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2019年广东省佛山二模试题(理数,Word精美版)

2019年广东省佛山二模试题(理数,Word精美版)
2019年广东省佛山二模试题(理数,Word精美版)

2019年佛山市普通高中高三教学质量检测(二)

数 学 (理科)

一、选择题:本大题共8小题,每小题5分,共40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.设全集{}1,2,3,4,5U =,集合{}1,2A =,{}2,3B =,则()U A B =e( )

A .{}4,5

B .{}2,3

C .{}1

D .{}1 2.设向量a 、b 满足:1=a ,2=b ,()0?-=a a b ,则a 与b 的夹角是( )

A .30?

B .60?

C .90?

D .120? 3.若0,0x y ≥≥,且21x y +=,则223x y +的最小值是( )

A .2

B .

34 C .2

3

D .0 4.已知,a b 为实数,则“||||1a b +<”是“1||2a <且1

||2

b <”的( )

A .充分不必要条件

B .必要不充分条件

C .充分必要条件

D .既不充分也不必要条件 5.函数x

y =,()(),00,x ππ∈-的图像可能是下列图像中的( )

A .

B .

C .

D .

6.已知直线m 、l 与平面α、β、γ满足l β

γ=,//l α,m α?,m γ⊥,则下列命题一定正确的是

( ) A .αγ⊥且 l m ⊥ B .αγ⊥且//m β

C .//m β且l m ⊥

D .//αβ且αγ⊥ 7.如图所示为函数()()2sin f x x ω?=+(0,0ω?π>≤≤)的部 分图像,其中,A B 两点之间的距离为5,那么()1f -=( ) A .2 B C . D .2- 8.已知函数()M f x 的定义域为实数集R ,满足()1,0,M x M

f x x M

∈?=???(M 是R 的非空真子集),在R 上有两

个非空真子集,A B ,且A

B =?,则()()()()1

1

A B A B f x F x f x f x +=

++的值域为( )

2012年4月18日

F

A

E

D

B

C

A .20,3?

? ??

? B .{}1 C .12,,123?????? D .1,13

??

????

二、填空题:本题共7小题,考生作答6小题,每小题5分,共30分 (一)必做题(9~13题)

9. 设i 为虚数单位,则()5

1i +的虚部为 .

10. 设,x y 满足约束条件0201x x y x y ≥??

-≥??-≤?

,则2z x y =+的最大值是 .

11. 抛掷一枚质地均匀的骰子,所得点数的样本空间为{}1,2,3,4,5,6S =,令事件{}2,3,5A =,事件

{}1,2,4,5,6B =,则()|P A B 的值为 .

12. 直线2y x =和圆221x y +=交于,A B 两点,以Ox 为始边,OA ,OB 为终边的角分别为,αβ,则

()sin αβ+的值为 .

13. 已知等比数列{}n a 的首项为2,公比为2,则

1

123n n

a a a a a a a a a a +=???

? .

(二)选做题(14、15题,考生只能从中选做一题,两题全答的,只计前一题的得分)

14.(坐标系与参数方程选做题)在极坐标系中,射线()03

π

θρ=

≥与曲线1C :4sin ρθ=的异于极点的交

点为A ,与曲线2C :8sin ρθ=的异于极点的交点为B ,则||AB =________. 15.(几何证明选做题)如图,已知圆中两条弦AB 与CD 相交于点F ,E 是

AB 延长线上一点,

且DF CF ==:::4:2:1AF FB BE ,若CE

与圆相切,则线段CE 的长为 .

三、解答题:本大题共6小题,满分80分,解答须写出文字说明、证明过程或演算步骤.

16.(本题满分12分)

在四边形ABCD 中,2AB =,4BC CD ==,6AD =,A C π∠+∠=. (Ⅰ)求AC 的长; (Ⅱ)求四边形ABCD 的面积.

P

C

E

F

B

A

17.(本题满分12分)

空气质量指数PM2.5(单位:3/g m μ)表示每立方米空气中可入肺颗粒物的含量,这个值越高,就代表空气污染越严重:

35 3575 75115 115150 150

250 一级 二级 三级

四级

五级 优

轻度污染 中度污染

重度污染

严重污染形图:

(Ⅰ)估计该城市一个月内空气质量类别为良的概率;(Ⅱ)在上述30个监测数据中任取2个,设X 为空气 质量类别为优的天数,求X 的分布列.

