Let - and - be the roots of ax2-bx-c- 0- then limx-1-cosax2-bx-c-x-2-

The correct option is B a^2(α β)^2/2 lim_x→α1 cos ( ax^2+bx+c )/ ( x α nbsp; )^2=lim_x→α2sin ^2 ( ax^2+bx+c/2 )/ ( x α nbsp; )^2 =lim_x→α ...

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Composite Function

Let α and ...

Question

Let $α$ and $β$ be the roots of $a x^{2} + b x + c = 0$ , then $lim x \to α \frac{1 - cos (a x^{2} + b x + c)}{(x - α)^{2}} =$

\frac{a^{2} (α - β)^{2}}{2}

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\frac{a^{2}}{2 (α - β)^{2}}

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\frac{a^{2}}{(α - β)^{2}}

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- \frac{a^{2}}{2 (α - β)^{2}}

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Solution

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Similar questions

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

x^{2} - p x - c - p = 0

\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c} =

lim x \to β \frac{1 - cos (a x^{2} + b x + c)}{(x - β)^{2}}

α, β

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

α, β

x^{2} - p (x + 1) - c = 0

\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c}

α, β

x^{2} - p (x + 1) - c = 0

(α + 1) (β + 1) = 1 - c

\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c} = 1

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