JEE Main 2024 April 5 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 Friday 05
th
 April, 2024)   TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let d be the distance of the point of intersection of
the lines 
x 6 y z 1
3 2 1
?K?K
?]?] and 
x 7 y 9 z 4
4 3 2
?M ?M ?M ?]?] from the point (7, 8, 9). Then
d
2
+ 6 is equal to :
(1) 72 (2) 69
(3) 75 (4) 78
Ans. (3) 
Sol. 
x 6 y z 1
3 2 1
?K?K
?] ?] ?] ?? …(1) 
 x = 3 ?? – 6, y = 2 ?? , z = ?? – 1  
x 7 y 9 z 4
µ
4 3 2
?M ?M ?M ?] ?] ?] …(2) 
x = 4µ + 7, y = 3µ + 9, z = 2µ + 4 
3 ?? – 6 = 4µ + 7 ?? 3 ?? – 4µ = 13 …(3) × 2 
2 ?? = 3µ + 9 ?? 2 ?? – 3µ = 9 …(4) × 3 
6 ?? – 8µ = 26 
6 ?? – 9µ = 27 
– + – .
       µ = –1  
?? 3 ?? – 4(–1) = 13
3 ?? = 9 
?? = 3
int. point  (3, 6, 2) ;  (7, 8, 9) 
d
2
 = 16 + 4 + 49 = 69  
Ans. d
2
 + 6 = 69 + 6 = 75 
2. Let a rectangle ABCD of sides 2 and 4 be inscribed
in another rectangle PQRS such that the vertices of
the rectangle ABCD lie on the sides of the
rectangle PQRS. Let a and b be the sides of the
rectangle PQRS when its area is maximum. Then
(a + b)
2
 is equal to :
(1) 72 (2) 60
(3) 80 (4) 64
Ans. (1 ) 
Sol. 
2cos(90– ?? )
 
P
 
B
 
Q
 
4
 
4
 
R
 
S
 
A
 
2 2
 
4cos(90– ?? )
 
2sin(90– ?? )
 
D
 
90– ?? 
?? 
?? 
Area = (4cos ?? + 2sin ?? ) (2cos ?? + 4sin ?? ) 
= 8cos
2
?? + 16sin ?? cos ?? + 4sin ?? cos ?? + 8sin
2
?? 
= 8 + 20 sin ?? cos ?? 
= 8 + 10 sin2 ?? ?@ Max Area =  8 + 10 = 18 (sin2 ?? ?@?] ?@?Q?I ?@ ?? ?@?] ?@ ?T?U?? ?@ (a + b)
2
 = (4cos ?? + 2sin ?? ?@ + 2cos ?? + 4sin ?? )
2
?@ = (6cos ?? + 6sin ?? )
2 
?@ = 36 (sin ?? + cos ?? )
2 
?@ ?]?@
2
36( 2) ?@ ?] ?@?W?R
Page 2


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Friday 05
th
 April, 2024)   TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let d be the distance of the point of intersection of
the lines 
x 6 y z 1
3 2 1
?K?K
?]?] and 
x 7 y 9 z 4
4 3 2
?M ?M ?M ?]?] from the point (7, 8, 9). Then
d
2
+ 6 is equal to :
(1) 72 (2) 69
(3) 75 (4) 78
Ans. (3) 
Sol. 
x 6 y z 1
3 2 1
?K?K
?] ?] ?] ?? …(1) 
 x = 3 ?? – 6, y = 2 ?? , z = ?? – 1  
x 7 y 9 z 4
µ
4 3 2
?M ?M ?M ?] ?] ?] …(2) 
x = 4µ + 7, y = 3µ + 9, z = 2µ + 4 
3 ?? – 6 = 4µ + 7 ?? 3 ?? – 4µ = 13 …(3) × 2 
2 ?? = 3µ + 9 ?? 2 ?? – 3µ = 9 …(4) × 3 
6 ?? – 8µ = 26 
6 ?? – 9µ = 27 
– + – .
       µ = –1  
?? 3 ?? – 4(–1) = 13
3 ?? = 9 
?? = 3
int. point  (3, 6, 2) ;  (7, 8, 9) 
d
2
 = 16 + 4 + 49 = 69  
Ans. d
2
 + 6 = 69 + 6 = 75 
2. Let a rectangle ABCD of sides 2 and 4 be inscribed
in another rectangle PQRS such that the vertices of
the rectangle ABCD lie on the sides of the
rectangle PQRS. Let a and b be the sides of the
rectangle PQRS when its area is maximum. Then
(a + b)
2
 is equal to :
(1) 72 (2) 60
(3) 80 (4) 64
Ans. (1 ) 
Sol. 
2cos(90– ?? )
 
P
 
B
 
Q
 
4
 
4
 
R
 
S
 
A
 
2 2
 
4cos(90– ?? )
 
2sin(90– ?? )
 
D
 
90– ?? 
?? 
?? 
Area = (4cos ?? + 2sin ?? ) (2cos ?? + 4sin ?? ) 
= 8cos
2
?? + 16sin ?? cos ?? + 4sin ?? cos ?? + 8sin
2
?? 
= 8 + 20 sin ?? cos ?? 
= 8 + 10 sin2 ?? ?@ Max Area =  8 + 10 = 18 (sin2 ?? ?@?] ?@?Q?I ?@ ?? ?@?] ?@ ?T?U?? ?@ (a + b)
2
 = (4cos ?? + 2sin ?? ?@ + 2cos ?? + 4sin ?? )
2
?@ = (6cos ?? + 6sin ?? )
2 
?@ = 36 (sin ?? + cos ?? )
2 
?@ ?]?@
2
36( 2) ?@ ?] ?@?W?R 3. Let two straight lines drawn from the origin  
O intersect the line 3x + 4y = 12 at the points P and 
Q such that ?d OPQ is an isosceles triangle and 
?? POQ = 90°. If l = OP
2
 + PQ
2
 + QO
2
, then the 
greatest integer less than or equal to l is :  
 (1) 44  (2) 48  
 (3) 46   (4) 42  
 Ans. (3) 
Sol.  
O
 
?? ?@ P(rcos ?? , rsin ?? )
 
Q(rcos(90+???I , rsin(90+ ?? ) = (–rsin ???L rcos ?? )
 
 
 3x + 4y = 12 
 3(rcos ?? ) + 4(rsin ?? ) = 12 
 r(3cos ?? + 4sin ?? ) = 12   ...(1) 
 3(–rsin ?? ) + 4(rcos ?? ) = 12 
 r(–3sin ?? + 4cos ?? ) = 12  ...(2) 
 
2 2
2 2
12 12
(3cos 4sin ) ( 3sin 4cos )
r r
?? ?? ?? ?? ?K ?] ?? ?K ?? ?K ?M ?? ?K ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
2
12
2 9 16
r
????
?]?K
????
????
 
