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

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

$$ = {\pi \over 2} + \sqrt 2 \left( {{1 \over {\sqrt 2 }}\sin 2x + {1 \over {\sqrt 2 }}\cos 2x} \right)$$

$$ = {\pi \over 2} + \sqrt 2 \left( {\cos {\pi \over 4}\sin 2x + \sin {\pi \over 4}\cos 2x} \right)$$

$$ = {\pi \over 2} + \sqrt 2 \,.\,\sin \left( {2x + {\pi \over 4}} \right)$$

f(x) is maximum when $$\sin \left( {2x + {\pi \over 4}} \right)$$ is maximum means $$x = {\pi \over 8}$$ or $$\sin \left( {2 \times {\pi \over 8} + {\pi \over 4}} \right) = \sin {\pi \over 2} = 1$$

$$\therefore$$ $${\left[ {f(x)} \right]_{\max }} = {\pi \over 2} + \sqrt 2 \,.\,1 = {\pi \over 2} + \sqrt 2 = M$$

f(x) is minimum when $$\sin \left( {2x + {\pi \over 4}} \right)$$ is minimum means $$x = 0$$ or $$\sin \left( {2 \times 0 + {\pi \over 4}} \right) = {1 \over {\sqrt 2 }}$$

$$\therefore$$ $${\left[ {f(x)} \right]_{\min }} = {\pi \over 2} + \sqrt 2 \,.\,{1 \over {\sqrt 2 }} = {\pi \over 2} + 1 = m$$

$$\therefore$$ $$m + M = {\pi \over 2} + \sqrt 2 + {\pi \over 2} + 1 = \pi + \sqrt 2 + 1$$