A) 2, -1
B) 0, -1, \[-\frac{1}{5}\]
C) \[-1,-\frac{1}{5}\]
D) None of these
Correct Answer: D
Solution :
\[x+\frac{1}{x}=2\,\,\Rightarrow \,\,x+\frac{1}{x}-2=0\]\[(\because x\ne 0)\] Þ \[{{x}^{2}}-2x+1=0\]Þ\[{{(x-1)}^{2}}=0\]Þ \[x=1,1\].
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