Let f :R--->R  is a function satisfying f(10-x) = f(x) and f(2-x) = f(2+x) for all x belonging to R.
If f(0) = 101, then find the minimum possible no. of values of x satisfying f(x) = 101 for x belonging to [0,30].

The correct answer is 11.

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{f}\left(10-\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)....\left(\mathrm{i}\right) \\ \mathrm{Put} \mathrm{x}\to 5-\mathrm{x} \\ \Rightarrow \mathrm{f}\left(10-5+\mathrm{x}\right)=\mathrm{f}\left(5-\mathrm{x}\right) \\ \Rightarrow \mathrm{f}\left(5+\mathrm{x}\right)=\mathrm{f}\left(5-\mathrm{x}\right) \\ \mathrm{Hence} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{symmetric} \mathrm{about} \mathrm{line} \mathrm{x}=5 \\ \mathrm{Put} \mathrm{x}=5 \\ \Rightarrow \mathrm{f}\left(5+5\right)=\mathrm{f}\left(0\right) \\ \Rightarrow \mathrm{f}\left(10\right)=101 \left[\mathrm{As} \mathrm{f}\left(0\right)=101\right] \\ \mathrm{Also} \mathrm{given} \mathrm{that} \\ \mathrm{f}\left(2+\mathrm{x}\right)=\mathrm{f}\left(2-\mathrm{x}\right) \left[\mathrm{Hence} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{symmetric} \mathrm{about} \mathrm{line} \mathrm{x}=2\right] \\ \mathrm{x}\to 2 \\ \Rightarrow \mathrm{f}\left(2+2\right)=\mathrm{f}\left(2-2\right) \\ \Rightarrow \mathrm{f}\left(4\right)=\mathrm{f}\left(0\right) \\ \Rightarrow \mathrm{f}\left(4\right)=101 \\ \mathrm{As} \mathrm{f} \mathrm{is} \mathrm{symmetric} \mathrm{about} \mathrm{x}=5 \mathrm{and} \mathrm{f}\left(4\right)=101, \mathrm{hence} \mathrm{f}\left(6\right)=101 \mathrm{is} \mathrm{also} \mathrm{true} \\ \mathrm{f}\left(2+\mathrm{x}\right)=\mathrm{f}\left(2-\mathrm{x}\right)....\left(\mathrm{ii}\right) \\ \mathrm{In} \left(\mathrm{ii}\right) \mathrm{put} \mathrm{x}-2\to \mathrm{x}-4 \\ \mathrm{you} \mathrm{will} \mathrm{get} \\ \mathrm{f}\left(4-\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)....\left(\mathrm{iii}\right) \\ \mathrm{Using} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{iii}\right) \\ \mathrm{we} \mathrm{can} \mathrm{find} \mathrm{that} \\ \mathbf{f}\left(\mathbf{x}\right)\mathbf{=}\mathbf{f}\left(\mathbf{6}\mathbf{+}\mathbf{x}\right) \\ \mathrm{And} \mathrm{then} \mathrm{again} \mathrm{use} \left(\mathrm{ii}\right) \mathrm{to} \mathrm{get} \\ \mathbf{f}\left(\mathbf{x}\right)\mathbf{=}\mathbf{f}\left(\mathbf{x}\mathbf{-}\mathbf{2}\right) \\ \mathrm{You} \mathrm{already} \mathrm{have} \\ \mathrm{f}\left(0\right)=\mathrm{f}\left(4\right)=\mathrm{f}\left(6\right)=\mathrm{f}\left(10\right)=101 \\ \mathrm{USe} \mathrm{them} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{answer}. \end{gathered}$$

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