If the complete set u is r, a = {x | 1 ≤ x ≤ 2}, if the complement of B and a in R = R, the complement of B intersection a in R = {x | 0}

If the complete set u is r, a = {x | 1 ≤ x ≤ 2}, if the complement of B and a in R = R, the complement of B intersection a in R = {x | 0}

∵ the complete set u is r, a = {x | 1 ≤ x ≤ 2},
∴CuA={x|x2}
∵ the complement of b-union a in R = R, the complement of b-intersection a in R = {x | 0

If the solution set of inequality ax ^ 2 + BX + C ＜ 0 is {x | - 1 / 2 ＜ x ＜ 1}, then the solution set of Cx ^ 2 + BX + a ＜ 0 is {x | - 1 / 2 ＜ x ＜ 1}

From the above, we can know that - 1 / 2 and 1 are the two roots of the equation, and a is greater than 0, so we have (according to the sum of the two, we can get) - 1 / 2 + 1 = - B / A; (- 1 / 2) * 1 = C / A, so we can use the formula containing a to express B and C respectively, B = - A / 2, C = - A / 2, and substitute these two formulas into CX ^ 2 + BX + a ＜

Let a, B, C be real numbers, f (x) = (x + a) (x2 + BX + C), G (x) = (AX + 1) (CX2 + BX + 1). Let s = {x | f (x) = 0, X ∈ r}, t = {x | g (x) = 0, X ∈ r}. If {s}, {t} are the number of elements of S, T & nbsp; respectively, then the following conclusion is impossible ()
A. {s} = 1 and {t} = 0b. {s} = 1 and {t} = 1C. {s} = 2 and {t} = 2D. {s} = 2 and {t} = 3

∵ f (x) = (x + a) (x2 + BX + C), when f (x) = 0, there is at least one root x = - A, when b2-4c = 0, there is another root x = - B2, as long as B ≠ 2a, f (x) = 0 has two roots; when B = 2A, f (x) = 0 is a root; when b2-4c < 0, f (x) = 0 has only one root; when b2-4c > 0, f (x) = 0 has two roots