A) \[(-2,1)\]
B) \[\left( -2,\frac{2}{5} \right)\]
C) \[\left( \frac{2}{5},1 \right)\]
D) None of these
Correct Answer: B
Solution :
Since, \[({{\lambda }^{2}}+\lambda -2){{x}^{2}}+(\lambda +2)x-1<0\] For all \[x\in R\] then D < 0 and a < 0. \[\therefore \]\[{{(\lambda +2)}^{2}}+4({{\lambda }^{2}}+\lambda -2)<0\] and\[{{\lambda }^{2}}+\lambda -2<0\]\[\Rightarrow \]\[5{{\lambda }^{2}}+8\lambda -4<0\] and\[{{\lambda }^{2}}+\lambda -2<0\] \[\Rightarrow \]\[(5\lambda -2)(\lambda +2)<0\]and \[(\lambda +2)(\lambda -1)<0\] \[\Rightarrow \]\[-2<\lambda <\frac{2}{5}\]and \[-2<\lambda <1\] \[\Rightarrow \]\[-2<\lambda <\frac{2}{5}\Rightarrow \lambda \in \left( -2,\frac{2}{5} \right)\]
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