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

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a

a2(α−β)22

b

a22(α−β)2

c

a2(α−β)2

d

−a22(α−β)2

answer is A.

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Detailed Solution

ax2+bx+c=a(x−α)(x−β)given limit is Limx→α2sin2(a(x−α)(x−β)2)2(x−α)2=a2(α−β)22

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