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江苏省苏北四市(徐州、淮安、连云港、宿迁)联考2017届高三(上)期末数学试卷(解析版)

江苏省苏北四市(徐州、淮安、连云港、宿迁)联考2017届高三(上)期末数学试卷(解析版)
江苏省苏北四市(徐州、淮安、连云港、宿迁)联考2017届高三(上)期末数学试卷(解析版)

2016-2017学年江苏省苏北四市(徐州、淮安、连云港、宿迁)

联考高三(上)期末数学试卷

一、填空题(本大题共14小题,每小题5分,共70分)

1.已知集合A={﹣2,0},B={﹣2,3},则A∪B=.

2.已知复数z满足(1﹣i)z=2i,其中i为虚数单位,则z的模为.

3.某次比赛甲得分的茎叶图如图所示,若去掉一个最高分,去掉一个最低分,则剩下4个分数的方差为.

4.根据如图所示的伪代码,则输出S的值为.

5.从1,2,3,4,5,6这六个数中一次随机地取2个数,则所取2个数的和能被3整除的概率为.

6.若抛物线y2=8x的焦点恰好是双曲线的右焦点,则实数a

的值为.

7.已知圆锥的底面直径与高都是2,则该圆锥的侧面积为.

8.若函数的最小正周期为,则的值为.

9.已知等比数列{a n}的前n项和为S n,若S2=2a2+3,S3=2a3+3,则公比q的值为.

10.已知函数f(x)是定义R在上的奇函数,当x>0时,f(x)=2x﹣3,则不等

式f(x)≤﹣5的解集为.

11.若实数x,y满足,则的最小值为.

12.已知非零向量满足,则与夹角的余弦值为.

13

.已知A,B是圆上的动点,,P是圆

上的动点,则的取值范围为.

14.已知函数,若函数f(x)的图象与直线y=x 有三个不同的公共点,则实数a的取值集合为.

二、解答题(本大题共6小题,共90分.解答应写出必要的文字说明、证明或演算步骤)

15.(14分)在△ABC中,角A,B,C的对边分别为a,b,c.已知2cosA(bcosC+ccosB)=a.

(1)求角A的值;

(2)若,求sin(B﹣C)的值.

16.(14分)如图,在四棱锥E﹣ABCD中,平面EAB⊥平面ABCD,四边形ABCD 为矩形,EA⊥EB,点M,N分别是AE,CD的中点.

求证:(1)直线MN∥平面EBC;

(2)直线EA⊥平面EBC.

17.(14分)如图,已知A,B两镇分别位于东西湖岸MN的A处和湖中小岛的

B处,点C在A的正西方向1km处,tan∠BAN=,∠BCN=,现计划铺设一条

电缆联通A ,B 两镇,有两种铺设方案:①沿线段AB 在水下铺设;②在湖岸MN 上选一点P ,先沿线段AP 在地下铺设,再沿线段PB 在水下铺设,预算地下、水下的电缆铺设费用分别为2万元∕km 、4万元∕km . (1)求A ,B 两镇间的距离;

(2)应该如何铺设,使总铺设费用最低?

18.(16分)在平面直角坐标系xOy 中,已知椭圆C : +=1(a >b >0)

的离心率为

,且右焦点F 到左准线的距离为6

(1)求椭圆C 的标准方程;

(2)设A 为椭圆C 的左顶点,P 为椭圆C 上位于x 轴上方的点,直线PA 交y 轴于点M ,过点F 作MF 的垂线,交y 轴于点N .

(i )当直线PA 的斜率为时,求△MFN 的外接圆的方程; (ii )设直线AN 交椭圆C 于另一点Q ,求△PAQ 的面积的最大值.

19.(16分)已知函数ax e

x x f -=2)(2

,ax x x g -=ln )(,R a ∈

(1)解关于x (x ∈R )的不等式f (x )≤0; (2)证明:f (x )≥g (x );

(3)是否存在常数a ,b ,使得f (x )≥ax +b ≥g (x )对任意的x >0恒成立?若存在,求出a ,b 的值;若不存在,请说明理由.

20.(16分)已知正项数列{a n }的前n 项和为S n ,且a 1=a ,(a n +1)(a n +1+1)=6(S n +n ),n ∈N *.

