内蒙古呼和浩特市2024届高三第一次质量数据监测理科数学试卷

2024-03-20 · U1 上传 · 10页 · 1.8 M

{#{QQABCYKAggAgQAAAAQhCAwH6CAIQkBACCKoOQAAIoAAASQFABCA=}#}{#{QQABCYKAggAgQAAAAQhCAwH6CAIQkBACCKoOQAAIoAAASQFABCA=}#}呼和浩特市高三理科数学一模参考答案一、选择题123456789101112BDCADBBABCAA二、填空题x2y2113.914.164n15.2n116.0,e三、解答题17.解:(1)∵2ac2bcosC.2sinAsinC2sinBcosC2sin(BC)sinC2sinBcosC……………………………………2分2cosBsinCsinC0∵sinC01cosB………………………………………………………分25又B(0,)2∴B………………………………………………………分36{#{QQABCYKAggAgQAAAAQhCAwH6CAIQkBACCKoOQAAIoAAASQFABCA=}#}(2)△ABD中,由余弦定理,|AD|2|AB|2|BD|22|AB||BD|cosB可得|BD|22|BD|30解得|BD|=1.………………………………………………………8分AD是角平分线ACAB2,DCDB设DCm,则AC2m…………………………………………10分14m24(1m)222(1m)2整理得3m24m707m37DC…………………………12分318.(1)证明:如图,取BD中点E,连接,在菱形ABCD中,∵A1ECE.AB=又AD,CB=CD,,111则∴AB=AD,CB=CD,∠DAB=60°,AB=2.所|以A1B|=|A1D|=|BD|=|CB|=|CD|=2A1E=CE=3又1222,即A1EE⊥BCDEA1CA1ECE…………………………2分又BD∩CE=E,∴A1E平面BCD又A1E平面A1BD所以平面平面.A1BDBCD…………………………5分(2)以E为原点,ED为x轴,EC为y轴,EA1为z轴建立空间直角坐标系,则E0,0,0,A10,0,3,D1,0,0,B1,0,0,C0,3,0,,,,,,,,A1D103A1B103A1C033…………………………7分{#{QQABCYKAggAgQAAAAQhCAwH6CAIQkBACCKoOQAAIoAAASQFABCA=}#}设平面A1BC的一个法向量为nx,y,z,则nAB01取,n(3,1,1)…………………………9分nA1C03315sin|cosA1D,n|||…………………………11分255所以与平面所成角的正弦值为…………………………12分15A1DA1BC5.解:()…………………………………………分191.P20111211P……………………………………3分33232211PP0(1P)……………………………………5分43324当时,1(2)n1Pn(1Pn1)……………………………………8分211112P(P),Pn32n13133n1n1121121P,即Pn332n332当n1时,符合上式…………………………………10分121故数列P是首项为,公比为的等比数列n332…………………………12分{#{QQABCYKAggAgQAAAAQhCAwH6CAIQkBACCKoOQAAIoAAASQFABCA=}#}20.(1)∵f(x)(ex2)(x1)exf(1)e2又∵f(1)0切线l的方程为y(e2)(x1)…………………………………2分下证:f(x)(e2)(x1),即证:(x1)(ex2)(e2)(x1),即(x1)(exe)0令g(x)(x1)(exe),则g(1)0,且g(x)xexe,g(1)0g(x)ex(x1),g(1)0,故g(x)在(,1)上单调递减且小于0,在(1,)上单调递增故g(x)在(,1)上单调递减,在(1,)上单调递增故g(x)g(1)0,所以f(x)(e2)(x1),即f(x)的图象在直线l的上方.…………………………………………6分(2)由题知f(x)0有两根x11,x2ln2,记f(x)在点(1,0)、点(ln2,0)处的切线方程分别为l1、l2,由(1)知l1:y(e2)(x1)∵f(x)(ex2)(x1)ex,f(ln2)2ln21的方程为l2y2ln21(xln2)………………………………7分下证:f(x)的图象在直线l2的上方,即证:f(x)2ln21(xln2),即(x1)(ex2)2ln21(xln2)令h(x)(x1)(ex2)2ln21(xln2),则h(ln2)0h(x)xex2ln2,则h(ln2)0h(x)(x1)ex,则h(1)0故h(x)在(,1)上单调递减且小于0,在(1,)上单调递增;故h(x)在(,ln2)上单调递减,在(ln2,)上单调递增故h(x)h(ln2)0,所以f(x)2ln21(xln2),即f(x)的图象在直线l2的上方.