JEE Main 2024 January 30 Shift 1 Paper & Solutions | JEE Main & Advanced Mock Test Series CY_Marker_0 PDF Download

 Page 1


 
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024 
(Held On Tuesday 30
th
 January, 2024)            TIME : 9 : 00 AM  to  12 : 00 NOON 
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
 
 
SECTION-A 
1. A line passing through the point A(9, 0) makes an angle 
of 30º with the positive direction of x-axis. If this line is 
rotated about A through an angle of 15º in the clockwise 
direction, then its equation in the new position is 
 (1) 9
3 2
?K?]
?M y
x (2) 9
3 2
?K?]
?M x
y 
 (3) 9
3 2
?K?]
?K x
y (4) 9
3 2
?K?]
?K y
x 
Ans. (1) 
Sol. 
 
Eq
n 
: y – 0 = tan15° (x – 9) ?? y = (2 ?M 3 ?I?@ (x ?M 9) 
2. Let S
a
 denote the sum of first n terms an arithmetic 
progression. If S
20
 = 790 and S
10
 = 145, then S
15
 – 
S
5
 is : 
 (1) 395 
 (2) 390 
 (3) 405 
 (4) 410 
Ans. (1) 
Sol.
 
?{ ?} 20
20
S 2a 19d 790
2
?] ?K ?] 
 2a + 19d = 79 …..(1) 
 
?{ ?} 10
10
S 2a 9d 145
2
?] ?K ?] 
 2a + 9d = 29  …..(2) 
 From (1) and (2) a = -8, d = 5 
 
?{ ?} ?{ ?} 15 5
15 5
S S 2a 14d 2a 4d
2 2
?M ?] ?K ?M ?K 
 
?{ ?} ?{ ?} 15 5
16 70 16 20
2 2
?] ?M ?K ?M ?M ?K 
 = 405-10 
 = 395  
3. If z = x + iy, xy ?? 0 , satisfies the equation 
2
z i z 0 ?K?] , then |z
2
| is equal to : 
 (1) 9 
 (2) 1 
 (3) 4 
 (4) 
1
4
 
Ans. (2) 
Sol.
 
2
2
z iz
z iz
?]?M
?] 
 |z
2
| = |z| 
 |z|
2 
– |z| = 0 
 |z|(|z| – 1) = 0 
 |z| = 0 (not acceptable) 
 ?| |z| = 1 
 ?| |z|
2 
= 1 
4. Let 
i 2 3
ˆ ˆ ˆ
a a i a j a k ?] ?K ?K and 
1 2 3
ˆ ˆ ˆ
b b i b j b k ?] ?K ?K be 
two vectors such that a 1; ?] a.b 2 ?] and b 4. ?] If 
?H ?I c 2 a b 3b ?] ?? ?M , then the angle between b and c
is equal to : 
 (1) 
1
2
cos
3
?M ????
????
????
 
 (2) 
1
1
cos
3
?M????
?M????
????
 
 (3) 
1
3
cos
2
?M????
?M????
????
????
 
 (4) 
1
2
cos
3
?M ????
????
????
 
Ans. (3) 
Page 2


 
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024 
(Held On Tuesday 30
th
 January, 2024)            TIME : 9 : 00 AM  to  12 : 00 NOON 
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
 
 
SECTION-A 
1. A line passing through the point A(9, 0) makes an angle 
of 30º with the positive direction of x-axis. If this line is 
rotated about A through an angle of 15º in the clockwise 
direction, then its equation in the new position is 
 (1) 9
3 2
?K?]
?M y
x (2) 9
3 2
?K?]
?M x
y 
 (3) 9
3 2
?K?]
?K x
y (4) 9
3 2
?K?]
?K y
x 
Ans. (1) 
Sol. 
 
Eq
n 
: y – 0 = tan15° (x – 9) ?? y = (2 ?M 3 ?I?@ (x ?M 9) 
2. Let S
a
 denote the sum of first n terms an arithmetic 
progression. If S
20
 = 790 and S
10
 = 145, then S
15
 – 
S
5
 is : 
 (1) 395 
 (2) 390 
 (3) 405 
 (4) 410 
Ans. (1) 
Sol.
 
?{ ?} 20
20
S 2a 19d 790
2
?] ?K ?] 
 2a + 19d = 79 …..(1) 
 
?{ ?} 10
10
S 2a 9d 145
2
?] ?K ?] 
 2a + 9d = 29  …..(2) 
 From (1) and (2) a = -8, d = 5 
 
?{ ?} ?{ ?} 15 5
15 5
S S 2a 14d 2a 4d
2 2
?M ?] ?K ?M ?K 
 
?{ ?} ?{ ?} 15 5
16 70 16 20
2 2
?] ?M ?K ?M ?M ?K 
 = 405-10 
 = 395  
3. If z = x + iy, xy ?? 0 , satisfies the equation 
2
z i z 0 ?K?] , then |z
2
| is equal to : 
 (1) 9 
 (2) 1 
 (3) 4 
 (4) 
1
4
 
Ans. (2) 
Sol.
 
2
2
z iz
z iz
?]?M
?] 
 |z
2
| = |z| 
 |z|
2 
– |z| = 0 
 |z|(|z| – 1) = 0 
 |z| = 0 (not acceptable) 
 ?| |z| = 1 
 ?| |z|
2 
= 1 
4. Let 
i 2 3
ˆ ˆ ˆ
a a i a j a k ?] ?K ?K and 
1 2 3
ˆ ˆ ˆ
b b i b j b k ?] ?K ?K be 
two vectors such that a 1; ?] a.b 2 ?] and b 4. ?] If 
?H ?I c 2 a b 3b ?] ?? ?M , then the angle between b and c
is equal to : 
 (1) 
1
2
cos
3
?M ????
????
????
 
 (2) 
1
1
cos
3
?M????
?M????
????
 
 (3) 
1
3
cos
2
?M????
?M????
????
????
 
