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Page 1
SECTION-A
1. A bag contains 8 balls, whose colours are either
white or black. 4 balls are drawn at random
without replacement and it was found that 2 balls
are white and other 2 balls are black. The
probability that the bag contains equal number of
white and black balls is:
(1)
2
5
(2)
2
7
(3)
1
7
(4)
1
5
Ans. (2)
Sol.
P(4W4B/2W2B) =
(4 4 ) (2 2 /4 4 )
(2 6 ) (2 2 /2 6 ) (3 5 ) (2 2 /3 5 )
............. (6 2 ) (2 2 /6 2 )
P W B P W B W B
P W B P W B W B P W B P W B W B
P W B P W B W B
?
? ? ?
? ? ?
=
44
22
8
4
2 6 3 5 6 2
2 2 2 2 2 2
8 8 8
4 4 4
CC 1
5C
C C C C C C 1 1 1
...
5 C 5 C 5 C
?
?
? ? ?
? ? ? ? ? ?
=
2
7
2. The value of the integral
4
44
0
:
sin (2 ) cos (2 )
xdx
equals
xx
?
?
?
(1)
2
2
8
?
(2)
2
2
16
?
(3)
2
2
32
?
(4)
2
2
64
?
Ans. (3)
Sol.
4
44
0
sin (2 ) cos (2 )
xdx
xx
?
?
?
Let 2xt ? then
1
2
dx dt ?
2
44
0
2
44
0
2
44
0
2
44
0
4 2
4
0
1
4 sin cos
1 2
4
sin cos
22
1
2
4 sin cos
2
8 sin cos
sec
2
8 tan 1
tdt
I
tt
t dt
I
tt
dt
II
tt
dt
I
tt
tdt
I
t
?
?
?
?
?
?
??
?
?
?
?
?
??
?
??
??
?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
??
?
?
?
?
?
?
?
?
?
?
Let tant = y then sec
2
t dt = dy
2
4
0
(1 )
2
81
?
?
?
?
?
y dy
I
y
?
2
2
0
2
1
1
y
dy
1
16
y
y
?
?
?
?
?
?
Put
1
yp
y
??
? ?
2
2
dp
I
16
p2
?
??
?
?
?
?
1
p
tan
16 2 2
?
?
??
?? ? ??
?
?? ??
?? ??
2
16 2
? I
?
Page 2
SECTION-A
1. A bag contains 8 balls, whose colours are either
white or black. 4 balls are drawn at random
without replacement and it was found that 2 balls
are white and other 2 balls are black. The
probability that the bag contains equal number of
white and black balls is:
(1)
2
5
(2)
2
7
(3)
1
7
(4)
1
5
Ans. (2)
Sol.
P(4W4B/2W2B) =
(4 4 ) (2 2 /4 4 )
(2 6 ) (2 2 /2 6 ) (3 5 ) (2 2 /3 5 )
............. (6 2 ) (2 2 /6 2 )
P W B P W B W B
P W B P W B W B P W B P W B W B
P W B P W B W B
?
? ? ?
? ? ?
=
44
22
8
4
2 6 3 5 6 2
2 2 2 2 2 2
8 8 8
4 4 4
CC 1
5C
C C C C C C 1 1 1
...
5 C 5 C 5 C
?
?
? ? ?
? ? ? ? ? ?
=
2
7
2. The value of the integral
4
44
0
:
sin (2 ) cos (2 )
xdx
equals
xx
?
?
?
(1)
2
2
8
?
(2)
2
2
16
?
(3)
2
2
32
?
(4)
2
2
64
?
Ans. (3)
Sol.
4
44
0
sin (2 ) cos (2 )
xdx
xx
?
?
?
Let 2xt ? then
1
2
dx dt ?
2
44
0
2
44
0
2
44
0
2
44
0
4 2
4
0
1
4 sin cos
1 2
4
sin cos
22
1
2
4 sin cos
2
8 sin cos
sec
2
8 tan 1
tdt
I
tt
t dt
I
tt
dt
II
tt
dt
I
tt
tdt
I
t
?
?
?
?
?
?
??
?
?
?
?
?
??
?
??
??
?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
??
?
?
?
?
?
?
?
?
?
?
Let tant = y then sec
2
t dt = dy
2
4
0
(1 )
2
81
?
?
?
?
?
y dy
I
y
?
2
2
0
2
1
1
y
dy
1
16
y
y
?
?
?
?
?
?
Put
1
yp
y
??
? ?
2
2
dp
I
16
p2
?
??
?
?
?
?
1
p
tan
16 2 2
?
?
??
?? ? ??
?
?? ??
