JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Continuity and Differentiability

$f(x)=e^{-|{\log }_{e}x|}$ on $(0,\infty )$.

For $0<x<1$: ${\log }_{e}x<0$, so $|{\log }_{e}x|=-{\log }_{e}x$, giving $f(x)=e^{{\log }_{e}x}=x$.

For $x\ge 1$: ${\log }_{e}x\ge 0$, so $f(x)=e^{-{\log }_{e}x}=\frac{1}{x}$.

Both pieces agree at $x=1$ (both give $1$), and each piece is continuous on its own domain, so $f$ is continuous everywhere on $(0,\infty )$: $m=0$.

Checking differentiability at $x=1$: the left-hand derivative (from $f(x)=x$) is $1$; the right-hand derivative (from $f(x)=1/x$) is $-1/x^2$ at $x=1$, i.e. $-1$. These are unequal, so $f$ is not differentiable at $x=1$; this is the only such point, so $n=1$.

$m+n=0+1=1$.