If alpha and beta are the roots of the equati | vedclass

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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

Vedclass · Accepted Answer

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

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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 $L = \lim_{x ightarrow \alpha} \frac{1-\cos(a(x-\alpha)(x-\beta))}{(x-\alpha)^2}$.Using the identity $1-\cos(	heta) = 2\sin^2(\frac{	heta}{2})$,we get:$L = \lim_{x ightarrow \alpha} \frac{2\sin^2(\frac{a(x-\alpha)(x-\beta)}{2})}{(x-\alpha)^2}$.Multiply and divide by $(\frac{a(x-\beta)}{2})^2$:$L = \lim_{x}$ ${ightarrow \alpha} 2 \left[ \frac{\sin(\frac{a(x-\alpha)(x-\beta)}{2})}{\frac{a(x-\alpha)(x-\beta)}{2}} ight]^2 \cdot \frac{a^2(x-\alpha)^2(x-\beta)^2}{4(x-\alpha)^2}$.Since $\lim_{	heta ightarrow 0} \frac{\sin(	heta)}{	heta} = 1$,the expression simplifies to:$L = 2 \cdot 1^2 \cdot \frac{a^2(\alpha-\beta)^2}{4} = \frac{a^2(\alpha-\beta)^2}{2}$.

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

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