JEE Main 2024 April 6 Shift 1 Paper & Solutions | JEE Main & Advanced Mock Test Series 2025 PDF Download

 Page 1


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Saturday 06
th
 April, 2024)   TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If 
3
1
x sin , x 0
f(x) = ,then
x
0 , x 0
?? ????
?? ?? ????
???? ?? ?? ?] ?? (1) f ''(0) 1 ?] (2) 
2
2 24 –
f ''
2
?? ????
?] ????
???? ????
(3) 
2
2 12 –
f ''
2
?? ????
?] ????
???? ????
(4) f ''(0) 0 ?] Ans. (2) 
Sol. f'(x) = 
2
1 1
3x sin xcos
x x
?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? ?? ?? 1
sin
1 1 1 x
f "(x) 6xsin 3cos cos
x x x x
????
????
?? ?? ?? ?? ?? ?? ????
?] ?M ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 2
2 12 24
f "
2 2
?? ?M ?? ????
?] ?M ?] ????
?? ?? ?? ????
2. If A(3,1,–1), 
5 7 1
B , ,
3 3 3
????
????
????
, C(2,2,1) and 
10 2 –1
D , ,
3 3 3
????
????
????
 are the vertices of a quadrilateral 
ABCD, then its area is 
(1)
4 2
3
(2) 
5 2
3
(3) 2 2 (4) 
2 2
3
Ans. (1) 
Sol. 
D
 
C
 
B
 
A
 
Area = 
1
BD AC
2
?? 5 5 2
ˆˆ ˆ
BD i j k
3 3 3
?] ?M ?M ˆˆ ˆ
AC i j 2k ?] ?M ?M 3. 
?H ?I /4 2 2
2
3 3
0
cos xsin x
dx
cos x sin x
?? ?K ?? is equal to 
(1) 1/12 (2) 1/9
(3) 1/6 (4) 1/3
Ans. (3) 
Sol. Divide Nr & Dr by cosx 
?H ?I /4 2 2
2
3
0
tan xsec xdx
dx
1 tan x
?? ?K ?? Let 1 + tan
3
x = t 
tan
2
x sec
2
x dx = 
dt
3
2
2
1
1 dt 1
3 6 t
?] ?? 4. The mean and standard deviation of 20 observations
are found to be 10 and 2, respectively. On
respectively, it was found that an observation by
mistake was taken 8 instead of 12. The correct
standard deviation is
(1) 3.86 (2) 1.8
(3) 3.96 (4) 1.94
Ans. (3) 
Sol. Mean (x) 10 ?] 
i
x
10
20
?s ???]
?s x
i
 = 10×20 = 200
If 8 is replaced by 12, then ?s x
i
 = 200 – 8 + 12 = 204 
Page 2


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Saturday 06
th
 April, 2024)   TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If 
3
1
x sin , x 0
f(x) = ,then
x
0 , x 0
?? ????
?? ?? ????
???? ?? ?? ?] ?? (1) f ''(0) 1 ?] (2) 
2
2 24 –
f ''
2
?? ????
?] ????
???? ????
(3) 
2
2 12 –
f ''
2
?? ????
?] ????
???? ????
(4) f ''(0) 0 ?] Ans. (2) 
Sol. f'(x) = 
2
1 1
3x sin xcos
x x
?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? ?? ?? 1
sin
1 1 1 x
f "(x) 6xsin 3cos cos
x x x x
????
????
?? ?? ?? ?? ?? ?? ????
?] ?M ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 2
2 12 24
f "
2 2
?? ?M ?? ????
?] ?M ?] ????
?? ?? ?? ????
2. If A(3,1,–1), 
5 7 1
B , ,
3 3 3
????
????
????
, C(2,2,1) and 
10 2 –1
D , ,
3 3 3
????
????
????
 are the vertices of a quadrilateral 
ABCD, then its area is 
(1)
4 2
3
(2) 
5 2
3
(3) 2 2 (4) 
2 2
3
Ans. (1) 
Sol. 
D
 
C
 
B
 
A
 
Area = 
1
BD AC
2
?? 5 5 2
ˆˆ ˆ
BD i j k
3 3 3
?] ?M ?M ˆˆ ˆ
AC i j 2k ?] ?M ?M 3. 
?H ?I /4 2 2
2
3 3
0
cos xsin x
dx
cos x sin x
?? ?K ?? is equal to 
(1) 1/12 (2) 1/9
(3) 1/6 (4) 1/3
Ans. (3) 
Sol. Divide Nr & Dr by cosx 
?H ?I /4 2 2
2
3
0
tan xsec xdx
dx
1 tan x
?? ?K ?? Let 1 + tan
3
x = t 
tan
2
x sec
2
x dx = 
dt
3
2
2
1
1 dt 1
3 6 t
?] ?? 4. The mean and standard deviation of 20 observations
are found to be 10 and 2, respectively. On
respectively, it was found that an observation by
mistake was taken 8 instead of 12. The correct
standard deviation is
(1) 3.86 (2) 1.8
(3) 3.96 (4) 1.94
Ans. (3) 
Sol. Mean (x) 10 ?] 
i
x
10
20
?s ???]
?s x
i
 = 10×20 = 200
If 8 is replaced by 12, then ?s x
i
 = 200 – 8 + 12 = 204 
 ?| Correct mean 
i
x
(x)
20
?s ?] 
 
