吉林省吉林市2024-2025学年高三下学期3月三模数学答案

吉林地区普通高中2024—2025学年度高三年级第三次模拟考试8.教学提示参考选择性必修第二册教材P553(3)，科赫雪花曲线.数学学科参考答案若原正三角形的边长为，则雪花曲线满足：1Pn一、单项选择题nn11①边数：n1；②边长：1；③周长：4；12345678an34bnLn333ABBCACDBn1n1348343④面积：SSSSS，其中S．6.教学提示n111111595594先找的外心，发现为线段的四等分点（靠近），ΔAFD1O1O1A1DA1二、多项选择题则球心在过且与平面垂直的直线上91011OO1AD1F.利用几何法或者坐标法均可，坐标法具体做法参考如下：ABABDACD以D为原点，建立如图所示的空间直角坐标系，则11.教学提示33，，，设球心33，O1(,0,)A(2,0,0)E(1,2,0)O(,m,)参考选择性必修第二册教材P8283，牛顿切线法───用导数方法求方程的近似解.222232227f(x)xx1，f(x)3x1，由OAOE，求出m1，从而求出ROA.2在横坐标为的点处的切线斜率为2，7.教学提示yf(x)xn13xn11yz在横坐标为的点处的切线方程为32．xlnxyeelnz5yf(x)xn1y(xn1xn11)(3xn11)(xxn1)，y，z，如图所示：3lnx5xe5ylnz5e2x1令y0，则xn1，C选项正确；n2由图知，yzx，故A，B选项错误；3xn11x（也可以通过yx，ye，ylnx图象的变化f(x)3x210在R上恒成立，f(x)在R上单调递增．速度判断x，y，z的大小关系）212f()，f(1)1，则f()f(1)0．yx553273由得，P(,)y5x22.2由零点存在性定理可知，f(x)在R上存在唯一零点r，且r,1．3yex和ylnx关于yx对称，5533A(y,ey)和B(x,lnx)关于P(,)对称，f(r)rr10，则1rr．22xy5且lnxy，lnxy2y5，故C选项错误；3323232xn112xn113rxn1r2xn13rxn1ryyxrrex，ezxzxy5，故D选项正确.n2223xn113xn113xn11第1页共5页(xr)2(2xr)四、解答题n1n1．2．【解析】3xn1115a1(a14d)64,a14,a116,2（Ⅰ）由题意得解得或（舍）要证，只需证2，xnrxn1r2xn1r3xn11a12d10.d3,d3.2an4(n1)33n1，即数列{an}的通项公式是an3n1．·······························4分只需证r3xn12xn11，即证rxn1(3xn12)1，（Ⅱ）①，当时，，得，2Sn2bn1n1b1S12b11b11x，x(3x2)11，n13n1n1当n≥2时，Sn2bn1②，由①②得，SnSn(2bn1)(2bn1)，21111r1rxn1(3xn12)1成立.3bn≥化简得，bn2bn1，即2(n2)，2b，选项正确．n1xnrxn1rD三、填空题n1数列{bn}是以1为首项，2为公比的等比数列，bn2．········································8分25812.13.214.(2分)；(3分)n1，337cnanbn3n1214.教学提示n[4(3n1)]1(12n)3n25n35T2n12nn2n1．·····················13分n212222|PF||AF|5511，，，n[4(3n1)]35由角平分线定理可知，|F1F2|2c|PF1|c（或n2）|PF||AF|34Tn2bn12nn122222．【解析】由双曲线定义可知，，，16|AF1||AF2|2a|AF1|5a|AF2|3a（Ⅰ）（法一）X的所有可能取值为0，1，2，且X服从超几何分布．在中，由余弦定理可得，2222π，ΔAF1F24c(5a)(3a)25a3acos3021120C3C31C3C33C3C312，，．P(X0)2P(X1)2P(X2)2222c4974c49a，e，e1，e，C65C65C65a242的分布列为连接，为内心，是的角平分线，X0F1IIΔAF1F2F1IAF1F2X12131AIAFaaP在中，由角平分线定理可知|||1|5448．ΔAF1P555|IP||PF|5ce71c4·····························································································································4分131X的数学期望E(X)0121．···························································5分555（法二）X服从超几何分布，且N6，M3，n2．扫码查看圆锥曲线的光学性质及动态演示过程第2页共5页k2kCCx2X的分布列为P(Xk)33，k0，1，2．