Page 1
SECTION-A
1. If in a G.P. of 64 terms, the sum of all the terms is
7 times the sum of the odd terms of the G.P, then
the common ratio of the G.P. is equal to
(1) 7 (2) 4
(3) 5 (4) 6
Ans. (4)
Sol.
2 3 63
a ar ar ar .... ar ? ? ? ? ?
=
2 4 62
7(a ar ar ..... ar ) ? ? ?
64 64
2
a(1 r ) 7a(1 r )
1r
1r
??
??
?
?
r = 6
2. In an A.P., the sixth terms a
6
= 2. If the
1 4 5
a a a is
the greatest, then the common difference of the
A.P., is equal to
(1)
3
2
(2)
8
5
(3)
2
3
(4)
5
8
Ans. (2)
Sol.
6
a 2 a 5d 2 ? ? ? ?
1 4 5
a a a a(a 3d)(a 4d) ? ? ?
= (2 5d)(2 2d)(2 d) ? ? ?
2 2 3
f(d) 8 32d 34d 20d 30d 10d ? ? ? ? ? ?
f'(d) 2(5d 8)(3d 2) ? ? ? ?
+ – –
8/5 2/3
8
d
5
?
3. If ? ?
2 2x , 1 x 0
f x ;
x
1 , 0 x 3
3
? ? ? ? ?
?
?
?
? ? ?
?
?
? ?
x , 3 x 0
gx
x, 0 x 1
? ? ? ? ?
?
?
??
?
,
then range of (fog(x)) is
(1) (0, 1] (2) [0, 3)
(3) [0, 1] (4) [0, 1)
Ans. (3)
Sol.
2 2g(x) , 1 g(x) 0 .....(1)
f(g(x))
g(x)
1 , 0 g(x) 3 .....(2)
3
? ? ? ? ?
?
?
?
? ? ?
?
?
By (1) x ??
And by (2) x [ 3,0] and x [0,1] ? ? ?
(–3,3)
(1,1)
y=f(x)
–3 O 1
1
2/3
y=f(g(x))
Range of f(g(x)) is [0, 1]
4. A fair die is thrown until 2 appears. Then the
probability, that 2 appears in even number of
throws , is
(1)
5
6
(2)
1
6
(3)
5
11
(4)
6
11
Ans. (3)
Sol. Required probability =
35
5 1 5 1 5 1
.....
6 6 6 6 6 6
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
=
5
1
6
25
6
1
36
?
?
=
5
11
Page 2
SECTION-A
1. If in a G.P. of 64 terms, the sum of all the terms is
7 times the sum of the odd terms of the G.P, then
the common ratio of the G.P. is equal to
(1) 7 (2) 4
(3) 5 (4) 6
Ans. (4)
Sol.
2 3 63
a ar ar ar .... ar ? ? ? ? ?
=
2 4 62
7(a ar ar ..... ar ) ? ? ?
64 64
2
a(1 r ) 7a(1 r )
1r
1r
??
??
?
?
r = 6
2. In an A.P., the sixth terms a
6
= 2. If the
1 4 5
a a a is
the greatest, then the common difference of the
A.P., is equal to
(1)
3
2
(2)
8
5
(3)
2
3
(4)
5
8
Ans. (2)
Sol.
6
a 2 a 5d 2 ? ? ? ?
1 4 5
a a a a(a 3d)(a 4d) ? ? ?
= (2 5d)(2 2d)(2 d) ? ? ?
2 2 3
f(d) 8 32d 34d 20d 30d 10d ? ? ? ? ? ?
f'(d) 2(5d 8)(3d 2) ? ? ? ?
+ – –
8/5 2/3
8
d
5
?
3. If ? ?
2 2x , 1 x 0
f x ;
x
1 , 0 x 3
3
? ? ? ? ?
?
?
?
? ? ?
?
?
? ?
x , 3 x 0
gx
x, 0 x 1
? ? ? ? ?
?
?
??
?
,
then range of (fog(x)) is
(1) (0, 1] (2) [0, 3)
(3) [0, 1] (4) [0, 1)
Ans. (3)
Sol.
2 2g(x) , 1 g(x) 0 .....(1)
f(g(x))
g(x)
1 , 0 g(x) 3 .....(2)
3
? ? ? ? ?
?
?
?
? ? ?
?
?
By (1) x ??
And by (2) x [ 3,0] and x [0,1] ? ? ?
(–3,3)
(1,1)
y=f(x)
–3 O 1
1
2/3
y=f(g(x))
Range of f(g(x)) is [0, 1]
4. A fair die is thrown until 2 appears. Then the
probability, that 2 appears in even number of
throws , is
(1)
5
6
(2)
1
6
(3)
5
11
(4)
6
11
Ans. (3)
Sol. Required probability =
35
5 1 5 1 5 1
.....
6 6 6 6 6 6
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
=
5
1
6
25
6
1
36
?
?
