Difference between revisions of "2010 AMC 12B Problems/Problem 24"
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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> | ||
− | 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. | + | 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 02:42, 2 February 2011
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
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
We shall say that
.
has three vertical asymptotes at
. 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
. The function intersects
at some point from
to
, and at some point from
to
, and at some point to the right of
. The intervals where the function is greater than
are between the points where the function equals
and the vertical asymptotes.
If ,
, and
are values of x where
, then the sum of the lengths of the intervals is
.
And now our job is simply to find the sum of the roots of .
and the sum is the negative of the coefficient on the
term, which is
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 |