MCQ
If $\frac{\big[\text{x} – 7\big]}{(\text{x} – 7)\geq 0}$ then:
- A$\text{x}\in\big[7,\infty)$
- B$\text{x}\in(7,\infty)$
- C$\text{x}\in(\infty, 7)$
- D$\text{x}\in(-\infty, 7)$
Solution:
Given,
$\frac{|\text{x}-7|}{(\text{x}-7)}\geq0$
This is possible when $\text{x}-7\geq0,$ and $\text{x}-7\neq0.$
Here, $\text{x}\geq7$ but $\text{x}\neq7$
Therefore, $\text{x}> 7, \text{i}.\text{e}. \text{x}\in(7,\infty).$
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