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If $\alpha$ is a repeated root of $ax^2+bx+c=0$, then $\lim\limits_{x\to \alpha}\large\frac{\sin(ax^2+bx+c)}{(x-\alpha)^2}$ =

$(a)\;0\qquad(b)\;a\qquad(c)\;b\qquad(d)\;c$

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Since $\alpha$ is a repeated root
$a\alpha^2+b\alpha+c=0$
$2a\alpha+b=0$
$\lim\limits_{x\to \alpha}\large\frac{(2ax+b) \cos (ax^2+bx+c)}{2(x-\alpha)}\qquad\big(\large\frac{0}{0}\big)$
$\lim\limits_{x\to \alpha}\large\frac{2a\cos(ax^2+bx+c)-(2ax+b)^2\sin(ax^2+bx+c)}{2}$
$\Rightarrow \large\frac{2a}{2}$$=a$
Hence (b) is the correct option.

 

answered Jan 6, 2014 by sreemathi.v
edited Mar 17, 2014 by balaji.thirumalai
 

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