MCQ
The solution set for $|3 x-2| \leq \frac{1}{2}$
  • A
    $\left[\frac{5}{6}, \frac{2}{3}\right]$
  • B
    $\left[\frac{2}{3}, \frac{2}{3}\right]$
  • $\left[\frac{1}{2}, \frac{5}{6}\right]$
  • D
    $\left[\frac{5}{6}, \frac{1}{2}\right]$

Answer

Correct option: C.
$\left[\frac{1}{2}, \frac{5}{6}\right]$
$|3 x-2| \leq \frac{1}{2}$
$\Rightarrow \frac{-1}{2} \leq 3 x-2 \leq \frac{1}{2}$
$\Rightarrow \frac{-1}{2}+2 \leq 3 x-2+2 \leq \frac{1}{2}+2$
$\Rightarrow \frac{3}{2} \leq 3 x \leq \frac{5}{2} \quad[\because|x| \leq a \Leftrightarrow-a \leq x \leq a]$
$\Rightarrow \frac{3}{2} \cdot \frac{1}{3} \leq 3 x \cdot \frac{1}{3} \leq \frac{5}{2} \cdot \frac{1}{3}$
$\Rightarrow \frac{1}{2} \leq x \leq \frac{5}{6}$
$\Rightarrow x \in\left[\frac{1}{2}, \frac{5}{6}\right]$

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