A) \[0,\,\,\pi \]
B) \[0,\,\frac{\pi }{4}\]
C) \[-\frac{\pi }{4},\frac{\pi }{4}\]
D) \[\frac{\pi }{4},\,\frac{\pi }{2}\]
Correct Answer: B
Solution :
We have, \[{{\tan }^{-1}}\,\left( \frac{1-x}{1+x} \right)={{\tan }^{-1}}1-{{\tan }^{-1}}x=\frac{\pi }{4}-{{\tan }^{-1}}x\] Since \[0\le x\le 1\,\,\Rightarrow \,\,0\le {{\tan }^{-1}}x\le \frac{\pi }{4}\] \[\Rightarrow \,\,0\ge -{{\tan }^{-1}}x\ge \frac{-\pi }{4}\Rightarrow \,\,\frac{\pi }{4}\ge \frac{\pi }{4}-{{\tan }^{-1}}x\ge 0\] \[\Rightarrow \,\,\frac{\pi }{4}\ge {{\tan }^{-1}}\left( \frac{1-x}{1+x} \right)\ge 0\].
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