Let `f(x+p)=1+{2-3f(x)+3(f(x))^(2)-(f(x))^(3)}^(1//3), forall x in R`. Where `p gt 0`, prove f(x) is periodic.

Options

A. p
B. 3p
C. 2p
D. `p^(2)`

Answer

C

Explanation

We have ,
`f(x+p)=1+[2-3f(x)+3(f(x))^(2)-(f(x))^(3)]^(1//3)`
`implies f(x+p)=1+[1+{1-f(x)}^(3)]^(1//3)`
`implies f(x+p)-1=[1-{f(X)-1}^(3)]^(1//3)`
` implies g(x+p)=[1-{g(x)}^(3)]^(1/3)" "`…..(i)
where g(x) =f(x)-1.
`implies g(x+2p)=[1-{g(x+p)}^(3)]^(1//3)" "`[ On replacing x by x+p]
`implies g(x+2p)=[1-{1-g(x)}^(3)]^(1//3)" "`[Using (i)]
`implies g(x+2p)=g(x)` for all ` x in R `
`implies g(x+2p)-1=f(x)-1`
`implies f(x+2p)=f(x)`
Hence , f(x) is a period function with period 2p.