# Difference between revisions of "1990 AHSME Problems/Problem 12"

## Problem

Let $f$ be the function defined by $f(x)=ax^2-\sqrt{2}$ for some positive $a$. If $f(f(\sqrt{2}))=-\sqrt{2}$ then $a=$

$\text{(A) } \frac{2-\sqrt{2}}{2}\quad \text{(B) } \frac{1}{2}\quad \text{(C) } 2-\sqrt{2}\quad \text{(D) } \frac{\sqrt{2}}{2}\quad \text{(E) } \frac{2+\sqrt{2}}{2}$

## Solution

If $f(u)=-\sqrt2$, then $au^2=0\implies u=0$. Therefore $f(\sqrt2)=0\implies 2a=\sqrt2$, so $\fbox{D}$