=
5
11
5. If
1
z 2i
2
?? , is such that
? ? |z 1| z 1 i ,i 1 ? ? ? ? ? ? ? ? and ,R ? ? ? , then
? ? ? is equal to
(1) –4 (2) 3
(3) 2 (4) –1
Ans. (2)
Sol.
1
z 2i
2
??
z 1 z (1 i) ? ? ? ? ? ?
3
2i 2 i i
22
?
? ? ? ? ? ? ? ?
? ?
3
2i 2 i
22
? ??
? ? ? ? ? ? ? ?
??
??
9
2 and 4
24
?
? ? ? ? ? ? ?
3 ? ? ? ?
6.
3
3
2
2 3
x
x
2
11
lim cos dt
t
x
2
? ??
??
??
?
?
??
??
??
??
??
??
? ? ? ??
?
?? ??
????
?
is equal to
(1)
3
8
?
(2)
2
3
4
?
(3)
2
3
8
?
(4)
3
4
?
Ans. (3)
Sol. Using L’hopital rule
=
2
x
2
0 cosx 3x
lim
2x
2
?
?
?
??
? ??
?
??
??
=
2
x
2
sin x
3 2
lim
4
2x
2
?
?
?
? ??
?
??
?
??
?
? ??
?
??
??
=
2
3
8
?
7. In a ABC, ? suppose y = x is the equation of the
bisector of the angle B and the equation of the side
AC is 2x –y =2. If 2AB = BC and the point A and
B are respectively (4, 6) and ? ? , ?? , then 2 ? ? ? is
equal to
(1) 42 (2) 39
(3) 48 (4) 45
Ans. (1)
Sol.
A(4,6)
B( , ) ?? C(–2,–6)
A’(6,4)
y=x
(2,2)
D
AD : DC = 1 : 2
4 10
68
??
?
??
? = ?
14 ?? and ? = 14
8. Let a, b and c be three non-zero vectors such that
b and c are non-collinear .ifa 5b ? is collinear
with c,b 6c ? is collinear with a and
a b c 0 ? ? ? ? ? , then ? ? ? is equal to
(1) 35 (2) 30
(3) – 30 (4)–25
Ans. (1)
Sol. a 5b c ? ? ?
b 6c a ? ? ?
Eliminating a
61
c 5b c b ? ? ? ?
??
?
1
, 30
5
?
? ? ? ? ?
5, 30 ? ? ? ?
Page 3
SECTION-A
1. If in a G.P. of 64 terms, the sum of all the terms is
7 times the sum of the odd terms of the G.P, then
the common ratio of the G.P. is equal to
(1) 7 (2) 4
(3) 5 (4) 6
Ans. (4)
Sol.
2 3 63
a ar ar ar .... ar ? ? ? ? ?
=
2 4 62
7(a ar ar ..... ar ) ? ? ?
64 64
2
a(1 r ) 7a(1 r )
1r
1r
??
??
?
?
r = 6
2. In an A.P., the sixth terms a
6
= 2. If the
1 4 5
a a a is
the greatest, then the common difference of the
A.P., is equal to
(1)
3
2
(2)
8
5
(3)
2
3
(4)
5
8
Ans. (2)
Sol.
6
a 2 a 5d 2 ? ? ? ?
1 4 5
a a a a(a 3d)(a 4d) ? ? ?
= (2 5d)(2 2d)(2 d) ? ? ?
2 2 3
f(d) 8 32d 34d 20d 30d 10d ? ? ? ? ? ?
f'(d) 2(5d 8)(3d 2) ? ? ? ?
+ – –
8/5 2/3
8
d
5
?
3. If ? ?
2 2x , 1 x 0
f x ;
x
1 , 0 x 3
3
? ? ? ? ?
?
?
?
? ? ?
?
?
? ?
x , 3 x 0
gx
x, 0 x 1
? ? ? ? ?
?
?
??
?
,
then range of (fog(x)) is
(1) (0, 1] (2) [0, 3)
(3) [0, 1] (4) [0, 1)
Ans. (3)
Sol.
2 2g(x) , 1 g(x) 0 .....(1)
f(g(x))
g(x)
1 , 0 g(x) 3 .....(2)
3
? ? ? ? ?
?
?
?
? ? ?
?
?
By (1) x ??
And by (2) x [ 3,0] and x [0,1] ? ? ?
(–3,3)
(1,1)
y=f(x)
–3 O 1
1
2/3
y=f(g(x))
Range of f(g(x)) is [0, 1]
4. A fair die is thrown until 2 appears. Then the
probability, that 2 appears in even number of
throws , is
(1)
5
6
(2)
1
6
(3)
5
11
(4)
6
11
Ans. (3)
Sol. Required probability =
35
5 1 5 1 5 1
.....
6 6 6 6 6 6
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
=
5
1
6
25
6
1
36
?
?
