The known set a = {x 2-3x-4 ≤ 0, X ∈ r}, B = {x 2-2mx + M2-4 ≤ 0, X ∈ R, m ∈ r} (1) If a ∩ B = [1,4], find the value of real number M. (2) if a is included in CRB, find the value range of real number M

The known set a = {x 2-3x-4 ≤ 0, X ∈ r}, B = {x 2-2mx + M2-4 ≤ 0, X ∈ R, m ∈ r} (1) If a ∩ B = [1,4], find the value of real number M. (2) if a is included in CRB, find the value range of real number M

（1）A={X|-1
solution
(1) From a: - 1 ≤ x ≤ 4
From B: m-2 ≤ x ≤ m + 2
A ∩ B = [1,4], then M + 2 ≥ and m-2 = 1, so m = 3
(2) A is contained in CRB, CRB: XM + 2
So m-2 ≥ 4 or M + 2 ≤ - 1
M ≥ 6 or m ≤ - 3
Known: set a = {x | x2-3x + 2 = 0}, B = {x | 2x-ax + A-1 = 0}. A ∪ B = A. find the value of real number a?
Predigesting sets a and B
According to the meaning of the title,
A=｛1,2｝ B=(1-a)/(2-a)
∵A∪B=A
B is contained in a
∴①(1-a)/(2-a)=1
The solution is not suitable for the purpose of the question
②(1-a)/(2-a)=2
The solution is a = 3
It is known that a = 2x-1 / x2 + 3x + 2 > 0, B = x2 + ax + B is less than or equal to 0, and a intersection B = 1 / 2
Is a = 2x-1 / x2 + 3x + 2 > 0, B = x2 + ax + b less than or equal to 0 (2x-1) / (x ^ 2) + 3x + 2 > 0, x ^ 2 + ax + b less than or equal to 0?
The known set a = [x | x is greater than or equal to 3 and less than or equal to 7], B = [x | x is greater than 2 and less than 10], C = [x | x is less than a], and the complete set is real number set R
Finding the intersection B of [CRA]
∵CRA={x7}
The [CRA] intersection B = {2
B
{x | x greater than 2 less than 3 or x greater than 7 less than 10}
Let f (x) whose domain is r satisfy f (x + y) = f (x) * f (y) for any real number x and y, and f (0) ≠ 0, prove f (x) & gt; 0
f(x)=f(x/2)*f(x/2)=[f(x/2)]²≥0
Suppose that there exists x satisfying: F (x) = 0
∵ f (0) = f (x-x) = f (x) * f (- x) = 0 * f (- x) = 0, which contradicts the known condition f (0) ≠ 0
The hypothesis does not hold, that is, f (x) ≠ 0
In conclusion: F (x) > 0
By substituting x = y = 0, we get the following result:
f(0)=[f(0)]²
The result is: F (0) = 1 or F (0) = 0
Then: F (0) = 1
By substituting y = x, we get the following result:
f(2x)=[f(x)]²
Namely:
f(x)=[f(x/2)]²>0
Then when x ∈ R, f (x) > 0
Let f (x) = 2x3 + ax and G (x) = bx2 + C pass through point P (2,0) and have the same tangent at point P. (I) find the value of real numbers a, B and C; (II) Let f (x) = f (x) + G (x) find the monotone interval of F (x)
(1) From the question, we know that: F (2) = 0g (2) = 0f ′ (2) = g ′ (2) {16 + 2A = 04B + C = 024 + a = 4B} a = − 8b = 4C = − 16. The values of real numbers a, B and C are: - 8, 4 and - 16, respectively（ 2) If f (x) = 2x3 + 4x2-8x-16f ′ (x) = 6x2 + 8x-8, if f ′ (x) = 6x2 + 8x-8 > 0, then x ＞ 23 or X ＜ - 2, if f ′ (x) = 6x2 + 8x-8 < 0, then − 2 < x ＜ 23. Therefore, the increasing range of F (x) is (− ∞, − 2) and the decreasing range of (23, + ∞) is (− 2, 23)
For any real number x, f (x-1) = f (3-x), and f (x-1) = f (x-3), if 1
From F (x-1) = f (3-x), and f (x-1) = f (x-3), we know that: F (3-x) = f (x-3), let 3-x = t, then, since x belongs to R, then t also belongs to R, so we have: for t belongs to R, f (T) = f (- t), so we preliminarily get that f (x) is even function; furthermore, since f (x-1) = f (x-3), let X-1 = s, then the above formula can be reduced to: F (s) = f (S-2), that is, f (s) = f (s + 2), so we know from the definition of periodic function that the period of the function is 2, So we can draw the graph of the function (it's not convenient to draw here, but we can draw it by ourselves through these properties), and then we can get the monotone interval. Your final answer is: monotone decreasing interval is [2n, 2n + 1], n belongs to R
Let f (x) = x ^ 3 + ax, G (x) = 2x ^ 2 + B, we know that their images have the same tangent at x = 1
① Find the analytic expressions of the functions f (x) and G (x). 2. If the function f (x) = f (x) - Mg (x) is a monotone decreasing function in the interval [0.5,3], find the value range of the real number M
① If f (x) and G (x) have the same tangent at x = 1, then f '(1) = g' (1), f (1) = g (1), i.e. 3 + a = 4,1 + a = 2 + B, the solution is a = 1, B = 0, f (x) = x + X, G (x) = 2x, ② f (x) = f (x) - Mg (x) = x + x-m (2x) = x-2mx + X in [0.5,3] monotone decreasing ∧ f '(x) = 3x-4mx + 1 ≥ 0 in [0.5,3], 3] If M ≤ (3x + 1) / 4x = 3x / 4 + 1 / (4x), let u = 3x / 4 + 1 / (4x) ≥ 2 √ [(3x / 4) × (1 / (4x))] = √ 3 / 2 if and only if 3x / 4 = 1 / (4x), then x = √ 3 / 3  u monotonically decreases in [0.5. √ 3 / 3], (√ 3 / 3,3] monotonically increases and UMIN = √ 3 / 2. To satisfy m ≤ u, we only need m ≤ UMIN = √ 3 / 2} m ∈ (- ∞, √ 3 / 2]
For any two unequal real numbers a and B, if f (a) - f (b) / A-B > 0 holds for the function f (x) over the domain R, then it must have
A. The function f (x) is an increasing then decreasing function
B. The function f (x) is a decreasing then increasing function
C. F (x) is an increasing function on R
D. F (x) is a decreasing function on R
[f(a)-f(b)]/(a-b)>0
When a > b
a-b>0
therefore
[f(a)-f(b)]>0
f(a)>f(b)
The function is an increasing function
When a > b
A-b
C. F (x) is an increasing function on R
f(a)-f(b)/a-b>0
When a > b, f (a) > F (b)
Given that f (x) = X3 + ax and G (x) = 2x2 + B have the same tangent at x = 1, then a + B = ()
A. -1B. 0C. 1D. 2
F ′ (x) = 3x2 + A, G ′ (x) = 4x. From the condition, we know that f (1) = g (1) f ′ (1) = g ′ (1), ∥ 1 + a = 2 + B3 + a = 4, ∥ a = 1b = 0 ∥ a + B = 1, so we choose C