JEE Main 2024 April 8 Shift 1 Paper & Solutions | JEE Main & Advanced Mock Test Series 2025 PDF Download

 Page 1


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Monday 08
th
 April, 2024)    TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. The value of k ?? ???? for which the integral
I
n
 = 
1
k n
0
(1 – x ) dx,
?? n ???@ ?? , satisfies 147 I
20
= 148 I
21
is :  
(1) 10 (2) 8
(3) 14 (4) 7
Ans. (4) 
Sol. I
n
 = 
1
k n
0
(1 x ) .1 dx ?M ?? I
n
 = (1 – x
k
)
n
.x – nk
1
k n 1 k 1
0
(1 x ) .x .dx
?M?M
?M ?? I
n
 = nk
1
k n k n 1
0
[(1 x ) (1 x ) ]dx
?M ?M ?M ?M ?? I
n
 = nkI
n
 – nkI
n
  
n
n 1
I nk
I nk 1
?M ?] ?K 21
20
I 21k
I 1 21k
?] ?K 147
148
?] ???@?@ k = 7
2. The sum of all the solutions of the equation
(8)
2x
 – 16 ?? (8)
x
 + 48 = 0 is :
(1) 1 + log
6
(8) (2) log
8
(6)
(3) 1 + log
8
(6) (4) log
8
(4)
Ans. (3) 
Sol. (8)
2x
 – 16 ?? (8)
x
 + 48 = 0 
Put 8
x
 = t 
t
2
 – 16 + 48 = 0 
?? t = 4 or t = 12
?? 8
x
 = 4 8
x
 = 12 
?? x = log
8
x x = log
8
12 
sum of solution = log
8
4 + log
8
12 
= log
8
48 = log
8
(6.8) 
= 1 + log
8
6 
3. Let the circles C
1
 : (x – ?? )
2
 + (y – ?? )
2
 =
2
1
r and 
C
2
 : (x – 8)
2
 +
2
15
y
2
????
?M????
????
 = 
2
2
r touch each other
externally at the point (6, 6). If the point (6, 6) 
divides the line segment joining the centres of the 
circles C
1
 and C
2
 internally in the ratio 2 : 1, then 
( ?? + ?? ) + 4 
?H ?I 2 2
1 2
r r ?K equals  
(1) 110 (2) 130
(3) 125 (4) 145
Ans. (2) 
Sol. 
r
1 
r
2
C
2
C
1
( ?? ?L?? ) 
P
2 : 1
 
C
2
( ?? ?L ?? ) C
1 P 
(6, 6) 
16
6
3
?K??
?] and
15
6
3
?K??
?] ?? ( ?? , ?? ) ?? (2, 3)
Also, C
1
C
2
 = r
1 
+
 
r
2
 
???@
2
2
2 2
15
(2 8) 3 2r r
2
????
?M ?K ?M ?] ?K ????
????
?? r
2
 =
5
2
?? r
1
 = 2r
2
 = 5
?| ( ?? ?@?K ?@ ?? ) + 4
2 2
1 2
(r r ) ?K = 5 + 4
25
25
4
????
?K ????
????
 = 130 ?@
Page 2


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Monday 08
th
 April, 2024)    TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. The value of k ?? ???? for which the integral
I
n
 = 
1
k n
0
(1 – x ) dx,
?? n ???@ ?? , satisfies 147 I
20
= 148 I
21
is :  
(1) 10 (2) 8
(3) 14 (4) 7
Ans. (4) 
Sol. I
n
 = 
1
k n
0
(1 x ) .1 dx ?M ?? I
n
 = (1 – x
k
)
n
.x – nk
1
k n 1 k 1
0
(1 x ) .x .dx
?M?M
?M ?? I
n
 = nk
1
k n k n 1
0
[(1 x ) (1 x ) ]dx
?M ?M ?M ?M ?? I
n
 = nkI
n
 – nkI
n
  
n
n 1
I nk
I nk 1
?M ?] ?K 21
20
I 21k
I 1 21k
?] ?K 147
148
?] ???@?@ k = 7
2. The sum of all the solutions of the equation
(8)
2x
 – 16 ?? (8)
x
 + 48 = 0 is :
(1) 1 + log
6
(8) (2) log
8
(6)
(3) 1 + log
8
(6) (4) log
8
(4)
Ans. (3) 
Sol. (8)
2x
 – 16 ?? (8)
x
 + 48 = 0 
Put 8
x
 = t 
t
2
 – 16 + 48 = 0 
?? t = 4 or t = 12
?? 8
x
 = 4 8
x
 = 12 
?? x = log
8
x x = log
8
12 
sum of solution = log
8
4 + log
8
12 
= log
8
48 = log
8
(6.8) 
= 1 + log
8
6 
3. Let the circles C
1
 : (x – ?? )
2
 + (y – ?? )
2
 =
2
1
r and 
C
2
 : (x – 8)
2
 +
2
15
y
2
????
?M????
????
 = 
2
2
r touch each other
externally at the point (6, 6). If the point (6, 6) 
divides the line segment joining the centres of the 
circles C
1
 and C
2
 internally in the ratio 2 : 1, then 
( ?? + ?? ) + 4 
?H ?I 2 2
1 2
r r ?K equals  
(1) 110 (2) 130
(3) 125 (4) 145
Ans. (2) 
Sol. 
r
1 
r
2
C
2
C
1
( ?? ?L?? ) 
P
2 : 1
 
