JEE Main 2025 January 22 Shift 1 Paper & Solutions | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


 1 
JEE–MAIN EXAMINATION – JANUARY 2025
MATHEMATICS TEST PAPER WITH SOLUTION 
(HELD ON WEDNESDAY 22
nd
 JANUARY 2025) TIME : 9:00 AM  TO  12:00 NOON
SECTION-A 
1. The number of non-empty equivalence relations on
the set {1,2,3} is :
(1) 6 (2) 7
(3) 5 (4) 4
Ans.  (3) 
Sol. Let R be the required relation 
A = {(1, 1) (2, 2), (3, 3)} 
(i) | R | = 3, when R = A
(ii) | R | = 5, e.g. R = A ? {(1, 2), (2, 1)}
Number of R can be [3] 
(iii) R = {1, 2, 3} × {1, 2, 3}
Ans. (5) 
2. Let ƒ : R ?R be a twice differentiable function
such that ƒ(x + y) = ƒ(x) ƒ(y) for all x, y ? R. If
ƒ'(0) = 4a and ƒ staisfies ƒ''(x) – 3a ƒ'(x) – ƒ(x) = 0,
a > 0, then the area of the region
R = {(x,y) | 0 ? y ? ƒ(ax), 0 ? x ? 2} is :
(1) e
2
 – 1 (2) e
4
 + 1
(3) e
4
 – 1 (4) e
2
 + 1
Ans.  (1) 
Sol. f(x + y) = f(x).f(y) 
? f(x) = e
?x
  f ?(0) = 4a
? f ?(x) = ?e
?x
  ? ? = 4a
So, f(x) = e
4ax
 
f ??(x) – 3af ?(x) – f (x) = 0 
? ?
2
 – 3a ? – 1 = 0
? 16a
2
 – 12a
2
 – 1 = 0 ? 4a
2
 = 1 ? ?
1
a
2
?
x = 0 
x = 2 
f(ax) = e
x 
F(x) = e
2x
 
2
x2
0
Area e dx e 1 ? ? ?
?
3. Let the triangle PQR be the image of the
triangle with vertices (1,3), (3,1) and (2, 4) in
the line x + 2y = 2. If the centroid of ?PQR is
the point ( ?, ?), then 15( ? – ?) is equal to :
(1) 24 (2) 19
(3) 21 (4) 22
Ans.  (4) 
Sol. Let ‘G’ be the centroid of ? formed by (1, 3) (3, 1) 
& (2, 4) 
8
G 2,
3
??
?
??
??
Image of G w.r.t. x + 2y – 2 = 0 
16
8
22
2 3
3
2
1 2 1 4
??
??
?? ??
??
??
? ? ?
?
2 16
53
? ??
?
??
??
? 
32 2
2
15 15
??
? ? ? ? , 
32 2 8 24
15 3 15
? ? ?
? ? ? ?
15( ? – ?) = – 2 + 24 = 22 
4. Let z
1
, z
2
 and z
3
 be three complex numbers on the circle
|z| = 1 with 
??
?
1
arg(z )
4
, arg(z
2
) = 0 and 
?
?
3
arg(z )
4
. 
If ? ? ? ? ? ?
2
1 2 2 3 3 1
z z z z z z 2 , ?, ? ? Z, then the 
value of ?
2
 + ?
2
 is : 
(1) 24 (2) 41
(3) 31 (4) 29
Ans.  (4) 
Sol. Z
1
 = e
–i ?/4
 , Z
2
 = 1, Z
3
 = e
i ?/4
2
1 2 2 3 3 1
z z z z z z ?? = 
2
i i i i
4 4 4 4
e 1 1 e e e
? ? ? ?
??
? ? ? ? ?
2
i i i
4 4 4
eee
? ? ?
??
??
2
i
4
2e i
?
?
?? = 
2
2 2i i ? ? ?
= 
? ? ? ?
22
2 1 2 ?? = 2 + 1 + 2 – 2 2 = 5 – 22
? = 5, ? = –2
? ?
2
 + ?
2
 = 29
Page 2


 1 
JEE–MAIN EXAMINATION – JANUARY 2025
MATHEMATICS TEST PAPER WITH SOLUTION 
(HELD ON WEDNESDAY 22
nd
 JANUARY 2025) TIME : 9:00 AM  TO  12:00 NOON
SECTION-A 
1. The number of non-empty equivalence relations on
the set {1,2,3} is :
(1) 6 (2) 7
(3) 5 (4) 4
Ans.  (3) 
Sol. Let R be the required relation 
A = {(1, 1) (2, 2), (3, 3)} 
(i) | R | = 3, when R = A
(ii) | R | = 5, e.g. R = A ? {(1, 2), (2, 1)}
Number of R can be [3] 
(iii) R = {1, 2, 3} × {1, 2, 3}
Ans. (5) 
2. Let ƒ : R ?R be a twice differentiable function
such that ƒ(x + y) = ƒ(x) ƒ(y) for all x, y ? R. If
ƒ'(0) = 4a and ƒ staisfies ƒ''(x) – 3a ƒ'(x) – ƒ(x) = 0,
a > 0, then the area of the region
R = {(x,y) | 0 ? y ? ƒ(ax), 0 ? x ? 2} is :
(1) e
2
 – 1 (2) e
4
 + 1
(3) e
4
 – 1 (4) e
2
 + 1
Ans.  (1) 
Sol. f(x + y) = f(x).f(y) 
? f(x) = e
?x
  f ?(0) = 4a
? f ?(x) = ?e
?x
  ? ? = 4a
So, f(x) = e
4ax
 
f ??(x) – 3af ?(x) – f (x) = 0 
? ?
2
 – 3a ? – 1 = 0
? 16a
2
 – 12a
2
 – 1 = 0 ? 4a
2
 = 1 ? ?
1
a
2
?
x = 0 
x = 2 
f(ax) = e
x 
F(x) = e
2x
 
2
x2
0
Area e dx e 1 ? ? ?
?
3. Let the triangle PQR be the image of the
triangle with vertices (1,3), (3,1) and (2, 4) in
the line x + 2y = 2. If the centroid of ?PQR is
the point ( ?, ?), then 15( ? – ?) is equal to :
(1) 24 (2) 19
(3) 21 (4) 22
Ans.  (4) 
Sol. Let ‘G’ be the centroid of ? formed by (1, 3) (3, 1) 
& (2, 4) 
8
G 2,
3
??
?
??
??
Image of G w.r.t. x + 2y – 2 = 0 
16
8
22
2 3
3
2
1 2 1 4
??
??
?? ??
??
??
? ? ?
?
2 16
53
? ??
?
??
??
? 
32 2
2
15 15
??
? ? ? ? , 
32 2 8 24
15 3 15
? ? ?
? ? ? ?
15( ? – ?) = – 2 + 24 = 22 
4. Let z
1
, z
2
 and z
3
 be three complex numbers on the circle
|z| = 1 with 
??
?
1
arg(z )
4
, arg(z
2
) = 0 and 
?
?
3
arg(z )
4
. 
If ? ? ? ? ? ?
2
1 2 2 3 3 1
z z z z z z 2 , ?, ? ? Z, then the 
value of ?
2
 + ?
2
 is : 
(1) 24 (2) 41
(3) 31 (4) 29
Ans.  (4) 
Sol. Z
1
 = e
–i ?/4
 , Z
2
 = 1, Z
3
 = e
i ?/4
2
1 2 2 3 3 1
z z z z z z ?? = 
2
i i i i
4 4 4 4
e 1 1 e e e
? ? ? ?
??
? ? ? ? ?
2
i i i
4 4 4
eee
? ? ?
??
??
2
i
4
2e i
?
?
?? = 
2
2 2i i ? ? ?
= 
? ? ? ?
22
2 1 2 ?? = 2 + 1 + 2 – 2 2 = 5 – 22
? = 5, ? = –2
? ?
2
 + ?
2
 = 29
 