(1)求数列{a n }的通项公式;

(2)若对于?n ∈N *,都有S n ≤n (3n +1)成立,求实数a 取值范围;

(3)当a=2时,将数列{a n }中的部分项按原来的顺序构成数列{b n },且b 1=a 2,

证明:存在无数个满足条件的无穷等比数列{b n}.

附加题[选做题]本题包括A、B、C、D四小题,请选定其中两题,并在相应的答题区域内作答.若多做,则按作答的前两题评分.解答时应写出文字说明、证明过程或演算步骤.A.[选修4-1:几何证明选讲](本小题满分0分)

21.如图,AB为半圆O的直径,D为弧BC的中点,E为BC的中点,求证:AB?BC=2AD?BD.

[选修4-2:矩阵与变换](本小题满分0分)

22.已知矩阵A=的一个特征值为2,其对应的一个特征向量为a=,求实数a,b的值.

[选修4-4:坐标系与参数方程](本小题满分0分)

23.在平面直角坐标系xOy中,以O为极点,x轴的正半轴为极轴建立极坐标系.直

线l:ρsin(θ﹣)=m(m∈R),圆C的参数方程为(t为参

数).当圆心C到直线l的距离为时,求m的值.

[选修4-5:不等式选讲](本小题满分0分)

24.已知a,b,c为正实数, +++27abc的最小值为m,解关于x的不等式|x+l|﹣2x<m.

【必做题】第22题、第23题,每题10分,共计20分.请在答题卡指定区域内作答,解答时应写出文字说明、证明过程或演算步骤.

25.甲、乙、丙分别从A,B,C,D四道题中独立地选做两道题,其中甲必选B

题.

(1)求甲选做D题,且乙、丙都不选做D题的概率;

(2)设随机变量X表示D题被甲、乙、丙选做的次数,求X的概率分布和数学期望E(X).

26.已知等式(1+x)2n﹣1=(1+x)n﹣1(1+x)n.

(1)求(1+x)2n﹣1的展开式中含x n的项的系数,并化简:++…+

(2)证明:()2+2()2+…+n()2=n.

2016-2017学年江苏省苏北四市(徐州、淮安、连云港、宿迁)联考高三(上)期末数学试卷

参考答案与试题解析

一、填空题(本大题共14小题,每小题5分,共70分)

1.已知集合A={﹣2,0},B={﹣2,3},则A∪B={﹣2,0,3} .

【考点】并集及其运算.

【分析】利用并集定义直接求解.

【解答】解:∵集合A={﹣2,0},B={﹣2,3},

∴A∪B={﹣2,0,3}.

故答案为:{﹣2,0,3}.

【点评】本题考查并集的求法,是基础题,解题时要认真审题,注意并集定义的合理运用.

2.已知复数z满足(1﹣i)z=2i,其中i为虚数单位,则z的模为.

【考点】复数代数形式的乘除运算.

【分析】由(1﹣i)z=2i,得,然后利用复数代数形式的乘除运算化简复数z,再由复数求模公式计算得答案.

【解答】解:由(1﹣i)z=2i,

得=,

则z的模为:.

故答案为:.

【点评】本题考查了复数代数形式的乘除运算,考查了复数模的求法,是基础题.

3.某次比赛甲得分的茎叶图如图所示,若去掉一个最高分,去掉一个最低分,则剩下4个分数的方差为14.

【考点】茎叶图.

【分析】求出剩下的4个分数平均数,代入方差公式,求出方差即可.

【解答】解:剩下的4个分数是:

42,44,46,52,

平均数是:46,

故方差是:(16+4+0+36)=14,

故答案为:14.

【点评】本题考查了读茎叶图问题,考查求平均数以及方差问题,是一道基础题.4.根据如图所示的伪代码,则输出S的值为20.

【考点】程序框图.

【分析】根据条件进行模拟计算即可.

