If the function$f\left(x\right)=$ $\left\{\begin{array}{l}x+a^2\sqrt{2}\sin x, 0\le x<\frac{\pi }{4}\\ x\cot x+b,\frac{\pi }{4}\le x<\frac{\pi }{2}\\ b\sin 2x-a\cos 2x,\frac{\pi }{2}\le x\le \pi \end{array}\right.$   is continuous in the interval $[0,\pi ],$ then the values of $(a,b)$ are

Since f is continuous at $x=\frac{\pi }{4}\therefore f\left(\frac{\pi }{4}\right)=\underset{{ }_{h\to 0}}{f}(\frac{\pi }{4}+h)=\underset{h\to 0}{f}(\frac{\pi }{4}-h)$

$\Rightarrow \frac{\pi }{4}\cot \frac{\pi }{4}+b=\underset{h\to 0}{f}(\frac{\pi }{4}+h)+a^2\sqrt{2}\sin (\frac{\pi }{4}+h)$ $\Rightarrow \frac{\pi }{4}\left(1\right)+b=(\frac{\pi }{4}+0)+a^2\sqrt{2}\sin (\frac{\pi }{4}+0)$

$\Rightarrow \frac{\pi }{4}+b=\frac{\pi }{4}+a^2\sqrt{2}\sin \frac{\pi }{4}$$\Rightarrow b=a^2\sqrt{2}\frac{1}{\sqrt{2}}\Rightarrow b=a^2$ Also as f is continuous at $x=\frac{\pi }{2}$

$\therefore f\left(\frac{\pi }{2}\right)=\underset{x\to \frac{\pi }{2}-0}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(\frac{\pi }{2}-h)$ 

$\Rightarrow bsin2\frac{\pi }{2}-acos2\frac{\pi }{2}=\underset{h\to 0}{lim}[(\frac{\pi }{2}-h)cot(\frac{\pi }{2}-h)+b]$

$\Rightarrow b.0-a(-1)=0+b\Rightarrow a=b$

Hence $\left(0,0\right) \left(1,1\right)$ satisfy the above relations.