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Difference between revisions of "2003 AMC 12A Problems/Problem 25"

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(I got somewhere, someone else go the rest of the way.)
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==Problem==
 
==Problem==
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Let <math>\displaystyle f(x)= \sqrt{ax^2+bx} </math>.  For how many [[real number | real]] values of <math>a</math> is there at least one [[positive number | positive]] value of <math> b </math> for which the [[domain]] of <math>f </math> and the [[range]] of <math> f </math> are the same [[set]]?
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Let <math>f(x)= \sqrt{ax^2+bx} </math>.  For how many [[real number | real]] values of <math>a</math> is there at least one [[positive number | positive]] value of <math> b </math> for which the [[domain]] of <math>f </math> and the [[range]] of <math> f </math> are the same [[set]]?
  
 
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} }  </math>
 
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} }  </math>
  
 
== Solution==
 
== Solution==
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{{solution}}
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The domain of this function is the range of the inverse function, and vice versa, so we find the inverse function:
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 +
<math>y=\sqrt{ax^2+bx}</math>
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<math>y^2=ax^2+bx</math>
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<math>x=\dfrac{-b\pm\sqrt{b^2+4ay^2}}{2a}</math>
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The domain of this is all real <math>y</math> such that <math>4ay^2\geq -b^2</math>
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The range of this function is the domain of the other function, which is all <math>x</math> such that <math>ax^2+bx\geq 0</math>. Thus we need to find all real <math>a</math> such that for all <math>x</math>, either both of those are true or neither are.
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{{incomplete|solution}}
  
 
==See Also==
 
==See Also==

Revision as of 10:09, 11 August 2008

Problem

Let $f(x)= \sqrt{ax^2+bx}$. For how many real values of $a$ is there at least one positive value of $b$ for which the domain of $f$ and the range of $f$ are the same set?

$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} }$

Solution

The domain of this function is the range of the inverse function, and vice versa, so we find the inverse function:

$y=\sqrt{ax^2+bx}$

$y^2=ax^2+bx$

$x=\dfrac{-b\pm\sqrt{b^2+4ay^2}}{2a}$

The domain of this is all real $y$ such that $4ay^2\geq -b^2$

The range of this function is the domain of the other function, which is all $x$ such that $ax^2+bx\geq 0$. Thus we need to find all real $a$ such that for all $x$, either both of those are true or neither are.

Template:Incomplete

See Also

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