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**Hint:**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{align}

& x,x\ge 0 \\

& -x,x<0 \\

\end{align} \right.$

So,

$f\left( x \right)=\left\{ \begin{align}

& \dfrac{x}{1+x},x\ge 0 \\

& \dfrac{x}{1-x},x<0 \\

\end{align} \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{align}

& \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{align}$

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{align}

& \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{align}$

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{align}

& \dfrac{x}{1+x},x\ge 0 \\

& \dfrac{x}{1-x},x<0 \\

\end{align} \right.$

For $x\ge 0,f\left( x \right)=\dfrac{x}{1+x}$

Let $f\left( x \right)=y$,

$\begin{align}

& \Rightarrow y=\dfrac{x}{1+x} \\

& \Rightarrow y\left( 1+x \right)=x \\

& \Rightarrow y+xy=x \\

& \Rightarrow x=\dfrac{y}{1-x} \\

\end{align}$

Let $f\left( x \right)=y$,

$\begin{align}

& \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{align}$

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.

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