If any two real genera A and B are taken in the interval [0,2], then the probability that the function f (x) = x ^ 3 + ax-b has and only has one zero point in the interval [- 1,1] is zero

If any two real genera A and B are taken in the interval [0,2], then the probability that the function f (x) = x ^ 3 + ax-b has and only has one zero point in the interval [- 1,1] is zero

F '(x) = 3x ^ 2 + a > = 0, so the function increases monotonically, and there is only one zero point at most
F (- 1) = - 1-a-b = 0, that is, when B-A > = 1, there is a zero in [- 1,1]
Take a as the horizontal axis and B as the vertical axis to make a square with side length of 2, and the vertices are (0,0), (2,0), (0,2), (2,2) respectively
Then only the part above the line B = a + 1 in this region makes the equation have a zero point at [- 1,1]
Probability is the area ratio: 1 / 2 * 1 * 1 / (2 ^ 2) = 1 / 8
Given that a is a real number, the function f (x) = - x ^ 2 + ax-3 has a zero point on the interval (0,1) and (2,4) respectively, and the value range of a is obtained
If the function f (x) = - x ^ 2 + ax-3 has a zero point in the interval (0,1) and (2,4), then there are two unequal real roots when - x ^ 2 + ax-3 = 0, namely a ^ 2-12 ＞ 0, then a ＜ - 2 √ 3, a ＞ 2 √ 3, and there is a zero point in the interval (0,1) and (2,4), namely X1 + x2 ＞ 0, then the distances from a ＞ 0x1 and X2 to the axis of symmetry are equal, X1 to the pair of
4 = < a0
f((0)0
f((2)>0
f((4)
If the function f (x) = x2-ax + 1 has zeros in the interval (1 / 2,3), then the value range of real number a is
A（2 ,5/2）B (2,10/3)
If f (x) = x2-ax + 1, the solution has zeros on the interval (1 / 2,3),
That is, x ^ 2-ax + 1 = 0 has a solution on the interval (1 / 2,3)
That is, ax = x ^ 2 + 1 has a solution on the interval (1 / 2,3)
That is, a = x + 1 / X has a solution on the interval (1 / 2,3)
Let g (x) = x + 1 / x, X belong to (1 / 2,3)
The function decreases on (1 / 2,1)
Increasing on (1,3)
So when x = 1, y = g (x) has a minimum value of 2
When x = 3, y = g (x) = 10 / 3
When x = 1 / 2, y = g (x) = 5 / 2
So the function g (x) = x + 1 / X is B (2,10 / 3) when x belongs to (1 / 2,3)
So choose B
If there is a root on (1 / 2,3), then f (1 / 2) * f (3) 1 / 2
a^2-4>0
f(1/2)>0 f(3)>0
2
Given that the function f (x) = ax + 1 has zeros in the interval (- 1,1), then the value range of real number a
(1) If a = 0, there is no zero
(2) If a ≠ 0
The zero point is - 1 / A
-1
A. B is a nonempty set. Define a × B = {x | x ∈ a ∪ B, and X ∉ a ∩ B}. If a = {x | y = x2 − 3x}, B = {y | y = 3x}, then a × B = {x | y = 3x}=______ .
A = {x | y = x2 − 3x} = x | x2-3x ≥ 0 = x | x ≤ 0 or X ≥ 3, B = {y | y = 3x} = {y | Y > 0}, so a ∪ B = R, a ∩ B = [3, + ∞), so a × B = (- ∞, 3)
If two of the equations 2x2-mx-4 = 0 are X1 and X2, and 1x1 + 1x2 = 2, then the value of real number m is equal to ()
A. 4B. -4C. 8D. -8
∵ x 1 + x 2 = − Ba, x 1 x 2 = Ca, ∵ 1x1 + 1x2 = 2 = X1 + x2x1x2 = M2 − 2 = 2, M2 = - 4, M = - 8
Let AB be a nonempty set, define a * b = {x | x ∈ a ∪ B and X &; a ∩ B, a = {x | 0 ≤ x ≤ 2} B = {y | Y > 1}, then a * B=
A ∪ B is x ≥ 0
A ∩ B is 12}
Given that M is a real number, the two roots of the equation 2x ^ 2-mx-30 = 0 are x1, X2, and 5x1 + 3x2 = 0. In the rectangular coordinate system, the parabola
