Q46 
 
46. If f(x) = sin^-1 (sin x) + cos^-1(cosx)  , then which of the following is/are correct​
$$\begin{gathered} \left(\text{A}\right) \text{Domain of f(x) is [-1 , 1] } \\ \text{(B) Range of f(x) is [0, π] } \\ \text{(C) Domain of f(x) is R} \\ \text{ (D) Range of f(x) is [0, 2π]} \end{gathered}$$

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\sin }^{-1}\left(\mathrm{sinx}\right)+{\cos }^{-1}\left(\mathrm{cosx}\right) \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{domain} \mathrm{of} \mathrm{sinx} \mathrm{and} \mathrm{cosx} \mathrm{is} \mathrm{R} \mathrm{and} \mathrm{their} \mathrm{range} \mathrm{is} \left[-1,1\right] \\ \mathrm{and} {\sin }^{-1}\mathrm{\theta } \mathrm{is} \mathrm{defined} \mathrm{when} \mathrm{\theta }\in \left[-1,1\right] \\ \mathrm{Hence} \mathrm{f}\left(\mathrm{x}\right) \mathrm{will} \mathrm{be} \mathrm{defined} \mathrm{for} \mathrm{all} \mathrm{real} \mathrm{values} \mathrm{of} \mathrm{x}. \\ \mathrm{Hence} \\ \mathrm{Range} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{R}. \\ \mathrm{Both} \mathrm{sinx} \mathrm{and} \mathrm{cosx} \mathrm{are} \mathrm{periodic} \mathrm{with} \mathrm{period} 2\mathrm{\pi }, \mathrm{hence} \mathrm{period} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right) \\ \mathrm{is} 2\mathrm{\pi }. \\ \mathrm{When} \mathrm{x}\in \left[0,\frac{\mathrm{\pi }}{2}\right] \\ {\sin }^{-1}\left(\mathrm{sinx}\right)=\mathrm{x} \mathrm{and} {\cos }^{-1}\left(\mathrm{cosx}\right)=\mathrm{x} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{x}=2\mathrm{x} \\ \mathrm{As} \mathrm{x}\in \left[0,\frac{\mathrm{\pi }}{2}\right] \\ \Rightarrow 2\mathrm{x}\in \left[0,\mathrm{\pi }\right] \\ \mathrm{Hence} \mathrm{f}\left(\mathrm{x}\right)\in \left[0,\mathrm{\pi }\right] \mathrm{when} \mathrm{x}\in \left[0,\frac{\mathrm{\pi }}{2}\right] \\ \mathrm{When} \mathrm{x}\in \left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right) \\ {\sin }^{-1}\left(\mathrm{sinx}\right)=\mathrm{\pi }-\mathrm{x} \mathrm{and} {\cos }^{-1}\left(\mathrm{cosx}\right)=\mathrm{x} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{\pi }-\mathrm{x}+\mathrm{x}=\mathrm{\pi } \\ \mathrm{Hence} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{\pi } \mathrm{for} \mathrm{all} \mathrm{x}\in \left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right) \\ \mathrm{Similarly} \mathrm{after} \mathrm{breaking} \mathrm{function} \mathrm{in} \left(\mathrm{\pi },2\mathrm{\pi }\right) \mathrm{you} \mathrm{will} \mathrm{see} \mathrm{that} \mathrm{range} \mathrm{comes} \mathrm{out} \\ \mathrm{to} \mathrm{be} \left[0,\mathrm{\pi }\right] \\ \mathrm{Hence} \mathrm{option}\left(\mathrm{B},\mathrm{C}\right) \mathrm{is} \mathrm{correct}. \end{gathered}$$

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