WBJEE Previous Year Question Paper - 2024 | WBJEE Sample Papers, Section Wise & Full Mock Tests 2026 PDF Download

 Page 1


Question 1
All values of a for which the inequality  is satisfied,
lie in the interval
Options:
A.
(1, 2)
B.
(0, 3)
C.
(0, 4)
D.
(1, 4)
Answer: C
-------------------------------------------------------------------------------------------------
Question 2
For any integer  has the value :
Options:
A.
B.
1
1
v a
a
?
1
(
3
2
v x + 1 -
1
v x
) d x < 4
n ,
p
?
0
e
c o s
2
x
· c o s
3
( 2 n + 1 ) x d x
p
WBJEE Paper - 2024
Mathematics
Page 2


Question 1
All values of a for which the inequality  is satisfied,
lie in the interval
Options:
A.
(1, 2)
B.
(0, 3)
C.
(0, 4)
D.
(1, 4)
Answer: C
-------------------------------------------------------------------------------------------------
Question 2
For any integer  has the value :
Options:
A.
B.
1
1
v a
a
?
1
(
3
2
v x + 1 -
1
v x
) d x < 4
n ,
p
?
0
e
c o s
2
x
· c o s
3
( 2 n + 1 ) x d x
p
WBJEE Paper - 2024
Mathematics
C.
0
D.
Answer: C
Solution:
To solve the given integral, let's consider the integral:
First, observe the integrand carefully. It includes the term  which can be expanded using the trigonometric
identity for cosine:
Applying this identity to the integrand, we have:
Substitute this back into the integral:
This can be broken down into two separate integrals:
Now, consider the properties of definite integrals and the cosine function over the interval from 0 to . For odd multiples of , the
integral over the symmetric limits will cancel out to zero due to the periodicity and symmetry of the cosine function.
Specifically, for any odd integer :
Therefore:
and
Thus, combining these results:
Therefore, the value of the integral is:
So, the correct option is:
Option C: 0
3 p
2
?
p
0
e
c o s
2
x
· c o s
3
( ( 2 n + 1 ) x ) d x
c o s
3
( ( 2 n + 1 ) x )
c o s
3
( ? ) =
1
4
( c o s ( 3 ? ) + 3 c o s ( ? ) )
c o s
3
( ( 2 n + 1 ) x ) =
1
4
( c o s ( 3 ( 2 n + 1 ) x ) + 3 c o s ( ( 2 n + 1 ) x ) )
?
p
0
e
c o s
2
x
·
1
4
( c o s ( 3 ( 2 n + 1 ) x ) + 3 c o s ( ( 2 n + 1 ) x ) ) d x
1
4
?
p
0
e
c o s
2
x
c o s ( 3 ( 2 n + 1 ) x ) d x +
3
4
?
p
0
e
c o s
2
x
c o s ( ( 2 n + 1 ) x ) d x
p x
k
?
p
0
c o s ( k x ) d x = 0
?
p
0
e
c o s
2
x
c o s ( 3 ( 2 n + 1 ) x ) d x = 0
?
p
0
e
c o s
2
x
c o s ( ( 2 n + 1 ) x ) d x = 0
1
4
× 0 +
3
4
× 0 = 0
0
Page 3


Question 1
All values of a for which the inequality  is satisfied,
lie in the interval
Options:
A.
(1, 2)
B.
(0, 3)
C.
(0, 4)
D.
(1, 4)
Answer: C
-------------------------------------------------------------------------------------------------
Question 2
For any integer  has the value :
Options:
A.
B.
1
1
v a
a
?
1
(
3
2
v x + 1 -
1
v x
) d x < 4
n ,
p
?
0
e
c o s
2
x
· c o s
3
( 2 n + 1 ) x d x
p
WBJEE Paper - 2024
Mathematics
C.
0
D.
Answer: C
Solution:
To solve the given integral, let's consider the integral:
First, observe the integrand carefully. It includes the term  which can be expanded using the trigonometric
identity for cosine:
Applying this identity to the integrand, we have:
Substitute this back into the integral:
This can be broken down into two separate integrals:
Now, consider the properties of definite integrals and the cosine function over the interval from 0 to . For odd multiples of , the
integral over the symmetric limits will cancel out to zero due to the periodicity and symmetry of the cosine function.
