Browse Questions

# 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

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

$\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)$