JEE ADVANCED 2017_21-05-17 (PAPER-II)_Code-0
[1]
MATHEMATICS
SECTION 1 (Maximum Marks : 21)

This section contains SEVEN questions

Each question has FOUR options (A), (B), (C) and (D), ONLY ONE of these four options is correct.

For each question, darken the bubble corresponding to the correct option in the ORS.

For each question, marks will be awarded in one of the following categories :
Full Marks
: +3
If only the bubble corresponding to the correct option is darkened.
Full Marks
:0
If none of the bubbles is darkened.
Negative Marks : –1
37.
In all other cases.
Let S = {1, 2, 3,...,9}. For k = 1, 2,....5, let Nk be the number of subsets of S, each containing five
elements out of which exactly k are odd. Then N1 + N2 + N3 + N4 + N5 =
[A] 125
[B] 210
[C] 252
[D] 126
Ans. (D)
Sol. In’s odd Nos, are five(s) and even number = 4 (four)
N1  4 C 4  5 C1  5
N2  4 C3  5 C2  4  10  40
N3  4 C2  5 C3  6  10  60
N4  4 C1  5 C4  4  5  20
N5  4 C 0  5 C 5  1

38.
N1 + N2 + N3 + N4 + N5 =5 + 40 + 60 + 20 + 1 = 126
The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y – 2z = 5
and 3x – 6y – 2z = 7, is
[A] 14x  2y  15z  3
[B] 14x  2y  15z  27
[C] 14x  2y  15z  1
[D] 14x  2y  15z  31
Ans. (D)
Equation of plane
(x  1)a  (y  1)b (z 1)c  0
Also, 2a  b  2c  0 and 3a  6b  2c  0 

....(i)
a b c
 
14 2 15
Equation of plane is 14x + 2y + 15z = 31
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[2]
39.
Let O be the origin and let PQR be an arbitrary triangle. The point S is such that
           
OP  OQ  OR  OS  OR  OP  OQ  OS  OQ  OR  OP  OS . Then the triangle PQR has S as its
[A] incentre
[B] circumcentre
[C] orthocentre
[D] centroid
Ans. (C)
   
Sol. (OQ  OR)  (OP  OS)  0
....(i)
   
(OP  OQ)  (OR  OS)  0
....(ii)
 
 
 QR  PS  0 and
 PQ  RS  0
 S is orthocentre of PQR
40.
How many 3  3 matrices M with entries from {0, 1, 2} are there, for which the sum of the diagonal entries
of MTM is 5?
[A] 162
[B] 135
[C] 126
[D] 198
Ans. (D)
a b c 


Sol. Let m  d e f 
g h i 
sum of diagonal elements of MT M is  a2  b2  c 2  d2  e2  f 2  g2  h2  i2  5
Here, any five entries are 1, rest zero
or any seven entries are zero and rest are 1 and 2.
so, 9 C5  9 C7  2  198
41.
Three randomly chosen nonnegative integers x,y and z are found to satisfy the equation x + y + z = 10.
Then the probability that z is even, is
[A]
6
11
36
55
[B]
[C]
1
2
[D]
5
11
Ans. (A)
Sol. x + y + z = 10
number of non negative solution =
12
C2  66
x + y = 10 – z, when z is even
number of solution = 11 + 9 + 7 + 5 + 3 + 1 = 36
required probability 
6
11
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42.
[3]
 1 1
If f : R  R is a twice differentiable function such that f ''(x)  0 for all x  R, and f    ,f(1)  1, then
2 2
[A]
1
 f '(1)  1
2
[B]
0  f '(1) 
1
2
[C]
f '(1)  0
[D] f '(1)  1
Ans. (D)
1 
Sol. since f(x) is continuous and differentiable in  ,1
2 
 1
f(1)  f  
2 1
1 
 f '(x) 
for at least one x   ,1
1
1
2 
2
But f ''(x)  0  f'(x) is increasing function
1 
 f '(1)  f'(x) for x   ,1
2 
 f '(1)  1
43.
If y = y(x) satisfies the differential equation 8 x

1


9  x dy   4  9  x  dx, x  0 and



y(0)  7, then y(256) 
[A] 80
[B] 9
[C] 16
Ans. (D)
Sol.
dy 
8 x

dx


9 x  4 9 x 



Let 4  9  x  t 2
dx
 t dt
8 x 9 x
 y   dt  4  9  x  C
y(0)  7  C  0
y(256) = 3
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JEE ADVANCED 2017_21-05-17 (PAPER-II)_Code-0
[4]
SECTION 2 (Maximum Marks : 28)

This section contains SEVEN questions.

