Question

The set of all real$x$ satisfying the inequality$\frac{3-\left|x\right|}{4-\left|x\right|}\ge 0$

• A. $[-3,3]\cup (-\infty ,–4)\cup (4,\infty )$
• B. $(-\infty ,–4)\cup (4,\infty )$
• C. $(-\infty ,–3)\cup (4,\infty )$
• D. $(-\infty ,–3)\cup (3,\infty )$
• E. $(-3,3)\cup (4,\infty )$

Answer 1

The correct option is A

$[-3,3]\cup (-\infty ,–4)\cup (4,\infty )$

Explanation for the correct option.

Given, $\frac{3-\left|x\right|}{4-\left|x\right|}\ge 0$

$\Rightarrow 3–\left|x\right|\le 0$ and $4–\left|x\right|<0$or $3–\left|x\right|\ge 0$ and $4–\left|x\right|>0$

$\Rightarrow \left|x\right|\ge 3$and $\left|x\right|>4$or $\left|x\right|\le 3$and$\left|x\right|<4$

$\Rightarrow \left|x\right|>4$or $\left|x\right|\le 3$

$\Rightarrow x\in (-\infty ,–4)\cup [-3,3]\cup (4,\infty )$

Hence, option A is the correct option.