JEE Main 2024 April 8 Shift 2 Paper & Solutions | JEE Main & Advanced Mock Test Series 2025 PDF Download

 Page 1


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Monday 08
th
 April, 2024) TIME : 3 : 00 PM  to  6 : 00 PM 
NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If the image of the point (-4, 5) in the line
x + 2y = 2 lies on the circle (x + 4)
2
 + (y-3)
2
 = r
2
,
then r is equal lo :
(1) 1 (2) 2
(3) 75 (4) 3
Ans. (2) 
Sol. Image of point (–4, 5) 
1 1 1 1
2 2
x x y y ax by c
2
a b a b
?M ?M ?K ?K ????
?] ?] ?M????
?K ????
Line : x + 2y – 2 = 0 
2 2
x 4 y 5 4 10 2
2
1 2 1 2
?K ?M ?M ?K ?M????
?] ?] ?M????
?K ????
    
8
5
?M ?] 8 28
x 4
5 5
?] ?M ?M ?] ?M 16 9
y 5
5 5
?] ?M ?K ?] Point lies on circle (x + 4)
2
 + (y – 3)
2
 = r
2
 
2
2
64 9
3 r
25 5
????
?K ?M ?] ????
????
2
100
r , r 2
25
?]?]
2. Let 
ˆˆ ˆ
a i 2j 3k ?] ?K ?K , 
ˆˆ ˆ
b 2i 3j 5k ?] ?K ?M and
ˆˆ ˆ
c 3i j k ?] ?M ?K ?? be three vectors. Let r be a unit
vector along b c ?K . If r . a 3 ?] , then 3 ?? is equal
to :
(1) 27 (2) 25
(3) 25 (4) 21
Ans. (2) 
Sol. ?H ?I r k b c ?]?K 
r . a 3 ?] 
?H ?I r.a k b.a c.a ?]?K
3 = k(2 + 6 – 15 + 3 – 2 + 3 ?? ) 
3 = k(–6 + 3 ?? ) …(1) 
ˆˆ ˆ
r k(5i 2j (5 )k) ?] ?K ?M ?M ?? 2
r k 25 4 25 10 1 ?] ?K ?K ?K ?? ?M ?? ?] …(2)
3 1
k
6 3 2
?]?]
?M ?K ?? ?M ?K ?? put in (2) 
4 + ?? 2
 – 4 ?? = 54 + ?? 2
 – 10 ?? 
6 ?? = 50 
3 ?? = 25 
3. If ???@ ?? ?@ a, ?? ?@ ?? ?@ b, ?? ?@ ?? ?@ c and
b c
a c 0
a b
?? ???]
?? , then 
a b
a b c
?? ?K?K
?? ?M ?? ?M ?? ?M  is equal to : 
(1) 2 (2) 3
(3) 0 (4) 1
Ans. (3) 
Sol. R
1
 ?? R
 1
– R
 2
, R
 2
?? ?@ R
 2
– R
 3
a b 0
0 b c 0
a b
?? ?M ?M ?? ?? ?M ?M ?? ?] ?? ( ?? –a) ( ?? ( ?? –b) – b(c– ?? )) – (b – ?? ) (– a(c - ?? )) = 0
?? ( ?? – a) ( ?? – b) – b( ?? – a) (c – ?? ) + a(b - ?? ) (c – ?? )
b a
0
c b a
?? ?K ?K ?] ?? ?M ?? ?M ?? ?M
Page 2


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Monday 08
th
 April, 2024) TIME : 3 : 00 PM  to  6 : 00 PM 
NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If the image of the point (-4, 5) in the line
x + 2y = 2 lies on the circle (x + 4)
2
 + (y-3)
2
 = r
2
,
then r is equal lo :
(1) 1 (2) 2
(3) 75 (4) 3
Ans. (2) 
Sol. Image of point (–4, 5) 
1 1 1 1
2 2
x x y y ax by c
2
a b a b
?M ?M ?K ?K ????
?] ?] ?M????
?K ????
Line : x + 2y – 2 = 0 
2 2
x 4 y 5 4 10 2
2
1 2 1 2
?K ?M ?M ?K ?M????
?] ?] ?M????
?K ????
    
8
5
?M ?] 8 28
x 4
5 5
?] ?M ?M ?] ?M 16 9
y 5
5 5
?] ?M ?K ?] Point lies on circle (x + 4)
2
 + (y – 3)
2
 = r
2
 
2
2
64 9
3 r
25 5
????
?K ?M ?] ????
????
2
100
r , r 2
25
?]?]
2. Let 
ˆˆ ˆ
a i 2j 3k ?] ?K ?K , 
ˆˆ ˆ
b 2i 3j 5k ?] ?K ?M and
ˆˆ ˆ
c 3i j k ?] ?M ?K ?? be three vectors. Let r be a unit
vector along b c ?K . If r . a 3 ?] , then 3 ?? is equal
to :
(1) 27 (2) 25
(3) 25 (4) 21
Ans. (2) 
Sol. ?H ?I r k b c ?]?K 
r . a 3 ?] 
?H ?I r.a k b.a c.a ?]?K
3 = k(2 + 6 – 15 + 3 – 2 + 3 ?? ) 
3 = k(–6 + 3 ?? ) …(1) 
ˆˆ ˆ
r k(5i 2j (5 )k) ?] ?K ?M ?M ?? 2
r k 25 4 25 10 1 ?] ?K ?K ?K ?? ?M ?? ?] …(2)
3 1
k
6 3 2
?]?]
?M ?K ?? ?M ?K ?? put in (2) 
4 + ?? 2
 – 4 ?? = 54 + ?? 2
 – 10 ?? 
6 ?? = 50 
3 ?? = 25 
3. If ???@ ?? ?@ a, ?? ?@ ?? ?@ b, ?? ?@ ?? ?@ c and
b c
a c 0
a b
?? ???]
?? , then 
a b
a b c
?? ?K?K
?? ?M ?? ?M ?? ?M  is equal to : 
(1) 2 (2) 3
(3) 0 (4) 1
Ans. (3) 
Sol. R
1
 ?? R
 1
– R
 2
, R
 2
?? ?@ R
 2
– R
 3
a b 0
0 b c 0
a b
?? ?M ?M ?? ?? ?M ?M ?? ?] ?? ( ?? –a) ( ?? ( ?? –b) – b(c– ?? )) – (b – ?? ) (– a(c - ?? )) = 0
?? ( ?? – a) ( ?? – b) – b( ?? – a) (c – ?? ) + a(b - ?? ) (c – ?? )
b a
0
c b a
?? ?K ?K ?] ?? ?M ?? ?M ?? ?M 4. In an increasing geometric progression ol positive 
terms, the sum of the second and sixth terms is 
70
3
and the product of the third and fifth terms is 
49. Then the sum of the 4
th
, 6
th
 and 8
th
 terms is :- 
 (1) 96 (2) 78 
 (3) 91  (4) 84 
 Ans. (3) 
Sol. 
2 6
70
T T
3
?K?] 
 ar + ar
5
 = 
70
3
 
 T
3
 . T
5
 = 49 
 ar
2
 . ar
4
 = 49 
 a
2
r
6
 = 49 
 ar
3
 = +7, 
3
7
a
r
?] 
 ar(1 + r
4
) = 
70
3
 
 
4 2
2
7 70
(1 r ) ,r t
3 r
?K ?] ?] 
 
