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Let $f (x)$ be non-constant differentiable function for all real $x$ and $f (x) = f (1 - x)$ Then Rolle's theorem is not applicable for $f (x)$ on

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$[0, \frac{2}{3}]$

$[0, 1]$

$[- 2, 3]$

$[- 1, 2]$

answer is D.

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Detailed Solution

$f (x) = f (1 - x) x \in [0, \frac{2}{3}] f (0) \neq f (\frac{2}{3})$

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