204
10.2
20
?]?] 
 ?q Standard deviation = 2 
 ?| Variance = (S.D.)
2
 = 2
2
 = 4 
 ?? 
2
2
i i
x x
4
20 20
????
?s?s
?M?]
????
????
????
 
 ?? 
?H ?I 2
2
i
x
10 4
20
?s ?M?] 
 ??  
2
i
x
104
20
?s ?] 
 ?? ?s x
i
2
 = 2080 
 Now, replaced '8' observations by '12' 
 Then, 
2 2 2
i
x 2080 8 12 2160 ?s ?] ?M ?K ?] 
 ?| Variance of removing observations  
 ?? 
2
2
i i
x x
20 20
????
?s?s
?M????
????
????
 
 ?? ?@ ?H ?I 2 2160
10.2
20
?M ?@ ?@ ?? ?@ 108 – 104.04 
 ?? 3.96 
 Correct standard deviation 
 = 3.96 
 
5. The function 
2
2
x 2x – 15
f(x)
x – 4x 9
?K ?] ?K ,  x?? R is  
 (1) both one-one and onto.  
 (2) onto but not one-one. 
 (3) neither one-one nor onto.   
 (4) one-one but not onto. 
 NTA Ans. (3) 
 Ans. Bonus 
Sol. 
2
(x 5)(x 3)
f(x)
x 4x 9
?K?M
?] ?M?K
 
 Let g(x) = x
2
 – 4x + 9 
 D < 0 
 g(x) > 0 for x ?? R 
  
x
 
 
?@ f( 5) 0
f(3) 0
?M?] ?? ?| ?? ?] ?? ?@ So, f(x) is many-one. 
 again,  
 yx
2
 – 4xy + 9y = x
2
 + 2x – 15 
 x
2
 (y – 1) – 2x(2y + 1) + (9y + 15) = 0 
 for ?B ?@ x ?? R ?? D ?? 0 
 D = 4(2y + 1)
2
 – 4(y –1) (9y + 15) ???@ 0 
 5y
2
 + 2y + 16 ?? 0 
 (5y – 8) (y + 2) ?? ?@ 0 
  
–2
 
8/5
 
?? ?@ ?? ?@ ?? 
 
 
8
y –2,
5
????
??????
????
range 
 Note : If function is defined from f : R ?? R then 
only correct answer is option (3) 
 ???@ Bonus  
 
6. Let A = {n ?? ?@ [100, 700] ???@ N : n is neither a 
multiple of 3 nor a multiple of 4}. Then the 
number of elements in A is  
 (1) 300 (2) 280 
 (3) 310  (4) 290 
 Ans. (1) 
Sol. n(3) ?? multiple of 3 
 102, 105, 108, ..... , 699 
 T
n
= 699 = 102 + (n – 1)(3) 
 n = 200 
 n(3) = 200 
 ?q n(4) ?? multiple of 4 
Page 3


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Saturday 06
th
 April, 2024)   TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If 
3
1
x sin , x 0
f(x) = ,then
x
0 , x 0
?? ????
?? ?? ????
???? ?? ?? ?] ?? (1) f ''(0) 1 ?] (2) 
2
2 24 –
f ''
2
?? ????
?] ????
???? ????
(3) 
2
2 12 –
f ''
2
?? ????
?] ????
???? ????
(4) f ''(0) 0 ?] Ans. (2) 
Sol. f'(x) = 
2
1 1
3x sin xcos
x x
?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? ?? ?? 1
sin
1 1 1 x
f "(x) 6xsin 3cos cos
x x x x
????
????
?? ?? ?? ?? ?? ?? ????
?] ?M ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 2
2 12 24
f "
2 2
?? ?M ?? ????
?] ?M ?] ????
?? ?? ?? ????
2. If A(3,1,–1), 
5 7 1
B , ,
3 3 3
????
????
????
, C(2,2,1) and 
10 2 –1
D , ,
3 3 3
????
????
????
 are the vertices of a quadrilateral 
ABCD, then its area is 
(1)
4 2
3
(2) 
5 2
3
(3) 2 2 (4) 
2 2
3
Ans. (1) 
Sol. 
D
 
C
 
B
 
A
 
Area = 
1
BD AC
2
?? 5 5 2
ˆˆ ˆ
BD i j k
3 3 3
?] ?M ?M ˆˆ ˆ
AC i j 2k ?] ?M ?M 3. 
?H ?I /4 2 2
2
3 3
0
cos xsin x
dx
cos x sin x
?? ?K ?? is equal to 
(1) 1/12 (2) 1/9
(3) 1/6 (4) 1/3
Ans. (3) 
Sol. Divide Nr & Dr by cosx 
?H ?I /4 2 2
2
3
0
tan xsec xdx
dx
1 tan x
?? ?K ?? Let 1 + tan
3
x = t 
tan
2
x sec
2
x dx = 
dt
3
2
2
1
1 dt 1
3 6 t
?] ?? 4. The mean and standard deviation of 20 observations
are found to be 10 and 2, respectively. On
respectively, it was found that an observation by
mistake was taken 8 instead of 12. The correct
standard deviation is
(1) 3.86 (2) 1.8
(3) 3.96 (4) 1.94
Ans. (3) 
Sol. Mean (x) 10 ?] 
i
x
10
20
?s ???]
?s x
i
 = 10×20 = 200
If 8 is replaced by 12, then ?s x
i
 = 200 – 8 + 12 = 204 
 ?| Correct mean 
i
x
(x)
20
?s ?] 
 
