JEE Main 2024 April 4 Shift 1 Paper & Solutions | JEE Main & Advanced Mock Test Series CY_Marker_0 PDF Download

 Page 1


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Thursday 04
th
 April, 2024)    TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ƒ : R ?? R be a function given by
?H ?I 2
1 cos2x
, x 0
x
ƒ x , x 0
1 cosx
, x 0
x
?? ?M ?\ ?? ?? ?? ?] ?? ?] ?? ?? ???M
?? ?^ ?? ?? , where ?? , ?? ?? R. If 
ƒ is continuous at x = 0, then ?? 2
 + ?? 2
 is equal to : 
(1) 48 (2) 12
(3) 3 (4) 6
Ans. (2) 
Sol. ƒ(0
–
) = 
2
2
x 0
2sin x
lim
x
?M ?? = 2 = ?? 
ƒ(0
+
) = 
x 0
x
sin
2
lim 2 2
x
2
2
2
?K ?? ?? ?? ?? ?] ?] ?? ?? = 2 2 
?? 2
+ ?? 2
 = 4 + 8 = 12
2. Three urns A, B and C contain 7 red, 5 black;
5 red, 7 black and 6 red, 6 black balls, respectively.
One of the urn is selected at random and a ball is
drawn from it. If the ball drawn is black, then the
probability that it is drawn from urn A is :
(1) 
4
17
(2) 
5
18
(3) 
7
18
(4) 
5
16
Ans. (2) 
Sol. 
A
7R, 5B
B
5R, 7B
C
6R, 6B
P(B) = 
1 5 1 7 1 6
. . .
3 12 3 12 3 12
?K?K
required probability = 
1 5
.
5
3 12
1 5 7 6 18
.
3 12 12 12
?] ????
?K?K
????
????
3. The vertices of a triangle are A(–1, 3), B(–2, 2) and
C(3, –1). A new triangle is formed by shifting the sides
of the triangle by one unit inwards. Then the equation
of the side of the new triangle nearest to origin is :
(1) x – y – 
?H ?I 2 2 ?K = 0
(2) –x + y – 
?H ?I 2 2 ?M = 0
(3) x + y – 
?H ?I 2 2 ?M = 0
(4) x + y + 
?H ?I 2 2 ?M = 0
Ans. (3) 
Sol. 
C
(–2,2) 
m = –1 
(3, –1) 
B 
A 
(–1,3) 
equation of AC ?? x + y = 2 
equation of line parallel to AC x + y = d 
?M ?] d 2
1
2
d 2 2 ?]?M
eq
n
of new required line
x y 2 2 ?K ?] ?M 4. If the solution y = y(x) of the differential equation
(x
4
 + 2x
3
 + 3x
2
 + 2x + 2)dy – (2x
2
 + 2x + 3)dx = 0
satisfies y(–1) = 
4
?? ?M , then y(0) is equal to :
(1) 
12
?? ?M (2) 0 
(3) 
4
?? (4) 
2
?? Ans. (3) 
Page 2


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Thursday 04
th
 April, 2024)    TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ƒ : R ?? R be a function given by
?H ?I 2
1 cos2x
, x 0
x
ƒ x , x 0
1 cosx
, x 0
x
?? ?M ?\ ?? ?? ?? ?] ?? ?] ?? ?? ???M
?? ?^ ?? ?? , where ?? , ?? ?? R. If 
ƒ is continuous at x = 0, then ?? 2
 + ?? 2
 is equal to : 
(1) 48 (2) 12
(3) 3 (4) 6
Ans. (2) 
Sol. ƒ(0
–
) = 
2
2
x 0
2sin x
lim
x
?M ?? = 2 = ?? 
ƒ(0
+
) = 
x 0
x
sin
2
lim 2 2
x
2
2
2
?K ?? ?? ?? ?? ?] ?] ?? ?? = 2 2 
?? 2
+ ?? 2
 = 4 + 8 = 12
2. Three urns A, B and C contain 7 red, 5 black;
5 red, 7 black and 6 red, 6 black balls, respectively.
One of the urn is selected at random and a ball is
drawn from it. If the ball drawn is black, then the
probability that it is drawn from urn A is :
(1) 
4
17
(2) 
5
18
(3) 
7
18
(4) 
5
16
Ans. (2) 
Sol. 
A
7R, 5B
B
5R, 7B
C
6R, 6B
P(B) = 
1 5 1 7 1 6
. . .
3 12 3 12 3 12
?K?K
required probability = 
1 5
.
5
3 12
1 5 7 6 18
.
3 12 12 12
?] ????
?K?K
????
????
3. The vertices of a triangle are A(–1, 3), B(–2, 2) and
C(3, –1). A new triangle is formed by shifting the sides
of the triangle by one unit inwards. Then the equation
of the side of the new triangle nearest to origin is :
(1) x – y – 
?H ?I 2 2 ?K = 0
(2) –x + y – 
?H ?I 2 2 ?M = 0
(3) x + y – 
?H ?I 2 2 ?M = 0
(4) x + y + 
?H ?I 2 2 ?M = 0
Ans. (3) 
Sol. 
C
(–2,2) 
m = –1 
(3, –1) 
B 
A 
(–1,3) 
equation of AC ?? x + y = 2 
equation of line parallel to AC x + y = d 
?M ?] d 2
1
2
d 2 2 ?]?M
eq
n
of new required line
x y 2 2 ?K ?] ?M 4. If the solution y = y(x) of the differential equation
(x
4
 + 2x
3
 + 3x
2
 + 2x + 2)dy – (2x
2
 + 2x + 3)dx = 0
satisfies y(–1) = 
4
?? ?M , then y(0) is equal to :
(1) 
12
?? ?M (2) 0 
(3) 
4
?? (4) 
2
?? Ans. (3) 
Sol. 
?H ?I 2
4 3 2
2x 2x 3
dy dx
x 2x 3x 2x 2
?K?K
?] ?K ?K ?K ?K ????
 
 
?H ?I ?H ?I ?H ?I 2
2 2
2x 2x 3
y dx
x 1 x 2x 2
?K?K
?] ?K ?K ?K ?? 
 
