Given the real number a > 0, the function f (x) = ax (X-2) 2 (x ∈ R) has a maximum 8. (I) find the monotone interval of function f (x); (II) find the value of real number a

Given the real number a > 0, the function f (x) = ax (X-2) 2 (x ∈ R) has a maximum 8. (I) find the monotone interval of function f (x); (II) find the value of real number a

(I) the derivative of the function is f '(x) = a (X-2) 2 + 2 (X-2) AX = 3ax2-8ax + 4A = 3A (x − 2) (x − 23), because a ＞ 0, then f' (x) ＞ 0, then x ＞ 2 or X ＜ 23, then the function increases monotonically, and from F '(x) ＜ 0, then 23 ＜ x ＜ 2, then the function decreases monotonically. That is to say, the monotone increasing interval of the function is (2, + ∞) and (- ∞, 23). The monotone decreasing interval of the function is (23, 2) (II) from (I), we know that when x = 23, the function has a maximum, so from F (23) = 8, f (23) = 23a (23 − 2) 2 = 32a27 = 8, the solution is a = 274
It is known that the inequality about X, the square of X - 2x + 3 ＞ a, holds for all real numbers
A.a≤2
B.a＞2
C.a＜2
D.a≥0
x²-2x+3＞a
（x-1）²>a-2
∵（x-1）²≥0
∴a-2≤0
a≤2
Option a
Given that a > 0, a ≠ 1, and loga 3 > loga 2, if f (x) = the maximum and minimum of Loga X in the interval [a, 2A]
(1) find the value of A;
(2) solve the inequality Log1 / 3 (x-1) > Log1 / 3 (A-X);
(3) find the range of function f (x) = | loga x | and point out its monotonicity
The difference is 1
1、
loga 3>loga 2
Increasing function
A>1
A
The square of equation x + AX-1 = 0 has several real roots
The discriminant = a ^ 2 + 4 is always greater than 0, so there are two unequal real roots
2
2
If the maximum value of the function f (x) = loga x (0 < a < 1) in the interval [a, 2A] is three times of the minimum value, then the value of a is?
Because 0 < a < 1
So f (x) = loga, X is a decreasing function
So f (a) is three times of F (2a)
That is, loga A is three times as big as loga 2A
1=3*（loga 2a）
1=3*（（loga 2）+1）
-2/3=loga 2
The - 2 / 3 power of a = 2
The square of a under 1 / cubic root = 2
a=√2/4
F (x) = loga x (0 ＜ a ＜ 1), where the interval is a monotone decreasing function, then the maximum value is
loga a=3loga 2a=3 +3 loga 2
So loga 2 = - 2 / 3
a^-2/3=2
So a = 1 / 8 under the root
If the square of the equation (AX + 1) about x = a + 1 has real roots, then the value range of a is
The equation (AX + 1) ^ 2 = a + 1 of X has real roots. Because (AX + 1) ^ 2 is a non negative number, it only needs to satisfy a + 1 ≥ 0, that is, a ≥ - 1. Another solution (AX + 1) ^ 2 = a + 1 is changed to AX ^ 2 + 2x-1 = 0 (1). When a = 0, the equation becomes 2x-1 = 0, x = 1 / 2, which satisfies the problem (2) when a ≠ 0, the equation AX ^ 2 + 2x-1 = 0 is unary binary
If the difference between the maximum value and the minimum value of the function f (x) = loga x (a > 0 and a ≠ 1) in the interval [2,8] is 2, the value of a is obtained
F (x) = loga x is monotone on interval [2,8]
So, | loga 2-loga 8 | = 2
|loga 4|=2
Loga 4 = 2 or loga 4 = - 2
So, a = 2 or a = 1 / 2
It is known that a and B are the real roots of the equation x square + ax + 2A + 1 = 0, and a square + b square
The relationship between the coefficient and the heel
a+b=-a,ab=2a+1.
From a ^ 2 + B ^ 2 = (a + b) ^ 2-2ab
=(-a)^2-2(2a+1)
If the function y = loga (x2-ax + 1) has a minimum value, then the value range of a is ()
A. 0＜a＜1B. 0＜a＜2，a≠1C. 1＜a＜2D. a≥2
Let g (x) = x2-ax + 1 (a ＞ 0, and a ≠ 1). ① when a ＞ 1, G (x) increases monotonically on R, ■ △ 0, ｜ 1 ＜ a ＜ 2; ② when 0 ＜ a ＜ 1, x2-ax + 1 has no maximum value, so the function y = loga (x2-ax + 1) cannot have a minimum value, which is not in line with the meaning of the problem
If the square + ax + 3 = 0 of the equation x has two real roots, one is greater than 1 and the other is less than 1, the value range of a is obtained
The answer is a
It is not enough to use discriminant alone, which can not reflect two roots, one is greater than 1, the other is less than 1
The algebraic method can be as follows:
Image processing using quadratic function
y=x²+ax+3
Let the equation have two real roots, one is greater than 1 and the other is less than 1,
The abscissa of the intersection of the image and the X axis corresponding to the function y = x & # 178; + ax + 3 is greater than 1 and less than 1
The opening of the image is upward, so only when x = 1, the value of the function
Let two be X1 and x2. (x1-1) (x2-1)