A) \[(-\infty ,0)\]
B) \[(-\infty ,\infty )\]
C) \[(-4,-2)\]
D) \[(-1,0)\cup (0,\infty )\]
Correct Answer: D
Solution :
[d] \[f(x)={{x}^{3}}+{{x}^{2}}+2x-1\] \[f'(x)=3{{x}^{2}}+2x+2>0\,\forall x\in R\] So, \[f(x)\] is an increasing function. Now \[f(f(x))>f(2x-1)\] \[\Rightarrow \,\,\,\,\,f(x)>2x-1\] \[\Rightarrow \,\,\,\,{{x}^{3}}+{{x}^{2}}+2x-1>2x-1\] \[\Rightarrow \,\,\,\,{{x}^{2}}(1+x)>0\] \[\Rightarrow \,\,\,\,x\in (-1,0)\cup (0,\infty )\]
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