Write the solution set of |\(x\) + \(\frac{1}{x}\)| > 2.|x + 1/x| > 2

1.	Write the solution set of \|\(x\) + \(\frac{1}{x}\)\| > 2.\|x + 1/x\| > 2.
Answer» \|x + 1/x\| > 2 ⇒ - 2 < \(x\) + \(\frac{1}{x}\) < 2 Part I : \(x\) + \(\frac{1}{x}\) > - 2 ⇒ \(\frac{x^2+2x+1}{x}\) > 0 ⇒ \(\frac{(x+1)^2}{x}\) > 0 Square term is always positive and (x + 1)² ≠ 0. So, x > 0 and x ≠ -1 Part II : \(x\) + \(\frac{1}{x}\)< 2 ⇒ \(\frac{x^2-2x+1}{x}\) < 0 ⇒ \(\frac{(x-1)^2}{x}\) < 0 Square term is always positive and (x – 1)² ≠ 0. So, x < 0 and x ≠ 1 So, by taking union of part I and part II We have final solution i.e., x ∈ R – {-1, 0, 1}

Write the solution set of |\(x\) + \(\frac{1}{x}\)| > 2.|x + 1/x| > 2.

Answer»

|x + 1/x| > 2

⇒ - 2 < \(x\) + \(\frac{1}{x}\) < 2

Part I : \(x\) + \(\frac{1}{x}\) > - 2

⇒ \(\frac{x^2+2x+1}{x}\) > 0

⇒ \(\frac{(x+1)^2}{x}\) > 0

Square term is always positive and (x + 1)² ≠ 0.

So, x > 0 and x ≠ -1

Part II : \(x\) + \(\frac{1}{x}\)< 2

⇒ \(\frac{x^2-2x+1}{x}\) < 0

⇒ \(\frac{(x-1)^2}{x}\) < 0

Square term is always positive and (x – 1)² ≠ 0.

So, x < 0 and x ≠ 1

So, by taking union of part I and part II

We have final solution i.e.,

x ∈ R – {-1, 0, 1}