JEE Advanced 2017 Paper 2

Section: Mathematics

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Question : 50 of 54

Marks: +1, -0

If g(x)=∫sinxsin(2x)sin−1dt,then :

[JEE Adv 2017 P2]

g′(
π
2
)=−2π
g′(−
π
2
)=−2π
g′(−
π
2
)=2π
g′(
π
2
)=2π
0

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