1. If the graph of $y=f(2 x)$ is symmetric about the lines $x=\frac{a}{2}$ and $x=\frac{b}{2}(b>a)$, then a period of $f(x)$ is
A. $\frac{a+b}{2}$
B. $2(b-a)$
C. $\frac{b-a}{2}$
D. $4(b-a)$
1. D Because $y=f(2 x)$ is symmetric about $x=\frac{a}{2}$. So $f(a+2 x)=f(a-2 x)$.

So $f(2 a-2 x)=f[a+(a-2 x)]=f[a-(a-2 x)]=f(2 x)$.
Similarly, $f(b+2 x)=f(b-2 x)$.
So $f(2 b-2 x)=f(2 x)$.
So $f(2 b-2 a+2 x)=f[2 b-(2 a-2 x)]=f(2 a-2 x)=f(2 x)$.
So a period of $f(2 x)$ is $2 b-2 a$.
Therefore, a period of $f(x)$ is $4(b-a)$, the option is $\mathrm{I}$.
Explanation The problem examines the symmetry and periodicity of functions. Analogous to the rules for solving periodic transformations and symmetry in trigonometric functions. If the graph of the function $y=f(x)$ is symmetric about the lines $x=a$ and $x=b$ $(a \neq b)$, then this function is a periodic function, with a period of $2(b-a)$.