1. For the quadratic function $f(x)=x^{2}-3 x+2$, the number of distinct real roots of the equation $f(f(x))=0$ is ( ). (A)1
(B) 2
(C) 3
(D) 4
Answer: Choose $D$.
Because $f(f(x))=\left(x^{2}-3 x+2\right)^{2}-3\left(x^{2}-3 x+2\right)+2=x^{4}-6 x^{3}+10 x^{2}-3 x$, therefore we have $x(x-3)\left(x^{2}-3 x+1\right)=0, x_{1}=0, x_{2}=3, x_{3,4}=\frac{3 \pm \sqrt{5}}{2}$, so the original equation has 4 distinct real roots.

Note: It is also possible to discuss the distribution of the roots of $f(x)=0$. Since when $x \leq \frac{3}{2}$, the function $f(x)$ is monotonically decreasing, and when $x>\frac{3}{2}$, the function $f(x)$ is monotonically increasing, and the two roots of $f(x)=0$ are 1 and 2, therefore when $x \leq \frac{3}{2}$, the function $f(x) \in\left[-\frac{1}{4},+\infty\right) \supset[1,2]$, so $f(f(x))=0$ has two distinct real roots; when $x>\frac{3}{2}$, the function $f(x) \in\left(-\frac{1}{4},+\infty\right) \supset[1,2]$, so $f(f(x))=0$ also has two distinct real roots. In summary, the original equation has 4 distinct real roots.