Question

Let ${}^{\prime }{f}^{\prime }$ be a real valued function such that $0\le f\left(x\right)\le \frac{1}{2}$ and for some fixed $a>0,f(x+a)=\frac{1}{2}-\sqrt{f\left(x\right)-\left(f\right(x){)}^2},\forall x\in R$, then the period of the function $f\left(x\right)$ is

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

Answer 1

Answer: (C), (D)

$4a$

$f(x+a)=\frac{1}{2}-\sqrt{f\left(x\right)-\left(f\right(x){)}^2}\dots \left(1\right)$
$f(x+a+a)=\frac{1}{2}-\sqrt{f(x+a)-\left(f\right(x+a){)}^2}$
$=\frac{1}{2}-\sqrt{\frac{1}{2}-\sqrt{f\left(x\right)-\left(f\right(x){)}^2}-(\frac{1}{4}+f(x)-(f\left(x\right){)}^2-\sqrt{f\left(x\right)-\left(f\right(x){)}^2})}$
$=\frac{1}{2}-\sqrt{\left(f\right(x){)}^2-f(x)+\frac{1}{4}}$
$=\frac{1}{2}-\sqrt{{\left(f\right(x)-\frac{1}{2})}^2}$
$=\frac{1}{2}+f\left(x\right)-\frac{1}{2}$$(\because \frac{1}{2}-f(x\left)\right)\ge 0$
$\Rightarrow f(x+2a)=f\left(x\right)$
Hence $f\left(x\right)$ will be periodic whose fundamental period is $2a$.
Thus period of this function will be in multiples of $2a.$