If the range of the function f(x)=5–xx2–3x+2, x≠1,2, is (–∞,α]∪[β,∞), then α2+β2 is equal to : [2025]
(2)
y=5–xx2–3x+2, x≠1,2
⇒ yx2–3xy+2y+x–5=0
⇒ yx2+(–3y+1)x+(2y–5)=0
Case I : If y = 0
⇒ x = 5
Case II : if y ≠ 0
For real solutions, D≥0
⇒ (–3y+1)2–4(y)(2y–5)≥0
⇒ 9y2+1–6y–8y2+20y≥0
⇒ y2+14y+1≥0
⇒ (y+7)2–48≥0
⇒ |y+7|≥43 ⇒ y+7≥43 or y+7≤–43
⇒ y≥43–7 or y≤–43–7
From Case I and Case II, we have
y∈(–∞,–43–7]∪[43–7,∞)
∴ α=–43–7 and β=43–7
⇒ α2+β2=(–43–7)2+(43–7)2=2(48+49)=194.