f′(x)=
2 |
x |
2a |
e |
2(e−ax) |
ex |
当a=0时,由f′(x)=
2 |
x |
当a>0时,由f′(x)=
2(e−ax) |
ex |
e |
a |
当a<0时,由f′(x)=
2(e−ax) |
ex |
e |
a |
所以当a=0时,函数f(x)的递增区间是(0,+∞);
当a>0时,函数f(x)的递增区间是(0,
e |
a |
当a<0时,函数f(x)的递增区间是(-∞,
e |
a |
(Ⅱ)因为f′(x)=
2 |
x |
2 |
e |
2(e−x) |
ex |
所以以p1(x1,f(x1))为切点的切线的斜率为
2(e−x1) |
ex1 |
以p2(x2,f(x2))为切点的切线的斜率为
2(e−x2) |
ex2 |
又因为切线过点p(0,t),
所以t−lnx12+
2x1 |
e |
2(e−x1) |
ex1 |
2x2 |
e |
2(e−x2) |
ex2 |
解得,x12=et+2,x22=et+2.则x12=x22.
由已知x1≠x2
所以,x1+x2=0.