For each x in [0,1], define
f(x)= \begin{cases}2 x, \text { if } 0 \leq x \leq \dfrac{1}{2} \\ 2-2 x, \text { if } \dfrac{1}{2}<x \leq 1\end{cases}
Let f^{[2]}(x)=f(f(x)), and f^{[n+1]}(x)=f^{[n]}(f(x)) for each integer n \geq 2. For how many values of x in [0,1] is f^{[2005]}(x)=1 / 2 ?
Answer Choices
A. 0
B. 2005
C. 4010
D. 2005^{2}
E. 2^{2005}