2016年福州市普通高中毕业班综合质量检测
理科数学能力测试
(完卷时间:120分钟;满分:150分)
本试卷分第Ⅰ卷和第Ⅱ卷两部分.第Ⅰ卷1至2页,第Ⅱ卷3至4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收回.
第Ⅰ卷
一、选择题:本大题共12小题,每小题5分,共60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、已知全集为R ,集合{1,1,2,4}M =-,2{|23}N x x x =->,则()M N =R e (A ){1,1,2}-
(B ){1,2}
(C ){4}
(D ){}12x x
-剟
2、复数z 满足(1i)|1i |z -=+,则复数z 的共轭复数在复平面内的对应点位于 (A )第一象限 (B )第二象限 (C )第三象限 (D )第四象限
3、函数()sin()f x A x ?=+(0A >)在π
3
x =处取得最小值,则
(A )π()3f x +是奇函数 (B )π
()3f x +是偶函数
(C )π()3f x -是奇函数 (D )π
()3
f x -是偶函数
4、在ABC ?中,5AB AC ?= ,4BA BC ?=
,则AB = (A )9 (B )3 (C )2 (D )1
5、已知某工程在很大程度上受当地年降水量的影响,施工期间的年降水量X (单位:mm )对工期延误天数Y 的影响及相应的概率P 如下表所示:
在降水量X 至少是100的条件下,工期延误不超过15天的概率为 (A )0.1 (B )0.3 (C )0.42 (D )0.5
6、若,x y 满足约束条件10,20,220,x x y x y +??
-+??++?
………且目标函数z ax y =-取得最大值的点有无数个,则z 的最小
降水量X 100
X < 100200X <… 20030X <… 300X …
工期延误天数Y 0 5 15 30 概率P
0.4
0.2
0.1
0.3
值等于 (A )2- (B )3
2
-
(C )1
2
-
(D )
12
7、执行右面的程序框图,若输入n 值为4,则输出的结果为 (A )8 (B )21 (C )34
(D )55
8、5
12x x ?
?++ ??
?的展开式中,2x 的系数为
(A )45 (B )60
(C )90
(D )120
9、正项等比数列{}n a 满足11a =,2635a a a a +=128,则下列结论正确的是 (A )n ?∈*N ,12n n n a a a ++…
(B )n ?∈*N ,212n n n a a a +++= (C )n ?∈*N ,1n n S a +<
(D )n ?∈*N ,312n n n n a a a a ++++=+
10、双曲线22
22:1(0,0)x y E a b a b
-=>>的左、右焦点分别为1F ,2F ,P 是E 左支上一点,112PF F F =,
直线2PF 与圆222x y a +=相切,则E 的离心率为 (A )
54
(B
(C )53
(D
11、一个三棱锥的三视图如图所示,则该三棱锥的体积等于 (A )2 (B
(C
(D )3
12、设m ∈R ,函数222()()(e 2)x f x x m m =-+-.若存在0x 使得
01
()5
f x …
成立,则m = (A )15
(B )
25
(C )35
(D )
45
俯视图
(1)A (2)D (3)B (4)B (5)D (6)C (7)C (8)D (9)C (10)C (11)A (12)B 第Ⅱ卷
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4小题,每小题5分,共20分.把答案填在答题卡相应位置.
13、知函数1,02,
()1,20.x x f x x -=?--?…剟若()()[],2,2g x f x ax x =+∈-为偶函数,则实数a = .
14、所有棱长均为2的正四棱锥的外接球的表面积等于 .
15、抛物线2:4C y x =的准线与x 轴交于点M ,过焦点F 作倾斜角为60?的直线与C 交于,A B 两点,则tan AMB ∠= .
16、数列{}n a 的前n 项和为n S .已知12a =,1(1)2n n n S S n ++-=,则100S =________.
(13)1
2
- (14)8π (15) (16)198
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)
ABC ?的内角A ,B ,C 所对的边分别为,,a b c ,已知tan 21tan A c
B b
+=
. (Ⅰ)求A ;
(Ⅱ)若BC 边上的中线AM =AH =ABC ?的面积.
(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:(Ⅰ)因为tan 21tan A c B b +=,所以sin cos 2sin 1sin cos sin A B C
B A B
+=
, ······················· 2分 即sin()2sin sin cos sin A B C B A B
+=
, 因为sin()sin 0A B C +=≠,sin 0B ≠,
所以1
cos 2
A =, ················································································ 4分
又因为(0,π)A ∈,所以π
3A =. ···························································· 5分
(Ⅱ)由M 是BC 中点,得1()2
AM AB AC =+
,
即2221(2)4
AM AB AC AB AC =++? ,
所以2232c b bc ++=,① ····································································· 7分
由11
sin 22S AH BC AB AC A =?=??,
=,即2bc a =,② ···························································· 9分 又根据余弦定理,有222a b c bc =+-,③ ··············································· 10分
联立①②③,得2()3222
bc
bc =-,
解得8bc =.
