JEE Main 2025 January 29 Shift 2 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 29
th
 JANUARY 2025) TIME : 3:00 PM  TO  6:00 PM
SECTION-A 
1. If the set of all a ? ?R, for which the equation 2x
2
 +
(a – 5)x + 15 = 3a has no real root, is the interval
(???), and X = {x ? Z : ? < x < ?}, then
2
xX
x
?
?
is
equal to  
(1) 2109 (2) 2129
(3) 2139 (4) 2119
Ans.  (3) 
Sol. (a – 5)
2
 – 8(15 – 3a) < 0 
a
2
 + 14a + 25 – 120 < 0 
a
2
 + 14a – 95 < 0 
(a + 19)(a – 5) < 0 
a ??(–19, 5) 
? –19 < x < 5
? 
2
xX
x
?
?
= (1
2
+ 2
2
+ ….+ 4
2
) + (1
2
+ 2
2
+ … + 18
2
) 
4 5 9 18 19 37
66
? ? ? ?
??
= 30 + 2109 
= 2139 
2. If sinx + sin
2
x = 1, x 0,
2
? ??
?
??
??
, then 
(cos
12
x + tan
12
x) + 3(cos
10
x +
 
tan
10
x + cos
8
x + tan
8
x) 
+ (cos
6
x + tan
6
x) is equal to
(1) 4 (2) 3
(3) 2 (4) 1
Ans.  (3) 
Sol. sinx + sin
2
x = 1 
??sinx = cos
2
x ? tanx = cosx
? Given expression
= 2cos
12
x + 6[cos
10
x + cos
8
x] + 2cos
6
x 
= 2[sin
6
x + 3sin
5
x + 3sin
4
x + sin
3
x] 
= 2sin
3
x[(sinx + 1)
3
] 
= 2[sin
2
x + sinx]
3 
= 2 
3. Let the area enclosed between the curves |y| = 1 –
x
2
 and x
2
 + y
2
 = 1 be ?. If 9 ? = ?? + ?; ? ? ? are
integers, then the value of | ? ? ?| equals
(1) 27 (2) 18
(3) 15 (4) 33
Ans.  (4) 
Sol. C
1
 : |y| = 1 – x
2
C
2
 : x
2
 + y
2
 = 1
(0, 1) 
(1, 0) 
??Required Area
= ? = 4[Area of circle in 1
st
 quad. – 
1
2
0
(1 x )dx ?
?
] 
1
3
0
x
4x
43
??
?? ?
?? ? ? ?
??
?? ??
??
 
8
3
? ? ? ?
??3 ? = 3 ? – 8
??9 ? = 9 ? – 24
? ? ? = 9, ? = – 24
? | ? – ?| = 33
4. If the domain of the function
? ?
2
5
log 18x x 77 ??
is ( ? ? ?) and the domain of the function 
? ?
2
x1 2
2x 3x 2
log
x 3x 4
?
??
??
??
??
??
??
 is ( ???), then ?
2
 + ?
2
 + ?
2
 
is equal to : 
(1) 195 (2) 174
(3) 186 (4) 179
Page 2


 1 
JEE–MAIN EXAMINATION – JANUARY 2025
MATHEMATICS TEST PAPER WITH SOLUTION 
(HELD ON WEDNESDAY 29
th
 JANUARY 2025) TIME : 3:00 PM  TO  6:00 PM
SECTION-A 
1. If the set of all a ? ?R, for which the equation 2x
2
 +
(a – 5)x + 15 = 3a has no real root, is the interval
(???), and X = {x ? Z : ? < x < ?}, then
2
xX
x
?
?
is
equal to  
(1) 2109 (2) 2129
(3) 2139 (4) 2119
Ans.  (3) 
Sol. (a – 5)
2
 – 8(15 – 3a) < 0 
a
2
 + 14a + 25 – 120 < 0 
a
2
 + 14a – 95 < 0 
(a + 19)(a – 5) < 0 
a ??(–19, 5) 
? –19 < x < 5
? 
2
xX
x
?
?
= (1
2
+ 2
2
+ ….+ 4
2
) + (1
2
+ 2
2
+ … + 18
2
) 
4 5 9 18 19 37
66
? ? ? ?
??
= 30 + 2109 
= 2139 
2. If sinx + sin
2
x = 1, x 0,
2
? ??
?
??
??
, then 
(cos
12
x + tan
12
x) + 3(cos
10
x +
 
tan
10
x + cos
8
x + tan
8
x) 
+ (cos
6
x + tan
6
x) is equal to
(1) 4 (2) 3
(3) 2 (4) 1
Ans.  (3) 
Sol. sinx + sin
2
x = 1 
??sinx = cos
2
x ? tanx = cosx
? Given expression
= 2cos
12
x + 6[cos
10
x + cos
8
x] + 2cos
6
x 
= 2[sin
6
x + 3sin
5
x + 3sin
4
x + sin
3
x] 
= 2sin
3
x[(sinx + 1)
3
] 
= 2[sin
2
x + sinx]
3 
= 2 
3. Let the area enclosed between the curves |y| = 1 –
x
2
 and x
2
 + y
2
 = 1 be ?. If 9 ? = ?? + ?; ? ? ? are
integers, then the value of | ? ? ?| equals
(1) 27 (2) 18
(3) 15 (4) 33
Ans.  (4) 
Sol. C
1
 : |y| = 1 – x
2
C
2
 : x
2
 + y
2
 = 1
(0, 1) 
(1, 0) 
??Required Area
= ? = 4[Area of circle in 1
st
 quad. – 
1
2
0
(1 x )dx ?
?
] 
1
3
0
x
4x
43
??
?? ?
?? ? ? ?
??
?? ??
??
 
