Quantitative Aptitude Ques 933

Question: 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

Options:

A) 0

B) $ -2 $

C) 2

D) $ -,5 $

Show Answer

Answer:

Correct Answer: B

Solution:

  • Given, $ x+\frac{1}{x}=-,2 $

$ \Rightarrow $ $ x+\frac{1}{x}+2=0 $

$ \Rightarrow $ $ {{( \sqrt{x}+\frac{1}{\sqrt{x}} )}^{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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