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JEE Advanced 2017 Paper 2
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© examsnet.com
Question : 50
Total: 54
If
g
(
x
)
=
∫
s
i
n
x
s
i
n
(
2
x
)
s
i
n
−
1
d
t
,then :
g
′
(
π
2
)
=
−
2
π
g
′
(
−
π
2
)
=
−
2
π
g
′
(
−
π
2
)
=
2
π
g
′
(
π
2
)
=
2
π
0
Validate
Solution:
g
(
x
)
=
s
i
n
(
2
x
)
∫
s
i
n
x
s
i
n
−
1
d
t
g
′
(
x
)
=
s
i
n
−
1
s
i
n
(
2
x
)
⋅
c
o
s
2
x
⋅
2
−
s
i
n
−
1
(
s
i
n
x
)
⋅
c
o
s
x
g
′
(
π
2
)
=
s
i
n
−
1
(
s
i
n
π
)
⋅
c
o
s
π
⋅
2
−
s
i
n
−
1
(
s
i
n
π
2
)
⋅
c
o
s
π
2
=
0
g
′
(
−
π
2
)
=
0
© examsnet.com
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