2 2
dx dx
y
x 2x 2 x 1
?]?K
?K ?K ?K ????
 
 y = tan
–1
(x + 1) + tan
–1
x + C 
 y(–1) = 
4
?M??
 
 0 C
4 4
?M ?? ?? ?] ?M ?K  ?? C = 0 
 ?? y = tan
–1
(x + 1) + tan
–1
x 
 y(0) = tan
–1
1 = 
4
?? 
5. Let the sum of the maximum and the minimum 
values of the function f(x) = 
2
2
2x 3x 8
2x 3x 8
?M?K
?K?K
 be 
m
n
, 
where gcd(m, n) = 1. Then m + n is equal to : 
 (1) 182 (2) 217 
 (3) 195  (4) 201 
 Ans. (4) 
Sol. 
2
2
2x 3x 8
y
2x 3x 8
?M?K
?] ?K?K
 
 x
2
(2y – 2) + x(3y + 3) + 8y – 8 = 0 
 use D ?? 0 
 (3y + 3)
2
 – 4(2y – 2) (8y – 8) ?? 0 
 (11y – 5) (5y – 11) ?? 0 
 
5 11
y ,
11 5
????
????
????
????
 
 y = 1 is also included 
6. One of the points of intersection of the curves  
y = 1 + 3x – 2x
2
 and y = 
1
x
 is 
1
,2
2
????
????
????
. Let the area 
of the region enclosed by these curves be 
?H ?I 1
5 m
24
?K – nlog
e ?H ?I 1 5 ?K , where ?? , m, n ?? 
N. Then ?? + m + n is equal to 
 (1) 32 (2) 30 
 (3) 29  (4) 31 
 Ans. (2) 
Sol. 
 
 
 
 
 
1 5
2
2
1
2
1
A 1 3x 2x dx
x
?K????
?] ?K ?M ?M ????
????
?? 
 
1 5
2 3
2
1
2
3x 2x
A x n x
2 3
?K ????
?] ?K ?M ?M ????
????
 
 
2 3
1 5 3 1 5 2 1 5 1 5
A n
2 2 2 3 2 2
?? ?? ?? ?? ?? ?? ?K ?K ?K ?K ?] ?K ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
1 3 1 2 1 1
n
2 2 4 3 8 2
?? ?? ?? ?? ?? ?? ?M ?M ?K ?K ?? ?? ?? ?? ????
?? ?? ?? ?? ????
 
 
1 5 3 3 15 4 2
A 5 5
2 2 8 4 8 3 3
?] ?K ?K ?K ?K ?M ?M 
 
?H ?I 1 3 1
n 1 5
2 8 12
?M ?M ?K ?M ?K 
 
?H ?I 1 3 2 15 4 1
5 n 1 5
2 4 3 8 3 12
????
?] ?K ?M ?K ?M ?K ?M ?K ????
????
 
 
?H ?I ?] ?K ?M ?K 14 15
5 n 1 5
24 24
 
7. If the system of equations 
 
?H ?I ?H ?I x 2 sin y 2 cos z 0 ?K ?? ?K ?? ?] 
 x + (cos ?? )y + (sin ?? )z = 0 
 x + (sin ?? )y – (cos ?? )z = 0 
 has a non-trivial solution, then 0,
2
?? ????
????
????
????
 is equal to : 
 (1) 
3
4
?? (2) 
7
24
?? 
 (3) 
5
24
??  (4) 
11
24
?? 
 Ans. (3) 
Page 3


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Thursday 04
th
 April, 2024)    TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ƒ : R ?? R be a function given by
?H ?I 2
1 cos2x
, x 0
x
ƒ x , x 0
1 cosx
, x 0
x
?? ?M ?\ ?? ?? ?? ?] ?? ?] ?? ?? ???M
?? ?^ ?? ?? , where ?? , ?? ?? R. If 
ƒ is continuous at x = 0, then ?? 2
 + ?? 2
 is equal to : 
(1) 48 (2) 12
(3) 3 (4) 6
Ans. (2) 
Sol. ƒ(0
–
) = 
2
2
x 0
2sin x
lim
x
?M ?? = 2 = ?? 
ƒ(0
+
) = 
x 0
x
sin
2
lim 2 2
x
2
2
2
?K ?? ?? ?? ?? ?] ?] ?? ?? = 2 2 
?? 2
+ ?? 2
 = 4 + 8 = 12
2. Three urns A, B and C contain 7 red, 5 black;
5 red, 7 black and 6 red, 6 black balls, respectively.
One of the urn is selected at random and a ball is
drawn from it. If the ball drawn is black, then the
probability that it is drawn from urn A is :
(1) 
4
17
(2) 
5
18
(3) 
7
18
(4) 
5
16
Ans. (2) 
Sol. 
A
7R, 5B
B
5R, 7B
C
6R, 6B
P(B) = 
1 5 1 7 1 6
. . .
3 12 3 12 3 12
?K?K
required probability = 
1 5
.
5
3 12
1 5 7 6 18
.
3 12 12 12
?] ????
?K?K
????
????
3. The vertices of a triangle are A(–1, 3), B(–2, 2) and
C(3, –1). A new triangle is formed by shifting the sides
of the triangle by one unit inwards. Then the equation
of the side of the new triangle nearest to origin is :
(1) x – y – 
?H ?I 2 2 ?K = 0
(2) –x + y – 
?H ?I 2 2 ?M = 0
(3) x + y – 
?H ?I 2 2 ?M = 0
(4) x + y + 
?H ?I 2 2 ?M = 0
Ans. (3) 
Sol. 
C
(–2,2) 
m = –1 
(3, –1) 
B 
A 
(–1,3) 
equation of AC ?? x + y = 2 
equation of line parallel to AC x + y = d 
?M ?] d 2
1
2
d 2 2 ?]?M
eq
n
of new required line
x y 2 2 ?K ?] ?M 4. If the solution y = y(x) of the differential equation
(x
4
 + 2x
3
 + 3x
2
 + 2x + 2)dy – (2x
2
 + 2x + 3)dx = 0
satisfies y(–1) = 
4
?? ?M , then y(0) is equal to :
(1) 
12
?? ?M (2) 0 
(3) 
4
?? (4) 
2
?? Ans. (3) 
Sol. 
?H ?I 2
4 3 2
2x 2x 3
dy dx
x 2x 3x 2x 2
?K?K
?] ?K ?K ?K ?K ????
 
