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2014-2015年福州市第一学期高三期末理科数学质量检查

福州市2014－2015学年度第一学期高三质量检查

理科数学试卷

（满分：150分;完卷时间：120分钟）

注意事项：

1．本科考试分试题卷和答题卷，考生须在答题卷上作答，答题前，请在答题卷的密封线内填写学校、班级、准考证号、姓名；

2．本试卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分，全卷满分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，共60分．在每小题所给的四个选项中有且只有一个选项是正确的．把正确选项涂在答题卡的相应位置上．） 1．如图，复平面上的点1234,,,Z Z Z Z 到原点的距离都相等．若复数z

所对应的点为1Z ，则复数z 的共轭复数所对应的点为（）． A ．1Z B ．2Z C ．3Z

D ．4Z

2．已知π

tan()34

+=α，则tan α的值是（）．

A ．2

B ．

C ．1-

D ．3-

3．已知A ?≠

B ，则“x A ∈”是“x 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

4．某班有49位同学玩“数字接龙”游戏，具体规则按如图所示的程序

框图执行（其中a 为座位号），并以输出的值作为下一个输入的值．若第一次输入的值为8，则第三次输出的值为（）． A ．8 B ．15 C ．29

D ．36

5．如图，若在矩形OABC 中随机撒一粒豆子，则豆子落在图中阴影部

分的概率为（）． A ．1

π B ．2π C ．

3π

D ．

第4题图

第5题图

6．已知函数()lg(1)=-f x x 的值域为(,1]-∞，则函数()f x 的定义域为（）．

A ．[9,)-+∞

B ．[0,)+∞

C ．(9,1)-

D ．[9,1)-

7．已知抛掷一枚质地均匀的硬币，正面朝上的概率为0.5．现采用随机模拟试验的方法

估计抛掷这枚硬币三次恰有两次正面朝上的概率：先由计算器产生0或1的随机数，用0表示正面朝上，用1表示反面朝上；再以每三个随机数做为一组，代表这三次投掷的结果．经随机模拟试验产生了如下20组随机数：

101 111 010 101 010 100 100 011 111 110 000 011 010 001 111 011 100 000 101 101 据此估计，抛掷这枚硬币三次恰有两次正面朝上的概率为（）． A ．0.30

B ．0.35

C ．0.40

D ．0.65

8． ABC △的三个内角,,A B C 所对的边分别为,,a b c .

若

cos cos A b

B a

==C 的大小为（）． A ．60?

B ． 75?

C ．90?

D ．120?

9．若双曲线22

22:1x y a b

Γ-=（0,0a b >>）的右焦点()4,0

到其渐近线的距离为，则

双曲线Γ的离心率为（）． A

C ．2

D ．4

10．定义运算“*”为：,0,2,0a b ab a a b a +

≥.若函数()(1)f x x x =+*，则该函数的图

象大致是（）．

11．已知ABC ?的三个顶点,,A B C 的坐标分别为())

()0,1,

,0,2-

，O 为坐标原点，动

点P 满足1CP =

，则OA OB OP ++的最小值是（）．

A ．4-

B 1

C 1

D 12．已知直线:l y ax b =+与曲线:Γ1

x y y

+没有公共点．若平行于l 的直线与曲线Γ有且只有一个公共点，则符合条件的直线l （）． A ．不存在

B ．恰有一条

C ．恰有两条

D ．有无数条

第Ⅱ卷（非选择题共90分）

二、填空题（本大题共4小题，每小题4分，共16分，把答案填在答题卡的相应位置上．） 13．若变量,x y 满足约束条件0,

0,2x y y x ??

??-?

≤≥≤，则z x y =+的最小值为 ★★★ ．

14．已知6234560123456(1)x a a x a x a x a x a x a x +=++++++，则016,,,a a a ???中的所有偶数．．

的和等于 ★★★ ．

15．已知椭圆2239x y +=的左焦点为1F ，点P 是椭圆上异于顶点的任意一点，O 为坐标

原点．若点D 是线段1PF 的中点，则1

FOD ?的周长为 ★★★ ． 16. 若数列{}n a 满足112n n n a a a +-+≥（2n ≥），则称数列{}n a 为凹数列．已知等差数列{}n b 的公差为d ，12b =，且数列n b n ??

????

