60 m
(d) 80 m
Here, from the figure we see that,
3
θ2 = tan−1
5
3
tan θ2 =
⇒
5
Since, median divides a triangle into two
triangles of equal area.
Q
 3
subtends an angle tan −1   at a point in the
 5
horizontal plane through its foot and at a
distance 40 m from the foot. A possible
height of the vertical pole is
[AIEEE 2003]
Exp. (b)
A
Given,
1
1
< |a | <
2
2
1
(c) | a | ≤
2
3
(a) 20 m
Exp. (c)
(a)
74. The upper   th portion of a vertical pole
 4
π
π
π
⇒ − ≤ 2 sin−1 a ≤
2
2
2
π
π
π
⇒ sin  −  ≤ a ≤ sin
 4
4
4
1
1
⇒ |a| ≤
2
2
…(i)
B
3h
4
h
C
1h
4
θ2
θ1
O
40 m
A
AC
AO
1
h
h
tan θ1 = 4 =
40 160
tan θ1 =
In ∆ AOC,
⇒
and in ∆ AOB,
…(ii)
AB
h
=
AO 40
tan θ1 + tan θ2
h
⇒
=
1 − tan θ1 tan θ2 40
3
h
+
160 5 = h [from Eqs. (i) and (ii)]
⇒
h
3 40
×
1−
160 5
5[h + 96]
h
=
⇒
800 − 3h 40
tan (θ1 + θ2 ) =
⇒
200[h + 96] = 800h − 3h2
⇒
3h2 − 600h + 19200 = 0
⇒
h2 − 200h + 6400 = 0
⇒
⇒
(h − 160)(h − 40) = 0
h = 160 or h = 40
Hence, height of the vertical pole is 40 m.
367
Trigonometry
75. cot −1 ( cos α ) − tan −1 ( cos α ) = x , then sin x
is equal to
[AIEEE 2002]
α
(a) tan 2  
 2
A+C= π−B
A− B+ C π
B
⇒
= − B adding − both sides
2
2
2


Exp. (a)
Given that,
cot −1 ( cos α ) − tan−1 ( cos α ) = x
…(i)
cot −1 ( cos α ) + tan−1 ( cos α ) =
π
2
…(ii)
Qcot −1 x + tan−1 x = π 

2 
On adding Eqs. (i) and (ii), we get
π
2 cot −1 ( cos α ) = + x
2
⇒

A − B +C 
 is equal to

2
(c) b − c − a
(d) c 2 − a 2 − b 2
2
is true, if and only if
(a) x − y ≠ 0
(c) x + y ≠ 0
(b) x = − y [AIEEE 2002]
(d) x ≠ 0, y ≠ 0
sin2 θ ≤ 1
4 xy
( x + y)2


4 xy
≤ 1 Qsin2 θ =
, given
2
(
+
)
x
y


⇒
x2 + y2 + 2 xy − 4 xy ≥ 0
⇒
( x − y)2 ≥ 0
which is true for all real values of x and y
4 xy
will be
provided x + y ≠ 0, otherwise
( x + y)2
meaningless.
(b) c 2 + a 2 − b 2
2
( x + y )2
⇒
(a) a 2 + b 2 − c 2
2
4xy
Q
On squaring both sides, we get
x
x
x
x
cos 2 + sin2 − 2 sin cos
2
2
2
2
⇒
cos α =
x
x
x
x
cos 2 + sin2 + 2 sin cos
2
2
2
2
1 − sin x
cos α =
⇒
1 + sin x
α
1 − tan2
2 = 1 − sin x
⇒
2 α
1 + sin x
1 + tan
2
Applying componendo and dividendo rule, we get
α
sin x = tan2  
2
76. In a ∆ABC ,2 ca sin 
77. sin 2 θ =
Exp. (c)
x
π
cos α = cot  + 
 4 2
x
cot − 1
2
cos α =
x
1 + cot
2
x
x
cos − sin
2
2
cos α =
x
x
cos + sin
2
2
⇒
A − B + C
π

∴ 2 ca sin 
 = 2 ca sin  − B
2



2
2
2
2
a + c − b 

= 2 ac cos B = 2 ac 
2 ac


= a2 + c 2 − b 2
We know that,
⇒
A+ B+ C = π
Since,
⇒
α
(b) cot 2  
 2
α
(d) cot  
 2
(c) tan α
Exp. (b)
[AIEEE 2002]
78. The value of
(a) 1
1 − tan 2 15°
1 + tan 2 15°
(b)
3
is
(c)
[AIEEE 2002]
3
2
(d) 2
Exp. (c)
We know that,
1 − tan2 15°
1 + tan2 15°
= cos 30°
=
4
3
79. If tan θ = − , then sin θ is
4
4
but not
5
5
4 4
(b) − or
5 5
4
4
(c) but not −
5
5
(d) None of the above
(a) −
3
2
[AIEEE 2002]
368
JEE Main Chapterwise Mathematics
Exp. (b)
4
3
BC 4
sin θ =
=
∴
AC 5
But tan θ is negative which is A
possible only, if θ lies in IInd
and IVth quadrants.
4
4
So, sin θ may be or − .
5
5
Q
C
tan θ = −
5
θ
3
4
(b) c 2 + 3c + 7 = 0
(c) c 2 − 3c + 7 = 0
(d) c 2 + 3c − 7 = 0
Exp. (a)
a = 4, b = 3 and
c 2 + 9 − 16
Now, cos 60° =
2 × 3×c
1
2
[AIEEE 2002]
⇒ c 2 − 3c − 7 = 0
A 5
C 2
= , tan = , then
2 6
2 5
(a) a , c and b are in AP
Exp. (a)
sin (α + β ) = 1
π
…(i)
α+β=
⇒
2
1
and
sin (α − β ) =
2
π
…(ii)
α −β =
⇒
6
On solving Eqs. (i) and (ii), we get
π
π
and β =
α=
3
6
∴ tan (α + 2 β ) tan (2 α + β )
2 π
 5π
= tan 
 tan 

 3 
 6 
π
π
= tan  π −  tan  π − 


3
6
π 
π

=  − cot   − cot 

3 
6
1
=
× 3 =1
3
Q
[AIEEE 2002]
(b) y ≤ 2
(d) y ≥ 2
[AIEEE 2002]
(b) a , b and c are in AP
(c) b , a and c are in AP
(d) a , b and c are in GP
Exp. (b)
Given that,
A 5
C 2
and tan =
=
2 6
2 5
A
C 5 2
Now, tan tan = ×
2
2 6 5
(s − b )(s − c ) (s − a)(s − b ) 1
⇒
=
⋅
3
s(s − a)
s(s − c )
s−b 1
=
⇒
s
3
⇒
2 s = 3b
[Q2 s = a + b + c]
⇒
a + c = 2b
So, a, b and c are in AP.
tan
84. The equation a sin x + b cos x = c, where
|c | > a 2 + b 2 has
(a)
(b)
(c)
(d)
[AIEEE 2002]
a unique solution
infinite number of solutions
no solution
None of the above
Exp. (c)
Exp. (d)
∠ A = 60°
1 c2 − 7
=
2 2 × 3c
⇒
83. In a ∆ABC , tan
(b) –1
(c) zero
(d) None of the above
81. If y = sin 2 θ + cosec2 θ, θ ≠ 0, then
[AIEEE 2002]
(a) c 2 − 3c − 7 = 0
Given that,
tan (α + 2 β )tan (2 α + β ) is equal to
(a) y = 0
(c) y ≥ − 2
the root of the equation
B
80. If sin (α + β ) = 1 and sin (α − β ) = , then
(a) 1
82. In a ∆ABC , a = 4,b = 3 and ∠ A = 60°, thenc is
Given that,
y = sin2 θ + cosec 2 θ
Since,
∴
y = (sin θ − cosec θ)2 + 2
⇒
⇒
y ≥ 2, θ ≠ 0
|c| >
c< −
a2 + b 2
a2 + b 2
369
Trigonometry
c>
and
But −
a2 + b 2
a2 + b 2 ≤ a sin x + b cos x
≤
a2 + b 2
…(i)
and a sin x + b cos x = c
…(ii)
From Eqs. (i) and (ii), we see that no solution
exists.
85. If α is a root of 25cos 2 θ + 5cos θ − 12 = 0,
π
< α < π , then sin 2 α is equal to
2
24
(a)
25
13
(c)
18
24
(b) −
25
13
(d) −
18
π
< α < π i.e., in second quadrant.
2
4
cos α = −
∴
5
3
sin α =
⇒
5
Now,
sin 2 α = 2 sin α cos α
3
4
24
= 2 × ×  −  = −
5  5
25
But
1
2
86. tan −1   + tan −1   is equal to
 9
 4
[AIEEE 2002]
Exp. (b)
25 cos 2 α + 5 cos α − 12 = 0
⇒
(5 cos α − 3)(5 cos α + 4) = 0
⇒
cos α = −
[AIEEE 2002]
1
3
(b) sin −1  
 5
2
1
(d) tan −1  
 2
Exp. (d)
Since, α is a root of
25 cos 2 θ + 5 cos θ − 12 = 0
∴
1
3
(a) cos−1  
 5
2
1
3
(c) tan −1  
 5
2
−1
Now, tan
3
4
and
5
5
 1  + tan−1
 
