Difference between revisions of "1971 AHSME Problems/Problem 30"

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== Solution 1 ==
 
== Solution 1 ==
<math>\boxed{\textbf{(D) }\frac{1}{1-x}}</math>.
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Extend the definition of <math>f_n(x)</math> to <math>n\leq0</math>: Let <math>f_{n-1}(x)</math> be the function such that <math>f_1(f_{n-1}(x))=f_n(x)</math>. From the problem, <math>f_5(x)=f_{35}(x)</math>, so the functions <math>f_1(x),f_2(x),\ldots</math> must repeat in a cycle whose length is a cycle which is a divisor of <math>35-5=30</math>. Thus, if <math>f_a(x)=f_b(x)</math> for integers <math>a</math> and <math>b</math>, we know that <math>a\equiv b</math> [[modulo]] <math>30</math>. Thus, because <math>28\equiv-2\pmod{30}</math>, we know that <math>f_{-2}(x)=f_{28}(x)</math>.
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It is clear that <math>f_0(x)=x</math>, because <math>f_1(f_0(x))=f_1(x)</math>.
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Let <math>f_{-1}(x)=y</math>. Then, we know that <math>f_1(y)=f_0(x)=x</math>, so we have the following equation we can solve for <math>y</math>:
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\begin{align*}
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\frac{2y-1}{y+1} &= x \
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2y-1 &= xy+x \
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y(2-x) &= x+1 \
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y &= \frac{x+1}{2-x}
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\end{align*}
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Let <math>f_{-2}(x)=z</math>. Then, we know that <math>f_1(z)=f_{-1}(x)=\frac{x+1}{2-x}</math>, so we have the following equation we can solve for <math>z</math>:
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\begin{align*}
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\frac{2z-1}{z+1} &= \frac{x+1}{2-x} \
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4z-2-2xz+x &= xz+x+z+1 \
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3z-3xz &= 3 \
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z(1-x) &= 1 \
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z &= \frac1{1-x}
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\end{align*}
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We derived earlier the fact that <math>z=f_{-2}(x)=f_{28}(x)</math>, so <math>f_{28}(x)=\boxed{\textbf{(D) }\frac{1}{1-x}}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
<math>\boxed{\textbf{(D) }\frac{1}{1-x}}</math>.
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Keep solving for the next function in the sequence of <math>f_n(x)</math> while being sure not to make silly algebra mistakes. This process reveals that <math>f_6(x)=x</math>, so <math>f_7(x)=f_1(x)</math>, and the cycle of functions repeats [[modulo]] <math>6</math>. Because <math>28\equiv4\pmod6</math>, we know that <math>f_{28}(x)=f_4(x)=\boxed{\textbf{(D) }\frac{1}{1-x}}</math>, which we calculated on the way to deducing that <math>f_6(x)=x</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AHSME 35p box|year=1971|num-b=29|num-a=31}}
 
{{AHSME 35p box|year=1971|num-b=29|num-a=31}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:53, 8 August 2024

Problem

Given the linear fractional transformation of $x$ into $f_1(x)=\dfrac{2x-1}{x+1}$. Define $f_{n+1}(x)=f_1(f_n(x))$ for $n=1,2,3,\cdots$. Assuming that $f_{35}(x)=f_5(x)$, it follows that $f_{28}(x)$ is equal to

$\textbf{(A) }x\qquad \textbf{(B) }\frac{1}{x}\qquad \textbf{(C) }\frac{x-1}{x}\qquad \textbf{(D) }\frac{1}{1-x}\qquad  \textbf{(E) }\text{None of these}$

Solution 1

Extend the definition of $f_n(x)$ to $n\leq0$: Let $f_{n-1}(x)$ be the function such that $f_1(f_{n-1}(x))=f_n(x)$. From the problem, $f_5(x)=f_{35}(x)$, so the functions $f_1(x),f_2(x),\ldots$ must repeat in a cycle whose length is a cycle which is a divisor of $35-5=30$. Thus, if $f_a(x)=f_b(x)$ for integers $a$ and $b$, we know that $a\equiv b$ modulo $30$. Thus, because $28\equiv-2\pmod{30}$, we know that $f_{-2}(x)=f_{28}(x)$.

It is clear that $f_0(x)=x$, because $f_1(f_0(x))=f_1(x)$.

Let $f_{-1}(x)=y$. Then, we know that $f_1(y)=f_0(x)=x$, so we have the following equation we can solve for $y$: 2y1y+1=x2y1=xy+xy(2x)=x+1y=x+12x

Let $f_{-2}(x)=z$. Then, we know that $f_1(z)=f_{-1}(x)=\frac{x+1}{2-x}$, so we have the following equation we can solve for $z$: 2z1z+1=x+12x4z22xz+x=xz+x+z+13z3xz=3z(1x)=1z=11x

We derived earlier the fact that $z=f_{-2}(x)=f_{28}(x)$, so $f_{28}(x)=\boxed{\textbf{(D) }\frac{1}{1-x}}$.

Solution 2

Keep solving for the next function in the sequence of $f_n(x)$ while being sure not to make silly algebra mistakes. This process reveals that $f_6(x)=x$, so $f_7(x)=f_1(x)$, and the cycle of functions repeats modulo $6$. Because $28\equiv4\pmod6$, we know that $f_{28}(x)=f_4(x)=\boxed{\textbf{(D) }\frac{1}{1-x}}$, which we calculated on the way to deducing that $f_6(x)=x$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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