Question:
The solution set of \(\left|x + \frac{1}{x}\right| > 2\) is
The solution set of \(\left|x + \frac{1}{x}\right| > 2\) is
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$x + \frac{1}{x} \geq 2$ or $\leq -2$ except at $x=\pm1$.
Updated On: Apr 30, 2026
- $\mathbb{R}$
- $\mathbb{R} - \{0\}$
- $\mathbb{R} - \{1,-1\}$
- $\mathbb{R} - \{-1\}$
- $\mathbb{R} - \{-1,0,1\}$
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The Correct Option is
Solution and Explanation
Concept:
Use inequality transformation.
Step 1: Square both sides
\[ \left(x + \frac{1}{x}\right)^2 > 4 \] \[ x^2 + \frac{1}{x^2} + 2 > 4 \Rightarrow x^2 + \frac{1}{x^2} > 2 \]
Step 2: Always true except special points
Equality holds when $x = \pm1$ Also $x \neq 0$ Final Conclusion:
\[ \mathbb{R} - \{-1,0,1\} \] Option (E)
Step 1: Square both sides
\[ \left(x + \frac{1}{x}\right)^2 > 4 \] \[ x^2 + \frac{1}{x^2} + 2 > 4 \Rightarrow x^2 + \frac{1}{x^2} > 2 \]
Step 2: Always true except special points
Equality holds when $x = \pm1$ Also $x \neq 0$ Final Conclusion:
\[ \mathbb{R} - \{-1,0,1\} \] Option (E)
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