Consider the function f: (0, ∞) → R defined by f(x) = e^−|logex|.

Question

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

(1) 0

(2) 3

(3) 1

(4) 2

Answer 1

Correct option is (3) 1

\(f:(0, \infty) \rightarrow R\)

\(f(x)=e^{-\left|\log _{e} x\right|}\)

\(f(x)=\frac{1}{e^{|\ln x|}}=\left\{\begin{array}{l}\frac{1}{e^{-\ln x}} ; 0<x<1 \\ \frac{1}{e^{\ln x}} ; x \geq 1\end{array}\right.\)

\(=\left\{\begin{array}{l}\frac{1}{\frac 1x}=x ; 0<x<1 \\ \frac{1}{x}, x \geq 1\end{array}\right.\)

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

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

\(\therefore \mathrm{m}+\mathrm{n}=1\)