Difference between revisions of "2001 AMC 12 Problems/Problem 9"
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== Solution 1 == | == Solution 1 == | ||
− | Letting <math>x = 500</math> and <math>y = \dfrac65</math> in the given equation, we get <math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, or <math>f(600) = \boxed{\textbf{C } \frac52}</math>. | + | Letting <math>x = 500</math> and <math>y = \dfrac65</math> in the given equation, we get <math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, or <math>f(600) = \boxed{\textbf{(C) } \frac52}</math>. |
== Solution 2 == | == Solution 2 == | ||
− | The only function that satisfies the given condition is <math>y = \frac{k}{x}</math>, for some constant <math>k</math>. Thus, the answer is <math>\frac{500 \cdot 3}{600} = \frac52</math>. | + | The only function that satisfies the given condition is <math>y = \frac{k}{x}</math>, for some constant <math>k</math>. Thus, the answer is <math>\frac{500 \cdot 3}{600} = \boxed{\textbf{(C) } \frac52}</math>. |
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==Solution 3== | ==Solution 3== | ||
− | Note that the equation given above is symmetric, so we have <math>x \cdot f(x)=y \cdot f(y)</math>. Plugging in <math>x=500</math> and <math>y=600</math> gives <math>f(y)= | + | Note that the equation given above is symmetric, so we have <math>x \cdot f(x)=y \cdot f(y)</math>. Plugging in <math>x=500</math> and <math>y=600</math> gives <math>f(y)=\boxed{\textbf{(C) } \frac52}</math>. |
== See Also == | == See Also == |
Latest revision as of 23:33, 29 December 2023
Problem
Let be a function satisfying for all positive real numbers and . If , what is the value of ?
Solution 1
Letting and in the given equation, we get , or .
Solution 2
The only function that satisfies the given condition is , for some constant . Thus, the answer is .
Solution 3
Note that the equation given above is symmetric, so we have . Plugging in and gives .
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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