数学（理科）参考答案

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(402240−4080)∴K2===5053.841··············5分(a+b)(c+d)(a+c)(b+d)80120801209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：Sn2.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Sn2．··························································································12分方法二由（1）得：n∴an=−21·························································································6分∴a2=321,a3=732,a4=1543····················································7分nn22n−n当n≥5时，恒有2=(11)+1C+nn+C=nn(−1)1+，故2−1nn(−1)0···········8分∴对任意n≥2时，恒有2n−1nn(−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−1nn(−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||PBPF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=|cosP