JEE Main 2024 April 9 Shift 2 Paper & Solutions | JEE Main & Advanced Mock Test Series 2025 PDF Download

 Page 1


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Tuesday 09
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. 
?H ?I 1
2x
x 0
e
1 2x
Lim
x
?? ?M ?K is equal to : 
(1) e (2) 
2
e
?M (3) 0 (4) e–e
2
Ans. (1) 
Sol. 
1
ln(1 2x)
2x
x 0
e e
Lim
x
?K ?? ?M = 
?H ?I ?H ?I ln 1 2x
1
2x
x 0
e 1
Lim( e)
x
?K ?M ?? ?M ?M = 
?H ?I 2
x 0
ln 2x
1 2x
Lim( e)
2x
?? ?M ?K ?M = (–e) × (–1) 
4
2 2 ?? = e 
2. Consider the line L passing through the points
(1, 2, 3) and (2, 3, 5). The distance of the point
11 11 19
, ,
3 3 3
????
????
????
from the line L along the line 
3x 11 3y 11 3z 19
2 1 2
?M ?M ?M ?]?] is equal to : 
(1) 3 (2) 5
(3) 4 (4) 6
Ans. (1) 
Sol. 
x 1 y 2 z 3
2 1 3 2 5 3
?M ?M ?M ?]?]
?M ?M ?M 
?? 
x 1 y 2 z 3
1 1 2
?M ?M ?M ?]?] = ?? 
B 
B(1 + ?? , 2 + ?? , 3 + 2 ?? ) 
D.R. of AB = <
3 8 3 5 6 10
, ,
3 3 3
?? ?M ?? ?M ?? ?M > 
B 
5 8 13
, ,
3 3 3
????
????
????
3 8 2
3 5 1
???M
?] ???M
?? 3 ?? – 8 = 6 ?? – 10
3 ?? = 2 
?? =
2
3
AB = 
36 9 36 9
3
3 3
?K?K
?]?]
3. Let ?H ?I ?H ?I x x
2
0 0
1 dt y(t)dt y'
t
?M?]
????
, 0 ?? x ?? 3, y ?? 0, 
y (0) = 0. Then at x = 2, y" + y + 1 is equal to : 
(1) 1 (2) 2
(3) 2 (4) 1/2
Ans. (1) 
Sol. ?H ?I ?H ?I 2
1 y y'(x)
x
?M?]
1 – 
2
2
dy
y
dx
????
?] ????
????
2
2
dy
1 y
dx
????
?]?M
????
????
Page 2


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Tuesday 09
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. 
?H ?I 1
2x
x 0
e
1 2x
Lim
x
?? ?M ?K is equal to : 
(1) e (2) 
2
e
?M (3) 0 (4) e–e
2
Ans. (1) 
Sol. 
1
ln(1 2x)
2x
x 0
e e
Lim
x
?K ?? ?M = 
?H ?I ?H ?I ln 1 2x
1
2x
x 0
e 1
Lim( e)
x
?K ?M ?? ?M ?M = 
?H ?I 2
x 0
ln 2x
1 2x
Lim( e)
2x
?? ?M ?K ?M = (–e) × (–1) 
4
2 2 ?? = e 
2. Consider the line L passing through the points
(1, 2, 3) and (2, 3, 5). The distance of the point
11 11 19
, ,
3 3 3
????
????
????
from the line L along the line 
3x 11 3y 11 3z 19
2 1 2
?M ?M ?M ?]?] is equal to : 
(1) 3 (2) 5
(3) 4 (4) 6
Ans. (1) 
Sol. 
x 1 y 2 z 3
2 1 3 2 5 3
?M ?M ?M ?]?]
?M ?M ?M 
?? 
x 1 y 2 z 3
1 1 2
?M ?M ?M ?]?] = ?? 
B 
B(1 + ?? , 2 + ?? , 3 + 2 ?? ) 
D.R. of AB = <
3 8 3 5 6 10
, ,
3 3 3
?? ?M ?? ?M ?? ?M > 
B 
5 8 13
, ,
3 3 3
????
????
????
3 8 2
3 5 1
???M
?] ???M
?? 3 ?? – 8 = 6 ?? – 10
3 ?? = 2 
?? =
2
3
AB = 
36 9 36 9
3
3 3
?K?K
?]?]
3. Let ?H ?I ?H ?I x x
2
0 0
1 dt y(t)dt y'
t
?M?]
????
, 0 ?? x ?? 3, y ?? 0, 
y (0) = 0. Then at x = 2, y" + y + 1 is equal to : 
(1) 1 (2) 2
(3) 2 (4) 1/2
Ans. (1) 
Sol. ?H ?I ?H ?I 2
1 y y'(x)
x
?M?]
1 – 
2
2
dy
y
dx
????
?] ????
????
2
2
dy
1 y
dx
????
?]?M
????
????
 
2
dy
dx
1 y
?] ?M OR 
2
dy
dx
1 y
?]?M
?M 
 ???@ ?@ sin
–1
y = x + c, sin
–1
y = –x + c 
 x = 0, y = 0    ??   c = 0 
 sin
–1
 y = x, as y ?? 0 
 sinx = y 
 ?? 
dy
cos x
dx
?] 
 
