The values of \(a\) and \(b\), so that the function
\(
f(x)=\left\{\begin{array}{l}
x+\mathrm{a} \sqrt{2} \sin x, 0 \leq x \leq \frac{\pi}{4} \
2 x \cot x+\mathrm{b}, \frac{\pi}{4} \leq x
\end{array}\right.
\) \(\leq \frac{\pi}{2} \
\operatorname{acos} 2 x-\mathrm{b} \sin x, \frac{\pi}{2} < x \leq \pi\)
is continuous for \(0 \leq x \leq \pi\), are respectively given by

As the given function is continuous at \(x=\frac{\pi}{4}\) and \(\frac{\pi}{2}\), we get
\(\lim _{x \rightarrow \frac{\pi^{-}}{4}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{4}} \mathrm{f}(x) \)
\( \therefore \lim _{x \rightarrow \frac{\pi}{4}}(x+\mathrm{a} \sqrt{2} \sin x)=\lim _{x \rightarrow \frac{\pi}{4}}(2 x \cot x+\mathrm{b}) \)
\( \therefore \frac{\pi}{4}+\mathrm{a}=\frac{2 \pi}{4}+\mathrm{b} \)
\( \therefore \mathrm{a}-\mathrm{b}=\frac{\pi}{4} \)
\( \text { Also, } \lim _{x \rightarrow \frac{\pi^{-}}{2}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} \mathrm{f}(x) \)
\( \lim _{x \rightarrow \frac{\pi}{2}} 2 x \cot x+\mathrm{b}=\lim _{x \rightarrow \frac{\pi}{2}} \mathrm{a} \cos 2 x-\mathrm{b} \sin x \)
\( \therefore 0+\mathrm{b}=-\mathrm{a}-\mathrm{b} . \)
\( \therefore \mathrm{a}+2 \mathrm{~b}=0 \)
Solving equations (i) and (ii), we get \(a=\frac{\pi}{6}\) and \(b=\frac{-\pi}{12}\)