 
?H ?I ?H ?I ?H ?I 2
2 2
2x 2x 3
y dx
x 1 x 2x 2
?K?K
?] ?K ?K ?K ?? 
 
2 2
dx dx
y
x 2x 2 x 1
?]?K
?K ?K ?K ????
 
 y = tan
–1
(x + 1) + tan
–1
x + C 
 y(–1) = 
4
?M??
 
 0 C
4 4
?M ?? ?? ?] ?M ?K  ?? C = 0 
 ?? y = tan
–1
(x + 1) + tan
–1
x 
 y(0) = tan
–1
1 = 
4
?? 
5. Let the sum of the maximum and the minimum 
values of the function f(x) = 
2
2
2x 3x 8
2x 3x 8
?M?K
?K?K
 be 
m
n
, 
where gcd(m, n) = 1. Then m + n is equal to : 
 (1) 182 (2) 217 
 (3) 195  (4) 201 
 Ans. (4) 
Sol. 
2
2
2x 3x 8
y
2x 3x 8
?M?K
?] ?K?K
 
 x
2
(2y – 2) + x(3y + 3) + 8y – 8 = 0 
 use D ?? 0 
 (3y + 3)
2
 – 4(2y – 2) (8y – 8) ?? 0 
 (11y – 5) (5y – 11) ?? 0 
 
5 11
y ,
11 5
????
????
????
????
 
 y = 1 is also included 
6. One of the points of intersection of the curves  
y = 1 + 3x – 2x
2
 and y = 
1
x
 is 
1
,2
2
????
????
????
. Let the area 
of the region enclosed by these curves be 
?H ?I 1
5 m
24
?K – nlog
e ?H ?I 1 5 ?K , where ?? , m, n ?? 
N. Then ?? + m + n is equal to 
 (1) 32 (2) 30 
 (3) 29  (4) 31 
 Ans. (2) 
Sol. 
 
 
 
 
 
1 5
2
2
1
2
1
A 1 3x 2x dx
x
?K????
?] ?K ?M ?M ????
????
?? 
 
1 5
2 3
2
1
2
3x 2x
A x n x
2 3
?K ????
?] ?K ?M ?M ????
????
 
 
2 3
1 5 3 1 5 2 1 5 1 5
A n
2 2 2 3 2 2
?? ?? ?? ?? ?? ?? ?K ?K ?K ?K ?] ?K ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
1 3 1 2 1 1
n
2 2 4 3 8 2
?? ?? ?? ?? ?? ?? ?M ?M ?K ?K ?? ?? ?? ?? ????
?? ?? ?? ?? ????
 
 
1 5 3 3 15 4 2
A 5 5
2 2 8 4 8 3 3
?] ?K ?K ?K ?K ?M ?M 
 
?H ?I 1 3 1
n 1 5
2 8 12
?M ?M ?K ?M ?K 
 
?H ?I 1 3 2 15 4 1
5 n 1 5
2 4 3 8 3 12
????
?] ?K ?M ?K ?M ?K ?M ?K ????
????
 
 
?H ?I ?] ?K ?M ?K 14 15
5 n 1 5
24 24
 
7. If the system of equations 
 
?H ?I ?H ?I x 2 sin y 2 cos z 0 ?K ?? ?K ?? ?] 
 x + (cos ?? )y + (sin ?? )z = 0 
 x + (sin ?? )y – (cos ?? )z = 0 
 has a non-trivial solution, then 0,
2
?? ????
????
????
????
 is equal to : 
 (1) 
3
4
?? (2) 
7
24
?? 
 (3) 
5
24
??  (4) 
11
24
?? 
 Ans. (3) 
 
Sol. 
????
?? ?M ?? ?] ????
1 2 sin 2 cos
1 sin cos 0
1 cos sin
 
 ?? 
?H ?I ?H ?I ?M ?? ?? ?K ?? ?K ?? ?? ?M ?? ?] 1 2 sin sin cos 2 cos cos sin 0
 
 ?? ?K ?? ?M ?? ?] 1 2 cos2 2 sin2 0 
 ?? ?M ?? ?] ?M 1
cos2 sin 2
2
 
 
?? ????
?? ?K ?] ?M ????
????
1
cos 2
4 2
 
 
????
?? ?K ?] ?? ?? 2
2 2n
4 3
 
 
????
?? ?K ?] ?? ?? n
8 3
 
 n = 0,  
 
5
x
3 8 24
?? ?? ?? ?]?M?] 
8. There are 5 points P
1
, P
2
, P
3
, P
4
, P
5
 on the side AB, 
excluding A and B, of a triangle ABC. Similarly 
there are 6 points P
6
, P
7
, …, P
11
 on the side BC and 7 
points P
12
, P
13
, …, P
18
 on the side CA of the triangle. 
The number of triangles, that can be formed using the 
points P
1
, P
2
, …, P
18
 as vertices, is : 
 (1) 776 (2) 751 
 (3) 796  (4) 771 
 Ans. (2) 
Sol. 
18
C
3
 – 
5
C
3
 – 
6
C
3
 – 
7
C
3
 
 = 751 
9. Let ƒ(x) = 
2, 2 x 0
x 2, 0 x 2
?M ?M ?? ?? ?? ?? ?M ?\ ?? ?? and h(x) = f(|x|) + |f(x)|. 
Then ?H ?I 2
2
h x dx
?M ?? is equal to : 
 (1) 2 (2) 4 
 (3) 1  (4) 6 
 Ans. (1) 
Sol. 
 
y=f(x
x 
2 
x–2 
–2 
–2 
 
  f(|x|) ??    |f(x)| 
 
x 
(2,0) 
x–2 
(0,–2) 
(–2,0) 
0 
–x–2 
    
 
x 
2 
–2 
0 
2 
 
 
?M ?K ?M ?] ?? ?? ?? ?] ?? ?M ?M ?K ?] ?M ?M ?? ?\ ?? x 2 2 x 0, 0 x 2
h(x)
x 2 2 x 2 x 0
 
