Let - and - be the distinct roots of ax2-bx-c-0 then x-lim1-cosax2-bx-c-x-2 is equal to -

The correct option is C a^2/2(α β)^2 lim_x→α1 cos ( ax^2+bx+c )/ ( x α #160; )^2=lim_x→α2sin ^2 ( ax^2+bx+c2 ) ( x+α #160; )^2 =lim_x→α4s ...

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Substitution Method to Remove Indeterminate Form

Let α and ...

Question

Let $α$ and $β$ be the distinct roots of $a x^{2} + b x + c = 0$ then $l i m x \to α \frac{1 - cos (a x^{2} + b x + c)}{(x - α)^{2}}$ is equal to -

\frac{1}{2} (α - β)^{2}

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\frac{- a^{2}}{2} (α - β)^{2}

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0

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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}}

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

α

β

a α^{2} + b α + c = 0

a β^{2} + b β + c = 0

α, β

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

\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 - 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} =

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

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

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