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Let $\mathrm{f}(\mathrm{x}) \begin{cases}=\mathrm{x}+\mathrm{a} \sqrt{2} \sin \mathrm{x}, & 0 \leq \mathrm{x} < \frac{\pi}{4} \\ =2 \mathrm{x} \cot \mathrm{x}+\mathrm{b}, & \frac{\pi}{4} \leq \mathrm{x} < \frac{\pi}{2} \\ =\mathrm{a} \cos 2 \mathrm{x}-\mathrm{b} \sin \mathrm{x}, & \frac{\pi}{2} \leq \mathrm{x} \leq \pi\end{cases}$ If $\mathrm{f}(\mathrm{x})$ is continuous for $0 \leq \mathrm{x} \leq \pi$, then
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$a=\frac{\pi}{6}, b=\frac{-\pi}{12}$
$\lim _{x \rightarrow \frac{\pi^{-}}{4}} f(x)=\lim _{x \rightarrow \frac{\pi^{-}}{4}} x+a \sqrt{2} \sin x$
$\lim _{x \rightarrow \frac{\pi^{+}}{4}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{4}} 2 x \cot x+b$
$\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{-}}{2}} 2 x \cot x+b$
$\begin{aligned} & \lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} a \cos 2 x-b \sin x \\ & =a \cos 2\left(\frac{\pi}{2}\right)-b \sin \left(\frac{\pi}{2}\right)\end{aligned}$
Since $f(x)$ is continuous at $\frac{\pi}{4}$ and $\frac{\pi}{2}$, we write
$\frac{\pi}{4}+\mathrm{a}=\frac{\pi}{2}+\mathrm{b} \quad \ldots[$ From (1) and (2) $]$
$\mathrm{b}=-\mathrm{a}-\mathrm{b} \quad \ldots[$ From (3) and (4) $]$
$\lim _{x \rightarrow \frac{\pi^{+}}{4}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{4}} 2 x \cot x+b$
$\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{-}}{2}} 2 x \cot x+b$
$\begin{aligned} & \lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} a \cos 2 x-b \sin x \\ & =a \cos 2\left(\frac{\pi}{2}\right)-b \sin \left(\frac{\pi}{2}\right)\end{aligned}$
Since $f(x)$ is continuous at $\frac{\pi}{4}$ and $\frac{\pi}{2}$, we write
$\frac{\pi}{4}+\mathrm{a}=\frac{\pi}{2}+\mathrm{b} \quad \ldots[$ From (1) and (2) $]$
$\mathrm{b}=-\mathrm{a}-\mathrm{b} \quad \ldots[$ From (3) and (4) $]$
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