Question

Which of the following functions is/are periodic?

• A. $f(x)=\{\begin{array}{l}1,x\text{ is integer }\\ 0,x\text{ is non - integer }\end{array}$
• B. $f(x)=\{\begin{array}{ll}x-[x]& ;2n\le x<2n+1\\ \frac{1}{2}& ;2n+1\le x<2n+2\end{array}$; where [. ] denotes the greatest integer function.
• C. $f(x)=(-1{)}^{[2x/\pi ]}$, where $[.]$ denotes the greatest integer function
• D. $f(x)=x-[x+3]+\tan \left(\frac{\pi x}{2}\right)$, where $[.]$ denotes the greatest integer function$

Answer 1

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

$f(x)=\{\begin{array}{l}1,x\text{ is integer }\\ 0,x\text{ is non-integer }\end{array}$
$\Rightarrow f(x+k)=\{\begin{array}{ll}1,& x+k\text{ is integer }\\ 0,& x+k\text{ is non -integer }\end{array}$
$\Rightarrow f(x+k)=f(x)$
$\Rightarrow f(x)$ is periodic function.
$\Rightarrow f(x)=\{\begin{array}{ll}x-[x],& 2n\le x<2n+1\\ 1/2,& 2n+1\le x<2n+2\end{array}$

From the graph it can be verified that period is 2 .
$\Rightarrow f(x)=(-1{)}^{\left[\frac{2x}{\pi }\right]}\Rightarrow f(x+\pi )=(-1{)}^{\left[\frac{2(\pi +x)}{\pi }\right]}=(-1{)}^{\left[\frac{2x}{\pi }\right]+2}=(-1{)}^{\left[\frac{2x}{\pi }\right]}$
$\Rightarrow f(x)=x-[x+3]+\tan \left(\frac{\pi x}{2}\right)=\{x\}-3+\tan \left(\frac{\pi x}{2}\right)$
Hence, $\{x\}$ is periodic with $1,\tan \left(\frac{\pi x}{2}\right)$ is periodic with period 2 .
Now, the LCM of 1 and 2 is 2 . Hence, the period of $f(x)$ is 2