数学（文科）参考答案

巴中市高2023届一诊考试文科数学参考答案一．选择题：本大题共12个小题，每小题5分，共60分．DCBABDACDBAB二．填空题：本大题共4个小题，每小题5分，共20分．13．x=−1．14．8．15．14．16．[532,−+532]．三．解答题：本大题共6小题，共70分，解答应写出文字说明，证明过程或演算步骤．17—21题为必考题，每个试题考生都要作答．22、23为选考题，考生按要求作答．（一）必考题：共60分17．（本小题满分12分）某中学为了解高中数学学习中抽象思维与性别的关系，随机频率组距抽取了男生120人，女生80人进行测试．根据测试成绩按[0,20)，0.025[20,40)，[40,60)，[60,80)，[80,100]分组得到右图所示的频率分布直方图，并且男生的测试成绩不小于60分的有80人．（1）求这200人测试成绩的中位数和平均数的估计值；（同0.01一区间的数据用该区间中点值作代表）0.0075（2）填写下面的2×2列联表，判断是否有95%的把握认为0.0050.0025高中数学学习中抽象思维与性别有关．020406080100成绩成绩小于60成绩不小于60合计男女合计附：22PKk()≥0.100.0500.010n()ad−bcK2=(a+b)(c+d)(a+c)(b+d)k2.7063.8416.635解：（1）设中位数的估计值为x0，则20(0.00250.00750.01)0.025(60)0.5+++−=x0·····················································2分化简得0.025(60)0.1x0−=，解得x0=64∴中位数的估计值为64············································································3分设平均数的估计值为x，则x=1020+0.0025+30200.0075+50+200.0170200.02590200.005··········5分=++++=0.54.51035959∴平均数的估计值为59············································································6分注：中位数估计值的另一算法20+−=0.0050.025(80)0.5x0，化简得0.025(80)0.4−=x0，解得（2）成绩小于60分的人数为：200++==[(0.00250.00750.01)20]2000.480··············7分由题意，得2×2列联表如下表：···································································9分成绩小于60成绩不小于60合计男4080120女404080合计80120200巴中市高2023届一诊考试文科数学参考答案共9页第1页n()200(40404080)adbc−−22∴K2===5053.841············11分()()()()80120801209abcdacbd++++故有95%的把握认为高中数学学习中抽象思维与性别有关·····························12分18．（本小题满分12分）已知数列{}an满足a1=1，aann+1=+21．（1）证明：数列{a1}n+是等比数列；n（2）设bn=，求数列{}bn的前n项和Sn．an+1解：（1）由aann+1=+21得：aann+1+=+12(1)·······························································2分由a1=1知：a1+=12a+1∴n+1=2··························································································4分an+1∴数列{an+1}是以2为首项，2为公比的等比数列·········································5分（2）方法一n由（1）得：an+=12················································································6分∴nn分bn==n····················································································7an+12∴S=+++123n①n222223n21S=++++23n②···································································8分n22221n−②−①得：S=++++−1111n························································10分n222221nn−11−n2nn+2分=−=−nn2···························································111−1222∴S=−2n+2．··················································································12分n2n方法二由（1）得：················································································6分∴····················································································7分∴①1121S=++++nn−②·························································8分2n2222231nn+①−②得：1111S=+++−n·························································10分22n22221nn+11(1)−n22nnn12+分=−=−−=−nnnn+++11111····································111−122222∴．··················································································12分19．（本小题满分12分）如图，正方形ABCD中，E，F分别是ADPAB，BC的中点，将△AED，△DCF分别沿DE，DF折起，使A，C两点重合于点P，过P作PH⊥BD，垂足为H．E（1）证明：PH⊥平面BFDE；（2）若四棱锥P−BFDE的体积为12，HB求正方形的边长．FC巴中市高2023届一诊考试文科数学参考答案共9页第2页解：（1）证明：方法一在正方形ABCD中，有ACB⊥D，由已知得EFA//C∴EFB⊥D···························································································1分由折叠的性质知：PDPEPDPFPEPFP⊥⊥=,,∴PD⊥平面PEF····················································································2分又EF平面PEF，故PDE⊥F····································································3分∵PDBDD=∴EF⊥平面PBD···················································································4分又PH平面PBD，故EF⊥PH···································································5分∵PHB⊥D，EFB,D平面BFDE，且EF与BD相交∴PH⊥平面BFDE·················································································6分方法二在正方形ABCD中，有，由已知得∴···························································································1分由折叠的性质知：∴平面PEF····················································································2分又平面PEF，故····································································3分∵∴平面PBD···················································································4分又平面BFDE，故平面BFDE⊥平面PBD················································5分∵PH平面PBD，平面BFDE平面PBD=BD，∴平面BFDE·················································································6分（2）方法一P连结EF，设EFBD=Q，正方形ABCD的边长为2aa(0)22则S四边形BFDE=4a−2S△ADE=2a···································································7分由已知得：PD=2a,PE=PF=a,EF=2a,DQ=32aD2E∴，2分PE⊥PFPQ=a································································QH···········82BF由（1）知PD⊥PQ，在直角△PDQ中PDPQ由PH⊥DQ得：PH==2a····························································10分DQ3又由（1）知：平面BFDE∴V=1SPH=12a232a=4a=12，解得a=3······················11分P−BFDE3四边形BFDE339∴正方形的边长6．··············································································12分方法二连结EF，设，正方形ABCD的边长为由已知得：PD=2,aPE=PF=a,EF=2,aDQ=3BQ，EF⊥BD····················7分∴，SS△DEF=3△BEF∴VVP−−DEF=3PBEF······································································