JEE Main 2024 April 6 Shift 2 Paper & Solutions | JEE Main & Advanced Mock Test Series CY_Marker_0 PDF Download

 Page 1


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Saturday 06
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM 
NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ABC be an equilateral triangle. A new triangle
is formed by joining the middle points of all sides
of the triangle ABC and the same process is
repeated infinitely many times. If P is the sum of
perimeters and Q is be the sum of areas of all the
triangles formed in this process, then:
(1)
2
P 36 3Q ?] (2) 
2
P 6 3Q ?] (3) 
2
P 36 3Q ?] (4) 
2
P 72 3Q ?] Ans. (1) 
Sol. 
a/2
 
a
 
a
 
a
 
Area of first 
2
3a
4
?d?]
Area of second 
2 2 2
3a a 3a
4 4 16
?d ?] ?] Area of third 
2
a
64
?S ?d?]
sum of area = 
2
3a 1 1
1 ....
4 4 16
????
?K?K
????
????
2 2
3a 1 a
Q
3
4
3
4
?]?] 
perimeter of 1
st
?d = 3a
perimeter of 2
nd
 ?d = 
3a
2
perimeter of 3
rd
 ?d =
3a
4
1 1
P 3a 1 ...
2 4
????
?] ?K ?K ?K????
????
P = 3a.2 = 6a 
P
a
6
?] 2
1 P
Q
36
3
?]??
2
P 36 3Q ?] 2. Let A = {1, 2, 3, 4, 5}. Let R be a relation on A
defined by xRy if and only if 4x ?? ?@ 5y. Let m be the
number of elements in R and n be the minimum
number of elements from A × A that are required
to be added to R to make it a symmetric relation.
Then m + n is equal to:
(1) 24 (2) 23
(3) 25 (4) 26
Ans. (3) 
Sol. Given :  4x ?? 5y 
then 
?H ?I ?? R 1,1 ,(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4)
(2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)}
?] i.e. 16 elements.
i.e. m = 16
Now to make R a symmetric relation add 
{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)} 
i.e. n = 9
So m + n = 25 
3. If three letters can be posted to any one of the 5
different addresses, then the probability that the
three letters are posted to exactly two addresses is:
(1) 
12
25
(2) 
18
25
(3) 
4
25
(4) 
6
25
Ans. (1) 
Sol. Total method = 5
3
 
faverable = 
?H ?I 5 3
2
C 2 – 2 60 ?] probability = 
60 12
125 25
?]
Page 2


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Saturday 06
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM 
NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ABC be an equilateral triangle. A new triangle
is formed by joining the middle points of all sides
of the triangle ABC and the same process is
repeated infinitely many times. If P is the sum of
perimeters and Q is be the sum of areas of all the
triangles formed in this process, then:
(1)
2
P 36 3Q ?] (2) 
2
P 6 3Q ?] (3) 
2
P 36 3Q ?] (4) 
2
P 72 3Q ?] Ans. (1) 
Sol. 
a/2
 
a
 
a
 
a
 
Area of first 
2
3a
4
?d?]
Area of second 
2 2 2
3a a 3a
4 4 16
?d ?] ?] Area of third 
2
a
64
?S ?d?]
sum of area = 
2
3a 1 1
1 ....
4 4 16
????
?K?K
????
????
2 2
3a 1 a
Q
3
4
3
4
?]?] 
perimeter of 1
st
?d = 3a
perimeter of 2
nd
 ?d = 
3a
2
perimeter of 3
rd
 ?d =
3a
4
1 1
P 3a 1 ...
2 4
????
?] ?K ?K ?K????
????
P = 3a.2 = 6a 
P
a
6
?] 2
1 P
Q
36
3
?]??
2
P 36 3Q ?] 2. Let A = {1, 2, 3, 4, 5}. Let R be a relation on A
defined by xRy if and only if 4x ?? ?@ 5y. Let m be the
number of elements in R and n be the minimum
number of elements from A × A that are required
to be added to R to make it a symmetric relation.
Then m + n is equal to:
(1) 24 (2) 23
(3) 25 (4) 26
Ans. (3) 
Sol. Given :  4x ?? 5y 
then 
?H ?I ?? R 1,1 ,(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4)
(2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)}
?] i.e. 16 elements.
i.e. m = 16
Now to make R a symmetric relation add 
{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)} 
i.e. n = 9
So m + n = 25 
3. If three letters can be posted to any one of the 5
different addresses, then the probability that the
three letters are posted to exactly two addresses is:
(1) 
12
25
(2) 
18
25
(3) 
4
25
(4) 
6
25
Ans. (1) 
Sol. Total method = 5
3
 
faverable = 
?H ?I 5 3
2
C 2 – 2 60 ?] probability = 
60 12
125 25
?] 4. Suppose the solution of the differential equation 
dy (2 )x y 2
dx x 2 y ( 4 )
?K ?? ?M ?? ?K ?] ?? ?M ?? ?M ?? ?? ?M ?? represents a circle 
passing through origin. Then the radius of this 
circle is : 
 (1) 17 (2) 
1
2
 
 (3) 
17
2
  (4) 2 
 Ans. (3) 
Sol. 
?H ?I ?H ?I dy 2 x – y 2
dx x – y 2 4
?K ?? ?? ?K ?] ?? ?? ?K ?? ?K ?? 
 ?H ?I xdy – (2 )ydy 4 dy 2 xdx – ydx 2dx ?? ?? ?K ?? ?K ?? ?] ?K ?? ?? ?K 
 ?H ?I (xdy ydx) – (2 )ydy 4 dy 2 xdx 2dx ?? ?K ?? ?K ?? ?K ?? ?] ?K ?? ?K 
 
?H ?I ?H ?I 2 2
2 y 2 x
xy – 4 y
2 2
?? ?K ?? ?K??
?? ?K ?? ?] 
 0 ?? ?? ?] for this to be circle  
 
?H ?I 2
2
x
2 y 2x – 4 y 0
2
?K ?? ?K ?? ?K ?? ?] 
  
