2024福建省福州市高三下学期4月末质量检测数学试题及答案

（在此卷上答题无效）20244月份质量检测数学试题（完卷时间120满分分）友情提示：请将所有答案填写到答题卡上！请不要错位、越界答题！一、单项选择题：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.ì1ü1.已知集合M=x„0，则ðM=ýRíx+1îþxx<-}xx„-}xx>-}D.xx-}aab2.设a,bÎR，则“ab<0”是“0”的b．充分不必要条件．充要条件．必要不充分条件D．既不充分也不必要条件3.等轴双曲线经过点-，则其焦点到渐近线的距离为．22．2．4D．2π144.若a+)=a，则sin2=4158787．-．．-．iD．88zz5.已知非零复数z满足z-1=z-i，则=．1．-1的展开式中x的系数为．-6D．-i6.-x)5+2x)42．-．34D．747.数列an}共有5项前三等差且公差为d后项成等比且公比为q.若第2项于2第14的和于10第35项和等于30d-q=．1．2．3D．48.四棱锥E-的顶点均在球O的球面上，底面为矩形，平面^平面，BC=5，CD==1，BE=2，则O到平面的距离为131425．．．D．48二、多项选择题：本题共3小题，每小题6分，共18分.在每小题给出的四个选项中，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.在一次射击比赛甲、乙两名选手射击环如下表，则下说法正确的是甲乙87909691869086928795．甲选手射击环极差大于乙选手射击环数的．甲选手射击环数的数等于乙选手射击环平均数．甲选手射击环数的方差大于乙选手射击环数的D．甲选手射击环第75百分位数大于乙选手射击环数的第75百分位数πè3öπè3öπ10.已知函数f(x)=sin(2x+j)满足f+x=f-x，且f()>f(π)，则ç÷ç÷øø2112．sinj=．sinj=-213．y=f(x)的图象关于点12π,0)对称πD．f(x)在区间(,π)单调递减211.已知函数f(x)=ax(e．x+x+x=0x+e-x)-e+e-x恰有三个x,x,x，且x<x<x，则x123123．实数a的取值范围为D．3+a>1123．1+1>0三、填空题：本大题共3小题，每小题5分，共15分.12.若向量a=-4)在向量b=(-上的投影向lb，则l等于__________.π13.倾斜角为的直线经过抛物线C：y2=12x的焦点C交于AB点，Q为线段3AB的中点，P为C上一点，则PF+的最小值为__________.D114.如图，ABCDACD的一个面是边长为2的111C1形，,,均垂直于平面且=1CC=2，11111A1则该六面体的体积于__________，表面积等_________.DCAB四、解答题：本大题共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.（13分）已知数列a}满足a=2，a=a+2n（n2）.n1nn-1（1）求数列n}的通项公式；1（2）记数列{}的前n项和为S，证明：S<1.annn16.（15分）甲企业生产线上的零件的误差X服从正分布XÎ(-0.6,0.6)N(0,0.2)规定XÎ(-0.2,0.2)2的零件为等品，（1）从该生产线上随机抽取100个零件，估计抽到合格品但非优等品的个数（精确到整数的零件为合格品．（）乙企业拟向甲企业购买这批零件，先对该批进行质量抽检，检测的案是：从这批零件中任取2个作这2个都是等品通这2零件中恰有1个为等品1为合格品但非优品零件中任取1作优等品，则通余情不通过检测.求通过检测时，检测了2个零件的概率（精确到0.01）.x~N(m,s2)P(m-s<x<m+s)=0.6827（附：若随机变量，则，P(m-s<x<m+s)=0.9545P(m-s<x<m+s)=0.9973，）17.（15分）如图以正方形的边AB所在直线为余三AF»边旋转120°形成的面围成一个体-BCEP是CEDHE上的一点，G，H分别为线段AP，EF的中点.（1）证明：GH∥平面；GB（2BP^求平面与平面夹角的.CP18.（17分）点P是椭圆E：x22y22+=a>b>右端除外一个点，F(-,0)F,0)12ab分别是E左、右a2|2|d（1）设点P到直线l:x=的距离为d，证明为定值并求个定值；c（2）△F的重心与内心内切圆圆心别为，I，已知直线IGx12（ⅰ）求E心率；（ⅱ椭圆E的长轴为求△F被直线分部分的图形面积之的取12值范围．19.（17分）记集合Lf(x),xÎD=l(x)=+b(xÎR)|"xÎD,f(x)„l(x$xÎD,f(x)=l(x}，集合000Tf(x),ÎD=l(x)=+b(xÎR)|"xÎD,f(x)l(x),且$xÎD,f(x)=l(x}.若l(x)ÎL，f(x),ÎD000则称直线y=l(x)为函数f(x)在D上的“最l(x)ÎTf(x),xÎD称直线y=l(x)为函数f(x)在D上的“最下界线”.（1）已知数f(x)=-x2+x，l(x)=+1.若l(x)ÎL，求k的值；f(x),xÎR00（2）已知g(x)=e+1.x（ⅰ直线y=l(x)是曲线y=g(x)的一条切线的充要条件是线y=l(x)是函数g(x)在R上的“佳下界线”；（ⅱh(x)=ln(x-直接写集合h(x),Î(1,+¥)Ig(x),ÎR中元素的个无需.20244月份质量检测参考答案与评分细则一、选择题：本大题考查基础知识和基本运算．每小题5分，满分40分.1．D2．C3．A4．C5．D．B7．B8．A二、选择题：本大题考查基础知识和基本运算．每小题6分，满分18分．全部选对的得6分，部分选对的得部分分，有选错的得0分．9．10．．三、填空题：本大题考查基础知识和基本运算．每小题5分，满分15分．12．-213．