厦门市2017届高中毕业班第一次质量检查
数学（理科）试题 2017.03
本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分．满分150分,考试时间120分钟．
第Ⅰ卷（选择题 共60分）
一、选择题：本大题共12小题，每小题5分，在每小题给出的四个选项中，只有一项是符合题目要
求的．
1. 已知集合{}
2560A x x x =--≤，11B x x ??
=>??-??
0，则A B 等于
A. [1
6]-， B. (16]， C. [1+)-∞， D. [23]， 2.已知复数i
i
a z -+=
1（其中i 为虚数单位），若z 为纯虚数，则实数a 等于 A. 1- B. 0 C. 1 D. 2
3. ABC ?的内角A ，B ，C 的对边分别为a ，b ，c ，若4523A a b =?==，，，则B 等于 A. 30? B. 60? C. 30?或150? D. 60?或120?
4. 若实数x y ，满足条件1
230x x y y x
≥??
-+≥??≥?
，则1y z x =+的最小值为
A.
1
3
B. 12
C. 34
D. 1
5.已知平面α⊥平面β，=l αβ ，直线m α?，直线n β?，且m n ⊥，有以下四个结论： ① 若//n l ，则m β⊥ ② 若m β⊥，则//n l
③ m β⊥和n α⊥同时成立 ④ m β⊥和n α⊥中至少有一个成立 其中正确的是
A ．①③
B ． ①④
C ． ②③
D ． ②④
6．已知Rt ABC ?，点D 为斜边BC 的中点，63AB = ，6AC = ，12
AE ED = ，则AE EB ?
等
于
A. 14-
B. 9-
C. 9
D.14
7.抛物线24y x =的焦点为F ，点(3,2)A ，P 为抛物线上一点，且P 不在直线AF 上，则PAF ?周长的最小值为
A. 4
B. 5
C. 4+22
D.5+5
8.某校高三年级有男生220人，学籍编号1，2，…，220；女生380人，学籍编号221，222，…，600.为了解学生学习的心理状态，按学籍编号采用系统抽样的方法从这600名学生中抽取10人进行问卷调查（第一组采用简单随机抽样，抽到的号码为10），然后再从这10位学生中随机抽取3人座谈，则3人中既有男生又有女生的概率是 A ．
15 B. 310 C. 710 D.45
9.二分法是求方程近似解的一种方法，其原理是“一分为二，无限逼近”.执行如图所示的程序框图，若输入12120.1x x d ===，，，则输出n 的值为
12()0?
x x d f m -<=或结束
1x m =1()()0?
f m f x <2x m
=12
2
x x m +=
2()2f x x =-12d x x 输入精确度和，的值
1
n =，是
否
是
否
1
n n =+n
输出开始
A.2
B.3
C.4
D. 5
10.已知定义在(0,)+∞上连续可导的函数()f x 满足'()()xf x f x x +=，且(1)1f =，则
A. ()f x 是增函数
B.()f x 是减函数
C. ()f x 有最大值1
D. ()f x 有最小值1
11.已知双曲线22
221(,0)x y a b a b
-=>，过x 轴上点P 的直线l 与双曲线的右支交于N M ，两点（M
在第一象限），直线MO 交双曲线左支于点Q （O 为坐标原点），连接QN .若60MPO ∠=?，
30MNQ ∠=?，则该双曲线的离心率为
A.
2 B.
3 C. 2 D. 4
12．已知P ，Q 为动直线2
(0)2
y m m =<<
与sin y x =和cos y x =在区间[0,]2π上的左，右两
个交点，P ，Q 在x 轴上的投影分别为S ，R .当矩形PQRS 面积取得最大值时，点P 的横坐标为0x ，则 A ．08
x π
< B. 08
x π
=
C.
