A) \[a<-2\]
B) \[a>-2\]
C) \[-3<a<0\]
D) \[-\infty <a<-3\]
Correct Answer: D
Solution :
For the function \[f(x)=(a+2){{x}^{3}}-3a{{x}^{2}}+9ax-1\] \[f'(x)=3(a+2){{x}^{2}}-6ax+9a\] \[\because \]Function is monotonically decreasing \[\forall x\in R\] \[\therefore \]\[f'(x)\le 0\,\forall x\in R\] \[3(a+2){{x}^{2}}-6ax+9a\le 0\,\forall x\in R\] \[(a+2){{x}^{2}}-2ax+3a\le 0\,\forall x\in R\] This is only possible when coefficient of x2 is less than zero and discriminate is negative. \[\Rightarrow \]\[a+2<0\]and\[4{{a}^{2}}-4(a+2)(3a)<0\] \[\Rightarrow \]\[a<-2\]and\[4{{a}^{2}}-12{{a}^{2}}-24a<0\] \[\Rightarrow \]\[a<-2\]and\[-8{{a}^{2}}-24\,a<0\] \[\Rightarrow \]\[a<-2\]and\[-8a(a+3)<0\] \[\Rightarrow \]\[a<-2\]and\[a(a+3)>0\] \[\therefore \]Required solution is \[(-\infty ,-3).\]You need to login to perform this action.
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