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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

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$(p, q)$

$(p / 3, q / 3)$

$(p + q, p - q)$

$(3 p, 3 q)$

answer is A.

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Detailed Solution

The centroid of the given triangle is the point

$(\frac{α + β + γ}{3}, \frac{\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}}{3}) = (\frac{3 p}{3}, \frac{α β + β γ + γ α}{3 α β γ})$

$= (p, q) [∵ α + β + γ = 3 p, α β + β γ + γ α = 3 q, α β γ = 1]$

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