C
2
( ?? ?L ?? ) C
1 P 
(6, 6) 
16
6
3
?K??
?] and
15
6
3
?K??
?] ?? ( ?? , ?? ) ?? (2, 3)
Also, C
1
C
2
 = r
1 
+
 
r
2
 
???@
2
2
2 2
15
(2 8) 3 2r r
2
????
?M ?K ?M ?] ?K ????
????
?? r
2
 =
5
2
?? r
1
 = 2r
2
 = 5
?| ( ?? ?@?K ?@ ?? ) + 4
2 2
1 2
(r r ) ?K = 5 + 4
25
25
4
????
?K ????
????
 = 130 ?@ 4.  Let P(x, y, z) be a point in the first octant, whose 
projection in the xy-plane is the point Q. Let  
OP = ?? ; the angle between OQ and the positive  
x-axis be ?? ; and the angle between OP and the 
positive z-axis be ?? , where O is the origin. Then 
the distance of P from the x-axis is :      
 (1) 
2 2
1 sin cos ?? ?M ?? ?? (2) 
2 2
1 cos sin ?? ?K ?? ?? 
 (3) 
2 2
1 sin cos ?? ?M ?? ?? (4) 
2 2
1 cos sin ?? ?K ?? ?? 
 Ans. (1) 
Sol. P(x, y, z), Q(x, y, O) ; x
2
 + y
2
 + z
2
= ?? 2
  
 
ˆˆ
OQ xi yj ?]?K  
 cos ?? = 
2 2
x
x y ?K 
 cos ?? = 
2 2 2
z
x y z ?K?K
  
 ???@ sin
2
???@ = 
2 2
2 2 2
x y
x y z
?K ?K?K
 
 distance of P from x-axis 
2 2
y z ?K 
 ???@
2 2
x ???M  ???@ 
2
2
x
1 ???M
??  
 = 
2 2
1 cos sin ?? ?M ?? ?? ?@ 
5. The number of critical points of the function  
f(x) = (x – 2)
2/3
 (2x + 1) is : 
 (1) 2 (2) 0 
 (3) 1  (4) 3 
 Ans. (1) 
Sol. f(x) = (x – 2)
2/3
 (2x + 1)  
 f'(x) = 
2
3
(x – 2)
–1/3
 (2x + 1) + (x – 2)
2/3
 (2) 
 f'(x) = 2 × 
1/3
(2x 1) (x 2)
3(x 2)
?K ?K ?M ?M 
 
1/3
3x 1
0
(x 2)
?M ?] ?M 
 Critical points x = 
1
3
 and x = 2  
 
6.  Let f(x) be a positive function such that the area 
bounded by y = f(x), y = 0 from x = 0 to x = a > 0 
is e
–a
 + 4a
2
 + a – 1. Then the differential equation, 
whose general solution is y = c
1
f(x) + c
2
 , where c
1
 
and c
2
 are arbitrary constants, is : 
 (1) 
2
x
2
d y dy
(8e 1) 0
dx dx
?M ?K ?] 
 (2) 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?M ?] 
 (3) 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?K ?] 
 (4) 
2
x
2
d y dy
(8e 1) 0
dx dx
?M ?M ?] 
 Ans. (3) 
Sol. 
a
a 2
0
f(x)dx e 4a a 1
?M ?] ?K ?K ?M ?? 
 f(a) = –e
–a
 + 8a + 1 
 f(x) = –e
–x
 + 8x + 1  
 Now y = C
1
f(x) + C
2
  
 
1
dy
C f '(x)
dx
?] = C
1
(e
–x 
+ 8) …..(1) 
 
2
2
d y
dx
 = –C
1
e
–x
  ?? –e
x
 
2
2
d y
dx
 
 Put in equation (1) 
 
2
x x
2
dy d y
e (e 8)
dx dx
?M ?] ?M ?K 
 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?K ?] 
 
Page 3


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Monday 08
th
 April, 2024)    TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. The value of k ?? ???? for which the integral
I
n
 = 
1
k n
0
(1 – x ) dx,
?? n ???@ ?? , satisfies 147 I
20
= 148 I
21
is :  
(1) 10 (2) 8
(3) 14 (4) 7
Ans. (4) 
Sol. I
n
 = 
1
k n
0
(1 x ) .1 dx ?M ?? I
n
 = (1 – x
k
)
n
.x – nk
1
k n 1 k 1
0
(1 x ) .x .dx
?M?M
?M ?? I
n
 = nk
1
k n k n 1
0
[(1 x ) (1 x ) ]dx
?M ?M ?M ?M ?? I
n
 = nkI
n
 – nkI
n
  
n
n 1
I nk
I nk 1
?M ?] ?K 21
20
I 21k
I 1 21k
?] ?K 147
148
?] ???@?@ k = 7
2. The sum of all the solutions of the equation
(8)
2x
 – 16 ?? (8)
x
 + 48 = 0 is :
(1) 1 + log
6
(8) (2) log
8
(6)
(3) 1 + log
8
(6) (4) log
8
(4)
Ans. (3) 
Sol. (8)
2x
 – 16 ?? (8)
x
 + 48 = 0 
Put 8
x
 = t 
t
2
 – 16 + 48 = 0 
?? t = 4 or t = 12
?? 8
x
 = 4 8
x
 = 12 
?? x = log
8
x x = log
8
12 
sum of solution = log
8
4 + log
8
12 
= log
8
48 = log
8
(6.8) 
= 1 + log
8
6 
3. Let the circles C
1
 : (x – ?? )
2
 + (y – ?? )
2
 =
2
1
r and 
C
2
 : (x – 8)
2
 +
2
15
y
2
????
?M????
????
 = 
2
2
r touch each other
externally at the point (6, 6). If the point (6, 6) 
divides the line segment joining the centres of the 
circles C
1
 and C
2
 internally in the ratio 2 : 1, then 
( ?? + ?? ) + 4 
?H ?I 2 2
1 2
r r ?K equals  
(1) 110 (2) 130
(3) 125 (4) 145
Ans. (2) 
Sol. 
r
1 
r
2
C
2
C
1
( ?? ?L?? ) 
P
2 : 1
 