08
6
x π
π
<<
D.06
x π>
第Ⅱ卷(非选择题 共90分) 二、填空题：本大题共4小题，每小题5分，共20分．
13．5
1(2+)x x 的展开式中，x 的系数为___________ 14．化简：
00
13
cos80sin80-=____________
15．某三棱锥的三视图如图所示，则其外接球的表面积为______ 16．若实数a ，b ，c 满足22(21)(ln )0a b a c c --+--=，则b c
-的最小值是_________
三、解答题：本大题共6小题，共70分，解答应写出文字说明、证明过程或演算步骤． 17．（本小题满分12分） 已知数列{}n a ，满足11a =，1323
n n n a a a +=
+，*
n N ∈.
（Ⅰ）求证：数列1n a ??
?
???
为等差数列； （Ⅱ）设212233445212221
111111
n n n n n T a a a a a a a a a a a a -+=
-+-++-
，求2n T .
18．（本小题满分12分）
为了响应厦门市政府“低碳生活，绿色出行”的号召，思明区委文明办率先全市发起“少开一天车，呵护厦门蓝”绿色出行活动，“从今天开始，从我做起，力争每周至少一天不开车，上下班或公务活动带头选择步行、骑车或乘坐公交车，鼓励拼车……”铿锵有力的话语，传递了低碳生活、绿色出
行的理念。某机构随机调查了本市500名成年市民某月的骑车次数，统计如下：
人数 次数 年龄 [0,10) [10,20) [20,30) [30,40) [40,50) [50,60] 18岁至30岁
6
14
20
32
40
48
31岁至44岁 4 6 20 28 40 42 45岁至59岁 22 18 33 37 19 11 60岁及以上
15
13
10
12
5
5
联合国世界卫生组织于2013年确定新的年龄分段：44岁及以下为青年人，45岁至59岁为中年人，60岁及以上为老年人．记本市一个年满18岁的青年人月骑车的平均次数为μ.以样本估计总体. （Ⅰ）估计μ的值；
（Ⅱ）在本市老年人或中年人中随机访问3位，其中月骑车次数超过μ的人数记为ξ，求ξ的分布列与数学期望.
19．（本小题满分12分）
在如图所示的六面体中，面ABCD 是边长为2的正方形，面ABEF 是直角梯形，90FAB ∠=
，
//AF BE ，24BE AF ==.
（Ⅰ）求证：AC //平面DEF ；
（Ⅱ）若二面角E AB D --为60
，求直线CE 和平面DEF 所成角的正弦值.
20.（本小题满分12分）
已知函数()ln 1()f x x kx k R =-+∈. （Ⅰ）讨论函数()f x 的零点个数；
（Ⅱ）当1k =时，求证：12()2x
f x x e -≤--恒成立.
21．（本小题满分12分）
已知椭圆22223:1(2)43x y C b b +=<< ，动圆P ：22004
()()3
x x y y -+-= （圆心P 为椭圆C 上异
于左右顶点的任意一点），过原点O 作两条射线与圆P 相切，分别交椭圆于M ，N 两点，且切线长
的最小值为
63
. （Ⅰ）求椭圆C 的方程；
（Ⅱ）求证：MON ?的面积为定值.
