Question

The interval(s) of $x$ which satisfies the inequality $\frac{1}{x+2}<\frac{1}{3x+7}$ is/are

• A. $(-\frac{7}{3},-2)$
• B. $(-2,\infty )$
• C. $(-\frac{5}{2},-\frac{7}{3})$
• D. $(-\infty ,-\frac{5}{2})$

Answer 1

Answer: (A), (D)

The correct options are
A $(-\frac{7}{3},-2)$
D $(-\infty ,-\frac{5}{2})$

$\frac{1}{x+2}<\frac{1}{3x+7}$
$\Rightarrow \frac{1}{x+2}-\frac{1}{3x+7}<0$
$\Rightarrow \frac{3x+7-x-2}{(x+2)(3x+7)}<0$
$\Rightarrow \frac{2x+5}{(x+2)(3x+7)}<0$

Critical points are $-\frac{5}{2},-\frac{7}{3},-2$


Therefore, $x\in (-\infty ,-\frac{5}{2})\cup (-\frac{7}{3},-2)$