18.(本题满分14分)

如图所示四棱锥P ABCD -中,PA ⊥底面ABCD ,四边形ABCD 中,AB AD ⊥,//BC AD ,2PA AB BC ===,4AD =,E 为PD 的中 点,F 为PC 中点.

(Ⅰ)求证:CD ⊥平面PAC ; (Ⅱ)求证://BF 平面ACE ;

(Ⅲ)求直线PD 与平面PAC 所成的角的正弦值;

19.(本题满分14分)

已知椭圆E :()222210x y a b a b +=>>

的一个交点为()

1F ,

而且过点12H ???.

(Ⅰ)求椭圆E 的方程;

(Ⅱ)设椭圆E 的上下顶点分别为12,A A ,P 是椭圆上异于 12,A A 的任一点,直线12,PA PA 分别交x 轴于点,N M ,若直线

OT 与过点,M N 的圆G 相切,切点为T .证明:线段OT 的长

为定值,并求出该定值.

20.(本题满分14分)

记函数()()(

)*

112,n

n f x x n n =+-≥∈N

的导函数为()n

f x ',函数()()n

g x f x nx =-.

(Ⅰ)讨论函数()g x 的单调区间和极值; (Ⅱ)若实数0x 和正数k 满足:()()()

()

0101n n

n n f x f k f x f k ++'=',求证:00x k <<.

21.(本题满分14分)

设曲线C :2

2

1x y -=上的点P 到点()0,n n A a 的距离的最小值为n d ,若00a =

,1n n a -,*

n ∈N

(Ⅰ)求数列{}n a 的通项公式; (Ⅱ)求证:

3

212124

35

2146

22

n n

n n a a a a a a a a a a a a -+++++

<+++

; (Ⅲ)是否存在常数M ,使得对*

n ?∈N ,都有不等式:33312

111

n

M a a a +++

<成立?请说明理由.

2019年佛山市普通高中高三教学质量检测(二)参考答案

数 学 (理科)

一、选择题:本题共8小题,每小题5分,共40分

2012年4月18日

A

B

C

D

P

D

E

F

A

O

G

二、填空题:本题共7小题,考生作答6小题,每小题5分,共30分

9.4-; 10.5; 11.

25; 12.45-; 13.4;

三、解答题:本大题共6小题,满分80分.解答应写出文字说明、证明过程和演算步骤

16.【解析】(Ⅰ)如图,连结AC ,依题意可知,B D π+=, 在ABC ?中,由余弦定理得2

2

2

24224cos AC B =+-?? 2016cos B =-

在ACD ?中,由余弦定理得2

2

2

64264cos AC D =+-?? 5248cos 5248cos D B =-=+

由2016cos 5248cos B B -=+,解得1

cos 2

B =-

从而2

2016cos 28AC B =-=,即AC =6分 (Ⅱ)由(Ⅰ)可知sin sin B D ==, 所以11

sin sin 22

ABCD ABC ACD S S S AB BC B AD CD D ??=+=

?+?==………12分 17.【解析】(Ⅰ)由条形统计图可知,空气质量类别为良的天数为16天, 所以此次监测结果中空气质量类别为良的概率为 168

3015

=.…………………4分 (Ⅱ)随机变量X 的可能取值为0,1,2,则

()2

222302310435C P X C ===,()118222301761435C C P X C ===,()282

3028

2435

C P X C === 所以X 的分布列为:

