Question

Let $f:R\to R$ be a periodic function such that
$f(T+x)=1+{[1-3f(x)+3f^2(x)-f^3(x\left)\right]}^{1/3},$ where $T$ is a fixed positive number. Then period of $f\left(x\right)$ is

• A. $T$
• B. $3T$
• C. $2T$
• D. $4T$

Answer 1

The correct option is C $2T$

Given, $f(T+x)=1+{[1-3f(x)+3f^2(x)-f^3(x\left)\right]}^{1/3},$
$\Rightarrow f(T+x)=1+{\left[\right\{1-f\left(x\right){\}}^3]}^{1/3}$
$\Rightarrow f(T+x)=1+1-f\left(x\right)$
$\Rightarrow f(T+x)+f\left(x\right)=2\cdots \left(1\right)$
substitute $x\to x+T,$ we get
$f(x+2T)+f(x+T)=2\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we get
$f(x+2T)-f\left(x\right)=0$
$\Rightarrow f(x+2T)=f\left(x\right)$

Hence, period of function is $2T$.