 
2
2 144
25
r
?? ?] ?? ?@ 288 = 25r
2 
?@ ?? ?@ 2
288
r
25
?] ?@ ?@ 12
2 r
5
????
???]
????
????
?@ ?@ ?? = OP
2
 + PQ
2
 + QO
2
 
?@ ???@ = r
2
 + r
2
?@ + r
2
(cos ?? + sin ?? )
2
 + r
2
(sin ?? + cos ?? )
2
 
 = 2r
2
 + r
2
(1 + sin2 ?? + 1 – 2sin2 ?? ) 
 = 2r
2
 + 2r
2
 
 = 4r
2 
 
1152 288
4 46.08
25 25
????
?] ?] ?] ????
????
 
 [ ?? ] = 46 
4. If y = y(x) is the solution of the differential 
equation 
dy
dx
 + 2y = sin (2x), y(0) = 
3
4
, then 
y
8
?? ????
????
????
 is equal to :  
 (1) e
– ?? /8
  (2) e
– ?? /4
 
 (3) e
?? /4
  (4) e
?? /8
 
 Ans. (2 ) 
Sol. 
dy
2y sin 2x
dx
?K?] , 
3
y(0)
4
?]  
 I.F = 
2dx
e
?? = e
2x
  
 
2x 2x
y.e e sin2xdx ?] ??  
 
2x
2x
e (2sin 2x 2cos2x)
y.e C
4 4
?M ?]?K
?K  
 x = 0, y = 
3
4
 ?? 
3 1(0 2)
.1 C
4 8
?M ?]?K  
 
3 1
C
4 4
?] ?M ?K  
 1 = C  
 
2x
2sin 2x 2cos2x
y 1.e
8
?M ?M ?]?K  
 x
8
?? ?] ,  
2
8
1
y 2sin 2cos e
8 4 4
?? ????
?M ????
????
???? ????
?] ?M ?K ????
????
  
 
4
y 0 e
?? ?M ?]?K  
5. For the function  
 f(x) = sinx + 3x – 
2
?? (x
2
 + x), where x ?? 0,
2
?? ????
????
????
, 
consider the following two statements :  
 (I)  f is increasing in 0,
2
?? ????
????
????
. 
 (II) f ?? is decreasing in 0,
2
?? ????
????
????
.  
 Between the above two statements,  
 (1) only (I) is true. 
 (2) only (II) is true.  
 (3) neither (I) nor (II) is true.  
 (4) both (I) and (II) are true.  
Page 3


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Friday 05
th
 April, 2024)   TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let d be the distance of the point of intersection of
the lines 
x 6 y z 1
3 2 1
?K?K
?]?] and 
x 7 y 9 z 4
4 3 2
?M ?M ?M ?]?] from the point (7, 8, 9). Then
d
2
+ 6 is equal to :
(1) 72 (2) 69
(3) 75 (4) 78
Ans. (3) 
Sol. 
x 6 y z 1
3 2 1
?K?K
?] ?] ?] ?? …(1) 
 x = 3 ?? – 6, y = 2 ?? , z = ?? – 1  
x 7 y 9 z 4
µ
4 3 2
?M ?M ?M ?] ?] ?] …(2) 
x = 4µ + 7, y = 3µ + 9, z = 2µ + 4 
3 ?? – 6 = 4µ + 7 ?? 3 ?? – 4µ = 13 …(3) × 2 
2 ?? = 3µ + 9 ?? 2 ?? – 3µ = 9 …(4) × 3 
6 ?? – 8µ = 26 
6 ?? – 9µ = 27 
– + – .
       µ = –1  
?? 3 ?? – 4(–1) = 13
3 ?? = 9 
?? = 3
int. point  (3, 6, 2) ;  (7, 8, 9) 
d
2
 = 16 + 4 + 49 = 69  
Ans. d
2
 + 6 = 69 + 6 = 75 
2. Let a rectangle ABCD of sides 2 and 4 be inscribed
in another rectangle PQRS such that the vertices of
the rectangle ABCD lie on the sides of the
rectangle PQRS. Let a and b be the sides of the
rectangle PQRS when its area is maximum. Then
(a + b)
2
 is equal to :
(1) 72 (2) 60
(3) 80 (4) 64
Ans. (1 ) 
Sol. 
2cos(90– ?? )
 
P
 
B
 
Q
 
4
 
4
 
R
 
S
 
A
 
2 2
 
4cos(90– ?? )
 
2sin(90– ?? )
 
D
 
90– ?? 
?? 
?? 
Area = (4cos ?? + 2sin ?? ) (2cos ?? + 4sin ?? ) 
= 8cos
2
?? + 16sin ?? cos ?? + 4sin ?? cos ?? + 8sin
2
?? 
= 8 + 20 sin ?? cos ?? 
= 8 + 10 sin2 ?? ?@ Max Area =  8 + 10 = 18 (sin2 ?? ?@?] ?@?Q?I ?@ ?? ?@?] ?@ ?T?U?? ?@ (a + b)
2
 = (4cos ?? + 2sin ?? ?@ + 2cos ?? + 4sin ?? )
2
?@ = (6cos ?? + 6sin ?? )
2 
?@ = 36 (sin ?? + cos ?? )
2 
?@ ?]?@
2
36( 2) ?@ ?] ?@?W?R 3. Let two straight lines drawn from the origin  
O intersect the line 3x + 4y = 12 at the points P and 
Q such that ?d OPQ is an isosceles triangle and 
?? POQ = 90°. If l = OP
2
 + PQ
2
 + QO
2
, then the 
greatest integer less than or equal to l is :  
 (1) 44  (2) 48  
 (3) 46   (4) 42  
 Ans. (3) 
Sol.  
O
 
