2025 AMC 12A Problems/Problem 25: Difference between revisions

Revision as of 02:25, 7 November 2025

Polynomials $P(x)$ and $Q(x)$ each have degree $3$ and leading coefficient $1$ , and their roots are all elements of $\{1,2,3,4,5\}$ . The function $f(x) = \tfrac{P(x)}{Q(x)}$ has the property that there exist real numbers $a < b < c < d$ such that the set of all real numbers $x$ such that $f(x) \leq 0$ consists of the closed interval $[a,b]$ together with the open interval $(c,d)$ . How many functions $f(x)$ are possible?

Solution 1 by Victor Zhang

\documentclass[12pt]{article} \usepackage{amsmath, amssymb, amsthm} \usepackage{enumitem} \begin{document} \section*{Solution}

\subsection*{Step 1: Sign chart analysis} From the structure $f(x) \le 0$ on $[a,b] \cup (c,d)$ and $f(x) > 0$ elsewhere, we deduce: \begin{itemize} \item $a$ and $b$ are zeros of $f$ (since we transition from $f > 0$ to $f \le 0$ at $a$ and from $f \le 0$ to $f > 0$ at $b$ ). \item $c$ and $d$ are poles or holes of $f$ (since on $(c,d)$ we have $f \le 0$ , at $c^+$ and $d^-$ the sign is negative, and immediately outside the interval $f > 0$ ). \end{itemize} Thus the sign pattern is:\\ $\textstyle \begin{array}{cccccccccc} & (-\infty, a) & a & (a,b) & b & (b,c) & c & (c,d) & d & (d,\infty) \\ f(x) & + & 0 & - & 0 & + & \text{pole} & - & \text{pole} & + \end{array}$ \section*{Step 2: Structure of $ f(x) $} According to the given conditions, $ f(x) $ can be expressed as: \[ f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}. \] Combining the sign analysis from Step 1, we can rewrite this expression as: \[ f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}. \] Furthermore, we can break down the expression into two parts: \[ f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}. \] Defining $ f(x) = f_1(x) \cdot f_2(x) $, we have: \[ f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \quad f_2(x) = \frac{x-p_3}{x-q_3}. \]

\section*{Step 3: Analysis of $ p_3 $ and $ q_3 $} \begin{itemize} \item By analyzing the sign changes of $ f_1(x) $, we find that it completely satisfies the sign chart for $ f(x) $. We can therefore conclude that $ f_2(x) $ must be positive on every interval; otherwise, it would introduce an extra sign change. This forces the condition $ p_3 = q_3 $, because if $ p_3 \neq q_3 $, the factor $ f_2(x) $ would be negative on the interval between $ p_3 $ and $ q_3 $. Let $ p_3 = q_3 = t $. \item Furthermore, $ t $ cannot lie within $ [a,b] \cup (c,d) $. If it did, it would contradict the given condition that the set $ \{x : f(x) \le 0\} $ consists of the closed interval $ [a, b] $ and the open interval $ (c, d) $. \item Additionally, $ t $ cannot be equal to $ a $ or $ b $. If it were, that point would be a hole (a point of discontinuity) in the graph of $ f(x) $, which contradicts the fact that $ [a, b] $ is a \textbf{closed} interval where $ f(x) \le 0 $. On a closed interval, the function must be defined at the endpoints. \end{itemize} In conclusion, $ t $ can only be equal to $ c $ or $ d $, or lie in an interval outside of $ [a,b] \cup (c,d) $. \section*{Step 4: The Possible Values of $a$ , $b$ , $c$ , $d$ , and $t$ } \begin{itemize} \item Given $a < b < c < d$ and the condition that $\{x: f(x) \leq 0\} = [a,b] \cup (c,d)$ , the number of ways to choose $a,b,c,d$ from $\{1,2,3,4,5\}$ is: \[ \binom{5}{4} = 5 \] The five cases are: $[1,2]\cup (3,4)$ ; $[1,2]\cup (3,5)$ ; $[1,2]\cup (4,5)$ ; $[1,3]\cup (4,5)$ ; $[2,3]\cup (4,5)$ \item \textbf{Case 1:} $[1,2]\cup (3,4)$ , $t$ can be 3, 4, or 5.\\ This gives us 3 combinations: \[ f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)} \] \[ f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4)} \] \[ f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5)} \] \item \textbf{Case 2:} $[1,2]\cup (3,5)$ , $t$ can be 3 or 5.\\

