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If $f$ is defined by $f(x)=\left\{\begin{array}{l}x \text { for } 0 \leq x < 1 \\ 2-x \text { for } x \geq 1\end{array}\right.$, then at $x=1, f(x)$ is
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The correct answer is:
continuous but not differentiable
Give function,
$\begin{aligned}
& f(x)=\left\{\begin{array}{lll}
x & \text { for } & 0 \leq x < 1 \\
2-x & \text { for } & x \geq 1
\end{array}\right. \\
& \text { At } x=1 \\
& \Rightarrow f(x)=2-x \\
& \Rightarrow f(1)=2-1=1 \\
& \text { LHL }=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} x=1 \\
& \mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(2-x)=2-1=1 \\
& \mathrm{LHL}=\mathrm{RHL}=f(1) \\
& \therefore \quad f(x) \text { is continuous at } x=1 \\
& \Rightarrow \text { To check the differentiability of } f(x) \text {, } \\
& f^{\prime}(x)=\left\{\begin{array}{lll}
1 & \text { for } & 0 \leq x < 1 \\
-1 & \text { for } & x \geq 1
\end{array}\right. \\
& \text { LHD }=\lim _{x \rightarrow 1^{-}} f^{\prime}(x)=\lim _{x \rightarrow 1^{-}} 1=1 \\
& \text { RHD }=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(-1)=-1 \\
& \mathrm{LHD} \neq \mathrm{RHD} \\
&
\end{aligned}$
$\therefore f(x)$ is not differentiable at $x=1$
$\begin{aligned}
& f(x)=\left\{\begin{array}{lll}
x & \text { for } & 0 \leq x < 1 \\
2-x & \text { for } & x \geq 1
\end{array}\right. \\
& \text { At } x=1 \\
& \Rightarrow f(x)=2-x \\
& \Rightarrow f(1)=2-1=1 \\
& \text { LHL }=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} x=1 \\
& \mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(2-x)=2-1=1 \\
& \mathrm{LHL}=\mathrm{RHL}=f(1) \\
& \therefore \quad f(x) \text { is continuous at } x=1 \\
& \Rightarrow \text { To check the differentiability of } f(x) \text {, } \\
& f^{\prime}(x)=\left\{\begin{array}{lll}
1 & \text { for } & 0 \leq x < 1 \\
-1 & \text { for } & x \geq 1
\end{array}\right. \\
& \text { LHD }=\lim _{x \rightarrow 1^{-}} f^{\prime}(x)=\lim _{x \rightarrow 1^{-}} 1=1 \\
& \text { RHD }=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(-1)=-1 \\
& \mathrm{LHD} \neq \mathrm{RHD} \\
&
\end{aligned}$
$\therefore f(x)$ is not differentiable at $x=1$
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