?? ?@ P(rcos ?? , rsin ?? )
 
Q(rcos(90+???I , rsin(90+ ?? ) = (–rsin ???L rcos ?? )
 
 
 3x + 4y = 12 
 3(rcos ?? ) + 4(rsin ?? ) = 12 
 r(3cos ?? + 4sin ?? ) = 12   ...(1) 
 3(–rsin ?? ) + 4(rcos ?? ) = 12 
 r(–3sin ?? + 4cos ?? ) = 12  ...(2) 
 
2 2
2 2
12 12
(3cos 4sin ) ( 3sin 4cos )
r r
?? ?? ?? ?? ?K ?] ?? ?K ?? ?K ?M ?? ?K ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
2
12
2 9 16
r
????
?]?K
????
????
 
 
2
2 144
25
r
?? ?] ?? ?@ 288 = 25r
2 
?@ ?? ?@ 2
288
r
25
?] ?@ ?@ 12
2 r
5
????
???]
????
????
?@ ?@ ?? = OP
2
 + PQ
2
 + QO
2
 
?@ ???@ = r
2
 + r
2
?@ + r
2
(cos ?? + sin ?? )
2
 + r
2
(sin ?? + cos ?? )
2
 
 = 2r
2
 + r
2
(1 + sin2 ?? + 1 – 2sin2 ?? ) 
 = 2r
2
 + 2r
2
 
 = 4r
2 
 
1152 288
4 46.08
25 25
????
?] ?] ?] ????
????
 
 [ ?? ] = 46 
4. If y = y(x) is the solution of the differential 
equation 
dy
dx
 + 2y = sin (2x), y(0) = 
3
4
, then 
y
8
?? ????
????
????
 is equal to :  
 (1) e
– ?? /8
  (2) e
– ?? /4
 
 (3) e
?? /4
  (4) e
?? /8
 
 Ans. (2 ) 
Sol. 
dy
2y sin 2x
dx
?K?] , 
3
y(0)
4
?]  
 I.F = 
2dx
e
?? = e
2x
  
 
2x 2x
y.e e sin2xdx ?] ??  
 
2x
2x
e (2sin 2x 2cos2x)
y.e C
4 4
?M ?]?K
?K  
 x = 0, y = 
3
4
 ?? 
3 1(0 2)
.1 C
4 8
?M ?]?K  
 
3 1
C
4 4
?] ?M ?K  
 1 = C  
 
2x
2sin 2x 2cos2x
y 1.e
8
?M ?M ?]?K  
 x
8
?? ?] ,  
2
8
1
y 2sin 2cos e
8 4 4
?? ????
?M ????
????
???? ????
?] ?M ?K ????
????
  
 
4
y 0 e
?? ?M ?]?K  
5. For the function  
 f(x) = sinx + 3x – 
2
?? (x
2
 + x), where x ?? 0,
2
?? ????
????
????
, 
consider the following two statements :  
 (I)  f is increasing in 0,
2
?? ????
????
????
. 
 (II) f ?? is decreasing in 0,
2
?? ????
????
????
.  
 Between the above two statements,  
 (1) only (I) is true. 
 (2) only (II) is true.  
 (3) neither (I) nor (II) is true.  
 (4) both (I) and (II) are true.  
 
 Ans. (4) 
Sol. f(x) = sinx + 3x – 
2
?? (x
2
 + x)  x 0,
2
?? ????
?? ????
????
  
 f ?? (x) = cosx + 3 – 
2
?? (2x + 1) > 0  f(x) ??  
 f ?? (x) = –sinx + 0 – 
2
?? (2)  
 = –sinx – 
4
?? < 0   f ?? (x)  ??  
 0 < x < 
2
??  
 ?? 
?H ?I 1 1 1
2
0 2x
?K ?K?K
?M ?\ ?\ ?? ??  
 
3 3 3
2 2 2
(2x 1) ( 1)
?K ?K ?K ?M ?M ?^ ?K ?^ ?M ?? ?K ?? ?? ??  
 
( ve) ( ve)
2 2 2
3 3 (2x 1) 3 ( 1)
?K?K
?M ?^ ?M ?K ?^ ?M ?? ?K ?? ?? ??  
6. If the system of equations  
 11x + y + ?? z = –5 
 2x + 3y + 5z = 3  
 8x – 19y – 39z = µ 
 has infinitely many solutions, then ?? 4
 – µ is equal 
to :  
 (1) 49  (2) 45  
 (3) 47   (4) 51  
 Ans. (3) 
Sol. 11x + y + ?? z = –5  
 2x + 3y + 5z = 3  
 8x – 19y – 39z = µ  
 for infinite sol.  
 
11 1
D 2 3 5 0
8 19 39
?? ?]?]
?M?M
  
 ?? 11(–117 + 95) – 1(–78 – 40) + ?? (–38 – 24)  
 ?? 11(–22) + 118 – ?? (62) = 0  
 ?? 62 ?? = 118 – 242  
 ?? 
124
2
62
?M ?? ?] ?] ?M  
 
1
5 1 2
D 3 3 5 0
µ 19 39
?M?M
?]?]
?M?M
  
 ?? –5(–117 + 95) – 1(–117 – 5µ) – 2(–57 – 3µ) = 0  
 ?? –5(–22) + 117 + 5µ + 114 + 6µ = 0  
 ?? 11µ = –110 – 231 = –341  
 ?? µ = –31  
 ?? 4
 – µ = (–2)
4
 – (–31) = 16 + 31 = 47 
7. Let A = {1, 3, 7, 9, 11} and B = {2, 4, 5, 7, 8, 10, 12}. 
Then the total number of one-one maps  
f : A ?? B, such that f (1) + f(3) = 14, is :  
 (1) 180  (2) 120  
 (3) 480   (4) 240  
 Ans. (4) 
Sol. 
1 
3 
7 
9 
11
 
