………………………………………11分4422uuuruuurxx2而QAx1a11QBx2a21442uuuuuurrxx2所以QAQBx1ax2a11214422xxxx22x1x2x1x2ax1x2a2121211644a2842a2a2110…………………………………13分411x21Ⅰf′x1……………………………………………………………1分xxy
在区间01上f′x0函数fx单调递增；在区间1∞上f′x0函数函数
fx单调递减∴当x1时fx取最大值f11…………………………………………………3分
Ⅱ直线PP2的斜率为k1
ax2l
x2ax1l
x1l
x2l
x1……………………4分ax2x1x2x1
f34515526docPage8of8
由Ⅰ的结论知fxxl
x≤1且仅当x1时取等号∴
x2xxxxxl
x2l
x11l
21l
221l
x2l
x121x1x1x1x1x1x2x1x1

x1xxxxxl
x2l
x11l
11l
111l
x2l
x121……7分x2x2x2x2x2x2x1x21l
x2l
x1111akax2x2x1x1x2x1111满足∈aa所以fx图象上存在点P0x0y0，xx2x1
∴
又在x1x2上f′xa
x1x0x2，且fx图象上以P0为切点的切线与直线PP2平行………………………8分1
Ⅲfx
33131xl
xf′x∴a
1…………………………………9分22x2a

a3
13a263131312a22a23a220a42a22a323123a222a2
2a21a220a22
3120a12……………………………11分2a1
下面我们证明当0a12时x2
2x2
且x2
2
∈N事实上当
1时0a12a2
3122a1
a4a2
13a2632a21a22a20a4a2结论成立…………12分23a2223a22
若当
k时结论成立即x2k2x2k且x2k2
∈N则
x2k1
31312x2k222x2k2x2k113a2k2632a2k21a2k22a2k20a2k4a2k223a2k2223a2k22
a2k4a2k2
由上述证明可知a1的取值范围是02……………………………………………14分
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