If f : R → A given by $f(x)=x^2-6 x+12$ is a surjective function, then the set A is:

$[3, \infty)$

The correct answer is Option (3) → $[3, \infty)$

$f(x)=x^2-6 x+12$

Since the coefficient of $x^2$ is positive, the parabola open upwards. The minimum value of $f(x)$ will occur at its vertex.

for vertex, $f'(c)=0$

$⇒2c-6=0$

$⇒c=\frac{6}{2}=3$

∴ f minimum = $f(3)=(3)^2-6(3)+12=3$

∴ Range (f(x)) = $[3, \infty)$