C
2
( ?? ?L ?? ) C
1 P 
(6, 6) 
16
6
3
?K??
?] and
15
6
3
?K??
?] ?? ( ?? , ?? ) ?? (2, 3)
Also, C
1
C
2
 = r
1 
+
 
r
2
 
???@
2
2
2 2
15
(2 8) 3 2r r
2
????
?M ?K ?M ?] ?K ????
????
?? r
2
 =
5
2
?? r
1
 = 2r
2
 = 5
?| ( ?? ?@?K ?@ ?? ) + 4
2 2
1 2
(r r ) ?K = 5 + 4
25
25
4
????
?K ????
????
 = 130 ?@ 4.  Let P(x, y, z) be a point in the first octant, whose 
projection in the xy-plane is the point Q. Let  
OP = ?? ; the angle between OQ and the positive  
x-axis be ?? ; and the angle between OP and the 
positive z-axis be ?? , where O is the origin. Then 
the distance of P from the x-axis is :      
 (1) 
2 2
1 sin cos ?? ?M ?? ?? (2) 
2 2
1 cos sin ?? ?K ?? ?? 
 (3) 
2 2
1 sin cos ?? ?M ?? ?? (4) 
2 2
1 cos sin ?? ?K ?? ?? 
 Ans. (1) 
Sol. P(x, y, z), Q(x, y, O) ; x
2
 + y
2
 + z
2
= ?? 2
  
 
ˆˆ
OQ xi yj ?]?K  
 cos ?? = 
2 2
x
x y ?K 
 cos ?? = 
2 2 2
z
x y z ?K?K
  
 ???@ sin
2
???@ = 
2 2
2 2 2
x y
x y z
?K ?K?K
 
 distance of P from x-axis 
2 2
y z ?K 
 ???@
2 2
x ???M  ???@ 
2
2
x
1 ???M
??  
 = 
2 2
1 cos sin ?? ?M ?? ?? ?@ 
5. The number of critical points of the function  
f(x) = (x – 2)
2/3
 (2x + 1) is : 
 (1) 2 (2) 0 
 (3) 1  (4) 3 
 Ans. (1) 
Sol. f(x) = (x – 2)
2/3
 (2x + 1)  
 f'(x) = 
2
3
(x – 2)
–1/3
 (2x + 1) + (x – 2)
2/3
 (2) 
 f'(x) = 2 × 
1/3
(2x 1) (x 2)
3(x 2)
?K ?K ?M ?M 
 
1/3
3x 1
0
(x 2)
?M ?] ?M 
 Critical points x = 
1
3
 and x = 2  
 
6.  Let f(x) be a positive function such that the area 
bounded by y = f(x), y = 0 from x = 0 to x = a > 0 
is e
–a
 + 4a
2
 + a – 1. Then the differential equation, 
whose general solution is y = c
1
f(x) + c
2
 , where c
1
 
and c
2
 are arbitrary constants, is : 
 (1) 
2
x
2
d y dy
(8e 1) 0
dx dx
?M ?K ?] 
 (2) 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?M ?] 
 (3) 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?K ?] 
 (4) 
2
x
2
d y dy
(8e 1) 0
dx dx
?M ?M ?] 
 Ans. (3) 
Sol. 
a
a 2
0
f(x)dx e 4a a 1
?M ?] ?K ?K ?M ?? 
 f(a) = –e
–a
 + 8a + 1 
 f(x) = –e
–x
 + 8x + 1  
 Now y = C
1
f(x) + C
2
  
 
1
dy
C f '(x)
dx
?] = C
1
(e
–x 
+ 8) …..(1) 
 
2
2
d y
dx
 = –C
1
e
–x
  ?? –e
x
 
2
2
d y
dx
 
 Put in equation (1) 
 
2
x x
2
dy d y
e (e 8)
dx dx
?M ?] ?M ?K 
 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?K ?] 
 
 
7. Let f(x) = 4cos
3
x + 3 3 cos
2
 x – 10. The number 
of points of local maxima of f in interval (0, 2 ?? ) is: 
 (1) 1 (2) 2 
 (3) 3  (4) 4 
 Ans. (2) 
Sol. f(x) = 4cos
3
(x) + 3 3 cos
2
(x) – 10 ; x ?? (0, 2 ?? ) 
 ???@ f'(x) = 12cos
2
x[–sin(x)] + 3 3 (2cos(x))[–sin(x)] 
 ???@ f'(x) = –6sin(x) cos(x)[2cos(x) + 3 ] 
 
 
 
–
 
+
 
–
 
+
 
–
 
+
 
–
 
 
 
 
 
2 ?? 0 
 
 local maxima at x = 
5 7
,
6 6
????
 
8.  Let A = 
2 a 0
1 3 1
0 5 b
????
????
????
????
. If A
3
 = 4A
2
 – A – 21I, where 
I is the identity matrix of order 3 × 3, then 2a + 3b 
is equal to :  
 (1) –10 (2) –13 
 (3) –9  (4) –12 
 Ans. (2) 
Sol. A
3
 – 4A
2
 + A + 21 I = 0 
 tr(A) = 4 = 5 + 6   ???@ b = –1 
 |A| = –21 
 –16 + a = –21   ??   a = –5 
 2a + 3b = –13  
9. If the shortest distance between the lines  
 
1
ˆˆ ˆ
L : r (2 )i (1 3 )j (3 4 )k, ?] ?K ?? ?K ?M ?? ?K ?K ?? ?? ?? 
 
2
ˆˆ ˆ
L : r 2(1 )i 3(1 )j (5 )k, ?] ?K ?? ?K ?K ?? ?K ?K ?? ?? ?? 
 
is 
m
n
, where gcd (m, n) = 1, then the value of  
m + n equals. 
 (1) 384 (2) 387 
 (3) 377  (4) 390 
 Ans. (2) 
Sol. 
 
