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Question

Let f:RR be a periodic function such that
f(T+x)=1+[13f(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 C 2T
Given, f(T+x)=1+[13f(x)+3f2(x)f3(x)]1/3,
f(T+x)=1+[{1f(x)}3]1/3
f(T+x)=1+1f(x)
f(T+x)+f(x)=2 (1)
substitute xx+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)

Hence, period of function is 2T.

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