​$$\text{Let } f \colon [0, \infty) \to [0, 2] \text{ be defined by } f(x) = \frac{2x}{1 + x}, \text{ then } f \text{ is:}$$​​

Correct option is A

Given function is

$$\begin{aligned}f(x) &= \frac{2x}{1 + x} \\f(x) &= f(y) \\\frac{2x}{1 + x} &= \frac{2y}{1 + y} \\2x + 2xy &= 2y + 2xy \\x &= y \\\text{So, } f \text{ is one-one.} \\\text{We have, } y &= \frac{2x}{x + 1} \\\Rightarrow 2x &= yx + y \\\Rightarrow x &= \frac{y}{2 - y} \\\text{Now, } x \in [0, \infty) \\\Rightarrow 0 \leq y &< 2 \\\Rightarrow \text{Range of } f(x) = [0, 2) \\\text{Since range } &\subset \text{ co-domain} \\\Rightarrow f &\text{ is not onto.}\end{aligned}$$​