1961 AMC 12 Problem 34

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