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Difference between revisions of "1981 AHSME Problems/Problem 28"

(Created page with "==Problem 28== Consider the set of all equations <math> x^3 + a_2x^2 + a_1x + a_0 = 0</math>, where <math> a_2</math>, <math> a_1</math>, <math> a_0</math> are real constants...")
 
(solution added)
 
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<math> \textbf{(A)}\ 1 < r < \dfrac{3}{2}\qquad \textbf{(B)}\ \dfrac{3}{2} < r < 2\qquad \textbf{(C)}\ 2 < r < \dfrac{5}{2}\qquad \textbf{(D)}\ \dfrac{5}{2} < r < 3\qquad \\ \textbf{(E)}\ 3 < r < \dfrac{7}{2}</math>
 
<math> \textbf{(A)}\ 1 < r < \dfrac{3}{2}\qquad \textbf{(B)}\ \dfrac{3}{2} < r < 2\qquad \textbf{(C)}\ 2 < r < \dfrac{5}{2}\qquad \textbf{(D)}\ \dfrac{5}{2} < r < 3\qquad \\ \textbf{(E)}\ 3 < r < \dfrac{7}{2}</math>
  
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[[1981 AHSME Problems/Problem 28|Solution]]
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==Solution==
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Since <math>x^3 = -(a_2x^2 + a_1x + a_0)</math> and <math>x</math> will be as big as possible, we need <math>x^3</math> to be as big as possible, which means <math>a_2x^2 + a_1x + a_0</math> is as small as possible. Since <math>x</math> is positive (according to the options), it makes sense for all of the coefficients to be <math>-2</math>.
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Evaluating <math>f(\frac{5}{2})</math> gives a negative number, <math>f(3)</math> 1, and <math>f(\frac{7}{2})</math> a number greater than 1, so the answer is <math>\boxed{D}</math>

Latest revision as of 12:22, 5 September 2021

Problem 28

Consider the set of all equations $x^3 + a_2x^2 + a_1x + a_0 = 0$, where $a_2$, $a_1$, $a_0$ are real constants and $|a_i| < 2$ for $i = 0,1,2$. Let $r$ be the largest positive real number which satisfies at least one of these equations. Then

$\textbf{(A)}\ 1 < r < \dfrac{3}{2}\qquad \textbf{(B)}\ \dfrac{3}{2} < r < 2\qquad \textbf{(C)}\ 2 < r < \dfrac{5}{2}\qquad \textbf{(D)}\ \dfrac{5}{2} < r < 3\qquad \\ \textbf{(E)}\ 3 < r < \dfrac{7}{2}$

Solution

Since $x^3 = -(a_2x^2 + a_1x + a_0)$ and $x$ will be as big as possible, we need $x^3$ to be as big as possible, which means $a_2x^2 + a_1x + a_0$ is as small as possible. Since $x$ is positive (according to the options), it makes sense for all of the coefficients to be $-2$.

Evaluating $f(\frac{5}{2})$ gives a negative number, $f(3)$ 1, and $f(\frac{7}{2})$ a number greater than 1, so the answer is $\boxed{D}$

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