Quantitative Aptitude Ques 1802

Question: If $ x+\frac{1}{x}=2, $ then what is the value of $ x-\frac{1}{x} $ ?

Options:

A) $ 0 $

B) $ 1 $

C) $ 2 $

D) $ -2 $

Show Answer

Answer:

Correct Answer: A

Solution:

  • Given that, $ x+\frac{1}{x}=2 $ …(i) On squaring both sides, we get $ {{( x+\frac{1}{x} )}^{2}}=4 $

$ \Rightarrow $ $ x^{2}+\frac{1}{x^{2}}+2=4 $
$ \Rightarrow $ $ x^{2}+\frac{1}{x^{2}}=2 $ Now, $ {{( x-\frac{1}{x} )}^{2}}=( x^{2}+\frac{1}{x^{2}} )-2 $

$ \Rightarrow $ $ {{( x-\frac{1}{x} )}^{2}}=2-2=0 $ [from Eq.(ii)]

$ \therefore $ $ x-\frac{1}{x}=0 $