Difference between revisions of "2003 AMC 12A Problems/Problem 25"
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<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} } </math> | <math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} } </math> | ||
− | == Solution== | + | == Solution 1== |
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>. | ||
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As the domain and the range are the same, we have that <math>\frac{-b}{a} = \sqrt{\frac{-b^2}{4a}} = \frac{b}{2\sqrt{-a}} \Longrightarrow a(a+4) = 0</math> (we can divide through by <math>b</math> since it is given that <math>b</math> is positive). Hence <math>a = 0, -4</math>, which both we can verify work, and the answer is <math>\mathbf{(C)}</math>. | As the domain and the range are the same, we have that <math>\frac{-b}{a} = \sqrt{\frac{-b^2}{4a}} = \frac{b}{2\sqrt{-a}} \Longrightarrow a(a+4) = 0</math> (we can divide through by <math>b</math> since it is given that <math>b</math> is positive). Hence <math>a = 0, -4</math>, which both we can verify work, and the answer is <math>\mathbf{(C)}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | If <math>f(x)=y</math>, then squaring both sides of the given equation and subtracting <math>ax^2</math> and <math>bx</math> yields <math>y^2-ax^2-bx=0</math>. Completing the square, we get <math>(x+\frac{b}{2a})^2-\frac{y^2}{a}=\frac{b^2}{4a^2}</math> where <math>y\geq 0</math>. Divide out by <math>\frac{b^2}{4a^2}</math> to put the equation in the standard form of an ellipse or hyperbola (depending on the sign of <math>a</math>) to get <math>\frac{(x+\frac{b}{2a})^2}{(\frac{b}{2a})^2}-\frac{y^2}{(\frac{b}{2\sqrt {\pm a}})^2}=1</math>. | ||
+ | |||
+ | Before continuing, it is important to note that because <math>f(x)=\sqrt{ax^2+bx}=\sqrt{ax(x+\frac{b}{a})}</math>, <math>f(x)</math> has roots 0 and <math>-\frac{b}{a}</math>. Now, we can use the function we deduced to figure out some of its properties when: | ||
+ | |||
+ | <math>a>0</math>: A semi-hyperbola above or on the x-axis. Therefore, no positive value of <math>a</math> allows the domain and range to be the same set because the range will always be <math>[0, \infty)</math> and the domain will always be undefined on some finite range between some value and zero. | ||
+ | |||
+ | <math>a=0</math>: 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 <math>[0, \infty )</math>. <math>a=0</math> is the only case where this happens. | ||
+ | |||
+ | <math>a<0</math>: A semi-ellipse in quadrant one. Since its roots are 0 and <math>-\frac{b}{a}</math>, its domain must be <math>[0, -\frac{b}{a}]</math>. To make its domain and range equal, the maximum value of the ellipse must then be <math>-\frac{b}{a}</math>. But we have another expression for the maximum value of the ellipse, which is <math>0+\frac{b}{2\sqrt {\pm a}}</math>. Setting these two expressions equal to each other will find us the final value of <math>a</math> that satisfies the question. | ||
+ | |||
+ | <math>\frac{b}{-a}=\frac{b}{2\sqrt {\pm a}}</math> | ||
+ | |||
+ | <math>-a=2\sqrt {\pm a}</math> | ||
+ | |||
+ | <math>a^2=\pm 4a</math> | ||
+ | |||
+ | <math>a=0,\pm 4</math> | ||
+ | |||
+ | We already knew 0 was a solution from earlier, so -4 is our only new solution (we already ruled out any positive value of <math>a</math> as a solution, so 4 does not work). Thus there are <math>\boxed {2\implies C}</math> values of <math>a</math> that make the domain and range of <math>f(x)</math> the same set. | ||
==See Also== | ==See Also== |
Revision as of 01:51, 28 October 2017
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. 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
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
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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