Question

$f:R\to R$ is defined by $f\left(x\right)=\frac{e^{x^2}-e^{-x^2}}{e^{x^2+e^{-x^2}}}\mathrm{is}$
(a) one-one but not onto
(b) many-one but onto
(c) one-one and onto
(d) neither one-one nor onto

Answer 1

(d) neither one-one nor onto

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=\frac{e^{x^2}-e^{-x^2}}{e^{x^2+e^{-x^2}}} \\ \mathrm{Here},-2,2\in R \\ \mathrm{Now},2\ne -2 \\ \mathrm{But},f\left(2\right)=f\left(-2\right) \\ \mathrm{Therefore},\mathrm{function}\mathrm{is}\mathrm{not}\mathrm{one}-\mathrm{one}. \\ \mathrm{And}, \\ \mathrm{The}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{is}0\mathrm{and}\mathrm{maximum}\mathrm{value}\mathrm{is}1 \\ \mathrm{That}\mathrm{is}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{is}\left[0,1\right]\mathrm{but}\mathrm{the}\mathrm{co}-\mathrm{domain}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{is}\mathrm{given}\mathrm{R}. \\ \mathrm{Therefore},\mathrm{function}\mathrm{is}\mathrm{not}\mathrm{onto}. \\ \therefore \mathrm{function}\mathrm{is}\mathrm{neither}\mathrm{one}-\mathrm{one}\mathrm{nor}\mathrm{onto}. \end{gathered}$$