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The function $f(x)=\left\{\begin{array}{l}x, \text { if } 0 \leq x \leq 1 \\ 1, \text { if } 1 \lt x \leq 2\end{array}\right.$ is
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Continuous at all $x, \quad 0 \leq x \leq 2$ and differentiable at all $x$, except $x=1$ in the interval $[0,2]$
$\begin{aligned}& f(x)= \begin{cases}x, & 0 \leq x \leq 1 \\
1, & 1 \lt x \leq 2\end{cases} \\
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(1-h)=1 \\
& \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=1\end{aligned}$
Hence function is continuous in $(0,2)$.
$\begin{aligned}& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=0=f(0) \\
& \text { Now } \\
& \lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)=1=f(2)\end{aligned}$
Hence function is continuous in $[0,2]$
Clearly, from graph it is not differentiable at $x=1$.
1, & 1 \lt x \leq 2\end{cases} \\
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(1-h)=1 \\
& \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=1\end{aligned}$
Hence function is continuous in $(0,2)$.
$\begin{aligned}& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=0=f(0) \\
& \text { Now } \\
& \lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)=1=f(2)\end{aligned}$
Hence function is continuous in $[0,2]$
Clearly, from graph it is not differentiable at $x=1$.
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