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