Show that function $f : R \rightarrow\{ x \in R :-1< x <1\}$ defined by $f ( x )=\frac{x}{1+|x|^{\prime}} x \in R$ is one-one and onto function.

It is given that $f : R \rightarrow\{ x \in R :-1< x <1\}$ is defined as $f ( x )=\frac{x}{1+|x|^{\prime}} x \in R$

For one-one

Suppose $f(x)=f(y),$ where $x, y \in R$

$\Rightarrow \frac{x}{1+|x|}=\frac{y}{1+|y|}$

It can be observed that if $x$ is positive and $y$ is negative,

Then, we have $\frac{x}{1+x}=\frac{y}{1+y} \Rightarrow 2 x y=x-y$

since, $x$ is positive and $y$ is negative

$x>y \Rightarrow x-y>0$

But, $2 x y$ is negative.

Then $2 x y \neq x-y$

Thus, the case of $x$ being positive and $y $ being negative can be ruled out.

Under a similar argument, $ x$ being negative and $y$ being positive can also be ruled out.

$\therefore $ $x$ and $y$ have to be either positive or negative.

When $x$ and $y$ are both positive, we have

$f(x)=f(y) \Rightarrow \frac{x}{1+x}$ $=\frac{y}{1+y} \Rightarrow x+x y=y+x y \Rightarrow x=y$

When $x$ and $y$ are both negative, we have

$f(x)=f(y) \Rightarrow \frac{x}{1-x}=\frac{y}{1-y} $ $\Rightarrow x-x y=y-x y \Rightarrow x=y$

$\therefore f$ is one $-$ one.

For onto

Now, let $y \in R$ such that $-1 < y < 1$

$(x)=f\left(\frac{y}{1+y}\right)$ $=\frac{\left(\frac{y}{1+y}\right)}{1+\left|\frac{y}{1+y}\right|}$ $=\frac{\frac{y}{1+y}}{1+\left(\frac{-y}{1+y}\right)}$ $=\frac{y}{1+y-y}=y$

If $y$ is positive, then, there exists $x=y 1-y \in R$ such that

$(x)=f\left(\frac{y}{1-y}\right)$ $=\frac{\left(\frac{y}{1-y}\right)}{1+\left|\frac{y}{1-y}\right|}$ $=\frac{\frac{y}{1-y}}{1+\left(\frac{-y}{1-y}\right)}$ $=\frac{y}{1-y+y}=y$

$\therefore f$ is onto.

Hence, $f$ is one $-$ one and onto.