1. Given $\mathbf{Z}$ is the set of integers, the set $A=\{x|| x-3 |<\pi, x \in \mathbf{Z}\}, B=\left\{x \mid x^{2}-11 x+5<0\right.$, $x \in \mathbf{Z}\}, C=\left\{x|| 2 x^{2}-11 x+10|\geqslant| 3 x-2 \mid, x \in \mathbf{Z}\right\}, \bar{C}$ is the complement of $C$ in $\mathbf{Z}$, then the number of proper subsets of $A \cap B \cap \bar{C}$ is
A. 7
B. 8
C. 15
D. 16
1. A Obviously, the sets are $A=\{0,1,2,3,4,5,6\} ; B=\{0,1,2,3,4,5,6,7,8,9,10\}$; for set $C$, $\left|2 x^{2}-11 x+10\right| \geqslant|3 x-2| \Leftrightarrow\left(2 x^{2}-11 x+10\right)^{2}-(3 x-2)^{2} \geqslant 0 \Leftrightarrow\left(2 x^{2}-8 x+8\right)\left(2 x^{2}-14 x+12\right)$ $\geqslant 0 \Leftrightarrow(x-2)^{2}(x-1)(x-6) \geqslant 0$, hence $C=\{x \mid x \leqslant 1$ or $x \geqslant 6$ or $x=2, x \in \mathbf{Z}\}$, thus $\bar{C}=\{3,4,5\}$, so $A \cap B \cap \bar{C}=\{3,4,5\}$, it has $2^{3}-1=7$ proper subsets.