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(MATHEMATICS)
[1]
PART-III : MATHEMATICS
SECTION – I
(Single Correct Answer Type)
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of
which ONLY ONE option is correct.
41.
In a triangle the sum of two sides is x and the product of the same two sides is y. If x2 – c2 = y, where c is the third
side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is
3y
2x(x  c)
(A)
(B)
3y
2c(x  c)
(C)
3y
4x(x  c)
(D)
3y
4c(x  c)
Sol. (B)
(a  b)2  c 2  ab
(a  b  c)(a  b  c)  ab
s(s  c) 
ab
4
s(s  c) 1

ab
4
sin
C 1
  C  120
2 2

1
3
ab sinC 
ab
2
4
s
abc x c
3y

 r
2
2
2(x  c)
2R 
c
c
2c
c


R
sinc
3
3
3
2
r
3y  3
3y


R 2(x  c)  c 2c(x  c)
42.
The common tangents to the circle x2 + y2 = 2 and the parabola y2 = 8x touch the circle at the points P, Q and the
parabola at the points R, S. Then the area of the quadrilateral PQRS is
(A) 3
(B) 6
(C) 9
(D) 15
Sol. (D)
y  mx  2 (1  m2 )
...(i)
2
m
...(ii)
y  mx 
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(MATHEMATICS)
[ 2 ]
2
m
2 1  m2 
2(1  m2 ) 
JEE ADVANCED 2014_25-05-14 (PAPER-II)_CODE-4
4
2
(2,4)
R
 m2 (1  m2 )  2  0
(-1,1)
P
m
m4  m2  2  0
m2  1, 2
Q
(-1,-1)
m  1
S
(2, -4)
y  x2 

Equation of tangents are
y   x  2
Area of quadrilateral 
43.
(4  1)
 3  2  15
2
The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has
(A) only purely imaginary roots
(B) all real roots
(C) two real and two purely imaginary roots
(D) neither real nor purely imaginary roots
Sol. (D)
Let p(x) be ax2 + b
a, b > 0
p(p(x)) = 0
 a(ax 2  b)2  b  0  (ax 2  b)2  
ax2  b  
b
i
a
ax2   b 
b
i
a
b
a
 p(p(x)) = 0 has neither purely imaginary nor real roots.
44.
Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each
envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover
the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is
(A) 264
(B) 265
(C) 53
(D) 67
Sol. (C)
Total derrangement of six things = 265
Where placement of card numbered 1 is equally likely in 2, 3, 4, 5 & 6.
Therefore required number of ways 
45.
265
 53
5
Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least
one more than the number of girls ahead of her, is
(A)
1
2
(B)
1
3
(C)
2
3
(D)
3
4
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(MATHEMATICS)
[3]
Sol. (A)
n(S)  5
Favourable ways n(A)  (2  3)  2  3  60
60
1

120 2
P(A) 
46.
Let f : [0,2]  R be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f(0) = 1. Let
x2
F(x) 
 f(
t )dt for x  [0,2] . If F'(x)  f '(x) for all x  (0,2), then F(2) equals
0
(A) e2  1
(B)
e4  1
(C) e – 1
(D) e4
Sol. (B)
x2
F(x) 
 f(
t )dt
0
F'(x)  f(x)2x  f '(x)  f(x)2x
For y  f(x),
dy
 2xy
dx
dy
 2x dx
y
ln(y)  x 2  logk
y  ke x
2
f(0)  1  k  1  f(x)  e x
2
2
2
2
F'(x)  e x 2x  F(x)   2x e x dx  e x  c
Now, F(0) = 0  c = – 1
2
 F(x)  ex  1  F(2)  e 4  1
47.
The function y = f(x) is the solution of the differential equation
dy
xy
x 4  2x


in (– 1, 1) satisfying f(0) = 0.
dx x2  1
1  x2
3
2

Then

(A)
f(x)dx is
3
2

3

3
2
(B)

3

3
4
(C)

3

6
4
(D)
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
3

6
2
(MATHEMATICS)
[ 4 ]
JEE ADVANCED 2014_25-05-14 (PAPER-II)_CODE-4
Sol. (B)
P
x
x2  1
;Q
x 4  2x
1 x2
Integrating factor  e  Pdx  1  x2
y  1  x 2   (x 4  2x) dx  c
f(x)  y 
x2
1  x2
3
2
Now,
f(x)dx  2


