Difference between revisions of "2010 AMC 12B Problems/Problem 24"
(Created page with '== Problem 24 == The set of real numbers <math>x</math> for which <cmath>\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1</cmath> is the union of intervals of the for…') |
|||
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> | <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>. |
Revision as of 17:36, 25 December 2010
Problem 24
The set of real numbers for which
is the union of intervals of the form . What is the sum of the lengths of these intervals?
Solution
First, we shift the graph so it will be easier to manipulate
We can see that this is equivalent to the original problem. We now want to solve this equation.
We first find where this will lead to equality. We have
.
Our answer is just the sum of the roots, and using Vieta's, we get that to be
.