8146，22四、解答题：本大题共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.逻辑推理等核心素养；基础性综合性.13分.解：（1n=n-1+2n,n2，所以n-n-1=2n，···································1分当n2时，a=(a-an-1)+(an-1-an-2)+L+(a-a)+a，nn211所以n=2n+2n-2+L+4+2，·························································3分n(2+2n)所以n=,n2，所以n=n+n,n2，··································4分22又因为1=2，···············································································5分所以n=n+,nÎN.······································································6分（2）由（）可知n=n+n=n(n+nÎN，·············································7分2*2*1111所以==-，····························································9分ann(n+)nn+11111所以Sn=++L++L++1´22´3(n-nn(n+1212131111=1-所以Sn=1-+--+-，·····································分n-1nnn+11，·········································································12分n+1又因为n1，所以Sn<1.·································································································13分参考答案第1页共9页16.【考查意图】本小题主要考查正态分布、全概率公式、条件概率等基础知识，考查数学建模能力、逻辑思维概率统计思想查数学建模据分核心素养基础性综性和应15分.解：（1题意得，m=s=，···························································1分所以零件为合格品概率为P(-<X<0.6)=P(m-s<X<m+s)=0.9973，···································································································2分零件为优品的P(-<X<0.2)=P(m-s<X<m+s)=0.6827，·····3分所以零件为合格品但非优等的概率为P=0.9973-0.6827=0.3146，···········5分所以从该生产线上随机抽取100个零件，估计抽到合格品但非优等品的个数为100´0.3146»31.·············································································6分（2）设从这批零件中任取2检测，2有2等品为事件A，恰有1个优等品1个为合等品为事件B从这批零件中任取1个是优等品为事件C，这批产品通过检测D，····························································8分则D=A+，且A与BC互斥，9分所以P(D)=P()+P(BC)10分=P()+P(B)PC|B)分=22´0.68272+21´0.6827´0.3146´0.6827=1.6292´0.68272，12分所以这批零件通过检测时，检测了2个零件的P(AD)P(A|D)=13分P(D)0.68272=1.6292´0.682721=1.6292»0.61.15分答：这批零件通过检测时，检测了2个零件的0.61.17.【考查意图】本小题主要考查直线与平面平行的判定定理、直线与平面垂直的判定与性质定理、与的夹角间向量、三概念等基础知识，考査直观想象能力、逻辑推理能力、运算求解能力等,考查数形结合、化归与转化等，考査直观想象、逻辑推理、数学运算等核心素养，基础性综性.满分15分.解法一（）在ABEF中，连接并延长，交BE的延长于点K，连接.···································································································2分参考答案第2页共9页因为G,H分别为线段AP,EF中点，所以=，AFDH所以△≌KEH，所以AH=，·····························4分GBEKCGH∥.································5P所以分GHË面BCE,PKÌ面BCE又因为，所以GH面BCE.···········································································7分（2）依题意得，AB^面BCE，又因为BPÌ面BCE，所以AB^BP.∩又因为BP^AE,AB=A,AB,AEÌ面ABEF，所以BP^面ABEF，········································································8分又BEÌ面ABEF，所以BP^BE，·····················································9分所以BP,BE,BA两两垂直.zx,y,zA以B为原点，BP,BE,BA所在分别为间直角坐标系，如所示.