………………………………9分假设l1与yt交于x3,l2与yt交于x4,则x3x4l1:y(e2)(x1)t联立得x31yte2l2:y2ln21(xln2)t联立得x4ln2yt2ln21ttt(e2ln2)所以x1x2x3x41ln21ln2.e22ln212(e2)(1ln2)………………………………12分{#{QQABCYKAggAgQAAAAQhCAwH6CAIQkBACCKoOQAAIoAAASQFABCA=}#}22ppp21.(1)解:令R(x,y),则RFx2y2pyy2pyy222pppp因为y0,所以y,RF的最小值为,即1,抛物线的方程为x24y22222分1(1)2(2)证明:令M(x,y),N(x,y),l:y1(xt),即y1(xt)1122PF0tt211211则PFPMPN11111(1)1y(1)1y(1)k2k21k222111(1)y1(1)y2(1)11yy2112y11y21y11y21y1y22y11y21y1y212lPF:y1(xt)16联立得2,由韦达定理得ty(22)y10y1y212tx4y211综上所述:PFPMPN7分{#{QQABCYKAggAgQAAAAQhCAwH6CAIQkBACCKoOQAAIoAAASQFABCA=}#}211(3)满足的关系为:.QGQEQHl:y13x联立AB得A234,743,B234,7432x4y假设抛物线C在A处的切线为l1:y74332x234在B处的切线为l2:y74332x234联立l1、l2得Q(23,1)令E(x3,y3),H(x4,y4),G(x0,y0),lQG:y1k(x23)211211则QGQEQH1111y(1)1y(1)1y(1)k20k23k24211y0(1)y3(1)y4(1)211(*)y01y31y41lQG:y1k(x23)7k32k3联立得y0,k3y14klAB:y13x0lQG:y1k(x23)联立得y2243k4k2y(123k)202x4y所以11y3y42y3y424kk3k32y31y41y31y41y3y4y3y4116k4k112211所以,即.y31y41y01QGQEQH----------12分22解:(1)设P点的极坐标为(,),---------------------1分3则OP,OMcosOMOP12312cos4cos------------------4分4cos0------------------5分2{#{QQABCYKAggAgQAAAAQhCAwH6CAIQkBACCKoOQAAIoAAASQFABCA=}#}(2)直线C2过半圆的圆心(2,0)所以直线的倾斜角为或2时满足题意33曲线C2的普通方程为y3(x2)----------------10分注:只写出一条直线给3分23、(1)当时��,=2�−1+2�,−解2得+�当�<0时−,5�+3≥2,解�<得011当0≤�<时2,−3�+,解3得≥20≤�≤31当2≤�<时1,�+1≥,2解得�∈∅综上�≥所1述,5�−3≥的2解集为�≥1或分1(2)�(�)≥2{�|�≤3�≥1}⋯⋯⋯⋯⋯⋯⋯5当时,的最小值为分13�=2��2⋯⋯⋯⋯⋯⋯⋯⋯⋯7,∴�+�=12�+�1当�+且�仅=当1,∴��≤取等=241�=�=令则21�=��,0<�≤4{#{QQABCYKAggAgQAAAAQhCAwH6CAIQkBACCKoOQAAIoAAASQFABCA=}#}211�+�−2��12225+�+�=+��+=��+−2=�+−2≥当且�仅当�取等,此时�������411�=�=�=42分⋯⋯⋯⋯⋯⋯⋯⋯10{#{QQABCYKAggAgQAAAAQhCAwH6CAIQkBACCKoOQAAIoAAASQFABCA=}#}

VIP会员专享最低仅需0.2元/天

VIP会员免费下载,付费最高可省50%

开通VIP

导出为Word

图片预览模式

文字预览模式
版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报
预览说明:图片预览排版和原文档一致,但图片尺寸过小时会导致预览不清晰,文字预览已重新排版并隐藏图片
相关精选
查看更多
更多推荐