$\lim\limits_{x \rightarrow0}(\mathrm{e}^x+ax^2+bx)^{\frac{1}{x^2}}=1$ $\therefore$ 左边= $\mathrm{e}^{\lim\limits_{x\rightarrow0}\frac{1}{x^2}\ln (\mathrm{e}^x+ax^2+bx)}=\mathrm{e}^{\lim\limits_{x \rightarrow 0}\frac{\mathrm{e}^x+ax^2+bx-1}{x^2}}=1$ $\therefore~\lim\limits_{x \rightarrow0}\frac{\mathrm{e}^x+ax^2+bx-1}{x^2}=0$ $\therefore$ 上式= $\lim\limits_{x \rightarrow 0}\frac{(\frac{1}{2}+a)x^2+(1+b)x+o(x^2)}{x^2} \therefore \begin{cases}1+b=0\\\\ \frac{1}{2}+a=0\end{cases} \therefore \begin{cases}a=-\frac{1}{2}\\\\ b=-1\end{cases},$ 选 $\rm B.$