【解答】解:第一次I=1,满足条件I≤5,I=1+1=2,S=0+2=2,

第二次I=2,满足条件I≤5,I=2+1=3,S=2+3=5,

第三次I=3,满足条件I≤5,I=3+1=4,S=5+4=9,

第四次I=4,满足条件I≤5,I=4+1=5,S=9+5=14,

第五次I=5,满足条件I≤5,I=5+1=6,S=14+6=20,

第六次I=6不满足条件I≤5,查询终止,

输出S=20,

故答案为:20

【点评】本题主要考查程序框图的识别和应用,根据条件进行模拟计算是解决本题的关键.

5.从1,2,3,4,5,6这六个数中一次随机地取2个数,则所取2个数的和能

被3整除的概率为.

【考点】列举法计算基本事件数及事件发生的概率.

【分析】基本事件总数n=,再用列举法求出所取2个数的和能被3整除包含的基本事件个数,由此能求出所取2个数的和能被3整除的概率.

【解答】解:从1,2,3,4,5,6这六个数中一次随机地取2个数,

基本事件总数n=,

所取2个数的和能被3整除包含的基本事件有:

(1,2),(1,5),(2,4),(3,6),(4,5),

共有5个,

∴所取2个数的和能被3整除的概率p=.

故答案为:.

【点评】本题考查概率的求法,是基础题,解题时要认真审题,注意列举法的合理运用.

6.若抛物线y2=8x的焦点恰好是双曲线的右焦点,则实数a 的值为1.

【考点】双曲线的简单性质.

【分析】求得抛物线的焦点,双曲线的右焦点,由题意可得方程,解方程即可得到a的值.

【解答】解:抛物线y2=8x的焦点为(2,0),

双曲线的右焦点为(,0),

由题意可得为=2,

解得a=1.

故答案为:1.

【点评】本题考查双曲线的方程和性质,同时考查抛物线的焦点,考查运算能力,属于基础题.

7.已知圆锥的底面直径与高都是2,则该圆锥的侧面积为.

【考点】旋转体(圆柱、圆锥、圆台).

【分析】首先根据底面半径和高利用勾股定理求得母线长,然后直接利用圆锥的侧面积公式代入求出即可.

【解答】解:∵圆锥的底面直径与高都是2,

∴母线长为:=,

∴圆锥的侧面积为:πrl=.

故答案为:.

【点评】本题考查了圆锥的侧面积的计算,正确理解圆锥的侧面展开图与原来的扇形之间的关系是解决本题的关键.

8.若函数的最小正周期为,则的值为﹣

【考点】正弦函数的图象.

【分析】利用正弦函数的周期性求得ω,再利用诱导公式求得的值.

【解答】解:∵函数的最小正周期为=,∴ω=10,

则=sin(10π?﹣)=sin=sin=﹣sin=﹣,

故答案为:.

【点评】本题主要考查正弦函数的周期性,利用诱导公式求三角函数的值,属于基础题.

9.已知等比数列{a n}的前n项和为S n,若S2=2a2+3,S3=2a3+3,则公比q的值为2.

【考点】等比数列的通项公式.

【分析】利用等比数列的通项公式与求和公式即可得出.

【解答】解:∵S2=2a2+3,S3=2a3+3,

∴a1=a1q+3,a1(1+q)=+3,

∴q2﹣2q=0,q≠0.

则公比q=2.

故答案为:2.

【点评】本题考查了等比数列的通项公式与求和公式,考查了推理能力与计算能力,属于中档题.

10.已知函数f(x)是定义R在上的奇函数,当x>0时,f(x)=2x﹣3,则不等式f(x)≤﹣5的解集为(﹣∞,﹣3] .

【考点】函数奇偶性的性质.

【分析】根据函数奇偶性的性质求出当x<0的解析式,讨论x>0,x<0,x=0,解不等式即可.

【解答】解:若x<0,则﹣x>0,

∵当x>0时,f(x)=2x﹣3,

∴当﹣x>0时,f(﹣x)=2﹣x﹣3,

∵f(x)是定义在R上的奇函数,

∴f(﹣x)=2﹣x﹣3=﹣f(x),

则f(x)=﹣2﹣x+3,x<0,

当x>0时,不等式f(x)≤﹣5等价为2x﹣3≤﹣5即2x≤﹣2,无解,不成立;当x<0时,不等式f(x)≤﹣5等价为﹣2﹣x+3≤﹣5即2﹣x≥8,

得﹣x≥3,即x≤﹣3;

当x=0时,f(0)=0,不等式f(x)≤﹣5不成立,

综上,不等式的解为x≤﹣3.

