JEE Main 2024 April 4 Shift 2 Paper & Solutions | JEE Main & Advanced Mock Test Series 2025 PDF Download

 Page 1


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Thursday 04
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM 
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If the function ?H ?I ?? ?M ?M ?K ?? ?? ?] ?M?K ?? ?? ?] ?? x x x
e e
72 9 8 1
, x 0
f x
2 1 cosx
a log 2log 3 , x 0
is continuous at x = 0, then the value of a
2
 is equal 
to  
(1) 968 (2) 1152
(3) 746 (4) 1250
Ans. (2) 
Sol. ?H ?I ?? ?] x 0
lim f x a n2 n3
?H ?I ?H ?I ????
?M?M
?M ?M ?K ?] ?M ?K ?M ?K x x
x x x
n 0 x 0
8 1 9 1
72 9 8 1
lim lim
2 1 cosx 2 1 cosx
?H ?I ?? ?? ?? ?? ?? ?? ?? ?M?M
?K?K
?? ?? ?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? x x 2
n 0
8 1 9 1 x
lim 2 1 cosx
x x 1 cosx
?| ?? ?? ?? ?] n8 n9 2 2 2 24 2 n2 n3
?| a = 24 2 , a
2
 = 576 × 2 = 1152
2. If ?? > 0, let ?? be the angle between the vectors
?] ?K ?? ?Mˆˆ ˆ
a i j 3k and ?] ?M ?K ˆˆ ˆ
b 3i j 2k . If the vectors
?K a b and ?M a b are mutually perpendicular, then
the value of (14 cos ?? )
2
 is equal to
(1) 25 (2) 20
(3) 50 (4) 40
Ans. (1) 
Sol. 
?H ?I ?H ?I ?K ?M ?] a b . a b 0 , ?? ?@?^ 0
?M?]
2
2
a b 0 ?? 1 + ?? 2
+ 9 = 9 + 1 + 4 
?| ?? = 2, 
?M ?M ?? ?M ?? ?] ?] a b 3 6
cos
a . b 14. 14
14cos ?? = 3 – 8 = –5 
?| (14 cos ?? )
2
 = 25
3. Let C be a circle with radius 10 units and centre
at the origin. Let the line x + y = 2 intersects the
circle C at the points P and Q. Let MN be a chord
of C of length 2 unit and slope –1. Then, a distance
(in units) between the chord PQ and the chord MN
is
(1) ?M 2 3 (2) ?M 3 2
(3) ?M 2 1 (4) ?K 2 1
Ans. (2) 
M 
N 
O 
A 
10
C : x
2
 + y
2
 = 10 
AN = ?] MN
1
2
?| In ?d OAN ?? (ON)
2
 = (OA)
2
 + (AN)
2 
10 = (OA)
2
 + 1 ?? ?@ OA = 3 
Perpendicular distance of center from 
?K?M
?]?]
0 0 2
PQ 2
2
Perpendicular distance between MN and 
?]?K PQ OA 2 or ?M OA 2
?]?K3 2 or ?M 3 2
Page 2


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Thursday 04
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM 
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If the function ?H ?I ?? ?M ?M ?K ?? ?? ?] ?M?K ?? ?? ?] ?? x x x
e e
72 9 8 1
, x 0
f x
2 1 cosx
a log 2log 3 , x 0
is continuous at x = 0, then the value of a
2
 is equal 
to  
(1) 968 (2) 1152
(3) 746 (4) 1250
Ans. (2) 
Sol. ?H ?I ?? ?] x 0
lim f x a n2 n3
?H ?I ?H ?I ????
?M?M
?M ?M ?K ?] ?M ?K ?M ?K x x
x x x
n 0 x 0
8 1 9 1
72 9 8 1
lim lim
2 1 cosx 2 1 cosx
?H ?I ?? ?? ?? ?? ?? ?? ?? ?M?M
?K?K
?? ?? ?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? x x 2
n 0
8 1 9 1 x
lim 2 1 cosx
x x 1 cosx
?| ?? ?? ?? ?] n8 n9 2 2 2 24 2 n2 n3
?| a = 24 2 , a
2
 = 576 × 2 = 1152
2. If ?? > 0, let ?? be the angle between the vectors
?] ?K ?? ?Mˆˆ ˆ
a i j 3k and ?] ?M ?K ˆˆ ˆ
b 3i j 2k . If the vectors
?K a b and ?M a b are mutually perpendicular, then
the value of (14 cos ?? )
2
 is equal to
(1) 25 (2) 20
(3) 50 (4) 40
Ans. (1) 
Sol. 
?H ?I ?H ?I ?K ?M ?] a b . a b 0 , ?? ?@?^ 0
?M?]
2
2
a b 0 ?? 1 + ?? 2
+ 9 = 9 + 1 + 4 
?| ?? = 2, 
?M ?M ?? ?M ?? ?] ?] a b 3 6
cos
a . b 14. 14
14cos ?? = 3 – 8 = –5 
?| (14 cos ?? )
2
 = 25
3. Let C be a circle with radius 10 units and centre
at the origin. Let the line x + y = 2 intersects the
circle C at the points P and Q. Let MN be a chord
of C of length 2 unit and slope –1. Then, a distance
(in units) between the chord PQ and the chord MN
is
(1) ?M 2 3 (2) ?M 3 2
(3) ?M 2 1 (4) ?K 2 1
Ans. (2) 
M 
N 
O 
A 
10
C : x
2
 + y
2
 = 10 
AN = ?] MN
1
2
?| In ?d OAN ?? (ON)
2
 = (OA)
2
 + (AN)
2 
10 = (OA)
2
 + 1 ?? ?@ OA = 3 
Perpendicular distance of center from 
?K?M
?]?]
0 0 2
PQ 2
2
Perpendicular distance between MN and 
?]?K PQ OA 2 or ?M OA 2
?]?K3 2 or ?M 3 2
4. Let a relation R on ?? be defined as :  
 (x
1
,y
1
) R(x
2
,y
2
) if and only if x
1
 < x
2
 or y
1
 < y
2
  