2 + a = 2a 
coeff. of  
x
2
 = y
2 
 
 2 ?? ?? ?] 
 i.e. 2x
2
 + 2y
2
 + 2x – 8y = 0 
 x
2
 + y
2
 + x – 4y = 0 
 
1 17
rd 4
4 2
?] ?K ?]  
5. If the locus of the point, whose distances from the 
point (2, 1) and (1, 3) are in the ratio 5 : 4, is  
ax
2
 + by
2
 + cxy + dx + ey + 170 = 0, then the value 
of a
2
 + 2b + 3c + 4d + e is equal to:  
 (1) 5 (2) –27 
 (3) 37  (4) 437 
 Ans. (3) 
Sol. let P(x, y) 
 
?H ?I ?H ?I ?H ?I 2
2
2 2
x – 2 (y – 1) 25
16
x – 1 y – 3
?K ?] ?K 
 9x
2
 + 9y
2
 + 14x – 118y + 170 = 0 
 a
2
 + 2b + 3c + 4d + e 
 = 81 + 18 + 0 + 56 – 118 
 = 155 – 118 
 = 37 
6.  
?H ?I 2 2
2
3 3 3 2 2 2
n
(1 1)(n 1) (2 2)(n 2) ..... · 1
(n 1) (n 1)
lim
(1 2 ..... n ) (1 2 ..... n )
????
?M ?M ?K ?M ?M ?K ?K ?M ?M ?M ?K ?K ?K ?M ?K ?K ?K is equal to: 
 (1) 
2
3
 (2) 
1
3
 
 (3) 
3
4
  (4) 
1
2
 
 Ans. (2) 
Sol. 
?H ?I ?H ?I n–1
2
r 1
n n
n
3 2
r 1 r 1
r – r n – r
lim
r – r
?] ????
?]?]
?? ????
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I n–1
3 2
r 1
2
n
–r r n 1 – nr
lim
n n 1 n n 1 2n 1
–
2 6
?] ????
?K?K
???? ?K ?K ?K ????
????
?? 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 2
2
n
n – 1 n n 1 n – 1 n(2n – 1) n (n – 1)
–
2 6 2
lim
n(n 1) n n 1 2n 1
–
2 2 3
????
????
?K ?K ????
????
???? ?K ?K ?K ????
????
 
 
?H ?I ?H ?I ?H ?I 2
n
n n – 1 –n n – 1 n 1 (2n – 1)
–n
2 2 3
lim
n(n 1) 3n 3n – 4n – 2
2 6
????
???? ?K ?K????
????
?K?K
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 2 2
2
n
n – 1 –3n 3n 2 2n n – 1 – 6
lim
n 1 3n – n – 2
????
?K ?K ?K ?K 
 
?H ?I ?H ?I ?H ?I ?H ?I 2
2
n
n –1 n 5n – 8 1
lim
3
n 1 3n – n – 2
????
?K ?] ?K 
7. Let 0 ?? r ?? n. If 
n+1
C
r+1
 : 
n
C
r
 : 
n–1
C
r–1
 = 55 : 35 : 21, 
then 2n + 5r is equal to: 
 (1) 60 (2) 62 
 (3) 50  (4) 55 
 Ans. (3) 
Ans. 
n 1
r
n
r
C 55
35 C
?K ?] 
 
?H ?I ?H ?I ?H ?I n 1 ! r! n – r ! 11
!
r 1 !(n – r) n! 7
?K ?] ?K 
 
?H ?I n 1 11
r 1 7
?K ?] ?K 
Page 3


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Saturday 06
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM 
NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ABC be an equilateral triangle. A new triangle
is formed by joining the middle points of all sides
of the triangle ABC and the same process is
repeated infinitely many times. If P is the sum of
perimeters and Q is be the sum of areas of all the
triangles formed in this process, then:
(1)
2
P 36 3Q ?] (2) 
2
P 6 3Q ?] (3) 
2
P 36 3Q ?] (4) 
2
P 72 3Q ?] Ans. (1) 
Sol. 
a/2
 
a
 
a
 
a
 
Area of first 
2
3a
4
?d?]
Area of second 
2 2 2
3a a 3a
4 4 16
?d ?] ?] Area of third 
2
a
64
?S ?d?]
sum of area = 
2
3a 1 1
1 ....
4 4 16
????
?K?K
????
????
2 2
3a 1 a
Q
3
4
3
4
?]?] 
perimeter of 1
st
?d = 3a
perimeter of 2
nd
 ?d = 
3a
2
perimeter of 3
rd
 ?d =
3a
4
1 1
P 3a 1 ...
2 4
????
?] ?K ?K ?K????
????
P = 3a.2 = 6a 
P
a
6
?] 2
1 P
Q
36
3
?]??
2
P 36 3Q ?] 2. Let A = {1, 2, 3, 4, 5}. Let R be a relation on A
defined by xRy if and only if 4x ?? ?@ 5y. Let m be the
number of elements in R and n be the minimum
number of elements from A × A that are required
to be added to R to make it a symmetric relation.
Then m + n is equal to:
(1) 24 (2) 23
(3) 25 (4) 26
Ans. (3) 
Sol. Given :  4x ?? 5y 
then 
?H ?I ?? R 1,1 ,(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4)
(2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)}
?] i.e. 16 elements.
i.e. m = 16
Now to make R a symmetric relation add 
{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)} 
i.e. n = 9
So m + n = 25 
3. If three letters can be posted to any one of the 5
different addresses, then the probability that the
three letters are posted to exactly two addresses is:
(1) 
12
25
(2) 
18
25
(3) 
4
25
(4) 
6
25
Ans. (1) 
Sol. Total method = 5
3
 
faverable = 
?H ?I 5 3
2
C 2 – 2 60 ?] probability = 
60 12
125 25
?] 4. Suppose the solution of the differential equation 
dy (2 )x y 2
dx x 2 y ( 4 )
?K ?? ?M ?? ?K ?] ?? ?M ?? ?M ?? ?? ?M ?? represents a circle 
passing through origin. Then the radius of this 
circle is : 
 (1) 17 (2) 
1
2
 
 (3) 
17
2
  (4) 2 
 Ans. (3) 
Sol. 
?H ?I ?H ?I dy 2 x – y 2
dx x – y 2 4
?K ?? ?? ?K ?] ?? ?? ?K ?? ?K ?? 
 ?H ?I xdy – (2 )ydy 4 dy 2 xdx – ydx 2dx ?? ?? ?K ?? ?K ?? ?] ?K ?? ?? ?K 
 ?H ?I (xdy ydx) – (2 )ydy 4 dy 2 xdx 2dx ?? ?K ?? ?K ?? ?K ?? ?] ?K ?? ?K 
 
?H ?I ?H ?I 2 2
2 y 2 x
xy – 4 y
2 2
?? ?K ?? ?K??
?? ?K ?? ?] 
 0 ?? ?? ?] for this to be circle  
 
?H ?I 2
2
x
2 y 2x – 4 y 0
2
?K ?? ?K ?? ?K ?? ?] 
  
