If $x^4 + 3\cos(ax^2 + bx +c) = 2(x^2-2) $ has two solutions with $a,b,c \in (2,5)$, then find the maximum value of $\frac{ac}{b^2} $

Question

The answer given is 1. i tried like this $3\cos(ax^2 + bx +c) = -x^4 +2x^2-4 = -(x^2 -1)^2 -3 $. The maximum value of $-x^4 +2x^2-4$ is $-3$ so $3\cos(ax^2 + bx +c) =-3$ and the two values of x are $1$ and $-1$. Now $\cos(a + b +c) =-1$ and $\cos(a - b +c) =-1$. How do i find maximum value of $ \frac{ac}{b^2} $

Answer 1

You are on the right track. The next step is to utilize the condition $a, b, c \in (2, 5)$.

Since $\cos(a + b + c) = \cos(a - b + c)$ and $a + b + c > a - b + c$, we have $$a + b + c = a - b + c + 2k\pi$$ for some positive integer $k$, implying $b = k\pi$. Since $b \in (2, 5)$, it can be seen that $k = 1$, and $b = \pi$. Additionally, $\cos(a + b + c) = -1$, $a \in (2, 5)$, $c \in (2, 5)$ and $b = \pi$ imply that $$a + b + c = (2\ell + 1)\pi$$ with $(2\ell + 1)\pi \in (4 + \pi, 10 + \pi)$, hence $\ell \in (2/\pi, 5/\pi)$, that is $\ell = 1$. It then follows that $a + c = 2\pi$.

By AM-GM inequality, when $a = c = \pi$: $$ac \leq \left(\frac{a + c}{2}\right)^2 = \pi^2.$$ Recall $b = \pi$, thus $$\frac{ac}{b^2} \leq \frac{\pi^2}{\pi^2} = 1.$$