First slide

Introduction to sets

Question

$Solution of 0 < |3 x + 1| < \frac{1}{3} is$

Easy

A

$(- \frac{4}{9}, - \frac{2}{9})$
B

$[- \frac{4}{9}, - \frac{2}{9}]$
C

$(- \frac{4}{9}, - \frac{2}{9}) - \{- \frac{1}{3}\}$
D

$[- \frac{4}{9}, - \frac{2}{9}] - \{- \frac{1}{3}\}$

Solution

$\begin{array}{l} Let us first solve | 3 x + 1 | < \frac{1}{3} \\ \Leftrightarrow - \frac{1}{3} < 3 x + 1 < \frac{1}{3} \Leftrightarrow - \frac{4}{3} < 3 x < - \frac{2}{3} \end{array}$
$\Leftrightarrow - \frac{4}{9} < x < - \frac{2}{9}$
$Also, 0 < |3 x + 1| is satisfied by each x except when$
$\begin{array}{l} 3 x + 1 = 0, i.e., x = - \frac{1}{3} \\ ∴ Solution is (- \frac{4}{9}, - \frac{2}{9}) - \{- \frac{1}{3}\} \end{array}$

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