 (4) 
1
2
cos
3
?M ????
????
????
 
Ans. (3) 
 
 
Sol. Given a 1, b 4, a.b 2 ?] ?] ?] 
 
?H ?I c 2 a b 3b ?] ?? ?M 
 Dot product with a on both sides 
 c.a 6 ?]?M …..(1) 
 Dot product with b on both sides 
 b.c 48 ?]?M …..(2) 
 
2 2
c.c 4 a b 9 b ?] ?? ?K 
 
?H ?I 2 2
2 2 2
c 4 a b a.b 9 b
????
?] ?M ?K ????
????
 
 ?H ?I ?H ?I ?H ?I ?H ?I 2 2
c 4 1 4 4 9 16
????
?] ?M ?K ????
????
 
 
?{ ?} 2
c 4 12 144 ?]?K 
 
2
c 48 144 ?]?K 
 
2
c 192 ?] 
 
b.c
cos
b c
?| ?? ?] 
 
48
cos
192.4
?M ?| ?? ?] 
 
48
cos
8 3.4
?M ?| ?? ?] 
 
3
cos
2 3
?M ?| ?? ?] 
 
1
3 3
cos cos
2 2
?M????
?M?M
?| ?? ?] ?? ?? ?] ????
????
????
 
5. The maximum area of a triangle whose one vertex 
is at (0, 0) and the other two vertices lie on the 
curve y = -2x
2
 + 54 at points (x, y) and (-x, y) 
where y > 0 is : 
 (1) 88 
 (2) 122 
 (3) 92 
 (4) 108 
Ans. (4) 
Sol. 
(x, y)
(-x, y)
(0, 0)
 
 Area of ?d ?@ ?@ 0 0 1
1
x y 1
2
x y 1
?] ?M ?@ ?@ ?H ?I 1
xy xy xy
2
?? ?K ?] ?@ ?@ ?H ?I ?H ?I 2
Area xy x 2x 54 ?d ?] ?] ?M ?K ?@ ?@ ?H ?I ?H ?I 2
d
d
6x 54 0
dx dx
?d ?d ?] ?M ?K ?? ?] ?@ at x = 3  
 Area = 3 (-2 × 9 + 54) = 108  
6. The value of 
?H ?I ?H ?I 3 n
2 2 2
n
k 1
n
lim
n k n 3k
?R ????
?] ?K?K
?? is : 
 (1) 
?H ?I 2 3 3
24
?K??
 
 (2) 
?H ?I 13
8 4 3 3
?? ?K 
 (3) 
?H ?I 13 2 3 3
8
?M??
 
 (4) 
?H ?I 8 2 3 3
?? ?K 
Ans. (2) 
Sol.
 
3 n
2 2
n
k 1 4
2 2
n
lim
k 3k
n 1 1
n n
????
?] ?? ?? ?? ?? ?K?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
3 n
2 2
n
k 1
2 2
1 n
lim
n
k 3k
1 1
n n
????
?] ?] ?? ?? ?? ?? ?K?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
?H ?I 1
2 2
0
dx
1
3 1 x x
3
?] ????
?K?K
????
????
?? 
Page 3


 
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024 
(Held On Tuesday 30
th
 January, 2024)            TIME : 9 : 00 AM  to  12 : 00 NOON 
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
 
 
SECTION-A 
1. A line passing through the point A(9, 0) makes an angle 
of 30º with the positive direction of x-axis. If this line is 
rotated about A through an angle of 15º in the clockwise 
direction, then its equation in the new position is 
 (1) 9
3 2
?K?]
?M y
x (2) 9
3 2
?K?]
?M x
y 
 (3) 9
3 2
?K?]
?K x
y (4) 9
3 2
?K?]
?K y
x 
Ans. (1) 
Sol. 
 
Eq
n 
: y – 0 = tan15° (x – 9) ?? y = (2 ?M 3 ?I?@ (x ?M 9) 
2. Let S
a
 denote the sum of first n terms an arithmetic 
progression. If S
20
 = 790 and S
10
 = 145, then S
15
 – 
S
5
 is : 
 (1) 395 
 (2) 390 
 (3) 405 
 (4) 410 
Ans. (1) 
Sol.
 
?{ ?} 20
20
S 2a 19d 790
2
?] ?K ?] 
 2a + 19d = 79 …..(1) 
 
?{ ?} 10
10
S 2a 9d 145
2
?] ?K ?] 
 2a + 9d = 29  …..(2) 
 From (1) and (2) a = -8, d = 5 
 
?{ ?} ?{ ?} 15 5
15 5
S S 2a 14d 2a 4d
2 2
?M ?] ?K ?M ?K 
 
?{ ?} ?{ ?} 15 5
16 70 16 20
2 2
?] ?M ?K ?M ?M ?K 
 = 405-10 
 = 395  
3. If z = x + iy, xy ?? 0 , satisfies the equation 
2
z i z 0 ?K?] , then |z
2
| is equal to : 
 (1) 9 
 (2) 1 
 (3) 4 
 (4) 
1
4
 
Ans. (2) 
Sol.
 
2
2
z iz
z iz
?]?M
?] 
 |z
2
| = |z| 
 |z|
2 
– |z| = 0 
 |z|(|z| – 1) = 0 
 |z| = 0 (not acceptable) 
 ?| |z| = 1 
 ?| |z|
2 
= 1 
4. Let 
i 2 3
ˆ ˆ ˆ
a a i a j a k ?] ?K ?K and 
1 2 3
ˆ ˆ ˆ
b b i b j b k ?] ?K ?K be 
two vectors such that a 1; ?] a.b 2 ?] and b 4. ?] If 
?H ?I c 2 a b 3b ?] ?? ?M , then the angle between b and c
is equal to : 
 (1) 
1
2
cos
3
?M ????
????
????
 
 (2) 
1
1
cos
3
?M????
?M????
????
 
 (3) 
1
3
cos
2
?M????
?M????
????
????
 
 (4) 
1
2
cos
3
?M ????
????
????
 