是凹数列，则d 的取值范围为 ★★★ ．

三、解答题（本大题共6小题，共74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）

已知等比数列{}n a 的公比1q >，1a ，2a 是方程2320x x -+=的两根．（Ⅰ）求数列{}n a 的通项公式；（Ⅱ）求数列{}2n n a ?的前n 项和n S ．

18.（本小题满分12分）

“ALS 冰桶挑战赛”是一项社交网络上发起的筹款活动，活动规定：被邀请者要么在24小时内接受挑战，要么选择为慈善机构捐款（不接受挑战），并且不能重复参加该活动.若被邀请者接受挑战，则他需在网络上发布自己被冰水浇遍全身的视频内容，然后便可以邀请另外3个人参与这项活动.假设每个人接受挑战与不接受挑战是等可能的，且互不影响.

（Ⅰ）若某被邀请者接受挑战后，对其他3个人发出邀请，则这3个人中至少有2个人接受挑战的概率是多少？

（Ⅱ）假定（Ⅰ）中被邀请到的3个人中恰有两人接受挑战.根据活动规定，现记X 为接下来被邀请到的6个人中接受挑战的人数，求X 的分布列和均值（数学期望）.

19.（本小题满分12分）

已知函数()4f x x π??

= ???

在同一半周期内的图象过点,,O P Q ，其中O 为坐标原

点，P 为函数()f x 图象的最高点，Q 为函数()f x 的图象与x

的正半轴的交点．

（Ⅰ）试判断OPQ ?的形状，并说明理由．

（Ⅱ）若将O P Q ?绕原点O 按逆时针方向旋转02ααπ?

?<< ??

?时，顶点,P Q ''恰好同时落在曲线k y x =()0x >（如图所示），求实数k 的值．

20.（本小题满分12分）

一种药在病人血液中的含量不低于2克时，它才能起到有效治疗的作用．已知每服

用m （14m ≤≤且m ∈R ）个单位的药剂，药剂在血液中的含量y （克）随着时间x （小时）变化的函数关系式近似为)(x f m y ?=，其中()10

,06,4.4,682x x

f x x x ?

≤≤≤

(Ⅰ)若病人一次服用3个单位的药剂，则有效治疗时间可达多少小时？

(Ⅱ)若病人第一次服用2个单位的药剂，6个小时后再服用m 个单位的药剂，要使接下来的2小时中能够持续有效治疗，试求m 的最小值．

21.（本小题满分12分）

已知抛物线Γ的顶点为坐标原点，焦点为(0,1)F ．（Ⅰ）求抛物线Γ的方程；

（Ⅱ）若点P 为抛物线Γ的准线上的任意一点，过点P 作抛物线Γ的切线PA 与PB ，切点分别为,A B ，求证：直线AB 恒过某一定点；

(Ⅲ)分析(Ⅱ)的条件和结论，反思其解题过程，再对命题(Ⅱ)进行变式和推广．请写出一个你发现的真命题．．．，不要求证明（说明：本小题将根据所给出的命题的正确性和一般性酌情给分）． 22.（本小题满分14分）

已知函数()()e sin cos ,cos x x f x x x g x x x =-=，其中e 是自然对数的底数．

（Ⅰ）判断函数()y f x =在π

(0,)2

内的零点的个数，并说明理由；

（Ⅱ）12ππ0,,0,22x x ????

?∈?∈????????

，使得不等式12()()f x g x m +≥成立，试求实数m 的

取值范围；

（Ⅲ）若1x >-，求证：()()0f x g x ->．

第19题图

福州市2014－2015学年度第一学期高三质量检查

理科数学试卷参考答案及评分细则

一、选择题：本大题共12小题，每小题5分，共60分． 1．C 2．B 3．A 4．A 5．B 6．D 7．B 8．C 9．C 10．D 11．B 12．C

二、填空题：本大题共4小题，每小题4分，共16分，

13．2- 14．32 15．316．(,2]-∞ 三、解答题：本大题共6小题，共74分．

17．本题主要考查一元二次方程的根、等比数列的通项公式、错位相减法求数列的和等基础知识，考查应用能力、运算求解能力，考查函数与方程思想．解：（Ⅰ）方程2320x x -+=的两根分别为1,2， ·························································· 1分依题意得11a =，22a =． ································································································ 2分所以2q =，······················································································································· 3分所以数列{}n a 的通项公式为12n n a -=． ·········································································· 4分（Ⅱ）由（Ⅰ）知22n n n a n ?=?， ··················································································· 5分所以212222n n S n =?+?+???+?， ············································ ①

23121222(1)22n n n S n n +?=?+?+???+-?+?， ························· ② 由①－②得

23222n S -=+++???122n n n ++-?， ·

·············································································· 8分即 1222

