Difference between revisions of "1997 AHSME Problems/Problem 27"
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+ | ==Problem== | ||
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+ | Consider those functions <math>f</math> that satisfy <math> f(x+4)+f(x-4) = f(x) </math> for all real <math>x</math>. Any such function is periodic, and there is a least common positive period <math>p</math> for all of them. Find <math>p</math>. | ||
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+ | <math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 32 </math> | ||
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+ | ==Solution== | ||
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+ | Recall that <math>p</math> is the fundamental period of function <math>f</math> iff <math>p</math> is the smallest positive <math>p</math> such that <math>f(x) = f(x + p)</math> for all <math>x</math>. | ||
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+ | In this case, we know that <math>f(x+ 4) + f(x - 4) = f(x)</math>. Plugging in <math>x+4</math> in for <math>x</math> to get the next equation in the recursion, we also get <math>f(x + 8) + f(x) = f(x + 4)</math>. Adding those two equations gives <math>f(x + 8) + f(x - 4) = 0</math> after cancelling out common terms. | ||
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+ | Again plugging in <math>x + 4</math> in for <math>x</math> in that last equation (in order to get <math>f(x)</math>), we find that <math>f(x) = -f(x + 12)</math>. Now, plugging in <math>x+12</math> for <math>x</math>, we get <math>f(x + 12) = -f(x + 24)</math>. This proves that <math>f(x) = f(x + 24)</math>, so there is a period of <math>24</math>, which gives answer <math>\boxed{D}</math>. We now eliminate answers <math>A</math> through <math>C</math>. | ||
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+ | Let <math>f(x) = x</math> for <math>x \in \{1, 2, 3, 4, 5, 6, 7, 8\}</math>. Plugging in <math>x=5</math> into the initial equation gives <math>f(9) + f(1) = f(5)</math>, which implies that <math>f(9) = 4</math>. Since <math>f(1) \neq f(1 + 8)</math>, the function does not have period <math>8</math>. | ||
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+ | Continuing, <math>f(10) = f(6) - f(2) = 4</math>, <math>f(11) = f(7) - f(3) = 4</math>, and <math>f(12) = f(8) - f(4) = 4</math>. Once we hit <math>f(13)</math>, we have <math>f(13) = f(9) - f(5) = 4 - 5 = -1</math>. Since <math>f(1) \neq f(1 + 12)</math>, the function does not have period <math>12</math>. | ||
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+ | Finally, <math>f(14) = f(10) - f(6) = 4 - 6 = -2</math>, <math>f(15) = f(11) - f(7) = 4 - 7 = -3</math>, <math>f(16) = f(12) - f(8) = 4 - 8 = -4</math>, and <math>f(17) = f(13) - f(9) = -1 - 4 = -5</math>. Since <math>f(1) \neq f(1 + 16)</math>, the function does not have period <math>16</math>. | ||
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+ | To confirm that our original period works, we may see that <math>f(18) = -6</math>, <math>f(19) = -7</math>, <math>f(20) = -8</math>, <math>f(21) = -4</math>, <math>f(22) = -4</math>, <math>f(23) = -4</math>, and <math>f(24) = -4</math>. Finally, <math>f(25) = 1</math>, which is indeed the same as <math>f(1)</math>. | ||
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== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=26|num-a=28}} | {{AHSME box|year=1997|num-b=26|num-a=28}} |
Revision as of 15:43, 21 August 2011
Problem
Consider those functions that satisfy for all real . Any such function is periodic, and there is a least common positive period for all of them. Find .
Solution
Recall that is the fundamental period of function iff is the smallest positive such that for all .
In this case, we know that . Plugging in in for to get the next equation in the recursion, we also get . Adding those two equations gives after cancelling out common terms.
Again plugging in in for in that last equation (in order to get ), we find that . Now, plugging in for , we get . This proves that , so there is a period of , which gives answer . We now eliminate answers through .
Let for . Plugging in into the initial equation gives , which implies that . Since , the function does not have period .
Continuing, , , and . Once we hit , we have . Since , the function does not have period .
Finally, , , , and . Since , the function does not have period .
To confirm that our original period works, we may see that , , , , , , and . Finally, , which is indeed the same as .
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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