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f(x)=e^x-1-x-ax^2,f'(x)=e^x-1-2ax當a≤1/2時,f'(x)=e^x-1≥0在x≥0時恆成立,f(x)在[0,+∞)單調遞增,且f(0)=0所以當a≤1/2時,有x≥0時f(x)≥0當a>1/2時,將f(x)=e^x-1-x-ax^2分解為h(x)=e^x-1和g(x)=x+ax^2即f(x)=...
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