Art of Problem Solving
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1978 AHSME Problems/Problem 17: Difference between revisions

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== Problem 17 ==
If <math>k</math> is a positive number and <math>f</math> is a function such that, for every positive number <math>x</math>, <math>\left[f(x^2+1)\right]^{\sqrt{x}}=k</math>;
then, for every positive number <math>y</math>, <math>\left[f(\frac{9+y^2}{y^2})\right]^{\sqrt{\frac{12}{y}}}</math> is equal to
<math>\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}  </math> 
We are given that <cmath>[f(x^2 + 1)]^{\sqrt(x)} = k</cmath>
We are given that <cmath>[f(x^2 + 1)]^{\sqrt(x)} = k</cmath>
We can rewrite <math>\frac{9+y^2}{y^2}</math> as <math>\frac{9}{y^2} + 1</math>
We can rewrite <math>\frac{9+y^2}{y^2}</math> as <math>\frac{9}{y^2} + 1</math>
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~JustinLee2017
~JustinLee2017
==See Also==
{{AHSME box|year=1978|num-b=16|num-a=18}}
{{MAA Notice}}

Revision as of 20:44, 13 February 2021

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


See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AHSME Problems and Solutions

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