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If $$\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}}$$ then the value of $$\frac{{x - 1}}{x}$$ is?
Answer & Solution
Correct Answer:
Option
B
$$\eqalign{
& {x^2} + \frac{{{\text{ }}1}}{{{x^2}}} = 2 \cr
& \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} + 2.x.\frac{1}{x} = 2 \cr
& \Rightarrow x - \frac{1}{x} = 2 - 2 \cr
& \Rightarrow x - \frac{1}{x} = 0 \cr} $$
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