Given,

4 + $${1 \over 2}$$ sin² 2x $$-$$ 2cos⁴ x

= 4 + $${1 \over 2}$$ (2sinx cosx)² $$-$$ 2cos⁴x

= 4 + $${1 \over 2}$$ $$ \times $$ 4 sin²x cos²x $$-$$ 2cos⁴ x

= 4 + 2 (1 $$-$$ cos²x) cos²x $$-$$ 2cos⁴ x

= 4 + 2 cos²x $$-$$ 4cos⁴x

= $$-$$ 4 $$\left\{ {\cos } \right.$$⁴x $$-$$ $$\left. {{{{{\cos }^2}x} \over 2} - 1} \right\}$$

= $$-$$ 4 $$\left\{ {\cos } \right.$$⁴x $$-$$ 2 . $${1 \over 4}$$ . cos²x + $$\left. {{1 \over {16}} - {1 \over {16}} - 1} \right\}$$

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

We know,

O $$ \le $$ cos²x $$ \le $$ 1

$$ \Rightarrow $$   $$ - {1 \over 4}$$ $$ \le $$cos²x $$ - {1 \over 4}$$ $$ \le $$ $${3 \over 4}$$

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

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

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

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

$$ \therefore $$   Maximum value, M = $${{17} \over 4}$$

      Minimum value, m = $$2$$

$$ \therefore $$   M $$-$$ m = $${{17} \over 4}$$ $$-$$ $$2$$ = $${{9} \over 4}$$