11. Let the function $f(x)=\mathrm{e}^{x}\left(x^{3}-3 x+3\right)-a \mathrm{e}^{x}-x \ (x \geqslant-2)$, if the inequality $f(x) \leqslant 0$ has a solution, then the minimum value of the real number $a$ is
A. $\frac{2}{\mathrm{e}}-1$
B. $2-\frac{2}{\mathrm{e}}$
C. $1+2 \mathrm{e}^{2}$
D. $1-\frac{1}{\mathrm{e}}$
Solve:
$$
a \mathrm{e}^{x} \geqslant \mathrm{e}^{x}\left(x^{3}-3 x+3\right)-x \Rightarrow a \geqslant x^{3}-3 x+3-\frac{x}{\mathrm{e}^{x}},
$$

Let $g(x)=x^{3}-3 x+3-\frac{x}{\mathrm{e}^{x}}, g^{\prime}(x)=3 x^{2}-3-\frac{1-x}{\mathrm{e}^{x}}=(x-1)\left[3(x+1)+\frac{1}{\mathrm{e}^{x}}\right]$,
then there exists $x_{0} \in(-2,-1)$ such that $g(x)$ is monotonically increasing on $\left(-2, x_{0}\right)$, monotonically decreasing on $\left(x_{0}, 1\right)$, and monotonically increasing on $(1,+\infty)$, thus $g(x)_{\min }=g(1)=1-\frac{1}{\mathrm{e}} \Rightarrow a \geqslant 1-\frac{1}{\mathrm{e}}$, so the answer is $D$.