Each question has FOUR options (A), (B), (C) and (D), ONE OR MORE THAN ONE of these four options(s)
is (are) correct.

For each question, darken the bubble(s) corresponding to all the correct options(s) in the ORS.

For each question, marks will be awarded in one of the following categories :
Full Marks
: +4
If only the bubble(s) corresponding to the correct option(s) is(are) darkened.
Partial Marks
: +1
For darkening a bubble corresponding to each correct option, provided NO
incorrect option is darkened
Zero Marks
: 0
Negative Marks : –2
If none of the bubbles is darkened
In all other cases

For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will
result in +4marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result
in –2 marks, as a wrong option is also darkened.
44.
If the line x =  divides the area of region R  {(x,y)  R 2 : x 3  y  x, 0  x  1} into two equal part, then
[A] 0   
1
2
[B]
24  42  1  0
[C]
 4  4 2  1  0
[D]
1
  1
2
Ans. (B, D)
1
Sol.
 (x  x
3
0
)dx 
1
4


3
So,  (x  x )dx 
0
or
1  x2 x4 
  
8 2
4 0
2  4 1


2
4 8
or 2 4  4 2  1  0,
Clearly
2  1 
1
2
,1 
1
2
(not valid)
1
   1.
2
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[5]
cos(2 x) cos(2 x) sin(2 x)
45.
If f(x)   cos x
sin x
cos x
 sin x , then
sin x
cos x
[A] f(x) attains its maximum at x = 0
[B] f(x) attains its minimum at x = 0
[C] f '(x)  0 at more than three points in (, )
[D]
f '(x)  0 at exactly three points in (, )
Ans. (A, C)
Sol.
f(x)  cos 2x(cos2 x  sin2 x)  cos 2 x(  cos 2 x  sin2 x)  sin 2 x(  sinx cosx  sinx cosx)
f(x)  cos 2 x  cos 2 2x  (sin2 2 x)
f(x)  cos 2 x  cos 4 x
f(x) is maximum at x  0, 
f '(x)  2 sin2x  4 sin 4x
 2sin2x  8 sin 2x cos 2x
f '(x)  2sin 2 x(1  4cos2 x)
f '(x)  0 at 7 points is x  ( , )
46.
If I 
98
k 1
k 1
k
 
[A] I 
k 1
dx , then
x  x  1
49
50
[B] I 
49
50
[C] I  loge 99
[D] I  loge 99
Ans. (A,C)
98 k 1
Sol. I 

k 1 k
98 k 1
I

k 1 k
47.
k 1
dx 
x  x  1
k 1
dx 
x  x  1
Let f  x  
[A]
[C]
98 k 1

k 1 k
98 k 1

k 1 k
k 1
dx 
x  k  1
2
dx 
1
  x dx  ln99
k 1 k
98
k 1
 x  1
98 k 1

k 1
98
1
1
49


k  2 K 1 100 50

1  x 1 | 1  x |
 1 
cos 
 for x  1. Then
| 1 x |
 1 x 
lim f  x   0
[B]
lim f  x  does not exist
[D]
x 1
x 1
lim f  x   0
x 1
lim f  x  does not exist
x 1
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[6]
Ans. (B,C)
Sol.
f  x 
1  x 1 | 1  x |
| 1 x |
cos
1
1 x
for x  1
lim f  x   lim
1  2x  x 2
1
cos
0
1 x
1 x
lim f  x   lim
1  x2
1
1
cos
 lim  1  x  cos
x 1
1  x x 1
1 x
x 1
x 1
x 1
x 1
which does not exist.
48.
If f :    is a differentiable function such that f '  x   2f  x  for all x   , and f(0) = 1, then
[A] f(x) is increasing in  0,  
[B]
f '  x   e2x in  0,  
[C] f  x   e2x in  0,  
[D]
f  x  is decreasing in  0,  
Ans. (A,C)
Sol.
f ' x  2 f  x  e
f ' x  2e
f x 0
f ' x  2 f x  e2x f '  x   2e2x f  x   0
.......(i)
d 2x
e f x  0  x  
dx


