(11) Set $A=\left\{x \left\lvert\, \frac{1}{4} \leqslant 2^{x} \leqslant \frac{1}{2}\right., x \in \mathbf{R}\right\}, B=\left\{x \mid x^{2}-2 t x+1 \leqslant 0\right\}$, if $A \cap B=A$, then the range of real number $t$ is $\qquad$ .
$11\left(-\infty,-\frac{5}{4}\right]$ Hint: Since $A=\{x \mid-2 \leqslant x \leqslant-1\}, A \subseteq B$, hence $x^{2}-2 t x+1 \leqslant 0$ always holds for $x \in[-2,-1]$. Also, because $x+\frac{1}{x} \geqslant 2 t$, and when $x \in[-2,-1]$, $x+\frac{1}{x} \in\left[-\frac{5}{2},-2\right]$, so $-\frac{5}{2} \geqslant 2 t$, which means $t \leqslant-\frac{5}{4}$. Therefore, the range of real number $t$ is $\left(-\infty,-\frac{5}{4}\right]$.