Question

lf $f:[0,\infty )\to [0,\infty )$, and $f\left(x\right)=\frac{x}{1+x}$ then $f$ is

• A. one - one and onto
• B. one - one but not onto
• C. onto but not one - one
• D. neither one - one nor onto

Answer 1

The correct option is B one - one but not onto

Given $f:[0,\infty )\to [0,\infty )$

Here, domain is $[0,\infty )$ and codomain is $[0,\infty )$.

Thus, to check one-one

$f\left(x\right)=\frac{x}{1+x}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{{(1+x)}^2}>0,\forall x\in [0,\infty )$

$\therefore f\left(x\right)$ is increasing in its domain.

Thus $f\left(x\right)$ is one-one in its domain.To check onto

Again $y=f\left(x\right)=\frac{x}{1+x}\Rightarrow y+yx=x$

$\Rightarrow x=\frac{y}{1-y}\Rightarrow \frac{y}{1-y}\ge 0$

Since $x\ge 0,$ therefore $0\le y<1$

i.e Range $\ne$ Codomain

$\therefore f\left(x\right)$ is one-one but not onto.