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 $
BETA
