JEE Exam > JEE Notes > JEE Main & Advanced Mock Test Series 2025 > JEE Main 2024 April 9 Shift 1 Paper & Solutions
JEE Main 2024 April 9 Shift 1 Paper & Solutions | JEE Main & Advanced Mock Test Series 2025 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
FINAL JEE –MAIN EXAMINATION – APRIL, 2024
(Held On Tuesday 09
th
April, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A
1. Let the line L intersect the lines
x – 2 = –y = z – 1, 2 (x + 1) = 2(y – 1) = z + 1
and be parallel to the line
x 2 y 1 z 2
3 1 2
?M ?M ?M ?]?] .
Then which of the following points lies on L ?
(1)
1
– , 1 , 1
3
????
????
????
(2)
1
– , 1 , 1
3
????
?M ????
????
(3)
1
– , 1 , 1
3
????
?M?M
????
????
(4)
1
– , 1 , 1
3
????
?M ????
????
Ans. (2)
Sol.
dr's of line (3, 1, 2)
M(2+ ?? , – ?@ ?? , 1+ ?@ ?? ?I
N 1 , 1 , –1
2 2
???? ????
?M ?K ?K ?K ?? ????
????
L
1
L
2
1
x 2 y z 1
L :
1 1 1
?M?M
?] ?] ?] ?? ?M 2
x 1 y 1 z 1
L :
1 1
1
2 2
?K ?M ?K ?] ?] ?] ?? dr of line MN will be
3 , 1 ,2
2 2
????
?\ ?K ?? ?M ?M ?M ?? ?M ?K ?? ?M ?? ?^ & it will be
proportional to <3, 1, 2>
3 1
2
2 2
3 1 2
????
?K ?? ?M ?M ?M ?? ?M ?K ?? ?M ?? ?| ?] ?] 4 ?? + ?? = –6 4 + 3 ?? = 0
?@ ?? ?@ 4 2
&
3 3
?? ?] ?M ?? ?] ?M ?|?@ Coordinate of M will be <
2 4 1
, ,
3 3 3
????
?M ????
????
and equation of required line will be.
2 4 1
x y z
3 3 3
k
3 1 2
?M ?M ?K ?] ?] ?] So any point on this line will be
2 4 1
3k, k, 2k
3 3 3
????
?K ?K ?M ?K ????
????
2 1
3k
3 3
?K ?] ?M ?? ?@ 1
k
3
?]?M
?|?@ Point lie on the line for
1
k
3
?]?M is
1
,1, 1
3
????
?M?M
????
????
?@ 2. The parabola y
2
= 4x divides the area of the circle
x
2
+ y
2
= 5 in two parts. The area of the smaller
part is equal to :
(1)
1
2 2
5sin
3 5
?M????
?K ????
????
(2)
1
1 2
5sin
3 5
?M????
?K ????
????
(3)
1
1 2
5 sin
3 5
?M????
?K ????
????
(4)
1
2 2
5 sin
3 5
?M????
?K ????
????
Ans. (1)
Sol. y
2
= 4x
x
2
+ y
2
= 5
?| ?@ Area of shaded region as shown in the figure
will be
Page 2
FINAL JEE –MAIN EXAMINATION – APRIL, 2024
(Held On Tuesday 09
th
April, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A
1. Let the line L intersect the lines
x – 2 = –y = z – 1, 2 (x + 1) = 2(y – 1) = z + 1
and be parallel to the line
x 2 y 1 z 2
3 1 2
?M ?M ?M ?]?] .
Then which of the following points lies on L ?
(1)
1
– , 1 , 1
3
????
????
????
(2)
1
– , 1 , 1
3
????
?M ????
????
(3)
1
– , 1 , 1
3
????
?M?M
????
????
(4)
1
– , 1 , 1
3
????
?M ????
????
Ans. (2)
Sol.
dr's of line (3, 1, 2)
M(2+ ?? , – ?@ ?? , 1+ ?@ ?? ?I
N 1 , 1 , –1
2 2
???? ????
?M ?K ?K ?K ?? ????
????
L
1
L
2
1
x 2 y z 1
L :
1 1 1
?M?M
?] ?] ?] ?? ?M 2
x 1 y 1 z 1
L :
1 1
1
2 2
?K ?M ?K ?] ?] ?] ?? dr of line MN will be
3 , 1 ,2
2 2
????
?\ ?K ?? ?M ?M ?M ?? ?M ?K ?? ?M ?? ?^ & it will be
proportional to <3, 1, 2>
3 1
2
2 2
3 1 2
????
?K ?? ?M ?M ?M ?? ?M ?K ?? ?M ?? ?| ?] ?] 4 ?? + ?? = –6 4 + 3 ?? = 0
?@ ?? ?@ 4 2
&
3 3
?? ?] ?M ?? ?] ?M ?|?@ Coordinate of M will be <
2 4 1
, ,
3 3 3
????
?M ????
????
and equation of required line will be.
2 4 1
x y z
3 3 3
k
3 1 2
?M ?M ?K ?] ?] ?] So any point on this line will be
2 4 1
3k, k, 2k
3 3 3
????
?K ?K ?M ?K ????
????
2 1
3k
3 3
?K ?] ?M ?? ?@ 1
k
3
?]?M
?|?@ Point lie on the line for
1
k
3
?]?M is
1
,1, 1
3
????
?M?M
????
????
?@ 2. The parabola y
2
= 4x divides the area of the circle
x
2
+ y
2
= 5 in two parts. The area of the smaller
part is equal to :
(1)
1
2 2
5sin
3 5
?M????
?K ????
????
(2)
1
1 2
5sin
3 5
?M????
?K ????
????
(3)
1
1 2
5 sin
3 5
?M????