Ans. (3) 
 
 
Sol. Given a 1, b 4, a.b 2 ?] ?] ?] 
 
?H ?I c 2 a b 3b ?] ?? ?M 
 Dot product with a on both sides 
 c.a 6 ?]?M …..(1) 
 Dot product with b on both sides 
 b.c 48 ?]?M …..(2) 
 
2 2
c.c 4 a b 9 b ?] ?? ?K 
 
?H ?I 2 2
2 2 2
c 4 a b a.b 9 b
????
?] ?M ?K ????
????
 
 ?H ?I ?H ?I ?H ?I ?H ?I 2 2
c 4 1 4 4 9 16
????
?] ?M ?K ????
????
 
 
?{ ?} 2
c 4 12 144 ?]?K 
 
2
c 48 144 ?]?K 
 
2
c 192 ?] 
 
b.c
cos
b c
?| ?? ?] 
 
48
cos
192.4
?M ?| ?? ?] 
 
48
cos
8 3.4
?M ?| ?? ?] 
 
3
cos
2 3
?M ?| ?? ?] 
 
1
3 3
cos cos
2 2
?M????
?M?M
?| ?? ?] ?? ?? ?] ????
????
????
 
5. The maximum area of a triangle whose one vertex 
is at (0, 0) and the other two vertices lie on the 
curve y = -2x
2
 + 54 at points (x, y) and (-x, y) 
where y > 0 is : 
 (1) 88 
 (2) 122 
 (3) 92 
 (4) 108 
Ans. (4) 
Sol. 
(x, y)
(-x, y)
(0, 0)
 
 Area of ?d ?@ ?@ 0 0 1
1
x y 1
2
x y 1
?] ?M ?@ ?@ ?H ?I 1
xy xy xy
2
?? ?K ?] ?@ ?@ ?H ?I ?H ?I 2
Area xy x 2x 54 ?d ?] ?] ?M ?K ?@ ?@ ?H ?I ?H ?I 2
d
d
6x 54 0
dx dx
?d ?d ?] ?M ?K ?? ?] ?@ at x = 3  
 Area = 3 (-2 × 9 + 54) = 108  
6. The value of 
?H ?I ?H ?I 3 n
2 2 2
n
k 1
n
lim
n k n 3k
?R ????
?] ?K?K
?? is : 
 (1) 
?H ?I 2 3 3
24
?K??
 
 (2) 
?H ?I 13
8 4 3 3
?? ?K 
 (3) 
?H ?I 13 2 3 3
8
?M??
 
 (4) 
?H ?I 8 2 3 3
?? ?K 
Ans. (2) 
Sol.
 
3 n
2 2
n
k 1 4
2 2
n
lim
k 3k
n 1 1
n n
????
?] ?? ?? ?? ?? ?K?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
3 n
2 2
n
k 1
2 2
1 n
lim
n
k 3k
1 1
n n
????
?] ?] ?? ?? ?? ?? ?K?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
?H ?I 1
2 2
0
dx
1
3 1 x x
3
?] ????
?K?K
????
????
?? 
 
 
?H ?I ?H ?I 2 2
1
2 2
0
1
x 1 x
1 3 3
dx
1 3 2
1 x x
3
????
?K ?M ?K????
????
?]??
????
?K?K
????
????
?? 
 
1
2 2
2 0
1 1 1
dx
2
1 x
1
x
3
????
????
????
?]?M
????
?K ????
????
?K????
????
????????
?? 
 
?H ?I ?H ?I 1 1
1 1
0 0
1 1
3 tan 3x tan x
2 2
?M?M
????
?]?M
????
 
 
3 1
2 3 2 4 8 2 3
?? ?? ?? ?? ?? ?? ?? ?? ?] ?M ?] ?M ?? ?? ?? ?? ?? ?? ?? ?? 
 ?H ?I 13
8. 4 3 3
?? ?] ?K 
7. Let g : R R ?? be a non constant twice 
differentiable such that 
1 3
g' g'
2 2
?? ?? ?? ?? ?] ?? ?? ?? ?? ?? ?? ?? ?? . If a real 
valued function f is defined as 
?H ?I ?H ?I ?H ?I 1
f x g x g 2 x
2
?] ?? ?K ?M ?? ????
, then 
 (1) f”(x) = 0 for atleast two x in (0, 2) 
 (2) f”(x) = 0 for exactly one x in (0, 1) 
 (3) f”(x) = 0 for no x in (0, 1) 
 (4) 
3 1
f ' f ' 1
2 2
?? ?? ?? ?? ?K?]
?? ?? ?? ?? ?? ?? ?? ?? 
Ans. (1) 
Sol. ?H ?I ?H ?I ?H ?I 3 1
g' g'
g' x g' 2 x 3 2 2
f ' x ,f ' 0
2 2 2
?? ?? ?? ?? ?M ?? ?? ?? ?? ?M?M
????
?? ?? ?? ?? ?] ?] ?] ????
????
 
 Also 
1 3
g' g'
1 1 2 2
f ' 0, f ' 0
2 2 2
?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?] ?] ?] ?? ?? ?? ?? ?? ?? ?? ?? 
 
3 1
f ' f ' 0
2 2
?? ?? ?? ?? ?? ?] ?] ?? ?? ?? ?? ?? ?? ?? ?? 
 
1 3
rootsin ,1 and 1,
2 2
?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 ?H ?I f " x ?? is zero at least twice in 
1 3
,
2 2
????
????
????
 
8. The area (in square units) of the region bounded by 
the parabola y
2
 = 4(x – 2) and the line y = 2x - 8 
 (1) 8 
 (2) 9 
 (3) 6 
 (4) 7 
Ans. (2) 
Sol. Let X = x – 2 
 y
2
 = 4x, y = 2 (x + 2) – 8 
 y
2
 = 4x, y = 2x – 4 
 
4
2
2
y y 4
A
4 2
?M ?K ?]?M
?? 
  
