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

Revision as of 13:06, 18 July 2018

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

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 == Solution ==
-<math>\fbox{D}</math>
+If <math>f(w)=-\sqrt2</math>, then <math>aw^2=0\implies w=0</math>. Therefore <math>f(\sqrt2)=0\implies 2a=\sqrt2</math>, so <math>\fbox{D}</math>
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Difference between revisions of "1990 AHSME Problems/Problem 12"

Revision as of 13:06, 18 July 2018

Problem

Solution

See also