JEE Main 12 Jan 2019 Evening Question 10
Question: $ \underset{x\to {1^{-}}}{\mathop{\lim }},\frac{\sqrt{\pi }-\sqrt{2{{\sin }^{-1}}x}}{\sqrt{1-x}} $ is equal to
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ \sqrt{\pi } $
B) $ \frac{1}{\sqrt{2\pi }} $
C) $ \sqrt{\frac{\pi }{2}} $
D) $ \sqrt{\frac{2}{\pi }} $
Show Answer
Answer:
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( \frac{\pi }{2}-{{\sin }^{-1}}x )}{\sqrt{1-x}( \sqrt{\pi }+\sqrt{2{{\sin }^{-1}}x} )} $ $ =\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 ( \frac{\theta }{2} )}.\frac{1}{2\sqrt{\pi }} $ [Putting $ x=\cos \theta $ ] $ =\frac{4}{2\sqrt{2\pi }}=\sqrt{\frac{2}{\pi }} $
