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Q.
The function $f: R \rightarrow\{x \in R:-1 < x < 1\}$ defined by $f(x)=\frac{x}{1+|x|}, x \in R$ is
Relations and Functions - Part 2
Solution:
It is given that $f: R \rightarrow\{x \in R:-1 < x < 1\}$ 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
$\frac{x}{1+x} =\frac{y}{1-y} (\text { as } x>0 \text { and } y < 0)$
$\Rightarrow 2 x y =x-y$
Since $x$ is positive and $y$ is negative, then
$x>y \Rightarrow x-y>0$
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, 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+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} $
$\Rightarrow x-x y =y-y x$
$\Rightarrow x =y$
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+\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 $x=\frac{y}{1-y} \in R$ such that
$f(x) =f\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+\left|\left(\frac{y}{1-y}\right)\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.