If the range of the function f(x) = 5-x/x²−3x+2, x ≠ 1,2, is (−∞, α] ∪ [β, ∞), then α² + β² is equal to:
Question If the range of the function f ( x ) = 5 − x x 2 − 3 x + 2 , x ≠ 1 , 2 is ( − ∞ , α ] ∪ [ β , ∞ ) (-\infty, \alpha] \cup [\beta, \infty) , then find the value of α 2 + β 2 \alpha^{2}+\beta^{2} . Question Image Short Solution Let y = 5 − x x 2 − 3 x + 2 y = \dfrac{5 - x}{x^{2} - 3x + 2} ' Multiply through to get: y ( x 2 − 3 x + 2 ) = 5 − x Rearrange to a quadratic in x x : y x 2 − 3 y x + 2 y − 5 + x = 0 or y x 2 + ( − 3 y + 1 ) x + ( 2 y − 5 ) = 0 For x x to be real, discriminant Δ \Delta must be non-negative: Δ = ( − 3 y + 1 ) 2 − 4 y ( 2 y − 5 ) ≥ 0 Simplify Δ \Delta : Δ = 9 y 2 − 6 y + 1 − 8 y 2 + 20 y = y 2 + 14 y + 1 Solve Δ = 0 \Delta =0 : y = − 14 ± 196 − 4 2 = − 14 ± 192 2 = − 14 ± 8 3 2 y = \frac{-14\pm \sqrt{196-4}}{2} = \frac{-14\pm \sqrt{192}}{2} = \frac{-14\pm 8\sqrt3}{2} y = − 7 ± 4 3 y = -7 \pm 4\sqrt3 Because the quadratic in x x is valid for Δ ≥ 0 \Delta\ge0 , the range of y y is: ( − ∞ ,...