If $m$ and $M$ are the minimum and the maximum values of $4+\frac{1}{2}{\sin }^{2}2x-2{\cos }^{4}x, x \in R,$ then $M-m$ is equal to:

Given expression is $4+\frac{1}{2}{\sin }^{2}2x-2{\cos }^{4}x$

$=$$4+2\left(1-{\cos }^{2}x\right){\cos }^{2}x-2{\cos }^{4}x$

Creating perfect squares, we get

$=-4 \left\{{\left({\cos }^{2}x-\frac{1}{4}\right)}^2-\frac{17}{16}\right\}$

Since, $0 \le {\cos }^{2}x \le 1$

$\Rightarrow$ $-\frac{1}{4} \le {\cos }^{2}x-\frac{1}{4} \le \frac{3}{4}$

$\Rightarrow$ $0 \le {\left({\cos }^{2}x-\frac{1}{4}\right)}^2 \le \frac{9}{16}$

$\Rightarrow$ $-\frac{17}{16} \le {\left({\cos }^{2}x-\frac{1}{4}\right)}^2-\frac{17}{16} \le \frac{9}{16}-\frac{17}{16}$

$\Rightarrow$ $\frac{17}{4} \ge -4 \left\{{\left({\cos }^{2}x-\frac{1}{4}\right)}^2-\frac{17}{16}\right\} \ge 2$

$\Rightarrow$ $M=\frac{17}{4}$ and $m=2$
$\Rightarrow$ $M-m=\frac{17}{4}-2=\frac{9}{4}$