It is known that f (x) is a differentiable function defined on (0, + ∞) and satisfies XF '(x) - f (x) ≥ 0. For any positive number a, B, if a > b, there must be () A. af（a）≤bf（b） B. bf（b）≤af（a） C. af（b）≤bf（a） D. bf（a）≤af（b）

It is known that f (x) is a differentiable function defined on (0, + ∞) and satisfies XF '(x) - f (x) ≥ 0. For any positive number a, B, if a > b, there must be () A. af（a）≤bf（b） B. bf（b）≤af（a） C. af（b）≤bf（a） D. bf（a）≤af（b）

F（x）=f(x)
x，
F '(x) = 1
x2[xf′（x）-f（x）]，
From XF ′ (x) - f (x) ≥ 0, it is discussed in two cases:
① XF '(x) - f (x) > 0, so f' (x) > 0, that is, f (x) is an increasing function,
That is, when a > b > 0, f (a) > F (b),
∴f(b)
b＜f(a)
a. Thus AF (b) < BF (a);
② XF '(x) - f (x) = 0, so f (x) is a constant function,
Yes f (b)
b=f(a)
a. That is, AF (b) = BF (a);
AF (b) ≤ BF (a);
Therefore, select C;

F (x) is a differentiable function defined on (0, + ∞) and satisfies XF '(x) - f (x) a

xf'(x)-f(x)=x ² [f(x)/x]'

The differentiable function f (x) defined on (0, + ∞) satisfies: X • f '(x) < f (x) and f (1) = 0, then f (x) The solution set of x < 0 is () A. （0，1） B. （0，1）∪（1，+∞） C. （1，+∞） D. ϕ

The definition field of function f (x) is x > 0, so f (x) < 0,
When f (x) ＜ 0,
xf'（x）＜f（x），
Then XF '(x) ＜ 0,
∵x＞0
∴f'（x）＜0
The function f (x) is a subtractive function on (0, + ∞),
∵f（1）=0
f（x）＜0=f（1）
The solution is x > 1,
Therefore, C

F X is a non positive differentiable function defined on 0 positive infinity and satisfies XF '(x) - f (x)

Let g (x) = f (x) / x, then G '(x) = [XF' (x) - f (x)] / X ² f(b)/b
Multiply both sides by A ²: af(a)>a ² f(b)/b
Because ABF (b)

The nonnegative differentiable function of F (x) defined on (0, + ∞) is known, and it satisfies XF '(x) - f (x) ≥ 0. For any positive numbers a and B, if a < B, ① AF (b) ≤ BF (a); ②af（b）≥bf（a）； ③af（a）≤bf（b）； ④ AF (a) ≥ BF (b). The correct one is () A. ③ B. ①③ C. ②④ D. ②③

The constructor g (x) = XF (x) ‡ g ′ (x) = XF '(x) + F (x) ∵ XF' (x) - f (x) ≥ 0, ∵ g ′ (x) ≥ 2F (x) ≥ 0 ∵ g (x) is a monotonic increasing function on (0, + ∞) ∵ a < B,

F (x) is a nonnegative differentiable function defined on (0, + ∞) and satisfies XF '(x) + F (x) ≤ 0. For any positive numbers a and B, if a

F (x) = f (x) / x, then f '(x) = [XF' (x) - f (x)] / x ^ 2 = [XF '(x) + F (x)] / x ^ 2-2f (x) / x ^ 2F (b) / b, equivalent to
bf(a)>af(b).
Your conclusion is also right and can be used to prove it
af(b)