Question

Which of the following statements are true?

• A. If f(x) be continuous and periodic with periodicity T, Then \(I=\int_{a}^{a+T}f(x)dx\) depend on 'a'.
• B. If f(x) be continuous and periodic with periodicity T, Then \(I=\int_{a}^{a+T}f(x)dx\) does not depend on 'a'.
• C. Let f(x)=\(\begin{Bmatrix} 1 &if \,x\, \,is \,rational \\   0&if \,x\, \,is \,irrational  \end{Bmatrix}\), then f is periodic of the periodicity T only if T is rational
• D. f defined in (C) is periodic for all T.

Answer 1

The Correct Option is B, C

Definition: A function f(x) is called periodic with period T if: $f(x + T) = f(x)$ for all $x$  

Part 1: About the integral We consider the integral: $I = \int_a^{a+T} f(x) \, dx$ 
Since $f(x)$ is periodic with period $T$, this integral over one full period is the same no matter where it starts: $\int_a^{a+T} f(x) \, dx = \int_0^T f(x) \, dx$ 
Therefore, I is independent of the choice of a So, Option (B) is correct 

Part 2: Consider the function $f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \notin \mathbb{Q} \end{cases}$ This is the Dirichlet function. 

Let us examine when it is periodic: - Suppose $T$ is rational, then for any $x$, $x + T$ is rational if $x$ is rational and irrational if $x$ is irrational. - Hence, $f(x + T) = f(x)$ So, function is periodic if T is rational 
If $T$ is irrational, this property fails. 

Therefore, Option (C) is also correct 

Final Answer: (B) and (C)