A. The position of two points B on the number axis is shown in the figure. Now two points a and B move to the left at the speed of 1 unit / s and 4 units / s respectively. (1) after a few seconds, the origin is just in the middle of two points? (2) A few seconds later, OA: OB = 1:2?

A. The position of two points B on the number axis is shown in the figure. Now two points a and B move to the left at the speed of 1 unit / s and 4 units / s respectively. (1) after a few seconds, the origin is just in the middle of two points? (2) A few seconds later, OA: OB = 1:2?

(1) Let the motion time be x seconds, according to the meaning of the question: x + 3 = 12-4x, the solution is: x = 1.8, a: after 1.8 seconds, the origin is just in the middle of two points; (2) let the motion time be x seconds, divided into two cases: ① before B and a meet: 12-4x = 2 (x + 3), the solution is: x = 1, ② after B and a meet: 4x-12 = 2 (x + 3)
If the system of inequalities x ＜ m + 1 (1) x ＞ 2m-1 (2) has a solution, then the range of M is?
emergency
x＜m+1①
X ＞ 2m-1, ② has solution
2m-1
The known functions f (x) = cos & # 178; X + cos & # 178; (x + α) + cos & # 178; (x + β)
Where α and β are constants and satisfy 0
To make f (x) independent of X, cos ^ 2 (x + a) + cos ^ 2 (x + b) = sin ^ 2 x + K (k is a constant) cos ^ 2 x cos ^ 2 A + sin ^ 2 x sin ^ 2 A + 2cosxsinx cosasina + cos ^ 2 x cos ^ 2 B + sin ^ 2 x sin ^ 2 B + 2cosx SiN x cosbsinb = sin ^ 2 x + KCOs ^ 2 x (C
Given the set a = {x | - 2 ≤ x ≤ 7}, B = {x | - M + 1 ≤ x ≤ 2m-1}, if AUB = B, then the value range of real number m is
From AUB = B: A is a subset of B, so m + 1 ≤ - 2 and 2m-1 ≥ 7, the solution is m ≤ - 3 and m ≥ 4
Given the set a = {x | - 2 ≤ x ≤ 7}, B = {x | - M + 1 ≤ x ≤ 2m-1}, if AUB = B, then the value range of real number m is an empty set
Given the set a = {x | - 2 ≤ x ≤ 7}, B = {x | - M + 1 ≤ x ≤ 2m-1}, if AUB = a, then the value range of real number m is {m | - 3 ≤ m ≤ 4}
If f (x) = A-B cosx (b > 0) has a maximum value of 3 / 2 and a minimum value of - 1 / 2, find the maximum value of G (x) = - 4asin3bx
Good points are great!
f(x)=-bcosx+a
b>0,-b
Given that the set a = {x | - 2 ≤ x ≤ 5}, B = {x | m + 1 ≤ x ≤ 2m + 1} satisfies that B is included in a, (1) then the value range of the real number m is? (2) the number of nonempty proper subsets of a when x belongs to Z? (3) when x belongs to R, there is no element x, so that x belongs to a and X belongs to B simultaneously, so the value range of M can be obtained?
(1) We know that M + 1 is greater than - 2 and 2m + 1 is less than 5, and M + 1 is less than or equal to 2m + 1, so we know that M is greater than or equal to 0 and less than 2. (2) we know that a = (, - 2-1,0,1,2,3,4,5), so the number of non empty subsets is: 2 ^ 8-1 = 255
-3≤m≥2
1) B included in a means that the range of B is less than or equal to the range of a, so the minimum value of B should be greater than or equal to the minimum value of a, and the maximum value of B should be less than or equal to the maximum value of a, so that B is included in a. Formula:
M + 1 ≥ - 2 and 2m + 1 ≤ 5, the solution is: - 3 ≤ m ≤ 2
2) Z is an integer. A non empty proper subset means that it cannot be an empty set (an empty set is a subset of any set) and the range of the subset is less than the range of the set (note that if it is less than or equal to, it becomes a subset instead of a proper subset).
There is another formula: if
1) B included in a means that the range of B is less than or equal to the range of a, so the minimum value of B should be greater than or equal to the minimum value of a, and the maximum value of B should be less than or equal to the maximum value of a, so that B is included in a. Formula:
M + 1 ≥ - 2 and 2m + 1 ≤ 5, the solution is: - 3 ≤ m ≤ 2
2) Z is an integer. A non empty proper subset means that it cannot be an empty set (an empty set is a subset of any set) and the range of the subset is less than the range of the set (note that if it is less than or equal to, it becomes a subset instead of a proper subset).
