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 a bijection
• B. $f$ is an injection only
• C. $f$ is a surjection
• D. $f$ is neither an injection nor a surjection

Answer 1

The correct option is D $f$ is neither an injection nor a surjection

Given equation is $f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
$\Rightarrow f\left(x\right)=\frac{e^{2x}-1}{e^{2x}+1}$
Let's check if $f\left(x\right)$ is injective .
Let $f\left(x\right)=f\left(y\right)$
$\Rightarrow \frac{e^{2x}-1}{e^{2x}+1}=\frac{e^{2y}-1}{e^{2y}+1}$
$\Rightarrow 2e^{2x}=2e^{2y}$
$\Rightarrow e^{2(x-y)}=0$
which does not implies $x=y$
Hence, $f\left(x\right)$ is not injective.
Now, we will check if $f\left(x\right)$ is onto.
Let $y=\frac{e^{2x}-1}{e^{2x}+1}$
$\Rightarrow e^{2x}=\frac{2}{1-y}$
$\Rightarrow x=\frac{1}{2}\log \left(\frac{2}{1-y}\right)$
$\Rightarrow \frac{2}{1-y}>0$
$\Rightarrow y<1$
Clearly , $y\ge 0$
Hence, range of f is $[0,1)$
Range of $f$ $\ne$ Co-domain $R$.
Hence, $f$ is not surjective.