江苏省无锡市2024-2025学年高二上学期期终教学质量调研测试数学答案

无锡市2024年秋学期高二期终教学质量调研测试数学参考答案2025.01一、选择题：本题共8小题，每小题5分，共40分。在每小题给出的四个选项中，只有一项是符合题目要求的。1．A2．D3．D4．B5．C6．A7．C8．A二、选择题：本题共3小题，每小题6分，共18分。在每小题给出的选项中，有多项符合题目要求。全部选对的得6分，部分选对的得部分分，有选错的得0分。9．ABD10．ACD11．ABD三、填空题：本题共3小题，每小题5分，共15分。3712．913．14．2a-b=333四、解答题：本题共5小题，共77分。解答应写出文字说明、证明过程或演算步骤。15．解：（1）因为圆心C在直线xy10上，所以设C(a,a1)，因为圆C经过A(5,1),B(1,7)两点，所以(5a)2(1a1)2(a1)2(a17)2，解得a2，·········································································································2分即C(2,3)，半径r(52)2(13)25，······························································4分所以圆C的标准方程为(x2)2(y3)225．···························································6分（2）因为过点A的直线l被圆C截得的弦长为8，8所以C到直线l距离d52()23，···································································8分2当直线l斜率不存在时，直线x5满足题意；····························································9分当直线l斜率存在时，设直线方程为yk(x5)1，即kxy5k10，2k35k17所以d3，解得k，k21247此时直线方程为y(x5)1，即7x24y110．··············································12分24综上所述，直线l的方程为x5或7x24y110．··················································13分16．解：（1）由题可设等比数列{an}的公比为q(qZ)，因为S28，6a1a330，32a1a1a2a1(1q)8a127所以，解得或（舍去）·········································4分a(6q2)30q331q4n1所以数列{an}的通项公式为an=23；····································································6分2(13n)（2）由（1）知S3n1，·····································································8分n13n所以bn(2n1)(1Sn)(2n1)3．·····································································10分23n234n1Tn133353(2n1)3，3Tn133353(2n1)3{#{QQABaQKQggioAAIAAAhCAQGSCgMQkBECAQoGBEAQoAABgBFABAA=}#}23nn1两式相减得2Tn13232323(2n1)318(13n1)3(2n1)3n13n16(2n1)3n1(22n)3n16··14分13n1所以Tn(n1)33．·····················································································15分17．解：以D为原点，DA，DC，DP所在直线分别为x轴、y轴、z轴，建立如图所示的空间直角坐标系，设DC1．11（1）依题意得A(1,0,0)，P(0,0,1)，B(1,1,0)，C0,1,0，E(0,,)．221111则PB(1,1,1)，DE(0,,)，故PBDE00，2222所以PBDE．····································································································（2）由（1）知PA(1,0,1)，PC(0,1,1)，BC(1,0,0)，设平面PBC的法向量为n1(x,y,z)，直线PA与平面PBC所成的角为，n1·PC0yz0所以，即，取x0，y1，z1，x0n1·BC0所以n1(0,1,1)是平面PBC的一个法向量，································································6分PAn111所以sincosPA,n1．·······················································8分PAn1222故直线PA与平面PBC所成的角大小30．································································9分（3）设平面PAC的法向量为n2(a,b,c)，平面PAC与平面PBC夹角为，n2·PA0ac0因为PA(1,0,1)，PC(0,1,1)，所以，即，取a1，则b1，c1，bc0n2·PC0所以n2(1,1,1)是平面PAC的一个法向量．······························································11分由（2）知n1(0,1,1)是平面PBC的一个法向量，·······················································12分所以n1n226．分coscosn1,n2··················································14n1n22336故平面PAC与平面PBC夹角的余弦值为．·························································15分318．解：（1）因为3Sn(n2)an①，所以3Sn1(n1)an1(n≥2)②，a①－②整理得，因为，所以，故nn1≥(n1)an(n1)an1a12an0(n2).an1n1ana3a2n1n432所以ana12nn.··································2分an1a2a1n1n221经检验n1也符合.······························································································3分2所以annn.····································································································4分{#{QQABaQKQggioAAIAAAhCAQGSCgMQkBECAQoGBEAQoAABgBFABAA=}#}11111（2）bn()，annn(n2)2nn21111111111所以Tbbb(1)n12n232435n1n1nn213113111()()，······················································6分22n1n242n1n23又nN*，所以T.···························································································8分n4n*n（3）因为an24对nN恒成立，且20，n2n4所以对nN*恒成立.·····································································10分2nn2n4设f(n)，2n(n1)2(n1)4n2n4n2n6(n3)(n2)则f(n1)f(n)，·············12分2n12n2n12n1所以f(1)f(2)f(3)f(4)f(5)，·······························································14分n2n4可得f(n)的最大值为f(3)f(4)1，······················································16分2n故实数≥1.······································································································17分19．解：（1）设双曲线的半焦距为c，cb由题意得，2，F(c,0)，双曲线的渐近线yx，即bxay0，aabc所以3，即b3，·············································································2分a2b2而c2a2b24a2，所以a21，y2所以双曲线的方程为x21．··········································································4分3（）易知，，设直线的方程为，，，2A(1,0)F(2,0)PQxmy2P(x1,y1)Q(x2,y2)xmy2由方程组2，得(3m21)y212my90，··················································6分2yx1311因为过焦点F的直线与C的右支交于P，Q两点，所以||3，即0m2，m312m9由根与系数关系得y1y2，y1y2．··················································8分3m213m2192y1y2y1y23m1（i）kAPkAQ9．x1x1m2yym(