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Page 1
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024
(Held On Thursday 01
st
February, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
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
?? ?? ?K ?? ?K ?K ??
=
4 4
2 2
8
4
2 6 3 5 6 2
2 2 2 2 2 2
8 8 8
4 4 4
C C 1
5 C
C C C C C C 1 1 1
...
5 C 5 C 5 C
?? ?? ?? ?? ?? ?? ?K ?? ?K ?K ??
=
2
7
2. The value of the integral
4
4 4
0
:
sin (2 ) cos (2 )
xdx
equals
x x
?? ?K ??
(1)
2
2
8
?? (2)
2
2
16
??
(3)
2
2
32
?? (4)
2
2
64
??
Ans. (3)
Sol.
4
4 4
0
sin (2 ) cos (2 )
xdx
x x
?? ?K ??
Let 2x t ?] then
1
2
dx dt ?]
2
4 4
0
2
4 4
0
2
4 4
0
2
4 4
0
4 2
4
0
1
4 sin cos
1 2
4
sin cos
2 2
1
2
4 sin cos
2
8 sin cos
sec
2
8 tan 1
tdt
I
t t
t dt
I
t t
dt
I I
t t
dt
I
t t
tdt
I
t
?? ?? ?? ?? ?? ?? ????
?? ?? ?? ?] ?K ????
?M ????
????
?] ?? ?? ?? ?? ?M ?K ?M ?? ?? ?? ?? ?? ?? ?? ?? ?]?M
?K ?] ?K ?] ?K ?? ?? ?? ?? ??
Let tant = y then sec
2
t dt = dy
2
4
0
(1 )
2
8 1
?? ?K ?] ?K ?? y dy
I
y
??
2
2
0
2
1
1
y
dy
1
16
y
y
?? ?K ?? ?] ?K ??
Put
1
y p
y
?M?]
?H ?I 2
2
dp
I
16
p 2
?? ?M??
?? ?] ?K ??
1
p
tan
16 2 2
?? ?M ?M??
???? ?? ????
?] ???? ????
???? ????
2
16 2
?] I
??
Page 2
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024
(Held On Thursday 01
st
February, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
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
?? ?? ?K ?? ?K ?K ??
=
4 4
2 2
8
4
2 6 3 5 6 2
2 2 2 2 2 2
8 8 8
4 4 4
C C 1
5 C
C C C C C C 1 1 1
...
5 C 5 C 5 C
?? ?? ?? ?? ?? ?? ?K ?? ?K ?K ??
=
2
7
2. The value of the integral
4
4 4
0
:
sin (2 ) cos (2 )
xdx
equals
x x
?? ?K ??
(1)
2
2
8
?? (2)
2
2
16
??
(3)
2
2
32
?? (4)
2
2
64
??
Ans. (3)
Sol.
4
4 4
0
sin (2 ) cos (2 )
xdx
x x
?? ?K ??
Let 2x t ?] then
1
2
dx dt ?]
2
4 4
0
2
4 4
0
2
4 4
0
2
4 4
0
4 2
4
0
1
4 sin cos
1 2
4
sin cos
2 2
1
2
4 sin cos
2
8 sin cos
sec
2
8 tan 1
tdt
I
t t
t dt
I
t t
dt
I I
t t
dt
I
t t
tdt
I
t
?? ?? ?? ?? ?? ?? ????
?? ?? ?? ?] ?K ????
?M ????
????
?] ?? ?? ?? ?? ?M ?K ?M ?? ?? ?? ?? ?? ?? ?? ?? ?]?M
?K ?] ?K ?] ?K ?? ?? ?? ?? ??
Let tant = y then sec
2
t dt = dy
2
4
0
(1 )
2
8 1
?? ?K ?] ?K ?? y dy
I
y
??
2
2
0
2
1
1
y
dy
1
16
y
y
?? ?K ?? ?] ?K ??
Put
1
y p
y
?M?]
?H ?I 2
2
dp
I
16
p 2
?? ?M??
?? ?] ?K ??
1
p
tan
16 2 2
?? ?M ?M??
???? ?? ????
?] ???? ????
???? ????
