5. If on 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 $f(x)$ on 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.
5. (B)

On $[1,2]$, $g(x)=x+\frac{1}{x^{2}}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^{2}} \geqslant 3 \sqrt[3]{\frac{1}{4}}=\frac{3}{2} \sqrt[3]{2}$.
$\because f(x)$ and $g(x)$ take equal minimum values at the same point, $\therefore-\frac{p}{2}=\sqrt[3]{2}$, $\frac{4 q-p^{2}}{4}=\frac{3}{2} \sqrt[3]{2}$
When $\frac{x}{2}=\frac{1}{x^{2}}$, i.e., $x=\sqrt[3]{2} \in[1,2]$, $g(x)$ is $=\frac{3}{2} \sqrt[3]{2}$.
Solving gives $p=-2 \sqrt[3]{2}, q=\frac{3}{2} \sqrt[3]{2}+\sqrt[3]{4}$. Thus, $f(x)=x^{2}-$ $2 \sqrt[3]{2 x}+\frac{3}{2} \sqrt[3]{2}+\sqrt[3]{4}$

Since $\sqrt[3]{2}-1<2-\sqrt[3]{2}$,
$\therefore f(x)$ is $=f(2)=4-\frac{5}{2} \sqrt[3]{2}+\sqrt[3]{4}$. Therefore, the answer is (B).