Problem:
Let f(x)=ax2+bxβ. 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?
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. infinitely many
Solution:
The domain of f is {xβ£ax2+bxβ₯0}. If a=0, then for every positive value of b, the domain and range of f are each equal to the interval [0,β), so 0 is a possible value of a.
If aξ =0, the graph of y=ax2+bx is a parabola with x-intercepts at x=0 and x=βb/a. If a>0, the domain of f is (ββ,βb/a]βͺ[0,β), but the range of f cannot contain negative numbers. If a<0, the domain of f is [0,βb/a]. The maximum value of f occurs halfway between the x-intercepts, at x=βb/2a, and