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. $4+\frac{11}{4} \sqrt[3]{2}+\sqrt[3]{4}$
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
Since $g(x)=x+\frac{1}{x^{2}}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^{2}} \geqslant 3 \sqrt[3]{\frac{1}{4}}=\frac{3 \sqrt[3]{2}}{2}$, equality holds when $x=\sqrt[3]{2}$. Thus, $f(x)=(x-\sqrt[3]{2})^{2}+\frac{3 \sqrt[3]{2}}{2} \Rightarrow f(x)_{\max }=f(2)=(2-\sqrt[3]{2})^{2}+\frac{3 \sqrt[3]{2}}{2}$ $=4-4 \sqrt[3]{2}+\sqrt[3]{4}+\frac{3 \sqrt[3]{2}}{2}=4-\frac{5}{2} \sqrt[3]{2}+\sqrt[3]{4}$. Therefore, the answer is $B$.