If the range of the function f(x)=5-xx^2-3x+2,x 1,2, is (-, ] [ ,), then ^2+ ^2 is equal t…

Question

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

Accepted Answer

$\begin{gathered} y=\frac{5-x}{{x}^{2}-3x+2} \ y{x}^{2}-3xy+2y+x-5=0 \ y{z}^{2}+(-3y+1)x+(2y-5)=0 \end{gathered}$

Case I : If $y=0$ (Accepted)

$\Rightarrow x=5$

Case II : If $y
e 0$

$\begin{gathered} D\ge 0 \ (-3y+1{)}^{2}-4(y)(2y-5)\ge 0 \ 9{y}^{2}+1-6y-8{y}^{2}+20y\ge 0 \ {y}^{2}+14y+1\ge 0 \ (y+7{)}^{2}-48\ge 0 \ |y+7|\ge 4\sqrt{3} \ \Rightarrow y+7\ge 4\sqrt{3}y+7\le -4\sqrt{3} \ \Rightarrow y\ge 4\sqrt{3}-7y\le -4\sqrt{3}-7 \end{gathered}$

From Case I and Case II

$y\in (-\infty,-4\sqrt{3}-7]\cup [4\sqrt{3}-7,\infty)$

$\begin{gathered} \alpha =-4\sqrt{3}-7 \ \, \beta =4\sqrt{3}-7 \ \Rightarrow {a}^{2}+{b}^{2} =(-4\sqrt{3}-7{)}^{2}+(4\sqrt{3}-7{)}^{2} \ =2(48+49) \ =194 \end{gathered}$

Solutions · 1