 2
0

0
3
2

3
48.
3
2

x5
5 1  x2
(as f(0)  0)
x2 dx
1  x2

3
2

3
sin 
 cos  d   (1  cos 2)d  
3 4
cos 
0
For x (0, ) , the equation sin x + 2 sin 2x – sin 3x = 3 has
(A) infinitely many solutions
(B) three solutions
(C) one solution
(D) no solution
Sol. (D)
sin x  4 sin x cos x  3 sin x  4 sin3 x  3
4 sin3 x  4 sin x cos x  2 sin x  3
2sin x [2  2cos2 x  2cos x  1]  3
2sin x[1  2cos2 x  2cos x]  3
2
3 
1 
4 sin x    cos x     3
2 
 4 

1

3 sin x  3  4 sin x  cos x  
2




LHS is  ve
2
RHS is  ve
Therefore no solution.
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
2
49.
The following integral
17
 (2cosec x)
(MATHEMATICS)
dx is equal to

4
log(1 2)
2(eu  e  u )16 du
(B)
0
log(1 2)
(eu  e u )17 du
(D)
0
(A)
0
(C)
0
log(1 2)
(eu  e u )17 du
log(1 2)
2(eu  e u )16 du
Sol. (A)

2
I   (2cosec x)17 dx

4
eu  cosecx  cot x ; e u  cosec x  cot x
For x 

 eu  2  1 ; u  ln( 2  1)
4

x
; u0
2
eu du   cosecx(cosecx  cot x)dx
dx 
1
 2du
du 
u
cosecx
e  eu
0
Now, I 

ln( 2 1)
50.
u
(e  e
u 17
)

ln( 2 1)
( 2du)
u
(e  e
u
2
)

(eu  eu )16 du
0
Coefficient of x11 in the expansion of (1  x 2 )4 (1  x3 )7 (1  x 4 )12 is
(A) 1051
(B) 1106
(C) 1113
(D) 1120
Sol. (C)
Coefficient of x11 in (1  x2 )4 (1  x3 )7 (1  x 4 )12 will be
 4 C0  7C1  12C2    4 C1  7C3  12C0    4 C2  7C1  12C1    4 C4  7 C1  12C0 
= 1113.
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[5]
(MATHEMATICS)
[ 6 ]
JEE ADVANCED 2014_25-05-14 (PAPER-II)_CODE-4
SECTION – II
Comprehension Type (Only One Option Correct)
This section contains 3 paragraphs, each describing theory, experiment, data etc. Six questions relate to the three
paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given
options (A), (B), (C) and (D).
Paragraph For Questions 51 and 52
Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and
box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be
the number on the card drawn from the ith box, i = 1, 2, 3.
51.
The probability that x1 + x2 + x3 is odd, is
(A)
29
105
(B)
53
105
(C)
57
105
(D)
1
2
Sol. (B)
Total cases = 3C1 × 5C1 × 7C1 = 105
For x1 + x2 + x3 to be odd following cases are possible.
Case -1 Every card bears odd number, the total ways of doing so will be 2C1 × 3C1 × 4C1 = 24 ways.
Case - 2 Two cards bear even number and one bears odd number. Hence the ways will be,
Box-1
Box-2
Box-3
Even
Even
Odd

1 × 2C1 × 4C1 = 8 ways
(ii) Even
Odd
Even

1 × 3C1 × 3C1 = 9 ways
(iii) Odd
Even
Even

2C
1
(i)
× 2C1 × 3C1 = 12 ways
Hence, favourable ways = 24 + 8 + 9 + 12 = 53 ways.
Therefore, required probability =
52.
53
.
105
The probability that x1, x2, x3 are in an arithmetic progression, is
(A)
9
105
(B)
10
105
(C)
11
105
(D)
7
105
Sol. (C)
for x1 + x3 = 2x2, we require x1 + x3 to be even for which x1 and x3 both should be odd or both should be even.
Now, x1  {1, 2, 3} and x3  {1, 2, 3, 4, 5, 6, 7}
x1 and x3 both even can be selected in 1C1 × 3C1 ways whereas both odd can be selected in 2C1 × 4C1 ways
hence, favourable cases = 3 + 8 = 11.
and required probability =
11
105
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JEE ADVANCED 2014_25-05-14 (PAPER-II)_CODE-4
(MATHEMATICS)
[7]
Paragraph For Questions 53 and 54
Let a, r, s, t be nonzero real numbers. Let P(at2, 2at); Q, R(ar2, 2ar) and S(as2, 2as) be distinct points on the parabola
y2 = 4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (2a, 0).
53.
The value of r is
(A) 
1
t
(B)
t2  1
t
(C)
1
t
(D)
t2  1
t
Sol. (D)
Let the co-ordinate of Q is (at12, 2at1)
T(, )
and PQ is focal chord.