2
2
d y
sin x
dx
?]?M 
 ?? – sinx + sinx + 1 = 1 
4. Let z be a complex number such that the real part 
of 
z 2i
z 2i
?M ?K is zero. Then, the maximum value of  
|z –(6+8i)| is equal to : 
 (1) 12 (2) ?? 
 (3) 10  (4) 8 
 Ans. (1) 
Sol. 
z 2i z 2i
0
z 2i z 2i
?M?K
?K?]
?K?M
 
 zz 2iz 2iz 4( 1) ?M ?M ?K ?M 
 zz 2zi 2zi 4( 1) 0 ?K ?K ?K ?K ?M ?] 
 ???@ 2|z|
2
 = 8 ?? |z| = 2  
 ?H ?I maximum
10 2 12 z
6 8i
?] ?K ?] ?M ?K 
5. The area (in square units) of the region enclosed by 
the ellipse x
2
 + 3y
2
 = 18 in the first quadrant below 
the line y = x  is :  
 (1) 
3
3
4
???K (2) 3 ?? 
 (3) 
3
3
4
???M (4) 3 1 ???K 
 Ans. (2) 
Sol. 
2 2
x y
1
18 6
?K?] 
 
 
 
y = x 
 
 
2 2
x 3x
1
18 18
?K?] ?? 4x
2
 = 18 ?? x
2
 = 
9
2
 
 
3 2 2
3
2
18 x
dx
3
?M ??  
 = 
3 2
2
1
3
2
1
x 18 x 18 x
sin
3 2 2 3 2
?M ????
?M ?K????
????
 
 = 
1
3 3 3
9 9
3 2 6 2 2 2
????
????
?? ?M ?? ?M ?? ????
????
 
 Required Area  
 
1 9 1
18 9 3
2 2 3 6 4
????
?? ?] ?? ?K ?M ????
????
 
 3 ?]??  
6. Let the foci of a hyperbola H coincide with the foci 
of the ellipse E : 
?H ?I ?H ?I 2
2
y 1
x 1
1
100 75
?M ?M ?K?] and the 
eccentricity of the hyperbola H be the reciprocal of 
the eccentricity of the ellipse E. If the length of the 
transverse axis of H is ?? and the length of its 
conjugate axis is ?? , then 3 ?? 2
 + 2 ?? 2
 is equal to :  
 (1) 242 
 (2) 225 
 (3) 237  
 (4) 205 
 Ans. (2) 
Page 3


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Tuesday 09
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. 
?H ?I 1
2x
x 0
e
1 2x
Lim
x
?? ?M ?K is equal to : 
(1) e (2) 
2
e
?M (3) 0 (4) e–e
2
Ans. (1) 
Sol. 
1
ln(1 2x)
2x
x 0
e e
Lim
x
?K ?? ?M = 
?H ?I ?H ?I ln 1 2x
1
2x
x 0
e 1
Lim( e)
x
?K ?M ?? ?M ?M = 
?H ?I 2
x 0
ln 2x
1 2x
Lim( e)
2x
?? ?M ?K ?M = (–e) × (–1) 
4
2 2 ?? = e 
2. Consider the line L passing through the points
(1, 2, 3) and (2, 3, 5). The distance of the point
11 11 19
, ,
3 3 3
????
????
????
from the line L along the line 
3x 11 3y 11 3z 19
2 1 2
?M ?M ?M ?]?] is equal to : 
(1) 3 (2) 5
(3) 4 (4) 6
Ans. (1) 
Sol. 
x 1 y 2 z 3
2 1 3 2 5 3
?M ?M ?M ?]?]
?M ?M ?M 
?? 
x 1 y 2 z 3
1 1 2
?M ?M ?M ?]?] = ?? 
B 
B(1 + ?? , 2 + ?? , 3 + 2 ?? ) 
D.R. of AB = <
3 8 3 5 6 10
, ,
3 3 3
?? ?M ?? ?M ?? ?M > 
B 
5 8 13
, ,
3 3 3
????
????
????
3 8 2
3 5 1
???M
?] ???M
?? 3 ?? – 8 = 6 ?? – 10
3 ?? = 2 
?? =
2
3
AB = 
36 9 36 9
3
3 3
?K?K
?]?]
3. Let ?H ?I ?H ?I x x
2
0 0
1 dt y(t)dt y'
t
?M?]
????
, 0 ?? x ?? 3, y ?? 0, 
y (0) = 0. Then at x = 2, y" + y + 1 is equal to : 
(1) 1 (2) 2
(3) 2 (4) 1/2
Ans. (1) 
Sol. ?H ?I ?H ?I 2
1 y y'(x)
x
?M?]
1 – 
2
2
dy
y
dx
????
?] ????
????
2
2
dy
1 y
dx
????
?]?M
????
????
 
2
dy
dx
1 y
?] ?M OR 
2
dy
dx
1 y
?]?M
?M 
 ???@ ?@ sin
–1
y = x + c, sin
–1
y = –x + c 
 x = 0, y = 0    ??   c = 0 
 sin
–1
 y = x, as y ?? 0 
 sinx = y 
 ?? 
dy
cos x
dx
?] 
 
