Question

Set of values of x satisfying the inequality $\frac{x^2+6x-7}{|x+4|}<0$ is / are

• A. $(-\infty ,-7)$
• B. $(-7,-4)$
• C. $(-7,-4)\cup (-4,1)$
• D. $(1,\infty )$

Answer 1

Answer: (C)

$\frac{x^2+6x-7}{|x+4|}<0\Rightarrow x^2+6x-7<0$ provided
$x+4\ne 0[\because |x+4|>0$ if $x\ne -4]$
$\Rightarrow (x+7)(x-1)<0,x\ne -4$
$\Rightarrow -7<x<1,x\ne -4$
$\therefore x\in (-7,-4)\cup (-4,1)$