JEE MAIN 2017
MATHEMATICS - PAPER CODE 'D'
1.
Solution : [Ans. (4)]
For no solution = 0
1 1 1
1 a 1  1( a  b )  1(1  a )  1(b  a 2 )  0
a b 1


a – b – 1 + a + b – a2 = 0

a2 – 2a + 1 = 0
a=1

For a = 1 and b = 1, given equations become x + y + z = 1, x + y + z = 1 and x + y + z = 0. Here first two
equations are same. Equations x + y + z = 1 and x + y + z = 0 represent two parallel planes and hence given
system of equations will have no solution when a = 1 and b = 1
 S = {1} which is a singleton set.
NOTE: When a = 1,
1 1 1
1 1 1
1 1 1
1  1 1 1  0 ;  2  1 1 1  0 ;  3  1 1 1  0
0 b 1
1 0 1
1 b 0
From this it is wrong to conclude that given system of equations will have infinitely many solutions for
all values of b when a = 1 and unique solution when a is not equal to 1.
This result can be used only when all the three equations are different.
2.
Solution : [Ans. (1)]
(p  q) + [(– p  q)  q]
3.
p
q
–p
pq
–pq
T
T
F
T
T
(– p  q) q
T
[(p  q) (– p  q)  q]
T
T
F
F
F
T
F
T
F
T
T
T
T
T
T
F
F
T
T
F
T
T
Solution : [Ans. (4)]
Given, 5 (sin2x – cos4x) = 2cos2x (2cos2x – 1) + 9cos2x

5 (1 – cos2x – cos4x) = 4cos4x + 7cos2x

9 cos4x + 12cos2x – 5 = 0

cos 2 x 
12  144  180 12  18 1


18
18
3
2
1 
7

 1
cos 4x = 2 cos2 2x – 1 = 2(2cos2. x – 1)2 – 1  2  2   1  1  2    1 




3
9
9
4.
Solution : [Ans. (2)]
P(A) + P(B) – 2P(A B) 
1
4
P(B) + P(C) – 2P (B C) 
1
4
1
P (A) + P (C) – 2P (A C) 
1
4
2[P(A) + P(B) + P(C) – P(A B) – P(B C) – P (A C) 

3
4
P (A B C) = P(A) + P(B) + P(C) – P(A B) – P(B C) – P (A C) + P (A B C) 
5.
Solution : [Ans. (1)]
2  1  3i   
1
1
1  3i
2
1
1  2  1  2  3k
Now,
1

2
3
1
0

0 2
7
1
 2  3k


3(2 – ) = 3k

1
3
1
3
 3i  k as  2   
i,  
i
2 2
2 2
6.
k = – z
Solution : [Ans. (4)]
Y
|k (k – 2) + 5 (2 + 3k) – k (– 4k)| = 56

3 1
7


8 16 16
( 2, 2)
C
5k2 + 13k + 10 = ± 56
2

5k + 13k – 46 = 0

k
L
2

5k + 13k + 66 = 0
13  169  920 13  169  1320 13  33


2
10
10
10
A (2, – 6), B (5, 2), C  (– 2, 2)
Equation of perpenditure from B to AC is y  2 
B (5, 2)
x=2
X
O
A (2,  6)
1
( x  5)
2
x – 2y – 1 = 0
 1
Clearly  2,  satisfies it.
2
7.
Solution : [Ans. (3)]
l = r
r
Given, r + 2r = 20

(+ 2) r = 20
Area of sector, y 
r
...(1)
1 2
1  20

r   r 2 .   2  10r  r 2

2
2  r
dy
= 10 – 2r = 0
dr


l
r=5
2
1  20

 2 = 25
ymax  .25 

2
5
8.
Solution : [Ans. (4)]
y  1 x

( y  1)2  x
Y
2
x =4y
(1,2)
(0,1)
A1
A2
y=1+ x
(2,1)
O
x+y=3
X
1
1


x2 
2
x3 
2 1 12  8  1 19
A1    1  x   dx   x  x3 / 2    1  


4
3
12
3
12
12
12




0
0
2
2


x2 
x 2 x3 
2 
1 1

A 2    (3  x)   dx   3 x 
   6  2    3  

 
4
2
12
3
2 12 




1
1

10 29 11


3 12 12
Required area A1 + A2 
9.
19 11 30 5



12 12 12 2
Solution : [Ans. (2)]
P(1, 2,3)
x 1 y  2 z  3


r
PQ :
1
4
5
Let L (r + 1, 4r – 2, 5r + 3)
L
L lies on plane, 2x + 3y – 4y + 22 = 0


