MCQ
If $g(x)=\int_{\sin x}^{\sin (2 x)} \sin ^{-1}(t) d t$, then
  • A
    $g^{\prime}\left(\frac{\pi}{2}\right)=-2 \pi$
  • B
    $g^{\prime}\left(-\frac{\pi}{2}\right)=2 \pi$
  • $g^{\prime}\left(\frac{\pi}{2}\right)=0 \pi$
  • D
    $g^{\prime}\left(-\frac{\pi}{2}\right)=-2 \pi$

Answer

Correct option: C.
$g^{\prime}\left(\frac{\pi}{2}\right)=0 \pi$
c
$g(x)=\int_{\sin x}^{\sin (2 x)} \sin ^{-1}(t) d t$

$g^{\prime}(x)=2(\cos 2 x) \sin ^{-1}(\sin 2 x)-(\cos x) \sin ^{-1}(\sin x)$

$g^{\prime}\left(\frac{\pi}{2}\right)=2(-1)(0)-\cos \left(\frac{\pi}{2}\right)(1)=0$

$g^{\prime}\left(-\frac{\pi}{2}\right)=0$

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