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

1978 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1978 AHSME Problems/Problem 17

Problem 17

Solution

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