$\text{ Find the range of }f\left(x\right)={\cot }^{-1}\left(2x-x^2\right)\text{ . }$

$\begin{array}{l}\text{ Let }\theta ={\cot }^{-1}\left(2x-x^2\right),\text{ where }\theta \in (0,\pi )\\ \text{⇒}\cot \theta =2x-x^2,\text{ where }\theta \in (0,\pi )\\ \text{⇒}\cot \theta =1-\left(1-2x+x^2\right),\text{ where }\theta \in (0,\pi )\\ \text{ ⇒}\cot \theta =1-(1-x{)}^2,\text{ where }\theta \in (0,\pi )\\ \text{⇒}\cot \theta \le 1,\text{ where }\theta \in (0,\pi )\\ \begin{array}{rl} & \text{⇒ }\frac{\pi }{4}\le \theta <\pi \\ & \text{ So, the range of }f\left(x\right)\text{ is }\left[\frac{\pi }{4},\pi \right)\text{ . }\end{array}\end{array}$