=
5
11
5. If
1
z 2i
2
?? , is such that
? ? |z 1| z 1 i ,i 1 ? ? ? ? ? ? ? ? and ,R ? ? ? , then
? ? ? is equal to
(1) –4 (2) 3
(3) 2 (4) –1
Ans. (2)
Sol.
1
z 2i
2
??
z 1 z (1 i) ? ? ? ? ? ?
3
2i 2 i i
22
?
? ? ? ? ? ? ? ?
? ?
3
2i 2 i
22
? ??
? ? ? ? ? ? ? ?
??
??
9
2 and 4
24
?
? ? ? ? ? ? ?
3 ? ? ? ?
6.
3
3
2
2 3
x
x
2
11
lim cos dt
t
x
2
? ??
??
??
?
?
??
??
??
??
??
??
? ? ? ??
?
?? ??
????
?
is equal to
(1)
3
8
?
(2)
2
3
4
?
(3)
2
3
8
?
(4)
3
4
?
Ans. (3)
Sol. Using L’hopital rule
=
2
x
2
0 cosx 3x
lim
2x
2
?
?
?
??
? ??
?
??
??
=
2
x
2
sin x
3 2
lim
4
2x
2
?
?
?
? ??
?
??
?
??
?
? ??
?
??
??
=
2
3
8
?
7. In a ABC, ? suppose y = x is the equation of the
bisector of the angle B and the equation of the side
AC is 2x –y =2. If 2AB = BC and the point A and
B are respectively (4, 6) and ? ? , ?? , then 2 ? ? ? is
equal to
(1) 42 (2) 39
(3) 48 (4) 45
Ans. (1)
Sol.
A(4,6)
B( , ) ?? C(–2,–6)
A’(6,4)
y=x
(2,2)
D
AD : DC = 1 : 2
4 10
68
??
?
??
? = ?
14 ?? and ? = 14
8. Let a, b and c be three non-zero vectors such that
b and c are non-collinear .ifa 5b ? is collinear
with c,b 6c ? is collinear with a and
a b c 0 ? ? ? ? ? , then ? ? ? is equal to
(1) 35 (2) 30
(3) – 30 (4)–25
Ans. (1)
Sol. a 5b c ? ? ?
b 6c a ? ? ?
Eliminating a
61
c 5b c b ? ? ? ?
??
?
1
, 30
5
?
? ? ? ? ?
5, 30 ? ? ? ?
9. Let
a
5,
4
??
??
??
, be the circumcenter of a triangle with
vertices ? ? ? ? A a, 2 , B a,6 ? and
a
C , –2
4
??
??
??
. Let ?
denote the circumradius, ? denote the area and ?
denote the perimeter of the triangle. Then ? ? ? ? ? is
(1) 60 (2) 53
(3) 62 (4) 30
Ans. (2)
Sol. A(a, –2), B(a, 6),
a
C , 2
4
??
?
??
??
,
a
O 5,
4
??
??
??
AO = BO
22
22
aa
(a 5) 2 (a 5) 6
44
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
a = 8
AB = 8, AC = 6, BC = 10
5, 24, 24 ? ? ? ? ? ?
10. For x,
22
??
??
??
??
??
, if
? ?
2
cosecx sinx
y x dx
cosecxsecx tanxsin x
?
?
?
?
and
? ?
x
2
lim y x 0
?
? ??
?
??
??
? theny
4
?
??
??
??
is equal to
(1)
1
1
tan
2
?
??
??
??
(2)
1
11
tan
2 2
?
??
??
??
(3)
1
11
tan
22
?
??
?
??
??
(4)
1
11
tan
2 2
???
?
??
??
Ans. (4)
Sol.
2
4
(1 sin x)cosx
y(x) dx
1 sin x
?
?
?
?
Put sinx = t
=
2
4
1t
dt
t1
?
?
?
=
1
1
t
1 t
tan C
22
?
??
?
??
??
?
x ,t 1
2
?
??
? C = 0
1
11
y tan
42
2
?
? ? ? ? ?
??
? ? ? ?
? ? ? ?
11. If ,
22
??
? ? ? ? ? is the solution of 4cos 5sin 1 ? ? ? ? ,
then the value of tan ? is
(1)
10 10
6
?
(2)
10 10
12
?
(3)
10 10
12
?
(4)
10 10
6
?
Ans. (3)
Sol. 4 5tan sec ?? ??
Squaring :
2
24tan 40tan 15 0 ?? ? ? ?
10 10
tan
12
?
??
?
and
10 10
tan
12
??
?
??
??
??
? is Rejected.
(3) is correct.
12. A function y = f(x) satisfies
? ? ? ? ? ?
2
f x sin2x sinx 1 cos x f' x 0 ? ? ? ? with condition
f(0) = 0 . Then f
2
?
??
??
??
is equal to
(1) 1 (2) 0 (3) –1 (4) 2
Ans. (1)
Sol.
2
dy sin 2x
y sin x
dx 1 cos x
??
??
??
?
??
