Question

Let $S$ be the set of values assumed by the fraction $\frac{2x+3}{x+2}$ when $x$ is any member of the interval $x\ge0.$ Let $M$ be the least upper bound of $S,$ and let $m$ be the greatest lower bound of $S.$ We may then say:

$m$ is in $S,$ but $M$ is not in $S$

$M$ is in $S,$ but $m$ is not in $S$

both $m$ and $M$ are in $S$

neither $m$ nor $M$ is in $S$

$M$ does not exist either in or outside $S$

Answer: A

Concepts:function inequality bounding to limit cases

Difficulty rating: 1500

Small Hint:

Rewrite the fraction as $2-\frac1{x+2}$

Big Hint:

Evaluate the lower endpoint and examine the limit as $x$ increases

Solution:

We have $\frac{2x+3}{x+2}=2-\frac1{x+2}.$ For $x\ge0,$ this increases from $\frac{3}{2}$ toward $2$ without reaching $2.$ Thus $S=[\frac32,2),$ so its greatest lower bound $m=\frac{3}{2}$ belongs to $S,$ while its least upper bound $M=2$ does not.

Therefore, the correct answer is A.

1961 AMC 12 Problem 34

Answer: A

Difficulty rating: 1500

Small Hint:

Big Hint:

Solution:

Problem 34 in Other Years