(IIT)(JEE)(Engineering)

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) x9 (2)  y9
32 32 3. If z = x + iy, xy  0 , satisfies the equation

(3)
x
(4)
y z 2  i z  0 , then |z2| is equal to :
 y9 x9
32 32 (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) cos1  
 2 
S10 
10
 2a  9d   145  
2
2
2a + 9d = 29 …..(2) (4) cos1  
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  412  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 33  
 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     cos1  (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