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Let a = {x | x2-5x + 4 ＞ 0} and B = {x | x2-2ax + A + 2 = 0}. If a ∩ B ≠ 0, the value range of a is obtained
From the inequality in the set a, we get: (x-1) (x-4) > 0, the solution is: x > 4 or x < 1, that is, a = (- ∞, 1) ∪ (4, + ∞); Let f (x) = x2-2ax + A + 2, from a ∩ B ≠ 0, we get that f (x) has no intersection with X axis or two intersections are in the interval [1,4], and ∩ a = 4a2-4 (a + 2) < 0 or ∩ B = 4a2 − 4
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