2012 AMC 12A 第 25 题

$f$ is a periodic triangular wave with $0 \le f(x) \le 1,$ so all solutions lie in $[0, n]$

On each unit interval the graph of $y = f(xf(x))$ oscillates a growing number of times; the total number of intersections with $y = \dfrac{x}{n}$ is $2n^2$

Since $0 \le f(x) \le 1,$ every solution lies in $[0,n].$ The function $f$ is a triangular wave of period $1.$ Put $g(x)=xf(x).$ For each integer $a\ge1,$ the function $g$ decreases from $a$ to $0$ on $[a,a+\tfrac12),$ while it increases from $0$ to $a+1$ on $[a+\tfrac12,a+1).$ The first interval $[0,\tfrac12)$ is exceptional, but it contributes no intersection with $y=\frac{x}{n}.$

Counting the oscillations, on the intervals $[a, a + \tfrac12)$ and $[a + \tfrac12, a+1)$ the curve $y = f(g(x))$ meets the line $y = \tfrac{x}{n}$ a total of $2a$ and $2(a+1)$ times. Summing over $a = 0, \ldots, n-1$ gives $$\sum_{a=0}^{n-1}\bigl(2a + 2(a+1)\bigr) = 2n^2$$ real solutions.

The smallest $n$ with $2n^2 \ge 2012$ is $n = 32,$ since $2 \cdot 31^2 = 1922$ and $2 \cdot 32^2 = 2048.$

Thus, the correct answer is C.