Question

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

Here, $f:[0,\infty )\to [0,\infty )$ i.e, domain is $[0,\infty )$ and codomain is $[0,\infty )$

For one-one

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

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

$\therefore f\left(x\right)$ is increasing in this domain. Thus $f\left(x\right)$ is one-one in its domain

For onto (we find range)

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

$\Rightarrow y+yx=x$

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

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

$\therefore 0\le y\ne 1$ and $y<1$

$\therefore$ Range $\ne$ Codomain

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