福建省泉州市2023-2024学年高三上学期质量监测（二）数学答案

保密★使用前泉州市2024届高中毕业班质量监测（二）2024.01高三数学四、解答题：本题共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．（10分）等差数列an和等比数列bn中，a1b12，a3b35，a52b20．（1）求a的公差d；n（2）记数列ab的前n项和为S，若a0，求S．nnnn20【命题意图】本小题主要考查等差数列、等比数列与数列求和等知识，考查运算求解等能力，考查函数与方程、化归与转化等思想，体现基础性，导向对发展数学运算等核心素养的关注．【试题解析】解法一：222d2q5,（1）设等比数列bn的公比为q，由题意，得·······························2分24d4q0,22q2d3,整理，得···········································································3分2q2d10,q1,q2,解得1或5d,d,2215故d或d．·················································································5分22q1,q2,（2）由（1），得1或5d,d,22q1,因为an0，所以d0，故1d,2n3n1从而a，b2(1)，·································································7分nn2高三数学试题第1页（共9页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} n1ab(n3)(1)，nn所以S(45)(67)(2223)10．············································10分20解法二：222d2q5,（1）设等比数列bn的公比为q，由题意，得·······························2分24d4q0,22q2d3,整理，得···········································································3分2q2d10,2消去q，得4d8d50，······································································4分15解得d或d．··············································································5分22q1,q2,（2）由（1），得1或5d,d,22q1,因为an0，所以d0，故1d,2n3n1从而a，b2(1)，·································································7分nn2n1ab(n3)(1)，nn10(422)10(523)所以S(46822)(57923)10．2022··············································································································10分高三数学试题第2页（共9页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} 18．（12分）教育部印发的《国家学生体质健康标准》，要求学校每学年开展全校学生的体质健康测试工作．某中学为提高学生的体质健康水平，组织了“坐位体前屈”专项训练．现随机抽取高一男生和高二男生共60人进行“坐位体前屈”专项测试．高一男生成绩的频率分布直方图如图所示，其中成绩在5,10的男生有4人．高二男生成绩（单位：cm）如下：10.212.86.46.614.38.316.815.99.717.518.618.319.423.019.720.524.920.525.117.5（1）估计高一男生成绩的平均数和高二男生成绩的第40百分位数；（2）《国家学生体质健康标准》规定，高一男生“坐位体前屈”成绩良好等级线为15cm，高二男生为16.1cm．已知该校高一年男生有600人，高二年男生有500人，完成下列22列联表，依据小概率值0.005的独立性检验，能否认为该校男生“坐位体前屈”成绩优良等级与年级有关？等级良好及以上良好以下合计年级高一高二合计22n(adbc)附：χ，其中nabcd．(ab)(cd)(ac)(bd)0.050.0100.0050.001x3.8416.6357.87910.828【命题意图】本题主要考查频率分布直方图、平均数、百分位数、独立性检验等知识；考查高三数学试题第3页（共9页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} 运算求解、数据处理等能力；考查数形结合等思想；体现基础性与应用性，导向对发展数据分析、数学运算、数学建模等核心素养的关注．【试题解析】（1）依题意得，抽取高二男生20人，所以抽取高一男生40人.·······························1分4因为高一男生成绩在5,10的男生有4人，所以a50.1，解得a0.02.40由(0.01a0.07b0.04)51，解得b0.06.···········································2分由样本估计总体，可估计高一年男生成绩的平均数x12.50.017.50.0212.50.0717.50.0622.50.045······················3分12.5100.0550.100.3550.3100.215.····························4分由200.48，可知样本数据的第40百分位数是第8项和第9项数据的均值，···············································································································5分高二年男生“坐位体前屈”成绩在5,15有7人，15,20有8人，15.916.8所以第40百分位数m在15,20中，故m16.35，2由样本估计总体，可估计高二年男生成绩的第40百分位数为16.35．·················6分（2）根据样本，知高一男生成绩良好及以上占50%，良好以下占50%，高二男生成绩良好128及以上占60%，良好以下占40%，由样本估计总体，可得22列联表如下：2020良好及以上良好以下合计高一300300600高二300200500合计6005001100···············································································································8分零假设为H：该校男生“坐位体前屈”成绩等级与年级之间无关.0221100(300200300300)根据列联表中的数据，得117.879x······11分0.005600500600500根据小概率值0.005的独立性检验，我们推断H不成立，即认为“坐位体前屈”成绩0等级与年级有关，此推断犯错误的概率不大于0.005.······································12分高三数学试题第4页（共9页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} 19．