A) \[x<-2\]
B) \[x>0\]
C) \[-3<x<0\]
D) \[-3<x<4\]
Correct Answer: B
Solution :
Given,\[x+2>\,\sqrt{x+4}\]\[\Rightarrow {{(x+2)}^{2}}\,\,>(x+4)\] \[\Rightarrow {{x}^{2}}+4x+4>x+4\]\[\Rightarrow {{x}^{2}}+3x>0\] \[\Rightarrow x(x+3)>0\] Þ x < - 3 or x > 0 \[\Rightarrow \,x>0\].
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