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(B) Given function,$f(x) = \begin{cases} x & 	ext{for } 0 \leq x < 1 \ 2-x & 	ext{for } x \geq 1 \end{cases}$.At $x=1$,$f(1) = 2-1 = 1$.Left Hand L

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continuous but not differentiable

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(B) Given function,$f(x) = \begin{cases} x & 	ext{for } 0 \leq x At $x=1$,$f(1) = 2-1 = 1$.Left Hand Limit $(LHL)$ = $\lim_{x ightarrow 1^-} f(x) = \lim_{x ightarrow 1^-} x = 1$.Right Hand Limit $(RHL)$ = $\lim_{x ightarrow 1^+} f(x) = \lim_{x ightarrow 1^+} (2-x) = 2-1 = 1$.Since $LHL$ = $RHL$ = $f(1)$,the function $f(x)$ is continuous at $x=1$.Now,to check differentiability,we find the derivative $f'(x)$:$f'(x) = \begin{cases} 1 & 	ext{for } 0 \leq x  1 \end{cases}$.Left Hand Derivative $(LHD)$ = $\lim_{x ightarrow 1^-} f'(x) = 1$.Right Hand Derivative $(RHD)$ = $\lim_{x ightarrow 1^+} f'(x) = -1$.Since $LHD$ $
eq$ $RHD$,the function is not differentiable at $x=1$.Therefore,$f(x)$ is continuous but not differentiable at $x=1$.

If $f$ is defined by $f(x) = \begin{cases} x & \text{for } 0 \leq x < 1 \\ 2-x & \text{for } x \geq 1 \end{cases}$,then at $x=1$,$f(x)$ is

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