2
16 2
?] I
??
3. If A =
2 1
1 2
????
????
?M????
????
, B =
1 0
1 1
????
????
????
, 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.
2 1
det( ) 3
1 2
1 0
det( ) 1
1 1
A A
B B
????
?] ?? ?] ????
?M????
????
????
?] ?? ?] ????
????
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 =
2 2
1
,tan
( 1) 1
?] ?K ?K ?K ?K x
B
x x x x x
and
?H ?I 1
3 2 1
2
tan ,0 , , ,
2
?M ?M ?M ?] ?K ?K ?\ ?\ C x x x A B C then
??
A + B is equal to :
(1) C
(2) C ?? ?M
(3) 2 C ?? ?M
(4)
2
C
?? ?M
Ans. (1)
Sol.
Finding tan (A + B) we get
?? tan (A + B) =
2 2
2
1
( 1) 1 tan tan
1
1 tan tan
1
1
x
x x x x x A B
A B
x x
?K ?K ?K ?K ?K ?K ?] ?M ?M ?K?K
?? tan (A + B) =
?H ?I ?H ?I ?H ?I ?H ?I 2
2
1 1 x x x
x x x
?K ?K ?K ?K
?H ?I ?H ?I ?H ?I ?H ?I 2
2
2
1 1
1
tan( ) tan
x x x
x x x
x x
A B C
x x
A B C
?K ?K ?K ?K ?K?K
?K ?] ?] ?K?]
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
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024
(Held On Thursday 01
st
February, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
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
?? ?? ?K ?? ?K ?K ??
=
4 4
2 2
8
4
2 6 3 5 6 2
2 2 2 2 2 2
8 8 8
4 4 4
C C 1
5 C
C C C C C C 1 1 1
...
5 C 5 C 5 C
?? ?? ?? ?? ?? ?? ?K ?? ?K ?K ??
=
2
7
2. The value of the integral
4
4 4
0
:
sin (2 ) cos (2 )
xdx
equals
x x
?? ?K ??
(1)
2
2
8
?? (2)
2
2
16
??
(3)
2
2
32
?? (4)
2
2
64
??
Ans. (3)
Sol.
4
4 4
0
sin (2 ) cos (2 )
xdx
x x
?? ?K ??
Let 2x t ?] then
1
2
dx dt ?]
2
4 4
0
2
4 4
0
2
4 4
0
2
4 4
0
4 2
4
0
1
4 sin cos
1 2
4
sin cos
2 2
1
2
4 sin cos
2
8 sin cos
sec
2
8 tan 1
tdt
I
t t
t dt
I
t t
dt
I I
t t
dt
I
t t
tdt
I
t
?? ?? ?? ?? ?? ?? ????
?? ?? ?? ?] ?K ????
?M ????
????
?] ?? ?? ?? ?? ?M ?K ?M ?? ?? ?? ?? ?? ?? ?? ?? ?]?M
?K ?] ?K ?] ?K ?? ?? ?? ?? ??
Let tant = y then sec
2
t dt = dy
2
4
0
(1 )
2
8 1
?? ?K ?] ?K ?? y dy
I
y
??
2
2
0
2
1
1
y
dy
1
16
y
y
?? ?K ?? ?] ?K ??
Put
1
y p
y
?M?]
?H ?I 2
2
dp
I
16
p 2
?? ?M??
?? ?] ?K ??
1
p
tan
16 2 2
?? ?M ?M??
???? ?? ????
?] ???? ????
???? ????
2
16 2
?] I
??
3. If A =
2 1
1 2
????
????
?M????
????
, B =
1 0
1 1
????
????
????
, 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.
2 1
det( ) 3
1 2
1 0
det( ) 1
1 1
A A
B B
????
?] ?? ?] ????
?M????
????
????
?] ?? ?] ????
????
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 =
2 2
1
,tan
( 1) 1
?] ?K ?K ?K ?K x
B
x x x x x
and
?H ?I 1
3 2 1
2
tan ,0 , , ,
2
?M ?M ?M ?] ?K ?K ?\ ?\ C x x x A B C then
??
