Question:
The function $f: R \rightarrow R$ defined by
$f(x)=(x-1)(x-2)(x-3)$ is
(a) one-one but not onto
(b) onto but not one-one
(c) both one and onto
(d) neither one-one nor onto
Solution:
$f(x)=(x-1)(x-2)(x-3)$
Injectivity:
$f(1)=(1-1)(1-2)(1-3)=0$
$f(2)=(2-1)(2-2)(2-3)=0$
$f(3)=(3-1)(3-2)(3-3)=0$
$\Rightarrow f(1)=f(2)=f(3)=0$
So, $f$ is not one-one.
Surjectivity:
Let y be an element in the co domain R, such that
$y=f(x)$
$\Rightarrow y=(x-1)(x-2)(x-3)$
Since $y \in R$ and $x \in R, f$ is onto.
So, the answer is (b).