1961 AHSME Problems/Problem 34

Problem

Let S be the set of values assumed by the fraction $\frac{2x+3}{x+2}$ . When $x$ is any member of the interval $x \ge 0$ . If there exists a number $M$ such that no number of the set $S$ is greater than $M$ , then $M$ is an upper bound of $S$ . If there exists a number $m$ such that such that no number of the set $S$ is less than $m$ , then $m$ is a lower bound of $S$ . We may then say:

$\textbf{(A)}\ \text{m is in S, but M is not in S}\qquad\\ \textbf{(B)}\ \text{M is in S, but m is not in S}\qquad\\ \textbf{(C)}\ \text{Both m and M are in S}\qquad\\ \textbf{(D)}\ \text{Neither m nor M are in S}\qquad\\ \textbf{(E)}\ \text{M does not exist either in or outside S}$

Solution

This problem is really finding the range of a function with a restricted domain.

Dividing $x+2$ into $2x+3$ yields $2 - \frac{1}{x+2}$ . Since $x \ge 0$ , as $x$ gets larger, $\frac{1}{x+2}$ approaches $0$ , so $\frac{2x+3}{x+2}$ approaches $2$ as $x$ gets larger. That means $M = 2$ . Since $\frac{1}{x+2}$ can never be $0$ , $\frac{2x+3}{x+2}$ can never be $2$ , so $M$ is not in the set $S$ .

For the smallest value, plug in $0$ in $\frac{2x+3}{x+2}$ to get $\frac{3}{2}$ , so $m = \frac{3}{2}$ . Since plugging in $0$ results in $m$ , $m$ is in the set $S$ .

Thus, the answer is $\boxed{\textbf{(A)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 33	Followed by Problem 35
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1961 AHSME Problems/Problem 34

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