I.F. = 1 + cos
2
x
? ? ? ?
2
y 1 cos x sin x dx ? ? ?
?
= – cosx + C
x = 0, C = 1
y1
2
? ??
?
??
??
13. Let O be the origin and the position vector of A
and B be
ˆ ˆ ˆ
2i 2j k ?? and
ˆ ˆ ˆ
2i 4j 4k ?? respectively. If
the internal bisector of AOB ? meets the line AB
at C, then the length of OC is
(1)
2
31
3
(2)
2
34
3
(3)
3
34
4
(4)
3
31
2
Page 4
SECTION-A
1. If in a G.P. of 64 terms, the sum of all the terms is
7 times the sum of the odd terms of the G.P, then
the common ratio of the G.P. is equal to
(1) 7 (2) 4
(3) 5 (4) 6
Ans. (4)
Sol.
2 3 63
a ar ar ar .... ar ? ? ? ? ?
=
2 4 62
7(a ar ar ..... ar ) ? ? ?
64 64
2
a(1 r ) 7a(1 r )
1r
1r
??
??
?
?
r = 6
2. In an A.P., the sixth terms a
6
= 2. If the
1 4 5
a a a is
the greatest, then the common difference of the
A.P., is equal to
(1)
3
2
(2)
8
5
(3)
2
3
(4)
5
8
Ans. (2)
Sol.
6
a 2 a 5d 2 ? ? ? ?
1 4 5
a a a a(a 3d)(a 4d) ? ? ?
= (2 5d)(2 2d)(2 d) ? ? ?
2 2 3
f(d) 8 32d 34d 20d 30d 10d ? ? ? ? ? ?
f'(d) 2(5d 8)(3d 2) ? ? ? ?
+ – –
8/5 2/3
8
d
5
?
3. If ? ?
2 2x , 1 x 0
f x ;
x
1 , 0 x 3
3
? ? ? ? ?
?
?
?
? ? ?
?
?
? ?
x , 3 x 0
gx
x, 0 x 1
? ? ? ? ?
?
?
??
?
,
then range of (fog(x)) is
(1) (0, 1] (2) [0, 3)
(3) [0, 1] (4) [0, 1)
Ans. (3)
Sol.
2 2g(x) , 1 g(x) 0 .....(1)
f(g(x))
g(x)
1 , 0 g(x) 3 .....(2)
3
? ? ? ? ?
?
?
?
? ? ?
?
?
By (1) x ??
And by (2) x [ 3,0] and x [0,1] ? ? ?
(–3,3)
(1,1)
y=f(x)
–3 O 1
1
2/3
y=f(g(x))
Range of f(g(x)) is [0, 1]
4. A fair die is thrown until 2 appears. Then the
probability, that 2 appears in even number of
throws , is
(1)
5
6
(2)
1
6
(3)
5
11
(4)
6
11
Ans. (3)
Sol. Required probability =
35
5 1 5 1 5 1
.....
6 6 6 6 6 6
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
=
5
1
6
25
6
1
36
?
?
=
5
11
5. If
1
z 2i
2
?? , is such that
? ? |z 1| z 1 i ,i 1 ? ? ? ? ? ? ? ? and ,R ? ? ? , then
? ? ? is equal to
(1) –4 (2) 3
(3) 2 (4) –1
Ans. (2)
Sol.
1
z 2i
2
??
z 1 z (1 i) ? ? ? ? ? ?
3
2i 2 i i
22
?
? ? ? ? ? ? ? ?
? ?
3
2i 2 i
22
? ??
? ? ? ? ? ? ? ?
??
??
9
2 and 4
24
?
? ? ? ? ? ? ?
3 ? ? ? ?
6.
3
3
2
2 3
x
x
2
11
lim cos dt
t
x
2
? ??
??
??
?
?
??
??
??
??
??
??
? ? ? ??
?
?? ??
????
?
is equal to
(1)
3
8
?
(2)
2
3
4
?
(3)
2
3
8
?
(4)
3
4
?
Ans. (3)
Sol. Using L’hopital rule
=
2
x
2
0 cosx 3x
lim
2x
2
?
?
?
??
? ??
?
??
??
=
2
x
2
sin x
3 2
lim
4
2x
2
?
?
?
? ??
?
??
?
??
?
? ??
?
??
??
=
2
3
8
?
7. In a ABC, ? suppose y = x is the equation of the
bisector of the angle B and the equation of the side
AC is 2x –y =2. If 2AB = BC and the point A and
B are respectively (4, 6) and ? ? , ?? , then 2 ? ? ? is
equal to
(1) 42 (2) 39
(3) 48 (4) 45
Ans. (1)
Sol.
A(4,6)
B( , ) ?? C(–2,–6)
A’(6,4)
y=x
(2,2)
D
AD : DC = 1 : 2
4 10
68
??
?
??
? = ?
