Question

Show that function

$f:R\to \{x\in R:-1<x<1\}$ defined by$f\left(x\right)=\frac{x}{1+|x|},x\in R$ is one-one and onto function.

Answer 1

Simplify given data.
$f:R\to \{x\in R:-1<x<1\}$

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

we know that

$|x|=\{\begin{array}{cc}x,& x\ge 0\\ -x,& x<0\end{array}$

So, $f\left(x\right)=\{\begin{array}{cc}\frac{x}{1+x}& x\ge 0\\ \frac{x}{1-x},& x<0\end{array}$

Solve for one-one
When $x\ge 0$

$f\left(x_1\right)=\frac{x_1}{1+x_1}$

$f\left(x_2\right)=\frac{x_2}{1+x_2}$

For one-one, putting $f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow \frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}$

$\Rightarrow x_1(1+x_2)=x_2(1+x_1)$

$\Rightarrow x_1+x_1{x}_2=x_2+x_2{x}_1$

$\Rightarrow x_1=x_2$

For $x<0$

$f\left(x_1\right)=\frac{x_1}{1-x_1}$

$f\left(x_2\right)=\frac{x_2}{1+x_2}$

For one-one, putting $f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow \frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}$

$\Rightarrow x_1(1-x_2)=x_2(1-x_1)$
$\Rightarrow x_1-x_1{x}_2=x_2-x_2{x}_1$

$\Rightarrow x_1=x_2$

Hence, if $f\left(x_1\right)=f\left(x_2\right),then{x}_1=x_2$

$\therefore$ f is one-one.

Solve for onto.

$f\left(x\right)=\{\begin{array}{cc}\frac{x}{1+x}& x\ge 0\\ \frac{x}{1-x},& x<0\end{array}$

For $x\ge 0$

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

$Letf\left(x\right)=y$

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

$\Rightarrow y(1+x)=x$

$\Rightarrow y+xy=x$

$\Rightarrow y=x-xy$

$\Rightarrow x-xy=y$

$\Rightarrow x(1-y)=y$

$\Rightarrow x=\frac{y}{1-y}$
For x < 0
$f\left(x\right)=\frac{x}{1-x}$
Let f(x)= y
$\Rightarrow y=\frac{x}{1-x}$
$\Rightarrow y(1-x)=x$
$\Rightarrow y-xy=x$
$\Rightarrow y=x+xy$
$\Rightarrow x+xy=y$
$\Rightarrow x(1+y)=y$
$\Rightarrow x=\frac{y}{1+y}$
Thus,
$x=\frac{y}{1-y}forx\ge 0\&x=\frac{y}{1+y},forx<0$

Here, $y\in \{y\in R:-1<y<1\}$

If $y=1,$

$x=\frac{y}{1-y}$ will be not defined,

If $y=-1$

$x=\frac{y}{1+y}$ will be not defined,

So, $y\in (-1,1)$
So here we see that codomain $=$ Range
$\therefore$ f is onto.
Hence, f is both one-one and onto
Proved.