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)

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