故不等式的解集为(﹣∞,﹣3].

故答案为:(﹣∞,﹣3].

【点评】本题主要考查不等式的解集的求解,根据函数奇偶性的性质求出函数的解析式是解决本题的关键.

11.若实数x,y满足,则的最小值为8.

【考点】基本不等式.

【分析】实数x,y满足,可得x=∈,解得y>3.则

=y+3+=y﹣3++6,利用基本不等式的性质即可得出.

【解答】解:∵实数x,y满足,

∴x=∈,解得y>3.

则=y+3+=y﹣3++6≥+6=8,当且仅当y=4

(x=)时取等号.

故答案为:8.

【点评】本题考查了基本不等式的性质,考查了推理能力与计算能力,属于中档题.

12.已知非零向量满足,则与夹角的余弦值为

【考点】平面向量数量积的运算.

【分析】利用两个向量的加减法的法则,以及其几何意义,余弦定理,数形结合

求得与夹角的余弦值.

【解答】解:非零向量满足,不妨设=1,

设与夹角为θ,如图所示:

设=,=,=+,则OA=0B=0C=1,设=2=2,则=2﹣,

∠ODA即为θ,△OAC和△OBC都是边长等于3的等边三角形.

利用余弦定理可得BD==,

cosθ==,

故答案为:.

【点评】本题主要考查两个向量的加减法的法则,以及其几何意义,余弦定理的应用,属于中档题.

13.已知A,B是圆上的动点,,P是圆

上的动点,则的取值范围为[7,13] .【考点】圆与圆的位置关系及其判定.

【分析】求出AB的中点的轨迹方程,即可求出的取值范围.

【解答】解:取AB的中点C,则=2||,C的轨迹方程是x2+y2=,|C1C2|=5

由题意,||最大值为5+1+=,最小值为5﹣1﹣=.

∴的取值范围为[7,13],

故答案为[:7,13].

【点评】本题考查圆与圆的位置关系,考查学生的计算能力,正确转化是关键.

14.已知函数,若函数f(x)的图象与直线y=x 有三个不同的公共点,则实数a的取值集合为[﹣20,﹣16] .

【考点】分段函数的应用.

【分析】因为y=sinx (x<1)与y=x无交点,故只需函数f(x)=x3﹣9x2+25x+a (x≥1)的图象与直线y=x有三个不同的公共点即可,只需g(x)=x3﹣9x2+24x+a (x≥1)与x轴有3个交点即可,

【解答】解:因为y=sinx (x<1)与y=x无交点,故只需函数f(x)=x3﹣9x2+25x+a (x≥1)的图象与直线y=x有三个不同的公共点即可,

令g(x)=x3﹣9x2+24x+a(x≥1),

g′(x)=3x2﹣18x+24=3(x2﹣6x+8)=2(x﹣2)(x﹣4),

当x∈(1,2),(4,+∞)时g(x)单调递增,当x∈(2,4)时g(x)单调递减,

依题意只需g(x)=x3﹣9x2+24x+a(x≥1)与x轴有3个交点即可,

及g(1)=16+a≤0,g(2)=20+a≥0,∴﹣20≤a≤﹣16.

故答案为[﹣20,﹣16]

【点评】题主要考查函数的图象的交点以及数形结合方法,数形结合是数学解题中常用的思想方法,属于基础题.

二、解答题(本大题共6小题,共90分.解答应写出必要的文字说明、证明或演算步骤)

15.(14分)(2016秋?淮安期末)在△ABC中,角A,B,C的对边分别为a,b,c.已知2cosA(bcosC+ccosB)=a.

(1)求角A的值;

(2)若,求sin(B﹣C)的值.

【考点】正弦定理;余弦定理.

【分析】(1)由正弦定理化简已知等式可得2cosAsinA=sinA,结合sinA≠0,可

求,结合范围A∈(0,π),可求A的值.

