If the function f (x) = x ^ 2-2mx is a decreasing function on X ≤ 1, find the range of M

If the function f (x) = x ^ 2-2mx is a decreasing function on X ≤ 1, find the range of M

This is a parabola with the opening up
So on the left side of the axis of symmetry x = m is a decreasing function
Now it's a decreasing function on X ≤ 1
Then the axis of symmetry is on the right side of the interval
So m ≥ 1
Given that the definition domain of function y = SiNx is [a, b], and the value domain is [- 1,1 / 2], then the value of B-A may be? For detailed explanation
2π/3
The range of SiNx on [- π / 2, π / 6] is [- 1,1 / 2]
A
If the function y = x2 + 2mx + 3 is a decreasing function on (- ∞, 2], then the value range of M is_____ .
Parabolic opening upward
Axis of symmetry x = - B / 2A = - M
In (- ∞, 2] is a decreasing function
To the left of the axis of symmetry is a decreasing function
It can be seen that the axis of symmetry moves with the function image
So make the axis of symmetry greater than or equal to 2
Namely
-m≥2
m≤-2
The most important problem is the combination of number and shape. If you can't think of it, you can draw pictures to help you understand it
Given that the function y = cos ^ 2x + asinx-2a + 5 has the maximum value 2, find the real number a
above.
Y = cos & sup2; X + asinx-2a + 5, = 1-sin & sup2; X + asinx-2a + 5 = - (sinx-a / 2) & sup2; + A & sup2 / / 4-2a + 6. If y is to have the maximum, then - (sinx-a / 2) & sup2; is to be the maximum, then sinx-a / 2 = 0, SiNx = A / 2, and | SiNx | ≤ 1, that is, | A / 2 | ≤ 1, - 2 ≤ a ≤ 2
Y = cos ^ 2x + asinx-2a + 5 = 1-2sin & sup2; X + asinx-2a + 5, where t = SiNx ∈ [- 1, 1]
y=-2t²+at+6-2a
When a / 4 ≥ 1, the maximum value is obtained when t = 1, that is, a = 2 does not satisfy the range rounding
When - 1 ＜ A / 4 ＜ 1, ymax = A & sup2 / 8-2a + 6 = 2, the solution is a = 8 ± 4 √ 2
When a / 4 ≤ - 1, t = - 1 takes the maximum value
Y = cos ^ 2x + asinx-2a + 5 = 1-2sin & sup2; X + asinx-2a + 5, where t = SiNx ∈ [- 1, 1]
y=-2t²+at+6-2a
When a / 4 ≥ 1, the maximum value is obtained when t = 1, that is, a = 2 does not satisfy the range rounding
When - 1 ＜ A / 4 ＜ 1, ymax = A & sup2 / 8-2a + 6 = 2, the solution is a = 8 ± 4 √ 2
When a / 4 ≤ - 1, the maximum value of T = - 1 is a = 2 / 3
To sum up, only a = 8-4 √ 2
PS: what else? Why is the sum of squares and multipliers always inconsistent now? I often read it wrong
Let f (x) be a decreasing function defined on (- 2,2) and satisfy the following conditions: F (- x) = - f (x), and f (m-1) + F (2m-1) > 0, then the value range of real number m is obtained
The inequality f (m-1) + F (2m-1) > 0, that is, f (m-1) > F (2m-1), ∵ f (- x) = - f (x), we can get - f (2m-1) = f (- 2m + 1) ∵ the original inequality is transformed into f (m-1) > F (- 2m + 1) and ∵ f (x) is a decreasing function defined on (- 2,2), ∵ - 2 < M-1 < - 2m + 1 < 2, the solution is - 12
It is known that the square + 2A + 5 function of y = cos square x + asinx-a has a maximum value of 2. Try to find the value of real number
Y = 1-sin ^ 2x + asinx-a ^ 2 + 2A + 5 = - (sinx-a / 2) ^ 2-3 / 4A ^ 2 + 2A + 6 when sinx-a / 2 = 0, y has the maximum value, that is - 3 / 4A ^ 2 + 2A + 6 = 2. Solving the equation, a = - 4 / 3, a = 4, rounding off, so a = - 4 / 3
Given the function y = (2m-3) x + (m-1), the value range of M is determined according to the following conditions
(1) The value of function y increases with the increase of the value of X
(2) The intersection of its image and Y axis is on the negative half axis
(1) According to the meaning 2m-3 > 0
So m > 3 / 2
So m ∈ (3 / 2, positive infinity)
(2) According to the meaning of the title, M-1
Given that the function y = cos2x + asinx-a2 + 2A + 5 has the maximum value 2, try to find the value of real number a
Let SiNx = t, then y = f (T) = - t (T) = - T 2 + at-a2 + 2A + 6, t ∈ [-1, 1, 1], the symmetry axis is t = A2, the symmetry axis is t = A2, when A2 ＜ − 1, that is, a ＜ - 2, ymax = f (- 1) = -a2 + A + A + 5 = 2, a = 1 ± 132 (rounding) when − 1 ≤ A2 ≤ 1, that is - 2 ≤ a ≤ 2 ≤ 2, that is - 2 ≤ 2 ≤ a ≤ 2, ymax = f (A2) = f (A2) = f (A2) = 34a2 + 2A + 2A + 2A + 2A + 2A + 2A + 6 + 6 = 2, that is, at this time, a = 4 (rounding) or a = 43. When A2, that is a \\\\\\\\\+ 5 = 2, a = 3 + 212 or a = 3 − 212, so a = 3 + 212 or a = − 43
Inequality (m ^ 2-2m-3) x ^ 2 - (M-3) X-1
∵ the solution set is r
Sub situation
Case 1:
M ^ 2-2m-3 = 0 and M-3 = 0
The solution is m = 3
Case 2:
When m ^ 2-2m-0 is not equal to 0
∴m^2-2m-3
The known function y = - cos ^ 2x + asinx + 1 / 2 (a is a constant, and a
y=sin^2x+asinx-1/2
=(sinx+a/2)^2-1/2-a^2/4.
Also a