2
2
d y
sin x
dx
?]?M 
 ?? – sinx + sinx + 1 = 1 
4. Let z be a complex number such that the real part 
of 
z 2i
z 2i
?M ?K is zero. Then, the maximum value of  
|z –(6+8i)| is equal to : 
 (1) 12 (2) ?? 
 (3) 10  (4) 8 
 Ans. (1) 
Sol. 
z 2i z 2i
0
z 2i z 2i
?M?K
?K?]
?K?M
 
 zz 2iz 2iz 4( 1) ?M ?M ?K ?M 
 zz 2zi 2zi 4( 1) 0 ?K ?K ?K ?K ?M ?] 
 ???@ 2|z|
2
 = 8 ?? |z| = 2  
 ?H ?I maximum
10 2 12 z
6 8i
?] ?K ?] ?M ?K 
5. The area (in square units) of the region enclosed by 
the ellipse x
2
 + 3y
2
 = 18 in the first quadrant below 
the line y = x  is :  
 (1) 
3
3
4
???K (2) 3 ?? 
 (3) 
3
3
4
???M (4) 3 1 ???K 
 Ans. (2) 
Sol. 
2 2
x y
1
18 6
?K?] 
 
 
 
y = x 
 
 
2 2
x 3x
1
18 18
?K?] ?? 4x
2
 = 18 ?? x
2
 = 
9
2
 
 
3 2 2
3
2
18 x
dx
3
?M ??  
 = 
3 2
2
1
3
2
1
x 18 x 18 x
sin
3 2 2 3 2
?M ????
?M ?K????
????
 
 = 
1
3 3 3
9 9
3 2 6 2 2 2
????
????
?? ?M ?? ?M ?? ????
????
 
 Required Area  
 
1 9 1
18 9 3
2 2 3 6 4
????
?? ?] ?? ?K ?M ????
????
 
 3 ?]??  
6. Let the foci of a hyperbola H coincide with the foci 
of the ellipse E : 
?H ?I ?H ?I 2
2
y 1
x 1
1
100 75
?M ?M ?K?] and the 
eccentricity of the hyperbola H be the reciprocal of 
the eccentricity of the ellipse E. If the length of the 
transverse axis of H is ?? and the length of its 
conjugate axis is ?? , then 3 ?? 2
 + 2 ?? 2
 is equal to :  
 (1) 242 
 (2) 225 
 (3) 237  
 (4) 205 
 Ans. (2) 
 
Sol.  
 
 
  
 
 F
12
 (1, 1) 
F
1
 
 
 e
1
 = 
75 5 1
1
100 10 2
?M ?] ?] 
 e
2
 = 2 
 F
1
 (6, 1), F
2
 (–4, 1) 
 2ae
2
 = 10 ?? a = 
5
2a 5
2
???] 
 ???@?? = 5 
 4 = 1 + 
2
2 2
2
b
b 3a
a
???] 
 b = 
5
3
2
?? 
 ?? ?@ = 5 3 
 3 ?? 2
 + 2 ?? 2
 = 3 × 25 + 2 × 25 × 3 
 = 225 
7. Two vertices of a triangle ABC are A(3, –1) and  
B (–2, 3), and its orthocentre is P(1, 1). If the 
coordinates of the point C are ( ?? , ?? ) and the centre 
of the circle circumscribing the triangle PAB is  
(h, k), then the value of ( ?? + ?? ) + 2 (h + k) equals : 
 (1) 51 (2) 81 
 (3) 5  (4) 15 
 Ans. (3) 
Sol.  
 
 
 
C ( ?? , ?? ) 
A(3,–1) 
B (–2, 3) 
D 
 
 M
AB
 = 
4
5 ?M ?? M
DP
 = 
5
4
 
 Equation of PC is y – 1 = ?H ?I 5
x 1
4
?M .......(1) 
 M
AP
 =
2
1
2
?]?M
?M ?? M
BC
 = + 1  
 Equation of BC is y – 3 = (x + 2) .........(2)  
 On solving (1) and (2) 
 x + 4 = ?H ?I 5
x 1
4
?M ?? 4x + 16 = 5x – 5 ?? ?? = 21  
?@ ?? ?? = y = x + 5 = 26 
 ?? + ?? = 47 
 Equation of ?~ bisector of AP 
 y – 0 = (x – 2) ............ (3) 
 Equation of ?~ bisector of AB 
 y – 1 = 
5 1
x
4 2
????
?M ????
????
 ............(4) 
 On solving (3) & (4)  
 (x – 3)4 = 5x – 
5
2
 
 x = 
19
h
2
?M ?] 
 y = 
23
k
2
?M ?] 
 ???@ 2(h + k) = –42  
8. If the variance of the frequency distribution is 160, 
then the value of c ?? N is   
x c 2c 3c 4c 5c 6c 
f 2 1 1 1 1 1 
 
