福建省2017届高三二模试卷(理)(含详细解析版)

福建省2017届高考数学二模试卷（理科）

一、选择题（共12小题，每小题5分，满分60分）

1．已知i为虚数单位，若复数z满足（1﹣i）z=1+i，则|z|=（）

A．1 B．

C．D．2

2．cos10°sin70°﹣cos80°sin20°=（）

A．B．

C．﹣D．﹣

3．“方程f′（x）=0有解”是“函数y=f（x）有极值”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．甲、乙两人各写一张贺年卡随意送给丙、丁两人中的一人，则甲、乙将贺年卡送给同一人的概率是（）

A．B．

C．D．

5．平面内动点P到两点A、B距离之比为常数λ（λ＞0，λ≠1），则动点P的轨迹叫做阿波罗尼斯圆，若已知A（﹣2，0），B（2，0），λ=，则此阿波尼斯圆的方程为（）A．x2+y2﹣12x+4=0 B．x2+y2+12x+4=0

C．x2+y2﹣x+4=0 D．x2+y2+x+4=0

6．设四边形ABCD为平行四边形，||=6，||=4，若点M、N满足，，则=（）

A．20 B．15

C．9 D．6

7．已知三棱锥O﹣ABC底面ABC的顶点在半径为的球O表面上,且AB=，AC=，BC=2，则三棱锥O﹣ABC的体积为（）

A．1 B．C．D．

8．执行如图所示的程序框图，则输出S的值为（）

A．40 B．38 C．32 D．20

9．已知函数f（x）=sinωx+cosωx（ω＞0），在区间（﹣，）上单调递增，则ω的取值范围为（）

A．（0，1] B．[1，2）C．[，2）D．（2，+∞）10．（2x﹣）5的展开式中各项系数的和为2，则该展开式中常数项为（）

A．﹣40 B．﹣20 C．20 D．40

11．设A、B、P是双曲线（a＞0，b＞0）上不同的三个点，且A、B连线经过坐标原点，若直线PA、PB的斜率之积为，则该双曲线的离心率为（）A．B．C．D．

12．定义在R上的函数f（x），f′（x）是其导函数，且满足f（x）+f′（x）＞2，f（1）=2+，则不等式e x f（x）＞4+2e x的解集为（）

A．（﹣∞，1）B．（1，+∞）C．（﹣∞，2）D．（2，+∞）

二、填空题（共4小题，每小题5分，满分20分）

13．已知f（x）是定义R上的偶函数，且当x＞0时，f（x）=2x，则f（log4）的值为．14．已知变量x，y满足约束条件，则目标函数z=3x﹣y的取值范围是．15．过点H（1，﹣1）作抛物线Γ：x2=4y的两条切线HA、HB，切点分别为A，B，则以线

段AB为直径的圆方程为．

16．在Rt△ABC中，A=，AB=2，AC=2，线段EF在斜边BC上运动，且EF=1，设∠EAF=θ，则tanθ的取值范围是．

三、解答题（共5小题，满分60分）

17．（12分）已知正项数列{a n}的前n项和为S n，且4S n=（a n+1）2（n∈N*）．

（1）求数列{a n}的通项公式；

（2）设b n=2n?a n，求数列{b n}的前n项和T n．

18．（12分）在多面体ABCDE中，平面ABC⊥平面BCE，四边形ABED为平行四边形，AB=AC=BC=2，CE=1，BE=，O为AC的中点．

（1）求证：BO⊥AE；

（2）求平面ABC与平面ACD所成锐二面角的大小．

19．（12分）为了解甲、乙两个教学班级的学生中任选两人，记其中甲班的学生人数为ξ，求ξ的概率分布列与数学期望．

20．（12分）左、右焦点分别为F1、F2的椭圆C：+=1（a＞b＞0）经过点Q（0，

），P为椭圆上一点，△PF1F2的重心为G，内心为I，IG∥F1F2．

（1）求椭圆C的方程；

（2）M为直线x﹣y=4上一点，过点M作椭圆C的两条切线MA、MB，A、B为切点，问直线AB是否过定点？若过定点，求出定点的坐标；若不过定点，请说明理由．

21．（12分）已知函数g（x）=ln x，f（x）=ag（x）+﹣2（a+1），（a∈R）．

（1）求函数f（x）的单调区间；

（2）将函数f（x）解析式中的g（x）改为g（x）的反函数得函数h（x），若x＞0时，

h（x）≥0．求a的取值范围．

四、选修题：[选修4－4：坐标系与参数方程]（请考生在第22、23二题中任选一题作答.注意：只能做所选定的题目.如果多做，则按所做第一个题目计分.）（共1小题，满分10分）22．（10分）在直角坐标系中，以原点为极点，x轴的正半轴为极轴建立极坐标系，直线l 的极坐标方程为ρcos（θ+）=，曲线C的参数方程为（θ为参数）．（1）求直线l的直角坐标方程和曲线C的普通方程；