14 ?? and ? = 14
8. Let a, b and c be three non-zero vectors such that
b and c are non-collinear .ifa 5b ? is collinear
with c,b 6c ? is collinear with a and
a b c 0 ? ? ? ? ? , then ? ? ? is equal to
(1) 35 (2) 30
(3) – 30 (4)–25
Ans. (1)
Sol. a 5b c ? ? ?
b 6c a ? ? ?
Eliminating a
61
c 5b c b ? ? ? ?
??
?
1
, 30
5
?
? ? ? ? ?
5, 30 ? ? ? ?
9. Let
a
5,
4
??
??
??
, be the circumcenter of a triangle with
vertices ? ? ? ? A a, 2 , B a,6 ? and
a
C , –2
4
??
??
??
. Let ?
denote the circumradius, ? denote the area and ?
denote the perimeter of the triangle. Then ? ? ? ? ? is
(1) 60 (2) 53
(3) 62 (4) 30
Ans. (2)
Sol. A(a, –2), B(a, 6),
a
C , 2
4
??
?
??
??
,
a
O 5,
4
??
??
??
AO = BO
22
22
aa
(a 5) 2 (a 5) 6
44
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
a = 8
AB = 8, AC = 6, BC = 10
5, 24, 24 ? ? ? ? ? ?
10. For x,
22
??
??
??
??
??
, if
? ?
2
cosecx sinx
y x dx
cosecxsecx tanxsin x
?
?
?
?
and
? ?
x
2
lim y x 0
?
? ??
?
??
??
? theny
4
?
??
??
??
is equal to
(1)
1
1
tan
2
?
??
??
??
(2)
1
11
tan
2 2
?
??
??
??
(3)
1
11
tan
22
?
??
?
??
??
(4)
1
11
tan
2 2
???
?
??
??
Ans. (4)
Sol.
2
4
(1 sin x)cosx
y(x) dx
1 sin x
?
?
?
?
Put sinx = t
=
2
4
1t
dt
t1
?
?
?
=
1
1
t
1 t
tan C
22
?
??
?
??
??
?
x ,t 1
2
?
??
? C = 0
1
11
y tan
42
2
?
? ? ? ? ?
??
? ? ? ?
? ? ? ?
11. If ,
22
??
? ? ? ? ? is the solution of 4cos 5sin 1 ? ? ? ? ,
then the value of tan ? is
(1)
10 10
6
?
(2)
10 10
12
?
(3)
10 10
12
?
(4)
10 10
6
?
Ans. (3)
Sol. 4 5tan sec ?? ??
Squaring :
2
24tan 40tan 15 0 ?? ? ? ?
10 10
tan
12
?
??
?
and
10 10
tan
12
??
?
??
??
??
? is Rejected.
(3) is correct.
12. A function y = f(x) satisfies
? ? ? ? ? ?
2
f x sin2x sinx 1 cos x f' x 0 ? ? ? ? with condition
f(0) = 0 . Then f
2
?
??
??
??
is equal to
(1) 1 (2) 0 (3) –1 (4) 2
Ans. (1)
Sol.
2
dy sin 2x
y sin x
dx 1 cos x
??
??
??
?
??
I.F. = 1 + cos
2
x
? ? ? ?
2
y 1 cos x sin x dx ? ? ?
?
= – cosx + C
x = 0, C = 1
y1
2
? ??
?
??
??
13. Let O be the origin and the position vector of A
and B be
ˆ ˆ ˆ
2i 2j k ?? and
ˆ ˆ ˆ
2i 4j 4k ?? respectively. If
the internal bisector of AOB ? meets the line AB
at C, then the length of OC is
(1)
2
31
3
(2)
2
34
3
(3)
3
34
4
(4)
3
31
2
Ans. (2)
Sol.
O
C
1 : 2
3
B A
(2, 4, 4) (2, 2, 1)
6
length of
136 2 34
OC
33
??
14. Consider the function
1
f : ,1 R
2
??
?
??
??
defined by
? ?
3
f x 4 2x 3 2x 1 ? ? ? . Consider the statements
(I) The curve y = f(x) intersects the x-axis exactly
at one point
(II) The curve y = f(x) intersects the x-axis at
x cos
12
?
?
Then
(1) Only (II) is correct
(2) Both (I) and (II) are incorrect
(3) Only (I) is correct
(4) Both (I) and (II) are correct
Ans. (4)
Sol. ? ?
2
f' x 12 2x 3 2 0 ? ? ? for
1
,1
2
??
??
??
1
f0
2
??
?
??
??
f(1) > 0 ? (A) is correct.
? ? ? ?
3
f x 2 4x 3x 1 0 ? ? ? ?
Let cos ? = x,
cos 3 ? = cos
4
?
? ? =
12
?
x cos
12
?
?
(4) is correct.
15. Let
1 0 0
A0
0
??
??
? ? ?
??
?? ??
??
and
3 21
|2A| 2 ? where ,Z ? ? ? ,
Then a value of ? is
(1) 3 (2) 5
(3) 17 (4) 9
Ans. (2)
Sol.
