The function $f : [0, ∞) → R$ given by $f(x) =\frac{x}{x+1}$, is

one-one and onto one-one but not onto onto but not one-one neither one-one nor onto

one-one but not onto

We have, $f(x) =\frac{x}{x+1}=1-\frac{1}{1+x}$ Let $x, y ∈ R$ be such that $f(x) = f(y)$. Then, $f(x) = f(y)$ $⇒1-\frac{1}{1+x}=1-\frac{1}{1+y}⇒1+x=1+y⇒x=y$ ∴ f is one-one. Clearly, f is not onto as f takes only values less than 1 i.e. Range f = (-∞, 1) ≠ co-domain of f.