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Difference between revisions of "2002 AMC 12P Problems/Problem 10"

(Created page with "== Problem == How many positive integers <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer? <math> \mathrm{(A) \ 0 } \qquad \mathrm{(B...")
 
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== Problem ==
 
== Problem ==
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How many positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
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Let <math>f_n (x) = \text{sin}^n x + \text{cos}^n x.</math> For how many <math>x</math> in <math>[0,\pi]</math> is it true that
  
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<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 }  </math>
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<cmath>6f_{4}(x)-4f_{6}(x)=2f_{2}(x)?</cmath>
  
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== Solution ==
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<math>
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If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
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\text{(A) }2
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\qquad
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\text{(B) }4
 +
\qquad
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\text{(C) }6
 +
\qquad
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\text{(D) }8
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\qquad
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\text{(E) more than }8
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</math>
 +
 
 +
== Solution 1 ==
 +
Divide by 2 on both sides to get <cmath>3f_{4}(x)-2f_{6}(x)=f_{2}(x)</cmath>
 +
Substituting the definitions of <math>f_{2}(x)</math>, <math>f_{4}(x)</math>, and <math>f_{6}(x)</math>, we may rewrite the expression as
 +
<cmath>3(\text{sin}^4{x} + \text{cos}^4{x}) - 2(\text{sin}^6{x} + \text{cos}^6{x}) = 1</cmath>
 +
We now simplify each term separately using some algebraic manipulation and the Pythagorean identity.
 +
 
 +
We can rewrite <math>3(\text{sin}^4 x + \text{cos}^4 x)</math> as
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<math>3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x</math>, which is equivalent to
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<math>3 - 6\text{sin}^2 x \text{cos}^2 x</math>.
 +
 
 +
As for <math>2(\text{sin}^6 x + \text{cos}^6 x)</math>, we may factor it as
 +
<math>2(\text{sin}^2 x + \text{cos}^2 x)(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)</math> which can be rewritten as
 +
<math>2(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)</math>, and then as
 +
<math>2(\text{sin}^2 x + \text{cos}^2)^2) - 6\text{sin}^2 x \text{cos}^2 x</math>, which is equivalent to  
 +
<math>2 - 6\text{sin}^2 x \text{cos}^2 x</math>.
 +
 
 +
Putting everything together, we have <math>(3 - 6\text{sin}^2 {x} \text{cos}^2 {x}) - (2 - 6\text{sin}^2 {x} \text{cos}^2 {x}) = 1</math> or <math>1 = 1</math>. Therefore, the given equation <math>3f_{4}(x)-2f_{6}(x)=f_{2}(x)</math> is true for all real <math>x</math>, meaning that there are more than <math>8</math> values of <math>x</math> that satisfy the given equation and so the answer is <math>\boxed{\text{(E) more than }8}</math>.
  
 
== See also ==
 
== See also ==
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{{AMC12 box|year=2000|num-b=6|num-a=8}}
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{{AMC12 box|year=2002|ab=P|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:16, 12 July 2024

Problem

Let $f_n (x) = \text{sin}^n x + \text{cos}^n x.$ For how many $x$ in $[0,\pi]$ is it true that

\[6f_{4}(x)-4f_{6}(x)=2f_{2}(x)?\]

$\text{(A) }2 \qquad \text{(B) }4 \qquad \text{(C) }6 \qquad \text{(D) }8 \qquad \text{(E) more than }8$

Solution 1

Divide by 2 on both sides to get \[3f_{4}(x)-2f_{6}(x)=f_{2}(x)\] Substituting the definitions of $f_{2}(x)$, $f_{4}(x)$, and $f_{6}(x)$, we may rewrite the expression as \[3(\text{sin}^4{x} + \text{cos}^4{x}) - 2(\text{sin}^6{x} + \text{cos}^6{x}) = 1\] We now simplify each term separately using some algebraic manipulation and the Pythagorean identity.

We can rewrite $3(\text{sin}^4 x + \text{cos}^4 x)$ as $3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x$, which is equivalent to $3 - 6\text{sin}^2 x \text{cos}^2 x$.

As for $2(\text{sin}^6 x + \text{cos}^6 x)$, we may factor it as $2(\text{sin}^2 x + \text{cos}^2 x)(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)$ which can be rewritten as $2(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)$, and then as $2(\text{sin}^2 x + \text{cos}^2)^2) - 6\text{sin}^2 x \text{cos}^2 x$, which is equivalent to $2 - 6\text{sin}^2 x \text{cos}^2 x$.

Putting everything together, we have $(3 - 6\text{sin}^2 {x} \text{cos}^2 {x}) - (2 - 6\text{sin}^2 {x} \text{cos}^2 {x}) = 1$ or $1 = 1$. Therefore, the given equation $3f_{4}(x)-2f_{6}(x)=f_{2}(x)$ is true for all real $x$, meaning that there are more than $8$ values of $x$ that satisfy the given equation and so the answer is $\boxed{\text{(E) more than }8}$.

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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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