JEE Main 2024 April 5 Shift 2 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 : 3 : 00 PM  to  6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ƒ: [–1, 2] ?? R be given by
ƒ(x) = 2x
2
 + x + [x
2
] – [x], where [t] denotes the
greatest integer less than or equal to t. The number
of points, where ƒ is not continuous, is :
(1) 6 (2) 3
(3) 4 (4) 5
Ans. (3) 
Sol. Doubtful points : –1, 0, 1, 2 , 3 , 2 
at x = 2 , 3
?H ?I 2 2
Cont. Cont.
f(x) 2x x [x] x Discount
?? ?? ????
?] ?K ?M ?K ?]????
at x = –1 : 
RHL f(x) (2 1 ( 1)) 0 2
Dis.
f( 1) 2 1 ( 1) 1 3
?? ?] ?M ?M ?M ?K ?] ?? ?? ?M ?] ?M ?M ?M ?K ?] ?? at x = 2 : 
LHL f(x) 8 2 1 3 12
Cont.
f(2) 8 2 2 4 12
?? ?] ?K ?M ?K ?] ?? ?? ?] ?K ?M ?K ?] ?? at x = 0 : 
LHL 0 0 ( 1) 0 1
Dis.
f(0) 0
?? ?K ?M ?M ?K ?] ?? ?? ?] ?? at x = 1 
LHL 2 1 0 0 3
f(1) 3 1 1 3 Cont.
RHL 2 1 1 1 3
?? ?K ?M ?K ?] ?? ?? ?] ?M ?K ?] ?? ?? ?? ?K ?M ?K ?] ?? 2. The differential equation of the family of circles
passing the origin and having center at the line
y = x is :
(1) (x
2
 – y
2
 + 2xy)dx = (x
2
 – y
2
 + 2xy)dy
(2) (x
2
 + y
2
 + 2xy)dx = (x
2
 + y
2
 – 2xy)dy
(3) (x
2
 – y
2
 + 2xy)dx = (x
2
 – y
2
 – 2xy)dy
(4) (x
2
 + y
2
 – 2xy)dx = (x
2
 + y
2
 + 2xy)dy
Ans. (3) 
Sol. C ?? x
2
 + y
2
 + gx + gy = 0        .....(1) 
2x + 2yy' + g + gy' = 0 
2x 2yy'
g
1 y'
???? ?K ?]?M
????
?K????
Put in (1) 
2 2
2x 2yy'
x y (x y) 0
1 y'
???? ?K ?K ?M ?K ?] ????
?K????
(x
2
 – y
2
 – 2xy)y' = x
2
 – y
2
 + 2xy 
3. Let S
1
 = {z ?? C : |z| ?? 5},
 
2
z 1 3 i
S z C : Im 0
1 3 i
???? ????
?K?M ????
?] ?? ?? ???? ????
????
?M ???? ???? ????
and  
S
3
 = {z ?? C : Re (z) ?? ?@?P }. Then 
(1) 
125
6
?? (2) 
125
24
?? (3) 
125
4
?? (4) 
125
12
?? Ans. (4) 
Sol. S
1
 : x
2
 + y
2
 ?? 25 .....(1) 
S
2
 : lm of 
?H ?I ?H ?I z 1 3 i
0
1 3 i
?K?M
?? ?M lm of 
x iy
1 0
1 3 i
????
?K ?K??
????
????
?M????
lm of 
?H ?I ?H ?I x iy 1 3 i
0
4
????
?K?K
????
?? ????
????
?? 3 x y 0 ?K?? ......(2) 
S
3
 : x ?? 0 ......(3) 
Area = 
?H ?I 2
5
(5)
12
??
Page 2


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Friday 05
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ƒ: [–1, 2] ?? R be given by
ƒ(x) = 2x
2
 + x + [x
2
] – [x], where [t] denotes the
greatest integer less than or equal to t. The number
of points, where ƒ is not continuous, is :
(1) 6 (2) 3
(3) 4 (4) 5
Ans. (3) 
Sol. Doubtful points : –1, 0, 1, 2 , 3 , 2 
at x = 2 , 3
?H ?I 2 2
Cont. Cont.
f(x) 2x x [x] x Discount
?? ?? ????
?] ?K ?M ?K ?]????
at x = –1 : 
RHL f(x) (2 1 ( 1)) 0 2
Dis.
f( 1) 2 1 ( 1) 1 3
?? ?] ?M ?M ?M ?K ?] ?? ?? ?M ?] ?M ?M ?M ?K ?] ?? at x = 2 : 
LHL f(x) 8 2 1 3 12
Cont.
f(2) 8 2 2 4 12
?? ?] ?K ?M ?K ?] ?? ?? ?] ?K ?M ?K ?] ?? at x = 0 : 
LHL 0 0 ( 1) 0 1
Dis.
f(0) 0
?? ?K ?M ?M ?K ?] ?? ?? ?] ?? at x = 1 
LHL 2 1 0 0 3
f(1) 3 1 1 3 Cont.
RHL 2 1 1 1 3
?? ?K ?M ?K ?] ?? ?? ?] ?M ?K ?] ?? ?? ?? ?K ?M ?K ?] ?? 2. The differential equation of the family of circles
passing the origin and having center at the line
y = x is :
(1) (x
2
 – y
2
 + 2xy)dx = (x
2
 – y
2
 + 2xy)dy
(2) (x
2
 + y
2
 + 2xy)dx = (x
2
 + y
2
 – 2xy)dy
(3) (x
2
 – y
2
 + 2xy)dx = (x
2
 – y
2
 – 2xy)dy
(4) (x
2
 + y
2
 – 2xy)dx = (x
2
 + y
2
 + 2xy)dy
Ans. (3) 
Sol. C ?? x
2
 + y
2
 + gx + gy = 0        .....(1) 
2x + 2yy' + g + gy' = 0 
2x 2yy'
g
1 y'
???? ?K ?]?M
????
?K????
Put in (1) 
2 2
2x 2yy'
x y (x y) 0
1 y'
???? ?K ?K ?M ?K ?] ????
?K????
(x
2
 – y
2
 – 2xy)y' = x
2
 – y
2
 + 2xy 
3. Let S
1
 = {z ?? C : |z| ?? 5},
 