2
1 10
(1 t )
t 3
?K?] 
 3t
2
 – 10t + 3 = 0 
 
1
t 3,
3
?] 
 Increasing G.P. r
2
 = 3, r 3 ?] 
 T
4
 + T
6
 + T
8
 
 = ar
3
 + ar
5
 + ar
7
 
 = ar
3
(1 + r
2
 + r
4
) 
 = 7(1 + 3 + 9) = 91 
5. The number of ways five alphabets can be chosen 
from the alphabets of the word MATHEMATICS, 
where the chosen alphabets are not necessarily 
distinct, is equal to : 
 (1) 175 (2) 181 
 (3) 177  (4) 179 
 Ans. (4) 
Sol. AA, MM, TT, H, I, C, S, E 
 (1) All distinct 
  
8
C
5
 ?? 56 
 (2) 2 same, 3 different 
  
3
C
?Q × 
7
C
3
 ?? 105 
 (3) 2 same I
st
 kind, 2 same 2
nd
 kind, 1 different 
  
3
C
2
 × 
6
C
1
 ?? 18 
  Total ?? 179 
6. The sum of all possible values of ???@?? [– ?@?? , 2 ?? ], for 
which 
1 icos
1 2icos
?K??
?M??
 is purely imaginary, is equal 
to 
 (1) 2 ?? (2) 3 ?? 
 (3) 5 ??  (4) 4 ?? 
 Ans. (2) 
Sol. 
1 icos
Z
1 2icos
?K??
?] ?M??
 
 Z Z ?]?M ??  
1 icos 1 icos
1 2icos 1 2icos
????
?K ?? ?K ?? ?]?M
????
?M ?? ?M ?? ????
 
 (1+ icos ?? ) ?H ?I 1 2icos ?M??
=–(1 – 2i cos ?? ) ?H ?I 1 icos ?K??
 
 (1+icos ?? ) (1 + 2icos ?? ) = –(1 – 2icos ?? ) (1 – icos ?? ) 
 1 + 3icos ?? – 2cos
2
?? = –(1 – 3icos ?? – 2cos
2
?? ) 
 2 – 4cos
2
?? = 0  
 ???@ cos
2
?? = 
1
2
?@?? 
3 3 5 7
, , , , ,
4 4 4 4 4 4
?? ?? ?? ?? ?? ?? ?? ?] ?M ?M 
 sum = 3 ?? 
  
7. If the system of equations x + 4y – z = ?? ,  
7x + 9y + ?? z = –3, 5x + y + 2z = –1 has infinitely 
many solutions, then (2 ?? . + 3 ?? ) is equal to : 
 (1) 2 (2) –3 
 (3) 3  (4) –2 
 Ans. (2) 
Sol. 
1 4 1
7 9 0
5 1 2
?M ?d ?] ?? ?] 
 ??  (18– ?? ) – 4(14–5 ?? ) – (7 – 45) = 0 ?? ?@?? = 0 
 ?d = ?d x
 = ?d y
 = ?d z
 = 0 (For infinite solution) 
 
x
4 1
3 9 0
1 1 2
???M
?d ?] ?M ?? ?] ?M 
 ?? (18 – ?? ) – 4(–6 + ?? ) –1(–3 + 9) = 0 
 18 ?? + 24 – 6 = 0  ?? ?? = –1 ?@
Page 3


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Monday 08
th
 April, 2024) TIME : 3 : 00 PM  to  6 : 00 PM 
NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If the image of the point (-4, 5) in the line
x + 2y = 2 lies on the circle (x + 4)
2
 + (y-3)
2
 = r
2
,
then r is equal lo :
(1) 1 (2) 2
(3) 75 (4) 3
Ans. (2) 
Sol. Image of point (–4, 5) 
1 1 1 1
2 2
x x y y ax by c
2
a b a b
?M ?M ?K ?K ????
?] ?] ?M????
?K ????
Line : x + 2y – 2 = 0 
2 2
x 4 y 5 4 10 2
2
1 2 1 2
?K ?M ?M ?K ?M????
?] ?] ?M????
?K ????
    
8
5
?M ?] 8 28
x 4
5 5
?] ?M ?M ?] ?M 16 9
y 5
5 5
?] ?M ?K ?] Point lies on circle (x + 4)
2
 + (y – 3)
2
 = r
2
 
2
2
64 9
3 r
25 5
????
?K ?M ?] ????
????
2
100
r , r 2
25
?]?]
2. Let 
ˆˆ ˆ
a i 2j 3k ?] ?K ?K , 
ˆˆ ˆ
b 2i 3j 5k ?] ?K ?M and
ˆˆ ˆ
c 3i j k ?] ?M ?K ?? be three vectors. Let r be a unit
vector along b c ?K . If r . a 3 ?] , then 3 ?? is equal
to :
(1) 27 (2) 25
(3) 25 (4) 21
Ans. (2) 
Sol. ?H ?I r k b c ?]?K 
r . a 3 ?] 
?H ?I r.a k b.a c.a ?]?K
3 = k(2 + 6 – 15 + 3 – 2 + 3 ?? ) 
3 = k(–6 + 3 ?? ) …(1) 
ˆˆ ˆ
r k(5i 2j (5 )k) ?] ?K ?M ?M ?? 2
r k 25 4 25 10 1 ?] ?K ?K ?K ?? ?M ?? ?] …(2)
3 1
k
6 3 2
?]?]
?M ?K ?? ?M ?K ?? put in (2) 
4 + ?? 2
 – 4 ?? = 54 + ?? 2
 – 10 ?? 
6 ?? = 50 
3 ?? = 25 
3. If ???@ ?? ?@ a, ?? ?@ ?? ?@ b, ?? ?@ ?? ?@ c and
b c
a c 0
a b
?? ???]
?? , then 
a b
a b c
?? ?K?K
?? ?M ?? ?M ?? ?M  is equal to : 
(1) 2 (2) 3
(3) 0 (4) 1
Ans. (3) 
Sol. R
1
 ?? R
 1
– R
 2
, R
 2
?? ?@ R
 2
– R
 3
a b 0
0 b c 0
a b
?? ?M ?M ?? ?? ?M ?M ?? ?] ?? ( ?? –a) ( ?? ( ?? –b) – b(c– ?? )) – (b – ?? ) (– a(c - ?? )) = 0
?? ( ?? – a) ( ?? – b) – b( ?? – a) (c – ?? ) + a(b - ?? ) (c – ?? )
b a
0
c b a
?? ?K ?K ?] ?? ?M ?? ?M ?? ?M 4. In an increasing geometric progression ol positive 
terms, the sum of the second and sixth terms is 
70
3
and the product of the third and fifth terms is 
49. Then the sum of the 4
th
, 6
th
 and 8
th
 terms is :- 
 (1) 96 (2) 78 
 (3) 91  (4) 84 
 Ans. (3) 
Sol. 
2 6
70
T T
3
?K?] 
 ar + ar
5
 = 
70
3
 
 T
3
 . T
5
 = 49 
 ar
2
 . ar
4
 = 49 
 a
2
r
6
 = 49 
 ar
3
 = +7, 
3
7
a
r
?] 
 ar(1 + r
4
) = 
70
3
 
 
4 2
2
7 70
(1 r ) ,r t
3 r
?K ?] ?] 
 