2 (r + 1) + 3 (4r – 2) – 4 (5r + 3) + 22 = 0
– 6r + 6 = 0

L (2, 2, 8)

Q  (3, 6, 13)

r=1
Q
2
2
2
PQ  2  8  10  2 1  16  25  2 42
10. Solution : [Ans. (1)]
3/ 2
 6 x. x 
1 2.3 x
y  tan 1 

tan

 1  9 x3 
1  3 x3 / 2


2
 2 tan 1 (3 x 3/ 2 )

dy
2
3
9

.3.
x
x  g ( x) x
dx 1  9 x 3 2
1  9 x3

g ( x) 
9
1  9 x3
3
11. Solution : [Ans. (1)]
(2  sin x)
dy
 ( y  1)cos x  0
dx
dy
  ( y  1)cos x
dx

(2  sin x )

 y  1    2  sin x dx

log (y + 1) = – log (2 + sinx) + log C

y +1=
dy
cos x
C
2  sin x
When x = 0, y = 1
From (1), y  1 
...(1)

2=
C
2

C=4
4
2  sin x
4
1
2  sin x

y

y () 
4
1
1
2 1
3
12. Solution : [Ans. (3)]
tan  
x 1

4x 4
tan(  ) 
2x 1

4x 2
tan = tan ()
B
x
C

x
A

P
4x
1 1
1

2
2
4

 4
1 1 9 9
1 .
2 4 8
13. Solution : [Ans. (2)]
 2 3
A
 4 1 

 2 3  2 3  16 9 
A2  

 4 1   4 1   12 13 
 48 27   24 36   72 63
(3A 2  12A)  


 36 39   48 12   84 51 
4

 51 63
adj A  84 72 


14. Solution : [Ans. (2)]
9(25a 2  b 2 )  25(c 2  3ac )  15b(3a  c)

225a 2  9b2  25c 2  75ac  45ab  15bc

225a 2  9b2  25c2  45ab  15bc  75ac  0

(15a ) 2  (3b) 2  (5c )2  15a.3b  3b  5c  15a  5c  0

 
Where = 15a, = 3b, = 5c

()2 + ()2 + ()2 = 0



15a = 3b = 5c = k (say)

a
k
k
k
,b  , c 
15
3
5

a
k
5k
3k
,b 
,c 
15
15
15

a, c, b are in A.P.

b, c, a are in A.P.
15. Solution : [Ans. (2)]
15 d.r's of given lines are 1, – 2, 3 and 2, – 1, – 1

d.r's of normal to plane are 6, 7, 3

Eqn of plane is 5(x – 1) + 7( y + 1) + 3(z + 1) = 0
or, 5x + 7y + 3z + 5 = 0

...(1)
Required distance of plane (1) from point (1, 3, – 7) 
| 5  1  7  3  3(7)  5 |
52  7 2  32

10
83
16. Solution : [Ans. (2)]
n
In I n   tan xdx

In   tan n  2 x (sec 2 x  1) dx   tan n  2 x sec 2 xdx   tan n  2 xdx 

In  In  2 

I6  I 4 

1
a  ,b  0
5
tan n 1
 In2
n 1
tan n 1 x
n 1
tan 5 x
 a tan 5 x  bx5  c
5
17. Solution : [Ans. (2)]
5
Y
a
4
e
a = 4e


a  4
2
2
Eqn. of ellipse is
A
O (0,0)
 1
b = a (1 – e )  4 1    3
4
2
4
1
2
2
x=4
x= 4
x2 y2

1
22 3
 3
Eqn. of tangent at P 1,  is
 2
x.1 y.3

1
4 32

x y
 1
4 2

x  2y  4

 3
Eqn of normal at 1,  will be
2
3

2x  y   2  1    0

2
2x  y 
or,
1
0
2
4x  2 y  1  0
18. Solution : [Ans. (2)]
Here, ae = 2. Centre of hyperbola is mid point of (2, 0) and (– 2, 0)

