If $$\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}}$$   then the value of $$\frac{{x - 1}}{x}$$  is?

A. -2

B. 0

C. 1

D. -1

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & {x^2} + \frac{{{\text{ }}1}}{{{x^2}}} = 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} + 2.x.\frac{1}{x} = 2 \cr & \Rightarrow x - \frac{1}{x} = 2 - 2 \cr & \Rightarrow x - \frac{1}{x} = 0 \cr} $$