2 + a = 2a 
coeff. of  
x
2
 = y
2 
 
 2 ?? ?? ?] 
 i.e. 2x
2
 + 2y
2
 + 2x – 8y = 0 
 x
2
 + y
2
 + x – 4y = 0 
 
1 17
rd 4
4 2
?] ?K ?]  
5. If the locus of the point, whose distances from the 
point (2, 1) and (1, 3) are in the ratio 5 : 4, is  
ax
2
 + by
2
 + cxy + dx + ey + 170 = 0, then the value 
of a
2
 + 2b + 3c + 4d + e is equal to:  
 (1) 5 (2) –27 
 (3) 37  (4) 437 
 Ans. (3) 
Sol. let P(x, y) 
 
?H ?I ?H ?I ?H ?I 2
2
2 2
x – 2 (y – 1) 25
16
x – 1 y – 3
?K ?] ?K 
 9x
2
 + 9y
2
 + 14x – 118y + 170 = 0 
 a
2
 + 2b + 3c + 4d + e 
 = 81 + 18 + 0 + 56 – 118 
 = 155 – 118 
 = 37 
6.  
?H ?I 2 2
2
3 3 3 2 2 2
n
(1 1)(n 1) (2 2)(n 2) ..... · 1
(n 1) (n 1)
lim
(1 2 ..... n ) (1 2 ..... n )
????
?M ?M ?K ?M ?M ?K ?K ?M ?M ?M ?K ?K ?K ?M ?K ?K ?K is equal to: 
 (1) 
2
3
 (2) 
1
3
 
 (3) 
3
4
  (4) 
1
2
 
 Ans. (2) 
Sol. 
?H ?I ?H ?I n–1
2
r 1
n n
n
3 2
r 1 r 1
r – r n – r
lim
r – r
?] ????
?]?]
?? ????
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I n–1
3 2
r 1
2
n
–r r n 1 – nr
lim
n n 1 n n 1 2n 1
–
2 6
?] ????
?K?K
???? ?K ?K ?K ????
????
?? 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 2
2
n
n – 1 n n 1 n – 1 n(2n – 1) n (n – 1)
–
2 6 2
lim
n(n 1) n n 1 2n 1
–
2 2 3
????
????
?K ?K ????
????
???? ?K ?K ?K ????
????
 
 
?H ?I ?H ?I ?H ?I 2
n
n n – 1 –n n – 1 n 1 (2n – 1)
–n
2 2 3
lim
n(n 1) 3n 3n – 4n – 2
2 6
????
???? ?K ?K????
????
?K?K
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 2 2
2
n
n – 1 –3n 3n 2 2n n – 1 – 6
lim
n 1 3n – n – 2
????
?K ?K ?K ?K 
 
?H ?I ?H ?I ?H ?I ?H ?I 2
2
n
n –1 n 5n – 8 1
lim
3
n 1 3n – n – 2
????
?K ?] ?K 
7. Let 0 ?? r ?? n. If 
n+1
C
r+1
 : 
n
C
r
 : 
n–1
C
r–1
 = 55 : 35 : 21, 
then 2n + 5r is equal to: 
 (1) 60 (2) 62 
 (3) 50  (4) 55 
 Ans. (3) 
Ans. 
n 1
r
n
r
C 55
35 C
?K ?] 
 
?H ?I ?H ?I ?H ?I n 1 ! r! n – r ! 11
!
r 1 !(n – r) n! 7
?K ?] ?K 
 
?H ?I n 1 11
r 1 7
?K ?] ?K 
 
 7n = 4 + 11r 
 
n
r
n–1
r–1
C 35
21 C
?] 
 
?H ?I n! (r –1)!(n – r)! 5
r! n – r ! (n –1)! 3
?]?] 
 
n 5
r 3
?] 
 3n = 5r 
 By solving r = 6  n = 10 
 2n + 5r = 50 
8. A software company sets up m number of 
computer systems to finish an assignment in  
17 days. If 4 computer systems crashed on the start 
of the second day, 4 more computer systems 
crashed on the start of the third day and so on, then 
it took 8 more days to finish the assignment. The 
value of m is equal to : 
 (1) 125 (2) 150 
 (3) 180  (4) 160 
 Ans. (2) 
Sol. 17m = m + (m – 4) + (m – 4 × 2)...+...(m – 4 × 24) 
 17 m = 25m – 4 (1 + 2 ...24) 
 
4 24 25
8m 150
2
????
?]?] 
9. If z
1
, z
2
 are two distinct complex number such that 
1 2
1 2
z 2z
2
1
z z
2
?M ?] ?M , then  
(1) either z
1
 lies on a circle of radius 1 or z
2
 lies on a 
circle of radius 
1
2
 
(2) either z
1
 lies on a circle of radius 
1
2
 or z
2
 lies on 
a circle of radius 1. 
(3) z
1
 lies on a circle of radius 
1
2
 and z
2
 lies on a 
circle of radius 1. 
(4) both z
1
 and z
2
 lie on the same circle. 
 Ans. (1) 
Sol. 
1 2 1 2
1 2 1 2
z – 2z z – 2z
4
1 1
– z z – z z
2 2
???] 
 
2 2
1 1 2 1 2 2
z 2z z – 2z z 4 z ?K 
 
2 2
1 2 1 2
1 2
z z z z 1
4 – – z z
4 2 2
????
?]?K
????
????
 
 
1 1 2 2 1 1 2 2
z z 2z 2z – z z 2z 2z – 1 0 ?K ?? ?] 
 ?H ?I ?H ?I 1 2 2
z,z –1 1– 2z 2z 0 ???] 
 ?H ?I ?H ?I 2 2
1 2
z –1 2z –1 0 ?] 
10. If the function 
2x
1
f(x)
x
????
?] ????
????
 ; x > 0 attains the 
maximum value at 
1
x
e
?] then : 
 (1) 
e
e
?? ?\?? (2) 
2 e
e (2 )
?? ?\?? 
 (3) 
e
e
?? ?^??  (4) 
(2e)
(2e)
?? ?^?? 
 Ans. (3) 
Sol. Let 
2x
1
y
x
????
?] ????
????
 
