Which of the following function is/are periodic?

Moderate

A

$f (x) = \{\begin{cases} 1, x is rational \\ 0, x is irrational \end{cases}$
B

$f (x) = \{\begin{array}{cc} x - [x]; & 2 n \leq x < 2 n + 1 \\ \frac{1}{2}; & 2 n + 1 \leq x < 2 n + 2 \end{array}$
where [.] denotes the greatest integer function, $n \in Z$
C

$f (x) = (- 1)^{[\frac{2 x}{π}]}$ .where [.] denotes the greatest integer function
D

$f (x) = x - [x + 3] + \tan (\frac{πx}{2})$ where [.] denotes the greatest integer function and a is a rational number

Solution

$\begin{array}{l} f (x) = \{\begin{matrix} 1, x is rational \\ 0, x is irrational \end{matrix} \\ or f (x + k) = \{\begin{matrix} 1, x + k is rational \\ 0, x + k is irrational \end{matrix} \end{array}$
where k is any rational number
$= \{\begin{array}{l} 1, x is rational \\ 0, x is irrational \end{array} = f (x)$
Therefore, f (x) is periodic function, but its fundamental period cannot be determined. Thus,
$f (x) = \{\begin{array}{lc} x - [x], & 2 n \leq x < 2 n + 1 \\ 1 / 2, & 2 n + 1 \leq x < 2 n + 2 \end{array}$
Draw the graph from which it can be verified that period is 2.

$f (x) = (- 1)^{[\frac{2 x}{π}]}$
or $f (x + π) = (- 1)^{[\frac{2 (π + x)}{π}]} = (- 1)^{[\frac{2 x}{π}] + 2} = (- 1)^{[\frac{2 x}{π}]} .$
Hence, the period is $π$ .
$f (x) = x - [x + 3] + \tan (\frac{πx}{2}) = {x} - 3 + \tan (\frac{πx}{2})$
{x} is periodic with period 1; tan $(\frac{πx}{2}) x$ is periodic with period 2.
Now, the LCM of I and 2 is 2. Hence, the period of f(x) is 2.

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