22
A ? ? ? ?
3
21
2A 2 ?
4
A2 ??
22
16 ? ? ? ?
? ? ? ? 16 ? ? ? ? ? ? ? 4 or 5 ? ? ?
16. Let PQR be a triangle with ? ? R 1,4,2 ? . Suppose
M(2, 1, 2) is the mid point of PQ. The distance of
the centroid of PQR ? from the point of
intersection of the line
x 2 y z 3 x 1 y 3 z 1
and
0 2 1 1 3 1
? ? ? ? ?
? ? ? ?
??
is
(1) 69 (2) 9
(3) 69
(4) 99
Ans. (3)
Sol. Centroid G divides MR in 1 : 2
G(1, 2, 2)
Point of intersection A of given lines is (2,–6, 0)
AG 69 ?
17. Let R be a relation on Z × Z defined by
(a, b)R(c, d) if and only if ad – bc is divisible by 5.
Then R is
(1) Reflexive and symmetric but not transitive
(2) Reflexive but neither symmetric not transitive
(3) Reflexive, symmetric and transitive
(4) Reflexive and transitive but not symmetric
Ans. (1)
Page 5
SECTION-A
1. If in a G.P. of 64 terms, the sum of all the terms is
7 times the sum of the odd terms of the G.P, then
the common ratio of the G.P. is equal to
(1) 7 (2) 4
(3) 5 (4) 6
Ans. (4)
Sol.
2 3 63
a ar ar ar .... ar ? ? ? ? ?
=
2 4 62
7(a ar ar ..... ar ) ? ? ?
64 64
2
a(1 r ) 7a(1 r )
1r
1r
??
??
?
?
r = 6
2. In an A.P., the sixth terms a
6
= 2. If the
1 4 5
a a a is
the greatest, then the common difference of the
A.P., is equal to
(1)
3
2
(2)
8
5
(3)
2
3
(4)
5
8
Ans. (2)
Sol.
6
a 2 a 5d 2 ? ? ? ?
1 4 5
a a a a(a 3d)(a 4d) ? ? ?
= (2 5d)(2 2d)(2 d) ? ? ?
2 2 3
f(d) 8 32d 34d 20d 30d 10d ? ? ? ? ? ?
f'(d) 2(5d 8)(3d 2) ? ? ? ?
+ – –
8/5 2/3
8
d
5
?
3. If ? ?
2 2x , 1 x 0
f x ;
x
1 , 0 x 3
3
? ? ? ? ?
?
?
?
? ? ?
?
?
? ?
x , 3 x 0
gx
x, 0 x 1
? ? ? ? ?
?
?
??
?
,
then range of (fog(x)) is
(1) (0, 1] (2) [0, 3)
(3) [0, 1] (4) [0, 1)
Ans. (3)
Sol.
2 2g(x) , 1 g(x) 0 .....(1)
f(g(x))
g(x)
1 , 0 g(x) 3 .....(2)
3
? ? ? ? ?
?
?
?
? ? ?
?
?
By (1) x ??
And by (2) x [ 3,0] and x [0,1] ? ? ?
(–3,3)
(1,1)
y=f(x)
–3 O 1
1
2/3
y=f(g(x))
Range of f(g(x)) is [0, 1]
4. A fair die is thrown until 2 appears. Then the
probability, that 2 appears in even number of
throws , is
(1)
5
6
(2)
1
6
(3)
5
11
(4)
6
11
Ans. (3)
Sol. Required probability =
35
5 1 5 1 5 1
.....
6 6 6 6 6 6
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
=
5
1
6
25
6
1
36
?
?
=
5
11
5. If
1
z 2i
2
?? , is such that
? ? |z 1| z 1 i ,i 1 ? ? ? ? ? ? ? ? and ,R ? ? ? , then
? ? ? is equal to
(1) –4 (2) 3
(3) 2 (4) –1
Ans. (2)
Sol.
1
z 2i
2
??
z 1 z (1 i) ? ? ? ? ? ?
3
2i 2 i i
22
?
? ? ? ? ? ? ? ?
? ?
3
2i 2 i
22
? ??
? ? ? ? ? ? ? ?
??
??
9
2 and 4
24
?
? ? ? ? ? ? ?
3 ? ? ? ?
6.
3
3
2
2 3
x
x
2
11
lim cos dt
t
x
2
? ??
??
??
?
?
??
??
??
??
??
??
? ? ? ??
?
?? ??
????
?
is equal to
(1)
3
8
?
(2)
2
3
4
?
(3)
2
3
8
?
(4)
3
4
?
Ans. (3)
Sol. Using L’hopital rule
=
2
x
2
0 cosx 3x
lim
2x
2
?
?
?
??
? ??
?
??
??
=
2
x
2
sin x
3 2
lim
4
2x
2
?
?
?
? ??
?
??
?
??
?
? ??
