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Difference between revisions of "1969 AHSME Problems/Problem 35"

(Created page with "== Problem == Let <math>L(m)</math> be the <math>x</math> coordinate of the left end point of the intersection of the graphs of <math>y=x^2-6</math> and <math>y=m</math>, where ...")
 
(Solution to Problem 35 (credit to soy_un_chemisto and tim9099xxzz) -- calculus problem on non-calculus test)
 
Line 9: Line 9:
 
\text{(E) undetermined} </math>
 
\text{(E) undetermined} </math>
  
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== Solution ==
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== Solutions ==
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<math>\fbox{B}</math>
 
  
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== See also ==
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===Solution 1===
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{{AHSME 35p box|year=1969|num-b=34|num-a=35}}   
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 +
Since <math>L(a)</math> is the <math>x</math> coordinate of the left end point of the intersection of the graphs of <math>y=x^2-6</math> and <math>y=a</math>, we can substitute <math>a</math> for <math>y</math> and find the lowest solution <math>x</math>.
 +
<cmath>x^2 - 6 = a</cmath>
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<cmath>x = \pm \sqrt{a+6}</cmath>
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That means <math>L(m) = -\sqrt{m+6}</math> and <math>L(-m) = -\sqrt{-m+6}</math>.  That means
 +
<cmath>r = \frac{-\sqrt{-m+6}+\sqrt{m+6}}{m}</cmath>
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Since plugging in <math>0</math> for <math>m</math> results in <math>0/0</math>, there is a removable discontinuity.  Multiply the fraction by <math>\frac{\sqrt{-m+6}+\sqrt{m+6}}{\sqrt{-m+6}+\sqrt{m+6}}</math> to get
 +
<cmath>r = \frac{m+6 - (-m+6)}{m(\sqrt{-m+6}+\sqrt{m+6})}</cmath>
 +
<cmath>r = \frac{2m}{m(\sqrt{-m+6}+\sqrt{m+6})}</cmath>
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<cmath>r = \frac{2}{(\sqrt{-m+6}+\sqrt{m+6})}</cmath>
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Now there wouldn't be a problem plugging in <math>0</math> for <math>m</math>.  Doing so results in <math>r = \tfrac{2}{2\sqrt{6}} = \tfrac{1}{\sqrt{6}}</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>.
 +
 
 +
===Solution 2===
 +
 
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From Solution 1, <math>L(m) = -\sqrt{m+6}</math> and <math>L(-m) = -\sqrt{-m+6}</math>, so
 +
<cmath>r = \frac{-\sqrt{-m+6}+\sqrt{m+6}}{m}</cmath>
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Since <math>m</math> is arbitrarily close to <math>0</math>, we wish to find
 +
<cmath>\lim_{m \rightarrow 0} \frac{-\sqrt{-m + 6} + \sqrt{m + 6}}{m}</cmath>
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Using [[L'Hopital's Rule]], the limit is equivalent to
 +
<cmath>\lim_{m \rightarrow 0} \frac{1}{2 \sqrt{-m + 6}} + \frac{1}{2 \sqrt{m + 6}}</cmath>
 +
Calculating the limit shows that <math>r</math> is <math>\boxed{\text{(B) arbitrarily close to }\frac{1}{\sqrt{6}}}</math>.
 +
 
 +
== See Also ==
 +
{{AHSME 35p box|year=1969|num-b=34|after=Last Problem}}   
  
 
[[Category: Intermediate Algebra Problems]]
 
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:19, 27 July 2018

Problem

Let $L(m)$ be the $x$ coordinate of the left end point of the intersection of the graphs of $y=x^2-6$ and $y=m$, where $-6<m<6$. Let $r=[L(-m)-L(m)]/m$. Then, as $m$ is made arbitrarily close to zero, the value of $r$ is:

$\text{(A) arbitrarily close to } 0\quad\\ \text{(B) arbitrarily close to }\frac{1}{\sqrt{6}}\quad \text{(C) arbitrarily close to }\frac{2}{\sqrt{6}} \quad \text{(D) arbitrarily large} \quad \text{(E) undetermined}$

Solutions

Solution 1

Since $L(a)$ is the $x$ coordinate of the left end point of the intersection of the graphs of $y=x^2-6$ and $y=a$, we can substitute $a$ for $y$ and find the lowest solution $x$. \[x^2 - 6 = a\] \[x = \pm \sqrt{a+6}\] That means $L(m) = -\sqrt{m+6}$ and $L(-m) = -\sqrt{-m+6}$. That means \[r = \frac{-\sqrt{-m+6}+\sqrt{m+6}}{m}\] Since plugging in $0$ for $m$ results in $0/0$, there is a removable discontinuity. Multiply the fraction by $\frac{\sqrt{-m+6}+\sqrt{m+6}}{\sqrt{-m+6}+\sqrt{m+6}}$ to get \[r = \frac{m+6 - (-m+6)}{m(\sqrt{-m+6}+\sqrt{m+6})}\] \[r = \frac{2m}{m(\sqrt{-m+6}+\sqrt{m+6})}\] \[r = \frac{2}{(\sqrt{-m+6}+\sqrt{m+6})}\] Now there wouldn't be a problem plugging in $0$ for $m$. Doing so results in $r = \tfrac{2}{2\sqrt{6}} = \tfrac{1}{\sqrt{6}}$, so the answer is $\boxed{\textbf{(B)}}$.

Solution 2

From Solution 1, $L(m) = -\sqrt{m+6}$ and $L(-m) = -\sqrt{-m+6}$, so \[r = \frac{-\sqrt{-m+6}+\sqrt{m+6}}{m}\] Since $m$ is arbitrarily close to $0$, we wish to find \[\lim_{m \rightarrow 0} \frac{-\sqrt{-m + 6} + \sqrt{m + 6}}{m}\] Using L'Hopital's Rule, the limit is equivalent to \[\lim_{m \rightarrow 0} \frac{1}{2 \sqrt{-m + 6}} + \frac{1}{2 \sqrt{m + 6}}\] Calculating the limit shows that $r$ is $\boxed{\text{(B) arbitrarily close to }\frac{1}{\sqrt{6}}}$.

See Also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Last Problem
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 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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