Question

If f : [0, $\infty )\to [0,\infty ),andf(x)=\frac{x}{1+x}$ , then f is

• A. one - one and onto
• B. one - one but not onto
• C. onto but not one-one
• D. nether one-one nor onto

Answer 1

The correct option is A one - one and onto

$f^{\prime }\left(x\right)=\frac{1+x-x}{(1+x{)}^2}=\frac{1}{(1+x{)}^2}$

Clearly $f\left(x\right)$ is an increasing function. Hence minimum and maximum values will be attained at lowest value of x and highest value of x respectively in given domain.

Also as the function is monotonic it is one-one.

$f\left(0\right)=0$

${lim}_{x\to \infty }f\left(x\right)=\frac{\frac{x}{x}}{\frac{1}{x}+\frac{x}{x}}=\frac{1}{1+\frac{1}{x}}=1$

Hence function is not onto.