陕西省西安中学2022届高三数学理科五月全仿真模拟考试（一）（PDF版含答案）

陕西省西安中学高2022届高三第一次仿真模拟考试理科数学试题(时间：120分钟满分：150分)第Ⅰ卷(选择题共60分)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.命题"所有实数的平方都是正数"的否定为()A.所有实数的平方都不是正数B.有的实数的平方是正数C.至少有一个实数的平方不是正数D.至少有一个实数的平方是正数2.已知M,N为集合I的非空真子集,且M,N不相等,若N∩(∁IM)=⌀,则M∪N=()A.MB.NC.ID.⌀3.某工厂利用随机数表对生产的700个零件进行抽样测试，先将700个零件进行编号001、002、…、699、700.从中抽取70个样本，下图提供随机数表的第4行到第6行，若从表中第5行第6列开始向右读取数据，则得到的第5个样本编号是()3321183429 7864560732  5242064438  122343567735789056428442125331  3457860736  2530073285  234578890723689608043256780843  6789535577  3489948375  22535578324577892345A.607B.328C.253D.0074.如图水平放置的一个平面图形的直观图是边长为1cm的正方形，则原图形的周长是()A.8cmB.6cmC.2(1+3)cmD.2(1+2)cm5.《九章算术》中盈不足章中有这样一则故事：“今有良马与驽马发长安，至齐．齐去长安三千里．良马初日行一百九十三里，日增一十二里；驽马初日行九十七里，日减二里．”为了计算每天良马和驽马所走的路程之和，设计框图如下图．若输出的S的值为350，则判断框中可填()A.i>6?B.i>7?C.i>8?D.i>9?高三年级数学试题第1页共4页\n22226.在平面直角坐标系xOy中，圆C1:x+y+2x−6y+6=0与圆C2:x+y−4x+2y+4=0,则两圆的公切线的条数是()A.4条B.3条C.2条D.1条7.下面是关于公差d>0的等差数列{an}的四个命题:p1:数列an是递增数列；p2:数列nan是递增数列；anp3:数列n是递增数列；p4:数列an+3nd是递增数列；其中的真命题为()A.p1,p2B.p1,p4C.p2,p3D.p3,p420228.设a∈Z，且0≤a<13，若51+a能被13整除，则a=()A.0B.1C.11D.129.设z1，z2是复数，则下列命题中的假命题是()A.若|z1-z2|=0，则z1=z2B.若z1=z2，则z1=z222C.若|z1|=|z2|，则z1·z1=z2·z2D.若|z1|=|z2|，则z1=z2π110.已知函数f(x)=sin(2x-),若方程f(x)=在（0,π）内的解为x1,x2(x1<x2),则sin(x1-33x2)=()22311A.-B.-C.-D.-322311.关于圆周率π，数学发展史上出现过许多很有创意的求法，如著名的浦丰实验和查理斯实验．受其启发，我们也可以通过设计下面的实验来估计π的值：先请全校m名同学每人随机写下两个都小于1的正实数x，y组成的实数对(x,y)；再统计两数能与1构成钝角三角形三边的数对(x,y)的个数n；最后再根据统计数n估计π的值，那么可以估计π的值约为()4nn+24n+2mn+2mA.B.C.D.mmmm312.已知双曲线C过点(3,2)且渐近线为y=±x，则下列结论错误的是()32x2A.曲线C的方程为-y=1；B.左焦点到一条渐近线距离为1；3C.直线x-2y-1=0与曲线C有两个公共点；D.过右焦点截双曲线所得弦长为23的直线只有三条。第Ⅱ卷(非选择题共90分)二、填空题(本题共4小题，每小题5分，共20分)π13.设0<θ<，向量a=sin2θ，cosθ，b=cosθ，1，若a//b，则tanθ=_______.2高三年级数学试题第2页共4页\n14.已知等比数列an中有a3a11=4a7，数列bn是等差数列，且a7=b7，则b5+b9=(2x+1)π15.函数f(x)=sin-ln|x-2|的所有零点之和为_________.216.平面α过正方体ABCD-A1B1C1D1的顶点A，α∥平面CB1D1，α∩平面ABCD=m，α∩平面ABB1A1=n，则m，n所成角的正切值为三、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)为了测量隧道口A、B间的距离，开车从A点出发，沿正西方向行驶4002米到达D点，然后从D点出发，沿正北方向行驶一段路程后到达C点，再从C点出发，沿东南方向行驶400米到达隧道口B点处，测得BD间的距离为1000米．(1)若隧道口B在点D的北偏东θ度的方向上，求cosθ的值；(2)求隧道口AB间的距离．18.(本小题满分12分)某企业为确定下一年投入某种产品的研发费用，需了解年研发费用x(单位：千万元)对年销售量y(单位：千万件)的影响，统计了近10年投入的年研发费用xi与年销售量yi(i=1,2⋯,10)的数据，得到散点图如图所示．