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Difference between revisions of "1991 AHSME Problems/Problem 21"

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== Problem ==
 
== Problem ==
  
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For all real numbers <math>x</math> except <math>x=0</math> and <math>x=1</math> the function <math>f(x)</math> is defined by <math>f(x/(1-x))=1/x</math>. Suppose <math>0\leq t\leq \pi/2</math>. What is the value of <math>f(\sec^2t)</math>?
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For all real numbers <math>x</math> except <math>x=0</math> and <math>x=1</math> the function <math>f(x)</math> is defined by <math>f(x/(x-1))=1/x</math>. Suppose <math>0\leq t\leq \pi/2</math>. What is the value of <math>f(\sec^2t)</math>?
  
 
<math>\text{(A) } sin^2\theta\quad
 
<math>\text{(A) } sin^2\theta\quad
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== Solution ==
 
== Solution ==
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Let <math>y=\frac{x}{x-1} \Rightarrow xy-y=x \Rightarrow x=\frac{y}{y-1}</math>
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<math>f(y)=\frac{1}{x}=\frac{y-1}{y}=1-\frac{1}{y}</math>
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<math>f(sec^2t)=sin^2t</math>
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<math>\fbox{A}</math>
 
<math>\fbox{A}</math>
  

Latest revision as of 20:56, 17 October 2016

Problem

For all real numbers $x$ except $x=0$ and $x=1$ the function $f(x)$ is defined by $f(x/(x-1))=1/x$. Suppose $0\leq t\leq \pi/2$. What is the value of $f(\sec^2t)$?

$\text{(A) } sin^2\theta\quad \text{(B) } cos^2\theta\quad \text{(C) } tan^2\theta\quad \text{(D) } cot^2\theta\quad \text{(E) } csc^2\theta$

Solution

Let $y=\frac{x}{x-1} \Rightarrow xy-y=x \Rightarrow x=\frac{y}{y-1}$

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

$f(sec^2t)=sin^2t$

$\fbox{A}$

See also

1991 AHSME (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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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