解：（I ）21()ln (0)2
f x ax x x =+> 1
'()'(1)1
f x ax x f a =+=+ ·
······················································································ 2分 依题意得11,2a a +=-=- ········································································ 4分
（Ⅱ）211'()ax f x ax x x
+=+= 0,'()0x f x >∴≥等价于210ax +≥ ·
······················································· 6分 ①当0a ≥时210ax +≥恒成立，
()f x ∴的单调递增区间为(0,)+∞ ·
······························································ 8分 ②当0a <时，由210ax +≥得a a x --≤≤- 0,0a x x a
->∴<≤- ()f x ∴的单调递增区间为(0,]a a
-- ·························································· 11分 综上所述：当0a ≥时()f x 的单调递增区间为(0,)+∞；
当0a <时，()f x 的单调递增区间为(0,]a a
-- ··········································· 12分 22．本题主要考查直线与椭圆的位置关系等基础知识；考查运算求解能力及化归与转化思想。满分14分。
解：（I ）设椭圆E 的方程为22
221()x y a b a b
+=> 由已知得：
112
a c c a -=???=?? ······························································································ 2分 21
a c =?∴?=? 2223
b a
c ∴=-=
∴椭圆E 的方程为22
143
x y += ··································································· 4分 （Ⅱ）设1122(,),(,)P x y Q x y ，线段PQ 中点T 的坐标为
00(,)x y ，则：
第6页 共8页 由22
143(1)x y y k x ?+=???=-?
得22234(1)12x k x +-= 化简得：
2222(34)8(412)0k x k x k +-+-=……5分
直线:(1)(0)l y k x k =-≠过点(1,0)
而点(1,0)在椭圆E 内，0∴?>
221212228412,4343
k k x x x x k k -+==++ ······························································ 6分 2002243,4343
k k x y k k -∴==++ 所以PQ 中垂直'l 的方程为：222143()4343
k k y x k k k =---++ 所以直线'l 在y 轴上的截距213
434k b k k k
==++ ·········································· 8分