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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