8
3
? ? ? ?
??3 ? = 3 ? – 8
??9 ? = 9 ? – 24
? ? ? = 9, ? = – 24
? | ? – ?| = 33
4. If the domain of the function
? ?
2
5
log 18x x 77 ??
is ( ? ? ?) and the domain of the function 
? ?
2
x1 2
2x 3x 2
log
x 3x 4
?
??
??
??
??
??
??
 is ( ???), then ?
2
 + ?
2
 + ?
2
 
is equal to : 
(1) 195 (2) 174
(3) 186 (4) 179
2 
Ans.  (3) 
Sol. f
1
(x) = log
5 
(18x – x
2
 – 77)
??18x – x
2
 – 77 > 0
x
2
 – 18x + 77 < 0
x ? (7, 11)  ? = 7, ? = 11 
f
2
(x) = 
2
(x 1) 2
2x 3x 2
log
x 3x 4
?
?? ??
??
??
??
? x – 1 > 0 , x – 1 ? 1,
2
2
2x 3x 2
0
x 3x 4
??
?
??
x > 1 , x ? 2 ,
(2x 1)(x 2)
0
(x 4)(x 1)
??
?
??
x > 1 , x ? 2 , 
4 1/2 –1 –2 
+ – + – + 
? x ? (4, ?)
? ? = 4
? ? ?
?
?? ? ?
?
? ? ? ?
?
?= 49 + 121 + 16
= 186
5. Let the function f(x) = (x
2
 – 1)|x
2
 – ax + 2| + cos|x|
be not differentiable at the two points x = ? = 2
and x = ?. Then the distance of the point ( ? ? ?)
from the line 12x + 5y + 10 = 0 is equal to :
2
2
 – ax + 2 = 0
One root is given, ? = 2
? 4 – 2a + 2 = 0
a = 3
? other root ? = 1
but for x = 1 f(x) is differentiable 
(Drop) 
6. Let a straight line L pass through the point
P(2,–1,3) and be perpendicular to the lines
x 1 y 1 z 3
2 1 2
? ? ?
??
?
and 
x 3 y 2 z 2
1 3 4
? ? ?
?? . 
If the line L intersects the yz-plane at the point Q, 
then the distance between the points P and Q is :  
(1) 2 (2) 10
(3) 3 (4)23
Ans.  (3) 
Sol. Vector parallel to 'L' 
ˆ ˆ ˆ
i j k
2 1 2
1 3 4
??
ˆ ˆ ˆ
10i 10j 5k ? ? ?
ˆ ˆ ˆ
5(2i 2j k) ? ? ?
Equation of 'L' 
x 2 y 1 z 3
(say)
2 2 1
? ? ?
? ? ? ?
?
Let   Q(2 ? + 2, –2 ? – 1, ? + 3) 
? 2? + 2 = 0  ? ? = –1 ?
??Q (0, 1, 2)
d(P, Q) = 3 
7. Let S = N ? ???? ??Define a relation R from S to R
by :
? ?
?? ??
? ? ? ?
??
??
?? ??
ee
2
x,y :log y xlog ,x S,y
5
RR . 
Then, the sum of all the elements in the range of R 
is equal to  
(1) 
3
2
(2) 
5
3
(3) 
10
9
(4) 
5
2
Ans.  (2) 
Sol. S = {0, 1, 2, 3 …..} 
log
e
y = xlog
e
2
5
??
??
??
??
x
2
y
5
??
?
??
??
(1)
– ax + 2|
? Not differentiable at
x
 3 (2) 4
(3) 2 (4) 5
 3 (2) 4
(3) 2 (4) 5
A Ans. ns.    NTA Ans. (1) 
 Sol. cos|x| is always differentiable
? we have to check only for |x
Page 3


 1 
JEE–MAIN EXAMINATION – JANUARY 2025
MATHEMATICS TEST PAPER WITH SOLUTION 
(HELD ON WEDNESDAY 29
th
 JANUARY 2025) TIME : 3:00 PM  TO  6:00 PM
SECTION-A 
1. If the set of all a ? ?R, for which the equation 2x
2
 +
(a – 5)x + 15 = 3a has no real root, is the interval
(???), and X = {x ? Z : ? < x < ?}, then
2
xX
x
?
?
is
equal to  
(1) 2109 (2) 2129
(3) 2139 (4) 2119
Ans.  (3) 
Sol. (a – 5)
2
 – 8(15 – 3a) < 0 
a
2
 + 14a + 25 – 120 < 0 
a
2
 + 14a – 95 < 0 
(a + 19)(a – 5) < 0 
a ??(–19, 5) 
? –19 < x < 5
? 
2
xX
x
?
?
= (1
2
+ 2
2
+ ….+ 4
2
) + (1
2
+ 2
2
+ … + 18
2
) 
4 5 9 18 19 37
66
? ? ? ?
??
= 30 + 2109 
= 2139 
2. If sinx + sin
2
x = 1, x 0,
2
? ??
?
??
??
, then 
(cos
12
x + tan
12
x) + 3(cos
10
x +
 
