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

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

Latest revision as of 14:47, 3 September 2024

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

See also

1990 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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