?K ????
????
(4)
1
2 2
5 sin
3 5
?M????
?K ????
????
Ans. (1)
Sol. y
2
= 4x
x
2
+ y
2
= 5
?| ?@ Area of shaded region as shown in the figure
will be
A(1,2)
B(1, –2)
C( 5, 0)
1 5
2
1
0 1
A 4x dx 5 x dx ?] ?K ?M ????
1
5
3
2 1
2
1
0
4 x 5 x
x 5 x sin
3 2 2 5
?M ????
????
?] ?? ?K ?M ?K ????
????
????
????
1
1 5 5 1
sin
3 4 2 5
?M ?? ????
?] ?K ?M ????
????
?@ ?| ?@ Required Area = 2 A
1
1
2 5 1
5sin
3 2 5
?M ?? ????
?] ?K ?M ????
????
1
2 1
5 sin
3 2 5
?M ??????
?] ?K ?M????
????
1
2 1
5cos
3 5
?M ?]?K
1
2 2
5sin
3 5
?M????
?]?K
????
????
3. The solution curve, of the differential equation
dy dy
2y 3 5
dx dx
?K?] , passing through the point
(0, 1) is a conic, whose vertex lies on the line :
(1) 2x + 3y = 9 (2) 2x + 3y = –9
(3) 2x + 3y = –6 (4) 2x + 3y = 6
Ans. (1)
Sol. ?H ?I dy
2y 5 3
dx
?M ?] ?M
?H ?I 2y 5 dy 3dx ?M ?] ?M
2
y
2 5y 3x
2
?? ?M ?] ?M ?K ??
Curve passes through (0, 1)
?@ ???@?? = – 4
Curve will be
2
5 3
y 3 x
2 4
?? ?? ?? ?? ?M ?] ?M ?M ?? ?? ?? ?? ?? ?? ?? ??
?| Vertex of parabola will be
3 5
,
4 2
????
????
????
2x + 3y = 9
4. A ray of light coming from the point P (1, 2) gets
reflected from the point Q on the x-axis and then
passes through the point R (4, 3). If the point S (h,
k) is such that PQRS is a parallelogram, then hk
2
is
equal to :
(1) 80 (2) 90
(3) 60 (4) 70
Ans. (4)
Sol.
Image of P wrt x-axis will be P'(1, –2) equation of
line joining P'R will be
?H ?I 5
y 3 x 4
3
?M ?] ?M
Above line will meet x-axis at Q where
11
y 0 x
5
?] ?? ?]
11
Q ,0
5
????
?| ????
????
PQRS is parallelogram so their diagonals will
bisects each other
Page 3
FINAL JEE –MAIN EXAMINATION – APRIL, 2024
(Held On Tuesday 09
th
April, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A
1. Let the line L intersect the lines
x – 2 = –y = z – 1, 2 (x + 1) = 2(y – 1) = z + 1
and be parallel to the line
x 2 y 1 z 2
3 1 2
?M ?M ?M ?]?] .
Then which of the following points lies on L ?
(1)
1
– , 1 , 1
3
????
????
????
(2)
1
– , 1 , 1
3
????
?M ????
????
(3)
1
– , 1 , 1
3
????
?M?M
????
????
(4)
1
– , 1 , 1
3
????
?M ????
????
Ans. (2)
Sol.
dr's of line (3, 1, 2)
M(2+ ?? , – ?@ ?? , 1+ ?@ ?? ?I
N 1 , 1 , –1
2 2
???? ????
?M ?K ?K ?K ?? ????
????
L
1
L
2
1
x 2 y z 1
L :
1 1 1
?M?M
?] ?] ?] ?? ?M 2
x 1 y 1 z 1
L :
1 1
1
2 2
?K ?M ?K ?] ?] ?] ?? dr of line MN will be
3 , 1 ,2
2 2
????
?\ ?K ?? ?M ?M ?M ?? ?M ?K ?? ?M ?? ?^ & it will be
proportional to <3, 1, 2>
3 1
2
2 2
3 1 2
????
?K ?? ?M ?M ?M ?? ?M ?K ?? ?M ?? ?| ?] ?] 4 ?? + ?? = –6 4 + 3 ?? = 0
?@ ?? ?@ 4 2
&
3 3
?? ?] ?M ?? ?] ?M ?|?@ Coordinate of M will be <
2 4 1
, ,
3 3 3
????
?M ????
????
and equation of required line will be.
2 4 1
x y z
3 3 3
k
3 1 2
?M ?M ?K ?] ?] ?] So any point on this line will be
2 4 1
3k, k, 2k
3 3 3
????
?K ?K ?M ?K ????
????
2 1
3k
3 3
?K ?] ?M ?? ?@ 1
k
3
?]?M
?|?@ Point lie on the line for
1
k
3
?]?M is
1
,1, 1
3
????
?M?M
????
????
?@ 2. The parabola y
2
= 4x divides the area of the circle
x
2
+ y
2
= 5 in two parts. The area of the smaller
part is equal to :
(1)
1
2 2
5sin
3 5
?M????
?K ????
????
(2)
1
1 2
5sin
3 5
?M????
?K ????
????
(3)
1
1 2
5 sin
3 5
?M????
?K ????
????
(4)
1
2 2
5 sin
3 5
?M????
?K ????
????
Ans. (1)
Sol. y
2
= 4x
x
2
+ y
2
= 5
?| ?@ Area of shaded region as shown in the figure
will be
A(1,2)
B(1, –2)
C( 5, 0)
1 5
2
1
0 1
A 4x dx 5 x dx ?] ?K ?M ????
1
5
3
2 1
2
1
0
4 x 5 x
x 5 x sin
3 2 2 5
?M ????
