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Q.
Which of the following functions is/are periodic?
Relations and Functions - Part 2
Solution:
$ f ( x )=\begin{cases}1, x \text { is integer } \\ 0, x \text { is non-integer }\end{cases}$
$\Rightarrow f ( x + k )=\begin{cases}1, & x + k \text { is integer } \\ 0, & x + k \text { is non -integer }\end{cases}$
$\Rightarrow f ( x + k )= f ( x )$
$\Rightarrow f ( x )$ is periodic function.
$\Rightarrow f ( x )=\begin{cases} x -[ x ], & 2 n \leq x <2 n +1 \\ 1 / 2, & 2 n +1 \leq x <2 n +2\end{cases}$
From the graph it can be verified that period is 2 .
$\Rightarrow f ( x )=(-1)^{\left[\frac{2 x }{\pi}\right]} \Rightarrow f ( x +\pi)=(-1)^{\left[\frac{2(\pi+ x )}{\pi}\right]}=(-1)^{\left[\frac{2 x }{\pi}\right]+2}=(-1)^{\left[\frac{2 x }{\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
