If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ be a function defined by   $\mathrm{f}\left(\mathrm{x}\right)={\log }_{\sqrt{\mathrm{m}}}\left\{\sqrt{2}\right(\sin \mathrm{x}-\cos \mathrm{x})+\mathrm{m}-2\}$ for some $m$, such that
the range of $f$ is $\left[0,2\right]$. Then the value of $m$ is ……

$\mathrm{f}\left(\mathrm{x}\right)={\log }_{\sqrt{\mathrm{m}}}\left(\sqrt{2}\right(\sin \mathrm{x}-\cos \mathrm{x})+\mathrm{m}-2)$

$\mathrm{range} \mathrm{of} \sqrt{2}(\sin \mathrm{x}-\cos \mathrm{x})=\sqrt{2}[-\sqrt{2},\sqrt{2}]=[-2,2]$

$\mathrm{range} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right)={\log }_{\sqrt{\mathrm{m}}}\left(\right[-2,2]+\mathrm{m}-2)=={\log }_{\sqrt{\mathrm{m}}}[-4+\mathrm{m}]\in [0,2]$

$\begin{array}{l}{\log }_{\sqrt{m}}(-4+\mathrm{m})=0\Rightarrow -4+\mathrm{m}=1\Rightarrow \mathrm{m}=5\end{array}$