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JEE Main Math Class 12 Integrals Questions Part 3 Questions
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© examsnet.com
Question : 2
Total: 100
Let
f
n
=
π
2
∫
0
(
n
∑
k
=
1
s
i
n
k
−
1
x
)
(
n
∑
k
=
1
(
2
k
−
1
)
s
i
n
k
−
1
x
)
cos
x
dx
,
n
∈
N
. Then
f
21
−
f
20
is equal to _________
[13-Apr-2023 shift 2]
Your Answer:
Validate
Solution:
f
n
=
π
∕
2
∫
0
(
n
∑
k
=
1
s
i
n
k
−
1
x
)
(
n
∑
k
=
1
(
2
k
−
1
)
s
i
n
k
−
1
x
)
cos
x
dx
s
i
n
x
=
t
cos
x
dx
=
d
t
f
n
1
∫
0
(
k
−
1
∑
k
=
1
n
t
)
(
k
−
1
∑
k
=
1
n
(
2
k
−
1
)
t
)
dt
=
1
∫
0
(
1
+
t
+
t
2
.
.
.
t
n
)
(
1
+
3
t
+
5
t
2
+
.
.
.
.
+
(
2
n
+
1
)
t
n
)
dt
=
1
∫
0
(
1
+
t
+
t
2
+
.
.
.
t
n
−
1
)
(
1
+
3
t
+
5
t
2
+
.
.
.
+
(
2
n
−
1
)
t
n
−
1
)
dt
+
1
∫
0
(
1
+
3
t
+
5
t
2
+
.
.
.
(
2
n
+
1
)
)
t
n
dt
+
1
∫
0
(
1
+
t
+
t
2
+
.
.
.
+
t
n
−
1
)
(
2
n
+
1
)
t
n
dt
f
n
+
1
−
f
n
=
1
∫
0
(
1
+
3
t
+
5
t
2
+
.
.
.
+
(
2
n
+
1
)
t
n
)
t
n
dt
+
1
∫
0
(
1
+
t
+
t
2
+
.
.
.
t
n
+
1
)
(
(
2
n
+
1
)
t
n
)
dt
put
n
=
20
f
21
−
f
20
=
1
∫
0
(
1
+
3
t
+
5
t
2
.
.
.
41
⋅
t
20
)
t
20
dt
+
1
∫
0
(
1
+
t
+
t
2
.
.
.
t
19
)
(
41
⋅
t
20
)
dt
=
(
1
21
+
3
22
+
5
23
+
.
.
.
+
39
40
+
41
41
)
+
(
41
21
+
41
22
+
41
40
)
=
1
+
41
21
+
3
+
41
22
+
.
.
.
+
39
+
41
40
+
1
=
40
+
1
=
41
© examsnet.com
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