Difference between revisions of "2010 AMC 12B Problems/Problem 24"

Revision as of 17:36, 25 December 2010

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?

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

@@ Line 7: / Line 7: @@
 <math>\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}</math>
-== Solution ==
-First, we shift the graph so it will be easier to manipulate
-<cmath> \dfrac{1}{x - 1} + \dfrac{1}{x} + \dfrac{1}{x+1} \ge 1 </cmath>
-We can see that this is equivalent to the original problem. We now want to solve this equation.
-<cmath> \frac{(x)(x + 1) + (x-1)(x+1) + (x)(x-1)}{(x-1)(x)(x+1)} \ge 1 </cmath>
-<cmath> \frac{(x^2+x+x^2-1+x^2-x)}{x^3-x} \ge 1 </cmath>
-<cmath> \frac{3x^2-1}{x^3-x}\ge 1 </cmath>
-We first find where this will lead to equality. We have <math>x^3 - 3x^2 - x + 1 = 0</math>.
-Our answer is just the sum of the roots, and using Vieta's, we get that to be <math>\boxed{(C)}</math>.

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Difference between revisions of "2010 AMC 12B Problems/Problem 24"

Revision as of 17:36, 25 December 2010

Problem 24