(5)
 
2
 
(7)
 
12
 
4
 
5
 
7
 
8
 
10
 
 
 A = {1, 3, 7, 9, 11} 
 B = {2, 4, 5, 7, 8, 10, 12} 
 f(1) + f(3) = 14 
 (i) 2 + 12 
 (ii) 4 + 10 
 2 × (2 × 5 × 4 × 3) = 240  
8. If the function 
3
sin3x sin x cos3x
f(x)
x
?K ?? ?M ?? ?] ,  
x ?? R, is continuous at x = 0, then f(0) is equal to :  
 (1) 2  (2) –2  
 (3) 4   (4) –4  
 Ans. (4) 
Sol. 
3
sin3x sin x cos3x
f(x)
x
?K ?? ?M ?? ?]  
 is continuous at x = 0 
3 3 2
3
x 0
(3x) x (3x)
3x ... x ... 1 ...
3 3 2
lim f(0)
x
?? ????????
?M ?K ?K ?? ?M ?? ?M ?M ????????
???? ????
?]?] 
Page 4


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Friday 05
th
 April, 2024)   TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let d be the distance of the point of intersection of
the lines 
x 6 y z 1
3 2 1
?K?K
?]?] and 
x 7 y 9 z 4
4 3 2
?M ?M ?M ?]?] from the point (7, 8, 9). Then
d
2
+ 6 is equal to :
(1) 72 (2) 69
(3) 75 (4) 78
Ans. (3) 
Sol. 
x 6 y z 1
3 2 1
?K?K
?] ?] ?] ?? …(1) 
 x = 3 ?? – 6, y = 2 ?? , z = ?? – 1  
x 7 y 9 z 4
µ
4 3 2
?M ?M ?M ?] ?] ?] …(2) 
x = 4µ + 7, y = 3µ + 9, z = 2µ + 4 
3 ?? – 6 = 4µ + 7 ?? 3 ?? – 4µ = 13 …(3) × 2 
2 ?? = 3µ + 9 ?? 2 ?? – 3µ = 9 …(4) × 3 
6 ?? – 8µ = 26 
6 ?? – 9µ = 27 
– + – .
       µ = –1  
?? 3 ?? – 4(–1) = 13
3 ?? = 9 
?? = 3
int. point  (3, 6, 2) ;  (7, 8, 9) 
d
2
 = 16 + 4 + 49 = 69  
Ans. d
2
 + 6 = 69 + 6 = 75 
2. Let a rectangle ABCD of sides 2 and 4 be inscribed
in another rectangle PQRS such that the vertices of
the rectangle ABCD lie on the sides of the
rectangle PQRS. Let a and b be the sides of the
rectangle PQRS when its area is maximum. Then
(a + b)
2
 is equal to :
(1) 72 (2) 60
(3) 80 (4) 64
Ans. (1 ) 
Sol. 
2cos(90– ?? )
 
P
 
B
 
Q
 
4
 
4
 
R
 
S
 
A
 
2 2
 
4cos(90– ?? )
 
2sin(90– ?? )
 
D
 
90– ?? 
?? 
?? 
Area = (4cos ?? + 2sin ?? ) (2cos ?? + 4sin ?? ) 
= 8cos
2
?? + 16sin ?? cos ?? + 4sin ?? cos ?? + 8sin
2
?? 
= 8 + 20 sin ?? cos ?? 
= 8 + 10 sin2 ?? ?@ Max Area =  8 + 10 = 18 (sin2 ?? ?@?] ?@?Q?I ?@ ?? ?@?] ?@ ?T?U?? ?@ (a + b)
2
 = (4cos ?? + 2sin ?? ?@ + 2cos ?? + 4sin ?? )
2
?@ = (6cos ?? + 6sin ?? )
2 
?@ = 36 (sin ?? + cos ?? )
2 
?@ ?]?@
2
36( 2) ?@ ?] ?@?W?R 3. Let two straight lines drawn from the origin  
O intersect the line 3x + 4y = 12 at the points P and 
Q such that ?d OPQ is an isosceles triangle and 
?? POQ = 90°. If l = OP
2
 + PQ
2
 + QO
2
, then the 
greatest integer less than or equal to l is :  
 (1) 44  (2) 48  
 (3) 46   (4) 42  
 Ans. (3) 
Sol.  
O
 
?? ?@ P(rcos ?? , rsin ?? )
 
Q(rcos(90+???I , rsin(90+ ?? ) = (–rsin ???L rcos ?? )
 
 
 3x + 4y = 12 
 3(rcos ?? ) + 4(rsin ?? ) = 12 
 r(3cos ?? + 4sin ?? ) = 12   ...(1) 
 3(–rsin ?? ) + 4(rcos ?? ) = 12 
 r(–3sin ?? + 4cos ?? ) = 12  ...(2) 
 
2 2
2 2
12 12
(3cos 4sin ) ( 3sin 4cos )
r r
?? ?? ?? ?? ?K ?] ?? ?K ?? ?K ?M ?? ?K ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
2
12
2 9 16
r
????
?]?K
????
????
 
 
2
2 144
25
r
?? ?] ?? ?@ 288 = 25r
2 
?@ ?? ?@ 2
288
r
25
?] ?@ ?@ 12
2 r
5
????
???]
????
????
?@ ?@ ?? = OP
2
 + PQ
2
 + QO
2
 
?@ ???@ = r
2
 + r
2
?@ + r
2
(cos ?? + sin ?? )
2
 + r
2
(sin ?? + cos ?? )
2
 
 = 2r
2
 + r
2
(1 + sin2 ?? + 1 – 2sin2 ?? ) 
 = 2r
2
 + 2r
2
 
 = 4r
2 
 
1152 288
4 46.08
25 25
????
?] ?] ?] ????
????
 
 [ ?? ] = 46 
4. If y = y(x) is the solution of the differential 
equation 
dy
dx
 + 2y = sin (2x), y(0) = 
3
4
, then 
y
8
?? ????
????
????
 is equal to :  
 (1) e
– ?? /8
  (2) e
– ?? /4
 
 (3) e
?? /4
  (4) e
?? /8
 
 Ans. (2 ) 
Sol. 
dy
2y sin 2x
dx
?K?] , 
3
y(0)
4
?]  
 I.F = 
2dx
e
?? = e
2x
  
 
2x 2x
y.e e sin2xdx ?] ??  
 
