Let α. and β be the roots of ax2+bx+c=0, then limx→a 1−cos⁡ax2+bx+c(x−α)2 is equal to

A
0
B
$\frac{1}{2} (α - β)^{2}$
C
$\frac{a^{2}}{2} (α - β)^{2}$
D
$- \frac{a^{2}}{2} (α - β)^{2}$

Solution:

It is given that $α_{,} β$ are roots of $a x^{2} + b x + c$

$∴ a x^{2} + b x + c = a (x - α) (x - β)$

Now, $lim_{x \to α} \frac{1 - \cos (a x^{2} + b x + c)}{(x - α)^{2}}$

$\begin{array}{l} = 2 lim_{x \to α} \frac{\sin^{2} \{\frac{(a x^{2} + b x + c)}{2}\}}{(x - α)^{2}} \\ = 2 lim_{x \to α} \frac{\sin^{2} \{\frac{a (x - α) (x - β)}{2}\}}{(x - α)^{2}}\} \\ = 2 lim_{x \to α} [\frac{{\sin \{\frac{a (x - α) (x - β)}{2}\}]}^{2}}{\frac{a (x - α) (x - β)}{2}}] \times \frac{a^{2}}{4} (x - β)^{2} \\ = 2 (1)^{2} \times \frac{a^{2}}{4} (α - β)^{2} = \frac{a^{2}}{2} (α - β)^{2} \end{array}$

Let $α$ . and $β$ be the roots of $a x^{2} + b x + c = 0$ , then $lim_{x \to a} \frac{1 - \cos (a x^{2} + b x + c)}{(x - α)^{2}}$ is equal to

Solution:

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