The function f (x) = xlnx is known 1. If the function g (x) = f (x) + x ^ 2 + ax + 2 has zeros, find the maximum of real number a 2. If any x is greater than 0 and f (x) / X is less than or equal to x-kx ^ 2-1, the value range of real number k is obtained

The function f (x) = xlnx is known 1. If the function g (x) = f (x) + x ^ 2 + ax + 2 has zeros, find the maximum of real number a 2. If any x is greater than 0 and f (x) / X is less than or equal to x-kx ^ 2-1, the value range of real number k is obtained

Given the function f (x) = xlnx1, if the function g (x) = f (x) + x ^ 2 + ax + 2 has zero, find the maximum value of the real number A. if any x is greater than 0, f (x) / X is less than or equal to x-kx ^ 2-1, find the value range of the real number K. (1) analytic: ∵ function f (x) = xlnx, Let f '(x) = LNX + 1 = 0 = = > x = 1 / E, f' '(x) = 1 / x > 0 ∵ function f (x) ∵

Given the function f (x) = xlnx, we discuss the number of solutions of the equation f (x) - M = (m ∈ R) about X
The first question is how to calculate the equation of minimum y = f (1 / E) and the second question is how to write it?

1) F '(x) = LNX + 1 = 0, x = 1 / E
F (1 / E) is the minimum, f (1 / E) = 1 / E * ln (1 / E) = - 1 / E
It is also the minimum value, that is, the minimum value is - 1 / E
2) From M = f (x)
00, f (x) increases monotonically, f (+ ∞) - > + ∞, so the value in this interval is (- 1 / E, + ∞)
Therefore, there are:
m

Given f (x) = xlnx, G (x) = - x ^ 2 + ax + x-3, if for all x ∈ (0, + ∞), 2f (x) ≥ g (x) is constant

2F (x) ≥ g (x), X ∈ (0, + ∞), that is, 2xlnx ≥ - X & # 178; + ax + x-3, ax ≤ 2x · LNX + X & # 178; - x + 3, a ≤ 2lnx + X - 1 + 3 / x, X ∈ (0, + ∞), let H (x) = 2lnx + X - 1 + 3 / x, so that a ≤ [H (x)] min, X ∈ (0, + ∞), obtain H '(x) = 2 / x + 1 - 3 / X & # 178; = (X & # 178; + 2x -

Let f (x) = xlnx + 4, if f (x) ≤ ax & sup2; - ax + 4 when x ≥ 1, the value range of a is obtained

F (x) ≤ ax & # 178; - ax + 4 is equivalent to xlnx ≤ ax & # 178; - ax. It is equivalent to LNX ≤ a (x-1). (because x ≥ 1) when x = 1, the above formula is 0 ≤ 0, which holds. When x > 1, X-1 > 0, the above formula is a ≥ LNX / (x-1), which only requires the maximum value on the right side