(2)由已知利用同角三角函数基本关系式可求sinB,利用倍角公式可求sin2B,

cos2B,由sin(B﹣C)=sin(2B﹣),利用两角差的正弦函数公式即可计算得解.

【解答】(本题满分为14分)

解:(1)由正弦定理可知,2cosA(sinBcosC+sinCcosB)=sinA,…(2分)

即2cosAsinA=sinA,

因为A∈(0,π),

所以sinA≠0,

所以2cosA=1,即,…(4分)

又A∈(0,π),

所以.…(6分)

(2)因为,B∈(0,π),

所以,…(8分)

所以,,…(10分)

所以

=…(12分)

==.…(14分)

【点评】本题主要考查了正弦定理,同角三角函数基本关系式,倍角公式,两角差的正弦函数公式在解三角形中的应用,考查了计算能力和转化思想,属于中档题.

16.(14分)(2016秋?淮安期末)如图,在四棱锥E﹣ABCD中,平面EAB⊥平面ABCD,四边形ABCD为矩形,EA⊥EB,点M,N分别是AE,CD的中点.求证:(1)直线MN∥平面EBC;

(2)直线EA⊥平面EBC.

【考点】直线与平面平行的判定;直线与平面垂直的判定.

【分析】(1)取BE中点F,连结CF,MF,证明四边形MNCF是平行四边形,所以MN∥CF,即可证明直线MN∥平面EBC;

(2)证明BC⊥平面EAB,得到BC⊥EA,又EA⊥EB,BC∩EB=B,EB,BC?平面EBC,即可证明直线EA⊥平面EBC.

【解答】证明:(1)取BE中点F,连结CF,MF,

又M是AE的中点,所以MF=AB,

又N是矩形ABCD边CD的中点,

所以NC=AB,所以MF平行且等于NC,

所以四边形MNCF是平行四边形,…(4分)

所以MN∥CF,

又MN?平面EBC,CF?平面EBC,

所以MN∥平面EBC.…(7分)

(2)在矩形ABCD中,BC⊥AB,

又平面EAB⊥平面ABCD,平面ABCD∩平面EAB=AB,BC?平面ABCD,

所以BC⊥平面EAB,…(10分)

又EA?平面EAB,所以BC⊥EA,

又EA⊥EB,BC∩EB=B,EB,BC?平面EBC,

所以EA⊥平面EBC.…(14分)

【点评】本题考查线面平行、线面垂直的证明,考查学生分析解决问题的能力,属于中档题.

17.(14分)(2016秋?淮安期末)如图,已知A,B两镇分别位于东西湖岸

MN的A处和湖中小岛的B处,点C在A的正西方向1km处,tan∠BAN=,∠

BCN=,现计划铺设一条电缆联通A,B两镇,有两种铺设方案:①沿线段AB 在水下铺设;②在湖岸MN上选一点P,先沿线段AP在地下铺设,再沿线段PB 在水下铺设,预算地下、水下的电缆铺设费用分别为2万元∕km、4万元∕km.(1)求A,B两镇间的距离;

(2)应该如何铺设,使总铺设费用最低?

【考点】导数在最大值、最小值问题中的应用.

【分析】(1)由tan∠BAN=,∠BCN=,得到|AD|,|DB|、|AB|间的关系,然后利用直角三角形的性质求解;

(2)方案①:总铺设费用为5×4=20(万元).

方案②:设∠BPD=θ,则,其中θ0=∠BAN,

在Rt△BDP中,,,

则总铺设费用为.

设,则,

,求出函数的极小值,即函数的最小值得答案.

【解答】解:(1)过B作MN的垂线,垂足为D,如图示:

在Rt△ABD中,,

所以,

在Rt△BCD中,,

所以CD=BD.

则,即BD=3,

所以CD=3,AD=4,

由勾股定理得,(km).

所以A,B两镇间的距离为5km.…(4分)

(2)方案①:沿线段AB在水下铺设时,总铺设费用为5×4=20(万元).…(6分)

方案②:设∠BPD=θ,则,其中θ0=∠BAN,

在Rt△BDP中,,,

所以.

则总铺设费用为.…(8分)

设,则,

令f'(θ)=0,得,列表如下:

所以f(θ)的最小值为.