 4
 1+ 2 
 2  = tan−1  4 9 


 
 9
1 − 1 × 2 

4 9
17
1
= tan−1   = tan−1  
 34 
2
16
Mathematical Reasoning
1. The contrapositive of the statement ‘‘If you
are born in India, then you are a citizen of
India’’, is
[JEE Main 2019, 8 April Shift-I]
(a) If you are not a citizen of India, then you are
not born in India.
(b) If you are a citizen of India, then you are
born in India.
(c) If you are born in India, then you are not a
citizen of India.
(d) If you are not born in India, then you are
not a citizen of India.
Exp. (a)
Given statement is ‘‘If you are born in India, then
you are a citizen of India’’.
Now, let statement p : you are born in India
and q : you are citizen of India.
Then, given statement, ‘‘If you are born in India
then you are a citizen of India’’ is equivalent to
p ⇒q.
QThe contrapositive of statement p ⇒ q is
~ q ⇒ ~ p.
∴ The contrapositive of the given statement is ‘‘If
you are not a citizen of India, then you are not born
in India.
2. Which one of the following statements is not
a tautology?
[JEE Main 2019, 8 April Shift-II]
(a) (p ∧ q ) → (~ p ) ∨ q
(b) (p ∧ q ) → p
(c) p → (p ∨ q )
(d) (p ∨ q ) → (p ∨ (~ q ))
Exp. (d)
(a) ( p ∧ q ) → (~ p) ∨ q
≡ (~( p ∧ q )) ∨ ((~ p) ∨ q )
≡ ((~ p) ∨ (~ q )) ∨ ((~ p) ∨ q )
≡ (~ p) ∨ (~ q ) ∨ q [Q(~ p) ∨ (~ p) ≡ ~ p]
≡ (~ p) ∨ T
[Q~ q ∨ q ≡ T]
≡T
So, it is a tautology
[Q ((~ q ) ∨ q ) is tautology]
(b) ( p ∧ q ) → p ≡ ( p ∧ q ) ∨ p
≡ ((~ p) ∨ (~ q )) ∨ p
[Q~( p ∧ q ) ≡ (~ p) ∨ (~ q )]
≡ (~ p ∨ p) ∨ (~ q ) is tautology.
[Q~ p ∨ p is a tautology and (~ q ) ∨ T ≡ T]
(c)Q p → ( p ∨ q ) ≡ (~ p) ∨ ( p ∨ q )
[Q p → q is equivalent to (~ p ∨ q )]
≡ (~ p ∨ p) ∨ q is tautology.
[Q(~ p ∨ p) is tautology and q ∨ T ≡ T]
(d) ( p ∨ q ) → ( p ∨ (~ q ))
≡ (~( p ∨ q )) ∨ ( p ∨ (~ q ))
≡ ((~ p) ∧ (~ q )) ∨ ( p ∨ (~ q ))
≡ ( p ∨ (~ q ) ∨ ((~ p) ∧ (~ q ))
≡ ( p ∨ (~ q ) ∨ (~ p)) ∧ ( p ∨ (~ q ) ∨ (~ q ))
≡ (T ∨ (~ q )) ∧ ( p ∧ (~ q ))
≡ T ∧ ( p ∧ (~ q ))
≡ p ∧ (~ q ), which is not a tautology.
3. For any
two statements p and q, the
negation of the expression p ∨ (~ p ∧ q ) is
[JEE Main 2019, 9 April Shift-I]
(a) ~ p ∧ ~ q
(c) p ∧ q
(b) ~ p ∨ ~ q
(d) p ↔ q
371
Mathematical Reasoning
Exp. (a)
6. The negation of the boolean expression
Q p ∨ ( (~ p) ∧ q )
= ( p ∨ (~ p)) ∧ ( p ∨ q ) [by Distributive law]
[Qp ∨(~ p) is tautology]
= p∨q
So negation of p ∨ ((~ p) ∧ q )
= ~ [ p ∨ (~ p) ∧ q ] = ~( p ∨ q )
[by Demorgan’s law]
= (~ p) ∧ (~ q )
4. If p ⇒ (q ∨ r ) is false, then the truth values of
p , q , r are respectively
[JEE Main 2019, 9 April Shift-II]
(a) T, T, F
(c) F, F, F
(b) T, F, F
(d) F, T, T
Exp. (b)
Given statement p ⇒(q ∨ r ) is false.
Q
p → (q ∨ r ) = (~ p) ∨ (q ∨ r )
Now, by trial and error method, if truth value of pis
T, q is F and r is F,
then truth value of (q ∨ r ) is F.
So, truth value of [(~ p) ∨ (q ∨ r )] is false.
Thus, if truth value of p, q , r are T, F, F, then the
statement p → (q ∨ r ) is false.
5. Which one of the following Boolean
expressions is a tautology ?
[JEE Main 2019, 10 April Shift-I]
(a)
(b)
(c)
(d)
(p ∨ q ) ∨ (p ∨ ~ q )
(p ∧ q ) ∨ (p ∧ ~ q )
(p ∨ q ) ∧ (p ∨ ~ q )
(p ∨ q ) ∧ (~ p ∨ ~ q )
Exp. (a)
Option (a) ( p ∨ q ) ∨ ( p ∨ (~ q ))
≡ p ∨ (q ∨ ~ q ) is tautology,
[Qq ∨ (~ q ) ≡ T and p ∨ T ≡ T]
Option (b)
( p ∧ q ) ∨ ( p ∧ (~ q )) ≡ p ∧ (q ∨ ~ q )
not a tautology,
[Qq ∨ ~ q ≡ T and p ∧ T ≡ p]
Option (c)
( p ∨ q ) ∧ ( p ∨ (~ q )) ≡ p ∨ (q ∧ ~ q )
not a tautology
[Qq ∧ ~ q ≡ F and p ∨ F ≡ p]
Option (d)
( p ∨ q ) ∧ ((~ p) ∨ (~ q )) ≡ ( p ∨ q ) ∧ (~( p ∧ q ))
not a tautology.
~ s ∨ (~ r ∧ s ) is equivalent to
[JEE Main 2019, 10 April Shift-II]
(a) s ∧ r
(b) ~ s ∧ ~ r (c) s ∨ r
(d) r
Exp. (a)
Key Idea Use De-morgan’s law, Distributive law
and Identity law as p ∨ F ≡ p
The given boolean expression is ~ s ∨ ((~ r ) ∧ s )
Now, the negation of given boolean expression is
~(~ s ∨ ((~ r ) ∧ s ))
= s ∧ (~((~ r ) ∧ s ))
[Q~( p ∧ q ) = ~ p∨ ~ q ]
= s ∧ (r ∨ (~ s ))
[Q~( p ∨ q ) = ~ p∧ ~ q ]
= (s ∧ r ) ∨ (s ∧ (~ s ))
[Q p ∧ (q ∨ r ) ≡ ( p ∧ q ) ∨ ( p ∧ r )]
= (s ∧ r )
[Q p∧ ~ p ≡ F ]
7. If the truth value of the statement
p → (~ q ∨ r ) is false (F), then the truth values
of the statements p , q and r are respectively
[JEE Main 2019, 12 April Shift-I]
(a) T, T and F
(c) T, F and T
(b) T, F and F
(d) F, T and T
Exp. (a)
Key Idea Use formula :
p → q = ~p ∨ q
Given statement is
p → (~ q ∨ r ) = ~ p ∨ (~ q ∨ r )
Now, from the options
(a) When p = T, q = T and r = F
then ~ p ∨ (~ q ∨ r ) = F ∨ (F ∨ F ) = F
(b) When p = T, q = F and r = F
then ~ p ∨ (~ q ∨ r ) = F ∨ (T ∨ F ) = T
(c) When p = T, q = F and r = T
then ~ p ∨ (~ q ∨ r ) = F ∨ (T ∨ T) = T
(d) When p = F, q = T and r = T
then ~ p ∨ (~ q ∨ r ) = T ∨ (F ∨ T) = T
8. The boolean expression ~ (p ⇒ (~ q )) is
equivalent to
(a) p ∧ q
[JEE Main 2019, 12 April Shift-II]
(b) q ⇒~ p (c) p ∨ q (d) (~ p ) ⇒ q
Exp. (a)
Given boolean expression is
~( p ⇒ (~ q ))
≡ ~((~ p) ∨ (~ q ))
[Q p ⇒ q ≡ ~ p ∨ q ]
≡ p∧q
372
JEE Main Chapterwise Mathematics
≡
≡
≡
≡
[p ∧ (~ q ∨ r )] ∧ (~q ∧ r )]
(distributive law)
(associative law)
p ∧ [(~ q ∨ r ) ∧ (~ q ∧ r )]
(commutative law)
p ∧ [(~ q ∧ r ) ∧ (~q ∨ r )]
p ∧ [{(~ q ∧ r ) ∧ (~ q )} ∧ {(~ q ∧ r ) ∧ r ]
(distributive law)
(idempotent law)
≡ p ∧ [(~ q ∧ r ) ∨ (~ q ∧ r )]
(idempotent law)
≡ p ∧ [~ q ∧ r ]
(associative law)
≡ p ∧ ~q ∧ r ≡ ( p ∧ r ) ∧ (~ q )
9. If the Boolean expression (p ⊕ q ) ∧(~ p ⋅ q ) is
equivalent to p ∧ q , where ⊕, ⋅∈{ ∧,∨}, then
the ordered pair(⊕, ⋅) is
[JEE Main 2019, 9 Jan Shift-I]
(a) (∧, ∨)
(b) (∧, ∧)
(c) (∨, ∧)
(d) (∨, ∨)
Exp. (a)
Let us check all the options
(a) Consider, ⊕ = ∧ and ⋅ = ∨.
In that case, we get
(p∧q)
∧
(∼ p ∨ q )
≡
(p∧q)
11. Consider the following three statements:
P : 5 is a prime number.
Q : 7 is a factor of 192.
R : LCM of 5 and 7 is 35.
Then, the truth value of which one of the
following statements is true ?
and
[JEE Main 2019, 10 Jan Shift-II]
[take ∨ ≈ ∪and ∧ ≈ ∩]
(b) Consider, ⊕ = ∧ and ⋅ = ∨.
In that case, we get
(p∧q) ∧
(∼ p ∧ q )
≡ null set
(a) ( P ∧ Q ) ∨ (~ R )
(c) (~ P ) ∨ (Q ∧ R )
Exp. (b)
Since, the statements
P : 5 is a prime number, is true statement.
Q : 7 is a factor of 192, is false statement
and R : LCM of 5 and 7 is 35, is true statement.
So, truth value of
P is T, Q is F, R is T
Now let us check all the options.
and
(c) Consider, ⊕ = ∨ and ⋅ = ∧.
In that case, we get
(p∨q) ∧
(∼ p ∧ q )
≡
(∼ p ∧ q )
and
(d) Consider, ⊕ = ∨ and u = ∨.
In that case, we get
(p∨q) ∧
(∼ p ∨ q )
≡
q
(b) P ∨ (~ Q ∧ R )
(d) (~ P ) ∧ (~ Q ∧ R )
P
Q
R
T
F
T
~ P ~Q ~ R
F
T
( P ∧ Q ) ∨ (~ R)
P ∨ (~Q ∧ R)
F
T
P ∧ Q Q ∧ R ~Q ∧ R
F
F
F
T
(~ P) ∨ (Q ∧ R) (~ P) ∧ (~Q ∧ R)
F
F
Clearly, the truth value of P ∨ (~ Q ∧ R ) is T.
and
12. Ifq is false andp ∧q ←→ r is true, then which
10. The logical statement
[~(~ p ∨ q) ∨ ( p ∧ r)] ∧ (~ q ∧ r)
is equivalent to
[JEE Main 2019, 9 Jan Shift-II]
(a) ~ p ∨ r
(c) (p ∧ r ) ∧ ~ q
(b) (p ∧ ~ q ) ∨ r
(d) (~ p ∧ ~ q ) ∧ r
Exp. (c)
Clearly, [~(~ p ∨ q ) ∨ ( p ∧ r )] ∧ (~ q ∧ r )
≡ [( p ∧ ~q ) ∨ ( p ∧ r )] ∧ (~ q ∧ r )
(Q~(~ p ∨ q ) ≡ ~(~ p) ∧ ~ q ≡ p ∧ ~ q
by De Morgan’s law)
one of the following statements is a
tautology?
[JEE Main 2019, 11 Jan Shift-I]
(a) p ∨r
(c) (p ∨r )→ (p ∧r )
(b) (p ∧r )→ (p ∨r )
(d) p ∧r
Exp. (b)
Given, ( p ∧ q ) ↔ r is true. This is possible under
two cases
Case I When both p ∧ q and r are true,
which is not possible because q is false.
Case II When both ( p ∧ q ) and r are false.
⇒
p ≡ T or F; q ≡ F, r ≡ F
373
Mathematical Reasoning
In this case,
(a) p ∨ r is T or F
(b) ( p ∧ r ) → ( p ∨ r ) is F→ (T or F ) ,
which always result in T.
(c) ( p ∨ r ) → ( p ∧ r ) is (T or F) → F,
which may be T or F.
(d) p ∧ r is F.
Exp. (d)
Since, the expression, p → q ≡ ~ p ∨ q
So,
~ p→ q ≡ p ∨ q
and
therefore ~(~ p → q ) ≡ ~( p ∨ q )
≡ (~ p) ∧ (~ q )
[by DeMorgan’s law]
13. Contrapositive of the statement “If two
16. The boolean expression~ (p ∨ q ) ∨ (~ p ∧ q ) is
equivalent to
numbers are not equal, then their squares
are not equal” is [JEE Main 2019, 11 Jan Shift-II]
(a) ~ p
We have, ~( p ∨ q ) ∨ (~ p ∧ q )
≡ (~ p ∧ ~ q ) ∨ (~ p ∧ q )
[Q By DeMorgan’s law ~( p ∨ q ) = (~ p ∧ ~ q )]
≡~ p ∧ (~ q ∨ q )[By distributive law]
[~q ∨ q = t ]
≡ ~p∧t
≡~ p
17. The statement (p → q ) → [(~ p → q ) → q ] is
[JEE Main 2017 (offline)]
(a) a tautology
(b) equivalent to ~ p → q
(c) equivalent to p → ~ q (d) a fallacy
Exp. (a)
The truth table of the given expression is given
below
14. The Boolean expression
(a) p ∧ q
(c) p ∧ (~ q )
(b) p ∨ (~ q )
(d) (~ p ) ∧ (~ q )
Exp. (d)
Let the given Boolean expression
(( p ∧ q ) ∨ ( p∨ ~ q )) ∧ (~ p∧ ~ q ) ≡ r
Now, let us construct the following truth table
~ p ~q
p ∧q
p ∨ ~q ~ p ∧ ~q
T
F
T
F
T
F
F
F
F
F
F
F
T
T
T
T
T
F
F
T
T
T
F
F
T
F
F
T
T
F
F
F
T
T
Clearly, r ≡ ~ p∧ ~ q
15. The expression ~ (~ p → q ) is logically
equivalent to
(a) p ∧ ~ q
[JEE Main 2019, 12 Jan Shift-II]
(b) p ∧ q
(c) ~ p ∧ q
T
T
F
T
T
T
T
F
F
F
T
F
T
F
T
T
T
T
T
T
F
F
T
T
F
T
T
(p ∧ ~ q ) ∨ q ∨ (~ p ∧ q ) is equivalent to
F
T
q
T
18. The Boolean expression
r
q
x ≡ p → q ~ p ~p → q y ≡ (~ p → q) → q x → y
p
Hence, it is a tautology.
(p ∧ q ) ∨
( p ∨ ~q )
p
(d) ~ q
Key Idea Use DeMorgan’s and distributive law.
Exp. (b)
((p ∧ q ) ∨ (p ∨ ~ q )) ∧ (~ p ∧ ~ q ) is equivalent
to
[JEE Main 2019, 12 Jan Shift-I]
(c) q
Exp. (a)
(a) If the squares of two numbers are not equal,
then the numbers are not equal.
(b) If the squares of two numbers are equal,
then the numbers are equal.
(c) If the squares of two numbers are not equal,
then the numbers are equal.
(d) If the squares of two numbers are equal,
then the numbers are not equal.
We know that, contrapositive of
p → q is ~q → ~p
Therefore, the contrapositive of the given
statement is
‘‘If the squares of two numbers are equal, then the
numbers are equal’’.
[JEE Main 2018]
(b) p
(d) ~ p ∧ ~ q
[JEE Main 2016 (offline)]
(a) ~ p ∧ q
(b) p ∧ q
(c) p ∨ q
(d) p ∨ ~ q
Exp. (c)
Consider, ( p ∧ ~ q ) ∨ q ∨ (~ p ∧ q )
≡ [( p ∧ ~ q ) ∨ q ] ∨ (~ p ∧q )
≡ [( p ∨ q ) ∧ (~ q ∨ q )] ∨ (~ p ∧ q )
≡ [( p ∨ q ) ∧ t ] ∨ (~ p ∧ q )
≡ ( p ∨ q ) ∨ (~ p ∧ q )
≡ ( p ∨ q ∨ ~ p) ∧ ( p ∨ q ∨ q )
≡ (q ∨ t ) ∧ ( p ∨ q ) ≡ t ∧ ( p ∨ q ) ≡ p ∨ q
374
JEE Main Chapterwise Mathematics
19. The negation of ~ s ∨ (~ r ∧ s ) is equivalent to
(a) s ∧ ~ r
[JEE Main 2015]
(b) s ∧ (r ∧ ~ s )
(c) s ∨ (r ∨ ~ s )
(d) s ∧ r
Exp. (d)
~(~ s ∨ (~ r ∧ s ))
≡ s ∧ (~(~ r ∧ s )) ≡ s ∧ (r ∨ ~ s ) ≡ (s ∧ r ) ∨ (s ∧ ~ s )
≡(s ∧ r ) ∨ F
≡s∧r
(Q s ∧ ~ s is false)
20. The statement ~ (p ↔ ~ q ) is
(a) equivalent to p ↔ q
[JEE Main 2014]
(b) equivalent to ~ p ↔ q (c) a tautology
(d) a fallacy
Exp. (a)
p
q
~p
~q
q↔q
p ↔ ~q
~p ↔ q
~ (p ↔ ~ q )
T
F
F
T
F
T
T
F
F
T
T
F
F
T
T
F
T
T
F
F
T
F
F
T
F
F
T
T
T
F
F
T
~( p ↔~ q ) is equivalent to ( p ↔ q ).
21. Consider
Statement I (p ∧ ~ q ) ∧ (~ p ∧ q ) is a fallacy.
Statement II (p → q ) ↔ (~ q → ~ p ) is a tautology.
[JEE Main 2013]
(a) Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
Exp. (b)
Statement II ( p → q ) ↔ (~ q → ~ p) ≡ ( p → q ) ↔ ( p → q )
which is always true, so Statement II is true.
Statement I ( p ∧ ~ q ) ∧ (~ p ∧ q ) ≡ p ∧ ~ q ∧ ~ p ∧ q ≡ p ∧ ~ p ∧ ~ q ∧ q ≡ f ∧ f ≡ f
Hence, it is a fallacy statement.
So, Statement I is true and statement II is not a correct explanation for statement I.
Alternate Solution
Statement II ( p → q ) ↔ (~ q → ~ p) ~ q → ~ p is contrapositive of p → q
Hence, ( p → q ) ↔ ( p → q ) will be a tautology.
Statement I ( p ∧ ~ q ) ∧ (~ p ∧ q )
q
~p
~q
p ∧~q
~p ∧ q
(p ∧ ~ q ) ∧ (~ p ∧ q )
T
T
F
F
F
F
F
T
F
F
T
T
F
F
F
T
T
F
F
T
F
F
F
T
T
F
F
F
p
Hence, it is a fallacy.
375
Mathematical Reasoning
22. The negation of the statement
“If I become a teacher, then I will open a school”, is
(a)
(b)
(c)
(d)
[AIEEE 2012]
I will become a teacher and I will not open a school
Either I will not become a teacher or I will not open a school
Neither I will become a teacher nor I will open a school
I will not become a teacher or I will open a school
Exp. (a)
Let us assume that p : ‘I become a teacher’ and
q : I will open a school.
Then, we can easily as certain that
Negation of ( p → q ) is ~( p → q ) = p ∧ ~ q
which means that ‘I will become a teacher and I will not open a school’.
23. Consider the following statements
P : Suman is brilliant.
R : Suman is honest.
Q : Suman is rich.
The negation of the statement. ‘‘Suman is brilliant and dishonest, if and only if Suman is rich’’ can
be expressed as
[AIEEE 2011]
(a) ~ [Q ↔ ( P ∧ ~ R )]
(b) ~ Q ↔ P ∧ R
(c) ~ ( P ∧ ~ R ) ↔ Q
(d) ~ P ∧ (Q ↔ ~ R )
Exp. (a)
Suman is brilliant and dishonest if and only if Suman is rich, is expressed as,
Q ↔ (P ∧ ~ R )
So, negation of it will be,
~ [Q ↔ (P ∧ ~ R )].
24. The only statement among the followings that is a tautology is
(a) B → [A ∧ ( A → B )]
(b) A ∧ ( A ∨ B )
[AIEEE 2011]
(c) A ∨ ( A ∧ B )
(d) [A ∧ ( A → B )]→ B
Exp. (d)
A
B
A∨B
A∧B
A ∧ ( A ∨ B)
A ∨ ( A ∧ B)
A→ B
A ∧ ( A → B)
A ∧ ( A → B )→ B
B → [ A ∧ ( A → B )]
T
T
F
F
T
F
T
F
T
T
T
F
T
F
F
F
T
T
F
F
T
T
F
F
T
F
T
T
T
F
F
F
T
T
T
T
T
T
F
T
Hence the truth value of all the elements of the column [ A ∧ ( A → B)] → B is T.
∴ A ∧ ( A → B) → B is tautology.
25. Let S be a non-empty subset of R. Consider the following statement
P : There is a rational number x ∈S such that
x > 0.
Which of the following statements is the negation of the statement P ?
(a) There is a rational number x ∈S such that x ≤ 0
(b) There is no rational number x ∈S such that x ≤ 0
(c) Every rational number x ∈S satisfies x ≤ 0
(d) x ∈S and x ≤ 0 ⇒ x is not rational
[AIEEE 2010]
376
JEE Main Chapterwise Mathematics
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation for
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation for
Statement I
(d) Statement I is true, Statement II is false
Exp. (c)
P : There is rational number x ∈ S such that x > 0.
~ P : Every rational number x ∈ S satisfies x ≤ 0.
26. Statement I ~ (p ↔ ~ q ) is equivalent to
p ↔ q.
Statement II ~ (p ↔ ~ q ) is a tautology.
Exp. (∗)
(a) Statement I is true, Statement II is true;
Here, p : x is an irrational number
q : y is a transcendental number
r : x is a rational number, iff y is a transcendental
number
⇒
r :~ p ↔q
But S1 : r ≡ q ∨ p, so statement I is not correct.
And S 2 : r ≡ ~( p ↔~ q )
Statement II is a correct explanation for
Statement I
(b) Statement I is true, Statement II is true;
Statement II is not a correct explanation for
Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
[AIEEE 2009]
p q~p~q r~p ↔ q
Exp. (c)
p
q
p↔q
~q
p ↔ ~q
~ (p ↔ ~ q )
T
T
F
F
T
F
T
F
T
F
F
T
F
T
F
T
F
T
T
F
T
F
F
T
From the table we see that, truth values of
column p ↔ q and ~( p ↔ ~ q ) are identical.
So, ~( p ↔ ~ q ) s equivalent to p ↔ q .
The values of the column ~( p ↔ ~ q ) are not all
‘T’. So, ~( p ↔ ~ q ) is not a tautology.
Hence, Statement I is true but Statement II is false.
T
T
F
F
Statement I r is equivalent to either q or p.
Statement II r is equivalent to ~ (p ↔ ~ q ).
[AIEEE 2008]
F
F
T
T
F
T
F
T
Statement II
q ∨ p (p ↔~ q ) ~ (p ↔~ q )
F
T
T
F
T
T
T
F
F
T
T
F
T
F
F
T
It is clear from the table that r is not equivalent to
either of the statements. Hence, none of the given
options is correct.
28. The statement p → (q → p ) is equivalent to
[AIEEE 2008]
(a) p → (p ↔ q )
(c) p → (p ∨ q )
27. Let p be the statement “x is an irrational
number”, q be the statement “y is a
transcendental number” and r be the
statement “x is a rational number, iff y is a
transcendental number”.
T
F
T
F
Statement I
(b) p → (p → q )
(d) p → (p ∧ q )
Exp. (c)
q
p
q→ p
p → (q → p )
p ∨q
p → (p ∨ q )
T
T
F
F
T
F
T
F
T
F
T
T
T
T
T
T
T
T
T
F
T
T
T
T
So, statement p → (q → p) is equivalent to
p → ( p ∨ q ).
Practice Set 1
Instructions
This test consists of 30 questions to be completed in 3 hrs.
Each question is allotted 4 marks for correct response.
1 marks will be deducted for indicating incorrect response of each question. No deduction from the total
score will be made if no response is indicated for an item in the answer sheet.
There is only one correct response for each question. Filling up more than one response in any question
1. The domain of the function
 2 x − 1
3 tan x
is
f ( x ) = 3 1 − 3x + 3 cos −1 
 +e
 3 
(a) (–1, 2)
(c) ( − ∞ , ∞ )
(b) [–1, 2]
(d) None of these
1/( x − α )
then lim (1 + ax + bx + c )
x→a
(a) log| a (α − β )|
(c) e
a( α − β )
is
(b) e a( β − α )
y = f ( x ), z = g ( x ), f ′ ( x ) = tan x
dy
is
g ′ ( x ) = sec x , then the value of
dz
3. If
(a)
(c)
3
⋅
tan x 3
5x 2 sec x 5
3x 2 tan x 3
⋅
5 sec x 5
5
(b)
and
5x 2 sec x 5
⋅
3 tan x 3
(d) None of these
be equal to
2 2n
0
3 ⋅ 42n , where n is multiple of 3
None of the above
5. The planes 3x − y + z + 1 = 0 and
5x + y + 3z = 0 intersect in the line PQ. The
equation of the plane through the point
(2 , 1, 4) and perpendicular to PQ is
(a) x + y − 2 z = 5
(c) x + y + 2 z = 5
(b) x + y − 2 z = − 5
(d) x + y + 2 z = − 5
(c) log 2
(d) – log 2
variable X are 2 and 1, respectively. The
probability that X takes values greater than
1, is
9
16
11
(c)
16
(a)
4. If z = − 2 + 2 3 i , then z 2n + 2 2n zn + 2 4n may
(a)
(b)
(c)
(d)
(b) 1
7. The mean and variance of a binomial
(d) None of these
3
to
(a) –1
2. If α and β are the roots of ax 2 + bx + c = 0,
2
 xx − x−x 
 , then f ′(1) is equal
2


6. If f ( x ) = cot −1 
(b)
5
16
(d) None of these
8. If the line ax + by + c = 0 is a tangent to the
curve xy = 4, then
(a) a > 0 and b > 0
(c) a < 0 and b > 0
(b) a > 0 and b < 0
(d) None of these
9. In a statistical investigation of 1003 families
of Kolkata, it was found that 63 families had
neither a radio nor a TV, 794 families had a
radio and 187 had a TV. The number of
families in that group having both a radio
and a TV is
(a) 36
(c) 41
(b) 32
(d) None of these
10. Three boys and three girls are to be seated
around a table in a circle. Among them, the
boy X does not want any girl neighbour and
the girlY does not want any boy neighbour.
The number of such arrangements po ssible is
(a) 6
(b) 4
(c) 8
(d) 9
Note All the Practice Sets (1-10) available for free online practice see detailed instructions back side of the title.
380
JEE Main Chapterwise Mathematics
11. If the system of linear equations
x + 2 ay + az = 0, x + 3by + bz = 0
and x + 4cy + cz = 0
has a non-zero solution, then a ,b and c
(a) are in AP
(b) are in GP
(c) are in HP
(d) satisfy a + 2 b + 3c = 0
Directions (Q. Nos. 12-14) Each of
these questions contains two statements :
Statement I (Assertion) and Statement II
(Reason). Each of these questions also
has four alternative choices, only one of
which is the correct answer. You have to
select one of the codes (a), (b), (c), and (d)
given below.
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation for
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation
for Statement I
(d) Statement I is true, Statement II is false
12. Statement I f ( x ) =
1
is decreasing in
x −5
x ∈( − ∞ , 5) ∪ (5, ∞ ).
Statement II
f ′ ( x ) < 0, ∀ x ≠ 5.
13. Statement I The degree of the
differential equation
 d 2y 
d 2y dy
+
= log  2  is 2.
2
dx
dx
 dx 
Statement II The degree of the
differential equation which can be written
as polynomial in the derivatives is the
degree of the derivative of the highest
order occuring in it.
14. Statement I Circles x 2 + y 2 = 144 and
x + y − 6x − 8y = 0 do not have any
common tangent.
2
2
Statement II If one circle lies completely
inside the other circle, then both have no
common tangent..
cos x
1
x
15. If f ( x ) = 2 sin x x 2 2 x , then lim
tan x
1
x
(a) − 2
(c) 2
x→ 0
f ′ (x )
is
x
(b) –1
(d) 1
16. AB is a vertical pole andC is the middle point.
The end A is on the level ground and P is any
point on the level ground other than A. The
portion CB subtends an angle β at P. If
AP : AB = 2 : 1, then β is equal to
1
(a) tan −1  
 9
4
(b) tan −1  
 9
5
(c) tan −1  
 9
2
(d) tan −1  
 9
17. a , b and c are three non-coplanar vectors and
r is any arbitrary vector, then
[ b c r ]a + [ c a r ] b + [a b r ] c is always equal to
(a) [a b c]r
(c) 3[a b c]r
(b) 2[a b c]r
(d) None of these
π π
18. In the interval − ,  , the equation
 2 2
log sin θ (cos 2 θ ) = 2 has
(a)
(b)
(c)
(d)
no solution
a unique solution
two solutions
infinitely many solutions
19. A parallelogram is constructed on 3a + b and
a − 4b, where |a| = 6 | b| = 8 and a and b are
anti-parallel, then the length of the longer
diagonal is
(a) 40
(b) 64
(c) 32
(d) 48
20. The image of the point (– 8, 12) with respect to
the line mirror 4x + 7y + 13 = 0 is
(a) (–16, 2)
(c) (16, 2)
21. tan
(b) (16, – 2)
(d) (–16, – 2)
4π
6π
π
− 33 tan
+ 27 tan 2 is equal to
9
9
9
(a) 0
(b)
3
(c) 3
(d) 9
381
Practice Set 1
22. If the angle between the straight lines
joining foci and the ends of minor axis of the
x2 y 2
ellipse
+
= 1 is 90°, then the
a2 b2
eccentricity is
1
2
1
(c)
2
(a)
(b)
26. If n C 0 , nC 1 , nC 2 ,... , nC n are the coefficients of
the expansion of (1 + x )n , then the value of
k n
C
∑ k +k1 is
0
3
2
(a) 0
(c)
(d) None of these
23. The portion of a tangent to a parabola
y 2 = 4ax cut off between the directrix and
the curve subtends an angle θ at the focus,
where θ is equal to
π
3
π
(c)
2
(a)
(b)
diameter is AB , where A and B
intersection points of circles
x 2 + y 2 − 4x − 4y + 4 = 0
1
and x 2 + y 2 − x − y + = 0 is
4
between the roots of the equation
x 2 + (1 − 2 k )x + (k 2 − k − 2 ) = 0 is given by
(a) k < 2
28.
are
2
3
3
7
(a)  x −  +  y −  =


