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(D) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,we can write $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.We need t

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$\frac{a^2}{2}(\alpha - \beta)^2$

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(D) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,we can write $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.We need to evaluate the limit $L = \lim_{x ightarrow \beta} \frac{1 - \cos(a(x - \alpha)(x - \beta))}{(x - \beta)^2}$.Using the identity $1 - \cos(	heta) = 2 \sin^2(\frac{	heta}{2})$,we get:$L = \lim_{x ightarrow \beta} \frac{2 \sin^2(\frac{a(x - \alpha)(x - \beta)}{2})}{(x - \beta)^2}$.Multiplying and dividing by $(\frac{a(x - \alpha)}{2})^2$,we have:$L = \lim_{x}$ ${ightarrow \beta} 2 \left[ \frac{\sin(\frac{a(x - \alpha)(x - \beta)}{2})}{\frac{a(x - \alpha)(x - \beta)}{2}} ight]^2 	imes \frac{a^2(x - \alpha)^2}{4}$.As $x ightarrow \beta$,the term in the bracket approaches $1$.Thus,$L = 2 	imes 1^2 	imes \frac{a^2(\beta - \alpha)^2}{4} = \frac{a^2(\alpha - \beta)^2}{2}$.

If $\alpha, \beta$ are the roots of the equation $ax^2 + bx + c = 0$,then $\lim_{x \rightarrow \beta} \frac{1 - \cos(ax^2 + bx + c)}{(x - \beta)^2}$ is

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