所以△ABC
的面积1
sin 2
S bc A == ··············································· 12分
18、(本小题满分12分)
为了研究某学科成绩是否与学生性别有关,采用分层抽样的方法,从高三年级抽取了30名男生和20名女生的该学科成绩,得到如下所示男生成绩的频率分布直方图和女生成绩的茎叶图,规定80分以
上为优分
(含80分
).
科成绩与性别有关”?
(Ⅱ)将频率视作概率,从高三年级该学科成绩中任意抽取3名学生的成绩,求至少2名学生的成绩为优分的概率.
)
)()()(()
(2
d b c a d c b a bc ad n K ++++-=
.
(18)本小题主要考查频率分布直方图、茎叶图、n 次独立重复试验、独立性检验等基础知识,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. 分 假设0H :该学科成绩与性别无关,
2
K 的观测值22
()50(991121) 3.125()()()()20302030
n ad bc k a b c d a c b d -?-?===++++???,
因为3.125 2.706>,所以能在犯错误概率不超过10%的前提下认为该学科成绩与性别有关. ·············································································································· 6分
(Ⅱ)由于有较大的把握认为该学科成绩与性别有关,因此需要将男女生成绩的优分频率
20
0.450
f =
=视作概率. ·
············································································· 7分 设从高三年级中任意抽取3名学生的该学科成绩中,优分人数为X ,则X 服从二项分布(3,0.4)B , ······························································································· 9分 所求概率223
333(2)(3)0.40.60.40.352P P X P X C C ==+==??+?=. ····································································································· 12分
19、(本小题满分12分)
如图所示,四棱锥P ABCD -的底面是梯形,且//AB CD ,AB ⊥平面PAD ,E 是PB 中点,
1
2
CD PD AD AB ===. (Ⅰ)求证:CE ⊥平面PAB ;
(Ⅱ)若CE =,4AB =,求直线CE 与平面PDC 所成角的大小.
(19)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.
(Ⅰ)证明:取AP 的中点F ,连结,DF EF ,如图所示. 因为PD AD =,所以DF AP ⊥. ·························································· 1分 因为AB ⊥平面PAD ,DF ?平面PAD , 所以AB DF ⊥.
又因为AP AB A = , 所以DF ⊥平面PAB . ······································································· 3分 因为点E 是PB 中点,
所以//EF AB ,且2AB
EF =. ······························································ 4分
又因为//AB CD ,且2
AB
CD =,
所以//EF CD ,且EF CD =, 所以四边形EFDC 为平行四边形,
所以//CE DF ,所以CE ⊥平面PAB . ·················································· 6分 (Ⅱ)解:设点O ,G 分别为AD ,BC 的中点,连结OG ,则//OG AB , 因为AB ⊥平面PAD ,AD ?平面PAD , 所以AB AD ⊥,所以OG AD ⊥. ·························································· 7分
因为EC =
DF = 又因为4AB =,所以2AD =,
所以22,AP AF === 所以APD ?为正三角形,所以PO AD ⊥, 因为AB ⊥平面PAD ,PO ?平面PAD , 所以AB PO ⊥.
又因为AD AB A = ,所以PO ⊥平面ABCD . ········································ 8分
故,,OA OG OP 两两垂直,可以点O 为原点,分别以,,OA OG OP
的方向为,,x y z 轴的正方向,建立空间直角坐标系O xyz -,如图所示.
P ,(1,2,0),(1,0,0)C D --
,1(,2E ,
E D
C B A P
设平面PDC 的法向量(,,)x y z =n ,
则0,0,PD PC ??=???=??n n
所以0,
20,x x y ?-=??-+=?? 取1z =,则(=n , ···········································
····················· 10分 设EC 与平面PDC 所成的角为α,
则1
sin |cos ,||2EC α=<>==n , ···················································· 11分 因为π
[0,]2
α∈,所以π6α=,
所以EC 与平面PDC 所成角的大小为π
6
. ·············································· 12分
20、(本小题满分12分)
在平面直角坐标系xOy 中,已知点,A B 的坐标分别为()()2,0,2,0-.直线,AP BP 相交于点P ,且它
们的斜率之积是1
4
-.记点P 的轨迹为Γ. (Ⅰ)求Γ的方程;
(Ⅱ)已知直线,AP BP 分别交直线:4l x =于点,M N ,轨迹Γ在点P 处的切线与线段MN 交于点Q ,
求MQ NQ
的值.