-2
4
 
 = 9 
 
9. Let y = y (x) be the solution of the differential 
equation sec x dy + {2(1 – x) tan x + x(2 – x)}  
dx = 0 such that y(0) = 2.Then y(2) is equal to : 
 (1) 2 
 (2) 2{1 – sin (2)} 
 (3) 2{sin (2) + 1} 
 (4) 1 
Ans. (1) 
Sol. ?H ?I ?H ?I 2
dy
2 x 1 sin x x 2x cos x
dx
?] ?M ?K ?M 
 Now both side integrate  
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 2
y x 2 x 1 sin xdx x 2x sin x 2x 2 sin x dx
????
?] ?M ?K ?M ?M ?M????
????
????
 
 ?H ?I ?H ?I 2
y x x 2x sin x ?] ?M ?K ?? 
 ?H ?I y 0 0 2 ?] ?K ?? ?? ?] ?? 
 ?H ?I ?H ?I 2
y x x 2x sin x 2 ?] ?M ?K 
 ?H ?I y 2 2 ?] 
Page 4


 
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024 
(Held On Tuesday 30
th
 January, 2024)            TIME : 9 : 00 AM  to  12 : 00 NOON 
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
 
 
SECTION-A 
1. A line passing through the point A(9, 0) makes an angle 
of 30º with the positive direction of x-axis. If this line is 
rotated about A through an angle of 15º in the clockwise 
direction, then its equation in the new position is 
 (1) 9
3 2
?K?]
?M y
x (2) 9
3 2
?K?]
?M x
y 
 (3) 9
3 2
?K?]
?K x
y (4) 9
3 2
?K?]
?K y
x 
Ans. (1) 
Sol. 
 
Eq
n 
: y – 0 = tan15° (x – 9) ?? y = (2 ?M 3 ?I?@ (x ?M 9) 
2. Let S
a
 denote the sum of first n terms an arithmetic 
progression. If S
20
 = 790 and S
10
 = 145, then S
15
 – 
S
5
 is : 
 (1) 395 
 (2) 390 
 (3) 405 
 (4) 410 
Ans. (1) 
Sol.
 
?{ ?} 20
20
S 2a 19d 790
2
?] ?K ?] 
 2a + 19d = 79 …..(1) 
 
?{ ?} 10
10
S 2a 9d 145
2
?] ?K ?] 
 2a + 9d = 29  …..(2) 
 From (1) and (2) a = -8, d = 5 
 
?{ ?} ?{ ?} 15 5
15 5
S S 2a 14d 2a 4d
2 2
?M ?] ?K ?M ?K 
 
?{ ?} ?{ ?} 15 5
16 70 16 20
2 2
?] ?M ?K ?M ?M ?K 
 = 405-10 
 = 395  
3. If z = x + iy, xy ?? 0 , satisfies the equation 
2
z i z 0 ?K?] , then |z
2
| is equal to : 
 (1) 9 
 (2) 1 
 (3) 4 
 (4) 
1
4
 
Ans. (2) 
Sol.
 
2
2
z iz
z iz
?]?M
?] 
 |z
2
| = |z| 
 |z|
2 
– |z| = 0 
 |z|(|z| – 1) = 0 
 |z| = 0 (not acceptable) 
 ?| |z| = 1 
 ?| |z|
2 
= 1 
4. Let 
i 2 3
ˆ ˆ ˆ
a a i a j a k ?] ?K ?K and 
1 2 3
ˆ ˆ ˆ
b b i b j b k ?] ?K ?K be 
two vectors such that a 1; ?] a.b 2 ?] and b 4. ?] If 
?H ?I c 2 a b 3b ?] ?? ?M , then the angle between b and c
is equal to : 
 (1) 
1
2
cos
3
?M ????
????
????
 
 (2) 
1
1
cos
3
?M????
?M????
????
 
 (3) 
1
3
cos
2
?M????
?M????
????
????
 
 (4) 
1
2
cos
3
?M ????
????
????
 
Ans. (3) 
 
 
Sol. Given a 1, b 4, a.b 2 ?] ?] ?] 
 
?H ?I c 2 a b 3b ?] ?? ?M 
 Dot product with a on both sides 
 c.a 6 ?]?M …..(1) 
 Dot product with b on both sides 
 b.c 48 ?]?M …..(2) 
 
2 2
c.c 4 a b 9 b ?] ?? ?K 
 
?H ?I 2 2
2 2 2
c 4 a b a.b 9 b
????
?] ?M ?K ????
????
 
 ?H ?I ?H ?I ?H ?I ?H ?I 2 2
c 4 1 4 4 9 16
????
?] ?M ?K ????
????
 
 
?{ ?} 2
c 4 12 144 ?]?K 
 
2
c 48 144 ?]?K 
 
2
c 192 ?] 
 
b.c
cos
b c
?| ?? ?] 
 
48
cos
192.4
?M ?| ?? ?] 
 
48
cos
8 3.4
?M ?| ?? ?] 
 
3
cos
2 3
?M ?| ?? ?] 
 
1
3 3
cos cos
2 2
?M????
?M?M
?| ?? ?] ?? ?? ?] ????
????
????
 
5. The maximum area of a triangle whose one vertex 
is at (0, 0) and the other two vertices lie on the 
curve y = -2x
2
 + 54 at points (x, y) and (-x, y) 
where y > 0 is : 
 (1) 88 
 (2) 122 
 (3) 92 
 (4) 108 
Ans. (4) 
Sol. 
(x, y)
(-x, y)
(0, 0)
 
 Area of ?d ?@ ?@ 0 0 1
1
x y 1
2
x y 1
?] ?M ?@ ?@ ?H ?I 1
xy xy xy
2
?? ?K ?] ?@ ?@ ?H ?I ?H ?I 2
Area xy x 2x 54 ?d ?] ?] ?M ?K ?@ ?@ ?H ?I ?H ?I 2
d
d
6x 54 0
dx dx
?d ?d ?] ?M ?K ?? ?] ?@ at x = 3  
 Area = 3 (-2 × 9 + 54) = 108  
6. The value of 
?H ?I ?H ?I 3 n
2 2 2
n
k 1
n
lim
n k n 3k
?R ????
?] ?K?K
?? is : 
 (1) 
?H ?I 2 3 3
24
?K??
 
