The function f (x) = xlnx is known if f (x) ≥ AX-1 holds for any x > 0 How to find the value range of real number a?

The function f (x) = xlnx is known if f (x) ≥ AX-1 holds for any x > 0 How to find the value range of real number a?

a≤1

If the function y = xlnx + A has zero, then the value range of real number a is______ ．

Since the domain of definition of function is (0, + ∞), let the derivative of function y '= LNX + 1 = 0, and get x = 1E. On (0, 1e), y' < 0, y is a decreasing function, on (1E, + ∞), y '> 0, y is an increasing function, so when x = 1E, the function gets the minimum value. In order to make the function have zero point, the minimum value of function should be less than or equal to zero, that is to say 1E · ln1e + a ≤ 0, ｜ a ≤ 1E, that is, the value range of real number a is (− ∞, 1e), so the answer is (− ∞, 1e]