2
1 10
(1 t )
t 3
?K?] 
 3t
2
 – 10t + 3 = 0 
 
1
t 3,
3
?] 
 Increasing G.P. r
2
 = 3, r 3 ?] 
 T
4
 + T
6
 + T
8
 
 = ar
3
 + ar
5
 + ar
7
 
 = ar
3
(1 + r
2
 + r
4
) 
 = 7(1 + 3 + 9) = 91 
5. The number of ways five alphabets can be chosen 
from the alphabets of the word MATHEMATICS, 
where the chosen alphabets are not necessarily 
distinct, is equal to : 
 (1) 175 (2) 181 
 (3) 177  (4) 179 
 Ans. (4) 
Sol. AA, MM, TT, H, I, C, S, E 
 (1) All distinct 
  
8
C
5
 ?? 56 
 (2) 2 same, 3 different 
  
3
C
?Q × 
7
C
3
 ?? 105 
 (3) 2 same I
st
 kind, 2 same 2
nd
 kind, 1 different 
  
3
C
2
 × 
6
C
1
 ?? 18 
  Total ?? 179 
6. The sum of all possible values of ???@?? [– ?@?? , 2 ?? ], for 
which 
1 icos
1 2icos
?K??
?M??
 is purely imaginary, is equal 
to 
 (1) 2 ?? (2) 3 ?? 
 (3) 5 ??  (4) 4 ?? 
 Ans. (2) 
Sol. 
1 icos
Z
1 2icos
?K??
?] ?M??
 
 Z Z ?]?M ??  
1 icos 1 icos
1 2icos 1 2icos
????
?K ?? ?K ?? ?]?M
????
?M ?? ?M ?? ????
 
 (1+ icos ?? ) ?H ?I 1 2icos ?M??
=–(1 – 2i cos ?? ) ?H ?I 1 icos ?K??
 
 (1+icos ?? ) (1 + 2icos ?? ) = –(1 – 2icos ?? ) (1 – icos ?? ) 
 1 + 3icos ?? – 2cos
2
?? = –(1 – 3icos ?? – 2cos
2
?? ) 
 2 – 4cos
2
?? = 0  
 ???@ cos
2
?? = 
1
2
?@?? 
3 3 5 7
, , , , ,
4 4 4 4 4 4
?? ?? ?? ?? ?? ?? ?? ?] ?M ?M 
 sum = 3 ?? 
  
7. If the system of equations x + 4y – z = ?? ,  
7x + 9y + ?? z = –3, 5x + y + 2z = –1 has infinitely 
many solutions, then (2 ?? . + 3 ?? ) is equal to : 
 (1) 2 (2) –3 
 (3) 3  (4) –2 
 Ans. (2) 
Sol. 
1 4 1
7 9 0
5 1 2
?M ?d ?] ?? ?] 
 ??  (18– ?? ) – 4(14–5 ?? ) – (7 – 45) = 0 ?? ?@?? = 0 
 ?d = ?d x
 = ?d y
 = ?d z
 = 0 (For infinite solution) 
 
x
4 1
3 9 0
1 1 2
???M
?d ?] ?M ?? ?] ?M 
 ?? (18 – ?? ) – 4(–6 + ?? ) –1(–3 + 9) = 0 
 18 ?? + 24 – 6 = 0  ?? ?? = –1 ?@ 
8. If the shortest distance between the lines 
x y 4 z 3
2 3 4
?M ?? ?M ?M ?]?] and 
x 2 y 4 z 7
4 6 8
?M ?M ?M ?]?] is 
13
29
, then a value 
of ?? is : 
 (1) –
13
25
 (2) 
13
25
 
 (3) 1  (4) –1 
 Ans. (3) 
Sol. 
?H ?I ?H ?I 1
1
2
2
ˆ ˆ
b 2i 3j 4k
ˆ ˆ ˆ ˆ ˆˆ
r i 4 j 3k (2i 3j 4k)
ˆˆ ˆ
a i 4 j 3k
ˆ ˆ ˆ ˆ ˆˆ
r 2i 4 j 7k (2i 3j 4k)
ˆˆ ˆ
a 2i 4 j 7k
?] ?K ?K ?? ?? ?] ?? ?K ?K ?K ?? ?K ?K ?K ?? ?K ?K ?? ?] ?K ?K ?K ?? ?K ?K ?? ?? ?]?K?K
 
 
2 1
b (a a )
13
Shortest dist.
b
29
???M
?]?]  
 
?H ?I ?H ?I ˆ ˆ ˆ ˆˆ
2i 3j 4k (2 )i 4k
13
29 29
?K ?K ?? ?M ?? ?K ?] 
 
ˆ ˆ ˆ ˆ
8j 3(2 )k 12i 4(2 ) j 13 ?M ?M ?M ?? ?K ?K ?M ?? ?] 
 
ˆˆ ˆ
12i 4 j (3 6)k 13 ?M ?? ?K ?? ?M ?] 
 144 + 16?@??
 2
 + (3 ?? – 6)
2
 = 169 
 16 ?? 2
 + (3 ?? – 6)
2
 = 25 = ?? ?@?? = 1 
9. If the value of 
3cos36 5sin18
5cos36 3sin18
?? ?K ?? ?? ?M ?? is 
a 5 b
c
?M , 
where a, b, c are natural numbers and gcd(a, c) = 1, 
then a + b + c is equal to : 
 (1) 50 (2) 40 
 (3) 52  (4) 54 
 Ans. (3) 
Sol. 
?H ?I 3 5 1 5 1
5
8 5 2
4 4
2 5 8 5 1 5 1
5 3
4 4
????
?K?M
?K????
?M ????
?] ?? ?? ?? ?? ?K ?K?M
?M ?? ?? ?? ?? ?? ?? ?? ?? 
 
4 5 1 5 4
5 4 5 4
?M?M
?]??
?K?M
 
 
20 16 5 5 4
11
?M ?M ?K ?] ?M 
 
17 5 24
11
?M ?] ?? a = 17, b = 27, c = 11 
 a + b + c = 52 
10. Let y = y(x) be the solution curve of the 
differential equation secy 
dy
dx
 + 2xsiny = x
3
cosy, 
y(1) = 0. Then 
?H ?I y 3 is equal to : 
 (1) 
3
?? (2) 
6
?? 
 (3) 
4
??  (4) 
12
?? 
 Ans. (3) 
Sol. sec
2
y 
dy
dx
 + 2xsiny secy = x
3
cosy secy 
 sec
2
y 
dy
dx
 + 2xtany = x
3
 
 tany = t ?@ ?? 
2
dy dt
sec y
dx dx
?] 
 
