Question

The function $f:R\to R$ defined by
$f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)$ is
(a) one-one but not onto
(b) onto but not one-one
(c) both one and onto
(d) neither one-one nor onto

Answer 1

$f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)$

Injectivity:
$$\begin{gathered} f\left(1\right)=\left(1-1\right)\left(1-2\right)\left(1-3\right)=0 \\ f\left(2\right)=\left(2-1\right)\left(2-2\right)\left(2-3\right)=0 \\ f\left(3\right)=\left(3-1\right)\left(3-2\right)\left(3-3\right)=0 \\ \text{⇒}f\left(1\right)=f\left(2\right)=f\left(3\right)=0 \end{gathered}$$
So, f is not one-one.

Surjectivity:
Let y be an element in the co domain R, such that
$$\begin{gathered} y=f\left(x\right) \\ \Rightarrow y=\left(x-1\right)\left(x-2\right)\left(x-3\right) \\ \text{Since}y\in R\text{and}x\in R,f\text{is onto.} \end{gathered}$$

So, the answer is (b).