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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 3"

 
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==Problem==
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A [[function]] <math>f(x)</math> is defined for all real numbers <math>x</math>. For all non-zero values <math>x</math>, we have
  
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<math>3.</math> A function <math>f(x)</math> is defined for all real numbers <math>x</math>. For all non-zero values <math>x</math>, we have
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<cmath>2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4</cmath>
  
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<math>2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4</math>
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Let <math>S</math> denote the sum of all of the values of <math>x</math> for which <math>f(x) = 2004</math>. Compute the integer nearest to <math>S</math>.
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==Solution==
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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>
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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):
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<cmath>\begin{align*}
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3f(x) &= 10x + 4 - \frac 5x \\
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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>.
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==See Also==
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{{Mock AIME box|year=Pre 2005|n=3|num-b=2|num-a=4}}
  
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Let <math>S</math> denote the sum of all of the values of <math>x</math> for which <math>f(x) = 2004</math>. Compute the integer nearest to <math>S</math>.
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 09:22, 4 April 2012

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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