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If $\alpha,\beta$ are 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

$\begin{array}{1 1}(a)\;(a-b)^2&(b)\;\large\frac{(\alpha-\beta)^2}{2}\\(c)\;\big(\large\frac{\alpha(\alpha-\beta)}{2}\big)^2&(d)\;\text{None of these}\end{array}$

1 Answer

$\lim\limits_{x\to \alpha}\large\frac{1-\cos(ax^2+bx+c)}{(x-\alpha)^2}$$\qquad\big(\large\frac{0}{0}\big)$
Using L Hospital rule
$\lim\limits_{x\to \alpha}\large\frac{\sin(ax^2+bx+c)(2ax+b)}{2(x-\alpha)}$
$\Rightarrow \lim\limits_{x\to \alpha}\large\frac{\sin(a(x-\alpha)(x-\beta))2ax+b}{2.a(x-\alpha)(x-\beta)}$$.a(x-\beta)$
$\Rightarrow \large\frac{a^2}{2}$$(2\alpha+\large\frac{b}{a})$$(\alpha-\beta)$
$\Rightarrow \large\frac{a^2}{2}$$(2\alpha-\alpha-\beta)(\alpha-\beta)$
$\Rightarrow \large\frac{a^2(\alpha-\beta)^2}{2}$
Hence (d) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

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