Question

Solve: ${\log }_{0.2}$ |x-3| $\ge$ 0

• A. x ∈ [2,4]
• B. x ∈ [2, 4] - {3}
• C. x ∈ [-2, 4]
• D. x ∈ [-2, 4] -{3}

Answer 1

The correct option is D

x ∈ [-2, 4] -{3}

${\log }_{0.2}$ |x-3| $\ge$ 0

Base of the logarithm lies between 0 to 1, then inequality is equivalent to

Then 0 < |x-3| $\le$ $(0.2{)}^{\circ }$

0 < |x-3| $\le$ 1

$\Rightarrow$ -1 $\le$ x - 3 $\le$ 1 and |x-3| ≠ 0

-1 + 3 $\le$ x $\le$ 1 + 3x - 3 ≠ 0

-2 $\le$ x $\le$ 4

x ∈ [-2, 4] - {3}