 Consider the two statements :  
 (I) R is reflexive but not symmetric. 
 (II) R is transitive  
Then which one of the following is true ?  
 (1) Only (II) is correct.  
 (2) Only (I) is correct. 
 (3) Both (I) and (II) are correct. 
 (4) Neither (I) nor (II) is correct. 
 Ans. (2) 
Sol. All ((x
1
y
1
), (x
1
,y
1
)) are in R where  
 x
1
, y
1
 ?? N ?| R is reflexive 
 ((1,1), (2,3)) ?? R but ((2,3), (1,1)) ?? R 
 ?| R is not symmetric  
 ((2,4), (3,3)) ?? R and ((3,3), (1,3)) ?? R but ((2,4), 
(1,3)) ?? R 
 ?| R is not transitive  
5. Let three real numbers a,b,c be in arithmetic 
progression and a + 1, b, c + 3 be in geometric 
progression. If a > 10 and the arithmetic mean of 
a,b and c is 8, then the cube of the geometric mean 
of a,b and c is  
 (1) 120 (2) 312 
 (3) 316  (4) 128 
 Ans. (1) 
Sol. 2b = a + c, b
2
 = (a + 1) (c + 3), 
 
?K?K
?] ?? ?] ?K ?] a b c
8 b 8,a c 16
3
 
 64 = (a + 1) (19 – a) = 19 + 18a – a
2
 
 a
2
 – 18a – 45 = 0 ?? (a – 15) (a + 3) = 0, (a > 10) 
 a = 15, c = 1, b = 8 
 ((abc)
1/3
)
3
 = abc = 120 
6. Let 
????
?]????
????
1 2
A
0 1
 and B = I + adj(A) + (adj A)
2
+…+ 
(adj A)
10
. Then, the sum of all the elements of the 
matrix B is :  
 (1) –110 (2) 22 
 (3) –88  (4) –124 
 Ans. (3) 
Sol. Adj(A) = 
?M ????
????
????
1 2
0 1
  
 (AdjA)
2
 = 
?M ????
????
????
1 4
0 1
 
    | 
    | 
 ?H ?I ?M ????
?]????
????
10
1 20
AdjA
0 1
 
 
?M ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?] ?K ?K ?K ?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 1 0 1 2 1 4 1 20
B ...
0 1 0 1 0 1 0 1
 
 
?M ????
?]????
????
11 110
B
0 11
 ?? sum of elements of B  
 = –88 
7. The value of 
?H ?I ?? ?K ?? ?K ?K ?? ?? ?K ?? ?K ?K ?? 2
2 2
2 2 2
1 2 2 3 ... 100 101
1 2 2 3 ... 100 101
 is  
 (1) 
306
305
 (2) 
305
301
 
 (3) 
32
31
  (4) 
31
30
 
 Ans. (2) 
Sol. 
?H ?I ?H ?I ?H ?I ?] ?] ?K ?? ?K ?? ?K ?K ?? ?] ?? ?K ?? ?K ?K ?? ?K ?? ?? 100
2
2
2 2
r 1
2 2 2 100
2
r 1
r r 1
1 2 2 3 ... 100 101
1 2 2 3 ... 100 101
r r 1
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?] ?] ????
?K ?K ?K ?K ?????K?K
?K?K
????
????
?]?]
???? ?K ?K ?K ?K ?K ????
????
?? ?? 2
100
3 2
r 1
100 2
3 2
r 1
n n 1 2.n n 1 2n 1 n n 1
r 2r r
2 6 2
n n 1 n n 1 2n 1
r r
2 6
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ???? ?K?K
?K ?K ?K ????
????
?] ???? ?K ?K ?K ?K ????
????
n n 1 n n 1
2
. 2n 1 1
2 2 3
n n 1 n n 1 2n 1
2 2 3
;Put n = 100 
 
?H ?I ?H ?I ?K?K
?] ?] ?] ?? ?K 100 101
2
201 1
5185 305
2 3
100 101 201
5117 301
2 3
 
Page 3


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Thursday 04
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM 
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If the function ?H ?I ?? ?M ?M ?K ?? ?? ?] ?M?K ?? ?? ?] ?? x x x
e e
72 9 8 1
, x 0
f x
2 1 cosx
a log 2log 3 , x 0
is continuous at x = 0, then the value of a
2
 is equal 
to  
(1) 968 (2) 1152
(3) 746 (4) 1250
Ans. (2) 
Sol. ?H ?I ?? ?] x 0
lim f x a n2 n3
?H ?I ?H ?I ????
?M?M
?M ?M ?K ?] ?M ?K ?M ?K x x
x x x
n 0 x 0
8 1 9 1
72 9 8 1
lim lim
2 1 cosx 2 1 cosx
?H ?I ?? ?? ?? ?? ?? ?? ?? ?M?M
?K?K
?? ?? ?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? x x 2
n 0
8 1 9 1 x
lim 2 1 cosx
x x 1 cosx
?| ?? ?? ?? ?] n8 n9 2 2 2 24 2 n2 n3
?| a = 24 2 , a
2
 = 576 × 2 = 1152
2. If ?? > 0, let ?? be the angle between the vectors
?] ?K ?? ?Mˆˆ ˆ
a i j 3k and ?] ?M ?K ˆˆ ˆ
b 3i j 2k . If the vectors
?K a b and ?M a b are mutually perpendicular, then
the value of (14 cos ?? )
2
 is equal to
(1) 25 (2) 20
(3) 50 (4) 40
Ans. (1) 
Sol. 
?H ?I ?H ?I ?K ?M ?] a b . a b 0 , ?? ?@?^ 0
?M?]
2
2
a b 0 ?? 1 + ?? 2
+ 9 = 9 + 1 + 4 
?| ?? = 2, 
?M ?M ?? ?M ?? ?] ?] a b 3 6
cos
a . b 14. 14
14cos ?? = 3 – 8 = –5 
?| (14 cos ?? )
2
 = 25
3. Let C be a circle with radius 10 units and centre
at the origin. Let the line x + y = 2 intersects the
circle C at the points P and Q. Let MN be a chord
of C of length 2 unit and slope –1. Then, a distance
(in units) between the chord PQ and the chord MN
is
(1) ?M 2 3 (2) ?M 3 2
(3) ?M 2 1 (4) ?K 2 1
Ans. (2) 
M 
N 
O 
A 
10
C : x
2
 + y
2
 = 10 
AN = ?] MN
1
2
?| In ?d OAN ?? (ON)
2
 = (OA)
2
 + (AN)
2 
10 = (OA)
2
 + 1 ?? ?@ OA = 3 
Perpendicular distance of center from 
?K?M
?]?]
0 0 2
PQ 2
2
Perpendicular distance between MN and 
?]?K PQ OA 2 or ?M OA 2
?]?K3 2 or ?M 3 2
4. Let a relation R on ?? be defined as :  
 (x
1
,y
1
) R(x
2
,y
2
) if and only if x
1
 < x
2
 or y
1
 < y
2
  
