Question

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

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

Answer 1

Answer: (C)

$f(x)=\frac{e^x-e^{|x|}}{e^x+e^{|x|}}=\{\begin{array}{ll}0,& x\ge 0\\ \frac{e^x-e^{|x|}}{e^x+e^{|x|}},& x<0\end{array}$
Clearly, $f(x)$ is identically xero if $x\ge 0\dots$ (1)
If $x<0$, let $y=f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$ or
$e^{2x}=\frac{1+y}{1-y}$
$\because x<0,e^{2x}<1$ or $0<e^{2x}<1$
$\therefore 0<\frac{1+y}{1-y}<1$ or $\frac{1+y}{1-y}>0$
and $\frac{1+y}{1-y}<1$
or $(y+1)(y-1)<0$
or $y>1$ or $-1<y<0$