 
 
C 
D 
 Shortes distance (CD) = 
AB p q
| p q |
????
?? 
 = 
ˆ ˆ ˆ ˆ ˆˆ
(0i 2j 2k).( 15i 7j 9k)
355
?K ?K ?M ?K ?K 
 = 
0 14 18
355
?K?K
 = 
32
355
 
 ?| m + n = 32 + 355 = 387  
10. Let the sum of two positive integers be 24. If the 
probability, that their product is not less than  
3
4
 times their greatest positive product, is 
m
n
, 
where gcd(m, n) = 1, then n – m equals : 
 (1) 9 (2) 11 
 (3) 8  (4) 10 
 Ans. (4) 
Sol. x + y = 24, x, y ?? ?@ N 
 AM > GM  ???@ xy ?? 144 
 xy ?? 108 
 Favorable pairs of (x, y) are 
 (13, 11), (12, 12), (14, 10), (15, 9), (16, 8),  
 (17, 7), (18, 6), (6, 18), (7, 17), (8, 16), (9, 15),  
 (10, 14), (11, 13) 
 i.e. 13 cases 
 Total choices for x + y = 24 is 23 
 Probability = 
13 m
23 n
?] 
 n – m = 10  
Page 4


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Monday 08
th
 April, 2024)    TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. The value of k ?? ???? for which the integral
I
n
 = 
1
k n
0
(1 – x ) dx,
?? n ???@ ?? , satisfies 147 I
20
= 148 I
21
is :  
(1) 10 (2) 8
(3) 14 (4) 7
Ans. (4) 
Sol. I
n
 = 
1
k n
0
(1 x ) .1 dx ?M ?? I
n
 = (1 – x
k
)
n
.x – nk
1
k n 1 k 1
0
(1 x ) .x .dx
?M?M
?M ?? I
n
 = nk
1
k n k n 1
0
[(1 x ) (1 x ) ]dx
?M ?M ?M ?M ?? I
n
 = nkI
n
 – nkI
n
  
n
n 1
I nk
I nk 1
?M ?] ?K 21
20
I 21k
I 1 21k
?] ?K 147
148
?] ???@?@ k = 7
2. The sum of all the solutions of the equation
(8)
2x
 – 16 ?? (8)
x
 + 48 = 0 is :
(1) 1 + log
6
(8) (2) log
8
(6)
(3) 1 + log
8
(6) (4) log
8
(4)
Ans. (3) 
Sol. (8)
2x
 – 16 ?? (8)
x
 + 48 = 0 
Put 8
x
 = t 
t
2
 – 16 + 48 = 0 
?? t = 4 or t = 12
?? 8
x
 = 4 8
x
 = 12 
?? x = log
8
x x = log
8
12 
sum of solution = log
8
4 + log
8
12 
= log
8
48 = log
8
(6.8) 
= 1 + log
8
6 
3. Let the circles C
1
 : (x – ?? )
2
 + (y – ?? )
2
 =
2
1
r and 
C
2
 : (x – 8)
2
 +
2
15
y
2
????
?M????
????
 = 
2
2
r touch each other
externally at the point (6, 6). If the point (6, 6) 
divides the line segment joining the centres of the 
circles C
1
 and C
2
 internally in the ratio 2 : 1, then 
( ?? + ?? ) + 4 
?H ?I 2 2
1 2
r r ?K equals  
(1) 110 (2) 130
(3) 125 (4) 145
Ans. (2) 
Sol. 
r
1 
r
2
C
2
C
1
( ?? ?L?? ) 
P
2 : 1
 
C
2
( ?? ?L ?? ) C
1 P 
(6, 6) 
16
6
3
?K??
?] and
15
6
3
?K??
?] ?? ( ?? , ?? ) ?? (2, 3)
Also, C
1
C
2
 = r
1 
+
 
r
2
 
???@
2
2
2 2
15
(2 8) 3 2r r
2
????
?M ?K ?M ?] ?K ????
????
?? r
2
 =
5
2
?? r
1
 = 2r
2
 = 5
?| ( ?? ?@?K ?@ ?? ) + 4
2 2
1 2
(r r ) ?K = 5 + 4
25
25
4
????
?K ????
????
 = 130 ?@ 4.  Let P(x, y, z) be a point in the first octant, whose 
projection in the xy-plane is the point Q. Let  
OP = ?? ; the angle between OQ and the positive  
x-axis be ?? ; and the angle between OP and the 
positive z-axis be ?? , where O is the origin. Then 
the distance of P from the x-axis is :      
 (1) 
2 2
1 sin cos ?? ?M ?? ?? (2) 
2 2
1 cos sin ?? ?K ?? ?? 
 (3) 
2 2
1 sin cos ?? ?M ?? ?? (4) 
2 2
1 cos sin ?? ?K ?? ?? 
 Ans. (1) 
Sol. P(x, y, z), Q(x, y, O) ; x
2
 + y
2
 + z
2
= ?? 2
  
 
ˆˆ
OQ xi yj ?]?K  
 cos ?? = 
2 2
x
x y ?K 
 cos ?? = 
2 2 2
z
x y z ?K?K
  
 ???@ sin
2
???@ = 
2 2
2 2 2
x y
x y z
?K ?K?K
 
 distance of P from x-axis 
2 2
y z ?K 
 ???@
2 2
x ???M  ???@ 
2
2
x
1 ???M
??  
 = 
2 2
1 cos sin ?? ?M ?? ?? ?@ 
5. The number of critical points of the function  
f(x) = (x – 2)
2/3
 (2x + 1) is : 
 (1) 2 (2) 0 
 (3) 1  (4) 3 
 Ans. (1) 
Sol. f(x) = (x – 2)
2/3
 (2x + 1)  
 f'(x) = 
2
3
(x – 2)
–1/3
 (2x + 1) + (x – 2)
2/3
 (2) 
 f'(x) = 2 × 
1/3
(2x 1) (x 2)
3(x 2)
?K ?K ?M ?M 
 
