| If \[x+\frac{1}{x}=-2,\]then the value of\[{{x}^{2n+1}}+\frac{1}{{{x}^{2n+1}}},\] where n is a positive integer, is |
A) 0
B) \[-\,\,2\]
C) 2
D) \[-\,5\]
Correct Answer: B
Solution :
| Given, \[x+\frac{1}{x}=-\,2\] |
| \[\Rightarrow \] \[x+\frac{1}{x}+2=0\] |
| \[\Rightarrow \]\[{{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{2}}=0\] |
| \[\Rightarrow \]\[\sqrt{x}+\frac{1}{\sqrt{x}}=0\]\[\Rightarrow \]\[x=-1\] |
| Then, \[{{x}^{2n+1}}+\frac{1}{{{x}^{2n+1}}}={{(-1)}^{2n+1}}+\frac{1}{{{(-1)}^{2n+1}}}\] |
| \[=(-1)+(-1)=-\,2\] |
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