Consider the function f:(0,) R defined by f(x)=e^-| _ex|.

Question

Consider the function $f:(0,\infty)	o R$ defined by $f(x)={e}^{-|{\log }_{e}x|}$. 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

Accepted Answer

$$\begin{aligned}
& f:(0, \infty) \rightarrow R \
& f(x)=e^{-\left|\log _e x\right|}
\end{aligned}$$

$$\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{e}^{|\ln \mathrm{x}|}}=\left\{\[\begin{gathered} \frac{1}{\mathrm{e}^{-\ln \mathrm{x}}} ; 0<\mathrm{x}<1 \ \frac{1}{\mathrm{e}^{\ln \mathrm{x}}} ; \mathrm{x} \geq 1 \end{gathered}$$\right.\]

$$\left\{\[\begin{gathered} \frac{1}{\frac{1}{x}}=x ; 0< x<1 \ \frac{1}{x}, x \geq 1 \end{gathered}$$\right.\]

$$\mathrm{m}=0$$ (No point at which function is not continuous)

$$\mathrm{n}=1$$ (Not differentiable)

$$\therefore \mathrm{m}+\mathrm{n}=1$$

Solutions · 1