数学-福建省莆田市2025届高中毕业班第二次教学质量检测试卷（莆田二检）试题和答案

莆田市2025届高中毕业班第二次教学质量检测数学试题参考解答及评分标准12345678CCADBBCB91011ACDACDABD 12.1213.2614.415．本小题主要考查等差数列、等比数列及数列求和等基础知识；考查运算求解能力，逻辑推理能力等；考查化归与转化思想，分类与整合思想，函数与方程思想等；考查逻辑推解1）解法一：依题意，设数列{an}的公差为d，因为S3=15，所以a1+a2+a3=15，则3a1+3d=15，即a1+d=5①,···············1分因为2a4=a3+14，即2(a1+3d)=a1+2d+14，所以a1+4d=14②,·················2分由①②,得a1=2，·················································································3分d=3.·················································································4分数学答案第1页（共11页）所以an=2+(n-1)×3.············································································5分所以an=3n-1.······················································································6分 ,50，···································8分其中5和17为数列{2n+1}的前5项与{an}的前17项的公共项， 10分=(3+5+9+17+33)+(a1+a2+…+a17)-5-17··············································11分=67+2×17+×3-5-17.·································································12分=487.··································································································13分解法二1）依题意，设数列{an}的公差因为S3=15，所以=3a2=15，所以a2=5，·····································1分又因为2a4=a3+14，所以a3+a5=a3+14，所以a5=14，································2分所以3d=a5-a2=9，解得d=3.·································································3分所以a1=a2-d=2.·················································································4分所以an=2+(n-1)×3.············································································5分=3n-1.······················································································6分（2）设cn=2n+1，因为a20=3×20-1=59，···········································································7分所以由cn=2n+1≤59可得n≤5，·································································8分因为c1=3,c2=5,c3=9,c4=17,c5=33,···························································9分其中c2=5=a2，c4=17=a6，···································································10分数学答案第2页（共11页）131217)································································11分=487.··································································································13分分.丄AB.····························1分所以AB1B.····················································3分B，所以AB1BC.················································································4分.···················································5分B1B，所以BC丄平面AA1B1B，···········································································7分轴的正方向，建立空间直角坐标系.·······························8分C(2,0,0)，D(1,1,1)，·················································9分数学答案第3页（共11页）取x=1，则z=-1，所以n=(1,0,-1)是平面ABD的一个法向量.···········----→ 所以直线A1C与平面ABD所成角的正弦值为.·········································15分丄BC.······························6分B1B，所以BC丄平面AA1B1B，即BC，BA，BB1两两垂直.······································7分 C=，·············································8分 所以△ABD的面积S=2.·······································································10分 所以CC,=2，·····················································································13分设直线A1C与平面ABD所成角为θ,则sinθ=.················································································14分数学答案第4页（共11页） 所以直线A1C与平面ABD所成角的正弦值为.···············································15分性．满分15分. 事件B=“测试结果语音识别成功”.····························································1分·······························································3分············································································4分=0.69.·····························································································5分条件下事件A发生的概率.··········································································6分（2）方案一的测试次数的数学期望为4.·····························································9分用X表示“方案二测试的次数”，由题意得X的可能取值为3，5.···················10分=0.512，·······································································11分P(X=5)=1P(X=3)=10.512=0.488.·····················································12分又因为E(X)<4，···················································································14分所以，以测试次数的期望值大小为决策依据，应选择方案二.····························15分数学答案第5页（共11页）解1）由已知，得b=1，············································································1分2，e==，···································································2分所以a=2.······························································································3分所以E1的方程为:+y2=1.······································································4分PyPAAOQBOQBx此时AP=BQ.·······················································································5分2所以弦AB中点的横坐标为22，········································································8分xPxP综上所述，AP=BQ.·············································································10分数学答案第6页（共11页）2，所以···········································12分又，·······································································14分 1所以km=②.····················································································15分 12由①②解得，经检验符合题意.·························································16分所以m=3.·····························································································17分AP=BQ.·····························································································5分数学答案第7页（共11页）BBB化简得xA+xB=xp+xQ.············································································9分22综上所述，AP=BQ.·············································································10分11则BPOR=PACS.··············11分11又因为BP=2PA，所以BP=2PA，所以OR=CS，·····················12分PTPTyySSAARROBCx所以O为Rt△CST斜边CT的中点.····························································13分2=2=m②,·····················16分所以m=3.·····························································································17分数学答案第8页（共11页）综合性和创新性．满分17分.···············································································2分············································3分故f(x)的单调递增区间为(0，e)，单调递减区间为(e，+∞).·····························4分因为f(2)=f(4).······················································································6分所以，当x∈[2，e]时，x+2∈[4，e+············································································································7分所以f(x+2)≤f(x)；················································································8分f(x+2)≤f(x).································9分f(x+2)≤f(x).即f(x)在[2，+∞)上具有性质P(2).···················