$\max (f(x))$

$\begin{aligned} f(x) & = 2\sin^2 x + 2\cos^4 x \\ & = 2\sin^2 x + 2(1-\sin^2 x)^2 \\ & = 2\sin^2 x + 2-4\sin^2 x + 2\sin^4 x \\ & = 2(\sin^4 x - \sin^2 x +1) \\ & = 2\left(\sin^2 x - \frac 12\right)^2 + \frac 32 \end{aligned}$

$\begin{aligned} \implies \max (f(x)) & = 2\left(\max(\sin^2 x) - \frac 12\right)^2 + \frac 32 \\ & = 2\left(1 - \frac 12\right)^2 + \frac 32 = \frac 12 + \frac 32 = \boxed 2 \end{aligned}$

The given function is $2+2(\sin^4 x-\sin^2 x)$ . Since $|\sin x|\leq 1$ , $\sin^4 x-\sin^2 x\leq 0$ for all $x$ . So, the required maximum is $2+0=\boxed 2$