题目
题型:解答题难度:一般来源:不详
(Ⅰ)当a=2时,证明函数f(x)在R上是增函数;
(Ⅱ)若a>2时,当x≥1时,f(x)≥
x2-2x+1 |
ex |
答案
f′(x)=e-x-xe-x+ex-2+(x-2)ex-2=(x-1)(ex-2-e-x)=e-x(x-1)(ex-1-1)(ex-1+1).
当x≥1时,x-1≥0,ex-1-1≥0,所以f′(x)≥0,
当x<1时,x-1<0,ex-1-1<0,所以f′(x)≥0,
所以对任意实数x,f′(x)≥0,
所以f(x)在R上是增函数;
(II)当x≥1时,f(x)≥
x2-2x+1 |
ex |
设h(x)=(x-2)e2x-a-x2+3x-1(x≥1),则h′(x)=(2x-3)(e2x-a-1),
令h′(x)=(2x-3)(e2x-a-1)=0,解得x1=
3 |
2 |
a |
2 |
(1)当1<
a |
2 |
3 |
2 |