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\[f(x) = \left\{ {\begin{array}{*{20}{c}}

{1,}&{x{\text{ is rational}}}\\

{0,}&{x{\text{ is irrational}}}

\end{array}\;\;\;\;\;is} \right.\]

A.1

B.2

C.non-periodic

D.periodic but having no fundamental period

Answer
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Given that:

\[f(x) = \left\{ {\begin{array}{*{20}{c}}

{1,}&{x{\text{ is rational}}}\\

{0,}&{x{\text{ is irrational}}}

\end{array}\;\;\;\;\;is} \right.\]

We know there will be infinite points between two numbers, whether the numbers are irrational or rational.

Hence there would be an irrational number in the immediate neighborhoods of a rational number and there would be rational numbers in the immediate neighborhood of an irrational number.

Hence the function will take the value of 1 and 0 periodically.

However, since there are infinite points on the numbers lines and still infinite points between any two numbers, the period of the function cannot be determined.

Hence, we also know that a constant function is a period function without any fundamental period.

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