?? ??
2
16 2
? I
?
3. If A =
21
12
??
??
???
??
, B =
10
11
??
??
??
, C = ABA
T
and X
= A
T
C
2
A, then det X is equal to :
(1) 243
(2) 729
(3) 27
(4) 891
Ans. (2)
Sol.
21
det( ) 3
12
10
det( ) 1
11
AA
BB
??
? ? ? ??
???
??
??
? ? ?
??
??
Now C = ABA
T
?det(C) = (dct (A))
2
x det(B)
9 C ?
Now |X| = |A
T
C
2
A|
= |A
T
| |C|
2
|A|
= |A|
2
|C|
2
= 9 x 81
= 729
4. If tanA =
22
1
,tan
( 1) 1
?
? ? ? ?
x
B
x x x x x
and
? ?
1
3 2 1
2
tan ,0 , , ,
2
? ? ?
? ? ? ? ? C x x x A B C then
?
A + B is equal to :
(1) C
(2) C ? ?
(3) 2 C ? ?
(4)
2
C
?
?
Ans. (1)
Sol.
Finding tan (A + B) we get
? tan (A + B) =
22
2
1
( 1) 1 tan tan
1
1 tan tan
1
1
x
x x x x x AB
AB
xx
?
? ? ? ? ?
?
?
?
??
? tan (A + B) =
? ?
? ?
? ? ? ?
2
2
11 x x x
x x x
? ? ?
?
? ?
? ?
? ? ? ?
2
2
2
11
1
tan( ) tan
x x x
x x x
xx
A B C
xx
A B C
? ? ?
?
??
? ? ?
??
5. If n is the number of ways five different employees
can sit into four indistinguishable offices where
any office may have any number of persons
including zero, then n is equal to:
(1) 47
(2) 53
(3) 51
(4) 43
Ans. (3)
Sol.
Total ways to partition 5 into 4 parts are :
5, 0, 0, 0 ?1 way
4, 1, 0, 0 ?
5!
4!
? 5 ways
3, 2, 0, 0,
5!
10
3!2!
?? ways
5!
2,2,0,1 15
2!2!2!
?? ways
3
5!
2,1,1,1 10
2!(1!) 3!
?? ways
5!
3,1,1,0 10
3!2!
?? ways
Total ? 1+5+10+15+10+10 = 51 ways
Page 3
SECTION-A
1. A bag contains 8 balls, whose colours are either
white or black. 4 balls are drawn at random
without replacement and it was found that 2 balls
are white and other 2 balls are black. The
probability that the bag contains equal number of
white and black balls is:
(1)
2
5
(2)
2
7
(3)
1
7
(4)
1
5
Ans. (2)
Sol.
P(4W4B/2W2B) =
(4 4 ) (2 2 /4 4 )
(2 6 ) (2 2 /2 6 ) (3 5 ) (2 2 /3 5 )
............. (6 2 ) (2 2 /6 2 )
P W B P W B W B
P W B P W B W B P W B P W B W B
P W B P W B W B
?
? ? ?
? ? ?
=
44
22
8
4
2 6 3 5 6 2
2 2 2 2 2 2
8 8 8
4 4 4
CC 1
5C
C C C C C C 1 1 1
...
5 C 5 C 5 C
?
?
? ? ?
? ? ? ? ? ?
=
2
7
2. The value of the integral
4
44
0
:
sin (2 ) cos (2 )
xdx
equals
xx
?
?
?
(1)
2
2
8
?
(2)
2
2
16
?
(3)
2
2
32
?
(4)
2
2
64
?
Ans. (3)
Sol.
4
44
0
sin (2 ) cos (2 )
xdx
xx
?
?
?
Let 2xt ? then
1
2
dx dt ?
2
44
0
2
44
0
2
44
0
2
44
0
4 2
4
0
1
4 sin cos
1 2
4
sin cos
22
1
2
4 sin cos
2
8 sin cos
sec
2
8 tan 1
tdt
I
tt
t dt
I
tt
dt
II
tt
dt
I
tt
tdt
I
t
?
?
?
?
?
?
??
?
?
?
?
?
??
?
??
??
?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
??
?
?
?
?
?
?
?
?
?
?
Let tant = y then sec
2
t dt = dy
2
4
0
(1 )
2
81
?
?
?
?
?
y dy
I
y
?
2
2
0
2
1
1
y
dy
1
16
y
y
?
?
?
?
?
?
Put
1
yp
y
??
? ?
2
2
dp
I
16
p2
?
??
?
?
?
?
1
p
tan
16 2 2
?
?
??
?? ? ??
?
?? ??
?? ??
