If a and b are roots of the equation px^2 + q | vedclass

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(D) Let $f(x) = px^2 + qx + r$. Since $a$ and $b$ are roots,$f(x) = p(x - a)(x - b)$.We need to evaluate $L = \lim_{x \rightarrow b} \frac{1 - \cos 2(

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$a^2 - 2ab + b^2$

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(D) Let $f(x) = px^2 + qx + r$. Since $a$ and $b$ are roots,$f(x) = p(x - a)(x - b)$.We need to evaluate $L = \lim_{x ightarrow b} \frac{1 - \cos 2(f(x))}{2(px - pb)^2}$.Using the identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$,we have $1 - \cos 2(f(x)) = 2 \sin^2(f(x))$.So,$L = \lim_{x ightarrow b} \frac{2 \sin^2(f(x))}{2 p^2(x - b)^2} = \frac{1}{p^2} \lim_{x ightarrow b} \left( \frac{\sin(f(x))}{x - b} ight)^2$.Since $f(x) = p(x - a)(x - b)$,we have $\frac{f(x)}{x - b} = p(x - a)$.As $x ightarrow b$,$p(x - a) ightarrow p(b - a)$.Thus,$L = \frac{1}{p^2} \lim_{x ightarrow b} \left( \frac{\sin(p(x - a)(x - b))}{p(x - a)(x - b)} \cdot p(x - a) ight)^2$.Since $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$,we get $L = \frac{1}{p^2} \cdot (p(b - a))^2 = \frac{p^2(b - a)^2}{p^2} = (b - a)^2$.Note: The provided options seem to have a typo. $(b - a)^2$ is equivalent to $a^2 - 2ab + b^2$,which is option $D$.

If $a$ and $b$ are roots of the equation $px^2 + qx + r = 0$,then $\lim_{x \rightarrow b} \frac{1 - \cos 2(px^2 + qx + r)}{2(px - pb)^2}$ is equal to

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