 
 
2 0 
–2 
 
 ?? ?] ?? 2
0
h(x)dx 0 and ?H ?I ?M ?] ?? 0
2
h x dx 2 
10. The sum of all rational terms in the expansion of 
?H ?I 15
1 1
5 3
2 5 ?K is equal to : 
 (1) 3133 (2) 633 
 (3) 931  (4) 6131 
 Ans. (1) 
Sol. T
r + 1
 = 
15
C
r
 
?H ?I ?H ?I ?M 15 r r
1 1
3 5
5 2 
 
?M ?] r 15 r
15
3 5
r
C 5 . 2 
 R = 3 ?? , 15µ 
 ?? r = 0, 15 
 2 rational terms 
 ?H ?I ???K
5
15 3 15
0 15
C 2 C 5 
 = 8 + 3125 = 3133 
Page 4


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Thursday 04
th
 April, 2024)    TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ƒ : R ?? R be a function given by
?H ?I 2
1 cos2x
, x 0
x
ƒ x , x 0
1 cosx
, x 0
x
?? ?M ?\ ?? ?? ?? ?] ?? ?] ?? ?? ???M
?? ?^ ?? ?? , where ?? , ?? ?? R. If 
ƒ is continuous at x = 0, then ?? 2
 + ?? 2
 is equal to : 
(1) 48 (2) 12
(3) 3 (4) 6
Ans. (2) 
Sol. ƒ(0
–
) = 
2
2
x 0
2sin x
lim
x
?M ?? = 2 = ?? 
ƒ(0
+
) = 
x 0
x
sin
2
lim 2 2
x
2
2
2
?K ?? ?? ?? ?? ?] ?] ?? ?? = 2 2 
?? 2
+ ?? 2
 = 4 + 8 = 12
2. Three urns A, B and C contain 7 red, 5 black;
5 red, 7 black and 6 red, 6 black balls, respectively.
One of the urn is selected at random and a ball is
drawn from it. If the ball drawn is black, then the
probability that it is drawn from urn A is :
(1) 
4
17
(2) 
5
18
(3) 
7
18
(4) 
5
16
Ans. (2) 
Sol. 
A
7R, 5B
B
5R, 7B
C
6R, 6B
P(B) = 
1 5 1 7 1 6
. . .
3 12 3 12 3 12
?K?K
required probability = 
1 5
.
5
3 12
1 5 7 6 18
.
3 12 12 12
?] ????
?K?K
????
????
3. The vertices of a triangle are A(–1, 3), B(–2, 2) and
C(3, –1). A new triangle is formed by shifting the sides
of the triangle by one unit inwards. Then the equation
of the side of the new triangle nearest to origin is :
(1) x – y – 
?H ?I 2 2 ?K = 0
(2) –x + y – 
?H ?I 2 2 ?M = 0
(3) x + y – 
?H ?I 2 2 ?M = 0
(4) x + y + 
?H ?I 2 2 ?M = 0
Ans. (3) 
Sol. 
C
(–2,2) 
m = –1 
(3, –1) 
B 
A 
(–1,3) 
equation of AC ?? x + y = 2 
equation of line parallel to AC x + y = d 
?M ?] d 2
1
2
d 2 2 ?]?M
eq
n
of new required line
x y 2 2 ?K ?] ?M 4. If the solution y = y(x) of the differential equation
(x
4
 + 2x
3
 + 3x
2
 + 2x + 2)dy – (2x
2
 + 2x + 3)dx = 0
satisfies y(–1) = 
4
?? ?M , then y(0) is equal to :
(1) 
12
?? ?M (2) 0 
(3) 
4
?? (4) 
2
?? Ans. (3) 
Sol. 
?H ?I 2
4 3 2
2x 2x 3
dy dx
x 2x 3x 2x 2
?K?K
?] ?K ?K ?K ?K ????
 
 
?H ?I ?H ?I ?H ?I 2
2 2
2x 2x 3
y dx
x 1 x 2x 2
?K?K
?] ?K ?K ?K ?? 
 
2 2
dx dx
y
x 2x 2 x 1
?]?K
?K ?K ?K ????
 
 y = tan
–1
(x + 1) + tan
–1
x + C 
 y(–1) = 
4
?M??
 
 0 C
4 4
?M ?? ?? ?] ?M ?K  ?? C = 0 
 ?? y = tan
–1
(x + 1) + tan
–1
x 
 y(0) = tan
–1
1 = 
4
?? 
5. Let the sum of the maximum and the minimum 
values of the function f(x) = 
2
2
2x 3x 8
2x 3x 8
?M?K
?K?K
 be 
m
n
, 
where gcd(m, n) = 1. Then m + n is equal to : 
 (1) 182 (2) 217 
 (3) 195  (4) 201 
 Ans. (4) 
Sol. 
2
2
2x 3x 8
y
2x 3x 8
?M?K
?] ?K?K
 
 x
2
(2y – 2) + x(3y + 3) + 8y – 8 = 0 
 use D ?? 0 
 (3y + 3)
2
 – 4(2y – 2) (8y – 8) ?? 0 
 (11y – 5) (5y – 11) ?? 0 
 
5 11
y ,
11 5
????
????
????
????
 
 y = 1 is also included 
6. One of the points of intersection of the curves  
y = 1 + 3x – 2x
2
 and y = 
1
x
 is 
1
,2
2
????
????
????
. Let the area 
of the region enclosed by these curves be 
?H ?I 1
5 m
24
?K – nlog
e ?H ?I 1 5 ?K , where ?? , m, n ?? 
N. Then ?? + m + n is equal to 
 (1) 32 (2) 30 
 (3) 29  (4) 31 
 Ans. (2) 
Sol. 
 
 
 
 
 
1 5
2
2
1
2
1
A 1 3x 2x dx
x
?K????
?] ?K ?M ?M ????
????
?? 
 
1 5
2 3
2
1
2
3x 2x
A x n x
2 3
?K ????
?] ?K ?M ?M ????
????
 
