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Question

If f(x)ϵ[1,2] when xϵR and for a fixed positive real number p,
f(x+p)=1+2f(x)|f(x)|2 for all xϵR then prove that f(x) is a periodic function.

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Solution

|f(x)|2=(f(x))2
So, f(x+p)=1+2f(x)(f(x))2

f(x+2p)=1+2f(x+p)(f(x+p))2=1+2(1+2f(x)(f(x))2)(1+2f(x)(f(x))2)2
f(x+2p)=1+2+22f(x)(f(x))2(1+(2f(x)(f(x))2)2+22f(x)(f(x))2)
f(x+2p)=1+12f(x)+(f(x)2)=1+(f(x)1)2=1+|f(x)1|

As f(x)[1,2], |f(x)1|=f(x)1

ie, f(x+2p)=f(x)
So, f(x) is a periodic function with a period of 2p

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