3
dt
2xt x
dx
?K?] , If 
2 2xdx
x
e e
???]?] 
 
2 2
x 3 x
te x .e dx c ?]?K
?? 
 x
2
 = Z  ??  
?{ ?} Z Z Z Z
1 1
t.e e .ZdZ e .Z e c
2 2
?] ?] ?M ?K ?? 
 
2
2 x
2tan y (x 1) 2ce
?M ?] ?M ?K 
 y(1) = 0  ?? c = 0 ??  y( 3)
4
?? ?] 
11. The area of the region in the first quadrant inside 
the circle x
2
 + y
2
 = 8 and outside the pnrabola   
y
2
 = 2x is equal to : 
 (1) 
1
2 3
?? ?M (2) 
2
3
???M 
 (3) 
2
2 3
?? ?M  (4) 
1
3
???M 
 Ans. (2) 
Page 4


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Monday 08
th
 April, 2024) TIME : 3 : 00 PM  to  6 : 00 PM 
NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If the image of the point (-4, 5) in the line
x + 2y = 2 lies on the circle (x + 4)
2
 + (y-3)
2
 = r
2
,
then r is equal lo :
(1) 1 (2) 2
(3) 75 (4) 3
Ans. (2) 
Sol. Image of point (–4, 5) 
1 1 1 1
2 2
x x y y ax by c
2
a b a b
?M ?M ?K ?K ????
?] ?] ?M????
?K ????
Line : x + 2y – 2 = 0 
2 2
x 4 y 5 4 10 2
2
1 2 1 2
?K ?M ?M ?K ?M????
?] ?] ?M????
?K ????
    
8
5
?M ?] 8 28
x 4
5 5
?] ?M ?M ?] ?M 16 9
y 5
5 5
?] ?M ?K ?] Point lies on circle (x + 4)
2
 + (y – 3)
2
 = r
2
 
2
2
64 9
3 r
25 5
????
?K ?M ?] ????
????
2
100
r , r 2
25
?]?]
2. Let 
ˆˆ ˆ
a i 2j 3k ?] ?K ?K , 
ˆˆ ˆ
b 2i 3j 5k ?] ?K ?M and
ˆˆ ˆ
c 3i j k ?] ?M ?K ?? be three vectors. Let r be a unit
vector along b c ?K . If r . a 3 ?] , then 3 ?? is equal
to :
(1) 27 (2) 25
(3) 25 (4) 21
Ans. (2) 
Sol. ?H ?I r k b c ?]?K 
r . a 3 ?] 
?H ?I r.a k b.a c.a ?]?K
3 = k(2 + 6 – 15 + 3 – 2 + 3 ?? ) 
3 = k(–6 + 3 ?? ) …(1) 
ˆˆ ˆ
r k(5i 2j (5 )k) ?] ?K ?M ?M ?? 2
r k 25 4 25 10 1 ?] ?K ?K ?K ?? ?M ?? ?] …(2)
3 1
k
6 3 2
?]?]
?M ?K ?? ?M ?K ?? put in (2) 
4 + ?? 2
 – 4 ?? = 54 + ?? 2
 – 10 ?? 
6 ?? = 50 
3 ?? = 25 
3. If ???@ ?? ?@ a, ?? ?@ ?? ?@ b, ?? ?@ ?? ?@ c and
b c
a c 0
a b
?? ???]
?? , then 
a b
a b c
?? ?K?K
?? ?M ?? ?M ?? ?M  is equal to : 
(1) 2 (2) 3
(3) 0 (4) 1
Ans. (3) 
Sol. R
1
 ?? R
 1
– R
 2
, R
 2
?? ?@ R
 2
– R
 3
a b 0
0 b c 0
a b
?? ?M ?M ?? ?? ?M ?M ?? ?] ?? ( ?? –a) ( ?? ( ?? –b) – b(c– ?? )) – (b – ?? ) (– a(c - ?? )) = 0
?? ( ?? – a) ( ?? – b) – b( ?? – a) (c – ?? ) + a(b - ?? ) (c – ?? )
b a
0
c b a
?? ?K ?K ?] ?? ?M ?? ?M ?? ?M 4. In an increasing geometric progression ol positive 
terms, the sum of the second and sixth terms is 
70
3
and the product of the third and fifth terms is 
49. Then the sum of the 4
th
, 6
th
 and 8
th
 terms is :- 
 (1) 96 (2) 78 
 (3) 91  (4) 84 
 Ans. (3) 
Sol. 
2 6
70
T T
3
?K?] 
 ar + ar
5
 = 
70
3
 
 T
3
 . T
5
 = 49 
 ar
2
 . ar
4
 = 49 
 a
2
r
6
 = 49 
 ar
3
 = +7, 
3
7
a
r
?] 
 ar(1 + r
4
) = 
70
3
 
 
4 2
2
7 70
(1 r ) ,r t
3 r
?K ?] ?] 
 
2
1 10
(1 t )
t 3
?K?] 
 3t
2
 – 10t + 3 = 0 
 
1
t 3,
3
?] 
 Increasing G.P. r
2
 = 3, r 3 ?] 
 T
4
 + T
6
 + T
8
 
 = ar
3
 + ar
5
 + ar
7
 
 = ar
3
(1 + r
2
 + r
4
) 
 = 7(1 + 3 + 9) = 91 
5. The number of ways five alphabets can be chosen 
from the alphabets of the word MATHEMATICS, 
where the chosen alphabets are not necessarily 
distinct, is equal to : 
 (1) 175 (2) 181 
 (3) 177  (4) 179 
 Ans. (4) 
Sol. AA, MM, TT, H, I, C, S, E 
 (1) All distinct 
  
8
C
5
 ?? 56 
 (2) 2 same, 3 different 
  
3
C
?Q × 
7
C
3
 ?? 105 
 (3) 2 same I
st
 kind, 2 same 2
nd
 kind, 1 different 
  
3
C
2
 × 
6
C
1
 ?? 18 
  Total ?? 179 
6. The sum of all possible values of ???@?? [– ?@?? , 2 ?? ], for 
which 
1 icos
1 2icos
?K??
?M??
 is purely imaginary, is equal 
to 
 (1) 2 ?? (2) 3 ?? 
 (3) 5 ??  (4) 4 ?? 
 Ans. (2) 
Sol. 
1 icos
Z
1 2icos
?K??
?] ?M??
 
 Z Z ?]?M ??  
1 icos 1 icos
1 2icos 1 2icos
????
?K ?? ?K ?? ?]?M
????
?M ?? ?M ?? ????
 