 (1) 5 (2) 8 
 (3) 7  (4) 6 
 Ans. (3) 
Sol.  
x C 2C 3C 4C 5C 6C 
f 2 1 1 1 1 1 
 
(2 2 3 4 5 6)C 22C
x
7 7
?K ?K ?K ?K ?K ?]?] 
Page 4


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Tuesday 09
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. 
?H ?I 1
2x
x 0
e
1 2x
Lim
x
?? ?M ?K is equal to : 
(1) e (2) 
2
e
?M (3) 0 (4) e–e
2
Ans. (1) 
Sol. 
1
ln(1 2x)
2x
x 0
e e
Lim
x
?K ?? ?M = 
?H ?I ?H ?I ln 1 2x
1
2x
x 0
e 1
Lim( e)
x
?K ?M ?? ?M ?M = 
?H ?I 2
x 0
ln 2x
1 2x
Lim( e)
2x
?? ?M ?K ?M = (–e) × (–1) 
4
2 2 ?? = e 
2. Consider the line L passing through the points
(1, 2, 3) and (2, 3, 5). The distance of the point
11 11 19
, ,
3 3 3
????
????
????
from the line L along the line 
3x 11 3y 11 3z 19
2 1 2
?M ?M ?M ?]?] is equal to : 
(1) 3 (2) 5
(3) 4 (4) 6
Ans. (1) 
Sol. 
x 1 y 2 z 3
2 1 3 2 5 3
?M ?M ?M ?]?]
?M ?M ?M 
?? 
x 1 y 2 z 3
1 1 2
?M ?M ?M ?]?] = ?? 
B 
B(1 + ?? , 2 + ?? , 3 + 2 ?? ) 
D.R. of AB = <
3 8 3 5 6 10
, ,
3 3 3
?? ?M ?? ?M ?? ?M > 
B 
5 8 13
, ,
3 3 3
????
????
????
3 8 2
3 5 1
???M
?] ???M
?? 3 ?? – 8 = 6 ?? – 10
3 ?? = 2 
?? =
2
3
AB = 
36 9 36 9
3
3 3
?K?K
?]?]
3. Let ?H ?I ?H ?I x x
2
0 0
1 dt y(t)dt y'
t
?M?]
????
, 0 ?? x ?? 3, y ?? 0, 
y (0) = 0. Then at x = 2, y" + y + 1 is equal to : 
(1) 1 (2) 2
(3) 2 (4) 1/2
Ans. (1) 
Sol. ?H ?I ?H ?I 2
1 y y'(x)
x
?M?]
1 – 
2
2
dy
y
dx
????
?] ????
????
2
2
dy
1 y
dx
????
?]?M
????
????
 
2
dy
dx
1 y
?] ?M OR 
2
dy
dx
1 y
?]?M
?M 
 ???@ ?@ sin
–1
y = x + c, sin
–1
y = –x + c 
 x = 0, y = 0    ??   c = 0 
 sin
–1
 y = x, as y ?? 0 
 sinx = y 
 ?? 
dy
cos x
dx
?] 
 
2
2
d y
sin x
dx
?]?M 
 ?? – sinx + sinx + 1 = 1 
4. Let z be a complex number such that the real part 
of 
z 2i
z 2i
?M ?K is zero. Then, the maximum value of  
|z –(6+8i)| is equal to : 
 (1) 12 (2) ?? 
 (3) 10  (4) 8 
 Ans. (1) 
Sol. 
z 2i z 2i
0
z 2i z 2i
?M?K
?K?]
?K?M
 
 zz 2iz 2iz 4( 1) ?M ?M ?K ?M 
 zz 2zi 2zi 4( 1) 0 ?K ?K ?K ?K ?M ?] 
 ???@ 2|z|
2
 = 8 ?? |z| = 2  
 ?H ?I maximum
10 2 12 z
6 8i
?] ?K ?] ?M ?K 
5. The area (in square units) of the region enclosed by 
the ellipse x
2
 + 3y
2
 = 18 in the first quadrant below 
the line y = x  is :  
 (1) 
3
3
4
???K (2) 3 ?? 
 (3) 
3
3
4
???M (4) 3 1 ???K 
 Ans. (2) 
Sol. 
2 2
x y
1
18 6
?K?] 
 
 
 
y = x 
 
 
2 2
x 3x
1
18 18
?K?] ?? 4x
2
 = 18 ?? x
2
 = 
9
2
 
 
3 2 2
3
2
18 x
dx
3
?M ??  
 = 
3 2
2
1
3
2
1
x 18 x 18 x
sin
3 2 2 3 2
?M ????
?M ?K????
????
 
 = 
1
3 3 3
9 9
3 2 6 2 2 2
????
????
?? ?M ?? ?M ?? ????
????
 
 Required Area  
 
1 9 1
18 9 3
2 2 3 6 4
????
?? ?] ?? ?K ?M ????
????
 
 3 ?]??  
6. Let the foci of a hyperbola H coincide with the foci 
of the ellipse E : 
?H ?I ?H ?I 2
2
y 1
x 1
1
100 75
?M ?M ?K?] and the 
eccentricity of the hyperbola H be the reciprocal of 
the eccentricity of the ellipse E. If the length of the 
transverse axis of H is ?? and the length of its 
conjugate axis is ?? , then 3 ?? 2
 + 2 ?? 2
 is equal to :  
 (1) 242 
 (2) 225 
 (3) 237  
 (4) 205 
 Ans. (2) 
 
Sol.  
 
