Jee advanced doubt... Please answer it as soon as possible

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{This} \mathrm{was} \mathrm{a} \mathrm{bonus} \mathrm{question} \mathrm{in} \mathrm{JEE}\hspace{0.17em}\mathrm{Adavanced} 2017 \mathrm{because} \mathrm{none} \mathrm{of} \mathrm{the} \\ \mathrm{options} \mathrm{were} \mathrm{matching}. \\ \mathrm{g}\left(\mathrm{x}\right)={\int }_{\mathrm{sinx}}^{\sin \left(2\mathrm{x}\right)} {\sin }^{-1}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \mathrm{using} \mathrm{Lebnitz} \mathrm{theorem}: \\ \mathrm{g}\text{'}\left(\mathrm{x}\right)= {\sin }^{-1}\left(\sin 2\mathrm{x}\right).\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin 2\mathrm{x}\right)-{\sin }^{-1}\left(\mathrm{sinx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sinx}\right) \\ \mathrm{g}\text{'}\left(\mathrm{x}\right)=2 {\sin }^{-1}\left(\sin 2\mathrm{x}\right)\cos 2\mathrm{x}-{\sin }^{-1}\left(\mathrm{sinx}\right)\mathrm{cosx} \\ \mathrm{Hence} \\ \mathrm{g}\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=2 {\sin }^{-1}\left(\mathrm{sin\pi }\right)\mathrm{cos\pi }-{\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{2}\right)\cos \frac{\mathrm{\pi }}{2} \\ =2{\sin }^{-1}\left(0\right)\times \left(-1\right)-{\sin }^{-1}\left(1\right)\times 0 \\ =0-0 \\ =0 \\ \mathrm{Similarly} \\ \mathrm{g}\text{'}\left(-\frac{\mathrm{\pi }}{2}\right) \mathrm{is} \mathrm{also} 0. \\ \mathrm{Hence} \mathrm{none} \mathrm{of} \mathrm{the} \mathrm{options} \mathrm{are} \mathrm{correct}. \end{gathered}$$

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