(IIT)(JEE)(Engineering)

FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Tuesday 30th January, 2024) TIME : 3 : 00 PM to 6 : 00 PM


MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A Sol. 3sin cos + 3sin cos
1. Consider the system of linear equations = 2sin cos – 2sin cos
x + y + z = 5, x + 2y +2z = 9, 5sin cos = –sin cos
x + 3y +z = , where ,   R. Then, which of 1
tan    tan 
the following statement is NOT correct? 5
(1) System has infinite number of solution if 1 tan = –5tan
and  =13 3. Let A(, 0) and B(0, ) be the points on the line
(2) System is inconsistent if 1 and   13 5x + 7y = 50. Let the point P divide the line
(3) System is consistent if 1 and  13 segment AB internally in the ratio 7 : 3. Let 3x –
(4) System has unique solution if 1 and   13 x2 y2
25 = 0 be a directrix of the ellipse E :  1
Ans. (4) a 2 b2
1 1 1 and the corresponding focus be S. If from S, the
Sol. 1 2  0 2
perpendicular on the x-axis passes through P, then
1 3  the length of the latus rectum of E is equal to
 22 –  – 1 = 0 25 32
(1) (2)
1 3 9
 1, 
2 25 32
(3) (4)
1 1 5 9 5
2  2
9  0    13 Ans. (4 )
3   A  (10, 0) 

Infinite solution  = 1 &  = 13 Sol.  50   P  (3, 5)
B   0, 
For unique sol  1 n
 7 
For no soln  = 1 &   13
If   1 and  13
1
Considering the case when    and   13 this
2 S
will generate no solution case
25
  x=
2. For ,   0, , let 3sin ( )  2sin( ) and a 3
 2 
ae = 3
real number k be such that tan   k tan  . Then the
a 25
value of k is equal to : 
e 3
2
(1)  (2) –5 a=5
3
b=4
2
(3) (4) 5 2b2 32
3 Length of LR  
Ans. (2 ) a 5

,4. Let a  ˆi ˆj  k,
ˆ  , R . Let a vector b be such 6. Let a and b be be two distinct positive real
 2 numbers. Let 11th term of a GP, whose first term is
that the angle between a and b is and b  6 ,
4 a and third term is b, is equal to pth term of another
If a.b  3 2 , then the value of  2 2  a  b
2
is GP, whose first term is a and fifth term is b. Then p

equal to is equal to
(1) 90 (2) 75 (1) 20 (2) 25
(3) 95 (4) 85 (3) 21 (4) 24
Ans. (1) Ans. (3)
  
Sol. | b |2  6 ; | a || b | cos   3 2 b
Sol. 1st GP  t1 = a, t3 = b = ar2  r2 =
 
a
| a |2 | b |2 cos2   18 5
b
 t11 = ar = a(r ) = a   
10 2 5

| a |2  6 a
Also 1 + 2 +  2 = 6 2nd G.P.  T1 = a, T5 = ar4 = b
2 + 2 = 5 b b
1/4

 r  r  
4
to find a a
 
(2 +  2) | a |2 | b |2 sin 2  p 1
b 4
Tp = ar p –1
 a 
1 a
= (5)(6)(6)  
2 p 1
5
= 90 b b 4
t11  Tp  a    a  
5. Let f(x)  (x  3)2 (x  2)3 , x  [ 4, 4] . If M and m are a a

the maximum and minimum values of f, p 1
 5  p  21
respectively in [–4, 4], then the value of M – m is : 4
(1) 600 (2) 392 7. If x2 – y2 + 2hxy + 2gx + 2fy + c = 0 is the locus of
(3) 608 (4) 108 a point, which moves such that it is always
Ans. (3) equidistant from the lines x + 2y + 7 = 0 and 2x – y
2 2 3
Sol. f'(x) = (x + 3) . 3(x – 2) + (x –2) 2(x + 3) + 8 = 0, then the value of g + c + h – f equals
2
= 5(x + 3) (x – 2) (x + 1) (1) 14 (2) 6
f'(x) = 0, x = –3, –1, 2
(3) 8 (4) 29
Ans. (1)
+ – + + Sol. Cocus of point P(x, y) whose distance from
–3 –1 2 Gives
f(–4) = –216
X + 2y + 7 = 0 & 2x – y + 8 = 0 are equal is
f(–3) = 0, f(4) = 49 × 8 = 392
x  2y  7 2x  y  8
M = 392, m = –216 
5 5
M – m = 392 + 216 = 608
Ans = '3' (x + 2y + 7)2 – (2x – y + 8)2 = 0

, Combined equation of lines dy
Sol. y = f(x)   f '(x)
(x – 3y + 1) (3x + y + 15) = 0 dx
3x2 – 3y2 – 8xy + 18x – 44y + 15 = 0 dy   1 1
8 44   f '(1)  tan   f '(1) 
x2 – y2 – xy  6x  y  5  0 dx (1,f (1)) 6 3 3
3 3
2 2
x – y + 2h xy + 2gx 2 + 2fy + c = 0 dy  
  f '(3)  tan  1  f '(3)  1
dx (3,f (3)) 4
4 22
h  , g  3, f   , c  5
3 3
 
3
27   f '(t)   1 f "(t)dt     3
2
4 22
gc h f 35   8  6  14 1
3 3
 
3
I    f '(t)   1 f "(t)dt
2
8. Let a and b be two vectors such that
1
2
| b |  1 and | b  a | 2 . Then (b  a)  b is equal
f'(t) = z  f"(t) dt = dz
to z = f'(3) = 1
(1) 3
1
(2) 5 z = f'(1) =
3
(3) 1
1
(4) 4 1
 z3 
I   (z  1)dz    z 
2
Ans. (2) 3 1/
1/ 3 3
  
Sol. | b | 1 & | b a | 2 1  1 1 1 
   1     
        3  3 3 3 3
 b  a   b  b  b  a   0
    4 10 4 10

2

2

2     3
   3 9 3 3 27
(b  a )  b  b  a  b
 4 10 
=4+1=5    3  27   3   36  10 3
 3 27 
9. Let y  f(x) be a thrice differentiable function in
= 36,  = – 10
(–5, 5). Let the tangents to the curve y=f(x) at
 +  = 36 – 10 = 26
 
(1, f(1)) and (3, f(3)) make angles and ,
6 4 x2 y2
10. Let P be a point on the hyperbola H :  1,
respectively with positive x-axis. If 9 4

 
3
in the first quadrant such that the area of triangle
27  f (t)   1 f (t)dt   3
2
where , are
1
formed by P and the two foci of H is 2 13 . Then,
integers, then the value of  +  equals
the square of the distance of P from the origin is
(1) –14
(1) 18
(2) 26
(3) –16 (2) 26
(4) 36 (3) 22
Ans. (2) (4) 20
Ans. (3)