Question

The set of solutions for $\frac{3+x}{3-x}\ge 0$ is

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

Answer 1

The correct option is B $[-3,3)$

$\frac{3+x}{3-x}\ge 0,(x\ne 3)$


$\therefore x\in [-3,3)$