1) If $g(x)=\int_{\sin x}^{\sin(2x)} \sin^{-1}(t)dt$ , then A) $g'(-\frac{\pi}{2})=2\pi$ B) $g'(-\frac{\pi}{2})=-2\pi$ C) $g'(\frac{\pi}{2})=2\pi$ D) $g'(\frac{\pi}{2})=-2\pi$ Answer: Option *Explanation:$g(x)=\int_{\sin x}^{\sin(2x)} \sin^{-1}(t)dt$ $g'(x)=2\cos 2x\sin^{-1}(\sin2x)-\cos x\sin^{-1}(\sin x)$ $g'(\frac{\pi}{2})=-2\sin^{-1}(0)=0$ $g'(-\frac{\pi}{2})=-2\sin^{-1}(0)=0$ no option is matching
