Let α and β be the distinct roots of ax2+bx+c=0 then the value of limx→α 1−cos⁡ax2+bx+c(x−α)2 is

limx→α 1−cos⁡ax2+bx+c(x−α)2as α and β are two distinct roots.∴ ax2+bx+c=a(x−α)(x−β) i.e. α,β=−b±b2−4ac2a =limx→α 1−cos⁡[(x−α)(x−β)a](x−u)2=limx→α 2sin2⁡[(x−α)(x−β)a]2(x−α)2=limx→α 2sin2⁡[(x−α)(x−β)a]2(x−α)(x−β)a22a2x−β42=limx→α 24a2(x−β)2=a22(α−β)2

Let α and β be the distinct roots of ax2+bx+c=0 then the value of limx→α 1−cos⁡ax2+bx+c(x−α)2 is

limx→α 1−cos⁡ax2+bx+c(x−α)2as α and β are two distinct roots.∴ ax2+bx+c=a(x−α)(x−β) i.e. α,β=−b±b2−4ac2a =limx→α 1−cos⁡[(x−α)(x−β)a](x−u)2=limx→α 2sin2⁡[(x−α)(x−β)a]2(x−α)2=limx→α 2sin2⁡[(x−α)(x−β)a]2(x−α)(x−β)a22a2x−β42=limx→α 24a2(x−β)2=a22(α−β)2

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Let $α and β$ be the distinct roots of ${ax}^{2} + bx + c = 0$ then the value of $lim_{x \to α} \frac{1 - \cos ({ax}^{2} + bx + c)}{(x - α)^{2}}$ is

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$\frac{a^{2}}{2} (α - β)^{2}$

$\frac{a^{2}}{2} (α + β)^{2}$

$\frac{1}{2} (α - β)^{2}$

answer is A.

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Detailed Solution

$lim_{x \to α} \frac{1 - \cos ({ax}^{2} + bx + c)}{(x - α)^{2}}$

as $α and β$ are two distinct roots.

$\begin{array}{l} ∴ {ax}^{2} + bx + c = a (x - α) (x - β) \\ i.e. α, β = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ = lim_{x \to α} \frac{1 - \cos [(x - α) (x - β) a]}{(x - u)^{2}} \end{array}$

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