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(D) Given $f(x) = \begin{cases} x, & 0 \leq x \leq 1 \ 2-x, & 1 < x \leq 2 \end{cases}$.For Rolle's theorem to be applicable,$f(x)$ must be continuou

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$f(x)$ is not differentiable on $(0, 2)$

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(D) Given $f(x) = \begin{cases} x, & 0 \leq x \leq 1 \ 2-x, & 1 For Rolle's theorem to be applicable,$f(x)$ must be continuous on $[0, 2]$,differentiable on $(0, 2)$,and $f(0) = f(2)$.First,check continuity at $x = 1$:$LHL = \lim_{x 	o 1^-} x = 1$.$RHL = \lim_{x 	o 1^+} (2-x) = 2 - 1 = 1$.$f(1) = 1$.Since $LHL = RHL = f(1)$,$f(x)$ is continuous at $x = 1$.Next,check differentiability at $x = 1$:$f'(1^-) = \frac{d}{dx}(x) = 1$.$f'(1^+) = \frac{d}{dx}(2-x) = -1$.Since $f'(1^-) 
eq f'(1^+)$,the function $f(x)$ is not differentiable at $x = 1$.Therefore,Rolle's theorem is not applicable because $f(x)$ is not differentiable on the interval $(0, 2)$.

If $f(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ 2-x, & 1 < x \leq 2 \end{cases}$,then Rolle's theorem is not applicable to $f(x)$ because

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