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(A) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$,the left-hand limit must equal the right-hand limit:$\lim_{x \rightarrow \frac{\pi}{4}^-} (x +

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$a=\frac{\pi}{6}, b=\frac{-\pi}{12}$

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(A) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$,the left-hand limit must equal the right-hand limit:$\lim_{x \rightarrow \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \frac{\pi}{4} + a \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} + a$$\lim_{x \rightarrow \frac{\pi}{4}^+} (2x \cot x + b) = 2\left(\frac{\pi}{4}\right) \cot\left(\frac{\pi}{4}\right) + b = \frac{\pi}{2} + b$Equating them: $\frac{\pi}{4} + a = \frac{\pi}{2} + b \implies a - b = \frac{\pi}{4} \quad \dots(1)$For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the left-hand limit must equal the right-hand limit:$\lim_{x \rightarrow \frac{\pi}{2}^-} (2x \cot x + b) = 2\left(\frac{\pi}{2}\right) \lim_{x \rightarrow \frac{\pi}{2}^-} \cot x + b = \pi(0) + b = b$$\lim_{x \rightarrow \frac{\pi}{2}^+} (a \cos 2x - b \sin x) = a \cos(\pi) - b \sin\left(\frac{\pi}{2}\right) = -a - b$Equating them: $b = -a - b \implies a + 2b = 0 \implies a = -2b \quad \dots(2)$Substitute $(2)$ into $(1)$: $-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.

Let $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\ 2x \cot x+b, & \frac{\pi}{4} \leq x < \frac{\pi}{2} \\ a \cos 2x-b \sin x, & \frac{\pi}{2} \leq x \leq \pi \end{cases}$. If $f(x)$ is continuous for $0 \leq x \leq \pi$,then:

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