Question 40: Find all pairs of real numbers $(a, b)$ such that the function $\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}$ satisfies for any real numbers $\mathrm{x}, \mathrm{y} \in$ $[0,1]$, we have $\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})+\mathrm{f}(\mathrm{x}+\mathrm{y}-\mathrm{xy}) \leq 0$.
Question 40, Solution: Let $F(x, y)=f(x) f(y)+f(x+y-x y)=a(a-1) x y+a(b+1)(x+$ $y)+b(b+1)$, then $F_{x}(x, y)$ and $F_{y}(x, y)$ are linear functions of $x$ and $y$, respectively, hence
$$
\begin{array}{l}
F(x, y) \leq 0(x, y \in[0,1]) \\
\Leftrightarrow\left\{\begin{array}{ll}
F(0, & 0) \leq 0 \\
F(1, & 1) \leq 0 \\
F(1, & 0) \leq 0 \\
F(0, & 1) \leq 0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{c}
b(b+1) \leq 0 \\
a(a-1)+2 a(b+1)+b(b+1) \leq 0 \\
a(b+1)+b(b+1) \leq 0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{c}
b(b+1) \leq 0 \\
(a+b)(a+b+1) \leq 0 \\
(a+b)(b+1) \leq 0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{c}
b \in[-1,0] \\
a \in[-(b+1),-b]
\end{array}\right. \\
\end{array}
$$

In summary, the set of $(a, b)$ is $\{(a, b) \mid -1 \leq b \leq 0, -(b+1) \leq a \leq -b\}$.