# Write the Solution of Set of ∣ ∣ ∣ X + 1 X ∣ ∣ ∣ > 2 - Mathematics

Write the solution of set of$\left| x + \frac{1}{x} \right| > 2$

#### Solution

$\text{ We have }:$
$\left| x + \frac{1}{x} \right| > 2$
$\Rightarrow \left| x + \frac{1}{x} \right| - 2 > 0$
$\text{ CASE 1: When } x + \frac{1}{x} > 0, \text{ then } \left| x + \frac{1}{x} \right| = x + \frac{1}{x}$
$\text{ Now }, \left| x + \frac{1}{x} \right| - 2 > 0$
$\Rightarrow x + \frac{1}{x} - 2 > 0$
$\Rightarrow \frac{x^2 + 1 - 2x}{x} > 0$
$\Rightarrow \frac{(x - 1 )^2}{x} > 0$
$\Rightarrow x > 0 \text{ and } x \neq 1$
$\Rightarrow x \in (0, 1)U(1, \infty ) . . . \left( i \right)$
$\text{ CASE 2: When } x + \frac{1}{x} < 0, \text{ then } \left| x + \frac{1}{x} \right| = - (x + \frac{1}{x})$
$\text{ Now }, \left| x + \frac{1}{x} \right| - 2 > 0$
$\Rightarrow - x - \frac{1}{x} - 2 > 0$
$\Rightarrow \frac{- x^2 - 1 - 2x}{x} > 0$
$\Rightarrow \frac{x^2 + 1 + 2x}{x} < 0 \left[ \text{ Multiplying both sides by } - 1 \right]$
$\Rightarrow \frac{(x + 1 )^2}{x} < 0$
$\Rightarrow x < 0 \text{ and } x \neq - 1$
$\Rightarrow x \in ( - \infty , - 1)U( - 1, 0) . . . \left( ii \right)$
$\text{ Thus, the solution set of the given inequation is the union of } \left( i \right) and \left( ii \right)$
$(0, 1)U(1, \infty ) \cup ( - \infty , - 1)U( - 1, 0) = R - \left\{ - 1, 0, 1 \right\}$
$\therefore x \in R - \left\{ - 1, 0, \right\}$


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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 15 Linear Inequations
Q 9 | Page 31