A) 1
B) - 2
C) 2
D) None of these
Correct Answer: B
Solution :
Let \[\alpha \]be a root of \[{{x}^{2}}-x+k=0,\]then \[2\alpha \]is a root of \[{{x}^{2}}-x+3k=0\]. \[\therefore \,\,\,{{\alpha }^{2}}-\alpha +k=0\] and \[4{{\alpha }^{2}}-2\alpha +3k=0\] Þ \[\frac{{{\alpha }^{2}}}{-3k+2k}=\frac{\alpha }{4k-3k}=\frac{1}{-2+4}\] Þ \[{{\alpha }^{2}}=-k/2\] and \[\alpha =k/2\] Now, \[{{\alpha }^{2}}={{(\alpha )}^{2}}\,\,\Rightarrow -k/2={{(k/2)}^{2}}\] \[\Rightarrow {{k}^{2}}+2k=0\Rightarrow k=0\] or - 2.
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