（12分）如图，两个棱长均等于2的正四棱锥拼接得到多面体PABCDQ．（1）求证：PA∥平面QBC；（2）求平面PCD与平面QBC的夹角的正弦值．【命题意图】本小题主要考查直线、平面间的平行，平面与平面的夹角等知识；考查空间想象能力、推理论证及运算求解等能力；考查数形结合、化归与转化等思想；体现基础性、综合性与应用性，导向对发展逻辑推理、数学运算、直观想象等核心素养的关注．【试题解析】解法一：（1）连结AC,BD，交于点O，连结PO,QO，易证PO平面ABCD，QO平面ABCD，所以P,O,Q三点共线，·············································································1分因为PQACO，所以PQ,AC确定平面PAQC，········································2分又因为PAPC2，AC22，222所以PAPCAC，故APC90，且PACPCA45，同理，可得△QAC中AQC90，且QACQCA45，·························3分故PAQAQC180，从而PA∥QC，····················································5分又PA平面QBC，QC平面QBC，所以PA∥平面QBC．···············································································6分高三数学试题第5页（共9页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} （2）分别取BC,CD的中点，ACBDO，由题意，可知OP,OM,ON两两垂直，···············································································································7分以O为原点，分别以OM,ON,OP的方向为x轴、y轴、z轴的正方向建立空间直角坐标系Oxyz，则P(0,0,2)，D(1,1,0)，C(1,1,0)，B(1,1,0)，Q(0,0,2)，从而PC(1,1,2)，PD(1,1,2)，QC(1,1,2)，QB(1,1,2)············8分nPC,xy2z0,1111设平面PCD的法向量n1(x1,y1,z1)，则即nPD,xy2z0,1111x0,1整理，得令z11，得n1(0,2,1)，············································9分y2z,11nQB,xy2z0,2222设平面QBC的法向量n2(x2,y2,z2)，则即nQC,xy2z0,2222x2z,22整理，得令z21，得n2(2,0,1)，······································10分y0,2nn1112cosn,n，···························································11分12|n1||n2|3331设平面PCD与平面QBC的夹角为，则coscosn,n，123222所以sin1cos．·································································12分3高三数学试题第6页（共9页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} 解法二：（1）连结AC,BD，交于点O，连结PO,QO，易证PO平面ABCD，QO平面ABCD，所以P,O,Q三点共线，易证得PQ,AC,BD两两垂直，且交于点O．···················1分以O为原点，分别以OB,OC,OP的方向为x轴、y轴、z轴的正方向建立空间直角坐标系Oxyz，···············································································································2分则P(0,0,2)，A(0,2,0)D(2,0,0)，C(0,2,0)，B(2,0,0)，Q(0,0,2)，从而PA(0,2,2)，QC(0,2,2)，故PAQC，························································································3分又AQC，所以PA∥QC，·······································································4分又PA平面QBC，QC平面QBC，所以PA∥平面QBC．···············································································6分（2）由（1），得PC(0,2,2)，PD(2,0,2)，QC(0,2,2)，QB(2,0,2)nPC,2y2z0,111设平面PCD的法向量n1(x1,y1,z1)，则即nPD,2x2z0,111yz,11整理，得令z1，得n1(1,1,1)，·················································8分xz,11高三数学试题第7页（共9页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} nQB,2x2z0,222设平面QBC的法向量n2(x2,y2,z2)，则即nQC,2y2z0,222xz,22整理，得令z21，得n2(1,1,1)，···········································10分yz,22nn111112cosn,n，···························································11分12|n1||n2|3331设平面PCD与平面QBC的夹角为，则coscosn,n，123222所以sin1cos．·································································12分3解法三：（1）连结AC,BD，交于点O，连结PO,QO，易证PO平面ABCD，QO平面ABCD，所以P,O,Q三点共线，·············································································1分因为PQACO，所以PQ,AC确定平面PAQC，········································2分又POOQ，AOOC，所以四边形PAQC平行四边形，·······························3分所以PA∥QC，·······················································································4分又PA平面QBC，QC平面QBC，························································5分所以PA∥平面QBC．