204
10.2
20
?]?] 
 ?q Standard deviation = 2 
 ?| Variance = (S.D.)
2
 = 2
2
 = 4 
 ?? 
2
2
i i
x x
4
20 20
????
?s?s
?M?]
????
????
????
 
 ?? 
?H ?I 2
2
i
x
10 4
20
?s ?M?] 
 ??  
2
i
x
104
20
?s ?] 
 ?? ?s x
i
2
 = 2080 
 Now, replaced '8' observations by '12' 
 Then, 
2 2 2
i
x 2080 8 12 2160 ?s ?] ?M ?K ?] 
 ?| Variance of removing observations  
 ?? 
2
2
i i
x x
20 20
????
?s?s
?M????
????
????
 
 ?? ?@ ?H ?I 2 2160
10.2
20
?M ?@ ?@ ?? ?@ 108 – 104.04 
 ?? 3.96 
 Correct standard deviation 
 = 3.96 
 
5. The function 
2
2
x 2x – 15
f(x)
x – 4x 9
?K ?] ?K ,  x?? R is  
 (1) both one-one and onto.  
 (2) onto but not one-one. 
 (3) neither one-one nor onto.   
 (4) one-one but not onto. 
 NTA Ans. (3) 
 Ans. Bonus 
Sol. 
2
(x 5)(x 3)
f(x)
x 4x 9
?K?M
?] ?M?K
 
 Let g(x) = x
2
 – 4x + 9 
 D < 0 
 g(x) > 0 for x ?? R 
  
x
 
 
?@ f( 5) 0
f(3) 0
?M?] ?? ?| ?? ?] ?? ?@ So, f(x) is many-one. 
 again,  
 yx
2
 – 4xy + 9y = x
2
 + 2x – 15 
 x
2
 (y – 1) – 2x(2y + 1) + (9y + 15) = 0 
 for ?B ?@ x ?? R ?? D ?? 0 
 D = 4(2y + 1)
2
 – 4(y –1) (9y + 15) ???@ 0 
 5y
2
 + 2y + 16 ?? 0 
 (5y – 8) (y + 2) ?? ?@ 0 
  
–2
 
8/5
 
?? ?@ ?? ?@ ?? 
 
 
8
y –2,
5
????
??????
????
range 
 Note : If function is defined from f : R ?? R then 
only correct answer is option (3) 
 ???@ Bonus  
 
6. Let A = {n ?? ?@ [100, 700] ???@ N : n is neither a 
multiple of 3 nor a multiple of 4}. Then the 
number of elements in A is  
 (1) 300 (2) 280 
 (3) 310  (4) 290 
 Ans. (1) 
Sol. n(3) ?? multiple of 3 
 102, 105, 108, ..... , 699 
 T
n
= 699 = 102 + (n – 1)(3) 
 n = 200 
 n(3) = 200 
 ?q n(4) ?? multiple of 4 
 
 
 100, 104, 108, ...., 700 
 T
n
 = 700 = 100 + (n – 1) (4) 
 n = 151 
 n(4) = 151 
 n(3 ?? 4) ?? multiple of 3 & 4 both 
 108, 120, 132, ....., 696 
 T
n
 = 696 = 108 + (n – 1)(12) 
 n = 50 
 n(3 ?? ?@ 4) = 50 
 n(3 ???@ 4) = n(3) + n(4) – n(3 ?? ?@ 4) 
        = 200 + 151 – 50 
        = 301 
 
?H ?I n 3 4 ?? = Total – n(3 ?? ?@ 4) = neither a multiple 
of 3 nor a multiple of 4 
  = 601 – 301 = 300 
 
7. Let C be the circle of minimum area touching the 
parabola y = 6 – x
2
 and the lines y = 3 x . Then, 
which one of the following points lies on the circle 
C ?  
 (1) (2, 4) (2) (1, 2) 
 (3) (2, 2)  (4) (1, 1) 
 Ans. (1) 
Sol.  
  
(0,6)
 
(0,6–r)
 
 
 
 Equation of circle 
 x
2
 + (y – (6 – r))
2
 = r
2
  
 touches 3 x y 0 ?M?] 
 p = r 
 
0 (6 r)
r
2
?M?M
?] 
 |r – 6| = 2r 
 r = 2 
 ?| Circle x
2
 + (y – 4)
2
 = 4 
 (2, 4) Satisfies this equation 
 
8. For ?? , ?? ?? R and a natural number n, let 
2
2
r
n
r 1
2
A 2r 2 n –
n(3n – 1)
3r – 2 3
2
?K??
?]?? . Then 2A
10
 – A
8
 is  
 (1) 4 ?? + 2 ?? (2) 2 ?? + 4 ?? 
 (3) 2n  (4) 0 
 Ans. (1) 
Sol. 
2
2
r
n
r 1
2
A 2r 2 n –
n(3n – 1)
3r – 2 3
2
?K??
?]?? 
   
2
2
10 8
n
20 1
2
2A – A 40 2 n –
n(3n – 1)
56 3
2
?K??
?]??
?H ?I 2
2
n
8 1
2
– 16 2 n –
n 3n – 1
22 3
2
?K??
?? 
 
