Find the maximum value of $\sin \left( x^{2}\right) +\cos x$ or prove that there is no maximum value

Question

I had proved this function <2. But I still couldn't find a way to prove the maximum value is inexistence or found it.

Answer 1

There are infinitely many solutions to its derivative

$$F'(x)=2x \cos(x^2)- \sin x =0 $$

Answer 2

In order to have $\sin(x^2) + \cos(x) = 2$, you'd need $x = n \pi$ for some even integer $n$ to make $\cos(x) = 1$, and $x^2 = (m+1/2) \pi$ for some even integer $m$. Then $n^2 \pi^2 = x^2 = (m+1/2) \pi$, which makes $\pi = (m+1/2)/n^2$. But that's impossible, because the right side is rational but $\pi$ is irrational.

EDIT: Now we want to prove that the function takes values arbitrarily close to $2$. For any $\epsilon > 0$, take $\delta = \arccos(1-\epsilon) > 0$. Then $\cos(x) > 1-\epsilon$ whenever $x$ is in one of the intervals $(n\pi - \delta, n\pi + \delta)$ for even integers $n$. I claim if $n$ is large enough, the interval $(n\pi- \delta, n\pi+ \delta)$ contains some $x$ for which $\sin(x^2) = 1$. Well, as $x$ goes from $n\pi- \delta$ to $n\pi + \delta$, $x^2$ goes from $(n\pi - \delta)^2$ to $(n\pi + \delta)^2$, a distance of $(n\pi + \delta)^2 - (n\pi - \delta)^2 = 4 n \pi \delta$.
If $4 n \pi \delta > 2 \pi$, that interval contains at least one number $y$ where $\sin(y) = 1$.