Difference between revisions of "1993 AHSME Problems/Problem 12"

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== Solution ==
 
== Solution ==
<math>\fbox{E}</math>
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As <math>f(2x)=\frac{2}{2+x}</math>, we have that <math>f(x)=\frac{2}{2+\frac{x}{2}}</math>. This also means that <math>2f(x)=\frac{2}{2(2+\frac{x}{2})}</math> which implies that the answer is \fbox{E}$
  
 
== See also ==
 
== See also ==

Revision as of 21:32, 4 April 2021

Problem

If $f(2x)=\frac{2}{2+x}$ for all $x>0$, then $2f(x)=$

$\text{(A) } \frac{2}{1+x}\quad \text{(B) } \frac{2}{2+x}\quad \text{(C) } \frac{4}{1+x}\quad \text{(D) } \frac{4}{2+x}\quad \text{(E) } \frac{8}{4+x}$

Solution

As $f(2x)=\frac{2}{2+x}$, we have that $f(x)=\frac{2}{2+\frac{x}{2}}$. This also means that $2f(x)=\frac{2}{2(2+\frac{x}{2})}$ which implies that the answer is \fbox{E}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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