2
z 1 3 i
S z C : Im 0
1 3 i
???? ????
?K?M ????
?] ?? ?? ???? ????
????
?M ???? ???? ????
and  
S
3
 = {z ?? C : Re (z) ?? ?@?P }. Then 
(1) 
125
6
?? (2) 
125
24
?? (3) 
125
4
?? (4) 
125
12
?? Ans. (4) 
Sol. S
1
 : x
2
 + y
2
 ?? 25 .....(1) 
S
2
 : lm of 
?H ?I ?H ?I z 1 3 i
0
1 3 i
?K?M
?? ?M lm of 
x iy
1 0
1 3 i
????
?K ?K??
????
????
?M????
lm of 
?H ?I ?H ?I x iy 1 3 i
0
4
????
?K?K
????
?? ????
????
?? 3 x y 0 ?K?? ......(2) 
S
3
 : x ?? 0 ......(3) 
Area = 
?H ?I 2
5
(5)
12
?? 
 
60° 
 
 
4. The area enclosed between the curves y = x|x| and 
y = x – |x| is : 
 (1) 
8
3
 (2) 
2
3
 
 (3) 1  (4) 
4
3
 
 Ans. (4) 
Sol.  
 
 
y = 2x 
y = x
2 
y = 0
 
(0,0)
 
(–2,–4)
 
y = –x
2 
 
 
0
2
2
A x 2x
?M ?] ?M ?M ?? 
4
3
?] 
5. 60 words can be made using all the letters of the 
word BHBJO, with or without meaning. If these 
words are written as in a dictionary, then the 50
th
 
word is : 
 (1) OBBHJ (2) HBBJO 
 (3) OBBJH  (4) JBBOH 
 Ans. (3) 
Sol. B  B  H  J  O 
 B _____4! 24 ?] 
 
4!
H _____ 12
2!
?] 
 
4!
J _____ 12
2!
?] 
 O  B  B  H  J 
 O  B  B  J  H ?? 50
th
 rank 
6. Let 
ˆˆ ˆ
a 2i 5j k ?] ?K ?M , 
ˆˆ ˆ
b 2i 2j 2k ?] ?M ?K  
 and c be  three vectors such that  
 
?H ?I ?H ?I ?H ?I ˆ ˆ ˆ
c i a b i a c i ?K ?? ?K ?K ?] ?? ?K 
. a.c 29 ?]?M ,  
 then 
?H ?I ˆˆ ˆ
c. 2i j k ?M ?K ?K is equal to : 
 (1) 10 (2) 5 
 (3) 15  (4) 12 
 Ans. (2) 
Sol. Let's assume 
ˆ
v a b i ?] ?K ?K 
       
ˆˆ ˆ
5i 3j k ?] ?K ?K 
 and 
ˆ
c i p ?K?] 
 So, 
 p v a p ?? ?] ?? 
 p v p a 0 ?? ?K ?? ?] 
 ?H ?I p v a 0 ?? ?K ?] 
 ?? 
?H ?I p v a ?] ?? ?K 
 
?H ?I ˆˆ
c i 7i 8j ?K ?] ?? ?K 
 
?H ?I ˆ ˆ ˆ
a.c a.i a. 7i 8j ?K ?] ?? ?K 
 
?H ?I 29 2 14 40 ?M ?K ?] ?? ?K 
 
1
2
?? ?] ?M 
?H ?I ?H ?I ?H ?I ?H ?I ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
c. 2i j k i. 2i j k 7i 8j . 2i j k ?M ?K ?K ?K ?M ?K ?K ?] ?? ?K ?M ?K ?K ?H ?I 1
14 8 2 5
2
?] ?M ?M ?K ?K ?] 
7. Consider three vectors a,b,c . Let a 2, b 3 ?]?] 
and a b c ?]?? . If 0,
3
?? ????
????
????
????
 is the angle between 
the vectors b and c , then the minimum value of 
2
27 c a ?M is equal to : 
 (1) 110 (2) 105 
 (3) 124  (4) 121 
 Ans. (3) 
Page 3


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Friday 05
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ƒ: [–1, 2] ?? R be given by
ƒ(x) = 2x
2
 + x + [x
2
] – [x], where [t] denotes the
greatest integer less than or equal to t. The number
of points, where ƒ is not continuous, is :
(1) 6 (2) 3
(3) 4 (4) 5
Ans. (3) 
Sol. Doubtful points : –1, 0, 1, 2 , 3 , 2 
at x = 2 , 3
?H ?I 2 2
Cont. Cont.
f(x) 2x x [x] x Discount
?? ?? ????
?] ?K ?M ?K ?]????
at x = –1 : 
RHL f(x) (2 1 ( 1)) 0 2
Dis.
f( 1) 2 1 ( 1) 1 3
?? ?] ?M ?M ?M ?K ?] ?? ?? ?M ?] ?M ?M ?M ?K ?] ?? at x = 2 : 
LHL f(x) 8 2 1 3 12
Cont.
f(2) 8 2 2 4 12
?? ?] ?K ?M ?K ?] ?? ?? ?] ?K ?M ?K ?] ?? at x = 0 : 
LHL 0 0 ( 1) 0 1
Dis.
f(0) 0
?? ?K ?M ?M ?K ?] ?? ?? ?] ?? at x = 1 
LHL 2 1 0 0 3
f(1) 3 1 1 3 Cont.
RHL 2 1 1 1 3
?? ?K ?M ?K ?] ?? ?? ?] ?M ?K ?] ?? ?? ?? ?K ?M ?K ?] ?? 2. The differential equation of the family of circles
passing the origin and having center at the line
y = x is :
(1) (x
2
 – y
2
 + 2xy)dx = (x
2
 – y
2
 + 2xy)dy
(2) (x
2
 + y
2
 + 2xy)dx = (x
2
 + y
2
 – 2xy)dy
(3) (x
2
 – y
2
 + 2xy)dx = (x
2
 – y
2
 – 2xy)dy
(4) (x
2
 + y
2
 – 2xy)dx = (x
2
 + y
2
 + 2xy)dy
Ans. (3) 
Sol. C ?? x
2
 + y
2
 + gx + gy = 0        .....(1) 
2x + 2yy' + g + gy' = 0 
2x 2yy'
g
1 y'
???? ?K ?]?M
????
?K????
Put in (1) 
2 2
2x 2yy'
x y (x y) 0
1 y'
???? ?K ?K ?M ?K ?] ????
?K????
(x
2
 – y
2
 – 2xy)y' = x
2
 – y
2
 + 2xy 
3. Let S
1
 = {z ?? C : |z| ?? 5},
 
2
z 1 3 i
S z C : Im 0
1 3 i
???? ????
?K?M ????
?] ?? ?? ???? ????
????
?M ???? ???? ????
and  
S
3
 = {z ?? C : Re (z) ?? ?@?P }. Then 
(1) 
125
6
?? (2) 
125
24
?? (3) 
125
4
?? (4) 
125
12
?? Ans. (4) 
Sol. S
1
 : x
2
 + y
2
 ?? 25 .....(1) 
S
2
 : lm of 
?H ?I ?H ?I z 1 3 i
0
1 3 i
?K?M
?? ?M lm of 
x iy
1 0
1 3 i
????
?K ?K??
????
????
?M????
lm of 
?H ?I ?H ?I x iy 1 3 i
0
4
????
?K?K
????
?? ????
????
?? 3 x y 0 ?K?? ......(2) 
S
3
 : x ?? 0 ......(3) 
Area = 
?H ?I 2
5
(5)
12
?? 
 
60° 
 
 
4. The area enclosed between the curves y = x|x| and 
y = x – |x| is : 
 (1) 
8
3
 (2) 
2
3
 
 (3) 1  (4) 
4
3
 
 Ans. (4) 
Sol.  
 
 
y = 2x 
y = x
2 
y = 0
 
(0,0)
 
(–2,–4)
 
y = –x
2 
 
 
0
2
2
A x 2x
?M ?] ?M ?M ?? 
4
3
?] 
5. 60 words can be made using all the letters of the 
word BHBJO, with or without meaning. If these 
words are written as in a dictionary, then the 50
th
 
word is : 
 (1) OBBHJ (2) HBBJO 
 (3) OBBJH  (4) JBBOH 
 Ans. (3) 
Sol. B  B  H  J  O 
 B _____4! 24 ?] 
 
4!
H _____ 12
2!
?] 
 
4!
J _____ 12
2!
?] 
 O  B  B  H  J 
 O  B  B  J  H ?? 50
th
 rank 
6. Let 
ˆˆ ˆ
a 2i 5j k ?] ?K ?M , 
ˆˆ ˆ
b 2i 2j 2k ?] ?M ?K  
 and c be  three vectors such that  
 
?H ?I ?H ?I ?H ?I ˆ ˆ ˆ
c i a b i a c i ?K ?? ?K ?K ?] ?? ?K 
. a.c 29 ?]?M ,  
 then 
?H ?I ˆˆ ˆ
c. 2i j k ?M ?K ?K is equal to : 
 (1) 10 (2) 5 
 (3) 15  (4) 12 
 Ans. (2) 
Sol. Let's assume 
ˆ
v a b i ?] ?K ?K 
       
ˆˆ ˆ
5i 3j k ?] ?K ?K 
 and 
ˆ
c i p ?K?] 
 So, 
 p v a p ?? ?] ?? 
 p v p a 0 ?? ?K ?? ?] 
 ?H ?I p v a 0 ?? ?K ?] 
 ?? 
?H ?I p v a ?] ?? ?K 
 
?H ?I ˆˆ
c i 7i 8j ?K ?] ?? ?K 
 
?H ?I ˆ ˆ ˆ
a.c a.i a. 7i 8j ?K ?] ?? ?K 
 
?H ?I 29 2 14 40 ?M ?K ?] ?? ?K 
 
1
2
?? ?] ?M 
?H ?I ?H ?I ?H ?I ?H ?I ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
c. 2i j k i. 2i j k 7i 8j . 2i j k ?M ?K ?K ?K ?M ?K ?K ?] ?? ?K ?M ?K ?K ?H ?I 1
14 8 2 5
2
?] ?M ?M ?K ?K ?] 
7. Consider three vectors a,b,c . Let a 2, b 3 ?]?] 
and a b c ?]?? . If 0,
3
?? ????
????
????
????
 is the angle between 
the vectors b and c , then the minimum value of 
2
27 c a ?M is equal to : 
 (1) 110 (2) 105 
 (3) 124  (4) 121 
 Ans. (3) 
 
 
Sol. 
2 2
c a c a 2a.c ?M ?] ?K ?M 
 
2
c 4 0 ?] ?K ?M 
 ?q a b c ?]?? 
 a b c ?]?? 
 2 3 c sin ?]?? 
 
2
c cosec
3
?]?? 0,
3
?? ????
????
????
????
 
 
min
2 2
c
3
3
?]?? 
2
cosec ,
3
????
?? ?? ?? ?? ??????
 
 ?? 
2
min
16
27 c a 27 4 124
27
????
?M ?] ?K ?] ????
????
 
8. Let A(–1, 1) and B(2, 3) be two points and P be a 
variable point above the line AB such that the area 
of ?d PAB is 10. If the locus of P is ax + by = 15, 
then 5a + 2b is : 
 (1) 
12
5
?M (2) 
6
5
?M 
 (3) 4  (4) 6 
 Ans. (1) 
Sol.  
 
 
P(h,k)
 
B(2,3)
 
A(–1,1)
 
 
 
h k 1
1
1 1 1 10
2
2 3 1
?M?] 
 –2x + 3y = 25 
 
6 9
x y 15
5 5
?M ?K ?] 
 a = 
6
5
?M , b = 
9
5
 
 5a = –6, 2b = 
18
5
 
9. Let ( ?? , ?? , ?? ) be the point (8, 5, 7) in the line 
x 1 y 1 z 2
2 3 5
?M ?K ?M ?]?] . Then ?? + ?? + ?? is equal to 
 (1) 16 (2) 18 
 (3) 14  (4) 20 
 Ans. (3) 
Sol.  
 