A + B is equal to :
(1) C
(2) C ?? ?M
(3) 2 C ?? ?M
(4)
2
C
?? ?M
Ans. (1)
Sol.
Finding tan (A + B) we get
?? tan (A + B) =
2 2
2
1
( 1) 1 tan tan
1
1 tan tan
1
1
x
x x x x x A B
A B
x x
?K ?K ?K ?K ?K ?K ?] ?M ?M ?K?K
?? tan (A + B) =
?H ?I ?H ?I ?H ?I ?H ?I 2
2
1 1 x x x
x x x
?K ?K ?K ?K
?H ?I ?H ?I ?H ?I ?H ?I 2
2
2
1 1
1
tan( ) tan
x x x
x x x
x x
A B C
x x
A B C
?K ?K ?K ?K ?K?K
?K ?] ?] ?K?]
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 ?? ?M ?] and
?H ?I ?H ?I ?H ?I 2 1 2 2 z z i z z ?M ?K ?M ?M ?] }. Let z
1
, z
2
S ?? be such that
1 2
max min
z s z s
z z and z z
?? ???]?] .
Then
2
1 2
2z z ?M equals :
(1) 1 (2) 4
(3) 3 (4) 2
Ans. (4)
Sol. Let Z = x + iy
Then (x - 1)
2
+ y
2
= 1 ?? (1)
&
?H ?I ?H ?I 2 1 2 (2 ) 2 2
( 2 1) 2 (2)
x i iy
x y
?M ?M ?] ?? ?M ?K ?] ??
Solving (1) & (2) we get
Either x = 1 or
1
(3)
2 2
x?]??
?M
On solving (3) with (2) we get
For x = 1 ?? y = 1 ?? Z
2
= 1 + i
& for
1
1 1 1
2 1
2 2 2 2 2
i
x y Z
????
?] ?? ?] ?M ?? ?] ?K ?K????
?M ????
Now
?H ?I 2
1 2
2
2
2
1
1 2 (1 )
2
2
2
z z
i i
?M ????
?] ?K ?K ?M ?K ????
????
?] ?]
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
7 7
a b ?K ?K ?K ?K ?K ?M ?K ?M ?]
?? a + b = 300
Mean =
170 125 230 190 210
175
7
a b ?K ?K ?K ?K ?K ?K ?]
Mean deviation
About mean =
50 175 175 5 15 35 55
7
a b ?K ?M ?K ?M ?K ?K ?K ?K = 30
8. Let
ˆˆ ˆ ˆ ˆ ˆ
5 3 , 2 4 a i j k b i j k ?] ?M ?K ?M ?] ?K ?M and
?H ?I ?H ?I ?H ?I ˆˆˆ
. c a b i i i ?] ?? ?? ?? ?? Then
?H ?I ˆ ˆˆ
c i j k ?? ?M ?K ?K is
equal to
(1) –12 (2) –10
(3) –13 (4) –15
Ans. (1)
Sol.
ˆ ˆ
5 3 ?] ?M ?K ?M a i j k
ˆ ˆˆ
2 4 ?] ?K ?M b i j k
?H ?I ?H ?I ˆ ˆ ˆ
( ) ?? ?? ?] ?? ?M ?? a b i a i b b i a
5 ?] ?M ?M b a
?H ?I ?H ?I ?H ?I ˆˆ
5 ?] ?M ?M ?? ?? b a i i
?H ?I ?H ?I ˆ ˆ ˆ ˆ
11 23 ?] ?M ?K ?? ?? j k i i
?H ?I ˆ ˆˆ
11 23 ?? ?K ?? k j i
?H ?I ˆ ˆ
11 23 ???M j k
?H ?I ˆ ˆˆ
. 11 23 12 ?M ?K ?K ?] ?M ?] ?M c i j k
Page 4
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024
(Held On Thursday 01
st
February, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
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
?? ?? ?K ?? ?K ?K ??
=
4 4
2 2
8
4
2 6 3 5 6 2
2 2 2 2 2 2
8 8 8
4 4 4
C C 1
5 C
C C C C C C 1 1 1
...
5 C 5 C 5 C
?? ?? ?? ?? ?? ?? ?K ?? ?K ?K ??
