1978 AHSME Problems/Problem 17

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

\[\boxed{D}\]

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