Question

Let $f:R\to (0,1)$ be a function defined by $f\left(x\right)=\frac{e^{|x|}-e^{-x}}{e^x+e^{-x}}$ Then

• A. $f$ is both one-one and onto.
• B. $f$ is one-one but not onto.
• C. $f$is onto but not one-one.
• D. $f$ is neither one-one nor onto.

Answer 1

The correct option is A $f$ is both one-one and onto.

Let f: R $\to (0,1)$

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

Let $x>0\Rightarrow \frac{e^x-e^{-x}}{e^x+e^{-x}}\&x<0\Rightarrow \frac{e^{-x}-e^{-x}}{e^x+e^{-x}}=0$

Function are many one

To find range

Let $x\ge =0$

$\Rightarrow f\left(x\right)=\frac{e^x+e^{-x}-2e^{-x}}{e^x+e^{-x}}=1-\frac{2e^{-x}}{e^x+e^{-x}}\times \frac{e^x}{e^x}\Rightarrow f\left(x\right)=1-\frac{2}{e^{2x}+1}$

We use $x\ge 0\Rightarrow 2x\ge 0\Rightarrow e^{2x}\ge 1\Rightarrow e^{2x}+1\ge 2$

$\Rightarrow \infty <e^{2x}+1\le 2\Rightarrow \frac{1}{\infty }<\frac{1}{e^{2x}+1}\le \frac{1}{2}\Rightarrow 0<\frac{1}{e^{2x}+1}\le \frac{1}{2}$

$\Rightarrow 0<\frac{2}{e^{2x}+1}\le \frac{1}{2}\times 2$

$\Rightarrow 0>\frac{-2}{e^{2x}+1}\ge -1$

$\Rightarrow 1>1-\frac{2}{e^{2x}+1}\ge 0$

$\Rightarrow$ [0, 1) = Range = co-Domain.

$\therefore$ f is both one-one and onto

$\therefore$ option A is correct.