 (2) 
?H ?I 13
8 4 3 3
?? ?K 
 (3) 
?H ?I 13 2 3 3
8
?M??
 
 (4) 
?H ?I 8 2 3 3
?? ?K 
Ans. (2) 
Sol.
 
3 n
2 2
n
k 1 4
2 2
n
lim
k 3k
n 1 1
n n
????
?] ?? ?? ?? ?? ?K?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
3 n
2 2
n
k 1
2 2
1 n
lim
n
k 3k
1 1
n n
????
?] ?] ?? ?? ?? ?? ?K?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
?H ?I 1
2 2
0
dx
1
3 1 x x
3
?] ????
?K?K
????
????
?? 
 
 
?H ?I ?H ?I 2 2
1
2 2
0
1
x 1 x
1 3 3
dx
1 3 2
1 x x
3
????
?K ?M ?K????
????
?]??
????
?K?K
????
????
?? 
 
1
2 2
2 0
1 1 1
dx
2
1 x
1
x
3
????
????
????
?]?M
????
?K ????
????
?K????
????
????????
?? 
 
?H ?I ?H ?I 1 1
1 1
0 0
1 1
3 tan 3x tan x
2 2
?M?M
????
?]?M
????
 
 
3 1
2 3 2 4 8 2 3
?? ?? ?? ?? ?? ?? ?? ?? ?] ?M ?] ?M ?? ?? ?? ?? ?? ?? ?? ?? 
 ?H ?I 13
8. 4 3 3
?? ?] ?K 
7. Let g : R R ?? be a non constant twice 
differentiable such that 
1 3
g' g'
2 2
?? ?? ?? ?? ?] ?? ?? ?? ?? ?? ?? ?? ?? . If a real 
valued function f is defined as 
?H ?I ?H ?I ?H ?I 1
f x g x g 2 x
2
?] ?? ?K ?M ?? ????
, then 
 (1) f”(x) = 0 for atleast two x in (0, 2) 
 (2) f”(x) = 0 for exactly one x in (0, 1) 
 (3) f”(x) = 0 for no x in (0, 1) 
 (4) 
3 1
f ' f ' 1
2 2
?? ?? ?? ?? ?K?]
?? ?? ?? ?? ?? ?? ?? ?? 
Ans. (1) 
Sol. ?H ?I ?H ?I ?H ?I 3 1
g' g'
g' x g' 2 x 3 2 2
f ' x ,f ' 0
2 2 2
?? ?? ?? ?? ?M ?? ?? ?? ?? ?M?M
????
?? ?? ?? ?? ?] ?] ?] ????
????
 
 Also 
1 3
g' g'
1 1 2 2
f ' 0, f ' 0
2 2 2
?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?] ?] ?] ?? ?? ?? ?? ?? ?? ?? ?? 
 
3 1
f ' f ' 0
2 2
?? ?? ?? ?? ?? ?] ?] ?? ?? ?? ?? ?? ?? ?? ?? 
 
1 3
rootsin ,1 and 1,
2 2
?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 ?H ?I f " x ?? is zero at least twice in 
1 3
,
2 2
????
????
????
 
8. The area (in square units) of the region bounded by 
the parabola y
2
 = 4(x – 2) and the line y = 2x - 8 
 (1) 8 
 (2) 9 
 (3) 6 
 (4) 7 
Ans. (2) 
Sol. Let X = x – 2 
 y
2
 = 4x, y = 2 (x + 2) – 8 
 y
2
 = 4x, y = 2x – 4 
 
4
2
2
y y 4
A
4 2
?M ?K ?]?M
?? 
  
-2
4
 
 = 9 
 
9. Let y = y (x) be the solution of the differential 
equation sec x dy + {2(1 – x) tan x + x(2 – x)}  
dx = 0 such that y(0) = 2.Then y(2) is equal to : 
 (1) 2 
 (2) 2{1 – sin (2)} 
 (3) 2{sin (2) + 1} 
 (4) 1 
Ans. (1) 
Sol. ?H ?I ?H ?I 2
dy
2 x 1 sin x x 2x cos x
dx
?] ?M ?K ?M 
 Now both side integrate  
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 2
y x 2 x 1 sin xdx x 2x sin x 2x 2 sin x dx
????
?] ?M ?K ?M ?M ?M????
????
????
 
 ?H ?I ?H ?I 2
y x x 2x sin x ?] ?M ?K ?? 
 ?H ?I y 0 0 2 ?] ?K ?? ?? ?] ?? 
 ?H ?I ?H ?I 2
y x x 2x sin x 2 ?] ?M ?K 
 ?H ?I y 2 2 ?] 
?? ?? be the foot of perpendicular from the 
point (1, 2, 3) on the line 
x 3 y 1 z 4
5 2 3
?K ?M ?K ?]?] . 
then ?H ?I 19 ?? ?K ?? ?K ?? is equal to : 
 (1) 102 
 (2) 101 
 (3) 99 
 (4) 100 
Ans. (2) 
Sol. 
(1, 2, 3)
P( ) ?? ?L?@?? ?L?@?? 
Let foot P (5k – 3, 2k + 1, 3k – 4) ?@ DR's AP: 5k 4, 2k 1 , 3k 7 ?? ?M ?M ?M ?@ DR's Line: 5, 2, 3 ?? ?@ Condition of perpendicular lines (25k-20) + (4k-2) + (9k – 21)=0 
Then 
43
k
38
?] ?@ Then ?H ?I 19 ?? ?K ?? ?K ?? =101 
11. Two integers x and y are chosen with replacement 
from the set {0, 1, 2, 3, ….., 10}. Then the 
probability that | x y| 5 ?M?^ is :   
 (1) 
30
121
 