=
2
7
2. The value of the integral
4
4 4
0
:
sin (2 ) cos (2 )
xdx
equals
x x
?? ?K ??
(1)
2
2
8
?? (2)
2
2
16
??
(3)
2
2
32
?? (4)
2
2
64
??
Ans. (3)
Sol.
4
4 4
0
sin (2 ) cos (2 )
xdx
x x
?? ?K ??
Let 2x t ?] then
1
2
dx dt ?]
2
4 4
0
2
4 4
0
2
4 4
0
2
4 4
0
4 2
4
0
1
4 sin cos
1 2
4
sin cos
2 2
1
2
4 sin cos
2
8 sin cos
sec
2
8 tan 1
tdt
I
t t
t dt
I
t t
dt
I I
t t
dt
I
t t
tdt
I
t
?? ?? ?? ?? ?? ?? ????
?? ?? ?? ?] ?K ????
?M ????
????
?] ?? ?? ?? ?? ?M ?K ?M ?? ?? ?? ?? ?? ?? ?? ?? ?]?M
?K ?] ?K ?] ?K ?? ?? ?? ?? ??
Let tant = y then sec
2
t dt = dy
2
4
0
(1 )
2
8 1
?? ?K ?] ?K ?? y dy
I
y
??
2
2
0
2
1
1
y
dy
1
16
y
y
?? ?K ?? ?] ?K ??
Put
1
y p
y
?M?]
?H ?I 2
2
dp
I
16
p 2
?? ?M??
?? ?] ?K ??
1
p
tan
16 2 2
?? ?M ?M??
???? ?? ????
?] ???? ????
???? ????
2
16 2
?] I
??
3. If A =
2 1
1 2
????
????
?M????
????
, B =
1 0
1 1
????
????
????
, 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.
2 1
det( ) 3
1 2
1 0
det( ) 1
1 1
A A
B B
????
?] ?? ?] ????
?M????
????
????
?] ?? ?] ????
????
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 =
2 2
1
,tan
( 1) 1
?] ?K ?K ?K ?K x
B
x x x x x
and
?H ?I 1
3 2 1
2
tan ,0 , , ,
2
?M ?M ?M ?] ?K ?K ?\ ?\ C x x x A B C then
??
A + B is equal to :
(1) C
(2) C ?? ?M
(3) 2 C ?? ?M
(4)
2
C
?? ?M
Ans. (1)
Sol.
Finding tan (A + B) we get
?? tan (A + B) =
2 2
2
1
( 1) 1 tan tan
1
1 tan tan
1
1
x
x x x x x A B
A B
x x
?K ?K ?K ?K ?K ?K ?] ?M ?M ?K?K
?? tan (A + B) =
?H ?I ?H ?I ?H ?I ?H ?I 2
2
1 1 x x x
x x x
?K ?K ?K ?K
?H ?I ?H ?I ?H ?I ?H ?I 2
2
2
1 1
1
tan( ) tan
x x x
x x x
x x
A B C
x x
A B C
?K ?K ?K ?K ?K?K
?K ?] ?] ?K?]
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 ?? ?M ?] and
?H ?I ?H ?I ?H ?I 2 1 2 2 z z i z z ?M ?K ?M ?M ?] }. Let z
1
, z
2
S ?? be such that
1 2
max min
z s z s
z z and z z
?? ???]?] .
Then
2
1 2
2z z ?M equals :
(1) 1 (2) 4
(3) 3 (4) 2
Ans. (4)
Sol. Let Z = x + iy
Then (x - 1)
2
+ y
2
= 1 ?? (1)
&
?H ?I ?H ?I 2 1 2 (2 ) 2 2
( 2 1) 2 (2)
x i iy
x y
?M ?M ?] ?? ?M ?K ?] ??
Solving (1) & (2) we get
Either x = 1 or
1
(3)
2 2
x?]??
?M
On solving (3) with (2) we get
For x = 1 ?? y = 1 ?? Z
2
= 1 + i
& for
1
1 1 1
2 1
2 2 2 2 2
i
x y Z
????
?] ?? ?] ?M ?? ?] ?K ?K????
