The value of limx-1-cos ax2-bx-c-x- 2 where - - - are the distinct roots of ax2-bx-c-0 is

The correct option is C (α β )^2/2 lim_x→β1 cos (ax^2+bx+c)(x β )^2=lim_x→ #160;β2sin ^2ax^2+bx+c2(x β )^2 =lim_x→βsin ^2(x α )(x β )2 ( #1 ...

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Cos(A+B)Cos(A-B)

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The value of $lim x \to β \frac{1 - cos (a x^{2} + b x + c)}{(x - β)^{2}}$ where $α, β$ are the distinct roots of $a x^{2} + b x + c = 0$ is

(α - β)^{2}

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\frac{(α - β)^{2}}{2}

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(α \frac{(α - β)^{2}}{2})

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none of these

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α, β

a x^{2} + b x + c = 0

L_{1} = L t x \to α \frac{1 - c o s ({a x}^{2} + b x + c)}{{(x - α)}^{2}}

L_{2} = L t x \to β \frac{1 - c o s ({a x}^{2} + b x + c)}{{(x - β)}^{2}}

L_{3} = L t (x - α) (x - β) \to 0 \frac{1 - c o s ({a x}^{2} + b x + c)}{{(x - α)}^{2} {(x - β)}^{2}}

α, β

a x^{2} + b x + c = 0

(\frac{α}{β} - \frac{α}{β})^{2}

α, β

a x^{2} + b x + c = 0

(α - β)^{2}, (α + β)^{2}

α, β

a x^{2} + b x + c = 0

\frac{α^{2} + β^{2}}{α^{- 2} + β^{- 2}}

α

β

a x^{2} + b x + c = 0,

a, b

c

\frac{a α^{2} + b α + 6 c}{a β^{2} + b β^{2} + 9 c} + \frac{a β^{2} + b β^{2} + 19 c}{a α^{2} + b α + 13 c}

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