Let f be a real valued function with domain R satisfying f (x + k) = 1 + [2 – 5 f (x) + 10 { f (x)}2 – 10 { f (x)}3 + 5 { f (x)}4 – { f (x)}5]1/5for all real x and some positive constant k, then the period of the function f (x) is

We have,f (x + k) = 1 + [1 + {1 – f (x)}5]1/5⇒ f (x + k) – 1 = [1 – ( f (x) – 1)5]1/5⇒ϕ(x+k)=1−{ϕ(x)}51/5 where  ϕ(x)=f(x)−1⇒ϕ(x+2k)=1−{ϕ(x+k)}51/5⇒ϕ(x+2k)=1−1−(ϕ(x))51/5=ϕ(x),∀x∈R⇒ f (x + 2k) – 1 = f (x) – 1⇒ f (x + 2k) = f (x), ∀ x ∈ R∴ f (x) is periodic with period 2k.