江苏2023高考数学仿真模拟试题


    

 
    

2023年江苏省高考仿真数学试卷
    

参考公式:柱体的体积,其中是柱体的底面积,是柱体的高.
    

一、填空题:本大题共14小题,每小题5分,共计70分.请把答案填写在答题卡相应位置上.
    

1.已知集合,则       .
    

2.已知是虚数单位,则复数的实部是       .
    

3.已知一组数据的平均数为4,则的值是       .
    

4.将一颗质地均匀的正方体骰子先后抛掷2次,观察向上的点数,则点数和为5的概率是       .
    

5.如图是一个算法流程图,若输出的值为,则输入的值是       .
    

F:\★★★2020\★★★★★★2020高考试卷转化\1231.tif
    

6.在平面直角坐标系xOy中,若双曲线的一条渐近线方程为,则该双曲线的离心率是       .
    

7.已知y=f(x)是奇函数,当x≥0时,,则的值是       .
    

8.已知=,则的值是       .
    

9.如图,六角螺帽毛坯是由一个正六棱柱挖去一个圆柱所构成的.已知螺帽的底面正六边形边长为2 cm,高为2 cm,内孔半轻为0.5 cm,则此六角螺帽毛坯的体积是       cm.
    

F:\★★★2020\★★★★★★2020高考试卷转化\1232.tif
    

10.将函数的图象向右平移个单位长度,则平移后的图象中与y轴最近的对称轴的方程是       .
    

11.设{an}是公差为d的等差数列,{bn}是公比为q的等比数列.已知数列{an+bn}的前n项和学科网,则d+q的值是       .
    

12.已知学科网,则学科网的最小值是       .
    

13.在△ABC中,学科网D在边BC上,延长AD到P,使得AP=9,若学科网(m为常数),则CD的长度是       .
    

F:\★★★2020\★★★★★★2020高考试卷转化\江苏13.tif
    

14.在平面直角坐标系xOy中,已知学科网,A,B是圆C:学科网上的两个动点,满足学科网,则△PAB面积的最大值是       .
    

二、解答题:本大题共6小题,共计90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.
    

15.(14分)在三棱柱ABC-A1B1C1中,AB⊥AC,B1C⊥平面ABC,E,F分别是AC,B1C的中点.
    

(1)求证:EF∥平面AB1C1;(2)求证:平面AB1C⊥平面ABB1.
    

                                               E:\2020年\2020年高考文档\定稿\未标题-2.tif
    

16.(14分)
    

在△ABC中,角A,B,C的对边分别为a,b,c,已知学科网
    

(1)求学科网的值;
    

(2)在边BC上取一点D,使得学科网,求学科网的值.
    

E:\2020年\2020年高考文档\定稿\未标题-1.tif
    

17.(14分)
    

某地准备在山谷中建一座桥梁,桥址位置的竖直截面图如图所示:谷底O在水平线MN上,桥AB与MN平行,为铅垂线(在AB上).经测量,左侧曲线AO上任一点D到MN的距离(米)与D到的距离a(米)之间满足关系式;右侧曲线BO上任一点F到MN的距离(米)与F到的距离b(米)之间满足关系式.已知点B到的距离为40米.
    

(1)求桥AB的长度;
    

(2)计划在谷底两侧建造平行于的桥墩CD和EF,且CE为80米,其中C,E在AB上(不包括端点)..桥墩EF每米造价k(万元)、桥墩CD每米造价(万元)(k>0),问为多少米时,桥墩CD与EF的总造价最低?
    


    

18.(16分)
    

在平面直角坐标系xOy中,已知椭圆的左、右焦点分别为F1,F2,点A在椭圆E上且在第一象限内,AF2⊥F1F2,直线AF1与椭圆E相交于另一点B.
    

E:\2016+2017真题2016年全国各省市真题精编荟萃·化学【终稿】\2020\江苏数学\18题图.tif
    

(1)求的周长;
    

(2)在x轴上任取一点P,直线AP与椭圆E的右准线相交于点Q,求的最小值;
    

(3)设点M在椭圆E上,记的面积分别为S1,S2,若,求点M的坐标.
    

19.(16分)
    

已知关于x的函数在区间D上恒有
    

(1)若,求h(x)的表达式;
    

(2)若,求k的取值范围;
    

(3)若eqWmf183GmgAAAAAAAIALIAIBCQAAAACwVwEACQAAA+QCAAACANAAAAAAAAUAAAACAQEAAAAFAAAAAQL/
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20.(16分)已知数列eqWmf183GmgAAAAAAAOAGQAIACQAAAACxWgEACQAAA4wCAAACALkAAAAAAAUAAAACAQEAAAAFAAAAAQL/
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AAQAAADwAQAAAwAAAAAA的首项a1=1,前n项和为Sn.设λ与k是常数,若对一切正整数n,均有成立,则称此数列为“λ~k”数列.
    

