Question

If $\frac{5x+8}{4-x}<2$, then

• A. $x\in (-\infty ,0)\cup (4,\infty )$
• B. $x\in (-\infty ,-1)\cup (3,\infty )$
• C. $x\in (-\infty ,0]\cup [4,\infty )$
• D. $x\in (-\infty ,4)$

Answer 1

The correct option is A $x\in (-\infty ,0)\cup (4,\infty )$

$\frac{5x+8}{4-x}-2<0$
$\Rightarrow \frac{5x+8-8+2x}{4-x}<0$
$\Rightarrow \frac{7x}{4-x}<0\Rightarrow \frac{x}{4-x}<0$

$\therefore$ Either $x<0$ and $4-x>0$ or $x>0$ and $4-x<0$

Case $1:$ When $x<0$ and $4-x>0$
$\Rightarrow x<0$ and $x<4$
$\Rightarrow x\in (-\infty ,0)$

Case $2:$ When $x>0$ and $4-x<0$
$\Rightarrow x>0$ and $4<x$
$\Rightarrow x\in (4,\infty )$

$\therefore x\in (-\infty ,0)\cup (4,\infty )$