Question

The function \( f:[0,\infty)\rightarrow [0,\infty) \) defined by} \[ f(x)=\frac{x}{1+x} \] \textbf{is

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

Hint:

For rational functions of the form \[ f(x)=\frac{x}{1+x}, \] first check injectivity by equating \(f(x_1)\) and \(f(x_2)\). To find the range, substitute \(y=f(x)\) and express \(x\) in terms of \(y\). The restrictions on \(x\) then give the range directly.

Answer 1

The Correct Option is B

Concept: A rational function is one-one if it produces equal outputs only for equal inputs. To check onto, compare its range with the given codomain.

Step 1: Given \(f(x)=\frac{x}{1+x},\;x\ge0\). Since \(f'(x)=\frac{1}{(1+x)^2}>0\), the function is strictly increasing on its domain and hence one-one.

Step 2: Also, \(\displaystyle \lim_{x\to0}f(x)=0\) and \(\displaystyle \lim_{x\to\infty}f(x)=1\). Therefore, the range is \([0,1)\).

Step 3: As the codomain is \([0,\infty)\), values greater than or equal to \(1\) are never attained. Hence, the function is not onto.

Step 4: Therefore, the function is \(\boxed{\text{One-one but not onto}}\).