2010 AMC 12B Problems/Problem 24

Problem 24

The set of real numbers $x$ for which

$\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1$

is the union of intervals of the form $a<x\le b$ . What is the sum of the lengths of these intervals?

$\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}$

Solution

Because the right side of the inequality is a horizontal line, the left side can be translated horizontally by any value and the intervals will remain the same. For simplicity of calculation, we will find the intervals where $\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1$ The left side of the inequality has three vertical asymptotes at $x=\{-1,0,1\}$ . Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positive infinity, and the function is decreasing at all intervals. In addition, the function has a horizontal asymptote at $y=0$ . The function intersects $1$ at some point from $x=-1$ to $x=0$ , and at some point from $x=0$ to $x=1$ , and at some point to the right of $x=1$ . The intervals where the function is greater than 1 are between the points where the function equals $1$ and the vertical asymptotes.

If $p$ , $q$ , and $r$ are values of x where $\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1$ , then the sum of the lengths of the intervals is $(p-(-1))+(q-0)+(r-1)=p+q+r$ .