The function

The given function is $f(x)=e^{|x|}$.

We know

If $f$ is continuous on its domain $D$, then $|f|$ is also continuous on $D$.

Now, the identity function $p(x)=x$ is continuous everywhere.

So, $g(x)=|p(x)|=|x|$ is also continuous everywhere.

Also, the exponential function $a^{x}, a>0$ is continuous everywhere.

So, $h(x)=e^{x}$ is continuous everywhere.

The composition of two continuous functions is continuous everywhere.

$\therefore f(x)=(h o g)(x)=e^{|x|}$ is continuous everywhere.

Now,

$g(x)=|x|= \begin{cases}x, & x \geq 0 \\ -x, & x<0\end{cases}$

$L g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{g(0-h)-g(0)}{-h}$

$\Rightarrow L g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{-(-h)-0}{-h}$

$\Rightarrow L g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{h}{-h}$

$\Rightarrow L g^{\prime}(0)=-1$

And

$R g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{h}$

$\Rightarrow R g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{h-0}{h}$

$\Rightarrow R g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{h}{h}$

$\Rightarrow R g^{\prime}(0)=1$

$\therefore L g^{\prime}(0) \neq R g^{\prime}(0)$

So, $g(x)=|x|$ is not differentiable at $x=0$.

We know

The exponential function $a^{x}, a>0$ is differentiable everywhere.

So, $h(x)=e^{x}$ is differentiable everywhere.

We know that, the composition of differentiable functions is differentiable.

Now, $e^{x}$ is differentiable everywhere, but $|x|$ is not differentiable at $x=0$.

$\therefore f(x)=(h o g)(x)=e^{|x|}$ is differentiable everywhere except at $x=0$

Thus, the function $f(x)=e^{|x|}$ is continuous every where but not differentiable at $x=0$.

Hence, the correct answer is option (a).