2010 AMC 12B Problem 24

Let $f(x)$ be the left-hand side. On each interval between consecutive asymptotes $2009, 2010, 2011,$ the function $f$ is decreasing, and $f\lt1$ for all $x\lt2009.$ On each of $(2009,2010),$ $(2010,2011),$ and $(2011,\infty),$ the solution is the part from the left asymptote up to a value $x_i$ where $f(x_i)=1.$ So the solution set consists of three intervals with left endpoints $2009, 2010, 2011$ and right endpoints $x_1, x_2, x_3.$ The total length is $(x_1-2009)$ $+(x_2-2010)$ $+(x_3-2011)$ $=x_1+x_2+x_3-6030.$ Clearing denominators in $f(x)=1$ gives $$\begin{aligned} &x^3 \\ &\quad \small{}-(2009+2010+2011+3)x^2 \\ &\quad {}+\cdots=0, \end{aligned}$$ whose roots are $x_1, x_2, x_3.$ By Vieta, $x_1+x_2+x_3=6033,$ so the sum of lengths is $6033-6030=3.$ Thus, the correct answer is C.