The solution set for |3 x-2| 1 2 - Vidyadip

MCQ

The solution set for $|3 x-2| \leq \frac{1}{2}$

A
$\left[\frac{5}{6}, \frac{2}{3}\right]$
B
$\left[\frac{2}{3}, \frac{2}{3}\right]$
C
$\left[\frac{1}{2}, \frac{5}{6}\right]$
D
$\left[\frac{5}{6}, \frac{1}{2}\right]$

✓

Answer

(c) $\left[\frac{1}{2}, \frac{5}{6}\right]$
Explanation: $|3 x-2| \leq \frac{1}{2}$
$\begin{array}{l}\Rightarrow \frac{-1}{2} \leq 3 x-2 \leq \frac{1}{2} \\ \Rightarrow \frac{-1}{2}+2 \leq 3 x-2+2 \leq \frac{1}{2}+2 \\ \Rightarrow \frac{3}{2} \leq 3 x \leq \frac{5}{2} \quad[\because|x| \leq a \Leftrightarrow-a \leq x \leq a] \\ \Rightarrow \frac{3}{2} \cdot \frac{1}{3} \leq 3 x \cdot \frac{1}{3} \leq \frac{5}{2} \cdot \frac{1}{3} \\ \Rightarrow \frac{1}{2} \leq x \leq \frac{5}{6} \\ \Rightarrow x \in\left[\frac{1}{2}, \frac{5}{6}\right]\end{array}$

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