Specifically, for any odd integer :
Therefore:
and
Thus, combining these results:
Therefore, the value of the integral is:
So, the correct option is:
Option C: 0
3 p
2
?
p
0
e
c o s
2
x
· c o s
3
( ( 2 n + 1 ) x ) d x
c o s
3
( ( 2 n + 1 ) x )
c o s
3
( ? ) =
1
4
( c o s ( 3 ? ) + 3 c o s ( ? ) )
c o s
3
( ( 2 n + 1 ) x ) =
1
4
( c o s ( 3 ( 2 n + 1 ) x ) + 3 c o s ( ( 2 n + 1 ) x ) )
?
p
0
e
c o s
2
x
·
1
4
( c o s ( 3 ( 2 n + 1 ) x ) + 3 c o s ( ( 2 n + 1 ) x ) ) d x
1
4
?
p
0
e
c o s
2
x
c o s ( 3 ( 2 n + 1 ) x ) d x +
3
4
?
p
0
e
c o s
2
x
c o s ( ( 2 n + 1 ) x ) d x
p x
k
?
p
0
c o s ( k x ) d x = 0
?
p
0
e
c o s
2
x
c o s ( 3 ( 2 n + 1 ) x ) d x = 0
?
p
0
e
c o s
2
x
c o s ( ( 2 n + 1 ) x ) d x = 0
1
4
× 0 +
3
4
× 0 = 0
0
-------------------------------------------------------------------------------------------------
Question 3
Let  be a differential function with . If ,
 then
Options:
A.
B.
C.
D.
Answer: B
-------------------------------------------------------------------------------------------------
Question 4
If , then  is
Options:
A.
B.
C.
D.
f l i m
x ? 8
f ( x ) = 0 y
'
+ y f
'
( x ) - f ( x ) f
'
( x ) = 0
l i m
x ? 8
y ( x ) = 0
y + 1 = e
f ( x )
+ f ( x )
y + 1 = e
- f ( x )
+ f ( x )
y + 2 = e
- f ( x )
+ f ( x )
y - 1 = e
- f ( x )
+ f ( x )
x y
'
+ y - e
x
= 0 , y ( a ) = b l i m
x ? 1
y ( x )
e + 2 a b - e
a
e
2
+ a b - e
- a
e - a b + e
a
Page 4


Question 1
All values of a for which the inequality  is satisfied,
lie in the interval
Options:
A.
(1, 2)
B.
(0, 3)
C.
(0, 4)
D.
(1, 4)
Answer: C
-------------------------------------------------------------------------------------------------
Question 2
For any integer  has the value :
Options:
A.
B.
1
1
v a
a
?
1
(
3
2
v x + 1 -
1
v x
) d x < 4
n ,
p
?
0
e
c o s
2
x
· c o s
3
( 2 n + 1 ) x d x
p
WBJEE Paper - 2024
Mathematics
C.
0
D.
Answer: C
Solution:
To solve the given integral, let's consider the integral:
First, observe the integrand carefully. It includes the term  which can be expanded using the trigonometric
identity for cosine:
Applying this identity to the integrand, we have:
Substitute this back into the integral:
This can be broken down into two separate integrals:
Now, consider the properties of definite integrals and the cosine function over the interval from 0 to . For odd multiples of , the
integral over the symmetric limits will cancel out to zero due to the periodicity and symmetry of the cosine function.
Specifically, for any odd integer :
Therefore:
and
Thus, combining these results:
Therefore, the value of the integral is:
So, the correct option is:
Option C: 0
3 p
2
?
p
0
e
c o s
2
x
· c o s
3
( ( 2 n + 1 ) x ) d x
c o s
3
( ( 2 n + 1 ) x )
c o s
3
( ? ) =
1
4
( c o s ( 3 ? ) + 3 c o s ( ? ) )
c o s
3
( ( 2 n + 1 ) x ) =
1
4
( c o s ( 3 ( 2 n + 1 ) x ) + 3 c o s ( ( 2 n + 1 ) x ) )
?