 ?? ny = 
1
2x n
x
????
????
????
 
 ny –2x nx ?] 
 
?H ?I 1 dy
–2 1 nx
y dx
?]?K 
 for 
1
x
e
?^ f
n
 is decreasing  
 so, e < ?? ?@ ?@ 2e 2
1 1
e
?? ?? ?? ?? ?? ?^ ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 e
?? > 
 
?? e 
11. Let 
ˆˆ ˆ
a 6i j k ?] ?K ?M and 
ˆˆ
b i j ?]?K . If c is a is vector 
such that 6, a.c 6 , 2 2
c c c a
?? ?] ?] ?M and the 
angle between a b ?? and c is 60°, then ?H ?I c
a b
?? ?? 
is equal to:  
 (1) 
?H ?I 9
6 6
2
?M (2) 
3
3
2
 
 (3) 
3
6
2
  (4) 
?H ?I 9
6 6
2
?K 
 Ans. (4) 
Page 4


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Saturday 06
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM 
NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ABC be an equilateral triangle. A new triangle
is formed by joining the middle points of all sides
of the triangle ABC and the same process is
repeated infinitely many times. If P is the sum of
perimeters and Q is be the sum of areas of all the
triangles formed in this process, then:
(1)
2
P 36 3Q ?] (2) 
2
P 6 3Q ?] (3) 
2
P 36 3Q ?] (4) 
2
P 72 3Q ?] Ans. (1) 
Sol. 
a/2
 
a
 
a
 
a
 
Area of first 
2
3a
4
?d?]
Area of second 
2 2 2
3a a 3a
4 4 16
?d ?] ?] Area of third 
2
a
64
?S ?d?]
sum of area = 
2
3a 1 1
1 ....
4 4 16
????
?K?K
????
????
2 2
3a 1 a
Q
3
4
3
4
?]?] 
perimeter of 1
st
?d = 3a
perimeter of 2
nd
 ?d = 
3a
2
perimeter of 3
rd
 ?d =
3a
4
1 1
P 3a 1 ...
2 4
????
?] ?K ?K ?K????
????
P = 3a.2 = 6a 
P
a
6
?] 2
1 P
Q
36
3
?]??
2
P 36 3Q ?] 2. Let A = {1, 2, 3, 4, 5}. Let R be a relation on A
defined by xRy if and only if 4x ?? ?@ 5y. Let m be the
number of elements in R and n be the minimum
number of elements from A × A that are required
to be added to R to make it a symmetric relation.
Then m + n is equal to:
(1) 24 (2) 23
(3) 25 (4) 26
Ans. (3) 
Sol. Given :  4x ?? 5y 
then 
?H ?I ?? R 1,1 ,(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4)
(2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)}
?] i.e. 16 elements.
i.e. m = 16
Now to make R a symmetric relation add 
{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)} 
i.e. n = 9
So m + n = 25 
3. If three letters can be posted to any one of the 5
different addresses, then the probability that the
three letters are posted to exactly two addresses is:
(1) 
12
25
(2) 
18
25
(3) 
4
25
(4) 
6
25
Ans. (1) 
Sol. Total method = 5
3
 
faverable = 
?H ?I 5 3
2
C 2 – 2 60 ?] probability = 
60 12
125 25
?] 4. Suppose the solution of the differential equation 
dy (2 )x y 2
dx x 2 y ( 4 )
?K ?? ?M ?? ?K ?] ?? ?M ?? ?M ?? ?? ?M ?? represents a circle 
passing through origin. Then the radius of this 
circle is : 
 (1) 17 (2) 
1
2
 
 (3) 
17
2
  (4) 2 
 Ans. (3) 
Sol. 
?H ?I ?H ?I dy 2 x – y 2
dx x – y 2 4
?K ?? ?? ?K ?] ?? ?? ?K ?? ?K ?? 
 ?H ?I xdy – (2 )ydy 4 dy 2 xdx – ydx 2dx ?? ?? ?K ?? ?K ?? ?] ?K ?? ?? ?K 
 ?H ?I (xdy ydx) – (2 )ydy 4 dy 2 xdx 2dx ?? ?K ?? ?K ?? ?K ?? ?] ?K ?? ?K 
 
?H ?I ?H ?I 2 2
2 y 2 x
xy – 4 y
2 2
?? ?K ?? ?K??
?? ?K ?? ?] 
 0 ?? ?? ?] for this to be circle  
 
?H ?I 2
2
x
2 y 2x – 4 y 0
2
?K ?? ?K ?? ?K ?? ?] 
  
2 + a = 2a 
coeff. of  
x
2
 = y
2 
 
 2 ?? ?? ?] 
 i.e. 2x
2
 + 2y
2
 + 2x – 8y = 0 
 x
2
 + y
2
 + x – 4y = 0 
 
1 17
rd 4
4 2
?] ?K ?]  
5. If the locus of the point, whose distances from the 
point (2, 1) and (1, 3) are in the ratio 5 : 4, is  
ax
2
 + by
2
 + cxy + dx + ey + 170 = 0, then the value 
of a
2
 + 2b + 3c + 4d + e is equal to:  
 (1) 5 (2) –27 
 (3) 37  (4) 437 
 Ans. (3) 
Sol. let P(x, y) 
 
?H ?I ?H ?I ?H ?I 2
2
2 2
x – 2 (y – 1) 25
16
x – 1 y – 3
?K ?] ?K 
 9x
2
 + 9y
2
 + 14x – 118y + 170 = 0 
 a
2
 + 2b + 3c + 4d + e 
 = 81 + 18 + 0 + 56 – 118 
 = 155 – 118 
 = 37 
6.  
?H ?I 2 2
2
3 3 3 2 2 2
n
(1 1)(n 1) (2 2)(n 2) ..... · 1
(n 1) (n 1)
lim
(1 2 ..... n ) (1 2 ..... n )
????
?M ?M ?K ?M ?M ?K ?K ?M ?M ?M ?K ?K ?K ?M ?K ?K ?K is equal to: 
 (1) 
2
3
 (2) 
1
3
 
 (3) 
3
4
  (4) 
1
2
 
 Ans. (2) 
Sol. 
?H ?I ?H ?I n–1
2
r 1
n n
n
3 2
r 1 r 1
r – r n – r
lim
r – r
?] ????
?]?]
?? ????
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I n–1
3 2
r 1
2
n
–r r n 1 – nr
lim
n n 1 n n 1 2n 1
–
2 6
?] ????
?K?K
???? ?K ?K ?K ????
????
?? 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 2
2
n
n – 1 n n 1 n – 1 n(2n – 1) n (n – 1)
–
2 6 2
lim
n(n 1) n n 1 2n 1
–
2 2 3
????
????
?K ?K ????
????
???? ?K ?K ?K ????
????
 