212

n n S n +-?-=-?-， ·

······················································································ 11分所以12(1)2n n S n +=+-?． ····························································································· 12分 18．本题主要考查离散型随机变量的概率、分布列、数学期望等基础知识，考查运算求解能力以及应用意识，考查必然与或然思想等．

解法一：（Ⅰ）这3个人接受挑战分别记为A 、B 、C ，则,,A B C 分别表示这3个人不接受挑战．

这3个人参与该项活动的可能结果为：{},,A B C ，{

},,A B C ，{},,A B C ，{},,A B C ，{}

,,A B C ，{},,A B C ，{},,A B C ，{},,A B C ．共有8种； ·

······························································· 2分其中，至少有2个人接受挑战的可能结果有：{},,A B C ，{},,A B C ，{},,A B C ，{},,A B C ，共有4种． ······················································································································ 3分

根据古典概型的概率公式，所求的概率为41

P ==． ·················································· 4分

（说明：若学生先设“用(),,x y z 中的,,x y z 依次表示甲、乙、丙三人接受或不接受挑战的情况”，再将所有结果写成

()

,,A B C ,

()

,,A B C ,

()

,,A B C ，

(),,A B C ,(),,A B C ,(),,A B C ,(),,A B C ,(),,A B C ，不扣分．

）（Ⅱ）因为每个人接受挑战与不接受挑战是等可能的，

所以每个人接受挑战的概率为12，不接受挑战的概率也为1

． ·

··································· 5分

所以()060

11102264P X C ????=== ? ?????，()5

1611631226432P X C ????==?== ? ?????

， ()2

426

111522264P X C ??

??==?= ?

???

??，()3

3611205

3226416P X C ????==?== ? ???

??， ()4

111542264P X C ??

??==?= ?

???

??，()5

5611635226432P X C ????

==?== ? ???

， ()6

661116.2264P X C ????=== ? ?????

······················································································· 9分故X

10分

所以()131551531

0123456364326416643264

E X =?+?+?+?+?+?+?=．

故所求的期望为3． ········································································································ 12分解法二：因为每个人接受挑战与不接受挑战是等可能的，

所以每个人接受挑战的概率为12，不接受挑战的概率也为1

． ·

··································· 1分（Ⅰ）设事件M 为“这3个人中至少有2个人接受挑战”，

则2

233

31111()2222P M C C ??????=?+= ? ? ???????

． ·

········································································· 4分（Ⅱ）因为X 为接下来被邀请的6个人中接受挑战的人数，

所以1~6,2X B ??

???

．

··········································································································· 5分所以()060

11102264P X C ????=== ? ?????，()5

1611631226432P X C ????==?== ? ?????

， ()2

426

111522264P X C ????==?= ? ???

??，()3

3611205

3226416P X C ????==?== ? ???

， ()4

46111542264P X C ??

??==?= ?

???

??，()5

5611635226432

P X C ????==?== ? ???

??， ()6

1116.2264P X C ????

=== ? ?????

······················································································· 9分

故X

10分

所以()1

632

E X =?=．

故所求的期望为3． ······································································································ 12分 19．本题主要考查反比例函数、三角函数的图象与性质、三角函数的定义、同角三角函数的基本关系式、二倍角公式、两角和的正弦公式等基础知识，考查运算求解能力，考查数形结合思想、化归与转化思想、函数与方程思想．解法一：（Ⅰ）OPQ ?为等边三角形． ············································································ 1分理由如下：

因为函数()4f x x π??

= ???

，

所以2π

T ==π，所以函数()f x 的半周期为4，所以4OQ =． ·················································································································· 2分又因为P 为函数()f x 图象的最高点，

所以点P

坐标为(2，，所以4OP =， ···································································· 4分又因为Q 坐标为(4,0)

，所以4PQ =，

所以OPQ ?为等边三角形． ··························································································· 6分（Ⅱ）由（Ⅰ）知，4OP OQ ==，

所以点P ',Q '的坐标分别为4cos 4sin 33αα??ππ???

?++ ? ? ??????

?，,(4cos 4sin )αα,， ················ 7分

代入k y x =，得216cos sin 8sin(2π)333k αααππ?

???=++=+ ? ??

???，

且16sin cos 8sin 2k ααα==， ························································································· 9分

所以2sin 2sin(2π)3αα=+，结合22sin (2)cos (2)1αα+=，02

απ

<<，

解得1

sin 22

α=，············································································································· 11分

所以4k =，所以所求的实数k 的值为4． ····································································· 12分解法二：（Ⅰ）OPQ ?为等边三角形． ·········································································· 1分理由如下：

因为函数()4f x x π??