This gives us 2 combinations: \[ f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)} \] \[ f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-5)(x-5)} \] \item \textbf{Case 3:} $[1,2]\cup (4,5)$ , $t$ can be 3, 4, or 5.\\ This gives us 3 combinations: \[ f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)} \] \[ f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4)} \] \[ f(x) = \frac{(x-1)(x-2)(x-5)}{(x-4)(x-5)(x-5)} \] \item \textbf{Case 4:} $[1,3]\cup (4,5)$ , $t$ can be 4 or 5.\\ This gives us 2 combinations: \[ f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)} \] \[ f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5)} \] \item \textbf{Case 5:} $[2,3]\cup (4,5)$ , $t$ can be 1, 4, or 5.\\ This gives us 3 combinations: \[ f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)} \] \[ f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)} \] \[ f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)} \] \end{itemize} \section*{Step 5: Final Answer} Summing up all possible cases from Step 4 gives us $3 + 2 + 3 + 2 + 3 = 13$ possible functions $f(x)$ . Thus the answer is $\boxed{E}$ . \end{document}

\section*{Step 4: The possible of a,b,c,d, and t } \begin{itemize} \item According to given $a < b < c < d$ and $ [a,b] \cup (c,d) $, the ways of choose a,b,c,d from {1,2,3,4,5}: $\binom{5}{4}=5$ List 5 cases as:\\$[1,2]\cup (3,4); [1,2]\cup (3,5); [1,2]\cup (4,5); [1,3]\cup (4,5); [2,3]\cup (4,5)$\item Case 1:$ (Error compiling LaTeX. Unknown error_msg)[1,2]\cup (3,4)$, t can be 3,4, or 5.\\ Give us 3 combnation: \[ f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)} \] \[ f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4 )} \] \[ f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5)}. \] \item Case 2:$ (Error compiling LaTeX. Unknown error_msg)[1,2]\cup (3,5)$, t can be 3, or 5.\\ Give us 2 combnation: \[ f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)} \] \[ f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5 )} \] \item Case 3:$ (Error compiling LaTeX. Unknown error_msg)[1,2]\cup (4,5)$, t can be 3,4, or 5.\\ Give us 3 combnation: \[ f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)} \] \[ f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4 )} \] \[ f(x) = \frac{(x-1)(x-2)(x-5)}{(x-4)(x-5)(x-5)}. \] \item Case 4:$ (Error compiling LaTeX. Unknown error_msg)[1,3]\cup (4,5)$, t can be 4, or 5.\\ Give us 2 combnation: \[ f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}

\] \[ f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5 )} \] \item Case 5:$ (Error compiling LaTeX. Unknown error_msg)[2,3]\cup (4,5)$, t can be 1,4, or 5.\\ Give us 3 combnation: \[ f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)} \] \[ f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)} \] \[ f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)} \]

\section*{Step 5: final answer } \Sum up all possible cases of step 4 give us 13 possible functions f(x).

Thus the answer is$ (Error compiling LaTeX. Unknown error_msg)\boxed{E}$. \end{document}

Video Solution 1 by OmegaLearn

https://youtu.be/SPbTyq3Dz_0

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2025 AMC 12A Problems/Problem 25: Difference between revisions

