Question

$f:R\to R$ is defined by $f\left(x\right)=\frac{e^{x^2}-e^{{-x}^2}}{e^{x^2}+e^{{-x}^2}}$ is

• A. one-one but onto
• B. many-one but not onto
• C. one-one and into
• D. Many-one and onto

Answer 1

The correct option is B many-one but not onto

Given, $f\left(x\right)=\frac{e^{x^2}-e^{-x^2}}{e^{x^2}+e^{-x^2}}$
$f\left(x\right)=\frac{e^{2x^2}-1}{e^{2x^2}+1}$
$f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow x_1=\pm x_2$
i.e $f\left(x\right)=f(-x)$
For example $f(-1)=f\left(1\right)$
$=f\left(x\right)=\frac{e^2-1}{e^2+1}$, many one function.

And $f\left(x\right)>0\forall x\in R$

Therefore, $f\left(x\right)$ is not onto.