2 
 
5. Using the principal values of the inverse 
trigonometric functions the sum of the maximum 
and the minimum values of 16((sec
–1
x)
2
 + (cosec
–1
x)
2
) 
is : 
 (1) 24 ?
2
 (2) 18 ?
2
 
 (3) 31 ?
2
  (4) 22 ?
2
 
Ans.  (4) 
Sol. 16(sec
–1
x)
2
 + (cosec
–1
x)
2
 
 Sec
–1
x = a ?[0, ?] – 
2
? ??
??
??
 
 cosec
–1
x = a
2
?
? 
 = 
2
2
22
16 a a 16 2a a
24
??
?? ?? ??
? ? ? ? ? ?
??
?? ??
??
????
??
 
 max]
a = 
?
 = 16[2 ?
2
 – ?
2 
+ 
2
4
? ] = 20 ?
2 
 
a
4
min]
?
?
 = 
2 2 2
2
2
16 2
16 4 4
?? ? ? ? ?
? ? ? ?
??
??
 
 Sum = 22 ?
2 
6. A coin is tossed three times. Let X denote the 
number of times a tail follows a head. If ? and ?
2
 
denote the mean and variance of X, then the value 
of 64( ? + ?
2
) is : 
 (1) 51 (2) 48 
 (3) 32  (4) 64 
Ans.  (2) 
Sol. HHH ? 0  
 HHT ? 0  
 HTH ? 1  
 HTT ? 0  
 THH ? 1  
 THT ? 1  
 TTH ? 1  
 TTT ? 0  
 Probability distribution  
 
i
i
x 0 1
11
P(x )
22
  
 
ii
1
xp
2
? ? ?
?
  
 
2 2 2
ii
x p µ ? ? ?
?
  
 
1 1 1
2 4 4
???  
 64(µ + ?
2
) = 
11
64
24
??
?
??
??
 = 48  
7. Let a
1
, a
2
, a
3
..... be a G.P. of increasing positive 
terms. If a
1
a
5
 = 28 and a
2 
+ a
4
 = 29, the a
6
 is equal to  
 (1) 628 (2) 526 
 (3) 784  (4) 812 
Ans.  (3) 
Sol. a
1
.a
5
 = 28 ? a.ar
4
 = 28 ? a
2
r
4
 = 28  …(1)  
 a
2
 + a
4
 = 29 ? ar + ar
3
 = 29  
 ? ar(1 + r
2
) = 29  
 ? a
2
r
2
(1 + r
2
)
2
 = (29)
2
   …(2)  
 By Eq. (1) & (2)  
 
2
22
r 28
(1 r ) 29 29
?
??
  
 ? 
2
r 28
1 r 29
?
?
 ? r 28 ?  
 ? a
2
r
4
 = 28 ? a
2
 × (28)
2
 = 28  
 ? 
1
a
28
?  
 ? a
6
 = ar
5
 = 
2
1
(28) 28 784
28
??  
8. Let 
? ? ?
??
1
x 1 y 2 z 3
L:
2 3 4
 and  
 
? ? ?
??
2
x 2 y 4 z 5
L:
3 4 5
 be two lines. Then which 
of the following points lies on the line of the 
shortest distance between L
1
 and L
2
 ? 
 (1) 
??
??
??
??
5
, 7,1
3
 (2) 
??
??
??
1
2,3,
3
 
 (3) 
??
?
??
??
81
, 1,
33
 (4) 
??
?
??
??
14 22
, 3,
33
 
Ans.  (4) 
Sol. 
 
P 
Q 
L
1
 
L
2
 
 
 P(2 ? + 1, 3 ? + 2, 4 ? + 3)   on L
1
  
 Q(3µ + 2, 4µ + 4, 5µ + 5)   on L
2
  
 Dr’s of PQ = 3µ – 2 ? + 1, 4µ – 3? + 2, 5µ – 4 ? + 2  
 PQ ? L
1
  
Page 3


 1 
JEE–MAIN EXAMINATION – JANUARY 2025
MATHEMATICS TEST PAPER WITH SOLUTION 
(HELD ON WEDNESDAY 22
nd
 JANUARY 2025) TIME : 9:00 AM  TO  12:00 NOON
SECTION-A 
1. The number of non-empty equivalence relations on
the set {1,2,3} is :
(1) 6 (2) 7
(3) 5 (4) 4
Ans.  (3) 
Sol. Let R be the required relation 
A = {(1, 1) (2, 2), (3, 3)} 
(i) | R | = 3, when R = A
(ii) | R | = 5, e.g. R = A ? {(1, 2), (2, 1)}
Number of R can be [3] 
(iii) R = {1, 2, 3} × {1, 2, 3}
Ans. (5) 
2. Let ƒ : R ?R be a twice differentiable function
such that ƒ(x + y) = ƒ(x) ƒ(y) for all x, y ? R. If
ƒ'(0) = 4a and ƒ staisfies ƒ''(x) – 3a ƒ'(x) – ƒ(x) = 0,
a > 0, then the area of the region
R = {(x,y) | 0 ? y ? ƒ(ax), 0 ? x ? 2} is :
(1) e
2
 – 1 (2) e
4
 + 1
(3) e
4
 – 1 (4) e
2
 + 1
Ans.  (1) 
Sol. f(x + y) = f(x).f(y) 
? f(x) = e
?x
  f ?(0) = 4a
? f ?(x) = ?e
?x
  ? ? = 4a
So, f(x) = e
4ax
 
f ??(x) – 3af ?(x) – f (x) = 0 
? ?
2
 – 3a ? – 1 = 0
? 16a
2
 – 12a
2
 – 1 = 0 ? 4a
2
 = 1 ? ?
1
a
2
?
x = 0 
x = 2 
f(ax) = e
x 
F(x) = e
2x
 