= ???

，

所以2π

T ==，所以函数()f x 的半周期为4，所以4OQ =， ··································· 2分因为P 为函数()f x 的图象的最高点，

所以点P

坐标为(2，，所以4OP =，所以OP OQ =． ······································ 4分又因为直线OP

的斜率k =

=60POQ ∠=?，所以OPQ ?为等边三角形． ··························································································· 6分

（Ⅱ）由（Ⅰ）知，4OP OQ ==，

所以点P ',Q '的坐标分别为4cos 4sin 33αα??ππ???

?++ ? ? ???????

,,(4cos 4sin )αα,

， ·················· 7分因为点P ',Q '在函数(0)k

y x x

=>的图象上，

所以16cos sin ,3316sin cos k k ?ππ?

???=++? ? ??

?????=?

αααα, ················································································ 8分所以28sin(2π),38sin 2k k ?

=+???=?

αα, ·

································································································· 9分消去k 得， 2

sin 2sin(2π)3αα=+，

所以22

sin 2sin 2cos πcos2sin π33

ααα=+，

所以3sin 222αα=

，所以tan 2α=，····························································· 10分

又因为 02απ<<，所以26απ=，所以1

sin 22

α=， ···················································· 11分

所以4k =．所以所求的实数k 的值为4． ····································································· 12分解法三：（Ⅰ）同解法一或同解法二；

（Ⅱ）由（Ⅰ）知，OPQ ?为等边三角形．

因为函数(0)k

y x x

=>的图象关于直线y x =对称， ························································ 8分

由图象可知，当12

απ=时，点P ',Q '恰在函数(0)k

y x x =>的图象上． ······················ 10分

此时点Q '的坐标为(4cos 4sin )1212

ππ

,， ·

········································································ 11分所以16sin cos 8sin 412126

k πππ

===，所以所求的实数k 的值为4． ···························· 12分

20．本题主要考查分段函数模型的应用问题、一元二次函数的最值、解不等式等基础知识，考查应用意识、运算求解能力，考查化归与转化思想、分类讨论思想等．

解：（I ）因为3m =，所以30

,06,4312,682x x

y x x ?

≤≤≤． ······················································ 1分

当06x <≤时，由30

24x

+≥，解得x ≤11，此时06x <≤； ·

······································ 3分当68x ≤≤时，由31222x -≥，解得203x ≤，此时20

63x ≤≤． ····························· 5分

综上所述，20

x ≤≤．

故若一次服用3个单位的药剂，则有效治疗的时间可达

小时． ······························ 6分（Ⅱ）当6x ≤≤8时，110102(4)[]824(6)2

y x m x x x =?-+=-++--， ······················· 8分

因为10822m

x x -+-≥对6x ≤≤8恒成立，

即2812

x x m -+≥对6x ≤≤8恒成立，

等价于2max 812

)10

x x m -+≥(，6x ≤≤8．

······································································ 9分令2812()10x x g x -+=，则函数2(4)4

()10

x g x --=在[6,8]是单调递增函数， ·············· 10分

当x =8时，函数2812

()10

x x g x -+=取得最大值为65， ················································ 11分

所以65

m ≥，所以所求的m 的最小值为6

5． ································································ 12分

解法二：（Ⅰ）同解法一；

（Ⅱ）当6x ≤≤8时，110102(4)[]824(6)2

y x m x x x =?-+=-++--， ······················· 8分

注意到18y x =-及2102

y x =-（14m ≤≤且m ∈R ）均关于x 在[6,8]上单调递减，

则1082

y x x =-+-关于x 在[6,8]上单调递减， ····························································· 10分

故10588823m m y -+=-≥，由523m

≥，得65

m ≥， ·

······················································ 11分所以所求的m 的最小值为6

． ······················································································· 12分

21．本题主要考查抛物线的标准方程与性质、直线与抛物线的位置关系、归纳推理等基础知识，考查推理论证能力、运算求解能力，考查化归与转化思想、数形结合思想、特殊与一般思想等．解：（Ⅰ）依题意可设抛物线Γ的方程为：22x py =（0p >）． ··································· 1分

由焦点为(0,1)F 可知12

=，所以2p =．······································································· 2分

所以所求的抛物线方程为24x y =． ················································································ 3分（Ⅱ）方法一：

设切点A 、B 坐标分别为22

1212,,,44x x x x ???? ? ?????，由（Ⅰ）知，1

2y x '=．

则切线PA PB 、的斜率分别为12

112211

,22

x x x x k y x k y x ==''==

==，故切线PA PB 、的方程分别为211111()42y x x x x -=-，222211

()42y x x x x -=-， ·

············ 4分

联立以上两个方程，得1212

,214x x x y x x +?=????=??

