If $α, β, γ$ are the real roots of the equation $x^{3} - 3 p x^{2} + 3 q x - 1 = 0$ , then the centroid of the triangle with vertices $(α, \frac{1}{α})$ , $(β, \frac{1}{β})$ and $(γ, \frac{1}{γ})$ is at the point

(p, q)

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(\frac{p}{3}, \frac{q}{3})

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(p + q, p - q)

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(3 p, 3 q)

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Solution

The correct option is A $(p, q)$
The centroid of the given triangle is the point $⎛ ⎜ ⎜ ⎜ ⎝ \frac{α + β + γ}{3}, \frac{\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}}{3} ⎞ ⎟ ⎟ ⎟ ⎠ = (\frac{3 p}{3}, \frac{α β + β γ + γ α}{3 α β γ}) = (p, q)$
$[∵ α + β + γ = 3 p, α β + β γ + γ α = 3 q, α β + β γ + γ α = 3 q, α β γ = 1]$

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