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