Given that the function f (x) = x + ax is an increasing function in the interval [2, + ∞), then the value range of real number a is______ ．

Given that the function f (x) = x + ax is an increasing function in the interval [2, + ∞), then the value range of real number a is______ ．

When a ≤ 0, the function f (x) = x + ax is an increasing function on R, satisfying the condition. When a > 0, ∵ x ∈ [2, + ∞), X2 ≥ 4, from F '(x) = 1-ax2 ≥ 0, i.e. a ≤ X2, 0 < A ≤ 4 can be obtained. In conclusion, a ≤ 4 can be obtained, so the answer is: {a | a ≤ 4}

If the function f (x) = x ^ 2 - (2 ^ a) x + 1 has an inverse function on [2,4], then the value range of real number a is

The inverse function is monotone
So the axis of symmetry x = 2 ^ A / 2 is not in the interval (2,4)
SO 2 ^ A / 2 = 4
2^a=8=2^3
So a = 3

If the function y = x + 4x has an inverse function on X ∈ (0, a], then the value range of real number a is______ ．

In this paper, the inverse function of function y = x + 4x on X ∈ (0, a], and the monotone function of function y = x + 4x on X ∈ (0, 2)]. The value range of real number a is (0, 2], so the answer is (0, 2]

Given that a and B are real numbers, the system of inequalities whose solution can be - 2 < x < 2 is ()
A. ax＞1bx＞1B. ax＞1bx＜1C. ax＜1bx＞1D. ax＜1bx＜1

A. If the solution set of the given inequality system is - 2 ＜ x ＜ 2, then a and B are positive and negative. If a ＞ 0, then B ＜ 0, the solution is x ＞ 1a, X ＜ 1b. The original inequality system has no solution. Similarly, if the sign of two numbers is changed, there is no solution, so it is wrong and does not conform to the meaning of the problem. B. if the solution set of the given inequality system is - 2 ＜ x ＜ 2, then