1/3
3x 1
0
(x 2)
?M ?] ?M 
 Critical points x = 
1
3
 and x = 2  
 
6.  Let f(x) be a positive function such that the area 
bounded by y = f(x), y = 0 from x = 0 to x = a > 0 
is e
–a
 + 4a
2
 + a – 1. Then the differential equation, 
whose general solution is y = c
1
f(x) + c
2
 , where c
1
 
and c
2
 are arbitrary constants, is : 
 (1) 
2
x
2
d y dy
(8e 1) 0
dx dx
?M ?K ?] 
 (2) 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?M ?] 
 (3) 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?K ?] 
 (4) 
2
x
2
d y dy
(8e 1) 0
dx dx
?M ?M ?] 
 Ans. (3) 
Sol. 
a
a 2
0
f(x)dx e 4a a 1
?M ?] ?K ?K ?M ?? 
 f(a) = –e
–a
 + 8a + 1 
 f(x) = –e
–x
 + 8x + 1  
 Now y = C
1
f(x) + C
2
  
 
1
dy
C f '(x)
dx
?] = C
1
(e
–x 
+ 8) …..(1) 
 
2
2
d y
dx
 = –C
1
e
–x
  ?? –e
x
 
2
2
d y
dx
 
 Put in equation (1) 
 
2
x x
2
dy d y
e (e 8)
dx dx
?M ?] ?M ?K 
 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?K ?] 
 
 
7. Let f(x) = 4cos
3
x + 3 3 cos
2
 x – 10. The number 
of points of local maxima of f in interval (0, 2 ?? ) is: 
 (1) 1 (2) 2 
 (3) 3  (4) 4 
 Ans. (2) 
Sol. f(x) = 4cos
3
(x) + 3 3 cos
2
(x) – 10 ; x ?? (0, 2 ?? ) 
 ???@ f'(x) = 12cos
2
x[–sin(x)] + 3 3 (2cos(x))[–sin(x)] 
 ???@ f'(x) = –6sin(x) cos(x)[2cos(x) + 3 ] 
 
 
 
–
 
+
 
–
 
+
 
–
 
+
 
–
 
 
 
 
 
2 ?? 0 
 
 local maxima at x = 
5 7
,
6 6
????
 
8.  Let A = 
2 a 0
1 3 1
0 5 b
????
????
????
????
. If A
3
 = 4A
2
 – A – 21I, where 
I is the identity matrix of order 3 × 3, then 2a + 3b 
is equal to :  
 (1) –10 (2) –13 
 (3) –9  (4) –12 
 Ans. (2) 
Sol. A
3
 – 4A
2
 + A + 21 I = 0 
 tr(A) = 4 = 5 + 6   ???@ b = –1 
 |A| = –21 
 –16 + a = –21   ??   a = –5 
 2a + 3b = –13  
9. If the shortest distance between the lines  
 
1
ˆˆ ˆ
L : r (2 )i (1 3 )j (3 4 )k, ?] ?K ?? ?K ?M ?? ?K ?K ?? ?? ?? 
 
2
ˆˆ ˆ
L : r 2(1 )i 3(1 )j (5 )k, ?] ?K ?? ?K ?K ?? ?K ?K ?? ?? ?? 
 
is 
m
n
, where gcd (m, n) = 1, then the value of  
m + n equals. 
 (1) 384 (2) 387 
 (3) 377  (4) 390 
 Ans. (2) 
Sol. 
 
 
 
C 
D 
 Shortes distance (CD) = 
AB p q
| p q |
????
?? 
 = 
ˆ ˆ ˆ ˆ ˆˆ
(0i 2j 2k).( 15i 7j 9k)
355
?K ?K ?M ?K ?K 
 = 
0 14 18
355
?K?K
 = 
32
355
 
 ?| m + n = 32 + 355 = 387  
10. Let the sum of two positive integers be 24. If the 
probability, that their product is not less than  
3
4
 times their greatest positive product, is 
m
n
, 
where gcd(m, n) = 1, then n – m equals : 
 (1) 9 (2) 11 
 (3) 8  (4) 10 
 Ans. (4) 
Sol. x + y = 24, x, y ?? ?@ N 
 AM > GM  ???@ xy ?? 144 
 xy ?? 108 
 Favorable pairs of (x, y) are 
 (13, 11), (12, 12), (14, 10), (15, 9), (16, 8),  
 (17, 7), (18, 6), (6, 18), (7, 17), (8, 16), (9, 15),  
 (10, 14), (11, 13) 
 i.e. 13 cases 
 Total choices for x + y = 24 is 23 
 Probability = 
13 m
23 n
?] 
 n – m = 10  
11. If sinx = –
3
5
, where ?? < x < 
3
2
?? ,  
 then 80(tan
2
x – cosx) is equal to : 
 (1) 109 (2) 108 
 (3) 18  (4) 19 
 Ans. (1) 
Sol. sinx = 
3
5
?M , ?? < x < 
3
2
?? 
 tanx = 
3
4
 cosx = 
4
5
?M 
 80(tan
2
x – cosx)  
 = 
9 4
80
16 5
????
?K ????
????
 = 45 + 64 = 109 
12.  Let I(x) = 
2 2
6
dx
sin x(1 cot x) ?M ?? . If I(0) = 3, then  
I
12
?? ????
????
????
 is equal to : 
 (1) 3 (2) 3 3 
 (3) 6 3  (4) 2 3 
 Ans. (2) 
Sol. I(x) = 
2
2 2 2
6dx 6cosec xdx
sin x(1 cot x) (1 cot x)
?] ?M?M
????
 
