# Let f : R→R be a function defined by

Question:

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

(a) f is a bijection
(b) f is an injection only
(c) f is surjection on only
(d) f is neither an injection nor a surjection

Solution:

(d) f is neither an injection nor a surjection

$f: R \rightarrow R$

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

For $x=-2$ and $-3 \in R$

$f(-2)=\frac{e^{|-2|}-e^{2}}{e^{-2}+e^{2}}$

$=\frac{e^{2}-e^{2}}{e^{-2}+e^{2}}$

$=0$

$\& f(-3)=\frac{e^{|-3|-e^{3}}}{e^{-3}+e^{3}}$'

$=\frac{e^{3}-e^{3}}{e^{-3}+e^{3}}$

$=0$

Hence, for different values of $x$ we are getting same values of $f(x)$

That means, the given function is many one.

Therefore, this function is not injective.

For $x<0$

$f(x)=0$

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

$=\frac{e^{x}+e^{-x}}{e^{x}+e^{-x}}-\frac{2 e^{-x}}{e^{x}+e^{-x}}$

$=1-\frac{2 e^{-x}}{e^{x}+e^{-x}}$

The value of $\frac{2 e^{-x}}{e^{x}+e^{-x}}$ is always positive.

Therefore, the value of $f(x)$ is always less than 1

Numbers more than 1 are not included in the range but they are included in codomain.

As the codomain is $\mathrm{R}$.

$\therefore$ Codomain $\neq$ Range

Hence, the given function is not onto.

Therefore, this function is not surjective .