 Consider the two statements :  
 (I) R is reflexive but not symmetric. 
 (II) R is transitive  
Then which one of the following is true ?  
 (1) Only (II) is correct.  
 (2) Only (I) is correct. 
 (3) Both (I) and (II) are correct. 
 (4) Neither (I) nor (II) is correct. 
 Ans. (2) 
Sol. All ((x
1
y
1
), (x
1
,y
1
)) are in R where  
 x
1
, y
1
 ?? N ?| R is reflexive 
 ((1,1), (2,3)) ?? R but ((2,3), (1,1)) ?? R 
 ?| R is not symmetric  
 ((2,4), (3,3)) ?? R and ((3,3), (1,3)) ?? R but ((2,4), 
(1,3)) ?? R 
 ?| R is not transitive  
5. Let three real numbers a,b,c be in arithmetic 
progression and a + 1, b, c + 3 be in geometric 
progression. If a > 10 and the arithmetic mean of 
a,b and c is 8, then the cube of the geometric mean 
of a,b and c is  
 (1) 120 (2) 312 
 (3) 316  (4) 128 
 Ans. (1) 
Sol. 2b = a + c, b
2
 = (a + 1) (c + 3), 
 
?K?K
?] ?? ?] ?K ?] a b c
8 b 8,a c 16
3
 
 64 = (a + 1) (19 – a) = 19 + 18a – a
2
 
 a
2
 – 18a – 45 = 0 ?? (a – 15) (a + 3) = 0, (a > 10) 
 a = 15, c = 1, b = 8 
 ((abc)
1/3
)
3
 = abc = 120 
6. Let 
????
?]????
????
1 2
A
0 1
 and B = I + adj(A) + (adj A)
2
+…+ 
(adj A)
10
. Then, the sum of all the elements of the 
matrix B is :  
 (1) –110 (2) 22 
 (3) –88  (4) –124 
 Ans. (3) 
Sol. Adj(A) = 
?M ????
????
????
1 2
0 1
  
 (AdjA)
2
 = 
?M ????
????
????
1 4
0 1
 
    | 
    | 
 ?H ?I ?M ????
?]????
????
10
1 20
AdjA
0 1
 
 
?M ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?] ?K ?K ?K ?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 1 0 1 2 1 4 1 20
B ...
0 1 0 1 0 1 0 1
 
 
?M ????
?]????
????
11 110
B
0 11
 ?? sum of elements of B  
 = –88 
7. The value of 
?H ?I ?? ?K ?? ?K ?K ?? ?? ?K ?? ?K ?K ?? 2
2 2
2 2 2
1 2 2 3 ... 100 101
1 2 2 3 ... 100 101
 is  
 (1) 
306
305
 (2) 
305
301
 
 (3) 
32
31
  (4) 
31
30
 
 Ans. (2) 
Sol. 
?H ?I ?H ?I ?H ?I ?] ?] ?K ?? ?K ?? ?K ?K ?? ?] ?? ?K ?? ?K ?K ?? ?K ?? ?? 100
2
2
2 2
r 1
2 2 2 100
2
r 1
r r 1
1 2 2 3 ... 100 101
1 2 2 3 ... 100 101
r r 1
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?] ?] ????
?K ?K ?K ?K ?????K?K
?K?K
????
????
?]?]
???? ?K ?K ?K ?K ?K ????
????
?? ?? 2
100
3 2
r 1
100 2
3 2
r 1
n n 1 2.n n 1 2n 1 n n 1
r 2r r
2 6 2
n n 1 n n 1 2n 1
r r
2 6
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ???? ?K?K
?K ?K ?K ????
????
?] ???? ?K ?K ?K ?K ????
????
n n 1 n n 1
2
. 2n 1 1
2 2 3
n n 1 n n 1 2n 1
2 2 3
;Put n = 100 
 
?H ?I ?H ?I ?K?K
?] ?] ?] ?? ?K 100 101
2
201 1
5185 305
2 3
100 101 201
5117 301
2 3
 
 
8. Let ?H ?I ?H ?I ?H ?I ?] ?K ?M ?? ?? x
t
0
f x t sin 1 e dt,x .  
 Then 
?H ?I ?? 3
x 0
f x
lim
x
is equal to  
 (1) 
1
6
 (2) ?M 1
6
 
 (3) ?M 2
3
  (4) 
2
3
 
 Ans. (2) 
Sol. 
?H ?I ?? 3
x 0
f x
lim
x
 
 Using L Hopital Rule. 
 
?H ?I ?H ?I ????
?K?M
?] x
2 2
x 0 x 0
x sin 1 e
f ' x
lim lim
3x 3x
 (Again L Hopital) 
 Using L.H. Rule 
 
?H ?I ?H ?I ?H ?I ??????
?M ?M ?M ?K ?M ????
?] x x x x x
x 0
sin 1 e e .e cos 1 e .e
lim
6
 
 ?]?M
1
6
 
9. The area (in sq. units) of the region described by 
{(x,y) : y
2
 < 2x, and y > 4x –1} is  
 (1) 
11
32
 (2) 
8
9
 
 (3) 
11
12
  (4) 
9
32
 
 Ans. (4) 
Sol. 
 
