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JEE Advanced 2020 Concept Recapitulation Test 2 Paper 1
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© examsnet.com
Question : 47
Total: 54
If
L
n
=
I
∕
4
∫
0
tan
n
x
d
x
, then
l
5
+
I
3
=
1
4
L
8
+
l
6
=
1
7
I
6
−
I
10
=
2
63
I
6
+
2
8
+
I
10
=
16
63
Validate
Solution:
∵
L
n
=
π
∕
4
∫
0
tan
n
−
2
x
(
sec
2
x
−
1
)
d
x
=
π
∕
4
∫
0
tan
n
−
2
x
⋅
sec
2
x
d
x
−
b
n
−
2
or
I
n
+
I
n
−
2
=
1
(
n
−
1
)
For
n
=
5
,
I
5
+
I
3
=
1
4
. . . (i)
For
n
=
8
,
I
8
+
l
s
=
1
7
. . . (ii)
For
n
=
10
,
l
10
+
L
g
=
1
9
. . . (iii)
Subtracting Eq. (iii) from (ii),
then
I
s
−
L
10
=
2
63
Adding Eqs. (ii) and (iii),
we get
I
5
+
21
8
+
L
10
=
16
63
© examsnet.com
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