所以方案②的总铺设费用最小为(万元),此时.…(12分)

而,

所以应选择方案②进行铺设,点P选在A的正西方向km处,总铺设费用最低.…(14分)

【点评】本题考查了简单的数学建模思想方法,考查了利用导数求函数的最值,

是中档题

18.(16分)(2016秋?淮安期末)在平面直角坐标系xOy中,已知椭圆C:

+=1(a>b>0)的离心率为,且右焦点F到左准线的距离为6.

(1)求椭圆C的标准方程;

(2)设A为椭圆C的左顶点,P为椭圆C上位于x轴上方的点,直线PA交y轴于点M,过点F作MF的垂线,交y轴于点N.

(i)当直线PA的斜率为时,求△MFN的外接圆的方程;

(ii)设直线AN交椭圆C于另一点Q,求△PAQ的面积的最大值.

【考点】椭圆的简单性质.

【分析】(1)由题意可知:离心率e==,则a=c,右焦点F到左准线的

距离c+=6,即可求得c和a的值,则b2=a2﹣c2=8,即可求得椭圆方程;(2)(i)设直线方程为:y=(x+4),求得M点,即可求得NF的方程和N的坐标,则丨MN丨=6,则以MN为圆心(0,﹣1),半径为3,即x2+(y+1)2=9;(ii)设直线方程为:y=k(x+4),代入椭圆方程,求得P点坐标,求得直线PF

方程,则求得N点坐标,则直线AN:y=﹣﹣,代入椭圆方程,求得M点坐

标,求得丨AM丨,△PAQ的面积S===

≤=10.

【解答】解:(1)由题意可知:椭圆C: +=1(a>b>0)焦点在x轴上,

由离心率e==,则a=c,

由右焦点F到左准线的距离c+=6,

解得:c=2,则a=4,

由b2=a2﹣c2=8,

∴椭圆的标准方程为:;

(2)(i)由(1)可知:椭圆的左顶点(﹣4,0),F(2,0),

设直线方程为:y=(x+4),即y=x+2,

则M(2,0),

==﹣,则k NF=,

k

MF

直线NF:y=(x﹣2)=﹣4,则N(0,﹣4),

丨MN丨=6,则以MN为圆心(0,﹣1),半径为3,即x2+(y+1)2=9,(ii)设直线方程为:y=k(x+4),

∴,整理得:(1+2k2)x2+16k2x+32k2﹣16=0,

解得:x1=4,x2=,则y2=,

则P(,),

∴k MF==﹣k,由M(0,4k),F(2,0),

∴k NF=,则NF:y=(x﹣2),

则N(0,﹣),

则直线AN:y=﹣﹣,

代入椭圆方程:整理得:(1+)x2+x+﹣16=0,

解得:x1=4,x2=,则y2=,则Q(,),

∴k PQ=,直线PQ:y﹣=(x﹣),

则x M =﹣=,

∴丨AM 丨=+4=,

△PAQ 的面积S==??=,

=

≤=10,

当且仅当2k=,即k=时,取最大值,

△PAQ 的面积的最大值10

【点评】本题考查椭圆的标准方程及简单几何性质,考查直线与椭圆的位置关系,考三角形的面积公式的应用,考查基本不等式的综合应用,属于难题.

19.(16分)已知函数ax e

x x f -=2)(2,ax x x g -=ln )(,R a ∈

(1)解关于x (x ∈R )的不等式f (x )≤0; (2)证明:f (x )≥g (x );

(3)是否存在常数a ,b ,使得f (x )≥ax +b ≥g (x )对任意的x >0恒成立?若存在,求出a ,b 的值;若不存在,请说明理由.

【考点】利用导数求闭区间上函数的最值;利用导数研究函数的单调性. 【分析】(1)通过讨论a 的范围,求出不等式的解集即可;

(2)设h (x )=f (x )﹣g (x ),求出函数的导数,解关于导函数的不等式,求出函数的单调区间,从而求出函数的最小值,证出结论即可;

(3)假设存在,得到对任意的x >0恒成立,根据函数的单调

性判断即可.

【解答】解:(1)当a=0时,,所以f (x )≤0的解集为{0};

当a ≠0时,

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