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 $f$ is

• A. a one one but not an onto function
• B. an onto but not a one-one function
• C. a bijective function
• D. neither an one-one nor an onto function

Answer 1

The correct option is D neither an one-one nor an onto function

$f\left(x\right)=\frac{e^{|x|}-e^{-x}}{e^x+e^{-x}}$

If $x\le 0,f\left(x\right)=\frac{e^{-x}-e^{-x}}{e^x+e^{-x}}=0$
$\Rightarrow f(-2)=f(-3)=0$
$\therefore f$ is not a one-one function.

If $x>0,f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
$\Rightarrow f\left(x\right)=\frac{e^{2x}-1}{e^{2x}+1}$
We know,
$e^{2x}-1<e^{2x}+1\forall x\in R$
$\Rightarrow \frac{e^{2x}-1}{e^{2x}+1}<1$
$\Rightarrow f\left(x\right)<1\forall x\in R$

So, pre image of $1$ does not exist.
$\therefore f$ is also not an onto function.