Mock AIME 3 Pre 2005 Problems/Problem 3

Revision as of 20:53, 19 June 2008 by Azjps (talk | contribs) (solution)

Problem

A function $f(x)$ is defined for all real numbers $x$. For all non-zero values $x$, we have

\[2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4\]

Let $S$ denote the sum of all of the values of $x$ for which $f(x) = 2004$. Compute the integer nearest to $S$.

Solution

Substituting $\frac{1}{x}$, we have \[2f\left(\frac 1x\right) + f\left(x\right) = \frac{5}{x} + 4\] This gives us two equations, which we can eliminate $f\left(\frac 1x\right)$ from (the first equation multiplied by two, subtracting the second): \begin{align*} 3f(x) &= 10x + 4 - \frac 5x \\ 0 &= x^2 - \frac{3 \times 2004 - 4}{10}x + \frac 52\end{align*} Clearly, the discriminant of the quadratic equation $\Delta > 0$, so both roots are real. By Vieta's formulas, the sum of the roots is the coefficient of the $x$ term, so our answer is $\left[\frac{3 \times 2004 - 4}{10}\right] = \boxed{601}$.

See also

Mock AIME 3 Pre 2005 (Problems, Source)
Preceded by
Problem 2
Followed by
Problem 4
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