 
  
 
 F
12
 (1, 1) 
F
1
 
 
 e
1
 = 
75 5 1
1
100 10 2
?M ?] ?] 
 e
2
 = 2 
 F
1
 (6, 1), F
2
 (–4, 1) 
 2ae
2
 = 10 ?? a = 
5
2a 5
2
???] 
 ???@?? = 5 
 4 = 1 + 
2
2 2
2
b
b 3a
a
???] 
 b = 
5
3
2
?? 
 ?? ?@ = 5 3 
 3 ?? 2
 + 2 ?? 2
 = 3 × 25 + 2 × 25 × 3 
 = 225 
7. Two vertices of a triangle ABC are A(3, –1) and  
B (–2, 3), and its orthocentre is P(1, 1). If the 
coordinates of the point C are ( ?? , ?? ) and the centre 
of the circle circumscribing the triangle PAB is  
(h, k), then the value of ( ?? + ?? ) + 2 (h + k) equals : 
 (1) 51 (2) 81 
 (3) 5  (4) 15 
 Ans. (3) 
Sol.  
 
 
 
C ( ?? , ?? ) 
A(3,–1) 
B (–2, 3) 
D 
 
 M
AB
 = 
4
5 ?M ?? M
DP
 = 
5
4
 
 Equation of PC is y – 1 = ?H ?I 5
x 1
4
?M .......(1) 
 M
AP
 =
2
1
2
?]?M
?M ?? M
BC
 = + 1  
 Equation of BC is y – 3 = (x + 2) .........(2)  
 On solving (1) and (2) 
 x + 4 = ?H ?I 5
x 1
4
?M ?? 4x + 16 = 5x – 5 ?? ?? = 21  
?@ ?? ?? = y = x + 5 = 26 
 ?? + ?? = 47 
 Equation of ?~ bisector of AP 
 y – 0 = (x – 2) ............ (3) 
 Equation of ?~ bisector of AB 
 y – 1 = 
5 1
x
4 2
????
?M ????
????
 ............(4) 
 On solving (3) & (4)  
 (x – 3)4 = 5x – 
5
2
 
 x = 
19
h
2
?M ?] 
 y = 
23
k
2
?M ?] 
 ???@ 2(h + k) = –42  
8. If the variance of the frequency distribution is 160, 
then the value of c ?? N is   
x c 2c 3c 4c 5c 6c 
f 2 1 1 1 1 1 
 
 (1) 5 (2) 8 
 (3) 7  (4) 6 
 Ans. (3) 
Sol.  
x C 2C 3C 4C 5C 6C 
f 2 1 1 1 1 1 
 
(2 2 3 4 5 6)C 22C
x
7 7
?K ?K ?K ?K ?K ?]?] 
 Var (x) = 
?H ?I 2
2 2 2 2 2
c
2 2 3 4 5 6
7
?K ?K ?K ?K ?K  
 –
2
22c
7
????
????
????
 
 = 
2
2
92c 484
c
7 49
?M?? 
 = 
?H ?I 2 2
c 60c
644 484
49 49
?Q ?M ?]  
 
2
160 c
160 c 7
49
?? ?] ?? ?] 
9. Let the range of the function 
 f (x) = 
1
2 sin3x cos3x ?K?K
, x ?? IR be [a, b].  
 If ?? and ?? are respectively the A.M. and the G.M. 
of a and b, then 
?? ?? is equal to : 
 (1) 2 (2) 2 
 (3) ??  (4) ?? 
 Ans. (1) 
Sol. f(x) 
1
2 sin3x cos3x ?K?K
 
 
1 1
,
2 2 2 2
????
????
?K?M ????
 
 
a b 1
a b
2 2 ab b a
???? ???K
?]?]
?K ????
?? ????
 
 = 
1
2 2 2 2
2
2 2 2 2
????
?M?K
???? ?K ????
?K?M
????
 
 = 
?H ?I ?H ?I 2 2 2 2
2 2
?K ?M?K
?? = 2 
10. Between the following two statements : 
 Statement-I : Let 
ˆ ˆ ˆ
a i 2j 3k ?] ?K ?M and 
ˆ ˆ ˆ
b 2i j k ?] ?K ?M . Then the vector r satisfying 
a r a b ?? ?] ?? and a.r 0 ?] is of magnitude 10 . 
 Statement-II : In a triangle ABC, cos2A + cos2B 
+ cos2C ?? –
3
2
. 
 (1) Both Statement-I and Statement-II are incorrect 
 (2) Statement-I is incorrect but Statement-II is 
correct 
 (3) Both Statement-I and Statement-II are correct   
(4) Statement-I is correct but Statement-II is 
incorrect 
 Ans. (2) 
Sol. 
ˆ ˆ ˆ
a i 2j 3k ?] ?K ?M 
 
