Question

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

• A. $\left(-\frac{4}{9},\,-\frac{2}{9}\right)$
• B. $\left[-\frac{4}{9},\,-\frac{2}{9}\right]$
• C. \left(-\frac{4}{9},\,-\frac{2}{9}\right) - \left\\{-\frac{1}{3}\right\\}
• D. \left[-\frac{4}{9},\,-\frac{2}{9}\right] - \left\\{-\frac{1}{3}\right\\}

Answer 1

Answer: (C)

\left(-\frac{4}{9},\,-\frac{2}{9}\right) - \left\\{-\frac{1}{3}\right\\}

Answer 2

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