A) \[(-\infty ,-7)\]
B) \[(-7,4)\]
C) \[(-4,1)\]
D) \[(1,\infty )\]
Correct Answer: C
Solution :
[c] \[\frac{{{x}^{2}}+6x-7}{\left| x+4 \right|}<0\] \[\Rightarrow {{x}^{2}}+6x-7<0,\] provided \[x+4\ne 0\] \[[\because \,\,\,\,\left| x+4 \right|>\,0\,\,if\,\,x\ne -\,4]\] \[\Rightarrow (x+7)(x-1)<0,x\ne -4\Rightarrow -7<x<1,\]\[x\ne -4\] \[\therefore x\in (-7,-4)\cup (-4,1)\]You need to login to perform this action.
You will be redirected in
3 sec