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 $\boxed{\textbf{(A)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 33
Followed by
Problem 35
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