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

Dear Student,
Please find below the solution to the asked query:

This was a bonus question in JEEAdavanced 2017 because none of theoptions were matching.gx=sinxsin2x sin-1tdtDifferentiate with respect to x using Lebnitz theorem:g'x= sin-1sin2x.ddxsin2x-sin-1sinxddxsinxg'x=2 sin-1sin2xcos2x-sin-1sinxcosxHenceg'π2=2 sin-1sinπcosπ-sin-1sinπ2cosπ2=2sin-10×-1-sin-11×0=0-0=0Similarly g'-π2 is also 0.Hence none of the options are correct.

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