2
16 2
? I
?
3. If A =
21
12
??
??
???
??
, B =
10
11
??
??
??
, C = ABA
T
and X
= A
T
C
2
A, then det X is equal to :
(1) 243
(2) 729
(3) 27
(4) 891
Ans. (2)
Sol.
21
det( ) 3
12
10
det( ) 1
11
AA
BB
??
? ? ? ??
???
??
??
? ? ?
??
??
Now C = ABA
T
?det(C) = (dct (A))
2
x det(B)
9 C ?
Now |X| = |A
T
C
2
A|
= |A
T
| |C|
2
|A|
= |A|
2
|C|
2
= 9 x 81
= 729
4. If tanA =
22
1
,tan
( 1) 1
?
? ? ? ?
x
B
x x x x x
and
? ?
1
3 2 1
2
tan ,0 , , ,
2
? ? ?
? ? ? ? ? C x x x A B C then
?
A + B is equal to :
(1) C
(2) C ? ?
(3) 2 C ? ?
(4)
2
C
?
?
Ans. (1)
Sol.
Finding tan (A + B) we get
? tan (A + B) =
22
2
1
( 1) 1 tan tan
1
1 tan tan
1
1
x
x x x x x AB
AB
xx
?
? ? ? ? ?
?
?
?
??
? tan (A + B) =
? ?
? ?
? ? ? ?
2
2
11 x x x
x x x
? ? ?
?
? ?
? ?
? ? ? ?
2
2
2
11
1
tan( ) tan
x x x
x x x
xx
A B C
xx
A B C
? ? ?
?
??
? ? ?
??
5. If n is the number of ways five different employees
can sit into four indistinguishable offices where
any office may have any number of persons
including zero, then n is equal to:
(1) 47
(2) 53
(3) 51
(4) 43
Ans. (3)
Sol.
Total ways to partition 5 into 4 parts are :
5, 0, 0, 0 ?1 way
4, 1, 0, 0 ?
5!
4!
? 5 ways
3, 2, 0, 0,
5!
10
3!2!
?? ways
5!
2,2,0,1 15
2!2!2!
?? ways
3
5!
2,1,1,1 10
2!(1!) 3!
?? ways
5!
3,1,1,0 10
3!2!
?? ways
Total ? 1+5+10+15+10+10 = 51 ways
6. LetS={ : 1 1 z C z ? ? ? and
? ?
? ? ? ? 2 1 2 2 z z i z z ? ? ? ? ? }. Let z
1
, z
2
S ? be such that
12
max min
zs zs
z z and z z
? ?
?? .
Then
2
12
2zz ? equals :
(1) 1 (2) 4
(3) 3 (4) 2
Ans. (4)
Sol. Let Z = x + iy
Then (x - 1)
2
+ y
2
= 1 ? (1)
&
? ?
? ? 2 1 2 (2 ) 2 2
( 2 1) 2 (2)
x i iy
xy
? ? ?
? ? ? ? ?
Solving (1) & (2) we get
Either x = 1 or
1
(3)
22
x??
?
On solving (3) with (2) we get
For x = 1 ?y = 1 ? Z
2
= 1 + i
& for
1
1 1 1
21
2 2 2 2 2
i
x y Z
??
? ? ? ? ? ? ? ?
??
? ??
Now
? ?
2
12
2
2
2
1
1 2 (1 )
2
2
2
zz
ii
?
??
? ? ? ? ?
??
??
?
?
7. Let the median and the mean deviation about the
median of 7 observation 170, 125, 230, 190, 210, a, b
be 170 and
205
7
respectively. Then the mean
deviation about the mean of these 7 observations is :
(1) 31
(2) 28
(3) 30
(4) 32
Ans. (3)
Sol. Median = 170 ?125, a, b, 170, 190, 210, 230
Mean deviation about
Median =
0 45 60 20 40 170 170 205
77
ab ? ? ? ? ? ? ? ?
?
?a + b = 300
Mean =
170 125 230 190 210
175
7
ab ? ? ? ? ? ?
?
Mean deviation
About mean =
50 175 175 5 15 35 55
7
ab ? ? ? ? ? ? ? ?
= 30
8. Let
ˆˆ ˆ ˆ ˆ ˆ
5 3 , 2 4 a i j k b i j k ? ? ? ? ? ? ? and
? ? ? ? ? ?
ˆˆˆ
. c a b i i i ? ? ? ? ? Then
? ?
ˆ ˆˆ
c i j k ? ? ? ? is
equal to
(1) –12 (2) –10
(3) –13 (4) –15
Ans. (1)
Sol.