 
(2 ?? +1, 3 ?? – 1, 5 ?? + 2) 
M 
A' 
A (8, 5, 7) 
 
 
 
?H ?I ˆˆ ˆ
AM. 2i 3j 5k 0 ?K ?K ?] 
 (2 ?? – 7)(2) + (3 ?? – 6)(3) + (5 ?? – 5)(5) = 0 
 38 ?? = 57 
 
3
2
???] 
 
7 19
M 4, ,
2 2
????
????
????
 
 A'(0,2,12) 
10. If the constant term in the expansion of 
12
5
3
3 2x
x
5
????
?K ????
????
????
, x ?? 0, is ?? ×2
8
×
5
3 , then 25 ?? is 
equal to : 
 (1) 639 (2) 724 
 (3) 693  (4) 742 
 Ans. (3) 
Sol. 
12 r
r
1/5
12
r 1 r
1/3
3 2x
T C
x 5
?M ?K ????
????
?] ????
????
????
????
 
 
?H ?I ?H ?I ?H ?I ?H ?I 12 r
r 2r 12
12
5
r
r 1
r/3
C 3 2 x
T
5
?M ?M ?K ?] 
 r = 6 
 
?H ?I ?H ?I 6/5 6
12
6 8 1/5
7
2
C 3 2
9 11 7
T 2 .3
25 5
???? ????
?]?]
????
????
 
 25 ?? = 693 
Page 4


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Friday 05
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ƒ: [–1, 2] ?? R be given by
ƒ(x) = 2x
2
 + x + [x
2
] – [x], where [t] denotes the
greatest integer less than or equal to t. The number
of points, where ƒ is not continuous, is :
(1) 6 (2) 3
(3) 4 (4) 5
Ans. (3) 
Sol. Doubtful points : –1, 0, 1, 2 , 3 , 2 
at x = 2 , 3
?H ?I 2 2
Cont. Cont.
f(x) 2x x [x] x Discount
?? ?? ????
?] ?K ?M ?K ?]????
at x = –1 : 
RHL f(x) (2 1 ( 1)) 0 2
Dis.
f( 1) 2 1 ( 1) 1 3
?? ?] ?M ?M ?M ?K ?] ?? ?? ?M ?] ?M ?M ?M ?K ?] ?? at x = 2 : 
LHL f(x) 8 2 1 3 12
Cont.
f(2) 8 2 2 4 12
?? ?] ?K ?M ?K ?] ?? ?? ?] ?K ?M ?K ?] ?? at x = 0 : 
LHL 0 0 ( 1) 0 1
Dis.
f(0) 0
?? ?K ?M ?M ?K ?] ?? ?? ?] ?? at x = 1 
LHL 2 1 0 0 3
f(1) 3 1 1 3 Cont.
RHL 2 1 1 1 3
?? ?K ?M ?K ?] ?? ?? ?] ?M ?K ?] ?? ?? ?? ?K ?M ?K ?] ?? 2. The differential equation of the family of circles
passing the origin and having center at the line
y = x is :
(1) (x
2
 – y
2
 + 2xy)dx = (x
2
 – y
2
 + 2xy)dy
(2) (x
2
 + y
2
 + 2xy)dx = (x
2
 + y
2
 – 2xy)dy
(3) (x
2
 – y
2
 + 2xy)dx = (x
2
 – y
2
 – 2xy)dy
(4) (x
2
 + y
2
 – 2xy)dx = (x
2
 + y
2
 + 2xy)dy
Ans. (3) 
Sol. C ?? x
2
 + y
2
 + gx + gy = 0        .....(1) 
2x + 2yy' + g + gy' = 0 
2x 2yy'
g
1 y'
???? ?K ?]?M
????
?K????
Put in (1) 
2 2
2x 2yy'
x y (x y) 0
1 y'
???? ?K ?K ?M ?K ?] ????
?K????
(x
2
 – y
2
 – 2xy)y' = x
2
 – y
2
 + 2xy 
3. Let S
1
 = {z ?? C : |z| ?? 5},
 
2
z 1 3 i
S z C : Im 0
1 3 i
???? ????
?K?M ????
?] ?? ?? ???? ????
????
?M ???? ???? ????
and  
S
3
 = {z ?? C : Re (z) ?? ?@?P }. Then 
(1) 
125
6
?? (2) 
125
24
?? (3) 
125
4
?? (4) 
125
12
?? Ans. (4) 
Sol. S
1
 : x
2
 + y
2
 ?? 25 .....(1) 
S
2
 : lm of 
?H ?I ?H ?I z 1 3 i
0
1 3 i
?K?M
?? ?M lm of 
x iy
1 0
1 3 i
????
?K ?K??
????
????
?M????
lm of 
?H ?I ?H ?I x iy 1 3 i
0
4
????
?K?K
????
?? ????
????
?? 3 x y 0 ?K?? ......(2) 
S
3
 : x ?? 0 ......(3) 
Area = 
?H ?I 2
5
(5)
12
?? 
 
60° 
 
 
4. The area enclosed between the curves y = x|x| and 
y = x – |x| is : 
 (1) 
8
3
 (2) 
2
3
 
 (3) 1  (4) 
4
3
 
 Ans. (4) 
Sol.  
 
 
y = 2x 
y = x
2 
y = 0
 
(0,0)
 
(–2,–4)
 
y = –x
2 
 
 
0
2
2
A x 2x
?M ?] ?M ?M ?? 
4
3
?] 
5. 60 words can be made using all the letters of the 
word BHBJO, with or without meaning. If these 
words are written as in a dictionary, then the 50
th
 
word is : 
 (1) OBBHJ (2) HBBJO 
 (3) OBBJH  (4) JBBOH 
 Ans. (3) 
Sol. B  B  H  J  O 
 B _____4! 24 ?] 
 
4!
H _____ 12
2!
?] 
 
