问题描述: 已知f(x)=2x-12 1个回答 分类:数学 2014-09-25 问题解答: 我来补答 (I)f′(x)=2−x,g′(x)=1xlna∵h(x)=f(x)-g(x)在定义域上为减函数∴h′(x)≤0在(0,+∞)上恒成立即1lna≥−x2+2x在(0,+∞)上恒成立即1lna≥ ( −x2+2x)maxx∈(0,+∞)令u(x)=-x2+2x=-(x-1)2+1≤1∴1lna≥1∵h′(x)存在零点∴x2−2x+1lna=0在(0,+∞)上有根∴△=4(1−1lna)≥0∴1lna≤1∴lna=1即a=e(II)∵g(x)=lnx,p(x)=ex令F(x)=ex(x−x2)−ex+ex2(x<x2)F′(x)=ex+exx-x2ex-ex=(x-x2)ex<0∴F(x)在(-∞,x2)上递减∴ex1(x1−x2)>ex1−ex2即ex1<ex1−ex2x1−x2同理ex1−ex2x1−x2<ex2所以有P(x1)<P(x0)<P(x2) 展开全文阅读