33 - 50 Let $a$ denote the fractional part of $\frac{1}{3-\sqrt{5}}$, then $\log _{2} a(2 a+1)$ equals
(A) -1.
(B) -2.
(C) 0.
(D) $\frac{1}{2}$.
(China Junior High School Mathematics Correspondence Competition, 1987)
[Solution] From $\frac{1}{3-\sqrt{5}}=\frac{3+\sqrt{5}}{4}$, and since $1<\frac{3+\sqrt{5}}{4}<2$, $\therefore \quad a=\frac{3+\sqrt{5}}{4}-1=\frac{\sqrt{5}-1}{4}$. Then $a(2 a+1)=\frac{\sqrt{5}-1}{4}\left(\frac{\sqrt{5}-1}{2}+1\right)$ $=\frac{(\sqrt{5}-1)(\sqrt{5}+1)}{8}=\frac{1}{2}$. $\therefore \quad \log _{2} a(2 a+1)=\log _{2} \frac{1}{2}=-1$. Therefore, the answer is $(A)$.