tan
10
x + cos
8
x + tan
8
x) 
+ (cos
6
x + tan
6
x) is equal to
(1) 4 (2) 3
(3) 2 (4) 1
Ans.  (3) 
Sol. sinx + sin
2
x = 1 
??sinx = cos
2
x ? tanx = cosx
? Given expression
= 2cos
12
x + 6[cos
10
x + cos
8
x] + 2cos
6
x 
= 2[sin
6
x + 3sin
5
x + 3sin
4
x + sin
3
x] 
= 2sin
3
x[(sinx + 1)
3
] 
= 2[sin
2
x + sinx]
3 
= 2 
3. Let the area enclosed between the curves |y| = 1 –
x
2
 and x
2
 + y
2
 = 1 be ?. If 9 ? = ?? + ?; ? ? ? are
integers, then the value of | ? ? ?| equals
(1) 27 (2) 18
(3) 15 (4) 33
Ans.  (4) 
Sol. C
1
 : |y| = 1 – x
2
C
2
 : x
2
 + y
2
 = 1
(0, 1) 
(1, 0) 
??Required Area
= ? = 4[Area of circle in 1
st
 quad. – 
1
2
0
(1 x )dx ?
?
] 
1
3
0
x
4x
43
??
?? ?
?? ? ? ?
??
?? ??
??
 
8
3
? ? ? ?
??3 ? = 3 ? – 8
??9 ? = 9 ? – 24
? ? ? = 9, ? = – 24
? | ? – ?| = 33
4. If the domain of the function
? ?
2
5
log 18x x 77 ??
is ( ? ? ?) and the domain of the function 
? ?
2
x1 2
2x 3x 2
log
x 3x 4
?
??
??
??
??
??
??
 is ( ???), then ?
2
 + ?
2
 + ?
2
 
is equal to : 
(1) 195 (2) 174
(3) 186 (4) 179
2 
Ans.  (3) 
Sol. f
1
(x) = log
5 
(18x – x
2
 – 77)
??18x – x
2
 – 77 > 0
x
2
 – 18x + 77 < 0
x ? (7, 11)  ? = 7, ? = 11 
f
2
(x) = 
2
(x 1) 2
2x 3x 2
log
x 3x 4
?
?? ??
??
??
??
? x – 1 > 0 , x – 1 ? 1,
2
2
2x 3x 2
0
x 3x 4
??
?
??
x > 1 , x ? 2 ,
(2x 1)(x 2)
0
(x 4)(x 1)
??
?
??
x > 1 , x ? 2 , 
4 1/2 –1 –2 
+ – + – + 
? x ? (4, ?)
? ? = 4
? ? ?
?
?? ? ?
?
? ? ? ?
?
?= 49 + 121 + 16
= 186
5. Let the function f(x) = (x
2
 – 1)|x
2
 – ax + 2| + cos|x|
be not differentiable at the two points x = ? = 2
and x = ?. Then the distance of the point ( ? ? ?)
from the line 12x + 5y + 10 = 0 is equal to :
2
2
 – ax + 2 = 0
One root is given, ? = 2
? 4 – 2a + 2 = 0
a = 3
? other root ? = 1
but for x = 1 f(x) is differentiable 
(Drop) 
6. Let a straight line L pass through the point
P(2,–1,3) and be perpendicular to the lines
x 1 y 1 z 3
2 1 2
? ? ?
??
?
and 
x 3 y 2 z 2
1 3 4
? ? ?
?? . 
If the line L intersects the yz-plane at the point Q, 
then the distance between the points P and Q is :  
(1) 2 (2) 10
(3) 3 (4)23
Ans.  (3) 
Sol. Vector parallel to 'L' 
ˆ ˆ ˆ
i j k
2 1 2
1 3 4
??
ˆ ˆ ˆ
10i 10j 5k ? ? ?
ˆ ˆ ˆ
5(2i 2j k) ? ? ?
Equation of 'L' 
x 2 y 1 z 3
(say)
2 2 1
? ? ?
? ? ? ?
?
Let   Q(2 ? + 2, –2 ? – 1, ? + 3) 
? 2? + 2 = 0  ? ? = –1 ?
??Q (0, 1, 2)
d(P, Q) = 3 
7. Let S = N ? ???? ??Define a relation R from S to R
by :
? ?
?? ??
? ? ? ?
??
??
?? ??
ee
2
x,y :log y xlog ,x S,y
5
RR . 
Then, the sum of all the elements in the range of R 
is equal to  
(1) 
3
2
(2) 
5
3
(3) 
10
9
(4) 
5
2
Ans.  (2) 
Sol. S = {0, 1, 2, 3 …..} 
log
e
y = xlog
e
2
5
??
??
??
??
x
2
y
5
??
?
??
??
(1)
– ax + 2|
? Not differentiable at
x
 3 (2) 4
(3) 2 (4) 5
 3 (2) 4
(3) 2 (4) 5
A Ans. ns.    NTA Ans. (1) 
 Sol. cos|x| is always differentiable
? we have to check only for |x
3 
(0, 1) 
Required 
Sum = 
1 2 3
2 2 2 1 5
1 .....
2
5 5 5 3
1
5
? ? ? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
?
8. Let the line x + y = 1 meet the axes of x and y at A
and B, respectively. A right angled triangle AMN
is inscribed in the triangle OAB, where O is the
origin and the points M and N lie on the lines OB
and AB, respectively. If the area of the triangle
AMN is 
4
9
of the area of the triangle OAB and 
AN : NB = ? : 1, then the sum of all possible 
value(s) of is ? : 
(1) 
1
2
 (2) 
13
6
(3) 
5
2
(4) 2
Ans.  (4) 
Sol. 
N 
M 
O 
B 
? ?
45º– ? ?
A 
Area of ?AOB = 
1
2
Area of ?AMN = 
4 1 2
9 2 9
??
Equation of AB is x + y = 1 
OA = 1,  AM = sec(45° – ?) 
AN = sec(45° – ?) cos ? 
MN = sec(45° – ?) sin ? ?
Ar( ?AMN) = 
1
2
? sec
2
(45° – ?)sin ??cos ?
2
9
? ?
??tan?? ? ????
1
2
?
tan ? = 2 is rejected 
AN
cot 2
NB 1
?
? ? ? ?
9. If ?x + ?y = 109 is the equation of the chord of the
ellipse 
22
xy
1
94
?? , whose mid point is 
51
,
22
??
??
??
, 
then ? + ? is equal to 
(1) 37 (2) 46
(3) 58 (4) 72
Ans.  (3) 
Sol. 
(0, 2) 
(3, 0) 
M 
51
,
22
??
??
??
Equation of chord T = S
1
 