 Put 1 – cotx = t 
 cosec
2
x dx = dt 
 I = 
2
6dt 6
c
t t
?M ?]?K
?? 
 I(x) = 
6
c
1 cot x
?M ?M , c = 3 
 I(x) = 
6
3
1 cot x
?M ?M , 
6
I 3
12
1 (2 3)
?? ????
?]?M
????
?M?K ????
 
 
6 6( 3 1)
I 3 3 3 3 2
12 2
3 1
???M ????
?] ?K ?] ?K ?] ????
?K ????
 
 
 
13. The equations of two sides AB and AC of a 
triangle ABC are 4x + y = 14 and 3x – 2y = 5, 
respectively. The point 
4
2,
3
????
?M ????
????
 divides the third 
side BC internally in the ratio 2 : 1. The equation 
of the side BC is : 
 (1) x – 6y – 10 = 0 (2) x – 3y – 6 = 0 
 (3) x + 3y + 2 = 0 (4) x + 6y + 6 = 0 
 Ans. (3) 
Sol.  
 
3x –2y = 5 
4x + y = 14 
A 
B (x
1
, 14 – 4x
1
) C 
2 : 1 
 
• 
 
2 1
2x x
2
3
?K ?] , 
?H ?I 2
1
3x 5
2 14 4x
2
3
?M ????
?K?M
????
????
= 
4
3
?M 
 2x
2
 + x
1
 = 6, 3x
2
 – 4x
1
 = – 13 
 x
2
 = 1, x
1
 = 4 
 So,  C(1, – 1) , B(4, –2) 
 m = 
1
3
?M 
 Equation of BC :  y + 1 = 
1
3
?M (x – 1) 
   3y + 3 = – x + 1 
   x + 3y + 2 = 0 
 
Page 5


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Monday 08
th
 April, 2024)    TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. The value of k ?? ???? for which the integral
I
n
 = 
1
k n
0
(1 – x ) dx,
?? n ???@ ?? , satisfies 147 I
20
= 148 I
21
is :  
(1) 10 (2) 8
(3) 14 (4) 7
Ans. (4) 
Sol. I
n
 = 
1
k n
0
(1 x ) .1 dx ?M ?? I
n
 = (1 – x
k
)
n
.x – nk
1
k n 1 k 1
0
(1 x ) .x .dx
?M?M
?M ?? I
n
 = nk
1
k n k n 1
0
[(1 x ) (1 x ) ]dx
?M ?M ?M ?M ?? I
n
 = nkI
n
 – nkI
n
  
n
n 1
I nk
I nk 1
?M ?] ?K 21
20
I 21k
I 1 21k
?] ?K 147
148
?] ???@?@ k = 7
2. The sum of all the solutions of the equation
(8)
2x
 – 16 ?? (8)
x
 + 48 = 0 is :
(1) 1 + log
6
(8) (2) log
8
(6)
(3) 1 + log
8
(6) (4) log
8
(4)
Ans. (3) 
Sol. (8)
2x
 – 16 ?? (8)
x
 + 48 = 0 
Put 8
x
 = t 
t
2
 – 16 + 48 = 0 
?? t = 4 or t = 12
?? 8
x
 = 4 8
x
 = 12 
?? x = log
8
x x = log
8
12 
sum of solution = log
8
4 + log
8
12 
= log
8
48 = log
8
(6.8) 
= 1 + log
8
6 
3. Let the circles C
1
 : (x – ?? )
2
 + (y – ?? )
2
 =
2
1
r and 
C
2
 : (x – 8)
2
 +
2
15
y
2
????
?M????
????
 = 
2
2
r touch each other
externally at the point (6, 6). If the point (6, 6) 
divides the line segment joining the centres of the 
circles C
1
 and C
2
 internally in the ratio 2 : 1, then 
( ?? + ?? ) + 4 
?H ?I 2 2
1 2
r r ?K equals  
(1) 110 (2) 130
(3) 125 (4) 145
Ans. (2) 
Sol. 
r
1 
r
2
C
2
C
1
( ?? ?L?? ) 
P
2 : 1
 
C
2
( ?? ?L ?? ) C
1 P 
(6, 6) 
16
6
3
?K??
?] and
15
6
3
?K??
?] ?? ( ?? , ?? ) ?? (2, 3)
Also, C
1
C
2
 = r
1 
+
 
r
2
 
???@
2
2
2 2
15
(2 8) 3 2r r
2
????
?M ?K ?M ?] ?K ????
????
?? r
2
 =
5
2
?? r
1
 = 2r
2
 = 5
?| ( ?? ?@?K ?@ ?? ) + 4
2 2
1 2
(r r ) ?K = 5 + 4
25
25
4
????
?K ????
????
 = 130 ?@ 4.  Let P(x, y, z) be a point in the first octant, whose 
projection in the xy-plane is the point Q. Let  
OP = ?? ; the angle between OQ and the positive  
x-axis be ?? ; and the angle between OP and the 
positive z-axis be ?? , where O is the origin. Then 
the distance of P from the x-axis is :      
 (1) 
2 2
1 sin cos ?? ?M ?? ?? (2) 
2 2
1 cos sin ?? ?K ?? ?? 
 (3) 
2 2
1 sin cos ?? ?M ?? ?? (4) 
2 2
1 cos sin ?? ?K ?? ?? 
 Ans. (1) 
Sol. P(x, y, z), Q(x, y, O) ; x
2
 + y
2
 + z
2
= ?? 2
  
