1. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $x - [x]$ is called the fractional part of $x$. If $t = \frac{1}{2 - \sqrt{3}}$, $a$ is the fractional part of $t$, and $b$ is the fractional part of $-t$, then $\frac{1}{2b} - \frac{1}{a} = (\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{3}}{2}$
(C) 1
(D) $\sqrt{3}$
- 1. A.

Notice,
$$
t=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}, 3<2+\sqrt{3}<4 \text {. }
$$

Thus, $a=t-3=\sqrt{3}-1$.
Also, $-t=-2-\sqrt{3}, -4<-2-\sqrt{3}<-3$, so $b=-t-(-4)=2-\sqrt{3}$.
Therefore, $\frac{1}{2 b}-\frac{1}{a}=\frac{1}{2(2-\sqrt{3})}-\frac{1}{\sqrt{3}-1}$
$$
=\frac{2+\sqrt{3}}{2}-\frac{\sqrt{3}+1}{2}=\frac{1}{2} \text {. }
$$