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

\(\lim _{x \rightarrow \frac{\pi^{-}}{4}} f(x)=\lim _{x \rightarrow \frac{\pi^{-}}{4}} x+a \sqrt{2} \sin x\)


\(\lim _{x \rightarrow \frac{\pi^{+}}{4}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{4}} 2 x \cot x+b\)

\(\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{-}}{2}} 2 x \cot x+b\)
\(\begin{aligned} & \lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} a \cos 2 x-b \sin x \\ & =a \cos 2\left(\frac{\pi}{2}\right)-b \sin \left(\frac{\pi}{2}\right)\end{aligned}\)

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) \(]\)