Let α and β be the roots of ax2+bx+c=0, then limx→α1−cos(ax2+bx+c)(x−α)2=

ax2+bx+c=a(x−α)(x−β)Given limit, limx→α 2sin2(a(x−α)(x−β)2)(x−α)2=2limx→α(sina(x−α)(x−β)2a(x−α)(x−β)2)2×(x−β2)2×a2=2(1)2×(α−β2)2×a2=2(1)2×(α−β2)2×a2=2a2(α−β)24.

Let α and β be the roots of ax2+bx+c=0, then limx→α1−cos(ax2+bx+c)(x−α)2=

ax2+bx+c=a(x−α)(x−β)Given limit, limx→α 2sin2(a(x−α)(x−β)2)(x−α)2=2limx→α(sina(x−α)(x−β)2a(x−α)(x−β)2)2×(x−β2)2×a2=2(1)2×(α−β2)2×a2=2(1)2×(α−β2)2×a2=2a2(α−β)24.

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

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$\frac{a^{2} {(α - β)}^{2}}{2}$

$\frac{a^{2}}{{(α - β)}^{2}}$

$\frac{- a^{2}}{2 {(α - β)}^{2}}$

$\frac{a^{2}}{2 {(α - β)}^{2}}$

answer is A.

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Detailed Solution

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

Given limit, $\lim_{x \to α} \frac{2 \sin^{2} (\frac{a (x - α) (x - β)}{2})}{{(x - α)}^{2}} = \underset{x \to α}{2 \lim} {(\frac{\sin \frac{a (x - α) (x - β)}{2}}{\frac{a (x - α) (x - β)}{2}})}^{2} \times {(\frac{x - β}{2})}^{2} \times a^{2}$