p
0
e
c o s
2
x
·
1
4
( c o s ( 3 ( 2 n + 1 ) x ) + 3 c o s ( ( 2 n + 1 ) x ) ) d x
1
4
?
p
0
e
c o s
2
x
c o s ( 3 ( 2 n + 1 ) x ) d x +
3
4
?
p
0
e
c o s
2
x
c o s ( ( 2 n + 1 ) x ) d x
p x
k
?
p
0
c o s ( k x ) d x = 0
?
p
0
e
c o s
2
x
c o s ( 3 ( 2 n + 1 ) x ) d x = 0
?
p
0
e
c o s
2
x
c o s ( ( 2 n + 1 ) x ) d x = 0
1
4
× 0 +
3
4
× 0 = 0
0
-------------------------------------------------------------------------------------------------
Question 3
Let  be a differential function with . If ,
 then
Options:
A.
B.
C.
D.
Answer: B
-------------------------------------------------------------------------------------------------
Question 4
If , then  is
Options:
A.
B.
C.
D.
f l i m
x ? 8
f ( x ) = 0 y
'
+ y f
'
( x ) - f ( x ) f
'
( x ) = 0
l i m
x ? 8
y ( x ) = 0
y + 1 = e
f ( x )
+ f ( x )
y + 1 = e
- f ( x )
+ f ( x )
y + 2 = e
- f ( x )
+ f ( x )
y - 1 = e
- f ( x )
+ f ( x )
x y
'
+ y - e
x
= 0 , y ( a ) = b l i m
x ? 1
y ( x )
e + 2 a b - e
a
e
2
+ a b - e
- a
e - a b + e
a
Answer: D
Solution:
Let's solve the given differential equation and find the limit .
The differential equation given is:
This is a first-order linear differential equation. We can rewrite it in the form:
We will solve this using the integrating factor method. The integrating factor (IF) is given by:
Since we are not given a specific interval for x, we assume x is positive:
Multiplying both sides of the differential equation by the integrating factor:
Recognize that the left side is the derivative of :
Integrate both sides with respect to x:
So we have:
Apply the initial condition :
Thus, the solution to the differential equation is:
Now, we need to determine:
Substituting  into the solution:
Therefore:
e + a b - e
a
, ( y
'
=
d y
d x
)
l i m
x ? 1
y ( x )
x y
'
+ y - e
x
= 0
y
'
+
1
x
y =
e
x
x
µ ( x ) = e
?
1
x
d x
= e
l n | x |
= | x |
µ ( x ) = x
x y
'
+ y = e
x
? x y
'
+ y = e
x
? x ( y
'
+
1
x
y ) = e
x
x · y
'
+ y = e
x
x y
d
d x
( x y ) = e
x
x y = ? e
x
d x = e
x
+ C
y =
e
x
+ C
x
y ( a ) = b
b =
e
a
+ C
a
C = a b - e
a
y =
e
x
+ a b - e
a
x
l i m
x ? 1
y ( x )
x = 1
y ( 1 ) =
e
1
+ a b - e
a
1
= e + a b - e
a
Page 5


Question 1
All values of a for which the inequality  is satisfied,
lie in the interval
Options:
A.
(1, 2)
B.
(0, 3)
C.
(0, 4)
D.
(1, 4)
Answer: C
-------------------------------------------------------------------------------------------------
Question 2
For any integer  has the value :
Options:
A.
B.
1
1
v a
a
?
1
(
3
2
v x + 1 -
1
v x
) d x < 4
n ,
p
?
0
e
c o s
2
x
· c o s
3
( 2 n + 1 ) x d x
p
WBJEE Paper - 2024
Mathematics
C.
0
D.
Answer: C
Solution:
To solve the given integral, let's consider the integral:
First, observe the integrand carefully. It includes the term  which can be expanded using the trigonometric
identity for cosine:
Applying this identity to the integrand, we have:
Substitute this back into the integral:
This can be broken down into two separate integrals:
Now, consider the properties of definite integrals and the cosine function over the interval from 0 to . For odd multiples of , the
integral over the symmetric limits will cancel out to zero due to the periodicity and symmetry of the cosine function.