The position relation between y = MX ^ 2 + (4 + k) x + K and X axis is ()
X1 + x2 = m / 2x1 * x2 = - 153x1 + 3x2 = 3m / 25x1 + 3x2 = 02x1 = - 3m / 2x1 = - 3m / 4x2 = m / 2 + 3m / 4 = 5m / 4x1 * x2 = - 15m ^ 2 / 16 = - 15m = + - 4 when m = 4 parabola △ = (4 + k) ^ 2-4mk = 16 + 8K + K ^ 2-16k = k ^ 2-8k + 16 = (K-4) ^ 2 > = 0 has intersection with X axis when m = - 4 parabola △ = (4 + k) ^ 2-4mk = 16
A:
The two roots of the equation 2x ^ 2-mx-30 = 0 are x1, x2,
According to Weida's theorem, the following conditions are satisfied: 5x1 + 3x2 = 0
x1*x2=-30/2=-15
The solution of the above two equations is as follows
X1 = - 3, X2 = 5 or X1 = 3, X2 = - 5
So: X1 + x2 = 2 or X1 + x2 = - 2
Thirdly, according to Weida's theorem:
x1+x2=m/2
So: M = 4 or M = - 4
1) When m = 4,... Expansion
A:
The two roots of the equation 2x ^ 2-mx-30 = 0 are x1, x2,
According to Weida's theorem, the following conditions are satisfied: 5x1 + 3x2 = 0
x1*x2=-30/2=-15
The solution of the above two equations is as follows
X1 = - 3, X2 = 5 or X1 = 3, X2 = - 5
So: X1 + x2 = 2 or X1 + x2 = - 2
Thirdly, according to Weida's theorem:
x1+x2=m/2
So: M = 4 or M = - 4
1) When m = 4, there is a parabola
y=mx^2+(4+k)x+k
=4x^2+(4+k)x+k
=(4x+1)(x+k)
So: the position relationship between Y and X axis of parabola is (at least one intersection (- 1 / 4,0))
2) When m = - 4, the parabola
y=mx^2+(4+k)x+k
=-4x^2+(4+k)x+k
Discriminant = (4 + k) ^ 2 + 4 * 4K
So: the position relationship between Y axis and X axis of parabola is (uncertain whether there is an intersection point) folded
The two roots of the equation 2x ^ 2-mx-30 = 0 are x1, X2, and 5x1 + 3x2 = 0,
So X1 + x2 = m / 2, x1x2 = - 30 / 2 = - 15
x1=-3x2/5
x1x2=-3x2^2/5=-15
x2^2=25
If x2 = (+ / -) 5, then X1 = (- / +) 3
m=2(x1+x2)=2*(+/-)2=(+/-)4
y=mx^2+(4+k)x+k
... unfold
The two roots of the equation 2x ^ 2-mx-30 = 0 are x1, X2, and 5x1 + 3x2 = 0,
So X1 + x2 = m / 2, x1x2 = - 30 / 2 = - 15
x1=-3x2/5
x1x2=-3x2^2/5=-15
x2^2=25
If x2 = (+ / -) 5, then X1 = (- / +) 3
m=2(x1+x2)=2*(+/-)2=(+/-)4
y=mx^2+(4+k)x+k
Discriminant = (4 + k) ^ 2-4mk
(1) M = 4, the above formula = 16 + 8K + K ^ 2-16k = k ^ 2 + 8K
(i) Dang-8
Let a = {x  2-ax + a2-19 = 0}, B = {x  2-5x + 6 = 0}, {x  2 + 2x-8 = 0}, (1) if a ∩ B = a ∪ B, find the value of a, (2) if ∅ is the value of a ∩ B
Let a = {x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\tofind the value of a
In this paper, we obtain 2 + 3 = a (2 + 3 = a) from the relationship between the roots and the coefficients, and 2 + 3 = a = 2 + 3 = a; 2 × 3 = 2 × 3 = 2-3 = a; 2 × 3 = a2-3 = a; 2 × 3 = 2 × 3 = a2-19 = 2-19, and the solution of a = 5 (2) (2) (2) (2) (2) (2) (2 × 3 = a2-19) is obtained by 2 × 3 = 2 × 3 = a2-2-19, and the solution of a = 5 (2) (2) (2) (2) (2) (2) (2) (2) (2) (2 × 3 = 2 × 3 = a2-19) a = 5 (2) (2) (2) (5 (2) (2) (2) (2) (2) \\\\\\\\\\\\\andb have a common
If one root of equation 2x2 + (a + 1) x + 2a-3 = 0 is less than - 1 and the other root is greater than 0, then the value range of real number a is___ .
Because one root of the equation 2x2 + (a + 1) x + 2a-3 = 0 is less than - 1, and the other root is greater than 0, the image of the corresponding function f (x) = 2x2 + (a + 1) x + 2a-3 is shown in the figure. From the figure, f (- 1) < 0 and f (0) < 0, {a < 32, that is, a < 32, so the answer is: a < 32