Difference between revisions of "2018 AMC 12A Problems/Problem 21"

Latest revision as of 05:58, 29 July 2024

1 Problem
2 Solution 1 (Intermediate Value Theorem, Inequalities, Graphs)
3 Solution 2 (Similar to Solution 1)
4 Solution 3 (Similar to Solution 1)
5 Solution 4
6 Solution 5 (Monotonic Function)
7 Solution 6 (Proof by Contradiction)
8 Solution 7 (Calculus)
9 Solution 8 (Calculus)
10 Video Solution by Richard Rusczyk
11 See Also

Problem

Which of the following polynomials has the greatest real root?

$\textbf{(A) } x^{19}+2018x^{11}+1 \qquad \textbf{(B) } x^{17}+2018x^{11}+1 \qquad \textbf{(C) } x^{19}+2018x^{13}+1 \qquad \textbf{(D) } x^{17}+2018x^{13}+1 \qquad \textbf{(E) } 2019x+2018$

Solution 1 (Intermediate Value Theorem, Inequalities, Graphs)

Denote the polynomials in the answer choices by $A(x),B(x),C(x),D(x),$ and $E(x),$ respectively.

Note that $A(x),B(x),C(x),D(x),$ and $E(x)$ are strictly increasing functions with range $(-\infty,\infty).$ So, each polynomial has exactly one real root. The real root of $E(x)$ is $x=-\frac{2018}{2019}\approx-1.000.$ On the other hand, since $A(-1)=B(-1)=C(-1)=D(-1)=-2018$ and $A(0)=B(0)=C(0)=D(0)=1,$ we conclude that the real root for each of $A(x),B(x),C(x),$ and $D(x)$ must satisfy $x\in(-1,0)$ by the Intermediate Value Theorem (IVT).

We analyze the polynomials for $x\in(-1,0):$

We have $\begin{align*} B(x)-A(x)=D(x)-C(x)&=x^{17}-x^{19} \\ &=x^{17}\left(1-x^2\right) \\ &<0. \end{align*}$ As the graph of $y=A(x)$ is always above the graph of $y=B(x)$ in this interval, we deduce that $B(x)$ has a greater real root than $A(x)$ does. By the same reasoning, $D(x)$ has a greater real root than $C(x)$ does.

We have $\begin{align*} B(x)-D(x)&=2018x^{11}-2018x^{13} \\ &=2018x^{11}\left(1-x^2\right) \\ &<0, \end{align*}$ from which $B(x)$ has a greater real root than $D(x)$ does.

Now, we are left with comparing the real roots of $B(x)$ and $E(x).$ Since $B\left(-\frac{1}{\sqrt2}\right)<0<B\left(-\frac{1}{2}\right),$ it follows that the real root of $B(x)$ must satisfy $x\in\left(-\frac{1}{\sqrt2},-\frac{1}{2}\right)$ by the IVT. Clearly, $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$ has the greatest real root.

~MRENTHUSIASM

Solution 2 (Similar to Solution 1)

We can see that our real solution has to lie in the open interval $(-1,0)$ . From there, note that $x^a < x^b$ if $a$ , $b$ are odd positive integers if $a<b$ , so hence it can only either be $\textbf{(B)}$ or $\textbf{(E)}$ (as all of the other polynomials will be larger than the polynomial $\textbf{(B)}$ ). Observe that $\textbf{(E)}$ gives the solution $x=-\frac{2018}{2019}$ . We can approximate the root for $\textbf{(B)}$ by using $x=-\frac 12$ : $\left(-\frac{1}{2}\right)^{17} - \frac{2018}{2048} + 1 \approx 0.$ Therefore, the root for $\textbf{(B)}$ is approximately $-\frac 12$ . The answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$ .

~cpma213

Solution 3 (Similar to Solution 1)

