【数学】陕西省西安中学2020届高三第四次模拟考试(理)

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【数学】陕西省西安中学2020届高三第四次模拟考试(理)

陕西省西安中学2020届高三第四次模拟考试(理)‎ 一、选择题:本大题共12小题,每小题5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,为自然数集,则中元素的个数为(  )‎ A.2 B.3 C.4 D.5‎ ‎2.复数满足,则复数的共轭复数是(  )‎ A. B. C. D.‎ ‎3.已知,,其中是互相垂直的单位向量,则(  )‎ A. B. C.28 D.24‎ ‎4.在等差数列中,首项,公差,是其前项和,若,则(  )‎ A. B. C. D.‎ ‎5.已知,则 (  )‎ A. B. C. D.‎ ‎6.已知函数,则=(  )[来源:学#科#网]‎ A. B. C. D.‎ ‎7.在正方体中,棱,的中点分别为,,则直线与平面所成角的余弦值为(  ) ‎ A. B. C. D.‎ ‎8.为得到函数的图像,可将函数的图像(  )‎ ‎ A.向左平移个单位 B.向左平移个单位 ‎ C.向右平移个单位 D.向右平移个单位 ‎9.图1是某高三学生14次模考数学成绩的茎叶图,第1次到第14次的考试成绩依次记为A1,A2,…,A14.将14次成绩输入图2的程序框图,则输出的结果是(  )‎ A.8 B.9 C.10 D.11‎ ‎10.已知抛物线与双曲线的渐近线交于A,B两点(异于原点O),若双曲线的离心率为,的面积为32,则抛物线的焦点坐标为(  )‎ A. B. C. D.‎ ‎11.为缓解城市道路交通压力,促进城市道路交通有序运转,减少机动车尾气排放对空气质量的影响,西安市人民政府决定:自2019年3月18日至2020年3月13日在相关区域实施工作日机动车尾号限行交通管理措施.已知每辆机动车每周一到周五都要限行一天,周末(周六和周日)不限行.某公司有A,B,C,D,E五辆车,每天至少有四辆车可以上路行驶.已知E车周四限行,B车昨天限行,从今天算起,A,C两辆车连续四天都能上路行驶,E车明天可以上路,由此可知下列推测一定正确的是(  )‎ A.今天是周四 B.今天是周六 C.A车周三限行 D.C车周五限行 ‎12.已知函数,若存在互不相等的实数,,,,满足 ‎,则 (  )‎ A.0 B.1 C.2 D.3‎ 二、填空题:本大题共4小题,每小题5分.‎ ‎13.的展开式中项的系数是__________.(用数字作答)‎ ‎14.若满足约束条件则的最小值为__________.‎ ‎15.已知三棱锥内接于球,,,则球的表面积为__________.‎ ‎16.已知函数为奇函数,,若函数与图像的交点为,,⋯,,则__________.‎ 三、解答题:共70分.解答应写出文字说明、证明过程或演算步骤.第17~21题为必考题,每个试题考生都必须作答.第22,23题为选考题,考生根据要求作答.‎ ‎(一)必考题:共60分.‎ ‎17.(本小题满分12分)‎ 在△ABC中,内角A,B,C所对的边长分别为a,b,c,且满足=.‎ ‎(Ⅰ)求角A;‎ ‎(Ⅱ)若a=,b=3,求△ABC的面积.‎ ‎18.(本小题满分12分)‎ ‎“绿水青山就是金山银山”的理念越来越深入人心,据此,某网站调查了人们对生态文明建设的关注情况,调查数据表明,参与调查的人员中关注生态文明建设的约占80%.现从参与调查的关注生态文明建设的人 员中随机选出200人,并将这200人按年龄(单位:岁)分组:第1组[15,25),第2组[25,35),第3组[35,45),第4组[45,55),第5组[55,65],得到的频率分布直方图如图3所示.‎ 图3‎ ‎(Ⅰ)求这200人的平均年龄(每一组用该组区间的中点值作为代表)和年龄的中位数(保留一位小数);‎ ‎(Ⅱ)现在要从年龄在第1,2组的人员中用分层抽样的方法抽取5人,再从这5人中随机抽取3人进行问卷调查,求抽取的3人中恰有2人的年龄在第2组中的概率;‎ ‎(Ⅲ)若从所有参与调查的人(人数很多)中任意选出3人,设这3人中关注生态文明建设的人数为X,求随机变量X的分布列与数学期望.‎ ‎19.(本小题满分12分)‎ 已知四棱锥P-ABCD(图4)的三视图如图5所示.‎ 图4 图5‎ ‎(Ⅰ)若点E为棱PC上的动点,求证:BD⊥AE;‎ ‎(Ⅱ)若点E为PC的中点,求钝二面角D-AE-B的大小.‎ ‎20.(本小题满分12分)‎ 已知点M,N分别是椭圆C:+=1(a>b>0)的左顶点和上顶点,F为其右焦点,,椭圆的离心率为.‎ ‎(Ⅰ)求椭圆C的方程;‎ ‎(Ⅱ)设不过原点O的直线l与椭圆C相交于A,B两点,若直线OA,AB,OB的斜率成等比数列,求△OAB面积的取值范围.‎ ‎21.(本小题满分12分)‎ 已知函数f(x)=xex+.‎ ‎(Ⅰ)求证:函数f(x)有唯一零点;‎ ‎(Ⅱ)若对任意x∈(0,+∞),xex-lnx≥1+kx恒成立,求实数k的取值范围.‎ ‎(二)选考题:共10分.请考生在第22,23题中任选一题作答.如果多做,那么按 所做的第一题计分.‎ ‎22.(本小题满分10分)[选修4—4:坐标系与参数方程]‎ 在直角坐标系xOy中,直线l1的参数方程为(t为参数),直线l2‎ 的参数方程为(m为参数).