MCQ
Solve: $\frac{-1}{(|\text{x}| – 2)}\geq1$ where $\text{x}\in\text{R}, \text{x}\neq\pm2$
- A$(-2, -1)$
- B$(-2, 2)$
- ✓$(-2, -1)\cup(1, 2)$
- D$\text{None of these}$
✓
Answer
Correct option: C.
$(-2, -1)\cup(1, 2)$
Given, $\frac{-1}{(|\text{x}| – 2)}\geq1$
$\Rightarrow\frac{-1}{(|\text{x}|-2) -1}\geq0$
$\Rightarrow\frac{-1 – (|\text{x}| – 2)}{(|\text{x}| – 2)}\geq0$
$\Rightarrow\frac{1 – |\text{x}|}{(|\text{x}| – 2)}\geq0$
$\Rightarrow\frac{-(|\text{x}| – 1)}{(|\text{x}| – 2)}\geq0$
Using number line rule:
$1\leq |\text{x}|<2$
$\Rightarrow\text{x}\in(-2, -1)\cup(1, 2)$
$\Rightarrow\frac{-1}{(|\text{x}|-2) -1}\geq0$
$\Rightarrow\frac{-1 – (|\text{x}| – 2)}{(|\text{x}| – 2)}\geq0$
$\Rightarrow\frac{1 – |\text{x}|}{(|\text{x}| – 2)}\geq0$
$\Rightarrow\frac{-(|\text{x}| – 1)}{(|\text{x}| – 2)}\geq0$
Using number line rule:
$1\leq |\text{x}|<2$
$\Rightarrow\text{x}\in(-2, -1)\cup(1, 2)$
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