Let $4{\mathrm{x}}^2-4(\mathrm{\alpha }-2)\mathrm{x}+\mathrm{\alpha }-2=0,(\mathrm{\alpha }\in \mathrm{R})$) be a quadratic equation. The set of values of α of which

$\begin{array}{l}\text{ A) }\frac{4(\mathrm{\alpha }-2)}{4}>\frac{1}{2}+\frac{1}{2}\Rightarrow \mathrm{\alpha }>3\Rightarrow \therefore \mathrm{\alpha }\in (3\propto )\\ \left(\mathrm{\Delta }=16(\mathrm{\alpha }-2{)}^2-16(\mathrm{\alpha }-2)\right.\\ =16(\mathrm{\alpha }-2)(\mathrm{\alpha }-2-1)=16(\mathrm{\alpha }-2)(\mathrm{\alpha }-3)\ge 0\text{; }\\ \mathrm{\alpha }\le 2\text{ or }\mathrm{\alpha }\ge 3\text{ ) }\\ \text{ B) }\therefore \mathrm{f}\left(0\right)<0,\mathrm{f}\left(\frac{1}{2}\right)>0\Rightarrow 1-4(\mathrm{\alpha }-2)\frac{1}{2}+\mathrm{\alpha }-2>0\\ \Rightarrow \mathrm{\alpha }<3\therefore \mathrm{\alpha }\in (-\mathrm{\infty },2]\\ \text{ C) }\frac{\mathrm{\alpha }-2}{4}<0\Rightarrow \mathrm{\alpha }<2\therefore \mathrm{\alpha }\in (-\mathrm{\infty },2)\end{array}$