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1969 AHSME Problems/Problem 35

Revision as of 13:19, 27 July 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 35 (credit to soy_un_chemisto and tim9099xxzz) -- calculus problem on non-calculus test)
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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)
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Problem 34 		Followed by
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All AHSME Problems and Solutions

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