(20)本小题考查椭圆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)设点P 坐标为(),x y ,则
直线AP 的斜率2AP y
k x =+(2x ≠-); 直线BP 的斜率2
BP y
k x =-(2x ≠). ···························································· 2分
由已知有1
224
y y x x ?=-+-(2x ≠±)
, ························································· 3分 化简得点P 的轨迹Γ的方程为2
214
x y +=(2x ≠±). ······································· 4分
(注:没写2x ≠或2x ≠-扣1分)
(Ⅱ)设()00,P x y (02x ≠±),则22
0014
x y +=. ·············································· 5分 直线AP 的方程为()0022y y x x =++,令4x =,得点M 纵坐标为0062M y
y x =+; ·
····· 6分 直线BP 的方程为()0022y y x x =
--,令4x =,得点N 纵坐标为0022
N y
y x =-; ·
······· 7分 设在点P 处的切线方程为()00y y k x x -=-,
由()0022
,44,y k x x y x y ?=-+?+=?
得()
()()2220000148440k x k y kx x y kx ++-+--=. ············· 8分 由0?=,得()()()22
22000064161410k y kx k y kx ??--+--=??
,
整理得222
20000214y kx y k x k -+=+.
将()2
222
00
001,414x y x y =-=-代入上式并整理得2
00202x y k ??+= ???,解得00
4x k y =-, ··· 9分 所以切线方程为()0000
4x
y y x x y -=--.
令4x =得,点Q 纵坐标为()()22
000000
000000
441441444Q x x x y x x x y y y y y y ---+-=-
===.
··········································································································· 10分
设MQ QN =
λ,所以()Q M N Q y y y y -=-λ, 所以
00000000162122x y y x y x x y ??
---=- ?+-??
λ. ·
······················································· 11分 所以()()()()()()
2
2
000000000012621222x x y y x x y x y x -+----=+-λ
. 将22
00
14x y =-代入上式,002+(2+)22
x x
-=-λ,
解得1=λ,即1MQ
NQ
=. ··········································································· 12分
解法二:(Ⅰ)同解法一.
(Ⅱ)设()00,P x y (02x ≠±),则22
0014
x y +=. ·············································· 5分 直线AP 的方程为()0022y y x x =++,令4x =,得点M 纵坐标为0062
M y
y x =+; ·
····· 6分 直线BP 的方程为()0022y y x x =
--,令4x =,得点N 纵坐标为0022
N y
y x =-; ·
······· 7分 设在点P 处的切线方程为()00y y k x x -=-,
由()0022
,44,y k x x y x y ?=-+?+=?
得()
()()2
220000148440k x k y kx x y kx ++-+--=. ············· 8分 由0?=,得()()()22
22000064161410k y kx k y kx ??--+--=??
,
整理得222
20000214y kx y k x k -+=+.
将()2
222
00
001,414x y x y =-=-代入上式并整理得2
00202x y k ??+= ??
?,解得004x k y =-, ··· 9分 所以切线方程为()0000
4x
y y x x y -=--.
令4x =得,点Q 纵坐标为()()22
000000
000000
441441444Q x x x y x x x y y y y y y ---+-=-
===.
··········································································································· 10分
所以()()0000000
22
00000
8181621222244M N Q x y x y y y x y y y x x x y y ---+=
+====+---, ············ 11分 所以Q 为线段MN 的中点,即
1MQ NQ
=. ······················································· 12分
21、(本小题满分12分)
已知a ∈R ,函数1()e x f x ax -=-的图象与x 轴相切. (Ⅰ)求()f x 的单调区间;
(Ⅱ)当1x >时,()(1)ln f x m x x >-,求实数m 的取值范围.
(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:(Ⅰ)()1e x f x a -'=-,设切点为0(,0)x , ·················································· 1分
依题意,00()0,()0,f x f x =??'=?即001
01e 0,
e 0,x x ax a --?-=??-=??
解得01,1,x a =??=?
························································································ 3分
所以()1e 1x f x -'=-.
当1x <时,()0f x '<;当1x >时,()0f x '>.
故()f x 的单调递减区间为(,1)-∞,单调递增区间为(1,)+∞. ························ 5分 (Ⅱ)令()()(1)ln g x f x m x x =--,0x >.
则11
()e (ln )1x x g x m x x
--'=-+-,
令()()h x g x '=,则1211
()e ()x h x m x x
-'=-+, ·············································· 6分
(ⅰ)若1
2
m …,
因为当1x >时,1e 1x ->,211
()1m x x
+<,所以()0h x '>,
所以()h x 即()g x '在(1,)+∞上单调递增.
