For constant number "a" consider the function $f(x)= ax + \cos 2x + \sin x + \cos x$ on $\mathbb R$ such that $f(u)<f(v)$ for $u< v$. If the range of a, for any real numbers $u, v$ is $[\dfrac{m}{n}, \infty)$. Find the minimum value of $(m+n)$
Attempt:
The problem is just an overly complicated way of saying "find the range for which f(x) is strictly increasing".
So, using $f'(x)>0$ I get:
$a> 2\sin 2x + \sin x - \cos x$
So basically we need to find the maximum of RHS for the range of a.
$g(x)= 2 \sin 2x + \sin x - \cos x $
For extremum, $g'(x)=0 $
$4 \cos 2x + \cos x+ \sin x= 0$
$\implies 16\cos^2 2x = 1+ \sin 2x$
$\implies 16\sin^2 2x + \sin 2x -15 =0$
$\implies \sin 2x = -1$ or $\sin 2x = \dfrac {15}{16}$
Then it seriously gets very complicated because we would have to extract $\sin x$ and $\cos x$ from $\sin 2x = \dfrac{15}{16}$. I tried but couldn't get it easily. After this we would have to plug in the values to g(x) to see which one of them gives maximum. (Just a side note: g is periodic with period $2\pi$)
Is it possible to do this problem using my method? If not, what are the other clever ways to solve it?
You want to maximize $$a=g(x)= 2\sin 2x + \sin x - \cos x=2\sin 2x + \sqrt{1-\sin 2x}$$ (I'm taking the plus sign on the square root because we only care about the maximum.) You've found the possible values of $\sin 2x$ that maximize $a,$ so it should be plain sailing from here.