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 domain | + | 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>. |
− | <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 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 |
− | <math> | + | <center><math>0 \le x \le \frac{-b}{a}.</math></center> |
− | <math>x=\ | + | [[Completing the square]], <math>f(x) = \sqrt{a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a}} \le \sqrt{\frac{-b^2}{4a}}</math> by the [[Trivial Inequality]] (remember that <math>a \le 0</math>). Since <math>f</math> is continuous and assumes this maximal value at <math>x = \frac{-b}{2a}</math>, it follows that the range of <math>f</math> is |
− | + | <center><math>0 \le f(x) \le \sqrt{\frac{-b^2}{4a}}.</math></center> | |
− | + | 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. | ||
+ | |||
+ | ==Solution 3 (more basic)== | ||
+ | *When numbers are referred to as positive and negative, it is understood that they are real numbers, as complex numbers don't have a definite positive/negative value. | ||
+ | |||
+ | First, notice that the square root of anything, whether it be a positive number, a negative number, or a complex, cannot yield a negative number. The square root of a positive number is a positive number, and the square root of a negative or complex number will yield a complex number. Complex numbers are a bit harder to address than real numbers, so I won't go into depth on complex domain/range (You don't need to, as this is a 2003 question, and the test writers most likely did not intend for test takers to take complex numbers into consideration. Also, even without considering complex domain/range, we can still get the answer). | ||
+ | |||
+ | Since the range cannot include negative numbers, the domain must be the same. This means that the graph of <math>f(x)</math> must only exist in the real coordinate plane for non-negative x-values. If any negative value of x yields a positive value under the square root, then <math>f(x)</math> will be positive, and then a negative x-value will be part the domain, and that contradicts what the first thing that we found. Now, first consider the graph of a quadratic equation/parabola. We know that a parabola always has a vertex, and always extends until <math>+</math> or <math>-</math> infinity, depending on the <math>a</math> value. Quite conveniently, we have an <math>a</math> value in the problem. | ||
+ | |||
+ | Consider the case where <math>a</math> is positive. Then, the parabola will "start" at the vertex, and extend upwards, hitting the x-axis at any roots. The y-values of a "positive" parabola are <math>[\infty, 0]</math> between x-values of <math>-\infty</math> and <math>r_1</math> (the leftmost root), negative between <math>r_1</math> and <math>r_2</math> (the two roots) and <math>[0, \infty]</math> between <math>r_2</math> and <math>\infty</math>. If we take the square root of such a parabola, the new function will exist only from <math>x=[-\infty, r_1]</math> and <math>x=[r_2, \infty]</math>. In order words, if <math>a</math> is positive, the graph of <math>f(x)</math> will exist from <math>-\infty</math> to <math>\infty</math> except between the two roots. This means, that regardless of where the roots are, some negative value of <math>x</math> will yield a positive value of <math>f(x)</math>. This means a positive value of <math>a</math> will not work. | ||
+ | |||
+ | Now we consider the case where <math>a</math> is negative. The parabola will do the opposite of the above; i.e. extend downwards. Now, either the entire graph of the parabola is below the x-axis (no roots), or some part including the vertex sticks out above it. The graph of the square root of the parabola will either be non-existent, or exist only between the 2 roots, respectively. This could work, as long as both roots are at least 0 (this way we will not have negative x-values in the domain). | ||
+ | |||
+ | Upon examination, we find that <math>ax^2+bx</math> can be rewritten <math>x(ax+b)</math>. The roots are then <math>x=0</math> and <math>x=-\frac{b}{a}</math>. Since <math>a</math> is negative, and <math>b</math> is positive, <math>-\frac{b}{a}</math> will always be positive. Therefore, the graph of <math>f(x)</math> will only exist from <math>x=\left[0, -\frac{b}{a}\right]</math> (the domain). | ||
+ | |||
+ | Notice that <math>ax^2+bx</math> can also be written in vertex form. The maximum value of <math>f(x)</math> (since <math>a</math> is negative) is the y-coordinate of the vertex, i.e. the range of <math>f(x)</math> is <math>[0, \text{vertex}]</math>. Converting <math>ax^2+bx</math> to vertex form yields <math>f(x) = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a}</math>. The y-coordinate of the vertex is then <math>-\frac{b^2}{4a}</math>. We want the square root of this, so the range of <math>f(x)</math> is <math>\left[0, \sqrt{-\frac{b^2}{4a}}\right]</math>. Since domain and range must be equivalent, we have <math>\frac{b}{-a} = \sqrt{\frac{b^2}{-4a}} \Rightarrow \frac{b}{-a} = \frac{b}{\sqrt{-4a}} \Rightarrow -a = \sqrt{-4a} \Rightarrow a^2 = -4a \Rightarrow a^2 + 4a = 0 \Rightarrow a(a+4) = 0</math>. So we have <math>a= 0, -4</math>, which is 2 values or <math>\boxed{\mathbf{(C)}}</math> | ||
==See Also== | ==See Also== | ||
− | + | {{AMC12 box|year=2003|ab=A|num-b=24|after=Last question}} | |
− | |||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:34, 10 July 2020
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
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.
Solution 3 (more basic)
- When numbers are referred to as positive and negative, it is understood that they are real numbers, as complex numbers don't have a definite positive/negative value.
First, notice that the square root of anything, whether it be a positive number, a negative number, or a complex, cannot yield a negative number. The square root of a positive number is a positive number, and the square root of a negative or complex number will yield a complex number. Complex numbers are a bit harder to address than real numbers, so I won't go into depth on complex domain/range (You don't need to, as this is a 2003 question, and the test writers most likely did not intend for test takers to take complex numbers into consideration. Also, even without considering complex domain/range, we can still get the answer).
Since the range cannot include negative numbers, the domain must be the same. This means that the graph of must only exist in the real coordinate plane for non-negative x-values. If any negative value of x yields a positive value under the square root, then will be positive, and then a negative x-value will be part the domain, and that contradicts what the first thing that we found. Now, first consider the graph of a quadratic equation/parabola. We know that a parabola always has a vertex, and always extends until or infinity, depending on the value. Quite conveniently, we have an value in the problem.
Consider the case where is positive. Then, the parabola will "start" at the vertex, and extend upwards, hitting the x-axis at any roots. The y-values of a "positive" parabola are between x-values of and (the leftmost root), negative between and (the two roots) and between and . If we take the square root of such a parabola, the new function will exist only from and . In order words, if is positive, the graph of will exist from to except between the two roots. This means, that regardless of where the roots are, some negative value of will yield a positive value of . This means a positive value of will not work.
Now we consider the case where is negative. The parabola will do the opposite of the above; i.e. extend downwards. Now, either the entire graph of the parabola is below the x-axis (no roots), or some part including the vertex sticks out above it. The graph of the square root of the parabola will either be non-existent, or exist only between the 2 roots, respectively. This could work, as long as both roots are at least 0 (this way we will not have negative x-values in the domain).
Upon examination, we find that can be rewritten . The roots are then and . Since is negative, and is positive, will always be positive. Therefore, the graph of will only exist from (the domain).
Notice that can also be written in vertex form. The maximum value of (since is negative) is the y-coordinate of the vertex, i.e. the range of is . Converting to vertex form yields . The y-coordinate of the vertex is then . We want the square root of this, so the range of is . Since domain and range must be equivalent, we have . So we have , which is 2 values or
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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