JEE Exam > JEE Notes > Mock Tests for JEE Main and Advanced 2025 > JEE Main 2024 February 1 Shift 1 Paper & Solutions
JEE Main 2024 February 1 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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)
Read More
|
357 docs|148 tests
|
FAQs on JEE Main 2024 February 1 Shift 1 Paper & Solutions - Mock Tests for JEE Main and Advanced 2025
| 1. What was the difficulty level of JEE Main 2024 February 1 Shift 1 Paper? |  |
Ans. The difficulty level of JEE Main 2024 February 1 Shift 1 Paper was moderate.
| 2. How many questions were there in JEE Main 2024 February 1 Shift 1 Paper? | |
Ans. There were a total of 90 questions in JEE Main 2024 February 1 Shift 1 Paper.
| 3. Were there any changes in the exam pattern for JEE Main 2024 February 1 Shift 1 Paper? | |
Ans. No, there were no changes in the exam pattern for JEE Main 2024 February 1 Shift 1 Paper compared to previous years.
| 4. What are the best preparation strategies for JEE Main 2024 February 1 Shift 1 Paper? | |
Ans. Some of the best preparation strategies for JEE Main include solving previous years' question papers, practicing mock tests, and focusing on strengthening concepts.
| 5. Are there any specific topics that were more heavily weighted in JEE Main 2024 February 1 Shift 1 Paper? | |
Ans. The distribution of topics in JEE Main 2024 February 1 Shift 1 Paper was balanced across all subjects, with no specific topic being more heavily weighted.
Related Exams
About this Document
|
4.76/5 Rating |
|
Dec 23, 2024 Last updated |
Document Description: JEE Main 2024 February 1 Shift 1 Paper & Solutions for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2025 preparation.
The notes and questions for JEE Main 2024 February 1 Shift 1 Paper & Solutions have been prepared according to the JEE exam syllabus. Information about JEE Main 2024 February 1 Shift 1 Paper & Solutions covers topics
like and JEE Main 2024 February 1 Shift 1 Paper & Solutions Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main 2024 February 1 Shift 1 Paper & Solutions.
Introduction of JEE Main 2024 February 1 Shift 1 Paper & Solutions in English is available as part of our Mock Tests for JEE Main and Advanced 2025
for JEE & JEE Main 2024 February 1 Shift 1 Paper & Solutions in Hindi for Mock Tests for JEE Main and Advanced 2025 course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: JEE Main 2024 February 1 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025
Description
Full syllabus notes, lecture & questions for JEE Main 2024 February 1 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 - JEE | Plus excerises question with solution to help you revise complete syllabus for Mock Tests for JEE Main and Advanced 2025 | Best notes, free PDF download
Information about JEE Main 2024 February 1 Shift 1 Paper & Solutions
In this doc you can find the meaning of JEE Main 2024 February 1 Shift 1 Paper & Solutions defined & explained in the simplest way possible. Besides explaining types of
JEE Main 2024 February 1 Shift 1 Paper & Solutions theory, EduRev gives you an ample number of questions to practice JEE Main 2024 February 1 Shift 1 Paper & Solutions tests, examples and also practice JEE
tests
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
study material
,mock tests for examination
,JEE Main 2024 February 1 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025
,Semester Notes
,video lectures
,ppt
,Viva Questions
,Objective type Questions
,past year papers
,JEE Main 2024 February 1 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025
,Summary
,Important questions
,Sample Paper
,shortcuts and tricks
,JEE Main 2024 February 1 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025
,practice quizzes
,Exam
,Previous Year Questions with Solutions
,Free
,MCQs
,Extra Questions
;
Additional Information about JEE Main 2024 February 1 Shift 1 Paper & Solutions for JEE Preparation
JEE Main 2024 February 1 Shift 1 Paper & Solutions Free PDF Download
The JEE Main 2024 February 1 Shift 1 Paper & Solutions is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the JEE Main 2024 February 1 Shift 1 Paper & Solutions now and kickstart your journey towards success in the JEE exam.
Importance of JEE Main 2024 February 1 Shift 1 Paper & Solutions
The importance of JEE Main 2024 February 1 Shift 1 Paper & Solutions cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
JEE Main 2024 February 1 Shift 1 Paper & Solutions Notes
JEE Main 2024 February 1 Shift 1 Paper & Solutions Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to JEE Main 2024 February 1 Shift 1 Paper & Solutions.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, JEE Main 2024 February 1 Shift 1 Paper & Solutions Notes on EduRev are your ultimate resource for success.
JEE Main 2024 February 1 Shift 1 Paper & Solutions JEE Questions
The "JEE Main 2024 February 1 Shift 1 Paper & Solutions JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study JEE Main 2024 February 1 Shift 1 Paper & Solutions on the App
Students of JEE can study JEE Main 2024 February 1 Shift 1 Paper & Solutions alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the JEE Main 2024 February 1 Shift 1 Paper & Solutions,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of JEE Main 2024 February 1 Shift 1 Paper & Solutions is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup