The function$f\left(x\right)=x^3+a{x}^2+bx+c,a^2\le 3b$ has

The correct option is C

no extreme value

Explanation for correct option

$f\left(x\right)=x^3+a{x}^2+bx+c$

$f\text{'}\left(x\right)=3x^2+2ax+b$

Now, $f\text{'}\left(x\right)=0$

$\Rightarrow 3x^2+2ax+b=0$

$\Rightarrow x=\frac{-2a\pm \sqrt{\left(4a^2-12b\right)}}{6}$

Since, $a^2\le 3b$

So, $x$ has imaginary values.

Hence, option C is correct.