ˆ ˆ ˆ
a 2i j k ?] ?K ?M 
 a r a b ?? ?] ?? ;  a.r 0 ?] 
 ?? ?H ?I a
r b
?? ?M 0 ?] 
 ?? ?H ?I a
r b
?]??
?M 
 ?H ?I a.a
a.r a.b
?]??
?M 
 14 = – 7 ??  ?? ?? = – 2 
 
a
r b
2
?M ?]?M ?? 
a
r b
2
?]?M 
 = 
ˆˆ
2b a 3i k
2 2
?M?K
?] 
 Statement (I) is incorrect 
 cos2A + cos 2B + cos 2c ?? – 
3
2
 
 2A + 2B + 2C = 2 ?? 
 cos2A + cos2B + cos 2C 
 = – 1 – 4 cosA·cosB·cosC 
 ?? –1 – 4 × 
1 1 1
2 2 2
???? 
 = – 
3
2
 
 Statement (II) is correct.  
Page 5


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Tuesday 09
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. 
?H ?I 1
2x
x 0
e
1 2x
Lim
x
?? ?M ?K is equal to : 
(1) e (2) 
2
e
?M (3) 0 (4) e–e
2
Ans. (1) 
Sol. 
1
ln(1 2x)
2x
x 0
e e
Lim
x
?K ?? ?M = 
?H ?I ?H ?I ln 1 2x
1
2x
x 0
e 1
Lim( e)
x
?K ?M ?? ?M ?M = 
?H ?I 2
x 0
ln 2x
1 2x
Lim( e)
2x
?? ?M ?K ?M = (–e) × (–1) 
4
2 2 ?? = e 
2. Consider the line L passing through the points
(1, 2, 3) and (2, 3, 5). The distance of the point
11 11 19
, ,
3 3 3
????
????
????
from the line L along the line 
3x 11 3y 11 3z 19
2 1 2
?M ?M ?M ?]?] is equal to : 
(1) 3 (2) 5
(3) 4 (4) 6
Ans. (1) 
Sol. 
x 1 y 2 z 3
2 1 3 2 5 3
?M ?M ?M ?]?]
?M ?M ?M 
?? 
x 1 y 2 z 3
1 1 2
?M ?M ?M ?]?] = ?? 
B 
B(1 + ?? , 2 + ?? , 3 + 2 ?? ) 
D.R. of AB = <
3 8 3 5 6 10
, ,
3 3 3
?? ?M ?? ?M ?? ?M > 
B 
5 8 13
, ,
3 3 3
????
????
????
3 8 2
3 5 1
???M
?] ???M
?? 3 ?? – 8 = 6 ?? – 10
3 ?? = 2 
?? =
2
3
AB = 
36 9 36 9
3
3 3
?K?K
?]?]
3. Let ?H ?I ?H ?I x x
2
0 0
1 dt y(t)dt y'
t
?M?]
????
, 0 ?? x ?? 3, y ?? 0, 
y (0) = 0. Then at x = 2, y" + y + 1 is equal to : 
(1) 1 (2) 2
(3) 2 (4) 1/2
Ans. (1) 
Sol. ?H ?I ?H ?I 2
1 y y'(x)
x
?M?]
1 – 
2
2
dy
y
dx
????
?] ????
????
2
2
dy
1 y
dx
????
?]?M
????
????
 
2
dy
dx
1 y
?] ?M OR 
2
dy
dx
1 y
?]?M
?M 
 ???@ ?@ sin
–1
y = x + c, sin
–1
y = –x + c 
 x = 0, y = 0    ??   c = 0 
 sin
–1
 y = x, as y ?? 0 
 sinx = y 
 ?? 
dy
cos x
dx
?] 
 
2
2
d y
sin x
dx
?]?M 
 ?? – sinx + sinx + 1 = 1 
4. Let z be a complex number such that the real part 
of 
z 2i
z 2i
?M ?K is zero. Then, the maximum value of  
|z –(6+8i)| is equal to : 
 (1) 12 (2) ?? 
 (3) 10  (4) 8 
 Ans. (1) 
Sol. 
z 2i z 2i
0
z 2i z 2i
?M?K
?K?]
?K?M
 
 zz 2iz 2iz 4( 1) ?M ?M ?K ?M 
 zz 2zi 2zi 4( 1) 0 ?K ?K ?K ?K ?M ?] 
 ???@ 2|z|
2
 = 8 ?? |z| = 2  
 ?H ?I maximum
10 2 12 z
6 8i
?] ?K ?] ?M ?K 
5. The area (in square units) of the region enclosed by 
the ellipse x
2
 + 3y
2
 = 18 in the first quadrant below 
the line y = x  is :  
 (1) 
3
3
4
???K (2) 3 ?? 
 (3) 
3
3
4
???M (4) 3 1 ???K 
 Ans. (2) 
Sol. 
2 2
x y
1
18 6
?K?] 
 
 
 
y = x 
 
 
2 2
x 3x
1
18 18
?K?] ?? 4x
2
 = 18 ?? x
2
 = 
9
2
 
 
3 2 2
3
2
18 x
dx
3
?M ??  
 = 
3 2
2
1
3
2
1
x 18 x 18 x
sin
3 2 2 3 2
?M ????
?M ?K????
????
 