Q 
1
 
P
 
y
 
x
 
0
 
–1/2
 
(0,–1)
 
y'
 
x'
 
(1/4,0
 
 Shaded area 
?H ?I ?M?]?M
?? 1
Right Left
1
2
x x dy 
 
?] ?]?M
?] ?] ?M 2
y 2x
y 4x 1 Solve
1
y 1,y
2
 
 Shaded area 
?M ????
?K ?]?M
????
????
?? 1 2
1
2
y 1 y
dy
4 2
 
 
?M ???? ????
?] ?K ?M ?] ???? ????
????
???? ????
1
2 3
1
2
1 y y 9
y
4 2 6 32
 
10. The area (in sq. units) of the region 
?H ?I ?H ?I ?H ?I ?? ?? ?] ?? ?M ?? ?K ?K ?M ?? ?? S z ; z 1 2; z z i z z 2,lm z 0
 is  
 (1) 
?? 7
3
 (2) 
?? 3
2
 
 (3) 
?? 17
8
  (4) 
?? 7
4
 
 Ans. (2) 
Sol. Put z = x + iy 
 ?H ?I ?M ?? ?? ?M ?K ?? 2
2
z 1 2 x 1 y 4    …(1) 
 
?H ?I ?H ?I ?H ?I ?K ?K ?M ?? ?? ?K ?? z z i z z 2 2x i 2iy 2 
 ?? ?M ?? x y 1    …(2) 
 Im(z) > 0 ?? y > 0   …(3) 
 
 
(–3,0) (1,0) (5,0) 
x–y=1 
?? /4 
 
 Required area  
 = Area of semi-circle – area  of sector A 
 ?H ?I ?? ???M
2 1
2
2 2
 
 
?? ?] 3
2
 
Page 4


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Thursday 04
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM 
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If the function ?H ?I ?? ?M ?M ?K ?? ?? ?] ?M?K ?? ?? ?] ?? x x x
e e
72 9 8 1
, x 0
f x
2 1 cosx
a log 2log 3 , x 0
is continuous at x = 0, then the value of a
2
 is equal 
to  
(1) 968 (2) 1152
(3) 746 (4) 1250
Ans. (2) 
Sol. ?H ?I ?? ?] x 0
lim f x a n2 n3
?H ?I ?H ?I ????
?M?M
?M ?M ?K ?] ?M ?K ?M ?K x x
x x x
n 0 x 0
8 1 9 1
72 9 8 1
lim lim
2 1 cosx 2 1 cosx
?H ?I ?? ?? ?? ?? ?? ?? ?? ?M?M
?K?K
?? ?? ?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? x x 2
n 0
8 1 9 1 x
lim 2 1 cosx
x x 1 cosx
?| ?? ?? ?? ?] n8 n9 2 2 2 24 2 n2 n3
?| a = 24 2 , a
2
 = 576 × 2 = 1152
2. If ?? > 0, let ?? be the angle between the vectors
?] ?K ?? ?Mˆˆ ˆ
a i j 3k and ?] ?M ?K ˆˆ ˆ
b 3i j 2k . If the vectors
?K a b and ?M a b are mutually perpendicular, then
the value of (14 cos ?? )
2
 is equal to
(1) 25 (2) 20
(3) 50 (4) 40
Ans. (1) 
Sol. 
?H ?I ?H ?I ?K ?M ?] a b . a b 0 , ?? ?@?^ 0
?M?]
2
2
a b 0 ?? 1 + ?? 2
+ 9 = 9 + 1 + 4 
?| ?? = 2, 
?M ?M ?? ?M ?? ?] ?] a b 3 6
cos
a . b 14. 14
14cos ?? = 3 – 8 = –5 
?| (14 cos ?? )
2
 = 25
3. Let C be a circle with radius 10 units and centre
at the origin. Let the line x + y = 2 intersects the
circle C at the points P and Q. Let MN be a chord
of C of length 2 unit and slope –1. Then, a distance
(in units) between the chord PQ and the chord MN
is
(1) ?M 2 3 (2) ?M 3 2
(3) ?M 2 1 (4) ?K 2 1
Ans. (2) 
M 
N 
O 
A 
10
C : x
2
 + y
2
 = 10 
AN = ?] MN
1
2
?| In ?d OAN ?? (ON)
2
 = (OA)
2
 + (AN)
2 
10 = (OA)
2
 + 1 ?? ?@ OA = 3 
Perpendicular distance of center from 
?K?M
?]?]
0 0 2
PQ 2
2
Perpendicular distance between MN and 
?]?K PQ OA 2 or ?M OA 2
?]?K3 2 or ?M 3 2
4. Let a relation R on ?? be defined as :  
 (x
1
,y
1
) R(x
2
,y
2
) if and only if x
1
 < x
2
 or y
1
 < y
2
  
 Consider the two statements :  
 (I) R is reflexive but not symmetric. 
 (II) R is transitive  
Then which one of the following is true ?  
 (1) Only (II) is correct.  
 (2) Only (I) is correct. 
 (3) Both (I) and (II) are correct. 
 (4) Neither (I) nor (II) is correct. 
 Ans. (2) 
Sol. All ((x
1
y
1
), (x
1
,y
1
)) are in R where  
 x
1
, y
1
 ?? N ?| R is reflexive 
 ((1,1), (2,3)) ?? R but ((2,3), (1,1)) ?? R 
 ?| R is not symmetric  
 ((2,4), (3,3)) ?? R and ((3,3), (1,3)) ?? R but ((2,4), 
(1,3)) ?? R 
 ?| R is not transitive  
5. Let three real numbers a,b,c be in arithmetic 
progression and a + 1, b, c + 3 be in geometric 
progression. If a > 10 and the arithmetic mean of 
a,b and c is 8, then the cube of the geometric mean 
of a,b and c is  
 (1) 120 (2) 312 
 (3) 316  (4) 128 
 Ans. (1) 
Sol. 2b = a + c, b
2
 = (a + 1) (c + 3), 
 
?K?K
?] ?? ?] ?K ?] a b c
8 b 8,a c 16
3
 
 64 = (a + 1) (19 – a) = 19 + 18a – a
2
 
 a
2
 – 18a – 45 = 0 ?? (a – 15) (a + 3) = 0, (a > 10) 
 a = 15, c = 1, b = 8 
 ((abc)
1/3
)
3
 = abc = 120 
6. Let 
????
?]????
????
1 2
A
0 1
 and B = I + adj(A) + (adj A)
2
+…+ 
(adj A)
10
. Then, the sum of all the elements of the 
matrix B is :  
 (1) –110 (2) 22 
 (3) –88  (4) –124 
 Ans. (3) 
Sol. Adj(A) = 
?M ????
????
????
1 2
0 1
  
 (AdjA)
2
 = 
?M ????
????
????
1 4
0 1
 
    | 
    | 
 ?H ?I ?M ????
?]????
????
10
1 20
AdjA
0 1
 
 
?M ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?] ?K ?K ?K ?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 1 0 1 2 1 4 1 20
B ...
0 1 0 1 0 1 0 1
 
