Given that f (x) = ax ^ 2 + BX + 3A + B is an even function, its domain of definition is [a-1,2a], then the locus of point (a, b) is A. Point B. line C. line ray

Given that f (x) = ax ^ 2 + BX + 3A + B is an even function, its domain of definition is [a-1,2a], then the locus of point (a, b) is A. Point B. line C. line ray

Since f (x) is an even function, then B = 0, and the domain of definition [a-1,2a] must be symmetric about the origin, then A-1 = - 2A gets a = 1 / 3, so point (a, b) is a fixed point (1 / 3,0). Is your topic correct

If the definition field is [2a-1, a ^ 2 + 1] function f (x) = ax ^ 2 + BX + 2a-b is even function, then the trajectory of point (a, b) is: A. one point B, two point C line segments DZ

B
The definition field is [2a-1, a ^ 2 + 1] function f (x) = ax ^ 2 + BX + 2a-b is even function
Then - (2a-1) = a ^ 2 + 1, then a = 0 or - 2
If a = 0, f is even, then B = 0 (a function of degree)
If a = 2, f is even, then B = 0 (quadratic function)

Given that the definition field of even function f (x) is {XLX ≠ 0}, and f (x) increases monotonically within (0, positive infinity), then the size relation of F (- 2), f (1), f (3) is——

F (x) increases monotonically in (0, + ∞)
So f (1) < f (2) < f (3)
And f (x) is even function
So f (- 2) = f (2)
So f (1) < f (- 2) < f (3)

If f (M + 1) < f (2m-1), find the value range of M
Thank you very much for the detailed process

Because f (x) is an even function
There is f (x) = f (- x),
There is f (x) decreasing monotonically on [2,0]
Because f (M + 1) < f (2m-1)
So if we have m + 1 | 2 or M = - 2, we get m > = - 1 / 2; if we have m + 1

It is known that f (x) is an even function and G (x) is an odd function. On the common domain {x | x is not equal to + - 1}, f (x) + G (x) = 1 / (x-1) is satisfied,
Finding the expression of F (x) and G (x)

Let - X be substituted into f (x) + G (x) = 1 / (x-1) (1), where f (x) is an even function and G (x) is an odd function
f（x）-g（x）=1/（-x-1）(2)
(1) (2) f (x) = 1 / [(x + 1) (x-1)] (x is not equal to + - 1)
Substituting (1) can get: G (x) = x / [(x + 1) (x-1)] (x is not equal to + - 1)

F (x) is an even function defined on R, whose image is symmetric with respect to the line x = 2, and if x ∈ (- 2,2), f (x) = - x2 + 1, then if x ∈ (- 6, - 2), f (x) is symmetric with respect to the line x = 2=______ ．

∵ f (x) is an even function defined on R ∵ f (- x) = f (x) ∵ its image is symmetric with respect to the line x = 2 ∵ f (4-x) = f (x) ∵ f (4-x) = f (- x) ∵ f (x) is a periodic function, and the period is 4. Let x ∈ (- 6, - 2), then x + 4 ∈ (- 2,2) so f (x + 4) = - (x + 4) 2 + 1 ∵ f (x) = - (x + 4) 2 + 1, so the answer is: - (x + 4) 2 + 1

Let f (x) be an even function defined on R, whose image is symmetric with respect to the line x = 1
wqer

If f (x) is an even function defined on R, its image is symmetric about y axis
If f (x) is an odd function defined on R, its image is symmetric about the origin

Given the quadratic function f (x) = X2 - (A-1) x + 5, if f (2-x) = f (2 + x), find the real number a
The second question is to find the value range of F (2) if f (x) is an increasing function in the interval (1 / 2,1)

According to the meaning of the title: (2-x) ^ 2 - (A-1) (2-x) + 5 = (2 + x) ^ 2 - (A-1) (2 + x) + 5
-4x-2(a-1)+(a-1)x=4x-2(a-1)-(a-1)x
2(a-1)x=8x
（a-1)=4
a=5
(2)
∵ f (x) is an increasing function in the interval (1 / 2,1)
∴f(1)-f(1/2)＞0
1-(a-1)+5-[(1/2)^2-(a-1)*(1/2)+5]＞0
1-(a-1)+5-1/4+1/2(a-1)-5＞0
3/4-1/2*(a-1）＞0
3＞2（a-1)...(1)
f(2)=4-2(a-1)+5=9-2(a-1)...(2)
From (1) and (2), f (2) > 6

The quadratic function f (x) = ax ^ 2 - (2 + 4a) x + 3A (a) is known

Δ=(2+4a)²-12a²>0
(1+2a)²-3a²>0
1+4a+4a²-3a²>0
a²+4a+1>0
Solve the equation A & # 178; + 4A + 1 = 0
A = - 2 - √ 3; a = - 2 + √ 3
So the solution of a & # 178; + 4A + 1 > 0 is A-2 + √ 3
So the value range of a is A-2 + √ 3

The minimum value of the quadratic function y = x2-4ax + 5a2-3a of the independent variable x is m, and a satisfies the inequality 0 ≤ a2-4a-2 ≤ 10, then what is the maximum value of M?

From 0 ≤ a2-4a-2 ≤ 0, the solution is: - 2 ≤ a ≤ 2-6 or 2 + 6 ≤ a ≤ 6. From y = x2-4ax + 5a2-3a, y = (x-2a) 2 + a2-3a is obtained, then the minimum value m = a2-3a = (A-32) 2-94, and the symmetry axis of its image is a = 32. Among the values of a, the distance between 6 and 32 is the largest. When a = 6, the minimum value m of the original function has the maximum value m = 62-3 × 6 = 18