It is known that f (x) is an even function, and it is a minus function on the interval [AB], (0

It is known that f (x) is an even function, and it is a minus function on the interval [AB], (0

Let x1, X2 belong to [- B, - A], and X1 therefore - x1, - x2 belong to [a, b] and - X1 > - x2
Since f is an even function and a minus function on the interval [AB], so
F (x1) = f (- x1), so f is an increasing function on [- B, - A]

If the function f (x) = (A-2) x2 + (A-1) x + 3 is even, then the monotone decreasing interval of function f (x) is______ .

∵ the function f (x) = (A-2) x2 + (A-1) x + 3 is an even function,
∴f（-x）=f（x）
∴（a-2）x2-（a-1）x+3=（a-2）x2+（a-1）x+3
ν - (A-1) = A-1, a = 1
∴f（x）=-x2+3
The monotone decreasing interval of function f (x) is [0, + ∞)
So the answer is: [0, + ∞)

If y = f (x) is an even function and a decreasing function on [0, positive infinity], then the monotone increasing interval of F (1-x ^ 2) is

∵ y = f (x) is an even function and a minus function on [0, + ∞)
ν y = f (x) is an increasing function on (- ∞, 0]
∵ y = f (x) is a decreasing function on [0, + ∞)
∴1-x^2∈[0,＋∞）,x∈[-1,1]
∵ y = 1-x ^ 2 is a minus function in [0, + ∞)
ν f (1-x ^ 2) increases monotonically on [0,1]
∵ y = f (x) is an increasing function on (- ∞, 0]
∴1-x^2∈（－∞,0],x∈（－∞,-1]∪[1,＋∞）
∵ y = 1-x ^ 2 is an increasing function on (- ∞, 0)
ν f (1-x ^ 2) increases monotonically on (- ∞, - 1]
The monotonic increasing interval of F (1-x ^ 2) is
（－∞,-1],[0,1]

Let y = f (x) be an even function on R and a minus function on the open interval from zero to positive infinity 1.f(-x1)>f(-x2) 2.f(-x1)=f(-x2) 3.f(-x1)

1、 Therefore - x1f (x2), i.e. f (- x1) > F (- x2)

If the function y = f (x) is an even function and a decreasing function on [0, + ∞), then the monotonic increasing interval of F (4-x2) is______ .

∵ the function y = f (x) is an even function and a minus function on [0, + ∞),
ν f (x) increases monotonically on (- ∞, 0)
Let t = 4-x2, then when t = 4-x2 ≥ 0, - 2 ≤ x ≤ 2, and the function T increases monotonically on X ∈ [- 2,0], and t decreases monotonically on X ∈ [0,2]
According to the same increase and different decrease of composite function, we know that the function f (4-x2) increases monotonically on [0,2]
Similarly, we can find that the function f (4-x2) increases monotonically on (- ∞, - 2]
So the answer is: (- ∞, - 2], [0, 2]

Given that the even function y = f (x) is a monotone decreasing function on [- 1,0], and α and β are the two inner angles of an acute triangle, then () A. f（sinα）＞f（cosβ） B. f（sinα）＜f（cosβ） C. f（sinα）＞f（sinβ） D. f（cosα）＞f（cosβ）

∵ even function y = f (x) is monotone decreasing function on [- 1, 0]
ν f (x) is a monotone increasing function on [0, 1]
α and β are the two inner angles of the acute triangle
∴α+β＞π
Two
∴π
2＞α＞π
2-β＞0
∴1＞sinα＞sin（ π
2−β）=cosβ＞0
∴f（sinα）＞f（cosβ）
Therefore, a

If the even function f (x) is a decreasing function in the interval [- 1,0], α and β are the two inner angles of an acute triangle, and α ≠ β, then the correct one in the following inequalities is () A. f（cosα）＞f（cosβ） B. f（sinα）＞f（cosβ） C. f（sinα）＞f（sinβ） D. f（cosα）＞f（sinβ）

∵ the even function f (x) is a minus function on the interval [- 1, 0],
ν f (x) is an increasing function on the interval [0,1]
In addition, α and β are the two inner angles of an acute triangle,
∴α+β＞π
2，α＞π
2-β，1＞sinα＞cosβ＞0．
∴f（sinα）＞f（cosβ）．
Therefore, B is selected

It is known that even function f (x) is monotone decreasing function on 〔 - 1.0] and α, β are two inner angles of acute triangle Because α and β are the two inner angles of an acute triangle, α + β > 90 degrees, so (1) when α "45 degrees", 45 degrees cos α (2) when 45 degrees

α "45 then 90 - α"

It is known that the function f (x) is an even function on R and a minus function on [- 1,0]. If α and β are the two inner angles of an acute triangle F (sin α), f (COS α), f (sin β), f (COS β) were compared

If f (x) is an even function on R, it is a decreasing function on [- 1,0], then it is an increasing function on [0,1]. Because a and B are the two inner angles of an acute triangle, a + b > π / 2, so a > π / 2-B > 0. Because y = SiNx is an increasing function on (0, π / 2), so Sina > sin (π / 2-B), that is sin

If the even function f (x) is a decreasing function in the interval [- 1,0], α and β are the two inner angles of an acute triangle, and α ≠ β, then the correct one in the following inequalities is () A. f（cosα）＞f（cosβ） B. f（sinα）＞f（cosβ） C. f（sinα）＞f（sinβ） D. f（cosα）＞f（sinβ）

∵ the even function f (x) is a minus function on the interval [- 1, 0],
ν f (x) is an increasing function on the interval [0,1]
In addition, α and β are the two inner angles of an acute triangle,
∴α+β＞π
2，α＞π
2-β，1＞sinα＞cosβ＞0．
∴f（sinα）＞f（cosβ）．
Therefore, B is selected