 
?M ????
?]????
????
11 110
B
0 11
 ?? sum of elements of B  
 = –88 
7. The value of 
?H ?I ?? ?K ?? ?K ?K ?? ?? ?K ?? ?K ?K ?? 2
2 2
2 2 2
1 2 2 3 ... 100 101
1 2 2 3 ... 100 101
 is  
 (1) 
306
305
 (2) 
305
301
 
 (3) 
32
31
  (4) 
31
30
 
 Ans. (2) 
Sol. 
?H ?I ?H ?I ?H ?I ?] ?] ?K ?? ?K ?? ?K ?K ?? ?] ?? ?K ?? ?K ?K ?? ?K ?? ?? 100
2
2
2 2
r 1
2 2 2 100
2
r 1
r r 1
1 2 2 3 ... 100 101
1 2 2 3 ... 100 101
r r 1
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?] ?] ????
?K ?K ?K ?K ?????K?K
?K?K
????
????
?]?]
???? ?K ?K ?K ?K ?K ????
????
?? ?? 2
100
3 2
r 1
100 2
3 2
r 1
n n 1 2.n n 1 2n 1 n n 1
r 2r r
2 6 2
n n 1 n n 1 2n 1
r r
2 6
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ???? ?K?K
?K ?K ?K ????
????
?] ???? ?K ?K ?K ?K ????
????
n n 1 n n 1
2
. 2n 1 1
2 2 3
n n 1 n n 1 2n 1
2 2 3
;Put n = 100 
 
?H ?I ?H ?I ?K?K
?] ?] ?] ?? ?K 100 101
2
201 1
5185 305
2 3
100 101 201
5117 301
2 3
 
 
8. Let ?H ?I ?H ?I ?H ?I ?] ?K ?M ?? ?? x
t
0
f x t sin 1 e dt,x .  
 Then 
?H ?I ?? 3
x 0
f x
lim
x
is equal to  
 (1) 
1
6
 (2) ?M 1
6
 
 (3) ?M 2
3
  (4) 
2
3
 
 Ans. (2) 
Sol. 
?H ?I ?? 3
x 0
f x
lim
x
 
 Using L Hopital Rule. 
 
?H ?I ?H ?I ????
?K?M
?] x
2 2
x 0 x 0
x sin 1 e
f ' x
lim lim
3x 3x
 (Again L Hopital) 
 Using L.H. Rule 
 
?H ?I ?H ?I ?H ?I ??????
?M ?M ?M ?K ?M ????
?] x x x x x
x 0
sin 1 e e .e cos 1 e .e
lim
6
 
 ?]?M
1
6
 
9. The area (in sq. units) of the region described by 
{(x,y) : y
2
 < 2x, and y > 4x –1} is  
 (1) 
11
32
 (2) 
8
9
 
 (3) 
11
12
  (4) 
9
32
 
 Ans. (4) 
Sol. 
 
Q 
1
 
P
 
y
 
x
 
0
 
–1/2
 
(0,–1)
 
y'
 
x'
 
(1/4,0
 
 Shaded area 
?H ?I ?M?]?M
?? 1
Right Left
1
2
x x dy 
 
?] ?]?M
?] ?] ?M 2
y 2x
y 4x 1 Solve
1
y 1,y
2
 
 Shaded area 
?M ????
?K ?]?M
????
????
?? 1 2
1
2
y 1 y
dy
4 2
 
 
?M ???? ????
?] ?K ?M ?] ???? ????
????
???? ????
1
2 3
1
2
1 y y 9
y
4 2 6 32
 
10. The area (in sq. units) of the region 
?H ?I ?H ?I ?H ?I ?? ?? ?] ?? ?M ?? ?K ?K ?M ?? ?? S z ; z 1 2; z z i z z 2,lm z 0
 is  
 (1) 
?? 7
3
 (2) 
?? 3
2
 
 (3) 
?? 17
8
  (4) 
?? 7
4
 
 Ans. (2) 
Sol. Put z = x + iy 
 ?H ?I ?M ?? ?? ?M ?K ?? 2
2
z 1 2 x 1 y 4    …(1) 
 
?H ?I ?H ?I ?H ?I ?K ?K ?M ?? ?? ?K ?? z z i z z 2 2x i 2iy 2 
 ?? ?M ?? x y 1    …(2) 
 Im(z) > 0 ?? y > 0   …(3) 
 
 
(–3,0) (1,0) (5,0) 
x–y=1 
?? /4 
 
 Required area  
 = Area of semi-circle – area  of sector A 
 ?H ?I ?? ???M
2 1
2
2 2
 
 
?? ?] 3
2
 
11. If the value of the integral 
?M ?? ?K ?? 1
x
1
cos x
dx
1 3
 is 
?? 2
. 
Then, a value of ?? is  
 (1) 
?? 6
 (2) 
?? 2
 
 (3) 
?? 3
  (4) 
?? 4
 
 Ans. (2) 
Sol. Let 
?K ?M ?? ?] ?K ?? 1
x
1
cos x
I dx
1 3
 …(I) 
 
?K ?M ?M ?? ?] ?K ?? 1
x
1
cos x
I dx
1 3
  
 ?H ?I ?H ?I ????
?] ?K ?M ????
????
????
????
b b
a a
using f x dx f a b x dx   …(II) 
 Add (1) and (II) 
 ?H ?I ?H ?I ?K ?M ?] ?? ?] ?? ????
1 1
1 0
2I cos x dx 2 cos x dx 
 ?H ?I ?? ?]?]
????
sin 2
I given 
 ?| 
?? ???]
2
 
