{y | y =-(x- 1) 2- 1, where x belongs to R}={y|y≤- 1}
∫a∩c≠φ∴ Equation x 2-(m- 1) x+2m = 0 has a solution on (1, +∞).
When a ∩ c = φ, the figure shows the symmetry axis (m- 1)/2≤ 1, f( 1)=m+2≥0, so -2≤m≤3.
When a ∩ c ≠ φ, m belongs to (-∞, -2)∩(3,+∞).
The equation x 2-(m- 1) x+2m = 0 has a solution in (-∞,-1).
When b ∩ c = φ, the figure shows the symmetry axis (m- 1)/2≥- 1, f (- 1) > 0, and m > 0.
When b ∴ c ≠ φ, m belongs to (-∞, 0].
∴m belongs to (-∞, -2)
∫A∪B =(-∞,-1]∨( 1, +∞) and ∫(A∪B)∩C =φ.
∴ The equation x 2-(m- 1) x+2m = 0 has no solution on (-∞,-1]∞( 1, +∞).
Drawing shows that f (- 1) = 3m > 0, and f (1) = m+2 ≥ 0 ∴ m > 0.
The existence that ∴m belongs to (0, +∞) makes (A ∪ B) ∩ C = φ hold.