 (1+ icos ?? ) ?H ?I 1 2icos ?M??
=–(1 – 2i cos ?? ) ?H ?I 1 icos ?K??
 
 (1+icos ?? ) (1 + 2icos ?? ) = –(1 – 2icos ?? ) (1 – icos ?? ) 
 1 + 3icos ?? – 2cos
2
?? = –(1 – 3icos ?? – 2cos
2
?? ) 
 2 – 4cos
2
?? = 0  
 ???@ cos
2
?? = 
1
2
?@?? 
3 3 5 7
, , , , ,
4 4 4 4 4 4
?? ?? ?? ?? ?? ?? ?? ?] ?M ?M 
 sum = 3 ?? 
  
7. If the system of equations x + 4y – z = ?? ,  
7x + 9y + ?? z = –3, 5x + y + 2z = –1 has infinitely 
many solutions, then (2 ?? . + 3 ?? ) is equal to : 
 (1) 2 (2) –3 
 (3) 3  (4) –2 
 Ans. (2) 
Sol. 
1 4 1
7 9 0
5 1 2
?M ?d ?] ?? ?] 
 ??  (18– ?? ) – 4(14–5 ?? ) – (7 – 45) = 0 ?? ?@?? = 0 
 ?d = ?d x
 = ?d y
 = ?d z
 = 0 (For infinite solution) 
 
x
4 1
3 9 0
1 1 2
???M
?d ?] ?M ?? ?] ?M 
 ?? (18 – ?? ) – 4(–6 + ?? ) –1(–3 + 9) = 0 
 18 ?? + 24 – 6 = 0  ?? ?? = –1 ?@ 
8. If the shortest distance between the lines 
x y 4 z 3
2 3 4
?M ?? ?M ?M ?]?] and 
x 2 y 4 z 7
4 6 8
?M ?M ?M ?]?] is 
13
29
, then a value 
of ?? is : 
 (1) –
13
25
 (2) 
13
25
 
 (3) 1  (4) –1 
 Ans. (3) 
Sol. 
?H ?I ?H ?I 1
1
2
2
ˆ ˆ
b 2i 3j 4k
ˆ ˆ ˆ ˆ ˆˆ
r i 4 j 3k (2i 3j 4k)
ˆˆ ˆ
a i 4 j 3k
ˆ ˆ ˆ ˆ ˆˆ
r 2i 4 j 7k (2i 3j 4k)
ˆˆ ˆ
a 2i 4 j 7k
?] ?K ?K ?? ?? ?] ?? ?K ?K ?K ?? ?K ?K ?K ?? ?K ?K ?? ?] ?K ?K ?K ?? ?K ?K ?? ?? ?]?K?K
 
 
2 1
b (a a )
13
Shortest dist.
b
29
???M
?]?]  
 
?H ?I ?H ?I ˆ ˆ ˆ ˆˆ
2i 3j 4k (2 )i 4k
13
29 29
?K ?K ?? ?M ?? ?K ?] 
 
ˆ ˆ ˆ ˆ
8j 3(2 )k 12i 4(2 ) j 13 ?M ?M ?M ?? ?K ?K ?M ?? ?] 
 
ˆˆ ˆ
12i 4 j (3 6)k 13 ?M ?? ?K ?? ?M ?] 
 144 + 16?@??
 2
 + (3 ?? – 6)
2
 = 169 
 16 ?? 2
 + (3 ?? – 6)
2
 = 25 = ?? ?@?? = 1 
9. If the value of 
3cos36 5sin18
5cos36 3sin18
?? ?K ?? ?? ?M ?? is 
a 5 b
c
?M , 
where a, b, c are natural numbers and gcd(a, c) = 1, 
then a + b + c is equal to : 
 (1) 50 (2) 40 
 (3) 52  (4) 54 
 Ans. (3) 
Sol. 
?H ?I 3 5 1 5 1
5
8 5 2
4 4
2 5 8 5 1 5 1
5 3
4 4
????
?K?M
?K????
?M ????
?] ?? ?? ?? ?? ?K ?K?M
?M ?? ?? ?? ?? ?? ?? ?? ?? 
 
4 5 1 5 4
5 4 5 4
?M?M
?]??
?K?M
 
 
20 16 5 5 4
11
?M ?M ?K ?] ?M 
 
17 5 24
11
?M ?] ?? a = 17, b = 27, c = 11 
 a + b + c = 52 
10. Let y = y(x) be the solution curve of the 
differential equation secy 
dy
dx
 + 2xsiny = x
3
cosy, 
y(1) = 0. Then 
?H ?I y 3 is equal to : 
 (1) 
3
?? (2) 
6
?? 
 (3) 
4
??  (4) 
12
?? 
 Ans. (3) 
Sol. sec
2
y 
dy
dx
 + 2xsiny secy = x
3
cosy secy 
 sec
2
y 
dy
dx
 + 2xtany = x
3
 
 tany = t ?@ ?? 
2
dy dt
sec y
dx dx
?] 
 
3
dt
2xt x
dx
?K?] , If 
2 2xdx
x
e e
???]?] 
 
2 2
x 3 x
te x .e dx c ?]?K
?? 
 x
2
 = Z  ??  
?{ ?} Z Z Z Z
1 1
t.e e .ZdZ e .Z e c
2 2
?] ?] ?M ?K ?? 
 
2
2 x
2tan y (x 1) 2ce
?M ?] ?M ?K 
 y(1) = 0  ?? c = 0 ??  y( 3)
4
?? ?] 
11. The area of the region in the first quadrant inside 
the circle x
2
 + y
2
 = 8 and outside the pnrabola   
y
2
 = 2x is equal to : 
 (1) 
1
2 3
?? ?M (2) 
2
3
???M 
 (3) 
2
2 3
?? ?M  (4) 
1
3
???M 
 Ans. (2) 
Sol.
  
(2 ,0) 
x=2 
 
 Required area = Ar(circle from 0 to 2) – 
 ar(para from 0 to 2) 
 
2
2
2
0
0
8 x dx 2x dx ?] ?M ?M ????
 
 
2
2
2 1
0
0
x 8 x x x
8 x sin 2
2 2 3 / 2
2 2
?M ????
????
?] ?M ?K ?M ????
????
????
????
 
 ?H ?I 1
2 8 2 2 2
8 4 sin 2 2 0
2 2 3
2 2
?M ?] ?M ?K ?M ?M 
 ??  
8 2
2 4 .
4 3 3
?? ?K ?M ?] ?? ?M 
12. If the line segment joining the points (5, 2) and  
(2, a) subtends an angle 
4
?? at the origin, then the 
absolute value of the product of all possible values 
of a is : 
 (1) 6 (2) 8 
 (3) 2  (4) 4 
 Ans. (4) 
Sol. 
 
B(2, a) 
?? /4 
A(5,2) 
O 
 
 
OA
2
m
5
?] 
 
OB
a
m
2
?] 
 
2 a
tan
5 2
4
?? ?M ?] 
 