 (2) 
62
121
 
 (3) 
60
121
 
 (4) 
31
121
 
Ans. (1) 
Sol. If x = 0, y = 6, 7, 8, 9, 10 
 If x = 1, y = 7, 8, 9, 10 
 If x = 2, y = 8, 9, 10 
 If x = 3, y = 9, 10 
 If x = 4, y = 10 
 If x = 5, y = no possible value 
        Total possible ways = (5 + 4 + 3 + 2 + 1) × 2 
 = 30 
 Required probability ?] 30 30
11 11 121
?] ?? ?@ 12. If the domain of the function 
?H ?I ?H ?I ?H ?I 1
1
e
2 x
f x cos log 3 x
4
?M ?M???? ?M ?] ?K ?M ????
????
 is 
?? ?? [ , ) y ?M?? ?? ?M , then ?? ?K ?? ?K ?? is equal to : 
 (1) 12 
 (2) 9 
 (3) 11 
 (4) 8 
Ans. (3) 
Sol.
 
2 x
1 1
4
?M ?M ?? ?? 
 
2 x
1
4
?M ???? 
 –4 ?? 2 – |x| ?? 4 
 –6 ?? – |x| ?? 2 
 –2 ?? |x| ?? 6 
 |x| ?? 6 
 ??  x ?? [–6, 6] …(1) 
 Now, 3 – x ?? 1 
 And x ?? 2  …(2) 
 and 3 – x > 0 
 x < 3  …(3) 
 From (1), (2) and (3) 
 ??  x ?? [–6, 3) – {2} 
 ?? = 6 
 ?? = 3 
 ?? = 2 
 ?? + ?? + ?? = 11 
13. Consider the system of linear equation x + y + z = 
4 ?? , x + 2y + 2 z ?? = 10 ?? , x + 3y + 4 ?? 2
z = ?? 2
 +15, 
where ?? , R ???? . Which one of the following 
statements is NOT correct ? 
 (1) The system has unique solution if 
1
2
???? and  
1 ???? , 15 
Page 5


 
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024 
(Held On Tuesday 30
th
 January, 2024)            TIME : 9 : 00 AM  to  12 : 00 NOON 
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
 
 
SECTION-A 
1. A line passing through the point A(9, 0) makes an angle 
of 30º with the positive direction of x-axis. If this line is 
rotated about A through an angle of 15º in the clockwise 
direction, then its equation in the new position is 
 (1) 9
3 2
?K?]
?M y
x (2) 9
3 2
?K?]
?M x
y 
 (3) 9
3 2
?K?]
?K x
y (4) 9
3 2
?K?]
?K y
x 
Ans. (1) 
Sol. 
 
Eq
n 
: y – 0 = tan15° (x – 9) ?? y = (2 ?M 3 ?I?@ (x ?M 9) 
2. Let S
a
 denote the sum of first n terms an arithmetic 
progression. If S
20
 = 790 and S
10
 = 145, then S
15
 – 
S
5
 is : 
 (1) 395 
 (2) 390 
 (3) 405 
 (4) 410 
Ans. (1) 
Sol.
 
?{ ?} 20
20
S 2a 19d 790
2
?] ?K ?] 
 2a + 19d = 79 …..(1) 
 
?{ ?} 10
10
S 2a 9d 145
2
?] ?K ?] 
 2a + 9d = 29  …..(2) 
 From (1) and (2) a = -8, d = 5 
 
?{ ?} ?{ ?} 15 5
15 5
S S 2a 14d 2a 4d
2 2
?M ?] ?K ?M ?K 
 
?{ ?} ?{ ?} 15 5
16 70 16 20
2 2
?] ?M ?K ?M ?M ?K 
 = 405-10 
 = 395  
3. If z = x + iy, xy ?? 0 , satisfies the equation 
2
z i z 0 ?K?] , then |z
2
| is equal to : 
 (1) 9 
 (2) 1 
 (3) 4 
 (4) 
1
4
 
Ans. (2) 
Sol.
 
2
2
z iz
z iz
?]?M
?] 
 |z
2
| = |z| 
 |z|
2 
– |z| = 0 
 |z|(|z| – 1) = 0 
 |z| = 0 (not acceptable) 
 ?| |z| = 1 
 ?| |z|
2 
= 1 
4. Let 
i 2 3
ˆ ˆ ˆ
a a i a j a k ?] ?K ?K and 
1 2 3
ˆ ˆ ˆ
b b i b j b k ?] ?K ?K be 
two vectors such that a 1; ?] a.b 2 ?] and b 4. ?] If 
?H ?I c 2 a b 3b ?] ?? ?M , then the angle between b and c
is equal to : 
 (1) 
1
2
cos
3
?M ????
????
????
 
 (2) 
1
1
cos
3
?M????
?M????
????
 
 (3) 
1
3
cos
2
?M????
?M????
????
????
 
 (4) 
1
2
cos
3
?M ????
????
????
 
Ans. (3) 
 
 
Sol. Given a 1, b 4, a.b 2 ?] ?] ?] 
 
?H ?I c 2 a b 3b ?] ?? ?M 
 Dot product with a on both sides 
 c.a 6 ?]?M …..(1) 
 Dot product with b on both sides 
 b.c 48 ?]?M …..(2) 
 
2 2
c.c 4 a b 9 b ?] ?? ?K 
 
?H ?I 2 2
2 2 2
c 4 a b a.b 9 b
????
?] ?M ?K ????
????
 
 ?H ?I ?H ?I ?H ?I ?H ?I 2 2
c 4 1 4 4 9 16
????
?] ?M ?K ????
????
 
 
?{ ?} 2
c 4 12 144 ?]?K 
 
2
c 48 144 ?]?K 
 
2
c 192 ?] 
 
b.c
cos
b c
?| ?? ?] 
 
48
cos
192.4
?M ?| ?? ?] 
 
