1978 AHSME Problems/Problem 17

Problem 17

If $k$ is a positive number and $f$ is a function such that, for every positive number $x$, $\left[f(x^2+1)\right]^{\sqrt{x}}=k$; then, for every positive number $y$, $\left[f(\frac{9+y^2}{y^2})\right]^{\sqrt{\frac{12}{y}}}$ is equal to

$\textbf{(A) }\sqrt{k}\qquad \textbf{(B) }2k\qquad \textbf{(C) }k\sqrt{k}\qquad \textbf{(D) }k^2\qquad  \textbf{(E) }y\sqrt{k}$

Solution

We are given that \[[f(x^2 + 1)]^{\sqrt(x)} = k\] We can rewrite $\frac{9+y^2}{y^2}$ as $\frac{9}{y^2} + 1$ Thus, our function is now \[[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{12}{y}}} = k\] \[\Rrightarrow[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y} \cdot 4}} = k\] \[\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{\sqrt{4}} = (k)^{\sqrt{4}}\] \[\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{2} = (k)^{2} = k^2\]

Thus, the answer is $k^2\implies \boxed{D}$.

~JustinLee2017

~clarification by Leon

See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png