If the range of the function $ f(x) = \frac{5-x}{x^2 - 3x + 2} , \ x \neq 1, 2, $ is $ (-\infty , \alpha] \cup [\beta, \infty) $, then $ \alpha^2 + \beta^2 $ is equal to :

<p>$$\begin{aligned} & y=\frac{5-x}{x^2-3 x+2} \\ & y x^2-3 x y+2 y+x-5=0 \\ & y z^2+(-3 y+1) x+(2 y-5)=0 \end{aligned}$$</p> <p>Case I : If $y=0$ (Accepted)</p> <p>$\Rightarrow x=5$</p> <p>Case II : If $y \neq 0$</p> <p>$$\begin{aligned} & \mathrm{D} \geq 0 \\ & (-3 y+1)^2-4(y)(2 y-5) \geq 0 \\ & 9 y^2+1-6 y-8 y^2+20 y \geq 0 \\ & y^2+14 y+1 \geq 0 \\ & (y+7)^2-48 \geq 0 \\ & |y+7| \geq 4 \sqrt{3} \\ & \Rightarrow y+7 \geq 4 \sqrt{3} \text { or } \mathrm{y}+7 \leq-4 \sqrt{3} \\ & \Rightarrow \mathrm{y} \geq 4 \sqrt{3}-7 \text { or } \mathrm{y} \leq-4 \sqrt{3}-7 \end{aligned}$$</p> <p>From Case I and Case II</p> <p>$y \in(-\infty,-4 \sqrt{3}-7] \cup[4 \sqrt{3}-7, \infty)$</p> <p>$$\begin{aligned} & \text { So } \alpha=-4 \sqrt{3}-7 \\ & \quad \beta=4 \sqrt{3}-7 \\ & \begin{aligned} \Rightarrow a^2+b^2 & =(-4 \sqrt{3}-7)^2+(4 \sqrt{3}-7)^2 \\ & =2(48+49) \\ & =194 \end{aligned} \end{aligned}$$</p>