Question

For a function, $y=f\left(x\right)=\frac{x}{1+|x|},x,y\in R$, which among the following is true :

• A. $f\left(x\right)$ is a one-one and an onto function
• B. $f\left(x\right)$ is an onto but not a one -one function
• C. $f\left(x\right)$ is a one-one but not an onto function
• D. $f\left(x\right)$ is neither a one-one nor an onto function

Answer 1

The correct option is C $f\left(x\right)$ is a one-one but not an onto function

$f\left(x\right)=\{\begin{array}{c}\frac{x}{1+x},x\ge 0\\ \frac{x}{1-x},x<0\end{array}$

For $x\ge 0$
$f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow \frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}$
$\Rightarrow x_1(1+x_2)=x_2(1+x_1)$
$\Rightarrow x_1+x_1{x}_2=x_2+x_2{x}_1$
$\Rightarrow x_1=x_2$

Similarly for $x<0$
we get $x_1=x_2$
Hence $f$ is a one-one function.

let $f\left(x\right)=1$
when $x\ge 0$
$1=\frac{x}{1+x}$
$1+x=x$
$1=0$ hence there is no such value for which $f\left(x\right)=1$

let $f\left(x\right)=1$
when $x<0$
$1=\frac{x}{1-x}$
$1-x=x$
$x=\frac{1}{2}\nless 0$ hence there is no such value for which $f\left(x\right)=1$
So range $\ne R$
Hence $f$ is an into function