Centre is (0, 0)
Let the eqn. of hyperbola be
x2 y 2

1
a2 b2
...(1)
Since hyperbola (1) passes through P( 2, 3)

2
3

1
a2 b2
...(2)
But b2 = a2e2 – a2 = 4 – a2
From (2),
2
3

1
2
a
4  a2

2
3

 1 [where t = a2]
t 4t

8  2t  3t  4t  t 2
6
t 2  9t  8  0

a2 = 1, 8


t = 1, 8
a = 1, 2 2

But when a = 2 2, ae = 2
e=

1
2
 1 (not possible)
a = 1, b2 = 4 – a2 = 3

Eqn. of hyperbola is
x2

2
1
y2
1
3
3x2 – y2 = 3
or
...(3)
Eqn of tangent at P( 2, 3) is
3.x 2  y 3  3
or 3 2 x  3 y  3

It passes through 2 2,3 3

19. Solution : [Ans. (3)]
y  f ( x) 
f '( x) 

x
1  x2
(1  x 2 ).1  x.2 x
1  x2

(1  x 2 ) 2
(1  x 2 )2
Sign of f ' (x) i.e. of (1 – x2) in R is
f is dec
 ve 1
f ( ) 0
inc
ve
dec
1 ve 
f ( ) 0
f is not one-one
f ( 1)  
1
1
, f (1) 
2
2
 1 1
Range    ,   codomain
 2 2
Hence f is onto function.
20. Solution : [Ans. (2)]
cot x  cos x
Lt
 Lt
 (  2 x )3
h0
x
2




cot   h  cos   h
2

2




   2  2  h 


3
 Lt
h0
 tan h  sin h
sin h(1  cos h)
 Lt
3
h

0
(  2 h)
8h3 .cos h
2 h
1 sin h 2sin 2 1
1
h2
1
1
 Lt
.
.
 1 2 
.

2
8 h
cos h 8
h
4  h 2 cos0 16
7
21. Solution : [Ans. (2)]
  
| (a  b )  c |  3

  
| a  b || c | sin 30º  3

  
| a  b || c |  6
...(1)
iˆ ˆj kˆ
 
a

b

2
1 2  2iˆ  2 ˆj  kˆ
Now,
1 1 0
 
| a  b |  2 2  ( 2)2  12  3

From (1), | c | 2
 
Now | c  a |  3
 
(c  a ) 2  9


c 2  a 2  2c .a  9


4  9  2c .a  9

c .a  2


22. Solution : [Ans. (2)]
Curve is y (x – 2) (x – 3) = x + 6
x=0

6y = 6

...(1)
y=1
Point where curve (1) intersects y-axis is (0, 1)

From (1), y 
x6
x  5x  6
2
dy ( x 2  5 x  6).1  ( x  6)(2 x  5)

dx
( x 2  5x  6) 2

At (0, 1),
dy 6  30

1
dx
62
Eqn of normal at (0, 1) is
y  1  1( x  0)
or,
x  y 1 0
...(2)
 1 1
Clearly line (2) passes through  , 
2 2
23. Solution : [Ans. (1)]
n(S)  11C2 
11  10
 55
2
E = Event that sum as well as absolute difference of the two selected numbers is a multiple of 4
= {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}
8
P (E) =
n (E)
6

n (S) 55
24. Solution : [Ans. (1)]
F rounds
X (Husband)  4L, 3M
Y (Wife)  3L, 4M
No. of ladies = 7, No. of men = 7
Husband's friends
Wife's friends
M
4
3
3
4
0
3
3
0
4
1
2
2
1
4
2
4
3
4
2
L
No. of ways
L
1
3
1
0
0
M
C0 × 3C3 × 3C3 × 4C0 = 1
C1 × 3C2 × 3C2 × 4C1 = 144
C2 × 3C1 × 3C1 × 4C2 = 324
C3 × 3C0 × 3C0 × 4C3 = 16
Required no. = 485.
25. Solution : [Ans. (4)]
Required sum = (21C1 + 21C2 + 21C3 + 21C4 + ... 21C10) – (10C1 + 10C2 + 10C3 + ... 10C10) = x – y
21
21
21
Now, C0 + C1 + ... C21 = 2