2x
2x
e (2sin 2x 2cos2x)
y.e C
4 4
?M ?]?K
?K  
 x = 0, y = 
3
4
 ?? 
3 1(0 2)
.1 C
4 8
?M ?]?K  
 
3 1
C
4 4
?] ?M ?K  
 1 = C  
 
2x
2sin 2x 2cos2x
y 1.e
8
?M ?M ?]?K  
 x
8
?? ?] ,  
2
8
1
y 2sin 2cos e
8 4 4
?? ????
?M ????
????
???? ????
?] ?M ?K ????
????
  
 
4
y 0 e
?? ?M ?]?K  
5. For the function  
 f(x) = sinx + 3x – 
2
?? (x
2
 + x), where x ?? 0,
2
?? ????
????
????
, 
consider the following two statements :  
 (I)  f is increasing in 0,
2
?? ????
????
????
. 
 (II) f ?? is decreasing in 0,
2
?? ????
????
????
.  
 Between the above two statements,  
 (1) only (I) is true. 
 (2) only (II) is true.  
 (3) neither (I) nor (II) is true.  
 (4) both (I) and (II) are true.  
 
 Ans. (4) 
Sol. f(x) = sinx + 3x – 
2
?? (x
2
 + x)  x 0,
2
?? ????
?? ????
????
  
 f ?? (x) = cosx + 3 – 
2
?? (2x + 1) > 0  f(x) ??  
 f ?? (x) = –sinx + 0 – 
2
?? (2)  
 = –sinx – 
4
?? < 0   f ?? (x)  ??  
 0 < x < 
2
??  
 ?? 
?H ?I 1 1 1
2
0 2x
?K ?K?K
?M ?\ ?\ ?? ??  
 
3 3 3
2 2 2
(2x 1) ( 1)
?K ?K ?K ?M ?M ?^ ?K ?^ ?M ?? ?K ?? ?? ??  
 
( ve) ( ve)
2 2 2
3 3 (2x 1) 3 ( 1)
?K?K
?M ?^ ?M ?K ?^ ?M ?? ?K ?? ?? ??  
6. If the system of equations  
 11x + y + ?? z = –5 
 2x + 3y + 5z = 3  
 8x – 19y – 39z = µ 
 has infinitely many solutions, then ?? 4
 – µ is equal 
to :  
 (1) 49  (2) 45  
 (3) 47   (4) 51  
 Ans. (3) 
Sol. 11x + y + ?? z = –5  
 2x + 3y + 5z = 3  
 8x – 19y – 39z = µ  
 for infinite sol.  
 
11 1
D 2 3 5 0
8 19 39
?? ?]?]
?M?M
  
 ?? 11(–117 + 95) – 1(–78 – 40) + ?? (–38 – 24)  
 ?? 11(–22) + 118 – ?? (62) = 0  
 ?? 62 ?? = 118 – 242  
 ?? 
124
2
62
?M ?? ?] ?] ?M  
 
1
5 1 2
D 3 3 5 0
µ 19 39
?M?M
?]?]
?M?M
  
 ?? –5(–117 + 95) – 1(–117 – 5µ) – 2(–57 – 3µ) = 0  
 ?? –5(–22) + 117 + 5µ + 114 + 6µ = 0  
 ?? 11µ = –110 – 231 = –341  
 ?? µ = –31  
 ?? 4
 – µ = (–2)
4
 – (–31) = 16 + 31 = 47 
7. Let A = {1, 3, 7, 9, 11} and B = {2, 4, 5, 7, 8, 10, 12}. 
Then the total number of one-one maps  
f : A ?? B, such that f (1) + f(3) = 14, is :  
 (1) 180  (2) 120  
 (3) 480   (4) 240  
 Ans. (4) 
Sol. 
1 
3 
7 
9 
11
 
(5)
 
2
 
(7)
 
12
 
4
 
5
 
7
 
8
 
10
 
 
 A = {1, 3, 7, 9, 11} 
 B = {2, 4, 5, 7, 8, 10, 12} 
 f(1) + f(3) = 14 
 (i) 2 + 12 
 (ii) 4 + 10 
 2 × (2 × 5 × 4 × 3) = 240  
8. If the function 
3
sin3x sin x cos3x
f(x)
x
?K ?? ?M ?? ?] ,  
x ?? R, is continuous at x = 0, then f(0) is equal to :  
 (1) 2  (2) –2  
 (3) 4   (4) –4  
 Ans. (4) 
Sol. 
3
sin3x sin x cos3x
f(x)
x
?K ?? ?M ?? ?]  
 is continuous at x = 0 
3 3 2
3
x 0
(3x) x (3x)
3x ... x ... 1 ...
3 3 2
lim f(0)
x
?? ????????
?M ?K ?K ?? ?M ?? ?M ?M ????????
???? ????
?]?] 
2
3
3
x 0
27 9 x
x(3 ) x ...
3 3 2
lim f(0)
x
?? ?M?? ?? ????
?M ?M ?? ?K ?K ?? ?K ?K ????
????
?]?] 
 for exist 
 ?? = 0, 3+ ?? = 0, 
27
f(0)
3 3
?? ?M ?M ?]  
 ?? ?@ = –3, 
27 ( 3)
f(0)
6 6
?M ?M ?M ?] 
 
27 3
f(0) 4
6
?M?K
?] ?] ?M 
9. The integral 
4
0
136sin x
dx
3sin x 5cosx
?? ?K ?? is equal to :  
 (1) 3 ?? – 50 log
e 
2 + 20 log
e 
5  
 (2) 3 ?? – 25 log
e 
2 + 10 log
e 
5 
 (3) 3 ?? – 10 log
e ?H ?I 2 2 + 10 log
e 
5 
 (4) 3 ?? – 30 log
e 
2 + 20 log
e 
5 
 Ans. (1) 
Sol. 
/4
0
136sin x
I dx
3sin x 5cosx
?? ?] ?K ??  
 136sinx = A(3sinx + 5cosx) + B(3cosx – 5sinx)  
 136 = 3A – 5B  …(1)  
 0 = 5A + 3B  …(2)  
 3B = –5A ?? 
5
B A
3
?]?M  
 
5
136 3A 5 A
3
????
?] ?M ?M????
????
 
