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Question

Let α and β be the distinct roots of ax2+bx+c=0, then limxα1cos(ax2+bx+c)(xα)2 is equal to

A
12(αβ)2
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B
α22(αβ)2
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C
0
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D
α22(αβ)2
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Solution

The correct option is B α22(αβ)2
We have, limxα1cos(ax2+bx+c)(xα)2

=limxα2sin2(ax2+bx+c2)(xα)2
[Since, α and β are the roots of ax2+bx+c=0, so it can be written as a(xα)(xβ)=0]

=limxα2sin2(a(xα)(xβ)2)(xα)2

=limxα2sin2(a2(xα)(xβ))(a2)2(xβ)2[(a2)(xα)(xβ)]2

=limxα2(a2)2(xβ)2

=limxαa22(xβ)2=a22(αβ)2

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