Question

Consider a function, f : R $\to$ R such that f(x + a) = $\frac{1}{2}+\sqrt{f\left(x\right)-f^2\left(x\right)}$, a is a real constant. If f(x) is periodic then its period can be

• A. $\frac{a}{2}$
• B. a
• C. 2a
• D. 8a

Answer 1

The correct option is C 2a

We have $f(x+a)\ge \frac{1}{2}$ and so $f\left(x\right)\ge \frac{1}{2}\forall xϵR$
Let's set, $g\left(x\right)=f\left(x\right)-\frac{1}{2}$ we have $g\left(x\right)\ge 0\forall xϵR$
$\Rightarrow g(x+a)=\sqrt{\frac{1}{4}-\left(g\right(x){)}^2}$
$\Rightarrow g^2(x+a)=\frac{1}{4}-g^2\left(x\right)\Rightarrow g^2$$(x+2a)=\frac{1}{4}-g^2(x+a)$
$\Rightarrow g^2(x+2a)=g^2\left(x\right)$
$\Rightarrow g(x+2a)=g\left(x\right)$
$\Rightarrow f(x+2a)=f\left(x\right)$