Question

Let $f(x+p)=1+{[2-3f\left(x\right)+3{\left(f\left(x\right)\right)}^2-{\left(f\left(x\right)\right)}^3]}^{1/3},\forall x\in R$, where $p>0$. Then, $f\left(x\right)$ is periodic with period

• A. $p$
• B. $2p$
• C. $4p$
• D. None of these

Answer 1

Answer: (B)

$2p$

We have, $f(x+p)=1+{[2-3f\left(x\right)+3{\left\{f\left(x\right)\right\}}^2-{\left\{f\left(x\right)\right\}}^2]}^{1/3}$

$\Rightarrow f(x+p)=1+{[1+{\{1-f\left(x\right)\}}^3]}^{1/3}$

$\Rightarrow f(x+p)-1={[1-{\{f\left(x\right)-1\}}^3]}^{1/3}$

$\Rightarrow g(x+p)={[1-{\left\{g\left(x\right)\right\}}^3]}^{1/3}$ ...(1)

where $g\left(x\right)=f\left(x\right)-1$ and $g(x+p)=f(x+p)-1$

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

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

From (1) and (2), we have

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

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

$\Rightarrow g(x+2p)={\left[{\left\{g\left(x\right)\right\}}^3\right]}^{1/3}$

$\Rightarrow g(x+2p)=g\left(x\right)$

which shows $f(x+2p)-1=f\left(x\right)-1$

$f(x+2p)=f\left(x\right)$

Hence, $f\left(x\right)$ is periodic with period $2p$.