When x ∈ (0,1], the function f (x) = x & # 8308; - 2aX & # 178; the slope of the tangent at any point of the image is less than 1, then the value range of the real number a How to be strict after obtaining 4x & # 179; - 4ax < 1?

When x ∈ (0,1], the function f (x) = x & # 8308; - 2aX & # 178; the slope of the tangent at any point of the image is less than 1, then the value range of the real number a How to be strict after obtaining 4x & # 179; - 4ax < 1?

Let g (x) = 4x ^ 3-4ax x ∈ (0,1] the original problem can be transformed into: X in (0,1], the value of G (x) is always less than 1, that is, the maximum value of G (x) is less than 1
Find the derivative of G (x), let H (x) = 12x ^ 2-4a
1) When A0
In this case, G (x) is an increasing function, and G (x) 3 / 4 does not conform to the condition
2) When a > 0, let H (x) = 0 to get x = √ 3A / 3, x = - √ 3A / 3 (rounding off does not conform to the range of x)
In this case, we need to discuss the size of √ 3A / 3 and 1, because it involves whether x ∈ (0,1) is the function g (x) decreasing first and increasing all the time or decreasing all the time
When you slowly analyze, the general idea is that the maximum value of G (x) in X ∈ (0,1) is less than 1

Draw the function f (x) = Xe ^ - X image (how to draw the function image) and solve the following problems: 1;
2. Find the monotone interval of function f (x);
3. How many roots does the equation x ^ 2E ^ - x = 0 have
4. If the equation x ^ 2E ^ - x-a = 0 has only one real root, what is the value of a?

F (x) is defined as R
1、f’(x)=e^-x-xe^-x=e^-x(1-x)
F '(0) = 1, that is, the tangent slope at the point (0, f (0)) is 1, and it passes through the origin. The tangent equation is y = X
2. Only if x = 1, f '(x) = 0, f (x) has only one extreme point, which is easily known as the maximum point
When x tends to positive infinity, Xe ^ - x tends to zero, and the image gradually approaches the X axis
When x tends to negative infinity, Xe ^ - x tends to negative infinity
Therefore, f (x) increases in (- ∞, 1) and decreases in [1, + ∞)
F (0) = 0, the function image passes through the origin
The function image can be drawn
3. The equation x ^ 2E ^ - x = 0 has only one root, that is, x = 0
4、g(x)= x^2e^-x.g’(x)=x(2-x)e^-x g’(0)=0 g’(2)=0
It is known that G (x) has two extremum points, and it is easy to know that x = 0 is the minimum point
And G (0) = 0, that is, x = 0 is both the zero point and the minimum point of G (x)
So there are two points of intersection between G (x) image and line y = a when a > 0, and one point of intersection when a = 0

Given the function f (x) = a (x-1) - 2lnx, G (x) = Xe ^ 1-x (where a ∈ R, e is the base of natural logarithm)
The function f (x) is always positive in the interval (0, + ∞), and the minimum value of a is obtained

A: F (x) = a (x-1) - 2lnx
F (x) > 0 is constant when x > 0
F (x) = a (x-1) - 2lnx > 0
a(x-1)>2lnx
If x = 1, the above inequality does not hold. If there is a problem with the title, please check and ask

Let f (x) = a (x-1) - 2lnx, G (x) = Xe superscript (1-x) (where a ∈ R, e is the base of natural logarithm)
(1) If the inequality f (x) > 0 holds for any x ∈ (0,1 / 2), find the value range of real number a
1) Let g (x0) hold, and find the value range of A

-Between 1 and 0