maximize$\frac{ac}{b^2}$

Question

The question goes like this :

Given the expression $\cos(ax^2 + bx + c) = -1$ has two distinct real roots $x=1$ and $x=-1$ , $a,b,c∈(2,5)$, find the maximum value of $\frac{ac}{b^2}$.

My approach:

Clearly $\cos(a+b+c) =-1$ and $\cos(a-b+c)=-1$.

and $a+b+c\in(6,15)$ and $(a-b+c)\in(-1,8)$.

Hence $a+b+c=2 \pi$, $3 \pi$, $4 \pi,$ and $a-b+c= \pi, 2\pi.$

Now we have $3$ variables and $2$ equations with different constants and I do not know how to proceed from here. Any help is appreciated.

Answer 1

You have a mistake in your last sentence: since the period of $\cos x$ is $2 \pi$, $a+b+c = 3 \pi$ and $a-b+c = \pi$.

Summing the two equations gives $2a + 2c = 4 \pi$ or $a + c = 2 \pi$, which means that $b = \pi$.

Can you take it from here and maximise the given expression?

Answer 2

Hint: You also know that $a+b+c = \cos^{-1}{-1} = \pi +2n\pi$ and equivalently that $a-b+c =\pi +2k\pi$ for some $n, k \in \Bbb Z$. This will eliminate the problem of your sums being equal to several constants.