Difference between revisions of "1974 AHSME Problems/Problem 12"

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==See Also==
 
==See Also==
 
{{AHSME box|year=1974|num-b=11|num-a=13}}
 
{{AHSME box|year=1974|num-b=11|num-a=13}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 12:43, 5 July 2013

Problem

If $g(x)=1-x^2$ and $f(g(x))=\frac{1-x^2}{x^2}$ when $x\not=0$, then $f(1/2)$ equals

$\mathrm{(A)\ } 3/4 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \  } 3 \qquad \mathrm{(D) \  } \sqrt{2}/2 \qquad \mathrm{(E) \  }\sqrt{2}$

Solution

Notice that if we can find a $y$ such that $g(y)=\frac{1}{2}$, then we can plug that into the second equation to get $f\left(\frac{1}{2}\right)=f(g(y))=\frac{1-y^2}{y^2}$. Thus, we let $1-y^2=\frac{1}{2}\implies y^2=\frac{1}{2}$. Notice that, since our final expression only involves $y^2$, we don't need to take the square root. Thus, $f\left(\frac{1}{2}\right)=\frac{1-y^2}{y^2}=\frac{1-\frac{1}{2}}{\frac{1}{2}}=1, \boxed{\text{B}}$.

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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