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

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{{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #20]] and [[2002 AMC 10P Problems|2002 AMC 10P #21]]}}
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== Problem ==
 
== Problem ==
 
Let <math>f</math> be a real-valued function such that
 
Let <math>f</math> be a real-valued function such that
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== Solution ==
 
== Solution ==
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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>.
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Setting <math>x = 2</math> gives <math> f(2) + 2f(1001) = 6</math>.
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Setting <math>x = 1001</math> gives <math> 2f(2) + f(1001) = 3003</math>.
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Adding these 2 equations and dividing by 3 gives
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<math>f(2) + f(1001) = \frac{6+3003}{3} = 1003</math>.
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Subtracting these 2 equations gives
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<math>f(2) - f(1001) = 2997</math>.
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Therefore, <math>f(2) = \frac{1003+2997}{2} = \boxed {\textbf{(B) }2000}</math>.
  
 
== See also ==
 
== 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}}
 
{{AMC12 box|year=2002|ab=P|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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.

Problem

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

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

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

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

Solution

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

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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