4!
J _____ 12
2!
?] 
 O  B  B  H  J 
 O  B  B  J  H ?? 50
th
 rank 
6. Let 
ˆˆ ˆ
a 2i 5j k ?] ?K ?M , 
ˆˆ ˆ
b 2i 2j 2k ?] ?M ?K  
 and c be  three vectors such that  
 
?H ?I ?H ?I ?H ?I ˆ ˆ ˆ
c i a b i a c i ?K ?? ?K ?K ?] ?? ?K 
. a.c 29 ?]?M ,  
 then 
?H ?I ˆˆ ˆ
c. 2i j k ?M ?K ?K is equal to : 
 (1) 10 (2) 5 
 (3) 15  (4) 12 
 Ans. (2) 
Sol. Let's assume 
ˆ
v a b i ?] ?K ?K 
       
ˆˆ ˆ
5i 3j k ?] ?K ?K 
 and 
ˆ
c i p ?K?] 
 So, 
 p v a p ?? ?] ?? 
 p v p a 0 ?? ?K ?? ?] 
 ?H ?I p v a 0 ?? ?K ?] 
 ?? 
?H ?I p v a ?] ?? ?K 
 
?H ?I ˆˆ
c i 7i 8j ?K ?] ?? ?K 
 
?H ?I ˆ ˆ ˆ
a.c a.i a. 7i 8j ?K ?] ?? ?K 
 
?H ?I 29 2 14 40 ?M ?K ?] ?? ?K 
 
1
2
?? ?] ?M 
?H ?I ?H ?I ?H ?I ?H ?I ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
c. 2i j k i. 2i j k 7i 8j . 2i j k ?M ?K ?K ?K ?M ?K ?K ?] ?? ?K ?M ?K ?K ?H ?I 1
14 8 2 5
2
?] ?M ?M ?K ?K ?] 
7. Consider three vectors a,b,c . Let a 2, b 3 ?]?] 
and a b c ?]?? . If 0,
3
?? ????
????
????
????
 is the angle between 
the vectors b and c , then the minimum value of 
2
27 c a ?M is equal to : 
 (1) 110 (2) 105 
 (3) 124  (4) 121 
 Ans. (3) 
 
 
Sol. 
2 2
c a c a 2a.c ?M ?] ?K ?M 
 
2
c 4 0 ?] ?K ?M 
 ?q a b c ?]?? 
 a b c ?]?? 
 2 3 c sin ?]?? 
 
2
c cosec
3
?]?? 0,
3
?? ????
????
????
????
 
 
min
2 2
c
3
3
?]?? 
2
cosec ,
3
????
?? ?? ?? ?? ??????
 
 ?? 
2
min
16
27 c a 27 4 124
27
????
?M ?] ?K ?] ????
????
 
8. Let A(–1, 1) and B(2, 3) be two points and P be a 
variable point above the line AB such that the area 
of ?d PAB is 10. If the locus of P is ax + by = 15, 
then 5a + 2b is : 
 (1) 
12
5
?M (2) 
6
5
?M 
 (3) 4  (4) 6 
 Ans. (1) 
Sol.  
 
 
P(h,k)
 
B(2,3)
 
A(–1,1)
 
 
 
h k 1
1
1 1 1 10
2
2 3 1
?M?] 
 –2x + 3y = 25 
 
6 9
x y 15
5 5
?M ?K ?] 
 a = 
6
5
?M , b = 
9
5
 
 5a = –6, 2b = 
18
5
 
9. Let ( ?? , ?? , ?? ) be the point (8, 5, 7) in the line 
x 1 y 1 z 2
2 3 5
?M ?K ?M ?]?] . Then ?? + ?? + ?? is equal to 
 (1) 16 (2) 18 
 (3) 14  (4) 20 
 Ans. (3) 
Sol.  
 
 
(2 ?? +1, 3 ?? – 1, 5 ?? + 2) 
M 
A' 
A (8, 5, 7) 
 
 
 
?H ?I ˆˆ ˆ
AM. 2i 3j 5k 0 ?K ?K ?] 
 (2 ?? – 7)(2) + (3 ?? – 6)(3) + (5 ?? – 5)(5) = 0 
 38 ?? = 57 
 
3
2
???] 
 
7 19
M 4, ,
2 2
????
????
????
 
 A'(0,2,12) 
10. If the constant term in the expansion of 
12
5
3
3 2x
x
5
????
?K ????
????
????
, x ?? 0, is ?? ×2
8
×
5
3 , then 25 ?? is 
equal to : 
 (1) 639 (2) 724 
 (3) 693  (4) 742 
 Ans. (3) 
Sol. 
12 r
r
1/5
12
r 1 r
1/3
3 2x
T C
x 5
?M ?K ????
????
?] ????
????
????
????
 
 
?H ?I ?H ?I ?H ?I ?H ?I 12 r
r 2r 12
12
5
r
r 1
r/3
C 3 2 x
T
5
?M ?M ?K ?] 
 r = 6 
 
?H ?I ?H ?I 6/5 6
12
6 8 1/5
7
2
C 3 2
9 11 7
T 2 .3
25 5
???? ????
?]?]
????
????
 
 25 ?? = 693 
11. Let ƒ, g : R ?? R be defined as : ƒ(x) = |x – 1| and 
x
e , x 0
g(x)
x 1, x 0
?? ?? ?? ?] ?? ?K?? ?? ?? . Then the function ƒ(g(x)) is  
 (1) neither one-one nor onto. 
 (2) one-one but not onto. 
 (3) both one-one and onto. 
 (4) onto but not one-one. 
 Ans. (1) 
Sol. f(g(x)) = |g(x) – 1| 
 
x
e 1 x 0
fog
x 1 1 x 0
?? ?M??
?? ?? ?K ?M ?? ?? 
 
x
e 1 x 0
fog
x x 0
?? ?M??
???M??
?? 
 
