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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.

user1442
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4 Answers4

3

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 $B$.

farruhota
  • 31,482
3

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 $\lambda\ne2$.

Bernard
  • 175,478
2

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.

Fly by Night
  • 32,272
1

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..