????
?] ?? ?K ?M ?K ????
????
????
????
1
1 5 5 1
sin
3 4 2 5
?M ?? ????
?] ?K ?M ????
????
?@ ?| ?@ Required Area = 2 A
1
1
2 5 1
5sin
3 2 5
?M ?? ????
?] ?K ?M ????
????
1
2 1
5 sin
3 2 5
?M ??????
?] ?K ?M????
????
1
2 1
5cos
3 5
?M ?]?K
1
2 2
5sin
3 5
?M????
?]?K
????
????
3. The solution curve, of the differential equation
dy dy
2y 3 5
dx dx
?K?] , passing through the point
(0, 1) is a conic, whose vertex lies on the line :
(1) 2x + 3y = 9 (2) 2x + 3y = –9
(3) 2x + 3y = –6 (4) 2x + 3y = 6
Ans. (1)
Sol. ?H ?I dy
2y 5 3
dx
?M ?] ?M
?H ?I 2y 5 dy 3dx ?M ?] ?M
2
y
2 5y 3x
2
?? ?M ?] ?M ?K ??
Curve passes through (0, 1)
?@ ???@?? = – 4
Curve will be
2
5 3
y 3 x
2 4
?? ?? ?? ?? ?M ?] ?M ?M ?? ?? ?? ?? ?? ?? ?? ??
?| Vertex of parabola will be
3 5
,
4 2
????
????
????
2x + 3y = 9
4. A ray of light coming from the point P (1, 2) gets
reflected from the point Q on the x-axis and then
passes through the point R (4, 3). If the point S (h,
k) is such that PQRS is a parallelogram, then hk
2
is
equal to :
(1) 80 (2) 90
(3) 60 (4) 70
Ans. (4)
Sol.
Image of P wrt x-axis will be P'(1, –2) equation of
line joining P'R will be
?H ?I 5
y 3 x 4
3
?M ?] ?M
Above line will meet x-axis at Q where
11
y 0 x
5
?] ?? ?]
11
Q ,0
5
????
?| ????
????
PQRS is parallelogram so their diagonals will
bisects each other
?? ?@ 11
h
4 1 2 3 k 0
5
&
2 2 2 2
?K ?K ?K ?K ?]?]
?@?? ?@ 14
h & k 5
5
?]?]
2 2
14
hk 5 70
5
?| ?] ?? ?] ?@ 5. Let ?? , ?? ?? R. If the system of equations
3x + 5y + ?? z = 3
7x + 11y –9z = 2
97x + 155y – 189z = ?? ?@ has infinitely many solutions, then ?? +2 ?? is equal
to :
(1) 25 (2) 24
(3) 27 (4) 22
Ans. (1)
Sol. 3x + 5y + ?? z = 3
7x + 11y –9z = 2
97x + 155y – 189z = ?? ?@ ?@ ?@?@?@ 93x + 155y + 31 ?? z = 93
97x + 155y – 189z = ?? ?@ ?@ – – + –
–4x + (31 ?? + 189)z = 93 – ?? ?@ 1085x + 1705y – 1395z = 310
1067x + 1705y – 2079z = 11 ??
?@ – – + –
18x + 684z = 310 – 11?? ?@ ?@ –36x + 9(31 ?? + 189)z = 9(93 – ?? )
36x + 1368z = 2 (310 – 11 ?? )
(279 ?? ?@ + 3069)z = 1457 – 31 ?? ?@ ?@ for infinite solutions -
3069 341
279 31
?M?M
?? ?] ?]
1457
31
???]
1457 682 775
2 25
31 31
?M ?? ?K ?? ?] ?] ?]
6. The coefficient of x
70
in x
2
(1 + x)
98
+ x
3
(1 + x)
97
+
x
4
(1 + x)
96
+ ........ +x
54
(1 + x)
46
is
99
C
p
–
46
C
q
.
Then a possible value to p + q is :
(1) 55 (2) 61
(3) 68 (4) 83
Ans. (4)
Sol.
?H ?I ?H ?I ?H ?I 98 96
2 3 97 4
x 1 x x 1 x x 1 x ....... ?K ?K ?K ?K ?K ?K
?H ?I 46
54
x 1 x ?K
Coeff. of
70 98 97 96
68 67 66
x : C C C .......... ?K ?K ?K
47 46
17 16
C C ?K
46 47 98
30 30 30
C C ........... C ?] ?K ?K
?H ?I 46 46 47 98 46
31 30 30 30 31
C C C .. .. .. .. .. . C – C ?] ?K ?K ?K
47 47 98 46
31 30 30 31
C C . . . . . . . . . . . C – C ?] ?K ?K
......
99 46 99 46
31 31 p q
C C C C ?] ?M ?] ?M
Possible values of (p + q) are 62, 83, 99, 46
?? p + q = 83
7. Let
?H ?I e
2 tan x 1
dx x log sin x cos x C
3 tan x 2
?M ?] ?? ?K ?? ?K ?? ?K ?K ?? , where C is the constant of integration.
Then
?? ???K
?? is equal to :
(1) 3 (2) 1
(3) 4 (4) 7
Ans. (3)
Sol.
2 tan x
dx
3 tan x
?M ?K ??
2cos x sin x
dx
3cos x sin x
?M ?] ?K ??
2 cosx – sinx = A(3cosx + sinx) + B(cosx – 3sinx)
3A + B = 2
A – 3B = –1
Page 4
FINAL JEE –MAIN EXAMINATION – APRIL, 2024
(Held On Tuesday 09
th
April, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A
1. Let the line L intersect the lines
x – 2 = –y = z – 1, 2 (x + 1) = 2(y – 1) = z + 1
and be parallel to the line
x 2 y 1 z 2
3 1 2
?M ?M ?M ?]?] .