 
2 3
1 5 3 1 5 2 1 5 1 5
A n
2 2 2 3 2 2
?? ?? ?? ?? ?? ?? ?K ?K ?K ?K ?] ?K ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
1 3 1 2 1 1
n
2 2 4 3 8 2
?? ?? ?? ?? ?? ?? ?M ?M ?K ?K ?? ?? ?? ?? ????
?? ?? ?? ?? ????
 
 
1 5 3 3 15 4 2
A 5 5
2 2 8 4 8 3 3
?] ?K ?K ?K ?K ?M ?M 
 
?H ?I 1 3 1
n 1 5
2 8 12
?M ?M ?K ?M ?K 
 
?H ?I 1 3 2 15 4 1
5 n 1 5
2 4 3 8 3 12
????
?] ?K ?M ?K ?M ?K ?M ?K ????
????
 
 
?H ?I ?] ?K ?M ?K 14 15
5 n 1 5
24 24
 
7. If the system of equations 
 
?H ?I ?H ?I x 2 sin y 2 cos z 0 ?K ?? ?K ?? ?] 
 x + (cos ?? )y + (sin ?? )z = 0 
 x + (sin ?? )y – (cos ?? )z = 0 
 has a non-trivial solution, then 0,
2
?? ????
????
????
????
 is equal to : 
 (1) 
3
4
?? (2) 
7
24
?? 
 (3) 
5
24
??  (4) 
11
24
?? 
 Ans. (3) 
 
Sol. 
????
?? ?M ?? ?] ????
1 2 sin 2 cos
1 sin cos 0
1 cos sin
 
 ?? 
?H ?I ?H ?I ?M ?? ?? ?K ?? ?K ?? ?? ?M ?? ?] 1 2 sin sin cos 2 cos cos sin 0
 
 ?? ?K ?? ?M ?? ?] 1 2 cos2 2 sin2 0 
 ?? ?M ?? ?] ?M 1
cos2 sin 2
2
 
 
?? ????
?? ?K ?] ?M ????
????
1
cos 2
4 2
 
 
????
?? ?K ?] ?? ?? 2
2 2n
4 3
 
 
????
?? ?K ?] ?? ?? n
8 3
 
 n = 0,  
 
5
x
3 8 24
?? ?? ?? ?]?M?] 
8. There are 5 points P
1
, P
2
, P
3
, P
4
, P
5
 on the side AB, 
excluding A and B, of a triangle ABC. Similarly 
there are 6 points P
6
, P
7
, …, P
11
 on the side BC and 7 
points P
12
, P
13
, …, P
18
 on the side CA of the triangle. 
The number of triangles, that can be formed using the 
points P
1
, P
2
, …, P
18
 as vertices, is : 
 (1) 776 (2) 751 
 (3) 796  (4) 771 
 Ans. (2) 
Sol. 
18
C
3
 – 
5
C
3
 – 
6
C
3
 – 
7
C
3
 
 = 751 
9. Let ƒ(x) = 
2, 2 x 0
x 2, 0 x 2
?M ?M ?? ?? ?? ?? ?M ?\ ?? ?? and h(x) = f(|x|) + |f(x)|. 
Then ?H ?I 2
2
h x dx
?M ?? is equal to : 
 (1) 2 (2) 4 
 (3) 1  (4) 6 
 Ans. (1) 
Sol. 
 
y=f(x
x 
2 
x–2 
–2 
–2 
 
  f(|x|) ??    |f(x)| 
 
x 
(2,0) 
x–2 
(0,–2) 
(–2,0) 
0 
–x–2 
    
 
x 
2 
–2 
0 
2 
 
 
?M ?K ?M ?] ?? ?? ?? ?] ?? ?M ?M ?K ?] ?M ?M ?? ?\ ?? x 2 2 x 0, 0 x 2
h(x)
x 2 2 x 2 x 0
 
 
 
2 0 
–2 
 
 ?? ?] ?? 2
0
h(x)dx 0 and ?H ?I ?M ?] ?? 0
2
h x dx 2 
10. The sum of all rational terms in the expansion of 
?H ?I 15
1 1
5 3
2 5 ?K is equal to : 
 (1) 3133 (2) 633 
 (3) 931  (4) 6131 
 Ans. (1) 
Sol. T
r + 1
 = 
15
C
r
 
?H ?I ?H ?I ?M 15 r r
1 1
3 5
5 2 
 
?M ?] r 15 r
15
3 5
r
C 5 . 2 
 R = 3 ?? , 15µ 
 ?? r = 0, 15 
 2 rational terms 
 ?H ?I ???K
5
15 3 15
0 15
C 2 C 5 
 = 8 + 3125 = 3133 
 
11. Let a unit vector which makes an angle of 60° with 
ˆˆ ˆ
2i 2j k ?K?M and an angle of 45° with 
ˆ ˆ
i k ?M be C . 
Then 
1 1 2
ˆˆ ˆ
C i j k
2 3
3 2
????
?K ?M ?K ?M????
????
 is : 
 (1) 
2 2 1 2 2
ˆˆ ˆ
i j k
3 3 2 3
????
?M ?K ?K ?K????
????
 
 (2) 
2 1 1
ˆˆ ˆ
i j k
3 2
3 2
?K?M 
 (3) 
1 1 1 1 1 2
ˆˆ ˆ
i j k
2 3
3 3 3 2 3
????
?? ?? ?? ?? ?K ?K ?M ?K ?K????
?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 (4) 
2 1
ˆ ˆ
i k
3 2
?M 
 Ans. (4) 
Sol. 
1 2 3
ˆˆ ˆ
C C i C j C k ?] ?K ?K 
 C
1
2
 + C
2
2
 + C
3
2
 = 1 
 
