Problem 29 · 2010 Math Kangaroo
Stretch
Algebra & Patterns
substitution
A function maps all positive real numbers to real numbers. For all \(x\in\mathbb{R}^{+}\) the following holds true: \(2f(x)+3f\!\left(\dfrac{2010}{x}\right)=5x\). Determine the value of f(6).
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Answer: A — 993
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Hint 1 of 2
Replace x with 2010/x to get a second equation in the same two unknowns.
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Hint 2 of 2
Solve the pair for f(x), then plug in x = 6.
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Approach: substitute x -> 2010/x and solve the system
- The given equation is 2f(x) + 3f(2010/x) = 5x.
- Swapping x and 2010/x gives 2f(2010/x) + 3f(x) = 5·2010/x.
- Eliminating f(2010/x) yields f(x) = 6030/x − 2x.
- Then f(6) = 1005 − 12 = 993.
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