Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 3"
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==Solution== | ==Solution== | ||
Substituting <math>\frac{1}{x}</math>, we have | Substituting <math>\frac{1}{x}</math>, we have | ||
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<cmath>2f\left(\frac 1x\right) + f\left(x\right) = \frac{5}{x} + 4</cmath> | <cmath>2f\left(\frac 1x\right) + f\left(x\right) = \frac{5}{x} + 4</cmath> | ||
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This gives us two equations, which we can eliminate <math>f\left(\frac 1x\right)</math> from (the first equation multiplied by two, subtracting the second): | This gives us two equations, which we can eliminate <math>f\left(\frac 1x\right)</math> from (the first equation multiplied by two, subtracting the second): | ||
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
3f(x) &= 10x + 4 - \frac 5x \\ | 3f(x) &= 10x + 4 - \frac 5x \\ | ||
0 &= x^2 - \frac{3 \times 2004 - 4}{10}x + \frac 52\end{align*}</cmath> | 0 &= x^2 - \frac{3 \times 2004 - 4}{10}x + \frac 52\end{align*}</cmath> | ||
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Clearly, the [[discriminant]] of the [[quadratic equation]] <math>\Delta > 0</math>, so both roots are real. By [[Vieta's formulas]], the sum of the roots is the coefficient of the <math>x</math> term, so our answer is <math>\left[\frac{3 \times 2004 - 4}{10}\right] = \boxed{601}</math>. | Clearly, the [[discriminant]] of the [[quadratic equation]] <math>\Delta > 0</math>, so both roots are real. By [[Vieta's formulas]], the sum of the roots is the coefficient of the <math>x</math> term, so our answer is <math>\left[\frac{3 \times 2004 - 4}{10}\right] = \boxed{601}</math>. | ||
Revision as of 08:06, 20 June 2008
Problem
A function is defined for all real numbers . For all non-zero values , we have
Let denote the sum of all of the values of for which . Compute the integer nearest to .
Solution
Substituting , we have
This gives us two equations, which we can eliminate from (the first equation multiplied by two, subtracting the second):
Clearly, the discriminant of the quadratic equation , so both roots are real. By Vieta's formulas, the sum of the roots is the coefficient of the term, so our answer is .
See also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |