It is known that the proposition p: ∃ m ∈ R, M + 1 ≤ 0, proposition q: ∀ x ∈ R, X2 + MX + 1 ＞ 0 is tenable. If P ∧ q is a false proposition and P ∨ q is a true proposition, then the value range of real number m is () A. M ≥ 2B. M ≤ - 2 or - 1 < m < 2C. M ≤ - 2 or m ≥ 2D. - 2 ≤ m ≤ 2

It is known that the proposition p: ∃ m ∈ R, M + 1 ≤ 0, proposition q: ∀ x ∈ R, X2 + MX + 1 ＞ 0 is tenable. If P ∧ q is a false proposition and P ∨ q is a true proposition, then the value range of real number m is () A. M ≥ 2B. M ≤ - 2 or - 1 < m < 2C. M ≤ - 2 or m ≥ 2D. - 2 ≤ m ≤ 2

∃ proposition p: ∃ m ∈ R, M + 1 ≤ 0, ∀ m ≤ - 1; proposition q: ∀ x ∈ R, X2 + MX + 1 ＞ 0, M2-4 ＜ 0, ∀ 2 ＜ m ＜ 2

Let P: "any x ∈ R, x2-2x ＞ a" and Q "exist x ∈ R, X2 + 2aX + 2-A = 0"; if "P or Q" is true and "P and Q" is false, the value range of a is obtained

From the proposition p: "any x ∈ R, x2-2x ＞ a", we can get that x2-2x-a ＞ 0 is tenable, so there is △ = 4 + 4A ＜ 0, a ＜ - 1. From the proposition q: "there is x ∈ R, X2 + 2aX + 2-A = 0", we can get △′ = 4a2-4 (2-A) = 4a2 + 4a-8 ≥ 0, and the solution is a ≤ - 2, or a ≥ 1. Then we can get P true, Q false, or P false, Q true by "P or Q" being true, so there is a ＜ - 1 − 2 ＜ a ＜ 1, or a ＜ 1 A ≥ − 1a ≤− 2 & nbsp;, or a ≥ 1. Find - 2 ＜ a ＜ - 1, or a ≥ 1, that is, a ＞ - 2. So the value range of a is (- 2, + ∞)

To resolve cosx (SiNx + cosx) - SiNx (cosx SiNx) / (SiNx + cosx) & #

cosx(sinx+cosx)-sinx(cosx-sinx)/(sinx+cosx)²=COSXSINX+COSX²-SINXCOSX+SINX²\(sinx+cosx)²=COSX²+SINX²\(sinx+cosx)²=（(sinx+cosx)²-2SINXCOSX）\(sinx+cosx)²=1 ...