7.97 If $k$ is a positive number, and for each positive number $x$, the function $f$ satisfies
$$
\left[f\left(x^{2}+1\right)\right]^{\sqrt{x}}=k \text {; }
$$

then for each positive number $y$, the expression $\left[f\left(\frac{9+y^{2}}{y^{2}}\right)\right]^{\sqrt{\frac{12}{y}}}$ equals
(A) $\sqrt{k}$.
(B) $2 k$.
(C) $k \sqrt{k}$.
(D) $k^{2}$.
(E) $k^{\sqrt{k}}$.
(29th American High School Mathematics Examination, 1978)
[Solution]
$$
\begin{aligned}
& {\left[f\left(\frac{9+y^{2}}{y^{2}}\right)\right]^{\sqrt{\frac{12}{y}}}=\left\{\left[f\left(\left(\frac{3}{y}\right)^{2}+1\right)\right] \sqrt{\frac{3}{y}}\right\}^{2} } \\
= & \left\{\left[f\left(t^{2}+1\right)\right]^{\sqrt{t}}\right\}^{2}=k^{2}
\end{aligned}
$$

Here, $\quad t=\frac{3}{y}$. Therefore, the answer is $(D)$.