(1)若等差数列是“λ~1”数列,求λ的值;
    

(2)若数列是“”数列,且,求数列的通项公式;
    

(3)对于给定的λ,是否存在三个不同的数列为“λ~3”数列,且?若存在,求λ的取值范围;若不存在,说明理由.
    

 
    

 
    

 
    

2023年江苏省高考仿真数学试卷答案
    

1.  2.3  3.2 4.   5. 6.  7.   8.  9.  10. 11.4  12.    13.或0  14.
    

15.证明:因为分别是的中点,所以.
    

平面平面
    

所以平面.
    


    

(2)因为平面平面,
    

所以.
    

,平面,平面,
    

所以平面.
    

又因为平面,所以平面平面.
    

16.解:(1)在中,因为
    

由余弦定理,得
    

所以.
    

中,由正弦定理
    


    

所以
    

(2)在中,因为,所以为钝角,
    

,所以为锐角.
    

.
    

因为,所以.
    

从而
    

.
    

17.解:(1)设都与垂直,是相应垂足.
    

由条件知,当时,
    

.
    


    

所以(米).
    


    

(2)以为原点,轴建立平面直角坐标系(如图所示).
    


    

.
    

因为所以.
    


    

所以
    

记桥墩的总造价为
    

 
    


    


    


    

所以当时,取得最小值.
    

答:(1)桥的长度为120米;
    

(2)当为20米时,桥墩的总造价最低.
    

18.解:(1)椭圆的长轴长为,短轴长为,焦距为
    

.
    

所以的周长为.
    

(2)椭圆的右准线为.
    


    


    


    

时取等号.
    

所以的最小值为.
    


    

(3)因为椭圆的左、右焦点分别为,点在椭圆上且在第一象限内,
    

.
    

所以直线
    

,因为,所以点到直线距离等于点到直线距离的3倍.
    

由此得
    

.
    

,此方程无解;
    

,所以.
    

代入直线,对应分别得.
    

因此点的坐标为.
    

19.解:(1)由条件,得
    

,得,所以
    

,得,此式对一切恒成立,
    

所以,则,此时恒成立,
    

所以
    

(2).
    

,则,得.
    


    

所以.则恒成立,
    

所以当且仅当时,恒成立.
    

另一方面,恒成立,即恒成立,
    

也即恒成立.
    

因为,对称轴为
    

所以,解得
    

因此,k的取值范围是
    

(3)①当时,
    

,得,整理得
    


    


    


    

恒成立,
    

所以上是减函数,则,即
    

所以不等式有解,设解为
    

因此
    

②当时,
    


    


    

学科网,得学科网
    

学科网时,学科网学科网是减函数;
    

学科网时,学科网学科网是增函数.
    

学科网学科网,则当学科网时,学科网
    

(或证:学科网.)
    

学科网,因此学科网
    

因为学科网,所以学科网
    

③当时,因为均为偶函数,因此也成立.
    

综上所述,
    

20.解:(1)因为等差数列是“λ~1”数列,则学科网,即
    

也即,此式对一切正整数n均成立.
    

,则恒成立,故,而
    

这与是等差数列矛盾.
    

所以.(此时,任意首项为1的等差数列都是“1~1”数列)
    

(2)因为数列是“”数列,
    

所以,即
    

因为,所以,则
    

,则,即
    

解得,即,也即
    

所以数列是公比为4的等比数列.
    

因为,所以.则
    

(3)设各项非负的数列为“”数列,
    

,即
    

因为,而,所以,则
    

,则,即.(*)
    

①若,则(*)只有一解为,即符合条件的数列只有一个.
    

(此数列为1,0,0,0,…)
    

②若,则(*)化为
    

因为,所以,则(*)只有一解为
    

即符合条件的数列只有一个.(此数列为1,0,0,0,…)
    

③若,则的两根分别在(0,1)与(1,+∞)内,
    

则方程(*)有两个大于或等于1的解:其中一个为1,另一个大于1(记此解为t).
    

所以
    

由于数列从任何一项求其后一项均有两种不同结果,所以这样的数列有无数多个,则对应的有无数多个.
    

综上所述,能存在三个各项非负的数列为“”数列,的取值范围是
    

21.【选做题】本题包括A、B、C三小题,请选定其中两小题,并在相应的答题区域内作答.若多做,则按作答的前两小题评分.解答时应写出文字说明、证明过程或演算步骤.
    

