A) \[\frac{2}{a}\]
B) \[\frac{2}{b}\]
C) \[\frac{2}{c}\]
D) \[\frac{-2}{a}\]
Correct Answer: D
Solution :
Since,\[\alpha \]and\[\beta \]are the roots of the equation \[a{{x}^{2}}+bx+c=0\] \[\therefore \] \[a{{\alpha }^{2}}+b\alpha +c=0\] and \[a{{\beta }^{2}}+b\beta +c=0\] \[\Rightarrow \] \[a{{\alpha }^{2}}+b\alpha =a{{\beta }^{2}}+b\beta =-c\] \[\therefore \] \[\frac{\alpha }{a\beta +b}+\frac{\beta }{a\alpha +b}+\frac{\alpha \beta }{a{{\beta }^{2}}+b\beta }+\frac{\alpha \beta }{a{{\alpha }^{2}}+b\alpha }\] \[=\frac{2\alpha \beta }{-c}=\frac{2}{-c}.\frac{c}{a}=-\frac{2}{a}\]
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