Let the real solution to $B$ be $a.$ It is easy to see that when $a$ is plugged in to $A,$ since $-1 < a < 0,$ it follows that $a^{19} < a^{17}$ thus making the real solution to $A$ more "negative", or smaller than $B.$ Similarly we can assert that $D > C.$ Now to compare $B$ and $D,$ we can use the same method to what we used before to compare $B$ to $A,$ in which it is easy to see that the smaller exponent $(11)$ "wins". Now, the only thing left is for us to compare $B$ and $E.$ Plugging $\frac{-2018}{2019}$ (or the solution to $E$ ) into $B$ we obtain $\frac{(-2018)^{17}}{2019^{17}} + 2018\cdot\frac{(-2018)^{11}}{2019^{11}} + 1,$ which is intuitively close to $-1 - 2018 + 1 = -2018,$ much smaller than the solution the required $0.$ (For a more rigorous proof, one can note that $\left(\frac{2018}{2019}\right)^{17}$ and $\left(\frac{2018}{2019}\right)^{11}$ are both much greater than $\left(\frac{2018}{2019}\right)^{2019} \approx \frac{1}{e},$ by the limit definition of $e.$ Since $- \frac{1}{e} - 2018 \cdot \frac{1}{e} + 1$ is still much smaller the required $0$ for the solution to $B$ to be a solution, our answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}.$

~fidgetboss_4000

Solution 4

Denote the polynomials in the answer choices by $A(x),B(x),C(x),D(x),$ and $E(x),$ respectively.

The real root of $E(x)$ is $x=-\frac{2018}{2019}\approx-1.000.$ , therefore, we can eliminate this polynomial first. Notice that all of the real roots for all answer choices are in $(-1,0)$ . Therefore, $x^{11}<0, \quad x^2<1, \quad x^{13}>x^{11}, \quad x^{19}+2018x^{13}+1>x^{19}+2018x^{11}+1, \quad C(x)>A(x)$ $\quad x^{17}+2018x^{13}+1>x^{17}+2018x^{11}+1, \quad D(x)>B(x)$ $x^{17}<0, \quad x^{19}>x^{17}, \quad x^{19}+2018x^{11}+1>x^{17}+2018x^{11}+1, \quad A(x)>B(x)$

Now compare the real roots of $A(x)$ and $D(x)$ . $A(x) - D(x) = x^{19}+2018x^{11}+1 - x^{17}-2018x^{13}-1 = x^{17}(x^2-1) - 2018x^{11}(x^2-1) = x^{11}(x^2-1)(x^6-2018)$ $\because x^{11}<0, \quad 0<x^2<1, \quad 0<x^6<1, \quad \therefore A(x) - D(x)<0, \quad A(x) < D(x)$

$C(x)>D(x)>A(x)>B(x)$

As $B(x)$ has the smallest value for the same $x$ , and $A(x),B(x),C(x),D(x),$ are all monotonically increasing functions, its real root must be the greatest, thus, the answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$

~isabelchen

Solution 5 (Monotonic Function)

Denote the polynomials in the answer choices by $A(x),B(x),C(x),D(x),$ and $E(x),$ respectively.

Notice that all of the real roots for all answer choices are in $(-1,0)$ . The real root of $E(x)$ is $x=-\frac{2018}{2019}\approx-1.000.$ , therefore, we can eliminate this polynomial first.

First compare the real roots of $A(x)$ and $C(x)$ :

Let the real root of $A(x)$ be $a$ , and the real root of $C(x)$ be $b$ . $a^{19}+2018a^{11}+1 = b^{19}+2018b^{13}+1, \quad a^{19}+2018a^{11} = b^{19}+2018b^{13}$ $\because a^2 < 1, \quad a^{11} < 0, \quad \therefore a^{13}>a^{11}$ $a^{19}+2018a^{11} < a^{19}+2018a^{13}, \quad b^{19}+2018b^{13} < a^{19}+2018a^{13}$ Let $f(x) = x^{19}+2018x^{13}$ . Notice that $f(x)$ is a monotonically increasing function. As $f(b)<f(a)$ , $b<a$ , $A(x)$ has a bigger real root than $C(x)$ .

Similarly, $B(x)$ has a bigger real root than $D(x)$ .

To determine the polynomial with the greatest real root, now we only need to compare $A(x)$ and $B(x)$ .

Let the real root of $A(x)$ be $a$ , and the real root of $B(x)$ be $b$ . $a^{19}+2018a^{11}+1 = b^{17}+2018b^{11}+1, \quad a^{19}+2018a^{11} = b^{17}+2018b^{11}$ $\because a^2 < 1, \quad a^{17} < 0, \quad \therefore a^{19}>a^{17}$ $a^{17}+2018a^{11} < a^{19}+2018a^{11}, \quad a^{17}+2018a^{11} < b^{17}+2018b^{11}$ Let $f(x) = x^{17}+2018x^{11}$ . Notice that $f(x)$ is a monotonically increasing function. As $f(a)<f(b)$ , $a<b$ , $B(x)$ has a bigger real root than $A(x)$ .