（2）曲线C交x轴于A、B两点，且点A的横坐标小于点B的横坐标，P为直线l上的动点，求△PAB周长的最小值．

[选修4－5：不等式选讲]

23．已知函数f（x）=|2x+1|﹣|x﹣4|．

（1）求不等式f（x）≥3的解集M；

（2）若a∈M，求证：|x+a|+|x﹣|≥．

参考答案

一、选择题

1．A

【解析】由（1﹣i）z=1+i，

得=，

则|z|=1．

故选：A．

2．B

【解析】cos10°sin70°﹣cos80°sin20°=sin80°cos20°﹣cos80°sin20°

=sin（80°﹣20°）

=sin60°

=．

故选：B．

3．B

【解析】“函数y=f（x）有极值”?“方程f′（x）=0有解”，反之不成立．

∴“方程f′（x）=0有解”是“函数y=f（x）有极值”的必要不充分条件．

故选：B．

4．A

【解析】甲、乙两人各写一张贺年卡随意送给丙、丁两人中的一人，

不同的送法有四种：甲送丙，乙送丙；甲送丙，乙送丁；甲送丁，乙送丙；甲送丁，乙送丁．甲、乙将贺年卡送给同一人的送法有两种：甲送丙，乙送丙；甲送丁，乙送丁．

∴甲、乙将贺年卡送给同一人的概率p=．

故选A．

5．D

【解析】由题意，设P（x，y），则=，

化简可得x2+y2+x+4=0，

故选：D．

6．C

【解析】∵四边形ABCD为平行四边形，点M、N满足，，

∴根据图形可得：=+=，

==，

∴=，

∵=?（）=2﹣，

2=22，

=22，

||=6，||=4，

∴=22=12﹣3=9

故选：C

7．C

【解析】∵三棱锥O﹣ABC底面ABC的顶点在半径为的球O表面上，且AB=，AC=，BC=2，

∴OA=OB=OC=AB=AC=，

∴OC2+OB2=BC2，AB2+AC2=BC2，∴∠BAC=∠BOC=90°，

取BC中点D，连结AD，OD，则AD⊥BC，OD⊥BC，

AD=OD===1，∴AD2+OD2=AO2，

∴OD⊥AD，∵BC∩AD=D，∴OD⊥平面ABC，

∴三棱锥O﹣ABC的体积为：

V O﹣ABC==

=．

故选：C．

8．B

【解析】分析程序中各变量、各语句的作用，

再根据流程图所示的顺序，可知：

该程序的作用是

累加S=4×5+3×4+2×3的值，

且S=4×5+3×4+2×3=38．

故选：B．

9．A

【解析】函数f（x）=sinωx+cosωx（ω＞0），

化简可得：f（x）=2sin（ωx+），

∵在区间（﹣，）上单调递增，

∴且，k∈Z，

解得：k∈Z，

∵ω＞0，

当k=0时，可得0＜ω≤1，

故选A

10．D

【解析】令x=1则有1+a=2，得a=1，故二项式为（x+）（2x﹣）5故其常数项为﹣22×C53+23C52=40．

故选：D．

11．A

【解析】根据双曲线的对称性可知A，B关于原点对称，

设A（x1，y1），B（﹣x1，﹣y1），P（x，y），

则，

∴k PA?k PB==﹣=﹣，

∴该双曲线的离心率e==．

故选：A．

12．B

【解析】令g（x）=e x f（x）﹣2e x﹣4，g′（x）=e x f（x）+e x f′（x）﹣2e x =e x[f（x）+f′（x）﹣2]；

∵f（x）+f′（x）＞2；

∴g′（x）＞0；

∴g（x）在R上单调递增；

；

∴；

∴x＞1时，g（x）＞0；

∴原不等式的解集为（1，+∞）．

故选B．

二、填空题

13．3

【解析】∵f（x）是R上的偶函数，

∴f（﹣x）=f（x），

∵当x＞0时，f（x）=2x，

∴f（log4）=f（log49）=f（log23）=3，

故答案为3．

14．[﹣，6]

