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1969 AMC 12 Problem 35

Attempt Problem 35 of the 1969 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1969 AMC 12 solutions, or check the answer key.

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35.

Let L(m)L(m) be the xx-coordinate of the left endpoint of the intersection of the graphs of y=x2−6y=x^2-6 and y=m,y=m, where −6<m<6.-6\lt m\lt6. Let r=[L(−m)−L(m)]m.r=\frac{[L(-m)-L(m)]}{m}. Then, as mm is made arbitrarily close to zero, the value of rr is:

arbitrarily close to zero

arbitrarily close to 16\dfrac1{\sqrt6}

arbitrarily close to 26\dfrac2{\sqrt6}

arbitrarily large

undetermined

Answer: B
Concepts:calculusradicalrationalizing denominator
Difficulty rating: 1970
Small Hint:

The left intersection coordinate is L(m)=−6+mL(m)=-\sqrt{6+m}

Big Hint:

Substitute into rr and rationalize the numerator

Solution:

The left intersection satisfies L(m)=−6+m. L(m)=-\sqrt{6+m}. Hence r=−6−m+6+mm=26+m+6−m. \begin{aligned} r&=\frac{-\sqrt{6-m}+\sqrt{6+m}}{m}\\ &=\frac{2}{\sqrt{6+m}+\sqrt{6-m}}. \end{aligned} As mm approaches 0,0, this approaches 226=16.\frac{2}{2\sqrt6}=\frac{1}{\sqrt6}.

Therefore, the correct answer is B.

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