 
?H ?I ?H ?I ?H ?I 2
n
n n – 1 –n n – 1 n 1 (2n – 1)
–n
2 2 3
lim
n(n 1) 3n 3n – 4n – 2
2 6
????
???? ?K ?K????
????
?K?K
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 2 2
2
n
n – 1 –3n 3n 2 2n n – 1 – 6
lim
n 1 3n – n – 2
????
?K ?K ?K ?K 
 
?H ?I ?H ?I ?H ?I ?H ?I 2
2
n
n –1 n 5n – 8 1
lim
3
n 1 3n – n – 2
????
?K ?] ?K 
7. Let 0 ?? r ?? n. If 
n+1
C
r+1
 : 
n
C
r
 : 
n–1
C
r–1
 = 55 : 35 : 21, 
then 2n + 5r is equal to: 
 (1) 60 (2) 62 
 (3) 50  (4) 55 
 Ans. (3) 
Ans. 
n 1
r
n
r
C 55
35 C
?K ?] 
 
?H ?I ?H ?I ?H ?I n 1 ! r! n – r ! 11
!
r 1 !(n – r) n! 7
?K ?] ?K 
 
?H ?I n 1 11
r 1 7
?K ?] ?K 
 
 7n = 4 + 11r 
 
n
r
n–1
r–1
C 35
21 C
?] 
 
?H ?I n! (r –1)!(n – r)! 5
r! n – r ! (n –1)! 3
?]?] 
 
n 5
r 3
?] 
 3n = 5r 
 By solving r = 6  n = 10 
 2n + 5r = 50 
8. A software company sets up m number of 
computer systems to finish an assignment in  
17 days. If 4 computer systems crashed on the start 
of the second day, 4 more computer systems 
crashed on the start of the third day and so on, then 
it took 8 more days to finish the assignment. The 
value of m is equal to : 
 (1) 125 (2) 150 
 (3) 180  (4) 160 
 Ans. (2) 
Sol. 17m = m + (m – 4) + (m – 4 × 2)...+...(m – 4 × 24) 
 17 m = 25m – 4 (1 + 2 ...24) 
 
4 24 25
8m 150
2
????
?]?] 
9. If z
1
, z
2
 are two distinct complex number such that 
1 2
1 2
z 2z
2
1
z z
2
?M ?] ?M , then  
(1) either z
1
 lies on a circle of radius 1 or z
2
 lies on a 
circle of radius 
1
2
 
(2) either z
1
 lies on a circle of radius 
1
2
 or z
2
 lies on 
a circle of radius 1. 
(3) z
1
 lies on a circle of radius 
1
2
 and z
2
 lies on a 
circle of radius 1. 
(4) both z
1
 and z
2
 lie on the same circle. 
 Ans. (1) 
Sol. 
1 2 1 2
1 2 1 2
z – 2z z – 2z
4
1 1
– z z – z z
2 2
???] 
 
2 2
1 1 2 1 2 2
z 2z z – 2z z 4 z ?K 
 
2 2
1 2 1 2
1 2
z z z z 1
4 – – z z
4 2 2
????
?]?K
????
????
 
 
1 1 2 2 1 1 2 2
z z 2z 2z – z z 2z 2z – 1 0 ?K ?? ?] 
 ?H ?I ?H ?I 1 2 2
z,z –1 1– 2z 2z 0 ???] 
 ?H ?I ?H ?I 2 2
1 2
z –1 2z –1 0 ?] 
10. If the function 
2x
1
f(x)
x
????
?] ????
????
 ; x > 0 attains the 
maximum value at 
1
x
e
?] then : 
 (1) 
e
e
?? ?\?? (2) 
2 e
e (2 )
?? ?\?? 
 (3) 
e
e
?? ?^??  (4) 
(2e)
(2e)
?? ?^?? 
 Ans. (3) 
Sol. Let 
2x
1
y
x
????
?] ????
????
 
 ?? ny = 
1
2x n
x
????
????
????
 
 ny –2x nx ?] 
 
?H ?I 1 dy
–2 1 nx
y dx
?]?K 
 for 
1
x
e
?^ f
n
 is decreasing  
 so, e < ?? ?@ ?@ 2e 2
1 1
e
?? ?? ?? ?? ?? ?^ ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 e
?? > 
 
?? e 
11. Let 
ˆˆ ˆ
a 6i j k ?] ?K ?M and 
ˆˆ
b i j ?]?K . If c is a is vector 
such that 6, a.c 6 , 2 2
c c c a
?? ?] ?] ?M and the 
angle between a b ?? and c is 60°, then ?H ?I c
a b
?? ?? 
is equal to:  
 (1) 
?H ?I 9
6 6
2
?M (2) 
3
3
2
 
 (3) 
3
6
2
  (4) 
?H ?I 9
6 6
2
?K 
 Ans. (4) 
Sol. 
?H ?I 3
a b c a b c
2
?? ?? ?] ?? 
 c – a 2 2 ?] 
 
2 2
c a – 2c a 8 ?K ?? ?] 
 
2
z 38 –12 z 8 ?K?] 
 
2
z –12 z 30 0 ?K?] 
 
12 144 –120
z
2
?? ?] 
 
12 2 6
2
?? ?] 
 z 6 6 ?]?K 
 
ˆˆ ˆ
j k
a b 6 1 –1
1 1 0
???] 
 
ˆˆ ˆ
– j 5k ?K 
 a b 27 ???] 
 
?H ?I ?H ?I 3
a b z 27 6 6
2
?? ?? ?] ?K 
 
?H ?I 9
6 6
2
?K 
12. If all the words with or without meaning made 
using all the letters of the word "NAGPUR" are 
arranged as in a dictionary, then the word at 315
th
 
position in this arrangement is : 
 (1) NRAGUP (2) NRAGPU 
 (3) NRAPGU (4) NRAPUG 
 Ans. (3) 
Sol. NAGPUR 
 A ?? 5! = 120 
 G ® 5! = 120  240 
 NA ® 4! = 24 264 
 NG ® 4! = 24 288 
 NP ® 4! = 24  312 
 NRAGPU = 1 313 
 NRAGUP  314 
 NRAPGU  315 
13. Suppose for a differentiable function h, h(0) = 0, 
h(1) = 1 and h'(0) = h'(1) = 2. If g(x) = h(e
x
) e
h(x)
, 
then g'(0) is equal to: 
 (1) 5 (2) 3 
 (3) 8  (4) 4 
 Ans. (4) 
Sol. 
?H ?I ?H ?I x h(x)
g x h e e ?]?? 
 
