If α and β are the roots of ax2+bx+c=0 and c≠0 then limx→1/α 1−cos⁡cx2+bx+a2(1−αx)2 is

If α , β are the roots of ax2 + bx + c = 0 then 1/α, 1/β will be the roots of cx2 + bx + a = 0.Thus , limx→1/a 1−cos⁡cx2+bx+a2(1−αx)2=limx→1/a sin⁡cx2+bx+a2|α||1/α−x|=limx→1/a sin⁡c2(x−1/α)(x−1/β)|α||c|2∣x−1/α∥x−1/β×|c|2|x−1/β|=c2α1α−1β

If α and β are the roots of ax2+bx+c=0 and c≠0 then limx→1/α 1−cos⁡cx2+bx+a2(1−αx)2 is

If α , β are the roots of ax2 + bx + c = 0 then 1/α, 1/β will be the roots of cx2 + bx + a = 0.Thus , limx→1/a 1−cos⁡cx2+bx+a2(1−αx)2=limx→1/a sin⁡cx2+bx+a2|α||1/α−x|=limx→1/a sin⁡c2(x−1/α)(x−1/β)|α||c|2∣x−1/α∥x−1/β×|c|2|x−1/β|=c2α1α−1β

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If $α$ and $β$ are the roots of $a x^{2} + b x + c = 0$ and $c \neq 0$ then
$lim_{x \to 1 / α} \sqrt{\frac{1 - \cos (c x^{2} + b x + a)}{2 (1 - α x)^{2}}}$ is

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$c (a - b)$

$\frac{c}{2 α} (\frac{1}{α} - \frac{1}{β})$

$\frac{1}{β} (\frac{1}{α} - \frac{1}{β})$

$|(\frac{1}{α} - \frac{1}{β}) \frac{c}{2 α}|$

answer is D.

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Detailed Solution

If $α, β$ are the roots of $a x^{2} + b x + c = 0$ then $1 / α, 1 / β$ will be the roots of $c x^{2} + b x + a = 0 .$

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

$= lim_{x \to 1 / a} \frac{|\sin \frac{c x^{2} + b x + a}{2}|}{| α | | 1 / α - x |}$

$= lim_{x \to 1 / a} \frac{|\sin \frac{c}{2} (x - 1 / α) (x - 1 / β)|}{| α | \frac{| c |}{2} ∣ x - 1 / α ∥ x - 1 / β} \times \frac{| c |}{2} | x - 1 / β |$