5 x 1 y 25 1
2 9 2 4 36 16
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
5x y 100 9 109
18 8 144 144
?
? ? ? ?
???40x + 18y = 109
???? = 40, ? = 18
???? + ? = 58
10. If all the words with or without meaning made
using all the letters of the word “KANPUR” are
arranged as in a dictionary, then the word at 440
th
position in this arrangement, is :
(1) PRNAKU (2) PRKANU
(3) PRKAUN (4) PRNAUK
Ans.  (3) 
Page 4


 1 
JEE–MAIN EXAMINATION – JANUARY 2025
MATHEMATICS TEST PAPER WITH SOLUTION 
(HELD ON WEDNESDAY 29
th
 JANUARY 2025) TIME : 3:00 PM  TO  6:00 PM
SECTION-A 
1. If the set of all a ? ?R, for which the equation 2x
2
 +
(a – 5)x + 15 = 3a has no real root, is the interval
(???), and X = {x ? Z : ? < x < ?}, then
2
xX
x
?
?
is
equal to  
(1) 2109 (2) 2129
(3) 2139 (4) 2119
Ans.  (3) 
Sol. (a – 5)
2
 – 8(15 – 3a) < 0 
a
2
 + 14a + 25 – 120 < 0 
a
2
 + 14a – 95 < 0 
(a + 19)(a – 5) < 0 
a ??(–19, 5) 
? –19 < x < 5
? 
2
xX
x
?
?
= (1
2
+ 2
2
+ ….+ 4
2
) + (1
2
+ 2
2
+ … + 18
2
) 
4 5 9 18 19 37
66
? ? ? ?
??
= 30 + 2109 
= 2139 
2. If sinx + sin
2
x = 1, x 0,
2
? ??
?
??
??
, then 
(cos
12
x + tan
12
x) + 3(cos
10
x +
 
tan
10
x + cos
8
x + tan
8
x) 
+ (cos
6
x + tan
6
x) is equal to
(1) 4 (2) 3
(3) 2 (4) 1
Ans.  (3) 
Sol. sinx + sin
2
x = 1 
??sinx = cos
2
x ? tanx = cosx
? Given expression
= 2cos
12
x + 6[cos
10
x + cos
8
x] + 2cos
6
x 
= 2[sin
6
x + 3sin
5
x + 3sin
4
x + sin
3
x] 
= 2sin
3
x[(sinx + 1)
3
] 
= 2[sin
2
x + sinx]
3 
= 2 
3. Let the area enclosed between the curves |y| = 1 –
x
2
 and x
2
 + y
2
 = 1 be ?. If 9 ? = ?? + ?; ? ? ? are
integers, then the value of | ? ? ?| equals
(1) 27 (2) 18
(3) 15 (4) 33
Ans.  (4) 
Sol. C
1
 : |y| = 1 – x
2
C
2
 : x
2
 + y
2
 = 1
(0, 1) 
(1, 0) 
??Required Area
= ? = 4[Area of circle in 1
st
 quad. – 
1
2
0
(1 x )dx ?
?
] 
1
3
0
x
4x
43
??
?? ?
?? ? ? ?
??
?? ??
??
 
8
3
? ? ? ?
??3 ? = 3 ? – 8
??9 ? = 9 ? – 24
? ? ? = 9, ? = – 24
? | ? – ?| = 33
4. If the domain of the function
? ?
2
5
log 18x x 77 ??
is ( ? ? ?) and the domain of the function 
? ?
2
x1 2
2x 3x 2
log
x 3x 4
?
??
??
??
??
??
??
 is ( ???), then ?
2
 + ?
2
 + ?
2
 
