Difference between revisions of "2023 AMC 12B Problems/Problem 6"
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==See Also== | ==See Also== |
Revision as of 20:41, 15 November 2023
- The following problem is from both the 2023 AMC 10B #12 and 2023 AMC 12B #6, so both problems redirect to this page.
Contents
[hide]Problem
When the roots of the polynomial
are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is positive?
Solution 1
is a product of or 10 terms. When , all terms are , but because there is an even number of terms. The sign keeps alternating . There are 11 intervals, so there are positives and 5 negatives.
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Solution 2
Denote by the interval for and the interval .
Therefore, the number of intervals that is positive is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3
We can use the turning point behavior at the roots of a polynomial graph to find out the amount of intervals that are positive.
First, we evaluate any value on the interval . Since the degree of is = = , and every term in P(x) is negative, multiplying 55 negatives gives a negative value. So is a negative interval.
We know that the roots of P(x) are at . When the degree of the term of each root is odd, the graph of P(x) will pass through the graph and change signs, and vice versa. So at , the graph will change signs; at , the graph will not, and so on.
This tells us that the interval is positive, is also positive, is negative, is also negative, and so on, with the pattern being .
The positive intervals are therefore , , , , , and , for a total of .
~nm1728
Video Solution 1 by OmegaLearn
See Also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 10 Problems and Solutions |
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.