 
 
12. Let the circle C
1
 : x
2
 + y
2
 – 2(x + y) + 1 = 0 and C
2
 
be a circle having centre at (–1, 0) and radius 2. If 
the line of the common chord of C
1
 and C
2
 
intersects the y-axis at the point P, then the square 
of the distance of P from the centre of C
1
 is : 
 (1) 2 (2) 1 
 (3) 6  (4) 4 
 Ans. (1) 
Sol. S
1
 : x
2
 + y
2
 – 2x – 2y + 1 = 0 
 S
2
 : x
2
 + y
2
 + 2x – 3 = 0 
 Common chord = S
1
 – S
2
 = 0 
 –4x – 2y + 4 = 0 
 2x + y = 2 ?? P(0, 2) 
 
2
(c,p)
d = (1 – 0)
2
 + (2 – 1)
2
 = 2 
 
13. Let the set S = {2, 4, 8, 16, ....., 512} be partitioned 
into 3 sets A, B, C with equal number of elements such 
that A ?? B ?? C = S and A ?? B = B ?? C = A ?? C = ?? . 
The maximum number of such possible partitions of S 
is equal to : 
 (1) 1680 (2) 1520 
 (3) 1710  (4) 1640 
 Ans. (1) 
Sol.  
 
 
A
 
B
 
C
 
 
 
?H ?I 9!
3!
3!3!3! 3!
?? 
14. The values of m, n, for which the system of 
equations 
 x + y + z = 4, 
 2x + 5y + 5z = 17, 
 x + 2y + mz = n 
 has infinitely many solutions, satisfy the equation : 
 (1) m
2
 + n
2
 – m – n = 46 
 (2) m
2
 + n
2
 + m + n = 64 
 (3) m
2
 + n
2
 + mn = 68 
 (4) m
2
 + n
2
 – mn = 39 
 Ans. (4) 
Sol. 
1 1 1
D 2 5 5 0
1 2 m
?]?] ?? m = 2 
 
3
1 1 4
D 2 5 17 0
1 2 n
?]?]
 
?? n = 7 
 
15. The coefficients a, b, c in the quadratic equation 
ax
2
 + bx + c = 0 are from the set {1, 2, 3, 4, 5, 6}. 
If the probability of this equation having one real 
root bigger than the other is p, then 216p equals :  
 (1) 57 (2) 38 
 (3) 19  (4) 76 
 Ans. (2) 
Page 5


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Friday 05
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ƒ: [–1, 2] ?? R be given by
ƒ(x) = 2x
2
 + x + [x
2
] – [x], where [t] denotes the
greatest integer less than or equal to t. The number
of points, where ƒ is not continuous, is :
(1) 6 (2) 3
(3) 4 (4) 5
Ans. (3) 
Sol. Doubtful points : –1, 0, 1, 2 , 3 , 2 
at x = 2 , 3
?H ?I 2 2
Cont. Cont.
f(x) 2x x [x] x Discount
?? ?? ????
?] ?K ?M ?K ?]????
at x = –1 : 
RHL f(x) (2 1 ( 1)) 0 2
Dis.
f( 1) 2 1 ( 1) 1 3
?? ?] ?M ?M ?M ?K ?] ?? ?? ?M ?] ?M ?M ?M ?K ?] ?? at x = 2 : 
LHL f(x) 8 2 1 3 12
Cont.
f(2) 8 2 2 4 12
?? ?] ?K ?M ?K ?] ?? ?? ?] ?K ?M ?K ?] ?? at x = 0 : 
LHL 0 0 ( 1) 0 1
Dis.
f(0) 0
?? ?K ?M ?M ?K ?] ?? ?? ?] ?? at x = 1 
LHL 2 1 0 0 3
f(1) 3 1 1 3 Cont.
RHL 2 1 1 1 3
?? ?K ?M ?K ?] ?? ?? ?] ?M ?K ?] ?? ?? ?? ?K ?M ?K ?] ?? 2. The differential equation of the family of circles
passing the origin and having center at the line
y = x is :
(1) (x
2
 – y
2
 + 2xy)dx = (x
2
 – y
2
 + 2xy)dy
(2) (x
2
 + y
2
 + 2xy)dx = (x
2
 + y
2
 – 2xy)dy
(3) (x
2
 – y
2
 + 2xy)dx = (x
2
 – y
2
 – 2xy)dy
(4) (x
2
 + y
2
 – 2xy)dx = (x
2
 + y
2
 + 2xy)dy
Ans. (3) 
Sol. C ?? x
2
 + y
2
 + gx + gy = 0        .....(1) 
2x + 2yy' + g + gy' = 0 
2x 2yy'
g
1 y'
???? ?K ?]?M
????
?K????
Put in (1) 
2 2
2x 2yy'
x y (x y) 0
1 y'
???? ?K ?K ?M ?K ?] ????
?K????
(x
2
 – y
2
 – 2xy)y' = x
2
 – y
2
 + 2xy 
3. Let S
1
 = {z ?? C : |z| ?? 5},
 
2
z 1 3 i
S z C : Im 0
1 3 i
???? ????
?K?M ????
?] ?? ?? ???? ????
????
?M ???? ???? ????
and  
S
3
 = {z ?? C : Re (z) ?? ?@?P }. Then 
(1) 
125
6
?? (2) 
125
24
?? (3) 
125
4
?? (4) 
125
12
?? Ans. (4) 
Sol. S
1
 : x
2
 + y
2
 ?? 25 .....(1) 
S
2
 : lm of 
?H ?I ?H ?I z 1 3 i
0
1 3 i
?K?M
?? ?M lm of 
x iy
1 0
1 3 i
????
?K ?K??
????
????
?M????
lm of 
?H ?I ?H ?I x iy 1 3 i
0
4
????
?K?K
????
?? ????
????
?? 3 x y 0 ?K?? ......(2) 
S
3
 : x ?? 0 ......(3) 
Area = 
?H ?I 2
5
(5)
12
?? 
 
60° 
 
 
4. The area enclosed between the curves y = x|x| and 
y = x – |x| is : 
 (1) 
8
3
 (2) 
2
3
 
 (3) 1  (4) 
4
3
 
 Ans. (4) 
Sol.  
 
 
y = 2x 
y = x
2 
y = 0
 
(0,0)
 
(–2,–4)
 
y = –x
2 
 
 
0
2
2
A x 2x
?M ?] ?M ?M ?? 
4
3
?] 
5. 60 words can be made using all the letters of the 
word BHBJO, with or without meaning. If these 
words are written as in a dictionary, then the 50
th
 
word is : 
 (1) OBBHJ (2) HBBJO 
 (3) OBBJH  (4) JBBOH 
 Ans. (3) 
Sol. B  B  H  J  O 
 B _____4! 24 ?] 
 
