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Difference between revisions of "1967 AHSME Problems/Problem 14"

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Since we know that <math>y=f(x)</math>, we can solve for <math>y</math> in terms of <math>x</math>. This gives us
 
Since we know that <math>y=f(x)</math>, we can solve for <math>y</math> in terms of <math>x</math>. This gives us
  
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<math>y=\frac{1}{1-x}</math>
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<math>y=\frac{x}{1-x}</math>
  
−
<math>\Rightarrow y(1-x)=1</math>
+
<math>\Rightarrow y(1-x)=x</math>
  
−
<math>\Rightarrow y-yx=1</math>
+
<math>\Rightarrow y-yx=x</math>
  
−
<math>\Rightarrow yx=y-1</math>
+
<math>\Rightarrow y=yx+x</math>
  
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<math>\Rightarrow x=\frac{y-1}{y}</math>
+
<math>\Rightarrow y=x(y+1)</math>
  
−
Therefore, we want to find the function with <math>y</math> that outputs <math>\frac{y-1}{y}</math>
+
<math>\Rightarrow x=\frac{y}{y+1}</math>
−
Listing out the possible outputs from each of the given functions we get
+
 
 +
Therefore, we want to find the function with <math>y</math> that outputs <math>\frac{y}{y+1}</math> Listing out the possible outputs from each of the given functions we get
 
<math>f\left(\frac{1}{y}\right)=\frac{1}{y-1}</math>
 
<math>f\left(\frac{1}{y}\right)=\frac{1}{y-1}</math>
  
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<math>f(-y)=\frac{-y}{y+1}</math>
 
<math>f(-y)=\frac{-y}{y+1}</math>
  
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<math>-f(y)=\frac{1}{y-1}</math>
+
<math>-f(y)=\frac{y}{y-1}</math>
  
 
<math>-f(-y)=\frac{y}{y+1}</math>
 
<math>-f(-y)=\frac{y}{y+1}</math>
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== See also ==
 
== See also ==
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{{AHSME box|year=1967|num-b=13|num-a=15}}   
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{{AHSME 40p box|year=1967|num-b=13|num-a=15}}   
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:42, 22 August 2023

Problem

Let $f(t)=\frac{t}{1-t}$, $t \not= 1$. If $y=f(x)$, then $x$ can be expressed as

$\textbf{(A)}\ f\left(\frac{1}{y}\right)\qquad \textbf{(B)}\ -f(y)\qquad \textbf{(C)}\ -f(-y)\qquad \textbf{(D)}\ f(-y)\qquad \textbf{(E)}\ f(y)$

Solution

Since we know that $y=f(x)$, we can solve for $y$ in terms of $x$. This gives us

$y=\frac{x}{1-x}$

$\Rightarrow y(1-x)=x$

$\Rightarrow y-yx=x$

$\Rightarrow y=yx+x$

$\Rightarrow y=x(y+1)$

$\Rightarrow x=\frac{y}{y+1}$

Therefore, we want to find the function with $y$ that outputs $\frac{y}{y+1}$ Listing out the possible outputs from each of the given functions we get $f\left(\frac{1}{y}\right)=\frac{1}{y-1}$

$f(y)=\frac{1}{1-y}$

$f(-y)=\frac{-y}{y+1}$

$-f(y)=\frac{y}{y-1}$

$-f(-y)=\frac{y}{y+1}$

Since $-f(-y)=\frac{y}{y+1}=x$ the answer must be $\boxed{C}$.

See also

1967 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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