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Paper of JEE exam which was held on 30th JAN 2024 1st session
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FINAL JEE–MAIN EXAMINATION – JANUARY, 2024(Held On Tuesday 30th January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A 15 5
S15 S5 2a 14d 2a 4d
1. A line passing through the point A(9, 0) makes an angle 2 2
of 30º with the positive direction of x-axis. If this line is 15 5
16 70 16 20
rotated about A through an angle of 15º in the clockwise 2 2
direction, then its equation in the new position is = 405-10
y x = 395
(1) x9 (2) y9
32 32 3. If z = x + iy, xy 0 , satisfies the equation
(3)
x
(4)
y z 2 i z 0 , then |z2| is equal to :
y9 x9
32 32 (1) 9
Ans. (1) (2) 1
(3) 4
1
Sol. (4)
4
Ans. (2)
Sol. z 2 iz
z 2 iz
|z2| = |z|
|z|2 – |z| = 0
|z|(|z| – 1) = 0
|z| = 0 (not acceptable)
Eqn : y – 0 = tan15° (x – 9) y = (2 3 (x 9) |z| = 1
|z|2 = 1
2. Let Sa denote the sum of first n terms an arithmetic
4. Let a a i ˆi a 2ˆj a 3kˆ and b b1ˆi b2ˆj b3kˆ be
progression. If S20 = 790 and S10 = 145, then S15 –
S5 is : two vectors such that a 1; a.b 2 and b 4. If
(1) 395
(2) 390
c 2 a b 3b , then the angle between b and c
is equal to :
(3) 405
2
(4) 410 (1) cos 1
3
Ans. (1)
1
20 (2) cos 1
Sol. S20 2a 19d 790
3
2
2a + 19d = 79 …..(1) 3
(3) cos1
2
S10
10
2a 9d 145
2
2
2a + 9d = 29 …..(2) (4) cos1
3
From (1) and (2) a = -8, d = 5
Ans. (3)
,Sol. Given a 1, b 4, a.b 2 (0, 0)
Sol.
c 2 a b 3b
Dot product with a on both sides
c.a 6 …..(1) (x, y) (-x, y)
Dot product with b on both sides Area of
b.c 48 …..(2) 0 0 1
1
2 2 x y 1
c.c 4 a b 9 b 2
x y 1
2 2
2 2 2
c 4 a b a.b 9 b 1
xy xy xy
2
c 4 1 4 4 9 16
2 2
Area xy x 2x 2 54
c 412 144
2
d
2
dx
6x 2 54 d
dx
0 at x = 3
c 48 144
Area = 3 (-2 × 9 + 54) = 108
2
c 192 n
n3
6. The value of lim is :
cos
b.c n k 1
n 2 k n 2 3k 2
b c
48 (1)
2 33
cos 24
192.4
13
48 (2)
cos
8 3.4
8 4 3 3
cos
3
(3)
13 2 3 3
2 3 8
3 3
cos cos1 (4)
2
2
8 2 3 3
5. The maximum area of a triangle whose one vertex Ans. (2)
is at (0, 0) and the other two vertices lie on the n
n3
curve y = -2x2 + 54 at points (x, y) and (-x, y) Sol. lim
n k 1 4 k 2 3k 2
where y > 0 is : n 1 2 1 2
n n
(1) 88
(2) 122 1 n n3
lim
n n k 1
(3) 92 k 2 3k 2
1 2 1 2
(4) 108 n n
Ans. (4) 1
dx
2 1 2
0 3 1 x x
3
, 1
1 3 x 2 1 x 2
1
3
8. The area (in square units) of the region bounded by
the parabola y2 = 4(x – 2) and the line y = 2x - 8
dx
0
3 2
3
1 x 2 2 1
x (1) 8
(2) 9
(3) 6
1 (4) 7
1 1 1
2 2 1 2
dx Ans. (2)
0 1 x2
x Sol. Let X = x – 2
3
y2 = 4x, y = 2 (x + 2) – 8
1 1
1 1 2
3 tan 1 3x tan 1 x y = 4x, y = 2x – 4
2 0 2 0
4
y2 y 4
3 1
A 4
2
2 3 2 4 2 3 8 2
13 4
8. 4 3 3
7. Let g : R R be a non constant twice
-2
1 3
differentiable such that g' g' . If a real
2 2
valued function f is defined as
=9
1
f x g x g 2 x , then 9. Let y = y (x) be the solution of the differential
2
equation sec x dy + {2(1 – x) tan x + x(2 – x)}
(1) f”(x) = 0 for atleast two x in (0, 2)
dx = 0 such that y(0) = 2.Then y(2) is equal to :
(2) f”(x) = 0 for exactly one x in (0, 1)
(1) 2
(3) f”(x) = 0 for no x in (0, 1)
(2) 2{1 – sin (2)}
3 1 (3) 2{sin (2) + 1}
(4) f ' f ' 1
2 2 (4) 1
Ans. (1) Ans. (1)
3 1
Sol. f ' x
g ' x g ' 2 x 3
g ' g '
,f '
2 2 0
Sol.
dy
dx
2 x 1 sin x x 2 2x cos x
2
2 2
Now both side integrate
1 3
y x 2 x 1 sin x dx x 2 2x sin x 2x 2 sin x dx
g ' g '
1 2 0, f ' 1 0
Also f '
2
2 2
2
y x x 2 2x sin x
3 1 y 0 0 2
f ' f ' 0
2 2
1 3
rootsin ,1 and 1,
y x x 2 2x sin x 2
2 2 y 2 2
1 3
f " x is zero at least twice in ,
2 2
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