Thus, the answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$

~isabelchen

Solution 6 (Proof by Contradiction)

Denote the polynomials in the answer choices by $A(x),B(x),C(x),D(x),$ and $E(x),$ respectively.

Notice that all of the real roots for all answer choices are in $(-1,0)$ . The real root of $E(x)$ is $x=-\frac{2018}{2019}\approx-1.000.$ , therefore, we can eliminate this polynomial first.

First compare the real roots of $A(x)$ and $C(x)$ :

Let the real root of $A(x)$ be $a$ , and the real root of $C(x)$ be $b$ . $a^{19}+2018a^{11}+1 = b^{19}+2018b^{13}+1$ $a^{11}(a^8+2018) = b^{11}(b^8+2018b^2)$ $\frac{ a^{11} }{ b^{11} } = \frac{ b^8+2018b^2 }{ a^8+2018 }$ As $b^2<1$ , $b^8+2018b^2<b^8+2018$ , $\frac{ a^{11} }{ b^{11} } = \frac{ b^8+2018b^2 }{ a^8+2018 } < \frac{ b^8+2018 }{ a^8+2018 }$

Assume that $-1<a<b<0$ . Thus, $-1<a^{11}<b^{11}<0, \quad 1<\frac{ a^{11} }{ b^{11} }, \quad 1< \frac{ b^8+2018 }{ a^8+2018 }$

$\because -1<a<b<0, \quad \therefore 1>a^8>b^8>0, \quad a^8 + 1 > b^8+1, \quad \therefore \frac{b^8+2018}{a^8+2018}<1$

$\text{ Contradicts with } 1<\frac{b^8+2018}{a^8+2018}$

$\therefore$ assumption $-1<a<b<0$ is false, $-1<b<a<0$ , $A(x)$ has a bigger real root than $C(x)$ .

Similarly, $B(x)$ has a bigger real root than $D(x)$ .

To determine the polynomial with the greatest real root, now we only need to compare $A(x)$ and $B(x)$ .

Let the real root of $A(x)$ be $a$ , and the real root of $B(x)$ be $b$ . $a^{19}+2018a^{11}+1 = b^{17}+2018b^{11}+1$ $a^{19} - b^{17} = 2018( b^{11} - a^{11} )$

Assume that $-1<b<a<0$ . Thus, $-1<b^{11}<a^{11}<0, \quad b^{11}-a^{11}<0, \quad 2018( b^{11} - a^{11} ) < 0, \quad a^{19} - b^{17} < 0$

$-1<b^{17}<a^{17}<0, \quad a^2<1, \quad a^{17}<a^{19}, \quad b^{17}<a^{19}, \quad a^{19} - b^{17} > 0$

$\text{ Contradicts with }a^{19} - b^{17} < 0$

$\therefore$ assumption $-1<b<a<0$ is false, $-1<a<b<0$ , $B(x)$ has a bigger real root than $A(x)$ .

Thus, the answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$

~isabelchen

Solution 7 (Calculus)

Note that $a(-1)=b(-1)=c(-1)=d(-1) < 0$ and $a(0)=b(0)=c(0)=d(0) > 0$ . Calculating the definite integral for each function in the interval $[-1,0]$ , we see that $B(x)\rvert^{0}_{-1}$ gives the most negative value. To maximize our real root, we want to maximize the area of the curve under the x-axis, which means we want our integral to be as negative as possible and thus the answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$ .

Solution 8 (Calculus)

Newton's Method is used to approximate the zero $x_{1}$ of any real valued function given an estimation for the root $x_{0}: \ x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}.$ After looking at all the options, $x_{0}=-1$ gives a reasonable estimate. For options $\textbf{(A)}$ to $\textbf{(D)},$ we have $f(-1) = -2018$ and the estimation becomes $x_{1}=-1+{\frac {2018}{f'(-1)}}.$ Thus we need to minimize the derivative, giving us $\textbf{(B)}$ . Now after comparing $\textbf{(B)}$ and $\textbf{(E)}$ through Newton's method, we see that $\textbf{(B)}$ has the higher root, so the answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$ .

