1. Given $\mathbf{Z}$ as the set of integers, the set $A=|x \quad| x-3|<\pi, x \in \mathbf{Z}|$
$$
\begin{array}{l}
B=\left\{x\left|x^{2}-11 x+5<0, x \in \mathbf{Z}\right|\right. \\
C=|x|\left|2 x^{2}-11 x+10\right| \geqslant|3 x-2|, x \in \mathbf{Z}
\end{array}
$$
$\bar{C}$ is the complement of $C$ in $\mathrm{Z}$. Then the number of true elements in $A \cap B \cap \bar{C}$ is ( ).
A. 7
B. 8
C. 15
D. 16
1. A

Given $A=\{0,1,2,3,4,5,6\}$
$$
B=\{1,2,3,4,5,6,7,8,9,10\}
$$

As for C. $\left|2 x^{2}-11 x+10\right| \geqslant|3 x-2| \Leftrightarrow\left(2 x^{2}-11 x+10\right)^{2}-(3 x-2)^{2} \geqslant 0$
$$
\Leftrightarrow\left(2 x^{2}-8 x+8\right)\left(2 x^{2}-14 x+12\right) \geqslant 0 \Leftrightarrow(x-2)^{2}(x-1)(x-6) \geqslant 0
$$

Therefore, $C=\{x \mid x \leqslant 1$, or $x \geqslant 6$ or $x=2, x \in \mathbf{Z}\}$
So $\bar{C}=\{3,4,5\}$
Thus, $A \cap B \cap \bar{C}=\{3,4,5\}$. It has $2^{3}-1=7$ subsets.