A) \[x\in (-\infty ,-a)\]
B) \[x\in (a,b)\]
C) \[x\in (-\infty ,a)\cup (b,\infty )\]
D) \[-b<x<-a\]
Correct Answer: D
Solution :
(d): \[{{x}^{2}}+\left( a+b \right)x+ab<0\Rightarrow \left( x+a \right)\left( x+b \right)<0\] \[\Rightarrow x+a<0,x+b>0\] or \[x+a>0,\] \[x+b<0,\Rightarrow x<-a,x>-b\] or \[x>-a,x<-b\Rightarrow x\in \left( -b,-a \right)\] or \[x\in \left( -a,-b \right)\] But since, \[a<b\therefore -a>-b,\] Hence, \[-b<x<-a\] holds true.
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