is equal to : 
(1) 195 (2) 174
(3) 186 (4) 179
2 
Ans.  (3) 
Sol. f
1
(x) = log
5 
(18x – x
2
 – 77)
??18x – x
2
 – 77 > 0
x
2
 – 18x + 77 < 0
x ? (7, 11)  ? = 7, ? = 11 
f
2
(x) = 
2
(x 1) 2
2x 3x 2
log
x 3x 4
?
?? ??
??
??
??
? x – 1 > 0 , x – 1 ? 1,
2
2
2x 3x 2
0
x 3x 4
??
?
??
x > 1 , x ? 2 ,
(2x 1)(x 2)
0
(x 4)(x 1)
??
?
??
x > 1 , x ? 2 , 
4 1/2 –1 –2 
+ – + – + 
? x ? (4, ?)
? ? = 4
? ? ?
?
?? ? ?
?
? ? ? ?
?
?= 49 + 121 + 16
= 186
5. Let the function f(x) = (x
2
 – 1)|x
2
 – ax + 2| + cos|x|
be not differentiable at the two points x = ? = 2
and x = ?. Then the distance of the point ( ? ? ?)
from the line 12x + 5y + 10 = 0 is equal to :
2
2
 – ax + 2 = 0
One root is given, ? = 2
? 4 – 2a + 2 = 0
a = 3
? other root ? = 1
but for x = 1 f(x) is differentiable 
(Drop) 
6. Let a straight line L pass through the point
P(2,–1,3) and be perpendicular to the lines
x 1 y 1 z 3
2 1 2
? ? ?
??
?
and 
x 3 y 2 z 2
1 3 4
? ? ?
?? . 
If the line L intersects the yz-plane at the point Q, 
then the distance between the points P and Q is :  
(1) 2 (2) 10
(3) 3 (4)23
Ans.  (3) 
Sol. Vector parallel to 'L' 
ˆ ˆ ˆ
i j k
2 1 2
1 3 4
??
ˆ ˆ ˆ
10i 10j 5k ? ? ?
ˆ ˆ ˆ
5(2i 2j k) ? ? ?
Equation of 'L' 
x 2 y 1 z 3
(say)
2 2 1
? ? ?
? ? ? ?
?
Let   Q(2 ? + 2, –2 ? – 1, ? + 3) 
? 2? + 2 = 0  ? ? = –1 ?
??Q (0, 1, 2)
d(P, Q) = 3 
7. Let S = N ? ???? ??Define a relation R from S to R
by :
? ?
?? ??
? ? ? ?
??
??
?? ??
ee
2
x,y :log y xlog ,x S,y
5
RR . 
Then, the sum of all the elements in the range of R 
is equal to  
(1) 
3
2
(2) 
5
3
(3) 
10
9
(4) 
5
2
Ans.  (2) 
Sol. S = {0, 1, 2, 3 …..} 
log
e
y = xlog
e
2
5
??
??
??
??
x
2
y
5
??
?
??
??
(1)
– ax + 2|
? Not differentiable at
x
 3 (2) 4
(3) 2 (4) 5
 3 (2) 4
(3) 2 (4) 5
A Ans. ns.    NTA Ans. (1) 
 Sol. cos|x| is always differentiable
? we have to check only for |x
3 
(0, 1) 
Required 
Sum = 
1 2 3
2 2 2 1 5
1 .....
2
5 5 5 3
1
5
? ? ? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
?
8. Let the line x + y = 1 meet the axes of x and y at A
and B, respectively. A right angled triangle AMN
is inscribed in the triangle OAB, where O is the
origin and the points M and N lie on the lines OB
and AB, respectively. If the area of the triangle
AMN is 
4
9
of the area of the triangle OAB and 
AN : NB = ? : 1, then the sum of all possible 
value(s) of is ? : 
(1) 
1
2
 (2) 
13
6
(3) 
5
2
(4) 2
Ans.  (4) 
Sol. 
N 
M 
O 
B 
? ?
45º– ? ?
A 
Area of ?AOB = 
1
2
Area of ?AMN = 
4 1 2
9 2 9
??
Equation of AB is x + y = 1 
OA = 1,  AM = sec(45° – ?) 
AN = sec(45° – ?) cos ? 
MN = sec(45° – ?) sin ? ?
Ar( ?AMN) = 
1
2
? sec
2
(45° – ?)sin ??cos ?
2
9
? ?
??tan?? ? ????
1
2
?
tan ? = 2 is rejected 
AN
cot 2
NB 1
?
? ? ? ?
9. If ?x + ?y = 109 is the equation of the chord of the
ellipse 
22
xy
1
94
?? , whose mid point is 
51
,
22
??
??
??
, 
then ? + ? is equal to 
(1) 37 (2) 46
(3) 58 (4) 72
Ans.  (3) 
Sol. 
(0, 2) 
(3, 0) 
M 
51
,
22
??
??
??
Equation of chord T = S
1
 
