Question

Let $f(x+p)=1+{\{2-3f(x)+3(f\left(x\right){)}^2-\left(f\right(x\left){)}^3\right\}}^{1/3},\forall x\in R$, where $p$ > 0, then

• A. $f\left(x\right)$ is periodic with period $p$
• B. $f\left(x\right)$ is periodic with period $2p$
• C. $f\left(x\right)$ is periodic with period $4p$
• D. $f\left(x\right)$ is not periodic

Answer 1

The correct option is B $f\left(x\right)$ is periodic with period $2p$

Since, $f(x+p)=1+{\{2-3f\left(x\right)+3{\left(f\left(x\right)\right)}^2-{\left(f\left(x\right)\right)}^3\}}^{\frac{1}{3}}$

$⟹f(x+p)=1+{\{1+{(1-f\left(x\right))}^3\}}^{\frac{1}{3}}$

$⟹f(x+p)-1={\{1+{(1-f\left(x\right))}^3\}}^{\frac{1}{3}}$

$⟹f(x+p)-1={\{1-{(f\left(x\right)-1)}^3\}}^{\frac{1}{3}}$

Now, take $f(x+p)-1=g\left(x\right),$ we have

$g(x+p)={(1-(g\left(x\right){)}^3)}^{\frac{1}{3}}$

$g(x+2p)={(1-(g(x+p){)}^3)}^{\frac{1}{3}}$

$⟹g(x+2p)=[1-1+(g\left(x\right){)}^3{]}^{\frac{1}{3}}$

$⟹g(x+2p)=g\left(x\right)$

$⟹f(x+2p)-1=f\left(x\right)-1$

$⟹f(x+2p)=f\left(x\right)$.