Revision as of 02:25, 7 November 2025

Solution 1 by Victor Zhang

Video Solution 1 by OmegaLearn

@@ Line 1: / Line 1: @@
-We begin by analyzing the sign chart. From the condition <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>f(x) > 0</imath> elsewhere, we deduce the following:
+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?
-- <imath>a</imath> and <imath>b</imath> are zeros of <imath>f</imath>, since we transition from <imath>f>0</imath> to <imath>f\le0</imath> at <imath>a</imath>, and from <imath>f\le0</imath> to <imath>f>0</imath> at <imath>b</imath>.
+==Solution 1 by Victor Zhang ==
-- <imath>c</imath> and <imath>d</imath> are poles or holes of <imath>f</imath>, since on <imath>(c,d)</imath> we have <imath>f\le0</imath>, while immediately outside that interval <imath>f>0</imath>.
+\documentclass[12pt]{article}
+\usepackage{amsmath, amssymb, amsthm}
+\usepackage{enumitem}
+\begin{document}
+\section*{Solution}
-Thus, the sign pattern of <imath>f(x)</imath> is
+\subsection*{Step 1: Sign chart analysis}
-<cmath>
+From the structure <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>f(x) &gt; 0</imath> elsewhere,
+we deduce:
+\begin{itemize}
+\item <imath>a</imath> and <imath>b</imath> are zeros of <imath>f</imath> (since we transition from <imath>f &gt; 0</imath> to <imath>f \le
+</imath> at <imath>a</imath> and from <imath>f \le 0</imath> to <imath>f &gt; 0</imath> at <imath>b</imath>).
+\item <imath>c</imath> and <imath>d</imath> are poles or holes of <imath>f</imath> (since on <imath>(c,d)</imath> we have <imath>f \le
+</imath>, at <imath>c^+</imath> and <imath>d^-</imath> the sign is negative, and immediately outside the
+interval <imath>f &gt; 0</imath>).
+\end{itemize}
+Thus the sign pattern is:\\
+<imath>\textstyle
 \begin{array}{cccccccccc}
- & (-\infty, a) & a & (a,b) & b & (b,c) & c & (c,d) & d & (d,\infty) \\
+&amp; (-\infty, a) &amp; a &amp; (a,b) &amp; b &amp; (b,c) &amp; c &amp; (c,d) &amp; d &amp; (d,\infty) \\
-f(x) & + & 0 & - & 0 & + & \text{pole} & - & \text{pole} & +
+f(x) &amp; + &amp; 0 &amp; - &amp; 0 &amp; + &amp; \text{pole} &amp; - &amp; \text{pole} &amp; +
 \end{array}
-</cmath>
+</imath>
+\section*{Step 2: Structure of \( f(x) \)}
----
+According to the given conditions, \( f(x) \) can be expressed as:
+\[
-**Step 2. Structure of <imath>f(x)</imath>**
-According to the given conditions, we can express
-<cmath>
 f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}.
-</cmath>
+\]
-Combining this with the sign analysis, we can write
+Combining the sign analysis from Step 1, we can rewrite this expression as:
-<cmath>
+\[
 f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}.
-</cmath>
+\]
-We can separate this into two factors:
+Furthermore, we can break down the expression into two parts:
-<cmath>
+\[
 f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}.
-</cmath>
+\]
-Define
+Defining \( f(x) = f_1(x) \cdot f_2(x) \), we have:
-<cmath>
+\[
-f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \qquad f_2(x) = \frac{x-p_3}{x-q_3}.
+f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \quad f_2(x) = \frac{x-p_3}{x-q_3}.
-</cmath>
+\]
----
+\section*{Step 3: Analysis of \( p_3 \) and \( q_3 \)}
+\begin{itemize}
+\item By analyzing the sign changes of \( f_1(x) \), we find that it completely
+satisfies the sign chart for \( f(x) \). We can therefore conclude that \( f_2(x) \)
+must be positive on every interval; otherwise, it would introduce an extra sign
+change. This forces the condition \( p_3 = q_3 \), because if \( p_3 \neq q_3 \),
+the factor \( f_2(x) \) would be negative on the interval between \( p_3 \) and \(
+q_3 \). Let \( p_3 = q_3 = t \).
+\item Furthermore, \( t \) cannot lie within \( [a,b] \cup (c,d) \). If it did, it
+would contradict the given condition that the set \( \{x : f(x) \le 0\} \) consists of
+the closed interval \( [a, b] \) and the open interval \( (c, d) \).