ˆ ˆ
53 ? ? ? ? a i j k
ˆ ˆˆ
24 ? ? ? b i j k
? ? ? ?
ˆ ˆ ˆ
() ? ? ? ? ? ? a b i a i b b i a
5 ? ? ? ba
? ? ? ? ? ?
ˆˆ
5 ? ? ? ? ? b a i i
? ? ? ?
ˆ ˆ ˆ ˆ
11 23 ? ? ? ? ? j k i i
? ?
ˆ ˆˆ
11 23 ? ? ? k j i
? ?
ˆ ˆ
11 23 ??jk
? ?
ˆ ˆˆ
. 11 23 12 ? ? ? ? ? ? ? c i j k
Page 4
SECTION-A
1. A bag contains 8 balls, whose colours are either
white or black. 4 balls are drawn at random
without replacement and it was found that 2 balls
are white and other 2 balls are black. The
probability that the bag contains equal number of
white and black balls is:
(1)
2
5
(2)
2
7
(3)
1
7
(4)
1
5
Ans. (2)
Sol.
P(4W4B/2W2B) =
(4 4 ) (2 2 /4 4 )
(2 6 ) (2 2 /2 6 ) (3 5 ) (2 2 /3 5 )
............. (6 2 ) (2 2 /6 2 )
P W B P W B W B
P W B P W B W B P W B P W B W B
P W B P W B W B
?
? ? ?
? ? ?
=
44
22
8
4
2 6 3 5 6 2
2 2 2 2 2 2
8 8 8
4 4 4
CC 1
5C
C C C C C C 1 1 1
...
5 C 5 C 5 C
?
?
? ? ?
? ? ? ? ? ?
=
2
7
2. The value of the integral
4
44
0
:
sin (2 ) cos (2 )
xdx
equals
xx
?
?
?
(1)
2
2
8
?
(2)
2
2
16
?
(3)
2
2
32
?
(4)
2
2
64
?
Ans. (3)
Sol.
4
44
0
sin (2 ) cos (2 )
xdx
xx
?
?
?
Let 2xt ? then
1
2
dx dt ?
2
44
0
2
44
0
2
44
0
2
44
0
4 2
4
0
1
4 sin cos
1 2
4
sin cos
22
1
2
4 sin cos
2
8 sin cos
sec
2
8 tan 1
tdt
I
tt
t dt
I
tt
dt
II
tt
dt
I
tt
tdt
I
t
?
?
?
?
?
?
??
?
?
?
?
?
??
?
??
??
?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
??
?
?
?
?
?
?
?
?
?
?
Let tant = y then sec
2
t dt = dy
2
4
0
(1 )
2
81
?
?
?
?
?
y dy
I
y
?
2
2
0
2
1
1
y
dy
1
16
y
y
?
?
?
?
?
?
Put
1
yp
y
??
? ?
2
2
dp
I
16
p2
?
??
?
?
?
?
1
p
tan
16 2 2
?
?
??
?? ? ??
?
?? ??
?? ??
2
16 2
? I
?
3. If A =
21
12
??
??
???
??
, B =
10
11
??
??
??
, C = ABA
T
and X
= A
T
C
2
A, then det X is equal to :
(1) 243
(2) 729
(3) 27
(4) 891
Ans. (2)
Sol.
21
det( ) 3
12
10
det( ) 1
11
AA
BB
??
? ? ? ??
???
??
??
? ? ?
??
??
Now C = ABA
T
?det(C) = (dct (A))
2
x det(B)
9 C ?
Now |X| = |A
T
C
2
A|
= |A
T
| |C|
2
|A|
= |A|
2
|C|
2
= 9 x 81
= 729
4. If tanA =
22
1
,tan
( 1) 1
?
? ? ? ?
x
B
x x x x x
and
? ?
1
3 2 1
2
tan ,0 , , ,
2
? ? ?
? ? ? ? ? C x x x A B C then
?
A + B is equal to :
(1) C
(2) C ? ?
(3) 2 C ? ?
(4)
2
C
?
?
Ans. (1)
Sol.
Finding tan (A + B) we get
? tan (A + B) =
22
2
1
( 1) 1 tan tan
1
1 tan tan
1
1
x
x x x x x AB
AB
xx
?
? ? ? ? ?
?
?
?
??
? tan (A + B) =
? ?
? ?
? ? ? ?
2
2
11 x x x
x x x
? ? ?
?
? ?
? ?
? ? ? ?
2
2
2
11
1
tan( ) tan
x x x
x x x
xx
A B C
xx
A B C
? ? ?
?
??
? ? ?
??
