Mock AIME 3 Pre 2005 Problems/Problem 3

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 $\left[\frac{3 \times 2004 - 4}{10}\right] = \boxed{601}$ .

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

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