1963 AHSME Problems/Problem 17

Problem

The expression $\dfrac{\dfrac{a}{a+y}+\dfrac{y}{a-y}}{\dfrac{y}{a+y}-\dfrac{a}{a-y}}$, $a$ real, $a\neq 0$, has the value $-1$ for:

$\textbf{(A)}\ \text{all but two real values of }y \qquad \\ \textbf{(B)}\ \text{only two real values of }y \qquad \\ \textbf{(C)}\ \text{all real values of }y\qquad \\ \textbf{(D)}\ \text{only one real value of }y\qquad \\ \textbf{(E)}\ \text{no real values of }y$

Solution

First, note that $y \ne \pm a$ because that would make the denominator $0$.

Create common denominators in the complex fraction. \[\frac{\frac{a^2-ay}{a^2-y^2} + \frac{ay+y^2}{a^2-y^2}}{\frac{ay-y^2}{a^2-y^2} - \frac{a^2+ay}{a^2-y^2}}\] \[\frac{\frac{a^2+y^2}{a^2-y^2}}{\frac{-a^2-y^2}{a^2-y^2}}\] Since $a \ne 0$, $a^2 + y^2 > 0$, so no other restrictions need to be made. \[\frac{a^2 + y^2}{a^2 - y^2} \cdot \frac{a^2 - y^2}{-a^2 - y^2}\] \[\frac{a^2 + y^2}{-(a^2 + y^2)}\] \[-1\] The expression (with suitable restrictions) simplifies to $-1$, so the answer is $\boxed{\textbf{(A)}}$.

See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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