 
25
136 3A A
3
?]?K  
 
34A
136
3
?]  
 ?? 
136 3
A 12
34
?? ?]?]  
 
5
B (12) –20
3
?M?]?]  
 
/4 /4
0 0
A(3sin x 5cosx) B(3cosx 5sin x)
I
3sin x 5cosx 3sin x 5cosx
????
?K?M
?]?K
?K?K
????
  
 ?H ?I ?H ?I /4 /4
0 0
A x B n 3sin x 5cosx
?? ?????? ?] ?K ?K????
  
 ?H ?I 3 5
12 20 n n 0 5
4
2 2
?? ???? ????
?] ?M ?K ?M ?K ???? ????
???? ????
  
 3 20 n4 2 20 n5 ?] ?? ?M ?K  
 
5
3 20 n2 20 n5
2
?] ?? ?M ?? ?K  
 3 50 n2 20 n5 ?] ?? ?M ?K 
10. The coefficients a, b, c in the quadratic equation  
ax
2
 + bx + c = 0 are chosen from the set  
 {1, 2, 3, 4, 5, 6, 7, 8}. The probability of this 
equation having repeated roots is :  
 (1) 
3
256
  (2) 
1
128
 
 (3) 
1
64
  (4) 
3
128
 
 Ans. (3) 
Sol. ax
2
 + bx + c = 0  
 a, b, c ?? {1, 2, 3, 4, 5, 6,7, 8}  
 Repeated roots D = 0  
 ?? b
2
 – 4ac = 0 ?? b
2
 = 4ac  
 Prob = 
8 1
8 8 8 64
?] ????
  
 ?? (a, b, c)  
 (1, 2, 1) ; (2, 4, 2) ; (1, 4, 4) ; (4, 4, 1) ; (3, 6, 3) ; 
(2, 8, 8) ; (8, 8, 2) ; (4, 8, 4)  
 8 case 
Page 5


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Friday 05
th
 April, 2024)   TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let d be the distance of the point of intersection of
the lines 
x 6 y z 1
3 2 1
?K?K
?]?] and 
x 7 y 9 z 4
4 3 2
?M ?M ?M ?]?] from the point (7, 8, 9). Then
d
2
+ 6 is equal to :
(1) 72 (2) 69
(3) 75 (4) 78
Ans. (3) 
Sol. 
x 6 y z 1
3 2 1
?K?K
?] ?] ?] ?? …(1) 
 x = 3 ?? – 6, y = 2 ?? , z = ?? – 1  
x 7 y 9 z 4
µ
4 3 2
?M ?M ?M ?] ?] ?] …(2) 
x = 4µ + 7, y = 3µ + 9, z = 2µ + 4 
3 ?? – 6 = 4µ + 7 ?? 3 ?? – 4µ = 13 …(3) × 2 
2 ?? = 3µ + 9 ?? 2 ?? – 3µ = 9 …(4) × 3 
6 ?? – 8µ = 26 
6 ?? – 9µ = 27 
– + – .
       µ = –1  
?? 3 ?? – 4(–1) = 13
3 ?? = 9 
?? = 3
int. point  (3, 6, 2) ;  (7, 8, 9) 
d
2
 = 16 + 4 + 49 = 69  
Ans. d
2
 + 6 = 69 + 6 = 75 
2. Let a rectangle ABCD of sides 2 and 4 be inscribed
in another rectangle PQRS such that the vertices of
the rectangle ABCD lie on the sides of the
rectangle PQRS. Let a and b be the sides of the
rectangle PQRS when its area is maximum. Then
(a + b)
2
 is equal to :
(1) 72 (2) 60
(3) 80 (4) 64
Ans. (1 ) 
Sol. 
2cos(90– ?? )
 
P
 
B
 
Q
 
4
 
4
 
R
 
S
 
A
 
2 2
 
4cos(90– ?? )
 
2sin(90– ?? )
 
D
 
90– ?? 
?? 
?? 
Area = (4cos ?? + 2sin ?? ) (2cos ?? + 4sin ?? ) 
= 8cos
2
?? + 16sin ?? cos ?? + 4sin ?? cos ?? + 8sin
2
?? 
= 8 + 20 sin ?? cos ?? 
= 8 + 10 sin2 ?? ?@ Max Area =  8 + 10 = 18 (sin2 ?? ?@?] ?@?Q?I ?@ ?? ?@?] ?@ ?T?U?? ?@ (a + b)
2
 = (4cos ?? + 2sin ?? ?@ + 2cos ?? + 4sin ?? )
2
?@ = (6cos ?? + 6sin ?? )
2 
?@ = 36 (sin ?? + cos ?? )
2 
?@ ?]?@
2
36( 2) ?@ ?] ?@?W?R 3. Let two straight lines drawn from the origin  
O intersect the line 3x + 4y = 12 at the points P and 
Q such that ?d OPQ is an isosceles triangle and 
?? POQ = 90°. If l = OP
2
 + PQ
2
 + QO
2
, then the 
greatest integer less than or equal to l is :  
 (1) 44  (2) 48  
 (3) 46   (4) 42  
 Ans. (3) 
Sol.  
O
 
?? ?@ P(rcos ?? , rsin ?? )
 
Q(rcos(90+???I , rsin(90+ ?? ) = (–rsin ???L rcos ?? )
 
 
 3x + 4y = 12 
 3(rcos ?? ) + 4(rsin ?? ) = 12 
 r(3cos ?? + 4sin ?? ) = 12   ...(1) 
 3(–rsin ?? ) + 4(rcos ?? ) = 12 
 r(–3sin ?? + 4cos ?? ) = 12  ...(2) 
 
2 2
2 2
12 12
(3cos 4sin ) ( 3sin 4cos )
r r
?? ?? ?? ?? ?K ?] ?? ?K ?? ?K ?M ?? ?K ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
2
12
2 9 16
r
????
?]?K
????
????
 