22.（本小题满分10分）选修4-4：坐标系与参数方程
在直角坐标系xOy 中，曲线1C ：27cos 7sin x y αα?=+??=??，
．
（α为参数）．以O 为极点，x 轴的正半轴为
极轴建立极坐标系，曲线2C 的极坐标方程为θρcos 8=，直线l 的极坐标方程为)(3
R ∈=ρπ
θ．
（Ⅰ）求曲线1C 的极坐标方程与直线l 的直角坐标方程；
（Ⅱ）若直线l 与1C ，2C 在第一象限分别交于A ，B 两点，P 为2C 上的动点， 求PAB ?面积的最大值．
23.（本小题满分10分）选修4-5：不等式选讲
已知函数()1(1)f x x x m m =-+-> ，若()4f x >的解集是{}
04x x x <>或. （Ⅰ）求m 的值；
（Ⅱ）若关于x 的不等式4)(2
-+ 厦门市2017届高中毕业班第一次质量检查 数学（理科）试题答案 一．选择题：本大题共12小题，每小题5分，在每小题给出的四个选项中，只有一项是符合题目要 求的． 1-6 BCDBBD 7-12 CDCDAA 12.解析：由题意知，P 与Q 关于直线4 x π = 对称，设(,sin )P x x ，则 (,sin )2 Q x x π -， ()(2)sin (0)24 S x x x x ππ ∴=-<<， '()2sin (2)cos 2S x x x x π∴=-+-，''()4cos (2)sin 2 S x x x x π ∴=---， 04x π << ，''()0S x ∴<，'()S x ∴在区间(0,)4π上单调递减，且'(0)02 S π =>， '()204S π=-<，'()S x ∴在区间(0,)4 π 存在唯一零点，即为0x . 令'0()0S x =得：0002sin ( 2)cos 2 x x x π =-，即00tan 4 x x π =-. 由不等式000tan (0)2x x x π><< 得： 004x x π ->，解得：08 x π < ，故选A. 命题意图：考查三角函数的图象与性质、导数、零点、不等式等，考查数形结合思想、转化与化归思想，考查逻辑推理能力. 二、填空题：本大题共4小题，每小题5分，共20分． 13. 40 14. 4 15. 323 π 16. 1 15.解析：由三视图可得三棱锥A BCD -的直观图如图所示，取BC 的中点E ，连接AE ， 设O 为ABC ?的外心，R 为三棱锥的外接球的半径，则O 在线段AE 上， 因为222OD OE DE =+，即222(6)(2)R R =-+，解得：26 3 R =， 所以，32.3 S π = 16.解析：（法一）由22(21)(ln )0a b a c c --+--=得：21, ln .a b a c c =+??=+? 在坐标系中考察函数()21f x x =+与()ln g x x x =+的图象， 所以，b c -的最小值等价于直线y a =与函数()21f x x =+， ()ln g x x x =+交点横坐标之间距离的最小值.设直线2:2l y x m =+与() g x 相切于点00(,)B x y ，则'0()2g x =，解得：01x =， 所以，(0,1),(1,1)A B ，故min 1b c -=. （法二）由22(21)(ln )0a b a c c --+--=得： 21,ln ,a b a c c =+??? =+? ln 1 2c c b +-=，则 ln 1ln 122c c c c b c c +----= -=，令ln 1()2c c f c --=，则'11 ()(1)2f c c =-， 当'()0f c =时，1c =；当'()0f c >时，01c <<；当'()0f c <时，1c >； 所以，()f c 在(0,1]单调递增，在[1,)+∞单调递减，故()(1)1f c f ≤=-?()1f c ≥， 所以，min min () 1.b c f c -== 三、解答题：本大题共6小题，共70分，解答应写出文字说明、证明过程或演算步骤． 17．本题以递推数列为背景考查等差数列的判定以及利用基本量的求和运算，（Ⅰ）重点考查利用数列递推形式构造等差或等比数列以及等差数列的判定方法；（Ⅱ）主要考查数列求和应首先探寻通项公式，通过分析通项公式的特征发现求和的方法. 满分12分． （Ⅰ）证明：法一：由1323n n n a a a += +，得 1231 12=33 n n n n a a a a ++=+ ····························· 3分 1112 =3 n n a a +∴ - ∴数列1n a ?????? 是首项为1，公差为2 3的等差数列 ··················································· 5分 法二：由1323 n n n a a a += +得 111=n n a a +-2313n n n a a a +- ························································································ 3分 1212 =( )=33 n n a a +- ·································································································· 4分 ∴数列1n a ?????? 是首项为1，公差为2 3的等差数列 ··················································· 5分 （Ⅱ）解：设2122212121211111 =()n n n n n n n n b a a a a a a a -+-+= - - ······························· 7分 由（Ⅰ）得，数列1n a ??? ? ?? 为公差为2 3的等差数列 ∴ 21 2111 4 =3n n a a -+- - 即21 21 2211141 =( ) 3n n n n n b a a a a -+-=-? ····································································· 8分 12224114416 ()3339n n n n b b a a ++-=--=-?=-， 且1 214141220 ()3339 b a a =-?=-?+=- {}n b ∴是首项1209b =- ，公差为16 9 -的等差数列 ············································ 10分 21220(1)16 ()929 n n n n T b b b n -∴=+++=-+?