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

Set \(y=f(x)=\frac{5−x}{{x}^{2}−3x+2}\).

Rearranging gives a quadratic in \(x\):

\(y{x}^{2}+(1−3y)x+(2y−5)=0\)

For real \(x\) to exist (note \(x=1,2\) are excluded as they don't satisfy this equation), the discriminant must be non-negative: \(D=(1−3y{)}^{2}−4y(2y−5)\geq 0\)

\(1−6y+9{y}^{2}−8{y}^{2}+20y\geq 0\)

\({y}^{2}+14y+1\geq 0\)

The roots of \({y}^{2}+14y+1=0\) are the boundary values \(\alpha\) and \(\beta\).

By Vieta's formulas: \(\alpha +\beta =−14,\ \alpha \beta =1\)

Therefore: \({\alpha }^{2}+{\beta }^{2}=(\alpha +\beta {)}^{2}−2\alpha \beta\)\(=(−14{)}^{2}−2(1)=196−2=194\)