?H ?I 2
2
n
12 1
2
24 2 n –
n 3n – 1
34 3
2
?K??
???? 
 
2
2
n
0 1
2
0 2 n –
n(3n – 1)
–2 3
2
?K??
???? 
 
?H ?I ?H ?I 2 2
– 2 (n – ) – n 2 ?? ?? ?K ?? 
 – 2(– – 2 ) ?? ?? ?? 4 2 ?? ?? ?K ?? 
 
 
 
Page 4


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Saturday 06
th
 April, 2024)   TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If 
3
1
x sin , x 0
f(x) = ,then
x
0 , x 0
?? ????
?? ?? ????
???? ?? ?? ?] ?? (1) f ''(0) 1 ?] (2) 
2
2 24 –
f ''
2
?? ????
?] ????
???? ????
(3) 
2
2 12 –
f ''
2
?? ????
?] ????
???? ????
(4) f ''(0) 0 ?] Ans. (2) 
Sol. f'(x) = 
2
1 1
3x sin xcos
x x
?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? ?? ?? 1
sin
1 1 1 x
f "(x) 6xsin 3cos cos
x x x x
????
????
?? ?? ?? ?? ?? ?? ????
?] ?M ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 2
2 12 24
f "
2 2
?? ?M ?? ????
?] ?M ?] ????
?? ?? ?? ????
2. If A(3,1,–1), 
5 7 1
B , ,
3 3 3
????
????
????
, C(2,2,1) and 
10 2 –1
D , ,
3 3 3
????
????
????
 are the vertices of a quadrilateral 
ABCD, then its area is 
(1)
4 2
3
(2) 
5 2
3
(3) 2 2 (4) 
2 2
3
Ans. (1) 
Sol. 
D
 
C
 
B
 
A
 
Area = 
1
BD AC
2
?? 5 5 2
ˆˆ ˆ
BD i j k
3 3 3
?] ?M ?M ˆˆ ˆ
AC i j 2k ?] ?M ?M 3. 
?H ?I /4 2 2
2
3 3
0
cos xsin x
dx
cos x sin x
?? ?K ?? is equal to 
(1) 1/12 (2) 1/9
(3) 1/6 (4) 1/3
Ans. (3) 
Sol. Divide Nr & Dr by cosx 
?H ?I /4 2 2
2
3
0
tan xsec xdx
dx
1 tan x
?? ?K ?? Let 1 + tan
3
x = t 
tan
2
x sec
2
x dx = 
dt
3
2
2
1
1 dt 1
3 6 t
?] ?? 4. The mean and standard deviation of 20 observations
are found to be 10 and 2, respectively. On
respectively, it was found that an observation by
mistake was taken 8 instead of 12. The correct
standard deviation is
(1) 3.86 (2) 1.8
(3) 3.96 (4) 1.94
Ans. (3) 
Sol. Mean (x) 10 ?] 
i
x
10
20
?s ???]
?s x
i
 = 10×20 = 200
If 8 is replaced by 12, then ?s x
i
 = 200 – 8 + 12 = 204 
 ?| Correct mean 
i
x
(x)
20
?s ?] 
 
204
10.2
20
?]?] 
 ?q Standard deviation = 2 
 ?| Variance = (S.D.)
2
 = 2
2
 = 4 
 ?? 
2
2
i i
x x
4
20 20
????
?s?s
?M?]
????
????
????
 
 ?? 
?H ?I 2
2
i
x
10 4
20
?s ?M?] 
 ??  
2
i
x
104
20
?s ?] 
 ?? ?s x
i
2
 = 2080 
 Now, replaced '8' observations by '12' 
 Then, 
2 2 2
i
x 2080 8 12 2160 ?s ?] ?M ?K ?] 
 ?| Variance of removing observations  
 ?? 
2
2
i i
x x
20 20
????
?s?s
?M????
????
????
 
 ?? ?@ ?H ?I 2 2160
10.2
20
?M ?@ ?@ ?? ?@ 108 – 104.04 
 ?? 3.96 
 Correct standard deviation 
 = 3.96 
 
5. The function 
2
2
x 2x – 15
f(x)
x – 4x 9
?K ?] ?K ,  x?? R is  
 (1) both one-one and onto.  
 (2) onto but not one-one. 
 (3) neither one-one nor onto.   
 (4) one-one but not onto. 
 NTA Ans. (3) 
 Ans. Bonus 
Sol. 
2
(x 5)(x 3)
f(x)
x 4x 9
?K?M
?] ?M?K
 
 Let g(x) = x
2
 – 4x + 9 
 D < 0 
 g(x) > 0 for x ?? R 
  
x
 
 
?@ f( 5) 0
f(3) 0
?M?] ?? ?| ?? ?] ?? ?@ So, f(x) is many-one. 
 again,  
 yx
2
 – 4xy + 9y = x
2
 + 2x – 15 
 x
2
 (y – 1) – 2x(2y + 1) + (9y + 15) = 0 
 for ?B ?@ x ?? R ?? D ?? 0 
 D = 4(2y + 1)
2
 – 4(y –1) (9y + 15) ???@ 0 
 5y
2
 + 2y + 16 ?? 0 
 (5y – 8) (y + 2) ?? ?@ 0 
  
–2
 
8/5
 
?? ?@ ?? ?@ ?? 
 