48
cos
8 3.4
?M ?| ?? ?] 
 
3
cos
2 3
?M ?| ?? ?] 
 
1
3 3
cos cos
2 2
?M????
?M?M
?| ?? ?] ?? ?? ?] ????
????
????
 
5. The maximum area of a triangle whose one vertex 
is at (0, 0) and the other two vertices lie on the 
curve y = -2x
2
 + 54 at points (x, y) and (-x, y) 
where y > 0 is : 
 (1) 88 
 (2) 122 
 (3) 92 
 (4) 108 
Ans. (4) 
Sol. 
(x, y)
(-x, y)
(0, 0)
 
 Area of ?d ?@ ?@ 0 0 1
1
x y 1
2
x y 1
?] ?M ?@ ?@ ?H ?I 1
xy xy xy
2
?? ?K ?] ?@ ?@ ?H ?I ?H ?I 2
Area xy x 2x 54 ?d ?] ?] ?M ?K ?@ ?@ ?H ?I ?H ?I 2
d
d
6x 54 0
dx dx
?d ?d ?] ?M ?K ?? ?] ?@ at x = 3  
 Area = 3 (-2 × 9 + 54) = 108  
6. The value of 
?H ?I ?H ?I 3 n
2 2 2
n
k 1
n
lim
n k n 3k
?R ????
?] ?K?K
?? is : 
 (1) 
?H ?I 2 3 3
24
?K??
 
 (2) 
?H ?I 13
8 4 3 3
?? ?K 
 (3) 
?H ?I 13 2 3 3
8
?M??
 
 (4) 
?H ?I 8 2 3 3
?? ?K 
Ans. (2) 
Sol.
 
3 n
2 2
n
k 1 4
2 2
n
lim
k 3k
n 1 1
n n
????
?] ?? ?? ?? ?? ?K?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
3 n
2 2
n
k 1
2 2
1 n
lim
n
k 3k
1 1
n n
????
?] ?] ?? ?? ?? ?? ?K?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 
?H ?I 1
2 2
0
dx
1
3 1 x x
3
?] ????
?K?K
????
????
?? 
 
 
?H ?I ?H ?I 2 2
1
2 2
0
1
x 1 x
1 3 3
dx
1 3 2
1 x x
3
????
?K ?M ?K????
????
?]??
????
?K?K
????
????
?? 
 
1
2 2
2 0
1 1 1
dx
2
1 x
1
x
3
????
????
????
?]?M
????
?K ????
????
?K????
????
????????
?? 
 
?H ?I ?H ?I 1 1
1 1
0 0
1 1
3 tan 3x tan x
2 2
?M?M
????
?]?M
????
 
 
3 1
2 3 2 4 8 2 3
?? ?? ?? ?? ?? ?? ?? ?? ?] ?M ?] ?M ?? ?? ?? ?? ?? ?? ?? ?? 
 ?H ?I 13
8. 4 3 3
?? ?] ?K 
7. Let g : R R ?? be a non constant twice 
differentiable such that 
1 3
g' g'
2 2
?? ?? ?? ?? ?] ?? ?? ?? ?? ?? ?? ?? ?? . If a real 
valued function f is defined as 
?H ?I ?H ?I ?H ?I 1
f x g x g 2 x
2
?] ?? ?K ?M ?? ????
, then 
 (1) f”(x) = 0 for atleast two x in (0, 2) 
 (2) f”(x) = 0 for exactly one x in (0, 1) 
 (3) f”(x) = 0 for no x in (0, 1) 
 (4) 
3 1
f ' f ' 1
2 2
?? ?? ?? ?? ?K?]
?? ?? ?? ?? ?? ?? ?? ?? 
Ans. (1) 
Sol. ?H ?I ?H ?I ?H ?I 3 1
g' g'
g' x g' 2 x 3 2 2
f ' x ,f ' 0
2 2 2
?? ?? ?? ?? ?M ?? ?? ?? ?? ?M?M
????
?? ?? ?? ?? ?] ?] ?] ????
????
 
 Also 
1 3
g' g'
1 1 2 2
f ' 0, f ' 0
2 2 2
?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?] ?] ?] ?? ?? ?? ?? ?? ?? ?? ?? 
 
3 1
f ' f ' 0
2 2
?? ?? ?? ?? ?? ?] ?] ?? ?? ?? ?? ?? ?? ?? ?? 
 
1 3
rootsin ,1 and 1,
2 2
?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 ?H ?I f " x ?? is zero at least twice in 
1 3
,
2 2
????
????
????
 
8. The area (in square units) of the region bounded by 
the parabola y
2
 = 4(x – 2) and the line y = 2x - 8 
 (1) 8 
 (2) 9 
 (3) 6 
 (4) 7 
Ans. (2) 
Sol. Let X = x – 2 
 y
2
 = 4x, y = 2 (x + 2) – 8 
 y
2
 = 4x, y = 2x – 4 
 
4
2
2
y y 4
A
4 2
?M ?K ?]?M
?? 
  
-2
4
 
 = 9 
 
9. Let y = y (x) be the solution of the differential 
equation sec x dy + {2(1 – x) tan x + x(2 – x)}  
dx = 0 such that y(0) = 2.Then y(2) is equal to : 
 (1) 2 
 (2) 2{1 – sin (2)} 
 (3) 2{sin (2) + 1} 
 (4) 1 
Ans. (1) 
Sol. ?H ?I ?H ?I 2
dy
2 x 1 sin x x 2x cos x
dx
?] ?M ?K ?M 
 Now both side integrate  
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 2
y x 2 x 1 sin xdx x 2x sin x 2x 2 sin x dx
????
?] ?M ?K ?M ?M ?M????
????
????
 
