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Question Number 119483 by Ar Brandon last updated on 24/Oct/20

Let a > 0 and f : R \to R a function satisfying

f (x + a) = 1 + {[2 - 3 f (x) + 3 f {(x)}^{2} - f {(x)}^{3}]}^{1 / 3}

for all x \in R . Then a period of f (x) is k a where k is

a positive integer whose value is

(A) 1 (B) 2 (C) 3 (D) 4

Answered by Olaf last updated on 25/Oct/20

f (x + a) = 1 + {[2 - 3 f (x) + 3 f {(x)}^{2} - f {(x)}^{3}]}^{1 / 3}

{[f (x + a) - 1]}^{3} = 2 - 3 f (x) + 3 f {(x)}^{2} - f {(x)}^{3}

{[f (x + a) - 1]}^{3} = 1 + [1 - 3 f (x) + 3 f {(x)}^{2} - f {(x)}^{3}]

{[f (x + a) - 1]}^{3} = 1 - {[f (x) - 1]}^{3}

{[f (x + 2 a) - 1]}^{3} = 1 - {[f (x + a) - 1]}^{3}

{[f (x + 2 a) - 1]}^{3} = 1 - (1 - {[f (x) - 1]}^{3})

{[f (x + 2 a) - 1]}^{3} = {[f (x) - 1]}^{3}

f (x + 2 a) - 1 = f (x) - 1

f (x + 2 a) = f (x)

\Rightarrow period is 2 a (k = 2)

Commented by Ar Brandon last updated on 25/Oct/20

Merci monsieur