2
x2
0
Area e dx e 1 ? ? ?
?
3. Let the triangle PQR be the image of the
triangle with vertices (1,3), (3,1) and (2, 4) in
the line x + 2y = 2. If the centroid of ?PQR is
the point ( ?, ?), then 15( ? – ?) is equal to :
(1) 24 (2) 19
(3) 21 (4) 22
Ans.  (4) 
Sol. Let ‘G’ be the centroid of ? formed by (1, 3) (3, 1) 
& (2, 4) 
8
G 2,
3
??
?
??
??
Image of G w.r.t. x + 2y – 2 = 0 
16
8
22
2 3
3
2
1 2 1 4
??
??
?? ??
??
??
? ? ?
?
2 16
53
? ??
?
??
??
? 
32 2
2
15 15
??
? ? ? ? , 
32 2 8 24
15 3 15
? ? ?
? ? ? ?
15( ? – ?) = – 2 + 24 = 22 
4. Let z
1
, z
2
 and z
3
 be three complex numbers on the circle
|z| = 1 with 
??
?
1
arg(z )
4
, arg(z
2
) = 0 and 
?
?
3
arg(z )
4
. 
If ? ? ? ? ? ?
2
1 2 2 3 3 1
z z z z z z 2 , ?, ? ? Z, then the 
value of ?
2
 + ?
2
 is : 
(1) 24 (2) 41
(3) 31 (4) 29
Ans.  (4) 
Sol. Z
1
 = e
–i ?/4
 , Z
2
 = 1, Z
3
 = e
i ?/4
2
1 2 2 3 3 1
z z z z z z ?? = 
2
i i i i
4 4 4 4
e 1 1 e e e
? ? ? ?
??
? ? ? ? ?
2
i i i
4 4 4
eee
? ? ?
??
??
2
i
4
2e i
?
?
?? = 
2
2 2i i ? ? ?
= 
? ? ? ?
22
2 1 2 ?? = 2 + 1 + 2 – 2 2 = 5 – 22
? = 5, ? = –2
? ?
2
 + ?
2
 = 29
 
2 
 
5. Using the principal values of the inverse 
trigonometric functions the sum of the maximum 
and the minimum values of 16((sec
–1
x)
2
 + (cosec
–1
x)
2
) 
is : 
 (1) 24 ?
2
 (2) 18 ?
2
 
 (3) 31 ?
2
  (4) 22 ?
2
 
Ans.  (4) 
Sol. 16(sec
–1
x)
2
 + (cosec
–1
x)
2
 
 Sec
–1
x = a ?[0, ?] – 
2
? ??
??
??
 
 cosec
–1
x = a
2
?
? 
 = 
2
2
22
16 a a 16 2a a
24
??
?? ?? ??
? ? ? ? ? ?
??
?? ??
??
????
??
 
 max]
a = 
?
 = 16[2 ?
2
 – ?
2 
+ 
2
4
? ] = 20 ?
2 
 
a
4
min]
?
?
 = 
2 2 2
2
2
16 2
16 4 4
?? ? ? ? ?
? ? ? ?
??
??
 
 Sum = 22 ?
2 
6. A coin is tossed three times. Let X denote the 
number of times a tail follows a head. If ? and ?
2
 
denote the mean and variance of X, then the value 
of 64( ? + ?
2
) is : 
 (1) 51 (2) 48 
 (3) 32  (4) 64 
Ans.  (2) 
Sol. HHH ? 0  
 HHT ? 0  
 HTH ? 1  
 HTT ? 0  
 THH ? 1  
 THT ? 1  
 TTH ? 1  
 TTT ? 0  
 Probability distribution  
 
i
i
x 0 1
11
P(x )
22
  
 
ii
1
xp
2
? ? ?
?
  
 
2 2 2
ii
x p µ ? ? ?
?
  
 
1 1 1
2 4 4
???  
 64(µ + ?
2
) = 
11
64
24
??
?
??
??
 = 48  
7. Let a
1
, a
2
, a
3
..... be a G.P. of increasing positive 
terms. If a
1
a
5
 = 28 and a
2 
+ a
4
 = 29, the a
6
 is equal to  
 (1) 628 (2) 526 
 (3) 784  (4) 812 
Ans.  (3) 
Sol. a
1
.a
5
 = 28 ? a.ar
4
 = 28 ? a
2
r
4
 = 28  …(1)  
 a
2
 + a
4
 = 29 ? ar + ar
3
 = 29  
 ? ar(1 + r
2
) = 29  
 ? a
2
r
2
(1 + r
2
)
2
 = (29)
2
   …(2)  
 By Eq. (1) & (2)  
 
2
22
r 28
(1 r ) 29 29
?
??
  
 ? 
2
r 28
1 r 29
?
?
 ? r 28 ?  
 ? a
2
r
4
 = 28 ? a
2
 × (28)
2
 = 28  
 ? 
1
a
28
?  
 ? a
6
 = ar
5
 = 
2
1
(28) 28 784
28
??  
8. Let 
? ? ?
??
1
x 1 y 2 z 3
L:
2 3 4
 and  
 
? ? ?
??
2
x 2 y 4 z 5
L:
3 4 5
 be two lines. Then which 
of the following points lies on the line of the 
shortest distance between L
1
 and L
2
 ? 
 (1) 
??
??
??
??
5
, 7,1
3
 (2) 
??
??
??
1
2,3,
3
 
 (3) 
??
?
??
??
81
, 1,
33
 (4) 
??
?
??
??
14 22
, 3,
33
 
Ans.  (4) 
Sol. 
 
P 
Q 
L
1
 
L
2
 
 
 P(2 ? + 1, 3 ? + 2, 4 ? + 3)   on L
1
  
 Q(3µ + 2, 4µ + 4, 5µ + 5)   on L
2
  
 Dr’s of PQ = 3µ – 2 ? + 1, 4µ – 3? + 2, 5µ – 4 ? + 2  
 PQ ? L
1
  
 
3 
 
 ? (3µ – 2? + 1)2 + (4µ – 3? + 2)3 + (5µ – 4? + 
2)4 = 0  
 38µ – 29 ? + 16 = 0  …(1)  
 PQ ? L
2
  
 ? (3µ – 2? + 1)3 + (4µ – 3? + 2)4 + (5µ – 4? + 
2)5 = 0  
 50µ – 38 ? + 21 = 0  …(2)  
 By (1) & (2)  
 
1
3
?? ;  
1
µ
6
?
?  
 ??
5 13
P ,3,
33
??
??
??
 & 
3 10 25
Q , ,
2 3 6
??
??
??
  
 Line PQ  
 
5
x
3
1
6
?
 
y3
1
3
?
?
 
13
z
3
1
6
?
  
 
5 13
xz
y3
33
1 2 1
??
?
??
?
  