Let g  x   e2x f  x 
for x  0, g  x   g  0 
e 2x f  x   1
f  x   e2x x  0,  
Now, from equation (i)
e2x f '  x   2e2x f  x 
 e2x f '  x   2  x  0,  
 f '  x   2e2x  f '  x   0  x  0,  
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49.
If g  x  

sin 2x 
sin x
[7]
sin1  t  dt , then
 
[A] g'     2
 2
[B]
 
g'     2
 2

[C] g'    2
2
[D]

g'    2
2
Ans. (None is correct)
50.
Let  and  be nonzero real numbers such that 2  cos   cos    cos  cos   1 . Then which of the
following is/are true?


[A] tan    3 tan    0
2
2
[B]


3 tan    tan    0
2
2


[C] tan    3 tan    0
2
2
[D]


3 tan    tan    0
2
2
Ans. (A,C)
Sol.
2  cos   cos    cos  cos   1 . Let tan
 1  y2 1  x 2
2 

2
1  x2
 1 y

  1  x2
  
2
  1 x

 


 y, tan  x
2
2
 1  y2 

 1  y 2   1



 2 1  y2 1  x 2  1  x 2 1  y 2
   1  x 1 y   1  x 1 y 
2


tan


  3 tan
2
2
tan




 3 tan  0 and tan  3 tan  0
2
2
2
2
2
2
2
2 x 2  y 2  x 2  y 2  x 2  3y 2  x   3 y
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[8]
SECTION – 3 (Maximum Marks : 12)

This section contains TWO paragraphs

Based on each paragraph, there will be TWO questions

Each question has FOUR option (A), (B), (C) and (D). ONLY ONE of these four option is correct.

For each question, darken the bubble corresponding to all the correct option in the ORS.

For each question, marks will be awarded in one of the following categories :
Full Marks
: +3
Zero Marks : 0
If only the bubble corresponding to the correct option is darkened.
In all other cases.
Paragraph-1
51.
  
  
Let O be the origin, and OX, OY, OZ be three unit vectors in the directions of the sides QR, RP, PQ ,
respectively, of a triangle PQR.
 
OX  OY 
[A] sin P  R 
[B]
sin2R
[C]
sin  P  Q
[D] sin  Q  R 
Ans. (C)
P
Sol. In  PQR
Since P  Q  
Now,
 
OX  OY  | 1| | 1| sin  P  Q  | nˆ |
 
OX  OY  sin P  Q 
52.
Q
R
If the triangle PQR varies, then the minimum value of cos  P  Q   cos  Q  R   cos R  P  is
[A] 
3
2
[B]
3
2
[C]
5
3
[D] 
5
3
Ans. (A)
Sol. Since, P  Q  R  
 cos  P  Q   cos  Q  R   cos R  P 
 cos    R   cos    P   cos    Q
   cosP  cosQ  cosR 
P
Q
R

  1  4 sin  sin  sin 
2
2
2

Since minimum value will be at
P QR 

3
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[9]
P
Q
R
1 1 1
3


  1  4 sin  sin  sin    1  4      
2
2
2
2
2
2
2




 Minimum value will be 
3
2
Paragraph-2
Let p, q be integers and let ,  be the roots of the equation, x 2  x  1  0 , where    . For n = 0, 1,
2,..., let an  pn  qn
FACT : If a and b are rational numbers and a  b 5  0 , then a = 0 = b.
53.
If a 4  28 , then p + 2q =
[A] 12
Ans. (A)
[B] 21
[C] 14
[D] 7
2
Sol. Given x  x  1 = 0

1 5
1 5
, 
2
2
Also, x 2  x  1
  2    1, 2    1
4
a 4  p 4  q 4
 1 5 
 1 5 
 28  p 
 q

 2 
 2 




4
 28  16   p  q   p  q  4 5   p  q 30  20 5  p  q   25  p  q
 28  16  56  p  q  24 5  p  q
 56  7  p  q   3 5  p  q   7  p  q  8   3 5 p  q  0
p  q  8  p  q  0
54.
p  q  4
 p  2q  12
[B]
[C]
a12 
[A] a11  2a10
a11  a10
a11  a10
Ans. (B)
Sol.
A12  p12  q12
 p10    1  q10   1
 x 2  x  1


 p11  p10  q11  q10  A11  A10
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[D] 2a11  a10