8
8
32
(b) 2 < k < 5
(c) ( x − 2 )2 + ( y − 2 )2 = 1
2
5
5
7
(d)  x −  +  y −  =


8
8
32
(d) k > 5
3
sin x dx
∫ (1 + cos2 x )
(a)
(b)
(c)
(d)
1 + cos 2 x + cos 4 x
is equal to
sec −1 (sec x + cos x ) + c
sec −1 (cos x − tan x ) + c
sec −1 ( sec x − cos x ) + c
None of the above
29. The area of one of the curvilinear triangles
formed by the curves y = sin x , y = cos x and
X-axis is
(a)
(b)
(c)
(d)
(b) ( x − 1 )2 + ( y − 1 )2 = 72
( 2 + 2 ) sq unit
( 2 − 2 ) sq unit
( 2 + 2 2 ) sq unit
None of these
30. If f and g are continuous functions on [0, π]
25. If p , q , r are simple propositions, with truth
values T , F and T, then the truth value of
(~ p ∨ q ) ∧ ~ r ⇒ p is
(a) true
(c) true, if r is false
(d) None of these
(c) 2 < k < 3
24. Equation of the smallest circle whose
2
2n + 1 − 1
n +1
2n − 1
n
27. The value of k for which the number 3 lies
π
4
(d) None of these
2
(b)
(b) false
(d) true, if q is true
satisfying
f ( x ) + f ( π − x ) = g ( x ) + g ( π − x ) = 1,
then ∫
π
0
[ f ( x ) + g ( x )] dx is equal to
(a) π
π
(c)
2
(b) 2 π
3π
(d)
2
Answers
1. (b)
11. (c)
21. (c)
2. (c)
12. (b)
22. (c)
3. (a)
13. (a)
23. (c)
4. (c)
14. (b)
24. (d)
5. (b)
15. (a)
25. (a)
6. (a)
16. (d)
26. (c)
7. (c)
17. (a)
27. (b)
8. (a)
18. (b)
28. (a)
9. (c)
19. (d)
29. (b)
10. (b)
20. (d)
30. (a)
Solutions
1. Given that,
2 x − 1
3 tan
f( x) = 3 1 − 3 x + 3cos 
 +e
 3 
2x − 1
is defined, if −1 ≤
≤1
3
⇒
− 3 ≤ 2 x − 1≤ 3
−1
x
⇒
−1 ≤ x ≤ 2
Domain of f( x) = [−1, 2 ]
2.
−x 
 x
6. Given that, f( x) = cot −1  x − x 

lim (1 + ax2 + bx + c )1/( x − α )
x→ α
=e
1
[( 1 + ax 2 + bx + c ) − 1]
lim
x→ α ( x − α)
lim
= ex→ α
3. Given that,
⇒
Also,
⇒
∴
( x − α)
= e a( α − β )
dy
[given, f ′( x) = tan x]
= 3 x2 tan x3
dx
z = g ( x5 )
dz
= g ′( x5 ) ⋅ 5 x4 = 5 x4 sec x5
dx
[given, g ′( x) = sec x]
3 x2 tan x3
/
dy dydx
= 4
=
/
dz dzdx
5 x sec x5
3
5x
 x2 x − 1

f( x) = cot −1 
x 
 2 x 
⇒
 tan2 θ − 1

y = f( x) = cot −1 
 2 tan θ 
∴
dy
= f ′( x3 ) ⋅ 3 x2
dx
y = f( x 3 ) ⇒
2
⋅

2
On putting x x = tan θ,
a ( x − α )( x − β )
=
4.
The plane perpendicular to PQ is
x + y − 2 z = λ.
Since, it passes through (2, 1, 4).
∴
2 + 1− 2 ⋅ 4 = λ
⇒
λ = −5
Hence, the required plane is
x + y − 2 z = − 5.
tan x3
sec x5
z = − 2 + 2 3 i = 4ω
z2 n + 2 2 n ⋅ zn + 2 4 n = 42 n ⋅ ω2 n + 2 2 n ⋅ 4n ⋅ ωn + 2 4 n
= 42 n [ω2 n + ωn + 1]
 0, if n is not a multiple of 3.
=
2n
3 ⋅ 4 , if n is a multiple of 3.
5. Let (l , m, n) be the direction cosines of PQ, then the
line is perpendicular to the given planes.
∴
3l − m + n = 0
and
5l + m + 3n = 0
l
m
n
=
=
⇒
−3 − 1 5 − 9 3 + 5
l m
n
=
=
⇒
1 1 −2
= cot −1 (− cot 2θ)
= π − cot −1 (cot 2 θ)
y = π − 2 θ = π − 2 tan−1 ( x x )
⇒
2
dy
=−
x x (1 + log x)
dx
1 + x2 x
⇒
⇒
2
 dy 
=−
(1 + 0) = − 1
 
 dx  ( x = 1)
1+ 1
mean, np = 2
7. Given,
and
⇒
…(i)
variance, npq = 1
1
q=
2
p= 1− q =
∴
From Eq. (i),
n×
1
2
1
=2
2
⇒
n=4
Now, P( X > 1) = P( X = 2 ) + P( X = 3) + P( X = 4)
2
2
1
3
1
1
1
1
1
= 4C 2     + 4 C 3     + 4 C 4  
2 2
2 2
2
6+ 4+1
=
16
11
=
16
4
383
Practice Set 1
xy = 4
dy
+ y⋅1= 0
x⋅
dx
dy
y
=−
dx
x
dy
4
=− 2
dx
x
8. Given that,
⇒
⇒
⇒
11. Given system of linear equations has a non-zero
solution, then
[Q xy = 4]
a
.
b
Since, the given line is a tangent to the curve.
a
4
a
>0
− 2 =−
∴
⇒
b
b
x
which is possible only when a > 0, b > 0 or
a < 0, b < 0.
Slope of the line ax + by + c = 0 is −
9. Let R be the set of families having a radio and
T be the set of families having a TV.
∴
n(R ∪ T ) = 1003 − 63 = 940,
n(R ) = 794
n(T ) = 187
n(R ∩ T ) = x
and
Let
U
R
T
794 – x x 187 – x
From Venn-diagram,
794 − x + x + 187 − x = 940
⇒
981 − x = 940
∴
x = 41
10. As shown in the figure, 1, 2 and X are the three boys
and 3, 4 and Y are the three girls. Boy X will have
neighbour as boys 1 and 2 and the girl Y will have
neighbour as girls 3 and 4. Boys 1 and 2 can be
arranged in P(2 , 2 ) ways.
= 2 ! = 2 ways
X
1
2
3
4
1 2a a
1 3b b = 0
1 4c c
Applying R 2 → R 2 − R1, R 3 → R 3 − R1, we get
a
1
2a
0 3b − 2 a b − a = 0
0 4c − 2 a c − a
⇒ (3 b − 2 a)(c − a) − (4c − 2 a)(b − a) = 0
⇒ 3 bc − 3 ba − 2 ac + 2 a2
= 4 bc − 2 ab − 4 ac + 2 a2
⇒
2 ac = bc + ab
2 1 1
⇒
= +
b a c
Hence, a, b and c are in HP.
12. Statement I Since, f( x) = 1
x−5
1
f ′ ( x) = −
< 0, ∀ x ∈ R − {5}
∴
( x − 5)2
⇒ f( x) is decreasing, ∀ x ∈ R − {5}.
Hence, option (b) is correct.
13. Since, the given equation cannot be written as a
polynomial in all the differential coefficients.
So, degree of equation is not defined.
Hence, Statement I is false and Statement II is true.
14. For circle x2 + y2 = 144,
centre, c1 = (0, 0) and radius, r1 = 12
For circle x2 + y2 − 6 x − 8 y = 0,
centre, c 2 = (3, 4) and radius, r2 = 5
Now, c1 c 2 = 5 and r1 − r2 = 7, thus c1 c 2 < r1 − r2 ,
hence one circle is completely lying inside other
without touching it. Hence, there is no common
tangent, therefore Statement I is true.
Hence, both the Statements are true and Statement II
is a correct explanation of Statement I.
cos x x 1
15. Given that, f( x) = 2 sin x x2 2 x
tan x
Y
Also, girls 3 and 4 can be arranged in P(2 , 2 ) ways
= 2 ! = 2 ways
Hence, required number of permutations
=2 ×2= 4
x
Applying R1 → R1 − R 3 , we get
cos x − tan x
=
2 sin x
tan x
1
0
0
x2 2 x
x
1
= (cos x − tan x)( x2 − 2 x2 )
⇒
f( x) = − x2 (cos x − tan x)
384
JEE Main Chapterwise Mathematics
On differentiating both sides w.r.t. x, we get
f ′( x) = − 2 x(cos x − tan x) − x2 (− sin x − sec 2 x)
f ′ ( x)
= lim [− 2 (cos x − tan x)
∴ lim
x→ 0
x→ 0
x
+ x (sin x + sec 2 x)]
= −2 × 1= −2
16. Let
AC = CB = h m and ∠ APC = α
AP 2
=
AB 1
Given that,
⇒
In ∆APC,
AP = 2 AB = 4 h
h
1
tan α =
⇒ tan α =
4h
4
B
18. Q θ ∈  − π , π 
 2 2 
∴
−1 ≤ sin θ ≤ 1
Now, we take
logsin θ cos 2 θ = 2
⇒
cos 2 θ = sin2 θ
⇒
1 − 2 sin2 θ = sin2 θ
⇒
3 sin2 θ = 1
⇒
sin θ = ±
∴
h
P
β
α
4h
Hence, given equation has a unique solution.
h
19. Consider (3a + b) ⋅ (a − 4b) = 3|a|2 − 11a ⋅ b − 4|b|2
A
= 3 ⋅ 36 − 11⋅ 6 ⋅ 8cos π − 4 ⋅ 64 > 0
So, angle between a and b is acute.
∴The longer diagonal is given by
2h 1
=
4h 2
tan α + tan β
1
=
1 − tan α ⋅ tan β 2
1
+ tan β
1
4
=
1
1 − tan β 2
4
1 + 4 tan β 1
=
4 − tan β
2
⇒
⇒
⇒
⇒
⇒
2 + 8 tan β = 4 − tan β
9 tan β = 2
2
β = tan−1  
 9
∴
1
3
[since, base of any log cannot be negative]
sin θ =
C
tan (α + β ) =
In ∆APB,
1
3
→
α = (3a + b) + (a − 4b) = 4a − 3b
Now,
|α|2 = |4a − 3b|2 = 16|a|2 + 9| b|2 − 24a ⋅ b
= 16 ⋅ 36 + 9 ⋅ 64 − 24 ⋅ 6 ⋅ 8cos πmn
= 16 × 144
∴|4a − 3b| = 48
20. Given equation of line is
4 x + 7 y + 13 = 0
…(i)
Let Q(α, β ) be the image of the point P(− 8, 12 ) w.r.t.
Eq.(i).
P (– 8, 12)
17. Let r = x1 a + x2 b + x3 c
⇒
r ⋅ (b × c ) = x1 a ⋅ (b × c )
x1 =
⇒
[r b c ]
[a b c ]
Also, r ⋅ (c × a ) = x2 b ⋅ (c × a )
[r c a ]
and r ⋅ (a × b) = x3 c ⋅ (a × b)
⇒
x2 =
[a b c ]
⇒
⇒
⇒
[r a b]
x3 =
[a b c ]
r=
[r a b]
[r b c ]
[r c a ]
a +
c
b+
[a b c ]
[a b c ]
[a b c ]
[b c r ]a + [c a r ]b + [a b r ]c = [a b c ]r
C
4x + 7y + 13 = 0
Q (α, β)
Then, PQ perpendicular to Eq. (i) and PC = CQ
Equation of the line PC is
7
y − 12 = ( x + 8)
4
…(ii)
⇒
7 x − 4 y + 104 = 0
On solving Eqs. (i) and (ii), we get
x = − 12 and y = 5
385
Practice Set 1
So, coordinates of C is (–12, 5).
Since, C is mid-point of PQ.
α−8
β + 12
and 5 =
∴
−12 =
2
2
and β = − 2
∴
α = − 16
23. The equation of the tangent at P(at 2 , 2 at ) to y2 = 4ax
is
3
θ
21. Since, tan 3 θ = 3 tanθ − tan
2
P (at2, 2at)
1 − 3 tan θ
tan
Q
X′
π
π
− tan3
9
9
π
1 − 3 tan2
9
3 tan
π
=
3
2
π
π
π
⇒ 3  1 − 3 tan2  =  3 tan − tan3 


9
9
9
π
π
π
tan6 − 33 tan4 + 27 tan2 = 3
⇒
9
9
9
b−0
b
22. Slope of line BS is m1 =
=−
0 − ae
ae
b−0
b
and slope of line BS is m2 =
=
0 + ae ae
2
X′
S′ (– ae, 0)
C
(ae, 0)S
X
X
Y′
and slope of QS is
a( t 2 − 1)
−0
(t 2 − 1)
t
=−
m2 =
−a − a
2t
Now,
90°
S (a, 0)

a( t 2 − 1)
Thus, Eq. (i) meets the directrix at Q  − a,
.
t


Now, slope of PS is
2 at − 0
2t
.
=
m1 = 2
at − a t 2 − 1
Y
B (0, b)
O
x+a=0
⇒
π
, we get
9
…(i)
It meets the directrix x = − a.
∴
ty = − a + at 2
a(t 2 − 1)
⇒
y=
t
Y
Hence, coordinates of Q is (–16, −2).
On putting θ =
ty = x + at 2
∴
 t 2 − 1
−
 = −1
×
2 t 
t 2 − 1 
π
θ=
2
m1m2 =
2t
24. Let the equations of circle be
B (0, – b)
and
Y′
⇒
b
b
×
= −1
ae ae
b 2 = a2e 2
⇒
a2 (1 − e 2 ) = a2e 2
⇒
1 − e2 = e2
⇒
2e2 = 1
⇒
e=
⇒
A
∠SBS ′ = 90°
m1m2 = − 1
Q
∴
−
1
2
[since, e cannot be negative]
S1 ≡ x2 + y2 − 4 x − 4 y + 4 = 0
1
S 2 ≡ x 2 + y2 − x − y + = 0
4
C1
(2, 2)
M
C2
(1/2, 1/2)
B
4x + 4y = 5
Equation of common chord AB is
S1 − S 2 = 0
⇒
4x + 4y = 5
4(2 ) + 4(2 ) − 5
11
Now,
=
C1M =
2
2
4
2
4 + 4
and
r1 =
4+ 4− 4 =2
…(i)
386
JEE Main Chapterwise Mathematics
Also, equation of line C1C 2 is
y= x
On solving Eqs. (i) and (ii), we get
5
5
and y =
x=
8
8
∴ Radius of required circle,
AM =
…(ii)
28. Let I = ∫
AC12 − C1M 2
121
= 4−
32
7
7
=
=
32 4 2
Hence, required equation of circle is
2
=
∫
=
∫
sin3 x
sin3 x
dx
cos
(sec
cos x) cos x
x
x
+




sec 2 x + 1 + cos 2 x 

sin3 x
dx

cos 2 x(sec x + cos x)



(sec x + cos x)2 − 1
Put sec x + cos x = z
sin3 x
dx = dz
⇒
cos 2 x
2
 x − 5 +  y − 5 = 7






8
8
32
I=
∴
Now, (~ p ∨ q ) ∧ ~ r means F.
∴
[(~ p ∨ q )∧ ~ r ] ⇒ p means T.
26. Let
Tr
+1
Cr
1
=
⋅
=
r+1 n+1
n+1
Cr
∑
0
Ck
1 n+1
(
=
C1 +
k+1 n+1
=
1
(2 n + 1 −
n+1
+1
Y
x = π/4
y = cos x
y = sin x
n+1
C2 + . . .
+
X′
X
O
n+1
Cn + 1)
n+1
C0 )
2n + 1 − 1
=
n+1
Y′
∴ Required area =
Since, the number 3 lies between the roots of the
given equation, if f(3) < 0.
= 10 − 7 k + k 2
Q
∴
⇒
⇒
f(3) < 0
k 2 − 7 k + 10 < 0
(k − 2 )(k − 5) < 0
2< k< 5
π/4
∫0
π/2
∫π/4
sin x dx +
= − [cos
x]π0 / 4
+ [sin
cos x dx
x]ππ // 24
1 
= 2  1 −
 = (2 − 2 ) sq unit

2
27. Let f( x) = x2 + (1 − 2 k )x + k 2 − k − 2
Now, f(3) = 9 + (1 − 2 k ) 3 + k 2 − k − 2
= sec −1 z + c
π 1 
y = cos x is  ,
.
 4 2
On putting r = 0, 1, 2 , . . . , n and then adding, we get
n
z2 − 1
z
29. The point of intersection of two curves y = sin x and
 n+1
n + 1n 
Q C r + 1 = r + 1 C r 


n
dz
∫
= sec −1 (sec x + cos x) + c
25. ~ p ∨ q means F ∨ F = F and ~ r means F.
n
dx
(1 + cos x) 1 + cos 2 x + cos 4 x
2
30. Now, ∫
π
=
∫0
=
∫0
=
∫0
π
⇒
π
π
0
f ( x) g ( x) =
π
∫0
f( π − x) g ( π − x) dx
[1 − f( x)] ⋅ [1 − g ( x)] dx
[1 − f( x) − g ( x) + f( x) ⋅ g ( x)] dx
1dx −
π
∫0
π
∫0
[f( x) + g ( x)] dx +
[f( x) + g ( x)] dx =
π
∫0
π
∫0
f( x) ⋅ g ( x) dx
1dx = π
Practice Set 2
Instructions
For instructions refer to Practice Set 1.
1. A survey shows that 63% of the Americans
like cheese whereas 76% like apples. If x % of
the Americans like both cheese and apples,
then
(a) x = 39
(c) 39 ≤ x ≤ 63
(b) x = 63
(d) None of these
2 (log 3 x )2 − |log 3 x | + a = 0 possess four real
solutions, is
1
8
(d) None of these
(b) 0 < a <
(c) 0 < a < 5

 x2 + 1
− ax − b  = 0 is
lim 
x→ ∞  x +1

(b) a = 2 and b = − 1
(c) a = − 1 and b = 1
(d) a = 1 and b = − 1
1 + 2a − a 2
(d)
1 − 2a + 2a 2
defective. An inspector takes out one
transistor at random, examines it for defects
and replace it. After it has been replaced
another inspector does the same thing and
then so does a third inspector. The
probability that atleast one of the inspectors
finds a defective transistor, is equal to
1
27
(b)
8
27
(c)
19
27
(d)
26
27
10. The function f ( x ) = x ax − x 2 , a > 0
1
1
1
1
n
= ( x )2
n
+ ( y )2
n
= ( x )2
n
− ( y )2
x−y
bn
(d)
and bn
an
all n ∈ N . Then, a1 a 2 a 3 ... an is equal to
for
xy
bn
6. If the tangent to the curve xy + ax + by = 0 at
(1, 1) is inclined at an angle tan −1 2 with
X-axis, then
(a) a = 1 and b = 2
(c) a = − 1 and b = − 2
(c)
(d) x 2 + y 2 + 3x − 6y − 45 = 0
5. If a n andbn are two sequences given by
(c)
1 + 2a + 2a 2
(a) x 2 + y 2 + 3x − 6y + 5 = 0
29
(b) x 2 + y 2 + 3x − 6y +
=0
4
(c) x 2 + y 2 + 3x − 6y − 31 = 0
f is not continuous
f ′ is differentiable for all x
f ′ is continuous
None of the above
x+y
bn
(b)
which rolls on the outside of the circle
x 2 + y 2 + 3x − 6y − 9 = 0 is
4. Let f ( x ) = ( x + | x |)| x |. Then, for all x
(b)
1 − 2a + a 2
9. The locus of the centre of a circle of radius 2,
(a) a = 0 and b = 0
(a) x − y
(a)
(a)
3. The values of constants a andb, so that
(a)
(b)
(c)
(d)
the curve 2 x 2 + y 2 − 2 x = 0 is given by
8. A box contains 15 transistors, 5 of which are
2. The value of a for which the equation
(a) −2 < a < 0
7. The longest distance of the point (a , 0) from
(b) a = 1 and b = − 2
(d) a = − 1 and b = 2
3a
(a) increases on the interval  0, 
 4
3a 
(b) decreases on the interval 
, 0
 4 
3a
(c) decreases on the interval  0, 
 4
3a 
(d) increases on the interval 
,a 
 4

