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Definition of Functions

Let r1,r2 a...

Question

Let $r_{1}, r_{2}$ and $r_{3}$ be the solution of the equation $x^{3} - 2 x^{2} + 4 x + 5074 = 0$ then the value of $(r_{1} + 2) (r_{2} + 2) (r_{3} + 2)$ is

5050

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5066

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- 5050

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- 5066

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Solution

The correct option is B $- 5050$
$x^{3} - 2 x^{2} + 4 x + 5074 = 0$ (given)
$r_{1} + r_{2} + r_{3} = \frac{- b}{a} = \frac{- (- 2)}{1} = 2$ ---(i)
$r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{c}{a} = \frac{4}{1} = 4$ ---(ii)
$r_{1} r_{2} r_{3} = \frac{- d}{a} = \frac{- 5074}{1} = - 5074$ ---(iii)
$(r_{1} + 2) (r_{2} + 2) (r_{3} + 2) = (r_{1} + 2) (r_{2} r_{3} + 2 r_{2} + 2 r_{3} + 4)$
$= r_{1} r_{2} r_{3} + 2 r_{1} r_{2} + 2 r_{3} r_{1} + 4 r_{1} + 2 r_{2} r_{3} + 4 r_{2} + 4 r_{3} + 8$
$= r_{1} r_{2} r_{3} + 2 (r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}) + 4 (r_{1} + r_{2} + r_{3}) + 8$
$= - 5074 + 2 (4) + 4 (2) + 8$ [from (i),(ii) and (iii)]
$= - 5074 + 8 + 8 + 8$
$= 24 - 5074$
$= - 5050$

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