Consider the function $f:\left(0,\infty \right)\to R$ defined by $f\left(x\right)=e^{-\left|{\log }_{e}x\right|}$. If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $m+n$ is

We know that, $\left|\log x\right|$ is continuous in $\left(0,\infty \right)$.

$\Rightarrow f\left(x\right)=e^{-\log \left|x\right|}$ is continuous in $\left(0,\infty \right)$

Thus, the number of points where $f\left(x\right)$ is discontinuous is $0$.

$\Rightarrow m=0$

$f\left(x\right)=\left\{\begin{array}{ll}{e}^{\log x},& 0<x<1\\ e^{-\log x},& x\ge 1\end{array}\right.$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{ll}x,& 0<x<1\\ \frac{1}{x},& x\ge 1\end{array}\right.$

$\Rightarrow f^{\text{'}}\left(x\right)=\left\{\begin{array}{ll}1,& 0<x<1\\ \frac{-1}{x^2},& x\ge 1\end{array}\right.$

$\Rightarrow f^{\text{'}}\left(1^{-}\right)=1 \& f^{\text{'}}\left(1^{+}\right)=-1$

So, the number of points where $f\left(x\right)$ is non-differentiable is $1$.

$\Rightarrow n=1$

$\Rightarrow m+n=1$