For $Δ A B C, Δ, R, r, r_{1}, r_{2}, r_{3}, s$ have the usual meanings, then if the cubic equation with roots $r_{1}, r_{2}, r_{3}$ is $x^{3} + l x^{2} + m x + n = 0$ , then $l =$

- (4 R + r)

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1 + \frac{r}{R}

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2 (R + r)

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(4 R + r)

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Solution

The correct option is D $- (4 R + r)$
As $r_{1}, r_{2}, r_{3}$ are roots of $x^{3} + l x^{2} + m x + n = 0$
$r_{1} + r_{2} + r_{3} = - l$
Using $r_{1} = \frac{Δ}{s - a}, r_{2} = \frac{Δ}{s - b}, r_{3} = \frac{Δ}{s - c}, r = \frac{Δ}{s}$
We get
$r_{1} + r_{2} + r_{3} - r = \frac{Δ}{s - a} + \frac{Δ}{s - b} + \frac{Δ}{s - c} - \frac{Δ}{s} = 4 R$
$\Rightarrow r_{1} + r_{2} + r_{3} = 4 R + r \Rightarrow l = - (4 R + r)$

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