1?2yM?21?4k23?8k2所以kBM?,............... ....................................10分 ???4kxM?04k?1?4k23?8k21???k?0?, 即?4kk12亦即k2?,所以k??, ............... ...........................................12分
84故EF的方程为?2x?4y?4?0. ............... ...........................................13分
14222的圆心O?0,0?到直线EF的距离为d?, ??23218所以直线EF与圆相离.
............... ...........................................14分
又圆x2?y2?
20.(I)解:f?x??a2x2?ax?lnx,
12a2x2?ax?1?ax?1??2ax?1?f??x??2ax?a????x?0?, xxx2 ............... ...........................................2 分 所以,a?0时,f?x?与f??x?的变化情况如下:
x ?1??0,? ?2a?1 2a?1?,???? 2a??f??x? f?x? - 0 + ↘ ↗ ?1??1?因此,函数f?x?的单调递增区间为?,???,单调递减区间为?0,?.
?2a??2a? ............... ...........................................4分 (II)证明:g?x??a2x2?f?x??lnx?ax,
g??x??1?a, x所以g??1??1?a,
所以l的斜率kl?1?a.
因为l?//l,且l?在y轴上的截距为1,
所以直线l?的方程为y??1?a?x?1................ ...........................................6分
令h?x??g?x?????1?a?x?1???lnx?x?1?x?0?,
则无论a取任何实数,函数g?x?的图像恒在直线l?的下方,等价于h?x??0??a?R,?x?0?, ............... ...........................................7分
11?xh??x???1?xx. 而
x??1,???h??x??0h?x?0时,??,当时,, hx0,11,???所以函数??的??上单调递增,在?上单调递减,
hxh1??2从而当x?1时,??取得极大值??,
0,???h?x?h1??2即在?上,取得最大值??,.....................................................8分
hx??2?0??a?R,?x?0?所以??,
gx因此,无论a取任何实数,函数??的图像恒在直线l?的下方.
............... ...........................................9分
当
x??0,1?
(III)因为所以
A?1,?a?,Q?x0,lnx0?ax0?,
kQA?lnx0?ax0?alnx0??ax0?1x0?1,
lnx0?a?2x?1x?1所以当0时,0, lnx0??a?2??x0?1??0即恒成立. ............... ...........................................10分 rx?lnx??a?2??x?1??x?1?令??,
1r??x????a?2?x则,
10??1x因为x?1,所以.
r?x?0(i)当a??2时,a?2?0,此时??,
rx1,???rx?r?1??0所以??在?上单调递增,有??不满足题意; (ii)当?2?a??1时,0?a?2?1,
1???1?x??1,x?,?????r??x??0a?2a?2????时,r??x??0, 所以当时,,当
1??t??1,?a?2??,使得r?t??r?1??0不满足题意; 所以至少存在
r?x?0(iii)当a??1时,a?2?1,此时??,
rx1,???rx?r1?0所以??在?上单调递减,????,满足题意.
综上可得a??1,
?1,???故所求实数a的取值范围是?.
............... ...........................................13分