# Solve 1 | X | − 3 ≤ 1 2 - Mathematics

Solve  $\frac{1}{\left| x \right| - 3} \leq \frac{1}{2}$

#### Solution

$\text{ As }, \frac{1}{\left| x \right| - 3} \leq \frac{1}{2}$
$\Rightarrow \frac{1}{\left| x \right| - 3} - \frac{1}{2} \leq 0$
$\Rightarrow \frac{2 - \left( \left| x \right| - 3 \right)}{2\left( \left| x \right| - 3 \right)} \leq 0$
$\Rightarrow \frac{2 - \left| x \right| + 3}{2\left( \left| x \right| - 3 \right)} \leq 0$
$\Rightarrow \frac{5 - \left| x \right|}{\left| x \right| - 3} \leq 0$
$\text{ Case I: When } x \geq 0, \left| x \right| = x,$
$\frac{5 - x}{x - 3} \leq 0$
$\Rightarrow \left( 5 - x \leq 0 \text{ and } x - 3 > 0 \right) \text{ or } \left( 5 - x \geq 0 \text{ and } x - 3 < 0 \right)$
$\Rightarrow \left( x \geq 5 \text{ and } x > 3 \right) \text{ or } \left( x \leq 5 \text{ and } x < 3 \right)$
$\Rightarrow x \geq 5 or x < 3$
$\Rightarrow x \in \left( 0, 3 \right) \cup [5, \infty )$
$\text{ Case II: When } x < 0, \left| x \right| = - x,$
$\frac{5 + x}{- x - 3} \leq 0$
$\Rightarrow \frac{x + 5}{x + 3} \geq 0$
$\Rightarrow \left( x + 5 > 0 \text{ and } x + 3 > 0 \right) or \left( x + 5 < 0 \text{ and } x + 3 < 0 \right)$
$\Rightarrow \left( x > - 5 \text{ and } x > - 3 \right) \text{ or } \left( x < - 5 \text{ and } x < - 3 \right)$
$\Rightarrow x > - 3 \text{ or } x < - 5$
$\Rightarrow x \in \left( - \infty , - 5 \right) \cup \left( - 3, \infty \right)$
$\text{ So, from both the cases, we get }$
$x \in \left( - \infty , - 5 \right) \cup \left( - 3, \infty \right) \cup \left( 0, 3 \right) \cup [5, \infty )$
$\therefore x \in ( - \infty , - 5] \cup \left( - 3, 3 \right) \cup [5, \infty )$

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Chapter 15: Linear Inequations - Exercise 15.3 [Page 22]

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 15 Linear Inequations
Exercise 15.3 | Q 10 | Page 22
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