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JEE(Advanced)-2016 Paper – II Mathematics Solutions
37.
x  3 y  1 z  7 2  6 
=
=
=
=4
1
1
1
3
x =  1, y = 5, z = 3
P (  1, 5, 3)
a(x+1)+b(y  5)+c (z  3)=0
a + 2b + c = 0 ...................(i)
a  5b  3c = 0
a b c
 
1 4 7
(x+1)+4 (y  5)  7 (z  3)=0
 x +4y  7z = 0
x  4y+7z = 0
38.
[c]
Graph given
3
1


 
Area a = Area of PQRS    x  3   Trapezium QABK   x  3 
4
3

 

=
3
2
[C]
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39.
Common difference d = 2 = logb2 – logb1 = log
b2
b1
 b2 = 2b1
 b1 b2 b3 …. GP
T = b1 + 2b1 + 4b1 … + 250b1 = b1 (251 -1)
S=
51
51
51
 a1  a51   b1  b51   b1 1  250 
2
2
2
S–T>0
b101 = 2100b1
a10 1 = a1 + 100d = 2a1 – a1 + 2(50d) = 2a51 – a1
= 2b51 – b1  b101 > a101 
[B]
sin 30
k 1 sin 
 45   k  1 30 sin  45  30k 
13
2
40.
13
2
k 1
sin  45  30k    45  (k  1)30
sin  45   k  1 30  sin  45  30k 
Use sin (A – B) = sinA CosB – CosA sinB
13
 2  cot(45  (k  1)30  cot(45  30k ) 
k 1
 2  cot 45  cot(45  30)   2
41.
 1 0 0
P 2   8 1 0  and P3 is


 48 8 1 

[C]

3 1
1
0 0
 1 0 0

12 1 0   P n   4n
1 0




96 12 1 
8( n 2  n) 4n 1 
0 0
 1
n  50  P50   200
1 0


 20400 200 1 
P50 – Q = I  200 – q21 = 0  q21 = 200
q32 = 200, q31 = 400×51 
q31  q32
 103
q21

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[B]
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42.
Replace x with –x and I + I

 2I 

43.
Lt
x 2
2



2
x 2 cos x d x  2  x 2 cos x 
0
2
2
4
2
f ( x) g ( x)
 1 Apply L’ Hospital Rule
f '( x) g '( x)
 f'' (2) = f(2)  [D]
Range of f(x)  (0, α)
f ' (x) = f ' (2) > 0

Gives minimum at x = 2
44.
[A]
C [2, 8], r = 2,
Normal equation y = mx – 2m – m2 passes through center [2, 8]
 8 = 2m -2m – m2  m = -2
Normal at P [am2, - 2am] = [4, 4]
Gives y = -2x + 12
Slope of Tangent =
45.
1
2
[A, C, D]
if a = 0, b = 1 f(x) = x sin(x3 + x) is differentiable
if a = 0, b = 1
f(x) = cos (x3 - x), differentiable at 1, 0
 [C, D] wrong
46.
f(x) = [x2] – 3, it is discontinues at x = 1, √2, √3, 2
g(x)
= 15x – 21
x<0
= 9x – 21,
0≤x≤1
= 6x – 14
1 ≤ x < √2
= 3x – 7
√2 ≤ x < √3
=0
√3 ≤ x < 2
=3
x=2
Thus at x = 0, 1, √2, √3
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A
B
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g(x) is not differentiable
 [B C]
47.
f (x)


1
1
1





n
n
 n  n 1  x / n   x / n   x / n   .... x   
2 
3 
n 

 
2
2

2n 
x  1 x  
1
n  n 2  1  2    2  .... 1  2 




 n 4 n   n 
1
2
 f    f   and f '(2)  0
3
3
x/n
 [B, C]
48.
ax + 2y = 
3x – 2y = µ

a 2
 2 a  6  0
3 2
 a  3
1 
 2
 2     
 2
2 
3 
 3     
3 
 [B, C, D]
49.
Perpendicular of u, v is u  v
w is perpendicular to u  v
w. (u ×v) = 1  infinitely many solutions.
[B, C]
50.
1
a  ibt

a  ibt a  ibt
a  ibt
x  iy  2
a  b 2t 2
a
bt
x 2
y 2
2 2
a b t
a  b 2t 2
z
2
x
1 

 1 
2
 x  y  x
  y 

a
2a 

 2a 
2
[A, C, D]
2
2
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1 1 1 1 1 1 5
     
2 2 2 6 6 2 12
51.
P(x > y) implies =
52.
P(x = y) =    2 
53.
a = 3, e 
1 1
 2 3
1 1 13
 
6 6 36
[B]
[C]
1
F1 = (-1, 0) F2 (1, 0)  parabola y2 = 4x
3
3
2


Point of intersection  ,  6 
 9 
,0  
 10 
 Or the center 
54.
Equation tangent
[A]
3x y 6

1
18
8
For x – axis, y = 0
 R [6, 0]
Normal at M
At y = 0
Area of le
3
3  3
x y2


2
2  2
2
7 
Q  ,0 
2 
1
7
5 6
6   6 
2
2
4
Quadrilateral area
= 2 F1 F2 M
1
 2    2
2
 6  2
6
5
6 5
4
  5 :8
Ratio =
2 6 8
[c]
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