$\mathrm{f}(x)$ is continuous in $0 \leq x \leq \pi$
$\therefore \quad \mathrm{f}(x)$ is continuous at $x=\frac{\pi}{4}$.
$\begin{aligned}
\therefore \quad & \lim _{x \rightarrow \frac{\pi^{-}}{4}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{4}} \mathrm{f}(x) \\
& \Rightarrow \lim _{x \rightarrow \frac{\pi}{4}^{-}}(x+\mathrm{a} \sqrt{2} \sin x)=\lim _{x \rightarrow \frac{\pi^{+}}{4}}(2 x \cot x+\mathrm{b}) \\
& \Rightarrow \frac{\pi}{4}+\mathrm{a} \sqrt{2} \times \sin \frac{\pi}{4}=2 \times \frac{\pi}{4} \cdot \cot \frac{\pi}{4}+\mathrm{b} \\
& \Rightarrow \frac{\pi}{4}+\mathrm{a}=\frac{\pi}{2}+\mathrm{b} \\
& \Rightarrow \mathrm{a}-\mathrm{b}=\frac{\pi}{4}... (i)
\end{aligned}$
Also, $\mathrm{f}(x)$ is continuous at $x=\frac{\pi}{2}$.
$\begin{aligned}
\therefore \quad & \lim _{x \rightarrow \frac{\pi^{-}}{2}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} \mathrm{f}(x) \\
& \Rightarrow \lim _{x \rightarrow \frac{\pi^{-}}{2}}(2 x \cot x+\mathrm{b})=\lim _{x \rightarrow \frac{\pi^{+}}{2}}(\mathrm{a} \cos 2 x-\mathrm{b} \sin x) \\
& \Rightarrow 2 \times \frac{\pi}{2} \times \cot \frac{\pi}{2}+\mathrm{b}=\mathrm{a} \cos \left(2 \times \frac{\pi}{2}\right)-\mathrm{b} \sin \frac{\pi}{2} \\
& \Rightarrow \mathrm{b}=-\mathrm{a}-\mathrm{b} \\
& \Rightarrow \mathrm{a}=-2 \mathrm{~b}
\end{aligned}$
Substituting $a=-2 b$ in equation (i), we get
$\begin{aligned}
& -2 b-b=\frac{\pi}{4} \\
& \Rightarrow b=\frac{-\pi}{12} \\
& \Rightarrow a=-2 \times \frac{-\pi}{12}=\frac{\pi}{6}
\end{aligned}$
Now, $\begin{aligned} 2 \mathrm{a}+3 \mathrm{~b} & =2\left(\frac{\pi}{6}\right)+3\left(\frac{-\pi}{12}\right) \\ & =\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}\end{aligned}$