5 x 1 y 25 1
2 9 2 4 36 16
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
5x y 100 9 109
18 8 144 144
?
? ? ? ?
???40x + 18y = 109
???? = 40, ? = 18
???? + ? = 58
10. If all the words with or without meaning made
using all the letters of the word “KANPUR” are
arranged as in a dictionary, then the word at 440
th
position in this arrangement, is :
(1) PRNAKU (2) PRKANU
(3) PRKAUN (4) PRNAUK
Ans.  (3) 
4 
Sol. A, K, N, P, R, U 
A …………………. = 5 = 120 
K …………………. = 5 = 120 
N …………………. = 5 = 120 
PA ………………. = 4 = 24 
PK ………………. = 4 = 24 
PN ………………. = 4 = 24 
P R A ……………. = 3 = 6 
P R K A N U          =  1 
P R K A U N          =  1 
Total = 440 
? 440
th
 word
11. Let ? ? ? ( ? ? ?) be the values of m, for which the
equations x + y + z = 1; x + 2y + 4z = m and
x + 4y + 10z = m
2
 have infinitely many solutions.
Then the value of 
? ?
10
n1
nn
??
?
?
?
 is equal to :
(1) 440 (2) 3080
(3) 3410 (4) 560
Ans.  (1) 
Sol. 
1 1 1
1 2 4
1 4 10
?? = 1(20 – 16) – 1(10 – 4) + 1(4 – 2) 
= 4 – 6 + 2 = 0 
For infinite solutions 
?
x
 = ?
y
 = ?
z
 = 0 
m
2
 – 3x + 2 = 0 
m = 1, 2 
? = 1, ? = 2
? 
10 10 10
12
n 1 n 1 n 1
(n n ) n n
??
? ? ?
? ? ?
? ? ?
= 
10(11) 10(11)(21)
26
?
= 55 + 385 
= 440 
12. Let A = [a
ij
] be a matrix of order 3 × 3, with
? ?
ij
ij
a2
?
? . If the sum of all the elements in the 
third row of A
2
 is 2 ? ? ? , ? ? ? , Z , then ? ?? ? ? is 
equal to   
(1) 280 (2) 168
(3) 210 (4) 224
Ans.  (4) 
Sol. 
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
234
3 4 5
4 5 6
2 2 2
A 2 2 2
2 2 2
??
??
??
?
??
??
??
??
??
2 2 2 4
A 2 2 4 4 2
4 4 2 8
??
??
???
??
??
??
22
1 2 2 1 2 2
A 2 2 2 2 2 2 2 2 2
2 2 2 4 2 2 2 4
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
4
(2 4 8) (2 2 4 2 8 2) (4 8 16)
??
? ? ?
??
? ? ? ?
??
??
? ? ? ? ? ?
??
Sum of elements of 3
rd
 row = 4(14 + 14 2 + 28) 
= 4(42 + 14 2 ) 
= 168 + 56 2 
? + ? 2
?? ? ?? ? ? = 168 + 56 = 224
13. Let P be the foot of the perpendicular from the
point (1, 2, 2) on the line L : 
x 1 y 1 z 2
1 1 2
? ? ?
??
?
. 
Let the line 
? ? ? ?
ˆ ˆ ˆ ˆ ˆ ˆ
r i j 2k i j k ? ? ? ? ? ? ? ? , ? ? ? ?R,
intersect the line L at Q. Then 2(PQ)
2
 is equal to:  
(1) 27 (2) 25
(3) 29 (4) 19
Ans.  (1) 
Page 5


 1 
JEE–MAIN EXAMINATION – JANUARY 2025
MATHEMATICS TEST PAPER WITH SOLUTION 
(HELD ON WEDNESDAY 29
th
 JANUARY 2025) TIME : 3:00 PM  TO  6:00 PM
SECTION-A 
1. If the set of all a ? ?R, for which the equation 2x
2
 +
(a – 5)x + 15 = 3a has no real root, is the interval
(???), and X = {x ? Z : ? < x < ?}, then
2
xX
x
?
?
is
equal to  
(1) 2109 (2) 2129
(3) 2139 (4) 2119
Ans.  (3) 
Sol. (a – 5)
2
 – 8(15 – 3a) < 0 
a
2
 + 14a + 25 – 120 < 0 
a
2
 + 14a – 95 < 0 
(a + 19)(a – 5) < 0 
a ??(–19, 5) 
? –19 < x < 5
? 
2
xX
x
?
?
= (1
2
+ 2
2
+ ….+ 4
2
) + (1
2
+ 2
2
+ … + 18
2
) 
4 5 9 18 19 37
66
? ? ? ?
??
= 30 + 2109 
= 2139 
2. If sinx + sin
2
x = 1, x 0,
2
? ??
?
??
??
, then 
(cos
12
x + tan
12
x) + 3(cos
10
x +
 
tan
10
x + cos
8
x + tan
8
x) 
+ (cos
6
x + tan
6
x) is equal to
(1) 4 (2) 3
(3) 2 (4) 1
Ans.  (3) 
Sol. sinx + sin
2
x = 1 
??sinx = cos
2
x ? tanx = cosx
? Given expression
= 2cos
12
x + 6[cos
10
x + cos
8
x] + 2cos
6
x 
= 2[sin
6
x + 3sin
5
x + 3sin
4
x + sin
3
x] 
= 2sin
3
x[(sinx + 1)
3
] 
= 2[sin
2
x + sinx]
3 
= 2 
3. Let the area enclosed between the curves |y| = 1 –
x
2
 and x
2
 + y
2
 = 1 be ?. If 9 ? = ?? + ?; ? ? ? are
integers, then the value of | ? ? ?| equals
(1) 27 (2) 18
(3) 15 (4) 33
Ans.  (4) 
Sol. C
1
 : |y| = 1 – x
2
C
2
 : x
2
 + y
2
 = 1
(0, 1) 
(1, 0) 
??Required Area
= ? = 4[Area of circle in 1
st
 quad. – 
1
2
0
(1 x )dx ?
?
] 
1
3
0
x
4x
43
??
?? ?
?? ? ? ?
??
?? ??
??
 