.故P 的坐标为12121(,)24x x x x +， ··························· 5分

因为点P 在抛物线Γ的准线上，所以121

x x =-，即124x x =-． ································· 6分

设直线AB 的方程为y kx m =+，代入抛物线方程24x y =，得2440x kx m --=，所以124x x m =-，即44m -=-，所以1m =． ································································· 7分故AB 的方程为1y kx =+，故直线AB 恒过定点(0,1)． ················································· 8分

方法二：设切点A 、B 坐标分别为22

12,,,44x x x x ???? ? ??

???，设(),1P m -，

易知直线PA PB 、斜率必存在，可设过点P 的切线方程为()1y k x m +=-．

由()21,4,y k x m x y ?+=-?=?

，消去y 并整理得()24410x kx km -++=． ····················· ①

因为切线与抛物线有且只有一个交点，

所以()2

416(1)0k km ?=-+=，整理得210k mk --=， ··································· ②

所以直线PA PB 、斜率12k k ,为方程②的两个根，故121k k ?=-， ·································· 4分另一方面，由0?=可得方程①的解为2x k =，所以11222,2x k x k ==． ···································································································· 5分假设存在一定点，使得直线AB 恒过该定点，则由抛物线对称性可知该定点必在y 轴上，设该定点为(0,)C c ， ································································································· 6分

则22

12(,),(,)44

x x CA x c CB x c =-=-．

所以//CA CB ，所以222112()()044x x x c c x ---=，整理得121221()()4

x x

c x x x x -=-

所以12x x ≠，

所以12124144

x x k k

c =-=-= ······························································································· 7分

所以直线AB 过定点()0,1． ····························································································· 8分（Ⅲ）结论一：若点P 为直线:l y t =（0t <）上的任意一点，过点P 作抛物线:Γ22x py =（0p >）的切线,PA PB ，切点分别为,A B ，则直线AB 恒过定点(0,)t -． ·············· 12分结论二：过点()0,Q m （0m >）任作一条直线交抛物线()2:20x py p Γ=>于,A B 两点，分别以点,A B 为切点作该抛物线的切线，两切线交于点P ，则点P 必在定直线y m =-上． ········································································································································· 12分结论三：已知点P 为直线:l y kx b =+上的一点，若过点P 可以作两条直线与抛物线:Γ22x py =（0p >）相切，切点分别为,A B ，则直线AB 恒过定点(),pk b -． ······ 12分

说明：①以上两结论只要给出其中一个即可或给出更一般性的结论； ②以上两结论中的抛物线开口方向均可改变；

考查推理论证能力、运算求解能力等，考查函数与方程思想、化归与转化思想、数形结合思想等．

解：（Ⅰ）函数()y f x =在π

(0,)2

上的零点的个数为1． ·············································· 1分

理由如下：

因为()e sin cos x f x x x =-，所以()e sin e cos sin x x f x x x x '=++． ······························· 2分

因为π

x <<

，所以()0f x '>，所以函数()f x 在π

(0,)2

上是单调递增函数． ··································································· 3分

因为(0)10f =-<，π

2π

()e 02

f =>，

根据函数零点存在性定理得

函数()y f x =在π

(0,)2

上的零点的个数为1．·································································· 4分

（Ⅱ）因为不等式12()()f x g x m +≥等价于12()()f x m g x -≥，

所以 12ππ

[0,],[0,]22

x x ?∈?∈，使得不等式12()()f x g x m +≥成立，等价于

()1min 2min ()()f x m g x -≥，即1min 2max ()()f x m g x -≥． ·················································· 6分当π[0,]2x ∈时，()e sin e cos sin 0x x f x x x x '=++>，故()f x 在区间π

[0,]2

上单调递增，所

以0x =时，()f x 取得最小值1-． ················································································· 7分

又()cos sin x g x x x x '=-，由于0cos 1,sin x x x x ≤≤≥

所以()g x '0<，故()g x 在区间π

[0,]2

上单调递减，

因此，0x =时，()g x 取得最大值 ······································································· 8分

