准考证号 姓名
(在此卷上答题无效)
机密★启用前
2016年福建省普通高中毕业班质量检查
理科数学
本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.第Ⅰ卷1至3页,第Ⅱ卷4至6页,满分 150分.
考生注意:
1.答题前,考生务必将自己的准考证号、姓名填写在答题卡上.考生要认真核对答题卡上粘贴
的条形码的“准考证号、姓名、考试科目”与考生本人准考证号、姓名是否一致.
2.第Ⅰ卷每小题选出答案后,用2B 铅笔把答题卡上对应题目的答案标号涂黑,如需改动,用
橡皮擦干净后,再选涂其他答案标号.第Ⅱ卷用0.5毫米的黑色墨水签字笔在答题卡上书写作答.若在试题卷上作答,答案无效.
3.考试结束,监考员将试题卷和答题卡一并交回.
第Ⅰ卷
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,a b ∈R ,i 是虚数单位,若i a +与2i b -互为共轭复数,则
2(i)a b +=
(A )34i - (B )34i +
(C )54i -
(D )54i +
(2)执行如图所示的程序框图,若要使输出的y 的值等于3,
则输入的x 的值可以是
(A )1 (B )2 (C )8 (D )9
(3)已知3cos 25απ?
?+= ??
?,22αππ-<<,则sin 2α的值等于
(A )
1225 (B )1225- (C )2425
(D )2425-
(4)已知0,0a b >>,则“1ab >”是“2a b +>”的
(A )充分不必要条件 (B )必要不充分条件 (C )充要条件 (D )既不充分也不必要条件
(5)若,x y 满足约束条件20,
20,20,
x y y x y -+≥??
+≥??++≤?
则11y x +-的取值范围为
(A )11,35??-???? (B )1,13??
-????
(C )11,,35????-∞-?+∞ ????
???
(D )[)1,1,3
??-∞-?+∞ ??
?
(6)已知等比数列{}n a 的各项均为正数且公比大于1,前n 项积为n T ,且243a a a =,则使得1n T >的n 的最小值为
(A )4 (B )5 (C )6 (D )7 (7)如图,网格纸上小正方形的边长为1,粗线画出的是某几何体的三视图,则该几何体的各个面的面积中,最小的值为
(A
) (B )8 (C
) (D
)(8)在ABC ?中,3
A π
=
,2AB =,3AC =,2CM MB =,则AM BC ?= (A )113
-
(B )43- (C )43 (D )11
3
(9)若椭圆上存在三点,使得这三点与椭圆中心恰好是一个正方形的四个顶点,则该椭圆的离心率为
(A
)
12 (B
)3 (C
)2 (D
)3
(11)已知12,F F 分别为双曲线()22
2210,0x y C a b a b
-=>>:的左、右焦点,若点P 是以12F F 为直径
的圆与C 右支的一个交点, 1PF 交C 于另一点Q ,且12PQ QF =,则C 的渐近线方程为
(12)已知)(x f 是定义在R 上的减函数,其导函数()f x '满足()
()
1f x x f x +<',则下列结论正确的是
(A )对于任意R ∈x , )(x f <0 (B )对于任意R ∈x , )(x f >0 (C )当且仅当()1,∞-∈x ,)(x f <0 (D )当且仅当()+∞∈,1x ,)(x f >0
第Ⅱ卷
注意事项:
第Ⅱ卷共3页,须用黑色墨水签字笔在答题卡上书写作答.若在试题卷上作答,答案无效.
本卷包括必考题和选考题两部分。第13题~第21题为必考题,每个试题考生都必须做答。第22题~第24题为选考题,考生根据要求做答。 二.填空题:本大题共4小题,每小题5分。 (13)若随机变量()2,X
N μσ,且()()510.2P X P X >=<-=,则()25P X <<= .
(14)若()5
112ax x x ?
?++ ??
?展开式中的常数项为40-,则a = .
(15)若数列{}n a 的各项均为正数,前n 项和为n S ,且111
1
1,n n n a S S a ++=+=
,则25a = . (16)已知点()53,1,,23A B ?? ???
