A) \[\frac{p}{r}\]
B) \[-\frac{p}{r}\]
C) \[\frac{q}{r}\]
D) \[-\frac{q}{r}\]
Correct Answer: A
Solution :
| [a] Since \[\alpha ,\beta ,\gamma \] are the zeroes of the polynomial |
| \[{{x}^{3}}+p{{x}^{2}}+qx+r.\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\alpha +\beta +\gamma =-p,\,\alpha \beta \gamma =-r\] |
| Now, \[\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha }=\frac{\alpha +\beta +\gamma }{\alpha \beta \gamma }=\frac{-p}{-r}=\frac{p}{r}\] |
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