~Qcumber

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2018amc12a/471

~ dolphin7

@@ Line 1: / Line 1: @@
-==Problem==
+== Problem ==
 Which of the following polynomials has the greatest real root?
 <math>\textbf{(A) }   x^{19}+2018x^{11}+1   \qquad        \textbf{(B) }   x^{17}+2018x^{11}+1   \qquad    \textbf{(C) }   x^{19}+2018x^{13}+1   \qquad   \textbf{(D) }  x^{17}+2018x^{13}+1 \qquad  \textbf{(E) }   2019x+2018 </math>
-==Solution 1==
+== Solution 1 (Intermediate Value Theorem, Inequalities, Graphs) ==
+Denote the polynomials in the answer choices by <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively.
+Note that <math>A(x),B(x),C(x),D(x),</math> and <math>E(x)</math> are strictly increasing functions with range <math>(-\infty,\infty).</math> So, each polynomial has exactly one real root. The real root of <math>E(x)</math> is <math>x=-\frac{2018}{2019}\approx-1.000.</math> On the other hand, since <math>A(-1)=B(-1)=C(-1)=D(-1)=-2018</math> and <math>A(0)=B(0)=C(0)=D(0)=1,</math> we conclude that the real root for each of <math>A(x),B(x),C(x),</math> and <math>D(x)</math> must satisfy <math>x\in(-1,0)</math> by the Intermediate Value Theorem (IVT).
+We analyze the polynomials for <math>x\in(-1,0):</math>
+<ol style="margin-left: 1.5em;">
+  <li>We have
+<cmath>\begin{align*}
+B(x)-A(x)=D(x)-C(x)&=x^{17}-x^{19} \\
+&=x^{17}\left(1-x^2\right) \\
+&<0.
+\end{align*}</cmath>
+As the graph of <math>y=A(x)</math> is always above the graph of <math>y=B(x)</math> in this interval, we deduce that <math>B(x)</math> has a greater real root than <math>A(x)</math> does. By the same reasoning, <math>D(x)</math> has a greater real root than <math>C(x)</math> does.</li><p>
+  <li>We have
+<cmath>\begin{align*}
+B(x)-D(x)&=2018x^{11}-2018x^{13} \\
+&=2018x^{11}\left(1-x^2\right) \\
+&<0,
+\end{align*}</cmath>
+from which <math>B(x)</math> has a greater real root than <math>D(x)</math> does.</li><p>
+</ol>
+Now, we are left with comparing the real roots of <math>B(x)</math> and <math>E(x).</math> Since <math>B\left(-\frac{1}{\sqrt2}\right)<0<B\left(-\frac{1}{2}\right),</math> it follows that the real root of <math>B(x)</math> must satisfy <math>x\in\left(-\frac{1}{\sqrt2},-\frac{1}{2}\right)</math> by the IVT. Clearly, <math>\boxed{\textbf{(B) }   x^{17}+2018x^{11}+1}</math> has the greatest real root.
+~MRENTHUSIASM
+== Solution 2 (Similar to Solution 1) ==
+We can see that our real solution has to lie in the open interval <math>(-1,0)</math>. From there, note that <math>x^a < x^b</math> if <math>a</math>, <math>b</math> are odd positive integers if <math>a<b</math>, so hence it can only either be <math>\textbf{(B)}</math> or <math>\textbf{(E)}</math> (as all of the other polynomials will be larger than the polynomial <math>\textbf{(B)}</math>). Observe that <math>\textbf{(E)}</math> gives the solution <math>x=-\frac{2018}{2019}</math>. We can approximate the root for <math>\textbf{(B)}</math> by using <math>x=-\frac 12</math>: <cmath>\left(-\frac{1}{2}\right)^{17} - \frac{2018}{2048} + 1 \approx 0.</cmath> Therefore, the root for <math>\textbf{(B)}</math> is approximately <math>-\frac 12</math>. The answer is <math>\boxed{\textbf{(B) }   x^{17}+2018x^{11}+1}</math>.
+~cpma213
+== Solution 3 (Similar to Solution 1) ==
