Track / Stage 3 / 153 of 260 #153 of 2000

Problem 153

AMC 10/12, early questions

Algebra Difficulty 3.5 Find the answer

Given the function $f(x)= \sqrt {ax^{2}+bx}$ , where $b > 0$ , find the value of the non-zero real number $a$ such that the domain and range of $f(x)$ are the same.

Your answer

Official solution

If $a > 0$ , since $ax^{2}+bx \geqslant 0$ , or $x(ax+b) \geqslant 0$ ,

For positive $b$ , the domain of $f(x)$ is: $D=(-\infty, -\frac{b}{a}] \cup [0, +\infty)$ ,

But the range of $f(x)$ is $A \subseteq [0, +\infty)$ , so $D \neq A$ , which does not meet the requirement.

If $a 0$ , thus $a=-4$ .

So the answer is: $\boxed{-4}$ .

This question mainly tests the understanding of the range of a function and the domain of a function for solving equations. It is a moderately difficult question.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.