 
ˆˆ
OQ xi yj ?]?K  
 cos ?? = 
2 2
x
x y ?K 
 cos ?? = 
2 2 2
z
x y z ?K?K
  
 ???@ sin
2
???@ = 
2 2
2 2 2
x y
x y z
?K ?K?K
 
 distance of P from x-axis 
2 2
y z ?K 
 ???@
2 2
x ???M  ???@ 
2
2
x
1 ???M
??  
 = 
2 2
1 cos sin ?? ?M ?? ?? ?@ 
5. The number of critical points of the function  
f(x) = (x – 2)
2/3
 (2x + 1) is : 
 (1) 2 (2) 0 
 (3) 1  (4) 3 
 Ans. (1) 
Sol. f(x) = (x – 2)
2/3
 (2x + 1)  
 f'(x) = 
2
3
(x – 2)
–1/3
 (2x + 1) + (x – 2)
2/3
 (2) 
 f'(x) = 2 × 
1/3
(2x 1) (x 2)
3(x 2)
?K ?K ?M ?M 
 
1/3
3x 1
0
(x 2)
?M ?] ?M 
 Critical points x = 
1
3
 and x = 2  
 
6.  Let f(x) be a positive function such that the area 
bounded by y = f(x), y = 0 from x = 0 to x = a > 0 
is e
–a
 + 4a
2
 + a – 1. Then the differential equation, 
whose general solution is y = c
1
f(x) + c
2
 , where c
1
 
and c
2
 are arbitrary constants, is : 
 (1) 
2
x
2
d y dy
(8e 1) 0
dx dx
?M ?K ?] 
 (2) 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?M ?] 
 (3) 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?K ?] 
 (4) 
2
x
2
d y dy
(8e 1) 0
dx dx
?M ?M ?] 
 Ans. (3) 
Sol. 
a
a 2
0
f(x)dx e 4a a 1
?M ?] ?K ?K ?M ?? 
 f(a) = –e
–a
 + 8a + 1 
 f(x) = –e
–x
 + 8x + 1  
 Now y = C
1
f(x) + C
2
  
 
1
dy
C f '(x)
dx
?] = C
1
(e
–x 
+ 8) …..(1) 
 
2
2
d y
dx
 = –C
1
e
–x
  ?? –e
x
 
2
2
d y
dx
 
 Put in equation (1) 
 
2
x x
2
dy d y
e (e 8)
dx dx
?M ?] ?M ?K 
 
2
x
2
d y dy
(8e 1) 0
dx dx
?K ?K ?] 
 
 
7. Let f(x) = 4cos
3
x + 3 3 cos
2
 x – 10. The number 
of points of local maxima of f in interval (0, 2 ?? ) is: 
 (1) 1 (2) 2 
 (3) 3  (4) 4 
 Ans. (2) 
Sol. f(x) = 4cos
3
(x) + 3 3 cos
2
(x) – 10 ; x ?? (0, 2 ?? ) 
 ???@ f'(x) = 12cos
2
x[–sin(x)] + 3 3 (2cos(x))[–sin(x)] 
 ???@ f'(x) = –6sin(x) cos(x)[2cos(x) + 3 ] 
 
 
 
–
 
+
 
–
 
+
 
–
 
+
 
–
 
 
 
 
 
2 ?? 0 
 
 local maxima at x = 
5 7
,
6 6
????
 
8.  Let A = 
2 a 0
1 3 1
0 5 b
????
????
????
????
. If A
3
 = 4A
2
 – A – 21I, where 
I is the identity matrix of order 3 × 3, then 2a + 3b 
is equal to :  
 (1) –10 (2) –13 
 (3) –9  (4) –12 
 Ans. (2) 
Sol. A
3
 – 4A
2
 + A + 21 I = 0 
 tr(A) = 4 = 5 + 6   ???@ b = –1 
 |A| = –21 
 –16 + a = –21   ??   a = –5 
 2a + 3b = –13  
9. If the shortest distance between the lines  
 
1
ˆˆ ˆ
L : r (2 )i (1 3 )j (3 4 )k, ?] ?K ?? ?K ?M ?? ?K ?K ?? ?? ?? 
 
2
ˆˆ ˆ
L : r 2(1 )i 3(1 )j (5 )k, ?] ?K ?? ?K ?K ?? ?K ?K ?? ?? ?? 
 
is 
m
n
, where gcd (m, n) = 1, then the value of  
m + n equals. 
 (1) 384 (2) 387 
 (3) 377  (4) 390 
 Ans. (2) 
Sol. 
 
 
 
C 
D 
 Shortes distance (CD) = 
AB p q
| p q |
????
?? 
 = 
ˆ ˆ ˆ ˆ ˆˆ
(0i 2j 2k).( 15i 7j 9k)
355
?K ?K ?M ?K ?K 
 = 
0 14 18
355
?K?K
 = 
32
355
 
 ?| m + n = 32 + 355 = 387  
10. Let the sum of two positive integers be 24. If the 
probability, that their product is not less than  
3
4
 times their greatest positive product, is 
m
n
, 
where gcd(m, n) = 1, then n – m equals : 
 (1) 9 (2) 11 
 (3) 8  (4) 10 
 Ans. (4) 
Sol. x + y = 24, x, y ?? ?@ N 
 AM > GM  ???@ xy ?? 144 
 xy ?? 108 
 Favorable pairs of (x, y) are 
 (13, 11), (12, 12), (14, 10), (15, 9), (16, 8),  
 (17, 7), (18, 6), (6, 18), (7, 17), (8, 16), (9, 15),  
 (10, 14), (11, 13) 
 i.e. 13 cases 
 Total choices for x + y = 24 is 23 
 Probability = 
13 m
23 n
?] 
 n – m = 10  
11. If sinx = –
3
5
, where ?? < x < 
3
2
?? ,  
 then 80(tan
2
x – cosx) is equal to : 
 (1) 109 (2) 108 
 (3) 18  (4) 19 
 Ans. (1) 
Sol. sinx = 
3
5
?M , ?? < x < 
3
2
?? 
 tanx = 
3
4
 cosx = 
4
5
?M 
 80(tan
2
x – cosx)  
 = 
9 4
80
16 5
????
?K ????
????
 = 45 + 64 = 109 
12.  Let I(x) = 
2 2
6
dx
sin x(1 cot x) ?M ?? . If I(0) = 3, then  
I
12
?? ????
????
????
 is equal to : 
 (1) 3 (2) 3 3 
 (3) 6 3  (4) 2 3 
 Ans. (2) 
Sol. I(x) = 
2
2 2 2
6dx 6cosec xdx
sin x(1 cot x) (1 cot x)
?] ?M?M
????
 