- 24 (23)9 n n =-+ ····························································································· 12分 18.本小题主要考查对频数分布表的理解与应用，古典概型、随机变量的数学期望等基础知识，考查运算求解能力、数据处理能力、应用意识，考查必然与或然思想、化归与转化思想．满分12分． 解：（Ⅰ）由已知可得下表 人数 次数 年龄 [0,10) [10,20) [20,30) [30,40) [40,50) [50,60] 合计 青年人 10 20 40 60 80 90 300 中年人 22 18 33 37 19 11 140 老年人 15 13 10 12 5 5 60 本市一个青年人月骑车的平均次数： 10204060809012000 5152535455540300300300300300300300 μ=? +?+?+?+?+?==. ················································································································································· 5分 （Ⅱ）本市老年人或中年人中月骑车时间超过40次的概率为 19115+51 140605 ++=+. ······ 7分 0,1,2,3ξ=，1~(3)5B ξ,，故()3314,0,1,2,355k k k P k C k ξ-?? ?? =? ? ??? ?? ＝=. · ··················· 9分 所以ξ的分布列如下： ξ 0 1 2 3 P 64 125 48125 12 125 1125 ··············································································································································· 11分 ()1 30.65 E ξ=?=. ············································································································ 12分 19．本题考查立体几何中的线面关系，空间角，空间向量在立体几何中的应用等基础知识，考查运算求解能力、空间想象能力、等价转化能力，考查数形结合思想、化归与转化、或然与必然等数学思想．满分12分． 证明：（1）法一：连接,AC BD 相交于点O ，取DE 的中点为G ,连接,FG OG . ABCD 是正方形，O ∴是BD 的中点，1 //,2 OG BE OG BE ∴= , 又因为1 //,2 AF BE AF BE = ,所以//OG AF 且OG AF =， 所以四边形AOGF 是平行四边形，· ·················································································· 3分 //AC FG ∴，又因为FG ?平面DEF ,AC ?平面DEF //AC ∴平面DEF · ·············································································································· 5分 法二：延长,BA EF 相交于点G ,连接GD . 因为1 //,2 AF BE AF BE = ， A ∴是BG 的中点，所以//DC GA 且DC GA =， 所以四边形ACDG 是平行四边形，· ················································································· 3分 //AC GD ∴，又因为GD ?平面DEF ,AC ?平面DEF //AC ∴平面DEF · ·············································································································· 5分 （2）ABCD 是正方形，ABEF 是直角梯形，90FAB ∠= ， ,DA AB FA AB ∴⊥⊥ AD AF A = ,AB ∴⊥平面AFD ,同理可得AB ⊥平面 EBC . 又AB ? 平面ABCD ，所以平面AFD ⊥平面ABCD , 又因为二面角E AB D --为60 ， 所以60FAD EBC ∠=∠= ,24BE AF ==,2BC =,由余弦定理得23EC =, 所以EC BC ⊥,又因为AB ⊥平面EBC ，EC AB ∴⊥,所以EC ⊥平面ABCD ， ················································································································································· 7分 法一：以C 为坐标原点，CB 为x 轴、CD 为y 轴、CE 为z 轴建立空间直角坐标系.则(0,0,0),(0,2,0),(0,0,23),(1,2,3)C D E F , ································································ 8分 所以(0,0,23), (1,0,3),(1,2,3)C E D F E F ===- ,设平面DEF 的一个法向量为 (,,)n x y z = ，则00n DF n EF ??=???=?? 即30 230 x z x y z ?+=??+-=??令3z =，则33x y =-??=?， 所以(3,3,3)n =- · ··········································································································· 11分 设直线CE 和平面DEF 所成角为θ， 则67 sin cos ,7 2321CE n θ=<>= = ? ································································ 12分 B A C D F E x y z G B A C D F E 法二：取AD 的中点为M ,BC 的中点为N ，连接,FM MN .