?H ?I ?H ?I x h(x) h(x) x x
g' x h(e ) e h'(x) e h' e e ?] ?? ?? ?K ?? 
 
h(0) h(0)
g'(0) h(1)e h'(0) e h'(1) ?]?K 
 = 2 + 2 = 4 
14. Let P ( ?? ?L?@ ?? ?L?@ ?? ) be the image of the point Q(3, –3, 1) 
in the line 
x 0 y 3 z 1
1 1 1
?M ?M ?M ?]?]
?M and R be the point 
(2, 5, –1). If the area of the triangle PQR is ?? and 
?? 2
 = 14K, then K is equal to: 
 (1) 36 (2) 72 
 (3) 18  (4) 81 
 Ans. (4) 
Sol.  
  
?? ?@ R(2,5,–1) 
S 
P( ?? , ?? , ?? ) 
Q(3,–3,1) 
 
 RQ 1 64 4 69 ?] ?K ?K ?] 
 
ˆˆ ˆ
RQ – 8j 2k ?]?K 
 
ˆˆ ˆ
RS j – k ?]?K 
 
RQ RS 1 – 8 – 2 9
cos
69 3 3 23 RQ RS
?? ?? ?] ?] ?] 
 
3 RS RS
cos
RQ
23 69
?? ?] ?] ?]          
 RS 3 3 ?] 
 
14 QS
sin
23 69
?? ?] ?] 
 QS 42 ?] 
 area = 
1
2QS RS 42 3 3
2
?? ?? ?] ?? 
 9 14 ???] 
 
2
81.14 14k ?? ?] ?] 
 k = 81 
Page 5


FINAL JEE –MAIN EXAMINATION – APRIL, 2024 
(Held On Saturday 06
th
 April, 2024)  TIME : 3 : 00 PM  to  6 : 00 PM 
NOON
MATHEMATICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Let ABC be an equilateral triangle. A new triangle
is formed by joining the middle points of all sides
of the triangle ABC and the same process is
repeated infinitely many times. If P is the sum of
perimeters and Q is be the sum of areas of all the
triangles formed in this process, then:
(1)
2
P 36 3Q ?] (2) 
2
P 6 3Q ?] (3) 
2
P 36 3Q ?] (4) 
2
P 72 3Q ?] Ans. (1) 
Sol. 
a/2
 
a
 
a
 
a
 
Area of first 
2
3a
4
?d?]
Area of second 
2 2 2
3a a 3a
4 4 16
?d ?] ?] Area of third 
2
a
64
?S ?d?]
sum of area = 
2
3a 1 1
1 ....
4 4 16
????
?K?K
????
????
2 2
3a 1 a
Q
3
4
3
4
?]?] 
perimeter of 1
st
?d = 3a
perimeter of 2
nd
 ?d = 
3a
2
perimeter of 3
rd
 ?d =
3a
4
1 1
P 3a 1 ...
2 4
????
?] ?K ?K ?K????
????
P = 3a.2 = 6a 
P
a
6
?] 2
1 P
Q
36
3
?]??
2
P 36 3Q ?] 2. Let A = {1, 2, 3, 4, 5}. Let R be a relation on A
defined by xRy if and only if 4x ?? ?@ 5y. Let m be the
number of elements in R and n be the minimum
number of elements from A × A that are required
to be added to R to make it a symmetric relation.
Then m + n is equal to:
(1) 24 (2) 23
(3) 25 (4) 26
Ans. (3) 
Sol. Given :  4x ?? 5y 
then 
?H ?I ?? R 1,1 ,(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4)
(2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)}
?] i.e. 16 elements.
i.e. m = 16
Now to make R a symmetric relation add 
{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)} 
i.e. n = 9
So m + n = 25 
3. If three letters can be posted to any one of the 5
different addresses, then the probability that the
three letters are posted to exactly two addresses is:
(1) 
12
25
(2) 
18
25
(3) 
4
25
(4) 
6
25
Ans. (1) 
Sol. Total method = 5
3
 
faverable = 
?H ?I 5 3
2
C 2 – 2 60 ?] probability = 
60 12
125 25
?] 4. Suppose the solution of the differential equation 
dy (2 )x y 2
dx x 2 y ( 4 )
?K ?? ?M ?? ?K ?] ?? ?M ?? ?M ?? ?? ?M ?? represents a circle 
passing through origin. Then the radius of this 
circle is : 
 (1) 17 (2) 
1
2
 
 (3) 
17
2
  (4) 2 
 Ans. (3) 
Sol. 
?H ?I ?H ?I dy 2 x – y 2
dx x – y 2 4
?K ?? ?? ?K ?] ?? ?? ?K ?? ?K ?? 
 ?H ?I xdy – (2 )ydy 4 dy 2 xdx – ydx 2dx ?? ?? ?K ?? ?K ?? ?] ?K ?? ?? ?K 
 ?H ?I (xdy ydx) – (2 )ydy 4 dy 2 xdx 2dx ?? ?K ?? ?K ?? ?K ?? ?] ?K ?? ?K 
 
?H ?I ?H ?I 2 2
2 y 2 x
xy – 4 y
2 2
?? ?K ?? ?K??
?? ?K ?? ?] 
 0 ?? ?? ?] for this to be circle  
 
?H ?I 2
2
x
2 y 2x – 4 y 0
2
?K ?? ?K ?? ?K ?? ?] 
  