11. If a , b and c are unit vectors such that
a + b + c = , then the value ofa ⋅ b + b ⋅ c + c ⋅ a
is
(a) 1
(b) 3
3
(c) −
2
(d) None of these
388
JEE Main Chapterwise Mathematics
12. Letα and β be any two positive values of x for
which 2 cos x ,|cos x |and 1 − 3 cos x are in
GP. The minimum value of |α − β | is
2
π
3
π
(c)
2
(b)
(a)
π
4
(d) None of these
1
(1 + an ), then
2

(1 − a 02 ) 
is equal to
cos 
 a1 a 2 a 3K to ∞ 


(b) −1
(c) a 0
(d)
(c) two
1
a0
p
is
2
(d) infinite
15. Axis of a parabola is y = x and vertex and
focus
are
at
a
distance
2l − m + 2n = 0,
π
4
π
(d)
2
(b)
x
A
∫0 [ x] dx = [ x]  2
(a) A = [x ] − 1
(c) A = [x ] + 1
tan −1 { x ( x + 1)} + sin −1 x 2 + x + 1 =
(b) one
π
6
π
(c)
3
(a)
by

+ B  , where [ ⋅]

denotes the greatest integer function, then
14. The number of real solutions of
(a) zero
cosines are given
lm + mn + nl = 0 is
19. If x > 0 and
13. If an + 1 =
(a) 1
18. The angle between the lines whose direction
2 and 2 2 ,
(b) B = − x − [x ]
(d) B = x + [x ]
20. The area above X-axis, bounded by the line
x = 4 and the curve y = f ( x ), where
f ( x ) = x 2 , 0 ≤ x ≤ 1 and f ( x ) = x , x ≥ 1, is
(a) 2 sq units
(c) 4 sq units
(b) 5 sq units
(d) 9 sq units
21. The differential equation of family of
parabolas with foci at the origin and axis
along the X-axis, is
2
respectively from the origin. Then, equation
of the parabola is
dy
dy
(a) y   + 2 x   − y = 0
 dx 
 dx 
(a) ( x − y )2 = 8( x + y − 2 )
(b) ( x + y )2 = 4( x + y − 2 )
dy
dy
(b) y   + 2 x   + y = 0
 dx 
 dx 
(c) ( x − y )2 = 4( x − y − 2 )
dy
dy
(c) x   + 2 y   − x = 0
 dx 
 dx 
2
2
(d) ( x − y )2 = 2 ( x − y + 2 )
(d) None of the above
16. If a > 2b > 0, then positive value of m for
which y = mx − b 1 + m 2 is a common
tangent to x 2 + y 2 = b 2 and ( x − a )2 + y 2 = b 2 ,
is
2b
(a)
(c)
a 2 − 4b 2
2b
a − 2b
(b)
(d)
b
a − 2b
a 2 − 4b 2
2b
17. If the axes are rectangular, the distance from
the point (3, 4, 5) to the point, where the line
x − 3 y − 4 z −5
meets the plane
=
=
1
2
2
x + y + z = 17, is
(a) 1
(c) 3
(b) 2
(d) None of these
Directions (Q. Nos. 22-24) Each of these
questions contains two statements :
Statement I (Assertion) and Statement II
(Reason). Each of these questions also has
four alternative choices, only one of which
is the correct answer. You have to select one
of the codes (a), (b), (c), and (d) given
below.
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation for
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation for
Statement I
(d) Statement I is true, Statement II is false
389
Practice Set 2
22. Statement I For n ∈ N ,2n > 1 + n 2n − 1 .
27. The minimum number of terms from the
1
2
+ 25 + ...,
3
3
so that the sum may exceed 1568, is
beginning of the series 20 + 22
GM > HM
Statement II
(AM)(HM) = (GM)2
and
(a) 25
23. Statement I The sum of the digits in the
tens place of all numbers with the help of 2,
3, 4, 5 taken all at a time is 84.
Statement II The sum of the digits in the
units place of all numbers formed with the
help of a1 , a 2 ,... , an taken all at a time is
(n − 1)!(a 1 + a 2 + ... + an ) (repetition of digits
not allowed).
n
Statement II
∑ ∑ (C i + C j ) = (n + 1)2n
j = 1i < j
25. If f ( x ) = ∫
x 2 + sin 2 x
1 + x2
f (1) is equal to
π
(a)
−1
4
(c) tan1 −
26.
π
4
sec 2x dx f (0) = 0, then
(d) 28
p ⇔~ q is true, when
(a) p is true and q is true
(b) p is false and q is true
(c) both p and q are false
(d) None of the above
xn
29.
n!
nπ
If f ( x ) = cos x cos
2
nπ
sin x sin
2
dn
dx n
2
4 , then the value of
8
[ f ( x )]x = 0 is equal to
(a) 0
(c) –1
(b) 1
(d) 2
30. The image of the pointP( 3 , 5) with respect to
π
(b) 1 −
4
(d) None of these
the line y = x is the pointQ and the image of
Q with respect to the line y = 0 is the point
R (a ,b ), then (a ,b ) is equal to
(a) (5, 3)
 12 x 
For all real values of x , 2 is
4x + 9
(a) ≤1
(c) >1
(c) 29
28. If p and q are simple propositions, then
24. Let (1 + x )n = C 0 + C 1x + C 2x 2 + ... + C n x n
Statement I
S = C 0 + (C 0 + C 1 ) + (C 0 + C 1 + C 2 )
+ ... + (C 0 + C 1 + ... + C n ) = n 2n −1
(b) 27
(b) (5, – 3)
(c) (– 5, 3)
(b) ≤ 2
(d) > 2
(d) None of the above
Answers
1. (c)
11. (c)
21. (a)
2. (b)
12. (d)
22. (b)
3. (d)
13. (c)
23. (b)
4. (c)
14. (c)
24. (d)
5. (c)
15. (a)
25. (c)
6. (b)
16. (a)
26. (a)
7. (d)
17. (c)
27. (c)
8. (c)
18. (d)
28. (b)
9. (c)
19. (a)
29. (a)
10. (a)
20. (b)
30. (b)
Solutions
1. Let A and B denote the set of Americans who like
cheese and apple, respectively.
∴
n( A) = 63, n(B) = 76
We know that, n( A ∪ B) = n( A) + n(B) − n( A ∩ B)
⇒
n( A ∪ B) = 63 + 76 − n( A ∩ B)
⇒
n( A ∩ B) = 139 − n( A ∪ B)
But
n( A ∪ B) ≤ 100
⇒
− n( A ∪ B) ≥ − 100
⇒
139 − n( A ∪ B) ≥ 139 − 100 = 39
…(i)
⇒
39 ≤ n( A ∩ B)
Again,
A ∩ B ⊆ A and A ∩ B ⊆ B
∴
n( A ∩ B) ≤ n( A) = 63
and
n( A ∩ B) ≤ n(B) = 76
…(ii)
∴
n( A ∩ B) ≤ 63
Then,
39 ≤ n( A ∩ B) ≤ 63
[from Eqs. (i) and (ii)]
∴
39 ≤ x ≤ 63
 x2 + 1
− ax −
lim 
 x+1
3. Given that,
⇒
x→ ∞
lim
x→ ∞
( x2 + 1) − (ax + b )( x + 1)
=0
x+1
x2 (1 − a) − (a + b )x − b + 1
=0
x→ ∞
x+1
(b − 1)
x(1 − a) − (a + b ) −
x =0
lim
⇒
1
x→ ∞
1+
x
which is possible, if
1 − a = 0 and a + b = 0
⇒
a = 1 and
b = −1
⇒
lim
4. Given that, f( x) = ( x + | x|)| x|
 ( x + x)x, if x ≥ 0
=
( x − x)(− x), if x < 0
log 3 x ≥ 0 ⇒ x ≥ 1
2. Case I If
2 x2 , if x ≥ 0
=
 0, if x < 0
From given equation
2(log 3 x)2 − log 3 x + a = 0
Y
For real solution
(−1)2 − 4 ⋅ 2 ⋅ a > 0
1
a<
∴
8
Case II If log 3 x < 0
∴
x<1
From given equation
2(log 3 x)2 + log 3 x + a = 0
For real solution
(1)2 − 4 ⋅ 2 ⋅ a > 0
1
a<
∴
8
y = 2x2 , ∀ x ≥ 0
…(i)
y = 0, ∀ x < 0
X′
log 3 x =
⇒
−1 ±
1 − 8a
4
<0 ⇒
Y′
…(ii)
Y
1 − 8a < 1
a> 0
1 − 8a
4
−1 ±
y = 4x
[here, log 3 x < 0]
1 − 8a < 0
[taking, + ve sign]
…(iii)
1
Hence, from Eqs. (i), (ii) and (iii), we get 0 < a < .
8
⇒
∴
X
O
It is clear from the figure that f( x) is continuous
everywhere.
4 x, if x ≥ 0
Now,
f ′ ( x) = 
 0, if x < 0
From equation
−1 ±

b  = 0

X′
O
X
Y′
It is clear from the figure that f ′( x) is continuous
everywhere but not differentiable at x = 0.
391
Practice Set 2
n
n
n
n
n
n
5. Given that, an = x1/ 2 + y1/ 2
bn = x1/ 2 − y1/ 2
and
n
n
Now,
an bn = ( x1/ 2 + y1/ 2 )( x1/ 2 − y1/ 2 )
⇒
an bn = ( x1/ 2 )2 − ( y1/ 2 )2
⇒
an bn = x 2
n
n
1
⇒
∴
1
n−1
− y2
n−1
…(i)
an bn = bn − 1
(a1a2 . . . an )
bn
a1a2 . . . an =
bn
(a a . . . an − 1 )(an bn )
= 1 2
bn
a1a2 . . . an − 1bn − 1
[from Eq. (i)]
=
bn
a a . . . bn − 2
= 1 2
bn
… … … … … …
… … … … … …
ab
b
x− y
⇒ a1a2 . . . an = 1 1 = 0 =
bn
bn
bn
6. The point (1, 1) lies on the curve
xy + ax + by = 0
∴
a + b = −1
On differentiating Eq. (i) w.r.t. x, we get
dy
dy
+ y+ a+ b =0
x
dx
dx

a
+ 1
 dy 
⇒
=−

 
 dx  ( 1, 1)
 b + 1
…(i)
…(ii)
On differentiating w.r.t. x, we get
dS
= − 2 x + 2 (1 − a)
2S
dx
dS
For S to be maximum,
=0
dx
⇒
−2 x + 2 (1 − a) = 0
⇒
x = 1− a
d 2S
It can be easily checked that 2 < 0
dx
for x = 1 − a.
Hence, S is maximum for x = 1 − a.
On putting x = 1 − a in Eq. (i), we get
S = 1 − 2 a + 2 a2
8. Probability of defective transistors = 5 = 1
15
3
and probability of non-defective transistors
1 2
= 1− =
3 3
∴ Required probability
= 1 − P (none of the defective transistors)
2 2 2
= 1− × ×
3 3 3
8
19
= 1−
=
27 27
9. Let C 2 (h, k ) be the coordinates of the centre of the
circle which rolls on the outside of the circle.
Then,
C1C 2 = r1 + r2
…(iii)
∴
Since, the tangent makes an angle tan−1 2 with
X-axis, therefore slope of the tangent is 2.
 a + 1
∴
2=−

 b + 1
⇒
a + 2b = − 3
On solving Eqs. (ii) and (iv), we get
a=1
and
b = −2
2
 h + 3  + (k − 3)2 = 9 + 2



2
2
r2
C2 (h, k)
r1
…(iv)
C1 (–3/2, 3)
x2 + y2 + 3x _ 6y _ 9 = 0
7. Let
( x, y) be the point on the curve
2 x2 + y2 − 2 x = 0. Then, its distance from (a, 0) is
given by
S = ( x − a)2 + y2
2
⇒
S = x2 − 2 ax + a2 + 2 x − 2 x2
[using, 2 x + y − 2 x = 0]
2
⇒
S = − x + 2 x(1 − a) + a
2
2
2
2
…(i)
⇒
⇒
9
13
+ 9 =  
2
4
h2 + k 2 + 3h − 6k − 31 = 0
h2 + k 2 + 3h − 6k +
Hence, locus of (h, k ) is
x2 + y2 + 3 x − 6 y − 31 = 0
2
392
JEE Main Chapterwise Mathematics
10. Given, f( x) = x ax − x2 , a > 0
f ′ ( x) = x ⋅
∴
=
1
2 ax − x
(3a − 4 x2 )
2 ax − x2
2
=
⋅ ( a − 2 x) +
ax − x2 ⋅ 1
− 4 x( x − 3a/4)
2 ax − x2
f ′ ( x) > 0
For
[increasing]
x ∈(0, 3a/4) and for f ′( x) < 0 [decreasing]
Then,
x ∈ (− ∞, 0) ∪ (3a/4, ∞ )
Then,
a +b+ c = 0
11. Given that,
∴
(a + b + c )2 = 0
⇒ |a| +|b| +|c |2 + 2 (a ⋅ b + b⋅ c + c ⋅ a ) = 0
2
2
⇒2 (a ⋅ b + b⋅ c + c ⋅ a ) = − [(1)2 + (1)2 + (1)2 ]
3
⇒ a ⋅b + b⋅c + c ⋅a = −
2
12. Since, 2cos x,|cos x| and 1 − 3cos 2 x are in GP.
∴
cos 2 x = 2 cos x(1 − 3cos 2 x)
⇒
cos x (6cos 2 x + cos x − 2 ) = 0
cos x = 0
(2 cos x − 1)(3cos x + 2 ) = 0
1
2
⇒
cos x = 0, and −
2
3
π
Hence, the two smallest positive values of x are
3
π
and .
2
π π π
|α − β| = − =
∴
2
3 6
Put
∴
1 + a0
a1 =
2
a0 = cosθ
θ
a1 = cos 
2
a2 =
∴








sinθ
 sinθ sinθ

= lim 
=
=
n→ ∞
θ


n
θ
 θ(1)
 2 sin 
 2n  θ 

⋅ n
  θ
2 

  2


2




2

 sinθ 
(1 − a0 ) 

= cos
∴ cos
 = cosθ = a0
 a a a K∞ 
 sinθ 
 1 2 3

 θ 
14. Q
⇒
and
13. Q
sin θ  
  2 r − 1  
= lim Π 

n→ ∞ r =1
 2 sin θ  

 2 r  
θ
θ
θ

sin n −1  
sin 2 
sin 
 sinθ
2 
2
2  
= lim 
⋅
...
⋅

n→ ∞
2 sin θ  2 sin θ  2 sin θ  2 sin θ  
n
3
2
2
 2  
2 
2 

n
tan−1 { x( x + 1)} + sin−1
1
⇒ cos −1
⇒ cos −1
⇒
( x2 + x)2 + 1
=
x2 + x + 1
x2 + x + 1
⇒ 1 = ( x2 + x + 1){( x2 + x)2 + 1}
⇒ ( x2 + x)3 + ( x2 + x)2 + ( x2 + x) + 1 = 1
⇒
( x2 + x)[( x2 + x)2 + ( x2 + x) + 1] = 0
⇒
x2 + x = 0
x = 0 and − 1
15. Since, distance of vertex from origin is 2 and focus
is 2 2, here a = 2.
Y
a1 a2 a3 K to ∞ = lim a1 a2 a3 K an
y=x
n
P
= lim Π ar
n→ ∞ r =1
a
= lim Π cos n 
n→ ∞ r =1
2 
2 sin θ  ⋅ cos θ
 r
 r
n 
2 
2

= lim Π 
n→ ∞ r =1
θ



2 sin r 

2 
π
− sin−1
2
= cos −1
( x2 + x)2 + 1
1
∴
1 + a1
θ
= cos 2 
2 
2
n→ ∞
=
( x2 + x)2 + 1
1
π
2
x2 + x + 1
x2 + x + 1 =
n
N






X′
V (1, 1)
F (2, 2)
M
x+y–2=0
O
Y′
∴ V(1, 1) and F(2 , 2 ) i.e., axis lies on y = x.
X
393
Practice Set 2
Length of latusrectum = 4a = 4 2
By definition of parabola, PM 2 = (4 a)(PN)
where, PN is length of perpendicular upon line NV.
x+ y−2 = 0
( x − y)2
x + y − 2
= 4 2 



2
2
⇒
∴
( x − y)2 = 8( x + y − 2 )
Alternate Solution
Given that the axis of parabola is y = x and vertex
and focus are at distance 2 and 2, respectively
form the origin.
P(x
, y)
Y
X′
C
X
Y′
⇒
⇒
⇒
( x + y)2 = 2 ( x − 2 )2 + 2 ( y − 2 )2
x + y + 2 xy = 2 x2 + 8 − 8 x + 2 y2 + 8 − 8 y
2
m2 + 1
|− b 1 + m2|
and
⇒
1 + m2
=b
=b
|ma − b 1 + m2| = | − b 1 + m2|
⇒ m2 a2 − 2 abm 1 + m2 + b 2 (1 + m2 )
= b 2 (1 + m2 )
ma − 2 b 1 + m2 = 0
m2 a2 = 4b 2 (1 + m2 )
2b
m=
2
a − 4b 2
17. Given line is
O
Now draw VC ⊥ OX, then since V is the vertex of
parabola whose axis is y = x.
∴Let the coordinates of V be (a, a) . Then, in right
angled ∆OVC, by applying Pythagoras theorem, we
get vertex of parabola is (1, 1.)
Similarly, coordinates of focus are F (2, 2 ).
Since, directrix is perpendicular to the axis of
parabola.
∴Equation of directrix is x + y + k = 0
Also, this equation will pass through the origin
(since, V is the mid-point of the focus and foot of the
directrix)
QEquation of directrix will become
x+ y=0
Let P ( x, y) be any point on the parabola
Now, by the definition of parabola,
|PM| = |PF|
x+ y
= ( x − 2 )2 + ( y − 2 )2
⇒
2
⇒
|ma − 0 − b 1 + m2|
⇒
√2
M
∴
⇒
F
2√
2
V
so its distance from centre = Radius of circle.
⇒
y=x
√2
16. Given, y = mx − b 1 + m2 touches both the circles,
2
8 x + 8 y − 16 = x2 + y2 − 2 xy
( x − y)2 = 8 ( x + y − 2 )
which is the required equation of parabola.
x−3 y−4 z−5
[say]
=
=
=r
1
2
2
Any point on the line is (r + 3, 2 r + 4, 2 r + 5), it lies
on the plane x + y + z = 17.
∴
(r + 3) + (2 r + 4) + (2 r + 5) = 17
⇒
r =1
Thus, the point of intersection of the plane and the
line is (4, 6, 7).
∴ Required distance
= Distance between (3, 4, 5) and (4, 6, 7)
= (4 − 3)2 + (6 − 4)2 + (7 − 5)2
= 1+ 4 + 4 = 3
18. The given equations are
2 l − m + 2 n = 01 and lm + mn + nl = 0
On eliminating m from the given equation, we get
2 (l + n)2 + nl = 0 [Q m = 2 l + 2 n]
⇒
(2 l + n) (l + 2 n) = 0
⇒
n = − 2l
⇒
m = − 2l
or
l = − 2n
⇒
m = − 2n
∴The direction ratios are(1, − 2, − 2 ) and (− 2, − 2, 1.)
Also, 1 (− 2 ) + (− 2 ) (− 2 ) + (− 2 ) × 1
= −2+ 4−2 = 0
∴Lines are perpendicular.
π
So, angle between them is .
2
394
JEE Main Chapterwise Mathematics
19. Let [ x] = l , then x = l + k,
x
∫0
∴
=
1
∫0
[ x] dx =
l + k
∫0
[ x] dx
2
[ x] dx + . . . +
∫1
[ x] dx +
0≤ k < 1
l
∫ l − 1 [ x] dx
+
∫l
l + k
[ x] dx
= 0 + 1 + 2 + . . . + (l − 1) + l ⋅ k
1
= (l − 1) l + lk
2
1
= [ x]([ x] − 1) + [ x]( x − [ x])
2
1
= [ x]  ([ x] − 1) + ( x − [ x])

2
∴ A = [ x] − 1 and B = x − [ x]
1/ n
 ( n − 1) n 
= 2 2 


⇒ (2 n − 1) > n ⋅ 2( n − 1)/ 2
1
4
∫1
0
23. Sum of the digits in the tens place
x dx
= Sum of the digits in the units place
= (4 − 1)! (2 + 3 + 4 + 5)
= 6 ⋅ 14 = 84
Hence, both the statements are true and Statement
II is the correct explanation for Statement I.
y = √x
(1, 1)
y = x2
X′
X
O
24. S can be written as
S = nC 0 + (n − 1) C1 + (n − 2 ) C 2
+ ... + 1⋅ C n −1 + 0C n …(i)
x=1 x=4
Y′
1
Using C r = C n − r , Eq. (i) can be written as
4
 x3 
 2 x3 / 2 
=  + 

 3  0  3 1
1 2
= + (8 − 1) = 5 sq units
3 3
S = 0C1 + 1⋅ C1 + 2C 2
21. Let the directrix be x = − 2 a and latusrectum be 4a.
Then, the equation of the parabola is distance from
focus = Distance from directrix
∴
x2 + y2 = (2 a + x)2
⇒
y2 = 4a (a + x)
…(i)
On differentiating w.r.t. x, we get
1 dy
dy
= 2a ⇒ a = y
y
2 dx
dx
On putting the value of a in Eq. (i), we get
dy  y dy
+ x
y2 = 2 y


dx  2 dx
⇒
2
2n > 1 + n 2n − 1
⇒
20. The required area = ∫ x2 dx +
Y
AM GM
=
>1
GM HM
∴
AM > GM
2
n −1
1 + 2 + 2 + ... + 2
> (1⋅ 2 ⋅ 2 2 ... 2 n − 1 )1/ n
⇒
n
1⋅ (2 n − 1)
2 −1
> {21 + 2 + 3 + . . . + ( n − 1) }1/ n
⇒
n
22. Q
dy
dy
y   + 2 x   − y = 0
 dx 
 dx 
+ ... + (n − 1)C n −1 + nC n …(ii)
On adding Eqs. (i) and (ii), we get
2S = n [C 0 + C1 + C 2 + ... + C n ] = n ⋅ 2 n

 n n
n
Q ∑ C r = 2 

 r = 0
⇒ S = n ⋅ 2 n −1
⇒ Statement I is true.
n
∑ ∑ (C i + C j ),
Also, in the expression
j =1 i < j
C i ⋅ (0 ≤ i ≤ n) occurs exactly n times.
n
n
Thus,
∑ ∑ (C i + C j ) = n ∑ C k
j =1 i < j
k =0
= n⋅2
⇒ Statement II is false.
n
each
395
Practice Set 2
2
2
25. Given that, f( x) = ∫ ( x + sin2 x)sec 2 x dx
28. p ⇔ ~ q is true, iff p, ~ q are both true or both false.
1+ x
=
=
∫
x2 + (1 − cos 2 x)
1 + x2
⋅ sec x dx
n!