 Point
14 22
, 3,
33
??
?
??
??
  
 lies on the line PQ  
9. The product of all solutions of the equation 
? ? ?
?
2
e
5 log x 3 8
ex , x > 0, is : 
 (1) e
8/5
 (2) e
6/5
 
 (3) e
2
  (4) e 
Ans.  (1) 
Sol. 
2
5( nx) 3 8
ex
?
?  
 ??
2
5( nx) 3 8
ne n x
?
? ??
 ? 5( ?nx)
2
 + 3 = 8?nx  
 ( ?nx = t)  
 ? 5t
2
 – 8t + 3 = 0  
  t
1
 + t
2
 = 
8
5
  
  ?nx
1
x
2
 = 
8
5
  
  x
1
x
2
 = e
8/5
  
10. If 
?
? ? ? ?
?
?
n
r
r1
(2n 1)(2n 1)(2n 3)(2n 5)
T
64
, then  
 
??
?
??
??
??
?
n
n
r1
r
1
lim
T
 is equal to : 
 (1) 1 (2) 0 
 (3) 
2
3
  (4) 
1
3
 
Ans.  (3) 
Sol. T
n
 = S
n
 – S
n–1
  
 ? T
n
 = 
1
8
(2n – 1)(2n + 1)(2n + 3)  
 ??
n
18
T (2n 1)(2n 1)(2n 3)
?
? ? ?
??
?
nn
nn
r 1 r 1 r
11
lim lim8
T (2n 1)(2n 1)(2n 3)
? ? ? ?
??
?
? ? ?
??
??
?
n
8 1 1
lim
4 (2n 1)(2n 1) (2n 1)(2n 3)
??
??
??
??
? ? ? ?
??
?
?
?
n
1 1 1 1
lim 2 ...
1.3 3.5 3.5 5.7
??
?? ? ? ? ?
? ? ? ? ?
? ? ? ? ??
? ? ? ? ??
??
?
2
3
? ??
11. From all the English alphabets, five letters are 
chosen and are arranged in alphabetical order. The 
total number of ways, in which the middle letter is 
‘M’, is : 
 (1) 14950 (2) 6084 
 (3) 4356  (4) 5148 
Ans.  (4) 
Sol. 
12 13
AB M N..........Z 
 
12 13
22
Selection of two Selection of two
letters before M letters after M
C C 5148 ? ? ? 
12. Let x = x(y) be the solution of the differential 
equation 
??
? ? ?
??
??
2
1
y dx x dy 0
y
. If x(1) = 1, then 
??
??
??
1
x
2
 is : 
 (1) ?
1
e
2
 (2) ?
3
e
2
 
 (3) 3 – e  (4) 3 + e 
Page 4


 1 
JEE–MAIN EXAMINATION – JANUARY 2025
MATHEMATICS TEST PAPER WITH SOLUTION 
(HELD ON WEDNESDAY 22
nd
 JANUARY 2025) TIME : 9:00 AM  TO  12:00 NOON
SECTION-A 
1. The number of non-empty equivalence relations on
the set {1,2,3} is :
(1) 6 (2) 7
(3) 5 (4) 4
Ans.  (3) 
Sol. Let R be the required relation 
A = {(1, 1) (2, 2), (3, 3)} 
(i) | R | = 3, when R = A
(ii) | R | = 5, e.g. R = A ? {(1, 2), (2, 1)}
Number of R can be [3] 
(iii) R = {1, 2, 3} × {1, 2, 3}
Ans. (5) 
2. Let ƒ : R ?R be a twice differentiable function
such that ƒ(x + y) = ƒ(x) ƒ(y) for all x, y ? R. If
ƒ'(0) = 4a and ƒ staisfies ƒ''(x) – 3a ƒ'(x) – ƒ(x) = 0,
a > 0, then the area of the region
R = {(x,y) | 0 ? y ? ƒ(ax), 0 ? x ? 2} is :
(1) e
2
 – 1 (2) e
4
 + 1
(3) e
4
 – 1 (4) e
2
 + 1
Ans.  (1) 
Sol. f(x + y) = f(x).f(y) 
? f(x) = e
?x
  f ?(0) = 4a
? f ?(x) = ?e
?x
  ? ? = 4a
So, f(x) = e
4ax
 
f ??(x) – 3af ?(x) – f (x) = 0 
? ?
2
 – 3a ? – 1 = 0
? 16a
2
 – 12a
2
 – 1 = 0 ? 4a
2
 = 1 ? ?
1
a
2
?
x = 0 
x = 2 
f(ax) = e
x 
F(x) = e
2x
 
2
x2
0
Area e dx e 1 ? ? ?
?
3. Let the triangle PQR be the image of the
triangle with vertices (1,3), (3,1) and (2, 4) in
the line x + 2y = 2. If the centroid of ?PQR is
the point ( ?, ?), then 15( ? – ?) is equal to :
(1) 24 (2) 19
(3) 21 (4) 22
Ans.  (4) 
Sol. Let ‘G’ be the centroid of ? formed by (1, 3) (3, 1) 
& (2, 4) 
8
G 2,
3
??
?
??
??
Image of G w.r.t. x + 2y – 2 = 0 
16
8
22
2 3
3
2
1 2 1 4
??
??
?? ??
??
??
? ? ?
?
2 16
53
? ??
?
??
??
? 
32 2
2
15 15
??
? ? ? ? , 
32 2 8 24
15 3 15
? ? ?
? ? ? ?
15( ? – ?) = – 2 + 24 = 22 
4. Let z
1
, z
2
 and z
3
 be three complex numbers on the circle
|z| = 1 with 
??
?
1
arg(z )
4
, arg(z
2
) = 0 and 
?
?
3
arg(z )
4
. 
If ? ? ? ? ? ?
2
1 2 2 3 3 1
z z z z z z 2 , ?, ? ? Z, then the 
value of ?
2
 + ?
2
 is : 
(1) 24 (2) 41
(3) 31 (4) 29
Ans.  (4) 
Sol. Z
1
 = e
–i ?/4
 , Z
2
 = 1, Z
3
 = e
i ?/4
2
1 2 2 3 3 1
z z z z z z ?? = 
2
i i i i
4 4 4 4
e 1 1 e e e
? ? ? ?
??
? ? ? ? ?
2
i i i
4 4 4
eee
? ? ?
??
??
2
i
4
2e i
?
?
?? = 
2
2 2i i ? ? ?
= 
? ? ? ?
22
2 1 2 ?? = 2 + 1 + 2 – 2 2 = 5 – 22
? = 5, ? = –2
? ?
2
 + ?
2
 = 29
 
2 
 
5. Using the principal values of the inverse 
trigonometric functions the sum of the maximum 
and the minimum values of 16((sec
–1
x)
2
 + (cosec
–1
x)
2
) 
is : 
 (1) 24 ?
2
 (2) 18 ?
2
 
 (3) 31 ?
2
  (4) 22 ?
2
 
Ans.  (4) 
Sol. 16(sec
–1
x)
2
 + (cosec
–1
x)
2
 
 Sec
–1
x = a ?[0, ?] – 
2
? ??
??
??
 
 cosec
–1
x = a
2
?
? 
 = 
2
2
22
16 a a 16 2a a
24
??
?? ?? ??
? ? ? ? ? ?
??
?? ??
??
????
??
 
 max]
a = 
?
 = 16[2 ?
2
 – ?
2 
+ 
2
4
? ] = 20 ?
2 
 
a
4
min]
?
?
 = 
2 2 2
2
2
16 2
16 4 4
?? ? ? ? ?
? ? ? ?
??
??
 
