## Tuesday, February 21, 2012

### this week's MGRE Math Beast Challenge

From here:

Go to it! My answer will appear in the comments. The official answer will appear next week.

_

#### 1 comment:

Kevin Kim said...

This one is a simple "plug-and-chug," as they say: simply plug in 2 for "x," and see what happens. If you really want to, you can plug in 10 for x as well, just to make sure you're on the right track.

For Quantity A, if you plug in a 2, then you get:

1/(1-(1/(1+(1/2))))

=

1/(1-(1/(3/2)))

=

1/(1-(2/3))

=

1/(1/3)

= 3

For Quantity B:

1/(1+(1/(1-(1/2))))

=

1/(1+(1/(1/2)))

= 1/(1+2)

= 1/3

Obviously, A is greater than B.

Just to be sure, plug in a 10 for x. In that case,

Quantity A = 11

Quantity B = 9/19

At this point, it's obvious that, as x gets larger, Quantity B remains a fraction less than 1, whereas Quantity A keeps increasing as x increases. A will always be greater than B.

For fun, I tried plugging in 10^6 for x. The result was still in favor of Quantity A. If you were to graph Quantity B as a function, you'd see that, as x increases in value, the function approaches 1/2 but can never reach it: 1/2 is, in fact, an asymptote! We can prove this by setting Quantity B equal to 1/2. If you do the math, you get the absurd result that -1 = 0. Such absurdities alert us to the presence of an asymptote.

So: f(x) can never equal or surpass 1/2.

f(x) < 1/2 as x approaches ∞.