A) \[abc>1\]
B) \[abc>-8\]
C) \[abc<-8\]
D) \[abc>-2\]
Correct Answer: B
Solution :
[b] \[\Delta >0\] \[abc+2>3{{\left( abc \right)}^{1/3}}\] Let \[{{\left( abc \right)}^{1/3}}=x\] \[{{x}^{3}}+1>3x\] \[{{\left( x-1 \right)}^{2}}\left( x+2 \right)>0\] \[\therefore x+2>0\] \[x>-2\] \[{{\left( abc \right)}^{1/3}}>-2\] \[abc>-8\]You need to login to perform this action.
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