所以(1m --≥，所以1m ≤-

．

所以实数m 的取值范围是(

,1-∞--．

······································································ 9分

（Ⅲ）当1x >-时，要证()()0f x g x ->，只要证()()f x g x >，

只要证e sin cos cos x x x x x x ->，

只要证(()e sin 1cos x x x x +>+，

由于sin 0,10x x +>+>，只要证e

1x x >+·············································· 10分下面证明1x >-时，不等式e

1x x >+成立．令()()e 11x

h x x x =>-+，则()()()()22

e 1e e 11x x x x x h x x x +-'==++，当()1,0x ∈-时，()0h x '<，()h x 单调递减；当()0,x ∈+∞时，()0h x '>，()h x 单调递增．

所以当且仅当0x =时，()h x 取得极小值也就是最小值为1．

令

k ，其可看作点()sin ,cos A x x 与点()

B 连线的斜率，

所以直线AB 的方程为：(y k x =，

由于点A 在圆221x y +=上，所以直线AB 与圆221x y +=相交或相切，

当直线AB 与圆221x y +=相切且切点在第二象限时，

直线AB 取得斜率k 的最大值为1． ··············································································· 12分

故0x =时，()10k h =<=；0x ≠时，()1h x k >≥． ··········································· 13分综上所述，当1x >-时，()()0f x g x ->成立． ························································ 14分

2018年高三数学模拟试题理科

黑池中学2018级高三数学期末模拟试题理科（四）一、选择题：本大题共12小题，每小题5分，共60分. 1.已知集合{}2,101，， -=A ，{} 2≥=x x B ，则A B =I A ．{}2,1,1- B.{ }2,1 C.{}2,1- D. {}2 2.复数1z i =-,则z 对应的点所在的象限为 A ．第一象限 B.第二象限 C.第三象限 D.第四象限 3 .下列函数中,是偶函数且在区间(0,+∞)上单调递减的函数是 A ．2x y = B ．y x = C ．y x = D ．2 1y x =-+ 4.函数 y=cos 2(x + π4 )－sin 2(x + π4 )的最小正周期为 A. 2π B. π C. π2 D. π 4 5. 以下说法错误的是（） A ．命题“若x 2 -3x+2=0,则x=1”的逆否命题为“若x≠1,则x 2 -3x+2≠0” B ．“x=2”是“x 2 -3x+2=0”的充分不必要条件 C ．若命题p:存在x 0∈R,使得2 0x -x 0+1<0,则﹁p:对任意x∈R,都有x 2 -x+1≥0 D ．若p 且q 为假命题,则p,q 均为假命题 6.在等差数列{}n a 中, 1516a a +=,则5S = A ．80 B ．40 C ．31 D ．-31 7.如图为某几何体的三视图，则该几何体的体积为 A ．π16+ B ．π416+ C ．π8+ D ．π48+ 8.二项式6 21()x x +的展开式中，常数项为 A ．64 B ．30 C ． 15 D ．1 9.函数3 ()ln f x x x =-的零点所在的区间是 A ．(1,2) B ．(2,)e C ． (,3)e D ．(3,)+∞ 10.执行右边的程序框图，若0.9p =，则输出的n 为 A. 6 B. 5 C. 4 D. 3 开始 10n S ==， S p

2020-2021高考理科数学模拟试题

高三上期第二次周练数学（理科）第Ⅰ卷（选择题，共60分）一、选择题：本大题共12个小题,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．设集合{}=0123A ，，，， {}=21B x x a a A =-∈，，则=( )A B ? A. {}12， B. {}13， C. {}01 ， D. {}13-， 2．已知i 是虚数单位，复数z 满足()12i z i +=，则z 的虚部是（） A. i - B. i C. 1- D. 1 3．在等比数列{}n a 中， 13521a a a ++=， 24642a a a ++=，则数列{}n a 的前9项的和9S =（） A. 255 B. 256 C. 511 D. 512 4．如图所示的阴影部分是由x 轴，直线1x =以及曲线1x y e =-围成，现向矩形区域OABC 内随机投掷一点，则该点落在阴影区域的概率是（） A. 1e B. 21 e e -- C. 11e - D. 11e - 5．在 52)(y x x ++ 的展开式中，含 2 5y x 的项的系数是（） A. 10 B. 20 C. 30 D. 60 6．已知一个简单几何体的三视图如右图所示，则该几何体的体积为 ( ) A. 36π+ B. 66π+ C. 312π+ D. 12 7．已知函数 ())2log(x a x f -= 在 )1,(-∞上单调递减，则a 的取值范围是( ) A. 11<<