We know that P: F (x) = 1 − X3, and | f (a) | 2, Q: set a = {x | x2 + (a + 2) x + 1 = 0, X ∈ r}, and a ≠ r}. If P or q are true propositions, P and Q are false propositions, we find the value range of real number a

We know that P: F (x) = 1 − X3, and | f (a) | 2, Q: set a = {x | x2 + (a + 2) x + 1 = 0, X ∈ r}, and a ≠ r}. If P or q are true propositions, P and Q are false propositions, we find the value range of real number a

If | f (a) | = | 1 − A3 | 2 holds, then - 6 | 1-A | 6, the solution is - 5 | a | 7, that is, when - 5 | a | 7, P is a true proposition; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;; if a ≠ 0, then the equation x2 + (a + 2) x + 1 = 0 has a real root, from △ = (a + 2) 2-4 ≥ 0, the solution is a ≤ - 4, or a ≥ 0, that is, when a ≤ - 4