Question

A function is defined as $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{e}^x;& x\le 0\end{array}\\ \begin{array}{cc}|x-1|;& x>0\end{array}\end{array}$ then $f\left(x\right)$ is

• A. continuous at $x=0,1$
• B. differentiable at $x=0$
• C. differentiable $x=1$
• D. none of these

Answer 1

The correct option is A continuous at $x=0,1$

$\Rightarrow \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}|x-1|=1[\because f\left(0\right)=1]$

$\Rightarrow \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}{e}^x=1$

$\Rightarrow f\left(x\right)$ is continuous at $x=0$

$\Rightarrow \underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(x-1)=0[\because f\left(1\right)=0]$

$\Rightarrow \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(1-x)=0$

$\Rightarrow f\left(x\right)$ is continuous at $x=1$

At $x=0$,

$\Rightarrow \underset{h\to 0^{+}}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{lim}\frac{1-h-1}{h}=-1$

$\Rightarrow \underset{h\to 0^{-}}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0^{-}}{lim}\frac{e^h-1}{h}=1$

Hence, it is not differentiable.