Let y = log2 (AX ^ 2-2x + 2) > 2 be constant on X, and find the value range of real number a

The function log2 (AX ^ 2-2x + 2) > 2 holds on X ∈ [1,2]
The function log2 (AX ^ 2-2x + 2) > log2 (4) holds on X ∈ [1,2]
∵ y = log2 (x) is an increasing function on (0, + ∞)
On X ∈ [1,2], ax ^ 2-2x + 2 > 4 is constant
That is, ax & # 178; > 2x + 2 is constant on X ∈ [1,2]
That is to say, a > 2 / x + 2 / X & # 178; it holds on X ∈ [1,2]
The maximum value of a > (2 / x + 2 / X & # 178;)
∵ f (x) = 2 / x + 2 / X & # 178; is a decreasing function on [1,2]
The maximum value of F (x) is f (1) = 4
∴ a>4
That is, the value range of a is a > 4. I hope it is useful for you! Please adopt it in time!

The proposition function f (x) = Log1 / 3 (x ^ 2-2ax + 3a) is a decreasing function on the interval (1, + infinity)

Let x ^ 2-2ax + 3A be t, then it is easy to know that log3 / 1 (T) is a decreasing function. Because of the composite function, it is necessary to let x ^ 2-2ax + 3A be an increasing function on (1, + infinity), so as long as - B / 2A (axis of symmetry) is less than or equal to 1 and G (1) is greater than or equal to 0, G (x) = x ^ 2-2ax + 3a

Let the proposition p: function f (x) = x2-2ax increase on (1, + ∞); the definition field of proposition q: function y = LG (ax2-x + a) is R. if P or q is true, P and Q is false, the value range of a is obtained

The image of function f (x) = x2-2ax is a parabola with the opening upward, and the axis of symmetry is x = A. to make function f (x) = x2-2ax increase on (1, + ∞), only a ≤ 1 is needed; the domain of definition of function y = LG (ax2-x + a) is r, that is, ax2-x + a > 0 holds for any x, so a > 0 △ = 1 − 4a2 < 0, and the solution is a > 12. When P or q is true, P and Q are false, P, q are true and false If P is true and Q is false, a ≤ 12 can be obtained from a ≤ 1a ≤ 12; if q is true and P is false, a > 1 can be obtained from a > 1A > 12, so the value range of a is a ≤ 12 or a > 1

Given proposition p: for any x ∈ [2,3], X & # 178; + 1 ≥ ax, proposition q: there exists x ∈ R, X & # 178; + 2aX + 2-A = 0, if proposition p and Q are true propositions
Then the value range of real number a is

Let y = log2 (AX ^ 2-2x + 2) > 2 be constant on X, and find the value range of real number a