 Put 1 – cotx = t 
 cosec
2
x dx = dt 
 I = 
2
6dt 6
c
t t
?M ?]?K
?? 
 I(x) = 
6
c
1 cot x
?M ?M , c = 3 
 I(x) = 
6
3
1 cot x
?M ?M , 
6
I 3
12
1 (2 3)
?? ????
?]?M
????
?M?K ????
 
 
6 6( 3 1)
I 3 3 3 3 2
12 2
3 1
???M ????
?] ?K ?] ?K ?] ????
?K ????
 
 
 
13. The equations of two sides AB and AC of a 
triangle ABC are 4x + y = 14 and 3x – 2y = 5, 
respectively. The point 
4
2,
3
????
?M ????
????
 divides the third 
side BC internally in the ratio 2 : 1. The equation 
of the side BC is : 
 (1) x – 6y – 10 = 0 (2) x – 3y – 6 = 0 
 (3) x + 3y + 2 = 0 (4) x + 6y + 6 = 0 
 Ans. (3) 
Sol.  
 
3x –2y = 5 
4x + y = 14 
A 
B (x
1
, 14 – 4x
1
) C 
2 : 1 
 
• 
 
2 1
2x x
2
3
?K ?] , 
?H ?I 2
1
3x 5
2 14 4x
2
3
?M ????
?K?M
????
????
= 
4
3
?M 
 2x
2
 + x
1
 = 6, 3x
2
 – 4x
1
 = – 13 
 x
2
 = 1, x
1
 = 4 
 So,  C(1, – 1) , B(4, –2) 
 m = 
1
3
?M 
 Equation of BC :  y + 1 = 
1
3
?M (x – 1) 
   3y + 3 = – x + 1 
   x + 3y + 2 = 0 
 
 
14.  Let [t] be the greatest integer less than or equal to 
t. Let A be the set of al prime factors of 2310 and  
f : A ???@ ?? be the function f(x) = 
3
2
2
x
log x
5
???? ???? ????
?K ???? ????
????
????
????
???? ???? ????
. 
The number of one-to-one functions from A to the 
range of f is : 
 (1) 20 (2) 120 
 (3) 25  (4) 24 
 Ans. (2) 
Sol. N = 2310 = 231 × 10 
  = 3 × 11 × 7 × 2 × 5 
 A = {2, 3, 5, 7, 11} 
 f(x) = 
3
2
2
x
log x
5
???? ???? ????
?K ???? ????
????
???? ???? ????
 
 f(2) = [log
2 
(5)] = 2  
 f(3) = [log
2 
(14)] = 3 
 f(5) = [log
2 
(25 + 25)] = 5 
 f(7) = [log
2 
(117)] = 6 
 f (11) = [log
2 
387] = 8 
 Range of f : B = {2, 3, 5, 6, 8} 
 No. of one-one functions = 5! = 120 
15. Let z be a complex number such that |z + 2| = 1 
and lm
z 1 1
z 2 5
?K ????
?] ????
?K ????
. Then the value of 
?H ?I Re z 2 ?K 
is : 
 (1) 
6
5
 (2) 
1 6
5
?K 
 (3) 
24
5
  (4) 
2 6
5
 
 Ans. (4) 
Sol. |z + 2| = 1, Im
z 1 1
z 2 5
?K ????
?] ????
?K ????
 
 Let z + 2 = cos ?? + isin ??  
 
1
cos isin
z 2
?] ?? ?M ?? ?K 
?@ ?? 
z 1 1
1
z 2 z 2
?K ?]?M
?K?K
 = 1 – (cos ?? – isin ?? ) 
 = (1 – cos ?? ) + isin ??   
 Im
z 1
z 2
?K ????
????
?K ????
 = sin ?? , sin ?? = 
1
5
 
 cos ?? = ±
1
1
25
?M = ±
2 6
5
 
 
2 6
Re(z 2)
5
?K?] 
 
16.  If the set R = {(a, b) ; a + 5b = 42, a, b ?? } 
has m elements and 
?H ?I m
n!
n 1
1 i
?] ?K ?? = x + iy, where  
I = 1 ?M , then the value of m + x + y is :  
 (1) 8 (2) 12 
 (3) 4  (4) 5 
 Ans. (2) 
Sol. a + 5b = 42, a, b ???@ N 
 a = 42 – 5b, b = 1, a = 37 
 b = 2, a = 32 
 b = 3, a = 27 
  
 b = 8, a = 2 
 R has "8" elements  ???@ m = 8 
 
8
n!
n 1
(1 i ) x iy
?] ?M ?] ?K ?? 
 for n ?? 4, i
n! 
= 1 
 ?? (1 – i) + (1 – i
2!
) + (1 – i
3!
) 
 = 1 – I + 2 + 1 + 1 
 = 5 – I = x + iy 
 m + x + y = 8 + 5 – 1 = 12