Difference between revisions of "2010 AMC 12A Problems/Problem 24"
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| − | == Problem | + | == Problem == |
Let <math>f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)</math>. The intersection of the domain of <math>f(x)</math> with the interval <math>[0,1]</math> is a union of <math>n</math> disjoint open intervals. What is <math>n</math>? | Let <math>f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)</math>. The intersection of the domain of <math>f(x)</math> with the interval <math>[0,1]</math> is a union of <math>n</math> disjoint open intervals. What is <math>n</math>? | ||
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First, let us find the number of zeros of the inside of the logarithm. | First, let us find the number of zeros of the inside of the logarithm. | ||
| − | + | <cmath>\begin{align*}\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x) &= 0\\ | |
| − | < | + | \sin(\pi x) &= 0\\ |
| − | + | x &= 0, 1\\ | |
| − | + | \sin(2 \pi x) &= 0\\ | |
| − | + | x &= 0, \frac{1}{2}, 1\\ | |
| − | + | \sin(3 \pi x) &= 0\\ | |
| − | + | x &= 0, \frac{1}{3}, \frac{2}{3}, 1\\ | |
| − | + | \sin(4 \pi x) &= 0\\ | |
| − | + | x &= 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1\\ | |
| − | + | &\cdots\end{align*}</cmath> | |
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After counting up the number of zeros for each factor and eliminating the excess cases we get <math>22</math> zeros. | After counting up the number of zeros for each factor and eliminating the excess cases we get <math>22</math> zeros. | ||
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And so, the total number of disjoint open intervals is <math>22 - 2\cdot{5} = \boxed{12\ \textbf{(B)}}</math> | And so, the total number of disjoint open intervals is <math>22 - 2\cdot{5} = \boxed{12\ \textbf{(B)}}</math> | ||
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| + | == See also == | ||
| + | {{AMC12 box|year=2010|num-b=23|num-a=25|ab=A}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 22:34, 25 February 2010
Problem
Let 
. The intersection of the domain of 
with the interval ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
is a union of 
disjoint open intervals. What is ?
Solution
The question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them.
We note that since all of the 
factors are inside a logarithm, the function is undefined where the inside of the logarithm is equal to or less than 
.
First, let us find the number of zeros of the inside of the logarithm.
After counting up the number of zeros for each factor and eliminating the excess cases we get 
zeros.
In order to find which intervals are negative, we must first realize that at every zero of each factor, the sign changes. We also have to be careful, as some zeros are doubled, or even tripled, quadrupled, etc.
The first interval 
is obviously positive. This means the next interval 
is negative. Continuing the pattern and accounting for doubled roots (which do not flip sign), we realize that there are 
negative intervals from to 
. Since the function is symmetric, we know that there are also negative intervals from to 
.
And so, the total number of disjoint open intervals is 
See also
| 2010 AMC 12A (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 | |