12. Let ?H ?I ?] ?M ?K ?M f x 3 x 2 4 x be a real valued 
function. If ?? and ?? are respectively the minimum 
and the maximum values of f, then ?? 2
 + 2 ?? 2
 is 
equal to  
 (1) 44 (2) 42 
 (3) 24  (4) 38 
 Ans. (2) 
Sol. f(x) = 3 x 2 4 x ?M ?K ?M 
 x – 2 ?? 0  &  4 – x ?? 0 
 ?| x ?? [2, 4] 
 Let x = 2sin
2
?? + 4cos
2
?? 
 ?| f(x) = 3 2 cos 2 sin ?? ?K ?? 
 ?| 2 3 2 cos 2 sin 9 2 2 ?? ?? ?K ?? ?? ?? ?K 
 2 3 2 cos 2 sin 20 ?? ?? ?K ?? ?? 
 ?| ?? = 2 ?? = 20 
 ?? 2
 + 2 ?? 2
 = 2 + 40 = 42 
13. If the coefficients of x
4
, x
5
 and x
6
 in the expansion 
of (1 + x)
n
 are in the arithmetic progression, then 
the maximum value of n is :  
 (1) 14 (2) 21 
 (3) 28  (4) 7 
 Ans. (1) 
Sol. Coeff. of x
4
 = 
n
C
4 
 Coeff. of x
5
 = 
n
C
5 
 
Coeff. of x
6
 = 
n
C
6 
 
n
C
4
, 
n
C
5
, 
n
C
6
 …. AP 
 2.
n
C
5
 = 
n
C
4
 + 
n
C
6
 
 
n n
6 4
n n
5 5
C C
2
C C
?]?K    
n
r
n
r 1
C n r 1
r C
?M????
?M?K ????
?]????
????
????
 
 
5 n 5
2
n 4 6
?M ?]?K
?M 
 12(n – 4) = 30 + n
2
 – 9n + 20 
 n
2
 – 21n + 98 = 0 
 (n – 14) (n – 7) = 0 
 n
max
 = 14 n
min
 = 7 
14. Consider a hyperbola H having centre at the origin 
and foci and the x-axis. Let C
1
 be the circle 
touching the hyperbola H and having the centre at 
the origin. Let C
2
 be the circle touching the 
hyperbola H at its vertex and having the centre at 
one of its foci. If areas (in sq. units) of C
1
 and C
2
 
are 36 ?? and 4 ?? , respectively, then the length (in 
units) of latus rectum of H is   
 (1) 
28
3
 (2) 
14
3
 
 (3) 
10
3
  (4) 
11
3
 
 Ans. (1) 
 
Page 5


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Thursday 04
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM 
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If the function ?H ?I ?? ?M ?M ?K ?? ?? ?] ?M?K ?? ?? ?] ?? x x x
e e
72 9 8 1
, x 0
f x
2 1 cosx
a log 2log 3 , x 0
is continuous at x = 0, then the value of a
2
 is equal 
to  
(1) 968 (2) 1152
(3) 746 (4) 1250
Ans. (2) 
Sol. ?H ?I ?? ?] x 0
lim f x a n2 n3
?H ?I ?H ?I ????
?M?M
?M ?M ?K ?] ?M ?K ?M ?K x x
x x x
n 0 x 0
8 1 9 1
72 9 8 1
lim lim
2 1 cosx 2 1 cosx
?H ?I ?? ?? ?? ?? ?? ?? ?? ?M?M
?K?K
?? ?? ?? ?? ?? ?? ?M ?? ?? ?? ?? ?? ?? x x 2
n 0
8 1 9 1 x
lim 2 1 cosx
x x 1 cosx
?| ?? ?? ?? ?] n8 n9 2 2 2 24 2 n2 n3
?| a = 24 2 , a
2
 = 576 × 2 = 1152
2. If ?? > 0, let ?? be the angle between the vectors
?] ?K ?? ?Mˆˆ ˆ
a i j 3k and ?] ?M ?K ˆˆ ˆ
b 3i j 2k . If the vectors
?K a b and ?M a b are mutually perpendicular, then
the value of (14 cos ?? )
2
 is equal to
(1) 25 (2) 20
(3) 50 (4) 40
Ans. (1) 
Sol. 
?H ?I ?H ?I ?K ?M ?] a b . a b 0 , ?? ?@?^ 0
?M?]
2
2
a b 0 ?? 1 + ?? 2
+ 9 = 9 + 1 + 4 
?| ?? = 2, 
?M ?M ?? ?M ?? ?] ?] a b 3 6
cos
a . b 14. 14
14cos ?? = 3 – 8 = –5 
?| (14 cos ?? )
2
 = 25
3. Let C be a circle with radius 10 units and centre
at the origin. Let the line x + y = 2 intersects the
circle C at the points P and Q. Let MN be a chord
of C of length 2 unit and slope –1. Then, a distance
(in units) between the chord PQ and the chord MN
is
(1) ?M 2 3 (2) ?M 3 2
(3) ?M 2 1 (4) ?K 2 1
Ans. (2) 
M 
N 
O 
A 
10
C : x
2
 + y
2
 = 10 
AN = ?] MN
1
2
?| In ?d OAN ?? (ON)
2
 = (OA)
2
 + (AN)
2 
10 = (OA)
2
 + 1 ?? ?@ OA = 3 
Perpendicular distance of center from 
?K?M
?]?]
0 0 2
PQ 2
2
Perpendicular distance between MN and 
?]?K PQ OA 2 or ?M OA 2
?]?K3 2 or ?M 3 2
4. Let a relation R on ?? be defined as :  
 (x
1
,y
1
) R(x
2
,y
2
) if and only if x
1
 < x
2
 or y
1
 < y
2
  
 Consider the two statements :  
 (I) R is reflexive but not symmetric. 
 (II) R is transitive  
Then which one of the following is true ?  
 (1) Only (II) is correct.  
 (2) Only (I) is correct. 
 (3) Both (I) and (II) are correct. 
 (4) Neither (I) nor (II) is correct. 
 Ans. (2) 
Sol. All ((x
1
y
1
), (x
1
,y
1
)) are in R where  
 x
1
, y
1
 ?? N ?| R is reflexive 
 ((1,1), (2,3)) ?? R but ((2,3), (1,1)) ?? R 
 ?| R is not symmetric  
 ((2,4), (3,3)) ?? R and ((3,3), (1,3)) ?? R but ((2,4), 
(1,3)) ?? R 
 ?| R is not transitive  
5. Let three real numbers a,b,c be in arithmetic 
progression and a + 1, b, c + 3 be in geometric 
progression. If a > 10 and the arithmetic mean of 
a,b and c is 8, then the cube of the geometric mean 
of a,b and c is  
 (1) 120 (2) 312 
 (3) 316  (4) 128 
 Ans. (1) 
Sol. 2b = a + c, b
2
 = (a + 1) (c + 3), 
 