1 
 sec 2 x −
 dx
1 + x2 

∫
i.e., iff either pis true and q is false or p is false and q
is true.
2
cos  x +
dn

f
x
[
(
)]
=
dxn
sin  x +

29.
= tan x − tan−1 x + c
Q
f(0) = 0 ⇒ c = 0
∴
f( x) = tan x − tan−1 x
⇒
f(1) = tan 1 − tan−1 1 = tan 1 −
12 x
26. Let
4 x2 + 9
∴
π
4
= y
4 yx2 − 12 x + 9 y = 0
D≥ 0
For real values of x,
∴
(12 )2 − 4 ⋅ 4 y ⋅ 9 y ≥ 0
⇒
1 − y2 ≥ 0 ⇒ y2 < 1
⇒
n
d
[f( x)]x = 0
dxn
a = 20,
Here,
d = 22
⇒
⇒
4
nπ 

2 
8
sin
nπ
2
30. Let Q( x1, y1 ) be the image of the point P(3, 5) with
respect to the line y = x.
Y
P (3, 5) y = x
Q (5, 3)
3
8
68
2
− 20 =
− 20 =
3
3
3
S n > 1568
n
8
40 + (n − 1)  > 1568
3 
2 
n 112 + 8n
⋅
> 1568
2
3
6
n2 + 14n > × 1568 = 1176
8
X′
⇒ n2 + 14n − 1176 > 0
⇒ (n + 42 )(n − 28) > 0
As n is positive, so we take n − 28 > 0 i.e., n > 28.
Hence, minimum value of n is 29.
X
O
R (a, b)
Since,
⇒
nπ 
nπ
 cos
2 
2
[since, C1 and C 2 are identical]
27. Given, series 20 + 22 2 + 25 1 + . . . is an AP.
3
2
2
n!
n!
nπ
nπ
= cos 2 cos 2 4
nπ
nπ
8
sin
sin
2
2
=0
 12 x 
| y| ≤ 1 ⇒  2
≤ 1
4 x + 9
⇒
n!
Y′
Then,
x1 = 5
and
y1 = 3
∴
Q = (5, 3)
Since, the image of the point Q(5, 3) w.r.t. the line
y = 0 is (a, b ).
∴
a=5
and
b= −3
∴
(a, b ) = (5, − 3)
Practice Set 3
Instructions
For instructions refer to Practice Set 1.
1. The range of the function f ( x ) = x 2 +
1
x +1
2
is
(d)
(a) [1, ∞ )
3
(c)  , ∞ 
 2 
(b) [2 , ∞ )
(d) None of these
2

2. If f ( x ) =  a2 x + b ,
b ≠ 0, x ≤ 1
, then f ( x )
bx
+
ax
+
c
,
x>1

is continuous and differentiable at x = 1, if
(a) c = 0, a = 2 b
(c) a = b , c = 0
(b) a = b , c ∈ R
(d) a = b , c ≠ 0
3. There is a point P (a , a , a )on the line passing
through the origin and equally inclined with
axes. The equation of the plane
perpendicular to OP and passing through P
cuts the intercepts on axes, the sum of
whose reciprocals is
2a
3
1
(c)
a
4
a
3
1
(d)
3a
cuts the circle x 2 + y 2 = 4 orthogonally, then
the equation of the locus of its centre is
(a) x 2 + y 2 − 3x − 8y + 1 = 0
1
2 (b − a )
log [ f ( x )] + C , then f ( x ) is equal to
(b)
a 2 sin 2 x − b 2 cos2 x
(a)
1
3
(b)
2
3
(c)
1
9
(d)
2
9
7. The points representing complex number z
for which| z − 3| = | z − 5| lie on the locus given
by
(a) an ellipse
(c) a straight line
(b) a circle
(d) None of these
8. The value of α, for which the equation
π
3
(b)
π
4
(c)
π
2
(d)
roots
π
6
9. For n ∈ N , 10n − 2 ≥ 81n, if
(a) n > 5
(c) n < 5
(b) n ≥ 5
(d) n > 8
of ( 3 + 2 x ) 74 whose coefficients are equal,
are
(c) 2 x + 4y − 9 = 0
(d) 2 x + 4y − 1 = 0
a 2 sin 2 x + b 2 cos2 x
1
2 2
is inscribed in
3
a circle and a point within the circle is
chosen at random. The probability that this
point lies outside the ellipse, is
10. The two consecutive terms in the expansion
(b) x 2 + y 2 − 2 x − 6y − 7 = 0
1
a 2 cos2 x + b 2 sin 2 x
6. An ellipse of eccentricity
(a)
4. If a circle passes through the point (1, 2) and
5. If ∫ f ( x ) sin x cos x dx =
1
a 2 cos2 x − b 2 sin 2 x
1
x 2 − (sin α − 2 ) x − (1 + sin α ) = 0 has
whose sum of square is least, is
(b)
(a)
(a)
(c)
2
2
(a) 11, 12
(c) 30, 31
11. Let Sn =
(b) 7, 8
(d) None of these
1
+
1+2
+ ...+
1 + 2 + ... + n
1
1 +2
1 + 2 3 + ... + n 3
n = 1, 2 , 3 ,... . Then, Sn is not greater than
(a)
1
2
3
3
(b) 1
3
(c) 2
3
(d) 4
;
397
Practice Set 3
12. The number of ways of choosing 10 balls
from infinite white, red, blue and green balls
is
(a) 84
(b) 90
(c) 100
(d) 286

13. If E (θ) =  cos θ
cos θ sin θ
 and θ and φ
sin 2 θ 
cos θ sin θ
π
differ by an odd multiple of ,then E (θ ) E ( φ )
2
is a
2
(a) unit matrix
(c) diagonal matrix
(b) null matrix
(d) None of these
14. A parabola is drawn with its focus at ( 3, 4)
and vertex at the focus of the parabola
y 2 − 12 x − 4y + 4 = 0. The equation of the
parabola is
(a) y 2 − 8x − 6y + 25 = 0 (b) y 2 − 6x + 8y − 25 = 0
(c) x − 6x − 8y + 25 = 0 (d) x + 6x − 8y − 25 = 0
2
2
15. If p , p ′ denote the lengths of the
perpendiculars from the focus and the
centre of an ellipse with semi-major axis of
length a, respectively on a tangent to the
ellipse and r denotes the focal distance of
the point, then
(a) ap ′ = rp + 1
(c) ap = rp ′ + 1
(b) rp = ap ′
(d) ap = rp ′
16. Area of a triangle with vertices (a ,b ),( x1 , y1 )
and( x 2 , y 2 ), wherea , x1 and x 2 are in GP with
common ratio r and b , y1 and y 2 are in GP
with common ratio s, is given by
(a)
(b)
(c)
(d)
1
ab(r − 1 )( s − 1 )( s − r )
2
1
ab(r + 1 )( s + 1 )( s − r )
2
ab(r − 1 )( s − 1 )( s − r )
None of the above
17. The equation of perpendicular bisectors of
sides AB and AC of a ∆ABC are x − y + 5 = 0
and x + 2 y = 0, respectively. If the
coordinates of vertex A are (1, –2), the
equation of BC is
(a) 14x + 23y − 40 = 0
(b) 14x − 23y + 40 = 0
(c) 23x + 14y − 40 = 0
(d) 23x − 14y + 40 = 0
3
3
and sin α = − , where θ does
2
5
not lie in the third quadrant, then
2 tan α + 3 tan θ
is equal to
cot 2 θ + cos α
18. If cos θ = −
7
22
9
(c)
22
5
22
22
(d)
5
(b)
(a)
19. The smallest positive x satisfying the
equation log cos x sin x + log sin x cos x = 2 is
π
2
π
(c)
4
π
3
π
(d)
6
(a)
(b)
20. A parallelogram is constructed on the
→
→
→
→
→
→
vectors a = 3α − β , b = α + 3 β , if | α | = | β | = 2
→
π
→
and angle between α and β is , then length
3
of a diagonal of the parallelogram is
(a) 4 5
(c) 4 17
(b) 4 3
(d) None of these
Directions (Q. Nos. 21-23) Each of these
questions contains two statements :
Statement I (Assertion) and Statement II
(Reason). Each of these questions also has
four alternative choices, only one of which
is the correct answer. You have to select one
of the codes (a), (b), (c), (d) given below.
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
21. Statement I If both roots of the equation
4x 2 − 2 x + a = 0, a ∈R lie in the interval
1
(–1, 1), then −2 < a ≤ .
4
If f ( x ) = 4x 2 − 2 x + a , then
1
D ≥ 0, f ( −1) > 0 and f (1) > 0 ⇒ −2 < a ≤ .
4
Statement II
398
JEE Main Chapterwise Mathematics
22. Statement I The equation
sin (cos x ) = cos (sin x ) does not possess real
roots.
Statement II If sin x > 0, then
2nπ < x < (2n + 1),n ∈I .
27. If f ( x ) = x 3 + bx 2 + cx + d and 0 < b 2 < C , then
in ( − ∞ , ∞ )
(a)
(b)
(c)
(d)
23. Statement I If a ,b , c are distinct and x , y , z
are not all zero and ax + by + cz = 0,
bx + cy + az = 0 and cx + ay + bz = 0, then
a + b + c = 0.
28.
1
1
−
2x 1 − x
1 − x2
1
1
(b)
−
2 x 1− x
1 − x2
24. The value of c, so that for all real x , the
1
1
+
2 x 1− x
1 − x2
1
1
(d) −
+
2 x 1− x
1 − x2
(c)
vectors c x i − 6 j + 3 k , x i + 2 j + 2 cx k make
an obtuse angle, is
(b) 0 < c <
(c) −
4
<c <0
3
4
3
(d) c > 0
29.
25. The solution of the equation
y −x
dy
=a
dx
 2 dy 
 y +  is

dx 
(a) y = c ( x + a )(1 − ay ) (b) y = c ( x + a )(1 + ay )
(c) y = c ( x − a )(1 + ay ) (d) None of these
26. The
area common to the circle
x 2 + y 2 = 16a 2 and the parabola y 2 = 6ax is
4a 2
(4π − 3)
3
2
4a
(c)
(8 π − 3)
3
(a)
(b)
4a 2
(4π +
3
3)
d
sin −1 ( x 1 − x + x 1 − x 2 ) is equal to
dx
(a) −
Statement II a 2 + b 2 + c 2 > ab + bc + ca , if
a ,b , c are distinct.
(a) c < 0
f ( x ) is strictly increasing function
f ( x ) has a local maxima
f ( x ) is strictly decreasing function
f ( x ) is bounded
sin θ
1 
 1

Let A = − sin θ
1
sin θ ,


− sin θ
1 
 −1
where 0 ≤ θ ≤ 2 π. Then, the range of | A | is
(a) 0
(c) [2, 4]
(b) {2, 4}
(d) None of these
30. The negation of the compound proposition
p ∨ (~ p ∨ q ) is
(a)
(b)
(c)
(d)
(p ∧ ~ q )∨ ~ p
(p ∧ ~ q )∧ ~ p
(p ∧ ~ q )∨ ~ p
None of the above
(d) None of these
Answers
1. (a)
11. (c)
21. (b)
2. (a)
12. (d)
22. (c)
3. (c)
13. (b)
23. (b)
4. (c)
14. (c)
24. (c)
5. (a)
15. (d)
25. (a)
6. (b)
16. (a)
26. (b)
7. (c)
17. (a)
27. (a)
8. (c)
18. (b)
28. (c)
9. (b)
19. (c)
29. (c)
10. (c)
20. (b)
30. (b)
Solutions
1. Q f( x) = x2 +
∴
1
4. Let the equation of circle be
x2 + 1
x2 + y2 + 2 gx + 2 fy + c = 0
 1

f( x) = x +  2
− 1 + 1
x +1 
2
 x2
= ( x2 + 1) −  2
x +


1

1
= 1 + x  1 − 2
x +



1
2
Since, this passes through (1, 2).
∴
12 + 2 2 + 2 g (1) + 2 f(2 ) + c = 0
⇒
∴
⇒
b ≠ 0, x ≤ 1
x>1
b ≠ 0, x ≤ 1
x>1
Since, f( x) is continuous at x = 1.
∴
lim f( x) = lim f( x)
x → 1−
x → 1+
⇒
a+ b=b+ a+c
5c = 0
⇒
Also, f( x) is differentiable at x = 1
LHD at x = 1 = RHD at x = 1
∴
⇒
2 a = 2 b (1) + a
⇒
a = 2b
3. Now, distance of P from origin,
OP =
DC’s of OP =
2
2 ( g ⋅ 0 + f ⋅ 0) = c − 4 ⇒ c = 4
On putting the value of c in Eq. (i), we get
2 g + 4f + 9 = 0
2. Given that,
 ax + b,
f ( x) =  2
bx + ax + c,
 2 ax,
f ′ ( x) = 
2 bx + a,
…(i)
Also, the circle x + y = 4 intersects the circle
x2 + y2 + 2 gx + 2 fy + c = 0 orthogonally.
≥ 1, ∀x ∈ R
Hence, range of f( x) is [1, ∞ ).
2
5 + 2g + 4f + c = 0
2
Hence, the locus of centre (− g , − f ) is
−2 x − 4y + 9 = 0
or
2 x + 4y − 9 = 0
5. Given that,
1
∫ f( x) sin x cos x dx = 2 (b2 − a2 ) log [f( x)] + C
On differentiating both sides, we get

d  log [f( x)]

f( x) sin x cos x =
+ C 
2
2
dx  2 (b − a )

1
1
f( x) sin x cos x =
⋅
f ′ ( x)
⇒
2 ( b 2 − a 2 ) f ( x)
f ′ ( x)
⇒ 2 (b 2 − a2 ) sin x cos x =
[f( x)]2
On integrating both sides, we get
2
2
∫ (2 b sin xcos x − 2 a sin x cos x) dx
a2 + a2 + a2 = a 3
a
a
a
1 1 1
,
,
,
,
=
a 3 a 3 a 3
3 3 3
Let equation of plane be
lx + my + nz = p
1
1
1
x+
y+
z=a 3
⇒
3
3
3
x
y
z
+
+
=1
⇒
3a 3a 3a
Intercepts on axes are 3a, 3a, 3a.
So, sum of their reciprocals is
1
1
1
1
S=
+
+
=
3a 3a 3a a
=
⇒ − b 2 cos 2 x − a2 sin2 x = −
∴ f ( x) =
f ′ ( x)
∫ [f( x)]2 dx
1
f ( x)
1
a2 sin2 x + b 2 cos 2 x
2
2
6. Let ellipse be x2 + y 2 = 1.
a
A′
[a > b ]
b
a
c
a
A
400
JEE Main Chapterwise Mathematics
∴Area of ellipse = π ab
= πa
74
1− e
2
= π a2 1 −
8 π a2
=
9
3

2 2
Qe = 3 


Area of circle = π a2
∴Required probability =
π a2
3 =2
2
3
πa
π a2 −
| z − 3| = | z − 5|
On squaring both sides, we get
( z − 3)( z − 3) = ( z − 5)( z − 5)
⇒
zz − 3 z − 3 z + 9 = zz − 5 z − 5 z + 25
⇒
2 z + 2 z = 16 ⇒ z + z = 8
⇒
2 x = 8 ⇒ x = 4 [putting z = x + iy]
Hence, locus of z is a straight line parallel to Y-axis.
7. Given that,
8. Let the roots of the equation be p and q.
Let
S = p2 + q 2
= ( p + q )2 − 2 pq
…(i)
Given equation is
x2 − (sin α − 2 ) x − (1 + sin α ) = 0
∴
p + q = (sin α − 2 ), pq = − (1 + sin α )
From Eq. (i), we get
S = (sin α − 2 )2 + 2 (1 + sin α )
= sin2 α − 4 sin α + 4 + 2 + 2 sin α
= sin2 α − 2 sin α + 6
= (sin α − 1)2 + 5
This is least when sin α − 1 = 0
π
α=
∴
2
For n = 4,102 ≥/ 81 × 4
For n = 5,103 ≥ 81 × 5
Hence, by mathematical induction for n ≥ 5, the
proposition is true.
10. General term of (3 + 2 x)74 is
+1
=
C r (3)74 − r 2 r xr
74
Let two consecutive terms be Tr + 1th and Tr
terms.
According to the given condition,
Coefficient of Tr + 1 = Coefficient of Tr + 2
⇒
74
C r 374 − r 2 r =
74
Cr
74 − ( r + 1) r + 1
2
+ 13
74
⇒
⇒
148 − 2 r = 3 r + 3
∴
r = 29
Hence, two consecutive terms are 30 and 31.
11. Given that, S n = 13 + 31 + 2 3
1
1 +2
1+ 2 + 3 + ... + n
+ ... + 3
1 + 2 3 + 33 + . . . + n3
1+ 2 + 3 + ... + n
Σn
Now, Tn = 3
=
1 + 2 3 + 33 + . . . + n3 Σn3
=
n(n + 1)/2
{n(n + 1)/2}2
=
1
2
1 
=2  −

n(n + 1)
 n n + 1
∴ S n = T1 + T2 + . . . + Tn
1
1 
1 1
1 1
= 2  −  + 2  −  + . . .+ 2  −

