If alpha beta 0 are the roots of the equa | vedclass

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(B) Given $ax^2 + bx + 1 = a(x - \alpha)(x - \beta)$,so $\alpha\beta = \frac{1}{a}$.Also,$x^2 + bx + a = a(1 - \alpha x)(1 - \beta x)$.Let $L = \lim_{

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$2\alpha$

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(B) Given $ax^2 + bx + 1 = a(x - \alpha)(x - \beta)$,so $\alpha\beta = \frac{1}{a}$.Also,$x^2 + bx + a = a(1 - \alpha x)(1 - \beta x)$.Let $L = \lim_{x}$ ${ightarrow \frac{1}{\alpha}} \left( \frac{1 - \cos(a(1 - \alpha x)(1 - \beta x))}{2(1 - \alpha x)^2} ight)^{\frac{1}{2}}$.Using $\lim_{	heta ightarrow 0} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$,we have:$L = \lim_{x}$ ${ightarrow \frac{1}{\alpha}} \left( \frac{1 - \cos(a(1 - \alpha x)(1 - \beta x))}{2(a(1 - \alpha x)(1 - \beta x))^2} \cdot a^2(1 - \beta x)^2 ight)^{\frac{1}{2}}$.$L = \left( \frac{1}{2} \cdot a^2 (1 - \frac{\beta}{\alpha})^2 ight)^{\frac{1}{2}} = \frac{a}{\sqrt{2}} \cdot \frac{1}{\alpha} (\alpha - \beta) = \frac{1}{\alpha\beta} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\alpha - \beta}{\alpha} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\alpha} (\frac{1}{\beta} - \frac{1}{\alpha})$.Wait,re-evaluating the limit: $L = \left( \frac{1}{2} \cdot \frac{a^2(1 - \beta x)^2}{1} ight)^{\frac{1}{2}} = \frac{a(1 - \beta/\alpha)}{\sqrt{2}} = \frac{1}{\alpha\beta} \cdot \frac{\alpha - \beta}{\alpha} \cdot \frac{1}{\sqrt{2}}$.Actually,the standard limit gives $L = \frac{a(1 - \beta/\alpha)}{2} = \frac{1}{2\alpha\beta} \cdot \frac{\alpha - \beta}{\alpha} = \frac{1}{2\alpha} (\frac{1}{\beta} - \frac{1}{\alpha})$.Comparing with $\frac{1}{k} (\frac{1}{\beta} - \frac{1}{\alpha})$,we get $k = 2\alpha$.

If $\alpha > \beta > 0$ are the roots of the equation $ax^2 + bx + 1 = 0$,and $\lim_{x}$ ${\rightarrow \frac{1}{\alpha}} \left( \frac{1 - \cos(x^2 + bx + a)}{2(1 - \alpha x)^2} \right)^{\frac{1}{2}} = \frac{1}{k} \left( \frac{1}{\beta} - \frac{1}{\alpha} \right)$,then $k$ is equal to

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