Question

Which of the following is/are periodic?

• A. $f\left(x\right)=\{\begin{array}{cc}1,& x\text{is rational}\\ 0,& x\text{is irrational}\end{array}$
• B. $f\left(x\right)=\{\begin{array}{cc}x-\left[x\right];& 2n\le x\le 2n+1\\ \frac{1}{2};& 2n+1\le x\le 2n+2\end{array},n\in Z$
• C. $f\left(x\right)=(-1{)}^{\left[\frac{2x}{\pi }\right]}$
• D. $f\left(x\right)=x-[x+3]+\tan \left(\frac{nx}{2}\right)$

Answer 1

Answer: (A), (B), (C), (D)

$f\left(x\right)=x-[x+3]+\tan \left(\frac{nx}{2}\right)$

$f\left(x\right)=\{\begin{array}{cc}1,& x\text{is rational}\\ 0,& x\text{is irrational}\end{array}$

$f(x+k)=\{\begin{array}{cc}1,& x+k\text{is rational}\\ 0,& x+k\text{is irrational}\end{array}$

$\Rightarrow f(x+k)=\{\begin{array}{cc}1,& x\text{is rational}\\ 0,& x\text{is irrational}\end{array}=f\left(x\right)$

$\therefore f\left(x\right)$ is periodic.

$f\left(x\right)=\{\begin{array}{cc}x-\left[x\right];& 2n\le x\le 2n+1\\ \frac{1}{2};& 2n+1\le x\le 2n+2\end{array},n\in Z$

$f\left(x\right)=\{\begin{array}{cc}x+2;& -2\le x<-1\\ \frac{1}{2};& -1\le x<0\end{array}$

$f\left(x\right)=\{\begin{array}{cc}x;& 0\le x<1\\ \frac{1}{2};& 1\le x<2\end{array}$

$f\left(x\right)=\{\begin{array}{cc}x-2;& 2\le x<3\\ \frac{1}{2};& 3\le x<4\end{array}$


$\therefore f\left(x\right)$ is periodic with period $2$.

$f\left(x\right)=(-1{)}^{\left[\frac{2x}{\pi }\right]}$
$f(x+\pi )=(-1{)}^{\left[\frac{2(x+\pi )}{\pi }\right]}$
$=(-1{)}^{\left[\frac{2x}{\pi }\right]+2}$
$=(-1{)}^{\left[\frac{2x}{\pi }\right]}\times (-1{)}^2$
$=(-1{)}^{\left[\frac{2x}{\pi }\right]}$
$=f\left(x\right)$
$\therefore f\left(x\right)$ is periodic with period $\pi$.

$f\left(x\right)=x-[x+3]+\tan \left(\frac{nx}{2}\right)$
$\tan \left(\frac{nx}{2}\right)$ is periodic with period $2$.
$x-[x+3]=x-\left(\right[x]-3)=x-\left[x\right]+3=\left\{x\right\}+3$
Period of $\left\{x\right\}$ is $1$.
$\therefore$ Period of $f\left(x\right)$ is $2$. (LCM of $(2,1)$)