 
8
y –2,
5
????
??????
????
range 
 Note : If function is defined from f : R ?? R then 
only correct answer is option (3) 
 ???@ Bonus  
 
6. Let A = {n ?? ?@ [100, 700] ???@ N : n is neither a 
multiple of 3 nor a multiple of 4}. Then the 
number of elements in A is  
 (1) 300 (2) 280 
 (3) 310  (4) 290 
 Ans. (1) 
Sol. n(3) ?? multiple of 3 
 102, 105, 108, ..... , 699 
 T
n
= 699 = 102 + (n – 1)(3) 
 n = 200 
 n(3) = 200 
 ?q n(4) ?? multiple of 4 
 
 
 100, 104, 108, ...., 700 
 T
n
 = 700 = 100 + (n – 1) (4) 
 n = 151 
 n(4) = 151 
 n(3 ?? 4) ?? multiple of 3 & 4 both 
 108, 120, 132, ....., 696 
 T
n
 = 696 = 108 + (n – 1)(12) 
 n = 50 
 n(3 ?? ?@ 4) = 50 
 n(3 ???@ 4) = n(3) + n(4) – n(3 ?? ?@ 4) 
        = 200 + 151 – 50 
        = 301 
 
?H ?I n 3 4 ?? = Total – n(3 ?? ?@ 4) = neither a multiple 
of 3 nor a multiple of 4 
  = 601 – 301 = 300 
 
7. Let C be the circle of minimum area touching the 
parabola y = 6 – x
2
 and the lines y = 3 x . Then, 
which one of the following points lies on the circle 
C ?  
 (1) (2, 4) (2) (1, 2) 
 (3) (2, 2)  (4) (1, 1) 
 Ans. (1) 
Sol.  
  
(0,6)
 
(0,6–r)
 
 
 
 Equation of circle 
 x
2
 + (y – (6 – r))
2
 = r
2
  
 touches 3 x y 0 ?M?] 
 p = r 
 
0 (6 r)
r
2
?M?M
?] 
 |r – 6| = 2r 
 r = 2 
 ?| Circle x
2
 + (y – 4)
2
 = 4 
 (2, 4) Satisfies this equation 
 
8. For ?? , ?? ?? R and a natural number n, let 
2
2
r
n
r 1
2
A 2r 2 n –
n(3n – 1)
3r – 2 3
2
?K??
?]?? . Then 2A
10
 – A
8
 is  
 (1) 4 ?? + 2 ?? (2) 2 ?? + 4 ?? 
 (3) 2n  (4) 0 
 Ans. (1) 
Sol. 
2
2
r
n
r 1
2
A 2r 2 n –
n(3n – 1)
3r – 2 3
2
?K??
?]?? 
   
2
2
10 8
n
20 1
2
2A – A 40 2 n –
n(3n – 1)
56 3
2
?K??
?]??
?H ?I 2
2
n
8 1
2
– 16 2 n –
n 3n – 1
22 3
2
?K??
?? 
 
?H ?I 2
2
n
12 1
2
24 2 n –
n 3n – 1
34 3
2
?K??
???? 
 
2
2
n
0 1
2
0 2 n –
n(3n – 1)
–2 3
2
?K??
???? 
 
?H ?I ?H ?I 2 2
– 2 (n – ) – n 2 ?? ?? ?K ?? 
 – 2(– – 2 ) ?? ?? ?? 4 2 ?? ?? ?K ?? 
 
 
 
9. The shortest distance between the lines  
 
x – 3 y 15 z – 9
2 –7 5
?K ?]?] and 
x 1 y –1 z – 9
2 1 –3
?K ?]?] is  
 (1) 6 3 (2) 4 3 
 (3) 5 3  (4) 8 3  
 Ans. (2) 
Sol. 
x – 3 y 15 z – 9
2 –7 5
?K ?]?] & 
x 1 y –1 z – 9
2 1 –3
?K ?]?]  
 S.D = 
?H ?I ?H ?I 2 1 1 2
1 2
a .a . b .b
b b ?? 
 a
1
 = 3, –15, 9 b
1
 = 2, –7, 5 
 a
2
 = –1, 1, 9  b
2
 = 2, 1, –3 
 a
2
 – a
1
 = –4, 16, 0 
 
1 2
ˆˆ ˆ
i j k
ˆˆ ˆ
b b 2 7 5 i(16) j( 16) k(16)
2 1 3
?? ?] ?M ?] ?M ?M ?K ?M 
 
ˆˆ ˆ
16(i j k) ?K?K 
 
1 2
b b 16 3 ???] 
 ?| ?H ?I ?H ?I ?{ ?} 2 1 1 2
a a . b b 16 4 16 (16)(12) ?M ?M ?] ?M ?K ?] 
 S.D. = 
(16)(12)
4 3
16 3
?] 
 
