Correct option is (C) -b/c
Given that \(\alpha\;and\;\beta\) are zeros of polynomial \(ax^2+bx+c.\)
\(\therefore\) \(\alpha+\beta=\frac{-b}a\) and \(\alpha\beta=\frac ca\)
\(\therefore\) \(\frac{\alpha+\beta}{\alpha\beta}=\cfrac{\frac{-b}a}{\frac ca}=\frac{-b}c\)
\(\Rightarrow\) \(\frac{1}{\alpha}+\frac{1}{\beta}\) \(=\frac{-b}c\)