(2)证明:(ⅰ)g(x)=f'(x)=ex(x-a+1)+(a-1),g'(x)=ex(x-a+2)------------------(5分)
当g'(x)<0时,x<a-2;当g'(x)>0时,x>a-2
因为a>2,所以函数g(x)在(0,a-2)上递减;在(a-2,+∞)上递增-----------------(7分)
又因为g(0)=0,g(a)=ea+a-1>0,
所以在(0,+∞)上恰有一个x0使得g(x0)=0.--------------------------------------------------(9分)
望采纳 谢谢
一般来说。第一题会了。第二题也能解
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