The smallest and the largest values of tan-11-x1+x,0≤x≤1 are

=tan-11-x1+1·x =tan-1tanπ4-x1+tanπ4·x =tan-1tanπ4-tan-1x given 0≤x≤1 x=1 ; π4-π4=0 x=0 ; π4-0=π4

The smallest and the largest values of tan-11-x1+x,0≤x≤1 are

=tan-11-x1+1·x =tan-1tanπ4-x1+tanπ4·x =tan-1tanπ4-tan-1x given 0≤x≤1 x=1 ; π4-π4=0 x=0 ; π4-0=π4

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

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$\frac{π}{4}, \frac{π}{2}$

$0, π$

$- \frac{π}{4}, \frac{π}{4}$

$0, \frac{π}{4}$

answer is B.

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Detailed Solution

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