【解析】∵变量x，y满足约束条件，

目标函数为：z=3x﹣y，

直线4x﹣y+1=0与x+2y﹣2=0交于点A（0，1），

直线2x+y﹣4=0与x+2y﹣2=0交于点B（2，0），

直线4x﹣y+1=0与2x+y﹣4=0交于点C（，3），

分析可知z在点C处取得最小值，z min=3×﹣1=﹣，

z在点B处取得最大值，z max=3×2﹣0=6，

∴﹣≤z≤6，

故答案为[﹣，6]；

15．

【解析】设A（x1，y1），B（x2，y2），

∵抛物线x2=4y，∴y′=x，

∴过点A的切线方程为y﹣y1=x1（x﹣x1），即x1x﹣2y﹣2y1=0．

H（1，﹣1）代入可得x1﹣2y1+2=0，

同理x2﹣2y2+2=0，

∴A（x1，y1），B（x2，y2）都满足方程x﹣2y+2=0，即为直线AB的方程，与抛物线Γ：x2=4y联立，可得x2﹣2x﹣4=0，∴AB的中点坐标为（1，），|AB|==5

∴以线段AB为直径的圆方程为，

故答案为．

16．[，]

【解析】如图建立直角坐标系，设BF=k，k∈[0，3]．

∴∠B=60°，∴F（2﹣，），E（，）．

∴tan∠EAB=，tan∠FAB=，．

tanθ=tan（∠EAB﹣∠FAB）=；

∵k∈[0，3]．∴，tanθ的取值范围是[] 故答案为[]．

三、解答题

17．解：（1）当n=1时，，∴a1=1

当n≥2时，，又，

两式相减得：，

即（a n+a n﹣1）（a n﹣a n﹣1﹣2）=0，

由a n＞0，∴a n﹣a n﹣1=2，

所以，数列{a n}是首项为1，公差为2的等差数列，即a n=2n﹣1．

（2）∵，

∴①

②

①﹣②得﹣T n=2+2（22+23+…+2n）﹣（2n﹣1）×2n+1=

=2﹣8+2n+2﹣（2n﹣1）×2n+1

=﹣6+2n+1（2﹣2n+1）=﹣6+2n+1（3﹣2n）

∴．

18．证明：（1）∵AB=AC=BC=2，

又O为AC中点，∴BO⊥AC

又，∴BC2+CE2=BE2，∴BC⊥CE

又∵平面ABC⊥平面BCE，且平面ABC∩平面BCE=BC，∴CE⊥平面ABC

∴CE⊥BO，又CE∩AC=C，∴BO⊥平面ACE

∵AE?平面ACE，∴BO⊥AE．

解：（2）以C为原点，CB为x轴，CE为y轴，建立空间直角坐标系C﹣xyz，

则

∵

由（1）知，是平面ABC的平面角，

设平面ACD的法向量为=（x，y，z），

，

∴，取x=1，得

设平面ABC与平面ACD所成锐二面角为θ，

则

∴，∴平面ABC与平面ACD所成锐二面角的大小为．

19．解：（Ⅰ）根据频率分布直方图可得甲班的成绩“优秀”的人数，（0.032+0.024）×10

×50=28，

“不优秀”的人数：50﹣28=22．

根据已知表格可得：乙班的成绩“优秀”的人数，12+8=20，

“不优秀”的人数：50﹣20=30．

可得以下表格：

根据列联表数据，．

所以，有85%的把握认为“成绩优秀”与所在教学班级有关．

（Ⅱ）由已知甲、乙两班级不及格人数分别是：4人、6人ξ的所有取值为：0，1，2 ，，

所求分布列的数学期望为：

20．解：（1）∵椭圆C：+=1（a＞b＞0）焦点在x轴上，且过点，∴

设△PF1F2内切圆的半径为r，点P的坐标为（x0，y0），

则△PF1F2重心G的坐标为，

∵IG∥F1F2，∴|y0|=3r．

由△PF1F2面积可得）r=，

即a=2c，，

则解得，

即所求的椭圆方程为则椭圆方程为

（2）设M（x1，y1），A（x2，y2），B（x3，y3）则切线MA，MB的方程分别为，．

∵点M在两条切线上，

∴，，

故直线AB的方程为．

又∵点M为直线x﹣y=4上，

∴y1=x1﹣4

即直线AB的方程可化为，整理得（3x+4y）x1=16y+12，

由解得，

因此，直线AB过定点．

21．解：（1）∵g（x）=ln x，

∴f（x）=ag（x）+﹣2（a+1）=a ln x+﹣2（a+1），（a∈R）；

∴f（x）定义域为（0，+∞），

且；

①当﹣1≤a≤0时，f'（x）＜0，即f（x）的单调减区间为（0，+∞）；

②当a＞0时，f（x）的单调增区间为，

单调减区间为；

③当a＜﹣1时，f（x）的单调增区间为，

单调减区间为；

（2）由题意得，

∵x＞0时，h（x）≥0，∴h（1）≥0，

则a（e﹣1）≥1，即；

则由，得，

即，x∈（0，+∞）；

设，

则；

令u'（x）=0，解得x=1或x=﹣?（0，+∞）舍去；

u'（x）＜0时，x∈（0，1）；u′（x）＞0 时，x∈（1，+∞）；

∴[u（x）]min=u（1）=，

