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FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Thursday 01st February, 2024) TIME : 9 : 00 AM to 12 : 00 NOON

MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A 
4
xdx
1. A bag contains 8 balls, whose colours are either Sol.  sin 4
(2 x)  cos 4 (2 x)
white or black. 4 balls are drawn at random 0

without replacement and it was found that 2 balls 1
Let 2x  t then dx  dt
are white and other 2 balls are black. The 2

probability that the bag contains equal number of
12 tdt
white and black balls is: I  4
4 0 sin t  cos 4 t
2 2
(1) (2)   
5 7 2   t  dt
1
I  2 
(3)
1
(4)
1 40    
7 5 sin 4   t   cos 4   t 
2  2 
Ans. (2)  
2 dt
Sol. 1
4 0 sin 4 t  cos 4 t
I 2 I
P(4W4B/2W2B) =
P(4W 4 B)  P(2W 2 B / 4W 4 B) 
P(2W 6 B)  P(2W 2 B / 2W 6 B)  P(3W 5 B)  P(2W 2 B / 3W 5 B)  2
dt
8  sin
.............  P(6W 2 B)  P(2W 2 B / 6W 2 B) 2I 
0
4
t  cos 4 t
1 4 C2  4C2 
 8
5 C4  2
sec 4 tdt
8 0 tan 4 t  1
= 2I 
1 C2  C2 1 C2  5C2
2 6 3
1 6 C  2C2
 8   8  ...   28
5 C4 5 C4 5 C4 Let tant = y then sec2t dt = dy
  (1  y 2 )dy
=
2 2I 
8  1 y4
7 0

1
 1
 y2
2. The value of the integral   dy
 16 0 y 2  1
4
xdx y2
 sin 4
(2 x)  cos 4 (2 x)
equals :
1
0 Put y   p
y
2 2 2 2 
(1) (2)  dp
8 16 I 
 
2
16  p2  2
2 2
2 2
(3) (4) 
32 64  1  p  

  tan  
Ans. (3) 16 2   2   
2
I
16 2

,  2 1  Sol.
1 0
3. If A =  , B =   , C = ABAT and X Finding tan (A + B) we get
 1 2  1 1  tan (A + B) =
= ATC2A, then det X is equal to : 1 x

(1) 243 tan A  tan B x( x  x  1)
2
x  x 1
2

(2) 729 1  tan A tan B 1
1
x  x 1
2
(3) 27
(4) 891
 tan (A + B) =
1  x   x2  x  1 
Ans. (2) x 2
x  x 
1  x   
Sol.
x2  x  1
 2 1 
A   det( A)  3 x 2
x  x 
 1 2 
1 0  x2  x  1
tan( A  B)   tan C
B   det( B)  1 x x
1 1 
A B  C
Now C = ABAT  det(C) = (dct (A))2 x det(B)

C 9 5. If n is the number of ways five different employees
can sit into four indistinguishable offices where
Now |X| = |ATC2A| any office may have any number of persons
= |AT| |C|2 |A| including zero, then n is equal to:
= |A|2 |C|2 (1) 47
(2) 53
= 9 x 81
(3) 51
= 729 (4) 43
1 x Ans. (3)
4. If tanA = , tan B  Sol.
x( x 2  x  1) x2  x  1
Total ways to partition 5 into 4 parts are :
and 5, 0, 0, 0  1 way
5!
1
 4, 1, 0, 0   5 ways

tan C  x  x  x3 2 1 2
 , 0  A, B, C 
2
, then 4!
5!
3, 2, 0, 0,   10 ways
A + B is equal to : 3!2!
(1) C 5!
2, 2,0,1   15 ways
(2)   C 2!2!2!
5!
(3) 2  C 2,1,1,1   10 ways
2!(1!)3 3!

(4) C 5!
2 3,1,1,0   10 ways
3!2!
Ans. (1) Total  1+5+10+15+10+10 = 51 ways

, 6. LetS={ z  C : z  1  1 and Sol. Median = 170  125, a, b, 170, 190, 210, 230

 
Mean deviation about
2  1  z  z   i  z  z   2 2 }. Let z1, z2
Median =
 S be such that z1  max z and z2  min z . 0  45  60  20  40  170  a  170  b 205
zs zs

2 7 7
Then 2z1  z2 equals :
 a + b = 300
(1) 1 (2) 4
(3) 3 (4) 2 Mean = 170  125  230  190  210  a  b  175
7
Ans. (4)
Mean deviation
Sol. Let Z = x + iy
Then (x - 1)2 + y2 = 1  (1) About mean =


&
 
2  1  2 x   i(2iy)  2 2
50  175  a  175  b  5  15  35  55
7
= 30

 ( 2  1) x  y  2  (2)
8. Let a  5iˆ  ˆj  3kˆ, b  iˆ  2 ˆj  4kˆ and
Solving (1) & (2) we get

Either x = 1 or x 
1
 (3)
c  a  b   iˆ  iˆ  iˆ. Then c   iˆ  ˆj  kˆ  is
2 2
equal to
On solving (3) with (2) we get
For x = 1  y = 1  Z2 = 1 + i
& for (1) –12 (2) –10
1 1  1  i (3) –13 (4) –15
x  y 2  Z1  1  
2 2 2  2 2 Ans. (1)
Now
2
Sol. a  5iˆ  j  3kˆ
2 z1  z2
b  iˆ  2 ˆj  4kˆ
2
 1 
  1 2  i  (1  i)
 2 
   
(a  b )  iˆ  a  iˆ b  b  iˆ a


 2  5b  a
2

2   5b  a   iˆ  iˆ
7. Let the median and the mean deviation about the
median of 7 observation 170, 125, 230, 190, 210, a, b    11 ˆj  23kˆ   iˆ   iˆ
205
be 170 and respectively. Then the mean
7 
 11kˆ  23 ˆj  iˆ 
deviation about the mean of these 7 observations is :
(1) 31
(2) 28

 11 ˆj  23kˆ 
(3) 30 c .  iˆ  ˆj  kˆ   11  23  12
(4) 32
Ans. (3)