···············································································6分（2）设平面平面PCD的法向量为m，平面QBC的法向量为n，平面QBC与平面PCD的夹角为，则coscosm,n．···································································7分高三数学试题第8页（共9页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} 由（1）知PA∥平面QBC，同理，可得PD∥平面QBC，又PA,PD平面PAD，且PAPDP，所以平面PAD∥平面QBC，所以平面QBC的法向量m是平面PAD的法向量，设平面PAD与平面PCD的夹角为，则coscosm,n，从而coscos，··················································································8分取PD的中点E，连结AE,CE，则AEPD，CEPD，又EPD，AE平面APD，CE平面CPD，所以AEC是二面角APDC的平面角，·················································10分在△ACE中，AC22，AECE3，222EAECAC3381由余弦定理，得cosAEC，即cosAEC，2EAEC2331从而coscosAEC，·····································································11分3222所以sin1cos．·································································12分3高三数学试题第9页（共9页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} 保密★使用前泉州市2024届高中毕业班质量监测（二）高三数学20．（12分）一个袋子中有10个大小相同的球，其中红球7个，黑球3个．每次从袋中随机摸出1个球，摸出的球不再放回．（1）求第2次摸到红球的概率；（2）设第1,2,3次都摸到红球的概率为P；第1次摸到红球的概率为P；在第1次摸到红12球的条件下，第2次摸到红球的概率为P；在第1,2次都摸到红球的条件下，第3次摸到红球3的概率为P．求P,P,P,P；41234（3）对于事件A,B,C，当P(AB)0时，写出P(A),P(B|A),P(C|AB),P(ABC)的等量关系式，并加以证明．【命题意图】本小题主要考查条件概率、全概率公式等知识；考查运算求解、推理论证等能力；考查化归与转化等思想．体现基础性、应用性和综合性，导向对发展数学运算、数学抽象等核心素养的关注．【试题解析】解法一：（1）记事件“第i次摸到红球”为Ai1,2,3,,10，则第2次摸到红球的事件为A，···1分i2于是由全概率公式，得PA2PA1PA2A1PA1PA2A1·························3分72377.········································4分103109103A77（2）由已知，得PPAAA，························································5分11233A24107P2PA1，·····················································································6分102PA1A2A7107102P3PA2A12，···········································7分PA1A1071573高三数学试题第1页（共11页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} PA1A2A37155P4PA3A1A2.····················································8分PA1A22478（3）由（2）可得PPPP，即PAAAPAPAAPAAA，1234123121312可猜想：PABCPAPBAPCAB，·················································9分证明如下：由条件概率及PA0,PAB0，PABPABC得PBA，PCAB，································10分PAPABPABPABC所以PAPBAPCABPAPABC.·······12分PAPAB解法二：（1）记事件“第i次摸到红球”为Ai1,2,3,,10，则第2次摸到红球的事件为A，········1分i2于是AAAAA，··············································································2分21212故PAPAAAAPAAPAA···············································3分2121212121111CCCC422177637······················································4分22AA9090101010（2）设事件“第1,2,3次都摸到红球”，事件“第1次摸到红球”，事件“第1次摸到红球的条件下第2次摸到红球”，事件“第1,2次均摸到红球的条件下第3次摸到红球”的四个事件的事件样本点个数和样本空间的样本点个数分别为n，Ni1,2,3,4，则由古典概型计ii算公式，得3nA717P，·················································································5分13NA24110n72P，························································································6分2N102n623P，····················································································7分3N933n54P.····························································································8分4N84（3）同解法一．····························································································12分高三数学试题第2页（共11页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} 21．（12分）sin(BC)bc△ABC的内角A,B,C所对的边分别为a,b,c．已知b3，．sinBsinCacπ（1）若C，求a；62（2）点D是△ABC外一点，AC平分BAD，且ADC，求△BCD的面积的取值3范围．【命题意图】本小题主要考查解三角形、三角恒等变换、导数的应用等知识，考查推理论证、运算求解等能力，考查数形结合和化归与转化等思想，体现综合性与应用性，导向对发展直观想象、逻辑推理及数学运算等核心素养的关注．