QJEEM22Ma165e: JEE Main 2022 Mathematics - Class 12 JEEM22Ma165e Solution: Let m and M respectively be the minimum and the maximum values…

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

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

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

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

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

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 $\sin \left( {2 \times 0 + {\pi \over 4}} \right) = {1 \over {\sqrt 2 }}$

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

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