1969 AMC 12 Problem 35

Question

Let

L(m)

be the

x

-coordinate of the left endpoint of the intersection of the graphs of

y=x^2-6

and

y=m,

where

-6\lt m\lt6.

Let

r=\frac{[L(-m)-L(m)]}{m}.

Then, as

m

is made arbitrarily close to zero, the value of

r

is:

arbitrarily close to zero

arbitrarily close to

\dfrac1{\sqrt6}

arbitrarily close to

\dfrac2{\sqrt6}

arbitrarily large

undetermined

The left intersection coordinate is

L(m)=-\sqrt{6+m}

Substitute into

r

and rationalize the numerator

The left intersection satisfies

L(m)=-\sqrt{6+m}.

Hence

\begin{aligned} r&=\frac{-\sqrt{6-m}+\sqrt{6+m}}{m}\\ &=\frac{2}{\sqrt{6+m}+\sqrt{6-m}}. \end{aligned}

As

m

approaches

0,

this approaches

\frac{2}{2\sqrt6}=\frac{1}{\sqrt6}.

Therefore, the correct answer is B.

Problem 35 in other years