+Let the real solution to <math>B</math> be <math>a.</math> It is easy to see that when <math>a</math> is plugged in to <math>A,</math> since <math>-1 < a < 0,</math> it follows that <math>a^{19} < a^{17}</math> thus making the real solution to <math>A</math> more "negative", or smaller than <math>B.</math> Similarly we can assert that <math>D > C.</math> Now to compare <math>B</math> and <math>D,</math> we can use the same method to what we used before to compare <math>B</math> to <math>A,</math> in which it is easy to see that the smaller exponent <math>(11)</math> "wins". Now, the only thing left is for us to compare <math>B</math> and <math>E.</math> Plugging <math>\frac{-2018}{2019}</math> (or the solution to <math>E</math>) into <math>B</math> we obtain <math>\frac{(-2018)^{17}}{2019^{17}} + 2018\cdot\frac{(-2018)^{11}}{2019^{11}} + 1,</math> which is intuitively close to <math>-1 - 2018 + 1 = -2018, </math> much smaller than the solution the required <math>0.</math> (For a more rigorous proof, one can note that <math>\left(\frac{2018}{2019}\right)^{17}</math> and <math>\left(\frac{2018}{2019}\right)^{11}</math> are both much greater than <math>\left(\frac{2018}{2019}\right)^{2019} \approx \frac{1}{e},</math> by the limit definition of <math>e.</math> Since <math>- \frac{1}{e} - 2018 \cdot \frac{1}{e} + 1</math> is still much smaller the required <math>0</math> for the solution to <math>B</math> to be a solution, our answer is <math>\boxed{\textbf{(B) }   x^{17}+2018x^{11}+1}.</math>
+~fidgetboss_4000
+==Solution 4==
+Denote the polynomials in the answer choices by <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively.
+The real root of <math>E(x)</math> is <math>x=-\frac{2018}{2019}\approx-1.000.</math>, therefore, we can eliminate this polynomial first. Notice that all of the real roots for all answer choices are in <math>(-1,0)</math>. Therefore,
+<cmath>x^{11}<0, \quad x^2<1, \quad x^{13}>x^{11}, \quad x^{19}+2018x^{13}+1>x^{19}+2018x^{11}+1, \quad C(x)>A(x)</cmath>
+<cmath>\quad x^{17}+2018x^{13}+1>x^{17}+2018x^{11}+1, \quad D(x)>B(x)</cmath>
+<cmath>x^{17}<0, \quad x^{19}>x^{17}, \quad x^{19}+2018x^{11}+1>x^{17}+2018x^{11}+1, \quad A(x)>B(x)</cmath>
+Now compare the real roots of <math>A(x)</math> and <math>D(x)</math>.
+<cmath>A(x) - D(x) = x^{19}+2018x^{11}+1 - x^{17}-2018x^{13}-1 = x^{17}(x^2-1) - 2018x^{11}(x^2-1) = x^{11}(x^2-1)(x^6-2018)</cmath>
+<cmath>\because x^{11}<0, \quad 0<x^2<1, \quad 0<x^6<1, \quad \therefore A(x) - D(x)<0, \quad A(x) < D(x)</cmath>
+<cmath>C(x)>D(x)>A(x)>B(x)</cmath>
+As <math>B(x)</math> has the smallest value for the same <math>x</math>, and <math>A(x),B(x),C(x),D(x),</math> are all monotonically increasing functions, its real root must be the greatest, thus, the answer is <math>\boxed{\textbf{(B) }   x^{17}+2018x^{11}+1}</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
+==Solution 5 (Monotonic Function)==
+Denote the polynomials in the answer choices by <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively.
+Notice that all of the real roots for all answer choices are in <math>(-1,0)</math>. The real root of <math>E(x)</math> is <math>x=-\frac{2018}{2019}\approx-1.000.</math>, therefore, we can eliminate this polynomial first.
+First compare the real roots of <math>A(x)</math> and <math>C(x)</math>:
+Let the real root of <math>A(x)</math> be <math>a</math>, and the real root of <math>C(x)</math> be <math>b</math>.
+<cmath>a^{19}+2018a^{11}+1 = b^{19}+2018b^{13}+1, \quad a^{19}+2018a^{11} = b^{19}+2018b^{13}</cmath>
+<cmath>\because a^2 < 1, \quad a^{11} < 0, \quad \therefore a^{13}>a^{11}</cmath>
+<cmath>a^{19}+2018a^{11} < a^{19}+2018a^{13}, \quad b^{19}+2018b^{13} < a^{19}+2018a^{13}</cmath>
+Let <math>f(x) = x^{19}+2018x^{13}</math>. Notice that <math>f(x)</math> is a monotonically increasing function. As <math>f(b)<f(a)</math>, <math>b<a</math>, <math>A(x)</math> has a bigger real root than <math>C(x)</math>.
+Similarly, <math>B(x)</math> has a bigger real root than <math>D(x)</math>.
