Discussion
Let \(f: \mathbb{R} \to \mathbb{R}\) be the function given by
\[ f(x) = \begin{cases} \cos x, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q}. \end{cases} \]
Which of the following statements is/are TRUE?
Answer: B
Only one statement is true: \(f(x)\) is continuous at \(\frac{\pi}{2}\). The final key accepts this option alone.
At \(\frac{\pi}{2}\), which is irrational, \(f\left(\frac{\pi}{2}\right) = 0\). For every \(x\), \(|f(x)| \leq |\cos x|\), and \(\cos x \to 0\) as \(x \to \frac{\pi}{2}\), so \(f(x) \to 0\): continuous.
At 0, which is rational, \(f(0) = \cos 0 = 1\), but irrational points arbitrarily close to 0 give \(f = 0\). So \(f\) is not continuous at 0.
On \([0,1]\), \(\cos x \geq \cos 1 \approx 0.54 \gt 0\). Every subinterval contains rationals, where \(f \geq \cos 1\), and irrationals, where \(f = 0\). So every upper Darboux sum is at least \(\cos 1\) while every lower sum is 0, and \(f\) is not Riemann integrable on \([0,1]\). Both integrability statements fail, whatever value they claim.
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