Question

The function $f:R\to \{x\in R:-1<x<1\}$ defined by $f(x)=\frac{x}{1+|x|},x\in R$ is

• A. one-one
• B. onto
• C. one-one and onto
• D. None of these

Answer 1

Answer: (C)

It is given that $f:R\to \{x\in R:-1<x<1\}$ is defined as $f(x)=\frac{x}{1+|x|},x\in R$
Suppose $f(x)=f(y)$, where $x,y\in R$
It can be observed that, if $x$ is positive and $y$ is negative, then we have
$\frac{x}{1+x}=\frac{y}{1-y}(\text{ as }x>0\text{ and }y<0)$
$\Rightarrow 2xy=x-y$
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, the case of $x$ 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
$f(x)=f(y)$
$\Rightarrow \frac{x}{1+x}=\frac{y}{1+y}$
$\Rightarrow x+xy=y+xy$
$\Rightarrow x=y$
When $x$ and $y$ are both negative, we have
$f(x)=f(y)$
$\Rightarrow \frac{x}{1-x}=\frac{y}{1-y}$
$\Rightarrow x-xy=y-yx$
$\Rightarrow x=y$
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
$f(x)=f\left(\frac{y}{1+y}\right)=\frac{\left(\frac{y}{1+y}\right)}{1+|\frac{y}{1+y}|}=\frac{\frac{y}{1+y}}{1+\left(\frac{-y}{1+y}\right)}$
$=\frac{y}{1+y-y}=y$
If $y$ is positive, then there exists $x=\frac{y}{1-y}\in R$ such that
$f(x)=f\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+|\left(\frac{y}{1-y}\right)|}=\frac{\frac{y}{1-y}}{1+\left(\frac{y}{1-y}\right)}$
$=\frac{y}{1-y+y}=y$
Also, for $y=0$, we have $x=0\in R$
Therefore, $f$ is onto. Hence, $f$ is one-one and onto.