?M ????
Now
?H ?I 2
1 2
2
2
2
1
1 2 (1 )
2
2
2
z z
i i
?M ????
?] ?K ?K ?M ?K ????
????
?] ?]
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
7 7
a b ?K ?K ?K ?K ?K ?M ?K ?M ?]
?? a + b = 300
Mean =
170 125 230 190 210
175
7
a b ?K ?K ?K ?K ?K ?K ?]
Mean deviation
About mean =
50 175 175 5 15 35 55
7
a b ?K ?M ?K ?M ?K ?K ?K ?K = 30
8. Let
ˆˆ ˆ ˆ ˆ ˆ
5 3 , 2 4 a i j k b i j k ?] ?M ?K ?M ?] ?K ?M and
?H ?I ?H ?I ?H ?I ˆˆˆ
. c a b i i i ?] ?? ?? ?? ?? Then
?H ?I ˆ ˆˆ
c i j k ?? ?M ?K ?K is
equal to
(1) –12 (2) –10
(3) –13 (4) –15
Ans. (1)
Sol.
ˆ ˆ
5 3 ?] ?M ?K ?M a i j k
ˆ ˆˆ
2 4 ?] ?K ?M b i j k
?H ?I ?H ?I ˆ ˆ ˆ
( ) ?? ?? ?] ?? ?M ?? a b i a i b b i a
5 ?] ?M ?M b a
?H ?I ?H ?I ?H ?I ˆˆ
5 ?] ?M ?M ?? ?? b a i i
?H ?I ?H ?I ˆ ˆ ˆ ˆ
11 23 ?] ?M ?K ?? ?? j k i i
?H ?I ˆ ˆˆ
11 23 ?? ?K ?? k j i
?H ?I ˆ ˆ
11 23 ???M j k
?H ?I ˆ ˆˆ
. 11 23 12 ?M ?K ?K ?] ?M ?] ?M c i j k
9. Let S = { :( 3 2) ( 3 2) 10}.
x x
x R ?? ?K ?K ?M ?]
Then the number of elements in S is :
(1) 4 (2) 0
(3) 2 (4) 1
Ans. (3)
Sol.
?H ?I ?H ?I x x
3 2 3 2 10 ?K ?K ?M ?]
Let
?H ?I x
3 2 t ?K?]
1
t 10
t
?K?]
t
2
– 10t + 1 = 0
10 100 4
t 5 2 6
2
???M
?] ?] ??
?H ?I ?H ?I x 2
3 2 3 2 ?K ?] ??
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 =
?H ?I 4
2
16
6
4
30 32ln2
x dx
x
?M???? ????
?M?M
???? ????
?K ???? ????
?]?M
??
11. Let f : R ?? R and g : R ?? R be defined as
f(x) =
e
x
log x , x 0
e , x 0
?M ?^ ?? ?? ?? ?? ?? ?? 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
f x
f x f x
e f x
?? ?? ?? ?\ ??
g(f(x)) =
?} ?H ?I ln
, ,0
,(0,1)
ln , 1,
x
x
e
e
x
?M ?? ?M??
?? ?? ?? ?? ?? ?? ?? ?? ??
(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
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024
(Held On Thursday 01
st
February, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
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
?? ?? ?K ?? ?K ?K ??
=
4 4
2 2
8
4
2 6 3 5 6 2
2 2 2 2 2 2
8 8 8
4 4 4
C C 1
5 C
C C C C C C 1 1 1
...
5 C 5 C 5 C
?? ?? ?? ?? ?? ?? ?K ?? ?K ?K ??
=
2
7
2. The value of the integral
4
4 4
0
:
sin (2 ) cos (2 )
xdx
equals
x x
?? ?K ??
(1)
2
2
8
?? (2)
2
2
16
??
(3)
2
2
32
?? (4)
2
2
64
??
Ans. (3)
Sol.
4
4 4
0
sin (2 ) cos (2 )
xdx
x x
?? ?K ??
Let 2x t ?] then
1
2
dx dt ?]
