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If $f(x)=\left\{\begin{array}{cc}x, & 0 \leq x \leq 1 \\ 2-x, & 1 \leq x \leq 2\end{array}, \quad\right.$ then Rolle's theorem is not applicable to $f(x)$ because
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Verified Answer
The correct answer is:
$f(x)$ is not differentiable on $(0,2)$
We have,
$$
\begin{aligned}
& f(x)=\left\{\begin{array}{cc}
x, & 0 \leq x \leq 1 \\
2-x, & 1 \leq x \leq 2
\end{array}\right. \\
& \text { LHL }=\lim _{x \rightarrow 1^{-}} x \\
& =\lim _{h \rightarrow 0}(1-h)=1 \\
& \text { RHL }=\lim _{x \rightarrow 1^{+}}(2-x)=\lim _{h \rightarrow 0^{+}}\{2-(1+h)\}=1 \\
& \text { and } \quad f(1)=2-1=1 \\
& \therefore \quad \mathrm{LHL}=\mathrm{RHL}=f(1) \\
&
\end{aligned}
$$
Hence, $f(x)$ is continuous.
Also, $f^{\prime}\left(1^{-}\right)=1$ and $f^{\prime}\left(1^{+}\right)=-1$
$$
\therefore \quad f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right)
$$
Hence, $f(x)$ is not differentiable at $x=1$.
So, Rolle's theorem is not applicable as $f(x)$ is not differentiable at $x=1$.
$$
\begin{aligned}
& f(x)=\left\{\begin{array}{cc}
x, & 0 \leq x \leq 1 \\
2-x, & 1 \leq x \leq 2
\end{array}\right. \\
& \text { LHL }=\lim _{x \rightarrow 1^{-}} x \\
& =\lim _{h \rightarrow 0}(1-h)=1 \\
& \text { RHL }=\lim _{x \rightarrow 1^{+}}(2-x)=\lim _{h \rightarrow 0^{+}}\{2-(1+h)\}=1 \\
& \text { and } \quad f(1)=2-1=1 \\
& \therefore \quad \mathrm{LHL}=\mathrm{RHL}=f(1) \\
&
\end{aligned}
$$
Hence, $f(x)$ is continuous.
Also, $f^{\prime}\left(1^{-}\right)=1$ and $f^{\prime}\left(1^{+}\right)=-1$
$$
\therefore \quad f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right)
$$
Hence, $f(x)$ is not differentiable at $x=1$.
So, Rolle's theorem is not applicable as $f(x)$ is not differentiable at $x=1$.
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