Then which of the following points lies on L ?
(1)
1
– , 1 , 1
3
????
????
????
(2)
1
– , 1 , 1
3
????
?M ????
????
(3)
1
– , 1 , 1
3
????
?M?M
????
????
(4)
1
– , 1 , 1
3
????
?M ????
????
Ans. (2)
Sol.
dr's of line (3, 1, 2)
M(2+ ?? , – ?@ ?? , 1+ ?@ ?? ?I
N 1 , 1 , –1
2 2
???? ????
?M ?K ?K ?K ?? ????
????
L
1
L
2
1
x 2 y z 1
L :
1 1 1
?M?M
?] ?] ?] ?? ?M 2
x 1 y 1 z 1
L :
1 1
1
2 2
?K ?M ?K ?] ?] ?] ?? dr of line MN will be
3 , 1 ,2
2 2
????
?\ ?K ?? ?M ?M ?M ?? ?M ?K ?? ?M ?? ?^ & it will be
proportional to <3, 1, 2>
3 1
2
2 2
3 1 2
????
?K ?? ?M ?M ?M ?? ?M ?K ?? ?M ?? ?| ?] ?] 4 ?? + ?? = –6 4 + 3 ?? = 0
?@ ?? ?@ 4 2
&
3 3
?? ?] ?M ?? ?] ?M ?|?@ Coordinate of M will be <
2 4 1
, ,
3 3 3
????
?M ????
????
and equation of required line will be.
2 4 1
x y z
3 3 3
k
3 1 2
?M ?M ?K ?] ?] ?] So any point on this line will be
2 4 1
3k, k, 2k
3 3 3
????
?K ?K ?M ?K ????
????
2 1
3k
3 3
?K ?] ?M ?? ?@ 1
k
3
?]?M
?|?@ Point lie on the line for
1
k
3
?]?M is
1
,1, 1
3
????
?M?M
????
????
?@ 2. The parabola y
2
= 4x divides the area of the circle
x
2
+ y
2
= 5 in two parts. The area of the smaller
part is equal to :
(1)
1
2 2
5sin
3 5
?M????
?K ????
????
(2)
1
1 2
5sin
3 5
?M????
?K ????
????
(3)
1
1 2
5 sin
3 5
?M????
?K ????
????
(4)
1
2 2
5 sin
3 5
?M????
?K ????
????
Ans. (1)
Sol. y
2
= 4x
x
2
+ y
2
= 5
?| ?@ Area of shaded region as shown in the figure
will be
A(1,2)
B(1, –2)
C( 5, 0)
1 5
2
1
0 1
A 4x dx 5 x dx ?] ?K ?M ????
1
5
3
2 1
2
1
0
4 x 5 x
x 5 x sin
3 2 2 5
?M ????
????
?] ?? ?K ?M ?K ????
????
????
????
1
1 5 5 1
sin
3 4 2 5
?M ?? ????
?] ?K ?M ????
????
?@ ?| ?@ Required Area = 2 A
1
1
2 5 1
5sin
3 2 5
?M ?? ????
?] ?K ?M ????
????
1
2 1
5 sin
3 2 5
?M ??????
?] ?K ?M????
????
1
2 1
5cos
3 5
?M ?]?K
1
2 2
5sin
3 5
?M????
?]?K
????
????
3. The solution curve, of the differential equation
dy dy
2y 3 5
dx dx
?K?] , passing through the point
(0, 1) is a conic, whose vertex lies on the line :
(1) 2x + 3y = 9 (2) 2x + 3y = –9
(3) 2x + 3y = –6 (4) 2x + 3y = 6
Ans. (1)
Sol. ?H ?I dy
2y 5 3
dx
?M ?] ?M
?H ?I 2y 5 dy 3dx ?M ?] ?M
2
y
2 5y 3x
2
?? ?M ?] ?M ?K ??
Curve passes through (0, 1)
?@ ???@?? = – 4
Curve will be
2
5 3
y 3 x
2 4
?? ?? ?? ?? ?M ?] ?M ?M ?? ?? ?? ?? ?? ?? ?? ??
?| Vertex of parabola will be
3 5
,
4 2
????
????
????
2x + 3y = 9
4. A ray of light coming from the point P (1, 2) gets
reflected from the point Q on the x-axis and then
passes through the point R (4, 3). If the point S (h,
k) is such that PQRS is a parallelogram, then hk
2
is
equal to :
(1) 80 (2) 90
(3) 60 (4) 70
Ans. (4)
Sol.
Image of P wrt x-axis will be P'(1, –2) equation of
line joining P'R will be
?H ?I 5
y 3 x 4
3
?M ?] ?M
Above line will meet x-axis at Q where
11
y 0 x
5
?] ?? ?]
11
Q ,0
5
????
?| ????
????
PQRS is parallelogram so their diagonals will
bisects each other
?? ?@ 11
h
4 1 2 3 k 0
5
&
2 2 2 2
?K ?K ?K ?K ?]?]
?@?? ?@ 14
h & k 5
5
?]?]
2 2
14
hk 5 70
5
?| ?] ?? ?] ?@ 5. Let ?? , ?? ?? R. If the system of equations
3x + 5y + ?? z = 3
7x + 11y –9z = 2
97x + 155y – 189z = ?? ?@ has infinitely many solutions, then ?? +2 ?? is equal
to :
(1) 25 (2) 24
(3) 27 (4) 22
Ans. (1)
Sol. 3x + 5y + ?? z = 3
7x + 11y –9z = 2
97x + 155y – 189z = ?? ?@ ?@ ?@?@?@ 93x + 155y + 31 ?? z = 93
97x + 155y – 189z = ?? ?@ ?@ – – + –
–4x + (31 ?? + 189)z = 93 – ?? ?@ 1085x + 1705y – 1395z = 310
1067x + 1705y – 2079z = 11 ??