+To determine the polynomial with the greatest real root, now we only need to compare <math>A(x)</math> and <math>B(x)</math>.
+Let the real root of <math>A(x)</math> be <math>a</math>, and the real root of <math>B(x)</math> be <math>b</math>.
+<cmath>a^{19}+2018a^{11}+1 = b^{17}+2018b^{11}+1, \quad a^{19}+2018a^{11} = b^{17}+2018b^{11}</cmath>
+<cmath>\because a^2 < 1, \quad a^{17} < 0, \quad \therefore a^{19}>a^{17}</cmath>
+<cmath>a^{17}+2018a^{11} < a^{19}+2018a^{11}, \quad a^{17}+2018a^{11} < b^{17}+2018b^{11}</cmath>
+Let <math>f(x) = x^{17}+2018x^{11}</math>. Notice that <math>f(x)</math> is a monotonically increasing function. As <math>f(a)<f(b)</math>, <math>a<b</math>, <math>B(x)</math> has a bigger real root than <math>A(x)</math>.
+Thus, the answer is <math>\boxed{\textbf{(B) }  x^{17}+2018x^{11}+1}</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
+==Solution 6 (Proof by Contradiction)==
+Denote the polynomials in the answer choices by <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively.
+Notice that all of the real roots for all answer choices are in <math>(-1,0)</math>. The real root of <math>E(x)</math> is <math>x=-\frac{2018}{2019}\approx-1.000.</math>, therefore, we can eliminate this polynomial first.
+First compare the real roots of <math>A(x)</math> and <math>C(x)</math>:
+Let the real root of <math>A(x)</math> be <math>a</math>, and the real root of <math>C(x)</math> be <math>b</math>.
+<cmath>a^{19}+2018a^{11}+1 = b^{19}+2018b^{13}+1</cmath>
+<cmath>a^{11}(a^8+2018) = b^{11}(b^8+2018b^2)</cmath>
+<cmath>\frac{ a^{11} }{ b^{11} } = \frac{ b^8+2018b^2 }{ a^8+2018 }</cmath>
+As <math>b^2<1</math>, <math>b^8+2018b^2<b^8+2018</math>,
+<cmath>\frac{ a^{11} }{ b^{11} } = \frac{ b^8+2018b^2 }{ a^8+2018 } < \frac{ b^8+2018 }{ a^8+2018 }</cmath>
+Assume that <math>-1<a<b<0</math>. Thus,
+<cmath>-1<a^{11}<b^{11}<0, \quad 1<\frac{ a^{11} }{ b^{11} }, \quad 1< \frac{ b^8+2018 }{ a^8+2018 }</cmath>
+<math>\because -1<a<b<0, \quad \therefore 1>a^8>b^8>0, \quad a^8 + 1 > b^8+1, \quad \therefore \frac{b^8+2018}{a^8+2018}<1</math>
+<cmath> \text{ Contradicts with } 1<\frac{b^8+2018}{a^8+2018}</cmath>
+<math>\therefore</math> assumption <math>-1<a<b<0</math> is false, <math>-1<b<a<0</math>, <math>A(x)</math> has a bigger real root than <math>C(x)</math>.
+Similarly, <math>B(x)</math> has a bigger real root than <math>D(x)</math>.
+To determine the polynomial with the greatest real root, now we only need to compare <math>A(x)</math> and <math>B(x)</math>.
+Let the real root of <math>A(x)</math> be <math>a</math>, and the real root of <math>B(x)</math> be <math>b</math>.
+<cmath>a^{19}+2018a^{11}+1 = b^{17}+2018b^{11}+1</cmath>
+<cmath>a^{19} - b^{17} = 2018( b^{11} - a^{11} )</cmath>
+Assume that <math>-1<b<a<0</math>. Thus,
+<cmath>-1<b^{11}<a^{11}<0, \quad b^{11}-a^{11}<0, \quad 2018( b^{11} - a^{11} ) < 0, \quad a^{19} - b^{17} < 0</cmath>
+<cmath>-1<b^{17}<a^{17}<0, \quad a^2<1, \quad a^{17}<a^{19}, \quad b^{17}<a^{19}, \quad a^{19} - b^{17} > 0</cmath>
+<cmath> \text{ Contradicts with }a^{19} - b^{17} < 0</cmath>
+<math>\therefore</math> assumption <math>-1<b<a<0</math> is false, <math>-1<a<b<0</math>, <math>B(x)</math> has a bigger real root than <math>A(x)</math>.
+Thus, the answer is <math>\boxed{\textbf{(B) }   x^{17}+2018x^{11}+1}</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
+== Solution 7 (Calculus) ==
