7.99 If the function $f$ is defined as $f(x)=a x^{2}-\sqrt{2}$, where $a$ is a positive constant, and if $f(f(\sqrt{2}))=-\sqrt{2}$, then $a$ equals
(A) $\frac{2-\sqrt{2}}{2}$.
(B) $\frac{1}{2}$.
(C) $2-\sqrt{2}$.
(D) $\frac{\sqrt{2}}{2}$.
(E) $\frac{2+\sqrt{2}}{2}$.
(41st American High School Mathematics Examination, 1990)
[Solution] Let $f(\sqrt{2})=a(\sqrt{2})^{2}-\sqrt{2}=2 a-\sqrt{2}$,
$$
\begin{array}{ll}
\text { Also } & f(f(\sqrt{2}))=a(2 a-\sqrt{2})^{2}-\sqrt{2}=-\sqrt{2}, \\
\therefore & a(2 a-\sqrt{2})^{2}=0 . \\
\because & a>0, \quad \therefore \quad a=\frac{\sqrt{2}}{2} .
\end{array}
$$

Therefore, the answer is $(D)$.