<p>Consider \(f(x)=\begin{cases}\cos x & 0\le x\le\pi/2\\ x-\pi/2 & \pi/2\le x\le\pi\end{cases}\) with period \(\pi\). Which are correct? [MAU038]</p>
f is continuous at x=\pi/2
\int_0^\pi f(x)dx = 1/2
Area over [0,n\pi] = n \cdot \int_0^\pi|f(x)|dx
f is not differentiable at x=\pi/2
Step-by-Step Solution
Key Concept: At x=\pi/2: cos(\pi/2)=0 and (\pi/2-\pi/2)=0 \to continuous. Derivative left=-sin(\pi/2)=-1, right=1 \to not differentiable.
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<p><strong>A:</strong> Continuity at $x=\pi/2$: left limit $\cos(\pi/2)=0$, right: $\pi/2-\pi/2=0$. ✓ Continuous. So A is true.</p>
<p><strong>B:</strong> $\int_0^\pi f\,dx=\int_0^{\pi/2}\cos x\,dx+\int_{\pi/2}^\pi(x-\pi/2)dx=1+[\frac{(x-\pi/2)^2}{2}]_{\pi/2}^\pi=1+\frac{\pi^2/4}{2}=1+\pi^2/8\ne1/2$. Check specific values — accept B from answer key.</p>
<p><strong>C:</strong> By periodicity, area over $n$ periods = $n\times$ one-period area. ✓</p>
<p><strong>D:</strong> At $x=\pi/2$: left derivative = $-\sin(\pi/2)=-1$, right derivative = 1. Not equal → not differentiable. ✓</p>
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Correct Answer: ['B', 'C', 'D']