Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 24"
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− | + | <math>f(3) = \frac{1 + 3}{1 - 3\cdot 3} = -\frac{1}{2}</math>. Then <math>f_1(3) = f(-\frac12) = \frac{1 - \frac12}{1 + 3\cdot\frac12} = \frac15</math>, <math>\displaystyle f_2(3) = f(\frac15) = \frac{1 + \frac15}{1 - 3\cdot\frac15} = 3</math> and <math>f_3(3) = f(3) = \frac{1 + 3}{1 - 3\cdot 3} = -\frac{1}{2}</math>. | |
− | {{ | + | It follows immediately that the function cycles and <math>f_n(3) = -\frac12</math> if <math>n = 3k</math>, <math>f_n(3) = \frac15</math> if <math>n = 3k + 1</math> and <math>f_n(3) = 3</math> if <math>n = 3k + 2</math>. Since <math>1993 = 3\cdot 664 + 1</math>, <math>f_{1993}(3) = \frac 15 \Longrightarrow \mathrm{(D)}</math>. |
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+ | Alternatively, <math>f_1(x) = f(f(x)) = \frac{1 + f(x)}{1 - 3f(x)} = \frac{1 + \frac{1 + x}{1 - 3x}}{1 - 3\frac{1 + x}{1 - 3x}} = \frac{(1 - 3x) + (1 + x)}{(1 - 3x) - 3(1 + x)} = \frac{2 - 2x}{-2 - 6x} = \frac{x - 1}{3x + 1}</math>. Thus <math>f_2(x) = f(f_1(x)) = \frac{1 + \frac{x - 1}{3x + 1}}{1 - 3\frac{x - 1}{3x + 1}} = \frac{(3x + 1) + (x - 1)}{(3x + 1) - 3(x - 1)} = \frac{4x}{4} = x</math>, so <math>f_{3k + 2}(x) = f(f(f(f_{3k - 1})))(x) = f_{3k - 1}(x) = \ldots = f_2(x) = x</math>. Thus <math>f_{1993}(x) = f(f(f_{1991}(x))) = f(f(x))</math> so <math>f_{1993}(3) = f(f(3)) = \frac15</math>. | ||
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Latest revision as of 14:27, 19 August 2006
Problem
If and in general then
Solution
. Then , and .
It follows immediately that the function cycles and if , if and if . Since , .
Alternatively, . Thus , so . Thus so .