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Difference between revisions of "2023 AMC 10B Problems/Problem 12"

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are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is <math>P(x)</math> positive?
 
are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is <math>P(x)</math> positive?
  
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==Solution==
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==Solution 1==
  
 
The interval of the alternating signs. <math>P(x)</math> is a product of <math>(x-r_n)</math> or 10 terms. When <math>x < 1</math>, all terms are <math>< 0</math>, but <math>P(x) > 0</math> because they are even number of terms. The sign keep alternates <math>+,-,+,-,....,+</math>.  There are 11 intervals, so there are 6 positives and 5 negatives.
 
The interval of the alternating signs. <math>P(x)</math> is a product of <math>(x-r_n)</math> or 10 terms. When <math>x < 1</math>, all terms are <math>< 0</math>, but <math>P(x) > 0</math> because they are even number of terms. The sign keep alternates <math>+,-,+,-,....,+</math>.  There are 11 intervals, so there are 6 positives and 5 negatives.

Revision as of 17:39, 15 November 2023

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

The interval of the alternating signs. $P(x)$ is a product of $(x-r_n)$ or 10 terms. When $x < 1$, all terms are $< 0$, but $P(x) > 0$ because they are even number of terms. The sign keep alternates $+,-,+,-,....,+$. There are 11 intervals, so there are 6 positives and 5 negatives.

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

Solution

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)

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