 = 
1
3 3 3
9 9
3 2 6 2 2 2
????
????
?? ?M ?? ?M ?? ????
????
 
 Required Area  
 
1 9 1
18 9 3
2 2 3 6 4
????
?? ?] ?? ?K ?M ????
????
 
 3 ?]??  
6. Let the foci of a hyperbola H coincide with the foci 
of the ellipse E : 
?H ?I ?H ?I 2
2
y 1
x 1
1
100 75
?M ?M ?K?] and the 
eccentricity of the hyperbola H be the reciprocal of 
the eccentricity of the ellipse E. If the length of the 
transverse axis of H is ?? and the length of its 
conjugate axis is ?? , then 3 ?? 2
 + 2 ?? 2
 is equal to :  
 (1) 242 
 (2) 225 
 (3) 237  
 (4) 205 
 Ans. (2) 
 
Sol.  
 
 
  
 
 F
12
 (1, 1) 
F
1
 
 
 e
1
 = 
75 5 1
1
100 10 2
?M ?] ?] 
 e
2
 = 2 
 F
1
 (6, 1), F
2
 (–4, 1) 
 2ae
2
 = 10 ?? a = 
5
2a 5
2
???] 
 ???@?? = 5 
 4 = 1 + 
2
2 2
2
b
b 3a
a
???] 
 b = 
5
3
2
?? 
 ?? ?@ = 5 3 
 3 ?? 2
 + 2 ?? 2
 = 3 × 25 + 2 × 25 × 3 
 = 225 
7. Two vertices of a triangle ABC are A(3, –1) and  
B (–2, 3), and its orthocentre is P(1, 1). If the 
coordinates of the point C are ( ?? , ?? ) and the centre 
of the circle circumscribing the triangle PAB is  
(h, k), then the value of ( ?? + ?? ) + 2 (h + k) equals : 
 (1) 51 (2) 81 
 (3) 5  (4) 15 
 Ans. (3) 
Sol.  
 
 
 
C ( ?? , ?? ) 
A(3,–1) 
B (–2, 3) 
D 
 
 M
AB
 = 
4
5 ?M ?? M
DP
 = 
5
4
 
 Equation of PC is y – 1 = ?H ?I 5
x 1
4
?M .......(1) 
 M
AP
 =
2
1
2
?]?M
?M ?? M
BC
 = + 1  
 Equation of BC is y – 3 = (x + 2) .........(2)  
 On solving (1) and (2) 
 x + 4 = ?H ?I 5
x 1
4
?M ?? 4x + 16 = 5x – 5 ?? ?? = 21  
?@ ?? ?? = y = x + 5 = 26 
 ?? + ?? = 47 
 Equation of ?~ bisector of AP 
 y – 0 = (x – 2) ............ (3) 
 Equation of ?~ bisector of AB 
 y – 1 = 
5 1
x
4 2
????
?M ????
????
 ............(4) 
 On solving (3) & (4)  
 (x – 3)4 = 5x – 
5
2
 
 x = 
19
h
2
?M ?] 
 y = 
23
k
2
?M ?] 
 ???@ 2(h + k) = –42  
8. If the variance of the frequency distribution is 160, 
then the value of c ?? N is   
x c 2c 3c 4c 5c 6c 
f 2 1 1 1 1 1 
 
 (1) 5 (2) 8 
 (3) 7  (4) 6 
 Ans. (3) 
Sol.  
x C 2C 3C 4C 5C 6C 
f 2 1 1 1 1 1 
 
(2 2 3 4 5 6)C 22C
x
7 7
?K ?K ?K ?K ?K ?]?] 
 Var (x) = 
?H ?I 2
2 2 2 2 2
c
2 2 3 4 5 6
7
?K ?K ?K ?K ?K  
 –
2
22c
7
????
????
????
 
 = 
2
2
92c 484
c
7 49
?M?? 
 = 
?H ?I 2 2
c 60c
644 484
49 49
?Q ?M ?]  
 
2
160 c
160 c 7
49
?? ?] ?? ?] 
9. Let the range of the function 
 f (x) = 
1
2 sin3x cos3x ?K?K
, x ?? IR be [a, b].  
 If ?? and ?? are respectively the A.M. and the G.M. 
of a and b, then 
?? ?? is equal to : 
 (1) 2 (2) 2 
 (3) ??  (4) ?? 
 Ans. (1) 
Sol. f(x) 
1
2 sin3x cos3x ?K?K
 
 
1 1
,
2 2 2 2
????
????
?K?M ????
 
 
a b 1
a b
2 2 ab b a
???? ???K
?]?]
?K ????
?? ????
 
 = 
1
2 2 2 2
2
2 2 2 2
????
?M?K
???? ?K ????
?K?M
????
 
