Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 13"

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m (<blank edit> solution incomplete, rigor necessary. Could someone provide?)
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== Problem ==
 
== Problem ==
Let <math>F(x)</math> be a polynomial such that <math>F(6) = 15</math> and  
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Let <math>F(x)</math> be a [[polynomial]] such that <math>F(6) = 15</math> and  
 
<cmath>\frac{F(3x)}{F(x+3)} = 9-\frac{48x+54}{x^2+5x+6}</cmath>
 
<cmath>\frac{F(3x)}{F(x+3)} = 9-\frac{48x+54}{x^2+5x+6}</cmath>
 
for <math>x \in \mathbb{R}</math> such that both sides are defined. Find <math>F(12)</math>.
 
for <math>x \in \mathbb{R}</math> such that both sides are defined. Find <math>F(12)</math>.
  
 
== Solution ==
 
== Solution ==
Combining denominators and simplifying,
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Combining [[denominator]]s and simplifying,
 
<cmath>\frac{F(3x)}{F(x+3)} = \frac{9(x^2+5x+6)-48x-54}{x^2+5x+6} = \frac{9x^2 - 3x}{x^2 + 5x + 6}= \frac{3x(3x-1)}{(x+3)(x+2)}</cmath>
 
<cmath>\frac{F(3x)}{F(x+3)} = \frac{9(x^2+5x+6)-48x-54}{x^2+5x+6} = \frac{9x^2 - 3x}{x^2 + 5x + 6}= \frac{3x(3x-1)}{(x+3)(x+2)}</cmath>
It becomes obvious that <math>F(x) = ax(x-1)</math>, for some constant <math>a</math>, matches the definition of the polynomial. To prove that <math>F(x)</math> must have this form, note that  
+
It becomes obvious that <math>F(x) = ax(x-1)</math>, for some [[constant]] <math>a</math>, matches the definition of the polynomial. To prove that <math>F(x)</math> must have this form, note that  
 
<cmath>(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)</cmath>
 
<cmath>(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)</cmath>
 
(rigor needed)
 
(rigor needed)

Revision as of 19:16, 2 April 2008

Problem

Let $F(x)$ be a polynomial such that $F(6) = 15$ and \[\frac{F(3x)}{F(x+3)} = 9-\frac{48x+54}{x^2+5x+6}\] for $x \in \mathbb{R}$ such that both sides are defined. Find $F(12)$.

Solution

Combining denominators and simplifying, \[\frac{F(3x)}{F(x+3)} = \frac{9(x^2+5x+6)-48x-54}{x^2+5x+6} = \frac{9x^2 - 3x}{x^2 + 5x + 6}= \frac{3x(3x-1)}{(x+3)(x+2)}\] It becomes obvious that $F(x) = ax(x-1)$, for some constant $a$, matches the definition of the polynomial. To prove that $F(x)$ must have this form, note that \[(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)\] (rigor needed)

By the given, $F(6) = a(6)(5) = 15 \Longrightarrow a = \frac 12$. Thus, $F(12) = \frac{1}{2}(12)(11) = \boxed{066}$.

Template:Incomplete

See also

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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