d(1)利用散点图判断y=a+bx和y=c·x(其中c，d均为大于0的常数)哪一个更适合作为年销售量y和年研发费用x的回归方程类型(只要给出判断即可，不必说明理由)；(2)对数据作出如下处理，令ui=lnxi，vi=lnyi，得到相关统计量的值如下表：101010102viui(ui-u(vi-v)(ui-ui=1i=1i=1i=1151528.2556.5根据第(1)问的判断结果及表中数据，求y关于x的回归方程；-39(3)已知企业年利润z(单位：千万元)与x，y的关系为z=18e4y-x(其中e≈2.71828)，根2据第(2)问的结果判断，要使得该企业下一年的年利润最大，预计下一年应投入多少研发费用？高三年级数学试题第3页共4页\n附：对于一组数据(u1,v1)，(u2,v2)，⋯，(un,vn)，其回归直线v=α+βu的斜率和截距的最小二n(ui-u(vi-v)i=1乘估计分别为β=，α=v-βu．n2(ui-ui=1∘19.(本小题满分12分)如图，在直角梯形ABCD中，AB⎳DC,∠ABC=90,AB=2DC=2BC，E为AB的中点，沿DE将△ADE折起，使得点A到点P位置，且PE⊥EB，M为PB的中点，N是BC上的动点(与点B,C不重合)．(1)证明：平面EMN⊥平面PBC；6(2)是否存在点N，使得二面角B-EN-M的余弦值为？若存在，确定N点位置；若不存6在，说明理由．20.(本小题满分12分)在平面直角坐标系xOy中，动点P到点F(1,0)的距离比到y轴的距离大1．(1)求点P的轨迹方程；2已知点Q在直线x=-1上，点P在第一象限，满足FP⊥FQ，记直线OP,OQ,PQ的斜率分别为k1,k2,k3，求k1k2k3的最小值．121.(本小题满分12分)已知函数f(x)=alnx+(x>0)．x(1)讨论函数f(x)的单调性；(2)若存在x1,x2满足0<x1<x2，且x1+x2=1，f(x1)=f(x2)，求实数a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]在平面直角坐标系xOy中，已知点P(1+cosα,sinα)，参数α∈[0,π]，直线l的方向向量为a=(1,1)，且过定点A(-1,0)．(1)在平面直角坐标系xOy中求点P的轨迹方程；(2)若直线l上有一点Q，求PQ的最小值．23.(本小题满分10分)[选修4-5：不等式选讲]设函数f(x)=x+3+x-1,x∈R，不等式f(x)<6的解集为M．(1)求M；22(2)当a∈M,b∈M时，证明：ab+2>2a+b．高三年级数学试题第4页共4页\n高三第一次仿真模拟考试理科数学答案１２３４５６７８９101112ＣＡＢＡＢＡＢＤＤＡＣＣ113.14.815.1216.32BDBC100040017.解：(Ⅰ)在ΔABC中，由正弦定理得=，即=，sinCsin∠CDBsin45°sin∠CDB2所以sin∠CDB=，··················································································(2分)5°由题可知，∠CDB<90，············································································(4分)2323所以cos∠CDB=，即cosθ=．······················································(6分)552(Ⅱ)由(Ⅰ)可知，cos∠ADB=sin∠CDB=，···································(8分)5222在ΔABD中，由余弦定理得AB=BD+AD-2⋅BD⋅AD⋅cos∠ADB222=1000+(4002)-2×1000×4002×=1000000，5所以AB=1000，故两隧道口AB间的距离为1000米．···························································(12分)d18.解：(1)由散点图知，选择回归类型y=c·x更适合；·················································(2分)d(2)对y=c·x两边取对数，得lny=lnc+dlnx，即v=lnc+du，······································(3分)328.251由表中数据可得u=v=，d==，256.5231333令lnc=m，则m=v-du=-×=，即c=e4，·····················································(7分)22243所以年销售量y和年研发费用x的回归方程为y=e4x；·····················································(8分)9(3)由(2)知，z(x)=18x-x，··················································································(9分)292令t=x，则f(t)=18t-t，2当t=2时f(t)取得最小值，·························································································(11分)所以当x=4千万元时，年利润z取最大值且最大值为z(4)=18千万元=1.8亿元故要使年利润取最大值，预计下一年要投入0.4亿元的研发费用．