4 5a
1
10 2a
?M ?] ?K 
 4 – 5a = ±(10 + 2a)  
 
 4 – 5a = 10 + 2a 4 – 5a = –10 – 2a 
 ?? 7a + 6 = 0 3a = 14 
 ?? 
6
a
7
?]?M  
14
a
3
?]?K 
 
6 14
4
7 3
?M ?? ?] ?M  
13. Let 
ˆˆ ˆ
a 4i j k ?] ?M ?K , 
ˆˆ ˆ
b 11i j k ?] ?M ?K and c be 
a vector such that 
 ?H ?I ?H ?I a b c c 2a 3b ?K ?? ?] ?? ?M ?K .  
 If ?H ?I 2a 3b .c 1670 ?K?] , then 
2
| c | is equal to :
 
 (1) 1627 (2) 1618 
 (3) 1600  (4) 1609 
 Ans. (2) 
Sol. ?H ?I ?H ?I a b c c 2a 3b 0 ?K ?? ?M ?? ?M ?K ?] 
 ?H ?I ?H ?I a b c 2a 3b c 0 ?K ?? ?K ?M ?K ?? ?] 
 ?? ?H ?I a b 2a 3b) c 0 ?K ?M ?K ?? ?] 
 ?? c (4b a) ?] ?? ?M 
 ??  ?H ?I ˆ ˆ ˆ ˆ ˆˆ
44i 4 j 4k 4i j k ?] ?? ?M ?K ?M ?K ?M 
 ?H ?I ˆˆ ˆ
40i 3j 3k ?] ?? ?M ?K 
 Now 
?H ?I ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
8i 2 j 2k 33i 3j 3k . (40i 3j 3k) 1670 ?M ?K ?K ?M ?K ?? ?M ?K ?] 
 ?? ?H ?I ˆ ˆ ˆ ˆ ˆˆ
41i 5j 5k .(40i 3j 3k) 1670 ?M ?K ?M ?K ?? ?? ?] ) 
 ??  (1640 + 15 + 15) ?? = 1670 ?? ?? = 1 
 so 
ˆˆ ˆ
c 40i 3j 3k ?] ?M ?M   
 ??  
2
c 1600 9 9 1618 ?] ?K ?K ?] 
Page 5


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Monday 08
th
 April, 2024) TIME : 3 : 00 PM  to  6 : 00 PM 
NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. If the image of the point (-4, 5) in the line
x + 2y = 2 lies on the circle (x + 4)
2
 + (y-3)
2
 = r
2
,
then r is equal lo :
(1) 1 (2) 2
(3) 75 (4) 3
Ans. (2) 
Sol. Image of point (–4, 5) 
1 1 1 1
2 2
x x y y ax by c
2
a b a b
?M ?M ?K ?K ????
?] ?] ?M????
?K ????
Line : x + 2y – 2 = 0 
2 2
x 4 y 5 4 10 2
2
1 2 1 2
?K ?M ?M ?K ?M????
?] ?] ?M????
?K ????
    
8
5
?M ?] 8 28
x 4
5 5
?] ?M ?M ?] ?M 16 9
y 5
5 5
?] ?M ?K ?] Point lies on circle (x + 4)
2
 + (y – 3)
2
 = r
2
 
2
2
64 9
3 r
25 5
????
?K ?M ?] ????
????
2
100
r , r 2
25
?]?]
2. Let 
ˆˆ ˆ
a i 2j 3k ?] ?K ?K , 
ˆˆ ˆ
b 2i 3j 5k ?] ?K ?M and
ˆˆ ˆ
c 3i j k ?] ?M ?K ?? be three vectors. Let r be a unit
vector along b c ?K . If r . a 3 ?] , then 3 ?? is equal
to :
(1) 27 (2) 25
(3) 25 (4) 21
Ans. (2) 
Sol. ?H ?I r k b c ?]?K 
r . a 3 ?] 
?H ?I r.a k b.a c.a ?]?K
3 = k(2 + 6 – 15 + 3 – 2 + 3 ?? ) 
3 = k(–6 + 3 ?? ) …(1) 
ˆˆ ˆ
r k(5i 2j (5 )k) ?] ?K ?M ?M ?? 2
r k 25 4 25 10 1 ?] ?K ?K ?K ?? ?M ?? ?] …(2)
3 1
k
6 3 2
?]?]
?M ?K ?? ?M ?K ?? put in (2) 
4 + ?? 2
 – 4 ?? = 54 + ?? 2
 – 10 ?? 
6 ?? = 50 
3 ?? = 25 
3. If ???@ ?? ?@ a, ?? ?@ ?? ?@ b, ?? ?@ ?? ?@ c and
b c
a c 0
a b
?? ???]
?? , then 
a b
a b c
?? ?K?K
?? ?M ?? ?M ?? ?M  is equal to : 
(1) 2 (2) 3
(3) 0 (4) 1
Ans. (3) 
Sol. R
1
 ?? R
 1
– R
 2
, R
 2
?? ?@ R
 2
– R
 3
a b 0
0 b c 0
a b
?? ?M ?M ?? ?? ?M ?M ?? ?] ?? ( ?? –a) ( ?? ( ?? –b) – b(c– ?? )) – (b – ?? ) (– a(c - ?? )) = 0
?? ( ?? – a) ( ?? – b) – b( ?? – a) (c – ?? ) + a(b - ?? ) (c – ?? )
b a
0
c b a
?? ?K ?K ?] ?? ?M ?? ?M ?? ?M 4. In an increasing geometric progression ol positive 
terms, the sum of the second and sixth terms is 
70
3
and the product of the third and fifth terms is 
49. Then the sum of the 4
th
, 6
th
 and 8
th
 terms is :- 
 (1) 96 (2) 78 
 (3) 91  (4) 84 
 Ans. (3) 
Sol. 
2 6
70
T T
3
?K?] 
 ar + ar
5
 = 
70
3
 
 T
3
 . T
5
 = 49 
 ar
2
 . ar
4
 = 49 
 a
2
r
6
 = 49 
 ar
3
 = +7, 
3
7
a
r
?] 
 ar(1 + r
4
) = 
70
3
 
 
4 2
2
7 70
(1 r ) ,r t
3 r
?K ?] ?] 
 
2
1 10
(1 t )
t 3
?K?] 
 3t
2
 – 10t + 3 = 0 
 
1
t 3,
3
?] 
 Increasing G.P. r
2
 = 3, r 3 ?] 
 T
4
 + T
6
 + T
8
 
 = ar
3
 + ar
5
 + ar
7
 
 = ar
3
(1 + r
2
 + r
4
) 
 = 7(1 + 3 + 9) = 91 
5. The number of ways five alphabets can be chosen 
from the alphabets of the word MATHEMATICS, 
where the chosen alphabets are not necessarily 
distinct, is equal to : 
 (1) 175 (2) 181 
 (3) 177  (4) 179 
 Ans. (4) 
Sol. AA, MM, TT, H, I, C, S, E 
 (1) All distinct 
  
8
C
5
 ?? 56 
 (2) 2 same, 3 different 
  
3
C
?Q × 
7
C
3
 ?? 105 
 (3) 2 same I
st
 kind, 2 same 2
nd
 kind, 1 different 
  
3
C
2
 × 
6
C
1
 ?? 18 
  Total ?? 179 
6. The sum of all possible values of ???@?? [– ?@?? , 2 ?? ], for 
which 
1 icos
1 2icos
?K??
?M??
 is purely imaginary, is equal 
to 
 (1) 2 ?? (2) 3 ?? 
 (3) 5 ??  (4) 4 ?? 
 Ans. (2) 
Sol. 
1 icos
Z
1 2icos
?K??
?] ?M??
 