+Note that <math>a(-1)=b(-1)=c(-1)=d(-1) < 0</math> and <math>a(0)=b(0)=c(0)=d(0) > 0</math>. Calculating the definite integral for each function in the interval <math>[-1,0]</math>, we see that <math>B(x)\rvert^{0}_{-1}</math> gives the most negative value. To maximize our real root, we want to maximize the area of the curve under the x-axis, which means we want our integral to be as negative as possible and thus the answer is <math>\boxed{\textbf{(B) }   x^{17}+2018x^{11}+1}</math>.
-We can see that our real solution has to lie in the open interval <math>(-1,0)</math>. From there, note that <math>x^a < x^b</math> if <math>a</math>, <math>b</math> are odd positive integers if <math>a<b</math>, so hence it can only either be B or E(as all of the other polynomials will be larger than the polynomial B). E gives the solution <math>x=-\frac{2018}{2019}</math>. We can approximate the root for B by using <math>x=-\frac 12</math>. <cmath> (- \frac {1}{2})  ^{17} - \frac{2018}{2048} + 1 \approx 0</cmath> therefore the root for B is approximately <math>-\frac 12</math>. The answer is <math>\fbox{B}</math>. (cpma213)
+== Solution 8 (Calculus) ==
+Newton's Method is used to approximate the zero <math>x_{1}</math> of any real valued function given an estimation for the root <math>x_{0}: \ x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}.</math> After looking at all the options, <math>x_{0}=-1</math> gives a reasonable estimate. For options <math>\textbf{(A)}</math> to <math>\textbf{(D)},</math> we have <math>f(-1) = -2018</math> and the estimation becomes <math>x_{1}=-1+{\frac {2018}{f'(-1)}}.</math> Thus we need to minimize the derivative, giving us <math>\textbf{(B)}</math>. Now after comparing <math>\textbf{(B)}</math> and <math>\textbf{(E)}</math> through Newton's method, we see that <math>\textbf{(B)}</math> has the higher root, so the answer is <math>\boxed{\textbf{(B) }   x^{17}+2018x^{11}+1}</math>.
-==Solution 2 (Calculus version of solution 1)==
+~Qcumber
-Note that <math>a(-1)=b(-1)=c(-1)=d(-1) < 0</math> and <math>a(0)=b(0)=c(0)=d(0) > 0</math>. Calculating the definite integral for each function on the interval <math>[-1,0]</math>, we see that <math>B(x)\rvert^{0}_{-1}</math> gives the most negative value. To maximize our real root, we want to maximize the area of the curve under the x-axis, which means we want our integral to be as negative as possible and thus the answer is <math>\fbox{B}</math>.
+== Video Solution by Richard Rusczyk ==
+https://artofproblemsolving.com/videos/amc/2018amc12a/471
-==Solution 3 (Alternate Calculus Version)==
+~ dolphin7
-Newton's Method is used to approximate the zero <math>x_{1}</math> of any real valued function given an estimation for the root <math>x_{0}</math>: <math>x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}\,.</math> After looking at all the options, <math>x_{0}=-1</math> gives a reasonable estimate. For options A to D, <math>f(-1) = -2018</math> and the estimation becomes <math>x_{1}=-1+{\frac {2018}{f'(-1)}}\,.</math> Thus we need to minimize the derivative, giving us B. Now after comparing B and E through Newton's method, we see that B has the higher root, so the answer is <math>\fbox{B}</math>. (Qcumber)
 ==See Also==
 {{AMC12 box|year=2018|ab=A|num-b=20|num-a=22}}
 {{MAA Notice}}

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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

Page

Toolbox

Search

Difference between revisions of "2018 AMC 12A Problems/Problem 21"

Latest revision as of 05:58, 29 July 2024

Contents

Problem

Solution 1 (Intermediate Value Theorem, Inequalities, Graphs)

Solution 2 (Similar to Solution 1)

Solution 3 (Similar to Solution 1)

Solution 4

Solution 5 (Monotonic Function)

Solution 6 (Proof by Contradiction)

Solution 7 (Calculus)

Solution 8 (Calculus)

Video Solution by Richard Rusczyk

See Also