···················································································(12分)19.证明:(1)∵PE⊥EB,PE⊥ED,EB∩ED=E,EB,ED⊂平面EBCD.∴PE⊥平面EBCD．·······················································(1分)又PE⊂平面PEB，∴平面PEB平面EBCD,而BC⊂平面EBCD,BC⊥EB,且平面PEB∩平面EBCD=EB,所以BC⊥平面PEB,··············(3分)而ME⊂平面PBE,∴BC⊥EM,由PE=EB,PM=MB知EM⊥PB,而BC∩PB=B,且PB,BC⊂平面PBC,可知EM⊥平面PBC,·················································(5分)又EM⊂平面EMN,∴平面EMN⊥平面PBC.·········(6分)(2)方法1：假设存在点N满足题意，过作MQ⊥EB于Q，由PE⊥EB知PE⎳MQ,由(1)得PE⊥平面EBCD,所以MQ⊥平面EBCD,··························································(7分)过Q作QR⊥EN于R,连接MR,由MQ⊥平面EBCD,且EN⊂平面EBCD,故MQ⊥EN,·1·\n而QR⊥EN,且QR∩QM=Q,QR,QM⊂平面MQR则EN⊥平面MQR,而MR⊂平面MQR,则EN⊥MR,即∠MRQ是二面角B-EN-M的平面角,·························································(9分)不妨设PE=EB=BC=2,则MQ=1,在Rt△EBN中,设BN=x(0<x<2),BNEN由Rt△EBN∼Rt△ERQ得,=RQEQ22x2+xx即=,得RQ=,········································································(10分)RQ122+x2MQx2+4∴tan∠MRQ==,RQx26x+4依题意知cos∠MRQ=,即tan∠MRQ==5,6x解得x=1∈(0,2),此时N为BC的中点.6综上知,存在点N,使得二面角B-EN-M的余弦值,此时N为BC的中点.···············(12分)6方法2：由(1)知，EB，ED，EP两两垂直，所以以E为原点，分别以EB，ED，EP所在直线为x轴、y轴、z轴建11立如图所示的空间直角坐标系，设EB=EP=ED=1，BN=t(0<t<1),则N(1,t,0)，M2,0,2，····(7分)平面EBN的法向量为m=(0,0,1)······················································································(8分)xz+=01设面EMN的一个法向量为n=(x,y,z)，则22，则可取n=1,−t,−1·····················(10分)x+ty=0161所以cosm,n==，解得t=，即N为BC的中点．····································(12分)1621+1+2t2220.解：(1)设P(x,y)，由已知得(x-1)+y=|x|+1，··························································(1分)2222当x≥0时，(x-1)+y=(x+1)，得y=4x，······································································(3分)222当x<0时，(x-1)+y=(-x+1)，得y=0;2所以点P的轨迹方程为y=4x或y=0(x<0)．···········································································(5分)2(2)设P(x0,y0)(x0>0,y0>0)，Q(-1,t)，则y0=4x0，因为FP⊥FQ，所以FP⋅FQ=-2(x0-1)+ty0=0，即ty0=2(x0-1)，y0y0-t因为k1=，k2=-t，k3=，··························································································(7分)x0x0+12-ty0(y0-t)-t(y0-ty0)-t[4x0-2(x0-1)]-2t所以k1k2k3====x0(x0+1)x0(x0+1)x0(x0+1)x02-4(x0-1)-4(y0-4)1441===-4-=4-，··································································(9分)x0y0y3y0y3y3y00003构造f(m)=4m-m(m>0)，23所以f'm=12m-1，令f'm≥0，得m≥，633所以f(m)在0,6上单调递减，在6,+∞上单调递增，33413所以f(m)min=f=-，即3-的最小值为-，69y0y0943所以k1k2k3的最小值为-．