?H ?I ?K ?M ?] ?? ˆˆ ˆ
C. 2i 2j k C 9 cos60 
 2C
1
 + 2C
2
 – C
3
 = 
3
2
 
 C
1
 – C
3
 = 1 
 C
1
 + 2C
2
 = 
1
2
 
 C
1
 = 
2 1
3 2
?K 
 C
2
 = 
1
3 2
?M 
 C
3
 = 
2 1
3 2
?M 
12. Let the first three terms 2, p and q, with q ?? 2, of a 
G.P. be respectively the 7
th
, 8
th
 and 13
th
 terms of an 
A.P. If the 5
th
 term of the G.P. is the n
th
 term of the 
A.P., then n is equal to 
 (1) 151 (2) 169 
 (3) 177  (4) 163 
 Ans. (4) 
Sol. p
2
 = 2q 
 2 = a + 6d ...(i) 
 p = a + 7d ...(ii)  
 q = a + 12d ...(iii) 
 p – 2 = d        ((ii) – (i)) 
 q – p = 5d        ((iii) – (ii))  
 q – p = 5(p – 2) 
 q = 6p – 10 
 p
2
 = 2(6p – 10) 
 p
2
 – 12p + 20 = 0 
 p = 10, 2 
 p = 10 ; q = 50 
 d = 8 
 a = –46 
 2, 10, 50, 250, 1250 
 ar
4
 = a + (n – 1)d 
 1250 = –46 + (n – 1)8 
 n = 163 
13. Let a, b ?? R. Let the mean and the variance of 6 
observations –3, 4, 7, –6, a, b be 2 and 23, 
respectively. The mean deviation about the mean 
of these 6 observations is : 
 (1) 
13
3
 (2) 
16
3
 
 (3) 
11
3
  (4) 
14
3
 
 Ans. (1) 
Sol. 
i
x
2
6
?? ?] and 
2
2 i
x
23
N
?? ?M ?? ?] 
 ?? + ?? = 10 
 ?? 2
 + ?? 2
 = 52 
 solving we get ?? = 4, ?? = 6 
 
???M
?K ?K ?K ?K ?K ?]?]
i
x x
5 2 5 8 2 4 13
6 6 3
 
 
 
Page 5


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Thursday 04
th
 April, 2024)    TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ƒ : R ?? R be a function given by
?H ?I 2
1 cos2x
, x 0
x
ƒ x , x 0
1 cosx
, x 0
x
?? ?M ?\ ?? ?? ?? ?] ?? ?] ?? ?? ???M
?? ?^ ?? ?? , where ?? , ?? ?? R. If 
ƒ is continuous at x = 0, then ?? 2
 + ?? 2
 is equal to : 
(1) 48 (2) 12
(3) 3 (4) 6
Ans. (2) 
Sol. ƒ(0
–
) = 
2
2
x 0
2sin x
lim
x
?M ?? = 2 = ?? 
ƒ(0
+
) = 
x 0
x
sin
2
lim 2 2
x
2
2
2
?K ?? ?? ?? ?? ?] ?] ?? ?? = 2 2 
?? 2
+ ?? 2
 = 4 + 8 = 12
2. Three urns A, B and C contain 7 red, 5 black;
5 red, 7 black and 6 red, 6 black balls, respectively.
One of the urn is selected at random and a ball is
drawn from it. If the ball drawn is black, then the
probability that it is drawn from urn A is :
(1) 
4
17
(2) 
5
18
(3) 
7
18
(4) 
5
16
Ans. (2) 
Sol. 
A
7R, 5B
B
5R, 7B
C
6R, 6B
P(B) = 
1 5 1 7 1 6
. . .
3 12 3 12 3 12
?K?K
required probability = 
1 5
.
5
3 12
1 5 7 6 18
.
3 12 12 12
?] ????
?K?K
????
????
3. The vertices of a triangle are A(–1, 3), B(–2, 2) and
C(3, –1). A new triangle is formed by shifting the sides
of the triangle by one unit inwards. Then the equation
of the side of the new triangle nearest to origin is :
(1) x – y – 
?H ?I 2 2 ?K = 0
(2) –x + y – 
?H ?I 2 2 ?M = 0
(3) x + y – 
?H ?I 2 2 ?M = 0
(4) x + y + 
?H ?I 2 2 ?M = 0
Ans. (3) 
Sol. 
C
(–2,2) 
m = –1 
(3, –1) 
B 
A 
(–1,3) 
equation of AC ?? x + y = 2 
equation of line parallel to AC x + y = d 
?M ?] d 2
1
2
d 2 2 ?]?M
eq
n
of new required line
x y 2 2 ?K ?] ?M 4. If the solution y = y(x) of the differential equation
(x
4
 + 2x
3
 + 3x
2
 + 2x + 2)dy – (2x
2
 + 2x + 3)dx = 0
satisfies y(–1) = 
4
?? ?M , then y(0) is equal to :
(1) 
12
?? ?M (2) 0 
(3) 
4
?? (4) 
2
?? Ans. (3) 
Sol. 
?H ?I 2
4 3 2
2x 2x 3
dy dx
x 2x 3x 2x 2
?K?K
?] ?K ?K ?K ?K ????
 
 
?H ?I ?H ?I ?H ?I 2
2 2
2x 2x 3
y dx
x 1 x 2x 2
?K?K
?] ?K ?K ?K ?? 
 
2 2
dx dx
y
x 2x 2 x 1
?]?K
?K ?K ?K ????
 
 y = tan
–1
(x + 1) + tan
–1
x + C 
 y(–1) = 
4
?M??
 
 0 C
4 4
?M ?? ?? ?] ?M ?K  ?? C = 0 
 ?? y = tan
–1
(x + 1) + tan
–1
x 
 y(0) = tan
–1
1 = 
4
?? 
5. Let the sum of the maximum and the minimum 
values of the function f(x) = 
2
2
2x 3x 8
2x 3x 8
?M?K
?K?K
 be 
m
n
, 
where gcd(m, n) = 1. Then m + n is equal to : 
 (1) 182 (2) 217 
 (3) 195  (4) 201 
 Ans. (4) 
Sol. 
2
2
2x 3x 8
y
2x 3x 8
?M?K
?] ?K?K
 
 x
2
(2y – 2) + x(3y + 3) + 8y – 8 = 0 
 use D ?? 0 
 (3y + 3)
2
 – 4(2y – 2) (8y – 8) ?? 0 
 (11y – 5) (5y – 11) ?? 0 
 
5 11
y ,
11 5
????
????
????
????
 
 y = 1 is also included 
6. One of the points of intersection of the curves  
y = 1 + 3x – 2x
2
 and y = 
1
x
 is 
1
,2
2
????
????
????
. Let the area 
of the region enclosed by these curves be 
?H ?I 1
5 m
24
?K – nlog
e ?H ?I 1 5 ?K , where ?? , m, n ?? 
N. Then ?? + m + n is equal to 
 (1) 32 (2) 30 
 (3) 29  (4) 31 
 Ans. (2) 
Sol. 
 
