JEE Main 2022 — Inverse Trigonometric Functions Question with Solution

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

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

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

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

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

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

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

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

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

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