2027届福建省福州市高三一检数学试题及答案

数学1．样本数据80，85，87，88，90，92，95的第50百分位数为A．87B．88C．89D．92C．3200D．3200A．729B．980C．100010．在四棱锥P−ABCD，，面面ABCD为正方形，面面PAD为正三角形，且平面PAD丄平面ABCD，则B．棱AD上存在点M，使得AD丄平面PMBC．异面直线AD与PB所成角的余弦值为D．二面角P−BC−A的大小为60。22=1被x，y轴截得的弦长分别为d1，d2，则下列说法正确的是A．a可以取任意实数D．若圆心C在直线y=x+1上，则d1+d2的最大值为232如图，在直三棱柱ABC−A1B1C1，，E是BC，点.（1）求证：A1B//平面AEC1；2.A.记△ABC的内角ABC的对边分别为abc．已知A=，sinB=.（1）求tanBtanC； （2）若△ABC的面积为，求b+c剩余罚点球的次数，则领先方提前获胜；否则，完成五轮时，进球数多的球队获胜.若完成五轮时双方进球数相同，则从第六轮开始进入“SuddenDeath”阶段，即双方继续各派新队员罚点球，直到某一轮，一方罚进而另一方罚失时决出胜负.某次高，足球联赛的淘汰赛，，甲、乙两队在常规时间和加时赛后以0:0战平，比球大战．假设每名队员是否罚进相互独立．已知甲队前五轮5名队员罚进点球的概率依次为----；乙队前五轮每名队员罚进点球的概率均为．甲队先罚.（2）前三轮罚点球后，比分达到3:3.进点球的个数，求X的分布列和数学期望.为a−拓展函数.（1）已知狄利克雷函数若函数f(x)=x.D(x)，x∈R，判断f(x)是否为2−拓展函数；2026-2027学年第一学期福州市高三年级第一次适应性练习 15.【考查意图】本小题主要考查直线与平间向量等基础知识，考查直观想象能力、逻辑推理能力、运算求解能力等，考查化转化、函数与方程思想等，考查直观想象、逻辑推理、数学运算等核心素养，体现则F是A1C的中点.···············································2分又E是BC中点，则EF//A1B.································3分因为EF平面AEC1，A1B丈平面AEC1，所以A1B//平面AEC1.············································································5分（2）由（1）知A1B//平面AEC1，所以直线A1B到平面AEC1的距离就是B到平面AEC1的距离，设为h.···············6分 因为CC1△ABE11()21 3 （2）由（1）知A1B//平面AEC1，所以直线A1B到平面AEC1的距离就是A1到平面AEC1的距离，设为h.··············6分 2 266(2)22(2,.····································因为点E为BC中点，所以点E到平面ACC1A1的距离为AB.·······················10分 3223 （2）以A为坐标原点，AB，AC，AA1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.·································则点A1到平面AEC1的距离········12分所以直线A1B到平面AEC1的距离就是A1到平面AEC1的距离， ·3····················································所以直线A1B到平面AEC1的距离为．·3····················································化归与转化思想、数形结合思想等，考查数学抽象、直观想象、逻辑推理、数学运F22222(x0+1y0)（2）设P(x0,y0)，因为A2(1,0)，所以A2P中点为N|(2,(x0+1y0)因为△MA2P是以A2P为底边的等腰三角形，所以MN丄A2P，所以MN.A2P=0，···························································9分0因为P(x0,y0)在椭圆上，所以4x+3y=4②·············································13分23233(2)(2)(3,(3,所以点P的坐标为|0,|或|0,(3,(3,(1)(1)(1)7(6,6，.(1)7(6,6，.所以点P在以M为圆心，为半径的圆上（异于点A2(1)22490(1)2249因为P(x0,y0)在椭圆上，所以4x+3y联立①、②,消去y0，化简得x0(x0−1)=0(x0≠1).····(2)(2)（2）设点P(x0,y0)，由题意P不与A2(1,0)重合，故y0≠0，2因为P(x0,y0)在椭圆上，所以4x+3y=4.①·····························因为△MA2P是以A2P为底边的等腰三角形(1)7(1)7(1)249(1)249(y0展开整理，得3m2=4m4．.········································· 由P在椭圆上，可设P.······················································8分因为△MA2P是以A2P为底边的等腰三角形，且A2(1)7(1)7辑推理、数学运算等核心素养，体现基础性和综合性．满分15因为故sinBsinC=，················382所以S△ABC=2bcsinA=2(2RsinB)(2RsinC)sinA=2R2sinBsinCsinA所以{， 解得{， 由余弦定理a2=b2+c2−2bccosA，····························································12分2222 22232所以S△ABC=1bcsinA=33，所以1bcsinτ=33，解得bc=6.················9分22232 abbbτ3babbbτ3bsinAsinBaa32a4c4c2a 由余弦定理a2=b2+c2−2bccosA，····························································12分2222 22232所以S△ABC=1bcsinA=33，所以1bcsinτ=3322232又因为S△ABC=2bcsinA=2×3(bsinA)(csinA)=3(asinB)(asinC) 由余弦定理a2=b2+c2−2bccosA·······························································12分22 分.（2i）由题意可知，在3:3后队、在第五轮时罚进点球就提前获胜，即第4轮队、罚进队、罚失，第5轮队、罚进.··················· 记所求事件为N，因为A4,B4,A5相互独立，P(A4（ii）X的可能取值为4,5.·····································································11W=F4WF5W，F4W,F5W互斥，且A4,A4,B4,B4,A5,A5,B5,B5,相互独立，(A4B4A5B5)PPAPPB4B5)P(N)则X的分列、为X45P 19.【考查意图】本小题主要考查函数单调性与奇偶性、函数与方程、导数与不等式等基所以D(2x)=D(x)；当x为无理数时，则2x也为无理数，故D(x)=0，D(2x)=0，所以D(2x)=D(x)；···············································································所以f(x)是2−拓展函数.·······················································因为f(x)为2−拓展函数，所以f(2x)=2f(x)，(x)(x)(x)(x)(x)(x