Given that f (x) = (6-A) x-4a, x < 1 ax, X ≥ 1 are increasing functions on R, then the value range of a is

Given that f (x) = (6-A) x-4a, x < 1 ax, X ≥ 1 are increasing functions on R, then the value range of a is

When x0, we get A1, f (x) increases monotonically with a > 0
If f (x) increases monotonically over R, then (6-A) x-4a ≤ ax when x = 1
6-a-4a ≤ a, a ≥ 1
So 1 ≤ a

Given the set a = {x | x2-2x-8 = 0}, B = {x | x2 + ax + a 2-12 = 0}, B ⊆ a, find the set composed of the value range of real number a

A = {x | x2-2x-8 = 0} = {x | (x-4) (x + 2) = 0} = {- 2,4}, when B = ∈, △ = A2-4 (a2-12) ＜ 0, the solution is a ＞ 4 or a ＜ - 4. When B ≠∞, if there is only one element in B, then △ = A2-4 (a2-12) = 0, the solution is a = ± 4, when a = 4, B = {- 2}, the condition is satisfied; when a = - 4, B = {2}, the condition is not satisfied. When B has two elements in B, a = - 2, and In conclusion, the value set of real number a is {a | a ＜ - 4, or a ≥ 4, or a = - 2}

If f (x) is an even function defined on R and a decreasing function defined on [2,6], then f (- 5)______ F (3) (fill in "<", ">" or "=")

∵ function f (x) is an even function defined on R, ∵ f (- 5) = f (5); ∵ f (x) is a decreasing function on [2,6]; ∵ f (5) < f (3); ∵ f (- 5) < f (3)

2. If the even function y = f (x) is a decreasing function on [2,4], then f (- 3)__ f(-π)

Even function y = f (x) is a decreasing function on [2,4]
Then f (x) is an increasing function on [- 4, - 2]
-4

When | ab | is the minimum, the equation of line L is ()
A. x-y+2=0B. x-y=0C. x+y-2=0D. x+y=0

∵ circle x2 + Y2 + 4x = 0, the standard equation is (x + 2) 2 + y2 = 4 ∵ the coordinate of the center of the circle is C (- 2, 0), so the slope of PC is k = 1 − 0 − 1 + 2 = 1 ∵ when the line L is perpendicular to PC, the slope k '= - 1K = - 1 of | ab | takes the minimum value ∵ L, the equation of line L is Y-1 = - (x + 1), and the simplification is x + y = 0

Let a be a real number, f (x) = x2 + | x-a | + 1, X ∈ R (1) discuss the parity of F (x); (2) find the minimum value of F (x)
A problem similar to the method and thinking of this problem/~

A is a real number, the function f (x) = x ^ 2 + | x-a | + 1, X ∈ R (1) discuss the parity of F (x); (2) find the minimum value of F (x). Analysis: the first question is to check the parity of the function, use the special value method to judge whether the function is odd or even; the second question is to find the maximum value, first judge the monotonicity of the function, and then find

The monotone decreasing interval of function f (x) = X2 - | x | is______ ．

∵ f (- x) = (- x) 2 - | - x | = X2 - | - x | = f (x), ∵ function f (x) = X2 - | - x | - is even function, and its image is symmetric about y-axis. The graph is as follows: ∵ monotone decreasing interval of function f (x) = X2 - | - x | - is (- ∞, - 12] and [0,12]. So the answer is: (- ∞, - 12] and [0,12]

Let f (x) defined on R satisfy f (x) · f (x + 2) = 13, if f (1) = 2, then f (99) = ()
A. 13B. 2C. 132D. 213

∵ f (x) · f (x + 2) = 13 and f (1) = 2 ∵ f (3) = 13F (1) = 132, f (5) = 13F (3) = 2, f (7) = 13F (5) = 132, f (9) = 13F (7) = 2, ∵ f (2n − 1) = 2 & nbsp; n is odd 132 & nbsp; n is even, f (99) = f (2 × 100 − 1) = 132, so C is selected

Let f (x) be differentiable on [0,1], and f (0) = 0, the absolute value of F '(x) is less than or equal to the absolute value of PF (x), and 0 is less than P and less than 1
Let f (x) be differentiable on [0,1], and f (0) = 0, the absolute value of F '(x) is less than or equal to the absolute value of PF (x), and 0 is less than P and less than 1. It is proved that f (x) on [0,1] is equal to 0

Let the maximum value of | f (x) | on [0,1] be | f (a) |, 0 ≤ a ≤ 1
Then | f (a) | = | [0 - > A] f '(T) DT | ≤ P ∫ [0 - > A] | f (T) |
≤p∫[0->a]|f(a)|dt=ap|f(a)|
However, if 0 ≤ AP ≤ P 0, then | f (a) | ≤ 0, that is | f (a) | = 0
When x ∈ [0,1], | f (x) | ≤| f (a) | = 0
That is, f (x) = 0, X ∈ [0,1]

When and where to build the new shopping mall
Which of the above is the correct answer
A.is not decided B.are not decided C.has not decided D.have not decided

A. when and where to build the new shopping mall is a whole. When and where to build the shopping mall is the subject, so the predicate verb is singular. CD is grammatically wrong, and the passive voice should be added