Let α. and β be the roots of ax2+bx+c=0, then limx→a 1−cosax2+bx+c(x−α)2 is equal to
0
12(α−β)2
a22(α−β)2
−a22(α−β)2
It is given that α,β are roots of ax2+bx+c
∴ ax2+bx+c=a(x−α)(x−β)
Now, limx→α 1−cosax2+bx+c(x−α)2
=2limx→α sin2ax2+bx+c2(x−α)2=2limx→α sin2a(x−α)(x−β)2(x−α)2=2limx→α sina(x−α)(x−β)22a(x−α)(x−β)2×a24(x−β)2=2(1)2×a24(α−β)2=a22(α−β)2