18.【解析】(Ⅰ)因为PA ⊥底面ABCD ,CD ?面ABCD ,

所以PA CD ⊥,又因为直角梯形面ABCD 中,AC CD == 所以222

AC CD AD +=,即AC CD ⊥,又PA

AC A =,所以CD ⊥平面PAC ;………4分

(Ⅱ)解法一:如图,连接BD ,交AC 于O ,取PE 中点G , 连接,,BG FG EO ,则在PCE ?中,//FG CE ,

又EC ?平面ACE ,FG ?平面ACE ,所以//FG 平面ACE ,

因为//BC AD ,所以BO GE

OD ED

=,则//OE BG , ……12分

P

C D

E

F

B A O G H

又OE ?平面ACE ,BG ?平面ACE ,所以//BG 平面ACE , 又BG

FG G =,所以平面//BFG 平面ACE ,

因为BF ?平面BFG ,所以//BF 平面ACE .………10分

解法二:如图,连接BD ,交AC 于O ,取PE 中点G , 连接FD 交CE 于H ,连接OH ,则//FG CE ,

在DFG ?中,//HE FG ,则12

GE FH ED HD ==,

在底面ABCD 中,//BC AD ,所以

1

2

BO BC OD AD ==, 所以

1

2

FH BO HD OD ==,故//BF OH ,又OH ?平面ACE ,BF ?平面ACE , 所以//BF 平面ACE .………10分

(Ⅲ)由(Ⅰ)可知,CD ⊥平面PAC ,所以DPC ∠为直线PD 与平面PAC 所成的角

,

在Rt

PCD ?中

,CD PD

==

=

所以sin CD DPC PD ∠=

=

=

, 所以直线PD 与平面PAC ………14分 19.【解析】(Ⅰ)解法一:由题意得22

3a b -=,223114a b

+=,解得224,1a b ==,

所以椭圆E 的方程为2

214

x y +=.

………………………………………………4分 解法二

:椭圆的两个交点分别为())

12,F F ,

由椭圆的定义可得1271

2||||422

a PF PF =+=

+=,所以2a =,21b =, 所以椭圆E 的方程为2

214

x y +=.………………………………………………4分 (Ⅱ)解法一:由(Ⅰ)可知()()120,1,0,1A A -,设()00,P x y , 直线1PA :0011y y x x --=,令0y =,得0

01

N x x y -=

-; 直线2PA :0011y y x x ++=

,令0y =,得001M x

x y =+; 设圆G 的圆心为00001,211x x h y y ????- ? ? ?+-????

, 则2r =2

2

22

0000000000112111411x x x x x h h y y y y y ??????--+=++?? ? ?+-++-??????,2

2200001411x x OG h y y ??=-+ ?+-??

22

222222

000002

00000114114111x x x x x OT OG r h h y y y y y ????=-=++---=

? ?+-+--???? 而

22

0014

x y +=,所以()220041x y =-,所以()2

02204141y OT y -==-, 所以||2OT =,即线段OT 的长度为定值2.…………………………………………14分

解法二:由(Ⅰ)可知()()120,1,0,1A A -,设()00,P x y , 直线1PA :0011y y x x --=,令0y =,得0

01N x x y -=

-; 直线2PA :0011y y x x ++=

,令0y =,得001

M x

x y =+; 则20002000||||111

x x x OM ON y y y -?=?=-+-,而220014x y +=,所以()22

0041x y =-,

所以2

02

0||||41

x OM ON y ?==-,由切割线定理得2||||4OT OM ON =?= 所以||2OT =,即线段OT 的长度为定值2.…………………………………………14分 20.【解析】(Ⅰ)由已知得()()11n

g x x nx =+--,所以()()

1

11n g x n x -??'=+-?

?

.………………2分

① 当2n ≥且n 为偶数时,1n -是奇数,由()0g x '>得0x >;由()0g x '<得0x <.

所以()g x 的递减区间为(),0-∞,递增区间为()0,+∞,极小值为()00g =.……………5分 ② 当2n ≥且n 为奇数时,1n -是偶数,

由()0g x '>得2x <-或0x >;由()0g x '<得20x -<<. 所以()g x 的递减区间为()2,0-,递增区间为(),2-∞-和()0,+∞,

此时()g x 的极大值为()222g n -=-,极小值为()00g =.……………8分

(Ⅱ)由()()()()0101n n

n n f x f k f x f k ++'='得()()()()()10101111111

n n

n n n x k n x k -+++-=+++-,

所以()()()10111111n n n k x n k +??+-??+=??++-??,()()()()0111111n

n

nk k x n k -++=??++-??

……………10分 显然分母()()1110n n k ??++->??

,设分子为()()()()1110n

h k nk k k =-++>

则()()()

()()()

1

1

111110n n n h k n k n k nk n n k k --'=+++-=++>

所以()h k 是()0,+∞上的增函数,所以()()00h k h >=,故00x >……………12分 又()()()(

)1

0111111n n

k n k x k n k +++-+-=

??

++-??

