Let α and β be the distinct roots of ax2+bx+c=0 then the value of limx→α 1−cosax2+bx+c(x−α)2 is
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a
a22(α−β)2
b
0
c
a22(α+β)2
d
12(α−β)2
answer is A.
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Detailed Solution
limx→α 1−cosax2+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