Express |3x + 1| < 2 in neighbourhood and interval form. - Vidyadip

Question

Express $|3x + 1| < 2$ in neighbourhood and interval form.

✓

Answer

$
\begin{aligned}
& |3 x+1|<2 \\
& \therefore\left|x+\frac{1}{3}\right|<\frac{1}{3} \\
& \therefore\left|x-\left(-\frac{1}{3}\right)\right|<\frac{2}{3} \\
& \therefore\left|x-\left(-\frac{1}{3}\right)\right|<\frac{2}{3}, \\
& \therefore a=-\frac{1}{3} \delta=\frac{2}{3}
\end{aligned}
$
In neighbourhood form: $N(a, \delta)$
$\therefore$ Neighbourhood form of $|3 \mathrm{x}+1|<2$
$
=\mathrm{N}\left(-\frac{1}{3}, \frac{2}{3}\right)
$
In Interval form: $(a-\delta, a+\delta)$
Interval form of $13 x+11<2$
$
\begin{aligned}
& =\left(-\frac{1}{3}-\frac{2}{3},-\frac{1}{3}+\frac{2}{3}\right) \\
& =\left(-1, \frac{1}{3}\right)
\end{aligned}
$

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