以M 为坐标原点，MD 为x 轴、MN 为y 轴、MF 为z 轴建立空间直角坐标系. 则(1,2,0),(1,2,23),(0,0,3),(1,0,0)C E F D · ····························································· 8分 以(0,0,23),(1,0,3),(1,2,3)CE DF FE ==-= , 设平面DEF 的一个法向量为(,,)n x y z = ， 则00n DF n FE ??=???=?? 即30230 x z x y z ?-+=??++=??令3z =，则33x y =??=-?， 所以(3,3,3)n =- · ··········································································································· 11分 设直线CE 和平面DEF 所成角为θ，则67 sin cos ,7 2321CE n θ=<>= = ? ··············································································································································· 12分 20.本题考查函数单调性、极值、零点的基础知识，考查学生运算求解与推理论证的能力，运用导数工具解决函数与方程、不等式综合问题的能力。考查数形结合，分类与整合，转化与化归的数学思想。 解：（1）法1：由已知ln 1 0x x k x +>∴= ， ··································································· 1分 令ln 1(x)x g x +=，2 21(lnx 1)ln (x)0x g x x -+-'=== ····················································· 2分 (0,1),(x)0,(x)x g g '∈>单调递增 (1,),(x)0x g '∈+∞<，(x)g 单调递减 max (x)=(1)=1g g ∴ ················································································································ 3分 (x)0x g →+∞→， 0(x)x g →→-∞， 综上：0k ≤或1k =时，有1个零点 01k <<时，有2个零点 1k >时，有0个零点 ··········································································································· 5分 法2：， 11()kx f x k x x -'= -=， ····················································································· 1分 0k ≤时，()0,(x)f x f '>单调递增 B E F D C A x z y M N 0(x)x f →→-∞，，(x)x →+∞→+∞，f ，所以，有1个零点 ···························· 2分 0k >时，11 ()0,0kx f x x x k -'= ==>， 1 (0,),(x)0,(x)x f f k ∈>单调递增 1(,),(x)0x f k '∈+∞<，(x)f 单调递减，max 11(x)()ln()f f k k == 1k >时，max 1 (x)ln()0f k =<，0个零点 1k =时，max 1 (x)ln()0f k ==，1个零点 · ········································································ 4分 01k <<时，max 1 (x)ln()0f k =>， 0(x)x f →→-∞，，(x)x →+∞→-∞，f ，所以，此时有2个零点 综上：0k ≤或1k =时，有1个零点 01k <<时，有2个零点 1k >时，有0个零点 · ·········································································································· 5分 （2）证明： 法1：要证12(x)2x f x e -≤-- 即证112(x)22ln x 0x x e f x x e --+--=-+≤ 令1(x)2ln x x g x e -=-+， 1122(x)-1x x x xe g e x x ----'=+-= ··················································································· 7分 1111(x)2,(x)11(x 1)x x x x h x xe h e xe e ----'=--=--+=-+-令 1(0,1),(x)1(x 1)0x x h e -'∈=-+-< 1 11 x 1(1,+),(x)1(x 1)x x x e x h e e -----'∈∞=-+-= 11(x)=x 1(x)=10x x m e m e --'---<令，9分 (x)0h '<即，(x)h ∴单调递减 ···························································································· 11分 12 (1)1=0,(0,1),(x)0,(x)x h e x h g x -=-+ -∈>单调递增，(x)(1)0,g g <= (1,+),(x)0,(x)x h g ∈∞<单调递减，(x)(1)0,g g <= P O M N y x 图1 综上：(x)(1)0g g ≤= ······································································································ 12分 法2要证12(x)2x f x e -≤-- 即证112(x)22ln x 0x x e f x x e --+--=-+≤ 令1(x)2ln x x g x e -=-+，12 (x)-1x g e x -'=+ - ······························································ 7分 令12(x)-1x h e x -=+-，211212212122(x)x x x x x e h e x e x x e -----'=-=-= 令211(x)2,(x)2x 2x x x e m e --'=-=-m 令111(x)2x 2,t (x)222(1)x x x e e e ---'=-=-=-t · ·························································· 9分 (0,1),t (x)0,(x)x m ''∈>单调递增，(1,+),t (x)0,(x)x m ''∈∞<单调递减 max (x)(1)0(x)0(x)m m m '''==∴≤∴m 单调递减，(0)=1m - (x)0(x)0(x)h h '∴<∴<∴m 单调递减 ··········································································· 11分 12 (1)1=0,(0,1),(x)0,(x)x h e x h g x -=-+ -∈>单调递增，(x)(1)0,g g <= (1,+),(x)0,(x)x h g ∈∞<单调递减，(x)(1)0,g g <= 综上：(x)(1)0g g ≤= · ····································································································· 12分 21．本题考查椭圆的标准方程的求法，直线与圆、直线与圆锥曲线的位置关系，考查推理运算和方程求解能力.运用化归转化手段. 将切线长最短问题转化为椭圆上的动点到定点距离最短问题；考查圆锥曲线中的有关定值问题，从变化中寻找不变量，并通过必要的推理和运算化简求值.考查转化化归思想、分类整合思想. （Ⅰ）解：如图1，因为2200214x y b +=，所以222 00(1)4 x y b =-， 由2323 b <<得：22 00x y +=2222204(1)43b x b b r -+≥>= 故点O 在圆P 外， 不妨设OM 与圆P 相切于T ，则有： 切线长 2 22 004433OT OP x y =- =+- · ········································································· 1分 P O M N y x 图2 A B O M N y x M 1 N 1 图3 代入得 2222044 (1)433 b OT x b b =-+-≥- · ················································································ 3分 由已知得：242 33 b - = ，解得：22b =， 所以椭圆的方程为：22 142 x y + = ··················································································· 4分 （Ⅱ）解：1°当切线OM 或ON 斜率不存在即圆P 与y 轴相切时，易得023 3 x r == ，代入椭圆方程得：023 3 y = ，说明圆P 同时也与x 轴相切（图2），此时M 、N 分别为长、短轴一个端点，则MON ?的面积为2. · ························································································································ 5分 2°当切线OM 、ON 斜率都存在时，设切线方程为：kx y =， 由r d =得： 3 2 12 00= +-k y kx ， 整理得：0436)43(2 0002 2 0=-+--y y kx k x （*）， · ······················································ 6分 由1°知：2 0340x -≠，即0233x ≠ ，此时023 3 y ≠，方程（*）必有两个非零根， 记为1212,()k k k k <,则12,k k 分别对应直线,OM ON 的斜率， 由韦达定理得：2012203434y k k x -?=-，将2 02024y x -=代入得：2 0122 0341862 y k k y -?==-- 7分 解法一：（求交点坐标） 由上知：210k k <<， 设点N 位于第一、三象限，点M 位于第二、四象限， 若点N 位于第一象限，点M 位于第二象限， 设OM ：x k y 1=与椭圆方程 12 4 22=+ y x 联立可得： )212,212(21 121 k k k M +- +- O M N y x 图4 设ON ：x k y 2=与椭圆方程 12 4 22=+ y x 联立可得： )212, 212( 22 222 k k k N ++ ··································································································· 9分 1 1 11 1111 ()()()()()2222 MON MOM MOM MM N N N M N M M M N N N M M N S S S S x x y y x y x y x y x y ???=--=-?+---=-梯形 ··············································································································································· 10分 代入坐标有： 122422)21)(21()(22121222 2122211 22 221222********* 2+++-+=++-=+?+-=?k k k k k k k k k k k k k k k k S MON 2) 1(21 22 2212 221=++++=k k k k 同理，当点M 、N 位于其它象限时，结论也成立 综上，MON ?的面积为定值2. ····················································································· 12分 解法二：（探寻直线MN 方程特征） （接上）设1122(,),(,)M x y N x y 由于点P 不与点A 、B 重合时，直线MN 的斜率存在，不妨设直线MN 的方程为：y kx m =+， 将MN 与椭圆方程联立可得：222 (12)4240k x kmx m +++-=， 22222216(816)(21)32168k m m k k m ?