 Z Z ?]?M ??  
1 icos 1 icos
1 2icos 1 2icos
????
?K ?? ?K ?? ?]?M
????
?M ?? ?M ?? ????
 
 (1+ icos ?? ) ?H ?I 1 2icos ?M??
=–(1 – 2i cos ?? ) ?H ?I 1 icos ?K??
 
 (1+icos ?? ) (1 + 2icos ?? ) = –(1 – 2icos ?? ) (1 – icos ?? ) 
 1 + 3icos ?? – 2cos
2
?? = –(1 – 3icos ?? – 2cos
2
?? ) 
 2 – 4cos
2
?? = 0  
 ???@ cos
2
?? = 
1
2
?@?? 
3 3 5 7
, , , , ,
4 4 4 4 4 4
?? ?? ?? ?? ?? ?? ?? ?] ?M ?M 
 sum = 3 ?? 
  
7. If the system of equations x + 4y – z = ?? ,  
7x + 9y + ?? z = –3, 5x + y + 2z = –1 has infinitely 
many solutions, then (2 ?? . + 3 ?? ) is equal to : 
 (1) 2 (2) –3 
 (3) 3  (4) –2 
 Ans. (2) 
Sol. 
1 4 1
7 9 0
5 1 2
?M ?d ?] ?? ?] 
 ??  (18– ?? ) – 4(14–5 ?? ) – (7 – 45) = 0 ?? ?@?? = 0 
 ?d = ?d x
 = ?d y
 = ?d z
 = 0 (For infinite solution) 
 
x
4 1
3 9 0
1 1 2
???M
?d ?] ?M ?? ?] ?M 
 ?? (18 – ?? ) – 4(–6 + ?? ) –1(–3 + 9) = 0 
 18 ?? + 24 – 6 = 0  ?? ?? = –1 ?@ 
8. If the shortest distance between the lines 
x y 4 z 3
2 3 4
?M ?? ?M ?M ?]?] and 
x 2 y 4 z 7
4 6 8
?M ?M ?M ?]?] is 
13
29
, then a value 
of ?? is : 
 (1) –
13
25
 (2) 
13
25
 
 (3) 1  (4) –1 
 Ans. (3) 
Sol. 
?H ?I ?H ?I 1
1
2
2
ˆ ˆ
b 2i 3j 4k
ˆ ˆ ˆ ˆ ˆˆ
r i 4 j 3k (2i 3j 4k)
ˆˆ ˆ
a i 4 j 3k
ˆ ˆ ˆ ˆ ˆˆ
r 2i 4 j 7k (2i 3j 4k)
ˆˆ ˆ
a 2i 4 j 7k
?] ?K ?K ?? ?? ?] ?? ?K ?K ?K ?? ?K ?K ?K ?? ?K ?K ?? ?] ?K ?K ?K ?? ?K ?K ?? ?? ?]?K?K
 
 
2 1
b (a a )
13
Shortest dist.
b
29
???M
?]?]  
 
?H ?I ?H ?I ˆ ˆ ˆ ˆˆ
2i 3j 4k (2 )i 4k
13
29 29
?K ?K ?? ?M ?? ?K ?] 
 
ˆ ˆ ˆ ˆ
8j 3(2 )k 12i 4(2 ) j 13 ?M ?M ?M ?? ?K ?K ?M ?? ?] 
 
ˆˆ ˆ
12i 4 j (3 6)k 13 ?M ?? ?K ?? ?M ?] 
 144 + 16?@??
 2
 + (3 ?? – 6)
2
 = 169 
 16 ?? 2
 + (3 ?? – 6)
2
 = 25 = ?? ?@?? = 1 
9. If the value of 
3cos36 5sin18
5cos36 3sin18
?? ?K ?? ?? ?M ?? is 
a 5 b
c
?M , 
where a, b, c are natural numbers and gcd(a, c) = 1, 
then a + b + c is equal to : 
 (1) 50 (2) 40 
 (3) 52  (4) 54 
 Ans. (3) 
Sol. 
?H ?I 3 5 1 5 1
5
8 5 2
4 4
2 5 8 5 1 5 1
5 3
4 4
????
?K?M
?K????
?M ????
?] ?? ?? ?? ?? ?K ?K?M
?M ?? ?? ?? ?? ?? ?? ?? ?? 
 
4 5 1 5 4
5 4 5 4
?M?M
?]??
?K?M
 
 
20 16 5 5 4
11
?M ?M ?K ?] ?M 
 
17 5 24
11
?M ?] ?? a = 17, b = 27, c = 11 
 a + b + c = 52 
10. Let y = y(x) be the solution curve of the 
differential equation secy 
dy
dx
 + 2xsiny = x
3
cosy, 
y(1) = 0. Then 
?H ?I y 3 is equal to : 
 (1) 
3
?? (2) 
6
?? 
 (3) 
4
??  (4) 
12
?? 
 Ans. (3) 
Sol. sec
2
y 
dy
dx
 + 2xsiny secy = x
3
cosy secy 
 sec
2
y 
dy
dx
 + 2xtany = x
3
 
 tany = t ?@ ?? 
2
dy dt
sec y
dx dx
?] 
 
3
dt
2xt x
dx
?K?] , If 
2 2xdx
x
e e
???]?] 
 
2 2
x 3 x
te x .e dx c ?]?K
?? 
 x
2
 = Z  ??  
?{ ?} Z Z Z Z
1 1
t.e e .ZdZ e .Z e c
2 2
?] ?] ?M ?K ?? 
 
2
2 x
2tan y (x 1) 2ce
?M ?] ?M ?K 
 y(1) = 0  ?? c = 0 ??  y( 3)
4
?? ?] 
11. The area of the region in the first quadrant inside 
the circle x
2
 + y
2
 = 8 and outside the pnrabola   
y
2
 = 2x is equal to : 
 (1) 
1
2 3
?? ?M (2) 
2
3
???M 
 (3) 
2
2 3
?? ?M  (4) 
1
3
???M 
 Ans. (2) 
Sol.
  
(2 ,0) 
x=2 
 
 Required area = Ar(circle from 0 to 2) – 
 ar(para from 0 to 2) 
 
2
2
2
0
0
8 x dx 2x dx ?] ?M ?M ????
 
