Question 6: Given $a, b \in \mathbb{R}, f(x) = x^2 + ax + b$, define $A = \{x \in \mathbb{R} \mid f(x) \leq 0\}, B = \{x \in \mathbb{R} \mid f(f(x)) \leq 3\}$, if $A = B \neq \emptyset$, try to find the range of values for $a$.
Question 6, Answer: Let the two roots of $f(x)$ be $x_{1} \leq x_{2}$, then $A=\left[x_{1}, x_{2}\right]$, hence $B=\left[x_{1}, x_{2}\right]$, so $x_{1}, x_{2}$ are also roots of $f(f(x))=3$. Thus, $3=f\left(f\left(x_{1}\right)\right)=f(0)=b$, hence
$$
\begin{array}{l}
f(x)=x^{2}+ax+3 \\
f(f(x))-3=\left(x^{2}+ax+3\right)^{2}+a\left(x^{2}+ax+3\right)=\left(x^{2}+ax+3\right)\left(x^{2}+ax+a+3\right)
\end{array}
$$

Therefore, the condition is equivalent to:
$$
\begin{array}{l}
\left\{\begin{array}{c}
x^{2}+a x+3=0 \text { has real roots } \\
x^{2}+a x+a+3=0 \text { has no real roots }
\end{array}\right. \text { } \\
\Leftrightarrow\left\{\begin{array}{c}
a^{2}-12 \geq 0 \\
a^{2}-4(a+3)<0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{r}
a \in(-\infty,-2 \sqrt{3}] \cup[2 \sqrt{3},+\infty) \\
a \in(-2,6)
\end{array}\right. \\
\Leftrightarrow a \in[2 \sqrt{3}, 6)
\end{array}
$$

In summary, the required $a \in[2 \sqrt{3}, 6)$.