Question

The range of the function $f\left(x\right)=\frac{e^x-e^{|x|}}{e^x+e^{|x|}}$ is

• A. $(-\infty ,\infty )$
• B. $[0,1)$
• C. $(-1,0]$
• D. $(-1,1)$

Answer 1

The correct option is C $(-1,0]$

$f\left(x\right)=\frac{e^x-e^{|x|}}{e^x+e^{|x|}}$
$e^x+e^{|x|}\ne 0$ for any real value of $x$
Hence, domain of $f$ is $R$
$|x|=\{\begin{array}{cc}x,& x\ge 0\\ -x,& x<0\end{array}$

$\therefore f\left(x\right)=\{\begin{array}{cc}0,& x\ge 0\\ \frac{e^x-e^{-x}}{e^x+e^{-x}},& x<0\end{array}$

Let $y=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
$=\frac{e^{2x}-1}{e^{2x}+1}$
$=1-\frac{2}{e^{2x}+1}$

Now, $-\infty <x<0$
$\Rightarrow -\infty <2x<0$
$\Rightarrow 0<e^{2x}<1$
$\Rightarrow 1<e^{2x}+1<2$
$\Rightarrow \frac{1}{2}<\frac{1}{e^{2x}+1}<1$
$\Rightarrow 1<\frac{2}{e^{2x}+1}<2$
$\Rightarrow -2<-\frac{2}{e^{2x}+1}<-1$
​​​​​​​​​​​​​​$\Rightarrow -1<1-\frac{2}{e^{2x}+1}<0$

For $x\ge 0,$ $f\left(x\right)=0$
For $x<0,$ $y\in (-1,0)$
$\therefore$ Range of $f$ is $(-1,0]$