A.[选修4-2:矩阵与变换](10分)
    

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FALCAokCCgAAADIKAAAAAAIAAADq+icEoAIFAAAAFAJiAwAFCQAAADIKAAAAAAEAAAAt+qACBQAA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对应的变换作用下得到点
    

(1)求实数的值;
    

(2)求矩阵eqWmf183GmgAAAAAAAIABgAECCQAAAAATXgEACQAAAy8BAAACAIwAAAAAAAUAAAACAQEAAAAFAAAAAQL/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的逆矩阵eqWmf183GmgAAAAAAACADwAEDCQAAAADyXAEACQAAA6IBAAACAJoAAAAAAAUAAAACAQEAAAAFAAAAAQL/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    

B.[选修4-4:坐标系与参数方程](10分)
    

在极坐标系中,已知点在直线eqWmf183GmgAAAAAAAGAMgAMBCQAAAADwUQEACQAAA5oCAAAIAL4AAAAAAAUAAAACAQEAAAAFAAAAAQL/
//8ABQAAAC4BGQAAAAUAAAALAgAAAAAFAAAADAKAA2AMCwAAACYGDwAMAE1hdGhUeXBlAACwABIA
AAAmBg8AGgD/////AAAQAAAAwP///7b///8gDAAANgMAAAgAAAD6AgAAAAAAAAAAAAIEAAAALQEA
AAUAAAAUAskBEAMFAAAAEwK2ATIDBQAAABMCOgKEAwUAAAATAt4A3AMFAAAAEwLeAJ4EBwAAAPwC
AAAAAAACAAAEAAAALQEBAAgAAAD6AgUAAQAAAAAAAAAEAAAALQECABoAAAAkAwsADgPFAToDrAGE
AxoC1wPWAJ4E1gCeBOYA4gPmAIwDOgJ8AzoCKgPBARMDzgEEAAAALQEAAAgAAAD6AgAAEAAAAAAA
AAIEAAAALQEDAAUAAAAUAswBzggFAAAAEwLMAaUJBQAAAAkCAAAAAgUAAAAUAiACXAIcAAAA+wKw
/gAAAAAAAJABAAAAAAACABBUaW1lcyBOZXcgUm9tYW4A3NcYAJSTbXWAAXF1XyJmswQAAAAtAQQA
EwAAADIKAAAAAAgAAAAyMnNpbigpMJAB3wCDAF4AqABhA8EBoAIFAAAAFAIoA+sIHAAAAPsCsP4A
AAAAAACQAQAAAAAAAgAQVGltZXMgTmV3IFJvbWFuANzXGACUk211gAFxdV8iZrMEAAAALQEFAAQA
AADwAQQACQAAADIKAAAAAAEAAAA0eaACBQAAABQCIAJFABwAAAD7ArD+AAAAAAAAkAEBAAABAAIA
EFN5bWJvbAB1UikKP0g/7gLc1xgAlJNtdYABcXVfImazBAAAAC0BBAAEAAAA8AEFAAoAAAAyCgAA
AAACAAAAcnFoBqACBQAAABQCSwHjCBwAAAD7ArD+AAAAAAAAkAEAAAABAAIAEFN5bWJvbAB1xCcK
YIg/7gLc1xgAlJNtdYABcXVfImazBAAAAC0BBQAEAAAA8AEEAAkAAAAyCgAAAAABAAAAcHGgAgUA
AAAUAiACWwEMAAAAMgoAAAAAAwAAACsrPT9tBqsCoAK+AAAAJgYPAHIBQXBwc01GQ0MBAEsBAABL
AQAARGVzaWduIFNjaWVuY2UsIEluYy4ABQEABglEU01UNgAAE1dpbkFsbEJhc2ljQ29kZVBhZ2Vz
ABEFVGltZXMgTmV3IFJvbWFuABEDU3ltYm9sABEFQ291cmllciBOZXcAEQRNVCBFeHRyYQATV2lu
QWxsQ29kZVBhZ2VzABEGy87M5QASAAghCl9Fj0QvQVD0EA9HX0FQ8h8eQVD0FQ9BAPRF9CX0j0Jf
QQD0EA9DX0EA9I9F9CpfSPSPQQD0EA9A9I9Bf0j0EA9BKl9EX0X0X0X0X0EPDAEAAQABAgICAgAC
AAEBAQADAAEABAAFAAoBAAIEhMEDcgIEhisAKwIAiDIAAwAKAAABAAIAiDIAAAsBAQAKAgKCcwAC
AIJpAAIAgm4AAgCCKAACBIS4A3ECBIYrACsDAAsAAAEACAIAAgR/wANwAAEAAgCINAAAAAIAgikA
AgSGPQA9AgCIMAAAAAoAAAAmBg8ACgD/////AQAAAAAACAAAAPoCAAAAAAAAAAAAAAQAAAAtAQQA
BwAAAPwCAAAAAAAAAAAEAAAALQEGABwAAAD7AhAABwAAAAAAvAIAAACGAQICIlN5c3RlbQAAXyJm
swAACgA4AIoBAAAAAP////8M4hgABAAAAC0BBwAEAAAA8AEFAAMAAAAAAA==上,点eqWmf183GmgAAAAAAAAAFgAMBCQAAAACQWAEACQAAA2oCAAAEAKgAAAAAAAUAAAACAQEAAAAFAAAAAQL/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在圆eqWmf183GmgAAAAAAAGAMgAMBCQAAAADwUQEACQAAA5oCAAAIAL4AAAAAAAUAAAACAQEAAAAFAAAAAQL/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上(其中).
    

