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Difference between revisions of "2010 AMC 12B Problems/Problem 24"
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== Solution ==
 
== 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 <cmath>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1</cmath>
 
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 <cmath>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1</cmath>
−
The left side of the inequality has three vertical asymptotes at <math>x=\{-1,0,1\}</math>. Values immediately to the left of each asymptote are hugely negative, and values immediately to the right of each asymptote are hugely positive. In addition, the function has a horizontal asymptote at <math>y=0</math>. The function intersects <math>1</math> at some point from <math>x=-1</math> to <math>x=0</math>, and from <math>x=0</math> to <math>x=1</math>, and at some point to the right of <math>x=1</math>. The intervals where the function is greater than 1 are between the points where <math>x=1</math> and the vertical asymptotes.
+
The left side of the inequality has three vertical asymptotes at <math>x=\{-1,0,1\}</math>. 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 <math>y=0</math>. The function intersects <math>1</math> at some point from <math>x=-1</math> to <math>x=0</math>, and at some point from <math>x=0</math> to <math>x=1</math>, and at some point to the right of <math>x=1</math>. The intervals where the function is greater than 1 are between the points where <math>x=1</math> and the vertical asymptotes.
  
 
If <math>p</math>, <math>q</math>, and <math>r</math> are values of x where <math>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1</math>, then the sum of the lengths of the intervals is <math>(p-(-1))+(q-0)+(r-1)=p+q+r</math>.
 
If <math>p</math>, <math>q</math>, and <math>r</math> are values of x where <math>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1</math>, then the sum of the lengths of the intervals is <math>(p-(-1))+(q-0)+(r-1)=p+q+r</math>.

Revision as of 23:33, 1 February 2011

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 $x=1$ 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 $(p-(-1))+(q-0)+(r-1)=p+q+r$.

\[\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1\] \[x(x-1)+(x-1)(x+1)+x(x+1)=x(x-1)(x+1)\] \[x^3-3x^2-x+1=0\]

And now our job is simply to find the sum of the roots of $x^3-3x^2-x+1$.

$x^3-3x^2-x+1=(x-p)(x-q)(x-r)$ $x^3-3x^2-x+1=x^3-(q+p+r)x^2+(qp+qr+pr)x-qpr$

and the sum $q+p+r$ is the negative of the coefficient on the $x^2$ term, which is $3\Rightarrow\boxed{C}$

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