5. (2007 Shaanxi Province College Entrance Examination, Science Subject, Question 11) $f(x)$ is a non-negative differentiable function defined on $(0,+\infty)$, and satisfies $x f^{\prime}(x)+f(x) \leqslant 0$. For any positive numbers $a$ and $b$, if $a<b$, then it must be true that
A. $a f(a) \leqslant f(b)$
B. $b f(b) \leqslant f(a)$
C. $a f(b) \leqslant b f(a)$
D. $b f(a) \leqslant a f(b)$
5. First, the problem has 3 conditions:
(1) $f(x)$ is a non-negative differentiable function defined on $(0,+\infty)$. Related to this, there are two mathematical concepts to clarify: one is a non-negative function $(f(x) \geqslant 0)$, and the other is a differentiable function ($f^{\prime}(x)$ exists).
(2) The inequality $x f^{\prime}(x)+f(x) \leqslant 0$. Its mathematical meaning can be understood from multiple perspectives, such as:
(1) $x f^{\prime}(x)+f(x) \leqslant 0 \Leftrightarrow f^{\prime}(x)=-\frac{f(x)}{x} \leqslant 0$, indicating that $f(x)$ is a non-increasing function.
(2) $[x f(x)]^{\prime}=x f^{\prime}(x)+f(x) \leqslant 0 \Rightarrow F(x)=x f(x)$ is a non-increasing function.
(3) The inequality $0<a<b$.
Second, the conclusion of the problem is one of 4 options, each of which is an inequality involving the comparison of $f(a)$, $f(b)$, $a f(a)$, $b f(b)$, $a f(b)$, and $b f(a)$.
Third, the basic connection between the conditions and the conclusion is to prove an inequality from another inequality, which is a "conditional inequality proof."
Thus, after reviewing the problem, at least two general methods can be derived from the understanding of $x f^{\prime}(x)+f(x) \leqslant 0$. (Special methods for multiple-choice questions are not discussed here)

Solution 1: From $x f^{\prime}(x)+f(x) \leqslant 0 \Rightarrow f^{\prime}(x)=-\frac{f(x)}{x} \leqslant 0$, we know that $f(x)$ is a non-increasing function on $(0,+\infty)$. For $0 < a < b$, we have $0 \leqslant f(b) \leqslant f(a)$.
Multiplying, we get $0 \leqslant a f(b) \leqslant b f(a)$. Choose $C$.
Solution 2: Define the function $F(x)=x f(x)$. From
$$
F^{\prime}(x)=[x f(x)]^{\prime}=x f^{\prime}(x)+f(x) \leqslant 0 .
$$
we know that the function $F(x)=x f(x)$ is a non-increasing function on $(0,+\infty)$. For $0 < a < b$, we have
$$
b f(b) \leqslant a f(a) \text{. }
$$
By shrinking the left side and expanding the right side, we get
$$
a f(b) \leqslant b f(b) \leqslant a f(a) \leqslant b f(a) . \text{ Choose } C .
$$