5. If n is the number of ways five different employees
can sit into four indistinguishable offices where
any office may have any number of persons
including zero, then n is equal to:
(1) 47
(2) 53
(3) 51
(4) 43
Ans. (3)
Sol.
Total ways to partition 5 into 4 parts are :
5, 0, 0, 0 ?1 way
4, 1, 0, 0 ?
5!
4!
? 5 ways
3, 2, 0, 0,
5!
10
3!2!
?? ways
5!
2,2,0,1 15
2!2!2!
?? ways
3
5!
2,1,1,1 10
2!(1!) 3!
?? ways
5!
3,1,1,0 10
3!2!
?? ways
Total ? 1+5+10+15+10+10 = 51 ways
6. LetS={ : 1 1 z C z ? ? ? and
? ?
? ? ? ? 2 1 2 2 z z i z z ? ? ? ? ? }. Let z
1
, z
2
S ? be such that
12
max min
zs zs
z z and z z
? ?
?? .
Then
2
12
2zz ? equals :
(1) 1 (2) 4
(3) 3 (4) 2
Ans. (4)
Sol. Let Z = x + iy
Then (x - 1)
2
+ y
2
= 1 ? (1)
&
? ?
? ? 2 1 2 (2 ) 2 2
( 2 1) 2 (2)
x i iy
xy
? ? ?
? ? ? ? ?
Solving (1) & (2) we get
Either x = 1 or
1
(3)
22
x??
?
On solving (3) with (2) we get
For x = 1 ?y = 1 ? Z
2
= 1 + i
& for
1
1 1 1
21
2 2 2 2 2
i
x y Z
??
? ? ? ? ? ? ? ?
??
? ??
Now
? ?
2
12
2
2
2
1
1 2 (1 )
2
2
2
zz
ii
?
??
? ? ? ? ?
??
??
?
?
7. Let the median and the mean deviation about the
median of 7 observation 170, 125, 230, 190, 210, a, b
be 170 and
205
7
respectively. Then the mean
deviation about the mean of these 7 observations is :
(1) 31
(2) 28
(3) 30
(4) 32
Ans. (3)
Sol. Median = 170 ?125, a, b, 170, 190, 210, 230
Mean deviation about
Median =
0 45 60 20 40 170 170 205
77
ab ? ? ? ? ? ? ? ?
?
?a + b = 300
Mean =
170 125 230 190 210
175
7
ab ? ? ? ? ? ?
?
Mean deviation
About mean =
50 175 175 5 15 35 55
7
ab ? ? ? ? ? ? ? ?
= 30
8. Let
ˆˆ ˆ ˆ ˆ ˆ
5 3 , 2 4 a i j k b i j k ? ? ? ? ? ? ? and
? ? ? ? ? ?
ˆˆˆ
. c a b i i i ? ? ? ? ? Then
? ?
ˆ ˆˆ
c i j k ? ? ? ? is
equal to
(1) –12 (2) –10
(3) –13 (4) –15
Ans. (1)
Sol.
ˆ ˆ
53 ? ? ? ? a i j k
ˆ ˆˆ
24 ? ? ? b i j k
? ? ? ?
ˆ ˆ ˆ
() ? ? ? ? ? ? a b i a i b b i a
5 ? ? ? ba
? ? ? ? ? ?
ˆˆ
5 ? ? ? ? ? b a i i
? ? ? ?
ˆ ˆ ˆ ˆ
11 23 ? ? ? ? ? j k i i
? ?
ˆ ˆˆ
11 23 ? ? ? k j i
? ?
ˆ ˆ
11 23 ??jk
? ?
ˆ ˆˆ
. 11 23 12 ? ? ? ? ? ? ? c i j k
9. Let S = { :( 3 2) ( 3 2) 10}.
xx
xR ? ? ? ? ?
Then the number of elements in S is :
(1) 4 (2) 0
(3) 2 (4) 1
Ans. (3)
Sol.
? ? ? ?
xx
3 2 3 2 10 ? ? ? ?
Let
? ?
x
3 2 t ??
1
t 10
t
??
t
2
– 10t + 1 = 0
10 100 4
t 5 2 6
2
??
? ? ?
? ? ? ?
x2
3 2 3 2 ? ? ?
x = 2 or x = –2
Number of solutions = 2
10. The area enclosed by the curves xy + 4y = 16 and
x + y = 6 is equal to :
(1) 28 – 30 log 2
e
(2) 30 – 28 log 2
e
(3) 30 – 32 log 2
e
(4) 32 – 30 log 2
e
Ans. (3)
Sol. xy + 4y = 16 , x + y = 6
y(x + 4) = 16 ____(1) , x + y = 6___(2)
on solving, (1) & (2)
we get x = 4, x = –2
–4
–2 4 (6,0)
(0,6)
Area =
? ?
