| If \[x+\frac{1}{x}=2,\]then \[{{x}^{2013}}+\frac{1}{{{x}^{2014}}}=?\] [SSC (CPO) 2014] |
A) \[0\]
B) \[1\]
C) \[-\,\,1\]
D) \[2\]
Correct Answer: D
Solution :
| \[x+\frac{1}{x}=2\]\[\Rightarrow \]\[{{x}^{2}}-2x+1=0\] |
| \[\Rightarrow \] \[{{(x-1)}^{2}}=0\]\[\Rightarrow \]\[x=1\] |
| \[\therefore \]\[{{x}^{2013}}+\frac{1}{{{x}^{2014}}}=1+1=2\] |
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