Let f (x)=x^3-6x^2 +12x-3, then at x=2, f(x) has:

The correct answer is Option (4) → neither a maximum nor a minimum

$f (x)=x^3-6x^2 +12x-3$

for maxima and minima, $f'(x)=0$

$⇒3x^2-12x+12=0$

$⇒x^2-4x+4=0$

$⇒(x-2)^2=0$

$⇒x=2\,or\,2$

Now, $f''(x)=2x-4$

$⇒f''(2)=0$

Hence at $x=2$, neither maxima nor minima.