 
2
2 144
25
r
?? ?] ?? ?@ 288 = 25r
2 
?@ ?? ?@ 2
288
r
25
?] ?@ ?@ 12
2 r
5
????
???]
????
????
?@ ?@ ?? = OP
2
 + PQ
2
 + QO
2
 
?@ ???@ = r
2
 + r
2
?@ + r
2
(cos ?? + sin ?? )
2
 + r
2
(sin ?? + cos ?? )
2
 
 = 2r
2
 + r
2
(1 + sin2 ?? + 1 – 2sin2 ?? ) 
 = 2r
2
 + 2r
2
 
 = 4r
2 
 
1152 288
4 46.08
25 25
????
?] ?] ?] ????
????
 
 [ ?? ] = 46 
4. If y = y(x) is the solution of the differential 
equation 
dy
dx
 + 2y = sin (2x), y(0) = 
3
4
, then 
y
8
?? ????
????
????
 is equal to :  
 (1) e
– ?? /8
  (2) e
– ?? /4
 
 (3) e
?? /4
  (4) e
?? /8
 
 Ans. (2 ) 
Sol. 
dy
2y sin 2x
dx
?K?] , 
3
y(0)
4
?]  
 I.F = 
2dx
e
?? = e
2x
  
 
2x 2x
y.e e sin2xdx ?] ??  
 
2x
2x
e (2sin 2x 2cos2x)
y.e C
4 4
?M ?]?K
?K  
 x = 0, y = 
3
4
 ?? 
3 1(0 2)
.1 C
4 8
?M ?]?K  
 
3 1
C
4 4
?] ?M ?K  
 1 = C  
 
2x
2sin 2x 2cos2x
y 1.e
8
?M ?M ?]?K  
 x
8
?? ?] ,  
2
8
1
y 2sin 2cos e
8 4 4
?? ????
?M ????
????
???? ????
?] ?M ?K ????
????
  
 
4
y 0 e
?? ?M ?]?K  
5. For the function  
 f(x) = sinx + 3x – 
2
?? (x
2
 + x), where x ?? 0,
2
?? ????
????
????
, 
consider the following two statements :  
 (I)  f is increasing in 0,
2
?? ????
????
????
. 
 (II) f ?? is decreasing in 0,
2
?? ????
????
????
.  
 Between the above two statements,  
 (1) only (I) is true. 
 (2) only (II) is true.  
 (3) neither (I) nor (II) is true.  
 (4) both (I) and (II) are true.  
 
 Ans. (4) 
Sol. f(x) = sinx + 3x – 
2
?? (x
2
 + x)  x 0,
2
?? ????
?? ????
????
  
 f ?? (x) = cosx + 3 – 
2
?? (2x + 1) > 0  f(x) ??  
 f ?? (x) = –sinx + 0 – 
2
?? (2)  
 = –sinx – 
4
?? < 0   f ?? (x)  ??  
 0 < x < 
2
??  
 ?? 
?H ?I 1 1 1
2
0 2x
?K ?K?K
?M ?\ ?\ ?? ??  
 
3 3 3
2 2 2
(2x 1) ( 1)
?K ?K ?K ?M ?M ?^ ?K ?^ ?M ?? ?K ?? ?? ??  
 
( ve) ( ve)
2 2 2
3 3 (2x 1) 3 ( 1)
?K?K
?M ?^ ?M ?K ?^ ?M ?? ?K ?? ?? ??  
6. If the system of equations  
 11x + y + ?? z = –5 
 2x + 3y + 5z = 3  
 8x – 19y – 39z = µ 
 has infinitely many solutions, then ?? 4
 – µ is equal 
to :  
 (1) 49  (2) 45  
 (3) 47   (4) 51  
 Ans. (3) 
Sol. 11x + y + ?? z = –5  
 2x + 3y + 5z = 3  
 8x – 19y – 39z = µ  
 for infinite sol.  
 
11 1
D 2 3 5 0
8 19 39
?? ?]?]
?M?M
  
 ?? 11(–117 + 95) – 1(–78 – 40) + ?? (–38 – 24)  
 ?? 11(–22) + 118 – ?? (62) = 0  
 ?? 62 ?? = 118 – 242  
 ?? 
124
2
62
?M ?? ?] ?] ?M  
 
1
5 1 2
D 3 3 5 0
µ 19 39
?M?M
?]?]
?M?M
  
 ?? –5(–117 + 95) – 1(–117 – 5µ) – 2(–57 – 3µ) = 0  
 ?? –5(–22) + 117 + 5µ + 114 + 6µ = 0  
 ?? 11µ = –110 – 231 = –341  
 ?? µ = –31  
 ?? 4
 – µ = (–2)
4
 – (–31) = 16 + 31 = 47 
7. Let A = {1, 3, 7, 9, 11} and B = {2, 4, 5, 7, 8, 10, 12}. 
Then the total number of one-one maps  
f : A ?? B, such that f (1) + f(3) = 14, is :  
 (1) 180  (2) 120  
 (3) 480   (4) 240  
 Ans. (4) 
Sol. 
1 
3 
7 
9 
11
 
(5)
 
2
 
(7)
 
12
 
4
 
5
 
7
 
8
 
10
 
 
 A = {1, 3, 7, 9, 11} 
 B = {2, 4, 5, 7, 8, 10, 12} 
 f(1) + f(3) = 14 
 (i) 2 + 12 
 (ii) 4 + 10 
 2 × (2 × 5 × 4 × 3) = 240  
8. If the function 
3
sin3x sin x cos3x
f(x)
x
?K ?? ?M ?? ?] ,  
x ?? R, is continuous at x = 0, then f(0) is equal to :  
 (1) 2  (2) –2  
 (3) 4   (4) –4  
 Ans. (4) 
Sol. 
3
sin3x sin x cos3x
f(x)
x
?K ?? ?M ?? ?]  
 is continuous at x = 0 
3 3 2
3
x 0
(3x) x (3x)
3x ... x ... 1 ...
3 3 2
lim f(0)
x
?? ????????
?M ?K ?K ?? ?M ?? ?M ?M ????????
???? ????
?]?] 
2
3
3
x 0
27 9 x
x(3 ) x ...
3 3 2
lim f(0)
x
?? ?M?? ?? ????
?M ?M ?? ?K ?K ?? ?K ?K ????
????
?]?] 
 for exist 
 ?? = 0, 3+ ?? = 0, 
27
f(0)
3 3
?? ?M ?M ?]  
 ?? ?@ = –3, 
27 ( 3)
f(0)
6 6
?M ?M ?M ?] 
 
