If alpha, beta are the roots of the quadratic | vedclass

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(C) Since $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we can write $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.We need to evaluate $L = \lim_

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

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(C) Since $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we can write $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.We need to evaluate $L = \lim_{x 	o \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}$.Using the identity $1 - \cos(	heta) = 2\sin^2(\frac{	heta}{2})$,we get:$L = \lim_{x 	o \alpha} \frac{2\sin^2(\frac{a(x - \alpha)(x - \beta)}{2})}{(x - \alpha)^2}$.Multiplying and dividing by $(\frac{a(x - \beta)}{2})^2$,we have:$L = \lim_{x 	o \alpha} 2 \cdot \left[ \frac{\sin(\frac{a(x - \alpha)(x - \beta)}{2})}{\frac{a(x - \alpha)(x - \beta)}{2}} ight]^2 \cdot \frac{a^2(x - \alpha)^2(x - \beta)^2}{4(x - \alpha)^2}$.Since $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,the limit becomes:$L = 2 \cdot (1)^2 \cdot \frac{a^2(\alpha - \beta)^2}{4} = \frac{a^2}{2}(\alpha - \beta)^2$.

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

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