The smallest and the largest values of $\tan^{- 1} (\frac{1 - x}{1 + x}), 0 \leq x \leq 1$ are

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0, π

\frac{π}{4}, \frac{π}{2}

0, \frac{π}{4}

\frac{π}{4}, \frac{π}{4}

answer is B.

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Detailed Solution

0 \leq x \leq 1

$\tan^{- 1} (\frac{1 - x}{1 + x}) = \tan^{- 1} 1 - \tan^{- 1} x = \frac{π}{4} - \tan^{- 1} x$

$0 \leq x \leq 1 \Rightarrow 0 \leq \tan^{- 1} x \leq \frac{π}{4}$

$- \frac{π}{4} \leq - \tan^{- 1} x \leq 0$

$0 \leq \frac{π}{4} - \tan^{- 1} x \leq \frac{π}{4}$

$0 \leq \tan^{- 1} (\frac{1 - x}{1 + x}) \leq \frac{π}{4}$

Smallest and longest values of $\tan^{- 1} (\frac{1 - x}{1 + x})$ are $0, \frac{π}{4}$

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