?K?K
?] ?? ?] ?K ?] a b c
8 b 8,a c 16
3
 
 64 = (a + 1) (19 – a) = 19 + 18a – a
2
 
 a
2
 – 18a – 45 = 0 ?? (a – 15) (a + 3) = 0, (a > 10) 
 a = 15, c = 1, b = 8 
 ((abc)
1/3
)
3
 = abc = 120 
6. Let 
????
?]????
????
1 2
A
0 1
 and B = I + adj(A) + (adj A)
2
+…+ 
(adj A)
10
. Then, the sum of all the elements of the 
matrix B is :  
 (1) –110 (2) 22 
 (3) –88  (4) –124 
 Ans. (3) 
Sol. Adj(A) = 
?M ????
????
????
1 2
0 1
  
 (AdjA)
2
 = 
?M ????
????
????
1 4
0 1
 
    | 
    | 
 ?H ?I ?M ????
?]????
????
10
1 20
AdjA
0 1
 
 
?M ?M ?M ?? ?? ?? ?? ?? ?? ?? ?? ?] ?K ?K ?K ?K ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? 1 0 1 2 1 4 1 20
B ...
0 1 0 1 0 1 0 1
 
 
?M ????
?]????
????
11 110
B
0 11
 ?? sum of elements of B  
 = –88 
7. The value of 
?H ?I ?? ?K ?? ?K ?K ?? ?? ?K ?? ?K ?K ?? 2
2 2
2 2 2
1 2 2 3 ... 100 101
1 2 2 3 ... 100 101
 is  
 (1) 
306
305
 (2) 
305
301
 
 (3) 
32
31
  (4) 
31
30
 
 Ans. (2) 
Sol. 
?H ?I ?H ?I ?H ?I ?] ?] ?K ?? ?K ?? ?K ?K ?? ?] ?? ?K ?? ?K ?K ?? ?K ?? ?? 100
2
2
2 2
r 1
2 2 2 100
2
r 1
r r 1
1 2 2 3 ... 100 101
1 2 2 3 ... 100 101
r r 1
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?] ?] ????
?K ?K ?K ?K ?????K?K
?K?K
????
????
?]?]
???? ?K ?K ?K ?K ?K ????
????
?? ?? 2
100
3 2
r 1
100 2
3 2
r 1
n n 1 2.n n 1 2n 1 n n 1
r 2r r
2 6 2
n n 1 n n 1 2n 1
r r
2 6
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ?H ?I ???? ?K?K
?K ?K ?K ????
????
?] ???? ?K ?K ?K ?K ????
????
n n 1 n n 1
2
. 2n 1 1
2 2 3
n n 1 n n 1 2n 1
2 2 3
;Put n = 100 
 
?H ?I ?H ?I ?K?K
?] ?] ?] ?? ?K 100 101
2
201 1
5185 305
2 3
100 101 201
5117 301
2 3
 
 
8. Let ?H ?I ?H ?I ?H ?I ?] ?K ?M ?? ?? x
t
0
f x t sin 1 e dt,x .  
 Then 
?H ?I ?? 3
x 0
f x
lim
x
is equal to  
 (1) 
1
6
 (2) ?M 1
6
 
 (3) ?M 2
3
  (4) 
2
3
 
 Ans. (2) 
Sol. 
?H ?I ?? 3
x 0
f x
lim
x
 
 Using L Hopital Rule. 
 
?H ?I ?H ?I ????
?K?M
?] x
2 2
x 0 x 0
x sin 1 e
f ' x
lim lim
3x 3x
 (Again L Hopital) 
 Using L.H. Rule 
 
?H ?I ?H ?I ?H ?I ??????
?M ?M ?M ?K ?M ????
?] x x x x x
x 0
sin 1 e e .e cos 1 e .e
lim
6
 
 ?]?M
1
6
 
9. The area (in sq. units) of the region described by 
{(x,y) : y
2
 < 2x, and y > 4x –1} is  
 (1) 
11
32
 (2) 
8
9
 
 (3) 
11
12
  (4) 
9
32
 
 Ans. (4) 
Sol. 
 
Q 
1
 
P
 
y
 
x
 
0
 
–1/2
 
(0,–1)
 
y'
 
x'
 
(1/4,0
 
 Shaded area 
?H ?I ?M?]?M
?? 1
Right Left
1
2
x x dy 
 
?] ?]?M
?] ?] ?M 2
y 2x
y 4x 1 Solve
1
y 1,y
2
 
 Shaded area 
?M ????
?K ?]?M
????
????
?? 1 2
1
2
y 1 y
dy
4 2
 
 
?M ???? ????
?] ?K ?M ?] ???? ????
????
???? ????
1
2 3
1
2
1 y y 9
y
4 2 6 32
 
10. The area (in sq. units) of the region 
?H ?I ?H ?I ?H ?I ?? ?? ?] ?? ?M ?? ?K ?K ?M ?? ?? S z ; z 1 2; z z i z z 2,lm z 0
 is  
 (1) 
?? 7
3
 (2) 
?? 3
2
 
 (3) 
?? 17
8
  (4) 
?? 7
4
 
 Ans. (2) 
Sol. Put z = x + iy 
 ?H ?I ?M ?? ?? ?M ?K ?? 2
2
z 1 2 x 1 y 4    …(1) 
 
?H ?I ?H ?I ?H ?I ?K ?K ?M ?? ?? ?K ?? z z i z z 2 2x i 2iy 2 
 ?? ?M ?? x y 1    …(2) 
 Im(z) > 0 ?? y > 0   …(3) 
 
 
(–3,0) (1,0) (5,0) 
x–y=1 
?? /4 
 
 Required area  
 = Area of semi-circle – area  of sector A 
 ?H ?I ?? ???M
2 1
2
2 2
 
 
?? ?] 3
2
 
11. If the value of the integral 
?M ?? ?K ?? 1
x
1
cos x
dx
1 3
 is 
?? 2
. 
Then, a value of ?? is  
 (1) 
?? 6
 (2) 
?? 2
 
 (3) 
?? 3
  (4) 
?? 4
 
 Ans. (2) 
Sol. Let 
?K ?M ?? ?] ?K ?? 1
x
1
cos x
I dx
1 3
 …(I) 
 
