A) \[\frac{2}{a}\]
B) \[\frac{2}{b}\]
C) \[\frac{2}{c}\]
D) \[-\frac{2}{a}\]
Correct Answer: D
Solution :
\[\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}\] and \[{{\alpha }^{2}}+{{\beta }^{2}}=\frac{({{b}^{2}}-2ac)}{{{a}^{2}}}\] Now \[\frac{\alpha }{\begin{align} & a\beta +b \\ & \\ \end{align}}+\frac{\beta }{a\alpha +b}\]\[=\frac{\alpha (a\alpha +b)+\beta (a\beta +b)}{(a\beta +b)(a\alpha +b)}\] \[=\frac{a({{\alpha }^{2}}+{{\beta }^{2}})+b(\alpha +\beta )}{\alpha \beta {{a}^{2}}+ab(\alpha +\beta )+{{b}^{2}}}=\frac{a\frac{({{b}^{2}}-2ac)}{{{a}^{2}}}+b\left( -\frac{b}{a} \right)}{\left( \frac{c}{a} \right){{a}^{2}}+ab\left( -\frac{b}{a} \right)+{{b}^{2}}}\] \[=\frac{{{b}^{2}}-ac-{{b}^{2}}}{{{a}^{2}}c-a{{b}^{2}}+a{{b}^{2}}}=\frac{-2ac}{{{a}^{2}}c}=-\frac{2}{a}\].
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