Mathematics - Functions Question with Solution | TestHub

Given $2\mathrm{f}(\sin \mathrm{x})+\mathrm{f}(\cos \mathrm{x})=\mathrm{x}$ ...(A) Replacing x by $\frac{\pi }{2}-x$, we have $2f(\cos x)+f(\sin x)=\frac{\pi }{2}-x$ ...(B) $(\mathrm{A})+(\mathrm{B})\Rightarrow 3\mathrm{f}(\sin \mathrm{x})+3\mathrm{f}(\cos \mathrm{x})=\frac{\pi }{2}\Rightarrow \mathrm{f}(\sin \mathrm{x})+\mathrm{f}(\cos \mathrm{x})=\frac{\pi }{6}$ ...(C) (A) - (C) $\Rightarrow \mathrm{f}(\sin \mathrm{x})=\mathrm{x}-\frac{\pi }{6}\Rightarrow \mathrm{f}(\mathrm{x})={\sin }^{-1}\mathrm{x}-\frac{\pi }{6}$.

Hence $D_f=[-1,1]$ and $R_f=\left[-\frac{\pi }{2}-\frac{\pi }{6},\frac{\pi }{2}-\frac{\pi }{6}\right]=\left[-\frac{2\pi }{3},\frac{\pi }{3}\right]$.]