,且平行四边形ABCD 的四个顶点都在函数()21
log 1
x f x x +=-的图象上,则四边形ABCD 的面积为 .
三.解答题: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)
在△ABC 中,3
B π
=
,点D 在边A B 上,1BD =,且DA DC =.
(Ⅰ)若△BCD CD ;
(Ⅱ)若AC =DCA ∠.
(18)(本小题满分12分)
如图,三棱柱111ABC A B C -中,底面ABC 为等腰直角三角形,1AB AC ==,12BB =,
160ABB ∠=.
(Ⅰ)证明:1AB B C ⊥;
(Ⅱ)若12B C =,求1AC 与平面1BCB 所成角的正弦值.
(19)(本小题满分12分)
甲、乙两家外卖公司,其送餐员的日工资方案如下:甲公司底薪70元,每单抽成2元;乙公司无底薪, 40单以内(含40单)的部分每单抽成4元,超出40单的部分每单抽成6元.假设同一公司
送餐员一天的送餐单数相同,现从两家公司各随机抽取一名送餐员,并分别记录其100天的送餐单数,得到如下频数表:
(Ⅰ)现从甲公司记录的这100天中随机抽取两天,求这两天送餐单数都大于40的概率; (Ⅱ)若将频率视为概率,回答以下问题:
(ⅰ)记乙公司送餐员日工资为X (单位:元), 求X 的分布列和数学期望;
(ⅱ)小明拟到甲、乙两家公司中的一家应聘送餐员,如果仅从日工资的角度考虑,请利
用所学的统计学知识为他作出选择,并说明理由.
(20)(本小题满分12分)
已知抛物线()2
:20E y px p =>的焦点为F ,过F 且垂直于x 轴的直线与抛物线E 交于,S T 两点,
以P ()3,0为圆心的圆过点,S T ,且90SPT ∠=. (Ⅰ)求抛物线E 和圆P 的方程;
(Ⅱ)设M 是圆P 上的点,过点M 且垂直于FM 的直线l 交E 于,A B 两点,证明:FA FB ⊥.
(21)(本小题满分12分)
已知函数()()ln 1f x ax x =-+,()e 1x g x x =--.曲线()y f x =与()y g x =在原点处的切线相同.
(Ⅰ)求()f x 的单调区间;
(Ⅱ)若0x ≥时,()()g x kf x ≥,求k 的取值范围.
请考生在第22、23、24题中任选一题做答,如果多做,则按所做的第一题计分,做答时请写清题号。
(22)(本小题满分10分)选修41-:几何证明选讲
如图,△ABC 的两条中线AD 和BE 相交于点G ,且,,,D C E G 四点共圆. (Ⅰ)求证:BAD ACG ∠=∠; (Ⅱ)若1GC =,求AB .
(23)(本小题满分10分)选修44-:坐标系与参数方程
在平面直角坐标系xOy 中,曲线C 的参数方程为3cos ,
sin x y αα=??=?
(α为参数),在以原点为极点,
x 轴正半轴为极轴的极坐标系中,直线l 的极坐标方程为sin 4ρθπ?
?-= ??
?.
(Ⅰ)求C 的普通方程和l 的倾斜角;
(Ⅱ)设点()0,2P ,l 和C 交于,A B 两点,求PA PB +.
(24)(本小题满分10分)选修45-:不等式选讲
已知函数()1f x x =+.
(Ⅰ)求不等式()211f x x <+-的解集M ;
(Ⅱ)设,a b M ∈,证明:()()()f ab f a f b >--.
2016年福建省普通高中毕业班质量检查
理科数学试题答案及评分参考
评分说明:
1.本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的主
要考查内容比照评分标准制定相应的评分细则.
2.对计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容
和难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应给分数的一半;如果后继部分的解答有较严重的错误,就不再给分.
3.解答右端所注分数,表示考生正确做到这一步应得的累加分数. 4.只给整数分数.选择题和填空题不给中间分.
一、选择题:本大题考查基础知识和基本运算.每小题5分,满分60分. (1)B (2)C (3)D (4)A (5)B (6)C (7)B (8)C (9)D (10)D (11)A (12)B 二、填空题:本大题考查基础知识和基本运算.每小题5分,满分20分.
