1965 AHSME Problems/Problem 18

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

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

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

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

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1965 AHSME Problems/Problem 18

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