Question

Let $f:R\to R$ be a function defined by $f\left(x\right)=\frac{e^{|x|}-e^{-x}}{e^x+e^{-x}}.$ Then

• A. $f$ is both one-one and onto
• B. $f$ is one-one but not onto
• C. $f$ is onto but not one-one
• D. $f$ is neither one-one nor onto

Answer 1

The correct option is D $f$ is neither one-one nor onto

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

For $x<0,f\left(x\right)=0$
$\Rightarrow f$ is not a one-one function.

For $x\ge 0,$
$f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$ $=\frac{e^{2x}-1}{e^{2x}+1}$
$f^{\prime }\left(x\right)=\frac{(e^{2x}+1)e^{2x}\cdot 2-e^{2x}\cdot 2\cdot (e^{2x}-1)}{(e^{2x}+1{)}^2}$
$=\frac{4e^{2x}}{(e^{2x}+1{)}^2}>0\forall x\ge 0$
$\Rightarrow f\left(x\right)$ is a strictly increasing function.

Also, $f\left(0\right)=0$ and $l=\underset{x\to \infty }{lim}\frac{e^{2x}-1}{e^{2x}+1}=\underset{x\to \infty }{lim}\frac{1-\frac{1}{e^{2x}}}{1+\frac{1}{e^{2x}}}=1$
$\therefore f\left(x\right)\in [0,l)=[0,1)$
$\Rightarrow f$ is not an onto function.
$\therefore f$ is neither one-one nor onto.