11. Given $A=\left\{x \mid x^{2}<x\right\}, B=\left\{x \mid x^{2}<\log _{a} x\right\}$, and $B \varsubsetneqq A$, find the range of real number $a$.

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$A=(0,1)$, when $0<a<1$, if $x^{2} \geqslant \log _{a} x=\frac{\ln x}{\ln a} \Rightarrow \ln a \leqslant \frac{\ln x}{x^{2}}$ for any $x \in[1,+\infty)$ to always hold,
Set $f(x)=\frac{\ln x}{x^{2}}, f^{\prime}(x)=\frac{1-2 \ln x}{x^{3}}$, thus $f(x)$ is monotonically increasing on $(1, \sqrt{\mathrm{e}})$ and monotonically decreasing on $(\sqrt{\mathrm{e}},+\infty)$, hence $f(x)_{\max }=f(\sqrt{\mathrm{e}})=\frac{1}{2 \mathrm{e}} \Rightarrow \ln a \leqslant \frac{1}{2 \mathrm{e}} \Rightarrow a \leqslant \mathrm{e}^{\frac{1}{2 e}}$.
When $a>1$, if $x^{2} \geqslant \log _{a} x=\frac{\ln x}{\ln a} \Rightarrow \ln a \geqslant \frac{\ln x}{x^{2}}$ for any $x \in[1,+\infty)$ to always hold,
Set $f(x)=\frac{\ln x}{x^{2}}, f^{\prime}(x)=\frac{1-2 \ln x}{x^{3}}$, thus $f(x)$ is monotonically increasing on $(1, \sqrt{\mathrm{e}})$ and monotonically decreasing on $(\sqrt{\mathrm{e}},+\infty)$, hence $f(x)_{\max }=f(\sqrt{\mathrm{e}})=\frac{1}{2 \mathrm{e}} \Rightarrow \ln a \geqslant \frac{1}{2 \mathrm{e}} \Rightarrow a \geqslant \mathrm{e}^{\frac{1}{2 e}}$.
In summary, the range of real number $a$ is $(0,1) \cup\left[\mathrm{e}^{\frac{1}{2 e}},+\infty\right)$.