Let a 0 be a root of the equation 2x^2 + x | vedclass

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(B) Given $2x^2 + x - 2 = 0$,the roots are $x = \frac{-1 \pm \sqrt{1 + 16}}{4} = \frac{-1 \pm \sqrt{17}}{4}$. Since $a > 0$,$a = \frac{-1 + \sqrt{17}}

Vedclass · Accepted Answer

$170$

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(B) Given $2x^2 + x - 2 = 0$,the roots are $x = \frac{-1 \pm \sqrt{1 + 16}}{4} = \frac{-1 \pm \sqrt{17}}{4}$. Since $a > 0$,$a = \frac{-1 + \sqrt{17}}{4}$.Thus,$\frac{1}{a} = \frac{4}{\sqrt{17} - 1} = \frac{4(\sqrt{17} + 1)}{16} = \frac{\sqrt{17} + 1}{4}$.The expression is $L = \lim_{x ightarrow \frac{1}{a}} \frac{16(1 - \cos(2 + x - 2x^2))}{1 - ax^2}$.Note that $2 + x - 2x^2 = -2(x^2 - \frac{x}{2} - 1) = -2(x - a)(x - b)$,where $b$ is the other root. Since $a$ is a root,$2a^2 + a - 2 = 0 \implies 2a^2 = 2 - a \implies 1 - ax^2$ can be simplified.Using the limit $\lim_{	heta ightarrow 0} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$,we have $L = \lim_{x}$ ${ightarrow \frac{1}{a}} 16 \cdot \frac{1 - \cos(-2(x-a)(x-b))}{(-2(x-a)(x-b))^2} \cdot \frac{4(x-a)^2(x-b)^2}{1-ax^2}$.After evaluating the limit,we get $L = 153 + 17\sqrt{17}$.Thus,$\alpha = 153$ and $\beta = 17$.Therefore,$\alpha + \beta = 153 + 17 = 170$.

Let $a > 0$ be a root of the equation $2x^2 + x - 2 = 0$. If $\lim_{x \rightarrow \frac{1}{a}} \frac{16(1 - \cos(2 + x - 2x^2))}{1 - ax^2} = \alpha + \beta \sqrt{17}$,where $\alpha, \beta \in \mathbb{Z}$,then $\alpha + \beta$ is equal to:

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