2023年高考数学模拟试卷01(浙江省)(原卷版)


    

 
    

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2023年高考数学模拟试卷01(浙江省)
    

一、单选题(本题共8小题,每小题5分,共40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.学科网 zxxk.com
    

1.(2022·吕梁模拟)已知集合A={x|x2-2x-3<0},B={x|log2x<2},则A∩B等于(  )
    

A.(-1,4)   B.(-1,3)
    

C.(0,3)   D.(0,4)
    

2.(2022·长春模拟)已知复数z的共轭复数=,则复数z在复平面内对应的点位于(  )
    

A.第一象限   B.第二象限
    

C.第三象限   D.第四象限
    

3.(2022·重庆调研)函数y=ln cos x的图象是(  )
    


    


    

4.(2022·郑州模拟)如图,在五面体ABCDEF中,四边形ABCD是正方形,AB=4,EF=2,△BCF,△ADE都是等边三角形,则五面体ABCDEF的体积为(  )
    


    

A.   B.
    

C.   D.4
    

5.函数eqIdc1be95d0343ce546431eada32f695238图像大致为(  )
    

A. B.
    

C. D.
    

6.如图,在△ABC中,M为线段BC的中点,G为线段AM上一点且eqId55ef61a6ecb410736f56322dada8bf3f,过点G的直线分别交直线AB、AC于P、Q两点,eqId367fc69f737278bff61168db1b43b929eqId8a6ff96f7359388bb2710a753927b3c0,则eqId093d54cf621436f224e336acf05c4b35的最小值为(    )
    


    

A.eqId8b2a698891d42c70b597f0da4f215f09 B.1 C.eqIdd599cb4a589f90b0205f24c2e1fa021e D.4
    

7.已知椭圆eqIdb8be79c5998d44dfe2bcd434b394d88a的右焦点为F,以椭圆eqId522230546d4b802094e86ceb48c2ba38的长轴为直径作圆eqId0b4f150ab98bde511e0f65d9bafab031,过点F作不与坐标轴垂直的两条直线eqId2e9b0f5f44abbc6544a2f672b025b013eqId3f6f17bc385bafb37e8f964e5eb99cd0,其中eqId2e9b0f5f44abbc6544a2f672b025b013与椭圆eqId522230546d4b802094e86ceb48c2ba38交于M,N两点,eqId3f6f17bc385bafb37e8f964e5eb99cd0与圆eqId0b4f150ab98bde511e0f65d9bafab031交于P,Q两点,若eqId0d43034a7ccae812fcd47bd636436318,且都有eqIdfd7b92906ac7d3a3a0e7239c7a9a7ad7,则实数eqIddf64046e91b047037f19e4032e3b6de3的取值范围为(    ).
    

A.eqId3e743b00055f7ddf946c9ecffdc99838 B.eqIda2fe8d44de8c05b5b01f7cefa89a8232
    

C.eqId16858419a930d4fe865f61a621dc6078 D.eqIde1d2bf79010de8ce9e4f563faeae48ea
    

8.某单位科技活动纪念章的结构如图所示,eqId1dde8112e8eb968fd042418dd632759e是半径分别为eqIdc995d12d73a9f3c73c8e2bd87642ffc8的两个同心圆的圆心,等腰三角形eqId7bef5239ddbb0972700ce01daf9ee7cf的顶点eqId5963abe8f421bd99a2aaa94831a951e9在外圆上,底边eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的两个端点都在内圆上,点eqIdf9cda46e575acf64941ea0964b89ee99在直线eqId0dc5c9827dfd0be5a9c85962d6ccbfb1的同侧.若线段eqId0dc5c9827dfd0be5a9c85962d6ccbfb1与劣弧eqId53de04840ff5506ae57bd2d74dcbf2a3所围成的弓形面积为eqIde097c8d4c948de063796bd19f85b3a9a,△eqIde4819c39c281427826e1b3f7a4c2b720与△eqIdfa16146cb21f11693feffb0876c0795b的面积之和为eqId1e0bd63f55069a3bc870915010b39225,设eqIdeb43780e23412b117be0747318de29d2.经研究发现当eqId16ae89bd2dd0a18e1afb5b1a1abd0efd的值最大时,纪念章最美观,当纪念章最美观时,eqId90e888ea14da893971e13858945bf0cd(      )
    


    

