The values of \(a\) and \(b\), so that the function
\(\mathrm{f}(x)= \begin{cases}x+\mathrm{a} \sqrt{2} \sin x & , 0 \leq x \leq \frac{\pi}{4} \ 2 x \cot x+\mathrm{b} , \end{cases}\) \(\frac{\pi}{4} \leq x \leq \frac{\pi}{2} \ \mathrm{a} \cos 2 x-\mathrm{b} \sin x , \frac{\pi}{2} \lt x \leq \pi\)
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
\(\lim _{x \rightarrow \frac{\pi^{-}}{4}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{4}} \mathrm{f}(x)\)
\(\begin{aligned}
& \therefore \quad \lim _{x \rightarrow \frac{\pi}{4}}(x+a \sqrt{2} \sin x)=\lim _{x \rightarrow \frac{\pi}{4}}(2 x \cot x+b) \\
& \therefore \quad \frac{\pi}{4}+a=\frac{2 \pi}{4}+b \\
& \therefore \quad a-b=\frac{\pi}{4}...(i)
\end{aligned}\)
Also, \(\lim _{x \rightarrow \frac{\pi^{-}}{2}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} \mathrm{f}(x)\)
\(\begin{aligned}
& \therefore \quad \lim _{x \rightarrow \frac{\pi}{2}}(2 x \cot x+b)=\lim _{x \rightarrow \frac{\pi}{2}}(a \cos 2 x-b \sin x) \\
& \therefore \quad 0+b=-a-b \\
& \therefore \quad a+2 b=0...(ii)
\end{aligned}\)
Solving equations (i) and (ii), we get \(a=\frac{\pi}{6}\) and \(b=\frac{-\pi}{12}\)