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

Solution:

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 unbounded but the range still starts at $0$ and grows without bound in a way that cannot match the domain, so no such $b$ exists.

If $a\lt0,$ the domain is $[0,-b/a]$ 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.

2003 AMC 12A Problem 25

Solution:

Problem 25 in Other Years