?@ – – + –
18x + 684z = 310 – 11?? ?@ ?@ –36x + 9(31 ?? + 189)z = 9(93 – ?? )
36x + 1368z = 2 (310 – 11 ?? )
(279 ?? ?@ + 3069)z = 1457 – 31 ?? ?@ ?@ for infinite solutions -
3069 341
279 31
?M?M
?? ?] ?]
1457
31
???]
1457 682 775
2 25
31 31
?M ?? ?K ?? ?] ?] ?]
6. The coefficient of x
70
in x
2
(1 + x)
98
+ x
3
(1 + x)
97
+
x
4
(1 + x)
96
+ ........ +x
54
(1 + x)
46
is
99
C
p
–
46
C
q
.
Then a possible value to p + q is :
(1) 55 (2) 61
(3) 68 (4) 83
Ans. (4)
Sol.
?H ?I ?H ?I ?H ?I 98 96
2 3 97 4
x 1 x x 1 x x 1 x ....... ?K ?K ?K ?K ?K ?K
?H ?I 46
54
x 1 x ?K
Coeff. of
70 98 97 96
68 67 66
x : C C C .......... ?K ?K ?K
47 46
17 16
C C ?K
46 47 98
30 30 30
C C ........... C ?] ?K ?K
?H ?I 46 46 47 98 46
31 30 30 30 31
C C C .. .. .. .. .. . C – C ?] ?K ?K ?K
47 47 98 46
31 30 30 31
C C . . . . . . . . . . . C – C ?] ?K ?K
......
99 46 99 46
31 31 p q
C C C C ?] ?M ?] ?M
Possible values of (p + q) are 62, 83, 99, 46
?? p + q = 83
7. Let
?H ?I e
2 tan x 1
dx x log sin x cos x C
3 tan x 2
?M ?] ?? ?K ?? ?K ?? ?K ?K ?? , where C is the constant of integration.
Then
?? ???K
?? is equal to :
(1) 3 (2) 1
(3) 4 (4) 7
Ans. (3)
Sol.
2 tan x
dx
3 tan x
?M ?K ??
2cos x sin x
dx
3cos x sin x
?M ?] ?K ??
2 cosx – sinx = A(3cosx + sinx) + B(cosx – 3sinx)
3A + B = 2
A – 3B = –1
?@ ??
1 1
A ,B
2 2
?]?]
2cos x sin x
dx
3cos x sin x
?M ?| ?K ??
x 1
ln 3cos x sin x C
2 2
?] ?K ?K ?K
?H ?I 1
x ln 3cos x sin x C
2
?] ?K ?K ?K
?H ?I 1
x ln sin x cos x C
2
?] ?? ?K ?? ?K ?? ?K
?? = 1, ?? = 1, ?? = 3
3
1 4
1
?? ?| ?? ?K ?] ?K ?] ??
8. A variable line L passes through the point (3, 5)
and intersects the positive coordinate axes at the
points A and B. The minimum area of the triangle
OAB, where O is the origin, is :
(1) 30 (2) 25
(3) 40 (4) 35
Ans. (1)
Sol.
x y
1
a b
?K?]
3 5
1
a b
?K?] ?? ?@ 5a
b ,a 3
a 3
?]?^
?M
(0, b)
O A
(a, 0)
(3, 5)
B
?H ?I 1 1 5a
A ab a
2 2 a 3
?]?]
?M 5 a
2 a 3
?R ?]??
?M
2
5 a 9 9
2 a 3
???? ?M?K
?]????
?M ????
5 9
a 3
2 a 3
?] ?K ?K????
?M ????
5 9
a 3 6 30
2 a 3
????
?] ?M ?K ?K ?? ????
?M ????
9. Let
?H ?I ?H ?I ?{ ?} 1
cos cos 60 cos 60 , 0,2
8
?? ?M ?? ?M ?? ?? ?? ?? ??
Then, the sum of all ?{ ?} 0, 2 ?? ?? ?? , where cos 3 ??
attains its maximum value, is :
(1) 9 ?? (2) 18 ?@??
(3) 6 ?@?? (4) 15 ?@??
Ans. (3)
Sol. We know that
(cos ?? ) (cos (60° – ?? ) (cos (60° + ?? )
1
cos3
4
?]??
So equation reduces to
1 1
cos3
4 8
????
?? ?@ 1
cos3
2
????
?? ?@ 1 1
cos3
2 2
?M ?? ?? ??
?? maximum value of cos3 ?? =
1
2
, here
?? 3 2n
3
?? ?? ?] ?? ??
2n
3 9
????
?? ?] ??
As ?? ?? [0, 2 ?? ] possible values are
5 7 11 13 17
, , , , ,
9 9 9 9 9 9
?? ?? ?? ?? ?? ?? ????
???]
????
????
Whose sum is
5 7 11 13 17
9 9 9 9 9 9
?? ?? ?? ?? ?? ?? ?K ?K ?K ?K ?K
54
6
9
?? ?] ?] ??
????
Page 5
FINAL JEE –MAIN EXAMINATION – APRIL, 2024
(Held On Tuesday 09
th
April, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A
1. Let the line L intersect the lines
x – 2 = –y = z – 1, 2 (x + 1) = 2(y – 1) = z + 1
and be parallel to the line
x 2 y 1 z 2
3 1 2
?M ?M ?M ?]?] .
Then which of the following points lies on L ?
(1)
1
– , 1 , 1
3
????
????
????
