A) \[x\ge 0\]
B) \[x\le 0\]
C) \[x\ge -1\]
D) \[-1\le x\le 1\]
Correct Answer: C
Solution :
\[{{x}^{3}}+1\ge {{x}^{2}}+x\] \[\Rightarrow \] \[{{x}^{3}}-{{x}^{2}}-x+1\ge 0\] \[\Rightarrow \] \[{{(x-1)}^{2}}(x+1)\ge 0\] \[\Rightarrow \] \[x\ge 1-1,x\ge 1\]
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