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| Polynomials <imath>P(x)</imath> and <imath>Q(x)</imath> each have degree <imath>3</imath> and leading coefficient <imath>1</imath>, and their roots are all elements of <imath>\{1,2,3,4,5\}</imath>. The function <imath>f(x) = \tfrac{P(x)}{Q(x)}</imath> has the property that there exist real numbers <imath>a < b < c < d</imath> such that the set of all real numbers <imath>x</imath> such that <imath>f(x) \leq 0</imath> consists of the closed interval <imath>[a,b]</imath> together with the open interval <imath>(c,d)</imath>. How many functions <imath>f(x)</imath> are possible? | | Polynomials <imath>P(x)</imath> and <imath>Q(x)</imath> each have degree <imath>3</imath> and leading coefficient <imath>1</imath>, and their roots are all elements of <imath>\{1,2,3,4,5\}</imath>. The function <imath>f(x) = \tfrac{P(x)}{Q(x)}</imath> has the property that there exist real numbers <imath>a < b < c < d</imath> such that the set of all real numbers <imath>x</imath> such that <imath>f(x) \leq 0</imath> consists of the closed interval <imath>[a,b]</imath> together with the open interval <imath>(c,d)</imath>. How many functions <imath>f(x)</imath> are possible? |
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| ==Solution 1 ==
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| Step 1: Sign Chart Analysis
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| From the given <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>f(x) > 0</imath> elsewhere, we deduce:
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| * <imath>a</imath> and <imath>b</imath> are zeros of <imath>f</imath> since we transition from <imath>f > 0</imath> to <imath>f \le 0</imath> at <imath>a</imath> and from <imath>f \le 0</imath> to <imath>f > 0</imath> at <imath>b</imath>.
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| * <imath>c</imath> and <imath>d</imath> are vertical asymptotes or holes of <imath>f</imath> since <imath>f(x)</imath> is undefined at <imath>c</imath> and <imath>d</imath>. We have <imath>f \le 0</imath> in <imath>(c,d)</imath> , and <imath>f> 0</imath> outside the interval.
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| Thus, the sign pattern is:
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| <imath>\textstyle
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| \begin{array}{cccccccccc}
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| & (-\infty, a) & a & (a,b) & b & (b,c) & c & (c,d) & d & (d,\infty) \\
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| f(x) & + & 0 & - & 0 & + & \text{pole} & - & \text{pole} & +
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| \end{array}
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| </imath>
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| Step 2: Structure of <imath>f(x)</imath>
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| According to the given conditions, \( f(x) \) can be expressed as:
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| <cmath>f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}.</cmath>
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| Combining the sign analysis from Step 1, we can rewrite this expression as:
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| <cmath>f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}.</cmath>
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| Furthermore, we can break down the expression into two parts:
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| <cmath>f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}.</cmath>
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| Defining \( f(x) = f_1(x) \cdot f_2(x) \), we have:
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| <cmath>f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \quad f_2(x) = \frac{x-p_3}{x-q_3}.</cmath>
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| Step 3: Analysis of \( p_3 \) and \( q_3 \)
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| * \( f_1(x) \) completely satisfies the sign chart for \( f(x) \). We can conclude that \( f_2(x) \) must be positive in every interval; otherwise, it would introduce extra sign changes. This forces the condition: \( p_3 = q_3 \), if \( p_3 \neq q_3 \), then \( f_2(x) \) would be negative in interval between <imath>p_3</imath> and <imath>q_3 </imath> . Let \( p_3 = q_3 = t \).
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| * Furthermore, \( t \) cannot lie within \( [a,b] \cup (c,d) \). If it did, there would be a hole in graph within that interval. This contradicts the given condition that the set \( \{x : f(x) \le 0\} \) consists of the closed interval \( [a, b] \) and the open interval \( (c, d) \).
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| * Additionally, \( t \) cannot be equal to \( a \) or \( b \), or that point would be a hole (a point of discontinuity) in the graph of \( f(x) \), which contradicts the closed interval \( [a, b] \).
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| Thus, \( t \) can only be equal to \( c \) or \( d \), or lie in an interval outside of \( [a,b] \cup (c,d) \).
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| Step 4: Ways to choose <imath>a,b,c,d</imath> from <imath>\{1,2,3,4,5\}</imath>
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| Given <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>a < b < c < d</imath>, the number of ways to choose <imath>a,b,c,d</imath> from <imath>\{1,2,3,4,5\}</imath> is:
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| <cmath>\binom{5}{4} = 5 </cmath>
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| Step 5: The possible value of <imath>a</imath>, <imath>b</imath>, <imath>c</imath>, <imath>d</imath>, and <imath>t</imath> for the five cases are:
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| * Case 1: <imath>[1,2]\cup (3,4)</imath>, <imath>t</imath> can be 3, 4, or 5
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| * Case 2: <imath>[1,2]\cup (3,5)</imath>, <imath>t</imath> can be 3 or 5
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| * Case 3: <imath>[1,2]\cup (4,5)</imath>, <imath>t</imath> can be 3, 4, or 5.
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| * Case 4: <imath>[1,3]\cup (4,5)</imath>, <imath>t</imath> can be 4 or 5.
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| * Case 5: <imath>[2,3]\cup (4,5)</imath>, <imath>t</imath> can be 1, 4, or 5.
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| Summing up all possible cases, <imath>3 + 2 + 3 + 2 + 3 = 13</imath> possible functions <imath>f(x)</imath>.
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| Thus, the answer is <imath>\boxed{\text{(E) }13}.</imath>
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| - Victor Zhang
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| ---------------------------------------------------------------------------------------------
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| Edit:
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| ~ABC09090927: I believe this solution is wrong. The actual answer is 8, which is not one of the answer choices. I didn't read this carefully, but there's definitely double-counts for this solution. Like for example:
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| <cmath>\frac{(x-1)(x-3)(x-4)}{(x-1)^2(x-2)} = \frac{(x-2)(x-3)(x-4)}{(x-1)(x-2)^2}.</cmath>
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| The question asks us how many different <imath>f</imath> are possible, not <imath>P,Q</imath>.
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| This video (Credited to Sohil Rathi - Omegalearn) deepdives into the error that MAA made themselves.
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| The answer is not 13. The answer is 8.
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| [https://www.youtube.com/watch?v=SPbTyq3Dz_0 The problem that no one got correct - AMC 12A 2025]
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| ==Video Solution 1 by OmegaLearn== | | ==Video Solution 1 by OmegaLearn== |
and 
each have degree 
and leading coefficient 
, and their roots are all elements of 
. The function 
has the property that there exist real numbers 
such that the set of all real numbers 
such that 
consists of the closed interval ![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
together with the open interval 
. How many functions 
are possible?