4
2
16
6
4
30 32ln2
x dx
x
?
?? ??
??
?? ??
?
?? ??
??
?
11. Let f : R ? R and g : R ? R be defined as
f(x) =
e
x
log x , x 0
e , x 0
?
? ?
?
?
? ?
?
and
g(x) =
x
x , x 0
e , x 0
? ?
?
?
? ?
?
. Then, gof : R ? R is :
(1) one-one but not onto
(2) neither one-one nor onto
(3) onto but not one-one
(4) both one-one and onto
Ans. (2)
Sol.
g(f(x)) =
()
( ), ( ) 0
, ( ) 0
fx
f x f x
e f x
? ?
?
?
?
g(f(x)) =
? ?
?
ln
, ,0
,(0,1)
ln , 1,
x
x
e
e
x
?
?
??
?
?
?
?
? ?
? ?
?
(1,0)
(0,1)
Graph of g(f(x))
g(f(x)) ?Many one into
12. If the system of equations
2x + 3y – z = 5
x + ?y + 3z = –4
3x – y + ?z = 7
has infinitely many solutions, then 13 ?? is equal
to
(1) 1110 (2) 1120
(3) 1210 (4) 1220
Ans. (2)
Page 5
SECTION-A
1. A bag contains 8 balls, whose colours are either
white or black. 4 balls are drawn at random
without replacement and it was found that 2 balls
are white and other 2 balls are black. The
probability that the bag contains equal number of
white and black balls is:
(1)
2
5
(2)
2
7
(3)
1
7
(4)
1
5
Ans. (2)
Sol.
P(4W4B/2W2B) =
(4 4 ) (2 2 /4 4 )
(2 6 ) (2 2 /2 6 ) (3 5 ) (2 2 /3 5 )
............. (6 2 ) (2 2 /6 2 )
P W B P W B W B
P W B P W B W B P W B P W B W B
P W B P W B W B
?
? ? ?
? ? ?
=
44
22
8
4
2 6 3 5 6 2
2 2 2 2 2 2
8 8 8
4 4 4
CC 1
5C
C C C C C C 1 1 1
...
5 C 5 C 5 C
?
?
? ? ?
? ? ? ? ? ?
=
2
7
2. The value of the integral
4
44
0
:
sin (2 ) cos (2 )
xdx
equals
xx
?
?
?
(1)
2
2
8
?
(2)
2
2
16
?
(3)
2
2
32
?
(4)
2
2
64
?
Ans. (3)
Sol.
4
44
0
sin (2 ) cos (2 )
xdx
xx
?
?
?
Let 2xt ? then
1
2
dx dt ?
2
44
0
2
44
0
2
44
0
2
44
0
4 2
4
0
1
4 sin cos
1 2
4
sin cos
22
1
2
4 sin cos
2
8 sin cos
sec
2
8 tan 1
tdt
I
tt
t dt
I
tt
dt
II
tt
dt
I
tt
tdt
I
t
?
?
?
?
?
?
??
?
?
?
?
?
??
?
??
??
?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
??
?
?
?
?
?
?
?
?
?
?
Let tant = y then sec
2
t dt = dy
2
4
0
(1 )
2
81
?
?
?
?
?
y dy
I
y
?
2
2
0
2
1
1
y
dy
1
16
y
y
?
?
?
?
?
?
Put
1
yp
y
??
? ?
2
2
dp
I
16
p2
?
??
?
?
?
?
1
p
tan
16 2 2
?
?
??
?? ? ??
?
?? ??
?? ??
2
16 2
? I
?
3. If A =
21
12
??
??
???
??
, B =
10
11
??
??
??
, C = ABA
T
and X
= A
T
C
2
A, then det X is equal to :
(1) 243
(2) 729
(3) 27
(4) 891
Ans. (2)
Sol.
21
det( ) 3
12
10
det( ) 1
11
AA
BB
??
? ? ? ??
???
??
??
? ? ?
??
??
Now C = ABA
T
?det(C) = (dct (A))
2
x det(B)
9 C ?
Now |X| = |A
T
C
2
A|
= |A
T
| |C|
2
|A|
= |A|
2
|C|
2
= 9 x 81
= 729
4. If tanA =
22
1
,tan
( 1) 1
?
? ? ? ?
x
B
x x x x x
and
? ?
1
3 2 1
2
tan ,0 , , ,
2
? ? ?
? ? ? ? ? C x x x A B C then
?
A + B is equal to :
(1) C
(2) C ? ?