S(as2, 2as)
we know t1t = –1.
P (at2, 2at)
According to question QR parallel to PK.
2ar  2at
1
2
2
ar  at
1

rt
2at


2
at  2a
(r
1
 t2 )
1
2

t
2
...(i)
F (0, 0) K
(2a, 0)
t 2
R
(ar2, 2ar)
1
t
1
t



2
2
1
rt
t 2
t 2
1
r
t
 rt  1  t 2  2  r 
Q(at12, 2at1)
t2  1
t
So correct Option is (D)
54.
If st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is
(A)
(t 2  1)2
(B)
2t 3
a(t 2  1)2
2t 3
(C)
a(t 2  1)2
t3
(D)
Sol. (B)
Given st = 1
 Equation of tangent at P is
yt = x + at2
and equation of normal at s is
xs + y = 2as + as3
Put s in equation (ii)
... (i)
... (ii)
x
2a a
y 

t
t
t3
xt2 + yt3 = 2at2 + a
Put x in equation (iii) then
t2(yt – at2) + yt3 = 2at2 + a
yt3 – at4 + yt3 = 2at2 + a
2yt3 = at4 + 2at2 + a
y
y
... (iii)
a(t 4  2t 2  1)
2t3
a(t 2  1)2
2t 3
.
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a(t 2  2)2
t3
(MATHEMATICS)
[ 8 ]
JEE ADVANCED 2014_25-05-14 (PAPER-II)_CODE-4
Paragraph For Questions 55 and 56
1h
a
a 1
 t (1  t) dt exists. Let this limit be g(a). In addition, it is given that the
Given that for each a  (0, 1), lim
h 0 
h
function g(a) is differentiable on (0, 1).
55.
 1
The value of g   is
2
(A) 
(B) 2

2
(C)
(D)

4
Sol. (A)
1h
 g(a)  Lt
t
h 0 
a
(1  t)a 1dt
h
 1
 g    Lt
 2  h0 
1h

1
1
t 2 (1  t) 2 dt
 Lt
h0 
h
1

t 
1 
2
 Lt sin 

h 0 
 1 
 2 
56.
1h
1h
dt

tt
h
 Lt
2
h0

h
dt
2
 1  1
  t  
2  2
2
1h

  
 
=
2  2 
h
 1
The value of g   is
 2
(A)

2
(B) 
(C)


2
(D) 0
Sol. (D)
1h
g(a)  Lt
h 0 
t
a
(1  t)a 1dt
h
1h
g(a)  Lt
h 0 
1h
 (t
 1
g    Lt
 2  h0
 1
g    Lt
 2  h0
a
a 1
(1  t)
ln(1  t)  t
a
a 1
ln t(1  t)
) dt
h
1h

ln(1  t)  ln(t)
t 1 t
h
1h

ln t  ln(1  t)
1 t
h
t
g(a)  Lt
h0 
 (t
a
(1  t)a 1[ln(1  t)  ln(t)] dt
h
dt ... (i)
dt
... (ii)
b
 f(x)dx 

a

b

f(a  b  x)dx 

a


(i) + (ii),
 1
2g    Lt
 2  h0 
1h
 0 dt = 0
h
Mentors Eduserv: Plot No.-136/137, Parus Lok Complex, Boring Road Crossing,
Patna-1, Ph. No. : 0612-3223701/02
JEE ADVANCED 2014_25-05-14 (PAPER-II)_CODE-4
(MATHEMATICS)
[9]
SECTION – III
Matching List Type (Only One Option Correct )
This section contains four questions, each having two matching list. Choices for the correct combination of elements
from List I and List II are given as options (A), (B), (C) and (D), out of which one is correct.
57.
P.
List I
The number of polynomials f(x) with non-negative integer coefficients
1.
List II
8
2.
2
3.
4
4.
0
of degree  2, satisfying f(0) = 0 and  1 f(x)dx  1, is
0
Q. The number of points in the interval [ 13, 13 ] at which
f(x) = sin(x2) + cos(x2) attains its maximum value, is
R.
S.
(A)
(B)
(C)
(D)
3x 2
2
 2









