Let f be a real valued function with domain R satisfying f(x+k)=1+[2−5f(x)+10{f(x)}2−10{f(x)3+5{f(x)}4−{f(x)}5}]1/5 For 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)}5]1/5,    where  ϕ(x)=f(x)−1 ⇒         ϕ(x+2k)=[1−{ϕ(x+k)5}]1/5 ⇒          ϕ(x+2k)=[1−{1−(ϕ(x))5}]1/5=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