If there is only one element in the set a = {x | x2 + ax + 1 = 0}, then the value of a is () A. 0b. 0 or 2C. 2D. - 2 or 2

If there is only one element in the set a = {x | x2 + ax + 1 = 0}, then the value of a is () A. 0b. 0 or 2C. 2D. - 2 or 2

If there is only one element in the set a = {x | x2 + ax + 1 = 0}, then the equation x2 + ax + 1 = 0 has and only one solution, then △ = A2-4 = 0, the solution is a = 2, or - 2, so the value of a satisfying the condition is 2 or - 2, so D is selected

Let f (x) = x2 + ax + B, a = {x | f (x) = x} = {a}, and the set composed of elements (a, b) be m

∵ a = {x | f (x) = x} = {a}, the two equal roots of the equation x2 + ax + B = x are a. ∵ x2 + (A-1) x + B = 0, both are a. ∵ a + a = - (a − 1) a · a = B, ∵ a = 13b = 19, and the set of ∵ elements (a, b) is m, ∵ M = {(13, 19)}

We know the set a = {x ∈ R | ax ^ 2 + 2x + 1 = 0}, where a ∈ r}
(1) If 1 ∈ a, a is represented by enumeration. (2) if there is only one element in a, find the set B composed of the values of A

（1）∵A={x ∈R|ax^2+2x+1＝0},a∈R
∵1∈A,a+3=0,a=-3
∴-3x^2+2x+1＝0
Let the other root of the equation be x1
∴x1=-1/3
∴A=｛1,-1/3｝
(2) If there is only one element in a
① When a = 0, there is only one element in a
∴a=0
② When a ≠ 0
∴△=4-4a=0
∴a=1
To sum up
∴B=｛0,1｝

It is known that the set a = {x | (x + 2) (x + 1) (2x-1) > 0}, B = {x | 2x ^ 2 + ax + B-1 ≤ 0}, and a ∪ B = {x | x + 2 > 0}, a ∩ B = {x | 1 / 2}

Because a = {x | (x + 2) (x + 1) (2x-1) > 0}, X ＞ 1 / 2 or - 2 ＜ x ＜ - 1
And because a ∪ B = {x | x + 2 > 0}, a ∩ B = {x | 1 / 2