······························································································(12分)9ax-121.解：(1)函数f(x)的定义域为(0,+∞)，f'(x)=．···························································(1分)2x当a≤0时，f'(x)<0，f(x)在(0,+∞)上单调递减；11当a>0时，令f'(x)<0，得0<x<，令f'(x)>0，得x>，aa11所以函数f(x)在0,a上单调递减，在a,+∞上单调递增，综上所述：当a≤0时，f(x)在(0,+∞)上单调递减；·2·\n11当a>0时，函数f(x)在0,a上单调递减，在a,+∞上单调递增．···································(5分)11x211(2)fx1=fx2⇒alnx1+=alnx2+⇒aln+-=0，x1x2x1x2x1x2x1x2又x1+x2=1，则aln+-=0．x1x2x1x21令t=>1，即方程alnt+-t=0在1,+∞上有解．································································(7分)x1t1令ht=alnt+-t，t∈1,+∞，t1-t2+at-1a-t+t1则h't==，t∈1,+∞，则t+>2，·······················································(8分)t2tt当a≤2时，h't<0，ht在1,+∞上单调递减，又h1=0，则ht<0在t∈1,+∞上恒成立，不合题意;22当a>2时，a-4>0，令-t+at-1=0，可知该方程有两个正根，2a+a-4因为方程两根之积为1且t>1，所以t=．222a+a-4a+a-4当t∈1,时，h't>0;当t∈,+∞时，h't<0;222a+a-4则t∈1,时，ht>h1=0，2a21a2a而he=a+a-e<a+1-ea>2．e2xx令φx=x+1-ex>2，则φ'x=2x-e，x令mx=φ'x，m'x=2-e<0，2则φ'x在2,+∞上单调递减，φ'x<φ'2=4-e<0，2a则φx在2,+∞上单调递减，φx<φ2=5-e<0，即he<0，2a+a-4a故存在t0∈,e，使得ht0=0，故a>2满足题意．2综上所述，实数a的取值范围是2,+∞．············································································(12分)x=1+cosα22.解：(1)由题知：点P的坐标满足y=sinα(α∈[0,π])22所以点P的轨迹方程为(x-1)+y=1(y≥0)···································································(5分)x=-1+t(2)直线l的方向向量为a=(1,1)，且过定点A(-1,0).转换为l的参数方程为y=t(t是参数)所以直线l的直角坐标方程为：y=x+1，············································································(8分)|1-0+1|所以|PQmin|=-1=2-1．············································································(10分)221+1-2x-2(x<-3),23.解：(Ⅰ)f(x)=|x+3|+|x-1|=4(-3≤x≤1),··················································(2分)2x+2(x>1),当x<-3时，-2x-2<6，解得-4<x<-3；当-3≤x≤1时，由4<6可得-3≤x≤1；当x>1时，2x+2<6，解得1<x<2．综上，不等式f(x)<6的解集M={x|-4<x<2}．···························································(5分)2222(Ⅱ)当a∈M，b∈M时，即0≤a<2，0≤b<2．22要证|ab+2|>|2|a+b|，只需证(ab+2)>2(a+b)，·····················································(7分)22222222只需证ab+4ab+4>2a+4ab+2b，只需证ab-2a-2b+4>0，22只需证(a-2)(b-2)>0，···································································································(9分)22又a-2<0，b-2<0，22故(a-2)(b-2)>0成立，即原不等式成立．··································································(10分)·3·