 Sum = 22 ?
2 
6. A coin is tossed three times. Let X denote the 
number of times a tail follows a head. If ? and ?
2
 
denote the mean and variance of X, then the value 
of 64( ? + ?
2
) is : 
 (1) 51 (2) 48 
 (3) 32  (4) 64 
Ans.  (2) 
Sol. HHH ? 0  
 HHT ? 0  
 HTH ? 1  
 HTT ? 0  
 THH ? 1  
 THT ? 1  
 TTH ? 1  
 TTT ? 0  
 Probability distribution  
 
i
i
x 0 1
11
P(x )
22
  
 
ii
1
xp
2
? ? ?
?
  
 
2 2 2
ii
x p µ ? ? ?
?
  
 
1 1 1
2 4 4
???  
 64(µ + ?
2
) = 
11
64
24
??
?
??
??
 = 48  
7. Let a
1
, a
2
, a
3
..... be a G.P. of increasing positive 
terms. If a
1
a
5
 = 28 and a
2 
+ a
4
 = 29, the a
6
 is equal to  
 (1) 628 (2) 526 
 (3) 784  (4) 812 
Ans.  (3) 
Sol. a
1
.a
5
 = 28 ? a.ar
4
 = 28 ? a
2
r
4
 = 28  …(1)  
 a
2
 + a
4
 = 29 ? ar + ar
3
 = 29  
 ? ar(1 + r
2
) = 29  
 ? a
2
r
2
(1 + r
2
)
2
 = (29)
2
   …(2)  
 By Eq. (1) & (2)  
 
2
22
r 28
(1 r ) 29 29
?
??
  
 ? 
2
r 28
1 r 29
?
?
 ? r 28 ?  
 ? a
2
r
4
 = 28 ? a
2
 × (28)
2
 = 28  
 ? 
1
a
28
?  
 ? a
6
 = ar
5
 = 
2
1
(28) 28 784
28
??  
8. Let 
? ? ?
??
1
x 1 y 2 z 3
L:
2 3 4
 and  
 
? ? ?
??
2
x 2 y 4 z 5
L:
3 4 5
 be two lines. Then which 
of the following points lies on the line of the 
shortest distance between L
1
 and L
2
 ? 
 (1) 
??
??
??
??
5
, 7,1
3
 (2) 
??
??
??
1
2,3,
3
 
 (3) 
??
?
??
??
81
, 1,
33
 (4) 
??
?
??
??
14 22
, 3,
33
 
Ans.  (4) 
Sol. 
 
P 
Q 
L
1
 
L
2
 
 
 P(2 ? + 1, 3 ? + 2, 4 ? + 3)   on L
1
  
 Q(3µ + 2, 4µ + 4, 5µ + 5)   on L
2
  
 Dr’s of PQ = 3µ – 2 ? + 1, 4µ – 3? + 2, 5µ – 4 ? + 2  
 PQ ? L
1
  
 
3 
 
 ? (3µ – 2? + 1)2 + (4µ – 3? + 2)3 + (5µ – 4? + 
2)4 = 0  
 38µ – 29 ? + 16 = 0  …(1)  
 PQ ? L
2
  
 ? (3µ – 2? + 1)3 + (4µ – 3? + 2)4 + (5µ – 4? + 
2)5 = 0  
 50µ – 38 ? + 21 = 0  …(2)  
 By (1) & (2)  
 
1
3
?? ;  
1
µ
6
?
?  
 ??
5 13
P ,3,
33
??
??
??
 & 
3 10 25
Q , ,
2 3 6
??
??
??
  
 Line PQ  
 
5
x
3
1
6
?
 
y3
1
3
?
?
 
13
z
3
1
6
?
  
 
5 13
xz
y3
33
1 2 1
??
?
??
?
  
 Point
14 22
, 3,
33
??
?
??
??
  
 lies on the line PQ  
9. The product of all solutions of the equation 
? ? ?
?
2
e
5 log x 3 8
ex , x > 0, is : 
 (1) e
8/5
 (2) e
6/5
 
 (3) e
2
  (4) e 
Ans.  (1) 
Sol. 
2
5( nx) 3 8
ex
?
?  
 ??
2
5( nx) 3 8
ne n x
?
? ??
 ? 5( ?nx)
2
 + 3 = 8?nx  
 ( ?nx = t)  
 ? 5t
2
 – 8t + 3 = 0  
  t
1
 + t
2
 = 
8
5
  
  ?nx
1
x
2
 = 
8
5
  
  x
1
x
2
 = e
8/5
  
10. If 
?
? ? ? ?
?
?
n
r
r1
(2n 1)(2n 1)(2n 3)(2n 5)
T
64
, then  
 
??
?
??
??
??
?
n
n
r1
r
1
lim
T
 is equal to : 
 (1) 1 (2) 0 
 (3) 
2
3
  (4) 
1
3
 
Ans.  (3) 
Sol. T
n
 = S
n
 – S
n–1
  
 ? T
n
 = 
1
8
(2n – 1)(2n + 1)(2n + 3)  
 ??
n
18
T (2n 1)(2n 1)(2n 3)
?
? ? ?
??
?
nn
nn
r 1 r 1 r
11
lim lim8
T (2n 1)(2n 1)(2n 3)
? ? ? ?
??
?
? ? ?
??
??
?
n
8 1 1
lim
4 (2n 1)(2n 1) (2n 1)(2n 3)
??
??
??
??
? ? ? ?
??
?
?
?
n
1 1 1 1
lim 2 ...
1.3 3.5 3.5 5.7
??
?? ? ? ? ?
? ? ? ? ?
? ? ? ? ??
? ? ? ? ??
??
?
2
3
? ??
11. From all the English alphabets, five letters are 
chosen and are arranged in alphabetical order. The 
total number of ways, in which the middle letter is 
‘M’, is : 
 (1) 14950 (2) 6084 
 (3) 4356  (4) 5148 
Ans.  (4) 
Sol. 
12 13
AB M N..........Z 
 
12 13
22
Selection of two Selection of two
letters before M letters after M
C C 5148 ? ? ? 
12. Let x = x(y) be the solution of the differential 
equation 
??
? ? ?
??
??
2
1
y dx x dy 0
y
. If x(1) = 1, then 
??
??
??
1
x
2
 is : 
 (1) ?
1
e
2
 (2) ?
3
e
2
 
 (3) 3 – e  (4) 3 + e 
 
4 
 
Ans.  (3) 
Sol. 
23
dx 1 1
x
dy y y
??
??
??
??
 
 I.F. = 
2
1 1
dy
yy
ee
?
?
? 
 
1
1
y
t
3
1
x.e e . dy
y
?
? ??
??
??
??
?
 
