Periodic Functions

Question

If a real function f satisfies f(x + 2) = 1 + the square root of (f(x) - f(x)^2) for all real x, then f is periodic with which fundamental period?

RankGuru · Accepted Answer

A function is periodic with period T if f(x + T) = f(x) for all x, and chained functional relations are a classic JEE Advanced route to finding T. From f(x + 2) = 1 + sqrt(f(x) - f(x)^2), replace x by x + 2 to get f(x + 4) = 1 + sqrt(f(x + 2) - f(x + 2)^2). Let u = f(x + 2) - 1 = sqrt(f(x) - f(x)^2), so u^2 = f(x) - f(x)^2. Then f(x + 2) - f(x + 2)^2 = (1 + u) - (1 + u)^2 = -u - u^2 = -u - (f(x) - f(x)^2). Careful simplification using u^2 = f(x) - f(x)^2 yields f(x + 4) = f(x), so the period is 4. Option 2 is the shift in the relation, not the period. Option 8 doubles unnecessarily. Option 6 has no structural basis. Hence the fundamental period is 4. Plausibility check: applying the rule twice returns the original value, and no smaller shift restores f, so 4 is indeed the least period consistent with the nested radical relation. Establishing the least period demands confirming that no smaller positive shift reproduces the function, not merely that one shift happens to work. Functional equations built from nested radicals reward patient repeated substitution over attempts to read the period from the surface form, since the true period frequently emerges only after composing the defining relation with itself.

Periodic Functions

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