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) $4+\frac{1113}{2} \sqrt[3]{2}+\sqrt[3]{4}$
(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 are correct
(B)
5.【Analysis and Solution】The function $f(x)$ reaches its minimum value on $\left(-\frac{p}{2}, \frac{4 q-p^{2}}{4}\right)$, and $g(x)=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^{2}} \geqslant 3$ $\left(\frac{x}{2} \frac{x}{2} \frac{1}{x^{2}}\right)^{\frac{1}{3}}=3 \times 4^{\frac{1}{3}}$. The equality holds when $\frac{x}{2}=\frac{1}{x^{2}}$, i.e., $x=2^{\frac{1}{3}}$, then we have $-\frac{p}{2}=2^{\frac{1}{3}}, \frac{4 q-p^{2}}{4}=3 \times 4^{\frac{1}{3}}$, solving for $p=-$ $2^{\frac{4}{3}}, q=3 \times 2^{-\frac{2}{3}}+2^{\frac{2}{3}}$. Since $2^{\frac{1}{3}}-1<2-2^{\frac{1}{3}}$, the maximum value of $f(x)$ on the interval $[1,2]$ is taken at $x=2$, i.e., $f(x)=f(2)=4-5 \times 2^{-\frac{2}{3}}+2^{\frac{2}{3}}$.