A) \[x>11\]or\[x<-\frac{3}{2}\]
B) \[x>11\]or \[x<-1\]
C) \[-\frac{3}{2}<x<-1\]
D) \[-1<x<11\]or \[x<-\frac{3}{2}\]
Correct Answer: D
Solution :
Given, \[\frac{{{x}^{2}}+2x+7}{2x+3}<6\] \[\Rightarrow \] \[\frac{{{x}^{2}}+2x+7}{2x+3}-6<0\] \[\Rightarrow \] \[\frac{{{x}^{2}}-10x-11}{2x+3}<0\] \[\Rightarrow \] \[\frac{(x-11)\,(x+1)}{2x+3}<0\] \[\Rightarrow \] \[\frac{(x-11)\,(x+1)\,(2x+3)}{{{(2x+3)}^{2}}}<0\] \[\Rightarrow \] \[(x-11)\,(x+1)\,(2x+3)<0\] \[\Rightarrow \] \[x\in \left( -\,\infty ,-\frac{3}{2} \right)\cup (-1,\,11)\]
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