1967 AHSME Problems/Problem 14

Revision as of 13:07, 10 August 2020 by Eulerntm (talk | contribs) (Problem 14)

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

Let $x^2 = a$ and $y^2 = b.$

then

        $a+4b=1$

and

        $4a+b=4$

By solving we find--

$a=1$

$b=0$

However \[a=x^2\] and \[b=y^2\]

Therefore $y=0$, and $x=1,-1$

Thus, the only solutions are $(0,1)$, and $(0,-1)$


So there are only 2 solutions


$\rightarrow$ $\fbox{C}$

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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