Specifically, for any odd integer :
Therefore:
and
Thus, combining these results:
Therefore, the value of the integral is:
So, the correct option is:
Option C: 0
3 p
2
?
p
0
e
c o s
2
x
· c o s
3
( ( 2 n + 1 ) x ) d x
c o s
3
( ( 2 n + 1 ) x )
c o s
3
( ? ) =
1
4
( c o s ( 3 ? ) + 3 c o s ( ? ) )
c o s
3
( ( 2 n + 1 ) x ) =
1
4
( c o s ( 3 ( 2 n + 1 ) x ) + 3 c o s ( ( 2 n + 1 ) x ) )
?
p
0
e
c o s
2
x
·
1
4
( c o s ( 3 ( 2 n + 1 ) x ) + 3 c o s ( ( 2 n + 1 ) x ) ) d x
1
4
?
p
0
e
c o s
2
x
c o s ( 3 ( 2 n + 1 ) x ) d x +
3
4
?
p
0
e
c o s
2
x
c o s ( ( 2 n + 1 ) x ) d x
p x
k
?
p
0
c o s ( k x ) d x = 0
?
p
0
e
c o s
2
x
c o s ( 3 ( 2 n + 1 ) x ) d x = 0
?
p
0
e
c o s
2
x
c o s ( ( 2 n + 1 ) x ) d x = 0
1
4
× 0 +
3
4
× 0 = 0
0
-------------------------------------------------------------------------------------------------
Question 3
Let  be a differential function with . If ,
 then
Options:
A.
B.
C.
D.
Answer: B
-------------------------------------------------------------------------------------------------
Question 4
If , then  is
Options:
A.
B.
C.
D.
f l i m
x ? 8
f ( x ) = 0 y
'
+ y f
'
( x ) - f ( x ) f
'
( x ) = 0
l i m
x ? 8
y ( x ) = 0
y + 1 = e
f ( x )
+ f ( x )
y + 1 = e
- f ( x )
+ f ( x )
y + 2 = e
- f ( x )
+ f ( x )
y - 1 = e
- f ( x )
+ f ( x )
x y
'
+ y - e
x
= 0 , y ( a ) = b l i m
x ? 1
y ( x )
e + 2 a b - e
a
e
2
+ a b - e
- a
e - a b + e
a
Answer: D
Solution:
Let's solve the given differential equation and find the limit .
The differential equation given is:
This is a first-order linear differential equation. We can rewrite it in the form:
We will solve this using the integrating factor method. The integrating factor (IF) is given by:
Since we are not given a specific interval for x, we assume x is positive:
Multiplying both sides of the differential equation by the integrating factor:
Recognize that the left side is the derivative of :
Integrate both sides with respect to x:
So we have:
Apply the initial condition :
Thus, the solution to the differential equation is:
Now, we need to determine:
Substituting  into the solution:
Therefore:
e + a b - e
a
, ( y
'
=
d y
d x
)
l i m
x ? 1
y ( x )
x y
'
+ y - e
x
= 0
y
'
+
1
x
y =
e
x
x
µ ( x ) = e
?
1
x
d x
= e
l n | x |
= | x |
µ ( x ) = x
x y
'
+ y = e
x
? x y
'
+ y = e
x
? x ( y
'
+
1
x
y ) = e
x
x · y
'
+ y = e
x
x y
d
d x
( x y ) = e
x
x y = ? e
x
d x = e
x
+ C
y =
e
x
+ C
x
y ( a ) = b
b =
e
a
+ C
a
C = a b - e
a
y =
e
x
+ a b - e
a
x
l i m
x ? 1
y ( x )
x = 1
y ( 1 ) =
e
1
+ a b - e
a
1
= e + a b - e
a
The correct answer is 
-------------------------------------------------------------------------------------------------
Question 5
The area bounded by the curves  and the Y-axis is
Options:
A.
16 sq. unit
B.
 sq. unit
C.
 sq. unit
D.
32 sq. unit
Answer: B
-------------------------------------------------------------------------------------------------
Question 6
 Then  is
Options:
A.
decreasing function
B.
increasing function
C.
neither increasing nor decreasing
D.
Option D
e + a b - e
a
x = 4 - y
2
3 2
3
1 6
3
f ( x ) = c o s x - 1 +
x
2
2 !
, x ? R f ( x )