=--+=+-， 由0?>得2242k m +>， 122 2 1224122412km x x k m x x k ? +=-??+?-??=?+? ·········································································· 8分 121212122122212121 22()()2(12)2()20 OM ON i y y k k x x y y x x kx m kx m x x k x x km x x m ?= =-?+=+++=++++= 代入有：222 2 222 248(12)201212m k m k m k k -+-+=++ 整理得：2221m k =+； · ···································································································· 9分 又222222 2 1222232168168111122212121 k m k k MN k x x k k k k k +-++=+-=+=+=+++ m 1 4 y O x 而原点O 到直线MN 的距离为22 2 1211m k d k k += = ++ ·················································· 11分 222211121 22222211 MON k k S MN d k k ?++∴=?=??=++. 所以MON ?的面积为定值2. · ························································································ 12分 22．本题考查学生对直角坐标方程、参数方程、极坐标方程之间的相互转化，利用极坐标方程求解 弦长问题，三角形最值问题，通过直角坐标方程、参数方程、极坐标方程之间的互化考查化归与转化、数形结合的思想. 解：（Ⅰ）依题意得，曲线1C 的普通方程为7)2(22=+-y x ， 曲线1C 的极坐标方程为03cos 42=--θρρ， · ····························································· 3分 直线l 的直角坐标方程为x y 3=． ·························································· 5分 （Ⅱ）曲线2C 的直角坐标方程为16)4(22=+-y x ，由题意设)3,(1π ρA ，)3 ,(2π ρB ， 则033 cos 412 1=--π ρρ，即032121=--ρρ，得31=ρ或11-=ρ（舍）， 43 cos 82==π ρ，则121=-=ρρAB ， ························································ 7分 2C )0,4(到l 的距离为324 34== d ． 以AB 为底边的PAB ?的高的最大值为324+． 则PAB ?的面积的最大值为 32)324(12 1 +=+??． ······························ 10分 23.本题考查学生对绝对值不等式的理解与运用，考查学生对绝对值函数的运算求解能力，考查分类与整合、函数与方程思想和数形结合等思想．本题以绝对值函数为背景，设置学生熟悉的绝对值函数化为分段函数以及不等式求解问题． 解：（Ⅰ）解法一：1m > 211 ()1121x m x f x m x m x m x m -++ ∴=-≤≤??-->? ，，， ················································································ 1分 作出函数)(x f 的图象 ……………………………………3分 由4)(>x f 的解集为{} 40> ? ??=--?=++?-41424 102m m 得3=m …………………………………………5分 解法二：1m > 211()1121x m x f x m x m x m x m -++ ∴=-≤≤??-->? ，，， ················································································ 1分 ① 12+14x x m ? 得?? ? ??-<<231 m x x 1m > 得 312m -<，32 m x -∴< ······································································· 2分 ②? ? ?>-≤≤411m m x 得1(5)x m m ≤≤>，不合题意 · ·················································· 3分 ③ ???>-->412m x m x 得?? ???+>>25m x m x 当5m ≥时，x m >，不符合4x >，舍去 当15m <<时，52 m x +> ····················································································· 4分 综上不等式的解集为3522m m x x x ?-+? < >???? 或 302 542 m m -?=??∴?+?=??， 3m ∴= · ····················································· 5分 （Ⅱ）解法一：由（Ⅰ）得421 ()213243x x f x x x x - =≤≤??->? ，， ， 2)(min =x f ··················································································· 6分 4)(2-+ 422-+-+a a ························································ 8分 0)2)(3(>-+a a 23>- 23>- 422-+-+a a ············································ 8分 0)2)(3(>-+a a 23>-