11. Given
$$
A=\left\{x \mid x^{2}<x\right\}, B=\left\{x \mid x^{2}<\log _{a} x\right\} \text {, }
$$

and $B \subset A$. Find the range of real number $a$.
II. 11. It is known that $A=\{x \mid 0<x<+\infty\}$ and $B=\{x \mid x^2 - \log_a x > 0\}$, where $a>0$ and $a \neq 1$.
When $0<a<1$, for $0<x<1$, $\log_a x > 0$; for $x>1$, $\log_a x < 0$.
When $a>1$, for $0<x<1$, $\log_a x < 0$; for $x>1$, $\log_a x > 0$.
Thus, $A \cap B=\varnothing$.
Since $B \subset A$, it follows that $B=\varnothing$.
Let $f(x)=x^{2}-\log _{a} x$. Then $B=\varnothing$ is equivalent to $f(x) \geqslant 0$ for all $x \in (0,+\infty)$.
Let $f^{\prime}(x)=2 x-\frac{1}{x \ln a}=0$, we get
$$
x_{0}=\sqrt{\frac{1}{2 \ln a}} .
$$

Then $f(x)$ is decreasing on the interval $\left(0, x_{0}\right)$ and increasing on the interval $\left(x_{0},+\infty\right)$.
Hence, the minimum value of $f(x)$ is $f\left(x_{0}\right)$.
To ensure that $f(x) \geqslant 0$ for all $x \in (0,+\infty)$, it is sufficient that $f\left(x_{0}\right) \geqslant 0$.
$$
\begin{array}{l}
\text { Also, } f\left(x_{0}\right) \geqslant 0 \\
\Leftrightarrow \frac{1}{2 \ln a}-\log _{a} \sqrt{\frac{1}{2 \ln a}} \geqslant 0 \\
\Leftrightarrow \frac{1}{2 \ln a}-\frac{\frac{1}{2} \ln \frac{1}{2 \ln a}}{\ln a} \geqslant 0 \\
\Leftrightarrow 1+\ln (2 \ln a) \geqslant 0 \\
\Leftrightarrow a \geqslant \mathrm{e}^{\frac{1}{x_{0}^{2}}} .
\end{array}
$$

In summary, $a \in(0,1) \cup\left[\mathrm{e}^{\frac{1}{x_{0}^{2}}},+\infty\right)$.