 = 
?H ?I ?H ?I 2 2 2 2
2 2
?K ?M?K
?? = 2 
10. Between the following two statements : 
 Statement-I : Let 
ˆ ˆ ˆ
a i 2j 3k ?] ?K ?M and 
ˆ ˆ ˆ
b 2i j k ?] ?K ?M . Then the vector r satisfying 
a r a b ?? ?] ?? and a.r 0 ?] is of magnitude 10 . 
 Statement-II : In a triangle ABC, cos2A + cos2B 
+ cos2C ?? –
3
2
. 
 (1) Both Statement-I and Statement-II are incorrect 
 (2) Statement-I is incorrect but Statement-II is 
correct 
 (3) Both Statement-I and Statement-II are correct   
(4) Statement-I is correct but Statement-II is 
incorrect 
 Ans. (2) 
Sol. 
ˆ ˆ ˆ
a i 2j 3k ?] ?K ?M 
 
ˆ ˆ ˆ
a 2i j k ?] ?K ?M 
 a r a b ?? ?] ?? ;  a.r 0 ?] 
 ?? ?H ?I a
r b
?? ?M 0 ?] 
 ?? ?H ?I a
r b
?]??
?M 
 ?H ?I a.a
a.r a.b
?]??
?M 
 14 = – 7 ??  ?? ?? = – 2 
 
a
r b
2
?M ?]?M ?? 
a
r b
2
?]?M 
 = 
ˆˆ
2b a 3i k
2 2
?M?K
?] 
 Statement (I) is incorrect 
 cos2A + cos 2B + cos 2c ?? – 
3
2
 
 2A + 2B + 2C = 2 ?? 
 cos2A + cos2B + cos 2C 
 = – 1 – 4 cosA·cosB·cosC 
 ?? –1 – 4 × 
1 1 1
2 2 2
???? 
 = – 
3
2
 
 Statement (II) is correct.  
 
11. 
?H ?I ?H ?I ?H ?I ?H ?I 3
3
/2
1/3 1/3
x
2
x
x
2
sin cos dt
2t t
lim
x
2
?? ??????
?K ????
????
?? ????
????
?M ???? ????
???? ????
?? is equal 
to : 
 (1) 
2
9
8
?? (2) 
2
11
10
?? 
 (3) 
2
3
2
??  (4) 
2
5
9
?? 
 Ans. (1) 
Sol. 
?? ?? 2
x
2
0 .3x sin(2x) cos(x)
lim
2 x
2
?? ?? ?M ?K ?? ????
?M ????
????
 
 = 
?? ?? 2
x
2
3x
2sin x cos x cos x
lim
2 x
2
?? ?? ?M ?K ?? ????
?M ????
????
 
 = 
2
x
2
2sin x sin sin x x
2 2
lim 3x
2 2 x x
2 2
?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?M?M
?? ?? ?? ?? ????
????
?? ?? ?? ?? ?K ????
???? ?? ?? ?? ?? ????
?M?M
?? ?? ?? ?? ???? ?? ?? ?? ?? ????
 
 = ?H ?I 2
1
3 1
1
2 2
?? ?? ?? ?? ?? ?K ?? ?? ?? ?? ?? ?? ?? ?? 
 = 
2
9
8
??  
12. The sum of the coefficient of x
2/3
 and x
–2/5
 in the 
binomial expansion of 
9
2/3 2/5
1
x x
2
?M ????
?K????
????
 is : 
 (1) 21/4 (2) 69/16 
 (3) 63/16  (4) 19/4 
 Ans. (1) 
Sol. T
r + 1
 = ?H ?I r
2/5
9 r 9
2/3
r
x
C
x
2
?M ?M????
????
????
 
 = ?H ?I 2r 2r
6
3 5
r
9
r
1
C
r
2
????
?M?M ????
????
????
????
????
 
 for coefficient of x
2/3
, put 
2r 2r 2
6
3 5 3
?M ?M ?] 
 ???@ r = 5  
 ?| Coefficient of x
2/3
 is = 
5
9
5
1
C
5
????
????
????
 
 For coefficient of x
–2/5
, put 
2r 2r 2
6
3 5 5
?M ?M ?] ?M 
 ?? r = 6 
 Coefficient of x
–2/5
 is 
9
C
6
 
6
1
2
????
????
????
 
 Sum = 
9
C
5
 
5
1
2
????
????
????
 + 
9
C
6
6
1
2
????
????
????
 = 
21
4
 
13. Let B = 
1 3
1 5
????
????
????
 and A be a 2 × 2 matrix such that 
AB
–1
 = A
–1
. If BCB
–1
 = A and C
4
 + ?? C
2
 + ?? I = O, 
then 2 ?? – ?? is equal to :  
 (1) 16 (2) 2 
 (3) 8  (4) 10 
 Ans. (4) 
Sol. BCB
–1
 = A  
 ?? (BCB
–1
) (BCB
–1
) = A.A 
 ?? BCI CB
–1
 = A
2
 
 ?? BC
2
B
–1
 = A
2
 
 ?? B
–1
(BC
2
B
–1
)B = B
–1
(A.A)B 
 From equation (1) 
 C
2
 = A
–1
.A.B  
 C
2
 = B 
 Also AB
–1
 = A
–1
 
 ?? AB
–1
.A = A
–1
 A = I 
 ?? A
–1
(AB
–1
A) =A
–1
I 
 B
–1
A = A
–1
  
 Now characteristics equation of C
2
 is 
 |C
2
– ?? I| = 0 
 |B – ?? I| = 0