(1  e x )
1
2
1

2

1
2
0
dx equals

 1 x  
cos2x log 
dx

 1 x  

 equals
 1 x  
cos 2xlog 
 dx
 1 x  

P
3
2
3
2
Q
2
3
2
3
R
4
4
1
1
S
1
1
4
4
Sol. (D)
(P) Let f(x) = ax2 + bx; a, b  W
1
 (ax
2
 bx) dx  1 
a b
  1  2a  3b  6
3 2
 (a,b)  (0,2); (3,0)
0
Hence (P)  (2)
(Q) sinx2 + cosx2 =


2 cos  x2  
4

whose maximum value is attained at x2 = 2n +
 x   2n 


4
x

; nI
4

9
,
4
4
Hence (Q)  (3)
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(MATHEMATICS)
[ 10 ]
2
(R) I 
3x 2
 1 e
x
2
dx 
2
3x 2
 1 e
x

0
3x 2
1  e x
JEE ADVANCED 2014_25-05-14 (PAPER-II)_CODE-4
2
dx  3x 2 dx  8.

0
Hence (R)  (1)
a
(S) Nr. is zero, because  f(x)dx  0 ; f(x) is odd
a
Also Dr. is a positive value.
Hence (S)  (4)
58.
List I
P.
List II
Let y(x) = cos(3cos–1x), x  [–1, 1], x  
3
. Then
2
1.
1
2.
2
1  2
d2 y(x)
dy(x) 
x
(x  1)

2
y(x) 
dx  equals
dx
Q. Let A1, A2, ..., An (n > 2) be the vertices of a regular polygon of n sides

with its centre at the origin. Let ak be the position vector of
the point Ak, k = 1, 2, ..., n. If
n  1  
a ) 
 k  1(a
k
k 1
n  1  
a ) ,
 k  1(a
k
k 1
then the minimum value of n is
x2 y2

 1 is
6
3
perpendicular to the line x + y = 8, then the value of h is
R.
If the normal from the point P (h, 1) on the ellipse
3.
8
S.
Number of positive solutions satisfying the equation
4.
9
 1 
1  1 
1  2 
tan1 
  tan 
  tan  2  is
 2x  1 
 4x  1 
x 
(A)
(B)
(C)
(D)
P
4
2
4
2
Q
3
4
3
4
R
2
3
1
1
S
1
1
2
3
Sol. (A)
(P) y(x) = cos(3cos–1x) = 4x3 – 3x
dy
 12x 2  3
dx
d2 y
dx2
 24x
1 2
1
[24x3  24x  12x3  3x]  9
Now, y [(x  1)y  xy ]  3
4x  3x
Hence (P)  (4)
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(MATHEMATICS)
(Q) Let OA1 = OA2 = ... = OAn = r
n 1
 
2
n
 
2
n
 (ak  ak 1)  (n  1)r 2 sin
k 1
n 1
 (ak  ak 1)  (n  1)r 2 cos
k 1
Hence sin

2
2
 cos
n
n
2
 
2 r 1
 r  ; r I 
 
n
2 4
n 2 4
 n
8
8.
2r  1
( nI)
Hence (Q)  (3)
(R) Let equation of normal be
 tan  
1
2
6x
3y

 3, it passes through (h, 1) and having slope 1.
cos  sin 
 h2
Hence (2, 1) is required point.
Hence (R)  (2)
1 

