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# 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.

Last updated date: 21st Jul 2024
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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.

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.