 
 
 
 
1 5
2
2
1
2
1
A 1 3x 2x dx
x
?K????
?] ?K ?M ?M ????
????
?? 
 
1 5
2 3
2
1
2
3x 2x
A x n x
2 3
?K ????
?] ?K ?M ?M ????
????
 
 
2 3
1 5 3 1 5 2 1 5 1 5
A n
2 2 2 3 2 2
?? ?? ?? ?? ?? ?? ?K ?K ?K ?K ?] ?K ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
1 3 1 2 1 1
n
2 2 4 3 8 2
?? ?? ?? ?? ?? ?? ?M ?M ?K ?K ?? ?? ?? ?? ????
?? ?? ?? ?? ????
 
 
1 5 3 3 15 4 2
A 5 5
2 2 8 4 8 3 3
?] ?K ?K ?K ?K ?M ?M 
 
?H ?I 1 3 1
n 1 5
2 8 12
?M ?M ?K ?M ?K 
 
?H ?I 1 3 2 15 4 1
5 n 1 5
2 4 3 8 3 12
????
?] ?K ?M ?K ?M ?K ?M ?K ????
????
 
 
?H ?I ?] ?K ?M ?K 14 15
5 n 1 5
24 24
 
7. If the system of equations 
 
?H ?I ?H ?I x 2 sin y 2 cos z 0 ?K ?? ?K ?? ?] 
 x + (cos ?? )y + (sin ?? )z = 0 
 x + (sin ?? )y – (cos ?? )z = 0 
 has a non-trivial solution, then 0,
2
?? ????
????
????
????
 is equal to : 
 (1) 
3
4
?? (2) 
7
24
?? 
 (3) 
5
24
??  (4) 
11
24
?? 
 Ans. (3) 
 
Sol. 
????
?? ?M ?? ?] ????
1 2 sin 2 cos
1 sin cos 0
1 cos sin
 
 ?? 
?H ?I ?H ?I ?M ?? ?? ?K ?? ?K ?? ?? ?M ?? ?] 1 2 sin sin cos 2 cos cos sin 0
 
 ?? ?K ?? ?M ?? ?] 1 2 cos2 2 sin2 0 
 ?? ?M ?? ?] ?M 1
cos2 sin 2
2
 
 
?? ????
?? ?K ?] ?M ????
????
1
cos 2
4 2
 
 
????
?? ?K ?] ?? ?? 2
2 2n
4 3
 
 
????
?? ?K ?] ?? ?? n
8 3
 
 n = 0,  
 
5
x
3 8 24
?? ?? ?? ?]?M?] 
8. There are 5 points P
1
, P
2
, P
3
, P
4
, P
5
 on the side AB, 
excluding A and B, of a triangle ABC. Similarly 
there are 6 points P
6
, P
7
, …, P
11
 on the side BC and 7 
points P
12
, P
13
, …, P
18
 on the side CA of the triangle. 
The number of triangles, that can be formed using the 
points P
1
, P
2
, …, P
18
 as vertices, is : 
 (1) 776 (2) 751 
 (3) 796  (4) 771 
 Ans. (2) 
Sol. 
18
C
3
 – 
5
C
3
 – 
6
C
3
 – 
7
C
3
 
 = 751 
9. Let ƒ(x) = 
2, 2 x 0
x 2, 0 x 2
?M ?M ?? ?? ?? ?? ?M ?\ ?? ?? and h(x) = f(|x|) + |f(x)|. 
Then ?H ?I 2
2
h x dx
?M ?? is equal to : 
 (1) 2 (2) 4 
 (3) 1  (4) 6 
 Ans. (1) 
Sol. 
 
y=f(x
x 
2 
x–2 
–2 
–2 
 
  f(|x|) ??    |f(x)| 
 
x 
(2,0) 
x–2 
(0,–2) 
(–2,0) 
0 
–x–2 
    
 
x 
2 
–2 
0 
2 
 
 
?M ?K ?M ?] ?? ?? ?? ?] ?? ?M ?M ?K ?] ?M ?M ?? ?\ ?? x 2 2 x 0, 0 x 2
h(x)
x 2 2 x 2 x 0
 
 
 
2 0 
–2 
 
 ?? ?] ?? 2
0
h(x)dx 0 and ?H ?I ?M ?] ?? 0
2
h x dx 2 
10. The sum of all rational terms in the expansion of 
?H ?I 15
1 1
5 3
2 5 ?K is equal to : 
 (1) 3133 (2) 633 
 (3) 931  (4) 6131 
 Ans. (1) 
Sol. T
r + 1
 = 
15
C
r
 
?H ?I ?H ?I ?M 15 r r
1 1
3 5
5 2 
 
?M ?] r 15 r
15
3 5
r
C 5 . 2 
 R = 3 ?? , 15µ 
 ?? r = 0, 15 
 2 rational terms 
 ?H ?I ???K
5
15 3 15
0 15
C 2 C 5 
 = 8 + 3125 = 3133 
 
11. Let a unit vector which makes an angle of 60° with 
ˆˆ ˆ
2i 2j k ?K?M and an angle of 45° with 
ˆ ˆ
i k ?M be C . 
Then 
1 1 2
ˆˆ ˆ
C i j k
2 3
3 2
????
?K ?M ?K ?M????
????
 is : 
 (1) 
2 2 1 2 2
ˆˆ ˆ
i j k
3 3 2 3
????
?M ?K ?K ?K????
????
 
