If f(x)=x2−1, then on the interval [0,π]

tan ⁡ f ( x ) and 1 f ( x ) are discontinuous

tan ⁡ f ( x ) and 1 f ( x ) are discontinuous

tan ⁡ f ( x ) and 1 f - 1 ( x ) are continuous

tan ⁡ f ( x ) is not continuous but 1 f ( x ) is continuous

tan ⁡ f ( x ) and 1 f ( x ) are discontinuous

tan ⁡ f ( x ) and 1 f ( x ) are continuous

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If $f (x) = \frac{x}{2} - 1,$ then on the interval $[0, π]$

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$\tan f (x) and \frac{1}{f^{- 1} (x)}$ are continuous

$\tan f (x)$ is not continuous but $\frac{1}{f (x)}$ is continuous

$\tan f (x) and \frac{1}{f (x)}$ are discontinuous

$\tan f (x) and \frac{1}{f (x)}$ are continuous

answer is C.

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$Range of f (x) is [- 1, \frac{π}{2} - 1]$

$\Rightarrow$ Hence tan f(x) is continuous

$\Rightarrow \frac{1}{f (x)} = \frac{2}{x - 2}, not continuous at x = 2$

$f^{- 1} (x) = 2 (x + 1), \frac{1}{f^{- 1} (x)} = \frac{1}{2 (x + 1)} is continuous$

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