$13 \cdot 33$ If in the interval $[1,2]$, the function $f(x)=x^{2}+p x+q$ and $g(x)$ $=x+\frac{1}{x^{2}}$ take the same minimum value at the same point, then the maximum value of $f(x)$ in this interval is
(A) $4+\frac{11}{2} \sqrt[3]{2}+\sqrt[3]{4}$.
(B) $4-\frac{5}{2} \sqrt[3]{2}+\sqrt[3]{4}$.
(C) $1-\frac{1}{2} \sqrt[3]{2}+\sqrt[3]{4}$.
(D) None of the above answers is correct.
(China High School Mathematics League, 1996)
[Solution] Now consider on the interval $[1,2]$,
$$
g(x)=x+\frac{1}{x^{2}}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^{2}} \geqslant 3 \sqrt[3]{\frac{1}{4}}=\frac{3}{2} \sqrt[3]{2}.
$$
$\because f(x)$ and $g(x)$ attain the same minimum value at the same point,
$$
\therefore \quad-\frac{p}{2}=\sqrt[3]{2}, \quad \frac{4 q-p^{2}}{4}=\frac{3}{2} \sqrt[3]{2} \text{. }
$$

When $\frac{x}{2}=\frac{1}{x^{2}}$, i.e., $x=\sqrt[3]{2} \in[1,2]$,
$$
g(x)_{\min }=\frac{3}{2} \sqrt[3]{2}
$$

Solving gives $\quad p=-2 \sqrt[3]{2}, \quad q=\frac{3}{2} \sqrt[3]{2}+\sqrt[3]{4}$.
Thus, $\quad f(x)=x^{2}-2 \sqrt[3]{2} x+\frac{3}{2} \sqrt[3]{2}+\sqrt[3]{4}$.
Since $\sqrt[3]{2}-1<2-\sqrt[3]{2}$,
$$
\therefore \quad f(x)_{\max }=f(2)=4-\frac{5}{2} \sqrt[3]{2}+\sqrt[3]{4} \text{. }
$$

Therefore, the answer is $(B)$.