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Question Number 119483 by Ar Brandon last updated on 24/Oct/20
Let a>0 and f:R→R a function satisfying       f(x+a)=1+[2−3f(x)+3f(x)^2 −f(x)^3 ]^(1/3)   for all x∈R. Then a period of f(x) is ka where k is  a positive integer whose value is  (A)1                    (B)2                    (C)3                    (D)4
$$\mathrm{Let}\:{a}>\mathrm{0}\:\mathrm{and}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{a}\:\mathrm{function}\:\mathrm{satisfying} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{x}+{a}\right)=\mathrm{1}+\left[\mathrm{2}−\mathrm{3}{f}\left(\mathrm{x}\right)+\mathrm{3}{f}\left(\mathrm{x}\right)^{\mathrm{2}} −{f}\left(\mathrm{x}\right)^{\mathrm{3}} \right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R}.\:\mathrm{Then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\left(\mathrm{x}\right)\:\mathrm{is}\:{ka}\:\mathrm{where}\:{k}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{whose}\:\mathrm{value}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\mathrm{4} \\ $$
Answered by Olaf last updated on 25/Oct/20
f(x+a) = 1+[2−3f(x)+3f(x)^2 −f(x)^3 ]^(1/3)   [f(x+a)−1]^3  = 2−3f(x)+3f(x)^2 −f(x)^3   [f(x+a)−1]^3  = 1+[1−3f(x)+3f(x)^2 −f(x)^3 ]  [f(x+a)−1]^3  = 1−[f(x)−1]^3     [f(x+2a)−1]^3  = 1−[f(x+a)−1]^3   [f(x+2a)−1]^3  = 1−(1−[f(x)−1]^3 )  [f(x+2a)−1]^3  = [f(x)−1]^3   f(x+2a)−1 = f(x)−1  f(x+2a) =  f(x)   ⇒ period is 2a (k = 2)
$${f}\left({x}+{a}\right)\:=\:\mathrm{1}+\left[\mathrm{2}−\mathrm{3}{f}\left({x}\right)+\mathrm{3}{f}\left({x}\right)^{\mathrm{2}} −{f}\left({x}\right)^{\mathrm{3}} \right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\left[{f}\left({x}+{a}\right)−\mathrm{1}\right]^{\mathrm{3}} \:=\:\mathrm{2}−\mathrm{3}{f}\left({x}\right)+\mathrm{3}{f}\left({x}\right)^{\mathrm{2}} −{f}\left({x}\right)^{\mathrm{3}} \\ $$$$\left[{f}\left({x}+{a}\right)−\mathrm{1}\right]^{\mathrm{3}} \:=\:\mathrm{1}+\left[\mathrm{1}−\mathrm{3}{f}\left({x}\right)+\mathrm{3}{f}\left({x}\right)^{\mathrm{2}} −{f}\left({x}\right)^{\mathrm{3}} \right] \\ $$$$\left[{f}\left({x}+{a}\right)−\mathrm{1}\right]^{\mathrm{3}} \:=\:\mathrm{1}−\left[{f}\left({x}\right)−\mathrm{1}\right]^{\mathrm{3}} \\ $$$$ \\ $$$$\left[{f}\left({x}+\mathrm{2}{a}\right)−\mathrm{1}\right]^{\mathrm{3}} \:=\:\mathrm{1}−\left[{f}\left({x}+{a}\right)−\mathrm{1}\right]^{\mathrm{3}} \\ $$$$\left[{f}\left({x}+\mathrm{2}{a}\right)−\mathrm{1}\right]^{\mathrm{3}} \:=\:\mathrm{1}−\left(\mathrm{1}−\left[{f}\left({x}\right)−\mathrm{1}\right]^{\mathrm{3}} \right) \\ $$$$\left[{f}\left({x}+\mathrm{2}{a}\right)−\mathrm{1}\right]^{\mathrm{3}} \:=\:\left[{f}\left({x}\right)−\mathrm{1}\right]^{\mathrm{3}} \\ $$$${f}\left({x}+\mathrm{2}{a}\right)−\mathrm{1}\:=\:{f}\left({x}\right)−\mathrm{1} \\ $$$${f}\left({x}+\mathrm{2}{a}\right)\:=\:\:{f}\left({x}\right)\: \\ $$$$\Rightarrow\:\mathrm{period}\:\mathrm{is}\:\mathrm{2}{a}\:\left({k}\:=\:\mathrm{2}\right) \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 25/Oct/20
Merci monsieur

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