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2 22cosa b cq- =

2 2

4 4cos 8cos 1q q- - =

2 3 3cos ,cos

4 2q q= =

As q in acute

6

pq =

Sol 8

Ans.(B)2 2 2(1 sin )(1 cos ) cos

R R R

q q q+ + =

2 210

1 11 sin sin (2 ) 1 1

4 4R Rq q+ + + +

9

4=

1010

1

9

4R

Rtp

=

Sol. 92 1

1 2 sin(cos ) 0x x y-

+ + = 1 1 2cos sin ( 1 )y y- -= -

2 1 21 2 sin(sin 1 ) 0x x y-+ + + =

2 21 2 1 ) 0x x y+ + - =

This equation is satisfy by only one solution i.e.

1,& 0x y= - =

Q.10

Ans. (A)

In a ABCD

tan tan tan tan tan tanA B C A B C+ + =

Hence 6 2 tan c=

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Hence (1) becomes 2y x x= +

This curve cut the x-axis at (0,0) and (-1,0) also it cut y axis at (0,0)

If meets the line x=1 at the print (1,2)Hence the curve is as Required area

0 1

1 0

ydx ydx-

= +

0 12 2

1 0

( ) ( )x x dx x x dx-

= + + +

0 13 2 3 2

1 03 2 3 2

x x x x

-

= + + +

1 1 1 1

3 2 3 2

- = - + + +

1 51

6 6= + =

Sol 18

Ans. (1)

If 1 2 3, ,z z z are the vertices of an equilateral triangle then we prove

2 2 2

1 2 3 1 2 2 3 3 1Z Z Z Z Z Z Z Z Z+ + = + +

Here 3 0Z =

2 21 2 1 2Z Z Z Z+ =

21 2 1 2( ) 3Z Z Z Z+ = ,

1 =3k/3 , k =1

Sol 19

Ans. (1)

The 2ndcurve in the ellipse2 24

125 16

x+ =

Any tangent to the ellipse is

225 16y mx m= + +

Let (x1y1) be a point on2 2 9x y- =

Then 2 21 1 1(25 ) 2 16 0m x mx y y- + + - =

X

o(0,0)

(-1,0 ) (1,0)

X

(1,2)

x=1

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Q.2 Given that x+ y = 5k where k is a constant and that all values of x between 0 and

5k are equally likely, then the probability that 29

4

xy k> is

(a)3

5(b)

4

5(c)

2

5(d)

1

2

Q.3 If , ,x y z be unit vectors, then the maximum value of

2 2 25 7 5 7 5 7x y y z z x- + - + -

(a) 179 (b) 3 (c) 176 (d) 178

Q.4 If , , , 0,2

p

andcos( )cos( ) sin (2sin sin )b g b g a g a + - = + then the

straight line

((sin ) cos ) in 0x y sa b g+ + = passes through the point

(a) (1, 1) (b) (1, -1) (c) (-1,1) (d) (-1, -1)

Q.5 If 2 sin (sin cos )z a iq q q= + , then the locus of z is

(a) a circle whose centre is a

(b) a circle whose center is a

(c) a circle whose radius is2

a

(d) a circle of radius 2a

Q.6 In a ABC, coordinates of A are (4, 6), AB=12 and BAC=

3

p. The midpoint of

AC is (4, 12) Then the radius of the in-circle of ABC is

(a) 3 3 (b) 2 3 (c) 4 3 (d) 3

Q.7 In ABC, the minimum value of 2 2 2tan tan tan2 2 2

A B C+ + is equal to

(a) 0 (b)1

2(c) 1 (d) 3

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Hence2

3(3 1)5 log2

x

d-

=

23

(3 3)10 log(3 1)

x

xd +=

+

( ) ( )( )

4x x3

2x 3

3 1 3 3

2 3 1

- += -

Which 2(3 1) 2(3 3)x x- = + 2 2 8 0y y- - =

Then, y =(3x-1)

( 4)( 2) 0y y- + =

3 1 4x - = or -2 (-2 is rejected)

3 5x =

3log 5x =

SECTION-2

Sol 9

Ans. (B)

C= 12 the possible values of a and b are below

a b

1 12

2 11 to 12

3 10 to 12

: :

: :

12 12

Total number of possible

=2(1+2+3+4+5+6)=42

Sol 10Ans. (D)

C=24 the possible values of a and b are

a b

13 13

14 14

: :

: :

23 23

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