+\item Additionally, \( t \) cannot be equal to \( a \) or \( b \). If it were, that
+point would be a hole (a point of discontinuity) in the graph of \( f(x) \), which
+contradicts the fact that \( [a, b] \) is a \textbf{closed} interval where \( f(x) \le 0
+\). On a closed interval, the function must be defined at the endpoints.
+\end{itemize}
+In conclusion, \( t \) can only be equal to \( c \) or \( d \), or lie in an interval
+outside of \( [a,b] \cup (c,d) \).
+\section*{Step 4: The Possible Values of <imath>a</imath>, <imath>b</imath>, <imath>c</imath>, <imath>d</imath>, and <imath>t</imath>}
+\begin{itemize}
+\item Given <imath>a &lt; b &lt; c &lt; d</imath> and the condition that <imath>\{x: f(x) \leq 0\} = [a,b]
+\cup (c,d)</imath>, the number of ways to choose <imath>a,b,c,d</imath> from <imath>\{1,2,3,4,5\}</imath> is:
+\[
+\binom{5}{4} = 5
+\]
+The five cases are:
+<imath>[1,2]\cup (3,4)</imath>; <imath>[1,2]\cup (3,5)</imath>; <imath>[1,2]\cup (4,5)</imath>; <imath>[1,3]\cup (4,5)</imath>;
+<imath>[2,3]\cup (4,5)</imath>
+\item \textbf{Case 1:} <imath>[1,2]\cup (3,4)</imath>, <imath>t</imath> can be 3, 4, or 5.\\
+This gives us 3 combinations:
+\[
+f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}
+\]
+\[
+f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4)}
+\]
+\[
+f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5)}
+\]
+\item \textbf{Case 2:} <imath>[1,2]\cup (3,5)</imath>, <imath>t</imath> can be 3 or 5.\\
-**Step 3. Determining <imath>p_3</imath> and <imath>q_3</imath>**
+This gives us 2 combinations:
+\[
+f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}
+\]
+\[
+f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-5)(x-5)}
+\]
+\item \textbf{Case 3:} <imath>[1,2]\cup (4,5)</imath>, <imath>t</imath> can be 3, 4, or 5.\\
+This gives us 3 combinations:
+\[
+f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)}
+\]
+\[
+f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4)}
+\]
+\[
+f(x) = \frac{(x-1)(x-2)(x-5)}{(x-4)(x-5)(x-5)}
+\]
+\item \textbf{Case 4:} <imath>[1,3]\cup (4,5)</imath>, <imath>t</imath> can be 4 or 5.\\
+This gives us 2 combinations:
+\[
+f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}
+\]
+\[
+f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5)}
+\]
+\item \textbf{Case 5:} <imath>[2,3]\cup (4,5)</imath>, <imath>t</imath> can be 1, 4, or 5.\\
+This gives us 3 combinations:
+\[
+f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)}
+\]
+\[
+f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)}
+\]
+\[
+f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)}
+\]
+\end{itemize}
+\section*{Step 5: Final Answer}
+Summing up all possible cases from Step 4 gives us <imath>3 + 2 + 3 + 2 + 3 = 13</imath>
+possible functions <imath>f(x)</imath>.
+Thus the answer is <imath>\boxed{E}</imath>.
+\end{document}
-By analyzing the sign changes of <imath>f_1(x)</imath>, we see that it already matches the required sign pattern for <imath>f(x)</imath>. Therefore, <imath>f_2(x)</imath> must remain positive on every interval; otherwise, it would introduce extra sign changes.
+\section*{Step 4: The possible of a,b,c,d, and t }
+\begin{itemize}
+\item According to given <imath>a &lt; b &lt; c &lt; d </imath> and \( [a,b] \cup (c,d) \), the ways
+of choose a,b,c,d from {1,2,3,4,5}:
+<cmath> \binom{5}{4}=5</cmath>
+List 5 cases as:\\$[1,2]\cup (3,4); [1,2]\cup (3,5); [1,2]\cup (4,5); [1,3]\cup
+(4,5); [2,3]\cup (4,5)<imath>
-This means <imath>p_3 = q_3</imath>, since if <imath>p_3 \ne q_3</imath>, the factor <imath>\frac{x-p_3}{x-q_3}</imath> would change sign between <imath>p_3</imath> and <imath>q_3</imath>.
+\item Case 1: </imath>[1,2]\cup (3,4)<imath>, t can be 3,4, or 5.\\
-Let <imath>p_3 = q_3 = t</imath>.
+Give us 3 combnation:
+\[
-Furthermore:
+f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}
