Find Your Co-ordinates

We can rewrite the above function as;

$\begin{aligned} f(x)&=2\dfrac{(\cos 2x+1)}2 + \sqrt 3 \sin 2x + 5\\ &=\cos 2x + 1 + \sqrt 3\sin 2x + 5\\ &=\cos 2x + \sqrt 3\sin 2x + 6\\ \end{aligned}$

This is now in the form of $a\cos x + b\sin x + c$ , and the range would be $\left[-\sqrt{a^2+b^2}+c, \sqrt{a^2+b^2}+c\right]$ . Proved here .

Here $a=1$ , $b=\sqrt 3$ and $c=6$ ; thus the range is; $\left[-\sqrt{1^2 + \sqrt{3} ^2}+6, \sqrt{1^2 + \sqrt{3} ^2} + 6 \right]=\left[4,8\right]$

Therefore $A+B = \boxed{12}$ .

Calculus method:

All we need to do is find the minimum and maximum values of the function in the domain $0 \leq x \leq \pi$ . This is because $\cos^2 x$ and $\sin 2x$ cycles once within this range. This cycle will be repeated in the exact same manner for $x > \pi$ and $x < 0$

Given that $f(x) = 2 \cos^2 x + \sqrt{3}\sin 2x+5$ , we have

$f'(x) = 4 \cos x (-\sin x) + 2\sqrt{3} \cos 2x\\ =-2\sin 2x + 2\sqrt{3} \cos 2x$

At turning points, $f'(x) = 0$

$-2\sin 2x + 2\sqrt{3}\cos 2x = 0\\ 2\sqrt{3}\cos 2x = 2\sin 2x\\ \dfrac{\sin 2x}{\cos 2x} = \dfrac{2\sqrt{3}}{2}\\ \tan 2x = \sqrt{3}$

Note that the reference angle, $\alpha = \tan ^{-1} \sqrt{3} = \dfrac{\pi}{3}$

The value of $\tan 2x$ is positive, therefore $2x$ is quadrant I and quadrant III. Also note that:

$0 \leq x \leq \pi\\ 0 \leq 2x \leq 2\pi$

Therefore, all values that satisfy the equation are:

$2x = \dfrac{\pi}{3}, \dfrac{4\pi}{3}\\ x=\dfrac{\pi}{6}, \dfrac{2\pi}{3}$

Calculate the values of $f(x)$ at these points (do it yourself!):

$x=\dfrac{\pi}{6} \implies f\left(\dfrac{\pi}{6}\right) = 8\\ x=\dfrac{2\pi}{3} \implies f\left(\dfrac{2\pi}{3}\right) = 4$

If you want a more solid proof, you can do the second derivative test to show that these are minimum and maximum points.

$f''(x) = -4\cos 2x -4\sqrt{3} \sin 2x$

$x=\dfrac{\pi}{6} \implies f''\left(\dfrac{\pi}{6}\right) = -8 < 0\\ x=\dfrac{2\pi}{3} \implies f''\left(\dfrac{2\pi}{3}\right) = 8 > 0$

Therefore, the minimum value of $f(x)$ is $4$ and the maximum value of $f(x)$ is $8$

The range of $f(x)$ is $[4,8]$

$A=4,\;B=8,\;A+B = 4+8 = \boxed{12}$

I honestly graphed the function and noticed it oscillates between 4 and 8 from the y-axis side, so I took the range $4\leq f\leq8$ $\implies\ 4+8=12$

write $5$ as $1+4$ and 1 as $(sinx)^2+cos^2x$ so, it becomes 4+(perfect square)