Inverse Trigonometric Functions Question 161
Question: The smallest and the largest values of $ {{\tan }^{-1}}( \frac{1-x}{1+x} )\text{ },0\le x\le 1 $ are
Options:
A) $ 0,\pi $
B) $ 0,\frac{\pi }{4} $
C) $ -\frac{\pi }{4},\frac{\pi }{4} $
D) $ \frac{\pi }{4},\frac{\pi }{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
We have, $ {{\tan }^{-1}}( \frac{1-x}{1+x} )={{\tan }^{-1}}1-{{\tan }^{-1}}x=\frac{\pi }{4}-{{\tan }^{-1}}x $
Since $ 0\le x\le 1\Rightarrow 0\le {{\tan }^{-1}}x\le \frac{\pi }{4} $
$ \Rightarrow 0\ge -{{\tan }^{-1}}x\ge \frac{-\pi }{4}\Rightarrow \frac{\pi }{4}\ge \frac{\pi }{4}-{{\tan }^{-1}}x\ge 0 $
$ \Rightarrow \frac{\pi }{4}\ge {{\tan }^{-1}}( \frac{1-x}{1+x} )\ge 0 $ .
BETA
