#### Explain solution rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 27

(a)

Hint: understand the concept of continuity and differentiability

Given: $f(x)=e^{|x|}$

Explanation:

$f(x)= \begin{cases}e^{x} & x>0 \\ e^{-x} & x<0\end{cases}$

\begin{aligned} &L H L=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} e^{-0}=1 \\ &R H L=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} e^{+0}=1 \end{aligned}

as LHL = RHL

$\therefore f(x)$ continuous at x

\begin{aligned} &L H D=\lim _{x \rightarrow 0^{-}}=\frac{f(x)-f(0)}{x-0} \\ \end{aligned}

\begin{aligned}&=\lim _{h \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{e^{-(-h)}-1}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{e^{h}-1}{-h}=-1 \quad\left[\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\right] \end{aligned}

\begin{aligned} &R H D=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1 \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{e^{h}}{1}=1 \end{aligned}

As LHD $\neq$ RHD

$f(x)$ is not differentiable

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support