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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 x3 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
zs zs
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)
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