设f(x)=x3+ax2+bx+1的导数f′(x)满足f′(1)=2a,f′(2)=﹣b,其中常数a,b∈R.(1)求曲线y=f(x)在点(1,f(1))处的切线方程.(2)设g(x)=f′(x)e﹣x.求函数g(x)的极值.
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