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