There is another formula: if B has n elements, then B has 2 ^ n subsets and (2 ^ n) - 2 nonempty proper subsets
Then the set a = {x | - 2 ≤ x ≤ 5, X belongs to Z} = {- 2, - 1,0,1,2,3,4}, a has n = 7 elements, then the nonempty proper subset has
(2 ^ 7) - 2 = 126
3) The meaning of the title is to take any element in a and not in B. in mathematical language, it is a ∩ B = empty set
You can think of it like this, with a as the center, B can't touch a, it can only be on the left and right sides of A
B is on the left side of A. if you want to disjoint, you can only have the maximum value of B less than the minimum value of A. the formula is 2m + 14
If both conditions are satisfied, the result is M4
Let f (x) = cosx + asinx-a / 4-1 / 2 (0 ≤ x ≤ π / 2)
(1) Let a denote the maximum value m (a) of F (x); (2) when m (a) = 2, find the value of a and the minimum value of F (x)
F (x) = cos ^ 2 * x asinx-a / 4-0.5 = 1-sin ^ 2x asinx-a / 4-1 / 2 = - (sinx-a / 2) ^ 2 1 / 2 (a ^ 2-A) / 40 ≤ SiNx ≤ 1 (1) 0 ≤ a ≤ 2, the maximum value of F (x) is 1 / 2 (a ^ 2-A) / 4 (2) when a > 2, the maximum value of F (x) is - (1-A / 2) ^ 2 1 / 2 (a ^ 2-A) / 4 (3) a
Given a = {x | - 2 ≤ x ≤ 7}, B = {x | - M + 1 ＜ x ＜ 2m-1}, if B ⊆ a, then the value range of M is______ .
In this case, the solution is: if a ≤ 2 m − 4 m, that is to say, if B ≤ 2 m ≠ 4 m − 1
Solution a {x | - 2
Given vector M = (cosx, 1-asinx), n = (cosx, 2), where a ∈ R, X ∈ R, Let f (x) = Mn, and the maximum value of function f (x) is g (a)
1. Find the analytic expression of the function g (a)
2, let 0 ≤ x < 2 π, find the maximum and minimum of G (2cosx + 1) and the corresponding x value
I want to do it by myself. Don't do it according to others. The answer I do is far from theirs. Let's think about it carefully
This is the answer given by them. Let's think about it
(1) f(x)=mn=(cosx)^2+2-2asinx=1-(sinx)^2+2-2asinx=-(sinx+a)^2+a^2+3
When a ∈ [- 1,1], G (a) = a ^ 2 + 3. When A1, G (a) = (A-1) ^ 2 + A ^ 2 + 3
(2) G (2cosx + 1) = ① (2cosx + 1) ^ 2 + 3, when x ∈ [2,3 π / 2]
② [(2cosx + 1) - 1] ^ 2, when x ∈ [0, π / 2) ∪ (3 π / 2, π)
The rest is worth your own calculation
1. Vector M = (cosx, 1-asinx), n = (cosx, 2)
F(x)=(cosx)^2+2-2asinx=3-[(sinx)^2+2asinx]=3-(sinx+a)^2+a^2
When - 1
This question is too familiar. I just did it
Wrong
(1) F (x) = Mn = (cosx) ^ 2 + 2-2asinx = 1 - (SiNx) ^ 2 + 2-2asinx = (SiNx + a) ^ 2 + A ^ 2 + 3 when a ∈ [- 1,1], G (a) = a ^ 2 + 3. When a & lt; - 1, G (a) = (a)
0-0
If the set a = {X / - 2 ≤ x ≤ 5}, B = {X / M + 1 ≤ x ≤ 2m-1}, and B is included in a, then the value range of real number m is?
My process is:
m+1≤2m-1,∴m≥2;
-2 ≤ m + 1 and 5 ≥ 2m-1, - 3 ≤ m ≤ 3
To sum up, m ≤ 3
But the correct answer in the reference book is m ≥ 2
It's wrong. The answer on the reference book is m ≤ 3
Sorry, now I think there are two possibilities. One is that B is a nonempty set, that is, 2 ≤ m ≤ 3;
The other is that B is an empty set, i.e. 2m-1 > m + 1, i.e. M
-2 ≤ m + 1 and 5 ≥ 2m-1, - 3 ≤ m ≤ 3
Or 2m-1
The building owner is right. You know, sometimes there are mistakes in teaching materials
B can be empty
What you have done is right. You may as well try to replace it. Let m = 0, then B becomes 1