Question

Let \( f: \mathbb{R} \to \mathbb{R} \) be the function given by \[ f(x) = \begin{cases} \cos x, & \text{if } x \in \mathbb{Q} \\ 0, & \text{if } x \notin \mathbb{Q} \end{cases} \]
Which of the following statements is/are TRUE?

• A. \( f(x) \) is continuous at 0.
• B. \( f(x) \) is continuous at \( \frac{\pi}{2} \).
• C. \( f(x) \) is Riemann integrable on \( [0, 1] \) and \( \int_0^1 f(x) \, dx = \sin 1 \).
• D. \( f(x) \) is Riemann integrable on \( [0, 1] \) and \( \int_0^1 f(x) \, dx = 0 \).

Hint:

A function can be Riemann integrable even if it is discontinuous at some points, as long as the set of discontinuities has measure zero.

Answer 1

The Correct Option is B

Step 1: Analyze continuity at \( x = 0 \).
The function \( f(x) \) is discontinuous at any point, as the limit from rationals and irrationals does not agree. Therefore, \( f(x) \) is not continuous at 0. Hence, option (A) is false.

Step 2: Analyze continuity at \( x = \frac{\pi}{2} \).
The function \( f(x) \) is discontinuous at \( x = \frac{\pi}{2} \) because the limit does not exist for both rationals and irrationals. Hence, option (B) is false.

Step 3: Analyze Riemann integrability on \( [0, 1] \).
The function is not continuous anywhere on \( [0, 1] \) but has a bounded set of discontinuities, which makes it Riemann integrable. Since the function is 0 almost everywhere except at rationals, the integral is 0. Hence, option (D) is true.

Step 4: Conclusion.
The correct answer is (D) since \( f(x) \) is Riemann integrable on \( [0, 1] \) and the integral equals 0.