Question

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

Answer 1

(d) f is neither an injection nor a surjection

$f:R\to R$

$$\begin{gathered} f\left(x\right)=\frac{e^{\left|x\right|}-e^{-x}}{e^x+e^{-x}} \\ \mathrm{For}x=-2\mathrm{and}-3\in R \\ f(-2)=\frac{e^{\left|-2\right|}-e^2}{e^{-2}+e^2} \\ =\frac{e^2-e^2}{e^{-2}+e^2} \\ =0 \\ \&f(-3)=\frac{e^{\left|-3\right|}-e^3}{e^{-3}+e^3} \\ =\frac{e^3-e^3}{e^{-3}+e^3} \\ =0 \\ \mathrm{Hence},\mathrm{for}\mathrm{different}\mathrm{values}\mathrm{of}x\mathrm{we}\mathrm{are}\mathrm{getting}\mathrm{same}\mathrm{values}\mathrm{of}f\left(x\right) \\ \mathrm{That}\mathrm{means},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{many}\mathrm{one}. \end{gathered}$$
Therefore, this function is not injective.

$$\begin{gathered} \mathrm{For}x<0 \\ f\left(x\right)=0 \\ \mathrm{For}x>0 \\ f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}} \\ =\frac{e^x+e^{-x}}{e^x+e^{-x}}-\frac{2e^{-x}}{e^x+e^{-x}} \\ =1-\frac{2e^{-x}}{e^x+e^{-x}} \\ \mathrm{The}\mathrm{value}\mathrm{of}\frac{2e^{-x}}{e^x+e^{-x}}\mathrm{is}\mathrm{always}\mathrm{positive}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}f\left(x\right)\mathrm{is}\mathrm{always}\mathrm{less}\mathrm{than}1 \\ \mathrm{Numbers}\mathrm{more}\mathrm{than}1\mathrm{are}\mathrm{not}\mathrm{included}\mathrm{in}\mathrm{the}\mathrm{range}\mathrm{but}\mathrm{they}\mathrm{are}\mathrm{included}\mathrm{in}\mathrm{codomain}. \\ \mathrm{As}\mathrm{the}\mathrm{codomain}\mathrm{is}\mathrm{R}. \\ \therefore \mathrm{Codomain}\ne \mathrm{Range} \\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{not}\mathrm{onto}. \end{gathered}$$
Therefore, this function is not surjective .