 2 3
1 2
 n n + 1

1
= 2 1 −
+
n



2
Q n + 1 ≤ 1



2
≤2
 =2 −
n+1
1
12. Required number of ways
= Coefficient of x10 in(1 + x + x2 + . . .)4
 1
= Coefficient of x10 in 
1 −


x
+ 2 th
4
=Coefficient of x10 in (1 − x)− 4
= Coefficient of x10 in
(1 + 4C1 x + 5C 2 x2 + . . . +
13 ⋅ 12 ⋅ 11
= 286
= 13C10 =
3⋅2 ⋅ 1
9. Let P (n) : 10n − 2 ≥ 81n
Tr
Cr
3
+1
=
2
Cr
74 − r 3
=
r+1 2
⇒
2
C10 x10 + . . .)
13
Alternate Solution
The required number of solution = Number of
non-negative integral solution of the equation
x1 + x2 + x3 + x4 = 10
where, x1 = number of white balls
x2 = number of red balls
x3 = number of blue balls
x4 = number of green balls
Now, the number of non-negative integral solution
of x1 + x2 + x3 + x4 = 10 =10 + 4 − 1 C 4 − 1
=13 C 3 =
= 286
13 ⋅ 12 ⋅ 11
3⋅2 ⋅1
401
Practice Set 3
cos θ sin θ

sin2 θ 
2

13. Given that, E(θ) =  cos θ
cos θ sin θ
∴
 cos θ
cos θ sin θ
E(θ) E(φ) = 

sin2 θ 
cos θ sin θ
∴
ap =
2
 cos 2 φ
cos φ sin φ
×

sin2 φ 
cos φsin φ
cos 2 θcos 2 φ + cos θ sin θ cos φ sin φ
=
2
2
 cos θ sin θcos φ + sin θ cos φ sin φ
cos 2 θ cos φ sin φ + cos θ sin θ sin2 φ

cos θ sin θ cos φ sin φ + sin2 θ sin2 φ 
cos θcos φ cos (θ − φ) cos θ sin φcos(θ − φ)
=

 cos φ sin θcos (θ − φ) sin θsin φ cos (θ − φ)
cos θ cos φ cos (2 n + 1) π cos θ sin φ cos(2 n + 1) π 
2
2
=

π
π
sin θ sin φ cos (2 n + 1) 
 cos φ sin θ cos (2 n + 1)
2
2
 0 0
=

 0 0
Qcos (2 n + 1) π = 0


2
14. Given equation can be rewritten as
( y − 2 )2 = 12 x
Here, vertex and foci are (0, 2) and (3, 2).
So, vertex of the required parabola is (3, 2) and focus
is (3, 4). The axis of symmetry is x = 3 and latusrectum
= 4 ⋅ 2 = 8.
Hence, required equation is
( x − 3)2 = 8( y − 2 )
⇒
Again, r = SP = a(1 − e cos θ)
x2 − 6 x − 8 y + 25 = 0
2
2
15. Tangent to the ellipse x2 + y 2 = 1 at (a cos θ, b sin θ)
a
b
cos 2 θ
a2
+
sin2 θ
= rp′
b2
16. Since, a, x1, x2 are in GP with common ratio r.
∴
x1 = ar,
x2 = ar 2
Also, b, y1, y2 are in GP with common ratio s.
∴
y1 = bs,
y2 = bs 2
The area of triangle is given by
a
a b 1
1
1
∆=
ar
x1 y1 1 =
2
2
ar 2
x2 y2 1
1
1
ab r
2
r2
=
1
s
⇒
∴
∆=
∆=
b 1
bs 1
bs 2 1
1
1
s2 1
Applying C1 → C1 − C 3 and C 2
we get
0
0
1
∆ = ab r − 1 s − 1
2
r 2 − 1 s2 − 1
1
ab(r − 1)(s − 1)
2
→ C2 − C3 ,
1
1
1
0
1
0
1
1
1
r+1 s+1 1
1
ab (r − 1)(s − 1)(s − r )
2
17. Let B( x1, y1 ) and C( x2 , y2 ) be two vertices and
P 

x1 + 1 y1 − 2 
,
 lies on perpendicular bisector
2
2 
x− y+ 5=0
is
x
y
…(i)
cos θ + sin θ = 1
a
b
∴ p = Perpendicular distance from focus
(ae, 0) to the Eq. (i)

ae cos θ + 0 − 1

1 − e cos θ
=
…(ii)
= a
cos 2 θ sin2 θ
cos 2 θ sin2 θ
+
+
a2
b2
a2
b2
Also, p′ = Perpendicular distance from centre
(0, 0) to the Eq. (i)
1
…(iii)
=
cos 2 θ sin2 θ
+
a2
b2
a − ae cos θ
x1 + 1 y1 − 2
−
= −5
2
2
x1 − y1 = − 13
∴
⇒
…(i)
A (1, –2)
Q
P
N
B
(x1, y1)
M
C
(x2, y2)
402
JEE Main Chapterwise Mathematics
⇒
⇒
⇒
Also, PN is perpendicular to AB.
y1 + 2
× 1= −1
x1 − 1
∴
⇒
⇒
y1 + 2 = − x1 + 1
x1 + y1 = − 1
⇒
…(ii)
On solving Eqs. (i) and (ii), we get
x1 = − 7,
logcos x sin x = 1
sin x = cos x
tan x = 1
π
x=
4
y1 = 6
20. Q
∴
AC = a + b
|AC| = |a + b|
So, the coordinates of B are (–7, 6). Similarly, the
11 2
coordinates of C are  ,  .
 5 5
Hence, equation of BC is
2
−6
( x + 7)
y−6= 5
11
+7
5
14
⇒
y−6=−
( x + 7)
23
⇒
14 x + 23 y − 40 = 0
18. Since, cos θ = − 3 < 0 and θ does not lie in third
2
quadrant.
So, θ must be lying in 2nd quadrant.
1
tan θ = −
⇒
3
and
cot θ = − 3
3
Also, α lies in 3rd quadrant andsin α = −
5
3
tan α =
∴
4
4
and
cos α = −
5
1
3
2⋅ − 3⋅
2 tan α + 3 tan θ
3
4
=
∴
4
cot 2 θ + cos α
3−
5
3
−1
5
= 2
=
4 22
3−
5
19. Given that, logcos x sin x + logsin x cos x = 2
⇒
logcos x sin x = y
1
y+ =2
y
( y − 1)2 = 0
⇒
y=1
Let
∴
D
C
a+b
b
b
A
B
a
⇒|AC|2 = |a|2 + |b|2 + 2 a ⋅ b
→
→
→
→
⇒|AC|2 = {|3 α − β|2 +|α + 3 β|2
→
→
→
→
+ 2 (3 α − β ) ⋅ ( α + 3 β )}
→
→
→ →
→
→
→ →
= 9 α 2 + β 2 − 6 α⋅ β + α 2 + 9 β 2 + 6 α⋅ β
→
→
→ →
+ 6 α 2 − 6 β 2 + 16 α⋅ β
→
→ 2
⇒ |AC|2 = 16 α 2 + 4 β
→ →
+ 16 α⋅ β
→
→
⇒ |AC|2 = 64 + 16 + 16|α ||β | cos
⇒ |AC|2 = 64 + 16 + 16 × 2 × 2 ×
π
3
1
2
⇒ |AC|2 = 112
|AC|= 4 7
∴
Similarly,
|BD| = |a − b| = 4 3
21. Let f( x) = 4 x2 − 2 x + a
Since, both roots of f( x) = 0 are lie in the interval
(−1, 1,) we can take
D ≥ 0, f(−1) > 0 and f(1) > 0
Consider D ≥ 0,
1
…(i)
(−2 )2 − 4 ⋅ 4 ⋅ a ≥ 0 ⇒ a ≤
4
Consider f(−1) > 0,
…(ii)
4(−1)2 − 2 (−1) + a > 0 ⇒ a > − 6
Consider f(1) > 0,
4(1)2 − 2 (1) + a > 0 ⇒ a > − 2
From Eqs. (i), (ii) and (iii), we get
1
−2 < a ≤
4
Hence, option (b) is correct.
…(iii)
403
Practice Set 3
Since, cos (sin x) = sin (cos x)
π
Consider cos (sin x) = cos  − cos x
2

π

sin x = 2 nπ ±  − cos
⇒
2
1

⇒
sin x ± cos x =  2 n ±  π

2
∴
x , n ∈ I

On squaring both sides, we get
⇒
1
1 ± sin 2 x =  2 n ± 

2
2
1
|sin 2 x| =  2 n ± 

2
2
2 n ± 1 



2
But
2
π2
⇒
3c 2 + 4c < 0
⇒
4
3c  c +  < 0

3
−
⇒
π2 − 1
π 2 > 2 for all n ∈ I
Since, sin x > 0
2 nπ < x < (2 n + 1) π, n ∈ I
4
<c< 0
3
dy
dy
= a  y2 + 

dx 
dx
y− x
dy
(a + x) = y − ay2
dx
dy
dx
∫ y(1 − ay) = ∫ a + x
⇒
Hence, the given equation does not possess real
roots.
⇒
(6c )2 − 4c (−12 ) < 0
25. Given that,
So,|sin 2 x| > 1 which is not possible.
Statement II
b 2 − 4ac < 0
Then,
22. Statement I
⇒
1

a
dx
∫  y + 1 − ay  dy = ∫ a +
⇒
⇒
⇒
⇒
x
log y − log (1 − ay) = log (a + x) + log c
log y = log (1 − ay)(a + x)c
y = c (1 − ay)(a + x)
26. The point of intersection of circle
x2 + y2 = 16 a2
Hence, option (c) is correct.
Y
a b c
∆= b c a
23. Let
c
P
a b
= − (a3 + b 3 + c 3 − 3abc )
X′
M
O
= − (a + b + c )(a + b + c − ab − bc − ca)
2
=0
2
2
Q
[Q a + b + c = 0]
a2 + b 2 + c 2 − ab − bc − ca
1
= {(a − b )2 + (b − c )2 + (c − a)2 } > 0
2
Hence, option (b) is correct.
Y′
and
24. For obtuse angle,
(cxi − 6 j + 3k ) ⋅ ( xi + 2 j + 2c xk ) < 0
⇒
c x2 − 12 + 6c x < 0
⇒
c x2 + 6c x − 12 < 0
We know that, if
ax2 + bx + c > 0 or < 0, ∀x
X
A (4a, 0)
and parabola, y2 = 6ax is
x2 + 6 ax − 16 a2 = 0
⇒
⇒
( x + 8 a)( x − 2 a) = 0
x = 2 a, y = ± 2 3 a
The required common area, A = 2[ APOA]
2a
= 2∫
0
= 2∫
0
2a
y dx + 2 ∫
4a
2a
y dx
6a x dx + 2 ∫
4a
2a
(4a)2 − x2 dx
404
JEE Main Chapterwise Mathematics
2a
28. d {sin−1 ( x 1 − x +
2
= 2 6a  x3 / 2 
 3
 0
dx
4a
x
x
1
(4a)2 − x2 + (4a)2 sin−1
+2
 2
2
4a  2 a
2
= 2 ⋅ 6a (2 a)3 / 2
3
1 

+ 2 (0 − 2 a 3a) + 8a2  sin−1 1 − sin−1 


2  
=
∴
π π 
4

= 2 ⋅ 2 3 ⋅ a2 + 2 −2 3a2 + 8a2  − 

2
3
6  
π
16
3a2 − 4 3a2 + 16a2
=
3
3
4 3 2 16 πa2 4a2
a +
( 3 + 4π)
=
=
3
3
3
27. Given, f( x) = x3 + bx2 + cx + d
⇒
f ′( x) = 3 x2 + 2 bx + c
[As we know that, if ax2 + bx + c > 0 for all
Now,
x ⇒ a > 0 and D < 0 and in above equation]
D = 4b 2 − 12 c = 4(b 2 − c ) − 8c
where, b 2 − c < 0 and c > 0
∴
D = (− ve ) or D < 0
⇒ f ′( x) = 3 x2 + 2 bx + c > 0 for all x ∈ (− ∞, ∞ )
[as D < 0 and a > 0]
Hence, f( x) is strictly increasing function.
x 1 − x2 )}
d
sin−1 {( x 1 − ( x )2 + x 1 − x2 )}
dx

Qsin−1 A + sin−1 B



= sin−1 ( A 1 − B2 + B 1 − A 2 )
d
{sin−1 ( x 1 − ( x )2 + x 1 − x2 )}
dx
d
(sin−1 x + sin−1 x )
=
dx
1
1
=
+
2
2
x
1− x
1− x
29. Since,
−1
⇒
sin θ
1
1
sin θ
− sin θ
1
1
| A| = − sin θ
| A| = 2 (1 + sin θ)
2
Now, 0 ≤ sin2 θ ≤ 1 for all θ ∈[0, 2 π )
⇒
2 ≤ 2 + 2 sin2 θ ≤ 4 for all θ ∈[0, 2 π )
Hence, the range of| A| is [2, 4].
30. ~[ p ∨ (~ p ∨ q )] = ~ p∧ ~(~ p ∨ q )
≡ ~ p ∧ [~(~ p)∧ ~ q ]
≡ ~ p ∧ ( p ∧ ~q )
Practice Set 4
Instructions
For instructions refer to Practice Set 1.
1. If y = |cos x | + |sin x |, then
(a) 0
(b) 1
(c)
dy
2π
at x =
is
dx
3
1− 3
2
(d)
3 −1
2

k
 ∑ x  − 100
k = 1 
is
The value of lim 
x→1
x −1
 100
2.
(a) − 5050
(c) 5050
(b) 0
(d) None of these
 sin x , x ≠ nπ
, where n ∈I and
x = nπ
 2,
3. Let f ( x ) = 
 x 2 + 1, x ≠ 2
,
g (x ) = 
x =2
 3,
then lim g [ f ( x )] is
(a) 1
(c) 3
x→ 0
4. The intercept made by the tangent to the
curve y = ∫
0
|t | dt , which is parallel to the
line y = 2 x , on X-axis is equal to
(b) − 2
(d) None of these
(a) 1
(c) 2
5.
1
1
If P = x 3 − 3 and Q = x − , x ∈(0, x ), then
x
x
P
minimum value of 2 is
Q
(a) 2 3
(c) Does not exist
(b) − 2 3
(d) None of these
6. Lines OA and OB are drawn from O with
direction
cosines
proportional
to
< 1, − 2 , − 1 > and < 3, − 2 , 3 >, respectively.
The direction ratios of the normal to the
plane AOB are
(a) < 4, 3 , 2 >
(c) < − 4, 3 , − 2 >
r = a + b and
non-collinear)
(b) < 4, − 3 , − 2 >
(d) < 4, 3 , − 2 >
r =a +s c
is
b, c
are
(b) b ⋅ c
(a) 0
b⋅c
(c)
|a |
(d) None of these
8. Locus of the point which divides double
ordinate of the ellipse
x2
a
2
y2
+
b2
= 1 in the ratio
1 : 2 internally, is
(a)
(c)
(b) 0
(d) Does not exist
x
7. The shortest distance between the lines
x2
2
−
a
9x 2
a2
9y 2
=
2
+
b
9y 2
b2
x2
9y 2
1
9
(b)
=1
(d) None of these
a
2
+
b2
=1
9. From any point on the hyperbola
x2
a2
−
y2
b2
= 1, tangents are drawn to the
x2
(a)
ab
2
2
−
y2
= 2 . The area cut off by
a
b2
the chord of contact on the asymptotes is
equal to
hyperbola
(b) ab
(c) 2ab
(d) 4ab
10. If A is a skew-symmetric matrix, then trace
of A is
(a) 1
(c) 0
(b) –1
(d) None of these
11. The arbitrary constant on which the value of
the determinant
1
α
α2
cos (p − d )a cos pa cos (p − d )a
sin (p − d )a
sin pa
sin (p − d )a
does not depend, is
(a) α
(b) p
(c) d
(d) a
406
JEE Main Chapterwise Mathematics
12. The total number of permutations of n
different things taken not more than r at a
time, where each thing may be repeated any
number of time, is
nr − 1
(a)
n −1
(c)
n(nr − 1 )
n −1
n(nn − 1 )
(b)
n −1
(d)
n(nr − 1 )
r −1
1 3 7 15
+ + +
+ ... is equal to
2 4 8 16
(b) 1 − 2 −n
(d) 2n − 1
14. The base of a cliff is circular. From the
extremities of a diameter of the base, angles
of elevation of the top of the cliff are 30° and
60°. If the height of the cliff is 500 m, then the
diameter of the base of the cliff is
2000
m
3
2000
(c)
m
2
(a)
(b)
1000
m
3
(d) 1000 3 m
15. If sin (θ + α ) = a and sin (θ + β ) =b, then
cos 2 (α − β ) − 4ab cos (α − β ) is equal to
(a) 1 − a 2 − b 2
(b) 1 − 2a 2 − 2 b 2
(c) 2 + a 2 − b 2
(d) 2 − a 2 − b 2
maximum
value
of
π
π

 π

sin  x +  + cos  x +  in the interval 0, 

 2

6
6
is attained at
π
12
π
(c) x =
3
17. If
sin 3 x sin 3x =
π
6
π
(d) x =
2
(b) x =
where
C 0 ,C 1 ,C 2 ,... ,C n are constants and C n ≠ 0,
then the value of n is
(b) 0
(d) 15
=3
x+
1
2
− 2 2x − 1 is
4
3
2
(c)
1
3
2
5
(d)
3
(b)
(a)
log 1/3 ( x 2 + x + 1) + 1 < 0 is
(a) ( − ∞ , − 2 ) ∪ (1, ∞ )
(c) (– 2, 1)
(b) [−1, 2 ]
(d) ( − ∞ , ∞ )
Directions (Q. Nos. 20-22) Each of these
questions contains two statements :
Statement I (Assertion) and Statement II
(Reason). Each of these questions also has
four alternative choices, only one of which
is the correct answer. You have to select one
of the codes (a), (b), (c), (d) given below.
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation of
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation of
Statement I
(d) Statement I is true, Statement II is false
2
where,
4( 2 − 1)
3
for 0 ≤ x < 1
∫ 0 f (x )dx =
 x2,
f (x ) = 
 x , for 1 ≤ x < 2
Statement II
f ( x ) is continuous in [0, 2].
divided by 7 is 4.
m =0
(a) 1
(c) 6
1
2
21. Statement I Remainder when 34562222 is
n
∑C m cos mx,
x−
20. Statement I
16. The
(a) x =
4x − 3
19. The solution set of the inequation
13. The sum of the first n terms of the series
(a) 2n − n + 1
(c) n + 2 − n − 1
18. The value of x in the given equation
Statement II Remainder when 52222 is
divided by 7 is 4.
22. Statement I If f ( x ) = max{| 6 − x 2 |,| x |} the
minimum value of f ( x ), in the interval
[ − 3, 3] is 2.
Statement II The minimum value of f ( x )
attains only at x = 2.
407
Practice Set 4
23. If zr = cos
rα
n2
+ i sin
rα
n2
, where r = 1, 2 , 3,... , n,
and y = x is
then lim z1 z2 ... zn is equal to
(a) cos α + i sin α
α
α
(b) cos   − i sin  
 2
 2
(c) e iα/2
(d)
3
e iα
corresponding weights n C 0 , nC 1 , nC 2 ,... , nC n
respectively is
2n
(b)
(n + 1 )
n
(d)
2
obtuse
angle
between
the
3x − 4y + 7 = 0 and −12 x − 5y + 2 = 0.
(a)
(b)
(c)
(d)
lines
21x + 77y − 101 = 0
99x − 27y + 81 = 0
21x − 77y + 101 = 0
None of the above
29. The equation of curve passing through the
25. Given that X is discrete random variable
which takes the values 0, 1, 2 and
144
1
, P ( X = 1) =
, then the value
P ( X = 0) =
169
169
of P ( X = 2 ) is
24
169
2
(c)
169
5
sq units
4
12
(d)
sq units
5
(b)
28. Find the equation of the bisector of the
24. The mean of the values 0, 1, 2, . . . . n with the
145
169
143
(d)
169
(a)
5
sq units
3
5
sq unit
(c)
12
(a)
n→∞
2n + 1
(a)
n(n + 1 )
n +1
(c)
2
27. The area enclosed between the curves y = x 3
 π
point 1,  and having slope of tangent at
 4
y
y
any point ( x , y ) as − cos 2   , is
 x
x
y
1 + tan  
x
(a) x = e
(b)
(c) x = e
26. Negation of “Paris is in France and London
is in England” is
(a) Paris is in England and London is in France
(b) Paris is not in France or London is not in
England
(c) Paris is in England or London is in France
(d) None of the above
(b) x = e
x
1 + tan  
 y
(d) x = e
y
1 − tan  
x
x
1 − tan  
 y
then
P (n ): 2 + 4 + 6 + ... + ( 2 n ), n ∈ N ,
implies
P (k ) = k (k + 1) + 2
P (k + 1) = (k + 1)(k + 2 ) + 2 is true for all
k ∈ N . So, statementP (n ) = n(n + 1) + 2 is true
for
30. If
(a) n ≥1
(c) n ≥ 3
(b) n ≥ 2
(d) None of these
Answers
1. (d)
11. (b)
21. (b)
2. (c)
12. (c)
22. (d)
3. (a)
13. (c)
23. (c)
4. (a)
14. (a)
24. (d)
5. (c)
15. (b)
25. (a)
6. (d)
16. (a)
26. (b)
7. (a)
17. (c)
27. (c)
8. (b)
18. (b)
28. (a)
9. (d)
19. (c)
29. (b)
10. (c)
20. (a)
30. (d)
Solutions
1. In the neighbourhood for x = 2 π , we have
3
cos x < 0 and sin x > 0
y = − cos x + sin x
∴
⇒
dy
= sin x + cos x
dx
⇒
2π
2π
 dy 
+ cos
=
  2 π = sin
 dx  x =
3
3
3 −1
2
x
4. Given that, y = ∫ |t| dt
0
On differentiating w.r.t. x, we get
dy
=| x|
dx
Since, the tangent is parallel to the line y = 2 x.
∴
| x|= 2 ⇒ x = ± 2
When x = 2, then y =
3
2
When x = − 2, then
( x + x2 + x3 + K + x100 ) − 100
= lim
x→ 1
( x − 1)
( x − 1) + ( x2 − 1) + ( x3 − 1) + K + ( x100 − 1)
= lim
x→ 1
( x − 1)
 x − 1  x2 − 1  x3 − 1
 + 
= lim 
 + 

x → 1  x − 1
 x−1   x−1 

 x100 − 1 
 
+ K+ 
 x − 1 
 x3 − 1
 x2 − 1
 x − 1

 + lim 
= lim 
 + lim 
x → 1 x − 1
x → 1 x − 1 
x → 1 x − 1 

 x100 − 1

+ K + lim 
x → 1 x − 1 

= 1 + 2 + 3 + K + 100
100 × 101
= 5050
2
[f( x)]2 + 1, f( x) ≠ 2
g [f( x)] = 
3,
f ( x) = 2

