Difference between revisions of "2010 AMC 12A Problems/Problem 24"
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<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36</math> | <math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36</math> | ||
− | == Solution == | + | == Solution 1== |
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. | 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. | ||
Revision as of 21:23, 26 April 2020
Problem
Let . The intersection of the domain of with the interval is a union of disjoint open intervals. What is ?
Solution 1
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 less than or equal to .
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 and intervals.
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
Solution 2 (cheap)
Note that the expression must be greater than zero, since logarithm functions are undefined for and negative numbers. Let temporarily be the dependent variables of the functions . It is easy to see that for to be positive for , must be even for . Since an even number of positives times an even number of negatives equals a positive, there can be or positive values of for for a given value of . (since is always positive on the range ) Since MAA allows rulers (and you should bring one to the actual exam), use it to your advantage and draw a larged scaled number line from to . (I recommend increments of at most .) If you don't have a ruler but have graph paper, you can use that instead. Then, designate rows for , respectively. Draw a large bar (label it with so you know it's positive) for all values of such that is even, and do that for all eight rows. Then, use your ruler (or another viable straightedge, such as the edge of another sheet of paper), place the straightedge perpendicular to the vertical line on your digram at , and slowly work your way to , marking all disjoint intervals in which your straightedge touches or boxes simultaneously. (If an interval excludes a value in that interval, you still have to count it as two disjoint intervals. Note that this will be important as to not undercounting disjoint intervals. ) If done correctly, you should obtain as your answer.
-fidgetboss_4000
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 |
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