10. A company has two plants A and B to manufacture 
motorcycles.  60% motorcycles are manufactured 
at plant A and the remaining are manufactured at 
plant B. 80% of the motorcycles manufactured at 
plant A are rated of the standard quality, while 
90% of the motorcycles manufactured at plant B 
are rated of the standard quality. A motorcycle 
picked up randomly from the total production is 
found to be of the standard quality. If p is the 
probability that it was manufactured at plant B, 
then 126p is     
 (1) 54 (2) 64 
 (3) 66  (4) 56 
 Ans. (1) 
Sol.  
 A B 
Manufactured 60% 40% 
Standard quality 80% 90% 
 
 P(Manufactured at B / found standard quality) = ? 
 A : Found S.Q 
 B : Manufacture B 
 C : Manufacture A 
 P(E
1
) = 
40
100
 
 P(E
2
) = 
60
100
 
 P(A/E
1
) = 
90
100
 
 P(A/E
2
) = 
80
100
 
 P(E
1
/A) = 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 1 1
1 1 2 2
P A / E P E
P A / E P E P A / E P E ?K 3
7
?] 
 ?| ?@ 126 P = 54 
 
11. Let, ?? , ?? be the distinct roots of the equation  
 
?H ?I 2 2
x – t – 5t 6 x 1 0, ?K ?K ?] t ?? R and a
n
 = ?? n
 
+ ?? n
. 
Then the minimum value of 
2023 2025
2024
a a
a
?K is    
 (1) 1/4 (2) –1/2 
 (3) –1/4  (4) 1/2 
 Ans. (3) 
Sol. by newton's theorem 
 a
n+2
 – (t
2
 – 5t + 6)a
n+1
 + a
n
 = 0 
 ?| ?@ a
2025
 + a
2023
 = (t
2
 – 5t + 6) a
2024 
 
?| ?@ 
2 2025 2023
2024
a a
t 5t 6
a
?K ?] ?M ?K 
 
2
2
1 5
t 5t 6 t
4 2
????
?M ?K ?] ?M ?M????
????
 
?@ ?| ?@ minimum value = 
1
4
?M 
Page 5


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Saturday 06
th
 April, 2024)   TIME : 9 : 00 AM  to  12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If 
3
1
x sin , x 0
f(x) = ,then
x
0 , x 0
?? ????
?? ?? ????
???? ?? ?? ?] ?? (1) f ''(0) 1 ?] (2) 
2
2 24 –
f ''
2
?? ????
?] ????
???? ????
(3) 
2
2 12 –
f ''
2
?? ????
?] ????
???? ????
(4) f ''(0) 0 ?] Ans. (2) 
Sol. f'(x) = 
2
1 1
3x sin xcos
x x
?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? ?? ?? 1
sin
1 1 1 x
f "(x) 6xsin 3cos cos
x x x x
????
????
?? ?? ?? ?? ?? ?? ????
?] ?M ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 2
2 12 24
f "
2 2
?? ?M ?? ????
?] ?M ?] ????
?? ?? ?? ????
2. If A(3,1,–1), 
5 7 1
B , ,
3 3 3
????
????
????
, C(2,2,1) and 
10 2 –1
D , ,
3 3 3
????
????
????
 are the vertices of a quadrilateral 
ABCD, then its area is 
(1)
4 2
3
(2) 
5 2
3
(3) 2 2 (4) 
2 2
3
Ans. (1) 
Sol. 
D
 
C
 
B
 
A
 
Area = 
1
BD AC
2
?? 5 5 2
ˆˆ ˆ
BD i j k
3 3 3
?] ?M ?M ˆˆ ˆ
AC i j 2k ?] ?M ?M 3. 
?H ?I /4 2 2
2
3 3
0
cos xsin x
dx
cos x sin x
?? ?K ?? is equal to 
(1) 1/12 (2) 1/9
(3) 1/6 (4) 1/3
Ans. (3) 
Sol. Divide Nr & Dr by cosx 
?H ?I /4 2 2
2
3
0
tan xsec xdx
dx
1 tan x
?? ?K ?? Let 1 + tan
3
x = t 
tan
2
x sec
2
x dx = 
dt
3
2
2
1
1 dt 1
3 6 t
?] ?? 4. The mean and standard deviation of 20 observations
are found to be 10 and 2, respectively. On
respectively, it was found that an observation by
mistake was taken 8 instead of 12. The correct
standard deviation is
(1) 3.86 (2) 1.8
(3) 3.96 (4) 1.94
Ans. (3) 
Sol. Mean (x) 10 ?] 
i
x
10
20
?s ???]
?s x
i
 = 10×20 = 200
If 8 is replaced by 12, then ?s x
i
 = 200 – 8 + 12 = 204 
 ?| Correct mean 
i
x
(x)
20
?s ?] 
 
204
10.2
20
?]?] 
 ?q Standard deviation = 2 
 ?| Variance = (S.D.)
2
 = 2
2
 = 4 
 ?? 
2
2
i i
x x
4
20 20
????
?s?s
?M?]
????
????
????
 
 ?? 
?H ?I 2
2
i
x
10 4
20
?s ?M?] 
 ??  
2
i
x
104
20
?s ?] 
 ?? ?s x
i
2
 = 2080 
 Now, replaced '8' observations by '12' 
 Then, 
2 2 2
i
x 2080 8 12 2160 ?s ?] ?M ?K ?] 
 ?| Variance of removing observations  
 ?? 
2
2
i i
x x
20 20
????
?s?s
?M????
????
????
 