 (2) 
2 1 1
ˆˆ ˆ
i j k
3 2
3 2
?K?M 
 (3) 
1 1 1 1 1 2
ˆˆ ˆ
i j k
2 3
3 3 3 2 3
????
?? ?? ?? ?? ?K ?K ?M ?K ?K????
?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 (4) 
2 1
ˆ ˆ
i k
3 2
?M 
 Ans. (4) 
Sol. 
1 2 3
ˆˆ ˆ
C C i C j C k ?] ?K ?K 
 C
1
2
 + C
2
2
 + C
3
2
 = 1 
 
?H ?I ?K ?M ?] ?? ˆˆ ˆ
C. 2i 2j k C 9 cos60 
 2C
1
 + 2C
2
 – C
3
 = 
3
2
 
 C
1
 – C
3
 = 1 
 C
1
 + 2C
2
 = 
1
2
 
 C
1
 = 
2 1
3 2
?K 
 C
2
 = 
1
3 2
?M 
 C
3
 = 
2 1
3 2
?M 
12. Let the first three terms 2, p and q, with q ?? 2, of a 
G.P. be respectively the 7
th
, 8
th
 and 13
th
 terms of an 
A.P. If the 5
th
 term of the G.P. is the n
th
 term of the 
A.P., then n is equal to 
 (1) 151 (2) 169 
 (3) 177  (4) 163 
 Ans. (4) 
Sol. p
2
 = 2q 
 2 = a + 6d ...(i) 
 p = a + 7d ...(ii)  
 q = a + 12d ...(iii) 
 p – 2 = d        ((ii) – (i)) 
 q – p = 5d        ((iii) – (ii))  
 q – p = 5(p – 2) 
 q = 6p – 10 
 p
2
 = 2(6p – 10) 
 p
2
 – 12p + 20 = 0 
 p = 10, 2 
 p = 10 ; q = 50 
 d = 8 
 a = –46 
 2, 10, 50, 250, 1250 
 ar
4
 = a + (n – 1)d 
 1250 = –46 + (n – 1)8 
 n = 163 
13. Let a, b ?? R. Let the mean and the variance of 6 
observations –3, 4, 7, –6, a, b be 2 and 23, 
respectively. The mean deviation about the mean 
of these 6 observations is : 
 (1) 
13
3
 (2) 
16
3
 
 (3) 
11
3
  (4) 
14
3
 
 Ans. (1) 
Sol. 
i
x
2
6
?? ?] and 
2
2 i
x
23
N
?? ?M ?? ?] 
 ?? + ?? = 10 
 ?? 2
 + ?? 2
 = 52 
 solving we get ?? = 4, ?? = 6 
 
???M
?K ?K ?K ?K ?K ?]?]
i
x x
5 2 5 8 2 4 13
6 6 3
 
 
 
 
14. If 2 and 6 are the roots of the equation ax
2
 + bx + 1 = 0, 
then the quadratic equation, whose roots are 
1
2a b ?K and 
1
6a b ?K , is : 
 (1) 2x
2
 + 11x + 12 = 0 (2) 4x
2
 + 14x + 12 = 0 
 (3) x
2
 + 10x + 16 = 0 (4) x
2
 + 8x + 12 = 0 
 Ans. (4) 
Sol. Sum = 8 = 
b
a
?M 
 Product = 12 = 
1
a
 ?? a = 
1
12
 
         b = 
2
3
?M 
 2a + b = 
2 2 1
12 3 2
?M ?] ?M 
 6a + b = 
6 2 1
12 3 6
?M ?] ?M 
 sum = –8 
 P = 12 
 x
2
 + 8x + 12 = 0 
15. Let ?? and ?? be the sum and the product of all the 
non-zero solutions of the equation 
?H ?I 2
z z 0 ?K?] , z ?? C. 
Then 4( ?? 2
 + ?? 2
) is equal to : 
 (1) 6 (2) 4 
 (3) 8  (4) 2 
 Ans. (2) 
 
Sol. z = x + iy 
 z x iy ?]?M 
 
2 2 2
z x y 2ixy ?] ?M ?M 
 
2 2 2 2
x y 2ixy x y 0 ?? ?M ?M ?K ?K ?] 
 ?? x = 0 or y = 0 
 –y
2
 + |y| = 0  x
2
 + |x| = 0 
 |y| = |y|
2
  ?? x = 0 
 y = 0, ±1 
 ?? i, –i  ?? ?? = i – i = 0 
 are roots  ?? = i(–i) = 1 
    4(0 + 1) = 4 
16. Let the point, on the line passing through the points 
P(1, –2, 3) and Q(5, –4, 7), farther from the origin 
and at a distance of 9 units from the point P, be  
( ?? , ?? , ?? ). Then ?? 2
 + ?? 2
 + ?? 2
 is equal to : 
 (1) 155 (2) 150 
 (3) 160  (4) 165 
 Ans. (1) 
Sol. PQ line 
 
x 1 y 2 z 3
4 2 4
?M ?K ?M ?]?]
?M 
 pt (4t + 1, –2t – 2, 4t + 3) 
 distance
2
 = 16t
2
 + 4t
2
 + 16t
2
 = 81 
 
3
t
2
?]?? 
 pt (7, –5, 9) 
 ?? 2
 + ?? 2
 + ?? 2
 = 155 
 option (1) 
17. A square is inscribed in the circle  
x
2
 + y
2
 – 10x – 6y + 30 = 0. One side of this square 
is parallel to y = x + 3. If (x
i
, y
i
) are the vertices of 
the square, then ?H ?I 2 2
i i
x y ?K ?? is equal to : 
 (1) 148 (2) 156 
 (3) 160  (4) 152 
 Ans. (4) 
Sol. 
 
(5,3) 
2 
 
 y = x + c & x + y + d = 0 
 
5 3 c
2
2
?M?K
?] 
8 d
2
2
?K ?] 
 |c + 2| = 2  8 + d = ±2 
 c = 0, – 4  d = –10, –6 
 pts (5, 5), (3, 3), (7, 3), (5, 1) 
 ?H ?I 2 2
i 1
x y 25 25 9 9 49 9 25 1 ?K ?] ?K ?K ?K ?K ?K ?K ?K ?? 
 = 152 
 Option (4)