If α and β are the roots of ax2+bx+c=0 then Ltx→α1+ax2+bx+c1x−α is

G⋅L=eliml→∞ α2+bx+cx−α=ellita(x−α)(x−β)x−α=ed(α−β)

If α and β are the roots of ax2+bx+c=0 then Ltx→α1+ax2+bx+c1x−α is

G⋅L=eliml→∞ α2+bx+cx−α=ellita(x−α)(x−β)x−α=ed(α−β)

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If $α$ and $β$ are the roots of $a x^{2} + b x + c = 0$ then $\underset{x \to α}{L t} {(1 + a x^{2} + b x + c)}^{\frac{1}{x - α}}$ is

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$a (α - β)$

$\log |a (α - β)|$

$e^{a (α - β)}$

$e^{a (α + β)}$

answer is C.

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Detailed Solution

$G \cdot L = e^{lim_{l \to \infty} \frac{α^{2} + b x + c}{x - α}} = e^{l^{l i t} \frac{a (x - α) (x - β)}{x - α}} = e^{d (α - β)}$

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If α and β are the roots of ax2+bx+c=0 then Ltx→α1+ax2+bx+c1x−α is

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If $α$ and $β$ are the roots of $a x^{2} + b x + c = 0$ then $\underset{x \to α}{L t} {(1 + a x^{2} + b x + c)}^{\frac{1}{x - α}}$ is