(3) 2 C ? ?
(4)
2
C
?
?
Ans. (1)
Sol.
Finding tan (A + B) we get
? tan (A + B) =
22
2
1
( 1) 1 tan tan
1
1 tan tan
1
1
x
x x x x x AB
AB
xx
?
? ? ? ? ?
?
?
?
??
? tan (A + B) =
? ?
? ?
? ? ? ?
2
2
11 x x x
x x x
? ? ?
?
? ?
? ?
? ? ? ?
2
2
2
11
1
tan( ) tan
x x x
x x x
xx
A B C
xx
A B C
? ? ?
?
??
? ? ?
??
5. If n is the number of ways five different employees
can sit into four indistinguishable offices where
any office may have any number of persons
including zero, then n is equal to:
(1) 47
(2) 53
(3) 51
(4) 43
Ans. (3)
Sol.
Total ways to partition 5 into 4 parts are :
5, 0, 0, 0 ?1 way
4, 1, 0, 0 ?
5!
4!
? 5 ways
3, 2, 0, 0,
5!
10
3!2!
?? ways
5!
2,2,0,1 15
2!2!2!
?? ways
3
5!
2,1,1,1 10
2!(1!) 3!
?? ways
5!
3,1,1,0 10
3!2!
?? ways
Total ? 1+5+10+15+10+10 = 51 ways
6. LetS={ : 1 1 z C z ? ? ? and
? ?
? ? ? ? 2 1 2 2 z z i z z ? ? ? ? ? }. Let z
1
, z
2
S ? be such that
12
max min
zs zs
z z and z z
? ?
?? .
Then
2
12
2zz ? equals :
(1) 1 (2) 4
(3) 3 (4) 2
Ans. (4)
Sol. Let Z = x + iy
Then (x - 1)
2
+ y
2
= 1 ? (1)
&
? ?
? ? 2 1 2 (2 ) 2 2
( 2 1) 2 (2)
x i iy
xy
? ? ?
? ? ? ? ?
Solving (1) & (2) we get
Either x = 1 or
1
(3)
22
x??
?
On solving (3) with (2) we get
For x = 1 ?y = 1 ? Z
2
= 1 + i
& for
1
1 1 1
21
2 2 2 2 2
i
x y Z
??
? ? ? ? ? ? ? ?
??
? ??
Now
? ?
2
12
2
2
2
1
1 2 (1 )
2
2
2
zz
ii
?
??
? ? ? ? ?
??
??
?
?
7. Let the median and the mean deviation about the
median of 7 observation 170, 125, 230, 190, 210, a, b
be 170 and
205
7
respectively. Then the mean
deviation about the mean of these 7 observations is :
(1) 31
(2) 28
(3) 30
(4) 32
Ans. (3)
Sol. Median = 170 ?125, a, b, 170, 190, 210, 230
Mean deviation about
Median =
0 45 60 20 40 170 170 205
77
ab ? ? ? ? ? ? ? ?
?
?a + b = 300
Mean =
170 125 230 190 210
175
7
ab ? ? ? ? ? ?
?
Mean deviation
About mean =
50 175 175 5 15 35 55
7
ab ? ? ? ? ? ? ? ?
= 30
8. Let
ˆˆ ˆ ˆ ˆ ˆ
5 3 , 2 4 a i j k b i j k ? ? ? ? ? ? ? and
? ? ? ? ? ?
ˆˆˆ
. c a b i i i ? ? ? ? ? Then
? ?
ˆ ˆˆ
c i j k ? ? ? ? is
equal to
(1) –12 (2) –10
(3) –13 (4) –15
Ans. (1)
Sol.
ˆ ˆ
53 ? ? ? ? a i j k
ˆ ˆˆ
24 ? ? ? b i j k
? ? ? ?
ˆ ˆ ˆ
() ? ? ? ? ? ? a b i a i b b i a
5 ? ? ? ba
? ? ? ? ? ?
ˆˆ
5 ? ? ? ? ? b a i i
? ? ? ?
ˆ ˆ ˆ ˆ
11 23 ? ? ? ? ? j k i i
? ?
ˆ ˆˆ
11 23 ? ? ? k j i
? ?
ˆ ˆ
11 23 ??jk
? ?
ˆ ˆˆ
. 11 23 12 ? ? ? ? ? ? ? c i j k
9. Let S = { :( 3 2) ( 3 2) 10}.
xx
xR ? ? ? ? ?
Then the number of elements in S is :
(1) 4 (2) 0
(3) 2 (4) 1
Ans. (3)
Sol.
? ? ? ?
xx
3 2 3 2 10 ? ? ? ?
