题目
题型:解答题难度:一般来源:不详
(Ⅰ)求实数a的值;
(Ⅱ)若函数g(x)=
f(x) |
x |
9 |
2(x+1) |
(Ⅲ)若f(x)>t(x-1)(t∈Z)对任意x>1恒成立,求t的最大值.
答案
故f′(e)=3,
即a+lne+1=3,
∴a=1.
(2)∵g(x)=
x+xlnx |
x |
9 |
2(x+1) |
=1+lnx+
9 |
2(x+1) |
∴g′(x)=
1 |
x |
9 |
2(x+1)2 |
(2x-1)(x-2) |
2x(x+1)2 |
令g′(x)=0,解得x=
1 |
2 |
列表如下
f(x) |
x |
9 |
2(x+1) |
x+xlnx |
x |
9 |
2(x+1) |
9 |
2(x+1) |
1 |
x |
9 |
2(x+1)2 |
(2x-1)(x-2) |
2x(x+1)2 |
1 |
2 |