The range of $\mathrm{f}\left(\mathrm{x}\right)={\cot }^{-1}\left({\log }_{4/5}^{(5{\mathrm{x}}^2-8\mathrm{x}+4)}\right) \mathrm{is}$

$$\begin{gathered} 5{\mathrm{x}}^2-8\mathrm{x}+4=5\left[\left({\mathrm{x}}^2-\frac{8\mathrm{x}}{5}\right)+\frac{4}{5}\right] \\ =5{\left(\mathrm{x}-\frac{4}{5}\right)}^2+\frac{4}{5} \\ \therefore \frac{4}{5}\le 5{\mathrm{x}}^2-8\mathrm{x}+5<\infty \\ \Rightarrow {\log }_{4/5}^{4/5}\ge {\log }_{4/5}\left(5{\mathrm{x}}^2-8\mathrm{x}+5\right)>\infty \\ \mathrm{taking} {\cot }^{-1} \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ \Rightarrow co{t}^{-1}\left(1\right)\le {\cot }^{-1}{\log }_{4/5}\left(5{\mathrm{x}}^2-8\mathrm{x}+5\right)<co{t}^{-1}(\infty ) \\ \frac{\mathrm{\pi }}{4}\le {\cot }^{-1}{\log }_{4/5}\left(5{\mathrm{x}}^2-8\mathrm{x}+5\right)<\mathrm{\pi } \end{gathered}$$