...(1)
21
21
C0 + 2(21C1 + ... + 21C10) + 21C21 = 221
21
C1 + 21C2 + ... + 21C10 = 2  2
2
x = 220 – 1
21
y = 10C1 + 10C2 + ... 10C10 = 210 – 1

x – y = 220 – 210
26. Solution : [Ans. (1)]
p = Prob. of drawing a green ball in any draw 
q = Prob. of not drawing a green ball 
15 3

25 5
2
5
Distribution is binomial

Variance = npq  10 

3 2
 [n = 10]
5 5
12
5
27. Solution : [Ans. (1)]
f ( x )  ax 2  bx  c
Given, f ( x  y )  f ( x)  f ( y )  xy
...(1)
...(2)
f (1)  a  b  c  3
Putting x = 1, y = 1 in (2), we get
9
f (2)  f (1)  f (1)  1  1  3  3  1  7
Putting x = 1, y = 2 in (2), we get
f (3)  f (1)  f (2)  1  2  3  7  2  12
Putting x = 1, y = 3, we get
f (4)  f (1)  f (3)  1  3  3  12  3  18
10
Now,
 f ( n) 
f (1)  f (2)  f (3)  f (4)  ...  f (10)  3  7  12  18  ... 10 term
n 1
Here, difference of terms form an A.P.
Let S = 3 + 7 + 12 + 18 + ... + tn
S =
3 + 7 + 12 + ...
...(3)
+ tn
...(4)
(3)–(4)  0 = 3 + 4 + 5 + 6 + ... to n term – tn
n
n
1 2 5
tn = 3 + 4 + 5 + ... to n terms  [6  (n  1).1]  (n  5)  n  n
2
2
2
2

n

Sn   
n 1

1 10 2 5 10
1
5
 n  2  n : S10  2 12  22  ...  102  2 (1  2  ...  10)
2 n 1
n 1


1  10  11  21 5  10  11 55  7 275 385  275 660



 330

  

2
6
2
2 
2
2
2
2
28. Solution : [Ans. (3)]
Y
AB=a
B (0,4)
A
(0, )
45º
AL=a
OA=
L
X
O
a = 4 – ;


cos 45º = 4 – 

 4
2


  4 2  2
4 2
 4 2( 2  1)  8  4 2
2 1

Radius  4  (8  4 2)  4 2  4  4


2 1
29. Solution : [Ans. (4)]
Given,
 ( x  n  1)( x  n)  10n

 ( x 2  (2n  1) x  n(n  1)  10n
n
n 1
10
n
nx 2  x  (2n  1)   n 2   n  10n

n 1
nx 2  xn 2 

n (n  1)(2n  1) n (n  1)

 10n
6
2
nx 2  n 2 x  10n 


nx 2  n 2 x 

x 2  nx 
n (n  1)(2n  1) n( n  1) 60 n  n(2n 2  3n  1)  3n( n  1)


6
2
6
n[60  2n 2  3n  1  3n  3] n( 2n 2  62) n( n 2  31)


6
6
3
n (n 2  31)
3
n 2  31
3
3 x 2  3nx  ( n 2  31)  0

n  10

3x2  30 x  69  0
n = 11

3x2  33x  90  0
[Product of roots should be integer and sum should also be integer]

x2  11x  30  0

( x  6)( x  5)  0

x = – 6, – 5
30. Solution : [Ans. (2)]
dx
 1  cos x  
3
4



4
1  cos x
dx = – cot x + cosec x
sin 2 x
3
dx
  cos ecx  cot x 4 
1  cos x

 
2 1 

2 1  2
4
ANSWERS
1.
5.
9.
13.
17.
21.
25.
29.
(4)
(1)
(2)
(2)
(2)
(2)
(4)
(4)
2.
6.
10.
14.
18.
22.
26.
30.
(1)
(4)
(1)
(2)
(2)
(2)
(1)
(2)
3.
7.
11.
15.
19.
23.
27.
(4)
(3)
(1)
(2)
(3)
(1)
(1)
4.
8.
12.
16.
20.
24.
28.
(2)
(4)
(3)
(2)
(2)
(1)
(3)
11