question_answer

The solution set of\[\frac{x+3}{x-2}\le 2\]is

A)  \[(-\infty ,\infty )\]                        

B)  \[(-\infty ,2]\cup [7,\infty )\]

C)  \[(-\infty ,2)\cup [7,\infty )\]

D)         \[[7,\infty )\]

E)  \[(-\infty ,2)\]  

Correct Answer: C

Solution :

\[\frac{x+3}{x-2}\le 2\] \[\Rightarrow \]               \[\frac{x+3}{x-2}-2\le 0\] \[\Rightarrow \]               \[\frac{x+3-2x+4}{x-2}\le 0\] \[\Rightarrow \]               \[\frac{-x+7}{x-2}\le 0\] \[\Rightarrow \]               \[\frac{x-7}{x-2}\ge 0\] \[\Rightarrow \]               \[\frac{(x-7)(x-2)}{{{(x-2)}^{2}}}\ge 0\] \[\Rightarrow \]               \[x\in (-\infty ,2)\cup [7,\infty )\]