(1) If the set $A=\left\{x \mid x^{2}-x-12 \leqslant 0\right\}, B=\left\{x \left\lvert\, \frac{x+1}{x-1}<0\right.\right\}, C=\{x \mid x \in A$ and $x \notin B\}$, then the set $C=$
(A) $[-3,-1) \cup(1,4]$
(B) $[-3,-1] \cup(1,4]$
(C) $[-3,-1) \cup[1,4]$
(D) $[-3,-1] \cup[1,4]$
Solution:
According to the problem, $A=\left\{x \mid x^{2}-x-12 \leqslant 0\right\}=[-3,4], \quad B=\left\{x \left\lvert\, \frac{x+1}{x-1}<0\right.\right\}=(-1,1)$.
Given $x \in A$, we know $-3 \leqslant x \leqslant 4$; and $x \notin B$, we know $x \leqslant-1$ or $x \geqslant 1$.
Therefore, $-3 \leqslant x \leqslant-1$ or $1 \leqslant x \leqslant 4$, which means $C=[-3,-1] \cup[1,4]$.
(D)