Question

The function defined by $f(x+\lambda )=1+\sqrt{2f\left(x\right)-f^2\left(x\right)},\forall \in R$ is

• A. not periodic
• B. periodic with period $2\lambda$
• C. periodic with period $4\lambda$
• D. periodic with period $\lambda$

Answer 1

Answer: (B)

periodic with period $2\lambda$

$f(x+\lambda )=1+\sqrt{2f\left(x\right)-f^2\left(x\right)}=1+\sqrt{1-{(f\left(x\right)-1)}^2}f(x+2\lambda )=1+\sqrt{1-{(f(x+\lambda )-1)}^2}=1+\sqrt{1-{(1-(1+\sqrt{1-{(f\left(x\right)-1)}^2}))}^2}=1+\sqrt{1-{(-\left(\sqrt{1-{(f\left(x\right)-1)}^2}\right))}^2}=1+\sqrt{1-{\left(\left(\sqrt{1-{(f\left(x\right)-1)}^2}\right)\right)}^2}=1+\sqrt{1-(1-{(f\left(x\right)-1)}^2)}=1+\sqrt{{(f\left(x\right)-1)}^2}=1+f\left(x\right)-1=f\left(x\right)$

$\therefore$ Period of the given function is $2\lambda$