4!
H _____ 12
2!
?] 
 
4!
J _____ 12
2!
?] 
 O  B  B  H  J 
 O  B  B  J  H ?? 50
th
 rank 
6. Let 
ˆˆ ˆ
a 2i 5j k ?] ?K ?M , 
ˆˆ ˆ
b 2i 2j 2k ?] ?M ?K  
 and c be  three vectors such that  
 
?H ?I ?H ?I ?H ?I ˆ ˆ ˆ
c i a b i a c i ?K ?? ?K ?K ?] ?? ?K 
. a.c 29 ?]?M ,  
 then 
?H ?I ˆˆ ˆ
c. 2i j k ?M ?K ?K is equal to : 
 (1) 10 (2) 5 
 (3) 15  (4) 12 
 Ans. (2) 
Sol. Let's assume 
ˆ
v a b i ?] ?K ?K 
       
ˆˆ ˆ
5i 3j k ?] ?K ?K 
 and 
ˆ
c i p ?K?] 
 So, 
 p v a p ?? ?] ?? 
 p v p a 0 ?? ?K ?? ?] 
 ?H ?I p v a 0 ?? ?K ?] 
 ?? 
?H ?I p v a ?] ?? ?K 
 
?H ?I ˆˆ
c i 7i 8j ?K ?] ?? ?K 
 
?H ?I ˆ ˆ ˆ
a.c a.i a. 7i 8j ?K ?] ?? ?K 
 
?H ?I 29 2 14 40 ?M ?K ?] ?? ?K 
 
1
2
?? ?] ?M 
?H ?I ?H ?I ?H ?I ?H ?I ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
c. 2i j k i. 2i j k 7i 8j . 2i j k ?M ?K ?K ?K ?M ?K ?K ?] ?? ?K ?M ?K ?K ?H ?I 1
14 8 2 5
2
?] ?M ?M ?K ?K ?] 
7. Consider three vectors a,b,c . Let a 2, b 3 ?]?] 
and a b c ?]?? . If 0,
3
?? ????
????
????
????
 is the angle between 
the vectors b and c , then the minimum value of 
2
27 c a ?M is equal to : 
 (1) 110 (2) 105 
 (3) 124  (4) 121 
 Ans. (3) 
 
 
Sol. 
2 2
c a c a 2a.c ?M ?] ?K ?M 
 
2
c 4 0 ?] ?K ?M 
 ?q a b c ?]?? 
 a b c ?]?? 
 2 3 c sin ?]?? 
 
2
c cosec
3
?]?? 0,
3
?? ????
????
????
????
 
 
min
2 2
c
3
3
?]?? 
2
cosec ,
3
????
?? ?? ?? ?? ??????
 
 ?? 
2
min
16
27 c a 27 4 124
27
????
?M ?] ?K ?] ????
????
 
8. Let A(–1, 1) and B(2, 3) be two points and P be a 
variable point above the line AB such that the area 
of ?d PAB is 10. If the locus of P is ax + by = 15, 
then 5a + 2b is : 
 (1) 
12
5
?M (2) 
6
5
?M 
 (3) 4  (4) 6 
 Ans. (1) 
Sol.  
 
 
P(h,k)
 
B(2,3)
 
A(–1,1)
 
 
 
h k 1
1
1 1 1 10
2
2 3 1
?M?] 
 –2x + 3y = 25 
 
6 9
x y 15
5 5
?M ?K ?] 
 a = 
6
5
?M , b = 
9
5
 
 5a = –6, 2b = 
18
5
 
9. Let ( ?? , ?? , ?? ) be the point (8, 5, 7) in the line 
x 1 y 1 z 2
2 3 5
?M ?K ?M ?]?] . Then ?? + ?? + ?? is equal to 
 (1) 16 (2) 18 
 (3) 14  (4) 20 
 Ans. (3) 
Sol.  
 
 
(2 ?? +1, 3 ?? – 1, 5 ?? + 2) 
M 
A' 
A (8, 5, 7) 
 
 
 
?H ?I ˆˆ ˆ
AM. 2i 3j 5k 0 ?K ?K ?] 
 (2 ?? – 7)(2) + (3 ?? – 6)(3) + (5 ?? – 5)(5) = 0 
 38 ?? = 57 
 
3
2
???] 
 
7 19
M 4, ,
2 2
????
????
????
 
 A'(0,2,12) 
10. If the constant term in the expansion of 
12
5
3
3 2x
x
5
????
?K ????
????
????
, x ?? 0, is ?? ×2
8
×
5
3 , then 25 ?? is 
equal to : 
 (1) 639 (2) 724 
 (3) 693  (4) 742 
 Ans. (3) 
Sol. 
12 r
r
1/5
12
r 1 r
1/3
3 2x
T C
x 5
?M ?K ????
????
?] ????
????
????
????
 
 
?H ?I ?H ?I ?H ?I ?H ?I 12 r
r 2r 12
12
5
r
r 1
r/3
C 3 2 x
T
5
?M ?M ?K ?] 
 r = 6 
 
?H ?I ?H ?I 6/5 6
12
6 8 1/5
7
2
C 3 2
9 11 7
T 2 .3
25 5
???? ????
?]?]
????
????
 
 25 ?? = 693 
11. Let ƒ, g : R ?? R be defined as : ƒ(x) = |x – 1| and 
x
e , x 0
g(x)
x 1, x 0
?? ?? ?? ?] ?? ?K?? ?? ?? . Then the function ƒ(g(x)) is  
 (1) neither one-one nor onto. 
 (2) one-one but not onto. 
 (3) both one-one and onto. 
 (4) onto but not one-one. 
 Ans. (1) 
Sol. f(g(x)) = |g(x) – 1| 
 
x
e 1 x 0
fog
x 1 1 x 0
?? ?M??
?? ?? ?K ?M ?? ?? 
 
x
e 1 x 0
fog
x x 0
?? ?M??
???M??
?? 
 