∴≥，解得a≥；

故a的取值范围是[，+∞）．

四、选修题

22．解：（1）∵直线l的极坐标方程为ρcos（θ+）=，

∴由直线l的极坐标方程，得=，即ρcosθ﹣ρsinθ=1，

∴直线l的直角坐标方程为x﹣y=1，即x﹣y﹣1=0，

∵曲线C的参数方程为（θ为参数），

∴由曲线C的参数方程得C的普通方程为：（x﹣5）2+y2=1．

（2）由（1）知曲线C表示圆心（5，0），半径r=1的圆，

令y=0，得x=4或x=6．

∴A点坐标为（4，0），B点坐标为（6，0）．

作A关于直线l的对称点A1得A1（1，3）．

由题设知当P为A1B与l的交点时，△PAB的周长最小，

∴△PAB周长的最小值为：|AP|+|PB|+|AB|=|A1B|+|AB|=．

23．（1）解：f（x）≥3可化为：|2x+1|﹣|x﹣4|≥3

即或或

解得x≤﹣8或x≥2，所以不等式的解集M为{x|x≤﹣8或x≥2} （2）证明：∵|x+a|+|x﹣|≥|a+|=|a|+||

令|a|=t，则t∈[2，+∞）

则y=y+是[2，+∞）上的增函数，

因此，y，故|x+a|+|x﹣|≥．

2017届上海市徐汇区高三英语二模卷(含听力文本和答案)

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北京市海淀区2017届高三英语二模试题答案(最新整理)

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长宁卷 72. 只有多练，你才能提高驾驶技术。(Only) 73. 在旅游旺季机票订得越早越便宜。(book) 74. 无论白天在学校发生了什么事情，晚上要尽量把负担卸下。(No matter) 75. 随着互联网的发展，海量信息唾手可得，或许你觉得在也没有必要去图书馆了。 (need n.) 72. Only through frequent practice can you improve your driving skills 73. The earlier you book a plane ticket in a tourist season , the lower its price will be . 74. No matter what happens at school during the day , as early in the evening as you can , put all your burden down. 或者..., burdens should be put down as many as possible in the evening. 75. With the development / growth of the Internet , an enormous amount of information is at hand./ at our fingertips / available so that you may think there is no need to go to the library . 杨浦卷 1. 新颁布的禁烟令得到了广大市民的支持。（ban） 2. 出乎我的意料，年轻人对中国古诗词显示出了极大的热情。（passion） 3. 共享单车不仅解决了最后一里路的问题，而且还有助于改善空气质量。 （Not only） 4. 一考定终身的日子已经一去不复返了，但不可否认的是考试越多，学生压力 越大。（denying） 72. The newly-issued ban on smoking has been supported by most citizens 73. Beyond my expectation, young people showed tremendous passion for ancient Chinese poetry. 74. Not only do shared bicycles solve the last-mile problem but also they help improve air quality 75. Gone are the days when one examination could determine students' fate, but there is no denying that the more examinations they have, the more stressful they will feel. 青浦卷 72. 演讲前，花点时间看一下提纲，你会更自信。(spend) 73. 两年后这个主题公园将对公众开放，但是它是否会盈利尚未确定。(remain)