If k is a positive number and f is a function such that, for every positive number x, [f(x^2+1)]^√ x=k, then, for every positive number y, [f((9+y^2)/(y^2))]^√(12)/(y) is equal to

1978 AMC 12 Problem 17

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17.

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\left(\frac{9+y^2}{y^2}\right)\right]^{\sqrt{\frac{12}{y}}}

is equal to

\sqrt k

2k

k\sqrt k

k^2

y\sqrt k

Choose

x

so that

x^2+1=\frac{9+y^2}{y^2}

Compare

\sqrt{\frac{12}{y}}

with

\sqrt x

after making the substitution

Set

x=\frac{3}{y},

which is positive. Then

\begin{aligned} x^2+1&=\frac{9+y^2}{y^2},\\ \sqrt{\frac{12}{y}}&=2\sqrt x. \end{aligned}

Therefore the requested expression is

\left(\left[f(x^2+1)\right]^{\sqrt x}\right)^2=k^2.

Therefore, the correct answer is D.