,由(Ⅰ)知,()()11n

g x x nx =+-- 是()0,+∞上的增函数,

故当0x >时,()()00g x g >=,即()11n

x nx +>+,所以()()1

111n k n k +++>+

所以00x k -<,从而0x k <. 综上,可知00x k <<.……………14分 21.【解析】(Ⅰ)设点(),P x y ,则2

2

1x y -=,所以

||n PA =

= 因为y R ∈,所以当2n a y =时,||n PA 取得最小值n d

,且n d =

又1n n a -,

所以1n n a +,

即1n n d +=

将1n n d +=

代入n d =

1n +=

两边平方得2212n n a a +-=,又00a =,2

12a =

故数列{}

2n a 是首项212a =,公差为2的等差数列,所以22n

a n =,

因为1n n a -0>,

所以n a ………………………………………6分

(Ⅱ)因为()()()222122120n n n n +--+=-<,所以()()()2221221n n n n +-<+

所以2221212n n n n a a a a +-+<

所以

2122122

n n n n a a a a -++<,所以321212434562122

,,

,

n n n n a a a

a a a

a a a a a a -++<<< 以上n 个不等式相加得3212124

35

2146

22

n n

n n a a a a

a a a a a a a a -+++++

<+++

.…………………

10分 (Ⅲ)

因为

31k a

=当2k

≥时

<=

=

=<=

<=

<

2

211

n

n

k k

==<=

<

所以3111142

n

n i k i

a ===<=+∑. 故存在常数14M =+

对*

n ?∈N ,都有不等式:33312111

n

M a a a +++

<成立. …………14分

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2019年全国新课标高考化学考试大纲 Ⅰ.考试性质 普通高等学校招生全国统一考试是合格的高中毕业生和具有同等学力的考生参加的选拔性考试。高等学校根据考生成绩,按已确定的招生计划,德、智、体全面衡量,择优录取。因此,高考应具有较高的信度、效度,必要的区分度和适当的难度。 Ⅱ.考试内容 根据普通高等学校对新生文化素质的要求,依据中华人民共和国教育部 2019年颁布的《普通高中课程方案(实验)》和《普通高中化学课程标准(实验)》,确定高考理工类化学科考核目标与要求。 2019年高考化学大纲考核目标与要求 化学科考试,为了有利于选拔具有学习潜能和创新精神的考生,以能力测试为主导,将在测试考生进一步学习所必需的知识、技能和方法的基础上,全面检测考生的化学科学素养。 化学科命题注重测量自主学习的能力,重视理论联系实际,关注与化学有关的科学技术、社会经济和生态环境的协调发展,以促进学生在知识和技能、过程和方法、情感态度和价值观等方面的全面发展。 (一)对化学学习能力的要求 1.接受、吸收、整合化学信息的能力 (1)能够对中学化学基础知识融会贯通,有正确复述、再现、辨认的能力。(2)能够通过对实际事物、实验现象、实物、模型、图形、图表的观察,以及对自然界、社会、生产、生活中的化学现象的观察,获取有关的感性知识和印象,并进行初步加工、吸收、有序存储的能力。 (3)能够从试题提供的新信息中,准确地提取实质性内容,并经与已有知识块整合,重组为新知识块的能力。 2.分析问题和解决(解答)化学问题的能力 (1)能够将实际问题分解,通过运用相关知识,采用分析、综合的方法,解决简单化学问题的能力。 (2)能够将分析解决问题的过程和成果,用正确的化学术语及文字、图表、模型、图形等表达并做出解释的能力。 3.化学实验与探究能力 (1)了解并初步实践化学实验研究的一般过程,掌握化学实验的基本方法和技能。 (2)在解决简单化学问题的过程中,运用科学的方法,初步了解化学变化规律,并对化学现象提出科学合理的解释。 (二)对知识内容的要求层次 为了便于考查,将高考化学命题对各部分知识内容要求的程度,由低到高分为了解、理解(掌握)、综合应用三个层次,高层次的要求包含低层次的要求。其含义分别为: 了解:对所学化学知识有初步认识,能够正确复述、再现、辨认或直接使用。 理解(掌握):领会所学化学知识的含义及其适用条件,能够正确判断、解释和说明有关化学现象和问题,即不仅“知其然”,还能“知其所以然” 综合应用:在理解所学各部分化学知识的本质区别与内在联系的基础上,运用所

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