设l1与l2的交点为P,当k变化时,P的轨迹为曲线C.‎ ‎(Ⅰ)写出C的普通方程;‎ ‎(Ⅱ)以坐标原点为极点,x轴正半轴为极轴建立极坐标系,设l3:ρ(cosθ+sinθ)-=0,M为l3与C的交点,求M的极径.‎ ‎23.(本小题满分10分)[选修4—5:不等式选讲]‎ 已知x,y∈R,且x+y=1.‎ ‎(Ⅰ)求证:x2+3y2≥;‎ ‎(Ⅱ)当xy>0时,不等式+≥|a-2|+|a+1|恒成立,求a的取值范围.‎ 参考答案 一、选择题:‎ 题号 ‎1‎ ‎2‎ ‎3‎ ‎4‎ ‎5‎ ‎6‎ ‎7‎ ‎8‎ ‎9‎ ‎10‎ ‎11‎ ‎12‎ 答案 D B A C A B D C D B A A 二、填空题:‎ ‎13.21 14.3 15. 16.‎ 三、解答题:‎ ‎17. 解:(Ⅰ)由= 及正弦定理可知,‎ ·=,··································································(4分)‎ 所以2cos A=1,又A∈(0,π),‎ 所以A=.·······················································································(6分)‎ ‎(Ⅱ)由余弦定理a2=b2+c2-2bccos A,‎ 得13=9+c2-3c,··············································································(8分)‎ 所以c2-3c-4=0,即(c-4)(c+1)=0,‎ 所以c=4.·······················································································(10分)‎ 从而.······································(12分)‎ ‎18.解:(Ⅰ)由10×(0.010+0.015+a+0.030+0.010)=1,得a=0.035,‎ 平均年龄为20×0.1+30×0.15+40×0.35+50×0.3+60×0.1=41.5(岁). ············(2分)‎ 设中位数为x岁,则10×0.010+10×0.015+(x-35)×0.035=0.5,解得x≈42.1,‎ 故这200人年龄的中位数为42.1岁.······················································(4分)‎ ‎(Ⅱ)易知从第1,2组中抽取的人数分别为2,3,‎ 设“抽取的3人中恰有2人的年龄在第2组中”为事件A,‎ 则P(A)=·········································(7分)‎ ‎(Ⅲ)从所有参与调查的人员中任意选出1人,则其关注生态文明建设的概率为.‎ 由题意知X的所有可能取值为0,1,2,3,‎ P(X=0)=C= P(X=1)=C= P(X=2)=C= P(X=3)=C= 所以X的分布列为 X ‎0‎ ‎1‎ ‎2‎ ‎3‎ P 因为X~B,所以E(X)=3×=.··············································(12分)‎ ‎19.证明:(Ⅰ)由三视图可知,四棱锥P-ABCD的底面是边长为1的正方形,‎ 侧棱PC⊥底面ABCD,且PC=2. ·····························································(2分)‎ 连接AC,∵ABCD是正方形,∴BD⊥AC.‎ ‎∵PC⊥底面ABCD,且BD⊂平面ABCD,∴BD⊥PC.‎ 又∵AC∩PC=C,∴BD⊥平面PAC. ··························································(4分)‎ ‎∵AE⊂平面PAC,‎ ‎∴BD⊥AE. ··························································································(5分)‎ ‎(Ⅱ)方法1. 在平面DAE内过点D作DF⊥AE于F,连接BF.‎ ‎∵AD=AB=1,DE=BE==,AE=,‎ ‎∴Rt△ADE≌Rt△ABE,从而△ADF≌△ABF,∴BF⊥AE.‎ ‎∴∠DFB为二面角D-AE-B的平面角. ····················································(7分)‎ 在Rt△ADE中,DF===,‎ ‎∴BF=.··························································································(9分)‎ 又BD=,在△DFB中,‎ 由余弦定理得cos∠DFB==-,······································(11分)‎ ‎∴∠DFB=,即二面角D-AE-B的大小为.······································(12分)‎ 方法2. 