Mark the correct

Given that function f : R → R where f(x) = (x – 1) (x – 2) (x – 3)

If f(x) = f(y)

Then

(x – 1) (x – 2) (x – 3) = (y – 1) (y – 2) (y – 3)

⇒ f(1) = f(2) = f(3) = 0

So, f is not one-one.

y = f(x)

∵ x ϵ R also y ϵ R so f is onto.