(13)0.3 (14)3- (15)5- (16)
26
3
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)本小题主要考查正弦定理、余弦定理、三角形面积公式及三角恒等变换等基础知识,考查运算求解能力,考查化归与转化思想、函数与方程思想等.满分12分.
解法一:(Ⅰ)因为BCD S △即1
sin 2
BC BD B ??= ····················· 2分 又因为3
B π
=
,1BD =,所以4BC = . ················································ 3分 在△BDC 中,由余弦定理得,2
2
2
2cos CD BC BD BC BD B =+-??, ·········· 5分
即2
1
161241132
CD =+-???
=,解得CD = ······························ 6分 (Ⅱ)在△ACD 中,DA DC =,可设A DCA θ∠=∠=,则ADC θ=π-2∠,
又AC =sin 2sin AC CD
θθ
=, ······································ 7分
所以2cos CD θ
=
. ·········································································· 8分
在△BDC 中, 22,23
BDC BCD θθπ
∠=∠=-, 由正弦定理得,sin sin CD BD
B BCD =
∠
,即12cos 2sin sin(2)33
θθ=ππ-, ············ 10分 化简得2cos sin(2)3
θθπ
=-,
于是2sin()sin(2)23θθππ
-=-. ·
······················································· 11分 因为02θπ<<,所以220,222333
θθπππππ
<-<-<-<
, 所以2223θθππ-=-或2+2=23θθππ--π,
解得==618θθππ或,故=618
DCA DCA ππ
∠∠=或. ······························ 12分
解法二:(Ⅰ)同解法一. (Ⅱ)因为DA DC =, 所以A DCA ∠=∠. 取AC 中点E ,连结DE ,
所以DE AC ⊥. ·············································································· 7分 设DCA A θ∠=∠=,
因为AC =
2
EA EC == 在Rt △CDE
中,cos CE CD DCA =
=∠ ····································· 8分
以下同解法一.
(18)本小题主要考查空间直线与直线、直线与平面的位置关系及直线与平面所成的角等基础知识,考查空间想象能力、推理论证能力、运算求解能力,考查化归与转化思想等.满分12分. 解法一:(Ⅰ)连结1AB ,在1ABB △中,111,2,60AB BB ABB ==∠=,
由余弦定理得,222
11112cos 3AB AB BB AB BB ABB =+-??∠=,
∴1AB =,…………………………………………1分
1
∴222
11BB AB AB =+,
∴1AB AB ⊥.………………………………………2分 又∵ABC △为等腰直角三角形,且AB AC =, ∴AC AB ⊥, 又∵1AC
AB A =,
∴AB ⊥平面1AB C . ········································································· 4分 又∵1B C ?平面1AB C ,
∴AB ⊥1B C . ················································································· 5分
(Ⅱ)∵11
1,2AB AB AC BC ===, ∴222
11B C AB AC =+,∴1AB AC ⊥. ················································ 6分
如图,以A 为原点,以1,,AB AC AB 的方向分别为x 轴,y 轴,z 轴的正方向建立空间直角坐标系,
······································································································ 7分
则(
)(()()1000,0,100010A B B C ,
,0,,,,,, ∴()
()11,0,3,1,1,0BB BC =-=-. ··················································· 8分 设平面1BCB 的法向量(),,x y z =n ,
由10,0,BB BC ??=?
??=??n n 得0,0,x x y ?-+=??
-+=?
?令1z =,得
x y ==
∴平面1BCB 的一个法向量为)
=n . (9)
分
∵
()((1110,1,0AC AC CC AC BB =+=+=+-=-,
……………………………………………………………………………10分
∴111cos ,||||AC AC AC ?<>=
==
n n n ,….……………11分
∴1AC 与平面1BCB ······································ 12分 解法二:(Ⅰ)同解法一.
