$\lambda$ for which the function $f(x)=2x^3-3(2+\lambda)x^2+12\lambda x$ has exactly one local maxima and exactly one local minima Let $S$ be the set of real values of parameter $\lambda$ for which the function $f(x)=2x^3-3(2+\lambda)x^2+12\lambda x$ has exactly one local maxima and exactly one local minima. Then the subset of $S$ is
$(A)(5,\infty)$ 
$(B)(-4,4)$
$(C)(3,8)$
$(D)(-\infty,-1)$

$$f'(x)=6x^2-6(2+\lambda)x+12\lambda=0$$gives $x=2,\lambda.$
The two local extrema means three roots.So applying the condition of three roots in a cubic polynomial.
$$f(2).f(\lambda)<0$$
gives $\lambda\in(\frac{6}{11},\frac{2}{3})$
But the subsets of $S$ are given $A,C,D$ in the answer.
 A: You found critical points correctly.
Take the second derivative: $$f''=12x-12-6\lambda.$$
Two cases:
$1) f''(2)>0, \ f''(\lambda)<0 \Rightarrow \lambda<2$ or
$2) f''(2)<0, \ f''(\lambda)>0 \Rightarrow \lambda>2.$
Hence: $\lambda\ne 2.$
Answer: all but $B$.
A: No need for considering the second derivative: as $f'(x)=6x^2-6(\lambda+2)x+12\lambda$, the condition is $f'(x)$ to have two critical values, and this happens if and only if the quadratic has a (reduced) discriminant
$$\Delta'=9(\lambda+2)^2-72\lambda=9\bigl((\lambda+2)^2-8\lambda\bigr)=9(\lambda-2)^2>0.$$
This condition is satisfied exactly when $\lambda\ne2$.
A: HINT:
The derivative of a cubic will be a quadratic.
A quadratic can have (at best) two distinct roots.
How does one decide on the number of roots a quadratic has?
If your cubic has a one local max, and one local min then it has two distinct turning points.
A: A cubic polynomial has between $0$ and $2$ critical points. You found the real solutions $2$ and $\lambda$ of the equation $f_\lambda'(x)=0$. If $\lambda\ne2$ then
$$f_\lambda''(2)f_\lambda''(\lambda)=-36(\lambda-2)^2<0\ ,$$
hence we have one local maximum and one local minimum. If $\lambda=2$ then we are talking about
$$f_2(x)=2x^3-12x^2+24x=2(x-2)^3+16\ .$$
This function is monotonically increasing. It follows that $S={\mathbb R}\setminus\{2\}$, hence  A), C), D) describe subsets of $S$, but B) doesn't..
