2005 AMC 12A 第 20 题

Let $g(n)$ be the number of solutions of $f^{[n]}(x) = \tfrac{1}{2};$ find a recursion for $g(n)$

Because $f$ maps each of $[0, \tfrac12]$ and $[\tfrac12, 1]$ onto all of $[0, 1],$ each solution splits into two, giving $g(n) = 2\,g(n-1)$

Let $g(n)$ count the solutions of $f^{[n]}(x) = \tfrac{1}{2}$ in $[0, 1].$ Since $f$ maps each of the two halves $[0, \tfrac12]$ and $[\tfrac12, 1]$ onto all of $[0, 1],$ every solution of $f^{[n-1]}(y) = \tfrac12$ comes from two values of $x$ (one in each half).

The boundary value $x = \tfrac12$ satisfies $f^{[n]}(\tfrac12) = f^{[n-1]}(1) = 0 \ne \tfrac12,$ so no solutions are lost, giving $g(n) = 2\,g(n-1).$

Since $g(1) = 2,$ we conclude $g(2005) = 2^{2005}.$

Thus, the correct answer is E.