https://artofproblemsolving.com/wiki/index.php?title=1974_AHSME_Problems/Problem_12&feed=atom&action=history
1974 AHSME Problems/Problem 12 - Revision history
2024-03-28T16:37:26Z
Revision history for this page on the wiki
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Nathan wailes at 16:43, 5 July 2013
2013-07-05T16:43:12Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:43, 5 July 2013</td>
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Nathan wailes
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1=2: added category
2012-05-30T14:18:49Z
<p>added category</p>
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1=2
https://artofproblemsolving.com/wiki/index.php?title=1974_AHSME_Problems/Problem_12&diff=47146&oldid=prev
Admin25: Created page with "==Problem== If <math> g(x)=1-x^2 </math> and <math> f(g(x))=\frac{1-x^2}{x^2} </math> when <math> x\not=0 </math>, then <math> f(1/2) </math> equals <math> \mathrm{(A)\ } 3/4 \..."
2012-05-26T22:24:16Z
<p>Created page with "==Problem== If <math> g(x)=1-x^2 </math> and <math> f(g(x))=\frac{1-x^2}{x^2} </math> when <math> x\not=0 </math>, then <math> f(1/2) </math> equals <math> \mathrm{(A)\ } 3/4 \..."</p>
<p><b>New page</b></p><div>==Problem==<br />
If <math> g(x)=1-x^2 </math> and <math> f(g(x))=\frac{1-x^2}{x^2} </math> when <math> x\not=0 </math>, then <math> f(1/2) </math> equals <br />
<br />
<math> \mathrm{(A)\ } 3/4 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } \sqrt{2}/2 \qquad \mathrm{(E) \ }\sqrt{2} </math><br />
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==Solution==<br />
Notice that if we can find a <math> y </math> such that <math> g(y)=\frac{1}{2} </math>, then we can plug that into the second equation to get <math> f\left(\frac{1}{2}\right)=f(g(y))=\frac{1-y^2}{y^2} </math>. Thus, we let <math> 1-y^2=\frac{1}{2}\implies y^2=\frac{1}{2} </math>. Notice that, since our final expression only involves <math> y^2 </math>, we don't need to take the square root. Thus, <math> f\left(\frac{1}{2}\right)=\frac{1-y^2}{y^2}=\frac{1-\frac{1}{2}}{\frac{1}{2}}=1, \boxed{\text{B}} </math>.<br />
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==See Also==<br />
{{AHSME box|year=1974|num-b=11|num-a=13}}</div>
Admin25