巴中市高2023届一诊考试理科数学参考答案一.选择题:本大题共12个小题,每小题5分,共60分.DCBABDACDABA二.填空题:本大题共4个小题,每小题5分,共20分.13.x=−1.14.15.15.14.16.[5−+32,532].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)填写下面的2×2列联表,判断是否有95%的把握认为0.01高中数学学习中抽象思维与性别有关;0.00750.005成绩小于60成绩不小于60合计0.0025男020406080100成绩女合计(2)规定成绩不小于60(百分制)为及格,按及格和不及格用分层抽样随机抽取10名学生进行座谈,再在这10名学生中选2名学生发言,设及格学生发言的人数为X,求X的分布列和期望.附:22PKk()≥0.100.0500.010n()adbc−K2=()()()()abcdacbd++++k2.7063.8416.635解:(1)成绩小于60分的人数为:200++==[(0.00250.00750.01)20]2000.480··············1分由题意,得2×2列联表如下表:···································································3分成绩小于60成绩不小于60合计男4080120女404080合计80120200n()200ad−bc(402240−4080)∴K2===5053.841··············5分(a+b)(c+d)(a+c)(b+d)80120801209故有95%的把握认为高中数学学习中抽象思维与性别有关·······························6分(2)由(1)知,200人中不及格的人数为80,及格人数为120∴用分层抽样随机抽取的10名学生中不及格有4人,及格有6人·····················7分由题意,X的所有可能取值为0,1,2,且X服从超几何分布CCkk2−64,即:分P(X==k)=2(k0,1,2)······························································8C10巴中市高2023届一诊考试理科数学参考答案共9页第1页C021120CCCCCPXPXPX(0),=========(1),(2)646464281························10分22215153CCC101010∴X的分布列为····················································································11分X012281P15153EX=++=0122816.······························································12分15153518.(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;1(2)设bn=,Snn=b12+b++b,证明:Sn2.an解:(1)由aann+1=+21得:aann+1+=1+2(1)·······························································2分由a1=1知:a1+=12a+1∴n+1=2··························································································4分an+1∴数列{a1}n+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−1∴an=−212≥,当且仅当n=1时取等号·················································8分∴11≤1分bn==nn−1············································································9an2−1211−n∴≤11122Snn=+++b12bb1++++21nn−==−222221−122即Sn2.··························································································12分方法二由(1)得:n∴an=−21·························································································6分∴a2=321,a3=732,a4=1543····················································7分nn22n−n当n≥5时,恒有2=(11)+1C+nn+C=nn(−1)1+,故2−1nn(−1)0···········8分∴对任意n≥2时,恒有2n−1nn(−1)01111∴当时,恒有bn==−·············································10分ann(n−−1)n1n∴S=+++bbb≤1+−+−++(11)(11)(1−=−1)212nn12223n−1nn即.··························································································12分说明:也可用数学归纳法证明:2nn−1≥2−1与2n−1nn(−1).19.(本小题满分12分如图,正方形ABCD中,E,F分别是AB,ADPBC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EPH⊥BD,垂足为H.(1)证明:PH⊥平面BFDE;(2)求PB与平面PED所成角的正弦值.HBC解:(1)证明:F方法一巴中市高2023届一诊考试理科数学参考答案共9页第2页在正方形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∴PDE⊥F···························································································3分∵PDBDD=∴EF⊥平面PBD···················································································4分∵PH平面PBD∴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不妨设正方形ABCD的边长为2,由已知得:PDPEPFEF====2,1,2∴PEPF⊥················································7分D故直线PD,PE,PF两两垂直以P为原点,PD,PE,PF所在直线分别为x,y,z轴Ey建立如图所示的空间直角坐标系QHBF则PDEF(0,0,0),(0,2,0),(1,0,0),(0,0,1)xz平面PED的一个法向量为PF=(0,0,1)··············8分设EFBDQ=,则Q(11,0,),且QB=1DQ=1(,1−2,1)=(,1−2,1)···········9分223322636∴PB=PQ+QB=(,10,1)+(,1−2,1)=(,2−2,2)··································10分226363332||PBPF设PB与平面PED所成角为,则sin|cos,=|===PBPF33·····11分||PBPF||2333∴与平面所成角的正弦值为3.·················································12分3方法二不妨设正方形ABCD的边长为,设EFBDQ=,与平面所成角为由已知得:,∴,PQBQ==2······································································7分2又PD⊥PF,故PF⊥平面PDE·····················································································8分222∴sin=|cosP
数学(理科)参考答案
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