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Question: The smallest and the largest values of \(\tan ^ { - 1 } \left( \frac { 1 - x } { 1 + x } \right) , ...

The smallest and the largest values of

tan⁡−1(1−x1+x),0≤x≤1\tan ^ { - 1 } \left( \frac { 1 - x } { 1 + x } \right) , 0 \leq x \leq 1are.

A

0,π0 , \pi

B

0,π40 , \frac { \pi } { 4 }

C

−π4,π4- \frac { \pi } { 4 } , \frac { \pi } { 4 }

D

π4,π2\frac { \pi } { 4 } , \frac { \pi } { 2 }

Answer

0,π40 , \frac { \pi } { 4 }

Explanation

Solution

We have,

tan⁡−1(1−x1+x)=tan⁡−11−tan⁡−1x=π4−tan⁡−1x\tan ^ { - 1 } \left( \frac { 1 - x } { 1 + x } \right) = \tan ^ { - 1 } 1 - \tan ^ { - 1 } x = \frac { \pi } { 4 } - \tan ^ { - 1 } x

Since 0≤x≤1⇒0≤tan⁡−1x≤π40 \leq x \leq 1 \Rightarrow 0 \leq \tan ^ { - 1 } x \leq \frac { \pi } { 4 }

⇒π4≥tan⁡−1(1−x1+x)≥0\Rightarrow \frac { \pi } { 4 } \geq \tan ^ { - 1 } \left( \frac { 1 - x } { 1 + x } \right) \geq 0.