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Difference between revisions of "2023 AMC 12B Problems/Problem 6"
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We know that the roots of P(x) are at <math>1,2,...,10</math>. When the degree of the term of each root is odd, the graph of P(x) will pass through the graph and change signs, and vice versa. So at <math>x=1</math>, the graph will change signs; at <math>x=2</math>, the graph will not, and so on.
 
We know that the roots of P(x) are at <math>1,2,...,10</math>. When the degree of the term of each root is odd, the graph of P(x) will pass through the graph and change signs, and vice versa. So at <math>x=1</math>, the graph will change signs; at <math>x=2</math>, the graph will not, and so on.
−
\newline
 
 
This tells us that the interval <math>(1,2)</math> is positive, <math>(2,3)</math> is also positive, <math>(3,4)</math> is negative, <math>(4,5)</math> is also negative, and so on, with the pattern being <math>+,+,-,-,+,+,-,-,...</math> .
 
This tells us that the interval <math>(1,2)</math> is positive, <math>(2,3)</math> is also positive, <math>(3,4)</math> is negative, <math>(4,5)</math> is also negative, and so on, with the pattern being <math>+,+,-,-,+,+,-,-,...</math> .
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\newline
+
The positive intervals are therefore <math>(1,2)</math>, <math>(2,3)</math>, <math>(5,6)</math>, <math>(6,7)</math>, <math>(9,10)</math>, and <math>(10,\infty)</math>, for a total of {\boxed{\textbf{(C) 6}}.
−
The positive intervals are therefore <math>(1,2)</math>, <math>(2,3)</math>, <math>(5,6)</math>, <math>(6,7)</math>, <math>(9,10)</math>, and <math>(10,\infty)</math>, for a total of \boxed{\textbf{(C) 6}}.
 
  
 
~nm1728
 
~nm1728

Revision as of 20:00, 15 November 2023

The following problem is from both the 2023 AMC 10B #12 and 2023 AMC 12B #6, so both problems redirect to this page.

Problem

When the roots of the polynomial

$P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}$

are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is $P(x)$ positive?

Solution 1

$P(x)$ is a product of $(x-r_n)$ or 10 terms. When $x < 1$, all terms are $< 0$, but $P(x) > 0$ because there is an even number of terms. The sign keeps alternating $+,-,+,-,....,+$. There are 11 intervals, so there are $\boxed{\textbf{6}}$ positives and 5 negatives. $\boxed{\textbf{(C) 6}}$

~$\textbf{Techno}\textcolor{red}{doggo}$

Solution 2

Denote by $I_k$ the interval $\left( k - 1 , k \right)$ for $k \in \left\{ 2, 3, \cdots , 10 \right\}$ and $I_1$ the interval $\left( - \infty, 1 \right)$.

Therefore, the number of intervals that $P(x)$ is positive is \begin{align*} 1 + \sum_{i=1}^{10} \Bbb I \left\{ \sum_{j=i}^{10} j \mbox{ is even} \right\} & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{\left( i + 10 \right) \left( 11 - i \right)}{2} \mbox{ is even} \right\} \\ & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{- i^2 + i + 110}{2} \mbox{ is even} \right\} \\ & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{i^2 - i}{2} \mbox{ is odd} \right\} \\ & = \boxed{\textbf{(C) 6}} . \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3

We can use the turning point behavior at the roots of a polynomial graph to find out the amount of intervals that are positive. \newline First, we evaluate any value on the interval $(-\infty, 1)$. Since the degree of the polynomial is $1+2+...+9$ = $\frac{9\times10}{2}$ = $45$, and every term in P(x) is negative, multiplying 45 negatives gives a negative value. So $(-\infty, 0)$ is a negative interval. \newline We know that the roots of P(x) are at $1,2,...,10$. When the degree of the term of each root is odd, the graph of P(x) will pass through the graph and change signs, and vice versa. So at $x=1$, the graph will change signs; at $x=2$, the graph will not, and so on. This tells us that the interval $(1,2)$ is positive, $(2,3)$ is also positive, $(3,4)$ is negative, $(4,5)$ is also negative, and so on, with the pattern being $+,+,-,-,+,+,-,-,...$ . The positive intervals are therefore $(1,2)$, $(2,3)$, $(5,6)$, $(6,7)$, $(9,10)$, and $(10,\infty)$, for a total of {\boxed{\textbf{(C) 6}}.

~nm1728

See Also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 10 Problems and Solutions
2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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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