JEE Main 201912 Jan, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002limx→1−π−2sin−1x1−x\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }}x→1−lim1−xπ−2sin−1x is equal to :AOption A. 2π\sqrt {{2 \over \pi }}π2BOption B. 12π{1 \over {\sqrt {2\pi } }}2π1COption C. π2\sqrt {{\pi \over 2}}2πDOption D. π\sqrt \piπCheckSkip · show solution
