2003 AMC 12A Problems/Problem 25
Problem
Let . For how many real values of
is there at least one positive value of
for which the domain of
and the range of
are the same set?
Solution
The function has a codomain of all non-negative numbers, or
. Since the domain and the range of
are the same, it follows that the domain of
also satisfies
.
The function has two zeroes at , which must be part of the domain. Since the domain and the range are the same set, it follows that
is in the codomain of
, or
. This implies that one (but not both) of
is non-positive. If
is positive, then
, which implies that a negative number falls in the domain of
, contradiction. Thus
must be non-positive,
is non-negative, and the domain of the function occurs when
, or
![$0 \le x \le \frac{-b}{a}.$](http://latex.artofproblemsolving.com/0/4/8/048dba23cb79813ba2028192a7ccc50fbe521655.png)
Completing the square, by the Trivial Inequality (remember that
). Since
is continuous and assumes this maximal value at
, it follows that the range of
is
![$0 \le f(x) \le \sqrt{\frac{-b^2}{4a}}.$](http://latex.artofproblemsolving.com/4/b/4/4b434ee413a6b27f6d7e0664398aae02a6f8d6c3.png)
As the domain and the range are the same, we have that (we can divide through by
since it is given that
is positive). Hence
, which both we can verify work, and the answer is
.
See Also
2003 AMC 12A (Problems • Answer Key • Resources) | |
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