【试题解析】解法一：abc（1）在△ABC中，由正弦定理，得2R，sinAsinBsinBabc所以sinA，sinB，sinC，2R2R2R又ABCπ，得BCπA，故sin(BC)sin(πA)sinA，sin(BC)sinAa所以，sinBsinCsinBsinCbcsin(BC)bcabc从而，可化为，············································1分sinBsinCacbcac222整理，得acbac，········································································2分222BABCAC1由余弦定理，得cosB，2BABC2π因为B(0,π)，所以B，3ππ又C，所以Aπ(BC)，62πb所以在Rt△ABC中，b3，C，且cosC，解得a2．·····················6分6a高三数学试题第3页（共11页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} π（2）由（1）得，B，32π因为AC平分BAD，且ADC，故设BACDAC,(0,)33ACBC3BC在△ABC中，由正弦定理，得，即，解得BC2sin，sinABCsinπsinsin3ACCD3CD在△ADC中，由正弦定理，得，即，解得CD2sin，sinADCsin2πsinsin3π2π又四边形ABCD中，B，ADC，BAD2，所以BCDπ2，331所以SBCCDsinBCD△BCD222sinsin(2)22sinsin2(1cos2)sin21sin2sin4·······································································9分21令f(x)sin2xsin4x,x[0,]，则23f(x)2cos2x2cos4x2(cos2x1)(2cos2x1)································································10分从而f(x)0，所以f(x)在[0,]单调递增，···············································11分3π33又f(0)0，f()，34ππ33所以当(0,)时，f(0)f()f()，即0f()，33433所以△BCD的面积取值范围为(0,)．····················································12分4解法二：（1）同解法一；····················································································6分高三数学试题第4页（共11页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} π2（2）四边形ABCD中，由（1）得B，又ADC，33所以BADC，从而A,B,C,D四点共圆，又因为AC平分BAD，所以BCCD，12设BCCDx(0x3)，所以SxsinBCD·································8分△BCD2ACBC1在△ABC中，由正弦定理，得，化简，得sinBACx，sinBsinBAC212故cosBAC1x，4π2又四边形ABCD中，BADC，所以BCDπBAD3，3所以sinBCDsinBAD，12又AC平分BAD，所以sinBADsin2BAC2sinBACcosBACx1x41212所以Sxx1x，··································································9分△BCD24216122所以Sx(1x)，其中0x3，△BCD441311314321312令f(t)t(1t)tt,t[0,3]，则f(t)ttt(3t)≥0，44416444所以f(t)在[0,3]上单调递增，27又f(0)0,f(3)，············································································11分16227故0S，△BCD1633所以△BCD的面积取值范围为(0,)．····················································12分4高三数学试题第5页（共11页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} 22．（12分）2222动圆C与圆C:(x5)y4和圆C:(x5)y4中的一个内切，另一个外切，12记点C的轨迹为E．（1）求E的方程；33（2）已知点M(1,t)（t），x轴与E交于A,B两点，直线AM与E交于另一点P，4249直线BM与E交于另一点Q，记△ABM,△PQM的面积分别为S,S．若SS，求直线122115PQ的方程．【命题意图】本小题主要考查双曲线的定义，双曲线的标准方程，圆与圆的位置关系、直线与双曲线的位置关系等知识；考查运算求解、逻辑推理等能力；考查数形结合、函数与方程等思想；体现基础性、综合性与创新性，导向对直观想象、逻辑推理、数学运算等核心素养的关注．【试题解析】解法一：（1）由题意，得两圆半径都为2，圆心分别为C(5,0),C(5,0).12设圆C的半径为R，由题意得RCC2CC2或RCC2CC2，1221故CCCC425CC，·······························································1分1212所以点C的轨迹是以C,C为焦点，实轴长为4的双曲线，·······························2分1222其中a2,c5,bca1，······························································3分【3分点：写出两个值可得此分；方程写对，这一步可不写】2x2所以轨迹E的方程为y1.·································································4分4t（2）由题意，得得A(2,0),B(2,0)，k，kt，AMBM3t所以直线AM的方程为y(x2)，直线BM的方程为yt(x2).················5分3设点P(x,y)，Q(x,y)．1122高三数学试题第6页（共11页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} ty(x2),32222由消去y，得(94t)x16tx16t360，···························6分2x2y1,42216t368t18由xx，x2，xx，得x，AP2AP11294t94t22t8t1812t8t1812t从而y(2)，故P(,)，·······························7分12222394t94t94t94tyt(x2),2222由x2消去y，得(14t)x16tx16t40，2y1,42216t48t2由xx，x2，xx，得x，BQ2BQ22214t4t1228t24t8t24t从而yt(2)，故Q(,)，·································8分222224t14t14t14t149因为△ABM,△PQM的面积分别为S,S，且SS，sinPMQsinAMB1221151MPMQsinPMQSMPMQx1x1(x1)(x1)221212所以S1MAMB1(2)1231MAMBsinAMB2228t188t2(1)(1)22222294t4t11(12t9)(4t3)(4t3)=，··················9分222233(94t)(4t1)(94t)(4t1)2249S49(4t3)492由SS，得即，212215S15(94t)(4t1)1512233(4t3)49又因为t，所以，2242(94t)(4t1)152化简，得t1，解得t1，·····································································10分2612104当t1时，P(,)，Q(,)，·························································11分5533所以直线PQ的方程为2xy80．