11. (20 points) Let $0<a<1$, and denote $b=1-a$. The inequality $\mathrm{e}^{x} \leqslant \frac{1+a x}{1-b x}$ holds for all $x \in[0,1]$. Find the maximum value of $a$.
11. Consider the inequality for $x \in (0,1]$, i.e., $\left(\mathrm{e}^{x} x - x\right) a \leqslant \mathrm{e}^{x} x - \mathrm{e}^{x} + 1$.
Thus, $a \leqslant \frac{\mathrm{e}^{x} x - \mathrm{e}^{x} + 1}{\mathrm{e}^{x} x - x}$ for $x \in (0,1]$.
Let $f(x) = \mathrm{e}^{x} x - \mathrm{e}^{x} + 1$, $g(x) = \mathrm{e}^{x} x - x$.
Then $f(0) = 0$, $g(0) = 0$.
Hence, $\lim _{x \rightarrow 0^{+}} \frac{f(x)}{g(x)} = \lim _{x \rightarrow 0^{+}} \frac{\frac{f(x) - f(0)}{x - 0}}{\frac{g(x) - g(0)}{x - 0}}$
$= \lim _{x \rightarrow 0^{+}} \frac{f^{\prime}(x)}{g^{\prime}(x)}$.
By $f^{\prime}(x) = \mathrm{e}^{x} x$, $f^{\prime}(0) = 0$,
$g^{\prime}(x) = \mathrm{e}^{x} x + \mathrm{e}^{x} - 1$, $g^{\prime}(0) = 0$,

we get
$\frac{\frac{f^{\prime}(x) - f^{\prime}(0)}{x - 0}}{\frac{g^{\prime}(x) - g^{\prime}(0)}{x - 0}}$
$= \lim _{x \rightarrow 0^{+}} \frac{f^{\prime \prime}(x)}{g^{\prime \prime}(x)}$.
Also, $f^{\prime \prime}(x) = \mathrm{e}^{x} x + \mathrm{e}^{x}$, $g^{\prime \prime}(x) = \mathrm{e}^{x} x + 2 \mathrm{e}^{x}$, so
$\lim _{x \rightarrow 0^{+}} \frac{f^{\prime \prime}(x)}{g^{\prime \prime}(x)} = \frac{f^{\prime \prime}(0)}{g^{\prime \prime}(0)} = \frac{1}{2}$.
Therefore, $a \leqslant \frac{1}{2}$.
Taking $a = \frac{1}{2}$, it is easy to prove that $\mathrm{e}^{x} \leqslant \frac{1 + \frac{1}{2} x}{1 - \frac{1}{2} x}$ for $x \in [0,1]$

always holds.
Thus, $a_{\text {max }} = \frac{1}{2}$.