1
(Ⅱ)过点A 作AH ⊥平面1BCB ,垂足为H ,连结1HC ,
则1AC H ∠为1AC 与平面1BCB 所成的角. ············································· 6分 由(Ⅰ) 知,1AB AB ⊥
,1AB ,1AB AC ==,12B C =,
∴222
11AB AC B C +=,∴1AB AC ⊥,
又∵AB
AC A =,∴1AB ⊥平面ABC , ·
··········································· 7分
∴111111332B ABC ABC V S AB AB AC AB -=
?=????=△. ······················· 8分 取BC 中点P ,连结1PB ,∵112BB B C ==,∴1PB BC ⊥. 又在Rt ABC △中,1AB AC ==
,∴BC =
2
BP =
,
∴12
PB ===,
∴1112B BC S BC B P =
?=△. ··························································· 9分 ∵11A BCB B ABC V V --=,
∴
113BCB S AH ?=△
,即1326AH ?=
,∴7
AH =. ············· 10分 ∵1AB ⊥平面ABC ,BC ?平面ABC ,∴1AB BC ⊥, 三棱柱111ABC A B C -中,11//BC B C ,112B C BC ==, ∴111AB B C ⊥
,∴1AC =
=····································· 11分
在1Rt AHC △
中,11sin 35AH AC H AC ∠===
, 所以1AC 与平面1BCB
································ 12分 (19)本小题主要考查古典概型、随机变量的分布列及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,考查分类与整合思想、必然与或然思想、化归与转化思想.满分12分.
1
解:(Ⅰ) 记“抽取的两天送餐单数都大于40”为事件M ,
则2
20210019
()495
C P M C ==. ···································································· 4分
(Ⅱ)(ⅰ)设乙公司送餐员送餐单数为a ,则 当38a =时,384152X =?=; 当39a =时,394156X =?=; 当40a =时,404160X =?=; 当41a =时,40416166X =?+?=; 当42a =时,40426172X =?+?=.
所以X 的所有可能取值为152,156,160,166,172. ······································ 6分 故X 的分布列为:
······································································································ 8分
11121
()1521561601661721621055510
E X =?
+?+?+?+?=所以. ·
····· 9分 (ⅱ)依题意, 甲公司送餐员日平均送餐单数为
380.2390.4400.2410.1420.139.5?+?+?+?+?=. ·
············· 10分 所以甲公司送餐员日平均工资为70239.5149+?=元. ·························· 11分 由(ⅰ)得乙公司送餐员日平均工资为162元.
因为149162<,故推荐小明去乙公司应聘. ·········································· 12分
(20)本小题考查圆与抛物线的标准方程及几何性质、直线与圆锥曲线的位置关系等基础知识,考查推理论证能力、运算求解能力,考查数形结合思想、函数与方程思想、分类与整合思想等.满分12分.
解法一:(Ⅰ)将2
p x =
代入2
2y px =,得y p =±,所以2ST p =, ································································
····························又因为90SPT ∠=,所以△SPT 是等腰直角三角形,
所以SF PF =,即32
p p =-, 解得2p =,
所以抛物线2
:4E y x =
,…………………………………………3分
此时圆P =
所以圆P 的方程为()2
2
38x y -+=. ··························································· 4分
(Ⅱ)设()()()001122,,,,,M x y A x y B x y ,
依题意()2
2
0038x y -+=,即2200061y x x =-+-. ·······································
··· 5分
(ⅰ)当直线l 斜率不存在时,()
3M ±, ①当3x =
+2
4y x =,得()
2y =±
.
不妨设()()
32,32A B ++-, 则1,1,1,AF BF AF BF k k k k ==-=-
即AF BF ⊥.
②当3x =-AF BF ⊥.………………….6分 (ⅱ)当直线l 斜率存在时,因为直线l 与抛物线E 交于,A B 两点,所以直线l 斜率不为零,01x ≠且00y ≠. 因为l MF ⊥,所以1l MF k k =-,
所以00
1l x k y -=,…………………………………………………..7分
直线()0
000
1:x l y x x y y -=
-+. 由()20
0004,1y x x y x x y y ?=?-?=-+??
得,22
2
0000004444011y x y x y y x x +--+=-- , ················ 8分 即2
00004204011y x y y x x --
+=--,所以00121200
4204
,11y x y y y y x x -+==--, ············· 9分
所以()()121211FA FB x x y y ?=--+=22
12121144y y y y ????
