2003 AMC 12A Problem 25

Question

Let

f(x) = \sqrt{ax^2 + bx}.

For how many real values of

a

is there at least one positive value of

b

for which the domain of

f

and the range of

f

are the same set?

0

1

2

3

infinitely many

If

a=0,

both the domain and range equal

[0,\infty)

for every

b\gt0

For

a\lt0

the domain is

[0,-\frac{b}{a}]

and the maximum of

f

is

\dfrac{b}{2\sqrt{-a}}

If

a=0,

then

f(x)=\sqrt{bx}

has domain and range both

[0,\infty),

so

a=0

works.

If

a\gt0,

the domain is

({-}\infty,-b/a]\cup[0,\infty),

while the range is

[0,\infty),

so no such

b

exists.

If

a\lt0,

the domain is

[0,-\frac{b}{a}]

and the range is

\left[0,\dfrac{b}{2\sqrt{-a}}\right].

Equating the right endpoints gives

-\dfrac ba=\dfrac{b}{2\sqrt{-a}},

so

2\sqrt{-a}=-a,

giving

a=-4.

Thus there are

2

values of

a,

and the correct answer is C.

Problem 25 in other years