?K ?M ?M ?? ?] ?K ?? 1
x
1
cos x
I dx
1 3
  
 ?H ?I ?H ?I ????
?] ?K ?M ????
????
????
????
b b
a a
using f x dx f a b x dx   …(II) 
 Add (1) and (II) 
 ?H ?I ?H ?I ?K ?M ?] ?? ?] ?? ????
1 1
1 0
2I cos x dx 2 cos x dx 
 ?H ?I ?? ?]?]
????
sin 2
I given 
 ?| 
?? ???]
2
 
12. Let ?H ?I ?] ?M ?K ?M f x 3 x 2 4 x be a real valued 
function. If ?? and ?? are respectively the minimum 
and the maximum values of f, then ?? 2
 + 2 ?? 2
 is 
equal to  
 (1) 44 (2) 42 
 (3) 24  (4) 38 
 Ans. (2) 
Sol. f(x) = 3 x 2 4 x ?M ?K ?M 
 x – 2 ?? 0  &  4 – x ?? 0 
 ?| x ?? [2, 4] 
 Let x = 2sin
2
?? + 4cos
2
?? 
 ?| f(x) = 3 2 cos 2 sin ?? ?K ?? 
 ?| 2 3 2 cos 2 sin 9 2 2 ?? ?? ?K ?? ?? ?? ?K 
 2 3 2 cos 2 sin 20 ?? ?? ?K ?? ?? 
 ?| ?? = 2 ?? = 20 
 ?? 2
 + 2 ?? 2
 = 2 + 40 = 42 
13. If the coefficients of x
4
, x
5
 and x
6
 in the expansion 
of (1 + x)
n
 are in the arithmetic progression, then 
the maximum value of n is :  
 (1) 14 (2) 21 
 (3) 28  (4) 7 
 Ans. (1) 
Sol. Coeff. of x
4
 = 
n
C
4 
 Coeff. of x
5
 = 
n
C
5 
 
Coeff. of x
6
 = 
n
C
6 
 
n
C
4
, 
n
C
5
, 
n
C
6
 …. AP 
 2.
n
C
5
 = 
n
C
4
 + 
n
C
6
 
 
n n
6 4
n n
5 5
C C
2
C C
?]?K    
n
r
n
r 1
C n r 1
r C
?M????
?M?K ????
?]????
????
????
 
 
5 n 5
2
n 4 6
?M ?]?K
?M 
 12(n – 4) = 30 + n
2
 – 9n + 20 
 n
2
 – 21n + 98 = 0 
 (n – 14) (n – 7) = 0 
 n
max
 = 14 n
min
 = 7 
14. Consider a hyperbola H having centre at the origin 
and foci and the x-axis. Let C
1
 be the circle 
touching the hyperbola H and having the centre at 
the origin. Let C
2
 be the circle touching the 
hyperbola H at its vertex and having the centre at 
one of its foci. If areas (in sq. units) of C
1
 and C
2
 
are 36 ?? and 4 ?? , respectively, then the length (in 
units) of latus rectum of H is   
 (1) 
28
3
 (2) 
14
3
 
 (3) 
10
3
  (4) 
11
3
 
 Ans. (1) 
 
 
Sol. Let H : 
2 2
2 2
x y
1
a b
?M?] (b
2
 = a
2
(e
2
 – 1)) 
 ?| eq
n
 of C
1
 = x
2
 + y
2
 = a
2
 
 Ar. = 36 ?? 
 ?? a
2
 = 36 ?? 
 a = 6 
 Now radius of C
2
 can be a(e – 1) or a(e + 1) 
 for r = a(e – 1) for r = a(e + 1) 
 Ar. = 4 ??  ?? r
2
 = 4 ?? 
 ?? a
2
(e – 1)
2
 = 4 ?? a
2
(e + 1)
2
 = 4 
 36 ?? (e – 1)
2
 = 4 ?? 36(e + 1)
2
 = 4 
 e – 1 = 
1
3
  e + 1 = 
1
3
 
 e = 
4
3
  
2
3
?M 
    Not possible 
 ?| b
2
 = 36
16
1
9
????
?M ????
????
 = 28 
 ?| LR = 
2
2b 2 28 28
a 6 3
?? ?]?] 
15. If the mean of the following probability 
distribution of a random variable X; 
 
?H ?I ?K X 0 2 4 6 8
P X a 2a a b 2b 3b
 
 is 
46
9
, then the variance of the distribution is  
 (1) 
581
81
 (2) 
566
81
 
 (3) 
173
27
  (4) 
151
27
 
 Ans. (2) 
Sol. 
i
P 1 ?] ?? 
 a + 2a + a + b + 2b + 3b = 1 
 4a + 6b = 1  …. (I) 
 E(x) = mean = 
46
9
 
 
i i
46
P X
9
?] ?? ?? 4a + 4a + 4b + 12b + 24b = 
46
9
 
 8a + 40b = 
46
9
 
 4a + 20b = 
23
9
 …. (II) 
 Subtract (I) from (II) we get 
 b = 
1
9
 & a = 
1
12
 
 Variance = E(x
i
2
) – E(x
i
)
2
 
 E(x
i
2
) = 0
2
 × 9
2
 + 2
2
 × 2a + 4
2
(a + b) + 6
2
(2b) + 8
2
(3b) 
 = 24a + 280b 
 Put a = 
1
12
 b = 
1
9
 
 E(x
i
2
) = 2 + 
280 298
9 9
?] 
 ?| ?@?? 2
 = E(x
i
2
) – E(x
i
)
2
 
 
2
298 46
9 9
????
?]?M
????
????
 
 
2
298 2116
9 81
?? ?] ?M 
 
566
81
?] 
16. Let PQ be a chord of the parabola y
2
 = 12x and the 
midpoint of PQ be at (4,1). Then, which of the 
following point lies on the line passing through the 
points P and Q ?  
 (1) (3,–3) (2) 
????
?M ????
????
3
, 16
2
 
 (3) ?M (2, 9)  (4) 
????
?M ????
????
1
, 20
2
 
 Ans. (4) 
Sol. 
 
Q 
(4,1
P
 
 
 T = S
1
 
 y – 6(x + 4) 
 = 1 – 48 
 6x – y = 23 
 Option 4 
????
?M ????
????
1
, 20
2
 will satisfy