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1 : (B)
Here,
E(x) = nP = 5 and Var (x) = npq = 2.5
2.5 1
5 2q
1
2p and n = 10
Now, ( 1) ( 0)P x P x
10
10
0
1( 1)
2P x C
2 : (D)
Correct option is (D)
3: (B)
As required circle touches y-axis at the origin.
Let Centre of the circle is d (a, 0) and radius is a
Equation of circle will be,
2 2 2( ) ( 0)x a y a
2 2 2 22x ax a y a
2 2 2 0x y ax …(i)
By differentiating above equation w.r.t. x, we get
101
( 1)2
P x
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2 2 2 0dy
x y adx
2 2 2dy
a x ydx
…(ii)
From (i) and (ii),
2 2 2 2 0dy
x y x y xdx
2 2 22 2 0dy
x y x xydx
2 2 2 0dy
x y xydx
4 : (B)
Here,
aij is stands for element of matrix A as ith row and jth column, and Aij stands for co-factor of element aij of
matrix A.
11 121, 1a a and 13 0a
And
2 1
21
1 0( 1) 1
2 1A
2 2
22
1 0( 1) 1
1 1A
2 3
23
1 1( 1) 1
1 2A
Therefore
11 21 12 22 13 23 1 ( 1) 1 (1) 0 ( 1) 0a A a A a A
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5 : (D)
Given, ( ) (sin cos )xf x e x x
'( ) [cos sin ] [sin cos ]x xf x e x x x x e
'( ) 2 sinxf x e x
To verify Rolle’s Theorem.
'( ) 0f c
2 sin 0ce c
sin 0c
c
6 : (C)
As given, both line passes through (0, 0), and 1 2,6 3
Equation of first line is.
0 tan ( 0)6
y x
1
3y x
3 0x y …(i)
Equation of second line is
0 tan ( 0)3
3
y x
y x
3 0x y …(ii)
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Hence, joint equation of these line is
( 3 )( 3 ) 0x y x y
2 23 3 3 0x xy xy y
2 23 3 4 0x y xy
7 : (B)
As given, 1 12 tan (cos ) tan (2 cos )x ec x
1 1 1tan (cos ) tan (cos ) tan (2 cos )x x ec x
1 1
2
2 costan tan (2 cos )
1 cos
xec x
x
2
2 cos2 cos
sin
xec x
x
2 cot 2x
cot 1x
4x
Hence,
sin cos sin cos4 4
x x
1 1
2 2
= 2
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8 : (A)
Option (A) is correct.
9 : (D)
Let 28 2
dxI
x x
29 2 1
dxI
x x
2 23 ( 1)
dxI
x
1 1sin
3
xI c
10 : (A)
As given,
3 2( ) 5 7 9f x x x x
(1.1) 8.6f
11 : (B)
As given,
1, 0 5
( ) 5
0,
xf x
otherwise
Now, probability of waiting time not more than 4 is 1
4 0.85
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12 : (B)
Let 2 2 2 2( ) cos ( ) sin
2 2
c cX a b a b
As we know,
( )( )sin
2
c s a s b
ab
and
( )cos
2
c s s c
ab
Where 2
a b cs
By substituting these value on above equation we will get
2X c
13 : (B)
Let 1 log (sec tan )y
21 1(sec tan sec )
sec tan
dya
d
1 sec [sec tan ]
[sec tan ]
dy
d
1 secdy
d
….(i)
Now,
Let 2 secy
2 sec tandy
d
…(ii)
1
2
seccot
sec tan
dy
dy
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1
24
cot 14
dy
dy
14 : (B)
We can say that both line passes through point (5, 3) and makes angle 45o and 135o with x axis
Equation of first line is ,
3 tan 45 ( 5)oy x
3 5y x
2 0y x …(i)
Similarly,
3 tan 135 ( 5)oy x
3 1( 5)y x
8 0y x ….(ii)
Joint equation of line is
( 2)( 8) 0y x y x
2 2 10 6 16 0x y x y
15 : (A)
As given, required point is on the
Curve 36 2y x
Therefore, only point (4, 11)
Satisfy the given equation,
Hence, option (A) is Correct.