Let
? ?
x
3 2 t ??
1
t 10
t
??
t
2
– 10t + 1 = 0
10 100 4
t 5 2 6
2
??
? ? ?
? ? ? ?
x2
3 2 3 2 ? ? ?
x = 2 or x = –2
Number of solutions = 2
10. The area enclosed by the curves xy + 4y = 16 and
x + y = 6 is equal to :
(1) 28 – 30 log 2
e
(2) 30 – 28 log 2
e
(3) 30 – 32 log 2
e
(4) 32 – 30 log 2
e
Ans. (3)
Sol. xy + 4y = 16 , x + y = 6
y(x + 4) = 16 ____(1) , x + y = 6___(2)
on solving, (1) & (2)
we get x = 4, x = –2
–4
–2 4 (6,0)
(0,6)
Area =
? ?
4
2
16
6
4
30 32ln2
x dx
x
?
?? ??
??
?? ??
?
?? ??
??
?
11. Let f : R ? R and g : R ? R be defined as
f(x) =
e
x
log x , x 0
e , x 0
?
? ?
?
?
? ?
?
and
g(x) =
x
x , x 0
e , x 0
? ?
?
?
? ?
?
. Then, gof : R ? R is :
(1) one-one but not onto
(2) neither one-one nor onto
(3) onto but not one-one
(4) both one-one and onto
Ans. (2)
Sol.
g(f(x)) =
()
( ), ( ) 0
, ( ) 0
fx
f x f x
e f x
? ?
?
?
?
g(f(x)) =
? ?
?
ln
, ,0
,(0,1)
ln , 1,
x
x
e
e
x
?
?
??
?
?
?
?
? ?
? ?
?
(1,0)
(0,1)
Graph of g(f(x))
g(f(x)) ?Many one into
12. If the system of equations
2x + 3y – z = 5
x + ?y + 3z = –4
3x – y + ?z = 7
has infinitely many solutions, then 13 ?? is equal
to
(1) 1110 (2) 1120
(3) 1210 (4) 1220
Ans. (2)
Sol. Using family of planes
2x + 3y –z – 5 = k
1
(x + ? y + 3z + 4) + k
2
(3x – y
+ ? z - 7)
2 = k
1
+ 3k
2
, 3 = k
1
? - k
2
, -1 = 3k
1
+ ? k
2
, -5 =
4k
1
– 7k
2
On solving we get
21
13 1 16
, , 70,
19 19 13
kk ??
??
? ? ? ? ?
13 ? ? = 13 (-70)
16
13
1120
? ??
??
??
?
13. For 0 < ? < ?/2, if the eccentricity of the hyperbola
x
2
– y
2
cosec
2
? = 5 is 7 times eccentricity of the
ellipse x
2
cosec
2
? + y
2
= 5, then the value of ? is :
(1)
6
?
(2)
5
12
?
(3)
3
?
(4)
4
?
Ans. (3)
Sol.
2
2
1 sin
1 sin
h
c
e
e
?
?
??
??
7
hc
ee ?
22
2
1 sin 7(1 sin )
63
sin
84
3
sin
2
3
??
?
?
?
?
? ? ?
??
?
?
14. Let y = y(x) be the solution of the differential
equation
dy
dx
= 2x (x + y)
3
– x (x + y) – 1, y(0) = 1.
Then,
2
11
y
22
?? ??
?
?? ??
?? ??
equals :
(1)
4
4e ?
(2)
3
3e ?
(3)
2
1e ?
(4)
1
2e ?
Ans. (4)
Sol.
3
2 ( ) ( ) 1 ? ? ? ? ?
dy
x x y x x y
dx
?? xyt
3
1 2 1 ? ? ? ?
dt
xt xt
dx
3
2
?
?
dt
xdx
tt
42
tdt
xdx
2t t
?
?
Let
2
? tz
? ?
2
dz
xdx
2 2z z
?
?
??
1
4
2
?
??
?
??
??
??
dz
xdx
zz
2
1
2
ln
?
??
z
xk
z
1
2
?
?
z
e
15. Let f : R ? R be defined as
f(x) =
2
2
a bcos2x
; x 0
x
x cx 2 ; 0 x 1
2x 1 ; x 1
? ?
?
?
?
?
? ? ? ?
?
?
??
?
?
?
If f is continuous everywhere in R and m is the
number of points where f is NOT differential then
m + a + b + c equals :
(1) 1 (2) 4
(3) 3 (4) 2
Ans. (4)
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The content of JEE Main 2024 February 1 Shift 1 Paper & Solutions is prepared as per the latest JEE syllabus.
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