2・106 Consider the set of all equations of the form:
$$
x^{3}+a_{2} x^{2}+a_{1} x+a_{0}=0
$$

where $a_{2}, a_{1}, a_{0}$ are real constants, and $\left|a_{i}\right| \leqslant 2,(i=0,1,2)$. Let $r$ be the largest positive real number that satisfies at least one such equation. Then
(A) $1 \leqslant r<\frac{3}{2}$.
(B) $\frac{3}{2} \leqslant r<2$.
(C) $2 \leqslant r<\frac{5}{2}$.
(D) $\frac{5}{2} \leqslant r<3$.
(E) $3 \leqslant r<\frac{7}{2}$.
(32nd American High School Mathematics Examination, 1981)
[Solution] Let $g(x)=x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$ be a special cubic polynomial that satisfies the given conditions. We know that when $x$ is sufficiently large, the value of $g(x)$ will depend on $x^{3}$, where the coefficient of the highest degree term of $g(x)$ is $1>0$. Therefore, for all $x$ greater than the largest root of $g(x)=0$, we have $g(x)>0$.

To maximize the largest real root of the equation, the terms $a_{2} x^{2}+a_{1} x+a_{0}$ should be as small as possible. Given the condition $\left|a_{i}\right| \leqslant 2$, we consider the extreme case:
$$
f(x)=x^{3}-2 x^{2}-2 x-2
$$

Let's assume the largest positive real root of this equation is $r_{0}$.
For all $x \geqslant 0$, we have
$$
-2 x^{2} \leqslant a_{2} x^{2}, \quad-2 x \leqslant a_{1} x, \quad-2 \leqslant a_{0},
$$

Adding these three inequalities and then adding $x^{3}$ to both sides, we get that for all $x \geqslant 0$:
$$
f(x)=x^{3}-2 x^{2}-2 x-2 \leqslant x^{3}+a_{2} x^{2}+a_{1} x+a_{0}=g(x).
$$

Therefore, for all $x>r_{0}$, we have $0<f(x) \leqslant g(x)$, so they cannot be roots of $g(x)$. This means that any number greater than $r_{0}$ will not be a root of $g(x)$, and thus $r_{0}$ is the desired $r$.

Sketching the graph of $f(x)$, we can see that it is a standard cubic curve, and its largest root is slightly less than 3.
In fact, $f(2)=-6, f(3)=1$, which indicates that the answer can only be $(\mathrm{C})$ or $(\mathrm{D})$.
To be certain, we calculate $f\left(\frac{5}{2}\right)=-\frac{31}{8}$, which is negative.
Therefore, the answer is $(D)$.