Question

$\frac{x + 4}{x -3} < 2 $ is satisfied when $x$ satisfies

• A. $\left(- \infty,\, 3\right) \cup \left(10,\, \infty\right)$
• B. $(3,\,10)$
• C. $\left(- \infty,\, 3\right) \cup [10,\, \infty)$
• D. None of these

Answer 1

Answer: (A)

We have, $\frac{x + 4}{x -3} < 2$ or $\frac{x + 4}{x -3} -2 < 0$
or $\quad\frac{-x + 10}{x -3} < 0$ or $\frac{x - 10}{x -3} > 0$
$\Rightarrow \quad$ $\{x - 10 > 0$ and $x-3>0\}$ or $\{ x - 10<0$ and $x-3<0\}$
$\Rightarrow \quad \{x > 10$ and $x > 3\}$ or $\{x < 10$ and $x < 3\}$
$\Rightarrow \quad x\in \left(-\infty,\,3\right) \cup \left(10,\, \infty\right)$