First slide

Introduction to sets

Question

$Solution of |x + \frac{1}{x}| > 2 is$

Easy

A

$R - {0}$
B

$R - {- 1, 0, 1}$
C

$R - {1}$
D

$R - {- 1, 1}$

Solution

$\begin{array}{l} |x + \frac{1}{x}| > 2 \\ (clearly, x \neq 0) \\ \Rightarrow |\frac{x^{2} + 1}{x}| > 2 \Rightarrow \frac{x^{2} + 1}{| x |} > 2 (∵ x^{2} + 1 > 0) \\ \Rightarrow x^{2} + 1 > 2 | x | \\ \Rightarrow | x |^{2} - 2 | x | + 1 > 0 \Rightarrow (| x | - 1)^{2} > 0 \\ \Rightarrow | x | \neq 1 \Rightarrow x \neq - 1, 1 \\ ∴ x \in R - {- 1, 0, 1} \end{array}$

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