(1)求eqWmf183GmgAAAAAAAKABAAICCQAAAACzXQEACQAAA2gBAAACAJMAAAAAAAUAAAACAQEAAAAFAAAAAQL/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的值;
    

(2)求出直线与圆的公共点的极坐标.
    

C.[选修4-5:不等式选讲](10分)
    

,解不等式
    

 【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.
    

 22.(10分)
    

在三棱锥A—BCD中,已知CB=CD=,BD=2,O为BD的中点,AO⊥平面BCD,AO=2,E为AC的中点.
    

C:\Users\蒋志华\Desktop\未标题-1.tif
    

(1)求直线AB与DE所成角的余弦值;
    

(2)若点F在BC上,满足BF=BC,设二面角F—DE—C的大小为θ,求sinθ的值.
    

23.(10分)
    

甲口袋中装有2个黑球和1个白球,乙口袋中装有3个白球.现从甲、乙两口袋中各任取一个球交换放入另一口袋,重复n次这样的操作,记甲口袋中黑球个数为Xn,恰有2个黑球的概率为pn,恰有1个黑球的概率为qn.
    

(1)求p1,q1和p2,q2;
    

(2)求2pn+qn与2pn-1+qn-1的递推关系式和Xn的数学期望E(Xn)(用n表示) .
    

 
    

 
    

 
    

 数学Ⅱ(附加题)参考答案
    

21.【选做题】
    

A.[选修4-2:矩阵与变换]
    

本小题主要考查矩阵的运算、逆矩阵等基础知识,考查运算求解能力.满分10分.
    

解:(1)因为学科网 ,所以
    

解得,所以学科网
    

(2)因为学科网,所以可逆,
    

从而学科网
    

B.[选修4-4:坐标系与参数方程]
    

本小题主要考查曲线的极坐标方程等基础知识,考查运算求解能力.满分10分.
    

解:(1)由,得,又(0,0)(即(0,))也在圆C上,
    

因此或0.
    

(2)由,所以
    

因为,所以
    

所以公共点的极坐标为
    

C.[选修4-5:不等式选讲]
    

本小题主要考查解不等式等基础知识,考查运算求解和推理论证能力.满分10分.
    

解:当x>0时,原不等式可化为,解得
    

时,原不等式可化为,解得
    

时,原不等式可化为,解得
    

综上,原不等式的解集为
    

22.【必做题】本小题主要考查空间向量、异面直线所成角和二面角等基础知识,考查空间想象能力和运算求解能力.满分10分.
    

解:(1)连结OC,因为CB =CD,O为BD中点,所以CO⊥BD.
    

又AO⊥平面BCD,所以AO⊥OB,AO⊥OC.
    

学科网为基底,建立空间直角坐标系O–xyz.
    

因为BD=2,学科网,AO=2,
    

所以B(1,0,0),D(–1,0,0),C(0,2,0),A(0,0,2).
    

因为E为AC的中点,所以E(0,1,1).
    

学科网=(1,0,–2),学科网=(1,1,1),
    

所以学科网
    

因此,直线AB与DE所成角的余弦值为学科网
    

(2)因为点F在BC上,学科网学科网=(–1,2,0).
    

所以学科网
    

学科网
    

学科网
    

学科网为平面DEF的一个法向量,
    

学科网学科网
    

学科网,得学科网学科网,所以学科网
    

学科网为平面DEC的一个法向量,又学科网=(1,2,0),
    

学科网学科网学科网,得学科网学科网
    

所以学科网
    

学科网
    

所以学科网
    

学科网
    

23.【必做题】本小题主要考查随机变量及其概率分布等基础知识,考查逻辑思维能力和推理论证能力.满分10分.
    

解:(1)学科网学科网
    

学科网
    

学科网
    

学科网
    

(2)当学科网时,
    

学科网,①
    

学科网
    

学科网,②
    

学科网,得学科网
    

从而学科网,又学科网
    

所以学科网学科网.③
    

由②,有学科网,又学科网
    

所以学科网学科网
    

由③,有学科网学科网
    

学科网学科网
    

学科网的概率分布
    

学科网
    

0
    

1
    

2
    

学科网
    

学科网
    

学科网
    

学科网
    

学科网