?
??
??
=
2
3
8
?
7. In a ABC, ? suppose y = x is the equation of the
bisector of the angle B and the equation of the side
AC is 2x –y =2. If 2AB = BC and the point A and
B are respectively (4, 6) and ? ? , ?? , then 2 ? ? ? is
equal to
(1) 42 (2) 39
(3) 48 (4) 45
Ans. (1)
Sol.
A(4,6)
B( , ) ?? C(–2,–6)
A’(6,4)
y=x
(2,2)
D
AD : DC = 1 : 2
4 10
68
??
?
??
? = ?
14 ?? and ? = 14
8. Let a, b and c be three non-zero vectors such that
b and c are non-collinear .ifa 5b ? is collinear
with c,b 6c ? is collinear with a and
a b c 0 ? ? ? ? ? , then ? ? ? is equal to
(1) 35 (2) 30
(3) – 30 (4)–25
Ans. (1)
Sol. a 5b c ? ? ?
b 6c a ? ? ?
Eliminating a
61
c 5b c b ? ? ? ?
??
?
1
, 30
5
?
? ? ? ? ?
5, 30 ? ? ? ?
9. Let
a
5,
4
??
??
??
, be the circumcenter of a triangle with
vertices ? ? ? ? A a, 2 , B a,6 ? and
a
C , –2
4
??
??
??
. Let ?
denote the circumradius, ? denote the area and ?
denote the perimeter of the triangle. Then ? ? ? ? ? is
(1) 60 (2) 53
(3) 62 (4) 30
Ans. (2)
Sol. A(a, –2), B(a, 6),
a
C , 2
4
??
?
??
??
,
a
O 5,
4
??
??
??
AO = BO
22
22
aa
(a 5) 2 (a 5) 6
44
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
a = 8
AB = 8, AC = 6, BC = 10
5, 24, 24 ? ? ? ? ? ?
10. For x,
22
??
??
??
??
??
, if
? ?
2
cosecx sinx
y x dx
cosecxsecx tanxsin x
?
?
?
?
and
? ?
x
2
lim y x 0
?
? ??
?
??
??
? theny
4
?
??
??
??
is equal to
(1)
1
1
tan
2
?
??
??
??
(2)
1
11
tan
2 2
?
??
??
??
(3)
1
11
tan
22
?
??
?
??
??
(4)
1
11
tan
2 2
???
?
??
??
Ans. (4)
Sol.
2
4
(1 sin x)cosx
y(x) dx
1 sin x
?
?
?
?
Put sinx = t
=
2
4
1t
dt
t1
?
?
?
=
1
1
t
1 t
tan C
22
?
??
?
??
??
?
x ,t 1
2
?
??
? C = 0
1
11
y tan
42
2
?
? ? ? ? ?
??
? ? ? ?
? ? ? ?
11. If ,
22
??
? ? ? ? ? is the solution of 4cos 5sin 1 ? ? ? ? ,
then the value of tan ? is
(1)
10 10
6
?
(2)
10 10
12
?
(3)
10 10
12
?
(4)
10 10
6
?
Ans. (3)
Sol. 4 5tan sec ?? ??
Squaring :
2
24tan 40tan 15 0 ?? ? ? ?
10 10
tan
12
?
??
?
and
10 10
tan
12
??
?
??
??
??
? is Rejected.
(3) is correct.
12. A function y = f(x) satisfies
? ? ? ? ? ?
2
f x sin2x sinx 1 cos x f' x 0 ? ? ? ? with condition
f(0) = 0 . Then f
2
?
??
??
??
is equal to
(1) 1 (2) 0 (3) –1 (4) 2
Ans. (1)
Sol.
2
dy sin 2x
y sin x
dx 1 cos x
??
??
??
?
??
I.F. = 1 + cos
2
x
? ? ? ?
2
y 1 cos x sin x dx ? ? ?
?
= – cosx + C
x = 0, C = 1
y1
2
? ??
?
??
??
13. Let O be the origin and the position vector of A
and B be
ˆ ˆ ˆ
2i 2j k ?? and
ˆ ˆ ˆ
2i 4j 4k ?? respectively. If
the internal bisector of AOB ? meets the line AB
at C, then the length of OC is
(1)
2
31
3
(2)
2
34
3
(3)
3
34
4
(4)
3
31
2
Ans. (2)
Sol.
O
C
1 : 2
3
B A
(2, 4, 4) (2, 2, 1)
6
length of
136 2 34
OC
33
??
14. Consider the function
1
f : ,1 R
2
??
?
??
??
defined by
? ?
3
f x 4 2x 3 2x 1 ? ? ? . Consider the statements
(I) The curve y = f(x) intersects the x-axis exactly
at one point
(II) The curve y = f(x) intersects the x-axis at
x cos
12
?
?
Then
(1) Only (II) is correct
(2) Both (I) and (II) are incorrect
(3) Only (I) is correct
(4) Both (I) and (II) are correct
Ans. (4)
Sol. ? ?