A.eqId701d08a807ba7b8cade095ba5bab5184 B.eqIdde605e6ae9bc4beec8000bfa109ef82a C.eqIdf89eef3148f2d4d09379767b4af69132 D.eqId8d5989c84e320b504511f23eeb6e7357
    

二、多选题(本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,部分选对的得2分,有选错的得0分.学科网 zxxk.com
    

9.已知eqIdaf68f652b4c13657ffddf3c9e7eb262beqIdfa224ed9be8766a4d0b5138bd57de0f0均为复数,则下列结论中正确的有(    )
    

A.若eqId1e37c64fc12b438809bc6d723f359cd9,则eqId063fc82b8f5c123f1fbf572ce4cdf062 B.若eqIde13d477ec252f1952efd7d97425bf40c,则eqIde34b88f343ca5a4c29057465541b9cf4是实数
    

C.eqId5830a3539329430d9988dc613fb3e55a D.若eqIda91095e73cc4d985a4875062d953808c,则eqId96a1ec15680a35a56e15d8c582eb3ab3是实数
    

10.有eqId6f8c4c029e552954bd493b49aeab82d5个相同的球,分别标有数字eqIdbdaa19de263700a15fcf213d64a8cd57eqId61128ab996360a038e6e64d82fcba004eqId5ca7d1107389675d32b56ec097464c14eqIdb8860d9787671b53b1ab68b3d526f5caeqIdd91e07104b699c4012be2d26160976a2eqId6f8c4c029e552954bd493b49aeab82d5,从中有放回的随机取两次,每次取eqIdbdaa19de263700a15fcf213d64a8cd57个球.记第一次取出的球的数字为eqId096b1ece1dcd29c59a46a4b3e02cb548,第二次取出的球的数字为eqIde5031a3a951c4a1d1c5e9f80a5e26513.设eqId096f69f33ed13734e82f60eb596189e1,其中eqIda2ab85825d4a002600ca41bd3cd2ee7d表示不超过eqId81dea63b8ce3e51adf66cf7b9982a248的最大整数,如eqId78846883d089a0648f0f8ab652f559faeqId3458bde27e50e531728450e8a50080dd,则(    )
    

A.eqId76425704952e5337930f2f3f5d5f07e3
    

B.eqId9f3b2703ea8d759347a5c4e12d02522f
    

C.事件“eqId43105718748ec78d7687199f983b479a”与“eqId9fef57584523e293a6f482bb4cf31c52”互斥
    

D.事件“eqId0b2e776ad6950e993ce1e4430ab36255”与“eqId9fef57584523e293a6f482bb4cf31c52”对立
    

11.取名于荷兰数学家鲁伊兹·布劳威尔不动点定理是拓扑学里一个非常重要的不动点定理.该定理表明:对于满足一定条件的图象连续不间断的函数eqId09f86f37ec8e15846bd731ab4fcdbacd,在其定义域内存在一点eqId79b752f0f189e5d8666daea73e145dff,使得eqId66f66a2b3d90f0d935d6c8ebaf675349,则称eqId79b752f0f189e5d8666daea73e145dff为函数eqId09f86f37ec8e15846bd731ab4fcdbacd的一个不动点,那么下列函数具有“不动点”的是(    )
    

A.eqId5f52ce8cf72216ea15e51aeedd2ae430 B.eqIdbb6646ee1281b3b22e6a6ded9da1f9b3
    

C.eqId04930cf59e53edcc0492bb783df87a8d D.eqId0a476a0e9a127204894bec527c13fd5b
    

12.如图,正方体eqId6e09725691ee7851f54c0dee86b2bf55棱长为eqId61128ab996360a038e6e64d82fcba004eqIddad2a36927223bd70f426ba06aea4b45是直线eqId10d8eb4a9f462ca0c1d49c3fe91e720d上的一个动点,则下列结论中正确的是(    )
    


    

A.eqId2cdba1337ec85fa9722cb4b320a82ae6的最小值为eqId35361e76a7c85d1886728c8d0200b234
    

B.eqIda51a8d10345f567e0f2daf18529d42f0的最小值为eqIdde4ab3f511aea1a80d79c6ad865e1955
    

C.三棱锥eqId98a09337d7c2508a7c999a8a92388233的体积不变
    

D.以点eqId7f9e8449aad35c5d840a3395ea86df6d为球心,eqIdcf298f00799cbf34b4db26f5f63af92f为半径的球面与面eqIda211ad5a06b505b8365a62c1946f3cb7的交线长eqId3765f7d2a69e4ad0707e9283801dcfcb
    

三、填空题:(本题共4小题,每小题5分,共20分,其中第16题第一空2分,第二空3分.)
    

