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

Question

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

Aicharya · Accepted Answer

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

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

Solution