8
3
? ? ? ?
??3 ? = 3 ? – 8
??9 ? = 9 ? – 24
? ? ? = 9, ? = – 24
? | ? – ?| = 33
4. If the domain of the function
? ?
2
5
log 18x x 77 ??
is ( ? ? ?) and the domain of the function 
? ?
2
x1 2
2x 3x 2
log
x 3x 4
?
??
??
??
??
??
??
 is ( ???), then ?
2
 + ?
2
 + ?
2
 
is equal to : 
(1) 195 (2) 174
(3) 186 (4) 179
2 
Ans.  (3) 
Sol. f
1
(x) = log
5 
(18x – x
2
 – 77)
??18x – x
2
 – 77 > 0
x
2
 – 18x + 77 < 0
x ? (7, 11)  ? = 7, ? = 11 
f
2
(x) = 
2
(x 1) 2
2x 3x 2
log
x 3x 4
?
?? ??
??
??
??
? x – 1 > 0 , x – 1 ? 1,
2
2
2x 3x 2
0
x 3x 4
??
?
??
x > 1 , x ? 2 ,
(2x 1)(x 2)
0
(x 4)(x 1)
??
?
??
x > 1 , x ? 2 , 
4 1/2 –1 –2 
+ – + – + 
? x ? (4, ?)
? ? = 4
? ? ?
?
?? ? ?
?
? ? ? ?
?
?= 49 + 121 + 16
= 186
5. Let the function f(x) = (x
2
 – 1)|x
2
 – ax + 2| + cos|x|
be not differentiable at the two points x = ? = 2
and x = ?. Then the distance of the point ( ? ? ?)
from the line 12x + 5y + 10 = 0 is equal to :
2
2
 – ax + 2 = 0
One root is given, ? = 2
? 4 – 2a + 2 = 0
a = 3
? other root ? = 1
but for x = 1 f(x) is differentiable 
(Drop) 
6. Let a straight line L pass through the point
P(2,–1,3) and be perpendicular to the lines
x 1 y 1 z 3
2 1 2
? ? ?
??
?
and 
x 3 y 2 z 2
1 3 4
? ? ?
?? . 
If the line L intersects the yz-plane at the point Q, 
then the distance between the points P and Q is :  
(1) 2 (2) 10
(3) 3 (4)23
Ans.  (3) 
Sol. Vector parallel to 'L' 
ˆ ˆ ˆ
i j k
2 1 2
1 3 4
??
ˆ ˆ ˆ
10i 10j 5k ? ? ?
ˆ ˆ ˆ
5(2i 2j k) ? ? ?
Equation of 'L' 
x 2 y 1 z 3
(say)
2 2 1
? ? ?
? ? ? ?
?
Let   Q(2 ? + 2, –2 ? – 1, ? + 3) 
? 2? + 2 = 0  ? ? = –1 ?
??Q (0, 1, 2)
d(P, Q) = 3 
7. Let S = N ? ???? ??Define a relation R from S to R
by :
? ?
?? ??
? ? ? ?
??
??
?? ??
ee
2
x,y :log y xlog ,x S,y
5
RR . 
Then, the sum of all the elements in the range of R 
is equal to  
(1) 
3
2
(2) 
5
3
(3) 
10
9
(4) 
5
2
Ans.  (2) 
Sol. S = {0, 1, 2, 3 …..} 
log
e
y = xlog
e
2
5
??
??
??
??
x
2
y
5
??
?
??
??
(1)
– ax + 2|
? Not differentiable at
x
 3 (2) 4
(3) 2 (4) 5
 3 (2) 4
(3) 2 (4) 5
A Ans. ns.    NTA Ans. (1) 
 Sol. cos|x| is always differentiable
? we have to check only for |x
3 
(0, 1) 
Required 
Sum = 
1 2 3
2 2 2 1 5
1 .....
2
5 5 5 3
1
5
? ? ? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
?
8. Let the line x + y = 1 meet the axes of x and y at A
and B, respectively. A right angled triangle AMN
is inscribed in the triangle OAB, where O is the
origin and the points M and N lie on the lines OB
and AB, respectively. If the area of the triangle
AMN is 
4
9
of the area of the triangle OAB and 
AN : NB = ? : 1, then the sum of all possible 
value(s) of is ? : 
(1) 
1
2
 (2) 
13
6
(3) 
5
2
(4) 2
Ans.  (4) 
Sol. 
N 
M 
O 
B 
? ?
45º– ? ?
A 
Area of ?AOB = 
1
2
Area of ?AMN = 
4 1 2
9 2 9
??
Equation of AB is x + y = 1 
OA = 1,  AM = sec(45° – ?) 
AN = sec(45° – ?) cos ? 
MN = sec(45° – ?) sin ? ?
Ar( ?AMN) = 
1
2
? sec
2
(45° – ?)sin ??cos ?
2
9
? ?
??tan?? ? ????
1
2
?
tan ? = 2 is rejected 
AN
cot 2
NB 1
?
? ? ? ?
9. If ?x + ?y = 109 is the equation of the chord of the
ellipse 
22
xy
1
94
?? , whose mid point is 
51
,
22
??
??
??
, 
then ? + ? is equal to 
(1) 37 (2) 46
(3) 58 (4) 72
Ans.  (3) 
Sol. 
(0, 2) 
(3, 0) 
M 
51
,
22
??
??
??
Equation of chord T = S
1
 