轴建立空FDHEG································································10分不妨设=1,ByCP31则PD(,-，x22ræ31öBP=1,0,0),=ç,-,1÷·························································，分ç÷22èøìm=ïm=(,y,z)设平面BPD的法向量为，则ím=ìx=ïy=2x=z=1，即í31取，得x-y+z=ïî22所以平面BPD的一个法向量是m=()，·········································13分分n=(0).················································14又平面的一个法向为mn225qcosq=<m,n>===.设平面BPD与平面的夹角为，则mn5´15参考答案第3页共9页255所以平面与平面夹角的余弦值为.····································15分解法二：（）证明：取的中点，连接QGQ,.·····1分AFD因为G,H分别为线段AP,EF的中点，HG1所以GQAB，GQ=AB，····························2分2BEQ又因为AB∥EF,=，CPGQ∥HE,=所以，·································3分······················································4GQEH所以四边形是平行边形，分分GH∥QE··············································································5所以又因为Ë,Ì，.············································································7，所以GH面分（2）同解.····················································································15分解法三：（）证明：取AB的中点I，连接,.········1分AFD因为G,H分别为线段AP,EF的中点，所以GI∥BP,∥，IHGBE又因为GIË面BCE,Ì面BCE，所以GI∥面BCE.············································3分因为Ë面BCE,Ì面BCE，CP所以HIBCE.············································································5分GI∩=I,GIÌ面GIH,Ì面GIH又因为，所以面GIH面BCE，·····································································6分又因为GHÌ面GIH，所以GH面.············································································7分（2）同解.····················································································15分18.【考查意图】本小题主要考查圆、椭圆的标准方程及简单几何性质，直线与椭圆的位置关系査直观想象逻辑,考查数形结思想、化归与转化思想思想逻辑核心素养体现基础性综合性创新性.17分.解法一（）依b2+c2=a．··························································1分2参考答案第4页共9页0a20b2设P(x,y)，则+=1，-a<0<a，00220a2b22所以|PF=(x-c)2+02=(0-c)2+b2-)=-)02-0+b2+c，2202ac2c所以|2|=02-20+a2|0-a|，············································3分a2aca2ca2又a>c，所以a>0，>x，所以|PF=a-0，d=-002acacca-0|2|dc|2|dca所以==，即为定值，且这个定为．··················4分a2aa-0c0y0（2ⅰ题意，G(,)，3303设直线与x轴交于点，因为IG⊥x，所以C(,0)，·······················5分03032所以|FC|-|FC|=(+c)-(c-)=0，··········································6分123因为△F的内切圆与x轴切于点，122所以|PF|-||FC|-|FC=0，·················································7分12123又因为|PF|+|PF=2a，1203y解得|2=a-，··········································8分Pc由（1）得|2=a-0，aGIc03xOC所以a-0=a-，ac13所以椭圆E的离心率e==．·························10分ac1（ⅱ2a=6a=3又=所以c=1b2=a2-c2=8，a3x2y2所以椭圆E的方程为+=1．