(2)
1
– , 1 , 1
3
????
?M ????
????
(3)
1
– , 1 , 1
3
????
?M?M
????
????
(4)
1
– , 1 , 1
3
????
?M ????
????
Ans. (2)
Sol.
dr's of line (3, 1, 2)
M(2+ ?? , – ?@ ?? , 1+ ?@ ?? ?I
N 1 , 1 , –1
2 2
???? ????
?M ?K ?K ?K ?? ????
????
L
1
L
2
1
x 2 y z 1
L :
1 1 1
?M?M
?] ?] ?] ?? ?M 2
x 1 y 1 z 1
L :
1 1
1
2 2
?K ?M ?K ?] ?] ?] ?? dr of line MN will be
3 , 1 ,2
2 2
????
?\ ?K ?? ?M ?M ?M ?? ?M ?K ?? ?M ?? ?^ & it will be
proportional to <3, 1, 2>
3 1
2
2 2
3 1 2
????
?K ?? ?M ?M ?M ?? ?M ?K ?? ?M ?? ?| ?] ?] 4 ?? + ?? = –6 4 + 3 ?? = 0
?@ ?? ?@ 4 2
&
3 3
?? ?] ?M ?? ?] ?M ?|?@ Coordinate of M will be <
2 4 1
, ,
3 3 3
????
?M ????
????
and equation of required line will be.
2 4 1
x y z
3 3 3
k
3 1 2
?M ?M ?K ?] ?] ?] So any point on this line will be
2 4 1
3k, k, 2k
3 3 3
????
?K ?K ?M ?K ????
????
2 1
3k
3 3
?K ?] ?M ?? ?@ 1
k
3
?]?M
?|?@ Point lie on the line for
1
k
3
?]?M is
1
,1, 1
3
????
?M?M
????
????
?@ 2. The parabola y
2
= 4x divides the area of the circle
x
2
+ y
2
= 5 in two parts. The area of the smaller
part is equal to :
(1)
1
2 2
5sin
3 5
?M????
?K ????
????
(2)
1
1 2
5sin
3 5
?M????
?K ????
????
(3)
1
1 2
5 sin
3 5
?M????
?K ????
????
(4)
1
2 2
5 sin
3 5
?M????
?K ????
????
Ans. (1)
Sol. y
2
= 4x
x
2
+ y
2
= 5
?| ?@ Area of shaded region as shown in the figure
will be
A(1,2)
B(1, –2)
C( 5, 0)
1 5
2
1
0 1
A 4x dx 5 x dx ?] ?K ?M ????
1
5
3
2 1
2
1
0
4 x 5 x
x 5 x sin
3 2 2 5
?M ????
????
?] ?? ?K ?M ?K ????
????
????
????
1
1 5 5 1
sin
3 4 2 5
?M ?? ????
?] ?K ?M ????
????
?@ ?| ?@ Required Area = 2 A
1
1
2 5 1
5sin
3 2 5
?M ?? ????
?] ?K ?M ????
????
1
2 1
5 sin
3 2 5
?M ??????
?] ?K ?M????
????
1
2 1
5cos
3 5
?M ?]?K
1
2 2
5sin
3 5
?M????
?]?K
????
????
3. The solution curve, of the differential equation
dy dy
2y 3 5
dx dx
?K?] , passing through the point
(0, 1) is a conic, whose vertex lies on the line :
(1) 2x + 3y = 9 (2) 2x + 3y = –9
(3) 2x + 3y = –6 (4) 2x + 3y = 6
Ans. (1)
Sol. ?H ?I dy
2y 5 3
dx
?M ?] ?M
?H ?I 2y 5 dy 3dx ?M ?] ?M
2
y
2 5y 3x
2
?? ?M ?] ?M ?K ??
Curve passes through (0, 1)
?@ ???@?? = – 4
Curve will be
2
5 3
y 3 x
2 4
?? ?? ?? ?? ?M ?] ?M ?M ?? ?? ?? ?? ?? ?? ?? ??
?| Vertex of parabola will be
3 5
,
4 2
????
????
????
2x + 3y = 9
4. A ray of light coming from the point P (1, 2) gets
reflected from the point Q on the x-axis and then
passes through the point R (4, 3). If the point S (h,
k) is such that PQRS is a parallelogram, then hk
2
is
equal to :
(1) 80 (2) 90
(3) 60 (4) 70
Ans. (4)
Sol.
Image of P wrt x-axis will be P'(1, –2) equation of
line joining P'R will be
?H ?I 5
y 3 x 4
3
?M ?] ?M
Above line will meet x-axis at Q where
11
y 0 x
5
?] ?? ?]
11
Q ,0
5
????
?| ????
????
PQRS is parallelogram so their diagonals will
bisects each other
?? ?@ 11
h
4 1 2 3 k 0
5
&
2 2 2 2
?K ?K ?K ?K ?]?]
?@?? ?@ 14
h & k 5
5
?]?]
2 2
14
hk 5 70
5
?| ?] ?? ?] ?@ 5. Let ?? , ?? ?? R. If the system of equations
3x + 5y + ?? z = 3
7x + 11y –9z = 2
97x + 155y – 189z = ?? ?@ has infinitely many solutions, then ?? +2 ?? is equal
to :
(1) 25 (2) 24
(3) 27 (4) 22
Ans. (1)
Sol. 3x + 5y + ?? z = 3
7x + 11y –9z = 2
97x + 155y – 189z = ?? ?@ ?@ ?@?@?@ 93x + 155y + 31 ?? z = 93
97x + 155y – 189z = ?? ?@ ?@ – – + –
–4x + (31 ?? + 189)z = 93 – ?? ?@ 1085x + 1705y – 1395z = 310
1067x + 1705y – 2079z = 11 ??
