Given, 4 + ${1 \over 2}$ sin 2 2x $-$ 2cos 4 x = 4 + ${1 \over 2}$ (2sinx cosx) 2 $-$ 2cos 4 x = 4 + ${1 \over 2}$ $\times$ 4 sin 2 x cos 2 x $-$ 2cos 4 x = 4 + 2 (1 $-$ cos 2 x) cos 2 x $-$ 2cos 4 x = 4 + 2 cos 2 x $-$ 4cos 4 x = $-$ 4 $\left\{ {\cos } \right.$ 4 x $-$ $\left. {{{{{\cos }^2}x} \over 2} - 1} \right\}$ = $-$ 4 $\left\{ {\cos } \right.$ 4 x $-$ 2 . ${1 \over 4}$ . cos 2 x + $\left. {{1 \over {16}} - {1 \over {16}} - 1} \right\}$ = $-$ 4 $\left\{ {{{\left( {{{\cos }^2}x - {1 \over 4}} \right)}^2} - {{17} \over {16}}} \right\}$ We know, O $\le$ cos 2 x $\le$ 1 $\Rightarrow$ $- {1 \over 4}$ $\le$cos 2 x $- {1 \over 4}$ $\le$ ${3 \over 4}$ $\Rightarrow$ 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 = ${{17} \over 4}$ $-$ $2$ = ${{9} \over 4}$