27 3
f(0) 4
6
?M?K
?] ?] ?M 
9. The integral 
4
0
136sin x
dx
3sin x 5cosx
?? ?K ?? is equal to :  
 (1) 3 ?? – 50 log
e 
2 + 20 log
e 
5  
 (2) 3 ?? – 25 log
e 
2 + 10 log
e 
5 
 (3) 3 ?? – 10 log
e ?H ?I 2 2 + 10 log
e 
5 
 (4) 3 ?? – 30 log
e 
2 + 20 log
e 
5 
 Ans. (1) 
Sol. 
/4
0
136sin x
I dx
3sin x 5cosx
?? ?] ?K ??  
 136sinx = A(3sinx + 5cosx) + B(3cosx – 5sinx)  
 136 = 3A – 5B  …(1)  
 0 = 5A + 3B  …(2)  
 3B = –5A ?? 
5
B A
3
?]?M  
 
5
136 3A 5 A
3
????
?] ?M ?M????
????
 
 
25
136 3A A
3
?]?K  
 
34A
136
3
?]  
 ?? 
136 3
A 12
34
?? ?]?]  
 
5
B (12) –20
3
?M?]?]  
 
/4 /4
0 0
A(3sin x 5cosx) B(3cosx 5sin x)
I
3sin x 5cosx 3sin x 5cosx
????
?K?M
?]?K
?K?K
????
  
 ?H ?I ?H ?I /4 /4
0 0
A x B n 3sin x 5cosx
?? ?????? ?] ?K ?K????
  
 ?H ?I 3 5
12 20 n n 0 5
4
2 2
?? ???? ????
?] ?M ?K ?M ?K ???? ????
???? ????
  
 3 20 n4 2 20 n5 ?] ?? ?M ?K  
 
5
3 20 n2 20 n5
2
?] ?? ?M ?? ?K  
 3 50 n2 20 n5 ?] ?? ?M ?K 
10. The coefficients a, b, c in the quadratic equation  
ax
2
 + bx + c = 0 are chosen from the set  
 {1, 2, 3, 4, 5, 6, 7, 8}. The probability of this 
equation having repeated roots is :  
 (1) 
3
256
  (2) 
1
128
 
 (3) 
1
64
  (4) 
3
128
 
 Ans. (3) 
Sol. ax
2
 + bx + c = 0  
 a, b, c ?? {1, 2, 3, 4, 5, 6,7, 8}  
 Repeated roots D = 0  
 ?? b
2
 – 4ac = 0 ?? b
2
 = 4ac  
 Prob = 
8 1
8 8 8 64
?] ????
  
 ?? (a, b, c)  
 (1, 2, 1) ; (2, 4, 2) ; (1, 4, 4) ; (4, 4, 1) ; (3, 6, 3) ; 
(2, 8, 8) ; (8, 8, 2) ; (4, 8, 4)  
 8 case 
 
11. Let A and B be two square matrices of order 3 
such that |A| = 3 and |B| = 2.  
 Then |A
T
 A(adj(2A))
–1
 (adj(4B))(adj(AB))
–1
AA
T
|  
 is equal to :  
 (1) 64 (2) 81 
 (3) 32  (4) 108 
 Ans. (1) 
Sol. |A| = 3, |B| = 2 
 |A
T
A(adj(2A))
–1 
(adj(4B)) (adj(AB))
–1
AA
T
| 
 = 3×3×|(adj(2A)
–1
| × |adj(4B)| × |(adj(AB))
–1
|×3×3 
           ??               ??                 ?? 
          
6
6 2
1
adj(2A)
1
2 adjA
1
2 · 3
?] ?]     
2
12
× 2
2
     
2 2
1
adj(AB)
1
adjB·adjA
1
2 · 3
?] ?] 
 
 
4 12 2
6 2 2 2
1 1
3 · ·2 ·2 · 64
2 · 3 2 · 3
?]?] 
12. Let a circle C of radius 1 and closer to the origin be 
such that the lines passing through the point (3, 2) 
and parallel to the coordinate axes touch it. Then 
the shortest distance of the circle C from the point 
(5, 5) is :  
 (1) 2 2 (2) 5 
 (3) 4 2  (4) 4 
 Ans. (4) 
Sol. 
O
 
R
 
R
 
P(3, 2)
 
Q(5, 5)
 
C(2, 1)
 
 
 Coordinates of the centre will be (2, 1) 
 Equation of circle will be 
 (x – 2)
2
 + (y – 1)
2
 = 1 
 QC = 
2 2
(5 2) (5 1) ?M ?K ?M 
 QC = 5 
 shortest distance 
 = RQ = CQ – CR 
 = 5 – 1 
 = 4 
13. Let the line 2x + 3y – k = 0, k > 0, intersect the  
x-axis and y-axis at the points A and B, 
respectively. If the equation of the circle having the 
line segment AB as a diameter is x
2
 + y
2
 – 3x – 2y = 0 
and the length of the latus rectum of the ellipse  
x
2
 + 9y
2
 = k
2
 is 
m
n
, where m and n are coprime, 
then 2m + n is equal to  
 (1) 10 (2) 11 
 (3) 13  (4) 12 
 Ans. (2) 
Sol. Centre of the circle = 
3
,1
2
????
????
????
  
 Equation of diameter = 2x + 3y – k = 0  
 
3
2 3(1) – k 0
2
????
?K?]
????
????
  
 ?? k = 6  
 Now, Equation of ellipse becomes  
 x
2
 + 9y
2
 = 36  
 
2 2
2 2
x y
1
6 2
?K?]