?@ – – + –
18x + 684z = 310 – 11?? ?@ ?@ –36x + 9(31 ?? + 189)z = 9(93 – ?? )
36x + 1368z = 2 (310 – 11 ?? )
(279 ?? ?@ + 3069)z = 1457 – 31 ?? ?@ ?@ for infinite solutions -
3069 341
279 31
?M?M
?? ?] ?]
1457
31
???]
1457 682 775
2 25
31 31
?M ?? ?K ?? ?] ?] ?]
6. The coefficient of x
70
in x
2
(1 + x)
98
+ x
3
(1 + x)
97
+
x
4
(1 + x)
96
+ ........ +x
54
(1 + x)
46
is
99
C
p
–
46
C
q
.
Then a possible value to p + q is :
(1) 55 (2) 61
(3) 68 (4) 83
Ans. (4)
Sol.
?H ?I ?H ?I ?H ?I 98 96
2 3 97 4
x 1 x x 1 x x 1 x ....... ?K ?K ?K ?K ?K ?K
?H ?I 46
54
x 1 x ?K
Coeff. of
70 98 97 96
68 67 66
x : C C C .......... ?K ?K ?K
47 46
17 16
C C ?K
46 47 98
30 30 30
C C ........... C ?] ?K ?K
?H ?I 46 46 47 98 46
31 30 30 30 31
C C C .. .. .. .. .. . C – C ?] ?K ?K ?K
47 47 98 46
31 30 30 31
C C . . . . . . . . . . . C – C ?] ?K ?K
......
99 46 99 46
31 31 p q
C C C C ?] ?M ?] ?M
Possible values of (p + q) are 62, 83, 99, 46
?? p + q = 83
7. Let
?H ?I e
2 tan x 1
dx x log sin x cos x C
3 tan x 2
?M ?] ?? ?K ?? ?K ?? ?K ?K ?? , where C is the constant of integration.
Then
?? ???K
?? is equal to :
(1) 3 (2) 1
(3) 4 (4) 7
Ans. (3)
Sol.
2 tan x
dx
3 tan x
?M ?K ??
2cos x sin x
dx
3cos x sin x
?M ?] ?K ??
2 cosx – sinx = A(3cosx + sinx) + B(cosx – 3sinx)
3A + B = 2
A – 3B = –1
?@ ??
1 1
A ,B
2 2
?]?]
2cos x sin x
dx
3cos x sin x
?M ?| ?K ??
x 1
ln 3cos x sin x C
2 2
?] ?K ?K ?K
?H ?I 1
x ln 3cos x sin x C
2
?] ?K ?K ?K
?H ?I 1
x ln sin x cos x C
2
?] ?? ?K ?? ?K ?? ?K
?? = 1, ?? = 1, ?? = 3
3
1 4
1
?? ?| ?? ?K ?] ?K ?] ??
8. A variable line L passes through the point (3, 5)
and intersects the positive coordinate axes at the
points A and B. The minimum area of the triangle
OAB, where O is the origin, is :
(1) 30 (2) 25
(3) 40 (4) 35
Ans. (1)
Sol.
x y
1
a b
?K?]
3 5
1
a b
?K?] ?? ?@ 5a
b ,a 3
a 3
?]?^
?M
(0, b)
O A
(a, 0)
(3, 5)
B
?H ?I 1 1 5a
A ab a
2 2 a 3
?]?]
?M 5 a
2 a 3
?R ?]??
?M
2
5 a 9 9
2 a 3
???? ?M?K
?]????
?M ????
5 9
a 3
2 a 3
?] ?K ?K????
?M ????
5 9
a 3 6 30
2 a 3
????
?] ?M ?K ?K ?? ????
?M ????
9. Let
?H ?I ?H ?I ?{ ?} 1
cos cos 60 cos 60 , 0,2
8
?? ?M ?? ?M ?? ?? ?? ?? ??
Then, the sum of all ?{ ?} 0, 2 ?? ?? ?? , where cos 3 ??
attains its maximum value, is :
(1) 9 ?? (2) 18 ?@??
(3) 6 ?@?? (4) 15 ?@??
Ans. (3)
Sol. We know that
(cos ?? ) (cos (60° – ?? ) (cos (60° + ?? )
1
cos3
4
?]??
So equation reduces to
1 1
cos3
4 8
????
?? ?@ 1
cos3
2
????
?? ?@ 1 1
cos3
2 2
?M ?? ?? ??
?? maximum value of cos3 ?? =
1
2
, here
?? 3 2n
3
?? ?? ?] ?? ??
2n
3 9
????
?? ?] ??
As ?? ?? [0, 2 ?? ] possible values are
5 7 11 13 17
, , , , ,
9 9 9 9 9 9
?? ?? ?? ?? ?? ?? ????
???]
????
????
Whose sum is
5 7 11 13 17
9 9 9 9 9 9
?? ?? ?? ?? ?? ?? ?K ?K ?K ?K ?K
54
6
9
?? ?] ?] ??
????
10. Let OA 2a,OB 6a 5b ?] ?] ?K and OC 3b ?] ,
where O is the origin. If the area of the
parallelogram with adjacent sides OA and OC is
15 sq. units, then the area (in sq. units) of the
quadrilateral OABC is equal to :
(1) 38 (2) 40
(3) 32 (4) 35
Ans. (4)
Sol.
C
(3b)
O A (2a)
B
(6a
T
5b)
Area of parallelogram having sides
OA & OC OA OC 2a 3b 15 ?] ?? ?] ?? ?]
6 a b 15 ???]
??
5
a b
2
???] ........(1)
Area of quadrilateral
1 2
1
OABC d d
2
?]??
