Let f (x) = x3-4x + A, 0 < a < 2. If the three zeros of F (x) are x1, X2, X3, and X1 < x2 < X3, then () A．x1＞－1 B．x2＜0 C．x2＞0 D．x3＞2 Which one should I choose? Specific ideas

Let f (x) = x3-4x + A, 0 < a < 2. If the three zeros of F (x) are x1, X2, X3, and X1 < x2 < X3, then () A．x1＞－1 B．x2＜0 C．x2＞0 D．x3＞2 Which one should I choose? Specific ideas

f'(x)=3x²-4
Let f '(x) ≥ 0
3x²-4≥0
3x²≥4
X ≥ 2 / √ 3 or X ≤ - 2 / √ 3
That is, the function increases monotonically in the interval (- ∞, - 2 / √ 3]; decreases monotonically in the interval [- 2 / √ 3,2 / √ 3]; increases monotonically in [2 / √ 3, + ∞)
F (- 1) = - 1 + 4 + a = a + 330, i.e. f (x) has no zero point on the interval (2, + ∞), D is wrong
To sum up, C. is correct
(x-x1)(x-x2)(x-x3)=0
(x^2 -x1x -x2x +x1x2)(x-x3)=0
(x^2 -x1x -x2x +x1x2)x-(x^2 -x1x -x2x +x1x2)x3=0
x^3 -x1x^2 - x2x^2 -x3x^2+x1x2x +x1x3x+x2x3x-x1x2x3=0
So X1 + x2 + X3 = 0, x1x2 + x2x3 + x1x3 = - 4, x1x2x3 = - A
If f (x) is an odd function and f (x) has three zeros x1, X2 and X3, then the value of X1 + x2 + X3 is zero______ .
∵ f (x) is an odd function, ∵ f (x) must pass through the origin. ∵ the equation f (x) = 0 has and only has three real roots x1, X2, X3, ∵ one of which is 0. Let x2 = 0. ∵ f (x) be an odd function. ∵ the two roots of the equation are symmetric about the origin, that is, X1 + X3 = 0. ∵ X1 + x2 + X3 = 0
Given the set a = {x | x2-ax + a2-19 = 0}, B = {x | x2-5x + 6 = 0}, C = {x | x2 + 2x-8 = 0} (1) if a ∩ B = a ∪ B, find the value of a; (2) if a ∩ B = a ∩ C ≠ 0, find the value of A
(1) ∵ B = {x | x2-5x + 6 = 0} = {2,3}, a ∩ B = a ∪ B, ∪ a = B. ∪ 2 and 3 are the two roots of the equation x2-ax + a2-19 = 0, ∩ 2 + 3 = a, ∩ a = 5. (2) a ∩ B = a ∩ C ≠∞, ∩ 2 ∈ a, ∩ 4-2a + a2-19 = 0, a = - 3, a = 5
It is known that the real number a makes at least one of the three univariate quadratic equations x * x-x + a = 0, X * x-2x + a = 0, X * x-4x + 2A = 0 have a solution. Do you have any simple method to get it
The range of a is given
First, find the value of a which makes them have no solution, then find the complement set, that is, the value of a which has at least one solution
therefore
1-4a
(1 / 2) 1 [set a = {the square of X - ax + the square of a - 19 = 0}, B = {the square of X - 5x + 6 = 0}, C = {the square of X + 2x-8 = 0} satisfies that a intersection B is not equal to an empty set
B = {2,3} into a, solution x = 5
B
x²-5x+6=0
X = 2 or x = 3
C
x²+2x-8=0
X = - 4 or x = 2
If a ∩ C is empty, then x ≠ - 4, X ≠ 2
So x = 3
3²-3a+a²-19=0
a²-3a-10=0
A = 5 or a = - 2
When a = 5, X & # 178; - 5x + 6 = 0, x = 2 or x = 3 (contradiction)
When a = - 2, X & # 178; + 2x-15 = 0
X = - 5 or x = 3, consistent with
∴a=-2
X & # 178; - 5x + 6 = 0 gives x = 2 or 3, so B = {2,3}
A ∩ B ≠ empty set
When a = {2}, 2 & # 178; - 2A + A & # 178; - 19 = 0, then a = 5 or - 3
If a = {3}, 3 & # 178; - 3A + A & # 178; - 19 = 0, then a = 5 or - 2
When a = {2,3}, a = 5 can be obtained from Veda's theorem 2 + 3 = a
2 * 3 = A & # 178; - 19 is a = ± 5 # a = 5
To sum up, a = 5, - 3, - 2
The intersection of a and B is not equal to an empty set, that is, at least one element of B set belongs to a set
Firstly, according to B = {X & # 178; - 5x + 6}, we get the set B = {2,3}
Let 2 be brought into a set, that is, 4-2a + A & # 178; - 19 = 0, and the solution is a = 5 or a = - 3
Let 3 be brought into a set, that is, 9-3a + A & # 178; - 19 = 0, and the solution is a = 5 or a = - 2
To sum up, a = {- 2, - 3,5} satisfies the condition
B={2,3}
,C={-4,2}
A intersection B is not equal to an empty set a intersection C is equal to an empty set
The square of 3 - 3A + the square of a - 19 = 0
The square of a-3a-10 = 0
Then a = 5 or a = - 2
A=5
A = B (rounding off)
a=-2
,x^2+2x-15=0
A = {3, - 5), consistent with
So a = - 2
May I ask an equation AX ^ 2-2 > = 2x ax (a belongs to all real numbers)
Such as the title, to explain in detail, urgent,
The original equation can be reduced to AX ^ 2 + (A-2) X-2 = 0
(1) When a = 0, the original equation is reduced to 2x + 2 = 0, and the solution is x = - 1
(2) When a is not equal to 0, the original equation is reduced to (x + 1) (AX-2) = 0
1. When a = - 2, the original equation is reduced to (x + 1) ^ 2 = 0, and the solution is x = - 1
2. When a is not equal to - 2, the original equation has two solutions: X1 = - 1, X2 = 2 / A
Change it into the general form of quadratic equation with one variable, and use the discriminant with to find a
Replace a into the equation and remove X
Given that the set a = {1}, the set B = {x | X & # 178; - 3x + a = 0}, and that a is really contained in B, find the value of the real number a
Set a = {1}, set B = {x | X & # 178; - 3x + a = 0}, and a is really contained in B, indicating that a is a proper subset of B
1 ∈ B, so 1-3 + a = 0, a = 2
When a = 2, x = 1 or x = 2 B + {1,2} satisfies the meaning of the problem
So a = 2
A is a real number. Write a necessary and sufficient condition for the equation AX ^ 2 + 2x + 1 = 0 to have at least one real root
If a = 0
=>X is solvable
On the contrary, the discriminant > = 0
=> a
Ax ^ 2 + 2x + 1 = 0 with at least one real root a
△>=0,
4-4a>=0,
A
Given the set a = (- 1,0), B = (x | X & # 178; + 2aX + 1 = 0), if a ∩ B = B, find the value range of real number a?
1. B is - # 8709;
4a²-4
Given x = 0 or - 1, if x = 0, it does not hold; if x = - 1, i.e. 2-2a = 0, a = 1; or B is an empty set, i.e. 4A ^ 2-4
The number of real number solutions of equation log2x + X & # 178; = 2
What do you mean by logarithm 2x 2 or logarithm 2x 10
If it is a logarithm of X with base 2, then
Let Y1 = log2x
Let y2 = 2-x & # 178;
Draw two functions on the same coordinate axis, the number of intersections is the number of real number solutions
Y1, Y2 have an intersection, so the equation has a real solution