Question

If $\alpha ,\beta ,\gamma$ are the real roots of the equation $x^3-3p{x}^2+3qx-1=0,(p,q)=(\pm 1,\pm 1)$ then the centroid of the triangle with vertices $({\alpha }^2,\frac{1}{{\alpha }^2}),({\beta }^2,\frac{1}{{\beta }^2})$ and $({\gamma }^2,\frac{1}{{\gamma }^2})$ can be

• A. $(1,1)$
• B. $(5,5)$
• C. $(1,5)$
• D. $(5,1)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(1,1)$
B $(1,5)$
C $(5,5)$
D $(5,1)$

$(\frac{{\alpha }^2+{\beta }^2+y^2}{3},\frac{{\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2}{3})=(\frac{{\left(3p\right)}^2-2\times 3q}{3},\frac{{\left(3q\right)}^2-2\times 3p}{3})=(3p^2-2q,3q^2-2p)$
Hence all options satisfies this