Question

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

Answer 1

It is given that $f:R\to \{x\in R:-1<x<1\}$ is defined as $f\left(x\right)=\frac{x}{x+|x{|}^{\prime }}x\in R.$
Suppose, $f\left(x\right)=f\left(y\right),wherex,y\in R\Rightarrow \frac{x}{1+|x|}=\frac{y}{1+|y|}$
It can be observed that if x is positive and y is negative, then we have
$\frac{x}{1+x}=\frac{y}{1-y}\Rightarrow 2xy=x-y$
It can be observed that if x is positive and y is negative, then we have
$\frac{x}{1+x}=\frac{y}{1-y}\Rightarrow 2xy=x-y$
Since, x is positive and y is negative, then
$x>y\Rightarrow x-y>0$
But, 2xy is negative. Then, $2xy\ne x-y.$
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
$f\left(x\right)=f\left(y\right)\Rightarrow \frac{x}{1+x}=\frac{y}{1+y}\Rightarrow x+xy=y+xy\Rightarrow x=y$
When x and y are both negative, we have
$f\left(x\right)=f\left(y\right)\Rightarrow \frac{x}{1-x}=\frac{y}{1-y}\Rightarrow x-xy=y-yx\Rightarrow x=y$
Therefore, f is one-one. Now, let $y\in R$ such that -1 < y < 1.
If y is negative, then there exists $x=\frac{y}{1+y}\in R$ such that
$f\left(x\right)=f\left(\frac{y}{1+y}\right)=\frac{\left(\frac{y}{1+y}\right)}{1+|\frac{y}{1+y}|}=\frac{\frac{y}{1+y}}{1+\left(\frac{-y}{1+y}\right)}=\frac{y}{1+y-y}=y$
If y is positive, then there exists $x=\frac{y}{1-y}\in R$ such that

$f\left(x\right)=f\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+|\left(\frac{y}{1-y}\right)|}=\frac{\frac{y}{1-y}}{1+\frac{y}{1-y}}=\frac{y}{1+\frac{y}{1-y}}=y$
Therefore, f is onto. Hence f is one-one and onto.