Difference between revisions of "2003 AMC 12A Problems/Problem 25"
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The function <math>f(x) = \sqrt{x(ax+b)}</math> has a [[codomain]] of all non-negative numbers, or <math>0 \le f(x)</math>. Since the domain and the range of <math>f</math> are the same, it follows that the domain of <math>f</math> also satisfies <math>0 \le x</math>. | The function <math>f(x) = \sqrt{x(ax+b)}</math> has a [[codomain]] of all non-negative numbers, or <math>0 \le f(x)</math>. Since the domain and the range of <math>f</math> are the same, it follows that the domain of <math>f</math> also satisfies <math>0 \le x</math>. | ||
− | The function has two zeroes at <math>x = 0, \frac{-b}{a}</math>, which must be part of the domain. Since the domain and the range are the same set, it follows that <math>\frac{-b}{a}</math> is in the codomain of <math>f</math>, or <math>0 \le \frac{-b}{a}</math>. This implies that one (but not both) of <math>a,b</math> is non-positive. | + | The function has two zeroes at <math>x = 0, \frac{-b}{a}</math>, which must be part of the domain. Since the domain and the range are the same set, it follows that <math>\frac{-b}{a}</math> is in the codomain of <math>f</math>, or <math>0 \le \frac{-b}{a}</math>. This implies that one (but not both) of <math>a,b</math> is non-positive. The problem states that there is at least one positive value of b that works, thus <math>a</math> must be non-positive, <math>b</math> is non-negative, and the domain of the function occurs when <math>x(ax+b) > 0</math>, or |
<center><math>0 \le x \le \frac{-b}{a}.</math></center> | <center><math>0 \le x \le \frac{-b}{a}.</math></center> |
Revision as of 16:48, 8 April 2019
Contents
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 1
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. The problem states that there is at least one positive value of b that works, 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
.
Solution 2
If , then squaring both sides of the given equation and subtracting
and
yields
. Completing the square, we get
where
. Divide out by
to put the equation in the standard form of an ellipse or hyperbola (depending on the sign of
) to get
.
Before continuing, it is important to note that because ,
has roots 0 and
. Now, we can use the function we deduced to figure out some of its properties when:
: A semi-hyperbola above or on the x-axis. Therefore, no positive value of
allows the domain and range to be the same set because the range will always be
and the domain will always be undefined on some finite range between some value and zero.
: We must refer back to the original function; this results in a horizontal semiparabola in the first quadrant, which satisfies that the domain and range of the function are equal. Specifically, both sets are
.
is the only case where this happens.
: A semi-ellipse in quadrant one. Since its roots are 0 and
, its domain must be
. To make its domain and range equal, the maximum value of the ellipse must then be
. But we have another expression for the maximum value of the ellipse, which is
. Setting these two expressions equal to each other will find us the final value of
that satisfies the question.
We already knew 0 was a solution from earlier, so -4 is our only new solution (we already ruled out any positive value of as a solution, so 4 does not work). Thus there are
values of
that make the domain and range of
the same set.
See Also
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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