Let alpha and beta be the roots of ax^2 + bx | vedclass

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(C) Given the quadratic equation $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.We need to evaluate $L = \lim_{x 	o \alpha} \frac{1 - \cos(ax^2 + bx + c)

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the quadratic equation $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.We need to evaluate $L = \lim_{x 	o \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}$.Using the identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$,we get:$L = \lim_{x 	o \alpha} \frac{2 \sin^2(\frac{a(x - \alpha)(x - \beta)}{2})}{(x - \alpha)^2}$.Multiply and divide by $(\frac{a(x - \beta)}{2})^2$:$L = 2 \lim_{x 	o \alpha} \left[ \frac{\sin(\frac{a(x - \alpha)(x - \beta)}{2})}{\frac{a(x - \alpha)(x - \beta)}{2}} ight]^2 \cdot \frac{a^2(x - \beta)^2}{4}$.Since $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,as $x 	o \alpha$,the term in the bracket approaches $1$.$L = 2 \cdot (1)^2 \cdot \frac{a^2(\alpha - \beta)^2}{4} = \frac{a^2}{2}(\alpha - \beta)^2$.

Let $\alpha$ and $\beta$ be the roots of $ax^2 + bx + c = 0$,then $\lim_{x \to \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}$ is equal to

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