1. If $x^{2}-12 x+1=0$, then the unit digit of $x^{4}+x^{-4}$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
- 1. B.

From $x^{2}-12 x+1=0$, we know $x \neq 0$.
Then $x+\frac{1}{x}=12 \Rightarrow x^{2}+x^{-2}=12^{2}-2=142$
$$
\Rightarrow x^{4}+x^{-4}=142^{2}-2 \text { . }
$$

Thus, the unit digit of $x^{4}+x^{-4}$ is $4-2=2$.