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 :
\[\frac{x+3}{x-2}\le 2\] \[\Rightarrow \] \[\frac{x+3}{x-2}-2\le 0\] \[\Rightarrow \] \[\frac{x+3-2x+4}{x-2}\le 0\] \[\Rightarrow \] \[\frac{-x+7}{x-2}\le 0\] \[\Rightarrow \] \[\frac{x-7}{x-2}\ge 0\] \[\Rightarrow \] \[\frac{(x-7)(x-2)}{{{(x-2)}^{2}}}\ge 0\] \[\Rightarrow \] \[x\in (-\infty ,2)\cup [7,\infty )\]You need to login to perform this action.
You will be redirected in
3 sec