Given the function f (x) = x2-2ax + 5, if f (x) is a decreasing function in the interval (- ∞, 2), and for any x1, X2 ∈ [1, a + 1], there is always | f (x1) - f (x2) | ≤ 4, the value range of real number a is obtained

Given the function f (x) = x2-2ax + 5, if f (x) is a decreasing function in the interval (- ∞, 2), and for any x1, X2 ∈ [1, a + 1], there is always | f (x1) - f (x2) | ≤ 4, the value range of real number a is obtained

The axis of symmetry of ∵ f (x) = x2-2ax + 5 is x = a, and f (x) is a decreasing function in the interval (- ∞, 2}, and ∵ a ≥ 2. For any x1, X2 ∈ [1, a + 1], there are always | f (x1) - f (x2) | ≤ 4, | f (1) - f (a) | ≤ 4, that is | a2-2a + 1 | ≤ 4, and the solution is - 1 ≤ a ≤ 3. In conclusion, the value range of a is {a | 2 ≤ a ≤ 3}

Quadratic inequality of one variable
ax2+ax+a+1＞0
Find the value range of A
That two is the square of X

Let the original formula = y, y = AX2 + ax + A + 1
Because y > 0 means that there is a minimum value, a > 0 (1-ary quadratic inequality, a cannot be equal to 0, in fact, if it is not 1-ary quadratic inequality, it can also be a = 0)
So using the formula method, when x = - 1 / 2, the minimum value of y = (3a ^ 2 + 4a) / 4A > 0
So 3A ^ 2 + 4A > 0 (a > 0)
So A0

0 ＜ B ＜ 1 + A, if there are exactly three integers in the solution set of inequality (X-B) ^ 2 ＞ (AX) ^ 2 about X, then ask a to get the value range?

Give me an idea
(X-B) ^ 2 - (AX) ^ 2 > 0, [(1 + a) X-B] [(1-A) X-B] > 0, because there are exactly three integers in the solution set
So the opening direction of the function corresponding to the quadratic inequality is downward, that is, the coefficient of the quadratic term is less than 0, so there is 1-a0, that is [(1 + a) X-B] [(A-1) x + b]