Let f:R→R be a periodic function such that f(T+x)=1+[1−3f(x)+3f2(x)−f3(x)]1/3, where T is a fixed positive number. Then period of f(x) is
A
T
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B
3T
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C
2T
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D
4T
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Solution
The correct option is C2T Given, f(T+x)=1+[1−3f(x)+3f2(x)−f3(x)]1/3, ⇒f(T+x)=1+[{1−f(x)}3]1/3 ⇒f(T+x)=1+1−f(x) ⇒f(T+x)+f(x)=2⋯(1)
substitute x→x+T, we get f(x+2T)+f(x+T)=2⋯(2)
From (1) and (2), we get f(x+2T)−f(x)=0 ⇒f(x+2T)=f(x)