1. Given $a+\frac{1}{b}=\frac{2}{a}+2 b \neq 0$. Then $\frac{a}{b}$ is ( ).
(A) -1
(B) 1
(C) 2
(D) cannot be determined
$-、 1 .(\mathrm{C})$.
From the condition, we get $a-2 b=\frac{2}{a}-\frac{1}{b}=\frac{2 b-a}{a b}$, $(a-2 b)\left(1+\frac{1}{a b}\right)=0$.

Since $\frac{1}{a}+b \neq 0$, i.e., $a b+1 \neq 0$,
and obviously $a b \neq 0$, hence $1+\frac{1}{a b} \neq 0$.
Therefore, $a-2 b=0$, i.e., $\frac{a}{b}=2$.