monksays studmonk the app facebook monksays studmonk – the app facebook 5 : (d) given, x) x

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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 Rolles 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 1sin3

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 xdy e mdx 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, 2log(1 2 )sin

0( )

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 12

sin3

26 : (B)

2

2

0

(2 )A x x dx

23

2

03

xA x

84

3A

4

3A sq unit.

27 : (A)

Given ( )