5 x 1 y 25 1
2 9 2 4 36 16
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
5x y 100 9 109
18 8 144 144
?
? ? ? ?
???40x + 18y = 109
???? = 40, ? = 18
???? + ? = 58
10. If all the words with or without meaning made
using all the letters of the word “KANPUR” are
arranged as in a dictionary, then the word at 440
th
position in this arrangement, is :
(1) PRNAKU (2) PRKANU
(3) PRKAUN (4) PRNAUK
Ans.  (3) 
4 
Sol. A, K, N, P, R, U 
A …………………. = 5 = 120 
K …………………. = 5 = 120 
N …………………. = 5 = 120 
PA ………………. = 4 = 24 
PK ………………. = 4 = 24 
PN ………………. = 4 = 24 
P R A ……………. = 3 = 6 
P R K A N U          =  1 
P R K A U N          =  1 
Total = 440 
? 440
th
 word
11. Let ? ? ? ( ? ? ?) be the values of m, for which the
equations x + y + z = 1; x + 2y + 4z = m and
x + 4y + 10z = m
2
 have infinitely many solutions.
Then the value of 
? ?
10
n1
nn
??
?
?
?
 is equal to :
(1) 440 (2) 3080
(3) 3410 (4) 560
Ans.  (1) 
Sol. 
1 1 1
1 2 4
1 4 10
?? = 1(20 – 16) – 1(10 – 4) + 1(4 – 2) 
= 4 – 6 + 2 = 0 
For infinite solutions 
?
x
 = ?
y
 = ?
z
 = 0 
m
2
 – 3x + 2 = 0 
m = 1, 2 
? = 1, ? = 2
? 
10 10 10
12
n 1 n 1 n 1
(n n ) n n
??
? ? ?
? ? ?
? ? ?
= 
10(11) 10(11)(21)
26
?
= 55 + 385 
= 440 
12. Let A = [a
ij
] be a matrix of order 3 × 3, with
? ?
ij
ij
a2
?
? . If the sum of all the elements in the 
third row of A
2
 is 2 ? ? ? , ? ? ? , Z , then ? ?? ? ? is 
equal to   
(1) 280 (2) 168
(3) 210 (4) 224
Ans.  (4) 
Sol. 
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
234
3 4 5
4 5 6
2 2 2
A 2 2 2
2 2 2
??
??
??
?
??
??
??
??
??
2 2 2 4
A 2 2 4 4 2
4 4 2 8
??
??
???
??
??
??
22
1 2 2 1 2 2
A 2 2 2 2 2 2 2 2 2
2 2 2 4 2 2 2 4
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
4
(2 4 8) (2 2 4 2 8 2) (4 8 16)
??
? ? ?
??
? ? ? ?
??
??
? ? ? ? ? ?
??
Sum of elements of 3
rd
 row = 4(14 + 14 2 + 28) 
= 4(42 + 14 2 ) 
= 168 + 56 2 
? + ? 2
?? ? ?? ? ? = 168 + 56 = 224
13. Let P be the foot of the perpendicular from the
point (1, 2, 2) on the line L : 
x 1 y 1 z 2
1 1 2
? ? ?
??
?
. 
Let the line 
? ? ? ?
ˆ ˆ ˆ ˆ ˆ ˆ
r i j 2k i j k ? ? ? ? ? ? ? ? , ? ? ? ?R,
intersect the line L at Q. Then 2(PQ)
2
 is equal to:  
(1) 27 (2) 25
(3) 29 (4) 19
Ans.  (1) 
5 
Sol. 
A(1, 2, 2) 
L 
P  (1, –1, 2) 
 (1, –1, 2) ? 
x 1 y 1 z 2
L:
1 1 2
? ? ?
? ? ? ?
?
P( ? + 1, – ?? – 1, 2? + 2) 
AP·d 0 ? ??(??, – ?? – 3, 2 ?)·(1, –1, 2) = 0 
??? + ? + 3 + 4 ? = 0 ??? = –
1
2
? 
1 1 1
P 1, 1, 2 2
2 2 2
?? ?? ??
? ? ? ?
?? ??
?? ??
11
P , ,1
22
? ??
??
??
Now general pt. on L
2
 is Q(–1 + ? , 1 – ?, –2 + ?) 
Equate it with general pt of L 
1 1 1 1 2 2 2
22
? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
2( ? – 2) + 2 = –2 + ?
2 ? – 4 + 2 = –2 + ?
? ? ??= – ??????? ? ? ???
? ? Q ? (–1 ,1 –2)
? ?
11
P , ,1 and Q 1,1, 2
22
? ??
??
??
??
22
2
11
PQ 1 1 (1 2)
22
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
9 9 54
9
4 4 4
? ? ? ?
? 2(PQ)
2
 = 
54
2 27
4
??
?
??
??
14. Let a circle C pass through the points (4, 2) and (0,
2), and its centre lie on 3x + 2y + 2 = 0. Then the
length of the chord, of the circle C, whose mid-
point is (1, 2), is:
(1) 3 (2) 23 
(3)42 (4)22
Ans.  (2) 
Sol. 
B(0, 2) 
A 
(4, 2) 
O 
 23
,
2
? ? ? ??
?
??
??
M
M
AB
 = 0  ??OM is vertical 
???? ?= 2
? ? Centre (0) ??(2, –4)
r = OA = 
22
(2 4) (2 4) 40 ? ? ? ?
mid point of chord is N ??(1, 2)  ? ON = 37
 ? length of chord = 
22
2 r (ON) ?
2 40 37 2 3 ? ? ?
15. Let A = [a
ij
] be a 2 × 2 matrix such that a
ij
 ? {0, 1}
for all i and j. Let the random variable X denote the
possible values of the determinant of the matrix A.
Then, the variance of X is:
(1) 
1
4
(2) 
3
8
(3) 
5
8
(4) 
3
4
Ans.  (2) 
Sol. 
11 12
21 22
aa
A
aa
?
= a
11 
a
22
 – a
21 
a
12 
= {–1, 0, 1} 
2
i i i 1 i
2
i i i i
x P P X P X
3 3 3
1
16 16 16
10
0 0 0
16
3 3 3
1
16 16 16
3
P X 0 P X
8
??
? ? ? ?
??var(x) =
22
i i i i
PX ( PX ) ? ? ?
= 
33
0
88
??