A) \[a<0\]
B) \[-1<a<0\]
C) \[-1<a<1\]
D) None of these
Correct Answer: D
Solution :
Given equation is \[2{{x}^{2}}+6x+a=0\] Now, \[\left( \frac{\alpha }{\beta } \right)+\left( \frac{\beta }{\alpha } \right)<2\] \[\Rightarrow \] \[\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }<2\] \[\Rightarrow \] \[\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{\alpha \beta }<2\] \[\Rightarrow \] \[\frac{9-a}{a/2}<2\]\[\Rightarrow \] \[9-a<a\] \[\left[ \begin{align} & \because \,\,\,\alpha +\beta =-3 \\ & and\,\,\alpha \beta =\frac{a}{2} \\ \end{align} \right]\] \[\Rightarrow \] \[\frac{9}{2}<a\]
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