Question

Let $f:R\to R$ is defined by $f\left(x\right)=\frac{x}{1+|x|}$. Then $f\left(x\right)$ is

• A. Injective but not surjective
• B. Surjective but not injective
• C. Injective as well as surjective
• D. Neither injective nor surjective

Answer 1

The correct option is A Injective but not surjective

For $x<0$

$f\left(x\right)=\frac{x}{1-x}$

$f\left(x_1\right)=f\left(x_2\right)$ implies

$\frac{x_1}{1-x_1}=\frac{x_2}{1-x_2}$

$x_1-x_1{x}_2=x_2-x_2{x}_1$

$x_1=x_2$ ...(i)

And for $x>0$

$f\left(x\right)=\frac{x}{1+x}$.

$f\left(x_1\right)=f\left(x_2\right)$

$\frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}$

$x_1.x_2+x_1=x_2+x_2{x}_1$

$x_1=x_2$ ...(ii)

Now

$|f\left(x\right)|<1$ ...(iii)

Hence from i, ii and iii,

$f\left(x\right)$ is injective but not surjective.