2 + a = 2a 
coeff. of  
x
2
 = y
2 
 
 2 ?? ?? ?] 
 i.e. 2x
2
 + 2y
2
 + 2x – 8y = 0 
 x
2
 + y
2
 + x – 4y = 0 
 
1 17
rd 4
4 2
?] ?K ?]  
5. If the locus of the point, whose distances from the 
point (2, 1) and (1, 3) are in the ratio 5 : 4, is  
ax
2
 + by
2
 + cxy + dx + ey + 170 = 0, then the value 
of a
2
 + 2b + 3c + 4d + e is equal to:  
 (1) 5 (2) –27 
 (3) 37  (4) 437 
 Ans. (3) 
Sol. let P(x, y) 
 
?H ?I ?H ?I ?H ?I 2
2
2 2
x – 2 (y – 1) 25
16
x – 1 y – 3
?K ?] ?K 
 9x
2
 + 9y
2
 + 14x – 118y + 170 = 0 
 a
2
 + 2b + 3c + 4d + e 
 = 81 + 18 + 0 + 56 – 118 
 = 155 – 118 
 = 37 
6.  
?H ?I 2 2
2
3 3 3 2 2 2
n
(1 1)(n 1) (2 2)(n 2) ..... · 1
(n 1) (n 1)
lim
(1 2 ..... n ) (1 2 ..... n )
????
?M ?M ?K ?M ?M ?K ?K ?M ?M ?M ?K ?K ?K ?M ?K ?K ?K is equal to: 
 (1) 
2
3
 (2) 
1
3
 
 (3) 
3
4
  (4) 
1
2
 
 Ans. (2) 
Sol. 
?H ?I ?H ?I n–1
2
r 1
n n
n
3 2
r 1 r 1
r – r n – r
lim
r – r
?] ????
?]?]
?? ????
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I n–1
3 2
r 1
2
n
–r r n 1 – nr
lim
n n 1 n n 1 2n 1
–
2 6
?] ????
?K?K
???? ?K ?K ?K ????
????
?? 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 2
2
n
n – 1 n n 1 n – 1 n(2n – 1) n (n – 1)
–
2 6 2
lim
n(n 1) n n 1 2n 1
–
2 2 3
????
????
?K ?K ????
????
???? ?K ?K ?K ????
????
 
 
?H ?I ?H ?I ?H ?I 2
n
n n – 1 –n n – 1 n 1 (2n – 1)
–n
2 2 3
lim
n(n 1) 3n 3n – 4n – 2
2 6
????
???? ?K ?K????
????
?K?K
 
 
?H ?I ?H ?I ?H ?I ?H ?I ?H ?I 2 2
2
n
n – 1 –3n 3n 2 2n n – 1 – 6
lim
n 1 3n – n – 2
????
?K ?K ?K ?K 
 
?H ?I ?H ?I ?H ?I ?H ?I 2
2
n
n –1 n 5n – 8 1
lim
3
n 1 3n – n – 2
????
?K ?] ?K 
7. Let 0 ?? r ?? n. If 
n+1
C
r+1
 : 
n
C
r
 : 
n–1
C
r–1
 = 55 : 35 : 21, 
then 2n + 5r is equal to: 
 (1) 60 (2) 62 
 (3) 50  (4) 55 
 Ans. (3) 
Ans. 
n 1
r
n
r
C 55
35 C
?K ?] 
 
?H ?I ?H ?I ?H ?I n 1 ! r! n – r ! 11
!
r 1 !(n – r) n! 7
?K ?] ?K 
 
?H ?I n 1 11
r 1 7
?K ?] ?K 
 
 7n = 4 + 11r 
 
n
r
n–1
r–1
C 35
21 C
?] 
 
?H ?I n! (r –1)!(n – r)! 5
r! n – r ! (n –1)! 3
?]?] 
 
n 5
r 3
?] 
 3n = 5r 
 By solving r = 6  n = 10 
 2n + 5r = 50 
8. A software company sets up m number of 
computer systems to finish an assignment in  
17 days. If 4 computer systems crashed on the start 
of the second day, 4 more computer systems 
crashed on the start of the third day and so on, then 
it took 8 more days to finish the assignment. The 
value of m is equal to : 
 (1) 125 (2) 150 
 (3) 180  (4) 160 
 Ans. (2) 
Sol. 17m = m + (m – 4) + (m – 4 × 2)...+...(m – 4 × 24) 
 17 m = 25m – 4 (1 + 2 ...24) 
 
4 24 25
8m 150
2
????
?]?] 
9. If z
1
, z
2
 are two distinct complex number such that 
1 2
1 2
z 2z
2
1
z z
2
?M ?] ?M , then  
(1) either z
1
 lies on a circle of radius 1 or z
2
 lies on a 
circle of radius 
1
2
 
(2) either z
1
 lies on a circle of radius 
1
2
 or z
2
 lies on 
a circle of radius 1. 
(3) z
1
 lies on a circle of radius 
1
2
 and z
2
 lies on a 
circle of radius 1. 
(4) both z
1
 and z
2
 lie on the same circle. 
 Ans. (1) 
Sol. 
1 2 1 2
1 2 1 2
z – 2z z – 2z
4
1 1
– z z – z z
2 2
???] 
 
2 2
1 1 2 1 2 2
z 2z z – 2z z 4 z ?K 
 
2 2
1 2 1 2
1 2
z z z z 1
4 – – z z
4 2 2
????
?]?K
????
????
 
 
1 1 2 2 1 1 2 2
z z 2z 2z – z z 2z 2z – 1 0 ?K ?? ?] 
 ?H ?I ?H ?I 1 2 2
z,z –1 1– 2z 2z 0 ???] 
 ?H ?I ?H ?I 2 2
1 2
z –1 2z –1 0 ?] 
10. If the function 
2x
1
f(x)
x
????
?] ????
????
 ; x > 0 attains the 
maximum value at 
1
x
e
?] then : 
 (1) 
e
e
?? ?\?? (2) 
2 e
e (2 )
?? ?\?? 
 (3) 
e
e
?? ?^??  (4) 
(2e)
(2e)
?? ?^?? 
 Ans. (3) 
Sol. Let 
2x
1
y
x
????
?] ????
????
 
 ?? ny = 
1
2x n
x
????
????
????
 
 ny –2x nx ?] 
 
?H ?I 1 dy
–2 1 nx
y dx
?]?K 
 for 
1
x
e
?^ f
n
 is decreasing  
 so, e < ?? ?@ ?@ 2e 2
1 1
e
?? ?? ?? ?? ?? ?^ ?? ?? ?? ?? ?? ?? ?? ?? ?? 
 e
?? > 
 
?? e 
11. Let 
ˆˆ ˆ
a 6i j k ?] ?K ?M and 
ˆˆ
b i j ?]?K . If c is a is vector 
such that 6, a.c 6 , 2 2
c c c a
?? ?] ?] ?M and the 
angle between a b ?? and c is 60°, then ?H ?I c
a b
?? ?? 
is equal to:  
 (1) 
?H ?I 9
6 6
2
?M (2) 
3
3
2
 
 (3) 
3
6
2
  (4) 
?H ?I 9
6 6
2
?K 
 Ans. (4) 
Sol. 
?H ?I 3
a b c a b c
2
?? ?? ?] ?? 
 c – a 2 2 ?] 
 
