If $g(x)=\int\limits_{\sin x}^{\sin 2 x} \sin ^{-1}(t) d t$, then

(a) $g'\left(\frac{\pi}{2}\right)=-2 \pi$
(b) $g'\left(-\frac{\pi}{2}\right)=2 \pi$
(c) $g'\left(\frac{\pi}{2}\right)=0$
(d) $g'\left(-\frac{\pi}{2}\right)=0$

(c), (d)

We have, $g(x)=\int\limits_{\sin x}^{\sin 2 x} \sin ^{-1}(t) d t$

Using Leibnitz's rule, we obtain

$g'(x)=0+\frac{d}{d x}(\sin 2 x) \sin ^{-1}(\sin 2 x)-\frac{d}{d x}(\sin x) \sin ^{-1}(\sin x)$

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

We know that

$\sin ^{-1}(\sin x)=\left\{\begin{aligned}-\pi-x,-\frac{3 \pi}{2} \leq x \leq-\frac{\pi}{2} \\ x,-\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \\ \pi-x, \frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}\end{aligned}\right.$

∴   $g'\left(\frac{\pi}{2}\right) =2 \cos \pi \sin ^{-1}(\sin \pi)-\cos \frac{\pi}{2} \sin ^{-1}\left(\sin \frac{\pi}{2}\right)$

$=-2(\pi-\pi)-0 \times \frac{\pi}{2}=0$

and, 

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

$=-2 \times(-\pi+\pi)-0 \times\left(-\frac{\pi}{2}\right)=0$