Let f (x) be an even function defined on R and increase monotonically in the interval (- ∞, 0), f (2a ^ 2 + A + 1)

Let f (x) be an even function defined on R and increase monotonically in the interval (- ∞, 0), f (2a ^ 2 + A + 1)

2a^2+a+1=2(a+1/4)^2+7/8>0
3a^2-2a+1=2a^2+(a^2-2a+1)=2a^2+(a-1)^2>0
Because f (x) is an even function, symmetric about the y-axis, it increases in the interval (- ∞, 0), and decreases in the interval (0, + ∞)
But f (3a ^ 2-2a + 1) > F (2a ^ 2 + A + 1)
Then 2A ^ 2 + A + 1 > 3A ^ 2-2a + 1
a^2-3a<0
a(a-3)<0
That is 0
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Given that the function f (x) = ax ^ 2 + BX + 3A + B is an even function defined in [a-1,2a], what is the value of a + B?

Even functions define domains that are symmetric about the origin
So A-1 is opposite to 2A
So A-1 + 2A = 0
a=1/3
If even function is symmetric, x = 0
Then x = - B / (2a) = 0
B=0
So, B = 1

Given that the definition domain of the function f (x) = ax ^ 2 + BX + 3A + B is even function of (a-1,2a), then the value of a + B is

Even function, then:
1. Excluding odd degree terms: B = 0
2. The definition domain is a symmetric interval about the origin: A-1 + 2A = 0, and a = 1 / 3
So, a + B = 1 / 3
Wish you happy! Hope to help you, if you do not understand, please ask, I wish you progress_ ∩)O

Given that the function f (x) = ax squared + BX + 3A + B is even function, and its definition domain is [a-1,2a], then a =? B =? B =? RT is not a definition field or a value field

B = 0, there is no first power term of X, even function definition domain symmetry, A-1 = - 2A

Given that f (x) = ax ^ 2 + BX + 3A + B is an even function and the domain is [a-1,2a], then a=_____ ,b=_____ ? (^ 2 is the square)

F (x) is an even function, so f (x) = f (- x) = a (- x) ^ 2-bx + 3A + B
So B = 0
Because f (x) is an even function, its domain must be symmetric about y-axis
Therefore, A-1 = - 2A. Therefore, a = 1 / 3, and its domain is [- 2 / 3,2 / 3]
Oh, I hope it helps

Given that the function f (x) = AX2 + 3a is even function, its definition domain is [A-1, 2A], find the maximum and minimum value of F (x)

∵ the definition domain of even functions [A-1, 2A] is symmetric about the origin,
∴a-1+2a=0
The solution is a = 1
Three
∴f（x）=1
3x2+1
Therefore, when x = 0, the function takes the minimum value of 0
When x = ± 2
When 3, the function takes the maximum value of 13
Nine

Given that the function FX = AX2 + BX + 3A + B is even function, its definition domain is [a-1,2a], find a and B It's a times x to the second power

Because the function is even, its definition domain is symmetric about the origin, so A-1 + 2A = 3a-1 = 0, a = 1 / 3;
FX = AX2 + BX + 3A + B is even function, quadratic function is even function, there is no first order term, so B = 0;

Even function f (x) defined on R decreases monotonically on [0, + ∞), and f (1) 2) If = 0, then f (Log1) is satisfied The set of 4x) < 0 is___ .

∵ even function f (x) defined on R decreases monotonically on [0, + ∞),
The even function f (x) increases monotonically on (- ∞, 0],
And ∵ f (1)
2）=0，
∴f（-1
2）=0，
If f (Log1
4x）＜0
Then Log1
4x＜-1
2, or Log1
4x＞1
Two
X > 2, or 0 < x < 1
Two
So the answer is: (0, 1
2）∪（2，+∞）

Even function y = f (x) defined on R decreases on [0, + ∞), and f (1) 2) If = 0, then f (Log1) is satisfied The set of X with 4x < 0 is () A. (−∞，1 2)∪(2，+∞) B. (1 2，1)∪(1，2) C. (1 2，1)∪(2，+∞) D. (0，1 2)∪(2，+∞)

Because the even function y = f (x) defined on R decreases on [0, + ∞), and f (1)
2) If = 0, then f (Log1) is satisfied
4x)＜0
⇔f(|log1
4x|)＜0＝f(1
2)⇔|log1
4x|＞1
2⇔
log1
4x≥0
log1
4x＞1
2 or
log1
4x＜0
−log1
4x＞1
2 ⇒0＜x＜1
2 or x > 2
Therefore, D

The even function y = f (x) defined on R increases on (- ∞, 0], and a zero point of the function y = f (x) is - 1 2. Satisfy f (Log1) 4x) ≥ 0

∵-1
2 is the zero point of the function, ν f (− 1)
2)＝0，… (1 point)
∵ f (x) is an even function, ᙽ f (1)
2)＝0，… (2 points)
∵ f (x) increases on (- ∞, 0], f (Log1)
4x)≥f(−1
2)… (4 points)
∴0≥log1
4x≥-1
2，∴1≤x≤2，… (7 points)
∵ f (x) is an even function, ᙽ f (x) is a single adjustment and subtraction on on [0, + ∞) (8 points)
F (Log1)
4x)≥f(1
2)，∴0≤log1
4x≤1
2，∴1
2≤x≤1，∴1
2≤x≤2．… (11 points)
Therefore, the value set of X is {x | 1
2≤x≤2}．… (12 points)