1
AC OB
2
?]??
?H ?I ?H ?I 1
3b 2a 6a 5b
2
?] ?M ?? ?K
1
1 8 b a – 1 0 a b 1 4 a b
2
?] ?? ?? ?] ??
5
14 35
2
?] ?? ?]
11. If the domain of the function
?H ?I 1
x 1
f x sin
2x 3
?M ?M ????
?] ????
?K ????
is R – ( ?? , ?? )
then 12???? is equal to :
(1) 36 (2) 24
(3) 40 (4) 32
Ans. (4)
Sol. Domain of
1
x 1
f (x) sin
2x 3
?M ?M ????
?] ????
?K ????
is
?H ?I x 1
3
2x 3 0 & x and 1
2 2x 3
?M ?M ?K ?? ?? ?? ?K
x 1 2x 3 ?M ?? ?K
y = |2x + 3|
y = |x – 1|
x = 1
x = – 4
For 2x 3 x 1 ?K ?? ?M
?H ?} 2
x – , 4 ,
3
????
?? ?? ?M ?? ?M ??????
????
2
4& :12 32
3
?? ?] ?M ?? ?] ?M ???? ?]
12. If the sum of series
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 1 1 1
.......
1 1 d 1 d 1 2d 1 9d 1 10d
?K ?K ?K ?? ?K ?K ?K ?K ?K
is equal to 5, then 50d is equal to :
(1) 20 (2) 5
(3) 15 (4) 10
Ans. (2)
Sol.
?H ?I ?H ?I ?H ?I 1 1
.............
1 1 d 1 d 1 2d
?K?K
?? ?K ?K ?K
?H ?I ?H ?I 1
5
1 9d 1 10d
?] ?K?K
Read More
|
1 videos|239 docs|217 tests
|
About this Document
Oct 03, 2025
Last updated
Related Exams
Document Description: JEE Main 2024 April 9 Shift 1 Paper & Solutions for JEE 2025 is part of JEE Main & Advanced Mock Test Series 2025 preparation.
The notes and questions for JEE Main 2024 April 9 Shift 1 Paper & Solutions have been prepared according to the JEE exam syllabus. Information about JEE Main 2024 April 9 Shift 1 Paper & Solutions covers topics
like and JEE Main 2024 April 9 Shift 1 Paper & Solutions Example, for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main 2024 April 9 Shift 1 Paper & Solutions.
Introduction of JEE Main 2024 April 9 Shift 1 Paper & Solutions in English is available as part of our JEE Main & Advanced Mock Test Series 2025
for JEE & JEE Main 2024 April 9 Shift 1 Paper & Solutions in Hindi for JEE Main & Advanced Mock Test Series 2025 course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: JEE Main 2024 April 9 Shift 1 Paper & Solutions | JEE Main & Advanced Mock Test Series 2025
Description
Full syllabus notes, lecture & questions for JEE Main 2024 April 9 Shift 1 Paper & Solutions | JEE Main & Advanced Mock Test Series 2025 - JEE | Plus excerises question with solution to help you revise complete syllabus for JEE Main & Advanced Mock Test Series 2025 | Best notes, free PDF download
Information about JEE Main 2024 April 9 Shift 1 Paper & Solutions
In this doc you can find the meaning of JEE Main 2024 April 9 Shift 1 Paper & Solutions defined & explained in the simplest way possible. Besides explaining types of
JEE Main 2024 April 9 Shift 1 Paper & Solutions theory, EduRev gives you an ample number of questions to practice JEE Main 2024 April 9 Shift 1 Paper & Solutions tests, examples and also practice JEE
tests
Related Searches
JEE Main 2024 April 9 Shift 1 Paper & Solutions | JEE Main & Advanced Mock Test Series 2025
,study material
,video lectures
,past year papers
,shortcuts and tricks
,Extra Questions
,Objective type Questions
,JEE Main 2024 April 9 Shift 1 Paper & Solutions | JEE Main & Advanced Mock Test Series 2025
,Previous Year Questions with Solutions
,Free
,Viva Questions
,Semester Notes
,MCQs
,Sample Paper
,Important questions
,mock tests for examination
,Summary
,ppt
,practice quizzes
,Exam
,JEE Main 2024 April 9 Shift 1 Paper & Solutions | JEE Main & Advanced Mock Test Series 2025
,
Additional Information about JEE Main 2024 April 9 Shift 1 Paper & Solutions for JEE Preparation
JEE Main 2024 April 9 Shift 1 Paper & Solutions Free PDF Download
The JEE Main 2024 April 9 Shift 1 Paper & Solutions is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the JEE Main 2024 April 9 Shift 1 Paper & Solutions now and kickstart your journey towards success in the JEE exam.
Importance of JEE Main 2024 April 9 Shift 1 Paper & Solutions
The importance of JEE Main 2024 April 9 Shift 1 Paper & Solutions cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
JEE Main 2024 April 9 Shift 1 Paper & Solutions Notes
JEE Main 2024 April 9 Shift 1 Paper & Solutions Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to JEE Main 2024 April 9 Shift 1 Paper & Solutions.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, JEE Main 2024 April 9 Shift 1 Paper & Solutions Notes on EduRev are your ultimate resource for success.
JEE Main 2024 April 9 Shift 1 Paper & Solutions JEE Questions
The "JEE Main 2024 April 9 Shift 1 Paper & Solutions JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study JEE Main 2024 April 9 Shift 1 Paper & Solutions on the App
Students of JEE can study JEE Main 2024 April 9 Shift 1 Paper & Solutions alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the JEE Main 2024 April 9 Shift 1 Paper & Solutions,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of JEE Main 2024 April 9 Shift 1 Paper & Solutions is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily