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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>
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We shall say that <math>f(x)=\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}</math>. <math>f(x)</math> has three vertical asymptotes at <math>x=\{-1,0,1\}</math>. As the sum of decreasing hyperbolas, the function is decreasing at all intervals. Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positive infinity. 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 the function equals <math>1</math> and the vertical asymptotes.
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We shall say that <math>f(x)=\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}</math>. <math>f(x)</math> has three vertical asymptotes at <math>x=\{-1,0,1\}</math>. As the sum of decreasing hyperbolas, the function is decreasing at all intervals. Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positive infinity. 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 <math>1</math> are between the points where the function equals <math>1</math> and the vertical asymptotes.
  
 
If <math>p</math>, <math>q</math>, and <math>r</math> are values of x where <math>f(x)=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>f(x)=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 01:42, 2 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 \[\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1\] We shall say that $f(x)=\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}$. $f(x)$ has three vertical asymptotes at $x=\{-1,0,1\}$. As the sum of decreasing hyperbolas, the function is decreasing at all intervals. Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positive infinity. 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 $f(x)=1$, 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 $3\Rightarrow\boxed{C}$

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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