高三文科数学普通高中毕业班质量检查模拟试卷及答案(4)

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

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