A) 1
B) \[\frac{2}{3}\]
C) 2
D) \[-1\]
Correct Answer: B
Solution :
For the nonreal roots of the equation \[2{{z}^{2}}+2z+\lambda =0\] ... (i) \[\operatorname{discriminant} <0.\] That is \[4-8\lambda <0\Rightarrow \,\,\,\lambda >\frac{1}{2}\] ... (ii) Let the roots of (i) be \[{{\operatorname{z}}_{1}} \And \,\,{{z}_{2}}\] Then \[{{\operatorname{z}}_{1}}+{{z}_{2}}=-\frac{2}{2}=-1,\,\,{{z}_{1}}{{z}_{2}}=\frac{\lambda }{2}\] \[{{\operatorname{z}}_{1}}\,and\,\,{{z}_{2}}\] with origin form equilateral triangle if \[{{\operatorname{z}}^{2}}+z_{2}^{2}\,-{{z}_{1}}{{z}_{2}}=0\] \[\Rightarrow \,\,\,\,{{({{z}_{1}}+{{z}_{2}})}^{2}}=\,\,3{{z}_{1}}{{z}_{2}}\] \[\Rightarrow \,\,\,{{(-1)}^{2}}=3\frac{\lambda }{2}\Rightarrow \lambda =\frac{2}{3}\] \[\lambda =\frac{2}{3}\left( >\frac{1}{2} \right)\] satisfies the condition (ii). Hence it is the required result.
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