-- <imath>t</imath> cannot lie within <imath>[a,b] \cup (c,d)</imath>, since that would alter the set <imath>\{x : f(x) \le 0\}</imath>.
+\]
-- <imath>t</imath> cannot be equal to <imath>a</imath> or <imath>b</imath>, since that would make those endpoints holes (discontinuities), contradicting that <imath>[a,b]</imath> is a *closed* interval.
+\[
+f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4
-Hence, <imath>t</imath> must either be equal to <imath>c</imath> or <imath>d</imath>, or lie outside <imath>[a,b] \cup (c,d)</imath>.
+)}
+\]
----
+\[
-**Step 4. Possible values of <imath>a,b,c,d,t</imath>**
-Given <imath>a<b<c<d</imath> and <imath>\{x : f(x) \le 0\} = [a,b] \cup (c,d)</imath>, we choose four distinct numbers from <imath>\{1,2,3,4,5\}</imath> for <imath>a,b,c,d</imath>:
-<cmath>
-\binom{5}{4} = 5.
-</cmath>
-The five possible cases are:
-<cmath>
-[1,2]\cup(3,4), \quad [1,2]\cup(3,5), \quad [1,2]\cup(4,5), \quad [1,3]\cup(4,5), \quad [2,3]\cup(4,5).
-</cmath>
----
-**Case 1:** <imath>[1,2]\cup(3,4)</imath>
-<imath>t</imath> can be <imath>3</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
-<cmath>
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}, \quad
-f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4)}, \quad
 f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5)}.
-</cmath>
+\]
+\item Case 2: </imath>[1,2]\cup (3,5)<imath>, t can be 3, or 5.\\
-**Case 2:** <imath>[1,2]\cup(3,5)</imath>
+Give us 2 combnation:
-<imath>t</imath> can be <imath>3</imath> or <imath>5</imath>, giving 2 functions:
+\[
-<cmath>
+f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}, \quad
+\]
-f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-5)(x-5)}.
+\[
-</cmath>
+f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5
+)}
-**Case 3:** <imath>[1,2]\cup(4,5)</imath>
+\]
-<imath>t</imath> can be <imath>3</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
+\item Case 3: </imath>[1,2]\cup (4,5)<imath>, t can be 3,4, or 5.\\
-<cmath>
+Give us 3 combnation:
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)}, \quad
+\[
-f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4)}, \quad
+f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)}
+\]
+\[
+f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4
+)}
+\]
+\[
 f(x) = \frac{(x-1)(x-2)(x-5)}{(x-4)(x-5)(x-5)}.
-</cmath>
+\]
+\item Case 4: </imath>[1,3]\cup (4,5)<imath>, t can be 4, or 5.\\
-**Case 4:** <imath>[1,3]\cup(4,5)</imath>
+Give us 2 combnation:
-<imath>t</imath> can be <imath>4</imath> or <imath>5</imath>, giving 2 functions:
+\[
-<cmath>
+f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}
-f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}, \quad
-f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5)}.
-</cmath>
-**Case 5:** <imath>[2,3]\cup(4,5)</imath>
+\]
-<imath>t</imath> can be <imath>1</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
+\[
-<cmath>
+f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5
-f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)}, \quad
+)}
-f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)}, \quad
+\]
-f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)}.
+\item Case 5: </imath>[2,3]\cup (4,5)<imath>, t can be 1,4, or 5.\\
-</cmath>
+Give us 3 combnation:
+\[
+f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)}
+\]
+\[
+f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)}
+\] \[
+f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)}
+\]
----
+\section*{Step 5: final answer }
+\Sum up all possible cases of step 4 give us 13 possible functions f(x).
-**Step 5. Final Answer**
+Thus the answer is </imath>\boxed{E}$.
+\end{document}
-Summing all possible cases:
+==Video Solution 1 by OmegaLearn==
-<cmath>
+https://youtu.be/SPbTyq3Dz_0
-+ 2 + 3 + 2 + 3 = 13.
-</cmath>
-Thus, there are <imath>\boxed{13}</imath> possible functions <imath>f(x)</imath>, corresponding to choice <imath>\boxed{E}</imath>.