A) \[0,\pi \]
B) \[1\]
C) \[5\]
D) \[10\]
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, \[0\le x\le 1\] Then, \[0\le {{\tan }^{-1}}x\le \frac{\pi }{4}\] \[\Rightarrow \] \[0\ge -{{\tan }^{-1}}x\ge -\frac{\pi }{4}\] \[\Rightarrow \] \[\frac{\pi }{4}\ge \frac{\pi }{4}-{{\tan }^{-1}}x\ge 0\] \[\Rightarrow \] \[\frac{\pi }{4}\ge {{\tan }^{-1}}\left\{ \frac{(1-x)}{(1+x)} \right\}\ge 0\]
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