 Put 
1
t
y
?? 
 
2
1
dy dt
y
?? 
 
1
t y
x.e t.e dt
?
??
?
 
 
1
tt y
x.e te e C
?
? ? ? ? 
 
1 1 1
y y y
1
x.e e e C
y
? ? ?
?
? ? ? 
 x = 1, y = 1 
 
1 1 1
C
e e e
? ? ? 
 ? 
1
C
e
?? 
 Put y = 
1
2
 
 
2 2 2
x 2 1 1
e e e e
? ? ? 
 x = 3 – e 
13. Let the parabola y = x
2
 + px – 3, meet the 
coordinate axes at the points P, Q and R. If the 
circle C with centre at (–1, –1) passes through the 
points P, Q and R, then the area of ?PQR is : 
 (1) 4 (2) 6 
 (3) 7  (4) 5 
Ans.  (2) 
Sol.  y = x
2
 + px – 3 
 Let P( ?, 0), Q( ?, 0), R(0, –3) 
 Circle with centre (–1, –1) is (x + 1)
2
 + (y + 1)
2
 = r
2
  
 Passes through (0, –3) 
 1
2
 + (–2)
2
 = r
2
 ] 
 r
2
 = 5 
 (x + 1)
2
 + (y + 1)
2
 = 5 
 Put y = 0 
 (x + 1)
2
 = 5 – 1 
 (x + 1)
2
 = 4 
 x + 1 = ±2 
 x = 1 or x = –3  
 ? P(1, 0) and Q(–3,0)  
 Area of ?PQR = 
1 0 1
1
3 0 1 6
2
0 3 1
??
?
  
14. A circle C of radius 2 lies in the second quadrant and 
touches both the coordinate axes. Let r be the radius 
of a circle that has centre at the point (2, 5) and 
intersects the circle C at exactly two points. If the set 
of all possible values of r is the interval ( ?, ?), then 
3 ? – 2 ? is equal to : 
 (1) 15 (2) 14 
 (3) 12  (4) 10 
Ans.  (1) 
Sol. 
 
(–2,2)  
 S
1
 : (x + 2)
2
 + (y – 2)
2
 = 2
2
  
 S
2
 : (x – 2)
2
 + (y – 5)
2
 = r
2
  
 Both circle intersect at two points  
 ? |r
1
 – r
2
| < c
1
c
2
 < r
1
 + r
2
  
 |r – 2| < 5 < 2 + r  
 ? 3 < r < 7  
 r ? (3, 7)  
 ? = 3, ? = 7  
 3 ? – 2? = 15  
15. Let for ƒ(x) = 7tan
8
x + 7tan
6
x – 3tan
4
x – 3tan
2
x, 
?
?
?
/4
1
0
I ƒ(x)dx and 
?
?
?
/4
2
0
I xƒ(x)dx . Then 7I
1
 + 12I
2
 
is equal to : 
 (1) 2? (2) ? 
 (3) 1  (4) 2 
Ans.  (3) 
 
Page 5


 1 
JEE–MAIN EXAMINATION – JANUARY 2025
MATHEMATICS TEST PAPER WITH SOLUTION 
(HELD ON WEDNESDAY 22
nd
 JANUARY 2025) TIME : 9:00 AM  TO  12:00 NOON
SECTION-A 
1. The number of non-empty equivalence relations on
the set {1,2,3} is :
(1) 6 (2) 7
(3) 5 (4) 4
Ans.  (3) 
Sol. Let R be the required relation 
A = {(1, 1) (2, 2), (3, 3)} 
(i) | R | = 3, when R = A
(ii) | R | = 5, e.g. R = A ? {(1, 2), (2, 1)}
Number of R can be [3] 
(iii) R = {1, 2, 3} × {1, 2, 3}
Ans. (5) 
2. Let ƒ : R ?R be a twice differentiable function
such that ƒ(x + y) = ƒ(x) ƒ(y) for all x, y ? R. If
ƒ'(0) = 4a and ƒ staisfies ƒ''(x) – 3a ƒ'(x) – ƒ(x) = 0,
a > 0, then the area of the region
R = {(x,y) | 0 ? y ? ƒ(ax), 0 ? x ? 2} is :
(1) e
2
 – 1 (2) e
4
 + 1
(3) e
4
 – 1 (4) e
2
 + 1
Ans.  (1) 
Sol. f(x + y) = f(x).f(y) 
? f(x) = e
?x
  f ?(0) = 4a
? f ?(x) = ?e
?x
  ? ? = 4a
So, f(x) = e
4ax
 
f ??(x) – 3af ?(x) – f (x) = 0 
? ?
2
 – 3a ? – 1 = 0
? 16a
2
 – 12a
2
 – 1 = 0 ? 4a
2
 = 1 ? ?
1
a
2
?
x = 0 
x = 2 
f(ax) = e
x 
F(x) = e
2x
 
2
x2
0
Area e dx e 1 ? ? ?
?
3. Let the triangle PQR be the image of the
triangle with vertices (1,3), (3,1) and (2, 4) in
the line x + 2y = 2. If the centroid of ?PQR is
the point ( ?, ?), then 15( ? – ?) is equal to :
(1) 24 (2) 19
(3) 21 (4) 22
Ans.  (4) 
Sol. Let ‘G’ be the centroid of ? formed by (1, 3) (3, 1) 
& (2, 4) 
8
G 2,
3
??
?
??
??
Image of G w.r.t. x + 2y – 2 = 0 
16
8
22
2 3
3
2
1 2 1 4
??
??
?? ??
??
??
? ? ?
?
2 16
53
? ??
?
??
??
? 
32 2
2
15 15
??
? ? ? ? , 
32 2 8 24
15 3 15
? ? ?
? ? ? ?
15( ? – ?) = – 2 + 24 = 22 
4. Let z
1
, z
2
 and z
3
 be three complex numbers on the circle
|z| = 1 with 
??
?
1
arg(z )
4
, arg(z
2
) = 0 and 
?
?
3
arg(z )
4
. 
If ? ? ? ? ? ?
2
1 2 2 3 3 1
z z z z z z 2 , ?, ? ? Z, then the 
value of ?
2
 + ?
2
 is : 
(1) 24 (2) 41
(3) 31 (4) 29
Ans.  (4) 
Sol. Z
1
 = e
–i ?/4
 , Z
2
 = 1, Z
3
 = e
i ?/4
2
1 2 2 3 3 1
z z z z z z ?? = 
2
i i i i
4 4 4 4
e 1 1 e e e
? ? ? ?
??
? ? ? ? ?
2
i i i
4 4 4
eee
? ? ?
??
??
2
i
4
2e i
?
?
?? = 
2
2 2i i ? ? ?
= 
? ? ? ?
22
2 1 2 ?? = 2 + 1 + 2 – 2 2 = 5 – 22
? = 5, ? = –2
? ?
2
 + ?
2
 = 29
 
2 
 
5. Using the principal values of the inverse 
trigonometric functions the sum of the maximum 
and the minimum values of 16((sec
–1
x)
2
 + (cosec
–1
x)
2
) 
is : 
 (1) 24 ?
2
 (2) 18 ?
2
 
 (3) 31 ?
2
  (4) 22 ?
2
 
Ans.  (4) 
Sol. 16(sec
–1
x)
2
 + (cosec
–1
x)
2
 
 Sec
–1
x = a ?[0, ?] – 
2
? ??
??
??
 
 cosec
–1
x = a
2
?
? 
 = 
2
2
22
16 a a 16 2a a
24
??
?? ?? ??
? ? ? ? ? ?
??
?? ??
??
????
??
 
 max]
a = 
?
 = 16[2 ?
2
 – ?
2 
+ 
2
4
? ] = 20 ?
2 
 
a
4
min]
?
?
 = 
2 2 2
2
2
16 2
16 4 4
?? ? ? ? ?
? ? ? ?
??
??
 
