Question

The function $f:[0,\infty )\to R\mathrm{given}\mathrm{by}f\left(x\right)=\frac{x}{x+1}\mathrm{is}$
(a) one-one and onto
(b) one-one but not onto
(c) onto but not one-one
(d) onto but not one-one

Answer 1

Injectivity:
Let x and y be two elements in the domain, such that
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow \frac{x}{x+1}=\frac{y}{y+1} \\ \Rightarrow xy+x=xy+y \\ \Rightarrow x=y \end{gathered}$$
So, f is one-one.

Surjectivity:
Let y be an element in the co domain R, such that
$$\begin{gathered} y=f\left(x\right) \\ \Rightarrow y=\frac{x}{x+1} \\ \Rightarrow xy+y=x \\ \Rightarrow x\left(y-1\right)=-y \\ \Rightarrow x=\frac{-y}{y-1} \\ \text{Range of}f=R-\left\{1\right\}\ne \text{co domain (}\mathit{\text{R}}\text{)} \end{gathered}$$
$\Rightarrow$f is not onto.
So, the answer is (b).