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let $\alpha$ and $\beta$ be the distinct roots of $ax^2+bx+c=0$ then $\lim\limits_{x\to \alpha}\large\frac{1-\cos(ax^2+bx+c)}{(x-\alpha)^2}$ is equal to

$\begin{array}{1 1}(a)\;-\large\frac{a^2}{2}\normalsize (\alpha-\beta)^2&(b)\;\large\frac{1}{2}\normalsize(\alpha-\beta)^2\\(c)\;\large\frac{a^2}{2}\normalsize(\alpha-\beta)^2&(d)\;0\end{array}$

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Since $\alpha,\beta$ are the roots of $ax^2+bx+c=0$
$\lim\limits_{x\to \alpha}\large\frac{1-\cos (x-\alpha)(x-\beta)}{(x-\alpha)^2}$
$\Rightarrow \lim\limits_{\large x=\alpha}\large\frac{2\sin^2\Large\frac{(x-\alpha)(x-\beta)}{2}}{\Large\frac{a^2(x-\alpha)^2(x-\beta)^2}{4}}\times \large\frac{a^2(x-\alpha)^2(x-\beta)^2}{4(x-\alpha)^2}$
$\Rightarrow \large\frac{2\times a^2(\alpha-\beta)^2}{4}$
$\Rightarrow \large\frac{ a^2(\alpha-\beta)^2}{2}$
Hence (c) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
edited Mar 7, 2016 by sharmaaparna1

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