Let's boycott Calculus this time! (Part 1)

$\begin{aligned} X & = \cos^2 x + \sin^4 x \\ & = \cos^2 x + \left(1-\cos^2 x\right)^2 \\ & = \cos^4 x - \cos^2 x + 1 \\ & = \color{#3D99F6}{\cos^4 x - \cos^2 x + \frac 14} + \frac 34 \\ & = \color{#3D99F6}{\left(\cos^2 x - \frac 12 \right)^2} + \frac 34 \end{aligned}$

Since $\left(\cos^2 x - \frac 12 \right)^2 \ge 0$ , $X \ge \frac 34$ , $X_{min} = \frac 34$ , when $\cos^2 x = \frac 12$ which is valid.

$X$ is maximum, when $\left(\cos^2 x - \frac 12 \right)^2$ is maximum, that is when $\cos^2 x = 1$ , then we have:

$\begin{aligned} X_{max} & = \left(1 - \frac 12 \right)^2 + \frac 34 = \frac 14 + \frac 34 = 1 \end{aligned}$

$\implies X_{max} + X_{min} = 1 + \dfrac 34 = \boxed{1.75}$

$\cos^2(x) + \sin^4(x)$

$=1 - \sin^2(x) + \sin^4(x)$

$=-\sin^2(x)(1-\sin^2(x)) + 1$

$=-\sin^2(x)\cos^2(x) +1$

$=- \frac{4 \sin^2(x)\cos^2(x)}{4} + 1$

$=- \frac{[2 \sin(x)\cos(x)]^2}{4} + 1$

$=- \frac{sin^2(2x)}{4} + 1$

Clearly, $-1 \leqslant -sin^2(2x) \leqslant 0$

$\implies -\frac{1}{4} \leqslant - \frac{sin^2(2x)}{4} \leqslant \frac{0}{4}$

$\implies -\frac{1}{4} + 1 \leqslant - \frac{sin^2(2x)}{4} + 1 \leqslant 0+1$

$\implies \boxed{\boxed{\frac{3}{4}} \leqslant - \frac{sin^2(2x)}{4} + 1 \leqslant \boxed{1}}$

Hence, the sum of the maximum and minimum values $= \frac{3}{4} + 1 = \frac{7}{4} = \boxed{1.75}$

convert everything to cos(2x)

X = $cos^2 x + sin^4 x = sin^4 x - sin^2 x + 1$ . Let $y = sin^2 x, then X = y^2 - y + 1 = (y - 1/2)^2 + 3/4$ . Now notice that $0 \leq y \leq 1, so 0 \leq (y - 1/2)^2 \leq 1/4 and therefore 3/4 \leq X \leq 1$ . The answer is then 3/4 + 1 = 1.75.