Question

If $f:R\to \left( -1,1 \right)$ be a function defined by $f\left( x \right)=\dfrac{x}{1+\left| x \right|}$, then $f$ is:
A. One-one but not onto.
B. Onto but not one-one.
C. Neither one-one nor onto.
D. Both one-one and onto.

Answer 1

For solving this question you should know about the general solution of a function and the terms one-one and onto for a function. In this problem we will check that the function is one-one or not and that is onto or not. We will check both these and come to the final decision.

Complete step by step answer:
According to our question it is asked to check if the function $f\left( x \right)=\dfrac{x}{1+\left| x \right|}$ is one-one or onto or not. So,
$f\left( x \right)=\dfrac{x}{1+\left| x \right|}$
We know that: \left| x \right|=\left\\{ \begin{aligned} & x,x\ge 0 \\\ & -x,x<0 \\\ \end{aligned} \right.
So,
f\left( x \right)=\left\\{ \begin{aligned} & \dfrac{x}{1+x},x\ge 0 \\\ & \dfrac{x}{1-x},x<0 \\\ \end{aligned} \right.
Now, checking for one-one for $x\ge 0$,
$f\left( {{x}_{1}} \right)=\dfrac{{{x}_{1}}}{1+{{x}_{1}}},f\left( {{x}_{2}} \right)=\dfrac{{{x}_{2}}}{1+{{x}_{2}}}$
Putting $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ , we get,
$\begin{aligned} & \dfrac{{{x}_{1}}}{1+{{x}_{1}}}=\dfrac{{{x}_{2}}}{1+{{x}_{2}}} \\\ & \Rightarrow {{x}_{1}}\left( 1+{{x}_{2}} \right)={{x}_{2}}\left( 1+{{x}_{1}} \right) \\\ & \Rightarrow {{x}_{1}}+{{x}_{1}}{{x}_{2}}={{x}_{2}}+{{x}_{2}}{{x}_{1}} \\\ & \Rightarrow {{x}_{1}}={{x}_{2}} \\\ \end{aligned}$
And for $x<0$, we have,
$f\left( {{x}_{1}} \right)=\dfrac{{{x}_{1}}}{1-{{x}_{1}}},f\left( {{x}_{2}} \right)=\dfrac{{{x}_{2}}}{1-{{x}_{2}}}$
Putting $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ , we get,
$\begin{aligned} & \dfrac{{{x}_{1}}}{1-{{x}_{1}}}=\dfrac{{{x}_{2}}}{1-{{x}_{2}}} \\\ & \Rightarrow {{x}_{1}}\left( 1-{{x}_{2}} \right)={{x}_{2}}\left( 1-{{x}_{1}} \right) \\\ & \Rightarrow {{x}_{1}}-{{x}_{1}}{{x}_{2}}={{x}_{2}}-{{x}_{2}}{{x}_{1}} \\\ & \Rightarrow {{x}_{1}}={{x}_{2}} \\\ \end{aligned}$
Hence if $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ , therefore $f$ is one-one.
Let us now check for onto. So,
f\left( x \right)=\left\\{ \begin{aligned} & \dfrac{x}{1+x},x\ge 0 \\\ & \dfrac{x}{1-x},x<0 \\\ \end{aligned} \right.
For $x\ge 0,f\left( x \right)=\dfrac{x}{1+x}$
Let $f\left( x \right)=y$,
$\begin{aligned} & \Rightarrow y=\dfrac{x}{1+x} \\\ & \Rightarrow y\left( 1+x \right)=x \\\ & \Rightarrow y+xy=x \\\ & \Rightarrow x=\dfrac{y}{1-x} \\\ \end{aligned}$
Let $f\left( x \right)=y$,
$\begin{aligned} & \Rightarrow y=\dfrac{x}{1-x} \\\ & \Rightarrow y\left( 1-x \right)=x \\\ & \Rightarrow y-xy=x \\\ & \Rightarrow x+xy=y \\\ & \Rightarrow x\left( 1+y \right)=y \\\ & \Rightarrow x=\dfrac{y}{1+y} \\\ \end{aligned}$
Thus, $x=\dfrac{y}{1-y}$ for $x\ge 0$ and $x=\dfrac{y}{1+y}$ for $x < 0$.
Here y\in \left\\{ x\in R:-1< x < 1 \right\\}
That is, the value of $y$ is from -1 to 1, $-1 < y < 1$.
If $y=1$, then $x=\dfrac{y}{1-y}$ will not be defined.
If $y=-1$, then $x=\dfrac{y}{1+y}$ will not be defined.
But here $-1 < y < 1$, so $x$ is defined for all values of $y$ and $x\in R$.
Therefore $f$ is onto.

So, the correct answer is “Option D”.

Note: While solving this type of questions you should be careful about checking the function $f\left( x \right)$ for one-one and for onto conditions. If both are satisfied, then both the options will be right otherwise the one that is valid will be right.