 ?? ?@ ?H ?I 2 2160
10.2
20
?M ?@ ?@ ?? ?@ 108 – 104.04 
 ?? 3.96 
 Correct standard deviation 
 = 3.96 
 
5. The function 
2
2
x 2x – 15
f(x)
x – 4x 9
?K ?] ?K ,  x?? R is  
 (1) both one-one and onto.  
 (2) onto but not one-one. 
 (3) neither one-one nor onto.   
 (4) one-one but not onto. 
 NTA Ans. (3) 
 Ans. Bonus 
Sol. 
2
(x 5)(x 3)
f(x)
x 4x 9
?K?M
?] ?M?K
 
 Let g(x) = x
2
 – 4x + 9 
 D < 0 
 g(x) > 0 for x ?? R 
  
x
 
 
?@ f( 5) 0
f(3) 0
?M?] ?? ?| ?? ?] ?? ?@ So, f(x) is many-one. 
 again,  
 yx
2
 – 4xy + 9y = x
2
 + 2x – 15 
 x
2
 (y – 1) – 2x(2y + 1) + (9y + 15) = 0 
 for ?B ?@ x ?? R ?? D ?? 0 
 D = 4(2y + 1)
2
 – 4(y –1) (9y + 15) ???@ 0 
 5y
2
 + 2y + 16 ?? 0 
 (5y – 8) (y + 2) ?? ?@ 0 
  
–2
 
8/5
 
?? ?@ ?? ?@ ?? 
 
 
8
y –2,
5
????
??????
????
range 
 Note : If function is defined from f : R ?? R then 
only correct answer is option (3) 
 ???@ Bonus  
 
6. Let A = {n ?? ?@ [100, 700] ???@ N : n is neither a 
multiple of 3 nor a multiple of 4}. Then the 
number of elements in A is  
 (1) 300 (2) 280 
 (3) 310  (4) 290 
 Ans. (1) 
Sol. n(3) ?? multiple of 3 
 102, 105, 108, ..... , 699 
 T
n
= 699 = 102 + (n – 1)(3) 
 n = 200 
 n(3) = 200 
 ?q n(4) ?? multiple of 4 
 
 
 100, 104, 108, ...., 700 
 T
n
 = 700 = 100 + (n – 1) (4) 
 n = 151 
 n(4) = 151 
 n(3 ?? 4) ?? multiple of 3 & 4 both 
 108, 120, 132, ....., 696 
 T
n
 = 696 = 108 + (n – 1)(12) 
 n = 50 
 n(3 ?? ?@ 4) = 50 
 n(3 ???@ 4) = n(3) + n(4) – n(3 ?? ?@ 4) 
        = 200 + 151 – 50 
        = 301 
 
?H ?I n 3 4 ?? = Total – n(3 ?? ?@ 4) = neither a multiple 
of 3 nor a multiple of 4 
  = 601 – 301 = 300 
 
7. Let C be the circle of minimum area touching the 
parabola y = 6 – x
2
 and the lines y = 3 x . Then, 
which one of the following points lies on the circle 
C ?  
 (1) (2, 4) (2) (1, 2) 
 (3) (2, 2)  (4) (1, 1) 
 Ans. (1) 
Sol.  
  
(0,6)
 
(0,6–r)
 
 
 
 Equation of circle 
 x
2
 + (y – (6 – r))
2
 = r
2
  
 touches 3 x y 0 ?M?] 
 p = r 
 
0 (6 r)
r
2
?M?M
?] 
 |r – 6| = 2r 
 r = 2 
 ?| Circle x
2
 + (y – 4)
2
 = 4 
 (2, 4) Satisfies this equation 
 
8. For ?? , ?? ?? R and a natural number n, let 
2
2
r
n
r 1
2
A 2r 2 n –
n(3n – 1)
3r – 2 3
2
?K??
?]?? . Then 2A
10
 – A
8
 is  
 (1) 4 ?? + 2 ?? (2) 2 ?? + 4 ?? 
 (3) 2n  (4) 0 
 Ans. (1) 
Sol. 
2
2
r
n
r 1
2
A 2r 2 n –
n(3n – 1)
3r – 2 3
2
?K??
?]?? 
   
2
2
10 8
n
20 1
2
2A – A 40 2 n –
n(3n – 1)
56 3
2
?K??
?]??
?H ?I 2
2
n
8 1
2
– 16 2 n –
n 3n – 1
22 3
2
?K??
?? 
 
?H ?I 2
2
n
12 1
2
24 2 n –
n 3n – 1
34 3
2
?K??
???? 
 
2
2
n
0 1
2
0 2 n –
n(3n – 1)
–2 3
2
?K??
???? 
 
?H ?I ?H ?I 2 2
– 2 (n – ) – n 2 ?? ?? ?K ?? 
 – 2(– – 2 ) ?? ?? ?? 4 2 ?? ?? ?K ?? 
 
 
 
9. The shortest distance between the lines  
 
x – 3 y 15 z – 9
2 –7 5
?K ?]?] and 
x 1 y –1 z – 9
2 1 –3
?K ?]?] is  
 (1) 6 3 (2) 4 3 
 (3) 5 3  (4) 8 3  
 Ans. (2) 
Sol. 
x – 3 y 15 z – 9
2 –7 5
?K ?]?] & 
x 1 y –1 z – 9
2 1 –3
?K ?]?]  
 S.D = 
?H ?I ?H ?I 2 1 1 2
1 2
a .a . b .b
b b ?? 
 a
1
 = 3, –15, 9 b
1
 = 2, –7, 5 
 a
2
 = –1, 1, 9  b
2
 = 2, 1, –3 
 a
2
 – a
1
 = –4, 16, 0 
 
1 2
ˆˆ ˆ
i j k
ˆˆ ˆ
b b 2 7 5 i(16) j( 16) k(16)
2 1 3
?? ?] ?M ?] ?M ?M ?K ?M 
 
ˆˆ ˆ
16(i j k) ?K?K 
 
1 2
b b 16 3 ???] 
 ?| ?H ?I ?H ?I ?{ ?} 2 1 1 2
a a . b b 16 4 16 (16)(12) ?M ?M ?] ?M ?K ?] 
 S.D. = 
(16)(12)
4 3
16 3
?] 
 