 ?H ?I ?H ?I 2
y x x 2x sin x ?] ?M ?K ?? 
 ?H ?I y 0 0 2 ?] ?K ?? ?? ?] ?? 
 ?H ?I ?H ?I 2
y x x 2x sin x 2 ?] ?M ?K 
 ?H ?I y 2 2 ?] 
?? ?? be the foot of perpendicular from the 
point (1, 2, 3) on the line 
x 3 y 1 z 4
5 2 3
?K ?M ?K ?]?] . 
then ?H ?I 19 ?? ?K ?? ?K ?? is equal to : 
 (1) 102 
 (2) 101 
 (3) 99 
 (4) 100 
Ans. (2) 
Sol. 
(1, 2, 3)
P( ) ?? ?L?@?? ?L?@?? 
Let foot P (5k – 3, 2k + 1, 3k – 4) ?@ DR's AP: 5k 4, 2k 1 , 3k 7 ?? ?M ?M ?M ?@ DR's Line: 5, 2, 3 ?? ?@ Condition of perpendicular lines (25k-20) + (4k-2) + (9k – 21)=0 
Then 
43
k
38
?] ?@ Then ?H ?I 19 ?? ?K ?? ?K ?? =101 
11. Two integers x and y are chosen with replacement 
from the set {0, 1, 2, 3, ….., 10}. Then the 
probability that | x y| 5 ?M?^ is :   
 (1) 
30
121
 
 (2) 
62
121
 
 (3) 
60
121
 
 (4) 
31
121
 
Ans. (1) 
Sol. If x = 0, y = 6, 7, 8, 9, 10 
 If x = 1, y = 7, 8, 9, 10 
 If x = 2, y = 8, 9, 10 
 If x = 3, y = 9, 10 
 If x = 4, y = 10 
 If x = 5, y = no possible value 
        Total possible ways = (5 + 4 + 3 + 2 + 1) × 2 
 = 30 
 Required probability ?] 30 30
11 11 121
?] ?? ?@ 12. If the domain of the function 
?H ?I ?H ?I ?H ?I 1
1
e
2 x
f x cos log 3 x
4
?M ?M???? ?M ?] ?K ?M ????
????
 is 
?? ?? [ , ) y ?M?? ?? ?M , then ?? ?K ?? ?K ?? is equal to : 
 (1) 12 
 (2) 9 
 (3) 11 
 (4) 8 
Ans. (3) 
Sol.
 
2 x
1 1
4
?M ?M ?? ?? 
 
2 x
1
4
?M ???? 
 –4 ?? 2 – |x| ?? 4 
 –6 ?? – |x| ?? 2 
 –2 ?? |x| ?? 6 
 |x| ?? 6 
 ??  x ?? [–6, 6] …(1) 
 Now, 3 – x ?? 1 
 And x ?? 2  …(2) 
 and 3 – x > 0 
 x < 3  …(3) 
 From (1), (2) and (3) 
 ??  x ?? [–6, 3) – {2} 
 ?? = 6 
 ?? = 3 
 ?? = 2 
 ?? + ?? + ?? = 11 
13. Consider the system of linear equation x + y + z = 
4 ?? , x + 2y + 2 z ?? = 10 ?? , x + 3y + 4 ?? 2
z = ?? 2
 +15, 
where ?? , R ???? . Which one of the following 
statements is NOT correct ? 
 (1) The system has unique solution if 
1
2
???? and  
1 ???? , 15 
 
 
 
 (2) The system is inconsistent if 
1
2
???] and 1 ???? 
 (3) The system has infinite number of solutions if 
1
2
???] and 15 ???] 
 (4) The system is consistent if 
1
2
???? 
Ans. (2) 
Sol. x + y + z = 4 ?? , x + 2y + 2 z ?? = 10 ?? , x + 3y + 4 ?? 2
z = ?? 2
 +15, 
 
2
1 1 1
1 2 2
1 3 4
?d ?] ?? ?? = ?H ?I 2
2 1 ???M ?@ ?@ For unique solution 
1
0, 2 1 0,
2
????
?d ?? ?? ?M ?? ?? ?? ????
????
 
 Let 
1
0,
2
?d ?] ?? ?] 
 
y x z
2
4 1 1
0, 10 2 1
15 3 1
?? ?d ?] ?d ?] ?d ?] ?? ???K
 
 ?H ?I ?H ?I 15 1 ?] ?? ?M ?? ?M 
 For infinite solution 
1
, 1 or 15
2
?? ?] ?? ?] 
  
14. If the circles ?H ?I ?H ?I 2 2 2
x 1 y 2 r ?K ?K ?K ?] and 
2 2
x y 4x 4y 4 0 ?K ?M ?M ?K ?] intersect at exactly two 
distinct points, then 
 (1) 5 < r < 9 
 (2) 0 < r < 7 
 (3) 3 < r < 7 
 (4) 
1
r 7
2
?\?\ 
Ans. (3) 
Sol. If two circles intersect at two distinct points 
 
1 2 1 2 1 2
r r C C r r ?? ?M ?\ ?\ ?K 
 r 2 9 16 r 2 ?M ?\ ?K ?\ ?K 
 |r – 2 | < 5 and r + 2 > 5 
 -5 < r – 2 < 5 r > 3 ……….(2) 
 -3 < r < 7  ………(1) 
 From (1) and (2) 
 3 < r < 7  
15. If the length of the minor axis of ellipse is equal to 
half of the distance between the foci, then the 
eccentricity of the ellipse is : 
 (1) 
5
3
 
 (2) 
3
2
 
 (3) 
1
3
 
 (4) 
2
5
 
Ans. (4) 
Sol. 2b = ae 
 
b e
a 2
?] 
 
2
e
e 1
4
?]?M 
 
2
e
5
?] 
16. Let M denote the median of the following 
frequency distribution. 
Class 0-4 4-8 8-12 12-16 16-20 
Frequency 3 9 10 8 6 
 Then 20 M is equal to : 
 (1) 416 
 (2) 104 
 (3) 52 
 (4) 208 
Ans. (4) 
Sol. 
Class Frequency Cumulative 
frequency 
0-4 3 3 
4-8 9 12 
8-12 10 22 
12-16 8 30 
16-20 6 36