Question 11. Let $\mathrm{R}$ be the set of all real numbers, and for the function
$$
f(x)=x^{2}+a x+b(a, b \in R),
$$

define the sets
$$
\begin{array}{l}
A=\{\mathbf{x} \mid \mathbf{x}=f(\mathbf{x}), \mathbf{x} \in \mathbf{R}\}, \\
B=\{\mathbf{x} \mid \mathbf{x}=\mathrm{f}(\mathrm{f}(\mathbf{x})), \mathbf{x} \in \mathbf{R}\} .
\end{array}
$$
(1) If $a=-1, b=-2$, find $A \cup B, A \cap B$;
(2) If $\mathbf{A}=\{-1,3\}$, find $B$.
(3) If $A=\{\alpha\}$, prove that $A \cap B=\{\alpha\}$.
Solve (1) From the given conditions, the function
$$
f(x)=x^{2}-x-2 .
$$

The equation $x=f(x)$ becomes $x^{2}-2 x-2=0$.
Its solution set is $\mathrm{A}$, so
$$
A=\{1-\sqrt{3},1+\sqrt{3}\} .
$$

Similarly, the equation $x=f(f(x))$ is
$$
x=\left(x^{2}-x-2\right)^{2}-\left(x^{2}-x-2\right)-2,
$$

Simplifying, we get $\left(x^{2}-x-2\right)^{2}-x^{2}=0$.
That is, $\left(x^{2}-2\right)\left(x^{2}-2 x-2\right)=0$.
Thus, $x^{2}-2=0$ or $x^{2}-2 x-2=0$.
The solution set is $\mathrm{B}$. Let the solution set of $\mathrm{x}^{2}-2=0$ be $\mathrm{C}$, then
$$
\mathrm{B}=\mathrm{A} \cup \mathrm{C} .
$$

Therefore,
$$
\begin{array}{l}
A \cup B=A \cup A \cup C=A \cup C \\
=\{1-\sqrt{3}, 1+\sqrt{3},-\sqrt{2}, \sqrt{2}\}
\end{array}
$$

Since $B \supseteq A$, it follows that
$$
\mathbf{A} \cap \mathrm{B}=\mathbf{A}=\{1-\sqrt{3}, 1+\sqrt{3}\} .
$$
(2) From $-1 \in \mathbf{A}, 3 \in \mathbf{A}$, we get
$$
\left\{\begin{array}{l}
f(-1)=0, \\
f(3)=0 . \\
1-a+b+1=0, \\
9+3 a+b-3=0 \\
a=-1, \\
b=-3 .
\end{array}\right.
$$

Thus, $f(x)=x^{2}-x-3$.
Therefore, the equation $\mathrm{f}(\mathrm{f}(\mathrm{x}))=\mathrm{x}$ should be
$$
\left(x^{2}-x-3\right)^{2}-\left(x^{2}-x-3\right)-3=x \text {. }
$$

Simplifying, we get $\left(x^{2}-x-3\right)^{2}-x^{2}=0$.
The solution set is
$$
\mathrm{B}=\{-\sqrt{3}, \sqrt{3},-1,3\} .
$$
(3) We only need to prove $\alpha \in B$. From $\alpha \in A$, we get $f(\alpha)=\alpha$.

Thus, $f(f(\alpha))=f(\alpha)=\alpha$.
Therefore, $\alpha \in B$. Since $A=\{\alpha\}$, we have
$$
A \cap B=\{\alpha\} .
$$
(Here, we ignore the multiplicity of the roots, although $\alpha$ may have different multiplicities in different equations, we consider $\alpha$ to be the same element.)