10. A company has two plants A and B to manufacture 
motorcycles.  60% motorcycles are manufactured 
at plant A and the remaining are manufactured at 
plant B. 80% of the motorcycles manufactured at 
plant A are rated of the standard quality, while 
90% of the motorcycles manufactured at plant B 
are rated of the standard quality. A motorcycle 
picked up randomly from the total production is 
found to be of the standard quality. If p is the 
probability that it was manufactured at plant B, 
then 126p is     
 (1) 54 (2) 64 
 (3) 66  (4) 56 
 Ans. (1) 
Sol.  
 A B 
Manufactured 60% 40% 
Standard quality 80% 90% 
 
 P(Manufactured at B / found standard quality) = ? 
 A : Found S.Q 
 B : Manufacture B 
 C : Manufacture A 
 P(E
1
) = 
40
100
 
 P(E
2
) = 
60
100
 
 P(A/E
1
) = 
90
100
 
 P(A/E
2
) = 
80
100
 
 P(E
1
/A) = 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 1 1
1 1 2 2
P A / E P E
P A / E P E P A / E P E ?K 3
7
?] 
 ?| ?@ 126 P = 54 
 
11. Let, ?? , ?? be the distinct roots of the equation  
 
?H ?I 2 2
x – t – 5t 6 x 1 0, ?K ?K ?] t ?? R and a
n
 = ?? n
 
+ ?? n
. 
Then the minimum value of 
2023 2025
2024
a a
a
?K is    
 (1) 1/4 (2) –1/2 
 (3) –1/4  (4) 1/2 
 Ans. (3) 
Sol. by newton's theorem 
 a
n+2
 – (t
2
 – 5t + 6)a
n+1
 + a
n
 = 0 
 ?| ?@ a
2025
 + a
2023
 = (t
2
 – 5t + 6) a
2024 
 
?| ?@ 
2 2025 2023
2024
a a
t 5t 6
a
?K ?] ?M ?K 
 
2
2
1 5
t 5t 6 t
4 2
????
?M ?K ?] ?M ?M????
????
 
?@ ?| ?@ minimum value = 
1
4
?M 
 
12. Let the relations R
1
 and R
2
 on the set  
 X = {1, 2, 3, ..., 20} be given by  
 R
1
 = {(x, y) : 2x – 3y = 2} and  
 R
2
 = {(x, y) : –5x + 4y = 0}. If M and N be the 
minimum number of elements required to be added 
in R
1
 and R
2
, respectively, in order to make the 
relations symmetric, then M + N equals 
 (1) 8 (2) 16 
 (3) 12  (4) 10 
 Ans. (4) 
Sol. x = {1, 2, 3, .......20} 
 R
1
 = {(x, y) : 2x – 3y = 2} 
 R
2
 = {(x, y) : –5x + 4y = 0} 
    R
1
 = {(4, 2), (7, 4), (10, 6), (13, 8), (16, 10), (19, 12)} 
    R
2
 = {(4, 5), (8, 10), (12, 15), (16, 20)} 
 in R
1
 6 element needed 
 in R
2
 4 element needed 
 So, total 6+4 = 10 element 
 
13. Let a variable line of slope m > 0 passing through 
the point (4, –9) intersect the coordinate axes at the 
points A and B. the minimum value of the sum of 
the distances of A and B from the origin is  
 (1) 25 (2) 30 
 (3) 15  (4) 10 
 Ans. (1) 
Sol. equation of line is  
 y + 9 = m (x – 4) 
 ?| 
9 4m
A ,0
m
?K????
?]????
????
  
  B = (0, – 9 – 4m) 
 ?|  OA + OB =
9 4m
9 4m
m
?K ?K?K 
 m 0 ?^ 
 
9
13 4m
m
?] ?K ?K 
 
9
4m
9
m
36 4m 12
2 m
?K ?? ?? ?K ?? 
 ?| OA + OB ?? 25  
 
14. The interval in which the function f(x) = x
x
, x > 0, 
is strictly increasing is 
 (1) 
1
0,
e
????
?? ?? ????
 (2) 
2
1
,1
e
????
?? ??????
 
 (3) (0, ?? )  (4) 
1
,
e
????
?? ?? ??????
 
 Ans. (4) 
Sol. f(x) = x
x
 ; x > 0 
 ?? ny = x ?? nx 
 
1 dy x
nx
y dx x
?]?K 
 
x
dy
x (1 nx)
dx
?]?K 
 for strictly increasing 
 
dy
0
dx
??   ?? ?@ x
x
 (1 + ?? nx) ?? 0 
 ?? ?@ ?? nx ?? –1 
 x ?? e
–1 
 x ?? 
1
e
 
 
1
x ,
e
????
????
?? ??????