Let $\alpha ,\beta ,\gamma$ be the roots of the equation $x^3-12x^2+44x-48=0.$ If the coordinates of the vertices of a triangle are $A(\alpha ,\frac{1}{\alpha }),$ $B(\beta ,\frac{1}{\beta })$ and $C(\gamma ,\frac{1}{\gamma }),$ then the centroid of the $△ABC$ is

The correct option is A $(4,\frac{11}{36})$
Given : $x^3-12x^2+44x-48=0$

Centroid $=(\frac{\alpha +\beta +\gamma }{3},\frac{\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }}{3})$

Now, $\frac{\alpha +\beta +\gamma }{3}=\frac{12}{3}=4$

$\frac{1}{3}(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })=\frac{1}{3}\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)=\frac{1}{3}\left(\frac{44}{48}\right)=\frac{11}{36}$

Therefore, centroid $=(4,\frac{11}{36})$