Aspire Faculty ID #16554 · Topic: NIMCET 2008 · Just now
NIMCET 2008

If $a,b,c$ are the roots of the equation $x^3-3px^2+3qx-1=0$, then the centroid of the triangle with vertices $\left(a,\frac1a\right),\left(b,\frac1b\right),\left(c,\frac1c\right)$ is the point

Solution

Centroid $=\left(\dfrac{a+b+c}{3},\dfrac{\frac1a+\frac1b+\frac1c}{3}\right)$ From the equation, $a+b+c=3p$ Also $\dfrac1a+\dfrac1b+\dfrac1c=\dfrac{ab+bc+ca}{abc}=\dfrac{3q}{1}=3q$ Hence centroid $=(p,q)$ Answer: $\boxed{(p,q)}$

Previous 10 Questions — NIMCET 2008

Nearest first

Next 10 Questions — NIMCET 2008

Ascending by ID
Ask Your Question or Put Your Review.

loading...