Question

Let $f$ be a real valued function defined by $f\left(x\right)=\frac{e^x-e^{-\left|x\right|}}{e^x+e^{\left|x\right|}}$, then range of $f\left(x\right)$ is:

• A. $R$
• B. $[0,1]$
• C. $(0,1)$
• D. $[0,\frac{1}{2}]$

Answer 1

The correct option is D $[0,\frac{1}{2}]$

$f\left(x\right)=\frac{e^x-e^{-\left|x\right|}}{e^x+e^{\left|x\right|}}$
$forx\ge 0$
$f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^x}=\frac{e^x-e^{-x}}{2e^x}=\frac{1}{2}-\frac{1}{2e^{2x}}$
$whenx=0$
$f\left(x\right)=\frac{1}{2}-\frac{1}{2}=0$
$whenx=\propto$
$f\left(x\right)=\frac{1}{2}-0=\frac{1}{2}$
$Hencef\left(x\right)\in [0,\frac{1}{2})$
$whenx<0$
$f\left(x\right)=\frac{e^x-e^{-(-x)}}{e^x+e^{-x}}=\frac{e^x-e^x}{e^x+e^{-x}}=0$
$HenceRange:[0,\frac{1}{2})$