2
f' x 12 2x 3 2 0 ? ? ? for
1
,1
2
??
??
??
1
f0
2
??
?
??
??
f(1) > 0 ? (A) is correct.
? ? ? ?
3
f x 2 4x 3x 1 0 ? ? ? ?
Let cos ? = x,
cos 3 ? = cos
4
?
? ? =
12
?
x cos
12
?
?
(4) is correct.
15. Let
1 0 0
A0
0
??
??
? ? ?
??
?? ??
??
and
3 21
|2A| 2 ? where ,Z ? ? ? ,
Then a value of ? is
(1) 3 (2) 5
(3) 17 (4) 9
Ans. (2)
Sol.
22
A ? ? ? ?
3
21
2A 2 ?
4
A2 ??
22
16 ? ? ? ?
? ? ? ? 16 ? ? ? ? ? ? ? 4 or 5 ? ? ?
16. Let PQR be a triangle with ? ? R 1,4,2 ? . Suppose
M(2, 1, 2) is the mid point of PQ. The distance of
the centroid of PQR ? from the point of
intersection of the line
x 2 y z 3 x 1 y 3 z 1
and
0 2 1 1 3 1
? ? ? ? ?
? ? ? ?
??
is
(1) 69 (2) 9
(3) 69
(4) 99
Ans. (3)
Sol. Centroid G divides MR in 1 : 2
G(1, 2, 2)
Point of intersection A of given lines is (2,–6, 0)
AG 69 ?
17. Let R be a relation on Z × Z defined by
(a, b)R(c, d) if and only if ad – bc is divisible by 5.
Then R is
(1) Reflexive and symmetric but not transitive
(2) Reflexive but neither symmetric not transitive
(3) Reflexive, symmetric and transitive
(4) Reflexive and transitive but not symmetric
Ans. (1)
Sol. (a, b)R(a, b) as ab – ab = 0
Therefore reflexive
Let (a,b)R(c,d) ?ad – bc is divisible by 5
? bc – ad is divisible by 5 ?(c,d)R(a,b)
Therefore symmetric
Relation not transitive as (3,1)R(10,5) and
(10,5)R(1,1) but (3,1) is not related to (1,1)
18. If the value of the integral
? ?
2023
22 2
x
sinx
2
x cosx 1 sin x
dx a 2
14
1e
?
?
?
?? ??
? ? ? ? ?
??
??
?
??
?
,
then the value of a is
(1) 3 (2)
3
2
? (3) 2 (4)
3
2
Ans. (1)
Sol.
2023
/2
22
x
sinx
/2
x cosx 1 sin x
I dx
1
1e
?
??
??
?
????
??
??
? ??
?
2023
/2
22
x
sin( x)
/2
x cosx 1 sin x
I dx
1
1e
?
?
?
??
??
?
????
??
??
? ??
?
On Adding, we get
? ?
/2
22
/2
2I x cosx 1 sin x dx
?
??
? ? ?
?
On solving
2
3
I2
44
??
? ? ?
a = 3
19. Suppose
? ?
? ? ? ?
? ?
x x 1 2
3
2
2 2 tanx tan x x 1
fx
7x 3x 1
??
? ? ?
?
??
,
Then the value of f '(0) is equal to
(1) ?
(2) 0
(3) ?
(4)
2
?
Ans. (3)
Sol.
h0
f(h) f(0)
f'(0) lim
h ?
?
?
=
h h 1 2
23
h0
(2 2 )tan h tan (h h 1) 0
lim
(7h 3h 1) h
??
?
? ? ? ?
??
= ?
20. Let A be a square matrix such that
T
AA I ? . Then
? ? ? ?
22
TT
1
A A A A A
2
??
? ? ?
??
??
is equal to
(1)
2
AI ?
(2)
3
AI ?
(3)
2T
AA ?
(4)
3T
AA ?
Ans. (4)
Sol.
TT
AA I A A ??
On solving given expression, we get
2 T 2 T 2 T 2 T
1
A A (A ) 2AA A (A ) 2AA
2
??
? ? ? ? ?
??
=
2 T 2
A[A (A ) ] ?
=
3T
AA ?
SECTION-B
21. Equation of two diameters of a circle are
2x 3y 5 ?? and 3x 4y 7 ?? . The line joining the
points
22
,4
7
??
??
??
??
and
1
,3
7
??
?
??
??
intersects the circle
at only one point ? ? P, ?? . Then 17??? is equal to
Ans. (2)
Sol. Centre of circle is (1, ?1)
C(1,–1)
A(–22/7,–4) B(–1/7,3) P( , ) ??
Equation of AB is 7x – 3y + 10 = 0 …(i)
Equation of CP is 3x + 7y + 4 = 0 …(ii)
Solving (i) and (ii)
41 1
,
29 29
?
? ? ? ?
17 2 ? ? ? ? ?
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