A) \[0\]
B) \[2\]
C) \[-2\]
D) \[-5\]
Correct Answer: C
Solution :
\[x+\frac{1}{x}=-2\] ? (i) \[\therefore \] \[{{\left( x-\frac{1}{x} \right)}^{2}}={{\left( x+\frac{1}{x} \right)}^{2}}-4\] \[={{(-2)}^{2}}-4=0\] \[\Rightarrow \] \[x-\frac{1}{x}=0\] ? (ii) Solving equations (i) and (ii), we have \[\therefore \] \[x=-1\] \[\therefore \] \[{{x}^{2n+1}}+\frac{1}{{{x}^{2n+1}}}\] \[={{(-1)}^{2n+1}}+\frac{1}{{{(-1)}^{2n+1}}}\] \[=-1-1=-2\]
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