2 2
c a – 2c a 8 ?K ?? ?] 
 
2
z 38 –12 z 8 ?K?] 
 
2
z –12 z 30 0 ?K?] 
 
12 144 –120
z
2
?? ?] 
 
12 2 6
2
?? ?] 
 z 6 6 ?]?K 
 
ˆˆ ˆ
j k
a b 6 1 –1
1 1 0
???] 
 
ˆˆ ˆ
– j 5k ?K 
 a b 27 ???] 
 
?H ?I ?H ?I 3
a b z 27 6 6
2
?? ?? ?] ?K 
 
?H ?I 9
6 6
2
?K 
12. If all the words with or without meaning made 
using all the letters of the word "NAGPUR" are 
arranged as in a dictionary, then the word at 315
th
 
position in this arrangement is : 
 (1) NRAGUP (2) NRAGPU 
 (3) NRAPGU (4) NRAPUG 
 Ans. (3) 
Sol. NAGPUR 
 A ?? 5! = 120 
 G ® 5! = 120  240 
 NA ® 4! = 24 264 
 NG ® 4! = 24 288 
 NP ® 4! = 24  312 
 NRAGPU = 1 313 
 NRAGUP  314 
 NRAPGU  315 
13. Suppose for a differentiable function h, h(0) = 0, 
h(1) = 1 and h'(0) = h'(1) = 2. If g(x) = h(e
x
) e
h(x)
, 
then g'(0) is equal to: 
 (1) 5 (2) 3 
 (3) 8  (4) 4 
 Ans. (4) 
Sol. 
?H ?I ?H ?I x h(x)
g x h e e ?]?? 
 
?H ?I ?H ?I x h(x) h(x) x x
g' x h(e ) e h'(x) e h' e e ?] ?? ?? ?K ?? 
 
h(0) h(0)
g'(0) h(1)e h'(0) e h'(1) ?]?K 
 = 2 + 2 = 4 
14. Let P ( ?? ?L?@ ?? ?L?@ ?? ) be the image of the point Q(3, –3, 1) 
in the line 
x 0 y 3 z 1
1 1 1
?M ?M ?M ?]?]
?M and R be the point 
(2, 5, –1). If the area of the triangle PQR is ?? and 
?? 2
 = 14K, then K is equal to: 
 (1) 36 (2) 72 
 (3) 18  (4) 81 
 Ans. (4) 
Sol.  
  
?? ?@ R(2,5,–1) 
S 
P( ?? , ?? , ?? ) 
Q(3,–3,1) 
 
 RQ 1 64 4 69 ?] ?K ?K ?] 
 
ˆˆ ˆ
RQ – 8j 2k ?]?K 
 
ˆˆ ˆ
RS j – k ?]?K 
 
RQ RS 1 – 8 – 2 9
cos
69 3 3 23 RQ RS
?? ?? ?] ?] ?] 
 
3 RS RS
cos
RQ
23 69
?? ?] ?] ?]          
 RS 3 3 ?] 
 
14 QS
sin
23 69
?? ?] ?] 
 QS 42 ?] 
 area = 
1
2QS RS 42 3 3
2
?? ?? ?] ?? 
 9 14 ???] 
 
2
81.14 14k ?? ?] ?] 
 k = 81 
 
 
15. If P(6, 1) be the orthocentre of the triangle whose 
vertices are A(5, –2), B(8, 3) and C(h, k), then the 
point C lies on the circle. 
 (1) x
2
 + y
2
 – 65 = 0 (2) x
2
 + y
2
 – 74 = 0 
 (3) x
2
 + y
2
 – 61 = 0 (4) x
2
 + y
2
 – 52 = 0 
 Ans. (1) 
Sol.  
  
B(8,3) 
D C(h, k) 
(6,1)P 
F E 
A(5,–2) 
 
 Slope of AD = 3 
 Slope of BC
1
–
3
?] 
 equation of BC = 3y + x – 17 = 0 
 slope of BE = 1 
 Slope of AC = –1 
 equation of AC is x + y – 3 = 0 
 point C is (–4, 7)  
 
16. Let 
1
f(x)
7 sin 5x
?] ?M be a function defined on R. 
Then the range of the function f(x) is equal to: 
 (1) 
1 1
,
8 5
????
????
????
 (2) 
1 1
,
7 6
????
????
????
 
 (3)
1 1
,
7 5
????
????
????
  (4) 
1 1
,
8 6
????
????
????
 
 Ans. (4) 
Sol. sin5x ?? [–1,1] 
 –sin5x ?? [–1, 1] 
 7 – sin5x ?? [6, 8] 
 
1 1 1
,
7 – sin 5x 8 6
????
?? ????
????
 
17. Let 
ˆˆ ˆ
a 2i j k ?] ?K ?M , 
?H ?I ?H ?I ?H ?Iˆˆ
ˆˆ
b a i i
i j
?] ?? ?? ?? ?K .  
 Then the square of the projection of a on b is : 
 (1) 
1
5
 (2) 2 
 (3) 
1
3
  (4) 
2
3
 
 Ans. (2) 
Sol. ?H ?I ˆˆ ˆ
i j k
ˆˆ
a i j 2 1 –1
1 1 0
?? ?K ?] 
 
ˆˆ ˆ
i – j k ?]?K 
 ?H ?I ?H ?I ˆ ˆ ˆ ˆ ˆ
a i j i k j ?? ?? ?? ?] ?K 
 ?H ?I ?H ?I ?H ?I ˆ ˆ ˆ ˆ ˆ
a i j i i j – k ?? ?? ?? ?? ?] 
 projection of a on 
ˆ
b
a b
b
?? ?] 
 
1 1
2
2
?K ?]?] 
18. If the area of the region  
 
2
a 1
(x,y) : y ,1 x 2, 0 a 1
x x
????
?? ?? ?? ?? ?\ ?\ ????
????
is  
(log
e
2) – 
1
7
then the value of 7a – 3 is equal to:  
 (1) 2 (2) 0 
 (3) –1  (4) 1 
 Ans. (3) 
Sol.  
  
a 1 2