························································12分高三数学试题第7页（共11页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} 解法二：（1）由题意，得两圆半径都为2，圆心分别为C(5,0),C(5,0).12设圆C的半径为R，由题意得RCC2CC2或RCC2CC2，1221故CCCC4即CCCC4.·······················································1分1212当CCCC4时，设C(x,y)，则122222(x5)y(x5)y4(x0)，···············2分【代入距离公式即给分】2222即(x5)y4(x5)y.222222两边平方，得(x5)y168(x5)y(x5)y，22整理，得5x42(x5)y，2x2上式两边再平方并整理得y1(x0)，················································2分42x2当CCCC4时，同理，可得y1(x0)，····································3分1242x2综上，轨迹E的方程为y1.···························································4分4t（2）由题意，得得A(2,0),B(2,0)，k，kt，AMBM331所以直线AM的方程为xy2，直线BM的方程为xy2．·················5分tt设点P(x,y)，Q(x,y)．1122高三数学试题第8页（共11页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} 3xy2,t9212由消去x，得(4)yy0，22x2tty1,4122t12t312t8t18由韦达定理，得y，x()2，12122994tt94t94t42t28t1812t所以P(,)；··········································································7分2294t94t1xy2,t124由消去x，得(4)yy0，22x2tty1,442t4t14t8t2从而y，x()2，2222214t1t4t14t142t28t24t所以Q(,)．···········································································8分224t14t149因为△ABM，△PQM的面积分别为S，S，且SS，sinPMQsinAMB1221151MPMQsinPMQSMPMQytyt2212所以S1MAMBt0t01MAMBsinAMB212t4ttt222294t4t1(4t3)=．·······································9分222t(94t)(4t1)2249S49(4t3)492由SS得即，212215S15(94t)(4t1)1512233(4t3)492因为t，所以，化简，得t1，解得t1，··········10分2242(94t)(4t1)152612104当t1时，P(,)，Q(,)，·························································11分5533所以直线PQ的方程为2xy80．························································12分解法三：高三数学试题第9页（共11页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} t（2）由题意，得得A(2,0),B(2,0)，k，kt，故k3k，AMBMBMAM3yy21设P(x,y)，Q(x,y)，则3.1122x2x2212x12y1y11y2y13由y1，得，所以.·······················5分14x2x24x2x241121设直线PQ的方程为xmyn，xmyn,222由x2，消去x，得(m4)y2mnyn40，2y1,422m40,m40,因为方程有两个不同的实数根，所以即2220,(2mn)4(m4)(n4)0,2m4,整理，得22mn4,22mnn4由韦达定理，得yy，yy············································6分122122m4m4yy3yy32121由得,x2x24myn2myn242121yy312即.22myym(n2)(yy)(n2)4121222mnn4将yy，yy，代入以上等式，122122m4m42n43整理，得，解得n4，24n16n164所以直线PQ的方程为xmy4，故直线PQ恒过点(4,0)，···························7分ttt从而yt(x2)t(x1)(my3)，1111333ytt(x2)tt(x1)t(my3)222249因为△ABM,△PQM的面积分别为S,S，且SS，sinPMQsinAMB1221151tMPMQsinPMQ(my3)t(my3)12S22MPMQy1ty2t3所以1002SMAMBttt1MAMBsinAMB2高三数学试题第10页（共11页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#} 22n42mn2m23m2922myy3m(yy)9m4m43m36m121212，22333m4m4··············································································································9分249S49m12492由SS，得，即，21215S115m41533因为直线PQ恒过点(4,0)，点M(1,t)（t），4222n444122所以yy==0，得m40，12222m4m4m4m212492m124921由，得，解得m，·····································10分22m415m415433411又因为t，所以yy，故m，所以m··························11分1P423221所以直线PQ的方程为xy4即2xy80.·······································12分2高三数学试题第11页（共11页）{#{QQABaYQAogCAABBAABgCUQVICkMQkACCAAoOBFAAsAAAAQFABCA=}#}