= Σ100 =
⇒
sin2 x + 1, x ≠ nπ
g [f( x)] = 
3,
x = nπ

∴
RHL = lim g [f(0 + h)]
h→ 0
= lim (sin h + 1) = 1
2
h→ 0
and
LHL = lim g [f(0 − h)]
h→ 0
= lim (sin2 h + 1) = 1
h→ 0
∴
= lim g [f( x)] = 1
x→ 0
2
t 2 
=   =2
 2 0
 100 k 
 ∑ x  − 100


2. lim  k = 1 
x→ 1
( x − 1)
3. Q
2
∫ 0 |t| dt = ∫ 0 t dt
y=
−2
∫0
|t| dt =
0
t 2 
∫ − 2 t dt =  2  = − 2
−2
0
Then, tangent at (2, 2) is
y − 2 = 2( x − 2)
⇒
2x − y − 2 = 0
For x-intercept, put y = 0 i.e., x = 1.
Also, tangent at (– 2, – 2) is
y + 2 = 2( x + 2)
⇒
2x − y + 2 = 0
For x-intercept, put y = 0 i.e., x = − 1.
5. Given that,
1
1
and Q = x −
x
x3
2
1
1
  x − 1 + 3
 x −   x2 + 1 +

 




P
x 
x
x2  = 
∴ 2 =
2
1

Q
 x − 1
x− 



x


x
1
3
=  x −  +

x   x − 1



x
P = x3 −
Clearly, the minimum does not exist.
6. Let < a, b, c > be the direction ratios of a normal to
the plane AOB, then
a − 2b − c = 0
and 3 a − 2 b + 3c = 0
a
b
c
=
=
⇒
− 6 − 2 − 3 − 3 −2 + 6
a b
c
= =
⇒
4 3 −2
409
Practice Set 4
7. The two lines are intersecting as both of these
contain the point A(a ).
Hence, the SD between these lines is 0.
8. Let P(acos θ, b sin θ), Q(acos θ, − b sin θ).
PR : RQ = 1 : 2
Given,
Y
P (a cos q, b sin q)
(h, k) R
X′
X
O
2a
−2 b
, y2 =
x′1
y′1
x′1
y′
+ 1
+
a
b
a
b
∴ Area of triangle


1
1  4ab × 2
= ( x1 y2 − x2 y1 ) =  2
2
2
2 x′
 1 − y′1
 2
 a
b2
and x2 =
10. Since, diagonal element of a skew-symmetric matrix
are all zero.
∴
tr( A) =
Q (a cos q, –b sin q)
9. Let P( x′1 , y′1 ) be a point on the hyperbola
y2
2
x′1
2
x1 =
−
y′1
= 1. Then, chord of
a2
b2
a2 b 2
contact of tangents from P to the hyperbola
xx′1 yy′1
x2
y2
…(i)
− 2 = 2 is
− 2 =2
2
a
b
a2
b
The equation of the asymptotes are
x y
− =0
a b
x
y
and
+ =0
a b
The point of intersection of Eq. (i) with the two
asymptotes are given by
= 1, then
i =1
sin ( p − d ) a
Let a point R(h, k ) divides the line joining the points
P and Q internally in the ratio 1 : 2.
∴
h = acosθ
h
…(i)
cosθ =
⇒
a
b
and
k = sinθ
3
3k
…(ii)
sinθ =
⇒
b
On squaring and adding Eqs. (i) and (ii), we get
9k 2
h2
+ 2 =1
2
a
b
9 y2
x2
Hence, locus of R is 2 + 2 = 1.
a
b
−
n
∑ aii = 0
1
α
α2
11. Let ∆ = cos ( p − d ) a cos pa cos ( p − d ) a
Y′
x2



 = 4ab



2a
2b
, y1 =
x′1 y′1
x′1 y′1
−
−
a
b
a
b
sin pa
sin ( p − d )a
Applying C 3 → C 3 − C1, we get
α
α2 − 1
1
∆ = cos ( p − d ) a cos pa
sin ( p − d ) a sin pa
0
0
= (α 2 − 1){− cos pa sin ( p − d )a
+ sin pa cos ( p − d ) a}
= (α 2 − 1)sin {−( p − d )a + pa}
⇒ ∆ = (α 2 − 1) sin da
which is independent of p.
12. One place can be filled in n ways by n things, 1st and
2nd places can be filled in n × n = n2 ways.
1st, 2nd and 3rd places can be filled in
n × n × n = n3 ways. Simiarly, 1st, 2nd, . . . . rth
places can be filled in
n × n × . . . × n (r times)= nr
∴ Required number of ways
= n + n2 + n3 + . . . + nr =
n(nr − 1)
n−1
13. Let S n be the sum of first n terms of the series
1 3
+ +
2 4
1
∴ S n =  1 −  +  1 −

2 
7 15
+
+ ...
8 16
1 
1
 + 1 −  + . . .
4 
8
1
+  1 − n 


2
1 − 1 

n

1 
1
2 = n − 1+
=n−
= n − 1 + 2− n
2 1 − 1 
2n



2
410
JEE Main Chapterwise Mathematics
14. In ∆AEC, tan 60° = 500 ⇒ d1 = 500 m
…(i)
3
d1
500 m
60°
30°
d 1 E d2
and in ∆BEC, tan 30° =
⇒
B
500
d2
d 2 = 500 3 m
…(ii)
Now, sin3 x sin 3 x
1
= (3sin x − sin 3 x) sin 3 x
4
1
3
= ⋅ 2 sin x sin 3 x − ⋅ 2 sin2 3 x
8
8
1
3
= (cos 2 x − cos 4 x) − (1 − cos 6 x)
8
8
3
1 3
= − + cos 2 x − cos 4 x
8
8 8
1
+ cos 6 x …(i)
8
∴ Required diameter,
500
2000
m
+ 500 3 =
AB = d1 + d 2 =
3
3
RHS =
⇒
θ + α = sin−1 a, θ + β = sin−1 b
⇒
α − β = sin−1 a − sin−1 b
π
π
= − cos −1 a − + cos −1 b
2
2
= cos −1 b − cos −1 a
⇒ cos (α − β ) = ab +
1 − b 2 ⋅ 1 − a2 )
1 − b 2 ⋅ 1 − a2
⇒ cos 2 (α − β ) = a2 b 2 + (1 − b 2 ) (1 − a2 )
+ 2 ab 1 − b 2 ⋅ 1 − a2
= a 2b 2 + 1 − a 2 − b 2 + a 2b 2
+ 2 ab 1 − b 2 − a2 + a2 b 2
∴ 2 cos (α − β ) − 4ab cos (α − β )
= 2 cos 2 (α − β ) − 1 − 4ab cos (α − β )
= 4a2 b 2 + 2 − 2 a2 − 2 b 2
+ 4ab 1 − a2 − b 2 + a2 b 2 − 1
− 4a2 b 2 − 4ab 1 − a2 − b 2 + a2 b 2
= 1 − 2 a2 − 2 b 2
16. Now, cos  x + π  + sin  x + π 




6
6
π
1
sin  x +
 +

6
2
π

 = 2 cos  x −

4
π
For maximum value, x =
12
 1
cos  x +
 2

π
= 2 cos  x +
−

6
= 2
π

6  
π

12 
n
∑ C m cos mx
m=0
= C 0 + C1 cos x + C 2 cos 2 x
+ C 3 cos 3 x + . . . + C n cos nx …(ii)
15. Given that, sin (θ + α ) = a, sin (θ + β ) = b
= cos −1 (ab +
n
∑ C m cos mx
m=0
C
A
17. Given that, sin3 x sin 3 x =
On comparing both sides of Eqs. (i) and (ii), we get
n=6
18. Given equation can be rewritten as
22 x + 22 x − 1 = 3
x+
1
2
+ 3
x−
1
2
1
⇒
x−
1
2 2 x  1 +  = 3 2 (3 + 1)


2
1
x−
3
= 3 2 ⋅4
2
⇒
22 x ⋅
⇒
22 x − 3 = 3
x−
3
2
Taking log on both sides, we get
3
( 2 x − 3 ) log 2 =  x −  log 3

2
3
⇒
2 x log 2 − 3 log 2 = x log 3 − log 3
2
3
x log 4 − x log 3 = 3log 2 − log 3
⇒
2
4
x log   = log 8 − log 3 3
⇒
 3
x
⇒
4
8
log   = log
 3
3 3
⇒
 4 = 8
 
 3
3 3
⇒
x
4
=  
 3
3
x=
2
3/ 2
411
Practice Set 4
19. Given inequality is
1
log1/ 3 ( x2 + x + 1) < − 1 = log1/ 3  
 3
1
x2 + x + 1 <  
 3
⇒
⇒
−1
−1
x2 + x − 2 < 0
⇒
⇒
( x + 2 )( x − 1) < 0
x ∈ (− 2 , 1)
23. Given that, zr = cos rα2 + i sin rα2
n
20. Statement I
2
1
2
∫ 0 f( x) dx = ∫ 0 f( x) dx + ∫ 1 f( x) dx
=
1 2
∫0 x
dx +
2
∫1
In the interval [– 3, 3],
−3 ≤ x < − 2
 | x|,

2
f( x) = |6 − x |, −2 ≤ x ≤ 2
 | x|,
2 < x≤ 3

Clearly, minimum value of f( x) obtain at x = − 2, 2
and is equal to 2.
Hence, option (d) is correct.
x dx
4 2 −1
1 2 3/ 2
+ (2
− 1) =
3
3 3
Hence, Statement I is false.
Statement II It is clear from the graph that f( x) is
continuous in [0, 2].
=
n
[where, r = 1, 2 , 3, ..., n]
α
α
z1 = cos 2 + i sin 2 ,
n
n
2α
2α
z2 = cos 2 + i sin 2 ,
n
n
M M M M M M M M M
nα
nα
zn = cos 2 + i sin 2
n
n
α
α

∴ lim ( z1 z2 . . . zn ) = lim  cos 2 + i sin 2 
n→ ∞ 
n→ ∞
n
n 
2α
2α
nα
nα
×  cos 2 + i sin 2  . . .  cos 2 + i sin 2 


n 
n
n
n 
α



= lim cos  2 (1 + 2 + 3 + . . . + n)
n→ ∞ 

n
α

+ i sin  2 (1 + 2 + 3 + . . . + n)

n

αn(n + 1)
αn(n + 1)
= lim cos
+ i sin

2
n→ ∞
2n
2 n2 

Y
X′
1
O
2
X
Y′
Hence, Statement II is ttrue.
iα
= cos
21. Consider
(3456)2222 = (7 × 493 + 5)2222
= (7 k + 5)2222
24. Required mean
= 7m + 5
2222
=
0 ⋅ nC 0 + 1 ⋅ nC1 + 2 ⋅ nC 2 + . . . + n ⋅ nC n
n
C 0 + nC1 + . . . + nC n
=
n ⋅ 2n − 1 n
=
2
2n
Now, consider 52222 = 52 (53 )740
= 25 (125)740
= 25 (126 − 1)740
= 25 (7 n + 1)
= 175n + 25
Thus, remainder when175 n + 25 is divided by 7 is 4.
Hence, both statements are correct and
Statement II is a correct explanation of Statement I.
⇒
144
169
1
P( X = 1) =
169
P( X = 0) =
25. Given that,
and
∴ P( X = 2 ) = 1 − P( X = 1) − P( X = 0)
= 1−
22. Let |6 − x2 | = | x|
⇒
α
α
+ i sin = e 2
2
2
6 − x2 = ± x
x = ± 2, ± 3
=
1
144
−
169 169
169 − 145 24
=
169
169
412
JEE Main Chapterwise Mathematics
26. Let p : Paris is in France
and q : London is in England
∴ We have, p ∧ q
Its negation is ~( p ∧ q ) = ~ p∨ ~ q
Hence, Paris is not in France or London is not in
England.
27. Solving the given curves
y=
y2 = x and y = x3
x or
Y
(1, 1)
y2 = x
y1 = x 3
X
X′
Y′
We get, the points of intersection (0, 0) and (1, 1.)
∴ Required area =
1
∫0 (
x − x3 ) dx
1
 x3 / 2
x4 
=
−

4 0
 3 /2
5
=
sq unit
12
28. Given, equations of lines are 3 x − 4 y + 7 = 0 and
−12 x − 5 y + 2 = 0.
Then,
a1a2 + b1b2 = 3 × (−12 ) + (−4)(−5)
= − 36 + 20
= − 16
⇒
a1a2 + b1b2 ≤ 0
Hence, obtuse angle bisector is
3x − 4y + 7
−12 x − 5 y + 2
=−
32 + (− 4)2
(−12 )2 + (−5)2
⇒
⇒
13(3 x − 4 y + 7 ) = − 5(−12 x − 5 y + 2 )
21x + 77 y − 101 = 0
29. Given that, dy = y − cos 2 y
dx x
x
x dy − y dx

2 y
⇒
= −  cos
 dx

x
x
x dy − y dx 
y
dx
⇒ sec 2   
 =−
 x 

x
x2
y
dx

2 y
⋅d   = −
sec
⇒
 x
x
x
y
[on integrating]
⇒
tan = − log x + c
x
π
When
x = 1, y = ,
4
c = 1 = log e
y
tan   = log e − log x
∴
 x
y
⇒
log x = 1 − tan  
 x
⇒
y
1 − tan  
 x
x=e
30. Here, P(1) = 2 and from the equation,
P(k ) = k(k + 1) + 2
⇒
P(1) = 4
So, P(1) is not true.
Hence, mathematical induction is not applicable.
Practice Set 5
Instructions
For instructions refer to Practice Set 1.
1. The value of
(a + h )2 sin (a + h ) − a 2 sin a
is
lim
h→0
h
(a) 2 a sin a + a 2 cos a
(b) 2 a sin a − a cos a
(c) 2 a cos a + a 2 sin a
(d) None of these
cos 2 x
dx = − log |cot x + cot 2 x − 1 |
sin x
+ A + C , then A is equal to
(a)
1
2
(b)
1
2

log



log


2 + 1 − tan 2 x


2 − 1 − tan 2 x
2 + 1 − sec 2 x


2 − 1 − sec 2 x
 2 + 1 − sin 2 x
1
(c)
log


2 
2
 2 − 1 − sin x
(d) None of the above
e2
log e x
dx is
 x 
3. The value of ∫ e −1 
3
2
(c) 3
(a)
between the vector c and the vector 3i + 4 j ,
then the unit vector in the direction of c is
1
(11 i − 10 j + 2 k )
15
1
(b)
(11 i + 10 j + 2 k )
15
1
(c) − (11 i + 10 j − 2 k )
15
1
(d) − (11 i + 10 j + 2 k )
15
(a) −
2
2. If ∫
5. If the vector − i + j − k bisects the angle
5
2
(d) 5
(b)
4. The area of the figure bounded by the
straight lines x = 0, x = 2 and the curves
y = 2 x , y = 2 x − x 2, is
 4
 4
8
8
(a) 
+  sq units (b) 
−  sq units
 log 2 3 
 log 2 3 
 8
 3
4
4
(c) 
−  sq units (d) 
−  sq units
 log 3 3 
 log 2 3 
6. A circle is drawn to pass through the
extremities of the latusrectum of the
parabola y 2 = 8x . It is given that, this circle
also touches the directrix of the parabola.
Radius of this circle is equal to
(a) 4
(b)
21
(c) 3
(d)
26
7. The equation of the ellipse whose axes are
coincident with the coordinate axes and
which
touches
the
straight
lines
3x − 2 y − 20 = 0 and x + 6y − 20 = 0, is
x2
+
5
x2
(c)
+
40
(a)
y2
=1
8
y2
=1
10
x2
+
40
x2
(d)
+
10
(b)
y2
= 10
10
y2
=1
40
8. The locus of the middle point of the chords
of the circle x 2 + y 2 = a 2 which subtend a
right angle at the centre, is
(a) 2 ( x 2 + y 2 ) − 2 a 2 = 0
(b) x 2 + y 2 − a 2 = 0
(c) 2 ( x 2 − y 2 ) + a 2 = 0
(d) 2 ( x 2 + y 2 ) − a 2 = 0
414
JEE Main Chapterwise Mathematics
Directions (Q. Nos. 9-11) Each of these
questions contains two statements :
Statement I (Assertion) and Statement II
(Reason). Each of these questions also has
four alternative choices, only one of which
is the correct answer you have to select one
of the codes (a), (b), (c) and (d) given below.
14. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12),
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true;
Statement II is a correct explanation for
Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation for
Statement I
(d) Statement I is true, Statement II is false
15. A five-digit number is formed by writing the
( 3, 9), (3, 12), (3, 6)} be a relation on the set
A = {3, 6, 9, 12}. The relation is
(a)
(b)
(c)
(d)
reflexive and symmetric only
reflexive only
an equivalence relation
reflexive and transitive only
digits 1, 2, 3, 4, 5 in a random order without
repetition. Then, the probability that the
number is divisible by 4, is
(a)
3
5
Statement II
Since, | sin x | ≤ 1.
10. Statement I If r ⋅ a = 0, r ⋅ b = 0 and r ⋅ c = 0
for some non-zero vector r, then a , b and c
are coplanar vectors.
Statement II If a , b and c are coplanar,
then a + b + c = .
(c)
1
5
(d)
6
5
x +1
x
,
( x + 1 )x
3x ( x − 1) x ( x − 1)( x − 2 ) ( x + 1)x ( x − 1)
2x
x ( x − 1)
then f (100) is equal to
(a) 0
(b) 1
 cos θ
(c) 100
(d) –100
sin θ
17. If A = 
 and A ⋅ (adj A ) = λ I ,
− sin θ cos θ
then the value of λ is
(a) 2
18. If (1 + x )n =
(b) 1
(c) 3
(d) 0
n
∑ nC r xr , then
r=0
11. Statement I If
dy
y
=−
dx
x
d
dy
y
Statement II
( xy ) = 0 ⇒
=−
dx
dx
x
e xy + log ( xy ) + cos ( xy ) + 5 = 0, then
12. The sum of n terms of three AP’s whose first
term is 1 and common differences are 1, 2
and 3 are S1 , S 2 and S 3 ,respectively. The true
relation is
(a) S 1 + S 3 = S 2
(b) S 1 + S 3 = 2 S 2
(c) S 1 + S 2 = 2 S 3
(d) S 1 + S 2 = S 3
13. If z1 , z2 , z3 and z4 are four complex numbers
represented by the vertices of a quadrilateral
taken in order such that z1 − z4 = z2 − z3 and
 z − z1  π
arg  4
 = , then the quadrilateral is
 z2 − z1  2
(a)
(b)
(c)
(d)
18
5
1
16. If f ( x ) =
9. Statement I The number of real solutions
of the equation sin x = 2 x + 2 − x is zero.
(b)
a square
a rectangle
a rhombus
a cyclic quadrilateral
 C1   C 2  
Cn 
 is equal to
1 +  1 +  ... 1 +
 C 0   C1   Cn − 1 
(a)
(n + 1 )n − 1
(n − 1 )!
(b)
nn − 1
(n − 1 )!
(c)
(n + 1 )n
n!
(d)
(n + 1 )n + 1
n!
19. Two events A and B have probabilities
0.25 and 0.50, respectively. The probability
that both A and B occurs simultaneously is
0.14. Then, the probability that neither A nor
B occurs, is
(a) 0.39
(c) 0.11
(b) 0.25
(d) None of these
20. If f ( x ) = sgn( x 3 ), then
(a)
(b)
(c)
(d)
f is continuous but not derivable at x = 0
f ′(0+ ) = 2
f ′(0− ) = 1
f is not derivable at x = 0
415
Practice Set 5
 2 x + 3
 , then
 3 − 2x
21. If f ′ ( x ) = sin (log x ) and y = f 
dy
at x = 1 is equal to
dx
(a) 6 sin log ( 5)
(c) 12 sin log ( 5)
(b) 5 sin log (6)
(d) 5 sin log (12 )


r = 2 i − 2 j + 3 k + λ( i − j + 4 k )
and the plane r ⋅ ( i + 5 j + k ) = 5 is
10
3
(b)
(c)
10
3 3
(a)
(d)
10
9
α 
 be any point on a line,

