The function f : R→R defined by

$f(x)=(x-1)(x-2)(x-3)$

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).