--+ ???????
·
······················ 10分 ()()()2
22
22
1212121212123
1116
41642y y y y y y y y y y y y ++=
-++=-++
()()
()
2
200
02
2
0005143061111x y x x x x --=-
++---()()()()()2
2
2
0000020514165111x y x x x x --+-+--=- ()
22
000
2
0244441x x y x ---=
-()
()
220002
046101x y x x -+-+=
=-,
所以AF BF ⊥. ···················································································· 12分 解法二:(Ⅰ)同解法一.
(Ⅱ)设()00,M x y ,依题意()2
2
0038x y -+=,即2200061y x x =-+-, (*) ······ 5分
设()22121212,,,44y y A y B y y y ????≠ ? ?????,则()2
2
2100211,,,4y y FM x y AB y y ??-=-=- ???,
2212010020,,,44y y MA x y y MB x y y ????
=--=-- ? ?????
, ·
······································ 6分 由于FM AB ⊥,//MA MB ,
所以()()()()22
21002122
1202001010,40.44y y x y y y y y x y y x y y ?--+-=??
??????-----= ? ???????
······························· 7分 注意到12y y ≠,()()()
()()
1200120120140,140.
2y y x y y y y y y x +-+=???
-++=?? ························ 8分 由(1)知,若01x =,则00y =,此时不满足(*),故010x -≠,
从而(1),(2)可化为00121200
4204
,11y x y y y y x x -+==--. ························ 9分 以下同解法一.
(21)本小题主要考查导数的几何意义、导数及其应用、不等式等基础知识,考查推理论证能力、运算求解能力、创新意识等,考查函数与方程思想、化归与转化思想、分类与整合思想、数形结合思想等.满分12分.
解法一:(Ⅰ)因为()()1
11
f x a x x '=-
>-+,()e 1x g x '=-, ·
···························· 2分 依题意,()()00f g ''=,解得1a =, ······················································· 3分 所以()111f x x '=-
+1
x
x =
+,当10x -<<时,()0f x '<;当0x >时,()0f x '>. 故()f x 的单调递减区间为()1,0-, 单调递增区间为()0,+∞. ··················· 5分 (Ⅱ)由(Ⅰ)知,当0x =时,()f x 取得最小值0.
所以()0f x ≥,即()ln 1x x +≥,从而e 1x x +≥. 设()()()()()e ln 111,x F x g x kf x k x k x =-=++-+- 则()()()e 11111
x k k
F x k x k x x '=+
-+++-+++≥, ·
··································· 6分 (ⅰ)当1k =时,因为0x ≥,所以()1
1201
F x x x '++
-+≥≥(当且仅当0x =时等号成立)
, 此时()F x 在[)0,+∞上单调递增,从而()()00F x F =≥,即()()g x kf x ≥. ······ 7分 (ⅱ)当1k <时,由于()0f x ≥,所以()()f x kf x ≥. ······························· 8分 由(ⅰ)知()()0g x f x -≥,所以()()()g x f x kf x ≥≥,故()0F x ≥,即()()g x kf x ≥. ······································································································ 9分
(ⅲ)当1k >时, 令()()e 11x k
h x k x =+
-++,则()()
2
e 1x k h x x '=-+,
显然()h x '在[)0,+∞上单调递增,又())
1
010,110h k h ''
=-<=->,
所以()h x '在()
1上存在唯一零点0x , ··········································· 10分 当()00,x x ∈时,()0,h x '<所以()h x 在[)00,x 上单调递减, 从而()()00h x h <=,即()0,F x '<所以()F x 在[)00,x 上单调递减,
从而当()00,x x ∈时,()()00F x F <=,即()()g x kf x <,不合题意. ·········· 11分 综上, 实数k 的取值范围为(],1-∞. ··················································· 12分 解法二:(Ⅰ)同解法一.
(Ⅱ)由(Ⅰ)知,当0x =时,()f x 取得最小值0.
所以()0f x ≥,即()ln 1x x +≥,从而e 1x x +≥.
