Explanation

(b)
Here, $f:\lbrack 0,\ \infty\rbrack \rightarrow \lbrack 0,\ \infty)ie,\$domain is $\lbrack 0,\ \infty)$ and codomain is $\lbrack 0,\ \infty).$
For one-one $f(x) = \frac{x}{1 + x}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f^{'}(x) = \frac{1}{(1 + x)^{2}} > 0,\ \forall\ x \in \lbrack 0,\ \infty)$
$\therefore f(x)$ is increasing in its domain. Thus, $f(x)$ is one-one in its domain.
For onto (we find range)
$f(x) = \frac{x}{1 + x}ie,\ y = \frac{x}{1 + x} \Rightarrow y + yx = x$
$\Rightarrow x = \frac{y}{1 - y} \Rightarrow \frac{y}{1 - y} \geq 0$ as $x \geq 0$ $\therefore 0 \leq y \neq 1$
$ie,$ Range$\ \neq \$Codomain
$\therefore f(x)$ is one-one but not onto.