2
4 4
0
2
4 4
0
2
4 4
0
2
4 4
0
4 2
4
0
1
4 sin cos
1 2
4
sin cos
2 2
1
2
4 sin cos
2
8 sin cos
sec
2
8 tan 1
tdt
I
t t
t dt
I
t t
dt
I I
t t
dt
I
t t
tdt
I
t
?? ?? ?? ?? ?? ?? ????
?? ?? ?? ?] ?K ????
?M ????
????
?] ?? ?? ?? ?? ?M ?K ?M ?? ?? ?? ?? ?? ?? ?? ?? ?]?M
?K ?] ?K ?] ?K ?? ?? ?? ?? ??
Let tant = y then sec
2
t dt = dy
2
4
0
(1 )
2
8 1
?? ?K ?] ?K ?? y dy
I
y
??
2
2
0
2
1
1
y
dy
1
16
y
y
?? ?K ?? ?] ?K ??
Put
1
y p
y
?M?]
?H ?I 2
2
dp
I
16
p 2
?? ?M??
?? ?] ?K ??
1
p
tan
16 2 2
?? ?M ?M??
???? ?? ????
?] ???? ????
???? ????
2
16 2
?] I
??
3. If A =
2 1
1 2
????
????
?M????
????
, B =
1 0
1 1
????
????
????
, 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.
2 1
det( ) 3
1 2
1 0
det( ) 1
1 1
A A
B B
????
?] ?? ?] ????
?M????
????
????
?] ?? ?] ????
????
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 =
2 2
1
,tan
( 1) 1
?] ?K ?K ?K ?K x
B
x x x x x
and
?H ?I 1
3 2 1
2
tan ,0 , , ,
2
?M ?M ?M ?] ?K ?K ?\ ?\ C x x x A B C then
??
A + B is equal to :
(1) C
(2) C ?? ?M
(3) 2 C ?? ?M
(4)
2
C
?? ?M
Ans. (1)
Sol.
Finding tan (A + B) we get
?? tan (A + B) =
2 2
2
1
( 1) 1 tan tan
1
1 tan tan
1
1
x
x x x x x A B
A B
x x
?K ?K ?K ?K ?K ?K ?] ?M ?M ?K?K
?? tan (A + B) =
?H ?I ?H ?I ?H ?I ?H ?I 2
2
1 1 x x x
x x x
?K ?K ?K ?K
?H ?I ?H ?I ?H ?I ?H ?I 2
2
2
1 1
1
tan( ) tan
x x x
x x x
x x
A B C
x x
A B C
?K ?K ?K ?K ?K?K
?K ?] ?] ?K?]
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 ?? ?M ?] and
?H ?I ?H ?I ?H ?I 2 1 2 2 z z i z z ?M ?K ?M ?M ?] }. Let z
1
, z
2
S ?? be such that
1 2
max min
z s z s
z z and z z
?? ???]?] .
Then
2
1 2
2z z ?M equals :
(1) 1 (2) 4
(3) 3 (4) 2
Ans. (4)
Sol. Let Z = x + iy
Then (x - 1)
2
+ y
2
= 1 ?? (1)
&
?H ?I ?H ?I 2 1 2 (2 ) 2 2
( 2 1) 2 (2)
x i iy
x y
?M ?M ?] ?? ?M ?K ?] ??
Solving (1) & (2) we get
Either x = 1 or
1
(3)
2 2
x?]??
?M
On solving (3) with (2) we get
For x = 1 ?? y = 1 ?? Z
2
= 1 + i
& for
1
1 1 1
2 1
2 2 2 2 2
i
x y Z
????
?] ?? ?] ?M ?? ?] ?K ?K????
?M ????
Now
?H ?I 2
1 2
2
2
2
1
1 2 (1 )
2
2
2
z z
i i
?M ????
?] ?K ?K ?M ?K ????
????
?] ?]
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
7 7
a b ?K ?K ?K ?K ?K ?M ?K ?M ?]
?? a + b = 300
Mean =
170 125 230 190 210
175
7
a b ?K ?K ?K ?K ?K ?K ?]