又因为(1)0g '=,所以当1x >时,()0g x '>, 从而()g x 在[1,)+∞上单调递增,
而(1)0g =,所以()0g x >,即()(1)ln f x m x x >-成立. ······························ 9分
(ⅱ)若1
2
m >,
可得1211
()e ()x h x m x x
-'=-+在(0,)+∞上单调递增.
因为(1)120h m '=-<,2
11
(1ln(2))2{
}01ln(2)[1ln(2)]h m m m m m '+=-+>++, 所以存在1(1,1ln(2))x m ∈+,使得1()0h x '=,
且当1(1,)x x ∈时,()0h x '<,所以()h x 即()g x '在1(1,)x 上单调递减,
又因为(1)0g '=,所以当1(1,)x x ∈时,()0g x '<,
从而()g x 在1(1,)x 上单调递减,
而(1)0g =,所以当1(1,)x x ∈时,()0g x <,即()(1)ln f x m x x >-不成立.
纵上所述,k 的取值范围是1
(,]2-∞. ······················································ 12分
请考生在第(22)、(23)、(24)三题中任选一题做答,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4-1:几何证明选讲
如图所示,ABC ?内接于圆O ,D 是 BAC 的中点,∠BAC 的平分线分别交BC 和圆O 于点E ,F .
(Ⅰ)求证:BF 是ABE ?外接圆的切线;
(Ⅱ)若3AB =,2AC =,求22DB DA -的值.
(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:(Ⅰ)设ABE ?外接圆的圆心为O ',连结BO '并延长交圆O '于G 点,连结GE , 则90BEG ∠=?,BAE BGE ∠=∠.
因为AF 平分∠BAC ,所以 =BF FC ,所以FBE BAE ∠=∠, ······················· 2分
所以18090FBG FBE EBG BGE EBG BEG ∠=∠+∠=∠+∠=?-∠=?, 所以O B BF '⊥,所以BF 是ABE ?外接圆的切线. ····································· 5分
(Ⅱ)连接DF ,则DF BC ⊥,所以DF 是圆O 的直径,
因为222BD BF DF +=,222DA AF DF +=, 所以2222
BD DA AF BF -=-. ··························································
因为AF 平分∠BAC ,所以ABF ?∽AEC ?,
所以AB AF AE AC
=,所以()AB AC AE AF AF EF AF ?=?=-?, 因为FBE BAE ∠=∠,所以FBE ?∽FAB ?,从而2BF FE FA =?,
所以22AB AC AF BF ?=-,
所以226BD DA AB AC -=?=. ····························································· 10分 23、(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 的参数方程为22cos ,
2sin x y αα=+??=?(α为参数).以O 为极点,x 轴正半轴
为极轴,并取相同的单位长度建立极坐标系.
(Ⅰ)写出1C 的极坐标方程;
(Ⅱ)设曲线222:14x C y +=经伸缩变换1
,
2x x y y
?'=???'=?后得到曲线3C ,
射线π3θ=(0ρ>)分别与1C 和3C 交于A ,B 两点,求||AB .
(23)选修44-;坐标系与参数方程
本小题考查极坐标方程和参数方程、伸缩变换等基础知识,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.
F
解:(Ⅰ)将22cos ,
2sin x y αα
=+??=?消去参数α,化为普通方程为22(2)4x y -+=,
即221:40C x y x +-=, ··············································································· 2分 将cos ,sin x y ρθρθ=??=?
代入221:40C x y x +-=,得24cos ρρθ=, ································· 4分
所以1C 的极坐标方程为4cos ρθ=. ······························································ 5分
(Ⅱ)将2,
x x y y '=??'
=?代入2C 得221x y ''+=,
所以3C 的方程为221x y +=. ······································································· 7分 3C 的极坐标方程为1ρ=,所以||1OB =.
又π
||4cos 23
OA ==,
所以||||||1AB OA OB =-=. ········································································ 10分 24、(本小题满分10分)选修4-5:不等式选讲 已知不等式|3|21x x +<+的解集为{|}x x m >. (Ⅰ)求m 的值;
(Ⅱ)设关于x 的方程1
||||x t x m t
-++=(0t ≠)有解,求实数t 的值.
(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:(Ⅰ)由|3|21x x +<+得, 3,(3)21,x x x -??-+<+? (3)
321,x x x >-??
+<+? ································································· 2分 解得2x >. 依题意2m =. ·························································································· 5分
(Ⅱ)因为()1111x t x x t x t t t t t t ?
?-++--+=+=+ ???…,
当且仅当()10x t x t ?
?-+ ??
?…时取等号, ···························································· 7分
因为关于x 的方程1
||||2x t x t
-++=(0t ≠)有实数根,
所以1
2t t
+…. ······················································································· 8分
另一方面,1
2t t
+…, 所以1
2t t
+
=, ························································································ 9分 所以1t =或1t =-. ·················································································· 10分