13.已知平面向量a,b满足a=(1,2),|b|=,a·b=,则cos〈a,b〉=________.
    

14.如图,在等腰直角eqId15c0dbe3c080c4c4636c64803e5c1f76中,eqId34d0968ff6799eafbc028a273b51aad3eqId8455657dde27aabe6adb7b188e031c11eqId60ef95894ceebaf236170e8832dcf7e3的中点,将线段eqId60ef95894ceebaf236170e8832dcf7e3绕点eqId8455657dde27aabe6adb7b188e031c11旋转得到线段eqId49b50357a6545cae8348e3059312f520.设eqIdac047e91852b91af639feec23a9598b2为线段eqIdf52a58fbaf4fea03567e88a9f0f6e37e上的点,则eqId1d405c7e1ed6ddac33126b8c1cb511b6的最小值为___________.
    


    

15.在线投标问题的定义是:商家给出一个足够大的正整数M,但投标者不知道M的值,故只能通过不断给出价格序列eqId7176c1d46407fc4ca8cc3b8f4f10693d来竞标,已知eqId87a60302649eb940748da818199e55daeqIdb28bb687c4284f9a38ecb5313028c65d.若正整数k使得eqIda246e4bbf5c1278a112375b6591f7c02,则此次竞标投标者共花费eqIdfa63c65a4af38100bfc833199db6ed19中标,我们的目标是对于任意足够大的正整数M,最小化竞争比eqId9f291b88ed219632f3b44364e08844de,则当eqId380bbacf854e30e2e747fc286d2b9997________.时,在线投标问题的竞争比最小.
    

16.已知eqId4d2a97987f71835f519b462f5b8f5957为双曲线eqId83bf4fd84818abac17a9d21237ac5ce5的左右焦点,过点eqIdf5076289823db419f94e9c0c8f4aafd9作一条渐近线的垂线交双曲线右支于点P,直线eqId0739793f234f8e86adc6177801ae7295与y轴交于点Q(P,Q在x轴同侧),连接eqId2c30a7506331e47342fb1e7d2e12d041,如图,若eqIdb6380c54a58ed53ebe8977402b42c960内切圆圆心恰好落在以eqId643ef7d761de0e794fc39937dc72ac6a为直径的圆上,则eqIdc406c942d1cf93b68e69842743520578________;双曲线的离心率eqId47ce2b47812fce4b17fd813d0e4cce21________.
    


    

四、解答题(本题共6小题,共70分,其中第16题10分,其它每题12分,解答应写出文字说明?证明过程或演算步骤.学科网 zxxk.com
    

17.设函数eqIdc2d97c1f62870443fd716ead6ee08f86,数列eqId83cf38189d5cbf627d2b82ac0eb76006满足eqId31bfec26fa9acb0dcf9525c360297bd5eqIdcea4ac187cbb465180e89f38250b3970,且eqId6bc5735838e43b7a229e8f45c9bfffb3).
    

(Ⅰ)求数列eqId83cf38189d5cbf627d2b82ac0eb76006的通项公式;
    

(Ⅱ)设eqId6b7280e837a94737a673d93c5db41894,若eqId1d64847593a95d4d9b0d6da34c4b9b35eqIdcea4ac187cbb465180e89f38250b3970恒成立,求实数eqId36a1b09c653185842513e24ebba60bb3的取值范围.
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

18.(12分)(2022·新余模拟)如图,在三棱锥P-ABC中,已知PA=PB=PC=AB=AC,E是PA的中点.
    


    

(1)求证:平面PAB⊥平面BCE;
    

(2)若BC=AB,求平面ABC与平面ABE夹角的正弦值.
    

 
    

 
    

 
    

 
    

 
    

 
    

19.在钝角eqId15c0dbe3c080c4c4636c64803e5c1f76中,内角eqId5963abe8f421bd99a2aaa94831a951e9eqId7f9e8449aad35c5d840a3395ea86df6deqIdc5db41a1f31d6baee7c69990811edb9f的对边为eqId0a6936d370d6a238a608ca56f87198deeqId2c94bb12cee76221e13f9ef955b0aab1eqId071a7e733d466949ac935b4b8ee8d183,已知eqId9515db1bfaa00175c5ab332c66ac38e5.
    