 1 1
1 2 
 tan1
(S) tan  tan

  tan  tan
2x

1
4x

1



x2 
1
1

2x

1
4x
1  2 

1
x2
1
(2x  1)(4x  1)
 3x3 – 7x2 – 6x = 0 
6x  2
2
8x  6x
x = 0, 3, 

2
x
2

3x3 + x2 = 8x2 + 6x.
2
3
But x = 3 is positive solution which satisfy the given equation.
Hence (S)  (1)
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[ 11 ]
(MATHEMATICS)
[ 12 ]
59.
JEE ADVANCED 2014_25-05-14 (PAPER-II)_CODE-4
Let f 1 :  R, f 2 : [0, )  R, f 3 : R  R and f4 : R  [0, ) be defined by
| x | if x  0,
f (x)   x
1
if x  0.
 e
f 2(x) = x2;
sin x if x  0,
f (x)  
3
if x  0.
 x
and
if x  0,
 f2 (f1(x))
f (x)  
4
 f2 (f1(x))  1 if x  0.
List I
P. f 4 is
Q. f 3 is
R.
S.
(A)
(B)
(C)
(D)
f 2 o f1 is
f 2 is
P
3
1
3
1
Q
1
3
1
3
1.
2.
3.
4.
R
4
4
2
2
List II
onto but not one-one
neither continuous nor one-one
differentiable but not one-one
continuous and one-one
S
2
2
4
4
Sol. (D)
 f2 (| x |), x  0  x 2 ,
x0


f (x)  

x
4
2x
f (e )  1, x  0 e  1, x  0
 2
which is clearly onto but not one one.
(P)  (1)
sin x if x  0,
f (x)  
3
if x  0.
 x
f3 : R  R
lim f  (0  h)  lim f  (0  h)  1
h 0 3
h0 3
hence f3 is differentiable, clearly f3 is many-one.
(Q)  (3)
 x 2 , x  0
f 2(f 1(x)) = 
e2x , x  0
which is discontinuous at x = 0.
(R)  (2)
f 2(x) = x2 ; [0, )  R
Clearly f 2 is continuous and one-one for x  [0, )
(S)  (4)
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60.
(MATHEMATICS)
 2k 
 2k 
 isin 
Let zk  cos 

 ; k  1,2, ...,9.
 10 
 10 
List I
P. For each zk there exists a zj such that
zk · z j = 1
Q. There exists a k  {1, 2, ..., 9} such that zj·z = zk
has no solution z in the set of complex numbers.
1.
List II
True
2.
False
3.
1
4.
2
| 1  z || 1  z | ... | 1  z |
1
R.
2
9
equals
10
S.
1
(A)
(B)
(C)
(D)
P
1
2
1
2

 2k 
9
cos 

k 1
 10 
equals
Q
2
1
2
1
R
4
3
3
4
S
3
4
4
3
Sol. (C)
(P) zk zj = 1
i
2k
10
i
2(k  j)
10
e
e
i
2j
 e 10  1
1
Since, k  {1, 2, 3, ..., 9}
If
k = 1, j = 9
k = 2, j = 8
k = 3, j = 7
and so on.. Hence True.
(P)  (1)
(Q) z1 z = zk  z 
z
k
z
i2
e
(k 1)
10
will have solution since k  {1, 2, 3, ... 9}
1
(Q)  (2)
(R) z  e
i2
k
10
k
z10  1
k
10
equation (i) zk  1  0
10
 (zk  1)  (zk  1)(zk  z1 )(zk  z 2 )...(zk  z9 )
where 1, z1, z2, ... z9 are the roots of z10  1.
k
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[ 13 ]
(MATHEMATICS)
[ 14 ]
(z10  1)
k
(z  1)
JEE ADVANCED 2014_25-05-14 (PAPER-II)_CODE-4
 (z  z )(z  z )...(z  z )
k
1
k
2
k
9
k
1  z  z 2  ...  z9  (z  z )(z  z )...(z  z )
k
k
k
k
1
k
2
k
9
| 1  z  z 2  ...  z 9 |  | (z  z )(z  z )...(z  z ) |
k
k
k
k
1
k
2
k
9
Put zk = 1  10 = |(1 – z1)(1 – z2) ... (1 – z9)|
| (1  z )(1  z )...(1  z ) |

1
2
10
9
1
(R)  (3)
9
 2k  
(S) 1   cos 

 10 
k 1
2
4
6
18 

1   cos
 cos
 cos
 ...  cos
= 1
10
10
10
10 

.
2 

sin  9 

2  
20  
 2

 8
 cos 

20  
 2  
 10
sin 

 20 
 9 
sin  
 10  cos 
= 1
= 1 + 1 = 2.
  
sin  
 10 
(R)  (4)
Mentors Eduserv: Plot No.-136/137, Parus Lok Complex, Boring Road Crossing,
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