A) \[\sqrt{\pi }\]
B) \[\frac{1}{\sqrt{2\pi }}\]
C) \[\sqrt{\frac{\pi }{2}}\]
D) \[\sqrt{\frac{2}{\pi }}\]
Correct Answer: D
Solution :
We have, \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{\sqrt{\pi }-\sqrt{2{{\sin }^{-1}}x}}{\sqrt{1-x}}\] \[=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{\sqrt{\pi }-\sqrt{2{{\sin }^{-1}}x}}{\sqrt{1-x}}\times \frac{\sqrt{\pi }+\sqrt{2{{\sin }^{-1}}x}}{\sqrt{\pi }+\sqrt{2{{\sin }^{-1}}x}}\] \[=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{2\left( \frac{\pi }{2}-{{\sin }^{-1}}x \right)}{\sqrt{1-x}\left( \sqrt{\pi }+\sqrt{2{{\sin }^{-1}}x} \right)}\] \[=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{2{{\cos }^{-1}}x}{\sqrt{1-x}}.\frac{1}{2\sqrt{\pi }}\] \[=\underset{\theta \to {{0}^{+}}}{\mathop{\lim }}\,\frac{2\theta }{\sqrt{2}\sin \left( \frac{\theta }{2} \right)}.\frac{1}{2\sqrt{\pi }}\] [Putting\[x=\cos \theta \]] \[=\frac{4}{2\sqrt{2\pi }}=\sqrt{\frac{2}{\pi }}\]
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