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

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== Solution ==
 
== Solution ==
<math>\fbox{D}</math>
+
If <math>f(u)=-\sqrt2</math>, then <math>au^2=0\implies u=0</math>. Therefore <math>f(\sqrt2)=0\implies 2a=\sqrt2</math>, so <math>\fbox{D}</math>
  
 
== See also ==
 
== See also ==

Revision as of 04:32, 4 February 2016

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 $\fbox{D}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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