设()()()()()e ln 111,x F x g x kf x k x k x =-=++-+- 则()()()e 11111x k k F x k x k x x '=+
-+++-+++≥()11
x
x k x =+-+, ·
············· 6分 (ⅰ)当1k ≤时,()0F x '≥在[)0,+∞恒成立,所以()F x 在[)0,+∞单调递增. 所以()()00F x F =≥,即()()g x kf x ≥. ·················································· 9分 (ⅱ)当1k >时,由(Ⅰ)知,当1x >-时,e
1x
x +≥
(当且仅当0x =时等号成立)
, 所以当01x <<时,e
1x
x ->-+,1
e 1x x
<
-. 所以1()e 1(1)e 111
x
x kx F x k x x '=---=--
++ 1111kx x x <
---+11
x kx
x x =--+()2
1
1()11k k x x k x -+-
+=-.················ 10分
于是当101k x k -<<
+时,()0,F x '<所以()F x 在10,1k k -??
??+??
上单调递减.
故当1
01
k x k -<<
+时,()(0)0F x F <=,即()()g x kf x <,不合题意. ······· 11分 综上, 实数k 的取值范围为(],1-∞. ··················································· 12分 解法三:(Ⅰ)同解法一.
(Ⅱ)(ⅰ)当0k ≤时,由(Ⅰ)知,当0x =时,()f x 取得最小值0. 所以()0f x ≥,即()ln 1x x +≥,从而e 1x x +≥,即()0g x ≥.
所以()0kf x ≤,()0g x ≥,()()g x kf x ≥. ················································ 6分 (ⅱ)当0k >时,
设()()()()()e ln 111,x F x g x kf x k x k x =-=++-+-则()()e 11
x k
F x k x '=+-++, 令()()h x F x '=,则()()
2
=e 1x k
h x x '-
+.
显然()h x '在[)0,+∞上单调递增. ·························································· 7分 ①当01k <≤时,()()'010h x h k '=-≥≥,所以()h x 在[)0,+∞上单调递增,()()00h x h =≥; 故()0F x '≥,所以()F x 在[)0,+∞上单调递增,()()00F x F =≥,即()()g x kf x ≥. ······································································································ 9分
②当1k >时,由于(
)
)
1
'010,'
110h k h =-<=->,
所以()h x '
在()
1上存在唯一零点0x , ··········································· 10分 当()00,x x ∈时,()0,h x '< ()h x 单调递减,
从而()()00h x h <=,即()0,F x '<()F x 在[)00,x 上单调递减,
从而当()00,x x ∈时,()()00F x F <=,即()()g x kf x <,不合题意. ·········· 11分 综上, 实数k 的取值范围为(],1-∞. ··················································· 12分
请考生在第(22),(23),(24)题中任选一题作答,如果多做,则按所做的第一题计分,作
答时请写清题号.
(22)选修41-:几何证明选讲
本小题主要考查圆周角定理、相似三角形的判定与性质、切割线定理等基础知识,考查推理论证能力、运算求解能力等,考查化归与转化思想等.满分10分.
解法一:(Ⅰ)连结DE ,因为,,,D C E G 四点共圆,则ADE ACG ∠=∠. ········ 2分 又因为,AD BE 为△ABC 的两条中线, 所以点,D E 分别是,BC AC 的中点,故DE
AB .·
··········································· 3分 所以BAD ADE ∠=∠, ················································································ 4分 从而BAD ACG ∠=∠. ················································································ 5分 (Ⅱ)因为G 为AD 与BE 的交点,
故G 为△ABC 的重心,延长CG 交AB 于F ,
则F 为AB 的中点,且2CG GF =. ······························································· 6分 在△AFC 与△GFA 中,因为FAG FCA ∠=∠,AFG CFA ∠=∠,
所以△AFG ∽△CFA , ······································································ 7分 所以
FA FG
FC FA
=
,即2FA FG FC =?.………………………………………………………9分 因为12FA AB =,12FG GC =,3
2
FC GC =, 所以
2213
44
AB GC =
,即AB =, 又1GC =
,所以AB =. ········································································ 10分 解法二:(Ⅰ)同解法一. ······································································ 5分 (Ⅱ) 由(Ⅰ) 知,BAD ACG ∠=∠,
F
A
B
C
D
E
G
因为,,,D C E G 四点共圆,所以ADB CEG ∠=∠, ········································· 6分
所以ABD △∽CGE △,所以
AB AD
CG CE
=
, ……………………………………………7分 由割线定理,AG AD AE AC ?=?, ····························································· 9分
又因为,AD BE 是ABC △的中线,所以G 是ABC △的重心, 所以2
3AG AD =
,又=2=2AC AE EC , 所以22
2=23AD EC
,所以AD CE =,
所以AB CG
=1CG =
,所以AB = ·
····································· 10分 (23)选修44-;坐标系与参数方程
本小题考查直线的极坐标方程和参数方程、椭圆的参数方程等基础知识,考查运算求解能力,考查数形结合思想、化归与转化思想等. 满分10分.
