1958 AMC 12 Problem 46

Question

For values of

x

less than

1

but greater than

-4,

the expression

\frac{x^2-2x+2}{2x-2}

has:

no maximum or minimum value

a minimum value of

1

a maximum value of

1

a minimum value of

-1

a maximum value of

-1

Set

t=x-1,

which is negative on the given interval

Rewrite the expression as

\tfrac12(t+\frac{1}{t})

and use the negative-

t

form of AM-GM

Set

t=x-1.

Then

-5\lt t\lt0,

and the expression becomes

\frac{t^2+1}{2t} =\frac12\left(t+\frac1t\right).

For

t\lt0,

t+\frac{1}{t}\le-2,

with equality when

t=-1.

That value is allowed and corresponds to

x=0.

Hence the expression is at most

-1,

and its maximum is

-1.

Thus, the correct answer is E.

Problem 46 in other years