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

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

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 $[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{12}{y}}} = k$ $\Rrightarrow[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y} \cdot 4}} = k$ $\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{\sqrt{4}} = (k)^{\sqrt{4}}$ $\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{2} = (k)^{2} = k^2$

$\boxed{D}$

~JustinLee2017

1978 AHSME (Problems • Answer Key • Resources)
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.

Page

Toolbox

Search

1978 AHSME Problems/Problem 17

Problem 17

See Also