2
2
then the range of values of t for which the
point P lies between the parallel line
x + 2 y − 1 and 2 x + 4y = 15 is
23. If P 1 +
α
3
10
,2+
−4 2
5 2
<α<
3
6
−4 2
(c)
<α<0
3
(a)
(b) 0 < α <
5 2
6
(d) None of these
population growth, we prosper’, is
(a) if we control population growth, we do not
prosper
(b) if we do not control population growth, we
prosper
(c) we control population growth and we do not
prosper
(d) we do not control population growth but we
prosper
25. lim sec
x→1
2x
⋅ log x is equal to
2
π log 2
−2
(c)
π log 2
(a)
(b)
2
−2  π 
 −  is
 7  2
2
3
4
(d)
5
(b)
3 5
1
(c)
5
28. A boat is being rowed away from a cliff of
150 m height. At the top of the cliff, the angle
of depression of boat changes from 60° to
45° in 2 min. Then, the speed of the boat
(in m/h) is
4500
3
4300
(c)
3
4500
( 3 − 1)
3
4500
(d)
( 3 + 1)
3
(a)
24. Negation of the preposition, ‘If we control
π
 4+ 7
(b) − 

 3 
1 − 7 
(d) 

 4 
 4− 7
(a) 

 3 
1 + 7 
(c) 

 4 
27. The value of tan cos −1 
22. The distance between the line
(a)
1
2
26. If 0 < x < π and cos x + sin x = , then tan x is
1
π log 2
(d) None of these
(b)
29. Let f : R → R be a function defined by
f (x ) =
x2 + 2x + 5
x2 + x + 1
is
(a) one-one and into (b) one-one and onto
(c) many-one and onto (d) many-one and into
30. The equation of a line of intersection of
planes
4x + 4y − 5z = 12
8x + 12 y − 13z = 32 can be written as
and
x −1 y + 2 z
=
=
−3
2
4
x −1 y − 2 z
(b)
=
=
2
3
4
x y +1 z − 2
(c) =
=
2
3
4
x y z−2
(d) = =
2 3
4
(a)
Answers
1. (a)
11. (b)
21. (c)
2. (a)
12. (b)
22. (c)
3. (b)
13. (b)
23. (a)
4. (d)
14. (b)
24. (c)
5. (d)
15. (c)
25. (a)
6. (a)
16. (a)
26. (b)
7. (c)
17. (b)
27. (a)
8. (d)
18. (c)
28. (b)
9. (b)
19. (a)
29. (d)
10. (c)
20. (d)
30. (b)
Solutions
2
2
1. lim (a + h) sin (a + h) − a sin a
h→ 0
 0 form
 0

h
3. Let
(a + h)2 cos (a + h) + 2 (a + h)sin (a + h)
= lim
h→ 0
1
[using L’Hospital’s rule]
= a2 cos a + 2 a sin a
sin x
=
cos x − sin x
2
∫
∫
=−
dx
=−
sec 2 θ − 1 ⋅
−1
− cosec 2 x
sec θ tan2 θ
1 + sec 2 θ
0
−1
z dx +
dθ = −
dθ
∫ cos θ + cos3 θ dθ
Y
(0, 1)
(2, 0)
X′
O
1 + cos 2 θ
x=2
sec 2 θ − 1|
cot 2 x − 1|
 2 + 1 − tan2
1
log 

2
2
 2 − 1 − tan
X
y = 2x – x2
d (sin θ)
x

+ C
x
cot 2 x − 1| + A + C
 2 + 1 − tan2 x 
1
A=
log 



2
2
 2 − 1 − tan x 
z dz
y = 2x
 2 + 1 − cos 2 θ
1
log 
+
+ C

2 
2
 2 − 1 − cos θ
but I = − log|cot x +
2
∫0
0
(1 + cos 2 θ) − 2 cos 2 θ
+
loge x
dx
x
4. Required area = ∫ ( y1 − y2 ) dx
1 − cos θ
= − log|cot x +
e2
∫1
2
sin2 θ
∫ cos θ (1 + cos2 θ) dθ
= − log|sec θ +

loge x
dx
 x 
2
2
∫
∫1
loge x
dx +
x
1
∫e
0
sec θ tan θ
= − ∫ sec θdθ + 2 ∫
∴
−1
 z2 
 z2 
= −
 +  
 2  −1  2  0
1
5
= +2=
2
2
dθ
cos θ (1 + cos 2 θ)
cos θ
= − ∫ sec θ dθ + 2 ∫
dθ
1 + cos 2 θ
=−
e2

loge x
dx +
 x 
1
I= −∫
∴
cot 2 x − 1 dx
∫
∫e
loge x
dx

 x 
−1
1
dx = dz
x
On putting cot x = sec θ and
−cosec 2 x dx = sec θ tan θ dθ, we get
I=
=
=−
⇒
2
sin2 x
∫
∫e
Put loge x = z
2. Let I = cos22 x dx
=
e2
I=
[given]
Y′
=
2
∫ 0 (2
x
− 2 x + x2 ) dx
2
 2x
x3 
=
− x2 +

3 0
 log 2
8
1
4
=
−4+ −
log 2
3 log 2
 3
4
=
−  sq units
 log 2 3 
5. Let xi + y j + zk be the unit vector along c. Since,
− i + j − k bisects the angle between c and 3i + 4 j.
Therefore,
417
Practice Set 5
λ (− i + j − k ) = ( xi + y j + zk ) +
is a tangent to the ellipse, therefore on comparing
with Eq. (i), we get
1
100
and a2 m2 + b 2 =
m= −
6
9
a2
100
2
⇒
+ b =
36
9
…(iii)
⇒
a2 + 36 b 2 = 400
3
= −λ
5
4
y+ =λ
5
z= −λ
x+
⇒
and
Now,
3i + 4 j
5
x2 + y2 + z2 = 1
[since, xi + y j + k is a unit vector]
2
2
3
4
⇒  − λ −  +  λ −  + λ2 = 1



5
5
2
⇒
λ = 0 or λ =
15
But
λ≠0
2
λ=
∴
15
11
10
and
x=− , y=−
⇒
15
15
1
∴ xi + y j + zk = − (11i + 10 j + 2 k )
15
On solving Eqs. (ii) and (iii), we get
a2 = 40 and b 2 = 10
Hence, required equation of the ellipse is
x2
y2
+
= 1.
40 10
8. Let P(h, k )be the mid-point of a chord ABof the circle
x2 + y2 = a2 . Then, the equation of AB is
z=−
2
15
hx + ky − a2 = h2 + k 2 − a2
⇒
6. Extremities of the latusrectum of the parabola are
(2, 4) and (2, – 4).
Since, any circle drawn with any focal chord at it’s
diameter touches the directrix, thus equation of
required circle is
( x − 2 )2 + ( y − 4)( y + 4) = 0
⇒
∴
O
π/2
A
 hx + ky 

x2 + y2 = a2  2
2
h + k 
Radius = (2 )2 + 12 = 4
⇒
7. Let the equation of the ellipse be
+
y2
=1
a2
b2
We know that, the general equation of the tangent to
the ellipse is
y = mx ±
a2 m2 + b 2
…(i)
Q 3 x − 2 y − 20 = 0
3
∴ y = x − 10 is a tangent to the ellipse.
2
Therefore, on comparing with Eq. (i), we get
3
m=
and a2 m2 + b 2 = 100
2
9
⇒
a2 ⋅ + b 2 = 100
4
…(ii)
⇒
9 a2 + 4 b 2 = 400
Similarly, line x + 6 y − 20 = 0
1
10
y=− x+
⇒
6
3
B
P (h, k)
The combined equation of OA and OB is
x2 + y2 − 4 x − 12 = 0
x2
[QT = S ′ ]
hx + ky = h2 + k 2
2
(h2 + k 2 )2 ( x2 + y2 ) − a2 (hx + ky)2 = 0
Since, OA and OB are perpendicular.
Coefficient of x2 + Coefficient of y2 = 0
∴
∴
(h2 + k 2 )2 − a2 h2 + (h2 + k 2 )2 − a2 k 2 = 0
⇒
2 (h2 + k 2 ) − a2 = 0
Hence, locus of (h, k ) is
2 ( x 2 + y2 ) − a 2 = 0
9. Given that,
sin x = 2 x + 2 − x
We know that, 2 + 2 − x ≥ 2
⇒
sin x ≥ 2
[impossible,Q| sin x| ≤ 1]
Hence, option (b) is correct.
x
10. We have,
r ⋅ a = 0 ⇒ r ⊥ a,
r ⋅b = 0 ⇒ r ⊥b
and
r ⋅c = 0 ⇒ r ⊥c
418
JEE Main Chapterwise Mathematics
So, the vectors a , b and c are coplanar vectors.
Also,
a + b+ c =0
∴
a ×b+ b×c + c ×a =0
⇒ a ⋅ (a × b) + a ⋅ (b × c ) + a ⋅ (c × a ) = 0
⇒
0 + [a b c ] + 0 = 0
⇒
[a b c ] = 0
Hence, a b and c are coplanar vectors.
Hence, option (c) is correct.
11. Since, e xy + log ( x y) + cos ( x y) + 5 = 0
1 d
d
( x y) +
( x y)
dx
xy dx
d
− sin ( x y) ( x y) = 0
dx


1
d
( x y) e x y +
− sin ( x y) = 0
⇒
dx
xy


1
xy
Q
e +
− sin ( x y) ≠ 0
xy
d
( x y) = 0
∴
dx
dy
x
+ y⋅1= 0
⇒
dx
dy
y
=−
∴
dx
x
Hence, option (b) is correct.
Then, e xy
⇒ Diagonals bisect each other.
 z − z1  π
Given that, arg  4
 =
 z2 − z1  2
π
Angle at z1 =
⇒
2
So, it form a rectangle.
14. Since, (3, 3), (6, 6), (9, 9), (12, 12) ∈ R
Hence, R is reflexive relation.
Now, (6, 12) ∈ R but (12, 6)∉R, so it is not a
symmetric relation.
15. The number is divisible by 4, if last two digits are
divisible by 4 i.e., 12, 24, 32 and 52.
Remaining three places can be filled by 3! ways.
∴
Required probability =
13. Given that, z1 − z4 = z2 − z3
⇒
z1 + z3
z + z4
= 2
2
2
3! × 4 1
=
5!
5
16. Let
1
f ( x) =
x+1
x
2x
x( x − 1)
3 x( x − 1) x( x − 1)( x − 2 )
( x + 1)x
( x + 1)x( x − 1)
Taking x( x + 1) common from C 2 , C 3 respectively,
we get
1
1
1
= x( x + 1)
12. Let a1, a2 , a3 and d1, d 2 , d 3 are the first term and
common difference of the three AP’s, respectively.
Given that,
a1 = a2 = a3 = 1
and
d1 = 1, d 2 = 2 , d 3 = 3
n
…(i)
∴
S1 = (n + 1)
2
n
…(ii)
S 2 = (2 n)
2
n
…(iii)
and
S 3 = (3 n − 1)
2
On adding Eqs. (i) and (iii), we get
n
S1 + S 3 = [(n + 1) + (3 n − 1)]
2
n
= 2  ( 2 n) = 2 S 2
 2

Favourable cases = 3! × 4
2x
( x − 1)
x
3 x( x − 1) ( x − 1)( x − 2 ) x( x − 1)
1
= x( x + 1)( x − 1) 2 x
3x
1
1
x−1 x
x−2
x
Applying C1 → C1 − C 3 and C 2 → C 2 − C 3
0 0 1
= x( x + 1)( x − 1) x −1 x
2 x −2 x
∴
⇒
= x( x + 1)( x − 1) (−2 x + 2 x) = 0
f ( x) = 0
f(100) = 0
17. Given that,
 cos θ sin θ 
A=

 − sin θ cos θ
419
Practice Set 5
∴
−1,

=  0,
 1,

cos θ − sin θ
adj ( A) = 

 sin θ cos θ 
Now,
 cos θ sin θ  cos θ − sin θ
A ⋅ adj ( A) = 


 − sinθ cos θ  sin θ cos θ 

cos 2 θ + sin2 θ
=
 − sin θcos θ + sin θ cos θ
− sin θ cos θ + sin θ cos θ

sin2 θ + cos 2 θ

 1 0
⇒ A ⋅ adj( A) = 
=I
 0 1
∴
Thus,
λ =1
n(n − 1) 


n 

2 !  . . .  1 + 1 
= 1 +  1 +



n 
n
1 


(1 + n) (1 + n) (1 + n)
(1 + n)
=
⋅
⋅
...
1
2
3
n
(n + 1)n
=
n!
= lim
∴
and
Now,
P( A ∪ B) = P( A) + P(B) − P( A ∩ B)
= 0.25 + 0.50 − 014
.
= 0.61
P( A ∩ B ) = P( A ∪ B)
= 1 − P( A ∪ B)
= 1 − 0.61
= 0.39
20. Here, f( x) = sgn ( x3 )
 x
, for x3 ≠ 0

= | x3 |
 0 , for x3 = 0

 x ,
for x ≠ 0

= | x|
 0 ,
for x = 0
3
is
neither
−1 − 0
→∞
−h
L f ′(0) ≠ R f ′(0)
Hence, f is not derivable at x = 0.


21. Given that, y = f  2 x + 3 
 3 − 2x
⇒
 2 x + 3 d  2 x + 3
dy
= f′ 



dx
 3 − 2 x  dx  3 − 2 x 

 2 x + 3    (3 − 2 x)(2 ) − (2 x + 3)(−2 )
= sin log 
 

(3 − 2 x)2
 3 − 2x 


=
19. Given that,
P( A) = 0.25
P(B) = 0.50
P( A ∩ B) = 014
.
which
continuous nor derivable at 0.
f(0 + h) − f(0)
Rf ′(0) = lim
h→ 0
h
(1 − 0)
= lim
→∞
h→ 0
h
f(0 − h) − f(0)
and
Lf ′(0) = lim
h→ 0
−h
h→ 0



C  
C 
18. Now,  1 + 1   1 + 2  . . .  1 + C n 

C0  
C1 
C n − 1 


∴
f( x) = sgn x3 = sgn x,
λI = I
∴
x< 0
x=0
x> 0
⇒

sin log
( 3 − 2 x)

12
2
 2 x + 3 


 3 − 2x
12
 dy 
=
sin log (5)
 
 dx  ( x = 1) (3 − 2 )2
= 12 sin log (5)
22. Line is parallel to plane as ( i − j + 4k ) ⋅ ( i + 5 j + k ) = 0
General point on the line is (λ + 2 , − λ − 2 , 4λ + 3).
For λ = 0 point on this line is (2, –2, 3) and distance
from
r ⋅ ( i + 5 j + k ) = 5 or x + 5 y + z = 5, is
|2 + 5 (−2 ) + 3 − 5|
d=
1 + 25 + 1
|− 10|
3 3
10
=
3 3
d=
⇒
23. QP  1 + α , 2 + α  lies between the parallel lines


2
2
x + 2 y = 1and 2 x + 4 y = 15, then
420
JEE Main Chapterwise Mathematics
⇒
⇒
⇒
1 + α  + 2 2 + α  − 1






2
2
<0
α 
α 

2  1 +
 + 4 2 +
 − 15


2
2
3α
4+
2 <0
6α
−5+
2

4 2
α +

3 <0


5 2
α −


6 
−4 2
5 2
<α<
3
6
24. Let p : We control population growth
q : We prosper.
We have, p ⇒q
Its negation is ~( p ⇒ q ) i.e., p ∧ ~ q
i.e., we control population growth and we do not
prosper.
25. lim sec πx log x = lim log x
x→ 1
2
π
2x
log[1 + ( x − 1)]
= lim
x→ 1
π
π
sin − x 
2 2 
x→ 1
cos
log[1 + ( x − 1)]
( x − 1)
x−1
= lim
x→ 1
π
π
sin − x 
2 2   π
π
⋅ −

 π − π  2 2x


x
2 2 
( x − 1)
= lim
x→ 1  π
π
 − x
2 2 
=
26. Since, cos x + sin x = 1 , 0 < x < π
⇒
⇒
⇒
On squaring Eq. (i), we get
1
(sin x + cos x)2 =  
2
1
⇒ sin2 x + cos 2 x + 2 sin x cos x =
4
1
1 + sin 2 x =
⇒
4
3
sin 2 x = −
⇒
4
2 tan x
3
⇒
=
−
4
1 + tan2 x
⇒
( x − 1)
 2 x −1 − 1

π 
x

 2
2
3 tan2 x + 8 tan x + 3 = 0
−4 ± 7
3
−4 − 7
tan x =
3
tan x =
⇒
⇒
[since, tan x is in II quadrant]
27. tan cos −1

− 2 − π



 7 2 
2
π
= tan  π − cos −1   − 


7
2


π

−1  2  
= tan  − cos   
7
2
2
= tan sin−1   
7

2 
= tan tan−1 

 3 5

2
=
3 5
x
π

sin − x 

2 2 
lim
= 1
x→ 1  π
π

 − x

2 2 
x→ 1
…(i)
2
1
1
1
cos x +
sin x =
2
2
2 2
π
1
cos  x −  =

4 2 2
π
3π
< x<
2
4
Hence, tan x lies in II quadrant.

log[1 + ( x − 1)]
= 1 and
Q xlim
→1
x−1

= lim
( x − 1)
2
2
=
lim
π x → 1 2 x −1 − 1 πlog 2
28. Let
In ∆APQ,
PQ = 150m
PQ
tan 60° =
AP
421
Practice Set 5
⇒
AP =
150
3
…(i)
x + x+1
Q
Since,
60°
and
150 m
45°
45°
B
P
and in ∆BPQ,
tan 45° =
PQ
AB + AP
150
= 150
3
150
AB =
( 3 − 1)
⇒
3
AB
∴ Speed of boat =
2
1 150
( 3 − 1) × 60 m/h
= ×
2
3
4500
( 3 − 1) m/h
=
3
⇒
AB +
x2 + x + 1 > 0
x + 2x + 5> 0
2
f( x) > 0, then it is into
( x2 + x + 1)(2 x + 2 ) − ( x2 + 2 x + 5)(2 x + 1)
Now, f ′( x) =
( x2 + x + 1)2
f ′( x) = 0 has real values of x, so function is many-one.
60°
A
2
29. Given that, f( x) = x 2 + 2 x + 5
30. Given equation of planes are
4 x + 4 y − 5 z = 12
and
8 x + 12 y − 13 z = 32
Let DR’s of required line be (l , m, n).
From Eqs. (i) and (ii), we get
4l + 4 m − 5 n = 0
…(i)
…(ii)
8l + 12 m − 13 n = 0
l
m
n
l m n
⇒
=
=
⇒
=
=
8 12 16
2
3 4
Now, we take intersection point with z = 0 is given by
…(iii)
4 x + 4 y = 12
and
…(iv)
8 x + 12 y = 32
On solving Eqs. (i) and (ii), we get the point (1, 2, 0).
x−1 y−2 z
Hence, the required line is
=
= ⋅
2
3
4
and