A) \[-1<x<1\]
B) \[0<x<2\]
C) \[x>3\]
D) \[1<x<2\]
E) \[1<x<3\]
Correct Answer: D
Solution :
\[\because \] \[f(x)=2{{x}^{3}}-9{{x}^{2}}+12x+4\] On differentiating w.r.t.\[x,\]we get \[f(x)=6{{x}^{2}}-18x+12\] For function to be decreasing \[f(x)<0\] \[\Rightarrow \]\[6({{x}^{2}}-3x+2)<0\] \[\Rightarrow \] \[({{x}^{2}}-2x-x+2)<0\] \[\Rightarrow \] \[(x-2)(x-1)<0\] \[\Rightarrow \] \[1<x<2\]You need to login to perform this action.
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