Question

The set of real numbers $x$ for which $\frac{1}{x-2009}+\frac{1}{x-2010}+\frac{1}{x-2011}\ge1$ is the union of intervals of the form $a\lt x\le b.$ What is the sum of the lengths of these intervals?

$\dfrac{1003}{335}$

$\dfrac{1004}{335}$

$3$

$\dfrac{403}{134}$

$\dfrac{202}{67}$

Solution:

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 $x^3-(2009+2010+2011+3)x^2+\cdots=0,$ 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.

2010 AMC 12B Problem 24

Solution:

Problem 24 in Other Years