1. Given real numbers $a, b$ satisfy $3^{a}=2007, 223^{b}=$ 2007. Then the value of $\frac{1}{a}+\frac{1}{2 b}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{4}$
(D) 1
- 1.A.

From $3^{a}=2007 \Rightarrow 9^{\frac{1}{2} a b}=2007^{b}$.
Also, $223^{b}=2007 \Rightarrow 223^{\frac{1}{2} a b}=2007^{\frac{a}{2}}$.
From the above two equations, we have
$$
\begin{array}{l}
\frac{1}{2} a b=\frac{a}{2}+b \Rightarrow a b=a+2 b \\
\Rightarrow \frac{1}{a}+\frac{1}{2 b}=\frac{2 b+a}{2 a b}=\frac{a b}{2 a b}=\frac{1}{2} .
\end{array}
$$