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.

Dear Student,
Please find below the solution to the asked query:

We havef10-x=fx....iPut x5-xf10-5+x=f5-xf5+x=f5-xHence fx is symmetric about line x=5Put x=5f5+5=f0f10=101 As f0=101Also given thatf2+x=f2-x Hence fx is symmetric about line x=2x2f2+2=f2-2f4=f0f4=101As f is symmetric about x=5 and f4=101, hence f6=101 is also truef2+x=f2-x....iiIn ii put x-2x-4you will getf4-x=fx....iiiUsing i and iiiwe can find thatfx=f6+xAnd then again use ii to getfx=fx-2You already havef0=f4=f6=f10=101USe them to find the answer.

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