 Sum = 22 ?
2 
6. A coin is tossed three times. Let X denote the 
number of times a tail follows a head. If ? and ?
2
 
denote the mean and variance of X, then the value 
of 64( ? + ?
2
) is : 
 (1) 51 (2) 48 
 (3) 32  (4) 64 
Ans.  (2) 
Sol. HHH ? 0  
 HHT ? 0  
 HTH ? 1  
 HTT ? 0  
 THH ? 1  
 THT ? 1  
 TTH ? 1  
 TTT ? 0  
 Probability distribution  
 
i
i
x 0 1
11
P(x )
22
  
 
ii
1
xp
2
? ? ?
?
  
 
2 2 2
ii
x p µ ? ? ?
?
  
 
1 1 1
2 4 4
???  
 64(µ + ?
2
) = 
11
64
24
??
?
??
??
 = 48  
7. Let a
1
, a
2
, a
3
..... be a G.P. of increasing positive 
terms. If a
1
a
5
 = 28 and a
2 
+ a
4
 = 29, the a
6
 is equal to  
 (1) 628 (2) 526 
 (3) 784  (4) 812 
Ans.  (3) 
Sol. a
1
.a
5
 = 28 ? a.ar
4
 = 28 ? a
2
r
4
 = 28  …(1)  
 a
2
 + a
4
 = 29 ? ar + ar
3
 = 29  
 ? ar(1 + r
2
) = 29  
 ? a
2
r
2
(1 + r
2
)
2
 = (29)
2
   …(2)  
 By Eq. (1) & (2)  
 
2
22
r 28
(1 r ) 29 29
?
??
  
 ? 
2
r 28
1 r 29
?
?
 ? r 28 ?  
 ? a
2
r
4
 = 28 ? a
2
 × (28)
2
 = 28  
 ? 
1
a
28
?  
 ? a
6
 = ar
5
 = 
2
1
(28) 28 784
28
??  
8. Let 
? ? ?
??
1
x 1 y 2 z 3
L:
2 3 4
 and  
 
? ? ?
??
2
x 2 y 4 z 5
L:
3 4 5
 be two lines. Then which 
of the following points lies on the line of the 
shortest distance between L
1
 and L
2
 ? 
 (1) 
??
??
??
??
5
, 7,1
3
 (2) 
??
??
??
1
2,3,
3
 
 (3) 
??
?
??
??
81
, 1,
33
 (4) 
??
?
??
??
14 22
, 3,
33
 
Ans.  (4) 
Sol. 
 
P 
Q 
L
1
 
L
2
 
 
 P(2 ? + 1, 3 ? + 2, 4 ? + 3)   on L
1
  
 Q(3µ + 2, 4µ + 4, 5µ + 5)   on L
2
  
 Dr’s of PQ = 3µ – 2 ? + 1, 4µ – 3? + 2, 5µ – 4 ? + 2  
 PQ ? L
1
  
 
3 
 
 ? (3µ – 2? + 1)2 + (4µ – 3? + 2)3 + (5µ – 4? + 
2)4 = 0  
 38µ – 29 ? + 16 = 0  …(1)  
 PQ ? L
2
  
 ? (3µ – 2? + 1)3 + (4µ – 3? + 2)4 + (5µ – 4? + 
2)5 = 0  
 50µ – 38 ? + 21 = 0  …(2)  
 By (1) & (2)  
 
1
3
?? ;  
1
µ
6
?
?  
 ??
5 13
P ,3,
33
??
??
??
 & 
3 10 25
Q , ,
2 3 6
??
??
??
  
 Line PQ  
 
5
x
3
1
6
?
 
y3
1
3
?
?
 
13
z
3
1
6
?
  
 
5 13
xz
y3
33
1 2 1
??
?
??
?
  
 Point
14 22
, 3,
33
??
?
??
??
  
 lies on the line PQ  
9. The product of all solutions of the equation 
? ? ?
?
2
e
5 log x 3 8
ex , x > 0, is : 
 (1) e
8/5
 (2) e
6/5
 
 (3) e
2
  (4) e 
Ans.  (1) 
Sol. 
2
5( nx) 3 8
ex
?
?  
 ??
2
5( nx) 3 8
ne n x
?
? ??
 ? 5( ?nx)
2
 + 3 = 8?nx  
 ( ?nx = t)  
 ? 5t
2
 – 8t + 3 = 0  
  t
1
 + t
2
 = 
8
5
  
  ?nx
1
x
2
 = 
8
5
  
  x
1
x
2
 = e
8/5
  
10. If 
?
? ? ? ?
?
?
n
r
r1
(2n 1)(2n 1)(2n 3)(2n 5)
T
64
, then  
 
??
?
??
??
??
?
n
n
r1
r
1
lim
T
 is equal to : 
 (1) 1 (2) 0 
 (3) 
2
3
  (4) 
1
3
 
Ans.  (3) 
Sol. T
n
 = S
n
 – S
n–1
  
 ? T
n
 = 
1
8
(2n – 1)(2n + 1)(2n + 3)  
 ??
n
18
T (2n 1)(2n 1)(2n 3)
?
? ? ?
??
?
nn
nn
r 1 r 1 r
11
lim lim8
T (2n 1)(2n 1)(2n 3)
? ? ? ?
??
?
? ? ?
??
??
?
n
8 1 1
lim
4 (2n 1)(2n 1) (2n 1)(2n 3)
??
??
??
??
? ? ? ?
??
?
?
?
n
1 1 1 1
lim 2 ...
1.3 3.5 3.5 5.7
??
?? ? ? ? ?
? ? ? ? ?
? ? ? ? ??
? ? ? ? ??
??
?
2
3
? ??
11. From all the English alphabets, five letters are 
chosen and are arranged in alphabetical order. The 
total number of ways, in which the middle letter is 
‘M’, is : 
 (1) 14950 (2) 6084 
 (3) 4356  (4) 5148 
Ans.  (4) 
Sol. 
12 13
AB M N..........Z 
 
12 13
22
Selection of two Selection of two
letters before M letters after M
C C 5148 ? ? ? 
12. Let x = x(y) be the solution of the differential 
equation 
??
? ? ?
??
??
2
1
y dx x dy 0
y
. If x(1) = 1, then 
??
??
??
1
x
2
 is : 
 (1) ?
1
e
2
 (2) ?
3
e
2
 
 (3) 3 – e  (4) 3 + e 
 
4 
 
Ans.  (3) 
Sol. 
23
dx 1 1
x
dy y y
??
??
??
??
 
