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

Latest revision as of 16:44, 15 July 2024

The following problem is from both the 2002 AMC 12P #20 and 2002 AMC 10P #21, so both problems redirect to this page.

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

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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 10 Problems and Solutions

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #20]] and [[2002 AMC 10P Problems|2002 AMC 10P #21]]}}
 == Problem ==
 Let <math>f</math> be a real-valued function such that
@@ Line 28: / Line 30: @@
 <math>f(2) - f(1001) = 2997</math>.
-Therefore, <math>f(2) = \frac{1003+2997}{2} = \boxed {\text{(B) }2000}</math>.
+Therefore, <math>f(2) = \frac{1003+2997}{2} = \boxed {\textbf{(B) }2000}</math>.
 == See also ==
+{{AMC10 box|year=2002|ab=P|num-b=20|num-a=22}}
 {{AMC12 box|year=2002|ab=P|num-b=19|num-a=21}}
 {{MAA Notice}}