JEE MAIN - Mathematics Hindi (2019 - 12th January Evening Slot - No. 5)
$$\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }}$$ का मान है :
$$\sqrt {{2 \over \pi }} $$
$${1 \over {\sqrt {2\pi } }}$$
$$\sqrt {{\pi \over 2}} $$
$$\sqrt \pi $$
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