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

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

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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
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