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Difference between revisions of "1965 AHSME Problems/Problem 18"

(Created page with "<math> \frac{1}{1 + y} = \frac{1 - y}{1 - y^2}</math> <math> \frac{1 - y}{\frac{1 - y}{1 - y^2}} = 1 - y^2</math> <math> \boxed{(B) y^2}</math>")
 
(Add problem statement)
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== Problem ==
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If <math>1 - y</math> is used as an approximation to the value of <math>\frac {1}{1 + y}, |y| < 1</math>, the ratio of the error made to the correct value is:
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<math>\textbf{(A)}\ y \qquad
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\textbf{(B) }\ y^2 \qquad
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\textbf{(C) }\ \frac {1}{1 + y} \qquad
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\textbf{(D) }\ \frac{y}{1+y}\qquad
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\textbf{(E) }\ \frac{y^2}{1+y}\qquad </math>
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== Solution ==
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<math> \frac{1}{1 + y} = \frac{1 - y}{1 - y^2}</math>
 
<math> \frac{1}{1 + y} = \frac{1 - y}{1 - y^2}</math>
  

Revision as of 13:51, 16 July 2024

Problem

If $1 - y$ is used as an approximation to the value of $\frac {1}{1 + y}, |y| < 1$, the ratio of the error made to the correct value is:

$\textbf{(A)}\ y \qquad \textbf{(B) }\ y^2 \qquad \textbf{(C) }\ \frac {1}{1 + y} \qquad \textbf{(D) }\ \frac{y}{1+y}\qquad \textbf{(E) }\ \frac{y^2}{1+y}\qquad$

Solution

$\frac{1}{1 + y} = \frac{1 - y}{1 - y^2}$

$\frac{1 - y}{\frac{1 - y}{1 - y^2}} = 1 - y^2$

$\boxed{(B) y^2}$

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