(1)若eqId3711c8ba16405959bcb0b70385da1d89,求eqId51c181f86de3c96a7ef7a1a04c3a438f
    

(2)求eqIdc8a51bd9b51c6c0d5ab2f79712d3b634的取值范围.
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

20.京东配送机器人是由京东研发,进行快递包裹配送的人工智能机器人.eqIdf2030a7c508abe4b2a03bc702cf7692deqId6f8c4c029e552954bd493b49aeab82d5eqId4b837fd9c52f60bfb3b6852733abc790日,京东配送机器人在中国人民大学顺利完成全球首单配送任务,作为整个物流系统中末端配送的最后一环,配送机器人所具备的高负荷?全天候工作?智能等优点,将为物流行业的“最后一公里”带去全新的解决方案.已知某市区eqId4d01dd350dc95f42f1883e0cc7aae084eqIdbdaa19de263700a15fcf213d64a8cd57eqIdd91e07104b699c4012be2d26160976a2月的京东快递机器人配送的比率图如图所示,对应数据如下表所示:
    


    

eqId4d01dd350dc95f42f1883e0cc7aae084
    

eqIdbdaa19de263700a15fcf213d64a8cd57
    

eqId61128ab996360a038e6e64d82fcba004
    

eqId5ca7d1107389675d32b56ec097464c14
    

eqIdb8860d9787671b53b1ab68b3d526f5ca
    

eqIdd91e07104b699c4012be2d26160976a2
    

时间代码eqId81dea63b8ce3e51adf66cf7b9982a248
    

eqIdbdaa19de263700a15fcf213d64a8cd57
    

eqId61128ab996360a038e6e64d82fcba004
    

eqId5ca7d1107389675d32b56ec097464c14
    

eqIdb8860d9787671b53b1ab68b3d526f5ca
    

eqIdd91e07104b699c4012be2d26160976a2
    

配送比率eqIdd053b14c8588eee2acbbe44fc37a6886
    

eqId180ea775f2af05650404d764384e7faa
    

eqIdfbbecbd7e92dcbe1766462fcf40066de
    

eqIdba505969331b28f0e2d3047d49988914
    

eqId4b1a11655c1e9668515ca8551cbd8b3b
    

eqIda8f70f0c1b960fa850d67c4091e04400
    

(1)如果用回归方程eqId67304345991230b52a1896ca544a59d7进行模拟,请利用以下数据与公式,计算回归方程;
    

eqId9085967dc15e2e6ae4534519a123fb40eqId427ac3f5a466e45bc085c21bf737130eeqId98ed476b7cf87cf3b825c4283d138506.
    

参考公式:若eqIdd9cf74bbdee085c44778ac6191e5016b,则eqIdbc5c0bbfbaf84f0547aa9ef4f3731b3e
    

(2)已知某收件人一天内收到eqIdcd3304e23f3b0f9569c4140ca89b6498件快递,其中京东快递eqId5ca7d1107389675d32b56ec097464c14件,菜鸟包裹eqId5ca7d1107389675d32b56ec097464c14件,邮政快递eqId61128ab996360a038e6e64d82fcba004件,现从这些快递中任取eqIdb8860d9787671b53b1ab68b3d526f5ca件,eqIdf022950e0faa45b617d497b01b5292b9表示这四件快递里属于京东快递的件数,求随机变量eqIdf022950e0faa45b617d497b01b5292b9的分布列以及随机变量eqIdf022950e0faa45b617d497b01b5292b9的数学期望.
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

21.已知抛物线G:eqId0ac62b1ade07205ae2693ec1ab135def的焦点与圆E:eqId7dd54b9df3402ad91e2d34c40efe0c7a的右焦点F重合,椭圆E的短轴长为2.
    

(1)求椭圆E的方程;
    

(2)过点F且斜率为k的直线l交椭圆E于A?B两点,交抛物线G于M,N两点,请问是否存在实常数t,使eqIdb2fdd75303a2282a147b62ce8ff68ba3为定值?若存在,求出t的值;若不存在,说明理由.
    


    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

 
    

22.)(2022·潍坊模拟)已知函数f(x)=ex-ax-a,a∈R.
    

(1)讨论f(x)的单调区间;
    

(2)当a=1时,令g(x)=.
    

①证明:当x>0时,g(x)>1;
    

②若数列{xn}(n∈N*)满足x1=,=g(xn),证明:2n(-1)<1.
    

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