 
2
2
2 1
0
0
x 8 x x x
8 x sin 2
2 2 3 / 2
2 2
?M ????
????
?] ?M ?K ?M ????
????
????
????
 
 ?H ?I 1
2 8 2 2 2
8 4 sin 2 2 0
2 2 3
2 2
?M ?] ?M ?K ?M ?M 
 ??  
8 2
2 4 .
4 3 3
?? ?K ?M ?] ?? ?M 
12. If the line segment joining the points (5, 2) and  
(2, a) subtends an angle 
4
?? at the origin, then the 
absolute value of the product of all possible values 
of a is : 
 (1) 6 (2) 8 
 (3) 2  (4) 4 
 Ans. (4) 
Sol. 
 
B(2, a) 
?? /4 
A(5,2) 
O 
 
 
OA
2
m
5
?] 
 
OB
a
m
2
?] 
 
2 a
tan
5 2
4
?? ?M ?] 
 
4 5a
1
10 2a
?M ?] ?K 
 4 – 5a = ±(10 + 2a)  
 
 4 – 5a = 10 + 2a 4 – 5a = –10 – 2a 
 ?? 7a + 6 = 0 3a = 14 
 ?? 
6
a
7
?]?M  
14
a
3
?]?K 
 
6 14
4
7 3
?M ?? ?] ?M  
13. Let 
ˆˆ ˆ
a 4i j k ?] ?M ?K , 
ˆˆ ˆ
b 11i j k ?] ?M ?K and c be 
a vector such that 
 ?H ?I ?H ?I a b c c 2a 3b ?K ?? ?] ?? ?M ?K .  
 If ?H ?I 2a 3b .c 1670 ?K?] , then 
2
| c | is equal to :
 
 (1) 1627 (2) 1618 
 (3) 1600  (4) 1609 
 Ans. (2) 
Sol. ?H ?I ?H ?I a b c c 2a 3b 0 ?K ?? ?M ?? ?M ?K ?] 
 ?H ?I ?H ?I a b c 2a 3b c 0 ?K ?? ?K ?M ?K ?? ?] 
 ?? ?H ?I a b 2a 3b) c 0 ?K ?M ?K ?? ?] 
 ?? c (4b a) ?] ?? ?M 
 ??  ?H ?I ˆ ˆ ˆ ˆ ˆˆ
44i 4 j 4k 4i j k ?] ?? ?M ?K ?M ?K ?M 
 ?H ?I ˆˆ ˆ
40i 3j 3k ?] ?? ?M ?K 
 Now 
?H ?I ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
8i 2 j 2k 33i 3j 3k . (40i 3j 3k) 1670 ?M ?K ?K ?M ?K ?? ?M ?K ?] 
 ?? ?H ?I ˆ ˆ ˆ ˆ ˆˆ
41i 5j 5k .(40i 3j 3k) 1670 ?M ?K ?M ?K ?? ?? ?] ) 
 ??  (1640 + 15 + 15) ?? = 1670 ?? ?? = 1 
 so 
ˆˆ ˆ
c 40i 3j 3k ?] ?M ?M   
 ??  
2
c 1600 9 9 1618 ?] ?K ?K ?] 
 
14. If the function f(x) = 2x
3
 – 9ax
2
 + 12a
2
x + 1, a > 0 
has a local maximum at x = ?? and a local 
minimum x = ?? 2
, then ?? and ?? 2
 are the roots of the 
equation : 
 (1) x
2
 – 6x + 8 = 0 (2) 8x
2
 + 6x – 8 = 0 
 (3) 8x
2
 – 6x + 1 = 0 (4) x
2
 + 6x + 8 = 0 
 Ans. (1) 
Sol.
  
f'(x) = 6x
2
 – 18ax + 12a
2
 = 0 
?? ?@ ?? 2
 
 
 ?? +  ?? 2
 = 3a & ?? × ?? 2
 = 2a
2
 
 ?@?@?@?@?@?? ?@ ( ?? + ?? 2
)
3
 = 27a
3
 
 ?? 2a
2
 + 4a
4
 + 3(3a) (2a
2
) = 27a
3
 
 ??  2 + 4a
2
 + 18a = 27a 
 ??  4a
2
 – 9a + 2 = 0 
 ??  4a
2
 – 8a – a + 2 = 0 
 ??  (4a – 1) (a – 2) = 0 ?? a = 2 
 so 6x
2
 – 36x + 48 = 0 
 ?? x
2
 – 6x + 8 = 0 (1) 
 If we take 
1
a
4
?] then 
1
2
???] which is not possible 
15. There are three bags X, Y and Z. Bag X contains 5 
one-rupee coins and 4 five-rupee coins; Bag Y 
contains 4 one-rupee coins and 5 five-rupee coins 
and Bag Z contains 3 one-rupee coins and  
6 five-rupee coins. A bag is selected at random and 
a coin drawn from it at random is found to be a 
one-rupee coin. Then the probability, that it came 
from bag Y, is : 
 (1) 
1
3
 (2) 
1
2
 
 (3) 
1
4
  (4) 
5
12
 
 Ans. (1) 
Sol. X                    Y       Z 
 5 one & 4 five    4 one & 5 five    3 one & 6 five 
 
4 / 9 4 1
P
5 / 9 4 / 9 3 / 9 12 3
?] ?] ?] ?K?K
 
16. Let 
log 4
x
e
dx
6
e 1
?? ?? ?] ?M ?? . Then e
?? and e
– ?? are the 
roots of the equation : 
 (1) 2x
2
 – 5x + 2 = 0 (2) x
2
 – 2x - 8 = 0 
 (3) 2x
2
 – 5x – 2 = 0 (4) x
2
 + 2x – 8 = 0 
 Ans. (1) 
Sol. 
log 4
x
e
dx
6
e 1
?? ?? ?] ?M ?? 
 Let e
x
 – 1 = t
2 
 
e
x
 dx = 2t dt 
 
2
2dt
t 1
?] ?K ?? 
 = 2 tan
–1
t 
 ?H ?I 4
e
log
1 x
2 tan e 1
?M ?? ?]?M 
 
1 1
2 tan 3 tan e 1
6
?M ?M ?? ?? ????
?] ?M ?M ?] ????
 
 
1
tan e 1
3 12
?M??
????
?] ?M ?M ?] 
 ?? 
1
tan e 1
4
?M??
?? ?M?] 
 e 2
?? ?]      
1
e
2
?M??
?] 
 
2
1
x 2 x 1 0
2
????
?M ?K ?K ?] ????
????
 
 2x
2
 – 5x + 2 = 0  
17. Let 
?? a if a x 0
(x)
x a if 0 x a
?M ?M ?? ?? ?] ?K ?\ ?? f  
 where a > 0 and g(x) = (f |x| ) – | f (x)| )/2.  
 Then the function g : [ –a, a] ?? [ –a, a] is 
 (1) neither one-one nor onto. 
 (2) both one-one and onto. 
 (3) one-one. 
 (4) onto 
 Ans. (1)