······················································分98根据椭圆称性设点P在第限或y轴正上即0„x<30<y„22，00又F(-0)，F0)，1200+1所以直线1的方程为y=(x+，03y(x+000+设直线与交于点D，因为x=，所以yD=，1D△F的面积S与△F的面积S之比为11211023y(x+(+´001S3(0+(0+2==，················································13分(x+x-118(0+´2´02(x+2令f(x)=（0„x<3，则f(x)=，18(x+18(x+2参考答案第5页共9页当xÎ，f(x)<0，当xÎ，f(x)>0，所以函数f(x)在单调递，在单调递增.1412又因为f(0)=，f=，f=，2941所以f(x)的值域是[,]，9241S1所以„„，··········································································15分9421所以„„1，·······································································16分5S-1根据对称，45△F被直线IG分成两部分的图形面积之比的值范围是[,]．········17分1254解法二：（）同解···········································································4分0y0（2ⅰ题意，G(,)，3303设直线与x轴交于点，因为IG⊥x，所以C(,0)，·······················5分03032所以|FC|-|FC|=(+c)-(c-)=0，··········································6分123因为△F的内切圆与x轴切于点，122所以|PF|-||FC|-|FC=0，·················································7分12123ìï030|PF1=a+,.ï又因为|PF|+|PF=2a，得···········································8分í12ï|2=a-ïî3ìï0303(0+c)22+022=a+,ï4所以两式平方后取，得40=ax对任意x成立，í003ï(0-c)+0=a-,ïîc13所以椭圆E的离心率e==．························································10分a（ⅱ）同法一···················································································17分解法三：（）同解···········································································4分0y003（2ⅰ题意，G(可求直线PF1方程为y=,)，因为IGx轴设点I坐标为(,t).··········5分3300+c(x+c)，00(+c)-t(0+c)3则点I到直线PF1的距离|t|，·································6分y20+(0+c)+(0+c)=0，①22æ03ö(y0即y(+c)-t(0+c)=t222)，化简得ç÷0èø030yt02+t(+c)(x+c)-y(+c)2003同理，由点I到直线2的距离等于|t|，可得参考答案第6页共9页0303yt02+t(-c)(x-c)-y(-c)2=0，②············································7分008404将式①式②，得t×=y×，则t=.·····································8分000330将t=代入式，得4y201x03+(0+c)(0+c)-(+c)2=0，1623x20y20化简得+=1，9c=a2c2得9c22，c13所以椭圆E的离心率e==．························································10分a（ⅱ）同法一···················································································17分19.【考查意图】本小题主要考查集合、导数、不等式等基础知识，考查逻辑推理能力、直观想象方程思想思想整合思想形结思想等，考查数学抽象、推理、直观想象、数学运算等核心素养，体现基础、综合与创新.满分17分.解：（1题意，0(x)ÎLf(x),xÎR所以"xÎR,-x+x„+1，且0R，-0令f(x)=-x+-k)x-1，D=(k-2-4，，22+x=+1，····················1分002则f(x)„0，且f(0)=0，ìD„所以í所以=0，···································································3分Dî即(k-2-4=0，解得k=3或-1.··············································································4分（2）先必要性.若直线y=l(x)是曲线y=g(x)的切线，设为x,e+1，(x0)0因为g(x)=e，所以切线方为y-ex0+)=ex0(x-0)，x即l(x)=e0x+-0)e0+1（）.························································5分一方面，g(x)=l(x)，所以$xÎR,g(x)=l(x)，································6分00000参考答案