Difference between revisions of "2002 AMC 12P Problems/Problem 20"

Latest revision as of 12:30, 28 March 2024

Let $f$ be a real-valued function such that

$f(x) + 2f(\frac{2002}{x}) = 3x$

for all $x>0.$ Find $f(2).$

$\text{(A) }1000 \qquad \text{(B) }2000 \qquad \text{(C) }3000 \qquad \text{(D) }4000 \qquad \text{(E) }6000$

Setting $x = 2$ gives $f(2) + 2f(1001) = 6$ . Setting $x = 1001$ gives $2f(2) + f(1001) = 3003$ .

Adding these 2 equations and dividing by 3 gives $f(2) + f(1001) = \frac{6+3003}{3} = 1003$ .

Subtracting these 2 equations gives $f(2) - f(1001) = 2997$ .

Therefore, $f(2) = \frac{1003+2997}{2} = \boxed {\text{(B) }2000}$ .

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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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 == Solution ==
-When <math>x = 2</math>, then we get <math> f(2) + 2f(1001) = 6</math>; we can also substitute <math>x</math> as <math>1001</math>, then we will get <math>f(1001) + 2f(2) =3003</math>. Solve this system of equations, then we get <math>f(2)= 2000</math> <math>\Longrightarrow \boxed{\mathrm{B}}</math>.
+Setting <math>x = 2</math> gives <math> f(2) + 2f(1001) = 6</math>.
+Setting <math>x = 1001</math> gives <math> 2f(2) + f(1001) = 3003</math>.
+Adding these 2 equations and dividing by 3 gives
+<math>f(2) + f(1001) = \frac{6+3003}{3} = 1003</math>.
+Subtracting these 2 equations gives
+<math>f(2) - f(1001) = 2997</math>.
+Therefore, <math>f(2) = \frac{1003+2997}{2} = \boxed {\text{(B) }2000}</math>.
 == See also ==
 {{AMC12 box|year=2002|ab=P|num-b=19|num-a=21}}
 {{MAA Notice}}