Mean deviation
About mean =
50 175 175 5 15 35 55
7
a b ?K ?M ?K ?M ?K ?K ?K ?K = 30
8. Let
ˆˆ ˆ ˆ ˆ ˆ
5 3 , 2 4 a i j k b i j k ?] ?M ?K ?M ?] ?K ?M and
?H ?I ?H ?I ?H ?I ˆˆˆ
. c a b i i i ?] ?? ?? ?? ?? Then
?H ?I ˆ ˆˆ
c i j k ?? ?M ?K ?K is
equal to
(1) –12 (2) –10
(3) –13 (4) –15
Ans. (1)
Sol.
ˆ ˆ
5 3 ?] ?M ?K ?M a i j k
ˆ ˆˆ
2 4 ?] ?K ?M b i j k
?H ?I ?H ?I ˆ ˆ ˆ
( ) ?? ?? ?] ?? ?M ?? a b i a i b b i a
5 ?] ?M ?M b a
?H ?I ?H ?I ?H ?I ˆˆ
5 ?] ?M ?M ?? ?? b a i i
?H ?I ?H ?I ˆ ˆ ˆ ˆ
11 23 ?] ?M ?K ?? ?? j k i i
?H ?I ˆ ˆˆ
11 23 ?? ?K ?? k j i
?H ?I ˆ ˆ
11 23 ???M j k
?H ?I ˆ ˆˆ
. 11 23 12 ?M ?K ?K ?] ?M ?] ?M c i j k
9. Let S = { :( 3 2) ( 3 2) 10}.
x x
x R ?? ?K ?K ?M ?]
Then the number of elements in S is :
(1) 4 (2) 0
(3) 2 (4) 1
Ans. (3)
Sol.
?H ?I ?H ?I x x
3 2 3 2 10 ?K ?K ?M ?]
Let
?H ?I x
3 2 t ?K?]
1
t 10
t
?K?]
t
2
– 10t + 1 = 0
10 100 4
t 5 2 6
2
???M
?] ?] ??
?H ?I ?H ?I x 2
3 2 3 2 ?K ?] ??
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 =
?H ?I 4
2
16
6
4
30 32ln2
x dx
x
?M???? ????
?M?M
???? ????
?K ???? ????
?]?M
??
11. Let f : R ?? R and g : R ?? R be defined as
f(x) =
e
x
log x , x 0
e , x 0
?M ?^ ?? ?? ?? ?? ?? ?? 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
f x
f x f x
e f x
?? ?? ?? ?\ ??
g(f(x)) =
?} ?H ?I ln
, ,0
,(0,1)
ln , 1,
x
x
e
e
x
?M ?? ?M??
?? ?? ?? ?? ?? ?? ?? ?? ??
(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
2 1
13 1 16
, , 70,
19 19 13
k k ????
?M?M
?] ?] ?] ?M ?]
13 ?? ?? = 13 (-70)
16
13
1120
?M????
????
????
?]
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
?? ?? ?]?K
?]?M
7
h c
e e ?]
2 2
2
1 sin 7(1 sin )
6 3
sin
8 4
3
sin
2
3
????
?? ?? ?? ?? ?K ?] ?M ?]?]
?] ?]
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
1 1
y
2 2
???? ????
?K ???? ????
???? ????
equals :
(1)
4
4 e ?K (2)
3
3 e ?M
(3)
2
1 e ?K (4)
1
2 e ?M
Ans. (4)
Sol.
3
2 ( ) ( ) 1 ?] ?K ?M ?K ?M dy
x x y x x y
dx
?K?] x y t
3
1 2 1 ?M ?] ?M ?M dt
xt xt
dx
3
2
?] ?M dt
xdx
t t
4 2
tdt
xdx
2t t
?] ?M
Let
2
?] t z
?H ?I 2
dz
xdx
2 2z z
?] ?M ????
1
4
2
?] ????
?M ????
????
????
dz
xdx
z z
2
1
2
ln
?M ?]?K
z
x k
z
1
2
?] ?M 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
?M ?? ?\ ?? ?? ?? ?K ?K ?? ?? ?? ???K?^
?? ?? ??
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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Additional Information about JEE Main 2024 February 1 Shift 1 Paper & Solutions for JEE Preparation
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