解法一:(Ⅰ)由3cos ,sin x y αα
=??=?消去参数α,得2
219x y +=, 即C 的普通方程为2
219
x y +=. ······························································ 2分
由sin 4ρθ?π?
-
= ???
,得sin cos 2ρθρθ-=,………(*) ·
··················· 3分 将cos ,sin x y ρθρθ=??=?
代入(*),化简得2y x =+, ·········································· 4分
所以直线l 的倾斜角为4
π
. ···································································· 5分
(Ⅱ)由(Ⅰ)知,点()0,2P 在直线l 上, 可设直线l 的参数方程为cos ,4
2sin 4
x t y t π?
=???π?=+??(t 为参数),
即,222x y ?=????=+??(t 为参数), ····························································· 7分 代入2
219
x y +=
并化简,得25270t ++=. ·································· 8分
(2
45271080?=-??=>.
设,A B 两点对应的参数分别为12,t t ,
则121227
0,05
t t t t +=<=>,所以120,0,t t << ·
·························· 9分
所以()1212PA PB t t t t +=+=-+= ·
··································· 10分 解法二:(Ⅰ)同解法一. ············································································ 5分 (Ⅱ)直线l 的普通方程为2y x =+.
由22
2,99y x x y =+??+=?
消去y 得2
1036270x x ++=, ····································· 7分 于是2
36410272160?=-??=>. 设1122(,),(,)A x y B x y ,则12180,5
x x +=-
<1227
010x x =>,所以120,0x x <<, ····································································································· 8分
故12120|0||PA PB x x x x +=--=+=
······· 10分 (24)选修45-:不等式选讲
本小题考查绝对值不等式的解法与性质、不等式的证明等基础知识,考查运算求解能力、推理论证能力,考查分类与整合思想、化归与转化思想等. 满分10分.
解法一:(Ⅰ)(ⅰ) 当1x -≤时,原不等式可化为122x x --<--,解得1x <-,
此时原不等式的解是1x <-; ······························································ 2分 (ⅱ)当1
12
x -<<-
时,原不等式可化为122x x +<--,解得1x <-, 此时原不等式无解; ··········································································· 3分
(ⅲ)当1
2
x -
≥时,原不等式可化为12x x +<,解得1x >, 此时原不等式的解是1x >; ································································ 4分
综上,{}
11M x x x =<->或. ·························································· 5分 (Ⅱ)因为()1f ab ab =+()()1ab b b =++- ····································· 6分
1ab b b +--≥
·
········································· 7分 11b a b =+--. ·
···································· 8分
因为,a b M ∈,所以1b >,10a +>, ··············································· 9分 所以()11f ab a b >+--,即()()()f ab f a f b >--. ······················ 10分 解法二:(Ⅰ)同解法一.
(Ⅱ)因为()()()1111f a f b a b a b a b --=+--++--+=+≤, ······ 7分 所以,要证()()()f ab f a f b >--,只需证1ab a b +>+,
即证22
1ab a b +>+, ····································································· 8分 即证2222
212a b ab a ab b ++>++,
即证2222
10a b a b --+>,即证()()
22110a b -->. ···························· 9分 因为,a b M ∈,所以2
2
1,1a b >>,所以(
)()
22110a b -->成立,
所以原不等式成立. ·········································································· 10分