Question

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. $g'\left(\frac{\pi}{2}\right)=2 \pi$
• D. $g'\left(-\frac{\pi}{2}\right)=-2 \pi$

Answer 1

Answer: (A), (D)

Given, $g(x)=\int\limits_{\sin x}^{\sin (2 x)} \sin ^{-1}(t) d t$
Applying Newton Leibniz Formula,
$g^{\prime}(x)=\sin ^{-1}[\sin (2 x)] \times 2 \cos 2 x-\sin ^{-1}(\sin x) \times \cos x$
$\Rightarrow g^{\prime}(x)=4 x \cos 2 x-x \cos x$
So, $g^{\prime}\left(\frac{\pi}{2}\right)=-2 \pi-0=-2 \pi$
And $g ^{\prime}\left(-\frac{\pi}{2}\right)=2 \pi-0=2 \pi$