A) \[4040\]
B) \[-2\]
C) 2
D) 3
Correct Answer: C
Solution :
\[x+\frac{1}{x}=2\] If we put the value of \[x=1,\]then the equation satisfies. So, \[{{x}^{2019}}+\frac{1}{{{x}^{2021}}}={{1}^{2018}}+\frac{1}{{{1}^{2021}}}\] \[=\,\,\,\,\,1+1=\underline{\mathbf{2}}\]You need to login to perform this action.
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