fiitjee mock paper
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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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