 I.F. = 
2
1 1
dy
yy
ee
?
?
? 
 
1
1
y
t
3
1
x.e e . dy
y
?
? ??
??
??
??
?
 
 Put 
1
t
y
?? 
 
2
1
dy dt
y
?? 
 
1
t y
x.e t.e dt
?
??
?
 
 
1
tt y
x.e te e C
?
? ? ? ? 
 
1 1 1
y y y
1
x.e e e C
y
? ? ?
?
? ? ? 
 x = 1, y = 1 
 
1 1 1
C
e e e
? ? ? 
 ? 
1
C
e
?? 
 Put y = 
1
2
 
 
2 2 2
x 2 1 1
e e e e
? ? ? 
 x = 3 – e 
13. Let the parabola y = x
2
 + px – 3, meet the 
coordinate axes at the points P, Q and R. If the 
circle C with centre at (–1, –1) passes through the 
points P, Q and R, then the area of ?PQR is : 
 (1) 4 (2) 6 
 (3) 7  (4) 5 
Ans.  (2) 
Sol.  y = x
2
 + px – 3 
 Let P( ?, 0), Q( ?, 0), R(0, –3) 
 Circle with centre (–1, –1) is (x + 1)
2
 + (y + 1)
2
 = r
2
  
 Passes through (0, –3) 
 1
2
 + (–2)
2
 = r
2
 ] 
 r
2
 = 5 
 (x + 1)
2
 + (y + 1)
2
 = 5 
 Put y = 0 
 (x + 1)
2
 = 5 – 1 
 (x + 1)
2
 = 4 
 x + 1 = ±2 
 x = 1 or x = –3  
 ? P(1, 0) and Q(–3,0)  
 Area of ?PQR = 
1 0 1
1
3 0 1 6
2
0 3 1
??
?
  
14. A circle C of radius 2 lies in the second quadrant and 
touches both the coordinate axes. Let r be the radius 
of a circle that has centre at the point (2, 5) and 
intersects the circle C at exactly two points. If the set 
of all possible values of r is the interval ( ?, ?), then 
3 ? – 2 ? is equal to : 
 (1) 15 (2) 14 
 (3) 12  (4) 10 
Ans.  (1) 
Sol. 
 
(–2,2)  
 S
1
 : (x + 2)
2
 + (y – 2)
2
 = 2
2
  
 S
2
 : (x – 2)
2
 + (y – 5)
2
 = r
2
  
 Both circle intersect at two points  
 ? |r
1
 – r
2
| < c
1
c
2
 < r
1
 + r
2
  
 |r – 2| < 5 < 2 + r  
 ? 3 < r < 7  
 r ? (3, 7)  
 ? = 3, ? = 7  
 3 ? – 2? = 15  
15. Let for ƒ(x) = 7tan
8
x + 7tan
6
x – 3tan
4
x – 3tan
2
x, 
?
?
?
/4
1
0
I ƒ(x)dx and 
?
?
?
/4
2
0
I xƒ(x)dx . Then 7I
1
 + 12I
2
 
is equal to : 
 (1) 2? (2) ? 
 (3) 1  (4) 2 
Ans.  (3) 
 
 
5 
 
Sol. f(x) = (7tan
6
x – 3tan
2
x)(sec
2
x)  
 
/4
6 2 2
1
0
I (7 tan x 3tan x)(sec x)dx
?
??
?
  
 Put tanx = t  
 
1
1
6 2 7 3
1
0
0
I (7t 3t )dt t t 0 ?? ? ? ? ? ?
?? ?
  
 
/4
6 2 2
2
I
0
II
I x (7 tan x 3tan x)(sec x)dx
?
??
?
  
 
? ?
/4
/4
7 3 7 3
0
0
x tan x tan x (tan x tan x)dx
?
?
??
? ? ? ?
??
?
 
 
? ? ? ?
/4
3 2 2
0
0 tan x tan x 1 1 tan x dx
?
? ? ? ?
?
  
 Put tanx = t  
 
? ?
1 64
53
0
t t 1
t t dt
6 4 12
??
? ? ? ? ? ? ?
??
??
?
  
 7I
1
 + 12I
2
 = 1  
16. Let ƒ(x) be a real differentiable function such that 
ƒ(0) = 1 and ƒ(x + y) = ƒ(x)ƒ'(y) + ƒ'(x) ƒ(y) for all 
x, y ? R. Then 
?
?
100
e
n1
log ƒ(n) is equal to : 
 (1) 2384 (2) 2525 
 (3) 5220  (4) 2406 
Ans.  (2) 
Sol. f(x + y) = f(x) f ?(y) + f ?(x) f(x) 
 Put = x = y = 0 
 f(0) = f(0)f ?(0) + f ?(0)f(0) 
 f ?(0) = 
1
2
 
 Put y = 0 
 f(x) = f(x) f ?(0) + f ?(x)f(0) 
 ? ? ? ? ? ?
1
f x f x f x
2
? ? ? 
 ? ?
? ? fx
fx
2
?? 
 
dy y dy dx
dx 2 y 2
? ? ?
??
 
 ? ?ny = 
x
c
2
? 
 ? f (0) = 1 ? C = 0 
 ?ny = 
2
?
? f(x) = 
x/2
e 
 ?n f (n) = 
n
2
 
 ? ?
100 100
n 1 n 1
1 5050
f n n
22
??
??
??
 
 = 2525 
17. Let A = {1,2,3,.......,10} and  
 
??
??
??
??
m
B : m, n A, m < n and gcd (m, n) =1
n
. 
Then n(B) is equal to : 
 (1) 31 (2) 36 
 (3) 37  (4) 29 
Ans.  (1) 
Sol. A = {1, 2, ….10} 
 B {
m
n
= m, n ??A, m < n, gcd (m, n) = 1} 
 n(B) 
 n = 2 
1
2
??
??
??
 
 n = 3 
12
,
33
??
??
??
 
 n = 4 
13
,
44
??
??
??
 
 n = 5 
1 2 3 4
, , ,
5555
??
??
??
 
 n = 6 
15
,
66
??
??
??
 
 n = 7 
1 2 3 4 5 6
,,,,,
777777
??
??
??
 
 n = 8 
1 3 5 7
, , ,
8 8 8 8
??
??
??
 
 n = 9 
1 2 4 5 7 8
, , , , ,
9 9 9 9 9 9
??
??
??
 
 n = 10 
1 3 7 9
,,,
10 10 10 10
??
??
??
 
 n(B) = 31