Let α. and β be the roots of ax2+bx+c=0, then limx→a 1−cos⁡ax2+bx+c(x−α)2 is equal to

Let α. and β be the roots of ax2+bx+c=0, then limxa1cosax2+bx+c(xα)2 is equal to

  1. A

    0

  2. B

    12(αβ)2

  3. C

    a22(αβ)2

  4. D

    a22(αβ)2

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    Solution:

    It is given that α,β are roots of ax2+bx+c

     ax2+bx+c=a(xα)(xβ)

    Now, limxα1cosax2+bx+c(xα)2

    =2limxαsin2ax2+bx+c2(xα)2=2limxαsin2a(xα)(xβ)2(xα)2=2limxαsina(xα)(xβ)22a(xα)(xβ)2×a24(xβ)2=2(1)2×a24(αβ)2=a22(αβ)2

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