# Get Answers to all your Questions

### Answers (1)

(a)

HINTS: Understand the  definition of continuity and differentiability

GIVEN: $f(x)=e^{-\left | x \right |}$

SOLUTION:

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

At x=0

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

Hence, f(x) is continuous everywhere

LHD at x=0

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

RHD  at x=0

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

LHD $\neq$ RHD

Hence, f(x) is  differentiable at x=0

View full answer

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

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support