Question

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

Answer: (D)

We have, If x < 0 |x| = -x
$\therefore f\left(x\right)=\frac{e^{-x}-e^{-x}}{e^x+e^{-x}}=0$
$\therefore f\left(x\right)=0\forall x<0$
$\therefore f\left(x\right)$ is not one-one
Next if $x\ge 0,\left|x\right|=x$
$\therefore f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
Let $y=\frac{e^x-e^{-x}}{e^x+e^{-x}}\Rightarrow y=\frac{e^{2x}-1}{e^{2x}+1}$
$\therefore e^{2x}=\frac{1+y}{1-y}$
For $x\ge 0,e^{2x}\ge 1$
$\therefore \frac{1+y}{1-y}\ge 1\Rightarrow \frac{2y}{1-y}\ge 0$
$\Rightarrow y\left(y-1\right)\le 0,y\ne 1\Rightarrow 0\le y<1$
$\therefore$ Range of f(x) = [0 1)
$\therefore$ f(x) is not onto