11. $[\mathbf{3 0}]$ Find all functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ such that
$$
\begin{aligned}
f(x) f(y) & =f(x)+f(y)-f(x y) \\
1+f(x+y) & =f(x y)+f(x) f(y)
\end{aligned}
$$
for all rational numbers $x, y$.
Answer: $f(x)=1 \forall x$, and $f(x)=1-x \forall x$
Solution. Considering the first equation, either side of the second equation is equal to $f(x)+f(y)$. Now write $g(x)=1-f(x)$, so that
$$
\begin{array}{l}
g(x y)=1-f(x y)=1-f(x)-f(y)+f(x) f(y)=(1-f(x))(1-f(y))=g(x) g(y) \\
g(x+y)=1-f(x+y)=1-f(x)+1-f(y)=g(x)+g(y)
\end{array}
$$

By induction, $g(n x)=n g(x)$ for all integers $n$, so that $g(p / q)=(p / q) g(1)$ for integers $p$ and $q$ with $q$ nonzero; i.e., $g(x)=x g(1)$. As $g$ is multiplicative, $g(1)=g(1)^{2}$, so the only possibilities are $g(1)=1$ and $g(1)=0$. These give $g(x)=x$ and $g(x)=0$, or $f(x)=1-x$ and $f(x)=1$, respectively. One easily checks that these functions are satisfactory.