A) \[\alpha \ne \beta \ne \gamma \ne \delta \]
B) \[\alpha \ne \beta and\,\gamma \ne \delta \]
C) \[a\alpha =a\beta =c\gamma =c\delta \]
D) \[\alpha =\beta \,and\,\gamma \ne \delta \]
E) \[\alpha \ne \beta \,and\,\gamma =\delta \]
Correct Answer: B
Solution :
Since\[\alpha \]and\[\beta \]are the roots of the equation\[a{{x}^{2}}+2bx+c=0\] \[\Rightarrow \]\[\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}\] And\[\gamma ,\delta \]are the roots of the equation \[c{{x}^{2}}+2bx+a=0\] \[\Rightarrow \] \[\gamma +\delta =\frac{-b}{c}\]and \[\delta \gamma =\frac{a}{c}\] Since, a, b, c are in GP\[\Rightarrow \]\[{{b}^{2}}=ac\] Now, if\[\alpha =\beta \Rightarrow {{b}^{2}}=4ac\therefore \alpha \ne \beta \] and if\[\gamma =\delta \Rightarrow {{b}^{2}}=4ac\therefore \gamma \ne \delta \]
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