16 : (A)
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1sin 0
( )
0
x for xf x x
k for x
0
1lim sinx
x kx
0 k
17 : (C)
Given, 1sinm xy e
1sin
2
1
1
m xdye m
dx x
2 2 2
21
dy m y
dx x
2
2 2 2(1 )dy
x m ydx
2A m
18 : (B)
Let 4 25
2 5
x
x
eI dx
e
10 25 6
2 5
x x
x
e eI dx
e
5(2 5) 6
2 5
x x
x
e eI dx
e
65
2 5
x
x
eI dx
e
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5 3 log(2 5)xI x e C
5A and 3B
19 : (B)
By solving we will get,
1 1
1 1
tan ( 3) sec ( 2) 4
1 5cos ( 2) cos
2ec
20 : (C)
As given, 2
log(1 2 )sin0
( )
0
x xfor x
f x x
k for x
Is continuous at x = 0
20
log(1 2 ) sinlimx
x xk
x
0 0
sin2 log(1 2 ) 180
lim lim2 180
180
x x
xx
xx
= k
2180
k
90k
21 : (A)
Given, 2 2
10 2 2log 2
x y
x y
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2 2
2 2100
x y
x y
2 2 2 2
2 2
100 1000
x y x y
x y
2 2
2 2
99 1010
x y
x y
2 299 101 0x y
(2 99) (2 101) 0dy
x ydx
99
101
dy x
dx y
22 : (D)
Let
/2
/2
2 sinlog
2 sin
xI dx
x
As given function is odd.
0I
23 : (C)
By using anti differentiation method,
We will get to know that, Option (C) is correct.
24 : (B)
Degree 3
Order 2
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25 : (B)
Acute angle is 1 2sin
3
26 : (B)
2
2
0
(2 )A x x dx
23
2
03
xA x
84
3A
4
3A sq unit.
27 : (A)
Given ( )
log [log sin ]log (sin )
f xdx x C
x
By using anti differentiation method,we will get
log [log sin ]d
x cdx
1 1cos
log (sin ) sinx
x x
cot
log (sin )
x
x
( ) cotf x x
28 : (A)
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Correct option is (A)
29 : (C)
3 2 2i k mi mj mk ni nj nk
3 ( 2 ) ( ) ( 2 )i k m n i m n i n n k
2 3m n …(i)
0m n ….(ii)
2 1n n …(iii)
As 0m n from (ii)
m n
3 3m
1m and n = 1
2m n
30 : (C)
Let
/2
0
sec
sec cos
n
n n
xI dx
x ec x
…(i)
/2
0
sec2
sec cos2 2
n
n n
x
I dx
x ec x
/2
0
cos
cos sec
n
n n
ec xI dx
ec x x
…(ii)
Adding equation (i) and (ii)
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/2
0
2I dx
/2
02I x
4I
31 : (C)
Correct option is (C)
32 : (A)
(1 log ) log 0dx
y x x xdy
1 log
log
x dydx
x x y
Integrating on both side
1 log
log
x dydx
x x y
log( log ) log logx x y C
log ( log ) log ( )x x y c
logx x y c …(i)
As 2,x e y e
2e e c
1c
e
Putting 1
ce
in eq (i) we get
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logy
x xe
logy ex x
33 : (C)
Correct option is (C)
34 : (D)
Given.
I.F of dy
py Qdx
is sin x
sinPdx
e x
ln(sin )P dx x
By anti-differentiation method, we will get
[ln(sin )]d
P xdx
1cos
sinx
x
cotP x
35 : (C)
Option (C) is the correct answer
36 : (A)
( 7) ( 7) ( 8) ( 9) ( 10)P x P x P x P x P x
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7 3 8 2 9 1 10
10 10 10 10
7 8 9 10
1 1 1 1 1 1 1
2 2 2 2 2 2 2C C C C
11
64
37 : (C)
Given,
sin 2 cos 2 0x x
Multiplying by 1
2 on both side,
1 1sin 2 cos 2 0
2 2x x
sin 2 04
x
24
x n
(4 1)
8
nx
11 15
8 8x and
38 : (A)
Correct option is (A)
39 : (A)
Given 2 2 0 1
3 2 1 0A and B
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1 12 2 0 11
3 2 1 010A and B
Now,
1 10 1 2 21
1 0 3 210B A
=3 21
2 210
1 1 12 2 2 210
( )2 3 2 310
B A
40 : (D)
p : Every square is a rectangle T
q : Every rhombus is a kite T
p q T
p q T
41 : (B)
Correct option is (B)
42 : (A)
Correct option is (A)
43 : (C)
2tan 1x
tan 1x
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4x n
44 : (B)
Correct option is (B)
45 : (B)
Correct option is (B)
46 : (C)
Given A x I
1x A
3 21
4 15x
3 21
4 15x
47 : (A)
1 1 1
2 1 10
1 1 4
1(4 1) 1(8 1) 1( 2 ) 10
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4 1 7 2 10
3 18
6
48 : (B)
Given
n = 5
1
3p
2
3q
(2 4) ( 3)p x p x
3 2
5
3
1 2
3 3C
5 4 1 4
2 27 9
40
243
49 : (C)
Correct option is (C)
50 : (D)
1s p
2s q
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( ) ( )p q p q
( )p q