如图,以点C为原点,CD,CB,CP所在的直线分别为x,y,z轴建立空间直角坐标系,则D(1,0,0),A(1,1,0),B(0,1,0),E(0,0,1),‎ 从而=(0,1,0),=(-1,0,1),=(1,0,0),=(0,-1,1).‎ 设平面ADE和平面ABE的法向量分别为n1=(x1,y1,z1),n2=(x2,y2,z2),‎ 由⇒取n1=(1,0,1);····································(7分)‎ 由⇒取n2=(0,-1,-1). ················· ··············(9分)‎ 设二面角D-AE-B的平面角为θ,则 cos θ===-, ······························································(11分)‎ ‎∴θ=,即二面角D-AE-B的大小为.·············································(12分)‎ ‎20. 解:(Ⅰ)设椭圆的半焦距为c,由题可知,,‎ ‎,,则.‎ 又e==,‎ 解得a=2,,c=1,‎ 所以椭圆C的方程为+=1. ···························································(4分)‎ ‎(Ⅱ)由题意可知,直线l的斜率存在且不为0.‎ 故可设直线l:y=kx+m(m≠0),A(x1,y1),B(x2,y2),‎ 联立直线和椭圆,消去y可得,(3+4k2)x2+8kmx+4m2-12=0,‎ 由题意可知,Δ=64(km)2-4(4k2+3)(4m2-12)=48(4k2-m2+3)>0,‎ 即4k2+3>m2,‎ 且x1+x2=-,x1x2=,···················································(6分)‎ 又直线OA,AB,OB的斜率依次成等比数列,所以·=k2,‎ 将y1,y2代入并整理得m2(4k2-3)=0,‎ 因为m≠0,k=±,00,‎ 因此ff(1)<0,即f(x)在区间(0,1)上恰有一个零点,‎ 由题可知f(x)>0在(1,+∞)上恒成立,即在(1,+∞)上无零点,‎ 所以f(x)在(0,+∞)上有唯一零点.··············································· ·······(5分)‎ ‎(Ⅱ)设f(x)的零点为x0,即x0e+=0.‎ 原不等式可化为≥k,‎ 令g(x)=,则g′(x)=,‎ 由(Ⅰ)可知g(x)在(0,x0)上单调递减,在(x0,+∞)上单调递增,‎ 故g(x0)为g(x)的最小值,····································································(7分)‎ 下面分析x0e+=0.‎ 设x0e=t,则=-t,‎ 可得,即x0(1-t)=ln t,··················································(9分)‎ 若t>1,等式左负右正不相等,‎ 若t<1,等式左正右负不相等,‎ 只能t=1.‎ 因此g(x0)=1,所以k≤1.·································(12分)‎ ‎22.解:(Ⅰ)消去参数t,得直线的普通方程为,‎ 消去参数m,得的普通方程为.·························· ····················(2分)设,由题设可得,消去k得 ,‎ 即C的普通方程为.······················································(5分)‎ ‎(Ⅱ)方法1.C的极坐标方程为,···(6分)‎ 联立,得,‎ 故,从而,.·········································(8分)‎ 代入,得,所以交点M的极径为.············ ··(10分)‎ 方法2.化为普通方程为 ························································(6分)‎ 联立 得 ···························································(8分)‎ ‎∴‎ ‎∴与C的交点M的极径为. ································································(10分)‎ ‎22.解:(Ⅰ)证明:由柯西不等式得[x2+(y)2][12+()2]≥(1·x+y·)2. ····(3分)‎ 所以(x2+3y2)×≥(x+y)2,当且仅当x=3y时取等号.‎ 所以x2+3y2≥. ············································································(5分)‎ ‎(Ⅱ)因为x,y∈R,且x+y=1,xy>0,‎ 所以+=(x+y)(+)=2++≥4,‎ 所以|a-2|+|a+1|≤4, ····································································(7分)‎ 当a≥2时,2a-1≤4,可得2≤a≤;‎ 当-1
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