 
 
12. Let the circle C
1
 : x
2
 + y
2
 – 2(x + y) + 1 = 0 and C
2
 
be a circle having centre at (–1, 0) and radius 2. If 
the line of the common chord of C
1
 and C
2
 
intersects the y-axis at the point P, then the square 
of the distance of P from the centre of C
1
 is : 
 (1) 2 (2) 1 
 (3) 6  (4) 4 
 Ans. (1) 
Sol. S
1
 : x
2
 + y
2
 – 2x – 2y + 1 = 0 
 S
2
 : x
2
 + y
2
 + 2x – 3 = 0 
 Common chord = S
1
 – S
2
 = 0 
 –4x – 2y + 4 = 0 
 2x + y = 2 ?? P(0, 2) 
 
2
(c,p)
d = (1 – 0)
2
 + (2 – 1)
2
 = 2 
 
13. Let the set S = {2, 4, 8, 16, ....., 512} be partitioned 
into 3 sets A, B, C with equal number of elements such 
that A ?? B ?? C = S and A ?? B = B ?? C = A ?? C = ?? . 
The maximum number of such possible partitions of S 
is equal to : 
 (1) 1680 (2) 1520 
 (3) 1710  (4) 1640 
 Ans. (1) 
Sol.  
 
 
A
 
B
 
C
 
 
 
?H ?I 9!
3!
3!3!3! 3!
?? 
14. The values of m, n, for which the system of 
equations 
 x + y + z = 4, 
 2x + 5y + 5z = 17, 
 x + 2y + mz = n 
 has infinitely many solutions, satisfy the equation : 
 (1) m
2
 + n
2
 – m – n = 46 
 (2) m
2
 + n
2
 + m + n = 64 
 (3) m
2
 + n
2
 + mn = 68 
 (4) m
2
 + n
2
 – mn = 39 
 Ans. (4) 
Sol. 
1 1 1
D 2 5 5 0
1 2 m
?]?] ?? m = 2 
 
3
1 1 4
D 2 5 17 0
1 2 n
?]?]
 
?? n = 7 
 
15. The coefficients a, b, c in the quadratic equation 
ax
2
 + bx + c = 0 are from the set {1, 2, 3, 4, 5, 6}. 
If the probability of this equation having one real 
root bigger than the other is p, then 216p equals :  
 (1) 57 (2) 38 
 (3) 19  (4) 76 
 Ans. (2) 
 
Sol. D > 0 
 b
2
 > 4ac 
 b = 3 : (a, c) = (1, 1)(1, 2)(2,1) 
 b = 4 : (a, c) = (1, 1)(1, 2)(2,1)(1,3)(3,1) 
 b = 5 : (a, c) = (1,1)(1,2)(2,1) (1,3)(3,1)(1,4)(4,1)  
       (1,5)(5,1)(1,6)(6,1)(2,3)(3,2)(2,2) 
 b = 6 : (a, c) = (1,1)(1,2)(2,1) (1,3)(3,1)(1,4)(4,1)  
    (1,5)(5,1)(1,6)(6,1)(2,3)(3,2)(2,4)(4,2)(2,2) 
 fav. cases = 38 
 Prob. : 
38
6 6 6 ????
 
16. Let ABCD and AEFG be squares of side 4 and  
2 units, respectively. The point E is on the line 
segment AB and the point F is on the diagonal AC. 
Then the radius r of the circle passing through the 
point F and touching the line segments BC and CD 
satisfies : 
 (1) r = 1 (2) r
2
 – 8r + 8 = 0 
 (3) 2r
2
 – 4r + 1 = 0 (4) 2r
2
 – 8r + 7 = 0 
 Ans. (2) 
Sol.  
  
A B 
E 
C D 
(0, 4) (4, 4) 
(4, 0) 
(2, 0) (0, 0) 
F(2,2) 
 
  
r 
r 
r 
o 
x = 4 
F 
(2,2) 
y = 4 
(4–r, 4–r ) 
 
 OF
2
 = r
2
 
 (2 – r)
2
 + (2 – r)
2
 = r
2
 
 r
2
 – 8r + 8 = 0 
17. Let ?? (m, n) = 
1
m 1 n 1
0
x (1 x) dx
?M?M
?M ?? , m, n > 0. If 
1
10 20
0
(1 x ) dx a (b,c) ?M ?] ?? ?? ?? , then 100(a + b + x) 
equals ______. 
 (1) 1021 (2) 1120 
 (3) 2012  (4) 2120 
 Ans. (4) 
Sol. 
?H ?I 1
20
10
0
I 1. 1 x dx ?]?M
?? 
 x
10
 = t 
 x = t
1/10
  
 ?H ?I 9/10 1
dx t dt
10
?M ?] 
 ?H ?I 1
9/10
20
0
1
I (1 t) t dt
10
?M ?]?M
?? 
 ?H ?I 1
20
9/10
0
1
I t 1 t dt
10
?M ?]?M
?? 
 a = 
1
10
 b = 
1
10
 c = 21 
18. Let ???? ?? 0 and 
3
A
2
????????
????
?] ?? ?? ?? ????
???? ?? ?M ?? ?? ????
.  
 If 
3 9 3
B 7 2
2 5 2
?? ?M ?? ????
????
?] ?M ?? ?M ?? ????
???? ?M ?? ?M ?? ????
 is the matrix of cofactors 
of the elements of A, then det(AB) is equal to : 
 (1) 343 (2) 125 
 (3) 64  (4) 216 
 Ans. (4) 
Sol. Equating co-factor fo A
21
  
 (2 ?? 2
 – 3 ?? ) = ?? 
 ?? = 0, 2 (accept) 
 Now, 2 ?? 2
 – ???? = 3 ?? 
 ?? = 2 ?? = 1 
 |AB| = |A cof (A)| = |A|
3
  
 
1 2 3
A 2 2 1 6 2(9) 3(6) 6
1 2 4
?] ?] ?M ?K ?] ?M