···················································4分y21,C2由4消去x，得(t24)y22mtym240.6xtymnM23X的数学期望E(X)1.······································································5分N62mtm2422Δ16(tm4)0，y1y22，y1y22，··········································8分t4t4（Ⅱ）（i）记C“每位员工经过培训合格”，A“每位员工第i轮培训达到优秀”(i1，2，3)，i8mx1x2ty1mty2mt(y1y2)2m2，t4，根据概率加法公式和事件相互独立定义得，CA1A2A3A1A2A3A1A2A3A1A2A322224m4tx1x2(ty1m)(ty2m)ty1y2mt(y1y2)m，t24P(C)P(A1A2A3)P(A1A2A3)P(A1A2A3)P(A1A2A3)，，MA(x12,y1)MB(x22,y2)MAMB，P(A)P(A)P(A)P(A)P(A)P(A)P(A)P(A)P(A)P(A)P(A)P(A)2221231231231234m4t16mm4MAMB(x2)(x2)yyxx2(xx)4yy4021111121121211212121212222．t4t4t4323323323323226化简得5m16m120，m，或m2（舍）15即每位员工经过培训合格的概率为．·······································································10分262111112112121直线AB过定点D,0.····················································································12分（注：记C“每位员工经过培训合格”，P(C)．53233233233232上述求解方法给满分.）（注：此处亦可按如下方法求m1（）记，两部门开展培训后合格的人数为，则，yyyyyyiiABDeepSeekYY~B(50,)1212122kMAkMB22x2x2(tym2)(tym2)tyyt(m2)(yy)(m2)112121212E(Y)5025．·····························································································12分2m24m26则253025205031100(万元)1，m）4(m2)24(m2)5即估计A，B两部门的员工参加DeepSeek培训后为公司创造的年利润为1100万元．········15分811设E为MD的中点，即E,0.（注：直接列式503050205031100(万元)给满分.）5221142若N与D不重合，则MD是RtΔMND的斜边，|NE||MD|；17.【解析】22551142c3若N与D重合，则|NE||MD|.（Ⅰ）e，a2，c3，b2a2c2431，2255a282x2综上所述，存在定点E,0，使得|NE|为定值.···················································15分即椭圆C的标准方程为y21.············································································4分554（Ⅱ）（法一）由题可知，直线AB的斜率不为0，（法二）当直线斜率不存在时，设，，，，1ABA(x0,y0)B(x0,y0)M(2,0)MA(x02,y0)设直线的方程为，设，，ABxtymA(x1,y1)B(x2,y2)1MB(x2,y)，MAMB，MAMB(x2)2y2x24x41x20，00000040第3页共5页65x216x120，解得x2，不符题意（舍），或x，符合题意.8200005综上所述，存在定点E,0，使得|NE|为定值.···················································15分556直线AB过点D,0.························································································6分18.【解析】5ACAB233（Ⅰ）ΔABC中，由正弦定理得，即，当直线斜率存在时，设直线的方程为，设，，sinABCsinACBsinABCπ2ABABykxm(k0)A(x1,y1)B(x2,y2)sin32故，又，π.分x2sinABC10ABCπABCBCAB····································2y1,2222由4消去y，得(14k)x8kmx4m40.又ΔACD为正三角形，CADACB，AD//BCADAB.·································4分ykxm又平面，平面分AA1ABCDADABCDAD